Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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proving the identity $b^3 = 6\binom{b}{3} + 6\binom{b}{2} + b$ for $b \in \mathbb{N}$, $b > 2$, by counting a set in different ways I'm trying to prove that $\forall b \in \mathbb{N}, b>2,$ $b^{3} = 6\binom{b} {3} +6 \binom{b}{2} +b$ without just using algebra. The idea I've thought about is that there are $b^{3}$ way... | Slight different approach (Case 2 below)!
Consider $b=3$ for simplicity.
There are $b^3=27$ ways to form a $3$-digit number with the digits $1,2,3$.
Case 1: All three digits are the same:
$${b\choose 1}={3\choose 1}=3 \Rightarrow 111,222,333.$$
Case 2: Two digits are the same:
$${3\choose 2} \Rightarrow \color{red}{**... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solving $\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5}$ $\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5}$
So, getting rid of the denominators I got to:
$(x+1)(x+5) + x(x+5) < 2x(x+1)$
$\rightarrow x < \frac{-5}{9}$
And also keeping in mind we can't have $-1,-5$ in the solution set, I got that the solution set was:
$\left(-\i... | To clear the fractions, you need to multiply by $|x(x+1)(x+5)|$ or $[x(x+1)(x+5)]^2.$ In these cases you can be sure that the inequality is preserved, since they are never negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that $2x^3+3x^2+x$ is always divisible by 6 if x is an integer. Prove that $2x^3+3x^2+x$ is always divisible by $6$ if $x$ is an integer.
I started by factoring the expression:
$x(2x^2+3x+1)=x(2x+1)(x+1)$
However I wasn’t sure how to progress from here to prove that it is always divisible by 6. Any ideas?
| You are almost done
$$x(2x^2+3x+1)=x(2x+1)(x+1)=x(2x-2+3)(x+1)
\\=2(x-1)x(x+1)+3x(x+1)$$
Now, among any three consecutive integers one is a multiple of 3, and among two consecutive integers one is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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Find the maximal value of $c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a}$ in terms of $c^3 + a^3 = m$ where $a, b$ and $c$ are positives.
Let $a, b$ and $c$ be positive real numbers such that $c^3 + a^3 = m$ and $$[(b^3 - abc) + ab(a + b) - c(a^2 + b^2)][(abc - b^3) - bc(b + c) + a(b^2 + c^2)] \le 0.$$ Find the maxi... | Constraint
$$((b^3 - abc) + ab(a + b) - c(a^2 + b^2))((abc - b^3) - bc(b + c) + a(b^2 + c^2)) \le 0,$$
or
$$(b^3-a^3)(b^3 - c^3)\ge 0,$$
is equivalent to
$$b^2-(a+c)b+ac\ge 0\tag1.$$
The goal function
$$f(a,b,c)=c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a}$$
under the conditions
$$a^3 + c^3 = m\tag2$$
can be pres... | {
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"timestamp": "2023-03-29T00:00:00",
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Knowing $\sin(45^\circ)={\sqrt2\over2}$, how do I find $\sin(135^\circ)$ and $\cos(135^\circ)$?
I just know about $$\sin(45^\circ)={\sqrt2\over2}$$
and I want understand how to find $\sin(135^\circ)$ and $\cos(135^\circ)$
Thank you.
| The cosine and sine of a directed angle with vertex at the origin and initial side on the positive $x$-axis are, respectively, the $x$- and $y$-coordinates of the points where the terminal side of the angle intersects the unit circle $x^2 + y^2 = 1$.
We are given the sine of a first-quadrant angle. Since $(\cos\theta... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How do I prove the limit $\frac{\sin(xy)}{\sqrt{x^2 + y^2}}$ as (x, y) approaches (0, 0) using $\epsilon - \delta$ I know that I can convert this limit to polar coordinates and solve the limit, but I want to see how I would do it using the $\epsilon - \delta$ definition of a limit.
This is my work so far:
We know that ... | Let $(x,y)\not =(0,0)$.
$|\sin (xy)| \le |xy|;$ $x^2+y^2 \ge |xy|$;
$0\le |\dfrac{\sin (xy)}{\sqrt{x^2+y^2}}| \le\dfrac{|xy|}{\sqrt{x^2+y^2}}\le$
$ \dfrac{x^2+y^2}{\sqrt{x^2+y^2}}= \sqrt{x^2+y^2}.$
Choose $\delta =\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373899",
"timestamp": "2023-03-29T00:00:00",
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Find all pairs of integers $(x,y)$ such that $x^{2}+y^{2}=(x-y)^{3}$.
Find all pairs of integers $(x,y)$ such that $x^{2}+y^{2}=(x-y)^{3}$.
I think that $(0,0)$ , $(1,0)$ and $(0,-1)$ are the only solutions to the above equation, but I'm unable to prove it.
I tried all sorts of things like working $\mod 9$ (but ther... | To extend your existing approach, multiply by $2$ to obtain :$$0=4y^2+4ay+2a^2-2a^3=(2y+a)^2+a^2(1-2a)$$
To obtain a factorisation with integer coefficients you need $2a-1=b^2$. For convenience, multiply through by $4$ to get $$0=(4y+2a)^2+4a^2(1-2a)=(4y+b^2+1)^2-(b^2+1)^2b^2$$ And the roots are $$4y=-(b^2+1)\pm b(b^2+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the maximum of the value $f(x)=\frac{n^4}{x}+4(n^2+3)x-4x^2-5n^2$
Let $n$ be a given positive integer, find the maximum of the value of
$$f(x)=\dfrac{n^4}{x}+4(n^2+3)x-4x^2-5n^2$$ where $x \le \dfrac{n^2}{4}$ and $x \in \mathbb N^{+}.$
If the $x$ is real positive number, the problem also not easy, because
$$f'... | $n=1$ makes no sense, while for $n=2$ the only allowed value is $x=2$. Thus assume $n>2$.
Using Descartes' rule of signs on $x^2f'(x)=-8x^3+4(n^2+3)x^2-n^4$ we see that $f'(x)$ has zero or two positive roots. Next, $f'(1) = -n^4+4 n^2+4<0$ while $f'\left(\frac{n^2}{4}\right) = 2 \left(n^2-2\right)>0$, so that $f$ has t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving $\cos \frac{165}{2}$ What are the steps to solve $$\frac{\sqrt{2\sqrt{2}(2\sqrt{2}-1-\sqrt3)}}{4}$$ into $$\frac{\sqrt{2-\sqrt{2+\sqrt{3}}}}{2}$$
Please explain. I got them from derivation of $$\cos(82.5^\circ)$$ in 2 different ways.
$$\cos(60^\circ+22.5^\circ)$$ and $$\cos\frac{(90^\circ+75^\circ)}{2}$$
Thank... | If we are allowed to solve analytically then
\begin{align}2\sqrt{2}(2\sqrt{2}-1-\sqrt3)&=
4\sqrt{4}-2\sqrt{2}-2\sqrt{6}\\&=
8-2(\sqrt{2}+\sqrt{6})\\&=
8-2\left(2\sqrt{2+\sqrt{3}}\right)\\&=
8-4\sqrt{2+\sqrt{3}}\\&=
4\left(2-\sqrt{2+\sqrt{3}}\right)
\end{align}
so that
$$\frac{\sqrt{4\left(2-\sqrt{2+\sqrt{3}}\right)}}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"Altered" Alternating Series Diverges or Converges? Consider the series
$$
1-\frac{1}{2} +\frac{1}{3} +\frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \cdots + \frac{1}{11} + \cdots
$$
which alternates between a block of positives and a block of negatives, with the block sizes,
$$
1, 1, 2, 3, 4, ... | Let: $$
S=1-\frac{1}{2} +\frac{1}{3} +\frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \cdots + \frac{1}{11} + \cdots
$$
For a given sequence of additions assume a constant denominator, specifically the smallest in the group. For a series of subtractions pick a common constant denominator, the larg... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find positive integers such that $2n^3 + 5|n^4 +n+1$ $$ 2n^3 + 5 | 2n^4 +2n +2 - 2n^4 - 5n$$
$$= 2n^3+5 | 2-3n$$
$$3n-2≥2n^3 + 5$$
is this correct? Is there a more efficient way?
| Mod $(2n^3+5)$, you have that $2n^3\equiv-5$. Now suppose $(2n^3+5)$ divides $n^4+n+1$. Then mod $(2n^3+5)$:
$$
\begin{align}
n^4&\equiv-n-1\\
2n^4&\equiv-2n-2\\
n(2n^3)&\equiv-2n-2\\
-5n&\equiv-2n-2\\
-3n&\equiv-2\\
2-3n&\equiv0
\end{align}
$$
So $2n^3+5$ must divide $2-3n$. You can then test values of $n$ to arrive a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral From Electricity and Magnetism In attempting to calculate the magnetic field due to a current loop at an arbitrary point, I ran into the following integrals
\begin{align}
&\int_0^{2\pi} \frac{z\cos t \ dt}{\left[ \cos^2 t+ (y - \sin t)^2 + z^2\right]^{\frac{3}{2}}}
\\&\int_0^{2\pi} \frac{z\sin t \ dt}{\left[ \... | I will work with integrals without normalization. Write $a$ for the diameter of the current loop. Then the first integral is simply
\begin{align*}
\int_{0}^{2\pi} \frac{a z \cos t}{(a^2 \cos^2 t + (y-a \sin t)^2 + z^2)^{3/2}} \, \mathrm{d}t
&= 0.
\end{align*}
The other two integrals involve elliptic functions, but its ... | {
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$f(x)$ is a quadratic function such that $f(0) = 1$ and $\int {f(x) \over x^2(x + 1)^3} dx $ is a rational function. Find $f(x)$.
$f(x)$ is a quadratic function such that $f(0) = 1$ and $$\int {f(x) \over x^2(x + 1)^3 } dx $$ is a rational function. Find $f(x)$.
Now first of all we don't even know what the degree of ... | Let $$f(x)=Z(x+1)(2x+1)~~~~(1)$$ where $Z$ is a constant then $$I=\int \frac{f(x)dx}{x^2(x+1)^3}=Z\int \frac{2x+1}{(x(x+1))^2} dx= \frac{-Z}{x^2+x}+C.$$
Choose $Z=1$ in (1) to have $f(0)=1$.
Explaination: Since $f(0)=1$, $x$ or $x^2$ cannot be ia factor of $f(x)$. Try to have $(x+1)$ as a factor of $f(x)$ so that your... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Maximum value of $(1 − x)(2 − y)^ 2 (x + y)$ Additional Info
$x < 1, y<2, x+y>0$
I tried doing AM-GM (since all of the terms are positive.
$(1 − x)(2 − y)^ 2 (x + y) = (2-2x)(2-y)(2-y)(2x+2y)*1/4$
this way we can cancel the x and y's
thus the maximum value is
$((2-2x+2-y+2-y+2x+2y+1/4)/5)^5$
but the answer is said to... | In your AM-GM the equality occurs for
$$2-2x=2-y=2x+2y=\frac{1}{4},$$ which is impossible.
I think, it's better to make the following.
By AM-GM
$$(1-x)(2-y)^2(x+y)=4(1-x)\left(1-\frac{y}{2}\right)^2(x+y)\leq$$
$$\leq4\left(\frac{1-x+2\left(1-\frac{y}{2}\right)+x+y}{4}\right)^4=\frac{81}{64}.$$
The equality occurs for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5... | Note that $$z=\sqrt{\frac{3 \pm \sqrt{5}}{2}}= \pm \left(\frac{1\pm \sqrt{5}}{2} \right).$$ So $$z_1=2\frac{1+\sqrt{5}}{4} =2 \cos 36^0$$ Hence the other three roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the digits of a number Consider the following statement!
$\overline{ABCD}+\overline{EFG}=8768$, and $\overline{ABC}+\overline{DEFG}=6005$.
If $A,\,B,\,C,\,D,\,E,\,F,$ and $G$ are different numbers, Then, what is $\overline{ABCD}$?
My idea is :
$$\begin{aligned}
&\overline{ABCD}+\overline{EFG}=8768\\
&1000(A)+100(B... | When you have $X+Y\le 9+9$ the most you can carry is $1$.
And if you have $X+Y+1\le 9+9+1$ the most you can carry is $1$.
So you can never carry more than one. Now if you have $X+Y\to 0$ that means $X+Y$ or $X+Y + 1=10$ and you do carry $1$ to the next column.
So
$\overline{ABC} + \overline{DEFG} = 6005$
has $A+E\to ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Inequality : $\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$ It's a charming problem :
Let $a,b,c>0$ such that $a+b+c=1$ then we have :
$$\frac{a}{\exp(a+b)}+\frac{b}{\exp(b+c)}+\frac{c}{\exp(c+a)}\leq \exp\Big(\frac{-2}{3}\Big)$$
I know the identity :
Let $a,b,c>0$ suc... | Fact 1: $\mathrm{e}^x \le \frac{2}{3}x^2 + x + 1, \quad \forall x \le \frac{2}{3}$.
(The proof is given at the end.)
The desired inequality is written as
$$a\mathrm{e}^{2/3-a-b} + b\mathrm{e}^{2/3-b-c} + c\mathrm{e}^{2/3-c-a} \le 1.$$
Let $f(x) \triangleq \frac{2}{3}x^2 + x + 1$. By Fact 1, it suffices to prove that
$$... | {
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Find $x,y,z$ for the given conditions $$4x^2+25y^2+9z^2-10xy-15yz-6zx=0$$
$$x+y+z=5$$
I tried two approaches
1) Substituting $x$ as $5-y-z$ in the first equation but didn't work out, I was getting $39y^2+19z^2-31yz-90y-70z+100=0$ which can't be factorized
2) First equation corresponds to $a^2+b^2+c^2-ab-bc-ca=0$, whic... | Just wanted to see. Here is how it looks if you make the Hessian matrix congruent to a diagonal matrix, the one that gives
$$ \frac{1}{4} \left(4x - 5 y - 3 z \right)^2 + \frac{3}{4} \left( 5y - 3 z\right)^2 = 4x^2 + 25 y^2 + 9 z^2 - 15 yz - 6 zx - 10 xy $$
is $ Q^T D Q = H $
The theorem is that the form does fac... | {
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"url": "https://math.stackexchange.com/questions/3389836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof by contradiction algebra Let n be an integer. Prove that if $n^2+2n-1$ is even, then $n$ is odd.
This is what i have tried so far
If $n$ is even then $n = 2k$
And $n^2+2n-1$ is odd
$$(2k)^2+2(2k)-1 = 4k^2+4k-1$$
| If $n^2 + 2n - 1$ is even, then
$(n + 1)^2 = n^2 + 2n + 1 = (n^2 + 2n - 1) + 2 \tag 1$
is also even; now, the square of an even is even, since
$n = 2k \Longrightarrow n^2 = 4k^2 = 2(2k^2) \Longrightarrow 2 \mid n^2; \tag 2$
likewise, the square of an odd is odd:
$n = 2k + 1 \Longrightarrow n^2 = 4k^2 + 4k + 1 = 2(2k^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391149",
"timestamp": "2023-03-29T00:00:00",
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$1^3+2^3+...+n^3= $? How I can find formula $1^3+2^3+...+n^3=$ has a formula which $(1+2+3+...+n)^2 $ or $[n(n + 1)/2]^2$ . We can verify with ınduction and I know how I can prove it. How we find what is the formula...
I tried like this
$1^3+2^3+...+n^3= An^4+ Bn^3+Cn^2+Dn+E$
after a long process i found it this way
$A... | $0=E, 1=A+B+C+D+E, 1^3+2^3=9=16A+8B+4C+2D+E,$
$ 1^3+2^3+3^3=36=81A+27B+9C+3D+E, $
and $1^3+2^3+3^3+4^3=100=256A+64B+16C+4D+E \implies$
$ E=0, 7=14A+6B+2C, 33=78A+24B+6C, $ and $96=252A+60B+12C \implies$
$12=36A+6B$ and $54=168A+24B\implies$
$24A=6\implies A=\dfrac14$. Can you solve now for $B, C, $ and $D$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Algebraic proof of a combinatoric question (Combinatoric proof is given) I had a IMO training about double counting. Then, there is a problem which I hope there is a combinatoric proof. Here comes the problem:
For every positive integer $n$, let $f\left(n\right)$ be the number of all positive integers with exactly $2... | Here is the generating function approach:
\begin{align}
\sum_{n=0}^\infty \left(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k}\binom{2k}{k} 2^{n-2k}\right) z^n
&= \sum_{k=0}^\infty \binom{2k}{k} 2^{-2k} \sum_{n=2k}^\infty \binom{n}{2k}(2z)^n \\
&= \sum_{k=0}^\infty \binom{2k}{k} 2^{-2k} \frac{(2z)^{2k}}{(1-2z)^{2k+1}} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Alternative way to calculate $\int_0^1(x^4(1-x)^4)/(1+x^2)dx$ $$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx$$
$$=\int_0^1(x^8-4x^7+6x^6-4x^5+x^4)/(x^2+1)dx$$
$$=\int_0^1(x^6-4x^5+5x^4-4x^2-4/(x^2+1)+4)dx$$
$$=[1/7x^7-2/3x^6+x^5-4/3x^3-4\tan^{-1}x+4x]_0^1$$
$$I=22/7-\pi$$
Any other method to solve this problem?
| Here is a systematic procedure. Let $\tan t=x$,
$$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx=\int_0^{\pi/4}\tan^4t(1-\tan t)^4dt$$
$$=I(8) -4I(7)+6I(6)-4I(5)+I(4)\tag{1}$$
where $I(n)=\int_0^{\pi/4}\tan^nt\>dt$ and it has the recursive relationship $I(n)=\frac{1}{n-1}-I(n-2)$.
Now, use the recursive equation to get
$$4I(7)+4I(... | {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$. Does $\sum_{n=1}^{\infty} \frac{2^n+5^n}{3^n+4^n}$ converge?
Dividing the top and bottom by $4^n$ gives
\begin{equation*}
\frac{2^n+5^n}{3^n+4^n} = \frac{\left(\frac{1}{2}\right)^n+\left(\frac{5}{4}\right)^n}{\left(\frac{3}{4}\right)^n+1}.
\end{equation*}
He... | Yes. :)
| {
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"url": "https://math.stackexchange.com/questions/3397130",
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"source": "stackexchange",
"question_score": "1",
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Given 2 red beads, 2 blue beads, 1 yellow bead, and 1 green bead, how many different necklaces can be made? I need to write a Matlab code to determine the answer (which was given as 16) and I need to utilize loops to remove flips and circular shifts
| We use the Polya Enumeration Theorem (PET).
The cycle index $Z(D_6)$ of the dihedral group $D_6$ is given by
$$Z(D_6) = \frac{1}{12}
\left(\sum_{d|6} \varphi(d) a_d^{6/d}
+ 3 a_2^3 + 3 a_1^2 a_2^2\right).$$
This is
$$\frac{1}{12}
\left(a_1^6 + a_2^3 + 2 a_3^2 + 2 a_6
+ 3 a_2^3 + 3 a_1^2 a_2^2\right)
\\ = \frac{1}{12}
\... | {
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Integrate $\int_0^x \frac{d\theta}{(1+\epsilon\cos{\theta})^2}$ for $\epsilon>0$ If I can solve this integral below in analytical form, then I will be able to provide the time dependant orbit for a two-body problem,
$$\int_0^x \frac{d\theta}{(1+\epsilon\cos{\theta})^2}$$
for $\epsilon>0$.
| Observe that,
$$\left(\frac{\sin\theta}{1+\epsilon\cos{\theta}} \right)'=
\frac{\cos\theta+\epsilon}{(1+\epsilon\cos{\theta})^2}
= \frac1\epsilon\left[\frac{1}{1+\epsilon\cos{\theta}}+\frac{\epsilon^2-1}{(1+\epsilon\cos{\theta})^2} \right]$$
and decompose the integral
$$I=\int_0^x \frac{d\theta}{(1+\epsilon\cos{\theta}... | {
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
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Trying to solve a recurrence relation by using generating functions: $a_n=3a_{n-1} + a_{n-2}$ I'm trying to solve the recurrence relation below by using generating function:
\begin{equation}
a_n=\begin{cases}
0, & \text{if $n<0$}\\
2, & \text{if $n=0$}\\
1, & \text{if $n=1$}\\
3a_{n-1} + a_{n-2}, & ... | I'm going to rewrite the problem as $$a_{n+2} = 3a_{n+1} + a_n; \; a_0 = 2,\; a_1=1.$$ Multiply by $x^n$, sum, and let $A(x) = \sum_{n\ge 0}a_nx^n$. So,
$$\sum_{n\ge 0}a_{n+2}x^n = 3\sum_{n\ge 0}a_{n+1}x^n + \sum_{n\ge 0}a_n x^n,$$
and we can see that $\sum_{n\ge 0}a_{n+2}x^n = a_2 + a_3x + \cdots = (1/x^2)(A(x)-a_0-a_... | {
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If $a$, $b$, $c$, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$, and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ are all integers, then $|a|=|b|=|c|$
Prove that if $a,b,c$ are integers and both $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ and $\frac{a}{c} + \frac{b}{a} + \frac{c}{b}$ are integers, then $|a|=|b|=|c|$.
Well this is wha... | I think what you've actually shown is that $a \mid b^5$, $b \mid c^5$ and $c \mid a^5$.
From this
you can argue as follows. Suppose $p$ is a prime factor of $a$. Then $a \mid b^5$ implies that
$p \mid b$, and similarly $b \mid c^5$ now implies that $p \mid c$.
Thus $p$ divides all of $a$, $b$, and $c$. Similarly, any p... | {
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"source": "stackexchange",
"question_score": "8",
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$f(x+1) + f(x-1) = x^2$ ; $f(x+4) + f(x-4) = 2\sin x$ , then $f(x) =$? Given $f$ is a complex valued function satisfying $$f(x+1) + f(x-1) = x^2 \\
f(x+4) + f(x-4) = 2\sin x$$
what is $f(x)$ ?
Here, only for the first part MathWolfram alpha is showing $f(x)$ to be of type $c_1(i)^x + c_2(-i)^x + \dfrac{x^2 - 1}{2}$ b... | The two conditions can be rewritten as
$$f(x+2)+f(x)=(x+1)^2$$
$$f(x+8)+f(x)=2\sin(x+4)$$
Iterating the first one gives
$$f(x+4)+f(x+2)=(x+3)^2$$
$$f(x+6)+f(x+4)=(x+5)^2$$
$$f(x+8)+f(x+6)=(x+7)^2$$
It follows
\begin{align}
2\sin(x+4) &= f(x+8)+f(x)\\
&= (x+7)^2-f(x+6)+f(x)\\
&= (x+7)^2-(x+5)^2+f(x+4)+f(x)\\
&= (x+7)^2-... | {
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Evaluate $\lim\limits_{n \to \infty}\sum\limits_{k=0}^n \dfrac{\sqrt{n}}{n+k^2}(n=1,2,\cdots)$ I tried to change it into a Riemann sum but failed, since
\begin{align*}
\lim_{n \to \infty}\sum_{k=0}^n \frac{\sqrt{n}}{n+k^2}=\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^n \frac{\sqrt{n}}{1+(k/\sqrt{n})^2}
,\end{align*}
which... | Integral bounds do the job: $$\int_k^{k+1} \frac{1}{n+t^2} dt \leq \frac{1}{n+k^2}\leq \int_{k-1}^{k} \frac{1}{n+t^2} dt$$
Summing the left-hand side inequalities for $k\in \{0,\ldots,n\}$ and the right-hand side inequalities for $k\in \{1,\ldots,n\}$ yields
$$\int_0^{n+1}\frac{1}{n+t^2} dt \leq \sum_{k=0}^n \frac{1}... | {
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Find $n$ for a $\sum_{x=1}^n \left[(x+1)^3-x^3\right]$ $$ \sum_{x=1}^n \left[(x+1)^3-x^3\right]$$
This is my sum, I tried simplfifying and got $3x^2+3x+1$ but Im stuck on how to resolve the sum for $n$.
| Note that
$$\sum_{x=1}^n \left[(x+1)^3-x^3\right] =\color{red}{2^3}-1^2+\color{red}{3^3}-\color{red}{2^3}+\ldots+\color{red}{n^3}-\color{red}{(n-1)^3}+(n+1)^3-\color{red}{n^3}$$
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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What is the coefficient of x^2 in the expansion of (x+2)^4(x+3)^5 I'm missing something here.
I've calculated the $x^2$ coefficient of $(x+2)^4$ as 24 with constant term 16. And $x^2$ term coefficient of $(x+3)^5$ as 270 with constant term 243.
if I'm correct here then the answer should be $(16*270)+(24*243)$? but this... | Using the binomial theorem we have:
$$(x+2)^4(x+3)^5 = \left( \sum_{k=0}^{4}\binom{4}{k}x^k2^{4-k}\right) \left(\sum_{j=0}^{5}\binom{5}{j}x^j3^{5-j} \right).$$
We get $x^2$ by taking an $x$ term from each sum or an $x^2$ and $x^0$ term from one or the other sum. So the coefficient of $x^2$ is $$\overbrace{\binom{4}{1}2... | {
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Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3) $ Solve equation $1+\sin^2\theta=3\sin\theta\cos\theta \text { (given } \tan71 ^{\circ}34^{\prime}=3) $
My attempt is as follows:-
$$1-2\sin\theta\cos\theta+\sin^2\theta-\sin\theta\cos\theta=0$$
$$\left(\sin\theta-\cos... | Use $3=\frac{1+\frac12}{1-1\cdot\frac12}=\tan(45^\circ+\theta)$.
| {
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"source": "stackexchange",
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How to show this inequality for complex numbers? Let $|y|<|z|$ then,
$$\left| \frac{y+z}{|y+z|} - \frac{z}{|z|}\right|\leq K |z|^{-1}|y|$$
for some constant $K.$
I want to use some Taylor expansion argument, but I am not sure if it would work for complex numbers. Any ideas will be much appreciated.
Edit: The expressio... | Let $y=\rho e^{i\phi}$ and $z=re^{i\theta}$. Then $|y|<|z|\implies\rho<r$ and $|z|^{-1}|y|=\rho/r$. Since \begin{align}|y+z|&=|(\rho\cos\phi+r\cos\theta)+i(\rho\sin\phi+r\sin\theta)|\\&=\sqrt{\rho^2\cos^2\phi+2\rho r\cos\theta\cos\phi+r^2\cos^2\theta+\rho^2\sin^2\phi+2\rho r\sin\theta\sin\phi+r^2\sin^2\theta}\\&=\sqrt... | {
"language": "en",
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"source": "stackexchange",
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using epsilon delta definition to proof limits
$$\lim_{\large x \to 1^+} \frac{2x+3}{x-1}=∞$$
let $$f(x)=\frac{2x+3}{x-1}$$ then
$\forall M∈ℝ>0, \exists\delta>0,\forall x\in D_f \left(
0 < \left|x-1 \right|=x-1<\delta\Longrightarrow \large\left|\frac{2x+3}{x-1}\right|\right)>M$
$$M<\left|\frac{2x+3}{x-1}\right|... | The computations you show here are usefull, but they aren't what should appear in your proof, eg let's prove
$$\lim_{\large x \to 1^+} \frac{2x+3}{x-1}=∞$$
We have to show that $\forall M>0, \ \exists \delta > 0, \ (x > 1 \textrm{ and } x-1 < \delta) \Longrightarrow \frac{2x+3}{x-1} >M$.
Take $M>0$. Your computation s... | {
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"source": "stackexchange",
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Cosine of a $2 \times 2$ non-diagonalisable matrix Given
$$A = \begin{bmatrix}
\pi-1 & 1\\
-1 & \pi+1
\end{bmatrix}$$
I need to calculate its cosine, $\cos(A)$. Typically, I use diagonalisation to approach this type of problems:
$$\cos(A) = P \cos(D) P^{-1}$$
However, in this problem, the eigen... | When you do the Jordan decomposition, you get $A = SJS^{-1}$ with
$$
S = \begin{pmatrix}1&1\\1&2\end{pmatrix}\quad\text{and}\quad J = \begin{pmatrix}\pi&1\\0&\pi\end{pmatrix}.
$$
You find that
$$
J^n = \begin{pmatrix}\pi^n&n\pi^{n-1}\\0&\pi^n\end{pmatrix}.
$$
Since $A^n = SJ^nS^{-1}$, this implies
\begin{align}
\cos(A)... | {
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Solve expression with unknown variables I am stuck in this next problem:
Given $x-y = 7$ and $xy = 5$, without solving for the values of $x$ and $y$, evaluate $x^2 + y^2$.
How can I solve this?
| First expand $(x-y)^2$ to give
\begin{align}(x-y)^2&=x^2+y^2-2xy\end{align}
We can then substitute in the values we have been given and solve the resulting equation
\begin{align}7^2&=x^2+y^2-2\times 5\\
49&=x^2+y^2-10\\
x^2+y^2&=59\end{align}
Most of the time questions like this involve you spotting an expansion of on... | {
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"source": "stackexchange",
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Computing double integral $\ \iint \sqrt{4-r^2}r \ dr d\theta $ I am trying to solve the following integral:
$$\ 2 \int_0^\pi \int_0^{2\sin\theta} \sqrt{4-r^2} \ r \ dr d\theta $$
My attempt:
$$\ 2 \int_0^\pi \int_0^{2\sin\theta} \sqrt{4-r^2} \ r \ dr \ d\theta \stackrel{t = 4-r^2}{=} -\int_0^\pi \int t^{\frac{1}{2}} ... | The error is due to assuming $(x^2)^{3/2}=x^3$ as opposed to $|x^3|$. This can be solved by noticing that $\cos^3x=|\cos^3x|$ for $x\in[0,\pi/2]$ and $-\cos^3x=|\cos^3x|$ for $x\in[\pi/2,\pi]$. Thus the integral should become $$2\times\left(-\frac{16}{3} \int_0^{\pi/2} \cos^3\theta - 1 \ d\theta\right)=\frac{16(3\pi-4)... | {
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"source": "stackexchange",
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"answer_count": 1,
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Inverse of polynomial over a finite field The question is the following:
Can you deduce if ${2x + 1}$ is invertible in $\mathbb{Z}_3[x]/(x^2 + 2x + 2)$? In case of a positive answer, give its inverse.
Following the Wilson's theorem, for $K[X]/(f)$, any polynomial of degree $1 \leq deg < n$ will admit an inverse of degr... | Note that $-\frac15 = 1$ in $\Bbb Z_3$, and $3 = 0$, so your inverse candidate is equal to $2x$. And we can just check whether this is in fact an inverse, using that $3 = 0$ and $x^2 = x+1$:
$$
2x\cdot (2x+1) = 4x^2 + 2x = 4(x+1) + 2x\\
= 6x + 4 = 1
$$
So yes, that is indeed the inverse you're looking for.
| {
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Is my Proof for the following problem correct? Prove that $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$ If $a,b,c$ are positive real numbers prove that $$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$$
I state the following:
Multiplying both sides by $abc$ yields $$(abc)(a+b+c)?{a^4+b^4+c^4}$$
By AM-G... | Using the inequality $$x^2+y^2+z^2\geq xy+yz+zx$$ twice we get
$$a^4+b^4+c^4\geq (ab)^2+(bc)^2+(ca)^2\geq abbc+abca+bcca=a^2bc+ab^2c+abc^2=abc(a+b+c)$$
| {
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"source": "stackexchange",
"question_score": "2",
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Prove that if p and q are both prime numbers, with p > q > 2, then $p^4− q^4$ is divisible by 16 Prove that if p and q are both prime numbers, with p > q > 2, then $p^4 − q^4$ is divisible by 16.
This is my attempt so far:
Since p and q are both prime numbers greater than 2, then they must be odd and hence can be writt... | You're off to a great start.
Rewrite $$p^4-q^4=(p^2+q^2)(p+q)(p-q).$$
Regarding $p^2+q^2$:
Since $p$ and $q$ are both odd, they can be written as $p=2m+1$ and $q=2n+1$. Then we have
\begin{align}
p^2+q^2&=(2m+1)^2+(2n+1)^2\\
&=4m^2+4m+1+4n^2+4n+1\\
&=2(2m^2+2m+2n^2+2n+1)
\end{align}
This tells us that $2|p^2+q^2$ (and... | {
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Determine the set of all values of $ x \in [0, 2 \pi] $
Determine the set of all values of $ x \in [0, 2 \pi] $ that simultaneously satisfy $ \frac {2 \sin ^ 2 x + \sin x-1} {\cos x-1} < 0 $ and $ \tan x + \sqrt {3} < (1+ \sqrt {3} \cot x) \cot x $
My ''solution'':
$\frac{2{{\sin }^{2}}x+\sin x-1}{\cos x-1}<0\Righta... | The first inequality,
$$ \frac {2 \sin ^ 2 x + \sin x-1} {\cos x-1}=\frac{(2\sin x -1)(\sin x+1)}{ \cos x-1}< 0 $$
Then,
$\sin x+1>0,\>\>\>\cos x-1<0 \implies \sin x > \frac12\implies x\in (\frac \pi6,\frac{5\pi}6)\tag{1}$
The second inequality,
$$\tan x+\sqrt{3}<\left( \tan x+\sqrt{3} \right)\cot^2x$$
Case 1.
$\tan x... | {
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Proving a Difficult Definite Integral in One Variable Let $t > 0 $, $N \in \mathbb{N}$, $q \geq 1 + \frac{1}{N}$.
Also let $r > 0 $, $r' = \frac{r}{r-1}$, such that $\frac{N}{2}(1 - \frac{1}{r}) + \frac{1}{2} < 1$.
We wish to prove the following integral equation:
$ \displaystyle \int^{t}_{0} \large (t-s)^{ -\frac{N}{2... | To show that the integral $$\int_0^1(1-x)^{-\frac{N}{2}(1-\frac{1}{r})-\frac{1}{2}}(1+x)^{-\frac{N}{2}(q-\frac{1}{r'})}\,dx$$ converges, note that $1\leq 1+x\leq 2$ in $(0,1)$, therefore the term $(1+x)^{-\frac{N}{2}(q-\frac{1}{r'})}$ is bounded above by a constant $C$. Therefore, \begin{align*}\int_0^1\left|(1-x)^{-\f... | {
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Prove $n+2(n-1)+3(n-2)+...+2(n-1)+n={n(n+1)(n+2)\over 6}$ by induction To prove:
$n+2(n-1)+3(n-2)+...+2(n-1)+n={n(n+1)(n+2)\over 6}$
I assume the statement was derived from:
$n+2(n-1)+3(n-2)+...+(n-1)(n-(n-2))+n(n-(n-1))={n(n+1)(n+2)\over 6}$
For $n=k+1$
$(k+1)+2(k)+3(k-1)+...+2(k)+(k+1)={(k+1)(k+2)(k+3)\over 6}$
Ho... | Hint: You can just extract $1$ from each product to use the induction hypothesis:
$$ \begin{aligned}
\phantom{=}&1\cdot(k+1) + 2\cdot k + 3\cdot(k-1) + \cdots + k\cdot 2 + (k+1) \\
=&(1\cdot k + \color{red}{1}) + (2\cdot (k-1) + \color{red}{2}) + \cdots + (k\cdot 1 + \color{red}{k}) + \color{red}{k+1}\\
=&\frac{(k-1)k(... | {
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Tough complex numbers problem Let $c$ be a complex number. Suppose there exist distinct complex numbers $r$, $s$, and $t$ such that for every complex number $z$, we have
$$(z - r)(z - s)(z - t) = (z - cr)(z - cs)(z - ct).$$
Compute the number of distinct possible values of $c$.
OK, so I thought $1, \omega,$ and $\ome... | Let $p(z)$ denote the cubic on the left-hand side; the right-hand side is then $c^3p(z/c)$. Write $p(z)=z^3+Az^2+Bz+C$ so$$z^3+Az^2+Bz+C\equiv z^3+cAz^2+c^2Bz+c^3C\\\implies(c-1)A=(c^2-1)B=(c^3-1)C=0.$$If $r,\,s,\,t$ are all nonzero, $C\ne0$ so $c^3=1$ as per your reasoning. For $c=\exp\frac{\pm 2\pi i}{3}$ to work, we... | {
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"source": "stackexchange",
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In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$
In a triangle, if $\tan\frac{A}{2}$, $\tan\frac{B}{2}$, $\tan\frac{C}{2}$ are in arithmetic progression, then show that $\cos A$, $\cos B$, $\cos C$ are in arithmetic progression.
$$2\tan\le... | For $A\ne B,A+B+C=\pi$
using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$$f(B,A)=\dfrac{\tan\dfrac B2-\tan\dfrac A2}{\cos B-\cos A}=-\dfrac1{2\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2}$$
By symmetry, $$f(B,A)=f(C,B)$$
Can you take it from here?
Similarly we can establish $$\dfrac{\cot\dfrac B2-\cot\dfrac ... | {
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Baldi - Stochastic Calculus - Exercise about construsction of Stratonovich integral I need to prove that $$ \lim_n \sum_{i=0}^{2^n -1} W(\frac{i}{2^n}) ( W(\frac{i+1}{2^n}) - W(\frac{i}{2^n}) ) \rightarrow \frac{1}{2}W_1^2 - \frac{1}{2}$$ in $L^2$.
where $W$ is a standard Wiener process.
So I started computing
$$E[( ... | You expectation numbers are incorrect, you could see that in your proof, your solution is independent of $n$...
First notice that
$$ W_1^2= \sum_{i=0}^{2^n -1} W(\frac{i+1}{2^n})^2- W(\frac{i}{2^n})^2$$
$$ W_1^2= \sum_{i=0}^{2^n -1}\left( W(\frac{i+1}{2^n})+ W(\frac{i}{2^n})\right)\left( W(\frac{i+1}{2^n})- W(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
For which integer values of $n$ does there exist an integer $m$ such that $n^{3} - m^{2} = -23$?
For which integer values of $n$ does there exist an integer $m$ such that $n^{3} - m^{2} = -23$?
I'm having a lot of trouble with this one, any help would be appreciated :)
So far, I've seen that if the expression were a ... | Suppose that $n^3+23=m^2$ for some integers $n,m$. Observe that
$$(n+3)(n^2-3n+9)=n^3+27=m^2+4.$$
If $m$ is odd, then $n$ is even. Since $m^2\equiv 1\pmod{8}$, we must get $n^3\equiv 1-23\equiv 2\pmod{8}$. However, this is impossible as $n^3\equiv 0\pmod{8}$ for every even integer $n$. Thus $m$ is even.
Since $m$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Identity about Fibonacci numbers If I note $(F_N)_N = \{0,1,1,2,3,5,...\}$ the Fibonacci sequence, I have proved the identity
$$ \forall N \geqslant 0,\,F_{N}^2 = F_{N} + 2\,\sum_{k=0}^{N-3} F_{k+1}\,F_{k+2}\,F_{N-k-2}. $$
This relation can be obtained by induction and by using the equality
$$F_{n+2}^2 =(F_{n+1}+F_n)^... | Define, for $n\ge3$, $\displaystyle S_n:=\sum_{k=0}^{n-3}F_{k+1}F_{k+2}F_{n-k-2}$. Then, the focus of your identity, $S_n=\frac{F_n^2-F_n}{2}$, is really on $S_n$ since the other terms are simpler.
Searching for the first few values of $S_n$ in the OEIS gives gives us sequence A191797, i.e. $S_n=\binom{F_n}{2}=\frac{2F... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solution attempt $xuu_x+yuu_y=u^2-1$ Solve $$
\begin{cases}
xuu_x+yuu_y=u^2-1\\
u(x,x^2)=x^3\\
\end{cases}
$$
I have got using Lagrange method:
$$F\left(\frac{x}{y}\right)=\frac{x^2}{u^2-1}$$
Applying $u(x,x^2)=x^3$:
$$u^2=\frac{y^6-x^6}{x^2y^2}+1$$
But plug in it to the PDE show that there is a mistake
| Calling $v = u^2$ we have
$$
\frac 12 xv_x +\frac 12 y v_y = v-1
$$
with solution
$$
v = x^2\phi\left(\frac yx\right)+1
$$
now
$$
v(x,x^2) = x^6\Rightarrow \phi\left(\frac yx\right) = \frac{y^6-x^6}{x^4y^2}
$$
and
$$
v(x,y) = x^2\left(\frac{y^6-x^6}{x^4y^2}\right)+1
$$
and finally
$$
u(x,y) = \sqrt{\frac{y^6-x^6}{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Rewriting a double summation. If I am given this function: $$f(x) = \sum_{i = 1}^{\infty} \frac{1}{i^x}$$
Is there a way to rewrite:
$$g(x) = \sum_{ j = 1}^{\infty} \sum_{i = j}^{\infty} \frac{1}{(i \cdot j)^x}$$
In terms of f(x). By simple arthimatic I know that $f(x)^2 = 2 \cdot g(x) - f(2x)$.
The question is if for ... | Let
\begin{eqnarray*}
f_1(x) &=& \sum_{i \geq 1} \frac{1}{i^x} \\
f_2(x) &=& \sum_{i >j \geq 1} \frac{1}{(ij) ^x} \\
f_3(x) &=& \sum_{i>j>k \geq 1} \frac{1}{(ijk)^x}. \\
\end{eqnarray*}
In your question you have calculated
\begin{eqnarray*}
\left( 1 + \frac{1}{2^x} + \frac{1}{3^x} + \cdots \right) ^2 &=& \sum_{i \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Understanding the equation $|x+1|=x^2 -1$ I want to understand the equation
$$|x+1|=x^2 -1$$
$$\Leftrightarrow x^2 - |x+1| - 1 = 0$$
Case $1$:
$$x+1 \geq 0 \Rightarrow x^2 - x-2 = 0$$
$$x_{1,2} = \frac{1}{2} \big( 1\pm \sqrt{1-4\cdot(-2)} \big) = \frac{1}{2}(1\pm3) \Rightarrow x_1 = 2, x_2 = -1$$
Case $2$:
$$x+1 < 0 \... | Suppose that $x$ satisfies
$$(*) \quad |x+1|=x^2 -1.$$
Then $|x+1|=(x -1)(x+1).$ Hence $|x+1|=|x-1| \cdot |x+1|.$
It is clear that $x=-1$ is a solution of $(*)$. Now we assume that $x \ne -1.$ Then we get $|x-1|=1.$ The last equation has the solutions $x=0$ and $x=2$. But only $x=2$ is a solutions of $(*)$.
Conseque... | {
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"url": "https://math.stackexchange.com/questions/3465480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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What fraction is shaded? Conrguent $\frac{5}{6}$ circles in a circle. What fraction is shaded?
]1
Solution:
Let $r$ be the radius of the small circles and $R$ the radius of the big one.
The colored section is three times five sixths of the area of one of the small circles.
Colored Section area= $3\times\dfrac{5}{6}\ti... |
Using your original drawing. Let the radius of the smaller circle $r=1$. The radius of the big circle is: $AD+AG=\frac{1}{\cos 30°}+\cos 30°=\frac{7\sqrt 3}{6}$. Thus, the ratio of the colored area to the area of the big circle is $$\frac{5 \over 2}{49 \over 12}=\frac{30}{49}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
$a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number? Let $a,b,c,d$ be natural numbers such that $a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number?
Now assume that $a+b+c+d = p > 2$ for some choice of $a,b,c,d$. Notice that we cannot have $a=b=c=... | We work in the principal ideal domain $\mathbb Z [\omega]$, where $\omega = \frac{-1 + \sqrt3 i}{2}$ (which we view as an element of $\mathbb C$).
Without loss of generality, assume that $a \geq b$ and $c \geq d$, and that we don't have equality at the same time.
Write the equation as $(a + b \omega)(a + b\overline\ome... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Find the value of $x$ in the figure
Attempt: $\frac{\sin5x}{\sin8x} = \frac{\sin2x}{\sin3x}$
$x= 6^\circ$
(The answer is $6^\circ$)
How to ensure there is no other solution?
| Let $t = e^{ix}$. We then have $\sin nx = \frac{1}{2i}(t^n - t^{-n})$.
Expanding everything in the identity $\frac{\sin5x}{\sin8x} = \frac{\sin2x}{\sin3x}$, we have an equation: $$t^{20} - t^{18} - t^{16} + t^{12} + t^8 - t^4 - t^2 + 1 = 0,$$ which after factorization becomes: $$(t - 1)^2 (t + 1)^2 (t^{16} + t^{14} -... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac12$ Let $a,b,c$ be a sides of triangle
Such that : $a+b+c=1$ Then
prove that : $a^{2}+b^{2}+c^{2}+4abc<\frac{1}{2}$
My effort :
Since $a+b+c=1$ $\implies$ $2S=sr=bc\sin A=\frac{abc}{2R}$
Also : $S=\sqrt{s(s-a)(s-b)(s-c)}$
Also :
$a^{2}+b^{2}+c^{2}+2(ab+ac+bc)=1$
But I don... | Using homogeneization and Ravi's substitution, the problem boils down to showing that
$$2(a+b+c)(a^2+b^2+c^2)+8abc \leq (a+b+c)^3 $$
holds for any triple $(a,b,c)$ of side lengths of a triangle, i.e. that
$$4(A+B+C)((B+C)^2+(A+C)^2+(A+B)^2)+8(A+B)(A+C)(B+C) \leq 8(A+B+C)^3 $$
holds for any triple $(A,B,C)$ of positive ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3468328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Compute $\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7})$
$\cos(\frac{2\pi}{7})\cos(\frac{4\pi}{7})+\cos(\frac{2\pi}{7})\cos(\frac{6\pi}{7})+\cos(\frac{4\pi}{7})\cos(\frac{6\pi}{7}) = -\frac12$
I tried showing the equation, but my attempts did... | Use the identity $\cos(a-b)+\cos(a+b) = 2\cos a\cos b$,
$$I=\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7} $$
$$=\frac12\left(\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}\right)
+\frac12\left(\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}\right)
+\frac12\left(\cos\frac{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solve a system by putting new variables
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{7}{25} \\ x^2+y^2=25 \end{array}$$
The first step is to determine the domain: $\begin{array}{|l} x \ne 0 \\ y \ne 0 \end{array}$
We can simplify the first equation of the system, and we get: $\begin{arr... | let $b=\frac{x}{y}$ (defined because $x\ne 0,y\ne 0$). Then by the first equation:
$b+\frac{1}{b}=\frac{7}{25}$
$\iff (b\ne0)$
$b^2-\frac{7}{25}b+1=0$
$\iff$
False (no real solutions)
so there are no solutions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Existence of $\lim_{k\to +\infty}\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\,\mathrm{d}x$
Prove that existence of limit $$\lim_{k\to +\infty}\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\,\mathrm{d}x$$ ... | How about this?
For fixed $x \in (0,\pi/2)$, the integrand
$$
\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}
$$
increases as $k$ increases. The integrand is nonnegative. So by the monotone convergence theorem,
$$
\lim_{k\to\infty}\int_0^{\pi/2}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Limit of $\frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$ as $n\to\infty$ I've tried to solve the limit
$$ \lim_{n \to \infty} \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1}$$
but I'm not sure.
$$ \frac { 2^{\sqrt{ (\ln n)^2+ \ln n^2}}}{n^2+1} = \frac { 2^{\sqrt{ (\ln n)^2+ 2\ln n}}}{n^2+1} = \frac { 2^{\ln n \sqr... | This is right, because it depends what is in your base. Consider
$ (e^2)^{\ln(n)} = (e^{\ln(n)})^2 = n^2 $
Since $10 > e^2$ or rather $\ln(10) = a > 2$
You have: $10^{\ln(n)}$ is equivalent of $ n^a $ and therefore
$ \frac { 10^{\ln(n)}}{n^2+1} \rightarrow +\infty$
In your initial example $2 < e < e^2$ so the limit i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving inequality $(a^2 + b^2)^3 \ge 32(a^3 + b^3)(ab - a - b)$
If $a, b \in \mathbb R$ and $a + b \geq 0$, then prove that $$(a^2 + b^2)^3 \geq 32(a^3 + b^3)(ab - a - b)$$
Since $a + b ≥ 0$, we can apply A.M.-G.M. inequality, I tried to apply the inequality, but wasn't able to reach a conclusive decision. How can I... | We need to prove that:
$$(a^2+b^2)^3+32(a+b)^2(a^2-ab+b^2)\geq32(a+b)(a^2-ab+b^2)ab$$ and since $a+b\geq0,$ it's enough to prove our inequality for $ab\geq0,$ which gives that $a$ and $b$ are non-negatives.
Now, by AM-GM
$$(a^2+b^2)^3+32(a+b)^2(a^2-ab+b^2)\geq2\sqrt{(a^2+b^2)^3\cdot32(a+b)^2(a^2-ab+b^2)}.$$
Thus, it's... | {
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"url": "https://math.stackexchange.com/questions/3480852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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value of $n$ in limits
If $\displaystyle \lim_{x\rightarrow 0}\frac{x^n\sin^{n}(x)}{x^{n}-(\sin x)^{n}}$ is a finite non zero number.
Then value of $n$ is equals
What I try:
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots $$
$$\lim_{x\rightarrow 0}\frac{x^n\bigg(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots... | $$\frac{\sin x}{x}\simeq 1-\frac{1}{6}x^2+\frac{1}{120}x^4+O(x^6)$$
$$\left(\frac{\sin x}{x}\right)^n\simeq 1-\frac{n}{6}x^2+\frac{n(5n-2)}{360}x^4+O(x^6)$$
$$\frac{\left(\frac{\sin x}{x}\right)^n}{1-\left(\frac{\sin x}{x}\right)^n}\simeq \frac{6}{n}x^{-2}-\frac{5n+2}{10n}+O(x^2)$$
$$\boxed{f(x)=\frac{x^n\sin^n x}{x^n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $x^4+2x^2y^2+y^4\geq2xy^3$ Suppose that $x,y$ are real numbers. I want to prove $$x^4+2x^2y^2+y^4\geq2xy^3.$$ I noticed that this is the same as $$(x^2+y^2)^2\geq 2xy^3.$$ Can we proceed from here?
| Consider :
$r>0$, $0\le \theta \lt 2π$.
Then obviously:
$r^4 \ge r\sin (2\theta) r^3 \sin^2 (\theta)$(Why?)
$r^4 \ge 2r \sin (\theta) \cos (\theta) r^3 \sin^2 (\theta)$
And with $x=r\cos (\theta)$, $y=r \sin (\theta)$:
$(x^2+y^2)^2\ge 2 xy^3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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$A (3,1)$ reflected to $y=2x$ what is area of AOA'? $y_1 = 2x$ = line of reflection
$x_1,y_1 = A(3,1) =$ reflected point
General formation of a line
$y = mx + k$
$Ax + By + C = 0$
Finding a line perpendicular to line of reflection
$m_2 = \frac{-1}{2}$
$y_2 = \frac{-1}{2}x + \frac{5}{2}$
Intersection of reflection line ... | Note that $|OA| = \sqrt{3^2+1}=\sqrt{10}$ and the angle $\theta$ between $y=2x$ and OA is given by
$$\tan\theta = \tan(\theta_2-\theta_1)
= \frac{\tan\theta_2 - \tan\theta_1}{1+\tan\theta_2\cdot \tan\theta_2 }
=\frac{2-\frac13}{1+2\cdot\frac13} = 1$$
which yields $\theta = 45^\circ$. Then, AOA' is an isosceles right ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\iint |xy|\,dx\,dy$ using polar coordinates If $R$ is the region bounded by $x^2+4y^2 \ge 1$ and $x^2+y^2 \le 1$. Then find the integral $$I=\iint_R |xy|\,dx\,dy.$$
I tried using Cartesian system and got the answer as $\frac{3}{8}$ using symmetry. Can we do this in Polar coordinates?
| First, use symmetry to only do the integral in the first quadrant (this is allowed because both the integrand and the region of integration share the same symmetry):
$$\iint_R |xy|\: dA = 4 \iint_{R_1} |xy|\:dA$$
Next, we'll have to figure out the bounds. The upper bounds aren't difficult, but the lower ones are given ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Derivative of $\sec^{-1}x$ and integral of $\frac{1}{x\sqrt{x^2-1}}$ My attempt is as follows:-
*
*Derivative of $\sec^{-1}x$
Let $\theta=\sec^{-1}x,$ where $\theta\in [0,\pi]-{\dfrac{\pi}{2}}.$
$$\sec\theta=x.$$
Differentiating both sides with respect to $x:$
$$\sec\theta\cdot\tan\theta\cdot\dfrac{\mathrm d\theta}{\... |
But many textbooks write that $\displaystyle\int\frac{1}{x\sqrt{x^2-1}}\,\mathrm dx=\sec^{-1}x+C.$
This is indeed wrong, since differentiating the equation at $x=-5$ gives $\displaystyle\frac{\sqrt6}{60}=-\frac{\sqrt6}{60}.$
Case $1:x>0$
the integral is definitely $\sec^{-1}x.$
Case $2: x<0$
the integral is $-\sec^{... | {
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"url": "https://math.stackexchange.com/questions/3485512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Let $x_1,x_2 \in \mathbb{R}$ be the roots of the equation $x^2+px+q=0$
Let $x_1,x_2 \in \mathbb{R}$ be the roots of the equation
$x^2+px+q=0$. Find $p$ and $q$ if it is known that $x_1+1$ and $x_2+1$
are the roots of the equation $x^2-p^2x+pq=0$.
The roots of $x^2+px+q=0$ satisfy $$x_1+x_2=-\dfrac{b}{a}=-p,x_1x... | Another way :
Like If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.
let $x_k+1=y_k; k=1,2$
$\implies x_k=y_k-1$
As $x_k$ is a root of $$x^2+px+q=0$$
$$(y_k-1)^2+p(y_k-1)+q=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluate $\int\frac{dx}{(1+\sqrt{x})(x-x^2)}$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}
$$
Set $\sqrt{x}=\cos2a\implies\dfrac{dx}{2\sqrt{x}}=-2\sin2a.da$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}=\int\frac{-4\sin2a\cos2a.da}{2\cos^2a.\cos^22a.\sin^22a}\\
=\int\frac{-2.\sec^2a.da}{\sin2a\cos2... | Following @J.G's subbing
$\sqrt{x}=y$ then $y=\frac{1-u}{1+u}$
$$I\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{2}{y(1+y)(1-y^2)}dy=\frac12\int\frac{(1+u)^2}{u(u-1)}du$$
$$=-\frac12\int\left(\frac1u-\frac4{u-1}-1\right)du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove that $R$ is reflexive, symmetric, and transitive. Define a relation $R$ on $\Bbb Z$ by declaring that $xRy$ if and only if $x^2\equiv y^2\pmod{4}$. Prove that $R$ is reflexive, symmetric, and transitive.
Suppose $x\in\Bbb Z$. Then $x^2\equiv x^2\pmod {4}$ means that $4\mid (x^2-x^2)$, so $x^2-x^2=4a$ where $a=0\i... | This is straight forward and I think you did ok. The transitivity follows immediately from transitivity of congruence. In fact all three follow from the fact that congruence mod $n$ is indeed an equivalence relation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int \frac{\cos^2x}{1+\tan x}dx$ Evaluate $\int \dfrac{\cos^2x}{1+\tan x}dx$
Here are my various unsuccessful attempts:-
Attempt $1$:
$$\tan x=t$$
$$\sec^2 x=\dfrac{dt}{dx}$$
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}$$
$$\ln(1+t)=y$$
$$\dfrac{1}{1+t}=\dfrac{dy}{dt}$$
$$\int \dfrac{dy}{\left(1+(e^y-1)^2\right)^2}$$
$$... | To find $A,B,C,D,E$, you should have:
$$A(1+t^2)^2+(Bt+C)(1+t)(1+t^2)+(Dt+E)(1+t)=1$$
Then equate coefficients.
Writing the integrand in partial fractions, we have
$$\int \dfrac{dt}{(1+t^2)^2(1+t)}=\int\left(\frac{1}{4(1+t)}+\frac{1-t}{4(1+t^2)}+\frac{1-t}{2(1+t^2)^2}\right)\,dt$$
$$=\frac{1}{4}\log(1+t)+\frac{1}{4}\ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Po-Shen Loh's new way of solving quadratic equation Quadratic equation, $ax^2+bx+c=0$ and its solution is quadratic equation, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Now setting $a=1$ then we have $x^2+bx+c=0$
$$x=\frac{-b\pm \sqrt{b^2-4c}}{2}$$ rewrite as
$$x=-\frac{b}{2}\pm \sqrt{\left(\frac{b}{2}\right)^2-c}$$
In thi... | Let us illustrate by example. Consider the equation
\begin{align}
x^2-2019x-2020 = (x-r_1)(x-r_2) = x^2-(r_1+r_2)x+r_1r_2=0.
\end{align}
The key observation is that the roots $r_1, r_2$ adds up to $2019$, which means the average of $r_1$ and $r_2$ is $\frac{2019}{2}$. Hence the roots have the form $r_\pm = \frac{2019}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 8,
"answer_id": 1
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How many ways can we get a number by addition if each part of the addition has to be smaller or equal to a set value? For example, if we need to get 5 with the largest number we can use being 3, we can use:
*
*3 + 2
*3 + 1 + 1
*2 + 2 + 1
*2 + 1 + 1 + 1
*1 + 1 + 1 + 1 + 1
Is there any way to find out the soluti... | It is a simple function call. In Mathematica, for example:
IntegerPartitions[5, {1, 5}, {1, 2, 3}]
(*
{{3, 2}, {3, 1, 1}, {2, 2, 1}, {2, 1, 1, 1}, {1, 1, 1, 1, 1}}
*)
If you merely seek the number of such partitions:
Length@IntegerPartitions[5, {1, 5}, {1, 2, 3}]
(* 5 *)
[This function asks for all integer partition... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How I can show that $\prod_{i=1}^{n}\frac{3^i(x+1)-2^i}{3^i(x+1)-3\cdot2^{i-1}}=\frac{1}{x}-\frac{1}{x}\left(\frac{2}{3}\right)^n+1$? How I can show that the following equality is true:
$$\prod_{i=1}^{n}\dfrac{3^i(x+1)-2^i}{3^i(x+1)-3\cdot2^{i-1}}=\dfrac{1}{x}-\dfrac{1}{x}\left(\dfrac{2}{3}\right)^n+1\,?$$
| Here we have a telescoping product.
We obtain for $n\geq 1$:
\begin{align*}
\color{blue}{\prod_{j=1}^n\frac{3^j(x+1)-2^j}{3^j(x+1)-3\cdot 2^{j-1}}}
&=\frac{1}{3^n}\prod_{j=1}^n\frac{3^j(x+1)-2^j}{3^{j-1}(x+1)-2^{j-1}}\tag{1}\\
&=\frac{1}{3^n}\,\frac{\prod_{j=1}^n \left(3^j(x+1)-2^j\right)}{\prod_{j=1}^{n}\left(3^{j-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to evaluate: $\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x) \, \mathrm dx$ $$\int_0^1 \frac{\frac{\pi^2}{6}-\operatorname{Li}_2(1-x)}{1-x}\cdot \ln^2(x)\,\mathrm dx=1.03693\ldots$$
This number looks like $\zeta(5)$ value.
We expand the terms
$$\int_0^1\frac{\frac{\pi^2}{6}}{1-x}\cdot ... | Here is an alternative approach through the harmonic sum thicket.
Starting with Euler's reflexion formula for the dilogarithm function, namely
$$\operatorname{Li}_2 (x) + \operatorname{Li}_2 (1 - x) = \zeta (2) - \ln x \ln (1- x),$$
your integral $I$ can be rewritten as
$$I = \int_0^1 \frac{\ln^3 x \ln (1 - x)}{1 - x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int\frac{dx}{(1+x^4)^{1/4}}$
Solve
$$
\int\frac{dx}{(1+x^4)^{1/4}}
$$
Set $t=\log x\implies x=e^t\implies dt=\dfrac{dx}{x}$
$$
\int\frac{dx}{(1+x^4)^{1/4}}=\int\frac{\dfrac{1}{x}dx}{\big(\dfrac{1}{x^4}+1\big)^{1/4}}=\int\frac{dt}{(e^{-4t}+1)^{1/4}}
$$
Set $e^{-4t}+1=y\implies-4e^{-4t}dt=dy$
$$
I=\int\fra... | Once you have
$$I=\int\frac{dx}{(x-1)x^{1/4}},$$
take $x=u^4$ so that $dx=4u^3du$. We have $$I=4\int\frac{u^2du}{u^4-1}.$$
This is
$$\begin{align}
\tfrac14I&=\int\frac{u^2du}{(u^2+1)(u^2-1)}\\
&=\frac12\int\frac{du}{u^2+1}+\frac14\int\frac{du}{u-1}-\frac14\int\frac{du}{u+1}.
\end{align}$$
The rest is easy from there.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3493494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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integrate $\int_{-\infty}^\infty e^{x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2}}dx $ How can we integrate $$\int_{-\infty}^\infty e^{x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2}}dx ~~?$$Is it integrable function over $\mathbb R$?, Can we use the fact $\int_{-\infty}^\infty e^{ax-x^2}dx = \sqrt{\pi}e^{\frac{a^2}{4}}$
| You can use the similar fact that
(i) $\int_{-\infty}^\infty e^{ax-bx^2}dx = \frac{\sqrt{\pi}}{\sqrt{b}}e^{\frac{a^2}{4b}}$
The expression in the exponent is:
$x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2} = x-\frac{x^2+1.9x + 0.95^2}{2\cdot (0.35)^2} = (1 + \frac{1.9x}{2\cdot (0.35)^2})x -\frac{1}{2\cdot (0.35)^2}x^2 + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Factoring $1 - y-x^2-y^2-yx^2+y^3$ I was struggling to factorize this expression, I need it cleaned to bracket form but I was struggling to understand the process to get the answer? Is there a special formula or technique for factorizing these sorts of expressions?
$$1 - y-x^2-y^2-yx^2+y^3$$
Answer: $$ \ ( 1+y)[(1-y)^2... | Once you gave us the answer, the factoring became easy. You just have to look for the factor $1+y$
$$1 - y-x^2-y^2-yx^2+y^3= (1+y^3)-(y+y^2)-x^2(1+y)$$
$$=(1+y)(1-y+y^2)-y(1+y)-x^2(1+y)=(1+y)(1-y+y^2-y-x^2)$$
$$=(1+y)[(1-y)^2-x^2]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve the irrational radical equation Solve equation:$$\frac{x + \sqrt{3}}{\sqrt{x} + \sqrt{x + \sqrt{3}}} + \frac{x - \sqrt{3}}{\sqrt{x} - \sqrt{x - \sqrt{3}}} = \sqrt{x}$$
answer: $S=\{2\}$
Attemp:
$$\frac{x+\sqrt{3}}{\sqrt{x}+\sqrt{x+\sqrt{3}}}+\frac{x-\sqrt{3}}{\sqrt{x}-\sqrt{x-\sqrt{3}}}=\sqrt{x}\Leftrightarrow \... | You have a mistake in your algebra (see the previous answer) but rationalizing the denominator is a good idea. Note that the domain is $x\geq \sqrt 3$. After you do that and multiply both sides by $\sqrt{3}$, you obtain
$$x\sqrt{x-\sqrt{3}}-\sqrt{3}\sqrt{x-\sqrt{3}}+x\sqrt{x+\sqrt{3}}+\sqrt{3}\sqrt{x+\sqrt{3}}=3\sqrt 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$
If real numbers $a$, $b$, $c$, $d$ satisfy
$$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$
then find $(a,b,c,d)$.
What I try:
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, pl... | Let $(p,q,r,s) = (1-a,a-b,b-c,c)$, we have $p+q+r+s = 1$.
We are given
$$p^2 + q^2 + r^2 + s^2 = \frac14$$
This leads to
$$\begin{align} &\;\left(p - \frac14\right)^2 +
\left(q - \frac14\right)^2 +
\left(r - \frac14\right)^2 +
\left(s - \frac14\right)^2\\
= &\; (p^2+q^2+r^2+s^2) - \frac12(p+q+r+s) + \frac14\\
= &\; \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Is this solution correct for equation $3x^2-4y^2=13$? Prove equation $3x^2-4y^2=13$ has no integer solution.
Solution: Suppose (y, 3)=1, We have:
$y≡( 1, 2) \mod (3)$
⇒ $4y^2≡( 1, 2) \ mod(3)$
$3x^2≡0 \mod (3)$
The common remainder between $3x^2$ and $4y^2$ is 0 , so we may write:
⇒ $3x^2-4y^2≡ 0 \ mod(3)$
But, $13≡1 \... | I think you are trying to say:
If you assume $\gcd(y,3) =1$ (Why are you assuming this? What if $\gcd(y,3) = 3$?) or in other words if $3\not\mid y$ then $y\equiv 1 \pmod 3$ or $y\equiv 2 \pmod 3$.
ANd therefore $y^2\equiv 1 \pmod 3$ or $y^2\equiv 4\equiv 1\pmod 3$. and so $4y^2 \equiv 1\pmod 3$. I'm not sure why you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Integration of a particular rational expression I am trying to solve the following integration, where $a,b,c,d,e$ and $f$ are constants:
$$I=\int\frac{x^4+ax^3+bx^2+cx+d}{x^3(x^3+ex+f)}dx$$
I tried to solve the integral using the following two methods, but both seemed to be very much complicated:
Method 1:
Using partia... | Taking into account what you wrote about the roots of the denominator, I should write
$$\frac{x^4+ax^3+bx^2+cx+d}{x^3(x^3+ex+f)}=\frac{x^4+ax^3+bx^2+cx+d}{x^3(x-r)(x^2+s x+t)}$$Now, partial fraction decomposition would give
$$-\frac{d}{r t }\frac 1 {x^2}+\frac{d r s-d t-c r t}{r^2 t^2 }\frac 1 {x}+\frac{a r^3+b r^2+c r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Chain rule partial derivative
Find the partial derivatives $\partial Z / \partial u $ and $\partial Z / \partial \nu$ for the following function:
$$
Z(x,y,z) = 2x^3 - 3xy^2 + 0.75~yu - 5u^2\quad \text{where}~x = \sqrt{u+\nu}~\text{and}~y=\nu^2
$$
No matter how hard I try I couldn't solve it and couldn't find anythi... | We calculate $\frac{\partial Z}{\partial u}$ in two ways.
First variant: We consider
\begin{align*}
Z(x(u,v),y(u,v),z(u,v))=2x^3-3xy^2+\frac{3}{4}yz-5z^2
\end{align*}
where
\begin{align*}
x=\sqrt{u+v},\qquad y=v^2,\qquad z=u
\end{align*}
According to the chain rule we obtain
\begin{align*}
\color{blue}{\frac{\partia... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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$50\cos^2 x + 5\cos x = 6\sin^2 x$, find $\tan x$
$50\cos^2 x + 5\cos x = 6\sin^2 x$
Find $\tan x$
I used $\cos^2 x + \sin^2 x = 1$ to get the equation $$56\cos^2 x + 5\cos x -6 = 0$$
I then solve this to get $\cos x = \dfrac27, -\dfrac38$
Then I used generic trig ratios to get $\tan x = \pm\dfrac{3\sqrt5}{2}, \pm\df... | Using the Weierstrass transformation, the equation rationalizes to
$$50\left(\frac{1-t^2}{1+t^2}\right)^2+5\frac{1-t^2}{1+t^2}=6\left(\frac{2t}{1+t^2}\right)^2$$
and gives, by solving the biquadratic
$$45t^4-100t^2+55=24t^2,$$
$$t^2=\frac 59,\frac{11}5.$$
The tangent follows, by
$$\pm\frac{2\dfrac{\sqrt5}3}{1-\dfrac 59... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Can we solve for $a$ and $b$ only knowing the value of $z$? We have the following equation for $z$ > 0
$ a^2b+ab^2 = z $
Only the value of $z$ is known. How can we solve this equation to get the values of $a$ and $b$?
Simplifying,
$ab(a+b) = z $
eg. $z=84$
then $a = 3$, $b=4$
Any approach or suggestions or trial... | If $a,b,z \in \mathbb{R}$ then there are an infinite number of solutions.
If we assume $a=b$ then we have $2a^3=z \Rightarrow a=b=\root 3 \of {\frac z 2}$.
But if we assume $2a=b$ then we have $6a^3=z \Rightarrow a=\root 3 \of {\frac z 6}, b=2\root 3 \of {\frac z 6}$.
And, in general, if $ka=b$ then $(k+k^2)a^3=z \Righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $C = \{(x, y, z) ∈ R^3: x + y + z = 1, x^2 + y^2 + z =7/4\}.$ compact? Is the following set compact? How can I show it? $C = \{(x, y, z) ∈ R^3: x + y + z = 1, x^2 + y^2 + z =7/4\}.$
Clearly it is closed as it contains its boundary, but I can not show that it is bounded..
| Substituting, we find that $x^2+y^2-x-y = (x-\frac{1}{2})^2 - \frac{1}{4} + (y-\frac{1}{2})^2 - \frac{1}{4} = \frac{3}{4}$, ie $(x-\frac{1}{2})^2 + (y-\frac{1}{2})^2 = \frac{1}{4}$, so the projection of $C$ on the z-plane is a circle, which is bounded as $|x|\le K$, $|y|\le K$ for some constant $K$.
Now, going back to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $\frac ab=\frac bc=\frac cd$, then $\frac ad=\sqrt{{a^5+b^2c^2+a^3c^2\over b^4c+d^4+b^2cd^2}}$
If
$$\frac ab=\frac bc=\frac cd$$
then prove that
$$\frac ad=\sqrt{{a^5+b^2c^2+a^3c^2\over b^4c+d^4+b^2cd^2}}$$
I don't want a full solution, just a little hint on how to start solving are welcome
| Note
$$\frac{d^2}{a^2}\cdot {a^5+b^2c^2+a^3c^2\over b^4c+d^4+b^2cd^2}
=\frac{a^3+\frac{b^2c^2}{a^2}+ac^2}{\frac{b^4c}{d^2}+d^2+b^2c}
=\frac{a^3+d^2+ac^2}{a^3+d^2+ac^2}=1$$
Thus,
$$\sqrt{{a^5+b^2c^2+a^3c^2\over b^4c+d^4+b^2cd^2}}=\sqrt{\frac{a^2}{d^2}}=\frac ad$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Does there exist a power of 2 whose last 100 digits are composed only of the digits 1 or 2? Does there exist a power of 2 whose last 100 digits are composed only of the digits 1 or 2? If so, what techniques could be used to prove its existence?
|
Let $q_n$ for $n>0$ be the recurring sequence of natural numbers defined by:
\begin{align}
&q_1=1&&q_{k+1}=\frac{q_k+5^k(2-q_k\bmod 2)}2
\end{align}
then for every $k>0$ the number $a_k=2^kq_k$ satisfy:
*
*$0<a_k<10^k$;
*the decimal digits of $a_k$ belongs to $\{1,2\}$;
*there exists $m\geq k$ such ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How do I evaluate the integral of the square root of a quartic equation? I'm currently trying to evaluate the integral
$$\int^1_0 \text{d}t\sqrt{(1-t^2)(1-k^2t^2)}$$
where $k\in(0,1)$. Is it the case that this can be expressed in terms of elliptic integrals? I'm struggling to see how this can be done as the elliptic in... | Rewriting the integrand and integrating by parts we find
\begin{align}f(k) &\equiv \int \limits_0^1 \sqrt{(1-x^2)(1-k^2 x^2)} \, \mathrm{d} x = \int \limits_0^1 \frac{\sqrt{1-x^2}}{x} x \sqrt{1-k^2 x^2} \, \mathrm{d} x \\
&= \left[\frac{\sqrt{1-x^2}}{x} \frac{1-(1-k^2 x^2)^{3/2}}{3 k^2}\right]_{x=0}^{x=1} - \int \limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$? The sum $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ is just a bit larger than $1$. Is there som... | For positive, unequal $a$ and $b$:
$\dfrac1a+\dfrac1b=\dfrac{a+b}{ab}>\dfrac4{a+b}$
because $(a+b)^2>4ab$ (the difference between these is $(a-b)^2$). So,
$\dfrac15+\dfrac17>\dfrac4{12}=\dfrac13$
$\dfrac19+\dfrac1{11}>\dfrac4{20}=\dfrac1{5}$
$\dfrac18+\dfrac1{12}>\dfrac4{20}=\dfrac1{5}$
When these inequalities are put... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
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Compute $\sigma$, given $\sigma^{11}$ in $S_{10}.$
Let $\sigma \in S_{10}$ with $\sigma^{11}=(3,1,4)(5,2,9,6,8,7,10)$. How to compute $\sigma$?
I tried:
$\sigma^{11}=(3,1,4)(5,2,9,6,8,7,10)=\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 4 & 9 & 1 & 3 & 2 & 8 & 10& 7 & 6 & 5\end{pmatrix}$
Now I tried to fin... | Let $\tau= (314)(529687a)$ (with $a=10$) & so $\tau^{21}=e$ & $\sigma^{21}=e$. We have $ \tau= \sigma^{11}$. Squaring gives $\sigma=\tau^2$. Which is easy to compute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Methods for evaluating $\sum_{n=1}^\infty \frac1{a+(n-1)n}$ I am interested in methods for evaluating the sum $$\sum_{n=1}^\infty \frac1{a+(n-1)n}.$$
Indeed I will give my own answer below using the Residue Theorem.
Please feel free to post other methods for the evaluation, such as Maclaurin series, methods from harmon... | Let $s$ and $t$ to be the roots of $n^2-n+a=0$. So
$$\frac{1}{a+n(n-1)}=\frac{1}{(n-s)(n-t)}=\frac{1}{s-t}\left(\frac 1{n-s}-\frac 1{n-t} \right)$$ Recalling that
$$\sum_{n-1}^p \frac 1{n-x}=\psi(p-x+1)-\psi (1-x)$$ we have that, if
$$S={\sqrt{1-4 a}}\sum_{n-1}^p \frac{1}{a+n(n-1)}$$
$$S=\psi \left(\frac{1}{2}+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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What is the theory behind a simple pattern involving binomial coefficients? These binomial equations are all true.
$\binom{7}{4} = \binom{4}{4} + 3[ \binom{4}{3} + \binom{4}{2}] + \binom{4}{1}$
$\binom{8}{4} = \binom{5}{4} + 3[ \binom{5}{3} + \binom{5}{2}] + \binom{5}{1}$
$\binom{9}{4} = \binom{6}{4} + 3[ \binom{6}... | By repeatedly applying Pascal's Theorem:
$$\begin{align*}
\binom{n}{4} &=\binom{n-1}{4}+\binom{n-1}{3}
\\ &=\left(\binom{n-2}{4}+\binom{n-2}{3}\right)+\left(\binom{n-2}{3}+\binom{n-2}{2}\right)
\\ &=\left(\left(\binom{n-3}{4}+\binom{n-3}{3}\right)+\left(\binom{n-3}{3}+\binom{n-3}{2}\right)\right)+\left(\left(\binom{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$ As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.
Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$
Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$
Therefore... | Your answer is extremely close to the correct derivation. The error occurs when you write
$$\frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=\sum_{\color{red}{n=0}}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$
where the last equality shou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Given a polynomial $P$ with integer coefficients, find $P(2)$ Question:
$P(x)$ be a polynomial with non-negative integer coefficients such that $P(0)=33$ , $P(1)=40$ and $P(9)=60000$.
Find $P(2)$.
My attempt:
Suppose $P(x) = a_nx^n + \dotsb + a_1x + a_0$.
Now, $P(0) = 33$ implies $a_0 = 33$.
$P(1)=40$ implies that the ... | The polynomial can be at most of degree $4$ because
$9^5=59409$
so
$p(x)=33+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5$
and
$a_1+a_2+a_3+a_4+a_5=7$
It is clear that $a_5=1$ otherwise
if $a_5=0$ then $p(9)<60000$
while if $a_5>1$ then $a_59^5>60000$
If $a_4\neq 0$ then $p(9)>60000$ so $a_4=0$. If $a_3=0$ we have
$a_1+a_2=6$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Combination of problem having identical objects without using generating function approach
A person goes in for an examination in which there are $4$ papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2m$ marks is
My Attempt
Number of ways of partitioning $n$ identical object... | We have to count the number of solutions of
$$x_1+x_2+x_3+x_4=2m$$
where $0\leq x_i\leq m$ are integers for $i=1,2,3,4$.
By Stars and Bars, if we consider just the condition $x_i\geq 0$ we have
$\binom{2m+3}{3}$ solutions. On the other hand, we can have at most one variable $x_i\geq m+1$. If $i=1$, we have to count t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Algebraic and Trigonometric expression is $>0$ for all real $x$
Prove that $$2x^2\sin x+2x\cos x+2x^2+1$$ is always positive for all real $x$.
From completing the square method
Write $1$ as $\sin^2 x+\cos^2 x$
$$x^4+2x^2\sin x+\sin^2 x+x^2+2x\cos x+\cos^2 x+x^2-x^4$$
$$(x^2+\sin x)^2+(x+\cos x)^2+x^2(1-x^2)$$
$\bulle... | By completing the square,
$$2(\sin x+1)x^2+2\cos x\,x+1\\
=2(\sin x+1)\left(x+\frac{\cos x}{2(\sin x+1)}\right)^2+1-\frac{\cos^2x}{2(\sin x+1)}\\
=2(\sin x+1)\left(\left(x+\frac{\cos x}{2(\sin x+1)}\right)^2+\frac{\sin x+1}2\right).$$
Obviously, $\sin x+1\ge0$. In case of equality, $\sin x=-1\implies \cos x=0$ and the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Let $X,Y$ be independent normally distributed random variables. Find the density of $\frac{X^2}{Y^2+X^2}$ Let $X,Y$ be independent standard normally distributed random variables and $X,Y\neq 0$. Find the density of $\frac{X^2}{Y^2+X^2}$
I was given the tip of first calculating the density of $(X^2,Y^2)$ and then calcul... | I’m on my phone so I can’t type it all out but X^2 and Y^2 are independent chi-squared r.v.s with one degree of freedom each.
These chi squared rvs are equivalent to Gamma(1/2,2) in distribution.
By a known fact about gamma random variables, if X and Y are independent gammas with the same location but different shape p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Computation of: $\lim_{n\to\infty}\left(\ln\left(1+\frac{1}{n^2+1}\right)^n+\ldots+\ln\left(1+\frac{1}{n^2+n}\right)^n\right)$
Evaluate:
$$\lim_{n\to\infty}\left(\ln\left(1+\frac{1}{n^2+1}\right)^n+\ldots+\ln\left(1+\frac{1}{n^2+n}\right)^n\right)\;n\in\mathbb N$$
My attempt:
Using the manual limit:
$$\lim_{x\to 0}... | You can squeeze it as follows:
$$n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2+n}\right)\leq n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2+k}\right)\leq n\sum_{k=1}^n\ln\left(1+\frac{1}{n^2}\right)$$
Hence,
$$\underbrace{n^2\ln\left(1+\frac{1}{n^2+n}\right)}_{\stackrel{n\to\infty}{\longrightarrow}1}\leq n\sum_{k=1}^n\ln\left(1+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx$
Evaluate
$$\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$$
I have tried substitution of $\sin x$ as well as $\cos x$ but it is not giving an answer.
Do not understand if there is a formula for this or not.
| Let $x=\tan ^{-1}(t)$ to make
$$I=\int \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx=\int \frac{dt}{t^{3/2}+t^2+\frac{1}{\sqrt{t}}+1}$$ Now, $t=u^2$
$$I=\int \frac{2 u^2}{u^5+u^4+u+1}\,du=\int \frac{du}{u+1}-\int \frac{u^3-u^2-u+1}{u^4+1}$$ For the second integral, use the roots of $u^4+1=0$ and partial fracti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Solving the inequality $1+1.2 + 1.2^2 + 1.2 ^3 + \cdots + 1.2^n < N/16?$ Given the inequality $$1+1.2 + 1.2^2 + 1.2 ^3 +\cdots + 1.2^n < \frac{N}{16}?$$
I need the value of $n$, or just an approximation. $N$ is known.
| Using the formula for the partial sum of a geometric series, rewrite this as
$$
\frac {1.2^{n+1}-1}{1.2-1} < \frac N{16} \implies
5(1.2^{n+1}-1) < \frac N{16} \implies
1.2^{n+1} < 1 + \frac N{80}.
$$
If we take a logarithm, we can solve this exactly to get
$$
n < \frac{\log(1 + \frac N{80})}{\log(1.2)} - 1.
$$
For a nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the radius of three equal circles, with radius $r$, inscribed in a rectangle. How can we find the radius of the small circles with the same given radius $r$. Three equal circles, with radius $r$, are inscribed in a rectangle in a way only one of them touches the others two, as the figure indicates. The circles cen... | If the white curve is a circle the you have three points of the big circle (assume symmetry: Big circle has $(-8,0),(0,6),(8,0)$) so the equation (and radius) can be figured.
Big Circle is center at $(0, -A)$ with radius $R$ and $8^2 + A^2 = R^2$ and $0^2 + (A+8)^2 = R^2$.
The three little circles have equations $(a_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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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.