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A binomial tail sum inequality Let $n$ be a positive even integer and $x_1,..,x_n$ be iid, $\mathop{Bernoulli}(\frac{1}{2})$ random variables. Let $S_n = x_1 + .. + x_n$. Let $0\leq k \leq \frac{n}{2}$ be an integer.
By Hoeffding's inequality, $$\text{Pr}\left[S_n-\frac{n}{2}\geq k\right] \leq e^{-\frac{2k^2}{n}}$$
Hence, we have
$$
\binom{n}{\frac{n}{2}+k} + \binom{n}{\frac{n}{2}+k+1} + .. + \binom{n}{n} \leq 2^{n}e^{-\frac{2k^2}{n}}
$$
Is there a direct proof for this inequality?
| $S_n$ is a sum of $n$ i.i.d Bernoulli random variables with each $x_i\sim Bernoulli(\frac{1}{2})$ (according to what you wanted to prove). Then $S_n\sim Binomial(n,\frac{1}{2})$ and $\mathbb E[S_n]=\frac{n}{2}$.
Now, let's evaluate $\mathbb P\left[S_n-\frac{n}{2}\geq k\right]$ :
\begin{align*}
\mathbb P\left[S_n-\frac{n}{2}\geq k\right]&=\mathbb P\left[S_n\geq k+\frac{n}{2}\right]\\
&=\sum_{k=0}^{n/2} \mathbb P\left[S_n= k+\frac{n}{2}\right]\\
&=\sum_{k=0}^{n/2} \binom{n}{k+\frac{n}{2}}\left(\frac{1}{2}\right)^{k+\frac{n}{2}}\left(\frac{1}{2}\right)^{n-k-\frac{n}{2}}\\
&=\frac{1}{2^n}\sum_{k=0}^{n/2} \binom{n}{k+\frac{n}{2}}\left(\frac{1}{2}\times2\right)^{k+\frac{n}{2}}\\
&=\frac{1}{2^n}\sum_{k=0}^{n/2} \binom{n}{k+\frac{n}{2}}\\
\end{align*}
On the other hand, using Markov's inequality $\forall s>0, \forall k\in\{0,...,\frac{n}{2}\}$, and the independence of $x_i$ :
\begin{align*}
\mathbb P\left[S_n-\frac{n}{2}\geq k\right]&=\mathbb P\left[e^{s(S_n-\frac{n}{2})}\geq e^{sk}\right]\\
&\leq \dfrac{\mathbb E\left[e^{s(S_n-\frac{n}{2})}\right]}{e^{sk}}\\
&\leq e^{-sk}e^{-\frac{sn}{2}}\mathbb E\left[e^{s(S_n)}\right]\\
&\leq e^{-sk}e^{-\frac{sn}{2}}\left(\frac{e^s+1}{2}\right)^n\\
&\leq \frac{1}{2^n}\left[e^{s(-k-\frac{n}{2})}(e^s+1)^n\right]
\end{align*}
Since $k\in\{0,...,\frac{n}{2}\}$, then $(-2k)\geq(-\frac{n}{2}-k)$
$$\frac{1}{2^n}\left[e^{s(-k-\frac{n}{2})}(e^s+1)^n\right]\leq \frac{1}{2^n}\left[e^{s(-2k)}(e^s+1)^n\right]$$
This is the best I could do. Hope it helps a bit.
There's always (first answer) the direct use of Hoeffding's inequality :
$$ \mathbb P\left[S_n-\frac{n}{2}\geq k\right]=\frac{1}{2^n}\sum_{k=0}^{n/2} \binom{n}{k+\frac{n}{2}}\leq e^{\frac{-2k^2}{n}}$$
Mulitplying by $2^n$ gives us
$$ \sum_{k=0}^{n/2} \binom{n}{k+\frac{n}{2}}=\binom{n}{\frac{n}{2}+k} + \binom{n}{\frac{n}{2}+k+1} + .. + \binom{n}{n} \leq 2^n e^{\frac{-2k^2}{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\frac{dx^5}{dx^2},$ the derivative of $x^5$ with respect to $x^2$
Let $f(x)=x^5.$ Find its derivative with respect to $x^2$ , i.e., find $\frac{dx^5}{dx^2}.$
We know that $\frac{dx^5}{dx}=5x^4.$
But what should I do when it is needed to take derivative with respect to $x^2 ,x^3$ or other weird things different from classical derivatives?
My approaches:
$1-)$ Say $x^2=a$ , then $x^5 =a^{5/2}$. Hence , find $$\frac{da^{5/2}}{da}=\frac{5}{2}a^{3/2}=\frac{5}{2}x^{3}$$
$2-)$ Find $$\lim_{h\to 0}\frac{f(x^2+h)-f(x^2)}{h}=\lim_{h\to 0}\frac{(x^2+h)^5-(x^2)^5}{h}=5x^8$$
My approaches conflict. Why do they give different results?
Addendum: Extra examples will be appreciated, for example, $dx^7 /dx^3,\;dx^6/dx^5.$
| In you 2nd attempt is wrong as stated
What is $\frac{dx^5}{dx^2}$
It is ratio of change in $ x^5$ to change in $x^2$
So $$\frac{dx^5}{dx^2}=\lim_{h\to 0}\frac{(x+h)^5-x^5}{(x+h)^2-x^2}
=\frac{5x^4h}{2xh}=\frac{5}{2}x^3$$
For general $\frac{dx^n}{dx^m}$
$$\frac{dx^m}{dx^n}=\lim_{h\to 0}\frac{(x+h)^m-x^m}{(x+h)^n-x^n}
=\frac{mx^{m-1}h}{nx^{n-1}h}=\frac{m}{n}x^{m-n}$$
And for your second approach $x^5$ is not a function of $x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How can I check if a single equation with two variables can be solved with only integers? How can I take a single equation, $k = \frac{132-n}{6n+1}$ and check if $k$ and $n$ can be solved as integers?
In this example, it's relatively easy to figure out by inspection that $k = 10$ and $n =2$, but for others, such as $\frac{100-n}{6n+1}$, there is no integer answers. Now I don't need to actually solve for $k$ and $n$, I just need to know if it's possible to get integer solutions for any given equation.
It also might be important to note that $k$ must be greater than or equal to $1$, so $n$ must be less than $132$ or $100$ depending on the example.
| Noodle:
$k = \frac {132-n}{6n+1}$
$6k = \frac {6\cdot 132 - 6n}{6n+1} =\frac {6\cdot 132 + 1 - (6n+1)}{6n+1}= \frac {793}{6n+1} -1$.
So $6n+1|793=13\cdot 61$ so we have $6n+1 = \pm 1,\pm 13,\pm 61,\pm 793$
So $6n = (\pm 1 - 1), (\pm 13 - 1), (\pm 61 - 1), (\pm 793 - 1)$ but $6|6n$ so we must have those terms all be multiples of $6$.
So $6n = 0, 12, 60, 792$ and $n = 0,2, 10, 132$.
And $6k= \frac {793}{6n+1} -1= \frac {793}{1,13,61,793}-1=792, 60,12,0$ all of which are divisible by $6$ so
If $n=0,2,10,132$ then $k = 132,10,2,0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why $\det\begin{pmatrix} 1&1&0&...&0 \\ -1 & 1 & 0 &...&0 \\ -1 & 0 & 1 &...&0 \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ -1&0&0&...&1\end{pmatrix}=2$? Let $$A = \begin{pmatrix}
1 & 1 & 0 & 0 &0 &... & 0
\\ -1 & 1 & 0 & 0 &0&...&0
\\ -1 & 0 & 1 & 0 &0&...&0
\\ -1 & 0 & 0 & 1 &0&...&0
\\ -1 & 0 & 0 & 0 &1&...&0
\\ \vdots & \vdots & \vdots & \vdots &\vdots&\ddots&\vdots
\\ -1 & 0 & 0 & 0 &0&...&1\end{pmatrix}$$
be an $n \times n$ matrix consisting of $1$'s on its diagonal, a $1$ in the entry located on row $1$ and column $2$, and $-1$'s from the second entry of the first column all the way to $n$. Prove that the determinant of $A$ is always $2$.
How should I begin this? I tried taking the determinants of the matrix when $n=2, 3, $ and $ 4$ and saw that they were all $2$, but am not sure how to proceed.
| Determinants are unchanged if you add a row, or a multiple thereof, to another.
Hence, add the first row to every other row.
$$\begin{pmatrix}
1 & 1 & 0 & 0 &0 & \cdots & 0
\\\ 0 & 2 & 0 & 0 &0& \cdots&0
\\\ 0 & 1 & 1 & 0 &0& \cdots&0
\\\ 0 & 1 & 0 & 1 &0& \cdots&0
\\\ 0 & 1 & 0 & 0 &1& \cdots&0
\\\ \vdots & \vdots & \vdots& \vdots &\vdots & \ddots &\vdots
\\\ 0 & 1 & 0 & 0 &0& \cdots&1\end{pmatrix}$$
Now add $-1/2$ times the second row to the first row:
$$\begin{pmatrix}
1 & 0 & 0 & 0 &0 & \cdots & 0
\\\ 0 & 2 & 0 & 0 &0& \cdots&0
\\\ 0 & 1 & 1 & 0 &0& \cdots&0
\\\ 0 & 1 & 0 & 1 &0& \cdots&0
\\\ 0 & 1 & 0 & 0 &1& \cdots&0
\\\ \vdots & \vdots & \vdots& \vdots &\vdots & \ddots &\vdots
\\\ 0 & 1 & 0 & 0 &0& \cdots &1\end{pmatrix}$$
This is a triangular matrix; the determinant of such a matrix is the product of its diagonal entries. Hence, the determinant is $2$.
Addendum:
As noted in the comments, a simpler solution in the same spirit is to just subtract the second row from the first. This gives
$$\begin{pmatrix}
2 & 0 & 0 & 0 &0 &... & 0
\\ -1 & 1 & 0 & 0 &0&...&0
\\ -1 & 0 & 1 & 0 &0&...&0
\\ -1 & 0 & 0 & 1 &0&...&0
\\ -1 & 0 & 0 & 0 &1&...&0
\\ \vdots & \vdots & \vdots & \vdots &\vdots&\ddots&\vdots
\\ -1 & 0 & 0 & 0 &0&...&1\end{pmatrix}$$
This too is lower-triangular, so the determinant is the product of the diagonal entries, which is clearly $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Two hard integrals: $\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx$ and $\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx$ I found two integrals that seem hard to evaluate:
$$I_1=\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx$$
$$I_2=\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx$$
I am just a beginner in logarithmic integral. So, I searched to find substitution like $x=\frac{1}{1+x}$, $x=\frac{1}{1-x}$, or $x=\frac{1-x}{1+x}$, but they didn't work.
Can I ask some ideas from every one? Thank you.
EDIT: After using Mathematica, with MZIntegrate paclet gives closed-form:
$$\begin{align}I_1&=\int_{0}^{1}\frac{\log{(x)}\log{(1-x)}\log{(1+x^2)}}{x}dx\\&=G^2+\frac{5 \text{Li}_4\left(\frac{1}{2}\right)}{4}+\frac{35}{32} \zeta (3) \log (2)-\frac{119 \pi ^4}{5760}+\frac{5 \log ^4(2)}{96}-\frac{5}{96} \pi ^2 \log ^2(2)\\I_2&=\int_{0}^{1}\frac{\log^2{(x)}\log{(1+x^2)}}{1-x}dx\\&=2 G^2+\frac{35}{16} \zeta (3) \log (2)-\frac{199 \pi ^4}{5760}\end{align}$$
where $G$ is Catalan's constant.
| Express the first integral via integration by parts in terms of the second and a third integral
$$
I_1=\int_{0}^{1}\frac{\ln x\ln(1-x)\ln (1+x^2)}{x}dx=\frac12I_2 -I_3\tag1$$
where
\begin{align}
I_2=&\int_{0}^{1}\frac{\ln^2x\ln (1+x^2)}{1-x}dx
= 2G^2 +\frac{35}{16}\ln2\zeta(3)-\frac{199\pi^4}{5760}\\
I_3=& \int_{0}^{1}\frac{x\ln^2 x\ln(1-x)}{1+x^2}dx
=-\frac54\text{Li}_4\left(\frac{1}{2}\right)+\frac{13 \pi ^4}{3840}+\frac{5\pi^2}{96} \ln ^22-\frac{5 }{96}\ln ^42
\end{align}
Similarly to $I_2$, $I_3$ is relatively manageable and is evaluated in the Appendix. Substitute above into (1) to obtain
$$I_1=\frac54\text{Li}_4\left(\frac{1}{2}\right)+ G^2+\frac{35}{32} \ln2\zeta (3) -\frac{119 \pi ^4}{5760}-\frac{5\pi^2}{96} \ln ^22+\frac{5 }{96}\ln ^42
$$
Appendix: (to be completed)
\begin{align}
I_3=\frac18 \int_{0}^{1}\frac{\ln^2 x\ln (1-x)}{1+x}\ \overset{x\to x^2}{dx}
- \int_{0}^{1}\frac{x\ln^2x\ln(1+x)}{1+x^2}dx\\
\end{align}
\begin{align}
\int_{0}^{1}\frac{\ln^2x\ln(1-x)}{1+x}dx=&\
-4\text{Li}_4\left(\frac{1}{2}\right) +\frac{\pi^4}{90}+\frac{\pi^2}6\ln^22-\frac16\ln^42\\
\int_{0}^{1}\frac{x\ln^2x\ln(1+x)}{1+x^2}dx=&\
\frac34 \text{Li}_4\left(\frac{1}{2}\right)-\frac{23\pi^4}{11520}-\frac{\pi^2}{32}\ln^22+\frac1{32}\ln^42
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $T_{200}$ of given series In a certain series, the $n$th term, $T_n=4T_{n-1} + n – 1$. If $T_1$ = 4, then find the value of $T_{200}$.
I tried the following:
$T_n-T_{n-1}=3T_{n-1} + n – 1$
$T_{n-1}-T_{n-2}=3T_{n-2} + n – 2$
$T_{n-2}-T_{n-3}=3T_{n-3} + n – 3$
Then I proceeded to calculate the difference of the first $2$ and the last $2$ to get:
$$T_n-3T_{n-1}+3T_{n-2}-T_{n-3}=3(T_{n-1} -T_{n-3})$$
Well $\dots$ basically I didn't reach anywhere (neither LHS nor RHS is in friendly form). Then I also tried taking $n^{th}$ order differences and then I got a GP at $3^{rd}$ order difference but still didn't know what to do from there (I'm only familiar with APs of $n^{th}$ order difference).
How to solve this problem? Please help.
Answer (and its format):
$$\frac {10(4^{200})-601}9$$
Edit:
Kindly suggest such that even if the question is tweaked a bit, I may be able to solve that too. For example:
$$T_n=3T_{n-1} + n – 1\text { and } T_1 = 3 \text { then } T_{100}=?$$
| Given
$$T_n = 4T_{n-1} + n - 1$$
by substitution we have
$$\begin{align*}
T_n &= 4 T_{n-1} + n - 1 \\[1ex]
&= 4^2 T_{n-2} + (n-1) + 4 (n-2) \\[1ex]
&= 4^3 T_{n-3} + (n-1) + 4 (n-2) + 4^2 (n-3) \\[1ex]
&~\vdots \\[1ex]
&= 4^k T_{n-k} + (n-1) + 4 (n-2) + 4^2 (n-3) + \cdots + 4^{k-1} (n-k) \\[1ex]
&= 4^k T_{n-k} + \sum_{i=1}^k 4^{i-1} (n-i)
\end{align*}$$
When $k=n-1$, we have
$$\begin{align*}
T_n &= 4^{n-1}T_{n-(n-1)} + \sum_{i=1}^{n-1} 4^{i-1} (n-i) \\[1ex]
&= 4^{n-1}T_1 + n \sum_{i=1}^{n-1} 4^{i-1} - \sum_{i=1}^{n-1} i \cdot 4^{i-1} \\[1ex]
&= 4^n + \frac{n(4^n-4)}{12} - \frac{4^{n+1} - 3n\cdot4^n - 4}{36} \\[1ex]
&= -\frac{4^{n+1}}{36} + 4^n - \frac n3 + 9
\end{align*}$$
and from here you can evaluate $T_{200}$. See here for methods on computing the geometric sums.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\lim _{n \rightarrow \infty} \sum_{k=n+1}^{2 n} \sin \frac{n}{n^{2}+k^{2}}$ Evaluate $$S=\lim _{n \rightarrow \infty} \sum_{k=n+1}^{2 n} \sin \frac{n}{n^{2}+k^{2}}$$
My try:
Since $\sin x$ is monotone increasing in $(0,1)$, we have:
$$n \sin \left (\frac{n}{n^2+(2n)^2}\right)\leq S \leq n\sin \left(\frac{n}{n^2+(n+1)^2}\right)$$
taking $\lim$ throughout we get:
$$0.2 \leq S \leq 0.5$$
I am stuck now?
| Another answer (just for the fun of it)
Using my favored $\large 1,400$ years old approximation
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ make $x=\frac{n}{k^2+n^2}$ and compute
$$S_n=\frac{16 n}{5 \pi }\sum _{k=n+1}^{2 n} \frac{k^2+\alpha }{\left(k^2+\beta \right) \left(k^2+\gamma \right)}$$ where
$$\alpha=n \left(n-\frac{1}{\pi }\right)$$
$$\beta=n^2-\frac{\left(\frac{2}{5}-\frac{4 i}{5}\right) n}{\pi }\qquad \qquad \gamma=n^2-\frac{\left(\frac{2}{5}+\frac{4 i}{5}\right) n}{\pi }$$ This will give a bunch of polylogarithms.
Expanding as a series for large values of $n$, the constant term is
$$\frac{4 \left(\pi -4 \cot ^{-1}(2)\right)}{5 \pi }$$ and comparing to the exact result
$$\frac{\frac{4 \left(\pi -4 \cot ^{-1}(2)\right)}{5 \pi }} {\frac{1}{4} \left(\pi -4 \cot ^{-1}(2)\right)}=\frac{16}{5 \pi }\sim \frac{56}{55}\qquad \text{using} \qquad \pi\sim \frac{22}{7}$$
The asymptotics is
$$\sum _{k=n+1}^{2 n} \sin \left(\frac{n}{k^2+n^2}\right)=\frac{4 \left(\pi -4 \cot ^{-1}(2)\right)}{5 \pi }-\frac{4-70 \pi +40 \cot ^{-1}(2)}{125 \pi ^2\,n}+$$ $$\frac{2 \left(5 \pi (85 \pi -207)+72 \left(16+75 \cot ^{-1}(2)\right)\right)}{9375
\pi ^3\,n^3}+O\left(\frac{1}{n^3}\right)$$ and notice again that
$$\frac{4-70 \pi +40 \cot ^{-1}(2)}{125 \pi ^2\,n}\times\frac {20}3\sim\frac {16}{15}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate integral: $\int\frac{C^2}{P^3-P^2C-QC^3} dC$, given $P,Q > 0$ I was trying to get a closed form solution of the integral
$$\int\frac{C^2}{P^3-P^2C-QC^3} dC$$
with $ P,Q > 0 $.
I went through the following steps:
$$\frac{C^2}{P^3-P^2C-QC^3} = \frac{-3QC^2}{-3Q(P^3-P^2C-QC^3)} \\+ \frac{-P^2}{-3Q(P^3-P^2C-QC^3)} + \frac{P^2}{-3Q(P^3-P^2C-QC^3)}$$
$$\int\frac{C^2}{P^3-P^2C-QC^3} \\= \int(\frac{-3QC^2-P^2}{-3Q(P^3-P^2C-QC^3)}+ \frac{P^2}{-3Q(P^3-P^2C-QC^3)}) dC$$
The first integral evaluates to:
$$\int(\frac{-3QC^2-P^2}{-3Q(P^3-P^2C-QC^3)}=\frac{-1}{3Q}\int(\frac{dz}{z}) =\frac{-1}{3Q}\log{z} $$
I am lost in evaluating the second integral :
$$\int\frac{P^2}{-3Q(P^3-P^2C-QC^3)} dC$$
Is there a closed form solution for this at all?
| Rewrite the integral as
$$I= -\frac1q \int\frac{x^2}{x^3 + \frac{p^2}q x-\frac{p^3}q}dx$$
and factorize $x^3 + \frac{p^2}q x-\frac{p^3}q=(x-r)(x^2+rx+\frac{p^3}{qr})$,
with
$$r=\frac p{\sqrt[3]{2q}} \bigg( \sqrt[3]{\sqrt{1+\frac4{27q}}+1}-\sqrt[3]{\sqrt{1+\frac4{27q}}-1}\bigg)
$$
Then, integrate per partial fractions to obtain the close-form below
\begin{align}
I=& -\frac1q \int\frac{x^2}{(x-r)(x^2+rx+\frac{p^3}{qr})}dx\\
=&\ -\frac1{p^3+2qr^3}
\int \frac{r^3}{x-r}+\frac{(p^3+qr^3)(2x+r)+ p^2r^2}{2q\ (x^2+rx+\frac{p^3}{qr})}\ dx\\
= &\ -\frac1{p^3+2qr^3}\bigg(
r^3\ln(x-r)+\frac{p^3+qr^3}{2q}\ln(x^2+rx+\frac{p^3}{qr})\\
&\hspace{35mm}+\frac{p r^2}q\sqrt{\frac{qr}{3p+r}}
\tan^{-1}\frac{p\sqrt{3p+r}\ (x+\frac r2)}{2\sqrt{qr}}
\bigg)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Pitchfork bifurcation Pitchfork breaking symmetry
I have this equation and I want to find when the breaking symmetry occurs:
$$
-x^3 + x^2(b - 1) + x(b - a)
$$
The roots are $x=0$ and $-x^2 + x(b - 1) + (b - a)=0$
The other solutions are
$$
\frac{b}{2} - \frac{1}{2} \pm
\frac{\sqrt{b^2 - 4a + 2b + 1}}{2}
$$
What I observe is, we have always the solution $x=0$, so if we try to find the breaking symmetry we need the other two solution be equal to zero.
The first condition is when $b=1$ and then the roots become
$\sqrt{1-a}$ and $-\sqrt{1-a}$.
here the picture becomes
Are my steps correct?
Edit:
I would like to thank @Sammy Black.
we want to find where the symmetry breaking occurs.
If we set (x)-->(-x) the equation is not symmetric.
see $f=−^3+^2(−1)+(−).$
Then $f(-x) = x^3+x^2(b-1)-x(b-a)$. The problem is with the quadratic term. So, can we say that the broken symmetry occurs when $b\neq 1$? If b=1, then we lose the quadratic term and we then have the normal form for pitchfork bifurcation.
Thank you.
| The two solutions to a quadratic equation
$$
Ax^2 + Bx + C = 0
$$
(with $A \neq 0$) become one solution where the discriminant
$$
\Delta = B^2 - 4AC
$$
is zero. In your equation, once you factor out $x$ (which you correctly associate with the solution $x=0$), the coefficients are
$$
\left\{
\begin{aligned}
A &= -1 \\
B &= b - 1 \\
C &= b - a
\end{aligned}
\right.
$$
Now,
$$
\Delta = (b - 1)^2 - 4(-1)(b - a)
= (b^2 - 2b + 1) + (4b - 4a)
= (b + 1)^2 - 4a
$$
so the bifurcation occurs where the parameters $a$ and $b$ conspire to make $\Delta = 0$, i.e. where
$$
a = \biggl( \frac{b+1}{2} \biggr)^2.
$$
In your bifurcation diagram, you've assumed that $b=1$, so you find the pitchfork bifurcation at $a=1$, but there are actually two parameters, so the bifurcation diagram consists of surfaces graphed over the $(a, b)$-plane. This is more difficult to picture.
For particular values of $b$ in the range $\{-15, \dots, 15\}$, the $x$-equilibrium values as a function of the parameter $a$ look like this:
On the other hand, for particular values of $a$ in the range $\{-15, \dots, 15\}$, the $x$-equilibrium values as a function of the parameter $b$ look like this:
| {
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"answer_id": 0
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How to compute derivative of $\sin(x^3)$ by definition? I am trying to proof that derivative of $\sin(x^3) = 3x^2\cos(x^3)$ by definition.
But I don't know an identity for $\sin(x^3)$ for getting $\cos(x^3)$.
Even I try to find a quantity similar to $\frac{\sin(x)}{x}$.
| $$\begin{align}
\frac{d}{dx}\sin(x^3)&=\lim_{h\to0}\frac{\sin{((x+h)^3)-\sin (x^3)}}{h}\\
\\
&=\lim_{h\to0}\frac{2\sin{\left(\frac{(x+h)^3-x^3}{2}\right)\cos \left(\frac{(x+h)^3+x^3}{2}\right)}}{h}\\
\\
&=\lim_{h\to0}\frac{2\sin{\left(\frac{(x+h)^3-x^3}{2}\right)}}{h}\cdot\lim_{h\to0}\cos \left(\frac{(x+h)^3+x^3}{2}\right)\\
\\
&=\left(\lim_{h\to0}\frac{2\sin{\left(\frac{(x+h)^3-x^3}{2}\right)}}{h}\right)\cdot\cos \left(\lim_{h\to0}\frac{(x+h)^3+x^3}{2}\right)\\
\\
&=\left(\lim_{h\to0}\frac{2\sin{\left(\frac{(x+h)^3-x^3}{2}\right)}}{h}\right)\cdot\cos \left(x^3\right)\\
\\
&=\cos \left(x^3\right)\cdot\lim_{h\to0}\left(\frac{2\sin{\left(\frac{(x+h)^3-x^3}{2}\right)}}{h}\cdot\frac{\frac{(x+h)^3-x^3}{2}}{\frac{(x+h)^3-x^3}{2}}\right)\\
\\
&=\cos \left(x^3\right)\cdot\lim_{h\to0}\frac{\sin{\left(\frac{(x+h)^3-x^3}{2}\right)}}{\frac{(x+h)^3-x^3}{2}}\cdot\lim_{h\to0}\frac{(x+h)^3-x^3}{h}\\
\\
&=\cos \left(x^3\right)\cdot 1\cdot\lim_{h\to0}\frac{(x+h)^3-x^3}{h}\\
\\
&=\cos \left(x^3\right)\cdot 3x^2\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove $\int_{0}^{2 \pi} \frac{x \sin (\theta)}{1 - 2 x \cos (\theta) + {x}^{2}} \mathcal{I}\phi(\theta)d \theta = \pi \phi (x)$ for $x \in (- 1 , 1)$. A function $\phi \left(z\right)$ is zero when $z = 0$, and is real when $z$ is real, and is analytic when $| z | \le 1$; if $f \left(x , y\right)$ is the imarginary component of $\phi \left(z\right)$, prove that if $x \in \left(- 1 , 1\right)$,
\begin{align}
\int_{0}^{2 \pi} \frac{x \sin \left(\theta\right)}{1 - 2 x \cos \left(\theta\right) + {x}^{2}} f \left(\cos \left(\theta\right) , \sin \left(\theta\right)\right) d \theta = \pi \phi \left(x\right) \\
\end{align}
Attempt:
I define $z = {e}^{i \theta}$ and $\mathrm{dz} = \frac{d \theta}{i z}$ s.t.
\begin{align}
& \int_{0}^{2 \pi} \frac{x \sin \left(\theta\right)}{1 - 2 x \cos \left(\theta\right) + {x}^{2}} f \left(\cos \left(\theta\right) , \sin \left(\theta\right)\right) d \theta & \\
= & \int_{C} \frac{x \frac{z - {z}^{- 1}}{2 i}}{1 - x \left(z + {z}^{- 1}\right) + {x}^{2}} f \left(z\right) \frac{\mathrm{dz}}{i z} \\
= & \int_{C} \frac{x \frac{z - {z}^{- 1}}{2 i}}{\left(1 - x z\right) \left(1 - x {z}^{- 1}\right)} f \left(z\right) \frac{\mathrm{dz}}{i z} \\
= & \int_{C} \frac{x \frac{{z}^{2} - 1}{2 i}}{\left(1 - x z\right) \left(z - x\right)} f \left(z\right) \frac{\mathrm{dz}}{i z} \\
\end{align}
Given that the imaginary component of $\phi \left(z\right) = \frac{\phi \left(z\right) - \overline{\phi \left(z\right)}}{2 i}$ and the complex conjugate is not holomorphic I am not sure how to compute the integral with $f$. I had hoped that I could decompose $f$ into its real and imaginary parts to quote the linearity of the integral, but I realized that I unfortunately do not know how to evalute the integral with the real component of $\phi \left(z\right)$. I feel that I shoudl be using one of the preconditions on $\phi \left(z\right)$, I was able to prove a restricted version of the problem when $\phi \left(z\right)$ is a linear function.
Anyway, I proceeded by evaluating the integral with $\phi \left(z\right)$ instead of $f \left(z\right)$.
\begin{align}
& \int_{C} \frac{x \frac{{z}^{2} - 1}{2 i}}{\left(1 - x z\right) \left(z - x\right)} \phi \left(z\right) \frac{\mathrm{dz}}{i z} \\
= & \frac{- 1}{2} \int_{C} \frac{x \left({z}^{2} - 1\right)}{\left(1 - x z\right) \left(z - x\right) z} \phi \left(z\right) \mathrm{dz} \\
= & - \pi \left[{\text{Res}}_{z = 0} \frac{x \left({z}^{2} - 1\right)}{\left(1 - x z\right) \left(z - x\right) z} + {\text{Res}}_{z = x} \frac{x \left({z}^{2} - 1\right)}{\left(1 - x z\right) \left(z - x\right) z}\right] \\
= & - \pi \left[1 - 1\right] = 0 \\
\end{align}
I believe therefore that the integral of the real and imaginary components of $\phi \left(z\right)$ should equal each other up to a possible constant of $i$ when $\phi \left(z\right)$ is analytic on $C$, but I am unsure how to best proceed from $f \left(z\right)$ to $\phi \left(z\right)$.
| Note that
$$ f \left(\cos \theta , \sin \theta\right)
= \Im \phi (e^{i\theta})
= \sum_{k=0}^\infty \frac{\phi^{(k)}(0)}{k!} \sin(k\theta)
$$
Then, integrate
\begin{align}
& \int_{0}^{2 \pi} \frac{x \sin \theta}{1 - 2 x \cos \theta + {x}^{2}} f \left(\cos \theta , \sin \theta\right) d \theta \\
=& \sum_{k=1}^\infty \frac{\phi^{(k)}(0)}{k!}
\int_{0}^{2 \pi} \frac{x \sin \theta \sin(k\theta)}{1 - 2 x \cos \theta + {x}^{2}} \ d \theta \\
=& \sum_{k=1}^\infty \frac{\phi^{(k)}(0)}{k!}
\int_{0}^{2 \pi} \sin(k\theta)\sum_{j=1}^\infty x^j\sin(j\theta)\ d \theta \\
=& \sum_{k=1}^\infty \frac{\phi^{(k)}(0)}{k!}\ \pi x^k=\pi \phi(x)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\lim\limits_{x \to \infty} \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi^2}{6}$? As we know, the result of some expressions and series is equal to $\frac{\pi^2}{6} $ that the most important of them is $\zeta(2)$
Now, I have founded an equation whose limit at point $x=\infty$ is equal to $\frac{\pi^2}{6} : $
$$\lim_{x \to \infty} \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi^2}{6}$$
I want to know how can we prove it?
| For simplicity let $x = \frac{1}{u^2}$. Then the limit becomes,
$$\lim_{u \to 0} (\frac{\pi u - \sin(\pi u)}{u^2 \sin(\pi u)})$$
This has indeterminate form of $\frac{0}{0}$, so we can apply l'hospitals rule,
$$=\lim_{u \to 0} (\frac{\pi - \pi \cos(\pi u)}{2u \sin(\pi u) + \pi u^2 \cos(\pi u)}) = \pi \lim_{u \to 0} (\frac{1 - \cos(\pi u)}{2u \sin(\pi u) + \pi u^2 \cos(\pi u)})$$
This has indeterminate form of $\frac{0}{0}$, so we can apply l'hospitals rule again,
$$=\pi \lim_{u \to 0} (\frac{\pi \sin(\pi u)}{2 \sin(\pi u) + 4 \pi u \cos(\pi u) - \pi^2u^2\sin(\pi u)}) = \pi^2 \lim_{u \to 0} (\frac{1}{2 + 4 \pi u \cot(\pi u) - \pi^2u^2})$$
Setting $v = \pi u$ simplifies the limit to,
$$=\pi^2 \lim_{v \to 0} (\frac{1}{2 + 4 v \cot(v) - v^2})$$
Now using the Maclaurin expansion of $\cot(v)$,
$$\pi^2 \lim_{v \to 0} (\frac{1}{2 + 4v(\frac{1}{v}-\frac{v}{3}-\frac{v^3}{45} - ...) - v^2}) = \pi^2 \lim_{v \to 0} (\frac{1}{2 + 4-\frac{4v^2}{3}-\frac{4v^4}{45} - ... - v^2})$$
Now setting $v=0$ the limit reduces to,
$$=\pi^2 (\frac{1}{2 + 4 - 0 - 0 - ... - 0}) = \frac{\pi^2}{6}$$
As required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving $\cos(2\theta)=\sin\left(\frac{\theta}{2}\right)$ Consider an acute angle $\theta$, which has the property that
$$\cos(2\theta)=\sin\left(\frac{\theta}{2}\right).$$
I am trying to find the value of $\theta$. Using the identity
$$\cos(2\theta)=\sin\left(\frac{\pi}{2}-2\theta\right),$$
I can write the first equation as
$$\sin\left(\frac{\pi}{2}-2\theta\right)=\sin\left(\frac{\theta}{2}\right).$$
Equating the argument of the $\sin$ functions, we see that
\begin{align}
\frac{\pi}{2}-2\theta&=\frac{\theta}{2} \\
\frac{5}{2}\theta&=\frac{\pi}{2} \\
\theta&=\frac{\pi}{5}.
\end{align}
While this yields the correct answer, I do not know how we can justify equating the arguments of the $\sin$ functions.
| $$
\begin{aligned}
\operatorname{cos} 2 \theta &=\sin \frac{\theta}{2}=\cos \left(\frac{\pi}{2}-\frac{\theta}{2}\right) \\
2 \theta &=2 n \pi \pm\left(\frac{\pi}{2}-\frac{\theta}{2}\right) \\
\frac{5 \theta}{2} &=2 n \pi+\frac{\pi}{2} \text { or } \frac{3 \theta}{2}=2 n \pi-\frac{\pi}{2} \\
\theta &=\frac{(4 n+1) \pi}{5} \text { or } \frac{(4 n-1) \pi}{3},
\end{aligned}
$$
where $n\in Z.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in which series Let $a^2, b^2$ and $c^2$ be three distinct numbers in AP. If $ab + bc + ca = 1$ then $(b + c), (c + a)$ and $(a + b)$ are in
(1)AP
(2) GP
(3) HP
(4) none of these
My approach is as follow
$2{b^2} = {a^2} + {c^2} \Rightarrow 2{b^2} + {b^2} = {a^2} + {c^2} + {b^2} \Rightarrow 3{b^2} = {a^2} + {c^2} + {b^2}$
${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ac} \right)$
${\left( {a + b + c} \right)^2} = 3{b^2} + 2 \Rightarrow {\left( {a + b + c} \right)^2} - {b^2} = 2{b^2} + 2$
$\Rightarrow \left( {a + b + c - b} \right)\left( {a + b + c + b} \right) = 2{b^2} + 2$
$\Rightarrow \left( {a + c} \right)\left( {a + c + 2b} \right) = 2{b^2} + 2 \Rightarrow {\left( {a + c} \right)^2} + 2b\left( {a + c} \right) = 2{b^2} + 2$
$ \Rightarrow {\left( {a + c} \right)^2} + 2\left( {ab + bc} \right) = 2{b^2} + 2 \Rightarrow {\left( {a + c} \right)^2} + 2\left( {1 - ac} \right) = 2{b^2} + 2$
$\Rightarrow {\left( {a + c} \right)^2} - 2ac = 2{b^2} \Rightarrow {\left( {a + c} \right)^2} = 2\left( {{b^2} + ac} \right)$
$ \Rightarrow \frac{{a + c}}{{{b^2} + ac}} = \frac{2}{{\left( {a + c} \right)}}$
Not able to proceed further.
| For multiple chase exercises, well, not my didactic or structural choice of giving exercises, is always a good idea to have a particular example that may rule out immediately some boxes. So let us take $A^2,B^2,C^2$ to be the squares $1,25,49$ in arithmetic progression. Well, $AB+BC+CA=1\cdot 5+5\cdot7+7\cdot1=5+35+7=47$, so we have to "rescale", i.e. to divide $A,B,C,$ by some number (each by $\sqrt{47}$) to obtain $a,b,c$ with $ab+bc+ca=1$.
Then $(b+c)$, $(c+a)$, $(a+b)$ are "rescaled" versions of $(B+C)=5+7=12$, $(C+A)=7+1=8$, and $(A+B)=1+5=6$, and to see if they are in AP, GP, and/or HP is the same equivalent game in the either original or "rescaled" version. So we used the "rescaled" version to test.
$(1)$ Are the numbers $12$, $8$, $6$ in AP? No, $8+ 8\ne 12+6$.
$(2)$ Are the numbers $12$, $8$, $6$ in GP? No, $8\cdot 8\ne 12\cdot6$.
$(3)$ Are the numbers $12$, $8$, $6$ in HP? Equivalently, are the reciprocals in AP? Yes, $\frac 18\cdot \frac 18= \frac 14=\frac 1{12}+\frac 16$.
So we look if this is the case in general, let $\Pi$ be the product $\Pi=(b+c)(c+a)(a+b)$:
$$
\begin{aligned}
&\frac 1{b+c} +\frac 1{a+b} - \frac2{c+a}
\\
&\qquad=\frac 1\Pi
\Big(\ (a+b)(a+c) +(c+a)(c+b) -2(b+a)(b+c)\ \Big)
\\
&\qquad=\frac 1\Pi
\Big(\ (a^2 +(ab+ac+bc)) + (c^2 + (ab+ac+bc)) -2(b^2+(ab+ac+bc))\ \Big)
\\
&\qquad=\frac 1\Pi
\Big(\ a^2 + c^2 -2b^2\ \Big)
\\
&\qquad=0\ .
\end{aligned}
$$
Yes, $(3)$ is in general valid. (There is no need for the norming $ab+bc+ca=1$ for $(3)$, but ok, the exercise wants to get arguments for the truth / failure of the possible choices under this supplementary condition.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4519209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $Au=v\sqrt{2}$ and $A^Tv=u\sqrt{2}$. Find value of $|u_1+2\sqrt{2}v_1|$ Let $$A=\begin{bmatrix}1&1\\0&1\\-1&1\end{bmatrix}$$ and let $A^T$ denote the transpose of A. Let $u=\begin{bmatrix}u_1\\u_2\end{bmatrix}$ and $v=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}$ be column vectors with entries in $\mathbb R$ such that
$u_1^2+u_2^2=v_1^2+v_2^2+v_3^2=1$.
Suppose $Au=v\sqrt{2}$ and $A^Tv=u\sqrt{2}$.
Find value of $|u_1+2\sqrt{2}v_1|$.
My Attempt
By equating and things like that I was able to obtain
$u_1+u_2=v_1\sqrt2$
$u_2=v_2\sqrt2$
$-u_1+u_2=v_3\sqrt2$
$v_1-v_3=u_1\sqrt2$
$v_1+v_2+v_3=u_2\sqrt2$
On solving the above equations I obtained
$u_2=v_2=0$
$v_3=-v_1$
$u_1=v_1\sqrt2$
and then by using $u_1^2+u_2^2=v_1^2+v_2^2+v_3^2=1$ I was able to obtain
$v_1^2=\frac{1}{2}$
$(u_1+2\sqrt2v_1)^2=u_1^2+8v_1^2+4\sqrt2u_1v_1=2v_1^2+8v_1^2+8v_1^2=9$
$\Rightarrow |u_1+2\sqrt2v_1|=3$
Is my answer correct .
Also can there be a shorter way.
My knowledge of matrices is very limited.
| What a crazy problem!
From the given equations $Au=\sqrt2 v$ and $A^\top v=\sqrt2 u$, we conclude that $(A^\top A)u = 2u$. But $A^\top A = \left[\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}\right]$, so $u=\pm\left[\begin{matrix} 1\\0\end{matrix}\right]$ and $v=\pm\dfrac1{\sqrt2}\left[\begin{matrix} 1\\0\\-1\end{matrix}\right]$. Then $|u_1+2\sqrt2 v_1| = 1+2 = 3$, as you said.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability question about choosing digits between 1 and 9 and that their product is a multiple of 3 Two digits between 1 and 9, inclusive, are selected at random. The same digit may be selected twice. What is the probability that their product is a multiple of 3?
I used this logic: if any one of the digits selected is 3, then the product would be a multiple of 3. There are 3 multiples of 3 (3, 6, 9) between 1-9, so $\frac{3}{9}$ = $\frac{1}{3}$, and since the second number can be anything, $\frac{1}{3} * \frac{1}{1} = \frac{1}{3}$. Therefore, my final answer is $\frac{1}{3}$. However, the answer key used complementary counting and got a different answer ($\frac{5}{9}$). What is wrong with my approach?
| Total number is: $9\cdot 9=81$ (9 possibilities for the first digit and 9 for the second).
Number of combinations that are not multiple of 3: $6\cdot 6=36$ (6 possibilities for the first digit and 6 for the second), 6 because from the set {1,2,4,5,7,8}.
Number of combinations that are multiple of 3: $81-36=45$.
Probability: $\frac{45}{81}=\frac{5}{9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proving a function is continuous using rigorous definition of a limit Im trying to prove that the function $$\begin{cases}f(x,y)=\dfrac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}, & (x,y)≠0\\ 0, & (x,y)=0\end{cases}$$ is continuous at point (0,0) using the rigorous defintion of a limit.
Attempting to find the upper limit of the function:
$$|f(x)-f(x_0)|= \left|\frac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}-0\right|$$
I see the denominator is always positive so this is equal to
$\dfrac{|2x^2y^4-3xy^5+x^6|}{(x^2+y^2)^2}$.
Using the triangle inequality i know that this is equal or less than
$\dfrac{|(2x^2y^4)-(3xy^5)|+|x^6|}{(x^2+y^2)^2}$.
From here I would like to continue finding expressions which are equal or greater than this, which allow me to cancel some terms against $((x^2+y^2)^2)$.
Im thinking i can write
$$x^6 = (x^2)^3 ≤ (x^2+y^2)^3 $$
for instance, but i am unsure of how to "handle" $|(2x^2y^4)-(3xy^5)$|.
Could someone give me any pointers?
| You can just use :
$x \leq \sqrt{x^2+y^2} =r$ and $y \leq \sqrt{x^2+y^2} =r $
Thus
$$ 0 \leq | f(x,y) | \leq \frac{2r^6+3r^6+r^6}{r^4} = 6r^2 $$
Then just make $r \to 0$ and it shows that $f(x,y) \to 0$ when ${(x,y) \to (0,0)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Trouble with the Binomial sum I guess the following sum is equal to $1$ since you can compute the baby cases for $t=1,2$ straightforward to get the desired result. So, in general, we should have the following sum to be $1$, I think. But, I am stuck in the last equality. Any comments or advice would be appreciated.
\begin{align*}
&\frac{t}{n}+t\left(\frac{\binom{n-t}{1}}{2\binom{n}{2}}+\frac{\binom{n-t}{2}}{3\binom{n}{3}}+\frac{\binom{n-t}{3}}{4\binom{n}{4}}+\cdots+\frac{\binom{n-t}{n-t}}{(n-t+1)\binom{n}{n-t+1}}\right)\\\\
=~~&\frac{t}{n}+t\left(\frac{\binom{n-t}{1}}{n\binom{n-1}{1}}+\frac{\binom{n-t}{2}}{n\binom{n-1}{2}}+\frac{\binom{n-t}{3}}{n\binom{n-1}{3}}+\cdots+\frac{\binom{n-t}{n-t}}{n\binom{n-1}{n-t}}\right)\\\\
=~~&\frac{t}{n}+\frac{t}{n}\left(\frac{\binom{n-t}{1}}{\binom{n-1}{1}}+\frac{\binom{n-t}{2}}{\binom{n-1}{2}}+\frac{\binom{n-t}{3}}{\binom{n-1}{3}}+\cdots+\frac{\binom{n-t}{n-t}}{\binom{n-1}{n-t}}\right)
\end{align*}
where the first equality is used $k\binom{n}{k}=n\binom{n-1}{k-1}$ and $t\in\{1,2,...,n\}.$
| Let us find the summation
$$S_{m,n}=\sum_{k=1}^{m} \frac{m \choose k}{(k+1){n\choose k+1}}$$
Use
${N\choose K}^{-1}=(N+1)\int_{0}^{1} x^K(1-x)^{N-K} dx$
$$S_{m,n}=(n+1)\sum_{k=1}^m\frac{m\choose k}{k+1} x^{k+1}(1-x)^{n-k-1}dx$$
Let $\frac{x}{1-x}=z$,
$$S_{m,n}=(n+1)\int_0^1 (1-x)^n \sum_{k=1}^{m} {m \choose k} \frac{z^{k+1}}{k+1}dx$$
By integration of $\sum_{k=1}^{m} {m\choose k} t^k=(1+t)^m-1$ from $t=0$ to $t=z$, we can re-write
$$S_{m,n}=(n+1)\int_{0}^{1}(1-x)^n \left(\frac{(1-x)^{-m-1}-1
}{m+1}-\frac{x}{1-x}\right)dx$$
$$S_{m,n}=(n+1)\int_{0}^1 \left[\frac{(1-x)^{n-m-1}}{m+1}-\frac{(1-x)^n}{m+1}-x(1-x)^{n-1}\right] dx=\frac{1}{n-m}$$
$$S_{m,n}=(n+1)\left[\frac{1}{(n-m)(m+1)}-\frac{1}{(n+1)(m+1)}\right]-\frac{1}{n}=\frac{m}{(n-m)n}.$$
Using this and $m=n-t$ the required expression is
$$\frac{t}{n}+tS_{m,n}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A precalculus solution for $x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 =0$
Using algebra (precalculus) and suggest the solution method for the polynomial $$x^6 - 3 x^4 + 2 x^3 + 3 x^2 - 3 x + 1 =0$$
I'm solving problems on polynomials. I'm stuck here.
My attempts.
First, I tried the Rational root theorem, then I failed.
Then I tried factorise the polynomial e.g. $(x^2+ax+b)(x^4+cx^3+dx^2+ex+f)$, but I failed again.
At the end I tried
$$P(x)/x^3=x^3-3x+2+\frac 3x-3\frac {1}{x^2}+\frac {1}{x^3}=x^3+\frac {1}{x^3}-3\bigg(x-\frac 1x\bigg)-\frac {3}{x^2}+2=0$$
I failed again.
| Let,
$$P(x)=x^6-3x^4+2x^3+3x^2-3x+1=0$$
We see that $x=0$ is not one of the roots of $P(x)$. Therefore, we can divide all terms of the polynomial $P(x)$ by $x^2:$
$$
\begin{align}
\frac {P(x)}{x^2}&=\color{red}{x^4}-\color{blue}{3x^2}+\color{red}{2x}+\color{green}{3}-\color{blue}{\frac {3}{x}}+\color{red}{\frac {1}{x^2}}\\
&=\color{red}{x^4+2x+\frac {1}{x^2}}-\color{blue}{3\left(x^2+\frac 1x\right)}+\color{green}{3}\\
&=\left(x^2+\frac 1x\right)^2-3\left(x^2+\frac 1x\right)+3=0.\end{align}
$$
Then substitute $x^2+\frac 1x=u$ , we obtain:
$$
\begin{align}u^2-3u+3=0\\
\implies u_{1,2}=\frac{3\pm i\sqrt 3}{2}\end{align}
$$
If you want to find the roots by radicals in exact form, you will need to solve the following cubic equation :
$$x^2+\frac 1x=u\implies x^3-ux+1=0$$
where, $u\in\left\{u_1,u_2\right\}$.
Thus, you have shown that the polynomial $P(x)$ is solvable by radicals.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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} |
Prove $\sum_{k=0}^{n} \binom{n}{k} (\tan \frac{x}{2})^{2k} (1 + \frac{2^{k}}{(1-\tan^{2} \frac{x}{2})^{k}} ) = \sec^{2n} \frac{x}{2} + \sec^{n} x$ (IMO SL) Prove the identity
\begin{equation}
\sum_{k=0}^{n} \binom{n}{k} \Bigl(\tan \frac{x}{2}\Bigl)^{2k} \Biggl(1 + \frac{2^{k}}{(1-\tan^{2} \frac{x}{2})^{k}} \Biggl) = \sec^{2n} \frac{x}{2} + \sec^{n} x
\end{equation}
for any natural number $n$ and any angle $x$.
*
*The solution probably uses the binomial theorem, but I don't see a clear way to use it.
*The identity $\tan^2 x = \sec^2 x - 1$ could also be potentially useful.
*Finally, I noticed that $\tan^{2} \frac{x}{2}$ appears twice on the left side, which might help.
| Put $\xi = \tan^2 \frac{x}2$, $\eta = \frac{2}{1-tan^2 \frac{x}2}$.
\begin{align*}
L.h.s. &= \sum \binom{n}{k} \xi^k (1 + \eta^k) \\
&= \sum \binom{n}{k} \xi^k + \sum \binom{n}{k} (\xi \eta)^k \\
&= (1+\xi)^n + (1+\xi \eta)^n \\
&= \Big(1+\tan^2 \frac{x}2\Big)^n + \Big(1+\frac{2\tan^2 \frac{x}2}{ 1 - \tan^2 \frac{x}2}\Big)^n \\
\end{align*}
From the identity $\tan^2 x = \sec^2 x - 1$, $1 + \tan^{2} \frac{x}{2} = \sec^{2} \frac{x}{2}$.
And,
\begin{equation}
\Big(1+\frac{2\tan^2 \frac{x}2}{ 1 - \tan^2 \frac{x}2}\Big)^n = \Big(\frac{1 + \tan^{2} \frac{x}{2}}{1 - \tan^{2} \frac{x}{2}} \Big)^n = \Big(\frac{\sec^{2} \frac{x}{2}}{1 - \tan^{2} \frac{x}{2}} \Big)^n = \Big(\frac{1}{\cos^2 \frac{x}{2} - \sin^{2} \frac{x}{2}} \Big)^n
\end{equation}
From the identity $\cos 2x = \cos^{2} x - \sin^{2} x$,
\begin{equation}
\Big(\frac{1}{\cos^2 \frac{x}{2} - \sin^{2} \frac{x}{2}} \Big)^n = (\frac{1}{\cos x})^n = \sec^{n} x
\end{equation}
Finally, we get
\begin{equation}
\Big(1+\tan^2 \frac{x}2\Big)^n + \Big(1+\frac{2\tan^2 \frac{x}2}{ 1 - \tan^2 \frac{x}2}\Big)^n = (\sec^{2} \frac{x}{2})^{n} + \sec^{n} x = \sec^{2n} \frac{x}{2} + \sec^{n} x = R.h.s
\end{equation}
| {
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} |
Is $\frac{1}{6}\ln(x+4)^6$ equivalent to $\ln(x+4)$? I was doing a question and was wondering if bringing the coefficient back into the exponent was allowed?
So, for example: $$\frac{1}{6}\ln(x+4)^6$$
Is it possible for me to do:
$$\ln(x+4)$$
Are those two equivalent?
I am struggling on what to do because I am doing a question where it is asking me to simplify to a single logarithm:
$$\frac{1}{6}\ln(x+4)^6 + \frac{1}{4}[\ln x-\ln(x^2+6x+8)^4]$$
If I can put them back into the exponent area, then I could do something such as:
$$\ln(x+4)+\ln x^\frac{1}{4}-\ln(x^2+6x+8)$$
After that, I could apply the basic adding and subtracting rules to get: $$\ln\left(\frac{(x+4)x^\frac{1}{4}}{x^2+6x+8}\right)$$
Finally, I could factor the bottom and reduce the numerator and denominator: $$\ln\left(\frac{(x+4)x^\frac{1}{4}}{(x+4)(x+2)}\right)$$
$$\ln\left(\frac{x^\frac{1}{4}}{x+2}\right)$$
Is this way of thinking right or am I not allowed to put the exponent back into the exponent area?
| Let's look at some examples. For the purposes of this discussion, all logarithms are natural (i.e., base $e$) and all logarithms are defined as functions from real numbers to the real numbers.
Consider $$f(x) = \frac{1}{6} \log (x+4)^6.$$ If $x = 0$, then we have $f(0) = \frac{1}{6} \log 4^6 = \log 4$. But we can also choose values for which $x+4 < 0$; e.g., $$f(-5) = \frac{1}{6} \log (-5+4)^6 = \frac{1}{6} \log(-1)^6 = \frac{1}{6} \log 1 = 0.$$ In fact, the only real number for which $f(x)$ is not defined is $x = -4$. This happens because raising $x+4$ to the sixth power always gives a nonnegative number, and in particular, a strictly positive number unless $x = -4$.
However, consider $$g(x) = \log (x+4).$$ Now if $x > -4$, we are okay because $x+4 > 0$. But we run into problems if $x < -4$, since without the benefit of raising $x+4$ to the sixth power, it can be negative; e.g., $$g(-5) = \log(-5 + 4) = \log(-1).$$
So clearly, $f(x) \ne g(x)$ in all cases. We do have equality if $x > -4$, but when $x < -4$, $f$ remains well-defined but $g$ does not. To remedy this problem, we can write instead
$$h(x) = \log |x+4|,$$
and here, $f(x) = h(x)$ for all real numbers $x$ except $x = -4$, in which case neither $f$ nor $h$ are well-defined.
This suggests a path forward for your problem: we have
$$\frac{1}{6} \log(x+4)^6 + \frac{1}{4} \left( \log x - \log (x^2 + 6x + 8)^4 \right) = \log|x+4| + \log x^{1/4} - \log |x^2 + 6x + 8|.$$ Note that we do not have to write $\log |x|^{1/4}$ (and in fact this would be incorrect), because the original function, $\frac{1}{4} \log x$, is defined if and only if $x > 0$ in the first place. But this also means that $x+4 > 0$ and $x^2 + 6x + 8 > 0$, so it is actually the restriction imposed by $\log x$ that allows us to remove the absolute values on the other terms without loss of generality. This gives us
$$\begin{align}\log (x+4) - \log(x^2 + 6x + 8) + \log x^{1/4} &= \log(x+4) - \log(x+4) - \log(x+2) + \log x^{1/4} \\&= \log x^{1/4} - \log (x+2) \\&= \log \frac{\sqrt[4]{x}}{x+2},\end{align}$$ where we require $x > 0$.
Keeping track of when expressions remain well-defined is an important part of simplifying algebraic expressions.
Plot of $\frac{1}{6} \log(x+4)^6 + \frac{1}{4}\left(\log x - log(x^2 + 6x + 8)^4\right)$:
Plot of $\log \frac{\sqrt[4]{x}}{x+2}$:
| {
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"answer_count": 1,
"answer_id": 0
} |
Find sequence from the generating function $ A(x)= \dfrac{2x^2-x-2}{3x^3+2x^2+x-1}$ I am working with the sequence $a_0=2, a_1=3$ and $a_2=5$ and
$$a_n=3a_{n-3}+2a_{n-2}+a_{n-1}$$
Using $$A(x)=\sum_{n=0}^\infty a_n x^n = 2+3x+5x^2 + \sum_{n=3}^\infty (3a_{n-3}+2a_{n-2}+a_{n-1})x^n$$
I found an equation for $A(x)$ and hence got the generating function
$$ A(x)= \dfrac{2x^2-x-2}{3x^3+2x^2+x-1}$$
Is there an easy way now to state the sequence explicitly after I have found the generating function
a) in this case or
b) in general?
| Given is the rational function
\begin{align*}
\color{blue}{A(x)=\frac{2x^2-x-2}{3x^3+2x^2+x-1}}\tag{1}
\end{align*}
We can derive from (1) a recurrence relation by recalling the following theorem:
Theorem: If a generating function has a representation as rational function of the form
\begin{align*}
A(x)=\sum_{n=0}^\infty a_n x^n=\frac{P(x)}{\color{blue}{Q(x)}}
\end{align*}
with $P(x), Q(x)$ polynomials, $\deg Q=q>\deg P$ and
\begin{align*}
\color{blue}{Q(x)=1+\alpha_1 x+\alpha_2 x^2+\cdots + \alpha_q x^q}
\end{align*}
then the coefficients $a_n$ follow the recurrence relation
\begin{align*}
\color{blue}{a_{n+q}+\alpha_1 a_{n+q-1}+\alpha_2 a_{n+q-2}+\cdots +\alpha_q a_{n}=0\qquad\qquad n\geq 0}
\end{align*}
See for instance theorem 4.1.1 in Enumerative Combinatorics, Vol. I by R. P. Stanley.
Thanks to this theorem we write $A(x)$ as
\begin{align*}
A(x)=\frac{2+x-2x^2}{1-x-2x^2-3x^3}\tag{2}
\end{align*}
and derive the recurrence relation from the denominator of (2) as
\begin{align*}
a_{n+3}-a_{n+2}-2a_{n+1}-3a_n=0\qquad\qquad n\geq 0
\end{align*}
resp. by shifting the indices we get
\begin{align*}
\color{blue}{a_n}&\color{blue}{=a_{n-1}+2a_{n-2}+3_{n-3}\qquad\qquad n\geq 3}\\
\end{align*}
Initial conditions:
We want to find starting values $a_0,a_1$ and $a_2$. We get from (1)
\begin{align*}
A(x)\left(3x^2+2x^2+x-1\right)&=2x^2-x-2\\
\left(a_0+a_1x+a_2x^2+\cdots\right)\left(3x^2+2x^2+x-1\right)&=2x^2-x-2\tag{3}
\end{align*}
*
*We obtain from (3) by inspection:
\begin{align*}
a_0(-1)=-2\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\to\qquad \color{blue}{a_0=2}
\end{align*}
*In the following we use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. Putting $a_0=2$ we obtain from (3)
\begin{align*}
[x^1](2+a_1x)(x-1)&=-1\\
2-a_1&=-1\qquad\qquad\qquad\qquad\qquad\qquad\to\qquad \color{blue}{a_1=3}
\end{align*}
*Putting $a_0=2$ and $a_1=3$ we obtain from (3)
\begin{align*}
[x^2](2+3x+a_2x^2)\left(3x^2+2x^2+x-1\right)&=2\\
2\cdot 2+3\cdot 1+a_2(-1)&=2\qquad\qquad\qquad\to\qquad\color{blue}{a_2=5}\\
\end{align*}
and get finally the recurrence relation
\begin{align*}
\color{blue}{a_n}&\color{blue}{=a_{n-1}+2a_{n-2}+3_{n-3}\qquad\qquad n\geq 3}\\
\color{blue}{a_0}&\color{blue}{=2, a_1=3, a_2=5}\\
\end{align*}
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Convergence of $\sum_{n=1}^{\infty}\frac{2n+1}{(n^{2}+n)^{2}}$ The series $\sum_{n=1}^{\infty}\frac{2n+1}{(n^{2}+n)^{2}}$
(a) converges to $1$
(b) converges to a number $>1$
(c) diverges to $\infty$
(d) has an oscillating sequence of partial sum
How to deal with convergence of the series $\sum_{n=1}^{\infty}\frac{2n+1}{(n^{2}+n)^{2}}$.
Here, I can use limit comparision test taking series $b_{n}=\sum\frac{1}{n^{3}}$ and got $\lim\frac{a_{n}}{b_{n}}=2$, so by limit comparison test, since the series $\frac{1}{n^{3}}$ is convergent, $a_n=\sum_{n=1}^{\infty}\frac{2n+1}{(n^{2}+n)^{2}}$ is also convergent.
But will it converge to $1$ or to a number greater than $1$? How should I proceed?
| The most important aspect of this problem is to notice that $2n+1=(n+1)^2-n^2$. From there, we have
$$\sum_{n=1}^\infty\frac{2n+1}{(n^2+n)^2} = \sum_{n=1}^\infty \frac{(n+1)^2-n^2}{n^2(n+1)^2}.$$
Also, note that $\frac{a-b}{ab}=\frac1b-\frac1a$. Then,
\begin{align*}\sum_{n=1}^\infty\frac{(n+1)^2-n^2}{n^2(n+1)^2} &= \sum_{n=1}^\infty\frac1{n^2}-\frac1{(n+1)^2}\\
&=1-\frac1{2^2}+\frac1{2^2}-\frac1{3^2}+\dots\end{align*}
I'm sure you can get the final answer from here. :)
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
What is the pattern in the powers of $\sqrt{2}-\sqrt{1}$? What is the pattern in this?
$$\begin{align}
\left(\sqrt{2}-\sqrt{1}\right)^1 &= \sqrt{2}-\sqrt{1}\\
\left(\sqrt{2}-\sqrt{1}\right)^2 &= \sqrt{9}-\sqrt{8}\\
\left(\sqrt{2}-\sqrt{1}\right)^3 &= \sqrt{50}-\sqrt{49}\\
\left(\sqrt{2}-\sqrt{1}\right)^4 &= \sqrt{289}-\sqrt{288}\\
\end{align}$$
I thought of applying the binomial theorem
| If $(\sqrt 2 - 1)^n = \sqrt {a_n } - \sqrt {a_n - 1}$, then $
(\sqrt 2 + 1)^n = \sqrt {a_n } + \sqrt {a_n - 1}$. Hence,
$$
a_n = \left( {\frac{{(\sqrt 2 + 1)^n + (\sqrt 2 - 1)^n }}{2}} \right)^2 = \frac{{2 + (3 + 2\sqrt 2 )^n + (3 - 2\sqrt 2 )^n }}{4}.
$$
This is A115599 in the OEIS. See also A055997. You can show using the binomial theorem that $a_n$ is always an integer. In fact $$
a_n = \frac{{3^n + 1}}{2} + \sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } \binom{n}{2k}2^{3k - 1} 3^{n - 2k} .
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Derivative of capital Pi product I wanted to find the derivative of this function at $x=6$
$$y= \prod_{i=1}^{10} (x-i) = (x-1)(x-2) \cdots (x-10) $$
without expanding all of the brackets, so I used the product rule to find a pattern. However, the resulting sum tells me that the derivative is zero at every whole number which is obviously not true. I've been over my solution and I can't see how I've gone wrong. Please could someone highlight where I went wrong? Thank you in advance.
\begin{align*}
\frac{\textit{d}y}{dx} &= (x-2)(x-3) \cdots (x-10) + (x-1) \frac{d}{dx} \biggl((x-2) \cdots (x-10) \biggr) \\
&= \prod_{i=2}^{10} (x-i) + (x-1) \frac{d}{dx} \biggl(\prod_{i=2}^{10} (x-i) \biggr) \\
&= \prod_{i=2}^{10} (x-i) + (x-1)\prod_{i=3}^{10} (x-i) + (x-1)(x-2)\frac{d}{dx} \biggl(\prod_{i=3}^{10} (x-i) \biggr) \\
&= \prod_{i=2}^{10} (x-i) + (x-1)\prod_{i=3}^{10} (x-i) + (x-1)(x-2)\biggl(\prod_{i=4}^{10} (x-i) \biggr) + \cdots \\
&= \frac{y}{x-1} + \frac{y}{x-2} + \frac{y}{x-3}+\cdots + \frac{y}{x-10} \\
&= \sum_{i=1}^{10} \biggl(\frac{y}{x-i}\biggr)
\end{align*}
| $y' = \sum_\limits{i=1}^{10} \frac {y}{x-i}$ is a fine way to express your derivative for all those values of $x\in\{1,2,3,4,5,6,7,8,9,10\}$ in which case $y'$ is undefined.
However, $\lim_\limits{x\to n} \sum_\limits{i=1}^{10} \frac {y}{x-i}$ with $n$ in the subset of the natural numbers above, does exist, and equals $y'(n)$
| {
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} |
To find the doubling map of the elliptic curve $y^2=x^3+1$ I want to verify that the doubling map of the elliptic curve $y^2=x^3+1$ is given by
\begin{align}
P=(x,y) &\to 2P \\
(x,y) &\mapsto \left(\frac{x^4-8x}{4x^3+4}, \frac{2x^6+40x^3}{8y^3} \right)
\end{align}
For calculation flexibility we denote $P=(x',y')$. We will find the tangent line at $P$ and make intersect with the curve to find $2P$.
Differentiating the curve with respect to $x$, we get $$\frac{dy}{dx}=\frac{3x^2}{2y}.$$
So the tangent line at $P$ is given by $$y-y'=\frac{3x'^2}{2y'}(x-x') \Rightarrow y=\frac{3x'^2x-3x'^3+2y'^2}{2y'}.$$
Substituting $y$ in the curve, we get
\begin{align}
&\left(\frac{3x'^2x-3x'^3+2y'^2}{2y'} \right)^2=x^3+1 \\
&\Rightarrow (3x'^2x-3x'^3+2y'^2)^2=4y'^2x^3+4y'^2 \\
&\Rightarrow 9x'^4x^2+9x'^6+4y'^4+18x'^5x+12x'^3y'^2-12x'^2y'^2x=4y'^2x^3+4y'^2 \\
& \Rightarrow 4y'^2x^3-9x'^4x^2+(12x'^2y'^2-18x'^5)x+(4y'^2-9x'^6-4y'^4-12x'^3y'^2)=0, \cdots (1)
\end{align}
The elliptic passes through $(0,1)$, and hence plugging it in $(1)$, we get
$$4y'^2-9x'^6-4y'^4-12x'^3y'^2=0, \cdots (2)$$
But now I am out of way, how to solve for $x', y'$ ?
what is the apropriate way to solve my question ?
| I tried to check optically the claimed formula on an obvious example of choice, taking $Q=(0,1)$, since the two denominators are divisible by $x$, thus getting $2Q\overset?=(0,0)$. Not a point on the curve. What happens here?
My way of computing is as follows. Consider the point $P=(u,v)$ on the elliptic curve with affine equation $Y^2 =X^3 +1$. Then the tangent line to the given curve in $(u,v)$ is $2Y\; dY=3X^2\; dX$, which taken in $(u,v)$ is $2v(Y-v) = 3u^2(X-u)$. We isolate and replace $Y$ from this equation, it is $Y=mx+n$ with $$m=\frac{3u^2}{2v}\ ,$$
into $-Y^2 + X^3 +\dots$ to get $0 = X^3 -m^2X^2+\dots$, and we already know two roots (with multiplicity counted), which are $u,u$, so the third root, $x$, is given by (Vieta) $u+u+x=m^2$. So
$$
x = -2u+m^2
= -2u + \frac {9u^4}{4v^2}
= \frac {-8uv^2 + 9u^4}{4v^2}
= \frac {-8u(u^3+1) + 9u^4}{4v^2}
= \frac {u^4-8u}{4(u^3+1)}
\ .
$$
To get the $Y$-component $y'$ of $-2P=(x,y')$, we use
$$
\begin{aligned}
y'&=mx+n=m(-2u+m^2)+n=-2mu+m^3+(v-mu)
=v-3mu+m^3
\\
&=\frac 1{8v^3}(8v^4-36u^3v^2+27u^6)\text{ which is either}
\\
&=\frac1{8v^3}(8v^4-36(v^2-1)v^2+27(v^2-1)^2)
=\frac1{8v^3}(-v^4-18v^2+27)\text{ or}
\\
&=\frac1{8v^3}(8(u^3+1)^2-36u^3(u^3+1)+27u^6)
=\frac1{8v^3}(-u^6-20u^3\color{red}{+8})\ .
\end{aligned}
$$
This gives for $(x,y)=2P=-(x,y')=(x,y)$ the formula:
$$
2P=(x,y)=
\left(\
\frac {u^4-8u}{4(u^3+1)}\ ,\
\frac{u^6+20u^3\color{red}{-8}}{8v^3}\ \right)\ .
$$
Chech using the point $Q=(0,1)$: Applying the above formula we get $2Q=(0,-1)$, which is a point on the curve.
Sage check (using the implemented sage formula for the field $K$, which is the fraction field of the ring $R=\Bbb Q[U,V]/(V^2 -(U^3+1))$):
F.<U,V> = PolynomialRing(QQ)
R.<u,v> = F.quotient(V^2 - U^3 - 1)
K = FractionField(R)
E = EllipticCurve(K, [0, 1])
P = E.point([u, v])
print(f'P = {P.xy()}')
print(f'2P = {(2*P).xy()}')
This gives:
P = (u, v)
2P = ((u*v^2 - 9*u)/(4*v^2), (v^4 + 18*v^2 - 27)/(8*v^3))
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Classify, up to similarity, real matrices of dimension $6\times6$ with minimal polynomial $(t-1)^2(t+1)(t-2)$ Classify, up to similarity, real matrices of dimension $6\times6$ with minimal polynomial $(t-1)^2(t+1)(t-2)$
My attempt:
If the minimal polynomial is $(t-1)^2(t+1)(t-2)$ then we only have the following invariant factors:
$(t-1)^2(t+1)(t-2)=t^4-3t^3+t^2+3t-2, (t-1)^2=t^2-2t+1$
So, every matrix of dimension $6\times6$ with minimal polynomial $(t-1)^2(t+1)(t-2)$ is similar to
$$A=\begin{bmatrix}
0 & 0 & 0 & 2 & 0 & 0 \\
1 & 0 & 0 & -3 & 0 & 0 \\
0 & 1 & 0 & -1 & 0 & 0 \\
0 & 0 & 1 & 3 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 \\
0 & 0 & 0 & 0 & 1 & 2 \\
\end{bmatrix}$$.
Is this reasoning correct?
Thanks in advance.
| Hint
The matrices having the desired minimal polynomial, are similar to the ones having for only non zero elements the elements $2,-1,1$ or $B=\begin{pmatrix}
1 & 1\\
0 & 1\end{pmatrix}$ on the diagonal, with at least one $2$, one $-1$ and one $B$ on the diagonal.
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
} |
What is the summation of $^nC_1 + ^nC_4 + ^nC_7+ \ldots$? I know how to find the following sum:
$$ S_0 = {}^nC_0 + ^nC_3 + ^nC_6 + ^nC_9 + \ldots \\= \sum {^nC_{3k}} $$
To solve it, we set: $$(1 + x)^n = \sum_{i=0}^{n} {^nC_{r}} {x^r} $$
Then we substitute $x$ with cube roots of unity to get 3 equations. We sum these 3 equations and then simplify them. Finally, we get $S_0 = \frac{2^n + 2\cos(\frac{n\pi}{3})}{3} $
Now, similarly, can we get the following sums as well?
$$ S_1 = {}^nC_1 + ^nC_4 + ^nC_7 + ^nC_{10} + \ldots \\ = \sum {^nC_{3k+1}} $$
$$ S_2 = {}^nC_2 + ^nC_5 + ^nC_8 + ^nC_{11} + \ldots \\ = \sum {^nC_{3k+2}} $$
I just can’t simplify the above expressions to get a nice and simplified result. Can you help me to find $S_1$ and $S_2$?
| $$(1 + x)^n = \sum_{i=0}^{n} {^nC_{r}} {x^r} $$
Let $w=e^{i\frac{2\pi}3}$. So, $w^3=1$ and $1+w+w^2=0$.
$1+w=-w^2=e^{i\pi+i\frac{4\pi}3}=e^{i\frac\pi3}$.
$1+w^2=-w=e^{i\pi+i\frac{2\pi}3}=e^{i\frac{-\pi}3}$.
$$(1 + 1)^n = \sum_{i=0}^{n} {^nC_{r}} =S_0 + S_1 + S_2\\
(1 + w)^n = \sum_{i=0}^{n} {^nC_{r}} {w^{r\%3}} =S_0 +S_1w+S_2w^2\\
(1 + w^2)^n = \sum_{i=0}^{n} {^nC_{r}} {w^{(2r)\%3}}=S_0 +S_1w^2+S_2w$$
$$3S_0=(1 + 1)^n + (1 + w)^n + (1 + w^2)^n\\=2^n + e^{i\frac{n\pi}3}+ e^{i\frac{-n\pi}3}=2^n + 2\cos(\frac{n\pi}3)\\
3S_1=(1 + 1)^n + (1 + w)^nw^2 + (1 + w^2)^nw\\=2^n + e^{i\frac{n\pi-2\pi}3}+ e^{i\frac{-n\pi+2\pi}3}=2^n + 2\cos(\frac{(n-2)\pi}3)\\
3S_2=(1 + 1)^n + (1 + w)^nw + (1 + w^2)^nw^2\\=2^n + e^{i\frac{n\pi-4\pi}3}+ e^{i\frac{-n\pi+4\pi}3}=2^n + 2\cos(\frac{(n-4)\pi}3)\\
$$
So for $r=0,1,2$, $$S_r=\frac{2^n+2\cos(\frac{(n-2i)\pi}3)}3$$
$$3S_r -2^n=\begin{cases}
2 &\text{if } (n-2r) \% 6 = 0 \\
1 &\text{if } (n-2r) \% 6 = 1 \\
-1 &\text{if } (n-2r) \% 6 = 2 \\
-2 &\text{if } (n-2r) \% 6 = 3 \\
-1 &\text{if } (n-2r) \% 6 = 4 \\
1 &\text{if } (n-2r) \% 6 = 5 \\
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving a property of a certain function $f:[0,1] \to R$ Consider a function $f:[0,1 ]\to R $ defined by $$ f(x)=\frac{1-x}{1+x}.$$ How can we show that for $x,y \in [0,1]$:
$$ f(x)+f(y) \geq f(xy).$$ I tried but get stuck in the end.We have the inequality holds iff
$$\frac{1-x}{1+x}+\frac{1-y}{1+y} \geq \frac{1-xy}{1+xy} \geq \frac{1-xy}{(1+x)(1+y)}.$$ That will be true if $$ (1-x)(1+y)+(1-y)(1+x) \geq 1-xy.$$
| Identically we have
$$
\frac{1-x}{1+x}
+
\frac{1-y}{1+y}
-
\frac{1-xy}{1+xy}
=
\frac
{(1-x)(1-y)(1-xy)}
{(1+x)(1+y)(1+xy)}
\ge
0
$$
Alternatively, continuing from where you left off
\begin{align*}
&
(1-x)(1+y)+(1-y)(1+x) \ge 1-xy
\\[4pt]
\iff\;&
2-2xy\ge 1-xy
\\[4pt]
\iff\;&
1\ge xy
\\[4pt]
\end{align*}
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $x-\frac{1}{x}$, given $x^3 - \frac{1}{x^3} = 108+76\sqrt{2}$
If $x^3 - \dfrac{1}{x^3} = 108+76\sqrt{2}$, find the value of $x-\dfrac{1}{x}$.
Here's what I've tried so far.
$$\begin{align}
\left(x-\dfrac{1}{x}\right)^3&=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right) \\
\rightarrow \quad \left(x-\dfrac{1}{x}\right)^3&=108+76\sqrt{2}-3\left(x-\dfrac{1}{x}\right) \\u:=x-\dfrac{1}{x} \quad\rightarrow \quad u^3+3u-108-76\sqrt{2}&=0
\end{align}$$
Got stuck here since I didn't know how to solve this cubic equation.
I also tried factorizing $x^3-\dfrac{1}{x^3}$.
$$\begin{align}x^3-\dfrac{1}{x^3}&=\left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+1\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}+2-1^2\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(\left(x+\dfrac{1}{x}\right)^2-1^2\right) \\
&= \left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}+1\right)\left(x+\dfrac{1}{x}-1\right)
\end{align}$$
Again, I didn't know what I could do with this.
| Note that
$$(x - \frac{1}{x})^3 = x^3 - 3x + \frac{3}{x} - \frac{1}{x^3}$$
$$(x - \frac{1}{x})^3 = x^3 - \frac{1}{x^3} - 3(x - \frac{1}{x})$$
$$x^3 - \frac{1}{x^3} = (x - \frac{1}{x})^3 + 3(x - \frac{1}{x})$$
So if you let $t = x - \frac{1}{x}$ (the value that you're ultimately planning to solve for), then your equation becomes:
$$t^3 + 3t = 108 + 76\sqrt{2}$$
This is a “depressed” cubic equation that you could solve with Cardano's Formula, but that gives you some ugly nested radicals. So instead, I'm going another route, by finding an equivalent polynomial equation with integer coefficients.
$$(t^3 + 3t - 108)^2 = (76\sqrt{2})^2$$
$$t^6 + 3t^4 - 108t^3 + 3t^4 + 9t^2 - 324t - 108t^3 - 324t + 11664 = 11552$$
$$t^6 + 6t^4 - 216t^3 + 9t^2 - 648t + 112 = 0$$
There's no general formula for a sixth-degree polynomial. And the Rational Root Theorem fails to provide us with any linear factors. But it turns out that a quadratic factor exists.
$$(t^2 - 6 t + 1) (t^4 + 6 t^3 + 41 t^2 + 24 t + 112) = 0$$
Solving the quadratic gives $t = 3 \pm 2\sqrt{2}$. Solving the remaining quartic gives 4 complex solutions:
$$t = -\frac{3}{2} + \sqrt{2} \pm \frac{3}{2} i \sqrt{7 - 4 \sqrt{2}}$$
$$t = -\frac{3}{2} - \sqrt{2} \pm \frac{3}{2} i \sqrt{7 + 4 \sqrt{2}}$$
But of these 6 roots, it turns out that only one of them is a valid solution to the original equation.
$$t = \boxed{3 + 2\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $
Finding anti-derivative of $f(x)= \sin^3 x \cos^2 x $
so, integrate $\int \sin^3 x \cos^2 x dx = \int \sin x (1- \cos^2 x) \cos^2 (x) dx $
let $u = \cos x$
$\int -u^2 (1-u^2) du = \int -u^2 dx + \int u^4 dx = \frac{-u^3}{3} + \frac{u^5}{5} + C $
Therefore $ \frac{\cos^3 x}{3}+\frac{\cos^5 x}{5} + C$
Why am I wrong? My teacher said the answer is $ \frac{-\cos x}{8} - \frac{\cos (3x)}{48} + \frac{\cos (5x)}{80} $
| To check the equivalence of different trigonometric forms, the simplest way is to use the following type of relation
$$
\cos5\theta=\Re (e^{5i\theta})
$$
and the binomial theorem to derive relations between multiples of angles. For example,
$$
\begin{align}
\left({e^{i\theta}}\right)^5&=(\cos\theta+i\sin\theta)^5\\
&=\cos^5\theta+5i\cos^4\theta \sin\theta-10\cos^3\theta \sin^2\theta\\
&-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^5\theta\\
\end{align}
$$
Discarding the complex part and writing $\sin^2=1-\cos^2$ gives us
$$
\begin{align}
\cos5\theta&=16\cos^5\theta-20\cos^3\theta+5\cos\theta
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve the Recurrence Relation $T(n) = 2T(n/2) +T(n/4) + n^2$. I am trying to find the best way to solve $T(n)=2T(\frac{n}{2})+T(\frac{n}{4})+n^2$. I tried to use the recurrence tree method, and after adding up all the levels of the tree i get the summation below:
$T(n)=n^2+\frac{9n^2}{16}+\frac{18n^2}{64}+\frac{36n^2}{256}+...$
When I try to find a closed form solution I get something like $288n^2$. Can I say that the recurrence is $\Theta(n^2)$?
| To make things simple, we denote that $n = 2^k$ ($k$ is not necessarily $\in \mathbb{N}$) and the recurrent sequence becomes
$$T(2^k) = 2T(2^{k-1}) + T(2^{k-2}) + 2^{2k} \tag{1}$$
Let's denote $f(k) = T(2^k)$, then
$$\begin{align}
(1)&\Longleftrightarrow f(k) = 2f(k-1) +f(k-2) + 4^{k} \\
&\Longleftrightarrow \left(f(k) - \frac{16}{7}4^k\right) = 2\left(f(k-1) - \frac{16}{7}4^{k-1}\right) +\left(f(k-2) - \frac{16}{7}4^{k-2}\right) \tag{2}\\
\end{align}$$
Denote $g(k) = f(k) - \frac{16}{7}4^k$, then
$$(2)\Longleftrightarrow g(k) = 2g(k-1) +g(k-2) \tag{3}$$
$(3)$ is a well-know recurrent sequence, so it's easy to prove that
$$g(k) = Ax_1^k+Bx_2^k \tag{4}$$
with $(x_1, x_2) = ( 1 + \sqrt{2}, 1 - \sqrt{2})$ are two roots of the quadratic equation $x^2=2x+1$ and $A,B$ can be determined from the initial values of the sequence $g(k)$ (for example, $g(0)$ and $g(1)$).
Finally, we have
$$T(2^k) =f(k) =g(k)+\frac{16}{7}4^k = Ax_1^k+Bx_2^k + \frac{16}{7}4^k$$
or with $k = \ln(n) / \ln(2)$
$$\begin{align}
T(n) &= Ax_1^\frac{\ln(n)}{\ln(2)}+Bx_2^\frac{\ln(n)}{\ln(2)} + \frac{16}{7}4^\frac{\ln(n)}{\ln(2)} \\
&=A n^\frac{\ln|x_1|}{\ln 2} + B n^\frac{\ln|x_2|}{\ln 2} + \frac{16}{7}n^2\tag{5}\\
\end{align} $$
We observe that $2 > \frac{\ln|x_1|}{\ln 2} > \frac{\ln|x_2|}{\ln 2}$, so the third term of $(5)$ is dominant for $n \longrightarrow + \infty$. Then
$$T(n) \sim \frac{16}{7}n^2 + \mathcal{o}(n^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
This problem can be solved graphically, but can it be solved mathematically?
I can solve it graphically, but not mathematically. Graphically, I found X and Y to be equal, X = Y = 1.33
| Let $ t = \angle ABC $
then
$ x = (y + 2) \sin t $
$ x = 1 \cdot \sin t + y \cos 45^\circ = \sin t + y \cos 45^\circ $
$ \dfrac{1}{x} = \dfrac{(y+1)}{(y + 2) \cos t } $
The last equation becomes
$ x(y + 1) = (y + 2) \cos t $
Use the first equation,
$ (y+1) (y+2) \sin t = (y+2) \cos t $
So
$\tan t = \dfrac{1}{y+1} $
It follows that
$ \cos t = \dfrac{y+1}{\sqrt{y^2 + 2 y + 2 }} $
$ \sin t = \dfrac{1}{\sqrt{y^2 + 2 y + 2 }} $
Substitute into the second equation,
$\dfrac{y + 2}{\sqrt{y^2 + 2 y + 2}} = \dfrac{1}{\sqrt{y^2+ 2 y + 2}} + \dfrac{y}{\sqrt{2}} $
And this simplifies to
$ y + 2 = 1 + \dfrac{y}{\sqrt{2}} \sqrt{y^2 + 2 y + 2} $
And further to
$ \sqrt{2} (y + 1) = y \sqrt{ y^2 + 2 y + 2} $
So that
$ 2 (y^2 + 2 y + 1) = y^2 (y^2 + 2 y + 2 ) $
And finally,
$ y^4 + 2 y^3 - 4 y - 2 = 0 $
whose solution is (from wolframalpha.com)
$ y = \dfrac{1}{2} (-1 + \sqrt{2} \sqrt[4]{3} + \sqrt{3} ) = 1.29663026289$
It follows that
$ t = \tan^{-1} \bigg( \dfrac{1}{y+1} \bigg) = 23.5292985676^\circ $
And
$ x = (y + 2) \sin t = 1.31607401295 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find slope of the tangent line of $4\sqrt x + 2e^\frac {3x-12}{x+2}$ at $ x_0$ Find the slope and the equation of the tangent line to the
graph $y = f(x)$ at $x_0=4$, $$4\sqrt x + 2e^\frac {3x-12}{x+2} $$
$$\lim_{h\to 0}\tfrac{4\sqrt {4+h} + 2e^\frac {12+3h-12}{4+h+2} - 10}{h} = \lim_{h\to 0}\frac{4\sqrt {4+h} + 2e^\frac {3h}{6+h} - 10}{h} $$ I am stuck at this part
| You can do this by using the direct calculus of derivatives, without passing through the incremental ratio.
The equation of the tangent line $r$ at $x = x_0$ for a certain $y = f(x)$ is given by
$$r: f(x_0) + f'(x_0)\cdot (x-x_0)$$
And its slope is just $f'(x_0)$.
The slope is given by $f'(x_0)$ where $x_0 = 4$ in your cases. Thence
$$f(x) = 4\sqrt{x} + 2e^{\frac{3x-12}{x+2}} \longrightarrow f'(x) = \frac{2}{\sqrt{x}} + 2 e^{\frac{3 x-12}{x+2}} \left(\frac{3}{x+2}-\frac{3 x-12}{(x+2)^2}\right)$$
Or arranging it a bit
$$f'(x) = \frac{2}{\sqrt{x}}+\frac{36 e^{\frac{3 (x-4)}{x+2}}}{(x+2)^2}$$
Now it's easy:
$$f'(4) = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Is there a way to represent $\frac{\int_0^1 x^{a-1} (1-x)^{b-1} dx}{\int_0^1 exp(-x^2)x^{a-1} (1-x)^{b-1} dx} $ in terms of a, b? We are asked to find the supremum of $\frac{exp(-x^2) \int_0^1 x^{a-1} (1-x)^{b-1} dx}{\int_0^1 exp(-x^2)x^{a-1} (1-x)^{b-1} dx}$ for $x \in (0, 1)$. I'm thinking since $\frac{\int_0^1 x^{a-1} (1-x)^{b-1} dx}{\int_0^1 exp(-x^2)x^{a-1} (1-x)^{b-1} dx}$ should be a fixed value, the supremum should just be $\frac{\int_0^1 x^{a-1} (1-x)^{b-1} dx}{\int_0^1 exp(-x^2)x^{a-1} (1-x)^{b-1} dx}$. But then I don't know how to simplify that.
I know that $\int_0^1 x^{a-1} (1-x)^{b-1} dx$ is just $B(a, b)$, and our instructor told us there is no need to calculate $\int_0^1 exp(-x^2)x^{a-1} (1-x)^{b-1} dx$, so I'm wondering is there anyway I can simplify that?
| By expanding the exponential into a series and integrating then it will be found that
$$ \int_{0}^{1} e^{-x^2} \, x^{a-1} \, (1-x)^{b-1} \, dx = B(a,b) \, {}_{2}F_{2}\left(\frac{a}{2}, \frac{a+1}{2}; \, \frac{a+b}{2}, \frac{a+b+1}{2}; \, -1 \right). $$
This leads to
$$\frac{ e^{-x^2} \, \int_{0}^{1} x^{a-1} \, (1-x)^{b-1} \, dx}{\int_{0}^{1} e^{-x^2} \, x^{a-1} \, (1-x)^{b-1} \, dx} = \frac{e^{-x^2}}{{}_{2}F_{2}\left(\frac{a}{2}, \frac{a+1}{2}; \, \frac{a+b}{2}, \frac{a+b+1}{2}; \, -1 \right)}. $$
The hypergeometric function is a constant which gives the form
$$\frac{ e^{-x^2} \, \int_{0}^{1} x^{a-1} \, (1-x)^{b-1} \, dx}{\int_{0}^{1} e^{-x^2} \, x^{a-1} \, (1-x)^{b-1} \, dx} = \frac{1}{c_{0}} \, e^{-x^2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Reduction of Order to Solve $y'' =-y^{3}$ I want to solve $y'' +y^3 = 0$ with the boundary conditions $y(0) = a$ and $y(k) = b$. My goal is to reduce this problem to $y' +y^2 = 0$ while solving but I'm not sure it can be done.
I tried reduction of order substitutions (ie. taking $y' = w$ and $y'' = \frac{dw}{dy}y'$) but that did not work. Then I tried to solve in the following way
$y'' y' = -y^3 y'$
$\frac{1}{2}[(y')^2]' = -[\frac{1}{4} y^4]'$
$\frac{1}{2}(y')^2 = -\frac{1}{4} y^4+C$
$(y')^2 = -\frac{1}{2} y^4+C$
$y' = \pm \sqrt{-\frac{1}{2} y^4+C}$
It seems to me if I take my original problem to be $y'' - 2y^3 = 0$ instead, I get
$y' = \pm \sqrt{y^4+C}$. If $C=0$, this would reduce to $y' - y^2 = 0$, which is close enough to what I want for my purposes. But I'm not sure how to get $C=0$ without a condition on the derivative, so maybe this was the wrong way to go.
*
*Can I reduce my original problem, $y'' +y^3 = 0$, to $y' +y^2 = 0$?
*Where do my boundary conditions come into play?
| Solve
\begin{gather*}
\boxed{y^{\prime \prime}+y^{3}=0}
\end{gather*}
Multiplying the ode by $y^{\prime}$ gives
$$
y^{\prime} y^{\prime \prime}+y^{3} y^{\prime} = 0
$$
Integrating the above w.r.t $x$ gives
\begin{align*}
\int \left(y^{\prime} y^{\prime \prime}+y^{3} y^{\prime}\right)d x &= 0 \\
\frac{\left(y^{\prime}\right)^{2}}{2}+\frac{y^{4}}{4} = c_2
\end{align*}
Which is now solved for $y$.
Solving for $y^{\prime}$ gives
\begin{align*}
y^{\prime}&=\frac{\sqrt{-2 y^{4}+8 c_{2}}}{2}\tag{1} \\
y^{\prime}&=-\frac{\sqrt{-2 y^{4}+8 c_{2}}}{2}\tag{2}
\end{align*}
These are separable which can be solved by integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4573004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
For what $x$ the expression of the equation has the same remainder after division. For what $x$ does the expression $(50x+7)(3x+11)$ have the same remainder as $(26x − 86)x$ when divided by $111$?
$(50x+7)(3x+11) \equiv y \mod 111$
$(26x − 86)x \equiv y \mod 111$
$(50x+7)(3x+11) - (26x − 86)x \equiv 0 \mod 111 $
$124x^2+657x+77 \equiv 0 \mod 111$
$111 = 3 \times 37$
There is a system
*
*$124x^2+657x+77 \equiv 0 \mod 3$
*$124x^2+657x+77 \equiv 0 \mod 37$
First:
$(11x+12)^2\equiv 1 \mod 3$
And $x\equiv -1 \mod 3$
But I don't know how to solve the second equation and what to do next.
How to combine these two equations and get at what $x$ these expressions have the same remainder from division.
| Simplifying the equation $$
(50 x+7)(3 x+11) \equiv(26 x-86) x \quad (\bmod 111)
$$
as
$$
13 x^2-9 x-34 \equiv 0 \Leftrightarrow (13 x+17)(x-2) \equiv 0 \quad (\bmod 111)
$$
We are going to investigate the equation case by case.
$$
\begin{aligned}(1)&:\left\{\begin{aligned}
13 x+17 \equiv 0 & \quad (\bmod 37) \\
x-2 \equiv 0 & \quad (\bmod 3)
\end{aligned}\right. \\\textrm{ or } (2)&:13x+7\equiv 0 \quad \pmod {111}\\ \textrm{ or } (3)&: x-2 \equiv 0 \quad(\bmod 111)\\ \textrm{ or } \ (4) &:\left\{\begin{aligned}
13 x+7 \equiv 0& \quad (\bmod 3) \\
x-2 \equiv 0 & \quad (\bmod 37)
\end{aligned}\right.\end{aligned}
$$
$(1)$ By the second equation, we have $x=3 k+2$ for some $k \in \mathbb{Z}\cdots (*).$
$$
\begin{array}{ll} &13(3 k+2)+17 \equiv 0 \quad (\bmod 37)\\
\Leftrightarrow & 2 k \equiv-6 \quad (\bmod 37)\\
\Leftrightarrow & k =-3+37m \text { for some }m \in Z.
\end{array}
$$
Putting back into $(*)$ yields $$
x=3(-3+37 m)+2=111 m-7\equiv 104 \quad \pmod {111}
$$
$(2)$ $$
\begin{array}{cc}
13 x+17 \equiv 0 & (\bmod 111) \\
13 x \equiv 104 & (\bmod 111) \\
x \equiv 67 & (\bmod 111)
\end{array}
$$
$(3)$
$$x\equiv 2 \quad \quad \pmod {111}$$
$(4)$
$$\begin{aligned}\left\{\begin{array}{tt}
13 x+17 \equiv 0& \quad (\bmod 3) \\
x-2 \equiv 0 & \quad (\bmod 37)\end{array}\right.\end{aligned}
$$
By the second equation, we have $x=37h+2$ for some $h \in \mathbb{Z}\cdots (**).$
Putting it into the first equation yields
$$
\begin{gathered}
13(37 h+2)+17 \equiv 0 \quad(\bmod 3) \\
h\equiv 2 \quad(\bmod 3) \\
h=3n+2 \textrm{ for some integer } n
\end{gathered}
$$
Putting back into $(**)$ yields
$$x=37(3n+2)+2\equiv 76 \quad \pmod {111}$$
Conclusively, the general solution to the equation is
$$x\equiv 104, 67, 2 \textrm{ or } 76 \quad \quad \pmod {111}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can someone please explain to me how to solve $\lim_{x \to -\infty} \frac {\sqrt{x^2 - 9}}{2x - 6}$ Can someone please explain to me how to solve $\lim_{x \to -\infty} \frac {\sqrt{x^2 - 9}}{2x - 6}$
In the following solution the only thing that I don't understand is the appearance of that minus sign before the limit after we include the divisor under root sign.
solution on wolframalpha
| When you push a positive number inside a square root, it squares itself:
$$3\sqrt{x+2} = \sqrt{9(x+2)} = \sqrt{9x+18}.$$
But when you try to push a negative number inside a square root, you have to leave its minus sign behind:
$$-3\sqrt{x+2} = -\sqrt{9(x+2)} = -\sqrt{9x+18},$$
because $\sqrt{9} = 3$ not $-3.$
When your number is a variable, it's not as obvious that there is a minus sign. If $x$ is negative, then $-x$ is positive. So in your problem the calculation is
$$\frac{1}{x}\sqrt{x^2-9} = -\frac{1}{-x}\sqrt{x^2-9} = -\frac{1}{\sqrt{(-x)^2}}\sqrt{x^2-9} $$
$$= -\frac{1}{\sqrt{x^2}}\sqrt{x^2-9}=-\sqrt{\frac{x^2-9}{x^2}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
} |
Calculate a certain integral Good day! Is this integral tabular? I calculated it in MatLab and am now trying to write down an analytical expression. How can I get a result?
\begin{align}
I &= \int\limits_{x=-\infty}^{-1} \frac{\mu \, dx}{2 \cdot (1+\mu^2 \cdot ((x-m) \cdot k)^2)^{3/2}} \\
&= \left[- \frac{\mu \cdot (m-x)}{2 k \mu \sqrt{(m-x)^2+1}} \right]_{x = -\infty}^{-1} \\
&= - \frac{\mu \cdot (m+1)}{2 k \mu \sqrt{(m+1)^2+1}} +
\lim_{x\to-\infty} \, \frac{\mu \cdot (m-x)}{2 k \mu \sqrt{(m-x)^2+1}} \\
&= - \frac{\mu \cdot (m+1)}{2 k \mu \sqrt{(m+1)^2+1}} +
\frac{\mu \cdot (m+\infty)}{2 k \mu \sqrt{(m+\infty)^2+1}} \\
&= \frac{1}{2 k} - \frac{\mu \cdot (m+1)}{2 k \mu \sqrt{(m+1)^2+1}}.
\end{align}
| The integral
$$ I = \int\limits_{x=-\infty}^{-1} \frac{\mu \, dx}{2 \cdot (1+\mu^2 \, k^2 \, (x-m)^2)^{3/2}} $$
can be evaluated as follows.
\begin{align}
I &= \int\limits_{x=-\infty}^{-1} \frac{\mu \, dx}{2 \cdot (1+\mu^2 \cdot ((x-m) \cdot k)^2)^{3/2}} \\
&= \frac{\mu}{2} \, \int_{1}^{\infty} \frac{dt}{(1 + \mu^2 \, k^2 \, (m+t)^2)^{3/2}} \hspace{5mm} \to x = -t \\
&= \frac{\mu}{2} \, \left[ \frac{m+t}{\sqrt{\mu^2 \, k^2 \, (m+t)^2 + 1}} \right]_{1}^{\infty} \\
&= \frac{\mu}{2} \, \left[ \lim_{t \to \infty} \frac{1}{\sqrt{\mu^2 \, k^2 + \frac{1}{(m+t)^2}}} - \frac{m+1}{\sqrt{\mu^2 \, k^2 \, (m+1)^2 + 1}} \right] \\
&= \frac{\mu}{2} \, \left[ \frac{1}{\mu \, k} - \frac{m+1}{\sqrt{\mu^2 \, k^2 \, (m+1)^2 + 1}} \right] \\
&= \frac{1}{2 \, k} - \frac{\mu \, (m+1)}{2 \, \sqrt{\mu^2 \, k^2 \, (m+1)^2 + 1}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Taylor series for $1/|R-r|$ I'm questioning all my math knowledge because somehow I don't end on the same textbook result.
Here is what I mean: Expanding
$$ \dfrac{1}{|R-r|} = \dfrac{1}{R}\,\dfrac{1}{\sqrt{1+ \frac{r^2}{R^2} - 2\,\frac{R\,r}{R^2}}}\quad \text{where} \quad R = [X,Y,Z] \quad r = [x,y,z]$$
Now for $r \ll R$ defining a small quantity $\delta = \frac{r^2}{R^2} - 2\,\frac{R\,r}{R^2}$ and expanding with respect to $\delta$ at $\delta = 0$:
$$\dfrac{1}{\sqrt{1 + \delta}} \approx 1 - \dfrac{1}{2}\,\delta + \dfrac{3}{8}\,\delta^2$$
Here is where the struggle begins: plugging $\delta$ back in, I receive:
$$\dfrac{1}{|R-r|} \approx\dfrac{1}{R}\,\left(1 - \dfrac{r^2}{2\,R^2} + \dfrac{R\,r}{R^2} + \dfrac{3}{8}\left(\,\dfrac{r^2}{R^2}- 2\,\dfrac{R\,r}{R^2}\right)^2\right)$$
At this point the textbook as well as all resources I sought for simplified this straight to:
$$\dfrac{1}{|R-r|} \approx\dfrac{1}{R}\,\left(1 - \dfrac{r^2}{2\,R^2} + \dfrac{R\,r}{R^2} + \dfrac{3}{2}\left(\,\dfrac{R\,r}{R^2}\right)^2\right)$$
How in the world of Mathematics is this possible?
| Hint:
You almost there! Just expand $\left(\frac{r^2}{R^2}-\frac{2Rr}{R^2}\right)^2$ and then the terms $\left(\frac{r}{R}\right)^3$ and $\left(\frac{r}{R}\right)^4$ may be negligible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Prove by induction $3^n\gt 5n^2$ for all $n \ge 4$ Prove by induction $3^n\gt 5n^2$ for all $n \ge 4$
Hello, I cannot seem to find this question, tried every search option (exact match), and not even wolframalpha could help me. Looking at a graph that is obviously true for all $n \ge 4$
My approach was trying to get a trinomial perfect square on the 5 side but it seems imposible with the tools of inequality.
| Base case: $n=4$
$3^4 = 81 > 80 = 5\cdot 16 = 5 \cdot 4^2$.
For the inductive step, we'll first prove a lemma by induction as well: $2 \cdot 3^n > 10 n + 5, \forall n \geq 4$
Base case': $n=4$
$2 \cdot 3^4 = 2 \cdot 81 = 162 > 45 = 10 \cdot 4 + 5$.
Inductive step':
Suppose $2 \cdot 3^n > 10 n + 5$ holds for some $n \in \mathbb{N}^{\geq 4}$.
$2 \cdot 3^n > 10 n + 5 \implies 3(2 \cdot 3^n) > 3(10 n + 5) \implies 2 \cdot 3^{n+1} > 30 n + 15$
Since $n \geq 4 > 0$,
$2 \cdot 3^{n+1} > 10 n + 15 = 10 n + 10 + 5 = 10(n+1) + 5$
Therefore, we have proved the lemma.
Let's procede with the inductive step for our original statement:
Inductive step:
Suppose $3^n > 5 n^2$ holds for some $n \in \mathbb{N}^{\geq 4}$
$3^n > 5 n^2 \implies 3^n + 10 n + 5 > 5 n^2 + 10 n + 5 = 5 (n^2 + 2n + 1) = 5 (n+1)^2$
This implies that
$3^{n+1} = 3 \cdot 3^n = 3^n + 2 \cdot 3^n > 3^n + 10n + 5 > 5n^2 + 10n + 5 = 5 (n+1)^2$
We got $3^{n+1} > 5(n+1)^2$. Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n$ Question:
$$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=?$$
My work:
$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=\lim_{n\to\infty}\left(\dfrac{n^2(1+5/n+3/n^2}{n^2(1+1/n+2/n^2)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)^n$
$\log L=n\log\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)$
Is this equal to 0? Then the answer would be $e^0=1$.
The answer was given as $e^4$ and I have no idea how to get to that.
| $$ y = \lim_{n\to\infty}\left(\frac{n^2+5n+3}{n^2+n+2}\right)^n$$
$$ y = \lim_{n\to\infty}\left(1 + \frac{4n+1}{n^2+n+2}\right)^n$$
$$\ln y = \lim_{n\to\infty}n \ln\left(1 + \frac{4n+1}{n^2+n+2}\right)$$
$$\ln y = \lim_{n\to\infty}n \left(\frac{4n+1}{n^2+n+2} - \left(\frac{4n+1}{n^2+n+2}\right)^2 + \left(\frac{4n+1}{n^2+n+2}\right)^3 \cdots\right)$$
$$\ln y = \lim_{n\to\infty}\left(\frac{4n^2+n}{n^2+n+2} + O(n^{-1})\cdots\right)$$
$$\ln y = \lim_{n\to\infty}\left(4 + \frac{-3n - 8}{n^2+n+2} + O(n^{-1})\cdots\right)$$
$$\ln y = 4$$
The expansion of $\ln(1 + z)$ around $z = 0$ only works because $\frac{4n+1}{n^2+n+2}$ approaches zero.
$\log L=n\log\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)$
Is this equal to 0? Then the answer would be $e^0=1$.
No because you have two different terms , the $n$ and the $\log\left(\dfrac{1+5/n+3/n^2}{1+1/n+2/n^2}\right)$, which can't be treated separately. The first goes to infinity, the second goes to zero, you have to figure out what the relative speed of them going to their limit is. If both went to zero or both went to infinity then you could conclude their product does as well, but that's not the case here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Does the imaginary part of a complex exponential function include the sign before it? I am deriving some equations, but encountered a problem.
We know that $$e^{i\theta}=\cos{\theta}+i\sin{\theta}$$
, where $\cos{\theta}$ is the real part, and $\sin{\theta}$ is the imaginary part.
However, for $$e^{-i\theta}=\cos{\theta}-i\sin{\theta}$$
What is the imaginary part?
Is it $\sin{\theta}$? Or $-\sin{\theta}$?
| Hint #1: Subsitute $-i \theta$ in the identity $e^{i \theta} = \cos \theta + i \sin \theta$ and note that $\cos (-\theta) = \cos (\theta)$ and $\sin (-\theta) = -\sin (\theta)$.
Hint #2 (using the series expansion for $e^x$)
Since $$e^{i\theta} = i^0 + i^1+\frac {i^2}{2!}+\frac {i^3}{3!}+\frac {i^4}{4!}+\frac {i^5}{5!}+\dots = 1 + i +\frac {1}{2!}+\frac {i}{3!}+\frac {1}{4!}+\frac {i}{5!} +\dots$$ we can separate the real and imaginary parts to get
$$(1 +\frac {1}{2!}+\frac {1}{4!} \dots) = \cos \theta$$ and $$ i(1 +\frac {1}{3!}+\frac {1}{5!} \dots) = i \sin \theta$$ thus $e^{i \theta} = \cos \theta + i \sin \theta$.
Doing the same for $$e^{-i\theta} = i^0 + i^{-1}+\frac {i^{-2}}{2!}+\frac {i^{-3}}{3!}+\frac {i^{-4}}{4!}+\frac {i^{-5}}{5!}\dots = 1 - i +\frac {1}{2!}-\frac {i}{3!}+\frac {1}{4!}-\frac {i}{5!} \dots$$ what do you notice when you separate the real and imaginary parts?
$$(1 +\frac {1}{2!}+\frac {1}{4!} \dots) = \cos \theta$$ and $$- i(1 +\frac {1}{3!}+\frac {1}{5!} \dots) = -i \sin \theta$$ thus $$e^{-i \theta} = \cos \theta - i \sin \theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4586075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Doubt in the solution of a competition combinatorial problem concerning the average number of times the letters of a sequence change I am currently going through Andreescu's 102 combinatorial problems. I couldn't solve number 28 after repeated attempts, so I checked the solution. The problem and solution go as follows:
However, I do not understand why summing the $N_i$'s and dividing over $20!$ is the average. As I see it, if $S_i$ denotes the number of rows with $i$ boy-girl or girl-boy appearances in the row, the average should be
$\dfrac{\sum_{i=2}^{13}iS_i}{\binom{20}{7}}$.
I was hoping that someone would throw some light on why my formula is incorrect and the logic behind the solution (I am relatively new to combinatorial problems in competitions.)
| Sometimes it is helpful to reduce the problem to smaller numbers so that it can be calculated manually. Here we take $5$ instead of $20$ persons and consider a group of two boys and three girls.
\begin{align*}
\{B_1,B_2,G_1,G_2,G_3\}\tag{1}
\end{align*}
In order to analyse the situation it is sufficient to consider the
\begin{align*}
\color{blue}{\binom{5}{2}=10}
\end{align*}
different rows
\begin{align*}
\begin{array}{l|c}
B\quad B\color{blue}{\bullet} G\quad G\quad G&1\\
B\color{blue}{\bullet} G\color{blue}{\bullet} B\color{blue}{\bullet} G\quad G&3\\
B\color{blue}{\bullet} G\quad G\color{blue}{\bullet} B\color{blue}{\bullet} G&3\\
B\color{blue}{\bullet} G\quad G\quad G\color{blue}{\bullet} B&2\\
G\color{blue}{\bullet} B\quad B\color{blue}{\bullet} G\quad G&2\\
G\color{blue}{\bullet} B\color{blue}{\bullet} G\color{blue}{\bullet} B\color{blue}{\bullet} G&4\\
G\color{blue}{\bullet} B\color{blue}{\bullet} G\quad G\color{blue}{\bullet} B&3\\
G\quad G\color{blue}{\bullet} B\quad B\color{blue}{\bullet} G&2\\
G\quad G\color{blue}{\bullet} B\color{blue}{\bullet} G\color{blue}{\bullet} B&3\\
G\quad G\quad G\color{blue}{\bullet} B\quad B&1
\end{array}
\end{align*}
Each of these rows represents the $2!\cdot 3!=12$ pairwise different configurations when we use the elements from (1) giving a total of $12\cdot 10=5!$ rows. But this is not so important, as we can only focus on $BG$ and $GB$ combinations which are marked with $\color{blue}{\mathrm{blue}}$ dots.
One approach:
We get the multiplicities from the right-most column of the table above and derive according to OPs approach
\begin{align*}
\color{blue}{\frac{1\cdot S_1+2\cdot S_2 + 3\cdot S_3+4\cdot S_4}{\binom{5}{2}}}=\frac{1\cdot 2+2\cdot 3+3\cdot 4+4\cdot 1}{10}
\color{blue}{=2,4}
\end{align*}
Another approach:
We focus on the combinations $BG$ and $GB$ for each of the four position pairs $(1,2), (2,3), (3,4), (4,5)$.
\begin{align*}
\begin{array}{ccccc}
B&G&\cdot&\cdot&\cdot\\
\cdot&B&G&\cdot&\cdot\\
\cdot&\cdot&B&G&\cdot\\
\cdot&\cdot&\cdot&B&G\\
G&B&\cdot&\cdot&\cdot\\
\cdot&G&B&\cdot&\cdot\\
\cdot&\cdot&G&B&\cdot\\
\cdot&\cdot&\cdot&G&B\\
\end{array}
\end{align*}
We fix a specific position $j, 1\leq j\leq 4$ and let's say the combination $BG$. We calculate the probability of the event that $BG$ occurs at a specific position $j$ by dividing the favorable number of possibilities by the total number which is as we have two boys and three girls:
\begin{align*}
\color{blue}{\frac{2\cdot 3}{5\cdot 4}}\tag{2}
\end{align*}
We note that (2) is strongly related with $N_j$ from OPs problem, since we can (2) write as
\begin{align*}
\frac{2\cdot 3}{5\cdot 4}=\frac{2\cdot 3\cdot 3!}{5\cdot 4\cdot 3!}=\frac{2\cdot 3\cdot 3!}{5!}\tag{3}
\end{align*}
Doubling (3) since we have to respect $BG$ as well as $GB$ at a specific position and using the linearity of expectation we calculate according to (2) and (3)
\begin{align*}
\color{blue}{\frac{N_1+N_2+N_3+N_4}{5!}}=4\cdot \frac{2 (2\cdot 3) 3!}{5!}=4\cdot\frac{ 2(2\cdot 3)}{5\cdot 4}\color{blue}{=2,4}
\end{align*}
and both results coincide.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fourier transform of $1/\sinh^n(x-\mathrm{i}\epsilon)$ I want to calculate
\begin{equation*}
\int_{-\infty}^{+\infty} \text{d}x \frac{ e^{-\mathrm{i}\omega x} }{ \left( \sinh{( x-\mathrm{i}\epsilon )} \right)^n }
\end{equation*}
with $n\in\text{N}$ and $\epsilon$ as an infinitesimal regularization factor. I tried several tricks and as a last resource I would try to Laurent-expand the $(\sinh(x))^{-n}$ function, but also this seems not possible. How can I calculate this integral?
Edit:
I may have found a way thanks to this article appendix D, but it's strongly non-trivial. I write it here because it can be useful to someone. First there is to do a substitution $x-\mathrm{i}\epsilon\to x$ and after that there is to consider the following path
inside which there is no singularity. The edges contributions are null for $r\to\pm\infty$ and we obtain
\begin{gather*}
e^{\omega\epsilon}
\int_{-\infty-\mathrm{i}\epsilon}^{+\infty-\mathrm{i}\epsilon} \text{d}x \frac{ e^{-\mathrm{i}\omega x} }{ \sinh^n{x} }
=
e^{\omega\epsilon}
\int_{-\infty - \mathrm{i} \frac{\pi}{2} }^{+\infty - \mathrm{i} \frac{\pi}{2}} \text{d}x \frac{ e^{-\mathrm{i}\omega x} }{ \sinh^n{x} }
\end{gather*}
Regularization parameter was useful to avoid any singularity on the path, but now is not useful anymore and we can put $\epsilon\to 0^+$, following variable substitution $x+\mathrm{i}\pi/2\to x$ leaving us with
\begin{gather*}
\mathrm{i}^n
e^{-\frac{\pi}{2}\omega}
\int_{-\infty}^{+\infty} \text{d}x \frac{ e^{-\mathrm{i}\omega x} }{ \cosh^n{x} }
\end{gather*}
that with variabile change $e^{2x}\to x$ becomes
\begin{gather*}
\mathrm{i}^n 2^{n-1}
e^{-\frac{\pi}{2}\omega}
B \left( \frac{n-\mathrm{i}\omega}{2}, \frac{n+\mathrm{i}\omega}{2} \right)
\end{gather*}
and we can write
\begin{gather*}
B \left( \frac{n-\mathrm{i}\omega}{2}, \frac{n+\mathrm{i}\omega}{2} \right)
=
\left( \Gamma (n) \right)^{-1}
\Gamma \left( \frac{n-\mathrm{i}\omega}{2} \right)
\Gamma \left( \frac{n+\mathrm{i}\omega}{2} \right)
\end{gather*}
Now, using just basic properties of the gamma function, it is not too difficult to show that
\begin{gather*}
\begin{cases}
\Gamma \left( \frac{n-\mathrm{i}\omega}{2} \right)
\Gamma \left( \frac{n+\mathrm{i}\omega}{2} \right)
=
\frac{\pi\omega}{\left( e^{\frac{\pi\omega}{2}} - e^{-\frac{\pi\omega}{2}} \right) }
\prod_{k=1}^{\frac{n}{2}-1} \left( \left( \frac{n}{2} - k \right)^2 + \frac{\omega^2}{4} \right)
& n \, \text{even}
\\
\Gamma \left( \frac{n-\mathrm{i}\omega}{2} \right)
\Gamma \left( \frac{n+\mathrm{i}\omega}{2} \right)
=
\frac{2\pi}{e^{\frac{\pi\omega}{2}} + e^{-\frac{\pi\omega}{2}} }
\prod_{k=1}^{\frac{n-1}{2}} \left( \left( \frac{n}{2} - k \right)^2 + \frac{\omega^2}{4} \right)
& n \, \text{odd}
\end{cases}
\end{gather*}
So that, in the end
\begin{equation*}
\lim_{\epsilon\to 0^+}
\int_{-\infty}^{+\infty} \text{d}x \frac{ e^{-\mathrm{i}\omega x} }{ \left( \sinh{( x-\mathrm{i}\epsilon )} \right)^n }
=
\begin{cases}
\frac{ \mathrm{i}^n 2^{n-1} }{\Gamma(n)}
\frac{\pi\omega}{\left( e^{\pi\omega} - 1 \right) }
\prod_{k=1}^{\frac{n}{2}-1} \left( \left( \frac{n}{2} - k \right)^2 + \frac{\omega^2}{4} \right)
& n \, \text{even}
\\
\frac{ \mathrm{i}^n 2^{n-1} }{\Gamma(n)}
\frac{2\pi}{e^{\pi\omega} + 1 }
\prod_{k=1}^{\frac{n-1}{2}} \left( \left( \frac{n}{2} - k \right)^2 + \frac{\omega^2}{4} \right)
& n \, \text{odd}
\end{cases}
\end{equation*}
| Hint $$\sinh(ix)=i\sin (x)$$
and then break down $\sin^n$. Not so difficult.
| {
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"timestamp": "2023-03-29T00:00:00",
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Positive Solution of Exponential Equation The equation is $2^{x+1}+2^{1/x^2}=6$.
By inspection I see that $1$ is a solution. However, after trying to algebraically isolate for $x$, I was unable to deduce that $1$ is a solution. Given the simplicity of the value of the solution, I was wondering if it would be possible to do so?
Also, I am only looking for the positive solution. However, when graphically analyzing the equation, I noticed that there exists a negative solution that Wolfram Alpha is incapable of giving an exact form for. Does there exist an exact form of the negative solution other than an infinite decimal?
| I don't believe it is actually possible to isolate for $x$ in this case, however I was able to "algebraically deduce" that 1 is a solution using the following rationale involving the AM-GM inequality twice:
\begin{align*}
2^{x+1}+2^{\frac{1}{x^2}}
& =
6
\\
2(2^x)+2^{\frac{1}{x^2}}
& =
6
\\
2^x+2^x+2^{\frac{1}{x^2}}
& =
6
\\
\implies\frac{2^x+2^x+2^{\frac{1}{x^2}}}{3}
& \ge
\sqrt[3]{2^x2^x2^{\frac{1}{x^2}}}
\\
\sqrt[3]{2^{x+x+\frac{1}{x^2}}}
& \le
2
\\
x+x+\frac{1}{x^2}
& \le
3
\\
\implies\frac{x+x+\frac{1}{x^2}}{3}
& \ge
\sqrt[3]{1}
\\
x+x+\frac{1}{x^2}
& \ge
3
\\
\therefore x+x+\frac{1}{x^2}
& =
3
\\
x=x
& =
\frac{1}{x^2}
\\
x
& =
1
\end{align*}
I tried deducing the negative solution. However, I was not able to, nor do I believe it is possible to do so.
| {
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Question about rational exponents From a book, the question goes like this:
This is how John simplified $\sqrt[3]{16x^5}$.
$\sqrt[3]{16x^5} = \sqrt[3]{8} \ \cdot \sqrt[3]{2x^5} = 2\sqrt[3]{2x^5}$
Explain the mistake in John's work and then correct it.
My answer: Isn't John correct? I mean,
\begin{align}
\sqrt[3]{16x^5} & = \sqrt[3]{(2\cdot 8)\cdot x^5} \\
& = (2\cdot 8 \cdot x^5)^{1/3}\\
& = 2^{1/3}\cdot8^{1/3}\cdot x^{5/3} \qquad \text{from $(xy)^n = x^ny^n$ }\\
& = 8^{1/3}\cdot2^{1/3}\cdot x^{5/3} \qquad \text{just rearranging}\\
& = \sqrt[3]{8} \cdot (2x^5)^{1/3}\\
& = \sqrt[3]{8} \cdot \sqrt[3]{2x^5} \qquad \text{which is the same with John's answer}\\
& = 2\sqrt[3]{2x^5}
\end{align}
Am I missing something here? Thanks for the help!
| I was an educator for 20 years, teaching mathematics at the secondary level throughout my career. Working from that context, and assuming the book in question was a high school mathematics textbook (perhaps Algebra 2), the John was mistake was leaving $x^5$ within the radical.
From a high school algebra student, the simplify for would be achieved as follows.
$$\sqrt[3]{16x^5}$$
$$\sqrt[3]{8\cdot 2\cdot x^3\cdot x^2}$$
$$\sqrt[3]{8x^3}\sqrt[3]{2x^2}$$
$$2x\sqrt[3]{2x^2}$$
While writing the expression using rational exponents would be correct, high school students and their textbooks generally do not use rational exponents.
| {
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} |
Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln(1-x)$? For $|x|<1,$ we have
$$
\begin{aligned}
& \frac{1}{1-x}=\sum_{k=0}^{\infty} x^k \quad \Rightarrow \quad \ln (1-x)=-\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x)}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k dx \\
& =-\sum_{k=0}^{\infty} \frac{1}{(k+1)^2} \\
& =- \zeta(2) \\
& =-\frac{\pi^2}{6}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln xdx \\
& =\sum_{k=0}^{\infty} \frac{1}{k+1}\cdot\frac{1}{(k+1)^2} \\
& =\zeta(3) \\
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^2 x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^2 xdx \\
& =-\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{2}{(k+1)^3} \\
& =-2 \zeta(4) \\
& =-\frac{\pi^4}{45}
\end{aligned}
$$
In a similar way, I dare guess that
$$\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x =(-1)^{n+1}\Gamma(n)\zeta(n+2),$$
where $n$ is a non-negative real number.
Proof:
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^k \ln ^n xdx \\
\end{aligned}
$$
Letting $y=-(k+1)\ln x $ transforms the last integral into a Gamma function as
$$
\begin{aligned}
\int_0^1 x^k \ln ^n x d x & =\int_{\infty}^0 e^{-\frac{k}{k+1}}\left(-\frac{y}{k+1}\right)^n\left(-\frac{1}{k+1} e^{-\frac{y}{k+1}} d y\right) \\
& =\frac{(-1)^n}{(k+1)^{n+1}} \int_0^{\infty} e^{-y} y^n d y \\
& =\frac{(-1)^n \Gamma(n+1)}{(k+1)^{n+1}}
\end{aligned}
$$
Now we can conclude that
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x & =(-1)^{n+1} \Gamma(n+1) \sum_{k=0}^{\infty} \frac{1}{(k+1)^{n+2}} \\
& =(-1)^{n+1} \Gamma(n+1)\zeta(n+2)
\end{aligned}
$$
Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln (1-x)$?
Your comments and alternative methods are highly appreciated?
| For natural number $n$, I can use differentiation of the integral
$$
\begin{aligned}
I(a) & =\int_0^1 x^a \ln (1-x) d x \\
& =-\int_0^1 x^a \sum_{k=0}^{\infty} \frac{1}{k+1} x^{k+1} d x \\
& =-\sum_{k=0}^{\infty} \frac{1}{k+1} \int_0^1 x^{a+k+1} d x \\
& =-\sum_{k=0}^{\infty}\left(\frac{1}{k+1} \cdot \frac{1}{a+k+2}\right)
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ by $n$ times at $a=-1$ yields our integral.
$$
\begin{aligned}
I & =-\left.\sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{(-1)^n n !}{(a+k+2)^{n+1}}\right|_{a=-1} \\
& =(-1)^{n+1} n ! \sum_{k=0}^{\infty} \frac{1}{k+1} \cdot \frac{1}{(k+1)^{n+1}} \\
& =(-1)^{n+1} n ! \zeta(n+2)
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
} |
How to prove identity $\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$? Looking at Jolley, Summation of Series, formula 445:
$\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$
How can one prove this?
Considering $\left( \sum^{n-1}_{j=0}Z_j\right)^2 = \sum^{n-1}_{j=0} Z_j^2 +\sum^{n-1}_{j=0}\sum^{n-1}_{i\neq j} Z_jZ_i$ from here
I thought one could take the log differential of the well known sine product:
$\begin{align}
\prod_{k=0}^{n-1}\sin\left(\theta + \frac{k\pi}{n}\right)
&= 2^{1-n} \sin(n\theta)
\end{align}$
to get
$n \cot(n\theta) = \sum_{k=0}^{n-1}\cot\left(\theta+\frac{k\pi}{n}\right)$
or
$-n \cot(\frac{n\pi}{2}+n\theta) = \sum_{k=0}^{n-1}\tan\left(\theta+\frac{k\pi}{n}\right)$
then square to get
$n^2 \cot(\frac{n\pi}{2}+n\theta)^2 = (\sum_{k=0}^{n-1}\tan\left(\theta+\frac{k\pi}{n})\right)^2$
but I cannot see how to show the
$\sum _{j=0}^{n-1} \sum _{k=0}^{n-1} \tan \left(\theta +\frac{\pi j}{n}\right) \tan \left(\theta +\frac{\pi k}{n}\right) [j\neq k]$
necessary for the $n(n-1)$ part of the identity.
Perhaps somebody could show a more successful method.
| So, here is my idea, it occurred to me when I saw this well known sine product, that it will appear if we integrate the function
$$f\left(\theta\right) = \sum_{k=0}^{n-1} \tan^{2}\left(\theta+\frac{k\pi}{n}\right).$$
Since we are going to differentiate in the end, I won't be bothering to write integration constant:
$$\int f\left(\theta\right)d\theta = \int \sum_{k=0}^{n-1} \left(\sec^{2}\left(\theta+\frac{k\pi}{n}\right) - 1\right) d\theta = \sum_{k=0}^{n-1} \tan\left(\theta+\frac{k\pi}{n}\right) - n\theta $$
$$\int f\left(\theta\right)d\theta = - n\theta - \sum_{k=0}^{n-1} \cot\left(\theta - \frac{\pi}{2}+\frac{k\pi}{n}\right) = - n\theta - \sum_{k=0}^{n-1} \frac{\frac{d}{d\theta}\sin\left(\theta - \frac{\pi}{2}+\frac{k\pi}{n}\right)}{\sin\left(\theta - \frac{\pi}{2}+\frac{k\pi}{n}\right)}$$
$$\int f\left(\theta\right)d\theta = - n\theta - \frac{\frac{d}{d\theta}\Pi_{k=0}^{n-1}\sin\left(\theta - \frac{\pi}{2}+\frac{k\pi}{n}\right)}{\Pi_{k=0}^{n-1}\sin\left(\theta - \frac{\pi}{2}+\frac{k\pi}{n}\right)} = -n\theta -\frac{\frac{d}{d\theta}2^{1-n}\sin\left(n\left(\theta-\frac{\pi}{2}\right)\right)}{2^{1-n}\sin\left(n\left(\theta-\frac{\pi}{2}\right)\right)}$$
$$
\int f\left(\theta\right)d\theta = - n\theta - n\frac{\cos\left(n\theta-\frac{n\pi}{2}\right)}{\sin\left(n\theta-\frac{n\pi}{2}\right)} = -n\theta -n\cot\left(n\theta-\frac{n\pi}{2}\right).
$$
If you sum $n\pi$ in the argument of $\cot$ it won't change its value, so:
$$\int f\left(\theta\right)d\theta = -n\theta -n\cot\left(n\theta+\frac{n\pi}{2}\right),$$
now differentiating we get:
$$f\left(\theta\right) = -n +n^{2}\csc^{2}\left(n\theta+\frac{n\pi}{2}\right) = -n +n^{2}\left(1+\cot^{2}\left(n\theta+\frac{n\pi}{2}\right)\right)$$
which give us the equality desired.
| {
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In any triangle $\triangle ABC$, show that $4R\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})=r$ This is a problem problem I found in a JEE examination prep textbook, it was a "starred" question which I believe implies that it is more challenging than usual. It goes as follows:
In any triangle $\triangle ABC$, show that $$4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=r$$
Hint: $$2R^2\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=\Delta$$
Here is my attempt at it. I want to know if this is correct and if there any better alternative approaches to achieve the same result, please do share them!
We know that:
$$\Delta=rs$$
Using the given hint:
$$2R^2\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=rs$$
$$16R^2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=r\left(\frac{a+b+c}{2}\right)$$
$$16R\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=r\left(\sin\left(A\right)+\sin\left(B\right)+\sin\left(C\right)\right)$$
Now, focusing on the equation of the right hand side for a bit, we know that:
$$A+B+C=\pi$$
$$\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$$
$$\sin\left(A\right)+\sin\left(B\right)+\sin\left(C\right)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)+2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)$$
$$2\cos\left(\frac{C}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{A+B}{2}\right)\right)$$
$$4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$$
Now substituting this back into the original problem:
$$16R\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=\left(4r\right)\left[\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)\right]$$
And that gives us:
$$4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=r$$
| Your proof is fine.
Another approach, with a little less trigonometry, uses Heron's formula and the law of cosines.
Square both sides and use $\sin^2(x/2)=\frac{1-\cos(x)}{2}$ so you want:
$$2R^2(1-\cos A)(1-\cos B)(1-\cos C)=r^2$$
The law of cosines says:
$$\cos A=\frac{b^2+c^2-a^2}{2bc},$$ so $$1-\cos A=\frac{a^2-(b-c)^2}{2bc}=\frac{(a+b-c)(a+c-b)}{2bc}$$ and likewise for $B,C.$
Starting with the hint:
$$\Delta =2R^2\sin A\sin B\sin C.$$
Multiply by $4R\Delta$, and use the law of sines:
$$4R\Delta^2=\Delta abc$$
Use Heron's formula on the left and $\Delta=rs$ on the right to get:
$$4Rs\frac{(a+b-c)(a+c-b)(b+c-a)}{8}=rsabc.$$
Squaring and re-arranging, we get:
$$2R^2\frac{(a^2-(b-c)^2)(b^2-(a-c)^2)(c^2-(a-b)^2)}{8a^2b^2c^2}=r^2$$
Then, by the law of cosines result above, this can be rewritten as:
$$2R^2(1-\cos A)(1-\cos B)(1-\cos C)=r^2$$
Which is what we wanted.
| {
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"answer_id": 1
} |
Evaluation of Trigonometric Limit having 5 terms Evaluation of
$\displaystyle \lim_{h\rightarrow 0}\bigg[\frac{\sin(60^\circ+4h)-4\sin(60^\circ+3h)+6\sin(60^\circ+2h)-4\sin(60^\circ+h)+\sin(60^\circ)}{h^4}\bigg]$
Here above limit is in $(0/0)$ form
So we have using D, L Hopital rule
$\displaystyle \lim_{h\rightarrow 0}\bigg[\frac{4\cos(60^\circ+4h)-12\cos(60^\circ+3h)+12\cos(60^\circ+2h)-4\cos(60^\circ+h)+0}{4h^3}\bigg]$
Again above limit is in $(0/0)$ form
So using D, L Hopital rule
$\displaystyle \lim_{h\rightarrow 0}\frac{-16\sin(60^\circ+4h)+36\sin(60^\circ+3h)-24\sin(60^\circ+2h)+4\sin(60^\circ+h)}{12h^2}$
Above limit is in $(0/0)$ form
So agian using D, L Rule
$\displaystyle \lim_{h\rightarrow 0}\frac{-64\cos(60^\circ+4h)+108\cos(60^\circ+3h)-48\cos(60^\circ+2h)+4\cos(60+h)}{24h}$
Again using D, L rule , We get
$\displaystyle \lim_{h\rightarrow 0}\frac{256\sin(60^\circ+4h)-324\sin(60^\circ+3h)+96\sin(60^\circ+2h)-4\sin(60+h)}{24}$
$\displaystyle \lim_{h\rightarrow 0}\frac{24\sin(60^\circ)}{24}=\frac{\sqrt{3}}{2}$
Above is very lengthy way
Please explain me some short way
Thanks
| The numerator can be restated as
$$16\sin^4\left(\frac{h}{2}\right) \sin(2h + 60^\circ).$$
Now, applying the first-order expansion and using radians instead of degrees (so $60^\circ$ transforms into $\frac{\sqrt{3}}{2}$), the limit is
$$\lim_{h\to 0} \frac{16\left(\frac{h}{2}\right)^4\left(\frac{\sqrt{3}}{2} + h\right)}{h⁴} = \frac{\sqrt{3}}{2}.$$
| {
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Showing $x^4 + 1$ is irreducible in $\mathbb{Q}[x]$. Clearly, none of the roots are in $\mathbb{Q}$ so $f(x) = x^4 + 1$ does not have any linear factors. Thus, the only thing left to check is to show that $f(x)$ cannot reduce to two quadratic factors.
My proposed solution was to state that $f(x) = x^4 + 1 = (x^2 + i)(x^2 - i)$ but $\pm i \not\in \mathbb{Q}$ so $f(x)$ is irreducible.
However, I stumbled across this post $x^4 + 1$ reducible over $\mathbb{R}$... is this possible? with a comment suggesting that $x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$ which turns out to be a case that I did not fully consider. It made me realize that $\mathbb{Q}[x]$ being a UFD only guarantees a unique factorization of irreducible elements in $\mathbb{Q}[x]$ (which $x^2 \pm i$ nor $x^2 \pm \sqrt{2} x + 1$ aren't in $\mathbb{Q}[x]$) so checking a single combination of quadratic products is not sufficient.
Therefore, what is the ideal method for checking that $x^4 + 1$ cannot be reduced to a product of two quadratic polynomials in $\mathbb{Q}[x]$? Am I forced to just brute force check solutions of $x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d)$ don't have rational solutions $(a,b,c,d) \in \mathbb{Q}^4$?
| You can indeed try a factorization of the form
$x^4+1=(x^2+ax+b)(x^2+cx+d).$
Expanding the right side and matching terms with like powers gives
$x^3$ terms: $a+c=0,c=-a$
$x^2$ terms: $ac+b+d=0,b+d=a^2$
$x^1$ terms: $ad+bc=a(d-b)=0,a=0$ or $d=b$
The case $a=0$ leads to $(x^2+i)(x^2-i)$ which fails to lie in $\mathbb Q[x]$. So we try $d=b$, which then means $d=b=a^2/2$ from the matching of $x^2$ terms. Then matching the $x^0$ terms gives:
$x^0$ terms: $bd=b^2=1.$
Then $b=a^2/2=\pm1$ and neither choice of the $\pm$ sign allows a rational value for $a$. In fact only $b=+1,a=\pm\sqrt2$ admits a quadratic-quadratic factorization even over the reals.
| {
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If $O$ is a point in square $ABCD$, then $\angle OAB+\angle OBC+\angle OCD+\angle ODA\ge\frac{3\pi}4$
$O$ is a point in square $ABCD$. Show that
$$\angle OAB+\angle OBC+\angle OCD+\angle ODA\ge\frac{3\pi}4$$
My solution is complex numbers. Let $-C=A=1$, $-D=B=i$. So
\begin{align*}&\mathrm{Arg}\left(\frac{B-A}{O-A}\cdot\frac{C-B}{O-B}\cdot\frac{D-C}{O-C}\cdot\frac{A-D}{O-D}\right)
\\={}&\mathrm{Arg}\left(\frac4{O^4-1}\right)=-\mathrm{Arg}(O^4-1).
\end{align*}
Here, $O$ is taken in the square, but $O^4$ is hard to manage. Firstly, when $|O|\le\dfrac{\sqrt2}2$, it can be in any direction then $O^4$’s locus is circle centered at $0$ with radius $\dfrac14$. However if $|O|$ is bigger, only some direction is in the square, namely $O$ moves on four arcs.
So my solution gets stuck here. You could actually change a method completely, but I’d like to see if my method would work.
| If someone wants to have an entirely algebraic trigonometric solution, not using complex numbers, nor having any geometric insight, below is one:
Let's assume $A=(0,0), B=(0,1), C=(1,1), D=(1,0)$, and $O=(a,b)$, where $0<a,b<1$.
It is easy to see that:
$$\angle OAB=\arctan \frac{a}{b}\\ \angle OBC=\arctan \frac{1-b}{a}\\\angle OCD=\arctan \frac{1-a}{1-b}\\\angle ODA=\arctan \frac{b}{1-a}.\\$$
Now, if $a=b$, or $a+b=1$, with the help of the relation $\arctan x+ \arctan \frac{1}{x}=\frac{\pi}{2}$, we get that:
$$A=\arctan \frac{a}{b}+\arctan \frac{1-b}{a}+\arctan \frac{1-a}{1-b}+\arctan \frac{b}{1-a}=\pi.$$
Hence four cases happen:
$$1>b+a, b>a \\ 1>b+a, a>b\\a+b>1, a>b\\a+b>1, b>a .$$
Assume $1>b+a, b>a$, or $a+b>1, a>b.$ In these cases, $\frac{1-b}{a}>\frac{1-a}{b}$ So,
$$A=\arctan \frac{a}{b}+\arctan \frac{1-b}{a}+\arctan \frac{1-a}{1-b}+\arctan \frac{b}{1-a}>\\\arctan \frac{a}{b}+(\arctan \frac{1-a}{b}+\arctan \frac{b}{1-a})+\arctan \frac{1-a}{1-b}=\\ \frac{\pi}{2}+\arctan \frac{a}{b}+\arctan \frac{1-a}{1-b}.$$
On the other hand, since either $a>b$ or $b>a$, $\arctan \frac{a}{b}+\arctan \frac{1-a}{1-b}> \frac{\pi}{4}$ (note that $\arctan 1= \frac{\pi}{4}$). As a result, $A>\frac{3\pi}{4}.$
If we assume $1>a+b, a>b$ or $a+b>1, b>a$, in these cases, $\frac {a}{b}>\frac{1-b}{1-a}$. So,
$$A=\arctan \frac{a}{b}+\arctan \frac{1-b}{a}+\arctan \frac{1-a}{1-b}+\arctan \frac{b}{1-a}>\\ (\arctan \frac{1-b}{1-a}+\arctan \frac{1-a}{1-b})+\arctan \frac{1-b}{a}+\arctan \frac{b}{1-a}=\\ \frac{\pi}{2}+\arctan \frac{1-b}{a}+\arctan \frac{b}{1-a};$$
since either $a+b>1$ or $1>a+b$, we have $\arctan \frac{1-b}{a}+\arctan \frac{b}{1-a}> \frac{\pi}{4}$. So, $A>\frac{3\pi}{4}.$
We are done.
Fig. 1 : Representation of the sum of the four angles as a surface $z=f(a,b)$ (for $0<a,b<1$). One notices a natural invariance by $k \pi/2$ rotations around point $(1/2,1/2)$.
| {
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Identity regarding roots of unity Let $\zeta$ be a primitive $n$-th root of unity and $m \in \{0,1,\dots,n-1\}$. I am interested in finding the value of the following expression:
$$\sum_{k=1}^{n-1}\frac{\zeta^{mk}}{1-\zeta^k}.$$
This has come up in a context where it should be a rational number (in fact it seems like it will be of the form $\frac{r}{2}$ where $r \in \mathbb Z$). For example for $m=0$ I can get the values $\frac{n-1}{2}$ by letting $f(x) = \frac{x^n-1}{x-1}$ and noticing that the desired sum equals $\frac{f'(1)}{f(1)}$. However I am not able to find such a "trick" when $m$ is nonzero.
| Note that with $\zeta$ being primitive the powers $\zeta^k$ with $0\le
k\le n-1$ just permute the powers of $\zeta = \exp(2\pi i/n)$ so we may
take that as our root. Next introduce
$$f(z) = \frac{z^m}{1-z} \frac{n/z}{z^n-1}.$$
We have for $1\le k\le n-1$
$$\mathrm{Res}(f(z); z = \zeta^k)
= \frac{\zeta^{km}}{1-\zeta^k}.$$
Residues sum to zero and with $m\lt n$ the residue at infinity is
zero. Hence the desired sum must be minus the residue at $z=1.$ We
write (the minus from the residue cancels the minus from the $1/(1-z)$
term):
$$- \mathrm{Res}(f(z); z=1) =
\mathrm{Res}\left(\frac{z^m}{z-1}
\frac{n/z}{(z-1)(1+z+z^2+\cdots+z^{n-1})}; z=1\right)
\\ = \left. \left( \frac{nz^{m-1}}{1+z+z^2+\cdots+z^{n-1}}
\right)' \right|_{z=1}
\\ = \left. \left( \frac{n(m-1)z^{m-2}}{1+z+z^2+\cdots+z^{n-1}}
- \frac{nz^{m-1} (1+2z+3z^2+\cdots+(n-1)z^{n-2})}
{(1+z+z^2+\cdots+z^{n-1})^2}
\right)\right|_{z=1}
\\ = \frac{n(m-1)}{n} - \frac{n \frac{1}{2} (n-1) n}{n^2}
\\ = m-1 - \frac{1}{2} (n-1).$$
| {
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"answer_id": 1
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Calculating $\int_{-\infty}^{\infty} \frac{\sin(t)}{t^2+2t+2} \ dt$ My attempt: We have $$ \int_{\Gamma} f(z) \ dt = \int_{\gamma} f(z) \ dz + \int_{\sigma} f(z) \ dz = 2 \pi i \sum_{z_0 \text{ pole inside } \Gamma} \mathrm{res}_{z_0}(f) $$ where $\Gamma$ is contour of the upper semi-circle of radius $r$, $\gamma$ is the real line segment going from $-r$ to $r$, $\sigma$ is the semi-circle, and $f(z) = \frac{sin(z)}{z^2 + 2z +2}$.
We have $$ \frac{\sin(t)}{t^2+2t+2} = \frac{\sin(t)}{(t+1 -i)(t+1+i)} $$ so the function has simple poles at $z = -1 \pm i$, and therefore
$$ \int_{-r}^r \frac{\sin(t)}{t^2+2t+2} \ dt = - \int_0^{ \pi} \frac{\sin(re^{i \theta} )ri e^{i \theta}}{(re^{i \theta})^2 + 2 re^{i \theta} + 2} \ d\theta + \pi \sin(-1+i). $$
By taking the limit on both sides we get
$$ \int_{-\infty}^{\infty} \frac{\sin(t)}{t^2+2t+2} \ dt = - \lim_{r \to \infty} \int_0^{ \pi} \frac{\sin(re^{i \theta}) ri e^{i \theta}}{(re^{i \theta})^2 + 2 re^{i \theta} + 2} \ d\theta + \pi \sin(-1+i) $$
but I don't know how to simplify the right hand side.
| \begin{align*}I & =\int_{-\infty}^\infty\frac{\sin x}{(x+1)^2+1}dx=\int_{-\infty}^{\infty}\frac{\sin(x-1)}{x^2+1}dx \\ & =\cos(1)\int_{-\infty}^\infty\frac{\sin x}{x^2+1}dx-\sin(1)\int_{-\infty}^\infty\frac{\cos x}{x^2+1}dx \\ &=-2\sin(1)\int_{0}^\infty\frac{\cos x}{x^2+1}dx=-2\sin(1)\frac{\pi}{2e}=\frac{-\pi\sin(1)}{e}. \end{align*}
(See here.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to compute $\lim_{ x\to 0} \frac{\sqrt{x+2}-\sqrt{2-x}}{\sqrt[3]{x+2}-\sqrt[3]{2-x}}$? How to compute the following limit?
$$\lim_{ x\to 0} \frac{\sqrt{x+2}-\sqrt{2-x}}{\sqrt[3]{x+2}-\sqrt[3]{2-x}}$$
In the context of the book I am reading, I must use some change of variables or multiply the expressions by some "conjugate". I tried to multiply it by $\left(\sqrt{2-x}+\sqrt{x+2}\right)$ to obtain:
$$\frac{2 x}{\left(\sqrt[3]{x+2}-\sqrt[3]{2-x}\right)
\left(\sqrt{2-x}+\sqrt{x+2}\right)}$$
And I tried to make the following change of variables $2+x =t^6$ and then I obtained:
$$\frac{t^3-\sqrt{4-t^6}}{t^2-\sqrt[3]{4-t^6}}$$
But I couldn't go much farther than this. Can you give me a hint?
| Use L'Hôpital's rule.
With
$$f(x){\mapsto }\sqrt{x+2}-\sqrt{2-x}$$
$$g(x){\mapsto }\left(x+2\right)^{\frac{1}{3}}-\left(2-x\right)^{\frac{1}{3}}$$
we compute the limit of the quotient of the first derivatives
$$\underset{x\to 0}{\text{lim}}\frac{f'(x)}{g'(x)}=\underset{x\to 0}{\lim }\, \frac{\frac{1}{2 \sqrt{x+2}}+\frac{1}{2 \sqrt{2-x}}}{\frac{1}{3 (x+2)^{2/3}}+\frac{1}{3 (2-x)^{2/3}}}=\frac{3 \sqrt[6]{2}}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Locus of points of intersection of some tangents of an ellipse The quadratic $x^2+4y^2-axy=0$, when $a>4$ represents two pair of lines through origin. Let them cut the ellipse $x^2+4y^2=4$ at A, B and C,D. The tangents at A and B or at C and D would be parallel which intersect at infinity. But, the tangents at A and C; A and B; C and B; B and D meet on some curve(s). Let us find these curves.
Let the point of intersection of two such tangents (e.g., at A and C) be E$(h,k)$, then line joining A and C is chord of contact of the ellipse namely $hx+4ky=4$. If we homogenize this with the equation of ellipse, we get the combined equation OA and OC as
$$x^2+4y^2=4(xh+4ky)^2/16 \implies (1-h^2/4)x^2+(4-4k^2)y^2-2hkxy=0$$
Now let us compare it with another given combined equation of OA and OC which is $x^2+4y^2-axy=0$, we get
$$\frac{1-h^2/4}{1}=\frac{4-4k^2}{4}=\frac{2hk}{a}\implies h^2=4k^2.$$
So we get the locus of point of intersection of these pair of tangents as $x^2=4y^2.$
The question is: How else this locus could be obtained?
| Let $y=mx$ be the lines OA and OC represented by $x^2+4y^2-axy=0$,then $m_1,m_2=1/4$ and let $|a|>4$. Further take $m_1=m, m_2=1/(4m)$.
The points A and C are $$A\left(\frac{2}{\sqrt{1+4m^2}},\frac{m}{\sqrt{1+4m^2}} \right),\quad C\left(\frac{4m}{\sqrt{1+4m^2}}, \frac{1}{\sqrt{1+4m^2}}\right).$$
Then equation of tangents at A and C are:
$$x+4my=2\sqrt{1+4m^2},\quad mx+y=\sqrt{1+4m^2},$$
Their point of intersection E is:
$$E\left(x=\frac{2\sqrt{1+4m^2}}{1+2m},y=\frac{\sqrt{1+4m^2}}{1+2m}\right)$$
By eliminating $m$. we get the required locus of point of intersection of tangents as $x^2=4y^2.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $x$ such that $\sqrt{x+1} - \sqrt{1-x} = 1$ To solve this equation, I started by putting the condition $x\in [-1, 1]$, then squared a few times: $\sqrt{x+1} - \sqrt{1-x} = 1 \iff x + 1 +1-x-2\sqrt{1-x^2} =1 \iff 2\sqrt{1-x^2}=1 \iff 4(1-x^2)=1 \iff 4x^2=3 \iff x=\pm \frac{\sqrt{3}}{2}$
This, however, is not the right solution, as $-\frac{\sqrt{3}}{2}$ returns $-1$, not $1$. My question is where did I miss a condition that excludes the negative "solution"? I expect somewhere along the line I squared where I wasn't allowed to square without an additional condition, hoping that I don't have to check these solutions every time.
| Since $\sqrt{x+1} - \sqrt{1-x} = 1$, we have
$\sqrt{x+1} - \sqrt{1-x}>0$, which gives
$$\tag{1}\sqrt{x+1}>\sqrt{1-x}.$$ Therefore, if $x=-\frac{\sqrt{3}}{2}$,
inequality $(1)$ is not satisfied.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(\sum x^2)^3\ge9\sum x^4yz$
Prove that $\displaystyle\left(x^2+y^2+z^2\right)^3\ge9\left(x^4yz+y^4xz+z^4xy\right)$, for $x$, $y$, $z\in\Bbb R_+$.
The $pqr$ method doesn't seem possible because the power is too high.
$$\iff\left(p^2-2q\right)^2\ge9r\left(p^3-3pq+3r\right).$$
Then expand the expression to get
$$\sum x^6+3\sum\left(x^4y^2+x^2y^4\right)+6x^2y^2z^2\ge9\sum x^4yz.$$
I wanted to use SOS but cannot find the weight of three squares, my progress:
$$3\sum x^4(y-z)^2=3\sum\left(x^4y^2+x^2y^4\right)-6\sum x^4yz.$$
Whats left is $\displaystyle\sum x^6+6x^2y^2z^2-3\sum x^4yz$. I have trouble dealing with it.
| This is a proof that continues your SOS attempt, which is to prove
$$
\sum x^6+6x^2y^2z^2\geqslant3\sum x^4yz
$$
By noticing the following identity, this inequality is obviously true by Schur's inequality
$$
\sum(x^6+6x^2y^2z^2-3x^4yz)=\sum x^4(x-y)(x-z)+4(x-y)^2(y-z)^2(z-x)^2+2\sum yz(x-y)^2(x-z)^2+2xyz\sum x(x-y)(x-z)
$$
This proof is by brute force using the Triangle of Coefficient Method. Though I've found a way to simplify this inequality, I didn't managed to solve it that way.
Namely, by dividing both sides with $(xyz)^2$ , we obtain
$$
\sum\left(\frac{x^2}{yz}\right)^2+6\geqslant3\sum\frac{x^2}{yz}
$$
Let $a=(x^2)/(yz),~b=(y^2)/(zx),~c=(z^2)/(xy)$ , we have to prove that
$$
a^2+b^2+c^2+6\geqslant 3(a+b+c)
$$
for positive real numbers $a,b,c$ such that $abc=1$ .
I hope someone would continue my attempt.
Edit. This edition contains a new proof of this inequality using SOS Method by LasterCircle
$$
\text{LHS}-\text{RHS}=\frac12\sum\left[\sum x^4+2xy(x^2+xy+y^2)-2xyz\sum x\right](x-y)^2
$$
It's apparent that $\sum x^4+2xy(x^2+xy+y^2)-2xyz\sum x>0$ , the inequality is true.
Edit2. River Li has given a nice proof in the comment.
And here's a simpler SOS
$$
\sum(x^6-3x^4yz)+6x^2y^2z^2=\frac1{18}\sum\left(x(x^2-yz)\right)^2+\frac12\sum x^2(x^2-y^2)(x^2+y^2)+\frac12\sum{x^2y^2(x-y)^2}\geqslant0
$$
| {
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How to show $k \cdot a + c \cdot b \ge \frac{1}{2} \cdot (k+c) \cdot (a+b)$? Assume k, c, a, b $\gt$ 0 and that $k \ge c$ and $a \ge b$. Prove $k \cdot a$ + $c \cdot b$ $\ge$ $\frac{1}{2}\cdot (k+c) \cdot (a+b)$. This is what I have so far:
$2\cdot k\cdot a +2\cdot c \cdot b$ = $(k\cdot a + k\cdot a)$ + $(c\cdot b + c\cdot b)=(k\cdot a + c\cdot b)$ + $(k\cdot a + c\cdot b)\ge(k\cdot a + c\cdot b)$ + $(k\cdot b + c\cdot b)$
| $(k-c)(a-b) \geq 0$
$ka+bc \geq bk+ac$
$2(ka+bc) \geq bk+ac+ka+bc$
$2(ka+bc) \geq (a+b)(k+c)$
| {
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"timestamp": "2023-03-29T00:00:00",
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The general solution of $2x^2y''-xy'+y=x$
Find the general solution of $2x^2y''-xy'+y=x$.
My attempt:
Consider its homogeneous version: $2x^2y''-xy'+y=0$. Letting $y=x^r$ and substituting this and its first- and second- order derivatives in the equation, I have the characteristic equation $2r^2-3r+1=0$. This gives $r=1,\frac{1}{2}$. It follows that the complementary solution is
$y_c(x)=C_1x+C_2x^{\frac{1}{2}}$
Now back to the original ODE. Dividing the whole equation by $2x^2$,
$y''-\frac{1}{2x}y'+\frac{1}{2x^2}y=\frac{1}{2x}=g(x)$
Then consider the Wronskian
$W= \begin{vmatrix} {y_1} & {y_2 }\\\ {y_1'} & {y_2'} \end{vmatrix}= \begin{vmatrix} x & x^{\frac{1}{2}}\\\ 1 & (2x^{\frac{1}{2}})^{-1} \end{vmatrix}=-\frac{x^{\frac{1}{2}}}{2}$
$W_1= \begin{vmatrix} 0 & y_2\\\ g(x) & y_2' \end{vmatrix}=\begin{vmatrix} 0 & x^\frac{1}{2}\\\ \frac{1}{2x} & (2x^\frac{1}{2})^{-1}\end{vmatrix}=(-2x^{\frac{1}{2}})^{-1}$
$u_1'=\frac{W_1}{W}=\frac{1}{x} \implies u_1=ln|x|+C$
$W_2=\begin{vmatrix} y_1 & 0\\\ y_1' & g(x)\end{vmatrix}=\begin{vmatrix} x & 0\\\ 1 & \frac{1}{2x}\end{vmatrix}=\frac{1}{2}$
$u_2'=\frac{W_2}{W}=-x^{-\frac{1}{2}} \implies u_2=-2x^{\frac{1}{2}}+C$
So, we have the particular solution
$y_p(x)=u_1y_1+u_2y_2=x\ln|x|-2x$
Hence, our solution is
$y(x)=y_c(x)+y_p(x)=C_1x+C_2x^\frac{1}{2}+x(\ln|x|-2)$
I think I've solved the problem, but I'm currently reviewing this material for computational PDE course, so I'm not sure. Is something wrong with this?
| Your solution looks correct here is another approach:
$$2x^2y''-xy'+y=x$$
$$2y''-\dfrac {y'}{x}+\dfrac y{x^2}=\dfrac 1x$$
$$2y''-\left (\dfrac {y}{x}\right)'=\dfrac 1x$$
Integrate.
$$2y'-\dfrac {y}{x}=\ln x+ C$$
$$\left (\dfrac {2y}{\sqrt x}\right )'=\dfrac 1 {\sqrt x}(\ln x+ C)$$
$$\dfrac {2y}{\sqrt x} =2C \sqrt x +C_2+ \int \dfrac {\ln x} {\sqrt x}dx$$
$$y( x) =C_0 x +C_1\sqrt x+ x {\ln x} $$
| {
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Proving this by quadratic function
Given are $x$, $y$ and $z\ge0$, if $x^2+y^2+z^2+2xyz=1$, show that
\[x+y+z\le\dfrac32.\]
I want to solve this by quadratic function.
Let $x+y+z=k$, so $z=k-x-y$. Plug in and expand to have
\[-2 x {{y}^{2}}+2 {{y}^{2}}-2 {{x}^{2}} y+2 k x y+2 x y-2 k y+2 {{x}^{2}}-2 k x+{{k}^{2}}-1=0.\]
See this as a quadratic function of $y$, so $\Delta\ge0$, which implies that
\begin{align}4 {{x}^{4}}-8 k {{x}^{3}}+8 {{x}^{3}}+4 {{k}^{2}} {{x}^{2}}-12 {{x}^{2}}+8 k x-8 x-4 {{k}^{2}}+8\ge0.\tag{1}\end{align}
We need to find the minimum of $k$, so that $\exists~x$ for $(1)$ to be true. When $k=\dfrac32$,
\[4 {{x}^{4}}-4 {{x}^{3}}-3 {{x}^{2}}+4 x-1=0\implies\left[ x=1\operatorname{,}x=-1\operatorname{,}x=\frac{1}{2}\right]\]
are all solutions of $\Delta=0$. How to prove $k\ge\dfrac32$ don't work?
| If $x = 1$, clearly the desired inequality is true.
In the following, assume that $x < 1$.
Proceeding along OP's approach,
we have
$$\Delta = 4(1-x^2)[3 - 2k -(x-k+1)^2] \ge 0$$
which results in
$k \le 3/2.$
| {
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What is the probability f(p) of A winning the match?
Here's my approach:
Let A denote the event "A wins the game" and B denote the event "B wins the game"
I am supposed to calculate the probability f(p)
There are two possible cases:-
i) A wins within 4 games
Possible cases: AAA,ABAA,BAAA,AABA
ii) A doesn't win within 4 games
Possible cases: AABB(A and B occur 2 times each) ABABABAB...(A and B occur k times each) AA
AABB can be arranged in 6 ways
ABABAB.... can be arranged in $(2!)^k$ ways (A and B are adjacent to one another)
Now :
f(p) = P(case i) + P(case ii)
P(case ii) = $(6 (p^2) (1-p)^2 ) \sum_{k=1}^∞ 2^k . p^k . (1-p)^k. p^2 $=$(6 (p^4) (1-p)^2 ) \sum_{k=1}^∞ 2^k . p^k . (1-p)^k$
= $(6. 2 . p^5 . (1-p)^3)/(1-2p+2p^2)$ (The denominator seems correct)
P(case i ) = $p^3 +3p^3(1-p)$
f(p) = $p^3 +3p^3(1-p) + \frac{(6. 2 . p^5 . (1-p)^3)}{(1-2p+2p^2)}$
The expected answer is not produced by solving the above equation.Where did I go wrong?
Edit:-
I finally figured out the mistake that I made in the above solution.
In P(case ii) = $(6 (p^4) (1-p)^2 ) \sum_{k=1}^∞ 2^k . p^k . (1-p)^k $, I replaced $\sum_{k=1}^∞$ with $\sum_{k=0}^∞$ . There has to be one case when B does not occur after the tie, i.e. A occurs only two times.
P(case ii) = $(6 (p^4) (1-p)^2 ) \sum_{k=0}^∞ 2^k . p^k . (1-p)^k $
= $(6 (p^4) (1-p)^2 ) .[1+2p(1-p)^1+2^2.p^2.(1-p)^2+...]$
=$(6 (p^4) (1-p)^2 ) .\frac{1}{1-2p(1-p)}$
f(p) = P(case i) + P(case ii)
=$p^3 +3p^3(1-p)+(6 (p^4) (1-p)^2 ) .\frac{1}{1-2p(1-p)}$
=$p^3[1+3(1-p)+\frac{6p(1-p)^2}{1-2p(1-p)}]$
=$p^3[1+(3-3p)+\frac{6p-12p^2+6p^3}{1-2p+2p^2)}]$
=$p^3[\frac{(1-2p+2p^2)+(3-6p+6p^2-3p+6p^2-6p^3)+(6p-12p^2+6p^3)}{1-2p+2p^2)}]$
=$p^3\frac{(4-5p+2p^2)}{1-2p+2p^2}$
| $A$ can win in three games with $3-0$ or $3-1$, else they must reach $2-2$ and then has to win by a difference of $2$
*
*To win directly, $A$ wins with $p^3 + \binom31 p^3(1-p) = 4p^3 -3p^4$
*To reach $2-2, Pr = \binom42p^2(1-p)^2$
*Let $P(A$ ultimately wins from $2-2$) be $d$
Then either A wins in straight two points or is back to equal scores, thus eqivalent to $2-2$, so $d = p^2 + 2p(1-p)d$ which gives $d = \frac{p^2}{2p^2-2p+1}$
*Putting the pieces together, $$P(A\; wins) = p^3+3p^3(1-p) + [6p^2(1-p)^2] \times \frac{p^2}{2p^2-2p+1}$$
Added
On giving it to Wolfram the answer I get is
$$P(A\; wins) = \frac{p^3(2p^2-5p+4)}{2p^2-2p+1}$$
which tallies with the book answer
| {
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Finding $x^3+y^3$, given $x + y = 5$ and $xy = 1$
Given
$$x + y = 5 \qquad xy = 1$$
Find $x^3 + y^3$.
To solve this, I tried this:
$y = \frac{1} {x}$
$x + \frac{1}{x} = 5$
$x^2 - 5x + 1 = 0$
What formula needs to be used to find the value of $x$?
| Use the binomial formula. We have
$$(x+y)^3=x^3 + 3x^2y + 3xy^2 + y^3,$$
which reduces to
$$125=x^3 + 3\cdot 5 + y^3$$
and thus
$$x^3 + y^3 = 110.$$
There is no need to solve for $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Alice plays a game of choosing numbers and replacing them with some other.
The following $100$ numbers are written on the board:$$2^1 - 1, 2^2 - 1, 2^3 - 1, \dots, 2^{100} - 1.$$
Alice chooses two numbers $a,b,$ erases them and writes the number $\dfrac{ab - 1}{a+b+2}$ on the board. She keeps doing this until a single number remains on the board. If the sum of all possible numbers she can end up with is $\dfrac{p}{q}$ where $p, q$ are coprime, then what
is the value of $\log_{2}(p+q)$?
Since many of these types of questions uses invariants, I was trying to find an invariant. But I'm unable to find it. I would like someone to provide hints.
| If we let $a_i$ be the numbers left on the board at any stage, then $$\Delta=\sum\limits_{i}\frac{1}{a_i+1}$$ is invariant.
Note that if $a,b$ are replaced at a stage by $\frac{ab-1}{a+b+2}$, then $\frac1{a+1}+\frac1{b+1}$ is changed to $\frac{1}{\frac{ab-1}{a+b+2}+1}=\frac1{\frac{(a+1)(b+1)}{a+b+2}}=\frac1{a+1}+\frac1{b+1}$.
What is this sum $\Delta$? It is equal to $\frac{2^{100}-1}{2^{100}}$, because it is invariant under this process and therefore, it shouldn't have changed from what we started with.
If the last no. on the board is $L$, then $$\Delta=\frac1{L+1}=\frac{2^{100}-1}{2^{100}}$$
$$\implies L=\frac{1}{2^{100}-1}=\frac p q.$$
We need $\mathrm{log}_2(p+q)$ which is $100$.
I noticed that applying $\frac{ab-1}{a+b+2}$ to $a=2^i-1,b=2^j-1$ always managed to clear the $-1$s and left me with a perfect power of $2$. So, I figured there had to be something involving $a+1,b+1$. So, I wrote the $\frac{ab-1}{a+b+2}$ as $\frac{ab-1}{(a+1)+(b+1)}$.
I also noticed that $ab-1=(a+1)(b+1)-((a+1)+(b+1))$. Using this, I wrote $\frac{ab-1}{a+b+2}+1=\frac{(a+1)(b+1)}{(a+1)+(b+1)}=\frac{1}{\frac{(a+1)+(b+1)}{(a+1)(b+1)}}=\frac{1}{\frac{1}{a+1}+\frac 1{b+1}}$.
I wrote $\frac{ab-1}{a+b+2}+1=\frac{1}{\frac{1}{\frac{ab-1}{a+b+2}+1}}$ and I knew I had the invariant. I guessed the $\Delta$ with this.
| {
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Approach on sum of $\binom{10+r}{r}\cdot\binom{20-r}{10-r}$ The problem: Find $$\sum_{r=0}^{10} \binom{10+r}{r}\cdot\binom{20-r}{10-r}$$
I do not even know how to approach this problem. None of the standard binomials or their combinations that I consider like $(1+x)^n$ or $(1-x)^{-n}$ seem to work. Would anyone please provide a hint on how to approach this?
This is expected to be solved by only using high school techniques and considering basic polynomials of the form $(1+x)^n$ and $(1+x)^{-n}$.
| Your sum is an instance of the Rothe-Hagen identity
$$\sum_{r=0}^n \frac{x}{x + rz} \binom{x + rz}{r} \frac{y}{y+(n-r)z} \binom{y+(n-r)z}{n-r} = \frac{x+y}{x+y+nz} \binom{x+y+nz}{n} \tag{1}$$ for the choice $$x = y = 11, \quad n = 10, \quad z = 1.$$ In particular, this yields
$$\begin{align}
\sum_{r=0}^{10} \binom{10+r}{r} \binom{20-r}{10-r}
&= \sum_{r=0}^{10} \frac{11}{11+r} \binom{11+r}{r} \frac{11}{21-r} \binom{21-r}{10-r} \\
&= \frac{22}{32} \binom{32}{10} \\
&= \binom{31}{10}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4637503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Probability of getting same number What is the probability of getting the same number when spinning a roulette wheel, tossing a die, and drawing a random card?
My approach:
Case A - Get a random number on the roulette wheel, now get the same number on the die and the card: $$1 \cdot \frac{1}{6} \cdot \frac{1}{13} = \frac{1}{78}$$
Case B - Get a random number on the die, then get the same number on the roulette wheel and the card: $$1 \cdot \frac{1}{6} \cdot \frac{1}{13} = \frac{1}{78}$$
Case C - Get a random number on the card, now get the same number on the die and the roulette wheel: $$\frac{6}{13} \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{78}$$ ($\frac{6}{13}$ for the card since we can only choose the numbers $1$–$6$)
Sum all $3$ cases: $$\frac{3}{78} = \bf{\frac{1}{26}}$$
Does this seem right or I am on the wrong approach here?
| Assuming it is a US roulette wheel, there are 38 numbers, the die has the most restrictive ($\{1,...,6\}$) and the cards have 52 possibilities.
The total number of selections is $38 \cdot 6 \cdot 52$ and the number of selections that match is $4 \cdot 6$ so the probability is ${1 \over 13 \cdot 38}$ (about $0.2 \%$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ How do I evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$ ?
Note that this is a Q&A post and I've presented my solution below.
| To evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$
$\Rightarrow\int \frac{-2x}{(x^2-1)^{\frac{3}{2}}\sqrt{x^2+1}}dx$
Now, one substitution that works but is rather tricky to find is,
Let $u=\sqrt{\frac{x^2+1}{x^2-1}}$
$\Rightarrow \frac{1}{2}(\frac{x^2-1}{x^2+1})^{\frac{1}{2}} \frac{(x^2-1)2x-(x^2+1)2x}{(x^2-1)^2}dx=du$
$\Rightarrow (\frac{x^2-1}{x^2+1})^{\frac{1}{2}} \frac{-2x}{(x^2-1)^2}dx=du$
$\Rightarrow \frac{-2x}{(x^2-1)^{\frac{3}{2}}\sqrt{x^2+1}}dx=du$
The integral reduces to
$\int du=u+c=\sqrt{\frac{x^2+1}{x^2-1}}+c$
Thus $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx=\sqrt{\frac{x^2+1}{x^2-1}}+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4640836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Question from MIT integration Bee 2023 final: Evaluate $\int^1_0 (\sum^\infty_{n=0}\frac{\left\lfloor 2^nx\right\rfloor}{3^n})^2{\rm d}x$ I am trying to evaluate the last question from MIT integration Bee 2023 Final.
$$\int^1_0 \left (\sum^\infty_{n=0}\frac{\left\lfloor 2^nx\right\rfloor}{3^n} \right )^2{\rm d}x$$
My approach is to divide $(0,1)$ into $1/2^n$ intervals and write the general term of the $y$-value. E.g. For $x \in (k/2^n, (k+1)/2^n)$,
$$f(x)=\left (\sum^{n-1}_{k=0}\frac{\left\lfloor k/2^k\right\rfloor}{3^{n-k}}\right)^2$$
I know that the final integral is just summing up the areas of all the infinite rectangles but I can't solve it. Please help. Thank you.
(The final answer of this question is $27/32$. Candidates were allowed to solve it within 4 minutes.)
| Here is a slightly advanced solution: Define $X_1, X_2, \ldots$ on $[0, 1]$ by
$$ X_k (x) := [\text{$k$th digit in the binary expansion of $x$}] = \lfloor 2^k x\rfloor - 2 \lfloor2^{k-1}x\rfloor. $$
Then
\begin{align*}
\sum_{n=0}^{\infty} \frac{\lfloor 2^n x \rfloor}{3^n}
= \sum_{n=1}^{\infty} \frac{1}{3^n} \sum_{k=1}^{n} 2^{n-k}X_k
= \sum_{k=1}^{\infty} \frac{X_k}{2^k} \sum_{n=k}^{\infty} \frac{2^n}{3^n}
= \sum_{k=1}^{\infty} \frac{X_k}{3^{k-1}} .
\end{align*}
Now by regarding $[0, 1]$ as a probability space with the probability measure $\mathrm{d}x$, we find that $X_1, X_2, \ldots$ are i.i.d. $\text{Bernoulli}(\frac{1}{2})$ variables. So,
\begin{align*}
\int_{0}^{1} \left( \sum_{n=0}^{\infty} \frac{\lfloor 2^n x \rfloor}{3^n} \right)^2 \, \mathrm{d}x
= \mathbf{E} \left[ \left( \sum_{k=1}^{\infty} \frac{X_k}{3^{k-1}} \right)^2 \right]
= \sum_{j,k=1}^{\infty} \frac{1}{3^{j+k-2}} \mathbf{E}[X_j X_k].
\end{align*}
Using the independence, we get $\mathbf{E}[X_j X_k] = \frac{1}{4} + \frac{1}{4} \mathbf{1}_{\{j = k\}}$. Hence, the expectation reduces to
\begin{align*}
\sum_{j,k=1}^{\infty} \frac{1}{3^{j+k-2}} \left( \frac{1}{4} + \frac{1}{4} \mathbf{1}_{\{j = k\}} \right)
= \frac{1}{4} \left( \sum_{k=1}^{\infty} \frac{1}{3^{k-1}} \right)^2 + \frac{1}{4} \left( \sum_{k=1}^{\infty} \frac{1}{9^{k-1}} \right)
= \boxed{\frac{27}{32}}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/4642139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
Finding all the minima and maxima within a range I'm not sure how to find all the maximas and minimas where the range is $1≤x≤18$ and the function is: $$20\sin \left(\fracπ6x-\frac {2\pi}3\right)+22$$ I already found the first derivative which is: $$\frac{10\pi}3\cos\left(\frac{\pi}{6}x-\frac{2\pi}{3}\right)=0$$ where $x$ is $7$ and using $f''(x)$ and subbing in my $x$, I get $-5.48$ which is a maximum.
| To simplify the problem, put $u = \dfrac{\pi}{6}x-\dfrac{2\pi}{3}$, then you have a much better function to deal with: $f(u) = 20\sin u + 22, -\dfrac{\pi}{2} \le u \le \dfrac{7\pi}{3}$. Since $-1 \le \sin u \le 1$, so $2 \le f(u) \le 42$, and the min and max are $2, 42$ respectively. The min occurs when $u = -\dfrac{\pi}{2} \implies \dfrac{\pi}{6}x - \dfrac{2\pi}{3} = -\dfrac{\pi}{2} \implies x = 1$, and the max occurs when $ u = \dfrac{\pi}{2} \implies \dfrac{\pi}{6}x - \dfrac{2\pi}{3} = \dfrac{\pi}{2} \implies x = 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Derivation of $\sin(15^\circ)$ geometrically This is my attempt:-
Let us consider a right $\triangle ABC$ such that angle $A$ is $15^\circ$ and $C$ is $75^\circ$.
On the line $AB$, let us assume a point $D$ such that $\frac{BC}{BD} =\frac{ 1}{\sqrt{3}}$ (Without Loss Of Generality). So $\angle BDC$ becomes $30^\circ$ and $\angle BCD$ becomes $60^\circ$. Then $\angle DCA$ becomes equal to $\angle BAC$, that is $15^\circ$; so $CD = DA$.
$CD$ will be $2$ times $BC$ (angle $BDC = 30^\circ$; $\sin 30^\circ$). On adding $BD$ and $AD$ we get $$AB = BC(2+\sqrt{3})$$
$$BC^2 + AB^2 = AC^2$$
$$\therefore AC = 2BC\sqrt{2 + \sqrt{3}}$$
$$\sin(15^\circ) = \frac{BC}{AC}$$
$$\sin(15^\circ) = \frac{1}{\sqrt{2 + \sqrt{3}}}$$
Rationalising the denominator 2 times we get:-
$$\sin(15^\circ) = \frac{(4-2\sqrt{3})(\sqrt{2+\sqrt{3}})}{4}$$
Further simplifying:-
$$\sin(15^\circ) = \frac{(2-\sqrt{3})(\sqrt{2+\sqrt{3}})}{2}$$
$$\sin(15^\circ) = \frac{(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}}) (\sqrt{2+\sqrt{3}})}{2}$$
$$\sin(15^\circ) = \frac{\sqrt{2-\sqrt{3}}}{2}$$
Is my answer correct? And is there any other method or way to get the value of $\sin(15^\circ)$ geometrically?
| You obtained the correct answer. However, not every step you wrote down is correct.
You found that
$$AC = 2BC\sqrt{2 + \sqrt{3}}$$
from which it follows that
$$\sin(15^\circ) = \frac{BC}{AC} = \frac{1}{2\sqrt{2 + \sqrt{3}}}$$
Multiplying the numerator and denominator by $\sqrt{2 - \sqrt{3}}$ yields
$$\sin(15^\circ) = \frac{\sqrt{2 - \sqrt{3}}}{2\sqrt{4 - 3}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$$
As for an alternative method, consider the diagram below.
Since $\angle AEC$ is a right angle, $\triangle ACE$ is a right triangle. Since the acute angles of a right triangle are complementary and $m\angle CAE = 45^\circ$, $m\angle ACE = 45^\circ$. Since $CE = 1$ and the ratio of the side lengths of a $45^\circ, 45^\circ, 90^\circ$ right triangle is $1: 1: \sqrt{2}$, $AE = 1$ and $AC = \sqrt{2}$.
Since $\angle AEC$ is a right angle, $\overline{AD} \perp \overline{CE}$. Thus, $\angle CED$ is also a right angle. Therefore, $\triangle CDE$ is a right triangle. Since the acute angles of a right triangle are complementary and $m\angle CDE = 30^\circ$, $m\angle ECD = 60^\circ$. Since the ratio of the side lengths in a $30^\circ, 60^\circ, 90^\circ$ right triangle is $1: \sqrt{3}: 2$ and $CE = 1$, $DE = \sqrt{3}$ and $CD = 2$.
By the Segment Addition Postulate, $AD = AE + ED = 1 + \sqrt{3}$.
Since $\angle ABD$ is a right angle, $\triangle ABD$ is a right triangle. Since the acute angles of a right triangle are complementary and $m\angle BDA = 30^\circ$, $m\angle BAD = 60^\circ$. Since the ratio of the side lengths in a $30^\circ, 60^\circ, 90^\circ$ right triangle is $1: \sqrt{3}: 2$ and $AD = 1 + \sqrt{3}$, $AB = \dfrac{1 + \sqrt{3}}{2}$ and $BD = \dfrac{3 + \sqrt{3}}{4}$.
By the Angle Addition Postulate, $m\angle BAD = m\angle BAC + m\angle CAE$. Hence, $m\angle BAC = m\angle BAD - m\angle CAE = 60^\circ - 45^\circ = 15^\circ$.
Since $\angle ABC$ is a right angle, $\triangle ABC$ is a right triangle. Since the acute angles of a right triangle are complementary and $m\angle BAC = 15^\circ$, $m\angle ACB = 75^\circ$.
By the Segment Addition Postulate, $BC + CD = BD$. Hence, $BC = BD - CD = \dfrac{3 + \sqrt{3}}{4} - 2 = \dfrac{-1 + \sqrt{3}}{2}$.
We summarize these results in the diagram below.
\begin{align*}
\sin(15^\circ) & = \frac{BC}{AC}\\
& = \frac{\frac{-1 + \sqrt{3}}{2}}{\sqrt{2}}\\
& = \frac{-1 + \sqrt{3}}{2\sqrt{2}}\\
& = \frac{-1 + \sqrt{3}}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\
& = \frac{-\sqrt{2} + \sqrt{6}}{4}\\
& = \frac{\sqrt{6} - \sqrt{2}}{4}
\end{align*}
which is equivalent to your answer since
\begin{align*}
\frac{\sqrt{2 - \sqrt{3}}}{2} & = \frac{\sqrt{2 - \sqrt{3}}}{2} \cdot \frac{2}{2}\\
& = \frac{\sqrt{8 - 4\sqrt{3}}}{4}\\
& = \frac{\sqrt{6 - 4\sqrt{3} + 2}}{4}\\
& = \frac{\sqrt{(\sqrt{6} - \sqrt{2})^2}}{4}\\
& = \frac{\sqrt{6} - \sqrt{2}}{4}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\int x \sqrt{x^2 - x}\ dx$
Problem is to integrate $ x \sqrt{x^2 - x}$.
My attempt:
I made it ready for a substitution $u = x^2 - x$
$$\begin{aligned} \int x \sqrt{x^2 -x}\ dx &= \int (2x-1)\sqrt{x^2 - x} \ dx - \int(x - 1)\sqrt{x^2 - x}\ dx\\& = \int \sqrt u\ du - \int(x - 1)\sqrt{x^2 - x}\ dx\\& = \frac23(x^2 -x)^{3/2} + C_1 - \int(x-1) \sqrt{x^2 -x}\ dx\end{aligned}$$
I don't know how to continue from here.
Alternatively I tried this:
$$\begin{aligned} \int x \sqrt{x^2 - x}\ dx &= \int x^2 \sqrt{1- \frac{1}{x}}dx\\ & \overset{1- \frac1x = t^2}{=} \int \frac{2t^2}{(1-t^2)^4}\ dt\\& \overset{t =\sin(\theta)}{=} \int \frac{2\sin^2(\theta) \cos(\theta)\ d\theta}{\cos^4(\theta)}\\& = \int 2 \tan(\theta)\tan(\theta) \sec(\theta) \ d\theta\\ & = \int 2\sqrt{\sec^2(\theta) - 1}\tan(\theta) \sec(\theta) \ d\theta\\& \overset{\sec(\theta) = u}{=} \int 2 \sqrt{\sec^2(\theta) - 1}\tan(\theta) \sec(\theta) \ d\theta\\& = \int 2 \sqrt{u^2 - 1}\ du\\& = u \sqrt{u^2 - 1} - \ln|u + \sqrt{u^2- 1}| + C\\& = \sqrt{x^2 - x} - \ln|x + \sqrt{x^2 - x}| + C\end{aligned}$$
This method is very tedious. Is there any easy way to do the original integral?
| $$
\begin{aligned}
I & =\int x \sqrt{x^2-x} d x \\
& =\int\left(\frac{2 x-1}{2}+\frac{1}{2}\right) \sqrt{x^2-x} d x \\
& =\frac{1}{2} \int \sqrt{x^2-x} d\left(x^2-x\right)+\frac{1}{2} \int \sqrt{x^2-x} d x \\
& =\frac{1}{3}\left(x^2-x\right)^{\frac{3}{2}}+\frac{1}{2} \underbrace{\int \sqrt{\left(x-\frac{1}{2}\right)^2-\frac{1}{4}} d x}_{J}
\end{aligned}
$$
Let $x-\frac{1}{2}=\frac{1}{2} \sec \theta$ and using integration by parts, then
$$
\begin{aligned}
J & =\frac{1}{4} \int \tan ^2 \theta \sec \theta d \theta \\
& =\frac{1}{8}(\tan \theta \sec \theta-\ln |\tan \theta+\sec \theta \mid)+C
\end{aligned}
$$
Plugging back $x$ yields
$$
I=\frac{1}{3}\left(x^2-x\right)^{\frac{3}{2}}+\frac{1}{16}\left[2(2 x-1) \sqrt{x^2-x}-\ln \left|2 \sqrt{x^2-x}+2 x-1\right|\right]+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4646982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Combinatorics: Amount of options for majority Say we have $2n$ people. Then the amount of options to form a majority (e.g. in a commission) are
$ \binom{2n}{n+1} + \binom{2n}{n+2} + \cdots + \binom{2n}{2n}$
I want to prove, that this is equal to $\frac{1}{2} \left[ 2^{2n} - \binom{2n}{n} \right]$
But I'm stuck. I have the this formula: $ \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n $
So I can say $ \binom{2n}{n+1} + \binom{2n}{n+2} + \cdots + \binom{2n}{n} = 2^{2n} - \left[ \binom{2n}{n} + \binom{2n}{n-1} + \cdots + \binom{2n}{n-n} \right] = 2^{2n} - \left[ \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n} \right] $
But I can't see where to go from there, even when writing the binomial coefficients as $ \frac{n!}{k!(n-k)!} $
Can anyone help me with it?
| By symmetry $\binom{2n}{k}=\binom{2n}{2n-k}$ for $k=0,\ldots,n-1.$
So the sum, call is $S$, you have is also equal to
$$
\binom{2n}{0}+\binom{2n}{1}+\cdots+\binom{2n}{n-1}.
$$
Now $2S+\binom{2n}{n}=2^{2n},$ does it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Help with integrating $\int \frac{dx}{1+\sqrt{\tan(x)}}$ Starting off with subbing $u^2 = \tan(x)$ to remove the square root, I got:
*
*$$\int \frac{2u}{(1+u)(1+u^4)} du$$ (Deriving that $\sec^2(x) = 1+u^4$)
Then by applying the partial fractions method, I get:
*$$\int \frac{-1}{1+u} du + \int \frac{u^3-u^2+u+1}{1+u^4} du$$
The first integral is manageable but for the second one I had to split the individual terms in the numerator into their own fractions to further obtain:
*$$\int \frac{u^3}{1+u^4}du +\int \frac{u}{1+u^4}du + \int \frac{1-u^2}{1+u^4}du $$
Now, the first two I could solve however it is the last one that I am unable to move forward with;
$$ \int \frac{1-u^2}{1+u^4}du $$
| Here there is slightly alternative way using a bit of brute force, pushing the decomposition into partial fractions. We can prepare the way for partial fractions by trying to factor the denominator in difference of squares; we can force this like
\begin{align*}1+u^4&=u^4+2u^2-2u^2+1\\
&=(u^4+2u^2+1)-(2u^2)\\
&=(u^2+1)^2-(\sqrt{2}u)^2\\
&=(u^2+1+\sqrt{2}u)(u^2+1-\sqrt{2}u)\\
&=(u^2+\sqrt{2}u+1)(u^2-\sqrt{2}u+1)
\end{align*}
Then we can use partial fraction in order to write
\begin{align*}\frac{1-u^2}{1+u^4}&=\frac{Au+B}{u^2+\sqrt{2}u+1}+\frac{Cu+D}{u^2-\sqrt{2}u+1}\\
&=\frac{\frac{1}{2\sqrt{2}}(2u+\sqrt{2})}{u^2+\sqrt{2}u+1}+\frac{-\frac{1}{2\sqrt{2}}(2u-\sqrt{2})}{u^2-\sqrt{2}u+1}.
\end{align*}
Substitution in each denominator reduce calculus to know $\int \frac{1}{t}dt=\ln|t|+K$. Substitution back give the answer
$$\frac{1}{2\sqrt{2}}\ln \left|\frac{u^2+\sqrt{2}u+1}{u^2-\sqrt{2}u+1} \right|+K$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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$A,B, C$ are independent $\implies (A \cup B)$ and $(B \cup C)$ are independent? Given that $A,B, C$ are independent events, I am trying to prove that $A \cup B$ and $B \cup C$ are independent.
$$P((A \cup B) \cap (B \cup C)) \\= P(B \cup (A \cap C)) \\= P(B) + P(A \cap C) - P(A \cap B \cap C) \\= P(B) + P(A)P(C) - P(A)P(B)P(C)\\ = P(B) + P(A)P(C) \left(1 - P(B) \right).$$
This clearly does not equal $P(A \cup B)P(B \cup C).$ How do I find an explicit counterexample to the claim that $A \cup B$ and $B \cup C$ are independent events? I am struggling because of the condition that $A,B,C$ must be independent.
| Consider an experiment of tossing two coins where each outcome is uniform. We have $\Omega =$ {H,T}$^2$. Label events:
$A =$ Event that first coin toss is heads. $P(A) = \frac{1}{2}$
$B =$ Event that second coin toss is heads $P(B) = \frac{1}{2}$
$C = \emptyset$ $P(C) = 0$
Check that $A,B,C$ are independent.
$P(A \cap B) = P({HH}) = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} $
$P(A \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot 0 $
$P(B \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot 0 $
$P(A \cap B \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot \frac{1}{2} \cdot 0 $
Now, check whether $A \cup B$ and $B \cup C$ are independent.
$P((A \cup B) \cap (B \cup C)) = P({HH,TH}) = \frac{1}{2}$. However, this is not equal to $P(A \cup B) \cdot P(B \cup C) = \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}$, so not independent.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$. Find the exact value of $\int_{0}^{2}x[\frac{1}{x}]dx$.
Let $[x]$ denote $\lceil{x-\frac{1}{2}}\rceil$.
Using Desmos, I got $2.46736022133$ and WolframAlpha does not give me a solution. My intuition tells me that it might be possible to find an exact value using Trapezoidal Reimann Sums but I am not really sure how to go about doing it. After my attempt, I got stuck but I was at a point where I could plug it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. Why did it come out so nicely?
My attempt:
Where $A_n$ denotes the area of the $nth$ trapezoid from the right:
$$A=\frac{h}{2}(a+b)$$
$$A_n=\frac{\frac{2}{2n-1}-\frac{2}{2n+1}}{2}(\frac{2n}{2n-1}+\frac{2n}{2n+1})$$
$$A_n=\frac{\frac{4n+2}{4n^{2}-1}-\frac{4n-2}{4n^{2}-1}}{2}\left(\frac{4n^{2}+2n}{4n^{2}-1}+\frac{4n^{2}-2n}{4n^{2}-1}\right)$$
$$A_n=\frac{2}{4n^{2}-1}\left(\frac{8n^{2}}{4n^{2}-1}\right)$$
$$A_n=\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
Then:
$$\int_{0}^{2}x[\frac{1}{x}]dx=\sum_{n=1}^{\infty}A_n=\sum_{n=1}^{\infty}\frac{16n^{2}}{\left(4n^{2}-1\right)^{2}}$$
I do not know how to solve this infinite summation so I plugged it into WolframAlpha and it gave me $\frac{\pi^2}{4}$. How did it get to this conclusion? Is there a more efficient way to solve this?
| Another way to get the sum of the series
$\displaystyle\sum_{n=1}^{\infty}\frac{16n^2}{\left(4n^2-1\right)^2}$
without double summations and without switching the order of the sums.
$\displaystyle\sum_{n=1}^N\frac{16n^2}{\left(4n^2-1\right)^2}=$
$=\displaystyle\sum_{n=1}^N\left[\frac{2n}{\left(2n-1\right)^2}-\frac{2n}{\left(2n+1\right)^2}\right]=$
$=\displaystyle2\sum_{n=1}^N\frac1{\left(2n-1\right)^2}+\sum_{n=1}^N\left[\frac{2(n-1)}{\left(2n-1\right)^2}-\frac{2n}{\left(2n+1\right)^2}\right]=$
$=\displaystyle2\left(\sum_{n=1}^{2N-1}\frac1{n^2}-\sum_{n=1}^{N-1}\frac1{(2n)^2}\right)+\left[0-\frac2{3^2}+\frac2{3^2}-\frac4{5^2}+\ldots+\\\quad+\frac{2(N-1)}{\left(2N-1\right)^2}-\frac{2N}{\left(2N+1\right)^2}\right]=$
$=\displaystyle2\left(\sum_{n=1}^{2N-1}\frac1{n^2}-\frac14\sum_{n=1}^{N-1}\frac1{n^2}\right)-\frac{2N}{\left(2N+1\right)^2}\,.$
Hence ,
$\displaystyle\sum_{n=1}^{\infty}\frac{16n^2}{\left(4n^2-1\right)^2}=$
$=\displaystyle\lim_{N\to\infty}\left[\sum_{n=1}^N\frac{16n^2}{\left(4n^2-1\right)^2}\right]=$
$=\displaystyle\lim_{N\to\infty}\left[2\left(\sum_{n=1}^{2N-1}\frac1{n^2}-\frac14\sum_{n=1}^{N-1}\frac1{n^2}\right)-\frac{2N}{\left(2N+1\right)^2}\right]=$
$=\displaystyle2\left(\sum_{n=1}^{\infty}\frac1{n^2}-\frac14\sum_{n=1}^{\infty}\frac1{n^2}\right)=$
$=\displaystyle2\left(\frac{\pi^2}6-\frac14\!\cdot\!\frac{\pi^2}6\right)=\displaystyle2\left(1-\frac14\right)\frac{\pi^2}6=$
$\displaystyle=2\!\cdot\!\frac34\!\cdot\!\frac{\pi^2}6=\frac{\pi^2}4\,.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Computing $\lim_{x\to 0}\left(\frac{\cot(x)}{x^3}-\frac{1}{x^4}+\frac{1}{3x^2}\right)$ I am trying to compute the following limit:
$$\lim_{x\to 0}\left(\frac{\cot(x)}{x^3}-\frac{1}{x^4}+\frac{1}{3x^2}\right)$$
I have tried by rewriting it as $\lim_{x\to 0}\left(\frac{3x+(x^3-3)\tan(x)}{3x^4 \tan(x)}\right)$ and applying De l'Hopital's Rule but the expression quickly becomes unmanageable:
$$\lim_{x\to 0}\left(\frac{3x+(x^3-3)\tan(x)}{3x^4 \tan(x)}\right)\overset{H}{=}\lim_{x\to 0}\frac{3+(x^2 - 3) \sec^2(x) + 2 x \tan(x)}{3 x^3 (4 \tan(x) + x \sec^2(x))}$$
so I then tried by using the Maclaurin expression for $\tan(x)$:
\begin{align*}
\lim_{x\to 0}\left(\frac{\cot(x)}{x^3}-\frac{1}{x^4}+\frac{1}{3x^2}\right)&=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{x-\tan(x)}{x^2\tan(x)}+\frac{1}{3}\right)=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{x-\left(x+\frac{x^3}{3}+\frac{2}{15}x^5\right)}{x\left(\frac{x^3}{3}+\frac{2}{15}x^5\right)}+\frac{1}{3}\right)\\
&=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{-\frac{1}{3}-\frac{2}{15}x^2}{1+\frac{x^2}{3}+\frac{2}{15}x^4}+\frac{1}{3}\right)=+\infty\cdot 0
\end{align*}
and I got an indeterminate form.
I am currently out of ideas so I would appreciate some help in figuring this out, thanks.
EDIT: It just occurred to me that
\begin{align*}
\lim_{x\to 0}\frac{1}{x^2}\left(\frac{-\frac{1}{3}-\frac{2}{15}x^2}{1+\frac{x^2}{3}+\frac{2}{15}x^4}+\frac{1}{3}\right)=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{-\frac{1}{3}-\frac{2}{15}x^2+\frac{1}{3}+\frac{1}{9}x^2+\frac{2}{45}x^4}{1+\frac{x^2}{3}+\frac{2}{15}x^4}\right)\\ \lim_{x\to 0} \frac{1}{x^2}\left(\frac{-\frac{1}{45}x^2+\frac{2}{45}x^4}{1+\frac{x^2}{3}+\frac{2}{15}x^4}\right)=\lim_{x\to 0}\frac{-\frac{1}{45}+\frac{2}{45}x^2}{1+\frac{x^2}{3}+\frac{2}{15}x^4}=-\frac{1}{45}.
\end{align*}
| The expression under limit can be written as $$\frac{3x-3\tan x+x^2\tan x}{3x^4\tan x}$$ Since $(\tan x) /x\to 1$ the desired limit is equal to the limit of $$\frac{3x-3\tan x+x^2\tan x}{3x^5}$$ Adding and subtracting $x^3$ in numerator we see that the expression can be rewritten as $$\frac{3x-3\tan x+x^3}{3x^5}+\frac{\tan x-x} {3x^3}$$ The last fraction tends to $1/9$ (via L'Hospital's Rule or Taylor expansion) and thus we need to evaluate the limit $L$ of first fraction above and get desired answer as $L+(1/9)$.
To get $L$ we can apply L'Hospital's Rule to get the expression $$\frac{1-\sec^2x+x^2} {5x^4} =\frac{x^2-\tan^2x}{5x^4}$$ which can be written as $$\frac{1}{5}\left(1+\frac{\tan x} {x} \right) \cdot\frac{x-\tan x} {x^3}$$ so that $$L=\frac{1}{5}\cdot 2\cdot \left(-\frac{1}{3}\right)=-\frac{2}{15}$$ The desired limit is $L+(1/9)=-1/45$. In the entire process L'Hospital's Rule has been used twice.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Linear Algebra - Linear transformation question Let $
b \in \mathbb{R}^4, \space A\in M_{4\times4} (\mathbb{R}).
$ Suppose
$$
\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1\end{bmatrix},\begin{bmatrix} 1 \\ 1 \\ 0 \\ 1\end{bmatrix},\begin{bmatrix} 1 \\ 1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1\end{bmatrix}$$
are all solutions of the equation $Ax=b$. Prove that $$b = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}.$$
This question appeared at my recent algebra exam. I had an intuiton that the given vectors are linear-dependent since there are five of them and the dim of the vector space is four. The professor solved that question after the exam using linear transformation which seemed much easier and clean solution however he didn't explain the intuition behind. I would like to get an explanation regarding that. Thanks
| $$\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1\end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ 1 \\ 1\end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 0 \\ 1\end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 1 \\ 0\end{bmatrix} - 3\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$
$$A\begin{bmatrix} 0 \\ 1 \\ 1 \\ 1\end{bmatrix} + A\begin{bmatrix} 1 \\ 0 \\ 1 \\ 1\end{bmatrix} + A\begin{bmatrix} 1 \\ 1 \\ 0 \\ 1\end{bmatrix} + A\begin{bmatrix} 1 \\ 1 \\ 1 \\ 0\end{bmatrix} - 3A\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1\end{bmatrix} = A\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}$$
$$b + b + b + b -3b = 0$$
$$b=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why are the differences between consecutive squares equal to the sequence of odd numbers? I was playing around with the squares and saw an interesting pattern in their differences.
$0^2 = 0$
+ 1
$1^2 = 1$
+ 3
$2^2 = 4$
+ 5
$3^2 = 9$
+ 7
$4^2 = 16$
+ 9
$5^2 = 25$
+ 11
$6^2 = 36$
etc.
(Also, in a very related question, which major Math Research Journal should I contact to publish my groundbreaking find in?)
| In general, $a^2-b^2=(a-b)(a+b)$ So $(x+1)^2-x^2=(x+1-x)(2x+1)=2x+1$ So all odd numbers are generated.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 5
} |
Explain why calculating this series could cause paradox? $$\ln2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots
= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - 2(\frac{1}{2} + \frac{1}{4} + \cdots)$$
$$= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) = 0$$
thanks.
| It's because the series for ln2 is conditionally convergent. (see also Riemann's rearrangement theorem)
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
How is this series in denominator converted to a series in numerator? How do we go from
\begin{equation*}
\frac{1}{1+(\frac{x}{2})+(\frac{x^2}{3})+(\frac{x^3}{4})+\dots}
\end{equation*}
to
\begin{equation*}
1-\left(\frac{x}{2}\right)-\left(\frac{x^2}{12}\right)-\dots?
\end{equation*}
If I could consolidate the series in denominator in some form, then I could use binomial expansion in numerator by raising it to power $-1$, but I don't see the one in denominator in any known form other than converting it to $\log.$
| Using $$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$$
We have that
$$ - \frac{x}{\log(1-x)} = \frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots}$$
Now the series for $$\frac{x}{\log(1-x)}$$ is well known.
See the question asked on this very site here: Formula for the harmonic series $H_n = \sum_{k=1}^n 1/k$ due to Gregorio Fontana
And the page here: http://en.wikipedia.org/wiki/Euler-Mascheroni_constant. (search the page for Gregory).
The series expansion is given by
$$\frac{x}{\log(1-x)} = \sum_{k=0}^{\infty} C_{k} x^{k} = -1 + \frac{x}{2} + \frac{x^2}{12} + \frac{x^3}{24} + \dots$$
The $C_{k}$ are called as Gregory coefficients. The wiki page I linked above tells you how they can be calculated using a recursive formula.
So to answer your question, we get
$$\frac{1}{1 + \frac{x}{2} + \frac{x^2}{3} + \dots} = - \sum_{k=0}^{\infty} C_{k} x^k = 1 - \frac{x}{2} - \frac{x^2}{12} - \frac{x^3}{24} - \dots$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
How to prove an identity in radicals? (4 / (3 - sqrt(5))) ^ 2 - ((6 - 5 * sqrt(6)) / (5 - sqrt(6))) ^ 2 = 2 * sqrt(61 + 24*sqrt(5))
$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$
How to prove it is right equality?
I come up with $\dfrac{16}{14-6\sqrt{5}}-6=2\sqrt{61+24\sqrt{5}}$, but still can't get it to the obvious equality.
Any ideas?
| You want to show:
$$\left(\frac{4}{3-\sqrt{5}}\right)^2 - \left(\frac{6 - 5 \sqrt{6}}{5 - \sqrt{6}}\right)^2 = 2\sqrt{61+24\sqrt{5}}$$
First multiply out the squares
$$\frac{16}{14-6\sqrt{5}} - \frac{186-60\sqrt{6}}{31-10\sqrt{6}} = 2\sqrt{61+24\sqrt{5}}$$
Then simplify it
$$\frac{16}{14-6\sqrt{5}} - 6 = 2\sqrt{61+24\sqrt{5}}$$
now we can start getting rid of the square roots on the right hand side by squaring both sides
$$\frac{256}{376-168\sqrt{5}} - \frac{192}{14-6\sqrt{5}} + 36 = 244 + 96 \sqrt{5}$$
at this point it makes sense to rationalize the denominators of the fractions
$$\left(\frac{256}{376-168\sqrt{5}}\right)\left(\frac{376+168\sqrt{5}}{376+168\sqrt{5}}\right) - \left(\frac{192}{14-6\sqrt{5}}\right)\left(\frac{14+6\sqrt{5}}{14+6\sqrt{5}}\right) + 36 = 244 + 96 \sqrt{5}$$
which simplifies to
$$376 + 168\sqrt{5} - 168 - 72\sqrt{5} + 36 = 244 + 96 \sqrt{5}$$
collecting like terms now gives
$$(376 - 168 + 36) + (168 - 72)\sqrt{5} = 244 + 96 \sqrt{5}$$
which is easily seem to be equal.
Another approach is, using the fact that algebraic numbers cannot be too close together - compute the first few digits of both sides of the original identity and compare them.
| {
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"source": "stackexchange",
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Simple Proof by induction: $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ I'm trying to prove using induction that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:
*
*Base case: $P(1) = (1) + (8) + (27) = 36, 36$ can be divided by $9$ so the base case is valid
*Inductive step: let $P(n)$ be the statement $9$ divides $n^3 + (n+1)^3 + (n+2)^3$. Assume $P(k)$ is true, so $9$ divides $k^3 + (k+1)^3 + (k+2)^3$.
And this is where I'm stuck. I'm not sure how to demonstrate that $9$ divides $P(n)$ when $n = k+1$. If someone could step me in the right direction that would be awesome. Thanks!
| $$n^3+(n+1)^3+(n+2)^3=n^3+(n+1)^3+(n+2)^3-3n(n+1)(n+2)+3n(n+1)(n+2)\equiv$$
$$\equiv n^3+(n+1)^3+(n+2)^3-3n(n+1)(n+2)=$$
$$=(n+n+1+n+2)(n^2+(n+1)^2+(n+2)^2-n(n+1)-n(n+2)-(n+1)(n+2))=$$
$$=3(n+1)\cdot3=9(n+1)\equiv0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating the improper integral $\int\limits_{0}^{\infty} \frac{x^{a-1} - x^{b-1}}{1-x} \ dx $ How does one evaluate the integral $$\int\limits_{0}^{\infty} \frac{x^{a-1} - x^{b-1}}{1-x} \ dx \quad \text{for} \ a,b \in (0,1)$$
| Here's a sketch of how we can evaluate the integral without complex analysis.
$$\text{Let} \quad I(a) = \int_0^\infty \frac{x^{a-1}}{1-x} dx.$$
Split the range into two intervals $(0,1)$ and $(1,\infty)$ then use the substitution $x=1/t$ in the latter part to obtain
$$I(a) = \int_0^1 \frac{x^{a-1}-x^{-a}}{1-x} dx.$$
Expand the integrand as a power series using $(1-x)^{-1} = \sum_{n=0}^\infty x^n$ and integrate to obtain
$$I(a) = \frac{1}{a} + \sum_{n=1}^\infty \left( \frac{1}{a+n} + \frac{1}{a-n} \right).$$
Now, by differentiating logarithmically the product formula for $\sin x,$
$$\sin \pi x = \pi x \prod_{n=1}^{\infty} \left( 1 - \frac{x^2}{n^2} \right),$$
we note that
$$\pi \cot \pi x = \frac{1}{x} + \sum_{n=1}^\infty \left( \frac{1}{x+n} + \frac{1}{x-n} \right).$$
Thus $$I(a) = \pi \cot(\pi a)$$ and the result follows since the integral in question is $I(a)-I(b).$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
| Here is an other one which is more or less what Euler did in one of his proofs.
The function $\sin x$ where $x\in\mathbb{R}$ is zero exactly at $x=n\pi$ for each integer $n$. If we factorized it as an infinite product we get
$$\sin x = \cdots\left(1+\frac{x}{3\pi}\right)\left(1+\frac{x}{2\pi}\right)\left(1+\frac{x}{\pi}\right)x\left(1-\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1-\frac{x}{3\pi}\right)\cdots =$$
$$= x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\left(1-\frac{x^2}{3^2\pi^2}\right)\cdots\quad.$$
We can also represent $\sin x$ as a Taylor series at $x=0$:
$$\sin x = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\quad.$$
Multiplying the product and identifying the coefficient of $x^3$ we see that
$$\frac{x^3}{3!}=x\left(\frac{x^2}{\pi^2} + \frac{x^2}{2^2\pi^2}+ \frac{x^2}{3^2\pi^2}+\cdots\right)=x^3\sum_{n=1}^{\infty}\frac{1}{n^2\pi^2}$$
or
$$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}.$$
Here are two interesting links:
*
*Euler's papers;
*Euler’s Solution of the Basel Problem – The Longer Story an essay on the subject written by Ed Sandifer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
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Funny identities Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?
| Facts about $\pi$ are always fun!
\begin{equation}
\frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\ldots\\
\end{equation}
\begin{equation}
\frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\ldots\\
\end{equation}
\begin{equation}
\frac{\pi^2}{6} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\ldots\\
\end{equation}
\begin{equation}
\frac{\pi^3}{32} = 1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\frac{1}{9^3}-\ldots\\
\end{equation}
\begin{equation}
\frac{\pi^4}{90} = 1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots\\
\end{equation}
\begin{equation}
\frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\ldots\\
\end{equation}
\begin{equation}
\pi = \cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\ldots}}}}}\\
\end{equation}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Determine the matrix relative to a given basis Question: (a) Let $f: V \rightarrow W$ with $ V,W \simeq \mathbb{R}^{3}$ given by: $$f(x_1, x_2, x_3) = (x_1 - x_3, 2x_1 -5x_2 -x_3, x_2 + x_3).$$
Determine the matrix of $f$ relative to the basis $\{(0,2,1),(-1,1,1),(2,-1,1)\}$ of $V$ and $\{(-1,-1,0),(1,-1,2),(0,2,0)\}$ of $W$.
(b) Let $n \in \mathbb{N}$ and $U_n$ the vector space of real polynomials of degree $\leq n$. The linear map $f: U_n \rightarrow U_n$ is given by $f(p) = p'$. Determine the matrix of $f$ relative to the basis $\{1,t,t^{2},...,t^{n}\}$ of $U_n$.
My attempt so far:
(a): First relative to the bases of $W$ I found the coordinates of an arbitrary vector: $\left( \begin{array}{r} a \\ b \\ c \end{array} \right) = x \left( \begin{array}{r} -1 \\ -1 \\ 0 \end{array} \right) + y \left( \begin{array}{r} 1 \\ -1 \\ 2 \end{array} \right) + z \left( \begin{array}{c} 0 \\ 2 \\ 0 \end{array} \right)$
$\begin{array}{l} a = -x + y \\ b = - x - y + 2z \\ c = 2y \end{array}$ or $\begin{array}{l} x = -a + \frac{1}{2}c \\ z = -\frac{1}{2}a + \frac{1}{2}b + \frac{1}{2}c \\ y = \frac{1}{2}c \end{array}$
At this point I believe I have the linear combinations of the given basis in $W$ for an arbitrary vector, so next I take the vectors from $V$ and send them to $W$ using the given function:
$\begin{array}{l} f(v_1) = f(0,2,1) = (-1,-11,3) = (1 + \frac{3}{2})w_1 + \frac{3}{2}w_2 + (\frac{1}{2} - \frac{11}{2} + \frac{3}{2})w_3 \\ f(v_2) = f(-1,1,1) = (-2,-8,2) = (2+1)w_1 + w_2 + (1 - 4 +1)w_3 \\ f(v_3) = f(2,-1,1) = (1,8,0) = w_1 + (-\frac{1}{2} + 4)w_3 \end{array}$
or $\left( \begin{array}{rrc} \frac{5}{2} & 3 & 1 \\ \frac{3}{2} & 1 & 0 \\ -\frac{7}{2} & -2 & \frac{7}{2}\end{array} \right)$
Was I taking the correct steps? I didn't really do anything differently based on the fact that $V,W$ were isometric... Is there a particular significance or interpretation for the resulting matrix?
(b): Not really sure here...
$f(p) = p'$
would it make sense to write something like:
$f(1,t,t^{2},\dots, t^{n}) = (0,1,2t, \dots, nt^{n-1})$?
and if a basis for $(1,t,t^{2},\dots, t^{n})$ would be $A = \left( \begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\0 & \cdots & \cdots & 0 & 1 \end{array} \right)$
could i write:
$A' = \left( \begin{array}{ccccc} 0 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\0 & 0 & 0 & 1 & 0 \end{array} \right)$?
| Here's a suggestion for seeing the bigger picture in your first question:
*
*The key idea below is that the column vectors of a matrix $M$ are the images of the standard ordered basis when the matrix $M$ is viewed as a transformation
$M: \mathbb{R}^n \to \mathbb{R}^n$.
*It is easy to find a matrix of f with respect to the standard ordered basis of $\mathbb{R}^3$: it is given by a matrix with columns $\{1, 2, 0\}$, $\{0, -5, 1\}$, and $\{-1, -1, 1\}$. Call this matrix $C$.
*The matrix which transforms the given ordered basis of $V$ to the standard basis of $\mathbb{R}^3$ is given by writing the basis vectors as columns of this matrix. Similarly, this can be done for the basis of $W$. Let's call these matrices $A$ and $B$.
*You desire a matrix which represents $f$. Interpreting the matrices above as transformations, consider the following diagram:
$$\begin{array}{ccc}
V & \to & \mathbb{R}^3\\
\downarrow & \qquad & \downarrow\\
W & \to & \mathbb{R}^3
\end{array}$$
where the horizontal maps are $A$ and $B$ and the right vertical map is $C$.
Your desired matrix is the left vertical arrow. It is now clear that it is given by the matrix $B^{-1} C A$.
| {
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Solve $x^3 \equiv 1 \pmod p$ for $x$ How can I find solution for $x^3 \equiv 1 \pmod p$ ($p$ a prime) efficiently?
Trivial root is $x_1 = 1$. I need to find other roots $x_2, x_3$.
| The integers modulo $p$ form a field. Since $x^3 - 1 = (x-1)(x^2+x+1)$, the problem is equivalent to solving $x^2+x+1\equiv 0 \pmod{p}$. For $p\neq 2$, the usual quadratic formula works; so you would need to find $y$ such that $y^2\equiv -3\pmod{p}$. If $p=2$, then $x^2+x+1=0$ has no solutions, and if $p=3$, then $x^3-1 = (x-1)^3$, so again there is no root other than $x=1$.
Let's consider the other primes, $p\gt 3$.
Using quadratic reciprocity, if $p\equiv 1\pmod{4}$, then $-1$ is a square modulo $p$ and we have:
$$\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right).$$
So if $p\equiv 1 \pmod{3}$ (hence $p\equiv 1 \pmod{12}$) then $-3$ is a square modulo $p$; if $p\equiv 2\pmod{3}$ (so $p\equiv 5\pmod{12}$), then $-3$ is not a square modulo $p$, so there is no other solution.
If $p\equiv 3\pmod{4}$, then $-1$ is not a square modulo $p$, and we have:
$$\left(\frac{-3}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right)$$
(since $\left(\frac{3}{p}\right) = -\left(\frac{p}{3}\right)$); again, if $p\equiv 1\pmod{3}$ (so $p\equiv 7\pmod{12}$) then $-3$ is a square modulo $p$; and if $p\equiv 2\pmod{3}$ (so $p\equiv 11\pmod{12}$) then $-3$ is not a square modulo $p$.
In summary, there are roots of $x^3-1$ modulo $p$ other than $x\equiv 1\pmod{p}$ if and only if $p\equiv 1\pmod{6}$. If $p=6k+1$ is such a prime, then the two roots of $x^3-1$ other than $x=1$ are precisely the roots of $x^2+x+1$, which are
$$\frac{-1\pm\sqrt{-3}}{2} = -3k(-1\pm a),$$
where $a$ is an integer such that $a^2\equiv -3\pmod{p}$. This means you still need to figure out the square root, which can be done using any of the methods suggested by Bill Dubuque, such as Tonelli's algorithm (say, in Shanks' version as described in Wikipedia).
Added: As Alex Bartel notes in comments and in his own answer, one can avoid the use of quadratic reciprocity above. Since the units of $\mathbb{Z}/p\mathbb{Z}$ form a cyclic group of order $p-1$, there are nontrivial solutions to $x^3=1$ if and only if $3$ divides the order, that is, if and only if $p\equiv 1\pmod{3}$. From there, it is an easy jump to solutions if and only if $p\equiv 1 \pmod{6}$, since otherwise we would have $p\equiv 4\pmod{6}$ which is impossible since $p$ is prime. Once one establishes that, the quadratic formula can be applied safe in the knowledge that these primes must have $-3$ as a quadratic residue.
| {
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Determine in How many ways $N!$ can be expressed as sum of consecutive numbers Determine in How many ways $N!$ can be expressed as sum of consecutive numbers.For example for $3!$ can be as $1+2+3$ only $1$ ways.
Please suggest an elegant approach.
| Let's deal with odd length and even length sequences separately.
A number $N$ is the sum of an odd number of consecutive integers, centered at $m$, whenever $m|N$ and $N/m$ is odd.
A number $N$ is the sum of an even number of consecutive integers, centered at $m+1/2$, whenever $2m+1|N$.
Now let $N = 2^k M$, where $M$ is odd. For each factor $m|M$ we get one odd-length sequence centered at $2^km$, and one even-length sequence centered at $m+1/2$.
Example: $N = 6 = 2^1 3$, so we get two odd-length sequences (centered at $2$ and $6$) and two even-length sequences (centered at $.5$ and $1.5$): $$6 = 1 + 2 + 3 = 6 = -5 -4 -3 -2 -1 + 0 + 1 + 2 + 3 + 4 + 5 + 6 = 0 + 1 + 2 + 3.$$
In order to obtain a formula for the number of representations of $n!$, all we need to do is to figure out how many odd divisors $n!$ has. I leave that to the reader.
| {
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Maximum likelihood and sufficient statistics $$f_T(t;B,C) = \frac{\exp(-t/C)-\exp(-t/B)}{C-B}$$
where our mean is $C+B$ and $t>0$.
so far i have found my log likelihood functions and differentiated them as follows:
$$dl/dB = \sum[t\exp(t/C) / (B^2(\exp(t/c)-\exp(t/B)))] +n/(C-B) = 0$$
i have also found a similar $dl/dC$.
I have now been asked to comment what you can find in the way of sufficient statistics for estimating these parameters and why there is no simple way of using Maximum Likelihood for estimation in the problem. I am simply unsure as to what to comment upon. Any help would be appreciated. Thanks, Rachel
Editor's Note: Given here is the probability density function
$$
f_T (t;B,C) = \frac{{e^{ - t/C} - e^{ - t/B} }}{{C - B}}, \;\; t > 0,
$$
where $B$ and $C$ are positive constants such that $C > B$. The mean is $C+B$.
For the log likelihood function, see the last equation in my answer to this related question, and differentiate accordingly (with respect to $B$ and $C$).
| Edit: Made a mistake in the MLE part
There is no MLE because the likelihood is maximized at at a parameter specification outside of the parameter range.
\begin{equation}
f_T (t;B,C) = \frac{{e^{ - \frac{t}{C}} - e^{ -\frac{t}{B}} }}{{C - B}}
\end{equation}
Taking the gradient we get:
\begin{equation}
\nabla f_T (t;B,C) = \{\frac{e^{\frac{-1}{x}}}{x^2(x-y)}-\frac{e^{\frac{-1}{x}}-e^{\frac{-1}{y}}}{(x-y)^2},\frac{e^{\frac{-1}{x}}-e^{\frac{-1}{y}}}{(x-y)^2}-\frac{e^{\frac{-1}{y}}}{y^2(x-y)}\}
\end{equation}
Setting this equal to 0 and re-arranging some terms gives us the following system of equations:
\begin{array}{lcl} \frac{e^{\frac{-1}{C}}}{C^2} & = & \frac{e^{\frac{-1}{C}}-e^{\frac{-1}{B}}}{(C-B)}\\
\frac{e^{\frac{-1}{B}}}{B^2} & = & \frac{e^{\frac{-1}{C}}-e^{\frac{-1}{B}}}{(C-B)}
\end{array}
\begin{array}{lcl} \frac{(C-B)}{C^2} & = & 1-e^{\frac{-1}{B}+\frac{1}{C}}\\
\frac{(C-B)}{B^2} & = & 1-e^{\frac{-1}{C}+\frac{1}{B}}
\end{array}
Which is maximized when $C = B$, which is not a valid parameterization.
If you have that $E[X] = C+B$ and $E[X^2] = 2(B^2 + BC + C^2)$ then:
\begin{equation}
Var[X] = 2B^2 + 2BC + 2C^2 - B^2 -2BC - C^2 = B^2 + C^2
\end{equation}
Using method of moments we get that $C = \bar{X} - B \Rightarrow \hat{\sigma^2} = B^2 + B^2 + \bar{X}^2 -2\bar{X}B \Rightarrow 0 = 2B^2 -2\bar{X}B + \bar{X}^2 - \hat{\sigma^2}$.
The roots of that quadratic get you $B$ (one is probably negative so take the positive one), and plugging that estimate for $B$ into $C = \bar{X} - B$ gives you the moment estimator for C.
Edit: Examining the methods of moment estimators
The root for the polynomial in $B$ is $\hat{B} = \frac{1}{2}(\sqrt{2\hat{\sigma}^2 - \bar{X}^2}+\bar{X})$. Plugging that into the expression for $C$ gives $\hat{C} = \frac{1}{2}(\bar{X} - \sqrt{2\hat{\sigma}^2 - \bar{X}^2})$.
Note that with this estimate $\hat{C} < \hat{B}$, but because in the estimating equations we didn't imply an ordering on $B$ or $C$, and because the formula for the mean and variance are symmetric in the parameters, we can just switch the estimators to make $\hat{B} = \frac{1}{2}(\bar{X} - \sqrt{2\hat{\sigma}^2 - \bar{X}^2})$ and $\hat{C} = \frac{1}{2}(\sqrt{2\hat{\sigma}^2 - \bar{X}^2}+\bar{X})$.
Noting that the MLE didn't converge to anything it might be interesting to examine whether these estimates are unbiased for the mean and variance and have asymptotic consistency.
\begin{equation}
E[\hat{C} + \hat{B}] = E[\frac{\bar{X} + \bar{X}}{2}] = E[\bar{X}] = \mu
\end{equation}
\begin{equation}
\begin{split}
E[\hat{C}^2 + \hat{B}^2]& = E[\frac{1}{4}((\bar{X}^2 +2\bar{X}\sqrt{2\hat{\sigma}^2 -\bar{X}^2} + 2\hat{\sigma}^2 - \bar{X}^2)) + (\bar{X}^2 -2\bar{X}\sqrt{2\hat{\sigma}^2 -\bar{X}^2} + 2\hat{\sigma}^2 - \bar{X}^2)] \\
& = E[\frac{2\hat{\sigma}^2 + 2\hat{\sigma}^2}{4}] \\
& = E[\hat{\sigma}^2] = \sigma^2
\end{split}
\end{equation}
To show convergence, note that by the law of large numbers convergence in probability to a constant $\bar{X}\overset{P}{\rightarrow}\mu$. Applying Slutsky's theorem to $2\hat{\sigma^2} - \bar{X}^2 \overset{P}{\rightarrow} 2\sigma^{2^*} - \mu^2$, where $\sigma^{2^*}$ is the limiting distribution of $\hat{\sigma}^2$.
Applying the continuous mapping theorem to the square root, then Slutsky's theorem again to the additional $\bar{X}$, we have that:
\begin{equation}
\hat{B} \overset{P}{\rightarrow} \frac{1}{2}(\mu- \sqrt{2\sigma^{2^*} - \mu^2})
\end{equation}
\begin{equation}
\hat{C} \overset{P}{\rightarrow} \frac{1}{2}(\sqrt{2\sigma^{2^*} - \mu^2} + \mu)
\end{equation}
By the law of large numbers $\sigma^{2^*} \overset{P}{\rightarrow} \sigma^2$. Plugging $\sigma^2 = B^2 + C^2$ and $\mu = B + C$ into these expressions gives us:
\begin{equation}
\begin{split}
\hat{B} & \overset{P}{\rightarrow} \frac{1}{2}(\mu- \sqrt{2\sigma^{2^*} - \mu^2})\\
& = \frac{1}{2}(B + C- \sqrt{B^2 -2BC + C^2})\\
& = \frac{1}{2}(B + C- \sqrt{(C-B)^2})\\
& = \frac{1}{2}(B + C- C + B)\\
& = B
\end{split}
\end{equation}
\begin{equation}
\begin{split}
\hat{C} & \overset{P}{\rightarrow} \frac{1}{2}( \sqrt{2\sigma^{2^*} - \mu^2} + \mu)\\
& = \frac{1}{2}(B + C + \sqrt{B^2 -2BC + C^2})\\
& = \frac{1}{2}(B + C + \sqrt{(C-B)^2})\\
& = \frac{1}{2}(B + C + C - B)\\
& = C
\end{split}
\end{equation}
| {
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Trying to derive an inverse trigonometric function I'd like to know how to derive these functions (I know the answers, I want to know how to get there)
\begin{align*}
f(x) &= \arcsin\left(\frac{x}{3}\right)\\
f(x) &= \arccos(2x+1)\\
f(x) &= \arctan(x^2)\\
f(x) &= \mathrm{arcsec}(x^7)\\
\end{align*}
etc.
| The formulas you need are the derivatives of $\arcsin(u)$, $\arccos(u)$, $\arctan(u)$, $\mathrm{arcsec}(u)$, and presumably $\mathrm{arccot}(u)$ and $\mathrm{arccsc}(u)$. Once you know these, you can apply the Chain Rule.
And how do you find these derivatives? Well, the Inverse Function Theorem is your first friend. If $y = g(x)$ has an inverse, is differentiable at $x=a$, $g(a)=b$, and $g'(a)\neq 0$, then $(g^{-1})'(b) = \frac{1}{g'(a)}$.
So, consider $y=\sin(\theta)$. Since the derivative of $\sin(\theta)$ is $\cos(\theta)$, you have that
$$\frac{d}{du}\arcsin(u) = \frac{1}{\cos(\arcsin(u))}.$$
But... what is $\cos(\arcsin(u))$? Suppose $\arcsin(u)=\theta$. That means that $\sin(\theta) = u$, and since $\sin^2(\theta)+\cos^2(\theta)=1$, then $\cos^2(\theta) = 1 - \sin^2(\theta) = 1-u^2$. Therefore, $|\cos(\theta)|=\sqrt{\cos^2\theta} = \sqrt{1-u^2}$; and because in order to talk about the inverse of $\sin \theta$ we must have $-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2}$, then $\cos\theta\geq 0$, so $|\cos\theta|=\cos\theta$. That is, $\cos\theta = \sqrt{1-u^2}$. So, plugging into the formula for the derivative of $\arcsin(u)$, we have:
$$\frac{d}{du}\arcsin(u) = \frac{1}{\cos(\arcsin u)} = \frac{1}{\sqrt{1-u^2}}.$$
Performing the same kind of analysis for $\arccos(u)$, we get
$$\frac{d}{du}\arccos(u) = \frac{1}{-\sin(\arccos u)} = -\frac{1}{\sqrt{1-u^2}}.$$
For $\arctan u$, using the fact that $(\tan\theta)' = \sec^2\theta$, we have
$$\frac{d}{du}\arctan u = \frac{1}{\sec^2(\arctan u)}.$$
Now, if $\arctan u = \theta$, then $\tan(\theta) = u$. Using the fact that $\tan^2\theta + 1 = \sec^2\theta$, we get that $sec^2(\arctan u) = \sec^2(\theta) = 1 + \tan^2(\theta) = 1+u^2$, so
$$\frac{d}{du}\arctan u = \frac{1}{\sec^2(\arctan u)} = \frac{1}{1+u^2}.$$
For $\mathrm{arccot u}$, the same analysis works, provided you remember that $(\cot\theta)' = -\csc^2\theta$ and that $1 + \cot^2\theta = \csc^2\theta$, so
$$\frac{d}{du}\mathrm{arccot}(u) = \frac{1}{-\csc^2(\mathrm{arccot}(u))} = -\frac{1}{1+u^2}.$$
With $\mathrm{arcsec}u$, we have $(sec\theta)' = sec\theta\tan\theta$, so
$$\frac{d}{du}\mathrm{arcsec}(u) = \frac{1}{\sec(\mathrm{arcsec} (u))\tan(\mathrm{arcsec} u)}.$$
Here, $\sec(\mathrm{arcsec} (u)) = u$; if $\mathrm{arcsec}(u)=\theta$, then $\sec\theta = u$, and from $\tan^2\theta + 1 = \sec^2\theta$, we get $|\tan\theta| = \sqrt{u^2 - 1}$. You get:
$$\frac{d}{du}\mathrm{arcsec}(u) = \frac{1}{\sec(\mathrm{arcsec}(u))\tan(\mathrm{arcsec}(u))} = \frac{1}{u\sqrt{u^2-1}}.$$
And finally, using the fact that $(\csc\theta)' = -\csc\theta\cot\theta$, you get
$$\frac{d}{du}\mathrm{arccsc}(u) = \frac{1}{-\csc(\mathrm{arccsc}(u))\cot(\mathrm{arccsc}(u))} = -\frac{1}{u\sqrt{u^2-1}}.$$
Once you have these formulas, the Chain Rule takes care of the rest.
So you have:
\begin{align*}
\frac{d}{du}\arcsin(u) &= \frac{1}{\sqrt{1-u^2}}, &\qquad \frac{d}{du}\arccos u &= -\frac{1}{\sqrt{1-u^2}},\\
\frac{d}{du}\arctan(u) &=\frac{1}{1+u^2}, &\frac{d}{du}\mathrm{arccot}(u) &= -\frac{1}{1+u^2},\\
\frac{d}{du}\mathrm{arcsec}(u) &=\frac{1}{u\sqrt{u^2-1}}, &\frac{d}{du}\mathrm{arccsc}(u) &= - \frac{1}{u\sqrt{u^2-1}}.
\end{align*}
| {
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Solve $\sin x = 1 - x$ How would you be solve sin x = 1 - x, without drawing the graph and manually measuring the point of intersection?
| As @Gerben answered, building the Taylor series around $x=\frac \pi 6$ gives
$$y=x+\sin(x)-1=\frac{\pi -3}{6} +\frac{2+\sqrt{3}}{2} x+\sum_{n=2}^\infty \frac{\sqrt{3} \sin \left(\frac{\pi n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)}{2 n!} \left(x-\frac{\pi }{6}\right)^n$$
Truncated to $O\left(\left(x-\frac{\pi }{6}\right)^6\right)$ for example and using series reversion, then
$$x=\frac{\pi }{6}+t+\left(1-\frac{\sqrt{3}}{2}\right)
t^2+\left(3-\frac{5}{\sqrt{3}}\right) t^3+\left(\frac{67}{6}-\frac{77}{4
\sqrt{3}}\right) t^4+\left(\frac{232}{5}-\frac{803}{10 \sqrt{3}}\right)
t^5+O\left(t^{6}\right)$$ where $t=\frac{y+\frac{3-\pi }{6}}{1+\frac{\sqrt{3}}{2}}$.
Making $y=0$ that is to say $t=-\frac{\left(2-\sqrt{3}\right) (\pi -3)}{3} $, this gives
$x=0.51097342938847$ while the exact solution is
$x=0.51097342938857$.
| {
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Techniques for forming square factorizations Say you have the polynomial
$$ x^4 + 2 + x^{-4} $$
Looking at it, you see you can do
$$\begin{align*}
x^4 + 1 + 1 + x^{-4} & =x^2( x^2 + x^{-2} ) + x^{-2}( x^2 + x^{-2} )\\
&= \left( x^2 + x^{-2} \right)^2.
\end{align*}$$
Another one is
$$\begin{align*}
x^2 + \frac{1}{2} + \frac{1}{16x^2} &=
x^2 + \frac{1}{4} + \frac{1}{4} + \frac{1}{16x^2} \\
&=x^2\left( 1 + \frac{1}{4x^2} \right) + \frac{1}{4}\left( 1 + \frac{1}{4x^2} \right)\\
&=\left( x^2 + \frac{1}{4}\right) \left( 1 + \frac{1}{4x^2} \right)\\
(x^2) \left( 1 + \frac{1}{4x^2} \right)^2.
\end{align*}$$
So the question is, I've been doing this by "inspection" - are there any techniques for recognizing when this type of factorization is possible or how to do more easily?
| For trinomials:
*
*If you have $(p(x))^2 + K + (q(x))^2$, then check to see if $K = \pm 2p(x)q(x)$. if so, then $(p(x))^2 + K + (q(x))^2 = (p(x)\pm q(x))^2$.
*
*In your first example, you have
$$x^4 + 2 + x^{-4} = (x^2)^2 + 2 + (x^{-2})^2.$$
Taking $p(x) = x^2$, $q(x) = x^{-2}$, we have $2p(x)q(x) = 2$, which is precisely the middle term, so $x^2+2+x^{-4} = (x^2 + x^{-2})^2$.
*In your second example, you have
$$x^2 + \frac{1}{2} + \frac{1}{16x^2} = (x)^2 + \frac{1}{2} + \left(\frac{1}{4x}\right)^2.$$
Taking $p(x) = x$ and $q(x) = \frac{1}{4x}$, you have $2p(x)q(x) = \frac{2x}{4x} = \frac{1}{2}$, again exactly the middle term, so you get
$$x^2 + \frac{1}{2} + \frac{1}{16x^2} = \left(x + \frac{1}{4x}\right)^2.$$
*More generally, if you have $(p(x))^2 + Kp(x) + L$, then see if you can find two expressions, $s(x)$ and $t(x)$, which when multiplied give $L$ and when added give $K$, $s(x)t(x) = L$, $s(x)+t(x) = K$. Then $(p(x))^2 + Kp(x) +L = (p(x)+s(t))(p(x)+t(x))$.
*
*For example,
$$2x^4 + (\sqrt{2}-1)x^2 + x.$$
You may notice that the leading term is $(\sqrt{2}x^2)^2$, and that you have a "middle term" of $\sqrt{2}x^2$. So rewriting a bit you get
$$(\sqrt{2}x^2)^2 + (\sqrt{2}x^2)1 + (x-x^2).$$
This suggests looking for $s(x)$ and $t(x)$ with $s(x)+t(x) = 1$ and $s(x)t(x)=x-x^2 = x(1-x)$. This in turn quickly suggests $s(x) = x$ and $t(x)=1-x$, which gives
$$2x^4 + (\sqrt{2}-1)x^2 + x = \left(\sqrt{2}x^2 + x\right)\left(\sqrt{2}x^2 + (1-x)\right).$$
But generally, there is some substantial amount of "inspect and notice" going on. The more of these you do, the more you will notice patterns and be able to "notice" the things that one needs to notice.
| {
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How does law of quadratic reciprocity work? From the book,
Suppose $p \equiv 1 \pmod{4}$, then by law of quadratic reciprocity, we have:
$$\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right) $$
Next, if $p \equiv 2 \pmod{3}$, then $p \equiv 5 \pmod{12}$
Hence, $$\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right) = -1$$
How do they get those Legendre fraction equal to $-1$?
From my understanding:
$$\left(\frac{q}{p}\right) \cdot \left(\frac{p}{q}\right) = (-1)^{\frac{p-1}{2}\cdot\frac{q-1}{2}}$$
So $q = 3 \implies \frac{q-1}{2} = \frac{3-1}{2} = 1$
For $p$, I take $p \equiv 5 \pmod{12} \implies p = 5 + 12k$, for some integers k.
Hence, $\frac{p-1}{2} = \frac{12k + 5 - 1}{2} = \frac{12k + 4}{2} = 6k + 2.$
And this $6k + 2$ is even :( ! How does $(-1)^{even} = -1$?
Any idea? I think I made some logic mistakes somewhere, but I couldn't find where.
Update
The problem was from Elementary Number Theory and Its Application - Kenneth H.Rosen 5th Edition.
Problem
Using the law of quadratic reciprocity, show that if $p$ is an odd prime, then
$$\left(\frac{3}{p}\right) = 1 \text{ if } p \equiv \pm 1 \pmod{12}$$
$$\left(\frac{3}{p}\right) = -1 \text{ if } p \equiv \pm 5 \pmod{12}$$
Solution
Thanks,
Now I'm even more confused :(!
Consider two cases:
Case 1
$$p \equiv 1 \pmod{4} \text{ and } p \equiv 1 \pmod{3}$$
Then,
$$p \equiv 1 \pmod{12} \implies p = 12k + 1$$
Hence,
$$\frac{p - 1}{2} \cdot \frac{3 - 1}{2} = \frac{12k}{2} = 6k = \text{ even }$$
Which implies
$$\left(\frac{3}{p}\right) \cdot \left(\frac{p}{3}\right) = 1$$
So both must be $1$ or $-1$.
Case 2
$$p \equiv 1 \pmod{4} \text{ and } p \equiv 2 \pmod{3}$$
Then,
$$p \equiv 5 \pmod{12} \implies p = 12k + 5$$
Hence,
$$\frac{p - 1}{2} \cdot \frac{3 - 1}{2} = \frac{12k + 4}{2} = 6k + 2 = 2(3k + 1) = \text{ even }$$
Which implies
$$\left(\frac{3}{p}\right) \cdot \left(\frac{p}{3}\right) = 1$$
So both must be $1$ or $-1$.
I don't see how these arguments can be deduced to the solution. Any suggestion?
| If $\rm\:q\:$ and $\rm\:p = 4\:k+1\:$ are distinct odd primes then by the law of quadratic reciprocity we have
$\displaystyle\rm\quad\quad\quad\quad\ \ \frac{p-1}{2}\ =\ 2\:k\ \ \Rightarrow\ \ \left(\frac{q}{p}\right)\ \times\ \left(\frac{p}{q}\right)\ =\ (-1)^{\frac{p-1}{2}\ \frac{q-1}{2}}\: =\ 1$
Therefore we deduce that $\rm\displaystyle\ \ \left(\frac{q}{p}\right)\ =\ \left(\frac{p}{q}\right)$
So $\displaystyle\rm\ q=3,\ p = 2+3\:n\ \Rightarrow\ \left(\frac{3}{p}\right)\ =\ \left(\frac{2+3\:n}{3}\right)\ =\ \left(\frac{2}{3}\right)\ =\:-1\ $ since $\rm\:2\:$ is not a square $\rm\:(mod\ 3)\:.$
| {
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Prove $\gcd(a+b, a-b) = 1$ or $2\,$ if $\,\gcd(a,b) = 1$
I want to show that for $\gcd(a,b) = 1$ $a,b \in Z$ $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ holds.
I think the first step should look something like this:
$d = \gcd(a+b, a-b) = \gcd(2a, a-b)$
From here I tried to proceed with two cases.
1: $a-b$ is even, which leads to $\gcd(a+b, a-b) = 2$
2: $a-b$ is odd, which leads to $\gcd(a+b, a-b) = 1$
My main problem I think is, that I do not know how I should include $\gcd(a,b) = 1$ in the proof.
Any help is appreciated. Thx in advance.
Cherio Woltan
| If $\gcd (a,b)=1$, then both $a$ and $b$ can't be even simultaneously. Now two cases arise,
Case $1$. Both $a$ and $b$ are odd
In this case, both $a+b$ and $a-b$ will be even, so their gcd must be an even number,
and so $\gcd (a+b, a-b) = \gcd (2a, a-b) = 2$, as $a-b$ is even.
Case $2$. One of the $a$ and $b$ is even and the other is odd
In this case, both $a+b$ and $a-b$ will be odd, so their $\gcd$ must be an odd number,
and in this case $\gcd (a+b, a-b) = \gcd (2a, a-b) = 1$, as $a-b$ is odd.
Note: $\gcd (a, a-b)$ can't exceed $1$ as $a$ and $b$ are relatively prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/32737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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sums of reciprocals Here is another tricky question: Suppose $m_1,\cdots,m_j$ and $m'_1,\cdots,m'_j$ are positive integers such that $\sum\limits_{k=1}^j m_k = \sum\limits_{k=1}^j m'_k$ and $\sum\limits_{k=1}^j m^2_k > \sum\limits_{k=1}^j m'^2_k$. What can we say about the sums of the reciprocals $\sum\limits_{k=1}^j \frac{1}{m_k}$ and $\sum\limits_{k=1}^j \frac{1}{m'_k}$. Which is larger than the other?
| Try some examples. $1 + 3 = 2 + 2$ with $1^2 + 3^2 > 2^2 + 2^2$ and $1/1 + 1/3 > 1/2 + 1/2$. On the other hand, $2 + 5 + 12 = 9 + 9 + 1$ with $2^2 + 5^2 + 12^2 > 9^2 + 9^2 + 1^2$ and $1/2 + 1/5 + 1/12 < 1/9 + 1/9 + 1/1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A simple way to evaluate $\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx$? I am currently trying to show that $\int_{-\infty}^\infty \cos(x^2) \, \mathrm dx = \sqrt{\frac{\pi}{2}}$ and the last integral I have to evaluate is
$$\int_{-a}^a \frac{x^2}{x^4+1} \, \mathrm dx.$$
Now of course I'm familiar with wolframalpha, however the way it solves this integral seems very awkward and also not elegant to me, even though the function to me looks quite simple. So, is there a simpler way to solve this integral or is the way described on wolframalpha already (one of) the simplest approach(es)? I ask this because often wolframalpha doesn't see tricks (occurred to me when I wanted to find a formula for the n-th derivative of some function) which a human eye might see.
Thanks for any answers in advance.
| I'm late to the party, but here's a simple way to derive the result shown in W|A without knowing before hand that $x^4+1$ factors as $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$. We have that $$\int_{-a}^a\dfrac{x^2}{x^4+1}\,\mathrm dx=\int_{-a}^a\dfrac{\tfrac1{x^2}\cdot x^2}{\tfrac1{x^2}\cdot \left(x^4+1\right)}\,\mathrm dx=\int_{-a}^a\dfrac{1}{x^2+\tfrac1{x^2}}\,\mathrm dx.$$ Now here's the trick, since $\left(x+\tfrac1x\right)^2=x^2+2+\tfrac1{x^2}$ it follows that $x^2+\tfrac1{x^2}=\left(x+\tfrac1x\right)^2-\sqrt{2}^2$. Hence the fraction decomposes nicely as follows, $$\begin{align} \frac{1}{\tfrac1{x^2}+x^2}&=\frac{1}{\left(\tfrac{1}x+x\right)^2-2} \\&=\dfrac{1} {\left (\tfrac{1}{x}+x+\sqrt{2}\right)\left(\tfrac{1}{x}+x-\sqrt{2}\right)}\\&= \dfrac{A}{\tfrac{1}{x}+x+\sqrt{2}}+\dfrac{B}{\tfrac{1}{x}+x-\sqrt{2}}. \end{align}$$ Using standard techniques we find that $A=\tfrac{-1}{2\sqrt{2}}$ and $B=\tfrac{1}{2\sqrt{2}}$, and so the last expression can be reduced to, $$\dfrac{x}{2\sqrt{2}\left(x^2+\sqrt{2}x+1\right)}-\dfrac{x}{2\sqrt{2}\left(x^2-\sqrt{2}x+1\right)}.$$ The rest will be messy but still pretty straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/37229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 7
} |
Two proofs involving Harmonic Mean If H is the harmonic mean between $a$ and $b$,then show that
$$\frac{1}{H-a}+\frac{1}{H-b} = \frac{1}{a} + \frac{1}{b}$$ and $$\frac{H+a}{H-a}+\frac{H+b}{H-b} = 2$$
I substituted $\displaystyle H = \frac{2ab}{a+b}$, then tried some algebraic manipulation, but I am not getting there.
Are they even valid? If yes, could somebody give me some ideas how to approach these?
| I don't know why you aren't getting this. You have
\begin{align*}
\frac{1}{H-a} + \frac{1}{H-b} &= \frac{1}{\frac{2ab}{a+b} -a} + \frac{1}{\frac{2ab}{a+b}-b} \\ &= \frac{a+b}{ab-a^{2}} + \frac{a+b}{ab-b^{2}} \\ &= (a+b) \cdot \biggl[ \frac{1}{a(b-a)} + \frac{1}{b(a-b)}\biggr] \\
\end{align*}
I think this should help you out. Try taking the factor $(a-b)$ common. For the next problem also again substitute the value of $H$ and try doing manipulations. Be patient, and be careful, with your calculations, you shall arrive at the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/38919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Calculating $a^p+b^p$ given $a^2+b^2=c$ If one has $a^2+b^2=c$ where $a$,$b$ and $c$ are real numbers, is there any way to calculate $a^p+b^p$ where $p$ may be any real number? If there is, would you please explain with an example with $p=3$?
| The other responses showing you cannot find it just from $a^2+b^2=c$ are correct. You may be interested in the fact that given $a^2+b^2=c, a+b=d$, you can form $(a^2+b^2)(a+b)=cd=a^3+a^2b+ab^2+b^3$ and $d^2-c=2ab$, so $cd-\frac{d^3-cd}{2}=a^3+b^3$. Given as many symmetric polynomials as variables you can find all the higher orders.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/39612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\sqrt{3}\tan\theta=2\sin\theta$ I am trying to solve $\sqrt{3}\tan\theta=2\sin\theta$ on the interval $[-\pi,\pi]$.
$$\sqrt{3}\tan\theta=2\sin\theta \Rightarrow \sqrt{3}=\frac{2\sin\theta}{\tan\theta}$$
$$\Rightarrow \sqrt{3}=2\sin\theta \cdot \frac{\cos\theta}{\sin\theta} \Rightarrow 3 = 4\cos^2\theta$$
I get $\displaystyle \cos \theta = \pm{\frac{\sqrt{3}}{2}}$; the cosine of $30^{\circ}$ and $150^{\circ}$ so arrived at the solutions $\displaystyle -\frac{5}{6}\pi,-\frac{1}{6}\pi,\frac{1}{6}\pi,\frac{5}{6}\pi$.
Looking in the back of the book (and checking with Wolfram), the answer is $\displaystyle -\pi,-\frac{1}{6}\pi,0,\frac{1}{6}\pi,\pi$.
Where am I going wrong please?
| When you cancelled the $\sin \theta$ factors, you lost some solutions. This is similar to starting with $x^2 = x$, cancelling an $x$ and concluding that $x=1$; the obvious, $x=0$ solution has been lost. The remedy for this is to factor instead. In the example I provided, that would amount to writing $x^2 - x = 0$, then factoring to get $x(x-1)=0$ then setting each factor equal to zero, as usual.
You also gained solutions when you squared. This is similar to starting with $x=1$, then squaring to get $x^2 = 1$, so $x= \pm 1$. Notice that in your problem, squaring was superfluous since a few steps later you took a square root that "undid" that squaring anyway.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/40077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Subsets and Splits
Fractions in Questions and Answers
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