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When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$... When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$. What is the denominator of the fraction?
I tried,
Let the numerator of the fraction be $x$ and the denominator be $y$.
Accordingly, $$\frac{x+4}y=\frac xy+\frac 23$$
I am not able to find the second equation.
| We have $\frac{x+4}y={\frac xy}+{\frac 23}=\frac{3x+2y}{3y}$ but: $$\frac{x+4}{y}=\frac{\color{red}3(x+4)}{\color{red}3y}=\frac{3x+12}{3y}$$ So $$\frac{3x+12}{3y}=\frac{3x+2y}{3y}$$ So if $y\neq 0$ then $3x+12=3x+2y$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the Legendre symbols $(\frac{503}{773})$ and $(\frac{501}{773})$ Evaluate the Legendre symbols $(\frac{503}{773})$ and $(\frac{501}{773})$.
My solution:
$(\frac{501}{773}) = (\frac{167 \cdot 3}{773}) = (\frac{167}{773}) \cdot (\frac{3}{773}) = (\frac{773}{167}) \cdot (\frac{773}{3})$
$= (\frac{105}{167}) \cdot (\frac{2}{3})$
$= (\frac{3}{167}) \cdot (\frac{5}{167}) \cdot (\frac{7}{167}) \cdot (\frac{2}{3})$
$=-1 \cdot (\frac{167}{3}) \cdot (\frac{167}{5}) \cdot -1 \cdot (\frac{167}{7}) \cdot (\frac{2}{3})$
$= (\frac{2}{3}) \cdot (\frac{2}{5}) \cdot (\frac{6}{7}) \cdot (\frac{2}{3})$
$= (\frac{2}{3}) \cdot (\frac{2}{5}) \cdot (\frac{2}{7}) \cdot (\frac{3}{7}) \cdot (\frac{2}{3})$
$= -1 \cdot -1 \cdot 1 \cdot -1 \cdot -1$
$=1$
Is it correct? I need confirmation if it is correct or not, and help in the other part.
| The calculation for $501$ is correct.
You should find that $503$ is a quadratic non-residue of $773$. The calculation is easier, because $503$ is prime and $270$ has lots of small factors. We have
$$(503/773)=(773/503)=(270/503)=(2/503)(3/503)(5/503).$$
Because $503$ is of the form $8k+7$, we have $(2/503)=1$. Easily, $(3/503)=-(503/3)=1$. And finally $(5/503)=(503/5)=(3/5)=-1$.
| {
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Algebraic proof of $\tan x>x$ I'm looking for a non-calculus proof of the statement that $\tan x>x$ on $(0,\pi/2)$, meaning "not using derivatives or integrals." (The calculus proof: if $f(x)=\tan x-x$ then $f'(x)=\sec^2 x-1>0$ so $f$ is increasing, and $f(0)=0$.) $\tan x$ is defined to be $\frac{\sin x}{\cos x}$ where these are defined by their infinite series. What I have so far:
$$|z|\le1\implies\left|\sum_{n=4}^\infty\frac{z^n}{n!}\right|<\sum_{n=0}^\infty\frac{|z|^4}{4!\,5^n}=\frac{5|z|^4}{4\cdot 4!}$$
$$\left|\sin x-\Big(x-\frac{x^3}6\Big)\right|=\Im\left[\sum_{n=4}^\infty\frac{(ix)^n}{n!}\right]<\frac{5x^4}{4\cdot 4!}<\frac{x^3}6$$
$$\left|\cos x-\Big(1-\frac{x^2}2\Big)\right|=\Re\left[\sum_{n=4}^\infty\frac{(ix)^n}{n!}\right]<\frac{5x^4}{4\cdot 4!}<\frac{x^2}6$$
Thus $\sin x>x-\frac{x^3}3$ and $\cos x<1-\frac{x^2}3$, so $\tan x>x$. However, this only covers the region $x\le1$, and I still need to bound $\tan x$ on $(1,\pi/2)$. My best approximation to $\pi$ is the very crude $2<\pi<4$, derived by combining the above bounds with the double angle formulas (note that $\pi$ is defined as the smallest positive root of $\sin x$), so I can't quite finish the proof with a bound like $\sin x>1/\sqrt 2$, $\cos x\le\pi/2-x$ (assuming now $x\ge1\ge\pi/4$) because the bound is too tight. Any ideas?
| (I should have added this as comment but don't have privilege yet)
Just in case OP haven't checked out yet, the infinite series expansion of $tan(x)$ is already available on Wikipedia. Quoting here:
$$tan(x) = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 + {...}$$
for $\lvert x \rvert < \frac{\pi}{2}$.
| {
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Finding the perimeter of the room If the length and breadth of a room are increased by $1$ $m$, the area is increased by $21$ $m^2$. If the length is increased by $1$ $m$ and breadth is decreased by $1$ $m$ the area is decreased by $5$ $m^2$. Find the perimeter of the room.
Let the length be $x$ and the breadth be $y$
Therefore, Area$=$$xy$ $m^2$
Accordingly, $(x+1) \cdot (y+1) \ = \ xy+21$ $m^2$
What should I do now? How should I find the second equation?
Should the second equation look like:
$(x+1) \ \cdot \ (y-1) \ = \ xy -5 \ $ $m^2$
| $(x+1)(y+1)=xy+21$, $(x+1)(y-1)=xy-5$, so by subtraction $(x+1)(y+1-y+1)=2x+2=26$. Solving gives $x=12$ and then using this to solve the first equation 13(y+1)=12y+21, so $y=8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} $ $$\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} dx$$
My approach is to calc
$$\int_{1}^{X} \frac{\ln{(2x-1)}}{x^2} dx$$ and then take the limit for the answer when $X \rightarrow \infty$
However, I must do something wrong. The correct answer should be $2\ln(2)$.
$$\int_{1}^{X} \frac{\ln{(2x-1)}}{x^2} dx = \left[-\frac{1}{x} \ln (2x-1) \right]_{1}^{X} + \int_{1}^{X} \frac{1}{x} \times \frac{2}{2x-1} dx = -\frac{1}{X}\ln(2X-1) + 2\int_{1}^{X} -\frac{1}{x} + \frac{2}{2x-1} dx = -1\frac{1}{X}\ln(2X-1)-2\ln X+2\ln(2X-1) $$
Am I wrong? If I'm not, how to proceed?
=== EDIT ===
After the edit I wonder if this is the correct way to proceed:
$$ - \frac{1}{X}\ln(2X-1)-2\ln X+2\ln(2X-1) $$ The first part will do to zero because of $\frac{1}{X} $ so we ignore that one, the second and third part:
$$ -2\ln X+2\ln(2X-1) = 2\ln \left( \frac{2X-1}{X}\right) = 2\ln \left( 2-\frac{1}{X} \right) = 2\ln (2)$$
| Your derivative is incorrect, it should be $$\frac{2}{2x-1}$$ instead of $$\frac{1}{2x-1}$$ Everything else seems correct.
| {
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Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$
If $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$
Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have tried so far:
Using Cauchy inequality I can write:$$\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)(a+b+c) \geq \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2$$
but I can't continue this.I tried expanded form:$$\sum \limits_{cyc} \frac{a^5c^2}{a^2b^2c^2} \geq \sum \limits_{cyc} \frac{a^3c}{abc}$$
Which proceeds me to this Cauchy:$$\sum \limits_{cyc} \frac{a^5c^2}{abc}\sum \limits_{cyc}a(abc)\geq \left(\sum \limits_{cyc}a^3c\right)^2$$
I can't continue this one too.
The main Challenge is $3$ fraction on both sides which all of them have different denominator.and it seems like using Cauchy from first step won't leads to anything good.
| $$\sum_{cyc}\frac{a^3}{b^2}-\sum_{cyc}\frac{a^2}{b}=\sum_{cyc}\left(\frac{a^3}{b^2}-\frac{a^2}{b}\right)=$$
$$=\sum_{cyc}\left(\frac{a^2(a-b)}{b^2}-(a-b)\right)=\sum_{cyc}\frac{(a-b)^2(a+b)}{b^2}\geq0.$$
Done!
| {
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What are the roots of $\sin(ax) + \sin((a + 2)x)$? I was playing around with $\sin(5x) + \sin(7x)$, wondering where the roots of the function are. I graphed it on wolframalpha and from the list of solutions I guessed that the solutions to $\sin(5x) + \sin(7x) = 0$ are exactly the roots of $\sin(6x)$.
After playing around with similar values, I have to wonder, is it true that solving
$$\sin(a\cdot x) + \sin((a + 2)\cdot x) = 0$$
is equivalent to
$$\sin((a + 1)\cdot x) = 0$$
This problem seems to want induction.
It is true that the roots of $\sin(x) + \sin(3x)$ are the same roots as $\sin(2x)$ (again checked via wolframalpha). So that takes care of the base case.
However, I have no idea how to get from the $\sin(a\cdot x) + \sin((a+2)\cdot x)$ to the $\sin((a+1)\cdot x)$ for any $a$. It's just been guesswork.
Is it true that $\sin(a\cdot x) + \sin((a + 2)\cdot x)$ has the same roots as $\sin((a + 1)\cdot x)$?
| Note the sum to product identity: $\sin A + \sin B = 2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$.
Hence, $\sin(ax) + \sin((a+2)x) = 2\sin((a+1)x)\cos(x)$.
So the roots of $\sin(ax) + \sin((a+2)x)$ are the roots of $\sin((a+1)x)$ along with the roots of $\cos(x)$.
If $x = \dfrac{\pi}{2}+\pi k$ (the roots of $\cos x$) are also roots of $\sin((a+1)x)$, then we are only left with the roots of $\sin((a+1)x)$. This happens iff $\sin\left((a+1)(\dfrac{\pi}{2}+\pi k)\right) = 0$, i.e. $a$ is odd (assuming $a \in \mathbb{Z}$).
Here are the graphs of $\sin(2x)+\sin(4x)$ and $\sin(3x)$. Note that $\sin(2x)+\sin(4x)$ has more roots.
| {
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Solving the logarithimic inequality $\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$ I tried solving the logarithmic inequality:
$$\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$$
several times but keeping getting wrong answers.
| $$\log_2 x - \log_2 2 + 2\log_2 x - \log_2 2 + \log_2 x \le 1$$
$$4\log_2 x - 2 \le 1$$
$$4\log_2 x \le 3$$
$$\log_2 x \le 3/4 $$
$$x \le 2 \cdot 3/4$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then $|z|=$? If z is a complex number and $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then find the value of $|z|$ .
I tried to put $ \frac {z^2 + z+ 1} {z^2 -z +1} =k $ then solve for $z$ and tried to find |z|, but it gets messy and I am stuck.
The answer given is |z|=1
| Suppose $|z|=r$, so that $z\bar{z}=r^2$ or $\bar{z}=r^2/z$.
We have that $$\overline{\frac{z^2+z+1}{z^2-z+1}}=\frac{\bar{z}^2+\bar{z}+1}{\bar{z}^2-\bar{z}+1}=\frac{r^4/z^2+r^2/z+1}{r^4/z^2-r^2/z+1}=\frac{z^2+r^2 z+r^4}{z^2-r^2z+r^4}. $$
But if the expression $$ \frac{z^2+z+1}{z^2-z+1}$$ is real, we must have
$$\overline{\frac{z^2+z+1}{z^2-z+1}}=\frac{z^2+z+1}{z^2-z+1}. $$
This should help you determining $r=|z|$.
| {
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Proving or disproving inequality $ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \ge x + y + z $
Given that $ x, y, z \in \mathbb{R}^{+}$, prove or disprove the
inequality
$$ \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z $$
I have rearranged the above to:
$$ x^2y(y - z) + y^2z(z - x) + z^2x(x - y) \ge 0 \\
\text{and, } \dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} \ge \dfrac{1}{xy} + \dfrac{1}{xz} + \dfrac{1}{yz} $$
What now? I thought of making use of the arithmetic and geometric mean properties:
$$
\dfrac{x^2 + y^2 + z^2}{3} \ge \sqrt[3]{(xyz)^2} \\
\text{and, } \dfrac{x + y + z}{3} \ge \sqrt[3]{xyz}
$$
but I am not sure how, or whether that'd help me at all.
| Use the rearrangement inequality:
$$xy={xyz\over z},$$
so WLOG(without loss of generality), if $0<x\le y\le z$, $$\begin{gather}\frac1x\ge\frac1y\ge\frac1z,\\ yz\ge zx\ge xy.\end{gather}$$
Then we have the result, since LHS of the given inequality is the maximum value when rearrange.
| {
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Find $ \int_0^2 \int_0^2\sqrt{5x^2+5y^2+8xy+1}\hspace{1mm}dy\hspace{1mm}dx$ I need the approximation to four decimals
Not sure how to start or if a closed form solution exists
All Ideas are appreciated
| Possible hint
It is not the most pleasant integral. However, using a CAS, the following result was obtained $$\int\sqrt{5x^2+5y^2+8xy+1}\hspace{1mm}dy=\sqrt{5 x^2+8 x y+5 y^2+1} \left(\frac{2 x}{5}+\frac{y}{2}\right)+\frac{\left(9
x^2+5\right) \log \left(\sqrt{5} \sqrt{5 x^2+8 x y+5 y^2+1}+4 x+5 y\right)}{10
\sqrt{5}}$$ So $$\int_0^2\sqrt{5x^2+5y^2+8xy+1}\hspace{1mm}dy=\frac{1}{50} \left(-20 \sqrt{5 x^2+1} x-\sqrt{5} \left(9 x^2+5\right) \log
\left(\sqrt{25 x^2+5}+4 x\right)+\sqrt{5} \left(9 x^2+5\right) \log \left(4
x+\sqrt{5} \sqrt{x (5 x+16)+21}+10\right)+10 (2 x+5) \sqrt{x (5 x+16)+21}\right)$$ Looking at this result over the range $0\leq x \leq 2$ reveals a function which looks very linear. So I hope that a simple quadrature method would lead to the required result.
| {
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Elementary algebra problem Consider the following problem (drawn from Stanford Math Competition 2014): "Find the minimum value of $\frac{1}{x-y}+\frac{1}{y-z}+ \frac{1}{x-z}$ for for reals $x > y > z$ given $(x − y)(y − z)(x − z) = 17.$"
Method 1 (official solution): Combining the first two terms, we have
$\frac{x−z}{(x-y)(y-z)} + \frac{1}{x-z}= \frac{(x-z)^2}{17}+ \frac{1}{x-z}.$
What remains is to find the minimum value of $f(a) = \frac{a^2}{17} + \frac{1}{a} = \frac{a^2}{17} + \frac{1}{2a}+ \frac{1}{2a}$ for positive values of $a.$ Using AM-GM, we get $f(a) \geq \frac{3}{68^{1/3}}$.
Method 2: Let $x-y:=a, \; y-z:=b, \; x-z:=c.$ Then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}= \frac{ab+bc+ac}{17}= \frac{3}{17}\frac{ab+bc+ac}{3} \geq \frac{3}{17} (a^2b^2c^2)^{1/3} =\frac{3}{17^{1/3}}.$
So Method 2 seems to give a sharper bound than the official solution. Have I done something wrong?
| The official solution is wrong. The equality happens when $$\frac{a^2}{17}=\frac1{2a}$$ or $a^3=\frac{17}2<a^3$. Note that in your solution, the equality does not happen either.
Edit: I would solve the problem as follows.
Let $a=x-y,b=y-z$ then $ab(a+b)=17$ and we need to minimize
$$\frac1a+\frac1b+\frac1{a+b}=\frac{(a+b)^2+ab}{ab(a+b)}=\frac{(a+b)^2+ab}{17}.$$
We then need to minimize $$f(t) = (a+b)^2+ab=t^2+\frac{17}{t},$$ where $a+b=t$ satisfying $$t^2\ge 4ab=\frac{68}{t},\text{ or } t\ge \sqrt[3]{68}.$$
It's easy to see that
$$f'(t)=2t-\frac{17}{t^2}=\frac{2t^3-17}{t^2}>0.$$
So $f(t)$ attains its minimum at $t=\sqrt[3]{68}$. That is $x-z=\sqrt[3]{68}$ and $x-y=y-z=\frac{\sqrt[3]{68}}{2}$ and we see that the minimum is $\displaystyle\color{red}{\frac{5}{\sqrt[3]{68}}}$.
| {
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Prove that $\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z} \ge \frac{(a+b+c)^3}{3(x+y+z)}$ a,b,c,x,y,z are positive real numbers. I stumbled upon it on some olympiad papers. Tried to AM>GM but didn't get any idea to move forward.
| Holder's Inequality gives:
$$\left( \frac{a^3}x + \frac{b^3}y + \frac{c^3}z \right)(x + y + z)(1+1+1) \ge (a+b+c)^3 $$
| {
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Explaining why $\sqrt {x^2+a} = x\sqrt{1+ \frac{a}{x^2}}$ For $x>0$. I understand the technical operation of extracting $x^2$ out of the root, but is there a way proving it?
$$\sqrt {x^2+a} = x\sqrt{1+ \frac{a}{x^2}}$$
| $$\sqrt{x^2+a}=\sqrt{x^2\left(1+\frac{a}{x^2}\right)}=\underbrace{\sqrt{x^2}}_{=|x|}\cdot \sqrt{1+\frac{a}{x^2}}\underset{(1)}{=}x\sqrt{1+\frac{a}{x^2}}$$
justification:
$(1):$ Because $x>0$.
| {
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Determine variables that fit this criterion... There is a unique triplet of positive integers $(a, b, c)$ such that $a ≤ b ≤ c$.
$$
\frac{25}{84} = \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc}
$$
Just having trouble with this Canadian Math Olympiad question. My thought process going into this, is:
Could we solve for $\frac{1}{a}$ in terms of the other variables? Then substitute that value in for each occurrence of $a$, to solve for $a$?
That's all I can really think of right now. It's a question I'm not exactly used to... It's sort of the first of these kinds that I've faced.
Thanks.
| Clearly $\displaystyle \frac{1}{a} < \frac{25}{84} < \frac{1}{3}$ so $a \geq 4$.
Then $b \geq 4$ and one of $a,b,c$ must be a multiple of $7$ so $c \geq 7$.
Hence $\displaystyle \frac{25}{84} \leq \frac{1}{a} + \frac{1}{4a} + \frac{1}{28a} = \frac{9}{7a}$ so $\displaystyle a \leq \frac{108}{25}<5$ so we must have $a = 4$.
Substitute $a=4$ into the original equation to obtain $\displaystyle \frac{c+1}{bc} = \frac{4}{21}.$ It follows that $4 \mid (c+1)$ and, since $c$ and $c+1$ are coprime, $c \mid 21$. Since we know that $c \geq 7$, we must have that $c = 7$ and so $b = 6$.
| {
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Inequality involving a finite sum this is my first post here so pardon me if I make any mistakes.
I am required to prove the following, through mathematical induction or otherwise:
$$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$$
I tried using mathematical induction through:
$Let$ $P(n) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$
$Since$ $P(1) = \frac{1}{\sqrt1} < 2{\sqrt{1}}, and$ $P(k) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{k}} < 2{\sqrt{k}},$
$P(k+1) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{k}}+ \frac{1}{\sqrt{k+1}} < 2{\sqrt{k+1}}$
Unfortunately, as I am quite new to induction, I couldn't really proceed from there. Additionally, I'm not sure how to express ${\sqrt{k+1}}$ in terms of ${\sqrt{k}}$ which would have helped me solve this question much more easily. I am also aware that this can be solved with Riemann's Sum (or at least I have seen it being solved in that way) but I do not remember nor quite understand it.
| We have a stronger inequality. Note that
$$\sqrt{x-\frac{1}{2}}+\sqrt{x+\frac{1}{2}}<2\sqrt{x}$$
for all $x\geq \dfrac{1}{2}$. Thus, for each $x>\dfrac12$, we get
$$\frac{1}{\sqrt{x}}< \frac{2}{\sqrt{x-\frac12}+\sqrt{x+\frac{1}{2}}}=2\,\left(\sqrt{x+\frac12}-\sqrt{x-\frac12}\right)\,.$$
That is, for every nonnegative integer $n$,
$$\sum_{k=1}^n\,\frac{1}{\sqrt{k}}\leq 2\,\sum_{k=1}^n\,\left(\sqrt{k+\frac12}-\sqrt{k-\frac12}\right)=2\,\left(\sqrt{n+\frac12}-\sqrt{\frac12}\right)\,.$$
The equality holds if and only if $n=0$. This is a sharper inequality than the required inequality as
$$\sqrt{x+y}\leq \sqrt{x}+\sqrt{y}$$
for any $x,y\geq 0$ (with equality iff $x=0$ or $y=0$).
On the other hand, you can show that
$$2\sqrt{x}\leq \sqrt{x-\frac{7}{16}}+\sqrt{x+\frac{9}{16}}$$
for every $x\geq 1$ (with equality case $x=1$). This gives
$$\sum_{k=1}^n\,\frac{1}{\sqrt{k}}\geq 2\,\left(\sqrt{n+\frac{9}{16}}-\frac{3}{4}\right)$$
for all integers $n\geq 0$ with equality cases $n=0$ and $n=1$. That is,
$$2\,(\sqrt{n+1}-1)\leq 2\,\left(\sqrt{n+\frac{9}{16}}-\frac{3}{4}\right)\leq \sum_{k=1}^n\,\frac{1}{\sqrt{k}}\leq 2\,\left(\sqrt{n+\frac12}-\sqrt{\frac12}\right)\leq 2\sqrt{n}$$
for every nonnegative integer $n$. If $n=0$, then every inequality is an equality. If $n=1$, then only the second inequality from the left becomes an inequality. For $n>1$, all inequalities are strict.
| {
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} |
Simplify the following compound fraction: $$\frac{2x+1}{\frac{3}{x^2}+\frac{2x+1}{x}}$$
My calculator says the final answer is $$\frac{x^2(2x+1)}{2x^2+x+3}$$
Please show the work. Thanks.
| You multiply at the numerator and at the denominator by $x^2$:
$$\frac{2x+1}{\frac{3}{x^2}+\frac{2x+1}{x}}=\frac{x^2(2x+1)}{x^2 \left ( \frac{3}{x^2}+\frac{2x+1}{x}\right)}=\frac{x^2(2x+1)}{3+x(2x+1)}=\frac{x^2(2x+1)}{2x^2+x+3}$$
| {
"language": "en",
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"source": "stackexchange",
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Consecutive Prime Gap Sum (Amateur) List of the first fifty prime gaps:
1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4.
My conjecture is that the sum of consecutive prime gaps is always prime whenever a prime gap of 2 is added.
$$ 1 + 2 = 3 $$
$$ 1 + 2 + 2 = 5 $$
$$ 1 + 2 + 2 + 4 + 2 = 11 $$
$$ 1 + 2 + 2 + 4 + 2 + 4 + 2 = 17 $$
$$ 1 + 2 + 2 + 4 + 2 + 4 + 2 + 4 + 6 + 2 = 29 $$
I don't know if this is meaningful or how to go about testing it completely (I've tested it up to 461) so I'll just leave this here and see what comes of it.
| The first gap is 3 - 2 = 1. The sum of gaps up to the first gap is 1, that is 2 less than 3.
The second gap is 5 - 3 = 2. The sum of gaps up to the second gap is 3, that is 2 less than 5.
The third gap is 7 - 5 = 2. The sum of gaps up to the third gap is 5, that is 2 less than 7.
The fourth gap is 11 - 7 = 4. The sum of gaps up to the fourth gap is 9, that is 2 less than 11.
You should be able to see a pattern now: The sum of the first n gaps is always two less than the n'th prime. And that's due to the way the gaps are calculated: The (n+1)st gap is the difference between the n+1st and the nth prime; adding this to the sum so far which is two less than the nth prime will give a result that is two less than the nth prime.
And of course every time the gap is two, two less than the nth prime happens to be the n-1st prime, which explains everything nicely.
| {
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Use the Cayley-Hamilton Theorem to find the inverse of this $3$-by-$3$ matrix $A=\begin{pmatrix}
2&0&0\\
1&3&0\\
-3&5&3
\end{pmatrix}.$
I can find the characteristic polynomial $C_a(A)= -A^3 + 8A^2-21A+18$.
How to continue from here to use the Cayley-Hamilton theorem to find the inverse matrix, $A^{-1}$?
| Since we know $p(A) = 0$, you can multiply by $A^{-1}$ both sides to have:
$$ -A^2 + 8 A - 21 \, I+ 18 A^{-1} = 0,$$
can you solve for $A^{-1}$?
Spoiler:
The answer is given by $A^{-1} = \frac{1}{18} \left( 21 I - 8 A + A^2 \right) =\left(\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -\frac{1}{6} & \frac{1}{3} & 0 \\ \frac{7}{9} & -\frac{5}{9} & \frac{1}{3}\end{array}\right)$
| {
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Evaluation of a sum of $(-1)^{k} {n \choose k} {2n-2k \choose n+1}$ I have some question about the paper of which name is Spanning trees: Let me count the ways. The question concerns about $\sum_{k=0}^{\lfloor\frac{n-1}{2} \rfloor} (-1)^{k}
{n \choose k} {2n-2k \choose n+1}$. Could you recommend me how to prove $\displaystyle \sum_{k=0}^{\lfloor\frac{n-1}{2} \rfloor} (-1)^{k} {n \choose k} {2n-2k \choose n+1}=n 2^{n-1}$?
| Suppose we seek to evaluate
$$\sum_{k=0}^n {n\choose k}(-1)^k {2n-2k\choose n+1}.$$
Start from
$${2n-2k\choose n+1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} (1+z)^{2n-2k} dz.$$
This yields the following expression for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{k=0}^n {n\choose k} (-1)^k
\frac{1}{z^{n+2}} (1+z)^{2n-2k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+2}}
\sum_{k=0}^n {n\choose k} (-1)^k
(1+z)^{-2k}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+2}}
\left(1-\frac{1}{(1+z)^2}\right)^n\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}}
\left(2z+z^2\right)^n\; dz.$$
It follows that the sum is given by
$$[z^{n+1}] \left(2z+z^2\right)^n
= [z^1] (2+z)^n = {n\choose 1} 2^{n-1} = n 2^{n-1}.$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
| {
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How prove $\frac{\sqrt{2}}{3}n^2<\sum_{k=1}^{n^2-1}\sqrt{1-\frac{\sqrt{k}}{n}}<\sqrt{2}n^2$
Show that
$$\dfrac{\sqrt{2}}{3}n^2<\sqrt{1-\dfrac{\sqrt{1}}{n}}+\sqrt{1-\dfrac{\sqrt{2}}{n}}+\sqrt{1-\dfrac{\sqrt{3}}{n}}+\cdots+\sqrt{1-\dfrac{\sqrt{n^2-1}}{n}}<\sqrt{2}n^2.$$
Maybe use
$$(1+x)^{\frac{1}{2}}<1+\dfrac{x}{2},x>-1$$
so
$$\sqrt{1-\dfrac{\sqrt{k}}{n}}<1-\dfrac{1}{2}\cdot\dfrac{\sqrt{k}}{n}$$
then how to prove it?
I don't want see without Mathematical induction,maybe can use integral to prove it?
| Note that $f(x)=\sqrt{1-\sqrt{x}}~$ is strictly decreasing on $[0,1]$. Thus
for $0\leq k<m$ we have
$$
\int_{k/m}^{(k+1)/m}f(t)dt<\frac{1}{m}f\left(\frac{k}{m}\right)\tag{1}
$$
and for for $0< k\leq m$ we have
$$
\int_{(k-1)/m}^{k/m}f(t)dt>\frac{1}{m}f\left(\frac{k}{m}\right)\tag{2}
$$
So, if
$$I_m=\frac{1}{m}\sum_{k=1}^{m-1}\sqrt{1-\sqrt{\frac{k}{m}}}$$ then,
from $(1)$ we get
$$
\int_{1/m}^1f(t)dt< I_m<\int_{0}^{1-1/m}f(t)dt<\int_{0}^{1}f(t)dt
$$
Now, it is not difficult to see that
$$\left(-\frac{4}{15}(1-\sqrt{x})^{3/2}(2+3\sqrt{x})\right)'=f(x)$$
Thus
$$
\frac{4}{15}\left(1-\frac{1}{\sqrt{m}}\right)^{3/2}\left(2+ \frac{3}{\sqrt{m}}\right)<I_m<\frac{8}{15}
$$
In particular, for $m=n^2$ we have
$$
\frac{4}{15}\left(1-\frac{1}{n}\right)^{3/2}\left(2+ \frac{3}{n}\right)<I_{n^2}<\frac{8}{15}
$$
Now, the left hand side of the previous inequality is an increasing function of $n$. So, for $n\geq 4$ we have
$$
\frac{\sqrt{2}}{3}<\frac{11\sqrt{3}}{40}<I_{n^2}<\frac{8}{15}
$$
Now, a direct check shows that the proposed inequality is valid also for $n=3$, but it is not for $n=1,2$. So, we have proved that
So, for $n\geq 3$ we have
$$
n^2\frac{\sqrt{2}}{3} <\sum_{k=1}^{n^2-1}\sqrt{1- \frac{\sqrt{k}}{n}}<\frac{8}{15}n^2
$$
which is the desired conclusion, with a stronger upper bound.
| {
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Find the sum of the multiples of $3$ and $5$ below $709$? I just cant figure this question out:
Find the sum of the multiples of $3$ or $5$ under $709$
For example, if we list all the natural numbers below $10$ that are multiples of $3$ or $5$, we get $3$, $5$, $6$ and $9$. The sum of these multiples is $23$.
| There are $\displaystyle\left\lfloor\frac{709}{3}\right\rfloor = 236$ multiples of $3$ below $709$, $\displaystyle\left\lfloor\frac{709}{5}\right\rfloor = 141$ multiples of $5$ below $709$ and $\displaystyle\left\lfloor\frac{709}{15}\right\rfloor = 47$ multiples of $15$ below $709$.
By the inclusion/exclusion principle:
$$\mathop{\sum_{(n|3\ \vee\ n|5)} n}_{ n \leq 709} = \mathop{\sum_{n|3} n}_{ n \leq 709} + \mathop{\sum_{n|5} n}_{ n \leq 709} - \mathop{\sum_{n|15} n}_{ n \leq 709} $$
Note that:
$$\mathop{\sum_{n|3} n}_{ n \leq 709} = 3 + 6 + \cdots + 3·236 = 3(1+2+\cdots + 236) = 3\frac{236·237}{2} = 83898$$
Similarly,
$$\mathop{\sum_{n|5} n}_{ n \leq 709} = 5 + 10 + \cdots + 5·141 = 5(1+2+\cdots + 141) = 5\frac{141·142}{2} = 50055$$
$$\mathop{\sum_{n|15} n}_{ n \leq 709} = 15 + 30 + \cdots + 15·47 = 15(1+2+\cdots + 47) = 15\frac{47·48}{2} = 16920$$
So the total sum is: $83898 + 50055 - 16920 = \boxed{117033}$
| {
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Is there any proof for this formula $\lim_{n \to \infty} \prod_{k=1}^n \left (1+\frac {kx}{n^2} \right) =e^{x/2}$? Some times ago, In a mathematical problem book I sow that this formula. I don't no whether it is true or not. But now I'm try to prove it. I have no idea how to begin it. Any hint or reference would be appreciate. Thank you.
$$\lim_{n \to \infty} \prod_{k=1}^n \left (1+\frac {kx}{n^2} \right) =e^{x/2}$$
| Here I use the method suggested by Daniel Fischer in the comment section:
Take the logarithm of $$\prod _{k=1}^n \left(1+\dfrac{kx}{n^2} \right).$$ Then we have
$$\ln\left(\prod_{k=1}^n\ \left(1+\dfrac{kx}{n^2 }\right) \right)=\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2 }\right).$$
By the Taylor series expansion of logarithms,
$$ \ln\left(1+\dfrac{kx}{n^2} \right)=\sum_{m=1}^\infty\dfrac{(-1)^{m-1}}{m}\left(\dfrac{ kx}{n^2 }\right)^m \\=
\left(\dfrac{kx}{n^2 }\right)-\dfrac{1}{2} \left(\dfrac{kx}{n^2 }\right)^2+\dfrac{1}{3} \left(\dfrac{kx}{n^2 }\right)^3-\dfrac{1}{4} \left(\dfrac{kx}{n^2 }\right)^4+\dfrac{1}{5} \left(\dfrac{kx}{n^2 }\right)^5-\dfrac{1}{6} \left(\dfrac{kx}{n^2 }\right)^6+ …$$
Now
$$\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2} \right) =\dfrac{x}{n^2 }\left( \sum_{k=1}^nk\right)-\dfrac{x^2}{2n^4 }\left( \sum_{k=1}^nk^2\right)+\dfrac{x^3}{3n^6 } \left(\sum_{k=1}^nk^3 \right)-\dfrac{x^4}{4n^8 }\left(\sum_{k=1}^nk^4 \right)+ …$$
Note that
$$\sum_{k=1}^nk^r =\dfrac{n^{r+1}}{r+1}+\dfrac{n^r}{2} + O(n^{r-1})$$
is a $(r+1)^{\text{th}}$ degree polynomial of $n.$
Therefore
$$\lim_{n\to\infty}\left(\dfrac{1}{n^{2r}} \sum_{k=1}^nk^r\right)=\left\{
\begin{array}{rl}
\frac{1}{2}:&r=1\\
0:&r>1
\end{array}
\right.
$$
Hence we can obtain $$\lim_{n\to\infty}\ln\left(\prod_{k=1}^n\left(1+\dfrac{kx}{n^2 }\right) \right)=\lim_{n\to\infty}\sum_{k=1}^n\ln\left(1+\dfrac{kx}{n^2 }\right)=\dfrac{x}{2}-0+0-0+ ...$$
$$\ln \left(\lim_{n\to\infty}\prod_{k=1}^n\left(1+\dfrac{kx}{n^2 }\right)\right)=\dfrac{x}{2}$$
$$\lim_{n\to\infty}\prod_{k=1}^n\left(1+\dfrac{kx}{n^2 }\right)=e^{x/2}$$
| {
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Analogue of $\zeta(2) = \frac{\pi^2}{6}$ for Dirichlet L-series of $\mathbb{Z}/3\mathbb{Z}$? Consider the two Dirichlet characters of $\mathbb{Z}/3\mathbb{Z}$.
$$
\begin{array}{c|ccr}
& 0 & 1 & 2 \\ \hline
\chi_1 & 0 & 1 & 1 \\
\chi_2 & 0 & 1 & -1
\end{array}
$$
I read the L-functions for these series have special values
*
*$ L(2,\chi_1) \in \pi^2 \sqrt{3}\;\mathbb{Q} $
*$ L(1,\chi_2) \in \pi \sqrt{3}\;\mathbb{Q} $
In other words, these numbers are $\pi^k \times \sqrt{3} \times \text{(rational number)}$. Is there a way to derive this similar to to the famous $\zeta(2) = \tfrac{\pi^2}{6}$ formula?
Here are 14 proofs of $\zeta(2) = \tfrac{\pi^2}{6}$ for reference
| After thinking about it for a while:
$$ L(\chi_1,2) = \tfrac{1}{1^2} + \tfrac{1}{2^2} + 0 + \tfrac{1}{4^2} + \tfrac{1}{5^2} + 0 + \dots = \sum \frac{1}{n^2} - \sum \frac{1}{(3n)^2}
= \left( 1 - \frac{1}{9}\right) \sum \frac{1}{n^2} = \frac{8}{9} \zeta(2)$$
| {
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In a triangle ABC, (b + c) cos A + (c + a) cos B + (a + b) cos C is equal to In a triangle $ABC$
$$(b + c)\cos A + (c + a)\cos B + (a + b)\cos C=?$$
| we know
$$a = b \cos C + c \cos B$$
$$b= a \cos C + c \cos A$$
$$c= b \cos A + a \cos B$$
adding together we get
$$(a+b) \cos C + (b+c) \cos A + (c+a) \cos B =a +b+c $$
| {
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Rationalize the denominator and simplify So the problem I have says rationalize the denominator and simplify.
$$ \frac{ \sqrt{15}}{\sqrt{10}-3}$$
My answer I got was $\frac{5 \sqrt6}{7}$.
Am I doing this wrong or is this the wrong answer I was told it was incorrect?
| It seems that you tried multiplying by $\frac{\sqrt{10}}{\sqrt{10}}$. Instead, you should try multiplying by the conjugate and take advantage of difference of squares:
$$
\frac{\sqrt{15}}{\sqrt{10} - 3}
= \frac{\sqrt{15}}{\sqrt{10} - 3} \cdot \frac{\sqrt{10} + 3}{\sqrt{10} + 3}
= \frac{\sqrt{150} + 3\sqrt{15}}{(\sqrt{10})^2 - 3^2}
= 5\sqrt{6} + 3\sqrt{15}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Summation of series of $2/(r-1)(r+1)$ using the method of differences Verify the identity
$$\frac{2r-1}{r(r-1)}-\frac{2r+1}{r(r+1)}=\frac{2}{(r-1)(r+1)}$$
Hence, using the method of differences, prove that
$$\sum_{r=2}^{n}\frac{2}{(r-1)(r+1)}=\frac{3}{2}-\frac{2n+1}{n(n+1)}$$
Deduce the sum of the infinite series
$$\frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+...+\frac{1}{(n-1)(n+1)}+...$$
I have done the first two parts and have recognized that half times the sum of the second part will equal the sum of the series in part three. How do I proceed?
| $$\begin{align}
S&=\frac 1{1\cdot 3}+\frac 1{2\cdot 4}+\cdots+\frac 1{(r-1)(r+1)}\cdots\\
&=\lim_{n\to\infty}\frac 12\sum_{r=2}^n \frac 2{(r-1)(r+1)}\\
&=\frac 12\lim_{n\to\infty}\sum_{r=2}^n \frac 2{(r-1)(r+1)}\\
&=\frac 12 \lim_{n\to\infty}\left[\frac 32-\frac{2n+1}{n(n+1)}\right]\\
&=\frac 12 \left[\frac 32-\lim_{n\to\infty}\frac{2n+1}{n(n+1)}\right]\\
&=\frac 34 -\frac 12\left[\lim_{n\to\infty}\frac{2n+1}{n(n+1)}\right]\\
&=\frac 34 -\frac 12\left[\lim_{n\to\infty}\frac{2n}{n^2}\right]\\
&=\frac 34 -\frac 12\underbrace{\left[\lim_{n\to\infty}\frac 1n\right]}_0\\
&=\frac 34\qquad \blacksquare \end{align}$$
| {
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Find $\sec \theta + \tan \theta$. If $\tan \theta=x-\frac{1}{x}$, find $\sec \theta + \tan \theta$.
This was the question ask in my unit test.
My Efforts:
$\tan^2 \theta=(x-\frac{1}{x})^2$
$\tan^2 \theta= (\frac {x^2-1}{x})^2$
Now we can use identity $\sec^2 \theta= 1 + \tan^2 \theta$.
But i am not able to get the answer using this.
I don't know the correct answer but I had got $2x\ or\ -\frac{2}{x} $, which was given wrong.
Also please tell me if there is better way to do this.
| You have the core of it. You are given that $ \tan \theta = x - \frac {1}{x} $, and you know that $ \sec^2 \theta = 1 + \tan^2 \theta $. Since $ \tan^2 \theta = x^2 + \frac {1}{x^2} - 2 $, you have $ \sec^2 \theta = x^2 + \frac {1}{x^2} - 1 $.
Therefore, $ \sec \theta = \sqrt {x^2 - \frac {1}{x^2} - 1} $ and $ \tan \theta = x - \frac {1}{x} $, so the answer is $$ \sec \theta + \tan \theta = x - \frac {1}{x} + \sqrt {x^2 + \frac {1}{x^2} - 1}. $$ I am not sure why you are saying the answer is $ 2x $ or $ - \frac {2}{x} $.
| {
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How do I solve $x^5 +x^3+x = y$ for $x$? I understand how to solve quadratics, but I do not know how to approach this question. Could anyone show me a step by step solution expression $x$ in terms of $y$?
The explicit question out of the book is to find $f^{-1}(3)$ for $f(x) = x^5 +x^3+x$
So far I have reduced $x^5 +x^3+x = y$ to $y/x - 3/4 = (x^2 + 1/2)^2$ or $y = x((x^2+1/2)^2 + 3/4)$ but Im still just as lost.
| In order to find $f^{-1}(3)=x$ for some $x$. We can apply the function to both sides giving us $3=f(a)$ now we need to find $x^5+x^3+x=3\implies x^5+x^3+x-3=0$. There are many ways to solve this however you can go ahead and guess that one solution would be $x=1$. Now we can perform polynomial division on $x^5+x^3+x-3$ by dividing by $x-1$ as that is a root and we get $x^4+x^3+2x^2+3$. The rest of your roots are there however it seems they are not real.
There is almost an art to guessing however one way to get a good feel is to try and plot the graph for different values. I guessed 1 because there were 3 terms with no coefficients and they evaluate to 3.
| {
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Closed-forms of infinite series with factorial in the denominator How to evaluate the closed-forms of series
\begin{equation}
1)\,\, \sum_{n=0}^\infty\frac{1}{(3n)!}\qquad\left|\qquad2)\,\, \sum_{n=0}^\infty\frac{1}{(3n+1)!}\qquad\right|\qquad3)\,\, \sum_{n=0}^\infty\frac{1}{(3n+2)!}\\
\end{equation}
Of course Wolfram Alpha can give us the closed-forms
\begin{align}
\sum_{n=0}^\infty\frac{1}{(3n)!}&=\frac{e}{3}+\frac{2\cos\left(\frac{\sqrt{3}}{2}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+1)!}&=\frac{e}{3}+\frac{2\sin\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+2)!}&=\frac{e}{3}-\frac{2\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}{3\sqrt{e}}
\end{align}
but how to get those closed-forms by hand? I can only notice that
\begin{equation}
\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{(3n)!}+\sum_{n=0}^\infty\frac{1}{(3n+1)!}+\sum_{n=0}^\infty\frac{1}{(3n+2)!}=e
\end{equation}
Could anyone here please help me? Any help would be greatly appreciated. Thank you.
PS: Please don't work backward.
| Let: $f(0)=1, f'(0)=0, f''(0)=0,f'''(x)=f(x)$
$$f(x) = \sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}$$
$$f(x) = \frac{e^x + e^{ux} + e^{vx}} {3} $$ $$u = -1/2+i\sqrt{3}/2; v= u^2 = -1/2-i\sqrt{3}/2$$
$$ f(x) = \frac{e^x}{3}+\frac{e^{-x/2}}{3}\cdot (e^{i\sqrt{3}/2}+e^{-i\sqrt{3}/2})$$
$$ f(x) = \frac{e^x}{3}+\frac{e^{-x/2}}{3}\cdot 2cos((\sqrt{3}/2) x)$$
$$ f(1) = \frac{e}{3}+\frac{2 cos(\frac{\sqrt{3}}{2})}{3 \sqrt{e}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/914176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equation*} \dfrac{5}{3} = 1 + \dfrac{2}{3}\text{.}
\end{equation*} This is because when 5 is divided by 3, 3 goes into 5
once (hence the $1$ term) and there is a remainder of $2$ (hence the
$\dfrac{2}{3}$ term). Note the following: every division problem can
be decomposed into an integer (the $1$ in this case) plus a fraction,
with the denominator being what you divide by (the $3$ in this
case).
So, when $n$ is divided by 14, the remainder is 10. This can be
written as \begin{equation*} \dfrac{n}{14} = a + \dfrac{10}{14}
\end{equation*} where $a$ is an integer.
We want to find the remainder when $n$ is divided by 7, which I'll
call $r$. So \begin{equation*} \dfrac{n}{7} = b + \dfrac{r}{7}\text{,}
\end{equation*} where $b$ is an integer.
Here's the key point to notice: notice that \begin{equation*}
\dfrac{n}{7} = \dfrac{2n}{14} = 2\left(\dfrac{n}{14}\right)\text{.}
\end{equation*} This is because $\dfrac{1}{7} = \dfrac{2}{14}$.
Thus, \begin{equation*} \dfrac{n}{7} = 2\left(\dfrac{n}{14}\right) =
2\left(a + \dfrac{10}{14}\right) = 2a + 2\left(\dfrac{10}{14}\right) =
2a + \dfrac{10}{7} = 2a + \dfrac{7}{7} + \dfrac{3}{7} = (2a+1) +
\dfrac{3}{7}\text{.} \end{equation*} So, since $a$ is an integer, $2a
+ 1$ is an integer, which is our $b$ from the original equation. Thus, $r = 3$.
To her, this method was not very intuitive. She did understand the explanation. Are there any suggestions for how I can explain this in another way?
| I wouldn't use fractions, instead use the usual division algorithm, note that every $7$ numbers, there is a multiple of $7$, ever $14$ a multiple of $14$, et cetera to motivate writing a number as
$$n=14q+r$$
with $0\le r < 14$ each time. Then say every number is also of the form
$$n=7q'+r'$$
with $0\le r'< 7$ and emphasize that clearly $r$ is unique. This is, of course, because you just count how many up you have to go from the nearest multiple of $7$, if you are $4$ more, then you are clearly not $3$ more.
If you like visuals you can demonstrate to the student with a simple list
$$\underbrace{\color{red}{0}}_{7\cdot 0},1,2,3,4,5,6,\underbrace{\color{red}{7}}_{7\cdot 1}, 8, 9, 10, 11, 12, 13, \underbrace{\color{red}{14}}_{7\cdot 2},\ldots$$
If the student knows enough about well-ordering, you can make this rigorous rather than simply intuitive since you can look at natural numbers of the form
$$\{n-7k: k\in\Bbb Z\}$$
and just define $r$ to be the minimal element of this set.
From either approach, you can write
$$14q+10=7(2q+1)+3$$
so that $q'=2q+1$ and $r'=3$.
Addendum: If you want to emphasize how things are evenly space for the other remainders, you can make the same list with different highlighting, here I'll do $14$ and highlight the related $7$ information
$$\underbrace{\color{red}{0}}_{14\cdot 0},1,2,\underbrace{\color{orange}{3}}_{7\cdot 1+3},4,5,6,7,8,9,\underbrace{\color{blue}{10}}_{14\cdot 0+10=7\cdot 1+3},11,12,13,\underbrace{\color{red}{14}}_{14\cdot 1},$$
$$15,16,\underbrace{\color{orange}{17}}_{7\cdot 2+3},18,19, 20,21,22,23,\underbrace{\color{blue}{24}}_{14\cdot 1+10=7\cdot 3+3},25,26,27,\underbrace{\color{red}{28}}_{14\cdot 2},$$
$$29,30,\underbrace{\color{orange}{31}}_{7\cdot 4+3},32,33,34,35,36,37,\underbrace{\color{blue}{38}}_{14\cdot 2+10=7\cdot 5+3},\ldots$$
This illustrates exactly how the $7q'+3$ numbers are distributed, and it's easy to see how they overlap with the $14q+10$ numbers every other one.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplifying the sum of powers of the golden ratio I seem to have forgotten some fundamental algebra. I know that:
$(\frac{1+\sqrt{5}}{2})^{k-2} + (\frac{1+\sqrt{5}}{2})^{k-1} = (\frac{1+\sqrt{5}}{2})^{k}$
But I don't remember how to show it algebraicly
factoring out the biggest term on the LHS gives
$(\frac{1+\sqrt{5}}{2})^{k-2}(1+(\frac{1+\sqrt{5}}{2}))$ which doesn't really help
| You are almost done. You have already found that
$$ ( \frac{1 + \sqrt{5}}{2} )^{k-2} + ( \frac{1 + \sqrt{5}}{2} )^{k-1} = ( \frac{1 + \sqrt{5}}{2} )^{k-2} (1 + \frac{1 + \sqrt{5}}{2} ) $$
You want to show that this quantity can be expressed as $ ( \frac {1 + \sqrt{5} }{2} )^k$.
Comparing what you have to what you need, you should be able to see that it would be sufficient to prove that
$ 1 + \frac{1 + \sqrt{5}}{2} = ( \frac {1 + \sqrt{5} }{2} )^2 $. This can be verified directly by simplifying both sides.
| {
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Prove $\frac{ab}{\sqrt{2c+a+b}}+\frac{bc}{\sqrt{2a+b+c}}+\frac{ca}{\sqrt{2b+c+a}}\le\sqrt\frac{2}{3}$ if $a+b+c=2$
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=2$. Prove: $$\frac{ab}{\sqrt{2c+a+b}}+\frac{bc}{\sqrt{2a+b+c}}+\frac{ca}{\sqrt{2b+c+a}}\le\sqrt\frac{2}{3}$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz, Hölder and AM-GM because I have background in them.
Things I have tried: I was thinking to making the denominator smaller using AM-GM.but I was not successful. My other idea was to re write LHS into this form.something like my idea on this question $$A-\frac{ab}{\sqrt{2c+a+b}}+B-\frac{bc}{\sqrt{2a+b+c}}+C-\frac{ca}{\sqrt{2b+c+a}}$$
But I was not able to observe something good.
I don't know this will lead to something useful but,here is my other idea. let $x=2c+a+b,y=2a+b+c,z=2b+c+a$. rewriting LHS:$$\sum\limits_{cyc}\frac{(3y-(x+z))(3z-(y+x))}{16\sqrt x} \le \sqrt\frac{2}{3}$$ $\sum\limits_{cyc}$denotes sums over cyclic permutations of the symbols $x,y,z$ . another thing I observed that $(x-y-z)^2-4(y-z)^2 = (3y-(x+z))(3z-(y+x))$
I looked at related problems and I think this and (Prove $\frac{a}{ab+2c}+\frac{b}{bc+2a}+\frac{c}{ca+2b} \ge \frac 98$) may have some common idea in proof with my question inequality.
Well,it seems like someone posted it a little after on AoPS.right now there is a solution there by $uvw$ and Cauchy-Schwarz. I post the starting part of solution that is with Cauchy (credits to arqady of AoPS). By Cauchy-Schwarz:$$\left(\sum_{cyc}\frac{ab}{\sqrt{2c+a+b}}\right)^2\leq(ab+ac+bc)\sum_{cyc}\frac{ab}{2c+a+b}$$
Hence, it remains to prove that: $$(ab+ac+bc)\sum_{cyc}\frac{ab}{2c+a+b}\leq\frac{(a+b+c)^3}{12}$$
I stuck here.
| Well I think figured out . I will write it down from first.
By Cauchy-Schwarz:
$$\left(\sum_{cyc}\frac{ab}{\sqrt{2c+a+b}}\right)^2\leq(ab+ac+bc)\sum_{cyc}\frac{ab}{2c+a+b}$$
Hence, it remains to prove that: $$(ab+ac+bc)\sum_{cyc}\frac{ab}{2c+a+b}\leq\frac{(a+b+c)^3}{12}$$
by the well known inequality:$$3(ab+bc+ca)\le(a+b+c)^2$$
So $$(ab+bc+ca)\le\frac{(a+b+c)^2}{3}$$
is true.it remain to show that $$\sum_{cyc}\frac{ab}{2c+a+b}\le \frac{(a+b+c)}{4}=\frac{1}{2}$$
by well known inequality $$\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}$$
So $$\frac{ab}{c+a}+\frac{ab}{c+b}\ge\frac{4ab}{2c+a+b} \rightarrow \frac{1}{4}(\frac{ab}{c+a}+\frac{ab}{c+b}) \ge\frac{ab}{2c+a+b} $$
so we can conclude that $$\frac{1}{4}\sum\limits_{cyc}(\frac{ab}{c+a}+\frac{ab}{c+b})=\frac{1}{4}(a+b+c)=\frac{1}{2}\ge\sum_{cyc}\frac{ab}{2c+a+b}$$
Done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Test the convergence of the series $\sum\limits_nn^p(\sqrt{n+1}-2\sqrt n + \sqrt{n-1})$ I am trying to test the convergence of the series
$$\sum_{n=1}^\infty n^p(\sqrt{n+1}-2\sqrt n + \sqrt{n-1})$$
$p$ is a fixed real number.
I tried the ratio test and the integral test without success. Now, I am stuck with the quantity between the round brackets. I have no idea on how to treat it in order to obtain something useful to work with.
| In essence the problem is to determine the order of
$$\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}$$
Tring the ususal trick of multiplying by the conjuguate we get
$$\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}
=\frac{2\sqrt{n^2-1}-2n}{\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}}$$ then doing the same again we get
$$\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}
=\frac{-2}{(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n})(\sqrt{n^2-1}+n)}$$
Now the order of $(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n})(\sqrt{n^2-1}+n)$ is $n^{\frac{3}{2}}$ as can be easily verified, since
$$\frac{\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}}{\sqrt{n}}\rightarrow 4$$ and
$$\frac{\sqrt{n^2-1}+n}{n}\rightarrow 2$$
So if we compare our series with the series $\sum n^{p-\frac{3}{2}}$
we see that we get a finite ratio. Thus the criteria for convergence is
$$p-\frac{3}{2}<-1$$
or $$p<\frac{1}{2}$$
| {
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Avoiding multiplication in inequality For $x,y,z\in \mathbb R^{+}$, prove that:
$$(x+y+z)^2(xy+yz+zx)^2\le3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+xz+x^2)$$
I have been trying to attack this problem by setting $a=x^2+xy+y^2, b=y^2+yz+z^2, c=y^2+yz+z^2$ and searching for those terms in the LHS using:
$$3(xy+yz+zx)\le 2(x^2+y^2+z^2)+xy+yz+zx=a+b+c$$
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\le a+b+c$$
But I think there is no hope in this method since I would be left with a $3abc$ in the RHS, which is almost never greater than something. But I haven't come up with another way to avoid multiplying the RHS(since using $x^2+y^2\ge2xy$ is not productive). Another idea is to use some sort of geometric argument by setting lengths $x,y,z$ with a $120°$ angle between them, but it hasn't worked so far. Any help would be greatly appreciated.
| Your geometry idea is nice, infact it works :) .
Consider a triangle with sides $ABC$ in which there is a point $H$ such that $AH=x, BH=y, CH=z$ and $\angle AHB = \angle BHC = \angle CHA = 120^{\circ}$. Then using cosine, rule gives $a^2= y^2+z^2+yz , b^2=x^2+z^2+xz , c^2 = x^2+y^2+xy $. Also since sum of area of these three triangles, equals area of $ABC$ this implies that, $xy+yz+xz= \frac{4A}{\sqrt{3}}$. Now, $(x+y+z)^2= \frac{a^2+b^2+c^2+4A\sqrt{3}}{2} $.
So given inequality is equivalent to, $$ \frac{a^2+b^2+c^2+4A\sqrt{3}}{2} \times \frac{16A^2}{3} \le 3a^2b^2c^2 $$
Writing, $abc = 4AR$ , this inequality is equivalent to,
$$a^2+b^2+c^2+4\sqrt{3}A \le 18R^2 $$
This is true, because distance between circumcentre of triangle and orthocentre of triangle is precisely $\sqrt{9R^2 -a^2-b^2-c^2}$ . And so, $9R^2 \ge a^2+b^2+c^2 $. And it is very well known that $a^2+b^2+c^2\ge 4\sqrt{3}A $ $\Box$
| {
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Proof by Induction - Algebra Problem (Steps included but not understood) I do not quite understand this proof, if anyone could explain the steps for me it would be greatly appreciated. It's probably something glaringly obvious I'm not seeing, thanks in advance.
Prove that for every integer $n \ge 0,$ the number $4^{2n+1}+3^{n+2} $ is a multiple of 13.
Proof. We use induction on n, starting with $n=0$
$P(0):4^{2(0)+1}+3^{0+2}=4+9=13=13\cdot1$
Assume $P(k):4^{2k+1}+3^{k+2}=13t$ for some integer $t$. We must prove
$P(k+1): 4^{2(k+1)+1}+3^{(k+1)+2}$ is a multiple of $13$.
We have
$4^{2(k+1)+1}+3^{(k+1)+2}=4^{(2k+1)+2}+3^{(k+2)+1}$
$=4^2(4^{2k+1})+4^2(3^{k+2}-3^{k+2})+3\cdot3^{k+2}$
$=4^2(4^{2k+1}+3^{k+2})+3^{k+2}(-4^2+3)$
$=16\cdot13t+3^{k+2}\cdot(-13)$ (by $P(k)$)
$=13(16t-3^{k+2})$, proven.
| We know something about $4^{2k+1}+3^{k+2}$, but in the expression
$$
4^2\cdot 4^{2k+1}+3\cdot3^{k+2}
$$
we have just $4^{2k+1}$. So we insert a term $4^2\cdot 3^{k+2}$ in order to apply the induction hypothesis. We add nothing provided we subtract the same term:
$$
4^2\cdot 4^{2k+1}+3\cdot3^{k+2}=4^2\cdot 4^{2k+1}+4^2\cdot 3^{k+2}-4^2\cdot 3^{k+2}+3\cdot3^{k+2}
$$
and go on with
$$
4^2\cdot 13t-3^{k+2}(16-3)
$$
which is a multiple of $13$.
A different way of doing the same thing is rewriting the induction hypothesis as
$$
4^{2k+1}=13t-3^{k+2}
$$
and substituting it in the expression for $P(k+1)$:
$$
4^2\cdot 4^{2k+1}+3\cdot3^{k+2}=4^2(13t-3^{k+2})+3\cdot3^{k+2}.
$$
| {
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$\int_0^{2\pi} \sqrt{a^2 +b^2+2ab \cos\varphi}\,\mathrm{d}\varphi$ $$\int_0^{2\pi} \sqrt{a^2 +b^2+2ab \cos\varphi}\,\mathrm{d}\varphi$$
Where $a$ and $b$ are constants.
I had to find the distance travelled by a point at a distance of $b$ from the centre of a rolling disc with radius $a$ in one full rotation. I got the final answer as this integral, but I was not able to solve this integral.
| First consider the integrand
\begin{align}
\sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)}
\end{align}
which can be seen as
\begin{align}
\sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)} &= \sqrt{(a+b)^{2} - 4ab \sin^{2}\left(\frac{\phi}{2}\right) } \\
&= (a+b) \sqrt{ 1 - \left( \frac{4ab}{(a+b)^{2}} \right) \, \sin^{2}\left( \frac{\phi}{2}\right) }.
\end{align}
Now,
\begin{align}
I &= \int_{0}^{2\pi} \sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)} \, d\phi \\
&= (a+b) \, \int_{0}^{2\pi} \sqrt{ 1 - \left( \frac{4ab}{(a+b)^{2}} \right) \, \sin^{2}\left( \frac{\phi}{2}\right) } \, d\phi \\
&= 2(a+b) \int_{0}^{\pi} \sqrt{1- k^{2} \sin^{2}(x) } \, dx
\end{align}
where $(a+b)^{2} k^{2} = 4ab$. The integral remaining if the complete elliptical integral of the second kind. The "solution" of the integral is
\begin{align}
I = 4(a+b) E\left(\frac{4ab}{(a+b)^{2}}\right).
\end{align}
Hence,
\begin{align}
\int_{0}^{2\pi} \sqrt{a^{2} + b^{2} + 2 a b \cos(\phi)} \, d\phi = 4(a+b) E\left(\frac{4ab}{(a+b)^{2}}\right).
\end{align}
| {
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Show that $\sum_{k=0}^{\infty} \binom{r}{k} \binom{s}{n+k} = \binom{r+s}{r+n}$. I can't resolve this exercise and I need some tips.
Let be $n$ a integer, $s$ a real number and $r \geq 0$ a integer. Show that
$$ \sum_{k=0}^{\infty} \binom{r}{k} \binom{s}{n+k} = \binom{r+s}{r+n} $$
| Suppose we seek to evaluate
$$\sum_{k=0}^r {r\choose k} {s\choose n+k}.$$
Start from
$${s\choose n+k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+k+1}} (1+z)^s \; dz.$$
This gives the following integral for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{k=0}^r {r\choose k}
\frac{1}{z^{n+k+1}} (1+z)^s \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^s}{z^{n+1}}
\sum_{k=0}^r {r\choose k}
\frac{1}{z^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^s}{z^{n+1}}
\left(1+\frac{1}{z}\right)^r \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^s}{z^{n+1}}
\frac{(1+z)^r}{z^r} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{r+s}}{z^{r+n+1}} \; dz.$$
This last integral can be evaluated by inspection and is given by
$$[z^{r+n}] (1+z)^{r+s} = {r+s\choose r+n}.$$
This obviously confirms the combinatorial proof, which is simple.
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
| {
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Finding asymptotes for $f(x)=\frac{x^2+3x-10}{3x^2+13x-10}$ $$f(x)=\frac{x^2+3x-10}{3x^2+13x-10}$$
I know that the horizontal asymptote is $1/3$. To find the vertical asymptotes, I set the denominator equal to zero and used the quadratic formula, and I got $-5$ and $2/3$, and this is wrong.
How do you find the vertical asymptotes for this problem?
thanks.
| $3x^2 + 13x - 10 = (3x - 2)(x + 5)$, and $x^2 + 3x - 10 = (x + 5)(x - 3)$. Thus only $x = \dfrac{2}{3}$ is the vertical asymptote since $\displaystyle \lim_{x \to -5} f(x) = \dfrac{8}{17} \neq \infty$
| {
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Summation of series Find $\sum_1^n$ $\frac {2r+1}{r^2(r+1)^2}$
Also, find the sum to infinity of the series.
I tried decomposing it into partial fractions of the form $\frac Ar$ + $\frac{B}{r^2}$ + $\frac{C}{(r+1)}$ + $\frac{D}{(r+1)^2}$ but it was getting too complicated and tedious. Is there some trick here that i'm missing?
| You're on the right track! Clearing the fractions, we have:
$$
2r + 1 = Ar(r + 1)^2 + B(r + 1)^2 + Cr^2(r + 1) + Dr^2
$$
Comparing coefficients, we have:
\begin{align*}
\boxed{r^3}:\qquad 0 &= A + C \\
\boxed{r^2}:\qquad 0 &= 2A + B + C + D \\
\boxed{r^1}:\qquad 2 &= A + 2B \\
\boxed{r^0}:\qquad 1 &= B \\
\end{align*}
Substituting, notice that $B = 1 \implies A = 0 \implies C = 0 \implies D = -1$. Hence, we obtain a telescoping series:
$$
\sum_{r=1}^\infty \frac {2r+1}{r^2(r+1)^2} = \sum_{r=1}^\infty \left[\frac{1}{r^2} - \frac{1}{(r + 1)^2} \right]
= \lim_{n\to\infty} \left[\frac{1}{1^2} - \frac{1}{(n + 1)^2} \right] = 1
$$
| {
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Solve system of equations with $\sin$ and $\cos$ Solve system of equations
$\begin{cases}
3x^2 + \sin 2y - \cos y - 3 = 0 \\
x^3 - 3x - \sin y - \cos 2y + 3 = 0
\end{cases}$
I tried to use substitution $x = \cos t$ or sth, but I get literally nothing
| Invoking the double-angle formulas $\sin 2 y = 2 \sin y \cos y$ and $\cos 2 y = 1 - 2 \sin^2 y$, and abbreviating "$\sin y$" and "$\cos y$" by "$s$" and "$c$", we can manipulate the equations into these forms
$$\begin{align}
c \;( 2 s - 1 ) &= - 3 ( x^2 - 1 ) &= (x-1)\;P \tag{1}\\
s \;( 2 s - 1 ) &= -(x - 1 )^2 ( x + 2 ) &= (x-1)\;Q \tag{2}
\end{align}$$
where $P := -3(x+1)$ and $Q := -(x-1)(x+2)$. From here, we see a family of solutions that make each side of $(1)$ and $(2)$ vanish simultaneously:
$x=1$, and all $y$ such that $\sin y = \frac12$ (in particular, $y = \frac\pi6$ and $y = \frac{5\pi}{6}$)
We also see that there are no other solutions with either $x=1$ or $\sin y = \frac12$. Moreover, there are no other solutions with either $P=0$ or $Q=0$: if $x=-1$, then $(2)$ gives an impossible relation, $c(2s-1)=-4$; likewise, if $x=-2$, then $(1)$ gives $c(2s-1) = -9$. Consequently, we can write
$$\frac{c}{P} = \frac{x-1}{2s-1} = \frac{s}{Q}$$
which implies
$$c = \pm\frac{P}{R}\qquad s = \pm\frac{Q}{R}$$
where $R := \sqrt{P^2+Q^2}$ and the "$\pm$"s match.
Substituting into $(1)$ (or $(2)$), then,
$$2 Q \mp R = ( x - 1 ) R^2 \quad\to\quad \left(\;2 Q - ( x - 1 )R^2\;\right)^2 = R^2$$
Rewriting $Q$ and $R^2 = P^2 + Q^2$ in terms of $x$, we have
$$x^{10}+ 2 x^9 + 9 x^8 + 28 x^7 + 38 x^6 + 48 x^5 + 73 x^4 - 118 x^3 - 345 x^2 - 48 x + 276 = 0$$
for which Mathematica finds two real roots
$$x_1 \approx 1.1352 \qquad x_2 \approx 0.848153$$
The corresponding values of $Q/P$ give the tangents of the associated $y$s:
$$\tan y_1 = \frac{-0.423884}{-6.4056} = \phantom{-}0.0661739$$
$$\tan y_2 = \frac{\phantom{-}0.432483}{-5.54446} = -0.0780028$$
One can check that
Only first-quadrant solutions $y_1$ and second-quadrant solutions $y_2$ actually satisfy the original equations.
Here's a graph of one period of the situation:
Equation $(1)$ defines the blue squiggles (right and middle); equation $(2)$ defines the red squiggle (left) and ovals. The line $x=1$ appears in gold.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The limit of $f(x)= \sqrt{x^2+4x+3} +x$ as $x\to\infty$ The problem is to find $\lim_{x\to\infty} \sqrt{x^2+4x+3} +x$.
Do I just divide everything by $x^2$ and get limit $= \sqrt{1}+0=1$?
| Since $x$ is large $$f(x)= \sqrt{x^2+4x+3} +x= \sqrt{x^2\Big(1+\frac{4}{x}+\frac{3}{x^2}\Big)}+x=x\sqrt{1+\frac{4}{x}+\frac{3}{x^2}}+x$$ Now consider $\sqrt{1+y}$ when $y$ is small; the Taylor expansion built at $y=0$ gives $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Relace now $y$ by $(\frac{4}{x}+\frac{3}{x^2})$; so $$\sqrt{1+\frac{4}{x}+\frac{3}{x^2}}=1+\frac{2}{x}-\frac{1}{2
x^2}+O\left(\left(\frac{1}{x}\right)^3\right)$$ So $$x\sqrt{1+\frac{4}{x}+\frac{3}{x^2}}+x \simeq 2x+2-\frac{1}{2
x}$$ and then $f(x$ behaves just as $(2x+2)$ which is an oblique asymptote for the curve which stays below its asymptote
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum area of Triangle $\Delta PAB$ In Complex plane $A$ and $B$ are two points given by $z_1=5-2i$ and $z_2=1+i$ and if $P(z)$ is any Point such that $$|z-z_1|=2|z-z_2|$$ Find the Maximum area of Triangle $\Delta PAB$.
I have done this problem algebraically using Heron's Formula taking $PA=2x$ and $PB=x$ and $AB=5$ . Is there any geometrical way of doing this?
| First, the area of a triangle $\Delta PAB$ is unchanged by translations or rotations of the complex plane. Since $|AB|=5$, we may therefore choose coordinates so that the points $A$ and $B$ are instead at $0$ and $5$ respectively. Writing the third point as $z=x+I y$, the area of $\Delta PAB$ is then $\frac{5}{2}y$ since the base and height are $5$ and $y$ respectively, and our goal becomes to maximize $y$.
To do so, note that the condition for $P(z)$ may be expressed as
$$|z|=2|z-5|\implies x^2+y^2=4(x-5)^2+4y^2\implies \left(x-\frac{20}{3}\right)^2+y^2=\left(\frac{10}{3}\right)^2$$
where we have squared both sides and then completed the squares. Thus $y^2=\left(\frac{10}{3}\right)^2-\left(x-\frac{20}{3}\right)^2\leq \left(\frac{10}{3}\right)^2,$ so $|y|\leq \frac{10}{3}$ (with equality iff $x=\frac{20}{3}$) and we conclude that $\Delta PAB$ has an area of at most $\frac{5}{2}\cdot \frac{10}{3}=\frac{25}{3}$.
| {
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Show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$ If $a^3+b^3+c^3=3abc$ and $a+b+c=0$ show that $\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b)^2}{3ab}=1$
| $$\frac{(b+c)^2} {3bc}+\frac{(c+a)^2}{3ac}+\frac{(a+b^2)}{3ab}\\=\frac{a(b+c)^2} {3abc}+\frac{b(c+a)^2}{3abc}+\frac{c(a+b^2)}{3abc}\\=\frac{a^3} {3abc}+\frac{b^3}{3abc}+\frac{c^3}{3abc}\\=\frac{a^3+b^3+c^3}{3abc}\\=\frac{3abc}{3abc}=1$$
| {
"language": "en",
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How to show that $a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$ How do you show that
$$a+b> \sqrt{a^2+b^2-ab}, \qquad a, b >0$$
I could write $\sqrt{a^2+b^2-ab}=\sqrt{(a+b)^2-3ab}$, but this seems to lead nowhere.
| Note that $2ab > -ab \Rightarrow a^2+b^2+2ab > a^2 + b^2 -ab \Rightarrow (a+b)^2 > a^2 + b^2 -ab \Rightarrow a+b > \sqrt{a^2 + b^2 -ab}$
I was able to do this by squaring the original inequality I was supposed to prove, and then I worked backwards.
| {
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How to solve the integral $\int \frac {(x^2 +1)}{x^4- x^2 +1} dx$ I have started this problem but I'm not completely sure I'm going down the right path with it.
So far I have completed the square in the denominator.
$x^4-x^2+1= (x^2-1/2)^2+\frac{3}{4}$
Then, let $u=x^2-\frac{1}{2}$ so $x=\sqrt(u+\frac{1}{2})$
$\int\frac{x^2+1}{x^4-x^2+1}dx = \int\frac{u+\frac{1}{2}+1}{u^2+\frac{3}{4}}du
=\int\frac{u}{u^2+\frac{3}{4}} +\frac{\frac{3}{2}}{u^2+\frac{3}{4}}du$
| Split it :
$\int \frac {(x^2 +1)}{x^4- x^2 +1} dx=\int\left(\frac{3 x^{2}}{x^{6} + 1} + \frac{1}{x^{2} + 1}\right)dx$
Then remember that $\arctan(x)'=\frac{1}{1+x^2}$
| {
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If $a+b+c=1$ and $abc>0$, then $ab+bc+ac<\frac{\sqrt{abc}}{2}+\frac{1}{4}.$ Question:
For any $a,b,c\in \mathbb{R}$ such that $a+b+c=1$ and $abc>0$, show that
$$ab+bc+ac<\dfrac{\sqrt{abc}}{2}+\dfrac{1}{4}.$$
My idea: let
$$a+b+c=p=1, \quad ab+bc+ac=q,\quad abc=r$$
so that
$$\Longleftrightarrow q<\dfrac{\sqrt{r}}{2}+\dfrac{1}{4}$$
Note this $a,b,c\in \mathbb{R}$, so we can't use schur inequality such
$$p^3-4pq+9r\ge 0, \quad pq\ge 9r$$
and so on
maybe can use AM-GM inequality to solve it.
| At least one of $a, b, c$ must be positive. Assume $a > 0$. Then we have $a^2 + ab + ac = a(a + b + c) = a$, so the inequality to be proved is equivalent to
$$bc + a - a^2 < \frac{1}{2}\sqrt{abc} + \frac{1}{4},$$
which can be rewritten as $f(\sqrt{bc}) < 0$, where
$$f(x) = x^2 - \frac{1}{2}\sqrt{a}x - a^2 + a - \frac{1}{4}.$$
Now $b$ and $c$ are subject only to the conditions $b + c = 1 - a$ and $bc > 0$. Thus $\sqrt{bc}$ varies in the interval $(0,\frac{1}{2}|1-a|]$.
Since $f(x)$ is a quadratic polynomial with positive leading coefficient, to check the inequality, it's enough to check it at its endpoints, namely, $f(0) \leq 0$ and $f(\frac{1}{2}|1-a|) < 0$. But $f(0) = -(a-\frac{1}{2})^2$, so the first of these inequalities is clear. Therefore, we need only prove that for all $a > 0$, we have
$$\frac{1}{4}(a-1)^2 -\frac{1}{4}\sqrt{a}|1-a| -a^2 +a - \frac{1}{4} < 0,$$
which we can rewrite as
$$-\frac{3}{4}a^2 +\frac{1}{2}a - \frac{1}{4}\sqrt{a}|1-a| < 0.$$
As the sum of the first two terms is already negative for $a \geq 1$, we need only consider the case $0 < a < 1$. Dividing by $t = \frac{1}{4}\sqrt{a}$, the inequality becomes
$$-3t^3 + t^2 + 2t - 1 < 0,$$
which must be proved for all $t \in (0,1)$ in order to conclude the proof.
This is straightforward to do with calculus.
| {
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If $x^2 + y^2 + Ax + By + C = 0 $. Find the condition on $A, B$ and $C$ such that this represents the equation of a circle.
If $x^2 + y^2 + Ax + By + C = 0 $. Find the condition on $A, B$ and $C$ such that this represents the equation of a circle.
Also find the center and radius of the circle.
Here's my solution, I'm not sure if it's correct or not (specifically the conditions on $A$, $B$ and $C$. I feel that my conditioning is invalid and that this may be some sort of triangle inequality case.
I grouped the $x^2$ and $Ax$ terms, and the $y^2$ and By terms together and equated them to -$C$.
After completing the square I am left with the following expression:
$(x + A/2)^2 + (y + B/2)^2 = -C + A^2/4 - B^2/4$
Conditions on A,B and C:
****CORRECTION MADE (Redundant inequalities need not be included)****
$C < (A^2 + B^2)/4$
Radius of the circle is $\sqrt{-C + A^2/4 + B^2/4}$
The center of the circle is $(-A/2, -B/2)$.
Correct?
| Completing the square is the right thing to do. We get
$$(x + A/2)^2 + (y + B/2)^2 = -C + A^2/4 + B^2/4$$
(I have corrected a little typo in the post).
For a circle, we need to have the right-hand side positive, or if you admit degenerate circles, non-negative. The condition I would use is
$$\frac{A^2}{4}+\frac{B^2}{4}-C\gt 0,$$
or something equivalent to that, such as $A^2+B^2\gt 4C$. No other condition is needed. (Note that in the post, the inequality runs the wrong way.)
The centre is right. The radius is $\sqrt{\frac{A^2}{4}+\frac{B^2}{4}-C}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\binom{2n}{n}\leq 4^n$ I would like to prove the following inequality, for $n=0,1,2,...$, $$ \binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof.
| A bit more can be proven with a bit more work. For $k\ge0$, we have the inequality
$$
\begin{align}
\left(\frac{k+\frac12}{k+1}\right)^2
&=\frac{k^2+k+\frac14}{k^2+2k+1}\\
&\le\frac{k+1}{k+2}\tag{1}
\end{align}
$$
because cross-multiplication gives $k^3+3k^2+\frac94k+\frac12\le k^3+3k^2+3k+1$.
Using $(1)$ yields
$$
\begin{align}
\frac{\binom{2k+2}{k+1}}{\binom{2k}{k}}
&=4\frac{k+\frac12}{k+1}\\
&\le4\sqrt{\frac{k+1}{k+2}}\tag{2}
\end{align}
$$
Multiplying $(2)$ for $k=0$ to $k=n-1$, we get
$$
\boxed{\bbox[5px]{\displaystyle\binom{2n}{n}\le\frac{4^n}{\sqrt{n+1}}}}\tag{3}
$$
As Olivier Oloa comments, Stirling's Formula tells us that
$$
\lim_{n\to\infty}\binom{2n}{n}\frac{\sqrt{\pi n}}{4^n}=1\tag{4}
$$
In fact, using inequalities similar to $(2)$, in this answer, it is shown that
$$
\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}}}\tag{5}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The area not covered by six pointed star In a circle with radius $r$, two equi triangles overlapping each other in the form of a six pointed star touching the circumference is inscribed! What is the area that is not covered by the star?
Progress
By subtracting area of the star from area of circle , the area of the surface can be found! But how to calculate the area of the star?
| Answer:
The required Area = Area of the Circle - (2*Area of the Equilateral Triangle - Area of the Hexagon that is formed by the superimposition of two equilateral trianle)
Area of the Circle $=\pi r^2$
Length of the equilateral triangle $\sqrt{3}r$
Area of the equilateral triangle then is = $\frac{3\sqrt{3}r^2}{4}$
Length of the hexagon (l) $= \frac{r}{\sqrt{3}}$
Area of the hexagon = $\frac{3\sqrt{3}l^2}{2}$
Thus Area of the hexagon = $\frac{3\sqrt{3}r^2}{6}$
The required area of the part of the circle uncovered by the six sided star = $(\pi r^2 - (2*\frac{3\sqrt{3}r^2}{4} - \frac{3\sqrt{3}r^2}{6}))$
which reduces to $$ = r^{2}(\pi -\sqrt{3})$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{1}{1^4}+\frac{1}{2^4}+\cdots+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ I have to show: $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ for natural $n$
I tried to show it by induction (but I think it could be possible to show it using some ineqaulity of means) so for $n=1$ we have $1=1$ so inequality holds then I assume it's true for $n$ and for $n+1$ my thesis is $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{(n+1)^4} \le 2-\frac{1}{\sqrt{n+1}}$
I know that:
$\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{(n+1)^4} \le 2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}$ but later I'm not sure if I have to show
$\displaystyle2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}\le2-\frac{1}{\sqrt{n+1}}$ or should be $\ge$
| If we are allowed to use tools from the calculus, we can show that the result holds at $n=1,2$, and then use the fact that
$$\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\cdots\lt \int_2^\infty x^{-4}\,dx.$$
From this we can conclude that our sum $S_n$ is always less than $1+\frac{1}{16}+\frac{1}{24}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$ Evaluation of $\displaystyle \int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$
$\bf{My\; Solution::}$ Given $\displaystyle \int\frac{1}{\sin^2 x\cdot (5+4\cos x)}dx = \int \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}dx$
Now Using Partial fraction for $\displaystyle \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}$
Now Let $\cos x= y\;,$ and Let $\displaystyle \frac{1}{(1-y)(1+y)(5+4y)} = \frac{A}{1+y}+\frac{B}{1-y}+\frac{C}{5+4y}$
after solving We Get $\displaystyle A = \frac{1}{2}$ and $\displaystyle B = -\frac{1}{18}$ and $\displaystyle C = -\frac{16}{9}$
So $\displaystyle \int \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}dx = \frac{1}{2}\int \frac{1}{1+\cos x}dx - \frac{1}{18}\int\frac{1}{1-\cos x}dx - \frac{16}{9}\int \frac{1}{5+4\cos x}dx$
And after that we can solve easily Like for
$\displaystyle \int\frac{1}{1+\cos x}dx = \int\frac{1-\cos x}{\sin^2 x}dx = \int \left(\csc^2 x-\csc x\cdot \cot x\right)dx = -\cot x +\csc x+\mathcal{C}$
My Question is , Is there is any other method by which we can solbe the above question.
OR without using partial fraction,
Thanks
| Yes there exists another one(OOps PF!):
$$\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx
\stackrel{t=\tan x/2}=\int\frac{t^6+3 t^4+3 t^2+1}{4 t^4+36 t^2}dx\\
= \int(t^2/4+128/(9 (t^2+9))+1/(36 t^2)-3/2)dx=...$$
Similiar to your method. Your's is best, why are you looking for other methods?
| {
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Find the value of $f(x)$ for $x = 2 + 2^{2/3} + 2^{1/3}$ If $x = 2 + 2^{2/3} + 2^{1/3}$, then find the value of $f(x)=x^3 - 6x^2 + 6x$.
I am unable to get to the answer - end up with more than one term. Please help me solve this!
| Letting $t=2^{1/3}\Rightarrow t^3=2$, we have
$$(x-2)^3=(t^2+t)^3$$$$\Rightarrow x^3-6x^2+12x-8=t^6+3t^5+3t^4+t^3=4+6t^2+6t+2.$$
Then, we have
$$\begin{align}x^3-6x^2+6x&=(6t^2+6t+6)+8-6x\\&=6t^2+6t+14-6(t^2+t+2)\\&=14-12\\&=2.\end{align}$$
| {
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Integration using exponent What could be the techniques we need to use to solve this integration
$\displaystyle \int\tan^2\theta\frac{\sin^2(\sec\theta\tan\theta)}{\sec^2\theta}d\theta \tag1$?
How do I convert this in to a solvable standard form ?
| $\int\tan^2\theta\dfrac{\sin^2(\sec\theta\tan\theta)}{\sec^2\theta}d\theta$
$=\int\sin^2\theta~\sin^2(\sec\theta\tan\theta)~d\theta$
$=\int\dfrac{(1-\cos^2\theta)(1-\cos(2\sec\theta\tan\theta))}{2}d\theta$
$=\int\dfrac{(\cos^2\theta-1)}{2}\sum\limits_{n=1}^\infty\dfrac{(-1)^n4^n\sec^{2n}\theta\tan^{2n}\theta}{(2n)!}d\theta$
$=\int\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}\sec^{2n-2}\theta\tan^{2n}\theta}{(2n)!}d\theta-\int\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}\sec^{2n}\theta\tan^{2n}\theta}{(2n)!}d\theta$
$=-\int\tan^2\theta~d\theta+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+3}\sec^{2n+2}\theta\tan^{2n+4}\theta}{(2n+4)!}d\theta+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}\sec^{2n+2}\theta\tan^{2n+2}\theta}{(2n+2)!}d\theta$
$=\int(1-\sec^2\theta)~d\theta+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+3}\sec^{2n}\theta\tan^{2n+4}\theta}{(2n+4)!}d(\tan\theta)+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}\sec^{2n}\theta\tan^{2n+2}\theta}{(2n+2)!}d(\tan\theta)$
$=\int(1-\sec^2\theta)~d\theta+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+3}(1+\tan^2\theta)^n\tan^{2n+4}\theta}{(2n+4)!}d(\tan\theta)+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}(1+\tan^2\theta)^n\tan^{2n+2}\theta}{(2n+2)!}d(\tan\theta)$
$=\int(1-\sec^2\theta)~d\theta+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+3}C_k^n\tan^{2k}\theta\tan^{2n+4}\theta}{(2n+4)!}d(\tan\theta)+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+1}C_k^n\tan^{2k}\theta\tan^{2n+2}\theta}{(2n+2)!}d(\tan\theta)$
$=\int(1-\sec^2\theta)~d\theta+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+3}n!\tan^{2n+2k+4}\theta}{(2n+4)!k!(n-k)!}d(\tan\theta)+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+1}n!\tan^{2n+2k+2}\theta}{(2n+2)!k!(n-k)!}d(\tan\theta)$
$=\theta-\tan\theta+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+3}n!\tan^{2n+2k+5}\theta}{(2n+4)!k!(n-k)!(2n+2k+5)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+1}n!\tan^{2n+2k+3}\theta}{(2n+2)!k!(n-k)!(2n+2k+3)}+C$
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Simplification of an expression containing $\operatorname{Li}_3(x)$ terms In my computations I ended up with this result:
$$\mathcal{K}=78\operatorname{Li}_3\left(\frac13\right)+15\operatorname{Li}_3\left(\frac23\right)-64\operatorname{Li}_3\left(\frac15\right)-102
\operatorname{Li}_3\left(\frac25\right)+126\operatorname{Li}_3\left(\frac35\right)\\+12\operatorname{Li}_3\left(\frac45\right)-89\operatorname{Li}_3\left(\frac16\right)-152\operatorname{Li}_3\left(\frac56\right)+63\operatorname{Li}_3\left(\frac38\right)+76\operatorname{Li}_3\left(\frac58\right).$$
I wonder if it's possible to simplify this expression somehow, e.g. to combine some trilogarithm terms to express them using logarithms, or at least to reduce the number of terms?
I tried to apply identities given at MathWorld and Wolfram Functions, but could not make the overall expression simpler. Mathematica could not simplify it either.
| Too long for a comment, but a related thing, really far from the solution.
$$\operatorname{Li}_3\left(\frac{3}{8}\right)+\operatorname{Li}_3\left(\frac{5}{8}\right)-\operatorname{Li}_3\left(\frac{3}{5}\right) = \frac{1}{6} \ln^3\left(\frac{5}{8}\right)-\frac{1}{2}\ln\left(\frac{3}{8}\right)\ln^2\left(\frac{5}{8}\right) + \frac{\pi^2}{6}\ln\left(\frac{5}{8}\right) + \zeta(3) - \frac{1}{4} \operatorname{Li}_3\left(\frac{9}{25}\right).$$
I get it by using this identitiy for $z:=3/8$.
An other one for $z:=1/6$ with the same identity.
$$\operatorname{Li}_3\left(\frac{1}{6}\right)+\operatorname{Li}_3\left(\frac{5}{6}\right)-\operatorname{Li}_3\left(\frac{1}{5}\right) = \frac{1}{6} \ln^3\left(\frac{5}{6}\right)-\frac{1}{2}\ln\left(\frac{1}{6}\right)\ln^2\left(\frac{5}{6}\right) + \frac{\pi^2}{6}\ln\left(\frac{5}{6}\right) + \zeta(3) - \frac{1}{4} \operatorname{Li}_3\left(\frac{1}{25}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/935366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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How do I solve $\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$? I'm having trouble finding this limit:
$$\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$$
I tried multiplying by the conjugate:
$$\lim_{x\to -\infty}(\frac{\sqrt{x^2 + x + 1} + x}{1} \times \frac{\sqrt{x^2 + x + 1} - x}{\sqrt{x^2 + x + 1} - x}) = \lim_{x\to -\infty}(\frac{x + 1}{\sqrt{x^2 + x + 1} - x})$$
And multiplying by $\frac{\frac{1}{x}}{\frac{1}{x}}$
$$\lim_{x\to -\infty}(\frac{x + 1}{\sqrt{x^2 + x + 1} - x} \times \frac{\frac{1}{x}}{\frac{1}{x}}) = \lim_{x\to -\infty}(\frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - 1})$$
That gives me $\frac{1}{0}$. WolframAlpha, my textbook, and my estimate suggest that it should be $-\frac{1}{2}$. What am I doing wrong?
(Problem from the 2nd chapter of Early Transcendentals by James Stewart)
| Things will be even clearer if we set $-\dfrac1x=h,$
$\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$
$=\lim_{h\to0^+}\dfrac{\sqrt{1-h+h^2}-1}h\ \ \ \ (1)$ as $h>0$
$=\lim_{h\to0^+}\dfrac{(1-h+h^2)-1}h\cdot\dfrac1{\lim_{h\to0^+}(1-h+h^2)+1}$
$=\lim_{h\to0^+}(h-1)\cdot\dfrac1{(1-0+0^2)+1}$ cancelling $h$ as $h\ne0$ as $h\to0^+$
$=\dfrac{0-1}{1+1}$
Alternatively, $(1)$ can also expressed $$\frac{d(\sqrt{1-x+x^2})}{dx}_{(\text{ at }x=0)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\dfrac{\tan(x+y)-\tan x}{1+\tan(x+y)\tan x}=\tan y$ Edit: got it, silly mistakes :)
I need to prove that $\dfrac{\tan(x+y)-\tan x}{1+\tan(x+y)\tan x}=\tan y$
$$=\frac{\tan x+\tan y-\tan x+\tan^2x\tan y}{1-\tan x\tan y+\tan^2x+\tan x\tan y}$$
$$=\frac{\tan y+\tan^2x\tan y}{1+\tan^2x}$$
$$=\frac{(\tan y)(1+\tan^2x)}{(1+\tan^2x)}$$
$$\boxed{=\tan y}$$
Any hints/help appreciated.
Thanks
| $$\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$$
$$a=\tan x\;\;and\;\;b=\tan y$$
$$\dfrac{\tan(x+y)-\tan x}{1+\tan(x+y)\tan x}=\frac{\frac{a+b}{1-ab}-a}{1+\frac{a+b}{1-ab}a}$$
$$=\frac{a+b-a(1-ab)}{1-ab+a(a+b)}$$
$$=\frac{b(a^2+1)}{a^2+1}$$
$$=b=\tan y$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/938432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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What is the limit of this specific function? Please evaluate the following limit for me:
$$\lim_{x \to -1} \frac{\sqrt{x^2+8}-3}{x+1} $$
I'd tried my best to solve this but unfortunately, it's too difficult for me. I tried multiplying by its conjugate and factoring the $x^2$ out but I can't get rid of that $x+1$ in the denominator so it always stay in the indeterminate form.
| Multiply the function by $$\frac{\sqrt{x^2 + 8} +3}{\sqrt{x^2 + 8}+3}$$
You'll have a difference of squares in the numerator of the form $(a -b)(a+ b)$ which, of course, is $a^2 - b^2$. You should have gotten, in the numerator: $$(\sqrt{x^2 + 8})^2 - (3^2) = x^2 + 8 - 9 = x^2 - 1 = (x+1)(x-1)$$
Simplifying, you'll get $$\lim_{x\to -1} \frac {\overbrace{x^2 -1}^{(x + 1)(x-1)}}{(x+1)(\sqrt{x^2 + 8} +3)} = \lim_{x\to -1} \frac {x-1}{\sqrt{x^2 + 8} +3} = \frac{-2}{6} = -\frac 13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
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How many functions can be constructed? How many functions $f:\left\{1, 2, 3, 4,5 \right\} \rightarrow \left\{ 1, 2, 3, 4, 5 \right\}$ satisfy the relation $f\left( x \right) =f\left( f\left( x \right) \right)$ for every $x\in \left\{ 1, 2, 3, 4, 5 \right\}$?
My book says the answer is 196.
| Let $D$ be the domain of $f$, $\{1,2,3,4,5\}$, and let $R\subset D$ be the range of $f$. We want $f(f(x)) = f(x)$ for all $x$, and since $f(x) = y$ if and only if $y\in R$, we have $f(y) = y$ if and only if $y\in R$.
Suppose $|R| = n$. There are $\binom 5n$ ways to pick $R$ itself. For each of these, $f(x)$ is completely determined when $x\in R$ (it is $x$) and when $x\notin R$, we have $n$ choices for $f(x)$. There are $5-n$ elements of $D\setminus R$, so $n^{5-n}$ choices for $f$, for each possible choice of $R$.
For example, suppose $|R| = 2$. There are $\binom52$ choices for $R$. For each of the three elements $x\in D\setminus R$, there are 2 choices for $f(x)$, for a total of $2^3=8$ functions for each choice of $R$. So there are $\binom52\cdot 8 = 80$ functions which have $|R|=2$.
The total number of functions $f$ is then $$\begin{align}
\sum_{n=1}^5 \binom5{n}n^{5-n} & =
\binom51\cdot 1^4 + \color{maroon}{\binom52\cdot 2^3} + \binom 53\cdot 3^2 + \binom 54\cdot 4^1 + \binom 55\cdot 5^0 \\
&= 5 \cdot 1 + \color{maroon}{10\cdot8} + 10\cdot 9+5\cdot4+ 1\cdot1 \\&= 196
\end{align}$$ where the colored term is the example from the previous paragraph.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show $\sum\limits_{k=1}^n{n-1\choose k-1} =2^{n-1}$
*
*Given $$\sum\limits_{k=1}^n k{n\choose k} = n\cdot 2^{n-1}$$
*
*I know that $$k\cdot{n\choose k}=n\cdot{n-1\choose k-1}=(n-k+1)\cdot{n\choose k-1}$$
Therefore $$\sum\limits_{k=1}^n k{n\choose k} = \sum\limits_{k=1}^n n{n-1\choose k-1} = n\cdot 2^{n-1}$$
So, $$n\cdot\sum\limits_{k=1}^n {n-1\choose k-1} = n\cdot 2^{n-1}$$
Therefore $$\sum\limits_{k=1}^n{n-1\choose k-1} =2^{n-1}$$
How is $\quad\sum\limits_{k=1}^n{n-1\choose k-1} =2^{n-1}\quad$?
| $$2^{n-1}=(1+1)^{n-1}=\sum_{k=0}^{n-1}\binom{n-1}{k}\cdot 1^{n-1-k}\cdot 1^k=\sum_{k=0}^{n-1}\binom{n-1}{k}=\sum_{k=1}^{n}\binom{n-1}{k-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/944422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Infinitely Many Circles in an Equilateral Triangle
In the figure there are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles.
I need to find the total area of the circles.
I know this is going to have something to do with summation as a value approaches infinity, but I'm not entirely sure how to approach the problem. Here's what I have so far:
I know that the radius of the central inscribed circle is $ \frac{\sqrt{3}}{6} $. As such, the area of the first circle is $$ A = \pi\left(\frac{\sqrt{3}}{6}\right)^2. $$ Because there are three "branches" of infinite circles, I'm assuming that the answer will look something like: $$ A = \pi\left(\frac{\sqrt{3}}{6}\right)^2 + 3 \sum_{1}^{\infty}\text{something}.$$
| Since the length of the equilateral triangle is 1, let its median length $x = \frac{\sqrt(3)}{2}$.
So, the radius of the first circle is $\frac{x}{3}$.
Hence, its area is $A_1 = {\pi}(\frac{x}{3})^2$
The radius of the next three identical circles will be $\frac{x}{9}$.
The radius of each progressing circle will be a third of the previous one.
So, the sum of the areas of these circles is
$A = A_1+3\sum_{n=1}^{\infty}\frac{\pi x^2}{3^{2n+2}}$
Now, considering the fact that $x=\frac{\sqrt(3)}{2}$,
$3\sum_{n=1}^{\infty}\frac{\pi x^2}{3^{2n+2}} = 3\sum_{n=1}^{\infty}\frac{9 \pi}{4*3^{2n+2}} = \frac{\pi}{4}\sum_{n=1}^{\infty}\frac{1}{9^n}$.
Considering the convergence of the geometric with r<1, the total area is given by
$A = A_1 + \frac{\pi}{4} \sum_{n=1}^{\infty} \frac{1}{9^n} = \frac{\pi}{12} + \frac{\pi}{32} = \frac{11 \pi}{96}$.
Thus, the total area of the circles is $\frac{11 \pi}{96}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/945123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
If the product $(x+2)(x+3)(x+4)\cdots(x+9)(x+10)$ expands to $a_9x^9+a_8x^8+\dots+a_1x+a_0$, then what is the value of $a_1+a_3+a_5+a_7+a_9$?
When expanded, the product $(x+2)(x+3)(x+4)\cdots(x+9)(x+10)$ can be written as $a_9x^9+a_8x^8+\dots+a_1x+a_0$. What is the value of $a_1+a_3+a_5+a_7+a_9$?
| On expansion of $$(x+2)(x+3)(x+4)...(x+9)(x+10)$$ we have
$$(x+2)(x+3)(x+4)...(x+9)(x+10) = x^9+54x^8+1186x^7+14084x^6+100045x^5+439850x^4+1182204x^3+1844936x^2+1458720x+403200$$
which is written in the form of $~~a_9x^9+a_8x^8+...+a_1x+a_0$.
Then, $$a_1+a_3+a_5+a_7+a_9~=~ 1458720~+~1182204~+~100045~+~1186~+~1~=~2742156$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/945590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that sum of elements of rows / columns of a matrix is equal to reciprocal of sum of elements of rows/colums of its inverse matrix Suppose $A=(a_{ij})_{n\times n}$ be a non singular matrix. Suppose sum of elements of each row is $k\neq 0$, then the sum of elements of rows of $A^{-1}$ is $\frac{1}{k}$.
Let $\,A^{-1}=(b_{ij})_{n\times n}$.
Then
\begin{align*}
&A\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]=k\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]\\
\implies &\frac{1}{k}A^{-1}A\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]=\frac{1}{k}A^{-1}k\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]\\
\implies &\frac{1}{k}I_n\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]=A^{-1}\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]\\
\implies &\frac{1}{k}I_n\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]=(b_{ij})\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]\\
\implies& (b_{ij})\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]=\frac{1}{k}I_n\left[ \begin{array}{c} 1\\ 1\\ 1\\\vdots \\1 \end{array}\right]\\
\implies &(\displaystyle\sum_{r=1}^n b_{1r}, \displaystyle\sum_{r=1}^n b_{2r},\cdots, \displaystyle\sum_{r=1}^n b_{nr})=(\frac{1}{k}, \frac{1}{k}\cdots, \frac{1}{k})\\
\implies & \displaystyle\sum_{r=1}^n b_{ir}=\frac{1}{k}, \forall i=1,2,\cdots,n
\end{align*}
| apply ${1\over k}A^{-1}$ to
$$
A
\begin{bmatrix}
1 \\
1 \\
\vdots \\
1 \\
\end{bmatrix}
=
k
\begin{bmatrix}
1 \\
1 \\
\vdots \\
1 \\
\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/946776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
finding a function from given function here is a function for:
$f(x-\frac{\pi}{2})=\sin(x)-2f(\frac{\pi}{3})$
what is the $f(x)$?
I calculate $f(x)$ as follows:
$$\begin{align}
x-\frac{\pi}{2} &= \frac{\pi}{3} \Rightarrow x= \frac{5\pi}{6} \\
f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6}-2f(\frac{\pi}{3}) \\
3f(\frac{\pi}{3}) &=\sin\frac{5\pi}{6} \\
f(\frac{\pi}{3}) &=(1/3)\sin\frac{5\pi}{6}
\end{align}$$
$f(x)=(1/3)\sin\frac{5x}{2}$
| $f(x)=f((x+\frac{1}{2}\pi)-\frac{1}{2}\pi)=\sin(x+\frac{1}{2}\pi)-2f(\frac{1}{3}\pi)=\cos(x)-2f(\frac{1}{3}\pi)$
To find $f(\frac{1}{3}\pi)$ we substitute $x=\frac{1}{3}\pi$:
$f(\frac{1}{3}\pi)=\cos(\frac{1}{3}\pi)-2f(\frac{1}{3}\pi)$
Then $f(\frac{1}{3}\pi)=\frac{1}{3}\cos(\frac{1}{3}\pi)=\frac{1}{6}$ so we end up with:
$$f(x)=\cos(x)-\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I show that $Var(Y) = n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}$? My original pdf is $f(y) = \frac{n (y_{n} - \theta_{1})^{n-1}}{(\theta_{2} - \theta_{1})^{n}}$ for $\theta_{1} < y < \theta_{2}$.
After using U-substitution, I obtain $E(Y) = \frac{n \theta_{2} + \theta_{1}}{(n+1)}$.
For variance of $Y$, I need to find $E(Y^2)$ first.
This is what I have with U-substitution: $E(Y^2) = \frac{n\theta_{2} - n \theta_{1} + 2n \theta_{1} \theta_{2} - n\theta_{1}^{2} + \theta_{1}^{2}}{(n+1)}$.
But with this the $Var(Y)= E(Y^2) - \left [E(Y) \right ]^{2} \neq n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}$
My mistake is definitely the $E(Y^2)$ part, so I would appreciate it if someone here can go over it with me.
Edit:
$$\begin{eqnarray}
\mathbb{E}\left(Y^2\right) &=& \mathbb{E}\left(\left(Y-\theta_1\right)^2\right) + 2 \theta_1 \cdot \mathbb{E}\left(\left(Y-\theta_1\right)\right) + \theta_1^2 \\ &=& \frac{n}{n+2} \left(\theta_2-\theta_1\right)^2 + 2 \theta_1 \cdot \frac{n}{n+1} \left(\theta_2-\theta_1\right) + \theta_1^2
\end{eqnarray}$$ from below and also my own work.
But now how do I simplify $$Var(Y) = \mathbb{E}\left(Y^2\right) - \left [ \mathbb{E}\left(Y\right) \right ]^{2}$$ to $$Var(Y) = n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}?$$
| Instead of $\theta_1$ and $\theta_2$ I will use $a$ and $b$.
For the expectation of $Y^2$, we want
$$\int_a^b \frac{ny^2(y-a)^{n-1}}{(b-a)^n}\,dy.$$
If we let $u=y-a$, then $y^2=u^2+2au+a^2$. So we want
$$\frac{n}{(b-a)^n}\int_{u=0}^{b-a} \left(u^{n+1}+2au^{n}+a^2u^{n-1}\right)\,du.$$
This is, after minor simplification, equal to
$$\frac{n}{n+2}(b-a)^{2}+\frac{2an}{n+1}(b-a)+a^2.$$
A similar but simpler calculation shows that $E(Y)=\frac{n}{n+1}(b-a)+a$.
The calculation of $E(Y^2)-(E(Y))^2$ is simple, since there is some nice cancellation. We get
$$(b-a)^2\left(\frac{n}{n+2}-\frac{n^2}{(n+1)^2}\right),$$
which simplifies to
$$(b-a)^2\frac{n}{(n+2)(n+1)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding cartesian equation for trigonometric parametric forms I'm trying to find the cartesian equation for these parameteric forms:
$$
x = sin\theta + 2 cos \theta \\
y = 2 sin\theta + cos\theta
$$
I tried:
$$\begin{align}
x^2 & = sin^2\theta + 4cos^2\theta \\
& = 1 - cos^2\theta + 4cos^2\theta \\
& = 1 + 3cos^2\theta \\
\\
y^2 & = 4sin^2\theta+cos^2\theta\\
& = 4(1 - cos^2\theta) + cos^2\theta \\
& = 4 - 3cos^2\theta \\
\\
\therefore & \space4 - y^2 = x^2 - 1\\
\space & x^2 + y^2 = 5
\end{align}$$
Which differs from the given answer of $5x^2 + 5y^2 - 8xy = 9$. Where am I going wrong?
| Note that $$x^2\not=\sin^2\theta+4\cos^2\theta$$
and that
$$x^2=\sin^2\theta+4\cos^2\theta+4\sin\theta\cos\theta$$
because
$$(a+b)^2=a^2+b^2+\color{red}{2ab}.$$
Since we can get
$$\sin\theta=\frac{-x+2y}{3},\ \ \ \cos\theta=\frac{2x-y}{3},$$
then use
$$\sin^2\theta+\cos^2\theta=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/949290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to proof the following function is always constant which satisfies $f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)\,dt} $? Suppose that $f(x)$ is a bounded continuous function on $\mathbb{R}$,and that there exists a positive number $a$ such that
$$f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)\,dt} $$
is constant. Can anybody show that $f$ is necessarily constant ?
| Starting with
\begin{align}
f(x) + a \int_{x-1}^{x} f(t) \, dt = c
\end{align}
then, by differentiation,
\begin{align}
f'(x) + a \left( f(x) - f(x-1) \right) = 0.
\end{align}
Now, for this equation to be satisfied consider it in the form of
\begin{align}
f'(x) = B = -a ( f(x) - f(x-1) ).
\end{align}
From the equation $f'(x) = B$ it is seen that $f(x) = Bx+c_{1}$. Now,
\begin{align}
f(x) - f(x-1) &= - \frac{B}{a} \\
(Bx+c_{1}) - (Bx - B + c_{1}) &= - \frac{B}{a} \\
B &= - \frac{B}{a}.
\end{align}
If this is to be satisfied then $B = 0$ which then implies that $$f(x) = c_{1}.$$
The alternative is that given the equation
\begin{align}
f'(x) + a (f(x) - f(x-1) ) = 0
\end{align}
then let $f(x) = a_{0} + a_{1} x + a_{2} x^{2} + \cdots$ to obtain
\begin{align}
0 &= [ a_{1} + 2 a_{2} x + \cdots ] + a [ a_{0} + a_{1} x + a_{2} x^{2} + \cdots ] - a[ a_{0} + a_{1} (x-1) + a_{2} (x-1)^{2} + \cdots] \\
&= [ a_{1} + 2 a_{2} x + \cdots] + a[ a_{1} + a_{2}(2x-1) +\cdots] \\
&= [ a_{1} - a(a_{2} - a_{3} + a_{4} - \cdots )] + [ 2 a_{2} + a(2 a_{2} - 3 a_{3} + \cdots) ] x + [ 3(1+a)a_{3} + \cdots ] x^{2} + \cdots .
\end{align}
All the coefficients of $x^{n}$, $0\leq n$, are zero which yields $a_{1} = a_{2} = \cdots = 0$. This then leaves $f(x) = a_{0}$ which is a constant.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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} |
Biggest set such that sum of any pair is perfect square What is the biggest set of positive integers such that the sum of any pair of them is a perfect square? (Or can we construct an infinite such set?)
One such set of size $3$ is $\{6,19,30\}$, which give sums $25,36,49$, but I can't find any larger yet.
| It's easy to do. The formula below. It is only necessary to take $a,b,c$.
You can even for 4 terms to write.
Do there exist four distinct integers such that the sum of any two of them is a perfect square? This is equivalent to solving the following system of equations: $$\left\{\begin{aligned}& b+a=x^2 \\&b+c=y^2\\&b+f=z^2\\&a+c=e^2\\&a+f=j^2\\&c+f=p^2\end{aligned}\right.$$ Let: $F,T,R,D$ - any asked us integers. For ease of calculation, let's make a replacement. $$q=(8F^2+4FT-T^2)R^2+2(T+2F)RD-D^2$$ $$k=(8F^2+8FT+2T^2)R^2+2(T+2F)RD$$ $$s=-T^2R^2+2(T+2F)RD-D^2$$ $$t=(8F^2+12TF+3T^2)R^2+2(T+2F)DR-D^2$$ Then the solutions are of the form: $$x=s^2+k^2-t^2+2(t-k-s)q$$ $$y=t^2+k^2-s^2+2ks-2tk$$ $$z=s^2+k^2-t^2$$ $$e=t^2+k^2+s^2-2kt-2ts$$ $$j=t^2+s^2-k^2+2ks-2ts$$ $$p=3s^2+3k^2+3t^2-6kt-6st+8ks+2(t-k-s)q$$ $$b=\frac{x^2+y^2-e^2}{2}$$ $$a=\frac{e^2+x^2-y^2}{2}$$ $$c=\frac{e^2+y^2-x^2}{2}$$ $$f=\frac{2z^2+e^2-x^2-y^2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Limit approaching negative infinity question?
I'm not sure how to solve this problem.. How do I get rid of the square root?
| $$\lim_{x\to -\infty}\frac{\sqrt{5x^2-2x^3+9x^6}}{-17+3x+8x^3}
=\lim_{x\to -\infty}\frac{-x^3\sqrt{5\frac{1}{x^4}-2\frac{1}{x^3}+9}}{-17+3x+8x^3}
=\lim_{x\to -\infty}\frac{\sqrt{5\frac{1}{x^4}-2\frac{1}{x^3}+9}}{\frac{17}{x^3}-\frac{3}{x^2}-8}=-\frac{3}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/954774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
For which natural n is $3^n+5^n$ divisible by $3^{n-1}+5^{n-1}$? For which natural n is $3^n+5^n$ divisible by $3^{n-1}+5^{n-1}$? I only got that $3^n+5^n=8k$ when $n$ is odd. How to solve this one?
| $$3(3^{n-1}+5^{n-1})=3^n+3*5^{n-1}<3^n+5^n<5*3^{n-1}+5^n=5(3^{n-1}+5^{n-1})$$
So the ratio is between $3$ and $5$, so it equals $4$.
$$4*3^{n-1}+4*5^{n-1}=3^n+5^n=3*3^{n-1}+5*5^{n-1}\\3^{n-1}=5^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x $ In a related question the following integral was evaluated
$$
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
=\int_0^{\pi} \frac{\mathrm{d}x/2}{1 + \cos x \sin x}
=\int_0^{2\pi} \frac{\mathrm{d}x/2}{2 + \sin x} \,\mathrm{d}x
=\int_{-\infty}^\infty \frac{\mathrm{d}x/2}{1+x+x^2}
$$
I noticed something interesting, namely that
$$
\begin{align*}
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
& = \int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\
& = \int_0^{\pi} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x
= \int_0^{\pi} \frac{(\sin x)^2}{1 - \cos x \sin x} \,\mathrm{d}x
\end{align*}
$$
The same trivially holds if the upper limits are changed to $\pi/2$ as well ($x \mapsto \pi/2 -u$).
But I had problems proving the first equality. Does anyone have some quick hints?
|
$$I=\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$
$$J=\int_0^{\pi/2} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$
Use $\int_a^b f(x)dx=\int_a^nf(a+b-x)dx$ and $\int_0^{2a}f(x)dx=\int_0^a f(x)dx+f(2a-x)dx$
$$I=\int_0^{\pi/2} \frac{(\cos (\pi-x))^2}{1 + \cos (\pi-x) \sin (\pi-x)} \,\mathrm{d}x+\frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$
$$I=\int_0^{\pi/2} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x+\frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$
$$I=\int_0^{\pi/2} \frac{(\cos (\pi/2-x))^2}{1 - \cos (\pi/2-x) \sin (\pi/2-x)} \,\mathrm{d}x+\frac{(\cos (\pi/2-x))^2}{1 + \cos (\pi/2-x) \sin (\pi/2-x)} \,\mathrm{d}x$$
$$I=\int_0^{\pi/2} \frac{(\sin x)^2}{1 - \sin x \cos x} \,\mathrm{d}x+\frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$
$$I=\int_0^{\pi/2} \frac{(\sin x)^2}{1 - \sin x \cos x} \,\mathrm{d}x+\frac{(\sin (\pi-x))^2}{1 - \cos (\pi-x) \sin(\pi-x)} \,\mathrm{d}x$$
$$I=\int_0^{\pi} \frac{(\sin x)^2}{1 - \sin x \cos x} \,\mathrm{d}x$$
Hope you can now show that:
$$J=\int_0^{\pi/2} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x=\int_0^{\pi/2} \frac{(\cos (\pi/2-x))^2}{1 + \cos (\pi/2-x) \sin (\pi/2-x)} \,\mathrm{d}x=\int_0^{\pi/2} \frac{(\sin x)^2}{1 + \sin x \cos x} \,\mathrm{d}x$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
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Solve for $x$ when $\sin 2x = \cos x$ where $ x$ is in the domain $ [0, 2\pi]$ Quick question on trig (which I haven't dealt with in a long time):
since $\sin 2x = 2\sin x\cos x $
$2\sin x\cos x = \cos x$
$2\sin x\cos x/\cos x = 1$
$\sin x = 1/2$
since $\sin x = 1/2$ in quadrants $1$ and $2$, $x = \pi/6$ and $x = 5\pi/6$
Is this correct? If not, hint please.
| After getting $$2\sin x\cos x=\cos x,$$
you cannot divide the both sides by $\cos x$ because $\cos x$ can be zero.
So, we have
$$2\sin x\cos x-\cos x=0\iff \cos x(2\sin x-1)=0\iff \cos x=0\ \text{or}\ 2\sin x-1=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/955971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Task "Inversion" (geometry with many circles) Incircle $\omega$ of triangle $ABC$ with center in point $I$ touches $AB, BC, CA$ in points $C_{1}, A_{1}, B_{1}$. Сircumcircle of triangle $AB_{1}C_{1}$ intersects second time circumcircle of $ABC$ in point $K$. Point $M$ is midpoint of $BC$, $L$ midpoint of $B_{1}C_{1}$. Сircumcircle of triangle $KA_{1}M$ intersects second time $\omega$ in point $T$.
Prove, that сircumcircles of triangle $KLT$ and triangle $LIM$ touch.
| Without loss of generality you can choose a coorinate system in such a way that $\omega$ is the unit circle and $A_1=(1,0)$. Then you can use a rational parametrization of the circle to describe $B_1$ and $C_1$ in terms of two parameters $b,c\in\mathbb R$, ending up with these coordinates:
\begin{align*}
I &= \begin{pmatrix}
0\\
0
\end{pmatrix}
&
A_1 &= \begin{pmatrix}
1\\
0
\end{pmatrix}
\\
B_1 &= \frac1{b^{2} + 1}\begin{pmatrix}
b^{2} - 1\\
2 b
\end{pmatrix}
&
C_1 &= \frac1{c^{2} + 1}\begin{pmatrix}
c^{2} - 1\\
2 c
\end{pmatrix}
\end{align*}
From there on, you can do a brute force coordinate computation, yielding these intermediate points:
\begin{align*}
A &= \frac1{b c + 1}\begin{pmatrix}
b c - 1\\
b + c
\end{pmatrix}
\\
B &= \frac1{c}\begin{pmatrix}
c\\
1
\end{pmatrix}
\\
C &= \frac1{b}\begin{pmatrix}
b\\
1
\end{pmatrix}
\\
M &= \frac1{2 b c}\begin{pmatrix}
2 b c\\
b + c
\end{pmatrix}
\\
L &= \frac1{b^{2} c^{2} + b^{2} + c^{2} + 1}\begin{pmatrix}
b^{2} c^{2} - 1\\
b^{2} c + b c^{2} + b + c
\end{pmatrix}
\\
K &= \frac1{b^{2} c^{2} + b^{2} + 4 b c + c^{2} + 1}\begin{pmatrix}
b^{2} c^{2} + b^{2} + 2 b c + c^{2} - 1\\
2 b + 2 c
\end{pmatrix}
\\
T &= {\scriptsize\frac1{16 b^{4} c^{4} + 9 b^{4} c^{2} + 18 b^{3} c^{3} + 9
b^{2} c^{4} + b^{4} + 6 b^{3} c + 10 b^{2} c^{2} + 6 b c^{3} + c^{4} +
b^{2} + 2 b c + c^{2}}}\cdot\\
&\phantom={\scriptsize\begin{pmatrix}
16 b^{4} c^{4} + 7 b^{4} c^{2} + 14 b^{3} c^{3} + 7 b^{2} c^{4} + b^{4}
+ 2 b^{3} c + 2 b^{2} c^{2} + 2 b c^{3} + c^{4} - b^{2} - 2 b c -
c^{2}\\
8 b^{4} c^{3} + 8 b^{3} c^{4} + 2 b^{4} c + 14 b^{3} c^{2} + 14 b^{2}
c^{3} + 2 b c^{4} + 2 b^{3} + 6 b^{2} c + 6 b c^{2} + 2 c^{3}
\end{pmatrix}}
\end{align*}
I'd like to paste equations for the circles as well, but I still have to work out how to format them since they are pretty big.
The circles $KLT$ and $LIM$ then have a common tangent:
\begin{align*}{\scriptsize
\bigl((b + c) \cdot (2 b^{4} c^{4} - 3 b^{4} c^{2} + 4 b^{3} c^{3} - 3 b^{2}
c^{4} - b^{4} - 8 b^{2} c^{2} - c^{4} - b^{2} - 4 b c - c^{2})\bigr)x}
\\
{\scriptsize+
\bigl(5 b^{5} c^{3} + 2 b^{4} c^{4} + 5 b^{3} c^{5} + b^{5} c + b^{4} c^{2} +
16 b^{3} c^{3} + b^{2} c^{4} + b c^{5} - b^{4} + 3 b^{3} c + 3 b c^{3}
- c^{4} - b^{2} - 2 b c - c^{2}\bigr)y}
\\
{\scriptsize= 2\cdot c \cdot b \cdot (b + c) \cdot (b c + 1)^{3}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 1
} |
How did we find the solution? In my lecture notes, I read that "We know that $$x^2 \equiv 2 \pmod {7^3}$$ has as solution $$x \equiv 108 \pmod {7^3}$$"
How did we find this solution?
Any help would be appreciated!
| Without referring to Hensel I'd make the "Ansatz"
$$x=a\cdot 7^2+b\cdot7+c$$
with $|a|$, $|b|$, $|c|$ integers $\leq3$. This is no restriction of generality. The condition $x^2=2\ (7^3)$
leads to
$$(b^2+2ac)\cdot7^2+2bc\cdot7+ c^2-2=0\quad(7^3)\ .\tag{1}$$
It follows that $c^2=2\ (7)$, which implies $c=\pm3$. Plugging this into $(1)$ gives
$$(b^2\pm 6a)\cdot7^2 \pm 6b\cdot7+7=0\quad(7^3)\ ,$$
and dividing by $7$ leads to
$$(b^2\pm 6a)\cdot7+(\pm 6b+1)=0\quad(7^2)\ .\tag{2}$$
This implies $b=\pm1$. Inserting this into $(2)$ and dividing by $7$ again we obtain
$$(1\pm 6a)+1=0\quad(7)\ ,$$
and this is solved by $a=\pm2$.
Therefore all solutions of the original equation have to be of the form
$$x=\pm(2\cdot7^2+1\cdot7+3)=\pm 108\quad(7^3)\ .$$
It is easily checked that $\pm108$ indeed does the job.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculation of $\int\frac1{\tan \frac{x}{2}+1}dx$ Calculation of $\displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$
$\bf{My\; Try}::$ Let $\displaystyle I = \displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$, Now let $\displaystyle \tan \frac{x}{2}=t\;,$ Then $\displaystyle dx=\frac{2}{1+t^2}dt$
So $\displaystyle I = 2\int\frac{1}{(1+t)\cdot (1+t^2)}dt$
Now Using Partial fraction,
$\displaystyle \frac{1}{(1+t)\cdot (1+t^2)} = \frac{A}{1+t}+\frac{Bt+C}{1+t^2}\Rightarrow 1=A(1+t^2)+(Bt+C)(1+t)$
Now put $(1+t)=0\Rightarrow t=-1\;,$ We get $\displaystyle 1=2A\Rightarrow A = \frac{1}{2}.$
Now Put $(1+t^2)=0\Rightarrow t^2=-1\;,$ We Get $1=Bt^2+(B+C)t+C$
So $\displaystyle 1=\left(-B+C\right)+(B+C)$. So Solving equation...$B+C=0$ and $-B+C=1$
So $\displaystyle B=-\frac{1}{2}$ and $\displaystyle C=\frac{1}{2}$
So $\displaystyle I = 2\int\frac{1}{(1+t)\cdot (1+t^2)}dt = \int\frac{1}{1+t}dt+\int\frac{-t+1}{1+t^2}dt$
So $\displaystyle I = \frac{1}{1+t}dt-\frac{1}{2}\int\frac{2t}{1+t^2}dt+\int \frac{1}{1+t^2}dt$
So $\displaystyle I = \ln \left|1+t\right|-\frac{1}{2}\ln \left|1+t^2\right|+\tan^{-1}(t)+\mathcal{C}$
So $\displaystyle I = \ln \left|1+\tan \frac{x}{2}\right|-\frac{1}{2}\ln \left|1+\tan^2 \frac{x}{2}\right|+\frac{x}{2}+\mathcal{C}$
Can we solve it without using partial fraction? If yes then please explain to me.
Thanks
| Thanks E W H Lee and shooting-squirrel
Another Try:: Let $\displaystyle I =\int\frac{1}{\tan \frac{x}{2}+1}dx$
Let $x=2y\;,$ Then $dx = 2dy$. So $\displaystyle I = 2\int\frac{1}{\tan y+1}dy = \int\frac{2\cos y}{\sin y+\cos y}dy$
So $\displaystyle I = \int\frac{\left(\cos y+\sin y\right)+\left(\cos y-\sin y\right)}{\sin y+\cos y}dy = y+\ln \left|\sin y+\cos y\right|+\mathcal{C}$
So $\displaystyle I =\int\frac{1}{\tan \frac{x}{2}+1}dx = \frac{x}{2}+\ln \left|\sin \frac{x}{2}+\cos \frac{x}{2}\right|+\mathcal{C}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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} |
Hard inequality $ (xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4} $ I need to prove or disprove the following inequality:
$$
(xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\ge\frac{9}{4}
$$
For $x,y,z \in \mathbb R^+$. I found no counter examples, so I think it should be true. I tried Cauchy-Schwarz, but I didn't get anything useful. Is it possible to prove this inequality without using brute force methods like Bunching and Schur?
This inequality was in the Iran MO in 1996.
| Each of the two expressions that are multiplying are symmetric with respect to the three variables x, y, z.
Making a change of varibales:
$$a = x + y\\
b = y + z\\
c = z + x$$
It causes the second expression to become another symmetric expression:
$$ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$$
And that the first expression also becomes another symmetrical expression:
$$ \frac{(a+b+c)^2-2(a^2+b^2+c^2)}{4}$$
Multiplying and eliminating the 4 of the denominator of the inequality:
$$ ((a+b+c)^2-2(a^2+b^2+c^2))(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) \geq 9 $$
Renaming $M_n = \sum_{k} a_k^n$, then:
$$ (M_1^2-2M_2)M_{-2} \geq 9 $$
$$ M_1^2-2M_2 \geq 9M_{-2}^{-1} $$
All this is related to statistics, arithmetic mean, variance, moments, etc.
$$\frac{M_1}{3} = \mu; \frac{M_2}{3}=\alpha_2; \frac{M_{-2}}{3}=\alpha_{-2} $$
$$ 9\mu^2-6\alpha_2 \geq 3\alpha_{-2}^{-1} $$
$$ 3\mu^2 \geq \alpha_{-2}^{-1}+2\alpha_2 $$
Continue...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Using the chain rule to find derivative Could you help me figure this one out?
$$f(x) = 6e^{x\sin x}$$
$$f'(x) = \,?$$
| Add logarithms to both sides of the equation $f(x) = 6e^{x \sin x}$
\begin{align}\ln f(x) &= \ln (6e^{x \sin x}) \\
&= \ln 6 + \ln e^{x \sin x} \\
&= \ln 6 + x \sin x,
\end{align}
and implicitly differentiate both sides to get
\begin{align}
\frac{f'(x)}{f(x)}=\sin x + x \cos x.
\end{align}
Finally, we get
\begin{align}
f'(x)&=f(x)(\sin x + x \cos x) \\ &= \boxed{6e^{x \sin x}(\sin x + x \cos x)}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
For what value of $x$ is this matrix invertible? I've been given the following matrix $X$:
$$X=
\begin{bmatrix}
1 & 4 & 8 & 1 \\
0 & 30 &1 & 0 \\
0 &2& 0& 0 \\
1 &2 & 9 & x \\
\end{bmatrix}
$$
and I have to determine for which value of $x$ the matrix is invertible.
Any hints/suggestions would be greatly appreciated.
So far I have substituted different values for $x$, but there seems to be an inverse for all values of $x$ above $1$.
| $$X=
\begin{pmatrix}
1 & 4 & 8 & 1 \\
0 & 30 &1 & 0 \\
0 &2& 0& 0 \\
1 &2 & 9 & x \\
\end{pmatrix}\stackrel{ R_4-R_1}\longrightarrow\;\;
\begin{pmatrix}
1 & 4 & 8 & 1 \\
0 & 30 &1 & 0 \\
0 &2& 0& 0 \\
0 &\!\!-2 & 1 & x-1 \\
\end{pmatrix}\stackrel{R_3-\frac1{15}R_2}{\stackrel{R_4+\frac1{15}R_2\;,\;R_2}\longrightarrow}
$$
$$\begin{pmatrix}
1 & 4 & 8 & 1 \\
0 & 30 &1 & 0 \\
0 &0& \!\! -\frac1{15}& 0 \\
0 &0 & \frac{16}{15} & x-1 \\
\end{pmatrix}\stackrel{R_4+16R_3}\longrightarrow\begin{pmatrix}
1 & 4 & 8 & 1 \\
0 & 30 &1 & 0 \\
0 &0& \!\! -\frac1{15}& 0 \\
0 &0 & 0 & x-1 \\
\end{pmatrix}$$
and the above matrix (and thus the original one as well) is singular iff the last row is all zeros, so...
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the maximum of $\frac{1}{1+|x|}+\frac{1}{1+|x-a|}$ Let $a>0$. Show that the maximum value of the function
$$f(x)= \frac{1}{1+|x|}+\frac{1}{1+|x-a|}$$
is $$\frac{2+a}{1+a}.$$
really need some help with this thing
| $f(a) = \dfrac{2+a}{1+a}$, and $f(x) = f(a-x)$, it suffices to consider $f$ on $[0,\frac{a}{2}]$, and on this interval $f(x) = \dfrac{1}{1+x} + \dfrac{1}{1+a-x}$, and $f'(x) = -\dfrac{1}{(1+x)^2} + \dfrac{1}{(1+a-x)^2} \leq 0$, thus $f(x) \leq f(0) = \dfrac{2+a}{1+a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to plot bivariate function involving modulus and floor functions? I need to plot:
*
*$\displaystyle\large|||x|-2|-1|+|||y|-2|-1|=1$
*$\displaystyle\large\left\lfloor\frac{|3x+4y|}{5}\right\rfloor+\left\lfloor\frac{|4x-3y|}{5}\right\rfloor=3$
either for finding area of some region or the length of the wire, if used to make the graph(maybe the line integral.) Only consider the plotting of these graphs:
and
| For the first :
The graph is symmetry both about $x$-axis and $y$-axis. So, we only need to consider the case when $x\ge 0$ and $y\ge 0$. Then, note that
$$|||a|-2|-1|=\begin{cases}-a+1&\text{if $0\le a\lt 1$}\\a-1&\text{if $1\le a\lt 2$}\\-a+3&\text{if $2\le a\lt 3$}\\a-3&\text{if $a\ge 3$}\end{cases}$$
For the second :
Since
$$\frac{|3x+4y|}{5}\ge 0\ \ \text{and}\ \ \frac{|4x-3y|}{5}\ge 0,$$
each term of the left-hand-side is a non-negative integer.
Hence, we have only four cases as the followings :
(1)
$$\left\lfloor\frac{|3x+4y|}{5}\right\rfloor=0\ \ \text{and}\ \ \left\lfloor\frac{|4x-3y|}{5}\right\rfloor=3.$$
(2)
$$\left\lfloor\frac{|3x+4y|}{5}\right\rfloor=1\ \ \text{and}\ \ \left\lfloor\frac{|4x-3y|}{5}\right\rfloor=2.$$
(3)
$$\left\lfloor\frac{|3x+4y|}{5}\right\rfloor=2\ \ \text{and}\ \ \left\lfloor\frac{|4x-3y|}{5}\right\rfloor=1.$$
(4)
$$\left\lfloor\frac{|3x+4y|}{5}\right\rfloor=3\ \ \text{and}\ \ \left\lfloor\frac{|4x-3y|}{5}\right\rfloor=0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove using induction principles $$\forall{n,a>1}:\;\sum\limits_{k=1}^{2^n-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{n(1-a)}}{1-2^{1-a}}\right)$$
For any fixed value of $a > 1$.
Induction step:
$$\sum_{k=1}^{2^{n+1} - 1} \frac{1}{k^a} = (\sum\limits_{k=1}^{2^n-1}\frac{1}{k^a}) + \frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \frac{1}{(2^n + 2)^a} + ... + \frac{1}{(2^{n+1} -1)^a} \leq\left(\frac{1-2^{n(1-a)}}{1-2^{1-a}}\right) + \frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \frac{1}{(2^n + 2)^a} + ... + \frac{1}{(2^{n+1} -1)^a}$$
I need help from Induction step and on. So if someone would help me, that would be greatly appreciated! People on this website keep putting this problem on hold even though I have clarified it as much as I can.
I need to prove P(n+1) is true:
$$\forall{n,a>1}:\;\sum\limits_{k=1}^{2^{n+1}-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{(n+1)(1-a)}}{1-2^{1-a}}\right)$$
| Hint. $$\sum\limits_{k=2^n}^{2^{n+1}-1} \frac{1}{k^a} \leq \sum\limits_{k=2^n}^{2^{n+1}-1}\frac{1}{2^{na}}=\frac{1}{2^{n(a-1)}}$$
and
$$
\frac{1 - 2^{n(1-a)}}{1-2^{1-a}} + \frac{1}{2^{n(a-1)}} = \cdots
$$
By the way, a circuitous route might by using the geometric series formula (you certainly don't need to do this, but it's interesting), you know that
$$
\sum\limits_{k=0}^{n-1} \frac{1}{2^{k(a-1)}} = \frac{1-2^{n(1-a)}}{1-2^{1-a}}
$$
and so
$$
\frac{1 - 2^{n(1-a)}}{1-2^{1-a}} + \frac{1}{2^{n(a-1)}} = \sum\limits_{k=0}^{n-1} \frac{1}{2^{k(a-1)}} + \frac{1}{2^{n(a-1)}} = \sum\limits_{k=0}^n \frac{1}{2^{k(a-1)}} = \cdots
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve this exponential equation? $2^{2x}3^x=4^{3x+1}$. I haven't been able to find the correct answer to this exponential equation:
$$\eqalign{
2^{2x}3^x&=4^{3x+1}\\
2^{2x} 3^x &= 2^2 \times 2^x \times 3^x\\
4^{3x+1} &= 4^3 \times 4^x \times 4\\
6^x \times 4 &= 4^x \times 256\\
x\log_6 6 + \log_6 4 &= x\log_64 + \log_6 256\\
x + \log_6 4 &= x\log_64 + \log_6 256\\
x-x\log_6 4 &= \log_6 256 - \log_6 4\\
x(1-\log_6 4) &= \log_6 256 - \log_6 4\\
x &= \dfrac{\log_6 256 - \log_6 4}{1-\log_6 4}\\
x &= 10.257}$$
so when I checked the answer I wasn't able to make them equal, I have tried variants of this method but I feel I'm missing something..
| You made a mistake in your first steps.
It should be :
$$4^{3x+1}= (4^3)^x \cdot 4$$
$$2^{2x} 3^x= 4^x 3^x$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of $\sqrt{n^2-4}, n\ge 3$ being irrational Is the proof of $n\ge 3$, $\sqrt{n^2-4} \notin \mathbb{Q} \ \text{correct}$?
$\sqrt{n^2-4} \in \mathbb{Q}
\\
\sqrt{n^2-4} = \frac{p}{q}
\\
(\sqrt{n^2-4})^2 = \left(\frac{p}{q}\right)^2
\\
q^2\left( n^2-4\right)=p^2
\\
\text{p is divisible by} \left (n^2-4 \right)
\Rightarrow p=k\left (n^2-4 \right)
\\
q^2 \left (n^2-4 \right)=k^2 \left(n^2-4 \right)^2
\\
q^2=k^2 \left (n^2-4 \right )\Rightarrow
\text{it follows that q is also divisible by} \left (n^2-4 \right)
\\
\text{Therefore p and q are not co-prime therefore} \Rightarrow \sqrt{n^2-4} \notin \mathbb{Q} \
\square
$
| You do not give sufficient reason why $p$ is divisible by $n^2-4$ should hold.
Consider this modified "proof" of $n\ge 3 \implies
\sqrt{n^2-5} \notin \mathbb{Q}$:
$$
\sqrt{n^2-5} \in \mathbb{Q}
\\
\sqrt{n^2-5} = \frac{p}{q}
\\
(\sqrt{n^2-5})^2 = \left(\frac{p}{q}\right)^2
\\
q^2\left( n^2-5\right)=p^2
\\
\text{p is divisible by} \left (n^2-5 \right)
\Rightarrow p=k\left (n^2-5 \right)
\\
q^2 \left (n^2-5 \right)=k^2 \left(n^2-5 \right)^2
\\
q^2=k^2 \left (n^2-5 \right )\Rightarrow
\text{it follows that q is also divisible by} \left (n^2-5 \right)
\\
\text{Therefore p and q are not co-prime therefore} \Rightarrow \sqrt{n^2-5} \notin \mathbb{Q} \ \square$$
This proof must be wrong because $n=3$ leads to $\sqrt{n^2-4}=2\in\mathbb Q$.
How can the proof be repaired?
First show that $n^2-4$ is not a perfect square (which would be a desaster for the claim, as $n^2-4=m^2$ implies $\sqrt{n^2-4}=m\in\mathbb Q$).
So assume $n^2-4=m^2$ with $0\le m<n$. Then $4=n^2-m^2=(n+m)(n-m)$. Compare this with the few possible factorizatons of $4$: The possibility $n+m=4$, $n-m=1$ leads to $n=\frac 52\notin\mathbb N$; the posssibility $n+m=n-m=2$ leads to $n=2$, which fails the important given condition $n\ge 3$.
So we conclude that $n^2-4$ is not a perfect square. Can you show (or do you already know) the important theorem:
If the positive integer $N$ is not a perfect square, then $\sqrt N\notin\mathbb Q$.
Hint: One can write $N=m^2k$ where $k$ is a product of one or more distic primes. Use one of these primes to show that $\sqrt k$ is irrational.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Taylor Expansion of complex function $ \frac{1}{1 - z - z^2} $ Knowing that
$$c_{n+2}=c_{n+1}+c_n, \quad n \geq 0,$$
the problem is to prove that we have
$$ \frac{1}{1 - z - z^2}=\sum^{+\infty}_{n=0} c_nz^n.$$
I have tried to transform the function into the form:
$$ A(\frac{1}{z-b_1}+\frac{1}{b_2-z})$$
Then expand $\frac{1}{z-b_1} \text{and} \frac{1}{b_2-z}$
However, the result seems having nothing to do with $c_{n+2}=c_{n+1}+c_n$
Am I wrong? How should I prove this in a elegant way?
| use this
$$\frac{8 x^2}{\sqrt{5} \left(\sqrt{5}+1\right)^2 \left(2 x+\sqrt{5}+1\right)}+\frac{8 x^2}{\sqrt{5} \left(\sqrt{5}-1\right)^2 \left(-2 x+\sqrt{5}-1\right)}+x+1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Trigonometry equation with arctan Solve the following equation: $\arctan x + \arctan (x^2-1) = \frac{3\pi}{4}$.
What I did
Let $\arctan x = \alpha, \arctan(x^2-1) = \beta$, $\qquad\alpha+\beta = \frac{3\pi}{4}$
$\tan(\alpha+\beta) = \tan(\frac{3\pi}{4}) = -1$
$$\frac{\tan\alpha + tan\beta}{1-\tan\alpha\tan\beta} = \frac{x+x^2-1}{1-x(x^2-1)} = -1$$
$\begin{align}
x^2+x-1 &= -(1-(x^3-x)) = -1+x^3-x \\
\iff x^2 + x &= x^3-x \\
\iff x(x+1) &= x(x^2-1) \qquad\implies \boxed{x_1 = 0}\\
\implies x+1 &= x^2-1 \\
\iff x^2-x-2 &= 0 \\
\end{align}$
$\therefore x_1 = 0,\quad x_2 = 2,\quad x_3 = -1$
However, the equation only works for $x=2$.
I wonder
Did I do this in an efficient manner? Is there any easy way to find $x$ where there's no fake solutions?
| Your three answers all look good to me. You did this in a pretty efficient manner, although I would've just jumped straight to the identity $$\arctan(A)+\arctan(B) = \arctan\left(\frac{A+B}{1-AB} \right)$$ without making the substitution $\arctan(x) = \alpha$, etc.
Anyway, you are confident that the equation works for $x=2$ (which it does.) But for $x=0$ note that a calculator will tell us that $$\arctan(0)+\arctan(0^2-1) = 0+\arctan(-1) = \frac{-\pi}{4}$$ However, this is because $$\tan\left( \frac{3\pi}{4}\right) =\tan\left( \frac{-\pi}{4} \right)$$ where there is a discrepancy with the calculator due to the fact that Cosine is negative and Sine is positive in the quadrant where $\frac{3\pi}{4}$ lies, and Cosine is positive while Sine is negative in the quadrant where $\frac{-\pi}{4}$ lies. Long story short, the calculator doesn't know the difference in the calculation and defaults to $\frac{-\pi}{4}$. The same exact thing happens for $x = -1$. All three of your answers are right though.
| {
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The common tangents to the circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$ form an equilateral triangle Problem :
Show that the common tangents to circles $x^2+y^2+2x=0$ and $x^2+y^2-6x=0$ form an equilateral triangle.
Solution :
Let $C_1 : x^2+y^2+2x=0$
here centre of the circle is $(-1,0) $ and radius 1 unit.
$C_2:x^2+y^2-6x=0$
here centre of the circle is $(3,0) $ and radius 3 units.
But how to proceed to prove that the tangents form equilateral triangle please suggest thanks.
| Clearly the $Y$ axis (i.e. $x=0$) is a common tangent. Let $y=mx+c$ be the equation of the other common tangent(s). Then we need both the quadratics
$$x^2+(mx+c)^2+2x=0 ; \qquad x^2+(mx+c)^2-6x=0$$
to have zero discriminant (why?). This gives us the conditions
$$(cm+1)^2=(m^2+1)c^2; \qquad (cm-3)^2=(m^2+1)c^2$$
Solving these, we have $\pm\sqrt3 y=x+3$ as the other tangents. Quite obviously the intersection points are then $(-3, 0), (0, \pm \sqrt3)$ and it easily follows that the distance between any two vertices is $\sqrt{3^2+3}=2\sqrt3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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If $x=t^2\sin3t$ and $y=t^2\cos3t$, find $\frac{dy}{dx}$ in terms of $t$ If $x=t^2\sin3t$ and $y=t^2\cos3t$, find $\frac{dy}{dx}$ in terms of $t$. This is how I tried solving it:
$$
\frac{dx}{dt} = 2t\sin3t + 3t^2\cos3t \\
\frac{dy}{dt} = 2t\cos3t - 3t^2\sin3t \\
\frac{dy}{dx} = \frac{2t\cos3t - 3t^2\sin3t}{2t\sin3t + 3t^2\cos3t}
$$
But the answer listed is:
$$
\frac{2-3t\tan3t}{2\tan3t+3t}
$$
Is my answer incorrect, or can I simplify it even more?
| Your answer is correct, $\frac{2t\cos3t - 3t^2\sin3t}{2t\sin3t + 3t^2\cos3t}=\frac{t\cos{3t}}{t\cos{3t}}\frac{2-3t\frac{\sin{3t}}{\cos{3t}}}{2\frac{\sin{3t}}{\cos{3t}}+3t}=\frac{2-3t\tan3t}{2\tan3t+3t}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trouble finding second implicit derivative I have trouble finding the second implicit derivative.
This is the question.
Find y'' in terms of x and y by implicit differentiation.
$x^5 +y^5 = 2^5$
The final answer I always get is $\displaystyle -\frac{(4x^3)(32)}{y^9}$.
I might be doing something wrong
| First, $\displaystyle x^5+y^5=2^5 \rightarrow 5x^4+5y^4\cdot{y'}=0 \Rightarrow y'=-\frac{x^4}{y^4}$.
Second, $\displaystyle 5x^4+5y^4\cdot{y'}=0 \rightarrow 20x^3+20y^3\cdot{y'}+5y^4\cdot{y''}=0 \Rightarrow y''=-\frac{4x^3+4y^3\cdot{y'}}{y^4}$.
Using $\displaystyle y'=-\frac{x^4}{y^4}$ we will get $\displaystyle y''=-\frac{4x^3+4y^3\cdot{(-\frac{x^4}{y^4})}}{y^4}=-\frac{4x^3-\frac{4x^4}{y}}{y^4}=\frac{4x^3(x-y)}{y^5}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$2^{50} < 3^{32}$ using elementary number theory How would you prove; without big calculations that involve calculator, program or log table; or calculus that
$2^{50} < 3^{32}$
using elementary number theory only?
If it helps you: $2^{50} - 3^{32} = -727120282009217$, $3^{32} \approx$ $2^{50.718800023077\ldots}$, $3^{32} $ $\div 2^{50}$ $=$ $1.6458125430068558$ (thanks to Henry).
| Note first that this is equivalent to $2^{25}\lt 3^{16}$ or $2\cdot 2^{24}\lt 3\cdot 3^{15}$ or $2\cdot 4^3\cdot 2^{24}\lt 3\cdot 4^3 \cdot 3^{15}$
Now $4\cdot 2^8=1024=10^3+24$ and $4\cdot 3^5=972=10^3-28$
So we can rewrite the last form of the inequality as equivalent to $$2\cdot(10^3+24)^3\lt 3\cdot (10^3-28)^3$$
Expanding the factors gives $$2\cdot 10^9+6\cdot24\cdot 10^6+6\cdot24^2\cdot10^3+24^3\lt 3\cdot 10^9-9\cdot28\cdot10^6+9\cdot 28^2\cdot10^3-28^3$$
Rearranging, this gives:
$$10^9+(9\cdot 28^2-6\cdot24^2)\cdot10^3\gt(9\cdot28+6\cdot24)\cdot10^6+28^3+24^3$$
Now it is obvious that this is true, because $9\cdot 28+6\cdot 24\lt 300+150\lt 500$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$ I would like some guidance regarding the following integral:
$$\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$$
EDIT: The upper problem was derived from the following integral $$\int\frac{\sqrt{x^2+2}}{x^2+1}dx$$
Where I rationalized the numerator which followed into: $$\int\frac{dx}{\sqrt{x^2+2}}+\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$$
| Kinda messy, but gets the answer.
Lets write $x^2+2=t \rightarrow x^2+1=t-1 \ and \ x=\pm\sqrt{t-2}$.
For the case $x=\sqrt{t-2}$ we will get $\displaystyle \frac{dt}{dx}=2x \rightarrow dx=\frac{dt}{2x}=\frac{dt}{2\sqrt{t-2}}$.
Now $$\int{\frac{dx}{(x^2+1) \sqrt{x^2+2}}}=\int\frac{\frac{dt}{2\sqrt{t-2}}}{(t-1)\sqrt{t}}=\frac{1}{2}\int\frac{1}{(t-1)\sqrt{t^2-2t}}dt=\frac{1}{2}\int\frac{1}{(t-1)\sqrt{(t-1)^2-1}}dt$$
Define $t-1=m \rightarrow dt=dm$ we get $\displaystyle \frac{1}{2}\int\frac{1}{m\sqrt{m^2-1}}dm$.
Define $\displaystyle p=\sqrt{m^2-1} \rightarrow \frac{dp}{dm}=\frac{2m}{2\sqrt{m^2-1}}=\frac{m}{p} \rightarrow dm=\frac{dp\cdot{p}}{m}$, hence $$\frac{1}{2}\int\frac{1}{m\sqrt{m^2-1}}dm=\frac{1}{2}\int\frac{1}{mp}\cdot{\frac{dp\cdot{p}}{m}}=\frac{1}{2}\int\frac{1}{m^2}dp=\frac{1}{2}\int\frac{1}{p^2+1}dp=\frac{1}{2}arctan(p)+C$$
We know that $p=\sqrt{m^2-1}=\sqrt{(t-1)^2-1}=\sqrt{(x^2+1)^2-1}$.
| {
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If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\begin{align*}
(a + ib)^3 &= 8\\
a + \sqrt{4-a^2} &= 2\\
\sqrt{4-a^2} &= 2 - a\\
2 - a &= 2 - a
\end{align*}
$$
Therefore if $(a + ib)^3 = 8$, then $a^2 + b^2 = 4$.
| Hint: $$
a^2 + b^2 = (a+ib)(a-ib)
$$
as $(a+ib)^3 = 8$, $(a-ib)^3 = 8$ and
$$
(a^2 + b^2)^3 = 8^2 = 64 \implies a^2 + b^2 = 4
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculus 1: Find the limit as x approaches 4 of $\frac{3-\sqrt{x+5}}{x-4}$ I understand how to find limits, but for some reason I cannot figure out the algebra of this problem. I tried multiplying by the conjugate and end up with 0/0. When I check on my calculator, or apply L'Hopital's rule I get -1/6. Is there an algebra trick that I am missing on this one?
$\displaystyle\frac{3-\sqrt{x+5}}{x-4}$
I have solved similar problems with the square root by multiplying by the conjugate, but it doesn't seem to work for this one.
| \begin{gathered}
\frac{{3 - \sqrt {x + 5} }}{{x - 4}} = \frac{{\left( {3 - \sqrt {x + 5} } \right)\left( {3 + \sqrt {x + 5} } \right)}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\
= \frac{{{3^2} - {{\sqrt {x + 5} }^2}}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\
= \frac{{9 - (x + 5)}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\
= \frac{{ - (x - 4)}}{{\left( {x - 4} \right)\left( {3 + \sqrt {x + 5} } \right)}} \\
= \frac{{ - 1}}{{\left( {3 + \sqrt {x + 5} } \right)}} \\
\end{gathered}
Take the limit of that as $x$ approaches $4$.
| {
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solve indefinite integral I have this indefinite integral $\int 3 \sqrt{x}\,dx$ to solve.
My attempt:
$$\int 3 \sqrt{x}\,dx = 3 \cdot \frac {x^{\frac {1}{2} + \frac {2}{2}}}{\frac {1}{2} + \frac {2}{2}}$$
$$\int 3 \sqrt{x}\,dx = 3 \frac{x^{\frac {3}{2}}}{\frac {3}{2}} = \frac{2}{3} \cdot \frac{9}{3} x^{\frac {3}{2}}$$
$$\int 3 \sqrt{x}\,dx = \frac{18}{3} x^{\frac{3}{2}} = 6 x^{\frac{3}{2}}$$
But according to wolframalpha the answer should be $2 x^{\frac {3}{2}}$
Where did I make a error in my calculation?
Thanks!
| $\frac{2}{3} \cdot \frac{9}{3}=2$. not $6$
| {
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Efficiently solving algebraic equation I would like to solve following equation:
$$15 (x+2)^{-4} = 11(x+2)^{-2} +4$$
I would first remove the negative power by adding $(x+2)^4$
Then I get
$$15 = 11(x+2)^2 + 4(x+2)^4\\
11(x+2)^2 + 4(x+2)^4 -15 = 0$$
should I do now a quadratic equation where $d=(x+2)^2$
$$11d + 4d^2 -15 =0$$
But this would be very much work
The other way is to expand them and factorize them. but this is even more work.
Isnt there are simple way?
| $$15(x+2)^{-4} = 11(x+2)^{-2} + 4$$
Multiply by $(x+2)^{4}$
$$15 = 11(x+2)^2 + 4(x+2)^4$$
Let $y = (x+2)^2$
$$15 = 11y + 4y^2$$
Subtract $15$
$$4y^2 + 11y - 15=0$$
Factor
$$(4y + 15)(y - 1)=0$$
Consider both cases:
$$(4y+15)=0$$
$$ y = -\frac{15}{4}$$
and
$$(y-1)=0$$
$$y=1$$
So, the two solutions for $y$ are $-\frac{15}{4}$ and $1$.
Substitute the values for $y$ into the expression containing $x$.
$$1 = (x+2)^2$$
$$\pm 1 = x+2$$
$$x = -2 \pm 1$$
$$x = -1, -3$$
and
$$-\frac{15}{4} = (x+2)^2$$
$$\pm \frac{\sqrt{15}i}{2} = (x+2)$$
$$x = -2 \pm \frac{\sqrt{15}i}{2}$$
So all of the solution for $x$ are:
$-1, -3, -2 + \dfrac{\sqrt{15}i}{2},$ and $-2 - \dfrac{\sqrt{15}i}{2}$
| {
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Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$. Also write the identity used Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$.
Also write the identity used
| Let $S=1^3+2^3+4^3-5^3-6^3.$ Then
$$S=3(1^3+2^3+3^3+4^3)-2(1^3+2^3+3^3)\\
-(1^3+2^3+3^3+4^3+5^3+6^3)+(1^3+2^3).$$
Now apply the formula $1^3+2^3+\cdots+k^3=(1+2+\cdots +k)^2.$
Then $$S=3\cdot 10^2-2\cdot 6^2 - 21^2 +3^2=-204.$$
This may not be what you want since we still need to square numbers to finish. [At least we don't have to cube any of them :)] But it seems there must be some computation necessary to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/985039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to find the roots of $(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$
Write down, in any form, all the roots of the equation $z^5 − 1 = 0$
Hence find all the roots of the equation
$$(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$$
and deduce that none of them is real
My Try:
I know how to do the first part:
$$z^5=1=cos 2\pi k + i sin 2\pi k$$
$$z= cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}$$
where $k=0,1,2,3,4$
Please help me to do the second part. Thanks.
Attempt:
$$z^5-1=(z-1)(z^4+z^3+z^2+1)=0$$
Have to find:
$$(w−1)^4 +(w−1)^3 +(w−1)^2 +w=0$$
Let $w=z+1$
$$z^4+z^3+z^2+z+1=0$$
Multiply by $(z-1)$ each side:
$$(z-1)(z^4+z^3+z^2+z+1)=0 (z-1)=0$$
$$z^5-1=(z-1)(z^4+z^3+z^2+z+1)=0$$
$$z^5-1=0$$ (back to Inital result)
Then
$$z=cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}$$
Since $w=z+1$ So $z=w-1$
$$w-1=cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}$$
$$w=cos \frac{2\pi k}{5}+ i sin \frac{2\pi k}{5}+1$$
This seems weird, is this correct?
| $$y:=w-1 \iff y^4+y^3+y^2+y+1=0 \stackrel{\cdot (y-1)}{\iff} (y-1)(y^4+y^3+y^2+y+1)=0 \iff y^5-1=0 \;\forall y\neq 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the coefficient of $x^{24}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^8$ I'm not sure how to go about doing this. Do I find the ways to add up to 24 using the exponents with repetition? Is the multinomial theorem useful here? I also have a feeling that generating functions might be useful here, but I can't see how. Any help would be appreciated.
| $$(1+x+\dots+x^5)^8=\left(\frac{1-x^6}{1-x}\right)^8=(1-x^6)^8(1-x)^{-8}$$
Using the binomial theoerem,
$$
(1-x^6)^8=\sum_{k\ge0}(-1)^k\binom{8}{k}x^{6k}
$$
and using the negative binomial theorem,
$$
(1-x)^{-8}=\sum_{k\ge0}(-1)^k\binom{-8}{k}x^k=\sum_{k\ge0}\binom{8+k-1}{k}x^k
$$
Thus, when we convolve the above two generating functions, the $x^{24}$ coefficient is
$$
\binom{8}{0}\binom{8+24-1}{24}-\binom{8}{1}\binom{8+18-1}{18}+\binom{8}{2}\binom{8+12-1}{12}\\
-\binom{8}{3}\binom{8+6-1}{6}+\binom{8}{4}\binom{8+0-1}{0}
$$
Addendum: If $a(x)=\sum_{n\ge0}a_nx^n$ and $b(x)=\sum_{n\ge 0}b_nx^n$, then the $x^{24}$ coefficient of $c(x)=a(x)b(x)$ is
$$
\sum_{k=0}^{24}a
_kb_{n-k}
$$
The final answer I wrote then comes from setting $a(x)=(1-x^6)^8$, $b(x)=(1-x)^{-8}$, and realizing that $a_k=0$ unless $k$ is a multiple of 6, so the above can be rewritten
$$
\sum_{\ell=0}^{4}a
_{6\ell}b_{n-6\ell}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested with various values of $a$ ( where $0<a<1$).
$$
D_1
\, =\,
\int_0^{2\pi}f_1\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
and
$$D_2\, =\,\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R$$
The hypothesis: $D_1$ = $D_2$ has been proved in a separate question Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ .
The remaining hypotheses $D_1$ = $R$ and $D_2$ = $R$ have not been proved. So the question is:-
Prove $D_1$ = $R$ or $D_2$ = $R$.
Only one proof is required because the other can then be obtained from $D_1$ = $D_2$.
For information
WolframAlpha computes expressions for the indefinite integrals $I_1,I_2$ as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
and
$$I_2
\, =\,
\int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
$$
$$constant2 -
\frac
{2a^2\sin\theta-sin\theta}
{2(a^2-1)^2(a\cos\theta-1)}
-\frac
{\sin\theta}
{2(a^2-1)(a\cos\theta-1)^2}
$$
$$
-\frac
{3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)}
{(a^2-1)^{5/2}}
$$
Note that the final terms of each expression ( i.e. the terms involving $\tanh^{-1} $ and $\tan$ ) are equivalent to each other.
Also, note that
$$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\frac{-\sin\theta}{(1-a\cos\theta)^3}
+\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta.
$$
Written with StackEdit.
UPDATE 20141028
I have accepted TenaliRaman's answer. I don't yet understand all the steps but his helpful exposition gives me confidence that with time I can understand it because the methods cited (binomials, series) are ones I have learned (at high school).
The answer of M.Strochyk also appears to give a good proof. But the residue method is too advanced for me to understand at present.
UPDATE 20220713
I have now accepted Quanto's answer (because it is simple enough for me to understand). I have also added an answer based on Quanto's but with the intermediate steps written out.
| NOTE: This is a continuation (the easy bit!) of M.Strochyk's answer.
using $$z = \frac{1-\sqrt{1-a^{2}}} {a}
\mathrm{and}\,\mathrm{ defining}\, Q=\sqrt{1-a^2}$$
so that $$\left(
z - \frac{1+\sqrt{1-a^{2}}} {a}
\right) = \frac{-2Q}{a}$$
We can analyze the three terms of the Residue separately and merge them later
$$$$
TERM 1
$$
\frac
{3 z}{{\left(
z - \frac{1+Q} {a}
\right)}^{3}}
= \frac
{3 \frac{1-Q} {a}}
{{\left(
\frac{-2Q}{a}
\right)}^{3}}
= \frac
{3 a^2 \left(1-Q \right) (-2Q)^{2}}
{{(-2Q)}^{5}}
$$TERM 2$$
- \frac{3 {\left(3 z^{2} + 1\right)}}{{\left(z - \frac{1+Q }{a}\right)}^{4}}
=- \frac{3a^4 {\left(3 (\frac{1-Q}{a})^{2} + 1\right) (-2Q) }}
{{\left(-2Q\right)}^{5}}
$$
TERM 3
$$
+ \frac{6 {\left((z)^{3} + z\right)}}{{\left(z - \frac{1+Q}{a}\right)}^{5}}
=+ \frac{6a^5 {\left((\frac{1-Q}{a})^{3} + (\frac{1-Q} {a})\right)}}
{{\left(-2Q\right)}^{5}}
$$
Summing together the three terms
$$
= \frac
{3 a^2 \left(4Q^2-4Q^3 \right)
+6Qa^2 {\left( 3(1-2Q+Q^2)+a^2\right) }
+6a^2 {\left((1-Q)^3+a^2(1-Q) \right)}
}
{{(-2Q)}^{5}}
$$
reducing to
$$
= \frac
{ \left(12a^2Q^2-12a^2Q^3 \right)
+{\left( 18a^2Q-36a^2Q^2+18a^2Q^3+6a^4Q\right) }
+6a^2 {\left((1-3Q+3Q^2-Q^3)+a^2-Qa^2 \right)}
}
{{(-2Q)}^{5}}
$$
and
$$
= \frac
{ \left(12a^2Q^2-12a^2Q^3 \right)
+{\left( 18a^2Q-36a^2Q^2+18a^2Q^3+6a^4Q\right) }
+ \left(6a^2-18a^2Q+18a^2Q^2-6a^2Q^3+6a^4-6a^4Q \right)
}
{{(-2Q)}^{5}}
$$
then
$$
= \frac
{
-6a^2Q^2
+ 6a^2+6a^4
}
{{(-2Q)}^{5}}
$$
So
$$
=\frac{6}{32} \frac
{(
a^2Q^2
- a^2-a^4 )
}
{Q^5}
$$
Multiplying by $-8\pi/a^3$ gives
$$
=-\frac{3\pi}{2a} \frac
{(
Q^2
- 1-a^2 )
}
{Q^5}
$$
but $Q^2$ = $1-a^2$ so
$$
=-\frac{3\pi}{2a} \frac
{(
-2a^2 )
}
{Q^5}
$$
giving
$$
= \frac
{3a\pi }
{(1-a^2)^{5/2}}
$$
which is the hypothesised result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/990813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Why can't prime numbers satisfy the Pythagoras Theorem? That is, why can't a set of 3 prime numbers be a Pythagorean triplet? Suppose $a$, $b$ and $c$ are three prime numbers.
How to prove that $a^2 + b^2 \neq c^2$?
| Primes are all odd expect $2$, so if $a, b, c$ don't contain $2$, $a^2 + b^2$ is even but $c^2$ is odd, then $a^2 + b^2 = c^2$ can't be true.
Of course if $c=2$, then $a^2 + b^2 = c^2$ can't be true.
If $a = 2$, then $c>b$ then $c^2 - b^2 \geq (b+2)^2 - b^2 = 4b + 4 > a^2 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/991947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 1
} |
How to express $z^8 − 1$ as the product of two linear factors and three quadratic factors
Verify
$$(z-e^{i \theta} ) (z - e^{-i \theta} ) ≡ z^2 - 2\cos \theta + 1$$
Hence express $z^8 − 1$ as the product of two linear factors and three quadratic factors, where all
coefficients are real and expressed in a non-trigonometric form.
For the first part I just expanded the LHS and showed its equal to RHS.
Roots of $z^8-1$:
$$z= e^{i \frac{\pi k}{4}}$$
Where $k=0,1,2,3,4,5,6,7$
Ok since they want non-Trignometric so :
I know two roots, which are obvious:
$$z=1,-1$$
$$z^8-1=(z-1)(z+1)(z^6+z^5+z^4+z^3+z^2+z+1)$$
They want two linear factors which I believe I have found: $(z+1)$ and $(z-1)$
How do I make
$$(z^6+z^5+z^4+z^3+z^2+z+1)$$
to three quadratic factors? And they mentioned Hence , so I have to use the identity I verified. Please help.
| Here is a method which make use of your identity.
$$z^8-1=(z-1)(z+1)(z^2+1)(z^4+1)$$
\begin{align*}
(z^4+1)&=(z^2+i)(z^2-i)\\
&=(z^2-e^{i-\frac{\pi}{2}})(z^2-e^{i\frac{\pi}{2}})\\&=(z+e^{-i\frac{\pi}{4}})(z-e^{-i\frac{\pi}{4}})(z+e^{i\frac{\pi}{4}})(z-e^{i\frac{\pi}{4}})\\&=(z-e^{-i\frac{5\pi}{4}})(z-e^{i\frac{5\pi}{4}})(z-e^{i\frac{\pi}{4}})(z-e^{-i\frac{\pi}{4}})\end{align*}
Then use your formula, you get $$z^4+1=(z^2+\sqrt2 z+1)(z^2-\sqrt2 z+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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