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Series expansion of $\frac{x^n-1}{x-1}$ at $ x=1$. $$\lim_{x\to 1}\left(\frac{x^n-1}{x-1}\right)=n$$
I thought that at $x=1$ the expansion is:
$$n+n(n-1)(x-1)+n(n-1)(n-2)(x-1)^2+...$$
but the answer is:
$$n+\frac{1}{2}n(n-1)(x-1)+\frac{1}{6}n(n-1)(n-2)(x-1)^2+...$$
Can you explain the origin of $$\frac{1}{2},\frac{1}{6},\frac{1}{24}, ...?$$
| By Binomial formula we have
$$x^n =((x-1)+1)^n = \sum_{j=0}^{n}{n\choose j}(x-1)^j=1+(x-1) \sum_{j=1}^{n}{n\choose j}(x-1)^{j-1}$$
Hence,
$$\begin{align}\frac{x^n-1}{x-1} &= \sum_{j=1}^{n}{n\choose j}(x-1)^{j-1} \\&= {n\choose 1}+{n\choose 2}(x-1)^{1}+{n\choose 3}(x-1)^{2} +{n\choose 4}(x-1)^{3}+\cdots
\\&= n +\color{blue}{\frac{n(n-1)}{2}}(x-1)^{1}+\color{blue}{\frac{n(n-1)(n-2)}{6}}(x-1)^{2} +\color{blue}{\frac{n(n-1)(n-2)(n-3)}{24}}(x-1)^{3}+\cdots \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find Laurent expansion of $f(z)=\frac{1}{(z+i)z^2}$ on $0<|z-i|<1$ I think the way is doing the decomposition as follows, but I just don't know how to deal with the quadratic term:
$f(z)=\frac{1}{(z+i)z^2}=-\frac{1}{z+i}+\frac{1}{z}-\frac{i}{z^2}$
where, $-\frac{1}{z+i}=-\frac{1}{2i}\frac{1}{1+\frac{z-i}{2i}}$ and $\frac{1}{z}=\frac{1}{1+\frac{z-i}{i}}$ can be expanded easily.
But how to deal with $-\frac{i}{z^2}$?
| Let $u = z+i$. Then $z = u - i$, so
\begin{align*}
\frac{1}{z^2} = \frac{1}{(i - u)^2} = \frac{1}{-(1 - u/i)^2} = -\frac{1}{(1 + iu)^2}
\end{align*}
Since $0 < |u| < 1$, then $0 < |iu|< 1$ as well, so this can be expanded as the derivative of a geometric series:
\begin{align*}
\frac{1}{1 - x} = \sum_{n \geq 0} x^n \implies \frac{1}{(1 - x)^2} = \frac{d}{dx} \frac{1}{1-x} = \frac{d}{dx} \sum_{n \geq 0} x^n = \sum_{n \geq 1} n x^{n-1} \, .
\end{align*}
| {
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Given $|x| = 1$ and $x \notin \mathbb R$ prove that $\Re\left(\frac{x-1}{x+1}\right) = 0$
Given $|x| = 1$ and $x \notin \mathbb R$ prove that $\Re\left(\frac{x-1}{x+1}\right) = 0$
I have tried to express $x$ in terms of $a$ and $b$, namely: $x = a+bi, |x|= \sqrt{a^2+b^2}$ and to evaluate this but this simply led me nowhere. I guess there must be a clever way to solve that I don't see. I'd be thankful if you could give me some hints (not the actual solution) on how to proceed with this problem.
| Hint 1: $\overline{x} + 1 = \overline{x+1}$.
Let $x=a+ib$. Then
$$ \overline{x} + 1 = \overline{a + ib} + 1 = a - ib + 1 = (a+1) - ib, \tag{1}$$
and
$$\overline{x+1} = \overline{a+ib+1} = \overline{(a+1) + ib} = (a+1) - ib. \tag{2}$$
Since (1) and (2) agree, we conclude that $\overline{x} + 1 = \overline{x+1}$.
Hint 2: $(x+1)(\overline{x+1}) = |x+1|^2$.
This is an application of the results that $z\overline{z} = |z|^2$ for any $z\in \mathbb{C}$. To see this, recall that if $z = a+ib$, then
$$|z| := \sqrt{a^2 + b^2}. $$
But
$$ z\overline{z} = (a+ib)(a-ib) = a^2 - aib + iba - i^2 b^2 = a^2 + b^2 = |z|^2, $$
which is the desired identity.
Hint 3: $(x-1)(\overline{x}+1) = |x|^2 + 2i\Im(x) - 1$.
Let $x = a+ib$. Then
\begin{align} (x-1)(\overline{x}+1) &= x\overline{x} + x - \overline{x} - 1 \\&= |x|^2 + (a+ib) - (a+ib) - 1 \\&= |x|^2 + 2ib - 1 \\&= |x|^2 + 2i \Im(x) - 1. \end{align}
If you can show each of these and put them together in the right order, they can be used to deduce a complete solution to your problem.
Putting the pieces together, we have
\begin{align}\frac{x-1}{x+1}&= \frac{x-1}{x+1} \cdot \frac{\overline{x+1}}{\overline{x+1}} \tag{"rationalize" the denominator}\\&= \frac{(x-1)(\overline{x}+1)}{(x+1)(\overline{x+1})} \tag{Hint 1} \\ &= \frac{|x|^2 + 2i\Im(x) - 1}{|x+1|^2}. \tag{Hints 2 and 3}\end{align}
Then the real part is given by
$$ \Re\left( \frac{|x|^2 + 2i\Im(x) - 1}{|x+1|^2} \right) = \frac{|x|^2 - 1}{|x+1|^2}. $$
But this is zero by the assumption that $|x|=1$.
| {
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Limit of $\frac{1}{x}\left( {1 - \frac{{\sin x}}{x}} \right)$ $\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left( {1 - \frac{{\sin x}}{x}} \right) = 0$
The result can be checked with Taylor series or with L'Hôpital's rule.
I wonder if it's possible to reach the same result with some standard algebraic manipulations (given that $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$)
| Following @labbhattacharjee hint here's my elaboration on Are all limits solvable without L'Hôpital Rule or Series Expansion.
I've added this "self-reply" in the hope it might be useful for others in this specific case. I've also added some more detail to the linked post derivation.
$L_0=\mathop {\lim }\limits_{x \to 0} \frac{1}{x}\left( {1 - \frac{{\sin x}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{{x^3}}}x$
We calculate the limit $L$ of the first factor inside $L_0$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{{x^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{3y - \sin 3y}}{{27{y^3}}}$ where we have set $y = 3x$
Now we must use the trigonometric identity
$\sin 3y = \sin \left( {2y + y} \right) = \sin 2y\cos y + \cos 2y\sin y = 2\sin y{\cos ^2}y + \left( {1 - 2{{\sin }^2}y} \right)\sin y = 2\sin y - 2{\sin ^3}y + \sin y - 2{\sin ^3}y$
that is
$\sin 3y = 3\sin y - 4{\sin ^3}y$
Then we have
$L = \mathop {\lim }\limits_{y \to 0} \frac{{3y - \sin 3y}}{{27{y^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{3y - 3\sin y + 4{{\sin }^3}y}}{{27{y^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{3y - 3\sin y}}{{27{y^3}}} + \mathop {\lim }\limits_{y \to 0} \frac{{4{{\sin }^3}y}}{{27{y^3}}} = \frac{1}{9}\mathop {\lim }\limits_{y \to 0} \frac{{y - \sin y}}{{{y^3}}} + \frac{4}{{27}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{{x^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{3y - \sin 3y}}{{27{y^3}}} = \mathop {\lim }\limits_{y \to 0} \frac{{y - \sin y}}{{{y^3}}} = \frac{1}{9}L + \frac{4}{{27}}$
So we have the equation
$L = \frac{1}{9}L + \frac{4}{{27}}$
whose solution is
$L = \frac{1}{6}$
The original limit is then $L_0=0$ as it is the product of a finite number and an infinitesimal quantity
| {
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Miller-Rabin: Showing non trivial divisors of $n$
Let's assume $n$ doesn't pass the Miller-Rabin test and $b$ is a witness. Meaning, $b^{\frac{n-1}{2^r}} \equiv 1 \pmod{n}$ where $\frac{n-1}{2^r}$ is even, but $c = b^{\frac{n-1}{2^{r+1}}} \not\equiv \pm1 \pmod{n}$. Show that $\gcd(c+1,n),\gcd(c-1, n)$ are non-trivial divisors of $n$.
So first of all, for my convenience:
*
*Denote $n-1 = 2^ls$, $s$ is odd.
*Denote $k=l-r$.
*
*$c = b^{2^{k-1}s}$
*$c^2 = b^{2^{k}s}$
*$2^ks$ is even
It is given that $c^2 \equiv 1 \pmod{n} \implies (c-1)(c+1) \equiv 0 \pmod {n} \implies n\mid (c-1)(c+1)$
Now, I think we can assume that $n$ passed Fermat's theorem test (that's part of the initial tests of the Miller-Rabin algorithm).
Hence,
$$c^{r+1} = b^{2^l s} = b^{n-1} \equiv 1 \pmod {n}$$
but that isn't revealing anything new other than $r$ is odd (since $2$ must divide $r+1$).
What am I missing?
| From $b^{\frac{n-1}{2^r}}\equiv 1\pmod n$, we see that there exists an integer $m$ such that
$$b^{\frac{n-1}{2^r}}=mn+1\tag1$$
From $b^{\frac{n-1}{2^{r+1}}}\not\equiv \pm 1\pmod n$, we see that there exist integers $k,r$ such that
$$b^{\frac{n-1}{2^{r+1}}}=kn+r+1\tag2$$
where $$\text{$1\le r\le n-1\quad$ with $\quad r\not=n-2$}\tag3$$
Since $\frac{n-1}{2^r}$ is even, we get, from $(1)(2)$,
$$mn+1=b^{\frac{n-1}{2^r}}=b^{\frac{n-1}{2^{r+1}}\times 2}=\left(b^{\frac{n-1}{2^{r+1}}}\right)^2=(kn+r+1)^2=(k^2n+2kr+2k)n+(r+1)^2$$
implying
$$1\equiv (r+1)^2\equiv r^2+2r+1\pmod n$$
which implies $$r(r+2)\equiv 0\pmod n\tag4$$
Here, supposing that $r+2=n+1$ implies $r\equiv 0\pmod n$ which is impossible.
So, from $(3)$, we have $1\lt r\lt n$ and $1\lt r+2\lt n$.
It follows from these and $(4)$ that both $r$ and $r+2$ are non-trivial divisors of $n$.
We have $b^{\frac{n-1}{2^{r+1}}}+1=kn+r+2$. Since $r+2$ is a non-trivial divisor of $n$, we see that $\gcd\left(b^{\frac{n-1}{2^{r+1}}} +1,n\right)$ is a non-trivial divisor of $n$.
We have $b^{\frac{n-1}{2^{r+1}}}-1=kn+r$. Since $r$ is a non-trivial divisor of $n$, we see that $\gcd\left(b^{\frac{n-1}{2^{r+1}}} -1,n\right)$ is a non-trivial divisor of $n$.
| {
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Expansion of $S=\frac{{nx^{n+1}-(n+1)x^{n}+1}}{(x-1)^2}$ Expand;
$S=\frac{{nx^{n+1}-(n+1)x^{n}+1}}{(x-1)^2}$
See solution, Also I am looking for an advance method if exists.
| Solution:
$S=\frac{{nx^{n+1}-(n+1)x^{n}+1}}{(x-1)^2}=\frac{{(n+1)x^{n+1}-(n+1)x^{n}+1}-x-x^{n+1}+x}{(x-1)^2}=\frac{(x-1)[(n+1)x^n -1]-(x^{n+1}-x)}{(x-1)^2}=(\frac{x^{n+1}-x}{x-1})^{'}=(x+x^2+x^3+ . . .x^n)^{'}=1+2x+3x^2+4x^3+ . . .nx^{n-1}$
| {
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Geometric interpretation for median formula $m_c = \sqrt{\frac{2(a^2 + b^2)- c^2}{4}}$ for $a + b < c$? For a triangular with side lengths $a, b, c$, the length of the median of $c$ is $m_c = \sqrt{\frac{2(a^2 + b^2) - c^2}{4}}$.
If $a + b < c$, the configuration is obviously not a triangle anymore.
However, there are values for $a, b, c$ with $a + b < c$ (e.g. $a = 7, b = 2, c = 10$) where the root is still positive. Is there any geometric interpretation for that?
| For the median, what we need is not just $m_c\ge0$. We actually need $\displaystyle m_c\ge\frac{|a-b|}{2}$.
When $a=7$ and $b=2$, $\displaystyle \frac{|a-b|}{2}=\frac{5}{2}$.
Now $\displaystyle \sqrt{\frac{2(a^2+b^2)-c^2}{4}}=\frac{\sqrt{6}}{2}<\frac{5}{2}$.
| {
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Proving $24\mid5^{2n}-1$ using modular arithmetic So, in general I am aware of how to use modular arithmetic to prove a divisibility. But I have the following problem:
Prove that $24\mid5^{2n}-1$ for all $n\in\mathbb Z$.
I know that theoretically, I could show 23 different cases that the expression is congruent to $0\bmod24$, but that seems like it might be excessive to me.
Is there a faster way to show this?
| 1)
Prove $8|5^{2n} - 1$ and $3|5^{2n} - 1$
a) $5^{2n} - 1 = (5^n -1)(5^n + 1)$. $5^n \pm 1 \equiv 1^n\pm 1 \equiv 0\mod 2$ so $2|5^2 - 1$ and $2|5^2 + 1$ and so $4|5^{2n}-1$. If $5^n - 1 = 2k$ then $5^{2n} + 1= 2k + 2 = 2(k+1)$. Either $k$ is even or $k+1$ is even so either $2k$ is even and $2(k+1)$ is divisible by $4$, or $2k$ is divisible by $4$ and $2(k+1)$ is even. Either way. $2k*2(k+1)= 5^{2n} -1$ is divisible by $8$.
b)$5^{2n}-1\equiv (-1)^{2n}-1 \equiv 1^n - 1 \equiv 0 \mod 3$.
So $3|5^{2n}-1$.
2)
Or notice $5^{2n} -1 = 25^n -1 \equiv 1^n -1 \equiv 0 \mod 24$.
(I will admit, I did not see the obvious right away.)
| {
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Prove that there does not exists RATIONAL numbers $x$ and $y$ such that $x^2 - y^2 = 1002$ I tried to prove this using contradiction.
Suppose there does exists rationals $x$ and $y$ such that $x^2 - y^2 = 1002$
Let $ x = a/b $ and $y = c/d$
Irreducible fractions.
I come up with $$ (ad)^2 - (bc)^2 = 1002 \cdot (bd)^2 $$
Thus
$$ (ad)^2 = (cb)^2 \pmod{1002}$$
And i'm stuck on this
Any tip or hint?
Thanks
This problem appears in the first chapter 'Contemporary Abstract Algebra"
| If $(x^2 - y^2) = 1002$
$(x + y)(x - y) = 1002$
Let $x + y = 2$ and $x - y = 501$
Solve for $x,y$. $x =\frac {503}2$ and $y = - \frac {499}{2}$.
That's a solution. $\frac {503^2}4 - \frac {499^2}{4} = \frac {253009-249001}4=\frac {4008}{4 }= 1002$.
So not true.
Now if $x$ and $y$ must be integers..
We have $(x+y)(x-y) = 2*3*167$.
$(x + y) + (x-y) = 2x$ so $x+y$ and $x-y$ must both be odd or both be even. But $2$ must divide one but only one of $x+y$ or $x-y$.
| {
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Integrate $\int \frac{\sin x \cos x}{\sin^4x + \cos^4x} \,dx$
Integrate $$\int \frac{\sin x \cos x}{\sin^4x + \cos^4x}dx$$
I solved the question by using the identity $\cos^4(x)+\sin^4(x) = \frac{1}{4}(\cos4x+3)$ and the substitution $u=\cos4x +3$, which turned it into a relatively familiar integral (see my answer below).
However, I'm pretty sure there are easier ways I am missing, so please feel free to post alternative answers.
There is a similar question here.
Problem Source: James Stewart Calculus, 6E
| $$ \frac{\sin x\cos x}{\sin^4 x + \cos^4 x} = \frac{\sin 2x}{2(1 - 2\sin^2x \cos^2 x)} = \frac{\sin 2x}{2 - (1-\cos 2x)(1 + \cos 2x)} $$
Substitute $u = \cos 2x$ to get
$$ -\frac{1}{2}\int\frac{du}{1+u^2} = -\frac{1}{2}\arctan u = \color{blue}{-\frac{1}{2}\arctan (\cos 2x)} $$
| {
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Evaluate $\int_{1}^{2} \int_{1}^{3} \frac{xy}{x^2 +y^2} \; dx\; dy$
How does one evaluate the following? $$\int_{1}^{2} \int_{1}^{3} \frac{xy}{x^2 +y^2} \, dx\, dy$$
It gets very tedious when I use a substitution method. Is there a short way to evaluate it?
| Notice that : $\int \frac{xy}{x^2+y^2}dx = \frac{y}{2}\ln(x^2+y^2) + C $
$$\int_{1}^{2} \int_{1}^{3} \frac{xy}{x^2 +y^2} \; dx\; dy = \int_1^2 \frac{y}{2}\ln(9+y^2)-\frac{y}{2}\ln(1+y^2)dy= \int_1^4 \frac{\ln(9+u)}{4}-\frac{\ln(1+u)}{4}du$$ where $u=y^2$ and $du=2ydy $
Setting $t=9+u$ and $dt=du$ and $s=1+u$ and $ds=du$ we have
$$\int_{10}^{13} \frac{\ln(t)}{4}dt-\int_2^5 \frac{\ln(s)}{4}ds$$
And then use integration by parts
| {
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Non negative real values of $a,b,c$ The question is to find out the minimum value of non negative real values of $a,b,c$ given that the equation $$x^4+ax^3+bx^2+cx+1=0$$ has real roots.
I tried dividing both sides by $x^2$ the resultant equation becomes $$x^2+ax+b+\frac{c}{x} + \frac{1}{x^2}=0$$ I couldn't transfer this to a quadratic equation.If I take a=c then a quadratic equation may be formed by substituting $x+1/x=t$ with this assumption the equation reduces to $$t^2+at+b-2=0$$ Now we know that $t \geq 2$ when $x$ is positive and $t \leq -2$ when $x$ is negative. So we can check that if the roots of $t$ be in the interval $(-2,2)$ the corresponding value of $x$ will be imaginary.For this we have three conditions $f(-2)>0$,$f(2)>0$ and $\frac{-a}{2} \in (-2,2)$ [This comes from the fact that the minimum must be in the interval of the roots] and the discriminant must be greater than or equal to zero.Using these conditions I get that (4,6,4) are the values of (a,b,c) However I am not sure how to proceed if $a\neq c$ .Any ideas?
| The idea followed here is, for $a \ne c$, to apply a shift to $x$ in order to transform the equation into a form for which the OP has already given a solution. From that, we can infer back to the values of $a,b,c$.
Applying a shift $x = y+q$ gives
$$
1 + c q + b q^2 + a q^3 + q^4 + (c + 2 b q + 3 a q^2+ 4 q^3) y + (b + 3 a q + 6 q^2) y^2 + (a + 4 q )y^3 + y^4 = 0
$$
Now you can select the $q = q(a,b,c)$ which solves
$$
c + 2 b q + 3 a q^2+ 4 q^3 = a' r\\
a + 4 q = a'\\
1 + c q + b q^2 + a q^3 + q^4 = r^2
$$
(coming back to that later).
Further, identify $b' = b + 3 a q + 6 q^2$. This gives
$$
0 = r^2+ a' r y + b'y^2 + a'y^3 + y^4 = y^2 (r^2/y^2+ a'r/ y + b' + a'y + y^2)
$$
Setting $z = y/\sqrt r$ gives
$$
0 = 1/z^2+ (a'/\sqrt r)/ z + b'/r + (a'/\sqrt r) z + z^2
$$
Now we can apply the OP's solution which gives
$a'/\sqrt r = 4$ and $b'/r = 6$. Plugging that back into the equations above gives
$$
c + 2 b q + 3 a q^2+ 4 q^3 = 4 \sqrt{r^3}\\
a + 4 q = 4 \sqrt r\\
1 + c q + b q^2 + a q^3 + q^4 = r^2\\
b + 3 a q + 6 q^2 = 6 r
$$
[One can check that for $a=c$, indeed $r=1$ and $q=0$ solve this system.]
Now here again fourth-order equations are present so it won't be easy. We need to identify $a,b,c$ from these four equations by eliminating $q$ and $r$. (is there a fifth equation hidden somewhere? This would generally be necessary for eliminating $r$ and $q$, however: wait for what follows.)
We get $r = (q + a/4)^2$ and hence $b/6 + a q/2 + q^2 = q^2 + a q/2 + a^2/16$ or $ b = 3a^2 / 8$.
Putting the results for $r$ and $b$ into the first system equation gives
$
c + 6 a^2 q / 8 + 3 a q^2+ 4 q^3 = a^3/16 + (3 a^2 q)/4 + 3 a q^2 + 4 q^3
$, so $c =a^3/16 $
Further, $r^2 = (q + a/4)^4 = a^4/256 + (a^3 q)/16 + (3 a^2 q^2)/8 + a q^3 + q^4 = 1 + c q + b q^2 + a q^3 + q^4 $. Plugging in $b$ and $c$ we have $a^4/256 = 1 $ which gives $a = 4$ no matter which shift $q$.
So collecting we have $(a,b,c) = (4,6,4)$ as the smallest values even if we started without the condition $a=c$.
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Calculate the domain of Rational function, which contains Trigonometric, Irrational and Log functions. I'm trying to find the domain of the following function:
$$f(x)={\frac{\displaystyle\ \sqrt{\frac 12-{\log_3}\biggl(\frac 12 \tan(x) + \sin(x)\biggr)} + \sqrt{\pi^2-4x^2}}{\displaystyle \arcsin\biggl(\sqrt{x^2-x} - |x|\biggr)}}$$
I have reasoned this way: since I have a Rational function, its denominator must be posed $\neq0$; the Irrational functions' argument need to be $\ge0$ and the Arcsin's argument must be $-1\le x\le 1$.
Hence I have to solve the following system:
$$
\left\{
\begin{array}{c}
\frac 12-{\log_3}\biggl(\frac 12 \tan(x) + \sin(x)\biggr)\ge0 \\
\frac 12 \tan(x) + \sin(x) >0\\
x\neq \frac \pi2 + k\pi\\
\pi^2-4x^2 \ge 0\\
\arcsin\biggl(\sqrt{x^2-x} - |x|\biggr) \neq 0 \\
-1\le\sqrt{x^2-x} - |x|\le1\\
x^2-x \ge0
\end{array}
\right.
$$
Now come the point in which I get blocked. I do not know how to solve the first inequality of the system. Specifically I proceed in this way:
I express the second member's inequality in logarithm and I obtain this situation:
$${\log_3}\biggl(\frac 12 \tan(x) + \sin(x)\biggr) \le \log_3\Bigl(\sqrt3 \Bigr)\quad\mathbf{(1)}$$
Next I apply logarithm inequalities rule, and then I express $\sin(x)$ in $\tan(x)$:
$$\frac 12 \tan(x)+\frac{\tan(x)}{\sqrt{1+\tan^2(x)}}\le\sqrt3$$
I try to simplify:
$$\frac{\tan(x)\sqrt{1+\tan^2(x)}+2\tan(x)-2\sqrt3\biggl(\sqrt{1+\tan^2(x)}\biggr)}{2\sqrt{1+\tan^2(x)}}\le0$$
From here, I do not know how to solve this inequality.
Someone could say me if my calculus is wrong or could give me another way (or hint) to determine the domain of f(x)? Thank you.
UPDATE:
The (1) became, by using logarithm inequalities rules:
$$\frac 12 \tan(x) + \sin(x) \le \sqrt3$$
$$\frac 12 \frac {\sin(x)}{\cos(x)} + \sin(x) \le \sqrt3 \quad \mathbf{(2)}$$
I do this sostitution: $$t=\tan\Bigl(\frac x2\Bigr); \quad \sin(x)=\frac{2t}{1+t^2};\quad \cos(x) = \frac{1-t^2}{1+t^2}.$$
So applying this to the (2):
$$\frac 12 \frac {2t}{1+t^2} \frac{1+t^2}{1-t^2} + \frac{2t}{1+t^2} \le \sqrt 3$$
$$\frac{t}{1-t^2} + \frac{2t}{1+t^2} \le \sqrt 3$$
$$\frac{t(1+t^2)+2t(1-t^2)-\sqrt{3}(1-t^4)}{(1-t^4)}\le0 \quad \mathbf{(3)}$$
I try to solve the (3), and so:
The numerator solution should be:
$$t+t^3+2t-2t^3-\sqrt 3 + \sqrt 3 t^4 \ge 0$$
$$\sqrt 3 t^4 -t^3 +3t -\sqrt 3 \ge 0$$
$$\biggl(t-\frac{\sqrt 3}{3}\biggr)\biggl(t^3+\sqrt 3\biggr)\ge0$$
$$t\le-\sqrt[6]{3} \quad\lor\quad t\ge\frac{\sqrt 3}{3}$$
$$\tan\Bigl(\frac x2\Bigr)\le-\sqrt[6]{3} \quad\lor\quad \tan\Bigl(\frac x2\Bigr)\ge\frac{\sqrt 3}{3}$$
$$\frac \pi2 < \frac x2 \le \arctan(\sqrt[6]{-3}),\ in \ k\pi \quad\lor\quad \frac \pi6 \le \frac x2 < \frac \pi2,\ in \ k\pi$$
$$\pi < x < -2 \arctan(\sqrt[6]{3}),\ in \ 2k\pi \quad\lor\quad \frac \pi3 \le x < \pi,\ in\ 2k\pi$$
The denominator solution should be:
$$1-t^4\ge0$$
$$-1\le t \le 1$$
$$-1\le \tan\Bigl(\frac x2\Bigr) \le 1$$
$$0\le \frac x2 \le \frac \pi4,\ in \ k\pi \quad\lor\quad \frac 34 \pi \le \frac x2 < \pi,\ in \ k\pi$$
$$0\le x\le \frac \pi2,\ in \ 2k\pi \quad\lor\quad \frac 32 \pi \le x < 2 \pi,\ in \ 2k\pi$$
But my numerator solution seems to not be right, someone has an idea or could say me where I'm wrong? Thank you.
| May be if we use this we might get something,
$$\frac{1}{2} - \log_3(\frac{1}{2}tan(x)+sin(x)) \geq 0$$
$$\Rightarrow \log_3(\frac{1}{2}tan(x)+sin(x))\leq \frac{1}{2}$$
$$\Rightarrow \frac{1}{2}tan(x)+sin(x) \leq 3^\frac{1}{2} $$
now, multiply $\sqrt{3}$ we get,
$$ \frac{\sqrt{3}}{2}tan(x)+{\sqrt{3}}sin(x) \leq 3$$
$$\Rightarrow sin(\frac{\pi}{3})tan(x) + tan(\frac{\pi}{3})sin(x)\leq3 $$
I am not able to solve further.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify: $\arctan \sqrt{\frac{1-\cos x}{1+\cos x}};$ $0$$\arctan \sqrt{ \frac{1-\cos x}{1+\cos x} } $$
There are two ways you can procced with this:
1.$$=\arctan{\sqrt{\left(\frac{2\sin^2{\frac{x}{2}}}{2\cos^2{\frac{x}{2}}}\right)}}$$
$$=\arctan{\left(\tan\frac{x}{2}\right)}$$
$$=\frac{x}{2}$$
2. $$=\arctan{\sqrt{\left(\frac{\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}-2\cos\frac{x}{2}\sin \frac{x}{2}}{\sin^2{\frac{x}{2}}+\cos^2{\frac{x}{2}}+2\cos{{\frac{x}{2}}}\sin{\frac{x}{2}}}\right)}}$$
$$=\arctan{
\sqrt{
\left(\frac
{\cos \frac{x}{2} -\sin \frac{x}{2} }{\cos \frac{x}{2} +\sin \frac{x}{2} } \right) ^2
}
}$$
$$=\arctan \left(\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\right)$$
Dividing by $\cos \frac{x}{2}$
$$=\arctan \left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right) $$
$$=\arctan\left[ \tan \left( \frac{\pi}{4}-\frac{x}{2} \right)\right] $$
$$=\frac{\pi}{4}-\frac{x}{2}$$
So, are both these answers correct? or one of them is incorrect?
| Your mistake follows from the following.
$$\sqrt{x^2}\neq x.$$
Take $x=-1.$
Actually,
$$\sqrt{x^2}=|x|.$$
| {
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Show that $n^2 \mod 5$ equals $0,1$, or $4$ for every integer $n$. Show that $n^2 \mod 5$ equals $0,1$, or $4$ for every integer $n$. Using divison in to cases.
Proof: let integer $n$ be given.
Case $1$: Suppose there exists an integer $k$ such that $n = 2k$
Case $2$: Suppose there exists an integer $k$ such that $n = 2k+1$
Do I have the right idea of having two cases for all integers, one that covers even numbers and one that covers odd, or am I not on the right track?
| You want to know the outcome modulo $5$ so if we are to break things into cases it makes sense to break things into cases of modulo $5$.
If $n \equiv 0,1,2,3,4 \mod 5$ then
$n^2 \equiv 0^2, 1^2, 2^2, 3^2, 4^2 \mod 5$ and ... the proof then practically writes itself.
It writes itself even better when you realize $3 \equiv -2 \mod 5$ and $4\equiv -1 \mod 5$ so
If $n \equiv 0,1,2,-2,-1 \mod 5$ then
$n^2 \equiv 0^2, 1^2, 2^2, (-2)^2, (-1)^2 \mod 5$.
======
P.S. If you want to do it the hard way we are trying to prove:
$n^2 = 5k + r$ and $0 \le r < 5$ then $r= 0,1$ or $4$ (and never $2$ or $3$).
We are dealing with remainders when divided by $5$ so if we let $n = 5j + m$ then
$n^2 = (5j + m)^2 = 25j^2 + 10jm + m^2 = 5(5j^2 + 2jm) + m^2 = 5k + r$ so we must have $r = m^2 + 5(5j^2 + 2jm - 1)$. And $m$ may be $0,1,2,3,4$ so $m^2$ may be $0,1,4,9,16$ and $r$ may be $0,1,2,3,4$
$r = 0 + 5(5j^2 + 2jm - 1)$ means $r= 0$
$r = 1 + 5(5j^2 + 2jm - 1)$ means $r = 1$.
$r = 4 + 5(5j^2 + 2jm - 1)$ means $r = 4$
$r = 9 + 5(5j^2 + 2jm - 1)= 4 + 5(5j^2 + 2jm)$ means $r =4$.
$r = 16 + 5(5j^2 + 2jm - 1)= 1 + 5(5j^2 + 2jm+2)$ means $r = 1$.
But that was the hard way.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $\alpha=3-\sqrt[5]{5}-\sqrt[5]{25}$ algebraic over $\mathbb{Q}$? How to determine whether the given real number $\alpha =3-\sqrt[5]{5}-\sqrt[5]{25}$ is algebraic or not. And, $[\mathbb{Q}(\alpha):\mathbb{Q}]=?$
Let $x=3-\sqrt[5]{5}-\sqrt[5]{25}$,
\begin{align*}
& x = 3-\sqrt[5]{5}-\sqrt[5]{25}\\
\implies & (x-3) = -\sqrt[5]{5}-\sqrt[5]{25}\\
\implies & (x-3)^5=(-\sqrt[5]{5}-\sqrt[5]{25})^5
\end{align*}
But, solving this is abit tedios process, is there any other way to do this?
| $$\alpha\pmatrix{1\\5^{1/5}\\5^{2/5}\\5^{3/5}\\5^{4/5}}
=\pmatrix{3&-1&-1&0&0\\0&3&-1&-1&0\\0&0&3&-1&-1\\-5&0&0&3&-1\\-5&-5&0&0&3}
\pmatrix{1\\5^{1/5}\\5^{2/5}\\5^{3/5}\\5^{4/5}}
$$
so that $\alpha$ is an eigenvalue of
$$\pmatrix{3&-1&-1&0&0\\0&3&-1&-1&0\\0&0&3&-1&-1\\-5&0&0&3&-1\\-5&-5&0&0&3}.
$$
This matrix has integer entries, and so its characteristic equation
has integer coefficients. Therefore $\alpha$ is an algebraic integer.
As $\alpha\in\Bbb Q(5^{1/5})$ and $\alpha\notin\Bbb Q$ and $|\Bbb Q(5^{1/5}):\Bbb Q|=5$ and $5$ is prime, then $\Bbb Q(\alpha)=\Bbb Q(5^{1/5})$.
| {
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Proving that $2^{2\cdot 3^{n-1}}\equiv 1+3^n\pmod{3^{n+1}}$ for every natural $n$
Prove that $2^{2\cdot 3^{n-1}}\equiv 1+3^n\pmod{3^{n+1}}$ for every natural $n$
Of course, should be done by induction. Base case ($n=1$) is easy.
I got stuck with the step:
Let's assume that it's right for $n$. Then,
$$2^{2\cdot 3^{n-1}} = 1 + 3^n \pmod{3^{n+1}}$$
We want to evaluate $$2^{2 \cdot 3^n} \pmod {3^{n+2}}$$
I think we can somehow utilize the fact that $2,3$ are coprime and reduce $3^{n+2}$ to $3^{n+1}$.
I'd be glad for help on that.
Thanks!
| $(1+3^nx)^3 = 1+3^{n+1}x+3^{2n+1}x^2+3^{3n}x^3\equiv 1+3^{n+1}x \pmod{3^{n+2}}$ if $n\ge 1$.
| {
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Need help to understand the following number theory proof Prove that there exist infinitely many integers n such that $n, n + 1, n + 2$ are each the sum of the squares of two integers:
Solution: First solution: Let a be an even integer such that $a^2 + 1$ is not prime. (For example, choose a ≡ 2 (mod 5), so that $a^2 + 1$ is divisible by 5.) Then we can write $a^2 + 1$ as a difference of squares $x^2 − b^2$, by factoring $a^2 + 1$ as rs with $r ≥ s > 1$, and setting x = (r + s)/2, b = (r − s)/2. Finally, put n = $x^2 − 1$, so that $n = a^2 + b^2$, $n + 1 = x^2$, n + 2 = $x^2 + 1$
I got lost in the following part:
Then we can write $a^2 + 1$ as a difference of squares $x^2 − b^2$, by factoring $a^2 + 1$ as rs with $r ≥ s > 1$. What is rs?. Can someone expand this proof, so it looks clearer?
| Let $a^2 + 1$ be odd and $a^2 + 1$ not be prime. The proof glibly assumes such numbers are easy to find by pointing out any $2 + 5k$ will be such a number.
Since $a^2 + 1$ is not prime nor then $a^2 + 1 = r*s$ for some odd $r, s$ and $s \le r$.
Let $x$ be the midpoint between $s$ and $r$. (In other words $\frac {r+s}2$). This midpoint is an integer because $r$ and $s$ are both odd.
Let $b = x-s = r-s$. This means $r = x+b$ and $s= x-b$. That's not surprising as $x$ is the midpoint and $b$ is the distance each is from the midpoint.
Consider the three consecutive numbers $n = x^2 -1$, $n+1 = x^2$ and $n + 2 = ^2 + 1$.
Obviously $n+1 = x^2 + 0^2$ is the sum of two squares. And so $n+2=x^2 + 1^2$.
But $n$ is as well because:
$a^2 + b^2 = (a^2 + 1) + b^2 - 1$
$= r*s + b^2 -1= (x+b)(x-b) + b^2 - 1$
$= x^2 -b^2 + b^2 -1 = x^2 - 1 = n$
So $n$ is the sum of $a^2+ b^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$. I'm looking for help with (b) and (c) specifically. I'm posting (a) for completeness.
(a) Show convergence for $a_n=\sqrt{n+1}-\sqrt{n}$ towards $0$ and test $\sqrt{n}a_n$ for convergence.
(b) Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$.
(c) For which $\alpha\in\mathbb{Q}_+$ does $n^\alpha b_n$ converge?
I'm pretty sure I solved (a). I have proven the convergence of $a_n$ by using the fact that $$\sqrt{n}<\sqrt{n+1}\leq\sqrt{n}+\frac{1}{2\sqrt{n}}$$ which holds true since $$(\sqrt{n}+\frac{1}{2\sqrt{n}})^2=n+1+\frac{1}{4n}\geq n+1\,.$$
This gives us $$0<\sqrt{n+1}-\sqrt{n}\leq\frac{1}{2\sqrt{n}}$$
and after applying the squeeze theorem with noting that $\frac{1}{2\sqrt{n}}\longrightarrow0$ we can tell that also $a_n\longrightarrow0$.
Now $x_n=\sqrt{n}a_n=\sqrt{n}(\sqrt{n+1}-\sqrt{n})$.
We have \begin{align*}\sqrt{n}(\sqrt{n+1}-\sqrt{n})&=\sqrt{n}\sqrt{n+1}-\sqrt{n}\sqrt{n}\\&=\sqrt{n(n+1)}-n\\&=\sqrt{n^2+n}-n\\&=\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n}\\&=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}\\&=\frac{n}{\sqrt{n^2+n}+n}\\&=\frac{n}{n\sqrt{1+\frac{1}{n}}+n}\\&=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\end{align*}
and hence since the harmonic sequence $\frac{1}{n}$ converges towards 0 we have $$\text{lim}_{n\rightarrow\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{1+1} = \frac{1}{2}\,._{\,\,\square}$$
| In order to solve (b), let $a=\sqrt[k]{n+1}$ and $b=\sqrt[k]n$. Then\begin{align}\sqrt[k]{n+1}-\sqrt[k]n&=a-b\\&=\frac{(a-b)(a^{k-1}+a^{k-2}b+\cdots+ab^{k-2}+b^{k-1})}{a^{k-1}+a^{k-2}b+\cdots+ab^{k-2}+b^{k-1}}\\&=\frac{a^k-b^k}{a^{k-1}+a^{k-2}b+\cdots+ab^{k-2}+b^{k-1}}\\&=\frac1{\sqrt[k]{n+1}^{k-1}+\sqrt[k]{n+1}^{k-2}\sqrt[k]n+\cdots+\sqrt[k]{n+1}\sqrt[k]n^{k-2}+\sqrt[k]n^{k-1}}\end{align}and therefore\begin{align}\lim_{n\to\infty}\sqrt[k]{n+1}-\sqrt[k]n&=\lim_{n\to\infty}\frac1{\sqrt[k]{n+1}^{k-1}+\sqrt[k]{n+1}^{k-2}\sqrt[k]n+\cdots+\sqrt[k]{n+1}\sqrt[k]n^{k-2}+\sqrt[k]n^{k-1}}\\&=0.\end{align}
| {
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$f(n)=\sqrt{2n+1+2\sqrt{n^2+n}}$, $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\dots+\frac{1}{f(99)}$ Given $f(n)=\sqrt{2n+1+2\sqrt{n^2+n}}$
Evaluate $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\dots+\frac{1}{f(99)}$
I have simplified the given to $f(n)=\sqrt{n}+\sqrt{n+1}$ but I am still not sure of how to solve this.
| Very simple Hint: Rationalize the denominator
$$\frac1{f(x)}=\frac1{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n}-\sqrt{n+1}}{(\sqrt{n}+\sqrt{n+1})(\sqrt{n}-\sqrt{n+1})}=\sqrt{n+1}-\sqrt{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prime numbers yield from Pythagoras triples Pythagoras theorem $$a^2+b^2=c^2$$
we got $$P_{prime}(a,b)={a^4+b^4+(a+b)^4\over a^2+b^2+(a+b)^2}$$
Where $(a,b,c)$ are Pythagoras theorem triples, this function $P_{prime}(a,b)$ always produce a prime number for the values of a and b.
Examples: $P_{prime}(3,4)=37$, $P_{prime}(5,12)=229$, $P_{prime}(68,285)=105229$ and so on...
I have checked a lot of values, it seem to be prime so far.
My question is: Does the function $P_{prime}(a,b)$ always produce prime numbers?
| Let $a=2xy,$ and $b=x^2-y^2$.
Hence, we obtain:
$$\frac{(2xy)^4+(x^2-y^2)^4+(2xy+x^2-y^2)^4}{(2xy)^2+(x^2-y^2)^2+(2xy+x^2-y^2)^2}=x^4+y^4+2x^3y-2xy^3+2x^2y^2.$$
We can get this expression by the following.
$$\frac{a^4+b^4+(a+b)^4}{a^2+b^2+(a+b)^2}=\frac{2(a^4+2a^3b+3a^2b^2+2ab^3+b^4)}{2(a^2+ab+b^2)}=$$
$$=\frac{(a^2+ab+b^2)^2}{a^2+ab+b^2}=a^2+ab+b^2=x^4+y^4+2x^3y-2xy^3+2x^2y^2.$$
Now, easy to find a counterexample: $x=10$ and $y=1$, which gives $a=20$ and $b=99.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Equation of circle after rolling Q: A circle of radius $5$ units touches both the axes and lies in the first quadrant. If the circle makes one complete roll on $+ x$-axis, then what will be its equation in the new position $?$
I got my answer as $x^2 +y^2 -x(10+20\pi)-10y+100\pi^2+100\pi+25=0.$
Is it correct?
| The circle travels a distance of $2\pi\cdot 5=10\pi$ units to the right after one full rotation. It's original equation is $(x-5)^2+(y-5)^2=5^2$. Hence, its new position is
$$(x-5-10\pi)^2+(y-5)^2=5^2\iff (x-5(1+2\pi))^2+(y-5)^2=25\\\iff x^2-10(1+2\pi)x+25(1+2\pi)^2+y^2-10y+25=25\\\iff x^2+y^2-x(10+20\pi)-10y+100\pi^2+100\pi+25=0$$
So yes, you're correct.
| {
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Prove that $\left(\frac {-1+\sqrt {-3}}{2}\right)^n + \left(\frac {-1-\sqrt {-3}}{2}\right)^n$ follows this pattern
Prove that:
$$\left(\dfrac {-1+\sqrt {-3}}{2}\right)^n + \left(\dfrac {-1-\sqrt {-3}}{2}\right)^n=\begin{cases}
2, & \textrm { if } n \textrm { is a multiple of 3},\\
-1, & \textrm { if } n \textrm { is any other integer}
\end{cases}$$
My Attempt:
$$\dfrac {-1+\sqrt {-3}}{2}=\dfrac {-1+i\sqrt {3}}{2}$$ which is a complex cube root of unity.
Let $\omega = \dfrac {-1+i\sqrt {3}}{2}$.
Similarly, $\omega^2=\dfrac {-1-i\sqrt {3}}{2}$
| Alternative solution: the two complex numbers $\frac{-1\pm\sqrt{3}i}{2}$ are the roots of $x^2+x+1=0.$ Consider a recursive formula
$$a_{n+2}+a_{n+1}+a_n=0,$$
with $a_0=2$ and $a_1=-1$ agreeing with $\left(\frac{-1+\sqrt{3}i}{2}\right)^n+\left(\frac{-1-\sqrt{3}i}{2}\right)^n$ at $n=0$ and $1$. Then from the characteristic root method, $a_n=\left(\frac{-1+\sqrt{3}i}{2}\right)^n+\left(\frac{-1-\sqrt{3}i}{2}\right)^n$. We can then prove the periodic-$3$ pattern using mathematical induction.
| {
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$A, B$ are $3 \times 3$ matrices such that $(A - B)^2 = 0$. Prove that $\operatorname{Tr}(AB - BA)^3 = 0$. I have been trying to solve this recent linear algebra problem:
Let $A, B$ be $3 \times 3$ matrices such that $(A-B)^2 = 0$. Prove that $\det (AB - BA) = 0$.
This was my approach:$\DeclareMathOperator{\Tr}{Tr}$
The following equality holds for any $3\times 3$ matrices $A, B$:
$$\det (AB - BA) = \frac13 \Tr(AB - BA)^3$$
It follows from the Hamilton-Cayley theorem applied on $AB - BA$. Therefore, it suffices to prove that $\Tr(AB - BA)^3 = 0$.
Expanding gives that $\Tr(AB - BA)^3$ is equal to:
\begin{align}
\Tr\left(\color{magenta}{ABABAB} - \color{blue}{ABABBA} - \color{purple}{ABBAAB} + \color{purple}{ABBABA} - \color{blue}{BAABAB} + \color{olive}{BAABBA} + \color{olive}{BABAAB} - \color{magenta}{BABABA}\right)\\
\end{align}
where the same-colored terms are cyclic permutations of each other so have the same trace.
So, $$\Tr(AB - BA)^3 = 2\Tr BAABBA - 2\Tr BAABAB = 2 \Tr BAAB(BA - AB)$$
I figured this was a good place to try to use the assumption $(A - B)^2 = 0$:
$$0 = (A - B)^2 = A^2 + B^2 - AB - BA \implies A^2 = AB + BA - B^2$$
So we have:
$$BAAB(BA - AB) = B(AB + BA - B^2)B(AB - BA) = $$
$$\color{OrangeRed}{BABBBA} + \color{green}{BBABBA} - BBBBBA - \color{green}{BABBAB} - \color{OrangeRed}{BBABAB} + BBBBAB$$
Again, the same-colored terms cancel out when taking the trace so:
$$\Tr BAAB(BA - AB) = \Tr BBBB(AB - BA)$$
A possible development is:
\begin{align}2 \Tr BAAB(BA - AB) &= \Tr (AB-BA)(BBBB - BAAB)\\
&= \Tr (AB-BA)(BBBB - BAAB)\\
&= \Tr (AB-BA)B(B^2 - A^2)B
\end{align}
But using $A^2 + B^2 = AB + BA$ here again gives $\Tr (BA - AB)BAAB$, so nothing new.
Is there a way to finish the proof?
| Let $N=A-B$. Then $N^2=0$ and $C=AB-BA=NB-BN$. Therefore
\begin{align}
C^2&=(NB-BN)(NB-BN)=NBNB-NBBN+BNBN,\\
C^3&=(NB-BN)(NBNB-NBBN+BNBN)\\
&=NBNBNB-NBNBBN+NBBNBN-BNBNBN.\tag{1}
\end{align}
Now, using the cyclic property of matrix trace $\operatorname{tr}(XY)=\operatorname{tr}(YX)$ and the assumption that $N^2=0$, we immediately see that the two middle summands on line $(1)$ are traceless and the traces of the two outer summands cancel out each other. Hence $\operatorname{tr}(C^3)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Selection of atleast 2 balls of each colour The question is:
A box contains $5$ different red and $6$ different white balls. In how
many ways can $6$ balls be selected so that there are atleast two balls
of each colour.
I am trying to solve it as follows:
To guarantee $2$ red balls, choosing $2$ red balls : $\binom{5}{2}$.
To guarantee $2$ white balls, choosing $2$ white balls : $\binom{6}{2}$.
For a total of $6$ balls, choosing $2$ from the remaining $7$ balls : $\binom{7}{2}$
So total number of ways = $\binom{5}{2} \times\binom{6}{2} \times \binom{7}{2}$
However, the answer is $425$. What am I doing wrong and why?
| Let's say red balls are $R_1$, $R_2$, $R_3$, $R_4$, $R_5$ and white balls are $W_1$, $W_2$, $W_3$, $W_4$, $W_5$, $W_6$.
Now, assume we are choosing the balls as in your method and we have chosen $R_1$ and $R_2$ from red balls, $W_1$ and $W_2$ from white balls first and other two balls are $R_3$ and $W_3$. But this is as same as choosing $R_1$ and $R_3$ from red balls, $W_1$ and $W_3$ from white balls first and other two balls are $R_2$ and $W_2$. So we are overcounting.
In order to solve this, we should choose all required balls at once like the following:
If we have at least $2$ red balls and $2$ white balls, we can choose them as $2$ red $4$ white balls or $3$ red $3$ white balls or $4$ red $2$ white balls, which is $$\binom{5}{2} \binom{6}{4} + \binom{5}{3}\binom{6}{3}+\binom{5}{4} \binom{6}{2} = 150 + 200 + 75 = 425$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $\int\limits_0^\infty x \left \lfloor{\frac1x}\right \rfloor \, dx$ This is an integral I computed but can't find the result online or on wolfram. So here's a proof sketch, please indulge this sanity check:
$$\int_0^\infty x \left \lfloor{\frac1x}\right \rfloor \ dx = \int_0^1 x \left \lfloor{\frac1x}\right \rfloor \ dx$$
$$= \sum_{n=1}^\infty \int_{1/(n+1)}^{1/n} nx \ dx =\sum_{n=1}^\infty\frac n2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right) $$
$$=
\sum_{n=1}^\infty\frac n2 \left(\frac{2n+1}{n^2(n+1)^2}\right)$$
$$= \sum_{n=1}^\infty\frac{1}{(n+1)^2} + \frac12 \sum_{n=1}^\infty \frac{1}{n(n+1)^2}$$
$$= \frac{\pi^2}{6} -1 + \frac12\left(\sum_{n=1}^\infty \frac1n - \frac{1}{n+1} - \frac{1}{(n+1)^2}\right)$$
$$=\frac{\pi^2}{6} -1 + \frac12\left(\sum_{n=1}^\infty \frac1n - \frac{1}{n+1}\right) -\frac12\left(\sum_{n=1}^\infty\frac{1}{(n+1)^2}\right)$$
$$= \left(\frac{\pi^2}{6} -1\right) + \left(\frac12\cdot 1\right) - \frac12\left(\frac{\pi^2}{6} -1\right)$$
$$= \frac{\pi^2}{12}.$$
Basically, I used the Basel sum several times, and the fifth line follows from a partial sum decomposition. The seventh follows from the known result for the Basel sum, as well as the fact that the first series in the 6th line telescopes.
I hope this is all correct.
| Use
$$n\left(\frac1{n^2}-\frac1{(n+1)^2}\right)=\frac n{n^2}-\frac{n+1-1}{(n+1)^2}=\frac1n-\frac1{n+1}+\frac1{(n+1)^2}.$$
The first two terms do telescope and the Basel series remains.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Fallacy in showing that $\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$ Given a cyclic quadrilateral $ABCD$, $AB=a, BC=b, CD=c, \angle ABC=120^\circ, \angle ABD=30^\circ$, then, show that $$\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$$
I tried to do it using some trig bashing.
What I did was basically assigned the angle $\angle BCA = \theta$ and used trigonometry.
It's not hard to see that the radius of the circle is $\frac{c}{2}$.
I've squared the equation on both sides and obtained: $$c+2a+2b=2\sqrt{(c+a)(c+b)}$$.
And using $\text {Sine Rule}$, the equation can be reduced further.
I used
*
*$\frac{a}{\sin\theta}=c$
*$\frac{b}{\sin(60^\circ - \theta)}=c$
And reduced the equation to: $$1+2\sin\theta+2\sin(60^\circ - \theta)=2\sqrt{(1+\sin\theta)(1+\sin(60^\circ - \theta))}$$.
Then I continued to reduce it using the addition-subtraction formulae of trigonometry.
And at the end of the day what I get is $$\boxed{\cos(30^\circ-\theta)=\frac{1}{2}}$$ after all those addition-subtraction of trigonometric equations.
And that's not true I believe.
May I get rectified?
| Noticing $\triangle{ADO}$ is an equilateral triangle where $O$ is the center of the circle should be a key.
Since the both sides of
$$\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$$
are non-negative, it is equivalent to
$$(\sqrt{c+a}-\sqrt{c+b})^2=c-a-b,$$
i.e.
$$c+2a+2b=2\sqrt{(c+a)(c+b)}$$
Since the both sides are positive, it is equivalent to
$$(c+2a+2b)^2=4(c+a)(c+b),$$
i.e.
$$\frac{3}{4}c^2=a^2+b^2+ab\tag1$$
So, all we need is to prove $(1)$.
Applying the law of cosines to $\triangle{ABC}$, we have
$$|\overline{AC}|^2=a^2+b^2-2ab\cos(120^\circ),$$
i.e.
$$|\overline{AC}|^2=a^2+b^2+ab\tag2$$
Applying the law of sines to $\triangle{ADC}$, we get
$$\frac{|\overline{CD}|}{\sin\angle{DAC}}=\frac{|\overline{AD}|}{\sin\angle{DCA}}$$
which implies
$$|\overline{AD}|=\frac c2$$
So, $\triangle{DAO}$ is an equilateral triangle where $O$ is the center of the circle.
Applying the law of sines to $\triangle{ADC}$ gives
$$\frac{|\overline{AC}|}{\sin\angle{ADC}}=\frac{|\overline{AD}|}{\sin\angle{DCA}}$$
which implies
$$|\overline{AC}|^2=\frac 34c^2\tag3$$
Now, $(1)$ follows from $(2)(3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all polynomials $f, g \in \mathbb{Z}[X]$ such that $f(g(X))=1 + X + ... + X^{p-1}$
Find all polynomials $f, g \in \mathbb{Z}[X]$ such that $f(g(X))=1 + X
+ ... + X^{p-1}$ where $p \gt 2$ is a prime.
One solution is $g = X, f =1 + X + ... + X^{p-1}$.
If $\deg(f)=m, \deg(g)=n$ then $f=\pm X^m + ... , g=\pm X^n + ... $ with $mn = p-1$
I have no idea how to get further.
UPDATE
$g = -X, f =1 - X + X^2 -X^3 + ... + X^{p-1}$ also $f = -X, g =- 1 - X - ... - X^{p-1}$ and $f = X, g =1 + X + ... + X^{p-1}$ are also solutions.
| Suppose $f(X) = \sum\limits_{k = 0}^m a_k X^k$ and $g(X) = \sum\limits_{j = 0}^n b_j X^j$. It is easy to see that $|a_m| = |b_n| = 1$.
First, consider the situation where $n = 1$. If $g(X) = X + b_0$, then$$
f(X + b_0) = 1 + X + \cdots + X^{p - 1} \Rightarrow f(X) = 1 + (X - b_0) + \cdots + (X - b_0)^m.
$$
If $g(X) = -X + b_0$, then$$
f(-X + b_0) = 1 + X + \cdots + X^{p - 1} \Rightarrow f(X) = 1 + (-X + b_0) + \cdots + (-X + b_0)^m.
$$
Next, suppose $n > 1$. Now focus on the coefficient of the $X^{p - 2} = X^{mn - 1}$ term. Expand $f(g(X))$ to get$$
\sum_{k = 0}^m a_k (g(X))^k = 1 + X + \cdots + X^{p - 1}.
$$
For $0 \leqslant k \leqslant m - 1$, $\deg(a_k (g(X))^k) = kn \leqslant (m - 1)n < mn - 1$, so the coefficient of $X^{mn - 1}$ in $f(g(X))$ is all contributed by $a_m(g(X))^m$. Note that $\deg g = n$, thus the coefficient of $X^{mn - 1}$ in $a_m(g(X))^m$ is $a_m \cdot \binom{m}{1} b_n^{m - 1} b_{n - 1} = m a_m b_{n - 1} b_n^{m - 1}$. Since the coefficient of $X^{mn - 1} = X^{p - 2}$ in $1 + X + \cdots + X^{p - 1}$ is $1$, then$$
m a_m b_{n - 1} b_n^{m - 1} = 1 \Longrightarrow m = 1.
$$
Now, if $f(X) = X + a_0$, then$$
g(X) + a_0 = 1 + X + \cdots + X^{p - 1} \Longrightarrow g(X) = 1 + X + \cdots + X^{p - 1} - a_0.
$$
If $f(X) = -X + a_0$, then$$
-g(X) + a_0 = 1 + X + \cdots + X^{p - 1} \Longrightarrow g(X) = -1 - X - \cdots - X^{p - 1} + a_0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Find: $ \lim_{x\to \infty}\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
Find: $\displaystyle \lim_{x\to \infty} f(x)=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
The answer provided in the book is 0 (also checked in Wolfram Alpha), but I can't find a good argument (without L'Hopital), to prove that. I end up in a $\infty \times 0$ situation which makes me uncomfortable.
Attempt: $$f(x)=(x-1)\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{x}(\sqrt{1+1/x}-1)}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{1+1/x}-1}{\sqrt{x+1}}$$
$$f(x)=x(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{x}\sqrt{1+1/x}}=\sqrt{x}(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{1+1/x}}.$$
Therefore the original limit is equivalent to
$$\lim_{x\to \infty} \sqrt{x}(1-1/x)\left(1-\frac{1}{\sqrt{1+1/x}}\right)$$
that appears to me a situation like $\infty\times 0$. How can I proceed to conclude that this limit is indeed $0$.
Hints and solutions are appreciated. Sorry if this is a duplicate.
| \begin{align*}
f(x)&=x\cdot\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x+1}}\right)+\dfrac{1}{\sqrt{x+1}}-\dfrac{1}{\sqrt{x}}\\
&=\dfrac{\sqrt{x}}{\sqrt{x+1}}\dfrac{1}{\sqrt{x+1}+\sqrt{x}}-\dfrac{1}{\sqrt{x+1}\sqrt{x}}\cdot\dfrac{1}{\sqrt{x}+\sqrt{x+1}}\\
&\rightarrow 1\cdot 0-0\\
&=0.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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A specific Jensen's inequality for a proof It's related to this Inequality : $\sum_{cyc}\frac{\sqrt{2}a^2b}{2a+b} \leq \sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$ more particulary to this passage :
Please suggest how to show that
$ \sqrt{2} \leq \displaystyle\sum_{cyc} \frac{\sqrt{a^2+b^2}}{2ab+1}$
My try
There is a simple method ,you can minimize each term like this but first you have:
$$\frac{\sqrt{a^2+b^2}}{2ab+1}$$
We can rewrite the expression like this :
$$\frac{\sqrt{(a+b)^2-2ab}}{2ab+1}$$
Now the idea is to remark that for a fixed sum of two numbers the product of this two numbers is maximal when each term is equal .
Here we use this method and remark that this method minimize the numerator and maximize the denominator so we have :
$a+b=x$ and $ab=0.25x^2$
$c+b=y$ and $bc=0.25y^2$
$a+c=z$ and $ac=0.25z^2$
We get :
$$\frac{\sqrt{x^2-0.5x^2}}{0.5x^2+1}+\frac{\sqrt{y^2-0.5y^2}}{0.5y^2+1}+\frac{\sqrt{z^2-0.5z^2}}{0.5z^2+1}$$
So we have to prove :
$$\frac{\sqrt{x^2-0.5x^2}}{0.5x^2+1}+\frac{\sqrt{y^2-0.5y^2}}{0.5y^2+1}+\frac{\sqrt{z^2-0.5z^2}}{0.5z^2+1}\geq \sqrt{2}$$
But I can't use directly Jensen because the function $f(x)=\frac{\sqrt{x^2-0.5x^2}}{0.5x^2+1}$ is not convex so my question is :
1)What's the kind of function $f(x)$ is ?
2)What's the specific Jensen's inequality we have to use ?
Thanks a lot .
| I think your reasoning is total wrong.
It remains to prove that
$$\sum\frac{a+b}{\frac{(a+b)^2}{2}+1}\geq2,$$ which is wrong for $c=3$ and $a=b\rightarrow0^+.$
I have a proof of your starting inequality by full expanding, by BW, by uvw and by SOS (which you saw).
Which way do you prefer?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x \to 0} \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}$
Find $\lim_{x \to 0} \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}$
By taylor polynomials we get: $e^{x^4}=1+x^4+\frac{x^8}{2}+\mathcal{O}(x^{12})$
$\sin(x^2)=x^2-\frac{x^6}{6}+\mathcal{O}(x^{10})$
$\cos(x^3)=1-\frac{x^6}{2}+\mathcal{O}(x^{12})$
so putting these together:
$$ \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)} = \frac{x^2+x^6+\frac{x^{10}}{2}-x^2+\frac{x^6}{6}-\mathcal{O}(x^{10})}{\frac{x^6}{2}-\mathcal{O}(x^{12})}=\frac{\frac{7}{6}x^6+\frac{1}{2}x^{10}-\mathcal{O}(x^{10})+\mathcal{O}({x^{12}})}{\frac{1}{2}x^6-\mathcal{O}(x^{12})}$$
Now I am not too familiar with the Big-Oh notation for limits so I am stuck here.
How does arithmetic work with them, can I simplify the oh's in the numerator and can I take $x$'s out?
| As $\;x\to\infty\;$ :
$$\frac{x^2e^{x^4}-\sin x^2}{1-\cos x^3}\ge\frac{x^2e^{x^4}-1}2\xrightarrow[x\to\infty]{}\infty$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $a^4+b^4+c^4$ The problem:
The sum of three numbers is $6$, the sum of their squares is $8$, and the sum of their cubes is $5$. What is the sum of their fourth powers?
Based on the above information, we have:
\begin{align}
a + b + c &= 6 \\
a^2 + b^2 + c^2 & = 8 \\
a^3 + b^3 + c^3 & = 5 \\
\end{align}
I had a feeling that this vaguely had to do with Viete's theorem, which states for a cubic polynomial $f(x) = x^3 - px^2 + qx - r$ which has roots $\alpha , \beta , \gamma$,
\begin{align}
p & = \alpha+\beta+\gamma \\
q & = \alpha \beta+\alpha\gamma+\beta\gamma \\
r & = \alpha\beta\gamma
\end{align}
Notice that we already have $p$, because $a+b+c=6=\alpha + \beta + \gamma = p$. Then to find $q$:
\begin{align}
2q& = 2\alpha\beta+2\alpha\gamma+2\beta\gamma \\
& = [(a+b+c)^2-(a^2+b^2+c^2)] \\
& = (a^2+ab+ac+b^2+ab+bc+c^2+ac+bc)-a^2-b^2-c^2 \\
& = 2ab+2ac+2bc
\end{align}
\begin{align}
q & = ab+ab+bc \\
& = \frac{1}{2}[(a+b+c)^2-(a^2+b^2+c^2)] \\
& = \frac{1}{2}[6^2-8] \\
& = 14
\end{align}
So now we have $f(x) = x^3-6x^2+14x-r$. And it follows that $f(\alpha)=f(\beta)=f(\gamma)=0$.
\begin{align}
f(\alpha) & = \alpha^3-6\alpha^2+14\alpha-r = 0\\
f(\beta) & = \beta^3-6\beta^2+14\beta-r = 0\\
f(\gamma) & = \gamma^3-6\gamma^2+14\gamma-r = 0\\
\end{align}
\begin{align}
0 & = f(\alpha) + f(\beta) + f(\gamma) \\
& = (\alpha^3+\beta^3+\gamma^3)-6(\alpha^2+\beta^2+\gamma^2)+14(\alpha+\beta+\gamma)-3r\\
& = 5-6(8)+14(6)-3r\\
3r& = 41\\
r & = \frac{41}{3} \\
\end{align}
Now we have that $f(x)=x^3-6x^2+14x-\frac{41}{3}$. Here's where I am stuck. Of course, the above working out was the culmination of hours of trying things out, and eventually we have this equation. If you are reading this now, I'd appreciate if you gave any hints as to how I should continue the problem.
I've had the idea of multiplying $f(x)$ by $x$ to get fourth powers, but I haven't tried that yet. Perhaps that might yield some results?
| Hint:
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\iff ab+bc+ca=?$$
$$a^3+b^3+c^3-3abc=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}\iff abc=?$$
Now $$a^4+b^4+c^4=(\underbrace{a^2+b^2+c^2})^2-2(a^2b^2+b^2c^2+c^2a^2)$$
$$a^2b^2+b^2c^2+c^2a^2=(\underbrace{ab+bc+ca})^2-2\underbrace{abc}(\underbrace{a+b+c})$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Taylor series for $f(z)=z\cos(z)/(z^2-2)$ Let $f(z)=z\cos(z)/(z^2-2)$, $z\in\mathbb{C}$.
Show that the Taylor series for $f$ around $0$ is given by
$$f(z)=\sum_{k=0}^\infty a_{2k+1} z^{2k+1} $$
with $a_1=-1/2$ and $a_{2k+1}=(a_{2k-1}+(-1)^{k+1}/(2k)!)/2$ for $k\geq 1$.
I tried to use the power series for $\cos(z)$ and the Taylor series for $z/(z^2-2)$ and somehow combine them, but got stuck.
This seems like it should be simple enough, but I can't get anywhere. Any help is appreciated
|
We observe since $\cos(z)=\sum_{j=0}^\infty \frac{(-1)^j}{(2j)!}z^{2j}$ is an even function, the function $f(z)$
is odd which makes the series representation
\begin{align*}
f(z)=\frac{z\cos(z)}{z^2-2}=\sum_{k=0}^\infty a_{2k+1}z^{2k+1}\tag{1}
\end{align*}
plausible.
It is convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ of a series. This way we can write for instance
\begin{align*}
[z^{2k+1}]f(z)&=[z^{2k+1}]\sum_{j=0}^\infty a_{2j+1}z^{2j+1}=a_{2k+1}
\end{align*}
We multipy (1) with $z^2-2$ and consider
\begin{align*}
f(z)(z^2-2)=z\cos(z)
\end{align*}
We obtain from (1) for $k\geq 0$:
\begin{align*}
[z^{2k+1}]f(z)(z^2-2)&=[z^{2k-1}]f(z)-2[z^{2k+1}]f(z)\\
&=a_{2k-1}-2a_{2k+1}
\end{align*}
Here we use the linearity of the coefficient of operator and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
On the other hand we obtain from (1) for $k\geq 0$:
\begin{align*}
[z^{2k+1}]z\cos(z)=[z^{2k}]\cos(z)=\frac{(-1)^k}{(2k)!}
\end{align*}
Equating both results gives
\begin{align*}
\color{blue}{a_{2k+1}=\frac{1}{2}\left(a_{2k-1}-\frac{(-1)^{k}}{(2k)!}\right)}
\end{align*}
and the claim follows.
We also obtain from (1)
\begin{align*}
[z^1]f(z)(z^2-2)&=-2[z^1]f(z)\\
&=\color{blue}{-2a_1}\\
[z^1]z\cos(z)&=[z^0]\cos(z)\\
&\color{blue}{=1}
\end{align*}
from which $\color{blue}{a_1=-\frac{1}{2}}$ follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Test $\sum_{n=0}^{\infty}\frac{n!}{2^{n^2}}$ for convergence. Applying the ratio test to $\sum_{n=0}^{\infty}\frac{n!}{2^{n^2}}$ with $x_n=\frac{n!}{2^{n^2}}$ we get
\begin{align*}
\bigg|\frac{x_{n+1}}{x_n}\bigg|&=\frac{\frac{(n+1)!}{2^{(n+1)^2}}}{\frac{n!}{2^{n^2}}}\\&=\frac{(n+1)!}{2^{(n+1)^2}}\frac{2^{n^2}}{n!}\\&=\frac{(n+1)n!\cdot2^{n^2}}{n!\cdot2^{n^2}\cdot2^{2n+1}}\\&=\frac{n+1}{2^{2n+1}}
\end{align*}
Quick question now: Is it sufficient here to state that $2^{2n+1}$ obviously grows faster than $n+1$ (and therefore our term converges towards $0$ and our series passed the test) or do I need to generally prove this first (e.g. by induction)?
| $$
0\le\frac{n!}{2^{n^2}} \le \frac{n^n}{2^{n^2}}
$$
by the root test for $n>1$:
$$
\sqrt[n]{\frac{n^n}{2^{n^2}}}=\frac{n}{2^n}\le \frac{1}{2}
$$
hence the series
$$\sum_{n=0}^{\infty}\frac{n^2}{2^{n^2}}$$
converges and by comparison
$$\sum_{n=0}^{\infty}\frac{n!}{2^{n^2}}$$
converges
| {
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"url": "https://math.stackexchange.com/questions/2593035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complex integral with residuals questions for complex integrals
*
*For $R>0$ let $\Gamma_R$ be the circle {$z\in \mathbb{C} : \lvert z \rvert =R $} equipped with counterclockwise direction. Figure out for which $R>0$, the function $f(z)=\frac{1}{\sin^2(z)}$ is continuous on $\Gamma_R$ and evaluate $$\int\limits_{\Gamma_R}\frac{dz}{\sin^2(z)}$$
for every such $R$ (the answer may depend on $R$)
*Evaluate $\int\limits_{0}^{\infty}\frac{\cos{x}}{1+x^6}dx$
I have no idea where to start from for any of the questions above, any help to get me started is greatly appreciated.
| Function $\sin{z}$ has zeros at $k\cdot \pi, k \in \mathbb{Z}$ and same applies to $\sin^2{z}$. As a result, function $\frac{1}{\sin^2{z}}$ has singularities at $k\cdot \pi, k \in \mathbb{Z}$. These singularities are poles of order $2$ since (using this fact)
$$\lim\limits_{z\rightarrow k\pi}\frac{(z-k\pi)^2}{\sin^2{z}}=\lim\limits_{z-k\pi\rightarrow 0}\frac{(z-k\pi)^2}{\sin^2{(z-k\pi)}}=\frac{1}{\lim\limits_{z-k\pi\rightarrow 0}\frac{\sin^2{(z-k\pi)}}{(z-k\pi)^2}}=1$$
and
$$\lim\limits_{z\rightarrow k\pi}\frac{(z-k\pi)^3}{\sin^2{z}}=0$$
Any $\Gamma_R$ where $R=k\pi, k\in\mathbb{N}\setminus\{0\}$ is of the form $\Gamma_R=\left\{k\pi e^{it}\mid 0\leq t < 2\pi\right\}$ and will contain $k\pi$ and $-k\pi$ where $\frac{1}{\sin^2{z}}$ is not continuous. In all the other cases ($R\ne k\pi, k\in\mathbb{N}\setminus\{0\}$), $\frac{1}{\sin^2{z}}$ is continuous on $\Gamma_R$. To calculate the integral on such a contour, residue theorem can be applied
$$\int\limits_{\Gamma_R}\frac{1}{\sin^2{z}}dz=2\pi i\sum\limits_{|k|<\frac{R}{\pi}\\ k\in \mathbb{Z}} \text{Res}\left(\frac{1}{\sin^2{z}}, k\pi\right)$$
We already know that the pole order at $k\pi$ is $2$, so calculating the residues shouldn't be a problem
$$\text{Res}\left(\frac{1}{\sin^2{z}},k\pi\right)=\lim\limits_{z\rightarrow k\pi}\left(\frac{(z-k\pi)^2}{\sin^2{z}}\right)'=\\
\lim\limits_{z\rightarrow k\pi}\left(\frac{2(z-k\pi)\sin^2{z}-2(z-k\pi)^2\sin{z}\cos{z}}{\sin^4{z}}\right)=\\
\lim\limits_{z\rightarrow k\pi}\left(\frac{2(z-k\pi)\left(\sin{z}-(z-k\pi)\cos{z}\right)}{\sin^3{z}}\right)\overset{\text{L'Hospital}}{=}\\
\lim\limits_{z\rightarrow k\pi}\left(\frac{2((z-k\pi)^2 \sin{z}-(z-k\pi) \cos{z} + \sin{z})}{3\sin^2{z}\cos{z}}\right)\overset{\text{L'Hospital}}{=}\\
\lim\limits_{z\rightarrow k\pi}\left(\frac{2(z-k\pi)\left((z-k\pi)\cos{z}+ 3\sin{z}\right) }{-\frac{3}{4}\left(\sin{z}-3\sin{3z}\right)}\right)\overset{\text{L'Hospital}}{=}\\
\lim\limits_{z\rightarrow k\pi}\left(\frac{ -2\left(\left(z^2-2 k\pi z-3+(k\pi)^2\right)\sin{z}-5(z-k\pi)\cos{z}\right) }{-\frac{3}{4}\left(\cos{z}-9\cos{3z}\right)}\right)=\\
\frac{0}{6\cdot (-1)^k}=0$$
As a result $$\int\limits_{\Gamma_R}\frac{1}{\sin^2{z}}dz=0$$
For the second one, function $\frac{\cos{x}}{1+x^6}$ is even, as a result
$$\int\limits_{0}^{\infty}\frac{\cos{x}}{1+x^6}dx= \lim\limits_{R\rightarrow\infty} \int\limits_{0}^{R}\frac{\cos{x}}{1+x^6}dx=\frac{1}{2} \lim\limits_{R\rightarrow\infty} \int\limits_{-R}^{R}\frac{\cos{x}}{1+x^6}dx=...$$
Now $\frac{\sin{x}}{1+x^6}$ is odd and as a result $\int\limits_{-R}^{R}\frac{\sin{x}}{1+x^6}dx=0$, thus
$$...=\frac{1}{2} \lim\limits_{R\rightarrow\infty} \int\limits_{-R}^{R}\frac{\cos{x}+i\sin{x}}{1+x^6}dx=\frac{1}{2} \lim\limits_{R\rightarrow\infty} \int\limits_{-R}^{R}\frac{e^{ix}}{1+x^6}dx=...$$
Now we will look at this contour
$$...=\frac{1}{2} \lim\limits_{R\rightarrow\infty} \left(\int\limits_{C_R}\frac{e^{iz}}{1+z^6}dz - \int\limits_{\Gamma_R}\frac{e^{iz}}{1+z^6}dz \right)=\\
\frac{1}{2} \lim\limits_{R\rightarrow\infty} \left(\int\limits_{C_R}\frac{e^{iz}}{(z-i)(z+i)\left(z-\frac{\sqrt{3}}{2}-\frac{i}{2}\right)\left(z+\frac{\sqrt{3}}{2}+\frac{i}{2}\right)\left(z-\frac{\sqrt{3}}{2}+\frac{i}{2}\right)\left(z+\frac{\sqrt{3}}{2}-\frac{i}{2}\right)}dz \\
- \int\limits_{\Gamma_R}\frac{e^{iz}}{1+z^6}dz \right)=
\frac{1}{2} \lim\limits_{R\rightarrow\infty} \left(\text{Res}\left(\frac{e^{iz}}{1+z^6},i\right)+
\text{Res}\left(\frac{e^{iz}}{1+z^6},\frac{\sqrt{3}}{2}+\frac{i}{2}\right)+\\
\text{Res}\left(\frac{e^{iz}}{1+z^6},-\frac{\sqrt{3}}{2}+\frac{i}{2}\right)
- \int\limits_{\Gamma_R}\frac{e^{iz}}{1+z^6}dz \right)=\\
\frac{1}{2}\left(\text{Res}\left(\frac{e^{iz}}{1+z^6},i\right)+
\text{Res}\left(\frac{e^{iz}}{1+z^6},\frac{\sqrt{3}}{2}+\frac{i}{2}\right)+
\text{Res}\left(\frac{e^{iz}}{1+z^6},-\frac{\sqrt{3}}{2}+\frac{i}{2}\right)\right)$$
It remains to show that
$$\lim\limits_{R\rightarrow\infty} \int\limits_{\Gamma_R}\frac{e^{iz}}{1+z^6}dz =0$$
which is
$$\left|\int\limits_{\Gamma_R}\frac{e^{iz}}{1+z^6}dz\right| \leq \text{length}(\Gamma_R) \max\limits_{z \in \Gamma_R}\left|\frac{e^{iz}}{1+z^6}\right|=\pi R \max\limits_{z \in \Gamma_R}\frac{|e^{iz}|}{\left|1+z^6\right|}\leq \pi R \max\limits_{z \in \Gamma_R}\frac{|e^{i(x+iy)}|}{\left||z^6|-|1|\right|}\leq\\
\pi R \frac{e^{-y}}{\left|R^6-1\right|}\leq \frac{\pi R}{R^6-1} \rightarrow 0, R \rightarrow \infty$$
I will leave you to finish the calculations of residues.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the intersection curve between a plane $x+y=1$ and a sphere $x^{2}+y^{2}+z^{2}=1$. Given two equations:
$\left\{\begin{matrix}
x^{2}+y^{2}+z^{2}=1\\
x+y=1
\end{matrix}\right.$
Find the set of points in $3$-space represented by this pair.
My work so far:
The equation $x^{2}+y^{2}+z^{2}=1$ represents a sphere with radius $1$ centered at the origin. The equation $x+y=1$ represents a plane in the $xy$-plane that has the points (1, 0, 0) and (0, 1, 0).
I am not sure how to find the equation of the curve (the circle) that is formed from the intersection of these two equations. Do I use substitution?
Let $x = 1 - y$.
$(1-y)^{2}+y^{2}+z^{2}=1$
$1-2y+y^{2}+y^{2}+z^{2}=1$
$2\left (y-\frac{1}{2} \right )^{2}+z^{2}=\frac{1}{2}$
I am close to the answer in the textbook, which is a circle with radius $\frac{1}{\sqrt{2}}$ centered at (1/2, 1/2, 0).
I know that the circle is in the $xy$-plane, so $z = 0$. However, how did they get the $x = \frac{1}{2}$ coordinate when I don't see $x$ in the equation...?
| Your derivation is correct but from here
$$2\left (y-\frac{1}{2} \right )^{2}+z^{2}=\frac{1}{2}$$
$$y = 1 - x$$
you find
$$\left (1-x-\frac{1}{2} \right )^{2}+\left (y-\frac{1}{2} \right )^{2}+z^{2}=\frac{1}{2}$$
$$\left (x-\frac{1}{2} \right )^{2}+\left (y-\frac{1}{2} \right )^{2}+z^{2}=\frac{1}{2}$$
which is the equation of a sphere centered in $C(\frac12,\frac12,0)$ and with radius $R=\frac{1}{\sqrt2}$.
Since the center belongs to the plane $x+y=1$ then the intersection between the sphere and the plane is a circle with center $C(\frac12,\frac12,0)$ and radius $R=\frac{1}{\sqrt2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that $\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$
prove that $$\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$$
My attempt:
C is semicircle in upper half complex plane
Simple poles = $e^{i\frac{\pi}{8}}, e^{i\frac{3\pi}{8}},e^{i\frac{5\pi}{8}},e^{i\frac{7\pi}{8}}$ lie in upper semi-circle C and real axis
Given integral value $= 2\pi i \cdot (\text{sum of residues}) = 2 \pi i \left(\frac{-1}{8}\right) \left[e^{i\frac{5\pi}{8}}+e^{i\frac{15\pi}{8}}+e^{i\frac{25\pi}{8}}+e^{i\frac{35\pi}{8}}\right] = 0.27059 \pi$
This is numerically equal to $\frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$. But without using calculator, how to get this expression.
| Just extending what you've got so far. Let's note $z=e^{i\frac{5\pi}{8}}$ and recall that $\cos{z}=\frac{e^{iz}+e^{-iz}}{2}$, then:
$$2 \pi i \left(\frac{-1}{8}\right) \left[e^{i\frac{5\pi}{8}}+e^{i\frac{15\pi}{8}}+e^{i\frac{25\pi}{8}}+e^{i\frac{35\pi}{8}}\right]=
2 \pi i \left(\frac{-1}{8}\right) \left[z+z^3+z^5+z^7\right]=\\
2 \pi i \left(\frac{-1}{8}\right) z \left[1+z^2+z^4+z^6\right] =
2 \pi i \left(\frac{-1}{8}\right) z \left[1+z^2+z^4(1+z^2)\right]=\\
2 \pi i \left(\frac{-1}{8}\right) z (1+z^2)(1+z^4)=
2 \pi i \left(\frac{-1}{8}\right) z^4 (z^{-1}+z)(z^{-2}+z^2)=\\
2 \pi i \left(\frac{-1}{8}\right) e^{i\frac{5\pi}{2}} 2\cos\left(\frac{5\pi}{8}\right)2 \cos\left(\frac{5\pi}{4}\right)=\pi i(-1)i \cos\left(\frac{\pi}{2}+\frac{\pi}{8}\right)\left(-\frac{1}{\sqrt{2}}\right)=\\
\frac{\pi}{\sqrt{2}}\sin\left(\frac{\pi}{8}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Integrate $\int \frac {x^4}{\sqrt {x^2-9}} \,dx$ Integrate $\int \dfrac {x^4}{\sqrt {x^2-9}} dx$
My Attempt:
Let $x=3\sec (\theta )$
$$dx=3\sec (\theta).\tan (\theta).d\theta$$
Then,
$$=\int \dfrac {x^4}{\sqrt {x^2-9}}$$
$$=\int \dfrac {81. \sec^4 (\theta)}{\sqrt {(3\sec (\theta))^2 - 9}} 3\sec (\theta).\tan (\theta).d\theta $$
$$=\int \dfrac {81 \sec^5 (\theta). 3\tan (\theta).d\theta}{3\tan (\theta)}$$
$$=81\int \sec^5 (\theta) d\theta$$
| For $x\ge 3$ let substitute $\begin{cases}x=3\cosh(u) & u\ge 0\\3\sinh(u)=\sqrt{x^2-9}\end{cases}$
$\begin{align}\displaystyle \int \dfrac{x^4}{\sqrt{x^2-9}}\mathop{dx}&=81\int \cosh(u)^4\mathop{du}\\&=\dfrac{81}8\bigg(2\cosh(u)^3\sinh(u)+3\cosh(u)\sinh(u)+3u\bigg)+C\\&=\frac 14x^3\sqrt{x^2-9}+\dfrac{27}{8}x\sqrt{x^2-9}+\dfrac{243}8\operatorname{argch}(\frac x3)+C\end{align}$
For $x\le-3$ just reverse the sign.
Your choice of substitution leads to $\sec^5(x)$ which is difficult to linearise.
With $\cosh^4(u)=\frac 18\cosh(4u)+\frac 12\cosh(2u)+\frac 38$ we can integrate and apply the inverse transformation to go back to powers of $(\cosh,\sinh)$, I wrote only the resulting anti-derivative.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Compute a division with integer and fractional part I have a problem that I don't know how to solve:
Compute $[\frac{\sqrt{7}}{frac(\sqrt{7})}]$
Here's what I've tried: $[\sqrt{7}]=2 \rightarrow frac(\sqrt{7}) = \sqrt{7}-2 \rightarrow [\frac{\sqrt{7}}{frac(\sqrt{7})}] =[\frac{\sqrt{7}}{\sqrt{7}-2}] = [\frac{\sqrt{7}(\sqrt{7}+2)}{7-4}] = [\frac{7+2\sqrt{7}}{3}]$
But from here I don't know what to do anymore.
| It is immediate that $2<\sqrt7<3$, which yields
$$\frac{11}3<\frac{7+2\sqrt7}3<\frac{13}3$$ and the answer is one of $3$ or $4$.
So let us evaluate
$$\frac{7+2\sqrt7}3-4=\frac{2\sqrt7-5}3.$$
It is clear that the numerator is positive, by
$$2\sqrt7>5\iff 28>25.$$
Hence, $4$.
You can also start with a tighter bracketing,
$$2.5=\frac52<\sqrt7<3,$$ justified by $$\frac{25}4<7.$$
Then
$$\frac{7+2\cdot\dfrac52}3=4<\frac{7+2\sqrt7}3<\frac{7+2\cdot3}3<5.$$
| {
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Find all pairs of integers (x,y) such that $x(x+1)(x^2+x+2)=2y^2$ A former Olympiad exercise is to find all pairs of integers $(x,y)$ such that
$$x(x+1)(x^2+x+2)=2y^2.$$
I could not solve this equation. How can one find all solutions?
I tried to rewrite the above equation in a different form, but it did not help. For example, the equation is equivalent to
\begin{align*} x(x+1)(x(x+1)+2)=2y^2&\iff (x(x+1))^2 + 2x(x+1) +1 =2y^2 +1\\
&\iff (x(x+1)+1)^2 =2y^2 +1\end{align*}
Can you tell me how to attack this problem?
Best wishes
| Since $gcd(x^2+x,x^2+x+2)=2$ we have $x^2+x=2a^2$ and $x^2 +x+2=b^2$ or $x^2+x = a^2$ and $x^2+x+2=2b^2$ where $a,b$ are relatively prime.
1st case: for $x>-2$ we have $$ x^2 <x^2+x+2<(x+1)^2$$ which is impossible. If $x<-1$ then we have $$(x+1)^2 <x^2+x+2<x^2$$ which is again impossible.
2nd case: for $x>0$ we have
$$x^2<x^2+x<(x+1)^2$$ which is impossible.
So we are left with the case if $x<-1$ we have $$(x+1)^2<x^2+x<x^2$$ impossible. So $x=0$ or $x=-1$...
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the greatest and least values of $(\sin^{-1}x)^2+(\cos^{-1}x)^2$ Find the upper and lower limit of
$$
(\sin^{-1}x)^2+(\cos^{-1}x)^2
$$
My Attempt:
$$
\frac{-\pi}{2}\leq\sin^{-1}x\leq \frac{\pi}{2}\quad\&\quad0\leq\cos^{-1}x\leq\pi\\(\sin^{-1}x)^2\leq\frac{\pi^2}{4}\quad\&\quad(\cos^{-1}x)^2\leq\pi^2\\
0\leq(\sin^{-1}x)^2+(\cos^{-1}x)^2\leq\frac{\pi^2}{4}+\pi^2=\frac{5\pi^2}{4}
$$
Here, I can see the upper limit is $\frac{5\pi^2}{4}$ which is fine. But, $0$ is one lower limit not the lower limit.
Why am I not getting the lower limit in my approach ?
How do I approach similar problems involving max and min, when you don't get the lower or upper limits ?
| By C-S $$\arcsin^2{x}+\arccos^2{x}=\frac{1}{2}(1^2+1^2)(\arcsin^2{x}+\arccos^2{x})\geq\frac{1}{2}(\arcsin{x}+\arccos{x})^2=\frac{\pi^2}{8}.$$
The equality occurs for $x=\frac{1}{\sqrt2}$, which says that we got a minimal value.
In another hand, for $x\geq0$ we obtain: $$\arcsin^2{x}+\arccos^2{x}\leq\left(\arcsin{x}+\arccos{x}\right)^2=\frac{\pi^2}{4}.$$
The equality occurs for $x=0$.
For $x<0$ let $t=-\arcsin{x}$.
Thus, $0<t\leq\frac{\pi}{2}$, $\arccos{x}=\pi-\arccos(-x)=\pi-\left(\frac{\pi}{2}-\arcsin(-x)\right)=\frac{\pi}{2}+t$ and
$$\arcsin^2{x}+\arccos^2{x}=t^2+\left(\frac{\pi}{2}+t\right)^2\leq\frac{5\pi^2}{4}$$ with equality for $x=-1$, which says that $\frac{5\pi^2}{4}$ is a maximal value.
| {
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Find largest possible value of $x^2+y^2$ given that $x^2+y^2=2x-2y+2$
Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$.
My attempt:
$x^2+y^2=2x-2y+2$
$(x^2-2x)+(y^2+1)=2$
$(x-1)^2+(y+1)^2=4$
I have no idea how to continue here. Any help?
| By Minkowski (the triangle inequality) we obtain:
$$\sqrt{x^2+y^2}\leq\sqrt{(x-1)^2+(y+1)^2}+\sqrt{1^2+(-1)^2}=2+\sqrt2.$$
Id est, $$x^2+y^2\leq(2+\sqrt2)^2=6+4\sqrt2.$$
The equality occurs for $(x,y)=(1+\sqrt2,-1-\sqrt2),$ which says that we got a maximal value.
| {
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McLaurin expansions help The problem:
Study the convergence/divergence of this limit, as $a \in \mathbb{R}$ changes
$$\lim_{x\to 0^+} x^a \biggl( \frac{1}{\log(2x+\sqrt{1+x^3})} -\frac{1}{\log(2x+1)} \biggr)$$
I did that:
$\sqrt{1+x^3}=1+\frac{x^3}{2} + o(x^3)$ so
$\log\left(1 +2x +\frac{x^3}{2} + o(x^3) \right) = 2x - \frac{x^2}{2} +\frac{x^3}{2} +o(x^3)$
and $\log\left(1 +2x \right) = 2x - \frac{x^2}{2} +\frac{8x^3}{3} +o(x^3)$ .
Now how can I expand $\frac{1}{ 2x - \frac{x^2}{2} +\frac{x^3}{2} +o(x^3)}$ and $\frac{1}{ 2x - \frac{x^2}{2} +\frac{8x^3}{3} +o(x^3)}$ ?
I don't find on internet some expansion for these fractions(!)
| Make them into one fraction.
$\frac {\ln (1+ 2x) - \ln (2x + \sqrt{1+x^3})}{\ln (1+ 2x) \ln (2x + \sqrt{1+x^3})}$
Expand the series for $\sqrt {1+x^3} = 1 + \frac 12 x^3 - \frac 18 x^6 \cdots$
$\frac {\ln (1+ 2x) - \ln ( 1 + 2x + \frac 12x^3\cdots)}{\ln (1+ 2x) \ln (1+2x + \frac 12 x^3 \cdots )}$
and then the logs.
$\ln(1+2x) = 2x - \frac 12 (2x)^2 + \frac 13 (2x)^3 \cdots\\
\ln(1+2x+\frac 12 x^3) = 2x + \frac 12 x^3 - \frac 12 (2x + \frac 12x^3)^2 \cdots $
the difference $\frac 12 x^3 - x^4\cdots$
The product $4x^2 - 8x^3 \cdots$
So it would appear that the numerator is of order $x^{a+3}$ and the denominator is of order $x^2$
The limit coverges to 0 if $x > -1$ and diverges if $a < -1$ and converges to $-\frac 18$ if $a = -1$
| {
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If $\sum\limits_{i=1}^na_i=\prod\limits_{i=1}^na_i$ for every $n$, identify $\lim\limits_{n\to \infty}a_n$
Let $\left(a_n\right)_{n \in\mathbb{N}} $ denote a sequence of real numbers such that, for every $n\geqslant1$, $$\sum_{i=1}^na_i=\prod_{i=1}^na_i$$ Identify the limit $$\lim_{n\to \infty}a_n$$
What I have done: $$a_1-a_1=0 \\a_1+a_2-(a_1a_2)=0 \to a_1\cdot a_2(\dfrac{a_1}{a_2}+\dfrac{a_2}{a_1}-1)=0\\.\\.\\.\\a_1\cdot a_2\cdot \cdot \cdot a_n(\dfrac{a_1}{a_2\cdot \cdot \cdot a_n}+\dfrac{a_2}{a_1\cdot \cdot \cdot a_n}+...+\dfrac{a_n}{a_2\cdot \cdot \cdot a_{n-1}}-1)=0$$
Now what do I do ?
| The answer is: if $a_1 > 1$, then the limit is 1; if $a_1 < 1$, then the limit is 0; and there is no sequence with $a_1 = 1$.
First, notice that if we let $S_n = \sum_{k=1}^n a_k$, then
$$S_n + a_{n+1} = \left( \sum_{k=1}^n a_k \right) + a_{n+1} = \sum_{k=1}^{n+1} a_k = \prod_{k=1}^{n+1} a_k = \left( \prod_{k=1}^n a_k \right) a_{n+1} = S_n a_{n+1}.$$
Solving $S_n + a_{n+1} = S_n a_{n+1}$ for $a_{n+1}$ gives $a_{n+1} = \frac{S_n}{S_n - 1}$ (and also implies that $a_1 = S_1 \ne 1$).
Now for the first case, suppose $a_1 > 1$. Then $a_{n+1} = \frac{S_n}{S_n - 1}$, so by induction, we can conclude that $a_n > 1$ and $S_n > 1$ for every $n$. Thus, $S_n = \sum_{k=1}^n a_k > n$, and so $a_{n+1} = 1 + \frac{1}{S_n - 1} < 1 + \frac{1}{n-1}$ for $n > 1$. By the squeeze theorem, we can conclude $a_n \to 1$ as $n \to \infty$.
Otherwise, suppose $a_1 < 1$. Then we get $S_{n+1} = S_n + \frac{S_n}{S_n - 1} = \frac{S_n^2}{S_n - 1}$. Now we can conclude by induction that $S_n \le 0$ for $n \ge 2$, and $S_n$ is nondecreasing for $n \ge 3$ since $a_n = \frac{S_{n-1}}{S_{n-1}-1} \ge 0$. It follows that $S_n$ must have some limit $L \le 0$, which must then satisfy $L = \frac{L^2}{L-1}$. This forces $L = 0$, so $a_n = S_n - S_{n-1} \to 0 - 0 = 0$ as $n \to \infty$.
| {
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"url": "https://math.stackexchange.com/questions/2612655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Evaluating $\lim\limits_{x\to \infty}\frac{x^2}{2^x-1}$
I would like to evaluate
$$\lim_{x\to \infty}\frac{x^2}{2^x-1}$$
Without using L'HOSPITAL's rule nor series.
I tried more than one
technique such the sub
$$x=\frac{1}{y}$$
But i could not get the solution ?
| Note that
$$\frac{x^2}{2^x-1}=\frac{2^x}{2^x-1}\cdot \frac{x^2}{2^x}\to1\cdot0=0$$
indeed
$$\frac{2^x}{2^x-1}=\frac{1}{1-\frac{1}{2^x}}\to 1$$
and since eventually, notably for $x>10$, we have that $2^x>x^3$
$$\frac{x^2}{2^x}<\frac{x^2}{x^3}=\frac1x\to0$$
EDIT
Proof by induction of $2^n>n^3$.
Base case
$n=10\implies2^{10}=1024>10^3=1000$
Induction step
We need to prove that
$2^n>n^3\implies 2^{n+1}>(n+1)^3$
Let observe that
$$2^{n+1}=2\cdot 2^n\stackrel{\text{inductive hypothesis}}\ge 2\cdot n^3 \stackrel{\text{?}}\ge (n+1)^3$$
and the last inequality is true since
$$2\cdot n^3 \ge (n+1)^3=n^3+3n^2+3n+1\iff n^3 \ge 3n^2+3n+1 \quad \square$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Polynomial vector space and an isomorphism. Let $V$ be the subspace of $P_3 (\Bbb R ) $ spanned by $(1+2x+x^2) , (x+x^2+x^3), (3+5x+2x^2 -x^3)$
Find $n=\dim(V) $ and write an explicit formula for an isomorphism $T: \Bbb R^n \to V $ and prove that it is an isomorphism.
I think that $P_3 (\Bbb R ) $ is spanned by 4 vectors $(1,x,x^2,x^3 )$
Looking at it $ -3(1+2x+x^2) +(x+x^2+x^3)+ (3+5x+2x^2 -x^3)=0$
So really all i need to consider is the span of the vectors $(1+2x+x^2), (3+5x+2x^2 -x^3)$ or $(1+2x+x^2), (1+x -x^3)$
This would seem to imply that $\dim(V) =2 $ as for the map from $\Bbb R^n $ i can't seem to imagine what that space looks like mapping into this one?
I meant id like to say take $T(1,0) \to(1+2x+x^2) $ and $T(0,1) \to(3+5x+2x^2 -x^3) $
where $T:\Bbb R^2 \to V $
| Write down the coefficients of the polynomial in a matrix, and row or column reduce it to an upper triangular form, to find the number of free variables which the span comprises of:
$$
\begin{pmatrix}
1 & 2 & 1 & 0\\
0 & 1 & 1 & 1 \\
3 & 5 & 2 & -1
\end{pmatrix}
\to
\begin{pmatrix}
1 & 0 & -1 & -2\\
0 & 1 & 1 & 1 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$
(I leave you to do this yourself. If you have trouble I can explain.)
Since there are precisely two non-zero rows present in the above matrix, the rank of the subspace is $2$.
Now, it is clear what the isomorphism should be as well. Simply take $T(1,0) = (1,0,1,2) = x + x^2 + 2x^3$, and $T(0,1) = (0,1,1,1) = x+x^2+x^3$, and extend $T$ linearly to the real plane. By the fact that the above are linearly independent, this is automatically an isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Binomial Coefficients: Summation and Floor Function Identities An exercise for my stochastic processes class is asking me to prove that
\begin{equation} \sum_{k=0}^{\left \lfloor \frac{n-1}{2} \right \rfloor } \begin{pmatrix} n \\ 2k+1 \end{pmatrix} 5^k \end{equation}
is always an integer multiple of $(2^{n-1})$, that is
\begin{equation} \sum_{k=0}^{\left \lfloor \frac{n-1}{2} \right \rfloor } \begin{pmatrix} n \\ 2k+1 \end{pmatrix} 5^k = p(2^{n-1}), \quad p \in \mathbb{N}\end{equation}
Some identities which seem helpful for this situation but which I have not quite figured out how to apply are the following:
\begin{equation}\begin{pmatrix} n \\ k \end{pmatrix} = \begin{pmatrix} \lfloor n/2 \rfloor \\ \lfloor k/2 \rfloor \end{pmatrix}\begin{pmatrix} n \mod 2 \\ k \mod 2 \end{pmatrix} \mod 2 \end{equation}
and
\begin{equation} \sum_{k=0}^n k\begin{pmatrix} n \\ k \end{pmatrix} = n2^{n-1}\end{equation}
Could someone please suggest some identities that may be helpful or point me to a resource where I might find some useful identities?
| The only identity you really need here is the binomial theorem:
$$
\binom n0 + \binom n1 x + \binom n2 x^2 + \dots + \binom{n}{n-1}x^{n-1} + \binom nn x^n = (1 + x)^n.
$$
By replacing $x$ with $-x$, we get:
$$
\binom n0 + \binom n1 (-x) + \binom n2 (-x)^2 + \dots + \binom{n}{n-1}(-x)^{n-1} + \binom nn (-x)^n = (1 - x)^n.
$$
You should ponder the relationship between these two sums and how you may use them to obtain the sum in your question, which only includes the odd terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$a_{n+2}=\frac{2}{a_{n+1}}+a_{n}.$ and $a_{1}=1,a_{2}=2$.Then value of $a_{2017}$
If ${a_{n}}$ be a sequence defined by $\displaystyle a_{n+2}=\frac{2}{a_{n+1}}+a_{n}.$ and $a_{1}=1,a_{2}=2$.Then value of $a_{2017}$
Try: $$a_{n+2}a_{n+1}-a_{n}a_{n+1}=2\frac{a_{n+1}}{a_{n-1}}$$
$$\sum^{2015}_{n=1}\left(a_{n+2}a_{n+1}-a_{n}a_{n+1}\right)=2\sum^{2015}_{n=1}\left(\frac{a_{n+1}}{a_{n-1}}\right)$$
$$a_{2017}a_{2016}-a_{2}a_{3}=a_{2015}a_{2016}$$
Could some help me to solve it, Thanks
| Hint: apply a telescoping sum to
$$
a_{n+2}a_{n+1}-a_{n+1}a_n=2
$$
The solution of the equation above is $a_{n+1}a_n=2n$
$$
\begin{align}
\frac{a_{2n+2}}{a_{2n}}
&=1+\frac{2}{a_{2n+1}a_{2n}}\\
&=1+\frac1{2n}
\end{align}
$$
Therefore,
$$
\begin{align}
a_{2n+2}
&=2\prod_{k=1}^n\frac{2n+1}{2n}\\
&=\frac{4n+2}{4^n}\binom{2n}{n}
\end{align}
$$
Since $a_{2n+2}a_{2n+1}=4n+2$, we get
$$
a_{2n+1}=\frac{4^n}{\binom{2n}{n}}
$$
Thus,
$$
a_{2017}=\frac{4^{1008}}{\binom{2016}{1008}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Deriving the fact that the approximation $\log(n!) \approx n\log(n) - n + \frac{1}{2}\log(2\pi n)$ is $O(1/n)$. I get to begin with Stirling's approximation, for any $C \in \mathbb{Z}_{\geq 0}$, there exists some $N \in \mathbb{Z}_{\geq 0}$ such that $N > C$ and for all $n > N$:
\begin{align*}
&\quad \left|n! - \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}\right| \leq C \left|\frac{1}{n}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right| \\
&\Rightarrow -C \frac{1}{n}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n \leq n! - \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} \leq C \frac{1}{n}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n \\
&\Rightarrow -C \frac{1}{n}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n + \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} \leq n! \leq C \frac{1}{n}\sqrt{2\pi n}\left(\frac{n}{e}\right)^n + \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n} \\
&\Rightarrow \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(-C \frac{1}{n} + 1\right) \leq n! \leq \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(C \frac{1}{n} + 1\right)
\end{align*}
Now we are in a position to make further manipulations:
\begin{align*}
&\quad \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(-C \frac{1}{n} + 1\right) \leq n! \leq \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(C \frac{1}{n} + 1\right) \\
&\{\text{$\log$ is monotonic}\} \\
&\Rightarrow \log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(-C \frac{1}{n} + 1\right)\right) \leq \log\left(n!\right) \leq \log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(C \frac{1}{n} + 1\right)\right) \\
&\{\text{take the average of the upper ($U(n)$) and lower bounds ($L(n)$)}\} \\
&\{\text{$\log(n!) \approx f(n)$ means $|f(n) - \log(n!)| \leq K \left[U(n) - L(n)\right]$, where $0 \leq K \leq 1$}\} \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(C \frac{1}{n} + 1\right)\right) + \frac{1}{2}\log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(-C \frac{1}{n} + 1\right)\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\left(\log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(C \frac{1}{n} + 1\right)\right) + \log\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(-C \frac{1}{n} + 1\right)\right)\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left((\sqrt{2\pi n})^2\left(\frac{n}{e}\right)^{2n}\left(C \frac{1}{n} + 1\right)\left(-C \frac{1}{n} + 1\right)\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left(2\pi n\left(\frac{n}{e}\right)^{2n}\left(\frac{-C^2}{n^2} + 1\right)\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left(2\pi n\right) + \frac{1}{2}\log\left(\frac{n^{2n}}{e^{2n}}\right) + \frac{1}{2}\log\left(1 - \frac{C^2}{n^2}\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left(2\pi n\right) + \frac{1}{2}\log\left(n^{2n}\right) - \frac{1}{2}\log\left(e^{2n}\right) + \frac{1}{2}\log\left(1 - \frac{C^2}{n^2}\right) \\
&\Rightarrow \log(n!) \approx \frac{1}{2}\log\left(2\pi n\right) + n\log\left(n\right) - n\log\left(e\right) + \frac{1}{2}\log\left(1 - \frac{C^2}{n^2}\right) \\
&\Rightarrow \log(n!) \approx n\log\left(n\right) - n + \frac{1}{2}\log\left(2\pi n\right) + \frac{1}{2}\log\left(1 - \frac{C^2}{n^2}\right)
\end{align*}
From here I could expand $\log(1 + x)$ about $x = 0$ (note that $0 < \frac{C^2}{n^2} < 1$, as $C < n$, and $0 < n, C$, and also $C^2/n^2$ is pretty close to $0$ most of the time, as $n > C$). The first term of the Taylor series expansion for $\log(1 + x)$ about $x = 0$ is simply $x$.
\begin{align*}
&\quad \log(n!) \approx n\log\left(n\right) - n + \frac{1}{2}\log\left(2\pi n\right) + \frac{1}{2}\log\left(1 - \frac{C^2}{n^2}\right) \\
&\Rightarrow \log(n!) \approx n\log\left(n\right) - n + \frac{1}{2}\log\left(2\pi n\right) - \frac{C^2}{2n^2}
\end{align*}
But this seems to suggest my error is $O(1/n^2)$? Isn't this approximation supposed to be $O(1/n)$? Please advise on where I went wrong.
| Nothing justifies that a function is equivalent to the average of the left and right-hand sides in an inequality.
$n^2<n^3<n^4$ but that doesn't make $n^3\sim \dfrac{n^2+n^4}{2}$.
Just after you use the monotonicity of $\log$ you can already conclude by saying that $\log(1+\frac{C}{n})\leq \frac{C}{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the limit containing $\arctan{x}$ and $\arcsin{x}$ Evaluate:
$$\lim_{x\to{0}}\bigg(\frac{2}{x^3}.(\arcsin{x}-\arctan{x})\bigg)^{2/x^2}$$
I can just expand $\arcsin{x}$ and $\arctan{x}$ using their taylor expansions, but is there any other method?
| Here is an approach which relies on trigonometric identities and certain standard limits. The approach involves some amount of labor in algebra and should be used only when more powerful tools like Taylor series or L'Hospital's Rule are forbidden. On the other hand it does show that simple tools can be used to tackle difficult problems if one applies them properly.
Let $t=\arcsin x$ so that $x=\sin t$ and then $\tan t=x(1-x^2)^{-1/2}$ so that $t=\arctan x(1-x^2)^{-1/2}$ and thus the expression $\arcsin x-\arctan x$ can be written as $$\arctan\frac{x} {\sqrt{1-x^2}}-\arctan x=\arctan\dfrac{x(1-\sqrt{1-x^2})}{x^2+\sqrt{1-x^2}} $$ which can be simplified as $$\arctan\frac{x^3}{(x^2+\sqrt{1-x^2})(1+\sqrt{1-x^2})}=\arctan u\text{ (say)} $$ where $u/x^3\to 1/2$ and further $$\frac{2u}{x^3}-1=\frac{1-(1+x^2)\sqrt{1-x^2}}{(x^2+\sqrt{1-x^2})(1+\sqrt{1-x^2})}$$ which can be written as $$\frac{-x^2+x^4+x^6}{(x^2+\sqrt{1-x^2})(1+\sqrt{1-x^2})(1+(1+x^2)\sqrt{1-x^2})}$$ and thus $$\lim_{x\to 0}\frac{1}{x^2}\left(\frac{2u}{x^3}-1\right)=-\frac{1}{4}\tag{1}$$ If $L$ is the desired limit in question and $f(x) $ is the function whose limit needs to be evaluated then
\begin{align}
\log L&=\log\lim_{x\to 0}f(x)=\lim_{x\to 0}\log f(x)\text{ (via continuity of log)} \notag\\
&=\lim_{x\to 0}\frac{2}{x^2}\log\left(\frac{2\arctan u} {x^3}\right)\notag\\
&=\lim_{x\to 0}\frac{2}{x^2}\cdot\dfrac{\log\left(1+\dfrac{2\arctan u-x^3}{x^3}\right)}{\dfrac{2\arctan u-x^3}{x^3}}\cdot\frac{2\arctan u-x^3}{x^3}\notag\\
&=\lim_{x\to 0}\frac{2}{x^2}\cdot\frac{2\arctan u-x^3}{x^3}\notag\\
&=\lim_{x\to 0}\frac{4\arctan u-4u}{x^5}+\frac{4u-2x^3}{x^5}\notag\\
&=\lim_{x\to 0}4\cdot\frac{\arctan u-u} {u^2}\cdot\frac{u^2}{x^6}\cdot x-\frac{1}{2}\text{ (via equation (1))}\notag\\
&=-\frac{1}{2}\notag\end{align}
The desired limit $L$ is thus $1/\sqrt{e}$. We have used the limit $$\lim_{x\to 0}\frac{\arctan x-x} {x^2}=\lim_{x\to 0}\frac{x-\tan x} {x^2}=\lim_{x\to 0}\frac{x\cos x-\sin x} {x^2}=0$$ which can be proved without Taylor series or L'Hospital's Rule. Also note that the factor $u^2/x^6=(u/x^3)^2\to (1/2)^2=1/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Proving complex series $1 + \cos\theta + \cos2\theta +... + \cos n\theta $ So I have this result
$1 + z + z^2 + ... + z^n = \frac{z^{n+1}-1}{z-1}$
which I proved already. Now I am supposed to use that result and De Moivre's formula to establish this identity
$1 + \cos\theta + \cos2\theta +... + \cos n\theta = \frac{1}{2} + \frac{\sin[(n+\frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})}$
Can anyone help me?
| Starting with (and using this midway)
$$1+\cos{\theta}+\cos{2\theta}+...+\cos{n\theta}=\\
\Re\left(1+\cos{\theta}+i\sin{\theta}+\cos{2\theta}+i\sin{2\theta}+...+\cos{n\theta}++i\sin{n\theta}\right)=\\
\Re\left(1+e^{i\theta}+e^{2i\theta}+...+e^{ni\theta}\right)=\\
\Re\left(\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}\right)=\Re\left(\frac{e^{\frac{(n+1)}{2}i\theta}}{e^{i\frac{\theta}{2}}}\cdot\frac{e^{\frac{(n+1)}{2}i\theta}-e^{\frac{-(n+1)}{2}i\theta}}{e^{i\frac{\theta}{2}}-e^{-i\frac{\theta}{2}}}\right)=\\
\Re\left(\frac{e^{\frac{(n+1)}{2}i\theta}}{e^{i\frac{\theta}{2}}}\cdot\frac{\sin{\frac{(n+1)\theta}{2}}}{\sin{\frac{\theta}{2}}}\right)=
\Re\left(e^{i\theta\frac{n}{2}}\cdot\frac{\sin{\frac{(n+1)\theta}{2}}}{\sin{\frac{\theta}{2}}}\right)=\\
\frac{\cos{\frac{n\theta}{2}}\cdot\sin{\frac{(n+1)\theta}{2}}}{\sin{\frac{\theta}{2}}}=
\frac{1}{2}\frac{\sin\left(\frac{n\theta}{2}+\frac{(n+1)\theta}{2}\right)-\sin\left(\frac{n\theta}{2}-\frac{(n+1)\theta}{2}\right)}{\sin{\frac{\theta}{2}}}=\\
\frac{1}{2}\frac{\sin\left(\left(n+\frac{1}{2}\right)\theta\right)+\sin{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}$$
and the final result follows.
| {
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"source": "stackexchange",
"question_score": "5",
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Evaluating: $\int \frac {1+\sin (x)}{1+\cos (x)} dx$
Evaluate: $\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$
My Attempt:
$$=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$$
$$=\int \dfrac {(\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2}))^2}{2\cos^2 (\dfrac {x}{2})} dx$$
$$=\dfrac {1}{2} \int (\dfrac {\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2})}{\cos (\dfrac {x}{2})})^2 dx$$
$$=\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx$$
How do I continue?
| From where you are you did a great job, note that
$$(\tan (\frac{x}{2}))' =\frac{1}{2}(1+\tan^2 (\frac{x}{2}))~~~and~~~ -2(\ln|\cos (\frac{x}{2})|)' = \tan (\frac{x}{2})$$
Therefore $$\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx = \dfrac {1}{2}\int (\tan^2 (\dfrac {x}{2}) +1) dx + \int \tan (\dfrac {x}{2}) dx \\=\tan (\frac{x}{2}) -2\ln|\cos (\frac{x}{2})|+c$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 6
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Number of non negative integer solution of the equation $x+y+z=n$
Total number of non negative integer solution of the equation $x+y+z=n$ subjected to the condition $x\leq y \leq z.$
Try:if $x=0$ Then we have $y+z=n$
If $x=1$ Then we have $y+z=n-1$
.....if $x=n$ Then we have $y+z=0$ so $x=y=0$
Could some help me how to calculate for $x+y=n,n-1,n-2$ ect.
Or any nice way to find solution of original equation.Thanks
| Let us define $A,B,C$ as follows :
$A$ : The number of non-negative solutions $(x,y,z)$ such that $x\lt y\lt z$.
$B$ : The number of non-negative solutions $(x,y,z)$ such that $x=y\lt z$ or $x\lt y=z$.
$C$ : The number of non-negative solutions $(x,y,z)$ such that $x=y=z$.
Then, the number of solutions such that $x\le y\le z$ is given by $A+B+C$.
The number of non-negative solutions for $x+y+z=n$ is given by $\binom{n+2}{2}=\frac{(n+2)(n+1)}{2}$, so we have
$$\frac{(n+2)(n+1)}{2}=3!A+\frac{3!}{2!}B+C\tag1$$
Since $C$ is given by
$$C=\begin{cases}1&\text{if $n\equiv 0\pmod 3$}\\\\0&\text{if $n\not\equiv 0\pmod 3$}\end{cases}$$
we can write
$$C=\left\lfloor\left\lfloor\frac n3\right\rfloor\div\frac n3\right\rfloor\tag2$$
To find $B$, let us consider $a+a+b=n$ where $a\not=b$, and then
$$b=n-2a\ge 0\implies a\le\frac n2\implies a=0,1,\cdots,\left\lfloor\frac n2\right\rfloor$$
So, we have
$$B=\left\lfloor\frac n2\right\rfloor+1-C\tag3$$
From $(1)$, we have
$$A=\frac 16\left(\frac{(n+2)(n+1)}{2}-3B-C\right)\tag4$$
Therefore, from $(2)(3)(4)$, the number of solutions with $x\le y\le z$, i.e. $A+B+C$ is given by
$$\frac{1}{12}\left((n+2)(n+1)+6\left\lfloor\frac n2\right\rfloor+6+4\left\lfloor\left\lfloor\frac n3\right\rfloor\div\frac n3\right\rfloor\right)$$
| {
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Prove the convergence of $\sum\limits_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$ Hints preferred:
I need to prove convergence and determine the limit of following series: $$ \sum_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$$
I need to use partial fraction decomposition to simplify the term, which I attempted with following:
$$
\frac 1 {(2k+1)(2k+5)} = \frac a {(2k+1)} + \frac b {(2k+5)} \\~\\
\frac 1 {(2k+1)(2k+5)} = \frac {a(2k+5)} {(2k+1)(2k+5)} + \frac {b(2k+1)} {(2k+1)(2k+5)} \\~\\$$
Then we have:
$$
k: 2a + 2b = 0 \\
k^0: 5a + b = 1 \\~\\
a = \frac 1 4 ~ \land b = -\frac 1 4
$$
So we have:
$$
\frac 1 {(2k+1)(2k+5)} = \frac { \frac 1 4 } {(2k+1)} - \frac { \frac 1 4} {(2k+5)} = \\ \frac { 1 } {4(2k+1)} - \frac { 1 } {4(2k+5)}
$$
Which can be defined as:
$$
\frac 1 4 (\frac { 1 } {2k+1} - \frac { 1 } {2k+5} )
$$
I'm not sure how to turn the result into a telescopic series (if the partial fraction decomposition is right in the first place).
| You are on the right track, this is a telescopic series. Thanks to your partial fraction decomposition (which is correct), we have that for $n\geq 2$,
$$
\begin{align}\sum_{k=2}^n \frac 1 {(2k+1)(2k+5)}&=
\frac 1 4 \sum_{k=2}^n\left(\frac { 1 } {2k+1} - \frac { 1 } {2k+5} \right)\\
&=\frac 1 4 \left(\sum_{k=2}^n\frac { 1 } {2k+1}-\sum_{k=2}^n\frac { 1 } {2k+5}\right)\\
&=\frac 1 4 \left(\sum_{k=2}^n\frac { 1 } {2k+1}-\sum_{k=4}^{n+2}\frac { 1 } {2(k-2)+5}\right)\\
&=\frac 1 4 \left(\sum_{k=2}^n\frac { 1 } {2k+1}-\sum_{k=4}^{n+2}\frac { 1 } {2k+1}\right)\\
&=\frac 1 4 \left(\sum_{k=2}^3\frac { 1 } {2k+1}-\sum_{k=n+1}^{n+2}\frac { 1 } {2k+1}\right)\\
\end{align}$$
where in the second sum we shifted the index.
Can you take it from here?
| {
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If $\exists n \in \mathbb{Z+}$,s.t. $2^n + 1$ is a prime, prove $\exists k \in \mathbb{Z+}, n = 2^k$ The actual description of the question is:
Suppose that $2^n + 1$ is a prime where $n$ is a positive integer. Prove that
$n = 2^k$ for some positive integer $k$.
There is only one approach that know of: To show that there is factorization in which at least one factor $(x-a)=1$, and the others are prime. A nice simplification would be to show that all except one root is having value such that the factor has value = $1$. For example, if given quadratic: $x^2 + 2x - 3 = (x+3)(x-1)$. As have two factors, so one of these factors to be $1$, and the other prime.
This leads to: $$x-1=1 \implies x=2 \implies (x+3)(x-1) = 5$$
else, $$x+3=1 \implies x = -2 \implies (x-1)(x+3) => -3$$
But, if consider positive integers then $x=2$ is the value.
if given quadratic: $x^2 - 5x +6 = (x-3)(x-2)$. As have two factors, so one of these factors to be $1$, and the other prime.
This leads to: $$x-3=1 \implies x=4 \implies (x-3)(x-2) = 2$$
else, $$x-2=1 \implies x = 3 \implies (x-3)(x-2) => 0$$
But, if consider that $0$ is not a prime, then $x=4$ is the value.
So, let us try to work out cases with value of $k$ starting form $0$.
$k=0, 2^0=1, 2^1+1 =3$
$k=1, 2^1=2, 2^2 +1=5$
So, weak induction will suffice here, as strong induction is not needed as no need for previous values to prove new case.
So, let the assumption be that it holds for $k=l$, need prove for $k=l+1$, but, $2^{2^{l+1}} = 2^{{2^l}^2} = (2^{2^l})^2$
So, if $2^{2^l} + 1$ is a prime, let p; then need prove $(p-1)^2+1$ is also.
i.e., need prove $p^2-2p+2$ is a prime, given $p$ is.
| Also from $2 \equiv -1 \pmod{3} \Rightarrow 2^{2k+1} \equiv (-1)^{2k+1} \equiv -1 \pmod{3}$. Since we ask for $2^n+1$ to be prime (thus it can't be divisible by 3, unless $n=1=2^0$), $n$ must be even or $n=2k_0$, then it becomes $4^{k_0}+1$ which is $4^{k_0} \equiv -1 \pmod{5}$ for odd $k_0$ so (unless $n=2^1$) it must be even $k_0=2k_1$ and $n=2^2k_1$ and so on by induction. This process stops for some $k_t$ s.t. $\frac{n}{2^{k_t}}\leq 1$.
| {
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Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{1}{6} - 1) (x^\frac{1}{6} + 1) } $$
$$\lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{2}{6} - 1) } $$
| Here's another solution depending on what you're allowed to use: substitute $u=x^{\frac{1}{6}}$.
Then we get
$$\lim_{u\rightarrow 1}\frac{u^{\frac{6}{5}}-1}{u-1}$$
and we can recognise this as the derivative of $u\mapsto u^{\frac{6}{5}}$ at 1, which is of course $\frac{6}{5}$.
| {
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"source": "stackexchange",
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Solving a biquadratic.
If $x$ is a real number and satisfies $$x+ \sqrt[4] {5-x^4}=2$$ then find the value of $$x\sqrt[4] {5-x^4}$$
My try :
The question is significantly asking for the value of $-x(x-2)$ if we get the root of the equation $$2x^4-8x^3+24x^2-32x+11=0$$
Using this I have reached till
$$-2x(x-2)(x^2-2x+8)=11$$
$$-x(x-2)=\frac {11}{2(x^2-2x-8)}$$
but I couldn't manipulate it further.
Also upon some second thought I want to ask whether could it be possible to form a quadratic polynomial with two roots $\alpha$ and $\beta$ such that $\alpha=x$ and $\beta = \sqrt[4] {5-x^4}$
but I still couldn't proceed further. Somebody please share some hints.
| Let $p=x\sqrt[4] {5-x^4}$, then by squaring both sides of $x+ \sqrt[4] {5-x^4}=2$ we get
$$x^2+ \sqrt{5-x^4}+2p=4$$
that is
$$x^2+ \sqrt{5-x^4}=4-2p.$$
Now take the square again and remember that $4-2p\geq 0$ (the l.h.s is non-negative), i.e. $p\leq 2$.
Hence
$$x^4+ (5-x^4)+2p^2=(4-2p)^2=16-16p+4p^2.$$
that is
$$2p^2-16p+11=0\implies p_1=4+\frac{\sqrt{42}}{2}\;\text{or}\; p_2=4-\frac{\sqrt{42}}{2}.$$
Since $p_1>2$, we have just ONE acceptable solution $p_2=4-\frac{\sqrt{42}}{2}$.
P.S. $p=4-\frac{\sqrt{42}}{2}$ works because the problem has at least a solution: $f(x):=x+ \sqrt[4] {5-x^4}$ is continuous in $[0,1]$, $f(0)=\sqrt[4] {5}<2$ and $f(1)=1+ \sqrt{2}>2$, so there is at least a real $x$ such that $f(x)=2$.
| {
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"source": "stackexchange",
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Calculate the limit $L=\lim_{x\to 0^+}\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac{1}{x}\right)\right)^x$. Question: Calculate the limit
$$L=\lim_{x\to 0^+}\left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac{1}{x}\right)\right)^x.$$
I'm thinking of using infinitesimal, but I'm not used to these kind of analysis arguments. Can someone explain how to deal with these kind of problems? Thanks in advance.
| Note that
$$ \left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac{1}{x}\right)\right)^x=e^{x\log \left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac{1}{x}\right)\right) }\to1$$
indeed
$$x\log \left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac{1}{x}\right)\right)\to0 $$
since
$$x\log \left(2\sin\left(\sqrt{x}\right)+\sqrt{x}\sin\left(\frac{1}{x}\right)\right)=x \left[\log \sqrt{x} +\log\left(2\frac{\sin\left(\sqrt{x}\right)}{\sqrt{x}}+\sin\left(\frac{1}{x}\right)\right)\right]=\sqrt{x} \left[ \sqrt{x} \log \sqrt{x} + \sqrt{x} \log\left(2\frac{\sin\left(\sqrt{x}\right)}{\sqrt{x}}+\sin\left(\frac{1}{x}\right)\right)\right]\to0\cdot(0+0)=0$$
| {
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How to prove that $ \Bigg(\sum^{n}\limits_{k=1}\sqrt{\frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}}}\Bigg)^{2} \le n\sqrt{\frac{n}{n+1}}$ for $n\ge1$
I need to prove Prove the inequality
$$ \Bigg(\sum^{n}_{k=1}\sqrt{\frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}}}\Bigg)^{2} \le n\sqrt{\frac{n}{n+1}}, $$
where $n$ is a positive integer.
Equivalently
$$\sum^{n}_{k=1}\sqrt{\frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}}} \le \sqrt[4]{\frac{n^3}{n+1}}, $$
Then proceed by induction assuming it is true for $n-1$ we have
$$\sum^{n}_{k=1}\sqrt{\frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}}} \le \sqrt[4]{\frac{(n-1)^3}{n}} + \sqrt{\frac{n-\sqrt{n^{2}-1}}{\sqrt{n(n+1)}}} $$
Then one need to show that the last term is not greater than $$\sqrt[4]{\frac{n^3}{n+1}}$$ which is not an obvious task.
Question; Is there an alternative way of proving this without appealing induction or can anyone help to prove this last step?
| It is a combination of Cauchy-Schwarz inequality and a telescope sum.
According to Cauchy-Schwarz inequality:
$$\Bigg(\sum^{n}_{k=1}\sqrt{\frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}}}\Bigg)^{2} \le \left( \sum_{k=1}^n 1^2 \right)\left( \sum_{k=1}^n \frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}} \right) = n \cdot \left( \sum_{k=1}^n \frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}} \right)$$
The sum on the right-hand side can now be transformed into a telescoping sum:
$$\left( \sum_{k=1}^n \frac{k-\sqrt{k^{2}-1}}{\sqrt{k(k+1)}} \right) = \left( \sum_{k=1}^n \frac{\sqrt{k}\sqrt{k}-\sqrt{(k-1)(k+1)}}{\sqrt{k(k+1)}} \right)\\ = \sum_{k=1}^n \left( \frac{\sqrt{k}}{\sqrt{k+1}} - \frac{\sqrt{k-1}}{\sqrt{k}} \right) = \sqrt{\frac{n}{n+1}}$$
Done.
| {
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Prove that the series $\sum_{n=1}^{\infty}\frac{1}{(n+x)(n+x+1)(n+x+2)}$ has a sum of $\frac{1}{2(x+1)(x+2)}$ I am trying to prove that the following series:
$$
\sum_{n=1}^{\infty}\frac{1}{(n+x)(n+x+1)(n+x+2)}
$$
has a sum of:
$$
\frac{1}{2(x+1)(x+2)}
$$
What I've tried:
Using the criteria that $S_n = b_1 - l$:
$$
\frac{1}{(n+x)} - \frac{1}{(n+x+1)(n+x+2)} = \frac{(n+x+1)(n+x+2)-(n+x)}{(n+x)(n+x+1)(n+x+2)} \\
\implies \frac{1}{(n+x)(n+x+1)(n+x+2)} = \frac{1}{(n+x+1)(n+x+2)-(n+x)}(\frac{1}{(n+x)} - \frac{1}{(n+x+1)(n+x+2)}) = \frac{1}{(n+x)(n+x+1)(n+x+2) - (n+x)^2} - \frac{1}{(n+x+1)^2(n+x+2)^2-(n+x)(n+x+1)(n+x+2)} = b_n - b_{n+1}
$$
Now that I have:
$$
b_n = \frac{1}{(n+x)(n+x+1)(n+x+2) - (n+x)^2} \\
l = \lim_{n \to \infty}{b_n} = \lim_{n \to \infty}{\frac{1}{(n+x)(n+x+1)(n+x+2) - (n+x)^2}} = \frac{1}{\infty} = 0 \\
b_1 = \frac{1}{(1+x)(1+x+1)(1+x+2) - (1+x)^2} = \frac{1}{(1+x)(2+x)(3+x) -
(1+x)^2}
$$
Therefore:
$$
S_n = b_1 - l \implies \\
S_n = \frac{1}{(1+x)(2+x)(3+x) -
(1+x)^2} - 0 \\
S_n = \frac{1}{(1+x)(2+x)(3+x) -
(1+x)^2} \\
= \frac{1}{(1+x)}.\frac{1}{(2+x)(3+x) -
(1+x)} \\
= \frac{1}{(1+x)}.\frac{1}{(6+3x+2x+x^2-1-x) -
(1+x)} \\
= \frac{1}{(1+x)}.\frac{1}{(x^2+4x+5)}
$$
Which is not the expected result. Am I mistaken somewhere or is this completely wrong?
| The given expression can be simplified as
$$\frac {1}{2} \left[ \frac {1}{(n+x)(n+x+1)} - \frac {1}{(n+x+1)(n+x+2)}\right ]$$
Which simply telescopes to
$$
\frac {1}{2(x+1)(x+2)}
$$
| {
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Proof of the generating function of $1 -2x +3x^2 -4x^3+5x^4-6x^5+\cdots$ The ordinary generating function for the sequence $\{a_n\}_{n\geq0}$ where $a_n = (-1)^n\,n$ is
$$1 -2x +3x^2 -4x^3+5x^4-6x^5+\cdots = \frac{1}{(x+1)^2}$$
I can see from the geometric formula, and even from long division that
$$1 -x +x^2 -x^3+x^4-x^5+\cdots = \frac{1}{1+x}$$
but I'm not seeing the coefficients to explain $\frac{1}{(x+1)^2}.$
| $$S=1 -x +x^2 -x^3+x^4-x^5+\cdots = \frac{1}{1+x}$$
Take the derivative
$$S'=0 -1 +2x -3x^2+4x^3-5x^4+\cdots = \frac{-1}{(1+x)^2}$$
Multiply by $-1$
$$-S'=1 -2x +3x^2-4x^3+5x^4+\cdots = \frac{1}{(1+x)^2}$$
$$\sum_{k=0}(-1)^k(k+1)x^k = \frac{1}{(1+x)^2}$$
| {
"language": "en",
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What is $\int \frac{ \sqrt{(a^2 - x^2)}}{x^2}$? I tried the following trig substitution:
$x = a\sin \theta$
$$ \int \frac{ \sqrt{(a^2 - x^2)}}{x^2} = \int \frac{\sqrt{a^2 - a^2 \sin^2 \theta}}{a^2 \sin^2 \theta} = \int \frac{a \cos \theta}{a^2 \sin^2}$$
Setting $u = \sin x$ yields:
$$\frac{1}{a} * \int u^{-2} = - \frac{1}{a \sin \theta}$$
However my textbook states that the answer should really be:
$$ -\frac{\sqrt{a^2 - x^2}+x\tan^{-1}(\frac{x}{\sqrt{a^2-x^2}})}{x}$$
Where did I mess up?
| $x = a\sin \theta \implies \color{blue}{ dx = a\cos \theta \,d\theta}$
$$ \int \frac{ \sqrt{(a^2 - x^2)}}{x^2}\,\color{blue}{dx} = \int \frac{\sqrt{a^2 - a^2 \sin^2 \theta}}{a^2 \sin^2 \theta}\,\color{blue}{(a \cos \theta\,d\theta)} = \int \frac{a^2 \cos^2 \theta\, d\theta }{a^2 \sin^2 \theta} = \int \frac {\cos^2 \theta}{\sin^2 \theta}\; d\theta = \int \cot^2 \theta \,d\theta $$
Or
$$=\int \frac{a^2 (1 -\sin^2 \theta)\, d\theta }{a^2 \sin^2 \theta} = \int \frac {1-\sin^2 \theta}{\sin^2 \theta}\; d\theta.$$
$$ = \int \frac 1{\sin^2 \theta} \, d\theta - \int \,d\theta$$
$$ = \left(\int \csc^2 \,d\theta\right) - \theta + c$$
$$= -\cot\theta - \theta + C$$
and recall that, since $x = a\sin \theta,$ then $\sin \theta = \frac xa$ and so $\theta = \sin^{-1}\left( \frac xa\right)$
That gives us $$= -\sin^{-1}\left(\frac xa\right) - \cot\left( \sin^{-1}\left(\frac xa\right)\right)+C\tag{1}$$
Now using that $$\cot \left(\sin^{-1}\frac xa\right) = \frac{\sqrt{1 - \left(\frac {x^2}{a^2}\right)}}{\frac xa},$$
we get that the integral is equal to $$- \frac{ a\sqrt{1 -\left(\frac xa\right)^2} + x \sin^{-1}\left(\frac xa\right)}{x}+C$$
And under appropriate restricted values for x and a, this is equivalent to
$$-\left(\frac{\sqrt{a^2 - x^2} + x \tan^{-1}\left(\frac x{\sqrt{a^2 - x^2}}\right)}{x}\right)+C\tag{form you seek}$$
Note that the result we obtain at $(1)$ is perfectly correct, as well.
| {
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Show that $f(x) = \sqrt{x}$ is continous at any $a > 0$ We first note that if $|x-a| < \delta$ and $\delta \leq 3a$, we have:
$|x-a|<3a \Leftrightarrow -3a < x-a < 3a \Leftrightarrow -2a < x < 4a \Leftrightarrow |x| < 4a \Leftrightarrow |\sqrt{x}| < 2\sqrt{a} $.
Let's also note that $|2\sqrt{a}-\sqrt{a}| \leq |\sqrt{x}-\sqrt{a}| $ and so, $ \frac{1}{|\sqrt{x}-\sqrt{a}|} \leq \frac{1}{|2\sqrt{a}-\sqrt{a}|} $.
Then, for all $ \varepsilon > 0 $ there exists $\delta = \min\{ 3a, \sqrt{a}\varepsilon \}$, if $|x-a|<\delta$:
\begin{align}
|f(x) - f(a)| &= |\sqrt{x} - \sqrt{a}| \\
&= \frac{|x-a|}{|\sqrt{x} - \sqrt{a}|} \\
&\leq \frac{|x-a|}{|2\sqrt{a}-\sqrt{a}|} \\
&=\frac{|x-a|}{\sqrt{a}} \\
&< \frac{\delta}{\sqrt{a}} \leq \frac{\sqrt{a}}{\sqrt{a}}\varepsilon = \varepsilon
\qquad \text{Q.E.D.}
\end{align}
Is this proof correct?
| If $x > 0$
and $h > 0$
then
$\begin{array}\\
|\sqrt{x+h}-\sqrt{x}|
&=|(\sqrt{x+h}-\sqrt{x})\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}|\\
&=|\dfrac{h}{\sqrt{x+h}+\sqrt{x}}|\\
&\lt|\dfrac{h}{2\sqrt{x}}|\\
\end{array}
$
Therefore,
if
$h < 2\epsilon \sqrt{x}$
then
$|\sqrt{x+h}-\sqrt{x}|
\lt \epsilon$.
| {
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Two approaches yield different results Here's the statement problem:
"From an ordinary deck of playing cards, cards are to be drawn successively at random and without replacement. The probability that the third spade appears on the sixth draw is computed as follows: Let A be the event of two spades in the first five cards drawn, and let B be the event of a spade on the sixth draw. Thus, the probability that we wish to compute is P(A ∩ B)."
The conditional probability approach: We have that $\frac{\left( \begin{array}{c} 13 \\ 2 \end{array} \right) \left( \begin{array}{c} 39 \\ 3 \end{array} \right)}{\left( \begin{array}{c} 52 \\ 5 \end{array} \right)}$ and $P(B|A) = \frac{11}{47}$, so the desired probability, $P(A \cap B) = P(A)P(B|A)= \frac{\left( \begin{array}{c} 13 \\ 2 \end{array} \right) \left( \begin{array}{c} 39 \\ 3 \end{array} \right) \times 11}{\left( \begin{array}{c} 52 \\ 5 \end{array} \right) \times47}$
The classical approach We have that $P(A \cap B) = \frac{N(A\cap B)}{N(S)}$, where $S$ is our sample space. We have that $N(A\cap B) = \left( \begin{array}{c} 13 \\ 2 \end{array} \right) \left( \begin{array}{c} 39 \\ 3 \end{array} \right){\left( \begin{array}{c} 11 \\ 1 \end{array} \right)}$, because we want to choose 2 spades out of 13 spades, then 3 non-spades of the remaining 39 cards, and then 6th position is one the spades in the 11 spades. On the other hand, $N(S) = \left( \begin{array}{c} 52 \\ 6 \end{array} \right)$, so we have $P(A \cap B) = \frac{\left( \begin{array}{c} 13 \\ 2 \end{array} \right) \left( \begin{array}{c} 39 \\ 3 \end{array} \right){\left( \begin{array}{c} 11 \\ 1 \end{array} \right)}}{\left( \begin{array}{c} 52 \\ 6 \end{array} \right)}$.
The two approaches should yield the same result, but here they' re not. Also, I noticed that while the numerators are the same, the denominators are different (though not too much). Where was I wrong?
| The first calculation is the correct one.
To convince you of this, you can take your attempts and try to apply the same formula to smaller examples. Briefly take a look at the related question of where we have a three-card deck: an ace of spades, a two of spades, and an ace of hearts. We ask the question of what is the probability that we draw our second spade as the third card.
By brute force listing out of the six possibilities, you should see that four of the six yield a spade as the last card.
Using the first approach, we would arrive at the same correct answer: $\frac{\binom{2}{1}\binom{1}{1}}{\binom{3}{2}}\cdot\frac{1}{1}=\frac{2}{3}$
Using the second approach, we would have arrived at a wrong answer: $\frac{\binom{2}{1}\binom{1}{1}\binom{1}{1}}{\binom{3}{3}}=2$, which happens to be off by a factor of three.
To fix your second calculation, it might be easiest to explain by instead of working in the scenario that order of cards in hand don't matter, that we work in the scenario where order of cards in hand do matter.
We approach the same way: $\binom{13}{2}$ ways to pick the early spades, $\binom{39}{3}$ ways to pick the other early cards, and then $\binom{11}{1}$ ways to pick the final spade. We then pick in what specific order the cards appeared in $5!$ ways, noting that the final spade must occur in the final position and nowhere else.
The bottom, similarly, will be $\binom{52}{6}6!$ as after picking what six cards apppear, we can arrange them in $6!$ ways.
This gives our corrected calculation at $$\frac{\binom{13}{2}\binom{39}{3}\binom{11}{1}5!}{\binom{52}{6}6!}=\frac{1}{6}\cdot\frac{\binom{13}{2}\binom{39}{3}\binom{11}{1}}{\binom{52}{6}}$$
which one can verify is equal to the answer given by your first attempt.
As discussed in comments elsewhere, that $6$ happens to be equal to $3!$ is merely a coincidence. We divide by $6$ since that is the number of cards in the hand, just as dividing by three and not by $2!$ corrected the calculation in the smaller three-card deck example above.
A different way of picturing the correction is that in the second attempt without the fix, we did not account for the final card actually appearing in the final position. The final card actually appears in the final position with probability $\frac{1}{6}$ (or more generally $\frac{1}{n}$ if referring to drawing $n$ cards), and so this should have been multiplied in as well.
Another example to try to convince that dividing by $n$ is the correct fix and not dividing by the factorial of the number of spades:
Consider the problem of asking what the probability that the first spade occurs in the $6$'th draw.
The first approach, again the correct one, would give an answer of $\frac{\binom{39}{5}}{\binom{52}{5}}\cdot\frac{13}{47}$
The second approach without the fix would give an answer of $\frac{\binom{39}{5}\binom{13}{1}}{\binom{52}{6}}$, an answer six times larger than the first, just as before, despite the number of spades having changed between the examples. Dividing by the factorial of the number of spades would not have corrected it.
| {
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"url": "https://math.stackexchange.com/questions/2635178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve an equation with cube roots How can I solve the following equation:
$$\frac{(34-x)\sqrt[3]{x+1}-(x+1)\sqrt[3]{34-x}}{\sqrt[3]{34-x}-\sqrt[3]{x+1}}=30$$
Attempts:
By substituting $\sqrt[3]{34-x}=a$ and $\sqrt[3]{x+1}=b$ we get:
$$\frac{a^3b-b^3a}{a-b}=\frac{ab(a^2-b^2)}{a-b} = ab(a-b)=30$$
Although it looks good, it doesn't lead to anything useful for me.
| Let $\sqrt[3]{34-x}=a$ and $\sqrt[3]{x+1}=b$.
Thus, $$a^3+b^3=35$$ and $$ab(a+b)=30,$$ which gives
$$(a+b)^3=125$$ or $$a+b=5.$$
Can you end it now?
I got $$\{26,7\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637568",
"timestamp": "2023-03-29T00:00:00",
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Is "$a^p + b^p = c^p + d^p \Rightarrow (a+b) \equiv (c+d) \mod p$" true?
Is $$a^p + b^p = c^p + d^p \implies (a+b) \equiv (c+d) \mod p$$ true?
Let $a,b,c,d$ be (distinct) positive integers, and $p$ be prime. My reasoning is roughly as follows:
$$(a^p + b^p) \equiv (a^p \mod p + b^p \mod p) \mod p$$
then by Fermat's Little Theorem
$$(a^p + b^p) \equiv (a \mod p + b \mod p) \mod p$$
$$\Rightarrow (a^p + b^p) \equiv (a + b) \mod p$$
and similarly
$$(c^p + d^p) \equiv (c+d) \mod p.$$
If $a^p+b^p=c^p+d^p$ then clearly $(a^p+b^p) \equiv (c^p+d^p) \mod p$, and hence (less clearly), $(a+b) \equiv (c+d) \mod p$.
EDIT: the motivation for this question is an efficient algorithm to search for Taxicab(5,2,2).
| Just found another nice proof of this fact (for n = 5, but can "easily" be generalised to all $p$): by the binomial theorem,
$$(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$$
which can be rearranged to show that
$$a^5+b^5 = (a+b)^5 - 5(a^4b + 2a^2b^3 + 2a^3b^2 + ab^4)$$
$$\Rightarrow a^5 + b^5 \equiv (a+b)^5 \mod 5$$
$$\therefore a^5 + b^5 \equiv c^5 + b^5 \mod 5 \Rightarrow (a+b)^5 \equiv (c+d)^5 \mod 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
If $\tan^{-1} \left(\frac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$, then prove $x^2=\sin(2\alpha)$ If $\tan^{-1} \left(\dfrac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$ then prove that: $x^2= \sin (2\alpha) $
My Attempt:
$$\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) =\alpha$$
$$\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}=\tan (\alpha )$$
$$\dfrac {1+x^2-2\sqrt {1+x^2}.\sqrt {1-x^2}+ 1 - x^2}{1+x^2-1+x^2}=\tan (\alpha)$$
$$\dfrac {1-\sqrt {1+x^2}.\sqrt {1-x^2}}{x^2}=\tan (\alpha)$$
| Since
\begin{align}
\tan\alpha = \frac{\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}} =\frac{\text{opposite}}{\text{adjacent}}
\end{align}
then it follows
\begin{align}
\sin \alpha =& \frac{\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt{(\sqrt {1+x^2} - \sqrt {1-x^2})^2+(\sqrt {1+x^2} + \sqrt {1-x^2})^2}}\\
=&\ \frac{\sqrt {1+x^2} - \sqrt {1-x^2}}{2} = \frac{\text{opposite}}{\text{hypotenuse}}
\end{align}
and
\begin{align}
\cos\alpha = \frac{\sqrt {1+x^2} + \sqrt {1-x^2}}{2} = \frac{\text{adjacent}}{\text{hypotenuse}}.
\end{align}
Then it follows
\begin{align}
2\sin\alpha \cos \alpha = \frac{(\sqrt {1+x^2} + \sqrt {1-x^2})(\sqrt {1+x^2} - \sqrt {1-x^2})}{2}=x^2.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2644104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the $x$ and $y$ part of the side of the triangle I need help finding x and y in this triangle.
Conditions:
It is not a right triangle;
there are no given angles;
$u$ doesn't bisect the corresponding angle;
$u$ doesn't split $c$ in two equal parts;
$c,b,a$ and $u$ are given;
Triangle example
| Check someone did suggest Stewart's Theorem as a comment!
$xb^2 + yc^2 = (x+y)(u^2+xy)=au^2+axy$
From here, $x$ can be found out this way:
\begin{align}
& x(b^2-ay)+yc^2=au^2 \\
& \implies x(b^2-ay-c^2)=a(u^2-c^2) \\
& \implies \boxed{x = \frac{a(u^2-c^2)}{b^2-ay-c^2}}
\end{align}
Similarly, $$\boxed{y=a-x =a- \frac{a(u^2-c^2)}{b^2-ay-c^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $\frac{b+c}{a+b+c+d}$ given $\frac{b}{a+b}, \frac{c}{c+d}, \frac{c}{c+a},\frac{b}{b+d}$ Given $\frac{b}{a+b}, \frac{c}{c+d}, \frac{c}{c+a},\frac{b}{b+d}$, where $a,b,c,d \geq 1$, can we solve for the value $\frac{b+c}{a+b+c+d}$? While it seems that we have 4 equations and 4 unknowns, due to the non-linearity I haven't been able to find a solution. I was thinking about inverting the equations, by saying $\frac{b}{a+b} = k \implies \frac{b+a}{b} = k \implies 1+\frac{a}{b} = k \implies \frac{a}{b}=k-1$. However, after doing this for all four equations, and trying to substitute, I haven't made any progress. I'd appreciate any help (or, if this isn't possible, I'd love to understand why)!
'Motivation': In my statistics class earlier today, we were talking about confusion matrices (https://en.wikipedia.org/wiki/Confusion_matrix), and I was curious to see if algebraically we can compute the 'total error' in a model from the error values. There really isn't any statistical importance of this question though (as far as I can tell)
| You don't really need the fourth equation.
Let $$\frac b{a+b}=A\implies b=\frac{Aa}{1-A}$$ $$\frac c{c+d}=B\implies d=\frac{1-B}Bc$$ $$\frac c{a+c}=C\implies c=\frac{Ca}{1-C}$$ so $$d=\frac{1-B}B\cdot\frac{Ca}{1-C}=\frac{C(1-B)}{B(1-C)}a$$ Hence $$\begin{align}\frac{b+c}{a+b+c+d}&=\frac{\frac{A}{1-A}+\frac{C}{1-C}}{1+\frac{A}{1-A}+\frac{C}{1-C}+\frac{C(1-B)}{B(1-C)}}\quad\text{cancel}\,a\\&=\frac{A(1-C)+C(1-A)}{(1-A)(1-C)+A(1-C)+C(1-A)+\frac CB(1-A)(1-B)}\\&=\frac{B(A+C-2AC)}{B(1-AC)+C(1-A)(1-B)}\end{align}$$
| {
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"url": "https://math.stackexchange.com/questions/2657526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the coefficient of $x^{18}$ in $(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 +\cdots)^5 $
Find the coefficient of $x^{18}$ in $$(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 +\cdots)^5 $$
This is the first time coming across a generating function question and am not quite sure how to solve this and am looking for some help, thanks!
| Your expression can be rewritten as $$x(1+x+x^2+x^3+x^4)\cdot (x^2(1+x+x^2+x^3+\cdots)) ^5$$
$$=x(1+x+x^2+x^3+x^4)\cdot x^{10}(1+x+x^2+x^3+\cdots) ^5$$
$$=x^{11}(1+x+x^2+x^3+x^4)\cdot (1+x+x^2+x^3+\cdots) ^5$$
The two expressions in the brackets are finite and an infinite GP respectively. Hence by formula for sum of GP we have
$$=x^{11}\cdot \left(\frac {1-x^5}{1-x}\right). \left(\frac {1}{1-x}\right) ^5$$
$$=x^{11}.\left (\frac {1-x^5}{(1-x)^6}\right) $$
$$=\left (\frac {x^{11}-x^{16}}{(1-x)^6}\right) $$
Now using the negative binomial series we need to find coefficient of $x^7$ and $x^2$ because we already have $x^{11} $ and $x^{16}$ in the numerator.
So using negative binomial series we get the answer as $$\binom {12}{5}- \binom {7}{2}=771$$
| {
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"url": "https://math.stackexchange.com/questions/2659565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate $π$ from first principles I would like to see a construction of the value of $\pi$ from first principles using e.g. a circle, without $\sin$ or $\cos$ and the usual relations for circumference or area.
I would consider e.g. a quarter of a circle, where you move in the $x$
direction from $0$ to $R$ (the radius of a circle), defining an infinitesimal section of the circumference, in order to get an integral expression for the circumference of $\frac14$ of a circle, corresponding to $\frac{\pi R}2$.
I tried to do so, but could not find a ‘simple’ solution for that. Any ideas?
Addendum:
*
*I do not want any ‘series’ or something to calculate $\pi$;
*I want somewhat the ‘proof’ that the circumference of a circle is indeed $2\pi r$ using geometric principles only (like Pythagoras and integration…)
| If you take on blind faith that the area of a circle with radius 1 is $\pi$, then you can calculate $\pi$ without trigonometry by taking the power series of $2\sqrt{1 - x^2}$ and integrating it from -1 to 1. The problem is any proof from first principles that the area of a circle with radius 1 is $\pi$ uses trigonometry. First I will calculate $\pi$ directly from its technical definition without using the assumption that the area of a circle with radius 1 is $\pi$. Next, I will prove from first principles that the area of a circle with radius 1 is $\pi$. I don't consider this image a proof from first principles that the area of a circle with radius 1 is $\pi$.
Here's a calculation of $\pi$ directly from the definition.
It's obvious that $\pi = 6 \times \sin^{-1}(\frac{1}{2})$. It turns out that the derivative of $\sin^{-1}$ is an elementary function. All we have to do is figure out the power series for the derivative of $\sin^{-1}$ centered at 0 and then integrate each term to get the power series for $\sin^{-1}$. In general, $\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$. So $\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))} = \frac{1}{\sqrt{1 - x^2}} = (1 - x^2)^{-\frac{1}{2}}$. To take the power series of this centered at 0, you first take the power series of $(1 + x)^{-\frac{1}{2}}$ centered at 0 and then substitute $-x^2$ for $x$. Now the first derivative of this is $-\frac{1}{2}(1 + x)^{-1\frac{1}{2}}$. Then the second derivative is $(-\frac{1}{2})(-1\frac{1}{2})(1 + x)^{-2\frac{1}{2}}$. Now to get its power series, you divide the derivatives by the factorials to get $(1 + x)^{-\frac{1}{2}} = 1 - \frac{1}{2}x + \frac{1}{2}(\frac{3}{4})x^2 - \frac{1}{2}(\frac{3}{4})(\frac{5}{6})x^3 ...$ Now substituting $-x^2$ for $x$, we get $(1 - x^2)^{-\frac{1}{2}} = 1 + \frac{1}{2}x^2 + \frac{1}{2}(\frac{3}{4})x^4 + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})x^6 ...$ Now finally, $\sin^{-1}(x) = \int_0^x(1 - t^2)^{-\frac{1}{2}}dt = x + \frac{1}{2}(\frac{1}{3})x^3 + \frac{1}{2}(\frac{3}{4})(\frac{1}{5})x^5 + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})(\frac{1}{7})x^7$. So $\pi = 6 \times \sin^{-1}(\frac{1}{2}) = 6(2^{-1} + \frac{1}{2}(\frac{1}{3})(2^{-3}) + \frac{1}{2}(\frac{3}{4})(\frac{1}{5})(2^{-5}) + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})(\frac{1}{7})(2^{-7}) ...)$.
Here's a proof that the area of a circle with radius 1 is $\pi$.
Using the substitution rule in reverse, we get $\int_{-1}^12\sqrt{1 - x^2} = 2\int_{-1}^1\sqrt{1 - x^2} = 2\int_{-1}^1\cos(\sin^{-1}(x)) = 2\int_{\sin(-\frac{\pi}{2})}^{\sin(\frac{\pi}{2})}\cos(\sin^{-1}(x)) = 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(\sin^{-1}(\sin(\sin'(x)))) = 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2(x) = 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos(2x) + 1}{2} = \pi$
Image source: Area of a circle $\pi r^2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Induction Proof for $\sum \limits_{i=1}^n \frac{2}{(i+1)(i+2)} = \frac{n}{(n+2)}$ I am asked: Prove, using induction, that for all $n \in \mathbb{N}$ the following equality holds.
\begin{align*}
\sum \limits_{i=1}^n \dfrac{2}{(i+1)(i+2)} = \dfrac{n}{(n+2)}
\end{align*}
Here's my attempt:
For all $n\in \mathbb{N}$, let $P(n)$: $\sum \limits_{i=1}^n \frac{2}{(i+1)(i+1)}=\frac{n}{n+2}$.
Base Case: When $n=1$, $P(1)$: $\sum \limits_{i=1}^1 \frac{2}{(i+1)(i+2)}=\frac{1}{1+2}\implies \frac{2}{(1+1)(1+2)}=\frac{1}{3}\implies \frac{2}{6}=\frac{1}{3}\implies \frac{1}{3}=\frac{1}{3}$. Therefore, the base case is true.
Inductive Assumption: Let $n\in \mathbb{N}$ be generic and assume $P(n)$ is true, i.e. $\sum \limits_{i=1}^n \frac{2}{(i+1)(i+2)}=\frac{n}{n+2}$.
Induction Step: Prove $P(n+1)$ is true. $\sum \limits_{i=1}^{n+1} \frac{2}{(i+1)(i+2)}=\frac{2}{((n+1)+1)((n+2)+2)}+\sum \limits_{i=1}^n \frac{2}{(i+1)(i+2)}=\frac{2}{(n+2)(n+3)}+\frac{n}{n+2}=\frac{2}{(n+2)(n+3)}+\frac{n(n+3)}{(n+2)(n+3}=\frac{2+n(n+3)}{(n+2)(n+3)}=\frac{n^2+3n+2}{(n+2)(n+3)}=\frac{(n+1)(n+2)}{(n+2)(n+3)}=\frac{n+1}{n+3}=\frac{n+1}{(n+1)+2}$.
Therefore, $P(n+1)$ is true. By induction, $P(n)$ is true. $\blacksquare$
Did I miss anything? Thanks!
| In the inductive step $$\sum \limits_{i=1}^{n+1} \frac{2}{(i+1)(i+2)}=\frac{2}{((n+1)+1)((n+2)+2)}+\sum \limits_{i=1}^n \frac{2}{(i+1)(i+2)}=$$
You meant $$\sum \limits_{i=1}^{n+1} \frac{2}{(i+1)(i+2)}=\frac{2}{((n+1)+1)((n+1)+2)}+\sum \limits_{i=1}^n \frac{2}{(i+1)(i+2)}=$$
The rest of your proof is flawless.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by mathematical induction that Prove by mathematical induction that
$$P(n)=3^{2n+1} + 2^{n-1}$$
is a multiple of 7.
My Attempt:
Here, $P(n)= 3^{2n+1}+2^{n-1}$
For $n=1$,
$$P(1)=3^3+2^0$$
$$=28$$
So, $P(1)$ is true.
Suppose, $P(m)$ is true for all $m\in N$
Now, we have to show that $P(m+1)$ is true,
$$P(m+1)=3^{2(m+1)+1}+2^{(m+1)-1}$$
$$=3^{2m+3}+2^m$$
| Inductive Step
$$
\begin{align}
P(m+1)
&=3^{2m+3}+2^m\\
&=9\cdot3^{2m+1}+2\cdot2^{m-1}\\
&=7\cdot3^{2m+1}+2\left(3^{2m+1}+2^{m-1}\right)\\
&=7\cdot3^{2m+1}+2P(m)
\end{align}
$$
Direct Proof
Note that since $2\cdot4\equiv1\pmod7$, we get
$$
\begin{align}
3^{2k+1}+2^{k-1}
&=3\cdot9^k+2^{-1}\cdot2^k\\
&\equiv3\cdot2^k+4\cdot2^k&\pmod7\\
&=7\cdot2^k\\
&\equiv0&\pmod7
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute $\lim_{x\to\frac{\pi}{2}} \frac{(1-\sin x)(1-\sin^2x)\dots(1-\sin^nx)}{\cos^{2n}x}$
$$\lim_{x\to\frac{\pi}{2}} \frac{(1-\sin x)(1-\sin^2x)\dots(1-\sin^nx)}{\cos^{2n}x}=?$$
Here's what I have done so far:
$y=\frac{\pi}{2}-x$ The limit becomes:$$\lim_{y\to 0}\frac{(1-\cos y)(1-\cos^2y)\dots(1-\cos^ny)}{\sin^{2n}y}=\lim_{y\to 0}\frac{1-\cos y}{\sin^2y}\frac{1-\cos^2y}{\sin^2y}\dots\frac{1-\cos^ny}{\sin^2y}$$
Also $1-\cos y=2\sin^2 \frac{x}{2}$
How should i write $1-\cos^2y ,1-\cos^3y,\dots1-\cos^ny$ in order to get the final answer.
| Note that
$$1-cos^ky=1-(1-\frac{y^2}{2}+o(y^2))^k=\frac{k}{2}y^2+o(y^2)$$
thus
$$\frac{(1-\cos y)(1-\cos^2y)\dots(1-cos^ny)}{\sin^{2n}y}=\frac{\frac{n!}{2^n}y^{2n}+o(y^{2n})}{y^{2n}+o(y^{2n})}=\frac{\frac{n!}{2^n}+o(1)}{1+o(1)}\to \frac{n!}{2^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of real $x$ satisfying the trigo equation $1+\sin 2x=\sin x+\sin^2 x$ Number of real $x$ satisfying the trigo equation $1+\sin 2x=\sin x+\sin^2 x$ in $x\in(0,2\pi)$
solution i try $1+\sin 2x =\sin x+\sin^2 x$
$5+4\sin 2x=4\sin^2 x+4\sin x+1$
$5+\sin 2x=(2\sin x+1)^2$ How i solve it after that point
| Do you remember the double angle formulas? I don't see you using that.
Now, consider writing it all again using this fact. I've attached my work that get you to a point where you need to reason out the angles, and I've hidden it so you can at least work out some more using the double angle formula! :)
$$ 1 + \sin(2x) = \sin(x) + \sin^2(x)\\1 + 2\sin(x)\cos(x) = \sin^2(x) + \sin(x)\\ 2\sin(x)\cos(x) = \sin^2(x) - 1 + \sin(x)\\ 2\sin(x)\cos(x) = -\cos^2(x) + \sin(x)\\ 2\sin(x)\cos(x) + \cos^2(x) = \sin(x)\\ 2\sin(x) + \cos(x) = \tan(x)\\ 2 + \cot(x) = \tan(x)\csc(x)\\ 2 + \cot(x) = \sec(x)\\ 2 = \sec(x) - \cot(x)\\ 2 = \frac{1}{\cos(x)} - \frac{\cos(x)}{\sin(x)} $$
And from there you should be able to reason out the angles which satisfy that equality.
Edit: Maybe you wrote the problem down wrong for us?
| {
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Finding the largest area of a right-angled triangle using Lagrange multipliers The area of a triangle of length a, b and c is given by $$A =\sqrt{s(s-a)(s-b)(s-c)}$$ where its perimeter is $2s$ such that $2s = a + b + c$
Consider a right-angled triangle with hypotenuse $a$ such that $a^2 = b^2 + c^2$. Using Lagrange multipliers find the largest area of a right angled triangle of perimeter $2s$ and express your answer in terms of the perimeter (i.e. in terms of $s$)
From this I understand that we have two constraints of $2s = a + b + c$ and $a^2 = b^2 + c^2$ on the equation $\sqrt{s(s-a)(s-b)(s-c)}$.
Therefore I constructed using Lagrange multipliers that $$L_a = \frac{-s(s-b)(s-c)}{2p} + \lambda - 2\mu a = 0$$ $$L_b = \frac{-s(s-a)(s-c)}{2p} + \lambda + 2\mu b = 0$$ $$L_c = \frac{-s(s-a)(s-b)}{2p} + \lambda + 2\mu c = 0$$ $$L_\lambda = 2s - a - b - c = 0 $$ $$L_\mu = a^2 - b^2 - c^2 = 0$$
where $p=\sqrt{s(s-a)(s-b)(s-c)}$ , $\lambda$ and $\mu$ being arbitrary constants.
I have attempted to solve this by multiplying $L_a$ by $(s-a)$, $L_b$ by $(s-b)$ and $L_c$ by $(s-c)$, adding them together and using the two constraints ($L_\lambda$ and $L_\mu$) to solve for $p$ (which represents the largest area). But it didn't seem to work.
What I am stuck on here is I'm not sure how I can solve for $p$ in terms of $s$ using the 5 equations above.
Thank you very much for your help.
| We have:
$$0 = L_a - L_b = -\frac{s(s-c)(b-a)}{2p} - 2\mu(a + b)$$
$$0 = L_b - L_c = -\frac{s(s-a)(c-b)}{2p} + 2\mu(b- c)$$
Multiply the first equation by $(b-c)$, the second one by $(a + b)$ and add them:
\begin{align}
0 &= -\frac{s}{2p}\Big((s-c)(b-a)(b-c) + (s-a)(c-b)(a+b)\Big)\\
&= -\frac{s(b-c)}{2p}\Big((s-c)(b-a)- (s-a)(a+b)\Big)\\
&= -\frac{s(b-c)}{4p}\Big((a+b-c)(b-a)- (b+c-a)(a+b)\Big)\\
&=-\frac{s(b-c)}{4p}\Big(-2bc\Big)\\
&=\frac{s(b-c)bc}{2p}\\
\end{align}
The only term which can be zero is $b - c$ so we conclude $b = c$.
Now we have $$a^2 = b^2 + c^2 = 2b^2 \implies a = \sqrt{2}b$$
Then $$2s = a + b + c = (2 + \sqrt{2})b \implies b = \frac{2s}{2 + \sqrt2}$$
Finally we get
$$p = \frac{bc}{2} = \frac{b^2}{2} = \frac{2s^2}{6 + 4\sqrt{2}}$$
| {
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"url": "https://math.stackexchange.com/questions/2669785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the remainder when $3^{29}$ is divided by $12$.
Find the remainder when $3^{29}$ is divided by $12$.
a) $2$ ; b) $3$ ; c) $7$ ; d) $9$ ; e) $12$
Since $\dfrac{3^{29}}{12} = \dfrac{3^{28}}{4} = \dfrac{9^{14}}{4}$, and $9 \equiv 1 \mod 4$, I thought I could do $9^{14} \equiv 1^{14} \mod 4$, so that the answer is $1$. However, that is not one of the answer choices... Where did I go wrong?
| Notice that we have
*
*$3^1 \equiv 3 \mod 12$
*$3^2 \equiv 9 \mod 12$
*$3^3 \equiv 3 \mod 12$
*$3^4 \equiv 9 \mod 12$
*...
So it goes like in this pattern. In general we have
*
*$3^{2k} \equiv 9 \mod 12,\ \ k > 0$
*$3^{2k+1} \equiv 3 \mod 12,\ \ k > 0$
And since $29$ is an odd number, the answer should be $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Jordan basis unique? Is Jordan basis unique? I have a 4x4 matrix, find eigenvectors and one generalized eigenvector, also trying different linearly independent eigenvectors but the matrix P so that PJP^-1 =A only works for certain vector.
\begin{pmatrix}-\frac{1}{2}&0&-\frac{1}{2}&0\\ \frac{3}{2}&0&\frac{1}{2}&0\\ -\frac{3}{2}&0&-\frac{3}{2}&-1\\ 1&1&1&1\end{pmatrix}\begin{pmatrix}1&-1&0&-1\\ 0&2&0&1\\ -2&1&-1&1\\ 2&-1&2&0\end{pmatrix}\begin{pmatrix}1&1&0&0\\ -1&0&1&1\\ -3&-1&0&0\\ 3&0&-1&0\end{pmatrix}=\begin{pmatrix}-1&0&0&0\\ 0&1&0&-1\\ 0&0&1&1\\ 0&0&0&1\end{pmatrix}
| Any basis is a Jordan basis for the identity.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\cos2\theta=0$, then $\Delta^2=$? I'll state the question from my textbook here:
If $\cos2\theta=0$, then $\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2=$?
This is how I solved the problem:
$\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2$
$= (\cos^3 \theta + \sin^3 \theta)^2$
$= (\cos \theta + \sin \theta)^2(\cos^2 \theta - \cos \theta \sin \theta + \sin^2 \theta)^2$
$= (1+ \sin2\theta)(1-\sin2\theta + \sin^2 \theta \cos^2 \theta)$
$= (1+ \sin2\theta)(1-\sin2\theta) + (1 + \sin2\theta)\sin^2\theta \cos^2 \theta$
$= \cos^2 2\theta + \frac 14 (1 + \sin2\theta)\sin^2 2\theta$
$= \frac 14 (1 + \sin2\theta)\sin^2 2\theta$
Now since $\cos2\theta=0$, $\sin2\theta = \pm 1$.
Therefore the above expression can take the values 0 and $\frac12$.
My textbook gives the answer as $\frac12$. I don't see any grounds on rejecting the other answer of 0. Have I made a mistake somewhere? Or am I forgetting something?
| You are right indeed by direct calculation we obtain
$$\cos 2\theta=0\iff \theta=\frac{\pi}4+k\frac{\pi}2$$
and since
$$\Delta^2=\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2= (-\cos^3 \theta - \sin^3 \theta)^2$$
*
*for $\theta=\frac{\pi}4 \implies \Delta^2=\left(-\frac{2\sqrt 2}{4}\right)^2=\frac12$
*for $\theta=\frac{3\pi}4\implies \Delta^2=0$
*for $\theta=\frac{5\pi}4\implies \Delta^2=\left(\frac{2\sqrt 2}{4}\right)^2=\frac12$
*for $\theta=\frac{7\pi}4\implies \Delta^2=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to simplify $\sec\tan^{-1}\pi$? How to simplify $\sec\tan^{-1}\pi$?
I've never dealt with inverse trigonometry before. Thanks in advance.
| Here is a right triangle.
In any right triangle, by definition, $\tan x = \frac ba$ and $\sec x = \frac ca$.
Also, of course, $a^2+b^2 = c^2$.
Since $\tan x =\frac ba $, we have $x = \tan^{-1}\frac ba$ and then: $$\begin{align}\sec x & = \frac ca \\ \sec\left(\tan^{-1} \frac ba\right) & = \frac ca \\ & = \frac {\sqrt{a^2+b^2}}a.\end{align}$$
Now take $b=\pi$ and $a=1$ and we have $$
\sec\tan^{-1} \frac \pi1 = \frac {\sqrt{1^2+\pi^2}}1.$$
Also, observe: since this works for any right triangle, we can also take $a=3, b=4$ and obtain $$\sec\tan^{-1} \frac 43 = \frac 53$$
without actually knowing what $\tan^{-1} \frac43$ is, and without even being able to calculate it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using polar co-ordinates to solve a dynamical system I have the system
$x^\prime = x-y-x(x^{2}+y^{2})$
$y^\prime = x+y-y(4x^{2}+y^{2})$.
I want to show with polar coordinates that I eventually end up with
$r^\prime = r-r^{3}(1+ \frac{3}{4}$sin$^{2}(2 \theta))$.
I have used the following substitutions:
$x = r$cos$(\theta)$, $y=r$sin$(\theta)$, $x^{2} + y^{2} = r^{2}$ and $rr^\prime = xx^\prime + yy^\prime $.
I have made a few attempts but my recent one is as follows:
\begin{align}
rr^\prime &= xx^\prime + yy^\prime
\\
&=x(x-y-x^{3} -xy^{2}) + y(x+y-4x^{2}y^{2}+y^{3})
\\
&=x^{2} -xy + x^{4} + x^{2}y^{2} + yx + y^{2} - 4x^{2}y^{2} + y^{4}
\\
&=r^{2} + (x^{4} + y^{4}) - 3x^{2}y^{2} .
\end{align}
I am not sure if I am heading in the right direction. I also have noted that $\sin(2\theta) = 2\sin(\theta)\cos(\theta) $.
Any help would be much appreciated.
| We are given
$$\tag 1 x' = x-y-x(x^2+y^2) \\y' = x+y-y(4x^2+y^2)$$
In polar coordinates
$$\tag 2 x^2+y^2 = r^2\\ x = r \cos \theta\\ y = r \sin \theta$$
Differentiating
$$\tag 3 2 x x' + 2 y y' = 2 r r' \implies r r' = x x' + y y' $$
Substituting $(1)$ and $(2)$ into $(3)$
$\begin{align}rr' &= x x' + y y'\\ &= r \cos \theta(r \cos \theta-r \sin \theta-r \cos \theta((r \cos \theta)^2+(r \sin \theta)^2)) +r \sin\theta ( r \cos \theta+r \sin \theta-r \sin \theta (4(r \cos \theta)^2+(r \sin \theta)^2))\\ &= \frac{3}{8} r^4 \cos (4 \theta )-\frac{11 r^4}{8}+r^2\end{align}$
Dividing by $r$
$$r' = \dfrac{r^3}{8}\left(3\cos (4 \theta )-11\right)+r$$
To find the angle
$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$
Using the quotient rule
$$\theta' = \dfrac{x y' - y x'}{r^2} = ...$$
Can you continue?
You should end up with
$$\theta' = r \left(\sin ^2(\theta )+\cos ^2(\theta )-3 r^2 \sin (\theta ) \cos ^3(\theta )\right) = \dfrac{r^3}{8} \left(-6 \sin (2 \theta )-3 \sin (4 \theta )\right) + r$$
Now you have $r'$ and $\theta'$, please continue.
| {
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"url": "https://math.stackexchange.com/questions/2674826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Contest Math Linear Algebra problem If $A\in M_2(\mathbb{R})$, prove that:
$$ \det(A^2 + A + I_2) \geq \frac{3}{4}(1-\det A)^2.$$
If $x,y$ are the eigenvalues of $A$ then we get the following inequality $$(x^2+x+1)(y^2+y+1)\geq \frac{3}{4}(1-xy)^2.$$ I was stuck after this step and it turns out that $$ (x ^2 + x +1)(y^2 +y +1) - \frac{3}{4} \left( xy-1 \right)^2 = \frac{1}{4}(2x +2 y +xy +1 )^2 \geq 0.$$ My question is how does one go about finding such a complete square? Are there some tricks that can help one to find such expressions? Perhaps someone knows a condition under which the equation of the a general conic
$$ax^2+hxy+by^2+cx+dy+f=0$$ is a perfect square?
| I would try like this. The following equation must be true for all $x,y$:
$$ (x ^2 + x +1)(y^2 +y +1) - \frac{3}{4} \left( xy-1 \right)^2 = (ax^2+hxy+by^2+cx+dy+f)^2$$
So what would we get if $y=0$:
$$ x ^2 + x +1 - \frac{3}{4} = (ax^2+cx+f)^2$$
so $$ (x+{1\over 2})^2 = (ax^2+cx+f)^2$$
Since this holds for all $x$ we have $a=0$, $c= 1$ and $f = {1\over 2}$.
With the same procedure for $x=0$ we get $b=0$ and $d=1$. We are left to calculate $h$.
For $x=y=-1$ we get then $$(h+{5\over 2})^2=9$$ and for $x=y=1$ we get then $$(h-{3\over 2})^2=1$$ so $h={1\over 2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Limit of a function tending to a finite number If
$$\lim_{x\to 0} \frac{ae^x - b\cos x +ce^{-x}}{x\sin x} = 2$$
then find the value of $a+b+c$.
My book has given the following solution to the above problem :-
We observe that as $x$ tends to zero , numerator tends to $a-b+c$ whereas the denominator tends to zero. Therefore for the limit to exist , we must have ,$a-b+c=0$
Now I am really confused at this point. Why would we want the numerator to attain the value of $0$ . Wouldn’t that give us an indeterminate answer? But actually it’s suposed to be two . Can you please explain ? Thank you for your help.
| If $a - b + c \ne 0$ then
$$\frac{ae^x-b\cos x+ce^{-x}}{x\sin x} \xrightarrow{x\to 0} \frac{a-b+c}{0} = \operatorname{sgn}(a-b+c) \cdot \infty \ne 2$$
Therefore $a - b + c = 0$. The limit is now an indeterminate form $\frac00$ so we can apply L'Hopital once. Differentiating gives
$$\frac{ae^x+b\sin x-ce^{-x}}{x\cos x+\sin x}$$
Again, if $a+b-c \ne 0$ then we would have:
$$\frac{ae^x+b\sin x-ce^{-x}}{x\cos x+\sin x} \xrightarrow{x\to 0} \frac{a+b-c}{0} = \operatorname{sgn}(a+b-c) \cdot \infty \ne 2$$
Therefore $a+b-c = 0$. The limit is now an indeterminate form $\frac00$ so we can again apply L'Hopital. Differentiating gives
$$\frac{ae^x+b\cos x+ce^{-x}}{-x\sin x+2\cos x}$$
Now the denominator is $\ne 0$ when $x\to 0$ so:
$$\frac{ae^x+b\cos x+ce^{-x}}{-x\sin x+2\cos x}\xrightarrow{x\to 0} \frac{a+b+c}{2}$$
This limit has to be $2$ so $$a+b+c = 4$$
$$a-b+c = 0$$
$$a+b-c = 0$$
The only solution is $a = 0, b = 2, c = 2$.
| {
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"source": "stackexchange",
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Let $a,b,c$ be positive reals. Find the minimum value of $P=\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c}{4a}$. This inequality problem is from Phan Boi Chau High School for the Gifted Grade 10 Selection Test, Nghe An, Vietnam.
Let $a,b,c$ be positive reals. Find the minimum value of $P=\dfrac{a^2}{(a+b)^2}+\dfrac{b^2}{(b+c)^2}+\dfrac{c}{4a}$.
I've done $\dfrac{c}{4a}=\dfrac{c^2}{4ac}\geq\dfrac{c^2}{(c+a)^2}$.
I think the minimum value is $\dfrac34$, so the problem is to prove $\displaystyle \sum_\text{cyc}\dfrac{a^2}{(a+b)^2}\geq\dfrac34$. But I still can't.
Can you help?
| Yes, you are right!
The minimal value is $\frac{3}{4}$ and occurs for $a=b=c$.
We can rewrite this inequality in the following form:
$$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{4z}\geq\frac{3}{4},$$
where $\frac{b}{a}=x$, $\frac{c}{b}=y$ and $\frac{a}{c}=z$.
Thus, $xyz=1$ and since $$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}\geq\frac{1}{1+xy}$$ it's $$xy(x-y)^2+(xy-1)^2\geq0,$$ it's enough to prove that
$$\frac{1}{1+\frac{1}{z}}+\frac{1}{4z}\geq\frac{3}{4},$$ which is
$$(z-1)^2\geq0.$$
Done!
It's interesting that your trying also helps.
Indeed, it's enough to prove that
$$\sum_{cyc}\frac{a^2}{(a+b)^2}\geq\frac{3}{4}.$$
Now, by C-S we obtain:
$$\sum_{cyc}\frac{a^2}{(a+b)^2}=\sum_{cyc}\frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a+b)^2(a+c)^2}.$$
Thus, it remains to prove that
$$4\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a+b)^2(a+c)^2$$ or
$$\sum_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2-8a^2bc)\geq0,$$
which is true by Muirhead because $(4,0,0)\succ(2,1,1)$, $(3,1,0)\succ(2,1,1)$ and $(2,2,0)\succ(2,1,1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$ I am in search of a correct and shortcut techniques to prove this.
Otherwise I have calculated each remainder according to the power of $20$ to prove this :-
$1 \equiv 1 \bmod 23$
$\Rightarrow 20 \equiv -3\bmod 23$
$\Rightarrow 20^2 \equiv (-3)^2 \equiv 9\bmod 23$
$20^{2n}\equiv (-3)^{2n}\bmod 23$
$20^{2n-1}\equiv (-3)^{2n-1}\bmod 23$.
Therefore
$1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$
For $\sum_{n=1}^{11} 20^{2n-1} \equiv r \bmod 23$
we have $$-3-4+10-2+5-1-9+11+7-6-8 \equiv 0 \bmod 23$$
For $\sum_{n=0}^{10} 20^{2n} \equiv s \bmod 23$
we have $$1+9+12+16+6+8+3+4+13+2-5 \equiv 0\bmod 23$$
Therefore $$\sum_{n=1}^{11} 20^{2n-1}+\sum_{n=0}^{10} 20^{2n} \equiv r+s \bmod 23$$
$$\Rightarrow 1+20+20^2+20^3+....+20^{21} \equiv 0+0 \equiv 0\bmod 23$$
If possible just show any short-cut correct way to prove this. Any help is appreciated.
| HINT
Note that by geometric series
$$1+20+20^2+20^3+....+20^{21} \equiv \sum_{k=0}^{21}(-3)^k=\frac{1- (-3)^{22} }{4} \pmod{23}$$
then use FLT for $(-3)^{22} \pmod{23}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is it wrong to factor in this way? I found an exercise for factoring:
$\frac{4x^2-8x+4}{x^2-1} * \frac{x+1}{12}$
I resolve like this:
= $\frac{(2x -2)^2}{(x-1)(x+1)} * \frac{(x+1)}{12}$
= $\frac{(2x -2)^2}{12(x-1)}$
Well up here, because here comes my problem:
$1-.$ If I factor like this:
= $\frac{2(x -1)^2}{12(x-1)}$
= $\frac{(x-1)}{6}$ <--- First answer
$2-.$ If I factor with this other way:
= $\frac{(2x-2)(2x-2)}{12(x-1)}$
= $\frac{4(x-1)(x-1)}{12(x-1)}$
= $\frac{(x-1)}{3}$ <-- Second answer
then, what is the wrong answer ? and why ?
| The first way is incorrect. You can't just factor a $2$ out of the squared expression just because you see the terms have one factor as $2$.
$$(2x-2)^2=\left[(2)(x-1)\right]^2=(2)^2(x-1)^2=4(x-1)^2\ne2(x-1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Magic sums with binomial coefficients (PART II) In these two questions (Question 1, Question 2) we proved that for a fixed $m\geq 1$ we have
$$a_n:=\sum_{k=1}^n\frac {(2n-k-1)!\cdot k}{n!(n-k)!} {{m+k}\choose{m}} =\frac{(m+1)(m+2n)!}{n!(m+n+1)!}$$
and
$$\sum_{n=0}^\infty a_n(1/4)^n=\sum_{n=0}^\infty\frac {(m+1)(m+2n)!}{n!(m+n+1)!}(1/4)^n=2^{m+1}.$$
I was wondering if there is a direct (and easier) proof of the fact that
$$\sum_{n=0}^\infty\sum_{k=1}^n\frac {(2n-k-1)!\cdot k}{n!(n-k)!} {{m+k}\choose{m}}(1/4)^n=2^{m+1}.$$
I know that defining $a_{n,k}=\frac {(2n-k-1)!\cdot k}{n!(n-k)!}$ then $\sum_{n,k\geq 0}a_{n,k}z^nx^k=\frac{2}{x\sqrt{1-4z}-x+2}.$ Can this help to show that $\sum_{n=0}^\infty\sum_{k=1}^na_{n,k} {{m+k}\choose{m}}(1/4)^n=2^{m+1}$?
| Two remarks seem to be required. Firstly, your expression for $a_n$ is ill-defined for $n=0$. Indeed by all rules of summation the left side of the equation is $0$ whereas the right side is $1$. With this in mind the equation which should be proved is:
$$\sum_{n=1}^\infty\sum_{k=1}^n\frac {(2n-k-1)!\cdot k}{n!(n-k)!} {{m+k}\choose{m}}\left(\frac{1}{4}\right)^n=2^{m+1}-1.$$
Secondly, the idea to prove the identity without intermediate steps is probably not the best one because instead of several interesting results only one is obtained. And indeed it is possible to find another interesting identity hidden in the expression:
$$
\sum_{n=k}^\infty\frac{1}{n}\binom{2n-k-1}{n-1}\left(\frac{x}{4}\right)^n
=\frac{1}{k}\left(\frac{1-\sqrt{1-x}}{2}\right)^k,\tag{1}
$$
where $k>0$ and $|x|\le1$ are assumed.
Differentiating the expression over $x$ and shifting indices one obtains:
$$
\sum_{n=k}^\infty\binom{2n-k}{n}\left(\frac{x}{4}\right)^n
=\frac{1}{\sqrt{1-x}}\left(\frac{1-\sqrt{1-x}}{2}\right)^k,\tag{2}
$$
which can be recognized as another form of the identity proved elsewhere.
To see it:
$$
\sum_{n=k}^\infty\binom{2n-k}{n}\left(\frac{x}{4}\right)^n
=\sum_{n=k}^\infty\binom{2n-k}{n-k}\left(\frac{\sqrt{x}}{2}\right)^{2n}
=\sum_{n'=0}^\infty\binom{2n'+k}{n'}\left(\frac{\sqrt{x}}{2}\right)^{2n'+2k}\\
=\left(\frac{\sqrt{x}}{2}\right)^{k}
\sum_{n'=0}^\infty\binom{2n'+k}{n'}\left(\frac{\sqrt{x}}{2}\right)^{2n'+k}
\stackrel{*}{=}\left(\frac{\sqrt{x}}{2}\right)^{k}
\frac{1}{\sqrt{1-x}}\left(\frac{1-\sqrt{1-x}}{\sqrt{x}}\right)^k
=\frac{1}{\sqrt{1-x}}\left(\frac{1-\sqrt{1-x}}{2}\right)^k,
$$
where in $\stackrel{*}{=}$ the aforementioned identity was used. Thus (2) and consequently (1) are proved.
Substituting $x=1$ in (1) one obtains:
$$
\sum_{n=k}^\infty\frac{1}{n}\binom{2n-k-1}{n-1}\left(\frac{1}{4}\right)^n
=\frac{1}{k}\left(\frac{1}{2}\right)^k,\tag{3}
$$
The rest is straightforward:
$$\sum_{n=1}^\infty\sum_{k=1}^n\frac {(2n-k-1)!\cdot k}{n!(n-k)!} {{m+k}\choose{m}}\left(\frac{1}{4}\right)^n
=\sum_{k=1}^\infty k{{m+k}\choose{m}}\sum_{n=k}^\infty\frac {(2n-k-1)!}{n!(n-k)!}\left(\frac{1}{4}\right)^n\\
\stackrel{(3)}{=}\sum_{k=1}^\infty {{m+k}\choose{m}}\left(\frac{1}{2}\right)^k
=2^{m+1}-1,$$
as required.
The last equality can be proved in various ways, as for example by induction. Indeed for $m=0$ the equality is obvious. Assume it is valid for $m-1$:
$$
S_m:=\sum_{k=1}^\infty {{m+k}\choose{m}}\left(\frac{1}{2}\right)^k=
\sum_{k=1}^\infty \left[{{m+k-1}\choose{m}}+{{m+k-1}\choose{m-1}}\right]
\left(\frac{1}{2}\right)^k
=\frac{1}{2}\left(1+S_m\right)+S_{m-1}\\
\Rightarrow S_m=2S_{m-1}+1
\stackrel{I.H.}{=}2(2^m-1)+1=2^{m+1}-1.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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polynomial long division Integration $$
\int \frac{3x^2 +6x +2}{x^2 + 3x +2}dx
$$
I've been trying to solve this integral, my gut tells me to use polynomial long division so I factored and got
$$
\int \frac{3x^2 +6x +2}{(2+x)(1+x)}dx
$$
I have never done a division with two factors in the denominator, would anybody be able to help out?
| $\int \frac{3x^2 +6x +2}{x^2 + 3x +2}dx
$
Since
$\dfrac1{x^2 + 3x +2}
=\dfrac1{(x+2)(x+1)}
=\dfrac1{x+1}-\dfrac1{x+2}
$
and
$\begin{array}\\
\dfrac{3x^2 +6x +2}{x^2 + 3x +2}
&=\dfrac{3(x^2+3x+2)-3(3x+2) +6x +2}{x^2 + 3x +2}\\
&=3-\dfrac{3x+4}{x^2 + 3x +2}\\
&=3-(3x+4)(\dfrac1{x+1}-\dfrac1{x+2})\\
&=3-\dfrac{3x+4}{x+1}-\dfrac{3x+4}{x+2}\\
&=3-\dfrac{3(x+1)+1}{x+1}+\dfrac{3(x+2)-2}{x+2}\\
&=3-3-\dfrac{1}{x+1}+3-\dfrac{2}{x+2}\\
&=3-\dfrac{1}{x+1}+\dfrac{2}{x+2}\\
\end{array}
$
and this you can
integrate by yourself.
| {
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Suppose $A \in M_{3,3}$ with eigenvalues $1,2,3$, eigenvectors $b_1, b_2, b_3$. Let $v = b_1 - 4b_2 + 3b_3$. Compute $A^5 v$. I think I did this right, but someone skim and double check?
Q: Suppose $A$ is a $3 \times 3$ matrix, with eigenvalues $1,2,3$ and corresponding eigenvectors $b_1, b_2, b_3$. Suppose that $v = b_1 - 4b_2 + 3b_3$. Compute $A^5 v$.
Let $\mathcal{B}$ be the eigenvector basis and $D$ be the diagonal eigenvalue matrix:
\begin{align*}
D &= \begin{pmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \\
\end{pmatrix} \\
P_{\mathcal{BE}} &= \begin{pmatrix} \begin{pmatrix} b_1 \end{pmatrix} & \begin{pmatrix} b_2 \end{pmatrix} & \begin{pmatrix} b_3 \end{pmatrix} \end{pmatrix} \\
[v]_{\mathcal{B}} &= (1,-4,3)^T \\
A &= P_{\mathcal{BE}}^{-1} D P_{\mathcal{BE}} \\
A^5 &= P_{\mathcal{BE}}^{-1} D^5 P_{\mathcal{BE}} \\
A^5 v &= P_{\mathcal{BE}}^{-1} D^5 P_{\mathcal{BE}} [v]_{\mathcal{E}} \\
[A^5 v]_{\mathcal{B}} &= D^5 [v]_{\mathcal{B}} \\
[A^5 v]_{\mathcal{B}} &= (1^5 \cdot 1, 2^5 \cdot -4, 3^5 \cdot 3)^T \\
[A^5 v]_{\mathcal{B}} &= (1, -128, 729)^T \\
A^5 v &= b_1 - 128 b_2 + 729 b_3 \\
\end{align*}
| $$ A^5v =A^5 b_1 - 4A^5b_2 + 3A^5b_3=$$
$$(1^5)b_1-4(2^5)b_2+3(3^5)b_3=$$
$$b_1-128b_2+729b_3$$
| {
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"answer_count": 3,
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} |
Finding the determinant of a matrix given determinants of other matrices Consider the following matrices:
$P$ = $\begin{pmatrix}a&2d&1\\ b&2e&-2\\ c&2f&-1\end{pmatrix}$
$U$ = $\begin{pmatrix}a&b&c\\ 2&3&2\\ d&e&f\end{pmatrix}$
$V$ = $\begin{pmatrix}a&b&c\\ d&e&f\\ 1&5&3\end{pmatrix}$
If you are given that $det(P)$ = $10$ and $det(U)$ = $-3$, then find the value of $det(V)$.
I personally am finding this problem very confusing. I know the general rules of how row operations and row changes affect the determinant of a matrix but I am not sure which ones are being applied here.
For the first matrix, we are given arbitrary values for the first two columns and for the last two we are given arbitrary values for the rows instead. So I am quite confused on how to work with this.
Any help?
| $$\mathbf P=\begin{pmatrix}a&2d&1\\b&2e&-2\\c&2f&-1\end{pmatrix},\,\mathbf U=\begin{pmatrix}a&b&c\\2&3&2\\d&e&f\end{pmatrix},\,\mathbf V=\begin{pmatrix}a&b&c\\d&e&f\\1&5&3\end{pmatrix}$$
For later use, we take the transpose of $\mathbf P$:
$$\mathbf P^\top=\begin{pmatrix}a&b&c\\2d&2e&2f\\1&-2&-1\end{pmatrix}$$
Recall that $\det\mathbf P=\det(\mathbf P^\top)$.
Denote by $\mathbf P_{i,j}$ the permutation matrix that, upon multiplication by a matrix $\mathbf A_{m\times n}$, swaps rows $i$ and $j$ in $\mathbf A$. So we can write
$$\mathbf P_{2,3}\mathbf U=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}\mathbf U=\begin{pmatrix}a&b&c\\d&e&f\\2&3&2\end{pmatrix}$$
Recall that the determinant of any square matrix with any pair of its rows/columns swapped negates the value of the new matrix, so that $\det\mathbf U=-\det(\mathbf P_{i,j}\mathbf U)$ when $i\neq j$.
Now, observe that we can write the last row of $\mathbf V$ as a linear combination of the last rows of $\mathbf P^\top$ and $\mathbf P_{2,3}\mathbf U$:
$$\begin{pmatrix}1&5&3\end{pmatrix}=(-1)\begin{pmatrix}1&-2&-1\end{pmatrix}+\begin{pmatrix}2&3&2\end{pmatrix}$$
The determinant has the property that it is linear with respect to any given row, which is to say: focusing our attention on a single row, if we can write it as a linear combination of other row vectors, then we can expand the determinant as the sum of two component determinants. To illustrate in practice, we can write
$$\begin{vmatrix}a&b&c\\d&e&f\\1&5&3\end{vmatrix}=\begin{vmatrix}a&b&c\\d&e&f\\(-1)1&(-1)(-2)&(-1)(-1)\end{vmatrix}+\begin{vmatrix}a&b&c\\d&e&f\\2&3&2\end{vmatrix}$$
Aside: I highly recommend watching this lecture from MIT if you ever feel the need to brush up on the properties of the determinant. Strang does a great job of explaining them.
Next, we can pull out a factor of $-1$ from the first determinant, and simultaneously distribute a factor of $2$ along the second row of the first matrix,
$$\begin{vmatrix}a&b&c\\d&e&f\\1&5&3\end{vmatrix}=-\frac12\begin{vmatrix}a&b&c\\2d&2e&2f\\1&-2&-1\end{vmatrix}+\begin{vmatrix}a&b&c\\d&e&f\\2&3&2\end{vmatrix}$$
and we see that we have written $\det\mathbf V$ in terms of known determinants. We get
$$\det\mathbf V=-\frac12\det(\mathbf P^\top)+\det(\mathbf P_{2,3}\mathbf U)=-\frac12\det\mathbf P-\det\mathbf U=-5-(-3)=-2$$
| {
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} |
Maximize the volume of a cone How do I maximize the volume of a cone which is inscribed inside a sphere of radius $r$. I know that $V=\pi r^2h$. But how do you inscribe that into a sphere with radius $r$.
| $V = \dfrac{π r^2 h}{3}$
And that of a sphere of radio $R$: $V = \dfrac{4π R^3}{3} $
The problem can be reduced to two dimensions considering the case of a circle centered on the origin of coordinates and radius $R$ that has inscribed a triangle of base $2r$ and height $R + a$, where $``a"$ is variable.
When both lines are rotated around the $X$-axis, they describe a sphere with an inscribed cone of radius $R$ and $r$ respectively and cone height $h = R + a$:
It is necessary to maximize the volume of the cone with the condition that it is inscribed in the sphere, that is, to maximize the function:
$V (a, r) = \dfrac{πr^2 (R + a)}{ 3}$.
Provided that the point $(a, r)$ belongs to the circle $x^2 + y^2 = R^2$, that is, that the relation is fulfilled:
$a^2 + r^2 = R^2$
$r^2 = R^2 - a^2 ........... (1)$
And substituting in the formula of the volume of the cone we have eliminated the variable $"r"$ and it remains:
$V (a) = \dfrac{π (R^2 - a^2) (R + a)}{3} = \dfrac{π (R^3 + aR^2 - a^2 R - a^2)}{3}$
Let's see for what value of "a" said volume is maximum. Deriving and equating to zero:
$V'(a) = \dfrac{π(R^2-2aR-3a^2)} { 3} = 0$
$R^2-2aR-3a^2 = 0\Rightarrow 3a^2+2aR-R^2 = 0$
Quadratic equation in "a", which by the resolvent leaves us
$a = \dfrac{-2R \pm \sqrt{4R^2 + 12 R^2}}{6} $
whose two roots are:
$a_1 = \dfrac{- 2R + 4R}{6} = \dfrac{R}{3}$
$a_2 = \dfrac{- 2R - 4R} {6} = - R$
Taking the only possible value, the positive one, i.e $a_1$, we substitute it in equation (1), we have left
$r^2 = R^2 - \dfrac{R^2}{9} = \dfrac{8R^2}{9}$
That is, the maximum volume of the cone inscribed in the sphere of radius R is obtained for a conical radius:
$R = \dfrac{\sqrt{8R^2}}{3} = \dfrac{2R \sqrt{2}}{3}$
And for a cone height of:
$h = R + a = R + \dfrac{R}{3} = \dfrac{4R}{3}$
The maximum volume of the inscribed cone will therefore be:
$V = π (\dfrac{8R^2}{9})(\dfrac{4R}{3}) = \dfrac{32 πR^3}{81}$
| {
"language": "en",
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Alternative for calculating the nth of quadratic sequence Given the quadratic sequence
$$f(n)=1, 7, 19, 37, \cdots$$
To calculate the $f(n)$ for $n\ge1$. $$f(n)=an^2+bn+c$$
We start with the general quadratic function, then sub in for $n:=1,2$ and $3$
$$f(1)=a+b+c$$
$$f(2)=4a+2b+c$$
$$f(3)=9a+3b+c$$
Now solve the simultaneous equations
$$a+b+c=1\tag1$$
$$4a+2b+c=7\tag2$$
$$9a+3b+c=19\tag3$$
$(2)-(1)$ and $(3)-(2)$
$$3a+b=6\tag4$$
$$5a+b=12\tag5$$
$(5)-(4)$
$$a=3$$
$$b=-3$$
$$c=1$$
$$f(n)=3n^2-3n+1$$
This method is very long. Is there another easy of calculating the $f(n)$?
| Another standard way is to calculate a difference scheme and then to work backwards:
$$\begin{matrix}
0 & & 1 & & 2 & & 3 & & 4 \\
& & 1 && 7 && 19 && 37 \\
&& &6& & 12 && 18 & \\
&&&& 6 && 6 && \\
\end{matrix} \Rightarrow
\begin{matrix}
& 0 & & 1 & & 2 & & 3 & & 4 \\
\color{blue}{c}= &\color{blue}{1} & & 1 && 7 && 19 && 37 \\
\color{blue}{a+b}= & &\color{blue}{0}&&6& & 12 && 18 & \\
\color{blue}{2a}= &&& \color{blue}{6}&& 6 && 6 && \\
\end{matrix}$$
$$\Rightarrow a = 3, \; b= -3, \; c = 1 \Rightarrow f(n) = 3n^2-3n+1$$
| {
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"url": "https://math.stackexchange.com/questions/2711621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Does such a operator exist? I have been looking for a matrix multiplier that is similar to a tensor product. The best way I can define the product is with the following example:
Suppose $A=\left[\begin{array}{cc}
1 & 2\\
3 & 4
\end{array}\right]$, and $B=\left[\begin{array}{cc}
0 & 5\\
6 & 7
\end{array}\right]$.
Does their exist a multiplier (call it $\#$) such that
$A\#B=\left[\begin{array}{cc}
1\left[\begin{array}{cc}
0 & 5\end{array}\right] & 2\left[\begin{array}{cc}
0 & 5\end{array}\right]\\
3\left[\begin{array}{cc}
6 & 7\end{array}\right] & 4\left[\begin{array}{cc}
6 & 7\end{array}\right]
\end{array}\right]=\left[\begin{array}{cccc}
0 & 5 & 0 & 10\\
18 & 21 & 24 & 28
\end{array}\right]$.
Does such a multiplier exist? If not, how could I use existing operators to attain my desired outcome?
| Can someone verify?
Suppose $A=\left[\begin{array}{cc}
1 & 2\\
3 & 4
\end{array}\right],$and $B=\left[\begin{array}{cc}
0 & 5\\
6 & 7
\end{array}\right]$.
By definintion of the Khatri-Rao product: $A*B=\left(A_{ij}\otimes B_{ij}\right)_{ij}$.
$A^{T}=\left[\begin{array}{cc}
1 & 3\\
2 & 4
\end{array}\right],$and $B^{T}=\left[\begin{array}{cc}
0 & 6\\
5 & 7
\end{array}\right]$
\begin{eqnarray*}
\left(A^{T}*B^{T}\right)^{T} & = & \left[A_{1}\otimes B_{1}|A_{2}\otimes B_{2}\right]^{T}\\
& = & \left[\begin{array}{cc}
1\cdot0 & 3\cdot6\\
1\cdot5 & 3\cdot7\\
2\cdot0 & 4\cdot5\\
2\cdot5 & 4\cdot7
\end{array}\right]^{T}\\
& = & \left[\begin{array}{cc}
1\cdot0 & 3\cdot6\\
1\cdot5 & 3\cdot7\\
2\cdot0 & 4\cdot6\\
2\cdot5 & 4\cdot7
\end{array}\right]^{T}\\
& = & \left[\begin{array}{cc}
0 & 18\\
5 & 21\\
0 & 24\\
10 & 28
\end{array}\right]^{T}\\
& = & \left[\begin{array}{cccc}
0 & 5 & 0 & 10\\
18 & 21 & 24 & 28
\end{array}\right]
\end{eqnarray*}
This seems related: Macedo, Hugo Daniel, and José Nuno Oliveira. 2015.
A linear algebra approach to OLAP. Formal Aspects of Computing 27,
(2) (03): 283-307
| {
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"source": "stackexchange",
"question_score": "3",
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Last three digits of $6^{2002}$ Find the last three digits of $6^{2002}$. I did some work and figured out that the last two digits is 36. Can anyone help me with the hundredth digit? By the way, I used modular arithmetic and the recursion method for the tens digit, but it fell short when I attempted to do the hundreds digit. Thank you in advance!
How I figured out the last two digits:
I used the formula $\frac {1}{10^k} [n-a_k (n)]$. The last digit is obviously 6. I obtained the tens digit this way: $\frac {1}{10}(6^{2002}-6)=\frac {6}{10}(6^{2001}-1)=\frac {3}{5} (6^{2001}-1)=\frac {3}{5}(6-1)(6^{2000}+6^{1999}+...+6^2+6+1)=3(6^{2000}+6^{1999}+...+6^2+6+1)\equiv3(6+6+...+6+6+1)=3(2000\cdot6+1)\equiv3 (mod 10)$
Therefore, the last two digits of $6^{2002}$ are $36$.
| I used the formula $\frac {1}{10^k} [n-a_k (n)]$. The last two digits is 36. I obtained the hundreds digit this way: $\frac {1}{100}(6^{2002}-36)=\frac {9}{25}(6^{2000}-1)=\frac {9}{25} (6^{2000}-1)=\frac {9}{25}((6^5)^{400}-1)=\frac {9}{25}((7776)^{400}-1)=\frac {9}{25}(7776-1)(7776^{399}+7776^{398}+...+7776^2+77766+1)\equiv9\cdot311\cdot(6+6+...+6+6+1)\equiv9\cdot1\cdot(399\cdot6+1)\equiv9\cdot1\cdot(9\cdot6+1)\equiv9\cdot1\cdot5\equiv5 (mod 10)$
Therefore, the last three digits of $6^{2002}$ are $536$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all pairs of $(x, y ,z)$ such that $x + y =\sqrt{z^{2} + 2018}, .... $ Find all pairs of $(x,y,z)$, of real numbers, such that
$$ x + y = \sqrt{z^{2} + 2018} $$
$$ x + z = \sqrt{y^{2} + 2018} $$
$$ y + z = \sqrt{x^{2} + 2018} $$
An attempt :
Squaring we get
$$ (x + y)^{2} = z^{2} + 2018 \implies (x + y)^{2} - z^{2} = 2018 $$
$$ (x + z)^{2} = y^{2} + 2018 \implies (x + z)^{2} - y^{2} = 2018 $$
$$ (y + z)^{2} = x^{2} + 2018 \implies (z + y)^{2} - x^{2} = 2018 $$
which also means
$$ (x + y - z)(x + y + z) = 2018 $$
$$ (x + z - y)(x + y + z) = 2018 $$
$$ (z + y - x)(x + y + z) = 2018 $$
so
$$ \frac{2018}{x+y-z} = \frac{2018}{x+z-y} = \frac{2018}{y+z-x} $$
$$(x+y-z) = (x+z-y) = (y+z-x)$$
$$y-z = z-y \implies z = y $$
$$x-y= y-x \implies x=y $$
so my answer is
$$(x, y, z), \:\:\: x=y=z $$
but with the 3 initial equations, we must also have
$$ (x+y) = \sqrt{z^{2} + 2018} \implies 4x^{2} = x^{2} + 2018 $$
or
$$ x^{2} = 2018/3 $$
so the solution is
$$(x, y, z), \:\:\: x=y=z = \sqrt{2018/3} $$
Is this sufficient already? are there better techniques?
| At the stage you have...
$$ (x + y - z)(x + y + z) = 2018 $$
$$ (x + z - y)(x + y + z) = 2018 $$
$$ (z + y - x)(x + y + z) = 2018 $$
This implies...
$$x+y-z=x+z-y=z+y-x=\dfrac{2018}{x+y+z}$$
...unless $x+y+z=0$.
Notice that $x+y+z=0$ implies $x+y=-z$ and so $-z=\sqrt{z^2+2018}$ which means $z^2=z^2+2018$ and so $z=0$. Likewise, $x=y=0$ (contradiction since $(0,0,0)$ isn't a solution).
Thus $x+y-z=x+z-y=z+y-x$ and so you got that $x=y=z$. But then $x+y=\sqrt{z^2+2018}$ implies that $2z=\sqrt{z^2+2018}$ and so $4z^2=z^2+2018$ and so $x=y=z = \pm \sqrt{\dfrac{2018}{3}}$. But $x+y $ etc. are square roots (thus non negative). So there is only one solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Showing $\int_{-\infty}^{\infty}\frac {\mathrm d x} {1+x+x^2}=3 \int_0^\infty \frac {\mathrm d x} {1+x+x^2}$ Is there an easy way to see why
$$\int_{-\infty}^{\infty}\frac {\mathrm d x} {1+x+x^2}=3 \int_0^\infty \frac {\mathrm d x} {1+x+x^2}$$
without having to evaluate the integrals explicitly? I was trying substitutions but nothing worked.
| Let $\displaystyle x=\frac{1}{v}$.
$$\int_{-1}^0\frac{dx}{1+x+x^2}=\int_{-1}^{-\infty}\frac{\frac{-1}{v^2}dv}{1+\frac{1}{v}+\frac{1}{v^2}}=\int_{-\infty}^{-1}\frac{dv}{1+v+v^2}$$
Let $\displaystyle x=-1-u$.
$$\int_{-\infty}^{-1}\frac{dx}{1+x+x^2}=\int_\infty^0\frac{-du}{1+(-1-u)+(-1-u)^2}=\int_0^{\infty}\frac{du}{1+u+u^2}$$
Therefore,
\begin{align*}
\int_{-\infty}^{\infty}\frac{dx}{1+x+x^2}&=\int_{-\infty}^{-1}\frac{dx}{1+x+x^2}+\int_{-1}^{0}\frac{dx}{1+x+x^2}+\int_{0}^{\infty}\frac{dx}{1+x+x^2}\\
&=3\int_{0}^{\infty}\frac{dx}{1+x+x^2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of an improper integral $\int_3^\infty \frac{\sin(x)}{x+2\cos(x)}dx$ I tried to check whether the following integral converges:
$$\int_3^\infty \dfrac{\sin(x)}{x+2\cos(x)}dx$$
Dirichlet criterion doesn't work here since the function
$$\dfrac{1}{x+2\cos(x)}$$
is not monotone.
| \begin{align*}
\int_{3}^{M}\dfrac{\sin x}{x+2\cos x}dx&=-\dfrac{\cos x}{x+2\cos x}\bigg|_{x=3}^{x=M}-\int_{3}^{M}\dfrac{-(2\sin(x)-1)\cos x}{-(x+2\cos x)^{2}}dx\\
&=-\dfrac{\cos M}{M+2\cos M}+\dfrac{\cos 3}{3+2\cos 3}-\int_{3}^{M}\dfrac{-\sin(2x)+\cos(x)}{(x+2\cos x)^{2}}dx.
\end{align*}
Now each of the integrals
\begin{align*}
\int_{3}^{\infty}\dfrac{-\sin(2x)}{(x+2\cos x)^{2}}dx.
\end{align*}
\begin{align*}
\int_{3}^{\infty}\dfrac{\cos(x)}{(x+2\cos x)^{2}}dx.
\end{align*}
absolutely converges, since
\begin{align*}
\int_{3}^{M}\dfrac{1}{(x+2\cos x)^{2}}dx\leq\int_{3}^{M}\dfrac{1}{(x-2)^{2}}dx.
\end{align*}
So the improper integral converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
solve the quintic non-linear system $x^5+y^5=33$ and $x+y=3$
I tried a variety of things here, all too ugly to post. the most promising was doing long division $(x^5+y^5)/(x+y), $but it still looked kind of ugly and not easily factorable. I am looking to solve this with simple high school level algebra techniques. Can someone give me a hint? Much appreciated.
| The first thing I try is plug some random small integers to the equations and see
whether they work. I already get two "obvious" solutions $(x,y) = (2,1)$ and $(1,2)$.
To see this exhaust all solutions, introduce a parameter $u$ such that $x = \frac{3+u}{2}$ and $y = \frac{3-u}{2}$.
In terms of $u$, we have
$$\begin{align}
& x^5 + y^5 = 33\\
\iff & (3+u)^5 + (3-u)^5 = 33\cdot 2^5\\
\iff & 2(3^5 + 10\cdot 3^3 u^2 + 5\cdot 3 u^4) = 33\cdot 2^5\\
\iff & 10\cdot 3^2 u^2 + 5 u^4 = \frac13\left(\frac12\cdot 33\cdot 2^5 - 3^5\right)
= 11\cdot16 - 81 = 85\\
\iff & u^4 + 18u^2 -19 = 0\\
\iff & (u^2-1)(u^2+19) = 0\\
\implies & u = \pm 1\end{align}$$
This means the equations have two and only two real solutions: $(x,y) = (2,1)$ and $(1,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric series sum involving tangents
$$\frac{1}{4}\tan\bigg(\frac{\pi}{8}\bigg)+\frac{1}{8}\tan\bigg(\frac{\pi}{16}\bigg)+\frac{1}{16}\tan\bigg(\frac{\pi}{32}\bigg)+\cdots\cdots \infty$$
Try: $$\cos x\cdot \cos(x/2)\cdot\cos(x/2^2)\cdots\cdots \cos(x/2^{n-1})=\frac{1}{2^{n-1}}\frac{\cos(x)}{\sin(x/2^{n-1})}$$
Could some help me how to solve ahead , thanks
| Using Viete's infinite product identity
$$\cos\frac{x}{2}\cdot\cos\frac{x}{4}\cdot...=\prod_{k\in\mathbb{N}}\cos\frac{x}{2^k}=\frac{\sin x}{x}$$
we obtain by taking logarithms on both sides and differentiating with respect to $x$ the following
$$-\frac{1}{2}\tan\frac{x}{2}-\frac{1}{4}\tan\frac{x}{4}-...=-\sum_{k\in\mathbb{N}}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{\cos x}{\sin x}-\frac{1}{x}$$
Therefore
$$\sum_{k\geqslant 2}\frac{1}{2^k}\tan\frac{x}{2^k}=\frac{1}{x}-\frac{\cos x}{\sin x}-\frac{1}{2}\tan\frac{x}{2}$$
Set $x=\pi/2$ to get
$$\frac{1}{4}\tan\frac{\pi}{8}+\frac{1}{8}\tan\frac{\pi}{16}+...=\sum_{k\geqslant 2}\frac{1}{2^k}\tan\frac{\pi}{2^{k+1}}=\frac{2}{\pi}-\frac{\cos (\pi/2)}{\sin(\pi/2)}-\frac{1}{2}\tan\frac{\pi}{4}=\frac{2}{\pi}-\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding nilpotent and diagonalizable operators for a linear transformation $T$
Let $T$ be the linear operator on $\mathbb{R}^3$ which is represented by the matrix:
$A=\begin{bmatrix}3 & 1 & -1\\ 2 & 2 & -1\\ 2 & 2 &0\end{bmatrix}$
in the standard ordered basis. Show that there is a diagonalizable operator $D$ on $R^3$ and a nilpotent operator $N$ on $\mathbb{R}^3$ such that $T=D+N$ and $DN=ND$. Find the matrices of $D$ and $N$ in the standard basis.
Here is what I've got so far.
$\det(xI-A)=\det\left(\begin{bmatrix}x-3 & -1 & 1\\ -2 & x-2 & 1\\ -2 & -2 &x\end{bmatrix}\right)=(x-1)(x-2)^2$
But now I've got to find polynomials $h_1(T)+h_2(T)=I$ where $E_i=h_i$. This was easy enough: $E_1=(x-2)^2$ and $E_2=-(x-1)(x-3)$. I've also computed the matricies $(T-2)^2$ and $(T-1)(T-3)$ beloe respectively:
$\begin{bmatrix}1 & -1 & 0\\ 0 & 0 & 0\\2 & -2 & 0\end{bmatrix}$
$\begin{bmatrix}0 & 1 & 0\\ 0 & 1 & 0\\-2 & 2 & 1\end{bmatrix}$
This is where I'm stuck. I'm not sure how to proceed from here in order to find $D$ and $N$.
| This can be accomplished as an extra step after doing the work for the Jordan Normal Form, but keeping careful account of the change of basis matrix and its inverse.
For $R^{-1} A R = J$ we get,
$$
\left(
\begin{array}{rrr}
1 & -1 & 0 \\
0 & 1 & 0 \\
2 & 0 & -1
\end{array}
\right)
\left(
\begin{array}{rrr}
3 & 1 & -1 \\
2 & 2 & -1 \\
2 & 2 & 0
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 1 & 0 \\
0 & 1 & 0 \\
2 & 2 & -1
\end{array}
\right) =
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{array}
\right)
$$
with the important reverse direction $RJR^{-1} = A$
$$
\left(
\begin{array}{rrr}
1 & 1 & 0 \\
0 & 1 & 0 \\
2 & 2 & -1
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & -1 & 0 \\
0 & 1 & 0 \\
2 & 0 & -1
\end{array}
\right) =
\left(
\begin{array}{rrr}
3 & 1 & -1 \\
2 & 2 & -1 \\
2 & 2 & 0
\end{array}
\right)
$$
Next we take
$$
D_0 =
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}
\right)
$$
and
$$
N_0 =
\left(
\begin{array}{rrr}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right)
$$
so that $D_0 N_0 = N_0 D_0$ and $D_0 + N_0 = J$
Then
$$
D = R D_0 R^{-1} =
\left(
\begin{array}{rrr}
1 & 1 & 0 \\
0 & 2 & 0 \\
-2 & 2 & 2
\end{array}
\right)
$$
while
$$
N = R N_0 R^{-1} =
\left(
\begin{array}{rrr}
2 & 0 & -1 \\
2 & 0 & -1 \\
4 & 0 & -2
\end{array}
\right)
$$
You can check easily enough, we get $D+N = A,$ also $DN=ND=2N$ and $N^2 = 0 \; .$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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