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Complex number inequality $|z-1| \ge \frac{2}{n-1}$
If $n \ge 3$ is an odd number and $z\in\mathbb{C}, z\neq -1$ such that $z^n=-1$, prove that
$$|z-1|\ge \frac{2}{n-1}$$
I was thinking that $-2=z^n-1=(z-1)(z^{n-1}+z^{n-2}+...+z^2+z+1)$ and because $z\neq -1$, the second factor can not be $0$:
$$|z-1|=\frac{2}{|z^{n-1}+z^{n-2}+...+z^2+z+1|}$$
Also $|z|=1$, so if I use triangle inequality, I find:
$$|z-1|\ge \frac{2}{|z^{n-1}|+...+|z|+1}=\frac{2}{n}$$
but this is lower than $\frac{2}{n-1}$. Does this help in any way?
| If $z^n = -1$, we have
$$
z = \exp\left(\frac{(2k + 1)i \pi}{n}\right),
\qquad k \in \{0, \ldots,n - 1\}
$$
Denote for $k \in \{0, \ldots,n - 1\}$
$$
a_k := \frac{(2k + 1) \pi}{n}
$$
Then
\begin{align}
| z - 1 |^2
& = | \cos(a_k) + i \sin(a_k) - 1 |^2
= (\cos(a_k) - 1)^2 + \sin^2(a_k) \\
& = \cos^2(a_k) + \sin^2(a_k) - 2 \cos(a_k) + 1 \\
& = 2 - 2 \cos(a_k),
\end{align}
therefore
$$
| z - 1 |
= \sqrt{2} \sqrt{1 - \cos(a_k)}.
$$
As $\sqrt{2} \left| \sin\left(\frac{x}{2}\right)\right| = \sqrt{1 - \cos(x)}$, we have
$$
| z - 1| = 2 \left| \sin\left(\frac{(2k + 1)\pi}{2n}\right) \right|.
$$
Therefore it remains to show that
$$
\left| \sin\left(\frac{\left(k + \frac{1}{2}\right)\pi}{n}\right) \right|
\ge \frac{1}{n - 1}, \qquad
k \in \{0, \ldots,n - 1\}.
$$
Also, we can exclude one $k$, which is the $k$ for which $a_k = 0$.
Notice that $\left| \sin\left(\frac{\left(k + \frac{1}{2}\right)\pi}{n}\right) \right|$ is smallest (exercise!) for $k = 0$ and $k = n - 1$ (they yield the same result since $\frac{\pi}{2n} = \frac{(n - 1/2) \pi}{n}$), so it suffices to show it for $k = 0$, which is to say
$$
\left| \sin\left(\frac{\pi}{2n}\right) \right|
\ge \frac{1}{n - 1}.
$$
But this follows from the concavity (similar to here) of $x \mapsto \sin(x)$ on $\left[0, \frac{\pi}{6}\right]$ and $n \ge 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of $n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$? I am having a hard time following this proof. Here is how it goes.
$$
(k+1)^3 = k^3+3k^2+3k+1\\
3k^2+3k+1 = (k+1)^3-k^3\\
$$
if $ k = 1, 2, 3, ... , n-1$ we add all the 5 formulas like this
$$
3(1)^2+3(1)+1 = ((1)+1)^3-(1)^3\\
3(2)^2+3(2)+1 = ((2)+1)^3-(2)^3\\
3(3)^2+3(3)+1 = ((3)+1)^3-(3)^3\\
3(4)^2+3(4)+1 = ((4)+1)^3-(4)^3\\
\vdots \\
3(n-1)^2+3(n-1)+1 = ((n-1)+1)^3-(n-1)^3\\
3n^2+3n+1 = (n+1)^3-(n-1)^3\\
$$
The result of adding these formulas is
$$
3[1^2+2^2 + ... + (n-1)^2+n^2] + 3[1 + 2 + ... + (n-1)+n] + n = (n+1)^3-1^3
$$
I am able to follow up-to this point easily. I don't understand how this last equation goes from that to this $ n^3/3 + n^2/2 + n/6 = [1^2 + 2^2 + ... + n^2]$ I know thee sum of arithmetic series to n is $n(n-1)/2$ and that replaces the second expression on the left side.
Can some one please show me the algebra step by step?
| $$ 3\left[1^2+2^2 + ... + (n-1)^2\right] + 3\left[1 + 2 + ... + (n-1)\right] + (n-1) = n^3-1^3 $$
$$3\left[1^2+2^2 + ... + (n-1)^2\right] + 3\left[1 + 2 + ... + (n-1)\right] + n = n^3 $$
$$3\left[1^2+2^2 + ... + (n-1)^2+n^2\right] + 3\left[1 + 2 + ... + (n-1)\right] + n = n^3+3n^2 $$
$$3\left[1^2+2^2 + ... + (n-1)^2+n^2\right] + 3\dfrac{(n-1)n}2 + n = n^3+3n^2 $$
$$3\left[1^2+2^2 + ... + (n-1)^2+n^2\right] = n^3+3n^2-n-3\dfrac{(n-1)n}2 $$
$$1^2+2^2 + ... + (n-1)^2+n^2 = \dfrac{n^3}3+n^2-\dfrac n3-\dfrac{(n-1)n}2 $$
$$1^2+2^2 + ... + (n-1)^2+n^2 = \dfrac{n^3}3+\dfrac{n^2}2 +\dfrac{n}6 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find a limit with sqrt $\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$ $$\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right)$$
I don't know how to rewrite or rationalize in order to find the limit.
| If you write the expression as $$x^4\left(1-\sqrt{1+6/x^2}+3/x^2\right)=\frac{1-\sqrt{1+6/x^2}+3/x^2}{1/x^4},$$ then you may be able to apply the Marquis de L'hopital's method.
After exactly two iterations, you get $$\frac92\frac{\frac{1}{\sqrt{1+6/x^2}}-1}{1/x^2},$$ where you may now take an elementary limit as $x\to+\infty.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Hard inequality problem (if $(a_2-a_1)^2 + (a_3-a_2)^2 + \ldots + (a_{2n}-a_{2n-1})^2 = 1$ ...) I have a hard problem :
If $$(a_2-a_1)^2 + (a_3-a_2)^2 + \ldots + (a_{2n}-a_{2n-1})^2 = 1$$ where $a_1,a_2...,a_{2n} \in \mathbb{R}$
What is the maximum of $$(a_{n+1}+a_{n+2}+…+a_{2n})−(a_1+a_2+…+a_n)?$$
I found that : $$1 \leq \sqrt{(a_1)^2+(a_2)^2 +...+ (a_{2n-1})^2} + \sqrt{(a_2)^2+(a_3)^2 +...+ (a_{2n})^2} $$
therefore : $$1 \leq (a_1)^2+(a_2)^2 +...+ (a_{2n-1})^2 + (a_2)^2+(a_3)^2 +...+ (a_{2n})^2 + 2\sqrt{\bigg((a_2)^2+(a_3)^2 +...+ (a_{2n})^2\bigg) \times \bigg((a_1)^2+(a_2)^2 +...+ (a_{2n-1})^2\bigg)}$$
Here we can use AM-GM but...
| Following achille hui's suggestion: Let
$$s_k = a_k - a_{k-1}, \qquad k \ge 2 $$
with $\sum\limits_{k=2}^{2n} s_{k}^2 = 1$. Now the target function is
$$
f = (a_{n+1}+a_{n+2}+…+a_{2n})−(a_1+a_2+…+a_n) \\
= (a_{n+2}+…+a_{2n})−(a_1+a_2+…+a_{n-1}) +a_{n+1} - a_{n} \\
= (a_{n+2}+…+a_{2n})−(a_1+a_2+…+a_{n-1}) +s_{n+1} \\
= (a_{n+3}+…+a_{2n})−(a_1+a_2+…+a_{n-2}) +s_{n+1} + a_{n+2} - a_{n-1}\\
= (a_{n+3}+…+a_{2n})−(a_1+a_2+…+a_{n-2}) +s_{n+1} + a_{n+2} - a_{n+1} +a_{n+1} - a_{n} + a_{n} - a_{n-1}\\
= (a_{n+3}+…+a_{2n})−(a_1+a_2+…+a_{n-2}) +s_{n+1} + (s_{n+2} +s_{n+1} + s_{n} )
$$
Continuing with that telescoping gives
$$
f = s_{n+1} + (s_{n+2} +s_{n+1} + s_{n}) + (s_{n+3} +s_{n+2} +s_{n+1} + s_{n}+ s_{n-1}) + \cdots + (\sum\limits_{k=2}^{2n} s_{k} )\\
= n s_{n +1} + \sum\limits_{k=1}^{n-1} k (s_{2n +1-k} + s_{k+1} )
$$
again with $\sum\limits_{k=1}^{2n-1} s_{k+1}^2 = 1$.
Now by the Cauchy-Schwarz inequality,
$$
f^2 = (n s_{n +1} + \sum\limits_{k=1}^{n-1} k (s_{2n +1-k} + s_{k+1} ))^2\\
\le (n^2 + 2 \sum\limits_{k=1}^{n-1} k^2 ) (s_{n +1}^2 + \sum\limits_{k=1}^{n-1} (s_{2n +1-k}^2 + s_{k+1}^2 )) \\
=
\frac{n (2 n^2 + 1)}{3} (\sum\limits_{k=1}^{2n-1} s_{k+1}^2 )
$$
Hence $$f \le \sqrt{\frac{n (2 n^2 + 1)}{3}}$$
This is indeed the maximum, as the Cauchy-Schwarz inequality is tight. Equality occurs when $c\cdot k = s_{2n +1-k} = s_{k+1}$ ($k = 1\cdots n$) with the constant $c$ given by the condition $\sum\limits_{k=1}^{2n-1} s_{k+1}^2 = 1$, i.e. $s_{2n +1-k} = s_{k+1} = k/ \sqrt{\frac{n (2 n^2 + 1)}{3}}$ (for $k = 1\cdots n$).
Just for getting a feeling, we can check that for the first two $n$.
For $n=1$, we have the obvious $s_2 = 1/ \sqrt{\frac{n (2 n^2 + 1)}{3}} = 1$. The $a_k$ can be recomputed from this: Fix some $a_1$ and let $a_2 = a_1 + 1$.
For $n=2$, we have $\sqrt{\frac{n (2 n^2 + 1)}{3}} = \sqrt{6}$, hence $s_2 = s_4 = 1/\sqrt{6}$ and $s_3 = 2/\sqrt{6}$. Indeed, $s_2^2 + s_3^2 + s_4^2 = 1$ and $f = s_2 + 2 s_3 + s_4 = 6/\sqrt{6} = \sqrt{6}$. Again, the $a_k$ can be recomputed from this: Fix some $a_1$ and let $a_2 = a_1 + 1/\sqrt{6}$, $a_3 = a_1 + 3/\sqrt{6}$, $a_4 = a_1 + 4/\sqrt{6}$. Then $f = (a_3 + a_4) - (a_1 +a_2) = 7/\sqrt{6} - 1/\sqrt{6} = \sqrt{6} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find general term of recursive sequences $ x_{n+1}=\frac{1}{2-x_n}, x_1=1/2,$ Please help to solve:
*
*$ x_{n+1}=\frac{1}{2-x_n}, x_1=1/2,$
*$x_{n+1}= \frac{2}{3-x_n}, x_1=1/2$
I know answers, but can't figure out the solution.
The first one is obvious if you calculate first 3-5 terms by hand. But how can I get the result not by guessing, but mathematically?
Answers are:
*
*$x_n = \frac{n}{n+1}$
*$x_n = \frac{3\cdot2^{n-1}-2}{3\cdot2^{n-1}-1}$
| A recurrence of the form:
$\begin{equation*}
w_{n + 1}
= \dfrac{a w_n + b}{c w_n + d}
\end{equation*}$
with $a d \ne b c$ and $c \ne 0$ is called a Ricatti recurrence. One way to solve them is to recognize the right hand side is a Möbius transform, and those can be composed like matrix products:
$\begin{align*}
A(z)
&= \frac{a_{1 1} z + a_{1 2}}{a_{2 1} z + a_{2 2}} \\
B(z)
&= \frac{b_{1 1} z + b_{1 2}}{b_{2 1} z + b_{2 2}} \\
C(z)
&= A(B(z)) \\
&= \frac{c_{1 1} z + c_{1 2}}{c_{2 1} z + c_{2 2}}
\end{align*}$
where the matrix of coefficients in $C$ is $B \cdot A$. Thus the solution (as a matrix) is given by $w_n = A^n(w_0)$.
Another way is due to Mitchell. Define a new variable $x_n = (1 + \eta w_n)^{-1}$, write the recurrence in terms of $x_n$:
$\begin{equation*}
x_{n + 1}
= \dfrac{(d \eta - c) x_n + c}
{(b \eta^2 - (a - d) \eta - c) x_n + a \eta + c}
\end{equation*}$
Selecting $\eta$ such that $b \eta^2 - (a - d) \eta - c = 0$ (both roots work fine) reduces the recurrence to linear of the first order.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$ When testing Mathematica, I accidentally found this equality:
$$\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}=\frac{2^{2 n+2} \Gamma \left(\frac{1}{2} (2 n+3)\right)}{\sqrt{\pi } (n+1)^3 \Gamma (n+1)}-\frac{1}{(n+1)^2}$$
Mathematica gives this closed-form directly, but I've found no proof. I'd like you to help to find the solution. Thanks!
| Use Binomial identity:
$$
(1+t)^n=\sum_{j=0}^{n} {n \choose j}t^n.
\tag{1}
$$
Integration of $(1)$ from $t=0$ to $t=x$ gives
$$
\frac{(1+x)^{n+1}-1}{n+1}= \sum_{j=0}^n {n \choose j}\frac{x^{j+1}}{j+1}.\tag{2}
$$
We can change $x$ to $1/x$ in $(2)$ to get
$$
\frac{(1+1/x)^{n+1}-1}{n+1}= \sum_{j=0}^n {n \choose j}\frac{x^{-j-1}}{j+1}
.
\tag{3}
$$
Multiplying $(2)$ and $(3)$ and collecting terms free of $x$ on RHS, we get
$$
\frac{x^{-n-1}\big[(1+x)^{2n+2}-(1+x)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}\big]}{(n+1)^2}=\sum_{j=0}^{n} \frac{{n \choose j}^2}{(j+1)^2} x^0+\dots
$$
Equating the coefficients yields
\begin{align}
S_n = \sum_{j=0}^{n} \frac{{n \choose j}^2}{(j+1)^2}
&= [x^{n+1}]~\frac{[(1+x)^{2n+2}-(1+x)^{n+1}-(1+x)^{n+1} x^{n+1}+x^{n+1}]}{(n+1)^2} \\
&= \frac{{2n+2 \choose n+1}-1}{(n+1)^2}.
\end{align}
It may be checked to be nothing but the RHS of OP.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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How to factor $x^6-4x^4+2x^3+1$ by hand? I generated this polynomial after playing around with the golden ratio. I first observed that (using various properties of $\phi$), $\phi^3+\phi^{-3}=4\phi-2$. This equation has no significance at all, I just mention it because the whole problem stems from me wondering: which other numbers does this equation hold for?
The six possible answers are the roots of $x^6-4x^4+2x^3+1=0$. Note that I am not interested in solving for $x$ itself as much as I am interested in a method which would allow me to completely factor out this polynomial into lowest degree factors which still have real coefficients. Note that I am treating this equation as if I had no clue that the golden ratio is one of the solutions. In other words, I am trying to factor this equation as if I never saw it before, so I can't just immediately factor out $(x^2-x-1)$ without a justifiable process, even though it is indeed one of the factors.
I first observed that the equation holds for $x=1$, so I was able to divide out $(x-1)$ to get the factorization of:
$$(x-1)(x^5+x^4-3x^3-x^2-x-1)$$
I tried making an assumption that the quintic reduces to a product of $(x^3+Ax^2+Bx+C)(x^2+Dx+E)$, multiplying out, and equalling coefficients, but I ended up with a system of two extremely convoluted equations which I had no idea how to solve. I also tried to turn the first five terms of the quintic into a palindromic polynomial and then perform the standard method of factoring palindromic polynomials, to no avail.
I am either missing something, or I don't know of a nice method that would let this expression be factored. I'm looking forward to being enlightened, thanks for any help.
| Your original method is tedious but it can be done.
You can show that $(x^3+Ax^2+Bx+C)(x^2+Dx+E)$ is equal to:
$$x^5+(D+A)x^4+(1+AD+B)x^3 + (AE+BD+C)x^2 + (BE+CD) + CE$$
so $A+D = 1, B+AD+1 = -3, AE+BD+C=-1, BE+CD=-1, CE=-1$.
Assuming $A,B,C,D,E$ are all integers, we either have $C=-1, E=1$ or $C=1, E=-1$.
If $C=-1, E=1$, then we have:
$$A+D=1 \tag{1}$$
$$B+AD=-4 \tag{2}$$
$$A+BD=0 \tag{3}$$
$$B-D=-1 \tag{4}$$
$(1)+(4)$ gives $A+B=0$ so $A=-B$, which gives:
$$-B+D=1 \tag{5}$$
$$B-BD=-4 \tag{6}$$
$$-B+BD=0 \tag{7}$$
$$B-D=-1 \tag{8}$$
and this is clearly impossible since $(6) + (7)$ gives $0=-4$.
Therefore we must have $C=1, E=-1$:
$$A+D=1 \tag{9}$$
$$B+AD=-4 \tag{10}$$
$$-A+BD=-2 \tag{11}$$
$$-B+D=-1 \tag{12}$$
This time $(9)-(12)$ gives $A+B=2$, so $A=2-B$:
$$-B+D=-1 \tag{13}$$
$$B+2D-BD=-4 \tag{14}$$
$$B+BD=0 \tag{15}$$
$$-B+D=-1 \tag{16}$$
$(14)+(15)$ gives $2B+2D = -4$, so $B+D=-2$. When we add this to $(16)$, $2D=-2$ so $D=-1$.
And the rest follows:
$$B - D = 1 \Rightarrow B+1=1, B=0$$
$$A=2-B \Rightarrow A=2$$
so the factorisation is $(x-1)(x^3+2x^2+1)(x^2-x-1)$.
I wouldn't wish this method on anybody.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $\sqrt{-5}$ a prime in $\mathbb{Z}{[\sqrt{-5}]}$? I know it is irreducible over $\mathbb{Z}{[\sqrt{-5}]}$ but since it is not even UFD, so we can't conclude primality from irreducibility.
My guess is yes since $N(\sqrt{-5})=5$ is prime. I started with $ab=c\sqrt{-5}$ where a, b and c are in $\mathbb{Z}{[\sqrt{-5}]}$, now we have to show one of $a$ or $b$ is divisible by $\sqrt{-5}$.
I got $N(a)N(b)=5N(c)$, hence $5$ divide either $N(a)$ or $N(b)$.
What should be my next step ?
| Say $5|N(a)$. Since $N(x+y\sqrt{-5}) = x^2+5y^2$, if $a=x+y\sqrt{-5}$ then $5|x$. So $a=5z+y\sqrt{-5} = \sqrt{-5}(y-z\sqrt{-5})$.
You can do this without invoking the norm, just low-tech: Note that an element $x+y\sqrt{-5}$ of $\mathbb{Z}[\sqrt{-5}]$ is a multiple of $\sqrt{-5}$ if and only if $x$ is a multiple of $5$ in $\mathbb{Z}$. Indeed, $5a+y\sqrt{-5} = \sqrt{-5}(y-a\sqrt{-5})$, and $\sqrt{-5}(m+n\sqrt{-5}) = 5(-n) + m\sqrt{-5}$.
Now say $\sqrt{-5}$ divides $(a+b\sqrt{-5})(x+y\sqrt{-5})$. Then it divides
$$(a+b\sqrt{-5})(x+y\sqrt{-5}) = (ax-5by) + (ay+bx)\sqrt{-5}$$
and therefore $ax-5by$ is a multiple of $5$; hence $ax$ is a multiple of $5$, so $5$ divides either $a$ or $x$, hence $\sqrt{-5}$ divides either $a+b\sqrt{-5}$ or $x+y\sqrt{-5}$ as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find sum of geometric-like series with binomial coefficients using complex analysis Studying analytic number theory, I stumbled across the problem of finding the sum of the series
$\sum_{n=0}^{\infty}\binom{2n}{n}\left(\frac{1}{5}\right)^n$
A professor gave me the hint of "using basic complex analysis" but honestly, I haven't been able to get anywhere. Appreciate any help/comments :)
| Consider the integral
$$\int_0^{2\pi}4^n\cos^{2n}(x)\:dx$$
Expanding out using Euler's formula and binomial expansion, we get that
$$\int_0^{2\pi}4^n\cos^{2n}(x)\:dx = \int_0^{2\pi}(e^{ix}+e^{-ix})^{2n}\:dx = \sum_{k=0}^{2n} {2n \choose k} \int_0^{2\pi}e^{i(2n-2k)x}\:dx$$
The integral on the right will always be $0$ unless the exponent is $0$, meaning the only term that survives the summation is $k=n$
$$\sum_{k=0}^{2n} {2n \choose k} \int_0^{2\pi}e^{i(2n-2k)x}\:dx = {2n \choose n}\cdot 2\pi$$
Now going backwards, we can substitute this value into the summation:
$$\sum_{n=0}^\infty {2n \choose n}\left(\frac{1}{5}\right)^n = \frac{1}{2\pi}\int_0^{2\pi} \:dx \sum_{n=0}^\infty \left(\frac{4}{5}\cos^2(x)\right)^n = \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{1-\frac{4}{5}\cos^2(x)}\:dx $$
Then rationalize and use trig identities to get the following expression:
$$= \frac{5}{2\pi}\int_0^{2\pi} \frac{1}{\cos^2(x) + 5\sin^2(x)}\:dx = \frac{\sqrt{5}}{2\pi}\int_0^{2\pi} \frac{\sqrt{5}\sec^2(x)}{1 + 5\tan^2(x)}\:dx$$
$$= \frac{\sqrt{5}}{2\pi}\tan^{-1}\left(\sqrt{5}\tan(x)\right)\Biggr|_0^{2\pi} = \frac{\sqrt{5}}{2\pi}\left(\frac{\pi}{2} + \pi + \frac{\pi}{2}\right) = \sqrt{5}$$
Leaving us with our final result
$$\sum_{n=0}^\infty {2n \choose n}\left(\frac{1}{5}\right)^n = \sqrt{5}$$
This only worked out so nicely because $5$ is the only number with the special property that $5-4=1$.
| {
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Simplify $(1+i^\frac{1}{2})^\frac{1}{2}$ How can I simplify $\sqrt{1+\sqrt{i}}$?
I thought about making $z^2=1+\sqrt{i}$ and then $w=z^2$
But I'm not really sure
| Using polar coordinates $\displaystyle \sqrt{1+\sqrt{i}} = \sqrt{1+\sqrt{e^{i \pi/2}}} = \sqrt{1 + e^{i \pi/4}} = \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} \, $ from this we have
$\displaystyle r^2 = \left( 1 + \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2 = 1 + \sqrt{2} + \frac{1}{2} + \frac{1}{2} = 2 + \sqrt{2} \implies r = \sqrt{2 + \sqrt{2}} $
and
$\displaystyle \tan \theta = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1 \implies \theta = \arctan(\sqrt{2} - 1)$
So, $\displaystyle \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}} = \sqrt{re^{i\theta}} = \sqrt{r} \, e^{i\theta/2}$
Finally, $\displaystyle \sqrt{1+\sqrt{i}} = \sqrt[4]{2 + \sqrt{2}} \, \left( \cos\left( \frac{\arctan(\sqrt{2} - 1)}{2} \right) + i\sin\left( \frac{\arctan(\sqrt{2} - 1)}{2} \right) \right)$
We can also obtain cartesian form:
$\displaystyle a + ib = \sqrt{1+\sqrt{i}} = \sqrt{1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}}$ yields
$\displaystyle a^2 + b^2 +i(-2ab) = 1 + \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \implies a^2 + b^2 = 1 + \frac{\sqrt{2}}{2}$ and $\displaystyle -2ab = \frac{\sqrt{2}}{2}$. So we have
$\displaystyle (a-b)^2 = 1 + \sqrt{2} \,$ and $\displaystyle (a+b)^2 = 1 $ which gives $\displaystyle a = \frac{1 + \sqrt{1 + \sqrt{2}}}{2} $ and $\displaystyle b = \frac{-1 + \sqrt{1 + \sqrt{2}}}{2}$
| {
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"answer_count": 3,
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} |
Let $U_1, U_2, \ldots$ be a sequence of independent random variables with PDF $f(u) = 1, \;\; 0 < u < 1$. Let $U_1, U_2, \ldots$ be a sequence of independent random variables with PDF
$$f(u) = 1, \;\; 0 < u < 1$$
Find $E[N]$ when $N = \min\left\{n\mid \sum\limits_{i = 1}^n U_i > 1\right\}$.
Hint: For $x\in (0,1)$, define $N(x) = \min\Big\{n\mid \sum\limits_{i = 1}^n U_i > x\Big\}$.
I am absolutely lost on this question. Any help would be greatly appreciated.
| For $x\ge 0$, let $N(x)=\min\left\{n:\sum_{i=1}^n U_i>x\right\}$ as in the hint, so that $N=N(1)$. We want to calculate $$p(x,n)=\Bbb P\big[N(x)=n\big]$$ when $n$ is a positive integer and $0\le x\le 1$.
For a positive integer $k$, let $S_k=U_1+U_2+\ldots+U_k$ with probability density $f_k$. For $0\le x\le 1$, observe that
$$f_k(x)=\frac{d}{dx}\int_0^x\int_0^{x-u_1}\ldots \int_0^{x-\sum_{j=1}^{k-2}u_j}\int_0^{x-\sum_{j=1}^{k-1}u_j}du_k\ du_{k-1}\ \ldots\ du_{2}\ du_1.$$
So $$f_k(x)=\frac{d}{dx}\left(\frac{x^k}{k!}\right)=\frac{x^{k-1}}{(k-1)!}$$ for $0\le x\le 1$.
If $N(x)=n$, then $S_{n-1}\le x<S_n$. We claim that $p(n,x)=\frac{x^{n-1}}{(n-1)!}-\frac{x^n}{n!}$ for every positive integer $n$. If $n=1$, then clearly $S_1=U_1$, so $$p(x,1)=\Bbb{P}[S_1>x]=\Bbb{P}[U_1>x]=1-x=\frac{1}{(1-1)!}-\frac{x^1}{1!}.$$ We now assume that $n>1$. Therefore
$$p(x,n)=\int_0^x \Bbb P\big[U_n>x-s\big]f_{n-1}(s)ds=\int_0^x\big(1-(x-s)\big)f_{n-1}(s)ds.$$
So
\begin{align}p(x,n)&=\int_0^x(1-x+s)\frac{s^{n-2}}{(n-2)!}ds=(1-x)\frac{x^{n-1}}{(n-1)!}+\frac{x^{n}}{n\cdot (n-2)!}\\&=\frac{(n-x)x^{n-1}}{n!}=\frac{x^{n-1}}{(n-1)!}-\frac{x^n}{n!}.\end{align}
(It should be verified that $\sum_{n=1}^\infty p(x,n)=1$.)
Specifically
$$\Bbb{P}[N=n]=p(1,n)=\frac1{(n-1)!}-\frac1{n!}=\frac{n-1}{n!}.$$
It follows that
$$\Bbb{E}\big[N(x)\big]=\sum_{n=1}^\infty n\Bbb{P}\big[N(x)=n\big]=\sum_{n=1}^\infty n\left(\frac{x^{n-1}}{(n-1)!}-\frac{x^n}{n!}\right)=e^x,$$
so $\Bbb{E}[N]=e$.
Write $p'(x,k)$ for $\frac{d}{dx}p(x,k)$. For $x>1$, we can show that
$$p(x,n)=-\int_{x-1}^x (1-x+s)\sum_{r=1}^{n-1}p'(s,r) ds.$$
That is, we have a recursion
$$p(x,n)=\sum_{r=1}^{n-1}\left(- p(x,r)+\int_{x-1}^x p(s,r)ds\right).$$
If $P(x,n)=\sum_{j=1}^np(x,n)=\Bbb{P}\big[N(x)\le n\big]$, then we get
$$P(x,n)=\int_{x-1}^x P(s,n-1) ds.$$
Since
$$P(x,1)=p(x,1)=\left\{\begin{array}{ll}1-x&\text{if }0\le x< 1\\
0&\text{if }x\ge1,
\end{array}\right.$$
and $$P(x,n)=1-\frac{x^n}{n!}$$ for $0\le x\le 1$, we can in principle find all $P(x,n)$ and $p(x,n)$ for every positive integer $n$ at any $x\ge 0$. For example,
$$P(x,2)=\left\{\begin{array}{ll}1-\frac{x^2}{2}&\text{if }0\le x<1\\
2-2x+\frac{x^2}{2}&\text{if }1\le x<2\\
0&\text{if }x\ge 2,
\end{array}\right.$$
$$p(x,2)=\left\{\begin{array}{ll}x-\frac{x^2}{2}&\text{if }0\le x<1\\
2-2x+\frac{x^2}{2}&\text{if }1\le x<2\\
0&\text{if }x\ge 2,
\end{array}\right.$$
$$P(x,3)=\left\{\begin{array}{ll}1-\frac{x^3}{6}&\text{if }0\le x<1\\
\frac{1}{2}+\frac{3x}{2}-\frac{3x^2}{2}+\frac{x^3}{3}&\text{if }1\le x<2\\
\frac{9}{2}-\frac{9x}{2}+\frac{3x^2}{2}-\frac{x^3}{6}&\text{if }2\le x<3\\
0&\text{if }x\ge 3,
\end{array}\right.$$
$$p(x,3)=\left\{\begin{array}{ll}\frac{x^2}{2}-\frac{x^3}{6}&\text{if }0\le x<1\\
-\frac{3}{2}+\frac{7x}{2}-2x^2+\frac{x^3}{3}&\text{if }1\le x<2\\
\frac{9}{2}-\frac{9x}{2}+\frac{3x^2}{2}-\frac{x^3}{6}&2\le x<3\\
0&\text{if }x\ge 3.
\end{array}\right.$$ I am not seeing a pattern to be able to find $P(x,n)$ or $p(x,n)$ for arbitrary $x$ and $n$, except: $$P(x,n)=p(x,n)=\frac{(n-x)^n}{n!}$$ if $n-1\le x\le n$, and trivially $P(x,n)=p(x,n)=0$ when $x>n$. Also, we have $$P(x,n)+P(n-x,n)=1$$ when $0\le x\le n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568937",
"timestamp": "2023-03-29T00:00:00",
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Multinomial theorem with imposed conditions
The number of ways in which 12 identical balls can be grouped in four marked non-empty sets $P, Q, R, S$ such that $n(P) < n(Q)$ is?
The answer to the above problem is the number of positive integral solutions of the expression $$P + Q + R + S = 12$$
when $P<Q$. I know that with no conditions imposed, the number of solutions is $^{12-1}C_{4-1} = ^{11}C_3$. How would one account for the condition when $P<Q$? Please explain the concept in detail in addition to providing a solution to the problem (which is given just as an example).
| Let $a,b,c,d\ge0$. Then set $|P|=a+1$, $|Q|=a+b+2$, $|R|=c+1$, $|S|=d+1$ where $2a+b+c+d+5=12$. We want to count the non-negative solutions to
$$
2a+b+c+d=7\tag1
$$
Considering the coefficients of $x^n$ in
$$
\overbrace{\left(1+x^2+x^4+x^6+\dots\right)\vphantom{{x^2}^3}}^a\overbrace{\left(1+x+x^2+x^3+\dots\right)^3}^{b,c,d}\tag2
$$
$(1)$ gives a generating function of
$$
\begin{align}
&\frac1{1-x^2}\left(\frac1{1-x}\right)^3\\
&=\frac1{1+x}\left(\frac1{1-x}\right)^4\\
&=\frac12\left(\frac1{1-x}+\frac1{1+x}\right)\left(\frac1{1-x}\right)^3\\
&=\frac12\left(\frac1{1-x}\right)^4+\frac14\left(\frac1{1-x}+\frac1{1+x}\right)\left(\frac1{1-x}\right)^2\\
&=\frac12\left(\frac1{1-x}\right)^4+\frac14\left(\frac1{1-x}\right)^3+\frac18\left(\frac1{1-x}+\frac1{1+x}\right)\frac1{1-x}\\
&=\frac12\left(\frac1{1-x}\right)^4+\frac14\left(\frac1{1-x}\right)^3+\frac18\left(\frac1{1-x}\right)^2+\frac1{16}\left(\frac1{1-x}+\frac1{1+x}\right)\tag3
\end{align}
$$
Therefore,
$$
\begin{align}
&\left[x^n\right]\frac1{1-x^2}\left(\frac1{1-x}\right)^3\\[3pt]
&=\textstyle\frac12(-1)^n\binom{-4}{n}+\frac14(-1)^n\binom{-3}{n}+\frac18(-1)^n\binom{-2}{n}+\frac1{16}(-1)^n\binom{-1}{n}+\frac1{16}\binom{-1}{n}\\[3pt]
&=\frac12\binom{n+3}{n}+\frac14\binom{n+2}{n}+\frac18\binom{n+1}{n}+\frac1{16}+\frac1{16}(-1)^n\\
&=\frac{4(n+3)(n+2)(n+1)+6(n+2)(n+1)+6(n+1)+3+3(-1)^n}{48}\\
&=\frac{(4n+14)(n+3)(n+1)+3\left(1+(-1)^n\right)}{48}\tag4
\end{align}
$$
For $n=7$, we get
$$
\frac{42\cdot10\cdot8}{48}=70\tag5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3571211",
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} |
$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$ Can I ask how to solve this type of equation:
$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)+\log_{zx}\left(\frac{y^2+4}{4\sqrt{zx}}\right)+\log_{xy}\left(\frac{z^2+4}{4\sqrt{xy}}\right)=0$$
It is given that $x,y,z>1$. Which properties of the logarithm have to be used?
I know that $\log_a b=\log a/\log b\to \log_{yz}((x^2+4)/(4\sqrt{yz}))=\log ((x^2+4)/(4\sqrt{yz}))/\log xy$
and $\log_a b/c=\log_a b/\log_a c\to \log_{yz}((x^2+4)/(4\sqrt{yz}))=\log_{yz} (x^2+4)/\log_{yz} (4\sqrt{yz})$
And how to solve this type of equation in general?
| Notice that we can use the following property $\log_a \frac{b}{c}=\log_a b-\log_a c$ to get:
$$\log_{yz} \left(\frac{x^2+4}{4\sqrt{yz}}\right)=\log_{yz}\frac{x^2+4}{4}-\log_{yz}\sqrt{yz}=\log_{yz}\frac{x^2+4}{4}-\frac{1}{2}$$
The equation can, thus, be written as:
$$\log_{yz}\frac{x^2+4}{4}+\log_{zx}\frac{y^2+4}{4}+\log_{xy}\frac{z^2+4}{4}=\frac{3}{2}$$
Notice that for any real number $a$, we have:
$$\frac{a^2+4}{4}\geq a \Leftrightarrow (a-2)^2\geq 0$$
with equality only if $a=2$. Therefore:
$$\log_{yz}\left(\frac{x^2+4}{4}\right)\ge \log_{yz} x=\frac{\ln x}{\ln y+\ln z}$$
Summing up, we arrive at:
$$\frac{3}{2}\geq \frac{\ln x}{\ln y+\ln z}+\frac{\ln y}{\ln x+\ln z}+\frac{\ln z}{\ln x+\ln y}$$
If we let $a=\ln x, b=\ln y, c=\ln z$, since $x,y,z>1$, we have $a,b,c>0$ and:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq \frac{3}{2}$$
However, Nesbitt's inequality states that for any positive real numbers $a,b,c$, we must have:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}$$
with equality only if $a=b=c$. This implies immediately that the only solution of the equation is $x=y=z=2$.
Note: As far as I know, there is no standard way to solve this sort of multivariable logarithm equations.
Most of the time, you have to use inequalities and show that the equation is a particular equality case. In fact, it given that $x,y,z>1$ so that the logarithm will be positive ($\log_a b$ is positive if $a,b>1$ or $a,b<1$).
| {
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Finding the joint distribution of two dependent random variables given conditions
Suppose $X \sim U(0,1)$ and $Y|X=x \sim U(0,1-x)$. Find the joint
distribution of $(X,Y)$ given that $X\le \frac{1}{2}$, $Y\le
\frac{1}{2}.$
My attempt:
The pdf of $X$ is $f_{X}(x) = 1$ for $x \in (0,1)$ and the pdf of $Y|X=x$ is $f_{Y|X}(y|x) = \frac{1}{1-x}$ for some $x\in (0,1)$ and $y\in (0,1-x)$.
Hence the (unconditional) joint density function should be $f_{XY}(x,y) = f_{Y|X}(y|x)f_X(x) = \frac{1}{1-x} $ where $x \in (0,1)$ and $y\in (0,1-x)$.
To find the distribution of $(X,Y)$ given the constraint let $A = \{(X,Y): \quad X\le \frac{1}{2} \cap Y\le \frac{1}{2} \}$.
Then, what I am trying to find is (using the CDF method) $$ \begin{align}\mathbb P (X\le x,Y\le y\vert (X,Y)\in A) &= \frac {\mathbb P\biggl(X\le x, Y\le y, X\le \frac 12, Y\le \frac {1}{2}\mathstrut \mathstrut \biggr) }{\mathbb P\Bigl(X\le \frac {1}{2},Y\le \frac 12\mathstrut \mathstrut \Bigr) }\mathstrut \\ &= \frac {\mathbb P\Bigl(X\le \text{min}\Bigl(x, \frac {1}{2}\mathstrut \Bigr), Y\le \text{min}(y, \frac {1}{2})\Bigr) }{\mathbb P(X\le \frac 12, Y\le \frac 12\mathstrut \mathstrut ) }\mathstrut \end{align} $$
at this point I'm not sure how to continue. For $(x,y) \in [0, \frac{1}{2}]^2$ the numerator becomes the (unconditional) CDF of just $(X,Y)$ but for e.g. $x > \frac{1}{2}$ and $y < \frac{1}{2}$ I'm not sure how to identify the distribution/simplify the expression.
| \begin{align}
P(Y\le y, X\le x)&= \int_{-\infty }^x P\left(Y\le y\ |\,X=t\right)f_X(t)dt\\
&=\cases{\displaystyle 0& if $\ x\le0\ $or$\ y\le0$\\
\displaystyle\int_0^x\frac{y}{1-t}dt&if $\ 0<x\le1-y,y< 1$\\
\displaystyle\int_0^{1-y} \frac{y}{1-t}dt+\int_{1-y}^xdt&if $\ 0<y<1, 1-y<x<1$\\
\displaystyle\int_0^{1-y} \frac{y}{1-t}dt+\int_{1-y}^1dt&if $\ 1\le x, 0<y<1 $\\
\displaystyle\int_0^xdt&if $\ 0<x<1, 1\le y$\\
\displaystyle 1&if $\ 1\le x,y$
}\\
&=\cases{
\displaystyle 0&if $\ x\le0\ $or$\ y\le0$\\
-y\ln(1-x)& if $\ 0<x\le1-y,y< 1$\\
-y\ln(y)+ x+y-1 &if $\ 0<y<1, 1-y<x<1$\\
-y\ln(y)+y&if $\ 1\le x, 0\le y<1$\\
x&if $\ 0<x<1, 1\le y$\\
1&if $\ 1\le x,y$}
\end{align}
Therefore $\ P\left(Y\le\frac{1}{2}, X\le\frac{1}{2}\right)=\frac{\ln(2)}{2}\ $, and
\begin{align}
P\left(X\le x, Y\le y\left|\ Y\le\frac{1}{2}, X\le\frac{1}{2}\right.\right)&=\frac{P\left(X\le\min\left(x,\frac{1}{2}\right), Y\le\min\left(y,\frac{1}{2}\right)\right)}{P \left(Y\le\frac{1}{2}, X\le\frac{1}{2}\right)}\\
&=\cases{
0&if $\ x\le0\ $ or $\ y\le 0\ $\\
\displaystyle-\frac{2y\ln(1-x)}{\ln(2)}&if $\ 0<x,y\le\frac{1}{2}, $\\
2y&if $\ \frac{1}{2}<x, 0<y\le\frac{1}{2} $\\
\displaystyle-\frac{\ln(1-x)}{\ln(2)}&if $\ 0<x\le \frac{1}{2}, \frac{1}{2}<y $\\
1&if $\ \frac{1}{2}<x,y$
}
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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At least one even number among $\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$
For any positive integer $n$, prove that the set
$$\{ \lfloor 2^{n}\sqrt{2} \rfloor, \lfloor 2^{n+1}\sqrt{2} \rfloor,..., \lfloor 2^{2n}\sqrt{2} \rfloor \}$$
contains at least one even number.
I tried to prove this question by contradiction, assuming that each element is odd. There exist the positive integers $k_1, k_2, ..., k_{n+1}$ such that
$$2k_1-1<2^n\sqrt{2}<2k_1$$
$$2k_2-1<2^{n+1}\sqrt{2}<2k_2$$
$$...$$
$$2k_{n+1}-1<2^{2n}\sqrt{2}<2k_{n+1}$$
But i can't find a contradiction among these inequalities.
| For the sake of contradiction assume that each element in the set is odd. Then, for some $m \geq 1$, we have:
$$2m-1 < 2^n\sqrt{2}<2m$$
and multiplying by $2$:
$$4m-2 < 2^{n+1}\sqrt{2}<4m$$
However, since $\lfloor 2^{n+1}\sqrt{2}\rfloor$ is odd, then
$$4m-1<2^{n+1}\sqrt{2}<4m$$
Repeating the process
$$2^{n+1}m-1<2^{2n}\sqrt{2}<2^{n+1}m\Rightarrow \frac{1}{2^{n+1}}>m-2^{n-1}\sqrt{2}=\frac{m^2-2^{2n-1}}{m+2^{n-1}\sqrt{2}}$$
Also, since $2m>2^n\sqrt{2}\Rightarrow m^2>2^{2n-1}\Rightarrow m^2\geq 2^{2n-1}-1$. Thus:
$$\frac{1}{2^{n+1}}>\frac{m^2-2^{2n-1}}{m+2^{n-1}\sqrt{2}}\geq \frac{1}{m+2^{n-1}\sqrt{2}}$$
Therefore:
$$2^{n+1}<m+2^{n-1}\sqrt{2}<\frac{2^{n}\sqrt{2}+1}{2}+2^{n-1}\sqrt{2}=2^n\sqrt{2}+\frac{1}{2}<2^n\sqrt{2}+\frac{2^n}{2}$$
and thus
$$2<\sqrt{2}+\frac{1}{2}$$
which is a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573684",
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"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Find the minimum value of $x^2+y^2$, where $x,y$ are nonnegative integers and $x+y=k$. Question: Let $k$ be a fixed odd positive integer. Find the minimum value of $x^2+y^2$, where $x,y$ are nonnegative integers and $x+y=k$.
My approach: After trying some examples I can conjecture that, the minimum value of $x^2+y^2$ is attained at $$x=\left\lceil \frac{k}{2}\right\rceil \text{and } y=\left\lfloor\frac{k}{2}\right\rfloor \text{and equivalently at } x=\left\lfloor\frac{k}{2}\right\rfloor \text{and }y=\left\lceil \frac{k}{2}\right\rceil.$$ This also implies that the minimum value of $x^2+y^2=\left\lceil \frac{k}{2}\right\rceil^2+\left\lfloor\frac{k}{2}\right\rfloor^2.$
But, how to prove the same?
Also, since $x+y=k$, this implies that $(x+y)^2=k^2\implies x^2+y^2=k^2-2xy.$ Therefore, we need to maximize $xy$ in order to minimize $x^2+y^2$.
But, again this is leading me nowhere.
| AM-GM: $x+y=k \geqslant 2 \sqrt{xy}$
${xy} \leqslant \frac {k^2} {4} $
$x^2+y^2=k^2-2xy\geqslant \frac {k^2} {2}$
If $k$ is even, then $minimum\; x^2+y^2=\frac {k^2} {2}$
If $k$ is odd, then
$minimum\; x^2+y^2=\frac {k^2+1} {2}$
| {
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How to show that $b_n(n) \neq 0$ for $b_n(x) = -\frac{16 x}{n} b_{n-1}(x) - \frac{48 x}{n} b_{n-2}(x)$? Lets define the recursive polynomial:
$b_0(x) = 1$
$b_1(x) = -16 x$
$b_n(x) = -\frac{16 x}{n} \left( b_{n-1}(x) + 3 b_{n-2}(x) \right) = -\frac{16 x}{n} b_{n-1}(x) - \frac{48 x}{n} b_{n-2}(x)$ ; with $x \in \mathbb{R}$ and $n \in \mathbb{N}$
Question 1: How can I show that $b_n(n) \neq 0$ ?
I can show that for the following recurrence relation $a_n \neq 0$.
$a_0 = 1$
$a_1 = -16$
$a_n = -16 \left( a_{n-1} + 3 a_{n-2} \right) = -16 a_{n-1} - 48 a_{n-2}$
The characteristic polynomial will be:
$\lambda^2 + 16\lambda + 48 = 0$
with the solutions $\lambda_1 = -12$ and $\lambda_2 = -4$.
Therefore I am able to express $a_n$ as:
$a_n = A_1 \lambda_1^n + A_2 \lambda_2^n = \frac{3}{2}(-12)^n - \frac{1}{2}(-4)^n = (-1)^n \frac{12^n - 4^n}{2} \neq 0$
My motivation was numerical evidence and that:
$b_n(n) = -16 \left( b_{n-1}(n) + 3 b_{n-2}(n) \right)$ which is like $a_n$ but
$b_{n-1}(n) = -16 \frac{n-1}{n} \left( b_{n-2}(n) + 3 b_{n-3}(n) \right)$ for example is only similar to $a_{n-1}$ in the long run when $n \to \infty$.
Question 2: Is it possible in any way to proof $b_n(n) \neq 0$ by using $a_n$ ?
| If we fix $n$ and define $a_k=b_k(n)$, then indeed we obtain a recursion
$$a_0=1,\quad a_1=-16n,\quad a_k=-16a_{k-1}-48a_{k-2}. $$
From here we infer that
$$\tag1 a_k=A(-12)^k+B(-4)^k$$
for suitable $A,B$.
The desired result is that this makes $a_n\ne0$.
Clearly, $a_0\ne 0$. Also, when $n=1$, we have $a_1=-16\ne 0$, so that we need only consider the cases with $n\ge 2$.
From $(1)$ and the given values for $a_0,a_1$, we infer $$\tag2A+B=1,\quad -12A-4B=-16n,$$
Or more explicitely
$$ \tag3A=2n-\frac12,\quad B=\frac32-2n.$$
Note that $a_n=0$ implies
$$\frac BA=-3^n\le -9, $$whereas $(3)$ leads to
$$\frac BA=\frac{\frac32-2n}{2n-\frac12}=-1+\frac1{2n-\frac12}>-1. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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For how many integers $n$ is $n^6+n^4+1$ a perfect square? QUESTION
For how many integers $n$ is $n^6+n^4+1$ a perfect square?
I am completely blank on how to start. Could anyone please provide tricks on how to get a start on such questions?
Thanks for any answers!
| $n^6 + n^4 + 1 = n^6 + n*n^3 + 1=n^6 + 2\frac n2*n^3 + \frac {n^2}4 +(1-\frac {n^2}4) =(n^3 + \frac n2)^2 + (1-\frac {n^2}4)$
So if $n\ge 2$ then $n^6 + n^4 + 1 \le (n^3 + \frac n2)^2$ with equality holding only if $n = 2$.
And $(n^3 +\frac {n}2 - 1)^2 =(n^3 +\frac {n-2}2)^2 = n^6 + (n-2)n^3 + \frac {n^2-4n -4}4=n^6 + n^4 - 2n^3 + \frac {n^2}4 -n -1<n^6+n^4-n-1 < n^6 + n^4 + 1$.
So if $n \ge 2$ then $(n^3 +\frac {n-2}2)^2 < n^6 + n^4 + 1 \le (n^3+\frac {n}2)^2$ (with equality holding only if $n =2$).
So if $n^6 + n^4 + 1$ is a perfect square then either $n=2$ and $n^6+n^4 + 1 = (n^3 +1)^2 = 9^2 = 81$....
or...
$n$ is odd and $n^6 + n^4 + 1 = (n^3 + \frac {n-1}2)^2$... which would mean $n^6 + n^4 + 1 = n^6 + (n-1)n^3 + (\frac{n-1}2)^2$ or in other words:
$n^3 -(\frac {n-1}2)^2 + 1=0$
$4n^3 - n^2 + 2n +3 = 0$
By rational root theorem the only integer larger than $2$ that could work would be $3$ and ... it doesn't.
Now $n^6 + n^6 + 1= (-n)^6 + (-n)^6 + 1$ so if $n$ is a solution if and only if $-n$ is a solution and $n = \pm 2$ is the only solution where $|n| \ge 2$.
So just need to check if $n = \pm 1, 0$.
$n=0$ yields $n^6 + n^4 + 1 = 0 = 1^2$ but $n=\pm 1$ yields $n^6 + n^4 + 1 = 3$ which is not a perfect square.
So $n=\pm 2$ and $n =0$ are the only three integers that yield perfect squares.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to obtain the asymptote formula of $\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx$? I want to obtain the formula in the form as follows:
$$ \int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx=A+\frac{B}{n}+\frac{C}{n^2}+\frac{D}{n^3}+o\left(\frac{1}{n^3}\right).$$
At least, it holds that $$\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx = \int_{0}^{\pi/2}\sqrt[n]{\sin x}\,dx = \int_{0}^{1}\frac{u^{1/n}}{\sqrt{1-u^2}}\,du = \frac{1}{2}\int_{0}^{1}(1-t)^{-\frac{1}{2}}t^{-\frac{1}{2}+\frac{1}{2n}}\,dt=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)}{2\,\Gamma\left(1+\frac{1}{2n}\right)},$$
but how to go on with this?
| You properly wrote
$$I_n=\int_{0}^{\pi/2}\sqrt[n]{\cos x}\,dx =\frac{\sqrt \pi}2\frac{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)}{\Gamma\left(1+\frac{1}{2n}\right)}$$ Now, take logarithms for the gamma functions, useStirling approximation of the gamma function and continue with Taylor expansions to get
$$\log \left(\frac{\Gamma \left(\frac{1}{2}+\frac{1}{2n}\right)}{\Gamma \left(1+\frac{1}{2
n}\right)}\right)=\frac{\log (\pi )}{2}-\frac{\log (2)}{n}+\frac{\pi ^2}{24 n^2}-\frac{\zeta (3)}{4
n^3}+O\left(\frac{1}{n^4}\right)$$
Take the exponential
$$\frac{\Gamma \left(\frac{1}{2}+\frac{1}{2n}\right)}{\Gamma \left(1+\frac{1}{2
n}\right)}=\sqrt{\pi }-\frac{\sqrt{\pi } \log (2)}{n}+\frac{\sqrt{\pi } \left(\pi ^2+12 \log
^2(2)\right)}{24 n^2}-\frac{\sqrt{\pi } \left(6 \zeta (3)+4 \log ^3(2)+\pi ^2
\log (2)\right)}{24 n^3}+O\left(\frac{1}{n^4}\right)$$
$$I_n=\frac{\pi }{2}-\frac{\pi \log (2)}{2 n}+\frac{\pi ^3+12 \pi \log ^2(2)}{48
n^2}-\frac{6 \pi \zeta (3)+4 \pi \log ^3(2)+\pi ^3 \log (2)}{48
n^3}+O\left(\frac{1}{n^4}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify $(1+x^2 )^{1/2}-(1-x^2)^{1/2}$ I need to simplify the following expression in a way that introduces minimal floating point cancellation errors.
$$(1+x^2 )^{\frac{1}{2}}-(1-x^2 )^{\frac{1}{2}}$$
The errors accumulate when numbers close together are subtracted from each other. I get
$$\sqrt{2}\left[1-\left(1-x^4 \right)^{1/2} \right]^{1/2}$$
But don't see how this helps, perhaps there is a better formula?
| I would try this:
$$\begin{align}(1+x^2 )^{\frac{1}{2}}-(1-x^2 )^{\frac{1}{2}}&=[(1+x^2 )^{\frac{1}{2}}-(1-x^2 )^{\frac{1}{2}}]\frac{(1+x^2 )^{\frac{1}{2}}+(1-x^2 )^{\frac{1}{2}}}{(1+x^2 )^{\frac{1}{2}}+(1-x^2 )^{\frac{1}{2}}}\\&=\frac{(1+x^2 )-(1-x^2 )}{(1+x^2 )^{\frac{1}{2}}+(1-x^2 )^{\frac{1}{2}}}\\&=\frac{2x^2}{(1+x^2 )^{\frac{1}{2}}+(1-x^2 )^{\frac{1}{2}}}\end{align}$$
| {
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The integral: $\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$ The integral: $$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$
has been encountered today while solving a longlish problem at MSE. The question here is: How would one evaluate it?
Addendum
For an interesting use of this integral see my Answer to: Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$
| $$I=\int_{0}^{\pi/2} \frac{\sin x \cos^5x}{(1-2\sin^2 x \cos^2 x)^2} dx$$
Next using $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx,$
we get $$I=\int_{0}^{\pi/2} \frac{\sin^5 x \cos x}{(1-2\sin^2 x \cos^2 x)^2} dx$$ Adding the two integrals, we get
So $$2I=\int_{0}^{\pi/2} \frac{\sin x \cos x(\sin^4 x+ \cos^4 x)}{(1-2\sin^2 x \cos^2 x)^2}=\int_{0}^{\pi/2} \frac{\sin x \cos x}{(1-2\sin^2 x \cos^2 x)}dx=\int_{0}^{\pi/2}\frac{1}{2} \frac{\sin 2x dx}{1+\cos^2 2x}$$
$$ \text{domain halved,}~\implies 2 I=\int_{0}^{\pi/4}\frac{2\sin 2x dx}{1+\cos^2 2x}=-\int_{1}^{0}\frac{dt}{1+t^2}=\frac{\pi}{4} \implies I=\frac{\pi} {8} $$
Lastly we have used $\cos 2x=t.$
| {
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The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all real values for $x\in R$ are: => The value of parameter $a$ for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$
takes all real values for $x\in R$ are:
My question is why we need to validate end points i.e. $1,7$ (Refer the last part of my attempt)
My attempt is as follows:-
$$y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$$
$$ya+3yx-4yx^2=ax^2+3x-4$$
$$x^2(-4y-a)+x(3y-3)+ya+4=0$$
As $x$ can be any real, so $D\ge0$
$$9y^2+9-18y-4(ya+4)(-4y-a)\ge0$$
$$9y^2+9-18y+4(4y^2a+ya^2+16y+4a)\ge0$$
$$y^2(9+16a)+y(4a^2+64-18)+9+16a\ge0$$
$$y^2(9+16a)+y(4a^2+46)+9+16a\ge0$$
As range is $R$, so discriminant of quadratic in $y$ should be less than equal to zero
$$4(2a^2+23)^2-4(9+16a)^2\le0$$
$$(2a^2+23-9-16a)(2a^2+23+9+16a)\le0$$
$$(2a^2-16a+14)(2a^2+16a+32)\le0$$
$$(a^2-8a+7)(a^2+8a+16)\le0$$
$$a\in[1,7]$$
But in such type of questions, we always check at endpoints like here we need to check at $a=1$ and $a=7$. But I don't understand what is so special about endpoints.
From the above calculation I can only say at $a=1,7$ discriminant of quadratic in $y$ is zero, but what is so special about this. Please help me in this.
| When $a=1$, $\displaystyle y=\frac{x^2+3x-4}{1+3x-4x^2}=\frac{-x-4}{1+4x}\ne-\dfrac14$ for all $x\in\mathbb{R}$.
When $a=7$, $\displaystyle y=\frac{7x^2+3x-4}{7+3x-4x^2}=\frac{7x-4}{7-4x}\ne-\dfrac74$ for all $x\in\mathbb{R}$.
| {
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show that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$ show that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$
Here is the algebric proof:
We have $a_n=\sqrt{n+1}-\sqrt{n}$, and we want to show that $\lim a_n=0$.
$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
So, when $n\to\infty$, we get $\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$.
Question: I am wondering does episilon-delta method work here as an alternative proof?
| Observe that $\sqrt{n+1} > \sqrt{n}$, so, $\sqrt{n+1} + \sqrt{n} > 2\sqrt{n}$ and then
$$\frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}} \ .$$
Now, let $\varepsilon>0$. Since $4\varepsilon^2$ is a positive real number, by the Archimedian property of $\mathbb{R}$ there exists $N \in \mathbb{N}$ such that
$$0< \frac{1}{N} < 4\varepsilon^2.$$
It follows that for $n \geq N$, $2\sqrt{n} \geq 2\sqrt{N}$ and then
$$a_n = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \dfrac{1}{2\sqrt{n}} \leq \dfrac{1}{2\sqrt{N}} < \varepsilon.$$
Since $\varepsilon$ was arbitrary, we have shown that $a_n$ can make arbitrarily small for sufficiently large $n$. Hence, $a_n \to 0$ as $n \to \infty$.
| {
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Particular generators of GL(2,R) I want to prove that $$ \text{GL}_{2}( \mathbb{R} ) = \left\langle \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & a \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}; \ a \in \mathbb{R} \setminus \{0\} \right\rangle $$ I know that $$ \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}^{k} = \begin{pmatrix} a^{k} & 0 \\ 0 & 1\end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & a \end{pmatrix}^{l} = \begin{pmatrix} 1 & 0 \\ 0 & a^{l} \end{pmatrix},$$$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{m} = \begin{pmatrix} 1 & m \\ 0 & 1\end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}^{h} = \begin{pmatrix} 1 & 0 \\ h & 1\end{pmatrix}. $$ Since the determinant of these matrices isn't zero $ \forall k,l,m,h \in \mathbb{Z}, a \neq 0$, the determinant of a product of any permutation of these matrices isn't zero. Therefore this generates invertible matrices and for $k=l=m=h=0$ we get identity matrix. Is there an efficient way to prove that this generates all invertible two by two matrices over $ \mathbb{R}$ other than checking all possible permutations?
| As noted by @LouisHainaut, these four matrices enable you to do the elementary row operations. Since it is possible to get to and from the identity matrix to any invertible matrix by doing elementary row operations, the result follows.
The first two allow you to multiply a row by a scalar. The last two, to add a multiple of a row to another.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of $\frac{a_{n+1}}{a_n}$, where $|a_{n+1}a_{n-1} - a_n^2| = 1 $ Let $(a_n)$ a non-decreasing sequence of positive real numbers such that $\lim a_n = \infty$ and $|a_{n+1}a_{n-1} - a_n^2| = 1 $. Prove that the sequence $(\frac{a_{n+1}}{a_n})$ converges.
My little ´´progress´´:
$|a_{n+1}a_{n-1} - a_n^2| = 1 $.
$\left|\dfrac{a_{n+1}a_{n-1}}{a_n^2} - 1\right| = \dfrac{1}{a_n^2} \to 0$
$\dfrac{a_{n+1}a_{n-1}}{a_n^2} \to 1$
$\dfrac{a_{n+1}}{a_n}\dfrac{a_{n-1}}{a_n} \to 1$
| We separate two cases.
Case 1: for every $n \geq 0$, we have $a_{n + 1} - a_n \leq 1$.
In this case, since the sequence $(a_n)_n$ tends to infinity by assumption, we have $\lim\limits_{n\rightarrow\infty}\frac{a_{n + 1} - a_n}{a_n} = 0$, and it follows that $\lim\limits_{n\rightarrow\infty}\frac{a_{n + 1}}{a_n} = 1$.
Case 2: there exists $m\geq 0$ such that $a_{m + 1} - a_m > 1$.
We then have $a_{m + 2} \geq \frac{a_{m + 1}^2 - 1}{a_m} > \frac{a_{m + 1}^2 - 1}{a_{m + 1} - 1} = a_{m + 1} + 1$. Hence by induction, we have $a_{n + 1} - a_n > 1$ for all $n \geq m$. This implies $a_n > a_m + (n - m)$ for all $n > m$.
Without loss of generality, we may assume that $m = 0$ (otherwise just consider the sequence $(b_k)_k$ with $b_k = a_{m + k}$). We then have $a_n > n$ for all $n \geq 1$.
Now rewrite the equation in the question as $\frac{a_{n + 1}}{a_n} - \frac{a_n}{a_{n - 1}} = \pm \frac 1{a_{n - 1}a_n}$. Thus for any $0 < n < N$, we have: $$\frac{a_{N + 1}}{a_N} - \frac{a_{n + 1}}{a_n} = \pm \frac 1{a_na_{n + 1}}\pm\frac1{a_{n + 1}a_{n + 2}}\pm \dotsc\pm \frac1{a_{N - 1}a_N}.$$ Taking absolute value, we get: $$\left|\frac{a_{N + 1}}{a_N} - \frac{a_{n + 1}}{a_n} \right| \leq \sum_{n < k \leq N}\frac1{a_{k - 1}a_k} <\sum_{n < k \leq N}\frac1{(k - 1)k} < \frac 1 n.$$
This shows that the sequence $\left(\frac{a_{n + 1}}{a_n}\right)_n$ is a Cauchy sequence and hence converges.
| {
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Calculate $\mathbb{E}(X-Y\mid 2X+Y).$ if $X\sim N(0,a)$ and $Y\sim N(0,b)$
Question: Given that $X$ and $Y$ are two random variables satisfying $X\sim N(0,a)$ and $Y\sim N(0,b)$ for some $a,b>0$. Assume that $X$ and $Y$ have correlation $\rho.$
Calculate
$$\mathbb{E}(X-Y \mid 2X+Y).$$
I tried to use the fact that if $A$ and $B$ are independent, then $\mathbb{E}(A\mid B) = \mathbb{E}(A)$ and uncorrelated implies independence in jointly normal distribution.
So, I attempted to express $X-Y$ as a linear combination of $2X+Y$ and $Z$ where $\operatorname{Cov}(2X+Y,Z) = 0.$
But I am not able to do so.
Any hint is appreciated.
| By @Kavi Rama Murthy answer (and me in other answer)
$$E(X-Y|2X+Y)=A(2X+Y)$$
Now By the Projection property ,$E(X-Y|2X+Y)$ minimized
$$E(X-Y-g(2X+Y))^2$$ conditional-expectation-as-best-predictor
I want to find $A$ by minimizing $E(X-Y-A(2X+Y))^2$
$$E(X-Y-A(2X+Y))^2=E((1-2A)X-(1+A)Y)^2$$
$$=E((1-2A)X)^2+E((1+A)Y)^2
-2E((1-2A)X (1+A)Y)2$$
$$=(1-2A)^2E(X)^2+(1+A)^2E(Y)^2
-2(1-2A)(1+A)E(X Y)$$
$$=(1-2A)^2a+(1+A)^2b
-2(1-2A)(1+A)cou(X Y)$$
$$=(1-2A)^2a+(1+A)^2b
-2(1-2A)(1+A)\rho \sqrt{a}\sqrt{b}$$
by derivation $\frac{d}{dA}$ and equal to $0$
$$\frac{d}{dA} E((1-2A)X-(1+A)Y)^2=0$$
$$\Leftrightarrow$$
$$0=
-4(1-2A)a+2(1+A)b-2(-2)(1+A)\rho \sqrt{a}\sqrt{b}-2(1-2A)\rho \sqrt{a}\sqrt{b}$$
$$\Leftrightarrow$$
$$0=\bigg(
-4a+2b+4\rho \sqrt{a}\sqrt{b}-2\rho\sqrt{a}\sqrt{b}
\bigg)+\bigg(
8a+2b+4\rho \sqrt{a}\sqrt{b}+4\rho \sqrt{a}\sqrt{b}
\bigg)A$$
$$\Leftrightarrow$$
$$0=\bigg(
-4a+2b+2\rho \sqrt{a}\sqrt{b}
\bigg)+\bigg(
8a+2b+8\rho \sqrt{a}\sqrt{b}
\bigg)A$$
$$\Leftrightarrow$$
$$A=\frac{2a-b-\rho \sqrt{a}\sqrt{b}}{4a+b+4\rho \sqrt{a}\sqrt{b}}$$
| {
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Solutions of the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ I have seen the equation $x=\sqrt{2 + \sqrt{2 +\sqrt 2+.....}}$ in many places and the answer is $x=2$ which is obtained by making the substitution for $x$ inside the radical sign i.e $x=\sqrt{2+x}$ which renders the quadratic $x^2 -x -2=0$
having the solutions $x=-1$ and $x=2$ and we just ignore $x=-1$ since it is negative.
But my question is, we could also have made the substitution for $x$ under the second radical sign i.e. $x=\sqrt{2 + \sqrt{2 +x}}$ and get the bi-quadratic $x^4-4x^2-x+2=0$. Which has the solutions
$$x=-1,2, \frac{-1-\sqrt{5}}{2} , \frac{-1+\sqrt{5}}{2}$$ Or we could also have substituted anywhere else in the infinite radical expression and it would have given us some real roots. So how do we decide that $x=2$ is the correct solution?
| Note that if that expression defines a number $x$, then $x$ satisfies $x^2-x-2$. That $x$ satisfies some other relations is not relevant. That some other numbers satisfy those other relations is even less relevant; those other numbers do not satisfy $x^2-x-2$.
As an aside; to prove that $x=2$ it is also necessary to prove that
$$x=\sqrt{2 + \sqrt{2 +\sqrt{2+.....}}},$$
actually defines a number. This comes down to showing that a certain infinite sequence converges.
| {
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Prove $2^{29}$ has exactly 9 distinct digits There is a problem that say: The number $2^{29}$ has exactly 9 distinct digits. Which digit is missing?
It is an Olympiad problem and I see it solved by using remainder modulo 9.
$2^{29}=536870912$ interesting!
My Question: Can we find any mathematically way to show $2^{29}$ has exactly 9 distinct digits?
| We know that $2^{10}=1024$, so we can calculate the value of $2^{29}$ by hand as follows
$2^{30} = 1024^3 = (10^3 + 24)^3 = 10^9 + 3.24.10^6 + 3.24^2.10^3 + 24^3$
Now $24=3.8$, so
$24^2 = 9.64 = 576 \\
\Rightarrow 3.24^2 = 1,728\\
24^3 = 24.576 = 3.8.576 = 3.4,608 = 13,824 \\
\Rightarrow 1024^3 = 10^9 + 72.10^6 + 1,728.10^3 + 13,824 \\
= 10^9 + 73.10^6 + 741.10^3 + 824 = 1,073,741,824 \\
\displaystyle \Rightarrow 10^{29} = \frac{1,073,741,824}{2} = 536,870,912$
| {
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Finding a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$ So the task is to find a coefficient of $x^{57}$ in a polynomial $(x^2+x^7+x^9)^{20}$
I was wondering if there is a more intelligible and less exhausting strategy in finding the coefficient, other than saying that $(x^2+x^7+x^9)^{20}=((x^2+x^7)+x^9)^{20}$ and then working with binomial expansion.
| $$
\begin{align}
\left[x^{57}\right]\left(x^2+x^7+x^9\right)^{20}
&=\left[x^{57}\right]\sum_{\substack{a+b+c=20\\a,b,c\ge0}}\frac{20!}{a!\,b!\,c!}x^{2a+7b+9c}\tag1\\
&=\sum_{\substack{2a+7b+9c=57\\a+b+c=20\\a,b,c\ge0}}\frac{20!}{a!\,b!\,c!}\tag2\\
&=\sum_{\substack{5b+7c=17\\b,c\ge0}}\frac{20!}{(20-b-c)!\,b!\,c!}\tag3\\[3pt]
&=\frac{20!}{(20-2-1)!\,2!\,1!}\tag4\\[15pt]
&=3420\tag5
\end{align}
$$
Explanation:
$(1)$: apply the Multinomial Theorem
$(2)$: extract the coefficient of $x^{57}$
$(3)$: substitute $a=20-b-c$
$(4)$: the only non-negative solution to $5b+7c=17$ is $(b,c)=(2,1)$
$(5)$: evaluate
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Euclid's algorithm to find multiplicative inverses over a polynomial field I'm trying to find the multiplicative inverse of $\overline{x+1}$ over the field $\mathbb{F}_3[x]/(x^3 + 2x + 1)$. I know I need to use Euclid's algorithm to do so, but I keep running into some difficulties.
I let $f(x) = x^3 + 2x + 1$ and $g(x) = x+1$. Then I should be able to compute
$$f(x) = q_1(x)g(x) + r_1(x)$$
$$g(x) = q_2(x)r_1(x) + r_2(x)$$
$$\vdots$$
$$ r_{m-1}(x) = q_{m+2}(x)r_m(x)$$
and then back substitute through the algorithm to solve for $a(x), b(x)$ in
$$ a(x)g(x) + b(x)f(x) = 1.$$
My problem is likely elementary, but it has me confused: I cannot find $q_1$, $q_2$ to make what should probably be a rather trivial iteration of the algorithm work at all. If I was solving for, say, the multiplicative inverse of $\overline{x ^2}$, I could let $f(x) = (x)(x^2) + (2x+1)$ with $g(x) = x^2 = (2x+1)(2x+2)+1$. The division follows nicely from there. However, I can't find out where I'm going wrong for $\overline{x+1}$... What am I missing?
| You could use polynomial long division, which as discussed & shown in the linked Wikipedia article is done similarly to base $10$ long division except rather than using powers of $10$, you use powers of $x$ instead. Also, similar to what the long division method does, you can notice by adding & subtracting various terms and regrouping them that
$$\begin{equation}\begin{aligned}
x^3 + 2x + 1 & = x^3 + (x^2 - x^2) + (- x + x) + 2x + (3 - 3) + 1 \\
& = (x^3 + x^2) - (x^2 + x) + (3x + 3) - 2 \\
& = x^2(x + 1) - x(x + 1) + 3(x + 1) - 2 \\
& = (x^2 - x + 3)(x + 1) - 2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
You thus have
$$q_1(x) = x^2 - x + 3 \tag{2}\label{eq2A}$$
$$r_1(x) = -2 \tag{3}\label{eq3A}$$
Also, note a fast way to determine $r_1(x)$ is to use that
$$x + 1 \equiv 0 \pmod{x + 1} \implies x \equiv -1 \pmod{x + 1} \tag{4}\label{eq4A}$$
Thus, this means
$$x^3 + 2x + 1 \equiv (-1)^3 + 2(-1) + 1 \equiv -2 \pmod{x + 1} \tag{5}\label{eq5A}$$
| {
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There are two straight lines passing through the point A(2,0) which make an angle of 45 deg with the tangent at $A$...
There are two straight lines passing through the point $A(2,0)$ intersecting the tangent from $A$ of the circle $x^2+y^2+4x-6y-12=0$ under $45^{\circ}$. Find the equation of the circles with radius $3$ units each, centred on these straight lines at a distance of $5\sqrt 2$ from $A$
The equation of the tangent at $A$ is $4x-3y-8=0$
Doing a fair bit of calculation, the equations of the two straight lines are
$$x-7y-2=0$$
$$7x+y-14=0$$
How do I find the centre of the circles?
Two ways came to my find
The Symmetric form
The formula $$\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r$$
Solving equations
Let the centre be $(h,k)$
Then
$$(h-2)^2 + k^2=50$$
And solving this with
the two obtained lines
Both methods are horribly tedious to solve. I am sure there is a more efficient way to solve.
| Observe that the two straight lines intersect the circles in two points $B,C$. Now, thanks to ortogonality of straight lines, you have $B\hat A C = 90^\circ$. So the points $A,B,C$ form a part of the square inscribed in the circle: moreover the diagonal of this square is the diameter of the circle that is $10$. Thanks to Pythagorean theorem you have $\overline{AC} = \overline{AB} = 5\sqrt{2}$ so the points $B$ and $C$ are what you required in your question. To find $B$ and $C$ we can rotate the point $A$ by $90^\circ$ around $O=(-2,3)$ center of your circle. The rotation is given by the formula (center is $(a,b)$ and the angle $\alpha$):
\begin{cases}
x' = (x-a)\cos\alpha - (y-b)\sin\alpha +a\\
y' = (x-a)\sin\alpha + (y-b)\cos\alpha +b\\
\end{cases}
So you have ($\alpha = 90^\circ$, $O$ the center)
\begin{cases}
x' = - (y-3) -2\\
y' = (x+2) +3\\
\end{cases}
and applying it to $A$ you have $B= (1,7)$. Doing the same thing with $\alpha = -90^\circ$ you have $C=(-5,-1)$.
For the other two points we can use also rotation: take $B$ and $C$ and rotate them around $A$ for $180^\circ$. Again you can use the above formula that become ($A$ is the center of rotation and $\alpha = 180^\circ$)
\begin{cases}
x' = 4-x\\
y' = -y\\
\end{cases}
And you finally obtain $B'=(3,-7)$ and $C'=(9,1)$.
Summarizing the four points are $B=(1,7)$, $B'=(3,-7)$, $C=(-5,-1)$, $C'=(9,1)$ and we did not resolve any equation.
| {
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Is $\sum_{n=0}^\infty \frac{(x^2-5)^n}{2^n}$ a power series? The series $\sum_{n=0}^\infty \frac{(x^2-5)^n}{2^n}$ can be expressed as the following geometric series:
$$\sum_{n=0}^\infty \left(\frac{x^2-5}{2}\right)^n.$$
This series should converge if $$\left|\frac{x^2-5}{2}\right|<1.$$
This gives the possible values of $x$ to lie in the range $$(-\sqrt7,-\sqrt3) \cup (\sqrt3,\sqrt7).$$ This would mean that the above series is not a power series as a power series cannot have any discontinuities in its interval of convergence. However the series can be rewritten in the following form: $$\sum_{n=0}^\infty \frac{((x-\sqrt5)(x+\sqrt5))^n}{2^n}$$
which can further be rewritten as
$$\sum_{n=0}^\infty \frac{((x-\sqrt5)^2+2\sqrt5(x-\sqrt5))^n}{2^n}.$$
This final series can be rewritten in the form
$$\sum_{n=0}^\infty a_n(x-\sqrt5)^n$$
by using the binomial expansion. According to the definition, a power series is any series of the form $$\sum_{n=0}^\infty a_n(x-c)^n.$$
This would mean that the above series is a power series with center $\sqrt5$. Thus there appears to be a contradiction which I am not able to resolve.
| A power series is any series of the form
$$\sum_{n} a_{n} \, (x-b)^n$$
which leads to saying the series in question is not a power series.
The two forms
$$\sum_{n=0}^{\infty} \left(\frac{x^2 - a^2}{b}\right)^n \quad \text{and} \quad \sum_{n=0}^{\infty} \frac{1}{b^n} \, ( (x - a)^2 + 2 a \, (x-a) )^n$$
provide the same result seen as follows:
$$ \sum_{n=0}^{\infty} \left(\frac{x^2 - a^2}{b}\right)^n = \frac{b}{b + a^2 - x^2}$$
and
\begin{align}
\sum_{n=0}^{\infty} \frac{1}{b^n} \, ( (x - a)^2 + 2 a \, (x-a) )^n &= \sum_{n=0}^{\infty} \sum_{k=0}^{n} \binom{n}{k} \, \frac{(2 a)^k}{b^n} \, (x-a)^{2n-k} \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \binom{n+k}{k} \, \frac{(2 a)^k}{b^{n+k}} \, (x-a)^{2n+k} \\
&= \sum_{n=0}^{\infty} \frac{(x-a)^{2n}}{b^n} \, \sum_{k=0}^{\infty} \binom{n+k}{k} \, \frac{(2 a (x - a))^k}{b^k} \\
&= \sum_{n=0}^{\infty} \frac{(x-a)^{2n}}{b^n} \, \frac{b^{n+1}}{(b + 2 a^2 - 2 a x)^{n+1}} \\
&= \frac{b}{b + 2 a^2 - 2 a x} \, \sum_{n=0}^{\infty} \left(\frac{(x-a)^2}{b + 2 a^2 - 2 a x}\right)^n \\
&= \frac{b}{b + a^2 - x^2}.
\end{align}
This is an indicator that both forms do not fit the definition of a power series.
This form brings into question the definitions of double sums where the coefficients are not strictly constants.
In general:
For any power series, one of the following is true:
*
*The series converges only for $x=0$
*The series converges absolutely for all $x=x_{0}$
*The series converges absolutely for all $x$ in some finite open interval $(-R,R)$ and diverges if $x<-R$ or $x>R$. At the points $x=R$ and $x=-R$, the series may converge absolutely, converge conditionally, or diverge.
Notes
Use was made of
\begin{align}
\sum_{n=0}^{\infty} \sum_{k=0}^{n} B(n,k) &= \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} B(n+k,k) \\
\sum_{k=0}^{\infty} \binom{n+k}{k} \, t^k &= \frac{1}{(1-t)^{n+1}}
\end{align}
to demonstrate the second series equals the first series when evaluated.
| {
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Cubic roots and difference of cubes in limits $\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$ Find the limit:
$$\lim\limits_{n\to\infty} (\sqrt[3]{n^6-6n^4+1} - n^2)$$
I applied the identity:
$$a^3 - b^3 = (a-b)(a^2+ab+b^2)$$
by multiplying the numerator and denominator by the complementary part.
$$\lim\limits_{n\to\infty} \frac{(\sqrt[3]{n^6-6n^4+1} - n^2)(\sqrt[3]{(n^6-6n^4+1)^2} + n^2\sqrt[3]{n^6-6n^4+1}+ n^4)}{(\sqrt[3]{(n^6-6n^4+1)^2} + n^2\sqrt[3]{n^6-6n^4+1}+ n^4)}=\lim\limits_{n\to\infty} \frac{n^6-6n^4+1 - n^6}{(\sqrt[3]{(n^6(1-6/n^2+1/n^6))^2} + n^2\sqrt[3]{n^6(1-6/n^2+1/n^6)}+ n^4)}=\lim\limits_{n\to\infty} \frac{-6n^4+1}{(n^4\sqrt[3]{(1-6/n^2+1/n^6)^2} + n^4\sqrt[3]{(1-6/n^2+1/n^6)}+ n^4)}=\lim\limits_{n\to\infty} \frac{-6+1/n^4}{(\sqrt[3]{(1-6/n^2+1/n^6)^2} + \sqrt[3]{(1-6/n^2+1/n^6)}+ 1)}=\frac{-6}{3}=-2$$
Is there any more elegant way to approach the problem?
| A possible way is turning it into a derivative using
$$n=\frac{1}{\sqrt t}$$
and consider $t\to 0^+$.
Hence,
$$\sqrt[3]{n^6-6n^4+1} - n^2 = \frac{\sqrt[3]{1-6t+t^3}-1}{t}$$
$$\stackrel{t\to 0^+}{\longrightarrow}\left.(\sqrt[3]{1-6t+t^3})'\right|_{t=0}=\frac{-6}{3}=-2$$
| {
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$f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y)dy$. Find $f(x)$ $f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y) dy$. Find $f(x)$
I've tried taking $\int_{0}^{1} (xy^2 + x^2y) f(y) dy$ to be $k(x)$ since it comes out to be a function of $x$. That transforms our equation to $f(x) = x + k(x)$.
$f(y) = y + k(y) \implies (xy^2 + x^2y)f(y) = (xy^2 + x^2y)y + (xy^2 + x^2y)k(y)$
Now I tried integrating on both sides against $dy$ from $0$ to $1$ in an attempt to find $k(x)$
$\int_{0}^{1} (xy^2 + x^2y)f(y).dy = k(x) = \int_{0}^{1} ((xy^2 + x^2y)y + (xy^2 + x^2y)k(y)).dy$
But I got stuck trying to integrate the right hand side. Any solutions or ideas are appreciated.
| $$f(x) = x + \int_{0}^{1} (xy^2 + x^2y) f(y) dy$$
$$f'(x) = 1 + \int_{0}^{1} (y^2 + 2xy) f(y) dy$$
$$f''(x) = \int_{0}^{1} 2y f(y) dy=\text{constant}$$
$$f(x)=ax^2+bx+c$$
$$ax^2+bx+c=x+ \int_{0}^{1} (xy^2 + x^2y) (ay^2+by+c) dy$$
After calculus of the integral and simplification :
$$(-\frac13 a+\frac13 b+\frac12 c)x^2+(1+\frac15 a-\frac34 b+\frac13 c)x-c=0$$
$$\begin{cases}
-\frac13 a+\frac13 b+\frac12 c=0 \\
1+\frac15 a-\frac34 b+\frac13 c=0 \\
c=0
\end{cases} \quad\implies\quad
\begin{cases}
a=\frac{80}{119} \\
b=\frac{180}{119}
\end{cases}$$
$$\boxed{f(x)=\frac{80}{119}x^2+\frac{180}{119}x}$$
| {
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Find the values of $x$ in terms of $a$ in $x^2+\frac{(ax)^2} {(x+a)^2} =3a^2$ The question is as follows.
Find the values of $x$ in terms of $a$ in
$x^2+\dfrac{(ax)^2}{(x+a)^2} =3a^2 $
My solution:
Multiply both sides by $(x+a)^2$ and expand. On rearranging we get
$x^4+2ax^3-a^2x^2-6a^3x-3a^4=0$
Now dividing by $a^4$ we and taking $\dfrac{x}{a}=y$, we get
$y^4+2y^3-y^2-6y^3-3=0$
Now we can write the above equation as
$(y^2+ay+b)(y^2+cy+d)=0$
Now expanding it and comparing coefficient with the equation we get
$a=-1;b=-1;c=3;d=3$
So we get
$(y^2-y-1)(y^2 +3y+3)=0$
The second bracket has no real roots and by solving 1st bracket we get
$y=\dfrac{x}{a} =\dfrac{1\pm\sqrt5}{2}$
So $x=\dfrac{a(1\pm\sqrt5) }{2}$
My question is
1: As you can see the method is long and tedious, is there any other or elegant way to solve it?
2:Can any step in my solution or the solution itself be improved upon or be replaced by another easier step (like but not limited to factorising the equation we got in $y$ directly into two quadratics or solving the quatic directly)
Thanks!
I got the question from a preparatory book for olympiads.
| Rewrite and factorize the equation as
$$\begin{align}
0 & = x^2(x+a)^2+(ax)^2 -3a^2 (x+a)^2\\
& = x^4+2x^3a+2x^2a^2 -3a^2 (x+a)^2 \\
& = x^4+2x^2a(x+a)-3a^2 (x+a)^2 \\
& = (x^2-a(x+a)) (x^2+3a(x+a))\\
\end{align}$$
| {
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Given: $\frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y},$ prove: a) $x^x\times y^y\times z^z =1,$ b) $x\times y \times z =1.$ Given: $ \frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y} $
Prove: $ a)\ x^x\times y^y\times z^z =1 \\ b) \ x\times y \times z =1 $
Here is a few step I tried to solve: $$ \log_{10}z^zy^yx^x => z\log_{10}z + y\log_{10}y + x\log_{10}x $$ and find each $\log_{10}z, \log_{10}y \ and \log_{10}x$ in the form of one given the equation by equalizing the denominator:
$(x-z)(y-x)\log_{10}z = (y-x)(z-y)\log_{10}y = (y-x)(x-z)log_{10}x $
and calculating $ \frac{\log_{10}z }{\log_{10}y}, \frac{\log_{10}z }{\log_{10}x} $
Somehow I couldn't figure out the how to get $ x^x\times y^y\times z^z =1... $
Can you help me to prove the equation? Thank you.
| Let $k$ be equal to our fractions:
$$ \frac{\log_{10}z}{y-x} = \frac{\log_{10}y}{x-z} = \frac{\log_{10}x}{z-y}=k.$$
Thus, $$\prod_{cyc}z=\prod_{cyc}10^{k(y-x)}=10^0=1.$$
Also, $$\prod_{cyc}z^z=\prod_{cyc}10^{kz(y-x)}=10^0=1.$$
I used $$\sum_{cyc}k(y-x)=k(y-x+z-y+x-z)=0$$ and
$$\sum_{cyc}kz(y-x)=k(zy-zx+xz-xy+yx-yz)=0.$$
| {
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For what $x$ and $y$ polynomial has maximum value?
For what $x,y\in\mathbb R$ does the polynomial
$$-5x^2-2xy-2y^2+14x+10y-1$$
attain a maximum?
My attempt:
I called $\alpha$ maximum value.
$$-5x^2-2xy-2y^2+14x+10y-1\leqslant\alpha$$
$$-5x^2-2xy-2y^2+14x+10y-1-\alpha\leqslant 0$$
$$5x^2+2xy+2y^2-14x-10y+1+\alpha\geqslant 0$$
$$(x+y)^2+(y-5)^2+3x^2+(x-7)^2-73+\alpha\geqslant0$$
$$\alpha\geqslant73$$
So the lowest maximum value turned out to be $73$, but after checking answers I was wrong-maximum is $16$, so my further plans to calculate from that $x$ and $y$ seemed purposless. I'd like to see solution using only high school knowledge.
Ans: $x=1$, $y=2$
| Your idea is quite correct, but you have to complete squares in a different way:
\begin{gather}
-(5x^2+2xy+2y^2-14x-10y+1) =\\
-\left[\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 +2y^2-10y+1-\frac{1}{5}y^2-\frac{49}{5}+\frac{14}{5}y\right]=\\
-\left[\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 +\frac{9}{5}y^2-\frac{36}{5}y-\frac{44}{5}\right]=\\
-\left[\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 +\left(\frac{3}{\sqrt 5}y-\frac{6}{\sqrt 5}\right)^2-\frac{36}{5}-\frac{44}{5}\right]=\\
-\left[\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 +\left(\frac{3}{\sqrt 5}y-\frac{6}{\sqrt 5}\right)^2-16\right]=\\
-\left(x\sqrt 5 + y\frac{1}{\sqrt 5} -\frac{7}{\sqrt 5}\right)^2 -\left(\frac{3}{\sqrt 5}y-\frac{6}{\sqrt 5}\right)^2+16 \leq 16
\end{gather}
In the first passage I complete the squares in order to eliminate the terms in $x^2$ and $xy$. Then complete in order to eliminate the term in $y^2$ and $y$. So we showed that our function is less or equal to $16$. Imposing the two squares equal to zero you obtain the maximum $16$ and if you do the computation you obtain the point $(1,2)$ that makes it.
This is a standard method to do this kind of exercise. You have to complete squares in a similar way as I did; then you obtain a inequality and putting the squares equal to zero you will find the minimum and the point that makes it.
| {
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Cambridge IGCSE Additional Mathematics Challenge Q This is a challenge question from my Cambridge IGCSE Additional Maths textbook. Bear with me on the drawing. The drawing consists of a square, circle, and quarter circle. The only measurement give is that the side length of the square is $10$cm. Can someone help me find the area of the shaded region? I am looking for an explanation of the answer as well.
| Sketch. Here is an elementary way to obtain the area of the lune. Join the points of intersection of the two arcs, which gives a common chord $C$ for the two involved circles. Thus the area we seek is the difference in area of the segment of the small circle and the big circle, cut off by $C.$ Let these areas respectively be $S$ and $T.$ Then we want $S-T.$ Now to get each of these, we subtract the area of the isosceles triangle defined by radii of the involved circle and $C$ from the area of the sector formed by this triangle and the segment. It follows that we need the length of $C,$ which I'll call $2y,$ and the angles subtended by the given arcs at their respective centres. Let the one for the small circle be $2\phi,$ and the other $2\psi.$ Finally let $x$ be the distance from the centre of the small circle to the line segment $C.$ If you represent all of this information on a diagram, you get a triangle defined by a half-diagonal of the given square, a radius of the small circle, and a radius of the large circle, with sides $5\sqrt2,5$ and $10$ respectively. The angles opposite these sides are an unnamed unknown (not needed to solve the problem), the angle $\psi,$ and the angle $180°-\phi.$ [All angles are measured in degrees.]
Thus applying the cosine rule to this triangle gives us that $$\cos\psi=\frac{5}{4\sqrt 2}.$$ Thus we obtain $$\sin\psi=\frac{\sqrt 7}{4\sqrt 2}.$$ Then using the sine rule gives us that $\sin\phi=2\sin\psi=\frac{\sqrt 7}{2\sqrt 2}.$ Thus we obtain that $\cos\phi=\frac{1}{2\sqrt 2}.$ This gives us $$x=5\cos\phi=\frac{5}{2\sqrt 2}$$ and $$y=\frac{5\sqrt 7}{2\sqrt 2}.$$
Hence we have that the area of the small triangle is $$xy=\frac{25}{8}\sqrt 7$$ and the area of the large triangle is $$(x+5\sqrt 2)y=xy+5y\sqrt 2=\frac{125}{8}\sqrt 7.$$ Therefore we have that the area $S$ of the small segment is given by $$\frac{2\phi}{360°}×π×5^2-xy=\frac54\left(\frac{π\phi}{9}-\frac58\sqrt 7\right)$$ and similarly that $$T=\frac{2\psi}{360°}×π×10^2-\frac{125}{8}\sqrt 7=5\left (\frac{π\psi}{9}-\frac{25}{8}\sqrt 7\right).$$
Therefore the area needed is given by $$S-T=\frac{5π}{9}\left(\frac{\phi}{4}-\psi\right)+\frac{425}{32}\sqrt 7,$$ where $$\cos\phi=\frac14\sqrt 2,\,\cos\psi=\frac58\sqrt 2$$ and $\phi,\,\psi$ are in degrees.
| {
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} |
How find the maximum of the value $t$ such $x^3+y^3+z^3+t(xy^2+yz^2+zx^2)\ge (1+t)(x^2y+y^2z+z^2x)$ let $x,y,z\ge 0$, find the maximum of the $t$, such
$$x^3+y^3+z^3+t(xy^2+yz^2+zx^2)\ge (1+t)(x^2y+y^2z+z^2x)$$
maybe this take $y=x+u,z=x+t$,But I can't get the answer,because this is very ugly, Thanks
| I assumed that it means that we need to find a maximal value of $t$, for which this inequality is true for any non-negatives $x$, $y$ and $z$.
If so, we can say that your way (BW) helps!
Let $x=\min\{x,y,z\}$, $y=x+u$, $z=x+v$ and $u=kv$.
Thus, $u$ and $v$ are non-negatives and $$\sum_{cyc}(x^2+txy^2-(t+1)x^2y)=$$
$$=2(u^2-uv+v^2)x+u^3-(t+1)u^2v+tuv^2+v^3\geq$$
$$\geq u^3-(t+1)u^2v+tuv^2+v^3=v^3(k^3-(t+1)k^2+tk+1).$$
Id est, it's enough to find a maximal value of $t$, for which the inequality $$k^3-(t+1)k^2+tk+1\geq0$$ is true for any non-negative value of $k$ or
$$k^3-k^2+1\geq t(k^2-k).$$
Now, it's enough to assume $k>1$ and it's remains to find$$\min_{k>1}\frac{k^3-k^2+1}{k^2-k}.$$
Can you end it now?
I got $$t_{max}=\frac{\sqrt{13+16\sqrt2}-1}{2}\approx2.484...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Identifying distributions
The pgf of a random variable $X$ has pgf
$$\frac{1}{7}(1+s+3s^2 +s^3 +s^4 )$$
What is the distribution of $X$ please?
I know that a pgf is defines as however the $z$ have just been replaced with $s$ in my example
$$
G(z)=E[z^X]=\sum_{x=0}^\infty p(x)z^x.
$$
| As in the definition, the PGF is the sum of probabilities times a variable to some power, where the probability is the probability that the power occurs. (Re-read this several times if that sentence doesn't make sense.)
So, we have a PGF of
$$
\begin{align}
\dfrac{1}{7}(1 + s + 3s^2 + s^3 + s^4) &= \dfrac{1}{7} \cdot 1 + \dfrac{1}{7} \cdot s + \dfrac{1}{7} \cdot 3s^2 + \dfrac{1}{7}\cdot s^3 + \dfrac{1}{7}\cdot s^4 \\
&= \dfrac{1}{7} \cdot s^0 + \dfrac{1}{7} \cdot s^1 + \dfrac{3}{7} \cdot s^2 + \dfrac{1}{7}\cdot s^3 + \dfrac{1}{7}\cdot s^4
\end{align}$$
Thus we have expressed the PGF as a sum of probabilities times a variable ($s$) to some power. Hence, the random variable $X$ is equal to $0$, $1$, $3$, and $4$ each with probability $\dfrac{1}{7}$, and equal to $2$ with probability $\dfrac{3}{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Checking the argument for calculation of the limit $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $\sqrt3 \over 2$
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $\sqrt3 \over 2$
In computing this limit, i have used following steps:
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\frac{\sqrt3}{2}(\sin(x)-\frac{1}{2})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\frac{\sqrt3}{2}\sin(x)-\frac{\sqrt3}{2}\frac{1}{2}} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)\cos(\frac{\pi}{6})-\cos(\frac{\pi}{6})\sin(\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
= $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)\cos(\frac{\pi}{6})-\cos(x)\sin(\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$ , because $\lim_{x\to \frac{\pi}{6}} \cos(x) = \cos(\frac{\pi}{6})$
= $\lim_{x\to \frac{\pi}{6}} \frac{\sin(x-\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$
from here by the use of $\lim_{\theta\to 0} \frac{\sin(\theta)}{\theta}$ and theorem for limits of composite functions, it can be proved that:
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)-\frac{1}{2}} {x-\frac{\pi}{6}}$ = $2 \over \sqrt3$, which is incorrect.
Can anyone tell me, what i did wrong in this computation?
It is likely that the assumption in this step is wrong:
$\lim_{x\to \frac{\pi}{6}} \frac{\sin(x)\cos(\frac{\pi}{6})-\cos(x)\sin(\frac{\pi}{6})} {\frac{\sqrt3}{2} (x-\frac{\pi}{6})}$ , because $\lim_{x\to \frac{\pi}{6}} \cos(x) = \cos(\frac{\pi}{6})$
If so, then can anyone tell me, what is wrong with this assumption?
| You cannot partially pass to the limit. See this example:
$$
1 = \lim_{x \to 1} 1 = \lim_{x \to 1} \frac{x-1}{x-1} = \lim_{x\to1} \frac{1-1}{x-1} = \lim_{x \to 1} 0 = 0.
$$
The error lies here:
$$
\lim_{x \to 1} \frac{x-1}{x-1} = \lim_{x\to1} \frac{1-1}{x-1}
$$
| {
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"url": "https://math.stackexchange.com/questions/3627825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Inequality with 4 variables For $a,b,c,d>0$ such that $abcd\ge 1$. Prove that $$\frac{1}{3a+2b+c+6}+\frac{1}{3b+2c+d+6}+\frac{1}{3c+2d+a+6}+\frac{1}{3d+2a+b+6} \leq \frac{1}{3}$$
My attempts:
By AM-GM $$3a+2b+c+6\geq12\sqrt[12]{a^3b^2c}$$ and I have to prove that
$$\frac{1}{\sqrt[12]{a^3b^2c}}+\frac{1}{\sqrt[12]{b^3c^2d}}+\frac{1}{\sqrt[12]{c^3d^2a}}+\frac{1}{\sqrt[12]{d^3a^2b}}\leq4,$$
But it's wrong.
| Let $a=kx$, $b=ky$, $c=kz$ and $d=kt$, where $k>0$ and $xyzt=1$.
Thus, $$k^4xyzt\geq1,$$ which gives $k\geq1$ and we obtain:
$$\sum_{cyc}\frac{1}{3a+2b+c+6}=\frac{1}{k(3x+2y+z)+6}\leq\sum_{cyc}\frac{1}{3x+2y+z+6}.$$
Id est, it's enough to prove that:
$$\sum_{cyc}\frac{1}{3x+2y+z+6}\leq\frac{1}{3}$$ or
$$\sum_{cyc}\left(\frac{1}{3x+2y+z+6}-\frac{1}{6}\right)\leq\frac{1}{3}-\frac{2}{3}$$ or
$$\sum_{cyc}\frac{3x+2y+z}{3x+2y+z+6}\geq2.$$
Now, by C-S we obtain:
$$\sum_{cyc}\frac{3x+2y+z}{3x+2y+z+6}=\sum_{cyc}\frac{(3x+2y+z)^2}{(3x+2y+z)^2+6(3x+2y+z)}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(3x+2y+z)\right)^2}{\sum\limits_{cyc}((3x+2y+z)^2+6(3x+2y+z))}=$$
$$=\frac{36(x+y+z+t)^2}{\sum\limits_{cyc}(14x^2+16xy+6xz)+36(x+y+z+t)}$$ and it's enough to prove that
$$9(x+y+z+t)^2\geq\sum_{cyc}(7x^2+8xy+3xz+18x)$$ or
$$9(x+y+z+t)^2+2(xz+yz)\geq\sum_{cyc}(7x^2+8xy+4xz+18x).$$
Now, let $x+y+z+t=4u$, $xy+xz+yz+xt+yz+zt=6v^2$, $xyz+xyt+xzt+yzt=4w^3$ and $xyzt=t^4$, where $t>0$.
Thus, since by AM-GM $$xt+yz\geq2\sqrt{xyzt}=2t^2,$$ it's enough to prove that
$$9\cdot16u^2+4t^2\geq7(16u^2-12v^2)+8\cdot6v^2+18\cdot4ut$$ or
$$8u^2+9v^2+t^2\geq18ut,$$ which is true by AM-GM:
$$8u^2+9v^2+t^2\geq18\sqrt[18]{u^{16}v^{18}t^2}=18\sqrt[18]{u^{16}v^{8}\cdot v^{10}t^2}\geq$$
$$\geq18\sqrt[18]{u^{16}\cdot u^2w^6\cdot t^{12}}\geq18\sqrt[18]{u^{18}t^{18}}=18ut$$ and we are done!
| {
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"url": "https://math.stackexchange.com/questions/3627943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3} \ -\sqrt{2x+2}}$ without l'hospital rule EDITED VERSION
find $$\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}}$$ without l'hospital rule.
using l'hospital rule, you'll have: $\lim_{x \to 1} \frac{2x -\sqrt{x^2 +3}}{\sqrt{x+3}
\ -\sqrt{2x+2}} = -6$.
I can show my attempt but it's pointless as the result i achieved $+\infty$, which is very wrong. is there an strategy trying to calculate a limit without l'hospital rule.
| Multiplying the top and the Bottom by $ \left(2x+\sqrt{x^2+3}\right)\left(\sqrt{x+3}+\sqrt{2x+2}\right) $, gives the following :
\begin{aligned} \lim_{x\to 1}{\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}}&=\lim_{x\to 1}{\frac{-3\left(x+1\right)\left(\sqrt{2x+2}+\sqrt{x+3}\right)}{\sqrt{x^{2}+3}+2x}}\\ &=\frac{-3\left(1+1\right)\times\left(\sqrt{2\times 1+2}+\sqrt{1+3}\right)}{\sqrt{1^{2}+3}+2\times 1}\\ \lim_{x\to 1}{\frac{2x-\sqrt{x^2+3}}{\sqrt{x+3}-\sqrt{2x+2}}}&=-6 \end{aligned}
| {
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"timestamp": "2023-03-29T00:00:00",
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To find the range of $f(x)=\frac {x^2-1}{x^2+3x+2}$, the "discriminant method" doesn't work—why and how can we fix it? Define $f: \mathbb R \setminus \{-1,-2\} \rightarrow \mathbb R $ by $$f(x)=\frac {x^2-1}{x^2+3x+2}$$
To find the range of $f$, we use the "discriminant method" (used in e.g. 1, 2, 3):
*
*Write $y=\frac {x^2-1}{x^2+3x+2}$.
*If $x^2+3x+2\neq0$ (or $x\neq -1,-2$), then we may cross-multiply to get: $$y(x^2+3x+2)=x^2-1.$$
*Rearrange: $(y-1)x^2+3yx+2y+1=0$.
*Check discriminant: $(3y)^2-4(y-1)(2y+1)=y^2+4y+4=(y+2)^2$ which is non-negative for all $y \in \mathbb R$.
*Conclude: The range of $f$ is $\mathbb R$.
The above though is incorrect. It turns out that the range of $f$ is $\mathbb R \setminus \{-2,1\}$. What went wrong above?
(Can we fix the above argument? In particular what do we need to add so that we can discover also that $-2$ and $1$ are not in the range of $f$ while every other real number is? Or is this argument always invalid?)
| With $y = 1$, note your equation has a $0$ coefficient for $x^2$, so it's not a quadratic any more. Instead, you have
$$\begin{equation}\begin{aligned}
3yx + 2y + 1 & = 0 \\
3x + 3 & = 0 \\
x & = -1
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
However, $x = -1$ is one of the excluded values, which means that $y = 1$ must be as well.
With $y \neq 1$, you get with the quadratic equation that
$$\begin{equation}\begin{aligned}
x & = \frac{-3y \pm (y + 2)}{2(y - 1)} \\
x(2(y - 1)) & = -3y \pm (y + 2)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Consider the case where $x = -2$. You then get
$$-4(y - 1) = -3y \pm (y + 2) \tag{3}\label{eq3A}$$
Thus you have that $-4y + 4 = -3y - y - 2 = -4y - 2$, which is not possible, or that $-4y + 4 = -3y + y + 2 = -2y + 2 \implies y = 1$, which has already been discounted. Also, consider the case where $x = -1$, you have $-2y + 2 = -3y + y + 2 = -2y + 2$, which is always true, or that $-2y + 2 = -3y - y - 2 = -4y - 2 \implies y = -2$. Actually, using $y = -2$ in \eqref{eq3A} gives $-6x = 6 \implies x = -1$, which confirms this value of $y$ is not allowed either, meaning since there are no other restrictions on the values of $y$, the range of $f$ is $\mathbb R \setminus \{-2,1\}$.
For another way to solve this, note you have
$$x^2 - 1 = (x + 1)(x - 1) \tag{4}\label{eq4A}$$
$$x^2 + 3x + 2 = (x + 1)(x + 2) \tag{5}\label{eq5A}$$
Thus, for $R \setminus \{-1,-2\}$, you get
$$\begin{equation}\begin{aligned}
f(x) & = \frac{x^2-1}{x^2+3x+2} \\
& = \frac{(x + 1)(x - 1)}{(x + 1)(x + 2)} \\
& = \frac{x-1}{x+2} \\
& = \frac{x + 2 - 2 - 1}{x + 2} \\
& = 1 + \frac{-3}{x + 2}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
Since $\frac{-3}{x + 2} \neq 0$, you have $f(x) \neq 1$. Also, since $x \neq -1$, you also have
$$\frac{-3}{x + 2} \neq \frac{-3}{-1 + 2} = -3 \tag{7}\label{eq7A}$$
which means $f(x) \neq 1 + (-3) = -2$ as well.
Since $\frac{-3}{x+2}$ has a range of all other real values (which I'll leave to you to show), this means that $f(x)$ does as well, apart from the previously mentioned $1$ and $-2$ so, once again, the range of $f$ is $\mathbb R \setminus \{-2,1\}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Evaluate $\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac{2}{45^{2^k}+45^{-2^k}}\right)$ Evaluate $$P=\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac{2}{45^{2^k}+45^{-2^k}}\right)$$
My try:
Let $a_k=45^{2^k}$ Then we have $45^{-2^k}=\frac{1}{a_k}$
So We get:
$$1+\frac{2}{a_k+\frac{1}{a_k}}=\frac{(a_k+1)^2}{a_k^2+1}$$
So
$$P=\lim_{n \to \infty}\prod_{k=0}^n\frac{(a_k+1)^2}{a_k^2+1}$$
Any way from here
| Note that $a_k^2=a_{k+1}$, so
$$
\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_k^2+1}=\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_{k+1}+1}
=\frac{a_0+1}{a_{n+1}+1}\cdot(a_0+1)(a_1+1)\ldots(a_n+1).
$$
Then,
$$
a_k+1=\frac{a_k^2-1}{a_k-1}=\frac{a_{k+1}-1}{a_k-1},~\text{so}~
$$
$$
\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_k^2+1}=\frac{a_0+1}{a_{n+1}+1}\cdot(a_0+1)(a_1+1)\ldots(a_n+1)=\frac{a_0+1}{a_{n+1}+1}\cdot\frac{a_{n+1}-1}{a_0-1}.
$$
Since $a_{n+1}\rightarrow +\infty$ when $n\rightarrow\infty$ we have
$$
\lim_{n\rightarrow\infty}\prod_{k=0}^{n}\frac{(a_k+1)^2}{a_k^2+1}=\frac{a_0+1}{a_0-1}=\frac{45+1}{45-1}=\frac{23}{22}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Inequality with the $2^n$ partial sum of the 'harmonic series' I would like to know whether $p_n:=\sum\limits_{k=1}^{2^n}\frac{1}{k}\le n+1$ for all $n\in\mathbb{N}\setminus\{0\}$ holds or not.
It holds for $n=1,2,3$. Furthermore computing $p_n$ for $n$ big the graph looks similar to the graph of the log-function. On the other hand on the left site of the inequality we have 2^n-summand whereas on the right hand site we have just n+1. Thus I'm not quite sure intuitively if this can be proven (then it can be done by induction) or disproven. Formally, the induction step: $$p_{n+1}=p_n+\frac{1}{2^n+1}+\frac{1}{2^n+2}+\cdots+\frac{1}{2^{n+1}}\le ( n+1)+ \frac{1}{2^n+1}+\frac{1}{2^n+2}+\cdots+\frac{1}{2^{n+1}} $$leads to check if $\frac{1}{2^n+1}+\frac{1}{2^n+2}+\cdots+\frac{1}{2^{n+1}} \le1$ holds. Here I'm stuck.
| Indeed, the induction step is correct. You have
$$\frac{1}{2^n +1} + \frac{1}{2^n +2}+ \ldots + \frac{1}{2^{n+1}} \leq \frac{1}{2^n +1} + \ldots + \frac{1}{2^n +1} \leq \frac{2^n}{2^n +1},$$ where in the last step we used the fact that there are $2^n$ terms in the sum. Since $\frac{2^n}{2^n +1}\leq 1,$ the proof folows by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim_{n \rightarrow \infty} \big( \frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n} \big)$ I need to compute
$$\lim_{n \rightarrow \infty} \big( \frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n} \big).$$
I've previously managed to show that
$$\lim_{n \rightarrow \infty} \Big( \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \Big) \rightarrow \log 2$$
by using the formula
$$\int_{0}^1 f(x) \mathrm{d}x = \lim_{n\rightarrow \infty} \sum_{i=1}^n f\Big( \frac{i}{n} \Big) \Big( \frac{1}{n}\Big)$$
but unfortunately I don't think the same trick works for what I need here, at least not without some adjustment that I can't think of. Or perhaps there is another way?
| $\begin{array}\\
s(n)
&= \big( \frac{1}{n+1}-\frac{1}{n+2}+...+\frac{(-1)^{n-1}}{2n} \big)\\
&= \sum_{k=1}^n\frac{(-1)^{k-1}}{n+k}\\
s(2n)
&= \sum_{k=1}^{2n}\frac{(-1)^{k-1}}{2n+k}\\
&= \sum_{k=1}^{n}\left(\frac{1}{2n-1+k}-\frac{1}{2n+k}\right)\\
&= \sum_{k=1}^{n}\frac{1}{(2n-1+k)(2n+k)}\\
&\le \sum_{k=1}^{n}\frac{1}{(2n)(2n+1)}\\
&= \frac{n}{(2n)(2n+1)}\\
&= \frac{1}{2(2n+1)}\\
&\to 0\\
s(2n+1)
&= \sum_{k=1}^{2n+1}\frac{(-1)^{k-1}}{2n+1+k}\\
&= \sum_{k=2}^{2n+2}\frac{(-1)^{k}}{2n+k}\\
&= \sum_{k=1}^{2n}\frac{(-1)^{k}}{2n+k}-\frac{(-1)^{1}}{2n+1}+\frac{(-1)^{2n+1}}{2n+2n+1}+\frac{(-1)^{2n+2}}{2n+2n+2}\\
&= -s(2n)+\frac{1}{2n+1}-\frac{1}{2n+2n+1}+\frac{1}{2n+2n+2}\\
&= -s(2n)+\frac{8 n + 1}{2 (2 n + 1) (4 n + 1)}\\
|s(2n+1)|
&= |-s(2n)+\frac{8 n + 1}{2 (2 n + 1) (4 n + 1)}|\\
&\le|-s(2n)|+\frac{8 n + 1}{2 (2 n + 1) (4 n + 1)}\\
&\le \frac{1}{2(2n+1)}+\frac{8 n + 1}{2 (2 n + 1) (4 n + 1)}\\
&\to 0\\
\end{array}
$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show that $f(x)=\sin(x)+\cos(x)$ is sinusoidal(alternative) There was this question in our trig homework; it was for plotting a graph but I found it far more interesting than that. When drawing the graph of $\sin(x)+\cos(x)$ (by hand, which I find rather pointless), I found that it looked like some sort of sine or cosine graph. So I set out with all my trig identities to prove this.
$f(x)=\sin(x)+\cos(x)$
$\begin{align} f^2(x)=\sin^2 (x) + \cos^2 (x)+2\sin(x)\cos(x)&=1+\sin(2x) \\ &= 1+\cos\left(\dfrac{\pi}{2} -2x\right) \\&= 1+\cos\left(2x-\dfrac{\pi}{2}\right) \\ &= 1+\cos\left(2\left(x-\dfrac{\pi}{4}\right) \right) \\ &= 1+2\cos^2 \left(x-\dfrac{\pi}{4}\right)-1 \\ &= 2\cos^2 \left(x-\dfrac{\pi}{4}\right) \end{align}$
So that means that $f(x)=\sqrt{2}\cos \left(x-\dfrac{\pi}{4}\right)$
Is there another shorter way to arrive at this result? Also, is there a geometric interpretation of this that can be explained to someone who doesn't know most of the identities I've used?
PS: If this question has already been asked please leave a link for it. I honestly tried to search for a similar question before asking this.
| Yes. Indeed all linear combinations of the form $a\cos x+b\sin x$ are sinusoids, as shown below:
Write this as $$\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}\left(a\cos x+b\sin x\right)=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\right)=R(\sin\phi\cos x+\cos\phi\sin x)=R\sin(\phi+x),$$ where $R=\sqrt{a^2+b^2}$ and $\tan\phi=a/b.$
In your case, you have $a=b=1,$ so that you have $$\sqrt 2\sin(π/4+x).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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evaluate $\int^{\infty}_1\frac{\ln(x)}{1+x^2}dx$ I need to calculate the integral: $$\int^{\infty}_1\frac{\ln(x)}{1+x^2}dx$$
My attempt:
IBP: $u=\ln(x) \Rightarrow du=\frac{1}{x}dx, vdv=\frac{1}{1+x^2}dx \Rightarrow v=\arctan(x)$
So, the integral becomes:
$$\ln(x)\arctan(x)\Big|^{\infty}_1-\int^{\infty}_1\frac{\arctan(x)}{x}dx=\ln(x)\arctan(x)\Big|^{\infty}_1-x+\frac{x^3}{3^2}-\frac{x^5}{5^2}+\frac{x^7}{7^2}-...\Big|^{\infty}_1$$
Since the integral $$\int^{\infty}_1\frac{\arctan(x)}{x}dx=\int^{\infty}_1\sum^{\infty}_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)x}dx=\int^{\infty}_1\sum^{\infty}_{n=0}\frac{x^{2n}}{2n+1}dx=\sum^{\infty}_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)^2}\Big|^{\infty}_1=x-\frac{x^3}{3^2}+\frac{x^5}{5^2}-\frac{x^7}{7^2}+...\Big|^{\infty}_1$$
Is this correct?
Also, this may help with the integral with lower bound $0$
https://www.youtube.com/watch?v=JCZFVQ9uFHg
| Convert the limits to $(0,1)$ to have the expansion converging,
$$\begin{align}
&\int^{\infty}_1\frac{\ln x}{1+x^2}dx
=- \int_{0}^1\frac{\ln t }{1+t^2}dt\\
& =-\ln t\arctan t\Big|_0^1+\int^{1}_0\frac{\arctan t}{t}dt \\
&=\int^{1}_0\frac1t (t-\frac{t^3}3 + \frac{t^5}5 - ...)dt
=(1-\frac{t^3}{3^2}+ \frac{t^5}{5^2}- ...)\bigg|^{1}_0\\
&= 1- \frac1{3^2} + \frac1{5^2} -\frac1{7^2} +...
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Necessary and sufficient conditions for a right triangle
Show that if for $\triangle ABC$ the equalities $h_c^2=a_1b_1$ and $b^2=b_1c$ are true where $h_c$ is the height, $AC=b, AB=c$ and $a_1$ and $b_1$ are the the projections of $BC$ and $AC$ on $AB$, then the triangle is right angled.
I want to demonstrate that together the conditions are sufficient without using the Law of Cosines or the Pythagorean theorem. The relations in a right triangle that we have studied and can use in the problem: $h_c^2=a_1b_1, a^2=ca_1$ and $b^2=cb_1$ where $BC=a, AC=b, AB=c$ and $a_1$ and $b_1$ are the projections of $AC$ and $BC$ on $AB$.
Counterexample:
$\triangle ABC$ $(AC<BC)$ is right triangle and then $h^2=a_1b_1$. Let $M$ lie on $BH$ and $HM=AH=b_1$. For $\triangle MBC$ $h^2=a_1b_1$ but it is NOT A RIGHT TRIANGLE.
| First Solution:
It's known that a triangle is right angled if and only if the three edges satisfies the Pythagorean theorem (this statement follows from the Low of Cosines). So we have only to verify the Pythagorean theorem for the triangle.
Named $a=BC$ we have:
\begin{gather}
a^2+b^2 = h_c^2 + a_1^2 + b_1 c = a_1b_1 + a_1^2 + b_1c=\\
a_1(b_1+a_1) + b_1 c = a_1c+b_1c=(a_1+b_1)c = c^2
\end{gather}
Where the first equality follows from Pythagorean theorem $a^2=h_c^2+a_1^2$, and we have used that $a_1+b_1=c$ because they are the projections.
Second solution:
Take a cartesian plane and put the triangle $ABC$ in it: $A=(-a_1,0)$, $B=(b_1,0)$, $C=(0,h_c)$.
Now take the line $AC$ and $CB$. Their angular coefficients are rispectively
\begin{gather}
AC:\quad \frac{h_c-0}{0-(-a_1)} = \frac{h_c}{a_1}\\
CB:\quad \frac{h_c-0}{0-b_1} = -\frac{h_c}{b_1}
\end{gather}
Multipling now the angular coefficients we obtain:
$$
\frac{h_c}{a_1} \cdot \left( - \frac{h_c}{b_1}\right) = \frac{h_c^2}{a_1b_1} = -1
$$
So the two lines are perpendicular and the angle in $C$ is a right angle.
Edit: I realize now that these arguments work for the angle in $A$ and $B$ acute.
| {
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"url": "https://math.stackexchange.com/questions/3648486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Theorem needed to prove a summation I know that the following relation holds:
$$\sum_{x=1}^y\frac{x(5x+6)}{45}=\frac{y(y+1)(10y+23)}{270}$$
But what theorem should I use to prove that relation?
| It is known that $\sum_{x=1}^y x=\frac{y(y+1)}{2}$ and $\sum_{x=1}^y x^2=\frac{y(y+1)(2y+1)}{6}$. Refer to here for example.
Your required sum will become:
$$\sum_{x=1}^y \frac{x^2}{9} + \frac{2}{15}x$$
$$=\frac{1}{9}\cdot\frac{y(y+1)(2y+1)}{6}+\frac{2}{15}\cdot\frac{y(y+1)}{2}$$
$$=y(y+1)\left[\frac{2y+1}{54}+\frac{1}{15}\right]$$
$$=y(y+1)\left[\frac{10y+5}{270}+\frac{18}{270}\right]$$
$$=y(y+1)\left[\frac{10y+23}{270}\right]$$
$$=\frac{y(y+1)(10y+23)}{270}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this Olympiad inequality proving technique (Isolated Fudging) work? A few years ago, I was in a math olympiad training camp and they taught us a technique to prove inequalities. I just came across it again recently. However, I am not able to understand why it works. So, here is how it goes. Suppose, you want to prove
$$ \frac{a}{b+c}+ \frac{b}{a+c}+ \frac{c}{b+a} \geq \frac{3}{2}.$$
What you do instead is to find an $\alpha$ such that
$$\frac{a}{b+c} \geq \frac{3}{2}\frac{a^\alpha}{a^\alpha+b^\alpha+c^\alpha}. \tag{1}\label{eq1}$$
The technique is primarily meant to find such an $\alpha$ (In an actual olympiad, this would be rough work and once you "know" $\alpha$, you would be supposed to prove the new inequality using standard techniques- Cauchy Schwarz, Hölder's...). To find $\alpha$, we set $b=c=1$. Now, we want to prove
$$\frac{a}{2} \geq \frac{3}{2} \frac{a^\alpha}{a^\alpha +2}$$
$$\Leftrightarrow a^{\alpha+ 1}- 3a^\alpha + 2 a \geq 0$$
Now, we differentiate (wrt a) the equation on the left-hand side and set it equal to zero for a=1. You get
$$\alpha + 1 - 3\alpha + 2 =0$$
$$\Rightarrow \alpha= 3/2$$
My question is why does this procedure work? When does it work? I understand that we are somehow setting the minima of Eq. \eqref{eq1}, but how does it all work out at $a=b=c=1$? I remember (maybe incorrectly) that for the inequality
$$ \sqrt{\frac{a}{b+c}}+ \sqrt{\frac{b}{a+c}}+ \sqrt{\frac{c}{b+a}} \geq 2$$
you need to use $b=1, c=0$. Why and what's the general rule here?
| It not always works.
More exactly, we not always can find this trick during a competition without computer.
For example, there is the following estimation (Ji Chen):
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$.
Prove that:
$$\frac{1}{2+a^2+b^2}\leq\frac{3(6a^2+b^2+c^2+2ab+2ac+4bc)}{32(a^2+b^2+c^2+ab+ac+bc)}.$$
This estimation gives $$\sum_{cyc}\frac{1}{2+a^2+b^2}\leq\sum_{cyc}\frac{3(6a^2+b^2+c^2+2ab+2ac+4bc)}{32(a^2+b^2+c^2+ab+ac+bc)}=\frac{3}{4}$$ and we proved that $$\sum_{cyc}\frac{1}{2+a^2+b^2}\leq\frac{3}{4}.$$
It seems as proof in one line, but we need to prove the Ji Chen's inequality and to find it, which is just impossible during the competition.
By the way, the last inequality we can prove by another ways (the best of them it's uvw, I think).
The inequality $$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq2$$ we can prove by your trick:
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq\sum_{cyc}\frac{2a}{a+b+c}=2,$$ but I think, much more better to get this estimation by AM-GM:
$$\sum_{cyc}\sqrt{\frac{a}{b+c}}=\sum_{cyc}\frac{2a}{2\sqrt{a(b+c)}}\geq\sum_{cyc}\frac{2a}{a+b+c}=2$$ without to look for $\alpha$, for which
$$\sqrt{\frac{a}{b+c}}\geq\frac{2a^{\alpha}}{a^{\alpha}+b^{\alpha}+c^{\alpha}}.$$
Also, we need to check $a=b=1$ and $c=0$ if we need to find some estimation, because the equality in the inequality $$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq2$$ occurs in this case.
Also, for any $n\geq2$ by Karamata we obtain:
$$\sqrt[n]{\frac{a}{b+c}}\geq\sqrt{\frac{a^{\frac{2}{n}}}{b^{\frac{2}{n}}+c^{\frac{2}{n}}}}\geq \frac{2a^{\frac{2}{n}}}{a^{\frac{2}{n}}+b^{\frac{2}{n}}+c^{\frac{2}{n}}},$$ which in the general gives:
$$\sum_{cyc}\sqrt[n]{\frac{a}{b+c}}\geq\sum_{cyc}\frac{2a^{\frac{2}{n}}}{a^{\frac{2}{n}}+b^{\frac{2}{n}}+c^{\frac{2}{n}}}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$ let $x,y,z\ge 0$ and such $x^2+y^2+z^2=3$ show that
$$\sum_{cyc}\dfrac{1}{5-2xy}\le 1$$
try:
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le 2$$
and
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le\sum_{cyc}\dfrac{(x+y)^2}{\frac{5}{3}z^2+\frac{2}{3}x^2+\frac{2}{3}y^2+(x-y)^2}\le\sum\dfrac{3(x+y)^2}{2(x^2+y^2)+5z^2}$$
following I want use C-S,But I don't Success
| We need to prove
$$ \sum \dfrac{xy}{5-2xy} \leqslant 1,$$
or
$$ \sum \dfrac{3xy}{5(x^2+y^2+z^2)-6xy} \leqslant 1,$$
Indeed, because
$$[13(x^2+y^2)+10z^2-18z(x+y)+36xy][5(x^2+y^2+z^2)-6xy]-108xy (x^2+y^2+z^2)$$
$$=\left[63(x^2+y^2)+\frac{109z^2}{2}+112xy-72z(x+y)\right](x-y)^2+\frac{(2x+2y-5z)^2(x+y-2z)^2}{2}$$
$$\geqslant 0.$$
Therefore
$$ \sum \dfrac{3xy}{5(x^2+y^2+z^2)-6xy} \leqslant \sum \frac{13(x^2+y^2)+10z^2-18z(x+y)+36xy}{36(x^2+y^2+z^2)}=1.$$
Done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Catalan triangle created by $[t^{n+1}](\frac{1-\sqrt{1-4t}}{2})^{k+1}=\frac{k+1}{n+1}{2n-k \choose n-k}$? I tried to get the catalan triangle from its Riordan array. I failed at $d_{n,k}=[t^{n+1}](\frac{1-\sqrt{1-4t}}{2})^{k+1}=\frac{k+1}{n+1}{2n-k \choose n-k}$. How do I get there?
| Working with the generating function $C(z)$ of the Catalan numbers
the claim is equivalent to
$$[z^{n-k}] C(z)^{k+1} = \frac{k+1}{n+1}
{2n-k\choose n-k}.$$
The generating function
$$C(z) = \frac{1-\sqrt{1-4z}}{2z}$$
has the functional equation
$$C(z) = 1 + z C(z)^2.$$
Note that $[z^0] C(z)^{k+1} = 1$ which matches the RHS of the claim
when $n=k,$ so we may suppose that $n\gt k.$ We then find
$$[z^{n-k}] C(z)^{k+1} =
\frac{1}{n-k} [z^{n-k-1}] (C(z)^{k+1})'
= \frac{k+1}{n-k} [z^{n-k-1}] C(z)^k C'(z).$$
We get from the Cauchy Coefficient Formula
$$\frac{k+1}{n-k} [z^{n-k-1}] C(z)^k C'(z)
= \frac{k+1}{n-k} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k}}
C(z)^k C'(z) \; dz.$$
Now we put $C(z) = w$ and get from the functional equation that
$(w-1)/w^2 = z.$ This yields the integral
$$\frac{k+1}{n-k} \frac{1}{2\pi i}
\int_{|w-1|=\gamma} \frac{w^{2n-2k}}{(w-1)^{n-k}} w^k \; dw
\\ = \frac{k+1}{n-k} \frac{1}{2\pi i}
\int_{|w-1|=\gamma} \frac{w^{2n-k}}{(w-1)^{n-k}} \; dw
\\ = \frac{k+1}{n-k} \frac{1}{2\pi i}
\int_{|w-1|=\gamma} \frac{1}{(w-1)^{n-k}}
\sum_{q=0}^{2n-k} {2n-k\choose q} (w-1)^q\; dw
\\ = \frac{k+1}{n-k} {2n-k\choose n-k-1}
= \frac{k+1}{n+1} {2n-k\choose n-k}.$$
This is the claim. Note that $C(z)$ is analytic in a neighborhood of
the origin. The above can also be obtained using Lagrange
Inversion,
in fact this entry treats the very same generating function.
| {
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Simplifying $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$
The expression $\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}$, where $0<x<1$, is equal to $x$ or $\sqrt{(1+x^2)}$ or $\frac1{\sqrt{(1+x^2)}}$ or $\frac x{\sqrt{(1+x^2)}}$? (one of these 4 is correct).
My attempt: $$0<x<1\implies 0<\arctan x<\frac{\pi}{4}\implies\frac{1}{\sqrt2}<\cos(\arctan x)<1$$ and $$0<x\sin(\arctan x)<\frac{x}{\sqrt2}$$$$\implies\frac{1}{\sqrt2}<\cos(\arctan x)+x\sin(\arctan x)<1+\frac{x}{\sqrt2}$$$$\implies\frac{1}{2}<(\cos(\arctan x)+x\sin(\arctan x))^2<(1+\frac{x}{\sqrt2})^2$$$$\implies\sqrt\frac{-1}{2}<\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}<\sqrt{(1+\frac{x}{\sqrt2})^2-1}$$$$\implies\sqrt\frac{-1}{2}<\sqrt{(\cos(\arctan x)+x\sin(\arctan x))^2-1}<\sqrt{\frac{x^2}{2}-\sqrt2x}$$ Not sure if I have reached anywhere.
| Your function at $x=0$ is $0$. Thus it can only be $x$ or $x/\sqrt{1+x^2}$. As $x\to 1$ it tends to $1$. Hence, it must be $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Conditional Probability: Papoulis/Pillai Ex 2-13 Question: A box contains white and black balls. When two balls are drawn without replacement, suppose the probability that both are white is 1/3. (A) Find the smallest number of balls in the box.
Solution from the text:
Let a and b denote the number of white and black balls in the box, and $W_k$ the event
$W_k=$ “a white ball is drawn at the kth draw”
We are given that $ P(W_1 \cap W_2) = 1/3 $.
It follows that $$\begin{align} P(W_1 \cap W_2) &= P(W_2 \cap W_1)\\ &= P(W_2 | W_1)P(W_1)\\&=\frac{a-1}{(a-1)+b}\frac{a}{a+b}\\&=1/3 \;\tag1\end{align}$$
The author goes on to state the following:
Because $b > 0$, $$\frac{a}{a+b} > \frac{a-1}{(a-1)+b}$$
We can rewrite $(1)$ as $$\left(\frac{a-1}{(a-1)+b}\right)^2 < 1/3 < \left(\frac{a}{a+b}\right)^2 \: \tag2$$
Which gives the inequalities $$ (\sqrt{3} + 1)b/2 < a < 1+ (\sqrt{3} +1)b/2 \: \tag3$$
And later goes to show that $a = 2$ for $b=1$.
It is the last two inequalities, $(2)$ and $(3)$, that are giving me trouble, as I can’t figure out how the author got there. I have attempted just looking at the denominators (to no avail), as well as multiplying it out (also to no avail). The only thing that got me close to an answer was focusing on $(2)$ without the two exterior terms squared, and looking solely at the left side of the inequality.
Thanks for your time.
| $$a^2 + b^2 + 2ab \lt 3a^2 \\
2a^2 - b^2 - 2ab \gt 0 \\
a \gt \frac{2b \ \pm \sqrt{4b^2 + 8b^2}}{4} \\
a \gt \frac{2b \ \pm 2b\sqrt{3}}{4} \\
\bbox[yellow]{a \gt \frac{( \sqrt{3} + 1)b}2} \\
3(a-1)^2 \lt (a-1)^2 + b^2 + 2b(a-1) \\
2(a-1)^2 - b^2 - 2b(a-1) \lt 0 \\
a-1 \lt \frac{2b \ \pm \sqrt{4b^2 +8b^2}}{4} \\
\bbox[yellow]{a \lt 1 + \frac{( \sqrt{3} + 1)b}2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given the Rodrigues' formula for Legendre's polynomials, show that it satisfies the ODE. The Rodrigues Formula for Legendre's Polynomials is $P_{l}(x)=\frac{1}{2^{l}l!}\frac{d^{l}}{dx^{l}}(x^2-1)^l$.
I wrote $P_{l}(x)=\frac{1}{2^{l}l!}\frac{d^l}{dx^l}\sum_{k=0}^l(-1)^{k-l}\frac{l!}{k!(l-k)!}x^{2k}=\sum_{k=0}^l\frac{(-1)^{k-l}}{2^l}\frac{(2k)!}{k!(l-k)!(2k-l)!}x^{2k-l}$ and put it into the ordinary Legendre's Equation :$\frac{d}{dx}[(1-x^2)\frac{d}{dx}P_{l}(x)]+l(l+1)P_{l}(x)=0$.
I got $\frac{d}{dx}[(1-x^2)\frac{d}{dx}P_{l}(x)]+l(l+1)P_{l}(x)=\sum_{k=0}^l\frac{(-1)^{k-l}}{2^l}\frac{(2k)!}{k!(l-k)!(2k-l-1)!}[(2k-l-1)x^{2k-l-2}+(2k-l+1)x^{2k-l}+\frac{l(l+1)}{2k-l}x^{2k-l}]$
But I can't see why it equals to zero.
Is there other way to do it?
Thanks
| This is another approach that does not explicitly use integration. Use $ D $ to stand for $ d/dx$.
First, apply Leibniz rule for the $n+2$ derivative of a product,
$$
\begin{align}
D^{n+2}(x^2-1)^{n+1} &= (x^2-1) D^{n+2}(x^2-1)^n \\
& \quad + \left( n+2 \atop 1 \right) 2x D^{n+1}(x^2-1)^n \\
&\quad\quad + 2 \left(n+2 \atop 2\right) D^n (x^2-1)^n \tag{1}
\end{align}
$$
Second, evaluate the same expression, but take one derivative first
$$
\begin{aligned}
D^{n+2}(x^2-1)^{n+1} &= D^{n+1} \Big( D(x^2-1)^{n+1} \Big) \\
&= D^{n+1} \Big( (n+1) (x^2-1)^n \cdot 2x \Big)
\end{aligned}
$$
And apply Leibniz again,
$$
\begin{align}
D^{n+2}(x^2-1)^{n+1} &= 2x(n+1) D^{n+1}(x^2-1)^n \\
&\quad + 2(n+1)\left(n+1 \atop 1 \right) D^n(x^2-1)^n \tag{2}
\end{align}
$$
Now divide through by $2^n n!$, subtract (1) from (2) and use Rodrigues's formula to obtain,
$$
\begin{align}
0 &= -(x^2-1) P''_n - 2 (n+2) x P'_n - (n+2)(n+1) P_n \\
&\quad \quad + 2(n+1)x P'_n + 2(n+1)(n+1) P_n \\
&= (1-x^2)P''_n -2x P'_n +n(n+1) P_n
\end{align}
$$
which is exactly what we wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
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Quadratic form on $\mathbb{C}^3$ Let
$$\Phi(x)= \frac{1}{4}x_1^2+\frac{3}{4}i(\bar{x_1}x_2-\bar{x_2}x_1)-\frac{1}{2\sqrt{2}}(\bar{x_1}x_3+\bar{x_3}x_1)+\frac{1}{4}x_2^2-\frac{1}{2\sqrt{2}}i(\bar{x_2}x_3+\bar{x_3}x_2)-\frac{1}{2}x_3^2$$
Show that $\Phi$ is hermitian and find its signature.
So in order to solve this I was thinking that, since we also have to show the signature, I could write the quadratic form in matrix form. Which is
$$A=\begin{pmatrix}\frac{1}{4}&\frac{3}{8}i&\frac{-1}{4\sqrt{2}}\\\frac{-3}{8}i&\frac{1}{4}&\frac{-1}{4\sqrt{2}}i\\\frac{-1}{4\sqrt{2}}&\frac{1}{4\sqrt{2}}i&\frac{-1}{2}\end{pmatrix}$$
It is easy to see that the matrix $A$ is hermitian since $A=A^{*}$.
My question is how do I find the signature? Do I have to diagonalise the matrix to see which eigenvalues it has?
| Your matrix is $\ast$-congruent to a diagonal real matrix.
We need not worry about the eigenvalues, those may or may not be nice.
$$
R =
\left(
\begin{array}{rrr}
1& -\frac{3i}{2} & - \sqrt 2 \\
0& 1 & - i \sqrt 2 \\
0& 0 & 1 \\
\end{array}
\right)
$$
$$
R^\ast A R =
\left(
\begin{array}{rrr}
\frac{1}{4} & 0 & 0 \\
0 & - \frac{5}{16} & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
$$
Without the bother of finding the eigenvalues, we know that one is positive, one zero, and one negative. Having one of them zero means finding them is not bad after all.
| {
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How to derive the condition? (I am not sure it is correct.) Given a polynomial with real coefficients $\alpha x^2 + \beta x + a^2 + b^2 + c^2 - ab- bc - ca $ has imaginary roots, how do we prove $ 2(\alpha - \beta) + ((a - b)^2 + (b - c)^2 + (c - a)^2) > 0 $?
Given all coefficients are real, roots are $ z $ and $\overline{z}$. Hence, $z \overline{z} > 0$. Hence, $ \dfrac{a^2 + b^2 + c^2 - ab - bc - ca}{\alpha} > 0 $. Thus $ \alpha > 0$ because numerator is greater than $ 0 $. Now $D < 0$. Therefore,
$ \begin{align}
\beta^2 - 2 \alpha ((a - b)^2 + (b - c)^2 + (c - a)^2) < 0\\
\end{align}$
How to proceed further?
| You can try with $2f(-1)>0$
| {
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Compute Taylor series $\frac{1}{x^2+4x+3}$ at $x = 2$ I was trying to solve a textbook exercise stated in the following:
Use completing the square and the geometric series to get the Taylor expansion about $x=2$ of $
\frac{1}{x^2+4x+3}$
My early attempt was
$\frac{1}{x^2+4x+3} = -1 (\frac{1}{1-(x+2)^2})$, even though the expression inside the parenthesis is of the form of geometric series $\frac{1}{1 - x}$. I realized that $x=2$ is not in the convergence domain of geometric series i.e. $|(x+2)^2| < 1$. So I should be wrong to proceed in this direction.
Could you please provide me some other directions to work with?
| Using partial fraction decomposition we get that
$$\frac{1}{x^2+4x+3} = \frac{1}{(x+3)(x+1)} = \frac{1}{2}\left(\frac{1}{x+1} - \frac{1}{x+3}\right) = \frac{1}{2}\left(\frac{1}{3}\frac{1}{1+\frac{x-2}{3}} - \frac{1}{5}\frac{1}{1+\frac{x-2}{5}} \right)$$
where we can now use geometric series to get
$$\frac{1}{2}\sum_{n=0}^\infty\left(\frac{1}{3^{n+1}}-\frac{1}{5^{n+1}}\right)(-1)^n(x-2)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3675629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int _0^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx$ with real methods I started like this:
$$\int _0^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx\:
=\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx+\overset {x=\frac{1}{x}}{\int _1^{\infty }\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx}$$
$$=2\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx+5G$$
where $G$ is Catalan's constant. But the integral left is very hard to calculate. As suggested by Zacky one can use the sub $x=\frac{1-t}{1+t}$ and get these integrals.
$$\ln 2\int _0^1\frac{1}{1+t^2}\:dt+\ln 5\int _0^1\frac{1}{1+t^2}\:dt+\int _0^1\frac{\ln \left(t^2+1-\frac{2}{\sqrt{5}}\right)}{1+t^2}\:dt$$
$$+\int _0^1\frac{\ln \left(t^2+1+\frac{2}{\sqrt{5}}\right)}{1+t^2}\:dt-5\int _0^1\frac{\ln \left(1+t\right)}{1+t^2}\:dt$$
To evaluate those one can use the identity
$$\int _0^1\frac{\ln \left(b+ax^2\right)}{1+x^2}\:dx=\frac{\pi }{2}\ln \left(\sqrt{a}+\sqrt{b}\right)+\text{Ti}_2\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)-G$$
In the end, the integral evaluates to
\begin{align}
\int _0^1\frac{\ln \left(x^5+1\right)}{x^2+1}\:dx
&= -\frac{3\pi }{8}\ln 2+\frac{\pi }{4}\ln 5-2G\\
&+\frac{\pi }{2}\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)+\text{Ti}_2\left(\frac{\sqrt{1-\frac{2}{\sqrt{5}}}-1}{\sqrt{1-\frac{2}{\sqrt{5}}}+1}\right)\\
&+\frac{\pi }{2}\ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)+\text{Ti}_2\left(\frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}\right)
\end{align}
But, I have no idea how to simplify the $\text{Ti}_2\left(z\right)$ terms, which seems possible because I found that a similar version can be expressed without these
$$\int _0^{\infty }\frac{\ln \left(1+x^5\right)}{\left(1+x^2\right)^2}\:dx=-\frac{5\pi }{8}-\frac{7\pi }{40}\ln \left(2\right)+\frac{\pi }{5}\ln \left(4+\sqrt{10-2\sqrt{5}}\right)
+\frac{\pi }{10}\ln \left(43+7\sqrt{5}+4\sqrt{130+38\sqrt{5}}\right)+\frac{G}{10}$$
NB:
$\displaystyle x\in\mathbb{R},\text{Ti}_2(x)=\int_0^x \frac{\arctan t}{t}dt$
| Continue to reduce the result in the post as follows
\begin{align}
& \text{Ti}_2\left(\frac{\sqrt{1-\frac{2}{\sqrt{5}}}-1}{\sqrt{1-\frac{2}{\sqrt{5}}}+1}\right)+\text{Ti}_2\left(\frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}\right)\\
=&\ \text{Ti}_2(\tan\frac\pi{20}) - \text{Ti}_2(\tan\frac{3\pi}{20})
=\int_{\tan\frac{3\pi}{20} }^{\tan\frac{\pi}{20} }\frac{\tan^{-1}x}xdx\\ \overset{ibp}=&
\ \frac\pi{20}\ln \tan\frac\pi{20}- \frac{3\pi}{20}\ln \tan\frac{3\pi}{20}+\int^{\frac{3\pi}{20} }_{\frac{\pi}{20} }\ln(\tan t )dt\\
=& \ \frac\pi{20}\ln \frac{\sqrt{1+\frac{2}{\sqrt{5}}}-1}{\sqrt{1+\frac{2}{\sqrt{5}}}+1}- \frac{3\pi}{20}\ln \frac{1-\sqrt{1-\frac{2}{\sqrt{5}}}}{1+\sqrt{1-\frac{2}{\sqrt{5}}}}-\frac25G\\
=& -\frac25G -\frac\pi5 \ln\frac2{\sqrt5 }
-\frac\pi5\ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)
+\frac{3\pi}5\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)
\end{align}
where $\int^{\frac{3\pi}{20} }_{\frac{\pi}{20} }\ln(\tan t )dt=-\frac25G$. Then, plug into the result to arrive at the elementary expression for the integral
\begin{align}
\int_0^\infty\frac{\ln(1+x^5)}{1+x^2}
&= \frac15G -\frac{19\pi }{20}\ln 2+\frac{3\pi }{5}\ln 5\\
&\hspace{5mm}+\frac{4\pi}5 \ln \left(1+\sqrt{1+\frac{2}{\sqrt{5}}}\right)+\frac{8\pi}5\ln \left(1+\sqrt{1-\frac{2}{\sqrt{5}}}\right)\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3675897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int \frac{\cos ^2x+\cos (\sin x)}{\sin x \sin (\sin x)+1} \, dx$ From software, I get
$$\int \frac{\cos ^2 x+\cos (\sin x)}{\sin x \sin (\sin x)+1} \, dx=-2 \tan ^{-1}\left(\cos \left(\frac{x}{2}-\frac{\sin x}{2}\right) \csc \left(\frac{x}{2}+\frac{\sin x}{2}\right)\right)$$
I tried $t=\sin x,x=\arcsin t$ and get
$$\int \frac{1-t^2+\cos t}{\sqrt{1-t^2} (1+t \sin t)} \, dt$$
Then, I can't figure out how to do it.
| Rewrite the denominator
\begin{align}
\sin x \sin (\sin x)+1 = &\frac12[\cos(x-\sin x) - \cos(x+\sin x)]+1
\\ = &\frac12[1-\cos(x+\sin x)] + \frac12[1+\cos(x-\sin x)]
\\ = & \sin^2t_+ + \cos^2t_-
\end{align}
where $t_\pm=\frac x2 \pm \frac{\sin x}2$. Also rewrite the numeritor
\begin{align}
\cos ^2x+\cos (\sin x)= &\cos x\cos(t_+ + t_-) + \cos(t_+ - t_-)
\\ = &(1+\cos x)\cos t_+ \cos t_- + (1-\cos x)\sin t_+ \sin t_-
\\ = & 2 \cos t_- \frac{d\sin t_+}{dx} - 2\sin t_+ \frac{d\cos t_-}{dx}
\end{align}
Substitute above into the integrand to obtain
\begin{align}
\int \frac{\cos ^2 x+\cos (\sin x)}{\sin x \sin (\sin x)+1} \, dx
& =\int \frac {2 \cos t_- \frac{d\sin t_+}{dx} - 2\sin t_+ \frac{d\cos t_-}{dx} }{ \sin^2t_+ + \cos^2t_-} \\
& =-2\int \frac {\sin^2 t_+ \frac{d}{dx}\left(\frac{\cos t_-}{\sin t_+}\right) }{ \sin^2t_+ + \cos^2t_-} \\
& =-2\int \frac {\frac{d}{dx}\left(\frac{\cos t_-}{\sin t_+}\right) }{ 1 + \left(\frac{\cos t_-}{\sin t_+}\right)^2}
= -2\tan ^{-1}\left(\frac{\cos t_-}{\sin t_+}\right) +C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\sum_{n,k} \binom{n}{k}^{-1} $
Evaluate $$\sum_{n,k} \frac{1}{\binom{n}{k}}, $$ where the summation ranges over all positive integers $n,k$ with $1<k<n-1$.
Thouhgts:
We are trying to evaluate $$\sum_{n=4}^{\infty} \sum_{k=2}^{n-2} \binom{n}{k}^{-1}$$
We may try to find a closed form of the inner summation which is of the form :
$$ \frac{1}{ {n \choose 2} } + \frac{1}{{n \choose 3} }+ \dotsb + \frac{1}{{n \choose n-2} }. $$
Notice that we may write $\frac{1}{ {n \choose 2} } = \frac{2! }{n(n-1)}$ and if keep doing the same for the other terms we obtain the following:
$$ \frac{ (n-2)! + (n-3)! (n-(n-2)) + (n-4)!(n-(n-2))(n-(n-3)) + \dotsb + 2! (n-3)! }{n!}, $$
which equals
$$ \frac{ (n-2)! + 2!(n-3)! + 3! (n-4)! + \dotsb + (n-3)! 2! }{n!} $$
and so this equals:
$$ \frac{1}{n(n-1)} + \frac{2}{n(n-1)(n-2)} + \dfrac{6}{n(n-1)(n-2)} + \dotsb + \dfrac{2}{n(n-1)(n-2) }. $$
But half of this terms are identical. Therefore, we are trying to sum up series of the form
$$\sum_{n \geq k} \frac{1}{(n-1)(n-2)(n-3)\dotso(n-k)} ,$$
which can be done by a telescoping trick, but it seems very formidable. Am I approaching this problem the right way? Any hints/suggestions?
| Let $\ell = n -k$, we have
$$\begin{align} \sum_{n=4}^\infty\sum_{k=2}^{n-2} \binom{n}{k}^{-1}
&= \sum_{k=2}^\infty\sum_{n=k+2}^\infty \binom{n}{k}^{-1}\\
&= \sum_{k=2}^\infty\sum_{\ell=2}^\infty \binom{k+\ell}{k}^{-1}
= \sum_{k=2}^\infty\sum_{\ell=2}^\infty \frac{k!\ell!}{(k+\ell)!}\\
&= \sum_{k=2}^\infty\sum_{\ell=2}^\infty (k+\ell+1)\frac{\Gamma(k+1)\Gamma(\ell+1)}{\Gamma(k+\ell+2)}\\
&= \sum_{k=2}^\infty\sum_{\ell=2}^\infty (k+\ell+1)\int_0^1 t^k (1-t)^\ell dt\\
&= \int_0^1 \sum_{k=2}^\infty\sum_{\ell=2}^\infty \left[ (k+\ell+1) t^k (1-t)^\ell\right] dt
\end{align}
$$
Notice when $s$ and $t$ are independent, we have
$$\begin{align} \sum_{k=2}^\infty\sum_{\ell=2}^\infty (k+\ell+1)t^k s^\ell
= & \sum_{k=2}^\infty\sum_{\ell=2}^\infty \left(t\frac{\partial}{\partial t} + s\frac{\partial}{\partial s} + 1 \right)t^k s^\ell
\\
= & \left(t\frac{\partial}{\partial t} + s\frac{\partial}{\partial s} + 1 \right)
\frac{t^2s^2}{(1-t)(1-s)}\\
= & \frac{s^2t^2(5-4(s+t)+3st)}{(1-s)^2(1-t)^2}
\end{align}$$
Substitute $s$ by $1-t$, we obtain
$$\sum_{k=2}^\infty\sum_{\ell=2}^\infty (k+\ell+1) t^k (1-t)^\ell = 1 + 3t(1-t)$$
As a result,
$$\sum_{n=4}^\infty\sum_{k=2}^{n-2} \binom{n}{k}^{-1} = \int_0^1 (1 + 3t(1-t)) dt = \frac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
$\sum(-1)^{k}k^{2}C_{n}^{k}$ I'm asked to compute $\sum^{n}_{k=1}(-1)^{k}k^{2}C_{n}^{k}$ for $n\geqslant 1$. I tried to use generating functions to solve the problem: $A(x):=\sum^{n}_{k=1}k^{2}C_{n}^{k}x^{k}$. So, I need to compute $A(-1)$.
For $n=1$ $A(-1)=-1$, for $n=2$ $A(-1)=2$, for $n=2$ $A(-1)=2$, for $n=3,4,5,6,7,...(?)$ $A(-1)=0$. It seems that the answer is $0$ for $n\geqslant 3$...
Thank you in advance!
| First note that
\begin{align}
\sum_{k=1}^\infty k^2 x^k
&=x^2\sum_{k=1}^\infty k(k-1) x^{k-2} + x\sum_{k=1}^\infty k x^{k-1}\\
&=x^2\frac{d^2}{dx^2}\sum_{k=0}^\infty x^k + x\frac{d}{dx} \sum_{k=0}^\infty x^k\\
&=x^2\frac{d^2}{dx^2}\frac{1}{1-x} + x\frac{d}{dx} \frac{1}{1-x}\\
&=x^2\frac{2}{(1-x)^3}+x\frac{1}{(1-x)^2}\\
&=\frac{x(1+x)}{(1-x)^3}. \tag1
\end{align}
Now
\begin{align}
\sum_{n=1}^\infty \sum_{k=1}^n (-1)^k k^2 \binom{n}{k} z^n
&=\sum_{k=1}^\infty (-1)^k k^2 \sum_{n=k}^\infty \binom{n}{k} z^n\\
&=\sum_{k=1}^\infty (-1)^k k^2 \frac{z^k}{(1-z)^{k+1}}\\
&=\frac{1}{1-z}\sum_{k=1}^\infty k^2 \left(\frac{-z}{1-z}\right)^k\\
&=\frac{1}{1-z}\frac{\frac{-z}{1-z}\left(1+\frac{-z}{1-z}\right)}{\left(1-\frac{-z}{1-z}\right)^3} &&\text{by $(1)$ with $x=\frac{-z}{1-z}$}\\
&=-z+2z^2.
\end{align}
Hence
$$\sum_{k=1}^n (-1)^k k^2 \binom{n}{k}=
\begin{cases}
-1 &\text{if $n=1$}\\
2 &\text{if $n=2$}\\
0 &\text{otherwise}
\end{cases}$$
Alternatively, @Phicar's approach yields
\begin{align}
\sum_{k=1}^n (-1)^k k^2 \binom{n}{k}
&=\sum_{k=1}^n (-1)^k \left(2\binom{k}{2}+\binom{k}{1}\right) \binom{n}{k} \\
&=2\sum_{k=2}^n (-1)^k \binom{k}{2} \binom{n}{k} +\sum_{k=1}^n (-1)^k \binom{k}{1} \binom{n}{k} \\
&=2\sum_{k=2}^n (-1)^k \binom{n}{2} \binom{n-2}{k-2} +\sum_{k=1}^n (-1)^k \binom{n}{1} \binom{n-1}{k-1} \\
&=2\binom{n}{2} \sum_{k=2}^n (-1)^{k-2} \binom{n-2}{k-2} -\binom{n}{1}\sum_{k=1}^n (-1)^{k-1} \binom{n-1}{k-1} \\
&=2\binom{n}{2} \sum_{k=0}^{n-2} (-1)^k \binom{n-2}{k} -\binom{n}{1}\sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} \\
&=2\binom{n}{2} (1-1)^{n-2} - \binom{n}{1}(1-1)^{n-1} \\
&=2\binom{n}{2} [n=2] - \binom{n}{1}[n=1] \\
&=
\begin{cases}
-1 &\text{if $n=1$}\\
2 &\text{if $n=2$}\\
0 &\text{otherwise}
\end{cases}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Compute $\int_0^1 \frac{\text{Li}_2(-x^2)\log (x^2+1)}{x^2+1} \, dx$ How can we evaluate: $$\int_0^1 \frac{\text{Li}_2\left(-x^2\right) \log \left(x^2+1\right)}{x^2+1} \, dx$$
Any help will be appreciated.
| Using double integration we have:
$$\scriptsize I=\frac{\pi ^2 C}{12}+2 C \log ^2(2)-16 \Im(\text{Li}_4(1+i))-\frac{21 \pi \zeta (3)}{8}+\frac{1}{6} \pi \log ^3(2)+\frac{5}{24} \pi ^3 \log (2)+\frac{11 \psi ^{(3)}\left(\frac{1}{4}\right)}{768}-\frac{11 \psi ^{(3)}\left(\frac{3}{4}\right)}{768}$$
And a corollary (thank to Shadhar)
$$\scriptsize \sum_{n=1}^\infty \frac{(-1)^nH_nH_n^{(2)}}{2n+1}=-\frac{\pi ^2 C}{12}+C \log ^2(2)-24 \Im(\text{Li}_4(1+i))+4 \log (2) \Im(\text{Li}_3(1+i))+\frac{5}{24} \pi ^3 \log (2)+\frac{5 \psi ^{(3)}\left(\frac{1}{4}\right)}{384}-\frac{5 \psi ^{(3)}\left(\frac{3}{4}\right)}{384}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum value of $x+2y$ given $\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$
Let $x$ and $y$ be positive real numbers such that
$$\frac{1}{x + 2} + \frac{1}{y + 2} = \frac{1}{3}.$$Find the minimum value of $x + 2y.$
I think I will need to use the Cauchy-Schwarz Inequality here, but I don't know how I should use it. Can anyone help?
Thanks!
| Hint: $y = \left(\dfrac{1}{3} - \dfrac{1}{x+2}\right)^{-1}-2= \dfrac{3x+6}{x-1}-2= \dfrac{x+8}{x-1}\implies x+2y=x+\dfrac{2x+16}{x-1}= \dfrac{x^2+x+16}{x-1}=f(x)$.From this point, you simply set $f'(x) = 0$ and solve for critical points and take it from there. It should be standard calculus problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find vertex coordinates of a square given their distances $p,\>s,\>q,\>r$ to an inner point
The figure represents a square, with P a point inside the square. The four segments are drawn from P to the four vertices of the square and they are named p, q, r, s.
If the bottom-left vertex is at origin and the sides of the square are parallel to the axes (square is drawn in the first quadrant), find the co-ordinates of other three vertices if the measures of p, q, r, s are given.
I'm seeking your help.
|
Let $P(x,y)$ and $a$ the side length of the square. Then,
$$x^2+y^2=p^2, \>\>\>
(a-x)^2+y^2=s^2,\>\>\> x^2+(a-y)^2=r^2
$$
The 2nd and 3rd equations leads to
$x= \frac{a^2+p^2-s^2 }{2a}$ and $y= \frac{a^2+p^2-r^2}{2a}$.
Substitute them into the 1st equation to get
$$ a^4 -(s^2+r^2)a^2 +\frac12[(s^2-p^2)^2+ (r^2-p^2)^2]=0
$$
Solve for $a$ to obtain the vertex coordinates $(0,a)$, $(a,0)$ and $(a,a)$ with
$$a= \left(\frac{s^2+r^2}2 +\left( p^2(s^2+r^2)-p^4 -\frac14(s^2-r^2)^2 \right)^{1/2} \right)^{1/2}
$$
Note that $q^2 = s^2+r^2-p^2$ can not be separately specified. For example, $p=3$, $s=7$ and $r=9$ result in $q=11$ and $a=\sqrt{65+7\sqrt{17}}=9.7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $C^{12}_{x} + C^{12}_{x+1} = C^{13}_{2x}$. $C^{12}_{x} + C^{12}_{x+1} = C^{13}_{2x}$
I did find by brute force the solutions $n=1$ and $n=4$, through the inequalities $2x \le 13, x \ge 0 \implies x \in \{0,1,2,3,4,5,6\}$
But is there a more analytical way to solve this?
Here is my attempt:
$C^{12}_{x} + C^{12}_{x+1} = C^{13}_{2x}$
$C^{13}_{x+1} = C^{13}_{2x}$
$\dfrac{13!}{(x+1)!(13-x-1)!}=\dfrac{13!}{(2x)!(13-2x)!}$
$\dfrac{(2x)!}{(x+1)!}=\dfrac{(13-x-1)!}{(13-2x)!}$
$\dfrac{(2x)(2x-1)...(2x-x)(2x-(x-1))!}{(x+1)!}=\dfrac{(13-x-1)!}{(13-2x)!}$
$(2x)(2x-1)...x=\dfrac{(13-x-1)(13-x-2)...(13-x-(x+1))(13-2x)!}{(13-2x)!}$
$(2x)(2x-1)...x=(13-x-1)(13-x-2)...(13-2x+1)$
Wolfram says these products can be written in terms of Pochhammer symbols as:
$x(x+1)_{x} = \dfrac{2(x-6)(13-2x)_{x+1}}{x-13}$
But I have no idea how to solve this.
Also, is by numerical methods the only way to get the non-natural solutions $-9,-8,-7,-6,-5,-4,-3,-2$ ?
Thanks.
| Using the property
$$C_r^n+C_{r-1}^n=C_r^{n+1}$$
On the LHS, with $n=12,r=x+1$ you have
$$C^{13}_{x+1}=C^{13}_{2x}$$
You now have either $x+1=2x$ or $2x+(x+1)=13$.
From here, it's fairly easy to get $x=1,4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculating a crazy trigonometric integral What is the fastest and best way to do this crAZy integral?
$$ \int\frac{1-\tan^4\theta d\theta}{\tan^{\frac{9}{6}}\theta(\sec^2 \theta+\tan\theta)^{\frac{1}{2}}+\tan^{\frac{10}{6}}\theta (\sec^2 \theta+ \tan\theta)^{\frac{1}{3}}}$$
I tried substituting:
$ \tan x = t$
but that comes out to something more ugly..
| $$ \int\frac{(1-\tan^4\theta) d\theta}{\tan^{\frac{9}{6}}\theta(\sec^2 \theta+\tan\theta)^{\frac{1}{2}}+\tan^{\frac{10}{6}}\theta (\sec^2 \theta+ \tan\theta)^{\frac{1}{3}}}$$
$$ \rightarrow \int \frac{ \sec^2 \theta (1- \tan^2 \theta) d\theta}{ \tan^{\frac{3}{2}} \theta \left[\sec^2 \theta + \tan \theta \right]^{\frac{1}{2}} + \tan^{ \frac{5}{3} } \theta \left[\sec^2 \theta + \tan \theta \right]^{\frac{1}{3}}}$$
$$ \rightarrow \int \frac{ \sec^2 \theta (1- \tan^2 \theta) d\theta}{ \tan^2 \theta \left[ ( \frac{\sec^2 \theta}{\tan \theta} +1 )^{\frac{1}{2}} + ( \frac{\sec^2 \theta}{\tan \theta} +1 )^{\frac{1}{3}} \right]}$$
$$ k= \frac{\sec^2 \theta}{\tan \theta} +1 $$
$$ dk = - \frac{ \sec^2 \theta (1- \tan^2 \theta)}{ \tan^2 \theta} d\theta$$
$$ \int \rightarrow -\frac{dk}{ k^{\frac{1}{2}} + k^{\frac{1}{3}}}$$
$$k^{\frac{1}{6} } = y$$
$$ dk = 6y^5 dy$$
$$ \rightarrow \int \frac{ - 6 y^5 dy}{y^3 + y^2}$$
$$=-2y^3 +3y^2-6y +6log(y+1) +C$$
And sub back $$ y= \left( \frac{\sec^2 \theta +1}{ \tan \theta} \right)^6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help to evaluate the integral $\iint_D\frac{y}{\sqrt{x^2+y^2}}dxdy$ I'm solving a problem about integrals in curves, and I got this integral:
$$\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dxdy.$$
I have been struggling to solve it. I'm sure i have to do some variable change to polar coordinates (to simplify the denominator expression), to be said,
$$x=r\cos\theta \phantom{a},\phantom{a}y=r\sin\theta.$$
$$\text{being: } \phantom{a}r=\sqrt{x^2+y^2}\phantom{a},\phantom{a}\theta=\arctan\frac{y}{x}$$
My problem is finding the new integration limits. The integration region is the square of vertices: $(1,1),(1,2),(2,1),(2,2)$. I'm not sure how is the square transforming to a polar coordinates region. How do i find the new integration limits?
| Use change of order of integration $$\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dxdy$$
$$=\int_1^2\int_1^2\frac{y}{\sqrt{x^2+y^2}}dydx$$
$$=\int_1^2\left(\frac12\int_1^2\frac{d(x^2+y^2)}{\sqrt{x^2+y^2}}\right)dx$$
$$=\int_1^2\left(\sqrt{x^2+y^2}\right)_1^2dx$$
$$=\int_1^2\left(\sqrt{x^2+4}-\sqrt{x^2+1}\right)dx$$
$$=\left(\frac x2\sqrt{x^2+4}+2\ln|x+\sqrt{x^2+4}|-\frac x2\sqrt{x^2+1}-\frac12\ln|x+\sqrt{x^2+1}|\right)_1^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3696069",
"timestamp": "2023-03-29T00:00:00",
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Finding the asymptotic of an integral I came across the following exercise on asymptotic behavior of integrals:
$$I(a) = \int_0^\infty\frac{\cos x}{x^a} \, dx, \text{ where } a\to0^+.$$
I have tried integrating by parts or replacing $\cos x$ with the first summands of its Taylor series, but I end up with something equal to infinity (independent of $a$ and that is not what we want).
I just thought about the substitution $x \leftrightarrow\frac{1}{1+x^2}$ and I get the following result:
$$I(a) = \frac{\cos(1/2)}{2^{2-a}} + \int_0^1\frac{2x^3 \cos\left(\frac{1}{1+x^2}\right)}{(1+x^2)^{3-a}} \, dx - \int_0^1 \frac{2x^3\sin\left(\frac{1}{1+x^2}\right)}{(1+x^2)^{2-2a}}\,dx$$
If I could prove that the last summands are $o(I(a))$ then it would be OK, but I think, in general my strategy won't work.
Apparently this is supposed to be an easy exercise, but I can't come up with anything right now.
| Integration by parts is a good idea (as it usually is when one factor of the integrand is oscillating): when $0<a<1$,
$$
I(a) = \int_0^\infty \frac{\cos x}{x^a} \,dx = \frac{\sin x}{x^a}\bigg|_0^\infty - \int_0^\infty \frac{-a\sin x}{x^{1+a}} \,dx = a \int_0^\infty \frac{\sin x}{x^{1+a}} \,dx.
$$
This is enough to prove convergence, but not yet enough to get an upper bound that goes to $0$ with $a$. However, let's integrate by parts again!—being careful to choose an antiderivative that still allows us to control functions' behavior at the lower endpoint $0$.
\begin{align*}
I(a) = a \int_0^\infty \frac{\sin x}{x^{1+a}} \,dx &= a\frac{1-\cos x}{x^{1+a}}\bigg|_0^\infty - a \int_0^\infty -(1+a)\frac{1-\cos x}{x^{2+a}}\,dx \\
&= a(1+a) \int_0^\infty \frac{1-\cos x}{x^{2+a}}\,dx.
\end{align*}
This new integrand is nonnegative (showing $I(a)\ge0$) and
\begin{align*}
I(a) &= a(1+a) \int_0^\infty \frac{1-\cos x}{x^{2+a}}\,dx \\
&= a(1+a) \int_0^1 \frac{1-\cos x}{x^{2+a}}\,dx + a(1+a) \int_1^\infty \frac{1-\cos x}{x^{2+a}}\,dx \\
&\le a(1+a) \int_0^1 \frac{x^2/2}{x^{2+a}}\,dx + a(1+a) \int_1^\infty \frac{1}{x^{2+a}}\,dx \\
&= \frac{a(1+a)}{2(1-a)} + 2a = a \cdot\frac{5-3a}{2(1-a)};
\end{align*}
since that last factor is continuous at $a=0$,
these inequalities are enough to show that $\lim_{a\to0+} I(a) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Measure of an angle "subtended by each pentagon" in a truncated icosahedron A soccer ball is a truncated icosahedron; it consists of 12 black regular pentagons and 20 white regular hexagons; the edge lengths of the pentagons and hexagons are congruent. What is the measure of the solid angle (in ste-radian) subtended by each pentagon of the soccer ball?
My Explanation:
If the edge length of the regular pentagon is $a$, according to mathworld.wolfram.com, the radius of the circle circumscribing the pentagon is
\begin{equation*}
R = \left(\frac{1}{10}\sqrt{50 + 10\sqrt{5}}\right)a \approx 0.8507 a ,
\end{equation*}
and the radius of the circle inscribed in the pentagon is
\begin{equation*}
r = \left(\frac{1}{10}\sqrt{25 + 10\sqrt{5}}\right)a \approx 0.6882 a.
\end{equation*}
So, the length of the line segment between a vertex of a pentagon and the midpoint of the side of the same pentagon across from this vertex is
\begin{equation*}
R + r
= \left(\frac{1}{10}\sqrt{50 + 10\sqrt{5}}\right)a + \left(\frac{1}{10}\sqrt{25 + 10\sqrt{5}}\right)a
\approx 1.5388 a.
\end{equation*}
According to wikipedia.org/wiki/Truncated_icosahedron, the radius of the sphere circumscribing the truncated icosahedron is
\begin{equation*}
\frac{a}{4} \sqrt{58 + 18\sqrt{5}} \approx 2.478 a.
\end{equation*}
An implementation of Heron's Formula yields the distance between the center of the sphere and the midpoint of a side of the pentagon of about $2.427a$. An implementation of the Law of Cosines shows that the angle with its vertex at the center of the icosahedron and its endpoints at a vertex of a pentagon and the midpoint of the side of the same pentagon across from this vertex is approximately $36.5^\circ$.
Is that correct? Is that what is meant by "the angle subtended by each pentagon"?
| The wikipedia about solid angle says that
The solid angle of a right $n$-gonal pyramid, where the pyramid base is a regular $n$-sided polygon of circumradius $r$, with a pyramid height $h$ is
$$2\pi -2n\arctan\bigg(\frac{\tan(\frac{\pi}{n})}{\sqrt{1+r^2/h^2}}\bigg)$$
Note here that our $r$ is the circumradius of the pentagon (not the radius of the sphere circumscribing the soccer ball). So, according to the wikipedia about regular pentagons, we get
$$r=a\sqrt{\frac{2}{5-\sqrt 5}}=a\sqrt{\frac{5+\sqrt 5}{10}}$$
Now, considering a right triangle $OAB$ where $O$ is the center of the soccer ball, $A$ is the center of the pentagon and $B$ is a vertex of the pentagon, we get
$$\small h=OA=\sqrt{OB^2-AB^2}=\sqrt{\bigg(\frac{a}{4}\sqrt{58+18\sqrt 5}\bigg)^2-\bigg(a\sqrt{\frac{5+\sqrt 5}{10}}\bigg)^2}=a\sqrt{\frac{41+25\sqrt 5}{8\sqrt 5}}$$
So, the solid angle we seek is
$$2\pi -10\arctan\bigg(\frac{\tan(\frac{\pi}{5})}{\sqrt{1+r^2/h^2}}\bigg)\approx 0.29507$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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identity for sum of binomial coefficients I am trying to prove that $$\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n}{2k+1}(-1)^k = \sqrt{2^n}\sin{}n\pi/4$$
This follows from the subtraction of $\sum_{k=0}^{...}\binom{n}{4k+1}$ and $\sum_{k=1}^{...}\binom{n}{4k-1}$, which according to wikipedia can be evaluated using the so called multisection of sums for binomial coefficients: https://en.wikipedia.org/wiki/Binomial_coefficient#Multisections_of_sums.
Is there a simple proof for the identity provided in the link?
Or is there a different proof for my identity of which I'm not aware of? Thanks
| Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$
we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$
Now take $a_k=\binom{n}{k+1}i^k$, where $i=\sqrt{-1}$, to obtain
\begin{align}
\sum_{k\ge 0} \binom{n}{2k+1}(-1)^k &= \sum_{k\ge 0} \binom{n}{2k+1}i^{2k} \\
&= \sum_{k\ge 0} \binom{n}{k+1}i^k \frac{1+(-1)^k}{2} \\
&= \frac{1}{2}\sum_{k\ge 0} \binom{n}{k+1}i^k + \frac{1}{2}\sum_{k\ge 0} \binom{n}{k+1} (-i)^k \\
&= \frac{1}{2i}\sum_{k\ge 0} \binom{n}{k+1}i^{k+1} - \frac{1}{2i}\sum_{k\ge 0} \binom{n}{k+1} (-i)^{k+1} \\
&= \frac{1}{2i}\sum_{k\ge 1} \binom{n}{k}i^k - \frac{1}{2i}\sum_{k\ge 1} \binom{n}{k} (-i)^k \\
&= \frac{1}{2i}\left((1+i)^n-1\right) - \frac{1}{2i}\left((1-i)^n-1\right) \\
&= \frac{(1+i)^n - (1-i)^n}{2i} \\
&= \frac{\left(\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4)\right)^n - \left(\sqrt{2}(\cos(-\pi/4)+i\sin(-\pi/4)\right)^n}{2i} \\
&= \frac{\sqrt{2}^n(\cos(n\pi/4)+i\sin(n\pi/4)) - \sqrt{2}^n(\cos(-n\pi/4)+i\sin(-n\pi/4))}{2i} \\
&= \frac{\sqrt{2}^n(\cos(n\pi/4)+i\sin(n\pi/4)) - \sqrt{2}^n(\cos(n\pi/4)-i\sin(n\pi/4))}{2i} \\
&= \sqrt{2}^n \sin(n\pi/4)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove or disprove that the ellipse of largest area (centered at origin) inscribed in $y=\pm e^{-x^2}$ has the equation $x^2+y^2=\frac12(1+\log2)$. I can show that $x^2+y^2=\frac12(1+\log2)$ is the equation of the circle of largest area inscribed in $y=\pm e^{-x^2}$:
The minimum distance $r$ (which will be the radius of the circle) between the origin and $y=e^{-x^2}$ can be found by finding the critical numbers of the derivative of the distance function.
\begin{align}
r&=\sqrt{x^2+(e^{-x^2})^2}\\
\frac{dr}{dx}&=\frac{2x-4xe^{-2x^2}}{2\sqrt{x^2+e^{-2x^2}}}\\
0&=2x(1-2e^{-2x^2})\\
x&=0\quad\text{(obviously inadmissible, or)}\\
x&=\sqrt{\frac12\log2}\\
\\
r&=\sqrt{\frac12\log2+e^{-\log2}}\\
r^2&=\frac12\log2+\frac12\\
\implies\quad x^2+y^2&=\frac12\left(1+\log2\right)
\end{align}
as desired.
The relationship between a circle and an ellipse is like that of a square and a rectangle: given a set perimeter, the more square the rectangle, the larger the area. But since this question is based on the curve $y=e^{-x^2}$ and not on any set perimeter, it does not seem like the same conclusion can be drawn. I would need something stronger (perhaps based on concavity?) to show whether $x^2+y^2=\frac12(1+\log2)$ is the largest ellipse or not.
| $$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \implies
\frac{dy}{dx}=-\frac{b^2}{a^2}\frac xy=-\frac{b}{a^2}\frac x{\sqrt{1-\frac{x^2}{a^2}}}.
$$
Hence we have the following system to find the tangent point:
$$\begin{cases}
e^{-x^2}=b\sqrt{1-\frac{x^2}{a^2}}\\
2xe^{-x^2}=\frac{b}{a^2}\frac x{\sqrt{1-\frac{x^2}{a^2}}}
\end{cases}\implies
1-\frac{x^2}{a^2}=\frac{1}{2a^2}\implies x^2=a^2-\frac12.\tag1
$$
Substituting this back into the equation of the tangent point one obtains equation to determine $b$:
$$
e^{\frac12-a^2}=\frac ba\sqrt{\frac12}\implies b=\sqrt{2e}\,ae^{-a^2}.
$$
The area is respectively:
$$
A=\pi ab=\pi \sqrt{2e}\,a^2e^{-a^2}.\tag2
$$
To find the extremum of the area we differentiate over $a$ to obtain:
$$
\frac1{\pi\sqrt{2e}}\frac{dA}{da}=2ae^{-a^2}-2a^3e^{-a^2}=2ae^{-a^2}(1-a^2),
$$
meaning that the largest value of area is achieved at $a=1$:
$$
A_\text{max}=\pi\sqrt{\frac2e}.
$$
Since $a\ne b$ the ellipse of the largest area is not a circle.
PS. In fact the system (1) has another solution: $x=0, b=1$. It can be however shown that the largest possible area of the inscribed ellipse in this case is $\frac\pi{\sqrt2}$ which corresponds to $a=\frac1{\sqrt2}$ in equation (2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find infimum of the sets of number $x + \frac{1}{x} $ Let $A = \{ z = x + \frac{1}{x} : x > 0 \} $ and $B = \{z = 2^x + 2^{1/x} : x > 0 \} $
I want to find $\inf A $ and $\inf B $.
Proof.
Clearly, by AM-GM inequality one has $x + \dfrac{1}{x} \geq 2 $ and $2^x + 2^{1/x} \geq 2 \sqrt{2^{x+1/x} } \geq 2 \sqrt{2^2} = 4 $.
Thus: claim $\inf A = 2 $ and $\inf B = 4 $
For the first one, we need to see that if $l$ is lower bound for $A$:
$$ x + \dfrac{1}{x} \geq l $$
for all $x$, then $2 \geq l$. Well, trivial: put $x=1$ then $1+1 \geq l $. So $\boxed{\inf A = 2 }$. Similarly, if $x=1$ in $B$ we see $2+2 \geq l$ so $\boxed{\inf B = 4}$ . QED
Is this enough work for the proof?
| Looks good to me. In fact, you prove not only that $\inf A=2$, but also that $\min A=2$. Similarly, you have proved that $\min B=4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Strategy to calculate $ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right) $. I am asked to calculate the following: $$ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right). $$
I simplify this a little bit, by moving the constant multiplicator out of the derivative:
$$ \left(\frac{1}{2}\right) \frac{d}{dx} \left(\frac{x^2-6x-9}{x^2(x+3)^2}\right) $$
But, using the quotient-rule, the resulting expressions really get unwieldy:
$$ \frac{1}{2} \frac{(2x-6)(x^2(x+3)^2) -(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} $$
I came up with two approaches (3 maybe):
*
*Split the terms up like this: $$ \frac{1}{2}\left( \frac{(2x-6)(x^2(x+3)^2)}{(x^2(x+3)^2)^2} - \frac{(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} \right) $$
so that I can simplify the left term to $$ \frac{2x-6}{x^2(x+3)^2}. $$
Taking this approach the right term still doesn't simplify nicely, and I struggle to combine the two terms into one fraction at the end.
*The brute-force-method. Just expand all the expressions in numerator and denominator, and add/subtract monomials of the same order. This definitely works, but i feel like a stupid robot doing this.
*The unofficial third-method. Grab a calculator, or computer-algebra-program and let it do the hard work.
Is there any strategy apart from my mentioned ones? Am I missing something in my first approach which would make the process go more smoothly?
I am looking for general tips to tackle polynomial fractions such as this one, not a plain answer to this specific problem.
| HINT
To begin with, notice that
\begin{align*}
x^{2} - 6x - 9 = 2x^{2} - (x^{2} + 6x + 9) = 2x^{2} - (x+3)^{2}
\end{align*}
Thus it results that
\begin{align*}
\frac{x^{2} - 6x - 9}{2x^{2}(x+3)^{2}} = \frac{2x^{2} - (x+3)^{2}}{2x^{2}(x+3)^{2}} = \frac{1}{(x+3)^{2}} - \frac{1}{2x^{2}}
\end{align*}
In the general case, polynomial long division and the partial fraction method would suffice to solve this kind of problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3707227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the distance between the points of two tangents along a circle I have the following problem: there is a circle with the $R = 5$ and center of the circle located on coordinate $(0, 0)$. I have two points $A(6, 8)$ and $B(-4, -6)$. From points, tangents to the circle were drawn. It is better illustrated as:
Let us denote points where tangents and circle intersect as $E, F, G, H$ (see the picture above for better the understandings). So we need to find the distance between E and F along the circle.
|
\begin{align}
|OE|=|OF|=
R&=5
,\quad
|OA|=10
,\quad
|OB|=2\sqrt{13}
,\quad
|AB|=2\sqrt{74}
,\\
\triangle AOE:\quad
|AE|&=5\sqrt3
,\\
\triangle BFO:\quad
|BF|&=3\sqrt3
.
\end{align}
\begin{align}
\angle EOF&=\angle AOB-\angle AOE-\angle FOB
,
\end{align}
\begin{align}
\angle AOB&=\arccos\frac{|OA|^2+|OB|^2-|AB|^2}{2\cdot|OA|\cdot|OB|}
=
\pi-\arccos(\tfrac{18}{65}\sqrt{13})
,\\
\angle AOE&=
\arccos\frac{|OE|}{|OA|}
=\tfrac\pi3
,\\
\angle FOB&=
\arccos\frac{|OF|}{|OB|}
=\arccos(\tfrac5{26}\sqrt{13})
,\\
\angle EOF&=
\tfrac{2\pi}3-\arccos(\tfrac{18}{65}\sqrt{13})
-\arccos(\tfrac5{26}\sqrt{13})
\approx 1.234262917
.
\end{align}
So,
the distance between $E$ and $F$ along the circle,
that is, the length of the arc $FE$ is
\begin{align}
R\cdot\angle EOF&=
5\cdot(\tfrac{2\pi}3-\arccos(\tfrac{18}{65}\sqrt{13})
-\arccos(\tfrac5{26}\sqrt{13}))
\approx 6.171314600
.
\end{align}
Expression for $\angle EOF$ can be simplified to
\begin{align}
\angle EOF&=
\arccos\frac{18+2\sqrt3}{65}
,
\end{align}
hence by the cosine rule
we can also find
\begin{align}
|EF|&=\tfrac1{13}\sqrt{6110-260\sqrt3}
\approx 5.78698130
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3712422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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To find $A$ given that $2I + A +A^2 = B$ where $B$ is given. How to find a matrix $A$ such that the following holds:
$$2I + A +A^2 = B,$$ where the matrix $B$ is given. I tried with char poly of $B$ but not getting any idea.
Note that it is also given that $B$ is invertible.
P.S. $B = \begin{pmatrix}-2&-7&-4\\ \:12&22&12\\ \:-12&-20&-10\end{pmatrix}$.
| $$A=SJS^{-1},\\ 2I+A+A^2=S(2I+J+J^2)S^{-1}=B\\
2I+J+J^2=S^{-1}BS$$
So incorporating (yeah, stealing) the idea of the Mostafa Ayaz's answer we get $$\left(J+\frac{1}{2}I\right)^2=S^{-1}BS-\frac{7}{4}I$$
Letting, for a second, that $S$ is the same for $$B=S\begin{pmatrix}2&0&0\\0&4&1\\0&0&4\end{pmatrix}S^{-1}$$
we feed this thing to wolframalpha obtaining
$$J+\frac{1}{2}I=\pm_{\small{1}}\begin{pmatrix}\frac12&0&0\\0&\pm_{\small{2}}\frac32&\pm_{\small{2}}\frac13
\\0&0&\pm_{\small{2}}\frac32\end{pmatrix}$$ and hence $A$. Can test here
| {
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"url": "https://math.stackexchange.com/questions/3713545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
$\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$ I was training for upcoming Olympiads, working on inequalities, and the following inequality came up:
$$\sqrt{a+b} (\sqrt{3a-b}+\sqrt{3b-a})\leq4\sqrt{ab}$$ with the obvious delimitations $3b\geq a;\: 3a\geq b.$
I've been pondering the question for quite a while, and tried the using CS among others but didn't find the solution, which, given the level of sophistication the problem should have, surprises me.
Any help would be appreciated.
| Presumably we also have the restriction $a,b\ge 0$.
With that assumption we can proceed as follows . . .
If $a+b=0$, then $a=b=0$, and for that case, the inequality clearly holds.
So assume $a+b > 0$.
Since the inequality is homogeneous, the truth of the inequality remains the same if $a,b$ are scaled by an arbitrary positive constant, hence without loss of generality, we can assume $a+b=1$.
Replacing $b$ by $1-a$, it remains to prove
$$
\sqrt{4a-1}+\sqrt{3-4a}\le 4\sqrt{a(1-a)}
\qquad\qquad\;\,
$$
for all $a\in \left[{\large{\frac{1}{4}}},{\large{\frac{3}{4}}}\right]$.
From here it's just routine algebra . . .
\begin{align*}
&
\sqrt{4a-1}+\sqrt{3-4a}\,\le 4\sqrt{a(1-a)}\\[4pt]
\iff\;&
\left(\sqrt{4a-1}+\sqrt{3-4a}\right)^2\le \left(4\sqrt{a(1-a)}\right)^2\\[4pt]
\iff\;&
2+2\sqrt{(4a-1)(3-4a)}\,\le -16a^2+16a\\[4pt]
\iff\;&
\sqrt{(4a-1)(3-4a)}\,\le -8a^2+8a-1\\[4pt]
\iff\;&
(4a-1)(3-4a)\le \left(-8a^2+8a-1\right)^2\\[4pt]
\iff\;&
-16a^2+16a-3\le 64a^4-128a^3+80a^2-16a+1\\[4pt]
\iff\;&
64a^4-128a^3+96a^2-32a+4\ge 0\\[4pt]
\iff\;&
16a^4-32a^3+24a^2-8a+1\ge 0\\[4pt]
\iff\;&
(2a-1)^4\ge 0\\[4pt]
\end{align*}
which is true.
Note:$\;$For the reverse implications, we need to have $-16a^2+16a\ge 0$ and $-8a^2+8a-1\ge 0$, both of which hold since $a\in \left[{\large{\frac{1}{4}}},{\large{\frac{3}{4}}}\right]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed{n\cdot2^{n-2}}.$ Am I on the right track?
| Sure. Notice that
$$\sum _{i = 0}^{n}2i\binom{n}{2i}=\sum _{i = 0}^{n}\binom{2i}{1}\binom{n}{2i}=\sum _{i = 1}^{n}\binom{n}{1}\binom{n-1}{2i-1}=n\sum _{i=1}^n\binom{n-1}{2i-1}=n\cdot 2^{n-2}.$$
Notice that the last step is because you are adding half of the binomials, and the odd half equals the even half by the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find the inverse... my idea is to find the domain of the inverse, and that would then be the range of the function. How to show otherwise what is the range here?
| Alternate way to find the range :
$$f(x) = y =\frac{x+2}{x^2+2x+1}$$
$$yx^2+(2y-1)x+(y-2)=0 $$
Now this quadratic has real roots (since real points exist belonging to the function for all values of x (we can remove the case of -1 later)) . So applying the condition for real roots : ($b^2-4ac \geq 0$)
$$(2y-1)^2-4(y)(y-2)\geq0$$
$$-4y+1 +8y\geq0$$
$$y\geq\frac{-1}{4}$$
$$y \in \left[ -\frac{1}{4},\infty \right)$$
Now the value of $y$ at $x=-1$ $\to \infty$ so no need to remove it from the range
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
} |
Name of the rule allowing the exchanging $\sin$ and $\cos$ in integrals with limits $0$ and $\pi/2$? As in $0$ to $\frac{\pi}{2}$ limits the area under curve of $\sin \theta$ and $\cos \theta$ are same, so in integration if the limits are from $0$ to $\frac{\pi}{2}$ we can replace $\sin \theta$ with $\cos \theta$ and vice versa. Example-
\begin{align*}
\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^3x-\cos x}{\cos^3x-\sin x} dx &=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^3x-\sin x}{\sin^3x-\sin x} dx\\
&=\int\limits_{0}^{\frac{\pi}{2}}dx\\
&=\frac{\pi}{2}
\end{align*}
I want to know what the name of this rule.
| This is not a specific rule. It is the property of definite integral: $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$ i.e. substitute $x=a+b-x$ everywhere in the integrand as follows
$$\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^3x-\cos x}{\cos^3x-\sin x} dx$$
$$=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^3\left(\frac{\pi}{2}-x\right)-\cos\left(\frac{\pi}{2}-x\right)}{\cos^3\left(\frac{\pi}{2}-x\right)-\sin \left(\frac{\pi}{2}-x\right)} dx$$
$$=\int\limits_{0}^{\frac{\pi}{2}} \frac{\cos^3x-\sin x}{\sin^3x-\cos x} dx$$
Using above property in this case is not useful because it's an improper integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 6
} |
Finding the number of solutions to $\sin^2x+2\cos^2x+3\sin x\cos x=0$ with $0\leq x<2\pi$
For $0 \leq x<2 \pi$, find the number of solutions of the equation
$$
\sin^2 x+2 \cos^2 x+3 \sin x \cos x=0
$$
I have dealed the problem like this
$\sin ^{2} x+\cos ^{2} x+\cos ^{2} x+3 \sin x \cos x=0$
LET, $\sin x=t ;\quad \sin ^{2} x+\cos ^{2} x=1$
$t^{2}+2-2 t^{2}+3 t \sqrt{1-t^{2}}=0$
$\left(t^{2}+2\right)^{2}=9 t^{2}\left(1-t^{2}\right)$
$t^{4}+4 t^{2}+h=9 t^{2}-9 t^{4}$
$10 t^{4}-5 t^{2}+4=0$
So the number of solution must be 4
P.s- Any other approach will be greatly appreciated!
correct me if I am wrong
| Other idea to solve $a\sin^2 x+b \cos^2x+ c \sin x \cos x =d $ is divide by $\cos^2x$ or $\sin^2x$ to turn to quadratic like equation of $\tan x$ function
$$\sin ^{2} x+2 \cos ^{2} x+3 \sin x \cos x=0 \div \sin ^{2} x \to \\ 1+2 \cot^2 x+3\cot x=0\\ and \div \cos^2x \to \tan^2x+2+3\tan x=0 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a+b+c=0, ab+bc+ca=1$ and $abc=1,$ then find the value of $\frac ab+\frac bc+\frac ca.$ Clearly $a, b, c$ are the roots of the cubic equation: $x^3+x-1=0\tag{1}.$ We have to find:
\begin{align}
\frac ab+\frac bc+\frac ca&=\frac{a^2c+b^2a+c^2b}{abc}\\\\
&=a^2c+b^2a+c^2b\\\\
&=p,\text{ say}.
\end{align} ($p$ is not a symmetric function of the roots.)
Now let: $q=ac^2+ba^2+cb^2.$ Then we have:
$0=\left(\sum ab\right)\left(\sum a\right)=p+s+3abc.$ This gives
$p+q=-3abc=-3.$
To find $p$ I multiplied $p$ and$q$ and obtained:
$pq=\sum a^3b^3+abc\left(\sum a^3\right)+3a^2b^2c^2.$
Now since $a, b, c$ are the roots of the Eq. $(1),$ so we can write:
$pq=\sum(1-a)(1-b)+abc\left[\sum(1-a)\right]+3(abc)^2=3-2\sum a+\sum ab+abc\left[3-\sum a\right]+3(abc)^2=3-0+1+1×(3-0)+3×1^2=10.$
This implies $p, q$ are the roots of the quadratic: $\color{green}{t^2+3t+10=0.}$ Which on solving gives
$\color{green}{t=\frac{-3\pm i\sqrt{31}}2\tag*{}.}$ Now my actual question is: between these two values of $t$ which one is $p$ and which one is $q$ ?
Please suggest. Thanks in advance.
| $p$ and $q$ are not symmetric to the roots of the polynomial, but they are both conjugates
They form the quadratic $t^2+3t+10=0$, as it happens the root of the quadratic are the values of both $p$ and $q$, both of them satisfy the quadratic, which in turn satisfies symmetric function of the polynomial
I can therefore write that $p^2+3p+10=0$ and also $q^2+3q+10=0$, this means the value of $p$ is $\frac{-3+\sqrt{-31}}{2}$ or $\frac{-3-\sqrt{-31}}{2}$ and also $q$ is $\frac{-3+\sqrt{-31}}{2}$ or $\frac{-3-\sqrt{-31}}{2}$
So the change affected here just depends on the arrangement of the root we take that forms their structure
$p = a^2c+b^2a+c^2b$
$q = a^2b+b^2c+c^2a$
So $p$ is any one of the solution of $t^2+3t+10=0$ while $q$ is the other
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
Integral $\int_{0}^{\frac{\pi}{2}} \sqrt{\cos^4{(x)} + \sin^2{(2x)}}\, dx.$
How to evaluate this integral:
$$\int_{0}^{\frac{\pi}{2}} \sqrt{\cos^4{(x)} + \sin^2{(2x)}}\, dx.$$
I can take the $\cos(x)$ out of the square root by expanding $\sin(2x)$ so it becomes $$\int_{0}^{\frac{\pi}{2}} \cos(x) \sqrt{\cos^2(x) +4\sin^2(x)} \, dx$$
but still I have no clue how to solve it.
| You can proceed as
$$\int_{0}^{\frac{\pi}{2}} \cos x\sqrt{\cos^2{(x)} + 4\sin^2{(x)}}\quad dx$$
$$=\int_{0}^{\frac{\pi}{2}} \cos x\sqrt{1-\sin^2{(x)} + 4\sin^2{(x)}}\quad dx$$
$$=\int_{0}^{\frac{\pi}{2}} \cos x\sqrt{1+ 3\sin^2{(x)}}\quad dx$$
Let $\sin x=\frac{t}{\sqrt 3}\implies \cos x\ dx=\frac{1}{\sqrt 3}dt$
$$=\frac{1}{\sqrt3}\int_{0}^{\sqrt 3} \sqrt{1+ t^2} dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Conditions for positive roots for cubic equation $$
\begin{array}{l}\text { The minimum value of ab if roots of the equation } x^{3}-a x^{2}+b x-2=0 \\ \text { are positive, is }\end{array}
$$
$$
\begin{array}{l}\text { Let } f(x)=x^{3}-a x^{2}+b x-2 \\ \therefore f^{\prime }(x)=3 x^{2}-2 a x+b \\ x_{-}=\frac{2 a-\sqrt{4 a^{2}-12 b}}{6} \\ \text { Also } D>0\end{array}
$$
I could not find the correct range of values of a and b.
Any help is appreciated!
| Starting with a generic, factored cubic equation
$$\begin{align*}f(x) &= (x-r_0)(x-r_1)(x-r_2) \\
\\
&= x^2 -(r_0 + r_1 +r_2)x^2 + (r_0r_1 + r_0r_2+r_1r_2)x - r_0r_1r_2 \\
\end{align*}$$
In your specific case, you have
$$\begin{align*}a &= (r_0 + r_1 +r_2) \\
b&= (r_0r_1 + r_0r_2+r_1r_2) \\
2 &= r_0r_1r_2 \\
\end{align*}$$
Computing the product $ab$
$$\begin{align*}ab &= (r_0 + r_1 +r_2)(r_0r_1 + r_0r_2+r_1r_2) \\
\\
&= 3r_0r_1r_2 +r_0^2(r_1+r_2)+r_1^2(r_0+r_2)+ r_2^2(r_0+r_1)\\
\end{align*}$$
And now with a little hand-waving, I'll say that since the product of the 3 roots is fixed to a constant, all 3 roots are positive and implied to be real, and since the expression for $ab$ has symmetry with respect to the 3 roots, that the expression for $ab$ is minimized when the three roots are equal. (Similar to a given regular, right rectangular volume has minimal total edge length when all the dimensions are equal [a cube]; or a given rectangular area has minimum perimeter when all the sides are equal [a square].)
$$\begin{align*}r &= r_0 = r_1 = r_2\\
\\
r^3 &= r_0r_1r_2 = 2\\
\\
\min{ab} &= 3 r^3 + 3r^2(2r)\\
\\
&= 9r^3\\
\\
\min{ab}&= 9(2) = 18
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Classifying Critical Point in 3D Question: $f(x, y, z) = px^2 +q(y^2 + z^2) +rxy + syz$ where $p,q,r,s \in \mathbb{R}$ has a critical point at $(0, 0, 0)$. Classify this critical point. You can assume the product of $p$ and $q$ is positive. Also, $r$ and $s$ cannot be both equal to zero (either one is zero and the other is not, or neither are zero).
Attempt: I've found the Hessian matrix evaluated at the critical point $H=\begin{bmatrix} 2p & r & 0 \\ r & 2q & s \\ 0 & s & 2q \\\end{bmatrix}$.
I've tried to find the eigenvalues ($\lambda$) of $H$ to assess whether the point is a local minimum, maximum or saddle point, but ended up with a long messy equation that cannot be factorised easily to solve for $\lambda$:
$$(2p - \lambda)((2q-\lambda)^2-s^2)-r^2(2q-\lambda)=0$$ which expands to $$-2r^2q+\lambda r^2 -2ps^2 +8q^2p-8\lambda qp +2\lambda ^2p+\lambda s^2 -4\lambda q^2 +4\lambda ^2 q-\lambda^3=0$$
Using Wolfram Alpha to solve this and find the eigenvalues gives these three solutions, which pushes me to consider another strategy.
So my next attempt was to see if I could classify the point via this method (see page 3) because I figured it would break it down into smaller more easier to manage equations with less unknowns, however it got pretty messy having to consider the two cases ($p,q>0$ and $p,q<0$) and then the three subcases (both nonzero $r$ and $s$, $r=0$ and nonzero $s$, nonzero $r$ and $s=0$):
*
*$f_{xx}(0, 0, 0) = 2p$
*$\det \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}=\det \begin{bmatrix} 2p & r \\ r & 2q \end{bmatrix} = 4pq-r^2$
*$\det H = \det \begin{bmatrix} 2p & r & 0 \\ r & 2q & s \\ 0 & s & 2q \\\end{bmatrix} = 2p(4q^2 - s^2)-2r^2q = 8pq^2 - 2ps^2 - 2r^2 q$
Dealing with the signs of the constants is really what's throwing me, as I am comfortable with the classification process. Any help would be greatly appreciated.
Edit: I'm pretty sure the type of critical point will be different depending on the different cases of what $p, q, r, s$ are (i.e. $p,q \gt 0$ or $p, q \lt 0$, and then subcases concerning $r$ and $s$ and whether they're zero or non-zero, remembering that they cannot be both zero).
| We have $f(x, y, z) = \frac{1}{2} u^\mathsf{T} H u$ where $u = [x, y, z]^\mathsf{T}$.
Results:
If $p, q > 0$,
then $(0, 0, 0)$ is a local minimizer iff $2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 \ge 0$.
If $p, q > 0$ and $2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 < 0$, then $(0, 0, 0)$ is a saddle point.
If $p, q < 0$, then $(0, 0, 0)$ is a local maximizer iff $-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 \ge 0$.
If $p, q < 0$ and $-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 < 0$, then
$(0, 0, 0)$ is a saddle point.
Details:
We apply the Schur complement. See https://www.cis.upenn.edu/~jean/schur-comp.pdf or https://en.wikipedia.org/wiki/Schur_complement
*
*$p, q > 0$:
We have
\begin{align}
H \succeq 0 \quad &\Longleftrightarrow \quad
\left(
\begin{array}{cc}
2q & s \\
s & 2q \\
\end{array}
\right) - \frac{1}{2p}\left(
\begin{array}{c}
r \\
0 \\
\end{array}
\right)\left(
\begin{array}{c}
r \\
0 \\
\end{array}
\right)^\mathsf{T} \succeq 0\\
\quad &\Longleftrightarrow \quad \left(
\begin{array}{cc}
2q - \frac{r^2}{2p} & s \\
s & 2q \\
\end{array}
\right) \succeq 0 \\
\quad &\Longleftrightarrow \quad
2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 \ge 0.
\end{align}
Also, $\det H = 4pq (2q - \frac{r^2}{2p} - \frac{1}{2q}s^2)$.
If $2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 < 0$, then $\det H < 0$.
Clearly, $H$ is not negative semidefinite since the diagonal entries are all positive.
Thus, if $2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 < 0$, then
$H$ has both positive and negative eigenvalues.
Thus, if $2q - \frac{r^2}{2p} - \frac{1}{2q}s^2 < 0$, then $(0, 0, 0)$ is a saddle point.
*$p, q < 0$:
We have
\begin{align}
-H \succeq 0 \quad &\Longleftrightarrow \quad
\left(
\begin{array}{cc}
-2q & -s \\
-s & -2q \\
\end{array}
\right) - \frac{1}{-2p}\left(
\begin{array}{c}
-r \\
0 \\
\end{array}
\right)\left(
\begin{array}{c}
-r \\
0 \\
\end{array}
\right)^\mathsf{T} \succeq 0\\
\quad &\Longleftrightarrow \quad \left(
\begin{array}{cc}
-2q + \frac{r^2}{2p} & -s \\
-s & -2q \\
\end{array}
\right) \succeq 0 \\
\quad &\Longleftrightarrow \quad
-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 \ge 0.
\end{align}
Also, $\det H = 4pq (2q - \frac{r^2}{2p} - \frac{1}{2q}s^2)$.
If $-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 < 0$, then $\det H > 0$.
Clearly, $H$ is not positive semidefinite since the diagonal entries are all negative.
Thus, if $-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 < 0$, then
$H$ has both positive and negative eigenvalues.
Thus, if $-2q + \frac{r^2}{2p} + \frac{1}{2q}s^2 < 0$, then $(0, 0, 0)$ is a saddle point.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find all possible values $m$ such that $f(x)=x^2-5mx+10m-4$ has 2 roots and one of them is twice as the other Consider the polynomial $f(x)=x^2-5mx+10m-4$, find $m$ such that there exist a number $a$ that satisfies $f(a)=f(2a)=0$. This was my attempt:
$f(a)=f(2a)$
$a^2-5ma+10m-4=4a^2-10ma+10m-4$
$a^2-5ma=4a^2-10ma$
$-3a^2+5ma=0$
$a*(-3a+5m)=0$
If $a=0$ then $f(a)=10m-4=0$ and $m=\frac{2}{5}$, else if $-3+5m=0$ then I can't solve.
Is my first answer correct? And if yes, how do I solve for the second one
| Let the roots be $\alpha, 2\alpha$.
By Vieta's formulas,
Sum of roots $=3\alpha = 5m$, product of roots $=2\alpha^2 = 10m-4$.
Hence, $2(\frac{5m}3)^2 = 10m-4 \\ \implies 25m^2 - 45m +18 = 0$, we get (by simple factorisation) $m=0.6$ or $1.2$.
Test by plugging these back into the original expression and getting $f(x) = x^2-3x+2$ and $f(x) = x^2 - 6x+8$ respectively, which each have two distinct roots with one twice the other.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Evaluating matrix equation A $2x2$ matrix $M$ satisfies the conditions $$M\begin{bmatrix}
-8 \\
1
\end{bmatrix} = \begin{bmatrix}
3 \\
8
\end{bmatrix}$$
and
$$M\begin{bmatrix}
1 \\
5
\end{bmatrix} = \begin{bmatrix}
-8 \\
7
\end{bmatrix}.$$
Evaluate $$M\begin{bmatrix}
1 \\
1
\end{bmatrix}.$$
What is the question essentially asking? Isn't this just a matrix equation?
| If you find two numbers $x,y\mathbb{R}$ such that $x\begin{bmatrix}-8\\1\end{bmatrix}+y\begin{bmatrix}1\\5\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$, then, make a left multiply by $M$:
$$xM\begin{bmatrix}-8\\1\end{bmatrix}+yM\begin{bmatrix}1\\5\end{bmatrix}=M\begin{bmatrix}1\\1\end{bmatrix}$$
finally,
$$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x\begin{bmatrix}3\\8\end{bmatrix}+y \begin{bmatrix}-8\\7\end{bmatrix} = M\begin{bmatrix}1\\1\end{bmatrix}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\cdots(I)$$
So, all your problem is reduced to $x\begin{bmatrix}-8\\1\end{bmatrix}+y\begin{bmatrix}1\\5\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$
which is equals to
$$\begin{cases} -8x + y & =1\\ x+5y & = 1\end{cases}
$$
with solution $x=-\frac{4}{41},~y=\frac{9}{41}$. Just reemplace $x,y$ in equation $(I)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the size of the Turan graph $T_r(n)$ is at least $(1 - \frac{1}{r}) \binom{n}{2}$. A Turan graph $T_r(n)$ is defined as the complete $r$-partite graph of order $n$ such that the number of vertices in each of the $r$ classes is either $\lfloor \frac{n}{r}\rfloor$ or $\lceil \frac{n}{r} \rceil$. For fixed $n$ and $r$, $T_r(n)$ is unique up-to isomorphism. The size of $T_r(n)$ can be simply counted as: $\binom{n}{2} - (n \bmod r) \binom{\lceil \frac{n}{r} \rceil}{2} - (r - (n\bmod r))\binom{\lfloor \frac{n}{r}\rfloor}{2}$.
Here is what I have: assume $n = kr + s,\ 0 \leq s \leq r-1$. Note that at least one class must have exactly $\lfloor \frac{n}{r} \rfloor$ vertices. Then, $\binom{n}{2} - (n \bmod r) \binom{\lceil \frac{n}{r} \rceil}{2} - (r - (n\bmod r))\binom{\lfloor \frac{n}{r}\rfloor}{2}$
$\geq$ $\binom{n}{2} -(r-1) \binom{\lceil \frac{n}{r} \rceil}{2} - \binom{\lfloor \frac{n}{r} \rfloor}{2}$
$=\binom{n}{2} - (r-1) \binom{k+1}{2} - \binom{k}{2}$
$= \frac{n(n-1)}{2} - \frac{(r-1)k(k+1)}{2} - \frac{k(k-1)}{2}$
$= \frac{n(n-1)}{2} - \frac{rk(k+1) - k(k+1)}{2} - \frac{k(k-1)}{2}$
$\geq \frac{n(n-1)}{2} - \frac{n(k+1) - k(k+1)}{2} - \frac{k(k-1)}{2}$
$= \frac{n(n-1)-n(k+1) + k(k+1) - k(k-1)}{2}$
$= \frac{n(n-1)-n(k+1) + 2k}{2}$
$= \binom{n}{2}(1 - \frac{k+1}{n-1} + \frac{2k}{n(n-1)})$.
But clearly, $(1 - \frac{k+1}{n-1} + \frac{2k}{n(n-1)})$ may be smaller than $1 - \frac{1}{r}$, as seen by taking $n=31$ and $r=5$.
| Let $e$ be the number of edges and $a$ the average degree of a vertex. We want to show that
$$e\ge\left(1-\frac1r\right)\binom n2.$$
Since $e=\frac{na}2$ and $\binom n2=\frac{n(n-1)}2$, that's the same as showing
$$\frac{na}2\ge\left(1-\frac1r\right)\frac{n(n-1)}2,$$
that is, we have to show that
$$a\ge\left(1-\frac1r\right)(n-1).$$
Since the degree of each vertex is either $n-\lfloor\frac nr\rfloor$ or $n-\lceil\frac nr\rceil$, and since $\lceil\frac nr\rceil\le\frac{n+r-1}r=1+\frac{n-1}r$, we have
$$a\ge n-\left\lceil\frac nr\right\rceil\ge n-\left(1+\frac{n-1}r\right)=\left(1-\frac1r\right)(n-1),$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find a matrix $B$ such that $B^2=A$.. Let $A$ be a $2 × 2$ matrix and $I$ be the identity matrix. Assume that the null spaces of $A − 4I$
and $A − I$ respectively are spanned by
$\begin{bmatrix}3\\2\end{bmatrix}$
and $\begin{bmatrix}1\\1\end{bmatrix}$
respectively. Find a matrix $B$ such that
$B^2 = A$.
How to approach this problem? Any hint. Thanks in advance.
| Observe that $A \dbinom{3}{2} = 4 \dbinom{3}{2}$ and that $A \dbinom{1}{1} = \dbinom{1}{1}$. So, if we define $P := \bigg( \begin{matrix} 3&1\\2&1 \end{matrix} \bigg)$ then
$$P^{-1}AP \dbinom{1}{0} = \dbinom{4}{0} \textrm{ and } P^{-1}AP \dbinom{0}{1} = \dbinom{0}{1},$$
that is, $P^{-1}AP = \bigg( \begin{matrix} 4&0\\0&1 \end{matrix} \bigg)$. Now, all that you need is a matrix $B$ such that $(P^{-1}BP)^2 = P^{-1}AP$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the solution of this summation? $$S(x) = \frac{x^4}{3(0)!} + \frac{x^5}{4(1)!}+\frac{x^6}{5(2)!}+.....$$
If the first term was $$x^3$$ and the next terms were $$x^{3+i}$$ then differentiating it would have given $$x^2.e^x$$ and then it was possible to integrate it. But how to solve this one?
| I'm assuming you've made a typo and you actually have
$$S(x) = \frac{x^4}{3(0)!} + \frac{x^5}{4(1)!}+\frac{x^{{\color{red}6}}}{5(2)!}+ \cdots.$$
Consider $P(x) = S(x)/x$. We have
\begin{align}
P(x) &= \frac{x^3}{3(0)!} + \frac{x^4}{4(1)!}+ \dfrac{x^5}{5(2)!} + \cdots\\
&=\int_0^x\left(\dfrac{t^2}{0!} + \dfrac{t^3}{1!} + \dfrac{t^4}{2!} + \cdots\right){\rm d}t\\
&= \int_0^x t^2e^t{\rm d}t\\
&= e^x(x^2 - 2x + 2) - 2.
\end{align}
Thus,
$$\boxed{S(x) = e^x(x^3 - 2x^2 + 2x) - 2x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Counting the number of strings with at least $2$ numbers Let $k$ and $n \ge 3$ be two natural numbers. How many strings in $\{1,...,n\}^k$ contain at least one occurrence of $1$ and $2$, or at least one occurrence of $2$ and $3$ or at least one occurrence of $1$ and $3$?
I tried to break it down and first count the number of strings that contain at least one $1$ which is $n^k - (n-1)^k$ and similarly to at least one $2$ and $3$, but how do we proceed from here?
| Use Inclusion/Exclusion.
Let $A$ be the subset of strings containing at least one $1$.
Let $B$ be the subset of strings containing at least one $2$.
Let $C$ be the subset of strings containing at least one $3$.
Then:
$$|(A\cap B)\cup (A\cap C)\cup (B\cap C)| = |A\cap B|+|A\cap C|+|B\cap C| - |(A\cap B)\cap(A\cap C)|-|(A\cap B)\cap (B\cap C)|-|(A\cap C)\cap (B\cap C)|+|(A\cap B)\cap (A\cap C)\cap (B\cap C)| = |A\cap B|+|A\cap C|+|B\cap C|-2|A\cap B\cap C|$$
And continuing to break this down:
$$|A\cup B| = |A|+|B|-|A\cap B|$$
$A\cup B$ is the subset of strings containing at least one $1$ or at least one $2$. So, the complement of that is a string that contains neither a $1$ nor a $2$. Thus:
$$|A\cup B| = n^k-(n-2)^k = 2(n^k-(n-1)^k)-|A\cap B|$$
This gives:
$$|A\cap B| = n^k-2(n-1)^k+(n-2)^k$$
Similarly, we can find the other intersections:
$$|A\cap B| = |A\cap C| = |B\cap C|$$
Next, we need to find $|A\cap B\cap C|$.
We have:
$$|A\cup B\cup C| = |A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$$
$$n^k-(n-3)^k = 3(n^k-(n-1)^k)-3(n^k-2(n-1)^k+(n-2)^k)+|A\cap B\cap C|$$
which yields:
$$|A\cap B\cap C| = n^k-3(n-1)^k+3(n-2)^k-(n-3)^k$$
Thus, the answer to your original question is:
$$|(A\cap B)\cup (A\cap C)\cup (B\cap C)| = 3(n^k-2(n-1)^k+(n-2)^k)-2(n^k-3(n-1)^k+3(n-2)^k-(n-3)^k) = n^k-3(n-2)^k+2(n-3)^k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is $P(\min\{X, Y\} = 1)$? If $y=1,2,3$ and $x=0,1,2$ where $P(X=x, Y=y) = k(2x+3y)$
I need to find $P(\min\{X, Y\} = 1)$.
I thought I need to use that the CMF of the minimum is $1-P(X)$, and maybe to find k by doing derivative on the equation and to sum it up to 1?
would love any direction on this.
| Given that $P(X=x, Y=y) = k(2x+3y)$
$\sum_{(x,y)}P(X=x,Y=y) = k(0+3) +k(2 + 3) + k(4 +3) + k(0+6) + k(2+6) + k(4+6) + k(0+9) + k(2+9) + k(4+9) = 72k = 1$
Hence $k = \frac{1}{72}$
Now, $P(\min(X,Y)=1) = P(X=1,Y=2,3) + P(X=2, Y=1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Gradient of squared norm I am looking for the gradient of the function
$$
f(x)= \dfrac{1}{2} \Vert A^Tx \Vert^2 - b^Tx \, .
$$
Well so far I came up with
$$
\nabla f(x) = AA^Tx-b
$$
because
$$
\begin{aligned}
f(x) &= \dfrac{1}{2} \Vert A^Tx \Vert^2 - b^Tx \\
&= \dfrac{1}{2} (A^Tx)^T(A^Tx) - b^Tx \\
&= \dfrac{1}{2} x^TAA^Tx - b^Tx \\
\end{aligned}
$$
and therefore
$$
\begin{aligned}
\nabla f(x) &= \dfrac{1}{2} (AA^T + (AA^T)^T)x - b \\
&= \dfrac{1}{2} (AA^T + AA^T)x - b \\
&= \dfrac{1}{2} \cdot 2 \cdot AA^T x - b \\
&= AA^T x - b \\
\end{aligned}
$$
Is my reasoning correct? Is there a shorter way to calculate the gradient?
| We can compute in the following way as well, using the identity $\|x-y\|^2 = \|x\|^2 - 2\langle x,y\rangle +\|y\|^2$:
$$\begin{align}f(x+h) - f(x) &= \frac{1}{2}\|A^T(x + h)\|^2 - \frac{1}{2}\|A^Tx\|^2 -b^T(x+h) + b^Tx\\
&= \langle A^Tx,A^T h\rangle - b^T h\\
&= \langle AA^Tx - b,h\rangle\end{align}$$
so the gradient $\nabla f(x)$ is $AA^Tx - b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
solving diophantine problem $a^3+b^3=2019(1+ab)$ for coprime $a$ and $b$ We're interested in solving
$$\begin{cases} a^3+b^3=2019(1+ab) \\ \gcd(a,b)=1 \end{cases}$$
I'm stuck with the deduction
*
*Say why $a^3 \equiv -b^3 \pmod{2019}$ . (done)
*Using Fermat, prove that $a^{672} \equiv b^{672} \pmod{2019}$ . (done)
*Deduce that $a \equiv -b \pmod{2019}$ (note that $673 = 1 + 244*3$) (stuck)
thanks for any help.
| Here's a method that will work, though you'll need a computer like Wolfram Alpha to do the computations... unless you really want to do it by hand, which is also doable.
The idea is to use the factorization $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$ Luckily, $$2019 = 3\cdot 673,$$ where the factor of $3$ is exceedingly important for the applicability of this technique. Then the equation $$a^3+b^3=2019(1+ab)$$ can be rewritten as $$a^3 + b^3+673^3 - 3ab\cdot 673 =3\cdot 673 +673^3$$ or equivalently $$(a+b+673)(a^2+b^2+673^2-ab-673a-673b)=2^2\cdot 673\cdot 113233.$$ The rest is easy, albeit computationally intensive. The number of positive divisors of the right side is $$(2+1)(1+1)(1+1)=12,$$ so there are $24$ ordered pairs of integers $(s,t)$ that can multiply to the right side, since we have to account for negative factors. So set the $a+b+673$ equal to each possible factor $s,$ isolate $b$ and substitute it into $$a^2+b^2+673^2-ab-673a-673b=t.$$ That will give you a quadratic in $a$ that you can solve using the quadratic formula. If you get an integer, find the corresponding $b$ using $b=s-a-673.$ Good luck with the details.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determining the closed form of $\sum\limits_{r = 2}^n \binom{n}{r} \binom{r}{2}$ Here is a sum in combinatorics, $\sum\limits_{r = 2}^n \binom{n}{r} \binom{r}{2}$ where $n>2$, does this have a closed form?
| $$\sum_{r=2}^n \binom{n}{r} \binom{r}{2}
=\sum_{r=2}^n \binom{n}{2} \binom{n-2}{r-2}
=\binom{n}{2} \sum_{r=2}^n \binom{n-2}{r-2}
=\binom{n}{2} 2^{n-2}
=n(n-1) 2^{n-3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Maximize $\boxed{\mathbf{x}+\mathbf{y}}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$ Maximize $\mathbf{x}+\mathbf{y}$ subject to the condition that $2 x^{2}+3 y^{2} \leq 1$
My approach
$\frac{x^{2}}{1 / 2}+\frac{y^{2}}{1 / 3} \leq 1$
Let $z=x+y$
$\mathrm{Now}, 4 \mathrm{x}+6 \mathrm{y} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{2 x}{3 y}$
$2 x^{2}+3 y^{2}=1$
What to do next? Any suggestion or Hint would be greatly appreciated!
| Use Schwarz inequality
$$(x+y)^2\le \left((\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{3}})^2\right)\left( (\sqrt{2}x)^2+(\sqrt{3}y)^2\right)\le 5/6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Symmetrical point with respect to a plane I have the point $A = (10,0,20)$, and I want the coordinates of its symmetrical point $B$ with respect to the plane $\pi = \begin{cases} x=2+3\alpha+\beta\\y=\alpha\\z=\alpha + 2\beta \end{cases}$, how can I do that? Look at the image:
image
These are the question's alternatives
$(2,20,24);\;(4,15,23);\;(-2,30,26);\;(0,25,25)$
| An equation of the plane is
\begin{equation}
x = 2 + 3 y + (z-y)/2 \Longleftrightarrow 2 x - 5 y -z -4 = 0
\end{equation}
Let $A = (x_0,y_0,z_0) = (10, 0, 20)$ and $B = (x_1, y_1, z_1)= A + t (2, -5, -1)$ because the vector $(2, -5, -1)$ is perpendicular to the plane. We have $2 x_0-5 y_0 -z_0 - 4 = -4$, hence we need $2 x_1-5 y_1 -z_1 - 4 = +4$ in order to have the midpoint $(A + B)/2$ on the plane. This implies $t = 8/30 = 4/ 15$, hence
\begin{equation}
x_1 = 10 + \frac{8}{15}=\frac{158}{15}\qquad
y_1 = -\frac{20}{15}\qquad z_1 = 20 -\frac{4}{15} = \frac{296}{15}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove that $\sum_{k=0}^n{(-1)^k{4n-2k\choose 2n}{2n\choose k}}=2^{2n}$? $$\sum_{k=0}^n{\left( -1 \right) ^k\left( \begin{array}{c} 4n-2k\\ 2n\\\end{array} \right) \left( \begin{array}{c} 2n\\ k\\\end{array} \right)}=2^{2n}$$
I know the correctness of this formula, but how can I prove it?
Thanks for your help.
|
Prove that$$\sum_{k=0}^n{\left( -1 \right) ^k\left( \begin{array}{c} 4n-2k\\ 2n\\\end{array} \right) \left( \begin{array}{c} 2n\\ k\\\end{array} \right)}=2^{2n}$$
Consider the following binomial expansion
\begin{align*}
(x^2-1)^{2n}&=\sum_{k=0}^{2n}(-1)^kx^{4n-2k}{2n\choose k}
\end{align*}
Put $x=1$ after differentiating both sides $2n$ times with respect to $x$ and you are done!
(Stop reading and try it yourself for a "confidence-booster")
\begin{align*}
2x(2n)(x^2-1)^{2n-1}&=\sum_k(-1)^k(4n-2k)x^{4n-2k-1}{2n\choose k}\\
2(2n)(x^2-1)^{2n-1}+4x^2(2n)(2n-1)(x^2-1)^{2n-2}&=\sum_{k}(-1)^k(4n-2k)x^{4n-2k-1}{2n\choose k}\\
\end{align*}
Let's neglect terms like the leftmost one in the above expression, for we are eventually going to put $x=1$ and since $2n-k>2n-k-1$. So,
\begin{align*}
\cdots+2^2x^2(2n)(2n-1)(x^2-1)^{2n-2}&=\sum_k(-1)^k(4n-2k)x^{4n-2k-1}{2n\choose k}\\
\cdots+2^3x^3(2n)(2n-1)(2n-2)(x^2-1)^{2n-3}&=\sum_k(-1)^k(4n-2k)(4n-2k-1)x^{4n-2k-2}{2n\choose k}\\
\end{align*}
and so on so that at last
\begin{align*}
2^{2n}x^{2n}(2n)!(x^2-1)^0&=\sum_k(-1)^k(4n-2k)(4n-2k-1)\cdots(2n-2k+1)x^{2n-2k}{2n\choose k}\\
\end{align*}
Now, what is the greatest value $k$ can take so that the latter terms aren't zero due to differentiation $\rightarrow2n-2k=0\Rightarrow k=n$.
So, after putting $x=1$, we get
\begin{align*}
2^{2n}(2n)!&=\sum_{k=0}^{n}(-1)^k(4n-2k)(4n-2k-1)\cdots(2n-2k+1){2n\choose k}\\
2^{2n}&=\sum_{k=0}^n{\left( -1 \right) ^k\left( \begin{array}{c} 4n-2k\\ 2n\\\end{array} \right) \left( \begin{array}{c} 2n\\ k\\\end{array} \right)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
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