Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proof of a summation of $k^4$ I am trying to prove $$\sum_{k=1}^n k^4$$
I am supposed to use the method where $$(n+1)^5 = \sum_{k=1}^n(k+1)^5 - \sum_{k=1}^nk^5$$
So I have done that and and after reindexing and a little algebra, I get $$(n+1)^5 = 1+ 5\sum_{k=1}^nk^4 + 10\sum_{k=1}^nk^3 + 10\sum_{k=1}^nk^2 + 5\sum_{k=1}... | Here's an alternative approach using binomial coefficients.
Firs we express $k^4$ as a linear combination of $\binom{k+a}{4}$ where $a=0,1,2,3$, i.e.
$$k^4={k+3\choose 4}+11{k+2\choose 4}+11{k+1\choose 4}+{k\choose 4}$$.
Summing this from $1$ to $n$ and using the hockey stick summation identity, i.e. $\sum_{r=0}^m {r\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Calculus 1: Find the limit as x approaches 4 of $\frac{3-\sqrt{x+5}}{x-4}$ I understand how to find limits, but for some reason I cannot figure out the algebra of this problem. I tried multiplying by the conjugate and end up with 0/0. When I check on my calculator, or apply L'Hopital's rule I get -1/6. Is there an alge... | $$\begin{array}{rcl}\lim_{x\to 4} \frac{3-\sqrt{x+5}}{x-4} & = & \lim_{x\to 4} \frac{(3-\sqrt{x+5})(3+\sqrt{x+5})}{(x-4)(3+\sqrt{x+5})}=\lim_{x\to 4} \frac{9-(x+5)}{(x-4)(3+\sqrt{x+5})} \\ & = & \lim_{x\to 4} \frac{4-x}{(x-4)(3+\sqrt{x+5})} =\lim_{x\to 4} \frac{-1}{3+\sqrt{x+5}}=- \frac{1}{6}.\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/981332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
solve indefinite integral I have this indefinite integral $\int 3 \sqrt{x}\,dx$ to solve.
My attempt:
$$\int 3 \sqrt{x}\,dx = 3 \cdot \frac {x^{\frac {1}{2} + \frac {2}{2}}}{\frac {1}{2} + \frac {2}{2}}$$
$$\int 3 \sqrt{x}\,dx = 3 \frac{x^{\frac {3}{2}}}{\frac {3}{2}} = \frac{2}{3} \cdot \frac{9}{3} x^{\frac {3}{2}}$$
... | In your penultimate step, $\frac 23$.$\frac 93$ $=\frac {18}{9}$ $=2$, not 6.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$. Also write the identity used Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$.
Also write the identity used
| You can use the identity that if $a + b + c=0$, then $a^3 + b^3 + c^3 = 3abc$. So we can say that:
*
*$2^3 + 4^3 + (-6)^3 = 3\cdot 2\cdot 4\cdot -6 = -144$.
*$1^3 + 4^3 + (-5)^3 = 3\cdot 1\cdot 4 \cdot -5 = -60$.
Adding both, we get:
$1^3 + 2^3 + 2 (4)^3 + (-5)^3 + (-6)^3 = -204$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $x^3 - x + 1 = 0$ Solve $x^3 - x + 1 = 0$, this cannot be done through elementary methods.
Although, this is way out of my capabilities, I would love to see a solution (closed form only).
Thanks!
| Let $x=u+v$ and note that $(u+v)^3=3uv(u+v)+(u^3+v^3)$
This has the required form if $3uv=1$ and $u^3+v^3=-1$
So $u^3v^3=\frac 1{27}$, and $u^3, v^3$ are roots of $y^2+y+\frac 1{27}=0$
$u,v= \sqrt[3] {\frac {-1\pm \sqrt{1-\frac 4{27}}}2 }$
$x=u+v=\sqrt[3] {\frac {-1+ \sqrt{\frac {23}{27}}}2 }+\sqrt[3] {\frac {-1- \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Find series expansion of 1/cosx Find the series expansion of 1/cosx from basic series expansions.
I tried to find 1/cosx from the expansion of cosx but was unsure how to continue. When I found 1/cosx from the basic formula for finding series expansions I didn't get the same answer.
| (going to the fifth term for an example purpose)
Using the basic expansions of cos(x) gives us
$$
\frac{1}{\cos(x)} = \frac{1}{1-\frac{x^2}{2}+\frac{x^4}{24} + \cdots}
$$
of the form $ \frac{1}{1-X} $ which has a known and easy expansion :
$$
1+X+X^2+X^3+X^4+X^5+\cdots
$$
where $ X = \frac{x^2}{2} + \frac{x^4}{24} $ (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Repeated substitution gone wrong It was an exam question.
$$ f(n)=
\begin{cases}
0 & \mbox{if } n \leq 1 \\
3 f(\lfloor n/5 \rfloor) + 1 & \mbox{if } n > 1 \\
\end{cases}$$
So by calculating some I have $f(5) = 1$, $f(10) = 4$, $f(50) = 13$.
I had to solve this recurrence. So to get rid of the floor operator I said let... | I didn't check your whole argument, but assuming it's correct, you can use the fact that if $n$ is $5^k$, then $k$ is $\log_5 n$ to express the result in terms of $n$.
(Side Question: are the $x$'s in the function definition supposed to be $n$'s?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A question on a nonnegative quadratic form Denote $x,y,z$ as variables, and $a,b,c$ as coefficients. Suppose $a\leq b\leq 0\leq c$ and $a+b+c=0$.
Could anyone help me prove whether the following quadratic form positive semi-definite?
\begin{equation*}
\begin{split}
I(x,y,z)=&(a^2+4b^2+4c^2)a^2x^2+(4a^2+b^2+4c^2)b^2y^2\... | Well, I showed you how to try to attack these problems a couple of days ago, and you can use exactly the same strategy here.
A problem on positive semi-definite quadratic forms/matrices
Once again using MATLAB Toolbox YALMIP to compute a sum-of-squares certificate. Ordering of the variables is not required, but the eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested w... | With
$$
\int_{0}^{2\pi} \frac{1}{1-a \cos \theta} d \theta=\frac{2\pi}{\sqrt{1-a^2}}
$$
evaluate the following integrals successively
\begin{align}\int_{0}^{2\pi} \frac{1}{(1-a \cos \theta)^2} d\theta
=&\frac{d}{da} \int_{0}^{2\pi} \frac{a}{1-a \cos \theta} d \theta=\frac{2\pi}{(1-a^2)^{3/2}}\\
\int_{0}^{2\pi} \frac{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Why can't prime numbers satisfy the Pythagoras Theorem? That is, why can't a set of 3 prime numbers be a Pythagorean triplet? Suppose $a$, $b$ and $c$ are three prime numbers.
How to prove that $a^2 + b^2 \neq c^2$?
| The sum of two odd numbers are even, so one of the numbers must be $2$.
If $a$ or $b$ are $2$ we have $a^2+4=c^2$ or $4=(c+a)(c-a)$ Since $c-a$ and $c+a$ have the same parity, this is impossible.
If $c=2$ we have $a^2+b^2=4$ but since $a$ and $b$ are positive, both must be $1$, but $1^2+1^2=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 0
} |
Show complex solutions exist Let A be a complex number and B a real number. Show that the equation
$\,\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B = 0\,$ has a solution iff $\,\lvert A^2\rvert \geq 4B$. If this is so, show that the solution set is a circle or a single point.
Well i am trying to do the first part first. So... | We have
$$
0=\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B =
z\overline{z}+\frac{1}{2}(Az+\overline{Az})+\frac{1}{4}\lvert A\rvert^2-\frac{1}{4}\lvert A\rvert^2+B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B
$$
If $4B>\lvert A\rvert^2>0$, then $\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B=\left\lvert z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to find the sum of the series $\sum_{k=2}^\infty \frac{1}{k^2-1}$? I have this problem :
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$
My solution
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\f... | Let $S_n=\displaystyle\sum_{k=1}^n\frac{1}{k^2-1}=\sum_{k=1}^n\frac{1}{2}\left[\frac{1}{k-1}-\frac{1}{k+1}\right]$
$\displaystyle\hspace{.2 in} =\frac{1}{2}\left[\bigg(\frac{1}{1}-\frac{1}{3}\bigg)+\bigg(\frac{1}{2}-\frac{1}{4}\bigg)+\bigg(\frac{1}{3}-\frac{1}{5}\bigg)+\cdots+\bigg(\frac{1}{n-2}-\frac{1}{n}\bigg)+\bigg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$ How to find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$
Let $y^2-xy-2x^2 =0...(1)$ and $y^2=x-2...(2)$
In equation (1) coefficient of $x^2 =-2; y^2=1, 2xy =\frac{-1}{2}$
We know that a second degree equation where $a... | $$y^2= 2x^2+xy \iff (y-2x)(x+y)=0 \iff y = -x \text{ or } y = 2x$$
so that is a couple of lines.
Hence you want to find the shortest distance from $y=-x$ or $y=2x$ to the parabola $y^2=x-2$. Let this be denoted by the line segment with end points $(a, -a)$ or $(a, 2a)$ and $(b^2+2, b)$.
Then we need the min of $(a-b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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An equation for the third powers of the roots of a given quadradic polynomial The roots of the equation $3x^2-4x+1=0$ are $\alpha$ and $\beta$. Find the equation with integer coefficients that has roots $\alpha^3$ and $\beta^3$.
GIVEN
SR: $\alpha + \beta = \frac43$
PR: $\alpha\beta = \frac13$
REQUIRED
SR: $\alpha^3 + \... | Using the quadratic formula, we have $$\alpha, \beta = \frac{-(-4) \pm \sqrt{(-4)^2-4(3)(1)}}{6} \\ = \frac{4 \pm \sqrt{16-12}}{6} \\ = \frac{4 \pm 2}{6} \\ = 1, \frac{1}{3} $$ Hence a polynomial you could use with roots at $\alpha^3 = 1^3 = 1$ and $\beta^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$ would be $$f(x) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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find common ratio of $\sum_{k=1}^\infty \frac{1}{k(k+1)}$ I have this problem, I need to find the sum.
$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{k(k+1)}$$
The problem is that the ratio is not conclusive, Any idea how to find the ratio?
Thanks!
| $$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{k}-\frac{1}{k+1})$$ Do you see how to do it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How would one solve this system $$12x^2=6z\\2y=-z\\6x-y=7$$
It's been many years since I've dealt with system equations, and now find myself in need to solve them. I am not quite sure what to do; I am interested in finding $x$ and $y$, so assuming that $z \neq 0$ (can I do this?), I isolate it and get
$$2x^2 = z = -2y ... | Given is $f(x,y)=4x^3+y^2$ with the constraint $y=6x-7$. The easiest way to solve this, is plugging in $y$ into the function, which gives
$$f(x)=4x^3+(6x-7)^2=4x^3+36x^2-84x+49$$
and we are asked to find the extrema. Differentiating:
$$f'(x)=12x^2+72x-84=0$$
Dividing by $12$ gives $f'(x)=x^2+6x-7=(x+7)(x-1)=0$. From th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/998124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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how to integrate $\mathrm{arcsin}\left(x^{15}\right)$? Integral by parts:
$$
I = x\sin^{-1}\left(x^{15}\right) - \int\frac{15x^{15}}{\sqrt{1-x^{30}}}dx
$$
then what?
The answer by wolfram gives an answer contains hypergeometric ${}_2F_1$ function,because it has no elementary answer. The question I want to know is, how ... | $\int\dfrac{15x^{15}}{\sqrt{1-x^{30}}}dx$
$=\int_0^x15t^{15}(1-t^{30})^{-\frac{1}{2}}~dt+C$
$=\int_0^{x^{30}}15t^\frac{1}{2}(1-t)^{-\frac{1}{2}}~d(t^\frac{1}{30})+C$
$=\dfrac{1}{2}\int_0^{x^{30}}t^{-\frac{7}{15}}(1-t)^{-\frac{1}{2}}~dt+C$
$=\dfrac{1}{2}\int_0^1(x^{30}t)^{-\frac{7}{15}}(1-x^{30}t)^{-\frac{1}{2}}~d(x^{30... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/999744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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derivative of $y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$ $y=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$
The answer is $\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5} \left(\dfrac{2}{x}+\dfrac{1}{2(x+1)}-\dfrac{1}{x+2}-\dfrac{5}{x-3}\right)$
I know that the quotient rule is used but I don't know how to do this problem. Would you multipl... | It's a long derivative. First use the quotient rule:
$$\frac{\frac{d}{dx}\left(x^2\sqrt{x+1}\right)(x+2)(x-3)^5- x^2\sqrt{x+1}\frac{d}{dx}\left((x+2)(x-3)^5\right)}{((x+2)(x-3)^5)^2}.$$
Then do the derivatives.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Question about recurrence relation problem.
solve the following recurrence relation, subject to given initial conditions.
$a_{n+1} = 6a_n -9,$
$a_0 = 0,$
$a_1 = 3.$
Here is what I have done.
$a_{n+1} - 6a_n +9 = 0$
$a_n = r^n$
$r^{n+1} - 6r^n + 9 = 0$
$r + 6 = 0$
$r = -6$
$a_n$ = $X(-6)^n$
$a_1 = 3 = X(-6)^1$
$X = -1/... | $$\begin{align}
a_{n+1}&=6a_n-9\\
a_{n+1}-\frac95&=6a_n-\frac{54}5\\
&=6\left(a_n-\frac95\right)\\
u_{n+1}&=6u_n\\
u_n&=6u_{n-1}=6^2u_{n-2}=\cdots=6^{n-1}u_1\\
&=6^{n-1}\left(a_1-\frac95\right)\\
&=6^{n-1}\left(3-\frac95\right)\\
a_n-\frac95&=\frac65(6^{n-1})\\
&=\frac{6^n}5\\
a_n&=\frac15\left(9+6^n\right)\qquad\black... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve $\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$ for $y$ I have the equation:
$\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$,
where $k$ and $c$ are arbitrary constants.
How do I go about simplifying this and solving for $y$ in terms of $x$, excluding the obvious solution $y=-x$
| Treat $x$ as a constant as well. We could use the standard quadratic formula, but since you noticed that $y = -x$ is a solution, let's try factoring instead. We obtain:
\begin{align*}
0
&= \tfrac{1}{2}ky^{2} - \tfrac{1}{2}kx^{2} + cy + cx \\
&= \tfrac{1}{2}k(y^{2} - x^{2}) + c(y + x) \\
&= \tfrac{1}{2}k(y - x)(y + x) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
taking the inverse of power series I am working with solution to near regular singular points.
I started with:
$$y_1(x)=x^\frac{1}{2}\left[1-\frac{3}{4}x+\frac{9}{64}x^2-\frac{3}{256}x^3+\cdots\right] $$
Then I squared it:
$$y_1^2(x) = x\left(1-\frac{3}{2}x+\frac{27}{32}x^2-\frac{15}{64}x^3+\cdots\right)$$
Why is the i... | If $$y=\sqrt{x} \left(1-\frac{3 x}{4}+\frac{9 x^2}{64}-\frac{3 x^3}{256}+\cdots\right)$$ then effectively $$y^2=x \left(1-\frac{3 x}{2}+\frac{27 x^2}{32}-\frac{15 x^3}{64} +\cdots\right)$$ Now you want to compute $\frac{1}{y^2 }$. You can write $$\frac{x}{y^2 }=\frac{1}{1-\frac{3 x}{2}+\frac{27 x^2}{32}-\frac{15 x^3}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question:
let $x,y,z>0$ and such $xyz=1$, show that
$$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$
My idea: use AM-GM inequality
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
so
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$
But this is not my inequality,so How pro... | Here is a possible solution: (although it is not the most elegant one)
I will employ Mixing Variables technique here. Since the inequality is symmetric, WLOG let $x=\min(x,y,z)$. Therefore $t^2:=yz \ge 1$.
Let
$$f(x,y,z)=x^3+y^3+z^3-2(x^2+y^2+z^2)$$
I wish to show
$$f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz}) = f(\frac1{t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
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Prove that: $\lim_{x\to 0}\frac{x}{\sin^2(x) + 1} = 0$ Prove
$$\displaystyle \lim_{x\to 0} \frac{x}{\sin^2(x) + 1} = 0$$
The proof:
Let $$|x| \le 1 \implies -1 \le x \le 1$$
$$\displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \epsilon\text{ for }\displaystyle |x| < \delta$$
$$-1 \le x \le 1
\\\implies \sin(-1) \le \sin(x) \... | This is way too complicated, don't you think?
Why not just say that
$$
\left| \frac{x}{1+\sin^2 x}
\right| \le |x| \le \epsilon
$$
as
soon as $|x|<\delta = \epsilon$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inequality proof by induction, what to do next in the step I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$
Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$
Step: assume holds for $n_0$.
$2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \f... | For induction step, it's enough to prove $\frac{1}{\sqrt{n+1}}>2(\sqrt{n+2}-\sqrt{n+1})$.
$$2(\sqrt{n+2}-\sqrt{n+1})=\frac{2}{\sqrt{n+2}+\sqrt{n+1}}<\frac{2}{2\sqrt{n+1}}=\frac{1}{\sqrt{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
$$
and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{... | Hint: $\frac 8 9 = \frac 1 2 + \frac 1 3 + \frac 1 {18}$ Now try using your second formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Trigonometric equation, missing some solutions I'm missing part of the answer, and I'm not quite sure why. The given answer doesn't even seem to hold...
Solve for x: $$\tan 2x = 3 \tan x $$
First some simplifications:
$$\tan 2x = 3 \tan x $$
$$\tan 2x - 3 \tan x = 0$$
$$\frac{\sin 2x}{\cos 2x} - \frac{3 \sin x}{\cos ... | Setting $\tan x=t$
we have $$\frac{2t}{1-t^2}=3t\iff2t=3t(1-t^2)\iff t(2-3+3t^2)=0$$
If $t=0,\tan x=0, x=n\pi$ where $n$ is any integer
$2-3+3t^2=0\iff 3t^2=1\implies\cos2x=\dfrac{1-t^2}{1+t^2}=\dfrac12=\cos\dfrac\pi3$
$\implies2x=2m\pi\pm\dfrac\pi3$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Linear independence of $\sin^2(x)$ and $\cos^2(x)$ The Wronskian for $\sin^2x, \cos^2x$ is
\begin{align}
& \left| \begin{array}{cc} \sin^2 x & \cos^2 x \\ 2\sin x\cos x & -2\cos x\sin x \end{array} \right| \\[8pt]
= {} & -2\sin^2x \cos x \sin x - 2 \cos^2 x \sin x \cos x,
\end{align}
with $x = \frac{π}{6},$ this is $=$... | It suffices to show that the Wronskian is not zero for a single value of $x$. We have: $$W(x) = \begin{vmatrix} \sin^2x & \cos^2x \\ 2\sin x \cos x & -2 \sin x \cos x\end{vmatrix} = -2\sin^3x \cos x - 2\sin x \cos^3 x$$
$$W(x) = -2\sin x \cos x = -\sin(2x)$$
Then, $W(\pi/4) = -1 \neq 0$, so the functions are linearly i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to integrate the dilogarithms? $\def\Li{\operatorname{Li}}$
How can you integrate $\Li_2$? I tried from $0 \to 1$
$\displaystyle \int_{0}^{1} \Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$
$$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$
$$(An + B)(n+1) + D(n^2) = 1$$
Let $n = -1, \implies D = 1... | Maybe you should look at your decomposition as
$$\frac1{n^2 (n+1)} = \frac1{n^2} - \frac1{n (n+1)}$$
The sum over the second term is easy, given that the indefinite sum is telescoping, i.e.,
$$\sum_{n=1}^N \frac1{n (n+1)} = 1-\frac1{N+1}$$
We take the limit as $N \to \infty$ and then we may view this as the infinite s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Clarification: Proof of the quotient rule for sequences My Problem
I am currently looking for a proof for the quotient rule for sequences:
$a_n$ and $b_n$ are two sequences with the limes a,b. So:
When
$ a_n \rightarrow a$
and
$ b_n \rightarrow b$
Then:
$\frac {a_n}{b_n} \rightarrow \frac{a}{b}$
Awesome stuff, but how... | If, for eny $\epsilon > 0 \in \mathbb{R}$, there exists an $n \in \mathbb{Z}$ such that $\forall i > n, |a_i - c| < \epsilon $, then $c$ is defined as the limit of the sequence of $a_i$.
Assuming the limit of sequence $a_i$ is $a > 0$, and the limit of $b_i$ is $b > 0$, we want to prove the limit of $\dfrac{a_i}{b_i} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve to find $y(x)$ of the $\frac{1}{\sum_{n=0}^{\infty }y^n}-\sum_{n=0}^{\infty }x^n=0$ Solve the equation to find the $y$ as a function to respect $x$ without $n$ $$\frac{1}{\sum_{n=0}^{\infty }y^n}-\sum_{n=0}^{\infty }x^n=0$$
| Assuming $|x| < 1$ and $|y| < 1$, the two geometric series simplify to:
$$\frac{1}{\frac{1}{1 - y}} = \frac{1}{1 - x}.$$
Consequently:
$$(1 - x)(1 - y) = 1.$$
Solving for $y$ in terms of $x$, you get:
$$1 - y = \frac{1}{1 - x}$$
which implies that
$$y(x) = 1 - \frac{1}{1 - x} = \frac{x}{x - 1}.$$
Note that $|y(x)| < 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of 6 cards being multiple of 6 I pick 6 cards from a set of 13 (ace-king). If ace = 1 and jack,queen,king = 10 what is the probability of the sum of the cards being a multiple of 6?
Tried so far:
I split the numbers into sets with values:
6n, 6n+1, 6n+2, 6n+3
like so:
{6}{1,7}{2,8}{3,9}{4,10,j,q,k}{5}
and then gro... | Your combinations are $4\cdot 4+1\cdot 0+1\cdot 2, 2\cdot 1+2\cdot2+2\cdot3, 4 \cdot 4+2\cdot 1,+$ something that doesn't make sense because there are three places you choose from $1$ and only $0,5$ qualify. The left number is the number of cards of that value $\pmod 6$ . You could also have $3\cdot 4+1 \cdot 5+1 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality of sides of triangle If a,b,c are the sides of a triangle ABC then prove $$ a/(b+c) + b/(a+c) + c/(a+b) < 2 $$I tried to solve this by triangle inequality but i was not able to get to the solution.
| Hint: let $x=a+b-c$, $y=a+c-b$, $z=b+c-a$. Clearly
$$a=\frac{x+y}2,\ b=\frac{x+z}{2},\ c=\frac{y+z}{2},\ a+b=\frac{2x+y+z}{2},\cdots$$
hence the original inequality reads
$$\frac{x+y}{x+y+2z}+ \frac{x+z}{x+2y+z}+\frac{y+z}{2x+y+z}<2.$$
Note that
$$\frac{x+y}{x+y+2z}+ \frac{x+z}{x+2y+z}+\frac{y+z}{2x+y+z}<\frac{x+y}{x+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How does does exponent property work on $\left(xe^{\frac{1}{x}}-x\right)$ How does
$\left(xe^{\frac{1}{x}}-x\right)$
become
$\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$
| With more intermediate steps :
$$\left(xe^{\frac{1}{x}}-x\right)=$$
$$=x\left(e^{\frac{1}{x}}-1\right)$$
Let $x=\frac{1}{t}$ Replace $x$ by $\frac{1}{t}$
$$=\frac{1}{t}\left(e^{\frac{1}{x}}-1\right)$$
$$=\frac{\left(e^{\frac{1}{x}}-1\right)}{t}$$
$t=\frac{1}{x}$ Replace $t$ by $\frac{1}{x}$
$$=\frac{\left(e^{\frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$
So Integral is $\displaystyle... | The answer in terms of elliptic integrals turns out to be very simple – but it requires a very sneaky trick.
As pointed out in Jack's answer, the integrals over $[0,1]$ and $[1,\infty]$ are the same, so
$$I=\int_0^\infty\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx=2\int_0^1\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx$$
The sneaky trick is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$ Then sum of all digits of $f(6)$
$(1):$ If $P(x)$ is a polynomial of Degree $4$ such that $P(-1) = P(1) = 5$ and
$P(-2)=P(0)=P(2)=2\;,$Then Max. value of $P(x).$
$(2):$ If $f(x)$ is a polynomial of degree $6$ with leading Coefficient $2009.$ Suppose
furthe... | For part 2, here's one way to simplify the differentiation. We have
$$f(x)-2x+1=2009\cdot (x-1)\cdot(x-2)\cdot (x-3)\cdot (x-4)\cdot (x-5)\cdot(x-r)$$
Substituting $x + 3 \to x$,
$$f(x + 3) - 2(x+3)+1 = 2009(x+2)(x+1)(x)(x-1)(x-2)(x+3-r)$$
$$f(x + 3) -2x - 5 = 2009x(x^2-1)(x^2-4)(x + 3 -r)$$
$$f(x + 3) - 2x - 5 = 2009(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$ What starting point would you recommend me for the one below?
$$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$
EDIT
Thanks to Felix Marin, we know the integral evaluates to
$$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$
| $\def\Li{{\rm{Li}_2\,}}$Denote the considered integral as $I$ and set $y=\pi x$, we have
\begin{equation}
I=\frac{1}{\pi^2}\int_0^\pi \frac{y\sin y}{\cos y}\,\ln(\sin y)\,dy
\end{equation}
Perform integration by parts by taking $u=y$, we have
\begin{align}
I&=-\left.\frac{y}{2\pi^2}\int\frac{\ln\left(1-\cos^2y\right)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 5,
"answer_id": 4
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How do I prove that a matrix is a rotation-matrix? I have to prove that this matrix is a rotation-matrix
$$\begin{pmatrix} \frac12 & 0 & \frac{\sqrt{3}}{2} \\
0 & 1 & 0 \\
\frac{\sqrt{3}}{2} & 0 & \frac12
\end{pmatrix}$$
How do I do this?
My idea is to multiplicate it with $\begin{pmatrix} x \\ y \\ z\end{pmatrix}$ an... | I think there is a minus sign missing. As it is, the determinant is not $1$. After fixing, this specific case is easy.
$$\begin{pmatrix} \frac12 & 0 & -\frac{\sqrt{3}}{2} \\
0 & 1 & 0 \\
\frac{\sqrt{3}}{2} & 0 & \frac12
\end{pmatrix} = \begin{pmatrix} \cos \frac{\pi}{3} & 0 & -\sin \frac{\pi}{3} \\ 0 & 1 & 0 \\ \sin \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Prove $ \lim_{x\to -\infty } \frac{x+8}{x+3} = 1 $ using only the definition of a limit I need to prove this limit:
$$ \lim_{x\to -\infty } \frac{x+8}{x+3} = 1 $$
I started with:
$|\frac{x+8}{x+3}-1|< ϵ $
$|\frac{x+8-x-3}{x+3}|=|\frac{5}{x+3}|=\frac{5}{|x+3|} <ϵ$
$\frac{5}{\epsilon} < |x+3|$
How do I proceed from he... | $$\begin{align*}
\frac5\epsilon &< \left|x+3\right|\\
\frac5\epsilon &< -\left(x+3\right)\\
x &< -\frac5\epsilon-3
\end{align*}$$
Hence, given $\epsilon>0$, we can take $M= -\frac5\epsilon-3$ such that $x<M$ implies $$\left|\frac{x+8}{x+3}-1\right|<\epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How does $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)}$ simplify? $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + ...$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... = 1$
My professor wrote this the other day. But I'm wondering...how does the series become $= 1 - \frac{1}{2} + \frac{1... | It is because $\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What's wrong with this argument? (Limits) In order to compute $\displaystyle \lim_{x \to \infty} \sqrt{9x^2 + x} - 3x$ we can multiply by the conjugate and eventually arrive at a limit value $1/6$.
But what about the line of reasoning below, what is wrong with the argument and why? I can't think of a simple explanatio... | The symbol $\approx $ is often misused. It can mean numerical approximation or it can mean asymptotically equal.
For example it is not true that (n+1) 2 $\approx $ n 2 in both senses. This you can see from the difference
${(n + 1)^2} - {n^2} = 1 + 2n$ .
You can observe that the difference is not small. Indeed it t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 6
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How can I evaluate $\lim_{n \to\infty}\left(1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)\right)/\left(1^2+2^2+3^2+\dots+n^2\right)^2$? How can I evaluate this limit? Give me a hint, please.
$$\lim_{n \to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)}{\left(1^2+2^2+3^2+\dots+n^2\right)^2}$$
| Hint:
$$1\cdot 2\cdot 3 + \dots +n(n+1)(n+2) \le n(n+2)^3 \le 8n^4$$
$$(1^2 + 2^2 + \dots + n^2)^2 = \left(\frac{n(n+1)(2n+1)}{6}\right)^2\ge \frac{1}{9}n^6$$
Hence
$$\frac{1\cdot 2\cdot 3 + \dots +n(n+1)(n+2)}{(1^2 + 2^2 + \dots + n^2)^2} \le \frac{8n^4}{1/9\cdot n^6}=\frac{72}{n^2}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$ How to solve this integral?
$$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$
my attempt
$$
\int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}=
\int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\
\r... | Both the function that you are integrating as the region over which you are integrating it get unchanged if you exchange $x$ with $y$. Therefore, your integral is equal to$$2\int_0^a\int_0^x\frac1{(a^2+x^2+y^2)^{3/2}}\,\mathrm dy\,\mathrm dx.$$You can compute this integral using polar coordinates: $\theta$ can take val... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Integral on sphere and ellipsoid Let $a,b,c \in \mathbb{R},$ $\mathbf{A}=\left[\begin{array}{*{20}{c}}
\mathbf{a}&{0}&{0}\\
{0}&\mathbf{b}&{0}\\
{0}&{0}&\mathbf{c}
\end{array}\right] , ~~\det A >1$
Let $~D = \{(x_1,x_2,x_3): x_1^2 + x_2^2 +x_3^2 \leq 1 \}~$ and
$~E = \left\{(x_1,x_2,x_3): \frac{x_1^2}{a^2} + \frac{x_2... | A function $~f~$ is said to be compactly supported if it is zero outside a compact set.
Let $~x=(x_1,x_2,x_3)\in\mathbb R^3~,$ be any arbitrary vector.
$$\therefore~~Ax=\left[\begin{array}{*{20}{c}}
\mathbf{a}&{0}&{0}\\
{0}&\mathbf{b}&{0}\\
{0}&{0}&\mathbf{c}
\end{array}\right]\left[\begin{array}{*{20}{c}}
{x_1}\\{x_2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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how many positive integer solutions to the following equation? $a^2 + b^2 + 25 = ab + 5a + 5b$
I have tried looking for a factorisation that could solve this question but couldn't find anything useful - found $(a+b+5)^2$ - don't know if this is useful
The equation does look similar to an equation of a circle - can you ... | The intuition is that the right-hand side is, with a few possible exceptins, smaller than the left-hand side.
Note that $(a-b)^2\ge 0$, so $ab\le \frac{a^2+b^2}{2}$. Thus the right-hand side is $\le \frac{a^2+b^2}{2}+5a+5b$.
It follows that
$$(a^2+b^2+25)-(ab+5a+5b)\ge \frac{a^2+b^2}{2}+25-5a-5b.$$
The right=hand sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Irrational number inequality : $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$ it is easy and simple I know but still do not know how to show it (obviously without simply calculating the sum but manipulation on numbers is key here.
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$$
| $2 \lt 3 \lt 4$ so $\sqrt{2} \lt \sqrt{3} \lt 2$ and
$$1+\frac1{\sqrt{2}}+\frac1{\
\sqrt{3}} \gt 1+\frac12+\frac12 = 2 \gt \sqrt{3}.$$
Of course that does not generalise in any nice way.
However $2 \lt \frac{100}{49}$ with $3 \lt \frac{49}{16}$ and $5 \lt \frac{81}{16}$
does yield $$1+\frac1{\sqrt{2}}+\frac1{\
\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
How to simpify $\cos x - \sin x$ How does one simplify
$$\cos x - \sin x$$
I tried multiplying by $\cos x + \sin x$, but that just gets me $$\cos x - \sin x = \frac{\cos 2x}{\cos x + \sin x}$$ which is worse.
Yet wolframalpha gives me $\cos x - \sin x = \sqrt{2}\sin\left(\dfrac{\pi}{4}-x\right)$. How does one obta... | $$
s = \cos x - \sin x \\
s^2 = \cos^2 x - 2 \cos x \sin x + \sin^2 x = 1 - \sin 2x \\
= 1 - \cos (\frac{\pi}2 -2 x)\\
= 1 - \left(1 - 2 \sin^2(\frac{\pi}4 - x)\right)\\
=2 \sin^2(\frac{\pi}4 - x)
$$
so
$$
s = \pm \sqrt{2} \sin(\frac{\pi}4 - x)
$$
and evaluating at $x=0$ shows that the positive sign must be taken
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Proving of $\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$ This is a homework for my son, he needs the proving.I tried to solve it by residue theory but I couldn't.
$$\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{14... | You can apply the residue theorem after a bit of playing with the sums:
\begin{align*}&\sum_{n=1}^\infty\frac{14}{576n^2-576n+95}-\sum_{n=1}^\infty\frac4{576n^2-576+140}=\\&\sum_{n=1}^\infty\left(\frac1{24n-19}-\frac1{24n-5}\right)-\sum_{n=1}^\infty\left(\frac1{24n-14}-\frac1{24n-10}\right)=\\&\sum_{n=1}^\infty\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$ This is the assignment I have:
Find a closed form for the equations
$1^3 = 1$
$2^3 = 3+5$
$3^3 = 7+9+11$
$4^3 = 13+15+17+19$
$5^3 = 21+23+25+27+29$
$...$
Hints. The equations are of the form $n^3 = a1 +a2 +···+an$, where
$a_{i+1} = a_... | $$n^3=\sum_{k=0}^{n-1}(n^2-(n-1)+2k)$$
Since
$\sum_{k=0}^{n-1}(n^2-(n-1)+2k)=n^3-n(n-1)+2(\frac{n(n-1)}{2})=n^3$
So it is the summation of $n$ consequitive odd number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Understanding a step in a double series proof I'm really confused, how do they get from the first line to the second line ?
$$\begin{align*}
S&=\frac12\left[\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2n}{3^m(n\cdot3^m+m\cdot3^n)}+\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{n^2m}{3^n(n\cdot3^m+m\cdot3^n)}\right]\\\\
&=\frac1... | Making a common denominator and factoring, observe that:
\begin{align*}
\frac{m^2n}{3^m(n \cdot 3^m + m \cdot 3^n)} + \frac{n^2m}{3^n(n \cdot 3^m + m \cdot 3^n)}
&= \frac{m^2n \cdot 3^n + n^2m \cdot 3^m}{3^m3^n(n \cdot 3^m + m \cdot 3^n)} \\
&= \frac{mn(m \cdot 3^n + n \cdot 3^m)}{3^{m+n}(n \cdot 3^m + m \cdot 3^n)} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate the following integral $\int_0^{\pi/2}\sin{x}\cos{x}\ln{(\sin{x})}\ln{(\cos{x})}\,dx$? How to evaluate the following integral
$$\int_0^{\pi/2}\sin{x}\cos{x}\ln{(\sin{x})}\ln{(\cos{x})}\,dx$$
It seems that it evaluates to$$\frac{1}{4}-\frac{\pi^2}{48}$$
Is this true? How would I prove it?
| Find this
$$I=\int_{0}^{\frac{\pi}{2}}\sin{x}\cos{x}\ln{(\cos{x})}\ln{(\sin{x})}dx$$
Solution
Since
$$\sin(2x) = 2\sin(x)\cos(x)$$
then
$$I=\dfrac{1}{8}\int_{0}^{\frac{\pi}{2}}\ln{(\sin^2{x})}
\ln{(\cos^2{x})}\sin{(2x)}dx$$
Let $\cos{(2x)}=y$, and since
$$\cos(2x) = 2\cos^2x - 1 = 1 - 2\sin^2x$$
we get
$$I=\dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
How can I differentiate this equation? $y = \sqrt[4]{\frac{(x^3+2\sqrt{x})^2(x-sinx)^5}{(e^{-2x}+3x)^3}}$ $y = \sqrt[4]{\frac{(x^3+2\sqrt{x})^2(x-sinx)^5}{(e^{-2x}+3x)^3}}$
I tried removing the root but that got me no where
| the solution of the given problem should be this here
$$\frac{\left(x^3+2 \sqrt{x}\right) (x-\sin (x))^4 \left(-3 \left(3-2 e^{-2
x}\right) \left(x^3+2 \sqrt{x}\right) (x-\sin (x))+5 \left(3 x+e^{-2 x}\right)
\left(x^3+2 \sqrt{x}\right) (1-\cos (x))+2 \left(3 x+e^{-2 x}\right) \left(3
x^2+\frac{1}{\sqrt{x}}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integral of $\frac{1}{\sqrt{x^2-x}}dx$ For a differential equation I have to solve the integral $\frac{dx}{\sqrt{x^2-x}}$. I eventually have to write the solution in the form $ x = ...$ It doesn't matter if I solve the integral myself or if I use a table to find the integral. However, the only helpful integral in an in... | $$\int \frac{dx}{\sqrt{x^2-x}} = \int \frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2 - \frac{1}{4}}}$$
Setting $ x - \frac{1}{2} = t $
$$ \int \frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2 - \frac{1}{4}}} = \int \frac{dt}{\sqrt{t^2 - \frac{1}{4}}} $$
It's possible to do this using a trig substitution, but if you want the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 5
} |
Show that $x^a+x-b=0$ must have only one positive real root and not exceed the $\sqrt[a]{b-1}$ If we take the equation $$x^3+x-3=0$$ and solve it to find the real roots, we will get only one positive real roots which is $(x=1.213411662)$. If we comparison this with $\sqrt[3]{3-1}=1.259921$, we will find that $x$ is les... | If $f(x)=x^a+x-b$ then $f'(x)=ax^{a-1}+1>0,$ since $a\in\mathbb{R}_+.$ Thus $f$ is strictly increasing and so it must take the value $0$ at most one time. Now, since $f(0)=-b<0$ and $f(b)>0$ it follows from the Intermediate Value Theorem that there exists a root in the interval $(0,b).$ So, $f$ has exactly one root.
Fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding Exact Value $7\csc(x)\cot(x)-9\cot(x)=0$ The values for $x$ on $[0,2\pi)$ solving $7\csc(x)\cot(x)-9\cot(x)=0$ are?
I think that $\dfrac{\pi}2$ is one but I can't find the others. what are the others?
| $7\csc x\cot x-9\cot x=0\implies7\cos x-9\cos x\sin x=0\implies (\cos x)(7-9\sin x)=0$,
so $\cos x=0\implies x=\frac{\pi}{2}$ or $x=\frac{3\pi}{2}$
and $\sin x=\frac{7}{9}\implies x=\sin^{-1}\frac{7}{9}$ or $x=\pi-\sin^{-1}\frac{7}{9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $ \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ Evaluation of $\displaystyle \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ where $\lfloor x \rfloor $ represent floor function of $x$.
My Try:: Here $\displaystyle f(x) = \frac{x^2}{\sin x\cdot \tan x}$ is an... | Let me continue where you stopped in your post.
Expand the denominator for a few terms and perform the long division. You should arrive to $$\frac{x^2}{\sin x\cdot \tan x} =1-\frac{x^2}{6}-\frac{7 x^4}{120}+O\left(x^5\right)$$ which is definitely smaller than $1$.
I am sure that you can take from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding matrix for given recurrence For the recurrence relation:
$f(0)=1$
$f(1)=1$
$f(2)=2$
$f(2n)=f(n)+f(n+1)+n$
$f(2n+1)=f(n)+f(n−1)+1$
How to find square matrices $M_0, M_1$ and vectors $u, v$ such that if the base-2 expansion of $n$ is given by $e_1 e_2 \cdots e_j$, then $$f(n) = u M_{e_1} \cdots M_{e_j} v.$$ ??
| This is how I would approach this question, and I guess my answer would not be unique.
For each $n = e_1e_2\cdots e_j$, there are $7$ numbers I should store (as a row vector) to calculate $f(2n)$ and $f(2n+1)$:
$$uM_{e_1}M_{e_2}\cdots M_{e_j} = w(n) = \pmatrix {1&n&f(n-2)&f(n-1)&f(n)&f(n+1)&f(n+2)}.$$
To calculate $f(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $2\cos^2 x-2\sin^2 x-2\cos x=0$ $$f(x) = 2\cos^2 x-2\sin^2 x-2\cos x$$
Need values of x that which make $f(x) = 0$
Tried $a^2-b^2 = (a+b)(a-b)$ with no luck
Really just need a hint that could bring me in the right direction
Thanks
EDIT: Solution thanks to everyones help! :D
$$f(x) = 2\cos^2 x-2\sin^2 x-2\cos x... | Hint:
Use $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\sin^2x=1-\cos^2x$
Update:
$$\cos x=-\frac{1}{2},1$$
$$\cos x=-\dfrac12=\cos\left(\frac{2\pi}3\right)\implies x=2k\pi\pm\dfrac{2\pi}3$$
$$\cos x=1\implies x=2k\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| Ill show an "tricky" method.
$\displaystyle \frac{x^6 - 1}{x-1}$
$= \displaystyle \frac{x^6 -x + x - 1}{x-1} = \frac{x^6 - x}{x-1} + 1 = \frac{x^6 - x^5 + x^5 - x}{x-1} + 1 = x^5 + 1 + \frac{x^5 - x}{x-1} = \frac{x^5 - x^4 + x^4 - x}{x-1} + x^5 + 1 = \frac{x^4(x - 1) + x^4 - x}{(x-1)}$
Do you see the pattern?
This i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 12,
"answer_id": 5
} |
Evaluate the integral by using Gauss divergence theorem. Evaluate $\int\int_SF.dS$ where $F=(xz,yz,x^2+y^2)$
by using the Gauss divergence theorem.
Where $S$ is the closed surface obtained from the surfaces $x^2+y^2\leq 4,z=2,x^2+y^2\leq 16,z=0$ on the top and the bottom and $z=4-\sqrt{x^2+y^2}$ on the side.
My calcula... | Anyway, this is what I get:
$$2 \int\int\int z dV = 2\int_0^2 \int_0^{4-z} \int_0^{2\pi} z r\ d\theta dr dz = 4\pi \int_0^2 \int_0^{4-z} zr\ dr dz$$
$$ = 2\pi \int_0^2 z(4-z)^2 dz = 2\pi \int_0^2 \bigg(z^3 - 8z^2 + 16z\bigg)dz$$
$$ = 2\pi \bigg( \frac{2^4}{4} - \frac{16(2^3)}{3} + \frac{16(2^2)}{2}\bigg) = \frac{88\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\lim_{x→0}\left(\frac{1+\tan x}{1+\sin x}\right)^{1/x^2} $ I have the following limit to evaluate:
$$ \displaystyle \lim_{x→0}\left(\frac{1+\tan x}{1+\sin x}\right)^{1/x^2} $$
What's the trick here?
| $$\lim_{x\to 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{x^2}}$$ $$=\exp\left(\lim_{x\to 0}{\frac{1}{x^2}}\left(\frac{1+\tan x}{1+\sin x}-1\right)\right)$$
$$=\exp\left(\lim_{x\to 0}{\frac{1}{x^2}}\left(\frac{\tan x-\sin x}{1+\sin x}\right)\right)$$
$$=\exp\left(\lim_{x\to 0}{\frac{1}{x^2}}\left(\frac{\frac{2tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Closed-form of sums from Fourier series of $\sqrt{1-k^2 \sin^2 x}$ Consider the even $\pi$-periodic function $f(x,k)=\sqrt{1-k^2 \sin^2 x}$ with Fourier cosine series $$f(x,k)=\frac{1}{2}a_0+\sum_{n=1}^\infty a_n \cos2nx,\quad a_n=\frac{2}{\pi}\int_0^{\pi} \sqrt{1-k^2 \sin^2 x}\cos 2nx \,dx.$$ This was considered in th... | We have:
$$ \frac{\pi}{2}\,a_n=\int_{0}^{\pi}\sqrt{1-k^2\sin^2\theta}\cos(2n\theta)\,d\theta =\frac{k^2}{4n}\int_{0}^{\pi}\frac{\sin(2\theta)\sin(2n\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta.\tag{1}$$
If we set:
$$ b_m = \int_{0}^{\pi}\frac{\cos(2m\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta,\qquad c_m = \int_{0}^{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such th... | Keep going!
$64\cdot 2^{3k-1} + 45\cdot 3^k = 9(2^{3k-1} + 5\cdot3^k) + 55\cdot2^{3k-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 1
} |
Evaluate $\int_0^{\pi/2}x\cot{(x)}\ln^4\cot\frac{x}{2}\,\mathrm dx$ How to evaluate the following integral ?:
$$
\int_{0}^{\pi/2}x\cot\left(\, x\,\right)\ln^{4}\left[\,\cot\left(\,{x \over 2}\,\right)\,\right]\,{\rm d}x
$$
It seems that evaluate to
$$
{\pi \over 16}\left[\,
5\pi^{4}\ln\left(\, 2\,\right) - 6\pi^{2}\zet... | Let $$J = \int_0^1 {\frac{{\arctan x{{\ln }^4}x}}{x}dx} \qquad K = \int_0^1 {\frac{x{\arctan x{{\ln }^4}x}}{{1 + {x^2}}}dx}$$
Then by M.N.C.E.'s comment, $$\tag{1}I = \int_0^{\pi /2} {x\cot x{{\ln }^4}\left( {\cot \frac{x}{2}} \right)dx} = 2J - 4K$$
Here is a symmetry of the integrand that we can exploit:
$$\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1063275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Diagonalization and find matrix that corresponds to the given condition Diagonalize the matrix
$$
A=
\begin{pmatrix}
1 & 2\\
0 & 3
\end{pmatrix}
$$
and find $B^3=A$.
I derived $A \sim \text{diag}(1,3)$ but I have problem finding any $B$. I tried to solve it by writing $B= \begin{pmatrix} 1 & x\\ 0 & 3\end{pmatrix}$, bu... | The eigen values are $1,3 $ clearly. So it is diagonalizabe(distinct eigen values).
And so, there exits $P$ such that
$$A=P\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)P^{-1}.$$
Now we need $B$ such that $B$ such that $B^3=A$
Supose there exists such a $B$ then, $$B^3=A=P\left(\begin{array}{cc}1& 0\\ 0& 3\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Probability of selecting a red ball first An urn contains 3 red and 7 black balls. Players A
and B withdraw balls from the urn consecutively
until a red ball is selected. Find the probability that
A selects the red ball. (A draws the first ball, then
B, and so on. There is no replacement of the balls
drawn.)
How do I c... | Here's what I was thinking.
$$\color{RED}R+BB\color{RED}R+BBBB\color{RED}R+BBBBBB\color{RED}R$$
$$P(A)=\frac{3}{10}+\frac{7}{10}\frac{6}{9}\frac{3}{8}+\frac{7}{10}\frac{6}{9}\frac{5}{8}\frac{4}{7}\frac{3}{6}+\frac{7}{10}\frac{6}{9}\frac{5}{8}\frac{4}{7}\frac{3}{6}\frac{2}{5}\frac{3}{4}=\frac{7}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Question about sines of angles in an acute triangle
Let $\triangle ABC$ be a triangle such that each angle is less than $ 90^\circ $.
I want to prove that $\sin A + \sin B + \sin C > 2$.
Here is what I have done:
Since $A+B+C=180^{\circ}$ and $0 < A,B,C < 90^\circ$, at least two of $A,B,C$ are in the range 45 < x <... | i am able to simplify $$
\sin A + \sin B + \sin (A + B) = \sin A + \sin B + \sin A \cos B + \sin B \cos A \\ = (1+\cos B)\sin A + (1 + \cos A)\sin B
\\ = 4\cos^2 B/2\sin A/2 \cos A/2 + 4\cos^2 A/2 \sin B/2 \cos B/2
\\ = 4\cos B/2 \cos A/2(\sin A/2 \cos B/2 + \sin B/2 \cos A/2)
\\ = 4\cos B/2 \cos A/2\sin (A/2 + B/2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this a legitimate proof? If not, how to prove? Question: Determine all natural numbers $n$ such that: $7 \mid \left(3^n - 2\right)
\implies3^{n}\equiv 2\pmod{7}$
Multiply both sides by 7
$7 \cdot 3^{n}\equiv 7\cdot2\pmod{7}$
Divide both sides by seven, since $\gcd(7,7) = 7$, we have to divide modulus by $7$
$\implie... | Noting that $n=1$ does not work, let $n \ge 2$. Then as $3^2 \equiv 2 \pmod 7$, we have the equivalent statement
$$2\cdot 3^{n-2} \equiv 2 \pmod 7 \iff3^{n-2}\equiv 1 \pmod 7$$
Now that has solutions $n = 6k+2$ as $3^6$ is the smallest positive power of $3$ that is $\equiv 1 \pmod 7$, so the solution is for natural nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How many positive integers of n digits chosen from the set {2,3,7,9} are divisible by 3? I'm preparing myself for math competitions. And I am trying to solve this problem from the Romanian Mathematical Regional Contest “Traian Lalescu’', $2003$:
Problem $\mathbf{7}$: How many positive integers of $n$ digits chosen fro... | Here is an alternative approach. Let $x_n,y_n$, and $z_n$ be as in the argument given in the question; clearly $x_1=2$, and $y_1=z_1=1$. For $n\ge 1$ let $X_n,Y_n$, and $Z_n$ be the sets of $n$-digit numbers using only the digits $2,3,7$, and $9$ and congruent modulo $3$ to $0,1$, and $2$, respectively (so that $x_n=|X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Calculate the distance between intersection points of tangents to a parabola
*
*Question
Tangent lines $T_1$ and $T_2$ are drawn at two points $P_1$ and $P_2$ on the parabola $y=x^2$ and they intersect at a point $P$. Another tangent line $T$ is drawn at a point between $P_1$ and $P_2$; it intersects $T_1$ at $Q_1$ a... | Suppose that the third tangent is drawn at a point $A$ with coordinates $A(a, a^2)$. Then its tangent intersects $T_{i}$ at
$$
Q_{i}\left(\frac{P_{ix} + a}{2}, P_{ix}a \right)
$$
using your equation $(3)$. In other words,
$$
2(Q_{1x} - Q_{2x})=P_{1x}-P_{2x}
$$
Therefore, in the argument of the square root in the numer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to solve a linear system in matrix form using Laplace transform? How to solve this linear system using Laplace transform?
$$\mathbf X'(t)=\left[\begin{array}{r,r,r}-3&0&2\\1&-1&0\\-2&-1&0\end{array}\right]\mathbf X(t); ~~~~~~~~\mathbf X(0)=\left[\begin{array}{r}4\\-1\\2\end{array}\right]$$
I am struggling with this... | We are given:
$$X'(t) = \begin{bmatrix} -3 & 0 & 2 \\ 1 & -1 & 0\\ -2 & -1 & 0\end{bmatrix} \begin{bmatrix} x(t) \\ y(t)\\ z(t)\end{bmatrix}, ~~ X(0) = \begin{bmatrix} 4 \\ -1\\ 2\end{bmatrix}$$
We can write this as:
$$\tag 1 \begin{align} x' &= -3x + 2z \\ y' &= x-y \\ z' &= -2x - y \end{align}$$
Taking the Laplace tr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating $\int_{0}^{\pi/2}\frac{x\sin x\cos x\;dx}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}$ How to evaluate the following integral
$$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$
For integrating I took $\cos^{2}x$ outside and applied integration by parts.
Given answer is $\dfrac{\pi}{4ab^... | let $x=\dfrac{t}{2}$, we have
$$I=\dfrac{1}{2}\int_{0}^{\pi}\dfrac{\dfrac{t}{2}\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}}{\left(a^2\sin^2{\dfrac{t}{2}}+b^2\cos^2{\dfrac{t}{2}}\right)^2}dt=\dfrac{1}{2}\int_{0}^{\pi}\dfrac{t\sin{t}}{[(a^2+b^2)+(a^2-b^2)\cos{t}]^2}dt$$
So
\begin{align*}I&=-\dfrac{1}{2(a^2-b^2)}\int_{0}^{\pi}t\... | {
"language": "en",
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"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Series sum $\sum 1/(n^2+(n+1)^2)$ In an exercise, I caculate the Fourier expansion of $e^x$ over $[0,\pi]$ is
$$e^x\sim \frac{e^\pi-1}{\pi}+\frac{2(e^\pi-1)}{\pi}\sum_{n=1}^\infty \frac{\cos 2nx}{4n^2+1}+\frac{4(1-e^\pi)}{\pi}\sum_{n=1}^\infty \frac{n\sin 2nx}{4n^2+1}.$$
From this, it is easy to deduce
$$\sum_{n=1}^\i... | We can approach such kind of series by considering logarithmic derivatives of Weierstrass products. For instance, from:
$$\cosh z = \prod_{n=1}^{+\infty}\left(1+\frac{4z^2}{(2n-1)^2\pi^2}\right)\tag{1}$$
we get:
$$\frac{\pi}{2}\tanh\frac{\pi z}{2} = \sum_{n=1}^{+\infty}\frac{2z}{z^2+(2n-1)^2}\tag{2},$$
so, evaluating i... | {
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"source": "stackexchange",
"question_score": "3",
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Proving $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}>2-\frac{2}{n}$ by induction for $n\geq 1$ I have the following inequality to prove with induction:
$$P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots\frac{1}{\sqrt{n}}>2-\frac{2}{n}, \forall n\in \mathbb{\:N... | Initial comment: Begin by noting that, for all $n\geq 1$, we have that
$$
n(\sqrt{n}-2)+2>0\Longleftrightarrow n\sqrt{n}-2n+2>0\Longleftrightarrow \color{red}{\sqrt{n}>2-\frac{2}{n}}.\tag{1}
$$
Thus, it suffices for us to prove the proposition $P(n)$ for all $n\geq 1$ where
$$
P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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In triangle ABC, Find $\tan(A)$.
In triangle ABC, if $(b+c)^2=a^2+16\triangle$, then find $\tan(A)$ . Where $\triangle$ is the area and a, b , c are the sides of the triangle.
$\implies b^2+c^2-a^2=16\triangle-2bc$
In triangle ABC, $\sin(A)=\frac{2\triangle}{bc}$, and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$,
$\implies \ta... | HINT:
$$16\triangle=(b+c+a)(b+c-a)$$
$$\iff16rs=2s(b+c-a)$$
$$8r=b+c-a$$
Using this and $a=2R\sin A$ etc.,
$$8\cdot4R\prod\sin\dfrac A2=2R\cdot4\cos\dfrac A2\sin\dfrac B2\sin\dfrac C2$$
$$\implies4\sin\dfrac A2=\cos\dfrac A2$$ as $0<B,C<\pi,\sin\dfrac B2\sin\dfrac C2\ne0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Inverse Laplace Transformation I have a question about laplace transformation.
$\frac{8s+4}{s^2+23}$
I tried to split them. $\frac{8s}{s^2+23}$ is the image of a cosine and $\frac{4}{s^2+23}$ is the image of a sine.
Here is what I did :
$\frac{8s}{s^2+(\sqrt{23})^2}$ is the image of $8\cos(\sqrt{23}t)$ and $\fr... | If you don't mind some Residue theory, we can check use that to check your solution.
\begin{align}
\mathcal{L}^{-1}\biggl\{\frac{8s+4}{s^2+23}\biggr\}&=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{8s+4}{s^2+23}e^{st}ds\\
&=\sum\text{Res}
\end{align}
The poles in the $s$-plane occur at $s=\pm i\sqrt{23}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073214",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^{2} \frac{dx}{\sqrt[3]{2x^2-x^3}}$ How to calculate this integral?
$$\int_0^{2} \frac{dx}{\sqrt[3]{2x^2-x^3}}$$
I suppose that it should be parted like this: $$\int_0^{1} \frac{dx}{\sqrt[3]{2x^2-x^3}} + \int_1^{2} \frac{dx}{\sqrt[3]{2x^2-x^3}}$$ but I have no idea how to calculate these two.
Thank yo... | $\sqrt[3]{2x^2-x^3} = x\sqrt[3]{\dfrac{2}{x}-1} \to u = \sqrt[3]{\dfrac{2}{x} - 1} \to u^3 = \dfrac{2}{x} - 1 \to x = \dfrac{2}{u^3+1}$. Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate this infinite product involving $a_k$ Let $a_0 = 5/2$ and $a_k = a_{k-1}^{2} - 2$ for $k \ge 1$ Compute:
$$\prod_{k=0}^{\infty} 1 - \frac{1}{a_k}$$
Off the bat, we can seperate $a_0$
$$= -3/2 \cdot \prod_{k=1}^{\infty} 1 - \frac{1}{a_k}$$
Lets see:
$a_0 = 5/2$
$a_1 = 25/4 - 2 = 17/4$
$a_2 = 289/16 - 2 = 257/16... | $\textbf{Hint:}$
show this by induction on $n\ge0$
$$a_n=2^{2^n}+2^{-2^n},\quad n=0,1,2\dots$$
and notice that $a_k+1=a_{k-1}^2-1=(a_{k-1}-1)(a_{k-1}+1),$
$$1-\frac{1}{a_k}=\frac{a_k-1}{a_k}=\frac{a_{k+1}+1}{a_k+1}\cdot\frac{1}{a_k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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3D coordinates of circle center given three point on the circle. Given the three coordinates $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $(x_3, y_3, z_3)$ defining a circle in 3D space, how to find the coordinates of the center of the circle $(x_0, y_0, z_0)$?
| In this formula
the center $O$ of
a circumscribed circle of $\triangle ABC$
is expressed as a convex combination of its vertices
in terms of
coordinates $A,B,C$ and corresponding side lengths $a,b,c$,
suitable for both 2d and 3d:
\begin{align}
O&=
A\cdot \frac{a^2\,(b^2+c^2-a^2)}{((b+c)^2-a^2)(a^2-(b-c)^2)}
\\
&+B\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 11,
"answer_id": 7
} |
Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$ in real numbers
Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$
The problem does not say it but I think solutions should be from $\mathbb{R}$. I tried to express the left sum as a sum of squares but that does not work out. Any suggestions?
| You can solve for $x$:
$(2)x^2+(6y-2)x+(5y^2-4y+1)=0\implies$
$x_{1,2}=\frac{-(6y-2)\pm\sqrt{(6y-2)^2-4\cdot2\cdot(5y^2-4y+1)}}{2\cdot2}=\frac{-6y+2\pm\sqrt{-4y^2+8y-4}}{4}=\frac{-6y+2\pm\sqrt{-4(y-1)^2}}{4}=\frac{-6y+2\pm2i(y-1)}{4}$
Then the only real solution is with $y=1$, hence $x=-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
No. of different real values of $x$ which satisfy $17^x+9^{x^2} = 23^x+3^{x^2}.$ Number of different real values of $x$ which satisfy $17^x+9^{x^2} = 23^x+3^{x^2}.$
$\bf{My\; Try::}$Using Hit and trial $x=0$ and $x=1$ are solution of above exponential equation.
Now we will calculate any other solution exists or not.
I... | Using derivatives, is studying functions
$ f, g: R \rightarrow R, f(x)= 9^{x^2}-3^{x^2}, g(x)=23^x-17^x$ and is found:
*
*$f$ has a minimum point in the interval $(0, 1)$ and limits to $+\infty$,$-\infty$ are equal with $+\infty$;
*$g$ has a negative minimum point and limited to $-\infty$ is $0$ and to $+\infty$ i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Prove $4^k - 1$ is divisible by $3$ for $k = 1, 2, 3, \dots$ For example:
$$\begin{align}
4^{1} - 1 \mod 3 &=
\\
4 -1 \mod 3 &=
\\
3 \mod 3 &=
\\3*1 \mod 3 &=0
\\
\\
4^{2} - 1 \mod 3 &=
\\
16 -1 \mod 3 &=
\\
15 \mod 3 &=
\\3*5 \mod 3 &= 0
\\
\\
4^{3} - 1 \mod 3 &=
\\
64 -1 \mod 3 &=
\\
21 \mod 3 &=
\\3*7 \mod 3 &=
... | Hint:
$$4 \equiv 1 \mod 3 \Rightarrow 4^n \equiv 1^n \mod 3 \Rightarrow 4^n \equiv 1 \mod 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 2
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Power series for the rational function $(1+x)^3/(1-x)^3$
Show that $$\dfrac{(1+x)^3}{(1-x)^3} =1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$$
I tried with the partial frationaising the expression that gives me
$\dfrac{-6}{(x-1)} - \dfrac{12}{(x-1)^2} - \dfrac{8}{(x-1)^3} -1$
how to proceed further on this havi... | The most simple way to prove your identity, IMHO, is to multiply both sides by $(1-x)^3$.
This leads to:
$$ 1+3x+3x^2+x^3\stackrel{?}{=}(1-3x+3x^2+x^3)\left(1+\sum_{n\geq 1}(4n^2+2)\,x^n\right).\tag{1}$$
If we set $a_n=(4n^2+2)$, for any $n\geq 4$ the coefficient of $x^n$ in the RHS is given by $a_n-3a_{n-1}+3a_{n-2}+... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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AlgebraII factoring polynomials equation: $2x^2 - 11x - 6$
Using the quadratic formula, I have found the zeros: $x_1 = 6, x_2 = -\frac{1}{2}$
Plug the zeros in: $2x^2 + \frac{1}{2}x - 6x - 6$
This is where I get lost. I factor $-6x - 6$ to: $-6(x + 1)$, but the answer says otherwise. I am also having trouble factoring ... | When you factor out the equation $2x^2-11x-6$, you get $(x-6)(2x+1)$ (David Peterson did the factoring process). This shows that the functions has two zeros in the graph. Thus, we have to sent $(x-6)(2x+1)$ equal to $0$: $$(x-6)(2x+1)=0.$$ Then we find the zeros:
$$(x-6)(2x+1)=0$$
$$x-6=0$$
$$\boxed{x=6}$$
$$2x+1... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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prove that $a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$ Good morning
help me to show the following inequality
for all $a$, $b$ two positive real numbers
$$a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$$
thanks you
| Let $a+b=2u$ and $ab=v^2$, where $v>0$. Hence, we need to prove that $2v^4u^2-(v^2-1)u-v^6-v^4\geq0$, for which it's enough to prove that $u\geq\frac{v^2-1+\sqrt{(v^2-1)^2+8v^4(v^4+v^6)}}{4v^4}$ or $(4v^5-v^2+1)^2\geq(v^2-1)^2+8v^4(v^4+v^6)$ because $u\geq v$, or $(v-1)^2(v+1)(v^2+v+1)\geq0$. Done!
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Simplification a trigonometric equation $$16 \cos \frac{2 \pi}{15} \cos\frac{4 \pi}{15} \cos\frac{8 \pi}{15} \cos\frac{14 \pi}{15}$$
$$=4\times 2 \cos \frac{2 \pi}{15} \cos\frac{4 \pi}{15} \times2 \cos\frac{8 \pi}{15} \cos\frac{14 \pi}{15}$$
I am intending in this way and then tried to apply the formula, $2\cos A \cos ... | Using the identity, $\cos\theta\cos2\theta\cos2^2\theta\cdots\cos2^{n-1}\theta=\dfrac{\sin2^n\theta}{2^n\sin\theta}$
By putting $n=4$ and $\theta=\dfrac{\pi}{15}$ you will get,
$- \cos\dfrac{\pi}{15}\cos \dfrac{2 \pi}{15} \cos\dfrac{4 \pi}{15} \cos\dfrac{8 \pi}{15}=-\dfrac{\sin2^4\dfrac{\pi}{15}}{2^4\sin\dfrac{\pi}{15... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Irrational number not ocurring in the period of rational numbers Write each rational number from $(0,1]$ as a fraction $a/b$ with $\gcd(a,b)=1$, and cover $a/b $ with the interval
$$
\left[\frac ab-\frac 1{4b^2}, \frac ab + \frac 1{4b^2}\right].
$$
Prove that the number $\frac 1{\sqrt{2}}$ is not covered. What I did w... | I got a good solution from one of my teachers...$$\left|\frac ab-\frac 1{\sqrt2}\right|\left(\frac ab+\frac 1{\sqrt2}\right)=\left|\frac {a^2}{b^2}-\frac 12\right|=\frac {|2a^2-b^2|}{2b^2}\gt\frac 1{2b^2}$$ Also we know that $\frac ab+\frac 1{\sqrt2}\lt2=>\left|\frac ab-\frac 1{\sqrt2}\right|\gt\frac 1{4b^2}$. Hence pr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the value of : $\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$ I need to calculate the limit of the function below:
$$\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$$
I tried multiplying by the conjugate, substituting $x=\frac{1}{t^4}$, and both led to nothing.
| Multiply both numerator and denominator by $\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}$
You will get $$\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}$$
Divide both numerator and denominator by $\sqrt{x}$
$$\dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{1}{x\sqrt{x}}}}+1}$$
On finding th... | {
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"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 0
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Functional inequalities Let x,y,z be the lengths of the sides of a triangle, and let$$f(x,y,z)=\left|\frac {x-y}{x+y}+\frac {y-z}{y+z}+\frac {z-x}{z+x}\right|.$$ Find the upper limit of $f(x,y,z)$. I simply used the fact that $|x-y|\le z$ and the other 3 to prove that $f(x,y,z)\le \frac 18=0.125$. But the answer given ... | Solution:let $x\ge y\ge z$, First Note
$$I=\dfrac{x-y}{x+y}+\dfrac{y-z}{y+z}+\dfrac{z-x}{z+x}=\dfrac{(x-y)(x-z)(y-z)}{(x+y)(y+z)(x+z)}$$
then let $$x=c+b,y=c+a,z=a+b,c\ge b\ge a>0$$
so
$$I=\dfrac{(c-b)(c-a)(b-a)}{(2a+b+c)(2b+a+c)(2c+a+b)}<\dfrac{(c-b)cb}{(b+c)(2b+c)(2c+b)}=\dfrac{1}{F}$$
then we only find $F$ minimum
l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Recurrence Relation Involving the gamma Function I'm having some doubts about my approach to the following problem. I am given that the function $k(z)$ is defined such that,
$$k(z)=\Gamma\left(\frac{1}{2}+z\right)\Gamma\left(\frac{1}{2}-z\right)\cos{\pi z}$$
I'm required to find the recurrence relation linking $k(z+1)$... | You made a sign error.
\begin{align}
\frac{k(n+1)}{k(n)}
&=\frac{\Gamma\left(\frac{1}{2}+n+1\right)\Gamma\left(\frac{1}{2}-n\color{red}{-1}\right)(-\cos(\pi n))}{\Gamma\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}-n\right)\cos(\pi n)}\\
&=\frac{\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}+n\right)\frac{\Gamma\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083502",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Is $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$? Plotting both functions $\ln(1+\frac{1}{x-1})$ and $\frac{1}{x}$ in $[2,\infty)$ gives the impression that $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$.
Is it possible to prove it?
| hint :
$$f(x)=ln(1+\frac{1}{x-1})-\frac{1}{x}\\x≥2\\ f'<0$$f(x) is decreasing function ,but f(x) is above the x axis $$\\f(2)>0,f(\infty)>0\\f(x)>0\\so\\ln(1+\frac{1}{x-1})-\frac{1}{x}≥0\\ln(1+\frac{1}{x-1})≥\frac{1}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How find the maximum possible length of OC, where ABCD is a square, and AD is the chord of the circle? Given a circle $o(O(0,0), r=1)$. How to find the maximum possible length of $OC$, where $ABCD$ is a square, and $AD$ is the chord of the circle?
I have no idea how to do this, can this be proved with simple geometry?
|
With the diagram as labeled, we see that
$$\begin{align}
|\overline{OC}|^2 &= \cos^2\theta + ( \sin\theta+2\cos\theta )^2 \\
&= \cos^2\theta + \sin^2\theta + 4 \cos\theta\sin\theta + 4 \cos^2\theta \\
&= 3+2 \sin 2\theta + 2 \cos 2\theta \\
&= 3+2\sqrt{2}\left( \sin 2\theta \cos45^\circ + \cos 2\theta \sin45^\circ \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove that $ \sum_{n=1}^\infty \ln\big(n\sin \frac{1}{n}\big)$ converges.
Prove that $\displaystyle \sum_{n=1}^\infty \;\ln\left(n\sin\frac{1}{n}\right)$ converges.
My Work:
$$\left|\ln \left(n \sin \frac{1}{n}\right)\right| \leq\left|\ln \left(n \sin \frac{1}{n^{2}}\right)\right| \leq\left|\ln \left(\sin \frac{1}{n^... | Let's amply use Taylor series to give the leading orders.
Note that $\sin\left( \frac{1}{n} \right) \approx \frac{1}{n} - \frac{1}{6n^3}$. So $n \sin \left( \frac{1}{n} \right) \approx 1 - \frac{1}{6n^2}$.
Note also that $\ln(1 - x) \approx x + x^2 + \dots$, so that
$$\ln\left( n \sin \frac{1}{n} \right) \approx \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$ Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$
I seem to struggle with this simple proof. All I managed to find is that ac+bd=-4 (which might not even be correct).
| 1st Method
$\begin{align}\left(a^2+b^2\right)\left(c^2+d^2\right)=1& \implies (ac+bd)^2+(ad-bc)^2=1\\&\implies (ac+bd)^2\le1\end{align}$
2nd Method
$\begin{align}\left(a^2+b^2\right)+\left(c^2+d^2\right)=2& \implies \left(a^2+c^2\right)+\left(b^2+d^2\right)=2\\&\implies 2(ac+bd)\le 2\qquad \text{(by A.M.- G.M. Inequal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
How do I integrate: $\int\sqrt{\frac{x-3}{2-x}} dx$? I need to solve:
$$\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$
What I did is:
Substitute: $x=2\cos^2 \theta + 3\sin^2 \theta$. Now:
$$\begin{align}
x &= 2 - 2\sin^2 \theta + 3 \sin^2 \theta \\
x &= 2+ \sin^2 \theta \\
\sin \theta &= \sqrt{x-2} \\
\theta &=\sin^{-1}\sqrt{... |
$$I=\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$
Integrating
Let $x=2\cos^2t+3\sin^2t$, $dx=\sin2tdt$
$$I=\int\sqrt{\frac{-\cos^2t}{-\sin^2t}}\sin2tdt=\int2\cos^2tdt=\int(1+\cos2t)dt=t+\frac12\sin2t+c\\I=\underbrace{\cos^{-1}\sqrt{3-x}}_{\pi/2-\sin^{-1}\sqrt{3-x}}+\sqrt{x-2}\sqrt{3-x}+c\\I=\underbrace{\sqrt{x-2}\sqrt{3-x}}_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Limit involving square roots, more than two "rooted" terms The limit is
$$\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right)$$
which has a value of $\dfrac{27}{4}$.
Normally, I would know how to approach a limit of the form
$$\lim_{x\to\infty}\left( \sqrt{a_1x^2+b_1x+c_1}\pm\sqrt{a_2x^2+b_2... | Rewrite it as:
$$\left(\sqrt{x^2+5x-2} - \left(x+\frac{5}{2}\right)\right) -\\ \left(\sqrt{4x^2-3x+7}-\left(2x-\frac{3}{4}
\right)\right) +\\
\left(\sqrt{x^2+7x+5}-\left(x+\frac{7}{2}\right)\right)+\left(\frac{5}{2}+\frac 34+\frac{7}2\right)$$
Or something like that. Each of the first three terms has limit zero...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How many $4\times 3$ matrices of rank three are there over a finite field of three elements?
Let $M$ be the space of all $4\times 3$ matrices with entries in the finite field of three elements. Then the number of matrices of rank three in $M$ is
A. $(3^4 - 3)(3^4 - 3^2)(3^4-3^3)$
B. $(3^4 - 1)(3^4 - 2)(3^4 - 3)$
C. $(... | Sorry for the previous post.the answer I am getting is c
$(3^4-1)(3^4-3)(3^4-3^2)$
For the $1$st column we have $4$ places and $3$ elements to fill it.So $3^4$ choices but the elements can't be all zero.S0 we have $3^4-1$ choices .For the second column we have $3^4$ choices but the second column cant be linearly depen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Factorise a matrix using the factor theorem Can someone check this please?
$$
\begin{vmatrix}
x&y&z\\
x^2&y^2&z^2\\
x^3&y^3&z^3\\
\end{vmatrix}$$
$$C_2=C_2-C_1\implies\quad
\begin{vmatrix}
x&y-x&z\\
x^2&y^2-x^2&z^2\\
x^3&y^3-x^3&z^3\\
\end{vmatrix}$$
$$(y-x)
\begin{vmatrix}
x&1&z\\
x^2&y+x&z^2\\
x^3&y^2+xy+x^2&z^3\\
\... | What you did is correct. But there is an easier way. Remember that for polynomial $p(x)$, if $p(a)=0$ then $(x-a)$ is a factor of $p(x)$.
Denote the determinant by $\Delta$. It is obviously a polynomial in $x,\ y$ and $z$. Now, note that:
*
*$x=0\implies \Delta = 0$, so $x$ is a factor of $\Delta$. Same for $y = 0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given $a+b+c$, Can I calculate $a^2+b^2+c^2$? I want to calculate $a^2 + b^2 + c^2$ when I am given $a+b+c$.
It is known that a,b,c are positive integers.
Is there any way to find that.
| If $a+b+c = k$ then $(a+b+c)^2 = k^2 \implies a^2+b^2+c^2 = k^2 - 2(ab + ac + bc)$, hence you would have to know the value of $ab + ac + bc$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Laplace tranform of $t^{5/2}$ It is asked to transform $t^{5/2}$.
I did $t^{5/2}=t^3\cdot t^{-1/2}$. Then followed the table result $$L\{{t^nf(t)}\}=(-1)^n\cdot\frac{d^n}{ds^n}F(s)$$
However i got $\frac{1}{2} \cdot\sqrt\pi \cdot s^{-7/2}$ instead of $\frac{15}{8} \cdot\sqrt\pi \cdot s^{-7/2}$.
Can you help me with the... | For every real number $r>-1$, we have,
$$L(t^r)(s)=\int_0^\infty e^{-st}t^rdt\\\hspace{60mm}=\int_0^\infty e^{-x}(\frac{x}{s})^r\frac{dx}{s}\hspace{10mm}\text{(Putting $x=st$)}\\\hspace{20mm}=\frac{1}{s^{r+1}}\int_0^\infty e^{-x}x^rdt\\\hspace{5mm}=\frac{\Gamma(r+1)}{s^{r+1}}.$$
So for $r=\frac{5}{2}$, we have, $$L(t^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to define an affine transformation using 2 triangles? I have $2$ triangles ($6$ dots) on a $2D$ plane.
The points of the triangles are: a, b, c and x, y, z
I would like to find a matrix, using I can transform every point in the 2D space.
If I transform a, then the result is x. For b the result is y, and for c the r... | There is a neat formula for your case
$$
\vec{P}(p_1; p_2) = (-1)
\frac{
\det
\begin{pmatrix}
0 & \vec{x} & \vec{y} & \vec{z} \\
p_1 & a_1 & b_1 & c_1 \\
p_2 & a_2 & b_2 & c_2 \\
1 & 1 & 1 & 1 \\
\end{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.