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Is this simple proof about symmetric sums correct? I'm asked to prove
$$x \gt 0, y \gt 0, z \gt 0 \rightarrow$$
$$\left(\frac{x+y}{x+y+z}\right)^\frac{1}{2}+\left(\frac{x+z}{x+y+z}\right)^\frac{1}{2} + \left(\frac{y+z}{x+y+z}\right)^\frac{1}{2} \le 6^\frac{1}{2}$$
I rewrite the summands and say that it is sufficient to prove:
$$A^\frac{1}{2} + B^\frac{1}{2} + C^\frac{1}{2} \le 6^\frac{1}{2} $$
$$ A +B +C = 2$$
$$ 0 \lt A, B, C \le 1$$
Now I just square both sides to get:
$$A + (AB)^\frac{1}{2} + (AC)^\frac{1}{2} + B + (BC)^\frac{1}{2} + C \le 6$$
This seems simple:
$$2 + (AB)^\frac{1}{2} + (AC)^\frac{1}{2} + (BC)^\frac{1}{2} \le$$
$$ 2 + 1 + 1 + 1 \le 5 \le 6$$
So I ended up proving that the original bounds were too loose. That makes me worry that I messed up my proof somewhere.
| You are almost there. Squaring both sides yields $$A+B+C+2(AB)^{1/2}+2(BC)^{1/2}+2(CA)^{1/2}\leq 6,$$ and so you only need to prove that $$(AB)^{1/2}+(BC)^{1/2}+(CA)^{1/2}\leq 2.$$ This inequality follows directly from the fact that $A+B+C=2$ along with Cauchy Schwarz.
Alternative Solution: Before squaring both sides, we can deduce desired inequality directly using Cauchy Schwarz. We have that $$\left(A^{1/2}+B^{1/2}+C^{1/2}\right)^2\leq (A+B+C)(1+1+1),$$ and so we see that $$A^{1/2}+B^{1/2}+C^{1/2}\leq 6^{1/2}.$$
| {
"language": "en",
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Calculate this sum $\sum \frac{1}{(4n+1)(4n+3)}$ I'm having troubles to calculate this sum: $\sum \frac{1}{(4n+1)(4n+3)}$. I'm trying to use telescopic series, without success:
$\sum \frac{1}{(4n+1)(4n+3)}=1/2\sum \frac{1}{(4n+1)}-\frac{1}{(4n+3)}$
I need help here
Thanks a lot
| $$\begin{align} \frac{1}{2} \sum_{k=0}^{\infty} \left ( \frac{1}{4 k+1} - \frac{1}{4 n+3} \right ) &= \frac{1}{2} \left ( 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots \right )\\ &= \frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \\ &= \frac{\pi}{8} \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Permutation question. $qpq^{-1}$, $q,p,r,s \in S_{8}$. Let $p,r,s,q \in S_{8}$ be the permutation given by the following products of cycles:
$$p=(1,4,3,8,2)(1,2)(1,5)$$
$$q=(1,2,3)(4,5,6,8)$$
$$r=(1,2,3,8,7,4,3)(5,6)$$
$$s=(1,3,4)(2,3,5,7)(1,8,4,6)$$
Compute $qpq^{-1}$ and $r^{-2}sr^{2}.$
thanks for your help.
I want to write the following permutations like :
$p=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\4&1&8&3&?&?&?&2\end{pmatrix}$
$q=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\2&3&1&5&6&8&7&4\end{pmatrix}$
$r=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&3&8&3&6&5&4&7\end{pmatrix}$
$s=\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 3&?&4&1&?&?&?&?\end{pmatrix}$
can you help me plese to fill the $?$ mark. Is there another method to compute $qpq^{-1}$?
thanks:)
| I assume right to left associativity, as is usually the case. Note that
$$(\alpha \beta)^{-1} = {\beta}^{-1} {\alpha}^{-1}, \text{ and}$$
$$ (a_1 a_2 \dots a_n)^{-1} = (a_1 a_n \dots a_2).$$
These are standard facts, and easily shown if you like to. Hence,
$$qpq^{-1} = (1 2 3)(4 5 6 8)(1 4 3 8 2)(1 2)(1 5)(4 8 6 5)(1 3 2).$$
I know no smart way to do this, and write the changes down one cycle at a time, right to left. For instance, after the first step, $(1 3 2)$, you get:
$$\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&3&1&4&5&6&7&8\end{pmatrix}.$$
After the next step, $(4 8 6 5)$, you get:
$$\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&3&1&5&6&8&7&4\end{pmatrix}.$$
If you keep going with the remaining cycles, you get
$$qpq^{-1} = \begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 5&6&4&3&2&8&7&1\end{pmatrix},$$
if I didn't mess a step up. You read off the resulting permutation, written as cycles:
$$qpq^{-1} = (15268)(34).$$
You calculate $r^{-2}sr^2$ similarly. It's grindwork mostly.
Edit: took it temporarily down to correct some typos.
Edit 2 - write $p$ (per comment below):
You apply the $3$ cycles of $p = (14382)(12)(15)$ one step at a time. Applying $(15)$ first, you get
$$\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 5&2&3&4&1&6&7&8\end{pmatrix}.$$
Applying $(12)$ next, we get
$$\begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 2&5&3&4&1&6&7&8\end{pmatrix}.$$
Finally applying the last cycle $(14382)$, we get
$$p = \begin{pmatrix}
1 & 2&3&4&5&6&7&8\\ 5&8&4&2&1&6&7&3\end{pmatrix}.$$
| {
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Least value of $a$ for which at least one solution exists? What is the least value of $a$ for which
$$\frac{4}{\sin(x)}+\frac{1}{1-\sin(x)}=a$$
has atleast one solution in the interval $(0,\frac{\pi}{2})$?
I first calculate $f'(x)$ and put it equal to $0$ to find out the critical points.
This gives
$$\sin(x)=\frac{2}{3}$$
as $\cos(x)$ is not $0$ in $(0,\frac{\pi}{2})$. I calculate $f''(x)$ and at $\sin(x)=\frac{2}{3}$, I get a minima. Now to have at least one solution, putting $\sin(x)=\frac{2}{3}$ in the main equation, I get $f=9-a$, which should be greater than or equal to $0$. I then get the 'maximum' value of $a$ as $9$.
Where did I go wrong?
[Note the function is $f(x)=LHS-RHS$ of the main equation.]
| One possible approach: Find a common denominator, then :
$$\frac{4}{\sin x}+\frac{1}{1-\sin x}=a\iff \frac{4(1- \sin x) + \sin x}{\sin x - \sin^2x} = a$$ $$ \iff 4-3\sin x = a(\sin x - \sin^2 x)\tag{$\sin x \neq 0$}$$
Now write the equation as a quadratic equation in $\sin x$:
$$a\sin^2 x - (3 + a)\sin x + 4 = 0 $$
You can solve for when the equation has a real solution (by determining when the discriminant is greater than or equal to 0). $$b^2 - 4ac \geq 0 \iff (3+a)^2 - 16 a \geq 0 \iff a^2 -10a + 9 \geq 0 \iff (a - 1)(a-9) \geq 0$$
Then determine which values of $a$ satisfy the inequality and give in the desired interval.
| {
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Verify that: $2\cot{x}/\tan{2}x = \csc^2x-2$ Verify the following:
$$\frac{2\cot{x}}{\tan{2}x} = \csc^2x-2\;.$$
| Hint:$$\tan 2x=\frac{2 \tan x}{1-\tan ^2x}$$
$$\frac{2\cot x}{\tan 2x}=\frac{1-\tan ^2x}{\tan ^2x}=\frac{1}{\tan ^2x}-1=\frac{cos^2x}{sin^2x}-1=$$$$=\frac{1-sin^2x}{sin^2x}-1= \csc^2x-2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Limit of a sequence with indeterminate form Let $\displaystyle u_n =\frac{n}{2}-\sum_{k=1}^n\frac{n^2}{(n+k)^2}$. The question is: Find the limit of the sequence $(u_n)$.
The problem is if we write $\displaystyle u_n=n\left(\frac{1}{2}-\frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$ and we use the fact that the limit of Riemann sum $\displaystyle \frac{1}{n}\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}$ is $\displaystyle \int_0^1 \frac{dx}{(1+x)^2}=\frac{1}{2}$ we find the indeterminate form $\infty\times 0$. How can we avoid this problem? Thanks for help.
| Not sure if this is correct, but here goes anyway,
$$u_n=n\left(\int_0^1\frac{1}{(1+x)^2}dx-\frac 1n\sum_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\right)$$
Let $$\begin{align} A_k
&=\int_{\frac{k-1}{n}}^{\frac{k}{n}}\frac{1}{(1+x)^2}\mathrm{d}x -\frac{1}{n\left(1+\frac{k}{n}\right)^2} \\
& =\frac{1}{n\left(1+\frac{k}{n}\right)}\cdot\left(\frac{1}{1+\frac{k-1}{n}}-\frac{1}{1+\frac{k}{n}}\right) \\
&=\frac{1}{n^2}\cdot\frac{1}{\left(1+\frac kn\right)^2}\frac{1}{1+\frac{k-1}{n}} \end{align}$$
Then
$$u_n=\sum_{k=1}^nnA_k=\int_0^1\frac{1}{(1+x)^3}dx=\frac 38$$
| {
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If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then find the largest possible value of $|a-b|$ I came across the following problem that says:
If $a,b \in \mathbb R$ satisfy $a^2+2ab+2b^2=7,$ then the largest possible value of $|a-b|$ is which of the following?
$(1)\sqrt 7$, $(2)\sqrt{7/2}$ , $(3)\sqrt {35}$ $(4)7$
My Attempt: We notice, $(a-b)^2=7-(4ab+b^2)$ and hence $|a-b|=\sqrt {7-(4ab+b^2)}$ and so $|a-b|$ will be maximum whenever $(4ab+b^2)$ will be minimum. But now I am not sure how to progress further hereon.Can someone point me in the right direction? Thanks in advance for your time.
| The previous solution was wrong, because I used $\frac {da}{db} = -1$ instead. I've added an explanation of why we should have used $\frac {da}{db} = 1$.
The simplest approach I can think of, is to realize that you have a conic section, which is an ellipse.
Because you are interested in extreme values of $a-b = K$, this would be lines of the form $a = b + K$. From calculus, it follows that the extremum occurs at the points of your ellipse where $1 = \frac {da}{db}$. By implicit differentiation,
$$ 2a \frac {da}{db} + 2a + 2b \frac {da}{db} + 4b = 0,$$
hence $a = -\frac {3b}{2}$. Plugging this back into the equation, we obtain $\frac {9b^2}{4} - 3b^2 + 2b^2 = 7$, or that $b = \pm 2 \sqrt{\frac {7}{5}}$. Check the corresponding values of $a$ are given by $ ( a, b) = ( -3 \sqrt{ \frac {7}{5} }, 2 \sqrt{ \frac {7}{5} }), ( 3\sqrt{\frac {7}{5}} , -2 \sqrt{ \frac {7}{5}})$. Hence, the maximum of $|a-b|$ is $\sqrt{ 35}$.
| {
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A simultaneous system of equations Solve for $a,b,c$:
\begin{align}
2ab+a+2b=24\\
2bc+b+c=52\\
2ac+2c+a=74\\
\end{align}
Solving them simultaneously is leading to very difficult situation. Plz help.
| \begin{align}
(a+1)(2b+1)&=(2ab+a+2b)+1=25\\
(2b+1)(2c+1)&=2(bc+b+c)+1=105\\
(2c+1)(a+1)&=(2ca+2c+a)+1=75
\end{align}
So put $u=a+1,v=2b+1,w=2c+1$ and I think you'll get the answer. (Hint: consider $u^2=(uv)(uw)/(vw)$, et cetera)
| {
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a elementary number theory problem please help me to find $ a,b \in \Bbb Z $ such that $a|b$ and $ \forall z \in \Bbb Z $,we have $a+z|b+z$.
| If $a = b$ the result is trivial.
So I will assume that $a \ne b$.
If $a$ and $b$ are positive,
we must have $a < b$.
If we choose $z$ such that
$2(a+z) > (b+z)$
(i.e., $z > b-2a$),
then $1 < \frac{b+z}{a+z} < 2$
which contradicts $(a+z)|(b+z)$.
So $a$ and $b$ cannot both be positive.
If $a$ and $b$ are both negative,
by looking at negative $z$
we come to the same conclusion.
So one of them must be positive and
the other negative.
Assume $a < 0$, and let
$c = -a$. Then
$(z-c)|(z+b)$ and $b$ and $c$ are positive integers.
Again by choosing $z$ large enough,
so $z > c$ and $z+b < 2(z-c)$
(or $z > b+2c$),
we get $1 < \frac{z+b}{z-c} < 2$,
a contradiction.
Therefore the only solution is $a = b$.
Note that this works if $a, b$, and $z$
are restricted to the positive integers.
In this case, we cannot set $z = 0$
to get $a|b$.
| {
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Introduction to Calculus equality I was reading through Apostol's Calculus where he has this equality, but I can't figure out how to prove this equality. I would appreciate a hint if you could help. Thank you.
$$
[(\sum_{k=1}^{n-1}k)+1]^3-(\sum_{k=1}^{n-1}k)^3=n^3-1^3
$$
Above shown to be incorrect should be:
$$
\sum_{k=1}^{n-1}(k+1)^3-(\sum_{k=1}^{n-1}k)^3=n^3-1^3
$$
Perhaps I am wrong in conveying the equation. Let me show you where I get this from. I am trying to obtain the equation to calculate any sum $(1^2+2^2+...+n^2)$ valid for $n\in Z , n\ge1$
We start by showing $3k^2 + 3k + 1 = (k+1)^3 - k^3.$ Taking k = 1, 2, . . . , n-1 we get the n-1 formulas:
$3\bullet1^2 + 3\bullet 1 + 1 = 2^3 - 1^3$
$3\bullet2^2 + 3\bullet 2 + 1 = 3^3 - 2^3$
$3(n-1)^2+3(n-1)+1=n^3-(n-1)^3$
Now, when we add the formulas we get:
$3[1^2+2^2+. . . + (n-1)^2]+3[1+2+. . . =(n-1)]+(n-1)=n^3-1^3.$
What I don't get is by how adding the right side of the n-1 formulas we obtain the right side of the last equality.
Perhaps I wrote the sums listed above incorrectly.
| I think there was a little problem with parentheses. You may intend
$$\sum_{k=1}^{n-1}(k+1)^3 -\sum_{k=1}^{n-1}k^3.$$
If so, it is just a question of expanding out the sums. The first is the sum of the cubes from $2^3$ to $n^3$, and the second is the sum of the cubes from $1^3$ to $(n-1)^3$. Equivalently, it is
$$\left(2^3+3^3+4^3+\cdots+(n-1)^3+n^3\right) -\left(1^3+2^3+3^3 +\cdots+(n-1)^3\right).$$
Note the wholesale cancellation when we remove the parentheses. The only things that survive are the $n^3$ from the first part and the $-1^3$ from the second part.
| {
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Integral $\int\limits_0^\infty \prod\limits_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx$ Does anybody know how to prove this identity?
$$\int_0^\infty \prod_{k=0}^\infty\frac{1+\frac{x^2}{(b+1+k)^2}}{1+\frac{x^2}{(a+k)^2}} \ dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(a+\frac{1}{2}\right)\Gamma(b+1)\Gamma \left(b-a+\frac{1}{2}\right)}{\Gamma(a)\Gamma \left(b+\frac{1}{2}\right)\Gamma(b-a+1)}$$
I found this on Wikipedia.
| Responding to the observation/comment by Random Variable from Thu Aug 8, what is missing above to turn this into a rigorous proof is an evaluation of the quantity
$$R(a,b) = \sum_{m=a}^b
\operatorname{Res}\left(\prod_{k=a}^b \frac{1}{k^2+x^2};x=im\right).$$
We put
$$g_1(x) = \prod_{k=a}^b \frac{1}{k+ix}
\quad \text{and} \quad
g_2(x) = \prod_{k=a}^b \frac{1}{k-ix}.$$
The partial fraction decompositions of $g_1(x)$ and $g_2(x)$ are
$$ g_1(x) = \frac{1}{(b-a)!}
\sum_{k=a}^b (-1)^{k+1-a} {b-a \choose k-a} \frac{i}{x-ki}$$
and
$$ g_2(x) = \frac{1}{(b-a)!}
\sum_{k=a}^b (-1)^{k-a} {b-a \choose k-a} \frac{i}{x+ki}.$$
This implies that
$$ R(a,b) = \sum_{m=a}^b \frac{i}{(b-a)!} (-1)^{m+1-a}
{b-a \choose m-a}
\left(\frac{1}{(b-a)!}
\sum_{k=a}^b (-1)^{k-a} {b-a \choose k-a} \frac{1}{m+k} \right)$$
which is
$$ \frac{i}{((b-a)!)^2}
\sum_{m=a}^b \sum_{k=a}^b (-1)^{m+1+k-2a} {b-a \choose m-a}
{b-a \choose k-a} \frac{1}{m+k}.$$
Now observe that
$$ \sum_{m=a}^b \sum_{k=a}^b
(-1)^{m+1+k-2a} {b-a \choose m-a} {b-a \choose k-a} x^{m+k-1}
\\ = - x^{2a-1}
\sum_{m=a}^b {b-a \choose m-a} (-1)^{m-a} x^{m-a}
\sum_{k=a}^b {b-a \choose k-a} (-1)^{k-a} x^{k-a} \\
= - x^{2a-1} (1-x)^{2(b-a)}.$$
Integrating we find that
$$ R(a,b) = - i \frac{1}{((b-a)!)^2}
\operatorname{B}(2a, 2(b-a)+1).$$
Returning to $I(a,b)$ from the other post, we get that
$$I(a,b) = \frac{1}{2} \frac{\Gamma(b+1)^2}{\Gamma(a)^2}
\times 2\pi i \times
\left(- i \frac{1}{((b-a)!)^2}
\operatorname{B}(2a, 2(b-a)+1) \right) \\=
\pi \frac{\Gamma(b+1)^2}{\Gamma(a)^2}
\frac{1}{((b-a)!)^2} \operatorname{B}(2a, 2(b-a)+1).$$
Switching to gamma functions, this becomes
$$ I(a,b) = \pi \frac{\Gamma(b+1)^2}{\Gamma(a)^2}
\frac{1}{\Gamma(b-a+1)^2} \frac{\Gamma(2a)\Gamma(2(b-a)+1)}{\Gamma(2b+1)}.$$
To conclude we apply the duplication formula several times, getting
$$ \pi \frac{\Gamma(b+1)^2}{\Gamma(a)^2 \Gamma(b-a+1)^2}
\frac{\frac{2^{2a-1}}{\sqrt{\pi}} \Gamma(a) \Gamma(a+1/2)
\frac{2^{2b-2a}}{\sqrt{\pi}} \Gamma(b-a+1/2) \Gamma(b-a+1)}
{\frac{2^{2b}}{\sqrt{\pi}} \Gamma(b+1/2) \Gamma(b+1)},$$
which is
$$ \sqrt{\pi} 2^{2a-1+2b-2a-2b}
\frac{\Gamma(b+1)}{\Gamma(a) \Gamma(b-a+1)}\Gamma(a+1/2)
\frac{ \Gamma(b-a+1/2)}{ \Gamma(b+1/2)},$$
which is indeed
$$ \frac{\sqrt{\pi}}{2}
\frac{\Gamma(b+1)\Gamma(a+1/2)\Gamma(b-a+1/2)}
{\Gamma(a) \Gamma(b-a+1) \Gamma(b+1/2)},$$
as claimed.
| {
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Help with following equality Could you please help me understand how this is?
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \,.$$
Thank you.
| $\text{L.H.S.}$
$=\displaystyle \frac 1 1-\frac 1 2+\frac 1 3 - \frac 1 4 +\cdots+\frac 1 {2n-1} - \frac 1 {2n}$
$\displaystyle=\frac 1 1+\frac 1 2+\frac 1 3 + \frac 1 4 +\cdots+\frac 1 {2n-1} + \frac 1 {2n}-2\Big(\frac 1 2+\frac 1 4+\cdots+\frac 1 {2n}\Big)$
$\displaystyle=\frac 1 1+\frac 1 2+\frac 1 3 + \frac 1 4 +\cdots+\frac 1 {2n-1} + \frac 1 {2n}-\Big(\frac 1 1+\frac 1 2+\cdots+\frac 1 {n}\Big)$
$=\displaystyle\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} $
$= \text{R.H.S.}$
Q.E.D.
| {
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An integral related to the beta function: $\int_{-1}^{1} \frac{(1+x)^{2m-1}(1-x)^{2n-1}}{(1+x^{2})^{m+n}} \, dx $ I came across an exercise in a textbook that says to show that $$ \int_{-1}^{1} \frac{(1+x)^{2m-1}(1-x)^{2n-1}}{(1+x^{2})^{m+n}} \, dx = 2^{m+n-2} B(m,n), \ (m,n >0),$$ and then deduce that $$ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \Big( \frac{\cos x + \sin x}{\cos x - \sin x} \Big)^{\cos \alpha} \, dx = \frac{\pi}{2 \sin \left( \pi \cos^{2} \frac{\alpha}{2}\right)}, $$ where $\alpha$ is not a multiple of $\pi$.
(Considering that $m$ and $n$ aren't necessarily integers here, using them for the parameters is perhaps a bit unconventional.)
I managed to figure out the second part of the exercise (which I'll show below), but not the first part.
I assume that with the right substitution, one can show that $$\int_{-1}^{1} \frac{(1+x)^{2m-1}(1-x)^{2n-1}}{(1+x^{2})^{m+n}} \, dx = 2^{m+n-2} \int_{0}^{1} u^{m-1} (1-u)^{n-1} \, du. $$
$$ \begin{align} \int_{-1}^{1} \frac{(1+x)^{2m-1}(1-x)^{2n-1}}{(1+x^{2})^{m+n}} \, dx &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{(1 + \tan u)^{2m-1}(1-\tan u)^{2n-1}}{\sec^{2(m+n)} (u)} \, \sec^{2} (u) \, du \\ &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\cos u + \sin u)^{2m-1}(\cos u - \sin u)^{2n-1} \ du \end{align}$$
If we then let $\displaystyle m = \frac{1 + \cos \alpha}{2}$ and $\displaystyle n= \frac{1- \cos \alpha}{2}$, we get
$$ \begin{align} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \Big( \frac{\cos x + \sin x}{\cos x - \sin x} \Big)^{\cos \alpha} \, dx &= \frac{1}{2} \frac{\Gamma \left(\frac{1 + \cos \alpha}{2} \right) \Gamma \left(\frac{1 - \cos \alpha}{2} \right)}{\Gamma (1)} \\ &= \frac{1}{2} \, \Gamma \left(\frac{1 + \cos \alpha}{2} \right) \Gamma \left( 1- \frac{ 1 + \cos \alpha}{2} \right) \\ &= \frac{1}{2} \, \frac{\pi}{\sin \, \left( \pi \frac{1+\cos \alpha}{2} \right)} \\ &= \frac{\pi}{2 \sin \left( \pi \cos^{2} \frac{\alpha}{2}\right)} . \end{align}$$
| Let's prove the first part, you need change the variables.
*
*$y:=\frac{(1+x)^2}{2(1+x^2)}$, then $1-y=\frac{(1-x)^2}{2(1+x^2)}$, and $$\mathrm{d}y=\frac{1-x^2}{(1+x^2)^2}=2\cdot y^{\frac{1}{2}}\cdot (1-y)^{\frac{1}{2}}\cdot \frac{1}{1+x^2}\mathrm{d}x$$
Theorfore,
\begin{eqnarray}
&&\int_{-1}^{1}\frac{(1+x)^{2m-1}(1-x)^{2n-1}}{(1+x^2)^{m+n}}\mathrm{d}x \\
&=&\int_0^1 (2y)^{m-\frac{1}{2}}\cdot (2(1-y))^{n-\frac{1}{2}}\cdot\frac{1}{1+x^2} \big(2\cdot y^{\frac{1}{2}}\cdot (1-y)^{\frac{1}{2}}\cdot\frac{1}{1+x^2}\big)^{-1}\mathrm{d}y\\
&=&2^{m+n-2}\cdot \int_{0}^{1}y^{m-1}\cdot(1-y)^{n-1}\mathrm{d}y\\
&=&2^{m+n-2}\cdot B(m,n)
\end{eqnarray}
| {
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Conditions that $\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$ is rational Motivation
I am working on one of the questions from Hardy's Course of Pure Mathematics and was wondering if I could get some assistance on where to go next in my proof. I have attempted rearranging the expression in numerous ways from the step I am at, but seem to get no-where.
Question
If $a^2-b>0$, then the necessary and sufficient conditions that $\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$ is rational are $a^2-b$ and $\dfrac{1}{2} (a+ \sqrt{a^2-b})$ be squares of rational numbers.
Attempt
Suppose that $\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$ is rational. Then it can be written as the ratio of two integers, p and q, that have no common factor. Write this as:
$\dfrac{p}{q}=\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}}$
Then by squaring both sides we have:
$\dfrac{p^2}{q^2} = (\sqrt{a+\sqrt{b}} + \sqrt{a-\sqrt{b}})^2=2a+ 2\sqrt{a^2-b}$
-Note sure where to go from here.
| Try to square $\sqrt{a+\sqrt b}+\sqrt{a-\sqrt b}$.
| {
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Number of non-negative integer solutions of an equation using generating functions Using convolution, find how many non-negative integer solutions of
$$\begin{aligned} x_1 + 5 x_2 &= n\\ x_1 + 5 x_2 &= 60\end{aligned}$$
Is anyone able to solve this problem or figure out the idea? Thank you so much.
| Here's a combinatorial interpretation: the number of solutions to the second equation is the number of ways to make 60 cents in change using 1 cent and 5 cent coins.
I assume that the second equation is a special case of the first, and that you want to know how to solve both. The method for $n=60$ illustrates the general case.
For a non-negative integer $a$, let $r_a$ be the number of non-negative integer solutions to $x_1=a$ and let $s_a$ be the number of non-negative integer solutions to $5x_2=a$. So $r_a$ is the number of ways to make $a$ cents in change using 1 cent coins alone, and $s_a$ is the number of ways of making $a$ cents in change using 5 cent coins alone. Can you characterize $r_a$ and $s_a$? The number of non-negative integer solutions to $x_1+5x_2=60$ is then given by the convolution
$$r_0s_{60}+r_1s_{59}+\ldots+r_{60}s_0.$$
This is the coefficient of $x^{60}$ in the product of the generating functions for the coefficients $r_a$ and $s_a,$
$$(r_0+r_1x+r_2x^2+\ldots)(s_0+s_1x+s_2x^2+\ldots).$$
You should have found that $r_a=1$ for all $a$ and that $s_a=1$ for $a$ divisible by 5 and $s_a=0$ otherwise. Then
$$r_0+r_1x+r_2x^2+\ldots=1+x+x^2+\ldots=\frac{1}{1-x}$$
and
$$s_0+s_1x+s_2x^2+\ldots=1+x^5+x^{10}+\ldots=\frac{1}{1-x^5}.$$
Here's one method for extracting the coefficient of $x^n$ in the generating function
$$\frac{1}{(1-x)(1-x^5)}.$$
Write
$$\frac{1}{1-x}=\frac{1+x+x^2+x^3+x^4}{1-x^5}.$$
Then
$$\frac{1}{(1-x)(1-x^5)}=\frac{1+x+x^2+x^3+x^4}{(1-x^5)^2}=(1+x+x^2+x^3+x^4)\sum_{r=0}^\infty\binom{-2}{r}(-x^5)^r,$$
where $\binom{-2}{r}=\frac{(-2)(-3)\ldots(-2-r+1)}{r!}$ is the generalized binomial coefficient. You can show that $(-1)^r\binom{-2}{r}=r+1$, and so the generating function is
$$(1+x+x^2+x^3+x^4)\sum_{r=0}^\infty(r+1)x^{5r}.$$
The $x^{60}$ term comes from multiplying the $1=x^0$ term in the polynomial with the $r=12$ term in the summation. Hence the answer is $1\cdot(12+1)=13$.
This method can be generalized to the case where you have additional coin values, say 10 cent coins.
| {
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How does $\frac{1}{2} \sqrt{4 + 4e^4} = \sqrt{1 + e^4}$ My understanding would lead me to believe that:
$$\frac{1}{2} \sqrt{4 + 4e^4} = \frac{1}{2}(2 + 2e^4) = 1 + e^4$$
But it actually equals: $\sqrt{1 + e^4}$
Can you explain why?
| Also you can use squaring both sides:
$$
\sqrt{4+4e^4}=2\sqrt{1+e^4}\\
4+4e^4=4(1+e^4)
$$
since $\sqrt{x^2}= \pm x$ and the expressions you start with is one of these two solutions (positive)
| {
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Column space of complex matrix Let $A=\begin{pmatrix}
1 & \alpha & \alpha^2\\
1 & \beta & \beta^2
\end{pmatrix}, \alpha,\beta\in\mathbb{C},\alpha\ne\beta$. I know that the column space of A is supposed to be $\mathbb{C}^2$, but I'm not sure how to get there.
My attempt:
The column space of $A$ is the set of all vectors of the form
$c_1\begin{pmatrix}1\\1\end{pmatrix}+c_2\begin{pmatrix}\alpha\\\beta\end{pmatrix}+c_3\begin{pmatrix}\alpha^2\\\beta^2\end{pmatrix}=\begin{pmatrix}c_1+\alpha c_2+\alpha^2 c_3\\c_1+\beta c_2+\beta^2 c_3\end{pmatrix}$
Is this the correct way to go about it?
| As $\alpha \neq \beta$, than the first two columns are linear independent, so the dimension of the image is at least $2$ that means your column space is $\mathbb{C}^2$.
If you like a constructive one more than make the following
$$ \begin{pmatrix} \alpha \\ \beta \end{pmatrix} + (-\alpha) \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ \beta - \alpha \end{pmatrix}$$
As $\alpha \neq \beta$ we can scale it to $$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$
Now we calculate $$\begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ \end{pmatrix}$$
and we have the canonical basis. The dimension argument would be enough anyway.
| {
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$ \forall n \in \Bbb N , 18\mid1^n+2^n+\ldots+9^n-3(1+6^n+8^n)$ How to prove:$ \forall n \in \Bbb N , 18\mid1^n+2^n+\ldots+9^n-3(1+6^n+8^n)$ ?
| It is enough to check divisibility by $2$ and by $9$. Divisibility by $2$ is trivial, since $1^n+\cdots+9^n$ and $3(1^n+6^n+8^n)$ are both odd.
To show divisibility by $9$, we first check separately the case $n=1$. This is easy, the expression in that case is $0$. For $n\ge 2$, the terms $3^n$, $6^n$, and $9^n$ are divisible by $9$, so we need to show that
$$1^n+2^n+4^n+5^n+7^n +8^n -3(1^n+8^n)$$
is divisible by $9$. Recall that if $a$ is relatively prime to $9$, then $a^6\equiv 1\pmod{9}$, since $\varphi(9)=6$, where $\varphi$ is the Euler $\varphi$-function.
Thus we will be finished if we can show that
$1^n+2^n+4^n+5^n+7^n+8^n-3(1^n+8^n)$ is divisible by $9$ for $n=0,1,2,3,4,5$. This is a finite computation, so in principle we are almost finished.
To cut down on the work, note that $8\equiv -1\pmod{9}$, $7\equiv -2\pmod{9}$, and $5\equiv -4\pmod{9}$. So if $n$ is odd, the terms $1^n$ and $8^n$, and $2^n$ and $7^n$, and $4^n$ and $5^n$ "cancel" modulo $9$.
Thus it only remains to check the cases $n=0$, $2$, and $4$. The case $n=0$ is easy. For the other $2$ cases, compute modulo $9$.
| {
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How to calculate the asymptotic expansion of $\sum \sqrt{k}$? Denote $u_n:=\sum_{k=1}^n \sqrt{k}$. We can easily see that
$$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + O(k^{-1/2}),$$
hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + O(n^{1/2})$, because $\sum_1^n O(k^{-1/2}) =O(n^{1/2})$.
With some more calculations, we get
$$ k^{1/2} = \frac{2}{3} (k^{3/2} - (k-1)^{3/2}) + \frac{1}{2} (k^{1/2}-(k-1)^{-3/2}) + O(k^{-1/2}),$$
hence $\sum_1^n \sqrt{k} = \frac{2}{3}n^{3/2} + \frac{1}{2} n^{1/2} + C + O(n^{1/2})$ for some constant $C$, because $\sum_n^\infty O(k^{-3/2}) = O(n^{-1/2})$.
Now let's go further. I have made the following calculation
$$k^{1/2} = \frac{3}{2} \Delta_{3/2}(k) + \frac{1}{2} \Delta_{1/2}(k) + \frac{1}{24} \Delta_{-1/2}(k) + O(k^{-5/2}),$$
where $\Delta_\alpha(k) = k^\alpha-(k-1)^{\alpha}$. Hence :
$$\sum_{k=1}^n \sqrt{k} = \frac{2}{3} n^{3/2} + \frac{1}{2} n^{1/2} + C + \frac{1}{24} n^{-1/2} + O(n^{-3/2}).$$
And one can continue ad vitam aeternam, but the only term I don't know how to compute is the constant term.
How do we find $C$ ?
| With Mathematica it very easy:
Series[Sum[Sqrt[k], {k, 1, n}], {n, Infinity, 2}]// TeXForm
$$\frac{2 n^{3/2}}{3}+\frac{\sqrt{n}}{2}+\zeta
\left(-\frac{1}{2}\right)+\frac{\sqrt{\frac{1}{n}}}{24}+O\left(\left(\frac{1}{n}\right)^{5/2}\right)$$
| {
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$\cos{(A-2B)}+\cos{(B-2C)}+\cos{(C-2A)}=\cos{(2A-B)}+\cos{(2B-C)}+\cos{(2C-A)}$ Let $A,B,C\in \mathbb{R}$ with $\sin{A}+\sin{B}+\sin{C}=0$. Prove that
$$\cos{(A-2B)}+\cos{(B-2C)}+\cos{(C-2A)}=\cos{(2A-B)}+\cos{(2B-C)}+\cos{(2C-A)}$$
| The first idea that comes to mind it to express everything in terms of $\sin A,\ \sin B,\ \sin C$ (and the cosines). It turns out that it works:
Begin with $\cos (A - 2B ) = \cos A \cos 2B + \sin A \sin 2B $ (by the formula for $\cos (x+y)$). Now, get rid of the double angles: $\cos 2B = 1 - 2 \sin^2 B$ and $\sin 2B = 2 \sin B \cos B$. This gives you:
$$\cos (A - 2B ) = \cos A - 2 \cos A \sin^2B + 2 \sin A \sin B \cos B $$
Now, the expression on the left side is (using cyclic summation):
$$ L = \sum_{\text{cyc}} \cos A - 2\sum_{\text{cyc}} \cos A \sin^2B + 2 \sum_{\text{cyc}}\sin A \sin B \cos B $$
Note that it follows from the assumption $\sin A + \sin B + \sin C = 0$ (by moving $\sin B$ to the other side, and multiplying by $\sin B$) that $- \sin^2 B = \sin A \sin B + \sin C \sin B$, so this is:
$$ L = \sum_{\text{cyc}} \cos A + 2\sum_{\text{cyc}} \cos A \sin B \sin A + 2\sum_{\text{cyc}} \cos A \sin B \sin C + 2 \sum_{\text{cyc}}\sin A \sin B \cos B $$
This has grown a little messy, but if you shift the summation in the last sum so that it becomes $ \sum_{\text{cyc}}\sin C \sin A \cos A$, you can rearrange it into something nicer:
$$ L = \sum_{\text{cyc}} \cos A + 2\sum_{\text{cyc}} \cos A (\sin A \sin B + \sin B \sin C + \sin C \sin A ) \\
= (1 + 2\sin A \sin B + 2\sin B \sin C + 2\sin C \sin A )\sum_{\text{cyc}} \cos A
$$
Now, look at the right side. Of course, we might redo the computations, and it would come out the same. There is a nicer way to see this: right side differs from the left side just by ordering of $A,B,C$ (in the sense that if you denote the expression on the left by $L(A,B,C)$, and the one on the right by $R(A,B,C)$, then $R(A,B,C) = L(C,B,A)$). But the formula we arrived at does not depend on the order of $A,B,C$! So $R(A,B,C) = L(C,B,A) = L(A,B,C)$, which is what we wanted.
To make sure the proposed solution is more readable, let me explain what I mean by $\sum_{\text{cyc}}$. If $f(A,B,C)$ is any expression involving $A,B,C$, by $\sum_{\text{cyc}} f(A,B,C)$ I mean the expression $f(A,B,C) + f(B,C,A) + f(C,A,B)$. For example, $\sum_{\text{cyc}} \sin A \cos B = \sin A \cos B + \sin B \cos C + \sin C \cos A$. The reasoning strongly relies on the fat that the problem does not change under cyclic rearrangement of $A,B,C$, so if I do some computation for, say, $\sin^2 B$, roughly the same can be done for $\sin^2 A $ and $\sin^2 C$. Note that $\sum_{\text{cyc}} f(A,B,C) = \sum_{\text{cyc}} f(B,C,A)$. Also, $\sum_{\text{cyc}} f(A,B,C)g(A,B,C) = g(A,B,C) \sum_{\text{cyc}} f(A,B,C)$ provided that $g(A,B,C)= g(B,C,A)=g(C,A,B)$
| {
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Prove inequality: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ Prove: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ with $a, b, c \ge0$
I can do this by: $EAT^2$ (expand all of the thing)
*
*$(x+y+z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y+4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{
x}^{2}{z}^{2}+4\,x{y}^{3}+4\,x{z}^{3}+4\,{y}^{3}z+6\,{y}^{2}{z}^{2}+4
\,y{z}^{3}+12\,x{y}^{2}z+12\,xy{z}^{2}+12\,{x}^{2}yz$
*$(x+y-z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y-4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{
x}^{2}{z}^{2}+4\,x{y}^{3}-4\,x{z}^{3}-4\,{y}^{3}z+6\,{y}^{2}{z}^{2}-4
\,y{z}^{3}-12\,x{y}^{2}z+12\,xy{z}^{2}-12\,{x}^{2}yz$
...
$$28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4\\
\iff a^4 + b^4 + c^4 \ge a^2b^2+c^2a^2+b^2c^2 \text{(clearly hold by AM-GM)}$$
but any other ways that smarter ?
| For non-negative variables also TL helps:
Let $a+b+c=3$ (we can assume it because our inequality is homogeneous).
Hence, we need to prove that
$$\sum_{cyc}(28a^4-(3-2a)^4-27)\geq0$$ or
$$\sum_{cyc}(a-1)(a^3+9a^2-9a+9)\geq0$$ or
$$\sum_{cyc}\left((a-1)(a^3+9a^2-9a+9)-10(a-1)\right)\geq0$$ or
$$\sum_{cyc}(a-1)^2(a^2+10a+1)\geq0.$$
Done!
| {
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Triple Integral and symmetry The problem is as follows: Compute the intergal
$$I=\iiint_B \frac{x^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz,$$ where $B$ is the unit ball defined by $B=\{(x,y,z) \mid x^2+y^2+z^2 \leq 1\}$.
The official solution is tricky: The change of variable $(x,y,z) \mapsto (z,y,x)$ transforms the integral into $$\iiint_B \frac{z^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz,\text{ hence }2I= \iiint_B \:dx\:dy\:dz= 4\pi /3,$$ which implies $I=2\pi/3$.
My question is: what is meant by $(x,y,z) \mapsto (z,y,x)$, isn't it ambiguous? and the jacobian $J= -1$, why the integral $$I =\iiint_B \frac{z^4+2y^4}{x^4+4y^4+z^4} \,dx\,dy\,dz$$ instead of $$-\iiint_B \frac{z^4+2y^4}{x^4+4y^4+z^4} \,dx\,dy\,dz\text{ ??}$$
thank you very much!!
| Forget about the Jacobian. From symmetry considerations it is obvious that
$$I:=\iiint_B \frac{x^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz=\iiint_B \frac{z^4+2y^4}{x^4+4y^4+z^4} \:dx\:dy\:dz\ .$$
Therefore $2 I=\int_B 1 \:dx\:dy\:dz= {\rm vol}(B)$ and $I={2\pi\over3}$.
| {
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Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble:
Monotonic:
The sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\leq a_{n+1}$
$$a_1=1\leq 1+\frac{1}{2^2}=a_2$$
Need to show that $a_{n+1}\leq a_{n+2}$
$$a_{n+1}=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=a_{n+2}$$
Thus the sequence is monotone and increasing.
Boundedness:
Since the sequence is increasing it is bounded below by $a_1=1$.
Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don't know what my thinking process should be to find an upper bound.
Can anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence?
Thanks so much in advance!
| This here should work with $n \geq 1 $ :
$$s_n = \sum\limits_{n=1}^\infty \frac{1}{n^2}= \frac{1}{1} + \frac{1}{4} +\frac{1}{9} + \frac{1}{16}+ ...+ \frac{1}{n²} $$$$
b_n = \sum\limits_{n=1}^\infty \frac{1}{2^{n-1}} = \frac{1}{1} + \frac{1}{2} +\frac{1}{4}+\frac{1}{8} + ...+ \frac{1}{2^{n-1}} $$
$b_n$ is directly compared greater than $s_n$ :
$$s_n < b_n$$
and $b_n$ converges, because of its ratio test :
$$\frac{1}{2^{n-1+1}} / \frac{1}{2^{n-1}} = \frac{2^{n-1}}{2^{n}} = \frac{1}{2} < 1$$
| {
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Evaluate $\sum_{k=1}^{n}(k^2 \cdot (k+1)!)$ We have to evaluate the following:
$$1^2 \cdot 2! + 2^2 \cdot 3! + \cdots + n^2 \cdot (n+1)! =\sum_{k=1}^{n} k^2 \cdot (k+1)!$$
Any hints ?
| Using $k^2 = (k+3)(k+2) - 5 (k+2) + 4$ we write $$f_k = k^2 (k+1)! = (k+3)! - 5 (k+2)! + 4 (k+1)! = G_{k+1} - G_k$$
where $G_k = (k+2)! - 4(k+1)!$, therefore
$$
\sum_{k=1}^n f_k = \sum_{k=1}^n \left(G_{k+1} - G_k\right) = G_{n+1} - G_1
$$
Since $G_1 = -2$, and $G_{n+1} = (n+3)! - 4(n+2)! = (n+2)! (n-1)$ we get
$$
\sum_{k=1}^n k^2 (k+1)! = (n-1) (n+2)! + 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Series expansion $(1-\cos{x})^{-1}$ How do i get the series expansion $(1-\cos{x})^{-1} = \frac{2}{x^2}+\frac{1}{6}+\frac{x^2}{120}+o(x^4)$ ?
| $$\frac{1}{1 - x} = 1 + x + x^2 + O(x^3)$$
and so
$$\frac{1}{1 - \cos x}= \frac{1}{\frac{x^2}{2} - \frac{x^4}{24} + \frac{x^6}{6!}} = \frac{2}{x^2} \frac{1}{1 - (\frac{x^2}{12} - \frac{x^4}{360} + O(x^6))}$$
$$ = \frac{2}{x^2} \left( 1 + \left(\frac{x^2}{12} - \frac{x^4}{360}\right) + \frac{x^4}{144} + O(x^6) \right) = \frac{2}{x^2} + \frac{1}{6} + \frac{x^2}{120} + O(x^4) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/334685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Euler's infinite product for the sine function and differential equation relation Euler's infinite product for the sine function
$$\displaystyle \sin( x) = x \prod_{k=1}^\infty \left( 1 - \frac{x^2}{\pi^2k^2} \right)$$
http://en.wikipedia.org/wiki/Basel_problem
We know that $\sin( x)$ satisfies $y''+y=0$ differential equation.
$$\displaystyle \frac{\sin'( x)}{\sin( x)} = \frac{1}{x}-2x \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}$$
$$\displaystyle \sin'( x) = \sin( x)\left(\frac{1}{x}-2x \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2} \right)$$
$$\displaystyle \sin''( x) = \sin'( x) \left(\frac{1}{x}-2x \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}\right)+ \sin( x) \left(-\frac{1}{x^2}-2\sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}-4x^2 \sum_{k=1}^\infty \frac{1}{(\pi^2k^2-x^2)^2}\right)$$
$$\displaystyle \sin''( x) = \left(\frac{\sin( x)}{x}-2x \sin( x) \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}\right) \left(\frac{1}{x}-2x \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}\right)+ \sin{x} \left(-\frac{1}{x^2}-2\sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2}-4x^2 \sum_{k=1}^\infty \frac{1}{(\pi^2k^2-x^2)^2}\right)$$
$$\displaystyle \sin''( x) = \sin x \left(+4x^2\left(\sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2} \right)^2-6 \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2} -4x^2\sum_{k=1}^\infty \frac{1}{(\pi^2k^2-x^2)^2} \right)$$
If $\sin( x)$ satisfies $y''+y=0$ differential equation.
Then $$ 4x^2\left(\sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2} \right)^2-6 \sum_{k=1}^\infty \frac{1}{\pi^2k^2-x^2} -4x^2\sum_{k=1}^\infty \frac{1}{(\pi^2k^2-x^2)^2}=-1$$ must be equal
I am stuck to prove the relation in another way. How can I prove that the last relation is equal to $-1$ ?
Note:
If
$x=0$
Easily we can see that
$$ -6 \sum_{k=1}^\infty \frac{1}{\pi^2k^2} =-1$$
$$ \sum_{k=1}^\infty \frac{1}{k^2} =\frac{\pi^2}{6}$$
This result famous basel problem result.
Thanks for answers
| The first sum is a known sum which I will not prove here:
$$\sum_{k=1}^{\infty} \frac{1}{\pi^2 k^2-x^2} = \frac{1}{2 x} \left ( \frac{1}{x}-\cot{x}\right)$$
The second sum, on the other hand, I could not find in a reference. You can, however, evaluate it using residues. That is,
$$\sum_{k=-\infty}^{\infty} \frac{1}{(\pi^2 k^2-x^2)^2} = -\sum_{\pm} \text{Res}_{z=\pm x/\pi} \frac{\pi \cot{\pi z}}{(\pi^2 z^2-x^2)^2}$$
I will spare you the residue calculation here; needless to say, the result for the sum is
$$\sum_{k=-\infty}^{\infty} \frac{1}{(\pi^2 k^2-x^2)^2} = \frac{\cot^2{x}}{2 x^2} + \frac{\cot{x}}{2 x^3}+ \frac{1}{2 x^2}$$
which means that
$$\sum_{k=1}^{\infty} \frac{1}{(\pi^2 k^2-x^2)^2} = \frac{\cot^2{x}}{4 x^2} + \frac{\cot{x}}{4 x^3}+ \frac{1}{4 x^2} - \frac{1}{2 x^4}$$
I also leave the algebra to the reader in plugging these expressions into the equation the OP has provided. In the end, yes, the relation is true.
| {
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"url": "https://math.stackexchange.com/questions/335789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a + b^3)(4)} \ge 3\sqrt{a} + \sqrt{b^3}$$
$$\sqrt{(3a + b^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2}$$
Also I get:
$$\sqrt{(3b + c^3)} \ge \frac{3\sqrt{b} + \sqrt{c^3}}{2}$$
$$\sqrt{(3c + a^3)} \ge \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
If I add add 3 inequalities I get:
$$\sqrt{(3a + b^3)} + \sqrt{(3b + c^3)} + \sqrt{(3c + a^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2} + \frac{3\sqrt{b} + \sqrt{c^3}}{2} + \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$
Now i need to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} + \frac{3\sqrt{b} + \sqrt{b^3}}{2} + \frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge 6 = 2(a+b+c)$$
It's enough now to prove that:
$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} \ge b+c = 3-a$$
$$\frac{3\sqrt{b} + \sqrt{b^3}}{2} \ge a+c = 3-b$$
$$\frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge b+a = 3-c$$
All three inequalities are of the form:
$$\frac{3\sqrt{x} + \sqrt{x^3}}{2} \ge 3-x$$
$$3\sqrt{x} + \sqrt{x^3} \ge 6-2x$$
$$(3\sqrt{x} + \sqrt{x^3})^2 \ge (6-2x)^2$$
$$9x + x^3 + 6x^2 \ge 36 - 24x + 4x^2$$
$$x^3 + 2x^2 + 33x - 36 \ge 0$$
$$(x-1)(x^2 + 3x + 33) \ge 0$$
Case 1:
$$(x-1) \ge 0 \ \ \ \ \text{ for any }\ x \geq 1$$
$$(x^2 + 3x + 33) \ge 0 \ \ \ \ \text{ for any x in R} $$
Case 2:
$$0 \ge (x-1) \ \ \ \ \text{ for any }\ 1 \geq x$$
$$0 \ge (x^2 + 3x + 33) \ \ \ \ \text{there are no solutions in R} $$
This proves that for $$x \geq 1$$ $$(x-1)(x^2 + 3x + 33) \ge 0$$ is true and so it is
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$$, but a, b, c can be every non-negative number. I proved it's true for $$a,b,c \geq 1$$, but i can't for $$a,b,c \geq 0$$
| maybe this following solution is by Vasc?
since use Cauchy-Schwarz inequality,we have
$$(a^3+3b)(a+3b)\ge (a^2+3b)^2$$
It suffices to show that
$$\sum_{cyc}\dfrac{a^2+3b}{\sqrt{a+3b}}\ge 6$$
By Holder
$$\left(\sum_{cyc}\dfrac{a^2+3b}{\sqrt{a+3b}}\right)^2[\sum_{cyc}(a^2+3b)(a+3b)]\ge[\sum_{cyc}(a^2+3b)]^3=[a^2+b^2+c^2+9]^3$$
it is enought to show that
$$\left(a^2+b^2+c^2+9\right)^3\ge 36\sum_{cyc}(a^2+3b)(a+3b)$$
let $p=a+b+c=3,q=ab+bc+ac\le 3$,we have
$$\sum_{cyc}(a^2+3b)(a+3b)=108-24q+3[abc+\sum_{cyc}a^2b]$$
use this well know：see inequality
$$abc+\sum_{cyc}a^2b\le 4$$
we get
$$\sum_{cyc}(a^2+3b)(a+3b)\le 24(5-q)$$
and
$$a^2+b^2+c^2+9=2(9-q)$$
It suffices to show that
$$(9-q)^3\ge 108(5-q)$$
which is true,because equivalent
$$(q-3)^2(21-q)\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 0
} |
How can find this sequence $ a_{n+1}=a_{n}+na_{n-1},$ let $a_{n+1}=a_{n}+na_{n-1},a_{1}=1，a_{2}=2$.
find the $a_{n}=?$
my ideas: $\dfrac{a_{n+1}}{(n+1)!}=\dfrac{1}{n+1}\dfrac{a_{n}}{n!}+\dfrac{1}{n+1}\dfrac{a_{n-1}}{(n-1)!},$
and let $b_{n}=\dfrac{a_{n}}{n!}$,then we have
$(n+1)b_{n+1}=b_{n}+b_{n-1},b_{1}=1,b_{2}=1$,but I failure,
so let $f(x)=\displaystyle\sum_{n=0}^{\infty}b_{n}x^n$,and assume that $b_{0}=0$
$f'(x)=\displaystyle\sum_{n=0}^{\infty}nb_{n}x^{n-1}=1+\displaystyle\sum_{n=1}^{\infty}(n+1)b_{n+1}x^{n-1}=1+\displaystyle\sum_{n=1}^{\infty}(b_{n}+b_{n-1})x^{n-1}=1+\dfrac{f(x)}{x}+f(x)$
then $f'(x)-(1+1/x)f(x)=1,f'(0)=1$
so
$f(x)=xe^x(\displaystyle\int \dfrac{e^{-x}}{x}dx+c)$
so $c=0$
and $f(x)=\displaystyle\sum_{n=0}^{\infty}\dfrac{x^{n+1}}{n!}\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{x^n}{(n+1)!}$
and my problem：How can prove that:
if:$f(x)=\displaystyle\sum_{n=0}^{\infty}b_{n}x^n=\displaystyle\sum_{n=0}^{\infty}\dfrac{x^{n+1}}{n!}\displaystyle\sum_{n=0}^{\infty}(-1)^n\dfrac{x^n}{(n+1)!}$
then we have
$b_{n}=\displaystyle\sum_{k=0}^{[n/2]}\dfrac{1}{（n-2k)!2^kk!}?$
| If $f(x)=\sum_{n\ge 1}a_nx^{n+1}$ then
$$
x^3f'(x)+(x-1)f(x)+x^3+x^2=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Relation between area of a triangle on a sphere and plane We know area of a plane triangle $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\frac{a+b+c}{2}$.
I was just thinking: let we have a triangle with arc length $a,b,c$ on a sphere of radius $r$, do we have any similar kind of formula for that spherical triangle? when radius of $r\to \infty$ we get the plane, so do we have any estimate of area of spherical triangle when $r\to\infty$?
Any reference and article link are also welcome! Thank you.
|
Source: This Dr. Math Article
Novice here, so please excuse any mistakes. The ratio should be:
$$
\frac{ 180 \cdot \sqrt{ s(s-a)(s-b)(s-c) } }
{ 4 \cdot \pi \cdot R^2 \cdot \arctan \left(
\sqrt{ \tan \left( \frac{s}{2} \right) \cdot \tan \left( \frac{s-a}{2} \right) \cdot \tan \left( \frac{s-b}{2} \right) \cdot \tan \left( \frac{s-c}{2} \right) }
\right) } $$
Still needs some simplification, though...
Basically, I just placed Heron's formula for the area of planar $\Delta$s above the $\frac{ \pi \cdot R^2 \cdot E}{180}$ formula for the area of a spherical $\Delta$. Since I didn't have the angular measures required to calculate $E$, I used this formula:
$$ \tan \left( \frac{E}{4} \right) = \sqrt{ \tan \left( \frac{s}{2} \right) \cdot \tan \left( \frac{s-a}{2} \right) \cdot \tan \left( \frac{s-b}{2} \right) \cdot \tan \left( \frac{s-c}{2} \right) } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
What are some useful tricks/shortcuts for verifying trigonometric identities? What "tricks" are there that could help verify trigonometric identities?
For example one is:
$$a\cos\theta+b\sin\theta = \sqrt{a^2+b^2}\,\cos(\theta-\phi)$$
| Expand the cosine of the difference of angles on the right.
$$ \cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi$$
Now, collect terms so that the expression is a linear combination of $\sin \theta$ and $\cos \theta$, as in the expression on the left.
$$ \begin{align}
\sqrt{a^2 + b^2}\cos(\theta - \phi) &= \sqrt{a^2 + b^2}(\cos \theta \cos \phi + \sin \theta \sin \phi)\\
&=\sqrt{a^2 + b^2}\cos \phi \cdot \cos \theta + \sqrt{a^2 + b^2}\sin \phi \cdot \sin \theta\\
&=a \cos \theta + b \sin \theta
\end{align}$$
This last equality holds as long as we choose $\phi$ such that
$$ \left\{\begin{align} \cos \phi &= \frac{a}{\sqrt{a^2 + b^2}}\\ \sin \phi &= \frac{b}{\sqrt{a^2 + b^2}} \end{align}\right.$$
This is always possible since those are the coordinates of a point on the unit circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Calculate:$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$ How to calculate following with out using L'Hospital rule
$$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$$
| Provided the following limits exist,
\begin{align*}
A &= \lim_{x \to -1^{+}}\left(\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{\sqrt{x+1}} \right)\\
&= \lim_{x\to -1^+}\left(\frac{\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{x + 1}}{\frac{\sqrt{x+1} - 0}{x + 1}} \right)\\
&= \frac{\lim_{x\to -1^+}\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{x + 1}}{\lim_{x\to -1^+}\frac{\sqrt{x+1} - 0}{x + 1}},
\end{align*}
which we recognize as the quotient of the "derivative from the right" of $\sqrt{\cos^{-1} x}$ and $\sqrt{ x + 1}$ at $x = -1$. So,
\begin{align*}
A &= \lim_{x\to -1^+}\frac{\frac{1}{2\sqrt{1 - x^2}\sqrt{\cos^{-1} x}}}{\frac{1}{2\sqrt{x + 1}}}\\
&= \lim_{x\to -1^+}\frac{\sqrt{x + 1}}{\sqrt{1 - x^2}\sqrt{\cos^{-1} x}}\\
&= \lim_{x\to -1^+}\frac{\sqrt{x + 1}}{\sqrt{1 - x}\sqrt{1 + x}\sqrt{\cos^{-1} x}}\\
&= \lim_{x\to -1^+}\frac{1}{\sqrt{1 - x}\sqrt{\cos^{-1} x}}\\
&= \frac{1}{\sqrt{2\pi}}.\\
\end{align*}
(in the first equation we flip the sign of the derivative because they're being taken in different orders: the top has the value first, limit second, bottom has limit first, value second.)
Admittedly, I'm playing pretty fast and loose here, but if you're ambitious and have a bit of tenacity (and free time), you can probably fill in the missing details.
| {
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"url": "https://math.stackexchange.com/questions/337603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| Such a sequence is called Arithemtico-Geomteric Progression.
$$S_n=\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i } } } $$
$$\frac{S_n}{2}=\sum _{ i=1 }^{ n }{ \frac { i }{ { 2 }^{ i+1 } } }=\sum _{ i=2 }^{ n+1 }{ \frac { i-1 }{ { 2 }^{ i } } }$$
Subtracting
$$\frac{S_n}{2}=\sum _{ i=1 }^{ n }{ \frac { 1 }{ { 2 }^{ i } } } -\frac { n }{ { 2 }^{ n+1 } } $$
as $n\rightarrow \infty$
It's easily seen that $S_{\infty}=2$
How to evaluate that limit
$$\sum _{ i=1 }^{ n }{ x^{i-1} } =\frac{1-x^n}{1-x}$$ If $|x|<1$ as $n\rightarrow \infty$
$$\sum _{ i=1 }^{ n }{ x^{i-1} } =\frac{1-x^n}{1-x}=\frac{1}{1-x}$$
Second one is directly from Taylor series.
Although there exist simpler proof , I have a rigorous proof of the second part of the limit
$$0<\log _{ 2 }{ x } =\log _{ 2 }{ e } \int _{ 1 }^{ x }{ \frac { 1 }{ t } } dt$$
When $x>1$ $\frac{1}{t}<\frac{1}{\sqrt{t}}$ is valid
$\log _{ 2 }{ e } \int _{ 1 }^{ x }{ \frac { 1 }{ t } } dt<\log _{ 2 }{ e } \int _{ 1 }^{ x }{ \frac { 1 }{ \sqrt { t } } } dt=2\log _{ 2 }{ e } \left( \sqrt { x } -1 \right) <2\log _{ 2 }{ e }\cdot \sqrt { x } $
$0<\log _{ 2 }{ x } <2\log _{ 2 }{ e } \sqrt { x }$ $$ \Rightarrow 0<\frac { \log _{ 2 }{ x } }{ x } <\frac { 2\log _{ 2 }{ e } }{ \sqrt { x } } \tag{1} $$
As $x\rightarrow \infty$ using $(1)$ and squeeze principle. We get
$$\lim_{x\rightarrow \infty}{\frac{\log_{2}{x}}{x}}=0\tag{2}$$
By continuity of $2^t$ making the sutbtituion $x=2^t$ and as $x\rightarrow \infty$ then.$t\rightarrow \infty$
Now $(2)$ is changed to $$\lim_{t\rightarrow \infty}{\frac{t}{2^t}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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"answer_id": 0
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Use the $\varepsilon$ - $\delta$ definition to prove $\lim_{x\to\,-1}\frac{x}{2x+1}=1$ Use the $\varepsilon$ - $\delta$ definition of limit to prove that $\displaystyle\lim_{x\to\,-1}\frac{x}{2x+1}=1$.
My working:
$\left|\frac{x}{2x+1}-1\right|=\left|\frac{-x-1}{2x+1}\right|=\frac{1}{\left|2x+1\right|}\cdot \left|x+1\right|$
First restrict $x$ to $0<\left|x+1\right|<\frac{1}{4}$ $\Rightarrow$ initial choice of $\delta=\frac{1}{4}$
$\left| 2x+1 \right|$ = $\left|2(x+1)-1\right|$
$\le$ $\left|2(x+1)\right|+\left|-1\right|$ = $2\left|x+1\right|+1$ $> 1$
Thus if $\left|x+1\right|<\frac{1}{4}$ , then
$\left|\frac{x}{2x+1}-1\right|=\frac{1}{\left|2x+1\right|}.\left|x+1\right|$
$<1.\left|x+1\right|$.
Therefore, $\delta = \min\{\frac{1}{4},\varepsilon\}$
$0<\left|x+1\right|<\delta$ $\Rightarrow$ $\left|\frac{x}{2x+1}-1\right| < 1\cdot\left|x+1\right| < 1\cdot\varepsilon = \varepsilon$
Thus, the limit is 1.
| To "discover" the proof, we typically work backwards. We "assume" ${|x+1|\over|2x+1|}<\epsilon$, and find what $\delta$ should be. Then we will have to rewrite the proof.
Let's think about the numerator and denominator separately. We want $|x+1|$ small, and intuitively, we want $|2x+1|$ large. So let's keep $2x+1$ away from $0$ first. I think that's the $\delta<{1\over4}$ part. Then
$$-{1\over 4}<x-(-1)<{1\over4},$$
so $-{5\over 4}<x<-{3\over4}$, so $-{5\over 2}<2x<-{3\over2}$ so $-{3\over 2}<2x+1<-{1\over2}$. So we use the fact that $|2x+1|>{1\over2}$. Thus,
$${|x+1|\over|2x+1|}<2|x+1|<\epsilon.$$
Then we notice
$$\delta<\min\bigg\{{\epsilon\over2},{1\over4}\bigg\}$$
Should do the trick. At this point, we've just "discovered" the proof, and $\delta(\epsilon)$, so now we proceed to write the logic out concisely. Fix $\epsilon$, blah blah blah
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $N$ be a $2× 2$ complex matrix such that $N^2=0$. how could I show $N=0$, or $N$ is similar over the matrix. Let $N$ be a $2× 2$ complex matrix such that $N^2=0$. how could I show $N=0$, or $N$ is similar over $\mathbb{C}$ to \begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}
| Elementarily, assume $$N = \begin{bmatrix}a & b\\c & d\end{bmatrix} $$
Then
$$0=N^2 = \begin{bmatrix}a^2+bc & b(a+d)\\c(a+d) & d^2+bc\end{bmatrix} $$
Now if $b=0$, you get
$$0=N^2 = \begin{bmatrix}a^2 & 0\\c(a+d) & d^2\end{bmatrix} $$
so that $a=d=0$, and the result follows.
Otherwise, you must conclude $a=-d$ so
$$0=N^2 = \begin{bmatrix}a^2+bc & 0\\0 & a^2+bc\end{bmatrix} $$
giving that $c=-\frac{a^2}{b}$ so the original matrix is
$$N = \begin{bmatrix}a & b\\-\frac{a^2}{b} & -a\end{bmatrix} $$
and then observe that
$$\begin{bmatrix}0 & 1\\\frac{1}{b} & -\frac{a}{b}\end{bmatrix}
\begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}
\begin{bmatrix}a & b\\1 & 0\end{bmatrix} = N$$
| {
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"url": "https://math.stackexchange.com/questions/339721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $a+b+c=0$, simplify the following. I am here again to ask a question about an exercise I saw around but i'm having a lot of trouble with. I know the answer is 3abc, but as in many of my questions, I am interested in the why and how.
Given $a+b+c=0$, simplify:
$$\frac{(a^3-abc)^3+(c^3-abc)^3+(b^3-abc)^3}{(c^2-ab)(b^2-ac)(a^2-bc)}$$
| If you just sub in $c=-a-b$ then each factor of the denominator reduces to $(a^2+ab+b^2)$.
Each term of the numerator has a factor that is one of these denominator factors as well. For instance the first term from the numerator is $a^3(a^2-bc)^3$, which is $a^3(a^2+ab+b^2)^3$.
So we have
$$\begin{align}
\frac{a^3(a^2+ab+b^2)^3+c^3(a^2+ab+b^2)^3+b^3(a^2+ab+b^2)^3}{(a^2+ab+b^2)^3}
\end{align}
$$
Now provided $a^2+ab+b^2\neq0$ (which is guaranteed the case if all our numbers are real), we have $$a^3+b^3+c^3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Integral check $\int \frac{6x+4}{x^4+3x^2+5} \ \text{dx}$ $$\int \frac{6x+4}{x^4+3x^2+5} \ \text{dx}$$ Partial fraction decomposition of $\frac{6x+4}{x^4+3x^2+5}$ is of the following form: $$\frac{Ax+B}{x^2+2}+\frac{Cx+D}{x^2+1}$$ We need to find $A,B,C$ and $D$ \ $$\frac{6x+4}{x^4+3x^2+5}=\frac{Ax+B}{x^2+2}+\frac{Cx+D}{x^2+1}$$ Or, $$6x+4=x^3(A+C)+x^2(B+D)+x(2A+C)+2B+D$$ By solving: $$\begin{cases} A+C=0 \\ B+D=0 \\ 2A+C=6 \\2B+D=4 \end{cases}$$ We get $$A=-6,B=-4,C=6,D=4$$ Hence $$\frac{6x+4}{x^4+3x^2+5}=-\frac{6x+4}{x^2+2}+\frac{6x+4}{x^2+1}$$ Now, by linearity, $$\int \frac{6x+4}{x^4+3x^2+5} \ \text{dx}=-\int\frac{6x+4}{x^2+2}\ \text{dx}+\int\frac{6x+4}{x^2+1}\ \text{dx}$$ Or,
$$-3\int\frac{2x}{x^2+2}\ \text{dx}-4\int\frac{1}{x^2+2}\ \text{dx}+3\int\frac{2x}{x^2+1}\ \text{dx}+4\int\frac{1}{x^2+1}\ \text{dx}$$ Which equals to:
$$-3\ln(x^2+2)+\frac{4}{\sqrt{2}}\text{arctan}(\frac{x}{\sqrt{2}})+3\ln(x^2+1)+4\text{arctan}(x)+C$$
| All the signs of all your letters are switched. This is because $A+A+C=A+0=A=6$, $B+B+D=B+0=B=4$, etc. Also, $$(x^2+1)(x^2+2)=x^4+3x^3+2\neq x^2+3x+5$$ Your polynomial has no real roots, so you'll have to do some manipulation in $\Bbb C$, as follows:
Let $x^2=u$. Then $$u^2+3u+5=0$$ has solutions $$u_1=\frac{-3+i\sqrt 11}{2}$$
$$u_2=\frac{-3-i\sqrt 11}{2}$$
Thus $$u^2+3u+5=\left(u-\frac{-3-i\sqrt 11}{2}\right)\left(u-\frac{-3+i\sqrt 11}{2}\right)$$
We can write then $${u^2} + 3u + 5 = \left( {u + {3 \over 2} + {{i\sqrt 1 1} \over 2}} \right)\left( {u + {3 \over 2} - {{i\sqrt 1 1} \over 2}} \right)$$ and using the difference of squares factorization $${u^2} + 3u + 5 = {\left( {u + {3 \over 2}} \right)^2} - {{{i^2}11} \over 4} = {\left( {u + {3 \over 2}} \right)^2} + {{11} \over 4}$$ Thus $${x^4} + 3{x^2} + 5 = {\left( {{x^2} + {3 \over 2}} \right)^2} + {{11} \over 4}$$
You can move on from that.
| {
"language": "en",
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} |
Show that $\displaystyle\lim_{x\rightarrow 0}\frac{5^x-4^x}{x}=\log_e\left({\frac{5}{4}}\right)$
*
*Show that $\displaystyle\lim_{x\rightarrow 0}\frac{5^x-4^x}{x}=\log_e\left({\frac{5}{4}}\right)$
*If $0<\theta < \frac{\pi}{2} $ and $\sin 2\theta=\cos 3\theta~~$ then find the value of $\sin\theta$
| For the first one:
$$\lim_{x\to 0}\frac{5^x-4^x}{x} = \frac{\mathrm{d}(5^x-4^x)}{\mathrm{d}x}(0) = \frac{\mathrm{d}5^x}{\mathrm{d}x}(0)-\frac{\mathrm{d}4^x}{\mathrm{d}x}(0) = \log_e\frac{5}{4}$$
For the second:
$$\sin 2\frac{\pi}{10} + \cos 3\frac{\pi}{10} = 0.$$
The golden-ratio triangle has angles $\frac{\pi}{5}$ and two $\frac{2\pi}{5}$, so $\cos\frac{\pi}{5} = \frac{1+\sqrt{5}}{4}$ and using $\cos 2\theta = 1-2\sin^2\theta$ we get
$$\sin\frac{\pi}{10} =
\sqrt{\frac{1-\cos\frac{\pi}{5}}{2}} =
\sqrt{\frac{3-\sqrt{5}}{8}} =
\sqrt{\frac{5-2\sqrt{5}+1}{16}} =
\frac{\sqrt{5}-1}{4}
.$$
I hope this helps ;-)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The minimum value of $a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}?$ I came across the following problem :
Let $a,b,c$ are non-zero real numbers .Then the minimum value of $a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}?$ This is a multiple choice question and the options are $0,6,3^2,6^2.$
I do not know how to progress with the problem. Can someone point me in the right direction? Thanks in advance for your time.
| $a^2+b^2+c^2+a^{-2}+b^{-2}+c^{-2}=(a-a^{-1})^2+(b-b^{-1})^2+(c-c^{-1})^2+6$, whence the minimum occurs when $a=a^{-1},b=b^{-1},c=c^{-1}$ and is $6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Calculate volume in a 3D sort of space using cartesian coordinates Find the volume bounded by the cylinder $x^2 + y^2 = 1$, the planes $x=0, z=0, z=y$ and lies in the first octant. (where x, y, and z are all positive)
| Use a triple integral. We're simply trying to find the volume$-$there's no density involved or anything$-$so the important question is just to find the correct bounds.
Let's examine the $x$ variable first. $x$ can move from $0$ to $1$, since it's constrained below by $0$ and cannot be greater than $1$ (otherwise $x^2 + y^2 > 1$).
Next, look at $y$. $y$ is bounded below by $0$ (must lie in the first octet) and above by $\sqrt{1-x^2}$ (so that $x^2 + y^2 \le 1$).
Finally, look at $z$. $z$ is bounded below by $0$ and above by $y$ (because of the plane $z=y$)
\begin{align*}
\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^y 1 dz dy dx &= \int_0^1 \int_0^{\sqrt{1-x^2}} ydydx \\
&= \int_0^1 \frac{1}{2} (1-x^2) dx \\
&= \frac{1}{2} - \frac{1}{6} \\
&= \frac{1}{3}
\end{align*}
| {
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"source": "stackexchange",
"question_score": "2",
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Find the value of : $\lim_{x\to\infty}x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$ How can I show/explain the following limit?
$$\lim_{x\to\infty} \;x\left(\sqrt{x^2-1}-\sqrt{x^2+1}\right)=-1$$
Some trivial transformation seems to be eluding me.
| A short calculation, using equivalents:
$$ x\Bigl(\sqrt{x^2-1}-\sqrt{x^2+1}\Bigr)=\frac{x\bigl((x^2-1)-(x^2+1)\bigr)}{\sqrt{x^2-12}+\sqrt{x^2+1}}= \frac{-2x}{\sqrt{x^2-1}+\sqrt{x^2+1}}.$$
Now, for $x>0$, we have
$$ \sqrt{x^2-1}+\sqrt{x^2+1}=x\biggl(\sqrt{1-\frac1{x^2}}+\sqrt{1+\frac 1{x^2}\biggr)}
\sim_{+\infty}2x, $$
since the contents of the parenthesis tends to $2$ at $\infty$, so
$$ x\Bigl(\sqrt{x^2-1}-\sqrt{x^2+1}\Bigr)\sim_{+\infty}\frac{-2x}{2x}=-1.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Struggling with an integral with trig substitution I've got another problem with my CalcII homework. The problem deals with trig substitution in the integral for integrals following this pattern: $\sqrt{a^2 + x^2}$. So, here's the problem:
$$\int_{-2}^2 \frac{\mathrm{d}x}{4 + x^2}$$
I graphed the function and because of symmetry, I'm using the integral: $2\int_0^2 \frac{\mathrm{d}x}{4 + x^2}$
Since the denominator is not of the form: $\sqrt{a^2 + x^2}$ but is basically what I want, I ultimately decided to take the square root of the numerator and denominator:
$$2 \int_0^2 \frac{\sqrt{1}}{\sqrt{4+x^2}}\mathrm{d}x = 2 \int_0^2 \frac{\mathrm{d}x}{\sqrt{4+x^2}}$$
From there, I now have, using the following: $\tan\theta = \frac{x}{2} => x = 2\tan\theta => dx = 2\sec^2\theta d\theta$
$$
\begin{array}{rcl}
2\int_{0}^{2}\frac{\mathrm{d}x}{4+x^2}\mathrm{d}x & = & \sqrt{2}\int_{0}^{2}\frac{\mathrm{d}x}{\sqrt{4+x^2}}\mathrm{d}x \\
& = & \sqrt{2}\int_{0}^{2}\frac{2\sec^2(\theta)}{\sqrt{4+4\tan^2(\theta)}}\mathrm{d}\theta \\
& = & \sqrt{2}\int_{0}^{2}\frac{2\sec^2(\theta)}{2\sqrt{1+\tan^2(\theta)}}\mathrm{d}\theta \\
& = & \sqrt{2}\int_{0}^{2}\frac{\sec^2(\theta)}{\sqrt{\sec^2(\theta)}}\mathrm{d}\theta \\
& = & \sqrt{2}\int_{0}^{2}\frac{\sec^2(\theta)}{\sec(\theta)}\mathrm{d}\theta \\
& = & \sqrt{2}\int_{0}^{2}\sec(\theta)\mathrm{d}\theta \\
& = & \sqrt{2}\left [\ln{\sec(\theta)+\tan(\theta)} \right|_{0}^{2}] \\
& = & \sqrt{2}\left [ \ln{\frac{\sqrt{4+x^2}}{2}+\frac{x}{2} } \right|_{0}^{2} ]
\end{array}
$$
I'm not sure if I've correctly made the integral look like the pattern it's supposed to have. That is, trig substitutions are supposed to be for $\sqrt{a^2 + x^2}$ (in this case that is, there are others). This particular problem is an odd numbered problem and the answer is supposed to be $\frac{\pi}{4}$. I'm not getting that. So, the obvious question is, what am I doing wrong? Also note, I had trouble getting the absolute value bars to produce for the ln: don't know what I did wrong there either.
Thanks for any help,
Andy
| Let $x=2\tan\theta$
and $dx=2 \sec^2\theta$
$2\tan\theta = -2, 2$
$\theta = \frac{-\pi}{4}, \frac{\pi}{4}$
$\int_{\frac{-\pi}{4}}^\frac{\pi}{4}\frac{2\sec^2\theta}{4+4\tan^2\theta}d\theta$
$\int_{\frac{-\pi}{4}}^\frac{\pi}{4}\frac{2\sec^2\theta}{4\sec^2\theta}d\theta$
$\int_{\frac{-\pi}{4}}^\frac{\pi}{4}\frac{1}{2} d\theta$
$\frac{\pi}{8}-\frac{-\pi}{8} = \frac{\pi}{4}$
Sorry about the formatting. I'm just learning latex.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Set of all triangles with two equal edges inscribed in a circle. Let $\Delta$ be the set of all triangles with two equal edges inscribed in a circle of radius $R$.
So, how do I show that:
1, The equilateral triangle in $\Delta$ is the one maximizing the area.
2, The equilateral triangle in $\Delta$ is the one maximizing the circumference?
Help greatly appreciated!
| We can prove that Isosceles Triangles have Two Equal Angles
Now using Law of Sines, $$\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}=2R$$
Let $A=B=x\implies C=\pi-2x,\sin C=\sin(\pi-2x)=\sin 2x$
So, $x=A>0$ and $\pi-2x=C>0\implies x<\frac\pi2\implies 0<x<\frac\pi2$
So, $a=b=2R\sin x,c=2R\sin2x$
So, the circumference will be $=a+b+c=2R(2\sin x+\sin2x)$ where $R$ is constant
and the area will be $\frac{abc}{4R}=\frac{(2R\sin x)^2(2R\sin2x)}{4R}=4R^2\sin^3x\cos x$ where $R$ is constant
We can apply Second Derivative Test w.r.t. $x$, to find the required extreme values as follows
Let $f(x)=2\sin x+\sin2x\implies f'(x)=2(\cos x+\cos2x)$
For the extreme values of $f(x),f'(x)=0$
$\implies \cos2x+\cos x=0\implies 2\cos^2x+\cos x-1=0$
$\implies \cos x=\frac{-1\pm\sqrt{1^2-4\cdot2\cdot(-1)}}{2\cdot2}=-1$ or $\frac12$
As $0<x<\frac\pi2,\cos x>0\implies \cos x=\frac12\implies x=\frac\pi3$
$$\text{ Now, show that }f''(\frac\pi3)<0$$
Let $g(x)=8\sin^3x\cos x$ which can be simplified to $2\sin2x-\sin4x$ (See the comment below)
For $g'(x)=0,$ $4(\cos2x-\cos4x)=0\implies 2\cos^22x-\cos2x-1=0$
$$\implies \cos2x=1\text{ or }-\frac12$$
As $0<x<\frac\pi2\implies 0<2x<\pi \implies \cos x\ne1\implies \cos 2x=-\frac12=\cos\frac{2\pi}3$
$\implies 2x=2n\pi\pm \frac{2\pi}3\implies 2x=\frac{2\pi}3\iff x=\frac\pi3$ as $0<2x<\pi$
Now, show that $g''(\frac\pi3)<0$
Some generalization can be found here and here
| {
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"source": "stackexchange",
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Simplify $\sum_{i=0}^n (i+1)\binom ni$ Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$
$$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$
| Just double your expression and regroup using $\binom{i}{n} = \binom{n-i}{n}$ $$ 1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= \\ =\frac 1 2 \left(
(1 + (n+1))\cdot\binom{n}{0}+ (2 + n)\cdot\binom{n}{1}+(3 + (n-1))\cdot\binom{n}{2}+ \cdots+((n+1)+1)\cdot\binom{n}{n}\right)=
\frac 1 2 (n + 2)\left(\binom{n}{0} + ... + \binom{n}{n}\right) = (n+2) \cdot 2^{n-1}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Convergence of the infinite series $ \sum_{n = 1}^\infty \frac{1} {n^2 - x^2}$ How can I prove that for every $ x \notin \mathbb Z$ the series
$$ \sum_{n = 1}^\infty \frac{1} {n^2 - x^2}$$
converges uniformly in a neighborhood of $ x $?
| If $x \notin \mathbb{Z}$, then there exists some $\delta>0$ such that $(x-\delta,x+\delta) \cap \mathbb{Z} = \emptyset$. Then $s(y) = \sum_{n = 1}^\infty \frac{1} {n^2 - y^2}$ converges uniformly for $y \in B(x,\delta)$.
To see this, let $\epsilon>0$ and choose $N$ such that $N^2 > 2 (|x|+\delta)^2$. Then if $n \ge N$ and $|x-y|< \delta$, we have $n^2-y^2 \ge \frac{1}{2} n^2$. Now choose $N' \ge N$, such that $\sum_{k=N'}^\infty \frac{1}{k^2} < \frac{1}{2} \epsilon$.
Then if $n \ge N'$, we have $0 \le \sum_{k = n}^\infty \frac{1} {k^2 - y^2} \le \sum_{k = N'}^\infty \frac{1} {k^2 - y^2} \le 2\sum_{k = N'}^\infty \frac{1} {k^2} < \epsilon$. Hence the convergence is uniform on $B(x,\delta)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/352339",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Darts on a ruler probability If two points are selected at random on an interval from 0 to 1.5 inches, what is
the probability that the distance between them is less than or equal to 1/4"?
| The square $S=\{(x,y) | x,y \in [0, \frac{3}{2}] \}$ has area $(\frac{3}{2})^2$. The area $\Delta= \{(x,y) \in S \, |\, |x-y| > \frac{1}{4} \}$ can be easily rearranged to be a square with area $(\frac{3}{2}-\frac{1}{4})^2$. Hence the chance of landing in $S \setminus \Delta$ is $\frac{(\frac{3}{2})^2-(\frac{3}{2}-\frac{1}{4})^2}{(\frac{3}{2})^2} = 1 - (1-\frac{1}{6})^2 = \frac{11}{36}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/352698",
"timestamp": "2023-03-29T00:00:00",
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Sum of the selected elements of matrix is $255$ A $5\times 10$ matrix is given:
$$\begin{pmatrix}
1 & 6 & 11 & 16 & 21 & 26 & 31 & 36 & 41 & 46\\
2 & 7 & 12 & 17 & 22 & 27 & 32 & 37 & 42 & 47\\
3 & 8 & 13 & 18 & 23 & 28 & 33 & 38 & 43 & 48\\
4 & 9 & 14 & 19 & 24 & 29 & 34 & 39 & 44 & 49\\
5 & 10& 15 & 20 & 25 & 30 & 35 & 40 & 45 & 50\\
\end{pmatrix}$$
If we select $10$ distinct elements such that
*
*exactly $2$ elements are chosen from one row and
*exactly $1$ element is chosen from each column then
we will have the sum of these $10$ elements as $255$.
I can recognize the pattern here but how to prove this in general?
| Your matrix is the sum of these two:
$$A = \begin{pmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2\\
3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3\\
4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 4 & 4\\
5 & 5 & 5 & 5 & 5 & 5 & 5 & 5 & 5 & 5\\
\end{pmatrix}$$
and
$$B = \begin{pmatrix}
0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45\\
0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45\\
0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45\\
0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45\\
0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 & 40 & 45\\
\end{pmatrix}$$
Condition 1 says that the contribution to the sum from $A$ is
$(1 + 1) + (2 + 2) + (3 + 3) + (4 + 4) + (5 + 5) = 30$
Condition 2 says that the contribution to the sum from $B$ is
$0 + 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 = 225$
| {
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Simplify $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots+ \frac{1}{\sqrt{24} + \sqrt{25}}$ Simplify$$\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{24} + \sqrt{25}}.$$
I know you can solve this using generating functions but I'm not totally sure.
| Hint: Multiply top and bottom of $\dfrac{1}{\sqrt{k+1}+\sqrt{k}}$ by $\sqrt{k+1}-\sqrt{k}$, and watch the house of cards collapse.
| {
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question about inverse functions Functions $f$ and $g$ are defined by
$$ f: x\mapsto 2x+1$$
$$g: x \mapsto \dfrac{2x +1}{x+3}$$
(i) Solve the equation $gf(x) = x $
(iii) Show that the equation $g^{-1} (x) = x$ has no solutions
I need help with these.
| $$ f(x)=2x+1,g(x)= \frac{2x +1}{x+3}$$
$$g(f(x)) =\frac{2(2x+1) +1}{2x+1+3}= x$$
$$\frac{4x+3}{2x+4}= x,x\neq-2\Rightarrow2x^2+4x=4x+3,x^2=3/2,x=\pm\sqrt{3/2}$$
$$g(x)=y= \frac{2x +1}{x+3}$$
$$x= \frac{2y +1}{y+3},xy+3x=2y+1,y(x-2)=1-3x,y=\frac{1-3x}{x-2},x\neq2$$
$$g^{-1}(x)=\frac{1-3x}{x-2}=x\Rightarrow x^2+x-1=0,x=\frac{-1\pm\sqrt{5}}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding all $\alpha$ such that a matrix is positive definite I have
$A = $
$
\left[\begin{array}{rrr}
2 & \alpha & -1 \\
\alpha & 2 & 1 \\
-1 & 1 & 4
\end{array}\right]
$
and I want to find all $\alpha$ such that $A$ is positive definite.
I tried
$ x^tAx = $
$
\left[\begin{array}{r}
x & y & z
\end{array}\right]
$
$
\left[\begin{array}{rrr}
2 & \alpha & -1 \\
\alpha & 2 & 1 \\
-1 & 1 & 4
\end{array}\right]
$
$
\left[\begin{array}{r}
x \\
y \\
z
\end{array}\right]
$
$=$
$
\left[\begin{array}{r}
2x + \alpha y - z & \alpha x + 2y + z & -x + y + 4z
\end{array}\right]
$
$
\left[\begin{array}{r}
x \\
y \\
z
\end{array}\right]
$
$= 2x^2 + \alpha xy - xz + \alpha xy + 2y^2 + yz - xz + yz + 4z^2$
$= 2 \alpha xy + 2x^2 + 2y^2 - 2xz + 2yz + 4z^2$
and I wanted to solve the inequality $2 \alpha xy + 2x^2 + 2y^2 - 2xz + 2yz + 4z^2 > 0$ for $\alpha$, but I wasn't sure what to do next.
Am I doing this correctly?
| Maybe this helps:
A hermitian matrix is positive definite $\Leftrightarrow$ all leading principal minors are positive.
So $\left|\begin{array}{r}
2
\end{array}\right|$, $\left|\begin{array}{rr}
2 & \alpha \\
\alpha & 2
\end{array}\right|$ and $\left|\begin{array}{rrr}
2 & \alpha & -1 \\
\alpha & 2 & 1 \\
-1 & 1 & 4
\end{array}\right|$ have to be positive.
This gives us
$$\begin{align}&\text{I:}\quad 2>0\\
&\text{II:}\quad 4-\alpha^2>0\\
&\text{III:}\quad -4\alpha^2 - 2\alpha +12>0
\end{align}
$$
Try to solve this!
| {
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When is this rational fraction bounded? Let $\Omega=(0,\infty)^2$. For $\alpha \gt 0, \beta \gt 0$, define a function
$f_{\alpha,\beta}$ on $\Omega$ by putting
$$
f_{\alpha,\beta}(x,y)=\frac{(xy)^{\alpha}(1+x^2)(1+y^2)}{\big(xy(x+y)+1\big)^{\beta}}
$$
For which pairs $(\alpha,\beta)$ is $f_{\alpha,\beta}$ bounded ?
Putting $x=y=t$ and letting $t\to \infty$, we see that $3\beta \geq 2(\alpha+2)$ is a necessary condition. I don’t know if it is sufficient.
| The necessary and sufficient condition is $3\beta\ge 2\alpha+4$ and $\alpha\ge 2$.
When $x,y\ge 1$, you have seen $3\beta\ge 2\alpha+4$ is necessary. In this case, it is also sufficient. Note that $xy(x+y)+1\ge 2(xy)^{\frac{3}{2}}$, and when $x,y\ge 1$, $1+x^2\le 2x^2$, $1+y^2\le 2y^2$. It follows that
$$f_{\alpha,\beta}(x,y)\le 2 (xy)^{\alpha+2-\frac{3}{2}\beta}\le 2.$$
When $x\le 1$ or $y\le 1$, an additional condition is $\alpha\ge 2$. By symmetry, we may assume $x\le 1$. When $y\le 1$, it it easy to see $f_{\alpha,\beta}(x,y)\le 4$, so let us also assume $y\ge 1$. Then $1\le 1+x^2\le 2$, $y^2\le 1+y^2\le 2y^2$ and $xy^2\le xy(x+y)\le 2xy^2$, and hence
$$\frac{x^\alpha y^{\alpha+2}}{(2xy^2+1)^\beta}\le f_{\alpha,\beta}(x,y)\le \frac{4x^\alpha y^{\alpha+2}}{(xy^2+1)^\beta}.$$
Let $z=xy^2$. If $z=1$,
$$\frac{x^\alpha y^{\alpha+2}}{(2xy^2+1)^\beta}=3^{-\beta}y^{2-\alpha}, $$
so $\alpha\ge 2$ is necessary.
Now let us show the sufficiency. If $z\le 1$, then
$$\frac{x^\alpha y^{\alpha+2}}{(xy^2+1)^\beta} \le x^\alpha y^{\alpha+2}=z^\alpha y^{\alpha-2}\le 1.$$
If $z\ge 1$,
$$\frac{x^\alpha y^{\alpha+2}}{(xy^2+1)^\beta}\le x^{\alpha-\beta}y^{\alpha+2-2\beta}=z^{\alpha-\beta}y^{2-\alpha}.$$
When $\alpha\ge \beta$, since $x\le 1$, $y\ge 1$ and $3\beta\ge 2\alpha+4$,
$$x^{\alpha-\beta}y^{\alpha+2-2\beta}\le y^{\frac{1}{2}(2\alpha+4-3\beta)}\le 1;$$
when $\alpha\le \beta$, since $z\ge 1$, $y\ge 1$ and and $\alpha\ge 2$,
$$z^{\alpha-\beta}y^{2-\alpha}\le 1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $z$ is a complex number of unit modulus and argument theta If $z$ is a complex number such that $|z|=1$ and $\text{arg} z=\theta$, then what is $$\text{arg}\frac{1 + z}{1+ \overline{z}}?$$
| $$ \frac{1+z}{1 + \bar z} = \frac{1+z}{1 + \bar z} \times \frac{1+z}{1 + z} = \frac{(1+z)^2}{|1+z|^2} $$
the argument should be $ 2 \arg (1 +z) = 2 \arctan \left( y \over x+1\right)$
Let $x = \cos \theta$ and $y = \sin \theta $, we have $\arctan \left( \frac{y}{x+1}\right) = \arctan \left( \frac{\sin \theta}{\cos \theta+1}\right) = \arctan \left( \frac{\sin \frac{\theta}{2}}{\cos {\theta\over 2 }}\right)$
$2 \arg (1+z)^2 = \arg (z)$
| {
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mixed permutations and combinations I have a problem that I am not too sure of. In a team of 16, there are 5 couples and 6 single people. In how many ways can at most 1 couple be chosen if 6 people are required to represent the team at a conference?
This is my solution: 6P6 (0 couples and 6 single people only) + 11C6 * 6! * 2! (choose 1 member of a couple and the other 6 single people, this can be done in 2!*6! ways) + 5C1 * 2! * 10C4 *2! * 4! (exactly one couple is chosen and any of the singles along with the other partners of the other couples)
Is this correct?
| The following approach is reasonably systematic. There will be lots of words, but at the end there will be a more or less compact formula.
We count first the teams that have no couple, then, in basically the same way, the teams that have $1$ couple. A couple, viewed as an entity, will be called a family. There are $5$ families.
No couples: Maybe we choose $6$ singles. That can be done in $\binom{6}{6}$ ways. Of course this is $1$, but we call it by the complicated name $\binom{6}{6}\binom{5}{0}2^0$. Soon that will look reasonable!
Or else we pick $5$ singles, $1$ family, and a representative of the family. This can be done in $\binom{6}{5}\binom{5}{1}2^1$ ways.
Or else we pick $4$ singles, $2$ families, and a representative of each family. This can be done in $\binom{6}{4}\binom{5}{2}2^2$ ways.
Or else we pick $3$ singles, $3$ families, and a representative of each family. This can be done in $\binom{6}{3}\binom{5}{3}2^3$ ways.
Or else we pick $2$ singles, $4$ families, and a representative of each family. This can be done in $\binom{6}{2}\binom{5}{4}2^4$ ways.
Or else, finally, we pick $1$ singles, $5$ families, and a representative of each family. This can be done in $\binom{6}{1}\binom{5}{5}2^5$ ways.
Add up. A number of cases, but only one idea.
One couple: The idea is the same. There are $\binom{5}{1}$ ways o pick the couple. We will count the number of ways to pick the remaining $4$ people, add them up, and multiply by $\binom{5}{1}$. But rom here on, we only count the ways of picking the $4$.
We could pick $4$ singles. This can be done in $\binom{6}{4}$ ways, but we call the number $\binom{6}{4}\binom{5}{0}2^0$.
Or else we pick $3$ singles, $1$ family, and a representative of the family. This can be done in $\binom{6}{3}\binom{5}{1}2^1$ ways.
Or else we pick $2$ singles, $2$ families, and a representative of each family. This can be done in $\binom{6}{2}\binom{5}{2}2^2$ ways.
Or else we pick $1$ single, $3$ families, and a representative of each family. This can be done in $\binom{6}{1}\binom{5}{3}2^3$ ways.
Or else, finally, we pick $0$ singles, $4$ families, and a representative of each family. This can be done in $\binom{6}{0}\binom{5}{4}2^4$ ways.
Final answer: We gather the whole thing into a compact formula.
$$\sum_{i=0}^5 \binom{6}{6-i}\binom{5}{i}2^i +\binom{5}{1}\sum_{i=0}^4 \binom{6}{4-i}\binom{5}{i}2^i.$$
| {
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Find minimum in a constrained two-variable inequation I would appreciate if somebody could help me with the following problem:
Q: find minimum
$$9a^2+9b^2+c^2$$
where $a^2+b^2\leq 9, c=\sqrt{9-a^2}\sqrt{9-b^2}-2ab$
| Another approach - the symmetry here suggests Purkiss Principle (conditions to be verified), so the extremum is attained when $a = b$.
So $c = (9-a^2) - 2a^2 = 9-3a^2$
and $9(a^2+b^2) + c^2 = 18a^2 + (9-3a^2)^2 = 9a^4 - 36a^2 + 81 = 9 (a^2 - 2)^2 + 45$
which is minimised when $a^2 = 2$ or $a = \pm \sqrt2$.
| {
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How can I prove that $xy\leq x^2+y^2$? How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?
| Technique 1:$$\sqrt{x^2y^2} = xy \le \dfrac{x^2 + y^2}{2}\le x^2 + y^2$$ Technique 2:$$(x - y)^2 \ge 0 \implies x^2 +y^2 - 2xy \ge 0 \implies x^2 + y^2 \ge 2xy\ge xy$$ Technique 3 (my favorite): There's a statement $\dfrac{y}{x} + \dfrac{x}{y} \ge 2$ with many classical proofs (which I'd not state here). We can write the inequality as follows:$$\dfrac{y}{x}+\dfrac{x}{y} \ge 1$$Divide both sides by $xy$.$$\dfrac{1}{x^2} + \dfrac{1}{y^2} \ge \dfrac{1}{xy}$$Rewrite.$$\dfrac{x^2 + y^2}{x^2y^2} \ge \dfrac{xy}{x^2y^2}$$And finally...$$x^2 + y^2 \ge xy$$
| {
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Diagonalizing symmetric 2x2 matrix If A is a $2\times2$ symmetric matrix ($A^T = A$) where $b$ does not equal zero ($a$'s are on the diagonal, $b$'s occupy the other $2$ spaces), find a matrix $X$ such that $X^T AX$ is diagonal.
What is the simple way to solve this problem (using orthogonal diagonalization intro linear algebra)?
| You can also just go with the computations ;)
You easily get the two eigenvalues : $\lambda_1=a-b$ and $\lambda_2=a+b$ and the corresponding eigenvectors:
$V_1=\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ and $V_2=\begin{pmatrix} 1 \\ 1 \end{pmatrix}$
you can also make them have norm 1. Then:
$V_1=\frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ and $V_2=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$
And thus you have your two matrices $P$ and $D$ such that $A=PDP^{-1}$:
$D=\begin{pmatrix} a-b & 0 \\0 & a+b \end{pmatrix}$ and $P=\frac{1}{\sqrt{2}}\begin{pmatrix} -1 & 1 \\1 & 1 \end{pmatrix}$
your matrix $P$ is clearly symmetric, and $P^{-1}=P^T$. And there you have it:
$A=PDP^{-1}=PDP^T$ which gives $D=P^TAP$
| {
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How find this value:$\dfrac{f(a_{1})+f(a_{3})}{f(a_{2})}$
Let $f(x)=\ln{x}-\dfrac{1}{x}+3$, and $a_{i}>0,i=1,2,3$, such that $a_{3}:a_{2}:a_{1}=e^2:e:1$.
Suppose $f(a_{1})+f(a_{2})+f(a_{3})=\dfrac{e^5-e^2}{1-e}$; what is the value of $\dfrac{f(a_{1})+f(a_{3})}{f(a_{2})}$? (where $e=2.718\cdots, \ln{x}=\log_{e}{x}$)
| $\dfrac{e^2+1}{e}\quad $ B:$\dfrac{e^2+3}{e+1}\quad$ C :$\dfrac{e^2+5}{e+2}\quad$ D :$\dfrac{e^3+e+2}{e^2+1}$
this problem have nice solution? Thank you
my idea:$a_{3}=e^2*a_{1},a_{2}=e*a_{1}$
since $f(a_{1})+f(a_{2})+f(a_{3})=3\ln{a_{1}}+3-\dfrac{1}{a_{1}}\left(1+\dfrac{1}{e}+\dfrac{1}{e^2}\right)+9=\dfrac{e^5-e^2}{1-e}$,
$$\Longrightarrow 3\ln{a_{1}}=\dfrac{1}{a_{1}}\left(1+\dfrac{1}{e}+\dfrac{1}{e^2}\right)+\dfrac{e^5-e^2}{1-e}-12$$
then find the value
$$\dfrac{f(a_{1})+f(a_{3})}{f(a_{2})}=\dfrac{2\ln{a_{1}}+8-\dfrac{1}{a_{1}}(1+\dfrac{1}{e^2})}{\ln{a_{1}}-\dfrac{1}{a_{1}}\dfrac{1}{e}+4}=\dfrac{\dfrac{1}{3a_{1}}\left(\dfrac{2}{e}-1-\dfrac{1}{e^2}\right)+\dfrac{2}{3}\dfrac{e^5-e^2}{1-e}}{-\dfrac{1}{3a_{1}}\left(\dfrac{2}{e}-1-\dfrac{1}{e^2}\right)+\dfrac{1}{3}\dfrac{e^5-e^2}{1-e}}$$
then we must find the $a_{1}$,Now I think follow is very ugly.someone have nice methods? Thank you
oh, I have solution:let $a_{2}=x$, It's easy have
$$3\ln{x}+9=(e^2+e+1)\left(\dfrac{1}{ex}-e^2\right)$$
so let
$$F(x)=3\ln{x}+9-(e^2+e+1)\left(\dfrac{1}{ex}-e^2\right),x>0$$
so
$F'(x)=\dfrac{3}{x}+\dfrac{e^2+e+1}{e}\dfrac{1}{x^2}>0$
and we have $F(e^{-3})=0$
so $a_{2}=e^{-3}$
and follwing is very easy
| {
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How do I evaluate $\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots}{n^3}$ How to find this limit : $$\lim_{n \rightarrow \infty} \frac{1\cdot2+2\cdot3 +3\cdot4 +4\cdot5+\ldots+n(n+1)}{n^3}$$ As, if we look this limit problem viz. $\lim_{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$ then we take the sum of numerator which is sum of first n natural numbers and we can write :
$$\lim_{x \rightarrow \infty} \frac{n(n+1)}{2n^2}$$ which gives after simplification :
$$ \frac{1}{2} $$ as other terms contain $\frac{1}{x}$ etc. and becomes zero.
| Squeeze principle can also help. Note that $$\lim_{n \to \infty}\frac{ 1\cdot 1 + 2\cdot 2 + 3 \cdot 3 + \cdots + n \cdot n}{n^{3}} \leq \lim_{n \to \infty}\frac{1\cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n\cdot(n+1)}{n^{3}}\leq \lim_{n\to\infty}\: \frac{2^{2}+3^{2}+4^{2}+\cdots +(n+1)^{2}}{n^{3}}$$
$$\Longrightarrow \frac{1}{3} \leq \lim_{n \to \infty}\frac{1\cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n\cdot(n+1)}{n^{3}} \leq \frac{1}{3}$$
| {
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Determine the limiting behaviour of $\lim_{x \to \infty}{\frac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$ Determine the limiting behaviour of $\lim_{x \to \infty}{\dfrac{\sqrt{x^4+1}}{\sqrt[3]{x^6+1}}}$
Used L'Hopitals to get $\;\dfrac{(x^6+1)^{\frac{2}{3}}}{x^2 \sqrt{x^4+1}}$ but not sure what more i can do after that.
| Have you tried dividing the original numerator and denominator each by $x^2 = \sqrt{x^4} = \sqrt[3]{x^6}\,$? This is a good example where algebraic manipulations are easier to use than is using L'Hopital.
$$
\lim_{x \to +\infty}\frac{\sqrt{x^4+1}}{\sqrt[\large 3]{x^6+1}}\cdot \frac {1/\sqrt{x^4}}{1/\sqrt[3]{x^6}} =\lim_{x\to+\infty} \frac{\sqrt{1+1/x^4}}{\sqrt[\large 3]{1+1/x^6}}=\frac{\sqrt 1}{\sqrt[\large 3]{1}}=1
$$
| {
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Irrational equation, How to solve it? The equation
$$\sqrt[3]{x^2-1} + x = \sqrt{x^3-2}$$ has a solution $x = 3.$ How to solve this eqution?
| You have to rewrite this as
$$
\sqrt[3]{x^2-1}= \sqrt{x^3-2}-x
$$
and take the 3rd power of both members obtaining
$$
x^2-1=(x^3-2)\sqrt{x^3-2}-3x(x^3-2)+3x^2\sqrt{x^3-2}-x^3.
$$
Now, you can isolate the square root obtaining
$$
3x^4+x^3+x^2-6x-1=(x^3+3x^2-2)\sqrt{x^3-2}.
$$
I think from here you can go on by yourself.
| {
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I need to find the value of $a,b \in \mathbb R$ such that the given limit is true I am given that $\lim_{x \to \infty} \sqrt[3]{8x^3+ax^2}-bx=1$ need to find the value of $a,b \in \mathbb R$ such that the given limit is true. I was able to work the whole thing out, but I have a question about one step in my work. There is a lot of rough work because I simplify by using the rule of the difference of cubes, so here is a condensed part of my work:
$$\begin{align} \lim_{x \to \infty} \sqrt[3]{8x^3+ax^2}-bx &=\lim_{x \to \infty} \frac{8x^3+ax^2-b^3x^3}{(\sqrt[3]{8x^3+ax^2})^2+bx\sqrt[3]{8x^3+ax^2}+b^2x^2} \\&= \lim_{x \to \infty} \frac{8+a\frac{1}{x}-b^3}{\frac{1}{x^3}(\sqrt[3]{8x^3+ax^2})^2+b\frac{1}{x^2}\sqrt[3]{8x^3+ax^2}+b^2\frac{1}{x}} \\&=\frac{\lim_{x \to \infty}8+\lim_{x \to \infty}a\frac{1}{x}-\lim_{x \to \infty}b^3}{\lim_{x \to \infty}\frac{1}{x^3}(\sqrt[3]{8x^3+ax^2})^2+\lim_{x \to \infty}b\frac{1}{x^2}\sqrt[3]{8x^3+ax^2}+\lim_{x \to \infty}b^2\frac{1}{x}} \\&= \frac{8-b^3}{0+0+0} \\&= \frac{8-b^3}{0}\end{align}$$ Thus $8-n^3$ must also equal $0$ which implies that $b=2$. (This is the part I am unsure about. Is what I said true? If $b=2$ then this would give me an indeterminate form, but other than that I'm not sure if what I said holds, and if it does hold why does it hold?) Regardless of my uncertainty, I went on and using this assumption I found that $a=12$ in a similar manner, and when I check $\lim_{x \to \infty} \sqrt[3]{8x^3+12x^2}-2x$ it does equal $1$ .
Any help as to why/why not my assumption is correct?
Thanks in advance!
(If anyone wants me to post the method as to how i got 12 for $a$, let me know and then I'll type it up).
| Your assumption is correct; if $b\neq 2$, then $8-b^3\neq 0$, and hence the limit would either not exist or be infinite. But you know the limit is $1$.
A shorter way to do the first part is: $\sqrt[3]{8x^3+ax^2}-bx=x(\sqrt[3]{8+\frac{a}{x}}-b)$. The cube root approaches 2 as $x\rightarrow \infty$, so if $b\neq 2$, the product approaches $\pm\infty$ (not $1$ as in the hypothesis).
| {
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Finding the limit of function - irrational function How can I find the following limit:
$$ \lim_{x \rightarrow -1 }\left(\frac{1+\sqrt[5]{x}}{1+\sqrt[7]{x}}\right)$$
| A useful principle when trying to find a limit is
"always expand around zero".
Since we want to see what happens
when $x \to -1$,
let $x = y-1$ so that
we can look at what happens when $y \to 0$.
The expression becomes
$\frac{1+\sqrt[5]{y-1}}{1+\sqrt[7]{y-1}}$.
For an odd value of $n$,
since $y-1 < 0$ (for small $y$),
$\sqrt[n]{y-1}
= - \sqrt[n]{1-y}
$.
By the binomial theorem,
as $y \to 0$,
$\sqrt[n]{1-y}
\approx 1-y/n$.
Putting this in for $n = 5$ and $n = 7$,
$\frac{1-\sqrt[5]{1-y}}{1-\sqrt[7]{1-y}}
\approx \frac{1-(1-y/5)}{1-(1-y/7)}
= \frac{y/5}{y/7}
= \frac{7}{5}
$.
| {
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Finding coefficient of generating functions I have the equation
$$(1+x+x^2+\ldots+x^k+\ldots)(1+x^2+x^4+\ldots+x^{2k}+\ldots)(x^2+x^3)$$
how of I find the coefficent of $x^{24}$. I know to condense this down to
$$\frac1{1-x}\cdot\frac1{1-x^2}\cdot x(1+x)$$
but I don't know what to do after that
| Here's another way: the coefficient of $x^{24}$ is the number of integer solutions to $$a+b+c=24$$ where $a\ge0$; $b\ge0$ is even; and $c$ is $2$ or $3$.
If $c=2$, then we have $a+b=22$. There are $12$ values of $b$ (namely, $0,2,\dots,22$), and for each, a value of $a$. So, $12$ solutions so far.
If $c=3$, then $a+b=21$, so there are $11$ values of $b$ ($0,2,\dots,20$); for each, there's a value of $a$, so, another $11$ solutions.
All told, $12+11=23$ solutions, so that's the coefficient you're looking for.
| {
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How to solve this equation by Fourier series? $$ y''+3y=\sin ^4 x ,\quad y=\frac{1}{8} +\frac{\cos2x}{2}-\frac{\cos4x}{104}.$$
Now the text book states the solution, but I don't know the process of solving this equation. I need your help!
| The first thing you should do is express $\sin^4{x}$ in terms of cosines:
$$\begin{align}\sin^4{x} &= \left ( \frac{1-\cos{2 x}}{2} \right )^2\\ &= \frac{1}{4} - \frac{1}{2} \cos{2 x} + \frac{1}{4} \cos^2{2 x}\\ &= \frac{1}{4} - \frac{1}{2} \cos{2 x} + \frac{1}{4} \left (\frac{1+\cos{4 x}}{2} \right )\\ &= \frac38 - \frac{1}{2} \cos{2 x} + \frac18 \cos{4 x}\end{align}$$
Because $y'' + 3 y$ is a linear differential expression, it follows that the solution to the above equation must be a linear combination of $1$, $\cos{2 x}$, and $\cos{4 x}$:
$$y(x) = a_0 + a_2 \cos{2 x} + a_4 \cos{4 x}$$
Plug this into the differential expression:
$$y'' + 3 y = 3 a_0 - a_2 \cos{2 x} - 13 a_4 \cos{4 x} = \frac38 - \frac{1}{2} \cos{2 x} + \frac18 \cos{4 x}$$
Equating the coefficients of $1$, $\cos{2 x}$, and $\cos{4 x}$, we get the stated result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove this inequality $a \sqrt{1-b^2}+b\sqrt{1-a^2}\le1 $ Prove that for $a,b\in [-1,1]$:
$$a\sqrt{1-b^2}+b\sqrt{1-a^2}\leq 1$$
| Use the AM-GM inequality: For nonnegative $x,y$, we have $\displaystyle\sqrt{xy}\le\frac{x+y}2$. This follows immediately from expanding and rearranging the obvious inequality $(\sqrt x-\sqrt y)^2\ge0$.
Here, we have $$\begin{array}{rl} a\sqrt{1-b^2}+b\sqrt{1-a^2}&\le|a|\sqrt{1-b^2}+|b|\sqrt{1-a^2}\\&\le\frac{a^2+(1-b^2)}2+\frac{b^2+(1-a^2)}2\\&=\frac22=1.\end{array}$$
Note that the AM-GM inequality is an equality iff $x=y$, so here we have $a=|a|$, $b=|b|$, and $a^2=1-b^2$, or $a^2+b^2=1$, so $a=\sin\theta$ and $b=\cos\theta$ for some $\theta$ in the first quadrant. This suggests the other solution, where more generally we let $a=\sin\alpha$, $b=\cos\beta$ for some $\alpha,\beta$ (or some variant thereof).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding equations of lines I'm sitting an exam soon and have been revising for a while but I sometimes get stuck with questions that I know are simple but still have no idea about how to do them at all! Please explain to me how this is done. (Note: this is all the info I have about the question):
Let $L$ be the line with equation $x + y = 0$, $F = (1,1)$ and $P = (a,b)$.
(a) Find the equation of the line perpendicular to $L$ which passes through $P$, the intersection of the two lines and hence the least distance between $P$ and $L$.
(b) Determine the Cartesian equation of the parabola with focus $F$ and directrix $L$.
(c) Show that the tangent to the parabola at the point $(x_1,y_1)$ has equation
$(x_1 - y_1 + 2)y + (y_1 - x_1 + 2)x - 2(x_1 + x_1y_1 + y_1) + x_1^2 + y_1^2 = 0$.
| a) The line $y_L=-x$ has a slope of $-1$, so a perpendicular line must have a slope of $-\frac{-1}{1}=1$; hence $y_P=x+C$, where $b=y_P(a)=a+C$ implies $C=b-a$. The lines intersect at $y_L(x_I)=-x_I=x_I+b-a=y_P(x_I)$; that is, at $x_I=\frac{a-b}{2}$. Thus the distance between the point $P=(a,b)$ and the line $L$ is given by $\sqrt{\left(a-\frac{a-b}{2}\right)^2+\left(b-\frac{b-a}{2}\right)^2}=\sqrt{2}\left(\frac{a+b}{2}\right)$.
b) Let $(x,y)$ be any point on the parabola. The distance between the point and the focus $F=(1,1)$ is $\sqrt{(x-1)^2+(y-1)^2}$ and the distance between the point and the directrix $L$ is $\sqrt{2}\left(\frac{x+y}{2}\right)$, which we calculated above. Now, we equate and square the two expressions:
$$\begin{align*} &\quad(x-1)^2+(y-1)^2=2\left(\frac{x+y}{2}\right)^2=\frac{(x+y)^2}{2}\\
&\implies x^2-2x+1+y^2-2y+1=\frac{x^2+2xy+y^2}{2}\\
&\implies \frac{x^2}{2}-2x+\frac{y^2}{2}-2y-xy+2=0 \end{align*}$$
This is the equation of the parabola with focus $F$ and directrix $L$.
c) We first find the slope of the tangent line. To do this, we consider $y$ as a function of $x$ and differentiate with respect to $x$:
$$\begin{align*} &\quad\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^2}{2}-2x+\frac{y(x)^2}{2}-2y(x)-xy(x)+2\right)=\\
&=x-2+y(x)y'(x)-2y'(x)-y(x)-xy'(x)=0\\
&\implies \frac{\mathrm{d}y}{\mathrm{d}x}=y'(x)=\frac{2-x+y}{y-x-2} \end{align*}$$
At the point $(x_1,y_1)$, we obtain the slope $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2-x_1+y_1}{y_1-x_1-2}$. Hence the equation of the tangent line is given by
$$y-y_1=\frac{2-x_1+y_1}{y_1-x_1-2}(x-x_1)\\
\implies yy_1-yx_1-2y-y_1^2+x_1y_1+2y_1=2x-xx_1+xy_1-2x_1+x_1^2-x_1y_1\\
\implies x(y_1-x_1+2)+y(x_1-y_1+2)-2(x_1+x_1y_1+y_1)+x_1^2+y_1^2=0$$
This is what we wanted to show.
| {
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"source": "stackexchange",
"question_score": "1",
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Fractional Trigonometric Integrands
*
*$$∫\frac{a\sin x+b\cos x+c}{d\sin x+e\cos x+f}dx$$
*$$∫\frac{a\sin x+b\cos x}{c\sin x+d\cos x}dx$$
*$$∫\frac{dx}{a\sin x+\cos x}$$
What are the relations between the numerator in the denominator, and what is the general pattern to solve these type of questions?
| You can use the following way to calculate all of them once. Let
$$ A=\int \frac{\sin xdx}{d\sin x+e\cos x+f}, B=\int \frac{\cos xdx}{d\sin x+e\cos x+f}, C=\int \frac{dx}{d\sin x+e\cos x+f}. $$
It is easy to check
$$ dA+eB+fC=1, dB-eB=\ln|d\sin x+e\cos x+f|+Const. $$
So
$$ A=\frac{d(1-fC)}{d^2+e^2}-\frac{e\ln|d\sin x+e\cos x|}{d^2+e^2}+Const, B=\frac{e(1-fC)}{d^2+e^2}+\frac{f\ln|d\sin x+e\cos x|}{d^2+e^2}+Const.$$
Thus we only need to get $C$. Let
$$\tan \frac{x}{2}=t,x=2\arctan t. $$
Then
$$\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2}{1+t^2}dt.$$
Then we have
$$ C=\int \frac{dx}{d\sin x+e\cos x+f}=2\int\frac{dt}{(f-e)t^2+2dt+(f+e)}$$
which is not hard to get.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Find all matrices $A$ of order $2 \times 2$ that satisfy the equation $A^2-5A+6I = O$
Find all matrices $A$ of order $2 \times 2$ that satisfy the equation
$$
A^2-5A+6I = O
$$
My Attempt:
We can separate the $A$ term of the given equality:
$$
\begin{align}
A^2-5A+6I &= O\\
A^2-3A-2A+6I^2 &= O
\end{align}
$$
This implies that $A\in\{3I,2I\} = \left\{\begin{pmatrix}
3 & 0\\
0 & 3
\end{pmatrix},
\begin{pmatrix}
2 & 0\\
0 & 2
\end{pmatrix}\right\}$.
Are these the only two possible values for $A$, or are there other solutions?If there are other solutions, how can I find them?
| Two matrices $A$ and $B$ are similar if there exists a matrix $P$ such that $A=PBP^{-1}$.
The solutions to your equation are $x=2,3$. Thus, all matrices which satisfy your equation must be similar to $B=\begin{bmatrix}v_1&0\\0&v_2\end{bmatrix}$, where $v_1$ and $v_2$ are either $2$ or $3$.
Choosing $P=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, all solutions to your equation are
$$
A=PBP^{-1}=\frac{1}{ad-bc}\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}v_1&0\\0&v_2\end{bmatrix}\begin{bmatrix}d&-b\\-c&a\end{bmatrix},
$$
for any choice of $a,b,c,d$ where $ad-bc\neq0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the limit of these sequences? Let $\{a_n\}$ be a real-valued sequence such that $a_1 \geq 0$ and $$a_{n+1}=\ln(a_{n}+1)$$ for all $n\ge1$. How can we find the following limits?
$$\lim_{n\to \infty}na_n=?,$$ $$\lim_{n\to \infty}\frac{n(na_n-2)}{\ln n}=?$$
Thanks in advance.
| If you use Stolz–Cesàro theorem, it is much easier to get the limit. Note that $\lim_{n\to\infty}a_n=0$. By using Stolz–Cesàro theorem,
\begin{eqnarray*}
\lim_{n\to\infty}na_n&=&\lim_{n\to\infty}\frac{n}{\frac{1}{a_n}}=\lim_{n\to\infty}\frac{1}{\frac{1}{a_{n+1}}-\frac{1}{a_n}}=\lim_{n\to\infty}\frac{a_na_{n+1}}{a_n-a_{n+1}}\\
&=&\lim_{n\to\infty}\frac{a_n\ln(a_n+1)}{a_n-\ln(a_n+1)}=\lim_{x\to 0}\frac{x\ln(x+1)}{x-\ln(x+1)}\\
&=&\lim_{x\to 0}\frac{x(x-\frac{1}{2}x^2+O(x^3))}{x-(x-\frac{1}{2}x^2+O(x^3))}\\
&=&2.
\end{eqnarray*}
Here $\ln(x+1)=x-\frac{1}{2}x^2+O(x^3)$. For the second part, we first note that
$$ a_n=\frac{2}{n}+o(1), \frac{1}{\ln(1+x)}=\frac{1}{x}+\frac{1}{2}-\frac{x}{12}+O(x^2) $$
and hence
\begin{eqnarray*}
\frac{1}{a_{n+1}}&=&\frac{1}{\ln(1+a_n)}=\frac{1}{a_n}+\frac{1}{2}-\frac{1}{12}a_n+O(a_n^2)=\frac{1}{a_n}+\frac{1}{2}-\frac{1}{6n}+O(\frac{1}{n^2}).
\end{eqnarray*}
So by using Stolz–Cesàro theorem
\begin{eqnarray*}
\lim_{n\to\infty}\frac{n(na_n-2)}{\ln n}&=&\lim_{n\to\infty}\frac{na_n(na_n-2)}{a_n\ln n}\\
&=&2\lim_{n\to\infty}\frac{na_n-2}{a_n\ln n}=4\lim_{n\to\infty}\frac{\frac{n}{2}-\frac{1}{a_n}}{\ln n}\\
&=&4\lim_{n\to\infty}\frac{(\frac{n+1}{2}-\frac{1}{a_{n+1}})-(\frac{n}{2}-\frac{1}{a_n})}{\ln (n+1)-\ln n}\\
&=&4\lim_{n\to\infty}\frac{\frac{1}{2}-\frac{1}{a_{n+1}}+\frac{1}{a_n}}{\ln (n+1)-\ln n}\\
&=&4\lim_{n\to\infty}\frac{\frac{1}{6n}+O(\frac{1}{n^2})}{\ln (n+1)-\ln n}\\
&=&\frac{2}{3}\lim_{n\to\infty}\frac{1}{n\ln(1+\frac{1}{n})}\\
&=&\frac{2}{3}.
\end{eqnarray*}
| {
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"url": "https://math.stackexchange.com/questions/379443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Inequality with mathematical expectations Let a random variable $X \ge 0.$ How to prove the inequality $EX^4EX^8 \le EX^3EX^9$?
| Let $p=\frac{6}{5}$ and $q=6$. Note that $\frac{1}{p}+\frac{1}{q}=1$, $X^4=X^{\frac{5}{2}}\cdot X^{\frac{3}{2}}$, $X^8=X^{\frac{1}{2}}\cdot X^{\frac{15}{2}}$, $\frac{5p}{2}=\frac{q}{2}=3$ and $\frac{3q}{2}=\frac{15p}{2}=9$. Therefore, by Hölder's inequality,
$$E X^4\le \left(E(X^{\frac{5}{2}})^p\right)^{\frac{1}{p}}\cdot \left(E(X^{\frac{3}{2}})^q\right)^{\frac{1}{q}}= \left(E X^3\right)^{\frac{1}{p}}\cdot\left(E X^9\right)^{\frac{1}{q}},\tag{1}$$
and
$$E X^8\le \left(E(X^{\frac{1}{2}})^q\right)^{\frac{1}{q}}\cdot \left(E(X^{\frac{15}{2}})^p\right)^{\frac{1}{p}}= \left(E X^3\right)^{\frac{1}{q}}\cdot\left(E X^9\right)^{\frac{1}{p}}.\tag{2}$$
The conclusion follows from $(1)\times(2)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the number of real roots of the given equation?
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
(A) $0$, (B) $1$, (C) $2$, (D) infinitely many.
Trial: $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-1} \left( \frac{2^x+2^{-x}}{2} \right)\end{align}$$ Then I can't proceed.
| There's a clever approach to this problem. In particular, $$(w-1)^2 \geq 0 \implies w^2 + 1 \geq 2w \implies w + \frac{1}{w} \geq 2$$
and this works for any $w$. Using this, you can show that one side is always at least $2$. Meanwhile, the other is at most $2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How does one get the Bernoulli numbers via the generating function? Here is the definition:
Bernoulli numbers arise in Taylor series in the expansion $$\frac{x}{e^x-1}=\sum_{k=0}^\infty B_k \frac{x^k}{k!}$$
I've tried to naively expand $\frac{x}{e^x-1}$ around $x_0=0$ and didn't quite understand what should be done with all the $(e^x-1)^{k+1}$ in the denominators of $f^{(k)}(0)$?
How does one get to $$1 - \frac{x}{2} + \frac{x^2}{12} - \frac{x^4}{720} + \ ...$$ directly from the definition (without rearranging the whole thing into the recursive formula)?
Thanks a lot!
| Expand $e^{x}$ around $x=0$ and then reexpand
\begin{align}\frac{1}{1+\frac{x}{2!}+\frac{x^2}{3!}+\ldots}=1-\left(\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}\ldots\right)+\left(\frac{x}{2!}+\frac{x^2}{3!}+\ldots\right)^2-\left(\frac{x}{2!}+\ldots\right)^3+\ldots=\\
=1-\left(\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}+\ldots\right)+\frac{x^2}{4}\left(1+\frac{x}{3}+\ldots\right)^2-\frac{x^3}{8}\left(1+\ldots\right)^3+\ldots=\\
=1-\left(\frac{x}{2}+\frac{x^2}{6}+\frac{x^3}{24}\right)+\frac{x^2}{4}\left(1+\frac{2x}{3}\right)-\frac{x^3}{8}+O\left(x^4\right)=\\
=1-\frac{x}{2}+\left(-\frac{1}{6}+\frac14\right)x^2+\left(-\frac{1}{24}+\frac{1}{6}-\frac{1}{8}\right)x^3+O\left(x^4\right)=\\
=1-\frac{x}{2}+\frac{x^2}{12}+O\left(x^4\right).\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing that $\mathbb Q(\sqrt{17})$ has class number 1 Let $K=\mathbb Q(\sqrt{d})$ with $d=17$. The Minkowski-Bound is $\frac{1}{2}\sqrt{17}<\frac{1}{2}\frac{9}{2}=2.25<3$.
The ideal $(2)$ splits, since $d\equiv 1$ mod $8$. So we get $(2)=(2,\frac{1+\sqrt{d}}{2})(2,\frac{1-\sqrt{d}}{2})$ and $(2,\frac{1\pm\sqrt{d}}{2})$ are two ideals of norm $2$.
Now if we can show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal ideals, then we know that every ideal class contains a principal ideal, which shows that the class number is $1$.
But how can we show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal?
| Here because of a duplicate. Though it does seem like the original asker does pop in from time to time.
I'm not sure if the duplicate asker is aware of algebraic integers such as $$\theta = \frac{1}{2} + \frac{\sqrt{17}}{2},$$ which is a solution to $x^2 - x - 4$, though the duplicate asker is aware of the Minkowski bound.
So this tells us $N(\theta) = -4$, while obviously $N(2) = 4$. This suggests that $$¿ N\left(\left\langle 2, \frac{1}{2} + \frac{\sqrt{17}}{2} \right\rangle\right) = 4 ?$$ However, if $\mathcal O_{\mathbb Q(\sqrt{17})}$ does indeed have class number 1, that would mean that 16 has one distinct factorization (ignoring multiplication by units) and so $$16 = 2^4 = \left(\frac{1}{2} - \frac{\sqrt{17}}{2}\right)^2 \left(\frac{1}{2} + \frac{\sqrt{17}}{2}\right)^2$$ represents incomplete factorizations, just like $16 = 4^2 = 2 \times 8$ in $\mathbb Z$.
It's not a given that this is a Euclidean domain even if it does have class number 1. However, it wouldn't hurt to try. And so we find by the Euclidean algorithm that $$\gcd\left(2, \frac{1}{2} + \frac{\sqrt{17}}{2}\right) = \frac{5}{2} + \frac{\sqrt{17}}{2},$$ and indeed $$2 + \frac{1}{2} + \frac{\sqrt{17}}{2} = \frac{5}{2} + \frac{\sqrt{17}}{2}.$$
Furthermore, since $$\frac{5}{2} - \frac{\sqrt{17}}{2} \in \left\langle \frac{5}{2} + \frac{\sqrt{17}}{2} \right\rangle,$$ it follows that $$\langle 2 \rangle = \left\langle \frac{5}{2} + \frac{\sqrt{17}}{2} \right\rangle^2.$$ That's a principal ideal after all.
Since $$\left(\frac{17}{3}\right) = -1$$ (that's the Legendre symbol), 3 is prime in this ring. But it's well over the Minkowski bound anyway, so we're done.
EDIT: Jyrki Lahtonen points out a mistake I made regarding $\langle 2 \rangle$. The correct factorization is $$\langle 2 \rangle = \left\langle \frac{5}{2} - \frac{\sqrt{17}}{2} \right\rangle \left\langle \frac{5}{2} + \frac{\sqrt{17}}{2} \right\rangle.$$ This does not detract from the point that these are all principal ideals.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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if $ab=cd$ then $a+b+c+d $ is composite Let $a,b,c,d$ be natural numbers with $ab=cd$.
Prove that $a+b+c+d$ is composite.
I have my own solution for this (As posted) and i want to see if there is any other good proofs.
| From $ab=cd$, We may assume $a=\frac{cd}{b}$. So $M=a+b+c+d = \frac{cd}{b}+b+c+d = \frac{(b+c)(b+d)}{b}$ and so $bM=(b+c)(b+d)$ and $M|(b+c)(b+d)$. We assume that $M$ is not composite, so it is prime. Now we may know that either $b+c$ or $b+d$ is divisible by $M$. So $M\leq b+c$ or $M\leq b+d$ which both result in contradiction because $M=a+b+c+d > b+c$ or $b+d$. So our assumption was wrong and $M$ is a composite number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/383394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Evaluating this integral : $ \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx $ The question :
$$ \int \frac {1-7\cos^2x} {\sin^7x \cos^2x} dx $$
I tried dividing by $\cos^2 x$ and splitting the fraction.
That turned out to be complicated(Atleast for me!)
How do I proceed now?
| Let $\displaystyle I = \int\frac{1-7\cos^2 x}{\sin^7 x\cos^2 x} = \int\frac{\sin^2 x-6\cos^2 x}{\sin^7 x\cos^2 x}dx$
$\displaystyle I = \int\frac{\sin^7 x-6\sin^5 x\cos^2 x}{\sin^{12}x\cos^2 x}dx = -\int \bigg[\frac{1}{(\sin^ 6 x\cos x)^2}\bigg]'dx = -\frac{1}{\sin^6 x\cos x}+\mathcal{C} $
| {
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"url": "https://math.stackexchange.com/questions/385636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Please correct my work, finding Eigenvector Determine whether matrix A is a Diagonalizable. if it is , determine matrix P that Diagnolizes it and compute $P^{-1}AP$.
$$A=
\begin{bmatrix}
+3 & +2\\
-2 & -3\\
\end{bmatrix}
$$
$$A-\ell I =
\begin{bmatrix}
+3-\ell & +2\\
-2 & -3-\ell\\
\end{bmatrix}
$$
Then Determinant should be zero :
$$
\begin{vmatrix}
+3-\ell & +2\\
-2 & -3-\ell\\
\end{vmatrix}=(3-\ell)(-3-\ell)+4=0 \to \ell^2-5=0 \to \ell_{1}=\sqrt 5,\ell_{2}=-\sqrt5 ,
$$
$$
\begin{bmatrix}
3-\ell_{1}& +2\\
-2 & -3-\ell{1}\\
\end{bmatrix}=
\begin{bmatrix}
3-\sqrt 5=0.76& +2\\
-2 & -3-\sqrt 5=-5.24\\
\end{bmatrix}\to{R_1\Leftarrow\Rightarrow R2 \mapsto}
\begin{bmatrix}
-2 & -5.24\\
0.76& +2\\
\end{bmatrix}\to{R_1=R1/{-2}\mapsto}
\begin{bmatrix}
1 & 2.62\\
0.76& +2\\
\end{bmatrix}\to{R_2=R2-0.76R1\mapsto}
\begin{bmatrix}
1 & 2.62\\
0 & 0.009\\
\end{bmatrix}\to{R_2=R2/0.009\mapsto}
\begin{bmatrix}
1 & 2.62\\
0 & 1\\
\end{bmatrix}\to{R_1=R1-2.62R1\mapsto}
\begin{bmatrix}
1 & 0 | 0\\
0 & 1 | 0\\
\end{bmatrix}
$$
$then$
$$ V_1= \begin{bmatrix}
0 \\
0 \\
\end{bmatrix}
$$
I am stuck here , matrix of $0,0$ is the right answer Eigenvector 1 ?
| The problem is that you rounded. The 0.009 in the second row is in fact a 0.
| {
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How find all this positive of $n$? Let $n$ be a number of the form $a^2+b^2$ with $a,b\in N^{+},(a,b)=1$, such that
every prime $P \lt \sqrt{n}$ satisfies $P|ab$. Find all such positive integers $n$?
I find $n=5$, and $n=13$,are there any others?
when $n=5$,then $a=2,b=1$, and $P=2<\sqrt{5}$
when $n=13$,then $a=3,b=2$,and $P=2$ or $3$
| First note that $\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \mid ab$. Thus $n=a^2+b^2 \geq 2ab \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p}$.
If $n \geq 26$, then $\sqrt{n}>5$, so $n \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \geq 2\prod\limits_{p \leq 5, p \text{prime}}{p}=60$. This implies that $\sqrt{n}>7$, so $n \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \geq 2\prod\limits_{p \leq 7, p \text{prime}}{p}=420$. This implies that $\sqrt{n}>19$, so $n \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \geq 2\prod\limits_{p \leq 19, p \text{prime}}{p}=199399380$. This implies that $\sqrt{n} \geq \sqrt{199399380}>\sqrt{100000000}=10000$.
Let $k=\lfloor \frac{\sqrt{n}}{16} \rfloor$, so $624=\frac{10000}{16}-1 \leq \frac{\sqrt{n}}{16}-1<k \leq \frac{\sqrt{n}}{16}$. We have $16k \leq \sqrt{n}$. Also, $\sqrt{n}<16(k+1)$ so $n<256(k+1)^2$.
By Bertrand's postulate, there exists primes $p, q, r, s$ with $k<p<2k<q<4k<r<8k<s<16k \leq \sqrt{n}$. Thus $256(k+1)^2 \geq n \geq 2\prod\limits_{p<\sqrt{n}, p \text{prime}}{p} \geq 2pqrs>2(k)(2k)(4k)(8k)=128k^4>64k^4$. Thus $4(k+1)^2>k^4$, so $2(k+1)>k^2$ so $k \leq 2$, a contradiction.
Therefore $n \leq 25$. Note that $25 \geq n=a^2+b^2>\max(a^2, b^2)$, so $a, b \leq 4$. If $n \geq 10$, then $\sqrt{n}>3$ so $2(3) \mid ab$. WLOG assume $3 \mid b$, so that $b=3$, so $2 \mid a$, so $a=2, 4$. This gives $n=2^2+3^2=13$ and $n=4^2+3^2=25$, which are both solutions.
Otherwise $n \leq 9$, so $9 \geq n=a^2+b^2>\max(a^2, b^2)$, so $a, b \leq 2$. $a, b$ cannot be both $2$, so we have $n=1^2+2^2=5$ and $n=1^2+1^2=2$, which are both solutions.
In conclusion, all solutions are given by $n=2=1^2+1^2, n=5=1^2+2^2, n=13=2^2+3^2, n=25=4^2+3^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx$ Prove that:
$(1)$$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \ dx=\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \ dx$$
$(2)$$$\int_0^{\infty } \frac{1}{\sqrt{8 x^3+x+7}} \ dx>1$$
What I do for $(1)$ is (something trival):
$$\int_0^{\infty } \frac{1}{\sqrt{6 x^3+6 x+9}} \, dx=\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx$$
$$\int_0^{\infty } \frac{1}{\sqrt{9 x^3+4 x+4}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$
so it remains to prove that
$$\int_0^{\infty } \frac{1}{\sqrt{x^3+36 x+324}} \, dx=\frac{\sqrt3}3\int_0^{\infty } \frac{1}{\sqrt{x^3+4 x+12}} \, dx$$
Thanks in advance!
| The second integral in closed form (by computer algebra) equals
$$ \frac1{\sqrt{2(a_2-a_1)}} F\left(\text{asin}(\sqrt{1-a_2/a_1}), \sqrt{\frac{a_1-a_3}{a_1-a_2}}\right), $$
where $a_1$ is the real root and $a_{2,3}$ are the complex roots of $8x^3+x+7=0$, and $F(\phi,k)$ is the incomplete elliptic integral of the first kind. This evaluates to
$$ 1.0001425023196181464480\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/387760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx$ I need help with calculating this integral:
$$\int_0^1\arctan\,_4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{2},\frac{3}{4},\frac{5}{4};\frac{x}{64}\right)\,\mathrm dx,$$
Where $_pF_q$ is a generalized hypergeometric function.
I was told it has a closed-form representation in terms of elementary functions and integers.
| This hypergeometric function is not an elementary function, but its inverse is - see Bring radical.
\begin{align}
I&= \int_0^1\arctan{_4F_3}\left(\frac15,\frac25,\frac35,\frac45;\frac12,\frac34,\frac54;\frac{x}{64}\right)\,dx \\
&=\frac{3125}{48}\left(5+3\pi+6\ln2-3\alpha^4+4\alpha^3+6\alpha^2-12\alpha\\-12\left(\alpha^5-\alpha^4+1\right)\arctan\frac1\alpha-6\ln\left(1+\alpha^2\right)\right)\\
&=0.7857194\dots
\end{align}
where $\alpha$ is the positive root of the polynomial $625\alpha^4-500\alpha^3-100\alpha^2-20\alpha-4$. It can be expressed in radicals as follows:
$$\alpha=\frac15+\sqrt\beta+\sqrt{\frac15-\beta +\frac1{25\sqrt\beta}},$$
where
$$\beta=\frac1{30}\left(\frac\gamma5-\frac4\gamma+2\right),$$
where
$$\gamma=\sqrt[3]{15\sqrt{105}-125}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
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Integral of $ \int_{-1}^{1} \frac{x^4}{x^2+1}\,dx $ Any suggestions how to solve it? by parts?
$$ \int_{-1}^{1} \frac{x^4}{x^2+1}dx$$
Thanks!
| $$\text{Note that, we have }\dfrac{x^4}{x^2+1} = \dfrac{x^4-1}{x^2+1}+\dfrac1{x^2+1} = x^2-1 + \dfrac1{x^2+1}$$
$$\text{Hence, }\int \dfrac{x^4}{x^2+1} dx = \int (x^2-1) + \int \dfrac1{x^2+1} = \dfrac{x^3}3 - x + \arctan(x) + \text{ constant}$$
| {
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"url": "https://math.stackexchange.com/questions/390169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solve the roots of a cubic polynomial? I have had trouble with this question - mainly due to the fact that I do not fully understand what a 'geometric progression' is:
"Solve the equation $x^3 - 14x^2 + 56x - 64 = 0$" if the roots are in geometric progression.
Any help would be appreciated.
| Let the roots be $a,a\cdot r,a\cdot r^2$
Using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies a(1+r+r^2)=14$
and $a(a\cdot r)+a\cdot r(a\cdot r^2)+a(a\cdot r^2)=56\implies a^2\cdot r(1+r+r^2)=56$
On division, $ar=4$ as $a\cdot r\ne0$
$$\implies a=\frac 4r\implies \frac{4(1+r+r^2)}r=14\implies 2r^2-5r+2=0\implies r=2\text{ or }\frac12$$
Alternatively,
using Vieta's formula $a\cdot(a\cdot r) \cdot (a\cdot r^2)=64\implies (ar)^3=64$
So, $a\cdot r$ can be one of $4,4w,4w^2$ where $w$ is the cube root of $1$
Using Polynomial Remainder Theorem, observe that $4$ is a root of the given equation
$\implies a\cdot r=4\iff a=\frac4r$
Again, using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies 2r^2-5r+2=0$
| {
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$x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$ I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.
Given the function
$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$
Prove that the taylor polynomial of degree $4$ of $f$ is equal to
$5+x^4+y^4$.
First, $4\cos(xy) = 4 - 2(xy)^2 + 4R_3 $
$(x^2+y^2+1)^2=x^4+2 x^2 y^2+2 x^2+y^4+2 y^2+1$
Therefore: $(x^2+y^2+1)^2 - 2(x^2+y^2)=x^4+2 x^2 y^2+y^4+1$
Therefore: $f(x,y)=x^4+y^4+5+4R_3$
I don't know exactly why I can now conclude that Taylor Polynomial of degree 4 must be $5+x^4+y^4$, but I don't know exactly why.
Now the second question is: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$
New edit
I understand this now thanks to hint of Hagen von Eitzen, thanks !
The third question is:
Determine what kind of stationary point you have in $(0,0)$.
| Write $x^4=2x^4-x^4$ and similarly $y^4=2y^4-y^4$
$$(x^2-y^2)^2 \ge 0$$
$$x^4+y^4-2x^2y^2 \ge 0$$
$$2x^4-x^4+2y^4-y^4-2x^2y^2 \ge0$$
$$2x^4+2y^4 \ge x^4+y^4+2x^2y^2$$
$$x^4+y^4 \ge \dfrac{(x^2+y^2)^2}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Power series of $\frac{\sqrt{1-\cos x}}{\sin x}$ When I'm trying to find the limit of $\frac{\sqrt{1-\cos x}}{\sin x}$ when x approaches 0, using power series with "epsilon function" notation, it goes :
$\dfrac{\sqrt{1-\cos x}}{\sin x} = \dfrac{\sqrt{\frac{x^2}{2}+x^2\epsilon_1(x)}}{x+x\epsilon_2(x)} = \dfrac{\sqrt{x^2(1+2\epsilon_1(x))}}{\sqrt{2}x(1+\epsilon_2(x))} = \dfrac{|x|}{\sqrt{2}x}\dfrac{\sqrt{1+2\epsilon_1(x)}}{1+\epsilon_2(x)} $
But I can't seem to do it properly using Landau notation
I wrote :
$ \dfrac{\sqrt{\frac{x^2}{2}+o(x^2)}}{x+o(x)} $
and I'm stuck... I don't know how to carry these o(x) to the end
Could anyone please show me what the step-by-step solution using Landau notation looks like when written properly ?
| Now
$$\frac{\sqrt{1 - \cos(x)}}{\sin(x)} = \frac{\sqrt{2 \sin^2(x/2)}}{\sin(x)}
= \sqrt{2} \frac{ |\sin(x/2)|}{\sin(x)} = \sqrt{2} \frac{|x/2| + O(x^3)}{x + O(x^3)}\\
= \sqrt{2} \frac{\text{sgn}(x)/2 + O(x^2)}{1 + O(x^2)} = \text{sgn}(x)/\sqrt{2} + O(x^2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I find all the solutions of $\sin^5x+\cos^3x=1$
Find all the solutions of $$\sin^5x+\cos^3x=1$$
Trial:$x=0$ is a solution of this equation. How can I find other solutions (if any). Please help.
| Use what you know about the magnitudes of $\sin x$ and $\cos x$.
$$
\begin{aligned}
\sin^5x+\cos^3x &\le |\sin^5x+\cos^3x| \\
&\le |\sin^5 x| + |\cos^3 x| \\
&\le |\sin^2 x| + |\cos^2 x| \\
&= \sin^2 x + \cos^2 x\\
&= 1
\end{aligned}
$$
The inequality where the exponents are changed is only satisfied if the individual terms are equal: $\sin x$ and $\cos x$ must both be $0$ or $1$. Put that into the original equation, and you get $\sin x = 1$ or $\cos x = 1$. So, $x= \frac\pi2 + 2n\pi$ or $x = 2m\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/394649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
Finding $y$ value of canonical ellipse. I have an ellipse:
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
This may be a simple question, but my mind plays tricks on me at the moment;
Which is the most efficient way if I have $x$, $a$ and $b$ and want to find the value of $y$?
Hope someone can help me - thanks in advance :)!
| You can rearrange $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ to $$y^2 = b^2(\frac{x^2}{a^2} - 1)$$
With knowledge of $a,b$ and $x$ you can evaluate $y^2$ and $$y_{1,2} = \pm \sqrt{y^2} = \pm \sqrt{b^2(\frac{x^2}{a^2} - 1)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Having trouble using eigenvectors to solve differential equations The question asked to solve $$\frac{dx}{dy} = \begin{pmatrix}
5 & 4 \\
-1 & 1\\
\end{pmatrix}x$$ ,where $$ x = \begin{pmatrix} x_1 \\
x_2 \\ \end{pmatrix}$$
I went ahead an found the determinant of matrix $$ |A - I\lambda| = \lambda^2 - 4\lambda + 9$$
And found $\lambda = 3$
Then the $\alpha$ matrices was found to be
$$ \begin{pmatrix}
5 & 4\\
-1 & 1\\
\end{pmatrix} \alpha = 3\alpha$$ where $$\alpha = \begin{pmatrix}
\alpha_1 \\
\alpha_2 \\
\end{pmatrix}$$
Ultimately $-\alpha_1 = 2\alpha_2$ so I write $$\alpha = \begin{pmatrix}
-1 \\
2 \\
\end{pmatrix}$$
Then because $\lambda$ is a repeated root I know the solution is supposed to look something like this:
$$x = c_1\alpha e^{\lambda t} + c_2( \alpha t + \beta) e^{\lambda t}t$$
And then this is where it gets tricky for me. I know we find the $\beta$ matrix by figuring this out:
$$(A - I\lambda)\beta = \alpha$$
Now when I multiply all that out I get
$$2\beta_1 + 4\beta_2 = -1$$
$$\beta_1 + 2\beta_2 = -2$$
This is the system of equations I can't seem to solve to get a suitable $\beta$. One option I have is to make $$\beta = \begin{pmatrix}
0\\
-1\\
\end{pmatrix}$$
But this doesn't work for the second system of equations. Help please.
| The characteristic polynomial is:
$$|A - \lambda I| = 0 \rightarrow \lambda^2-6 \lambda+9 = 0 \rightarrow \lambda_{1,2} = 3$$
Substituting in the first eigenvalue to find the first eigenvector:
$$[A - \lambda I]v_1 = 0 \rightarrow \begin{pmatrix}
2 & 4 \\-1 & -2\\\end{pmatrix}v_1 = 0$$
After RREF, for the first eigenvector, I would have chosen:
$$a + 2b = 0 \rightarrow b = 1 \rightarrow a = -2$$
So, the first eigenvector is $v_1 = (-2, 1)$.
Since we have a repeated eigenvalue, we need a generalized eigenvector and you did the right approach, we have:
$$[A - \lambda I]v_2 = v_1$$
$$\begin{pmatrix}
2 & 4 \\-1 & -2 \end{pmatrix}v_2 = \begin{pmatrix}-2\\1 \end{pmatrix}$$
The RREF is:
$$\begin{pmatrix}1 & 2 \\ 0 & 0 \end{pmatrix}v_2 = \begin{pmatrix}-1\\0 \end{pmatrix} $$
This yields:
$$a + 2b = -1 \rightarrow b = 0 \rightarrow a = -1$$
From this, the second eigenvector is $v_2 = (-1, 0)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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find out the value of $\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$ If $(x-3)^2+(y-5)^2+(z-4)^2=0$,then find out the value of $$\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$$
just give hint to start solution.
| Hint: $(x-3)^3\geq0,(y-5)^2\geq0,(z-4)^2\geq0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/396284",
"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$? The answer to a question I asked earlier today hinged on the fact that
$$\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}} = 7$$
How does one evaluate such expressions? And, is there a way to evaluate the general expression
$$\sqrt[3]{a + ib} + \sqrt[3]{a - ib}$$
| Here's one way for square roots:
$$(\sqrt{x+y}+\sqrt{x-y})^2 = 2x + 2\sqrt{x^2-y^2}.$$
Perhaps this is easier to evaluate. For cube roots:
$$
(\sqrt[3]{x+y}+\sqrt[3]{x-y})^3 = 2x + 3\sqrt[3]{(x+y)(x^2-y^2)}+3\sqrt[3]{(x-y)(x^2-y^2)}.
$$
Of course, to get the coefficients in all of these I am using the binomial theorem. There really is no guarantee that these are easier to deal with in general.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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Find min of $IA + IB + IC +ID$ in tetrahedron $ABCD$ Let the point $I$ in tetrahedron $ABCD$. Find $\min\{IA + IB + IC + ID\}$.
I can't solve this problem, even in the case ABCD regular. Please help
|
I only can solve regular case, here is the solution:
$H$ and $F$ are one the plane $BCD$, $H,F$ is the foot of $A,I$,
$G$ is on $AH$ and $GH=IF=q,IG=FH=p,\angle FHB=\theta,BH=HD=HB=b,AH=h $
$AI=\sqrt{p^2+(h-q)^2},BF^2=p^2+b^2-2pbcos(\theta),FD^2=p^2+b^2-2pbcos(\dfrac{2\pi}{3}-\theta),FC^2=p^2+b^2-2pbcos(\dfrac{2\pi}{3}+\theta)$
$AI+BI+CI+DI=\sqrt{p^2+(h-q)^2}+\sqrt{p^2-2pbcos(\theta)+b^2+q^2}+\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}-\theta)+q^2}+\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}+\theta)+q^2}=f(p,q,\theta)$
we note that $p,q$ is independent, so we take $p$ first to let $f_{p}=f_{p_{min}}$
$f'_{p}=\dfrac{2p}{\sqrt{p^2+(h-q)^2}}+\dfrac{2p-2bcos(\theta)}{\sqrt{p^2-2pbcos(\theta)+b^2+q^2}}+\dfrac{2p-2bcos(\dfrac{2\pi}{3}-\theta)}{\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}-\theta)}}+\dfrac{2p-2bcos(\dfrac{2\pi}{3}+\theta)}{\sqrt{p^2+b^2-2pbcos(\dfrac{2\pi}{3}+\theta)+q^2}}=0$
here we proof that $g(x)=\dfrac{2x-2m}{\sqrt{x^2-2mx+m^2+q^2}}$ will be mono increasing function. because:
$g'(x)=\dfrac{q^2}{(x^2-2mx+m^2+q^2)^{\frac{3}{2}}}>0 \to f'_{p}$is mono increasing function also . luckily $f'_{p}(0)=0$, so we have only one root. it is trivial that $\sqrt{p^2-tp+s}$ will be increasing function when $p$ is big, so $ p=0 \to f_{p_{min}}$ and again we are lucky to clean $\theta$ also.
now $f=3\sqrt{b^2+q^2}+h-q$, $b,h$ is known so you can find final result.
| {
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Show that $8 \mid (a^2-b^2)$ for $a$ and $b$ both odd
If $a,b \in \mathbb{Z}$ and odd, show $8 \mid (a^2-b^2)$.
Let $a=2k+1$ and $b=2j+1$. I tried to get $8\mid (a^2-b^2)$ in to some equivalent form involving congruences and I started with
$$a^2\equiv b^2 \mod{8} \Rightarrow 4k^2+4k \equiv 4j^2+4j \mod{8}$$
$$\Rightarrow k^2+k-j^2-j=2m$$
for some $m \in \mathbb{Z}$ but I am not sure this is heading anywhere that I can tell.
Second attempt: Use Euler's Theorem and as $\gcd(a,8)=\gcd(b,8)=1$ and $\phi(8)=4$, $a^4 \equiv b^4 \equiv 1 \mod 8$ so $a^4-b^4\equiv 0 \mod{8}$.
I haven't gotten too much further are there any hints?
| HINT:
$k^2-j^2+k-j=(k-j)(k+j+1)$
As $(k+j+1)-(k-j)=2j+1$ which is odd, they must be of opposite parity, exactly one of them must be divisible by $2$
Method 2:
If $a,b$ are odd, observe that one of $(a-b),(a+b)$ is divisible by $4,$ the other by $2$
Method 3:
$(2a+1)^2=4a^2+4a+1=8\frac{a(a+1)}2+1\equiv 1\pmod 8$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\to \pi/2} \;\frac 1{\sec x+ \tan x}$ how to solve it answer is $0$, but $\frac 1{\infty + \infty}$ is indeterminate form
$$\lim_{x \to \pi/2} \frac 1{\sec x + \tan x}$$
| We have for $x$ with $\cos x \ne 0$
\begin{align*}
\frac 1{\tan x + \sec x} &= \frac 1{\frac{\sin x}{\cos x} + \frac 1{\cos x}}\\
&= \frac{\cos x}{1 + \sin x}
\end{align*}
And hence
\[ \lim_{x \to \frac \pi 2} f(x) = \lim_{x\to\frac\pi 2} \frac{\cos x}{1 + \sin x} = \frac{\cos \frac \pi 2}{1 + \sin \frac \pi 2} = \frac{0}{1 + 1} = 0. \]
| {
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Generating functions combinatorical problem In how many ways can you choose $10$ balls, of a pile of balls containing $10$ identical blue balls, $5$ identical green balls and $5$ identical red balls?
My solution (not sure if correct, would like to have input):
Define generated function:
$$\begin{align}
A(x) & =(x^0+x^1+x^2+...+x^{10})(x^0+x^1+x^2+...+x^5)(x^0+x^1+x^2+...+x^5) \\
& ={1-x^{11} \over 1-x}\cdot \left({1-x^6 \over 1-x}\right)^2 \\
& =(1-x^{11})(1-2x^6+x^{12}) \cdot {1 \over (1-x)^3} \\
& =(1-2x^6-x^{11}+x^{12}+2x^{17}-x^{23})\cdot \sum_{n=0}^∞ {n+2 \choose 2} \cdot x^n.
\end{align}$$
We look for the coefficient of $x^{10}$, so we get:
$$a_{10}=1 \cdot {10+2 \choose 2}-2\cdot {4+2 \choose 2}=36.$$
This seems incorrect (sadly I'm terrible in 'ordinary combinatorics' so I'm not sure how to calculate this 'regularly'). I would love to get input and hints.
Thanks in advance.
| Here’s one elementary approach. As you see, it confirms your result.
Let $b,g$, and $r$ be the numbers of blue, green, and red balls chosen to make up a set of $10$ balls. You’re looking for the number of solutions in non-negative integers to the equation $$b+g+r=10\;,\tag{1}$$ subject to the condition that $g\le 5$ and $r\le 5$. (You also have to have $b\le 10$, but that imposes no additional constraint when the sum is to be $10$.) Without the upper bounds this is a standard stars-and-bars problem whose solution is
$$\binom{10+3-1}{3-1}=\binom{12}2\;.$$
However, this count includes solutions with too many green or red balls. Let $g'=g-6$; then there is a bijection between solutions to $$b+g'+r=4\tag{2}$$ in non-negative integers and solutions to $(1)$ for which $g>5$. Thus, we need only count solutions to $(2)$ to get the number of solutions to $(1)$ with $g>5$. This is another stars-and-bars problem, and the answer is
$$\binom{4+3-1}{3-1}=\binom62\;.$$
Similarly, there are $\dbinom62$ solutions to $(1)$ that have $r>5$. There are no solutions that exceed the upper limits on both $g$ and $r$, so the number of solutions to $(1)$ that satisfy all of the conditions is
$$\binom{12}2-2\binom62=36\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/398415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
A problem on matrices: Find the value of $k$
If $
\begin{bmatrix}
\cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\
\sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7} \\
\end{bmatrix}^k
=
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}
$, then the least positive integral value of $k$ is?
Actually, I got no idea how to solve this.
I did trial and error method and got 7 as the answer.
how do i solve this? Can you please offer your assistance? Thank you
| I founded out that $$A^k=
\begin{bmatrix}
\cos \frac{k.2 \pi}{7} & -\sin \frac{k.2 \pi}{7} \\
\sin \frac{k.2 \pi}{7} & \cos \frac{k.2 \pi}{7} \\
\end{bmatrix}
$$
using appropriate trigonometric formulae. Now for $A=I$, $k$ should be equal to $7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Algebraic expression in its most simplified form I am trying to simplify the algebraic expression:
$$\bigg(x-\dfrac{4}{(x-3)}\bigg)\div \bigg(x+\dfrac{2+6x}{(x-3)}\bigg)$$
I am having trouble though. My current thoughts are:
$$=\bigg(\dfrac{x}{1}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x}{1}+\dfrac{2+6x}{(x-3)}\bigg)$$
$$=\bigg(\dfrac{x(x-3)}{1(x-3)}-\dfrac{4}{(x-3)}\bigg)\div \bigg(\dfrac{x(x-3)}{1(x-3)}+\dfrac{2+6x}{(x-3)}\bigg)$$
$$=\bigg(\dfrac{x(x-3)+(-4)}{(x-3)}\bigg)\div \bigg(\dfrac{x(x-3)+2+6x}{(x-3)}\bigg)$$
$$=\dfrac{x(x-3)+(-4)}{(x-3)}\times \dfrac{(x-3)}{x(x-3)+2+6x}$$
$$=\dfrac{x(x-3)+(-4)(x-3)}{(x-3)x(x-3)+2+6x} $$
$$\boxed{=\dfrac{-4(x-3)}{2(1+3x)} }$$ Which does not appear is not the answer. Am I close? Where exactly did I go wrong? I have tried this question multiple times.
Edit: Figured it out!
$\dfrac{x(x-3) - 4}{x(x - 3) + 2(1 + 3x)}\implies\dfrac{x^2-3x-4}{x^2+3x+2}\implies \dfrac{(x-4)(x+1)}{(x+2)(x+1)}$
$(x+1)$'s cancel leaving us with: $\boxed{\dfrac{x-4}{x+2}}$
| You did not distribute the term $(x - 3)$ in the denominator when you wrote:
$$\begin{align} & =\dfrac{x(x-3)+(-4)}{(x-3)}\times \dfrac{(x-3)}{x(x-3)+2+6x} \\ \\ & =\dfrac{x(x-3)+(-4)(x-3)}{(x-3)x(x-3)+2+6x} \end{align}$$
What would be correct is the following denominator:
$$\begin{align} & \quad\color{blue}{(x-3)}[x(x-3)+2+6x] \\ \\ & = \color{blue}{(x-3)}x(x-3)+\color{blue}{(x-3)}(2+6x)\end{align} $$
But note
$$\dfrac{x(x-3)+(-4)}{\color{blue}{\bf (x-3)}}\times \dfrac{\color{blue}{\bf(x-3)}}{x(x-3)+2+6x}$$
The highlighted terms cancel, leaving you:
$$\begin{align} & =\dfrac{x(x-3)+(-4)}{1}\times \dfrac{1}{x(x-3)+2+6x} \\ \\
& = \frac{x(x-3) - 4}{x(x - 3) + 2 + 6x} \\ \\
& = \dfrac{x^2-3x-4}{x^2+3x+2} \tag{$\diamondsuit$}
\end{align} $$
Now, all both the numerator and denominator of $\diamondsuit$ factor very nicely, and in fact, share a common factor, and hence, can be further simplified.
Recall: $$\frac{[b + c]a}{a[d+e]} = \frac{a[b+c]}{a[d+e]} = \frac{b+c}{d+e}$$
Or: $$\frac{a[b+c]}{a[d+e]} = \frac{ab+ac}{ad+ae}= \frac{b+c}{d+e}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Show that $n \ge \sqrt{n+1}+\sqrt{n}$ (how) Can I show that:
$n \ge \sqrt{n+1}+\sqrt{n}$ ?
It should be true for all $n \ge 5$.
Tried it via induction:
*
*$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.
*$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$
Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)
Is this the right way?
| You can replace the $\sqrt{n}$ on the RHS with another $\sqrt{n+1}$. Therefore you have $n\ge2\sqrt{n+1}$, or $n^2\ge4n+4$. $n^2-4n+4\ge8$, or $(n-2)^2\ge8$. The lowest integer solution to this is $5$, so $n\ge5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 5
} |
The Laplace operator. What will be the value of $\Delta\left(\frac 1{r^2}\right)$ if $r=|x|=\sqrt{x_1^2+x_2^2+x_3^2}$?
May be this can be determined using Green's formula.
| If we don't wanna use coordinate transform, it can be also done in a "hands-on" way:
$$\Delta u= \sum_{i}\frac{\partial^2 u}{\partial x_i^2},$$
where $$u = \frac{1}{r^2} = \frac{1}{\sum\limits_i x_i^2}= \frac{1}{x_1^2 + x_2^2 + x_3^2}.$$
We can compute it term by term:
$$
\frac{\partial u}{\partial x_1} = -\frac{1}{(\sum_i x_i^2)^2}\cdot 2 x_1 = -\frac{ 2 x_1}{(x_1^2 + x_2^2 + x_3^2)^2} ,
$$
and
$$
\frac{\partial^2 u}{\partial x_1^2} = -\frac{2 (\sum_i x_i^2) ^2 - 2 x_1 \cdot2(\sum_i x_i^2) 2 x_1 }{(\sum_i x_i^2)^4} = -\frac{2 (x_1^2 + x_2^2 + x_3^2)^2- 8 x_1^2 (x_1^2 + x_2^2 + x_3^2) }{(x_1^2 + x_2^2 + x_3^2)^4} ,
$$
Hence:
$$
\Delta u = -\frac{6 (x_1^2 + x_2^2 + x_3^2)^2- 8 (x_1^2 + x_2^2 + x_3^2)^2 }{(x_1^2 + x_2^2 + x_3^2)^4} = \frac{2}{(x_1^2 + x_2^2 + x_3^2)^2} = \frac{2}{r^4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate : $\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$ Evaluate: $$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$$
First approach :
$$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4(1-\cos^2x)}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{\cos^2xdx}{4 - 3\cos^2x}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3}\{\frac{4-3\cos^2x-4}{4 - 3\cos^2x}\}\,dx$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3}\{ 1- \frac{4}{4 - 3\cos^2x}\}\,dx$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3} 1\,dx- \int^{\frac{\pi}{2}}_0 \frac{1}{3} \frac{4\sec^2x}{4\sec^2x - 3}\,dx$$
$$=\int^{\frac{\pi}{2}}_0 \frac{1}{3} 1\,dx- \int^{\frac{\pi}{2}}_0 \frac{1}{3} \frac{4
\sec^2x}{4(1+\tan^2x) - 3}\,dx$$
Now I can easily put $\tan x = t$ and I get $\sec^2x \,dx =dt$
Second approach :
$$\int^{\frac{\pi}{2}}_0 \frac{\cos^2x\,dx}{\cos^2x+4\sin^2x}$$
Dividing numerator and denominator by $\cos^2x$ we get :
$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{1 +4\tan^2x}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{(4)\{\frac{1}{4} +\tan^2x\}}$$
$$=\int^{\frac{\pi}{2}}_0 \frac{dx}{(4)\{\{\frac{1}{2}\}^2 +(\tan x)^2\}}$$
Can we apply this formula of integral here :
$$\int \frac{1}{a^2+x^2}dx = \frac{1}{a}\tan^{-1}\frac{x}{a}$$
I tried but its not working here, I think doing some manipulation we can implement this here.. Please suggest thanks...
| You multiply and divide by $\sec^{4}(x)$ and see if it's working. The denominator becomes $$\sec^{2}(x) + 4 \tan^{2}(x) \sec^{2}(x)$$ Now use $1+\tan^{2}(x)=\sec^{2}(x)$.
So
*
*In numerator you have $\sec^{2}(x)$
*In denominator you have $\sec^{2}(x) \cdot \left(1+4\tan^{2}(x)\right) = (1+\tan^{2}(x))\cdot(1+4\tan^{2}(x))$.
*Now put $t=\tan(x)$ and your integral becomes $$\int_{0}^{\infty} \frac{dt}{(1+t^{2}) \cdot (1+4t^{2})} = \int_{0}^{\infty} \left\{ \frac{1}{1+t^2} -\frac{4}{1+4t^2}\right\} \ dt$$
Oh. I missed a trick here. Note that you had $$\int_{0}^{\pi/2} \frac{dx}{1+4\tan^{2}(x)}$$ Now put $t = \tan(x)$. Then you have $dt = \sec^{2}(x) \ dx $ and so $dx = \frac{dt}{(1+t^{2})}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Algebra simplification in mathematical induction . I was proving some mathematical induction problems and came through an algebra expression that shows as follows:
$$\frac{k(k+1)(2k+1)}{6} + (k + 1)^2$$
The final answer is supposed to be:
$$\frac{(k+1)(k+2)(2k+3)}{6}$$
I walked through every possible expansion; I combine like terms, simplify, factor, but never arrived at the answer.
Could someone explain the steps?
| When we are given an expression of the form:
$$\frac{k(k+1)(2k+1)}{6} + (k + 1)^2$$
We should recognize that this is a case of simply adding fractions which can also be referred to as rational expressions.
The first thing I would suggest is to rewrite this as a case of adding fractions:
$$\dfrac{k(k+1)(2k+1)}{6} + \dfrac{(k + 1)^2}{1}$$
Notice the denominators are different. In order to add fractions or rational expressions we need a common denominator. In this case, a common denominator could be 6.
What we will do is multiply the numerator and the denominator $\dfrac{(k + 1)^2}{1}$ by 6. We should also expand $(k+1)^2=(k+1)(k+1)$.
Rewriting the equation and having a common denominator of 6:
$$\dfrac{k(k+1)(2k+1)}{6} + \dfrac{6(k + 1)(k+1)}{1(6)}$$
Now that we have a common denominator of 6 we can simply add the fraction and simplify.
$$\dfrac{k(k+1)(2k+1)+6(k + 1)(k+1)}{6} $$
$$=\dfrac{k(2k^2+3k+1)+6(k^2+2k+1)}{6} $$
$$=\dfrac{(2k^3+3k^2+k)+(6k^2+12k+6)}{6} $$
$$=\dfrac{2k^3+9k^2+13k+6}{6} $$
Factoring again:
$$=\dfrac{2k^3+9k^2+13k+6}{6} $$
Giving us our final answer simplified:
$$\boxed{\dfrac{(k+1)(k+2)(2k+3)}{6}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$ Background: Evaluation of $\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx$
We can prove using the Beta-Function identity that
$$\int_0^\infty \frac{1}{(1+x^2)^\lambda}dx=\sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \quad \lambda>\frac{1}{2}$$
Differentiating the above equation with respect to $\lambda$, we obtain an expression involving the Digamma Function $\psi_0(z)$.
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^\lambda}dx = \sqrt{\pi}\frac{\Gamma \left(\lambda-\frac{1}{2} \right)}{\Gamma(\lambda)} \left(\psi_0(\lambda)-\psi_0 \left( \lambda-\frac{1}{2}\right) \right)$$
Putting $\lambda=2$, we get
$$\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx = -\frac{\pi}{4}+\frac{\pi}{2}\log(2)$$
Question:
But, does anybody know how to evaluate $\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx$ and $\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx$?
Mathematica gives the values
*
*$\displaystyle \int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx = -\frac{G}{6}+\pi \left(-\frac{3}{8}+\frac{1}{8}\log(2)+\frac{1}{3}\log \left(2+\sqrt{3} \right) \right)$
*$\displaystyle \int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx = -\frac{\pi}{2}+\frac{\pi \log \left( 6+4\sqrt{2}\right)}{4}$
Here, $G$ denotes the Catalan's Constant.
Initially, my approach was to find closed forms for
$$\int_0^\infty \frac{1}{(1+x^2)^2(1+x^3)^\lambda}dx \ \ , \int_0^\infty \frac{1}{(1+x^2)^2(1+x^4)^\lambda}dx$$
and then differentiate them with respect to $\lambda$ but it didn't prove to be of any help.
Please help me prove these two results.
| The generalized results for the even and odd cases of $$I_n=\int_0^\infty \frac{\log(1+x^n)}{(1+x^2)^2}dx
$$
are respectively as follows
\begin{align}
I_{2m} =& -\frac{m\pi}4+\frac{m\pi}2\ln2+\pi\sum_{k=1}^{[\frac m2]}\ln \cos\frac{(m-2k+1)\pi}{4m}\\
I_{2m+1} =& -\frac{(2m+1)\pi}8+\frac{(4m+1)\pi}8\ln2+\frac{(-1)^m G}{2(2m+1)}\\
&\ +\frac\pi2\sum_{k=0}^{m-1}\left[\ln \cos\frac{(2k+1)\pi}{4(2m+1)}+\frac{(-1)^{m+k}(2k+1)}{2(2m+1)}\ln\tan\frac{(2k+1)\pi}{4(2m+1)}
\right]\\
\end{align}
Specifically
\begin{align}
\int_0^\infty \frac{\log(1+x)}{(1+x^2)^2}dx
=& -\frac\pi8+\frac\pi8\ln2 +\frac12G\\
\int_0^\infty \frac{\log(1+x^2)}{(1+x^2)^2}dx
=& -\frac\pi4+\frac\pi2\ln2 \\
\int_0^\infty \frac{\log(1+x^3)}{(1+x^2)^2}dx
=& -\frac{3\pi}8+\frac{\pi}8\ln2
+\frac\pi3\ln(2+\sqrt3) -\frac16G\\
\int_0^\infty \frac{\log(1+x^4)}{(1+x^2)^2}dx
=& -\frac\pi2+\frac\pi2\ln(2+\sqrt2)\\
\int_0^\infty \frac{\log(1+x^5)}{(1+x^2)^2}dx
=& -\frac{5\pi}8-\frac{3\pi}8\ln2+\frac\pi2\ln\left(1+\sqrt5+\sqrt{2(5+\sqrt5)}\right)\\
&\ +\frac\pi{20}\ln\tan\frac\pi{20}-\frac{3\pi}{20}\ln\tan\frac{3\pi}{20} +\frac1{10}G\\
\int_0^\infty \frac{\log(1+x^6)}{(1+x^2)^2}dx
=& -\frac{3\pi}4+\frac\pi2\ln6\\
\int_0^\infty \frac{\log(1+x^7)}{(1+x^2)^2}dx
=& -\frac{7\pi}8+\frac{13\pi}8\ln2+\frac\pi2\ln\left(\cos\frac\pi{28} \cos\frac{3\pi}{28} \cos\frac{5\pi}{28}\right)\\
& -\frac\pi{28}\ln\tan\frac\pi{28}+\frac{3\pi}{28}\ln\tan\frac{3\pi}{28} -\frac{5\pi}{28}\ln\tan\frac{5\pi}{28} -\frac1{14}G\\
\int_0^\infty \frac{\log(1+x^8)}{(1+x^2)^2}dx
=& -\pi+\pi\ln\left(\sqrt2+\sqrt{2+\sqrt2}\right)\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/414642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 6,
"answer_id": 2
} |
How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't help as it leaves $\frac{1+x-2x^2}{1-4x^2}$ any ideas?
| Hint:$f^{-1}(f(x))=x$ where $f(x)=\frac{1-x}{1-2x}$,$(x\ne 1/2)$
Solution:Here $f^{-1}(x)=\frac{1-x}{1-2x}$.We know that $f^{-1}(f(x))=x$ and as $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}=f^{-1}(f(x))$ so it follows that $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}=f^{-1}(f(x))=x$ So we dont need to simplify.
(Here $f'(x)=\frac{1}{(1-2x)^2}>0$ so $f^{-1} $ exists.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/415304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
How to find out X in a trinomial How can I find out what X equals in this?
$$x^2 - 2x - 3 = 117$$
How would I get started? I'm truly stuck.
| A different way if you have not seen the quadratic formula yet.
Recall that $(x-1)^2 = x^2-2x+1$ and so
$$
x^2-2x-3 = \left(x^2-2x+1\right)-4=(x-1)^2-4
$$
and your equation $120=x^2-2x-3$ becomes equivalent to
$$
117=(x-1)^2-4
$$
so $(x-1)^2 = 121 = 11^2$. Therefore, $x-1 = 11$ or $x-1 = -11$.
In the first case, $x = 11+1 = 12$.
In the second case, $x = -11+1 = -10$.
Check
In the first case, $x=12$ so $x^2-2x-3 = 144-24-3 = 117$.
In the second case, $x=-10$, so $x^2-2x-3 = 100+20-3 = 117$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/416818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
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