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Small angle approximations - different answers I would like to approximate $$\frac{\cos^2{x}}{\sin(x) \tan(x)}$$ using the small angle approximations.
Throughout I will use $\sin(x) \approx x$, $\tan(x) \approx x$, $\cos(x) \approx 1 - \frac{x^2}{2}$.
Method 1:
$$\frac{\cos^2{x}}{\sin(x) \tan(x)} \approx \frac{\left(1-\frac{x^2}{2}\right)^2}{x^2} = \frac{1}{4}(x^2 - x^{-2} - 1)$$
Method 2:
$$\frac{\cos^2{x}}{\sin(x) \tan(x)} = \frac{\cos^3{x}}{\sin^2(x)} \approx \frac{1}{x^2} \left(1 - \frac{x^2}{2}\right)^3$$
which when expanded out gives for example $-\frac{3}{2} \neq - \frac{1}{4}$ as the constant term.
Why the differences? Which is the better way to approximate this and why?
| Because you are using less terms in your first expansion. What happens if you take $\tan(x) \approx \frac{x}{1 -x^2/2}$? It gives the second expansion. So the first one is a weaker approximation. In the first your are not keeping all terms of order $x^2$, but you do in the second. Note that the leading order terms for small $x$ are $x^{-2}$ which are the same.
| {
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"url": "https://math.stackexchange.com/questions/3200995",
"timestamp": "2023-03-29T00:00:00",
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If $\det\left(\begin{smallmatrix}a & 1 & 1\\1 & b & 1\\1 & 1 & c\end{smallmatrix}\right) > 0 $ then prove that $abc> -8$
If the value of the determinant $$\begin{vmatrix}
a & 1 & 1\\
1 & b & 1\\
1 & 1 & c
\end{vmatrix} > 0 $$ then prove that $abc> -8$
I have calculated its determinant and got
$$abc - (a+b+c) > -2$$
Please tell me how to proceed further to the given answer form.
| Another counterexample to this claim is $(a,b,c) = (-1,-1,-k)$ for any $k\ge 8$. Then
$$
\begin{vmatrix}
a & 1 & 1\\
1 & b & 1\\
1 & 1 & c
\end{vmatrix} = \begin{vmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -k
\end{vmatrix} = 4
$$
for any choice of $k$, but $abc = -k$ can be made as small as we want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202121",
"timestamp": "2023-03-29T00:00:00",
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Minimum value of $\frac{2-\cos(x)}{\sin(x)}$ without differentiation I have to find the minimum value of the expression
$$\frac{2 - \cos x}{ \sin x}$$
Also $x$ lies between $0$ to $\pi$. One way is to find the minima using differentiation. But it is not taught in my grade so my teacher asked me to do it without differentiation.
Here's what I did
Let$$\frac{2 - \cos x }{ \sin x} = y~,$$
so that $$ (2-y \sin x)^2 = 1 - \sin^2(x) \\
\implies \sin^2(x) \cdot (y^2 +1) - 4 y \sin x +3=0$$
Now I am struck. I tried using Discriminant $\ge{0}$ but no use as our variable $\sin x$ lies between $0$and $1$. Please help.
| Let $$k=\frac{2-\cos x}{\sin x}\Rightarrow 2-\cos x=k\sin x$$
So we have $$k\sin x+\cos x=2$$
Using $$|a\sin x+b\cos x|\leq \sqrt{a^2+b^2}$$
So we have $$|k\sin x+\cos x|\leq \sqrt{k^2+1}$$
$$2\leq \sqrt{k^2+1}\Rightarrow k^2+1\geq 4\Rightarrow k\geq \sqrt{3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prime numbers of the form $p=m^2+n^2$ such that $p \mid m^3+n^3-4$
Find all prime numbers $p$, for which there are positive integers $m$ and $n$ such that $p=m^2+n^2$ and $p \mid m^3+n^3-4$.
I simplified this a little bit.
$$m^2+n^2 \mid m^3+n^3-4 =(m+n)(m^2+n^2-mn)-4 \\ \Longrightarrow m^2+n^2 \mid (m+n)mn+4 $$
The only case that $m$ and $n$ can both be odd is $m=n=1$, which leads to $p=2$. If one of $m$ and $n$ is $1$ (For example $n=1$), then $p=m^2+1$ and
$$m^2+1 \mid m^2+m+4 \Longrightarrow m^2+1 \mid m+3 \Longrightarrow m=2$$
Which gives $p=5$.
For $m,n>1$ with different parity, I could not find a strong argument. Can you guys give it a try?
| This problem is from 2004 Silkroad Mathematical Competition. By the way, if you are into this problems, I think you should really visit AoPS forums, audience here at stack exchange are not very experienced with olympiadic problems.
Here is a short reasoning (similar to above). Check that, $p=2,5$ indeed works. Suppose now $p>5$. Note that, $m^3+n^3=(m+n)(m^2-mn+n^2)$. Notice that, $m^2-mn+n^2\equiv -mn\pmod{p}$. This yields, $p\mid mn(m+n)+4$. Next, note also that, $p=m^2+n^2$ implies, $(m+n)^2\equiv 2mn\pmod{p}$, and thus, $mn\equiv \frac{(m+n)^2}{2}\pmod{p}$. This yields, $p\mid (m+n)^3+8=(m+n+2)((m+n)^2+2(m+n)+4)$. Now, if $p\mid m+n$, we have $m^2+n^2\mid m+n$ with $m,n$ not simultaneously one, which is not sound. Similarly, if $p\mid m^2+2mn+n^2+2(m+n)+4$, then note that, $p\mid 2mn+2(m+n)+4$, or equivalently, $p\mid mn+m+n+2$, since $p>2$. This yields $m^2+n^2\leq mn+m+n+2$. Now, using AM-GM inequality, we also have $m^2+n^2\geq 2mn$, which implies, $mn-m-n+1\leq 3$, that is, $(m-1)(n-1)\leq 3$. In particular, either $(m-1)(n-1)=1$, which is not possible, or $(m-1)(n-1)=2$, yielding $m=3,n=2$, and $p=13$, for which the condition does not hold, or $(m-1)(n-1)=3$, yielding $m=4,n=2$, for which, again, $m^2+n^2$ is not prime.
| {
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infinitely many negative and infinitely many positive numbers Suppose that
$$x_1=\frac{1}{4}, \ x_{n+1}=x_{n}^3-3x_n.$$
Show that the sequence has infinitely many negative and infinitely many positive numbers.
My idea: Suppose that it has finitely many negative numbers. then all the numbers after some index, must be larger than $\sqrt{3}$. I want to show that the sequence cannot escape some interval.
| The desired claim follows from the following two observations:
Claim. If $x_n \in \{-1/4, 1/4\}$, then $x_n \neq 0$ for all $n \geq 1$.
Proof. Let $p_1 = \operatorname{sign}(x_1)$ and $p_{n+1} = p_n^3 - 3p_n 4^{2 \cdot 3^{n-1}}$. Then we inductively check that $p_n$ is always odd and $ x_n = p_n / 4^{3^{n-1}}$. Since the numerator is always odd integer, it cannot vanish, and the claim follows.
Claim 2. If either $x_n \geq 0$ for any sufficiently large $n$ or $x_n \leq 0$ for any sufficiently $n$, then we actually have $x_n = 0$ for any sufficiently large $n$.
Proof. Write $x_n = 2\cos(2 \pi f_n)$. Then
$$ \cos(2 \pi f_{n+1}) = \frac{x_{n+1}}{2} = \frac{x_n^3 - 3x_n}{2} = 4\cos^3(2 \pi f_n) - 3\cos(2 \pi f_n) = \cos(2 \pi \cdot 3f_n). $$
So it follows that $\cos(2\pi f_{N+n}) = \cos(2\pi \cdot 3^n f_N)$. Now, replacing $(x_n)$ by $(-x_n)$ if necessary, we may assume that $x_n \geq 0$ for all sufficiently large $n$. In other words, there exists $N$ so that $x_{N+n} \geq 0$ for all $n \geq 0$. Then each $3^n f_N$ must avoid the sets $(\frac{1}{4}, \frac{3}{4}) + \mathbb{Z}$. So it follows that
$$ f_N \in \mathbb{R} \setminus \bigcup_{n=0}^{\infty} \bigcup_{k\in\mathbb{Z}} \left( \frac{4k+1}{4 \cdot 3^n}, \frac{4k+3}{4 \cdot 3^n} \right) = \left\{ k \pm \frac{1}{4\cdot 3^n} : n \geq 0 \right\}. $$
This implies that $3^{n_0} f_N = \pm 1$ for some $n_0 \geq 0$, and hence $x_{N+n} = 0$ for all $n \geq n_0$. This proves the desired claim.
| {
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"timestamp": "2023-03-29T00:00:00",
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Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$
If $$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}= \frac{3\sqrt{3}}{4}$$ for $a,b,c,d>0$
Then
Value of $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}+\frac{1}{\sqrt{1+d^2}}$ is
Try: using A.M G.M Inequality
$$1+a^2\geq 2a\;, 1+b^2\geq 2b\;,1+c^2\geq 2c\; 1+d^2\geq 2d$$
$$\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}\geq 4\sqrt{abcd}$$
$$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}\leq \frac{a+b+c+d}{4\sqrt{abcd}}$$
I have edited it, please have a look
Could some help me to solve it , Thanks
| We'll prove that
$$\frac{a+b+c+d}{\sqrt{(1+a^2)(1+b^2)(1+c^2)(1+d^2)}}\leq\frac{3\sqrt3}{4},$$ where the equality occurs for $a=b=c=d=\frac{1}{\sqrt3}$ only.
Indeed, let $a=\frac{x}{\sqrt3},$ $b=\frac{y}{\sqrt3},$ $c=\frac{z}{\sqrt3}$ and $d=\frac{t}{\sqrt3}.$
Thus, we need to prove that
$$(x^2+3)(y^2+3)(z^2+3)(t^2+3)\geq16(x+y+z+t)^2.$$
Consider five cases.
*
*$x\geq1\geq y\geq z\geq t$.
Thus, by C-S $$\prod_{cyc}(x^2+3)=(x^2+3)\prod_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1+4)=$$
$$=(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)+4\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)(z^2-1)+\prod_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)\right)=$$
$$=(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)+4\sum_{y\rightarrow z\rightarrow t\rightarrow y}(1-y^2)(1-z^2)-\prod_{y\rightarrow z\rightarrow t\rightarrow y}(1-y^2)\right)\geq$$
$$\geq(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)+4\sum_{y\rightarrow z\rightarrow t\rightarrow y}(1-y^2)(1-z^2)-(1-y^2)(1-z^2)\right)\geq$$
$$\geq(x^2+3)\left(4^3+4^2\sum_{y\rightarrow z\rightarrow t\rightarrow y}(y^2-1)\right)=16(x^2+1+1+1)(1+y^2+z^2+t^2)\geq$$
$$\geq16(x+y+z+t)^2;$$
2. $x\geq y\geq1\geq z\geq t.$
Thus, by C-S again we obtain:
$$\prod_{cyc}(x^2+3)=\prod_{cyc}(4+(x^2-1))\geq(16+4(x^2+y^2-2))(16+4(z^2+t^2-2))=$$
$$=16(x^2+y^2+1+1)(1+1+z^2+t^2)\geq16(x+y+z+t)^2;$$
The cases
*$x\geq y\geq z\geq1\geq t$,
*$1\geq x\geq y\geq z\geq t$ and
*$x\geq y\geq z\geq t\geq1$ for you.
Can you end it now?
| {
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How do I determine the limit of this sequence? I need a refresher.
I would assume that the answer would be $\infty$, but I am not quite sure.
The problem reads:
Determine the limit of this sequence? $c_n$ = $\sqrt {n^2+n} - \sqrt {n^2 - n}$ .
I would need a refresher on this. That's all thank you.
| $$\begin{align*}
c_n &= \frac{\left(\sqrt{n^2+n}-\sqrt{n^2-n}\right)\left(\sqrt{n^2+n}+\sqrt{n^2-n}\right)}{\sqrt{n^2+n}+\sqrt{n^2-n}}\\
&= \frac{n^2+n-n^2+n}{\sqrt{n^2+n}+\sqrt{n^2-n}}\\
&= \frac{2n}{\sqrt{n^2+n}+\sqrt{n^2-n}}\\
&= \frac{2}{\sqrt{1+1/n}+\sqrt{1-1/n}}\\
&\to \frac{2}{\sqrt{1+0}+\sqrt{1-0}}\\
&= 1
\end{align*}$$
| {
"language": "en",
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What is all N that make $2^N + 1$ divisible by 3? When $N = 1$, $2^N + 1 = 3$ which is divisible by $3$.
When $N = 7$, $2^N + 1 = 129$ which is also divisible by $3$.
But when $N = 2$, $2^N + 1 = 5$ which is not divisible by $3$.
A quick lookout can determine that when $N$ is an odd number, $2^N + 1$ is divisible by $3$. If is that correct, how can it be proven? Otherwise, what is the counterexample and the correct $N$?
Can this be made into a more general form, for example $2^N + C$ or something similar?
| You can easily prove by induction on $k$ that $2\times 4^k+1$ is divisible by $3$: It is for $k=1$, and if it is for some $k$, then
$$2\times 4^{k+1}+1=2\times 4^k\times 4+1=4(2\times 4^k+1)-3$$
Therefore $2^{2m+1}+1$ is divisible by $3$ for every natural $m$.
| {
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Solve $4+\frac{1}{x}-\frac{1}{x^2}$ using quadratic formula I am to solve for x using the quadratic formula:
$$4+\frac{1}{x}-\frac{1}{x^2}=0$$
The solution provided in the answers section is: $\dfrac{-1\pm\sqrt{17}}{8}$ whereas I arrived at something entirely different: $$\dfrac{\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
Here's my working:
Start with $$4+\frac{1}{x}-\frac{1}{x^2}=0$$
Rearranging into standard form:
$$-\frac{1}{x^2}+\frac{1}{x}+4=0$$
Multiply by $-1$ to get a positive leading coefficient $a$:
$$\frac{1}{x^2}-\frac{1}{x}-4=0$$
I'm not sure how to determine my inputs $a,b$ and $c$ with these fractions but I guess $a=\dfrac{1}{x^2}$, $b=\dfrac{1}{x}$ and $c=-4$.
Plugging into quadratic function:
$$x = \frac{-\frac{1}{x}\pm\sqrt{\frac{1}{x^2}+\frac{16}{x}}}{\frac{2}{x}}$$
I find this challenging due to the coefficients $a$ and $b$ being fractions.
How can I apply the quadratic formula to $4+\dfrac{1}{x}-\dfrac{1}{x^2}=0$ to arrive at $\dfrac{-1\pm\sqrt{17}}{8}$?
| Other answers followed my suggestion in the comments.
Here's an alternative:
Let $z=\dfrac1x.$ Then we have $-z^2+z+4=0$, so, using the quadratic formula, $z=\dfrac{-1\pm\sqrt{17}}{-2}.$
Therefore $x=\dfrac1z=\dfrac{-2}{-1\pm\sqrt{17}}=\dfrac{2(-1\mp\sqrt{17})}{16}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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An algebra problem: Let $a+b+c=0$. Prove that $a^2+b^2+c^2 =6/5$ Let $a,b,c$ be nonzero real numbers such that $a+b+c=0$ and $a^3+b^3+c^3 = a^5 +b^5 +c^5$. Prove that $a^2+b^2+c^2 =6/5$
I tried to expand $(a+b+c)^5$ but I can't get term of $a^2+b^2+c^2$.
| When you see symmetric equations, try the following substitution $$a+b+c=s=0\qquad ab+bc+ac=q\qquad abc=p$$ And rewrite the equations in terms of $s,q,p$
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\implies a^3+b^3+c^3=3p$$
$$a^5+b^5+c^5\stackrel{(*)}=-5(ab+bc+ac)(abc)=-5qp$$
(*) means: in this case.
If you're wondering how I obtained this, use Newton's Identities. It isn't hard, but long.
Finally $$a^3+b^3+c^3=a^5+b^5+c^5\implies -5qp=3p\implies q=-\frac35$$ Now $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=s^2-2q=\frac65$$
| {
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Finding the Inverse Laplace Transform of $-\frac{1}{s(s+1)}+\frac{(s+1)(s+2)}{s\left((s+1)^2+1\right)}$
I am trying to find the inverse Laplace transform of $$F(s)=-\frac{1}{s(s+1)}+\frac{(s+1)(s+2)}{s\left((s+1)^2+1\right)}.$$
I proceeded as follows:
\begin{align}
F(s)&=-\frac{1}{s(s+1)}+\frac{(s+1)(s+2)}{s\left((s+1)^2+1\right)} \\
&=-\frac{1}{s}+\frac{1}{s+1}+\frac{s+1}{(s+1)^2+1}+\frac{4}{s\left((s+1)^2+1\right)} \\
f(t)&=-1+e^{-t}+e^{-t}\cos(t)+\mathcal{L}^{-1}\left(\frac{4}{s}\times\frac{1}{(s+1)^2+1}\right) \\
&=-1+e^{-t}+e^{-t}\cos(t)+4\int_0^{\infty}e^{-u}\sin(u) \ du \ \ \ \ \ \text{(convolution theorem)} \\
&=1+e^{-t}(1+\cos(t))
\end{align}
But the answer is apparently $f(t)=e^{-t}(1+\sin(t)).$ I have looked over my worked and agree with my solution.
edit:
\begin{align}
\frac{(s+1)(s+2)}{s\left((s+1)^2+1\right)}&=\frac{s^2+3s+2}{s\left((s+1)^2+1\right)} \\
&=\frac{s^2+3s}{s\left((s+1)^2+1\right)}+\frac{2}{s\left((s+1)^2+1\right)} \\
&=\frac{s(s+3)}{s\left((s+1)^2+1\right)}+\frac{2}{s\left((s+1)^2+1\right)} \\
&=\frac{s+3}{(s+1)^2+1}+\frac{2}{s\left((s+1)^2+1\right)} \\
&=\frac{s+1}{(s+1)^2+1}+\frac{4}{s\left((s+1)^2+1\right)}
\end{align}
| Your partial fractions are not correct for the second term:
\begin{equation}
\frac{(s+1)(s+2)}{s((s+1)^2+1)}=\frac{1}{(s+1)^2+1}+\frac{1}{s}.
\end{equation}
Then,
\begin{equation}
F(s) = \frac{1}{(s+1)}+\frac{1}{(s+1)^2+1}.
\end{equation}
Thus, you get the desired result!
| {
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The number of positive integers of 5 digits such that each digit is 1, 2 or 3, and all three of the digits appear atleast once. The number of positive integers of 5 digits such that each digit is 1, 2 or 3, and all three of the digits appear at least once.
Working:
Three 1, one 2, one 3, number of numbers = 5!/3! = 20 Two 1, one 2, two 3, number of numbers = 5!/(2!*2!) = 30 Two 1, two 2, one 3, number of such numbers = 5!/(2!*2!) = 30 One 1, one 2, three 3, number of such numbers = 5!/3! = 20 One 1, two 2, two 3, number of such numbers = 5!/(2!*2!) = 30 One 1, three 2, one 3, number of such numbers = 5!/(3!) = 20 Therefore, total number of such numbers = 20 + 30 + 30 + 20 + 30 + 20 = 150
Question: Is there any other possible easier, and simpler approach?
| Inclusion Exclusion is somewhat easier.
The total number of such integers, ignoring the requirement that all three digits appear, is $3^5$. We then subtract off the instances in which one specified digit does not appear, that's $3\times 2^5$. Then we add back the $3$ instances in which two specified digits do not appear. Thus we get $$3^5-3\times 2^5+3=150$$
| {
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Remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$? I need to remove the discontinuity of $f(x) = \frac{x^2-11x+28}{x-7}$ at $f(7)$, and find out what $f(7)$ equals. I am not sure what I've done wrong, but I'm getting 33, which the website I'm using tells me is wrong.
Please help me figure out what I'm doing wrong here.
My Steps:
$$\frac{x^2-11x+28}{x-7} \cdot \frac{x+7}{x+7}$$
$$=\frac{x^3+7x^2-11x^2-77x+28x+196}{x^2-49}$$
$$=\frac{x^3-4x^2-49x+196}{x^2-49}$$
$$=\frac{x^2(x-4)-49(x+4)}{x^2-49}$$
$$=(x-4)(x+4)$$
plug in $7$ for $x$.
$$(7-4)(7+4)=7^2-16=33$$
| Hint: $x^2-11x+28=(x-7)(x-4)$. Can you finish?
| {
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Use numerical and graphical evidence to guess the value of the limit $\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}$?
Use numerical and graphical evidence to guess the value of the limit $$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}$$
I am a Calculus 1 student, and I'm not sure what this problem in my textbook means when it says, "use numerical and graphical evidence". I worked around with it a bit and found that the answer is 6. However, I'm not sure if this is what my professor wants.
The textbook is "Calculus: Early Transcendentals", 8th Edition, by James Stewart. This is problem 55a in section 2.2.
What does this question mean by, "Use numerical and graphical evidence"?
$$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}$$
$$=\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}\cdot \frac{\sqrt{x} + 1}{\sqrt{x} + 1}$$
$$=\underset{x \to 1} \lim \frac{(x^3-1)(\sqrt{x} + 1)}{x-1}$$
$$=\underset{x \to 1} \lim \frac{(x-1)(x^2+x+1)(\sqrt{x} + 1)}{x-1}$$
$$=\underset{x \to 1} \lim \;\ (x^2+x+1)(\sqrt{x} + 1)$$
$$= ((1)^2+(1)+1)(\sqrt{1}+1)=(3)(2)=6$$
| You did it well but it could be done faster.
Let $x=y^2$
$$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}=\underset{y \to 1} \lim \frac{y^6-1}{y-1}=1+y+y^2+y^3+y^4+y^5\,\, \to 6$$
If you want to also see how the limit is approached, let $x=1+t$
$$\underset{x \to 1} \lim \frac{x^3-1}{\sqrt{x}-1}=\underset{t \to 0} \lim \frac{(1+t)^3-1}{\sqrt{1+t}-1}$$
Using the binomial expansion or Taylor series
$$\frac{(1+t)^3-1}{\sqrt{1+t}-1}=\frac{3t+3t^2+t^3 } {\left( 1+\frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+O\left(t^4\right)\right)-1 }=6+\frac{15 t}{2}+O\left(t^2\right)$$ that is to say
$$\frac{x^3-1}{\sqrt{x}-1}=6+\frac{15}{2}(x-1)+O\left((x-1)^2\right)$$
Using the examples given by D.B., you get $6+\frac{15}{2}(0.99-1)=5.925$ and $6+\frac{15}{2}(1.01-1)=6.075$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Prove $\sum_{k=0}^\infty \binom{2n+k+1}{n}/2^{2n+k+1}=1$. I was trying to find a closed form for the sum
$\sum_{k=0}^\infty \binom{2n+k+1}{n}/2^{2n+k+1}$.
According to Wolfram
https://www.wolframalpha.com/input/?i=sum+(2n%2Bk%2B1)!%2F(n!*n%2Bk%2B1)!*2%5E(2n%2Bk%2B1))+from+k%3D0+to+infinity
this sum evaluates to 1, but I can't figure out how to prove this.
Any hints.
| Preliminary
$$
\begin{align}
a_n
&=\sum_{k=0}^n\frac{\binom{k+n}{k}}{2^k}\tag{1a}\\
&=\sum_{k=0}^n\frac{\binom{k+n-1}{k-1}+\binom{k+n-1}{k}}{2^k}\tag{1b}\\
&=\sum_{k=0}^{n-1}\frac{\binom{k+n}{k}}{2^{k+1}}+\sum_{k=0}^n\frac{\binom{k+n-1}{k}}{2^k}\tag{1c}\\
&=\frac12a_n-\frac{\binom{2n}{n}}{2^{n+1}}+a_{n-1}+\frac{\binom{2n-1}{n}}{2^n}\tag{1d}\\[3pt]
&=\frac12a_n+a_{n-1}\tag{1e}\\[9pt]
&=2a_{n-1}\tag{1f}
\end{align}
$$
Explanation:
$\text{(1a)}$: define $a_n$
$\text{(1b)}$: Pascal Identity
$\text{(1c)}$: substitute $k\mapsto k+1$ in the left sum
$\text{(1d)}$: apply $\text{(1a)}$
$\text{(1e)}$: cancel terms
$\text{(1f)}$: $2$ times $\text{(1e)}$ minus $\text{(1a)}$
Since $a_0=1$, we get
$$
\sum_{k=0}^n\frac{\binom{k+n}{k}}{2^k}=2^n\tag2
$$
Answer
$$
\begin{align}
\sum_{k=0}^\infty\frac{\binom{2n+k+1}{n}}{2^{2n+k+1}}
&=\sum_{k=0}^\infty\frac{\binom{2n+k+1}{n+k+1}}{2^{2n+k+1}}\tag{3a}\\
&=\frac1{2^n}\sum_{k=0}^\infty(-1)^{n+k+1}\frac{\binom{-n-1}{n+k+1}}{2^{n+k+1}}\tag{3b}\\
&=\frac1{2^n}\sum_{k=n+1}^\infty(-1)^k\frac{\binom{-n-1}{k}}{2^k}\tag{3c}\\
&=\frac1{2^n}2^{n+1}-\frac1{2^n}\sum_{k=0}^n(-1)^k\frac{\binom{-n-1}{k}}{2^k}\tag{3d}\\
&=2-\frac1{2^n}\sum_{k=0}^n\frac{\binom{k+n}{k}}{2^k}\tag{3e}\\[9pt]
&=1\tag{3f}
\end{align}
$$
Explanation:
$\text{(3a)}$: symmetry of Pascal's Triangle
$\text{(3b)}$: negative binomial coefficient
$\text{(3c)}$: substitute $k\mapsto k-n-1$
$\text{(3d)}$: Binomial Theorem
$\text{(3e)}$: negative binomial coefficient
$\text{(3f)}$: apply $(2)$
| {
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"url": "https://math.stackexchange.com/questions/3215020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Algebra question/geometry solving the system 3x + 2y = 21, (7 - x)² + y² = 2².
what does this mean ?
How to get the answer (7−4/√13,6/√13)
| This means that you have a linear graph 3x+2y=21 and a circle, with origin at (7,0) and a radius of 2. To solve this system, you need to find the intersection point of these 2 graphs.
Edit:
My algebraic solution:
$3x+2y=21\\
2y=21-3x\\
y=\frac{21-3x}{2}\\
(7-x)^2+\frac{9}{4} \cdot (7-x)^2=4\\
\frac{13}{4} \cdot (7-x)^2=4\\
(7-x)^2=\frac{16}{13}\\
x-7=\pm \frac{4}{\sqrt{13}}\\
x=7 \pm \frac{4}{\sqrt{13}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factorize the polynomial in the ring $\mathbb{Z}_5[x]$ I need to factor the polynomial $x^5+3x^4+x^3+x^2+3$ into the product of the irreducible ones in the ring $\mathbb{Z}_5[x]$.
The problem is I don't see any whole roots (I tried every possible divisor of 3) in the given polynomial. Does it mean that it is already irreducible?
| It is easy to check that $-1$ is a root, and by Horner's algorithm, you obtain
$$x^5+3x^4+x^3+x^2+3=(x+1)(x^4+2x^3-x^2+2x-2).$$
Similarly, the quotient has $2$ as a root and
$$ x^4+2x^3-x^2+2x-2=(x-2)(x^3-x^2+2x+1).$$
The latter quotient has $-2$ as a root, and
$$x^3-x^2+2x+=(x+2)(x^2+2x+3),$$
so that $$x^5+3x^4+x^3+x^2+3=(x+1)(x-2)(x+2)(x^2+2x+3). $$
Now the quadratic factor is irreducible because it has no root. Indeed
$$ x^2+2x+3=(x+1)^2+2,$$
and the non-zero squares in $\mathbf Z/5\mathbf Z$ are $\pm 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Any easy way to evaluate this double integral? $$ \iint_S \frac{x^2+xy+3y^2}{2x^2+xy+2y^2}dxdy$$
where
$$S=\left\{(x,y)\vert x^2+y^2\leq1 \right\}$$
I tried
$$x=r\cos \theta, \quad y=r\sin\theta$$ substitution but it still not clear.
Is polar-coordinates the only solution for this?
Or how can I solve this easier?
|
I tried
$$x=r\cos \theta, \quad y=r\sin\theta$$ substitution but it still not clear.
This is a good idea. After simplifying (see below), you get:
$$\frac{x^2+xy+3y^2}{2x^2+xy+2y^2} \xrightarrow{\color{red}{(\star)}} 1 - \frac{2 \cos(2 \theta)}{4 + \sin(2 \theta)}$$
Don't forget the extra $\color{blue}{r}$ when converting to polar coordinates and you can split the variables:
$$\begin{align}
\iint_S \frac{x^2+xy+3y^2}{2x^2+xy+2y^2} \,\mbox{d}x\,\mbox{d}y
& = \iint_S \left(1 - \frac{2 \cos(2 \theta)}{4 + \sin(2 \theta)}\right) \color{blue}{r} \,\mbox{d}r\,\mbox{d}\theta \\[6pt]
& = \underbrace{\int_0^1 r \,\mbox{d}r}_{\frac{1}{2}} \; \underbrace{\int_0^{2\pi} \left(1 - \frac{2 \cos(2 \theta)}{4 + \sin(2 \theta)}\right) \,\mbox{d}\theta}_{2\pi \; + \; 0}
\end{align}$$
$$$$
$$\begin{align}
\color{red}{(\star)} \quad\frac{x^2+xy+3y^2}{2x^2+xy+2y^2}
& = \frac{2x^2+xy+2y^2+y^2-x^2}{2x^2+xy+2y^2} \\[5pt]
& = 1+\frac{y^2-x^2}{2\left(x^2+y^2\right)+xy} \\[5pt]
& = 1+\frac{r^2\sin^2\theta-r^2\cos^2\theta}{2r^2+r^2\cos\theta\sin\theta} \\[5pt]
& = 1-\frac{2\cos(2\theta)}{4+\sin(2\theta)}
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Number of possible integer values of $x$ for which $\frac{x^3+2x^2+9}{x^2+4x+5}$ is integer How many integer numbers, $x$, verify that the following
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5}
\end{equation*}
is an integer?
I managed to do:
\begin{equation*}
\frac{x^3+2x^2+9}{x^2+4x+5} = x-2 + \frac{3x+19}{x^2+4x+5}
\end{equation*}
but I cannot go forward.
| You're off to a good start. Now note that the denominator $x^2+4x+5$ is quickly larger than the numerator $3x+19$; you can quickly reduce the problem to only finitely many values for $x$ to check.
More details: (Hover to show)
The fraction is certainly not an integer if the denominator is greater than the numerator, i.e. if $$x^2+4x+5>3x+19,$$ unless perhaps the numerator is zero, but that is not possible in this case. The quadratic formula shows that the inequality above holds if $x\leq5$ or $x\geq4$. Then it remains to check whether the fraction is an integer for $x$ in the range $-4\leq x\leq3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Definite integral containing 2 trig functions and a square root function $$\int_{-\pi/4}^{\pi/4}\bigl(\cos x+ \sqrt {1+x^2}\sin^3x \cos^3x \bigr)\, dx $$
This question is from a math GRE practice test
I've tried to solve this integral for 2 days... starting to think it is a typo
The answer is $\sqrt2$ but I need to know how to solve with steps
Some useful trig identities and derivatives:
$$\frac{d}{dx} \sin^2x = 2\cos x \sin x$$
$$\frac{d}{dx} \tan^{-1}x = \frac{1}{1+x^2}$$
$$\sin^2x = \frac{1}{2} - \frac{1}{2} \cos 2x$$
$$\cos^2x = \frac{1}{2} + \frac{1}{2} \cos 2x$$
$$\sin 2x = 2\cos x \sin x$$
| Let $$I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\cos(x)+\sqrt{1+x^2}\sin^3(x)\cos^3(x)\right)dx$$ $$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos(x)dx+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\sqrt{1+x^2}\sin^3(x)\cos^3(x)\right)dx$$
For the second integral, notice $$f(x)=\sqrt{1+x^2}\sin^3(x)\cos^3(x)$$ is an odd function since $$f(-x)=\sqrt{1+(-x)^2}\sin^3(-x)\cos^3(-x)=-\left(f(x)\right)$$
So recall that $$\int_{-b}^bf(x)dx=0$$ if $f$ is odd and the integral exists over the interval. Conclude that $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sqrt{1+x^2}\sin^3(x)\cos^3(x)dx=0.$
Then $$I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos(x)dx+0=2\int_0^{\frac{\pi}{4}}\cos(x)dx=2\sin(x)\bigg|^{\frac{\pi}{4}}_0=2\sin\left(\frac{\pi}{4}\right)-2\sin(0)=\sqrt2$$
| {
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Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(1-\frac{x}{2}+\frac{x^2}{3}+\cdots \right)=e\cdot\exp \left(-\frac{x}{2}+\frac{x^2}{3}+\cdots \right) \\
&=e\left[1+\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e\left(1-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\\
&=e-\frac{ex}{2}+\frac{11}{24}ex^2+\cdots
\end{align*}
Likewise, expanding $e^{(1+x)^{\frac{1}{x}}}$ at $x=0$, we obtain
\begin{align*}
e^{(1+x)^{\frac{1}{x}}}&=(e^e)^{1-\frac{x}{2}+\frac{11}{24}x^2-\cdots}=e^e\cdot (e^e)^{-\frac{x}{2}+\frac{11}{24}x^2+\cdots}\\
&=e^e\cdot\left[1+\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)\ln e^e+\frac{1}{2!}\left(-\frac{x}{2}+\frac{11}{24}x^2+\cdots\right)^2\ln^2 e^e+\cdots\right]\\
&=e^e \cdot\left[1-\frac{ex}{2}+\frac{1}{24}(11e+3e^2)x^2+\cdots\right]
\end{align*}
Expanding $(1+x)^{\frac{e}{x}}$ at $x=0$, it follows that
\begin{align*}
(1+x)^{\frac{e}{x}}&=\exp\left[\frac{e\ln(1+x)}{x}\right]=\exp \left(e\cdot\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left(e-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)=e^e\cdot\exp \left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right) \\
&=e^e\left[1+\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)+\frac{1}{2!}\left(-\frac{ex}{2}+\frac{ex^2}{3}+\cdots \right)^2+\cdots\right]\\
&=e^e\left[1-\frac{ex}{2}+\frac{1}{24}e(8+3e)x^2+\cdots\right]
\end{align*}
Therefore
\begin{align*}
&\lim_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}\\
=&\lim_{x \to 0}\frac{e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(11+3e)x^2+\cdots\right]-e^e\cdot\left[1-\dfrac{ex}{2}+\dfrac{1}{24}e(8+3e)x^2+\cdots\right]}{x^2}\\
=&e^e\cdot\frac{1}{8}e\\
=&\frac{1}{8}e^{e+1}
\end{align*}
Please check. Is there any more simpler solution?
| In response to reading some of the comments, I just want to say that the L'hospital calculation is not horrible, if one works a little strategically. Of course, this is not entriely tedium free...
Let $$g(x) = \frac{\log(1+x)}{x}.$$ We have $g \to 1,$ and $g' \to -1/2$ as $x \to 0.$ (you'll have to show this for $g'$ but this is not hard).
Now let the limit in the question be written as $$L = \lim \frac{f_1(x) - f_2(x)}{x^2}, $$ where
\begin{align}
f_1(x) &:= e^{e^{g(x)}} \\ f_2(x) &:= e^{eg(x)} \end{align}
(I'll now start dropping the $(x)$ in $g(x)$ for legibility.)
By the chain rule, \begin{align} f_1' &= \left(e^{g(x)}\right)' e^{e^{g}} = g' e^{g} e^{e^{g}} \\ f_2' &= eg' e^{eg}\end{align}
So, by L'Hospital, $$ L = \lim \frac{g'e^g \left(e^{e^g} -e e^{(e-1)g }\right) }{2x}$$
But, $g'e^{g} \to -e/2$, so $$ L = -\frac{e}{4} \lim \frac{e^{e^g} - e^{(e-1)g + 1} }{x} $$ (assuming of course that the latter exists - this should be easy to show).
The point of doing this is get back to something that looks (almost) like $f_1 - f_2$ on the numerator. In particular, I don't want the second derivative of $f_1$ to enter my expressions - that's too much work.
Applying L'Hospital to this again, we get $$ -\frac{4L}{e} = \lim g'e^g e^{e^g} - (e-1)g' e^{(e-1)g + 1} = -\frac{1}{2}e\cdot e^e + \frac{(e-1)}{2}e^e = -\frac{1}{2}e^e,$$ where I've simply evaluated using $g \to 1, g' \to -1/2,$ giving again $$ L = e \cdot e^e/8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $x, y, z$ are three distinct positive integers, where $x + y + z = 13$ and $xy, xz, yz$ form an arithmetic... If $x, y, z$ are three distinct positive integers, where $x + y + z = 13$ and $xy, xz, yz$ form an increasing arithmetic sequence, what is the value of $(x + y)^z$ ?
I've been trying to solve this by forming equations based off of formulas ($yz - xz = xz - xy$) and by trying to clean it up by systems of equations (I did $(13 - z)^2 = 13$), but I can't seem to find the solution.
| This is actually not too bad to brute force since $x + y + z = 13$ and they are all positive integers. There are actually only $11$ possible values for $x$: $1, 2, 3, \ldots, 11$.
However, with a bit of analysis, we can reduce the number of cases.
Since $xz - xy = yz - xz$, we have that $2xz = xy + yz = y(x+z) = y(13-y)$. Since
\begin{align*}
(x+z)^2 \ge 4xz = 2y(13-y),
\end{align*}
we have that
\begin{align*}
(13-y)^2 \ge 2y(13-y).
\end{align*}
Since $0 < y < 13$ (otherwise $x = z = 0$ which is impossible since $x$ and $z$ are positive integers), then $13 - y > 0$ so we can divide both sides by $13-y$ without changing the sign to get
\begin{align*}
13-y \ge 2y,
\end{align*}
so this implies $y \le 4$. We have 4 cases to consider.
*
*If $y = 1$, then $x + z = 12$ and $xz = 6$. No solution.
*If $y = 2$, then $x + z = 11$ and $xz = 11$. No solution.
*If $y = 3$, then $x + z = 10$ and $xz = 15$. No solution.
*If $y = 4$, then $x + z = 9$ and $xz = 18$. This gives $(x,z) = (6,3)$ or $(3,6)$.
Hence, $(x+y)^z$ is either $(6+4)^3 = 1000$ or $(3+4)^6 = 117649$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that the largest eigenvalue of $A$ lies in the given interval
Show that if the given matrix $A$ is positive semi-definite then the largest eigen value of $A$ lies in the interval $(6,7)$.
$$A=\begin{bmatrix}
5&1&1&1&1&1\\
1&2&1&0&0&0\\
1&1&2&0&0&0\\
1&0&0&1&0&0\\
1&0&0&0&1&0\\
1&0&0&0&0&1
\end{bmatrix}$$
My try:
Since the matrix is given to be positive semidefinite so the spectral radius $\rho(A)$ must be an eigen value of $A$.
Also $\rho (A)=\max_{||x||=1}x^TAx$
I considered the vector $x=(1,0,0,0,0,0)$ then $\rho(A)\ge 5$
So I tried various $x$ such that $||x||=1$ but I find largest lower bound to be $5$.
Is there any way I can show that $\rho\ge 6$
| There exist $2\times 2$ orthogonal matrix $Q_2$ and $3\times 3$ orthogonal matrix $Q_3$ such that
$$Q_2\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}\sqrt{2}\\0\end{bmatrix}\quad\text{and}\quad Q_3\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}\sqrt{3}\\0\\0\end{bmatrix}\,,$$
e.g. Householder reflections. Define $6\times6$ orthogonal block diagonal matrix $Q$ with $Q=\operatorname{diag}(1,Q_2,Q_3)$ and note that matrix $A$ is similar to the matrix
$$QAQ^T=\begin{bmatrix}5&\sqrt{2}&0&\sqrt{3}&0&0\\\sqrt{2}&3&0&0&0&0\\0&0&1&0&0&0\\\sqrt{3}&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\end{bmatrix}\,.$$
Now we know that eigenvalues of $A$ different from $1$ must be eigenvalues of the matrix $B$ defined as
$$B=\begin{bmatrix}5&\sqrt{2}&\sqrt{3}\\\sqrt{2}&3&0\\\sqrt{3}&0&1\end{bmatrix}\,.$$
Also,
$\rho(A)=\max\{1,\rho(B)\}\,.$
Characteristic polynomial of matrix $B$ is
$$k(\lambda) = \lambda^3-9\lambda^2+18\lambda-4\,.$$
Its derivative is a quadratic polynomial whose zeros are $3\pm\sqrt{3}$. From here we conclude that $k'(\lambda)>0$ for $\lambda\in(3+\sqrt{3},+\infty)$. This means that $k(\lambda)$ is strictly increasing on that interval and has at most one zero in it.
Since $k(6)=-4<0$ and $k(7)=24>0$, we know that zero is actualy in $(6,7)$.
This shows that the largest eigenvalue of $B$, $\rho(B)$, satisfies $6<\rho(B)<7$, from where $6<\rho(A)<7$ follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can the number $2^n + n^2$ be divisible by $5$ for some natural number $n$? Can the number $2^n + n^2$ be divisible by $5$ for some natural number $n$?
I found some solutions, i.e., $n=6, n=8, n=12, n=14$ but not 18 or 20 by trial and error and somewhere I realized that the last digit of either of $2^n$ or $n^2$ should be either (6 or 4) or (9 or 1). How can I prove that the expression is divisible by 5?
| Hint:
First observe that if $n\equiv 0\mod 5$, there are no solutions. Next, the non-zero squares modulo $5$ are congruent to $\pm 1$, hence if $2^n+n^2\equiv 0\bmod 5$, $\;2^n\equiv\mp 1$.
Now, by lil' Fermat, $\;2^n\equiv 2^{n\bmod 4}\mod 5$, so that
*
*either $n\equiv \pm 1\mod 5$, which implies $n^2\equiv 1\mod 5$. The equation has a solution if $2^n\equiv -1\mod 5\iff n\equiv 2\mod 4$.
*or $n\equiv\pm 2\mod 5$, which implies $n^2\equiv -1\mod 5$. The equation has a solution if $2^n\equiv 1\mod 5\iff n\equiv 0\mod 4$.
This amounts to solving the systems of congruences
$$\begin{cases}n\equiv 2\mod 4,\\n\equiv \pm 1 \mod 5\end{cases}\qquad\qquad
\begin{cases}n\equiv 0\mod 4,\\n\equiv \pm 2 \mod 5\end{cases}$$
which are solved with the Chinese remainder theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Different methods give different answers. Let A,B,C be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Let $A,B,C$ be three angles such that $ A=\frac{\pi}{4} $ and $ \tan B \tan C=p $. Find all possible values of $p$ such that $A,B$ and $C$ are angles of a triangle.
case 1- discriminant
We can rewrite the following equation
$ f(x) = x^2 - (p-1)x + p $
As we know the sum and product of $ \tan C $ and $ \tan B $
Settings discriminant greater than equal to zero.
$ { (p-1)}^2 - 4p \ge 0 $
This gives $ p \le 3 - 2\sqrt2 $. Or $ p \ge 3 + 2\sqrt2 $
solving both equation
$ A + B + C = \pi $
$ C + B + \frac{\pi}{4} = \pi $
$ C + B = \frac{3\pi}{4} $
Using this to solve both the equation give $ p \in $ real
I found this on Quora.
https://www.quora.com/Let-A-B-C-be-three-angles-such-that-A-frac-pi-4-and-tan-B-tan-C-p-What-are-all-the-possible-value-of-p-such-that-A-B-C-are-the-angles-of-the-triangle
the right method
$ 0 \lt B , C \lt \frac{3\pi}{4} $
Converting tan into sin and cos gives
$ \dfrac {\sin B \sin C}{\cos B \cos C} = p $
Now using componendo and dividendo
$ \frac{\cos (B-C) }{- \cos(B+C) } = \frac{p+1}{p-1} $
We know $ \cos (B+C) = 1/\sqrt2 $
We know the range of $B$ and $C$ $(0, 3π/4)$
Thus the range of $B - C$. $(0, 3π/4 )$
Thus range of $\cos(B+C)$ is $ \frac{ -1}{\sqrt2} $ to $1$
Thus using this to find range gives
$ P \lt 0 $ or $ p \ge 3+ 2\sqrt2 $
| 1) The second method is wrong because of a silly mistake.
2) The first method is wrong because apart from discriminant, it is also important to note that there are restrictions on the values of $ B $ and $ C $ , and thus their are restrictions on $ tan C $, and $ tan B $, and thus their are restrictions on $ p $.
When both $B$ and $C$ are acute angles, both the roots of the above equations are positive. thus $ p \gt 1 $.
When one of them is obtuse, $$ \tan B \tan C \lt 0 .
$$
thus $$ p \lt 0 .$$
This with the intersection of non-negative discriminant, gives the correct answer.
Which give the range obtained in the third answer.
| {
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"source": "stackexchange",
"question_score": "6",
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$S=\left \{z\in\mathbb{C}| (z+i)^{n}=(z-i)^{n} \right \}$ $S=?$ If $n\in\mathbb{N}, n\geq2$ and $S=\big \{z\in\mathbb{C}| (z+i)^{n}=(z-i)^{n} \big \}$ then $S=?$ The right answer is
$$
S=\left \{\operatorname{ctg}\frac{k\pi}{n} |1\leq k\leq n-1;k\in\mathbb{N}\right \}
$$
I started like this $\left(\frac{z+i}{z-i}\right)^{n}=1$. How to continue? Some ideas?
| $$\frac{z+i}{z-i} = \frac{x + (y+1)i}{x + (y-1)i} = \frac{x^2 - y^2 + 1}{x^2 + (y-1)^2} + \frac{2xy}{x^2 + (y-1)^2}i = r e^{i\theta}$$
so
$$r = \vert \frac{z+i}{z-i} \vert = \sqrt{ \frac{(x^2 - y^2 + 1)^2}{(x^2 + (y-1)^2)^2} + \frac{4x^2y^2}{(x^2 + (y-1)^2)^2} } = 1$$
which is
$$ (x^2 - y^2 + 1)^2+4x^2y^2 = (x^2 + (y-1)^2)^2 \tag{1}$$
Another equation could be obtained from the phase, that is
$$n\theta = 2k\pi \tag{2}$$
where $\theta = \arctan \frac{2xy}{x^2-y^2 + 1} $
| {
"language": "en",
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"source": "stackexchange",
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How can this binomial expansion result in two different approximations of root 2? I have been working on a problem on approximating $\sqrt{2}$ using the first three terms of the following binomial expansion, and a substitution of $x = -\frac{1}{10}$ :
$$(4 - 5x)^.5 = 2 - \frac{5x}{4} - \frac{25x^2}{64}$$
After substituting, I got to the stage:
$ \frac{3}{\sqrt{2}}= \frac{543}{256}$
Now what is bizarre, is that if I solve this equation for $\sqrt{2}$, there are two routes I could take, but they give slightly different answers.
Route 1:
Reciprocate both sides, and multiply both sides by 3 to get:
$ \sqrt{2}= \frac{256}{181} $
Route 2:
Rationalize the left hand side to make the equation:
$ \frac{3\sqrt{2}}{2}= \frac{543}{256}$
Divide both sides by $\frac{3}{2}$:
$ \sqrt{2}= \frac{181}{128}$
So we end up with two approximations of $ \sqrt{2}$: $\frac{256}{181}$ and $\frac{181}{128}$
I am really struggling to understand how this has happened?
Why is the same equation leading to two different solutions?
| Let $f(x) = 2 - \frac{5x}{4} - \frac{25x^2}{64}$.
Let $c$ denote the value $f(-\frac{1}{10}) = \frac{543}{256}$.
We can write
$$\tag 1 \frac{3}{\sqrt 2} = c + \varepsilon$$
If we take the OP's route 1 (but keeping $\varepsilon$), then
$$\tag 2 \sqrt 2 = \frac{3}{c+\varepsilon}$$
We can check that the 'route 1 approximation', $\frac{3}{c}$, is strictly greater than $\sqrt 2$. With $\varepsilon$ playing a part in the denominator of the rhs of $\text{(2)}$, to 'fix things up' it must be true that $\varepsilon \gt 0$.
If we take the OP's route 2 (but keeping $\varepsilon$), then
$$\tag 3 \sqrt 2 = \frac{2}{3}\,(c + \varepsilon)$$
Since $\varepsilon \gt 0$, the 'route 2 approximation' $\frac{2c}{3}$ is strictly less than $\sqrt 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Knowing that $a+b\equiv 1 \pmod{7^{n+1}}$ show that $a^7+b^7\equiv 1 \pmod{7^{n+2}}$ Knowing that $a,b$ are prime integers and $a+b\equiv 1 \pmod{7^{n+1}}$ show that $a^7+b^7\equiv 1 \pmod{7^{n+2}}$
I used $a^7+b^7=(a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)$ and tried to show that $(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)\equiv 1 \pmod 7$ but after some trial and error I figured that it might not help as much as I thought. I also tried to show that $ a^7+b^7-1\equiv 0 \pmod{7^{n+2}}$, but I got stuck.
How should I solve this?
| It's not true. Take $a=7, b=43$. Then $a+b\equiv 1\bmod 49$, but $a^7+b^7\equiv 295\bmod 343$.
In fact, I would think that the primality or otherwise of $a$ and $b$ is irrelevant, given that we are only interested in their values mod $7^{n+1}$; but perhaps if you require that they be co-prime to $7$, your result might hold.
Edited to add: No, that doesn't work either. Take $a=53,b=193$ (both prime, and both co-prime to $7$). Then $a+b\equiv 1\bmod 49$, but $a^7+b^7\equiv 134\bmod 343$. So it seems your claim is simply false.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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let $a_{k}=\frac{k^4-17k^2+16}{k^4-8k^2+16}$ real number for $5≤k\in N$ If
$a_{k}=\frac{k^4-17k^2+16}{k^4-8k^2+16}$ real
number for $5≤k\in N$
Then find :
$\lim_{n\to +\infty} a_{5}a_{6}a_{7}...a_{n}$
My try :
$k^4-17k^2+16=(k^2-1)(k^2-16)$
And
$k^4-8k^2+16=(k-4)^2$
But how I complete
Answer is $\frac{1}{14}$
| Let $a_k = p_k/q_k$, where $p_k = k^4 - 17k^2 + 16 = (k-4)(k-1)(k+1)(k+4)$, and $q_k = k^4 - 8k^2 + 16 = (k-2)(k-2)(k+2)(k+2)$. Then $$\prod_{k=5}^n a_k = \prod_{k=5}^n \frac{(k-4)(k-1)(k+1)(k+4)}{(k-2)(k-2)(k+2)(k+2)}.$$ This is a telescoping product; for instance, $$\prod_{k=5}^n \frac{k-4}{k-2} = \frac{\prod_{k=1}^{n-4} k}{\prod_{k=3}^{n-2} k} = \frac{1 \cdot 2}{(n-3)(n-2)}.$$ Treat the other fractions similarly and you will find your answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k=4^{-n}~{2n \choose n}.$ I have happened to have proved this sum while attempting to prove another summation. Let
$$S_n=\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k$$
${2k \choose k} $ is the coefficient of $x^0$ in $(x+1/x)^{2k}$.
Consequently, $S_n$ is the coefficient of $x^0$ in $$f(x)= \sum_{k=0}^{2n} {2n \choose k}~\left (x+\frac{1}{x}\right)^{2k}~\left ( \frac{-1}{2}\right)^{k}=\left(1-\left(\frac{x+1/x}{\sqrt{2}}\right)^2 \right)^{2n} = 4^{-n} ~ \left(x^2+\frac{1}{x^2}\right)^{2n}.$$ Finally, the coefficient of $x^0$
in $f(x)$ is $$4^{-n}~{2n \choose n}=S_n.$$
I hope that you will find it interesting and prove it in some other way. Do try!
| We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. Recalling the generating function of the central binomial coefficients we can write
\begin{align*}
[z^n]\frac{1}{\sqrt{1-4z}}=\binom{2n}{n}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{2n}}&\color{blue}{\binom{2n}{k}\binom{2k}{k}\left(-\frac{1}{2}\right)^k}\\
&=\sum_{k=0}^{2n}\binom{2n}{k}[z^k]\frac{1}{\sqrt{1+2z}}\tag{2}\\
&=[z^0]\frac{1}{\sqrt{1+2z}}\sum_{k=0}^{2n}\binom{2n}{k}z^{-k}\tag{3}\\
&=[z^0]\frac{1}{\sqrt{1+2z}}\left(1+\frac{1}{z}\right)^{2n}\tag{4}\\
&=[z^{-1}]\frac{(1+z)^{2n}}{z^{2n+1}\sqrt{1+2z}}\tag{5}\\
&=[t^{-1}]\frac{\left(1+\frac{t}{1-t}\right)^{2n}}{\left(\frac{t}{1-t}\right)^{2n+1}\sqrt{1+\frac{2t}{1-t}}}\cdot\frac{1}{(1-t)^2}\tag{6}\\
&=[t^{2n}]\frac{1}{\sqrt{1-t^2}}\tag{7}\\
&\,\,\color{blue}{=\frac{1}{4^n}\binom{2n}{n}}\tag{8}
\end{align*}
Comment:
*
*In (2) we apply the coefficient of operator according to (1).
*In (3) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (4) we apply the binomial theorem.
*In (5) we write the expression using formal residual by applying again the rule from comment (3).
*In (6) we use the substitution $z=\frac{t}{1-t}, dz=\frac{1}{(1-t)^2}dt$.
*In (7) we do some simplifications.
*In (8) we select the coefficient of $t^{2n}$ by taking (1) evaluated at $z=\frac{1}{4}t^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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Prove $\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$ if $3\leq aI don't really know if I should use brute force or some kind of theorem, it comes on a calculus past exam and it says:
suppose: $3≤a<b≤8$
prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$
| By the mean value theorem and since $(\sqrt{1+x})'=\frac{1}{2\sqrt{1+x}},$ we obtain: $$\frac{\sqrt{1+b}-\sqrt{1+a}}{b-a}=\frac{1}{2\sqrt{1+c}},$$ where $3<c<8$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate the derivative of $\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ at $x=0$? Let $F(x):=\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ ,where $[\frac{1}{t}]$ is the largest integer no more than $\frac{1}{t}$.Prove $F'(0)=\frac{1}{2}$.
I have tried in this way:
\begin{equation}
\begin{aligned}
\lim_{n\to\infty}nF\left(\frac{1}{n}\right)&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-\left[\frac{1}{t}\right]\right)dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty\int_\frac{1}{k+1}^\frac{1}{k}\left(\frac{1}{t}-k\right) dt\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\ln(1+\frac{1}{k})-\frac{1}{k+1}\right)\\&=\lim_{n\to\infty}n\sum_{k=n}^\infty \left(\frac{1}{k}-\frac{1}{k+1}\right)\\&=1.
\end{aligned}
\end{equation}
Please give me some ideas,thank you!
| i) Consider $$ A_n :=\int^{\frac{1}{n-1}}_{ \frac{1}{n} } \
\frac{1}{x}\ dx - (n-1) \bigg[ \frac{1}{n-1} - \frac{1}{n}\bigg] $$
Hence $F(\frac{1}{n-1})=A_n +A_{n+1}+\cdots $
So $$ A_n\bigg[ \frac{1}{n-1} - \frac{1}{n}\bigg]^{-1} \rightarrow
\frac{1}{2}$$
so that $\frac{F(\frac{1}{n-1}) }{ \frac{1}{n-1} } \rightarrow
\frac{1}{2} $
ii) $ \frac{F(\frac{1}{n} )}{\frac{1}{n-1}}\leq \frac{F(x)}{x} \leq
\frac{F(\frac{1}{n-1} )}{\frac{1}{n}} $ where $\frac{1}{n}\leq
x<\frac{1}{n-1} $
| {
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Prove by mathematical induction that $\sum_{i=0}^n (2i-1)^2 = \frac{n(2n+1)(2n-1)}{3}$ One of my homework problems is to prove that $\sum_{i=1}^n (2i-1)^2 = \frac{n(2n+1)(2n-1)}{3}$
I already completed the basis step
$[2(1)-1]^2 = 1 $
$\frac{(1)[2(1)+1][2(1)-1]}{3} = 1$
Then I assumed that the proposition was true for n=k+1 which leaves me with
$\frac{k(2k+1)(2k-1)}{3}$ + $[2(k+1)-1]^2$ = $\frac{k[2(k+1)+1][2(k+1)-1]}{3}$
However, I am stuck here, as the term $[2(k+1)-1]^2$ produces a constant term and $\frac{k[2(k+1)+1][2(k+1)-1]}{3}$ (when the numerator is multiplied out) has no constants. Any ideas, guys?
Edits:
After reading some of the comments I found my original mistake. I should have written $\frac{(k+1)[2(k+1)+1][2(k+1)-1]}{3}$ instead of $\frac{k[2(k+1)+1][2(k+1)-1]}{3}$.
This leaves me with $\frac{k(2k+1)(2k-1)}{3}$ + $[2(k+1)-1]^2$ = $\frac{(k+1)[2(k+1)+1][2(k+1)-1]}{3}$
However, I am still lost on what to do next. I'm not having any luck combining these fractions, anyone got good ideas?
| What you have to prove is $$\frac{k(2k+1)(2k-1)}{3}+(2k+1)^2=\frac{(k+1)(2k+3)(2k+1)}{3}$$
The left-hand side is $$\frac{k(2k+1)(2k-1)+3(2k+1)^2}{3}=\frac{(2k+1)(2k+3)(k+1)}{3}$$
$$(2k+1)(2k^2-k+6k+3)$$ and $$(2k+3)(k+1)=2k^2-k+6k+3$$
| {
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"question_score": "1",
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Find the condition on $p$ such that equation $x^2+3 \equiv 0 \pmod{4p^2}$ has roots
Problem: Find the condition on $p$ such that equation $x^2+3 \equiv 0 \mod 4p^2$ has roots with $p$ is a prime and find the number of roots of this equation.
My solution: $x^2+3 \equiv 0 \mod 4p^2 \Rightarrow x^2 \equiv -3 \mod 4p^2 \Rightarrow x^{4p^2-1} \equiv (-3)^{(4p^2-1)/2}$. By little Fermat theorem we have $x^{4p^2-1} \equiv 1 \mod 4p^2 \Rightarrow (-3)^{(4p^2-1)/2} = 1$. I didn't knew how to continue.
Please give some hint, I'm not sure it right, I think there some doubt in my solution. Thank all!
| Here we go. I leave the $p=2$ case to you, and will assume $p>2$.
Now, $x^2+3\equiv 0\pmod{4p^2}$ if and only if $x^2\equiv 1\pmod{4}$ and $x^2\equiv -3\pmod{p^2}$. For the former, it suffices to have $x$ to be odd, that is, $x\equiv 1\pmod{2}$.
Now, we turn our attention to the latter. If $p=3$, observe that we don't get a solution. Hence, $p>3$. Now, $-3$ is a quadratic residue modulo $p$. It is well-known that this holds if and only if $p\equiv 1\pmod{6}$. Now, you want more, you want modulo $p^2$. I'll show that, there is a $q$ such that $q^2+3\equiv 0\pmod{p^2}$ (and note that, there are precisely two values of $q$, $q$ and $p^2-q$). To see this, let us first suppose $q_0^2+3\equiv 0\pmod{p}$, and $0<q_0<p$. Now, consider the numbers, $a_\ell=q_0+\ell p$, where $\ell$ is to be determined. You get $a_\ell^2+3 = q_0^2+3+2q_0\ell p + p^2\equiv q_0^2+3+2q_0\ell p\pmod{p^2}$. Let $q_0^2+3=kp$ for some $k$. Then, $a_\ell^2+3 = p(k+2q_0\ell)$. In particular, you want to ensure, $p\mid k+2q_0\ell$. Since $q_0$ and $k$ are fixed, it follows that $\ell\equiv -k(2q_0)^{-1}\pmod{p}$ (and these lines also justify the uniqueness up to a sign flip).
We are done.
| {
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Showing that generating function coefficients agree with recurrence relation I have the recurrence relation $a_n = 3a_{n-1} + 4a_{n-2}$ with $a_0 = 1, a_1 = 0$. The solution to this is $a_n = \frac{4^n}{5} + \frac{4(-1)^n}{5}$. The generating function is $g(x) = \frac{1-3x}{1-3x-4x^2} = (1-3x)(\sum_{i=0}^{\infty}(-1)^ix^i)(\sum_{k=0}^{\infty}4^kx^k$). I need to show that, for any non-negative integer $n$, the coefficient of $x^n$ is equal to $a_n$ (the solution to the recurrence relation). But how can I find a general formula for this $N$th coefficient, given that there are so many possibilities for what $i$ and $k$ could be?
| One way is expanding $\sum_{n=0}^\infty a_nx^n$. Note that $a_n=\frac{4^n}{5} + \frac{4(-1)^n}{5}$ evaluated at $n=0$ and $n=1$ coincides with the stated $a_0=1, a_1=0$, so that we can focus at
\begin{align*}
a_n=\frac{4^n}{5} + \frac{4(-1)^n}{5}\qquad\qquad n\geq 0
\end{align*}
Recalling the geometric series formula we obtain
\begin{align*}
\color{blue}{\sum_{n=0}^\infty a_n x^n}&=\sum_{n=0}^\infty\left(\frac{1}{5}4^n+\frac{4}{5}(-1)^n\right)x^n\\
&=\frac{1}{5}\sum_{n=0}^\infty 4^nx^n+\frac{4}{5}\sum_{n=0}^\infty(-1)^nx^n\\
&=\frac{1}{5}\sum_{n=0}^\infty(4x)^n+\frac{4}{5}\sum_{n=0}^\infty(-x)^n\\
&=\frac{1}{5}\cdot\frac{1}{1-4x}+\frac{4}{5}\cdot\frac{1}{1+x}\\
&=\frac{1}{5}\cdot\frac{1+x+4(1-4x)}{(1-4x)(1+x)}\\
&\,\,\color{blue}{=\frac{1-3x}{1-3x-4x^2}}
\end{align*}
| {
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Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$
Solve the following system of equations. $$\large \left\{ \begin{aligned} x^2 + y^2 &= 2x^2y^2\\ (x + y)(1 + xy) &= 4x^2y^2 \end{aligned} \right.$$
From the system of equations, we have that
$$\left\{ \begin{align*} (x + y)^2 \le 2(x^2 + y^2) = 4x^2y^2\\ 4x^2y^2 = (x + y)(1 + xy) \le \frac{[(x + 1)(y + 1)]^2}{4} \le \frac{(x + y + 2)^4}{4^3} \end{align*} \right.$$
$$\implies (x + y)^2 \le \frac{(x + y + 2)^4}{4^3} \implies |x + y| \le \left|\frac{x + y + 2}{2^3}\right|$$
I don't know what to do next. Please help me solve this problem.
| The first equation is equivalent to $(x+y)^2=2xy(1+xy)$.
Now multiplying the second equation by the previous relation yields: $$(x+y)^3=(2xy)^3$$ thus $x+y=2xy$ and then: $$(x+y)^2=2\times 2x^2y^2=2(x^2+y^2)\iff (x-y)^2=0$$
| {
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$(X^2-1)^2+(X^2-1)-12=0$ has 2 solutions only or 3? I am asked to solve for X using substitution:
$(X^2-1)^2+(X^2-1)-12=0$
Let $U = X^2-1$
$U^2+U-12=0$
$U^2+4U-3U-12=0$
$U(U+4)-3(U+4)=0$
$(U+4)(U-3)$
Then:
$U-3=0$
$U=3$
$X^2-1=3$
$X^2=4$
$X=\pm2$
This is the provided solution by my textbook.
However:
$U+4=0$
$U=-4$
$X^2-1=-4$
$X^2=-3$
$X=i\sqrt{3}$
My textbook does not provide $X=i\sqrt{3}$ as a solution. Why is $X=i\sqrt{3}$ not a solution in addition to $\pm2$?
| Note that if $X^2=-3$, this does not mean that $X=i\sqrt 3$. There is another solution $X=-i\sqrt 3$ that you have neglected. In either case, the reason your textbook might not have included $X=\pm 2,\pm i\sqrt 3$ as all four solutions is because it was only seeking for real solutions, of which there are only $\pm 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find $\lim\limits_{n \to \infty}n\int_{0}^{1}(\cos x-\sin x)^ndx$? I want to compute
$$\lim\limits_{n \to \infty}n\int_{0}^{1}(\cos x-\sin x)^ndx$$
Someone helped me find the limit of the integral, which is $0$, but now I can't figure out this one. Also tried squeeze theorem but I only get only one side of it to converge to $1$. Moreover, I'm bound to using elementary calculus.
| Alternative solution
For $n\ge 4$, we have
\begin{align*}
I_n &= n\int_0^{1/\sqrt{n}} (\cos x - \sin x)^n \mathrm{d} x
+ n\int_{1/\sqrt{n}}^{\pi/4} (\cos x - \sin x)^n \mathrm{d} x\\
&\quad + n\int_{\pi/4}^1 (\cos x - \sin x)^n \mathrm{d} x\\
&= I_{n, 1} + I_{n, 2} + I_{n, 3}.
\end{align*}
Let us prove that $\lim_{n\to \infty} I_{n, 1} = 1$, $\lim_{n\to \infty} I_{n, 2} = 0$,
and $\lim_{n\to \infty} I_{n, 3} = 0$.
*
*First, since $|\cos x - \sin x| = \sin x - \cos x \le \sin x \le \sin 1$ for all $x$ in $[\pi/4, 1]$, we have
$$|I_{n, 3}| \le n\int_{\pi/4}^1 |\cos x - \sin x|^n \mathrm{d} x
\le n (\sin 1)^n (1 - \pi/4).$$
Since $\lim_{n\to \infty} n (\sin 1)^n (1 - \pi/4) = 0$, by the squeeze theorem, we have
$$\lim_{n\to \infty} I_{n, 3} = 0.$$
*Second, it is easy to prove that $0\le \cos x - \sin x
\le \cos \frac{1}{\sqrt{n}} - \sin \frac{1}{\sqrt{n}} \le 1 - \frac{1}{\sqrt{n}}$ for
all $n\ge 4$ and all $x$ in $[1/\sqrt{n}, \pi/4]$. Thus, we have
$$0 \le I_{n, 2} \le n\left(1 - \tfrac{1}{\sqrt{n}}\right)^n\frac{\pi}{4}.$$
Since $\lim_{n\to \infty} n\left(1 - \tfrac{1}{\sqrt{n}}\right)^n = 0$, by the squeeze theorem, we have
$$\lim_{n\to \infty} I_{n, 2} = 0.$$
*Third, with the substitution $x = \frac{y}{\sqrt{n}}$, using $(\cos x - \sin x)^2 = 1 - \sin 2x$, we have
$$I_{n, 1}
= \int_0^1 \sqrt{n}\left(1 - \sin \frac{2y}{\sqrt{n}}\right)^{n/2} \mathrm{d} y.$$
It is easy to prove that, for all $y$ in $[0, 1]$ and all $n\ge 4$,
$$\left(1 - \frac{1}{n}\right)\frac{2y}{\sqrt{n}} \le \sin \frac{2y}{\sqrt{n}} \le \frac{2y}{\sqrt{n}}.$$
Thus, we have
$$\int_0^1 \sqrt{n}\left(1 - \frac{2y}{\sqrt{n}}\right)^{n/2} \mathrm{d} y \le I_{n, 1} \le \int_0^1 \sqrt{n}\left(1 - \left(1 - \frac{1}{n}\right)\frac{2y}{\sqrt{n}}\right)^{n/2} \mathrm{d} y$$
which results in
$$\frac{n \left(1 - \left(1 - \frac{2}{\sqrt{n}}\right)^{n/2 + 1}\right)}{n+2}
\le I_{n, 1} \le \frac{n^2 \left(1 - \left(1 - \left(1 - \frac{1}{n}\right)\frac{2}{\sqrt{n}}\right)^{n/2 + 1}\right)}{(n+2)(n-1)}.$$
Since
\begin{align*}
\lim_{n\to \infty} \frac{n \left(1 - \left(1 - \frac{2}{\sqrt{n}}\right)^{n/2 + 1}\right)}{n+2} &= 1, \\
\lim_{n\to \infty} \frac{n^2 \left(1 - \left(1 - \left(1 - \frac{1}{n}\right)\frac{2}{\sqrt{n}}\right)^{n/2 + 1}\right)}{(n+2)(n-1)} &= 1,
\end{align*}
by the squeeze theorem, we have $\lim_{n\to \infty} I_{n, 1} = 1$.
Thus, we have $\lim_{n\to \infty} I_n = 1$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
IMO 1984: Prove that $0 ≤ yz + zx +xy −2xyz ≤ \frac {7}{27}$, where $x,y$ and $z$ are non-negative real numbers for which $x + y + z = 1.$ I tried to solve this inequality because I find exotic. Actually, I didn't look at the right solution. Because before I look at the right solution, I want to know if my solution is right or not.
Prove that $0 ≤ yz + zx +xy -2xyz≤\frac{7}{27}$, where $x,y$ and $z$ are non-negative real numbers for which $x + y + z = 1.$
Attempts:
If $x=0$. The left side of the inequality is correct. So, I can accept $x,y,z≠0$.
It is enough to prove $\frac 1x+\frac1y+\frac1z≥2$
We have,
$x+y+z≥3\sqrt[3]{xyz}\Longrightarrow xyz≤\frac1{27}$
$\frac 1x+\frac1y+\frac1z≥\frac3{\sqrt[3]{xyz}}≥9≥2$
The left side proved.
It is obvious at least one of the numbers is less than $\frac 12.$
So, we can choose $y$, such that $y≤\frac 12$.
Therefore, we have
$yz + zx +xy −2xyz ≤ \frac7{27}$
$x(y+z)+yz(1-2x)-\frac7{27}≤0$
$x(1-x)+y(1-x-y)(1-2x)-\frac7{27}≤0$
$x-x^2+y-xy-y^2-2xy+2x^2y+2xy^2-\frac7{27}≤0$
$x^2-x-y+xy+y^2+2xy-2x^2y-2xy^2+\frac{7}{27}≥0$
$x^2(1-2y)+x(3y-2y^2-1)+(y^2-y+\frac7{27})≥0$
$2(\frac12-y)\left(x+\frac{y-1}{2}\right)^2+\frac{1}{108} (3y-1)^2(6y+1)≥0$
Of course, I'm not sure the solution is correct. Can you verify the solution?
Thank you.
| Assume $x= \max\{\,x,\,y,\,z\,\}\,\therefore\,3\,x\geqq 1$.
For $z= 1- x- y$, we need to prove
$$x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}\geqq 0$$
We have
$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$
$$= \left \{ 4(\,1- 2\,x\,)y^{\,2}- 4(\,x- 1\,)(\,2\,x- 1\,)y+ \frac{1}{27}(\,5- 9\,x\,)^{\,2} \right \}+$$
$$+ \frac{1}{3}(\,2\,x- 1\,)(\,3\,y- 1\,)(\,3\,x+ 3\,y- 2\,)$$
$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$
$$= \frac{1}{27}(\,3\,x- 1\,)^{\,2}- \frac{1}{9}(\,2\,x- 1\,)(\,3\,y- 1\,)(\,3\,x+ 3\,y- 2\,)$$
$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$
$$= \frac{1}{108}(\,1- 3\,x\,)^{\,2}(\,6\,x+ 1\,)+ \frac{1}{4}(\,1- 2\,x\,)(\,x+ 2\,y- 1\,)^{\,2}$$
Because
$$4(\,1- 2\,x\,)y^{\,2}- 4(\,x- 1\,)(\,2\,x- 1\,)y+ \frac{1}{27}(\,5- 9\,x\,)^{\,2}\geqq 0$$
$$\because\,{\rm discriminant}[\,4(\,1- 2\,x\,)y^{\,2}- 4(\,x- 1\,)(\,2\,x- 1\,)y+ \frac{1}{27}(\,5- 9\,x\,)^{\,2},\,y\,]=$$
$$= \frac{32}{27}(\,2\,x- 1\,)(\,3\,x- 1\,)^{\,3}\leqq 0$$
So
$$x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}\geqq 0$$
$\lceil$ ANOTHER!way $\rfloor$
As @Elvin's above, we can choose $1- 2\,x\geqq 0$ (or $\because\,(\,1- 2\,x\,)(\,1- 2\,y)\leqq 0\,\because\,x+ y< 1$).
$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$
$$= \frac{1}{108}(\,1- 3\,x\,)^{\,2}(\,6\,x+ 1\,)+ \frac{1}{4}(\,1- 2\,x\,)(\,x+ 2\,y- 1\,)^{\,2}\geqq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Theorem of Pythagoras - Incorrect Derivation I am trying to solve below question from Coursera Intro to Calculus (link)
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and
$2ab$ units respectively, where $a$ and $b$ are positive real numbers
such that $a$ is greater than $b$. Find an exact expression for the
length of the hypotenuse (in appropriate units).
Below are the choices
$(a - b)^2$
$\sqrt(a^4 + 4a^2b^2 -b^4)$
$a^2 + b^2$
$\sqrt(a^2 + 2ab -b^2)$
$(a + b)^2$
When I attempt to work out the solution (and I even got a 2nd pair of eyes to look at it, but he arrived at the same conclusion), I get this:
$(a^2 - b^2)^2 + (2ab)^2 = x^2$
$a^4 -2a^2b^2 + b^4 + (2ab)^2 = x^2$
$a^4 -2a^2b^2 + b^4 + 4a^2b^2 = x^2$
$a^4 + 2a^2b^2 + b^4 = x^2$
$\sqrt(a^4 + 2a^2b^2 + b^4) = x$
Please help! How to get the correct solution?
| $$c^2=(a^2-b^2)^2+(2ab)^2$$ $$=(a^4-2a^2b^2+b^4)+(4a^2b^2)$$ $$=a^4+2a^2b^2+b^4=(a^2+b^2)^2$$
$$\Rightarrow c=a^2+b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Determinant of a specific block matrix Let $A$ be the following block matrix
$$A = \begin{pmatrix}
0 & A_{12} & A_{13} \\
A_{21} & A_{22} & 0 \\
A_{31} & 0 & A_{33}\\
\end{pmatrix}$$
I am finding difficulty to show that
$$\det A = \det \begin{pmatrix}
0 & A_{12} \\
A_{21} & A_{22} \\
\end{pmatrix} \times \det(A_{33}) + \det \begin{pmatrix}
0 & A_{13} \\
A_{31} & A_{33} \\
\end{pmatrix} \times \det(A_{22})$$
I tried using the Schur complement but got nowhere. Any hints/suggestions?
Diagonal blocks are square and $A_{21} = A_{12}^t$, $A_{13} = A_{31}^t$, where $A_{21},A_{13}$ are column vectors.
| This is not true. Counterexample:
$$
A=\left[\begin{array}{cc|cc|cc}
0&0&1&0&0&0\\
0&0&0&0&0&1\\
\hline
1&0&0&0&0&0\\
0&0&0&1&0&0\\
\hline
0&0&0&0&1&0\\
0&1&0&0&0&0
\end{array}\right].
$$
We have $\det(A)\ne0$ because $A$ is a permutation matrix, but as both $A_{22}$ and $A_{33}$ are singular,
$$
\det\pmatrix{0&A_{12}\\ A_{21}&A_{22}}\det(A_{33})
+\det\pmatrix{0&A_{13}\\ A_{31}&A_{33}}\det(A_{22})=0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Closed form for $\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}$ By Mathematica, we find $$\sum_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\pi^2\log(2)-\frac{7}{2}\zeta(3).$$
How to find the closed form for general series:
$$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}? \ \ (p\ge 3)$$
| We can make use of the following representation
$$\sf 2\arcsin^2z=\sum\limits_{n\geq1}\frac {(2z)^{2n}}{n^2\binom {2n}n}, \ z\in[-1,1]$$
Which gives integrating once with respect to $\sf z$ from $\sf 0$ to $\sf x$:
$$\sf 4\int_0^x \frac{\arcsin^2 z}{z}dz =\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^3 \binom{2n}{n}}$$
So the sum can be written as
$$\sf S_3=\sum_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=4\int_0^1 \frac{\arcsin^2 t}{t}dt$$
Now let $\sf t=\sin x$ and integrate by parts in order to get:
$$\sf S_3=4\int_0^\frac{\pi}{2} x^2\cot xdx=-8\int_0^\frac{\pi}{2} x\ln(\sin x)dx$$
Also we can use the fourier series of log sine
$$\sf S_3=8\ln 2 \int_0^\frac{\pi}{2} xdx+8\sum_{n=1}^\infty \frac{1}{n}\int_0^\frac{\pi}{2}x\cos(2nx)dx$$
The second integral is easily doable integrating by parts, thus:
$$\sf S_3=\pi^2 \ln 2+2\sum_{n=1}^\infty \frac{(-1)^n-1}{n^3} = \boxed{\pi^2\ln 2 -\frac72\zeta(3)}$$
For higher $p$ things will get quite complicate, but the approach is the same. For case $p=4$ we have:
$$\sf \frac{4}{x}\int_0^x \frac{\arcsin^2 z}{z}dz =\sum_{n=1}^\infty \frac{4^{n}x^{2n-1}}{n^3 \binom{2n}{n}}$$
And integrating once again produces
$$\sf 8\int_0^t\frac{1}{x}\int_0^x \frac{\arcsin^2 z}{z}dzdx =\sum_{n=1}^\infty \frac{4^{n}t^{2n}}{n^4 \binom{2n}{n}}$$
$$\sf \Rightarrow S_4=\sum_{n=1}^\infty \frac{4^{n}}{n^4 \binom{2n}{n}}=8\int_0^1\frac{1}{x}\int_0^x \frac{\arcsin^2 z}{z}dzdx$$
$$\sf =8\int_0^1\int_z^1 \frac{1}{x}\frac{\arcsin^2 z}{z}dx dz=-8\int_0^1 \frac{\arcsin^2 z \ln z}{z}dz$$
Set $z=\sin x$ and integrate by parts to get
$$\sf S_4=-8\int_0^\frac{\pi}{2} x^2 \ln(\sin x)\cot x dx=8\int_0^\frac{\pi}{2} x\ln^2(\sin x)dx$$
$$=\boxed{8\operatorname{Li}_2\left(\frac12\right)+\frac13\ln^42 +4\zeta(2)\ln^2 2-\frac{19}{4}\zeta(4)}$$
See here for the above integral.
Or for $p=5$ we have by the same approach:
$$\sf 8\int_0^y\frac{1}{t}\int_0^t\frac{1}{x}\int_0^x \frac{\arcsin^2 z}{z}dzdxdt =\sum_{n=1}^\infty \frac{4^{n}y^{2n}}{n^5 \binom{2n}{n}}$$
$$\sf \sum_{n=1}^\infty \frac{4^{n}}{n^5 \binom{2n}{n}}=8\int_0^1 \int_z^1\int_z^1 \frac{\arcsin^2 z}{xtz}dxdtdz=8\int_0^1 \frac{\arcsin^2 z\ln^2 z}{z}dz$$
$$\sf \overset{z=\sin x}=8\int_0^\frac{\pi}{2}x^2\ln^2(\sin x)\cot x dx \overset{IBP}=-\frac{16}3\int_0^\frac{\pi}{2} x\ln^3(\sin x)dx$$
Furthermore this paper may be useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Jensen gap for a real-valued random variable $X \geq 0$ For a real-valued random variable $X \geq 0$,
We have $1 - \frac{1}{1+E \left[x\right]} \geq E \left[ 1 - \frac{1}{1+x} \right]$ (Jensen's inequality).
We want to get a tight constant gap between $1 - \frac{1}{1+E \left[x\right]}$ and $E \left[ 1 - \frac{1}{1+x} \right]$, i.e., $ \lvert 1 - \frac{1}{1+E \left[x\right]} - E \left[ 1 - \frac{1}{1+x} \right] \rvert \leq \epsilon_0$.
Any hints for this inequality?
| Taylor expanding $\frac{1}{1+x}$ around $x = E[x]$ gives us
\begin{align*}
\text{Gap} &= \left|1 - \frac{1}{1+E[x]} - E\left[1 - \frac{1}{1+x}\right] \right| \\&= \left|E\left[\frac{1}{1+x}\right] - \frac{1}{1+E[x]} \right| \\
&= \left|\frac{1}{1+E[x]} - E\left[\frac{x - E[x]}{(1 + E[x])^2}\right] + E\left[\frac{(x - E[x])^2}{(1 + E[x])^3}\right] - \cdots - \frac{1}{1+E[x]}\right| \\
&\le \frac{\text{Var}(x)}{(1 + E[x])^3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Advanced algebra-precalculus Find the value of $\frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{2}}{w+x+z}+\frac{z^{2}}{w+x+y}$ when $\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y}$ = 1, where $w,x,y,z \in \mathbb R $ .
I have tried setting one variable equal to zero, two variables equal to zero, and many other combinations to no avail of mine. I am training for a math olympiad, and this question has been boggling my head. A solution to this would be appreciated, but not as much as resources I can use to find a definitive answer to this problem.
| Let's try your method! Set $z=w=0$, then:
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y}=1 \Rightarrow \\
\frac{x}{y}+\frac{y}{x}=1 \Rightarrow x^2+y^2=xy\Rightarrow x^2+y^2\ge 2|xy|>xy$$
So, there is no real solution.
However, for complex numbers:
$$\frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{2}}{w+x+z}+\frac{z^{2}}{w+x+y}=\\
\frac{x^2}{y}+\frac{y^2}{x}=\frac{x^3+y^3}{xy}=\frac{(x+y)(x^2+y^2-xy)}{xy}=0.$$
Alternatively, lucky combination $w+z=0,x+y=-y$:
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{w+x+z}+\frac{z}{w+x+y}=1 \Rightarrow \\
\frac{-z}{z-y}+\frac{-2y}{y}+\frac{y}{-2y}+\frac{z}{-z-y}=1 \Rightarrow \\
\frac{z}{y-z}-\frac{z}{y+z}=\frac72 \Rightarrow \\
11z^2=7y^2\Rightarrow \\
z=\pm \sqrt{\frac{7}{11}}y$$
So:
$$y=1,x=-2,z=\sqrt{\frac7{11}}=-w\\
\frac{w^{2}}{x+y+z}+\frac{x^{2}}{w+y+z}+\frac{y^{2}}{w+x+z}+\frac{z^{2}}{w+x+y}=\\
\frac{z^2}{z-1}+\frac{4}{1}+\frac{1}{-2}+\frac{z^2}{-z-1}=\\
\frac{2z^2}{z^2-1}+\frac72=\\
\frac{14}{11}\cdot \left(-\frac{11}{4}\right)+\frac72=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Let $A, B$ and $C$ be the angles of an acute triangle. Show that: $\sin A+\sin B +\sin C > 2$. Let $A, B$ and $C$ be the angles of an acute triangle. Show that: $\sin A+\sin B +\sin C > 2$.
I started from considering
$$\begin{align}\sin A+\sin B+\sin (180^o-A-B) &= \sin A+\sin B+\sin(A+B)
\\&=\sin A+\sin B+\sin A\cos B + \cos A\sin B
\\&=\sin A(1+\cos B)+\sin B(1+\cos A).\end{align}$$
How to proceed?
| Also, we can make the following.
Let $b^2+c^2-a^2=x$, $a^2+c^2-b^2=y$ and $a^2+b^2-c^2=z$.
Thus, $x$, $y$ and $z$ are positives, $a=\sqrt{\frac{y+z}{2}},$ $b=\sqrt{\frac{x+z}{2}},$ $c=\sqrt{\frac{x+y}{2}}$ and in the standard notation
$$\sum_{cyc}\sin\alpha=\sum_{cyc}\frac{2S}{bc}=\sum_{cyc}\frac{\frac{1}{2}\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}}{bc}=\sum_{cyc}\frac{\sqrt{xy+xz+yz}}{\sqrt{(x+z)(x+y)}}=$$
$$=\sqrt{\frac{(xy+xz+yz)\left(\sum\limits_{cyc}\sqrt{x+y}\right)^2}{\prod\limits_{cyc}(x+y)}}=\sqrt{\frac{(xy+xz+yz)\left(2(x+y+z)+2\sum\limits_{cyc}\sqrt{(x+y)(x+z)}\right)}{\prod\limits_{cyc}(x+y)}}>$$
$$>\sqrt{\frac{(xy+xz+yz)\left(2(x+y+z)+2(x+y+z)\right)}{\prod\limits_{cyc}(x+y)}}=2\sqrt{\frac{(xy+xz+yz)(x+y+z)}{(x+y)(x+z)(y+z)}}>2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Elementary Second partial derivative If $V=\frac{xy} {(x^2+y^2)^2}$ and $x=r\cos\theta$, $y=r\sin\theta$ show that $\frac{\partial^2V} {\partial{r^2}}+\frac{1}{r}\frac{\partial{V}} {\partial{r}}+\frac{1} {r^2} \frac{\partial^2{V}}{\partial{\theta^2}}=0$
My attempt :
$\frac{\partial {V}} {\partial{x}} =y\cdot\frac{y^2-3x^2} {(x^2+y^2)^3}\\$
$\frac{\partial {V}} {\partial{y}} =x\cdot\frac{y^2-3x^2} {(x^2+y^2)^3}$
$\frac{\partial{V} } {\partial{r}}= \frac{\partial {V}} {\partial{x}}\frac{\partial {x}} {\partial{r}}+\frac{\partial {V}} {\partial{y}}\cdot\frac{\partial {y}} {\partial{r}}\\\frac{\partial {V}} {\partial{r}}=\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\}$
$\frac{\partial^2{V} } {\partial{r}^2} =\frac{\partial} {\partial{r}}\biggr\{\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\} \biggr\}\\=$
$\frac{\partial} {\partial{\theta}}\biggr\{\frac{y^2-3x^2} {(x^2+y^2)^3}\{y\cos\theta+x\sin\theta\} \biggr\} \frac{\partial{\theta} }{\partial{r}}$
At this point it is getting clumsy. Any hint on this.
| Hint: Write $$V=\frac{\sin2\theta}{2r^2}.$$Now take partial derivative with respect to $r$ and $\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3251673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Do I have a chance to get a closed form for this integral? I conjecture that
$$\int_{-z}^z \frac{\sin \left(\frac \pi {2z} x + \frac \pi 2 \right)} z \int_z^\infty \exp\left({- \frac {(y-x)^2 \pi^2}{16}}\right) \,d y d x \sim \frac 1 {z^2}$$
when $z\to \infty$.
Maybe there should be also equality.
| As Paul Enta already answered, the integral can be exactly computed.
$$J=\int_z^\infty \exp\left({- \frac { \pi^2}{16}(y-x)^2}\right) \,d y=\frac{2 }{\sqrt{\pi }}\left(1+\text{erf}\left(\frac{\pi}{4} (x-z)\right)\right)$$
$$I=\int \frac{\cos \left(\frac{\pi x}{2 z}\right)}{z}J \,dx=\frac{2 e^{-\frac{1}{z^2}}}{\pi ^{3/2}}\left(\text{erf}\left(\frac{\pi z (z-x)+4 i}{4 z}\right)-\text{erf}\left(\frac{\pi z (x-z)+4 i}{4 z}\right) \right)+\frac{4 \sin
\left(\frac{\pi x}{2 z}\right)}{\pi ^{3/2}} \left(1+\text{erf}\left(\frac{\pi (x-z)}{4} \right)\right)$$ Now, using the bounds, the result of the given definite integral is
$$\frac{2 \left(4-2 \text{erf}\left(\frac{\pi z}{2}\right)-i e^{-\frac{1}{z^2}}
\left(\text{erfi}\left(\frac{1}{z}-\frac{i \pi
z}{2}\right)-\text{erfi}\left(\frac{1}{z}+\frac{i \pi
z}{2}\right)\right)\right)}{\pi ^{3/2}}$$ and expanding for large values of $z$, this gives
$$\sim \frac{4}{\pi ^{3/2} z^2}\left(1-\frac{1}{2 z^2}+\frac{1}{6 z^4}+O\left(\frac{1}{z^6}\right) \right)+\frac{32 e^{-\frac{1}{4} \pi ^2 z^2}}{\pi ^5 z^5}\left(1-\frac{12}{\pi ^2 z^2}+\frac{4 \left(45-\pi ^2\right)}{\pi ^4
z^4}+O\left(\frac{1}{z^6}\right) \right)$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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Is it possible to have 2 different but equal size real number sets that have the same mean and standard deviation? By inspection I notice that
*
*Shifting does not change the standard deviation but change mean.
{1,3,4} has the same standard deviation as {11,13,14} for example.
*Sets with the same (or reversed) sequence of adjacent difference have the same standard deviation. For example, {1,3,4}, {0,2,3}, {0,1,3} have the same standard deviation. But the means are different.
My conjecture: There are no two distinct sets with the same length, mean and standard deviation.
Question
Is it possible to have 2 different but equal size real number sets that have the same mean and standard deviation?
| You seem to like the set $\{1,3,4\}$. Here are six more three element sets having the same mean and standard deviation as your given set. \begin{align*}
&\frac{202}{171} & &\frac{1}{171} \left(583+\sqrt{19842}\right) & &\frac{1}{171}
\left(583-\sqrt{19842}\right) \\
&\frac{688}{171} & &\frac{1}{171} \left(340+\sqrt{27861}\right) & &\frac{1}{171}
\left(340-\sqrt{27861}\right) \\
&\frac{544}{171} & &\frac{1}{171} \left(412-\sqrt{62421}\right) & &\frac{1}{171}
\left(412+\sqrt{62421}\right) \\
&\frac{32}{19} & &\frac{1}{19} \left(60-\sqrt{581}\right) & &\frac{1}{19}
\left(60+\sqrt{581}\right) \\
&\frac{27}{19} & &\frac{1}{38} \left(125-\sqrt{1689}\right) & &\frac{1}{38}
\left(125+\sqrt{1689}\right) \\
&-\frac{2}{3} \left(-4+\sqrt{7}\right) & &\frac{1}{6} \left(16+2 \sqrt{7}\right) &
&\frac{1}{3} \left(8+\sqrt{7}\right)
\end{align*}
and, more generally, let $(x,y)$ be any point on the ellipse given by the equation
$$ x^2 + xy + y^2 -8x -8y + 19 = 0 $$
and set $z = 8 - x - y$. This triple of values has the same mean and standard deviation as does $\{1,3,4\}$. (This is found by eliminating $z$ from the system mean$(x,y,z) = {}$mean$(1,3,4)$ and stddev$(x,y,z) = {}$stddev$(1,3,4)$, i.e., $x+y+z = 8, x^2 + y^2 + z^2 - xy - xz - yz = 7$.)
| {
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"source": "stackexchange",
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Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ I have just started learning sums. I need to evaluate the following sum:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)}$$
$$a_{1} = \frac{1}{2}, a_{2} = \frac{1}{6}, a_{3} = \frac{1}{12}, a_{4} = \frac{1}{20}, a_{5} = \frac{1}{30}, a_{6} = \frac{1}{42}, ...$$
I tried splitting it to partial fractions:
$$\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}$$
$$1 = A(k+1) + Bk$$
$$1 = Ak + A + Bk$$
$$\begin{cases} A = 1 \\ 0 = A + B \quad \Rightarrow B = -1\end{cases}$$
$$\Longrightarrow S_{n} = \sum_{k=1}^{n} \Big( \frac{1}{k} - \frac{1}{k+1} \Big)$$
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k+1}$$
Wolframalpha gives answer $S_{n} = \frac{n}{n+1}$
Help's appreciated. P.S: Are the tags okay?
| We have
\begin{split}\sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \frac1k - \frac1{k+1}&=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\dots+\left(\frac1n-\frac1{n+1}\right)\\&=1-\frac1{n+1}=\frac n{n+1}.\end{split}
| {
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Prove that $HK$ passes through a fixed point.
Point $A$ lies on the perpendicular bisector of $BC$. $M$ and $N$ are points respectively in line segments $AB$ and $AC$ such that $MN$ is tangent to the incircle of $ABC$ at point $H$. $MP, NQ \perp BC$ ($P,Q \in BC$). The intersection of $MQ$ and $NP$ is point $K$. Prove that $HK$ passes through a fixed point.
It's the midpoint of $BC$, obviously, but how to prove it is more important, which I failed to do. This problem is adapted from a recent competition.
| We can analytically prove that the straight line spanned by $HK$ passes through the midpoint of $BC$ as follows. Let the side $BC$ of the triangle lies along $x$-axis, and the origin $O$ is the midpoint of $BC$. Let ther vertex $A$ is a point $(0,a)$ and the vertices $B$ and $C$ are points $(-b,0)$ and $(b,0)$ respectively ($a,b>0$). Let $I=(0,r)$ be the center of the incircle and $H=(x_h,y_h)=I+r(\cos\varphi,\sin\varphi)$. Let $R$ (resp. $S$) be the tangent point of the incircle to the side $AB$ (resp. $AC$) of the triangle $ABC$. Since the point $H$ belongs to the upper arc of the incircle spanned by the points $R$ and $S$, $\theta\le\varphi\le\pi-\vartheta$, where $\tan\theta=\tfrac ba$. Let $M=(m,h_m)$, $P=(0,h_m)$, $N=(n,h_n)$, $Q=(0,h_n)$. Collinearity of points $A$, $B$, and $M$ implies $h_m=a\left(\tfrac {m}{b}+1\right)$. Similarly, collinearity of points $A$, $C$, and $N$ implies $h_n=a\left(\tfrac {-n}{b}+1\right)$. Let $K=(x_k,y_k)$. Similarity of triangles $KMP$ and $KQN$ implies $\tfrac{x_k-m}{h_m}=\tfrac{n-x_k}{h_n}$, so $x_k=\tfrac {nh_m+mh_n}{h_m+h_n}.$ Considering an angle $NPQ$, we obtain $\tfrac{x_k-m}{y_k}=\tfrac{|PQ|}{|NQ|}=\tfrac{n-m}{h_n}$. Thus $$y_k=\frac {h_n(x_k-m)}{n-m}=\frac{h_n}{n-m}\left(\frac {nh_m+mh_n}{h_m+h_n}-m\right)=\frac {h_mh_n}{h_m+h_n}.$$ It remains to show that the points $O$, $K$, and $H$ are collinear, that is $$\frac{b}{a}\left(\frac{n}{b-n}+\frac{m}{m+b}\right)= \frac{n}{h_n}+\frac{m}{h_m}=\frac{x_k}{y_k}=\frac{x_h}{y_h}=\tfrac{\cos\varphi}{1+\sin\varphi}.$$
Since $MH\perp IH$ we have
$(r\cos\varphi-m)r\cos\varphi+\left(r(1+\sin\varphi)- a\left(\tfrac {m}{b}+1\right)\right)r\sin\varphi =0$
$(r\cos\varphi-m)\cos\varphi+\left(r(1+\sin\varphi)- a\left(\tfrac {m}{b}+1\right)\right)\sin\varphi =0$
$br-bm\cos\varphi+rb\sin\varphi-am\sin\varphi-ab\sin\varphi=0$
If $ b\cos\varphi+ a\sin\varphi =0$ then $HI\perp AB$, so $H=R$. Then $N=A$, so $Q=O$ and formally $M$ can be any point of $AB$. But in order to have the straight line spanned by $HK$ passes through $O$, we restrict ourselves to $M=R$ in this case. Then $H=M$ and $K$ belongs to $MO$.
Below we shall assume that $a\sin\varphi+ b\cos\varphi \ne 0$. Then
$m=b\frac{r(1+\sin\varphi)-a\sin\varphi }{b\cos\varphi+ a\sin\varphi}.$
It follows
$\frac{m}{m+b}=1-\frac{ a\sin\varphi+ b\cos\varphi}{b\cos\varphi+ r(1+\sin\varphi)}.$
Similarly, since $NH\perp IH$ we have
$(r\cos\varphi-n)r\cos\varphi+\left(r(1+\sin\varphi)-a\left(\tfrac{-n}{b}+1\right)\right)r\sin\varphi =0$
$br-bn\cos\varphi+rb\sin\varphi+an\sin\varphi-ab\sin\varphi=0$
If $b\cos\varphi-a\sin\varphi=0$ then $HI\perp AC$, so $H=S$. Then $M=A$, so $P=O$ and formally $N$ can be any point of $AC$. But in order to have the straight line spanned by $HK$ passes through $O$, we restrict ourselves to $M=S$ in this case. Then $H=M$ and $K$ belongs to $NO$.
Below we shall assume that $b\cos\varphi-a\sin\varphi\ne 0$. Then
$n=b\frac{r(1+\sin\varphi) -a\sin\varphi}{b\cos\varphi-a\sin\varphi }.$
It follows
$\frac{n}{b-n}=-1+\frac{b\cos\varphi-a\sin\varphi }{b\cos\varphi-r(1+\sin\varphi)}.$
Thus
$\frac{m}{m+b}+\frac{n}{b-n}=\frac{b\cos\varphi-a\sin\varphi }{b\cos\varphi-r(1+\sin\varphi)}-
\frac{ a\sin\varphi+ b\cos\varphi}{b\cos\varphi+ r(1+\sin\varphi)}=$
$\frac{2b\cos\varphi(r(1+\sin\varphi)-a\sin\varphi)}{b^2\cos^2\varphi-r^2(1+\sin\varphi)^2}$.
So it remains to verify that
$\frac ba\frac{2b\cos\varphi(r(1+\sin\varphi)-a\sin\varphi)}{b^2\cos^2\varphi-r^2(1+\sin\varphi)^2}=\frac{\cos\varphi}{1+\sin\varphi}$
It suffices to check that
$\frac{2b^2(r(1+\sin\varphi)-a\sin\varphi)}{b^2\cos^2\varphi-r^2(1+\sin\varphi)^2}=\frac{a}{1+\sin\varphi}$
$\frac{2b^2(r(1+\sin\varphi)-a\sin\varphi)}{b^2(1+\sin\varphi)(1-\sin\varphi)-r^2(1+\sin\varphi)^2}=\frac{a}{1+\sin\varphi}$
$2b^2(r(1+\sin\varphi)-a\sin\varphi)=a(b^2(1-\sin\varphi)-r^2(1+\sin\varphi))$
$(2b^2r-ab^2+ar^2)(1-\sin\varphi)=0$
That is, we need to show that $2b^2r-ab^2+ar^2=0$. That is $r=\tfrac{-2b^2\pm \sqrt{4b^4+4a^2b^2}}{2a}=\tfrac{-b^2\pm b\sqrt{b^2+a^2}}{a}$. Indeed, the radius $r$ equals the area $ab$ of the triangle $ABC$ divided by its semiperimeter $b+\sqrt{a^2+b^2}$. It follows $r=\frac ba\left(\sqrt{a^2+b^2}-b \right)$.
| {
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"url": "https://math.stackexchange.com/questions/3257634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Investigate $\sum_{n\geq1}\frac{3^nx^n\left(\frac{1}{3}-x\right)}{n}$ for uniform convergence on $x\in [0, \frac{1}{3}]$: is my solution correct?
Exercise:
Investigate $$\sum_{n\geq1}\frac{3^nx^n\left(\frac{1}{3}-x\right)}{n}$$ for uniform convergence on $x\in [0, \frac{1}{3}].$
Graphs:
From a few graphs, it seems to be uniformly convergent:
My solution:
in $x=0$ and $x=\frac{1}{3}$, the limit function to which the graphs try to reach out takes value $0$.
Every other $x$ has form $\frac{1}{a}$, where $a>3$, $a\in\mathbb{R}$.
$$\frac{\left(\frac{3}{a}\right)^n\left(\frac{1}{3}-\frac{1}{a}\right)}{n},$$
in terms of convergence, is equivalent to
$$\frac{3^n}{a^n\cdot n}=\frac{(1+2)^n}{a^n\cdot n}=\frac{1+n2+\cdots+2^n}{a^n\cdot n}<\frac{1+n2^n}{a^n \cdot n}=\frac{1}{a^n\cdot n}+\frac{2^n}{a^n\cdot n}.$$
$$\lim_{n\to\infty}\frac{a^n\cdot n}{a^{n+1}\cdot (n+1)}=\frac{1}{a}<1,$$
$$\lim_{n\to\infty}\frac{2^n\cdot 2\cdot a^n\cdot n}{a^{n+1}\cdot (n+1)\cdot 2^n}=\frac{2}{a}<1.$$
On the right-hand side, there's a sum of two converging entities, which is a converging entity. By the Weierstrass test, the original series is uniformly convergent for $0\leq x\leq \frac{1}{3}$.
Is my solution correct?
If not, how to do it correctly?
If possible, I would also love to know if there's a way to find the limit function.
Thank you.
| Your solution shows pointwise convergence, but not uniform. To prove uniform convergence, we can use some calculus and the Weierstrass M-test. Define
$$g_n(x) = \frac{3^nx^n(\frac13-x)}{n}.$$
The function $g_n$ is continuous and non-negative on $[0,\frac13]$, and $g_n(0)=g_n(\frac13)=0$, so $g_n$ will take a maximum somewhere in $(0,1)$. To find this maximum, we differentiate:
$$g_n'(x) = \frac{3^n}n\Big[x^n\cdot(-1) + nx^{n-1}(\tfrac13 - x)\Big] = 3^nx^{n-1}\Big[\tfrac13 - (1+\tfrac1n)x\Big]. $$
Hence, for $x\in(0,\frac13)$, one has $g_n'(x) = 0$ if and only if $x=\frac1{3(1+\frac1n)}=\frac{n}{3(n+1)}$. This must be where $g_n$ takes its maximum, and its value is
$$M_n:= g\big(\tfrac{1}{3(1+\frac1n)}\big) = \frac{3^n(\frac{1}{3(1+\frac1n)})^n(\frac13-\frac{1}{3(1+\frac1n)})}{n} = \frac1{3(1+\frac1n)^nn(n+1)}.$$
Now since $M_n \le \frac1{n^2}$ for all $n\ge1$, and $0 \le g_n(x) \le M_n$ for all $x\in[0,\frac13]$ and $n\ge1$, it follows from the Weierstrass M-test that the series
$$\sum_{n=1}^\infty g_n(x)$$
converges absolutely and uniformly, giving you the desired conclusion.
Note also that one can also fairly easily evaluate the limit function. First, show $g_n(x) = (1-3x)\int_0^x(3t)^{n-1}$, and so since the convergence is uniform we have
$$\sum_{n=1}^\infty g_n(x) = (1-3x)\int_0^x\sum_{n=1}^\infty(3t)^{n-1} dt = (1-3x)\int_0^x\frac1{1-3t} dt = -\tfrac13(1-3x)\log(1-3x).$$
| {
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If $x+y$ is a factor of $x^2$ prove it's also a factor of $y^2$ If $X+Y$ is a factor of $X^2$ prove $X+Y$ is a factor of $Y^2$.
I have tried the rmainder theorem, attempted factorisation but those don't work.
| $x,y$ and $k$ are integers.
$x^2=k(x+y)$;
$((x+y)-y)^2=$
$ (x+y)^2-2y(x+y)+y^2=k(x+y);$
$y^2=k(x+y) -(x+y)^2 +2(x+y)y;$
Hence?
| {
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Integral $\ 4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{x}\ dx$ How to prove that
$$4\int_0^1\frac{\chi_2(x)\operatorname{Li}_2(x)}{x}\ dx+\int_0^1\frac{\log(1-x)\log^2(x)\log(1+x)}{x}\ dx=\frac{29}4\zeta(2)\zeta(3)-\frac{91}8\zeta(5)$$
Where $\chi_2(x)=\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}$ is the Legendre Chi function and $ \operatorname{Li}_2(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$ is the Dilogarithm function.
This integral was proposed by Cornel.
| This approach is pretty identical to Cornel's solution posted on his FB page.
using the fact that $\quad\displaystyle \sum_{n=1}^\infty a_{2n}=\frac12\left(\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty (-1)^na_n\right),\ $ we have
\begin{align}
\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}&=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)^2}=\frac12\left(\sum_{n=0}^\infty\frac{x^{n+1}}{(n+1)^2}+\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{(n+1)^2}\right)\\
&=\frac12\left(\sum_{n=1}^\infty\frac{x^n}{n^2}-\sum_{n=1}^\infty(-1)^n\frac{x^n}{n^2}\right)=\frac12\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)\right)
\end{align}
then, the first integral:
\begin{align}
I_1&=4\int_0^1\left(\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)^2}\right)\frac{\operatorname{Li}_2(x)}{x}\ dx\\
&=2\sum_{n=1}^\infty\left(\frac1{n^2}-\frac{(-1)^n}{n^2}\right)\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\
&=2\sum_{n=1}^\infty\left(\frac1{n^2}-\frac{(-1)^n}{n^2}\right)\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)\\
&=\zeta(2)\zeta(3)-2\zeta(2)\operatorname{Li}_3(-1)-2\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}\\
&\boxed{=\frac72\zeta(2)\zeta(3)-2\sum_{n=1}^\infty\frac{H_n}{n^4}+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}}
\end{align}
and the second integral:
using the following identity proved by Cornel and can be found in his book, (Almost) Impossible Integrals, Sums and Series. $\quad\displaystyle\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n}$.
multiply both sides by $\displaystyle\frac{\ln^2x}{x}$ then integrate from $0$ to $1$, we get
\begin{align}
I_2&=\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n-1}\ln^2x\ dx\\
&=\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(\frac{2}{(2n)^3}\right)\\
&=-4\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^4}+\frac14\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)\\
&=-2\sum_{n=1}^\infty\frac{H_n}{n^4}-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}+\frac14\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)\\
&\boxed{=-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^4}-\frac74\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac18\zeta(5)}
\end{align}
Finally
\begin{align}
I&=I_1+I_2\\
&=\frac72\zeta(2)\zeta(3)-\frac18\zeta(5)-\frac{15}4\sum_{n=1}^\infty\frac{H_n}{n^4}\\
&=\frac72\zeta(2)\zeta(3)-\frac18\zeta(5)-\frac{15}4\left(3\zeta(5)-\zeta(2)\zeta(3)\right)\\
&\boxed{=\frac{29}{4}\zeta(2)\zeta(3)-\frac{91}{8}\zeta(5)}
\end{align}
| {
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Counterexample to the primality test This is a generalization of this claim .
Can you provide a counterexample to the following claim?
Let $n$ be a natural number greater than two . Let $r$ be the smallest odd prime number such that $r \nmid n$ and $n^2 \not\equiv 1 \pmod r$ . Let $P_n^{(a)}(x)=\left(\frac{1}{2}\right)\cdot\left(\left(x-\sqrt{x^2+a}\right)^n+\left(x+\sqrt{x^2+a}\right)^n\right)$ , where $a$ is a nonzero integer coprime to $n$ . Then $n$ is a prime number if and only if $P_n^{(a)}(x) \equiv x^n \pmod {x^r-1,n}$ .
You can run this test here.
I have tested this claim for many random values of $n$ and $a$ and there were no counterexamples .
EDIT
Algorithm implementation in PARI/GP without directly computing $P_n^{(a)}(x)$.
I have verified this claim for $n \in [3,100000]$ with $a \in [-100,100]$ .
| The claim is true.
It is true that if $n$ is a prime number, then $P_n^{(a)}(x)\equiv x^n\pmod{x^r-1,n}$.
Proof :
We have, by the binomial theorem,
$$\begin{align}P_n^{(a)}(x)&=\frac 12\left(\left(x-\sqrt{x^2+a}\right)^n+\left(x+\sqrt{x^2+a}\right)^n\right)
\\\\&=\frac 12\sum_{i=0}^{n}\binom nix^{n-i}\bigg(\bigg(-\sqrt{x^2+a}\bigg)^i+\bigg(\sqrt{x^2+a}\bigg)^i\bigg)
\\\\&=\sum_{j=0}^{(n-1)/2}\binom{n}{2j}x^{n-2j}(x^2+a)^j
\\\\&=x^n+\sum_{j=1}^{(n-1)/2}\binom{n}{2j}x^{n-2j}(x^2+a)^j\end{align}$$
Since $\binom nm\equiv 0\pmod n$ for $1\le m\le n-1$, there exists a polynomial $f$ with integer coefficients such that
$$P_n^{(a)}(x)=x^n+0\times (x^r-1)+nf$$
from which
$$P_n^{(a)}(x)\equiv x^n\pmod{x^r-1,n}$$
follows.$\quad\blacksquare$
It is true that if $P_n^{(a)}(x)\equiv x^n\pmod{x^r-1,n}$, then $n$ is a prime number.
Proof :
Suppose that $n$ is an even number. Then, there exist a polynomial $f$ with integer coefficients and an integer $s$ such that$$P_n^{(a)}(x)=\sum_{i=0}^{n/2}\binom{n}{2i}x^{n-2i}(x^2+a)^i=x^n+s(x^r-1)+nf$$
Considering $[x^{n}]$ where $[x^k]$ denotes the coefficient of $x^k$ in $P_n^{(a)}(x)$, we get
$$\sum_{i=0}^{n/2}\binom{n}{2i}\equiv 1\pmod n,$$
i.e.
$$2^{n-1}\equiv 1\pmod n$$which is impossible.
So, $n$ has to be an odd number.
There exist a polynomial $\displaystyle g=\sum_{i=0}^{n}a_ix^i$ where $a_i$ are integers and an integer $t$ such that
$$P_n^{(a)}(x)=\sum_{j=0}^{(n-1)/2}\binom n{2j}x^{n-2j}(x^2+a)^j=x^n+t(x^r-1)+ng$$
Considering $[x^0]$, we have
$$0=-t+na_0\implies t=na_0$$
So, we see that there exists a polynomial $h$ with integer coefficients such that
$$P_n^{(a)}(x)=\sum_{j=0}^{(n-1)/2}\binom n{2j}x^{n-2j}(x^2+a)^j=x^n+nh\tag1$$
It follows that $[x^k]\equiv 0\pmod n$ for all $k$ such that $0\le k\le n-1$.
Now, $(1)$ can be written as
$$P_n^{(a)}(x)=\sum_{j=0}^{(n-1)/2}\sum_{k=0}^{j}\binom n{2j}\binom jkx^{n-2(j-k)}a^{j-k}=x^n+nh$$
So, we see that
$$\begin{align}&[x^3]\equiv 0\pmod n
\\\\&\implies \left(\binom n{n-3}\binom {(n-3)/2}0+\binom n{n-1}\binom {(n-1)/2}1\right)a^{(n-3)/2}\equiv 0\pmod n
\\\\&\implies \binom n{n-3}\equiv 0\pmod n\end{align}$$
since $\gcd(a,n)=1$.
Also, we have
$$\begin{align}&[x^5]\equiv 0\pmod n
\\\\&\implies \bigg(\binom n{n-5}\binom {(n-5)/2}0+\binom n{n-3}\binom {(n-3)/2}1
\\&\qquad\qquad+\binom n{n-1}\binom {(n-1)/2}2\bigg)a^{(n-5)/2}\equiv 0\pmod n
\\\\&\implies \binom n{n-5}\equiv 0\pmod n\end{align}$$
So, we can get (one can prove by induction)
$$\begin{align}&[x^3]\equiv [x^5]\equiv [x^7]\equiv\cdots\equiv [x^{n-2}]\equiv 0\pmod n
\\\\&\implies\binom n{n-3}\equiv\binom n{n-5}\equiv\binom n{n-7}\equiv\cdots\equiv\binom{n}{2}\equiv 0\pmod n
\\\\&\implies\binom{n}{2}\equiv \binom{n}{3}\equiv \binom n4\cdots \equiv\binom n{n-2}\equiv 0\pmod n\tag2\end{align}$$
Suppose here that $\displaystyle n=\prod_{i=1}^mp_i^{b_i}$ is a composite number where $p_1\lt p_2\lt\cdots\lt p_m$ are primes and $b_i$ are positive integers.
Let $[[N]]$ be the number of prime factor $p_i$ in $N$.
Then, we have the followings :
*
*$[[1!]]=[[2!]]=\cdots =[[(p_i-1)!]]=0$
*$[[p_i!]]=1$
*$[[(n-1)!]]=[[(n-2)!]]=\cdots =[[(n-p_i)!]]$
Using these, we see that
$$\binom n1=\frac{n!}{1!(n-1)!}=n,\binom n2=\frac{n!}{2!(n-2)!},\cdots, \binom{n}{p_i-1}=\frac{n!}{(p_i-1)!(n-(p_i-1))!}$$
are divisible by $p_i^{b_i}$, and that
$$\binom n{p_i}=\frac{n!}{p_i!(n-p_i)!}$$
is not divisible by $p_i^{b_i}$.
Therefore, we see that $$\binom n1=\frac{n!}{1!(n-1)!}=n,\binom n2=\frac{n!}{2!(n-2)!},\cdots, \binom{n}{p_1-1}=\frac{n!}{(p_1-1)!(n-(p_1-1))!}$$
are divisible by $n$, and that $$\binom{n}{p_1}=\frac{n}{p_1!(n-p_1)!}$$ is not divisible by $n$, which contradicts $(2)$.
It follows that $n$ is a prime number.$\quad\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Find remainder of division of $x^3$ by $x^2-x+1$ I am stuck at my exam practice here.
The remainder of the division of $x^3$ by $x^2-x+1$ is ..... and that of $x^{2007}$ by $x^2-x+1$ is .....
I tried the polynomial remainder theorem but I am not sure if I did it correctly.
By factor theorem definition, provided by Wikipedia,
the remainder of the division of a polynomial $f(x)$ by a linear polynomial $x-r$ is equal to $f(r)$.
So I attempted to find $r$ by factorizing $x^2-x+1$ first but I got the complex form $x=\frac{1\pm\sqrt{3}i}{2}=r$.
$f(r)$ is then $(\frac{1+\sqrt{3}i}{2})^3$ or $(\frac{1-\sqrt{3}i}{2})^3$ which do not sound right.
However, the answer key provided is $-1$ for the first question and also $-1$ for the second one. Please help.
| (1) Since $x^3=(x^3+1)-1=(x+1)(x^2-x+1)-1$, the remainder is $-1$.
(2) $\displaystyle x^{2007}=(x^3)^{669}=[(x+1)(x^2-x+1)-1]^{669}=-1+\sum_{k=1}^{669}(x+1)^k(x^2-x+1)^k(-1)^{669-k}$
The remainder is $-1$.
| {
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"source": "stackexchange",
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Proving that $\iint\frac{x-y}{(x+y)^3}\,dx\,dy$ does not exist over $0 \leq x \leq 1 , 0 \leq y \leq 1$ To prove that double integral does not exist:
$$\iint\frac{x-y}{(x+y)^2}\,dx\,dy$$ over region $0 \leq x \leq 1 , 0 \leq y \leq 1$.
I put $x - y = u$ and $x+y = v$ and I got integral as
$$\iint \frac{u}{v^3}\,du\,dv$$
Limits of $u$ are from $2-v \leq u\leq v-2$ and $v$ are from $0$ to $1$. I am not sure about my new limits and how to prove it further.
EDIT : The correct expression is $\iint\frac{x-y}{(x+y)^3}\,dx\,dy$
| Given Integral $$ \begin{align} I &= \int^1_0 \int^1_0 \cfrac{x-y}{(x+y)^2} dx dy \\
& = \int^1_0 \int^1_0 \cfrac{x+y-2y}{(x+y)^2} dx dy \\ & = \int^1_0 \int^1_0 \cfrac{1}{(x+y)} dx dy - 2 \int^1_0 \int^1_0 \cfrac{y}{(x+y)^2} dx dy \\ & = \int^1_0 \ln(x+y) \bigg|_0^1 dy + 2 \int^1_0 \cfrac{y}{x+y} \bigg |^1_0 dy \\ & = \int^1_0\ln (y+1) dy - \int^1_0\ln(y) dy - 2 \int^1_0 \cfrac{dy}{y+1} \\& = (y+1)(\ln(y+1)-1 )\bigg|_0^1 - y(\ln(y)-1) \bigg|_0^1 - 2 \ln(y+1) \bigg |^1_0 \end{align}$$
Which is easy to evaluate, and I think the only slight problem would be in evaluating $y \ln(y) $ at $x = 0$, which is just $0$.
I don't see any problem with this.
After changing the question
Given integral:
$$ \begin{align} I &= \int^1_0 \int^1_0 \cfrac{x-y}{(x+y)^3} dx dy \\
& = \int^1_0 \int^1_0 \cfrac{x+y-2y}{(x+y)^3} dx dy \\& = \int^1_0 \int^1_0 \cfrac{dx dy}{(x+y)^2} - 2 \int^1_0 \int_0^1 \cfrac{y}{(x+y)^3} dx dy \\& = -\int^1_0 \cfrac{ dy}{(x+y)} \bigg |^1_0 + \cfrac{2}{2} \int^1_0 \cfrac{y}{(x+y)^2} \bigg |^1_0 dy \\& = - \int^1_0 \bigg( \cfrac{1}{y+1} - \cfrac{1}{y} \bigg) dy + \int^1_0 \bigg( \cfrac{y}{(1+y)^2} - \cfrac{1}{y} \bigg ) dy \\ & = - \int^1_0 \cfrac{dy}{(1+y)^2} \\& = {- \cfrac{1}{2}}\end{align}$$
But $$ \int^1_0 \int^1_0 \cfrac{x-y}{(x+y)^3} dy dx = \cfrac{1}{2} $$
Then from converse of Fubini's theorem this integral doesn't exists.
| {
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"url": "https://math.stackexchange.com/questions/3266784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do you come up with great and ingenious questions In mathematics how are the professionals and great authors able to come up with such ingenious questions which which are so difficult yet elegant.
Like in integral calculus it involve a substitution or adding and subtracting sometimes or a trick which would make you say WOW! and you cannot understand like how anyone could come up with that. Like this or this or this
Or questions on co-ordinate geometry when they ask about making a tangent or something and then, something and to prove that it's always true for some conditions like how could you come up with that.
I am posing this question to understand what goes on in the minds of a paper setter who tries to make a difficult question how does he come up with new ideas how is it that he can come off the new and ingenious your difficult questions every time be it trigonometry,calculus,algebra and geometry
| I can show, how it can work in creating of inequalities.
We know that the $uvw$ method (see here: https://artofproblemsolving.com/community/c6h278791 or here http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mo&paperid=194&option_lang=rus ) is very useful.
Sometimes this method gives a proof without great effort, but I want that my students will a bit of think.
We know that if our inequality is equivalent to $f\geq0,$ where $f$ is monotone function of $w^3$ or $f$ is a concave function of $w^3$, so we can use the $uvw$.
All this we can say if $f$ is a function of $u$ or of $v^2$.
We want to create an inequality, which very easy to write in the $uvw$'s language, but the method does not work or gives a very hard and ugly solution.
For which we can take an easy expression, where $u$, $v^2$ and $w^3$ are involved and fixture it.
Id est, we got a condition in the inequality, which we want to create.
We know that
$$(a+b+c)^3=\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)=(a^3+b^3+c^3)+3(a+b)(a+c)(b+c).$$
Let $(a+b)(a+c)(b+c)$ be this expression, where $a$, $b$ and $c$ are positives.
If $(a+b)(a+c)(b+c)=8$, so
$$(a+b+c)^3=a^3+b^3+c^3+24=a^3+b^3+c^3+8\cdot3.$$
It's obvious that $a^3+b^3+c^3\geq3,$ but in the sum $a^3+b^3+c^3+8\cdot3$
the expression $a^3+b^3+c^3$ does not play a major role,
which says that after using AM-GM we'll get a very strong inequality:
$$(a+b+c)^3=a^3+b^3+c^3+8\cdot3\geq9\sqrt[9]{(a^3+b^3+c^3)\cdot3^8},$$ which gives
$$\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^3+b^3+c^3}{3}}.$$
Since $(a+b)(a+c)(b+c)=9uv^2-w^3$ and $a^3+b^3+c^3=27u^3-27uv^2+3w^3,$ we see that our inequality is equivalent to $$u\geq\sqrt[27]{9u^3-9uv^2+w^3}$$ or $F(u,v^2,w^3)\geq0,$ where
$$F(u,v^2,w^3)=u^{27}-(9u^3-9uv^2+w^3)\left(\frac{9uv^2-w^3}{8}\right)^8.$$
We see that $uvw$ gives a very hard solution, but we got a nice inequality, which we can prove in one line!
| {
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"url": "https://math.stackexchange.com/questions/3267221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Development of the sine function into the power series I would like to check my solution for the development of the real sine function, $f(x) = \sin x$ into the power series. Here is my solution:
First, we have that $\mathcal{D}_f = \mathbb{R}$. Now, we have to check if the given function is infinitely diferentiable and where:
$$f'(x) = \cos x = \sin (x + \frac{\pi}{2}),$$
$$f''(x) = (\sin(x + \frac{\pi}{2}))' = \cos(x + \frac{\pi}{2}) = \sin (x + \frac{2\pi}{2}),$$
$$f'''(x) = (\sin(x + \frac{2\pi}{2}))' = \cos(x + \frac{2\pi}{2}) = \sin(x + \frac{3\pi}{2}),$$
$$\vdots$$
$$f^{(n)}(x) = \sin(x + \frac{n\pi}{2}).$$
So, we, have that given function is infinitely diferentiable on the set $\mathbb{R}$.
Now, notice the point $x_0 = 0$, in which we are going to try to develop the given function.
Because $f$ is infinitely diferentiable, from that follows that $f$ is also continuous with all of its derivatives up to the $n$ - th order, in the arbitrarily neighborhood $U$ of the point $x_0 = 0$, and it has the derivative of the $n + 1$ - st order in $U$. Now, we have that $f$ fulfills the conditions of the theorem from here:
Checking my understanding of the process of developing function into power series
Now, we have that the $f$ can be represented with Maclaurin series:
$$f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + R_n(x) =$$
$$= \sum\limits_{n = 0}^{\infty} \frac{\sin \frac{n \pi}{2}}{n!} x^n + R_n(x).$$
Next, I need to examine if $\lim\limits_{n \to \infty} R_n(x) = 0$:
We can notice that $\forall n \in \mathbb{N}$, $\forall x \in \mathbb{R}$, $|f^{(n)}(x)| \le 1$.
Further, we can notice that for $\forall h \in \mathbb{R}$, $f$ is infinitely diferentiable in $[x_0 - h, x_0 + h]$ and that there is a constant $M = 1$, such that $\forall n \in \mathbb{N}$ and $\forall x \in [x_0 - h, x_0 + h]$ it is $|f^{(n)}(x)| \le M$.
So, We have that conditions of the lema from the question I already linked, are fulfilled. Because of that, we can conclude that $\lim\limits_{n \to \infty} R_n(x) = 0$ for $\forall x \in [x_0 - h, x_0 + h]$, or in this case $\forall x \in \mathbb{R}$.
Conclulsion:
We can conclude that the function $f(x) = \sin x$, can be developed into the power series
$$\sum\limits_{n = 0}^{\infty} \frac{\sin \frac{n \pi}{2}}{n!} x^n$$
in the neighborhood of the point $x_0 = 0$.
Edit:
I tried to simplify the power series I got, to try and get the shape suggested in the answer by Chris Custer. I think I did it:
$$\sum\limits_{n = 0}^{\infty} \frac{\sin \frac{n \pi}{2}}{n!} x^n = \frac{0 x^0}{0!} + \frac{1 x^1}{1!} + \frac{0 x^2}{2!} + \frac{(-1) x^3}{3!}+\frac{0 x^4}{4!} + \frac{1 x^5}{5!} + \frac{0 x^6}{6!} + \frac{(-1) x^7}{7!} + \cdots =$$
$$= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} + \frac{x^{13}}{13!} - \frac{x^{15}}{15!} + \cdots = $$
$$= (-1)^0 x^{2\cdot 0 + 1} + (-1)^1 \frac{x^{2\cdot 1 + 1}}{(2\cdot 1 + 1)!} + (-1)^2 \frac{x^{2\cdot 2 + 1}}{(2\cdot 2 + 1)!} + (-1)^3 \frac{x^{2\cdot 3 + 1}}{(2\cdot 3 + 1)!} + (-1)^4 \frac{x^{2\cdot 4 + 1}}{(2\cdot 4 + 1)!} + (-1)^5 \frac{x^{2\cdot 5 + 1}}{(2\cdot 5 + 1)!} + \cdots = $$
$$= \sum\limits_{n = 0}^{\infty} (-1)^n \frac{x^{2n + 1}}{(2n + 1)!}$$
Please, could you tell me if my solution is correct, and if it isn't could you tell me where I made a mistake?
| Using Taylor's theorem: $f(x)=\sum_{n=0}^{\infty}\dfrac {f^{(n)}(0)}{n!}x^n$, as you appear to have done, you should get $\sin x=\sum_{n=0}^\infty(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}$.
Note that $\sin n\frac{\pi}2$ takes values $0,1,0,-1$ in succession.
| {
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"source": "stackexchange",
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Convergence and divergence of three series Which of the following is true.
A)$ \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n}$ does not converge.
B) $ \sum_{n=1}^{\infty} \dfrac{1}{n}$ converges.
C) $ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{(n+m)^2}$ converges.
D) $ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \dfrac{1}{(n+m)^2}$ diverges.
I know that 1st is true but i am confused with 3rd and 4th means is it possible that series is neither convergent nor divergent.
Any one explain how it is possible.
| Note that
\begin{align*}
\sum_{m = \mathbf{0}}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(n + m)^2} &= \sum_{n=1}^{\infty} \frac{1}{(n + 0 )^2} + \sum_{n=1}^{\infty} \frac{1}{(n + 1)^2} + \sum_{n=1}^{\infty} \frac{1}{(n + 2 )^2} + \ldots \\\\ &= \sum_{n=1}^{\infty} \frac{1}{(n )^2} + \sum_{n=2}^{\infty} \frac{1}{(n )^2} + \sum_{n=3}^{\infty} \frac{1}{(n )^2} + \ldots \\ &= \sum_{n = 1}^{\infty} \frac{n}{n^2}
\end{align*}
so
$$ \sum_{m = \mathbf{1}}^{\infty} \sum_{n=1}^{\infty} \frac{1}{(n + m)^2} = \sum_{n = 1}^{\infty} \frac{1}{n} - \sum_{n = 1}^{\infty} \frac{1}{n^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sum\limits_{cyc}\sqrt{a+11bc+6}\geq9\sqrt2.$
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc+abc=4.$ Prove that:
$$\sqrt{a+11bc+6}+\sqrt{b+11ac+6}+\sqrt{c+11ab+6}\geq 9\sqrt2.$$
The equality occurs for $(a,b,c)=(1,1,1)$ and again for $(a,b,c)=(2,2,0)$ and for the cyclic permutations of the last.
I tried Holder:
$$\sum_{cyc}\sqrt{a+11bc+6}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{a+11bc+6}\right)^2\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}{\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}}\geq$$
$$\geq \sqrt{\frac{\left(\sum\limits_{cyc}(a+11bc+6)(3a+4)\right)^3}{\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}}$$ and it's enough to prove that
$$\left(\sum\limits_{cyc}(a+11bc+6)(3a+4)\right)^3\geq162\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3,$$
which is true for $(a,b,c)=(2,2,0)$, $(a,b,c)=(1,1,1)$, but it's wrong for $(a,b,c)=\left(8,\frac{1}{2},0\right).$
Thanks to River Li for this counterexample.
Also, I tried a substitution $a=\frac{2x}{y+z},$ $b=\frac{2y}{x+z}$, $c=\frac{2z}{x+y}$ and SOS, but it seems very complicated.
Also, I tried the following estimation. By Minkowcki:
$$\sqrt{a+11bc+6}+\sqrt{b+11ac+6}\geq\sqrt{(\sqrt{a}+\sqrt{b})^2(1+11c)+24}.$$
Now, for $c=\min\{a,b,c\}$ it's enough to prove that
$$\sqrt{(\sqrt{a}+\sqrt{b})^2(1+11c)+24}+\sqrt{c+11ab+6}\geq9\sqrt2,$$ which not so helps.
Also, LM does not help.
Thank you!
Update
Also, there is the following.
We need to prove that:
$$\sum_{cyc}\sqrt{\frac{2x}{y+z}+\frac{44yz}{(x+y)(x+z)}+6}\geq9\sqrt2$$ or
$$\sum_{cyc}\sqrt{\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3}\geq9,$$ where $x$, $y$ and $z$ are non-negatives such that $xy+xz+yz\neq0.$
Now, by Holder
$$\left(\sum_{cyc}\sqrt{\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3}\right)^2\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(kx^2+y^2+z^2+myz+nxy+nxz)^3\geq$$
$$\geq\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(kx^2+y^2+z^2+myz+nxy+nxz)\right)^3,$$ where $k$, $m$ and $n$ are reals such that the expression $kx^2+y^2+z^2+myz+nxy+nxz$
is non-negative for all non-negatives $x$, $y$ and $z$.
Thus, it's enough to choose values of $k$, $m$ and $n$ for which the following inequality is true.
$$\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(kx^2+y^2+z^2+myz+nxy+nxz)\right)^3\geq$$
$$\geq81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(kx^2+y^2+z^2+myz+nxy+nxz)^3$$
From the equality case we can get that should be $$2k-5m+2n=8.$$
For $k=1$, $m=0$ and $n=3$ we need to prove that:
$$\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(x^2+y^2+z^2+3xy+3xz)\right)^3\geq$$
$$\geq81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(x^2+y^2+z^2+3xy+3xz)^3,$$
which is true for $y=z$ and it's true for $z=0$, but I have no a proof for all non-negative variables.
| Michael Rozenberg actually gave a proof. I do a little bit by the Buffalo Way to prove that
\begin{align}
&\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(x^2+y^2+z^2+3xy+3xz)\right)^3\\
\geq\ & 81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(x^2+y^2+z^2+3xy+3xz)^3.
\end{align}
It suffices to prove that $f(x,y,z)\ge 0$ where $f(x,y,z)$ is a polynomial (a long expression).
WLOG, assume that $z = \min(x,y,z).$ There are two possible cases:
1) If $z \le y \le x$, let $y=z+s, \ x = z+s+t; \ s,t \ge 0$.
Note that $f(z+s+t, z+s, z)$ is a polynomial in $z, s, t$ with non-negative coefficients.
It is true.
2) If $z \le x\le y$, let $x = z+s, \ y = z+s+t; \ s,t \ge 0.$
Note that $f(z+s, z+s+t, z)$ is a polynomial in $z, s, t$ with non-negative coefficients.
It is true. We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+7=(1-\sqrt{x+2})^2$ # square both sides
(Use perfect square formula on right hand side $a^2-2ab+b^2$)
$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula
$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify
$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying
$2x+4=2\sqrt{x+2}$ # simplify across both sides
$(2x+4)^2=(2\sqrt{x+2})^2$
$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again
$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x
For use int he quadratic function, my parameters are: a=4, b=12 and c=14:
$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$
$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$
$x=\frac{-12\pm{\sqrt{-80}}}{8}$
$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1
This is as far as I get:
$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$
I must have gone of course somewhere further up since the solution is provided as x=-2.
How can I arrive at -2?
| Your first step is wrong. It should be $$\sqrt{3x+7}-\sqrt{x+2}=1\implies\sqrt{3x+7}=1+\sqrt{x+2}$$ so we have $$3x+7=(1+\sqrt{x+2})^2$$ from which I think you can continue.
Note that as a check to your textbook solution, at $x=-2$, we get $$\sqrt{3(-2)+7}-\sqrt{-2+2}$$ which is indeed equal to $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\iint_A \frac{x^2}{(4x^2+3y^2)^{3/2}} \,dx\,dy$ where $A=\{(x,y)|4x^2+3y^2\leq1\}$
Find $\displaystyle\iint_A \frac{x^2}{(4x^2+3y^2)^{3/2}} \,dx\,dy$ where $A=\{(x,y)|4x^2+3y^2\leq1\}.$
I got an answer, however the computer marks it as wrong and I can't find my mistake (if there is one, because sometimes it deems correct answers incorrect).
First of all I will use the substitution:
\begin{align*}
x&=\frac{1}{2}\,r\cos(\theta)\\
y&=\frac{1}{\sqrt{3}}\,r\sin(\theta).
\end{align*}
The Jacobian I get is $\dfrac{r}{\sqrt{3}}$.
Now, I will use this sequence of domains:
$D_n=\left[\frac{1}{n},1\right]\times[0,2\pi]$.
Now I tried to calculate the integral:
\begin{align*}
\iint_A \frac{x^2}{(4x^2+3y^2)^{3/2}} \,dx\,dy
&=\iint_{D_n} \frac{\frac{1}{4}r^2\cos^2(\theta)}{(r^2)^{3/2}}\frac{1}{\sqrt{3}}\,r \,dr\,d\theta \\
&=\int_{0}^{2\pi}\left(\int_{\frac{1}{n}}^{1} \frac{1}{4\sqrt{3}}\,\cos^2(\theta)\, dr \right)d\theta\\
&=\int_{0}^{2\pi}\left(1-\frac{1}{n}\right)\frac{\cos^2(\theta)}{4\sqrt{3}} \,d\theta.
\end{align*}
From here this is a simple integral that gets the value $\dfrac{\pi}{4\sqrt{3}}$ when $n \rightarrow \infty$, however this is not the correct answer.
Where is my mistake?
| There is an error in the computation of the Jacobian: it is equal to $\frac r{2\sqrt3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proofs involving strict inequalities let $a,b \in R$
Prove that if $3 \lt a \lt 5$ and $b= 2 + \sqrt{a-2}$ then, $3 \lt b \lt a$
My approach was simply to start with the first inequality and transform it into b and see what happens.
Subtracting 2 gives: $1 \lt a-2 \lt 3$
Taking the square root gives: $1 \lt \sqrt{a-2} \lt \sqrt{3}$
Adding two gives: $3 \lt 2 + \sqrt{a-2} \lt \sqrt{3}+2$
thus $3 \lt b \lt \sqrt{3}+2$
So the interval on b is in fact smaller than the interval on a but that just doesn't seem like it is enough. That's not a very convincing argument. Isn't it plausible that b could equal 3.11 and a could equal 3.10 or something like that. Those numbers fall in the inside the allowed intervals.
Plus this method didn't work on the other 4 questions in the problem.
Suppose $ a \gt 2$ and $b = 1 + \sqrt{a-1}$ using the same technique lands
$b$ as $b \gt 2$
| Let $b=f(a)=2+\sqrt{a-2}$, where $a\in (3,5)$. You have proved that $3<f(a)=b$. Now consider $$g(a)=a-f(a)=a-2-\sqrt{a-2}.$$
Then $g(a)$ is a increasing function since $g'(a)=1-\frac{1}{2\sqrt{a-2}}>1-\frac{1}{2\sqrt{3-2}}=\frac{1}{2}>0$. So $$a-b=a-f(a)=g(a)> g(3)= 3-2-\sqrt{3-2}=0.$$
More precisely, we have $0 < a-b < g(5)=5-2-\sqrt{5-2}=3-\sqrt{3}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding A point $Q$ on a surface such that Line $PQ$ get tangent to that surface
Consider the surface $f: x^2 + y^2 - 2z^2 = 1$ and Point $P= (1,1,1)$.
We want to find all points $Q$ on the surface such that line $PQ$ get
tangent to the surface. Also we want to find the point $Q_0$ with
above conditions such that $PQ_0$ gets smallest.
For the first part; I solved the problem like this: $\nabla f = (2x,2y,-4z) , Q=(x,y,z) \Rightarrow PQ : (x-1 , y-1 , z-1)$ and $PQ \perp \nabla f$ by using the dot product we get at last $2x^2 + 2y^2 -4z^2 -2x-2y +4 =0$ and using the equation of surface we get: $x+y-2z =1$ and $x^2 + y^2 -2z^2 = 1$.(Which is a curve - probably and elliptic curve)
After these, I don't know how to find Q such that PQ gets smallest. I know that I must make $(x-1)^2 + (y-1)^2 + (z-1)^2$ smallest but I don't know how to use equations I got to do this.
| Given point $P=(1,1,1)$, find set $S$ of all $Q=(x,y,z)$ on the surface $x^2+y^2-2z^2=1$
subject to the constraint that the line $PQ$ is tangent to the surface at $Q$. Then locate the $Q$ in that set closest to $P$.
OP has already found that the set $S$ satisfies the equations
$$x+y-2z =1 $$
$$x^2 + y^2 -2z^2 = 1 $$
So it remains to minimize
$$ R(x,y,z)=(x-1)^2+(y-1)^2+(z-1)^2 $$
subject to constraints (1) and (2).
Using the method of Lagrange multipliers, let
$$f(x,y,z)=x^2+y^2-2z^2-1 \tag{1}$$
$$g(x,y,z)=x+y-2z-1 \tag{2}$$
Then solve
$$ \begin{cases}
\nabla R(x,y,z)-\lambda\nabla f(x,y,z)-\mu\nabla g(x,y,z)=0&\\
x+y-2z =1&\\
x^2 + y^2 -2z^2 = 1
\end{cases}$$
The gradient equation leads to
$$ x=y=\frac{\mu+2}{2(1-\lambda)} $$
which reduces the other two equations to
$$ \begin{cases}
x=z+\frac{1}{2}&\\x^2=z^2+\frac{1}{2}
\end{cases} $$
giving $x=y=\frac{3}{4},\,z=\frac{1}{4}$. So the nearest $Q$ to $P$ is $Q=\left(\frac{3}{4},\frac{3}{4},\frac{1}{4}\right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......$
Prove that $$\frac{9}{1!}+\frac{19}{2!}+\frac{35}{3!}+\frac{57}{4!}+\frac{85}{5!}+......=12e-5$$
$$
e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+......
$$
I have no clue of where to start and I am not able to find the general expression of the nth term.
| You noticed a pattern about the denominators, they are straight forward. How do you figure out what's going on with the numerators?
It's good to check ratios between consecutive terms if they are increasing, but that doesn't get you anywhere here. Anther good rule of thumb, take consecutive differences, perhaps differences in those differences, and so on. If your iterated sequences of differences ends up with 0s, then you know you are dealing with a polynomial of degree $n-1$ where $n$ is the number of differences you needed to take.
So we have 9,19, 35, 57, 85.
First layer of differences is 10, 16, 22, 28.
Second Layer: 6,6,6.
Third Layer: 0,0.
So we needed three layers to get a layer of zeros, so our numerators are formed by a quadratic function in n.
With some algebra, you can determined that given a polynomial of $f(x)=ax^2+bx+c$, layers of $f(x+1)-f(x)$, the second layer will be a constant,term by term, and specifically $2a$. Six is our iterated constant and $2a=6$, so the coefficient of $n^2$ must be 3.
We then know that $3+b+c=9$ and $12+2b+c=19$ . So, subtracting the first from the second, we get that $9+b=10$. So $b=1$. Subbing $b=1$ into the first equation and solving for $c$, we find $c=5$.
So we get to Azif00's numerator $3n^2+n+5$.
Now for the sum.
We can break it up.
$$\sum_{n=1}^\infty \frac{3n^2+n+5}{n!}=\sum_{n=1}^\infty\frac{3n^2}{n!}+\sum_{n=1}^\infty\frac{n}{n!}+\sum_{n=1}^\infty\frac{5 }{n!}$$
Note for $e$ , $n$ must start with 0 and not 1, so we need to modify our terms.
The third sum is $5(e-1/0!)=5(e-1).$
The second sum is of $1/(n-1)!$ with $n$ starting at 1. But the $(n-1)$ makes this start at 0. So the second sum is $e$.
Now the first part of the sum is $\sum_{n=1}^\infty\frac{3n}{(n-1)!}$
We can change the index to k starting with k=0 and letting $n=k+1$.
So the sum is $\sum_{k=0}^\infty \frac{3(k+1)}{k!}=\sum_{k=0}^\infty \frac{3k}{k!}+\sum_{k=0}^\infty\frac{3}{k!}$. The second sum is $3e$.
The first term in the left hand sum is zero, so we can start it at k=1 instead and simply to sum over $3/(k-1)!$ But by similar arguments to the above, this is just $3e$.
Put it all together we get $12e-5$
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of Newton Method for monotonic polynomials Consider a polynomial $p : \mathbb R \to \mathbb R$ with $p'(x) > 0$ for all $x \in \mathbb R$. The function $p$ has exactly one real zero. Will the Newton method
$$x_{n+1} = x_n - \frac{p(x_n)}{p'(x_n)}$$
converge for all $x_0 \in \mathbb R$?
Intuitively I think there still might be a counterexample - but I couldn't find one, so is it maybe possible that it indeed converges for every $x_0$?
| No, not necessarily. In particular, consider the polynomial,
$$p(x) = \frac{7}{2}x - \frac{5}{2}x^3 + x^5.$$
Note that
$$p'(x) = \frac{7}{2} - \frac{15}{2}x^2 + 5x^4,$$
a positive quadratic in $x^2$ with a negative discriminant $-\frac{55}{4}$, and hence is strictly positive everywhere. This means $p$ is strictly increasing, as required.
Take an initial iterate $x_0 = 1$. Then,
\begin{align*}
x_1 &= 1 - \frac{p(1)}{p'(1)} = 1 - \frac{2}{1} = -1 \\
x_2 &= -1 - \frac{p(-1)}{p'(-1)} = -1 - \frac{-2}{1} = 1,
\end{align*}
and so the iterates repeat.
Method
It's not difficult to form a cycle with Newton's method. I wanted a polynomial that passes through $(-1, -2)$ and $(1, 2)$, both with derivative $1$. Following the tangent at $(-1, -2)$ to the $x$-axis will yield $x = 1$, so if we take $x_n = -1$, then $x_{n+1} = 1$. Following the tangent at $(1, 2)$ yields an $x$-intercept of $x = -1$, so $x_{n+2} = -1$, and so the iterates repeat.
I tried for a degree $5$ polynomial. Let our polynomial be
$$p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5.$$
Our restrictions turn into
\begin{align*}
p(-1) = a_0 - a_1 + a_2 - a_3 + a_4 - a_5 &= -2 \\
p'(-1) = a_1 - 2a_2 + 3a_3 - 4a_4 + 5a_5 &= 1 \\
p(1) = a_0 + a_1 + a_2 + a_3 + a_4 + a_5 &= 2 \\
p'(-1) = a_1 + 2a_2 + 3a_3 + 4a_4 + 5a_5 &= 1.
\end{align*}
Solving this, we get a general solution in terms of $s$ and $t$:
$$p(x) = \frac{5}{2}x - \frac{1}{2}x^3 + t(1 - 2x^2 + x^4) + s(x - 2x^3 + x^5).$$
In particular, I needed to choose $s$ and $t$ so that the derivative
\begin{align*}
p'(x) &= \frac{5}{2} - \frac{3}{2}x^2 + t(3x^3 - 4x) + s(1 - 6x^2 + 5x^4) > 0
\end{align*}
Clearly, we required $s > 0$. I also decided to chose $t = 0$; it may not have been necessary, but it made things simpler. Now $p'(x)$ is now a cubic in $x^2$:
$$p'(x) = \left(\frac{5}{2} + s\right) - \left(\frac{3}{2} + 6s\right)x^2 + 5sx^4.$$
I wanted the discriminant to be negative, which is to say
$$\left(\frac{3}{2} + s\right)^2 - 20s\left(\frac{5}{2} + s\right) < 0.$$
Choosing $s = 1$ did the trick, and gave us the previously presented polynomial.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $f(x) =x^3+x+1 \in \mathbb{Q} [x] $. If ideal $I=(f(x)) $, find the inverse of $x^2+x+1$ in quotient $\mathbb{Q} [x] /I$. Let $f(x) =x^3+x+1 \in \mathbb{Q} [x] $. If ideal $I=(f(x)) $, find the inverse of $x^2+x+1$ in quotient $\mathbb{Q} [x] /I$.
I am having trouble with this. What exactly is the quotient $\mathbb{Q} [x] /I$ here? I guess it's polynomials with degree 2 or less which in multiplication satisfy the relation $ x^3+x+1=0$. How do I find the inverse? I guess I have to find a polynomial of degree 2 or less so that when multiplied by $x^2+x+1$ I get something divisible by $ x^3+x+1$.
| A polynomial in the quotient ring can be represented by $ax^2+bx+c.$
We want $(ax^2+bx+c)(x^2+x+1)\equiv 1 \pmod {x^3+x+1}$.
Note that $\color{blue}{x^3\equiv-x-1}$ and $\color{green}{x^4\equiv -x^2-x}$.
Thus $(ax^2+bx+c)(x^2+x+1)\equiv \color{green}{ax^4}+\color{blue}{(a+b)x^3}+(a+b+c)x^2+(b+c)x+c$
$\equiv \color{green}{-ax^2-ax}\color{blue}{-(a+b)x-(a+b)}+(a+b+c)x^2+(b+c)x+c$
$\equiv (b+c)x^2+(c-2a)x+(c-a-b)\equiv1,$
so we want $b+c=0, c-2a=0, $ and $c-a-b=1$.
Can you solve this system of equations for $a, b, $ and $c$?
Hint: $c=-b=2a$.
| {
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$bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$ My book has used the equality $$bc(b-c) + ca(c-a) + ab(a-b) = -(b-c)(c-a)(a-b)$$ But its proof is not given.
When I open the LHS, I get $b^2c-bc^2+c^2a-ca^2+a^2b-ab^2$. I do not know how to proceed next.
| $$bc(b-c)+ca(c-a)+ab(a-b)=b^2c-a^2c-bc^2+ac^2+ab(a-b)=$$
$$=c^2(a-b)-c(a-b)(a+b)+ab(a-b)=(a-b)(c^2-ac-bc+ab)=$$
$$=(a-b)(c-a)(c-b).$$
| {
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Prove that all multiples of 4 can appear in a Primitive Pythagorean Triple I have attempted by using the fact that the even numbers $b$ in a PPT $(a,b,c)$ can be generated using $ b = \frac{s^2 - t^2}{2} $ for any odd integers $s >t \ge 1$ that share no common factors.
We can express any odd numbers as $2k+1$ for some non-negative integers $k$, so:
$$\frac{(2n+1)^2 - (2m^2 + 1)}{2} = \frac{4(n^2 + n - m^2 - m)}{2} \\(2n+1)^2 - (2m^2 + 1)= 2(n(n+1)-m(m+1)) \\ \text{So }\frac{ s^2 - t^2}{2} = 2(2k)=4k \text{ for some integer k }$$
But I think all I did was that I proved that all even numbers $b$ in PPT must be a multiple of 4, rather than proving that all multiples of 4 can appear as b, how would I proceed from this?
| $(4n^2-1)^2+(4n)^2=(4n^2+1)^2.$
All Pyth. triplets are of the form $(\,k(a^2-b^2),\,2kab,\, k(a^2+b^2)\,)$ with positive integers $k,a,b.$
All primitive triplets have $k=1$, with $a,b$ co-prime and such that $a-b$ is odd. These conditions are necessary but are also sufficient. (Along with, obviously, $a>b.)$
If $2ab=4n$ then $ab=2n,$ and we see that the above conditions on $a,b$ for a primitive triplet are met if $a=2n$ and $b=1.$
| {
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Solving simple equation I can't seem to solve/simplify step by step to get from equation 3) to 4) as they do in this paper. As per the paper:
3) $p = \frac{p' + (r-R)p}{1+r}$
Because both sides of equation 3) involve the current price, $p$, the equation can be solved as follows:
4) $p = \frac{p'}{1+R}$
Could somebody please step by step show how I'd solve this step please?
| We start with the original equation:
\begin{align*}
p = \frac{p' + (r-R)p}{1+r}
\end{align*}
We want to obtain $p$ as a function of the other values. So, we first want to collect all terms with $p$ on the same side. We do this by splitting the fraction and bringing one term over.
\begin{align*}
p &= \frac{p'}{1+r} + \frac{(r-R)p}{1+r} \\
p - \frac{(r-R)p}{1+r} &= \frac{p'}{1+r}
\end{align*}
Now, we can factor out the $p$ on the left side, and put everything on the same denominator.
\begin{align*}
p \left(
1 - \frac{(r-R)}{1+r}
\right) &= \frac{p'}{1+r} \\
p \left(
\frac{1+r}{1+r} - \frac{r-R}{1+r}
\right) &= \frac{p'}{1+r} \\
p \left(
\frac{1+R}{1+r}
\right) &= \frac{p'}{1+r}
\end{align*}
(Note of course that $(1+r)-(r-R) = (1+r) + (-r +R) = 1+R$.). Finally, we multiply both sides of the equation by $(1+r)/(1+R)$ to get:
\begin{align*}
p \left( \frac{1+R}{1+r} \right) \cdot \frac{1+r}{1+R}
&= \frac{p'}{1+r} \cdot \frac{1+r}{1+R} \\
p = \frac{p'}{1+R}
\end{align*}
And this is the desired result.
| {
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Compute the integral in a closed form : $\int_0^{1}\operatorname{Li}_2(1-x)dx$ How I can find the closed form of the following integration :
$I=\int_0^{1}\operatorname{Li}_2(1-x)dx$
$J=\int_0^{1}\operatorname{Li}_2(1-x)\operatorname{Li}_2(1-\frac{1}{x})dx$
$K=\int_0^{1}\ln(x)\operatorname{Li}_2(1-\frac{1}{x})dx$
Wolfram alpha give me for $I=$ see here
For $J=$ see here
For $K=$ see here
My attempt first integral is :
$\operatorname{Li}_2(1-x)=\sum_{n=1}^{\infty}\frac{(1-x)^n}{n^2}$
So :
$I=\sum_{n=1}^{\infty}\frac{\int_0^{1}(1-x)^{n}dx}{n^2}$
$=\sum_{n=1}^{\infty}\frac{\beta(1,n+1)}{n^2}$
$=\sum_{n=1}^{\infty}\frac{1}{(n+1)n^2}=\zeta (2)-1$
But what about from second and last integral ?
We can use same method ?
| The integrals $J$ and $K$ can be computed using the polylogarithm identity
$$ \operatorname{Li}_2 \left(1 - \frac{1}{x}\right) = - \operatorname{Li}_2(1-x) - \frac{1}{2} \log^2(x) \, , \, x > 0 \, , \tag{1} $$
and the integrals
$$ \int \limits_0^1 \frac{[-\log(x)]^n}{1-x} \, \mathrm{d} x = n! \zeta(n+1) \, , \, n \in \mathbb{N} \, . \tag{2}$$
We have
\begin{align}
K &\stackrel{(1)}{=} \int \limits_0^1 \left[\frac{- \log^3(x)}{2} - \log(x) \operatorname{Li}_2(1-x)\right]\, \mathrm{d} x \\
&= 3 + \left[\left(x - 1 - x \log(x)\right)\operatorname{Li}_2(1-x)\right]_{x = 0}^{x=1} + \int \limits_0^1 \frac{[1 - x + x \log(x)] \log(x)}{1-x} \, \mathrm{d} x \\
&= 3 + \zeta(2) - 1 - 2 + \int \limits_0^1 \frac{\log^2(x)}{1-x} \, \mathrm{d} x \stackrel{(2)}{=} \zeta(2) + 2 \zeta(3) = \frac{\pi^2}{6} + 2\zeta(3) \, .
\end{align}
Similarly, we can write
$$ - J \stackrel{(1)}{=} \int \limits_0^1 \operatorname{Li}_2^2(1-x) \, \mathrm{d} x + \frac{1}{2} \int \limits_0^1 \log^2(x) \operatorname{Li}_2(1-x) \, \mathrm{d} x \equiv M + \frac{1}{2} N \, .$$
The remaining integrals can be evaluated using integration by parts:
\begin{align}
M &= \zeta^2(2) - 2 \int \limits_0^1 - \log(x) \operatorname{Li}_2(1-x) \, \mathrm{d} x \\
&= \zeta^2(2) - 2 \zeta(2) + 2 \int \limits_0^1 \frac{\log(x)[x - 1 - x \log(x)]}{1-x} \, \mathrm{d} x \\
&= -\zeta(2)(2-\zeta(2)) + 2 + 4 - 2 \int \limits_0^1 \frac{\log^2(x)}{1-x} \, \mathrm{d} x \stackrel{(2)}{=} 6 - \zeta(2)(2-\zeta(2)) - 4 \zeta(3)
\end{align}
and
\begin{align}
N &= \int \limits_0^1 \frac{- x \log(x) [\log^2(x) - 2 \log(x) + 2]}{1-x} \, \mathrm{d} x \stackrel{(2)}{=} 6 \zeta(4) + 4 \zeta(3) + 2 \zeta(2) - 12 \, .
\end{align}
Therefore,
$$ - J = 3 \zeta(4) - 2 \zeta(3) + \zeta(2)(\zeta(2)-1) = \frac{11 \pi^4}{180} - \frac{\pi^2}{6} - 2 \zeta(3) \, . $$
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minimum value of of $(5+x)(5+y)$
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$
What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
| The symmetry of the problem suggests to substitute $x = u+v$, $y = u-v$. Then the given constraint is
$$
3 = x^2 + xy + y^2 = 3u^2 + v^2
$$
and in particular $-1 \le u \le 1$.
We look for the minimal value of
$$
(5+x)(5+y) = (5+u)^2 - v^2 = (5+u)^2 - (3-3u^2) \\
= 4u^2 + 10 u + 22 \, .
$$
This expression is increasing for $u \in [-1, 1]$, and therefore
$$
\ge 4 (-1)^2 + 10(-1) + 22 = 16 \, .
$$
Equality holds for $u=-1, v=0$, corresponding to $x=-1, y=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Let $x, y, z$ be real numbers. If $x + y + z = 1$ and $x^2 + y^2 + z^2 = 1$, then what is the minimum value of $x^3 + y^3 + z^3$? Through manipulation, I got as far as $x^3 +y^3 + z^3 = 1 + 3xyz$
I tried solving it using AM - GM inequality but it only gave me the maximum value.
GM - HM does not to help either.
| You can use Cauchy-Schwarz inequality. $X+y=1-z$, thus $(1-z)^2 = (x+y)^2$ is less than or equal to $(1+1)(x^2+y^2)=2 (1-z^2)$. I get $z$ more than or equal to $-1/3$. $x^3+y^3+z^3= (x^3+y^3)+z^3=(x+y)(x^2-xy+y^2)+z^3= (3(x^2+y^2)-(x+y)^2)/2 + z^3$ doing the necessary substitutions I get $1-3 z^2 +3 z^3$, equals $5/9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if someone has a more straightforward solution.
So applying the sum formula for sine and doing simple algebra we have:
$$\lim_{x\to 0} \frac{\cos{2} \,(\sin{x^2}-\sin{x})}{x} - \frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
The first limit is easy to evaluate and is equal to $-\cos{2}$. However, the second limit is harder, as it follows:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
I came across a solution by using the following sum-to-product identity:
$$\cos{A}-\cos{B}=-2\sin{\Big(\frac{A+B}{2}\Big)} \sin{\Big(\frac{A-B}{2}\Big)}$$
Setting $A=x^2$ and $B=x$, we have that
$$\cos{x^2}-\cos{x}=-2\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}$$
This is my only point of concern whether I applied the identity correctly. The rest of it flows more easily:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} = \lim_{x\to 0} \frac{-2\sin{2}\,\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}}{x}$$
$$= -2 \sin{2} \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} \lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}\Big)} $$
The first limit can be solved as it follows:
$$\lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} = \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)} \Big(\frac{x^2+x}{2}\Big)}{x \Big(\frac{x^2+x}{2}\Big)} = 1 \cdot \lim_{x \to 0} \frac{x^2 + x}{2x} = \frac{1}{2} $$
The second limit is equal to zero
$$\lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}}\Big)=0$$
| On applying Taylor series, and binomial expansion and taking linear degree terms in numerator to evaluate the limit,(all constant terms are powers of 2, they cancel out as they are same for both terms in numerator)
We get the limit as $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}. = -1 + \frac{{3} \choose {1} * 2^2}{3!} - \frac{{5} \choose {1} *2^4}{5!} \ldots =-(1-\frac{2^2}{2!}+\frac{2^4}{4!} \ldots)=-\cos(2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
$ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$ I'm currently working on a definite integral and am hoping to find alternative methods to evaluate. Here I will to address the integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx
\end{equation}
Where $n \in \mathbb{N}$. We first observe that when $n = 2k + 1$ ($k\in \mathbb{Z}, k \geq 0$) that,
\begin{equation}
I_{2k + 1} = \int_0^\frac{\pi}{2}\ln^{2k + 1}\left(\tan(x)\right)\:dx = 0
\end{equation}
This can be easily shown by noticing that the integrand is odd over the region of integration about $x = \frac{\pi}{4}$. Thus, we need only resolve the cases when $n = 2k$, i.e.
\begin{equation}
I_{2k} = \int_0^\frac{\pi}{2}\ln^{2k}\left(\tan(x)\right)\:dx
\end{equation}
Here I have isolated two methods.
Method 1:
Let $u = \tan(x)$:
\begin{equation}
I_{2k} = \int_0^\infty\ln^{2k}\left(u\right) \cdot \frac{1}{u^2 + 1}\:du = \int_0^\infty \frac{\ln^{2k}\left(u\right)}{u^2 + 1}\:du
\end{equation}
We note that:
\begin{equation}
\ln^{2k}(u) = \frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}
\end{equation}
By Leibniz's Integral Rule:
\begin{align}
I_{2k} &= \int_0^\infty \frac{\frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}}{u^2 + 1}\:du = \frac{d^{2k}}{dy^{2k}} \left[ \int_0^\infty \frac{u^y}{u^2 + 1} \right]_{y = 0} \nonumber \\
&= \frac{d^{2k}}{dy^{2k}} \left[ \frac{1}{2}B\left(1 - \frac{y + 1}{2}, \frac{y + 1}{2} \right) \right]_{y = 0} =\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \Gamma\left(1 - \frac{y + 1}{2}\right)\Gamma\left( \frac{y + 1}{2} \right) \right]_{y = 0} \nonumber \\
&=\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \frac{\pi}{\sin\left(\pi\left(\frac{y + 1}{2}\right)\right)} \right]_{y = 0} = \frac{\pi}{2}\frac{d^{2k}}{dy^{2k}} \left[\operatorname{cosec}\left(\frac{\pi}{2}\left(y + 1\right)\right) \right]_{y = 0}
\end{align}
Method 2:
We first observe that:
\begin{align}
\ln^{2k}\left(\tan(x)\right) &= \big[\ln\left(\sin(x)\right) - \ln\left(\cos(x)\right) \big]^{2k} \nonumber \\
&= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)
\end{align}
By the linearity property of proper integrals we observe:
\begin{align}
I_{2k} &= \int_0^\frac{\pi}{2} \left[ \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right) \right]\:dx \nonumber \\
&= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \int_0^\frac{\pi}{2} \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)\:dx \nonumber \\
& = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j F_{n,m}(0,0)
\end{align}
Where
\begin{equation}
F_{n,m}(a,b) = \int_0^\frac{\pi}{2} \ln^n\left(\cos(x)\right)\ln^{m}\left(\sin(x)\right)\:dx
\end{equation}
Utilising the same identity given before, this becomes:
\begin{align}
F_{n,m}(a,b) &= \int_0^\frac{\pi}{2} \frac{d^n}{da^n}\big[\sin^a(x) \big] \cdot \frac{d^m}{db^m}\big[\cos^b(x) \big]\big|\:dx \nonumber \\
&= \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[ \int_0^\frac{\pi}{2} \sin^a(x)\cos^b(x)\:dx\right] = \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{1}{2} B\left(\frac{a + 1}{2}, \frac{b + 1}{2} \right)\right] \nonumber \\
&= \frac{1}{2}\frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]
\end{align}
Thus,
\begin{equation}
I_{2k} = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \frac{1}{2}\frac{\partial^{2k }}{\partial a^j \partial b^{2k - j}}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]_{(a,b) = (0,0)}
\end{equation}
So, I'm curious, are there any other Real Based Methods to evaluate this definite integral?
| Make an exponential generating function with $I_k$,
$$I_k: = \frac{1}{2} (1 + (-1)^k) \int_0^\infty \frac{ \log^{k}(u) }{u^2+1} du $$
Then
$$I(x)=\sum_{k=0}^\infty I_k\,\frac{x^k}{k!} = \frac{1}{2}\int_0^\infty \frac{ u^x + u^{-x} }{u^2+1} du $$
where an interchange of $\sum$ and $\int$ has been made. The integral can be solved in closed form, $I(x) = \pi/2 \cdot \sec{(\pi x/2)}.$ Expanding the sec in a power series will give the last answer that Stafan Lafon gave, in terms of Euler numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Real values of $k$ such that $kx^2+(k+3)x+k-3=0$ has $2$ positive integer roots
There is only one real values of $k$ for which the quadratic equation $kx^2+(k+3)x+k-3=0$ has $2$ positive integer roots. Then the product of these two solutions is
What i tried.
$$kx^2+(k+3)x+k-3=0$$
for $k=0,$ i am getting $x=1$(only one integer roots)
for $\displaystyle k\neq 0, x=\frac{-k\pm \sqrt{(k+3)^2-4k(k-3)}}{2k}$
for integer roots, $(k+3)^2-4k(k-3)=p^2$
How do i solve it , Help me please
| There is no such $k.$
Since the product we're looking for is $$\frac {k-3}{k},$$ first we must have $k\ne 0.$ Also, since the roots are positive integers, it follows that their product is also a positive integer. Thus, $k\ne 3$ and we also have that $k-3$ and $k$ must have the same sign.
Having noticed this, we find that the discriminant $$(k+3)^2-4k(k-3)$$ must be a nonzero square. Thus we must have $$0<(k-3)^2<12.$$ It follows that $(k-3)^2$ is either $1,4,$ or $9.$ Thus, $k-3$ must be some of $\pm 1,\pm 2, \pm 3,$ so that correspondingly, $k$ is $4,2;5,1;6,0.$
Now using the fact that $k$ is nonzero and has the same sign as $k-3$ leaves just $k$ being one of $4,5,6.$ Finally using the condition that $(k-3)/k$ be integral leaves us no such value of $k.$
Thus, the claim that there exists such a $k$ is false. Your conditions are too restrictive. Completely killed off the $k$'s.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that if $a$ and $b$ are real numbers with $0 < a < b$ then $\frac{1}{b} < \frac{1}{a}$
Suppose a and b are real numbers. Prove that if $0 < a < b$
then $\frac{1}{b} < \frac{1}{a}$.
My attempt:
Given that $0<<$
We can write $b$ as $b = an$, where $n>1$
$$\tag1\frac{1}{a} = \frac{n}{an} = \frac{n}{b} $$
$$\tag2\frac{n}{b} > \frac{1}{b} \implies \frac{1}{a} > \frac{1}{b}$$
Is it correct?
| Yes, your solution is correct.
Of course you could have multiplied both sides of your inequality by $$\frac {1}{ab}$$ to get the result in one shot.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Determine polynomial of $3$ degree real coefficients $f(x)$ such that $f(x) \vdots (x-2)$ and $f(x)$ dividing by $x^2-1$ have remainder $2x$ I need help in this problem.
Problem: Determine polynomial of $3$ degree real coefficients $f(x)$ such that $f(x) \vdots (x-2)$ and $f(x)$ divided by $x^2-1$ has remainder $2x$ and suppose $f(x)=x^3+ax^2+bx+c$.
| Hint:
$f(x)=(x^2-1)(x-a)+2x$ and $f(2)=0$.
Solve for $a$.
$f(x)=x^3-ax^2+x+a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find modulus and argument of $\omega = {\frac {\sin (P + Q) + i (1 - \cos (P + Q))} {(\cos P + \cos Q) + i (\sin P - \sin Q) }} $ A past examination paper had the following question that I found somewhat difficult. I tried having a go at it but haven't come around with any possible double angle identities. How would one go about tackling it?
Given:
$$\omega = {\frac {\sin (P + Q) + i (1 - \cos (P + Q))} {(\cos P + \cos Q) + i (\sin P - \sin Q) }} $$
To prove:
$$|\omega| = \tan \frac {P + Q} {2} \qquad\text{and}\qquad \arg(\omega) = Q $$
A guideline on how/ which identity to use would be greatly appreciated.
To give an idea how one would start it is by;
Proof:
$$|\omega| = {\frac {\sqrt{\sin^2 (P + Q) + (1 - \cos (P + Q))^2}}
{\sqrt{(\cos P + \cos Q)^2 + (\sin P - \sin Q)^2 }}} $$
I'm still unsure about the above or how the square root come about
| We have
\begin{align}
N
& := \sin^2(P+Q) + (1-\cos(P+Q))^2 = \sin^2(P+Q) + \cos^2(P+Q) + 1 - 2\cos(P+Q) \\
& = 2 (1-\cos(P+Q))
= 2\cdot2\sin^2\frac{P+Q}{2}
= 4\sin^2\frac{P+Q}{2}
\end{align}
and
\begin{align}
D
& = \cos^2P +\cos^2Q + \sin^2P + \sin^2Q + 2(\cos P\cos Q - \sin P \sin Q) \\
&= 2 +2(\cos(P+Q))
= 2(1+\cos(P+Q)) = 4\cos^2\frac{P+Q}{2}
\end{align}
Now, $$|\omega| = \sqrt{\frac{N}{D}} = \tan\frac{P+Q}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $\alpha$ and $\beta$ are the solutions of $a\cos \theta+ b\sin \theta= c$ show that If $\alpha$ and $\beta$ are the solutions of $a\cos \theta+ b\sin \theta= c$ show that
1) $\sin \alpha + \sin \beta = \dfrac{2bc}{a^2 + b^2}$
2) $\sin \alpha \sin \beta = \dfrac{c^2-a^2}{a^2+b^2}$
I couldn't even start the problem, and I generally have a lot of difficulty in compound angles so please help me with this.
Thanks!
| $$a\cos\theta = c - b\sin\theta \implies a^2\cos^2\theta = c^2-2bc\sin\theta+b^2\sin^2\theta \\ \implies (a^2-a^2\sin^2\theta) = c^2-2bc\sin\theta+b^2\sin^2\theta $$
$$(a^2+b^2)\sin^2\theta - (2bc)\sin\theta+(c^2-a^2)=0 \equiv Ax^2 + Bx +C = 0$$
As $\alpha$ and $\beta$ are the solutions of the given equation, $\sin\alpha \equiv x_1$ and $\sin\beta \equiv x_2$ satisfy the above equation.
If $x_1$ and $x_2$ are the solutions of $Ax^2 + Bx +C = 0$, then ,
$$x_1+x_2 = -\frac{B}{A} \implies \sin\alpha + \sin\beta = - \frac{-2bc}{a^2+b^2} = \frac{2bc}{a^2+b^2}$$
and
$$x_1 \cdot x_2 = \frac{C}{A} \implies \sin\alpha\cdot\sin\beta=\frac{c^2-a^2}{a^2+b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
1025th term of the sequence $ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $ Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^{th}\, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it We can observe the following -
$1 - 1$
$2 - 2 $
$3 - 2$
$4 - 4$
$5 - 4$
. . .
$8 - 8$
$9 - 8$
. . .
We can notice that $ 4^{th}$ term is 4 and similarly, the $ 8^{th}$ term is 8.
So the $ 1025^{th}$ term must be 1024 as $ 1024^{th} $ term starts with 1024.
So the value of $ 1025^{th}$ term is $ 2^{10} $ .
Is there any other method to solve this question?
| A higher brow way of writing the same thing is to say the $n^{th}$ term is
$2^{\lfloor \log_2 n\rfloor}$, then to evaluate that at $n=1025$. The base $2$ log of $1025$ is slightly greater than $10$, so the term is $2^{10}=1024$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
} |
How to derive this Bessel function relationship? $$J_n(x) = \frac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt$$
| Show that
$J_n(x) = \dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt
$
I did my usual
naive plugging in of definitions
and, with some help from Wolfy,
it worked out.
A pleasant surprise.
Since
$J_a(x)
=\sum_{m=0}^{\infty} \dfrac{(-1)^m}{m!(m+a)!}\dfrac{x^{2m+a}}{2^{2m+a}}
$
and,
according to Wolfy,
$\int_0^1 (1 - x^2)^a x^b dx
= \dfrac{Γ(a + 1) Γ((b + 1)/2)}{2 Γ(a + b/2 + 3/2)}
$
for
$Re(b)>-1 ∧ Re(a)>-1
$,
$\begin{array}\\
U_{n,m}(x)
&=\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt\\
&=\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt\\
&=\int_0^1(1-t^2)^{n-m-1}t^{m+1}\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k+m}t^{2k+m}}{2^{2k+m}}dt\\
&=\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k+m}}{2^{2k+m}}\int_0^1(1-t^2)^{n-m-1}t^{m+1}t^{2k+m}dt\\
&=\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k+m}}{2^{2k+m}}\int_0^1(1-t^2)^{n-m-1}t^{2k+2m+1}dt\\
&=\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k+m}}{2^{2k+m}}\dfrac{Γ(n-m) Γ((2k+2m+2)/2)}{2 Γ(n-m-1 + (2k+2m+1)/2 + 3/2)}\\
&=\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k+m}}{2^{2k+m}}\dfrac{Γ(n-m) Γ(k+m+1}{2 Γ(n-m-1 + k+m+2)}\\
&=\dfrac{x^{m}}{2^{m}}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k}}{2^{2k}}\dfrac{Γ(n-m) Γ(k+m+1}{2 Γ(n+k+1)}\\
&=\dfrac{x^{m}\Gamma(n-m)}{2^{m}}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k}}{2^{2k}}\dfrac{ Γ(k+m+1)}{2 Γ(n+k+1)}\\
&=\dfrac{x^{m}\Gamma(n-m)}{2^{m}}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(k+m)!}\dfrac{x^{2k}}{2^{2k}}\dfrac{ (k+m)!}{2(n+k)!}\\
&=\dfrac{x^{m}\Gamma(n-m)}{2^{m}}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!2(n+k)!}\dfrac{x^{2k}}{2^{2k}}\\
&=\dfrac{\Gamma(n-m)}{2}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(n+k)!}\dfrac{x^{2k+m}}{2^{2k+m}}\\
&=\dfrac{\Gamma(n-m)x^{m-n}}{2^{n-m+1}}\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!(n+k)!}\dfrac{x^{2k+n}}{2^{2k+n}}\\
&=\dfrac{\Gamma(n-m)x^{m-n}}{2^{n-m+1}}J_n(x)\\
\end{array}
$
so
$\begin{array}\\
\dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}U_{n, m}(x) &=\dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\int_0^1(1-t^2)^{n-m-1}t^{m+1}J_m(xt)dt\\
&= \dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\dfrac{\Gamma(n-m)x^{m-n}}{2^{n-m+1}}J_n(x)\\
&= \dfrac{2x^{n-m}}{2^{n-m}\Gamma(n-m)}\dfrac{\Gamma(n-m)x^{m-n}}{2^{n-m+1}}J_n(x)\\
&= J_n(x)\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find range of $m$ such that $|x^2-3x+2 | = mx$ has 4 distinct solutions
$|x^2-3x+2 | = mx$
I think that $(x^2-3x+2) = mx$ or $-(x^2-3x+2) = mx$
For $-(x^2-3x+2) = mx$, $D > 0$,
$m = 3 \pm \sqrt{8}$
For $(x^2-3x+2) = mx$, $D > 0$,
$m = -3 \pm \sqrt{8}$
Now. What do you think it's the range?
Note that $ |x^2-3x+2 | = mx$ which means $mx > 0$ isn't it?
| Solving $|x^2-3x+2|=mx$ is same as finding the intersection of the graph of $y=|x^2-3x+2|$ and the lines $y=mx$ passing through the origin. Here is a figure to better illustrate the idea.
By changing the value of $m$ we are changing the slope of the line passing through the origin. Unless this line is tangent or passes through the middle cusp in the graph there cannot be $4$ solutions. This means the first instance the line will intersect the graph at $4$ points (tangent point is counted as double point) is when this line is tangent to the graph.
Ques: what is the point of tangency?
Let $(h,k)$ be the point on (the middle cusp) the curve such that the line $y=mx$ is tangent. Then using the fact that slope of the line $\frac{k}{h}$ should be the same as the slope of the curve (in particular the middle cusp which is given by $y=-x^2+3x-2$), we get
$$m=\frac{k}{h}=\frac{dy}{dx}\Big|_{(h,k)}=-2h+3.$$
But $k=-h^2+3h-2$, so this gives $h=\pm \sqrt{2}$. Note that the negative value is not possible. So $\color{red}{h=\sqrt{2}}$.
Observe that the slope $m=-2h+3=3-2\sqrt{2}$.
Since you have asked for $4$ distinct solutions, so
$$\color{red}{0 <m < 3-2\sqrt{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$a, b$ are the integer part and decimal fraction of $\sqrt7$ find integer part of $\frac{a}{b}$
$a, b$ are the integer part and decimal fraction of $\sqrt7$
find integer part of $\frac{a}{b}$
using calculator :
$\sqrt 7$ = 2.645
$\frac{a}{b} = \frac{2}{0.6} = \frac{20}{6} = 3.333$
integer part of $\frac{a}{b} = 3$
how to do it without calculator? if i don't know $\sqrt7$ part
| My solution without a calculator:
Since $2^2 = 4$ and $3^2=9$, we know that $\sqrt{7} = 2 +b $ and therefore
$$
\frac{a}{b} = \frac{2}{b} = \frac{2}{\sqrt{7}-2} = \frac{2}{(\sqrt{7}-2)} \frac{(\sqrt{7}+2)}{(\sqrt{7}+2)} = \frac{2(\sqrt{7}+2)}{7-4} = \frac{2}{3} \left(\sqrt{7}+2 \right)
$$
And since $2 < \sqrt{7} < 3$, we know that (plugging these into the expression)
$$
\frac{8}{3} < \frac{2}{3} \left(\sqrt{7}+2 \right) < \frac{10}{3}
$$
which indicates that the integer part of $a/b$ is $3$.
Edit: the lower bound is not tight enough. But this can be fixed.
First observation:
$
4\cdot 7 = 28 > 25 \Rightarrow \sqrt{7}> \frac{5}{2}
$
Then:
$$
\left[ \frac{2}{3}\left( \sqrt{7}+2 \right) \right]^2 = \frac{4}{9} (11+4 \sqrt{7}) > \frac{4}{9} (11+4 \cdot \frac{5}{2}) = \frac{4}{9}\cdot 21 = \frac{28}{3}> \frac{27}{3} = 9
$$
so therefore $\frac{2}{3} \left(\sqrt{7}+2 \right) > 3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the Range of $y=\frac{x+a}{x^2+bx+c^2}$ Given that $x^2-4cx+b^2 \gt 0$ $\:$ $\forall$ $x \in \mathbb{R}$ and $a^2+c^2-ab \lt 0$
Then find the Range of $$y=\frac{x+a}{x^2+bx+c^2}$$
My try:
Since $$x^2-4cx+b^2 \gt 0$$ we have Discriminant
$$D \lt 0$$ $\implies$
$$b^2-4c^2 \gt 0$$
Also
$$x^2+bx+c^2=(x+a)^2+(b-2a)(x+a)+a^2+c^2-ab$$
Hence
$$y=\frac{1}{(x+a)+b-2a+\frac{a^2+c^2-ab}{x+a}}$$
But Since $a^2+c^2-ab \lt 0$ We can't Use AM GM Inequality
Any way to proceed?
| This is not a complete solution. Just my opinion on the possible solution.
Cross multiplying and subtracting $x+a$ gives an equation:
$x^2y+x(by-1)+(yc^2-a)=0$
The discriminant turns out to be:
$y^2(b^2-4c^2)+y(4a-2b)+1$
Further the discriminant of the above equation is:
$16(a^2+c^2-ab)$
Which is certainly less than 0 according to info provided. I believe in a similar manner the range shall pop up. But certainly, I might be wrong...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$
I would like a little help, so that I can finish solving this exercise.
So far, I've got
$$
\frac{\frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h)^2}{(x-a-h)^2(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h)^2}{(x-a-h)^2(x-a)^2}}{ \frac{h}{1}}
= \frac{(x-a)^2 - (x-a-h)^2}{h(x-a-h)^2(x-a)^2}
$$
In the numerator maybe I can continue with a difference of squares, but I'm a little confused.
| The top simplifies
$$ (x-a)^2 -[(x-a)-h]^2 =$$
$$(x-a)^2 -[(x-a)^2-2h(x-a)+h^2]$$
$$=2h(x-a)-h^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Minimizing $9 \sec^2{x} + 16 \csc^2{x}$ Find the minimum value of $$9 \sec^2{x} + 16 \csc^2{x}$$
My turn :
Using AM-GM
$$9\sec^2{x} + 16\csc^2{x} \geq 2 \sqrt{144 \sec^2{x} \csc^2{x}}$$
$$9 \sec^2{x} + 16\csc^2{x} \geq 24 \sec{x} \csc{x} $$
But the equality sign holds iff
$$9 \sec^2{x} = 16\csc^2{x}$$
Then $$ \tan{x} = \frac{4}{3}$$
Then the minimum value is $$24 \times \frac{5}{4} \times \frac{5}{3} = 50$$
Is there any mistake with the solution ?
| By differentiating and setting to $0$ or plotting the graph, the minimum value is actually $49$, so there must be something wrong in your solution.
From $9 \sec^2 x = 16 \csc^2 x$ you cannot deduce $\tan x = \frac{4}{3}$. You have $\tan^2 x = \frac{16}{9}$, but $\tan x$ can either be $\frac{4}{3}$ or $-\frac{4}{3}$.
In addition, $24 \sec x \csc x$ is actually negative for some values of $x$, such as $x = 2$. Therefore you have not actually proven that $50$ is the minimum value, but rather that the function is always greater than some negative number.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a function with given conditions I want to find the equation of a function $f(x)$ such that the following conditions are met:
$(i)\ f(3) = 0, f(0) =1$.
$(ii)\ f(x)$ is an even function.
$(iii)\ f(x)$ has vertical asymptodes at $x=\pm 4$.
$(iv)\ f(x)$ has horizontal asymptode at $ y=2$.
Now since the function is even, I get $x=-3$ as another root and then I can graph the function. But how do I find its equation? Any hints? Thank you.
In general is there some method or software to find the equation of a function given its graph?
| We use building blocks having one or more of the wanted properties and put them together conveniently so that the final function $f$ has all wanted properties.
*
*We start with (iii) and take as function with asymptote at $x=\pm 4$ the rational function
\begin{align*}
x\to\frac{1}{(x-4)(x+4)}=\frac{1}{x^2-16}\tag{1}
\end{align*}
which is already even.
To get a zero at $x=3$ we multiply with $x-3$ and to keep the function even we multiply with $(x-3)(x+3)=x^2-9$.
*
*We obtain
\begin{align*}
x\to\frac{x^2-9}{x^2-16}\tag{2}
\end{align*}
Note that $\frac{x^2-9}{x^2-16}$ evaluated at $x=0$ is $\frac{9}{16}$.
*
*We multiply the function (2) with $\frac{16}{9}$ and obtain with
\begin{align*}
x\to\frac{16}{9}\cdot\frac{x^2-9}{x^2-16}\tag{3}
\end{align*}
an even function which fulfills (i) and (ii).
Finally we extend (3) to get a function with horizontal asymptote at $y=2$. Note that multiplying the function with $e^{-x^2}$ gives
\begin{align*}
\lim_{x\to\pm\infty}\frac{x^2-9}{x^2-16}e^{-x^2}=0
\end{align*}
Since we also have
\begin{align*}
\lim_{x\to\pm\infty}\frac{x^2-9}{x^2-16}=1
\end{align*}
we use the Ansatz
\begin{align*}
f(x)=\frac{16}{9}\cdot\frac{x^2-9}{x^2-16}\left(ae^{-x^2}+b\right)
\end{align*}
solve
\begin{align*}
f(0)&=a+b=1\\
\lim_{x\to\pm\infty}f(x)&=\frac{16}{9}b=2
\end{align*}
and find $a=-\frac{1}{8}$ and $b=\frac{9}{8}$.
*
*We finally obtain a function
\begin{align*}
&\color{blue}{f:\mathbb{R}\to\mathbb{R}\setminus\{\pm 4\}}\\
&\color{blue}{f(x)=\frac{2}{9}\cdot\frac{x^2-9}{x^2-16}\left(-e^{-x^2}+9\right)}
\end{align*}
which fulfills (i) to (iv).
Note: To find an even function is usually not a problem. When we have a function $g$ fulfilling all other properties we can take $f(x)=\frac{1}{2}\left(g(x)+g(-x)\right)$ to get an even function with the wanted properties.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x\sin A=y\sin(A+2π/3) =z\sin(A+4π/3),$ derive a relation among $x, y, z$ by eliminating $A$
If $x\sin A=y\sin(A+2π/3) =z\sin(A+4π/3),$ derive a relation among $x, y, z$ by eliminating $A$
This problem was bothering me for a while, and I finally could not solve it. I tried taking the whole equation as $k$ but the calculation was a mess. Would someone please help me to find a solution using a simpler approach?
| Consider (first & second terms) $$x\sin A=y\sin\left(A+\frac{2\pi}{3}\right)$$
$$x\sin A=y\left(\sin A\cos \frac{2\pi}{3}+\cos A\sin\frac{2\pi}{3}\right)$$
$$x=y\left(\frac{\sin A\cos \frac{2\pi}{3}}{\sin A}+\frac{\cos A\sin\frac{2\pi}{3}}{\sin A}\right)$$
$$x=y\left(-\frac12+\cot A \frac{\sqrt3}{2}\right)$$
$$\cot A=\frac{2}{\sqrt3}\left(\frac xy+\frac12\right)\tag 1$$
Similarly, consider (first & third terms) $$x\sin A=z\sin\left(A+\frac{4\pi}{3}\right)$$
$$x\sin A=z\left(\sin A\cos \frac{4\pi}{3}+\cos A\sin\frac{4\pi}{3}\right)$$
$$x=z\left(\frac{\sin A\cos \frac{4\pi}{3}}{\sin A}+\frac{\cos A\sin\frac{4\pi}{3}}{\sin A}\right)$$
$$x=z\left(-\frac12+\cot A \left(-\frac{\sqrt3}{2}\right)\right)$$
$$\cot A=-\frac{2}{\sqrt3}\left(\frac xz+\frac12\right)\tag 2$$
Equating (1) & (2), we get
$$\frac{2}{\sqrt3}\left(\frac xy+\frac12\right)=-\frac{2}{\sqrt3}\left(\frac xz+\frac12\right)$$
$$\frac xy+\frac xz+1=0$$
Hope above is the required relation among $x, y$ & $z$
| {
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The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two partss. Find the area of smaller part. Question: The Curve $y = \frac{x^2}2$ divides the curve $x^2 + y^2 = 8$ in two parts. Find the area of smaller part.
This question is from Basic Mathematics. Please explain how I can solve it according to class 11th student.
My work is below.
$\implies y = \frac{x^2}2 \\
\implies 2y = x^2 \\
\implies -x^2 +2y = 0 \ \ \ \ (1)$
and
$x^2 + y^2 = 8 \ \ \ \ (2) $
Adding (1) and (2)
$y^2 + 2y = 8 \\
(y-2)(y+4) =0 \\
\therefore y = 2 or -4$
But when I plot it in the plain with help of https://www.desmos.com/ value of y can only be 2 for the point of intersections.
I'm unable to understand how I can eliminate -4 from the results.
| Because $y$ should be positive.
By your work it's:
$$2\int\limits_0^2\left(2-\frac{x^2}{2}\right)dx+\frac{1}{4}\cdot\pi(2\sqrt2)^2-\frac{1}{2}(2\sqrt2)^2.$$
I got $\frac{4}{3}+2\pi.$
| {
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Can i factor this expression: $x^3+y^3+z^3$ I have the following numerical expression, which is exactly equal to $1$
Text version:
(-(7 2^(2/3))/38 + 3^(2/3)/19 + (23 5^(2/3))/95 + (5 6^(1/3))/19 + 3/38 (5^(1/3) 6^(2/3)) - 10^(1/3)/19 - 3/190 (3^(1/3) 10^(2/3)) - (4 15^(1/3))/19 - 6/95 (2^(1/3) 15^(2/3)))*(2^1/3+3^1/3+5^1/3)
Image version:
according to this fact, $x^3+y^3+z^3$ would be factorized at all and first factor of it should be $x+y+z$ if we set
$
x^3=2,y^3=3,z^3=5
$
If so then how to factor $x^3+y^3+z^3$ ?
PS. I tried to divide $x^3+y^3+z^3$ by $x+y+z$ and failed (got remainder).
| Your expression can be verified by the following identity:
\begin{align}
& \; \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c} \\
= & \:
\dfrac
{(a+b+c)^3-27abc}
{\left(
\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}-
\sqrt[3]{bc}-\sqrt[3]{ca}-\sqrt[3]{ab} \,
\right)
\left[
(a+b+c)^2+
3(a+b+c)\sqrt[3]{abc}+
9\sqrt[3]{a^2 b^2 c^2}
\right]}
\end{align}
Note that $$(a,b,c)=(2,3,5) \implies (a+b+c)^3-27abc=190$$
Your first factor is just rationalization of the denominator for
$$\frac{1}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{5}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find limit of $2x+ (4x^2 +3x - 2)^{1/2}$ when $x\to-\infty$ I have to find a limit $\lim _{x \to - \infty}{(2x + \sqrt{4x^2 +3x - 2})}$. I know, that the correct answer is $-3/4$, I've checked it in a graphing tool.
But could someone explain it to me, how to find it analytically, preferably without the Lhopital rule, since the one uses derivatives, and in my textbook I've not yet reached the derivatives chapter, so I'm sure the exercise should and can be solved without the Lhopital rule.
I have a hint in the textbook, that I should multiply by the conjugate. I did:
$$
\lim _{x \to - \infty}{(2x + \sqrt{4x^2 +3x - 2})} = \\
\lim _{x \to - \infty}{(\frac{4x^2 - 4x^2 -3x + 2}{2x - \sqrt{4x^2 +3x - 2}})} = \\
\lim _{x \to - \infty}{(\frac{-3x + 2}{2x - x\sqrt{4 +3/x - 2/x^2}})} = \\
\lim _{x \to - \infty}{(\frac{-3 + 2/x}{2 - \sqrt{4 +3/x - 2/x^2}})} = \\
\lim _{x \to - \infty}{(\frac{-3 - 0}{2 - \sqrt{4 - 0 + 0}})} = \\
\lim _{x \to - \infty}{(\frac{-3}{0})}
$$
That is not correct. Could anyone please, point where I'm wrong and what should I do?
| $\begin{array}\\
ax-\sqrt{a^2x^2+bx+c}
&=(ax-\sqrt{a^2x^2+bx+c})\dfrac{ax+\sqrt{a^2x^2+bx+c}
}{ax+\sqrt{a^2x^2+bx+c}}\\
&=\dfrac{a^2x^2-(a^2x^2+bx+c)
}{ax+\sqrt{a^2x^2+bx+c}}\\
&=\dfrac{-bx-c}{ax+\sqrt{a^2x^2+bx+c}}\\
&=\dfrac{-b-c/x}{a+\sqrt{a^2+b/x+c/x^2}}\\
&=\dfrac{-b-c/x}{a+a\sqrt{1+b/(a^2x)+c/(a^2x^2)}}\\
&\to\dfrac{-b}{2a}\\
\end{array}
$
When $a=2, b=3$ this is
$-\dfrac34$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating Decimal points of modulo operations I am not too sure of the step that has occurred here (I found this form a cryptography book but I do not understand how he finds the second 9).
The function found in the image is
$(2 \cdot 1)^{-1}(3 \cdot 5^2 + 2) = 2^{-1} \cdot 9 \equiv 9 \cdot 9 \equiv 13\,\mod\,17$
My calculations are that:
$(2 \cdot 1)^{-1}(3 \cdot 5^2 + 2) = 2^{-1} \cdot 9 = 4.5 \not\equiv 9 \cdot 9 \equiv 13\,\mod\,17$
Equation from Book
| In modulo arithimetic then expression $(2^{-1})$ does not mean $0.5$ which is completely out of the purview of modular arithmetic (which is about equivalence classes of integers).
Instead $(2^{-1})$ refers to the the solution to $2x \equiv 1\pmod n$
And $2x \equiv 1 \pmod {17}\iff x \equiv 9 \pmod {17}$[1]
So $2^{-1}\equiv 9 \pmod {17}$.
======
[1] $2x \equiv 1 \pmod {17} \implies \exists k\in \mathbb Z$ so that $2x = 1 + 17k$ so $2|1+17k$ but $2\not \mid 1$ so $2\not \mid 17k$ so $k$ is odd. If $k = 2m + 1$ then $2x = 1 + (2m+1)17 = 18 + 34m$ which means $x = 9 + 17m$ or $x \equiv 9 \pmod {17}$ is the only solution ($\pmod {17}$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the area under a curve when the area is bounded by 3 curves. The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney.
Problem:
Find the area of the region bounded by the given curves and lines.
$$ y = x, y = \frac{1}{ \sqrt{x} }, x = 2 $$
Answer:
Let $A$ be the area we seek. The first step in the process is find where the first two curves
interset.
\begin{align*}
y &= x \\
y &= \frac{1}{ \sqrt{x} } \\
x &= \frac{1}{ \sqrt{x} } \\
x^2 &= \frac{1}{x} \\
x^3 &= 1 \\
x &= \pm 1 \\
\end{align*}
\begin{align*}
A &= \int_1^2 x^{-\frac{1}{2}} \, dx = 2x^{\frac{1}{2}} \Big|_1^2 \\
A &= 2(2)^{\frac{1}{2}} - 2 = 2(\sqrt{2} - 1) \\
\end{align*}
The book's answer is $\frac{7 - 4\sqrt{2}}{2}$. Where did I go wrong?
One thought on what I might have done wrong. It appears to me, the problem only gives three sides to the region
you are trying to find the area of. I did not include a graph in the solution but I did produce a graph on paper.
| You have the bounds correct, but since the area is bounded above by $x$ and below by $\frac{1}{\sqrt{x}}$ the integral should be
$$\int_1^2 (x-\frac{1}{\sqrt{x}})dx $$
$$=(\frac{1}{2}x^2-2\sqrt{x} )|^2_1$$
$$=2 - 2\sqrt{2} - (\frac{1}{2}-2)$$
$$=\frac{7}{2}-2\sqrt{2}$$
| {
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equation simplification. $(5y-1)/3 + 4 =(-8y+4)/6$ Simplification of this equation gives two answers when approched by two different methods.
Method 1 Using L.C.M( least common multiple)
$(5y-1)/3 + 4 =(-8y+4)/6$
$(5y-1+12)/3 = (-8y+4)/6$
$5y-11 = (-8y+4)/2$
$(5y-11)2= (-8y+4)$
$10y-22 = -8y+4$
$18y=26$
$y = 26/18=13/9$
Method 2 multiplying every term by 3
$3(5y-1)/3 + 4*3 = 3(-8y+4)/6$
$5y-1 + 12 = (-8y+4)/2$
$2(5y-1 + 12) = -8y+4$
$10y-2+24 = -8y+4$
$18y + 22 = 4$
$18y = -18$
$y = -1$
The correct method is method 2 and the correct answer is y = -1
Why is method 1 is incorrect? Could anyone explain why the answer is wrong when using the L.C.M( method 1)?
| Well, in the first case, there is a sign error:
$$\frac{5y-1+12}{3} = \frac{-8y+4}{6}$$
$$5y-11 = \frac{-8y+4}{2}$$
It should be
$$5y+11 = \frac{-8y+4}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Choosing two numbers from set $\{10,11...,99\}$ that satisfy the given conditions Given the set of numbers $\{10,11,...,99\}$, with no repetitions and no order significance.
Let $A$ be the set of options of choosing pairs with the same tens number.
Let $B$ be the set of options of choosing only two even numbers.
Let $C$ be the set of options which the difference between $2$ numbers satisfies $-2 \leq x \leq 2$.
How do I calculate the sizes of $A$, $B$, $C$?
For $A$: Ss we know we can choose $2$ numbers from $\{10, \ldots, 99\}$, we have two positions that we need to fill. Therefore, for the first position, we have $90$ possibilities. And after choosing the first number, the second one will only have $9$ options from a group with the same tens number. So that gives us $$\frac{90 \cdot 9}{2}$$
For $B$: In total, we have $45$ even numbers out of $\{10, \ldots, 99\}$.
For the first position, we have $45$ possibilities. And for the second position, we remain with only $44$ even numbers to choose from.
So that gives us $$\frac{45 \cdot 44}{2}$$
For $C$: For each given number chosen from $\{12, \ldots,97\}$, we can pair it with $4$ different numbers that will fulfill the condition (ex. <12, 10\11\13\14> the subtraction of 12 and all those numbers will give as a difference that is $-2 \leq x \leq 2$.)
And for the numbers $11$ and $98$, there are only $3$ numbers to choose from.
And for the numbers $10$ and $99$, there are only $2$ numbers to choose from. In total: $85 \cdot 4 + 2 \cdot 3 + 2 \cdot 2$.
Is this calculation is right?
| In your attempt, you used the Multiplication Principle. Since the order of selection does not matter in the first two parts, I will use combinations.
The number of ways of choosing a subset with $k$ elements from a set with $n$ elements is
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
where $n!$, read "$n$ factorial", is the product of the first $n$ positive integers if $n$ is a positive integer and $0!$ is defined to be $1$.
In how many ways can two elements be selected from the set $S = \{10, 11, 12, \ldots, 99\}$ that have the same tens digit?
There are nine possible choices for the tens digit. For each such choice, there are ten numbers with that tens digit, of which we must choose two. Therefore, the number of ways two elements of $S$ with the same tens digit can be selected is
$$\binom{9}{1}\binom{10}{2} = \frac{9!}{1!8!} \cdot \frac{10!}{2!8!} = \frac{9 \cdot 8!}{1!8!} \cdot \frac{10 \cdot 9 \cdot 8!}{2 \cdot 1 \cdot 8!} = 9 \cdot 45 = 405$$
Notice that this agrees with your answer since
$$\frac{90 \cdot 9}{2} = 45 \cdot 9 = 405$$
In how many ways can two even numbers be selected from the set $S = \{10, 11, 12, \ldots, 99\}$?
The set $S$ contains $99 - 9 = 90$ elements. Since the elements of $S$ are consecutive integers, half of them are even. Hence, set $S$ contains $45$ even numbers. We can select two of those $45$ even numbers in
$$\binom{45}{2} = \frac{45!}{2!43!} = \frac{45 \cdot 44 \cdot 43}{2 \cdot 1 \cdot 43!} = \frac{45 \cdot 44}{2} = 45 \cdot 22 = 990$$
as you found.
In how many ways can two elements be selected from set $S$ such that the difference of the two numbers satisfies $-2 \leq d \leq 2$?
Given the phrasing of the question, I will assume we are selecting ordered pairs so that $(10, 12)$ has difference $-2$ while $(12, 10)$ has difference $2$. I will also assume that we are choosing two different elements of $S$.
If the first number is $10$, the second number must be one of the two numbers $11$, or $12$.
If the first number if $11$, the second number must be one of the three numbers $10$, $12$, or $13$.
If the first number is $m$, where $12 \leq m \leq 97$, there are four possibilities for the second number. They are $m - 2, m - 1, m + 1, m + 2$.
If the first number is $98$, the second number must be one of the three numbers $96$, $97$, or $99$.
If the first number is $99$, the second number must be one of the two numbers $97$ or $98$.
Hence, there are $$2 \cdot 2 + 2 \cdot 3 + 86 \cdot 4 = 4 + 6 + 344 = 354$$
ordered pairs of two different numbers in the set $S$ whose difference has absolute value at most $2$.
The only error you made was counting the integers that satisfy the inequalities $12 \leq m \leq 97$. As indicated in the comments, there are $86$ numbers in the subset $\{12, 13, 14, \ldots, 97\}$ since $97 - 11 = 86$, where we subtract the $11$ positive integers that are not in the subset from the $97$ positive integers that are at most $97$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Having problems with Exponents $1^n =1$ where n is a positive integer I understand.
But $1^{1/2}$ or $1^{0.5}$ is also $1$, I have difficulty in understanding that.
And does that mean $1^n$ where n is any positive real no equals $1$?
$2^{-5}$ = $1 / 2^5$ = $1/32 =0.03125 $
I can compute this but I really don't understand this law.How can you raise something to negative power and if you can how is it obtained by removing the $-$ sign and dividing it by $1$ ?
| $1^N = 1$ for any real value of N.
You are OK with the case of N being integer. This is good start.
The case where $N$ is not integer, as in your example: $1^{0.5}$ (also referred to as $\sqrt{1}$) could be justified as follows:
What is the number $x$ that if you multiply it by itself, you get 1?
This calls for solving the equation $$x^2=1$$ Taking the square root of both sides, we get two numbers that satisfy the equation, namely the number $+1$ and the number $-1$.
That is for the $+1$ case: $$\sqrt{x^2}=\sqrt{1}=1^{0.5}=1$$
I hope that his is somewhat convincing.
As for the negative powers part, think of how you would do $\frac{4^3}{4^5}$, you may want to write it down as where the "." is a multiplication sign:
$$K=\frac{4^3}{4^5} = \frac{4.4.4}{4.4.4.4.4}=\frac{1}{4.4}$$
If you notice the powers of the numerator (3) and the denominator (5) and compare them with the power of the denominator (2), you may observe that $5-3=2$. That is we could write:
$$K=\frac{4^3}{4^5}=\frac{1}{4^{5-3}}=\frac{1}{4^{2}}$$
Suppose now we want to calculate $Z$ by multiplying $K$ by $4^7$:
$$Z=4^{7}\frac{1}{4^{2}}$$
Using the above approach:
$$Z=4^{7}\frac{1}{4^{2}}=\frac{4.4.4.4.4.4.4}{4.4}=4.4.4.4.4=4^5$$
Observe the powers here:
$$Z=4^{7}\frac{1}{4^{2}}=4^5$$
Do you see that $7-2=7+(-2)=5$?
The (-2) here is what you use the negative power for.
That is why we can do:
$$Z=4^{7}.4^{-2}=4^{7-2}=4^5$$
and we can do $$K=\frac{4^3}{4^5}=4^{3-5}=4^{-2}=\frac{1}{4^2}$$
This may be a good read: What do Fractional Exponents mean
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of products of combinatorials In the proof of some proposition, it appears that the following statement should hold:
$$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\beta} =2^{2n}. $$
However, using the definition of combinatorials does not help:
$$ \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} C^{n+1}_r C^n_{\beta} = \sum_{r=1}^{n+1} \sum_{\beta=0}^{r-1} \frac{(n+1)!}{(n+1-r)!r!} \frac{n!}{(n-\beta)! \beta!}.$$
I suspect that this has something to do with the binomial expansion. Any ideas?
| We have
$$\sum_{r=1}^{n+1} {n+1\choose r}
\sum_{\beta=0}^{r-1} {n\choose \beta}
\\ = \sum_{r=1}^{n+1} {n+1\choose r}
\sum_{\beta\ge 0} {n\choose \beta} [[0\le \beta\le r-1]]
\\ = \sum_{r=1}^{n+1} {n+1\choose r}
\sum_{\beta\ge 0} {n\choose \beta}
[z^{r-1}] \frac{z^{\beta}}{1-z}
\\ = \sum_{r=1}^{n+1} {n+1\choose r}
[z^{r-1}] \frac{1}{1-z}
\sum_{\beta\ge 0} {n\choose \beta} z^{\beta}
\\ = \mathrm{Res}_{z=0} \frac{1}{1-z} (1+z)^n
\sum_{r=1}^{n+1} {n+1\choose r}
\frac{1}{z^r}
\\ = \mathrm{Res}_{z=0} \frac{1}{1-z} (1+z)^n
\left(-1 + \left(1+\frac{1}{z}\right)^{n+1}\right)
\\ = \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{1-z} (1+z)^{2n+1}
\\ = \sum_{q=0}^n {2n+1\choose q} = \frac{1}{2} 2^{2n+1}
= 2^{2n}.$$
| {
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"url": "https://math.stackexchange.com/questions/3324706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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If $a+b+c = 4, a^2+b^2+c^2=7, a^3+b^3+c^3=28$ find $a^4+b^4+c^4$ and $a^5+b^5+c^5$ I have tried to solve it but cannot find any approach which would lead me to the answer
| $$a+b+c=4\,\Rightarrow\,(a+b+c)^2=16$$
$$(a^2+b^2+c^2)+2(ab+bc+ac)=16\Rightarrow2(ab+bc+ac)=9$$
Now:
$$(a+b+c)^3=(a^3+b^3+c^3)+3\left[(a+b+c)(ab+ac+bc)-abc\right]$$
$$64=28+3\left[4\times9-abc\right]\therefore\,abc=24$$
Now you need to try and simplify:
$${256=(a^4+b^4+c^4)+6(a^2b^2+a^2c^2+b^2c^2)+4(ab^3+ac^3+ba^3+bc^3+ca^3+cb^3)+12(abc^2+acb^2+bca^2)}$$
By using other methods like the fact that:
$$(a^2+b^2+c^2)^2=(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+a^2c^2)$$
| {
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"url": "https://math.stackexchange.com/questions/3325068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Pair of straight lines parallel vs coincident
The equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of parallel
lines if $\dfrac{a}{h}=\dfrac{h}{b}=\dfrac{g}{f}$ and the distance between the parallel lines is $2\sqrt{\dfrac{g^2-ac}{a(a+b)}}$ or $2\sqrt{\dfrac{f^2-bc}{b(a+b)}}$
This is given in many references and have been in discussion with Deriving conditions for a pair of straight lines to be parallel
and Condition for the pair of lines $^2+2ℎ+^2+2+2+=0$ to be parallel.
First part of the statement is attempted to prove as follows:
Let, $l=Ax+By+C=0,k=Dx+Ey+F=0$ be the pair of lines
$$
lk=0
$$
Taking partial derivative w.r.t $x$ resulting in the line:
$$
L_x=lk'_x+l'_xk=0
$$
Taking partial derivative w.r.t $y$ resulting in the line:
$$
L_y=lk'_y+l'_yk=0
$$
Now, $l=0$ and $k=0\implies L_x=0$ and $L_y=0$ : the point of intersection of $l=0,k=0$ are the same as that of $L_x=0,L_y=0$.
$\implies$ when $l=0,k=0$ are coincident, the resulting lines $L_x=0,L_y=0$ are also coincident. ie, all 4 coincide.
$\implies$$\dfrac{a}{h}=\dfrac{h}{b}=\dfrac{g}{f}$
And when $l=0,k=0$ are parallel I don't think we can say much about $L_x=0,L_y=0$ from the equations. So it seems like the condition $\dfrac{a}{h}=\dfrac{h}{b}=\dfrac{g}{f}$ is for coincident lines ?
But plotting an example, check desmos, thanx @ganeshie8 for the help, it seems like if $l=0,k=0$ parallel you'll still get coincident $L_x=0,L_y=0$.
It'd be very helpful if someone could help me clarify this ?
Thanx @David K for the post in Deriving conditions for a pair of straight lines to be parallel. There it is proved that
The equation $^2+2ℎ+^2+2+2+=0$ is an equation of two parallel lines if and only if $ℎ^2=, =ℎ, =ℎ$, and $^2≥$
But, I think all these conditions do not say whether the lines are coincident or just parallel ?, please correct me if I am wrong ?
And how do I prove the distance between the lines to be $2\sqrt{\dfrac{g^2-ac}{a(a+b)}}$ or $2\sqrt{\dfrac{f^2-bc}{b(a+b)}}$ ?
| I should have read your links and their comments before starting an answer.
That would have saved a big waste of time.
I have posted an alternative proof in an answer to
Deriving conditions for a pair of straight lines to be parallel..
I do not think the approach using partial derivatives in that question is a good one.
The condition $\frac ah=\frac hb=\frac gf$ tells us that the equation
$ax^2+2hxy+by^2+2gx+2fy+c=0$
is either the equation of two parallel lines, the equation of
one line (which could be regarded as "two parallel lines" that are coincident),
or the equation of nothing.
The question of whether $ax^2+2hxy+by^2+2gx+2fy+c=0$ is satisfied by any points at all cannot be answered merely by looking at $a,$ $b,$ $f,$ $g,$ and $h.$
You also have to look at $c.$
If $\frac ah=\frac hb=\frac gf,$ a necessary additional condition to have a solution at all is either $g^2 \geq ac$ or $f^2 \geq bc.$
This could be guessed by looking at the formulas for the distance between the lines, realizing that $a$ and $b$ must have the same sign (because $ab = h$),
and realizing that the expressions inside the square roots are non-negative
only if $g^2 - ac$ or $f^2 - bc$ are non-negative.
An additional condition to have one line instead of two is
either $g^2 = ac$ or $f^2 = bc,$ which puts a zero inside the square root.
Regarding the second part of the question, namely, how we can show that the distance between the parallel lines is $2\sqrt{\frac{g^2-ac}{a(a+b)}}$
or $2\sqrt{\frac{f^2-bc}{b(a+b)}}$:
We suppose in all of the following that $ax^2+2hxy+by^2+2gx+2fy+c=0$ is the equation of one line or two parallel lines.
For simplicity, let's just consider $2\sqrt{\frac{g^2-ac}{a(a+b)}}$ at first.
This formula obviously works only if $a \neq 0.$ Otherwise you would be dividing by zero. So assume $a\neq 0.$ It follows that $h\neq 0.$
Then as I showed in my answer to Deriving conditions for a pair of straight lines to be parallel.,
we can rewrite $ax^2+2hxy+by^2+2gx+2fy+c=0$ as
$$
a(x+By)^2 + 2g(x+By) + c = 0
$$
where $B=\frac ha = \frac bh.$
The two lines are parallel to the line $x+By=0$ and perpendicular to the line $Bx-y=0.$ To find the distance between the lines we can take the distance between their intersection points with that perpendicular line.
To find the intersection points of the two lines with $Bx-y=0,$ we can set $y=Bx$ in $a(x+By)^2 + 2g(x+By) + c = 0,$ which gives us the equation
$$
a(x+B^2x)^2 + 2g(x+B^2x) + c = 0.
$$
Solving this as a quadratic in $x+B^2x,$
$$
x+B^2x = \frac{-g \pm \sqrt{g^2 - ac}}{a}.
$$
Pulling the factor $1+B^2 = 1 + \frac ba$ out of each side, we get
$$
x = \frac{-g \pm \sqrt{g^2 - ac}}{a + b}.
$$
Since we have assumed the equation has a solution,
this gives us $x$ coordinates separated by a horizontal distance
$$\Delta x = 2\frac{\sqrt{g^2 - ac}}{a + b}.$$
Meanwhile, since $y = Bx = \left(\sqrt{\frac ba}\right)x$ at the two intersection points with the perpendicular line,
the $y$ coordinates of these points are separated by a vertical distance
$$\Delta y = 2\left(\sqrt{\frac ba}\right)\frac{\sqrt{g^2 - ac}}{a + b}.$$
The total distance between the two intersection points is therefore
\begin{align}
\sqrt{(\Delta x)^2 + (\Delta y)^2}
&= \sqrt{4\frac{g^2 - ac}{(a + b)^2}
+ 4\left({\frac ba}\right)\frac{g^2 - ac}{(a + b)^2}} \\
&= \sqrt{4\left({1+\frac ba}\right)\frac{g^2 - ac}{(a + b)^2}} \\
&= 2\sqrt{\left(\frac{a+b}{a}\right)\frac{g^2 - ac}{(a + b)^2}} \\
&= 2\sqrt{\frac{g^2 - ac}{a(a + b)}}.
\end{align}
The proof that the distance is $2\sqrt{\frac{f^2-bc}{b(a+b)}}$ when $b\neq 0$
is similar.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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