Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-... | $f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$ is false !
Correct is:
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(-x^2)-\sqrt{3-x^2}}{(3+x)^2}$.
Now proceed !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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Probability of winning chocolates in throwing game In urn there are 12 balls from which 7 are green.John draws at random 4 balls.Then he hits a target with the green balls he drew(if there are any).His probability of hitting successfully is $\frac{1}{4}$ in every shot and if he succeeds he wins a chocolate for every su... | Note that John can only win 2 chocolates if he drew at least 2 green balls from the urn. So the first step is to find the probs of drawing 2, 3 and 4 greens from the urn.
We have $$p(2G)= 6\frac{7}{12}\frac{6}{11}\frac{5}{10}\frac{4}{9}=\frac{42}{99}$$ Similarly $$p(3G)=\frac{35}{99},p(4G)=\frac{7}{99}$$
We also need t... | {
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"url": "https://math.stackexchange.com/questions/3526086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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ordered pair of $(p,q)$ in $20!$
A rational number given in the form
$\displaystyle \frac{p}{q},\;\;p,q\in \mathbb{Z}^{+}\;,\frac{p}{q}\in(0,1)$ and $p,q$ are coprime to each other.If $pq=20!.$ Then number of ordered pair of $p,q$ are
what i try
$20!=2^{18}\cdot 3^{8}\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19... | Since $p.q=20!$ hence I can treat this like there are eight distinct objects i.e $2^{18},3^8,5^4,7^2,11,13,17,19$ need to be distributed into two different groups $p , q$ where
$$ $$
Total number of ways are $2^8$ since each object can go in any one of the group. And since $20!$ is not a perfect square hence $p\neq q$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve a system of equation: $\cos(2x) + \cos(y) = 1$, $\sin(2x) + \sin(y) = 1$ Solve a system of equation:
$$\cos(2x) + \cos(y) = 1$$
$$\sin(2x) + \sin(y) = 1$$
My idea:
Let's see what is product of this two equations.
$$\cos(2x)\sin(2x) + \cos(2x)\sin(y) + \cos(y)\sin(2x) + \cos(y)\sin(y) = 1$$
$$\cos(2x)\sin(2x) + \s... | HINT.-We have from the two given equations $$2\cos\left(\dfrac{2x+y}{2}\right)\cos\left(\dfrac{2x-y}{2}\right)=1\\2\sin\left(\dfrac{2x+y}{2}\right)\cos\left(\dfrac{2x-y}{2}\right)=1$$ so, by division $$\tan\left(\dfrac{2x+y}{2}\right)=1\Rightarrow\dfrac{2x+y}{2}=\dfrac{(2n+1)\pi}{4}$$ Each of the two given equations is... | {
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"url": "https://math.stackexchange.com/questions/3533104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complicated question, please help me out I'm not able to get a proper defined function and not able to determine it's nature, pls help
Let the function $f:[0,1] \longrightarrow \Bbb R$ be defined by
$$f(x) = \max\left\{\frac{|x-y|}{x+y+1}\, :\, 0\leqslant y \leqslant 1\right\}.$$
Then what are the intervals on which t... | For each $x\in [0,1]$, the $y\in [0,1]$ which maximizes $\frac{|x-y|}{x+y+1}$ also maximizes ${\left(\frac{|x-y|}{x+y+1}\right)}^2$.
Because $|x-y|^2 = (x-y)^2$, we may apply calculus techniques to the second one.
We could also not square, and consider when $y\leqslant x$ and when $y>x$, but squaring does away with cas... | {
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"url": "https://math.stackexchange.com/questions/3535441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=... | Let $\dfrac1x=h$ where $h>0$
$$\sqrt[3]{x^3+2x}= \dfrac{\sqrt[3]{1+2h^2}}h$$ and $$\sqrt{x^2-2x}=\dfrac{\sqrt{1-2h}}h$$
Now for $F=\lim_{y\to0}\dfrac{\sqrt[n]{1+my}-1}y,$ set $\sqrt[n]{1+my}-1=z,my=(1+z)^n-1$
$F=\lim_{z\to0}\dfrac{mz}{nz+O(z^2)}=\dfrac mn$
$$\implies\lim_{y\to0}\sqrt[n]{1+my}=\dfrac{my}n+1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Is the function $f$ continuous at $(0,0)$
Is the function $f$ continuous at $(0,0)$?
$f(x, y)$ := $\frac{xy}{|x|+|y|}$, if $(x, y)$ $\not= (0, 0)$
and $f(0, 0) := 0$
My attempt:
$f(x, y)$ = $\frac{xy}{x+y}$ if $(x,y) > (0,0)$ and
$f(x, y)$ = $\frac{xy}{-(x+y)}$ if $(x,y) < (0,0)$
So using polar coordinates:
$f(x, y)$... | If $x=r\cos\theta$ and $y=r\sin\theta$, then\begin{align}\left\lvert\frac{xy}{\lvert x\rvert+\lvert y\rvert}\right\rvert&=\frac{r^2\left\lvert\cos(\theta)\sin(\theta)\right\rvert}{r\bigl(\lvert\cos\theta\rvert+\lvert\sin\theta\rvert\bigr)}\\&\leqslant\frac r2,\end{align}since $\left\lvert\cos(\theta)\sin(\theta)\right\... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$
Prove that
$$\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$$
According to Wolfram the above holds. Could someone show me the steps for this?
| The $n$-th Catalan number is defined as
$$ C_n := \frac 1 {n+1}\binom{2n}n.$$
By inspecting the family of definite integrals
$$ J_n(\alpha) := \frac 1 \pi \int_0^\alpha \xi^{2n} \sqrt{\alpha^2-\xi^2} d\xi, \qquad \alpha > 0,\ n\geq 0, $$
it can be shown, via standard integration techniques, that $C_n = J_n(2)$:
equival... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
Find all possible integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
what i try $y^2-(x+2)y+x^2-2x=0$
$$y=\frac{x+2\pm\sqrt{(x+2)^2-4(x^2-2x)}}{2}$$
$$y=\frac{x+2\pm \sqrt{-3x^2+12x+4}}{2}$$
How do i solve it Help me please
| The equation is equivalent to $(x-y)^{2}+(x-2)^{2}+(y-2)^{2}=8$. However due to symmetry we can assume $x-y\geq 0$ but then there are only three possible choices $x-y=0,1,2$. If $x-y=0$ $(x,y)=(0,0),(4,4)$ if $x-y=1$ then $7=(y-1)^{2}+(y-2)^{2}=0+7=1+6=2+5=3+4$ so no solution.
If $x-y=2$ then $y^{2}+(y-2)^{2}=4=0+4=3+1... | {
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Definite integral with irrational function I wonder if it's possible solve this definite integral in terms of elemental or special functions:
$$\int_0^1\sqrt{\sqrt{x^4+a}-x^2}\,dx$$
with $a>0$. I've tried with Wolfram Mathematica but doesn't give me anything.
| Let's rewrite the integral as
$$\int_0^1 \sqrt{\sqrt{x^4+b^2}-x^2}\:dx = \int_0^1 \frac{1}{2}\sqrt{\sqrt{1+\frac{b^2}{y^2}}-1}\:dy$$
for $a = b^2$ and letting $x^2 = y$. Now use the substitution $y = b\operatorname{csch}t$ to get
$$\int_{\sinh^{-1}(b)}^\infty \frac{b\cosh t}{2\sinh^2 t}\sqrt{\cosh t - 1}\:dt$$ $$ =
\... | {
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"url": "https://math.stackexchange.com/questions/3545629",
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If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$... |
Your calculation is fine, except the interpretation. Note that,
$$h(-\sqrt{2-\sqrt3}) = h(\sqrt{2+\sqrt3}) = 2\sqrt2$$
As seen from the plot, $2\sqrt2 $ at $-\sqrt{2-\sqrt3}$ and $\sqrt{2+\sqrt3}$ are the two local minima, while $-2\sqrt2 $ are the local maxima.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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recurrence initial conditions I'm working on a homework assignment involving recursion and I'm having trouble finding an easy way to determine the initial conditions. Heres the problem:
We want to tile ann×1 strip with tiles of three types: 1×1 tiles that are dark-blue, light-blue,and red; 2×1 green tiles, and 3×1 sky... | Let $a_n$ and $b_n$ the number of $n$-tilings that start with red or green, and blue, respectively. Also, let $c_n$ (your $B_n$) be the total number of $n$-tilings. Then $a_0=a_1=b_0=c_0=1$, $b_1=b_2=2$, $c_1=3$, and, by conditioning on the next tile, we see that
\begin{align}
a_n &= c_{n-1} + c_{n-2} &&\text{for $n ... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\int_{0}^{1}\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} dx = 2\pi\frac{\sqrt{3}}{27}$ The given question is from Complex Variables written by Levinson and Redheffer, Problem 7 in Chapter 4. We want to show that $\int_{0}^{1}\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}\,dx = 2\pi\frac{\sqrt{3}}{27}$. Below is... | Let $f(z)$ be the branch of $\sqrt[\leftroot{-3}\uproot{3}3]{z^2(1 - z)}$ that is real positive at $z$ on $P_1$. Then the value of $f(x)$ for $x>1$ is $\sqrt[\leftroot{-3}\uproot{3}3]{x^2(x-1)}\cdot e^{-\frac{\pi i}{3}}$ and $f(x)=\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1-x)}\cdot e^{-\frac{2\pi i}{3}}$ on $P_2$. Therefor... | {
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"timestamp": "2023-03-29T00:00:00",
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Describe the set of points $z$ in the complex plane that satisfies the following equation. $|2z−i|=4$
I have tried solving it multiple times but I cannot. Here are my steps. I hope somebody can catch my mistakes.
$|2z−i|=4$
$|2(a+bi)−i|=4$
$|(2a+2bi)−i|=4$
$|(2a+2bi)^2+(0\cdot-i)^2|=4$ (because every number is a co... | At the end of step 3) you have
$|2a + 2bi - i|=4$. Try to combine the $i$ terms:
$|2a + (2b-1)i| = 4$. What you have here is the complex number $w = 2z -i = 2a + (2b-1)i$ so that $Re(w) = 2a (= 2Re(z))$ and $Im(w) = 2b -1(= 2Re(z) - 1)$.
Now you can do the formula that $|a+bi| = \sqrt{a^2 + b^2}$ or in this case
$|2a... | {
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"timestamp": "2023-03-29T00:00:00",
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Does the following series (inspired by harmonic things) converge? $$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right) \right) ^ {\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right)} \right] + $$
$$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}}... | This series diverges, but it's somewhat instructive in why - this is an instance where we can just "peel back" all the layers of the expression by applying linear approximations and eventually reduce it to something trivial.
In particular, I assume that you know
$$\lim_{x\rightarrow 0}(1+x)^{1/x} = e$$
and thought to l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552054",
"timestamp": "2023-03-29T00:00:00",
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If $f(x)$ is a polynomial of degree three leaves remainder $1$ when divided by $(x−1)^2$ and leaves remainder $–1$ when divided by $(x+1)^2$
$f(x)$ is a polynomial of degree three which leaves remainder $1$ when divided by $(x−1)^2$ and leaves remainder $–1$ when divided by $(x+1)^2$.
If $f(x)=0$ has roots $\alpha,\be... | Hint
Let $f(x)=1+(x-1)^2(ax+b)=-1+(x+1)^2(cx+d)$
$ax^3+x^2(b-2a)+x(a-2b)+b-1=cx^3+x^2(d+2c)+x(c+2d)+d+1$
Now compare the coefficients of the different exponent of $x$
$\implies c=a$
$b-1=d+1\iff d=b-2$
$b-2a=d+2c\iff b=d+2(c+a)=d+4a$
$a-2b=c+2d\implies b=-d=2-b$
Hope you can take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Help with induction using divisibility. Need some help with this question for induction using divisibility: $8*19^n-2*3^{2n+2}$ is divisible by 10 for any positive integer.
What I have so far is proving the first value and $P_k$ and $P_{k+1}$
$P_1$ = $8*19^1-2*3^{2+2}$=-10; $P_1$ is true.
$P_k$ = $8*19^k-2*3^{2k+2}$= 1... | Let $P(n)$ be the statement that $10 \mid 8 \cdot 19^n - 2 \cdot 3^{2n + 2}$.
If we wish to establish that the statement holds for all nonnegative integers, we have to verify that $P(0)$ holds since that will be our base case.
Since $8 \cdot 19^0 - 2 \cdot 3^2 = 8 - 18 = -10 = 10 \cdot (-1)$, $P(0)$ holds.
If we wish t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On conjectured continued fractions and $e$ Playing around with numbers, I conjectured three incredibly interesting things:
$$9+\cfrac{1}{18+0\times 12\cfrac{1}{18+1\times 12+\cfrac{1}{18+2\times 12+\cfrac{1}{18+3\times 12+\ddots}}}}=\frac{4e^{1/3}-2}{e^{1/3}-1}$$
$$6+\cfrac{1}{9+0\times 6+\cfrac{1}{9+1\times 6+\cfrac{... | First of all, the terms on the right are of the form
$$
\frac{4e^z-2}{e^z-1} = 4 + \frac2{e^z-1}
$$
Let's examine the last of your cases which is for $z=1$. It is known that
$$
e = 1 + \frac2{[1;6\;10\;14\;18\cdots]}
$$
hence
$$
4+\frac2{e-1} = 4+[1;6\;10\;14\;18\cdots] = [5;6\;10\;14\;18\cdots]
$$
which is your 3rd f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Revisiting Ahmed Integral in $(0,\infty)$ A recent post in MSE:
Evaluate $\int_0^{\infty } \frac{\tan ^{-1}\left(a^2+x^2\right)}{\left(x^2+1\right)\sqrt{a^2+x^2}} \, dx$
re-emphasizes that when Ahmed Integral is converted to a two-dimensional integral in $x,y$, then the sameness of domain the $(0,1)$ of $x$ and $y$ mak... | \begin{align}
&\int_{0}^{\infty} \frac{\tan^{-1}\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}~ dx\\
=& \int_{0}^{\infty}\int_0^1 \frac{1}{(1+x^2)(1+y^2(2+x^2))}dy~dx\\
= & \int_{0}^{1}\int_0^\infty \frac{1}{1+y^2}
\bigg( \frac{1}{1+x^2}-\frac{y^2}{1+y^2(2+x^2) }\bigg) dx~dy\\
= & \ \frac\pi2\int_{0}^1 \frac{1}{1+y^2}
\bigg( 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the open intervals where $x(x-6)^3$ is concave upwards and concave downwards. Find the open intervals where $f(x)=x(x-6)^3$ is concave upwards and concave downwards.
Solution: We want to find the concavity of $f(x)=x(x-6)^3$, so we have to take the second derivative, find the critical points (where it equals zero... | Set $z=x-6$;
This amounts to a translation of the coordinate system and does not change the characteristics of the function.
Then $Y=(z+6)z^3=z^4+6z^3;$
$Y'=4z^3+18z^2$;
$Y''=12z^2+36z=12z(z+3)$;
This is a parabola.
$Y'' \gt 0$: Convex;
$12z(z+3)\gt 0$;
1) $z>0$; 2) $z <-3$;
$Y'' <0$: Concave: $-3 <z <0$.
Reset to $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}$
Hence or otherwise show that for all positive integers $n$
$$
n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}
$$
This is related to another part of this question where I proved $\ln \left(\frac{4-t}{t}\right) \geq 2-t$ for $0<t \leq 2$ by int... | Integrating
$\dfrac1{1+t}
=\sum_{k=0}^{m-1}(-1)^k t^k+\dfrac{(-1)^mt^m}{1+t}
$,
we get
$\ln(1+x)
=\sum_{k=0}^{m-1}(-1)^k \dfrac{x^{k+1}}{k+1}+\int_0^x \dfrac{(-1)^mt^m}{1+t}dt
$
for any $m \ge 1$ and $x \ge 0$.
Therefore
$n\ln(1+1/n)
=\sum_{k=0}^{m-1}(-1)^k \dfrac1{(k+1)n^k}+n\int_0^{1/n} \dfrac{(-1)^mt^m}{1+t}dt
$.
Pu... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the indefinite integral $\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$ $$\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$$
My Attempt:
$$I = \int \frac{\sin^2\frac\theta2\cdot\sin\frac\theta2\... | Use $s = \frac{t-1}{t+1} \implies t = \frac{1+s}{1-s}$
$$I = \frac{1}{2}\int \frac{2s}{\sqrt{3-2s^2-s^4}}\:ds = \frac{1}{2}\int\frac{2s}{\sqrt{4-(s^2+1)^2}}\:ds = \frac{1}{2}\sin^{-1}\left(\frac{s^2+1}{2}\right)$$
Then keep reversing the substitutions
$$I = \frac{1}{2}\sin^{-1}\left(\frac{\left(\frac{t-1}{t+1}\right)^2... | {
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Why does $3x^4 + 16x^3 + 20x^2 - 9x - 18$ = $(x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3$? $$3x^4 + 16x^3 + 20x^2 - 9x - 18 $$
When simplified I arrive to:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6}) $$
But the math book wrote:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3 $$
with that extra 3 at the end. The graph calculator s... | Let $ a\ne 0$.
If the equation $$ax^2+bx+c=0$$ has two roots $ x_1 $ and $ x_2 $, then
$$ax^2+bx+c=a(x-x_1)(x-x_2)$$
do not forget the coefficient $ a$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Modular arithmetic and inverse functions The question is shown below:
Suppose $S=\{0,1,2,3,4,5,6,7,8,9,10\}$ and that the function $f:S\rightarrow S$ is given by:
$f(x)=6x^2+3x+8$ (mod 11)
Let $T=\{0,5\}$.
Find $f^{-1}\left(T\right)$.
Alright,
so my initial approach with this question was to find the inverse function,
... | Considering the quadratic $f(x)=6x^2 + 3x + 8$ we have a discriminant (in $\Bbb F_{11}$) of
$$D=3^2 - 4 \cdot 6 \cdot 8 \equiv 4 \pmod{11}$$
which is a square (whether it is $\ge 0$ or not is irrelevant, and we don't use $\sqrt{D}$ in finite fields. We just have the two roots $2, -2 \equiv 9$. So $f(x)$ has two roots ... | {
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Finding the MacLaurin series of $\frac{x+3}{2-x}$ I did:
$$\frac{x+3}{2-x} = \frac{x}{2-x}+\frac{3}{2-x} = x(\frac{1}{2-x})+3(\frac{1}{2-x}) = \\
= \frac{x}{2}(\frac{1}{1-\frac{x}{2}})+\frac{3}{2}(\frac{1}{1-\frac{x}{2}}) = \\
= \frac{x+3}{2}\sum(\frac{x}{2})^n = ...?$$
The answer my professor got was $\frac{(3 + x)}{(... | $\begin{align}\frac{3 + x}{2 - x} &=\frac{-x-3}{x-2}\end{align}$
Use long division to find the quotient and remainder of $\frac{-x-3}{x-2}$ and then rewrite it as the quotient plus the remainder over the denominator:
$\begin{align}\frac{-x-3}{x-2}
&= -1-\frac{5}{x-2}=-1+\frac{5}{2-x}\\
&=-1-\frac{\frac52}{1-\frac x... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Prove $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$ $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$
I need to solve this trigonometric identity for a trianlge.
I'm not allowed to use the formula $a+b+c=s$ where 's' is per... | Start from the RHS and use the following formulas:
Step 1:
$\cot \frac{A}{2} = \frac{\Delta}{(s-b)(s-c)}$ etc. in the numerator
$\cot A = \frac{R}{abc} (b^2 + c^2 - a^2)$ etc. in the denominator
Step 2:
$R = \frac{abc}{4 \Delta}$
Step 3:
$\Delta^2 = s(s-a)(s-b)(s-c)$
You are done!
| {
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"source": "stackexchange",
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Divisibility by $33^{33}$ Let $P_n=(19+92)(19^2+92^2)\cdots(19^n+92^n)$ for each positive integer $n$. Determine , with proof the least positive integer $n$, if it exists , for which $P_n$ is divisible by $33^{33}$.
I have made no progress concrete enough to show .
| Modulo $3$, we have
$$19^k+92^k\equiv 1^k+(-1)^k\pmod 3 $$
and this is $0$ iff $k$ is odd. So for each odd $k$, the factor $19^k+92^k$ adds (at least) one factor $3$; and for even $k$, it doesn't.
This alone gives us
$3^m\mid P_{2m-1}$, or equivalently
$$3^{\lfloor\frac{n+1}2\rfloor}\mid P_n. $$
There may be higher p... | {
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"source": "stackexchange",
"question_score": "2",
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Finding xy+yz+zx such that the given determinant = 0
$x≠y≠z$
$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix} = 0$
Then xy+yz+zx = | A. x+y+z | B. $xyz$ | C. $xyz\over(x+y+z)$ | D. $(x+y+z)\over xyz$ |
Given Ans - D
What I did first was R1->R1-R3 & R2->R2-R3 and throwing (x-z) and (y-z) to the 0... | Hint:
$$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix}=\begin{vmatrix}x&x^3&x^4\\y&y^3&y^4\\z&z^3&z^4\end{vmatrix}-\begin{vmatrix}x&x^3&1\\y&y^3&1\\z&z^3&1\end{vmatrix}=xyz\begin{vmatrix}1&x^2&x^3\\1&y^2&y^3\\1&z^2&z^3\end{vmatrix}-\begin{vmatrix}1&x&x^3\\1&y&y^3\\1&z&z^3\end{vmatrix}$$
Can you end i... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Multinomial expansion sum Given,
$$ \frac{x^2+x+1}{1-x}= a_0+a_1x +a_2x^2+\cdots $$
then, find the sum:
$$ \sum^{50}_{r=1}a_r $$
I knew using multinomial expansion, that
$$ (1-x)^{-1} = 1+ x+x^2+\cdots$$
Hence, $$ \frac{x^2+x+1}{1-x}=1\times(1+x+x^2+\cdots) +x\times(1+x+x^2+\cdots)+x^2\times(1+x+x^2+\cdots) $$
Hence,... | Let's start from here:
$$1\times(1+x+x^2+\cdots) +x\times(1+x+x^2+\cdots)+x^2\times(1+x+x^2+\cdots)$$
Let us rewrite this in a somewhat more friendly way as well by writing out what the sum would be, in effect:
$$\begin{array}{rrrrr}
1 & +x & +x^2 & +x^3 & \cdots \\
& x & +x^2 & +x^3 & \cdots \\
& & x^2 &+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3581419",
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"source": "stackexchange",
"question_score": "3",
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How many subsets are there with exactly 8 elements in a set of 16 elements? Below is a problem I did which I believe I did correctly. I would like somebody to confirm that I did, or tell me where I went wrong.
Problem:
Consider a set with $16$ elements in it. How many subsets does it have with exactly $8$ elements?
Ans... | The number of subsets with exactly k elements ${ n \choose k}$
Your answer:
${ 16 \choose 8}$
R code
> choose(16,8)
[1] 12870
| {
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"source": "stackexchange",
"question_score": "2",
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Solving binomial summation $\sum_{k=0}^{\lfloor{n/2}\rfloor} \binom{n-k}{k} 2^{n-k}$ How can we solve the sum
$$\sum_{k=0}^{\lfloor{n/2}\rfloor} \binom{n-k}{k} 2^{n-k}$$
The problem arose from a counting question, but I am unable to solve this sum.
Edit:
The counting problem was similar to what @Phicar has written, ie,... | Here's the "snake-oil" approach that uses generating functions:
\begin{align}
\sum_{n=0}^\infty \sum_{k=0}^{\lfloor{n/2}\rfloor} \binom{n-k}{k} 2^{n-k} z^n
&= \sum_{k=0}^\infty \sum_{n=2k}^\infty \binom{n-k}{k} 2^{n-k} z^n\\
&= \sum_{k=0}^\infty 2^{-k} \sum_{n=2k}^\infty \binom{n-k}{k} (2z)^n\\
&= \sum_{k=0}^\infty 2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Comparing $2$ infinite continued fractions
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$
I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{... |
Are there any more ways to solve this type of problem?
We can get some loose bounds by truncating the fractions.
$A = 1 +\dfrac{1}{1 + \left[\frac{1}{1 + \frac{1}{\ddots}}\right]}$
The quantity in the square brackets is clearly smaller than 11 (we have 11 divided by something larger than 11); it follows that $A>1+\fr... | {
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"source": "stackexchange",
"question_score": "5",
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Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$ Evaluate of this sum:
$$S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$$
Expand out the sum:
$$S=\prod_{k=1}^{1}\frac{2k}{k+2}+\prod_{k=1}^{2}\frac{2k}{k+3}+\prod_{k=1}^{3}\frac{2k}{k+4}+\cdots$$
$$S=\frac{2}{3}+\frac{2}{4}\cdot\frac{4}{5}+\frac{2}{... | Amazing is to recognize some series.
Consider
$$S=\sum_{j=1}^{\infty} \frac{j!\,(j+1)!\, 2^j}{(2j+1)!}x^{2j}$$ Let $x=y \sqrt 2$ to make
$$S=\sum_{j=1}^{\infty}\frac{4^j\, j!\, (j+1)! }{(2 j+1)!}y^{2 j}=-\frac{y^2}{y^2-1}+\frac{1}{2 \left(y^2-1\right)}-\frac{\sin ^{-1}(y)}{2
\sqrt{1-y^2} \left(y^2-1\right) y}$$
Mak... | {
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"source": "stackexchange",
"question_score": "1",
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Evaluate: $S=\sum_{n=1}^{\infty}\frac{(2n)!}{(2n+1)!!^2}$ How to evaluate this sum?
$$S=\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-1/2)}{(j+1/2)^2}$$
$$\frac{j(j-1/2)}{(j+1/2)^2}=\frac{2j(2j-1)}{(2j+1)^2}$$
$$S=\prod_{j=1}^{1}\frac{2j(2j-1)}{(2j+1)^2}+\prod_{j=1}^{2}\frac{2j(2j-1)}{(2j+1)^2}+\prod_{j=1}^{3}\frac{2j(2j... | I suppose that there are several ways to arrive to the result.
I am not very happy with the following
$$S(x)=\sum_{n=1}^{\infty}\frac{(2n)!}{\big[(2n+1)!!\big]^2}x^{n-1}=\frac{2}{9} \, _3F_2\left(1,\frac{3}{2},2;\frac{5}{2},\frac{5}{2};x\right)$$
$$S(1)=2 C-1$$ where $C$ is Catalan constant.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Exact expression of a trigonometric integral Let $a>2$ be a real number and consider the following integral
$$
I(a)=\int_0^\pi\int_0^\pi \frac{\sin^2(x)\sin^2(y)}{a+\cos(x)+\cos(y)} \mathrm{d}x\,\mathrm{d}y
$$
My question. Does there exist a closed-form expression of $I(a)$?
Some comments. Since $a-2<a+\cos(x)+\cos(y... | Partial answer:
As the integrand is an even function, write $$I(a)=\frac14\int_{-\pi}^\pi\int_{-\pi}^\pi\frac{\sin^2x\sin^2y}{a+\cos x+\cos y}\,dx\,dy$$ and let $w=e^{ix}$. Then for $C_1=\{w:|w|=1\}$, we have \begin{align}\int_{-\pi}^\pi\frac{\sin^2x}{a+\cos x+\cos y}\,dx&=\oint_{C_1}\frac{\frac{(w-1/w)^2}{-4}}{b+\frac... | {
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"source": "stackexchange",
"question_score": "10",
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Integration through partial fractions with complex roots In integrating the following: $\frac{1}{(x^2+2x+3)^2}$ I am trying to use partial fraction decomposition as follows:
$\frac{1}{(x^2+2x+3)^2} = \frac{Ax + B}{x^2+2x+3} + \frac{Cx + D}{(x^2+2x+3)^2}$
Which gives me:
$1 = (Ax + B)(x^2 +2x +3) + (Bx + C)$
And that ge... | You don't need he partial fractions decomposition: as already observed, the substitution $u=x+1$ results in having to compute the integral $\;\int\frac{\mathrm du}{(u^2+2)^2}$.
Now this integral pertains to a well-known type:
$$I_n=\int\frac{\mathrm d x}{(x^2+a^2)^n},$$
which is easily computed though a recursion formu... | {
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"source": "stackexchange",
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Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$...
Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$. Then $\sum_{n=1}^{10} f(n)$ is
From the first equation
$$a(x^2+y^2+2xy)+b(x+y)+c=ax^2+bx+c+ay^2+by+c+... | You have that
$$f(x+y)=f(x)+f(y)+xy \tag{1}\label{eq1A}$$
for all $x,y\in R$. Your simplification of \eqref{eq1A} gives
$$2axy=c+xy \implies (2a-1)xy = c \tag{2}\label{eq2A}$$
Since this is true for every real $x$ and $y$, this means that $2a - 1 = 0 \implies a = \frac{1}{2}$ and $c = 0$. This thus gives that $b = 3 - ... | {
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"source": "stackexchange",
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Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers
I am trying to prove
$$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$
for all positive integers.
My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}... | An elementary way. Let $x=n^{1/3}\geq 1$ then it suffices to show that
$$(x^3+1)^{2/3} -x^2 <\frac{2}{3x}$$
that is
$$(x^3+1)^2<\left(x^2+\frac{2}{3x}\right)^3$$
or
$$x^6+2x^3+1<x^6+2x^3+\frac{4}{3}+\frac{8}{27x^3}$$
which trivially holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Question on sum of digits of squares Let $D$ be the function define as $D(b,n)$ be the sum of the base-$b$ digits of $n$.
Example: $D(2,7)=3$ means $7=(111)_2\implies D(2,7)=1+1+1=3$
Define $S_m(a)=1^m+2^m+3^m+...+a^m$ where $a,m\in\mathbb{Z}_+$
Can it be shown that
(1)$$D(a,S_2(a))\le 2(a-1)?$$
(2) $$D(a,S_2(a))< a\i... | We can continue your casework on $ n \pmod{6}$.
Here's a sketch of it. Fill in the rest of the details yourself.
If $ n \equiv 0 \pmod{6}$, then $ \frac{ 2n^3 + 3n^2 + n } { 6} = \frac{2n}{6} \times n^2 + \frac{3n}{6} \times n + \frac{n}{6} \times 1 $,
so $D_2 = \frac{2n}{6} + \frac{3n}{6} + \frac{n}{6} = n $.
If ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Roots of the equation $(x – 1)(x – 2)(x – 3) = 24$ The equation $(x – 1)(x – 2)(x – 3) = 24$ has the real root equal to 'a' and the complex roots 'b' and 'c'. Then find the value of $\frac{bc}{a}$
My approach is as follow $y=f(x)=(x – 1)(x – 2)(x – 3) - 24=0$
$y'=3x^2-12x+11=0$
Solving we get $x=2\pm\sqrt{\frac{1}{3}}$... | Noticing that $24=2\cdot3\cdot4$, $5$ must be a root. Then after long division by $x-5$, $x^2-x+6=0$.
Using Vieta,
$$\frac{bc}a=\frac65.$$
Now for a "general" solution, you first deplete the cubic by setting $z:=x-2$ and the equation is
$$(z+1)z(z-1)=z^3-z=24.$$
Then with $z:=\dfrac2{\sqrt 3}\cosh u$,
$$4\cosh^3u-3\c... | {
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Proving $\def\n#1{\left(\frac12+\sum\limits_{k=1}^n{#1}^{k^2}\right)}\n{a}\n{b}\ge{\n{(ab)}}^2$
Let $n$ be an even postive integer, and $a,b\in (-1,1)$, $a+b\ge0$. Show that
$$\left(\frac12+\sum_{k=1}^na^{k^2}\right)\left(\frac12+\sum_{k=1}^nb^{k^2}\right)\ge\left(\frac12+\sum_{k=1}^n(ab)^{k^2}\right)^2\tag{1}$$
It... | Partial answer
WLOG, assume that $a \ge b$. The inequality is written as
$$\frac{1}{2}(\sum a^{k^2} + \sum b^{k^2}) + \sum a^{k^2} \sum b^{k^2}
\ge \sum (ab)^{k^2} + (\sum (ab)^{k^2})^2. \tag{1}$$
If $b > 0$, the proof is easy. Indeed, since $0\le ab \le 1$, we have (by AM-GM)
$$\sum a^{k^2} + \sum b^{k^2} \ge \sum 2(\... | {
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"source": "stackexchange",
"question_score": "11",
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If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find m+n. If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find $m+n$.
I did and I got
$$(x+2)^3(x-2)^{3+m}(x+1)^n(x^2-x+1)^n(x-3)^m=0$$, so I find $3+3+m+n+2n+m=18\implies 2m+3n=12$, the answer is m+n=5. What I have to do now?
| $2×3+3×n+2×m = 18$
Find $m+n$
$3*n+2*m = 12$
( Diophantine equation )
Offcourse $m,n € Z$ were $Z → (0,1,2,3,4,5,............)$
If we construct a table,
Put $n=0$, $m=6$
Put $n=1$, $m$ ≠ $Z$
Put $n=2$, $m=3$
Put $n=3$, $m$ ≠ $Z$
Put $n=4$, $m=0$
So there are only there values that works here $n=0$, $m=6$ and $n=2$, $m... | {
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"question_score": "1",
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Show that there do not exist any distinct natural numbers a,b,c,d such that
Show that there do not exist any distinct natural numbers a,b,c,d such that $a^3+b^3=c^3+d^3$ and $a+b=c+d$.
| Suppose $a+b=c+d$ and $a^3+b^3=c^3+d^3$.
$$a+b=c+d$$
$$(a+b)^3=(c+d)^3$$
$$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$
$$3ab(a+b)=3cd(c+d)$$
$$ab=cd$$
Let $a+b=c+d=m$ and $ab=cd=n$
a and b are the roots of the quadratic equation
$$x^2-mx+n=0$$
by Vieta's relations because a+b=m and ab=n.
But c and d are also roots of the equat... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is the integral $ \int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1} $ converges or diverges. Determine if the following integral diverges/converges, if it converges, is it absolutely or conditionally converges.
$$
\int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1}
$$
What i tried:
In that interval we know that:
$$
0 < x
$$
Therefore ... | It diverges.
Multiplying by the conjugate:
$\frac{1}{\sqrt{x^2-x+1}-1} = \frac{\sqrt{x^2-x+1}+1}{x(x-1)}$.
A partial fraction decomposition yields:
$\frac{1}{x(x-1)}=\frac{1}{x}-\frac{1}{x-1}$ and therefore
$\frac{\sqrt{x^2-x+1}+1}{x(x-1)}=\frac{\sqrt{x^2-x+1}+1}{x}-\frac{\sqrt{x^2-x+1}+1}{x-1}$.
The right-hand side eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding number of integral solutions to an equation.
Find the number of integral solutions to:
$$x^2+y^2-6x-8y=0.$$
My attempt:
The equation can be rewritten as:
$$x^2+y^2-6x-8y+9+16=25,$$
basically adding 25 to both sides, or equivalently,
$$(x-3)^2+(y-4)^2=25.$$
This is a Pythagorean triplet. The only triplet of ... | You are correct; there are solutions corresponding to $0^2+(\pm5)^2=25$, and there are $12$ integral solutions in total. Another way to see this, is to note that
$$(x-3)^2+(y-4)^2=25,$$
forces $|y-4|\leq5$, leaving $11$ possible values for $y$ and yielding $6$ different quadratics in $x$. A quick check then shows that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Establish a lower bound for the generalized alternating harmonic series I'm given the following series:$$\sum_{n=1}^{\infty}{(-1)^{n-1}\over n^p}$$
I need to show that the sum is greater than $1/2$ for every $p > 0$. For $p \ge1$ this is obvious, as it follows by grouping terms $2$ by $2$, and the first sum is greater ... | let $S_p(N)=\sum_{k=1}^{N}{k^{-p}}$; for notational simplicity we fix for now $0<p<1$ and let $S_p(N)=S(N)$.
We note that $\eta(p)=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n^p} > S(2N)-2^{1-p}S(N)$ for any $N \ge 1$ as the remainder is an alternating sum with decreasing terms that starts at the positive $\frac{1}{(2N+1)^p}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove that the series $\sum\limits_{n=1}^{\infty} \log x_n$ is convergent?
Consider the sequence $\{x_n \}$ defined by $$x_n = e \left (\frac {n} {n+1} \right )^{n + \frac 1 2},\ n \geq 1.$$
Prove that the series $\sum\limits_{n=1}^{\infty} \log x_n$ is convergent.
My attempt: I find that \begin{align*} \sum\... | Notice, that:
$$\sum_{i=1}^n \log x_i=
\log \prod_{i=1}^n x_i =
\log \left( e\left(\frac{1}{2}\right)^{\frac{3}{2}} e\left(\frac{2}{3}\right)^{\frac{5}{2}} \dots e\left(\frac{n}{n+1}\right)^{\frac{2n+1}{2}} \right)=
\log \left( e^n \frac{n!}{(n+1)^{n+\frac{1}{2}}}\right)$$
Using Stirling's approximation:
$$\lim_{n \to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why doesnt this trignometric substituition work? $$\frac 1 5\int\frac{x+1+5}{(x+1)^2+5}\,\mathrm dx$$
where i substitute $x+1 = \sqrt{5}\tan(\theta)$
After Substituition :
$$\frac15\int\frac{\sqrt5\tan(\theta)+5}{\tan^2(\theta)+1}\sqrt5\sec^2(\theta)\,\mathrm d\theta$$
but if i do $u = x +1$
and then further substitute... | Where is your trig substitution failing?
$\int \frac{(x+1)+5}{(x+1)^2 + 5} \ dx$
$x+1 = \sqrt 5 \tan \theta\\
dx = \sqrt 5 \sec^2 \theta$
$\int \frac{(\sqrt 5 \tan \theta +5)(\sqrt 5 \sec^2 \theta)}{5\sec^2 \theta} \ d\theta\\
\int \frac{\tan \theta}{\sqrt 5} + \sqrt 5 \ d\theta$
That looks relatively simple from here.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving modulus equations with multiple unknowns I have the following equations:
$$8\equiv-(3a+7b)\mod11$$
and
$$9\equiv-(5a+4b)\mod11$$
How do I solve them?
I've tried to subtract one equation from the other, but I end up with the wrong answer.
It is supposed to be:
$a=2$ and $b=9$.
| By rearranging the first congruence equation $8 \equiv -3a -7b \pmod{11}$, we have $4b \equiv 8 + 3a \pmod{11}$. Since $\gcd(4, 11) = 1$, there is an inverse of $4$, which is $3$. We thus have
\begin{equation}
b \equiv 24 + 9a \equiv 2 - 2a \pmod{11}
\end{equation}
Plugging this into the other congruence equation $9 \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the equation of a plane that contains a given line and is 2 units away from a given point. The given line has a vector of (2,-1,4) and contains the points A(-1,1,-1) and B(1,0,3). The given point is P(2,3,4). The problem asks you to find a plane such that its distance to P is 2 units and it contains line AB, and ... | Let $T$ be a point of the sphere $s$ centered at $P$ withe radius $2$ such that the distance from the plane $p$ defined by $T$, $A$ and $B$ to $P$ is equal to $2$. Then $p$ is tangent to $s$ and therefore the vectors $TA$ and $TB$ are orthogonal to $TP$. So, you can find the point $T$ by solving the system$$\left\{\beg... | {
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"url": "https://math.stackexchange.com/questions/3620094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When can a sum of 3 identical squares be written as a sum of 3 non-identical squares? Consider $3n^2$ for some positive integer $n$. When do there exist positive integers $n_1,n_2,n_3$, not all equal to $n$, such that
$$
3n^2 = n_1^2 + n_2^2 + n_3^2
$$
An example is $27$:
$$
27 = 3^2 + 3^2 + 3^2 = 1^2 + 1^2 + 5^2
$$
| I have one hint:
$(n-\alpha)^2 + (n- \alpha)^2 + (n+\alpha)^2 = n^2 -2n\alpha + \alpha^2 + n^2 -2n\alpha + \alpha^2 + n^2 +2n\alpha + \alpha^2 = \\ n^2 + n^2 + n^2 + 3\alpha^2 - 2n\alpha.$
So, if we want for the sum to have $3n^2$, then it is mandatory that we have $3\alpha^2-2n\alpha =0 \iff 3\alpha = 2n$, i.e. $\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$ $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$
Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+... | You can only use $t$ going to $\sqrt[6]{2}$. This is because you have $t^3 = \sqrt{x} \gt 0$ for $x \to 2$, but $t = -\sqrt[6]{2}$ means $t^3 \lt 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$. Use a Taylor series expansion to find $f^{(80)}(27)$.
I tried the following:
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
%
f''(x)
&= \frac{1}{3}(x... | We first expand $f(x)$ at $x=27$. Because it already has a factor $(x-27)$,we just need to expand $(x+3)^{1/3}:=g(x)$.
Do the Taylor expansion:
$$
g(x) = g(27)+g'(27)(x-27)+\cdots+\frac{1}{79!}g^{(79)}(27)(x-27)^{79}+o((x-27)^{79})
$$
Then the Taylor expansion of $f(x)$ should be (according to the uniqueness of Taylor... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Calculating determinant of a symmetric matrix where the $k$th row is given by $[a_{k-1},a_k,...,a_0,a_1,...,a_{n-(k-1)}]$
For $j = 0,...,n$ set $a_{j} = a_{0} + jd$, where $a_{0}, d$ are fixed real numbers. Calculate the determinant of the $(n+1)\times (n+1)$ matrix
$$A = \begin{pmatrix}
a_{0} & a_{1} & a_{2} ... | Subtracting from each row the one above it, we shall obtain
$$ \begin{pmatrix}
a_{0} & a_{1} & a_{2} & ... & a_{n}\\
d & -d & -d & ... & -d\\
d & d & -d & .... & -d\\
\ldots & \ldots & \ldots & \ldots & \ldots\\
d & d & d & ... & -d
\end{pmatrix}$$
Now, subtracting from each column the one be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Using that $1 + z + z^{2} + ... + z^{n} = \frac{1-z^{n+1}}{1-z}$ and taking the real parts, prove that: $$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = \frac12+\frac{\sin[(n + \frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})} $$
for $0 < \theta < 2\pi$.
Alright. What I have done is this, using the De Moivre's Formu... | $$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = \frac{\sin[(n + \frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})} $$
Using De moivre's theorem you arrived that
$$ \operatorname{Re}(1 + e^{i\theta} + e^{2i\theta} + ... e^{ni\theta}) = \operatorname{Re} \biggl(\frac{1 - e^{(n+1)i\theta}}{1 - e^{i\theta}}\biggr)$$
Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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(Different) derivatives of $f(x) = \arcsin\left(\left(5 x + 12 \sqrt{1-x^2}\right)/13\right)$ via two different substitutions? We have been given a function to differentiate:
$$f(x) = \arcsin \left(\frac{5x + 12\sqrt{1-x^2}}{13}\right)$$
My teacher told me the method to substitute $ x= \sin\vartheta$ which would sim... | HINT.-$(5,12,13)$ is a Pythagorean triple so what is the relation between $\arctan\left(\frac{12}{5}\right)$ and $\arctan\left(\frac{5}{12}\right)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3636162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Proof existence of $\int_0^\infty \frac{1}{\sqrt{x}} \frac{1}{1+x^2} \,\mathrm dx$ - is this proof correct? I have to show that the integral
$$ \int_0^\infty \frac{1}{\sqrt{x}} \frac{1}{1+x^2} \,\mathrm dx $$
exists.
My approach: Let $b > 1$, then the integral
$$ \int_1^b \frac{1}{\sqrt x} \frac{1}{1+x^2} \,\mathrm dx ... | Looks fine, here is another approach :
\begin{aligned}&\bullet \ f:x\mapsto\frac{1}{\sqrt{x}\left(1+x\right)}\textrm{ is continuous on }\left(0,+\infty\right)\cdot\\ &\bullet \ \lim_{x\to 0}{\frac{x^{\frac{3}{4}}}{\sqrt{x}\left(1+x\right)}}=0\textrm{, and thus }f\left(x\right)=\underset{\overset{x\to 0}{}}{\mathcal{o}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3636826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim_{n\to\infty}a_n=\frac{\sum_{i=1}^k2ia_i}{k(k+1)}$
Given that a sequence $(a_n)$ satisfies $a_{n+k} = \dfrac{a_n + a_{n+1} + \cdots + a_{n+k-1}}{k}$ for $n\geq 1,$ where $k\in\mathbb{N},$ prove that $\lim\limits_{n\to\infty} a_n = \dfrac{2a_1}{k(k+1)}+\dfrac{4a_2}{k(k+1)}+\cdots +\dfrac{(2k)a_k}{k(k+1)... | Based on the comment by Paramanand Singh, we see that we have the relation
$kx_{n+k} = x_{n+k-1} + x_{n+k-2} + \cdots + x_n.$
Hence $kx_{n+k}+(k-1)x_{n+k-1} +\cdots + 2x_{n+2} +x_{n+1} = kx_{n+k-1}+(k-1)x_{n+k-2}+\cdots + 2x_{n+1}+x_n.$
Repeating this process $n-1$ more times, we see that $kx_{n+k}+(k-1)x_{n+k-1}+\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3638306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$
for $a,b \in \rm{I\!R}_{+}$. Applying ... | Let us denote by $\pi - x$ the angle between the sides $a$ and $b$ from a triangle with third side $c(x)$, the application of the cosine's law results into
\begin{align*}
c^{2}(x) = a^{2} + b^{2} - 2ab\cos(\pi - x)
\end{align*}
Thus the given expression can be rewrriten as
\begin{align*}
f(x) = \frac{ab\sin(x)}{2\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Determine the probability that the chosen coin is coin $2.$
Coin $1$ is a fair coin and coin $2$ is an unfair coin such that the probability of getting heads is $0.6.$ One of the coins is chosen at random and flipped repeatedly until the first head is obtained. Suppose that the first head is observed in the fifth flip... | Yes, your approach and your result are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $ Question -
Let $a, b, c$ be positive real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ . Prove that
$$
\frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1
$... | It is enough to show that $\sqrt a + \sqrt b + \sqrt c = 3 \implies (ab)^{2/3} + (bc)^{2/3} + (ca)^{2/3} \leqslant 3$. For ease, let us replace $a, b, c$ with $x^6, y^6, z^6$, so we need to show for positives, $x^3+y^3+z^3=3 \implies (xy)^4+(yz)^4+(zx)^4\leqslant 3$. This one is in fact a known old chestnut.
We note ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof $\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt2} + \frac{\sqrt{2 + \sqrt2}}{\sqrt2 \sqrt{2 - \sqrt{2 - \sqrt2}}} =\sqrt{2-\sqrt{2+\sqrt2}}$ I want to prove the following equation
$$\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt{2}} + \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2} \sqrt{2 - \sqrt{2 - \sqrt{2}}}}$$
I know this m... | Note that the second term of the LHS can be expressed as
$$\begin{align}
\frac{\sqrt{2 + \sqrt{2} }}{\sqrt{2} \sqrt{2 - \sqrt{2 - \sqrt{2}}}}
=\frac{\sqrt{2 + \sqrt{2 - \sqrt{2}}}}
{\sqrt{2}}
\end{align}$$
which can be verified by cross multiply. Then,
$$LHS = \sqrt{ \frac{2 + \sqrt{2 - \sqrt{2}}} {2} }
- \sqrt{ \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha... | Given three roots $\alpha, \beta, \gamma$ of a polynomial, it can generally be written as
$$(x-\alpha)(x-\beta)(x-\gamma)=0$$
Note how each of the roots contribute in making the equality work. On expanding the braces,
$$(x-\alpha)(x-\beta)(x-\gamma)=0$$
$$[(x^2-x(\beta)-x(\alpha)+(\alpha\beta)](x-\gamma)=0$$
$$(x^3-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\righ... | Let $a+b+c=x(a^2+b^2+c^2).$
Thus, $$\sum_{cyc}(a-xa^2)^2\geq0=\left(\sum_{cyc}(a-xa^2)\right)^2$$ or
$$\sum_{cyc}(a^2-2xa^3+x^2a^4)\geq\sum_{cyc}(a^2+2ab)-2x\sum_{cyc}(a^3+a^2b+a^2c)+x^2\sum_{cyc}(a^4+2a^2b^2)$$ or
$$\sum_{cyc}ab-x\sum_{cyc}(a^2b+a^2c)+x^2\sum_{cyc}a^2b^2\leq0.$$
Now, let $a^2b^2+a^2c^2+b^2c^2=A$, $\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3649363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Proving two binomial identities I would like to show that
\begin{align}
&\sum_{j=n-k}^n\binom nj(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&\qquad=\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&\qquad=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k-1}k(-x)^k
\end{al... | The first Identity follows by induction. When $k=0$
\begin{eqnarray*}
\binom{n}{n} \frac{x^{n-1}}{1-x} (n-nx) = \binom{n}{n} nx^{n-1}.
\end{eqnarray*}
Now assume the sum for $k$ and add the $(k+1)^{th}$ term ($j=n-k-1$)
\begin{eqnarray*}
\binom{n}{n-k}(n-k)(1-x)^kx^{n-k-1}+\binom{n}{n-k-1}(1-x)^{k}x^{n-k-2}(n-k-1-nx) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3650275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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n-th power of a matrix using the division of polynomials. Consider the matrix
$$
A=\begin{pmatrix}
0 & 0 & 0\\
-2 & 1 & -1\\
2 & 0 & 2
\end{pmatrix}
$$
*
*Calculate $A^3-3A^2+2A$.
*What is the remainder of the division of the polynomial $X^n$ by the polynomial $X^3-3X^2+2X$.
*Calculate $A^n$ for every natural num... | You could use induction.
For $n=1$, your statement is true.
Assuming
$$
A^n=\begin{pmatrix}
0 & 0 & 0\\
-2^n & 1 & 1-2^n\\
2^n & 0 & 2^n
\end{pmatrix}
$$
Then
$$
A^{n+1} = AA^n =
\begin{pmatrix}
0 & 0 & 0\\
-2 & 1 & -1\\
2 & 0 & 2
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0\\
-2^n & 1 & 1-2^n\\
2^n & 0 & 2^n
\end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Cauchy-Schwarz inequality problem with four variables Let a,b,c,d, are positive numbers. How can I prove
$a/(b+c) + b/(c+d) + c/(d+a) + d/(a+b) \ge 2$
I am unable to find a transformation which will lead me to the result. Any help would be very much appreciated.
I was proceeding as follows:
$E = a/(b+c) + b/(c+d) + c/... | Applying Titu's lemma, it suffices to show that (fill in the slight gap)
$$ (a+b+c+d)^2 \geq 2(ab+ac+bc+bd+cd+ac+da+bd ).$$
This is obviously true by expansion, since it becomes
$$ ( a -c)^2 + (b-d) ^2 \geq 0 .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving an ode of first order I want to solve this first order ode:
$$ x'(1+t^2) \sin(x)-2t \cos(x)=0 $$
$x(1)= \frac{ \pi}{3} $
I want to use the separation method, so :
$ x' (1+t^2) \sin(x)= 2t \cos(x) $
$\leftrightarrow x'= \frac{2t \cos (x)}{(1+t^2) \sin(x) }$
$ \frac{dx}{dy} = \frac{2t \cos (x)}{(1+t^2) \sin(x... | From your last line you get $$\log\left(|\cos x |\cdot(t^2+1)\right)= \log c \hspace{1 cm} \text{(for some c)} \\ \implies \cos x\cdot(t^2+1)= \pm c =C \\ \implies \cos x = \frac{C}{t^2+1}$$
Using $x(1)=\frac{\pi}{3}$, $$\cos x =\frac{1}{t^2+1} \\ x= \cos^{-1}\left( \frac{1}{t^2+1} \right) + 2n\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability problem on umbrellas
Two absent-minded roommates, mathematicians, forget their umbrellas in some way or another. $A$ always takes his umbrella when he goes out, while $B$ forgets to take his umbrella with probability $1/2$. Each of them forgets his umbrella at a shop with probability $1/4$. After visiting ... | You want the probability that only one of the two remembers their umbrella at every point.
At the end of the trip, $A$ will have an umbrella if it is remembered at each of the three stores. This has a probability of $(3/4)^3$
At the end of the trip, $B$ will have an umbrella if it is remembered at home and each of the... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove this algebraic version of the sine law? How to solve the following problem from Hall and Knight's Higher Algebra?
Suppose that
\begin{align}
a&=zb+yc,\tag{1}\\
b&=xc+za,\tag{2}\\
c&=ya+xb.\tag{3}
\end{align}
Prove that
$$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}.\tag{4}$$
(I suppose tha... | Let $a=0$.
Thus, $$xc=b$$ and $$xb=c,$$ which gives $$x^2bc=bc$$ or $$(x^2-1)bc=0$$ and since $x^2\neq1,$ we obtain $bc=0$ and from here $$a=b=c=0,$$ which gives that our statement is true.
Let $abc\neq0$.
Thus, $$\frac{zb}{a}+\frac{yc}{a}=1$$ and $$\frac{xc}{b}+\frac{za}{b}=1,$$ which gives
$$z^2+\frac{xyc^2}{ab}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 3
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Show that the derivative of a series $\sum_{n=1}^\infty \frac{-2x}{(2+x^2)^{n+1}}$ converges uniformly Consider the derivative series
$$
D = \sum_{n=1}^\infty \frac{-2x}{(2+x^2)^{n+1}}
$$
of
$$
S = \sum_{n=1}^\infty \frac{1}{n} \frac{1}{(2+x^2)^n}
$$
I have shown that $S$ converges by the use of Weiterstrass' M-test by... | If $f(x)=\frac{-2x}{(2+x^2)^{n+1}}$, then$$f'(x)=2 \left(x^2+2\right)^{-n-2} \left((2 n+1) x^2-2\right).$$So, the maximum of $|f|$ is attained when $x=\pm\sqrt{\frac2{2n+1}}$ and$$\left|f\left(\pm\sqrt{\frac2{2n+1}}\right)\right|=2 \sqrt{2} \sqrt{\frac{1}{2 n+1}} \left(\frac{2}{2 n+1}+2\right)^{-n-1}.$$Since$$\lim_{n\t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$ From Mr. Michael Rozenberg solution:
For $a,b,c>0$$,$ prove that$:$
$$(a+b+c)^2\prod_{cyc}(a+b)\geq4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab,$$
I found two SOS proof:
1) $$\text{LHS-RHS}={\frac { \left( a-b \right) ^{2}\cdot \text{M}+ab \left( {a}^{... | I think, the shortest way here it's $uvw$. See here: https://artofproblemsolving.com/community/c6h278791
Indeed, this inequality is a linear inequality of $w^3$ and for the proof it's enough to consider two cases:
*
*$c=0$;
*$b=c$, which very easy to make.
Due to the Nguyenhuyen_AG's post your inequality we can re... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of finite series using partial fraction I'm quite stuck with the following problem. I have seen on this forum that there is already an answer for the infinite sum to the problem but I can't seem to find how to find the sum for a finite value.
The first part of the questions asks to transform the given series using ... | For instance, with $n = 10$ we have
$$
\sum_{k=1}^n \frac1k - \frac 1{k+2} = \\
\left(1 - \frac 13 \right) +
\left(\frac12 - \frac 14 \right) +
\left(\frac13 - \frac 15 \right) + \cdots +
\left(\frac 18 - \frac 1{10} \right) +
\left(\frac19 - \frac 1{11} \right) +
\left(\frac1{10} - \frac 1{12} \right) =\\
1 + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3678339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cubic Discriminant Uses The discriminant for the cubic equation $ax^3+bx^2+cx+d=0$ is
$Δ\:=b^2c^2−4ac^3−4b^3d−27a^2d^2+18abcd$
And I am aware that you can determine the number of roots a cubic has using method shown below -
$Δ\:>0$: the equation has three distinct real roots
$Δ\:=0$: the equation has a repeated root... | When a monic cubic has square discriminant but no rational roots, what we expect is real roots that can be written as (doubled) cosines, or sums of them.
$$ x^3 + x^2 - 2x - 1 $$
has $$ 2 \cos \frac{2 \pi}{7} \; , \; \; 2 \cos \frac{4 \pi}{7} \; , \; \; 2 \cos \frac{8 \pi}{7} \; , \; \; $$
more in a minute
$$ x^3 ... | {
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"source": "stackexchange",
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Finding volume of solid in one quadrant - divide total volume by 4? 8? 2? I want to find the volume of the solid produced by revolving the region enclosed by $y=4x$ and $y=x^3$ in the first quadrant. The wording about the first quadrant confuses me but here's my work so far:
I know the volume unrestrained by quadrant i... | The first quadrant is defined by $\{(x,y)\in \mathbb R^2 | x,y\ge 0\}$. The $\ge$ instead of $>$ is debatable, but is of no consequence for this problem.
Thus your intersection points are $(0,0)$ and $(2,2)$. So, the volume integral would be $\int_0^2$.
To answer your other question, it would be twice the volume of the... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Olympiad Inequality Question I am new to the Olympiad-style questions and I hope someone could correct my proof for this question as I do not have the answer for it. Please leave some constructive criticism if possible so I could improve. Thanks in advance!
Let $a,b,c$ be positive real numbers. Prove that:
$$a^3 +b^... | Hint: $a^3+a^3+b^3 \ge 3a^2b$ by AM-GM. Do it $2$ more times with the pairs $(b,c)$ and $(c,a)$. Then add up the $3$ inequalities, and divide both sides by $3$ to complete the proof.
| {
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Double integral over $x^2+y^2 \le 1$ I am trying to calculate the double integral
$\displaystyle \iint_{x^2+y^2\leq 1} (\sin x+y+3)\,dA$
Here are my attempts so far:
1) I used polar coordinates
$ x= r \sin(\theta)$
$y= r \cos (\theta)$
where $\theta \in [0,2 \pi]$ and $r \in [0,1]$ which gives
$\displaystyle \int_0^... | The integral of a sum is equal to the sum of the integrals of the summands (assuming all exist, and finitely many summands).
$$\iint_{x^2+y^2=1}\sin(x)+y+3\,dA = \iint_{x^2+y^2=1}\sin(x)\,dA + \iint_{x^2+y^2=1}y\,dA + \iint_{x^2+y^2=1}3\,dA$$
Hint: For any odd function $f$ (a function such that $f(-x) = -f(x)$ for all ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to solve this problem with $P(Q(n))\equiv n\pmod p$ for all integers $n$, the degrees of $P$ and $Q$ are equal. $p$ is a prime. Let $K_p$ be the set of all polynomials with coefficients from the set $\{0,1,\dots ,p-1\}$ and degree less than $p$. Assume that for all pairs of polynomials $P,Q\in K_p$ such that $P(Q(n... | To complete Peter's answer, which shows that $P=x^a$ and $Q=x^b$ works for some $(a,b)$ unless $p\in\{2,3,5,7,13\}$, here is a full explication in these cases:
If $p=2$ then, since $P$ and $Q$ can clearly not be constant, they must be linear, and so $p=2$ works.
If $p=3$, then $\deg P(Q(x))=(\deg P)(\deg Q)$ must be ei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine the convergence of the series $\sum_{n=1} ^{\infty} \frac{5^{n}-2^{n}}{7^{n}-6^{n}}$ Does
$$\sum_{n=1} ^{\infty} \frac{5^{n}-2^{n}}{7^{n}-6^{n}}$$
converge?
I tried the ratio test but I failed.
| Ratio test:
$$
\frac{\displaystyle\frac{5^{n+1}-2^{n+1}}{7^{n+1}-6^{n+1}}}{\displaystyle\frac{5^{n}-2^{n}}{7^{n}-6^{n}}}
=\frac{7^{n}-6^{n}}{{7^{n+1}-6^{n+1}}}\,\frac{5^{n+1}-2^{n+1}}{5^{n}-2^{n}}
=\frac{1-(6/7)^n}{7-(6/n)^{n+1}}\,\frac{5-(2/5)^{n+1}}{1-(2/5)^n}\to\frac57<1,
$$
so the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ahmed integral revisited $\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx$ How to prove
$$\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} dx=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{... | Let
$$ I(a)=\int_0^1 \frac{\arctan(a\sqrt{x^2+4})}{(x^2+2)\sqrt{x^2+4}}dx,I(0)=0,I=I(1)$$
\begin{align}
I'(a)&=\int_0^1 \frac{1}{(x^2+4)[1+a^2(x^2+4)]}dx \\
&=\int_0^1 \frac{1}{x^2+4}dx-a^2\int_0^1 \frac{1}{1+a^2(x^2+4)}dx \\
&=\frac{1}{2} \arctan \frac{1}{2}-\frac{a}{\sqrt{1+4a^2}}\arctan \frac{a}{\sqrt{1+4a^2}} \\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3693547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Finding the Domain? The set ${{x:\left|x+\frac{1}{x}\right|>6}}$ equals the set
My Approach
We know,IF $|f(x)| \geqslant a, \quad ;a>0$
then $f(x) \geqslant a \quad \cup \quad f(x) \leqslant-a$
$\frac{x^{2}+1}{x}> 6$ & $\frac{x^{2}+1}{x}<-6$
$\frac{x^{2}-6 x+1}{x}>0..........(1)$
$\frac{x^{2}+6 x+1}{x}<0$............(... | We have $x\neq0$ and $$x^2+1>6|x|$$ or
$$|x|^2-6|x|+1>0,$$ which is
$$|x|>3+2\sqrt2$$ or
$$|x|<3-2\sqrt2,$$ which gives the answer:
$$(-\infty,-3-2\sqrt2)\cup(-3+2\sqrt2,0)\cup(0,3-2\sqrt2)\cup(3+2\sqrt2,+\infty)$$
| {
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Where does the equation of asymptotes of a hyperbola come from? It's known that the asymptotes of a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is given by $y=\pm\frac{b}{a}x$ if $a>b$.
I tried to find a proof of the fact that why the equations of these asymptotes are like that,however the only reference (Th... | Edited to do it properly -- see below
Original post:
We have
$$y=b\sqrt{\frac{x^2}{a^2}-1}=\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}$$
And as $x\to\pm\infty$, $\sqrt{1-\frac{a^2}{x^2}}\to 1$.
End of original post
But as mentioned in the comments, it is not enough to show that $\frac{y}{bx/a}\to 1$. We have to show that $y-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to solve equation with multiple trigonometric functions? Solve for $x$:
$\arccos( \cos(x) y + z) = \arcsin( \sin(x) a+b)$.
| $$\cos^{-1}(\cos(x)y+z)=\sin^{-1}(\sin(x)a+b)$$
Taking Sine of both the sides,
$$\sin(\cos^{-1}(\cos(x)y+z))=\sin(\sin^{-1}(\sin(x)a+b))$$
$$\sin\left(\sin^{-1}\sqrt{1-(\cos(x)y+z)^2}\right)=\sin(x)a+b$$
$$\sqrt{1-(\cos(x)y+z)^2}=a\sqrt{1-\cos^2(x)}+b$$
$$1-(\cos(x)y+z)^2=\left(a\sqrt{1-\cos^2(x)}+b\right)^2$$
$$\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a, b, c\in\mathbb R^+, $ then prove that $a^3b+b^3c+c^3a\ge abc(a+b+c) .$ While trying to prove it, I proved the following two inequalities:
$a^4+b^4+c^4\ge abc(a+b+c)$ and
$(a^2b+b^2c+c^2a)(ab+bc+ca)\ge abc(a+b+c)^2.$
The later one, on some simplification gives
$a^3b+b^3c+c^3a\ge abc(ab+bc+ca).$
But we can't claim... | Suppose $c=\min\{a,b,c\}.$ We have
$$\begin{aligned}a^3b+b^3c+c^3a-abc(a+b+c)&=c(a^3+b^3-a^2b-ab^2)+a(c^3+a^2b-bc^2-ca^2)\\&=c(a+b)(a-b)^2+a(c+a)(a-c)(b-c) \geqslant 0.\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Simplify $4^3\sin^4(20^\circ)\sin^2(70^\circ)-4\sqrt3\sin^3(20^\circ)\sin(70^\circ)+3$ I was trying to solve a question where the two sides of a triangle were $$\frac{a\sin(20^\circ)}{\sin(70^\circ)}$$and $$\frac{a\sin(60^\circ)\sin(30^\circ)}{\sin(70^\circ)\sin(40^\circ)}$$ and the ange between them was $70^\circ$ I u... | We have the following theorem:
$$
\prod_{0<k<n}2\sin\frac{k\pi}n=n.
$$
Particularly:
$$\begin{align}
\prod_{0<k<9}{\sin\frac{k\pi}9}=[2^4\sin(20^\circ)\sin(40^\circ)\sin(60^\circ)\sin(80^\circ)]^2=9\\
\implies \sin(20^\circ)\sin(40^\circ)\sin(60^\circ)\sin(80^\circ)=\frac{3}{16}.\tag1
\end{align}
$$
Let us apply this t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$ Reducing this whole expression i finally came to this
$$\sum \limits_{cyc}\left(ab^4+a^4b+a^2b^2c\right)\geq \sum \limits_{cyc}\left(a^3b^2+a^2b^3+a^3bc\right)$$
Here I am stuck. I ca... | Another way.
Afte using your $uvw$'s substitution we see that our inequality is a linear inequality of $w^3$,
which by $uvw$ says that it's enough to prove our inequality in the following cases.
*
*$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$ and $b=1$.
We obtain:
$$a(a+1)(a-1)^2\geq0;$$
2. Two variables are equal.
... | {
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"url": "https://math.stackexchange.com/questions/3705652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$? I want to solve the below:
$$\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$$
I know for trig substitution, if I have something in the form of $\sqrt{x^2-a^2}$, I can use $x = a\sec{u}$; it just so happens my integral has a numerator in t... | Given $\int \frac{\sqrt{16x^2}-9}{x}dx$ and that $\sqrt{x^2-a^2} \Rightarrow x=a \sec \theta \wedge a \sec \theta \tan \theta d\theta =dx$
Then,
$$\int \frac{\sqrt{16x^2}-9}{x}dx \Rightarrow \int \frac{\sqrt{16(3 \sec\theta)^2}-3^2}{3 \sec \theta} 3 \sec \theta \tan \theta d\theta $$
$$ = 12\int \tan \theta \sqrt{\sec^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 6
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If $d \equiv 3 \pmod 4 $ then $x^2 − dy^2 = −1$ has no solutions in positive integers $x, y$. In this case, I have come to the conclusion that $x^2 \equiv 0,1 \pmod 4$, $y^2 \equiv 0,1 \pmod 4$ but I am not sure what this means for $dy^2$ and how this won't have any solutions for the above equation. Can someone clarify... | $d\equiv3\equiv-1\pmod4$, so $x^2-dy^2\equiv x^2+y^2\pmod 4$.
Since $x^2\equiv0$ or $1$ and $y^2\equiv 0$ or $1$, $x^2+y^2\equiv 0, 1, $ or $2$, but not $3$ ($\equiv-1$).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Deriving the expansion of $\sin (\alpha - \beta)$ using $\sin x = \sqrt{1-\cos^2 x}$ I was deriving the expansion of the expansion of $\sin (\alpha - \beta)$ given that $\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$
Now, my textbook has done it in a different manner but I thought of doing it... | What is your question?
At start when you accepted
$$\sin x =\pm \sqrt{1-\cos^2 x}, \tag1$$
at the end why don't you accept
$$\sin (\alpha - \beta) = \pm \sqrt{1-(\cos \alpha \cos \beta + \sin \alpha \sin \beta)^2}\tag2$$
in that very same sense? What is extra in (2) ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3707342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\int \frac{x^2\,dx}{(a-bx^2)^2}$ How do I integrate $\int \frac{x^2\,dx}{(a-bx^2)^2}$
I've tried substitution and partial fraction decomposition, but I'm not getting anywhere.
| Almost without any substitution.
Let $x=\frac{\sqrt{a} }{\sqrt{b}}y$
$$I=\int \frac{x^2}{(a-b\,x^2)^2}\,dx=\frac{1}{\sqrt{a}\,\, b^{3/2}}\int \frac{y^2}{\left(1-y^2\right)^2}\,dy$$ Now, using partial fraction decomposition
$$\frac{y^2}{\left(1-y^2\right)^2}=\frac{1}{4 (y-1)}-\frac{1}{4 (y+1)}+\frac{1}{4 (y+1)^2}+\frac{... | {
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"timestamp": "2023-03-29T00:00:00",
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Eigenvectors of $\begin{pmatrix}2&-2\\ -4&-2\end{pmatrix}$ - What am I doing wrong? Eigenvalues are $2\sqrt{3}$ and $-2\sqrt{3}$, I'll calculate the eigenvector for $2\sqrt{3}$ here
We've got:
$\begin{pmatrix}2-2\sqrt{3}&-2\\ -4&-2-2\sqrt{3}\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pma... | You did nothing wrong. What you got were that the eigenvectors corresponding to the eigenvalue $2\sqrt3$ are those of the form $\left(1,1-\sqrt3\right)^Ty$, with $y\ne0$. That calculator got the vectors of the form $\left(-\frac{\sqrt3+1}2,1\right)^Ty$ with $y\ne0$. But$$\left(-\frac{\sqrt3+1}2\right)\begin{pmatrix}1\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$f(x - f(y))= f(f(y)) - 2xf(y) + f(x)$
Let $x,y \in \mathbb R$. Determine all $f : \mathbb R \to \mathbb R$ that satisfies $$f(x - f(y)) = f(f(y)) - 2xf(y) + f(x)$$
My solution: $f(x) = 0$ is obvious. So, I’ll consider other functions.
$x = f(y)$ $\to$ $f(0) = f(x) - 2x^2 + f(x)$
Which implies $f(x) = 2x^2 + \frac{f(... | Here is a solution: Suppose that $f$ solves the functional equation, and write $g(x) = f(x) - x^2$.
Claim. $ g(x-f(y)) = g(x) + \frac{f(0)}{2} $ holds for all $x, y \in \mathbb{R}$.
Proof. Using the functional equation, we get
\begin{align*}
g(x-f(y))
&= \bigl( f(f(y)) - 2xf(y) + f(x) \bigr) - \bigl( x - f(y) \bigr)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $ \int_{1}^{ \infty} \frac{1}{ (1+x^3)^3 } dx$ is convergent? I am trying to determine whether the following improper integral is convergent or not.
$$ \int_{1}^{ \infty} \frac{1}{ (1+x^3)^3 } dx$$
I tried the following:
$l = \lim_{x \to a} ((x-a)^k)f(x)$, if $l \in [0, \infty)$ and $k < 1$, then the in... | If $x \in [1,\infty)$, then $\frac{1}{(1+x^3)^3} < \frac{1}{x^6}<\frac{1}{x^2}$. So for every $n \in \Bbb N$, we have that $\int_{1}^n \frac{1}{(1+x^3)^3}<\int_{1}^n\frac{1}{x^2}$. But $\int_{1}^n\frac{1}{x^2} = 1 - \frac{1}{n}$. Therefore
\begin{equation}
\int_{1}^n \frac{1}{(1+x^3)^3}<1-\frac{1}{n}
\end{equation}
Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to computer $f(\frac{1}{2})$ given $f(f(x)) = x^2 + \frac{1}{4}$? I have observed that $f(f(\frac{1}{2})) = \frac{1}{2}$ and $f(f(f(x))) = f(x^2 + \frac{1}{4})$, and when $x = \frac{1}{2}$, we have $f(\frac{1}{2}^2 + \frac{1}{4}) = f(\frac{1}{2})$. But I don't know how to proceed, or if any of these observations ar... | Denote
$
c = f\left(\frac{1}{2}\right).
$
We have
$$
f\left(f\left( \frac{1}{2}\right)\right) = f(c) = \frac{1}{2}.
$$
$$
f\left(f\left( c\right)\right) = f\left(\frac{1}{2}\right) = c^2 + \frac{1}{4} = c.
$$
So
$$
\left(c - \frac{1}{2} \right)^2 = 0.
$$
This means that the only possible value for $c$ (if such function... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find eigenvalues of a Pauli matrix resolved direct product matrix? For a $2 \times 2$ hermitian matrix one can resolve the matrix
in terms of Pauli matrices like this
\begin{align}
H &= \begin{pmatrix}
a & b \\
b & -a
\end{pmatrix} \\
&= a \sigma_z + b \sigma_x \\
\end{align}
Here, I've assumed $ a,b \in... | The problem is your assumption that $H^2 = E^2 I_{2n}$.
In the $2 \times 2$ case, we use the equation $H^2 = E^2 I_2$ in order to exploit the fact that any trace-zero $2 \times 2$ matrix $H$ will be such that $H^2$ is a multiple of the identity. However, this trick no longer works in the generalized setting since it i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find integers $1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$ Root numbers Problem (Math Quiz Facebook):
Consider the following equation:
$$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$
Where $a,\,b,\,c,\,d$ are integers. Find $a+b+c+d$
I've tried it like this:
Let $w=\sqrt6,\, x=\sqrt3, \... | First an answer $$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{21+\sqrt{413+\sqrt{4656+ \sqrt{16588800}}}}.$$
Then an explanation.
Everything takes place inside the field $L=\Bbb{Q}(\sqrt2,\sqrt3)$. By elementary Galois theory the quadratic subfields of $L$ are $\Bbb{Q}(\sqrt2)$, $\Bbb{Q}(\sqrt3)$ and $\Bbb{Q}(\sqrt6)$. The number $c+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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How to show $\sum_{k=0}^{\infty} \frac{1}{k!} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k =x$? There are a lot of ways to show that $e^x$ and $\ln(x)$ are inverse functions of each other depending on how you define them. I am trying to show that given the definitions
$$ e^x:= \sum_{k=0}^{\infty} \frac{x^k}{k!} \qq... | By taking into account the comment by @Pythagoras, my attempt can be corrected to get the following:
\begin{align}
\frac{d^2}{dx^2} e^{\ln(x)} & = \frac{d^2}{dx^2}\left(\underbrace{1}_{k=\color{blue}{0}} + \underbrace{\int_{1}^{x} \frac{1}{t} \ dt}_{k=\color{blue}{1}} +\sum_{k={\color{blue}{2}}}^{\infty} \frac{1}{k!} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding all real $a$ such that $16x^4-(a)x^3+(2a+17)x^2−(a)x+16=0$ has four distinct roots in geometric progression
Determine all real values of the parameter, $a$, for which the equation $$16x^4−(a)x^3+(2a+17)x^2−(a)x+16=0$$ has exactly four distinct real roots that form a geometric progression?
I noticed that the ... | Let's do like derivation of Vieta's formulas rather than the formulas themselves.
Let $b,\,bq,\,bq^2,\,bq^3$ be the roots of $$p(x)=(x-b)(x-bq)(x-bq^2)(x-bq^3)$$ $$\hbox{and }f(x)=x^4−\frac{a}{16}x^3+\frac{2a+17}{16}x^2−\frac{a}{16}x+1.$$
It can be verified that $$p(x)=b^4 q^6 - b^3 (q^3 + q^2 + q + 1) q^3 x + \\
b^2 (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solving polynomials using De Moivre's theorem Given $\cos 4\theta=8\cos^4\theta−8\cos^2\theta+1$, solve $16x^4-16x^2+1=0$.
The textbook's answers are $x=\pm\cos\dfrac{\pi}{12},\pm\cos\dfrac{5\pi}{12}$. I managed to get two of the four answers and i can not figure out what i did wrong.
My Attempt
Let $x=\cos\theta$
$16\... | You wrote:
$$\cos4θ=\cfrac{1}{2}$$
$$\cos4θ=cos\cfrac{\pi}{3}$$
$$\therefore4θ=\cfrac{\pi}{3}+k\pi$$
This contains some errors
The $\cos$ and the $\sin$ function take each value to times in the interval $[0,2\pi]$ except $1$ and $-1$.
So the following holds:
$$\cos \phi=\cos (2\pi-\phi)$$
The period of these trigonomet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Divisibility Number Theory problem, explanation needed I can't understand the solution of the following problem:
$x$,$y$,$z$ are pairwise distinct natural numbers show that $(x-y)^5$ + $(y-z)^5$ + $(z-x)^5$ is divisible by $5(x-y)(y-z)(z-x)$. No need to explain the div. by 5.
The sol. says:
$(x-y)^5$ + $(y-z)^5$ + $(z-... | Let $x-y=a$ and $y-z=b$.
Thus, $z-x=-(a+b)$ and $$\sum_{cyc}(x-y)^5=a^5+b^5-(a+b)^5=-5a^4b-10a^3b^2-10a^2b^3-5ab^4=$$
$$=-5ab(a^3+2a^2b+2ab^2+b^3)=-5ab(a+b)(a^2-ab+b^2+2ab)$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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.