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how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Possible Duplicate:
Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
Summation of natural number set with power of $m$
How to get to the formula for the sum of squares of first n numbers?
how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Let,
$T_{2}(n)=1^2+2^2+3^2+\cdots+n^2$
$T_{2}(n)=(1^2+n^2)+(2^2+(n-1)^2)+\cdots$
$T_{2}(n)=((n+1)^2-2(1)(n))+((n+1)^2-2(2)(n-1))+\cdots$
| Hint: $(n+1)^3-n^3=3n^2+3n+1$ and use telescopic sum.
| {
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Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$ I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can be found in that pdf but I will write here to be more explicitly.
Exercise 1.3.4(a) Let $a,b,c$ be positive real numbers. Prove that
$$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4.$$
(b) For real numbers $a,b,c \gt0$ and $n \leq3$ prove that:
$$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$
| By C-S
$$\sum_{cyc}\frac{a}{b}=\sum_{cyc}\frac{a^2}{ab}\geq\frac{(a+b+c)^2}{ab+ac+bc}.$$
Thus, it's enough to prove that
$$\frac{(a+b+c)^2}{ab+ac+bc}+\frac{3\sqrt[3]{abc}}{a+b+c}\geq4.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $f(v^2)\geq0,$ where $f$ decreases, which says that it's enough to prove the last inequality for a maximal value of $v^2$, which happens for equality case of two variables.
Since the last inequality is homogeneous, we can assume $b=c=1$. Also, let $a=x^3$.
Id est, we need to prove that
$$\frac{(x^3+2)^2}{2x^3+1}+\frac{3x}{x^3+2}\geq4$$ or
$$(x-1)^2(x^6+2x^5+3x^4+2x^3+x^2+6x+3)\geq0.$$
Done!
The following inequality is also true.
Let $a$, $b$ and $c$ be positives. Prove that:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{24\sqrt[3]{abc}}{a+b+c}\geq11.$$
| {
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How to solve $x^3=-1$? How to solve $x^3=-1$? I got following:
$x^3=-1$
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
| $x^3+1=0\implies (x+1)(x^2-x+1)=0$
If $x+1=0,x=-1$
else $x^2-x+1=0$ then $x=\frac{1±\sqrt{1-4}}{2\cdot1}=\frac{1±\sqrt3i}{2}$ using the well known Quadratic Formula.
Alternatively, using Euler's identity and Euler's formula like other two solutions,
for $x^n=-1=e^{i(2m+1)\pi}$ as $e^{i\pi}=-1$ where $m$ any any integer, $n$ is a natural number .
We know $n$-degree equation has exactly $n$ roots, so the roots of $x^n+1=0$
are $e^{\frac{i(2m+1)\pi}{n}}=\cos\frac{(2m+1)\pi}{n}+i\sin \frac{(2m+1)\pi}{n}$ where $m=0,1,2,...n-1$. It's just customary, not mandatory that we have defined $m$ to assume this range of values, in fact it can assume any $n$ in-congruent values (for example, consecutive values like $r,r+1,...,r+n-1,$ where r is any integer), the reason is explained below.
(1)Using Repeated Root Theorem, let $f(x)=x^n+R$ where $n>1$ and $R ≠ 0\implies x≠0$
So, $\frac{df}{dx}=nx^{n-1}, \frac{df}{dx}=0$ does not have any non-zero root.
Clearly, $f(x)=x^n+R$ can not have any repeated root unless $R=0$
(2)Let $e^{\frac{i(2a+1)\pi}{n}}=e^{\frac{i(2b+1)\pi}{n}}$
$\implies e^{\frac{2i(a-b)\pi}{n}}=1=e^{2k\pi i}$ where $k$ is any integer.
$\implies a-b=kn\implies a≡b\pmod n$, so any $n$ in-congruent values of $m$ will give us essentially the same set of n distinct solutions.
Here $n=3$ so, let's take $m=0,1,2$.
$m=0\implies \cos\frac{\pi}{3}+i\sin \frac{\pi}{3}=\frac{1+i\sqrt3}{2} $
$m=1\implies \cos \pi+i\sin\pi=-1 $
$m=2\implies \cos\frac{5\pi}{3}+i\sin \frac{5\pi}{3}=\frac{1-i\sqrt3}{2} $
| {
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ Please help me for prove this inequality:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$
| It's a convexity inequality: the function $f(x) = \frac{1}{x}$ is convex on $\mathbb{R}^{+*}$ (its second derivative is positive), so for any $a,b,c\in\mathbb{R}^{+*}, \frac{1}{3}(f(a)+f(b)+f(c))\geq f(\frac{a+b+c}{3})$, which is $\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq\frac{3}{a+b+c}$, or finally $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{9}{a+b+c}$
| {
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Solving $\sqrt{(x-2)^2 + (y-1)^2} = \sqrt 2$ What is the answer of this:
$\sqrt{(x-2)^2 + (y-1)^2} = \sqrt 2$
| $\textbf{Hint}$ : First, try to relate this to the pythagorean theorem. Also note.
$\sqrt{(x - 2)^2 + (y - 1)^2} = \sqrt{2}$
$(x - 2)^2 + (y - 1)^2 = 2$
Think about circles.
| {
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Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$? Playing around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?
| Note that
$$ \tan \left(\arctan(1+z) + \arctan\left(2 + z + z^2 \right) + \arctan \left( 3+3\,z+4\,{z}^{2}+2\,{z}^{3}+{z}^{4} \right) \right)=z
$$
so that $$\arctan(1+z) + \arctan\left(2 + z + z^2 \right) + \arctan \left( 3+3\,z+4\,{z}^{2}+2\,{z}^{3}+{z}^{4} \right) = \arctan(z) + n \pi $$
for the appropriate integer $n$.
For integers $z$ we get interesting arctan identities from this.
$$\eqalign{ \arctan(1) + \arctan\left(2\right)+ \arctan\left(3\right) &= \pi \cr
\arctan(2) + \arctan(4) + \arctan(13) &= \arctan(1) + \pi \cr
\arctan(3) + \arctan(8) + \arctan(57) &= \arctan(2) + \pi \cr
\arctan(4) + \arctan(14) + \arctan(183) &= \arctan(3) + \pi \cr}$$
etc.
| {
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Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!
| Let $x=\sqrt{1+i\sqrt3}+\sqrt{1-i\sqrt3}$. Then
$$x^2=(1+i\sqrt3)+2\sqrt{1+i\sqrt3}\sqrt{1-i\sqrt3}+(1-i\sqrt3)=2+2\sqrt{1+i\sqrt3}\sqrt{1-i\sqrt3}$$
so
$$(x^2-2)^2=4(1+i\sqrt3)(1-i\sqrt3)=4(1+3)=16$$
so
$$x^4-4x^2-12=(x^2-6)(x^2+2)=0$$
Thus $x\in\{\sqrt6,-\sqrt6,i\sqrt2,-i\sqrt2\}$, so it remains to determine which of these four roots is meant.
The answer depends on which convention you decide to use when computing the square root of a complex number, $\sqrt{a+ib}$ with $b\not=0$. There are two common conventions: require $\sqrt{a+ib}$ with $b\not=0$ to have positive real part, or require it to have positve imaginary part. (In terms of polar coordinates, this amounts to saying $\sqrt{re^{i\theta}}=\sqrt re^{i\theta/2}$, but with $-\pi\lt\theta\le\pi$ for the first convention and $0\le\theta\lt2\pi$ for the second.)
If we assume the first convention, then $x$, being the sum of two square roots, must have positive real part, hence must equal $\sqrt6$. If we assume the second convention, then $x$ must have positive imaginary part, hence must equal $i\sqrt2$. Since the desired answer is $\sqrt6$, we see that the problem is tacitly assuming the first convention.
| {
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Solving for x with exponents (algebra) So I am trying to help a friend do her homework and I am a bit stuck.
$$8x+3 = 3x^2$$
I can look at this and see that the answer is $3$, but I am having a hard time remembering how to solve for $x$ in this situation.
Could someone be so kind as to break down the steps in solving for $x$.
Thanks in advance for replies.
| Here's a way without the formula. Using "completing the square"
\begin{align*}
3x^2-8x-3&=0\\
x^2-\frac{8}{3}x-1&=0\\
\left(x-\frac{4}{3}\right)^2-\frac{16}{9}-1&=0\\
x-\frac{4}{3}&=\pm \sqrt{\frac{25}{9}}\\
x &= \frac{4}{3}\pm \frac{5}{3}\\
x = \frac{9}{3} = 3\ \ &\mathrm{or}\ \ x = -\frac{1}{3}.
\end{align*}
| {
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Identity proof $(x^{n}-y^{n})/(x-y) = \sum_{k=1}^{n} x^{n-k}y^{k-1}$ In a proof from a textbook they use the following identity (without proof):
$(x^{n}-y^{n})/(x-y) = \sum_{k=1}^{n} x^{n-k}y^{k-1}$
Is there an easy way to prove the above? I suppose maybe an induction proof will be appropriate, but I would really like to find a more intuitive proof.
| Yes there is. First let's look at the series:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3} + xy^{n-2} + y^{n-1} \, . $$
This is a geometric series with $n$-terms, whose first term is $x^{n-1}$ and whose common ratio is $y/x$. Applying the standard formula we obtain:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3} + xy^{n-2} + y^{n-1} = x^{n-1}\left( \frac{1-(y/x)^n}{1-(y/x)} \right) . $$
Simplifying the numerator and denominator and then cancelling common factors gives:
$$x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \cdots + x^2y^{n-3} + xy^{n-2} + y^{n-1} = \frac{x^n-y^n}{x-y} \, . $$
| {
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Proving $\sum\limits_{i=1}^{n}{\frac{i}{2^i}}=2-\frac{n+2}{2^n}$ by induction.
Theorem (Principle of mathematical induction):
Let $G\subseteq \mathbb{N}$, suppose that
a. $1\in G$
b. if $n\in \mathbb{N}$ and $\{1,...,n\}\subseteq G$, then $n+1\in G$
Then $G=\mathbb{N}$
Proof by Induction
Prove that
$$\sum_{i=1}^{n}{\frac{i}{2^i}}=2-\frac{n+2}{2^n},\quad \quad \forall n\in \mathbb{N} \tag{1}$$
$\color{darkred}{\mbox{Proof:}}$
let $P(n)$ be the statement $\sum_{i=1}^{n}{\frac{i}{2^i}}=2-\frac{n+2}{2^n}$
$$\sum_{i=1}^{1}{\frac{i}{2^i}}=\frac{1}{2}=2-\frac{1+2}{2^1}\tag{Basis step $P(1)$}$$
Thus $P(1)$ is true
We assume that $P(n)$ is true and prove that $P(n+1)$ is true
$$\sum_{i=1}^{n+1}{\frac{i}{2^i}}\stackrel{?}{=}2-\frac{(n+1)+2}{2^{n+1}} $$
We know that
$$
\begin{align*}
\sum_{i=1}^{n+1}{\frac{i}{2^i}}&=\color{darkred}{\underbrace{\sum_{i=1}^{n}{\frac{i}{2^i}}}_\text{P(n) is true}}+\frac{n+1}{2^{n+1}} \tag{LHS} \\
&=\color{darkred}{2-\frac{n+2}{2^n}} + \frac{n+1}{2^{n+1}}\\
&=\frac{2^{n+1}-n-2}{2^n} + \frac{n+1}{2^{n+1}}\\
&=\frac{2\cdot 2^{n+1}-2n-4+n+1}{2^{n+1}}\\
&=\frac{2\cdot 2^{n+1}-n-3}{2^{n+1}}\tag{2}\\
\end{align*}
$$
And
$$
\begin{align*}
2-\frac{(n+1)+2}{2^{n+1}}&=\frac{2\cdot 2^{n+1}-n-3}{2^{n+1}} \tag{RHS}
\end{align*}
$$
So
$$\sum_{i=1}^{n+1}{\frac{i}{2^i}}\stackrel{}{=}2-\frac{(n+1)+2}{2^{n+1}} \tag{3}$$
Hence, if $P(n)$ is true, then $P(n+1)$ is true.
Therefore $\sum\limits_{i=1}^{n}{\frac{i}{2^i}}=2-\frac{n+2}{2^n},\quad \quad \forall n\in \mathbb{N} $
$\hspace{8cm}$ ${\Large ▫}$
My Question
$(i)$ does my solution make sense (is it correct?)
$(ii)$ If so, is this an correct approach to proving formulas by induction?
Can someone give me some hints/tips?
| You can also notice that $$\sum\limits_{i=1}^n \frac{i}{2^i} = \sum\limits_{i=1}^n \frac{1}{2^i}+ \sum\limits_{i=2}^{n} \frac{1}{2^i} + \dots + \sum\limits_{i=n}^n \frac{1}{2^i}= \sum\limits_{j=1}^n \sum\limits_{i=j}^n \frac{1}{2^i}$$ So you can find the result without knowing it in advance; this is only a geometric series.
| {
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Solving trigonometric equations of the form $a\sin x + b\cos x = c$ Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.
How do you solve this equation without using the method that moves $b\cos x$ to the right side and squaring left and right sides of the equation?
And how does solving $\sqrt{3}\sin x + \cos x = 2$ equal to solving $\sin (x+ \frac{\pi}{6}) = 1$
| Using complex numbers, and setting $z=e^{i\theta}$,
$$a\frac{z-z^{-1}}{2i}+b\frac{z+z^{-1}}2=c,$$
or
$$(b-ia)z^2-2cz+(b+ia)=0.$$
The discriminant is $c^2-b^2-a^2:=-d^2$, assumed negative, then the solution
$$z=\frac{c\pm id}{b-ia}.$$
Taking the logarithm, the real part $$\ln\left(\dfrac{\sqrt{c^2+d^2}}{\sqrt{a^2+b^2}}\right)=\ln\left(\dfrac{\sqrt{c^2+a^2+b^2-c^2}}{\sqrt{a^2+b^2}}\right)=\ln(1)$$ vanishes as expected, and the argument is
$$\theta=\pm\arctan\left(\frac dc\right)+\arctan\left(\frac ab\right).$$
The latter formula can be rewritten with a single $\arctan$, using
$$\theta=\arctan\left(\tan(\theta)\right)=\arctan\left(\frac{\pm\dfrac dc+\dfrac ab}{1\mp\dfrac dc\dfrac ab}\right)=\arctan\left(\frac{\pm bd+ac}{bc\mp ad}\right).$$
| {
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The intersection of a line with a circle Get the intersections of the line $y=x+2$ with the circle $x^2+y^2=10$
What I did:
$y^2=10-x^2$
$y=\sqrt{10-x^2}$ or $y=-\sqrt{10-x^2}$
$ x+ 2 = y=\sqrt{10-x^2}$
If you continue, $x=-3$ or $x=1$ , so you get 2 points $(1,3)$, $(-3,-1)$
But then, and here is where the problems come:
$x+2=-\sqrt{10-x^2}$
I then, after a while get the point $(-3\dfrac{1}{2}, -1\dfrac{1}{2}$) but this doesn't seem to be correct. What have I done wrong at the end?
| If you do everything correct, the solution of $x+2=-\sqrt{10-x^2}$ would be almost the same - $(1,-3)$, $(-3,1)$. It's refer to intersaction of $y = -(x+2)$ and original circle.
By the way, your answer is correct.
I have no idea how you'd get the point $(-3\dfrac{1}{2}, -1\dfrac{1}{2})$.
| {
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Basic Integral question Prove that $$ \int_0^\pi \frac{dx}{\alpha - \cos(x)}\ = \frac{\pi}{\sqrt{\alpha^2 - 1}}$$ for $\alpha > 1$
| Put $z = e^{ix}$ so that $dz = i e^{ix} dx$ and $dx = 1/z/i \,dz$. The integral becomes
$$\frac{1}{2} \int_{|z|=1} \frac{2}{2\alpha - z - 1/z} \frac{1}{iz} dz $$
Simplifying this yields
$$\frac{1}{2i} \int_{|z|=1} \frac{2}{2\alpha z - z^2 - 1} dz$$
The poles are at
$$z = \alpha \pm \sqrt{\alpha^2-1},$$
Of these, only the smaller one is inside the unit circle and the residue is
$$ \text{Res}_{z=\alpha - \sqrt{\alpha^2-1}} \frac{2}{2\alpha z - z^2 - 1} =
\frac{1}{\sqrt{\alpha^2-1}}.$$
Hence by the Cauchy Residue Theorem the integral is
$$ 2\pi i \frac{1}{2i} \frac{1}{\sqrt{\alpha^2-1}} =
\frac{\pi}{\sqrt{\alpha^2-1}}.$$
Here we have used the fact that
$$\text{Res}_{z=\rho_0} \frac{1}{(z-\rho_0)(z-\rho_1)} =
\lim_{z\rightarrow \rho_0} \frac{1}{z-\rho_1} = \frac{1}{\rho_0-\rho_1}.$$
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Show $x^2 +xy-y^2 = 0$ is only true when $x$ & $y$ are zero. Show that it is impossible to find non-zero integers $x$ and $y$ satisfying $x^2 +xy-y^2 = 0$.
| Let $g=\gcd(x,y)$. Let $x=rg$ and $y=sg$. Then
$$g^2r^2+g^2rs-g^2s^2=0$$
$$r^2+rs-s^2=0$$
$$s^2-r^2=rs$$
$$(s+r)(s-r)=rs$$
We've divided out all common factors, so $\gcd(r,s)=1$. For any prime $p$, if $p|r$, $p$ does not divide $s$ and therefore divides neither $s+r$ nor $s-r$. The same argument applies if $p|s$. Therefore, $r$ and $s$ contain no prime factors and $r=s=1.$ Since $r=s$, $x=y$. So
$$x^2+xy-y^2=x^2+x^2-x^2=x^2=0$$
$$x=0$$
| {
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Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$ Given $f(x+1/x) = x^2 +1/x^2$, find $f(x)$. Please show me the way you find it.
The answer in my textbook is $f(x)=\frac{1+x^2+x^4}{x\cdot \sqrt{1-x^2}}$
| Let $y=x+\frac{1}{x}$, try now to express $x$ as a function of $y$.
We have
$$x^2-xy+1=0$$
$$x=\frac{y \pm \sqrt{y^2 -4}}{2}$$
Substitute this value for x in your expression for $f$.
$$f(y)=\left(\frac{y \pm \sqrt{y^2 -4}}{2}\right)^2+\left(\frac{2}{y \pm \sqrt{y^2 -4}}\right)^2$$
$$f(y)=y^2-2$$
| {
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Implicit solution to an ODE I have come accross a problem which leads to the following ODE:
$\frac{dy}{dx}= \frac{y}{x}-\frac{1}{h}\frac{\sqrt{x^2+y^2}}{x}$,
where $h>0$ is a parameter. I am not able to solve it, however maple gives an implicit
solution:
$\frac{x^{1/h}y}{x}+\frac{x^{1/h}\sqrt{x^2+y^2}}{x}=$constant.
I want to understand, how to $\textit{find}$ this implicit solution. (Of course one can differentiate it and find back the ODE.)
| The important observation is that th equation is homogeneous. Indeed
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\frac{{\sqrt {{x^2} + {y^2}} }}{x} \cr
& \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\sqrt {\frac{{{x^2} + {y^2}}}{{{x^2}}}} \cr
& \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} \cr} $$
Thus, let $y=vx$. Then $y'=v'x+v$; and we get
$$\eqalign{
& \frac{{dv}}{{dx}}x + v = v - \frac{1}{h}\sqrt {1 + {v^2}} \cr
& \frac{{dv}}{{dx}}x = - \frac{1}{h}\sqrt {1 + {v^2}} \cr
& \frac{{dv}}{{\sqrt {1 + {v^2}} }} = - \frac{{dx}}{x}\frac{1}{h} \cr} $$
Can you move on? The complete solution would be:
$$\begin{eqnarray*}
{\sinh ^{ - 1}}v = - \frac{1}{h}\left( {\log x + C} \right)\\\log \left( {v + \sqrt {1 + {v^2}} } \right) = \log \frac{1}{{\root h \of x }} - \frac{C}{h}\\ \log \left( {\frac{y}{x} + \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} } \right) = \log \frac{1}{{\root h \of x }} - \frac{C}{h} \\\frac{y}{x} + \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} = \frac{k}{{\root h \of x }}\\y + \sqrt {{x^2} + {y^2}} = k{x^{1 - \frac{1}{h}}}
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/221672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Why $\cos^2 (2x) = \frac{1}{2}(1+\cos (4x))$? Why: $$\cos ^2(2x) = \frac{1}{2}(1+\cos (4x))$$
I don't understand this, how I must to multiply two trigonometric functions?
Thanks a lot.
| Recall the formula $$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ This gives us $$\cos^2(\theta) = \dfrac{1+\cos(2 \theta)}{2}$$ Plug in $\theta = 2x$, to get what you want.
EDIT
The identity
$$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ can be derived from $$\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B)$$
Setting $A = B = \theta$, we get that
$$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \cos^2(\theta) - (1-\cos^2(\theta)) = 2 \cos^2(\theta) - 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/222404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Derivative of a split function We have the function:
$$f(x) = \frac{x^2\sqrt[4]{x^3}}{x^3+2}.$$
I rewrote it as
$$f(x) = \frac{x^2{x^{3/4}}}{x^3+2}.$$
After a while of differentiating I get the final answer:
$$f(x)= \frac{- {\sqrt[4]{\left(\frac{1}{4}\right)^{19}} + \sqrt[4]{5.5^7}}}{(x^3+2)^2}$$(The minus isn't behind the four)
But my answer sheet gives a different answer, but they also show a wrong calculation, so I don't know what is the right answer, can you help me with this?
| Although collecting powers is a good idea as suggested, things are made clearer after taking formal logarithm before differentiating:
$$\ln f=2\ln x+\frac{3}{4}\ln x -\ln(x^3+2)$$
$$\frac{f'}{f}=\frac{2}{x}+\frac{3}{4x}-\frac{3x^2}{x^3+2}=\frac{11}{4x}-\frac{3x^2}{x^3+2}=-\frac{x^3-22}{4x(x^3+3)}$$
$$f'=-\frac{x^\frac{11}{4}(x^3-22)}{4x(x^3+2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/225438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Adding sine waves of different phase, sin(π/2) + sin(3π/2)? Adding sine waves of different phase, what is $\sin(\pi/2) + \sin(3\pi/2)$?
Please could someone explain this.
Thanks.
| $\sin(\pi+x)=\sin \pi\cos x+\cos\pi\sin x=-\sin x$ as $\sin \pi=0,\cos\pi=-1$
This can be achieved directly using "All-Sin-Tan-Cos" formula or using $\sin 2A+\sin 2B=2\sin(A+B)\cos(A-B)$, $\sin(\pi+x)+\sin x=2\sin\left(x+\frac \pi 2\right)cos\left(\frac \pi 2\right)=0$
So, $\sin(x)+\sin(\pi+x)=0$
More generally, $\sin(x+c)+\sin(x+d)=2\sin\left(x+\frac{c+d}2\right)\cos\left(\frac{c-d}2\right)$
So, the resultant phase is the arithmetic mean of the phases of the original waves provided $\cos\left(\frac{c-d}2\right)\ne 0,$ i.e., $\frac{c-d}2\ne \frac{(2n+1)\pi}2$ or $c-d\ne(2n+1)\pi$ where $n$ is any integer.
Here, $c-d=0-\pi=-\pi\implies \cos\left(\frac{c-d}2\right)=0 $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using induction to prove $3$ divides $\left \lfloor\left(\frac {7+\sqrt {37}}{2}\right)^n \right\rfloor$ How can I use induction to prove that $$\left \lfloor\left(\cfrac {7+\sqrt {37}}{2}\right)^n \right\rfloor$$ is divisible by $3$ for every natural number $n$?
| Since
$$
\left(x-\frac{7+\sqrt{37}}2\right)\left(x-\frac{7-\sqrt{37}}2\right)=\left(x-\frac72\right)^2-\frac{37}4=x^2-7x+3\;,
$$
the recurrence relation
$$x_n=7x_{n-1}-3x_{n-2}$$
has solutions
$$
x_n=c_1\left(\frac{7+\sqrt{37}}2\right)^n+c_2\left(\frac{7-\sqrt{37}}2\right)^n\;.
$$
With $c_1=c_2=1$ we have $x_0=2$ and $x_1=7$. Since $0\lt\frac{7-\sqrt{37}}2\lt1$ and $x_n$ is an integer,
$$
\left\lfloor\left(\frac{7+\sqrt{37}}2\right)^n\right\rfloor=x_n-1\;.
$$
Now you just need to use the recurrence relation to show that the property that $x_1$ has remainder $1$ mod $3$ is inherited by the remaining terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/226177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Number of squares in a rectangle Given a rectangle of length a and width b (as shown in the figure), how many different squares of edge greater than 1 can be formed using the cells inside.
For example, if a = 2, b = 2, then the number of such squares is just 1.
| Without loss of generality we may assume that that the width $b$ is $\ge$ to the height $a$. Note that there are $b+1$ vertical lines and $a+1$ horizontal lines in the picture.
If $a\ge 2$, the number of $2\times 2$ squares is $(a-1)(b-1)$. This is because the top left corner of a $2\times 2$ square can be any corner that is not in the last two columns or the last two rows.
If $a\ge 3$, the number of $3\times 3$ squares is $(a-2)(b-2)$. This is because the top left corner of a $3\times 3$ square can be any corner which is not in the last three columns or the last three rows.
If $a\ge 4$, the number of $4\times 4$ squares is $(a-3)(b-3)$.
Continue. Finally, the number of $a\times a$ squares is (a-(a-1))(b-(a-1))$.
Add up. To get a "nice" expression for the sum, note that
$$(a-k)(b-k)=ab-k(a+b)+k^2.\tag{$1$}$$
When we add up the expressions in $(1)$, we get $a-1$ terms with value $ab$, for a total of $(a-1)ab$.
The "middle" terms add up to $-(a+b)(1+2+\cdots+(a-1))$. The sum is
$-(a+b)\dfrac{(a-1)a}{2}$.
Finally, we need $1^2+2^2+\cdots +(a-1)^2$. By the usual formula for the sum of consecutive squares, this is equal to $\dfrac{(a-1)(a)(2a-1)}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/229182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Solve $2a + 5b = 20$ Is this equation solvable? It seems like you should be able to get a right number!
If this is solvable can you tell me step by step on how you solved it.
$$\begin{align}
{2a + 5b} & = {20}
\end{align}$$
My thinking process:
$$\begin{align}
{2a + 5b} & = {20} & {2a + 5b} & = {20} \\
{0a + 5b} & = {20} & {a + 0b} & = {20} \\
{0a + b} & = {4} & {a + 0b} & = {10} \\
{0a + b} & = {4/2} & {a + 0b} & = {10/2} \\
{0a + b} & = {2} & {a + 0b} & = {5} \\
\end{align}$$
The problem comes out to equal:
$$\begin{align}
{2(5) + 5(2)} & = {20} \\
{10 + 10} & = {20} \\
{20} & = {20}
\end{align}$$
since the there are two different variables could it not be solved with the right answer , but only "a answer?"
What do you guys think?
| Note that $2a$ must have as its unit digit $0, 2, 4, 6,$ or $8$ (because it's even!).
Similarly, note that $5b$ must have as its unit digit $0$ or $5$.
With a bit of thinking, you can see that for $2a + 5b$ to be $20$, we need to ensure that $2a$ has a unit digit of $0$ (hence $a$ is a multiple of $5$).
So let $a$ be a multiple of $5$. That is, let $a = 5k$, for some integer $k$.
Then $2a + 5b = 20$ becomes $10k + 5b = 20$, so that $5b = 20 - 10k$.
Dividing both sides of our last equation by $5$, we have $b = 4 - 2k$.
This gives you all the possible answers for $(a, b)$, namely, $(5k, 4-2k)$.
For example, when $k = 1$ you get $(5 \cdot 1, 4 - 2 \cdot 1) = (5, 2)$, which is the answer you came to.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Mixing things- ratios Suppose I take two things, A and B. C is made from a $1/4$ ratio of $A$ to $B$, while D is made from a $4/3$ ratio. If I want to know what ratio of $C$ to $D$ will give a $5/6$ ratio of $A$ to $B$, do I just solve the system
$A+4B=C, 4A+3B=D, 5A+6B=z$
to get
$z= \frac{9C+14D}{13} \rightarrow \frac{5A}{6B} = \frac{9C}{14D}?$ If not what am I doing wrong? What is the right method?
| The mixing ratios give
$$
\begin{bmatrix}
\frac15&\frac45\\
\frac47&\frac37
\end{bmatrix}
\begin{bmatrix}A\\B\end{bmatrix}=\begin{bmatrix}C\\D\end{bmatrix}
$$
That is, $1$ can of $C$ is $\frac15$ can of $A$ and $\frac45$ can of $B$, $1$ can of $D$ is $\frac47$ can of $A$ and $\frac37$ can of $B$.
Because
$$
\begin{align}
\begin{bmatrix}\frac5{11}&\frac6{11}\end{bmatrix}
\begin{bmatrix}
\frac15&\frac45\\
\frac47&\frac37
\end{bmatrix}^{-1}
&=\begin{bmatrix}\frac5{11}&\frac6{11}\end{bmatrix}
\begin{bmatrix}
-\frac{15}{13}&\frac{28}{13}\\
\frac{20}{13}&-\frac7{13}
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}\frac{45}{143}&\frac{98}{143}\end{bmatrix}
\end{align}
$$
we get
$$
\begin{bmatrix}\frac5{11}&\frac6{11}\end{bmatrix}
\begin{bmatrix}A\\B\end{bmatrix}
=\begin{bmatrix}\frac{45}{143}&\frac{98}{143}\end{bmatrix}
\begin{bmatrix}C\\D\end{bmatrix}
$$
Therefore, a $45:98$ mixture of $C$ and $D$ requires a $5:6$ ratio of $A$ and $B$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
convergence tests for series $p_n=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n)}$ If the sequence:
$p_n=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n)}$
Prove that the sequence
$((n+1/2)p_n^2)^{n=\infty}_{1}$ is decreasing.
and that the series $(np_n^2)^{n=\infty}_{1}$ is convergent.
Any hints/ answers would be great.
I'm unsure where to begin.
| Firstly, it should be noted that $\dfrac{p_{n+1}}{p_n}=\dfrac{2n+1}{2n+2}=\dfrac{n+\frac{1}{2}}{n+1},$ therefore,
$$\frac{a_{n+1}}{a_n}=\frac{\left(n+1+\frac{1}{2} \right)p_{n+1}^2}{\left(n+\frac{1}{2} \right)p_{n}^2}=\frac{\left(n+\frac{3}{2} \right)\left(n+\frac{1}{2} \right)^2}{\left(n+\frac{1}{2} \right)\left(n+1 \right)^2}=\frac{\left(n+\frac{3}{2} \right)\left(n+\frac{1}{2} \right)}{\left(n+1 \right)^2}=\frac{n^2+2n+\frac{3}{4}}{n^2+2n+1}<1.$$
Next, $p_n$ can be rewritten as $$p_n=\frac{3}{2}\cdot \frac{5}{4}\cdot \ldots \cdot \frac{2n-1}{2n-2}\cdot \frac{1}{2n} > \frac{3}{2}\cdot \frac{1}{2n}=\frac{3}{4n}, $$
which implies
$$
p_n^2>\frac{9}{16}\cdot\frac{1}{n^2}
$$
and $$np_n^2>n\cdot\dfrac{9}{16}\cdot\dfrac{1}{n^2}=\dfrac{9}{16}\cdot\dfrac{1}{n},$$
so $\sum\limits_{n=1}^{\infty}{np_n^2}$ diverges.
| {
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"url": "https://math.stackexchange.com/questions/232092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric Identity - $\tan(\pi/2 -x) - \cot(3\pi/2 -x) + \tan(2\pi-x) - \cot(\pi-x) ...$ $$\tan\left(\frac{\pi}{2} -x\right) - \cot\left(\frac{3\pi}{2} -x\right) + \tan(2\pi-x) - \cot(\pi-x) = \frac{4-2\sec^{2}x}{\tan{x}}$$
L.S.
$= \cot{x} - \tan{x} - \tan{x} + \cot{x}$
$= 2\cot{x} - 2\tan{x}$
$= 2\left(\frac{\cos{x}}{\sin{x}} - \frac{\sin{x}}{\cos{x}}\right)$
$= 2\left(\frac{\cos^{2}x - \sin^{2}x}{\sin{x}\cos{x}}\right)$
$= 2\left(\frac{1-2\sin^{2}x}{\sin{x}\cos{x}}\right)$
$= \frac{4 - 2\sin^{2}x}{\sin{x}\cos{x}}$
*
*Not sure where to go from here.
| You have simplified your initial expression to $2 \cot(x) - 2 \tan(x)$. Make use of the following trigonometric identities: $$\cot(x) = \dfrac1{\tan(x)}$$ $$\sec^2(x) - \tan^2(x) = 1$$ and simplify to get what you want.
Move your mouse over the gray area below for a complete solution.
$$2 \cot(x) - 2 \tan(x) = 2 \left( \dfrac1{\tan(x)} - \tan(x)\right) = 2 \left( \dfrac{1 - \tan^2(x)}{\tan(x)}\right)$$ Making use of this identity we have that $\tan^2(x) = \sec^2(x) - 1$. Plug this in your numerator and you that $$2 \cot(x) - 2 \tan(x) = 2 \left( \dfrac{1 - \sec^2(x) + 1}{\tan(x)}\right) = \dfrac{4 - 2 \sec^2(x)}{\tan(x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/232590",
"timestamp": "2023-03-29T00:00:00",
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Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Prove that
$$
\frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n,
\quad
\forall k\in\mathbb{N}.
$$
I tried already by induction over $k$ but i have problems showing the statement holds for $k=0$ or $k=1$.
| If we define
$$
f_n(t)=\sum_{k=0}^\infty\binom{2k+n}{k}t^k\tag{1}
$$
then we have
$$
\begin{align}
f_n'(t)
&=\sum_{k=0}^\infty\binom{2k+n}{k}kt^{k-1}\\
&=\sum_{k=0}^\infty\binom{2k+n-1}{k-1}(2k+n)t^{k-1}\\
&=\sum_{k=0}^\infty\binom{2k+n+1}{k}(2k+n+2)t^k\\[6pt]
&=(n+2)f_{n+1}(t)+2tf_{n+1}'(t)\tag{2}
\end{align}
$$
If we define
$$
\begin{align}
g_n(t)
&=\frac1{\sqrt{1-4t}}\left(\frac{1-\sqrt{1-4t}}{2t}\right)^n\\
&=\frac1{\sqrt{1-4t}}\left(\frac2{1+\sqrt{1-4t}}\right)^n\tag{3}
\end{align}
$$
then we have
$$
g_n'(t)
=\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+1}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\tag{4}
$$
and therefore
$$
\begin{align}
&(n+2)g_{n+1}(t)+2tg_{n+1}'(t)\\[9pt]
&=\frac{n+2}{\sqrt{1-4t}}\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\
&+2t\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+2}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+2}\\
&=\frac{n+2}{\sqrt{1-4t}}\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\
&+(1-\sqrt{1-4t})\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+2}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\
&=\left(\frac{1}{\sqrt{1-4t}^3}+\frac{n+1}{1-4t}\right)\left(\frac2{1+\sqrt{1-4t}}\right)^{n+1}\\[9pt]
&=g_n'(t)\tag{5}
\end{align}
$$
Equations $(2)$ and $(5)$ ensure that
$$
\begin{align}
f_n'(t)&=(n+2)f_{n+1}(t)+2tf_{n+1}'(t)=2t^{-\frac n2}\left[t^{\frac{n+2}{2}}f_{n+1}(t)\right]'\\
g_n'(t)&=(n+2)g_{n+1}(t)+2tg_{n+1}'(t)=2t^{-\frac n2}\left[t^{\frac{n+2}{2}}g_{n+1}(t)\right]'
\end{align}\tag{6}
$$
Furthermore, it follows from $(1)$ and $(3)$ that
$$
f_n(0)=g_n(0)=1\tag{7}
$$
The generalized binomial theorem yields
$$
\begin{align}
(1-4t)^{-1/2}
&=1+\frac124\frac{t}{1!}+\frac12\frac324^2\frac{t^2}{2!}+\frac12\frac32\frac524^3\frac{t^3}{3!}+\dots\\
&=\sum_{k=0}^\infty\frac{(2k-1)!!}{k!}2^kt^k\\
&=\sum_{k=0}^\infty\frac{(2k)!}{2^kk!k!}2^kt^k\\
&=\sum_{k=0}^\infty\binom{2k}{k}t^k\tag{8}
\end{align}
$$
Equation $(8)$ ensures that $f_0(t)=g_0(t)$.
Equations $(6)$, $(7)$, and $(8)$ ensure that
$$
f_n(t)=g_n(t)\tag{9}
$$
for all $n\ge0$. That is,
$$
\frac1{\sqrt{1-4t}}\left(\frac{1-\sqrt{1-4t}}{2t}\right)^n
=\sum_{k=0}^\infty\binom{2k+n}{k}t^k\tag{10}
$$
| {
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"url": "https://math.stackexchange.com/questions/237810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 6,
"answer_id": 0
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numbers' pattern It is known that
$$\begin{array}{ccc}1+2&=&3 \\ 4+5+6 &=& 7+8 \\
9+10+11+12 &=& 13+14+15 \\\
16+17+18+19+20 &=& 21+22+23+24 \\\
25+26+27+28+29+30 &=& 31+32+33+34+35 \\\ldots&=&\ldots
\end{array}$$
There is something similar for square numbers:
$$\begin{array}{ccc}3^2+4^2&=&5^2 \\ 10^2+11^2+12^2 &=& 13^2+14^2 \\ 21^2+22^2+23^2+24^2 &=& 25^2+26^2+27^2 \\ \ldots&=&\ldots
\end{array}$$
As such, I wonder if there are similar 'consecutive numbers' for cubic or higher powers.
Of course, we know that there is impossible for the following holds (by Fermat's last theorem):
$$k^3+(k+1)^3=(k+2)^3 $$
| There is the famous $3^3+4^3+5^3=6^3$. But I don't know if there are others.
At least not where the whole sequence is consecutive...
One gets the problem of three triangular numbers T1 T2 T3 for which one wants to solve $T1^2+T2^2=2 \cdot T3^2$.
| {
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"url": "https://math.stackexchange.com/questions/238532",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Find the sum of the first $n$ terms of $\sum^n_{k=1}k^3$ The question:
Find the sum of the first $n$ terms of $$\sum^n_{k=1}k^3$$
[Hint: consider $(k+1)^4-k^4$]
[Answer: $\frac{1}{4}n^2(n+1)^2$]
My solution:
$$\begin{align}
\sum^n_{k=1}k^3&=1^3+2^3+3^3+4^3+\cdots+(n-1)^3+n^3\\
&=\frac{n}{2}[\text{first term} + \text{last term}]\\
&=\frac{n(1^3+n^3)}{2}
\end{align}$$
What am I doing wrong?
| $$(k+1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1$$
Take sum from k=1 to n
$$\sum_{k=1}^n(k+1)^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n $$
$$\sum_{k=2}^{n+1}k^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n $$
Add 1 on both sides
$$\sum_{k=1}^{n}k^4 + (n+1)^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n + 1$$
Cancel $\sum_{k=1}^{n}k^4$
$$ (n+1)^4 = \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n + 1$$
You need $\sum_{k=1}^n k^2, \sum_{k=1}^n k$ to solve it. Hope you can do the rest.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Finding eigenvalues and eigenvectors, $2\times2$ matrices
Find all eigenvalues and eigenvectors:
a.) $\pmatrix{i&1\\0&-1+i}$
b.) $\pmatrix{\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta}$
For a I got:
$$\operatorname{det} \pmatrix{i-\lambda&1\\0&-1+i-\lambda}= \lambda^{2} - 2\lambda i + \lambda - i - 1
$$
For b I got:
$$\operatorname{det} \pmatrix{\cos\theta - \lambda & -\sin\theta \\ \sin\theta & \cos\theta - \lambda}= \cos^2\theta + \sin^2\theta + \lambda^2 -2\lambda \cos\theta = \lambda^2 -2\lambda \cos\theta +1$$
But how can I find the corresponding eigenvalues for a and b?
| Basic tools
For $2 \times 2$ matrices the characteristic polynomial is:
$$
p(\lambda) = \lambda^{2} - \lambda\, \text{tr }\mathbf{A} + \det \mathbf{A}
$$
The roots of this function are the eigenvalues, $\lambda_{k}$, k=1,2$.
The eigenvectors solve the eigenvalue equation
$$
\mathbf{A} u_{k} = \lambda_{k} u_{k}
$$
Case 1
$$
\mathbf{A} =
%
\left(
\begin{array}{cc}
i & 1 \\
0 & -1+i \\
\end{array}
\right)
$$
The trace and determinant are
$$
\text{tr } \mathbf{A} = -1 + 2 i, \qquad \det \mathbf{A} = -1 - i
$$
The characteristic polynomial is
$$
p(\lambda) = \lambda ^2+(1-2 i) \lambda +(-1-i)
$$
The roots of this polynomial are the eigenvalues:
$$ \lambda \left( \mathbf{A} \right) = \left\{
i, -1 + i
\right\}
$$
First eigenvector:
$$
\begin{align}
\left( \mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) u_{1} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{cc}
i & 1 \\
0 & -1+i \\
\end{array}
\right) -
\left( -1 + i \right)
%
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\left(
\begin{array}{cc}
u_{x} + u_{y}\\
0 \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right)
%
\end{align}
$$
The result is
$$
u_{1} =
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
=
\alpha
\left(
\begin{array}{r}
1 \\
-1 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$
Second eigenvector:
$$
\begin{align}
\left( \mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) u_{2} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{cr}
0 & 1 \\
0 & -1 \\
\end{array}
\right)
%
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\end{align}
$$
The result is
$$
u_{2} =
\alpha
\left(
\begin{array}{c}
1 \\
0 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$
Case 2
$$
\mathbf{A} =
%
\left(
\begin{array}{cr}
\cos (\theta ) & -\sin (\theta ) \\
\sin (\theta ) & \cos (\theta ) \\
\end{array}
\right)
$$
The trace and determinant are
$$
\text{tr } \mathbf{A} = 2 \cos \theta, \qquad \det \mathbf{A} = \cos^{2} \theta + \sin^{\theta} = 1
$$
The characteristic polynomial is
$$
p(\lambda) = \lambda ^2+(1-2 i) \lambda +(-1-i)
$$
The roots of this polynomial are the eigenvalues:
$$ \lambda \left( \mathbf{A} \right) = \left\{
\cos \theta -i \sin \theta ,\cos \theta +i \sin \theta
\right\}
$$
First eigenvector:
$$
\begin{align}
\left( \mathbf{A} - \lambda_{1} \mathbf{I}_{2} \right) u_{1} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{cr}
i \sin \theta & -\sin \theta \\
\sin \theta & i \sin \theta \\
\end{array}
\right)
%
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\left(
\begin{array}{cc}
-u_{y} \sin \theta +i u_{x} \sin \theta \\
u_{x} \sin \theta +i u_{y} \sin \theta
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right)
%
\end{align}
$$
The result is
$$
u_{1} =
\left(
\begin{array}{cc}
u_{x} \\
u_{y} \\
\end{array}
\right)
=
\alpha
\left(
\begin{array}{r}
i \\
-1 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$
Second eigenvector:
$$
\begin{align}
\left( \mathbf{A} - \lambda_{2} \mathbf{I}_{2} \right) u_{2} &= \mathbf{0} \\[3pt]
%
\left(
\begin{array}{rr}
-i \sin \theta & -\sin \theta \\
\sin \theta & -i \sin \theta \\
\end{array}
\right)
%
\left(
\begin{array}{c}
u_{x} \\
u_{y} \\
\end{array}
\right)
&=
\left(
\begin{array}{c}
0 \\
0 \\
\end{array}
\right) \\
%
\end{align}
$$
The result is
$$
u_{2} =
\alpha
\left(
\begin{array}{c}
i \\
1 \\
\end{array}
\right), \quad \alpha \in \mathbb{C}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/241612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Value of $\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ I remember that a couple of years ago a friend showed me and some other people the following expression:
$$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$
As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:
What is the value of this limit and how do we prove it rigorously?
(Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)
I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.
(1) The sequence $a_n$ satisfies $a_n<1$ for all $n\ge 2$.
Note that $a_n<1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n<n^n$. I prove this second statement by induction. Observe that $1^2=1<4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n<n^n$ for some integer $n\ge 2$. Then
$$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}<(n-1)n^n+n^{n+1}<(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$
(2) The sequence $a_n$ is increasing for all $n\ge 2$.
We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)
(2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.
We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By Bernoulli's Inequality, we may find:
$$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$
A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.
We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:
$$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$
By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:
$$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}<\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$
(In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.
I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.
Thanks!
| We can prove this by simple squeezing.
Using the fact that $A>B>0$ implies $A^n-B^n \leq n(A-B)A^{n-1}$, and the facts that $e^{-x}\leq\frac{1}{1+x},e^x\geq 1+x+\frac{x^2}{2}$, we have:
$$ e^{-k} - \left(1-\frac{k}{n}\right)^n \leq n\left(e^{-k/n}-1+\frac{k}{n}\right)e^{-k\,\frac{n-1}{n}}\leq e\cdot\frac{k^2 e^{-k}}{n+k}\leq \frac{2e}{n+k},$$
so:
$$\sum_{k=1}^{n}\left(e^{-k}-\left(1-\frac{k}{n}\right)^n\right)\leq \frac{2e}{n}\sum_{k=1}^{n}\frac{1}{1+k/n}\leq\frac{2e}{n}\int_{1}^{2}\frac{dx}{1+x}<\frac{9}{4n},$$
then:
$$\lim_{n\to+\infty}\sum_{k=1}^{n}\left(1-\frac{k}{n}\right)^n=\lim_{n\to+\infty}\sum_{k=1}^{n}e^{-k} = \frac{e}{e-1}, $$
no need to use any advanced technique.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/244657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Find an angle in a given triangle $\triangle ABC$ has sides $AC = BC$ and $\angle ACB = 96^\circ$. $D$ is a point in $\triangle ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$. What is the measure (in degrees) of $\angle ACD$?
| $\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$
As $\angle DAB=18^\circ, \angle DAC=(42-18)^\circ=24^\circ, $
Similarly, $\angle CBD=18^\circ$
Let $\angle ACD=x,$ so, $\angle DCB=96^\circ-x$
So, in $\triangle ADC, \angle ADC=180^\circ-(24^\circ+x)=156^\circ-x$
Similarly, from $\triangle BCD, \angle BDC=72^\circ-x$
Applying sine law in $\triangle BCD,$ $$\frac {CD}{\sin 12^\circ}=\frac {BC}{\sin(72^\circ+x)}$$
Similarly, from $\triangle ADC,$ $$\frac {CD}{\sin 24^\circ}=\frac {AC}{\sin(156^\circ-x)}$$
On division, $$\frac{\sin 24^\circ}{\sin 12^\circ}=\frac{\sin(156^\circ-x)}{\sin(72^\circ+x)}$$ as $AC=BC$
But $\sin(156^\circ-x)=\sin(180^\circ-(24^\circ+x))=\sin(24^\circ+x)$ and $\sin 24^\circ=2\sin 12^\circ\cos 12^\circ$
So, $$2\cos 12^\circ=\frac{\sin(24^\circ+x)}{\sin(72^\circ+x)}$$
Applying $2\sin A\cos B=\sin(A+B)+\sin(A-B),$
$$\sin(84^\circ+x)+\sin(60^\circ+x)=\sin(24^\circ+x)$$
or applying $\sin(A+B)$ formula and separating sine and cosine, $$\sin x (\cos24^\circ-\cos84^\circ-\cos60^\circ)=\cos x(\sin84^\circ-\sin24^\circ+\sin60^\circ)$$
$$\tan x=\frac{\sin84^\circ-\sin24^\circ+\sin60^\circ}{\cos24^\circ-\cos84^\circ-\cos60^\circ}=\frac{2\sin30^\circ\cos54^\circ+\sin60^\circ}{2\sin30^\circ\sin54^\circ-\cos60^\circ}$$ (applying $\sin C-\sin D,\cos C-\cos D$ formula)
$$=\frac{\cos54^\circ+\cos30^\circ}{\sin54^\circ-\sin30^\circ}=\frac{2\cos42^\circ\cos12^\circ}{2\cos42^\circ\sin12^\circ}$$
(applying $\sin C-\sin D,\cos C+\cos D$ formula)
$$=\cot 12 ^\circ=\tan(90-12)^\circ$$
$\implies x=78^\circ$ as $0<x<180^\circ$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/245608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Arc Length Problem: Assimilation of A Constant The problem I am working on is,
I understand the motivation of the step, but I am not quite sure how the author the solution manual absorbs the constant into the other term. What is the process?
EDIT:
Here is another one:
Here is my attempt to try and figure out what they are doing:
$1+[y']^2=[\frac{1}{2}(e^x-e^-x)]+1$
$1+[y']^2=[\frac{1}{2}(e^x-e^-x)+1]^2$
$1+[y']^2=[\frac{1}{2}(e^x-e^-x+2)]^2$ This is where I get stuck, because I can't truly see any sort of manipulation that would get rid of the 2 and put a plus sign between the exponential functions.
| $\quad\quad\quad$First question: $$1 + \frac14\left(x^8 - 2 +\frac{1}{x^8}\right)\tag{1.1}$$
$$=\frac14\cdot 4 + \frac14\left(x^8 - 2 + +\frac{1}{x^8}\right)\quad=\quad\frac14\left(4 + x^8 -2 +\frac{1}{x^8}\right)\tag{1.2}$$
$$=\frac14\left(x^8 + 2 + \frac{1}{x^8}\right)\tag{1.3}$$
$$=\frac14\left(x^4 + \frac{1}{x^4}\right)^2\tag{1.4}$$
$$(1.3) \to (1.4):\quad\quad\text{Note that}$$
$$\frac14 \left(x^4 + \frac{1}{x^4}\right)^2\tag{1.4}$$ $$= \frac14\left(x^{2\cdot 4} + 2\frac{x^4}{x^4} + \frac{1}{x^{2\cdot 4}}\right) = \frac14\left(x^8 + 2 + \frac{1}{x^8}\right)\tag{1.3}$$
For your second question: in your work you have:
$$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})]+1\tag{a}$$
$$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})+1]^2\tag{b}$$
$$1+[y']^2=[\frac{1}{2}(e^x-e^{-x}+2)]^2\tag{c}$$
How did you get from $(a)\to(b)$? Did you forget the exponent in $(a)$? Shouldn't $(a)$ be $$1+[y']^2=[\frac{1}{2}(e^x-e^{-x})]^2+1\quad?$$And if so, you cannot bring the constant term $1$ into an expression that is exponentiated without first making appropriate computations on the exponentiation expression. Failing that, your move from $(b)\to (c)$ is affected.
In this case, first expand $\left[\dfrac{1}{2}(e^x-e^{-x})\right]^2$, then worry about adding the $1$ term:
$$1+[y']^2=\left[\frac{1}{2}(e^x-e^{-x})\right]^2+1\tag{2.1}$$
$$=\frac14\left(e^{2x} -2\frac{e^{2x}}{e^{2x}} + \frac{1}{e^{2x}}\right) + 1\tag{2.2}$$
$$=\frac14\left(e^{2x} -2 + \frac{1}{e^{2x}}\right) + \frac14\cdot4
\quad=\quad\frac14\left(e^{2x} -2 +\frac{1}{e^{2x}}+4\right)\tag{2.3}$$ $$= \left(\frac12\right)^2\left(e^{2x} +2 +\frac{1}{e^{2x}}\right)\quad=\quad\left[\frac12(e^x + e^{-x})\right]^2\tag{2.4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/245658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An upper bound for $\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i}$? Does anyone know of a reasonable upper bound for the following:
$$\sum_{i = 1}^m \frac{\binom{i}{k}}{2^i},$$
where we $k$ and $m$ are fixed positive integers, and we assume that $\binom{i}{k} = 0$ whenever $k > i$.
One trivial upper bound uses the identity $\binom{i}{k} \le \binom{i}{\frac i2}$, and the fact that $\binom{i}{\frac{i}{2}} \le \frac{2^{i+1}}{\sqrt{i}}$, to give a bound of
$$2\sum_{i = 1}^m \frac{1}{\sqrt{i}},$$
where $\sum_{i = 1}^m \frac{1}{\sqrt{i}}$ is upper bounded by $2\sqrt{m}$, resulting in a bound of $4\sqrt{m}$. Can we do better?
Thanks!
Yair
| I like Pot's argument better, but here's another approach for those who may be interested.
I get an upper bound of $2$, independent of $k$ and $m$.
Applying summation by parts, we have, for $k \geq 2$,
$$\begin{align*}
\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i} &= \binom{m}{k} \sum_{i=1}^m \frac{1}{2^i} - \sum_{i=1}^{m-1} \left(\binom{i+1}{k} - \binom{i}{k}\right) \sum_{j=1}^i \frac{1}{2^j} \\
&= \binom{m}{k} \left(1 - \frac{1}{2^m}\right) - \sum_{i=1}^{m-1} \binom{i}{k-1} \left(1 - \frac{1}{2^i}\right)\\
&= \binom{m}{k} - \binom{m}{k}\frac{1}{2^m} - \sum_{i=1}^{m-1} \binom{i}{k-1} + \sum_{i=1}^{n-1} \binom{i}{k-1} \frac{1}{2^i}\\
&= \binom{m}{k} - \binom{m}{k}\frac{1}{2^m} - \binom{m}{k} + \sum_{i=1}^{m-1} \binom{i}{k-1} \frac{1}{2^i}\\
&= \sum_{i=1}^{m-1} \binom{i}{k-1} \frac{1}{2^i}- \binom{m}{k}\frac{1}{2^m},
\end{align*}$$
where, in the second-to-last step, we use the upper summation identity for the binomial coefficients, $\displaystyle \sum_{i=0}^m \binom{i}{k} = \binom{m+1}{k+1}$.
Letting $\displaystyle F(m,k) = \sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i}$, this means we have the recurrence $$F(m,k) = F(m-1,k-1) - \binom{m}{k}\frac{1}{2^m},$$ valid for $k \geq 2$.
Unrolling the recurrence is easy, and with $F(m-k+1,1) = 2 - \frac{m-k}{2^{m-k+1}} - \frac{3}{2^{m-k+1}},$ we have
$$\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i} = 2 - \frac{m-k}{2^{m-k+1}} - \frac{3}{2^{m-k+1}} - \sum_{i=0}^{k-2} \binom{m-i}{k-i} \frac{1}{2^{m-i}}.$$
Therefore,
$$\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i} \leq 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/247261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Rationalizing the denominator of $\frac {\sqrt{10}}{\sqrt{5} -2}$ I have the expression
$$\frac {\sqrt{10}}{\sqrt{5} -2}$$
I can't figure out what to do from here, I can't seem to pull any numbers out of either of the square roots so it appears that it must remain as is.
| To rationalize the denominator i. e. turn the denominator rational multiply both numerator and denominator by
the conjugate of the denominator $-\sqrt{5}-2$ or its symmetric $\sqrt{5}+2$, expand both and
simplify
$$\begin{eqnarray*}
\frac{\sqrt{10}}{\sqrt{5}-2} &=&\frac{\sqrt{10}\left( \sqrt{5}+2\right) }{
\left( \sqrt{5}-2\right) \left( \sqrt{5}+2\right) }=\frac{\sqrt{10}\sqrt{5}+\sqrt{10}\times 2}{\left( \sqrt{5}\right) ^{2}-2^{2}} \\
&=&\frac{\sqrt{50}+2\sqrt{10}}{5-4}=\frac{\sqrt{50}+2\sqrt{10}}{1}=\sqrt{50}%
+2\sqrt{10}.
\end{eqnarray*}.
$$
In general [Edited to correct] $$\frac{1}{a+\sqrt{b}}=\frac{a-\sqrt{b}}{\left( a+\sqrt{b}\right)
\left( a-\sqrt{b}\right) }=\frac{a-\sqrt{b}}{a^{2}-b}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/249563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to simplify polynomials I can't figure out how to simplify this polynominal $$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$
I tried combining like terms
$$5x^2+3x^4-7x^3+5x+8+2x^2-4x+9-6x^2+7x$$
$$(5x^2+5x)+3x^4-(7x^3+7x)+2x^2-4x-6x^2+(8+9)$$
$$5x^3+3x^4-7x^4+2x^2-4x-6x^2+17$$
It says the answer is $$3x^4-7x^3+x^2+8x+17$$ but how did they get it?
| Observe the magical power of color:
$$\color{blue}{5}x^\color{blue}{2}+3x^4-7x^3+\color{green}{5}x+\color{orange}{8}+\color{blue}{2}x^\color{blue}{2}+(\color{green}{-4})x+\color{orange}{9}+(\color{blue}{-6})x^\color{blue}{2}+\color{green}{7}x.$$
Instead of Color-Me-Elmo, we have Color-Me-Like-Terms-And-Combine (not as catchy, I know):
$$3x^4-7x^3+(\color{blue}{5}+\color{blue}{2}+(\color{blue}{-6}))x^\color{blue}{2}+(\color{green}{5}+(\color{green}{-4})+\color{green}{7})x+(\color{orange}{8}+\color{orange}{9}).$$
Presto-simplification-o!
Combining Like Terms
In a polynomial $p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$ and $q(x)=b_nx^n+b_{n-1}x^{n-1}+\dots+b_1x+b_0$, they are added thusly:
$$
\begin{align}
p(x)+q(x)&=a_nx^n+b_nx^n+a_{n-1}x^{n-1}+b_{n-1}x^{n-1}+\cdots+a_1x+b_1x+a_0+b_0\\
&=(a_n+b_n)x^n+(a_{n-1}+b_{n-1})x^{n-1}+\cdots+(a_1+b_1)x+(a_0+b_0).
\end{align}
$$
In other words, add the coefficients of terms with the same power.
| {
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"url": "https://math.stackexchange.com/questions/251172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Sum of infinite series with arctan: $\sum_{n=1}^{\infty}\left(\arctan\left(\frac{1}{4}-n\right)-\arctan\left(-\frac{1}{4}-n\right)\right)$ I'm trying to find the value of the following sum (if exist):
$$\sum_{n=1}^{\infty}\left(\arctan\left(\frac{1}{4}-n\right)-\arctan\left(-\frac{1}{4}-n\right)\right)$$
where, $\arctan$ represent the inverse tangent function - $\tan^{-1}$.
I tried to use the telescoping series idea and the sequence of partial sums but I couldn't cancel any terms!
| Consider
$$\tag{1}f(x):=\sum_{n=1}^\infty \arctan\left(\left(\frac 14-n\right)x\right)-\arctan\left(\left(-\frac 14-n\right)x\right)$$
and let's rewrite the derivative of $f$ :
\begin{align}
\tag{2}f'(x)&=\frac 4{x^2}\sum_{n=1}^\infty\frac{1-4n}{(4n-1)^2+\bigl(\frac 4x\bigr)^2}-\frac{-1-4n}{(4n+1)^2+\bigl(\frac 4x\bigr)^2}\\
\tag{3}f'(x)&=\frac 4{x^2}\left(\frac {-1}{1^2+\left(\frac 4x\right)^2}+\sum_{k=1}^\infty\frac{k\sin\bigl(k\frac {\pi}2\bigr)}{k^2+\bigl(\frac 4x\bigr)^2}\right)\\
\end{align}
(since the $k=1$ term didn't appear in $(2)$)
But the series in $(3)$ may be obtained from $\,\frac d{d\theta} C_a(\theta)\,$ with :
$$\tag{4}C_a(\theta)=\frac {\pi}{2a}\frac{\cosh((\pi-|\theta|)a)}{\sinh(\pi a)}-\frac 1{2a^2}=\sum_{k=1}^\infty\frac{\cos(k\,\theta)}{k^2+a^2}$$
which may be obtained from the $\cos(zx)$ formula here (with substitutions $\ x\to\pi-\theta,\ z\to ia$).
The replacement of the series in $(3)$ by $\,\frac d{d\theta} C_a(\theta)\,$ applied at $\,\theta=\frac {\pi}2$ gives us :
\begin{align}
f'(x)&=\left(-\arctan\left(\frac x4\right)\right)'-\frac 4{x^2}C_{\frac 4x}\left(\theta\right)'_{\theta=\frac {\pi}2}\\
&=\left(-\arctan\left(\frac x4\right)\right)'+\frac{4\pi}{2x^2}\frac{\sinh\left(\frac{\pi}2 \frac 4x\right)}{\sinh\left(\pi\frac 4x\right)}\\
&=\left(-\arctan\left(\frac x4\right)\right)'+\frac{\pi}{x^2}\frac 1{\cosh\left(\frac{2\pi}x\right)}\\
\end{align}
Integrating both terms returns (with constant of integration $\frac {\pi}2$ since $f(0)=0$) :
$$f(x)=\frac {\pi}2-\arctan\left(\frac x4\right)-\arctan\left(\tanh\left(\frac {\pi}x\right)\right)\quad\text{for}\ \ x>0$$
i.e. the neat :
$$\tag{5}\boxed{\displaystyle f(x)=\arctan\left(\frac 4x\right)-\arctan\left(\tanh\left(\frac {\pi}x\right)\right)}\quad\text{for}\ \ x>0$$
So that your solution will be (for $x=1$) :
$$\boxed{\displaystyle \arctan(4)-\arctan\left(\tanh(\pi)\right)}$$
| {
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"url": "https://math.stackexchange.com/questions/252780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Show that the expression, $\frac{(n)(n+1)}{2}$ will never yield a multiple of $n$ for even values of $n$ where $n \neq 0$. $F(n) =\frac{(n)(n+1)}{2}$
Show that $F(n)$ will never yield a multiple of $n$ for even values of $n$ where $n \neq 0$.
Let, $n = 2$
$F(2) =\frac{2(2+1)}{2}$
$F(2) = 3$
Let, $n = 4$
$F(4) =\frac{4(4+1)}{2}$
$F(4) = (2)(5)$
| $n$ and $n+1$ are relatively prime, or perhaps more simply, if $n$ is even, then $n+1$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/256912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\lim\limits_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)$ Evaluate
$$
\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)
$$
The answer is $\frac{1}{2}$, have no idea how to arrive at that.
| Alternatively:
$$ \sqrt{\frac{x^3}{x-1}} - x = x\sqrt{\frac{x}{x-1}} - x = x\left(\sqrt{1+\frac{1}{x-1}}-1\right) $$
For small $\alpha$ we have $\sqrt{1+\alpha}\approx 1+\alpha/2$, so for large $x$ we get
$$\cdots \approx x\left(1+\frac{1}{2x-2}-1\right) = \frac{x}{2x-2} \to \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to find $(-64\mathrm{i}) ^{1/3}$? How to find $$(-64\mathrm{i})^{\frac{1}{3}}$$
This is a complex variables question.
I need help by show step by step.
Thanks a lot.
| If you transform $-64i$ to polar form, you get $r=\sqrt{0+(-64)^2}=64$ and $\theta=-\pi/2$.
Then you have $$(-64i)^{1/3} = r^{1/3}\cdot (\cos(\theta*\frac{1}{3})+i\sin(\theta*\frac{1}{3})) = 64^{1/3}\cdot (\cos((-\pi/2)*\frac{1}{3})+i\sin((-\pi/2)*\frac{1}{3})$$
$$= 4\cdot (\cos(-\pi/6)+i\sin(-\pi/6))$$
Given that
$$\cos(-\pi/6)=\frac{\sqrt{3}}{2}$$
and
$$\sin(-\pi/6) = -\frac{1}{2}$$
We have:
$$4\cdot (\cos(-\pi/6)+i\sin(-\pi/6)) = 4\cdot (\frac{\sqrt{3}}{2}-\frac{1}{2}i) = 2\sqrt{3}-2i$$
The other roots can be found by adding $2\pi$ and $4\pi$ to $\theta$.
So,
$$4\cdot (\cos((\theta+2\pi)\cdot \frac{1}{3})+i\sin((\theta+2\pi)\cdot \frac{1}{3})) =4i$$
and
$$4\cdot (\cos((\theta+4\pi)\cdot \frac{1}{3})+i\sin((\theta+4\pi)\cdot \frac{1}{3})) = -2\sqrt{3}-2i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to integrate this $\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}} $ without trigonometric substitution? I have been looking for a possible solution but they are with trigonometric integration..
I need a solution for this function without trigonometric integration
$$\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}}$$
| $$\int\frac{dx}{(a^2+x^2)^{\frac n2}}=\int1\cdot \frac1{(a^2+x^2)^{\frac n2}}dx$$
$$=\frac x{(a^2+x^2)^{\frac n2}}-\int\left(\frac{-n}2\frac{2x\cdot x}{(a^2+x^2)^{\frac n2+1}} \right) dx$$
$$=\frac x{(a^2+x^2)^{\frac n2}}+n\int \left(\frac{(a^2+x^2-a^2)}{(a^2+x^2)^{\frac n2+1}}\right)dx $$
$$=\frac x{(a^2+x^2)^{\frac n2}}+n\left(\int\frac{dx}{(a^2+x^2)^{\frac n2}}-a^2\int\frac{dx}{(a^2+x^2)^{\frac n2+1}}\right)+c $$ where $c$ is the constant for indefinite integration.
or, $$na^2\int\frac{dx}{(a^2+x^2)^{\frac n2+1}}=\frac x{(a^2+x^2)^{\frac n2}}+(n-1)\int\frac{dx}{(a^2+x^2)^{\frac n2}}+c$$
Putting $n=1,$ $$a^2\int\frac{dx}{(a^2+x^2)^{\frac 32}}=\frac x{(a^2+x^2)^{\frac 12}}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/260831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Using product and chain rule to find derivative. Find the derivative of
$$y =(1+x^2)^4 (2-x^3)^5$$
To solve this I used the product rule and the chain rule.
$$u = (1+x^2)^4$$
$$u' = 4 (1+x^2)^3(2x)$$
$$v= (2-x^3)^5$$
$$v' = 5(2-x^3)^4(3x^2)$$
$$uv'+vu'$$
$$((1+x^2)^4)(5(2-x^3)^4(3x^2)) + ((2-x^3)^5 )(4 (1+x^2)^3(2x))$$
The answer I got is: $$(15x^2)(1-x^2)^4(2-x^3)^4 + 8x(2-x^3)^5(1+x^2)^3$$.
Why is the answer $$8x(x^2 +1)^3(2-x^3)^5-15x^2(x^2)(X^2+1)^4(2-x^3)4$$?
How did the $15x^2$ become negative?
| The problem is in your differentiation of $$v= (2-x^3)^5$$
You have: $$v'= 5(2-x^3)^4(3x^2)$$
However, the derivative of $2-x^3$ is $-3x^2$. Thus, $$v' = 5(2-x^3)^4(-3x^2)=-15x^2(2-x^3)^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/261011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Working out digits of Pi. I have always wondered how the digits of π are calculated. How do they do it?
Thanks.
| From one of my favorite mathematicians,
$$
\frac{1}{\pi}=\frac{2\sqrt{2}}{9801}\sum_{k \ge 0} \frac{(4n)!(1103+26390n)}{(n!)^4396^{4n}}.
$$
Looking here, we see some many formulae. Some of particular interest are:
$$
\pi=\sum_{k \ge 0}\frac{3^k-1}{4^k}\zeta(k+1),\quad \zeta(s)=\sum_{k\ge 0}k^{-s}.
$$
$$
\frac{1}{6}\pi^2=\sum_{k \ge 1}\frac{1}{k^2} \quad \text{via the Basel problem}.
$$
$$
\pi =\frac{3}{4}\sqrt{3}+24\int_0^{\frac{1}{4}}\sqrt{x-x^2}dx.
$$
$$
\frac{\pi}{5\sqrt{\phi+2}}=\frac{1}{2}\sum_{k \ge 0}\frac{(k!)^2}{\phi^{2k+1}(2k+1)!},\quad \text{where } \phi \text{ is the golden ratio.}
$$
$$
\pi=\frac{22}{7}-\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx.
$$
$$
\pi=4\sum_{1 \le k \le n}\frac{(-1)^{j+1}}{2j-1}+\frac{(-1)^n}{(2n-1)!}\sum_{i \ge 0}\frac{1}{16^i}\left( \frac{8}{(8i+1)_{2n}}-\frac{4}{(8i+3)_{2n}}-\frac{4}{(8i+4)_{2n}}-\frac{2}{(8i+5)_{2n}}+\frac{1}{(8i+7)_{2n}}+\frac{1}{(8i+8)_{2n}} \right),\quad \text{where } n \in \mathbb{Z}_{>0}, \text{ and } (x)_{n} \text{ represents the Pochhammer symbol.}
$$
$$
\pi = 2\left[\prod_{n \ge 0}\left(1+\frac{\sin\left(\frac{1}{2}p_n\right)}{p_n}\right) \right]^{-1},\quad p_n \text{ is the } n\text{th} \text{ prime number}.
$$
$$
\frac{2}{\pi}=\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}\sqrt{\frac{1}{2}+ \frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}}\cdots
$$
$$
\frac{\pi}{2}=\prod_{n \ge 1}\frac{(2n)^2}{(2n-1)(2n+1)}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What are the A,B,C parameters of this ellipse formula? I am looking at
$$A(x − h)^2 + B(x − h)(y − k) + C(y − k)^2 = 1$$
This is a rotating ellipse formula, where $h,k$ are the centroid of the ellipse. I have tried looking around for $A,B,C$ parameters, and I see that they are from Quadratic formula. But to be frank, I want to see how $A,B,C$ impact the orientation of the ellipse and I am looking for a more visual explanation for this question.
| I would like to explain the transforms on a picture.
$$\frac{x'^2}{a^2}+\frac{y'^2}{b^2}=1$$
$$x=x'.\cos \alpha - y'. \sin \alpha $$
$$y=x'.\sin \alpha + y'. \cos \alpha $$
We can get easily the result above from the picture
$$\cos \alpha. x=\cos \alpha (x'.\cos \alpha - y'. \sin \alpha) $$
$$\sin \alpha. y=\sin \alpha (x'.\sin \alpha + y'. \cos \alpha) $$
$$x'=x.\cos \alpha + y. \sin \alpha $$
$$-\sin \alpha. x=\cos \alpha (x'.\cos \alpha - y'. \sin \alpha) $$
$$\cos \alpha. y=\sin \alpha (x'.\sin \alpha + y'. \cos \alpha) $$
$$y'=-x.\sin \alpha + y. \cos \alpha $$
$$\frac{(x.\cos \alpha + y. \sin \alpha)^2}{a^2}+\frac{(-x.\sin \alpha + y. \cos \alpha)^2}{b^2}=1$$
$$(\frac{\cos^2 \alpha }{a^2}+\frac{\sin^2 \alpha }{b^2})x^2+2(\frac{\cos \alpha \sin \alpha }{a^2}-\frac{\cos \alpha \sin \alpha }{b^2})xy+(\frac{\sin^2 \alpha }{a^2}+\frac{\cos^2 \alpha }{b^2})y^2=1$$
$$(\frac{\cos^2 \alpha }{a^2}+\frac{\sin^2 \alpha }{b^2})x^2+ \sin 2\alpha (\frac{1 }{a^2}-\frac{1}{b^2})xy+(\frac{\sin^2 \alpha }{a^2}+\frac{\cos^2 \alpha }{b^2})y^2=1$$
$$Ax^2+ Bxy+Cy^2=1$$
$x=X-h$
$y=Y-k$
$$A(X-h)^2+ B(X-h)(Y-k)+C(Y-k)^2=1$$
$$A=\frac{\cos^2 \alpha }{a^2}+\frac{\sin^2 \alpha }{b^2}$$
$$B=\sin 2\alpha (\frac{1 }{a^2}-\frac{1}{b^2})$$
$$C=\frac{\sin^2 \alpha }{a^2}+\frac{\cos^2 \alpha }{b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/264440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for
$a,b,c>0$
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$
What I tried is using substitution:
$p=a+b+c$
$q=ab+bc+ca$
$r=abc$
But I cannot reduce $a^2(b+c)(c+a)+b^2(a+b)(c+a)+c(a+b)(b+c) $ interms of $p,q,r$
| Cauchy-Swartz:
$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+ \cdots +\frac{a_n^2}{b_n} \geq \frac{(a_1+a_2+ \cdots +a_n)^2}{b_1+b_2+ \cdot +b_n}$
So,
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} \geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/264931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simple integral help How do I integrate $$\int_{0}^1 x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\}\, dx$$
Where $\lceil x \rceil $ is the ceiling function, and $\left\{x\right\}$ is the fractional part function
| The main idea is to divide $(0,1)$ into "good" intervals. I'lll give only the main steps of computation
$$
\int\limits_{(0,1)} x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\} dx
=\sum\limits_{n=1}^\infty\int\limits_{n\leq \frac{1}{x}<n+1} x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\} dx
=\sum\limits_{n=1}^\infty\int\limits_{\frac{1}{n+1}< x\leq \frac{1}{n}} x (n+1) \left(\frac{1}{x}-n\right) dx
=\sum\limits_{n=1}^\infty\frac{1}{2n^2+2n}=\frac{1}{2}\sum\limits_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/266110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
how many number like $119$ How many 3-digits number has this property like $119$:
$119$ divided by $2$ the remainder is $1$
119 divided by $3$ the remainderis $2 $
$119$ divided by $4$ the remainder is $3$
$119$ divided by $5$ the remainder is $4$
$119$ divided by $6$ the remainder is $5$
| Exercises:
Try dividing $420 - 1$ by integers $2, 3, ..., 7$ and note the respective remainders are $1, 2, 3, 4, 5, 6$.
Then note that $420 = \textrm{lcm}\,(2, 3, 4, 5, 6, 7)$
Any $n = k\cdot 420 - 1$, $k\in \mathbb{Z}$, will yield the same remainders when divided by $2, 3, ..., 7$.
Try dividing $840 - 1 = 839\,$ by $\,2,\, 3, \,4, \,5, \,6, \,7,\,8\,$ $\implies$ respective remainders of $\;$...?...
Then note that $840 = \textrm{lcm}\,(2, 3, 4, 5, 6, 7, 8)$.
So the same respective remainders are obtained, when $k\cdot 840 - 1$ is divided by each of $\{2, 3, ..., 8\}$.
Now: What\s the least positive number, that when divided by each of $\{2, 3, 4, 5, 6, 7, 8, 9\}$ gives a corresponding remainder of $\{1, 2, 3, 4, 5, 6, 7, 8\}$?
Find the $\text{lcm}\,(2, 3, 4, 5, 6, 7, 8, 9\} = 2^3\cdot 3^2\cdot 5 \cdot 7 = 840\cdot 3 = 2520.\;$
Subtract $1:\;\;$ $2520 - 1 = 2519 = n_9$.
Test by dividing $n_9$ by each of $2, 3, 4, 5, 6, 7, 8, 9$, to confirm...
There you go.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I find two independent solution for this ODE? Please help me find two independent solutions for $$3x(x+1)y''+2(x+1)y'+4y=0$$ Thanks from a new beginner into ODE's.
| We suppose that the $\displaystyle y=\sum_{n=0}^{\infty}a_{n}x^n$ is a solution.
Then
$$y=a_{0}+a_{1}x+...+a_{n}x^n+... $$
$$y'=a_{1}+2a_{2}x+3a_{3}x^2+4a_{4}x^3+...+(n+1)a_{n+1}x^n+...$$
$$ y''=2a_{2}+3.2a_{3}x+4.3a_{4}x^2+...+(n+2)(n+1)a_{n+2}x^n+... $$
replacing this in the DE,
$3x^2y''+3xy''+2xy'+2y'+4y=0$,
we have
general term
$$3(n+2)(n+1)a_{n+2}x^{n+2}+3(n+3)(n+2)a_{n+3}x^{n+2}+2(n+2)a_{n+2}x^{n+2}+2(n+3)a_{n+3}x^{n+2}+4a_{n+2}x^{n+2}$$
Then we must have
$$3(n+2)(n+1)a_{n+2}+3(n+3)(n+2)a_{n+3}+2(n+2)a_{n+2}+2(n+3)a_{n+3}+4a_{n+2}=0$$
or
$$[3(n+3)(n+2)+2(n+3)]a_{n+3}+[3(n+2)(n+1)+2(n+2)+4]a_{n+2}=0$$
Now you solve this difference equation of the first order to $a_{0}=0$,$a_{1}=1$ and $a_{0}=1, a_{1}=0$.
This give you two solution LI.
Note that $a_{0}$ and $a_{1}$ are the initial condition $y(0)$ and $y'(0)$
| {
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"timestamp": "2023-03-29T00:00:00",
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$ \displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$ $$ \ X_n=\frac{1}{\sqrt{n^3+1}}+\frac{2}{\sqrt{n^3+2}}+\cdots+\frac{n}{\sqrt{n^3+n}}$$ Find $\displaystyle\lim_{n\to\infty} X_n$ using the squeeze theorem
I tried this approach:
$$
\frac{1}{\sqrt{n^3+1}}\le\frac{1}{\sqrt{n^3+1}}<\frac{n}{\sqrt{n^3+1}}
$$
$$
\frac{1}{\sqrt{n^3+1}}<\frac{2}{\sqrt{n^3+2}}<\frac{n}{\sqrt{n^3+1}}$$
$$\vdots$$
$$\frac{1}{\sqrt{n^3+1}}<\frac{n}{\sqrt{n^3+n}}<\frac{n}{\sqrt{n^3+1}}$$
Adding this inequalities:
$$\frac{n}{\sqrt{n^3+1}}\leq X_n<\frac{n^2}{\sqrt{n^3+1}}$$
And this doesn't help me much. How should i proced?
| Hint:
Use the fact that:
$$\sum_{i=1}^n{i}=\frac{n^2 +n}{2}$$
And:
$$\frac{1}{\sqrt{n^3+1}}\le\frac{i}{\sqrt{n^3+i}}\le\frac{i}{\sqrt{n^3+1}}$$
| {
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"url": "https://math.stackexchange.com/questions/270216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If both roots of the Quadratic Equation are similar then prove that If both roots of the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that $2a=b+c$.
Things should be known:
*
*Roots of a Quadratic Equations can be identified by:
The roots can be figured out by:
$$\frac{-b \pm \sqrt{d}}{2a},$$
where
$$d=b^2-4ac.$$
*When the equation has equal roots, then $d=b^2-4ac=0$.
*That means $d=(b-c)^2-4(a-b)(c-a)=0$
| The expression for $d$ factors as $(2a-b-c)^2$, so we must have $2a-b-c=0$.
The easiest way to see this is to let $x=a-b,y=c-a$ and note that $b-c=-x-y$: $d=(-x-y)^2-4xy=x^2+y^2+2xy-4xy=(x-y)^2$.
Alternative solution: Note that $(a-b)+(b-c)+(c-a)=0$, i.e. $x=1$ is a root of the equation. If both roots are equal, it means that the other root is also $1$, showing that $(a-b)x^2+(b-c)x+(c-a)=(a-b)(x-1)^2$.
Equating the constant term, we find: $c-a=a-b$, which means $2a=b+c$. (This can be done with the coefficient of $x$ too).
| {
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"timestamp": "2023-03-29T00:00:00",
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Recurrence relation by substitution I have an exercise where I need to prove by using the substitution method the following
$$T(n) = 4T(n/3)+n = \Theta(n^{\log_3 4})$$
using as guess like the one below will fail, I cannot see why, though, even if I developed the substitution
$$T(n) ≤ cn^{\log_3 4}$$
finally, they ask me to show how to substract off a lower-order term to make a substitution proof work. I was thiking about using something like
$$T(n) ≤ cn^{\log_3 4}-dn$$
but again, I cannot see how to verify this recurrence.
What I did:
$$T(n) = 4T(n/3)+n$$
$$\qquad ≤ \frac{4c}{3}n^{\log_3 4} + n$$
and then from here, how to proceed and conclude that the first guess fails?
The complete exercise says:
Using the master method, you can show that the solution to the recurrence $T(n) = 4T(n/3)+n$ is $T(n)=\Theta(n^{log_3 4})$. Show that a substitution proof with the assumption $T(n) ≤ cn^{log_3 4}$ fails. Then show how to subtract off a lower-order term to make a substitution proof work.
| This recurrence has the nice property that we can give an explicit value for all $n$, not just powers of three. Let $$ n = \sum_{k=0}^{\lfloor \log_3 n \rfloor} d_k 3^k$$ be the representation of $n$ in base three. With $T(0) = 0$, we have by inspection
$$ T(n) = \sum_{j=0}^{\lfloor \log_3 n \rfloor} 4^j
\sum_{k=j}^{\lfloor \log_3 n \rfloor} d_k 3^{k-j} =
\sum_{j=0}^{\lfloor \log_3 n \rfloor} 4^j
\sum_{k=0}^{\lfloor \log_3 n \rfloor - j } d_{k+j} 3^k
.$$
For a lower bound, consider those $n$ that consist of a leading one digit followed by zeroes. This gives
$$ T(n) \ge \sum_{j=0}^{\lfloor \log_3 n \rfloor} 4^j 3^{\lfloor \log_3 n \rfloor - j} =
3^{\lfloor \log_3 n \rfloor} \sum_{j=0}^{\lfloor \log_3 n \rfloor}
\left(\frac{4}{3}\right)^j $$ which is $$ 3^{\lfloor \log_3 n \rfloor}
\frac{(4/3)^{1+ \lfloor \log_3 n \rfloor}-1}{4/3-1} =
4^{1+ \lfloor \log_3 n \rfloor} - 3^{1+ \lfloor \log_3 n \rfloor}.$$
For an upper bound, consider those $n$ that consist entirely of digits with value two.
$$ T(n) \le 2 \sum_{j=0}^{\lfloor \log_3 n \rfloor} 4^j
\sum_{k=0}^{\lfloor \log_3 n \rfloor - j } 3^k =
2 \sum_{j=0}^{\lfloor \log_3 n \rfloor} 4^j
\frac{3^{1+\lfloor \log_3 n \rfloor - j}-1}{3-1} <
\sum_{j=0}^{\lfloor \log_3 n \rfloor} 4^j 3^{1+\lfloor \log_3 n \rfloor - j}
$$ which is
$$3^{1+ \lfloor \log_3 n \rfloor} \sum_{j=0}^{\lfloor \log_3 n \rfloor}
\left(\frac{4}{3}\right)^j =
3^{1+\lfloor \log_3 n \rfloor}
\frac{(4/3)^{1+ \lfloor \log_3 n \rfloor}-1}{4/3-1} =
3 \times 4^{1+ \lfloor \log_3 n \rfloor} - 3 \times 3^{1+ \lfloor \log_3 n \rfloor}$$
What we have shown here is that for all $n$,
$$ T(n) \in
\Theta\left(4^{1+ \lfloor \log_3 n \rfloor} - 3^{1+ \lfloor \log_3 n \rfloor}\right) =
\Theta\left(4^{1+ \lfloor \log_3 n \rfloor}\right) =
\Theta\left(4^{\lfloor \log_3 n \rfloor}\right).$$
But $$4^{\lfloor \log_3 n \rfloor} \le
4^{\log_3 n} = 4^{\frac{\log_4 n}{\log_4 3}} =
n^{\frac{1}{\log_4 3}} = n^{\log_3 4}$$
and similarly
$$4^{\lfloor \log_3 n \rfloor} >
4^{\log_3 n -1} = \frac{1}{4} 4^{\frac{\log_4 n}{\log_4 3}} =
\frac{1}{4}n^{\frac{1}{\log_4 3}} = \frac{1}{4}n^{\log_3 4}$$
so that finally
$$ T(n) \in \Theta\left(n^{\log_3 4}\right).$$
| {
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How do I prove by induction that, for $n≥1, \sum_{r=1}^n \frac{1}{r(r+1)}=\frac{n}{n+1}$? Hi can you help me solve this:
I have proved that $p(1)$ is true and am now assuming that $p(k)$ is true. I just don't know how to show $p(k+1)$ for both sides?
| If you want to do it by a conventional "blind" induction, suppose that for a certain $k$ we have
$$\sum_1^k \frac{1}{r(r+1)}=\frac{k}{k+1}.\tag{$1$}$$
We want to prove that
$$\sum_1^{k+1} \frac{1}{r(r+1)}=\frac{k+1}{k+2}.\tag{$2$}$$
Note that the left-hand side of $(2)$ is the left-hand side of $(1)$, plus $\dfrac{1}{(k+1)(k+2)}$.
So we want to prove that
$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{k+2}.\tag{$3$}$$
It seems reasonable to manipulate the left-hand side of $(3)$ and see whether we get the right-hand side. A common manipulation is to bring the expression to the common denominator $(k+1)(k+2)$. We get
$$\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}.$$
This is equal to $\dfrac{k^2+2k+1}{(k+1)(k+2)}$.
But the numerator is equal to $(k+1)^2$. Cancel a $k+1$.
Remark: The algebra at the end is neater, and closer to the informal "telescoping" argument, if we observe that $\dfrac{1}{(k+1)(k+2)}=\dfrac{1}{k+1}-\dfrac{1}{k+2}$. Thus
$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+ \frac{1}{k+1}-\frac{1}{k+2}=1-\frac{1}{k+2}=\frac{k+1}{k+2}.$$
| {
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Check convergence $\sum\limits_{n=1}^\infty\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$ Please help me to check convergence of $$\sum_{n=1}^{\infty}\left(\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}}\right)$$
| Hint: Consider the Taylor expansion of the summand about $1/n$ for $n$ large:
$$\sqrt{1+\frac{7}{n^{2}}}-\sqrt[3]{1-\frac{8}{n^{2}}+\frac{1}{n^{3}}} \approx \left (1 + \frac{7}{2 n^2} \right ) - \left (1 - \frac{8}{3 n^2} \right ) = \frac{37}{16 n^2}$$
Use the comparison test.
| {
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"timestamp": "2023-03-29T00:00:00",
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If x,y,z are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what If $x,y, z$ are positive reals, then the minimum value of $x^2+8y^2+27z^2$ where $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ is what?
$108$ , $216$ , $405$ , $1048$
| The function $t\mapsto t^3$ is convex on ${\mathbb R}_{>0}$. Given that $${1\over x}+{1\over y}+{1\over z}=1\qquad(*)$$ we therefore have by Jensen's inequality
$$x^2+8y^2+27 z^2={1\over x} x^3 + {1\over y}(2y)^3+{1\over z}(3z)^3 \geq\left({1\over x} x+ {1\over y}2y +{1\over z} 3z\right)^3 =216\ ,$$
where equality holds iff in addition to $(*)$ we have $x=2y=3z$. It follows that the minimum value we are looking for is $216$ and that it is taken at the point $(6,3,2)$.
| {
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Calculate $\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$ Please help me calculate this:
$$\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$$
Here I've tried multiplying by $\sqrt[4]{x+9}+2$ and few other method.
Thanks in advance for solution / hints using simple methods.
Edit
Please don't use l'Hosplital rule. We are before derivatives, don't know how to use it correctly yet. Thanks!
| One thing you should learn, is that analysts like to think of functions as power series (or at worst Laurent series) In this sense, L'Hopital's rules is essentially saying that "When we have a function $ \frac {(x-7)f(x)}{(x-7)g(x)}$, then we can 'fill in' the hole and carry along our own merry way".
So, if we don't have L'Hopital, and we know we want to use it, we simply force it out.
For example, notice that
$$(\sqrt[4]{x+9} - 2)(\sqrt[4]{x+9} + 2)= \sqrt[2]{x+9} -4,$$
which I'm sure you did. Does this help us? No, not yet, because we haven't forced out the troublesome $x-7$. So let's try again, and we use $$(\sqrt{x+9}-4)(\sqrt{x+9}+4) = x+9 - 16 = x-7.$$ Are we done with the denominator? You bet!
How about the numerator? It is likely giving us problems with $x-7$, so let's force it out. Try
$$(\sqrt{x+2} - \sqrt[3]{x+20})(\sqrt{x+2} + \sqrt[3]{x+20}) = x+2 - (x+20)^{2/3}.$$
Are we done? No not yet, I don't see a $x-7$. So let's use
$$ [(x+2) - (x+20)^{2/3} ][(x+2)^2 + (x+2)(x+20)^{2/3} + (x+20^{4/3} ] = (x+2)^3 - (x+20)^2.$$
Are we done? I most certainly hope so, and you can check that we can factor out an $(x-7)$, since $(7+2)^3 - (7+20)^2 = 0$.
What's the moral of the story?
$$\frac {\sqrt{x+2} - \sqrt[3]{x+20}} {\sqrt[4]{x+9} - 2} \times \frac {\mbox{stuff}} {\mbox{same stuff}} = \frac {(x-7) \times \mbox {something}}{(x-7) \times \mbox {more something}}.$$
And now we rejoice and wait for the cows to come home.
| {
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Is my limit evaluation correct? I'm studying for my calculus exam and I have the following limit:
$$\lim\limits_{n \to \infty} \left ( \frac{1}{\sqrt{n^3 +3}}+\frac{1}{\sqrt{n^3 +6}}+ \cdots +\frac{1}{\sqrt{n^3 +3n}} \right )$$
My solution is:
$$\begin{align*}
&\lim\limits_{n\ \to \infty} \left ( \frac{1}{\sqrt{n^3 +3}}+\frac{1}{\sqrt{n^3 +6}}+ \cdots +\frac{1}{\sqrt{n^3 +3n}} \right )\\&= \lim_{n \to \infty}\left ( \frac{1}{\sqrt{n^3 +3}} \right ) + \lim_{n \to \infty}\left ( \frac{1}{\sqrt{n^3 + 6}} \right ) + \cdots + \lim_{n \to \infty} \left ( \frac{1}{\sqrt{n^3 +3n}} \right )\\
&=0 + 0 + \cdots + 0\\
&= 0
\end{align*}$$
It turned out to be suspiciously easy to solve.
Is this correct? If it isn't, what is wrong and how can I solve it correctly?
| As $n$ grows, the number of terms increases. So you need extra care.
The sum is equal to $\sum_{k=1}^n \frac{1}{\sqrt{n^3+3k}}$.
We have
$\left | \sum_{k=1}^n \frac{1}{\sqrt{n^3+3k}}\right| \leq n \cdot \frac{1}{\sqrt{n^3}}=\frac{1}{\sqrt{n}}$.
Now you can take a limit and the limit is $0$.
| {
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How to show that $\frac{x^2}{x-1}$ simplifies to $x + \frac{1}{x-1} +1$ How does $\frac{x^2}{(x-1)}$ simplify to $x + \frac{1}{x-1} +1$?
The second expression would be much easier to work with, but I cant figure out how to get there.
Thanks
| We have that
$$
x^2=(x-1+1)^2=(x-1)^2+1+2(x-1)
$$
and so for $x\neq 1$:
$$
\frac{x^2}{x-1}=\frac{(x-1)^2+1+2(x-1)}{x-1}=x-1+\frac{1}{x-1}+2=x+1+\frac{1}{x-1}.
$$
| {
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Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2}$ How to prove the following inequalities without using Bernoulli's inequality?
*
*$$\prod_{k=1}^{n}{\sqrt[k+1]{k}} \leq \frac{2^n}{n+1},$$
*$$\sum_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq \frac{n^2+3n}{2n+2}.$$
My proof:
*
*
\begin{align*}
\prod_{k=1}^{n}{\sqrt[k+1]{k}} &= \prod_{k=1}^{n}{\sqrt[k+1]{k\cdot 1 \cdot 1 \cdots 1}} \leq \prod^{n}_{k=1}{\frac{k+1+1+\cdots +1}{k+1}}\\
&=\prod^{n}_{k=1}{\frac{2k}{k+1}}=2^n \cdot \prod^{n}_{k=1}{\frac{k}{k+1}}=\frac{2^n}{n+1}.\end{align*}
*
$$\sum_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}} \geq n \cdot \sqrt[n]{\prod_{k=1}^{n}{\frac{1}{\sqrt[k+1]{k}}}} \geq n \cdot \sqrt[n]{\frac{n+1}{2^n}}=\frac{n}{2}\cdot \sqrt[n]{n+1}.$$
It remains to prove that
$$\frac{n}{2}\cdot \sqrt[n]{n+1} \geq \frac{n^2+3n}{2n+2}=\frac{n(n+3)}{2(n+1)},$$
or
$$\sqrt[n]{n+1} \geq \frac{n+3}{n+1},$$
or
$$(n+1) \cdot (n+1)^{\frac{1}{n}} \geq n+3.$$
We apply Bernoulli's Inequality and we have:
$$(n+1)\cdot (1+n)^{\frac{1}{n}}\geq (n+1) \cdot \left(1+n\cdot \frac{1}{n}\right)=(n+1)\cdot 2 \geq n+3,$$
which is equivalent with:
$$2n+2 \geq n+3,$$ or
$$n\geq 1,$$ and this is true becaue $n \neq 0$, $n$ is a natural number.
Can you give another solution without using Bernoulli's inequality?
Thanks :-)
| The inequality to be shown is
$$(n+1)^{n+1}\geqslant(n+3)^n,
$$
for every positive integer $n$. Introduce the function $u$ defined by
$$
u(x)=(x+1)\log(x+1)-x\log(x+3),
$$
then standard computations yield
$$
u'(x)=\frac3{3+x}+\log\left(\frac{1+x}{3+x}\right),\qquad
u''(x)=\frac{3-x}{(1+x)(3+x)^2},
$$
hence $u'$ increases on $(0,3)$ and decreases on $(3,+\infty)$. Since $u'(1)=\frac34-\log2\gt0$ and $u'(+\infty)=0$, $u$ is increasing on $(1,+\infty)$. Since $u(1)=0$, this yields $u(n)\geqslant0$ for every positive integer, QED.
Edit: Equivalently, one wants to prove that $(k+2)^{k-1}\leqslant k^k$ for every $k\geqslant2$, that is, $k+2\geqslant\left(1+\frac2k\right)^k$. If one knows that the RHS is increasing and converges to $\mathrm e^2\lt8$, this yields the result for every $k\geqslant6$. The cases $2\leqslant k\leqslant5$ can be checked manually.
| {
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Showing that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left (\sqrt{a^2+1}-1\right)$. How can I show that $\displaystyle\int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2}dx = \pi\left(\sqrt{a^2+1}-1\right)$?
| A slight variant of lab bhattacharjee's method provides a simpler solution:
Let
$$ I(a) = \int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx = 2\int_{0}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx. $$
Then by a simple application of multivariable calculus,
\begin{align*}
I'(a)
&= 2 \left. \frac{\sqrt{a^2-x^2}}{1+x^2} \right|_{x=a} + 2 \int_{0}^{a} \frac{d}{da} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx \\
&= 2 \int_{0}^{a} \frac{a}{(1+x^2)\sqrt{a^2-x^2}} \, dx.
\end{align*}
Then with the change of variable $x = a \sin\theta$, we have
\begin{align*}
I'(a)
&= 2 \int_{0}^{\frac{\pi}{2}} \frac{a}{1+a^2\sin^2\theta} \, d\theta \\
&= 2 \int_{0}^{\frac{\pi}{2}} \frac{a \sec^2\theta}{1+(a^2+1)\tan^2\theta} \, d\theta \\
&= 2 \int_{0}^{\infty} \frac{a}{1+(a^2+1)t^2} \, dt \qquad (t = \tan\theta) \\
&= \frac{\pi a}{\sqrt{a^2+1}}.
\end{align*}
Thus by integrating, we must have
$$ I(a) = \pi\sqrt{a^2+1} + C $$
for some constant $C$. But
$$ I(0+) = \lim_{a\to0} \int_{-a}^{a} \frac{\sqrt{a^2-x^2}}{1+x^2} \, dx = \lim_{a\to0} \int_{-1}^{1} \frac{\sqrt{1-x^2}}{(1/a)^2+x^2} \, dx = 0 $$
and we must have $C = -\pi$. This proves the identity.
| {
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Let $a,b$ and $c$ be real numbers.evaluate the following determinant: |$b^2c^2 ,bc, b+c;c^2a^2,ca,c+a;a^2b^2,ab,a+b$| Let $a,b$ and $c$ be real numbers. Evaluate the following determinant:
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$
after long calculation I get that the answer will be $0$. Is there any short processs? Please help someone thank you.
| \begin{align*}
\begin{vmatrix}b^2c^2&bc&b+c\\c^2a^2&ca&c+a\\a^2b^2&ab&a+b\end{vmatrix}
\stackrel{\large R_1-R_3\,,\,R_2-R_3}{\longrightarrow}
&(c-a)(c-b)
\,\begin{vmatrix}b^2(c+a)&b&1\\a^2(c+b)&a&1\\a^2b^2&ab&a+b\end{vmatrix}\\
\stackrel{\large C_1\,-\,ab\,C_2}{\longrightarrow}
&(c-a)(c-b)c
\,\begin{vmatrix}b^2&b&1\\a^2&a&1\\0&ab&a+b\end{vmatrix}\\
\stackrel{\large R_1-R_2}{\longrightarrow}
&(c-a)(c-b)c(b-a)
\,\begin{vmatrix}b+a&1&0\\a^2&a&1\\0&ab&a+b\end{vmatrix}\\
\stackrel{\large R_2-aR_1}{\longrightarrow}
&(c-a)(c-b)c(b-a)
\,\begin{vmatrix}b+a&1&0\\-ab&0&1\\0&ab&a+b\end{vmatrix}=0.
\end{align*}
| {
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How many binomial coefficients are equal to a specific integer ($\binom{n}{r} = 2013$ or $\binom{n}{r} = 2014$)?
*
*Find the number of ordered pairs $(n,r)$ which satisfy $\binom{n}{r} = 2013$.
*Find the number of ordered pairs $(n,r)$ which satisfy $\binom{n}{r} = 2014$.
My Attempt for $(1)$:
By simple guessing, we can find two solutions:
$$
\binom{n}{r} = \binom{2013}{1}=\binom{2013}{2013-1}=\binom{2013}{2012}
$$
So two solutions are $(2013,1),(2013,2012)$.
We also know that $\binom{n}{r} = 3\times 11 \times 61$.
How can I calculate the remaining ordered pairs $(n,r)$ from this point?
| $$\binom n {r+1}\ge \binom nr \iff \frac{\binom n {r+1}}{\binom nr}\ge 1\iff\frac{n-r}{r+1}\ge1\iff r\le \frac{n-1}2$$
So, $$\binom n r\le \binom n {r+1}\iff r\le \frac{n-1}2$$ and $$\binom n r\ge\binom n {r+1}\iff r\ge\frac{n-1}2$$
For any integer $u,\binom n1=u\implies n=u$ will always have a solution in integers.
For $r=2,\binom n2=\frac{n(n-1)}{2}=2013\iff n^2-n-2\cdot2013=0$ but the discriminant $1+4\cdot2\cdot2013=16105$ is not a perfect square, hence we don't have any rational solution here.
For $r=3,\binom n3=\frac{n(n-1)(n-2)}{1\cdot2\cdot3},$
one of the term in the numerator $n-s$ (say,) where $0\le s\le 2$ is divisible by $61$
So, $n-s=61m,n=61m+s$ for some integer $m$ then $n-t\text{( where $0\le s\le 2$)}\ge 61m-2\ge 59m$ for $m\ge 1$
So, $\binom n3\ge \frac{(59m)^3}{1\cdot2\cdot3}>2013$ for $m\ge1$
Now, $\binom n{r+1}\ge \binom n3$ for $\frac{n-1}2\ge r\ge 3\implies \binom nr>2013$for $\frac{n-1}2\ge r\ge 3$
also $\binom n{n-3}\le \binom nr$ for $\frac{n-1}2\le r\le n-3\implies \binom nr>2013$ for $\frac{n-1}2\le r\le n-3$
$\implies \binom nr>2013$ for $3\le r\le n-3$
As $\binom nr=\binom n{n-r},$ the only other solution is $r=n-1$ corresponding to $r=1$
| {
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Question about theta of $T(n)=4T(n/5)+n$ I have this recurrence relation $T(n)=4T(\frac{n}{5})+n$ with the base case $T(x)=1$ when $x\leq5$. I want to solve it and find it's $\theta$. I think i have solved it correctly but I can't get the theta because of this term $\frac{5}{5^{log_{4}n}}$ . Any help?
$T(n)=4(4T(\frac{n}{5^{2}})+\frac{n}{5})+n$
$=4^{2}(4T(\frac{n}{5^{3}})+\frac{n}{5^{2}})+4\frac{n}{5}+n$
$=...$
$=4^{k}T(\frac{n}{5^{k}})+4^{k-1}\frac{n}{5^{k-1}}+...+4\frac{n}{5}+n$
$=...$
$=4^{m}T(\frac{n}{5^{m}})+4^{m-1}\frac{n}{5^{m-1}}+...+4\frac{n}{5}+n$ Assuming $n=4^{m}$
$=4^{m}T(\lceil(\frac{4}{5})^{m}\rceil)+((\frac{4}{5})^{m-1}+...+1)n$
$=n+\frac{1-(\frac{4}{5})^{m}}{1-\frac{4}{5}}n=n+5n-n^{2}\frac{5}{5^{log_{4}n}}$
$=6n-n^{2}\frac{5}{5^{log_{4}n}}$
| Hint :$\frac {\log{n}}{\log4}=\log_4{n}$
So $5^{\log_4{n}}=n^{\frac {1}{\log_5{4}}}$
Use this.
| {
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Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$? Consider a function $f(x) = x^4+x^3+x^2+x+1$, where x is an integer, $x\gt 1$. What will be the remainder when $f(x^5)$ is divided by $f(x)$ ?
$f(x)=x^4+x^3+x^2+x+1$
$f(x^5)=x^{20}+x^{15}+x^{10}+x^5+1$
| $f_1(x)=x+1$
$f_1(x^5)=x^5+1$
$f_1(x^5)$, when divided by $f_1(x)$ leaves a remainder 0.
$f_2(x)=x^2+x+1$
$f_2(x^5)=x^{10}+x^5+1$
$f_2(x^5)$, when divided by $f_2(x)$ leaves a remainder 0.
| {
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What is the formula for $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$ How can I find the formula for the following equation?
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}$$
More importantly, how would you approach finding the formula? I have found that every time, the denominator number seems to go up by $n+2$, but that's about as far as I have been able to get:
$\frac12 + \frac16 + \frac1{12} + \frac1{20} + \frac1{30}...$ the denominator increases by $4,6,8,10,12,\ldots$ etc.
So how should I approach finding the formula? Thanks!
| You can use these formulas for Problems like this , $$\frac{1}{a.b}=\frac{1}{(b-a)}\left(\frac{1}{a}-\frac{1}{b}\right)$$ $$\frac{1}{a.b.c}=\frac{1}{(c-a)}\left(\frac{1}{a.b}-\frac{1}{b.c}\right)$$
Hope it'll help you .
| {
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Evaluate integral with quadratic expression without root in the denominator $$\int \frac{1}{x(x^2+1)}dx = ? $$
How to solve it? Expanding to $\frac {A}{x}+ \frac{Bx +C}{x^2+1}$ would be wearisome.
| It is not hard to express it as partial fraction, if you write $1$ in the numerator as $x^2+1 - x^2$.
Hence, $$\dfrac1{x(x^2+1)} = \dfrac{x^2+1-x^2}{x(x^2+1)} = \dfrac{x^2+1}{x(x^2+1)} - \dfrac{x^2}{x(x^2+1)} = \dfrac1x - \dfrac{x}{x^2+1}$$
Hence,
$$\int \dfrac{dx}{x(x^2+1)} = \int \dfrac{dx}x - \int\dfrac{x}{x^2+1}dx = \log(x) - \dfrac12 \log(x^2+1) + \text{ constant}$$
| {
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Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$. So I came up with $b= a+1$ $\Rightarrow$ $ab=a(a+1) = a^2 + a$
So that:
$a^2+b^2 -1$ = $a^2 + (a+1)^2 -1$ = $2a^2 + 2a$ = $2(a^2 + a)$ $\Rightarrow$
$(a,b) = (a,a+1)$ are solutions.
My motivation is for this follow up question:
(b) With $a$ and $b$ as above, what are the possible values of:
$$
\frac{a^2 +b^2 −1}{ab}
$$
Update
With Will Jagy's computations, it seems that now I must show that the ratio can be any natural number $m\ge 2$, by the proof technique of vieta jumping.
Update
Via Coffeemath's answer, the proof is rather elementary and does not require such technique.
| If you want the ratio $(a^2+b^2-1)/(ab)$ to be equal to $a$, choose $b=a^2-1$.
Then the numerator is $$a^2+b^2-1=a^2+(a^2-1)^2-1=a^4-a^2=a^2(a^2-1),$$
which divided by the denominator $ab=a(a^2-1)$ gives the result $a$.
This answers in the affirmative the question of the "update" of the OP, since here $a$ can be any $m \ge 2$.
I noticed after posting this that Will Jagy mentions the choice $a=r,b=r^2-1$ in a comment. I'll delete this if asked to...
| {
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Show that $(a+b+c)^3 = a^3 + b^3 + c^3+ (a+b+c)(ab+ac+bc)$ As stated in the title, I'm supposed to show that $(a+b+c)^3 = a^3 + b^3 + c^3 + (a+b+c)(ab+ac+bc)$.
My reasoning:
$$(a + b + c)^3 = [(a + b) + c]^3 = (a + b)^3 + 3(a + b)^2c + 3(a + b)c^2 + c^3$$
$$(a + b + c)^3 = (a^3 + 3a^2b + 3ab^2 + b^3) + 3(a^2 + 2ab + b^2)c + 3(a + b)c^2+ c^3$$
$$(a + b + c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + (3a^2b + 3a^2c + 3abc) + (3ab^2 + 3b^2c + 3abc) + (3ac^2 + 3bc^2 + 3abc) - 3abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3a(ab + ac + bc) + 3b(ab + bc + ac) + 3c(ac + bc + ab) - 3abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3(a + b + c)(ab + ac + bc) - 3abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3[(a + b + c)(ab + ac + bc) - abc]$$
It doesn't look like I made careless mistakes, so I'm wondering if the statement asked is correct at all.
| $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b+c)(ab+ac+bc) - 3abc$ is the right factorisation
| {
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Limit of a Function: $ \lim_{x \to 0}\ (e^x + x)^ {\large \frac {1} {x}}$ What is the limit of the following function which consists of an exponential and algebraic expression? $$ \lim_{x \to 0}\ (e^x + x)^ {\large \frac {1} {x}}\;\;?$$
| $$\begin{align}\lim\limits_{x \to 0}\ (e^x + x)^ \frac {1} {x} &=\lim\limits_{x \to 0}\ (1+x+\frac{x^2}{2!}+\dots + x)^ \frac {1} {x}\\ &=\lim\limits_{x \to 0}\ (1+2x+\frac{x^2}{2!}+\dots )^ \frac {1} {x}=e^2\end{align}$$Note that since $x \to 0$ we can assume $\frac{x^2}{2!}+ \frac{x^3}{3!}+\cdots =0$ and $$\lim_{x \to 0}(1+2x)^{\frac{1}{x}}=\lim_{u \to 0}(1+u)^{\frac{1}{u}{2}}=e^2$$ where $u=2x.$
Alternatively note that $$\lim_{x \to 0}(1+2x+\frac{x^2}{2!}+\dots )^{\frac{1}{x}}=\lim_{u \to 0}(1+u)^{\frac{1}{u}[\lim_{x \to 0}\frac{u}{x}]} = \exp{\lim_{x \to 0}\frac{u}{x}}=e^2$$
where $u=2x+\frac{x^2}{2!}+\cdots$ and
$$\begin{align}\lim_{x \to 0}\frac{u}{x} &=\lim_{x \to 0}\frac{2x+\frac{x^2}{2!}+\cdots}{x}\\ &=\lim_{x \to 0}[2+\frac{x}{2!}+\frac{x^2}{3!}+\cdots]\\ &= 2\end{align}$$
I personally like the first method because with some logical fact the given limit reduces to a smooth limit. Some people dislike the first method though it is not wrong. If anybody want to avoid the logical fact and wanna solve the problem only with mathematical tools he/she may solve the problem in the second method.
| {
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Finding $n$ such that $\frac{3^n}{n!} \leq 10^{-6}$ This question actually came out of a question. In some other post, I saw a reference and going through, found this, $n>0$.
Solve for n explicitly without calculator:
$$\frac{3^n}{n!}\le10^{-6}$$
And I appreciate hint rather than explicit solution.
Thank You.
| Note that, for $n=3m$, $$3^{-3m}{(3m)!}=\left[m\left(m-\frac{1}{3}\right)\left(m-\frac{2}{3}\right)\right]\cdots\left[1\cdot\frac{2}{3}\cdot\frac{1}{3}\right] <\frac{2}{9}\left(m!\right)^3.$$
So you have to go at least far enough so that
$$
\frac{2}{9}\left(m!\right)^3>10^{6},
$$
or $m! > \sqrt[3]{4500000} > 150$. So $m=5$ (corresponding to $n=15$) isn't far enough; the smallest $n$ satisfying your inequality will be at least $16$.
Similarly, for $n=3m+1$,
$$
3^{-3m-1}(3m+1)!=\left[\left(m+\frac{1}{3}\right)m\left(m-\frac{1}{3}\right)\right]\cdots \left[\frac{4}{3}\cdot1\cdot\frac{2}{3}\right]\cdot\frac{1}{3} < \frac{1}{3}(m!)^3,$$
so you need $m!>\sqrt[3]{3000000}> 140$, and $m=5$ (that is, $n=16$) is still too small.
Finally, for $n=3m+2$,
$$
3^{-3m-2}(3m+2)!=\left[\left(m+\frac{2}{3}\right)\left(m+\frac{1}{3}\right)m\right]\cdots \left[\frac{5}{3}\cdot\frac{4}{3}\cdot1\right]\cdot\frac{2}{3}\cdot\frac{1}{3} > \frac{560}{729}(m!)^3,
$$
where the coefficient comes from the last eight terms, so it is sufficient that $m! > 100\cdot\sqrt[3]{729/560}.$ To show that $m=5$ is large enough, we need to verify that $(12/10)^3=216/125 > 729/560$. Carrying out the cross-multiplication, you can check without a calculator that $216\cdot 560 =120960$ is larger than $729\cdot 125=91125$, and conclude that $m=5$ (that is, $n=17$) is large enough. The inequality therefore holds for exactly all $n\ge 17$.
| {
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half-angle trig identity clarification I am working on the following trig half angle problem. I seem to be on the the right track except that my book answer shows -1/2 and I didn't get that in my answer. Where did I go wrong?
$$\sin{15^{\circ}} = $$
$$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \cos{30^{\circ}} }{ 2 } } $$
$$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \frac {\sqrt 3 }{ 2 } }{ 2 } } $$
$$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \frac {\sqrt 3 }{ 2 } }{ 2 } (\frac {2} {2}) } $$
$$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { 2 - \sqrt {3} } $$
Book Answer
$$\sin \frac { 30^{\circ} }{ 2 } = -\frac {1} {2} \sqrt { 2 - \sqrt {3} } $$
| In the $4^{th}$ equation, you should have $$\pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2} \cdot \frac{2}{2}}=\pm\sqrt{\frac{2-\sqrt{3}}{4}}=\pm\frac{1}{2}\sqrt{2-\sqrt{3}}.$$ Since $30/2$ is in the first quadrant, the answer should be the positive.
| {
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Sum of $\sum_{n=1}^\infty \frac{(-1)^n}{n^2+(n^2-1)^2}$ I came across these two series $$\sum_{n=1}^\infty \frac{(-1)^n}{n^2+(n^2-1)^2}$$ and $$\sum_{n=1}^\infty \frac{(-1)^n(n^2-1)}{n^2+(n^2-1)^2}.$$ Does anyone recognize the sums of these series?
| Using the Residue Theorem, we can show that, for sufficiently well-behaved functions $f(z)$:
$$\sum_{n=-\infty}^{\infty} (-1)^n f(n) = - \sum_{k=1}^m \mathrm{Res}_{z=z_k} \pi \csc{(\pi z)} f(z)$$
where the $z_k$ are the poles of $f$. I will not show this for now, only that I note that $f(z) = (z^2 + (z^2-1)^2)^{-1}$ is sufficiently well-behaved so that we may use the above formula to evaluate the sum.
To find the $z_k$, we solve $ z^2 + (z^2-1)^2 = z^4-z^2+1 = 0$, or
$$z_{(\pm \pm )} = \pm \exp{\left (\pm i \frac{\pi}{6} \right )}$$
We now evaluate the residues associated with the poles and add them up. The result (after some algebra) is
$$\sum_{k=1}^m \mathrm{Res}_{z=z_k} \pi \csc{(\pi z)} f(z) = \pi \csc{\left ( \frac{\pi}{3} \right )} \Im{\left [\exp{\left (-i \frac{\pi}{6} \right )} \csc{\left (\pi \exp{\left (i \frac{\pi}{6} \right )}\right )}\right ]} $$
The computation boils down to the evaluation of $\csc{\left (\pi \exp{\left (i \frac{\pi}{6} \right )}\right )}$. Again, I will spare the reader for now. The result is
$$S = \sum_{n=-\infty}^{\infty} (-1)^n f(n) = \frac{2 \pi}{\sqrt{3}} \frac{\sqrt{3} \cos{\left ( \sqrt{3} \frac{\pi}{2} \right )} \sinh{\left ( \frac{\pi}{2} \right )}+\sin{\left ( \sqrt{3} \frac{\pi}{2} \right )} \cosh{\left ( \frac{\pi}{2} \right )}}{\cosh{\pi} - \cos{(\sqrt{3} \pi)}}$$
However, this is not what we were asked for; rather, the result we seek is
$$\sum_{n=1}^{\infty} (-1)^n f(n) = \frac{1}{2} (S-1) $$
I verified this with a numerical and analytical result in Mathematica.
| {
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Two Problem: find $\max, \min$; number theory: find $x, y$
*
*Find $x, y \in \mathbb{N}$ such that $$\left.\frac{x^2+y^2}{x-y}~\right|~ 2010$$
*Find max and min of $\sqrt{x+1}+\sqrt{5-4x}$ (I know $\max = \frac{3\sqrt{5}}2,\, \min = \frac 3 2$)
| The second problem can be solved using routine calculus. There are also various non-calculus solutions. These in principle involve only basic machinery from algebra or trigonometry, but are in practice more complicated than the calculus solution.
The derivative of our function is
$$\frac{1}{2}\left(\frac{1}{\sqrt{x+1}}-\frac{4}{\sqrt{5-4x}}\right).$$
When you set this equal to $0$, and manipulate a little, you will get a linear equation for $x$. Our candidates are the solution of this equation, and the natural endpoints $x=-1$ and $x=\frac{5}{4}$. So we have $3$ candidates. Substitution will determine the winner and the loser.
For the first problem, it is not clear whether you want one solution, or a characterization of all solutions. We will find a couple of solutions, and you can do the rest.
Let's see if we can make the divisor $\frac{x^2+y^2}{x-y}$ of $2010$ equal to $2010$. We get the equation $\frac{x^2+y^2}{x-y}=2010$. Rewrite this equation as $x^2+y^2=2010(x-y)$, and then, completing the square, as
$$(x-1005)^2+(y+1005)^2=2(1005)^2.$$
So we want to express $2(1005)^2$ as a sum of two squares. We also want to make sure our solution produces positive $x$ and $y$.
Note that $2(1005)^2=50(201)^2= (201)^2+ 7^2(201)^2$.
Set $x-1005=201$ and $y+1005=7(201)$, and solve.
To find other solutions, repeat the calculation with divisors $d$ of $2010$ different from $2010$. By if necessary interchanging the roles of $x$ and $y$, we can confine attention to positive divisors.
Note that $2010$ has few divisors. The work is also made easier by the fact that if $q$ is a prime of the form $4k+3$, and the prime factorization of $m$ contains $q$ to an odd power, then the equation $s^2+t^2=m$ has no solutions. And if the prime factorization of $m$ contains $q$ to an even power, and $s^2+t^2=m$, then $q$ divides $s$ and $t$. Since $2010=(2)93)(5)(67)$, two of the prime factors of $m$ are of the form $4k+3$. This greatly simplifies the search for solutions.
The easiest thing to deal with is the divisor $10$. We end up looking at the equation $(x-5)^2+(y+5)^2= 50$, which gives $x=6$, $y=2$.
Slightly harder is the odd divisor $5$. When we complete the square, we get
$(x-5/2)^2+(y+5/2)^2=25/2$. Multiply through by $4$.
There are not many other cases to consider. Your turn!
| {
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Product of numbers Pair of numbers whose product is $+7$ and whose sum is $-8$.
Factorise $x^{2} - 8x + 7$.
I can factorise but it's just I can't find any products of $+7$ and that is a sum of
$-8$. Any idea? Thanks guys!
Thanks.
| $(x-a)(x-b)=x^2-(a+b)x+ab$
So, you are looking for the roots of $x^2-(+2)x+(-3)$, that is, $x^2-2x-3$, which is $=x^2-2x+1-4=(x-1)^2-4$, and this is $0$ iff $(x-1)^2=4$, that is, $x-1=\pm 2$, i.e., $-1,3$ is your solution.
Similarly for the other one: $x^2-8x+7=(x-4)^2-16+7=\dots$
Or, the other way: $7+1=8$ and $7\cdot 1=7$, so the roots of $x^2-8x+7$ are $1$ and $7$.
| {
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Proving that for any odd integer:$\left\lfloor \frac{n^2}{4} \right\rfloor = \frac{(n-1)(n+1)}{4}$ I'm trying to figure out how to prove that for any odd integer, the floor of:
$$\left\lfloor \frac{n^2}{4} \right\rfloor = \frac{(n-1)(n+1)}{4}$$
Any help is appreciated to construct this proof!
Thanks guys.
| Let $n$ be an odd integer.
Then there exists an integer $k$, such that:
$$n=2k+1$$
It follows that:
$$\begin{align}
\left\lfloor\frac{n^2}{4}\right\rfloor &= \left\lfloor\frac{(2k+1)^2}{4}\right\rfloor\\
&=\left\lfloor\frac{(4k^2+4k+1)}{4}\right\rfloor\\
&=\left\lfloor\frac{(4k^2+4k)}{4}+\frac{1}{4}\right\rfloor\\
&=\left\lfloor(k^2+k)+\frac{1}{4}\right\rfloor
\end{align}
$$
Because $k^2+k$ is an integer, we can now say:
$$\left\lfloor\frac{n^2}{4}\right\rfloor = k^2+k$$
It also follows that:
$$\begin{align}
\frac{(n-1)(n+1)}{4} &= \frac{n^2-1}{4}\\
&= \frac{(2n+1)^2-1}{4}\\
&= \frac{(4k^2+4k+1)-1}{4}\\
&= \frac{4k^2+4k}{4}\\
&= k^2+k\\
\end{align}$$
Therefore:
$$\left\lfloor\frac{n^2}{4}\right\rfloor=\frac{(n-1)(n+1)}{4}$$
Q.E.D.
| {
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Solving Recurrence $T(n) = T(n − 3) + 1/2$; I have to solve the following recurrence.
$$\begin{gather}
T(n) = T(n − 3) + 1/2\\
T(0) = T(1) = T(2) = 1.
\end{gather}$$
I tried solving it using the forward iteration.
$$\begin{align}
T(3) &= 1 + 1/2\\
T(4) &= 1 + 1/2\\
T(5) &= 1 + 1/2\\
T(6) &= 1 + 1/2 + 1/2 = 2\\
T(7) &= 1 + 1/2 + 1/2 = 2\\
T(8) &= 1 + 1/2 + 1/2 = 2\\
T(9) &= 2 + 1/2
\end{align}$$
I couldnt find any sequence here. can anyone help!
| I believe this is the right answer:
$$
\\
T(n) = T(n - 3) + \frac{1}{2}
\\
T(n) = T(n - 6) + \frac{1}{2} + \frac{1}{2} = T(n - 6) + \frac{2}{2}
\\
T(n) = T(n - 9) + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = T(n - 9) + \frac{3}{2}
\\
T(n) = T(n - 12) + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = T(n - 12) + \frac{4}{2}
\\
...
\\
T(n) = T(n - k) + \frac{k}{3} \times \frac{1}{2}
\\
\text{When }n - k = 0 \Rightarrow n = k
\\
T(n) = T(0) + \frac{n}{3} \times \frac{1}{2}
\\
\text{Hence, } T(n)\ \epsilon \ O(n)
$$
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evaluate $\int_0^\infty \dfrac{dx}{1+x^4}$ using $\int_0^\infty \dfrac{u^{p-1}}{1+u} du$ evaluate $\int_0^\infty \dfrac{dx}{1+x^4}$using $\int_0^\infty \dfrac{u^{p-1}}{1+u} du = \dfrac{\pi}{\sin( \pi p)}$. I am having trouble finding what is $p$. I set $u = x^4$, I figure $du = 4x^3 dx$, I am unsure though how to find $p$ though. Could someone tell me what I am missing? Thanks.
| If $u=x^4$ and $du=4x^3 dx$ then:
$$dx = \frac{1}{4x^3} du = \frac{1}{4{(x^4)}^{3/4}} du =
\frac{1}{4{u}^{3/4}} du$$
So your integral is:
$$\int_0^\infty \dfrac{dx}{1+x^4} = \frac{1}{4}\int_0^\infty \frac{u^{-3/4}du}{1+u} = \frac{\pi}{4\sin(\pi/4)}=\frac{\pi }{2 \sqrt{2}}$$
| {
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"url": "https://math.stackexchange.com/questions/301010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0Evaluate by complex methods
$$\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0<a<b<1$$
Sis.
| This integral is $1/2$ the integral over $[0,2 \pi)$. Let $z=e^{i \theta}$, $d\theta = dz/(i z)$; the result is
$$\frac{1}{2 i}\oint_{|z|=1} \frac{dz}{z} \frac{-\frac{1}{4} (z^2-1)^2}{(a z^2-(1+a^2)z+a)(b z^2-(1+b^2)z+b)}$$
which can be rewritten as
$$\frac{i}{8} \oint_{|z|=1} \frac{dz}{z} \frac{(z^2-1)^2}{(a z-1)(z-a)(b z-1)(z-b)}$$
There are 5 poles, although because $0<a<b<1$, only 3 of them fall within the contour. This integral is then $i 2 \pi$ times the sum of the residues of these poles. The residues of these poles are actually straightforward:
$$\mathrm{Res}_{z=0}\frac{i}{8} \frac{(z^2-1)^2}{z(a z-1)(z-a)(b z-1)(z-b)} = \frac{i}{8 a b}$$
$$\mathrm{Res}_{z=a}\frac{i}{8} \frac{(z^2-1)^2}{z(a z-1)(z-a)(b z-1)(z-b)} = \frac{i}{8 a} \frac{a^2-1}{(a b-1)(a-b)}$$
$$\mathrm{Res}_{z=b}\frac{i}{8} \frac{(z^2-1)^2}{z(a z-1)(z-a)(b z-1)(z-b)} = -\frac{i}{8 b} \frac{b^2-1}{(a b-1)(a-b)}$$
There is vast simplification from adding these pieces together, which I leave to the reader. The result is
$$\int_0^{\pi} d\theta \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)} = \frac{\pi}{2} \frac{1}{1-a b}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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how to solve, $(x^2-2x+2y^2)dx+2xydy=0$? solve the following differential equation.
$\tag 1(x^2-2x+2y^2)\,dx+2xy\,dy=0$
$\frac{dy}{dx}=\frac{2x-x^2-2y^2}{2xy}$
dividing (1) throughout by $y^2$ we have,
$\tag 2 \left(\frac{x^2}{y^2}+2-2\frac{x}{y^2}\right)dx+2(\frac{x}{y})dy=0$
| Illustrating @DavidMitra's idea: multiply through by $x$:
$$(x^3-2x^2+2xy^2)\,dx+2x^2y\,dy=0$$
We want to find a function $F(x,y)$ such that
$$\frac{\partial F}{\partial x} = x^3-2x^2+2xy^2$$
$$\frac{\partial F}{\partial y} = 2x^2y$$
From the former equation, we integrate with respect to $x$ and get
$$F(x,y) = \frac{1}{4} x^4 - \frac{2}{3} x^3 + x^2 y^2 + g(y)$$
From the latter, integrate with respect to $y$ and get:
$$F(x,y) = x^2 y^2 + f(x)$$
Comparing the two equations, we see that $f(x) = \frac{1}{4} x^4 - \frac{2}{3} x^3 $ and $g(y)=0$. Therefore
$$F(x,y) = \frac{1}{4} x^4 - \frac{2}{3} x^3 + x^2 y^2$$
Solutions of your differential equation satisfy $F(x,y) = C$, a constant.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$ for $x$ Is there any smart way to solve the equation: $$\frac{(b-c)(1+a^2)}{x+a^2}+\frac{(c-a)(1+b^2)}{x+b^2}+\frac{(a-b)(1+c^2)}{x+c^2}=0$$
Use Maple I can find $x \in \{1;ab+bc+ca\}$
| Starting with pipi's simplification of the mess:
$$
\frac{b - c}{x + a^2} + \frac{c - a}{x + b^2} + \frac{a - b}{x + c^2} = 0 \\
(b - c) (x + b^2) (x + c^2) + (c - a) (x + a^2) (x + c^2) + (a - b) (x + a^2) (x + b^2) = 0
$$
Suprisingly, this turns out a linear equation for $x$, with solution:
$$
x = a b + a c + b c
$$
(Many thanks to Maxima for help with algebra)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/306154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet
First, I showed that $(m^2 - n^2, 2mn, m^2 + n^2)$ is in fact a Pythagorean triplet.
$$\begin{align*} (m^2 - n^2)^2 + (2mn)^2 &= (m^2 + n^2)^2 \\
&= m^4 -2m^2n^2 + n^4 + 4m^2n^2 \\
&= m^4 + 2m^2n^2 + n^4 \\
&= 1\end{align*}$$ which shows that it respect $a^2+b^2 = c^2$
let p be a prime number, $ p|(m^2 + n^2) \text { and } p|(m^2 - n^2) $
if $gcd(m^2 + n^2, (m^2 - n^2)) = 1$
$p | (m^2 + n^2) , \text { so, } p |m^2 \text { and } p |n^2$
that means $ (m^2 + n^2) \text { and } (m^2 - n^2) $ are prime together
I'm kind of lost when I begin to show the gcd = 1... I think I know what to do, just not sure how to do it correctly.
Thanks
| You don't know that $\gcd(m^2-n^2, m^2+n^2)=1$, you need to prove it.
Now if $p|m^2-n^2$ and $p|m^2+n^2$ then $p$ divides their sum and their difference. Use this.
And to proceed further you need extra information, which you probably left out but it is the key: $m,n$ are relatively prime and of opposite parity....
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}.$ Find
$$\lim_{n\to \infty}\frac{1}{\ln n}\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}\;.$$
| I will be very non-rigorous here.
Letting $m = j+k$,
$\begin{align}
\sum_{j,k=1}^{n}\frac{j+k}{j^3+k^3}
&\approx \frac{1}{2}\sum_{m=2}^{2n} \sum_{j=1}^{m-1}\frac{m}{j^3+(m-j)^3}\\
&= \frac{1}{2}\sum_{m=2}^{2n} m\sum_{j=1}^{m-1}\frac{1}{(j+(m-j))(j^2-j(m-j)+(m-j)^2)}\\
&= \frac{1}{2}\sum_{m=2}^{2n} m\sum_{j=1}^{m-1}\frac{1}{m(j^2-jm+j^2+m^2-2jm+j^2)}\\
&= \frac{1}{2}\sum_{m=2}^{2n} \sum_{j=1}^{m-1}\frac{1}{3j^2-3jm+m^2}\\
&= \frac{1}{2}\sum_{m=2}^{2n} \frac{1}{m^2}\sum_{j=1}^{m-1}\frac{1}{3(j/m)^2-3(j/m)+1}\\
&\approx \frac{1}{2}\sum_{m=2}^{2n} \frac{1}{m}\int_{0}^{1}
\frac{dx}{3x^2-3x+1}\\
&= \frac{1}{2}\int_{0}^{1}\frac{dx}{3x^2-3x+1}
\sum_{m=2}^{2n} \frac{1}{m}\\
&\approx \frac{\ln(2n)}{2} \int_{0}^{1}\frac{dx}{x^2-x+1/3} \\
\end{align}$.
To evaluate the integral,
$\begin{align}
I = \int_{0}^{1}\frac{dx}{x^2-x+1/3}
&=\int_{0}^{1}\frac{dx}{x^2-x+1/4-1/4+1/3}\\
&=\int_{0}^{1}\frac{dx}{(x-1/2)^2+1/12}\\
&=\int_{-1/2}^{1/2}\frac{dx}{x^2+1/12}\\
&=2\int_{0}^{1/2}\frac{dx}{x^2+1/12}\\
\end{align}
$.
Letting $x = y/\sqrt{12}= y/(2\sqrt{3})$
\begin{align}
I &= 2\int_{0}^{1/2}\frac{dx}{x^2+1/12}\\
&= 2\int_{0}^{\sqrt{3}}\frac{dy}{2\sqrt{3}((y^2)/12+1/12)}\\
&= \frac{12}{\sqrt{3}}\int_{0}^{\sqrt{3}}\frac{dy}{y^2+1}\\
&= \frac{12}{\sqrt{3}}\tan^{-1}(\sqrt{3})\\
&= \frac{12}{\sqrt{3}}(\pi/3)\\
&=\frac{4\pi}{\sqrt{3}}\\
&=\frac{4\sqrt{3}\pi}{3}\\
\end{align}
so the sum is about
$$ \frac{\ln(2n)}{2} \frac{4\sqrt{3}\pi}{3}
=\frac{4\sqrt{3}\pi\ln(2n)}{6}
\approx \frac{2\sqrt{3}\pi\ln(n)}{3}
$$
so the limit is
$\dfrac{2\sqrt{3}\pi}{3} $.
As usual, this is done off the top of my head,
doing math as $\LaTeX$,
but at least I got a reasonable limit.
Who knows, this might even be correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving for x with radicals and negative exponents How do I go about solving for $x$ in this equation?
$$\displaystyle -x^{-\large\frac{3}{4}} + \frac{15^{\large\frac{1}{4}}}{15} = 0$$
| Hint: $$-x^{-3/4} + \frac{15^{1/4}}{15} = 0 \iff -\frac {1}{x^{3/4}} + \frac{15^{1/4}}{15^{4/4}} = 0$$
$$\iff -\frac {1}{x^{3/4}} + \frac{1}{15^{3/4}} = 0$$
$$\iff \dfrac{x^{3/4}}{15^{3/4}} = 1 \iff \left(\frac{x}{15}\right)^{3/4} = 1$$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Domain of analyticity for $\operatorname{Log}(z^4 - 1)^{1/2}$ How do I solve for the analyticity of $\operatorname{Log}(z^4 - 1)^{1/2}$?
So I factored out $$z^4 - 1 = x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4 - 1$$
Now the $$\operatorname{Re}(z^4 - 1) = x^4 - 6x^2y^2 + y^4 - 1 < 0 \implies x^4 - 6x^2y^2 + y^4 < 1$$ but I'm stuck here on how to find the set where $\operatorname{Re}(z^4 - 1)$ is nonnegative.
$$\operatorname{Im}(z^4 - 1) = 4x^3y - 4xy^3 = 4xy(x^2 - y^2) = 0 \implies x=0, y=0$$ but unsure how to interpret that either, is it only $z=0$ where it is not analytic or on $x=0$ and $y=0$ axis?
If anyone knows a textbook reference that clearly explains these concepts that would be great as well.
| You're doing fairly well. As for the inequality, find that polynomial's roots first:
$$
x^4-6x^2y^2+y^4-1 = 0 \\
x^2 = 3y^2 \pm \sqrt{9y^4-y^4+1} = 3y^2 \pm \sqrt{8y^4+1} \\
x_{1,2,3,4} = \pm \sqrt { 3y^2 \pm \sqrt{8y^4+1}}
$$
So inequality can be rearranged as
$$
(x-x_1)(x-x_2)(x-x_3)(x-x_4) > 0
$$
As for the second region with $\text{Im} (z^4-1) \ne 0$ you didn't find all roots of that equation
$$
4x^3y-4xy^3=0 \\
4xy(x^2-y^2) = 0 \\
\left [ \begin{array}{l}
xy = 0 \\
x^2-y^2 =0
\end{array}
\right . \Longrightarrow \left [ \begin{array}{l}
\left [ \begin{array}{l}
x = 0, y \in \mathbb R \\
y = 0, x \in \mathbb R
\end{array}\right . \\
|x| = |y|
\end{array}
\right .
$$
So final regions looks like filled region below (white color means excluded points)
| {
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Probability of choosing two equal bits from three random bits Given three random bits, pick two (without replacement). What is the probability that the two you pick are equal?
I would like to know if the following analysis is correct and/or if there is a better way to think about it.
$$\Pr[\text{choose two equal bits}] = \Pr[\text{2nd bit} = 0 \mid \text{1st bit} = 0] + \Pr[\text{2nd bit} = 1 \mid \text{1st bit} = 1]$$
Given three random bits, once you remove the first bit the other two bits can be: 00, 01, 11, each of which occurring with probability $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$. Thus,
$$\Pr[\text{2nd bit} = 0] = 1\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{4} + 0\cdot\frac{1}{4} = \frac{3}{8}$$
And $\Pr[\text{2nd bit} = 1] = \Pr[\text{2nd bit} = 0]$ by the same analysis.
Therefore,
$$\Pr[\text{2nd bit}=0 \mid \text{1st bit} = 0] = \frac{\Pr[\text{1st and 2nd bits are 0}]}{\Pr[\text{1st bit}=0]} = \frac{1/2\cdot3/8}{1/2} = \frac{3}{8}$$
and by the same analysis, $\Pr[\text{2nd bit} = 1 \mid \text{1st bit} = 1] = \frac{3}{8}$.
Thus, $$\Pr[\text{choose two equal bits}] = 2\cdot\frac{3}{8} = \frac{3}{4}$$
| Clearly the intuition tells us that the answer should be $\frac{1}{2}$, because choosing 2 random bits at random without replacing is just like simply choosing two random bits which are equal with probability $\frac{1}{2}$.
There are 3 way to choose two bits out of 3, i.e. choosing $B_1$ and $B_2$ or $B_2$ and $B_3$ or $B_1$ and $B_3$ which are equally likely, thus:
$$
\Pr\left(\text{two same bits}\right) = \frac{1}{3} \Pr\left(B_1 = B_2\right) + \frac{1}{3} \Pr\left(B_2 = B_3\right) + \frac{1}{3} \Pr\left(B_1 = B_3\right)
$$
since each bit is independent and equally likely to be either 0 or 1, $\Pr\left(B_1 = B_2\right) = \Pr\left(B_2 = B_3\right) = \Pr\left(B_1 = B_3\right) = \frac{1}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.
Given $a,b,c>0$, prove that $\displaystyle \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$.
I expanded the LHS, and realized I have to prove $\displaystyle\frac a b +\frac a c +\frac b c +\frac b a +\frac c a +\frac c b \geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$, but I don't know how. Please help. Thank you.
| I know it is a very late to answer this, but i came across a very nice method to prove this.
$$\displaystyle \left(1+\frac a b \right) \left(1+\frac b c \right)\left(1+\frac c a \right) \geq 2\left(1+ \frac{a+b+c}{\sqrt[3]{abc}}\right)$$
$$\displaystyle \iff (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right) \geq 3+2\left(\frac{a+b+c}{\sqrt[3]{abc}}\right)$$
$$\displaystyle \iff \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right) \geq \frac{3}{a+b+c}+\frac{2}{\sqrt[3]{abc}}$$
Replacing by their inverses
$$\iff 3AM\ge HM+2GM$$
Which is evident.
| {
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"timestamp": "2023-03-29T00:00:00",
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Statistical Probability We make sweets of $5$ different flavors. The sweets are made in batches of $1.8$ million sweets per batch and then combined/mixed together into a total random batch of $9$ million sweets.
We then pack into tubes of $14$ sweets and we would like to know the probability of getting all $5$ flavors in every pack. If the sweets are fed to the packing machine completely randomly from the mixed batch, how do we calculate this?
| The details depend on whether you want a theoretically exact answer, or an excellent approximation. Since this is packaged as an applied problem, we work with the approximation. So we will assume that each sweet has one of $5$ values, independently of the values of the previously chosen sweets. The number $9,000,000$ is so large that "with replacement" is essentially equivalent to "without replacement."
Let the flavours be called $1,2,3,4,5$. We find the probability that some flavour(s) is (are) missing. The method we use is Inclusion/Exclusion.
First we find the probability that flavour $1$ is missing. That is the probability all $14$ sweets are not of type $1$. Since the probability of being of type $1$ is $\frac{1}{5}$, the probability the sweets are all of type other than $1$ is $\left(\frac{4}{5}\right)^{14}$.
Now add together the probabilities for the $5$ flavours. We get a first estimate
$5\left(\frac{4}{5}\right)^{14}$ for the probability that some flavour(s) is (are) missing.
But for every pair $\{i,j\}$ of flavours, we have counted twice the probability that flavours $i$ and $j$ are both missing. The probability $i$ and $j$ are both missing is $\left(\frac{3}{5}\right)^{14}$. We must add up over the $\binom{5}{2}=10$ possible pairs, and subtract from our first estimate. So our next estimate is $5\left(\frac{4}{5}\right)^{14}-10\left(\frac{3}{5}\right)^{14}$.
But we have subtracted too much, for we have subtracted twice the probability that three flavours are missing. The probability that the flavours $i,j,k$ are missing is $\left(\frac{2}{5}\right)^{14}$. Add over the $\binom{5}{3}=10$ triples. So our next estimate is
$5\left(\frac{4}{5}\right)^{14}-10\left(\frac{3}{5}\right)^{14}+10\left(\frac{2}{5}\right)^{14}$.
We have added a little too much, for we have added one too many times the probabilities that $4$ flavours are missing. Thus the probability at least one flavour is missing is $5\left(\frac{4}{5}\right)^{14}-10\left(\frac{3}{5}\right)^{14}+10\left(\frac{2}{5}\right)^{14}-5\left(\frac{1}{5}\right)^{14}$. Thus the probability no flavour is missing is
$$1-\left(\left(\frac{4}{5}\right)^{14}-10\left(\frac{3}{5}\right)^{14}+10\left(\frac{2}{5}\right)^{14}-5\left(\frac{1}{5}\right)^{14}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do you find the inverse of a polynomial in Laurent series division ring? In the Laurent series division ring, how can we find the inverse of a given polynomial? For example how can you find the inverse of $3x^{-2} +x^{-1}+5x+7x^4$. Is there a certain formula to find the inverse?
I've tried to find the inverse for the above polynomial with a formula used in proving the existence of the units, but it gives me a lot of unknowns.
| $$ \frac{1}{7 x^4 + 5 x + x^{-1} + 3 x^{-2}} = \frac{x^2}{7 x^6 + 5 x^3 + x + 3}$$
The polynomial $P(x) = 7 x^6 + 5 x^3 + x + 3$ is irreducible over the rationals, with six
distinct complex roots $r_1$ to $r_6$ (none of them real, by the way). We have
a partial fraction decomposition
$$ \frac{x^2}{7 x^6 + 5 x^3 + x + 3} = \sum_{j=1}^6 \frac{r_j^2}{P'(r_j) (x - r_j)} $$
since $r_j^2/P'(r_j)$ is the residue of $x^2/P(x)$ at $x=r_j$.
Since $\dfrac{1}{x -r_j} = \dfrac{-r_j^{-1}}{1 - r_j^{-1} x} = - \sum_{k=0}^\infty r_j^{-k-1} x^k$, we obtain
$$ \frac{1}{7 x^4 + 5 x + x^{-1} + 3 x^{-2}} = -\sum_{k=0}^\infty \sum_{j=1}^6
\frac{r_j^{1-k}}{P'(r_j)} x^k$$
(the $k=0$ and $k=1$ terms will be $0$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the asymptote of a function Find the asymptote of $f(x)$ as $x \rightarrow \infty$ where
$$f(x) = \frac{x^3 -x^2 +2}{x-1}$$
The answer key mentioned that the answer is $h(x)=x^2$ where $h(x)$ is the equation of the asymptote of $f(x)$. I beg to differ with this solution:
$$
\begin{align*}
f(x) = \frac{x^3 -x^2 +2}{x-1}
&= \frac{x^2 (x-1) +2}{x-1}\\
&=x^2+\frac{2}{x-1}
\end{align*}
$$
With arrangement, we attempt to fit it in the form $f(x) + h(x) =\frac{2}{x-1}$ and so we have
$h(x) = -x^2$. So the asymptote of $f(x)$ as $x \rightarrow \infty$ is $y=-x^2$
Any ideas to solve this?
| With a rearrangement,
$$
\begin{align*}
f(x) -x^2
&=\frac{2}{x-1}\\
f(x) -h(x)
&=\frac{2}{x-1}\\
\lim_{x\rightarrow\infty}f(x) -h(x)
&=\lim_{x\rightarrow\infty}\frac{2}{x-1}\\
&=0
\end{align*}
$$
If you think about it, the idea is that as $x \rightarrow \infty$, the difference between $f(x)$ and $h(x)$ tends to $0$. So the functions are getting closer and closer to each other, but don't touch. It is no wonder $h(x)$ is considered the asymptote.
Hence, indeed the asymptote is $y=x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/316571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $x^{11}+x^8+5\equiv 0\pmod{49}$ Solve $x^{11}+x^8+5\pmod{49}$
My work
$f(x)=x^{11}+x^8+5$
consider the polynomial congruence $f(x) \equiv 0 \pmod {49}$
Prime factorization of $49 = 7^2$
we have $f(x) \equiv 0 \mod 7^2$
Test the value $x\equiv0,1,2,3,4,5,6$ for $x^{11}+x^8+5 \equiv 0\pmod 7$
It works for $x\equiv1$, $x\equiv1\pmod7$ is a solution.
We proceed to lift the solution $\mod 7^2$
$f'(x) = 11x^{10} + 8x^7 +5$, we have $f(1) = 7$ , $f'(x) = 24$
Since $7$ can not divide $f'(1)$, we need to solve $24t \equiv 0 \mod 7$
we get $t \equiv 0 \pmod 5$
and then ..................... ?????
| Why not use Hensel's Lemma ?
First, let us find a solution modulo $\,7\,$ of $\,f(x)=x^{11}+x^8+5\pmod 7\,$.
Clearly, $\,r=1\,$ is a root. Now, define
$$t:=-\frac{f(1)}{7}f'(1)^{-1}=-(11+8)^{-1}=(-19)^{-1}=30^{-1}=18\pmod{49}\Longrightarrow $$
$$s:=1+18\cdot 7=127=29\pmod{49}\,\,\,\text{is a root of}\,\,\,f(x)\pmod{49}$$
What is cool in this method is that it allows you find roots modulo $\,7^2\,,\,7^3\,\ldots\,,\,7^k\,$ , for any $\,k\in\Bbb N\,$ , as long as you have one simple root modulo $\,7\,$ (why does it have to be "simple"?)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Conditional Probability with balls in an urn
Two balls, each equally likely to be colored either red or blue, are put in an urn. At each stage one of the balls is randomly chosen, its color is noted, and it is then returned to the urn. If the first two balls chosen are colored red, what is the probability that
(a) both balls in the urn are colored red; (b) the next ball chosen will be red?
I'm wondering if my method for part (a) is correct:
Let $P(R)$ be the probability of picking a red ball: $P(R)=\frac{1}{2}$
Let $P(B)$ be the probability of picking a blue ball: $P(B)=\frac{1}{2}$
Let $P(C)$ be the probability of the condition, i.e., picking two red balls consecutively: $P(C)=P(C|H_1)P(H_1)+P(C|H_2)P(H_2)+P(C|H_3)P(H_3) $
where $P(H_1)$ is the probability both balls in the urn are red: $P(H_1)=(P(R))^2=(\frac{1}{2})^2=\frac{1}4$
$P(H_2)$ is the probability that both balls in the urn are blue: $P(H_2)=(P(B))^2=(\frac{1}{2})^2=\frac{1}4$
$P(H_3)$ is the probability that one ball is red and one is blue inside the urn: $P(H_3)=1-(P(H_1)+P(H_2))=1-\frac{1}2=\frac{1}2$ since the sum of the mutually exclusive hypotheses or events must sum to 1.
Thus $P(C)=1\times\frac{1}4+\frac{1}4\times0+\frac{1}2\times\frac{1}4=\frac{3}8$
and $P(H_1|C)= \frac{P(C \bigcap H_1)}{P(C)}=\frac{P(C|H_1)P(H_1)}{P(C)}=\frac{1\times\frac{1}4}{\frac{3}8}=\frac{2}3$
For part (b), I know that $P(R|C)$, the probability of picking a red ball given that the first two balls picked were red is to be found
$P(R|C)= \frac{P(C \bigcap R)}{P(C)}=\frac{P(C|R)P(R)}{P(C)}$
How can I find $P(C|R)$?
| The probability that the contents of the urn are two red is indeed $\frac{1}{4}$, as is the probability of two blue, and the probability of mixed is therefore $\frac{1}{2}$. The derivation could have been done more quickly.
Question (a) asks for the probability both are red given that the two drawn balls are red. Let $R$ be the event both are red, and $D$ be the event both drawn balls are red. We want $\Pr(R|D)$. By the usual formula this is $\frac{\Pr(R\cap D)}{\Pr(D)}$.
To find $\Pr(D)$, note that if both balls are red (probability $\frac{1}{2}$), then the probability of $D$ is $1$, while if one ball is red and the other is not (probability $\frac{1}{2}$) then the probability of $D$ is $\frac{1}{4}$. Thus the probability of $D$ is $\left(\frac{1}{4}\right)(1)+\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)$. This is $\frac{3}{8}$.
The probability of $R\cap D$ is $\frac{1}{4}$. So the ratio is indeed $\frac{2}{3}$.
For (b), you can use the calculation of (a). With probability $\frac{2}{3}$ we are drawing from a double red, and we will get red with probability $1$. With probability $\frac{1}{3}$ the urn is a mixed one, and the probability of drawing a red is $\frac{1}{2}$, for a total of $\frac{2}{3}\cdot 1+\frac{1}{3}\cdot\frac{1}{2}$.
One can also solve (b) without using the result of (a). With I hope self-explanatoru notation, we want $\Pr(RRR|RR)$. The probabilities needed in the conditional probability formula are easily computed. We have $\Pr(RRR\cap RR)=\Pr(RRR)=\frac{1}{4}\cdot 1+\frac{1}{2}\cdot\frac{1}{8}=\frac{5}{16}$. Divide by $\frac{3}{8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$2005|(a^3+b^3) , 2005|(a^4+b^4 ) \implies2005|a^5+b^5$ How can I show that if $$2005|a^3+b^3 , 2005|a^4+b^4$$ then $$2005|a^5+b^5$$
I'm trying to solve them from $a^{2k+1} + b^{2k+1}=...$ but I'm not getting anywhere.
Can you please point in me the correct direction?
Thanks in advance
| a triffle answer;we have $2005=401.5$ so $p=5,401$ also: $$a^5+b^5=(a+b)(a^4+b^4-ab(a^2-ab+b^2)) \tag I$$.also we have
$$a^3+b^3=(a+b)(a^2-ab+b^2) \tag {II} $$ from the $p|a^3+b^3$ we have 2 cases : 1) if $p|a+b$ then from $(I)$ we have $p|a^5+b^5$
2)if $p|(a^2-ab+b^2)$ by attention to $(II)$,$p|(-ab)(a^2-ab+b^2)$and therefore $p|a^5+b^5$ by(I) and $p|a^4+b^4$.$$$$
next subsequent step is using definition of LCD:$2005=[401,5]|a^5+b^5$ .
| {
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"timestamp": "2023-03-29T00:00:00",
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Express $\sin 4\theta$ by formulae involving $\sin$ and $\cos$ and its powers. I have an assignment question that says "Express $\sin 4\theta$ by formulae involving $\sin$ and $\cos$ and its powers."
I'm told that $\sin 2\theta = 2 \sin\theta \cos\theta$ but I don't know how this was found.
I used Wolfram Alpha to get the answer but this is what I could get :
$$
4\cos^3\theta\sin\theta- 4\cos\theta \sin^3\theta
$$
How can I solve this problem?
| (Note that this method is pretty explanatory and slow, you can do it faster).
Let $u = 2\theta$, then we have:
$$ \sin 4\theta = \sin 2u $$
We know that:
$$ \sin 2u = 2\sin u\cos u$$
Now put $u = 2\theta$ back in:
$$ \sin (2 \cdot 2\theta) = 2\sin 2\theta \cos 2\theta $$
$$ \sin (4\theta) = 2\sin 2\theta \cos 2\theta $$
We know that $\sin 2\theta = 2\sin\theta\cos\theta$, so:
$$ \sin (4\theta) = 4\sin \theta \cos \theta \cos 2\theta $$
Still, we must get rid of that pesky $\cos 2\theta$. You should know the other double angle sum formula for $\cos$:
$$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$$
So:
$$ \sin (4\theta) = 4\sin \theta \cos \theta \left( \cos^2 \theta - \sin^2 \theta \right)$$
$$ \sin (4\theta) = 4\sin \theta \cos^3 \theta - 4\sin^3 \theta \cos \theta$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do we integrate, $\int \frac{1}{x+\frac{1}{x^2}}dx$? How do we integrate the following integral?
$$\int \frac{1}{x+\large\frac{1}{x^2}}\,dx\quad\text{where}\;\;x\ne-1$$
| $$\quad\int \frac{1}{x+\large\frac{1}{x^2}}\,dx \quad = \quad\int \frac{1}{\Large\frac {x^3+ 1}{x^2}}\,dx\quad= \quad\int \frac {\color{blue}{\bf x^2\,dx}}{\color{red}{\bf x^3 + 1}}\quad(x \neq -1)$$
Let us substitute $\;\;\color{red}{\bf u} = \color{red}{\bf x^3 + 1} \;\implies\; du = 3x^2\,dx \;\implies \;\color{blue}{\bf\dfrac 13 du} = \color{blue}{\bf x^2\,dx}$.
Substituting equivalent expressions gives us:
$$\int \color{blue}{\bf \frac 13} \frac{\color{blue}{\bf du}}{\color{red}{\bf u}} \;\;= \;\;\frac 13 \ln|u| + C \;\;= \;\;\frac 13 \ln|x^3 + 1| + C, \;\;x\neq -1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Common basis for subspace intersection
Let $ W_1 = \textrm{span}\left\{\begin{pmatrix}1\\2\\3\end{pmatrix}, \begin{pmatrix}2\\1\\1\end{pmatrix}\right\}$, and $ W_2 = \textrm{span}\left\{\begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}3\\0\\-1\end{pmatrix}\right\}$. Find a basis for $W_1 \cap W_2$
I first thought of solving the augmented matrix:
$ \begin{pmatrix}1 && 2 && 1 && 3\\2 && 1 && 0 && 0\\3 && 1 && 1 && -1\end{pmatrix}$
But this matrix can have 3 pivots and so it's column space dimension can be at most 3 (which doesn't make sense since the basis I'm looking for must have dimension 2.
So, what is the correct way to solve these exercise.
| Let's try to make it easier: if you take a minute to read carefully what is $\,W_2\,$ , you'll see that
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\in W_2\iff y=0$$
Now, since an element of $\,w_1\in W_1\,$ has the general form
$$w_1=a\begin{pmatrix}1\\2\\3\end{pmatrix}+b\begin{pmatrix}2\\1\\1\end{pmatrix}=\begin{pmatrix}a+2b\\2a+b\\3a+b\end{pmatrix}$$
we see that $\,w_1\in W_2\iff 2a+b=0\,$
Since $\,\dim(W_1\cap W_2)=1\,$ (why?) , we get that the above condition completely determines the elements of the intersection, and thus
$$W_1\cap W_2=\left\{\,\begin{pmatrix}a+2b\\2a+b\\3a+b\end{pmatrix}\;;\;2a+b=0\,\right\}=\left\{\,\begin{pmatrix}-3a\\0\\a\end{pmatrix}\;;\;a\in\Bbb R\,\right\}=\operatorname{Span}\left\{\,\begin{pmatrix}-3\\0\\1\end{pmatrix}\,\right\}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Intersection points of a Triangle and a Circle How can I find all intersection points of the following circle and triangle?
Triangle
$$A:=\begin{pmatrix}22\\-1.5\\1 \end{pmatrix} B:=\begin{pmatrix}27\\-2.25\\4 \end{pmatrix} C:=\begin{pmatrix}25.2\\-2\\4.7 \end{pmatrix}$$
Circle
$$\frac{9}{16}=(x-25)^2 + (y+2)^2 + (z-3)^2$$
What I did so far was to determine the line equations of the triangle (a, b and c):
$a : \overrightarrow {OX} = \begin{pmatrix}27\\-2.25\\4 \end{pmatrix}+ \lambda_1*\begin{pmatrix}-1.8\\0.25\\0.7 \end{pmatrix} $
$b : \overrightarrow {OX} = \begin{pmatrix}22\\-1.5\\1 \end{pmatrix}+ \lambda_2*\begin{pmatrix}3.2\\-0.5\\3.7 \end{pmatrix} $
$c : \overrightarrow {OX} = \begin{pmatrix}22\\-1.5\\1 \end{pmatrix}+ \lambda_3*\begin{pmatrix}5\\-0.75\\3 \end{pmatrix} $
But I am not sure what I have to do next...
| You want to find the intersection points of the lines and the cycle. You can to this by substituting.
For instance: line $a$.
$$9 = (27-1.8\lambda_1 -25)^2 + (-2.25 + 0.25\lambda_1+2)^2 + (4+0.7\lambda_1)^2$$
Now you can calculate the value of $\lambda_1$, using the Quadratic formula.
You can ran into 3 cases: the discriminant is negative, in this case you have no intersection points, the discriminant is 0, there is one intersection point (note: the line is a tangent to the cycle in the found point), of discriminant positive, therefore 2 intersection points.
You get the coordinates of the points, by substituting $\lambda_1$ into the line equation $a$.
Now you have to check, if the points are indeed part of the triangle, or just part of the extension of the line. You can do this by making the creating the vectors $AB$ and $AP$. Look if they look both in the same direction and look, if $AB$ is longer than $AP$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $x$ such that $\sum_{k=1}^{2014} k^k \equiv x \pmod {10}$ Find $x$ such that $$\sum_{k=1}^{2014} k^k \equiv x \pmod {10}$$
I knew the answer was $3$.
| This is asking for the last digit of that sum. So we may ignore everything but the last digit of each number. Now eliminate the few last terms for a minute and consider:
$$(1^1+\cdots +9^9)+(1^{11}+\cdots +9^{19})+\cdots+(1^{2001}+\cdots +9^{2009})\equiv \,?\pmod{10}$$
Notice that we can now pair each digit and form geometric progressions:
$$\begin{align*}&\quad(1^1+\cdots +1^{2001})+(2^{2}+\cdots +2^{2002})+\cdots+(9^{9}+\cdots +9^{2009})\\[7pt]&=1(1^0+1^{10}+\cdots+1^{2000})+2^2(2^0+2^{10}+\cdots+2^{2000})+9^9(9^0+9^{10}+\cdots 9^{2000})\\[7pt]\end{align*}$$
Rather than writing out the geometric series compactly, let's take it each separately:
$$1^0+1^{10}+\cdots+1^{2000}\equiv 201\equiv 1\pmod{10}$$
$$2^0+2^{10}+\cdots+2^{2000}\equiv 1+\underbrace{4+6+4+\cdots +6}_{200}\equiv 1\pmod{10}$$
$$3^0+3^{10}+\cdots+3^{2000}\equiv 1+\underbrace{9+1+9+\cdots +1}_{200}\equiv 1\pmod{10}$$
$$\cdots$$
$$9^0+9^{10}+\cdots+9^{2000}\equiv 1+\underbrace{1+1+1+\cdots +1}_{200}\equiv 1\pmod{10}$$
Now back to the original expression, we are left with:
$$1+2^2+\cdots+9^9\equiv 7\pmod{10}$$
We have a few more terms to consider, namely:
$$1^{2011}+2^{2012}+3^{2013}+4^{2014}\equiv 6\pmod{10}$$
Hence:
$$\sum_{k=1}^{2014} k^k\equiv 3\pmod{10}$$
(I have left out some of the more trivial computations to avoid clogging this answer with tedious modular arithmetic)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need help proving this integration If $a>b>0$, prove that :
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \frac{2\pi}{b^2} \left(a-\sqrt{a^2-b^2} \right) $$
| First step: $\cos(\theta+\pi)=-\cos(\theta)$ and $\sin^2(\theta+\pi)=\sin^2(\theta)$ hence the integral $I$ to be computed is
$$
I=\int_0^\pi\frac{\sin^2\theta}{a+b\cos\theta}\mathrm d\theta+\int_0^\pi\frac{\sin^2\theta}{a-b\cos\theta}\mathrm d\theta=2aJ,
$$
with
$$
J=\int_0^\pi\frac{\sin^2\theta}{a^2-b^2\cos^2\theta}\mathrm d\theta.
$$
Second step: the transformation $\theta\to\theta+\pi$ leaves the integrand unchanged hence one uses the change of variable $t=\tan\theta$, $\mathrm dt=(1+t^2)\mathrm d\theta$, which yields
$$
J=\int_{-\infty}^{+\infty}\frac{t^2}{a^2(1+t^2)-b^2}\frac{\mathrm dt}{1+t^2}.
$$
Third step: the fraction with argument $t^2$ can be decomposed as
$$
\frac{x}{(a^2(1+x)-b^2)(1+x)}=\frac1{b^2}\left(\frac{1}{1+x}-\frac{a^2-b^2}{a^2(1+x)-b^2}\right),
$$
hence
$b^2J=K-(a^2-b^2)L$ with
$$
K=\int_{-\infty}^{+\infty}\frac{\mathrm dt}{1+t^2},\qquad L=\int_{-\infty}^{+\infty}\frac{\mathrm dt}{a^2t^2+a^2-b^2}.
$$
Fourth step: the change of variable $at=\sqrt{a^2-b^2}s$ yields
$$
L=\frac{\sqrt{a^2-b^2}}a\int_{-\infty}^{+\infty}\frac{\mathrm ds}{(a^2-b^2)(1+s^2)}=\frac{K}{a\sqrt{a^2-b^2}},
$$
and the integral $K$ is classical, its value is $K=\pi$.
Conclusion:
$$
I=2aJ=\frac{2a}{b^2}\left(K-(a^2-b^2)\frac{K}{a\sqrt{a^2-b^2}}\right)=\frac{2\pi}{b^2}\left(a-\sqrt{a^2-b^2}\right)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Fibonacci Sequence in $\mathbb Z_n$. Consider a Fibonacci sequence, except in $\mathbb Z_n$ instead of $\mathbb Z$:
$$F(1) = F(2) = 1$$ $$F(n+2) = F(n+1) + F(n)$$
It is easy to see that each of these sequences must cycle through some sequence and repeat. Consider, for example, the sequence in $\mathbb Z_4$:
$$1,1,2,3,1,0,1,1,2,3,1,0,\dots$$
Is there a nice way to describe the length of the sequence that results before repeating? Clearly, it must be less than $n^2$, since that would cycle through all possible combinations of $a,b$ with $a,b \in \mathbb Z_n$, and once you repeat two combinations, you have a cycle.
\begin{align*}
\mathbb Z_2: 1,1,0,\dots & 3 \\
\mathbb Z_3: 1,1,2,0,2,2,1,0,\dots & 8 \\
\mathbb Z_4: 1,1,2,3,1,0,\dots & 6 \\
\mathbb Z_5: 1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0,\dots & 20
\end{align*}
Examining the sequences above, it seems "obvious" that there's some pattern (just look at how $0$ occurs every $5$th number). Is there a "nice" way to describe these sequences (such as a way to calculate the $k^{th}$ term in a sequence), and, in particular, a rule for the lengths of the cycles?
| I found a way to do this with prime $n.$ Not sure how well it extends to prime powers.
We want to know what happens to the pair of consecutive entries $(x,y),$ shifting over one to get $(y,x+y).$ So we define the matrix
$$ A \; = \;
\left( \begin{array}{rr}
0 & 1 \\
1 & 1
\end{array}
\right) ,
$$ and note
$$
\left( \begin{array}{rr}
0 & 1 \\
1 & 1
\end{array}
\right)
\left( \begin{array}{c}
x \\
y
\end{array}
\right) \; = \;
\left( \begin{array}{c}
y \\
x+ y
\end{array}
\right)
$$
We find $A^2 = A + I,$ as well as $A^{-1} = A - I.$ As a result, the set of sums of powers of $A$ make a commutative ring; we calculate
$$ A^3 = 2A + I, $$
$$ A^4 = 3A + 2I, $$
$$ A^5 = 5A + 3I, $$
$$ A^6 = 8A + 5I, $$
$$ A^7 = 13A + 8I, $$
$$ A^8 = 21A + 13I, $$
and so on.
We do not know for sure that we have a repetition until we have repeated the pair $(x,y).$ Next, what if we regard the coefficients in this ring as elements of the field $\mathbf Z / p \mathbf Z ?$
We would like to know the smallest $m$ such that
$$ A^m \equiv I \pmod p. $$ There are two parts to this. We illustrate first.
$$ A^3 \equiv I \pmod 2. $$
$$ A^4 \equiv 2I \pmod 3, 2^2 \equiv 1 \pmod 3, A^8 \equiv I \pmod 3. $$
$$ A^5 \equiv 3I \pmod 5, 3^4 \equiv 1 \pmod 5, A^{20} \equiv I \pmod 5. $$
$$ A^8 \equiv 6I \pmod 7, 6^2 \equiv 1 \pmod 7, A^{16} \equiv I \pmod 7. $$
For each prime, the second exponent is a factor of $p-1,$ as
$$ \lambda^{p-1} \equiv 1 \pmod p. $$
I forgot one item earlier. If $p \equiv 2,3 \pmod 5,$ then there are no solutions to $x^2 - x - 1 \pmod p,$ the thing is irreducible, and our ring of matrices is a field with $p^2$ elements. The set of nonzero elements is a multiplicative group with $p^2 - 1$ elements. It follows that, with our $A^m \equiv I \pmod p,$ that $m | (p^2 - 1).$
For $p = 5$ or $p \equiv 1,4 \pmod 5,$ the ring has zero divisors, given explicitly by the factors of $x^2 - x - 1 \pmod p$ and scalar multiples of those. Annoying. On the other hand, for $p \equiv 1,4 \pmod 5$ it appears $A^{p-1} \equiv I \pmod p,$ true for $p=11,19,29,31,41,$ and perhaps provable.
From the link given in comments above about Pisano periods, I finally understand how to do $p \equiv 1,4 \pmod 5.$ Let $\phi$ be either of the two distinct roots of $\phi^2 - \phi - 1 \equiv 0 \pmod p. $ Then, with the numbering $F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5,$ we find
$$ F_n = \frac{\phi^n - (1-\phi)^n}{2 \phi - 1}. $$
Using $a^{p-1} \equiv 1 \pmod p$ and $\phi^2 \equiv \phi + 1 \pmod p$ and $\frac{1}{\phi} \equiv \phi - 1 \pmod p,$
we find both
$$ F_{p-1} \equiv 0 \pmod p, \; \; \; F_{p-2} \equiv 1 \pmod p. $$
From our earlier discussion of the matrix $A,$ we see that
$$ A^n = F_n A + F_{n-1} I, $$ pointed out below by awllower. So
$$ A^{p-1} = F_{p-1} A + F_{p-2} I \equiv I \pmod p. $$
As a result, any such sequence, Fibonacci or Lucas or what have you, repeats with period $p-1$ or some divisor thereof.
| {
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How do I transform the left side into the right side of this equation? How does one transform the left side into the right side?
$$
(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2
$$
| Use the method of foil on the right hand side and then foil the left hand side. It's easier doing right hand side to the left hand side.
$$(ac-bd)^2=(a^2c^2-2abcd+b^2d^2)$$
$$(ad+bc)^2=(a^2d^2+2abcd+b^2c^2)$$
Then simplify.
For the right hand side:
$$(ac-bd)^2+(ad+bc)^2= (a^2c^2-2abcd-b^2c^2)+(a^2d^2+2abcd+b^2c^2)$$
$$=a^2c^2+a^2d^2+b^2d^2+b^2c^2=RHS$$
Then take the left hand side and foil.
$$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2=LHS$$
Therefore
$RHS=LHS$ Does this make sense?
| {
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"source": "stackexchange",
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"answer_count": 7,
"answer_id": 4
} |
How to solve this system of equations? How to solve this system of equations?
$$\begin{cases}
1+\sqrt{2 x+y+1}=4 (2
x+y)^2+\sqrt{6 x+3 y},\\
(x+1) \sqrt{2 x^2-x+4}+8 x^2+4
x y=4.
\end{cases}$$
| The trick here is to not use both equations at the same time.
The first equation screams to use the substitution $k=2x+y$. Substituting that, we get $1+\sqrt{k+1}=4k^2+\sqrt{3k}$. So we transform it to $(4k^2-1)+(\sqrt{3k}-\sqrt{k+1})=0$. So $(4k^2-1)(\sqrt{3k}+\sqrt{k+1})+(2k-1)=0$.
(this is because we multiplied both sides by $\sqrt{3k}+\sqrt{k+1}$.
Thus we have either $2k-1=0$ or $(2k+1)(\sqrt{3k}+\sqrt{k+1})+1=0$. For the quantity $(2k+1)(\sqrt{3k}+\sqrt{k+1})$ to be real, we need $k$ nonnegative. However, if $k$ is nonnegative, then the quantity is clearly positive, so no roots.
Thus we need $k=\frac{1}{2}$. $8x^2+4xy=4x(2x+y)=2x$. So $(x+1)\sqrt{2x^2-x+4}+2x=4$. $(x+1)\sqrt{2x^2-x+4}=4-2x$, and squaring we get $(x+1)^2(2x^2-x+4)=(4-2x)^2$. After expanding, we get $2x^4+3x^3+23x-12=0$, so it factors as $(2x-1)(x+3)(x^2-x+4)=0$. Since the discriminant of $x^2-x+4$, which is $1-4(4)$, is negative, it doesn't have a real root.
Hence the possible values of $x$ are $\frac{1}{2}$ and $-3$, giving values of $y$ as $\frac{-1}{2}$ and $\frac{-11}{2}$. We could just "check" them to see that both work. (just plug back in)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/332059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int \frac{1}{{x^4+1}} dx$ I am trying to evaluate the integral
$$\int \frac{1}{1+x^4} \mathrm dx.$$
The integrand $\frac{1}{1+x^4}$ is a rational function (quotient of two polynomials), so I could solve the integral if I can find the partial fraction of $\frac{1}{1+x^4}$. But I failed to factorize $1+x^4$.
Any other methods are also wellcome.
| $$I =\int \frac{1}{{x^4+1}} dx$$
If we add and subtract $2x^2$ to $x^4 + 1$, we get:
$$\int \frac{1}{{x^4 + 2x^2 + 1 - 2x^2}}$$
We know $x^4 + 2x^2 + 1 = (x^2 + 1)^2$
$$\int \frac{1}{{(x^2 + 1)^2 - 2x^2}}$$
We know $a^2 - b^2 = (a - b)(a + b)$
Hence, $(x^2 + 1)^2 - 2x^2 = (x^2 + 1 - \sqrt{2}x)(x^2 + 1 + \sqrt{2}x)$
$$\int \frac{1}{{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)}}$$
Now using partial fraction decomposition:
$$\frac{1}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)} = \frac{Ax + B}{x^2-\sqrt{2}x+1} + \frac{Cx + D}{x^2+\sqrt{2}x+1}$$
After you have found A, B, C and D, its just a basic $\frac{linear}{quadratic}$ type integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
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