Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find $\sin^3 a + \cos^3 a$, if $\sin a + \cos a$ is known
Given that $\sin \phi +\cos \phi =1.2$, find $\sin^3\phi + \cos^3\phi$.
My work so far:
(I am replacing $\phi$ with the variable a for this)
$\sin^3 a + 3\sin^2 a *\cos a + 3\sin a *\cos^2 a + \cos^3 a = 1.728$. (This comes from cubing the already given state... | Squaring
$\sin a + \cos a = b$,
$b^2
=\sin^2a+2\sin a \cos a + \cos^2 a
= 1+2\sin a \cos a
$,
so
$\sin a \cos a
=(b^2-1)/2
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 1
} |
Finding value of 1 variable in a 3-variable $2^{nd}$ degree equation The question is: If $a,b,\space (a^2+b^2)/(ab-1)=q$ are positive integers, then prove that $q=5$. Also prove that for $q=5$ there are infinitely many solutions in $\mathbf N$ for $a$ and $b$. I simplified the equation as follows:-$$\frac {a^2+b^2}{ab-... | For such equations:
$$\frac{x^2+y^2}{xy-1}=-t^2$$
Using the solutions of the Pell equation. $$p^2-(t^4-4)s^2=1$$
You can write the solution.
$$x=-4tps$$ $$y=t(p^2+2t^2ps+(t^4-4)s^2)$$
It all comes down to the Pell equation - as I said. Considering specifically the equation:
$$\frac{x^2+y^2}{xy-1}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Help on proving a trigonometric identity involving cot and half angles Prove: $\cot\frac{x+y}{2}=-\left(\frac{\sin x-\sin y}{\cos x-\cos y}\right)$.
My original idea was to do this:
$\cot\frac{x+y}{2}$ = $\frac{\cos\frac{x+y}{2}}{\sin\frac{x+y}{2}}$, then substitute in the formulas for $\cos\frac{x+y}{2}$ and $\sin\fra... | Try using
$ \displaystyle \sin{x} - \sin{y} = \sin{\frac{2x}{2}} - \sin{\frac{2y}{2}} =
2 \frac{\sin \left( {\frac{x+y}{2} + \frac{x-y}{2}} \right) - \sin \left( {\frac{x+y}{2} - \frac{x-y}{2}} \right)}{2}$
and then the product to sum-formula for sine, i.e.
$ \displaystyle \frac {\sin \left({x + y}\right) - \sin \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
What is the coefficient of $x^{18}$ in the expansion of $(x+x^{2}+x^{3}+x^{4}+x^{5}+x^{6})^{4}$? How to approach this type of question in general?
*
*How to use binomial theorem?
*How to use multinomial theorem?
*Are there any other combinatorial arguments available to solve this type of question?
| We really seek the coefficient of $x^{14}$, factoring out an $x$ from each term in the generating function. Then observe that:
$(1 + x + x^{2} + x^{3} + x^{4} + x^{5}) = \frac{1-x^{6}}{1-x}$
Now raise this to the fourth to get: $f(x) = \left(\frac{1-x^{6}}{1-x}\right)^{4}$.
We have the identities:
$$(1-x^{m})^{n} = \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1096069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$ If $x,y,z \geq 1/2, xyz=1$, showing that $2(1/x+1/y+1/z) \geq 3+x+y+z$
I tried Schturm's method for quite some time, and Cauchy Schwarz for numerators because of the given product condition.
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality is equivalent to $f(v^2)\geq0$, where $f$ is a linear increasing function.
Hence, $f$ gets a minimal value, when $v^2$ gets a minimal value, which happens for equality of two variables or maybe one of them equal to $\frac{1}{2}$.
*
*$y=x$, $z=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How does one expression factor into the other? How does $$(k+1)(k^2+2k)(3k+5)$$ factor into $$(k)(k^2-1)(3k+2) + 12k(k+1)^2$$
| Well, $RHS=k(k^2-1)(3k+2) + 12k(k+1)^2 = k(k+1)((k-1)(3k+2)+12(k+1))=k(k+1)(3k^2+2k-3k-2+12k+12)=k(k+1)(3k^2+11k+10)=k(k+1)(k+2)(3k+5)=LHS$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I show the two limits How can I show the two limits
$$
\displaylines{
\mathop {\lim }\limits_{x \to + \infty } \frac{{x^2 e^{x + \frac{1}{x}} }}{{e^{ - x} \left( {\ln x} \right)^2 \sqrt x }} = \mathop {\lim }\limits_{x \to + \infty } \left( {\frac{{x^{\frac{3}{4}} e^{\left( {x + \frac{1}{{2x}}} \right)} }}{... | $$\frac{x^2e^x}{e^{-x} (\ln x)^2\sqrt x} < \frac{x^2e^{x + \frac1x}}{e^{-x} (\ln x)^2\sqrt x} \text{ for large $x$ since } e^{\frac1x} > 1$$
$$$$
$$\lim_{x \to \infty} \frac{x^2e^x}{e^{-x} (\ln x)^2\sqrt x} = e^{2x}\frac{x^{1.5}}{(\ln x)^2} = +\infty$$
$$$$
$$\text{Thus, the right hand side at the top should diverge to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
The roots of a certain recursively-defined family of polynomials are all real Let $P_0=1 \,\text{and}\,P_1=x+1$ and we have $$P_{n+2}=P_{n+1}+xP_n\,\,n=0,1,2,...$$
Show that for all $n\in \mathbb{N}$, $P_n(x)$ has no complex root?
| Interlacing is a good hint, but let we show a brute-force solution. By setting:
$$ f(t) = \sum_{n\geq 0}P_n(x)\frac{t^n}{n!} \tag{1}$$
we have that the recursion translates into the ODE:
$$ f''(t) = f'(t) + x\, f(t) \tag{2}$$
whose solutions are given by:
$$ f(t) = A \exp\left(t\frac{1+\sqrt{1+4x}}{2}\right) + B\exp\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{n \rightarrow \infty} \frac{1-(1-1/n)^4}{1-(1-1/n)^3}$ Find
$$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}$$
I can't figure out why the limit is equal to $\dfrac{4}{3}$ because I take the limit of a quotient to be the quotient of their limits.
I'm taking th... | $$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}
\stackrel{\mathscr{L}}{=}\lim_{n \rightarrow \infty} \dfrac{4\left(1-\dfrac{1}{n}\right)^3 \dfrac{1}{n^2}}{3\left(1-\dfrac{1}{n}\right)^2\dfrac{1}{n^2}}
=\lim_{n \rightarrow \infty} \dfrac{4\left(1-\dfrac{1}{n}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1102159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Finding determinant of $n \times n$ matrix I need to find a determinant of the matrix:
$$
A = \begin{pmatrix}
1 & 2 & 3 & \cdot & \cdot & \cdot & n \\
x & 1 & 2 & 3 & \cdot & \cdot & n-1 \\
x & x & 1 & 2 & 3 & \cdot & n-2 \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot... | Multiply the last row by $\frac{1-x}{x}$; this means that the determinant you want will be the determinant of the changed matrix times $-\frac{x}{x-1}$. Now subtract $r_1$ from $r_n$
leaving
$$r_n = (0, -x, -x, -x, \cdots, -x, \frac{(x-1)^2 - x^2}{x})
$$
where I have intentionally written
$$
\frac{1-x}{x} -1 = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1104569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $ \int^\infty_0\int^\infty_0 x^a y^{1-a} (1+x)^{-b-1}(1+y)^{-b-1} \exp(-c\frac{x}{y})dxdy $
Evaluate
$$ \int^\infty_0\int^\infty_0 x^a y^{1-a} (1+x)^{-b-1}(1+y)^{-b-1} \exp(-c\frac{x}{y})dxdy $$
under the condition $a>1$, $b>0$, $c>0$. Note that none of $a$, $b$ and $c$ is integer.
Mathematica found the ... | $\int_0^\infty\int_0^\infty x^ay^{1-a}(1+x)^{-b-1}(1+y)^{-b-1}e^{-c\frac{x}{y}}~dx~dy$
$=\Gamma(a+1)\int_0^\infty y^{1-a}(1+y)^{-b-1}U\left(a+1,a-b+1,\dfrac{c}{y}\right)dy$ (according to http://en.wikipedia.org/wiki/Confluent_hypergeometric_function#Integral_representations)
$=\dfrac{\Gamma(a+1)\Gamma(b-a)}{\Gamma(b+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Number theory: prove that if $a,b,c$ odd then $2\gcd(a,b,c) = \gcd(a+b,b+c, c+a)$ Please help! Am lost with the following:
Prove that if $a,b,c$ are odd integers, then $2 \gcd(a,b,c) = \gcd( a+b, b+c, c+a)$
Thanks a lot!!
| Let $d=\gcd(a,b,c)$. Then solve $d=ax+by+cz$.
Use that $2a=(a+b)+(a+c)-(b+c)$, $2b=(a+b)+(b+c)-(a+c)$ and $2c=(a+c)+(b+c)-(a+b)$. Then
$$2d=2ax+2by+2cz = (a+b)(x+y-z) + (a+c)(x-y+z) + (b+c)(y+z-x)$$
So we have a solution to:
$$2d = (a+b)X+(a+c)Y + (b+c)Z$$
So we know $\gcd(a+b,a+c,b+c)\mid 2\gcd(a,b,c)$. The other di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
why is $\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}}$=-3? Exercise taken from here: https://mooculus.osu.edu/textbook/mooculus.pdf (page 42, "Exercises for Section 2.2", exercise 4).
Why is $\lim_{x\to -\infty} \frac{3x+7}{\sqrt{x^2}}$=-3*? I always find 3 as the solution. I tried two approaches:
Approach 1:
$$
\lim_{x\... | For $x \to -\infty$, we have that $\sqrt{x^2} = |x| = -x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Considering $ (1+i)^n - (1 - i)^n $, Complex Analysis I have been working on problems from Complex Analysis by Ahlfors, and I got stuck in the following problem:
Evaluate:
$$
(1 + i)^n - (1-i)^n
$$
I have just "reduced" to:
$$
(1 + i)^n - (1-i)^n = \sum_{k=0} ^n i^k(1 - (-1)^k)
$$
by using expansion of each term.
Th... | There are a number of spiffy techniques one could use on this problem, but Ahlfors doesn't get to conjugation and modulus until 1.3 and geometry of the complex plane until Section 2 (of Chapter 1), while this is still in 1.1. [I dug up my copy of the third edition to see how much was discussed to that point.]
abel sh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
If $G$ is a group whereby $(a\cdot b)^{i} =a^i\cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show $G$ is abelian. If $G$ is a group in which $(a\cdot b)^{i} =a^i\cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show that $G$ is abelian.
Proof: Let $x$ be the smallest of the 3 cons... | Use $\backslash$overbrace{below}^{above}, as in (right click and select to see LaTeX commands):
$$
\overbrace{a\ldots a}^{27}
$$
The proof looks fine to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Which arrangement produces the largest number? I learnt that the power tower $2\uparrow3\uparrow4\uparrow...\uparrow n$ is larger
than any power tower with a different order of the numbers $2,3,4,...,n$.
Is this also true for conway-chains and for bowers array notation ?
Are $$2\rightarrow 3\rightarrow 4\rightarrow...... | No for both.
For example, for $n = 4$ we have $2 \rightarrow 3 \rightarrow 4 = 2 \rightarrow 4 \rightarrow 3 = 2 \uparrow \uparrow 65536$, whereas $3 \rightarrow 2 \rightarrow 4 = 3 \uparrow \uparrow 3^{27}$.
We can show by induction that $2 \rightarrow 3 \rightarrow n < 3 \rightarrow 2 \rightarrow n$, as
$$
2 \rightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\frac{(n-a)^2}{n}$ can be written as $1-\left(\frac{n}{a}\right)^2\cdot\frac{n}{(n/a)^2}$ \left(\frac{n}{a}\right)^2\cdot\frac{n}{(n/a)^2}$.
I have got so far to $(a^2/n)-2a+n$
But I can not see how to proceed. Can anyone help please?
| $$\frac{(n-a)^2}{n} =(n-a)^2\left(\frac{1}{n}\right) =(n-a)^2\left(\frac{1}{n}\right)\frac{a^2}{a^2} =\left(\frac{n-a}{a}\right)^2 \frac{a^2}{n}= \left(\frac{n}{a}-1\right)^2 \frac{a^2}{n} = \left(\frac{n}{a}-1\right)^2 \frac{n^2}{n^2} \frac{a^2}{n} = (-1)^2\left(1-\frac{n}{a}\right)^2 \frac{n}{(\frac{n}{a})^2}=\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Summation of an infinite series The sum is as follows:
$$
\sum_{n=1}^{\infty} n \left ( \frac{1}{6}\right ) \left ( \frac{5}{6} \right )^{n-1}\\
$$
This is how I started:
$$
= \frac{1}{6}\sum_{n=1}^{\infty} n \left ( \frac{5}{6} \right )^{n-1} \\
= \frac{1}{5}\sum_{n=1}^{\infty} n \left ( \frac{5}{6} \right )^{n}\\\\... | hint: differentiate the identity
$$\sum_{k=0}^{\infty} x^k = \frac{1}{1-x} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int \frac{x+2}{x^3-x} dx$ I need to solve this integral but I get stuck, let me show what I did:
$$\int \frac{x+2}{x^3-x} dx$$
then:
$$\int \frac{x}{x^3-x} + \int \frac{2}{x^3-x}$$
$$\int \frac{x}{x(x^2-1)} + 2\int \frac{1}{x^3-x}$$
$$\int \frac{1}{x^2-1} + 2\int \frac{1}{x^3-x}$$
now I need to resolve one... | The Heaviside cover-up method for solving partial fraction decompositions deserves to be more widely known. We want to find $A,B,C$ in this equation:
$$\frac{x+2}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$$
To find $A$, multiply by $x$:
$$\frac{x+2}{(x-1)(x+1)}=A+\frac{Bx}{x-1}+\frac{Cx}{x+1}$$
Now put $x=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Does $\int_0^\infty \sin(x^{2/3}) dx$ converges? My Try:
We substitute $y = x^{2/3}$. Therefore, $x = y^{3/2}$ and $\frac{dx}{dy} = \frac{2}{3}\frac{dy}{y^{1/3}}$
Hence, the integral after substitution is:
$$ \frac{3}{2} \int_0^\infty \sin(y)\sqrt{y} dy$$
Let's look at:
$$\int_0^\infty \left|\sin(y)\sqrt{y} \right| d... | $\sin x^{2/3}$ remains above $1/2$ for $x$ between $[(2n+1/6)\pi]^{3/2}$ and $[2n+5/6]^{3/2}$, so the integral rises by more than $\left([2n+5/6]^{3/2}-[2n+1/6]^{3/2}\right)\pi^{3/2}/2$ during that time.
$$[2n+5/6]^{3/2}-[2n+1/6]^{3/2}=\frac{[2n+5/6]^3-[2n+1/6]^3}{[2n+5/6]^{3/2}+[2n+1/6]^{3/2}}\\
>\frac{8n^2}{2[2n+1]^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
(Infinite) Nested radical equation, how to get the right solution? I've been tasked with coming up with exam questions for a high school math contest to be hosted at my university. I offer the following equation,
$$\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}=2$$
and ask for the solution for $x$.
Here's what I attempted ... | You know that:
$$\sqrt{x+\sqrt{x-\sqrt{x+\sqrt{x-\cdots}}}}=A = 2$$
and hence
$$A = \sqrt{x+\sqrt{x-A}} = 2$$
or equivalently
$$\sqrt{x+\sqrt{x-2}} = 2$$
Clearly, $\sqrt{x-2}$ is well defined when $$x \geq 2. ~~~(1)$$
Then, squaring both side, you get:
$$x+ \sqrt{x-2} = 4 \Rightarrow \sqrt{x-2} = 4-x ~~~(2).$$
Since $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Four different positive integers a, b, c, and d are such that $a^2 + b^2 = c^2 + d^2$ Four different positive integers $a, b, c$, and $d$ are such that $a^2 + b^2 = c^2 + d^2$
What is the smallest possible value of $abcd$?
I just need a few hints, nothing else. How should I begin? Number theory?
| The number of representations of a positive integer as a sum of two squares depends on the number of prime divisors of the form $4k+1$ (see Cox, Primes of the form $a^2+k b^2$). If we take the first two primes of such a form, $5$ and $13$, we have that $5\cdot 13$ can be represented as:
$$ 65 = 1^2+8^2 = 4^2+7^2 $$
so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can the cube of every perfect number be written as the sum of three cubes? I found an amazing conjecture: the cube of every perfect number can be written as the sum of three positive cubes. The equation is
$$x^3+y^3+z^3=\sigma^3$$
where $\sigma$ is a perfect number
(well it holds good for the first three perfect numbe... | This is more of a comment as opposed to an answer
There is a formula for finding the values of $a, b, c, d$ in the following equation: $$a^3 + b^3 + c^3 = d^3$$ Where $$\forall x, y\in \mathbb{Z}, \ \begin{align} a &= 3x^2 + 5y(x - y), \ b = 2\big(2x(x - y) + 3y^2\big) \\ c &= 5x(x - y) - 3y^2, \ d = 2\big(3x^2 - 2y(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 1
} |
Use integration by substitution I'm trying to evaluate integrals using substitution. I had
$$\int (x+1)(3x+1)^9 dx$$
My solution: Let $u=3x+1$ then $du/dx=3$
$$u=3x+1 \implies 3x=u-1 \implies x=\frac{1}{3}(u-1) \implies x+1=\frac{1}{3}(u+2) $$
Now I get $$\frac{1}{3} \int (x+1)(3x+1)^9 (3 \,dx) = \frac{1}{3} \int \fr... | $\frac{19683 x^{11}}{11}+\frac{39366 x^{10}}{5}+15309 x^9+17496 x^8+13122 x^7+6804 x^6+\frac{12474 x^5}{5}+648 x^4+117 x^3+14 x^2+x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx$ How do I evaluate the definite integral $$\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\,dx ?$$ I used trig substitution, and then a u substitution for $\sec\theta$.
I tried doing it and got an answer of: $-\sqrt{125}+12\sqrt{5}-16$, but apparently its wrong.
Can someone hel... | A way to compute this is as follows:
\begin{align*}
\int_0^1 \frac {x^3}{\sqrt {4+x^2}}\mathrm d x &=\int_0^1\frac{4x+x^3-4x}{\sqrt{4+x^2}}\mathrm d x\\
&=\int_0^1x\sqrt{4+x^2}\mathrm d x -2\int_0^1\frac{2x}{\sqrt{4+x^2}}\mathrm d x\\
&=\left.\frac{1}{3}(4+x^2)^{3/2}\right|_0^1-4\left.\left(4+x^2\right)^{1/2}\right|_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Chinese Remainder Theorem for non prime-numbers. Let's say I want to find x such that x leaves remainder 2 when divided by 3 and x leaves remainder 3 when divided by 5.
x % 3 = 2
x % 5 = 3
We break down the problem to:
x % 3 = 1
x % 5 = 0
Therefore,
5k % 3 = 1
2k % 3 = 1
k = 2
10, when remainder = 1
20, when remainder ... | Exactly the same way. The equation $x\equiv 3 \mod 7$ tells you that $x=3+7y$ .
Plugging this into the second equation gives you $3+7y\equiv 2 \mod 4$, that is $-y\equiv -1 \mod 4$, so $y=1+4z$, and $x=10+28z$, i.e. $x\equiv 10 \mod 28$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to prove this: $a^4+b^4+2 \ge 4ab$? How to prove this: $a^4+b^4+2 \ge 4ab$?
$a$ and $b$ are reals.
| Another possible method is generation of a function and calculating its minimum:
$$f(a,b)=a^4+b^4+2-4 a b$$
$$D_a f(a,b)=4 a^3-4 b = 0$$
$$ a^3=b$$
$$ f(a)=a^4+a^{12}+2-4a^4=a^{12}+2-3a^4$$
$$D_a f(a)=12a^{11}-12a^3=0$$
$$a^3 (a^8-1)=0$$
Minimum must be at one of:
$$f(0,0)=2$$
$$f(1,1)=0$$
$$f(-1,-1)=0$$
so
$$f(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Number theory: show that ${ 1^2, 2^2, 3^2,... , m^2}$ cannot be a complete residue system Is this an acceptable answer?
Question: show that ${ 1^2, 2^2, 3^2,... , m^2}$ cannot be a complete residue system.
Since the above has $m$ elements, one must show it cannot be a complete residue system
modulo $m$. Consider the... | Let's start with an example. Below is a congruence table modulo &11&.
\begin{array}{rrr}
k & k^2 & k^2 \pmod{11} \\
\hline
0 & 0 & 0\\
1 & 1 & 1\\
2 & 4 & 4\\
3 & 9 & 9\\
4 & 16 & 5\\
5 & 25 & 3\\
6 & 36 & 3\\
7 & 49 & 5\\
8 & 64 & 9\\
9 & 81 & 4\\
10 & 100 & 1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to solve this ordinary differential equation? I am just trying to find general solution
$$\frac{dy}{dx} = 1 + \sqrt{1 - xy}$$
| Let $u=-\sqrt{1-xy}$ ,
Then $y=\dfrac{1-u^2}{x}$
$\dfrac{dy}{dx}=\dfrac{u^2-1}{x^2}-\dfrac{2u}{x}\dfrac{du}{dx}$
$\therefore\dfrac{u^2-1}{x^2}-\dfrac{2u}{x}\dfrac{du}{dx}=1-u$
$\dfrac{2u}{x}\dfrac{du}{dx}=u-1+\dfrac{u^2-1}{x^2}$
Approach $1$:
$\dfrac{2u}{x}\dfrac{du}{dx}=\dfrac{(u-1)x^2+(u+1)(u-1)}{x^2}$
$(x^2+u+1)\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
$ A_n $ =$[\frac{n}{n+1},\frac{n+1}{n+2}] $ be closed subsets find $\bigcup_{n=1}^\infty A_n $ Let $ A_n =\frac{n}{n+1},\frac{n+1}{n+2}] $, $n=1,2,3...$ be closed subsets of real line R. Then $\bigcup_{n=1}^\infty A_n $ is
*
*(1/2,1)
*[1/2,1)
*(1/2,1]
*[1/2,1]
My attempt : think it could be [1/2,1) since $\lim ... | Hint: Since $f(x)=\frac{x}{x+1}$ is monotonically increasing, we have that for any $0\le x\lt1$,
$$
n=\left\lfloor\frac{x}{1-x}\right\rfloor\iff n\le\frac{x}{1-x}\lt n+1\iff\frac{n}{n+1}\le x\lt\frac{n+1}{n+2}
$$
Thus, for any $\frac12\le x\lt1$, if $n=\left\lfloor\frac{x}{1-x}\right\rfloor$, then $n\ge1$ and $x\in\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Differential calculus: integrate $\frac{1}{x \log^3 (x)}$ I would like a step by step description of how to integrate $$\frac{1}{x \log^3 (x)}$$
*
*I know that the answer is - $\frac{1}{2\log^2(x)}$
*and that the integral of $\frac{1}{\log^2(x)}$ is $\frac{1}{\log(x)}$
*and that the integral of $\frac{1}{\log(x)}... | For this problem:$$\int\frac{1}{x\log^3(x)}dx\tag{1}$$we can use the substitution:$$u=\log^3(x)\tag{2}$$$$\therefore \log(x)=u^{\frac{1}{3}}\tag{3}$$which leads to:$$du=3\log^2(x)\times\frac{1}{x}dx$$$$=3(u^{\frac{1}{3}})^2\times\frac{1}{x}dx=\frac{3u^{\frac{2}{3}}}{x}dx$$$$\therefore \frac{dx}{x}=\frac{du}{3u^{\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of digits $d$ in $d^k$ The title says it all really. For example, how many occurences of $6$ are there in $6^k$? It starts $6, 36, 216, \dots$ so $1, 1, 1,\dots$ The question can now be generalized into any digit or group of digits.
| Here is a general formula for base $10$ and $d\in[1,10-1]$:
$$\sum\limits_{n=1}^{k}1-\left\lceil\left(\frac{\left\lfloor\frac{d^k}{10^{n-1}}\right\rfloor-10\left\lfloor\frac{d^k}{10^n}\right\rfloor-d}{\left\lfloor\frac{d^k}{10^{n-1}}\right\rfloor-10\left\lfloor\frac{d^k}{10^n}\right\rfloor+d}\right)^2\right\rceil$$
He... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Show that the series $\sum_{k=0} ^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$ Show that the series $\sum_{k=0} ^\infty (-1)^k \dfrac{x^{2k+1}}{2k+1}$ converges for $|x|<1$ and that it converges to $\arctan x$
I tried using the ratio test but I got that it equals 1, so... | Modifying Jack D'Aurizio's answer,
since
$\sum_{k=0}^n x^k = \dfrac{1-x^{n+1}}{1-x}$,
putting $-x^2$ for $x$ we get
$\sum_{k=0}^n (-1)^kx^{2k}
= \dfrac{1-(-1)^{n+1}x^{2n+2}}{1+x^2}
= \dfrac1{1+x^2}+\dfrac{(-1)^{n}x^{2n+2}}{1+x^2}
$
or
$\dfrac1{1+x^2}
=\sum_{k=0}^n (-1)^kx^{2k} -\dfrac{(-1)^{n}x^{2n+2}}{1+x^2}
$.
Integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Let $x,y,z>0,xyz=1$. Prove that $\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\ge \frac34$ Let $x,y,z>0$ and $xyz=1$. Prove that $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{y^3}{(1+x)(1+z)}+\dfrac{z^3}{(1+x)(1+y)}\ge \dfrac34$
My attempt:
Since it is given that $xyz=1$, I tried substituting $x=\dfrac{a}{... | As another approach, you could prove it as follows.
Due to Hölder's inequality, we have:
$$
\left(\sum_{cyc} \frac{x^3}{(1+y)(1+z)}\right)\cdot\left(\sum_{cyc} (1+y)\right)\cdot\left(\sum_{cyc} (1+z)\right)\ge(x+y+z)^3\iff \sum_{cyc} \frac{x^3}{(1+y)(1+z)}≥\frac{(x+y+z)^3}{\left(\sum_{cyc} (1+y)\right)\cdot\left(\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove $\int_t^{\infty} e^{-x^2/2}\,dx > e^{-t^2/2}\left(\frac{1}{t} - \frac{1}{t^3}\right)$ How to formally prove the following inequality -
$$\int_t^{\infty} e^{-x^2/2}\,dx > e^{-t^2/2}\left(\frac{1}{t} - \frac{1}{t^3}\right)$$
| For a better lower-bound you may use the following proof by @robjohn:
$$x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t \le \int_x^\infty e^{-t^2/2}\,t\,\mathrm{d}t =e^{-x^2/2}$$
Integrate both sides of the preceding:
$$
\begin{align}
\int_s^\infty e^{-x^2/2}\,\mathrm{d}x
&\ge\int_s^\infty x\int_x^\infty e^{-t^2/2}\,\mathr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Sum $\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}$ I want to evaluate the sum $$\large\sum_{n=2}^{\infty} \frac{n^4+3n^2+10n+10}{2^n(n^4+4)}.$$ I did partial fraction decomposition to get $$\frac{1}{2^n}\left(\frac{-1}{n^2+2n+2}+\frac{4}{n^2-2n+2}+1\right)$$ I am absolutely stuck after this.
| Note that
$$\dfrac{n^4+3n^2+10n+10}{2^n(n^4+4)}=\dfrac{1}{2^n}+\dfrac{3n^2+10n+6}{2^n[(n^2+2)^2-(2n)^2]}$$
Then let's find constants $A,B$ suct that
$$\dfrac{3n^2+10n+6}{(n^2+2n+2)(n^2-2n+2)}=\dfrac{A(n+1)+B}{(n+1)^2+1}-4\Big[\dfrac{A(n-1)+B}{(n-1)^2+1}\Big]$$ to obtain the form $$f(n+1)-f(n-1).$$
For $n=-1,$ we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Derivative of $\frac { y }{ x } +\frac { x }{ y } =2y$ with respect to $x$ $$\frac { y }{ x } +\frac { x }{ y } =2y$$
Steps I took:
$$\frac { d }{ dx } \left[yx^{ -1 }1+xy^{ -1 }\right]=\frac { d }{ dx } [2y]$$
$$\frac { dy }{ dx } \left(\frac { 1 }{ x } \right)+(y)\left(-\frac { 1 }{ x^{ 2 } } \right)+(1)\left(\frac {... | $\text{ Assuming your y' is correct... } \\ \text{ then we should get rid of the compound fractions.. } \\ y'=\frac{\frac{-y}{x^2}+\frac{1}{y}}{2-\frac{1}{x}+\frac{x}{y^2}} \\ \text{ now we need to multiply top and bottom by } \\ x^2y^2 \text{ this is the lcm of the bottoms of the mini-fractions } \\ y'=\frac{-y(y^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In how many ways can you distribute 100 lemons between Dana, Sara and Lena so that Lena will get more lemons than Dana? Assume Dana has 0 lemons, so Lena must have 1 lemon. Now all i need to distribute is
$$x_1 + x_2 = 99 \text{ // because Lena already has 1 and Dana has 0}$$
The answer to above is 100.
Now assume Dana... | Alternatively, let $D, S$ and $L$ be the values, and let $L=D+1+L_0$. Then you want non-negative integer solutions to $1+D+S+L_0=100$, or $2D+S+L_0=99$.
You are doing $$\sum_{D=0}^{49} \sum_{S=0}^{99-2D} 1=\sum_{D=0}^{49} (100-2D)$$ which is a correct way to count this value.
A generating function solution would be to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$\alpha = \frac{1}{2}(1+\sqrt{-19})$ I have asked a similar question here before, which was about ring theory, but it is slightly different today and very trivial.
$\alpha = \frac{1}{2}(1+\sqrt{-19})$
Here, $\alpha$ is a root of $\alpha^2 - \alpha + 5$ and $\alpha^2 = \alpha - 5$, but I can't seem to understand this.... | $\alpha^2-\alpha+5=\frac{1}{4}(1+\sqrt{-19})^2-\frac{1}{2}(1+\sqrt{-19})+5=\frac{1}{4}(1+2\sqrt{-19}+(-19))-\frac{1}{4}(2+2\sqrt{-19})+\frac{1}{4}\cdot 20=\frac{1}{4}(1+2\sqrt{-19}+(-19)-2-2\sqrt{-19}+20)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\frac{n^n}{e^{n-1}} 2$. I am trying to prove
$$\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^{n}} \text{ for all }n > 2.$$
Here is the original source (Problem 1B, on page 12 of PDF)
Can this be proved by induction?
The base step $n=3$ is proved: $\frac {27}{e^2} < 6 < \frac{256}{e^3}$ (since $e^2 > 5$ and $... | Proof: We will prove the inequality by induction. Since $e^2 > 5$ and $e^3 < 27$, we have
$$\frac {27}{e^2} < 6 < \frac{256}{e^3}.$$ Thus, the statement for $n=3$ is true. The base step is complete.
For the induction step, we assume the statement is true for $n=k$. That is, assume $$\frac{k^k}{e^{k-1}}<k!<\frac{(k+1)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 2
} |
How are first digits of $\pi$ found? Since Pi or $\pi$ is an irrational number, its digits do not repeat. And there is no way to actually find out the digits of $\pi$ ($\frac{22}{7}$ is just a rough estimate but it's not accurate). I am talking about accurate digits by either multiplication or division or any other ope... | In the 18th century, Leonard Euler discovered an elegant formula:
$$\frac{π^4}{90}=\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\dots$$
The more terms you add, the more accurate the calculation of $π$ gets.
$$\frac{π^4}{90}=\frac{1}{1^4}=1.000$$
then $π=3.080$
$$\frac{π^4}{90}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 9,
"answer_id": 2
} |
Find The Equation This is the question:
Find the equations of the tangent lines to the curve $y = x − \frac 1x + 1$ that are parallel to the line $x − 2y = 3$.
There are two answers: 1) smaller y-intercept; 2) larger y-intercept
The work:
The slope of the line is (1/2).
$y' = ((x + 1) - (x - 1))/(x + 1)^2 = 2/(x + 1)^2... | there are no points on the graph of $y = x - \frac1x + 1$ has a tangent that is parallel to $x - 2y = 3.$
i will explain why.
you will get a better idea of the problem if you can draw the graph of $y = x - \frac1x + 1$ either on your calculator or by hand. the shape of the graph is called a hyperbola, similar looking t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$, where numbers $a$ and $b$ are rational If $a$ and $b$ are rational numbers such that $\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$
Then what is the value of $a$? The answer is $-1$.
$$\sqrt{4 - 2\sqrt{3}} = a + b\sqrt{3}$$
$$4 - 2\sqrt{3} = 2^2 - 2\sqrt{3}$$ Let $u =2$ hence,
$$\sqrt{u... | Using formula
$$\sqrt{a\pm\sqrt{b}}=\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}$$
you will get
$$\sqrt{4-2\sqrt3}=\sqrt{4-\sqrt{12}}=\sqrt{\dfrac{4+\sqrt{16-12}}{2}}-\sqrt{\dfrac{4-\sqrt{16-12}}{2}}=-1+\sqrt3$$
So, $a=-1$ and $b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
how to compute this limits given these conditions.
if $f(1)=1$ and $f'(x)=\frac{1}{x^2+[f(x)]^2}$ then compute $\lim\limits_{x\to+\infty}f(x)$
i tried to write it was
$$\frac{dy}{dx}=\frac{1}{x^2+y^2}\\
(x^2+y^2)\frac{dy}{dx}=1\\
(x^2+y^2)dy=dx$$
by the help
$$\begin{align}
f(x)&\le1+\int_1^x\frac{dt}{1+t^2}\\
&\le1+... | Since $f'(x)>0 $ and $f(1)=1$ then we have
$$ \frac{1}{x^2+f(x)^2}\leq \frac{1}{x^2+1}. $$
Also we will have
$$ \int_{1}^{x}f'(t)dt = \int_{1}^{x}\frac{dt}{t^2+f(t)^2} \leq \int_{1}^{x}\frac{dt}{t^2+1} $$
$$ \implies f(x) \leq 1+ \int_{1}^{x}\frac{dt}{t^2+1} . $$
Try to finish the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Question in relation to completing the square In description of "completing the square" at http://www.purplemath.com/modules/sqrquad.htm the following is given :
I'm having difficulty understanding the third part of the transformation.
Where is
$ -\frac{1}{4}$ derived from $-\frac{1}{2}$ ?
Why is $ -\frac {1}{4}$ squ... | Let's begin with the equation
$$x^2 - \frac{1}{2}x = \frac{5}{4}$$
What the author wants to do is to create a perfect square on the left hand side. That is, the author wants to transform the expression on the left hand side into the form $(a + b)^2 = a^2 + 2ab + b^2$. Assume that
$$a^2 + 2ab = x^2 - \frac{1}{2}x$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove $\lim\limits_{n \to \infty} \sup \left ( \frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!)} \right ) ^ {\frac 1 n} = \frac {e^2} 4$ This is a problem in Heuer (2009) "Lerbuch der Analysis Teil 1" on page 366. I assume that the proof should use $e = \sum\limits_{k = 0}^{\infty} \frac 1 {k!}$, but I cannot come further.
| If $a_n=\frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!}$. Compute the limit
$$\begin{align}\lim \frac{a_{n+1}}{a_n}&=\lim \frac{\frac{(2n + 1)^{2n + 1}}{2^{2n+2} (2n+2)!}}{\frac{(2n - 1)^{2n - 1}}{2^{2n} (2n)!}}\\&=\lim\frac{(2n+1)(2n+1)}{4(2n+1)(2n+2)}\left(\frac{2n+1}{2n-1}\right)^{2n-1}\\&=\frac{1}{4}\lim\left[\left(1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1157493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to calculate derivative of $f(x) = \frac{1}{1-2\cos^2x}$? $$f(x) = \frac{1}{1-2\cos^2x}$$
The result of $f'(x)$ should be equals
$$f'(x) = \frac{-4\cos x\sin x}{(1-2\cos^2x)^2}$$
I'm trying to do it in this way but my result is wrong.
$$f'(x) = \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} =
\frac {1-2\cos... | The problem is in this step, from here
$$f'(x) = \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} $$
to here
$$\frac {1-2\cos^2x-(1-(2\cos^2x)')}{(1-2\cos^2x)^2}$$
because $$1'=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Limit of an unknown cubic expression Let $f(x)=ax^3+bx^2+cx+d$ and $g(x)=x^2+x-2$. If $$\lim_{x \to 1}\frac{f(x)}{g(x)}=1$$ and $$\lim_{x \to -2}\frac{f(x)}{g(x)}=4$$ then find the value of $$\frac{c^2+d^2}{a^2+b^2}$$ Since the denominator is tending to $0$ in both cases, the numerator should also tend to $0$, in order... | What about? $$f(x)=-x^3+x^2+4x-4$$
So:
$$\frac{c^2+d^2}{a^2+b^2}=32/2=16$$
Let us make ansatz that $$f(x)=a(x+\alpha)(x-1)(x+2)$$
So:
$$a(1+\alpha)=1,a(\alpha-2)=4\implies a=-1,\alpha=-2$$
so:
$$f(x)=-(x-1)(x+2)(x-2)=-x^3+x^2+4x-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1162864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Mathematical induction for inequalities: $\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1$ Prove by induction: $$\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} > 1$$
adding $1/(3m+4)$ as the next $m+1$ value proves pretty fruitless. Can I make some simplifications in the inequality that because the $m$ step is t... | More generally
(one of my favorite phrases),
let
$s_k(n)
=\sum\limits_{i=n+1}^{kn+1} \frac1{i}
$.
I will show that
$s_k(n+1)>s_k(n)$
for $k \ge 3$.
In particular,
for $n \ge 1$
$s_3(n)
\ge s_3(1)
=\frac1{2}+\frac1{3}+\frac1{4}
=\frac{6+4+3}{12}
=\frac{13}{12}
> 1
$.
$\begin{array}\\
s_k(n+1)-s_k(n)
&=\sum\limits_{i=n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1164493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How can I demonstrate that $x-x^9$ is divisible by 30? How can I demonstrate that $x-x^9$ is divisible by $30$ whenever $x$ is an integer?
I know that $$x-x^9=x(1-x^8)=x(1-x^4)(1+x^4)=x(1-x^2)(1+x^2)(1+x^4)$$
but I don't know how to demonstrate that this number is divisible by $30$.
| Let's factor $x^9-x$ like you have done:
$$
x^9-x=(x-1)x(x+1)(x^2+1)(x^4+1).\tag{$*$}
$$
Let's look at the RHS. The product of the first 2 terms is divisible by $2$ because it consists of 2 consecutive integers. Similarly, the product of the first 3 terms is divisible by $3$. Now, if you had
$$
(x-2)(x-1)x(x+1)(x+2)
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proof of an Limit Using the formal definition of convergence, Prove that $\lim\limits_{n \to \infty} \frac{3n^2+5n}{4n^2 +2} = \frac{3}{4}$.
Workings:
If $n$ is large enough, $3n^2 + 5n$ behaves like $3n^2$
If $n$ is large enough $4n^2 + 2$ behaves like $4n^2$
More formally we can find $a,b$ such that $\frac{3n^2+5n}{... | Perhaps simpler:
With the Squeeze Theorem:
$$\frac34\xleftarrow[x\to\infty]{}\frac{3n^2}{4n^2}\le\frac{3n^2+5n}{4n^2+2}\le\frac{3n^2+5n}{4n^2}=\frac34+\frac54\frac1{n}\xrightarrow[n\to\infty]{}\frac34+0=\frac34$$
With arithmetic of limits:
$$\frac{3n^2+5n}{4n^2+2}=\frac{3n^2+5n}{4n^2+2}\cdot\frac{\frac1{n^2}}{\frac1{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $ f(x \cdot f(y) + f(x)) = y \cdot f(x) + x $, then $f(x)=x$ Let $ f : \mathbb{Q} \rightarrow \mathbb{Q} $ be a function which has the following property:
$$ f(x \cdot f(y) + f(x)) = y \cdot f(x) + x \;,\; \forall \; x, y \in \mathbb{Q} $$
Prove that $ f(x) = x, \; \forall \; x, y \in \mathbb{Q} $.
So far, I've foun... | If you have $f(f(x))=x$ it means that the function is onto
Furthermore, assume that $f(y_1) = f(y_2)$ for some $y_1 \neq y_2$
$$ f(-f(y_1)-1 ) =f(-f(y_1)-1 ) $$
$$\Rightarrow -y_1 -1 = -y_2 -1 $$
$$\Rightarrow y_1 = y_2 !! $$
Thus the function is $1$ $to$ $1$, its inverse $f^{-1}$ exists,
in addition to $f(f(x))=x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Find two linearly independent solutions of the differential equation $(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$ I want to find two linearly independent solutions of the differential equation
$$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$$
Previously I have seen that the following holds for the... | Hint
I do not know how much this could help you; so, please forgive me if this is off-topic.
If you define first $y=(3x-1)u$, $$(3x-1)^2 y''+(9x-3)y'-9y=0$$ rewrite $$(3 x-1)^2 \Big((3 x-1) u''+9 u'\Big)=0$$ for which order can be reduced and integration seems simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Proving that $7^n(3n+1)-1$ is divisible by 9 I'm trying to prove the above result for all $n\geq1$ but after substituting in the inductive hypothesis, I end up with a result that is not quite obviously divisible by 9.
Usually with these divisibility induction problems, it falls apart nicely and we can easily factorise... | First, show that this is true for $n=1$:
$7^1\cdot(3\cdot1+1)-1=9\cdot3$
Second, assume that this is true for $n$:
$7^n\cdot(3n+1)-1=9k$
Third, prove that this is true for $n+1$:
$7^{n+1}\cdot(3(n+1)+1)-1=$
$7^{n+1}\cdot(3n+3+1)-1=$
$7^{n+1}\cdot(3n+1+3)-1=$
$7^{n+1}\cdot(3n+1)+7^{n+1}\cdot(3)-1=$
$7\cdot\color{red}{7^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
Prove that if $\lim\limits_{x\to 0}f\big(x\big(\frac{1}{x}-\big\lfloor\frac{1}{x}\big\rfloor\big)\big)$, then $\lim\limits_{x\to 0}f(x)=0$ Prove that if $\lim\limits_{x\to 0}f\bigg(x\bigg(\dfrac{1}{x}-\bigg\lfloor\dfrac{1}{x}\bigg\rfloor\bigg)\bigg)$, then $\lim\limits_{x\to 0}f(x)=0$
My attempt:
If I can show that... | Use the Squeeze Theorem. First note that
$$\frac{1}{x}-1<\left\lfloor\frac{1}{x}\right\rfloor\le\frac{1}{x}$$
from which we obtain
$$ \left|x\right|\ge \left|x\right|\left( \frac{1}{x} -\left\lfloor\frac{1}{x}\right\rfloor\right)\ge 0$$
Therefore, as $x\to 0$, the Squeeze Theorem guarantees that
$$\lim_{x\to 0} x\,\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integration: Area between curves Let $f(x)=x^2−c^2$ and $g(x)=c^2−x^2.$ Find $c>0$ such that the area of the region enclosed by the parabolas $f(x)$ and $g(x)$ is 9.
The question above is what I am having trouble with. In order to solve this problem I use the formula given as:
$\int_a^b f(x) - g(x) dx$
Here is what I ... | Draw a picture. I would use symmetry and say that the area is $4$ times the integral from $0$ to $c$ of $c^2-x^2$. So we want
$$\int_0^c (c^2-x^2)\,dx=\frac{9}{4}.$$
The integral is equal to $\frac{2}{3}c^3$. Now solve for $c$.
Remark: Your approach will work, except that we want $\int_{-c}^c (c^2-x^2)\,dx=\frac{9}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Symplectic lie algebra Can anyone explain me why, in the symplectic lie algebra, which is defined as $ sp(n)=\{X \in gl_{2n}:X^tJ+JX=0\}$ where $J=\begin{pmatrix}
0 & I \\
-I & 0 \\
\end{pmatrix} $ we can write its elements, in block form $X=\begin{pmatrix}
A & B \\
C & -A^... | Hint: The block decomposition used for $J$ (as well as the answer) suggest writing a generic matrix $X \in \mathfrak{gl}_{2n}$ as
$$X = \begin{pmatrix} A & B \\ C & D \end{pmatrix}.$$
To produce the block matrix description of $\mathfrak{sp}_{2n}$, simply substitute the block expressions for $X$ and $J$ in the definiti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1177513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $p$ is prime and congruent to $1$, then show $((\frac{p-1}{2})!)^2 \equiv -1 \pmod p$ I got another one.
Quadratic residues are completely new to me...
Thanks!
| Take the congruences, $p-1 \equiv -1 \pmod p$, $\text{ }\text{ } p-2 \equiv -2 \pmod p$
and so on upto, $\frac{p+1}2 \equiv -\frac{p-1}2 \pmod p$.
Multiplying and rearranging, $$(p-1)!\equiv 1\cdot (-1)\cdot 2 \cdot (-2) ...\frac{p-1}2 \cdot (-\frac{p-1}2) \equiv (-1)^{\frac{p-1}2}[(\frac{p-1}{2})!]^2 \pmod p$$
Thus, b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
square root / factor problem $(A/B)^{13} - (B/A)^{13}$ Let
$A=\sqrt{13+\sqrt{1}}+\sqrt{13+\sqrt{2}}+\sqrt{13+\sqrt{3}}+\cdots+\sqrt{13+\sqrt{168}}$ and
$B=\sqrt{13-\sqrt{1}}+\sqrt{13-\sqrt{2}}+\sqrt{13-\sqrt{3}}+\cdots+\sqrt{13-\sqrt{168}}$.
Evaluate $(\frac{A}{B})^{13}-(\frac{B}{A})^{13}$.
By Calculator, I have $\frac... | Let
$$A=\sum_{n=1}^{168}\sqrt{13+\sqrt{n}},B=\sum_{n=1}^{168}\sqrt{13-\sqrt{n}}$$
since
$$\sqrt{2}A=\sum_{n=1}^{168}\sqrt{26+2\sqrt{n}}=\sum_{n=1}^{168}\left(\sqrt{13+\sqrt{169-n}}+\sqrt{13-\sqrt{169-n}}\right)=A+B$$ so we have $x=\dfrac{A}{B}=\sqrt{2}$,then we have
$$x=\sqrt{2}+1,\dfrac{1}{x}=\sqrt{2}-1\Longrightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 0
} |
Need clarification on a Taylor polynomial question $$f(x) = 5 \ln(x)-x$$
second Taylor polynomial centered around $b=1$ is $-1 + 4(x-1) - (5/2)(x-1)^2$
let $a$ be a real number : $0 < a < 1$
let $J$ be closed interval $[1-a, 1+a]$
find upper bound for the error $|f(x)-T_2(x)|$ on interval $J$
answer in terms of a
so i ... | We are given:
$$f(x) = 5 \ln(x) - x$$
The second Taylor polynomial centered around $b=1$ is given by:
$$T_2(x) = -\frac{1}{2} 5 (x-1)^2+4 (x-1)-1$$
We are told to let $a$ be a real number such that $0 \lt a \lt 1$ and let $J$ be the closed interval $[1 −a, 1 +a]$.
We are then asked to use the Quadratic Approximation Er... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For $x, y \in \Bbb R$ such that $(x+y+1)^2+5x+7y+10+y^2=0$. Show that $-5 \le x+y \le -2.$ I have a problem:
For $x, y \in \Bbb R$ such that $(x+y+1)^2+5x+7y+10+y^2=0$. Show that
$$-5 \le x+y \le -2.$$
I have tried:
I write $(x+y+1)^2+5x+7y+10+y^2=(x+y)^2+7(x+y)+(y+1)^2+10=0.$
Now I'm stuck :(
Any help will be a... | Since $(y+1)^2\ge 0$ we must have $(x+y)^2+7(x+y)+10=(x+y+5)(x+y+2)\leq 0$ then, solving this inequality for $x+y$, we get $$-5\le x+y \le -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to do multiplication in $GF(2^8)$? I am taking an Internet Security Class and we received some practice problems and answers, but I do not know how to do these problems , an explanation would be greatly appreciated
Try to compute the following value:(the number is in hexadecimal and each represents a polynomial in ... | To carry out the operation, we need to know the irreducible polynomial that is being used in this representation. By reverse-engineering the answer, I can see that the irreducible polynomial must be $x^8+x^4+x^3+x+1$ (Rijndael's finite field). To carry out a product of any two polynomials then, what you want to do is m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the initial value problem $u_{xx}+2u_{xy}-3u_{yy}=0,\ u(x,0)=\sin{x},\ u_{y}(x,0)=x$
Solve the partial differential equation $$u_{xx}+2u_{xy}-3u_{yy}=0$$ subjet to the initial conditions $u(x,0)=\sin{x}$, $u_{y}(x,0)=x$.
What I have done
$$
3\left(\frac{dx}{dy}\right)^2+2\frac{dx}{dy}-1=0
$$
implies
$$\frac{dx... | We use Laplace transform method and free CAS Maxima
http://maxima.sourceforge.net/
Answer:
$$u=\frac{\sin(x+y)}{4}+\frac{y^2}{3}+xy+\frac34\sin\left(x-\frac{y}{3}\right)$$
2 method
*
*$D_x^2+2D_xD_y-3D_x^2=(D_x-D_y)(D_x+3D_y)$
*General solution of $u_x-u_y=0$ is $u_1=f(x+y)$
*General solution of $u_x+3u_y=0$ is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Finding the order of an element in the dihedral group of order 4. How do I find the order of
$$S_1=\left({\begin{array}{cc}
\cos\frac{\pi}{3} & \sin\frac{\pi}{3}\\
\sin\frac{\pi}{3} & -\cos\frac{\pi}{3}\\
\end{array} }\right)$$
I know that $S_1$ is a dihedral group and is a reflection of the line that makes an... | $$S_1^2=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$$
In more detail
$$\begin{align}S_1^2=&\begin{matrix} \cos^2{\frac{\pi}{3}}+\sin^2{\frac{\pi}{3}}&\cos{\frac{\pi}{3}}\sin{\frac{\pi}{3}}-\cos{\frac{\pi}{3}}\sin{\frac{\pi}{3}}\\\cos{\frac{\pi}{3}}\sin{\frac{\pi}{3}}-\cos{\frac{\pi}{3}}\sin{\frac{\pi}{3}}&\cos^2{\frac{\pi}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
integration of definite integral involving sinx and cos x Evaluate $\int_0^{\pi}\frac{dx}{a^2\cos^2x +b^2 \sin^2x}$
I got numerator $\sec^2 x$ and denominator $b^2 ( a^2/b^2 + \tan^2x)$.
I made substitution $u= \tan x$. That way $\sec^2 x$ got cancelled and the answer was of form $1/ab$ ($\tan^{-1} (bu/a)$)
And then if... | Are you talking about this?:
$$\small\int\frac{dx}{a^2\cos^2x +b^2 \sin^2x}=\int\frac{\sec^2xdx}{a^2+b^2 \tan^2x}\stackrel{u=\tan x}=\frac1{b^2}\int\frac{du}{a^2/b^2+u^2}=\frac1{b^2}\frac1{a/b}\arctan\frac{\tan x}{a/b}$$
So:
$$\int_0^{\pi}\frac{dx}{a^2\cos^2x +b^2 \sin^2x}=\frac1{ab}\arctan\frac{b\tan x}a\Bigg|_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ Let $a_1=2$ and we define $a_{n+1}=a_n+\sqrt {a_n},n\geq 1$.
Is it possible to get a good aproximation of the $n$th term $a_n$?
The first terms are $2,2+\sqrt{2}$, $2+\sqrt{2}+\sqrt{2+\sqrt{2}}$ ...
Thanks in advance!
| For the third term, we have the following. Let $c_n = \sqrt{a_n} -\frac{1}{2}n + \frac{1}{4}\ln n$, then $c_{n+1} -c_{n} = -\frac{1}{2} \cdot \frac{\sqrt{a_n}}{(\sqrt{a_n}+\sqrt{\sqrt{a_n}+a_n})^2} + \frac{1}{4} \ln(1+ \frac{1}{n})\quad (1)$. Note that $a_n = \frac{1}{4}n^2 - \frac{1}{4}n\ln n + o(n \ln n)$, the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
there exsit postive integer $x,y$ such $p\mid(x^2+y^2+n)$ For any give the postive integer $n$,and for any give prime number $p$.
show that
there exsit postive integer $x,y$ such
$$p\mid(x^2+y^2+n)$$
My approach is the following:
Assmue that $n=1,p=2$,we choose$(x,y)=(1,2)$
$$2\mid6=1^2+2^2+1$$
Assmue that $n=1,p=3$, ... | The result is easy to prove if $p=2$, so we can assume from now on that $p$ is odd.
Modulo $p$, there are $\frac{p+1}{2}$ squares, namely the $\frac{p-1}{2}$ quadratic residues of $p$, and $0$.
So modulo $p$ there are $\frac{p+1}{2}$ distinct values of $x^2$. There are also (for fixed $n$) $\frac{p+1}{2}$ distinct valu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding for what $x$ values the error of $\sin x\approx x-\frac {x^3} 6$ is smaller than $10^{-5}$
Find for what $x$ values the error of $\sin x\approx x-\frac {x^3} 6$ is smaller than $10^{-5}$
I thought of two ways but got kinda stuck:
*
*Since we know that $R(x)=f(x)-P(x)$ then we could solve: $\sin x-x+\frac {... | Consider
$$
\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots.
$$
This series is alternating, and the terms are strictly decreasing in magnitude if $|x| < 1$. So we get
\begin{align*}
\text{estimate} - \text{ reality} &= x - \frac{x^3}{6} - \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots\right) \\
&=\frac{x^5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the following indefinite integral Evaluate the integral : $$\int\frac{1-x}{(1+x)\sqrt{x+x^2+x^3}}\,dx$$
I tried through putting $x=\tan \theta$ as well as $x=\tan^2\theta$ .but I am unable to remove the square root. I also tride by putting $x+x^2+x^3=z^2$. But I could not proceed anyway...Please help...
Updat... | Hint 1: $t \mapsto (1-x)/(1+x)$
$$\begin{equation}\displaystyle\int\frac{x-1}{(x+1)\sqrt{x+x^2+x^3}}\,\mathrm{d}x = 2\arccos\left(\frac{\sqrt{x}} {x+1}\right) + \mathcal{C}\end{equation}$$
Hint 2: One can show that $t = (1-x)/(1+x)$ is it's own inverse. In other words $x = (1-t)/(1+t)$. Hence the derivative becomes.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x) $ $f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x)$
Determine the greatest and least values of $\frac{39}{f(x)+14}$ and state a value of x at which greatest values occurs.
Do I just use a graphing calculator for this? Is there a way I could do this without a graphing calc... | HINT:
$$f(x)=17\cos^2x+7\sin^2x-24\sin x\cos x=\dfrac{17(1+\cos2x)+7(1-\cos2x)-24\sin2x}2$$
$$=12+5\cos2x-12\sin2x=12+\sqrt{12^2+5^2}\cos\left(2x+\arctan\dfrac{12}5\right)$$
Now for real $x,-1\le\cos\left(2x+\arctan\dfrac{12}5\right)\le1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Divergence of $\prod_{n=2}^\infty(1+(-1)^n/\sqrt n)$. Looking looking for a verification of my proof that the above product diverges.
$$\begin{align}
\prod_{n=2}^\infty\left(1+\frac{(-1)^n}{\sqrt n}\right) & =\prod_{n=1}^\infty\left(1+\frac1{\sqrt {2n}}\right)\left(1-\frac1{\sqrt{2n+1}}\right)\\
& =\prod_{n=1}^\infty\l... | I would like for some approval to my answer, please.
Show that $$\prod_2^\infty(1+\frac{(-1)^k}{\sqrt k})$$ diverges even though$$\sum_2^\infty \frac{(-1)^k}{\sqrt k}$$ converges.
$\prod_2^\infty(1+\frac{(-1)^k}{\sqrt k})$ diverges if and only if $\sum _2^\infty\log(1+\frac{(-1)^k}{\sqrt k})$ diverges.
Indeed, by Lei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $A,B>0$ and $A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$. If $A,B>0$ and $\displaystyle A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$.
$\bf{My\; Try::}$ Given $$\displaystyle A+ B = \frac{\pi}{3}$$ and $A,B>0$.
So we can say $$\displaystyle 0< A,B<\frac{\pi}{3}$$. No... | $$
\tan A\tan B=\tan A\tan\big(\frac{\pi}{3}-A\big)=\tan A.\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}=1-1+\frac{\sqrt{3}\tan A-\tan^2A}{1+\sqrt{3}\tan A}\\
=1+\frac{-1-\sqrt{3}\tan A+\sqrt{3}\tan A-\tan^2A}{1+\sqrt{3}\tan A}=1-\frac{1+\tan^2A}{1+\sqrt{3}\tan A}=1-\frac{\sec^2A}{1+\sqrt{3}\tan A}\\
=1-\frac{2}{2\cos^2A+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Inequality and Induction: $\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$ I needed to prove that
$\prod_{i=1}^n\frac{2i-1}{2i}$ $<$ $\frac{1}{\sqrt{2n+1}}$, $\forall n \geq 1$ .
I've atempted by induction.
I proved the case for $n=1$ and assumed it holds for some $n$.
The left-side of the n+1 case... | $$\begin{align}\frac{1}{2}.\frac{3}{4}. ... .\frac{2n-1}{2n}.\frac{2n+2-1}{2n+2} &<\frac{2n+1}{\sqrt{2n+1}(2n+2)}\\~\\&=\frac{\sqrt{2n+1}}{(2n+2)}\\~\\&=\frac{\sqrt{(2n+1)(2n+3)}}{(2n+2)\sqrt{2n+3}}\\~\\&=\frac{\sqrt{4n^2+8n+3}}{(2n+2)\sqrt{2n+3}}\\~\\&\lt \frac{\sqrt{4n^2+8n+3+\color{blue}{1}}}{(2n+2)\sqrt{2n+3}}\\~\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Taylor Expansion of Inverse of Difference of Vectors I am trying to derive the multipole moment of a gravitational potential, but I'm getting stuck on some math I believe. So basically the problem is finding the Taylor Expansion for $$\frac{1}{|\mathbf{x}-\mathbf{x'}|}=\frac{1}{\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}}.$$ T... | The multivariable Taylor expansion is probably most easily written as
$$ f(\mathbf{x+h}) = \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{i_1,i_2,\dotsc,i_k=1}^n (\partial_{x_{i_1}} \partial_{x_{i_2}} \dotsm \partial_{x_{i_k}} f(\mathbf{x}) ) h_{i_1} h_{i_2} \dotsm h_{i_k} $$
(which keeps all terms of the same order together),... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Another optimization problem I am having trouble figuring out a next step in an optimization problem
the question is to find the max and min values of $f(x,y)=\frac{x+y}{2+x^2+y^2}$
I calculated $f_x$ and $f_y$ and set both them equal to zero, and the only possibility you get is x=y. I dont know how else to find it aft... | Setting $f_x=0$ and $f_y=0$ gives $-x^2-2xy+y^2+2=0$ and $-y^2-2xy+x^2+2=0$, so
subtracting these equations gives $2y^2-2x^2=0, \;\;y^2=x^2,\; $ and so $y=\pm x$.
1) If $y=x$, substituting into the first equation gives $x^2=1$ so $x=\pm 1$.
2) If $y=-x$, substituting into the first equation gives $x^2=-1$, so there is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1211844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit as $x$ tend to zero of: $x/[\ln (x^2+2x+4) - \ln(x+4)]$ Without making use of LHôpital's Rule solve:
$$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)}$$
$ x^2+2x+4=0$ has no real roots which seems to be the gist of the issue.
I have attempted several variable changes but none seemed to work.
| An approach without L'Hopital's rule.
$$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)}=\lim_{x\to 0} {1\over {1\over x}\ln {x^2+2x+4\over x+4}}=\lim_{x\to 0} {1\over \ln \big ({x^2+2x+4\over x+4}\big)^{1\over x}}$$
but
$$({x^2+2x+4\over x+4}\big)^{1\over x}=({x^2+x+x+4\over x+4}\big)^{1\over x}=\big(1+{x^2+x\over x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Problem when $x=\cos (a) +i\sin(a),\ y=\cos (b) +i\sin(b),\ z=\cos (c) +i\sin(c),\ x+y+z=0$ Problem: If $$x=\cos (a) +i\sin(a),\ y=\cos (b) +i\sin(b),\ z=\cos (c) +i\sin(c),\ x+y+z=0$$ then which of the following can be true:
1) $\cos 3a + \cos 3b + \cos 3c = 3 \cos (a+b+c)$
2) $1+\cos (a-b) + \cos (b-c) =0$
3) $\cos 2... | HINT:
For $(1),$
Using If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.,
we have $x^3+y^3+z^3=3xyz$
Now $x^3=e^{i(3a)}=\cos3a+i\sin3a$
and $xy=e^{i(a+b)}=\cos(a+b)+i\sin(a+b),xyz=\cdots$
For $(3),(4)$
$x+y+z=0\implies\cos a+\cos b+\cos c=\sin a+\sin b+\sin c=0$
So, $x^{-1}+y^{-1}+z^{-1}=?$
Now $x^2+y^2+z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1215672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that equation $x^6+x^5-x^4-x^3+x^2+x-1=0$ has two real roots Prove that equation
$$x^6+x^5-x^4-x^3+x^2+x-1=0$$ has two real roots
and $$x^6-x^5+x^4+x^3-x^2-x+1=0$$
has two real roots
I think that:
$$x^{4k+2}+x^{4k+1}-x^{4k}-x^{4k-1}+x^{4k-2}+x^{4k-3}-..+x^2+x-1=0$$
and
$$x^{4k+2}-x^{4k+1}+x^{4k}+x^{4k-1}-x^{... | HINT: you can write this equation:
$x^6+x^5-x^4-x^3=-x^2-x+1$
$y=x^6+x^5-x^4-x^3$
$y=-x^2-x+1$
and draw two functions noting that there are two points where two fnctions intersect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Quadratic Integers in $\mathbb Q[\sqrt{-5}]$ Can someone tell me if $\frac{3}{5}$, $2+3\sqrt{-5}$, $\frac{3+8\sqrt{-5}}{2}$, $\frac{3+8\sqrt{-5}}{5}$, $i\sqrt{-5}$ are all quadratic integers in $\mathbb Q[\sqrt{-5}]$. And if so why are they in $\mathbb Q[\sqrt{-5}]$.
| Only one of them is. There are a number of different ways to tell, and of these the easiest are probably the minimal polynomial and the algebraic norm.
In a quadratic extension of $\mathbb{Q}$, the minimal polynomial of an algebraic number $z$ looks something like $a_2 x^2 + a_1 x + a_0$, and if $a_2 = 1$, we have an a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Infinite Product - Seems to telescope Evaluate
$$\left(1 + \frac{2}{3+1}\right)\left(1 + \frac{2}{3^2 + 1}\right)\left(1 + \frac{2}{3^3 + 1}\right)\cdots$$
It looks like this product telescopes: the denominators cancel out (except the last one) and the numerators all become 3.
What would my answer be?
| we have the following identity (which affirms that the product telescopes):
$$\left (1+\frac{2}{3^n+1}\right)=3\cdot\frac{3^{n-1}+1}{3^n+1}=\frac{1+3^{-(n-1)}}{1+3^{-n}}$$
(as denoted in the comment by Thomas Andrews)and as a result:
$$\prod_{k=1}^n \left (1+\frac{2}{3^k+1}\right)=\frac{2\cdot 3^n}{3^n+1}=\frac{2}{1+3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculation of real root values of $x$ in $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4x-1}.$
Calculation of x real root values from $ y(x)=\sqrt{x+1}-\sqrt{x-1}-\sqrt{4x-1} $
$\bf{My\; Solution::}$ Here domain of equation is $\displaystyle x\geq 1$. So squaring both sides we get
$\displaystyle (x+1)+(x-1)-2\sqrt{x^2-1}=(4x-1)$.
$... | For $x\ge1$ we have $$\sqrt{4x-1}\ge \sqrt {3x} $$
and $$\sqrt{x+1}\le \sqrt {2x}$$
hence
$$\sqrt{x+1}-\sqrt{x-1}\le \sqrt{2x}<\sqrt{3x}\le\sqrt{4x-1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
differentiation of a matrix function In statistics, the residual sum of squares is given by the formula
$$ \operatorname{RSS}(\beta) = (\mathbf{y} - \mathbf{X}\beta)^T(\mathbf{y} - \mathbf{X}\beta)$$
I know differentiation of scalar functions, but how to I perform derivatives on this wrt $\beta$? By the way, I am tryi... | First you can remove the transposition sign from the first bracket:
$RSS=(\mathbf{y}^T - \beta ^T \mathbf{X} ^T)(\mathbf{y} - \mathbf{X}\beta)$
Multiplying out:
$RSS=y^Ty-\beta ^T \mathbf{X} ^Ty-y^TX\beta+\beta^TX^T X\beta $
$\beta ^T \mathbf{X} ^Ty$ and $y^TX\beta$ are equal. Thus
$RSS=y^Ty-2\beta ^T \mathbf{X} ^Ty+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1224163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
All real values $a$ for a $2$-dimensional vector? Find all real numbers $a$ for which there exists a $2D$, nonzero vector $v$ such that:
$\begin{pmatrix} 2 & 12 \\ 2 & -3 \end{pmatrix} {v} = a {v}$.
I substituted $v$ with $\begin{pmatrix} c \\ d \end{pmatrix}$ and multiplied to obtain the system of equations:
$2x+12y ... | you can go from $$2x+12y = kx,\, 2x-3y = ky $$ to $$\frac{y}{x} = \frac{k-2}{12} = \frac2{k+3}.$$ therefore $k$ satisfies the characteristic equation $$0=(k+3)(k-2) - 24 = k^2+k-30 = (k+6)(k-5).$$ therefore $$k = 5, -6 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1224750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Expressing ${}_2F_1(a, b; c; z)^2$ as a single series Is there a way to express
$${}_2F_1\bigg(\frac{1}{12}, \frac{5}{12}; \frac{1}{2}; z\bigg)^2$$
as a single series a la Clausen? Note that Clausen's identity is not applicable here.
| Using Maple, I get
$$
\sum_{n=0}^\infty \frac{\Gamma \left( {\frac {7}{12}} \right) \Gamma \left( {\frac {11}{12}}
\right)
{\mbox{$_4$F$_3$}(\frac{1}{12},{\frac {5}{12}},-n,\frac{1}{2}-n;\,1/2,-n+{\frac {7}{12}},-n+{\frac {11}{12}};\,1)\;(4z)^n }}
{16\,\Gamma \left( -n+{\frac {11}{12}} \right) \Gamma \left( -n+{
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$ I came across this strange limit whilst showing convergence of a series:
$$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$$
How can I calculate this limit?
| Squeeze theorem gives you the proof that the limit is $\frac{3}{4}$. Since you mentioned you were looking for another way to verify that the limit is correct, here is one way (although not rigorous like the squeeze theorem) $$\begin{align}\frac{2^n+3^n}{3^n+4^n} = \frac{2^n}{3^n+4^n}+\frac{3^n}{3^n+4^n} \\ = \left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to integrate $\int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx$ How would I do the following integral?
$$\int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx$$
Where $x > 0$ and $k$ is a constant greater than $0$
| Consider the integral
\begin{align}
I = \int \frac{x^{\frac{k}{2}-1}}{1+x^k}dx
\end{align}
Let $t = x^{k/2}$ for which $x = t^{2/k}$ and $dx = (2/k) t^{(2/k) - 1} \, dt$ for which the integral becomes
\begin{align}
I = \frac{2}{k} \int \frac{dt}{1 + t^{2}}.
\end{align}
This is the integral for $\tan^{-1}(t)$ leading t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Stoppage time for sequence of uniform random numbers with a recursively shrinking domain Define $x_n = U(x_{n-1})$ where $U(x)\in\lbrace 0,1,\ldots,x\rbrace$ is a uniformly distributed random integer. Given $x_0$ as some large positive integer, what is the expected value of $n$ for which $x_n=0$?
The answer I came up w... | The expected waiting time $T_k$ to get down from $k$ to $0$ is $T_0 = 0$ for the base case, and otherwise it is $T_k = 1 + H_k$, where $H_k$ is the $k$th harmonic number $1 + 1/2 + 1/3 + \cdots + 1/k$. For large $k$ this is approximately $1 + \gamma + \ln k$, where $\gamma \doteq 0.57722$ is the Euler-Mascheroni const... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tangent line of a lemniscate at (0,0) I need to find the tangent line of the function $y=g(x)$ implicitly defined by
$(x^2+y^2)^2-2a^2(x^2-y^2)=0$
at $(0,0)$, but I don't know how.
I can't use implicit differentiation and evaluate at $(0,0)$, because when $y=0$ I can't use the Implicit Function Theorem to calculate the... | $$(x^2+y^2)^2-2a^2(x^2-y^2)=0$$
Solving for $y$ we do substitution $t=y^2$
$$x^4+x^2t+t^2-2a^2x^2+2a^2t=0$$
$$t^2+t(2x^2+2a^2)+x^4-2a^2x^2=0$$
$$t=\pm a\sqrt{4x^2+a^2}-x^2-a^2$$
As $t=-a\sqrt{4x^2+a^2}-x^2-a^2$ is not positive we get solutions
$$y=\pm \sqrt{a\sqrt{4x^2+a^2}-x^2-a^2}$$
Let $f(x)=\sqrt{a\sqrt{4x^2+a^2}-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of m. Problem :
If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of $m$.
My Approach:
$|x^2+4x+3|-mx+2m=0$
Case I : $x^2+4x+3-mx+2m=0$
$\Rightarrow x^2+ x (4-m) + 3+2m=0 $
Discriminant of above qudratic ... | $$m(x-2)=|(x+3)(x+1)|\ge0$$
If $m=0,$ there are two real solutions
Else $m(x-2)=|(x+3)(x+1)|=0$ has no solution
So, $$m(x-2)=|(x+3)(x+1)|>0$$
Now $|(x+3)(x+1)|=-(x+3)(x+1)$ if $-3\le x\le-1$
$=+(x+3)(x+1)$ otherwise
If $m>0,x-2>0\iff x>2\implies m(x-2)=x^2+4x+3$ which has exactly two solutions
If $m<0,x-2<0\iff x<2$
If... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
How to integrate $\frac{1}{(1+a\cos x)}$ from $-\pi$ to $\pi$ How to solve the following integration?$$\int_{-\pi}^\pi\frac{1}{1+a \cos x}$$
| You can look at my earlier answer here (on my previous avatar).
Below is another method. We have
$$I = \int_{-\pi}^{\pi} \dfrac{du}{1+a\cos(u)} = 2 \int_0^{\pi} \dfrac{du}{1+a \cos(u)} = 2\int_0^{\pi/2} \dfrac{du}{1+a\cos(u)} + 2\int_{\pi/2}^{\pi} \dfrac{du}{1+a\cos(u)}$$
Hence,
$$\dfrac{I}2 = \int_0^{\pi/2} \dfrac{du}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1235200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Express $\ln(3+x)$ and $\frac{1+x}{1-x}$ as Maclaurin series its probably a lot to ask.. but how can I obtain the Maclaurin series for the two functions $f(x)=\ln(3+x)$ and $g(x)=\frac{1+x}{1-x}$ ? as far as I know I cant use any commonly known series to help me with this one ? so finding the derivatives at 0:
$f(0) = ... | for $(x+1)/(1-x)$ write it as
$$
x \frac{1}{1-x} + \frac{1}{1-x}
$$
substitute the series for $1/(1-x)$ and proceed. To get a single series, add the terms one by one.
Also
$$
\log(3 -x) = \log(3) + \log(1 - x/3)
$$
so substitute $x/3$ in the series for $\log(1-y)$ which can be found be integrating the series for $1/(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Giving integer images (bis) Prove the statement below (the restriction $x < y < z$ is to avoid apparent uncertainties but the property is valid for all $x, y, z$ really).
$$F(x,y,z) = \frac{(y+z)x^n}{(z-x)(y-x)} +\frac{(z+x)y^n}{(z-y)(x-y)}+\frac{(x+y)z^n}{(x-z)(y-z)} \in \mathbb Z.$$
| If $n = 1$, then the whole expression is equal to $-1$. If $n=2$, it is equal to $0$. Let us therefore assume that $n > 2$.
We have:
$$F(x,y,z) = \frac{y^nz^2 - y^2z^n + x^n(y-z)(y+z) + x^2(z^n-y^n)}{(x-y)(x-z)(y-z)} = \frac{f(x,y,z)}{(x-y)(x-z)(y-z)}.$$
What we want to know is if the numerator is divisible by $x-y$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How many non prime factors are in the number $N=2^5 \cdot 3^7 \cdot 9^2 \cdot 11^4 \cdot 13^3$. to find non prime factors in the number
$N=2^5 \cdot 3^7 \cdot 9^2 \cdot 11^4 \cdot 13^3$.
First I tried finding all the factors by adding 1 to each of the exponents and then multiplying them
and then finding the prime facto... | Write it as $2^5 * 3^{11} * 11^4 * 13^3$. Thus, as you said, adding one to each exponent and multiplying, we get the total number of divisors, which is $6* 12 * 5* 4 = 1440$. Subtracting the four prime factors ($2,3,11,13$) leaves us with 1436.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Real Numbers are Roots $r, s$.
Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$, and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$. Find the sum of all possible values of $|b|$.
Using Vieta's Formulas,
$r+s+x_1$ $=0$ $\Rightarrow x_1$ $=-r-s$, where $x_1$ is the third root. Similarly, $x_2=-r-s-1$ $=x_1-1$, w... | Since $r,s$ are two roots of $x^3+ax+b=0$, we have
$$ r^3+ar+b=0,s^3+as+b=0 \tag{1}$$
and hence
$$ (r^3-s^3)+a(r-s)=0. $$
Assuming $r-s\neq0$, we have
$$ r^2+rs+s^2+a=0.\tag{2}$$
Similarly since $r+4,s-3$ are two roots of $x^3+ax+b+240=0$, we have
$$ (r+4)^3+a(r+4)+b+240=0,(s-3)^3+a(s-3)+b+240=0\tag{3}$$
and hence
$$ [... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many words with letters from the word ABRACADABRA if they must end in a consonant and $d$ must be after $r$. How many words with letters from the word ABRACADABRA if they must end in a consonant and $d$ must be after $r$.
What I did:
I have
$A:5$
$B:2$
$R:2$
$C:1$
$D:1$
If the words must end in a consonant and d mu... | Since D must appear after both R's, the last letter can be a B, C, or D.
Case 1: The last letter is D.
We have ten places to fill with five A's, two B's, two R's, and one C. We can fill five of the ten places with A's in $\binom{10}{5}$ ways. We can fill two of the remaining five places with B's in $\binom{5}{2}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Basis for the vector space P2 I am trying to wrap my head around vector spaces of polynomials in P2. If I represent the polynomial
$
ax^2 + bx + c
$
with the matrix
$
A = \begin{bmatrix}
1,0,0 \\
0,1,0 \\
0,0,1 \\
\end{bmatrix}
$
and the vector
$
\begin{bmatrix}
1 \\
x \\
x^2 \\
\end{bmatrix}
$
what corresponds to $... | I think you need to be clear about what you mean by "representing" the polynomial. You can if you like make the assignments
$$
x^2 \;\; \to \;\; \left [ \begin{array}{c}
0\\
0\\
1\\
\end{array} \right ] \hspace{2pc} x \;\; \to \;\; \left [ \begin{array}{c}
0\\
1\\
0\\
\end{array} \right ] \hspace{2pc} 1 \;\; \to \;\; ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $x$ has order $5$. let $ x \in G$ such that $(a^{-1})*(x^2)*(a) = x^3$ for some self inverse $a.$ Prove that $x$ has order $5.$
I don't know how to start this proof. Seems really difficult.
| $a^{-1}x^2a=x^3 \implies a^{-1}x^4a=x^6 \implies a^{-1}x^6a=x^9$ but $x^6=x^3x^3=(a^{-1}x^2a)(a^{-1}x^2a)=a^{-1}x^4a \implies a^{-1}(a^{-1}x^4a)a=x^9 \implies a^{-1}a^{-1}x^4aa=x^9$ now using $a^2=a^{-2}=e$ we have $x^4=x^9$ so $x^5=e$ and because $5$ is prime $x=e$ or $x$ has order $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Show that: $\sinh^{-1}(x) = \ln(x + \sqrt{x^2 +1 } )$ could someone Please give me some hint of how to do this question thanks
| Hint:
Assuming we already know these hyperbolic functions are invertible:
$$\sinh(\log(x+\sqrt{x^2+1})):=\frac12\left(e^{\log(x+\sqrt{x^2+1})}-e^{-\log(x+\sqrt{x^2+1})}\right)=$$
$$=\frac12\left(x+\sqrt{x^2+1}-\frac1{x+\sqrt{x^2+1}}\right)=\frac12\left(\frac{x^2+x^2+1+2x\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}}\right)=$$
$$=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How to find a function that is the upper bound of this sum? The Problem
Consider the recurrence
$ T(n) =
\begin{cases}
c & \text{if $n$ is 1} \\
T(\lfloor(n/2)\rfloor) + T(\lfloor(n/4)\rfloor) + 4n, & \text{if $n$ is > 1}
\end{cases}$
A. Express the cost of all levels of the recursion tree as a sum over the cost of e... | Suppose we start by solving the following recurrence:
$$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + 4n$$
where $T(1) = c$ and $T(0) = 0.$
Now let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
be the binary representation of $n.$
We unroll the recursion to obtain an exact formula for $n\ge 2$
$$T(n)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Epsilon and Delta proof of $\lim_{x\to0} \frac{2-\sqrt{4-x}}{ x}$ I need to prove $\lim_{x\to0} \frac{2-\sqrt{4-x}}{ x}$
I first found the limit to be $\frac{1}{4}$ by using l'hopital's rule.
By definition i need to find a $\delta > 0$ for every $\epsilon >0$
Then i will have $|x-0|<\delta$ and
$$|\frac{2-\sqrt{4-x}}... | It gets much simple when you write
$$
\frac{2−\sqrt{4−x}}x = \frac{(2-\sqrt{4−x})(2+\sqrt{4−x})}{x(2+\sqrt{4−x})}
=\frac1{2+\sqrt{4−x}}
$$
then:
\begin{align}
\left|
\frac1{2+\sqrt{4−x}} - \frac 14
\right|
&= \frac{|2 - \sqrt{4-x}|}{4(2+\sqrt{4−x})}
\le \frac{|2 - \sqrt{4-x}|}8
\\ &=
\frac{|2 - \sqrt{4-x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.