Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Converting Trig equation in Maple $$\frac{\sin^4(3x)}{18\cos^6(3x)} + \frac{\sin^4(3x)}{36\cos^4(3x)} = \frac{\tan^6(3x)}{18} + \frac{\tan^4(3x)}{12}$$
The answer on the left is what I got when I punched in the problem in Maple: $\operatorname{int}(\tan^3(3x)\cdot\sec(3x)^4, x)$
The answer on the right is my own using ... | Let's do this manually, since it's actually not that bad! I'll write in a condensed notation with $s=\sin(3x)$ and $c=\cos(3x)$ so I can type this more easily.
\begin{align*}
\frac{s^4}{18c^6} + \frac{s^4}{36c^4} &=\frac{s^4}{18c^6}-\frac{s^4}{18c^4} + \frac{s^4}{18c^4} + \frac{s^4}{36c^4} \\
&=\frac{s^4(c^2+s^2)}{18c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How many numbers are there which only contain digits $4$ and $7$ in them? I wanna know how many numbers $n$ are there which only contain digits $4$ and $7$ in them, where $1 ≤ n ≤ 10^9$.
Ex: $4, 7, 44, 47, 74, 77, ...$
I am trying to find a general equation to compute the numbers, given how many digits, which is $2$ in... | There are $2$ with length $1$, $2^2$ with length $2$, $2^3$ with length $3$, et cetera.
That observation leads to a total of:$$2+2^2+2^3+\cdots+2^9=2^{10}-2$$
The last equation on base of: $$(2-1)(2+2^2+2^3+\cdots+2^9)=(2^2+2^3+\cdots+2^{10})-(2+2^2+2^3+\cdots+2^9)=2^{10}-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Given $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$, where $a,b$ are positive constants $(a>b)$, prove that $f(x)$ is a periodic function
Given $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$, where $a,b$ are positive constants $(a>b)$, prove that $f(x)$ is a periodic function
I have done the following:
$f(a-x)=f(a+x)$....(1)
or, $f(-(... | I guess that $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$ hold for every $x$.
In particular, evaluating at $x+a$ instead of $x$,
$$
f(-x)=f(x+2a)
$$
Similarly $f(-x)=f(x+2b)$. Therefore
$$
f(x+2a)=f(x+2b)
$$
for every $x$. Evaluating at $x-2b$ instead of $x$, we obtain
$$
f(x+2a-2b)=f(x)
$$
and, setting $T=2a-2b$,
$$
f(x)=f(x+T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Tricks for estimating $ \lim_{x\rightarrow 0} \frac{d}{dx} \bigl(-\frac{1}{x} \ln\bigl(1 + \frac{(e^{-xu}-1) (e^{-xv}-1)}{e^{-x}-1} \bigr) \bigr)$ I'm trying to find a Taylor approximation of $ f(x) =-\frac{1}{x} \ln\left(1 + \frac{(e^{-xu}-1) (e^{-xv}-1)}{e^{-x}-1} \right) $ at $x = 0$.
For the derivation part Wolfra... | $\begin{array}\\
f(x)
&=-\frac{1}{x} \ln\left(1 + \dfrac{(e^{-ax}-1) (e^{-bx}-1)}{e^{-x}-1} \right)\\
&=-\frac{1}{x} \ln\left(1 + \dfrac{(-ax+O(x^2)) (-bx+O(x^2))}{-x+O(x^2)} \right)\\
&=-\frac{1}{x} \ln\left(1 + \dfrac{abx^2+O(x^3)}{-x+O(x^2)} \right)\\
&=-\frac{1}{x} \ln\left(1 - abx+O(x^2) \right)\\
&=\frac{1}{x}(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that gcd$(n^2+1, (n+1)^2+1)$ is either 1 or 5 My try to solve this question goes as follows:
$g=gcd(n^2+1, (n+1^2)+1) = gcd(n^2+1, 2n+1) = gcd(n^2-2n, 2n+1)$.
By long division:
$$n^2-2n = -2n(2n+1) + 5n^2$$
Since $g$ divides $n^2-2n$ and $g$ divides $(2n+1)$, $g$ divides $5n^2$. However, I need to show that $g$ ... | Clearly $$\gcd(n^ 2+1,(n+1)^2+1)\mid \bigl((n+1)^2+1\bigr)-\bigl(n^ 2+1\bigr)=2n+1,$$
hence also
$$\gcd(n^ 2+1,(n+1)^2+1)\mid (2n+1)n-2(n^2+1)=n-2 $$
as well as
$$\gcd(n^ 2+1,(n+1)^2+1)\mid (2n+1)n-2(n^2+1)=2n+1-2(n-2)=5. $$
This leaves only the possibilites $1$ and $5$ (which are both possible as can be seen by che... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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$ (5+2\sqrt6)^{\sin x} +(5-2\sqrt6)^{\sin x} = 2\sqrt3 $ , where $ 0 ≤ x≤ 360 $ There is something I haven't picked up on, a hint would be appreciated
Given that $(\sqrt3+\sqrt2)^2 = (5+2\sqrt6)$ and $ (\sqrt3-\sqrt2)^2 = (5-2\sqrt6)$
Find the values of x
for which$
(5+2\sqrt6)^{\sin x}
+(5-2\sqrt6)^{\sin x}
= 2\sqrt3... | Hint: Use that $$(5+2\sqrt{6})(5-2\sqrt{6})=1$$
Then you will get $$(5+2\sqrt{6})^{\sin(x)}+\frac{1}{(5+2\sqrt{6})^{\sin(x)}}=2\sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve recurrence relation! I was working with this recurrence relation :
$$\begin{cases}A(n,k) = A(n-1, k-1)+A(n-2, k-1)+A(n-1, k)\\
A(n, 0) = 1\\
A(n, 1) = 2n
\end{cases}$$
Generating function : $(1+x)/(1-x-x*y-x^2*y)$
Now since this involves two parameters, I tried changing it to a single parameter by fixing $k$ ... | We look at the generating function $\frac{1+x}{1-x-xy-x^2y}$ and derive the coefficients $A(n,k)$. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain
\begin{align*}
\color{blue}{A(n,k)}&=[x^ny^k]\frac{1+x}{1-x-xy-x^2y}\\
&=[x^ny^k]\frac{1+x}{1-x-x(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve $9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}$ $$9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}.$$
The equation states solve for $x$.
What I first did was put like bases together.
$$3^{2x}+3^{2x-1}= 2^{x+\frac{7}{2}}+ 2^{x+\frac{1}{2}}.$$
Then I factored $3^{2x}$ and $2^x$
$$3^{2x}(1+\frac{1}{3})=2^x(2^{\fr... | $$9^x-2^{x+\frac12}=2^{x+\frac72}-3^{2x-1}$$
$$\to\frac43(9^x)=9(2^{x+\frac12})$$
Hence we form the two iterates:
$$x_{n+1}=\log_9{\bigg[\frac{27}{4}(2^{x_n+\frac12})\bigg]}$$
and
$$x_{n+1}=\log_2{\bigg[\frac{4}{27}(9^{x_n})\bigg]}$$
The first gives the solution $x=\frac32$, the second diverges to $-\infty$, and thus t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $c$ and $n$ such that $\frac{x^3 \arctan x}{x^4 + \cos x +3} \sim cx^n$ as $x \to 0$ Where is my mistake in the below:
$$\frac{1}{c} \lim_{x \to 0} \frac{x^3 \arctan x}{x^{4+n} + x^n \cos x + 3x^n} = \frac{1}{c} \lim_{x \to 0} \frac{x^4}{x^{4+n}+x^n \cos x + 3x^n} \\ =\frac{1}{c} \lim_{x \to 0} \frac{1}{x^n + x^{n... | We have that
$$\frac{x^3 \arctan x}{x^4 + \cos x +3}\sim\frac{x^4}{x^4+4}\sim \frac14 x^4$$
and the guess is correct indeed
$$\frac{\frac{x^3 \arctan x}{x^4 + \cos x +3}}{\frac14 x^4}\to 1$$
As an alternative
$$\frac{\frac{x^3 \arctan x}{x^4 + \cos x +3}}{c x^n}=\frac1c\frac{x^3 \arctan x}{x^{4+n} + x^n\cos x +3x^n}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all pairs of positive integers $(a,b)$ such that $2^a+5^b$ is a perfect square. How do you solve such questions when they appear?
I know that this problem involves quadratic residues. Moreover, I also know that a=2,b=1 is possible. It may also be the only solution
I tried to take $\pmod{5}$ of this equation but i... | If $2^a+5^b = m^2$, then $m^2\equiv 0, 1, 4\pmod{5}$. $0$ is impossible since $2^a$ is never divisible by 5, so $2^a+5^b\equiv 2^a \equiv 1, 4\pmod{5}$. This implies that $a$ is even, say $a=2k$. This gives $5^b = m^2-2^{2k} = (m-2^k)(m+2^k)$, which implies that both $m-2^k$ and $m+2^k$ are powers of $5$, say
$$m-2^k =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Partial Fraction of $\int \frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)}$ If
$$\int\frac{ \left( \cos x + \sin 2x \right) \ \mathrm{d}x}{( 2 - \cos^2 x)(\sin x)} = \int \frac{A \ \mathrm{d}x}{(\sin x)} + B \int\frac{\sin x \ \mathrm{d}x}{ 1 + \sin^2 x} + C \int \frac{\mathrm{d}x}{1 + \s... | Write your Integrand in the form
$$\frac {\csc(x)(\cos(x)+2\sin(x)\cos(x))}{2-\cos^2(x)}$$
and then multiply numerator and denominator by
$$\frac{csc(x)(\cos(x)+2\sin(x)\cos(x))}{2-\cos^2(x)}$$
and we get
$$\newcommand{\dx}{\; \mathrm{d}}x\int \frac{\cot(x)(\csc(x)+2)\csc(x)}{\csc^2(x)+1}\dx$$
now Substitute
$$u=\csc(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$ A while ago I encountered this integral $$I=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$$ To be fair I spent some time with it and solved it in a heuristic way, I want to avoid that way so I won't show that approach, but the result I got is $\frac{\pi}{6} \ln(2+... | I thought about the comment you left the other day, and it has led me to this alternative approach, which is more direct than my other answer. Thanks a lot, I have learned so much from you!
Let $\omega=\sqrt[3]{-1}$, and let Ti2 denote the inverse tangent integral function:
\begin{align*}
I&=\int_0^1\frac{\left(1+x^2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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Finding the centroid of a tetrahedron I have four points to form a tetahedron
$$A=(0,-\frac{1}{2},-\frac{1}{4}\sqrt{\frac{1}{2}})
\\B=(0,\frac{1}{2},-\frac{1}{4}\sqrt{\frac{1}{2}})
\\C=(-\frac{1}{2},0,\frac{1}{4}\sqrt{\frac{1}{2}})
\\D=(\frac{1}{2},0,\frac{1}{4}\sqrt{\frac{1}{2}})$$
it looks like this :
I'm asked to ... | Your equation for the plane that passes through $\{A, B, C\}$ should be
$\displaystyle \frac12\sqrt{\frac12} x + \frac12 z = -\frac18 \sqrt{ \frac12 }$.
Namely, you were off by one negative sign starting from there.
As for the integrals, guessing your intention from the way you set them up, it looks like you also need ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
how is that expression is generated
I have no idea that how is red arrowed mammoth term is generated from the yellow arrowed term
Please explain
| This is just a substitution. You have from Taylor Series
$$
\sin(t) = t- \dfrac{t^3}{3!} + \dfrac{t^5}{5!}- \dfrac{t^7}{7!} + \cdots= t- \dfrac{t^3}{6} + \dfrac{t^5}{120}- \dfrac{t^7}{5040} + \cdots
$$
But you know also the Taylor Series for $e^x-1$:
$$
e^x-1 = x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots
$$
which th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Solving $z^{1+i}=4$. Solution:
Let $z=re^{i\theta}=r(\cos \theta + i\sin \theta)$. Then $z^{1+i}=e^{i\theta(1+i)}=e^{i\theta -\theta}=4$. So $e^{i\theta-\theta}=e^{-\theta}e^{i\theta}=e^{-\theta}(\cos \theta + i\sin \theta)=4(\cos 0 +i\sin 0) \Longleftrightarrow e^{-\theta}=4 \quad \text{and}\quad \theta =0$. But, $e^0... | $$4 = 2^2 = z^{1 + i}$$
$$2 \log 2 = (1+i) \log z$$
$${2 \over 1+i} \log 2 = \log z$$
$$2^{2 \over 1 + i} = z$$
$$2^{2(1-i) \over (1+i)(1-i)}$$
$$2^{2(1-i) \over 2} = z$$
$$2^{1 - i} = z$$
Check:
$$\left( 2^{1-i}\right)^{1+i} = 2^{1+1} = 2^2 = 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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$S(r,s) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+r+1)(2k+s+1)}$ for $r>s\geq0$ and even integers $S$ generates $\pi$ or rational values. \begin{align}
S(r,s) = \sum_{k=0}^{\infty} = \frac{(-1)^k}{(2k+r+1)(2k+s+1)}
\end{align}
Assuming $r>s\geq 0$ and both $r,s$ even integers.
A few values, using wolfram:
\begin{align}
&... | Write
$$H_z=\psi(z)+\gamma$$
where $\psi(z)$ is the digamma function. Then
$$H_{(z+1)/2}-H_{z/2}=2\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+z}$$
$$H_{z+1}=\frac{1}{z}+H_z$$
for $z$ nonzero and not a negative integer (DLMF 5.7.7, DLMF 5.5.2). Additionally, if $0<p/q<1$ is a rational number in lowest terms, then
$$H_{p/q}=-\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivati... | Your calculations are likely to be correct when $x>1$. When $x <-1$ write $\sqrt {\frac {x+1} {x-1}}$ as $\sqrt {\frac {-x-1} {-x+1}}=\frac {\sqrt {-x-1}}{\sqrt {-x+1}}$ and now compute the derivative as before. You will get the correct answer now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Using the Residue Theorem to Prove that $\int^{2\pi}_{0} \frac{1}{(a+\cos\theta)^{2}} d \theta=\frac{2\pi a}{(a^{2}-1)^{3/{2}}}.$ How do you evaluate the following integral? Here we take $a>1$.
$$\int^{2\pi}_{0} \frac{1}{(a+\cos\theta)^{2}} d \theta=\frac{2\pi a}{(a^{2}-1)^{\frac{3}{2}}}.$$
I know I have to use the Res... | Use the contour $|z| = 1$
First change the cosines into exponential forms.
$$\large\int \frac {1}{(a+\frac{e^{it}}{2} + \frac{e^{-it}}{2})^2} \ d\theta$$
$$z = e^{i\theta}\\ d\theta = \frac {1}{iz} dz$$
$$\large \oint_{|z| = 1} \frac {1}{iz(a+\frac{z}{2} + \frac{z^{-1}}{2})^2} \ dz$$
Which simplifies to:
$$\large \oint... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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For any integer $p \ge 3$, the largest integer $r$ such that $(x-1)^r$ is a factor of the polynomial $2x^{p+1}-p(p+1)x^2+2(p^2-1)x-p(p-1)$
For any integer $p \ge 3$, the largest integer $r$ such that $(x-1)^r$ is a factor of the polynomial $2x^{p+1}-p(p+1)x^2+2(p^2-1)x-p(p-1)$
I have tried the following:
If $(x-1)^r$... | For $(x-1)^r$ to be a divisor of $f(x)$, we should have $f^{(r-1)}(1)=0$ (the $r-1$-st derivative).
So consider
$$f'(x)=2(p+1)x^{p}-2p(p+1)x+2(p^2-1).$$
Then $f'(1)=2(p+1)-2p(p+1)+2(p^2-1)=0.$ This means $(x-1)^2$ is a factor of $f$.
Now consider,
$$f''(x)=2p(p+1)x^{p-1}-2p(p+1).$$
Thus $f''(1)=2p(p+1)-2p(p+1)=0$. Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2971803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Convergence or Divergence of $\sum_{n=1}^{\infty}(3^{1/n}-1)\sin(\pi/n)$ How to determine convergence of this series. $$\sum_{n=1}^{\infty}(3^{1/n}-1)\sin(\pi/n)$$
I've tried using comparison test:
$\sin(\pi/n) \leq \pi/n $, so:
$$(3^{1/n}-1)\sin(\pi/n)<(3^{1/n}-1)\pi/n < (3^{\frac{1}{n}})\frac{\pi}{n}$$
By comparis... | Your series is absolutely convergent by asymptotic comparison with $\sum_{n\geq 1}\frac{\log 3}{n}\cdot\frac{\pi}{n}=\frac{\pi^3}{6}\log(3)$.
If you like explicit bounds, you may notice that
$$ \frac{3}{2} = \frac{2n+1}{2n}\cdot\frac{2n+2}{2n+1}\cdot\ldots\cdot\frac{2n+n}{3n-1} ,\qquad 2=\frac{n+1}{n}\cdot\frac{n+2}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
find the sum to n term of $\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $ $$\frac{1}{1\cdot2\cdot3} + \frac{3}{2\cdot3\cdot4} + \frac{5}{3\cdot4\cdot5} + \frac{7}{4\cdot5\cdot6 } + ... $$
$$=\sum \limits_{k=1}^{n} \frac{2k-1}{k\cdot(k+1)\cdot(k+2)}$$ $$= ... | You are almost there. You can merge the parts of the series for which the denominator is similar and you will see they cancel each other. Then you are left with the terms for which the denominator is either smaller than $3$ or greater than $n$.
$$
\begin{aligned}
& \sum_{k=1}^n\frac{-1}{2k} + \sum_{k=1}^n\frac{3}{k+1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Fractional Linear Transformation of the Image of the Line $y=4-x$
I am trying to find the image of the line $y=4-x$ under the fractional linear transformation $$w=\frac{8}{z-2-2i}.$$
My method is as follows:
Rearranging yields $$z=\frac{8}{w}+2+2i.$$
Now, \begin{align}
z=x+iy&=\frac{8}{u+iv}+2+2i \\
&=\frac{8+(u+iv)(... | Your calculation is correct.
Here is a shorter way using the properties of the given mapping:
*
*Note that your mapping is a so called Möbius transformation. So, it maps generalized circles (circles and lines) onto generalized circles.
*To find the image of the line $y= 4-x$ which is the same as $z=x + (4-x)i$ you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2975939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Telescoping $\sum_{n\ge2} \ln(1-\frac1{n^2})$ leads to wrong result I know that this series converges to $-\ln2$. I'm not looking for someone to show me that here. I've used a way of telescoping the series that gives a result of 0. I want to know what's wrong about my process, I don't care (I already know) about other ... | Your argument should work if you carefully analyze the partial sum:
\begin{align*}
\sum_{k=1}^{2n} \log\left(1-\frac{1}{k^2}\right)
&= \log\left(\frac{1\cdot3\cdots(2n-1)}{3\cdot5\cdots(2n+1)}\right) + \log\left(\frac{2\cdot3\cdots(n+1)}{1\cdot2\cdots n}\right) \\
&= -\log(2n+1) + \log(n+1) \\
&= \log\left(\frac{n+1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
calculate $\lim_{x\to\infty} x + \sqrt[3]{1-x^3}$ So I multiplied by the conjugate and got $$\lim_{x\to\infty} \frac{x^2-(1-x^3)^\frac{2}{3} + x(1-x^3)^\frac{1}{3}-(1-x^3)}{x-(1-x^3)^\frac{2}{3}}$$
and this is where I got stuck.
| Here's a completely elementary solution.
For $x > 1$
$\begin{array}\\
d(x)
&=x + \sqrt[3]{1-x^3}\\
&=x - \sqrt[3]{x^3-1}\\
&=x(1 - \sqrt[3]{1-1/x^3})\\
&\gt 0\\
\end{array}
$
Also,
since
$(1-z)^3 < 1-z$
for $0 < z < 1$,
$\sqrt[3]{1-z} > 1-z$
so that
$\sqrt[3]{1-1/x^3}
\gt 1-\frac1{x^3}
$
for $x > 1$.
Therefore
$\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$2 < (1 + \frac{1}{n})^n < 3$ for all $n > 2$ Show by induction that $2 < (1 + \frac{1}{n})^n < 3$ for all natural n > 2.
*
*Induction base. For $n = 3$ the inequality is obviously true.
*Assume that for $n = k$ inequality is true, than i can prove $(1 + \frac{1}{n+1})^{n +1} < 3$.
$$\frac{(n + 1)^{n}}{n^{n}} < 3 ... | I'd like to point out an error in the proof for $\left(1+\frac1n\right)^n < 3$.
The step that says $\frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}} < 3$ is wrong (I plugged $n=3$ into the windows calculator and got a term slightly above 3). That term is equal to
$$\frac{3(n^2 + 2n)^n(n+2)}{(n^2+2n+1)^n(n+1)} = 3\frac{(n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find a locus of points Given a triangle $ABC$, $A'$ and $B'$ halves $BC$ and $AC$. We have a variable point on a line $AB$. Parallel to $AA'$ and $BB'$ through $P$ cuts $AC$ in $E$ and $BC$ in $F$. Now line $EF$ cuts $AA'$ in $M$ and $BB'$ in $N$. Lines $A'N$ and $B'M$ meet at $Q$. What is a locus of $Q$?
When playing... | Since the construction uses only affine-independent properties (incidence, collinearity, parallelism, midpoints) we can choose a convenient form of $\triangle ABC$. If we can show that the locus is the Steiner Inellipse for that form, then the locus is the Inellipse for every triangle.
We'll position vertex $C$ at the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Permutation of the string Ronald Mcdonald I am checking to see if I am on the right track for my Discrete Math class.
Given the string: RONALDMCDONALD
Vowels: AAOO
How many permutations do NOT have consecutive vowels in them?
I first visualized the problem: _R_N_L_D_M_C_D_N_L_D_
I answered this question by taking the ... |
How many permutations of the string RONALDMCDONALD do not have consecutive vowels in them?
The ten consonants consist of $1$ R, $2$ Ns, $2$ Ls, $3$ Ds, $1$ M, $1$ C. The four vowels consist of $2$ As and $2$ Os.
Your idea of arranging the consonants, then separating the vowels by placing them between the consonant... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determinate of symmetric $82\times 82$ matrix. I am trying to calculate the determinant of the matrix:
$
\left(\begin{matrix} -6 & -5 & & \dots & -5\\
-5 & -6 & \dots & & -5 \\
\vdots & & \ddots & & \vdots \\
-5 & \dots & -5 & -6 & -5\\
-5 & \dots & & -5 & -6
\end{matrix}\right)\in \mathbb{R}^{82\times 82}
$
How am I s... | Let $M = \left(\begin{matrix} -6 & -5 & & \dots & -5\\
-5 & -6 & \dots & & -5 \\
\vdots & & \ddots & & \vdots \\
-5 & \dots & -5 & -6 & -5\\
-5 & \dots & & -5 & -6
\end{matrix}\right)$.
Then $M=\left(\begin{matrix} -5 & -5 & & \dots & -5\\
-5 & -5 & \dots & & -5 \\
\vdots & & \ddots & & \vdots \\
-5 & \dots & -5 & -5 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Combinatorial proof of $\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3$?
Give a combinatorial proof of the following identity: $$\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3.$$
I've been working on this proof for hours, however I'm not able to show LHS = RHS-
I completely understand binomial theorem and few c... | Brute force:
$27n^{3} - 27n^{2}+ 6n = 27n^{3} - 27n^{2} + 6n$
$27n^{3} - 9n^{2} - 18n^{2} + 6n = 3n^{3} - 3n^{2} - 6n^{2} + 6n + 18n^{3} - 18n^{2} + 6n^{3}$
$(9n^{2} - 3n)(3n - 2) = (3n^{2} - 3n)(n - 2) + 18n^{2}(n-1) + 6n^{3}$
$3n(3n - 1)(3n - 2) = 3n(n - 1)(n - 2) + 3(6n)n(n-1) + 6n^{3}$
$3n(3n - 1)(3n - 2)/6 = 3n(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
} |
Find the value of $(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ Find the value of
$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$ where $\alpha,\beta,\gamma$ are roots of the equation $x^3+2x^2+3x+3=0$.
I tried to use the formula which is... | I think your way is not really practical. Here it is a shortcut:
$$ \begin{array}{|c|c|}\hline \text{Roots}&\text{Polynomial}\\
\hline \alpha,\beta,\gamma & x^3+2x^2+3x+3\\
\hline \frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma} & x^3+x^2+\frac{2}{3}x+\frac{1}{3}\\
\hline 1+\frac{1}{\alpha},1+\frac{1}{\beta},1+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$ It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$?
Here's my reasoning:
(1) For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2) For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3) Fo... | We can also proceed as follow
$$x(1 - ax) \le \frac{1}{4a} \iff ax^2- x+\frac{1}{4a}\ge 0$$
which holds when $a>0$ since
$$\Delta=1-4\cdot \frac{1}{4a}=1-1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Prove $\sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n$ without the Beta function I know how to prove
$$\sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n$$
by tackling it with the beta function.
I was actually wondering if there is a proof of this fact without using the property of the Beta function $$B(x,y) = \f... |
We have
\begin{align}
f_n&=\color{blue}{\sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\frac{1}{k}}\\
&=\sum_{k=1}^{n}(-1)^{k-1}\left(\binom{n-1}{k}+\binom{n-1}{k-1}\right)\frac{1}{k}\\
&=f_{n-1}+\sum_{k=1}^{n}(-1)^{k-1}\binom{n-1}{k-1}\frac{1}{k}\\
&=f_{n-1}-\frac{1}{n}\sum_{k=1}^{n}(-1)^k\binom{n}{k}\\
&=f_{n-1}-\frac{1}{n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
If $a=b+c$ prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer
If $a=b+c$, and $a$,$b$,$c\in \Bbb N$, prove that $S=a^4+b^4+c^4$ is twice the square of a positive integer.
Source: a list of problems used in the preparation to math contests.
My attempt:
By making the substitution $a=b+c$ in $S$ and de... | We have:
$$S=(b+c)^4+b^4+c^4=2b^4+4b^3c+6b^2c^2+4bc^3+2c^4$$
$$=2(b^4+2b^3c+3b^2c^2+2bc^3+c^4)=2 (b^2+bc+c^2)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2985553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Simultaneous real solution of $x^3+y^3+1+6xy=0$ & $xy^2+y+x^2=0$ I am trying to solve the following system of non-linear equations in real numbers:
$x^3+y^3+1+6xy=0$ & $xy^2+y+x^2=0$, with $x,y$ real.
I can only see that $xy\ne 0$.
I have no clue whether a solution exists or not and how to find any solution. I cannot... | Not a very satisfactory answer since it involves the cubic formula...
Set $u = xy$. Multiply the first equation by $x^3$ and the second by $x$. Then we get
$x^6+u^3+x^3+6ux^3 = 0$
$u^2+u+x^3 = 0$
Solving for $x^3$ in the second and substituting into the first, we get
$u^4+2u^3+u^2+u^3-u^2-u+6u(-u^2-u)=0$
$u^4+2u^3+u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
ّFind $x$ such that $ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$.
ّFind $x$ such that
$$ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}\,.$$
My attempt:
After clearing the denominators, I obtain this quartic equation
$$104 x^{4} -624 x^{3} +886 x^{2} +150x-225=0.$$
I don't know how to proceed from here.
| One can see that $1/2$ is a solution of $\frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$. Then it is easy to see that also $5/2$ is a solution.
Can you proceed ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
What value does $\frac{1}{n} \sqrt[n]{n(n+1)(n+2)\cdots(2n)}$ tend to? I need to find where this sequence tends to:
$$\frac{1}{n} \sqrt[n]{n(n+1)(n+2)...(2n)}$$
My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $\dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence... | By ratio root criterion we have
$$a_n=\frac{n(n+1)(n+2)...(2n)}{n^n} \quad b_n=\frac{1}{n} \sqrt[n]{n(n+1)(n+2)...(2n)}=\sqrt[n] a_n$$
then
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)(n+2)...(2n+2)}{n(n+1)(n+2)...(2n)}\frac{n^n}{(n+1)^{n+1}}=\frac{(2n+2)(2n+1)}{n(n+1)}\frac{1}{\left(1+\frac1n\right)^n}\to \frac 4 e$$
which impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $\det\left[\begin{smallmatrix}1&\cos a&\cos b\\ \cos a&1&\cos(a+b) \\ \cos b&\cos(a+b)&1 \end{smallmatrix}\right]=0$ I am unable to show - without expanding, by using determinant properties - that
$$\det\begin{bmatrix} 1 &\cos a &\cos b\\
\cos a &1 ... | To flesh out my comment from earlier,$$\Delta:=\left|\begin{array}{ccc}
1 & \cos a & \cos b\\
\cos a & 1 & \cos\left(a+b\right)\\
\cos b & \cos\left(a+b\right) & 1
\end{array}\right|=\left|\begin{array}{ccc}
1 & \cos a & \cos b\\
\cos a & 1 & \cos\left(a+b\right)\\
0 & -\sin a\sin b & \sin^{2}b
\end{array}\right|=\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What am I doing wrong finding the limit of $\left(\frac{3x^2-x+1}{2x^2+x+1}\right)^\left(\frac{x^3}{1-x}\right)$? $$\lim_{x\to\infty}\left(\frac{3x^2-x+1}{2x^2+x+1}\right)^\left(\frac{x^3}{1-x}\right)$$
Divide by $x^2$, get
$$\lim_{x\to\infty}(1,5)^\infty=\infty$$
The answer in the book is $0$.
I've also tried substitu... | Because the natural logarithm is injective, it turns out that: $\lim_{x \to \infty} \frac{3x²-x+1}{2x²+x+1}^{\frac{x^3}{1-x}}=(\lim_{x \to \infty} \frac{3x²-x+1}{2x²+x+1})^{(\lim_{x \to \infty} \frac{x^3}{1-x})}$. Thus the limit is $(\frac{3}{2})^{-\infty}=\frac{1}{1.5^{\infty}}=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the general solution to the ODE $x\frac{dy}{dx}=y-\frac{1}{y}$ I have been working through an ODE finding the general solution and following the modulus through the equation has left me with four general solutions, as shown below. Online ODE solvers, however, have only calculated the two (positive and negative), o... | If $-1<y<1$ then $|y^2-1|=1-y^2$.
Another approach: Write the equation as
$$\dfrac{x\ dy-y\ dx}{x^2}=\dfrac{-1}{x^2y}\ dx$$
or
$$\left(\dfrac{y}{x}\right)d\left(\dfrac{y}{x}\right)=\dfrac{-1}{x^3}\ dx$$
and
$$\left(\dfrac{y}{x}\right)^2=\dfrac{1}{x^2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\int_0^1\frac{\ln(1+x-x^2)}xdx$ without using polylogarithms.
Evaluate $$I=\int_0^1\frac{\ln(1+x-x^2)}xdx$$ without using polylogarithm functions.
This integral can be easily solved by factorizing $1+x-x^2$ and using the values of dilogarithm at some special points.
The motivation of writing this post i... | $$I=\int_0^1 \frac{\ln(1+x-x^2)}{x}dx\overset{x\to 1-x}=\int_0^1 \frac{\ln(1+x-x^2)}{1-x}dx$$
Averaging the two integrals from above gives us:
$$I=\frac12 \int_0^1 \frac{\ln(1+x-x^2)}{x-x^2}dx=\frac12I(1)$$
Where we considered, in order to apply Feynman's trick, the following integral:
$$I(a)=\int_0^1 \frac{\ln(1+a(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Seeking methods to solve $\int_{0}^{\infty} \frac{x - \sin(x)}{x^3\left(x^2 + 4\right)} \:dx$ I recently asked for definite integrals that can be solved using the Feynman Trick. One of the responses is the following integral:
$$I = \int_{0}^{\infty} \frac{x - \sin(x)}{x^3\left(x^2 + 4\right)} \:dx$$
I employed Laplace ... | My approach:
Let
$$I(t) = \int_{0}^{\infty} \frac{xt - \sin(xt)}{x^3\left(x^2 + 4\right)} \:dx$$
Where $I = I(1)$
Taking the first derivative:
$$ \frac{dI}{dt} = \int_{0}^{\infty} \frac{x - x\cos(xt)}{x^3\left(x^2 + 4\right)} \:dx = \int_{0}^{\infty} \frac{1 - \cos(xt)}{x^2\left(x^2 + 4\right)} \:dx$$
Taking the se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $\alpha$,$\beta$ if $\lim_{x→∞}[ \sqrt {ax^2+2bx+c} - \alpha x -\beta] = 0$ Here is my approach.
Consider;
$$ax^2 + 2bx + c = 0$$
or;
$$ x_{±} = \frac {-b±\sqrt {b^2-ac}}{a} = \frac {-b±\sqrt D}{a}$$
Hence;
$$\sqrt {ax^2+2bx+c} = \sqrt {(x+\frac {b-√D}{a})(x+\frac {b+√D}{a})}$$
$$=x\sqrt {(1+\frac {b-√D}{ax})(1+\f... | You want $$ \lim _{x\to \infty }ax^2+2bx+c -(\alpha x +\beta )^2=0$$
That implies $$ a= \alpha ^2, c=\beta ^2, b=\alpha \beta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3007306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving $\sin^2 x +1=2x$ How do I solve this equation?
$$\sin^2 x +1=2x$$
I have no idea how to attack the problem.
Thanks!
| Just for the fun of the approximation.
Using the double angle formula Rewrite the equation as $$\cos(2x)+4x=3$$ Now, let $t=2x$ to make the equation
$$\cos(t)+2t=3$$
Now, using the approximation
$$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad (-\frac \pi 2 \leq t\leq\frac \pi 2)$$ we get the cubic equation
$$2 t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Integrate $\frac{1}{ \sqrt{T^2 - \tau^2}}\exp\left(-\frac{a^2}{4 (T + \tau )} - \frac{b^2}{4 (T-\tau )}\right)$ I want to compute the integral
$$
\int_t^T \int_{-\infty}^\infty \frac{1}{ \sqrt{\tau - t} (T-\tau)} \exp\left(-\frac{(z-x)^2}{2(\tau - t)} -\frac{(z-v)^2 + (z-w)^2}{2(T-\tau)} \right) d z d \tau.
$$
First in... | $$I=\underbrace{\int_0^{T}\frac{d\tau}{\sqrt{T^2-\tau^2}}\exp\left(-\frac{a^2}{4(T+\tau)}-\frac{b^2}{4(T-\tau)}\right)}\\{\tau\rightarrow \frac{T(1-y^2)}{1+y^2}}$$
$$=\exp\left(-\frac{a^2+b^2}{8T}\right)\int_0^{1} \frac{dy}{1+y^2}\exp\left(-\frac{a^2y^2+\frac{b^2}{y^2}}{8T}\right)$$
Let's parametrize the integral in or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$ If $x \ge 5$, is $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{x} < \log_2 x$
I believe the answer is yes.
Here is my thinking:
(1) $\log_2{5} > 2.32 > 2.284 > 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}$
(2) Assume up to $x$ that $\... | We can use induction as well. If we assume that it's true for $n$:
$$1+\frac{1}{2}+...+\frac{1}{n}+\frac{1}{n+1}<\log_2(n)+\frac{1}{n+1}$$
So we need to prove that
$$\log_2(n)+\frac{1}{n+1} < \log_2(n+1)$$
$$\frac{1}{n+1}< \log_2\left(1+\frac{1}{n}\right)$$
$$2^{\frac{1}{n+1}}<1+\frac{1}{n}$$
$$2<\left(1+\frac{1}{n}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solution to infinite product $\prod_{p-primes}^{\infty} \frac{p}{p-1}$ I want to find the $\prod_{p-primes}^{\infty} \frac{p}{p-1}$. This question stems from a question from amc 12a 2018. It goes as follows: Let $A$ be the set of positive integers that have no prime factors other than $2$, $3$, or $5$. The infinite sum... | ${p\over p-1}=1+{1\over p-1}>1+\frac1p$ and the sum of the reciprocals of the primes diverges, so the given product also diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the remainder of the division of polynomials $x^{2007}$ divided by $x^2-x+1$.
I consider to solve this problem, should I break the $x^{2007}$ to find the formula $x^2-x+1$?
| Doing long division:
$$\frac{x^{2007}}{x^2-x+1}=\frac{x^{2008}+x^{2007}}{x^3+1}=\\
\frac{(x^3+1)(x^{2005}+x^{2004}-x^{2002}-x^{2001}+x^{1999}+x^{1998}-\cdots-x^4-x^3+x+1)-x-1}{x^3+1}=\\
a(x)-\frac{x+1}{x^3+1}=\\
a(x)+\frac{\color{red}{-1}}{x^2-x+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Closed form of $\int{\lfloor{x}\rfloor}dx$ I calculated $\int{\lfloor{x}\rfloor}dx$ and i got this result:
$$\int{\lfloor{x}\rfloor}dx = \frac{x^2-x}{2}+\sum_{k=1}^{\infty}\left(\frac{\sin(k\pi x)}{k\pi}\right)^2+c$$
Do you know if this series have a closed form?
We found a nice identity!
$$\int{\lfloor{x}\rfloor}dx = ... | Interpret the integral as a Riemann-Stieltjes integral, you can integrate it by part.
For $y > 0$, you get something like
$$\begin{align}\int_0^y \lfloor x \rfloor dx
&=
\int_{0-}^{y+} \lfloor x \rfloor dx =
y \lfloor y \rfloor - \int_{0-}^{y+} x d\lfloor x \rfloor\\
&= y \lfloor y \rfloor - \sum_{k=0}^{\lfloor y\rf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
System of 3 equations problem
If
$$x+y+z=1\\x^2+y^2+z^2={3\over2}\\x^3+y^3+z^3=1$$
Then how much is
$$x^4+y^4+z^4$$
So what I'm pretty sure about, this shouldn't be solved for $x$,$y$, and $z$, but I should try to make $x^4+y^4+z^4$ from these equations.
I did this:
$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+... | Hint:
$$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$
$$(x+y+z)^2=?$$
$$x^3+y^3+z^3-3xyz=(x+y+z)((x^2+y^2+z^2-(xy+yz+zx))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there another mathematical way to approach this problem?
$$f(x)-f(x-1) = x-5$$
$$f(16) = 74$$
Compute $f(1)$.
So, this is basically a linear function. I, however, calculated it without taking the easier way to approach this problem.
If
$$f(16) - f(15) = 11$$
Then
$$74-f(15) = 11 \implies f(15)=63$$
This will take ... | First, note that $f$ isn't linear at all (if it were linear, then $f(x)-f(x-1)$ would be constant. However, that difference does suggest something quadratic: its second difference will be constant. Thus, we'll try to find $a$, $b$, and $c$ such that $f(x) = ax^2 + bx + c$ satisfies this. For this purpose, we garner som... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Proving a three variables inequality Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)\ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c \ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ getting that $a+b+c\ge9$.
Then I al... | Since $ab+bc+ca= abc$ and $$ab+bc+ca\geq 3\sqrt[3]{a^2b^2c^2}\implies abc\geq 27$$
Now $$(a+1)(b+1)(c+1)=2abc+a+b+c+1\geq 55+3\sqrt[3]{27} = 64$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3026163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
How to factor equations with the form $x^2 + xy + y^2$? Title says it all. I'm having a hard time factoring anything looking like the equation $x + xy + y$. Especially trig ones, like $9\sin^2x + 12\sin x\cos x + 4\cos^2x$.
Thanks!
| If you want to factor expressions of the form $\alpha x^2+\beta xy+\gamma y^2$, observe that $$\begin{align*}\alpha x^2+\beta xy+\gamma y^2&=\alpha y^2\left((xy^{-1})^2+\beta\alpha^{-1} (xy^{-1})+\gamma\alpha^{-1}\right)\\&=\alpha y^2\left(xy^{-1}-\frac{-\beta+ \sqrt{\beta^2-4\alpha\gamma}}{2\alpha}\right)\left(xy^{-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Conjecture: if $a$, $b$ and $c$ have no common factors, dividing each of them by their sum yields at least one irreducible fraction
Let $a$, $b$ and $c$ be $3$ integers with no common factors.
I conjecture that at least one of the three fractions:
$$\frac{a}{a+b+c},\quad\frac{b}{a+b+c},\quad\frac{c}{a+b+c}$$
is irredu... | The smallest counterexample is $(2,3,25)$. An infinite family of counterexamples is $(2,3,25+30n)$:
$$
\begin{align*}
\frac{a}{a+b+c} &= \frac{2}{30+30n} = \frac{1}{15+15n}
\\ \frac{b}{a+b+c} &= \frac{3}{30+30n} = \frac{1}{10+10n}
\\ \frac{c}{a+b+c} &= \frac{25+30n}{30+30n} = \frac{5+6n}{6+6n}
\end{align*}
$$
There ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find limit $\lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}} -\sqrt{2x^{4}}\right)$ $\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^{4} +1}}\displaystyle -\sqrt{2x^{4}}\right)$
$\displaystyle \lim\limits _{x\rightarrow \infty }\left(\sqrt{x^{4} +x^{2}\sqrt{x^... | Now, use that
$$\sqrt{x^4+1}-x^2=\frac{1}{\sqrt{x^4+1}+x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove: $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2} < \infty$ without L'Hôpital's Given $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}$, prove that it converges.
I tried to use the Ratio test.
I got a terrible algebraic expression: $$\lim_{n \to \infty} \frac{a_{n+1}}... | Another approach using the Root test.
Let $L = \lim_{n\to\infty} \sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}} $
$$ \begin{align}
L &= \lim_{n\to\infty} \exp \log \sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}} \\
&= \exp \lim_{n\to\infty} \frac{\log \left(\left( \frac{n^2+1}{n^2+n+1} \right)^{n^2}\right)}{n} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $Q(x)$ at $x= -1$, knowing some of its properties.
Given polynomial $Q$ with real coefficients such that $Q(1)=1$ and
$$
\frac{Q(2x)}{Q(x+1)}= 8 -\frac{56}{x+7},
\quad \forall x \ne -7 \text{ and } Q(x+1)\neq 0\,.
$$
Find $Q(-1)$.
| Rewrite the equation as
$$(x+7)\,Q(2x)=8x\,Q(x+1)\,.\tag{*}$$
Thus, $2x\mid Q(2x)$ or $x\mid Q(x)$, and $x+7\mid Q(x+1)$ or $x+6\mid Q(x)$. That is, $Q(x)=x(x+6)\,R(x)$ for some polynomial $R(x)$. Show that
$$(x+3)\,R(2x)=2(x+1)\,R(x+1)\,.$$
Ergo, $2x+2\mid R(2x)$ or $x+2\mid R(x)$, and $x+3\mid R(x+1)$ or $x+2\mid R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving this equation: $3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}$
Solve this equation:
$$3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}\qquad (1)$$
I tried to make both sides of the equation have a same base and I started:
$$(1)\Leftrightarrow 3^{\log_{4}x}.\sqrt{3}+ \frac{3^{\lo... | First of all impose the necessary existence conditions, that is: $x > 0$ for the logarithms and $x \geq 0$ for the square root.
That is, eventually,
$$x > 0$$
for the whole equation.
Then follow Siong Thye Goh reasoning, obtaining the final equation he wrote.
At that point:
$$1 + \log_4(3)\log_4(x) = \frac{\log_4(3)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Limit of sum as definite integral I don't understand why $$\displaystyle \sum_{k=1}^n \dfrac{n}{n^2+kn+k^2} < \lim_{n\to \infty}\sum_{k=1}^n \dfrac{n}{n^2+kn+k^2}$$
whereas
$$\displaystyle \sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2} > \lim_{n\to \infty}\sum_{k=0}^{n-1} \dfrac{n}{n^2+kn+k^2} $$
I know that $$\lim_{n\to \inft... | Consider the function
$$f(x)=\frac{1}{1+x+x^2}$$
and note that the your inequality holds because $f$ is strictly decreasing.
Indeed for $n\geq 1$, $k\geq 0$, and $x\in [\frac{k}{n},\frac{k+1}{n}],$
$$f(\frac{k}{n})> f(x)> f(\frac{k+1}{n}).$$
By integrating over the interval $[\frac{k}{n},\frac{k+1}{n}]$, we get
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding matrix $A^n,$ when $\lim_{n \to \infty}$
Finding $\lim_{n\rightarrow \infty}\begin{pmatrix}
1 & \frac{x}{n}\\ \\
-\frac{x}{n} & 1
\end{pmatrix}^n$ for all $x\in \mathbb{R}$
Try:
Let $$ A = \begin{pmatrix}1&\frac{x}{n}\\\\-\frac{x}{n}&1\end{pmatrix}.$$
Then $$ A^2 = \begin{pmatrix}1-\frac{x^2}{n^2}&\frac{2x}... | We have
$$A = \sqrt{1 + \frac {x^2}{n^2}}\begin{pmatrix}\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}&\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}\\\\-\frac {\frac x n} {\sqrt{1 + \frac {x^2}{n^2}}}&\frac 1 {\sqrt{1 + \frac {x^2}{n^2}}}\end{pmatrix}$$
So if $\theta_n$ is such that $\cos \theta_n=\frac 1 {\sqrt{1 + \frac {x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3037992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve the equation $2z^{2}+i=-2$?
How to solve the equation $2z^{2}+i=-2$?
Assuming that $k=-1-i/2$ we have $|k|=\sqrt{5}/2$ and
$$
\cos \phi=(-2\sqrt5)/5,\quad \sin \phi= -\sqrt5/5.
$$
How should I find $\phi$?
| You won't find the exact value of $\phi$.
Here is a general method for solving $z^2=p+qi$, where $p$ and $q$ are real.
If $z=a+bi$ where $a$ and $b$ are real numbers, then $z^2=a^2-b^2+2abi=p+qi$, hence $p=a^2-b^2$ and $q=2ab$. You can also notice that $\vert z\vert^2=a^2+b^2=\sqrt{p^2+q^2}$, so you can find $a^2$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proof verification: For $a$, $b$, $c$ positive with $abc=1$, show $\sum_{\text{cyc}}\frac{1}{a^3(b+c)}\geq \frac32$ I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've use... | I think it's better to end your proof by C-S and AM-GM:
$$\sum_{cyc}\frac{x^2}{y+z}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(y+z)}=\frac{1}{2}(x+y+z)\geq\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving $\int\limits_{-\infty}^\infty \frac{1}{x^8+1}dx$ through Glasser's Master Theorem Trying to find a way to solve
$$\int_{-\infty}^\infty \frac{1}{x^8+1}dx$$
through Glasser's Master Theorem, more specifically the Cauchy–Schlömilch substitution. Preferably, I'm looking for the closed form solution, and I am a... | Decompose the integrand to express the integral as
\begin{align}
I=&\int_{-\infty}^\infty \frac1{1+x^8}dx \\
=& \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{\sqrt2+x^2}{x^4+\sqrt2 x^2+1}dx
+ \frac1{2\sqrt2}\int_{-\infty}^\infty \frac{\sqrt2-x^2}{x^4+\sqrt2 x^2+1}dx\\
\overset{x\to \frac1x}=&
\frac1{2\sqrt2}\int_{-\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
If $xy$ divides $x^2 + y^2$ show that $x=\pm y$ Problem statement : Let $x,y$ be integers, show that if $xy$ divides $x^2 + y^2$ then $x=\pm y.$
What I have tried:
I can reduce this to the case where $\gcd(x,y)=1$, since if $x$ and $y$ have a common factor, $d$ say, then $d^2$ divides through both $xy$ and $x^2 + y^... | Rational Algebraic Integer Approach
Suppose that
$$
\frac{x^2+y^2}{xy}=\frac xy+\frac yx\in\mathbb{Z}\tag1
$$
Note that if $q=\frac xy\in\mathbb{Q}$ and $q+\frac1q=n\in\mathbb{Z}$, then
$$
\left(q-\frac1q\right)^2=n^2-4\in\mathbb{Z}\tag2
$$
This means that $z=q-\frac1q$ is a rational solution to $z^2-(n^2-4)=0$; that i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Determining properties of a polynomial $f$ satisfying $f(x^2)-xf(x) = x^4(x^2-1)$ for $x \in\Bbb R^+$
Let $f$ be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x \in\Bbb R^+$. Then which of the following is correct?
A) $f$ is an even function
B) $f$ is an odd function
C) $\displaystyle\lim_{x\to \infty} \frac{f(x... | You don't need to find $f(x)$. For instance, statements A and B can be dealt with by considering
$$
f(x)=\frac{f(x^2)-x^4(x^2-1)}{x} \tag{*}
$$
and so
$$
f(-x)=\frac{f(x^2)-x^4(x^2-1)}{-x}=-f(x)
$$
showing that $f$ is an odd function. One can object that (*) only holds for $x>0$, but the right-hand side is a polynomial... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$
With $z\in \mathbb C$ find the maximum value for |z| such that
$$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
Source: List of problems for math-contest training.
My attempt: it is easy to see that the given c... | The reverse triangle inequality implies that $|z^2+1|\geq ||z^2|-1|$ so that if $|z|^2\geq 1$ we see that
$$|z|\geq |z|^2-1$$
By solving this inequality using the quadratic formula, we see that $|z|\leq \frac{1}{2}(1+\sqrt{5})$, so the maximum must be less than $\frac{1}{2}(1+\sqrt{5})$. We see that this maximum is at... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^... | Sketch of a solution rewriting the given system in terms of linear algebra:
*
*Set $\boxed{a = \binom{x}{y}}$. Then the given system looks as follows:
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases} \Longleftrightarrow \boxed{a +\frac{1}{||a||^2}\begin{pmatrix}3 & -1 \\ -1 & -3 \en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
We also know that
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have
$$\sin a + \sin b = 2 \sin\fra... | Use that $$\sin(A)+\sin(B)=2\cos\left(\frac{A-B}{2}\right)\sin\left(\frac{A+B}{2}\right)$$ and $$\sin(A+B)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Proof of polynomial divisibility without using complex numbers? My question is the same as polynomial of degree n and its divisor except I want a solution that does not make use of complex numbers
Problem: Find all positive integers $n$ such that $x^2+x+1\mid (x+1)^n+x^n+1$
Using wolframalpha, I can see that a numbe... | First I will prove that: $$n \in \{2,4\} \pmod{6}$$ is a necessary condition:
Just substitute x for the value 2 so that the equation results:
$$7 | 3 ^ n + 2 ^ n + 1$$
$$3 ^ n + 2 ^ n + 1 \equiv 0 \pmod{7}$$
whose all solutions are:
$$n \in \{2,4\} \pmod{6}$$
Now I am going to prove that these two conditions are suffic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Radius of largest circle in an ellipse Consider an ellipse whose major and minor axis of length $10$ and $8$ unit respectively. Then the radius of the largest circle which can be inscribed in such an ellipse if the circle's center is one focus of the ellipse.
What I tried:
Assuming that major axis and minor axis of an ... | You want to minimize $r^2=(x-3)^2+y^2$ subject to $\frac{x^2}{25}+\frac{y^2}{16}=1$:
$$L(x,y,\lambda)=(x-3)^2+y^2+\lambda\left(1-\frac{x^2}{25}-\frac{y^2}{16}\right)\\
\begin{cases}L_x=2x-6-\frac{2x\lambda}{25}=0\\ L_y=2y-\frac{y\lambda}{8}=0\\ L_{\lambda}=1-\frac{x^2}{25}-\frac{y^2}{16}=0\end{cases} \Rightarrow (x,y)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $-1\leq x, y \leq 1$ and $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1$, find $x^2+y^2$ Let $x,\, y\in\mathbb
R,\ -1\leq x,\, y\leq 1$ such that $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1.$
Find the sum $S = x^2+y^2.$
| Let $x=\sin\alpha$ and $y=\sin\beta$, where $\{\alpha,\beta\}\subset[-\frac{\pi}{2},\frac{\pi}{2}].$
Thus, the condition gives $\sin(\alpha+\beta)=1.$
Can you end it now?
Also, we can make squaring twice.
Indeed, the condition gives
$$x^2(1-y^2)+y^2(1-x^2)+2xy\sqrt{(1-x^2)(1-y^2)}=1$$ or
$$2xy\sqrt{(1-x^2)(1-y^2)}=1-x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the interval of convergence of the series $ \sum^{\infty}\limits_{k=0} ((-1)^k+3)^k(x-1)^k $ I wish to find the interval of convergence of the following series
\begin{align} \sum^{\infty}_{k=0} ((-1)^k+3)^k(x-1)^k \end{align}
PROOF
Wittingly,
\begin{align} \left[(-1)^k+3\right]^k= \begin{cases} 0,&\text{if}\;j=0;\... | Your justification of the equality$$\limsup_{k\to\infty}\sqrt[k]{((-1)^k+3)^k(x-1)^k}=\lvert x-1\rvert\limsup_{k\to\infty}\sqrt[k]{ 2^{2k+1}}$$doesn't hold. You have $\bigl((-1)^{2k+1}+3\bigr)^{2k+1}=2^{2k+1}$. Therefore, what you should consider here is$$\sqrt[2k+1]{2^{2k+1}}=2.$$But, since $\bigl((-1)^{2k}+3\bigr)^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the $\frac mn$ if $T=\sin 5°+\sin10°+\sin 15°+\cdots+\sin175°=\tan \frac mn$ It's really embarrassing to be able to doesn't solve this simple-looking trigonometry question.
$$T=\sin(5^\circ) +\sin(10^\circ) + \sin(15^\circ) + \cdots +\sin(175^\circ) =\tan \frac mn$$
Find the $\frac mn=?$, where $m$ and $n$ are p... | By geometric series,
$$T=\sin 5°+\sin10°+\sin 15°+...+\sin175° = \\
\operatorname{Im} (\sum_{n=1}^{35}\exp(i n 5 \pi/180)) =\\
\operatorname{Im} \exp(i 5 \pi/180) \frac{\exp(i 35 \cdot 5 \pi/180)-1}{\exp(i 5 \pi/180)-1} =\\
\operatorname{Im} \exp(i 5 \pi/180) \exp(i 34 \cdot 5 \pi/360) \frac{\sin( 35 \cdot 5 \pi/360)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
In $\Delta ABC$, find $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ if $b+c=3a$
If in a triangle ABC, $b+c=3a$, then $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ is equal to ?
My reference gives the solution $2$, but I have no clue of where to start ?
My Attempt
$$
\cot\dfrac{B}{2}.\cot\dfrac{C}{2}=\frac{\cos\frac{B}{2}.\cos\frac{C}{2}... | In the standard notation we obtain: $$\sin\beta+\sin\gamma=3\sin(\beta+\gamma)$$ or
$$2\sin\frac{\beta+\gamma}{2}\cos\frac{\beta-\gamma}{2}=6\sin\frac{\beta+\gamma}{2}\cos\frac{\beta-\gamma}{2}$$ or
$$\cos\frac{\beta}{2}\cos\frac{\gamma}{2}+\sin\frac{\beta}{2}\sin\frac{\gamma}{2}=3\left(\cos\frac{\beta}{2}\cos\frac{\ga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
prove $a,b,c$ in A.P if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$
In $\Delta ABC$, if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
\sin A=\frac{2.5}{6}.\frac{36}{61}=\frac{60}{61}\\
\sin C=\frac{2.2}{5}.\frac... | It is $$\tan(\alpha/2)=\frac{r}{s-a}$$ and $$\tan(\gamma/2)=\frac{r}{s-c}$$ where $$s=\frac{a+b+c}{2}$$ so we get $$\frac{5}{6}(s-a)=\frac{2}{5}(s-c)$$
Can you finish now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Proving prime divisibility relation between $a^2-a+3$ and $b^2-b+25$. Let $p$ be a given prime number. Prove that there exists an integer $a$ such that $p|a^2-a+3$ if and only if there exists an integer $b$ such that $p|b^2-b+25$.
I've managed to prove that if $p|a^2-a+3$ for some $a$, then $p|b^2-b+25$ for some $p$ l... | $a^2-a+3\equiv0\pmod p$
$\iff(2a-1)^2\equiv-11$ for odd $p$
$4(b^2-b+25)=(2b-1)^2+99$
So, we need $(2b-1)^2\equiv-99\pmod p$
$\implies(2b-1)^2\equiv3^2(2a-1)^2\pmod p$
$\iff2b-1\equiv\pm3(2a-1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to divide ${2k^3+3k^2+k-2j^3+3j^2-j}$ with $(k+1-j)$? The question I had was calculating $$\displaystyle\frac{1}{k+1-j}\sum_{i=j}^k i^2$$
Because I didn't know how to do a variable change, I did
$$\frac{1}{k+1-j}\sum_{i=j}^k i^2 = \frac{1}{k+1-j}\left(\sum_{i=1}^k i^2 - \sum_{i=1}^{j-1} i^2\right) = \frac{2k^3+3k^... | You could use synthetic division fairly easily here.
First, we consider $2k^3+3k^2+k-2j^3+3j^2-j$ as a polynomial in $k,$ and write the $4$ coefficients $$\begin{array}{c|cccc} & 2 & 3 & 1 &-2j^3+3j^2-j\\ & & & &\\\hline & & & &\end{array}$$ Next, we invert the coefficients of the divisor $k+1-j$ to get $-1k+j-1,$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Super hard system of equations
Solve the system of equation for real numbers
\begin{split}
(a+b) &(c+d) &= 1 & \qquad (1)\\
(a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\
(a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\
(a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\
\end{split}
First I used the identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)... | We are going to show that $(a,b)$ belongs to one of the two lines with equations $b=\sqrt{a}$ and $b=\frac{1}{\sqrt{a}}$ as displayed on the following figure. It will give the answer, due to the symmetry of the system of equations with respect to the group of variables $(a,b)$ vs. $(c,d)$. Moreover, we will establish ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 4,
"answer_id": 1
} |
What is $\int_0^{\pi/2}\sin^7(\theta)\cos^5(\theta)d\theta$ I have to integrate the following:
$\int_0^\limits\frac{\pi}{2}\sin^7(\theta)\cos^5(\theta)d\theta$
I decided to use a $u$ substitution of $u=\sin^2(\theta)$, and $\frac{du}{2}=\sin(\theta)\cos(\theta)$
and arrived at this integral
$\int_\limits{0}^{1}u^3(1-u)... | I think you complicated the last part, after all you are integrating a polynomial.
$\displaystyle \int_0^1 u^3(1-u)^2\mathop{du}=\int_0^1 (u^3-2u^4+u^5)\mathop{du}=\left[\frac{u^4}4-2\frac{u^5}5+\frac{u^6}6\right]_0^1=\frac 14-\frac 25+\frac 16=\frac 1{60}$
Also you dropped the coeff $\dfrac 12$ from $\dfrac{du}2$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Integral over recurrence relationship I'm interested in evaluating the following definite integral
\begin{equation}
I_n = \int_0^{\gamma} F_n(x)\:dx
\end{equation}
Where $\gamma \gt 0$ and $F_n(x)$ is based on the recurrence relationship:
\begin{equation}
F_{n + 1}(x) = \frac{1}{1 + F_n(x)}
\end{equation}
Here $F_0(x)... | This is copied from
another answer of mine:
Let
$f_1 = \frac{1}{1 + g(x) }
$
where
$g(x) > 0,
$,
and let
$f_n(x)
=\frac{1}{1+f_{n-1}(x)}
$.
Then
$f_n(x)
\to \dfrac{\sqrt{5}-1}{2}
$.
Note:
I doubt that any of this
is original,
but this was all done
just now by me.
Proof.
$\begin{array}\\
f_n(x)
&=\frac{1}{1+\frac{1}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3064562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Maximizing $f$ in $\mathbb{R}^3$
Find the domain and the maximum value that the function
$$f(x,y,z)=\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}$$
may attain in its domain.
I have found the domain of the function to be $\mathbb{R^3\backslash\mathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=\frac{-... | By C-S $$(x^2+y^2+z^2)(1^2+2^2+3^2)\geq(x+2y+3z)^2,$$
which gives
$$-\sqrt{14}\leq\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}\leq\sqrt{14}.$$
The equality occurs for $(x,y,z)||(1,2,3),$ which gives that we got a maximal value and the minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find maximum value of $\frac xy$
If $x^2-30x+y^2-40y+576=0$, find the maximum value of $\dfrac xy$.
First I completed the squares and got $(x-15)^2+(y-20)^2=7^2$, which is the equation of a circle.
I think I need to use some properties but I don't know what to do next.
| An alternative method, uses calculus. Similar to Sauhard Sharma's answer.
First note, this is the equation of a circle centred on $(15,20)$ of radius $7$, so it does not reach the $x$ axis (otherwise you'd be able to get arbitrarily large values for $x/y$). The value of $x/y$ is obviously going to be maximised on the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3065769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
How to solve pell type equation Example $x^2-7y^2=2$
Find $x,y$ in general when $x,y$ are integer
I don't know how to solve it , I need the method to help to solve it
| Apply the PQa algorithm for D=7, determining that $\sqrt{7}=[2,\overline{1,1,1,4}]$
$
\begin{array}{c|c|c|c|c|c}
j & P_j & Q_j & a_j & A_j & B_j \\
\hline
0 & 0 & 1 & 2 & 2 & 1\\
1 & 2 & 3 & 1 & \textbf{3} & \textbf{1}\\
2 & 1 & \textbf{2} & 1 & 5 & 2\\
3 & 1 & 3 & 1 & \textbf{8} & \textbf{3}\\
4 & 2 & \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the limit using Taylor seris with Big-O notation I have a limit
$$
\lim_{x \to 0} \frac{1}{\sin x} - \frac{1}{x^2}
$$
I've tried to solve it like this:
\begin{align}
&\lim_{x \to 0} \frac{1}{\sin x} - \frac{1}{x^2} =\\
&\lim_{x \to 0} \frac{x^2 - \sin x}{x^2\sin x} = \\
&\lim_{x \to 0} \frac{ x^2 - x + \frac{x^3}... | Just as you did
$$y=\frac{1}{\sin x} - \frac{1}{x^2}=\frac{x^2 - \sin x}{x^2\sin x}=\frac{x^2-\left( x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)\right) } {x^2\left(x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right) \right) }$$ that is to say
$$y=\frac{-x+x^2+\frac{x^3}{6}-\frac{x^5}{120}+O\left(x^7\right) } {x^3-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality in 3 variables with a constraint condition To prove : a(b+c)/bc + b(c+a)/ca + c(a+b)/ab > 2/(ab+bc+ca) where a+b+c=1 and a,b,c are positive real numbers
Here's my way : Add 1 to each summand in the LHS and subtract 3 (=1+1+1) and after some algebraic manipulation I got (ab+bc+ac)/abc > 2/(ab+bc+ac) +3 which ... | If we arrange
$$
\frac{ab+bc+ac}{abc} \ge \frac{2}{ab+bc+ac} +3
$$ then we have
$$
(ab+bc+ac)^2 =\sum_{\text{cyc}}a^2b^2 +2abc\ge abc(2+3(ab+bc+ac))=2abc +3\sum_{\text{cyc}}a^2b^2c.
$$ We need to show
$$
\sum_{\text{cyc}}a^2b^2\ge 3\sum_{\text{cyc}}a^2b^2c.
$$ This is equivalent to
$$
\sum_{\text{cyc}}a^2b^2(a+b+c)\ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all integer solutions to $x^2+xy+y^2=((x+y)/3 +1)^3$
Find all ordered pairs of integers $(x,y)$, that satisfy the following relation:
$$x^2+xy+y^2=((x+y)/3 +1)^3$$
I tried bashing at first. Then I assumed $x+y = 3k$ for some integer $k$ so that $y=3k-x$, substituted in the given equation and got a cubic polynomi... | Put $a = x+y, b = x -y \implies x^2+y^2 +xy = \dfrac{(x+y)^2+(x-y)^2}{2}+ \dfrac{(x+y)^2-(x-y)^2}{4}= \dfrac{a^2+b^2}{2}+\dfrac{a^2-b^2}{4}= \dfrac{3a^2+b^2}{4}$. The equation becomes: $\dfrac{3a^2+b^2}{4} = \dfrac{(a+3)^3}{27}\implies 81a^2+ 27b^2=4(a+3)^3$. Observe that $3 \mid a$, thus put $a = 3n$, and we have a n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3068288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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What is the maximum value of $(a+ b+c)$ if $(a^n + b^n + c^n)$ is divisible by $(a+ b+c)$ where the remainder is 0?
The ‘energy’ of an ordered triple $(a, b, c)$ formed by three
positive integers $a$, $b$ and $c$ is said to be n if the following $c$
$\ge b\geq a$, gcd$(a, b, c) = 1$, and $(a^n + b^n + c^n)$ is
divisib... | The question is to find the largest sum of $a, b, c$, given they're all relatively prime to each other and divide
$$a^n + b^n + c^n \text{ } \forall \text{ } n \ge 1 \tag{1}\label{eq1}$$
Since $a, b, c \ge 1$, then $a + b + c \ge 3$, so it consists of one or more prime factors. Call one of these prime factors $d$. Thu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Compute $S = \sum_{k=0}^{m} \left\lfloor \frac{k}{2}\right\rfloor$ I want to compute the following sum
$$S = \sum_{k=0}^{m} \left\lfloor \frac{k}{2}\right\rfloor.$$
Here is what I tried:
$$ S = \sum_{k\geq 0, 2|k}^{m} \left\lfloor \frac{k}{2}\right\rfloor + \sum_{k\geq 0, 2\not |k}^{m} \left\lfloor \frac{k}{2}\right\rf... | Yes, you are correct. You may also write the result as a more compact formula:
$$\sum_{k=0}^{m} \left\lfloor \frac{k}{2}\right\rfloor=
\begin{cases}
t^2&\text {if $m=2t$}\\
t(t+1)&\text {if $m=2t+1$}\\
\end{cases}=\left\lfloor \frac{m^2}{4}\right\rfloor.$$
Indeed, if $m=2t$ then
$$\left\lfloor \frac{m^2}{4}\right\rflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding maxima of a function $f(x) = \sqrt{x} - 2x^2$ without calculus My question is how to prove that $f(x) = \sqrt x - 2x^2$ has its maximum at point $x_0 = \frac{1}{4}$
It is easy to do that by finding its derivative and setting it to be zero (this is how I got $x_0 = \frac{1}{4}$). But the task is to do that witho... | $f\left(\frac{1}{4}\right)=\frac{3}{8}.$
We'll prove that we got a maximal value.
Indeed, we need to prove that
$$\sqrt{x}-2x^2\leq\frac{3}{8},$$ which is true by AM-GM:
$$2x^2+\frac{3}{8}=2x^2+3\cdot\frac{1}{8}\geq4\sqrt[4]{2x^2\left(\frac{1}{8}\right)^3}=\sqrt{x}.$$
We can get a value $\frac{3}{8}$ by the following w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 1
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Solving with L'Hôpital's rule. What is wrong? L'Hôpital's rule can be used infinitely many times as the limit remains $0/0$. However, this problem does not work with L'Hôpital's rule (I may have counted wrong).
Here it is:
$$\lim _{x\to 0}\left(\frac{\left(e^x+\sin x\right)x+e^x-\cos x}{x^2}\right)$$
We use L'Hôpital's... | You can't apply l'Hôpital on a non indeterminate form.
Let's see what happens with a Taylor expansion at degree $2$:
\begin{align}
(e^x+\sin x)x+e^x-\cos x
&=(1+x+x+o(x))x+1+x+\frac{x^2}{2}-1+\frac{x^2}{2}+o(x^2) \\
&=x+2x^2+x+x^2+o(x^2)\\
&=2x+3x^2+o(x^2)
\end{align}
Thus we see that the given limit cannot be finite (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Explicit calculation of $\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x$ Is it possible to confirm the value of this integral using the methods of complex analysis or similar?
$$
\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x=\frac{\pi^2-9}{12}
$$
Of course, on... | This will require some calculations but no special functions. Making the change of variables $x = -x$ on $\mathbb R^-$, we obtain
$$\int_{\mathbb R} \frac x {(x^2 + 1)^2 (e^{2 \pi x} - 1)} dx =
\underbrace {\int_{\mathbb R^+} \frac {2 x} {(x^2 + 1)^2 (e^{2 \pi x} - 1)} dx}_
{= I} +
\int_{\mathbb R^+} \frac x {(x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Using the R method for finding all solutions to $\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$. My solution differs from official answer.
How many solutions does
$$\sin(2a) - \cos(2a) = \frac{\sqrt{6}}{2}$$
have between $-90^\circ$ and $90^\circ$?
I used the R method and got
$$2a-45^\circ = \arcsin\left(\frac{\sqrt... | $\sin(2a-45) = \frac{\sqrt{3}}{2}$
$2a-45 = 60 + 360n \Rightarrow a = 52.5 + 180n$
$2a-45 = 120+ 360n \Rightarrow a = 82.5 + 180n$
Either $a = 52.5 $ or $a = 82.5$. If you square
$\sin(4a) = \frac{-1}{2}$
$4a = 210 + 360n \Rightarrow a = 52.5 + 90n $
$4a = 330 + 360n \Rightarrow a = 82.5 + 90n $
First two soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$ evaluation using expansion series $$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$$
This is the limit which I got from the book of Joseph Edwards' Differential Calculus for B... | You are on the right track. By using little-o notation you may control the expansions in a better way. We have that as $x\to 0^+$,
$$\frac{x^3(1 + \frac{x^4}{4} + o(x^4)) - (x^2 - \frac{x^6}{3!} + o(x^6))^{\frac 3 2} }{x^7}=\frac{1 + \frac{x^4}{4} + o(x^4) - (1 - \frac{x^4}{3!} + o(x^4))^{\frac 3 2} }{x^4}$$
Now all yo... | {
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"url": "https://math.stackexchange.com/questions/3073052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If a sports team is down by 1-2 in a best of 7 series, what is their chance of winning (if both teams have a 1/2 chance of winning every game)? I know that the team has to win 3 of the 4 remaining games. So I thought that from the 4 remaining games, there are a total of $2^4 = 16$ total different outcomes.
Could someo... | Let $p(a,b)$ be the probability that team A is the first to win 4 games, given that team A has already won $a$ games and team B has already won $b$ games. By conditioning on the outcome of the next game, we see that $p$ satisfies the following recurrence relations:
\begin{align}
p(4,b) &= 1 \\
p(a,4) &= 0 \\
p(a,b) &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073796",
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Evaluate $\lim\limits_{x \to 1}\dfrac{x-x^x}{1-x+\ln x}$. Problem
Evaluate $$\lim\limits_{x \to 1}\frac{x-x^x}{1-x+\ln x}$$.
Solution
Consider using Taylor's formula. Expand $x^x $ and $\ln x$ at $x=1$. We obtain
$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$\ln x=(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
\begin{align*}... | Why no?
$$\lim_{x\rightarrow1}\frac{x-x^x}{1-x+\ln{x}}=\lim_{x\rightarrow1}\frac{1-x^x(1+\ln{x})}{-1+\frac{1}{x}}=\lim_{x\rightarrow1}\frac{-x^x(1+\ln{x})^2-x^x\cdot\frac{1}{x}}{-\frac{1}{x^2}}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3075787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is an integer. Prove that $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer using mathematical induction.
I tried using mathematical induction but using binomial formula also it becomes little bit complicated.
Please show me ... | We have:
$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}
=\frac{15k^7+21k^5+70k^3-k}{3\cdot 5\cdot 7}
$$
To prove this is an integer we need that:
$$15k^7+21k^5+70k^3-k\equiv 0 \pmod{3\cdot 5\cdot 7}$$
According to the Chinese Remainder Theorem, this is the case iff
$$\begin{cases}15k^7+21k^5+70k^3-k\equiv 0... | {
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"source": "stackexchange",
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How do I find the order of a bijection? Question: List all bijections (permutations) from $\{1, 2, 3\}$ onto $\{1, 2, 3\}$. Find their order and sign.
I understand there will be n! permutations, namely:
$
\begin{Bmatrix}
1 & 2 & 3 \\
1 & 2 & 3 \\
\end{Bmatrix}
$,
$
\begin{Bmatrix}
1 & 2 & 3 \\
1 & 3 & 2... | The identity is not something you choose; the identity means the permutation that fixes each element, so the first one you considered is the identity permutation. Secondly, the order of the identity permutation is $1$, because $\epsilon^1=\epsilon$ (the order must be positive). The order of $
\sigma_1=\begin{Bmatrix}
... | {
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"source": "stackexchange",
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Evaluating $\lim_{x\to0^+}\frac{2x(\sin x)^2+\frac{2x^7+x^8}{3x^2+x^4}-\arctan(2x^3)}{\ln(\frac{1+x^2}{1-x^2})-2x^2+xe^{-{1\over x}}}$ To start with, the term $xe^{-{1\over x}}$ can be ignored.Then splitting the term $\ln(\frac{1+x^2}{1-x^2})=\ln(1+x^2)-\ln(1-x^2)$ and expanding both of them up to order 3 we have in th... | The items below are called Taylor Series. This is a good way to reduce errors, in cases where L'Hospital's method leads to messy derivatives. One may take time and get each series correct. In this case, all the series derive from that for $$ \frac{1}{1+t} = 1 - t + t^2 - t^3 + t^4 - t^5 \cdots, $$ which leads fairly qu... | {
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"url": "https://math.stackexchange.com/questions/3079733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I ... | Prove: for all $n$: $a^{2n+1}+b^{2n+1}=c^{2n+1}+d^{2n+1}$
If $a+b\ne 0$ then we get $$a^2-ab+b^2 =c^2-cd+d^2\implies (a+b)^2-3ab = (c+d)^2-3cd$$
so $$ab = cd$$
Induction step: $n-1,n\to n+1$
By I. H. we have $$a^{2n-1}+b^{2n-1} = c^{2n-1}+d^{2n-1}\;\;\; $$
now we multiply this with $a^2+b^2 = c^2+d^2$ we get $$ a^{2n+1... | {
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"timestamp": "2023-03-29T00:00:00",
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Differential Equations - Is the solution to $(e^y+1)^2e^{-y}dx+(e^x+1)^6e^{-x}dy=0$, $y=\ln(-5(e^x+1)^5+c)$? This is the problem. It is question 4. I typed the answer in explicit form, but my professor had said it is incorrect. Could someone check over my work, because I think I did have it correct. My Work
Because som... | The step from:
$$ -\frac{1}{e^y + 1} = \frac{1}{5\left(e^x+1\right)^5} +c_1 \tag{1} $$
to:
$$ -5(e^x +1)^5-c_2 = e^y +1 \tag{2}$$
is wrong. Because if what you say is true then it implies that (1) and (2) are a tautology, but if you replace a term of (2) in (1):
$$ -\frac{1}{e^y + 1} = \frac{1}{5\left(e^x+1\right)^5} +... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Relation between Roots and Coefficients Question If the roots of the equation $x^3+px^2+qx+r=0$ are consecutive terms of a geometric series, prove that $q^3 = p^3r$.
Show that this condition is satisfied for the equation $8x^3-100x^2+250x-125=0$ and solve this equation.
| The roots can be written as $a, ab, ab^2$. Then $(x-a)(x-ab)(x-ab^2) = x^3 + px^2 + qx +r$. Multiplying the left hand side and comparing yields $p = -a(1+b+b^2), q = a^2b(1+b+b^2), r = -a^3b^3$. This shows that $q^3 = p^3r$.
Since $r = -a^3b^3$, we get $ab = -\sqrt[3]{r}$ which is one of the three roots. Thus, if $q^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3083856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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