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Polynomial with "symmetrical" coefficients, and $\gcd(p,p')$ Consider the following polynomial:
$p(x) = x^6+x^5+2x^4+x^3+2x^2+x+1$. I would like to calculate $\gcd(p,p')$.
With Euclid's GCD algorithm, the calculation becomes very complicated, very quickly. I was wondering if the fact that the coefficient vector is "palindromic" could be of some use, but I didn't notice any interesting pattern while I was doing the calculation. Is there anything I'm missing?
| $p(x)=x^6 + x^5 + x^4 +x^4+ x^3 + x^2 +x^2+ x + 1$
$p(x)=x^4(x^2+x+1)+x^2(x^2+x+1)+1(x^2+x+1)=(x^2+x+1)(x^4+x^2+1)=(x^2+x+1)[(x^4+2x^2+1)-x^2]=(x^2+x+1)[(x^2+1)^2-x^2]=(x^2+x+1)(x^2+x+1)(x^2-x+1)=(x^2+x+1)^2(x^2-x+1)$
$p'(x)=2(x^2+x+1)(2x+1)(x^2-x+1)+(x^2+x+1)^2(2x-1)=(x^2+x+1)(6 x^3-x^2+3 x+1)$
Let $q(x)=6x^3-x^2+3x+1$
$q(\pm1)\ne 0;\;q\left(\pm\frac12\right)\ne 0;\;q\left(\pm\frac13\right)\ne 0;\;q\left(\pm\frac16\right)\ne 0$
$q(x)$ has no rational roots therefore it is irreducible in $\mathbb{Q}$
$\text{GCD}(p(x),p'(x))=x^2+x+1$
Hope this helps
| {
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Partial fractions decomposition of ${\frac{2x}{(x+2)^2}}$ Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g. $$2x = A\cdot (x+2) + B$$
Substitute $x=-2$
$2\cdot(-2)$ = $A\cdot (2-2) +B$
$-4 = B$
In other questions there is always another factor to multiply by at this stage.
| it must be
$$\frac{2x}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$$
| {
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A small trouble understanding the AMGM inequality usage for some basic minimization problems? I'm reading Byrne's "A first course in optimization". After the presentation and proof of the AMGM inequality, the author gives two examples:
Here's the first:
I guess I understand what was done, it seems he writes in a convenient form with the AMGM inequality:
$$\left(\frac{3\cdot12}{x} \cdot\frac{3\cdot18}{y} \cdot 3\cdot xy\right)^{\frac{1}{3}}\leq \frac{1 }{3} \left( \frac{3\cdot12}{x} +\frac{3\cdot18}{y} + 3\cdot xy\right)$$
Which in turn, yields:
$$\left(3^3\cdot 12 \cdot 18\right)^{\frac{1}{3}}=\left(3^3 2^3 3^3\right)^{\frac{1}{3}}=18\leq \frac{12}{x} +\frac{18}{y} + xy$$
From here, I know that for all possible choices of $x,y$, then $18$ is the smaller value, but I'm not sure on how to find the actual $x,y$. Given that he mentions that the smaller value for $\frac{1}{3} f(x,y)$ is $6$, then I guess he did the following:
$$\frac{1}{3} f(x,y)+\frac{1}{3} f(x,y)+\frac{1}{3} f(x,y) \leq 6+6+6=18$$
$$\overbrace{\left(\frac{4}{x} +\frac{6}{y} + \frac{xy}{3}\right)}^{= \;6}+\overbrace{\left(\frac{4}{x} +\frac{6}{y} + \frac{xy}{3}\right)}^{=\;6}+\overbrace{\left(\frac{4}{x} +\frac{6}{y} + \frac{xy}{3}\right)}^{=\;6}\leq 6+6+6=18$$
$$\overbrace{\left(\frac{4}{x} +\frac{4}{x} + \frac{4}{x}\right)}^{= \;6}+\overbrace{\left(\frac{6}{y} +\frac{6}{y} + \frac{6}{y}\right)}^{= \;6}+\overbrace{\left(\frac{xy}{3} +\frac{xy}{3} + \frac{xy}{3}\right)}^{= \;6}=\frac{12}{x} +\frac{18}{y} + xy\leq 18$$
And it seems that solving each of them individually yields the desired result, I am confused at why reorganizing $3$ sums whose total is $6$ in this way (supposing it is what actually was done) yields exactly the the minimal solution.
My doubt in the following example may stem from the same doubt I had before:
Here, he talks about a "constant sum" which is perhaps something very similar to what was done before, but I don't understand what is this "constant sum" nor why it is needed, I see that the second centered equation is the same as the first but I don't understand the workings of this "constant sum". I have tried to rewrite the equation in several other ways (on paper) to check if I could uncover something but I couldn't. Could you clarify?
| In both cases it was used:
$$a+b+c\ge 3\sqrt[3]{abc},$$
the equality occurs when $a=b=c$.
Example 1:
$$\frac{12}{x}+\frac{18}{y}+xy\ge 3\sqrt[3]{\frac{12}{x}\cdot \frac{18}{y}\cdot xy}=18,$$
$$\frac{12}{x}=\frac{18}{y}=xy \Rightarrow x=2, y=3$$
Example 2:
$$\frac{1}{12}\cdot (3x)(4y)(72-3x-4y)\le \frac{1}{12}\left(\frac{3x+4y+72-3x-4y}{3}\right)^3=1152, $$
$$3x=4y=72-3x-4y \Rightarrow x=8, y=6.$$
| {
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Stationary points and linearisation of non-linear system So, the problem is:
Find and discuss the behavior of the stationary points of the system :
$$ x'=-y+x\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2} =f(x,y)$$
$$ y'=x+y\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}=g(x,y)$$
So in the beggining I Linearised the non-linear system, using the limits:
$$ \lim_{ r\to 0}\frac{f_1(x,y)}{r}=0 $$ and $$\lim_{r \to 0}\frac{g_1(x,y)}{r}=0$$
where $$f_1(x,y)=x\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}$$ and
$$g_1(x,y)=y\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}$$
I linearise the non-linear system and reach it to the form:
$$x'=-y$$ and $$y'=x$$
but the problem is that i don't know how to calculate the stationary points from this non-linear system so to proceed with the next question of my problem .I also know that when i find the stationary points i have to take the jacobian so to characterise my stationary points.I would really, really appreciate any thorough help or hints/tips.
Thanks in advance!
| In order to get the critical points, we need to solve the system :
$$\begin{cases} -y+x\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2} =0 \\ \space \space\space x+y\cdot (x^2+y^2)\cdot \sin\sqrt{x^2+y^2}=0\end{cases}$$
The obvious solution is $x=0,y=0$, which means that $O(0,0)$ is a Stanionary Point of your system.
To prove that it's unique, let's assume $x,y\neq 0$ and multiply the first equation by $-y$ and the second one by $x$. We then get :
$$\begin{cases} y^2 -xy(x^2+y^2)\sin(\sqrt{x^2+y^2}) = 0 \\ x^2 + xy(x^2 + y^2)\sin(\sqrt{x^2 + y^2}) = 0\end{cases}$$
By adding these two, we obviously get :
$$ x^2 + y^2 = 0 $$
which can only hold for $x=y=0$, which is not compatible with our initial hypothesis.
Now, let's assume $x\neq 0,y=0$. We then get (through the initial given system) :
$$\begin{cases} x^3\cdot \sin\sqrt{x^2}=0\\x=0\end{cases}$$
which is also not compatible, since we had let $x\neq 0$.
Letting $x=0,y\neq 0$ gets us to :
$$\begin{cases} y=0 \\ y^3\sin\sqrt{y^2}=0\end{cases}$$
which is as well not compatible, since we had let $y\neq0$.
Thus, the only solution is : $\begin{cases} x=0 \\ y=0 \end{cases}$ and your only stationary point is the $O(0,0)$.
| {
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Proving $ \lim\limits_{n\to\infty} \frac{8n^2-5}{4n^2+7} = 2$ using $\epsilon-\delta-$ definition. I'm trying to prove the limit of this sequence using the formal definition. I've looked at other questions on the site but from the ones I've seen, the $n^2$ term always seems to cancel out, making it simpler.
Show that $$ \lim_{n\to\infty} \frac{8n^2-5}{4n^2+7} = 2$$
So this is how I started:
$$ \left|\frac{8n^2-5}{4n^2+7} -2 \right| = \left|\frac{-19}{4n^2+7}\right| = \frac{19}{4n^2+7} \leq \frac{19}{4n^2} = \epsilon$$
Let $\epsilon > 0 \implies n = \sqrt{\frac{19}{4\epsilon}}$
By Archimedian Property: $ N > \sqrt{\frac{19}{4\epsilon}} $
If $ n \geq N \geq \sqrt{\frac{19}{4\epsilon}} $
$$ \left|\frac{8n^2-5}{4n^2+7} -2 \right| \leq \frac{19}{4n^2} < \frac{19}{4(\frac{19}{4\epsilon})} = \epsilon $$
| we have $$\left|\frac{8n^2-5}{4n^2+7} -2 \right| = \left|\frac{-19}{4n^2+7}\right| < \frac{19}{4n^2} = $$
Now let such that $$\frac{19}{4n^2} < \epsilon\Longleftrightarrow n>\sqrt{\frac{19}{4\epsilon}} $$ Now choosing,
$$N =\left\lfloor \sqrt{\frac{19}{4\epsilon}}\right\rfloor +1$$ then we have $$n > N=\left\lfloor \sqrt{\frac{19}{4\epsilon}}\right\rfloor +1\Longrightarrow n>\sqrt{\frac{19}{4\epsilon}}\\\Longrightarrow \epsilon>\frac{19}{4n^2} > \left|\frac{8n^2-5}{4n^2+7} -2 \right|$$
that is we have $$n > N=\left\lfloor \sqrt{\frac{19}{4\epsilon}}\right\rfloor +1\Longrightarrow \left|\frac{8n^2-5}{4n^2+7} -2 \right|< \epsilon$$
| {
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Value of $e$ given that $\frac{a+b+c+d+e}{5} = 2 = \sqrt{\frac{a^2 + b^2 + c^2 + d^2 + e^2}{5}}$ I have five real numbers $a,b,c,d,e$ and their arithmetic mean is $2$. I also know that the arithmetic mean of $a^2, b^2,c^2,d^2$, and $e^2$ is $4$. Is there a way by which I can prove that the range of $e$ (or any ONE of the numbers) is $[0,16/5]$. I ran across this problem in a book and am stuck on it. Any help would be appreciated.
| By C-S $$(1^2+1^2+1^2+1^2)(a^2+b^2+c^2+d^2)\geq(a+b+c+d)^2$$ or
$$4(a^2+b^2+c^2+d^2)\geq(a+b+c+d)^2$$ or
$$4(20-e^2)\geq(10-e)^2$$ or
$$(e-2)^2\leq0$$ or
$$e=2.$$
This method works in the general case.
Given: $a+b+c+d+e=k$ and $a^2+b^2+c^2+d^2+e^2=l$.
Find the range of $e$.
| {
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How to simplify: $f(x)=(2^a-1)(3^a-1)(4^a-1)...(x^a-1)\mod {m}$ How to simplify $f(x)$ for positive integers $a,m,x$ where $x>1$ and $m$ is a composition of two very large primes, such that factoring $m$ is not an option. $a$ is relatively small number that doesn't have common divider with $m$
$f(x)=(2^a-1)(3^a-1)(4^a-1)...(x^a-1)\mod {m}$
I have been trying to play with small numbers to see how this expression looks like and come with below examples, my goal is to be able to solve this for big numbers of $x$, at the moment I can only solve this recursively.
$$f(3)=(2^a-1)(3^a-1)=$$
$$-2^a - 3^a + 6^a + 1$$
$$f(4)=(2^a-1)(3^a-1)(4^a-1)=$$
$$2^a + 3^a + 4^a - 6^a - 8^a - 12^a + 24^a - 1$$
$$f(5)=(2^a-1)(3^a-1)(4^a-1)(5^a-1)=$$
$$-2^a - 3^a - 4^a - 5^a + 6^a + 8^a + 10^a + 12^a + 15^a + 20^a - 24^a - 30^a - 40^a - 60^a + 120^a + 1$$
| This expression will be $0 \mod m\;$ in specific cases, for example if $x\ge m,$ because then one of the terms $n+i$ will be 1 $\pmod m,$ or if $a$ is a multiple of the Euler phi function $\varphi(m),$ because then $(n+i)^a\equiv 1 \pmod m$ for all $i.$
| {
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Evaluating $\int_1^\infty \frac{1}{x(x^2+1)}\ dx$ $$I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx$$
I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.
$$\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i}\right)$$
So
$$I = \frac{1}{2}\int_1^\infty\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i} \ dx \\ = \frac{1}{2}\left[\log |x| - \log(|x+i|) - \log(|x-i|)\right]_1^\infty
$$
which evaluates to be $\infty$.
| Why complicate it with complex coefficients?
$$\int_1^\infty\frac{1}{x(x^2+1)}dx=\int_1^\infty\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx=\left(\ln(x)-\frac{1}{2}\ln(x^2+1)\right)\bigg\vert_1^\infty\\=\ln\frac{x}{\sqrt{x^2+1}}\bigg\vert_1^\infty=\ln\frac{1}{\sqrt{1+1/x^2}}\bigg\vert_1^\infty=\ln 1-\ln\frac{1}{\sqrt{2}}=\ln\sqrt 2$$
| {
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Sketching $(x^2-2)^2+(y^2-2)^2=4$ I was recently asked to sketch $(x^2-2)^2+(y^2-2)^2=2$, which did not prove to be too problematic, for establishing the range and domain of the expression gives nearly all of it away.
I then asked myself what would happen as I change the constant on the R.H.S. I recommend you try this yourself:
https://www.desmos.com/calculator/tlksmrpzxu
The constant that stood out as yielding the most intriguing result is $4$. The graph looks like this:
I then wondered how on earth I could deduce this just by analysing $(x^2-2)^2+(y^2-2)^2=4$.
One of my approaches involved spotting that the two 'orbits' are in the shape of ellipses with equations (derived partly experimentally, partly using the ellipse formula) $x^2-xy\sqrt{2}+y^2=2$ and $x^2+xy\sqrt{2}+y^2=2$. Can I derive these just from $(x^2-2)^2+(y^2-2)^2=4$?
| More generally,
$(x^2-a)^2+(y^2-a)^2-b = (x^2-Axy+y^2-B)(x^2+Axy+y^2-B)$
iff
$A^2=2, B=a, b=a^2$
but like @Lubin I can't see this at once.
| {
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$a,b,c$ such that $ax^2+bx+c \equiv 0$ (mod $3$) holds for all $x$? Today I wrote a computer program that finds whole positive $a,b,c$ such that $ax^2+bx+c \equiv 0$ (mod $n$) holds for all $x$. For $n=2$ I get several results, such as $x^2+x$, $3x^2+x$ and more. However, I was not able to find $a,b,c$ for the case $n=3$.
(Why) Are there no $a,b,c$ for $n=3$?
Remark: Of course $\text{gcd}(a,b,c) = 1$ and $a,b,c \in \mathbb{Z}$ and $a \not = 0$ or $b \not = 0$ or $c \not = 0$.
| I assume $a,b,c$ and $x$ are all restricted to integers.
Considering the three cases $x = 0,1,2 \mod 3$ we have:
$x=0 \mod 3 \Rightarrow c=0 \mod 3$
$x=1 \mod 3 \Rightarrow a+b+c=0 \mod 3$
$x=2 \mod 3 \Rightarrow a +2b+c = 0 \mod 3$
These three simultaneous equations are only all satisfied if $a=b=c=0 \mod 3$ i.e. if $a$, $b$ and $c$ are all multiples of $3$.
| {
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Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$
Let $\log x=\log 2+\log y$ and $2^x+8^y=4$ then find the $x$
My Try :
$$\log x=\log 2+\log y \\ \log x=\log 2y \\x=2y $$
So we have :
$$2^x+8^y=4\\2^{2y}+2^{3y}=4 \\t=2^y \\t^2+t^3=4$$
now what ?
| I will assume you want a real solution. $t^2+t^3=4$ can be rewritten as $t^3+t^2-4=0$. Using the cubic formula, the real solution for $t$ is $$t=\frac{1}{3} \left(-1+\sqrt[3]{53-6\sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}\right)$$
We also know that $y=log_2(t)$ and that $x=2y$. So, using that,
$$x=2\cdot log_2\left(\frac{1}{3} \left(-1+\sqrt[3]{53-6\sqrt{78}}+\sqrt[3]{53+6 \sqrt{78}}\right)\right)$$
| {
"language": "en",
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Taylor expansion of arctan x How can I prove that the tailor expansion of arctan x converges only at [-1,1]? I was able to prove that it converges at the given interval but I don't know how to prove that it does not converge anywhere else . Can someone help?
| The derivative of $\arctan x$ is
$$\frac d{dx} \arctan x =\frac 1{1+x^2}$$
From the formula for geometric series, we have that
$1+y+y^2+y^3+...=\frac 1{1−y}$ if and only if $|y|<1$.
Plugging in $-x^2$ for $y$, we get that
$$\begin{align*}
\frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\
&= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\
&= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots
\end{align*}$$
provided that $|-x^2| \lt 1$; i.e., provided that $|x|\lt 1$.
And if $|x|\lt 1$, then $x \in [-1,1]$. If $x$ isn't in this interval, then we do not get convergence.
| {
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Series expansion of $\frac{x^n-1}{x-1}$ at $ x=1$. $$\lim_{x\to 1}\left(\frac{x^n-1}{x-1}\right)=n$$
I thought that at $x=1$ the expansion is:
$$n+n(n-1)(x-1)+n(n-1)(n-2)(x-1)^2+...$$
but the answer is:
$$n+\frac{1}{2}n(n-1)(x-1)+\frac{1}{6}n(n-1)(n-2)(x-1)^2+...$$
Can you explain the origin of $$\frac{1}{2},\frac{1}{6},\frac{1}{24}, ...?$$
| Let $f(x)=\frac{x^n-1}{x-1}=\sum_{k=0}^{n-1}x^k$ for $x\ne 1$. $\lim_{x->1}f(x)=n$.
| {
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How to find Laurent expansion of $f(z)=\frac{1}{(z+i)z^2}$ on $0<|z-i|<1$ I think the way is doing the decomposition as follows, but I just don't know how to deal with the quadratic term:
$f(z)=\frac{1}{(z+i)z^2}=-\frac{1}{z+i}+\frac{1}{z}-\frac{i}{z^2}$
where, $-\frac{1}{z+i}=-\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}$ and $\frac{1}{z}=\frac{1}{1+\frac{z-i}{i}}$ can be expanded easily.
But how to deal with $-\frac{i}{z^2}$?
| With $z-i=w$ and geometric series
$$\dfrac{1}{z}=\dfrac{1}{w+i}=\dfrac{1}{i}\dfrac{1}{1+\frac{w}{i}}=\dfrac{1}{i}\left(\sum_{n=0}^\infty(-\frac{w}{i})^n\right)$$
then it's derivative
$$\dfrac{1}{(w+i)^2}=\sum_{n=0}^\infty (n+1)\left(-\dfrac{w}{i}\right)^n$$
valid for $\Big|\dfrac{w}{i}\Big|<1$. Also
$$\dfrac{1}{z+i}=\dfrac{1}{w+2i}=\dfrac{1}{2i}\dfrac{1}{1+\frac{w}{2i}}=\dfrac{1}{2i}\left(\sum_{m=0}^\infty(-\frac{w}{2i})^m\right)$$
valid for $\Big|\dfrac{w}{2i}\Big|<1$. So
\begin{align}
\dfrac{1}{z^2(z+i)}
&= \dfrac{1}{(w+i)^2(w+2i)} \\
&= \sum_{n=0}^\infty (n+1)\left(-\dfrac{w}{i}\right)^n\dfrac{1}{2i}\left(\sum_{m=0}^\infty(-\frac{w}{2i})^m\right) \\
&= \dfrac{i}{2}\sum_{n=0}^\infty\sum_{m=0}^\infty (n+1)\left(-\dfrac{w}{i}\right)^n\left(-\frac{w}{2i}\right)^m \\
&= \dfrac{i}{2}-\dfrac54w+\cdots \\
&= \dfrac{i}{2}-\dfrac54(z-i)+\cdots
\end{align}
valid for $|z-i|<1$.
| {
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Almost-half iterate of $x^2+1$ Because I can't find a function $h:\mathbb R\mapsto\mathbb R$ with the property
$$h^{\circ 2}(x)=x^2+1$$
I'm looking for a function that almost has that property - that is, I would like to find a closed-form (and preferably elementary) function $h:\mathbb R\mapsto\mathbb R$ satisfying
$$\lim_{x\to\infty} (x^2+1-h(h(x)))=0$$
or, equivalently,
$$h(h(x))=x^2+1+\mathcal O(\epsilon(x))$$
where $\lim_{x\to\infty} \epsilon(x)=0$.
But I haven't been able to do this either. I've tried functions in the form $$|x|^{\sqrt 2}+C$$
but none of them have worked. Can anybody find such a function $h$?
| Starting with
$$
g(x) = x^\sqrt{2} + C
$$
we have
$$
\begin{align}
g(g(x)) &= \left(x^\sqrt{2} + C\right)^\sqrt{2} + C \\
&= x^2 \left(1 + Cx^{-\sqrt{2}}\right)^\sqrt{2} + C \\
&\approx x^2 \left( 1 + C\sqrt{2}x^{-\sqrt{2}} \right) + C \qquad \text{(binomial theorem)} \\
&= x^2 + C\sqrt{2}x^{2-\sqrt{2}} + C.
\end{align}
$$
It looks like we can get what we want if we replace $C$ with $x^{\sqrt{2}-2}/\sqrt{2}$.
Indeed, if
$$
h(x) = x^{\sqrt{2}} + \frac{1}{\sqrt{2}} x^{\sqrt{2}-2} \tag{$*$}
$$
then
$$
\begin{align}
h(h(x)) &= \left( x^{\sqrt{2}} + \frac{1}{\sqrt{2}} x^{\sqrt{2}-2} \right)^\sqrt{2} + \frac{1}{\sqrt{2}} \left( x^{\sqrt{2}} + \frac{1}{\sqrt{2}} x^{\sqrt{2}-2} \right)^{\sqrt{2}-2} \\
&= \left( x^{\sqrt{2}} + \frac{1}{\sqrt{2}} x^{\sqrt{2}-2} \right)^\sqrt{2} + O\!\left(x^{2(1-\sqrt{2})}\right) \\
&= x^2 \left( 1 + \frac{1}{\sqrt{2}} x^{-2} \right)^\sqrt{2} + O\!\left(x^{2(1-\sqrt{2})}\right) \\
&= x^2 \left[ 1 + x^{-2} + O\!\left(x^{-4}\right) \right] + O\!\left(x^{2(1-\sqrt{2})}\right) \\
&= x^2 + 1 + O\!\left(x^{2(1-\sqrt{2})}\right)
\end{align}
$$
as $x \to +\infty$.
| {
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"source": "stackexchange",
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Given $|x| = 1$ and $x \notin \mathbb R$ prove that $\Re\left(\frac{x-1}{x+1}\right) = 0$
Given $|x| = 1$ and $x \notin \mathbb R$ prove that $\Re\left(\frac{x-1}{x+1}\right) = 0$
I have tried to express $x$ in terms of $a$ and $b$, namely: $x = a+bi, |x|= \sqrt{a^2+b^2}$ and to evaluate this but this simply led me nowhere. I guess there must be a clever way to solve that I don't see. I'd be thankful if you could give me some hints (not the actual solution) on how to proceed with this problem.
| So let’s evaluate the expression by plugging in $x = a + bi$. $\frac{x-1}{x+1} = \frac{(a-1)+bi}{(a+1)+bi}$.
Multiply by the conjugate of the denominator, which is $(a+1)-bi$, to get $\dfrac{((a-1)+bi)((a+1)-bi)}{((a+1)+bi)((a+1)-bi)}$.
We know that the resulting denominator must be real and equal to $(a+1)^2+b^2$. We can expand the numerator to $(a-1)(a+1)+(a+1)bi-(a-1)bi+b^2$. Discard the imaginary middle terms, and we're left with $(a-1)(a+1)+b^2 = a^2 - 1 + b^2$. We know that $a^2+b^2=1$, and $1-1=0$. Thus, the real part of the numerator is zero, and since the denominator is also real, we know that the real part of the fraction is zero.
| {
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Inequality with angles of a triangle Let $\alpha$, $\beta$ and $\gamma$ be the three angles in a non-right triangle. How can i prove the following inequality?
$\dfrac{1+\sin^2(\alpha)}{\cos^2(\alpha)}+\dfrac{1+\sin^2(\beta)}{\cos^2(\beta)}+\dfrac{1+\sin^2(\gamma)}{\cos^2(\gamma)}\ge \dfrac{1+\sin(\alpha)\sin(\beta)}{1-\sin(\alpha)\sin(\beta)}+\dfrac{1+\sin(\beta)\sin(\gamma)}{1-\sin(\beta)\sin(\gamma)}+\dfrac{1+\sin(\alpha)\sin(\gamma)}{1-\sin(\alpha)\sin(\gamma)}$
| $$\sum_{cyc}\frac{(2bc)^2+16\Delta^2}{(b^2+c^2-a^2)^2}\geq \sum_{cyc}\frac{4 R^2+ab}{4R^2-ab}$$
is equivalent to
$$ 16\Delta^2\sum_{cyc}\frac{1}{(b^2+c^2-a^2)^2} \geq \sum_{cyc}\frac{ab}{4R^2-ab} $$
or to
$$\sum_{cyc}\frac{1}{(b^2+c^2-a^2)^2} \geq \sum_{cyc}\frac{1}{(b^2+c^2-a^2)^2+4(a-b)bc^2}$$
which can be proved by bashing. Although I suspect there are more efficient ways, based on the convexity of $f(x)=\frac{1+x}{1-x}$ over $(-1,1)$.
| {
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Miller-Rabin: Showing non trivial divisors of $n$
Let's assume $n$ doesn't pass the Miller-Rabin test and $b$ is a witness. Meaning, $b^{\frac{n-1}{2^r}} \equiv 1 \pmod{n}$ where $\frac{n-1}{2^r}$ is even, but $c = b^{\frac{n-1}{2^{r+1}}} \not\equiv \pm1 \pmod{n}$. Show that $\gcd(c+1,n),\gcd(c-1, n)$ are non-trivial divisors of $n$.
So first of all, for my convenience:
*
*Denote $n-1 = 2^ls$, $s$ is odd.
*Denote $k=l-r$.
*
*$c = b^{2^{k-1}s}$
*$c^2 = b^{2^{k}s}$
*$2^ks$ is even
It is given that $c^2 \equiv 1 \pmod{n} \implies (c-1)(c+1) \equiv 0 \pmod {n} \implies n\mid (c-1)(c+1)$
Now, I think we can assume that $n$ passed Fermat's theorem test (that's part of the initial tests of the Miller-Rabin algorithm).
Hence,
$$c^{r+1} = b^{2^l s} = b^{n-1} \equiv 1 \pmod {n}$$
but that isn't revealing anything new other than $r$ is odd (since $2$ must divide $r+1$).
What am I missing?
| The general statement is that whenever we have a nontrivial square root of $1$ modulo $n$, a value $c$ such that
$$
c^2 \equiv 1 \pmod{n}, \qquad c \not\equiv \pm1 \pmod{n},
$$
then we obtain a factorization of $n$.
If $c^2 \equiv 1 \pmod n$, then $(c+1)(c-1) \equiv 0 \pmod n$.
If $c+1$ were relatively prime to $n$, then it would have an inverse modulo $n$, and we could multiply by $(c+1)^{-1}$ to conclude that $c-1 \equiv 0 \pmod n$. But this is ruled out by our assumption that $c \not\equiv 1 \pmod n$.
If $c+1$ were divisible by $n$, then we would have $c+1 \equiv 0 \pmod n$. But this is ruled out by our assumption that $c \not\equiv -1 \pmod n$.
So the only remaining possibility is that $c+1$ shares some, but not all, prime factors with $n$: that $\gcd(c+1,n)$ is a nontrivial factor of $n$.
The same goes for the other factor $c-1$.
More generally, whenever we find $a$ and $b$ such that $a^2 \equiv b^2 \pmod n$ but $a \not\equiv \pm b \pmod n$, the same argument shows that $\gcd(a+b,n)$ and $\gcd(a-b,n)$ are nontrivial factors of $n$. This is the basis for many integer factorization algorithms starting from Fermat's factorization method and ending with the quadratic sieve.
| {
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How to find $\int \frac{e^{-x^2}}{x^2 + 1} dx$? I have a question about improper integrals:
How can we find $\lim_{n \rightarrow +\infty}\int_{-n}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx$?
$\textbf{Some effort:}$
$\lim_{n \rightarrow +\infty}\int_{-n}^{n} \frac{1}{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx = \lim_{n \rightarrow +\infty} \frac{2}{n}\int_{0}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx $ $~~~~~~~~~\textbf{(1)}$
By setting $nx^2 = u$, we have $dx = \frac{1}{2\sqrt{n}} \times \frac{1}{\sqrt{u}}$ and $x = \frac{\sqrt{u}}{\sqrt{n}}$. So by substituting these in $\textbf{(1)}$ we have (I will not put bounds and at the end will come back to the initial bounds and also I will drop the constant in integrals)
$\textbf{(1)} = \lim_{n \rightarrow +\infty} \frac{2}{n} \int \frac{1 - e^{-u}}{\frac{u}{n}(1+u)} \times \frac{1}{2 \sqrt{n}}\times \frac{1}{ \sqrt{u}} du = \lim_{n \rightarrow +\infty} \frac{1}{\sqrt{n}} \int \frac{1 - e^{-u}}{ u \sqrt{u}(1+u)} du $
$= \lim_{n \rightarrow +\infty} \frac{1}{\sqrt{n}} \int ( \frac{1 }{ u \sqrt{u}(1+u)} - \frac{e^{-u} }{ u \sqrt{u}(1+u)}) du$ $~~~~~~~~~\textbf{(2)}$
By setting $\sqrt{u} = v$, we have $\frac{1}{2\sqrt{u}}du = dv$. So by substituting these in $\textbf{(2)}$ we have
$\textbf{(2)} = \lim_{n \rightarrow +\infty} \frac{2}{\sqrt{n}} \int (\frac{1}{v^2(1+v^2)} - \frac{e^{-v^2}}{v^2(1+v^2)} dv) $ $~~~~~~~~~\textbf{(3)}$
$\textbf{(3)}= \lim_{n \rightarrow +\infty} \frac{2}{\sqrt{n}} \int ( \frac{1}{v^2} - \frac{1}{1+v^2} - \frac{e^{-v^2}}{v^2} + \frac{e^{-v^2}}{v^2 + 1}) dv$
Now we will calculate each term separately.
First part:
For to find $\int \frac{1}{v^2} dv $, by setting $k_1=-\frac{1}{v}$, we have $dk_1 =\frac{1}{v^2} dv$ and so we have $\int \frac{1}{v^2} dv = \int dk_1= k_1= -\frac{1}{v}= -\frac{1}{\sqrt{u}}= -\frac{1}{\sqrt{nx^2}} = -\frac{1}{\sqrt{n}|x|} $
Second part:
For to find $-\int \frac{1}{1+v^2} dv$, by setting $k_2 = \arctan(v)$, we have $dk_2 = \frac{1}{1 + v^2}dv$ and so we have $-\int \frac{1}{1+v^2} dv = -int dk_2= -k_2= -\arctan(v) = -\arctan(\sqrt{u}) = -\arctan(\sqrt{nx^2}) = -\arctan(\sqrt{n}|x|) $
Third part:
For to find $-\int \frac{e^{-v^2}}{v^2} dv $, by setting $\begin{cases}
k_3=e^{-v^2}\\
-\frac{1}{v^2}=dk_4
\end{cases}$ we will have $\begin{cases}
dk_3=-2ve^{-v^2}\\
k_4= \frac{1}{v}
\end{cases}$ and our integral will transform to $-\int \frac{e^{-v^2}}{v^2} dv = \frac{e^{-v^2}}{v} - 2 \int e^{-v^2} dv = \frac{e^{-v^2}}{v} - 2(\frac{\sqrt{\pi}}{2}) = \frac{e^{-v^2}}{v} - \sqrt{\pi} = \frac{e^{-(\sqrt{u})^2}}{\sqrt{u}} - \sqrt{\pi} = \frac{e^{-u}}{\sqrt{u}} - \sqrt{\pi} =\frac{e^{-nx^2}}{\sqrt{nx^2}} - \sqrt{\pi}=\frac{e^{-nx^2}}{\sqrt{n}|x|} - \sqrt{\pi}$
Forth part:
For to find $\int \frac{e^{-v^2}}{v^2 + 1} dv$, I cannot find it!
Can someone please help me to find $\int \frac{e^{-v^2}}{v^2 + 1} dv$?
Thanks!
| Unfortunately, this integral has no closed form in terms of known functions, elementary or non-elementary. You'd have to integrate it numerically if you want to do anything practical with it.
The series expansion of the integral (at $x=0$), while by no means a closed form, is, according to W|A:
$$\int \frac{e^{-v^2}}{v^2 + 1} dv \approx x - \frac{2 x^3}3 + \frac{x^5}2 - \frac{8x^7}{21} + \frac{65 x^9}{216} - \frac{163 x^11}{660} + O(x^{13}) $$
And of your original integral:$$\int \frac{e^{-nx^2}}{x^2(1 + nx^2)} dx \approx -\frac 1x - 2 n x + \frac{5 n^2 x^3}6 - \frac{8 n^3 x^5}{15} + O(x^6) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Expansion of $S=\frac{{nx^{n+1}-(n+1)x^{n}+1}}{(x-1)^2}$ Expand;
$S=\frac{{nx^{n+1}-(n+1)x^{n}+1}}{(x-1)^2}$
See solution, Also I am looking for an advance method if exists.
| We can also see that $$S =\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2} $$ $$\implies S(1-x)=\frac{(n+1)x^n-nx^{n+1}-1}{x-1}=\frac{(x-1)[1+x+x^2…x^{n-1}-nx^n]}{x-1}=[1+x+x^2+…x^{n-1}-nx^n]$$ $$S=\frac{1+x+x^2…x^{n-1}-nx^n}{1-x}=[1+x+x^2+…][1+x+x^2+…x^{n-1}-nx^n] = ?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $f(x)=x^3-x^2-x+a$ only have a real negative root then what is range of $a$
Let $f(x)=x^3-x^2-x+a$ only have a real negative root then what is range of $a$
I got it : $$g(x)=x^3-x^2-x \ \ \ h(x)=-a$$
Now we have :
so must $-a<-1 \to a>1$
But I want to be algebraic please help me !
| Discriminant $\Delta$ of the cubic function
\begin{align}
f(x)&=ax^3+bx^2+cx+d
\end{align}
is known to be
\begin{align}
\Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2
,
\end{align}
and the cubic polynomial has only one real root
when its $\Delta<0$.
For
\begin{align}
f(x)=x^3-x^2-x+a
\end{align}
we have
\begin{align}
\Delta&=-27a^2+22a+5
,\\
\Delta&<0 \quad\text{when}\quad a\in(-\infty,-\tfrac5{27})
\cup(1,\infty)
\tag{1}\label{1}
.
\end{align}
Let the three roots be
$x_1=-r$, $x_2=u+v\,i$ and $x_3=u-v\,i$
($r>0$, $u,v\in\mathbb{R}$, $i^2=-1$).
We also know that
\begin{align}
-a&=x_1\,x_2\,x_3
,\\
-a&=-r\,(u^2+v^2)
,\\
a&=r\,(u^2+v^2)>0
,
\end{align}
that is, for the real negative root $x_1=-r$
we must have $a>0$.
Hence, together with \eqref{1}, the answer is: $a>1$.
| {
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Geometric interpretation for median formula $m_c = \sqrt{\frac{2(a^2 + b^2)- c^2}{4}}$ for $a + b < c$? For a triangular with side lengths $a, b, c$, the length of the median of $c$ is $m_c = \sqrt{\frac{2(a^2 + b^2) - c^2}{4}}$.
If $a + b < c$, the configuration is obviously not a triangle anymore.
However, there are values for $a, b, c$ with $a + b < c$ (e.g. $a = 7, b = 2, c = 10$) where the root is still positive. Is there any geometric interpretation for that?
| If $a+b>c$, then we can draw circles of radius $a$ and $b$ about endpoints of segment $\overline{AB}$ of length $c$, and these circles will meet at distinct points $C$ and $C^\prime$. The circle about $M$, the midpoint $\overline{AB}$, with radius $m:=\frac12 \sqrt{2(a^2+b^2)-c^2}$, meets those points, as well. Thus, $\overleftrightarrow{CC^\prime}$ is a common secant line, and therefore also the common "radical axis" or "power line", of the circles. (When $a+b=c$, we can replace "common secant line" with "common tangent line".) This allows us to characterize the length $m$ thusly:
The circle coaxal with $\bigcirc A$ and $\bigcirc B$, centered at the midpoint of $\overline{AB}$, has radius $m$.
(The diagram illustrates the property that, for any point $P$ on the radical axis of a pair (or family) of circles, the tangent segments from $P$ to each of those circles are all congruent.)
As it turns out, the same description holds for $a+b<c$.
Here's a quick coordinate proof (valid whether $a$, $b$, $c$ make a triangle or not) that leverages a convenient property of radical axis equations.
Take $A = (-c/2, 0)$, $B = (c/2, 0)$, $M= (0,0)$. Then the equations of the circles are
$$
\bigcirc A: \left(x + \frac{c}{2}\right)^2 + y^2 = a^2 \qquad
\bigcirc B: \left(x - \frac{c}{2}\right)^2 + y^2 = b^2 \qquad
\bigcirc C: x^2 + y^2 = m^2
$$
Now, the equation of the radical axis of two circles is obtained by eliminating the $x^2$ (and $y^2$) terms from the equations of those circles. With our equations, a simple subtraction will do:
$$\begin{align}
\text{radical axis of $\bigcirc{A}$ and $\bigcirc{B}$}\; &\;:\; \bigcirc{A} - \bigcirc{B}\;\quad\to\quad \phantom{-}2cx = a^2 - b^2 \\[4pt]
\text{radical axis of $\bigcirc{A}$ and $\bigcirc{M}$} &\;:\; \bigcirc{A} - \bigcirc{M} \quad\to\quad \phantom{-2}cx = a^2 - m^2 - \frac{c^2}{4} \\[4pt]
\text{radical axis of $\bigcirc{B}$ and $\bigcirc{M}$} &\;:\; \bigcirc{B} - \bigcirc{M} \quad\to\quad \;-cx = b^2 - m^2 - \frac{c^2}{4}
\end{align}$$
These three (vertical) lines will coincide precisely when
$$m^2 = \frac{1}{4}\left(\; 2 ( a^2 + b^2 ) - c^2 \;\right)$$
$\square$
| {
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Large Remainders with Mod Determine the remainder of $5^{2017}$ when divided by 7.
I know that we need to use mod 7 to find all of the different remainders but I am not sure what specific steps to take and how to finish it out.
Please use Mod 7 as this question is very specific in the way it needs to be worked out.
| The powers of $5$ modulo $7$ obviously form a finite set. So one can try and look at a few values in order to get the idea:
\begin{alignat}{2}
5^0&\equiv 1&&\pmod{7}\\
5^1&\equiv 5&&\pmod{7}\\
5^2&\equiv 5\cdot 5\equiv 25\equiv 4&&\pmod{7}\\
5^3&\equiv 4\cdot 5\equiv 20\equiv 6&&\pmod{7}\\
5^4&\equiv 6\cdot 5\equiv 30\equiv 2&&\pmod{7}\\
5^5&\equiv 2\cdot 5\equiv 10\equiv 3&&\pmod{7}\\
5^6&\equiv 3\cdot 5\equiv 15\equiv 1&&\pmod{7}
\end{alignat}
Oh, good! Each time we have power $6$ we can reduce it to $1$ (modulo $7$). Since
$$
2017=6\cdot 336+1
$$
we are done! Indeed, $5^{2017}=(5^6)^{336}\cdot 5\equiv 1^{336}\cdot 5\equiv 5\pmod{7}$.
Fermat's little theorem, or, better, the Euler-Fermat theorem, is a generalization:
Theorem (Euler-Fermat). If $\gcd(a,n)=1$, then $a^{\varphi(n)}\equiv 1\pmod{n}$. In particular, if $p$ is prime and $p\nmid a$, then $a^{p-1}\equiv 1\pmod{p}$.
Here $\varphi$ is Euler's totient function: $\varphi(n)$ is the number of integers in the range $[0,n-1]$ which are coprime to $n$. For $p$ a prime, we obviously have $\varphi(p)=p-1$.
In case the prime is big, it would be a nuisance to compute all the powers like I did above for $7$.
Little Fermat's theorem is
Theorem (Fermat). If $p$ is prime, then $a^p\equiv a\pmod{p}$, for every integer $a$.
| {
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Find values of $p$ such that $p^2-p+1$ is cubic? Find all prime numbers such that $p^2-p+1$ is a cubic number. I've tried substituting values of $x$ for $p^2-p+1 = x^3$.
This gave me a few values, which were small. As values of $x$ ranged higher and higher, it was tedious to find values of $p$ that satisfied this condition. To my imminent horror, I realized that most of these values of $p$ where in fact NOT prime numbers.
Can someone guide me through the solution, as I prefer to use a slick solution, rather than using brute force? After the solution, can someone point me towards resources for solving problems like this?
| Here is a solution using simpler tools.
Our starting point is
$$p^2 - p + 1 = x^3$$
with $p$ prime, and $x$ is an integer.
As the smallest prime is 2, note that
$$x^3 = p^2 - p + 1 \ge 2^2 - 2 + 1 = 3 \quad \rightarrow \quad x \ge 3^{1/3} > 1$$
$$p^2 > p^2 - p + 1 = x^3 \quad \rightarrow \quad p > x^{3/2} > x $$
Since $p$ being prime is important, this hints we should try to rearrange the equation to factor as much as possible.
$$
\begin{eqnarray}
p^2 - p + 1 &=& x^3 \\
p^2 - p &=& x^3 - 1 \\
p(p-1) &=& (x-1)(x^2+x+1)
\end{eqnarray}
$$
Because $p$ is prime and divides the left hand side, it must divide the right hand side as well. Since $p>x$, this means $p$ cannot divide $(x-1)$ and therefore $(x^2+x+1)$ must be a multiple of $p$. Let's call this multiple $m$.
So we now have two equations
$$
\begin{eqnarray}
mp &=& (x^2+x+1) \\
(p-1) &=& m(x-1)
\end{eqnarray}
$$
Because $p>x$ the second equation makes it clear that $m>1$, and since it is an integer $m \ge 2$. Note also that $(x^2 + x +1)$ cannot be even, thus raising the limit further: $m \ge 3$.
Eliminating $p$ to get a quadratic on $x$
$$
\begin{eqnarray}
p &=& mx - m + 1 \\
m(mx - m + 1) &=& (x^2+x+1) \\
0 &=& x^2 + (1-m^2)x + (m^2-m+1)
\end{eqnarray}
$$
For the quadratic equation to have an integer solution, the discriminant must be a square value
$$
\begin{eqnarray}
(1-m^2)^2 - 4(m^2-m+1) &=& (1 - 2m^2 + m^4) - 4(m^2-m+1) \\
&=& m^4 - 6m^2 +4m-3
\end{eqnarray}
$$
Requiring this to be a square integer is quite restrictive, which is easier to see if we denote the square as $(m^2 - k)^2 = m^4 -2km^2 + k^2$. Using this along with the fact that $m \ge 3$ (thus $m^2 \ge 3m$), we can see
$$ m^4 - 6m^2 +4m-3 < m^4 - 4m^2 - 3 < (m^2 - 2)^2 $$
and
$$ m^4 - 6m^2 +4m-3 \ge m^4 - 6m^2 -3m + 18 > m^4 -8m^2 + 18 > (m^2 - 4)^2 $$
therefore
$$m^4 - 6m^2 +4m-3 = (m^2 - 3)^2 = m^4 - 6m^2 + 9$$
$$4m = 12 \quad \rightarrow \quad m=3$$
Now putting this back in our quadratic equation for $x$
$$ 0 = x^2 + (1-m^2)x + (m^2-m+1) = x^2 - 8x + 7 = (x-1)(x-7) $$
As $x>1$, this means the only solution is
$$x=7 \quad \rightarrow \quad p=m(x-1)+1=19. $$
| {
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} |
Solving: $(x-a)(x-b)=x-c$, $(x-c)(x-b)=x-a$ and $(x-c)(x-a)=x-b$ Given three distinct real numbers $a$, $b$, and $c$, show that at least two of the three following equations
$$(x-a)(x-b)=x-c$$
$$(x-c)(x-b)=x-a$$
$$(x-c)(x-a)=x-b$$
have real solutions.
My attempt: I tried to multiply all the equations side by side to obtain
$$(x-a)^2(x-b)^2(x-c)^2=(x-a)(x-b)(x-c)$$
then what next?
| Let (x-a) (x-b) / (x-c) = y
Solving, $ x^2 -(a+b+y)x + ab + cy = 0 $
D>0
$(a+b+y)^2 -4(ab+cy) > 0 $
$a^2 + b^2 + y^2 + 2ab + 2by + 2ay - 4ab - 4cy >0$
$ y^2 + (2b+2a-4c)y + (a-b)^2 > 0 $
Now for this Quadratic, D<0 gives the quadratic the whole Real values as range.
$ 4(b+a-2c)^2 - 4(a-b)^2 < 0$
$(a-c)(b-c)<0$
$Either, a<c<b, or ,b<c<a.$ Which means, when c is the middle value, it will have a solution for y=1.
Now when in the RHS, it is the smallest value(assume a here). The value of LHS at x=b(the largest value supposedly) is 0 and RHS is b-a which is positive. This makes sure that this equation have real solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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L'Hospital Rule with Trigonometry I've been working on this question for the final exam, yet I could not come to a closure. I'd appreciate if you can help me.
What is $$\lim_{x\to 0} \frac{\sin(\sin(\sin x)) - x}{x^3}\qquad?$$
| $$\lim_{x\rightarrow0}\frac{\sin\sin\sin{x}-x}{x^3}=\lim_{x\rightarrow0}\frac{\cos\sin\sin{x}\cos\sin{x}\cos{x}-1}{3x^2}=$$
$$=\lim_{x\rightarrow0}\tfrac{-\sin\sin\sin{x}\cos^2\sin{x}\cos^2x-\cos\sin\sin{x}\sin\sin{x}\cos^2x-\cos\sin\sin{x}\cos\sin{x}\sin{x}}{6x}=$$
$$=\lim_{x\rightarrow0}\tfrac{-\tfrac{\sin\sin\sin{x}}{\sin\sin{x}}\cdot\tfrac{\sin\sin{x}}{\sin{x}}\cdot\tfrac{\sin{x}}{x}\cdot\cos^2\sin{x}\cos^2x-\cos\sin\sin{x}\cdot\tfrac{\sin\sin{x}}{\sin{x}}\cdot\tfrac{\sin{x}}{x}\cdot\cos^2x-\cos\sin\sin{x}\cos\sin{x}\cdot\tfrac{\sin{x}}{x}}{6}=$$
$$=\frac{-1-1-1}{6}=-\frac{1}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate last four digits of $2^{2017}$? can u suggest me any short trick of solving these kind of problems. How to find last four digits of any number raised to some power ?
| Since $10^4 = 2^4 \times 5^4$, my first reaction is one should use Chinese remainder theorem.
For $2^4$, it is clear $2^{2017} \equiv 0 \pmod {2^4}$.
For $5^4$, we use following theorem
For $a,b \in \mathbb{Z}_{+}$ such that $\gcd(a,b) = 1$, we have $$a^{\varphi(b)} \equiv 1 \pmod b$$ where $\varphi(\cdot)$ is the Euler totient function.
For $a = 2$ and $b = 5^4$, we have $\varphi(5^4) = 4(5)^3 = 500$. This leads to
$$2^{2017} = 2^{4(500)+17} \equiv 2^{17} \equiv 65536\times 2 \equiv 1072\pmod {5^4}$$
It turns out we are lucky. Since $1072 \equiv 0\pmod {2^4}$ already, we don't need CRT to conclude $2^{2017} \equiv 1072 \pmod {10^4}$.
As a result, the last $4$ digits of $2^{2017}$ is $1072$.
| {
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$X\sim N(1,1)$. Find $\operatorname{Var}(X^2)$
Let $X\sim N(1,1)$. Find $\operatorname{Var}(X^2)$.
I know since, $X-1\sim N(0,1)$, then $(X-1)^2\sim \chi^2_1$, hence $\operatorname{Var}(X-1)^2=2$. But don't understand how to find $\operatorname{Var}(X^2)$. Any help appreciated. Thanks.
| $\newcommand{\e}{\operatorname E}$
$$
\operatorname{var}(X^2) = \e((X^2)^2) - (\e(X^2))^2.
$$
$$
\e(X^2) = \operatorname{var}(X) + (\e(X))^2 = 1 + 1^2 = 2.
$$
Let $Z=X-1$ so that $Z\sim N(0,1).$ Then
\begin{align}
\e(X^4) & = \e((Z+1)^4) = \e(Z^4) + 4\e(Z^3) + 6\e(Z^2) + 4\e(Z) + 1. \\[10pt]
\e(Z^2) & = \operatorname{var}(Z) = 1. \\[10pt]
\e(Z^3) & = 0 \text{ by symmetry.} \\[10pt] {}
\end{align}
\begin{align}
\e(Z^4) & = \int_{-\infty}^\infty z^4 \varphi(z)\, dz = 2\int_0^\infty z^4\varphi(z)\, dz = 2\int_0^\infty z^4 \frac 1 {\sqrt{2\pi}} e^{-z^2/2} \, dz \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty z^3 e^{-z^2/2} (z\,dz) = \sqrt{\frac 2 \pi} \int_0^\infty (2u)^{3/2} e^{-u} \, du \\[10pt]
& = \sqrt{\frac 2 \pi} \cdot 2^{3/2} \Gamma\left(\frac 5 2\right) = \frac 4 {\sqrt\pi} \cdot \frac 1 2 \cdot \frac 3 2 \cdot \Gamma\left( \frac 1 2 \right) \\[10pt]
& = 3.
\end{align}
| {
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Need help to understand the following number theory proof Prove that there exist infinitely many integers n such that $n, n + 1, n + 2$ are each the sum of the squares of two integers:
Solution: First solution: Let a be an even integer such that $a^2 + 1$ is not prime. (For example, choose a ≡ 2 (mod 5), so that $a^2 + 1$ is divisible by 5.) Then we can write $a^2 + 1$ as a difference of squares $x^2 − b^2$, by factoring $a^2 + 1$ as rs with $r ≥ s > 1$, and setting x = (r + s)/2, b = (r − s)/2. Finally, put n = $x^2 − 1$, so that $n = a^2 + b^2$, $n + 1 = x^2$, n + 2 = $x^2 + 1$
I got lost in the following part:
Then we can write $a^2 + 1$ as a difference of squares $x^2 − b^2$, by factoring $a^2 + 1$ as rs with $r ≥ s > 1$. What is rs?. Can someone expand this proof, so it looks clearer?
| Just follow the instructions:
*
*Pick some even $a$ such that $a^2+1$ is not a prime, for instance some $a\equiv 2\pmod{5}$.
Fine, let us pick $a=22$;
*Write $a^2+1$ as $r\cdot s$ with $r\geq s >1$.
Fine, $22^2+1 = 97\cdot 5$;
*Write $a^2+1$ as a difference of squares through the previous decomposition.
Fine, $22^2+1 = (51+46)(51-46) = 51^2-46^2$.
This gives that $51^2-1$ is the sum of two squares, $22^2+46^2$, so $51^2-1,51^2$ and $51^2+1^2$ are three consecutive numbers belonging to $\square+\square$. Is it a bit clearer now?
| {
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$f(n)=\sqrt{2n+1+2\sqrt{n^2+n}}$, $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\dots+\frac{1}{f(99)}$ Given $f(n)=\sqrt{2n+1+2\sqrt{n^2+n}}$
Evaluate $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\dots+\frac{1}{f(99)}$
I have simplified the given to $f(n)=\sqrt{n}+\sqrt{n+1}$ but I am still not sure of how to solve this.
| The hint.
Use $$\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}$$ and the telescopic sum.
Finally, we obtain $$\sqrt{100}-1=9.$$
| {
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$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ is true if and only if $x+y=0$ Prove that $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ is true if and only if $x+y=0$
I believe x and y could both be 0 as that satisfies the equations. Beyond that, I do not know how to prove this.
| By continuous rearranging and squaring:
$$\text{Equation} \Rightarrow x+\sqrt{x^2+1}=\frac{1}{y+\sqrt{y^2+1}} \Rightarrow x+y=\sqrt{y^2+1}-\sqrt{x^2+1} \Rightarrow$$
$$1-xy=\sqrt{x^2y^2+x^2+y^2+1} \Rightarrow (x+y)^2=0.$$
| {
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find the value of $a\in\Bbb{Z}$ such that $2+\sqrt{3}$ is a root of the polynomial? Find the value of $a\in\Bbb{Z}$ such that $2+\sqrt{3}$ is a root of the polynomial
$$x^3-5x^2 +ax -1$$
I got the answer and value of $a = -10 -5\sqrt{3} -1$.
Is my answer is correct or not?
| If $a \in \mathbb Z$, then if $2+\sqrt3$ is a root, then so is $2-\sqrt3$.
Therefore, $x^3-5x^2 +ax -1$ is divisible by $x^2 - 4 x + 1$, which has $2\pm\sqrt3$ as roots.
Now, $x^3-5x^2 +ax -1 = (x^2 - 4 x + 1)(x-1)+(a-5)x$ and so $a=5$.
| {
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Reverse continued fractions Basic continued fractions arise from recurrence relations such as:
$$
n = a + \frac{b}{n},
$$
This gives rise to the continued fraction:
$$
n = a + \frac{b}{a+\frac{b}{a+\frac{b}{a+...}}}.
$$
What about relations such as:
$$
n = a + \frac{n}{b}?
$$
Do these give rise to things like:
$$
n = a + \frac{a+\frac{a+\frac{a+\frac{a+\frac{a+...}{b}}{b}}{b}}{b}}{b}?
$$
Of course, one could just solve the original equation as:
\begin{align}
n &= a + \frac{n}{b} \\
n(1-b^{-1}) &= a \\
n &= \frac{a}{1-b^{-1}}.
\end{align}
But couldn't the above "reverse continued fraction" be generalized like the usual one to create interesting structures and definitions of known constants? For instance, one can obtain this strange result for $n=a + \frac{n}{b²}$ by setting $a=1$ and $b=\sqrt{2}^{-1}$:
$$
-1 = 1 + \frac{1+\frac{1+\frac{1+\frac{1+\frac{1+...}{\sqrt{2}^{-1}}}{\sqrt{2}^{-1}}}{\sqrt{2}^{-1}}}{\sqrt{2}^{-1}}}{\sqrt{2}^{-1}}?
$$
Is this known/of any particular interest?
| Note that $$n = a + \frac{a+\frac{a+\frac{a+\frac{a+\frac{a+...}{b}}{b}}{b}}{b}}{b}=a+\frac ab + \frac a{b^2}+ \frac a{b^3} +\ldots$$
gives you a geometric series. If the constants vary it is still just another way of writing an infinite sum, so doesn't give anything new.
| {
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Is this a valid method of solving $3^x + 2^x = 35$? I want to know if this is a valid method of solving this equation:
$3^x + 2^x = 35$
$3^x+2^x = (7)(5)$
$3^x+2^x = (3+2^2)(2+3)$
$3^x+2^x = 3^2 + 2^3 + 18$
$3^x+2^x = 9 + 18 + 2^3 $
$3^x+2^x = 27 + 2^3 $
$3^x+2^x = 3^3 + 2^3 $
And now comes my problem. Is it correct to say that this last equation implies $x=3$? I don't think I can just compare terms just like it was a polynomial...
If this is not correct, is there a way of solving this equation algebraically?
| As others have noted, $x=3$ is the only solution in the real numbers. However, there are infinitely many complex solutions. For example, there is a solution near $3.40258654642902+5.90727700642090 i$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit without l'hopital $\lim_{x \to 0} \left ( \frac{1}{x} -\frac{1}{e^x-1}\right )$ I want to find the limit without L'hopital's rule and without Taylor series:
$
\lim_{x \to 0} \left ( \frac{1}{x} -\frac{1}{e^x-1}\right )
$
Is it possible?
Hints are more than welcome!
Thanks!
| Let us rewrite the expression in the limit as follows
$$ \frac{1}{x} - \frac{1}{e^x-1} = \frac{e^x - 1 - x}{xe^x - x} $$
Using the Taylor expansion of $e^x$, which we recall is
$$ e^x = \sum_{i = 0}^{\infty} \frac{x^i}{i!} \ , $$
we further rewrite the numerator and denominator as
$$ e^x - 1 - x = \frac{x^2}{2} + \mathcal{O} (x^3) = x^2(\frac{1}{2} + \mathcal{O} (x) )\\ xe^x - x = x^2 + \mathcal{O} (x^3) = x^2( 1 + \mathcal{O}(x) ) $$
where $ \mathcal{O} (x^3) $ denotes the terms of order higher then $2$ in the variable $x$. Since $\lim_{x \rightarrow 0} \mathcal{O}(x) = 0$ we can compute, using the theorem for product of limits, in the following manner
$$ \lim_{x \rightarrow 0} (\frac{1}{x} - \frac{1}{e^x-1} ) = \lim_{x \rightarrow 0} (\frac{e^x - 1 - x}{xe^x - x} ) = \lim_{x \rightarrow 0} \frac{x^2(\frac{1}{2} + \mathcal{O} (x) )}{x^2(1 + \mathcal{O} (x) )} = \lim_{x \rightarrow 0} \frac{\frac{1}{2} + \mathcal{O} (x)}{1 + \mathcal{O} (x)} = \frac{1}{2} $$
| {
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Inequality with integer numbers Let $a,b$ be two positive integer numbers such that $a\sqrt{3}>b\sqrt{7}$.
Prove that $(a\sqrt{3}-b\sqrt{7})(a+b)>1$.
Attempt. From the assumption we have $3a^2-7b^2\geq 1$. We try to argue that $3a^2-7b^2\geq 3$ to get
$$
a\sqrt{3}-b\sqrt{7}=\frac{3a^2-7b^2}{a\sqrt{3}+b\sqrt{7}}\geq\frac{3}{a\sqrt{3}+b\sqrt{7}}>\frac{1}{a+b}.
$$
| $3a^2=7b^2+1$ is impossible for $b$ by modulo $3$.
$3a^2-2=7b^2$ is impossible for $a$ by modulo $7$.
Thus, $3a^2-7b^2\geq3$ and we are done!
The equality in the last inequality occurs for $a=55$ and $b=36$, which gives
$$S\geq91(55\sqrt3-36\sqrt7)=1.43....$$
Since $(55,36)$ is a minimal solution of the Pell $3a^2-7b^2=3$,
it seems that we got a minimal value of $S$. I have no a proof for this statement.
| {
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Equation of circle after rolling Q: A circle of radius $5$ units touches both the axes and lies in the first quadrant. If the circle makes one complete roll on $+ x$-axis, then what will be its equation in the new position $?$
I got my answer as $x^2 +y^2 -x(10+20\pi)-10y+100\pi^2+100\pi+25=0.$
Is it correct?
| The circle of radius $5$ has equation $(x-5)^2+(y-5)^2=5^2$.
One complete revolution covers a distance of $2\pi r$, or $10\pi$ in this case.
So the centre also shifts by $10\pi$ to the right.
So the new equation is
$$(x-5-10\pi)^2+(y-5)^2=5^2$$
So:
$$x^2-2(5+10\pi)x+(5+10\pi)^2+y^2-10y+25=25$$
$$x^2-(10+20\pi)x+25+100\pi+100\pi^2+y^2-10y=0$$
Thus:
$$x^2+y^2-x(10+20\pi)-10y+100\pi^2+100\pi+25=0$$
So your proof is correct.
| {
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sequence of functions converging uniformly on compact sets but not on the real line Consider the sequence of functions
$$
f_n(x)=n\sin((4\pi^2 n^2+x^2)^{1/2}), 0\leq x\leq C, C>0
$$
I cannot prove $(f_n)$ converges uniformly on $[0,C]$ by finding f(x) such that
$$
\lim_n \sup_{0\leq x\leq C}||f_n(x)-f(x)||=0
$$
But is the sequence uniformly convergent in $\mathbb{R}?$
I think not, because it oscillates to infinity for large x.
But how to prove it?
| Let's study pointwise convergence first. For fixed $x$ you have
$$\lim_{n \to \infty} 2 \pi n \left( \sqrt{1 + \frac{x^2}{(2 \pi n)^2}} -1\right) = \lim_{n \to \infty} 2 \pi n \left( \frac{x^2}{2(2 \pi n)^2}\right) = 0 $$
where we used the asymptotic approximation $\sqrt{1+t^2} \sim \frac{t^2}{2}$. Hence
$$n \sin \left( \sqrt{(2 \pi n)^2 + x^2} \right) = n \sin \left( 2 \pi n \sqrt{1 + \frac{x^2}{(2 \pi n)^2}} \right) = n \sin \left( 2 \pi n \sqrt{1 + \frac{x^2}{(2 \pi n)^2}} - 2 \pi n\right) =$$
$$= n \sin \left( 2 \pi n \left( \sqrt{1 + \frac{x^2}{(2 \pi n)^2}} -1 \right) \right)$$
which is of the form $\infty \cdot 0$ as $n \to \infty$. Using again the asymptotics above and $\sin t \sim t$, we can see that this limit equals
$$\lim_{n \to \infty} n \left( 2 \pi n \left( \frac{x^2}{2(2 \pi n)^2}\right) \right) = \frac{x^2}{4 \pi}$$
So we can define $f(x)= x^2/4 \pi$, and say that this is the pointwise limit of the $f_n$s.
Clearly $f$ is not bounded on $\Bbb R$, and all the $f_n$ have the trivial bound $\sup |f_n| \le n$, hence there is no uniform convergence on $\Bbb R$ (because for all $n$ we have $\sup |f_n - f| = \infty$).
| {
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Constrained stars and bars. I have been racking my brain to understand the solutions of constrained stars and bars problems like:
$a+b+c+d = 19$; $a,b,c,d \in \{0,1,2,3,4,5,6,7\}$
Finding the total number of solutions is really simple $\dbinom{22}{3}$.
What to do after that? I want to understand the entire procedure step by step.
Please do not mark it as a duplicate of this.
Reasons:
The accepted answer's trick doesn't always work.
Another person (the answer with two upvotes) has just listed out the binomial coefficients and I can't even remotely understand what he has tried to do.
The second answer isn't my cup of tea.
| Here's another method which applies generally to many such problems. You have the problem equivalent to searching for coefficient of $x^{19}$ in expansion of $(1+x+x^2 ... x^7)^4$
Each power of $x$ in each of the $4$ factors represents value of particular variable ( say, $a$) and power of $x$ in each term of expansion represents sum of value of all four variables. Note that in each term of expansion, some power of $x$ must come from each of $4$ factors.
We can rewrite this series as:
$$\begin{align}
(1+x+x^2 + ... x^7)^{4} &= \left(\frac{1-x^8}{1-x}\right)^4 \\
&= (1-x^8)^4(1-x)^{-4} \\
&= \left(\binom{4}{0}-\binom{4}{1}x^8 + \binom{4}{2}x^{16} ...\right) \left(1-x\right)^{-4}
\end{align}$$
Thus for $x^{19}$ we only need coefficients of $x^{19}$, $x^{11}$ and $x^{3}$ from second factor, ie,$(1-x)^{-4}$. Now coefficient of $x^{r}$ in $(1-x)^{-n}$ is $\binom{n+r-1}{r}$, so we have the following answer:
$$\binom{4}{0} \binom{22}{3}-\binom{4}{1} \binom{14}{3}+\binom{4}{2} \binom{6}{3}$$
Which matches with Mike's answer found using Principal of Inclusion Exclusion.
| {
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Intersection points of two functions Find all intersection points (with integer coordinates) of $$y=p$$ and $$y=|x^2+2x-15|$$ if p is prime number.
Now what I tried:
$$|x^2+2x-15|=p$$
then
$$|(x+1)^2-16|=p$$
then
$$(x+1)^2=16-p$$ $$or$$ $$(x+1)^2=p+16$$
So $p+16$ and $16-p$ must be square numbers. I found that if $p=7$ $$(x+1)^2=9$$ so $x=2$ or $x=-4$, therefore intersection points, which I found are $(2;7)$ and $(-4;7)$. But how to find other intersection points or prove that there are no others?
| You can instead write $y=x^2 + 2x - 15$ as $$y=(x+5)(x-3).$$ Then, we are searching for integers $x$ such that $$p=|(x+5)(x-3)| = |x+5| \cdot |x-3|$$ where $p$ is a prime. Thus, $$\begin{align}\text{either} \quad &|x+5| = 1 \quad\\ \text{or} \quad &|x-3|=1.\end{align}$$
Now,
$$
\begin{align}
|x+5| = 1 &\Leftrightarrow x = -6 \quad \text{or} \quad x=-4 \\
|x-3| = 1 &\Leftrightarrow x = \phantom{-}2 \quad \text{or} \quad x=\phantom{-}4.
\end{align}
$$
Thus, these are the only possible integer values that $x$ can take. Evaluating $|(x+5)(x-3)|$ at these values, we can check whether or not the value is prime.
$$
\begin{align}
x = -6 &\Rightarrow |(x+5)(x-3)| = 9 \leftarrow \text{NOT prime} \\
x = -4 &\Rightarrow |(x+5)(x-3)| = 7 \leftarrow \text{prime} \\
x = \phantom{-}2 &\Rightarrow |(x+5)(x-3)| = 7 \leftarrow \text{prime} \\
x = \phantom{-}4 &\Rightarrow |(x+5)(x-3)| = 9 \leftarrow \text{NOT prime}
\end{align}
$$
Thus, $(x,p) = (-4,7)$ and $(x,p) = (2,7)$ are the only possible solutions.
| {
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Find: $ \lim_{x\to \infty}\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
Find: $\displaystyle \lim_{x\to \infty} f(x)=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
The answer provided in the book is 0 (also checked in Wolfram Alpha), but I can't find a good argument (without L'Hopital), to prove that. I end up in a $\infty \times 0$ situation which makes me uncomfortable.
Attempt: $$f(x)=(x-1)\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{x}(\sqrt{1+1/x}-1)}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{1+1/x}-1}{\sqrt{x+1}}$$
$$f(x)=x(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{x}\sqrt{1+1/x}}=\sqrt{x}(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{1+1/x}}.$$
Therefore the original limit is equivalent to
$$\lim_{x\to \infty} \sqrt{x}(1-1/x)\left(1-\frac{1}{\sqrt{1+1/x}}\right)$$
that appears to me a situation like $\infty\times 0$. How can I proceed to conclude that this limit is indeed $0$.
Hints and solutions are appreciated. Sorry if this is a duplicate.
| Consider $$A=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}=\sqrt{x}-\frac{1}{\sqrt{x}}-\frac{x-1}{\sqrt{x}\sqrt{1+\frac 1x}}=\sqrt{x}-\frac{1}{\sqrt{x}}-\frac{x-1}{\sqrt{x}}\left(1+\frac 1x \right)^{-1/2}$$ Now, by Taylor (or generalized binomial theorem
$$\left(1+\frac 1x \right)^{-1/2}=1-\frac{1}{2 x}+\frac{3}{8 x^2}-\frac{5}{16
x^3}+O\left(\frac{1}{x^4}\right)$$ making
$$A=\sqrt{x}-\frac{1}{\sqrt{x}}-\frac{x-1}{\sqrt{x}}\left(1-\frac{1}{2 x}+\frac{3}{8 x^2}-\frac{5}{16
x^3}+O\left(\frac{1}{x^4}\right)\right)$$ Expand and simplify to get
$$A=\frac 1{2 \sqrt x}-\frac 7 {8 x \sqrt x}+O\left(\frac{1}{x^{5/2}}\right)$$ which for sure shows the limit.
But it ia also a good approximation of the function value. Suppose $x=100$; the exact value would be $\frac{99}{10}-\frac{99}{\sqrt{101}}\approx 0.0491318$; the first term of the expansion would give $\frac 1{20}= 0.05$ while including the second term would give $\frac{393}{8000}=0.049125$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$
$P,Q,R,S$ are polynomials such that: $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$
I thought a lot on this but no result!!
By the way,one idea is to insert some values for $x$ and try to produce a system of equations for the given polynomials,but I'm not sure it works.
| Firstly it is easy to see that the roots of $x^4+x^3+x^2+x+1 =0$ are all different. Also if $a$ is a root for $ x^4+x^3+x^2+x+1=0$ then $a^5=1$. The same holds for the rest of three roots: $b,c,d$.
Now if we plug in each root in
$P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$
we get:
$$P(1)+aQ(1)+a^2R(1)= 0$$
$$P(1)+bQ(1)+b^2R(1)= 0$$
$$P(1)+cQ(1)+c^2R(1)= 0$$
$$P(1)+dQ(1)+d^2R(1)= 0$$
So quadratic function $f(x) = P(1)+xQ(1)+x^2R(1)$ has 4 different roots and so $f(x)=0$ for all $x$, so $P(1)=Q(1)=R(1)=0$ and thus the conclusion.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $a^4+b^4+c^4$ The problem:
The sum of three numbers is $6$, the sum of their squares is $8$, and the sum of their cubes is $5$. What is the sum of their fourth powers?
Based on the above information, we have:
\begin{align}
a + b + c &= 6 \\
a^2 + b^2 + c^2 & = 8 \\
a^3 + b^3 + c^3 & = 5 \\
\end{align}
I had a feeling that this vaguely had to do with Viete's theorem, which states for a cubic polynomial $f(x) = x^3 - px^2 + qx - r$ which has roots $\alpha , \beta , \gamma$,
\begin{align}
p & = \alpha+\beta+\gamma \\
q & = \alpha \beta+\alpha\gamma+\beta\gamma \\
r & = \alpha\beta\gamma
\end{align}
Notice that we already have $p$, because $a+b+c=6=\alpha + \beta + \gamma = p$. Then to find $q$:
\begin{align}
2q& = 2\alpha\beta+2\alpha\gamma+2\beta\gamma \\
& = [(a+b+c)^2-(a^2+b^2+c^2)] \\
& = (a^2+ab+ac+b^2+ab+bc+c^2+ac+bc)-a^2-b^2-c^2 \\
& = 2ab+2ac+2bc
\end{align}
\begin{align}
q & = ab+ab+bc \\
& = \frac{1}{2}[(a+b+c)^2-(a^2+b^2+c^2)] \\
& = \frac{1}{2}[6^2-8] \\
& = 14
\end{align}
So now we have $f(x) = x^3-6x^2+14x-r$. And it follows that $f(\alpha)=f(\beta)=f(\gamma)=0$.
\begin{align}
f(\alpha) & = \alpha^3-6\alpha^2+14\alpha-r = 0\\
f(\beta) & = \beta^3-6\beta^2+14\beta-r = 0\\
f(\gamma) & = \gamma^3-6\gamma^2+14\gamma-r = 0\\
\end{align}
\begin{align}
0 & = f(\alpha) + f(\beta) + f(\gamma) \\
& = (\alpha^3+\beta^3+\gamma^3)-6(\alpha^2+\beta^2+\gamma^2)+14(\alpha+\beta+\gamma)-3r\\
& = 5-6(8)+14(6)-3r\\
3r& = 41\\
r & = \frac{41}{3} \\
\end{align}
Now we have that $f(x)=x^3-6x^2+14x-\frac{41}{3}$. Here's where I am stuck. Of course, the above working out was the culmination of hours of trying things out, and eventually we have this equation. If you are reading this now, I'd appreciate if you gave any hints as to how I should continue the problem.
I've had the idea of multiplying $f(x)$ by $x$ to get fourth powers, but I haven't tried that yet. Perhaps that might yield some results?
| $f(\alpha)=0$ implies that $\alpha f(\alpha)=0$.
So, $\alpha^4-6\alpha^3+14\alpha^2-\frac{41}{3}\alpha=0$
We have similar results for $\beta$ and $\gamma$.
So, $\alpha^4+\beta^4+\gamma^4=6(\alpha^3+\beta^3+\gamma^3)-14(\alpha^2+\beta^2+\gamma^2)+\frac{41}{3}(\alpha+\beta+\gamma)$
| {
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Manual calculation doesn't match Wolfram Alpha. Why? Say I want to evaluate this sum:
$$\sum_{x=2}^\infty \ln(x^3+1)-\ln(x^3-1)$$
We can rewrite the sum as
$$\ln\left(\prod_{x=2}^\infty \frac{x^3+1}{x^3-1}\right)$$
We can split the product into two products:
$$\ln\left(\prod_{x=2}^\infty \frac{x+1}{x-1}\right)+\ln\left(\prod_{x=2}^\infty \frac{x^2-x+1}{x^2+x+1}\right)$$
These are both telescopic products! We can rewrite them as
$$\ln\left(\prod_{x=2}^\infty \frac{(x+2)-1}{x-1}\right)+\ln\left(\prod_{x=2}^\infty \frac{(x-1)^2+(x-1)+1}{x^2+x+1}\right)$$
Plugging in numbers makes this pattern more obvious;
$$\ln\left(\prod_{x=2}^\infty \frac{\color{green}{3}}{1}\frac {\color{red}{4}}{2} \frac{\color{blue}{5}}{\color{green}{3}} \frac{\color{bluedark}{6}}{\color{red}{4}} \right) +
\ln\left(\prod_{x=2}^\infty \frac{3}{\color{green}{7}} \frac{\color{green}{7}}{\color{red}{13}} \frac{\color{red}{13}}{\color{blue}{21}} \right)$$
Simplifying, we get that the sum is equal to
$$\ln(1/2)+\ln(3)=\ln\left(\frac 32 \right) \approx 0.405465108108164381978013115464349136571990423462494197614$$
However, when I put the sum in Wolfram Alpha directly, I get the following number:
$$0.4054588737136331726677370820628648457601892466568342890929$$
Why are these two numbers different? It's not a small difference either; it's on the 5th number after the decimal point! How can Wolfram make such an error?
| When you split the product into two products, the first diverges to $\infty$ and the second goes to zero. This gives you the indeterminate $\infty-\infty$. All bets are off.
| {
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Show that $L^{-1}\{(s^2+a^2)^{-\frac{3}{2}}\}=\frac{t}{a}J_1(ax)$ Question:
Assume that $J_\alpha(x)$ is defined as below:
$$J_\alpha(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{n!\, \Gamma(n+\alpha+1)}\left(\frac{x}{2}\right)^{2n+\alpha}.$$
(a) Prove that $\frac{d}{dx}(x^{-\alpha} J_\alpha(x))=-x^{-\alpha}J_{\alpha+1}(x)$;
(b) If $L\{J_0(ax)\}=\frac{1}{\sqrt {s^2+a^2}}$, show that
$L^{-1}\{(s^2+a^2)^{-\frac{3}{2}}\}=\frac{x}{a}J_1(ax)$ .
I proved part (a). But about part (b), what I have is:
$$L^{-1}\{(s^2+a^2)^{-\frac{3}{2}}\}=L^{-1}\{({(s^2+a^2)^{-\frac{1}{2}}})^{3}\}=J_0(x)*J_0(x)*J_0(x).$$
How should I reach $\frac{t}{a}J_1(ax)$ from $J_0(x)*J_0(x)*J_0(x)$ ?
| Well, let's write:
$$\mathscr{L}_x\left[\frac{x}{\text{a}}\cdot\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}=\frac{1}{\text{a}}\cdot\mathscr{L}_x\left[x\cdot\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}\tag1$$
Using the 'frequency-domain derivative' property of the Laplace transform:
$$\frac{1}{\text{a}}\cdot\mathscr{L}_x\left[x\cdot\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}=-\frac{1}{\text{a}}\cdot\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}\right\}\tag2$$
Using the 'time scaling' property of the Laplace transform:
$$-\frac{1}{\text{a}}\cdot\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\mathscr{J}_\alpha\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}\right\}=-\frac{1}{\text{a}}\cdot\frac{\partial}{\partial\text{s}}\left\{\frac{1}{\text{b}}\cdot\mathscr{L}_x\left[\mathscr{J}_\alpha\left(x\right)\right]_{\left(\frac{\text{s}}{\text{b}}\right)}\right\}=$$
$$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\mathscr{J}_\alpha\left(x\right)\right]_{\left(\frac{\text{s}}{\text{b}}\right)}\right\}\tag3$$
Using the definition of the Bessel function:
$$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\mathscr{J}_\alpha\left(x\right)\right]_{\left(\frac{\text{s}}{\text{b}}\right)}\right\}=$$
$$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\frac{\partial}{\partial\text{s}}\left\{\mathscr{L}_x\left[\left(\frac{x}{2}\right)^{\alpha+2\text{n}}\right]_{\left(\frac{\text{s}}{\text{b}}\right)}\right\}=$$
$$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{1}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\frac{\partial}{\partial\text{s}}\left\{\frac{1}{2^{\alpha+2\text{n}}}\cdot\frac{\Gamma\left(1+\alpha+2\text{n}\right)}{\left(\frac{\text{s}}{\text{b}}\right)^{1+\alpha+2\text{n}}}\right\}=$$
$$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{1}{2^{\alpha+2\text{n}}}\cdot\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{\Gamma\left(1+\alpha+2\text{n}\right)}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\frac{\partial}{\partial\text{s}}\left\{\frac{1}{\left(\frac{\text{s}}{\text{b}}\right)^{1+\alpha+2\text{n}}}\right\}=$$
$$-\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{1}{2^{\alpha+2\text{n}}}\cdot\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{\Gamma\left(1+\alpha+2\text{n}\right)}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\left(-\frac{1+\alpha+2\text{n}}{\text{s}}\cdot\frac{1}{\left(\frac{\text{s}}{\text{b}}\right)^{1+\alpha+2\text{n}}}\right)=$$
$$\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{1}{2^{\alpha+2\text{n}}}\cdot\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{\Gamma\left(2+\alpha+2\text{n}\right)}{\Gamma\left(1+\alpha+\text{n}\right)}\cdot\frac{1}{\text{s}}\cdot\frac{1}{\left(\frac{\text{s}}{\text{b}}\right)^{1+\alpha+2\text{n}}}\tag4$$
When $\Re\left(\alpha+2\text{n}\right)>-1\space\wedge\space\Re\left(\text{s}\right)>0$
Now, when $\alpha=1$:
$$\mathscr{L}_x\left[\frac{x}{\text{a}}\cdot\mathscr{J}_1\left(\text{b}\cdot x\right)\right]_{\left(\text{s}\right)}=$$
$$\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{1}{2^{1+2\text{n}}}\cdot\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{\Gamma\left(2+1+2\text{n}\right)}{\Gamma\left(1+1+\text{n}\right)}\cdot\frac{1}{\text{s}}\cdot\frac{1}{\left(\frac{\text{s}}{\text{b}}\right)^{1+1+2\text{n}}}=$$
$$\frac{1}{\text{a}}\cdot\frac{1}{\text{b}}\cdot\sum_{\text{n}=0}^\infty\frac{1}{2^{1+2\text{n}}}\cdot\frac{\left(-1\right)^\text{n}}{\text{n}!}\cdot\frac{\Gamma\left(3+2\text{n}\right)}{\Gamma\left(2+\text{n}\right)}\cdot\frac{1}{\text{s}}\cdot\frac{1}{\left(\frac{\text{s}}{\text{b}}\right)^{2+2\text{n}}}\tag5$$
Now, when $\text{a}=\text{b}$:
$$\mathscr{L}_x\left[\frac{x}{\text{a}}\cdot\mathscr{J}_1\left(\text{a}\cdot x\right)\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}^3}\cdot\frac{1}{\left(1+\left(\frac{\text{a}}{\text{s}}\right)^2\right)^\frac{3}{2}}\tag6$$
Now, I let you prove:
$$\frac{1}{\text{s}^3}\cdot\frac{1}{\left(1+\left(\frac{\text{a}}{\text{s}}\right)^2\right)^\frac{3}{2}}=\frac{1}{\left(\text{s}^2+\text{a}^2\right)^\frac{3}{2}}\tag7$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The integral of $\frac{x+1}{\sqrt {x}-1}$ Can anyone help me with this? I have been racking my brain for days trying to solve this. I tried splitting the equation into two integrals and multiplying by $\frac{\sqrt{x}}{\sqrt{x}}$ for the differential but nothing works. The only hint the book gives me is that $u = \sqrt{x} - 1$.
Thanks for checking.
-Lester
| $$\int\frac{x+1}{\sqrt{x}-1}dx=\int\frac{x\sqrt{x}+\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}dx=\int\frac{x\sqrt{x}+\sqrt{x}-2+2}{\sqrt{x}(\sqrt{x}-1)}dx=$$
$$=\int\frac{x\sqrt{x}-x+x-\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}(\sqrt{x}-1)}dx+\int\frac{4}{2\sqrt{x}(\sqrt{x}-1)}dx=$$
$$=\int\frac{x+\sqrt{x}+2}{\sqrt{x}}dx+4\ln|\sqrt{x}-1|=\frac{2}{3}\sqrt{x^3}+x+4\sqrt{x}+4\ln|\sqrt{x}-1|+C.$$
| {
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A triangle in ellipse Coordinates of the vertices B and C of a triangle ABC are (2,0) and (8,0)respectively.The vertex A is varing in such a way that $4\tan(B/2)\tan(C/2)=1$.then the locus of A has to be find.
Now , i didnt get any idea about the same.
Pls help me solving this.
| Applying Componendo-Dividendo we get $\dfrac{\cos \left ( \dfrac{B+C}{2}\right)}{\cos \left ( \dfrac{B-C}{2}\right)} = \dfrac{3}{5}$
We can easily prove that $\dfrac{a}{b+c} = \dfrac{\cos \left ( \dfrac{B+C}{2}\right)}{\cos \left ( \dfrac{B-C}{2}\right)}$
Thus $b+c = 10$ and hence locus of $A$ is an ellipse with focii at $(2,0)$ and $(8,0)$ and major axis $=10$. Then the center is at $(5,0)$ and we get semi-minor axis is $4$.
So the equation of the ellipse is $\dfrac{(x-5)^2}{25}+\dfrac{y^2}{16}=1$
| {
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A counterintuitive problem in conditional probability/combinatorics Please help me to finish/validate my solution for the following problem:
Three players $A,B,C$ play a game. A game is played between two
players. After a game the winner plays against player who did not play
last time. Everyone has equal winning chance of $\frac12$. To become a
champion a player have to win 2 games in a row. Find the probabilities
of being a champion for each player, if the first game is played
between $A$ and $B$.
Ok, my attempt:
Let ${PL}_n$ is the probability of an event "player $PL$ became a champion after game $n$". Then it is easy to get that
$A_1=B_1=C_1 =C_2=0,A_2=B_2=\frac14$, $C_3=\frac14$.
With the probability $\frac14$ $4$-th game is played between $A$ and $B$, so the game tree with winning leaves cut off will repeat itself with a period $3$.
So $A_{3n}=B_{3n}= {\frac14(\frac14)}^{n-1}={(\frac14)}^{n}$,$C_{3n+1}= {\frac14(\frac14)}^{n-1}={(\frac14)}^{n}$. Then $A_{\infty}=B_{\infty}=$ sum of infinite sequence with first term $\frac14$ and factor $\frac14$, so $A_{\infty}=B_{\infty}=\frac13$. Since the probability the no one will will win the game after game $3n+1$ is equal to ${(\frac14)}^{n}\to0$ for large $n$, we get $C_{\infty}=1-A_{\infty}-B_{\infty}=\frac13$.
I do not see any errors in my solution and I like it and believe it is correct, but the result looks counterintuitive - it looks like players $A$ and $B$ should have better chance of being a champion.
Is my solution correct? Thanks a lot for your help!
| The game tree won't repeat itself after 3 games because even though the 4th game is between $A$ and $B$, one of them will be coming in with one win under her belt and will now win with probability $\frac{1}{2}$, not $\frac{1}{4}$.
However, your solution idea works with a little modification, now that we understand that the first game is special. (Spoiler alert: don't read on yet if you don't want to try it out yourself!)
Suppose $A$ wins the first game. Now $A$ wins with probability $\frac{1}{2}$ (by winning the second game), $C$ wins with probability $\frac{1}{4}$ (by winning the second and third games), $B$ wins with probability $\frac{1}{8}$ (by $A$ losing the second game, and then $B$ winning the next two), and with probability $\frac{1}{8}$ ($C$ wins, $B$ wins, then $A$ wins) the fifth game is between $A$ and $C$ and is in exactly the same situation as the second game. So, given that $A$ wins the first game, $A$ wins with probability $\frac{1}{2}(1 + \frac{1}{8} + \left(\frac{1}{8}\right)^2 + \dots) = \frac{1}{2} \cdot \frac{8}{7} = \frac{4}{7}$, $C$ with probability $\frac{1}{4}(1 + \frac{1}{8} + \left(\frac{1}{8}\right)^2 + \dots) = \frac{1}{4} \cdot \frac{8}{7} = \frac{2}{7}$, and $B$ with probability $\frac{1}{8}(1 + \frac{1}{8} + \left(\frac{1}{8}\right)^2 + \dots) = \frac{1}{8} \cdot \frac{8}{7} = \frac{1}{7}$.
Given that $B$ wins the first game, the roles of $A$ and $B$ are reversed: $B$ wins with probability $\frac{4}{7}$, $C$ with probability $\frac{2}{7}$, and $A$ with probability $\frac{1}{7}$.
So $A$ wins the competition with probability $\frac{1}{2}\cdot\frac{4}{7}+\frac{1}{2}\cdot\frac{1}{7} = \frac{5}{14}$, $B$ with probability $\frac{1}{2}\cdot\frac{1}{7}+\frac{1}{2}\cdot\frac{4}{7} = \frac{5}{14}$, and $C$ with probability $\frac{1}{2}\cdot\frac{2}{7}+\frac{1}{2}\cdot\frac{2}{7} = \frac{2}{7}$. This confirms your intuititon: $A$ and $B$ each have a slightly better chance than $C$.
| {
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Show that any integer can be expressed as $S^{k}_{n}$ Consider the following sequence: $a_1=6$, $a_2=4$, $a_3=1$, $a_4=2$, and $a_n=a_{n-4}$ for $n \geqslant 5$.
Let $S^{k}_{n}$ be the sum of $k + 1$ elements from $a_n$ of the previous sequence. For example:
$
S^{0}_{3} = 1
\\
S^{0}_{4} = 2
\\
S^{1}_{3} = 2 + 1 = 3
\\
S^{0}_{2} = 4
\\
S^{1}_{2} = 4 + 1 = 5
\\
…
\\
S^{6}_{2} = 4 + 1 + 2 + 6 + 4 + 1 + 2 = 20
\\
…
\\
$
Show that any positive integer can be expressed as $S^{k}_{n}$.
This question was on an entrance exam for graduation. I wrote it from my memory, as the exam is not available online yet, maybe the wording is not 100% precise but I hope the examples are clear.
| As you notice, the sequence $a_n$ is by definition periodic: $6,4,1,2,6,4,1,2,6,4,1,2,...$
The sum of a whole "period" is $6+4+1+2=13$.
You can easily check that every number between $1$ and $12$ can be obtained as a sum of CONSECUTIVE numbers in the sequence $a_n$ (for instance, $9=1+2+6$). Then you just write any desired number $k$ (choose for example $k=139$) as a multiple of $13$ plus a number between $0$ and $12$ (in the example, $139 = 13 \cdot 10 + 9$). Then it just suffices to choose the right starting point and the right number of elements (in the example: to get $9$ you need to start at some $1$ in the sequence, so that you get $1+2+6=9$; for example, $S_3^{42} = 1+2+6+4+1+2+6+4+...+1+2+6= 13 \cdot 10 + 9 = 139$).
| {
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"url": "https://math.stackexchange.com/questions/2598836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve $(x^4+y^4)\,\text{d}x-xy\,\text{d}y=0$ I was trying to solve
$$(x^4+y^4)\,\text{d}x-xy\,\text{d}y=0.$$
I ended up to the point of obtaining $$x^4\, \text{d}x+xy^4\frac{\text{d}(e^{1/2y^2}x)}{e^{1/2y^2}x}=0.$$ Couldn't proceed further. Can someone help me?.
| $$x^4+y^4=x^2\frac{2ydy}{2xdx}$$
Let $\quad\begin{cases}x^2=X\\y^2=Y\end{cases}$
$$X^2+Y^2=X\frac{dY}{dX}$$
$$\frac{dY}{dX}=X+\frac{Y^2}{X}$$
Riccati ODE. Let : $\quad Y(X)= -X\frac{F'(X)}{F(X)} \quad\implies\quad Y'=-\frac{F'}{F} -X\frac{F''}{F}+X\frac{(F')^2}{F^2} $
$\frac{dY}{dX}=X+\frac{Y^2}{X}=-\frac{F'}{F} -X\frac{F''}{F}+X\frac{(F')^2}{F^2}=X+\frac{\left(-X\frac{F'}{F} \right)^2}{X}$
$-\frac{F'}{F} -X\frac{F''}{F}=X$
$$F''(X)+\frac{1}{X}F'(X)+F(X)=0$$
Bessel equation :
$$F(X)=c_1J_0(X)+c_2Y_0(X)$$
$J_0(X)$ and $Y_0(X)$ are the Bessel functions of first and second kind, of order $0$.
$\frac{dJ_0(X)}{dX}=-J_1(X)$ and $\frac{dJ_0(X)}{dX}=-Y_1(X)$
$J_1(X)$ and $Y_1(X)$ are the Bessel functions of first and second kind, of order $1$.
$\frac{F'(X)}{F(X)}=-\frac{c_1J_1(X)+c_2Y_1(X)}{c_1J_0(X)+c_2Y_0(X)} = -\frac{C\,J_1(X)+Y_1(X)}{C\,J_0(X)+Y_0(X)} $ where $C=\frac{c_1}{c_2}$
$Y(X)=-X\frac{F'(X)}{F(X)}=X\frac{C\,J_1(X)+Y_1(X)}{C\,J_0(X)+Y_0(X)}$
$$y(x)=\pm \,x\,\sqrt{\frac{C\,J_1(x^2)+Y_1(x^2)}{C\,J_0(x^2)+Y_0(x^2)}}$$
$C$ is an arbitrary constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Continuity And Derivative Of A Function
Let $
f(x)=
\begin{cases}
\sin^2x \sin(\frac{1}{x}), x\neq 0\\
0, x=0\\
\end{cases}
$
a. Find all the points where $f(x)$ is continuous
b. Find $f'(x)$ for all the points $f(x)$ is continuous
c. Is $f'(x)$ continuous where it is defined? if not where and which discontinuity type is it?
So for a we look at $\sin^2x \sin(\frac{1}{x})$ at $x\neq 0$ this is a product of two continuous functions and therefore continuous for $x=0$ we look at $\lim_{x\to 0}\sin^2x \sin(\frac{1}{x})\leq \lim_{x\to 0}|\sin^2x \sin(\frac{1}{x})|=\lim_{x\to 0}\sin^2x |\sin(\frac{1}{x})|\leq 0\cdot 1=0$ so $$
f(x)=
\begin{cases}
\sin^2x \sin(\frac{1}{x}), x\neq 0\\
0, x=0\\
\end{cases}
$$ is continuous for all $x$
b. For $x\neq 0$ we can take $$[\sin^2x \sin(\frac{1}{x})]'=2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$$ (not sure why can we do it)
At $x=0$ we look at $$lim_{\Delta h\to 0}\frac{\sin^2(x+\Delta h) \sin(\frac{1}{x+\Delta h})-\sin^2x \sin(\frac{1}{x})}{\Delta h}\mid_{x=0}=lim_{\Delta h\to 0}\frac{\sin^2(\Delta h) \sin(\frac{1}{\Delta h})-0}{\Delta h}=lim_{\Delta h\to 0}\frac{\sin(\Delta h)}{\Delta h} \cdot sin(\Delta h)\cdot\sin(\frac{1}{\Delta h})\leq lim_{\Delta h\to 0}\mid\frac{\sin(\Delta h)}{\Delta h} \cdot sin(\Delta h)\cdot\sin(\frac{1}{\Delta h})\mid=lim_{\Delta h\to 0}\mid\frac{\sin(\Delta h)}{\Delta h}\mid \cdot \mid \sin(\Delta h)\mid \cdot \mid\sin(\frac{1}{\Delta h})\mid=1\cdot 0 \cdot 1=0$$
(When can I take the limit of a product of functions? just when I am sure the the limit of each function is finite?)
c. Now we have to check that $\lim_{x\to 0}2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$
(how do I find if this limit exist?)
In general are a and b are valid?
| Here you check whether this limit exists: $\lim_{x\to 0}2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$. The first term approaches zero, so we care about second (forget $-$ sign for some minuts):
$$\lim_{x\to 0}\frac{\cos(\frac{1}{x})\sin^2x}{x^2} \equiv \lim_{x\to 0}\cos(\frac{1}{x})$$
which surely does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
prove that $\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$
prove that $$\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$$
My attempt:
C is semicircle in upper half complex plane
Simple poles = $e^{i\frac{\pi}{8}}, e^{i\frac{3\pi}{8}},e^{i\frac{5\pi}{8}},e^{i\frac{7\pi}{8}}$ lie in upper semi-circle C and real axis
Given integral value $= 2\pi i \cdot (\text{sum of residues}) = 2 \pi i \left(\frac{-1}{8}\right) \left[e^{i\frac{5\pi}{8}}+e^{i\frac{15\pi}{8}}+e^{i\frac{25\pi}{8}}+e^{i\frac{35\pi}{8}}\right] = 0.27059 \pi$
This is numerically equal to $\frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$. But without using calculator, how to get this expression.
| A couple of alternative approaches to Euler's Beta function or contour integration.
1) The reflection formula for the digamma function
By parity we have $I=\int_{-\infty}^{+\infty}\frac{x^4}{1+x^8}\,dx = 2\int_{0}^{+\infty}\frac{x^4}{1+x^8}\,dx $ and by splitting $\mathbb{R}^+$ as $(0,1]\cup(1,+\infty)$, then enforcing the substitution $x\mapsto \frac{1}{x}$ on the second interval,
$$ I = 2\int_{0}^{1}\frac{x^2+x^4}{1+x^8}\,dx = 2\sum_{n\geq 0}\left[\frac{1}{16n+3}+\frac{1}{16n+5}-\frac{1}{16n+11}-\frac{1}{16n+13}\right] $$
where the RHS is a sort of BBP-type formula. Since $\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}$ and $\psi(x)-\psi(1-x)=-\pi\cot(\pi x)$,
$$ \sum_{n\geq 0}\left[\frac{1}{16n+3}-\frac{1}{16n+13}\right]=\frac{\pi}{16}\cot\frac{3\pi}{16},$$
$$ \sum_{n\geq 0}\left[\frac{1}{16n+5}-\frac{1}{16n+11}\right]=\frac{\pi}{16}\cot\frac{5\pi}{16},$$
and finally
$$ I = \frac{\pi}{8}\left[\cot\frac{3\pi}{16}+\cot\frac{5\pi}{16}\right] = \color{blue}{\frac{\pi}{\sqrt{2}}\sin\frac{\pi}{8}}=\frac{\pi}{2}\sqrt{1-\frac{1}{\sqrt{2}}}.$$
2) Glasser's Master Theorem
We have
$$ I = \int_{\mathbb{R}}\frac{dx}{x^4+\frac{1}{x^4}}=\int_{\mathbb{R}}\frac{dx}{\left(x^2+\frac{1}{x^2}\right)^2-2}=\int_{\mathbb{R}}\frac{dx}{\left[\left(x-\frac{1}{x}\right)^2+2\right]^2-2} $$
hence, by the residue theorem and de l'Hopital rule,
$$ I = \int_{\mathbb{R}}\frac{dz}{(z^2+2)^2-2}=2\pi i\sum_{\zeta\in Z}\operatorname*{Res}_{z=\zeta}\frac{1}{z^4+4z^2+2}\stackrel{\text{d.H.}}{=}\frac{\pi i}{2}\sum_{\zeta\in Z}\frac{1}{z^3+2z} $$
where $Z=\{i\sqrt{2-\sqrt{2}},i\sqrt{2+\sqrt{2}}\}$. Of course the outcome is the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Integrate $\int \frac {x^4}{\sqrt {x^2-9}} \,dx$ Integrate $\int \dfrac {x^4}{\sqrt {x^2-9}} dx$
My Attempt:
Let $x=3\sec (\theta )$
$$dx=3\sec (\theta).\tan (\theta).d\theta$$
Then,
$$=\int \dfrac {x^4}{\sqrt {x^2-9}}$$
$$=\int \dfrac {81. \sec^4 (\theta)}{\sqrt {(3\sec (\theta))^2 - 9}} 3\sec (\theta).\tan (\theta).d\theta $$
$$=\int \dfrac {81 \sec^5 (\theta). 3\tan (\theta).d\theta}{3\tan (\theta)}$$
$$=81\int \sec^5 (\theta) d\theta$$
| Let $ x + t = \sqrt{x^2 - 9}$
$$I = \dfrac{-1}{16}\int \dfrac{(t^2 + 9)^4}{t^5} dt = -\dfrac{486\ln(|t|)+\dfrac{t^4+72t^2}{4}-\dfrac{5832t^2+6561}{4t^4}}{16}+C$$
Then, $$\dfrac1{16}\left(\dfrac{t^4+72t^2}{4}-\dfrac{5832t^2+6561}{4t^4}\right) =- \left(\dfrac{27}8 x \sqrt{x^2 - 9} + x^3\dfrac{1}4 \sqrt{x^2 - 9}\right) $$
$$I = \dfrac{27}8 x \sqrt{x^2 - 9} + x^3\dfrac{1}4 \sqrt{x^2 - 9} - \dfrac{243}8\ln(\sqrt{x^2 - 9} - x) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find all pairs of integers (x,y) such that $x(x+1)(x^2+x+2)=2y^2$ A former Olympiad exercise is to find all pairs of integers $(x,y)$ such that
$$x(x+1)(x^2+x+2)=2y^2.$$
I could not solve this equation. How can one find all solutions?
I tried to rewrite the above equation in a different form, but it did not help. For example, the equation is equivalent to
\begin{align*} x(x+1)(x(x+1)+2)=2y^2&\iff (x(x+1))^2 + 2x(x+1) +1 =2y^2 +1\\
&\iff (x(x+1)+1)^2 =2y^2 +1\end{align*}
Can you tell me how to attack this problem?
Best wishes
| Set $z=x^2+x$, then we want $z(z+2) = 2y^2$. So $z$ is even and with $w=z/2 \in\Bbb{Z}$, we get the new Diophantine equation
$$2w(w+1) =y^2.$$
Notice that since $x^2+x=z$ has no real solution for $z<\frac{1}{4}$, we can assume $w\ge0$.
*
*If $2|w$, then since $gcd(2w, w+1)=1$ we have that $w+1$ is a square. Then $2w=z$ is a square, and so is $4z$. But since $x = -\frac{1}{2} \pm\sqrt{z+\frac{1}{4}}$ is an integer, $4z+1$ is a square. So $z=w=0$ which means $y=0$, and the two solutions for $x$ are $0$ and $-1$.
*If $2$ does not divide $w$, then $gcd(w, 2(w+1))=1$ implies that $w$ is a square, and this implies that $w+1 = 2k^2$ for $k\in \Bbb{Z}$. So $$x^2 +x = 2w = 4k^2-2$$
and since $x = -\frac{1}{2} \pm\sqrt{4k^2-2+\frac{1}{4}} \in \Bbb{Z}$, we have that $16k^2-7$ is a square. So is $16k^2$, and the only integer squares with difference $7$ are $16$ and $9$, so $k=1$, so $w=1$, so $y= \pm2$ and the possibilities for $x$ are $1$ and $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Find largest possible value of $x^2+y^2$ given that $x^2+y^2=2x-2y+2$
Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$.
My attempt:
$x^2+y^2=2x-2y+2$
$(x^2-2x)+(y^2+1)=2$
$(x-1)^2+(y+1)^2=4$
I have no idea how to continue here. Any help?
| Hint
You found
$$(x-1)^2+(y+1)^2=4$$
which is the equation of a circle centered in $(1,-1)$ and with radius $2$.
The function $x^2+y^2$ is the square of the distance of $(x,y)$ to the origin. Which point $(x,y)$ on this circle is located the furthest from the origin? It helps to make a simple sketch.
Alternatively, you can handle this as a constrained optimization problem and use for example the method of Lagrange multipliers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Does $x+\sqrt{x}$ ever round to a perfect square, given $x\in \mathbb{N}$? I'll define rounding as $$R(x)=\begin{cases} \lfloor x \rfloor, & x-\lfloor x \rfloor <0.5 \\ \lceil x \rceil, & else\end{cases}$$
Does $x+\sqrt{x}$ ever round (to the nearest integer) to a perfect square, given $x\in \mathbb{N}$?
For example, $7+\sqrt{7}=9.646...$ which rounds up, and $57+\sqrt{57}=64.549...$ which also round up. Also, $6+\sqrt{6}$ and $57+\sqrt{57}$ both round down.
I think the positive integers $x$ such that $\lfloor x+\sqrt{x} \rfloor =k^2, k\in \mathbb{Z}$ are all of the form $n^2+n+1, n\in \mathbb{Z}^+$. The set of all $x$ begins as: $\{3, 7, 13, 21, 31, 43, 57, \dots \}$ and all those numbers are of the form $n^2+n+1$
I tried to find a pattern for whether the decimal part of $\sqrt{n^2+n+1}$ is less than $0.5$ or not, and I tried to modify $\sqrt{n^2+n+1}$ to $\sqrt{n^2+2n+1}=(n+1)^2$ but that didn't lead anywhere.
Is there an algebraic proof/disproof of my above claim?
Thanks.
| $n+\sqrt{n}$ never rounds to a square
By way of introduction, note first that $N^{2}+N$ can never be a
perfect square (when $N$ is a whole number) -- this is because perfect squares
of whole numbers must differ by (at least) $2N+1$. If we let $n=N^{2}$ then
$N^{2}+N=n+\sqrt{n}$ and this can never be a perfect square; however, we can
ask if $n+\sqrt{n}$ can ever round to a perfect square -- i.e. is
there a perfect square which is within $\frac{1}{2}$ of it?
Let $G(n)=$ the nearest integer to $n+\sqrt{n}$, so that $G(n)$ lies within
$\frac{1}{2}$ of $n+\sqrt{n}$.$.$
Let $K$ be the nearest integer to $\sqrt{n}$, so
\begin{align}
K-\frac{1}{2} & <\sqrt{n}<K+\frac{1}{2}\text{ and}\tag{A}\\
\left( K-\frac{1}{2}\right) ^{2} & <n<\left( K+\frac{1}{2}\right)
^{2}\text{ and}\nonumber\\
K^{2}-K+\frac{1}{4} & <n<K^{2}+K+\frac{1}{4}\text{. Adding:}\tag{B}\\
K^{2}-\frac{1}{4} & <n+\sqrt{n}<K^{2}+2K+\frac{3}{4}\text{ or}\nonumber
\end{align}
\begin{equation}
K^{2}-\frac{1}{4}<n+\sqrt{n}<\left( K+1\right) ^{2}-\frac{1}{4}\tag{C}%
\end{equation}
Thus: $n+\sqrt{n}$ lies between $K^{2}-\frac{1}{4}$ and $\left( K+1\right)
^{2}-\frac{1}{4}$. It follows that $G(n)=K^{2}$ or $G(n)=\left( K+1\right)
^{2}$ if $G(n)$ is to be a perfect square. We eliminate each case separately.
CASE 1: Suppose $G(n)=K^{2}$ Then $n+\sqrt{n}<K^{2}+\frac{1}{2}$ so (from line C):
\begin{equation}
K^{2}-\frac{1}{4}<n+\sqrt{n}<K^{2}+\frac{1}{2}\text{.}\tag{D}%
\end{equation}
Multiplying line (A) above by $-1$ and adding:
\begin{align*}
-K-\frac{1}{2} & <-\sqrt{n}<-K+\frac{1}{2}\text{; adding to line (D):}\\
\left( K^{2}-K\right) -\frac{3}{4} & <n<\left( K^{2}-K\right) +1\text{.}%
\end{align*}
Since $n$ and $K$ are whole numbers, we must have $n=K^{2}-K$. But from line
(B) above this gives $\ n+\frac{1}{4}<n$ or $\frac{1}{4}<0$: a contradiction.
CASE 2: Suppose $G(n)=\left( K+1\right) ^{2}$. Then
\begin{align*}
\left( K+1\right) ^{2}-\frac{1}{2} & <n+\sqrt{n}<\left( K+1\right)
^{2}+\frac{1}{2}\text{ and from line (A):}\\
-K-\frac{1}{2} & <-\sqrt{n}<-K+\frac{1}{2}\text{. Adding:}\\
\left( K+1\right) ^{2}-K-1 & <n<\left( K+1\right) ^{2}-K+1\text{ or:}\\
\left( K+1\right) ^{2}-(K+1) & <n<\left( K+1\right) ^{2}-\left(
K+1\right) +2\text{.}%
\end{align*}
Since $n$ and $K$ are whole numbers and these are strict inequalities, we must
have $n=\left( K+1\right) ^{2}-(K+1)+1=(K+1)^{2}-K$. But from line (B) above:
\begin{align*}
\left( K+1\right) ^{2}-K & <K^{2}+K+\frac{1}{4}\text{ or:}\\
K^{2}+2K+1-K & <K^{2}+K+\frac{1}{4}\text{ or}\\
1 & <\frac{1}{4}\text{.}%
\end{align*}
This contradiction shows that $G(n)$ can never be a perfect (integer) square.
| {
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"url": "https://math.stackexchange.com/questions/2613253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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How can one show that $\int_{0}^{1}{x\ln{x}\ln(1-x^2)\over \sqrt{1-x^2}}\mathrm dx=4-{\pi^2\over 4}-\ln{4}?$ How can one show that
$$\int_{0}^{1}{x\ln{x}\ln(1-x^2)\over \sqrt{1-x^2}}\mathrm dx=4-{\pi^2\over 4}-\ln{4}?\tag1$$
I have tried IBP but it is seem too complicate. I am not sure what to do or how to tackle this problem. Please any help? Thank!
| Enforcing a substitution of $x \mapsto \sqrt{x}$ the integral becomes
$$I = \frac{1}{4} \int_0^1 \frac{\ln x \ln (1 - x)}{\sqrt{1 - x}} \, dx.$$
From the definition for Euler's beta function, namely
$$\text{B}(x,y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt,$$
we observe that
$$\lim_{x \to 1} \lim_{y \to 1/2} \partial_x \partial_y \text{B}(x,y) = \int_0^1 \frac{\ln t \ln (1 - t)}{\sqrt{1 - t}} \, dt.$$
On finding the required derivatives for the Beta function we obtain
$$\partial_x \partial_y \text{B}(x,y) = \text{B}(x,y) \left [\{\psi(x) - \psi(x + y)\}\{\psi(y) - \psi (x + y)\} - \psi^{(1)}(x + y) \right ].$$
Here $\psi(x)$ is the digamma function while $\psi^{(1)}(x)$ is the trigamma function or polygamma function of order one.
On taking the required limits we are left with
$$\lim_{x \to 1} \lim_{y \to 1/2} \partial_x \partial_y \text{B}(x,y) = \text{B} \left (1, \frac{1}{2} \right ) \left [\{\psi(1) - \psi(3/2)\}\{\psi (1/2) - \psi(3/2)\} - \psi^{(1)}(3/2) \right ].$$
Each of the values for the digamma and polygamma functions are well known. They are
\begin{align*}
\psi \left (\frac{1}{2} \right ) = -\gamma - \ln (4), & \quad \psi (1) = -\gamma,\\
\psi \left (\frac{3}{2} \right ) = 2 - \gamma - \ln (4),& \quad
\psi^{(1)} \left (\frac{3}{2} \right ) = 3 \zeta (2) - 4 = \frac{\pi^2}{2} - 4.
\end{align*}
As
$$\text{B} \left (1, \frac{1}{2} \right ) = \frac{\Gamma (1) \Gamma (1/2)}{\Gamma (3/2)} = 2,$$
we have
$$\lim_{x \to 1} \lim_{y \to 1/2} \partial_x \partial_y \text{B}(x,y) = 16 - 4 \ln (4) - \pi^2,$$
yielding
$$\int_0^1 \frac{x \ln x \ln (1 - x)}{\sqrt{1 - x^2}} \, dx = 4 - \ln (4) - \frac{\pi^2}{4}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Deriving the fact that the approximation $\log(n!) \approx n\log(n) - n + \frac{1}{2}\log(2\pi n)$ is $O(1/n)$. I get to begin with Stirling's approximation, for any $C \in \mathbb{Z}_{\geq 0}$, there exists some $N \in \mathbb{Z}_{\geq 0}$ such that $N > C$ and for all $n > N$:
\begin{align*}
&\quad \left|n! - \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}\right| \leq C \left|\frac{1}{n}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right| \\
&\Rightarrow -C \frac{1}{n}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n \leq n! - \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} \leq C \frac{1}{n}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n \\
&\Rightarrow -C \frac{1}{n}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n + \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} \leq n! \leq C \frac{1}{n}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n + \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} \\
&\Rightarrow \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(-C \frac{1}{n} + 1\right) \leq n! \leq \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(C \frac{1}{n} + 1\right)
\end{align*}
Now we are in a position to make further manipulations:
\begin{align*}
&\quad \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(-C \frac{1}{n} + 1\right) \leq n! \leq \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(C \frac{1}{n} + 1\right) \\
&\{\text{$\log$ is monotonic}\} \\
&\Rightarrow \log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(-C \frac{1}{n} + 1\right)\right) \leq \log\left(n!\right) \leq \log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(C \frac{1}{n} + 1\right)\right) \\
&\{\text{take the average of the upper ($U(n)$) and lower bounds ($L(n)$)}\} \\
&\{\text{$\log(n!) \approx f(n)$ means $|f(n) - \log(n!)| \leq K \left[U(n) - L(n)\right]$, where $0 \leq K \leq 1$}\} \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(C \frac{1}{n} + 1\right)\right) + \frac{1}{2}\log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(-C \frac{1}{n} + 1\right)\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\left(\log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(C \frac{1}{n} + 1\right)\right) + \log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(-C \frac{1}{n} + 1\right)\right)\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left((\sqrt{2\pi n})^2\left(\frac{n}{e}\right)^{2n}\left(C \frac{1}{n} + 1\right)\left(-C \frac{1}{n} + 1\right)\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left(2\pi n\left(\frac{n}{e}\right)^{2n}\left(\frac{-C^2}{n^2} + 1\right)\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left(2\pi n\right) + \frac{1}{2}\log\left(\frac{n^{2n}}{e^{2n}}\right) + \frac{1}{2}\log\left(1 - \frac{C^2}{n^2}\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left(2\pi n\right) + \frac{1}{2}\log\left(n^{2n}\right) - \frac{1}{2}\log\left(e^{2n}\right) + \frac{1}{2}\log\left(1 - \frac{C^2}{n^2}\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left(2\pi n\right) + n\log\left(n\right) - n\log\left(e\right) + \frac{1}{2}\log\left(1 - \frac{C^2}{n^2}\right) \\
&\Rightarrow \log(n!) \approx n\log\left(n\right) - n + \frac{1}{2}\log\left(2\pi n\right) + \frac{1}{2}\log\left(1 - \frac{C^2}{n^2}\right)
\end{align*}
From here I could expand $\log(1 + x)$ about $x = 0$ (note that $0 < \frac{C^2}{n^2} < 1$, as $C < n$, and $0 < n, C$, and also $C^2/n^2$ is pretty close to $0$ most of the time, as $n > C$). The first term of the Taylor series expansion for $\log(1 + x)$ about $x = 0$ is simply $x$.
\begin{align*}
&\quad \log(n!) \approx n\log\left(n\right) - n + \frac{1}{2}\log\left(2\pi n\right) + \frac{1}{2}\log\left(1 - \frac{C^2}{n^2}\right) \\
&\Rightarrow \log(n!) \approx n\log\left(n\right) - n + \frac{1}{2}\log\left(2\pi n\right) - \frac{C^2}{2n^2}
\end{align*}
But this seems to suggest my error is $O(1/n^2)$? Isn't this approximation supposed to be $O(1/n)$? Please advise on where I went wrong.
| Everything you've written here is 'correct' — but let's take a closer look at what it means. Your definition of $\approx$ says: "$\log(n!)\approx f(n)$ means that $\left|f(n)-\log(n!)\right|\leq K (U(n)-L(n))$ for some $0\leq K\leq 1$". And we have $L(n) = n\log n-n+\log(2\pi n)+\log(1-\frac Cn)$ and $U(n)=n\log n-n+\log(2\pi n)+\log(1+\frac Cn)$. So the difference between these terms — the bound you've put on the 'quality' of $\approx$ — is $\log(1+\frac Cn)-\log(1-\frac Cn)$. But the best you can say about this is that it's $O(\frac1n)$, so that's what your result is: $\log(n!)\in n\log n-n+\log(2\pi n)-\frac{C^2}{2n^2}+O(\frac1n)$. And now there's no point in including the $-\frac {C^2}{2n^2}$ term, because the $O(\frac1n)$ term 'overwhelms' it.
| {
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"url": "https://math.stackexchange.com/questions/2616658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Evaluating: $\int \frac {1+\sin (x)}{1+\cos (x)} dx$
Evaluate: $\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$
My Attempt:
$$=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$$
$$=\int \dfrac {(\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2}))^2}{2\cos^2 (\dfrac {x}{2})} dx$$
$$=\dfrac {1}{2} \int (\dfrac {\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2})}{\cos (\dfrac {x}{2})})^2 dx$$
$$=\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx$$
How do I continue?
| I may contribute by multiplying $(1-\cos(x))/(1-\cos(x)) $ :
$$ \int \frac{1 + \sin(x)}{1 + \cos(x)} dx= \int \frac{1 + \sin(x) - \cos(x) - \sin(x)\cos(x)}{1 - \cos^{2}(x)} dx = \int \frac{1 + \sin(x) - \cos(x) - \sin(x)\cos(x)}{\sin^{2}(x)} dx = \int \frac{\sin^{2}(x) + \cos^{2}(x) + \sin(x) - \cos(x) - \sin(x)\cos(x)}{\sin^{2}(x)} dx $$
$$ = \int \left(1 + \cot^{2}(x) + \frac{1}{\sin(x)}\right) dx - \int \frac{d(\sin(x))}{\sin^{2}(x)} - \ln | \sin(x)|$$
$$= x + \int \cot^{2}(x)dx + \int \frac{1}{\sin(x)}dx + \frac{1}{\sin(x)}-\ln|\sin(x)| $$
Now for one integral, multiply by $(1+\cos(x))/(1+\cos(x)) $ :
$$ \int \frac{1}{sin(x)} dx = \int \frac{1+\cos(x)}{\sin(x)(1+\cos(x))} dx = \int \frac{\sin^{2}(x) + \cos^{2}(x) + \cos(x)}{\sin(x)(1+\cos(x))} dx $$
$$ \int \frac{\sin(x)}{1+\cos(x)}dx + \int \frac{\cos(x)}{\sin(x)} dx = - \ln|1+\cos(x)| + \ln|\sin(x)| $$
so ...
$$ \int \frac{1 + \sin(x)}{1 + \cos(x)} dx= x - \ln|1+\cos(x)| + \frac{1}{\sin(x)} + \int \cot^{2}(x)dx$$
The last part may be continued..
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Integrating $\int (t^2+7)^{\frac{1}{3}}$dt
I have to integrate $\int (t^2+7)^{\frac{1}{3}}$dt
I tried some trigonometric substitution and then rationalisation but didn't get anything
Please provide a hint on how to proceed with this question
Please help!!!
| Positioning your integral:
\begin{equation}
\int \left(t^2 + 7 \right)^{\frac{1}{3}}\:dt = \int_0^t \left(w^2 + 7 \right)^{\frac{1}{3}}\:dw = \int_0^t \frac{1}{ \left(w^2 + 7 \right)^{\frac{1}{3}}}\:dw
\end{equation}
We can employ the solution that I've spoken to here:
\begin{equation}
\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}, \frac{1}{1 + ax^n} \right)\right]
\end{equation}
Here $a = 7$, $m = -\frac{1}{3}$, $k = 0$, and $n = 2$. Thus,
\begin{align}
\int \left(t^2 + 7 \right)^{\frac{1}{3}}\:dt &= \frac{1}{2}\cdot 7^{\frac{0 + 1}{2} --\frac{1}{3}}\left[B\left(-\frac{1}{3} - \frac{0 + 1}{2}, \frac{0 + 1}{2}\right) - B\left(-\frac{1}{3}- \frac{0 + 1}{2}, \frac{0 + 1}{2}, \frac{1}{1 + 7t^2} \right)\right] \\
&= \frac{7^{\frac{5}{6}}}{2}\left[B\left(-\frac{5}{6}, \frac{1}{2}\right) - B\left(-\frac{5}{6}, \frac{1}{2}, \frac{1}{1 + 7t^2} \right)\right]
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Number of non-negative integral solutions of $2x+3y+z=19$ The question is to find the number of solutions of $2x+3y+z=19$.
There are posts like this with answers arrived by counting the possibilities. But I decided to approach it in a different way.
My Approach:
I learnt a theorem stating that
The number of integral solutions of $$a_1x_1+a_2x_2+a_3x_3...+a_nx_n=r$$ is the coefficient of $x^r$ in $$(1+{x}^{a_1}+{x}^{2\cdot{a_1}}+...+{x}^{\left \lfloor{\frac{r}{a_1}}\right \rfloor\cdot{a_1}})\cdot(1+{x}^{a_2}+{x}^{2\cdot{a_2}}+...+{x}^{\left \lfloor{\frac{r}{a_2}}\right \rfloor\cdot{a_2}})...(1+{x}^{a_n}+{x}^{2\cdot{a_n}}+...+{x}^{\left \lfloor{\frac{r}{a_n}}\right \rfloor\cdot{a_n}})$$ (I don't know the name of this theorem)
Substituting and reducing the equation, I landed up with finding coefficient of $x^{19}$ in $${\frac{1-x^{20}}{1-x^2}}\cdot{\frac{1-x^{21}}{1-x^3}}\cdot{\frac{1-x^{20}}{1-x}}$$
But, I had no clue of finding that.
My doubts:
*
*Is there any other relatively elegant way of finding it? (Other than counting or stars and bars - as in here)
*If no, how to solve further to get the answer?
Thanks in advance...
| You can try directly. We see that $0\leq y\leq 6$
case 1: $y=6$ then we have $2x+z=1$, so $z=1$ and $x=0$;
case 2: $y=5$ then we have $2x+z=4$, so $x=2,z=0$ or $x=1,z=2$ or $x=0, z=4$;
case 3: $y=4$ then we have $2x+z=7$, so $0\leq x\leq 3$ so we have 4 solution
case 4: $y=3$ then we have $2x+z=10$, so $0\leq x\leq 5$ so we have 6 solution
case 5: $y=2$ then we have $2x+z=13$, so $0\leq x\leq 6$ so we have 7 solution
case 6: $y=1$ then we have $2x+z=16$, so $0\leq x\leq 8$ so we have 9 solution
case 7: $y=0$ then we have $2x+z=19$, so $0\leq x\leq 9$ so we have 10 solution
Thus we have total $1+3+4+6+7+9+10 = 40$ solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
$3$ primitive pythagorean triples from 6 integers. Do there exist $3$ different primitive Pythagorean triples $(a,d,w), (a,b,z)$ and $(c,d,z)$?
Explicitly, we want $6$ different integers $a,b,c,d,w,z$ such that...
(1) $a^2 + d^2 = w^2$
(2) $a^2 + b^2 = c^2 + d^2 = z^2$
| This is surprisingly simple given a function derived from Euclid's formula:
$$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2$$
To find a triple matching side-C,
$$C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor$$
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$, for example
$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$
$$33^2+56^2=63^2+16^2=65^2$$
The larger the value of $C$ the greater the chances there are more solutions, for example $C=1105$:
$f(24,23)=(47,1104,1105)\\
f(31,12)=(817,744,1105)\\
f(32,9)=(943,576,1105)\\
f(33,4)=(1073,264,1105)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Ratio of volumes of $n$-dimensional unit balls
Prove that $$\left( \frac{nS_n}{S_{n-1}}\right)^{1/n} \le 2$$ where $S_i$ is the volume of the $i$th dimensional unit ball and $n\ge 2$.
I think we can use the fact that an $n$-dimensional unit ball is contained in a hypercube whose edge measures $2$ units, and the ball contains a hypercube whose edge measures $\sqrt{2}$ units. So we have $\sqrt{2}^n\le S_n \le 2^n$ and thus $$\left( \frac{nS_n}{S_{n-1}}\right)^{1/n} \le \left( \frac{n2^n}{\sqrt{2}^{n-1}} \right)^{1/n} = n^{1/n}2^{\frac{n+1}{2n}} = n^{1/n}\sqrt{2}\sqrt{2}^{1/n}.$$
So if we take $n^{1/n} \le 2$ and $\sqrt{2}^{1/n}\le \sqrt{2}^{1/2}$, then an upper bound to my problem is $2^{1+3/4}$ which isn't good enough.
Do we need another approach or do we get better bounds on $n^{1/n}$?
| Since the volume of a $n$-dimensional unit ball is (see volume of an $n$- ball)$$
S_n = \frac{π^{\frac{n}{2}}}{Γ\left( n + \frac{1}{2} \right)},
$$
then\begin{align*}
\left( \frac{nS_n}{S_{n - 1}} \right)^{\frac{1}{n}} \leqslant 2 &\Longleftrightarrow \frac{n \sqrt{π} \cdot Γ\left( n - \frac{1}{2} \right)}{Γ\left( n + \frac{1}{2} \right)} \leqslant 2^n\\
&\Longleftrightarrow \frac{n \sqrt{π}}{n - \frac{1}{2}} \leqslant 2^n.
\end{align*}
Because$$
2^n \geqslant 4 > 2 \sqrt{π} \geqslant \frac{n \sqrt{π}}{n - \frac{1}{2}},
$$
then $\displaystyle \left( \frac{nS_n}{S_{n - 1}} \right)^{\frac{1}{n}} \leqslant 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove the convergence of $\sum\limits_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$ Hints preferred:
I need to prove convergence and determine the limit of following series: $$ \sum_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$$
I need to use partial fraction decomposition to simplify the term, which I attempted with following:
$$
\frac 1 {(2k+1)(2k+5)} = \frac a {(2k+1)} + \frac b {(2k+5)} \\~\\
\frac 1 {(2k+1)(2k+5)} = \frac {a(2k+5)} {(2k+1)(2k+5)} + \frac {b(2k+1)} {(2k+1)(2k+5)} \\~\\$$
Then we have:
$$
k: 2a + 2b = 0 \\
k^0: 5a + b = 1 \\~\\
a = \frac 1 4 ~ \land b = -\frac 1 4
$$
So we have:
$$
\frac 1 {(2k+1)(2k+5)} = \frac { \frac 1 4 } {(2k+1)} - \frac { \frac 1 4} {(2k+5)} = \\ \frac { 1 } {4(2k+1)} - \frac { 1 } {4(2k+5)}
$$
Which can be defined as:
$$
\frac 1 4 (\frac { 1 } {2k+1} - \frac { 1 } {2k+5} )
$$
I'm not sure how to turn the result into a telescopic series (if the partial fraction decomposition is right in the first place).
| Yes, this sum telescopes. We have: $$S =\sum_{k=2}^{\infty}\frac14\left(\frac1{2k+1}-\frac1{2k+5}\right)$$ giving us $$4S =\left(\frac15-\frac19\right) +\left(\frac17-\frac1{11}\right)+\left(\frac19-\frac1{13}\right)+ \ldots$$ $$\implies 4S=\frac15+\frac17$$ $$\boxed{\therefore S=\frac3{35}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $\exists n \in \mathbb{Z+}$,s.t. $2^n + 1$ is a prime, prove $\exists k \in \mathbb{Z+}, n = 2^k$ The actual description of the question is:
Suppose that $2^n + 1$ is a prime where $n$ is a positive integer. Prove that
$n = 2^k$ for some positive integer $k$.
There is only one approach that know of: To show that there is factorization in which at least one factor $(x-a)=1$, and the others are prime. A nice simplification would be to show that all except one root is having value such that the factor has value = $1$. For example, if given quadratic: $x^2 + 2x - 3 = (x+3)(x-1)$. As have two factors, so one of these factors to be $1$, and the other prime.
This leads to: $$x-1=1 \implies x=2 \implies (x+3)(x-1) = 5$$
else, $$x+3=1 \implies x = -2 \implies (x-1)(x+3) => -3$$
But, if consider positive integers then $x=2$ is the value.
if given quadratic: $x^2 - 5x +6 = (x-3)(x-2)$. As have two factors, so one of these factors to be $1$, and the other prime.
This leads to: $$x-3=1 \implies x=4 \implies (x-3)(x-2) = 2$$
else, $$x-2=1 \implies x = 3 \implies (x-3)(x-2) => 0$$
But, if consider that $0$ is not a prime, then $x=4$ is the value.
So, let us try to work out cases with value of $k$ starting form $0$.
$k=0, 2^0=1, 2^1+1 =3$
$k=1, 2^1=2, 2^2 +1=5$
So, weak induction will suffice here, as strong induction is not needed as no need for previous values to prove new case.
So, let the assumption be that it holds for $k=l$, need prove for $k=l+1$, but, $2^{2^{l+1}} = 2^{{2^l}^2} = (2^{2^l})^2$
So, if $2^{2^l} + 1$ is a prime, let p; then need prove $(p-1)^2+1$ is also.
i.e., need prove $p^2-2p+2$ is a prime, given $p$ is.
| We have:
$$2^{ab} + 1 = (2^{a} + 1)(2^{a(b-1)} - 2^{a(b-2)} + \cdots + 1)$$
where $b$ is an odd integer. It's obvious that both of the factors are bigger than 1 and so if $n$ has an odd factor we $2^n + 1$ has a nontrivial factor.
If you're more interested in primes of that form, they are called Fermat's Primes and till now only five of them are known.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Solving a biquadratic.
If $x$ is a real number and satisfies $$x+ \sqrt[4] {5-x^4}=2$$ then find the value of $$x\sqrt[4] {5-x^4}$$
My try :
The question is significantly asking for the value of $-x(x-2)$ if we get the root of the equation $$2x^4-8x^3+24x^2-32x+11=0$$
Using this I have reached till
$$-2x(x-2)(x^2-2x+8)=11$$
$$-x(x-2)=\frac {11}{2(x^2-2x-8)}$$
but I couldn't manipulate it further.
Also upon some second thought I want to ask whether could it be possible to form a quadratic polynomial with two roots $\alpha$ and $\beta$ such that $\alpha=x$ and $\beta = \sqrt[4] {5-x^4}$
but I still couldn't proceed further. Somebody please share some hints.
| $2x^4-8x^3+24^2-32x+11=0$
Let $u = x -1 $
$11-32(u+1)+24(u+1)^2-8(u+1)^3+2(u+1)^4=0$
$2u^4+12u^2-3=0$
Let $v = u^2$
$2v^2+12v-3=0$
Can you continue from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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} |
differential equation $y''-2y'+y=\frac{2(1-x^2) e^x}{(1+x^2)^2}$ using variation of parameter $y''-2y'+y=\frac{2(1-x^2) e^x}{(1+x^2)^2}$
so i got the homogeneous part
$y=Ae^x+Be^x . x$
i tried to find the non homogeneous part by using variation of parameter:
$W = \begin{vmatrix}
e^x& x.e^x \\
e^x &e^x+x.e^x \\
\end{vmatrix}$ =$e^{2x}$
$W_1=u_1'= \begin{vmatrix}
0& x.e^x \\
\frac{2(1-x^2) e^x}{(1+x^2)^2}&e^x+x.e^x \\
\end{vmatrix}$ =$\frac{-2x (1-x^2)}{(1+x^2)^2}$
$W_2=u_2' = \begin{vmatrix}
e^x& 0 \\
e^x&\frac{2(1-x^2) e^2x}{(1+x^2)^2} \\
\end{vmatrix}$ =$\frac{.2(1-x^2).}{(1+x^2)^2}$
$\int_{}^{} u_1'$$=\ln(x^2+1)+\frac{2}{x^2+1}$
however is my approach correct? for $\int u_2'$ the integral quite hard and i doubt i made mistake or maybe there is an easier method than this? thanks so much!!
| I think the idea is correct, but
$$u'_2=\frac{gy_1}W=\frac{2(1-x^2)e^x\cdot e^x}{e^{2x}(1+x^2)^2}=2\frac{1-x^2}{(1+x^2)^2}$$
and the integral of the above doesn't look so daunting.
Also
$$u'_1=-\frac{gy_2}W=-\frac{2(1-x^2)e^x\cdot xe^x}{e^{2x}(1+x^2)^2}=-2\frac{x(1-x^2)}{(1+x^2)^2}$$
and again: not so terrible an integral.
| {
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"url": "https://math.stackexchange.com/questions/2628010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Dividing polynomials such that we get the desired remainder?
Find $a,b,c$ such that the remainder of
$$\frac{a(x^{11}-x^4+1)^{15}-x^2+bx-c}{x^3+x}$$ is equal to
$x^2-4x+1$.
I can solve it by writing $$a(x^{11}-x^4+1)^{15}-x^2+bx-c=(x^3+x)q(x)+x^2-4x+1$$
And solve the system of linear equations by looking at $x\in\{0,i,-i\}$ since those are the roots of $x^3+x$ making the term $(x^3+x)q(x)=0$ in the above expression. This leaves us with three simple linear equations where I got $(a,b,c)=(2,-6,1)$.
But what if the roots of the denominator weren't so easily obtainable because of a high degree, or the denominator had less unique roots than the unknown parameters? Then this method wouldn't work.
Are there other ways a problem like this could be solved relatively easily, if this was the case?
Or would that make things more complicated; what would be best way to solve something like that then?
| Not sure if its easier, but here is another way:
Note we have $x^{11}-x^4+1 \equiv 1 \pmod x$ and $x^{11}-x^4+1 \equiv -x \pmod {x^2+1} $
So we have $(x^{11}-x^4+1)^{15} \equiv 1 \pmod x $ and $(x^{11}-x^4+1)^{15} \equiv x \pmod {x^2+1} $
Also as $(x^2+1)^{-1} \equiv 1 \pmod x$ and $x^{-1} \equiv -x \pmod {x^2+1}$, we apply CRT:
Thus, $\pmod {x^3+x}$ we must have
$(x^{11}-x^4+1)^{15} \equiv (1)(x^2+1)(1) + (x)(x)(-x) = x^2+1-x^3 \equiv x^2+x+1$
$$\implies a(x^2+x+1) -x^2+bx-c \equiv x^2-4x+1$$
$$\iff (a-1)x^2+ (a+b)x+(a-c) \equiv x^2-4x+1$$
$\therefore a = 2, b = -6, c = 1$ is a solution.
| {
"language": "en",
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} |
If $\sin (y)=x\sin (a+y)$, prove that: If $\sin (y)=x\sin (a+y)$, prove that:
$$\dfrac {dy}{dx}=\dfrac {\sin^2 (a+y)}{\sin (a)}$$
My Attempt:
$$\sin (y)=x\sin (a+y)$$
$$\dfrac {d}{dy} \sin (y)=\dfrac {d}{dy} (x\sin (a+y))$$
$$\cos (y)=x\cdot \cos (a+y)+\sin (a+y)\cdot\dfrac {dx}{dy}$$
$$\cos (y)-x\cdot\cos (a+y)=\sin (a+y)\cdot\dfrac {dx}{dy}$$
| By implicit differentiation, you should get
\begin{align}
\cos(y) y' = \sin(a+y) + x\cos(a+y)y' \ \ \implies \ \ y' = \frac{\sin(a+y)}{\cos(y)-x\cos(a+y)}.
\end{align}
However, since
\begin{align}
x= \frac{\sin(y)}{\sin(a+y)} \ \ \implies \ \ y' = \frac{\sin^2(a+y)}{\sin(a+y)\cos(y) - \sin(y)\cos(a+y)} = \frac{\sin^2(a+y)}{\sin(a)}
\end{align}
where the last equality uses the difference angle formula.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $A_{n+1}=3A_n+2^n$ Let's say I want to find a formula for the following expression given $n$ number of threes
$$\ldots(3(3(3(3(3(3+1)+2)+4)+8)+16)+\ldots$$
If $A_0=1$, then
$$A_{n+1}=3A_n+2^n$$
Plugging in values to see the pattern,
$$A_2 = 3+1$$
$$A_3 = 3^2+3+2^1$$
$$A_4 = 3^3+3^2+3\cdot2+2^2$$
But I don't know how to condense something like this into an explicit formula.
| One way to see the correct answer is to use that:
$$x^n-y^n=(x-y)\left(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}\right)$$
Putting in $x=3,y=2$ you get that:
$$3^n-2^n = 3^{n-1}+3^{n-2}\cdot2+\cdots+3\cdot 2^{n-2}+2^{n-1}$$
Now add $3^n$ to both sides, and you get:
$$2\cdot 3^n -2^n = 3^{n}+3^{n-1}+3^{n-2}\cdot2+\cdots+3\cdot 2^{n-2}+2^{n-1}$$
There are more advanced techniques to solve this sort of equation generally, but this is a good "eyeball" solution without appeal to generating functions.
The generating function approach is to write:
$$f(z)=\sum_{n=0}^{\infty} A_nz^n = A_0 + z\sum_{n=1}^{\infty} (3A_{n-1}+2^{n-1})z^{n-1} = 1+z\left(3f(z)+\frac{1}{1-2z}\right)$$ Solving for $f(z)$ gives us $$f(z)=\frac{1}{1-3z}\left(1+\frac{z}{1-2z}\right)=\frac{1-z}{(1-2z)(1-3z)}$$
You can then use partial fractions to get that:
$$f(z)=\frac{2}{1-3z}-\frac{1}{1-2z}$$
Thus giving $A_n=2\cdot 3^n-2^n.$
| {
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Prove that the series $\sum_{n=1}^{\infty}\frac{1}{(n+x)(n+x+1)(n+x+2)}$ has a sum of $\frac{1}{2(x+1)(x+2)}$ I am trying to prove that the following series:
$$
\sum_{n=1}^{\infty}\frac{1}{(n+x)(n+x+1)(n+x+2)}
$$
has a sum of:
$$
\frac{1}{2(x+1)(x+2)}
$$
What I've tried:
Using the criteria that $S_n = b_1 - l$:
$$
\frac{1}{(n+x)} - \frac{1}{(n+x+1)(n+x+2)} = \frac{(n+x+1)(n+x+2)-(n+x)}{(n+x)(n+x+1)(n+x+2)} \\
\implies \frac{1}{(n+x)(n+x+1)(n+x+2)} = \frac{1}{(n+x+1)(n+x+2)-(n+x)}(\frac{1}{(n+x)} - \frac{1}{(n+x+1)(n+x+2)}) = \frac{1}{(n+x)(n+x+1)(n+x+2) - (n+x)^2} - \frac{1}{(n+x+1)^2(n+x+2)^2-(n+x)(n+x+1)(n+x+2)} = b_n - b_{n+1}
$$
Now that I have:
$$
b_n = \frac{1}{(n+x)(n+x+1)(n+x+2) - (n+x)^2} \\
l = \lim_{n \to \infty}{b_n} = \lim_{n \to \infty}{\frac{1}{(n+x)(n+x+1)(n+x+2) - (n+x)^2}} = \frac{1}{\infty} = 0 \\
b_1 = \frac{1}{(1+x)(1+x+1)(1+x+2) - (1+x)^2} = \frac{1}{(1+x)(2+x)(3+x) -
(1+x)^2}
$$
Therefore:
$$
S_n = b_1 - l \implies \\
S_n = \frac{1}{(1+x)(2+x)(3+x) -
(1+x)^2} - 0 \\
S_n = \frac{1}{(1+x)(2+x)(3+x) -
(1+x)^2} \\
= \frac{1}{(1+x)}.\frac{1}{(2+x)(3+x) -
(1+x)} \\
= \frac{1}{(1+x)}.\frac{1}{(6+3x+2x+x^2-1-x) -
(1+x)} \\
= \frac{1}{(1+x)}.\frac{1}{(x^2+4x+5)}
$$
Which is not the expected result. Am I mistaken somewhere or is this completely wrong?
| I would suggest that you try to make what i mention below instead. I think it's more simple ( and it's a general method to find the sum of this type of series, later in this post i give you the general form of it)
The Trick:
To find the terms of telescoping sum we can add in the numerator the following difference: the greatest term (in your case $(n+x+2)$ ) minus the lowest term $(n+x)$, and to not alter the fraction divide by that difference (in your case $2$).
To find that sum.
Take $f(n)=\frac{1}{2(n+x)(n+x+1)}$. Apply the trick
$$ \frac{1}{(n+x) (n+x+1)(n+x+2)}= \frac{(n+x+2 ) -(n+x)} {2(n+x)(n+x+1)(n+x+2)} $$
$$=\frac{1}{2(n+x)(n+x+1)}-\frac{1}{2(n+x+1)(n+x+2)}=f(n)-f(n+1)$$
$$=- \left(f(n+1)-f(n) \right) .$$
You can see that $\lim\limits_{n \to \infty} f(n)=0,$
and $f(1)=\frac{1}{2(x+1)(x+2)}$
Then, use the telescopic sum
$$\sum_{n=1}^{\infty}\frac{1}{(n+x) (n+x+1)(n+x+2)}=- \lim_{n \to \infty} [f(n+1)-f(1)]= \frac{1}{2(x+1)(x+2)}.$$
Extra:
Let's see the general case
In general we calculate the following sum (then take $n \to \infty$
$$\sum^{n}_{k=0}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }. $$
The trick (again)
Every time we have this kind of summation we can use the telescoping sum. To find the terms of telescoping sum in a easy way we can sum add in the numerator the following term: sum $(ak+b+sa)$ the greatest term", subtract thelowest term" $(ak+b)$, and to not alter the fraction divide by $\frac{1}{sa}$ the resultant difference of those terms.
$$\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }=\frac{1}{sa}\frac{(ak+b+sa) -(ak+b)}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }$$
$$=\frac{1}{sa}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+(s-1)a) } -\frac{1}{sa}\frac{1}{(ak+b+a)\ldots (ak+b+sa) }. $$
Take $f(k)=\frac{1}{sa}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+(s-1)a) }.$
Then the summation is
$$\frac{-1}{sa}\sum^{n}_{k=0} \left(f(k+1) -f(k)\right) $$
by the telescopic summation it's
$$=\frac{-1}{sa}\left(f(n+1) -f(0)\right)=\frac{-1}{sa}\left(\frac{1}{(an+b)\ldots (an+b+(s-1)a) } -\frac{1}{(b)(b+a)\ldots (+b+(s-1)a) }\right). $$
Take $n \to \infty$, then
$$\sum^{\infty}_{k=0}\frac{1}{(ak+b)(ak+b+a)\ldots (ak+b+sa) }=\frac{1}{sa(b)(b+a)\ldots (+b+(s-1)a) }.$$
Example:
To get your case we make $a=1$, $b=x+1$ and $s=2$
$$\sum^{\infty}_{k=0}\frac{1}{(k+x+1)(k+x+2) (k+x+3) }=\frac{1}{2(x+1)(x+2) }.$$
| {
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Finite difference approximation of $u''(x)+u(x)=0, u(0)=1, u(\pi)=-1$ Finite difference approximation of $u''(x)+u(x)=0, u(0)=1, u(\pi)=-1$.
For the approximation for this, do I get:
$$\frac{1}{h^2}(U_{j-1}-2U_j+U_{j+1})+U_j$$
$$=\frac{1}{h^2}(U_{j-1}+(h^2-2)U_j+U_{j+1})$$
Then the matrix A would have $h^2-2$ on the main diagonal with ones on the diagonal above and below?
What I am confused about, is that I am trying to show that the eigenvalues go to 0 as h goes to 0 of the ODE, but this doesn't hold for this matrix, so I feel like I made a mistake in the derivation of the matrix.
Thanks.
| If you let $h^2-2-\lambda=x$, then the eigenvalue equation becomes $P_n(x)=0$, where $P_n(x)$ is the determinant of the $n\times n$ matrix with $x$ on the main diagonal and $1$ on the adjacent diagonals. For example,
$$P_4(x)=\det\begin{bmatrix}x&1&0&0\\
1&x&1&0\\
0&1&x&1\\
0&0&1&x\end{bmatrix}$$
If we expand the determinant along the first column we get the difference equation $P_n(x)=xP_{n-1}(x)-P_{n-2}(x)$ with initial conditions $P_1(x)=x$ and $P_2(x)=x^2-1$. Let $P_n(x)=Cr^n$. Then $r^2-xr+1=0$ or
$$r=\frac{x\pm\sqrt{x^2-4}}2$$
So
$$P_n(x)=C_1\left(\frac{x+\sqrt{x^2-4}}2\right)^n+C_2\left(\frac{x-\sqrt{x^2-4}}2\right)^n$$
Applying initial conditions,
$$P_1(x)=x=C_1\left(\frac{x+\sqrt{x^2-4}}2\right)+C_2\left(\frac{x-\sqrt{x^2-4}}2\right)$$
$$P_2(x)=x^2-1=C_1\left(\frac{x^2-2+x\sqrt{x^2-4}}2\right)+C_2\left(\frac{x^2-2-x\sqrt{x^2-4}}2\right)$$
$$xP_1(x)-P_2(x)=1=C_1+C_2$$
$$\frac{\sqrt{x^2-4}}2\left(C_1-C_2\right)+\frac x2\left(C_1+C_2\right)=\frac{\sqrt{x^2-4}}2\left(C_1-C_2\right)+\frac x2=P_1(x)=x$$
$$C_1-C_2=\frac x{\sqrt{x^2-4}}$$
Adding and subtracting we arrive at
$$C_1=\frac{x+\sqrt{x^2-4}}{2\sqrt{x^2-4}}\text{; }C_2=-\frac{x-\sqrt{x^2-4}}{2\sqrt{x^2-4}}$$
So
$$P_n(x)=\frac1{\sqrt{x^2-4}}\left[\left(\frac{x+\sqrt{x^2-4}}2\right)^{n+1}-\left(\frac{x-\sqrt{x^2-4}}2\right)^{n+1}\right]$$
All the roots are within $-2<x<2$, so
$$\left|\frac{x+\sqrt{x^2-4}}2\right|=\left|\frac{x+i\sqrt{4-x^2}}2\right|=\sqrt{\frac{x^2+4-x^2}4}=1$$
So if $P_n(x)=0$ then
$$\left(\frac{x+\sqrt{x^2-4}}2\right)^{2n+2}=1=e^{2\pi ik}$$
$$\frac{x+\sqrt{x^2-4}}2=\exp\left(\frac{\pi ik}{n+1}\right)=\cos\left(\frac{\pi k}{n+1}\right)+i\sin\left(\frac{\pi k}{n+1}\right)$$
Taking real parts,
$$x=2\cos\left(\frac{\pi k}{n+1}\right)=\frac{\pi^2}{(n+1)^2}-2-\lambda$$
So
$$\lambda=\frac{\pi^2}{(n+1)^2}-2\left(1+\cos\left(\frac{\pi k}{n+1}\right)\right)=\frac{\pi^2}{(n+1)^2}-4\cos^2\left(\frac{\pi k}{2(n+1)}\right)$$
The smallest eigenvalue is when $k=n$ and is about $\frac{\pi^4}{12(n+1)^4}$.
But what do you need the eigenvalues for? The difference equation for $U_j$ is easy enough to solve:
$$U_j=A\cos\left(j\theta-\delta\right)$$
Where
$$\theta=\text{atan2}\left(h\sqrt{4-h^2},2-h^2\right)$$
and $h=\frac{\pi}{n+1}$, $\delta=(n+1)\theta/2-\pi/2$, and $A=\sec\delta$
| {
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Proof of the generating function of $1 -2x +3x^2 -4x^3+5x^4-6x^5+\cdots$ The ordinary generating function for the sequence $\{a_n\}_{n\geq0}$ where $a_n = (-1)^n\,n$ is
$$1 -2x +3x^2 -4x^3+5x^4-6x^5+\cdots = \frac{1}{(x+1)^2}$$
I can see from the geometric formula, and even from long division that
$$1 -x +x^2 -x^3+x^4-x^5+\cdots = \frac{1}{1+x}$$
but I'm not seeing the coefficients to explain $\frac{1}{(x+1)^2}.$
| $$S_n = 1 -2x +3x^2 -4x^3+5x^4-6x^5+\cdots$$ $$\implies
xS_n = x - 2x^2 + 3x^3 -\cdots$$
$$(1+x)S_n=1-x+x^2-x^3\cdots=\frac{1}{1+x}$$
| {
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What is $\int \frac{ \sqrt{(a^2 - x^2)}}{x^2}$? I tried the following trig substitution:
$x = a\sin \theta$
$$ \int \frac{ \sqrt{(a^2 - x^2)}}{x^2} = \int \frac{\sqrt{a^2 - a^2 \sin^2 \theta}}{a^2 \sin^2 \theta} = \int \frac{a \cos \theta}{a^2 \sin^2}$$
Setting $u = \sin x$ yields:
$$\frac{1}{a} * \int u^{-2} = - \frac{1}{a \sin \theta}$$
However my textbook states that the answer should really be:
$$ -\frac{\sqrt{a^2 - x^2}+x\tan^{-1}(\frac{x}{\sqrt{a^2-x^2}})}{x}$$
Where did I mess up?
| You forgot to substitute dx. dx = acosθdθ
| {
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How to prove $\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}$ Prove
$$
\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}
$$
and justify why $\frac{4+\sqrt{7}}{3}$ is ignored.
My Attempt:
$$
\tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}\implies\tan x-\tan x\tan^2\frac{x}{2}=2\tan\frac{x}{2}\\
\implies \tan^2\frac{x}{2}(\tan x)+2\tan\frac{x}{2}-\tan x=0\\
\implies \tan\frac{x}{2}=\frac{-2\pm2\sec x}{2\tan x}=\frac{-1\pm\sec x}{\tan x}
$$
Using this,
$$
\tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-1\pm\frac{4}{\sqrt{7}}}{\frac{3}{\sqrt{7}}}=\frac{-\sqrt{7}\pm4}{3}
$$
As $0\leq\frac{\sin^{-1}3/4}{2}\leq\frac{\pi}{4}\implies0\leq\tan(\frac{\sin^{-1}3/4}{2})\leq1$
$$
\implies \tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-\sqrt{7}+4}{3}
$$
Is my attempt correct and where is the value $\frac{4+\sqrt{7}}{3}$ to be excluded ?
| The method you used to eliminate $\frac{-4-\sqrt7}{3}$ is correct. Here I want to explain why it comes out to be a root of the equation so that things become more clear.
Consider the following
$$\tan(n\pi +2\theta)=\tan2\theta$$
$$\frac{\tan(\frac{n\pi}{2}+\theta)}{1-\tan^2(\frac{n\pi}{2}+\theta)}=\tan2\theta$$
Now it follows that not only $\tan\theta$ but also $\tan(\frac{n\pi}{2}+\theta)$ is a root of $$\tan2\theta=\frac{x}{1-x^2}$$ and to decide which of the two roots of the quadratic equation is of $\tan\theta$ is to be determined by the value of $2\theta$ and not by merely knowing the value of $\tan2\theta$
| {
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Given $\triangle ABC$ with $\angle C = 60^\circ$, show $a^2+b^2-ab=c^2$ without trigonometry I have the following problem:
In $\triangle ABC$, the angle at vertex $C$ is $60^{\circ}$. Prove that $a^2+b^2-ab=c^2$.
Of course, it is easy if you use cosine rule. I believe there exists a beautiful proof as well without using trigonometry. Anyone?
| Here's a proof using Ptolemy's Theorem:
$$\begin{align}
|\overline{AB}| |\overline{A^\prime B^\prime}| &= |\overline{AA^\prime}||\overline{BB^\prime}|+|\overline{AB^\prime}||\overline{A^\prime B}| \tag{1} \\[4pt]
c^2 &= a b + ( a - b )^2 \tag{2} \\[4pt]
&= a^2 + b^2 - a b \tag{3}
\end{align}$$
Incidentally, the Wikipedia entry mentions that one can use Ptolemy to prove the Law of Cosines generally. However, the description there isn't as clear as it might be. All one needs to do is observe that, in general,
$$|\overline{AA^\prime}| = 2b\sin\frac{C}{2} \qquad |\overline{BB^\prime}| = 2a\sin\frac{C}{2} \tag{4}$$
so that the counterpart of $(2)$ is
$$c^2 = 4ab\sin^2\frac{C}{2} + ( a - b )^2 \tag{5}$$
and we have
$$c^2 = a^2 + b^2 - 2 a b \left( 1 - 2\sin^2\frac{C}{2} \right) = a^2 + b^2 - 2 a b \cos C \tag{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632935",
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"source": "stackexchange",
"question_score": "12",
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Differential equation $y''=y^2$ I must solve the differential equation $y''=y^2$. Clearly, the function $y=0$ is a solution. So, assume that $y$ is not identically zero. Unless I'm not mistaken, there is a trick to solve equations of this kind. If I multiply for $2y'$, the left-hand member is $2y'y''$, that is the derivative of $(y')^2$, while the right-hand member is $2y^2 y'$. At this point, how can I continue this exercise?
| We must solve $y'' = y^2$. Assuming the independent variable is $x$, let us write this as:
$\frac{d^2y}{dx^2} = y^2$. Then, using the chain rule, we can write the left hand side as:
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{dy} \left(\frac{dy}{dx}\right) \frac{dy}{dx}$.
But, this is equivalent to:
$\frac{d}{dy} \left[ \left(\frac{dx}{dy}\right)^{-1}\right] \left( \frac{dx}{dy}\right)^{-1}$.
Computing this derivative, we get:
$-\left(\frac{dx}{dy}\right)^{-2} \frac{d^2 x}{dy^2} \left( \frac{dx}{dy}\right)^{-1} = -\left(\frac{dx}{dy}\right)^{-3} \frac{d^2x}{dy^2}$.
But, this is simply:
$\frac{d}{dy} \left[ \frac{1}{2} \left( \frac{dx}{dy}\right)^{-2}\right]$.
Now, we go back to your original ODE:
$\frac{d^2 y}{dx^2} = y^2$
we can write as:
$\frac{d}{dy} \left[ \frac{1}{2} \left( \frac{dx}{dy}\right)^{-2}\right] = y^2$
Integrating both sides wrto $y$, we get:
$\frac{1}{2} \left( \frac{dx}{dy}\right)^{-2} = 2 \int y^2 dy + A$, where $A$ is just a constant.
Solving for $dx/dy$, we get that:
$\boxed{x + B = \pm \int \frac{dy}{\sqrt{2 \int y^2 dy + A}}}$.
The right side is a very messy integral, you get some elliptic functions, in fact, Mathematica gives that the right-hand-side is:
$\frac{\sqrt[6]{-1} 2^{2/3} \sqrt[12]{3} \sqrt[3]{A} \sqrt{(-1)^{5/6} \left(\frac{\sqrt[3]{-\frac{2}{3}} y}{\sqrt[3]{A}}-1\right)} \sqrt{\frac{\left(-\frac{2}{3}\right)^{2/3} y^2}{A^{2/3}}+\frac{\sqrt[3]{-\frac{2}{3}} y}{\sqrt[3]{A}}+1} F\left(\sin ^{-1}\left(\frac{\sqrt{-\frac{i y \sqrt[3]{-\frac{2}{3}}}{\sqrt[3]{A}}-(-1)^{5/6}}}{\sqrt[4]{3}}\right)|\sqrt[3]{-1}\right)}{\sqrt{A+\frac{2 y^3}{3}}}$,
where $F$ denotes the elliptic integral of the first kind.
In general, the boxed equation is the general implicit solution to this problem. These types of ODEs occur all the time in classical mechanics, and most often are solved numerically or using dynamical systems techniques.
Hope this is helpful!
| {
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Finding the Sum of $a,b,c,d$ that satisfy the following conditions Let $a,b,c,d$ represent $4$ different non-zero integers such that the absolute value of each integer is less than $11$. If $c$ and $d$ are the solutions for $x$ of $x^2+ax+b=0$ and if $a$ and $b$ are the solutions for $x$ of $2x^2-cx-20d=0$, find the value of $a+b+c+d$.
I know the answer to this problem should be $6$.
| From Vieta's formulas we know that:
$$
\left\{ \begin{array}{rrrrrr}
c+d &= &-\dfrac{a}{1}\\
cd &= &\dfrac{b}{1}\\
a+b &= &-\dfrac{-c}{2}\\
ab &= &\dfrac{-20d}{2}\\
\end{array} \right.
$$
Then
$$
\left\{ \begin{array}{rrrrrr}
a &= &-c -d\\
b &= &cd\\
-c-d+cd &= &\dfrac{c}{2}\\
-(c+d)cd &= &-10d\\
\end{array} \right.
$$
Next
$$
\left\{ \begin{array}{rrrrrr}
d &= &\dfrac{3c}{2(c-1)}\\
c^2 + cd &= &10\\
\end{array} \right.
$$
Therefore
$$ c^2 +c \dfrac{3c}{2(c-1)} =10 $$
Thus
$$ 2c^3 + c^2 -20c + 20 = 0 $$
After solving this equation we will get $c=2$ (and two other solutions which are not integers).
So $a = -5, b = 6, d = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int^{\pi}_{0}\frac{1}{\sqrt{a-\cos x}}dx$
Finding $$\int^{\pi}_{0}\frac{1}{\sqrt{a-\cos x}}dx$$
Try: using $$\cos x =\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$$
$$\int^{\pi}_{0}\frac{\sec \frac{x}{2}}{\sqrt{(a-1)+(a+1)\tan^2\frac{x}{2}}}dx$$
Could some help me to solve it , thanks
| Starting from
$$I(a)=\int^{\pi}_{0}\frac{\sec \frac{x}{2}}{\sqrt{(a-1)+(a+1)\tan^2\frac{x}{2}}}\mathrm{d}x$$
Let us consider the map $x\mapsto 2x$ and factor out the $(a-1)$ term, so that
$$I(a)=\frac{2}{\sqrt{a-1}}\int_{0}^{\pi/2}\frac{\sec x}{\sqrt{1+\dfrac{a+1}{a-1}\tan^2 x}}\mathrm{d}x \\
=\frac{2}{\sqrt{a-1}}\int_{0}^{\pi/2}\frac{1}{\sqrt{\cos^2x+\dfrac{a+1}{a-1}\sin^2 x}}\mathrm{d}x \\
=\frac{2}{\sqrt{a-1}}\int_{0}^{\pi/2}\frac{1}{\sqrt{1-\sin^2x+\dfrac{a+1}{a-1}\sin^2 x}}\mathrm{d}x \\
=\frac{2}{\sqrt{a-1}}\int_{0}^{\pi/2}\frac{1}{\sqrt{1-\dfrac{2}{1-a}\sin^2 x}}\mathrm{d}x $$
By defining $m=\dfrac{2}{1-a}$, the last integral reads
$$\int_{0}^{\pi/2}\frac{1}{\sqrt{1-m\sin^2 x}}\mathrm{d}x = K(m)$$
where $K(m)$ is the complete elliptic integral of the first kind.
Finally
$$ I(a) = \frac{2}{\sqrt{a-1}}K\left(\frac{2}{1-a}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $x+y=1$ then $x^{n+1}\sum\limits_{k=0}^n \binom{n+k}{k}y^k+y^{n+1}\sum\limits_{k=0}^n \binom{n+k}{k}x^k=1$ $$x^{n+1}\sum_{k=0}^n \binom{n+k}{k}y^k+y^{n+1}\sum_{k=0}^n \binom{n+k}{k}x^k=1$$
with one property such that x+y=1
I tried using the binomial equation to replace $y^k$ with $(1-x)^k$ but that will not cancel out anything! so I tried to use
$$\sum_{k=0}^n \binom{n}{k} x^k=(1+x)^n$$
which when $x=1$ gives the following
$$\sum_{k=0}^n \binom nk=2^n\\
\sum_{k=1}^n\binom nk=2^n-1$$
so I hope to combine the two such that $x+y=1$ and I can use the above equation? thanks
| I just copy my answer of this question.
$$\forall m,n\in\Bbb N,\ \forall x\in\Bbb R,\ \sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}$$
Let $m,n\in\Bbb N$, $f:x\mapsto\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}$, we have $f(0)=0$ and for all $x\in\Bbb R$,
\begin{align}
f'(x)&=(n+1)x^{n}\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^k-x^{n+1}\sum_{k=0}^{m}\binom{n+k}{k}k(1-x)^{k-1}\\
&=x^{n}(1-x)^{m}\cdot A
\end{align}
where
\begin{align}
A&=\sum_{k=0}^{m}\binom{n+k}{k}(n+1)(1-x)^{k-m}-\sum_{k=0}^{m}\binom{n+k}{k}kx(1-x)^{k-m-1}\\
&=\sum_{k=0}^{m}\binom{n+k}{k}(n+k+1)(1-x)^{k-m}-\sum_{k=0}^{m}\binom{n+k}{k}k(1-x)^{k-m-1}\tag1\\
&=\sum_{k=1}^{m+1}\binom{n+k}{k}k(1-x)^{k-m-1}-\sum_{k=1}^{m}\binom{n+k}{k}k(1-x)^{k-m-1}\tag2\\
&=\binom{n+m+1}{m+1}(m+1)\\
&=\frac{(n+m+1)!}{n!\ m!}
\end{align}
$(1):-x(1-x)^{k-m-1}=(1-\frac1{1-x})(1-x)^{k-m}=(1-x)^{k-m}-(1-x)^{k-m-1}$
$(2):\binom{n+k}{k}(n+k+1)=\frac{(n+k+1)!}{n!k!}=\binom{n+k+1}{k+1}(k+1)$
Thus,
\begin{align}
\forall x\in \Bbb R,\ f(x)&=A\int_0^xt^n(1-t)^mdt\\
&=A\int_{1-x}^1(1-s)^ns^mds\tag{$s=1-t$}\\
&=A\int_0^1(1-s)^ns^mds-A\int_0^{1-x}(1-s)^ns^mds\\
&=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}\tag3
\end{align}
$(3):$ integration by parts proves that $\int_0^1(1-s)^ns^mds=\frac{n!\ m!}{(n+m+1)!}=A^{-1}$, and $m\leftrightarrow n$ and $x\leftrightarrow 1-x$ in the expressions of $f$ give the second term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Differentiating w.r.t x Please help me understand this question.
\begin{gathered}
{(\sqrt x - \frac{1}{{\sqrt x }})^2} = y \hfill \\
{\text{Differentiating w}}{\text{.r}}{\text{.t x}} \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = \frac{d}{{dx}}{\left[ {(\sqrt x ) - \frac{1}{{\sqrt x }}} \right]^2} \hfill \\
\hfill \\
= 2\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)\frac{d}{{dx}}{\left[ {(\sqrt x ) - \frac{1}{{\sqrt x }}} \right]^{}}{\text{From Where 2 Come?}} \hfill \\
\hfill \\
\frac{{dy}}{{dx}} = 2\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right){\left[ {\frac{d}{{dx}}({x^{\frac{1}{2}}}) - \frac{d}{{dx}}({x^{ - \frac{1}{2}}})} \right]^{}} \hfill \\
\hfill \\
= 2\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right){\left[ {\frac{1}{2}({x^{\frac{1}{2} - 1}}) - \left( { - \frac{1}{2}} \right)({x^{ - \frac{1}{2} - 1}})} \right]^{}} \hfill \\
\hfill \\
= 2\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right){\left[ {\frac{1}{2}{x^{ - \frac{1}{2}}}) + \frac{1}{2}{x^{ - \frac{3}{2}}}} \right]^{}} \hfill \\
\hfill \\
= 2\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right){\text{x}}\frac{1}{2}{\left[ {\frac{1}{{\sqrt x }} + {{\frac{1}{{x\sqrt x }}}^{}}} \right]^{}} \hfill \\
\hfill \\
= \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)\left[ {\frac{{x + 1}}{{x\sqrt x }}} \right]{\text{How this equation come?}} \hfill \\
\hfill \\
= \left[ {\frac{{x - 1}}{{\sqrt x }}} \right]\left[ {\frac{{x + 1}}{{x\sqrt x }}} \right]{\text{How this equation come?}} \hfill \\
= \frac{{{x^2} - 1}}{{{x^2}}}Answer \hfill \\
\end{gathered} ]
| it is by the power and Quotient rule
$$2\left(\frac{x-1}{\sqrt{x}}\right)\left(\frac{\sqrt{x}-(x-1)\cdot \frac{1}{2}x^{-1/2}}{x}\right)$$
it can be simplified ton $$1-\frac{1}{x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Arithmetic progression and polynomials Suppose the quadratic polynomial $$p(x)=ax^2 + bx + c$$ has positive coefficients $a, b, c$ such that these are in AP in the given order. If $m$ and $n$ are the integer zeros of the polynomial then what is the value of $m+n+mn$? I have tried quadratic formula but ain't getting the answer.
| Since $m$ and $n$ are roots of the polynomial we have
\begin{align*}
ax^2+bx+c&=a\left(x-m\right)\left(x-n\right)\\
&=ax^2-a(m+n)x+amn
\end{align*}
Then, $$m+n=-\frac ba\quad\text{and}\quad mn=\frac ca$$
So, since $a,b$ and $c$ are in AP we get
$$m+n+mn=-\frac ba+\frac ca=\frac{c-b}a=\frac{b+b-a-b}a=\frac{b-a}a=\frac ba-1$$
and $\frac ba =-(m+n)$, then
$$m+n+mn=-(m+n)-1\quad \implies \quad mn+2(m+n)+4=3$$
So $$(m+2)(n+2)=3$$
Now, since $3$ is a prime we have $(m,n)\in\left\{(-1,1),(1,-1),(-3,-5),(-5,-3)\right\}$. Notice that $(m,n)=(-1,1)$ or $(m,n)=(1,-1)$ implies $b=0$, but we know that $a,\,b$ and $c$ are positive. So, $(m,n)=(-3,-5)$ or $(-5,-3)$, and then
$$\boxed{m+n+mn=-8+15=7}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the real part of the following analytic function Given $v = \frac{-y}{x^2+y^2}$, find the real component of the function $u$.
I'm not sure why I'm not getting this. Here's my work so far:
$$v_x = \frac{2xy}{(x^2+y^2)^2} = -u_y \Rightarrow u(x,y) = \frac{x}{x^2+y^2} + \phi(x) $$
where $\phi(x)$ is (as of yet) undetermined. Continuing on, I get
$$ u_x = \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2} + \phi'(x) = \frac{2y^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} = v_y $$
simplifying
$$ \phi'(x) = \frac{2x^2+2y^2}{(x^2+y^2)^2} - \frac{2}{x^2+y^2}$$
multiplying the rightmost fraction by $x^2+y^2$ (top and bottom) we find that
$$\phi'(x) = 0 \Rightarrow \phi(x) = C \in \mathbb{R}$$
so
$$ u(x,y) = \frac{x}{x^2+y^2} + C$$
This satisfies first first of the Cauchy-Reimann conditions, but it does not seem to satisfy the second. Explicitly the condition that $u_x = v_y$ is
$$ \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2} = \frac{2y^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} $$
I don't believe that this is generally true… The only idea that I have is that perhaps $\phi(x) = C$ is a constant with respect to $x$, but not $y$ but I think that would contradict my first line of reasoning above.
Any ideas?
| Hint:
$$ \frac{1}{x^2+y^2} - \frac{2x^2}{(x^2+y^2)^2} = \frac{2y^2}{(x^2+y^2)^2} - \frac{1}{x^2+y^2} \iff$$
$$\iff \frac{2}{x^2+y^2} = \frac{2(x^2+y^2)}{(x^2+y^2)^2} $$
So your work seems correct !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\tan^{-1} \left(\frac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$, then prove $x^2=\sin(2\alpha)$ If $\tan^{-1} \left(\dfrac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$ then prove that: $x^2= \sin (2\alpha) $
My Attempt:
$$\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) =\alpha$$
$$\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}=\tan (\alpha )$$
$$\dfrac {1+x^2-2\sqrt {1+x^2}.\sqrt {1-x^2}+ 1 - x^2}{1+x^2-1+x^2}=\tan (\alpha)$$
$$\dfrac {1-\sqrt {1+x^2}.\sqrt {1-x^2}}{x^2}=\tan (\alpha)$$
| Continuing from where you stopped
We get
$$1-x^2\tan\alpha= \sqrt {1+x^2} .\sqrt {1-x^2}$$
Squaring both sides we get
$$1+x^4\tan^2\alpha-2x^2\tan\alpha=1-x^4$$
$$x^4(1+\tan^2\alpha)=2x^2\tan\alpha$$
$$x^2=\frac {2\tan\alpha}{1+\tan^2\alpha}$$
Hence $$x^2=\sin(2\alpha)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Prove that: $\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$
Prove that:
$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$
My work so far:
$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$
$$\frac{(n+0)!}{n!0!}+\frac{(n+1)!}{n!1!}+...+\frac{(n+n)!}{n!n!}=\frac{(2n+1)!}{n!(n+1)!}$$
$$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+...+\binom{2n}{n}=\binom{2n+1}{n}$$
How to prove the last equality?
| You can calculate directly:
$$\begin{align} & \frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}=\frac{n!(1+n+1)}{1!}; \\
& \frac{n!(n+2)}{1!}+\frac{(n+2)!}{2!}=\frac{n!(n+2)(2+n+1)}{2!}; \\
& \cdots \\
& \frac{n!(n+2)(n+3)\cdots ((n-1)+n+1)}{(n-1)!}+\frac{(n+n)!}{n!}= \\
& \frac{n!(n+2)(n+3)\cdots (n+n+1)}{n!}=\frac{(2n+1)!}{(n+1)!}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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How to show $\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}$ I am trying to show $\displaystyle{\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}}.$
Any help?
(I am having troubles using the half circle infinite contour)
Or more specifically, what is the residue $\text{res} \left(\frac{1}{z^3+1},z_0=e^\frac{\pi i}{3} \right )$
Thanks!
| Hint
Substitute $u=1/x \ $ then $ \ \text{d}u=(-1/x^2)\text{d}x=-u^2\text{d}x$
$$
I=\int_{0}^{+\infty}\frac{\text{d}x}{1+x^3}=-\int_{+\infty}^{0}\frac{1}{1+(1/u)^3}\left(\frac{\text{d}u}{u^2}\right)
$$
Then
$$
\int_{0}^{+\infty}\frac{\text{d}x}{1+x^3}=\int_{0}^{+\infty}\frac{u}{1+u^3}\text{d}u$$
What is beautiful here is that you can sum the two different expressions
$$
2I=\int_{0}^{+\infty}\frac{1+u}{1+u^3}\text{d}u=\int_{0}^{+\infty}\frac{\text{d}u}{1-u+u^2}
$$
Then you can write that
$$
1-u+u^2=\left(u-\frac{1}{2}\right)^2+\frac{3}{4}=\frac{3}{4}\left(\frac{4}{3}\left(u-\frac{1}{2}\right)^2+1\right)$$
Then you can write $\displaystyle \frac{4}{3}$ as $\displaystyle \left(\frac{2}{\sqrt{3}}\right)^2$
So far we have
$$
2I=\frac{4}{3}\int_{0}^{+\infty}\frac{\text{d}u}{\displaystyle 1+\left(\frac{2}{\sqrt{3}}\left(u-\frac{1}{2}\right)\right)^2}=\frac{2}{\sqrt{3}}\int_{0}^{+\infty}\frac{2/\sqrt{3}}{\displaystyle 1+\left(\frac{2}{\sqrt{3}}\left(u-\frac{1}{2}\right)\right)^2}\text{d}u
$$
There we have the arctangente and
$$
2I=\frac{2}{\sqrt{3}}\left[\text{arctan}\left(\frac{2}{\sqrt{3}}\left(u-\frac{1}{2}\right)\right)\right]^{+\infty}_{0}=\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}-\text{arctan}\left(-\frac{1}{\sqrt{3}}\right)\right)
$$
Then with $\displaystyle \text{arctan}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}$ we finally have
$$
2I=\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}+\frac{\pi}{6}\right)=\frac{2}{\sqrt{3}}\frac{4 \pi}{6}$$
Here we have the claimed
$$
I=\frac{2 \pi}{3\sqrt{3}}
$$
Beautiful equality
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Real part of complex root of $X^5-X-k^5+k+1$ is $\frac{1}{23k}$ away from any integer ($k\geq 2$) Let $k\geq 2$ be an integer. The polynomial $P_k=X^5-X-k^5+k+1$ is easily
seen to have exactly one real root and two pair of conjugate non-real roots. Is it true
that if $\alpha_k + i \beta_k$ any non-real root of $P_k$, then $\lbrace \lbrace \alpha_k \rbrace \rbrace \geq \frac{1}{23k}$, where $\lbrace \lbrace \rbrace \rbrace$ denotes distance to the nearest integer ? I have checked this for $2 \leq k \leq 10^4$.
My thoughts : $\alpha_k$'s minimal polynomial can be shown to be of degree $10$, so Liouville's approximation theorem would suggest that $\lbrace \lbrace \alpha_k \rbrace \rbrace \geq \frac{C}{k^{10}}$ which is much weaker.
Motivation : I noticed that the real root of $P_k$ is very close to its nearest integer while the real parts of the other roots are comparatively much farther. If quantified successfully, this would yield a proof of the irreducibility of $P_k$ over the rationals.
| Not a full solution, but for large $k$ this should explain it.
Let $X = r e^{i \phi}$. Then
$$
P_k=r^5 e^{5 i \phi}-r e^{i \phi}- k^5+k+1
$$
Let $a = r^{-4}$.Then
$$
P_k=r^5 (e^{5 i \phi}-a e^{i \phi})- k^5+k+1
$$
For the roots, the imaginary part of this has to vanish, so we have
$$
0 = \sin(5 \phi)-a \sin(\phi)
$$
Using the trig. identity $\sin(5 \phi) = \sin(\phi) (16\cos^4(\phi) - 12\cos^2(\phi) + 1)$ this gives
$$
0 = 16\cos^4(\phi) - 12\cos^2(\phi) + 1 -a
$$
Again, let $y = \cos^2(\phi)$, reducing this to
$$
0 = 16 y^2 - 12 y + 1 -a
$$
This gives
$$
y = \cos^2(\phi) = \frac{3\pm \sqrt{4a + 5}}{8}
$$
Turning now to the real part of the root, we have
$$
0 =r^5 (\cos(5 \phi)-a \cos(\phi))- k^5+k+1
$$
With another trig. identity,
$$\cos(5 \phi) = \cos(\phi) (16\cos^4(\phi) - 20\cos^2(\phi) + 5)
$$
we have, using results from above,
$$
0 =r^5 \cos(\phi) (16\cos^4(\phi) - 20\cos^2(\phi) + 5 -a)- k^5+k+1 \\
= r^5 \cos(\phi) (16y^2 - 20y + 5 -a)- k^5+k+1 \\
=r^5 \cos(\phi) ( - 8y + 4 )- k^5+k+1 \\
=r^5 \cos(\phi) ( 1\pm \sqrt{4a + 5})- k^5+k+1
$$
But also
$$
0 =r^5 \cos(\phi) ( - 8y + 4 )- k^5+k+1 \\
= 4 r^5 \cos(\phi) (1 - 2 \cos^2(\phi))- k^5+k+1 \\
= - 4 r^5 \cos(\phi) \cos(2 \phi)- k^5+k+1 \\
$$
and from above
$$
- 4 \cos(2 \phi) = 1\pm \sqrt{5 + 4/r^4}
$$
For very large $k$, we have also very large $r$, which gives approximately
$$
- 4 \cos(2 \phi) = 1\pm \sqrt{5} \rightarrow \phi = \pm \frac{2 \pi}{5}
$$
So $- 4 \cos(\phi) \cos(2 \phi) = 1$ and
$$
0 = r^5 - k^5+k+1 \rightarrow r = k
$$
Hence
$$
a_k = r \cos(\phi) = \frac{k}{4} (\sqrt5 -1)
$$
is the real part of the root for very large $k$. Now, for large $k$, we have that $a_k$ will not be close enough ($< 1/(23 k)$) to any integer, say $n$, since this would approximately require $(4 n +k)^2 = 5 k^2 $.
The same can be done for the other solution pair.
It remains to be estimated how large $k$ has to become such that the derived bounds hold. For smaller $k$, then, validity can be shown manually.
| {
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"url": "https://math.stackexchange.com/questions/2646089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$? How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$?
| It's true even for all reals $x$, $y$ and $z$.
Indeed, we need to prove that
$$(x^2+y^2+z^2)^2\geq3xyz(x+y+z)$$ or
$$\sum_{cyc}(x^4+2x^2y^2-3x^2yz)\geq0$$ or
$$\sum_{cyc}(2x^4+4x^2y^2-6x^2yz)\geq0$$ or
$$\sum_{cyc}(x^4-2x^2y^2+y^4+3z^2x^2-6z^2xy+3z^2y^2)\geq0$$ or
$$\sum_{cyc}(x-y)^2((x+y)^2+3z^2)\geq0,$$
which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Compute the dihedral angle of a regular pyramid Given a regular pyramid, defined as a right pyramid with a base which is a regular polygon, with the vertex above the centroid of the base, I would like to compute the dihedral angle between adjacent faces, as well as the angle between a lateral face and the base, in terms of the number of sides and the slant edge length. Can someone walk through the calculation?
Using Google I found this helpful article about computing dihedral angles by Greg Egan.
We have isosceles triangle $CAD$ above the plane of triangle $CBD$, making a dihedral angle of $\epsilon$. Triangle $CBD$ is the orthogonal projection of $CAD$, so also isosceles. The apex angle between adjacent edges in triangle $CAD$ is $\alpha$. The angle of triangle $CBD$ is $\beta.$ Edge $AC$ makes a slant angle of $\gamma$ with the projected edge $BC$.
There is a plane perpendicular to slant edge $AC$ through the opposite vertex $D$, which intersects edge $AC$ at point $F$, and $BC$ at point $G$. Angle $GFD$ is $\delta.$
If triangle $CAD$ is the side of a regular pyramid, and $CBD$ its base, then $2\delta$ will be the dihedral angle between adjacent faces, and $\epsilon$ will be the dihedral angle between slant face and base.
I can see that plane $DFG$ is perpendicular to slant edge $AC$, hence why triangles $AFD$ and $AFG$ are right triangles. But why are $DGC$ and $FGD$ right angles?
How do I arrive at $$\sin\delta=\cos(\beta/2)/\cos(\alpha/2)$$?
| An alternate approach to obtaining the formula for $\delta$ is to take a unit normal $\underline{\mathbf{u}}$ for a given slope and then rotate this by the angle $\beta$ about the $z$-axis, ie about the vector $\underline{\mathbf{k}}$, to produce a unit normal $\underline{\mathbf{v}}$ for the adjacent slope, as shown in Fig. 1. This approach avoids the complication which arises when $\beta > 90^\circ$, for then $G$ is no longer on the line segment $BC$, but extends out to the left beyond $B$ (eg. in the case of the tetrahedron we have $\alpha = 60^\circ$ and $\beta = 120^\circ$, giving dihedral angle $2\delta = \arccos (1/3)$).
Since $\underline{\mathbf{u}}$ and $\underline{\mathbf{v}}$ are both outward pointing, we have :
$$
\underline{\mathbf{u}} \cdot \underline{\mathbf{v}} = \cos(180 - 2\delta) = -\cos 2\delta.
$$
From Fig. 1 we have :
$$
\underline{\mathbf{u}} = \sin \epsilon \; \underline{\mathbf{i}} + \cos \epsilon \; \underline{\mathbf{k}}.
$$
Rotating $\underline{\mathbf{u}}$ about the $z$-axis by angle $\beta$ leaves the $\underline{\mathbf{k}}$ component unchanged and changes the component $\sin \epsilon \; \underline{\mathbf{i}}$ to $\sin \epsilon \cos \beta \; \underline{\mathbf{i}} + \sin \epsilon \sin \beta \; \underline{\mathbf{j}}$ so that :
$$
\underline{\mathbf{v}} = \sin \epsilon \cos \beta \; \underline{\mathbf{i}} + \sin \epsilon \sin \beta \;\underline{\mathbf{j}} + \cos \epsilon \; \underline{\mathbf{k}}
$$
Thus :
\begin{eqnarray}
2 \sin^2 \frac{1}{2} \delta -1 = -\cos 2\delta & = & \underline{\mathbf{u}} \cdot \underline{\mathbf{v}} \nonumber \\
& = & \sin^2 \epsilon \cos \beta + \cos^2 \epsilon \label{eq:Rodrigues} \tag{1}
\end{eqnarray}
We require an expression that involves $\alpha$ and $\beta$ only. Denote the altitude of the regular pyramid by $h$, and the radius of its regular polygonal base by $r$. Then :
\begin{equation}
\sin \epsilon = \frac{h}{s \cos \frac{1}{2} \alpha}, \hspace{1em} \mbox{and} \hspace{1em} \cos \epsilon = \frac{r \cos \frac{1}{2} \beta}{s \cos \frac{1}{2} \alpha} \label{eq:angle-epsilon} \tag{2}
\end{equation}
with :
\begin{equation}
\frac{h}{s} = \sin \gamma \hspace{1em} \mbox{and} \hspace{1em} \frac{r}{s} = \cos \gamma. \label{eq:angle-gamma} \tag{3}
\end{equation}
As in the article we can readily establish by computing the length of EC in two ways that :
\begin{equation}
\cos \gamma = \frac{\sin \frac{1}{2} \alpha}{\sin \frac{1}{2} \beta}. \label{eq:angle-gamma-formula} \tag{4}
\end{equation}
Thus putting equations (\ref{eq:Rodrigues}) - (\ref{eq:angle-gamma-formula}) together we now have :
\begin{eqnarray*}
2 \sin^2 \frac{1}{2} \delta -1 & = & \frac{\sin^2 \gamma}{\cos^2 \frac{1}{2} \alpha} \cdot \cos \beta + \cos^2 \gamma \cdot \frac{\cos^2 \frac{1}{2} \beta}{\cos^2 \frac{1}{2} \alpha} \\
& = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \left( 1 - \frac{\sin^2 \frac{1}{2} \alpha}{\sin^2 \frac{1}{2} \beta} \right) \cdot (\cos^2 \frac{1}{2}\beta - \sin^2 \frac{1}{2}\beta) + \left( \frac{\sin^2 \frac{1}{2} \alpha}{\sin^2 \frac{1}{2} \beta} \right) \cdot \cos^2 \frac{1}{2} \beta \right] \\
& = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \cos^2 \frac{1}{2} \beta - \sin^2 \frac{1}{2} \beta + \sin^2 \frac{1}{2} \alpha \right] \\
\Rightarrow 2 \sin^2 \delta & = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \cos^2 \frac{1}{2} \beta - \sin^2 \frac{1}{2} \beta + 1 \right] \\
& = & \frac{2 \cos^2 \frac{1}{2} \beta}{\cos^2 \frac{1}{2} \alpha} \\
\Rightarrow \mbox{since } \alpha, \beta, \gamma \in (0, 180^\circ), \hspace{1.5em} \sin \delta & = & \frac{\cos \frac{1}{2} \beta}{\cos \frac{1}{2} \alpha}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $\arccos (x) +\arccos (2x)=\frac{3\pi}{4}$ $\arccos (x) +\arccos (2x)=\frac{3\pi}{4}$
$\implies \arccos (x)=\frac{3\pi}{4}-\arccos (2x)$
$\implies \cos(\arccos (x))=\cos(\frac{3\pi}{4}-\arccos (2x))$
$\implies x=\cos(\frac{3\pi}{4})\cos(\arccos (2x))+\sin(\frac{3\pi}{4})\sin(\arccos(2x))$
$\implies x=\cos(\frac{3\pi}{4})\cos(\arccos (2x))+\sin(\frac{3\pi}{4})\sin(\arcsin(\sqrt{1-4x^2}))$
$\implies x=-\frac{2x}{\sqrt{2}}+\frac{\sqrt{1-4x^2}}{\sqrt{2}}$
$\implies \sqrt{2}x=-2x+\sqrt{1-4x^2}$
$\implies x(\sqrt{2}+2)=\sqrt{1-4x^2}$
$\implies x^2(2+4\sqrt{2}+4)=1-4x^2$
$\implies x^2(4\sqrt{2}+10)=1$
$\implies x^2=\frac{1}{4\sqrt{2}+10}$
$x=\pm\sqrt{\frac{1}{4\sqrt{2}+10}}$
But $x=-\sqrt{\frac{1}{4\sqrt{2}+10}}$ does not satisfy the given equation.
$\therefore x= \sqrt{\frac{1}{4\sqrt{2}+10}}$
I am looking for other methods to solve this equation. I applied the formula to combine two $\arccos(x)$ functions into one $\arccos(x)$ function and on simplifying I obtained a complicated equation.
| use that $$\arccos(x)+\arccos(y)=\arccos(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$ for $$x+y\geq 0$$ and
$$2\pi -\arccos(xy-\sqrt{1-x^2}\sqrt{1-y^2})$$ for $$x+y<0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that the sequence is bounded and subsequences converge Sequence is defined in the following way:
$x_1 = \frac{2}{3}$, $x_{2n+1} = \frac{x_{2n}}{3} + \frac{2}{3} $, $x_{2n} = \frac{x_{2n-1}}{3}$
I need to show that the sequence is bounded, i.e $0<x_{n}<1$, and it does not converge.
This is what I have done so far:
I have separated odd and even elements as the following subsequences:
$x_{2n+1} = \frac{x_{2n-1}}{9} + \frac{2}{3} $
$x_{2n} = \frac{x_{2n-2}}{9} + \frac{2}{9} $
And I have found that $x_{2n+1} \to \frac{3}{4} $ and $x_{2n} \to \frac{1}{4} $
I know that $x_n$ is divergent, as the subsequential limits are not equal, but I have trouble showing that:
*
*$x_n$ is bounded from above. Do I need to look at subsequences separately and show that they are monotone or do I need to somehow estimate $x_n$?
*Showing that $x_{2n}$ and $x_{2n+1}$ are convergent
| We know that $$x_1=\dfrac{2}{3}\\x_{2n+1}=\dfrac{x_{2n}}{3}+\dfrac{2}{3}\qquad\qquad (I)\\x_{2n-1}=\dfrac{x_{2n-2}}{3}+\dfrac{2}{3}\qquad\qquad (II)\\x_{2n}=\dfrac{x_{2n-1}}{3}\qquad\qquad (III)$$by substituting $(II)$ in $(III)$ and $(III)$ in $(I)$ we respectively obtain$$x_{2n}=\dfrac{x_{2n-2}+2}{9}\\x_{2n+1}=\dfrac{x_{2n-1}+6}{9}$$which yields to $x_{2n}\to \dfrac{1}{4}$ and $x_{2n+1}\to \dfrac{3}{4}$ using $\epsilon-\delta$ approach (check it!) and $x_n$ is divergent.
For showing boundedness using induction, note that $0<x_1<1$. Also if $0<x_{2n}<1$ then $x_{2n+1}=0<\dfrac{2}{3}<\dfrac{x_{2n}+2} {3}<1$ and if $0<x_{2n+1}<1$ then $0<x_{2n+2}=0<\dfrac{x_{2n+1}}{3}<\dfrac{1}{3}<1$ which yields to $0<x_n<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Collinearity when $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$
Let $\mathbf{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\mathbf{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\mathbf{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if
$\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}.$
Since the cross product of two vectors gives an area, and for two vectors to give an area of $0$ they need to be on the same line (or they can be a point, but I'm assuming both are not $\mathbf 0$). However, in this problem, each of the cross products need not necessarily be $0$ since it's their sum that is $0$, and now I'm not sure what to do.
| If the vectors are collinear, they are of the form
$$
\mathbf{a}=\mathbf{u}+\alpha\mathbf{v},
\quad
\mathbf{b}=\mathbf{u}+\beta\mathbf{v},
\quad
\mathbf{c}=\mathbf{u}+\gamma\mathbf{v}
$$
Then
$$
\mathbf{a}\times\mathbf{b}=
\mathbf{u}\times\mathbf{u}+\alpha\mathbf{v}\times\mathbf{u}+
\beta\mathbf{u}\times\mathbf{v}+\alpha\gamma\mathbf{v}\times\mathbf{v}=
(\beta-\alpha)\mathbf{u}\times\mathbf{v}
$$
Similarly $\mathbf{b}\times\mathbf{c}=(\gamma-\beta)\mathbf{u}\times\mathbf{v}$ and $\mathbf{c}\times\mathbf{a}=(\alpha-\gamma)\mathbf{u}\times\mathbf{v}$, so easily $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$.
Conversely, suppose $\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} = \mathbf{0}$ and set $\mathbf{b}-\mathbf{a}=\mathbf{x}$. Without loss of generality, we can assume $\mathbf{a}\ne\mathbf{b}$ (or there would be just one or two vectors, which are obviously collinear). Then
\begin{align}
\mathbf{0}
&=\mathbf{a}\times(\mathbf{a}+\mathbf{x})
+(\mathbf{a}+\mathbf{x})\times\mathbf{c}
+\mathbf{c}\times\mathbf{a} \\[4px]
&=\mathbf{a}\times\mathbf{x}
+\mathbf{a}\times\mathbf{c}
+\mathbf{x}\times\mathbf{c}
+\mathbf{c}\times\mathbf{a} \\[4px]
&=\mathbf{x}\times(\mathbf{c}-\mathbf{a})
\end{align}
which implies $\mathbf{c}-\mathbf{a}=\delta\mathbf{x}$. Thus
$$
\mathbf{a}=\mathbf{a}+0\mathbf{x},
\quad
\mathbf{b}=\mathbf{a}+1\mathbf{x},
\quad
\mathbf{c}=\mathbf{a}+\delta\mathbf{x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Homogeneous fourth degree inequality : $\sum x_i^2x_j^2 +6x_1x_2x_3x_4 \geq\sum x_ix_jx_k^2$ Let $x_1,x_2,x_3,x_4$ be four real numbers. The inequality
$$
\sum_{i,j} x_i^2x_j^2 +6x_1x_2x_3x_4 \geq \sum_{i,j,k} x_ix_jx_k^2
$$
(where the sums are over unordered uples of distinct indices) is true because if one views the difference as a trinomial in $x_4$, this trinomial has negative discriminant $\frac{-3}{16}((x_2-x_1)(x_3-x_1)(x_3-x_2))^2$. The drawback of this method are (1) it is not very symmetrical, (2) it involves the somewhat painful computation of the discriminant, and (3) the equality case is not very easy to deduce from it.
Are there other methods, improving on (1) and/or (2) and/or (3) ?
| It's $$(x_1-x_2)^2(x_3-x_4)^2+(x_1-x_3)^2(x_2-x_4)^2+(x_1-x_4)^2(x_2-x_3)^2\geq0.$$
Because if $x_1=a$, $x_2=b$, $x_3=c$ and $x_4=d$ then:
$$(x_1-x_2)^2(x_3-x_4)^2+(x_1-x_3)^2(x_2-x_4)^2+(x_1-x_4)^2(x_2-x_3)^2=$$
$$=(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2=$$
$$=(a^2-2ab+b^2)(c^2-2cd+d^2)+$$
$$+(a^2-2ac+c^2)(b^2-2bd+d^2)+$$
$$+(a^2-2ad+d^2)(b^2-2bc+c^2)=$$
$$=a^2c^2+a^2d^2+b^2c^2+b^2d^2-2a^2cd-2b^2cd-2c^2ab-2d^2ab+4abcd+$$
$$+a^2b^2+a^2d^2+b^2c^2+c^2d^2-2a^2bd-2c^2bd-2b^2ac-2d^2ac+4abcd+$$
$$+a^2b^2+a^2c^2+b^2d^2+c^2d^2-2a^2bc-2d^2bc-2b^2ad-2c^2ad+4abcd=$$
$$=2(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)+12abcd-2\sum_{cyc}a^2(bc+bd+cd).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simpiflying Exponential Expressions
Note that $x = 5$
$$\frac {x^{x+3}+ x^{x+2}}{x^{x+1}+x^{x+2}} = ?$$
This is what I'm struggling with. There's my attempt as seen below.
$$\frac {x^x.x^3+x^x.x^2}{x^x . x + x^x .x^2}$$
Factoring $x^x$ and we get
$$\frac {x^x (x^3 + x^2)}{x^x(x . x^2)} = \frac {x^3}{x} = x^2 = (5)^2 = 25$$
According to my textbook, I've found wrong answer. Correct answer seems $5$, why?
| write $$\frac{x^{x+2}(1+x^{x+3-x-2})}{x^{x+1}(1+x^{x+2-x-1})}$$
can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integer solutions for $x^3=y^2+5$ I have been able to solve the equation $x^3=y^2+a$ for integers where $1 \leq a \leq 4$ by splitting in $\Bbb Z [i\sqrt a]$. However as a natural continuation I would like to know whether the equation $x^3=y^2+5$ can be solved using elementary methods. Seems this case is much tougher though .I would like some hints on how to proceed.
| Write $$ (x-1)(x^2+x+1) = y^2+4$$
If $x\equiv_4 0$ then $x-1\equiv_4 3$ so there is a prime $p\equiv_4 3$ such that $p\mid x-1$. So $p\mid y^2+2^2$ so $p\mid 2$, a contradiction!
If $x\equiv_4 1$ then $x^2+x+1\equiv_4 3$ so there is a prime $p\equiv_4 3$ such that $p\mid x^2+x+1$. So $p\mid y^2+2^2$ so $p\mid 2$, a contradiction!
If $x\equiv_4 2$ then $x^2+x+1\equiv_4 3$ so there is a prime $p\equiv_4 3$ such that $p\mid x^2+x+1$. So $p\mid y^2+2^2$ so $p\mid 2$, a contradiction!
We are left with the case $x\equiv_4 3$ ....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $ax^2+bx+c=0$ has $2$ roots, one root is twice the other, find expression for $b$ $ax^2+bx+c=0$, has two roots with one being twice the other, so $x$ and $2x$
I need to find an expression for $b$ (in terms of $a$ and $c$)
I know $b = \dfrac{-ax^2-c}x$, but I don't know how to use the roots and how it would affect the answer
| Without using the Viète-formulas:
$$x_1=2x_2$$
$$\frac{-b+\sqrt{b^2-4ac}}{2a}=2\frac{-b-\sqrt{b^2-4ac}}{2a}$$
$$\frac{-b+\sqrt{b^2-4ac}}{2}=-b+\sqrt{b^2-4ac}$$
$$\frac{b}{2}=-\frac{3}{2}\sqrt{b^2-4ac}$$
$$b^2=9(b^2-4ac)$$
$$8b^2=36ac$$
$$b=\pm \sqrt{\frac{36ac}{8}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculating the Inverse of a matrix. $A)$
Solve:
$x_1-x_2=1$
$-x_1+2x_2-x_3=0$
$\vdots$
$-x_{99}+2x_{100}=0$
$B$)Deduce the inverse of
A=\begin{pmatrix}
1 & -1 & 0 & \cdots& \cdots & 0 \\
-1 & 2 & -1 & \ddots& \cdots &\vdots\\
0 & -1 & 2 &-1& \ddots& \vdots\\
\vdots & \ddots & \ddots & \ddots &\ddots&0\\
\vdots & \cdots & \ddots & \ddots &2&-1\\
0 & \cdots & \cdots & 0 &-1&2\\
\end{pmatrix}
I've solved the first part and got:
$x_1=100$
$x_2=99$
$\vdots$
$x_{100}=1$ (Correct me if I'm wrong)
For the part B , I'm not sure how to approach it although I'm quite sure the first column of $A^{-1}$ should contain $x_1$ till $x_{100}$ in this order , however I have no idea on how to fill the other columns.
If anyone could help me or give me hints , I would be grateful.
Thanks in advance.
| $\begin{pmatrix}
1 & -1 \\
-1 & 2
\end{pmatrix}^{-1}=\begin{pmatrix}
2 & 1 \\
1 & 1
\end{pmatrix}$
$\begin{pmatrix}
1 & -1 & 0\\
-1 & 2 & -1\\
0 & -1 & 2
\end{pmatrix}^{-1}=\begin{pmatrix}
3 & 2 & 1\\
2 & 2 & 1\\
1 & 1 & 1
\end{pmatrix}$
$\begin{pmatrix}
1 & -1 & 0 & 0\\
-1 & 2 & -1 & 0\\
0 & -1 & 2 & -1\\
0 & 0 & -1 & 2
\end{pmatrix}^{-1}=\begin{pmatrix}
4 & 3 & 2 & 1\\
3 & 3 & 2 & 1\\
2 & 2 & 2 & 1\\
1 & 1 & 1 & 1
\end{pmatrix}$
Can you guess the inverse of $A$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
complex number written in $re^{ix}$ form where $r<0$
The complex number $\left(\sqrt{\frac{1}{2} + \frac{\sqrt{3}}{2}i}\right)^{1/3}$ can be written in the polar form as $r(\cos x + i\sin x)$. If $r<0$, find the smallest positive value of $x$ in degrees.
$\frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i\pi/3}$. Thus $\left(\sqrt{\frac{1}{2} + \frac{\sqrt{3}}{2}i}\right)^{1/3} = e^{i\pi/18}$. At this point, I feel like the answer is $\pi/18$ or 20 degrees, but I'm not sure about the $r<0$ part. How does that affect the answer (if it does)?
| Just do it.
$z= re^{i\theta} = r(\cos \theta + i\sin \theta)$ where $r = \pm |z|$.
As $z = e^{i\frac {\pi}{18}}$ and $|z| = 1$ then $r = -1$ and
$e^{i\frac{\pi}{18}} = -1(\cos \frac{\pi}{18} + i \sin \frac{\pi}{18})$.
We are asked to find the smallest positive $x$ so that -$1(\cos \frac{\pi}{18} + i \sin \frac{\pi}{18})=\cos x + i\sin x$.
In other words words $-\cos x = \cos \frac{\pi}{18}$. And $-\sin x = \cos \frac{\pi}{18}$.
Now basic trig identities $-\cos x = \cos(\pm\pi \pm x)$ and $-sin x = \sin(-x) = \sin(x \pm \pi)$. (All modulo $2\pi$ of course)
So the only case were $-\cos x = \cos y$ and $-\sin x = \sin y$ is when $y = x \pm \pi$, or vice versa, $x = y \pm \pi$. (All modulo $2\pi$ of course)
So $x = \frac{\pi}{18} \pm \pi$ (All modulo $2\pi$ of course).
At for $x$ to be the least positive value $x = \frac {\pi}{18} + \pi = \frac {19}{18}\pi$.
.....
Note: This reasoning leads to the identity that $e^{i\theta + i2k\pi} = -e^{i(\theta + \pi)+ i2k\pi}$ which is very easy to prove. [$e^{i(\theta + \pi)} = \cos (\theta + \pi) + i\sin(\theta + \pi) = -\cos\theta - i\sin \theta = -e^{i\theta}$]
Has we simply kept that in mind the solution would be simply:
$e^{i\frac {\pi}{18}}= -e^{i(\frac {\pi}{18}+\pi)} = \cos x + i\sin x$ so $x = \frac {\pi}{18}+\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Minimum value of slope of line tangent to ellipse I would like to solve the following problem.
Given that the point $p$ lies on the ellipse $$\frac{x^2}{100} + \frac{(y - 12)^2}{75} = 1$$ and the line $y = ax$ passes through $p$, let $n$ be the smallest positive value of $a$. Compute $n$.
What first came to my mind was finding the slope of the line tangent to the ellipse. This would require me to set the discriminant of the resulting quadratic in $x$ to be set to $0$.
However, the numbers in the denominator make the calculations a bit annoying. Is there a better way to solve the problem? Thanks!
| Not the best in math way answer because it is not general solution but it is based on my comment:
My idea is (for easy solution without hard work) to find the line
$y=αx$ that is tangent in the circle that contains the ellipse and
then test next few values with the real ellipse. [I know it is a hack
but also know that the numbers seems to be a little bit silly for this
question... Sure the solution is close to $1$ if not $1$]
The circle that contains the ellipse (don't know English name) is:
$x^2+(y-12)^2=10^2$
The tangent on this circle that can pass through $(0,0)$ is a line of function $y=\alpha \cdot x$ that has distance d=10 from the point $(0,12)$.
line is:
$\alpha \cdot x -y=0$ and point (0,12) have a distance:
$d=\dfrac{|\alpha \cdot 0 -12|}{\sqrt{\alpha^2+1}}=\dfrac{12}{\sqrt{\alpha^2+1}}=10$
Thus:
$100\alpha^2+100=144\Longrightarrow \alpha^2=\frac{44}{100}$.
We are looking for integer values of $\alpha$, and so we will start from value $1$ and test a few integers with the real ellipse [Possibly $1$ could already passes through].
Test value $\alpha=1$
$\begin{cases}y=x\\
\dfrac{x^2}{100}+\dfrac{(x-12)^2}{75}=1
\end{cases}\hspace{10pt}
\begin{cases}y=x\\
x^2+\dfrac{4}{3}(x-12)^2=100
\end{cases}\hspace{10pt}
\begin{cases}y=x\\
x^2+\dfrac{4}{3}x^2 +\dfrac{4}{3}144-\dfrac{4}{3}\cdot 24 \cdot x=100
\end{cases}$
$\begin{cases}\dfrac{7}{3}x^2-32\cdot x+16\cdot 12-100=0\\\end{cases}$
The last one becomes:
$7x^2- 96 x +(192-100)\cdot 3=0$
$7x^2-96 x+276=0$ with
$\Delta=(-96)^2-4\cdot 7\cdot 276\approx 10,000-7000 >0$
Solution =1
Edit: Visualizing why I found the numbers silly:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find the roots of $\mathrm{ \overline{Z} + 1 = iZ^2 + {|Z|}^2}$ I attempted the problem in two ways
Method 1
Resolved $\mathrm{ Z = X + iY}$ this led to a very big cubic equation and handling that became too difficult for me .
Method 2
I believe factorizing the equation is a far better method, I don't know how to go about it .
EDIT
The answer given is $\mathrm{Z = \frac{-i}{2} , i}$
As pointed out by @gimsui $\mathrm{\frac{-i}{2} }$ is not a solution most likely a printing error.
| \begin{align}
z &= x + iy \\
\hline
\overline z &= x - iy \\
z^2 &= (x^2-y^2)+2ixy \\
|z|^2 &= x^2+y^2
\end{align}
\begin{align}
\overline z + 1 &= iz^2 + |z|^2 \\
(x-iy) + 1 &= i(x^2-y^2 + 2ixy)+(x^2+y^2) \\
(x+1) - iy &= (x^2-2xy+y^2) + i(x^2-y^2) \\
x+1-x^2+2xy-y^2 &= i(x^2-y^2+y) \\
\hline
x^2-2xy+y^2 - x - 1 &= 0 \\
x^2 &= y^2-y \\
\hline
2y^2-2xy-x-y-1 &= 0 \\
2y^2 -(2x+1)y-(x+1)&=0 \\
y &= \dfrac{(2x+1)\pm \sqrt{(2x+1)^2+8(x+1)}}{4} \\
y &= \dfrac{(2x+1)\pm \sqrt{4x^2+12x+9}}{4} \\
y &= \dfrac{(2x+1)\pm |2x+3|}{4} \\
y &= \dfrac{(2x+1)\pm (2x+3)}{4} \\
y &\in \left\{ x+1, -\frac 12 \right\}
\end{align}
If $y=x+1$ and $x^2=y^2-y$, then
\begin{align}
x^2 &= (x+1)^2 - (x+1) \\
x^2 &= x^2 + x \\
(x,y) &= (0,1)
\end{align}
If $y=-\dfrac 12$ and $x^2=y^2-y$, then
\begin{align}
x^2 &= \dfrac 14 + \dfrac 12 \\
(x,y) &= \left( \pm \dfrac{\sqrt 3}{2}, -\dfrac 12 \right)
\end{align}
$$z \in \left\{ i,
-\dfrac{\sqrt 3}{2} - i\dfrac 12,
\dfrac{\sqrt 3}{2} - i\dfrac 12
\right\} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.