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Prove by induction that $21 | 4^{n+1} + 5^{2n-1}$ The problem that I have is: Prove by induction that 21 divides 4n+1 + 52n-1
So far I have:
Base Case:
$n = 1$
$4^{1+1} + 5^{2-1} = 4^2 + 5^1 = 16 + 5 = 21$
Inductive Step:
Assume: $4^{k+1} + 5^{2k-1} = 3m$
$4^{(k+1)+1} + 5^{2(k+1)-1} = 3m$
$4^{k+2} + 5^{2k+1} = 3m$
I'm pretty sure I am far off on this one and not sure where to go.
Thanks
| Consider this:
$$4^{n+2} + 5^{2n+1} = 4\cdot 4^{n+1} + (21+4)\cdot 5^{2n-1} = 21\cdot5^{2n-1} + 4\cdot (4^{n+1}+5^{2n-1})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is this sum equal to $0$? While solving a differential equation problem involving power series, I stumbled upon a sum (below) that seemed to be always equal to $0$, for any positive integer $s$.
$$
\sum_{k=0}^s \left( \frac{ \prod_{r=1}^k (-4r^2+10r-3) \prod_{r=1}^{s-k} (-4r^2+6r+1)}{2^s (2k)! (2s-2k+1)!} \times (2s-4k+1) \right)
$$
Why is this sum always equal to $0$?
The simplified version of this equation would be:
$$
\frac{1}{(-2)^s (2s+1)!} \sum_{k=0}^s \left( \binom{2s+1}{2k} (2s-4k+1) \prod_{r=1}^k (4r^2-10r+3) \prod_{r=1}^{s-k} (4r^2-6r-1) \right)
$$
or
$$
\frac{1}{(-2)^s (2s)!} \sum_{k=0}^s \left( \left( \binom{2s}{2k} - \binom{2s}{2k-1} \right) \prod_{r=1}^k (4r^2-10r+3) \prod_{r=1}^{s-k} (4r^2-6r-1) \right)
$$
Update:
$4r^2-10r+3$ and $4r^2-6r-1$ were derived from $n^2-5n+3$ by replacing $n$ with $2r$ and $2r+1$, respectively, which means this could be rewritten as:
$$
\frac{1}{(-2)^s (2s)!} \sum_{k=0}^s \left( \left( \binom{2s}{2k} - \binom{2s}{2k-1} \right) \prod_{r=1}^k a_{2r} \prod_{r=1}^{s-k} a_{2r+1} \right)
$$
where $a_n=n^2-5n+3$
If the sum on the first line is indeed equal to $0$, then this would also be true:
$$
\sum_{k=0}^s \left( \binom{2s}{2k} \prod_{r=1}^k a_{2r} \prod_{r=1}^{s-k} a_{2r+1} \right) = \sum_{k=0}^s \left( \binom{2s}{2k-1} \prod_{r=1}^k a_{2r} \prod_{r=1}^{s-k} a_{2r+1} \right)
$$
Then, I changed the coefficients of $a_n$ into unknown constants to check whether this equality is true in other cases: $a_n=n^2-bn+c$
It seemed that this equality stands when $b$ is an odd integer greater than or equal to $5$ and $s$ is an integer greater than or equal to $(b-3)/2$; when these conditions are met, the value of $c$ does not seem to affect anything.
How could one mathematically derive or prove the above conclusion?
| Let
$$
\alpha=\frac{5+\sqrt{13}}{4},\beta=\frac{5-\sqrt{13}}{4}
$$ be the roots of equation $4r^2-10r+3=0$, and
$$
\gamma=\frac{3+\sqrt{13}}{4},\delta=\frac{3-\sqrt{13}}{4}
$$
be the roots of equation $4r^2-6r-1=0$. Note that $\alpha-\gamma=\beta-\delta=\frac{1}{2}$,$\ \alpha+\beta=\frac{5}{2}$ (this will be used below). Using the Pochhammer symbol
$$
(x)_n=x(x+1) ... (x+n-1)
$$
and the identities
$$
(x)_{n-k}=(-1)^k\frac{(x)_n}{(-x-n+1)_k},
$$
$$
(2k)!=4^kk!\left(\frac{1}{2}\right)_k,\quad (2s-2k+1)!=2^{2s-2k}\left(1\right)_{s-k}\left(\frac{3}{2}\right)_{s-k}
$$
one can rewrite the sum as
$$
A(s)\cdot\sum_{k=0}^s\frac{(1-\alpha)_k(1-\beta)_k(-s)_k\left(-s-\frac{1}{2}\right)_k}{k!\left(\frac{1}{2}\right)_k(\gamma-s)_k(\delta-s)_k}\left(s-2k+\frac{1}{2}\right)
$$
where $A(s)$ is a non-essential constant that had absorbed all prefactors that depend only on $s$ (but not on $k$). Up to a prefactor this is equal to
$$
_7F_6\left({\textstyle a,1+\frac{a}{2},b,c,d,e,-s \atop \textstyle \frac{a}{2},1+a-b,1+a-c,1+a-d,1+a-e,1+a+s}\,;1\right)
$$
where $a=-s-\frac{1}{2}$,$\ b=1-\alpha$,$\ c=1-\beta$,$\ d=e=\frac{1+a}{2}$. One can check that $1+2a=b+c+d+e-s$, so the series can be summed using Dougall's formula
To show that this series under consideration is $0$ it is enough to note that (one can see this from the link above) this sum is proportional to the Pochhammer symbol $(1+a-b-c)_s=(1-s)_s$ which is $0$ for positive $s$.
| {
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"url": "https://math.stackexchange.com/questions/1525361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Evaluate the Integral: $\int\frac{e^{2x}}{e^{2x}+3e^x+2}\mathrm{d}x$ $$\int\frac{e^{2x}}{e^{2x}+3e^x+2}\mathrm{d}x$$
Step one- Factor the denominator
${e^{2x}+3e^x+2}$
$e^x\quad\quad\quad +1$ $=1e^x$
$e^x\quad\quad\quad +2$ $=2e^x$
$\quad\quad\quad\quad\quad\quad\quad3e^x$
$(e^x+1)(e^x+2)$
Step two: Further Decomposition:
$\frac{A}{e^x+1}+\frac{B}{e^x+2}$
$\frac{A(e^x+2)+B(e^x+1)}{(e^x+1)(e^x+2)}$
$e^2x=A(e^x+2)+B(e^x+1)$
I am stuck at this point
The book says I should use u substitution.
Also, how is $e^{2x}=(e^x)^2$
| Set $e^x=t$, we then have that $e^xdx = dt$. This gives us that
$$I=\int \dfrac{e^{2x}}{e^{2x}+3e^x+2} dx = \int \dfrac{t}{t^2+3t+2}dt = \int\dfrac{2dt}{t+2} - \int\dfrac{dt}{t+1}=2\ln(t+2)-\ln(t+1) + c$$
Hence,
$$I = 2\ln(2+e^x) - \ln(1+e^x) + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525751",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $ \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x} $. I need to solve the following integral:
$$
I = \int_{1}^{\infty} \frac{\sqrt{x - 1}}{(x + 1)^{2}} ~ \mathrm{d}{x}.
$$
Wolfram Alpha gives the answer as $ \dfrac{\pi}{2 \sqrt{2}} $.
I think it’s achievable by complex analysis, but I really have no idea how. Also, is there a special name for this integral, i.e., does it have some known physical significance?
| I think I have a different answer.
Let $\frac{x+1}{2} = u$, so that $dx = 2 du$ and $x-1 = 2(u-1)$. Then
$$\begin{align}
\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx
&= \int_1^\infty \frac{\sqrt{2} (u-1)^{\frac{1}{2}}}{4u^2} 2 du \\
&= \frac{1}{\sqrt{2}} \int_1^\infty u^{-2} (u-1)^\frac{1}{2} du
\end{align}$$
Now let $u = \frac{1}{t}$, so that $du = \frac{-1}{t^2} dt$. Continuing.
$$\begin{align}
\phantom{\int_1^\infty \frac{\sqrt{x-1}}{(x+1)^2} dx}
&= \frac{1}{\sqrt{2}} \int_0^1 t^2 (1-t)^\frac{1}{2} \frac{1}{\sqrt{t}} \frac{1}{t^2} dt \\
&= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{-1}{2}} (1-t)^\frac{1}{2} dt \\
&= \frac{1}{\sqrt{2}} \int_0^1 t^{\frac{1}{2}-1} (1-t)^{\frac{3}{2} - 1} dt \\
&= \frac{1}{\sqrt{2}} B ( \frac{1}{2} . \frac{3}{2} ) \\
&= \frac{\Gamma(\frac{1}{2}) \Gamma(\frac{3}{2})} {\sqrt{2}\Gamma(2)} \\
&= \frac{\pi}{2\sqrt{2}}.
\end{align}$$
Original image of work by hand.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The series of reciprocals of the least common multiples of integers $X_1
Let $X_1, X_2,\dots$ be a sequence of strictly increasing positive integers.
For each $n\ge1$, let $W_n$ be the least common multiple of the first $n$ terms $X_1, X_2,\dots, X_n$ of the sequence.
I need to prove the following statement:
The series $1/W_1+1/W_2+\dots+1/W_n\;(n\to\infty)$ is a
convergent series.
I tried several ways, including the hint given below, but I have no luck to overcome this problem. I'd like to learn the methods that can be used to prove that such a series converges.
| $\newcommand{\lcm}{\operatorname{lcm}}$
Note that for $n>1$ the least common multiple $W_n$ of natural numbers $1\leq X_1<X_2<\ldots<X_n$ fulfills
\begin{align*}
\lcm(X_{n-1},X_n) \leq W_n
\end{align*}
The following theorem was originally conjectured by P. Erdös:
Theorem: Let $X_0,\ldots,X_n$ be integers satifying $1\leq X_0<X_1<\ldots<X_n$ then
\begin{align*}
\frac{1}{\lcm(X_0,X_1)}+\frac{1}{\lcm(X_1,X_2)}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\leq 1-\frac{1}{2^n}\tag{1}
\end{align*}
From (1) we obtain by noting that $W_1=\lcm(1,X_1)$
\begin{align*}
\frac{1}{W_1}&+\frac{1}{W_2}+\frac{1}{W_3}+\cdots+\frac{1}{W_n}\\
&\leq \frac{1}{W_1}+\frac{1}{\lcm(X_1,X_2)}+\frac{1}{\lcm(X_2,X_3)}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\\
&\leq 1-\frac{1}{2^{n}}
\end{align*}
and OPs claim follows.
A proof of the theorem was given by D. Borwein in the paper A sum of reciprocals of least common multiples.
Proof (D. Borwein): For $j=1,2,\ldots,n$ let $$S_j=\frac{1}{\lcm(X_0,X_1)}+\cdots+\frac{1}{\lcm(X_{j-1},X_j)}$$
then $\lcm(X_{j-1},X_j)=u_jX_{j-1}=v_jX_{j}$ where $u_j>v_j\geq 1$. Hence
\begin{align*}
\frac{1}{\lcm(X_{j-1},X_j)}\leq \frac{1}{X_j}\tag{2}
\end{align*}
It follows from (2) that
\begin{align*}
\frac{1}{\lcm(X_{j-1},X_j)}\leq(u_j-v_j)\frac{1}{\lcm(X_{j-1},X_j)}=\frac{1}{X_{j-1}}-\frac{1}{X_j}\tag{3}
\end{align*}
We obtain from (3)
\begin{align*}
S_j\leq\frac{1}{X_0}-\frac{1}{X_j}\tag{4}
\end{align*}
Now we show (1) is valid by considering all possible conditions on $X_0,X_1,\ldots,X_n$
Case 1: $X_n\leq 2^n$. It follows from (4),
\begin{align*}
S_n\leq 1 - \frac{1}{X_n}\leq 1-\frac{1}{2^n}
\end{align*}
Case 2: $X_j>2^j$ for $1\leq j \leq n$. We obtain by (2)
\begin{align*}
S_n\leq\frac{1}{X_1}+\frac{1}{X_2}+\cdots+\frac{1}{X_n}<\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^n}=1-\frac{1}{2^n}
\end{align*}
Case 3: $X_k\leq2^k$ for some positive integer $k<n$ and $X_j>2^j$ for $k+1\leq j\leq n$. Then, by (2) and (4)
\begin{align*}
S_n&=S_k+\frac{1}{\lcm(X_{k},X_{k+1})}+\cdots+\frac{1}{\lcm(X_{n-1},X_n)}\\
&<1-\frac{1}{2^k}+\frac{1}{2^{k+1}}+\cdots+\frac{1}{2^n}\\
&=1-\frac{1}{2^n}
\end{align*}
We conclude (1) holds in all cases and the theorem follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the green function for the following BVP by using dirac delta function and solve the BVP using Green function Here's the question:
$y'' + y = x^2 + 1, y(0) = 5, y(1) = 0$
I manage to get a solution to be this:
$y = -6\cot(1)\sin(x) + 6\cos(x) + x^2 - 1 - 2\cos(1)\sin(x) + 2\cot(1)\sin(1)\sin(x)$
Can somebody help me check if my $y_p$ is correct?
Thanks!
| First of all, the last two terms of your solution cancel out since
$\cot(1)\sin(1) = \cos(1)$.
Then, $$y_p = -6\cot(1)\sin(x) + 6\cos(x) + x^2 - 1$$
$$y'_p = -6\cot(1)\cos(x) + 6\sin(x) + 2x$$
$$y''_p = 6\cot(1)\sin(x) - 6\cos(x) + 2$$
Thus, $y''_p + y_p = x^2 +1 $, and the boundary conditions are also satisfied,
$$y(0) = -6\cot(1)\sin(0) + 6\cos(0) + x^2 - 1 = 5$$
$$y(1) = -6\cot(1)\sin(1) + 6\cos(1) + 1^2 - 1 = 0$$
So yes, it works.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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The equation of the plane which passes through the point of intersection of two space lines and at greatest distance from the point $(0,0,0)$ The equation of the plane which passes through the point of intersection of lines $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ and at greatest distance from the point $(0,0,0)$ is
$(A)4x+3y+5z=25\hspace{1cm}(B)4x+3y+5z=50\hspace{1cm}(C)3x+4y+5z=49\hspace{1cm}(D)x+7y-5z=2$
I found the point of intersection of two lines as $(4,3,5)$.Let the equation of the required plane is $ax+by+cz+d=0$.Squared distance of plane $ax+by+cz+d=0$ from $(0,0,0)$ is $\frac{d^2}{a^2+b^2+c^2}$
We need to maximize $\frac{d^2}{a^2+b^2+c^2}$ under the constraint $4a+3b+5c+d=0$.
I cannot solve it further.I am stuck here.Please help me.Thanks.
| An alternative to your approach is to this geometrically. To be the furthest distance then the plane must be perpendicular to the vector $(4,3,5)$. So to find the points of the plane $(x,y,z)$ we have:
$$(4,3,5)\cdot((x,y,z)-(4,3,5))=0$$
$$4x+3y+5x-16-9-25=0$$
$$4x+3y+5x=50$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Evaluating $\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}$ Find the following limit
$$
\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}
$$
I have used natural logarithm to get
$$
\exp\lim_{x\to0}\frac1{x^2}\ln\left(\frac{1+x2^x}{1+x3^x}\right)
$$
After this, I have tried l'opital's rule but I was unable to get it to a simplified form.
How should I proceed from here? Any here is much appreciated!
| Solution Procedure Using Taylor Series and L'Hoptial
Try to understand or prove each of the following steps:
1) $\ln \left( {\frac{{1 + x{2^x}}}{{1 + x{3^x}}}} \right) = \ln \left( {1 + x\frac{{{2^x} - {3^x}}}{{1 + x{3^x}}}} \right)$
2) $\ln (1 + x) = x + o({x^2})$
3) $o\left( {{x^2}{{\left( {{{{2^x} - {3^x}} \over {1 + x{3^x}}}} \right)}^2}} \right) = o\left( {o\left( {{x^2}} \right)} \right) = o({x^2})$
4) $\ln \left( {1 + x{{{2^x} - {3^x}} \over {1 + x{3^x}}}} \right) = x{{{2^x} - {3^x}} \over {1 + x{3^x}}} + o({x^2})$
5) $\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\ln \left( {1 + x{{{2^x} - {3^x}} \over {1 + x{3^x}}}} \right) = \mathop {\lim }\limits_{x \to 0} \frac{1}{x^2} \times x{{{2^x} - {3^x}} \over {\left( {1 + x{3^x}} \right)}}$
6) $\mathop {\lim }\limits_{x \to 0} {{{2^x} - {3^x}} \over {x\left( {1 + x{3^x}} \right)}} = \mathop {\lim }\limits_{x \to 0} {{{2^x}\ln 2 - {3^x}\ln 3} \over {\left( {1 + x{3^x}} \right) + x\left( {1 + x\ln 3} \right){3^x}}} = \ln 2 - \ln 3 = \ln \left( {{2 \over 3}} \right)$
7) $ {\exp\lim_{x\to0}\frac1{x^2}\ln\left(\frac{1+x2^x}{1+x3^x}\right)} = {\exp(\ln{2 \over 3}})= {2 \over 3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531646",
"timestamp": "2023-03-29T00:00:00",
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Solve modular arithmetic equation $\frac{1}{24} \cdot n(n+1)(n+2)(n+3) \equiv 1 \pmod{10}$ From another problem, I have reduced it to: This is the last step in solving:
Solve $\frac{1}{24} \cdot n(n+1)(n+2)(n+3) \equiv 1 \pmod{10}$
How should I begin, a major problem is the $1/24$
| You want to find $n$ such that
$$n(n+1)(n+2)(n+3)\equiv 24\pmod{240}\tag1$$
By the way, $$\begin{align}(1)&\Rightarrow n(n+1)(n+2)(n+3)\equiv 4\pmod{10}\\\\&\Rightarrow n\equiv 1,6\pmod{10}\end{align}$$
So, you only need to check if each of the following 48 numbers satisfies $(1)$ :
$$1,11,21,\cdots, 231$$
$$6,16,26,\cdots, 236$$
Hence, the answer is
$$\small n\equiv 1,11,26,36,41,51,66,76,81,91,106,116,121,131,146,156,161,171,186,196,201,211,226,236\pmod{240}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$ using the half angle identities. Given is the sequence $x_1=0,\; x_{n+1}=\sqrt{2+x_n}$. Prove: $$\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$$
Hint:
Use the following formulas:
$$\cos\left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos x}{2}}$$
$$\sin\left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos x}{2}}$$
Any idea how to solve this problem?
| By mathematical induction it can be proved that $x_n=2\cos(\frac{\pi}{2^n})$.
for base of induction $x_0=0=2\cos(\frac{\pi}{2})$. Also we have $x_{n+1}=\sqrt{2+x_n}=\sqrt{2+2\cos(\frac{\pi}{2^n})}=2\sqrt{\frac{1+\cos(\frac{\pi}{2^n})}{2}}=2\cos(\frac{\pi}{2^{n+1}})$. Therefore $\sqrt{2-x_n}=2\sqrt{\frac{1-\cos(\frac{\pi}{2^n})}{2}}=2\sin(\frac{\pi}{2^{n+1}})$, so $\lim_{n\rightarrow \infty}2^n\sqrt{2-x_n}=\lim_{n\rightarrow \infty}2^{n+1}\sin(\frac{\pi}{2^{n+1}})=\lim_{n\rightarrow \infty}\pi( \frac{\sin(\frac{\pi}{2^{n+1}})}{\frac{\pi}{2^{n+1}}})=\pi$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $f(x)=x^3+3x+2$ and $g(x)$ is the inverse of it, find the area bounded by $g(x)$ and $x$-axis and ordinates $x=-2$ and $x=6$. Is there any way to solve it without finding inverse of the function because finding I can't find inverse?
| Here are the graphs of $f(x) = x^3 + 3 x + 2$ (red) and $g(x)$ (green) and the lines $x = -2$ and $x = 6$.
(Large version)
We see that the area $A$ between $g$ and the $x$-axis for $x \in [-2,6]$ is the same as the area between $f$ and the $y$-axis for $y \in [-2, 6]$.
The complication is the change of sign of $f$ at $x = g(0)$.
We could express $A$ as
\begin{align}
A
= &
A_1 - \left(- \int\limits_{g(-2)}^{g(0)} f(x) \,dx \right) + A_2 +
\int\limits_{g(0)}^{0} f(x) \,dx +
A_3 - \int\limits_0^{g(6)} f(x) \, dx \\
&= A_1 + A_2 + A_3
+ \int\limits_{g(-2)}^0 f(x) \,dx
- \int\limits_0^{g(6)} f(x) \, dx
\end{align}
with $A_1 = (g(0) - g(-2))\,(-(-2))$, $A_2 = (0 - g(0))\,(-(-2))$ and $A_3 = (g(6)-0)\,6$.
Their sum reduces to
$$
A_1 + A_2 + A_3 = -2 \, g(-2) + 6 \, g(6)
$$
So we need to know only two values of $g$ and can avoid the complicated $g(0)$:
\begin{align}
g(-2):& -2 = f(x) = x^3 + 3 x + 2 \iff
x = -1 \\
g(6):& 6 = f(x) = x^3 + 3 x + 2 \iff
4 = x^3 + 3 x \iff
x = 1 \\
\end{align}
This results in:
\begin{align}
A
&= 8 + \int\limits_{-1}^0 f(x) \, dx - \int\limits_0^1 f(x) \, dx \\
&= 8
+ \left[\frac{1}{4} x^4 + \frac{3}{2} x^2 + 2 x \right]_{-1}^0
- \left[\frac{1}{4} x^4 + \frac{3}{2} x^2 + 2 x \right]_0^1 \\
&= 8 - (7/4 - 2) - (7/4 + 2) \\
&= 8 - 14/4 \\
&= 9/2
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/1533604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sum up a function series $f(1/9)+f(2/9)+\dots+f(26/9)$ for $f(x)=\frac{9^{x}}{9^{x}+27}$ Given $f(x)=\dfrac{9^{x}}{9^{x}+27}$.
Find:
$$S=f\left(\frac{1}{9}\right)+f\left(\frac{2}{9}\right)+\dotsb+ f\left(\frac{26}{9}\right)$$
Teacher did not allow us to use calculator...Use sigma notation???
| Observe that $f(x)=\frac{1}{1+\frac{27}{9^x}}$ and $f(3-x)=\frac{1}{1+\frac{9^x}{27}}$. Adding these two expressions gives $f(x)+f(3-x)=1$. (It may be easier to see this by making a substitute such as $a=\frac{27}{9^x}$ and $a^{-1}=\frac{9^x}{27}$). Therefore, $$S=f\left(\frac{1}{9}\right)+f\left(\frac{2}{9}\right)+\dotsb+ f\left(\frac{26}{9}\right)=13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \sum_{k=0}^n k4^k = \frac 49((3n-1)4^n + 1) $ by induction Prove that for every position integer $n$ that
$$ \sum_{k=0}^n k4^k = \frac 49((3n-1)4^n + 1) $$
Proof: Let $P(n)$ denote the above statement.
Base case: $n=1$ : Note that $$ \sum_{k=1}^1 k4^k = \frac 49((3(1)-1)4^{(1)} + 1) $$
$\frac 49((3(1)-1)4^{(1)} + 1) = \frac49((2)4+1) = \frac49(8+1) = \frac 49(9) = 4$
$k4^k = (1)4^{(1)} = 4$
So P(1) holds.
Inductive Step: Let $s\ge1$. Assume P(s), so
$$ \sum_{k=1}^s k4^k = \frac 49((3s-1)4^s + 1) $$
Note
$$ \sum_{k=1}^{s+1} k4^k = \sum_{k=1}^{s} k4^k + (s+1)4^{s+1} $$
and by inductive hypothesis:
**
$$ \frac 49((3s-1)4^s + 1) + (s+1)4^{s+1} $$
**
I'm afraid I'm stuck after this point. I know my endpoint needs to be:
$$ \sum_{k=1}^{s+1} k4^k = \frac 49((3(s+1)-1)4^{s+1} + 1) $$
but I don't know how to get from the asterisks to the above. Any help would be greatly appreciated.
| All you need to do is expand the expression you got after using the inductive hypothesis (in general, if you don't know where to go, at least do something you know you can do!). You get:
$$
\begin{split} \frac{4}{9}((3s-1)4^s+1)+(s+1)4^{s+1}&= \frac{1}{9}(3s-1)4^{s+1}+\frac{4}{9}+s\cdot 4^{s+1}+4^{s+1} \\ &=\frac{1}{3}s\cdot 4^{s+1}-\frac{1}{9}4^{s+1}+\frac{4}{9}+s4^{s+1}+4^{s+1} \\ &=4^{s+1}\left[\frac{s}{3}-\frac{1}{9}+s+1 \right]+\frac{4}{9} \\ &= 4^{s+1} \left[ \frac{3s-1+9s+9}{9}\right]+\frac{4}{9} \\ &= 4^{s+1} \left[\frac{12s+8}{9} \right] \\ &= \frac{4}{9}[(3s+2)4^{s+1}+1] \\ &=\frac{4}{9}[(3(s+1)-1)4^{s+1}+1] \end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Simplify $2 \sqrt[3]{50x^2 z^5} × 3 \sqrt[3]{15 y^3 z}$ Image of problem
As you can see the answer is $30 y z^2 \sqrt[3]{6x^2}$ . I understand mostly everything in the problem, but one thing that I am having confusion on is where the "30" came from in the answer. I tried many ways to find out how the "30" got there.
| You have:
$$6 \sqrt[3]{5^2\cdot 2 \cdot x^2 \cdot z^5 \cdot 3 \cdot 5 \cdot y^3 \cdot z} = 6 \sqrt[3]{5^3\cdot 6 \cdot x^2 \cdot z^6 \cdot y^3} = 30 \cdot y \cdot z^2 \sqrt[3]{ 6 \cdot x^2} $$
I hope its clear now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Double integral over the region enclosed by 4 curves. I'm a bit stuck on a problem that involves trying to integrate the area between 4 curves.
Problem:
$\int \int_D x^2+y^2 dxdy$
Where D is the region enclosed by the curves $xy =2, xy=7, y= 2x^2$ and $y=5x^2$.
For this, I set $u=y/x^2$, $v=xy$ and then work through the problem. In the end, I find that I can't fully rewrite the end integral in terms of $u$ and $v$, so am likely making a mistake somewhere.
Any help in solving this would be much appreciated.
Many thanks.
| With $u=y/x^2$, $v=xy$, we first need to write $x^2+y^2$. Solving for $y$ on both equalities we obtain $x^2u=v/x$, so $x^3=v/u$. Then $x^2=(v/u)^{2/3}$. Squaring the second equality, $v^2=x^2y^2$, so $y/u=v^2/y^2$, or $y^3=v^2/u$. Then
$$
x^2+y^2=\left(\frac{v}{u}\right)^{2/3}+\left(\frac{v^2}{u}\right)^{2/3}.
$$
The Jacobian is
$$
\begin{vmatrix}
x_u&y_u\\ x_v&y_v
\end{vmatrix}
=\begin{vmatrix}
-\dfrac13\,\dfrac{v^{1/3}}{u^{4/3}}&-\dfrac13\,\dfrac{v^{2/3}}{u^{4/3}}\\
\dfrac13\,\dfrac{v^{-2/3}}{u^{1/3}}&\dfrac23\,\dfrac{v^{-1/3}}{u^{1/3}}
\end{vmatrix}
=-\dfrac19\left(\frac{2}{u^{5/3}} -\frac{v^{4/3}}{u^{5/3}}\right)
$$
since $x=(v/u)^{1/3}$, $y=(v^2/u)^{1/3}$. Now you can form the integral
$$
-\frac19\int_2^7\int_2^5\left(\left(\frac{v}{u}\right)^{2/3}+\left(\frac{v^2}{u}\right)^{2/3}\right)\,\left(\frac{2}{u^{5/3}} -\frac{v^{4/3}}{u^{5/3}}\right)\,dv\,du.
$$
This is trivial after distributing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1536901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Real polynomials with $P(x^3-2)=P(x)^3-2$ Find all real polynomials $P(x)$ such that $P(x^3-2)=P(x)^3-2$.
Clearly $P(x)=x$ works. If $P(x)=ax+b$ is linear, then $P(x^3-2)=ax^3-2a+b$ and $P(x)^3-2=a^3x^3+3a^2bx^2+3ab^2x+b^3-2$, so $ab=0$ and $a^3=a$ and $-2a+b=b^3-2$.
If $b=0$, then the last equation implies that $a=1$.
If $a=0$, then the last equation implies that $b^3-b-2=0$, which has a real root and yields a constant solution.
| This is not a complete answer as well but I hope the following result will help. This result guarantees there exist exactly one solution for polynomials of degree which is a power of $3$. If one can prove there exist no solution for polynomials with degree not a power of $3$ then the whole proof can be completed.
Lemma: If $P(x)$ and $Q(x)$ has same degree and $P(x^3-2)=P(x)^3-2$ and $Q(x^3-2)=Q(x)^3-2$ then $P(x)=Q(x)$.
Let $P(x)=\Sigma_0^n a_ix^i$ and let $Q(x)=\Sigma_0^nb_ix^i$.
Then $P(x^3-2)-Q(x^3-2)=P(x)^3-Q(x)^3$.
Base case: consider the coefficient of $x^{3n}$ on both side, left side has $a_n-b_n$ and right side has ${a_n}^3-{b_n}^3$ so $a_n-b_n=(a_n-b_n)({a_n}^2+a_nb_n+{b_n}^2)\implies a_n=b_n$.
Now suppose $a_i=b_i$ for all $i=n,n-1,n-2,...,k$. We look at the case $a_{k-1}$ and $b_{k-1}$.
First all the power of $x$ which is not a multiple of $3$ on the left side is obviously $0$. Now we look at the coefficient of $x^{3k-3}$ on the left side. By binomial theorem expansion it equals ${k-1\choose 0}(a_{k-1}-b_{k-1})+{k\choose 1}(a_k-b_k)+{k+1\choose2}(a_{k+1}-b_{k+1})+...+{n\choose n-k-1}(a_n-b_n)$.
By our induction hypothesis all terms except the first term is $0$ so it is equal to ${k-1\choose 0}(a_{k-1}-b_{k-1})$.
Suppose $a_{k-1}-b_{k-1}\neq 0$ then the left side has degree $3k-3$
Now right side is $P(x)^3-Q(x)^3=(P(x)-Q(x))(P(x)^2+P(x)Q(x)+Q(x)^2)$ has degree $(k-1)+2n=2n+k-1\geq 3k-1>3k-3$ contradiction.
Hence $a_{k-1}=b_{k-1}$ and by induction, we have all $a_i=b_i$ and hence $P(x)=Q(x)$.
| {
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"url": "https://math.stackexchange.com/questions/1538125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to find matrix from general solution?
Find a $2 \times 3$ system (two equations with three unknowns) such that its general solution has form $\begin{pmatrix}1\\1\\0 \end{pmatrix}+s\begin {pmatrix}1\\2\\1\end{pmatrix},\ s \in \Bbb R.$
I tried thinking that $s\begin {pmatrix}1\\2\\1\end{pmatrix}$ is solution to kernel of asked matrix, and tried matrix $\begin {pmatrix}1&-1&1\\1&-1&1\end{pmatrix}$, but it also includes $\begin{pmatrix}1\\1\\0 \end{pmatrix}$ in its kernel!
| One can translate the first two components of $u= (x,y,z)^T= (s+1, 2s+1, s)^T$ into the following system $Au=b$, the third component stays free and acts as $z=s$:
$$
x = s + 1 = z + 1 \iff x - z = 1 \\
y = 2s + 1 = 2 z + 1 \iff y - 2 z = 1 \\
$$
Or written as augmented matrix
$$
(A\mid b) =
\left(
\begin{array}{rrr|r}
1 & 0 & -1 & 1 \\
0 & 1 & -2 & 1 \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Double radical proof I'm trying to prove that
$$
\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
$$
With
$$
C=\sqrt{A^2 - B}
$$
How can I handle this?
Edit: obviously is easy that this holds when you know the r.h.s., but my question is: how to get the r.h.s. when you only know the l.h.s.
| In what follows, I will show how to start from the l.h.s. written as
\begin{equation*}
\sqrt{A+\sqrt{B}}
\end{equation*}
and arrive to the r.h.s written as
\begin{equation*}
\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}},\ \ \ \ with\ \ \ C=\sqrt{A^{2}-B}.
\end{equation*}
To this end, make use of the following standard identities of algebra
\begin{eqnarray*}
x^{2}-y^{2} &=&(x+y)(x-y) \\
(x+y)^{2} &=&x^{2}+y^{2}+2xy
\end{eqnarray*}
Note that this second identity can be written as follows
\begin{equation}
xy=\frac{1}{2}(x+y)^{2}-\frac{1}{2}x^{2}-\frac{1}{2}y^{2} \tag{K}
\end{equation}
Consider a rectangle triangle with the longest side $A$ and the others are $
\sqrt{B,}$ and $C.$ So, by the Pythagorean theorem one has,$\ $
\begin{equation}
A^{2}=B+C^{2}, \tag{P}
\end{equation}
then$\ \sqrt{B}=\sqrt{A^{2}-C^{2}},$ therefore,
\begin{eqnarray*}
\sqrt{A+\sqrt{B}} &=&\sqrt{A+\sqrt{A^{2}-C^{2}}} \\
&=&\sqrt{A+\sqrt{(A+C)(A-C)}} \\
&=&\sqrt{A+\sqrt{A+C}\sqrt{A-C}},\ let\ x=\sqrt{A+C},\ and\ y=\sqrt{A-C},\
and\ apply\ (K) \\
&=&\sqrt{A+\frac{1}{2}\left( \sqrt{A+C}+\sqrt{A-C}\right) ^{2}-\frac{1}{2}%
\left( \sqrt{A+C}\right) ^{2}-\frac{1}{2}\left( \sqrt{A-C}\right) ^{2}} \\
&=&\sqrt{A+\frac{1}{2}\left( \sqrt{2}\sqrt{\frac{A+C}{2}}+\sqrt{2}\sqrt{%
\frac{A-C}{2}}\right) ^{2}-\frac{1}{2}(A+C)-\frac{1}{2}\left( A-C\right) } \\
&=&\sqrt{A+\frac{1}{2}(\sqrt{2})^{2}\left( \sqrt{\frac{A+C}{2}}+\sqrt{\frac{%
A-C}{2}}\right) ^{2}-A} \\
&=&\sqrt{\left( \sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right) ^{2}} \\
\sqrt{A+\sqrt{B}} &=&\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}},\ \ \ with\ C=%
\sqrt{A^{2}-B},\ from\ (P).
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
Proofs of basilar powers identities We all know that a simple and intuitive way to show what $2^n$ is (for $n$ an integer number) is to write it as
$$2^n = \underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}$$
My question is: what is the most intuitive method to show the meaning of these two identities?
$$2^{-n} = \frac{1}{2^n}$$
$$2^{\frac{1}{n}} = \sqrt[n]{2}$$
| By the first relation
$$2^{n}=\underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}=\underbrace{2\times 2\times 2\times \cdots \times 2}_{n-1\ \text{times}}\times2=2^{n-1}\times2.$$
Then $$2^{n-1}=\frac{2^n}2,\\
2^{n-2}=\frac{2^n}{2^2},\\
\cdots\\
2^{n-m}=\frac{2^n}{2^m}.$$
By the first relation
$$2^{nm} = \underbrace{2\times 2\times 2\times \cdots \times 2}_{nm\ \text{times}}=\underbrace{\underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}\times \underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}\cdots \underbrace{2\times 2\times 2\times \cdots \times 2}_{n\ \text{times}}}_{m\text{ times}}\\
=(2^n)^m$$
and
$$\sqrt[m]{2^{nm}}=2^n=2^{nm/m},$$ or
$$\sqrt[m]{2^p}=2^{p/m}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
What's the expression $( \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 ) / ( \cos 5x + 5 \cos 3x + 10 \cos x ) $ equal to? $$\frac{ \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 }{ \cos 5x + 5 \cos 3x + 10 \cos x }$$
My approach so far : Tried to represent the denominator as a factor of numerator by manipulating numerator's $\cos 6x = \cos (5x+x)$ , $\cos 4x = \cos (3x+x)$ , so on .. but then $\sin x$ come up which make it more complex to solve .
The options for the answer are:
A) $\cos 2x$.
B) $2 \cos x$.
C) $\cos^2 x$.
D) $1 + \cos x$
| In a concept similar to the Chebyshev Polynomial,
For a fixed angle $x$, We let $T(n)=\cos nx$. Note the relation of $T(n+1)=2T(1)T(n)-T(n-1)$.
Now we have the numerator as $$T(6)+6T(4)+15T(2)+10=2T(1)T(5)-T(4)+6T(4)+15T(2)+10 = 2T(1)T(5)+5T(4)+15T(2)+10 = 2T(1)T(5)+5(2T(1)T(3)-T(2))+15T(2)+10=2T(1)T(5)+10T(1)T(3)+10T(2)+10=2T(1)T(5)+10T(1)T(3)+20T(1)T(1)-10T(0)+10=2T(1)T(5)+10T(1)T(3)+20T(1)T(1)=2T(1)(T(5)+5T(3)+10T(1))$$
Since the denominator is just $T(5)+5T(3)+10T(1)$, we have the answer as $2T(1)$, or $2 \cos x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the strategy to finding $\sum_{k=2}^{300} k^k \pmod 7$? I am stuck on this modular arithmetic problem for homework practice:
$$\sum_{k=1}^{300} k^k \pmod 7$$
I am not quite sure how to approach this problem. I've tried finding a pattern between the sums but i do not think there is one. I know how to find mods of a^b (mod c) but for this question i am clueless. Any hints and help appreciated.
| It does repeat, but only after $7*6 = 42$ terms (the bases repeat in cycles of 7 while the exponents repeat in cycles of 6). Try breaking it up like this
$$\sum_{k=1}^{43} 1^{1 + 7(k-1)} + \sum_{k=1}^{43} 2^{2+7(k-1)} + \sum_{k=1}^{43} 3^{3+7(k-1)} + \sum_{k=1}^{43} 4^{4+7(k-1)} + \sum_{k=1}^{43} 5^{5+7(k-1)} + \sum_{k=1}^{43} 6^{6+7(k-1)} + \sum_{k=1}^{42} 7^{7+7(k-1)}$$
Now, the exponents can be reduced $\bmod{6}$, since $a^{6} \equiv 1 \bmod{7}$ for every $a \neq 0$. This gives
$$\sum_{k=1}^{43} \left( 1^{1 + (k-1)} + 2^{2+(k-1)} + 3^{3+(k-1)} + 4^{4+(k-1)} + 5^{5+(k-1)} + 6^{(k-1)}\right)$$
or equivalently
$$7\sum_{k=1}^{6} \left(1^{1 + (k-1)} + 2^{2+(k-1)} + 3^{3+(k-1)} + 4^{4+(k-1)} + 5^{5+(k-1)} + 6^{(k-1)}\right) + 1^{1} + 2^{2} + 3^{3} + 4^{4} + 5^{5} + 6^{0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $2f\left(\frac{x}{x^2+x+1}\right)=\frac{x^2}{x^4+x^2+1}$ then what is $f(x)$? assume that:
$$2f\left(\frac{x}{x^2+x+1}\right)=\frac{x^2}{x^4+x^2+1}$$
Then what is $f(x)$?
| Assume that $u=\frac{x}{x^2+x+1}=\frac{1}{1+x+\frac{1}{x}}$. We have $\frac{1}{u}-1=x+\frac{1}{x}$, and by squaring:
$$ \frac{1}{u^2}-\frac{2}{u} = x^2+\frac{1}{x^2}+1,$$
hence:
$$ \frac{u^2}{1-2u} = \frac{1}{1+x^2+\frac{1}{x^2}} = \frac{x^2}{1+x^2+x^4} $$
and:
$$ f(u) = \color{red}{\frac{u^2}{2-4u}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solution to a complex equation I'm asked to solve this equation :
$$z^4+7(1+i)z^2+25i=0$$
by taking $z^2 = u $ :
$$u^2+7(1+i)u+25i=0$$
this turns out to be a quadratic equation. solving it gives:
$$u_1=-6-8i$$
$$u_2=-8-6i$$
so how are we supposed to solve $z^2=-6-8i$ and $z^2=-8-6i$ ?
answeres are :
$$z=\pm \sqrt5(sin\beta/2-icos\beta/2)$$
$$z=\pm \sqrt5(sin\alpha/2-icos\alpha/2)$$
where $\alpha = Arctan(4/3)$ and $\beta = Arctan(3/4)$
I have no idea how to get this answers.
| $$z^4+7(1+i)z^2+25i=0\Longleftrightarrow$$
$$z^4+(7+7i)z^2+25i=0\Longleftrightarrow$$
Substitute $x=z^2$:
$$x^2+(7+7i)x+25i=0\Longleftrightarrow$$
$$(x+(3+4i))(x+(4+3i))=0\Longleftrightarrow$$
$$x+(3+4i)=0\Longleftrightarrow\space\space\vee\space\space x+(4+3i)=0\Longleftrightarrow$$
$$x=-3-4i\Longleftrightarrow\space\space\vee\space\space x=-4-3i\Longleftrightarrow$$
$$z^2=-3-4i\Longleftrightarrow\space\space\vee\space\space z^2=-4-3i\Longleftrightarrow$$
$$z^2=|-3-4i|e^{\arg(-3-4i)i}\Longleftrightarrow\space\space\vee\space\space z^2=|-4-3i|e^{\arg(-4-3i)i}\Longleftrightarrow$$
$$z^2=5e^{\left(\arctan\left(\frac{4}{3}\right)-\pi\right)i}\Longleftrightarrow\space\space\vee\space\space z^2=5e^{\left(\arctan\left(\frac{3}{4}\right)-\pi\right)i}\Longleftrightarrow$$
$$z=\left(5e^{\left(\arctan\left(\frac{4}{3}\right)-\pi+2\pi k\right)i}\right)^{\frac{1}{2}}\Longleftrightarrow\space\space\vee\space\space z=\left(5e^{\left(\arctan\left(\frac{3}{4}\right)-\pi+2\pi l\right)i}\right)^{\frac{1}{2}}\Longleftrightarrow$$
$$z=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{4}{3}\right)-\pi+2\pi k\right)i}\space\space\vee\space\space z=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{3}{4}\right)-\pi+2\pi l\right)i}$$
With $k,l\in\mathbb{Z}$ and $k,l:0-1$
So the solutions are:
$$z_0=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{4}{3}\right)-\pi+2\pi \cdot 0\right)i}=1-2i$$
$$z_1=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{4}{3}\right)-\pi+2\pi \cdot 1\right)i}=-1+2i$$
$$z_2=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{3}{4}\right)-\pi+2\pi \cdot 0\right)i}=\sqrt{-4-3i}$$
$$z_3=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{3}{4}\right)-\pi+2\pi \cdot 1\right)i}=-\sqrt{-4-3i}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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To prove the limit of $\frac{x^2+2\cos x-2}{x\sin^3 x}$ at zero is $1/12$ To Prove $$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{1}{12}$$
I tried with L'Hospital rule but in vain.
| L'Hopital's rule may be used repeatedly until you find an answer:
$$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\lim_{x \to 0}\frac{2x-2\sin x}{\sin^2 x(\sin x+3x\cos x)}$$
$$= \lim_{x \to 0}\frac{2-2\cos x}{ 6\sin ^2(x)\cos (x) - 3x(\sin^3(x)-2\sin(x)\cos^2(x))}$$
$$= \lim_{x \to 0}\frac{2\sin (x)}{ 3(-3\sin^3 (x) + 6\cos^2(x)\sin(x) + x(2\cos^3(x)-7\cos(x)\sin^2(x)))}$$
$$= \lim_{x \to 0}\frac{-2\cos(x)}{ -24\cos^3(x)+84\sin^2(x)\cos(x)+x(60\sin(x)\cos^2(x)-21\sin^3(x))}$$
$$= \frac{-2\cos(0)}{ -24\cos^3(0)+84\sin^2(0)\cos(0)+0(60\sin(0)\cos^2(0)-21\sin^3(0))}$$
$$=\frac{1}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find two solutions of $299x+247y=13$ Applying Euclidean algorithm I got
$299=247+52$
$247=4\cdot 52+39$
$52=39+13$
$39=3\cdot 13+0$
Applying this in reverse order I got
$52-39=13$
$52-(247-4\cdot 52)=13$
$5(299-247)-247=13$
$5\cdot 299-6\cdot 247=13$
$x=5, y=-6$, this is one solution, what is the other solution?
| As $13|(299,247)=13$
from this, we shall have $13$ in-congruent solutions $\pmod{13}$
$299x+247y=13\iff23x+19y=1=23\cdot5-19\cdot6$
$\iff23(x-5)=-19(y+6)$
As $\dfrac{19(y+6)}{23}=5-x$ which is an integer,
$\implies23\mid(y+6)$ as $(19,23)=1$
$\iff y\equiv-6\pmod{23}$
and similarly, $x\equiv5\pmod{19}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Closed form of $\sum\limits_{p = k}^{n - k + 1} C^{k - 1}_{p - 1} C^{k - 1}_{n - p}$ Does anyone has an idea how to prove that $$\sum\limits_{p = k}^{n - k + 1} \binom{p - 1}{k - 1} \binom{n - p}{k - 1} = \frac{n!}{(2 k - 1)! (n - 2 k + 1)!}?$$
| Suppose we seek to evaluate
$$\sum_{p=k}^{n-k+1} {p-1\choose k-1} {n-p\choose k-1}.$$
Introduce
$${n-p\choose k-1} = {n-p\choose n-k+1-p} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+2-p}} (1+z)^{n-p} \; dz.$$
Observe that this controls the range being zero when $p\gt n-k+1$ so
we may extend $p$ to infinity to obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+2}} (1+z)^{n}
\sum_{p\ge k} {p-1\choose k-1} \frac{z^p}{(1+z)^p} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k+2}} (1+z)^{n} \frac{z^k}{(1+z)^k}
\sum_{p\ge 0} {p+k-1\choose k-1} \frac{z^p}{(1+z)^p} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k+2}} (1+z)^{n} \frac{1}{(1+z)^k}
\frac{1}{(1-z/(1+z))^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k+2}} (1+z)^{n}
\frac{1}{(1+z-z)^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k+2}} (1+z)^{n} \; dz
\\ = {n\choose n-2k+1} = {n\choose 2k-1}.$$
This completes the proof.
| {
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"source": "stackexchange",
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Locus of vertice of equilateral triangle. Here is a question:
On two mutually perpendicular lines, two points $A$ and $B$ are taken such that $A$ is a variable point and $B$ is a fixed point. An equilateral triangle $ABC$ is completed. Find the locus of point $C$.
I'm getting a quadratic equation but the back answer is showing a linear equation. Can someone please help me with this question?
| The midpoint between A and B is $$M=(\frac{a}{2},\frac{b}{2})$$ if $A=(a,0)$ and $B=(0,b)$
The direction normal to the line AB is $$\vec{n} = (\frac{b}{\sqrt{a^2+b^2}},\frac{a}{\sqrt{a^2+b^2}})$$
The height of an equilateral triangle of side $\ell = \sqrt{a^2+b^2}$ is $$h = \frac{\sqrt{3}}{2} \sqrt{a^2+b^2}$$
Point C is located a distance $h$ away from the midpoint M along the normal $\vec{n}$
$$ C = M + \vec{n} h = ( \frac{a}{2}, \frac{b}{2} ) + \sqrt{3} ( \frac{b}{2}, \frac{a}{2} ) = ( \frac{a+b\sqrt{3}}{2}, \frac{b+a\sqrt{3}}{2} ) $$
The locus is $C=(x,y)$ which from the above $a=2x - b\sqrt{3}$ and
$$ y = \frac{b+(2x - b\sqrt{3})\sqrt{3}}{2} $$
$$\boxed{y = x\sqrt{3} - b}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the following integral $\int_{-1/2}^{1/2}\big(\frac{\sin(n\pi f)}{\sin(\pi f)}\big)^4 df$ There are similar questions out there, but I was hoping someone could show how to would evaluate the following integral
$$\int_{-1/2}^{1/2}\bigg(\frac{\sin(n\pi f)}{\sin(\pi f)}\bigg)^4 df$$
I've seen the approach of using Euler's formula quite a bit, but I've never been able to wrap my head around it. Most of the answers that I've use that use it simply state it and skip some of the details. Any help?
| Assuming $n \gt 0$. I would rewrite as
$$\frac1{2 \pi} \int_{-\pi}^{\pi} dt \frac{\sin^4{n t}}{\sin^4{t}} $$
We can express this as an integral over the unit circle in the complex plane and use the residue theorem. Thus, the integral is equal to, upon subbing $z=e^{i t}$:
$$\frac1{i 2 \pi} \oint_{|z|=1} \frac{dz}{z} \left (\frac{z^n-z^{-n}}{z-z^{-1}}\right )^4 = \frac1{i 2 \pi} \oint_{|z|=1} dz \frac1{z^{4 n-3}} \left (\frac{z^{2 n}-1}{z^2-1} \right )^4$$
The only pole to speak of is at $z=0$. The residue at this pole is the coefficient of $z^{-1}$ in the Laurent expansion of the integrand. Note that
$$\frac1{z^{4 n-3}} \left (\frac{z^{2 n}-1}{z^2-1} \right )^4 = \frac1{z^{4 n-3}} \left (1+z^2+z^4+\cdots+z^{2 n-2} \right )^4$$
So, really, the residue is simply the coefficient of $z^{4 (n-1)}$ in the expansion of the fourth power of the sum in the parentheses. This is not as hard as it looks. Note that
$$\left (1+z^2+z^4+\cdots+z^{2 n-2} \right )^2 = 1+2 z^2 + 3 z^4 + \cdots + n z^{2 (n-1)}+\cdots+3 z^{4 n-8} + 2 z^{4 n-6} + z^{4 n-4}$$
So, therefore, the coefficient of $z^{4 (n-1)}$ in the square of the above sum is
$$n^2 + 2 \sum_{k=1}^{n-1} k^2 = n^2 + \frac13 (n-1)(n)(2 n-1) = \frac13 n (2 n^2+1)$$
By the residue theorem, the integral we seek is equal to that residue. Thus, for $n \gt 0$,
$$\frac1{2 \pi} \int_{-\pi}^{\pi} dt \frac{\sin^4{n t}}{\sin^4{t}} = \frac13 n (2 n^2+1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How do I simplify $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ )$? $\sqrt{3}(\cot 70^\circ + 4 \cos 70^\circ ) =$ ?
The answer is $3$.
My progress so far:
\begin{align}
\sqrt{3}(\cot 70^\circ+4\cos 70^\circ)&= \sqrt{3}(\tan 20^\circ+4\sin 20^\circ) \\
&= \sqrt{3}(\sin 20^\circ)\left(\frac{1}{\cos 20^\circ}+4\right)\\
&= \sqrt{3}\frac{(\sin 20^\circ)(1+4\cos 20^\circ)}{\cos 20^\circ}\\
&= \sqrt{3}\frac{(\sin 20^\circ+2\sin 40^\circ)}{\cos 20^\circ}\\
\end{align}
| To continue from your method:
$$\sqrt{3}\frac{(sin20+2sin40)}{cos20}=2\left(\frac{\frac{\sqrt{3}}{2}sin20+2\frac{\sqrt{3}}{2}sin40}{cos20}\right)$$ Using $\frac{\sqrt{3}}{2}=cos30$ we get
$$2\left(\frac{\frac{\sqrt{3}}{2}sin20+2\frac{\sqrt{3}}{2}sin40}{cos20}\right)=\left(\frac{2cos30sin20+2 \times 2cos30sin40}{cos20}\right)=\frac{sin50-sin10+2(sin70+sin10)}{cos20}=\frac{sin50+sin10+2sin70}{cos20} $$
But $sin50+sin10=cos20$ and $sin70=cos20$ so
$$\frac{sin50+sin10+2sin70}{cos20}=\frac{cos20+2cos20}{cos20}=3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find out the number of solutions to the equation- $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 25$ under certain constraints How many solutions are there to the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 25$ where each $x_i$ is a non-negative integer, $3 \leq x_1 \leq 10, \ 2 \leq x_2 \leq 7$ and $x_3 \geq 5$
I have been able to do all my counting problems but this one. I can not find an equation for max number per variety.
I know that I will subtract the min for each $x$ from the total, but I can not do $\begin{pmatrix} 15 \\ 6\end{pmatrix}$ cause that would not factor in the max per variety.
What is the equation I would use for this?
| The generating function would be
$$(x^3+x^4+...+x^{10})(x^2+x^3+...+x^7)(x^5+x^6+...)(1+x+x^2+...)^3$$
Simplify,
$$x^{10}(1-x^8)(1-x^6)({1\over 1-x})^6$$
Now we are finding the coefficient of $x^{25}$ of the above, which is same as $x^{15}$ in
$$(1-x^8)(1-x^6)({1\over 1-x})^6=(1-x^8-x^6+x^{14})({1\over 1-x})^6$$
So the coefficient of $x^{15}$ is
(coefficient of $x^{15}$ in $({1\over 1-x})^6$)$-$(coefficient of $x^{7}$ in $({1\over 1-x})^6$)$-$(coefficient of $x^{9}$ in $({1\over 1-x})^6$)$+$(coefficient of $x^{1}$ in $({1\over 1-x})^6$)
Which is basically
$${15+6-1\choose6-1}-{7+6-1\choose6-1}-{9+6-1\choose6-1}+{1+6-1\choose6-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559281",
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"source": "stackexchange",
"question_score": "3",
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$n^{th}$ determinant Find determinant $D_n$ of matrix $$
\begin{bmatrix}
1 & 1 & \cdots & 1 & -n \\
1 & 1 & \cdots & -n & 1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
1 & -n & \cdots & 1 & 1 \\
-n & 1 & \cdots & 1 & 1
\end{bmatrix}
$$
After multiplying first row by $-1$ and adding to $n-1$ rows:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & 0 & \cdots & -n-1 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & -n-1 & \cdots & 0 & n+1 \\
-n & 1 & \cdots & 1 & 1
\end{vmatrix}
$$
Second and $n-1$ row are changing place:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
-n & 1 & \cdots & 1 & 1
\end{vmatrix}
$$
Multiplying first row by $n$ and adding it to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & n+1 & \cdots & n+1 & 1-n^2
\end{vmatrix}
$$
Adding second to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & 0 & \cdots & n+1 & (1+n)(2-n)
\end{vmatrix}
$$
Adding $n-1$ to $n$-th row:
$$
\begin{vmatrix}
1 & 1 & \cdots & 1 & -n \\
0 & -n-1 & \cdots & 0 & n+1 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & -n-1 & n+1 \\
0 & 0 & \cdots & 0 & (1+n)(3-n)
\end{vmatrix}
$$
Product on main diagonal gives $$D_n=(n+1)(3-n)(-n-1)^{n-2}$$
This is not correct. What is wrong with this upper triangular transformation?
| Let $D_m$ be the determinant of the $m\times m$ matrix like this with $-n$ in the antidiagonal and $1$ everywhere else. You are seeking $D_n$.
Note for $k\leq n$, we can find a relation on $D_k$ as follows. Multiply the top row by $n-k+2$, thus scaling the determinant by $n-k+2$.
Then add rows $2$ through $k$ to the first row, which leaves the determinant unchanged, and you have a top row that is mostly $0$ except $-n(n-k+2)+k-1$ in the last entry.
Compute the determinant $D_n$ by expanding across this top row, and you have that $$\begin{align}
D_n&=(-1)^n(-n(n-n+2)+n-1)D_{n-1}/(n-n+2)\\
&=(-1)^n(-n-1)D_{n-1}/2\\
&=(-1)^n(-n-1)(-1)^{n-1}(-n(n-(n-1)+2)+n-1-1)D_{n-2}/[2(n-(n-1)+2)]\\
&=(-1)^{n+(n-1)}(-n-1)(-2n-2)D_{n-2}/[2(3)]\\
&=(-1)^{n+(n-1)}(-n-1)(-2n-2)(-1)^{n-2}(-n(n-(n-2)+2)+n-2-1)D_{n-3}/[2(3)(n-(n-2)+2)]\\
&=(-1)^{n+(n-1)+(n-2)}(-n-1)(-2n-2)(-3n-3)D_{n-3}/[2(3)(4)]\\
&\cdots\\
&=(-1)^{n+(n-1)+\cdots+2}(-n-1)(-2n-2)\cdots(-(n-1)n-(n-1))D_{1}/n!\\
&=(-1)^{n(n+1)/2-1}(n+1)(2n+2)\cdots((n-1)n+(n-1))(-n)/n!\\
&=(-1)^{n(n+1)/2}(n+1)^{n-1}(n-1)!(n)/n!\\
&=(-1)^{n(n+1)/2}(n+1)^{n-1}\\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$... Prove by induction that $3^n +7^n −2$ is divisible by $8$ for all positive integers $n$.
So far I have the base case completed, and believe I am close to completing the proof itself.
Base case:$(n=1)$
$3^1 + 7^1 - 2 = 8/8 = 1 $
Inductive Hypothesis: Assume that $3^n +7^n −2$ is divisible by 8 for all positive integers n.
Induction step $(n+1)$ case:
$$ 3^{n+1} + 7^{n+1} - 2 $$
$$3(3^{n}) + 7(7^{n}) - 2$$
$$3^n + 7^n = 8x $$
-It seems to me that this could be the end of the proof because whatever the answer is would be a multiple of 8: but I am unsure, any help is appreciated.
| We have that $3^n+7^n-2=8k$ for some $k$. Now we substitute this expression into $3(3^n)+7(7^n)-2$ to get:
$3(8k-7^n+2)+7(7^n)-2=24k+4(7^n)+4=24k+4(7^n+1)$ where $7^n+1$ is even so we may rewrite $24k+4(7^n+1)=24k+4(2m)=24k+8m$ which is divisible by $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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find all integers $a,b,c$ such that $a^2=bc+1,$ $b^2=ca+1.$ This was one of the problems in a math contest in India. This is how I tried it:
Subtracting the second equation from the first one gives $a^2-b^2=bc-ca$ or $(a-b)(a+b)=c(b-a)$. $a-b=-(b-a)$. Therefore a+b has to be -c. Thus all integers satisfying the condition $a+b=-c$ are such that $a^2=bc+1$ and $b^2=ca+1$. Also if $a-b=b-a,$ it means $a=b.$ Then both $b-a$ and $a-b$ are $0.$ Hence $c$ can thus be any integer. Hence, there are two groups of integers satisfying the required condition $-a=b$ with any $c$ and $a+b=-c.$
Please let me know whether my solution is correct or not.
| A partial counterexample $a=-3,b=-1,c=-4$ thus $b^2=1$ while ca+1=$-11$ thus i think the condition should be $a=<b=<c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit problem with roots I'm struggling solving the following limit problem:
$$\lim_{x \to \infty} \left(\sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\right)$$
At first I thought I could
Multiply by:
$$\frac{(x^9+x^8)^{\frac{9}{8}} + (x^9-x^8)^{\frac{9}{8}}}{(x^9+x^8)^{\frac{9}{8}} + (x^9-x^8)^{\frac{9}{8}}}$$
But that doesn't seem to take me anywhere closer to an answer.
Some help would be appreciated.
| Alternatively:
\begin{align}
&\sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\\
=&\frac{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}\left(\sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\right)\\
=&\frac{\left({x^9+x^8}\right)-\left({x^9-x^8}\right)}{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}\\
=&\frac{2x^8}{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}\\
\end{align}
Therefore,
\begin{align}
&\lim_{x\to\infty} \sqrt[9]{x^9+x^8} - \sqrt[9]{x^9-x^8}\\
=&\lim_{x\to\infty}\frac{2x^8}{\sum_{k=0}^8\left(\sqrt[9]{x^9+x^8}\right)^{8-k}\left(\sqrt[9]{x^9-x^8}\right)^k}\\
=&\lim_{x\to\infty}\frac{2x^8}{\sum_{k=0}^8x^{8-k}x^k}\\
=&\lim_{x\to\infty}\frac{2x^8}{9\cdot x^8}\\
=&\frac{2}{9}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this sum $\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$ Show that
$$\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$$
| We can compute the generating function:
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k2^{2n-2k}\binom{2n-k+1}{k}x^{2n+1}
&=\frac12\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k2^{2n-2k+1}\binom{2n-k+1}{2n-2k+1}x^{2n+1}\tag{1}\\
&=\frac12\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{k+1}2^{2n-2k+1}\binom{-k-1}{2n-2k+1}x^{2n+1}\tag{2}\\
&=\frac12\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^{k+1}2^{2n-2k+1}\binom{-k-1}{2n-2k+1}x^{2n+1}\tag{3}\\
&=\frac12\sum_{k=0}^\infty\sum_{n=0}^\infty(-1)^{k+1}2^{2n+1}\binom{-k-1}{2n+1}x^{2n+2k+1}\tag{4}\\
&=\frac14\sum_{k=0}^\infty(-1)^k\left(\frac{x^{2k}}{(1-2x)^{k+1}}-\frac{x^{2k}}{(1+2x)^{k+1}}\right)\tag{5}\\
&=\frac14\left(\frac1{1-2x+x^2}-\frac1{1+2x+x^2}\right)\tag{6}
\end{align}
$$
Explanation:
$(1)$: $\binom{n}{k}=\binom{n}{n-k}$
$(2)$: $\binom{-n}{k}=(-1)^k\binom{k+n-1}{k}$
$(3)$: switch the order of summation
$(4)$: substitute $n\mapsto n+k$
$(5)$: the sum in $n$ is the odd part of $(-1)^{k+1}\frac{x^{2k}}{(1+2x)^{k+1}}$
$(6)$: sum the geometric series
$(6)$ is the odd part of $\frac12\frac1{(1-x)^2}=\frac12\sum\limits_{n=0}^\infty(n+1)x^n$; that is,
$$
\frac12\sum_{n=0}^\infty((2n+1)+1)x^{2n+1}=\sum_{n=0}^\infty(n+1)x^{2n+1}\tag{7}
$$
Equating coefficients of $x^{2n+1}$, we get
$$
\sum_{k=0}^n(-1)^k2^{2n-2k}\binom{2n-k+1}{k}=n+1\tag{8}
$$
| {
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"url": "https://math.stackexchange.com/questions/1567183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the difference between the two real numbers that satisfy this equation?
What is the absolute difference between the two real numbers $x$ for which $(x+1)(x-1)(x-2) = (x+2)(x+3)(x-3)$? Express your answer in simplest radical form
I tried guessing solutions but seeing how there are no common zeroes to both the left- and right-hand sides I don't know what to do.
| We have
$$(x^2-1)(x-2) = (x^2-9)(x+2) \implies x^3-2x^2-x+2 = x^3+2x^2-9x-18$$
which simplifies to
$$4x^2-8x-20 = 0 \implies x^2-2x-5 = 0 \implies (x-1)^2 = 6$$
Hence, the roots are $1\pm\sqrt6$, which means that the difference between the roots is $2\sqrt6$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Euclidean Algorithm for Polynomials in $\mathbb F_7$
I have two polynomials $x^4+5x+1$ and $x^2-1$ and I'm trying to find their GCD over $\mathbb F_7$.
My attempt is:
$x^4+5x+1$ = $(x^2+1)(x^2-1)+5x+2$
$x^2-1 =\ ?$
I get lost on the second step...
| $x^4+5x+1=(x^2+1)(x^2-1)+5x-5=(x-1)(x^3+x^2+x+6)$
$x^2-1=(x-1)(x+1)$
$x^3+x^2+x+6$ has no $-1$ as a root
$\gcd(x^4+5x+1,x^2-1)=x-1$
Using the Euclidean algorithm: $x^2-1=(5x+2)(3x+1)+3(x-1)$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate this definite double integral? $$\int\int_D 6x\sqrt{y^2-x^2}dA, D=\{(x,y)|0\leq y\leq 2, 0 \leq x \leq y\}$$
I tried: $$\int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy$$
But that is incorrect.
| Notice, $$\int_{0}^2\int_0^y 6x\sqrt{y^2-x^2} \ dxdy=\int_{0}^2\left(\int_0^y 6x\sqrt{y^2-x^2} \ dx\right)dy $$
$$=\int_{0}^2\left(-3\int_0^y (y^2-x^2)^{1/2} \ d(y^2-x^2)\right)dy $$
$$=\int_{0}^2dy\left(-3\frac{2(y^2-x^2)^{3/2}}{3}\right)_{0}^{y}$$
$$=\int_{0}^2\left(2y^3\right)\ dy$$
$$=\left(2\frac{y^4}{4}\right)_{0}^{2}=\left(\frac{2^4}{2}-0\right)=\color{red}{8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The set of real solutions to an equation $$4^x - 7(2^{\frac{x-3}{2}}) = 2^{-x}$$
Set of real solutions is in which interval:
*
*$(-9, -2)$
*$(0, 3]$
*$(-2, 0]$
*$(7, 12]$
*$(3, 7]$
I tried the following. Dividing by $2^{-x}$ I get $2^{3x} - \frac{7\sqrt(2)}{4}2^{\frac{3x}{2}} - 1 = 0$. Substituing
$$
2^{3x} = c
$$ and solving I get $c_1= 2\sqrt(2)$ and $c_2=-\frac{\sqrt(2)}{4}$ and for $x$ I get $\frac{1}{2}$. That's not even in the given line. What is wrong with the calculations. What have I missed?
| Notice, $$4^x-7\cdot 2^{\frac{x-3}{2}}=2^{-x}$$
$$(2\sqrt2)2^{3x}-7\cdot 2^{\frac{3x}{2}}-2\sqrt2=0$$
let $2^{\frac{3x}{2}}=t$,
$$(2\sqrt 2)t^2-7t-2\sqrt 2=0$$
$$t=\frac{-(-7)\pm\sqrt{(-7)^2-4(2\sqrt 2)(-2\sqrt 2)}}{2(2\sqrt2)}=\frac{7\pm9}{4\sqrt 2}$$ but $t(=2^{\frac{3x}{2}})>0$ hence, one should get
$$2^{\frac{3x}{2}}=\frac{7+9}{4\sqrt 2}=2\sqrt 2=2^{\frac{3}{2}}$$
$$\implies \frac{3x}{2}=\frac{3}{2}$$ $$\color{red}{x=1}$$
| {
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Given a polynomial find the minimum value of the variable. If $x^5 - x^3 + x = a. $
Then we have to find the minimum value of $x^6$ in terms of a.
The answer given is $2a - 1$ if that gives any idea.
I have no idea how to approach this problem.
A hint would do fine.
| I reinterprete the problem a bit differently (based upon the proposed answer) to show that
$$\tag1x^5-x^3+x=a\implies x^6\ge 2a-1. $$
Note that $(1)$ is trivially true for $a\le \frac12$. Hence we may assume that $a$ is positive.
We have $$(x^2+1)a=(x^2+1)(x^5-x^3+x)=x^7+x $$
so that $x$ must be positive.
Divide by $x$ and subtract $1$ to arrive at
$$x^6=\frac{(x^2+1)a}{x}-1\ge\frac{2xa}{x}-1=2a-1, $$
where we used $x>0$ and $x^2+1\ge 2x$ (from $x^2+1-2x=(x-1)^2\ge 0$).
| {
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Evaluating limit by sandwich theorem: $\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$ $$\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$$
For using Sandwich theorem I need two functions such that $g(x)<f(x)<h(x)$
$$\frac{1}{n+n^2} +\frac{2}{n+n^2}+\cdots+\frac{n}{n+n^2} \leq \frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2} $$
But I can't find function greater than the given which will help me evaluate the limit.
| Hint:
$$(1+2+3+\cdots+n)\frac{1}{n^2+n} \leq \frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}\leq (1+2+3+\cdots+n)\frac{1}{n^2+1} $$
Limit is $\frac{1}{2}$
| {
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Suppose $n|a^2-1$ Show that $n=$gcd$(a-1,n)$gcd$(a+1,n)$ Suppose $n|a^2-1$ where $a>1$ and n is odd. Show that $n=$gcd$(a-1,n)$gcd$(a+1,n)$.
Part 2
Show that if $a<n-1$ then this gives a nontrivial factorization of n
What I did:
I found the gcd$(a-1,a+1)$ which is $2$ if $a$ is odd and $1$ if $a$ is even. And that's it. I'm stuck
| Since $n|(a^2-1)$, we have $\gcd(a^2-1, n)=n$.
First assume that $a$ is even. Then $\gcd(a-1, a+1)=1$. Therefore
$$\gcd(a^2-1, n)=\gcd(a-1, n)\gcd(a+1, n)$$
and we are done.
Now suppose $a$ is odd.
Then $\gcd(a^2-1, n)=\gcd((a^2-1)/4, n)$. This is because $n$ is odd.
Since $\gcd((a-1)/2, (a+1)/2)=1$, we get that
$$\gcd((a-1)/2, n)\gcd((a+1)/2, n)=n$$
Again using the fact that $n$ is odd to write $\gcd((a-1)/2, n)=\gcd(a-1, n)$ and $\gcd((a+1)/2, n)=\gcd(a+1, n)$, we are done.
| {
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Generating Functions To Deal With I've been working on producing a closed-form generating function for the
coefficients $a_n = \binom{n}{2}.$ I was wondering what might be a good
procedure to start on this. I get that
$$\sum_{n=0}^{\infty} x^n = x^0 + x^1 + x^2 + ... = \frac{1}{1-x}$$
by the convergence of geometric series. However, I'm wondering what might be
the proper steps to get LHS to look like $\sum_{n=0}^{\infty} \binom{n}{2}x^n.$
Unfortunately, I have a general lack of familiarity with discrete math, so any
help would be greatly appreciated.
| Differentiate term-by-term twice:
$$
\begin{aligned}
\frac{d}{dx} \sum_{n=0}^\infty x^n
&= \sum_{n=1}^\infty n x^{n-1} \\
\frac{d^2}{dx^2} \sum_{n=0}^\infty x^n
&= \sum_{n=2}^\infty n(n-1) x^{n-2}.
\end{aligned}
$$
Next, since $a_n = \binom{n}{2} = \frac{n(n-1)}{2}$ and $a_0 = a_1 = 0$, we have
$$
\begin{aligned}
\sum_{n=0}^\infty a_n x^n
&= a_0 + a_1 x + \sum_{n=2}^\infty \frac{n (n - 1)}{2} x^n \\
&= \frac{x^2}{2} \sum_{n=2}^\infty n (n-1) x^{n-2} \\
&= \frac{x^2}{2} \frac{d^2}{dx^2} \sum_{n=0}^\infty x^n \\
&= \frac{x^2}{2} \frac{d^2}{dx^2} \frac{1}{1-x} \\
&= -\frac{x^2}{\left(1 - x\right)^3}.
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Roots of complex numbers - true/false Here's another one of the problems I'm having trouble with. Below are two statements and the book is asking me to tell for each whether it is true or false:
*
*All the solutions to the equation $\ \ z^3=i\ \ $ are:
$z_1=-i\ ,\ z_2=\frac{\sqrt3}{2}+\frac{i}{2}\ ,\ z_3=-\frac{\sqrt3}{2}+\frac{i}{2}$
*All the solutions to the equation $\ \ z^2=i-\sqrt3\ \ $ are:
$z_1=\sqrt2\left( \cos\frac{\pi}{3}+i\sin\frac{\pi}{3} \right)\ ,\ z_2=\sqrt2\left( \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} \right)$
I know that the three answers listed for the first one are indeed roots of $i$ but I'm not so sure about the second statement.
| $$z^2=i-\sqrt{3}\Longleftrightarrow$$
$$z^2=|i-\sqrt{3}|e^{\arg(i-\sqrt{3})i}\Longleftrightarrow$$
$$z^2=2e^{\frac{5\pi i}{6}}\Longleftrightarrow$$
$$z=\left(2e^{\left(2\pi k+\frac{5\pi}{6}\right)i}\right)^{\frac{1}{2}}\Longleftrightarrow$$
$$z=\sqrt{2}e^{\frac{1}{2}\left(2\pi k+\frac{5\pi}{6}\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-1$
So the solutions are:
$$z_0=\sqrt{2}e^{\frac{1}{2}\left(2\pi\cdot0+\frac{5\pi}{6}\right)i}=\sqrt{2}e^{\frac{5\pi i}{12}}=\sqrt{2}\left(\cos\left(\frac{5\pi}{12}\right)+\sin\left(\frac{5\pi}{12}\right)i\right)$$
$$z_1=\sqrt{2}e^{\frac{1}{2}\left(2\pi\cdot1+\frac{5\pi}{6}\right)i}=\sqrt{2}e^{-\frac{7\pi i}{12}}=\sqrt{2}\left(\cos\left(-\frac{7\pi}{12}\right)+\sin\left(-\frac{7\pi}{12}\right)i\right)$$
Notice with $a,b\in\mathbb{R}$:
$$a+bi=|a+bi|e^{\arg(a+bi)i}=|a+bi|\left(\cos(\arg(a+bi))+\sin(\arg(a+bi))i\right)$$
With $e$ is the base of the natural logarithm.
| {
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I want to calculate the limit of: $\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $ I want to calculate the limit of: $$\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $$
or prove that it does not exist. Now I know the result is $4$, but I am having trouble getting to it. Any ideas would be greatly appreciated.
| Let $f(x) = \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $. For $x \to 0$, We have
$$\begin{align*}
\log f(x) &= \frac{1}{x} \log \left[ \frac{1}{2}\left(e^{x \log 2} + e^{x \log 8}\right) \right] \\
&= \frac{1}{x}\log\left(1 + x \log 4 + o(x)\right) \\
&= \frac{1}{x}\left(x \log 4 + o(x)\right) \\
&= \log 4 + o(1).
\end{align*}$$
Thus $f(x) \to 4$.
| {
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Find the Range of $f(a,b)=2a+b-3ab$ Let $a,b>0$,and such $$a^2+b^2-ab=4$$ Find the range
$$f(a,b)=2a+b-3ab$$
I try let $a=x+y,b=x-y$,then
$$a^2+b^2-ab=4\Longrightarrow x^2+3y^2=16,x>y,x>-y$$
so we Let
$$\begin{align}
x &=4\cos{t}, \qquad y=\dfrac{4}{\sqrt{3}}\sin{t} \\
f(a,b) &= 2a+b-3ab \\
&= \dfrac{3x}{2}+\dfrac{y}{2}-\frac{3}{4}x^2+\dfrac{3}{4}y^2\\
&=6\cos{t}-\dfrac{2}{\sqrt{3}}\sin{t}-12\cos^2{t}+4\sin^2{t}
\end{align}$$
Then I stuck
| I think you have made a mistake in your computations as I did the arithmetic with my CAS. However, if we make the substitutions
$$\begin{align}
a &= 2 \cos t + \frac{2}{\sqrt{3}} \sin t = \frac{4}{\sqrt{3}} \sin(t+\frac{\pi}{3}) \\
b &= 2 \cos t - \frac{2}{\sqrt{3}} \sin t = \frac{4}{\sqrt{3}} \sin(t+\frac{2\pi}{3})
\end{align}$$
then the constraint equation
$$a^2+b^2-ab=4$$
will be satisfied identically $4=4$. Also, to satisfy $a \gt 0$ and $b \gt 0$ we require that
$$\begin{cases}
0 \lt t+\frac{\pi}{3} \lt \pi \\
0 \lt t+\frac{2\pi}{3} \lt \pi
\end{cases}
\to
\begin{cases}
-\frac{\pi}{3} \lt t \lt \frac{2\pi}{3} \\
-\frac{2\pi}{3} \lt t \lt \frac{\pi}{3}
\end{cases}
\to
-\frac{\pi}{3} \lt t \lt \frac{\pi}{3}$$
So, it remains to work on $f(a,b)=2a+b- 3ab$. Now, if we do the substitution in $f(a,b)$ we will get
$$\begin{align}g(t) &= f(a(t),b(t)) \\
&=\frac{2}{\sqrt{3}}\sin t + 6 \cos t - 16 \cos^2 t +4 \\
\end{align}$$
Finally, we can use any method in calculus to find the Maximum and Minimum of this continuous function on the interval $(-\frac{\pi}{3},\frac{\pi}{3})$. So the range of $f(a,b)$ with the desired constraints on $a$ and $b$ will be obtained. In fact, we just reduced a constrained multi-variable optimization problem to a usual single variable optimization problem in calculus.
| {
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Find $\lim\limits_{x\to 0}\frac{e^{x^2}-\cos x^{\sqrt{2}}}{x^2}$
Find $\lim\limits_{x\to 0}\frac{e^{x^2}-\cos x^{\sqrt{2}}}{x^2}$
Using Taylor series:
$$\lim\limits_{x\to 0}\frac{e^{x^2}-\cos x^{\sqrt{2}}}{x^2}=\lim\limits_{x\to 0}\frac{(1+x^2+O(x^{2n}))-(1-\frac{x^{2\sqrt{2}}}{2}+O(x^{\sqrt{2}(2n+1)})}{x^2}=\lim\limits_{x\to 0}\frac{\frac{2x^2+x^{2\sqrt{2}}}{2}}{x^2}=\lim\limits_{x\to 0}\frac{2x^2+x^{2\sqrt{2}}}{2x^2}$$
How to evaluate this limit?
| Starting from where you left off, we divide the numerator and denominator by $x^2$ we obtain $\frac{2+x^{2\sqrt{2}-2}}{2}$. Now $\lim_{x\to 0}{\frac{2+x^{2\sqrt{2}-2}}{2}}=\frac{2+0}{2}=1$
| {
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$S1 = 1 + {x^3 \over 3!} + {x^6 \over 6!} + ...$ In one of my lecturer's problem sheets we were asked to evaluate the following sums:
$$S1 = 1 + {x^3 \over 3!} + {x^6 \over 6!} + \dots $$
$$S2 = {x^1 \over 1!} +{x^4 \over 4!} +{x^7 \over 7!} + \dots$$
$$S3 = {x^2 \over 2!} +{x^5 \over 5!} +{x^8 \over 8!} + \dots$$
In case it's relevant, we were previously required to solve $z^3 -1 =0$, which is simple. And show that if ω is one of the complex roots of the above equation, $ω^2 + ω +1 = 0$.
How would one go about solving these sums? Is there an obvious method which I am missing?
| Hint: We have
$$e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots$$ and
$$e^{\omega x}=1+\omega x+\frac{1}{2!}\omega^2 x^2+\frac{1}{3!}x^3+\cdots$$ and $$e^{\omega^2 x}=1+\omega^2 x+\frac{1}{2!}\omega x^2+\frac{1}{3!}x^3+\cdots.$$
Add. After finding $S_1(x)$ we can get the others by differentiating.
| {
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Calculate $\int_{-\infty}^{\infty}\;\left( \frac{x^2}{1+4x+3x^2-4x^3-2x^4+2x^5+x^6}\right) \;dx$
Calculate $$\displaystyle \int_{-\infty}^{\infty}\;\left(
\frac{x^{2}}{1+4x+3x^{2}-4x^{3}-2x^{4}+2x^{5}+x^{6}}\right) \;dx$$
The answer given is $\pi$. How does one calculate this?
| Let $F(x) = \frac{x^2}{P(x)}$ where
$$P(x) = x^6+2x^5-2x^4-4x^3+3x^2+4x+1 = (x^3+x^2-2x-1)^2 + (x^2+x)^2$$
Change variable to $u = \frac{1}{x+1} \iff x = \frac{1}{u}-1$. The integral at hand becomes
$$\int_{-\infty}^\infty F(x) dx
= \left(\int_{-\infty}^{-1^{-}} + \int_{-1^{+}}^\infty\right) F(x) dx
= \left(\int_{0^{-}}^{-\infty} + \int_{+\infty}^{0^{+}}\right) F\left(\frac{1}{u} - 1\right)\left(-\frac{du}{u^2}\right)\\
= \int_{-\infty}^\infty \frac{1}{u^2} F\left(\frac{1}{u}-1\right) du
$$
By direct substitution, we have
$$\frac{1}{u^2}F\left(\frac{1}{u}-1\right)
= \frac{(u^2-u)^2}{u^6-2u^5-2u^4+4u^3+3u^2-4u+1}
= \frac{(u^2-u)^2}{(u^3-u^2-2u+1)^2+(u^2-u)^2}$$
Notice the function defined by
$$g(u) \stackrel{def}{=} \frac{u^3-u^2-2u+1}{u^2-u} = u - \frac{1}{u}-\frac{1}{u-1}$$
has the form where
Glasser's Master Theorem applies, we get
$$\int_{-\infty}^\infty F(x) dx = \int_{-\infty}^\infty \frac{du}{g(u)^2 + 1}
= \int_{-\infty}^\infty \frac{dx}{x^2+1} = \pi
$$
NOTE
Please note that the statement about Glasser's Master theorem in above link is slightly off. The coefficient $|\alpha|$ in front of $x$ there need to be $1$. Otherwise, there will be an extra scaling factor on RHS of the identity. When in doubt, please consult the original paper by Glasser,
Glasser, M. L. "A Remarkable Property of Definite Integrals." Math. Comput. 40, 561-563, 1983.
and an online copy of that paper can be found here.
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Minimum integer solution of $\ \ \ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{55^2} < \frac{N}{500}$ If $\ \ \ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{55^2} < \frac{N}{500}$
find minimum integer N such that true for this inquality.
or maybe in the other cases for general term of $\ \ \ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{k^2} < \frac{N}{500}$
P.S. I try to bring them all to $RMS \geqslant AM \geqslant GM \geqslant HM $ ,but cannot find the way to solve.
P.S.2 the answer for $\ \ \ \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{55^2} < \frac{N}{500}$ is$\ \ N=314$
thank you so much for every advices and comments
| The RMS-AM-GM-HM inequality is very unlikely to help with this problem as the terms are not close to each other. Here is an alternative approach:
The solution to the Basel problem tells us that $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.
You can bound $\dfrac{1}{n(n+1)} \le \dfrac{1}{n^2} \le \dfrac{1}{(n-1)n}$ for all positive integers $n$.
Therefore, $\displaystyle\sum_{n = 56}^{\infty}\dfrac{1}{n^2} \le \sum_{n = 56}^{\infty}\dfrac{1}{(n-1)n} = \sum_{n = 56}^{\infty}\left(\dfrac{1}{n-1}-\dfrac{1}{n}\right) = \dfrac{1}{55}$.
Similarly, $\displaystyle\sum_{n = 56}^{\infty}\dfrac{1}{n^2} \ge \sum_{n = 56}^{\infty}\dfrac{1}{n(n+1)} = \sum_{n = 56}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) = \dfrac{1}{56}$.
Since $\displaystyle\sum_{n = 1}^{\infty}\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$ and $\dfrac{1}{56} \le \displaystyle\sum_{n = 56}^{\infty}\dfrac{1}{n^2} \le \dfrac{1}{55}$, we have $\dfrac{\pi^2}{6}-\dfrac{1}{55} \le \displaystyle\sum_{n = 1}^{55}\dfrac{1}{n^2} \le \dfrac{\pi^2}{6}-\dfrac{1}{56}$.
Subtract off the $n = 1$ term to get $\dfrac{\pi^2}{6}-1-\dfrac{1}{55} \le \displaystyle\sum_{n = 2}^{55}\dfrac{1}{n^2} \le \dfrac{\pi^2}{6}-1-\dfrac{1}{56}$.
Can you finish the problem from here?
| {
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"url": "https://math.stackexchange.com/questions/1589147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\frac{4\sqrt{7}}{3}\cos{\left(\frac{1}{3}\arccos{\frac{1}{\sqrt{28}}}\right)}+\frac{1}{3}$ If $\dfrac{2\sqrt{19}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{7}{\sqrt{76}}}\right)}-\dfrac{1}{3}$ can be simpified to $2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right)$.
How to simplify $\dfrac{4\sqrt{7}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{1}{\sqrt{28}}}\right)}+\dfrac{1}{3}$ ?
edit :
Now, I have get the answer :
$$\dfrac{4\sqrt{7}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{1}{\sqrt{28}}}\right)}+\dfrac{1}{3}=2\left(\cos{\dfrac{\pi}{7}}+\cos{\dfrac{2\pi}{7}}+\cos{\dfrac{3\pi}{7}}\right)$$
How to prove it?
| Call $\theta=\frac{1}{3}\arccos \frac{1}{\sqrt {28}}$. We can use the formula
$\cos (3\theta)=4\cos^3\theta -3\cos \theta$
to deduce
$$\frac{1}{\sqrt {28}}=4\cos^3\theta -3\cos \theta. \tag{1}$$
The question is asking us to evaluate the quantity
$$ y=\frac{4 \sqrt 7}{3}\cos \theta+\frac{1}{3}$$
with $\cos \theta$ satisfying $(1)$. Applying the substitution $\cos \theta=\frac{3y-1}{4 \sqrt 7}$we get
$$y^3-y^2-9y=-1. \tag{2}$$
The last step is to show that
$$y=2\left(\cos{\dfrac{\pi}{7}}+\cos{\dfrac{2\pi}{7}}+\cos{\dfrac{3\pi}{7}}\right)$$
satisfies $(2)$. This can be done by a straightforward computation, using Werner's formula to deal with the product of cosines. In details we get
$$y^2=2\Big(\cos{\dfrac{2\pi}{7}}+\cos{\dfrac{4\pi}{7}}+2\cos{\dfrac{2\pi}{7}}-\cos{\dfrac{6\pi}{7}}\Big)+6$$
and
$$y^3=y^2\cdot y=24\cos{\frac{\pi}{7}}+24\cos{\frac{2\pi}{7}}+18\cos{\frac{3\pi}{7}}+4\cos{\frac{4\pi}{7}}+2\cos{\frac{5\pi}{7}}+2\cos{\frac{6\pi}{7}}+6.$$
Now we can compute
\begin{align}
y^3-y^2-9y &=2\cos{\frac{\pi}{7}}+4\cos{\frac{2\pi}{7}}+2\cos{\frac{4\pi}{7}}+2\cos{\frac{5\pi}{7}}+4\cos{\frac{6\pi}{7}} \\
&= 2\cos{\frac{2\pi}{7}}-2\cos{\frac{3\pi}{7}}+ 2\cos{\frac{6\pi}{7}}=-1
\end{align}
from which follows
$$y^3-y^2-9y=-1.$$ The last cosines equality is a well known result and it's the only non trivial step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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} |
Derive the generating function: $\frac1{(1-x)^3}= \sum_{n=0}^\infty \binom{n+2}{2}x^n$
$$\frac1{(1-x)^3} = \sum_{n=0}^\infty \binom{n+2}2x^n$$
I am not sure how to begin.
$$\begin{aligned}\frac1{1-x}&= \sum_{n=0}^{\infty} x^n\\\frac1{(1-x)^2}&=\sum_{n=0}^{\infty} nx^{n-1}\end{aligned}$$
Then you would get:
$$\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} nx^{2n - 1}$$
Which is nowhere near the same...?
| Take the derivative of both sides
$$\frac{d}{dx} \left (\frac{1}{(1-x)^2} \right) = \frac{d}{dx} \left ( \sum_{n=0}^{\infty} n x^{n-1} \right)$$
$$-(-2)\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} n (n-1) x^{n-2}$$
$$\frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} \frac{n (n-1)}{2} x^{n-2} = \sum_{n=0}^{\infty} \frac{(n +2) (n+1)}{2} x^{n}.$$
Look at the definition of the binomial coefficient and you'll see that this is the same as the desired formula.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $\left (\frac{1}{a}+1 \right)\left (\frac{1}{b}+1 \right)\left (\frac{1}{c}+1 \right) \geq 64.$
Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$
Attempt
Expanding the LHS we obtain $\left (\dfrac{1+a}{a} \right)\left (\dfrac{1+b}{b} \right)\left (\dfrac{1+c}{c} \right)$. We are given that $a+b+c = 1$, so substituting that in we get $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right)$. Then do I say $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right) \geq 64$ and see if I can get a true statement from this?
| Use the AM GM inequality : $$a+a+b+c\geq 4(a^2bc)^\frac{1}{4}$$ $$a+b+b+c\geq 4(ab^2c)^\frac{1}{4}$$ $$a+b+c+c\geq 4(abc^2)^\frac{1}{4}$$ Multiply the three inequalities and then divide $abc$ on both sides to get the desired inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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simple proof that $\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$ It is well known that for $x>0$ that $\left(1+\frac{1}{x}\right)^x\le e\le\left(1+\frac{1}{x}\right)^{x+1}$ (see wikipedia). However, one can obtain the stronger inequality
$$
\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e\le\sqrt{1+\frac{1}{x}}\left(1+\frac{1}{x}\right)^{x}
$$
The second inequality can be found in Proposition B.3 of "Randomized Algorithms", by Raghaven and Motwani (which itself refers to the book "Analytic Inequalities" by Mitrinović) , and can be proven straight-forwardly by calculus (showing a first derivative is non-negative and such).
While I can also prove the first inequality using familiar calculus methods, it is a bit messy (ultimately requiring that $\frac{1}{y+2}+\frac{1}{3y+2}\ge \frac{1}{y+1}$ for $y\ge 0$).
Does anyone know a "simple" proof of $\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e$?
| I found a proof that seems well-motivated enough to satisfy me. The key lemma is the Hermite-Hadamard inequality (see wikipedia, or Old and new on the Hermite-Hadamard inequality).
Lemma (Hermite-Hadamard): Let $f:[a,b]\to\mathbb{R}$ be convex. Then $$
f\left(\frac{a+b}{2}\right)
\le\frac{1}{b-a}\int_a^b f(x)dx
\le \frac{f(a)+f(b)}{2}.
$$
This can be seen as an instantiation of the trapezoid rule for approximating integrals. The lower bound is Jensen's inequality, the upper bound follows from seeing (via convexity) that the curve $(x,f(x))$ is below the line $(a,f(a))\to(b,f(b))$ on the interval $[a,b]$. That Hermite-Hadamard is useful for the desired inequalities is in retrospect not surprising, as these inequalities can be used to prove Stirling's formula (up to the exact constant), and some proofs of Stirling's formula use Hermite-Hadamard/trapezoid-rule.
I'll first give a weaker result using Hermite-Hadamard which is pretty straightforward, then a proof of the exact result I wanted.
Applying Hermite-Hadamard to $\frac{1}{x}$ on $[x,x+1]$, and simplifying, we get that
$$
\frac{1}{x+\frac{1}{2}}
\le \ln\left(1+\frac{1}{x}\right)
\le \frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right)
$$
Manipulating the lower bound, we see that
$$1\le \left(x+\frac{1}{2}\right)\ln\left(1+\frac{1}{x}\right)$$
which is equivalent to
$$e\le \sqrt{1+\frac{1}{x}}\left(1+\frac{1}{x}\right)^x.$$
Turning to the other direction, we want to show
$$
\sqrt{1+\frac{1}{x+1}}\left(1+\frac{1}{x}\right)^x\le e
$$
Taking logarithms,
$$
\frac{1}{2}\ln\left(1+\frac{1}{x+1}\right)+x\ln\left(1+\frac{1}{x}\right)\le 1
$$
Using that $\ln(1+y)\le y$, and the above upper bound via Hermite-Hadamard, we get
$$
\frac{1}{2}\ln\left(1+\frac{1}{x+1}\right)+x\ln\left(1+\frac{1}{x}\right)
\le \frac{1}{2}\cdot\frac{1}{x+1}+x\cdot \frac{1}{2}\left(\frac{1}{x}+\frac{1}{x+1}\right)
=\frac{1}{2}\left(\frac{x}{x}+\frac{x+1}{x+1}\right)=1
,$$
as desired.
To get the tighter bound, we can apply Hermite-Hadamard to $\ln(x)$ (which is concave, so the inequality is reversed) on $[x,x+1]$, so that (after integrating),
$$
\frac{1}{2}(\ln(x+1)+\ln(x))
\le \ln(x+1)+x\ln\left(1+\frac{1}{x}\right)-1
\le \ln\left(x+\frac{1}{2}\right)
$$
Manipulating the upper bound yields
$$
1
\ge x\ln\left(1+\frac{1}{x}\right)+\ln\left(\frac{x+1}{x+\frac{1}{2}}\right)
=x\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{2}\cdot\frac{1}{x+\frac{1}{2}}\right).
$$
Exponentiating, this is
$$
e
\ge \left(1+\frac{1}{2}\cdot \frac{1}{x+\frac{1}{2}}\right)\left(1+\frac{1}{x}\right)^x
$$
Appealing to Bernoulli's inequality, $1+y/2\ge \sqrt{1+y}$
$$
e\ge \sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x,
$$
as desired. (One can also use the lower bound for Hermite-Hadamard on $\ln(x)$ to again get $e\le (1+1/x)^{x+1/2}$.)
| {
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"url": "https://math.stackexchange.com/questions/1592345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Simple question on trigonometry identities of sec and tan Please, I want to know different methods to prove following identity
$$\frac{\tan \theta + \sec\theta - 1}{\tan\theta-\sec\theta + 1}=\frac{1+\sin\theta}{\cos\theta}$$
| $$LHS = \frac{\sin\theta+1-\cos\theta}{\sin\theta - 1+ \cos\theta}$$
$$\Longleftrightarrow\frac{\sin\theta+1-\cos\theta}{\sin\theta - 1+ \cos\theta} = \frac{1+\sin\theta}{\cos\theta}$$
$$\Longleftrightarrow(\sin\theta+1-\cos\theta)\cos\theta = (\sin\theta - 1+ \cos\theta)(1+\sin\theta)$$
$$\Longleftrightarrow\sin\theta\cos\theta+\cos\theta-\cos^2\theta = \sin\theta - 1 + \cos\theta + \sin^2\theta-\sin\theta+\cos\theta\sin\theta$$
$$\Longleftrightarrow-\cos^2\theta = -1 + \sin^2\theta$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$? I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$
| $$\lim _{ x\to -\infty } \left( \frac { x\left( \sqrt { x^{ 2 }-x } -\sqrt { x^{ 2 }-1 } \right) \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 } }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 } } \right) =\\ =\lim _{ x\to -\infty } \left( \frac { x\left( x^{ 2 }-x-x^{ 2 }+1 \right) }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 } } \right) =\lim _{ x\to -\infty } \left( \frac { x\left( 1-x \right) }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 } } \right) =\\ \\ =\lim _{ x\to -\infty } \frac { x\left( 1-x \right) }{ \left| x \right| \left( \sqrt { 1-\frac { 1 }{ x } } +\sqrt { 1-\frac { 1 }{ { x }^{ 2 } } } \right) } =\\ \lim _{ x\to -\infty } \frac { x-1 }{ \left( \sqrt { 1-\frac { 1 }{ x } } +\sqrt { 1-\frac { 1 }{ { x }^{ 2 } } } \right) } =-\infty $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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For all nonnegative real numbers $x,y$ and $z$, prove that $\dfrac{(x+y+z)^2}{3} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}.$
For all nonnegative real numbers $x,y$ and $z$, prove that $$\frac{(x+y+z)^2}{3} \geq x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}.$$
It seems like AM-GM works here. We have $\dfrac{(x+y+z)^2}{3} \geq \dfrac{(2\sqrt{xy}+z)(2\sqrt{xz}+y)(2\sqrt{xy}+z)}{3}$. Then I get stuck and don't know what to do next.
| Using AM-GM inequality we have:
$\displaystyle \sum_{\text{cyclic}} x\sqrt{yz}\leq \displaystyle \sum_{\text{cyclic}}\dfrac{x(y+z)}{2}=\displaystyle \sum_{\text{cyclic}} xy\leq \dfrac{2\displaystyle \sum_{\text{cyclic}} xy + \displaystyle \sum_{\text{cyclic}} x^2}{3}=\dfrac{\displaystyle \left(\sum_{\text{cyclic}} x\right)^2}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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If $\alpha= \frac{\pi}{7}$ and also if $\frac{3- \tan^2 (\alpha)}{1 - \tan^2 (\alpha)} = \lambda \cos(\alpha)$. Then find $\lambda$. Given that
$$\frac{3- \tan^2 \left(\dfrac{\pi}{7}\right)}{1 - \tan^2 \left(\dfrac{\pi}{7}\right)} = \lambda \cos\left(\dfrac{\pi}{7}\right)$$
Then find the value of $\lambda$.
| Notice, setting the value of $\alpha=\pi/7$, one can simplify LHS as follows $$LHS=\frac{3-\tan^2(\alpha)}{1-\tan^2(\alpha)}$$
$$=\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}$$
$$=\frac{1-\tan^2\frac{\pi}{7}+2}{1-\tan^2\frac{\pi}{7}}$$
$$=1+\frac{2}{1-\tan^2\frac{\pi}{7}}$$
$$=1+\frac{2\tan\frac{\pi}{7}}{\left(1-\tan^2\frac{\pi}{7}\right)\tan\frac{\pi}{7}}$$
$$=1+\left(\frac{2\tan\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}\right)\frac{1}{\tan\frac{\pi}{7}}$$
using $\color{blue}{\frac{2\tan A}{1-\tan^2A}=\tan2A}$,
$$=1+\frac{\tan\frac{2\pi}{7}}{\tan\frac{\pi}{7}}$$
$$=1+\frac{\sin\frac{2\pi}{7}\cos\frac{\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+\sin\frac{2\pi}{7}\cos\frac{\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
using $\color{blue}{\sin A\cos B+\sin B\cos A=\sin(A+B)}$,
$$=\frac{\sin\frac{3\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{\sin\left(\pi-\frac{3\pi}{7}\right)}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{\sin\frac{4\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}}{\sin\frac{\pi}{7}\cos\frac{2\pi}{7}}$$
$$=\frac{2\sin\frac{2\pi}{7}}{\sin\frac{\pi}{7}}$$
$$=\frac{2\cdot 2\sin\frac{\pi}{7}\cos\frac{\pi}{7}}{\sin\frac{\pi}{7}}$$
$$=4\cos \frac{\pi}{7}$$
now, $RHS=\lambda\cos(\alpha)=\lambda\cos\frac{\pi}{7}$,
Now, equating LHS & RHS, we have $$4\cos\frac{\pi}{7}=\lambda\cos\frac{\pi}{7}$$
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\lambda=4}}$$
| {
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Solve $ \int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx $ I don’t know how to solve this integral:
$$\int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx$$
I know this is a rational integral but I don’t know how to write it in a different way
|
$$ \int{\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}}dx$$
Partial fractions:
You need to solve the equation: $$ \color{green}7x^2 +\color{red}0\cdot x +\color{blue}1 = A(x-1)(x+3) + B(x+1)(x+3) + C(x+1)(x-1) $$
$$=A(x^2+2x-3)+B(x^2+4x+3)+C(x^2-1)\\=\color{green}{(A+B+C)}x^2+\color{red}{(2A+4B)}x+\color{blue}{(-3A+3B-C)} $$
To find A, B and C you need to solve an simultaneous equation:
$ \color{green}{A + B + C} = \color{green}7$
$ \color{red}{2A + 4B} = \color{red}0$
$\color{blue}{3A+3B-C} = \color{blue}1$
After solving it you can find out that $ A = -2$, $B=1$ and $ C = 8$
So our integral can be written as:
$$= \int\bigg(-\frac{2}{x+1}+\frac{8}{x+3}+\frac{1}{x-1}\bigg) dx\\ \Longrightarrow=\boxed{\color{teal}{\ln|x-1|-2\ln|x+1|+8\ln|x+3|+C}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Proof of Multiple angle trigonometri ratios If $\tan^2{A}=1+2\tan^2B$ then prove that $\cos2B=1+2\cos2A$
I could not relate from $\tan$ to $\cos$
| Notice, given that $$\tan^2A=1+2\tan^2B$$
$$\tan^2A+1=2+2\tan^2B$$
$$1+\tan^2A=2(1+\tan^2B)$$
$$\frac{1}{1+\tan^2B}=\frac{2}{1+\tan^2A}$$
$$\frac{1-\tan^2 B}{1+\tan^2B}=\frac{2(1-\tan^2 B)}{1+\tan^2A}$$
$$\cos 2B=\frac{2-2\tan^2B}{1+\tan^2A}$$
setting $2\tan^2B=\tan^2A-1, $
$$\cos 2B=\frac{2-\tan^2A+1}{1+\tan^2A}$$
$$\cos 2B=\frac{(1+\tan^2A)+2(1-\tan^2A)}{1+\tan^2A}$$
$$\cos 2B=1+2\frac{(1-\tan^2A)}{1+\tan^2A}$$
$$\cos 2B=1+2\cos 2A$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Positive definite binary quadratic forms Please help me to solve this question or introduce references that help me:
Let $f(x,y) = ax^2 + bxy + cy^2$ be a reduced positive definite form.
Suppose that $\gcd(x, y) = 1$ and that $f(x, y) ≤ a + |b| + c$. Show
that $f(x,y)$ must be one of the numbers $a, c, a - |b| + c$ or $a +
|b| + c$.
| This answer breaks the problem into three cases: $y = 0$, $y = \pm 1$, and $|y| \ge 2$. Recall that because $f$ is reduced and positive definite, $|b| \le a \le c$.
Assume first that $y = 0$. Then because $\gcd(x, y) = \gcd(x, 0) = 1$, $x$ must equal $\pm 1$. In this case, $f(x, y) = f(\pm 1, 0) = a$, which satisfies the condition $f \le a + |b| + c$.
Next, assume that $y = \pm 1$. If $x = 0$, then $f(0, \pm 1) = c$. If $x = \pm 1$, then $f(\pm 1, \pm 1) = a \pm b + c = a \pm |b| + c$. Each of those three representations meets the conditions $\gcd(x, y) = 1$ and $f \le a + |b| + c$. If $|x| \ge 2$, we first derive the following inequality starting with the triangle inequality:
\begin{align}
|2ax + by| & \ge |2ax| - |by|\\
& \ge 4a - b &&\text{because } |x| \ge 2 \text{ and } y = \pm 1\\
& \ge 4a - a &&\text{because } |b| \le a\\
& = 3a
\end{align}
We now find a lower bound for $f(x, y)$:
\begin{align}
f(x, y) & = ax^2 + bxy + cy^2\\
& = \frac{1}{4a} \left( 4a^2x^2 + 4abxy + 4acy^2 \right)\\
& = \frac{1}{4a} \left( 4a^2x^2 + 4abxy + b^2y^2 - b^2y^2 + 4acy^2 \right)\\
& = \frac{1}{4a} \left( (2ax + by)^2 + (4ac - b^2)y^2 \right)\\
& \ge \frac{1}{4a} \left( (3a)^2 + 4ac - b^2 \right) &&\text{because } |2ax + by| \ge 3a \text{ and } y = \pm 1\\
& \ge \frac{1}{4a} \left( 9a^2 + 4ac - a^2 \right) &&\text{because } |b| \le a\\
& = 2a + c\\
& = a + a + c\\
& \ge a + |b| + c &&\text{because } |b| \le a
\end{align}
Due to the upper-bound condition $f \le a + |b| + c$, the only possible representation here is $f(x, \pm 1) = a + |b| + c$.
Finally, assume that $|y| \ge 2$.
\begin{align}
f(x, y) & = \frac{1}{4a} \left( (2ax + by)^2 + (4ac - b^2)y^2 \right) &&\text{as above}\\
& \ge \frac{4ac - b^2}{4a}y^2\\
& \ge \frac{4ac - b^2}{4a} \cdot 4 &&\text{because } |y| \ge 2\\
& = \frac{4ac - b^2}{a}\\
& \ge \frac{4ac - ac}{a} &&\text{because } |b| \le a \le c\\
& = 3c
\end{align}
Then
\begin{align}
3c \le f(x, y) & \le a + |b| + c &&\text{as just shown and using a required condition}\\
& \le 3c &&\text{because } |b| \le a \le c
\end{align}
Because both $f(x, y)$ and $a + |b| + c$ are sandwiched between $3c$ and $3c$, the only possible representation here is $f(x, y) = a + |b| + c$.
Thus, for all pairs of integers $(x, y)$ such that $\gcd(x, y) = 1$ and $f(x, y) \le a + |b| + c$, $f(x, y)$ must be one of the numbers $a$, $c$, $a - |b| + c$, or $a + |b| + c$.
Remark: This is Problem 7 of Section 3.7 and is similar to Lemma 3.24 in I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991. That book is a helpful reference for questions like this.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to factor $x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1$ by hand? I know that
$x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1 = (x^4-x^3+x+1)(x^2+2x-1)$
but I would not know how to do that factoring without a software.
Some idea? Thank you!
| The equation is palindromic (well, almost), so:
We can write it as $$x^3\left[x^3+x^2-3x+2+\frac 3x+\frac{1}{x^2}-\frac{1}{x^3}\right]$$
$$=x^3\left[(x^3-3x+\frac 3x-\frac{1}{x^3})+\left((x-\frac 1x)^2+2\right)+2\right]$$
$$=x^3\left[u^3+u^2+4\right],$$ where $u=x-\frac 1x$
And hence the factorization is $$x^3(u+2)(u^2-u+2)$$ which will give us the expected answer.
| {
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How to solve $\lim\limits_{x \to +\infty} \left(\frac{(\ln(x+1))^x}{(2x)^{x/3}}\right)$? I have a problem with this limit, I don't know what method to use.
Can you show a method for the resolution with asymptotic approximations, or with variable change(so without Hopital)? Thanks
$$\lim\limits_{x \to +\infty} \left(\frac{(\ln(x+1))^x}{(2x)^{x/3}}\right)$$
| Using Taylor expansions:
$$
\ln(x+1) = \ln x + \ln\left(1+\frac{1}{x}\right) = \ln x + \frac{1}{x} + o\left(\frac{1}{x}\right)
$$
and therefore
$$
\ln(x+1)^x = e^{x\ln\left(\ln x + \frac{1}{x} + o\left(\frac{1}{x}\right)\right) }
= e^{x\ln\ln x + x\ln\left(1 + \frac{1}{x\ln x} + o\left(\frac{1}{x\ln x}\right)\right) }
= e^{x\ln\ln x + \frac{1}{\ln x} + o\left(\frac{1}{\ln x}\right) } = e^{x\ln\ln x + o(1)}.
$$
The denominator, however, becomes
$$
(2x)^{x/3} = e^{\frac{x}{3}\ln(2x)} = e^{\frac{1}{3}x\ln x + \frac{\ln 2}{3}x}
$$
so that overall, the quantity considered is, at infinity,
$$
e^{x\ln\ln x - \frac{1}{3}x\ln x - \frac{\ln 2}{3}x + o(1)}.
$$
But the limit of $x\ln\ln x - \frac{1}{3}x\ln x - \frac{\ln 2}{3}x + o(1)$ is $-\infty$ (can you see why? The term that dominates is $\frac{1}{3}x\ln x$), so by continuity of $\exp$ the limit is $0$.
| {
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How to compute $\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)$? I have a problem with this limit, I don't know what method to use.
Can you show a method for the resolution ?
$$\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)$$ Thanks
| Let $\frac{1}{x}=t\implies t\to 0$ as $x\to \infty$, $$\lim_{x\to +\infty}\frac{x^4+x^5\sin\left(\frac 1x\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x}=\lim_{x\to +\infty}\frac{\frac 1x+\sin\left(\frac 1x\right)}{\frac 1x\ln\left(\frac{1}{2+\frac1x}\right)-\frac 1{x^4}}$$
$$=\lim_{t\to 0}\frac{t+\sin t}{-t\ln(2+t)-t^4}$$
$$=\lim_{t\to 0}\frac{1+\frac{\sin t}{t}}{-\ln(2+t)-t^3}$$
$$=\frac{1+1}{-\ln 2-0}=\color{red}{-\frac{2}{\ln 2}}$$
| {
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How to prove this inequality given the equation Given $ a>0 $ and $ a^5-a^3+a=3$, how can I prove the inequality: $ a^6 \ge 5 $ ?
I have tried factorizing the equation, solving for $ x, x^2, x^3 $ and then adding equations made together. Last but not least I tried, to solve for $x$ and then raise it to the power of $ x^6 $, but none of these seem to work
| Given $a^5 - a^3 + a = 3$, we can factorise the left hand side to write $a(a^4-a^2+1)=3$.
The second factor here is reminiscent of cyclotomic polynomials/finite geometric series - in any case, we can write it as $\frac{a^6+1}{a^2+1}$, and clear fractions to get to $a(a^6+1)=3(a^2+1)$.
Now, we spot the $a^6$ we're looking for on the left hand side, and rearrange to see that $a^6 = 3\frac{a^2+1}{a}-1=3(a+\frac{1}{a})-1$.
There are a number of ways of showing $a+\frac{1}{a}\geq2$ for $a>0$ - the easiest is probably differentiating with respect to $a$ to find a minimum at $a=1$, but you can equally use the arithmetic-geometric inequality or a number of other methods. Substituting $a+\frac{1}{a}\geq2$ back into the previous inequality gets us to $a^6\geq3(2)-1=5$ and we're done.
We also could have, upon spotting the $a^6$, subtracted 5 and looked for a reason for it to be positive - in this case, we'd see that $a^6 - 5 = 3(a+\frac{1}{a})-6=3(a+\frac{1}{a}-2)$, and pushing through this a bit takes us to $\frac{3(a^2-2a+1)}{a}=\frac{3(a-1)^2}{a}$.
It's then clear that $a^6=5+\frac{3(a-1)^2}{a}$, which displays fairly immediately that the inequality holds.
| {
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Find the length of latus rectum of the conic $7x^2+12xy-2y^2-2x+4y-7=0$. Find the length of latus rectum of the conic $7x^2+12xy-2y^2-2x+4y-7=0$.
The given conic $7x^2+12xy-2y^2-2x+4y-7=0$ is a hyperbola because when i compare it with $ax^2+2hxy+by^2+2gx+2hy+c=0$ and found $h^2-ab>0$,
Then i tried to change $7x^2+12xy-2y^2-2x+4y-7=0$ into the standard form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ so that length of latus rectum is $\frac{2b^2}{a}$
$7x^2+12xy-2y^2=7x^2-2(-6xy+y^2)=7x^2-2(9x^2-2.3x.y+y^2)+18x^2$
$=25x^2-2(3x-y)^2$
$7x^2+12xy-2y^2-2x+4y-7=25x^2-2(3x-y)^2-2x+4y-7$
Let $x=X,3x-y=Y$ so that $y=3X-Y$
$25x^2-2(3x-y)^2-2x+4y-7=0$ becomes $25X^2-2Y^2-2X+4(3X-Y)-7=0$
$25X^2-2Y^2+10X-4Y-7=0$
$(25X^2+10X+1)-2(Y^2+2Y+1)+2-1-7=0$
$(5X+1)^2-2(Y+1)^2=6$
$\frac{(5X+1)^2}{6}-\frac{(Y+1)^2}{3}=1$
$\frac{(5x+1)^2}{6}-\frac{(3x-y+1)^2}{3}=1$
Now this is in the standard form,So $a^2=6,b^2=3$ and length of latus rectum is $\frac{2b^2}{a}=\frac{6}{\sqrt6}=\sqrt6$
But the answer given is $\sqrt{\frac{48}{5}}$I dont know where i am wrong?
| All of your algebraic work is correct, but you will want a different approach. The problem is in your conclusion, when you change variables back to $x,y$:
$$\frac{(5x+1)^2}{6}-\frac{(3x-y+1)^2}{3}=1$$
Now this is in the standard form, so $\ldots$
This is not in standard form because the numerator of the second fraction has both $x$ and $y$ terms. Furthermore, the change of variables will change the foci, coördinates of the latera recta, and so on, and these coördinates do not scale linearly with the variables.
Instead, try a change of variables that consists of a rotation without scaling:
$$x = X\cos\theta - Y\sin\theta,\quad y=X\sin\theta + Y\cos\theta$$
where
$$\cot2\theta = \frac{A-C}B.$$
You can read more details about this transformation in this excerpt from Stewart. Note that you don't actually need to solve for $\theta$ since we only care about $\cos\theta$ and $\sin\theta$; in your case, you should end up with
$$x = 2\sqrt\frac15X - \sqrt\frac15Y, \quad y = \sqrt\frac15X + 2\sqrt\frac15Y.$$
You should be able to take it from here!
| {
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Prove that $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}} \leq \frac{3}{2}.$
Let $a,b,$ and $c$ be positive real numbers such that $a+b+c = abc$. Prove that $$\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}} \leq \dfrac{3}{2}.$$
This question seems tricky because how do we incorporate the $a+b+c = abc$ condition to solve this? I was thinking of solving for the variables and substituting like $a = \dfrac{b+c}{1-bc}$ etc., but I am not sure how that will help.
| Let $a=\tan(A)$, $b=\tan(B)$ and $c=\tan(C)$. We then have $$\tan(A)+\tan(B)+\tan(C) = \tan(A)\tan(B)\tan(C) \,\,\,\, (\spadesuit)$$
Recall that $(\spadesuit)$ is true iff $A+B+C = \pi$, i.e., $A$, $B$ and $C$ are angles of a triangle. (Infact the part we need is trivial, since we have $\tan(C) = -\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)} = \tan(\pi-(A+B))$.)
Further, since $a,b,c > 0$, we have that $0 \leq A,B,C < \dfrac{\pi}2$. We then have that
$$\dfrac1{\sqrt{1+a^2}} + \dfrac1{\sqrt{1+b^2}} + \dfrac1{{\sqrt{1+c^2}}} = \underbrace{\cos(A) + \cos(B) + \cos(C) \leq 3\cos\left
(\dfrac{A+B+C}3\right)}_{\text{By Jensen}} = \dfrac32$$
| {
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How to find the possible square roots of the two rowed unit matrix I?
How to find the possible square roots of the two rowed unit matrix I ?
I took a matrix like this $$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
and wrote,$A.A=I$
and got a set of equations
$a^2+bc=1$
$ab+bd=0$
$ac+cd=0$
$cb+d^2=1$
After that what to do?
| (Elaborating on Did's comment:) You already got that for
$$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
$A^2 = I$ is equivalent to the four equations
$$
\begin{aligned}
a^2+bc &=1 \\
(a+d)b &=0 \\
(a+d)c &= 0 \\
d^2 + bc &= 1
\end{aligned}
$$
Now distinguish two cases:
Case 1: $a+d \ne 0$. Then $b=c=0$ must hold. It follows that $a^2 = d^2 =1$ and therefore $a = d = \pm 1$. The solutions in this case
are
$$
A = I \text{ or } A = -I \, .
$$
Case 2: $a+d = 0$. Then $a^2 + bc=1$ must hold, i.e. the
solutions are
$$
A=\begin{pmatrix} a & b \\ c & -a \end{pmatrix}
\text{ for any $a, b, c$ such that $a^2 + bc=1$} \, .
$$
| {
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Proving that $\sec\frac{\pi}{11}\sec\frac{2\pi}{11}\sec\frac{3\pi}{11}\sec\frac{4\pi}{11}\sec\frac{5\pi}{11}=32$
If $11 \gamma = \pi$ then prove that
$
\sec(\gamma) \sec(2\gamma) \sec(3\gamma) \sec(4\gamma) \sec(5\gamma) = 32.
$
I could not use the relation $11 \gamma = \pi$.
| $$\sec\frac{\pi}{11}\sec\frac{2\pi}{11}\sec\frac{3\pi}{11}\sec\frac{4\pi}{11}\sec\frac{5\pi}{11}=\frac{1}{\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}}=$$
$$=\frac{1}{\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\left(-\cos\frac{8\pi}{11}\right)\cos\frac{4\pi}{11}\left(-\cos\frac{16\pi}{11}\right)}=\frac{32\sin\frac{\pi}{11}}{32\sin\frac{\pi}{11}\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{4\pi}{11}\cos\frac{8\pi}{11}\cos\frac{16\pi}{11}}=$$
$$=\frac{32\sin\frac{\pi}{11}}{\sin\frac{32\pi}{11}}=\frac{32\sin\frac{\pi}{11}}{\sin\left(3\pi-\frac{\pi}{11}\right)}=32$$
| {
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Solve the congruence $3x^2+x+8\equiv 0 \pmod{11}$ How to find the solutions of this congruence? $$3x^2+x+8\equiv 0 \pmod{11}$$
I need to find the inverse of $3$, and there I have a problem.
| In general, if $p$ is an odd prime and $p\nmid a$, then $$ax^2+bx+c\equiv 0\pmod{p}\iff 4a\left(ax^2+bx+c\right)\equiv 0\pmod{p}$$
$$\iff (2ax+b)^2\equiv b^2-4ac\pmod{p}$$
At this point, you're just left with finding the square roots of $b^2-4ac$ mod $p$.
In your case, $$3x^2+x+8\equiv 0\pmod{11}\iff (6x+1)^2\equiv 4\pmod{11}$$
In this case, the square roots of $4$ mod $p$ are simply $\pm 2$, so:
$$\iff 6x+1\equiv \pm 2\pmod{11}$$
$$\iff x\equiv \left\{\frac{2-1}{6},\frac{-2-1}{6}\right\}\equiv \left\{\frac{1}{6},\frac{-3}{6}\right\}\pmod{11}$$
$$\iff x\equiv \left\{\frac{12}{6}, -\frac{1}{2}\right\}\equiv \left\{2,-\frac{12}{2}\right\}\equiv \{2,-6\}\equiv \{2,5\}\pmod{11}$$
I'll explain about the general solution.
As I said, you're just left with finding the square roots of $b^2-4ac$ mod $p$ (after that it's simple).
So you're left with solving $X^2\equiv A\pmod{p}$ to find $X$,
where $X\equiv 2ax+b\pmod{p}$ and $A\equiv b^2-4ac\pmod{p}$.
A solution exists if and only if $\left(\frac{A}{p}\right)\in\{0,1\}$. This denotes the Legendre Symbol. Finding it might require Quadratic Reciprocity.
If a solution exists, then see this paper and Wikipedia for how to find the solutions.
If $p\equiv 3\pmod{4}$, then $X\equiv \pm A^{\frac{p+1}{4}}\pmod{p}$.
If $p\equiv 5\pmod{8}$, then either $X\equiv \pm A^{\frac{p+3}{8}}\pmod{p}$ or $X\equiv \pm 2^{\frac{p-1}{4}}A^{\frac{p+3}{8}}\pmod{p}$.
If $p\equiv 1\pmod{8}$, then it's complicated. You may need to apply either Cipolla's Algorithm or Tonelli-Shanks algorithm, or see the paper I linked.
| {
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Calculating: $I=\int_{-1}^1{(1+x)^{m+n}P_n(x)}dx$ Where $P_n$ is a Legendre Polynomial. Calculating:
$$I=\int_{-1}^1{(1+x)^{m+n}P_n(x)}dx$$
Where $P_n$ is a Legendre Polynomial.
My progress:
Using Rodrigues formula:
$$I=\dfrac{1}{2^n n!} \int_{-1}^1{(1+x)^{m+n}\large(\dfrac{d}{dx}\large)^n(x^2-1)^n}dx$$
A first interation by parts leads to:
$$I=\dfrac{1}{2^n n!}[(1+x)^{m+n} \large(\dfrac{d}{dx}\large)^{n-1}(x^2-1)^n -\int_{-1}^1{(m+n)(1+x)^{m+n-1}(\dfrac{d}{dx})^{n-1}(x^2-1)^n}dx]$$
The integrated terms vanishes due to $x^2-1$
A second integration with the vanishing terms is:
$$I=\dfrac{1}{2^n n!}[\int_{-1}^1{(m+n)(m+n-1)(1+x)^{m+n-2}(\dfrac{d}{dx})^{n-2}(x^2-1)^n}dx]$$
And here is where i have problems, continuing the process $n$-times to vanish the differntial inside the integrand.
$$I=\dfrac{1}{2^n n!}\int_{-1}^1{\dfrac{(m+n)!}{m!}(x^2-1)^n}dx$$
Can´t assure the coefficient is $\dfrac{(m+n)!}{m!}$
How can i finish the evaluation?
| I can follow your derivation up to the point where
$$ I= \frac{1}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m+n} \frac{d^n}{dx^n} (x^2-1)^n.$$
I assume for the following that $m,n\in \mathbb{N}_0$ (as was probably intended).
Starting from there, your idea is to integrate by parts ($n$-times). Let us observe what happens after the first application ($n>0$ assumed)
$$I = \frac{1}{2^n n!} (1+x)^{m+n} \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n \Bigl|_{x=-1}^1
-\frac{m+n}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m+n-1} \frac{d^{n-1}}{dx^{n-1}} (x^2-1)^n .$$
Following your argument, we want to discard the boundary term. Now it is easy to check that $d^{n-1}(x^2-1)^n/dx^{n-1}=(x^2-1) p(x)$ with $p(x)$ a polynomial. So the boundary term indeed vanishes.
So given $n>0$, we have that
$$I =
-\frac{m+n}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m+n-1} \frac{d^{n-1}}{dx^{n-1}} (x^2-1)^n .$$
We continue with the next application of integration by parts (assuming $n>1$)
$$I = - \frac{m+n}{2^n n!} (1+x)^{m+n-1} \frac{d^{n-2}}{dx^{n-2}} (x^2-1)^n \Bigl|_{x=-1}^1+
\frac{(m+n)(m+n-1)}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m+n-2} \frac{d^{n-2}}{dx^{n-2}} (x^2-1)^n .$$
This time it is important that $d^{n-2}(x^2-1)^n/dx^{n-2}=(x^2-1)^2 \tilde p(x)$ such that the boundary term vanishes once more.
So we are reasonably sure that after the $n$-th application of integration by parts we have that
$$ I = (-1)^n \frac{(m+n)(m+n-1)\cdots (m+1)}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m} (x^2-1)^n\\
= \frac{(-1)^n}{2^n} \binom{m+n}{n}\int_{-1}^1\!dx\,(1+x)^{m} (x^2-1)^n. $$
Of course this result can be proven by induction.
So the remaining task is to evaluate
$$J= \int_{-1}^1\!dx\,(1+x)^{m} (x^2-1)^n.$$
Also this is not completely trivial. However, observe that after substitution $2 y= 1+x$, we obtain
$$J = 2\int_0^1\!dy\, (2y)^m (-4 (1-y) y)^n
= (-1)^n 2^{m+2n+1} \int_0^1\!dy \,y^{m+n} (1-y)^n.$$
The last integral is the definition of the Beta function. Thus we obtain
$$J = (-1)^n 2^{m+2n+1} B(m+n+1, n+1) = (-1)^n 2^{m+2n+1} \frac{(m+n)! n!}{(m+2n +1)!}.$$
In conclusion, we have the result
$$I =\frac{(-1)^n}{2^n} \binom{m+n}{n}(-1)^n 2^{m+2n+1} \frac{(m+n)!\, n!}{(m+2n +1)!} = \frac{2^{m+n+1} (m+n)!^2}{m! \,(m + 2 n+!)!}. $$
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$(x+y)(x+1)(y+1) = 3$ and $x^3 + y^3 = \frac{45}{8}$ I've came across this problem :
If $x$ and $y$ are real numbers such that
$(x+y)(x+1)(y+1) = 3$ and
$x^3 + y^3 = \frac{45}{8}$,
find $xy$.
This is what I've tried so far:
$$x^3 + y^3 = (x+y)(x^2-xy+y^2) =\frac{45}{8}$$
So $$\frac{45}{8(x^2 - xy + y^2)} = \frac{3}{(x+1)(y+1)}$$
$$24x^2 - 24xy + 24y^2 = 45xy + 45(x+y+1)$$
$$24x^2 + 24y^2 - 45(x+y+1) = 69xy$$
But this doesn't seem to help. I tend to think it is just a matter of factorization and/or substitution, but I can't get it right.
A piece of advice would be apreciated.
Thanks!
| See, the whole thing is symmetric, so it would be natural to switch to the symmetric polynomials. Say, $xy=a$ and $x+y=b$. Then $(x+1)(y+1)=xy+x+y+1=a+b+1$, and $x^3+y^3=(x+y)^3-3x^2y-3xy^2=b^3-3ab$.
So $b(a+b+1)=3$ and $b^3-3ab={45\over8}$. From the first equation we have $a={3\over b}-b-1$. Plugging this into the second equation, we get $b^3-3({3\over b}-b-1)b=b^3+3b^2+3b-9=(b+1)^3-10={45\over8}$. Well, the rest is kinda obvious.
| {
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Perfect square is $0$ or $1$ modulo $4$ . Prove that for every integer $n$ either
*
*$n^2 \equiv 0\pmod{4}$ or
*$n^2\equiv 1\pmod{4}$
| $n$ can be even or odd.
When $n=2k$ we get
$(2k)^2=4k^2 \equiv 0 \pmod 4$.
When $n=2k+1$ we get
$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 \equiv 1 \pmod 4$.
| {
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Factorials/Binomial Coefficients (Finding Integer Solutions)
Question
There are many integer solutions to the equation $\begin{pmatrix}n\\r\\ \end{pmatrix} = \begin{pmatrix}n+1\\r-1\\ \end{pmatrix}$ including $n = r = 1$.
Find an expression for $n$ in terms of $r$, and hence find another of the integer solutions.
What I have attempted:
$$\begin{pmatrix}n\\r\\ \end{pmatrix} = \begin{pmatrix}n+1\\r-1\\ \end{pmatrix}$$
$$ {n!\over (n-r)!r!} = {(n+1)!\over (n+1-(r-1))!(r-1)!} $$
$$ {n!\over (n-r)!r!} = {(n+1)!\over (n-r+2)!(r-1)!} $$
$$ {(n-r+2)!\over (n-r)!r!} = {(n+1)!r!\over n!(r-1)!} $$
Most of the factorials cancel so
$$ (n-r+2)(n-r+1) = (n+1)r $$
$$ n^2 - nr + n - nr + r^2 - r + 2n - 2r + 2 = nr + r $$
$$ n^2 - 3nr + 3n + r^2 - 4r + 2 = 0 $$
$$ n^2 + (3-3r)n + (r^2 - 4r + 2) = 0 $$
$$ n = {-b\pm\sqrt{b^2-4ac} \over 2a} $$
$$ n = {3r-3\pm\sqrt{5r^2 - 2r + 1} \over 2} $$
Now I am trying to find another set of integer solutions how do I do that? (I want to solve it algebraically not guess and check)
| Notice that the last equation implies that you have to find another solution to $5r^2-2r+1=z^2$, where $r,z$ are both naturals.
Multiplying by 5 gives us $25r^2-10r+5=5z^2$. $\therefore (5r-1)^2+4=5z^2$.
Since you are merely trying to find another set, say
$5r-1=2k,z=2m$. The question becomes, $k^2-5m^2=-1$.
However, note that the solution you found, $(k,m)=(2,1)$, implies that $(2+\sqrt {5})(2- \sqrt{5})=-1$. Then this implies for any odd n,
$(2+\sqrt {5})^n(2- \sqrt{5})^n=-1$. If you put $n=5$,
$(2+\sqrt {5})^5(2- \sqrt{5})^5=(682+305\sqrt {5})(682-305\sqrt{5})=682^2-5*305^2=-1$
Therfore, $5r-1=1364, z=610$. Therefore $r=273, z=610$. Therefore $n=713$.
Another set of $(n,r)=(713,273)$. This is possibly the most algebraic way to do it. However, $(14,6)$ is also a set. This is largely because of the assumption that $5r-1$ is a even number. Further information can be found here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $6^{1000} \mod 23$
Find $6^{1000} \mod 23 $
Having just studied Fermat's theorem I've applied $6^{22}\equiv 1 \mod 23 $, but now I am quite clueless on the best way to proceed.
This is what I've tried:
Raising everything to the $4th$ power I have $$6^{88} \equiv 1 \mod 23 $$ $$6^{100} \equiv 6^{12} \mod 23 $$ $$6^{1000}\equiv 6^{120} \mod 23 $$ $$6^{1000} \equiv 6^{10} \mod 23$$ How do I simplify now the right hand side of the congruence ?
| We may exploit the fact that $1000\equiv 10\pmod{22}$ to get:
$$ 6^{1000}\equiv 6^{10} \equiv (6^5)^2 \equiv 2^2 \equiv\color{red}{4}\pmod{23}.$$
As an alternative, we may notice that $a^{11}\pmod{23}$ is the Legendre symbol $\left(\frac{a}{23}\right)$, so:
$$ 6^{11} \equiv \left(\frac{2}{23}\right)\cdot\left(\frac{3}{23}\right) \equiv 1\cdot 1\equiv 1\pmod{23} $$
gives $6^{1001}\equiv 1\pmod{23}$ and by multiplying both sides by the inverse of $6$ we get $4$ as above.
| {
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Find all complex roots of $T^4-{1/2}T^2-\sqrt{15}T+{69/16}$ I want to find all complex roots of $T^4-{1/2}T^2-\sqrt{15}T+{69/16}$.
The only way I can think to do it is to find 1 complex root, $\alpha$, by inspection, so we can rearrange the polynomial to be $(T-\alpha)(T^3+pT^2+qt+r) $then show the cubic equation is irreducible by Eisenstein's criterion. But I do not think this is going to work, so any ideas would be greatly appreciated
| It is better, I think, to try with $x^4-2x^2-8\sqrt{15}\space x+69$ where $x=2T$.
By some tedious calculation of undetermined coefficients we have
$x^4-2x^2-8\sqrt{15}\space x+69=[x^2-2\sqrt {5}\space x +(9-2\sqrt 3)]\cdot[x^2+2\sqrt {5}\space x +(9+2\sqrt 3)]$
Hence $x_{1,2}=\sqrt 5\pm\sqrt{4-2\sqrt 3}\space i$ and $x_{3,4}=-\sqrt 5\pm \sqrt{4+2\sqrt 3}\space i$
Thus $$\color{red}{\begin{cases}2T_{1,2}=\sqrt 5\pm\sqrt{4-2\sqrt 3}\space i\\ 2T_{3,4}=-\sqrt 5\pm \sqrt{4+2\sqrt 3}\space i\end{cases}}$$
The verification can be made by
$[(x-\sqrt 5)^2+(\sqrt 3-1)^2][ x+\sqrt 5)^2+(\sqrt 3+1)^2]=x^4-2x^2-8\sqrt{15}\space x+69$.
| {
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Prove that for all positive real numbers $a,b,$ and $c$ we have $a^5+b^5+c^5 \geq a^3bc+ab^3c+abc^3$
Prove that for all positive real numbers $a,b,$ and $c$ we have $a^5+b^5+c^5 \geq a^3bc+ab^3c+abc^3$.
This question reminds me of rearrangement, but I can't really find two sequences that fit. Maybe there is a way using the triangle inequality, but I am unsure.
| Since $f(a,b,c)=LHS-RHS$ is homogeneous, we can asumme $abc=1$. Thus, inequalitie is equivalent to $a^5+b^5+c^5\ge a^2+b^2+c^2$ subject to $abc=1$.
Now, if $a\ge b\ge c$, by Tchebyshev's inequality,
$$\begin{eqnarray}\dfrac{a^5+b^5+c^5}{3}&=&\dfrac{a^3a^2+b^3b^2+c^3c^2}{3}\\&\ge&\dfrac{a^3+b^3+c^3}{3}\dfrac{a^2+b^2+c^2}{3}\\&\ge&\sqrt[3]{a^3b^3c^3}\dfrac{a^2+b^2+c^2}{3}\\&=&\dfrac{a^2+b^2+c^2}{3}\end{eqnarray}$$
and the result follows.
Tchebyshev Inequality: If $a_1\le a_2\le...\le a_n$ and $b_1\le b_2\le...\le b_n$, then $$\dfrac{a_1b_1+a_2b_2+\cdots+a_nb_n}{n}\ge\dfrac{a_1+a_2+\cdots+a_n}{n}\dfrac{b_1+b_2+\cdots+b_n}{n}.$$
Proof Applying rearrangement inequality:
$$\begin{array}{rcl}
a1b_1+a_2b_2+...+a_nb_n&=&a_1b_1+a_2b_2+...+a_nb_n\\
a1b_1+a_2b_2+...+a_nb_n&\ge&a_1b_2+a_2b_3+...+a_nb_1\\
a1b_1+a_2b_2+...+a_nb_n&\ge&a_1b_3+a_2b_4+...+a_nb_2\\
\vdots&\vdots&\vdots\\
a1b_1+a_2b_2+...+a_nb_n&\ge&a_1b_n+a_2b_1+...+a_nb_{n-1}\\
\end{array}$$
Adding all the expressions give us the desired result. Note that equality is done iff $a_1=a_2=...=a_n$ or $b_1=b_2=...=b_n$
PD: I don't know if this Tchebyshev is the same of Statistics Theory
| {
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How to compute $\cos(\arctan(2)) = 1/\sqrt{5}$ I'm doing matrices and I rotated a line about an angle. The gradient of my line I'm rotating to the $x$-axis is $2$, from $y=2x$. So obviously the angle that the line $y=2x$ makes with the $x$-axis is $\arctan(2)$.
My question is how do I arrive at $\cos(\arctan(2)) = 1/\sqrt{5}$. <---
The $1/\sqrt{5}$ is what I'm confused with, how do I get this?
| Well..
Let $\arctan 2 = x$.
$\arctan (2) = x \implies \tan x = \sin x/\cos x = 2 \implies \sin x = 2 \cos x \implies \sin^2 x + \cos^2 x = 4\cos^2 x + \cos^2 x = 1 \implies \cos^2 x = 1/5 \implies \cos x = \pm 1/\sqrt{5}$.
By convention $\arctan$ takes values in (-$\pi/2, \pi/2$) where $\cos$ is presumed to be positive. So $\cos x = 1/\sqrt{5}$.
====
In general if $\tan x = b$ then $\sin x = b \cos x$ so $\sin^2 x + \cos^2 x = (b^2 + 1) \cos^2 x$ so $\cos x = 1/\sqrt{b^2 + 1}$ and hence the trig identity $\cos x = 1/\sqrt{\tan^2 x + 1}$ which is, confession time, one of those trig identities I absolutely can not remember and derive every single time.
| {
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To evaluate the sum $\frac{1}{5}-\frac{1 \cdot 4}{5 \cdot 10}+\frac{1 \cdot 4 \cdot 7}{5 \cdot 10 \cdot 15}-\ldots$ Right now I am working through archived papers of a math aptitude quiz. For some reason I seem to be haveing a hard time with these series problems. I have managed to write the above series in a compact form but thats as far as I got.
$$\frac{1}{5}-\frac{1 \cdot 4}{5 \cdot 10}+\frac{1 \cdot 4 \cdot 7}{5 \cdot 10 \cdot 15}-\ldots = \sum\limits_{i=0}^\infty (-1)^i\frac{\prod\limits_{j=0}^i (3j+1)}{5^{i+1}(i+1)!}$$
Help please!
| Lab Bhattacharjee is right. This is just an instance of the usual binomial series. More precisely,
$$
(1+x)^{-1/3}=1-\frac13x+\frac{1\cdot4}{3^2\cdot 2!}x^2-\frac{1\cdot4\cdot7}{3^3\cdot 3!}x^3+\cdots,
$$
that converges and has the prescribed sum, whenever $-1<x< 1$.
Plugging in $x=3/5$ gives
$$\frac12\root 3\of5=1-\frac15+\frac{1\cdot4}{5\cdot10}-\frac{1\cdot4\cdot7}{5\cdot10\cdot15}+\cdots.$$
This yields $1-\dfrac12\root3\of5$ as the value of the sum.
| {
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Why it's $2^8-127$, not $2^8-(-127)$ when "two's complement of $a$" is $2^n-a$ in binary notation? "Table 2.5.1 Powers of 2
$$\begin{array} {|c|c|c|c||c|}
\hline
\text {Power of 2} & 2^{10}& 2^9 & 2^8 & 2^7& 2^6&2^5 &2^4 &2^3 &2^2 &2^1 &2^0 \\ \hline
2 & 1024 & 512 & 256 & 128 & 64 & 32 & 16 & 8 & 4 & 2 & 1\\ \hline
\end{array} $$"
"Definition Given a positive integer a, the two's complement of a relative to a fixed bit length n is the n-bit binary representation of $2^n-a$"
p. 84
I don't understand the last five rows in Table 2.5.2, the above definition says the two's complement of a positive integer, but the below shows the complement of a negative integer, and I don't understand why negative integer $a$ becomes $-a$ unlike the definition $2^n-a$, for example one of the five rows, why it's $2^8-127$, not $2^8-(-127)$?
"Table 2.5.2 $$\begin{array} {|c|c|c|c||c|}
\hline
Integer & \begin{array}{c}\text {8-Bit Representation}\\\text{(ordinary 8-bit binary}\\\text{ if nonnegative or 8-bit two's}\\\text{complement of absolute value if negative)}\end{array} & \begin{array}{c}\text {Decimal form of}\\\text{Two's complement}\\\text{for Negative Integers)}\end{array} \\ \hline
127 & 01111111 & \\
126 & 01111110 & \\
. & . & \\
. & . & \\
. & . & \\
2 & 00000010 & \\
1 & 00000001 & \\
0 & 00000000 & \\
-1 & 11111111 & 2^8-1 \\
-2 & 11111110 & 2^8-2 \\
. & . & \\
. & . & \\
. & . & \\
-127 & 10000001 & 2^8-127\\
-128 & 10000000 & 2^8-128\\ \hline
\end{array} $$ p.87"
Source: Discrete Mathematics with Applications by Susanna Epp
| Taking two's complement implies inversing a number ie binary digits ie if $0$ then $1$ and vice -versa and adding $1$ to LSB so binary $8$ -bit representation of $1$ is $00000001$ so taking inverse we get $11111110$ and adding $1$ we get $11111111$ if on subtraction the answer is negative then its in the form of 2's complement and if extra bit is generated its positive and we discard it
| {
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How many non-negative integer solutions does the equation $3x + y + z = 24$ have? If the equation is $x + y + z = 24$ then it is solvable with stars and bars theorem. But what to do if it is $3x + y + z = 24$?
| Observe that $3x + y + z = 24$ implies that $y + z = 24 - 3x$. For a fixed value of $x$, a particular solution of the equation $y + z = 24 - 3x$ corresponds to where we place an addition sign in a row of $24 - 3x$ ones. For instance, if $x = 5$, $y + z = 9$ and
$$+ 1 1 1 1 1 1 1 1 1$$
corresponds to the solution $y = 0$ and $z = 9$, while
$$1 1 1 1 1 1 + 1 1 1$$
corresponds to the solution $y = 6$ and $z = 3$. The number of such solutions is the number of ways we can insert one addition sign in a row of $24 - 3x$ ones, which is $$\binom{24 - 3x + 1}{1} = \binom{25 - 3x}{1} = 25 - 3x$$ since we must select which of the $25 - 3x$ symbols ($24 - 3x$ ones and one addition sign) will be an addition sign. Since $x$ is a non-negative integer satisfying $0 \leq 24 - 3x \leq 24$, $x$ can assume the nine values $0, 1, 2, 3, 4, 5, 6, 7, 8$. Hence, the number of solutions of the equation $3x + y + z = 24$ in the non-negative integers is
$$\sum_{k = 0}^{8} (25 - 3x) = \frac{9(25 + 1)}{2} = \frac{9 \cdot 26}{2} = 9 \cdot 13 = 117$$
since $$\sum_{k = 0}^{8} (25 - 3x)$$ is an arithmetic series consisting of nine terms with initial term $25$ and final term $1$.
| {
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Evaluate $\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}dx$ Evaluate $$\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}d x$$
What substitution should I use ? $\sqrt\frac{1+x}{1-x}$ suggests $x=cos2\theta $ but its not useful in $\frac{2-3x}{2+3x}$
| HINT:
Setting $x=\cos2y$
$$I=\int\dfrac{2-3x}{2+3x}\sqrt{\dfrac{1+x}{1-x}}dx=-2\int\dfrac{2-3\cos2y}{2+3\cos2y}\cdot\cot y\sin2y\ dy$$
$$=-2\int\dfrac{2(1+\tan^2y)-3(1-\tan^2y)}{2(1+\tan^2y)+3(1-\tan^2y)}\cos^2y\ dy$$
$$=-2\int\dfrac{5\tan^2y-1}{(5-\tan^2y)(1+\tan^2y)^2}\sec^2y\ dy$$
Set $\tan y=u$ to find $$I=-2\int\dfrac{5u^2-1}{(5-u^2)(1+u^2)^2}du$$
Now use Partial Fraction Decomposition,
$$\dfrac{5v-1}{(5-v)(1+v)^2}=\dfrac A{5-v}+\dfrac B{1+v}+\dfrac C{(1+v)^2}$$
| {
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Quartic with $4 $ equidistant roots Today I got the problem $(x^2 -1)(x^2 -4)=k$, and I have no idea how to prove this algebraically. $K$ is a real, non-zero number that makes the equation have $4$ distinct real equidistant roots. Solve for $k$ algebraically.
What I have attempted:
I have tried solving for x and then in turn attempting to solve for k
$$(x^2 -1)(x^2 -4)=k$$
$$ (x^4 -4x^2-1x^2+4) = k$$
$$ x^4 - 5x^2 + 4 - k = 0 $$
$$ x^2 = {-b\pm\sqrt{b^2-4ac} \over 2a} $$
$$ x^2 = {5\pm\sqrt{25-4(4-k)} \over 2} $$
$$ x^2 = {5\pm\sqrt{9+4k} \over 2} $$
$$ x^2 = {5+\sqrt{9+4k} \over 2} $$ and $$ x^2 = {5-\sqrt{9+4k} \over 2} $$
Therefore
$$ x_1 = \sqrt {5+\sqrt{9+4k} \over 2} $$
$$ x_2 = -\sqrt {5+\sqrt{9+4k} \over 2} $$
$$ x_3 = \sqrt {5-\sqrt{9+4k} \over 2} $$
$$ x_4 = -\sqrt {5-\sqrt{9+4k} \over 2} $$
Now I am stuck trying to solve for k.. The answer is $k$ = 7/4 but having trouble solving for it algebraically..
| As the roots are equidistant, let the roots be $a\pm d,a\pm3d$
using Vieta's formula, immediately we have $$a+d+a-d+a+3d+a-3d=0\iff a=0$$
and $5=(-d)(d)+(-d)(3d)+(-d)(-3d)+d(3d)+(d)(-3d)+(3d)(-3d)$
Now $4-k= d\cdot-d\cdot3d\cdot-3d$
Can you take it from here?
In fact $$x^4-5x^2+4-k=(x^2-d^2)(x^2-9d^2)=x^4-10d^2x^2+9d^4$$
$\implies10d^2=5$ and $4-k=9d^4$
| {
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Proof of value of infinite series, using definition of the integral How do I prove that $$\lim_{n \to \infty }n^2\left ( \frac{1}{(n^2+1)^2} +\frac{2}{(n^2+2^2)^2}+...+\frac{n}{(n^2+n^2)^2} \right )=\frac{1}{4}$$ using the definition of the integral?
| $$
n^2\sum_{k=1}^n\frac{k}{(n^2+k^2)^2}=n^2\sum_{k=1}^n\frac{k}{n^4(1+(k/n)^2)^2}=\frac{1}{n}\sum_{k=1}^n\frac{k/n}{(1+(k/n)^2)^2}\ ,
$$
which is a Riemann sum, yielding
$$
\int_0^1 dx \frac{x}{(1+x^2)^2}=1/4
$$
in the limit.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Finding roots of cubic (trig)
The question is
By putting $x$ $=$ $\frac 23 \cos (\theta)$ Find the exact roots of the equation in terms of $\pi$
$$ 27x^3 - 9x = 1 $$
What I have attempted:
$$ 27x^3 - 9x = 1 $$
$$x=\frac 23 \cos (\theta)$$
$$ x^3=\frac 8{27} \cos^3 (\theta)$$
$$ \therefore 27\left(\frac 8{27} \cos^3 (\theta)\right) - 9\left(\frac 23 \cos (\theta)\right) = 1$$
$$ 8\cos^3(\theta) - 6\cos(\theta) - 1 = 0 $$
Now I tried letting $\cos(\theta)$ = $z$ and try solving the cubic but the solutions aren't exactly nice looking (rational). Is there a special trig identity that I can reduce this equation to?
| Recall that
$$\cos^3 \theta = \frac 3 4 \cos \theta + \frac 1 4 \cos 3\theta$$
Hence you can rewrite your equation as
\begin{align*}
0 &= 8 \cos^3 \theta - 6 \cos \theta - 1 \\
&= 6 \cos \theta + 2 \cos 3\theta - 6 \cos \theta - 1 \\
&= 2 \cos 3\theta + 1
\end{align*}
and hence reduce to $\cos 3\theta = -\frac 1 2$. After a little bit more work, you can write the root of the original polynomial as $\cos r\pi$ with $r$ a reasonably nice fraction.
| {
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How can I calculate the sum $\sum_{n=1}^{\infty} \frac{1}{16n^2-8n-3}$? I have to calculate the partial sum for an equation. How can I calculate the sum for
$$\sum_{n=1}^{\infty} \frac{1}{16n^2-8n-3}\ \text{?}$$
Thanks.
| Notice, use partial fractions: $\frac{1}{16n^2-8n-3}=\frac{1}{4}\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)$ as follows $$\sum_{n=1}^{\infty}\frac{1}{16n^2-8n-3}$$
$$=\frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)$$
$$=\frac{1}{4}\lim_{n\to \infty}\left(\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{13}\right)+\ldots+\left(\frac{1}{4n-7}-\frac{1}{4n-3}\right)+\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)\right)$$
$$=\frac{1}{4}\lim_{n\to \infty}\left(1-\frac{1}{4n+1}\right)$$
$$=\frac{1}{4}\left(1-0\right)=\color{red}{\frac 14}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of the surface of revolution of the ellipsoid I need to find the surface area of an ellipsoid using the equation of an ellipse. I believe my calculations are correct but the formulas I meet on the Internet are complex and have $\arcsin$ or $\arctan$ in the equation, which makes me believe that there is something wrong in my calculations.
Can anyone help me if there is something wrong? The integration limits are $0$ to $a$ to calculate half of the area but multiplied by $2$ to calculate the whole area.
\begin{align*}
y(x) &= \sqrt{b^2 a^2-x^2} \\
y'(x) &= -\frac{x}{\sqrt{b^2a^2-x^2}} \\
S &= 2\pi \int_a^b f(x)\sqrt{1 + y'^2} dx \\
&= 4\pi\int_0^a \sqrt{b^2a^2-x^2}
\sqrt{1 + \left( \frac{-x}{\sqrt{b^2a^2-x^2}} \right)^2} dx \\
&= 4\pi\int_0^a \sqrt{b^2a^2-x^2}
\sqrt{1 + \left( \frac{x^2}{b^2a^2-x^2} \right)} dx \\
&= 4\pi\int_0^a \sqrt{b^2a^2-x^2}
\sqrt{\frac{b^2a^2}{b^2a^2-x^2}}dx \\
&= 4\pi\int_0^a \sqrt{b^2a^2} dx \\
&= 4\pi\sqrt{b^2a^2} \left[ x \right]_0^a \\
&= 4\pi a\sqrt{b^2a^2}
\end{align*}
| The graph $y = \sqrt{a^{2} b^{2} - x^{2}}$ is the upper half of the circle
$$
x^{2} + y^{2} = a^{2} b^{2}
$$
of radius $|ab|$. Revolving gives a portion of a sphere, so it's not entirely a surprise the area element came out to be independent of $x$.
The correct area integral for a spheroid comes from
$$
y = \frac{b}{a}\sqrt{a^{2} - x^{2}},\qquad
y' = -\frac{bx}{a\sqrt{a^{2} - x^{2}}};
$$
this gives
$$
dS = 2\pi y\sqrt{1 + (y')^{2}}\, dx
= 2\pi b\sqrt{1 - \frac{a^{2} - b^{2}}{a^{4}} x^{2}}.
$$
How you proceed next depends on the sign of $a^{2} - b^{2}$, but in either case you'll need something like a trig or hyperbolic substitution. The result is, as you found on the Internet, a more complicated function of $a$ ad $b$ than your polynomial, though the integral does turn out to be elementary.
By contrast, the area of the ellipsoid
$$
\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1,\qquad
0 < a < b < c,
$$
is not an elementary function of $a$, $b$, and $c$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I compare the numbers $2^{39}$, $5^{19}$ and $52^7$? I have to compare the numbers $2^{39}$, $5^{19}$ and $52^7$. I don't know how to do that because their exponents don't have anything in common.
| $$ 52^7 = (26^7 \cdot 2^7) > ((2^{4.7})^7 \cdot 2^7) = 2^{39.9} > 2^{39} $$
As $26 = 2^{\frac{\ln 26}{ \ln 2}} \approx 2^{4.7004}$.
Also,
$$ 52^7 < 6^7 \cdot 10^7 = 6^7 5^7 2^7 < 6^7 5^7 5^4 = 5^{7 \cdot \frac{\ln 6}{\ln 5}} 5^7 5^4 = 5^{7.793 + 7 + 4} < 5^{19}$$
Using $2^7 < 5^4$.
Thus,
$$ 2^{39} < 52^7 < 5^{19}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Level Curves in multiple variabes I want to sketch some level curves to the function $f(x,y) = 3x^2 + 4xy + 3y^2$. I have tried to set the equation to C but I'm not sure how to proceed.
I've gotten hints to substitute $u = x + y$ and $v = x - y$ but it doesn't seem to make sense to me.
Thanks in advance!
| The curves are clearly elliptical.
To be more precise, notice that
$$
f(x,y) = (x,y)A\binom xy
\quad\text{where}\quad
A = \begin{bmatrix}3 & 2\\ 2& 3\end{bmatrix}
$$
Infinite many such $A$ exist, the one above is the one which is symmetric and hence diagonalizable.
Its eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 5$, with respective eigenversors $v_1 = \frac{1}{\sqrt 2}(1,-1)$ and $v_2 = \frac{1}{\sqrt 2}(1,1)$.
This means that
\begin{align}
f(x,y)
= &
(x,y)
\begin{bmatrix}3 & 2\\ 2& 3\end{bmatrix}
\binom xy
\\
= &
(x,y)
\frac{1}{\sqrt 2}
\begin{bmatrix}1 & 1\\ -1& 1\end{bmatrix}
\begin{bmatrix}1 & 0\\ 0& 5\end{bmatrix}
\sqrt 2
{\begin{bmatrix}1 & 1\\ -1& 1\end{bmatrix}}^{-1}
\binom xy
\\
= &
(x,y)
\begin{bmatrix}\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2}\\ -\frac{1}{\sqrt 2}& \frac{1}{\sqrt 2}\end{bmatrix}
\begin{bmatrix}1 & 0\\ 0& 5\end{bmatrix}
\begin{bmatrix}\frac{1}{\sqrt 2} & -\frac{1}{\sqrt 2}\\ \frac{1}{\sqrt 2}& \frac{1}{\sqrt 2}\end{bmatrix}
\binom xy
\end{align}
Letting
$$\tag{1}
\begin{cases}
\hat x & = \displaystyle\frac{x-y}{\sqrt 2} \\
\hat y & = \displaystyle\frac{x+y}{\sqrt 2} \\
\end{cases}
$$
you have
$$
f(\hat x,\hat y) = \hat x^2 + 5\hat y^2
$$
therefore the level curves of $f$ in $(\hat x,\hat y)$ coordinates are solutions to
$$
\hat x^2 + 5\hat y^2 = C
$$
For $C<0$ you have no solutions, for $C=0$ you have only the origin, while for $C>0$
$$
\frac{\hat x^2}{(\sqrt C)^2} + \frac{\hat y^2}{(\sqrt{C/5})^2} = 1
$$
which are ellipses centered in the origin and with semiaxes $\sqrt C$ and $\sqrt{C/5}$ parallel to $\hat x$ and $\hat y$ respectively.
Basically they are all concentrical ellipses with ratio horizonal/vertical axis equal to $1/\sqrt 5$.
Notice that the change of variable $(1)$ is simply a rotation of $\frac{\pi}{4}$, so that in the original coordinates $(x,y)$ your curves are such ellipses rotated by $-\frac{\pi}{4}$.
| {
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Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$ Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$.
I begin by letting $n=1$ then $\frac{1}{2}<\frac{1}{\sqrt{3}}$. Then assume $\frac{1\cdot 3\cdots(2k-1)}{2\cdot 4 \cdots(2k)}<\frac{1}{\sqrt{2k+1}}$, for some $k\ge 1$. Then if I multiply both sides by $\frac{(2(k+1)-1)}{2(k+1)}$ I get $\frac{1\cdot 3\cdots(2(k+1)-1)}{2\cdot 4 \cdots2(k+1)}<\frac{1}{\sqrt{2k+1}}\cdot \frac{(2(k+1)-1)}{2(k+1)}$. So I have what I need on the left hand side, but I'm not sure how to continue on the right hand side. I realized that if I show that $\frac{(2(k+1)-1)}{2(k+1)}<\frac{\sqrt{2k+1}}{\sqrt{2(k+1)+1}}$ then I would be done. This however leads to the same dilemma in that I can easily make the left the way I need, but not the right. So there must be a trick that I am missing.
| First observe that
$$\frac{2n-1}{2n}<\frac{2n}{2n+1}\qquad\text{since }(2n-1)(2n+1)=4n^2-1<4n^2=(2n)(2n)$$
Then
\begin{align}
\left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)^2&<\left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)\left(\prod_{k=1}^{n}\frac{2k}{2k+1}\right)\\
&=\prod_{k=1}^{n}\frac{2k-1}{2k+1}\qquad\text{this product is "telescopic"}\\
&=\frac{1}{2n+1}
\end{align}
So
$$\prod_{k=1}^{n}\frac{2k-1}{2k}<\frac{1}{\sqrt{2n+1}}$$
| {
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Bound on sum of combinations I came across the following inequality $\sum_{i=0}^D \binom N i \le N^D+1$.
I am not sure how to prove this. I tried to do it by induction on $D$, and started with observing the values of sum for different values of $N$, and $D$. The values are as follows:
$$
\begin{matrix}
N \downarrow D \rightarrow &0 &1& 2 &3 &4 &5& 6\\
1 &1 &2& 2& 2& 2& 2& 2\\
2& 1& 3& 4& 4& 4& 4& 4\\
3 &1 &4 &7 &8 &8 &8 &8\\
4 &1 &5 &11 &15 &16 &16 &16\\
5 &1 &6 &16 &26 &31 &32 &32\\
\end{matrix}
$$
Though form the base cases of $N=1, D=0$, and $N=1, D=2$, and so on, the inequality holds, but I am not sure how to proceed further. Any pointer on this would be great.
| We shall prove the statement
$$\sum \limits _{i=0} ^D \binom N i \le N^D + 1, \ \forall N \ge 1, \forall \ 0 \le D \le N$$
using induction on $N$.
For $N = 1$, if $D = 0$ then $\binom 1 0 = 1 \le 1 ^0 + 1$, and if $D = 1$ then $\binom 0 0 + \binom 1 0 = 1 + 1 = 2 \le 1^1 + 1$. Thus, for $N = 1$ the statement is verified.
Let us assume it true for $N$ and let us prove it for $N + 1$.
If $D = N + 1$, then it is clear that
$$\sum \limits _{i = 0} ^D \binom {N + 1} i = \sum \limits _{i = 0} ^{N + 1} \binom {N + 1} i 1^i 1^{N + 1 - i} = (1 + 1) ^{N + 1} = 2^{N + 1} \le (N+1) ^{N+1} + 1 .$$
If $D=0$, the conclusion is trivial: $\binom {N+1} 0 = 1 \le (N+1)^0 + 1$.
For $1 \le D \le N$, using Pascal's formula $\binom {N + 1} i = \binom N i + \binom N {i-1}$ we have that
$$\sum \limits _{i = 0} ^D \binom {N + 1} i = 1 + \sum \limits _{i = 1} ^D \binom {N + 1} i = 1 + \sum \limits _{i = 1} ^D \left( \binom N i + \binom N {i-1}\right) = \\ 1 + \sum \limits _{i = 1} ^D \binom N i + \sum \limits _{i = 1} \binom N {i-1} = \sum \limits _{i = 0} ^D \binom N i + \sum \limits _{j = 0} ^{D-1} \binom N j \le N^D + 1 + N^{D-1} + 1 \le \\ N^D + D N^{D-1} + 1 + 1 \le (N+1) ^D + 1$$
as desired (we have used that $(N+1)^D = N^D + D N^{D-1} + \dots + 1$).
| {
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The integral $\int\frac{{\rm d}x}{(x+3)^{\frac87}(x-2)^{\frac 67}}$ I need to solve the integral
$$\int\frac{{\rm d}x}{(x+3)^{\frac87}(x-2)^{\frac 67}}$$
I can't seem to be able to set it in a suitable form; and I know it has to do something with the fact that $\frac87+\frac67=2$ which here would be a negative integer; but I still can't integrate it. The question is from a book and Answer was "None of these" So I don't know the answer;
Thank you!
The other options are:
$$\frac{7}{5}\left(\frac{x+3}{x-2}\right)^{\frac{1}{7}}$$
$$\frac{7}{5}\left(\frac{x-2}{x+3}\right)^{\frac{-1}{7}}$$
$$\frac{5}{7}\left(\frac{x-2}{x+3}\right)^{\frac{1}{7}}$$ + constant of integration
| Let $$I = \int\frac{1}{(x+3)^{\frac{8}{7}}\cdot (x-2)^{\frac{6}{7}}}dx = \int\frac{1}{\left(\frac{x-2}{x+3}\right)^{\frac{6}{7}}\cdot (x+3)^2}dx$$
Now Put $$\left(\frac{x-2}{x+3}\right) = t\Rightarrow 1-\frac{5}{x+3} =t$$
So we get $$\frac{1}{(x+3)^2}dx = \frac{dt}{5}$$
So we get $$I = \frac{1}{5}\int t^{-\frac{6}{7}}dx = \frac{7}{5}t^{\frac{1}{7}}+\mathcal{C} = \frac{7}{5}\cdot \left(\frac{x-2}{x+3}\right)^{\frac{1}{7}}+\mathcal{C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve in integers $b^{11}-1=a^{2016}+a^{2015}+\dots+1$
Find all integers $(a,b)$ satisfying $$b^{11}-1=a^{2016}+a^{2015}+\dots+1.$$
Obviously, we can get the factorisation $(b-1)(b^{10}+\dots+1)=a^{2016}+a^{2015}+\dots+1$, but I'm not sure how to proceed from there.
| Note that $a^{2016}+a^{2015}+\dots+a+1$ has only prime factors that are $0,1$ modulo $2017$.
This implies that $b^{11} \equiv 1,2 \pmod {2017}$.
Therefore, $b^{2013} \equiv 1,2^{183} \pmod {2017}$ or $b^3,2^{183}b^3 \equiv 1 \pmod {2017}$.
This implies that $b \equiv 1 \pmod {2017}$ or $b^2+b+1 \equiv 0 \pmod {2017}$ or $2^{61}b \equiv 1 \pmod {2017}$ or $2^{122}b^2+2^{61}b+1 \equiv 0 \pmod {2017}$.
Since $b^{10}+b^9+\dots+b+1 \equiv 0,1 \pmod {2017}$, it is not difficult to see that $b \not \equiv 1 \pmod {2017}$ and $b^2+b+1 \not \equiv 0 \pmod {2017}$.
Also, note that if $2^{61}b \equiv 1 \pmod {2017}$ or $2^{122}b^2+2^{61}b+1 \equiv 0 \pmod {2017}$, $b-1 \not \equiv 1,0 \pmod {2017}$. Therefore, no such solutions exist.
| {
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Proof for $e^{\frac{1}{n+1}}-1-\frac{1}{n}\leq0$ I'm looking for a proof of $e^{\frac{1}{n+1}}-1-\frac{1}{n}\leq0$, optionally $\ln(n)+\frac{1}{n+1}\leq \ln(n+1)$
| If $x \ge 0$,
$$e^x = 1 + \int_0^x e^t\, dt \le 1 + \int_0^x e^x\, dt = 1 + xe^x.$$
So $(1 - x)e^x \le 1$, and thus
$$e^x \le \frac{1}{1 - x},\quad 0 \le x < 1.$$
Letting $x = \frac{1}{n+1}$, we obtain
$$e^{\frac{1}{n+1}} \le \frac{1}{1 - \frac{1}{n+1}} = \frac{n+1}{(n+1)-1} = \frac{n+1}{n} = 1 + \frac{1}{n}.$$
Equivalently,
$$e^{\frac{1}{n+1}} - 1 - \frac{1}{n} \le 0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ways to create a quadrilateral by joining vertices of regular polygon with no common side to polygon
How many ways are there to create a quadrilateral by joining vertices of a $n$- sided regular polygon with no common side to that polygon?
It's quite easy to solve for triangles for the same question, logic remains same, we need to choose $4$ vertices with none of them being consecutive, what I did is broke them into cases i.e.: {$n$ sided polygon}
$1)$ All $4$ consecutive - $n$ ways
$2)$ $3$ consecutive - $n (n-5)$ ways (draw a diagram and don't take the dots adjacent to chosen pair)
$3)\; 2+2$ consecutive - $n$ ways to choose the first pair, to choose the second, ways are $n-4-1=n-5$ so, total ways $\Rightarrow n(n-5)$
$4)\; 2$ consecutive strictly - $n$ ways, now to choose the other two, leaving the adjacent ones, we have $n-4$ left, so, $(n-6)(n-10)+2(n-8)$
According to me the answer is $\binom n4 - \{ n + n (n-5) + n(n-5) + ((n-6)(n-10)+(2)(n-8))n \}$
but for $n=20$, our teacher told us that answer is near to $200$ can somebody please confirm my or my teacher's answer?
| $1)$ All $4$ consecutive $=n$ is correct
$2)$ $3$ consecutive $=n(n-5)$ is also correct
$3)$ $2\times2$ consecutive $=\frac{n(n-5)}{2}$ You were double-counting in this case $(a,b,d,e) = (d,e,a,b)$
$4)$ $2\times1\times1=\frac{n(n-6)(n-5)}{2}$ [edited - this expression is cleaner]
Collectively:
\begin{align}&=n+2n\frac{(n-5)}{2}+n\frac{(n-5)}{2}+n(n-6)\frac{(n-5)}{2}\\&=n\left(1+(2+1+(n-6))\frac{(n-5)}{2}\right)\\&=n\left(1+(n-3)\frac{(n-5)}{2}\right)\\&=n\left(\frac{2+(n-3)(n-5)}{2}\right)\\&=n\left(\frac{n^2-8n+17}{2}\right)\\&=\frac{n^3-8n^2+17n}{2}\end{align}
The final answer is:$${n\choose4}-\frac{n^3-8n^2+17n}{2}$$
Confirm: $n=8\implies2; n=20\implies2275$ (Your teacher said near $2000$, right?)
| {
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Complex number ( prove ) Let
$$
{x-yi\over{x+yi}}=a+bi\;\;.
$$
Prove $a^2+b^2=1$
I don't know how to start prove it, can anyone help me?
| First off, the problem posed is flawed, because it forgot to specify that $x$ and $y$ are real. If not, try $x=i$ and $y=1$ to see that the relation does not hold.
Now you may want to prove this without knowledge of deMoivre's theorem (complex exponentials). Here is how:
$$\frac{x+iy}{x-iy}=\frac{x+iy}{x-iy}\frac{x+iy}{x+iy} = \frac{(x+iy)^2}{(x-iy)(x+iy)} = \frac{x^2 -y^2 + 2ixy}{x^2+y^2}$$ so $$ a = \frac{x^2 -y^2 }{x^2+y^2}\\ b = \frac{2xy}{x^2+y^2}
$$
Then
$$ a^2+b^2 = \frac{(x^2-y^2)^2+(2xy)^2}{(x^2+y^2)^2}=\frac{x^4 - 2x^2y^2+y^4 + 4x^2y^2)}{x^4 +2x^2y^2+y^4} = \frac{x^4 +2x^2y^2+y^4}{x^4 +2x^2y^2+y^4} = 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633926",
"timestamp": "2023-03-29T00:00:00",
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$11^{-1}$ modulo $91$ is $58$. Why? I am reading wiki article about Quadratic Sieve and it says
$11^{-1}$ modulo $91$ is $58$
Why? How is it been calculated?
| The reason $58 = 11^{-1} \pmod{91}$ is
$$58 \cdot 11 = 638 = 7 \cdot 91 + 1 \implies 58 \cdot 11 \equiv 638 \equiv 1 \pmod{91}$$
The reason $11$ has an inverse $\pmod{91}$ is that $\gcd(11, 91) = 1$.
To calculate the inverse, we use the extended Euclidean algorithm. First, we use the Euclidean Algorithm to solve for $\gcd(11, 91)$.
\begin{align*}
91 & = 8 \cdot 11 + 3\\
11 & = 3 \cdot 3 + 2\\
3 & = 1 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Working backwards, we express $1$ as a linear combination of $11$ and $91$.
\begin{align*}
1 & = 3 - 1 \cdot 2\\
& = 3 - 1(11 - 3 \cdot 3)\\
& = 4 \cdot 3 - 1 \cdot 11\\
& = 4(91 - 8 \cdot 11) - 1 \cdot 11\\
& = 4 \cdot 91 - 33 \cdot 11
\end{align*}
Hence,
$$-33 \cdot 11 = 1 - 4 \cdot 91 \implies -33 \cdot 11 \equiv 1 \pmod{91}$$
Thus, $-33 \equiv 11^{-1} \pmod{91}$. Moreover, any number of the form $-33 + 91t$ is an inverse of $11 \pmod{91}$ since
$$
11(-33 + 91t) = -363 + 11t \cdot 91 = 1 - 364 + 11t \cdot 91 = 1 + 91(11t - 4)$$
Thus, $11(-33 + 91t) \equiv 1 \pmod{91} \implies -33 + 91t \equiv 11^{-1} \pmod{91}$.
In particular, if $t = 1$, we obtain
$$11^{-1} \equiv -33 + 91 \equiv 58 \pmod{91}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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solving rectangle Diagonal of a rectangle is $13$ cm. If we extend the length of the rectangle for $4$ cm and width for $7$ cm, then diagonal will be longer for $7$ cm as well. Find sides (length and width) of the rectangle.
So $d^2=a^2+b^2 \implies b^2=d^2-a^2 \implies b^2=13^2-a^2 \implies b^2=169-a^2$ and $a^2+b^2=169$
\begin{align*}
d'^2 & =a'^2+b'^2\\
(d+7)^2 & =(a+4)^2+(b+7)^2\\
20^2 & = a^2+8a+16+b^2+14b+49\\
400 & = 8a+16+14b+49+169\\
\frac{166}{2} & = \frac{8a+14b}{2}\\
83 & =4a+7b
\end{align*}
...
That's what I've done. What should I do next?
| You can substitute in $a=\sqrt{169-b^2}$ and do the algebra. I don't think the numbers will be very pretty.
| {
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Why is the solution to $\sqrt{6-5x}=x$ only $x=1$ and not $x=-6$? I solved the equation $\sqrt{6-5x}=x$ as follows:
$$(\sqrt{6-5x})^2=x^2$$
$$6-5x=x^2$$
$$0=x^2+5x-6=(x+6)(x-1)$$
$$x=-6 \quad \text{or} \quad x=1$$
If I plug in $x=-6$ into the original equation, I get $\sqrt{6+30}=\sqrt{36}=\pm 6$ and if I plug in $x=1$, I get $\sqrt{6-5}=\sqrt{1}=\pm 1$.
To me it seems that both values satisfy the original equation. I am using an online education system for my class called MyMathLab and the solution is only $x=1$. Why is that?
Thank you in advance.
| You may not select all solutions from an expression $B$ derived (calculated) from expression $A$ when $B$ is not equivalent to $A$, and have to refer to the initial expression to consider the valid values.
Here (notice the / within the $\Leftrightarrow$)
$$\sqrt{6-5x}=x\quad\not\Leftrightarrow\quad6-5x=x^2$$
because removing the $\sqrt{ }$ removes a condition on the expression as well, as $\sqrt{x}$ is valid for $x \geqslant 0$, thus $\sqrt{\text{expression}} = x$ implies $x \geqslant 0$. And you may solve
$$6-5x \geqslant 0$$
that gives $x \leqslant \frac{6}{5}$.
From now on, you know the requirements: $x \geqslant 0$ and $x \leqslant \frac{6}{5}$.
$$\sqrt{6-5x}=x\quad\implies\quad6-5x=x^2,\quad 0\leqslant x \leqslant \frac{6}{5}$$
Then after solving $6-5x=x^2$ that gives
$$x=-6 \quad \text{or} \quad x=1$$
you refer to the requirements to remove the invalid $x = -6$, and may safely give the only solution
$$x = 1$$
| {
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Find the two values of $k$ for which $2x^3-9x^2+12x-k$ has a double real root.
Find the two values of $k$ for which $2x^3-9x^2+12x-k$ has a double
real root.
I've found one method which is to equate $$2x^3-9x^2+12x-k=2(x-r)^2(x-c)$$
Expanding and equating coefficients I get the system of equations :
\begin{array}
\space 2(c+2r) &=9 \\
2r(2c+r) &=12 \\
-2r^2c&=k \\
\end{array}
Solving this I've found the solutions $k=4,5$
However, I would like to know if there's some easier solution than the
one I've already found, or in general, if there's any other solution to the
problem.
| From $2x^3-9x^2+12x-k=2(x-r)^2(x-c)$, using Vieta's formulas:
$$-\frac{-9}{2} = r + r + c \implies c = \frac{9}{2} - 2r$$
$$\frac{12}{2} = r \cdot r + r \cdot c + c \cdot r \implies6=r^2 + 2(9/2 - 2r)r$$
hence $6 = r^2 + 9r - 4r^2$ or $3(r^2 - 3r + 2) = 0 \implies r = 1, 2$. Using the remainder theorem, $2(1)^3 - 9(1)^2 + 12(1) - k = 0$ i.e $k =5$, or $2(2)^3 - 9(2)^2 + 12(2) - k = 0$ i.e $k =4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.