Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$. Problem: Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.
My work: So I think I have to do a proof by induction and I just wanted some help editing my proof.
My attempt:
Let $P(n)=1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.
Then $$P(1)=1^2=\frac{1(1+1)(2+1)}{6}$$
$$1=\frac{6}{6}.$$
So $P(1)$ is true.
Next suppose that $P(k)=1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}$ for $k \in \mathbb{N}$. Then adding $(k+1)^2$ to both sides of $P(k)$ we obtain the following:
$$1^2+2^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$
$$=\frac{2k^3+3k^2+k+6(k^2+2k+1)}{6}$$
$$=\frac{2k^3+9k^2+13k+6}{6}$$
$$=\frac{(k^2+3k+2)(2k+3)}{6}$$
$$=\frac{(k+1)(k+2)(2k+3)}{6}$$
$$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$
$$=P(k+1).$$
Thus $P(k)$ is true for $k \in \mathbb{N}$.
Hence by mathematical induction, $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ is true for $n \in \mathbb{N}$.
| Consider any natural number $r$. You have $$r^3-(r-1)^3=3r^2-3r+1.$$
Now telescope it:
$$
1^3-0^3=3-3+1
$$
$$2^3-1^3=3\cdot2^2-3\cdot2+1
$$
$$\vdots
$$
$$
n^3-(n-1)^3=3n^2-3n+1
$$ Now add, and see them cancel out. You are left with $$n^3=3(1^2+2^2+\cdots+ n^2)-3(1+2+3+\cdots+n)+n$$ You must know
$$
1+2+3+\cdots+n=\frac{n(n+1)}{2}.
$$
Plug it in, and you get the answer. Also, please see that this method works even for $\sum r^4,r^5,\cdots$. I have tried it out. All you need is the sum of its previous powers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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} |
How to decompose the polynomial $x^{21} + 1$ into a product of irreducible factors over $\mathbb Z/2\mathbb Z$? This is a problem from a past abstract algebra exam, the degree $21$ was too high for me to solve it. No Wolfram Mathematica please!
| $$x^{21} + 1 = x^{21} - 1 = (x^7)^3 - 1 = (x^7 - 1)(x^{14} + x^7 + 1) = (x - 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)(x^{14} + x^7 + 1) = (x -1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)(x^2 + x + 1)(x^{12} - x^{11} + x^9 - x^8 + x^6 - x^4 + x^3 - x +1)$$
Now: The $2$nd one can be shown to be irreducible in $\mathbb Z[x]$; I don't know about $\frac{\mathbb Z}{2\mathbb Z}[x]$. The $3$rd one can be shown to be irreducible in $\frac{\mathbb Z}{2\mathbb Z}[x]$. Good luck with the last one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Use $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$ Is it possible to use the fact that $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$?
| Note
$$\sum_{n=1}^\infty\frac{1}{n^4}+\sum_{n=1}^\infty\frac{(-1)^n}{n^4}=2\left(\frac{1}{2^4}+\frac{1}{4^4}+\dots+\right)=\frac{1}{8}\left(\frac{1}{1^4}+\frac{1}{2^4}+\dots\right)=\frac{1}{8}\sum_{n=1}^\infty\frac{1}{n^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the limit of $ (1+1/n)^n$ to find the limit of $((n^2+2)/(n^2+1)) ^ {3n^2+1/n}$ Use: $ \lim_{n \to \infty} (1+1/n)^n = e$
To find: $ \lim_{n \to \infty} (\frac{n^2+2}{n^2+1}) ^ {3n^2+1/n}$
Any help is appreciated. Not looking for the complete answer since it's looked down upon here. So strong hints would be appreciated. I am lost.
| Hint: ${n^2 +2 \over n^2 +1}=1+{1 \over n^2 +1}$
It is convenient to write the exponent as $3n^2 + 1/n= 3(n^2 + 1) -3 + 1/n$ to get:
$${n^2 +2 \over n^2 +1}^{3n^2 + 1/n}=\left (\left (1+{1 \over n^2 +1} \right )^{n^2+1} \right )^3\left (1+{1 \over n^2 +1} \right )^{-3}\left (1+{1 \over n^2 +1} \right )^{1/n}$$
Now $n^2 +1 \rightarrow +\infty$ as $n \rightarrow +\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what values of k is the vector u in the span of the columns of A? Let $$u = \begin{pmatrix} -1 \\1 \\-1 \end{pmatrix}$$
and let$$A = \begin{pmatrix} 1&0&1 \\1&1&0 \\0&1&k \end{pmatrix}$$
For what values of k is the vector u in the span of the columns of A?
Here is my work so far:
$$Au = \begin{pmatrix} 1&0&1&-1 \\1&1&0&1 \\0&1&k&-1 \end{pmatrix}$$$$Au = \begin{pmatrix} 1&0&1&-1 \\0&1&-1&2 \\0&1&k&-1 \end{pmatrix}$$$$Au = \begin{pmatrix} 1&0&1&-1 \\0&1&-1&2 \\0&0&k+1&-3 \end{pmatrix}$$
I've reduced it as much as I can, but I'm stuck at this point. Does this mean that k can be equal to -4?
| $(-1, 1, -1) = x (1, 1, 0) + y (0, 1, 1) + z (1, 0, k) $
\begin{equation*}
\left\{\begin{aligned}
-1 = x 1 + y 0 + z 1 \\
1 = x 1 + y 1 + z 0 \\
-1 = x 0 + y 1 + z k
\end{aligned}
\right.
\end{equation*}
\begin{equation*}
\left\{\begin{aligned}
-1 - x &= z(x) \\
1 - x &= y(x) \\
-1 &= y + z k = 1-x + (-1-x) k
\end{aligned}
\right.
\end{equation*}
$- \frac{x-2}{x + 1} = k(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$
| Another way is as follows:
$$
(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 = (x^3 - y^3) -3xy(x - y) \quad \Rightarrow
$$
$$
x^3 - y^3 = (x - y)^3 + 3xy(x - y) = (x - y)[(x - y)^2 + 3xy] \quad \Rightarrow
$$
$$
x^3 - y^3 = (x - y)(x^2 + xy + y^2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Does the series of $\sum_{n=2}^{\infty} \frac{(-1)^n(n+1)}{n-1}$ converge? Does the following sum converge?
$$
\sum_{n=2}^{\infty} \frac{(-1)^n(n+1)}{n-1}
$$
Have tried ratio test and root test, inconclusive. alternating series test not useful.
| Let
$s(m)
=\sum_{n=2}^{m} \frac{(-1)^n(n+1)}{n-1}
$.
$\begin{array}\\
s(2m+1)
&=\sum_{n=2}^{2m+1} \frac{(-1)^n(n+1)}{n-1}\\
&=\sum_{n=1}^{m} (\frac{(-1)^{2n}(2n+1)}{2n-1}+\frac{(-1)^{2n+1}(2n+1+1)}{2n+1-1})\\
&=\sum_{n=1}^{m} (\frac{(2n+1)}{2n-1}-\frac{(2n+2)}{2n})\\
&=\sum_{n=1}^{m} (\frac{2n-1+2}{2n-1}-\frac{n+1}{n})\\
&=\sum_{n=1}^{m} (1+\frac{2}{2n-1}-(1+\frac{1}{n}))\\
&=\sum_{n=1}^{m} (\frac{2}{2n-1}-\frac{1}{n})\\
&=\sum_{n=1}^{m} \frac{2n-(2n-1)}{n(2n-1)}\\
&=\sum_{n=1}^{m} \frac1{n(2n-1)}\\
&\to c
\qquad\text{for some real }c\text{ since the sum converges}\\
s(2m+2)
&=s(2m+1)+ \frac{(-1)^{2m+2}(2m+2+1)}{2m+2-1}\\
&=s(2m+1)+ \frac{(2m+3)}{2m+1}\\
&=s(2m+1)+ 1+\frac{2}{2m+1}\\
&\to c+1\\
\end{array}
$
Therefore the even terms and odd sums
approach different limits,
so the series itself
does not have a limit.
Note:
this also works for
$\sum_{n=2}^{m} \frac{(-1)^n(n+a)}{n+b}
$
with $a \ne b$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trying to determine the Inverse function of $\sinh$ and $\cosh$ I'm trying to find out how to determine the inverse function in order to develop the $$ \sinh(x).$$
I tried to expand to its exponential form $$\sinh(x) = \frac{1}{2} (e^x-e^{-x}) .$$
So I wrote
$$ \frac{1}{2}(e^x-e^{-x}) = y . $$
So we find ourselves with this equation
By substituting $ e^x = t $,
we have $$ t^2 -2yt +1 = 0 .$$
Noting that $\sinh$ changes its sign from negative to positive as shown below.
What to do next because $$ \delta = 4y^2-4 .$$
| If you multiply both sides of the equation
$$y = \frac{e^x - e^{-x}}{2}$$
by $2e^x$, you obtain
\begin{align*}
2e^{x}y & = e^{2x} - 1\\
0 & = e^{2x} - 2e^{x}y - 1
\end{align*}
so you should have obtained the equation $t^2 - 2ty - 1 = 0$ when you made the substitution $y = e^x$.
To find the inverse, we interchange $x$ and $y$ in the equation $e^{2x} - 2e^{x}y - 1 = 0$, then solve for $y$.
\begin{align*}
e^{2y} - 2xe^{y} - 1 & = 0\\
e^{2y} - 2xe^{y} & = 1\\
e^{2y} - 2xe^{y} + x^2 & = 1 + x^2 && \text{complete the square}\\
(e^y - x)^2 & = 1 + x^2\\
e^y - x & = \pm \sqrt{1 + x^2}\\
e^y & = x \pm \sqrt{1 + x^2}
\end{align*}
Since the range of $e^y$ is the set of all positive real numbers, we discard the negative root since $x - \sqrt{1 + x^2} < 0$. Thus,
\begin{align*}
e^y & = x + \sqrt{1 + x^2}\\
y & = \log_e (x + \sqrt{1 + x^2})
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180824",
"timestamp": "2023-03-29T00:00:00",
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Maximum number of prime sums Zach thinks of six different positive integers, and for each pair of numbers, he adds them. This gives him $\binom{6}{2}=15$ sums. Among these sums, find the maximum number that can be prime numbers. I deducted that there has to be atleast one even number among the $6$ numbers. But will there be one even number, or two evens. Then what will the other numbers be.
| The sum of two different positive integers is always at least $3$, and so in order for it to be prime, it has to be odd. So one summand has to be even, the other odd. If there are $a$ even numbers and $b = 6-a$ odd numbers, then the potential number of primes is at most $ab \leq 9$, attained when $a=b=3$. Can this bound be attained? Here is an example:
$$
2,3,4,8,9,15.
$$
All of the following are primes:
$$
2+3 = 5 \\
3+4 = 7 \\
2+9 = 3 + 8 = 11 \\
4+9 = 13 \\
2+15 = 8+9 = 17 \\
4+15 = 19 \\
8+15 = 23
$$
If you don't like the repeated sums, you can try instead:
$$
2,3,4,8,39,99.
$$
All of the following are prime:
$$
2+3 = 5 \\
3+4 = 7 \\
3+8 = 11 \\
2+39 = 41 \\
4+39 = 43 \\
8+39 = 47 \\
2+99 = 101 \\
4+99 = 103 \\
8+99 = 107
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of no primes such that $x^2 + y^2 = z^2$ I'm in a pretty simple "CS Math" course for year 1 Comp Sci, and I came across this:
Disprove, $x^2 + y^2 = z^2$, such that $x, y, z$ are primes
I thought of this as, if n is a prime, then prime factorization of n must be:
$n = z*1$
$n^2 = (z*1)(z*1) = (z)(z)$
So I tried to derive a contradiction:
Assume $x, y, z$ are primes such that $x^2 + y^2 = z^2$
Then: $x^2 = z^2 - y^2$
Then: $x^2 = (z-y)(z+y)$ #Contradiction, prime factorization of $x^2$ must be uniquely represented.
But then I noticed that this statement is true if $z-y = 1$ and $z+y = x^2$, So:
Then: $y = -1+z = y = z-1$
Because: $x^2 + y^2 = z^2 $
Then: $(z-1)(z-1) + z+y = z^2$
Then: $z^2 - 2z + 1 + z + y = z^2$
Then: $z^2 - z + y + 1 = z^2$ # subtract $z^2$ from both sides
Then: $-z + y + 1 = 0$
Then: $y = -1 + z$ #Loop??
I remember reducing this to $x + y = z$, which I think is false?
EDIT: Thought of this:
z - y = 1
Then: z = y + 1
Then z must be 3, y must be 2
Because: z is greater than y by 1, since z is a prime and y is a prime, z and y must be 3 and 2 respectively since no other prime number is 1 apart because one of them would be even if greater than 2, and any even number has at least 3 divisors (2, the even number, and 1) thus not a prime.
Then: $x^2 + 2^2 = 3^2$ ,Because $2^2 = 4, 3^2 = 9$
Then: $x^2 = 9 - 4$
Then: $x = \sqrt5$
Contradiction: x is a prime, but sqrt(5) is not a prime!
Is this a solid evidence proof in that can I say that only prime numbers 1 apart are 2 and 3? I'm pretty sure there is a more simpler and intuitive way to prove this, are there any other ways to prove this via contradiction?
| I think this proof is sufficient:
If x, y, z are primes, $x^2 + y^2 \ne z^2$
There exists x, y, z that are primes such that $x^2 + y^2 = z^2$, derive contradiction
Then: $x^2 = z^2 - y^2$
Then: $x^2 = (z-y)(z+y)$
Then: $z - y = 1$ and $z + y = x^2$, because by the Fundamental Theorem of Arithmetic, the prime factorization of x which is a prime is:
$x = (x)(1)$
Then: $x^2 = (x)(1)(x)(1) = (x)(x)$
Then: $x^2$ has only three factors, $x^2, 1, x$. Since $x - y \ne x + y$, then by elimination process, $x-y = 1$ and $z + y = x^2$
Then: $x - y = 1$
Then: $x = y + 1$
Then: $x = 3, y = 2$
Because: x and y are primes, x is one greater than y implies that one of x or y must be even as they both cannot be odd (odd number - 1 cannot equal odd). The only even prime number is 2, so by deduction, $y = 2, z = 3$
Also: $z + y = x^2$, as shown above
Then: $x^2 + 2^2 = 3^2$
Then: $x^2 = 9 - 4$
Then: $x = \sqrt5$
Contradiction! Prime Numbers must be natural numbers, $\sqrt5$ is not a whole number or natural number
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Normalizer and Centralizer of Upper Triangular Matrix Consider the group $H_3(\mathbb{Z})=\left\{\begin{pmatrix}1 & a & c\\0 & 1 &
b\\0 & 0 & 1\end{pmatrix}:a, b, c\in\mathbb{Z}\right\}$.
Find $Z\left(\begin{pmatrix}1 & 1 & 0\\0 & 1 & 1\\0 & 0 &
1\end{pmatrix}\right)$.
Find $N_{H_3(\mathbb{Z})}\left(\left\{\begin{pmatrix}1 & a & 0\\0 & 1
& 0\\ 0 & 0 & 1\end{pmatrix}\right\}\right)$.
So to find the centralizer of \begin{pmatrix}1 & 1 & 0\\0 & 1 & 1\\0 & 0 &
1\end{pmatrix}
I need to find things that commute with each other right?
| To find the centralizer of
$$
A = \begin{bmatrix}1 & 1 & 0\\0 & 1 & 1\\0 & 0 &
1\end{bmatrix}
$$
you need to solve the equation
$$
A X = X A,
$$
where
$$
X = \begin{bmatrix}1 & a & c\\0 & 1 & b\\0 & 0 &
1\end{bmatrix}.
$$
This will yield three linear equations which, as you will see, admit the immediate solution $a = b$.
To find the normalizer, you might first note that your subgroup
$$
\left\{
\begin{bmatrix}1 & y & 0\\0 & 1 & 0\\0 & 0 &
1\end{bmatrix}
:
y \in \mathbb{Z}
\right\}
$$
is generated by
$$
B = \begin{bmatrix}1 & 1 & 0\\0 & 1 & 0\\0 & 0 &
1\end{bmatrix},
$$
so it is enough to find all $X$ as above such that
$$
B X = X B',
$$
where
$$
B' = \begin{bmatrix}1 & y & 0\\0 & 1 & 0\\0 & 0 &
1\end{bmatrix},
$$
for some $y$. You will find out that the condition is $b = 0$ here.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\int\frac{dx}{\sqrt{4-x^2}}$ with trigonometric substitution? The integral
$$\int\frac{dx}{\sqrt{4-x^2}}$$
I've found the variable
$$x=2\sin\theta$$
$$x^2=4\sin^2\theta$$
$$dx=2\cos\theta\,d\theta$$
Which gave me by substitution
$$\int\frac{2\cos\theta}{4-\sqrt{4\sin^2\theta}}\,d\theta$$
$$\int\frac{\cos\theta}{2-\sin\theta}\,d\theta$$
$$\frac{1}{2}\int \cos\theta \,d\theta-\int\frac{\cos\theta}{\sin\theta}\,d\theta$$
$$\frac{1}{2}\int \cos\theta \,d\theta-\int\cot\theta \,d\theta$$
$$= \frac{\sin\theta}{2}-\ln|\sin\theta| + C$$
Now if I look at the expected answer, it should be
$$\arcsin\left(\frac{x}{2}\right)+C$$
What am I missing ?
| Substitution of $x=2\sin\theta$ implies that $\sqrt{4-x^2}=\sqrt{4-4\sin^2\theta}=2\cos\theta$. Then,
$$\int \frac{dx}{\sqrt{4-x^2}}=\int \frac{\cos\theta d\theta}{\cos\theta}=\theta+C$$
where $C$ is a constant of integration. But, $\theta=\arcsin\left(\frac{x}{2}\right)$. So,
$$\int \frac{dx}{\sqrt{4-x^2}}=\arcsin\left(\frac{x}{2}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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differential equation; how to find the general solution. how do i get the general solution of $$\frac{dy}{dx} = \frac{(10x^2-1)y}{ x(10x^2+7x+1)}$$ please? i have been trying for a while now. I have been doing it as a separable DE and I get $$\frac {dy}y = \frac{(10x^2-1)\, dx}{x(10x^2+7x+1)} $$I can not seem to carry on.
| The solution is as follows
\begin{align}
\frac{dy}{dx} = \frac{(10x^2-1)y}{x(10x^2+7x+1)}
\end{align}
can be seen to be of the form
\begin{align}
\frac{dy}{dx} = y \left( \frac{2}{2x+1} + \frac{5}{5x+1} - \frac{1}{x} \right)
\end{align}
or
\begin{align}
\frac{dy}{y} = \left(\frac{2}{2x+1} + \frac{5}{5x+1} - \frac{1}{x} \right) dx
\end{align}
for which
\begin{align}
\ln(y) = \ln(2x+1) + \ln(5x+1) - \ln(x) + \ln(c_{0}) = \ln\left( \frac{c_{0} (2x+1)(5x+1)}{x} \right)
\end{align}
and the general solution for $y(x)$ becomes
\begin{align}
y(x) = \frac{c_{0} (2x+1)(5x+1)}{x}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Definition of multiplication of complex numbers I know that given two complex numbers $z_1 = a + bi$ and $z_2 = c + di$, the multiplication of these two numbers is defined as
$$
z_1*z_2 = (ac - bd) + i(ad + cb)
$$
I also know that I can easily derive this formula by applying the distributive property of multiplication and considering $i^2 = -1$.
But $i^2 = -1$ is a consequence of the definition of multiplication.
So, the question is: How mathematicians defined the multiplication of complex numbers whether they didn't know the value of $i^2$?
| Indeed, $i^2=-1$ is coherent with the definition of multiplication, but things appeared in the reverse order.
While working on the resolution of algebraic equations (roots of polynomials), the need arose to consider square roots of negative numbers. Hence the "imaginary" symbol $i=\sqrt{-1}$, which is enough to represent the square root of any negative number $\sqrt{-a}$ as $\sqrt ai$.
From there, the algebra of complex numbers easily follows, treating $i$ as a variable and using the rule $i^2=-1$:
$$(a+bi)+(c+di)=a+bi+c+di=(a+b)+(c+d)i,$$
$$(a+bi)(c+di)=ac+adi+bci+bdi^2=ac+adi+bci-bd=(ac-bd)+(ad+bc)i.$$
Division is not much more difficult to invent, using a trick to turn the denominator to a real number,
$$\frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{(a+bi)(c-di)}{c^2+d^2}.$$
Now, what about the square root $\sqrt{a+bi}$ ?
Well, let us try and solve $a+bi=(c+di)^2=c^2-d^2+2cdi$, or $$a=c^2-d^2\\b=2cd.$$
Then,
$$a^2+b^2=c^4-2c^2d^2+d^4+4c^2d^2=c^4+2c^2d^2+d^4=(c^2+d^2)^2,$$
so that
$$c^2=\frac{\sqrt{a^2+b^2}+a}2,\\d^2=\frac{\sqrt{a^2+b^2}-a}2.$$
As you can see, $c$ and $d$ are always real numbers, so that no new symbol needs to be introduced to take square roots. From the last formulas, we have $$\sqrt i=\frac{1+i}{\sqrt2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 3
} |
Evaluation of $\int\frac{1}{x^4-5x^2+16}dx$
Evaluation of $\displaystyle \int\frac{1}{x^4-5x^2+16}\,dx$
$\bf{My\; Try::}$ Given $$\displaystyle \int\frac{1}{x^4-5x^2+16}dx = \frac{1}{8}\int\frac{\left(x^2+4\right)-\left(x^2-4\right)}{x^4-5x^2+16}\,dx$$
So We get $$\displaystyle = \frac{1}{8}\int\frac{x^2+4}{x^4-5x^2+16}\,dx-\frac{1}{8}\int\frac{x^2-4}{x^4-5x^2+16}\,dx$$
So We get $$\displaystyle = \frac{1}{8}\int\frac{1+\frac{4}{x^2}}{\left(x-\frac{4}{x}\right)^2+\left(\sqrt{3}\right)^2}dx-\frac{1}{8}\int\frac{1-\frac{4}{x^2}}{\left(x+\frac{4}{x}\right)^2-\left(\sqrt{13}\right)^2}\,dx$$
Now Using $$\displaystyle \left(x-\frac{4}{x}\right)=t$$ and $$\displaystyle \left(1+\frac{4}{x^2}\right)dx = dt$$ in First Integral and
Using $$\displaystyle \left(x+\frac{4}{x}\right)=u$$ and $$\displaystyle \left(1-\frac{4}{x^2}\right)\,dx = du$$ in Second Integral.
So We Get $$\displaystyle = \frac{1}{8}\int\frac{1}{t^2+\left(\sqrt{3}\right)^2}dt-\frac{1}{8}\int\frac{1}{u^2-\left(\sqrt{13}\right)^2}\,du$$
So We Get $$\displaystyle = \frac{1}{8\sqrt{3}}\tan^{-1}\left(\frac{x^2-4}{\sqrt{3}x}\right)-\frac{1}{16\sqrt{13}}\ln \left|\frac{x^2-\sqrt{13}x+4}{x^2+\sqrt{13}x+4}\right|+{C}$$
My question is can we solve the above question any other Method, If yes then
please explain here. Thanks
| Use hyperbolic functions. The integrand can be stated as $ \frac {4}{4x^4 - 20x^2 + 64} = \frac {4}{(2x^2 - 5)^2 + 39} $
Let $ x = \sqrt{ \frac{5}{2} }\cosh y $, then $ \mathrm{d} x = \sqrt{ \frac {5}{2} } \sinh y \space \mathrm{d} y $
The integral becomes
$ \displaystyle 4 \cdot \sqrt{ \frac {5}{2} } \int \frac {\sinh y}{5 \sinh^2 y + 39 } \space \mathrm{d} y $
Using the identity $ \sinh^2 A - \cosh^2 A = -1 $, apply another substitution $ \cosh y = \frac {1}{\sqrt 5} z $
Then apply $ \int \frac {1}{x^2+a^2} \mathrm{d}x = \frac {1}{a} \ \arctan \left ( \frac {x}{a} \right ) $ for $ x > 0 $
Back substitute everything and you get your answer
| {
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"url": "https://math.stackexchange.com/questions/1186356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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How can we solve $y^2=x^3+23$ without trial and error? $$y^2=x^3+23$$
Are there any easy ways to solve this problem with number theory, abstract algebra, etc.?
(trial and error for mods by the way)
| $$ y^2 + 4 = x^3 + 27 = (x+3)(x^2 - 3 x + 9)$$
If $x$ were even, we would have $y^2 \equiv 3 \pmod 4,$ impossible.
So, $x$ is odd, $y$ is even. Note that $y^2 + 4 $ is divisible by $4.$ Since $x^2 - 3 x + 9$ is odd, this means $x+3$ is divisible by $4.$ That is $x \equiv 1 \pmod 4.$
As a result, since $-3 \equiv 1 \pmod 4,$ we have $-3 x \equiv 1 \pmod 4,$ so
$$ x^2 - 3 x + 9 \equiv 3 \pmod 4. $$
Finally, this means that there is some prime $q \equiv 3 \pmod 4$ such that
$q | (x^2 - 3 x + 9).$
The contradiction comes because that means $q |(y^2 + 4),$ this being impossible. Indeed, the lemma is that, given prime $q \equiv 3 \pmod 4,$ if $q |(u^2 + v^2),$ then both $q|u$ and $q |v.$ In this case, we have $v^2 = 4$ and $v = 2,$ and we cannot have $q | 2.$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Center of mass of Right angle Trapezoid Given bases $a$, $b$, and the height $h$.
Get the $M(x,y)$ coordinates formula from point $O(0,0)$, where $M$ is center of mass. Wiki has a formula for $M(y) = \frac{h}{3}\frac{2a+b}{a+b}$. And I'm interested in how to find that formula (also for $M(x)$).
The median $c$ is dividing the Trapezoid into 2 equal surfaces. I need to draw $m$ in right place $y$, so all 4 surfaces stay equal each other.
Here's my try:
$y(\frac{a+m}{2})=(h-y)(\frac{m+b}{2})$
$\frac{a+m}{2}=(\frac{h}{y}-1)(\frac{m+b}{2})$
$y=\frac{h(m+b)}{a+2m+b}$
on the other look
$2y(\frac{a+m}{2})=h(\frac{a+b}{2})$
$y=\frac{h(a+b)}{2(a+m)} = \frac{h(m+b)}{a+2m+b}$
From the last equality we have: $m^2 = \frac{a^2+b^2}{2}$
But I still cant get that formula back for $y$
| The formula for the distance from the base $b$ to the center of mass of a trapezoid
is
$$\bar y = \frac{b+2a}{3(a+b)} h.$$
You can find this in many on-line sources such as Wolfram Mathworld.
You can prove this by integrating $\int_0^h y(b + (a-b)\frac yh) dy$
and dividing by the area of the trapezoid. But here's a derivation without calculus,
using the fact that the distance from a side of a triangle to the triangle's
centroid is $\frac13$ the height of the triangle.
Let $T$ be a trapezoid with bases $a$ and $b$.
For the case $a < b$, extend the non-parallel sides of the trapezoid
until they intersect. The base $b$ of the trapezoid and the two extended
sides form a triangle $B$; the base $a$ divides this triangle into two pieces,
one of which is $T$ and the other of which is a triangle which we'll call $A$.
If the height of the trapezoid $T$ is $h$, the height of $B$ is $\frac{b}{b-a}h.$
The centroid of $B$ is at a distance $\frac{b}{3(b-a)}h$ from base $b$.
But another way to find the centroid of $B$ is to balance the two figures
$A$ and $T$ that compose $B$.
The triangle $A$ has height $\frac{a}{b-a}h$, so its area is
$\frac{a^2}{2(b-a)}h$ and its centroid is a distance $\frac{a}{3(b-a)}h + h$
from the base $b$ of the trapezoid.
Trapezoid $T$ has area $\frac{a+b}{2} h$ and a centroid at the unknown
distance $\bar y$ from base $b$.
To "balance" the two regions, we take a weighted average
of the distance of their centroids from base $b$.
The "weights" in this average are just the areas of the two regions.
This weighted average is the same as the distance of the centroid of $B$
(the combined figure) from base $b$.
That is,
$$\begin{eqnarray}
\frac{b}{3(b-a)}h
&=&\frac{\mathop{Area}(A) \cdot \left(\frac{a}{3(b-a)}h + h\right)
+ \mathop{Area}(T) \cdot \bar y}{\mathop{Area}(A) + \mathop{Area}(T)}\\
&=&\frac{\frac{a^2}{2(b-a)}h \cdot \left(\frac{a}{3(b-a)} + 1\right)h
+ \frac{a+b}{2} h \cdot \bar y}{\frac{a^2}{2(b-a)}h + \frac{a+b}{2} h}\\
\end{eqnarray}$$
Solve for $\bar y$. This looks messy, but it can be simplified if you
realize that
$$\frac{\mathop{Area}(T)}{\mathop{Area}(A)} = \frac{b^2 - a^2}{a^2}.$$
If you divide both the numerator and denominator on the
right-hand side of the weighted average by $\mathop{Area}(A)$, you get
$$
\frac{b}{3(b-a)}h
= \frac{ \frac{3b - 2a}{3(b-a)} h
+ \frac{b^2 - a^2}{a^2} \cdot \bar y}{1 + \frac{b^2 - a^2}{a^2}}
$$
After you finish collecting all the terms in
$a$, $b$, and $h$ on the left side of this equation, and factor
$(b-a)^2$ out of $b^3 - 3ba + 2a^3$, it all simplifies to the formula
for $\bar y$ shown above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Having trouble solving $\int\frac{5x^2+3x+2}{x(x+1)^2}$ I've first transformed the integral to
$$\int\frac{5x^2+3x+2}{x(x^2+2x+1)}dx$$
Which gave me
$$\frac{A}{x}+\frac{Bx+C}{x^2+2x+1}$$
$$=\frac{A(x^2+2x+1)+Bx^2+Cx}{x(x^2+2x+1)}$$
$$\frac{5x^2+3x+2}{x(x^2+2x+1)}=\frac{(A+B)x^2+(2A+C)x+A)}{x(x^2+2x+1)}$$
So I've found the corresponding variables
$$A = 2$$
$$A+B = 5, B = 3$$
$$2A+C=3, C=-1$$
So the final integral is
$$2\int\frac{dx}{x}+3\int\frac{xdx}{x^2+2x+1}-\int\frac{dx}{x^2+2x+1}$$
$$=2ln(x) -ln(x^2+2x+1)+\frac{3}{x+1}+3ln(x+1)$$
However, the expected answer is
$$2ln(x)+3ln(x+1)+\frac{4}{x+1}$$
What is my error ?
| you can see that $$\frac{5x^2 + 3x+2}{x(x+1)^2} = \frac 2 x + \frac{B}{x+1} - \frac 4{(x+1)^2}$$ by looking at the behavior of the two sides near the singularities $x = 0, x = -1.$ to find $B,$ put $x = 1.$ you will find $B = 3.$
you can now integrate $$ \int \frac{5x^2 + 3x+2}{x(x+1)^2} \, dx = 2 \ln x + 3 \ln(x+1) + \frac 4{x+1} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Indefinite integral of $ \int \frac{x^3}{\sqrt{x^2+1}} \,\text{d}x$. Can you please provide any sort of hint or suggestion on how to find the following indefinite integral?
$$\int\frac{x^3}{\sqrt{x^2+1}}\text{d}x$$
I tried substituting everything but it didn't work. I also tried trigonometric substitution but I couldn't find any valid trigonometric identity for $ \sqrt{\cos^2(x)+1} $ or $ \sqrt{\sin^2(x)+1} $.
| $$
\begin{aligned}
\int \frac{x^{3}}{\sqrt{x^{2}+1}} d x &=\int x^{2} d\left(\sqrt{x^{2}+1}\right) \\
&=x^{2} \sqrt{x^{2}+1}-2 \int x \sqrt{x^{2}+1} d x \\
&=x^{2} \sqrt{x^{2}+1}-\frac{2}{3}\left(x^{2}+1\right)^{\frac{3}{2}}+C \\
&=\frac{\sqrt{x^{2}+1}}{3}\left[3 x^{2}-2\left(x^{2}+1\right)\right]+C \\
&=\frac{\sqrt{x^{2}+1}}{3}\left(x^{2}-2\right)+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Locus of a midpoint Let $Γ_1$ be a circle of radius $4$, and let $Γ_2$ be a circle of radius $14$. The distance between the centers of $Γ_1$ and $Γ_2$ is $25$. Let $A$ be a variable point on $Γ_1$, let $B$ be a variable point on $Γ_2$, and let $M$ be the midpoint of $AB$. Let $S$ be the set of all possible locations of $M$. Then find the area of $S$. I am getting $81\pi$. See the figure
With one of my friends, I got this.Let C1 be centered at $(0,0)$ and let C2 be $(25, 0)$. The points that will lie on the boundary of S are
1. Midpoint of (-4, 0) and (11, 0) i.e. (3.5, 0)
2. Midpoint of (4, 0) and (39, 0) i.e. (21.5, 0)
3. Midpoint of the tangents joining C1 and C2
Let the angle that the point of intersection of the tanget to circle C1 and C2 makes with the x axis be $\theta $ then
The point of intersection on C1 and C2 are
$$(x_1, y_1) = (4\cos{\theta}, 4\sin{\theta})$$
$$(x_2, y_2) = (25+14\cos{\theta}, 14\sin{\theta})$$
The equation of the tangent is
$$ y = -\frac{x}{\tan{\theta}} + c$$
Putting the above two points in the line equation and eliminating c gives
$$ 10\sin{\theta} = -\frac{25 + 10\cos{\theta}}{\tan{\theta}}$$
$$ 10\sin{\theta}\tan{\theta} = -25 - 10\cos{\theta}
\frac{2}{\cos{\theta}} = -5
\cos{\theta} = -\frac{2}{5}$$
This gives
$$ \sin{\theta} = \pm \frac{\sqrt{21}}{5}$$
The two pairs of points of intersection in C1 and C2 are
$$ (x_1, y_1) = (-\frac{8}{5}, \frac{4\sqrt{21}}{5})$$
$$ (x_2, y_2) = (\frac{97}{5}, \frac{14\sqrt{21}}{5})$$
and
$$(x_1, y_1) = (-\frac{8}{5}, -\frac{4\sqrt{21}}{5})$$
$$ (x_2, y_2) = (\frac{97}{5}, -\frac{14\sqrt{21}}{5})$$
This gives the other two mid points as $(\frac{89}{10}, \frac{9\sqrt{21}}{5})$, $(\frac{89}{10}, -\frac{9\sqrt{21}}{5})$
Using all the mid points obtained and putting them into the ellipse equation
$$ \frac{(x - x_1)^2}{a^2} + \frac{(y - y_1)^2}{b^2}$$
following are obtained
$$ x_1 = \frac{25}{2}$$ and $$ y_1 = 0$$
$ a = 9$ and $b=9$.
Where is it wrong.($81\pi$ is wrong!) Please help, thanks.
| We can write
$$A = A_0 + 2 a (\cos\theta, \sin\theta) \qquad B = B_0 + 2 b (\cos\phi, \sin\phi)$$
where $A_0$ and $B_0$ are the centers of the circles, and $2a$ and $2b$ are the radii. Then
$$M = C_0 + (a\cos\theta + b \cos\phi, a\sin\theta+b\sin\phi)$$
where $C_0 = \frac12(A_0+B_0)$. Note that
$$\begin{align}
|\overline{MC_0}|^2 &= (a\cos\theta+b\cos\phi)^2+(a\sin\theta+b\sin\phi)^2 \\[4pt]
&= a^2 + b^2 + 2 a b \cos(\theta-\phi)
\end{align}$$
so that
$$| a - b | \;\leq\; |\overline{MC_0}| \;\leq\; a+b$$
Therefore, the locus of $M$ is definitely confined to the annulus with outer radius $a+b$ and inner radius $|a-b|$. Once you show that $M$ accounts for all points in that annulus, then the area is
$$\pi (a+b)^2 - \pi(a-b)^2 = 4 a b \pi = \pi\cdot(\text{product of radii})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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$ \sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6} $ then $ \sum _{n=1}^{\infty} \frac 1 {(2n -1)^2} $ If $ \sum _{n=1}^{\infty} \frac 1 {n^2} =\frac {\pi ^2}{6}
$ then $ \sum _{n=1}^{\infty} \frac 1 {(2n -1)^2} $
Dont know what kind of series is this. Please educate.
How to do such problems?
| $$\sum_{n=1}^\infty \frac1{n^2} =\sum_{n=1}^\infty \frac1{(2n)^2} + \sum_{n=1}^\infty \frac1{(2n-1)^2}=\frac14 \sum_{n=1}^\infty \frac1{n^2} + \sum_{n=1}^\infty \frac1{(2n-1)^2}$$
So, $\sum\limits_{n=1}^\infty \frac1{(2n-1)^2}= \frac34 \sum\limits_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there another way to solve this integral? My way to solve this integral. I wonder is there another way to solve it as it's very long for me.
$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$
Let $$u=\tan (\frac{x}{2})$$
$$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$
By Weierstrass Substitution
$$\sin (x)=\frac{2u}{u^2+1}$$
$$\cos (x)=\frac{1-u^2}{u^2+1}$$
$$dx=\frac{2du}{u^2+1}$$
$$=\int_{0}^{\infty }\frac{2(1-\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+1)}du$$
$$=\int_{0}^{\infty }\frac{2(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$
$$=2\int_{0}^{\infty }\frac{(u-1)^2}{(u+1)^2(u^2+1)}du $$
$$=2\int_{0}^{\infty }(\frac{2}{(u+1)^2}-\frac{1}{u^2+1})du $$
$$=-2\int_{0}^{\infty }\frac{1}{u^2+1}du+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du $$
$$\lim_{b\rightarrow \infty }\left | (-2\tan^{-1}(u)) \right |_{0}^{b}+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=(\lim_{b\rightarrow \infty}-2\tan^{-1}(b))+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
$$=-\pi+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$
Let $$s=u+1$$
$$ds=du$$
$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$
$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$
$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$
$$=4-\pi$$
$$\approx 0.85841$$
| The question asked "is there another way", so here's another one. Start by removing the trig from the numerator:
$$ \int_0^{\pi} \frac{1-\sin{x}}{1+\sin{x}} \, \mathrm{d}x =
\int_0^{\pi} \frac{2 - (1 +\sin{x})}{1+\sin{x}} \, \mathrm{d}x =
\int_0^{\pi} \left( \frac{2}{1+\sin{x}} - 1 \right) \, \mathrm{d}x =
-\pi + \int_0^{\pi} \frac{2 \, \mathrm{d}x}{1+\sin{x}} $$
There are many ways to solve this but I would like to demonstrate the substitution $u = 1 + \sin x$. It would be much more convenient if the limits ran from $0$ to $\frac{\pi}{2}$, so that $x = \sin^{-1} (u-1)$ is a continuous, monotonic function from $[1,2]$ to $[0,\frac{\pi}{2}]$, and also because $\cos x$ and $\sin x$ are both positive for $0 \leq x \leq \frac{\pi}{2}$. The latter property is often useful so we can write $\cos x=\sqrt{1-\sin^2 x}$ and $\sin x=\sqrt{1-\cos^2 x}$. Fortunately symmetry considerations allow a change of limits:
$$\int_0^{\pi} \frac{2 \, \mathrm{d}x}{1+\sin{x}} =
\int_0^{\pi/2} \frac{4 \, \mathrm{d}x}{1+\sin{x}}$$
We have $\frac{\mathrm{d}u}{\mathrm{d}x} = \cos x = \sqrt{1-\sin^2 x} = \sqrt{1-(u-1)^2} = \sqrt{2u - u^2}$ so the integral becomes:
$$\int_0^{\pi/2} \frac{4 \, \mathrm{d}x}{1+\sin{x}} =
\int_1^2 \frac{4 \, \mathrm{d}u}{u \sqrt{2u - u^2}} =
\int_1^2 \frac{4 \, \mathrm{d}u}{u \sqrt{-u(u-2)}}$$
This might look fearsome but is actually very amenable to the third Euler substitution, since we have a factorised quadratic expression inside the square root. The computation is very similar to this answer. In general the substitution is $\sqrt{au^2 + bu + c} = \sqrt{a(u-\alpha)(u-\beta)} = (u-\alpha)t$ which gives $u = \frac{a\beta-\alpha t^2}{a-t^2}$; in our case we can take $a=-1$, $b=2$, $c=0$, $\alpha=0$, and $\beta=2$ with $\sqrt{-u(u-2)} = ut$.
Since $u=\frac{(-1)(2)-0t^2}{-1-t^2}=2(t^2+1)^{-1}$ we obtain $\frac{\mathrm{d}u}{\mathrm{d}t} = -4t(t^2+1)^{-2}$ and to change the limits we set $t=\frac{\sqrt{-u(u-2)}}{u} = \sqrt{\frac{2-u}{u}}$. The remaining integral becomes:
$$\int_1^2 \frac{4 \, \mathrm{d}u}{u \sqrt{-u(u-2)}} =
\int_1^0 \frac{4 (-4t)(t^2+1)^{-2} \, \mathrm{d}t}{u^2 t} =
\int_0^1 \frac{4 (4t)(t^2+1)^{-2} \, \mathrm{d}t}{4(t^2+1)^{-2} t} =
\int_0^1 4 \, \mathrm{d}t =
4$$
This was not the most straightforward way to find the result $4 - \pi$, but I just wanted to draw out the similarity between this integral and one the original poster had asked about before (the main differences being the trig in the numerator - which is easily removed - and the limits).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$ I recently ran into this series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$
Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.
I had given the following solution:
$$\begin{aligned}
1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\
&\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\
&=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\
&= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\
&=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4} \right ] \\
&=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32}
\end{aligned}$$
where I used polygamma identities and made use of the absolute convergence of the series at $(*)$ in order to re-arrange the terms.
Any other approach using Fourier Series, or contour integration around a square, if that is possible?
| Method by Fourier Series
Consider the function $f(x) = x(1 - x)$, $0 \le x \le 1$. It has Fourier sine series expansion
$$f(x) = \frac{8}{\pi^3}\sum_{n = 1}^\infty \frac{1}{(2n-1)^3}\sin{(2n-1)\pi x}.$$
Setting $x = \frac{1}{2}$ results in
$$\frac{1}{4} = \frac{8}{\pi^3}\sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3},$$
or
$$\frac{\pi^3}{32} = \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}.$$
By reindexing the sum we can write
$$\frac{\pi^3}{32} = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$
Method by Contour Integration
Let $g(z) = \frac{1}{(2z - 1)^3}$. Then $g$ has only one pole of order $3$ at $z = \frac{1}{2}$. Let $N$ be a positive integer, and consider the contour integral
$$\frac{1}{2\pi i}\int_{\Gamma_N} \pi\csc \pi z\, g(z)\, dz,$$
where $\Gamma_N$ is a positively oriented square with vertices at $\left(N + \frac{1}{2}\right)(\pm 1 \pm i)$. The residue theorem gives
\begin{align}\frac{1}{2\pi i}\int_{\Gamma_N} \pi \csc \pi z\, g(z)\, dz &= \sum_{n = -N}^N \operatorname{Res}\limits_{z = n} \pi \csc \pi z\, g(z) + \operatorname{Res}\limits_{z = \frac{1}{2}} \pi \csc \pi z\, g(z)\\
&= \sum_{n = -N}^N (-1)^n g(n) + \frac{\pi^3}{16}.
\end{align}
For $|z| \ge 1$, $|g(z)| \le |z|^{-3}$. Thus, $$\frac{1}{2\pi i}\int_{\Gamma_N} \pi \csc \pi z\, g(z)\, dz \to 0 \quad \text{as} \quad N \to \infty.$$
Hence
$$0 = \sum_{n = -\infty}^\infty (-1)^n g(n) + \frac{\pi^3}{16}$$
that is,
$$\frac{\pi^3}{16} = \sum_{n = -\infty}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}.$$
Now
\begin{align}\sum_{n = -\infty}^\infty \frac{(-1)^{n-1}}{(2n-1)^3} &= \sum_{n = -\infty}^0 \frac{(-1)^{n-1}}{(2n-1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\
& = \sum_{n = 0}^\infty \frac{(-1)^{n}}{(2n+1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\
& = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.
\end{align}
Thus
$$\frac{\pi^3}{16} = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$
Finally, we have
$$\frac{\pi^3}{32} = \sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
A geometric proof about decagons. Let us have a regular decagon. Prove that the radius of the circle that can be drawn around it (if I have drawn correctly, that's the blue line) equals $|ad|-|ab|$.
How can I prove this statement, any ideas? :)
| WLOG you can take the radius to be unity. Using the Cosine Law we can easily say that $AD=2\sin\dfrac{3\pi}{10}$ and $AB=2\sin\dfrac{\pi}{10}$ . Then the difference $AD-AB:$
$$2\left(\sin\dfrac{3\pi}{10}-\sin\dfrac{\pi}{10}\right)=4\cos\dfrac{\pi}{5}\sin\dfrac{\pi}{10}={4\cos\dfrac{\pi}{5}\sin\dfrac{\pi}{10}\cos\dfrac{\pi}{10}\over\cos\dfrac{\pi}{10}}$$
$$={2\cos\dfrac{\pi}{5}\sin\dfrac{\pi}{5}\over\cos\dfrac{\pi}{10}}={\sin\dfrac{2\pi}{5}\over \cos\dfrac{\pi}{10}}=1$$
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Finf $f(x)$ which is a second degree polynomial, such that $f(1)=0$ and $f(x) = f(x-1)$ I must find a function $f(x) = ax^2+bx+c$ such that:
$$f(1) = a+b+c=0\\f(x)=f(x-1)\implies ax^2+bx+c = a(x-1)^2+b(x-1)+c\implies\\ax^2+bx+c = ax^2+(-2a+b)x+a-b+c\implies\\a = a, b = -2a+b, c = a-b+c$$
but this results fo $a=b=c=0$. What am I doing wrong?
| If $f$ is a polynomial and $f(1)=0$ together with $f(x)=f(x-1)$, then $$0=f(1)=f(2)=f(3)=f(4)=\ldots $$
and $f$ has an infinite number of real roots. That implies $f\equiv 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Finding the limit $\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$ I am trying to find this limit
$$\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$$
My so far method is this
*
*$f(x)>0.$
*$f^{\prime}(x)=\frac{1}{3\sqrt[3]{(x+1)^2}}-\frac{1}{3\sqrt[3]{x^2}}<0.$
*For every $0<y<1$ the equation $f(x)=y$ has a unique solution(found in maple):
$$x=\frac{1}{3}\cdot \frac{\frac{1}{3}y^2(3y^2+\sqrt{-3y^4+12y})+\frac{1}{6}\frac{3y^2+\sqrt{-3y^4+12y}}{y}-y^4-2y}{y}$$
The previous facts implies that the limit is $0$.
I am wondering if there is any way easier than this to find the limite.
thanks in advance.
| use the binomial theorem to expand $$(x+1)^{1/3} = x^{1/3} + \frac 13 x^{-2/3}+\cdots $$ so $$ (x+1)^{1/3} - x^{1/3} = \frac 13 x^{-2/3}+\cdots \rightarrow 0\text{ as } x \to \infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Solve this system of equation Solve this system of equations for real $x$ and $y$:
*
*$5x\left(1+\dfrac{1}{x^2+y^2}\right)=12$
*$5y\left(1-\dfrac{1}{x^2+y^2}\right)= 4$
I juggled with those equations and got $x-y+\dfrac{x+y}{x^2+y^2}=\dfrac{8}{5}$, from where I guessed a solution $(2,1)$.
But I don't know how to approach mathematically.
Please help me.
| \begin{align}
5x\left(1+\frac{1}{x^2+y^2}\right)&=12\\
5y\left(1-\frac{1}{x^2+y^2}\right)&=4\\
\end{align}
Obviously, $x\neq0$ and $y\neq0$.
\begin{align}
\left(1+\frac{1}{x^2+y^2}\right)&=\frac{12}{5x}\\
\left(1-\frac{1}{x^2+y^2}\right)&=\frac{4}{5y}\\
1&=\frac{6}{5x}+\frac{2}{5y}&=\frac{6y+2x}{5xy}\\
\frac{1}{x^2+y^2}&=\frac{6}{5x}-\frac{2}{5y}&=\frac{6y-2x}{5xy}\\
x^2+y^2&=\frac{6y+2x}{6y-2x}\\
5xy&=6y+2x&=(x^2+y^2)(6y-2x)\\
\end{align}
Now, consider $x=y.\alpha$. Remember that $y\neq 0$
\begin{align}
y(5\alpha) &=6+2\alpha&=y^2(1+\alpha^2)(6-2\alpha)
\end{align}
$$y=\frac{6+2\alpha}{5\alpha}$$
$$25\alpha^2=(1+\alpha^2)(36-4\alpha^2)$$
$$4\alpha^4-7\alpha^2-36=0$$
$$\alpha=\pm 2$$ (There are two other imaginary roots for $\alpha$ that gives two more solutions in $\mathbb C$)
Hence two solutions : $(x,y)=(2,1)$ or $(x,y)=(\frac{2}{5},-\frac{1}{5})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that the algebraic expressions are equivalent.
$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+2}-1}{2}$$
with steps make left hand side = right hand side by modifying one or both expressions
Thanks for your help guys, I solved it like this:
$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+1}-1}{2} + \frac{2*3^{k+1}}{2} = \frac{3^{k+2}-1}{2} $$
| Hint. $$
2\cdot 3^{k+1}=(3-1)\cdot 3^{k+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Proof by Induction Problem I need to prove that for natural numbers $a$, and positive integers $n$, the number $a^{2n+1}-a$ is divisible by $6$.
I have proved the case when $n=1$, that $a^3-a$ is divisible by $6$. I'm having trouble proving that if $a^{2n+1}-a$ is divisible by $6$, then $a^{2(n+1)+1}-a$ is also divisible by $6$.
| hint: $a^{2n+1} - a = a(a^{2n}-1) = a(a^2-1)(a^{2n-2} + a^{2n-4} + \cdots + 1) = a(a-1)(a+1)(.....)$ is divisible by $6$ since it has a product of $3$ consecutive integers: $a-1,a,a+1$.
To do an induction proof.
The base case is done. Assume $6\mid a^{2n+1} - a \Rightarrow a^{2n+3} - a = (a^{2n+3} - a^{2n+1}) + (a^{2n+1} - a) = a^{2n}\cdot a\cdot (a-1)(a+1) + (a^{2n+1}-a)$ is divisible by $6$ since the first term has again a product of $3$ consecutive integers, and the expression in the parentheses is divisible by $6$ by inductive step so the sum is divisible by $6$, complete the induction proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
calculating an integral $\int\frac{x \ln x}{(1+x^{2})^{2}}dx $ I tried to do subtition and to integrate by parts, unsuccessfully.
$\int\frac{x \ln x}{(1+x^{2})^{2}}dx
$
Thanks in advanced.
| $$\int \frac{x}{(1+x^2)^2}=\frac{1}{2}\int\frac{2x}{(1+x^2)^2}=\frac{1}{2}\int\frac{1}{(1+y)^2}=-\frac{1}{2}\frac{1}{1+y}=-\frac{1}{2}\frac{1}{1+x^2},$$ with $y=x^2$, was easy to integrate and its integral is rational.
On the other hand, $(\ln(x))'=\frac{1}{x}$ is rational.
Therefore, integration by parts rationalizes the integrand.
$$\int\frac{x}{(1+x^2)^2}\ln(x)=-\frac{1}{1+x^2}\ln(x)+\frac{1}{2}\int\frac{1}{1+x^2}\frac{1}{x}$$
The last integral you can do by expanding in simple fractions and all of that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this limits with $\sin{(\tan{x})}-\tan{(\sin{x})}$ How Find limit
$$\lim_{x\to 0}\dfrac{\sin{(\tan{(\sin{(\tan{x})})})}-\tan{(\sin{(\tan{(\sin{x})})})}}
{\sin{(\tan{x})}-\tan{(\sin{x})}}$$
My approach is the following: I use wolframalpha found this limits is $2$
| For this kind of problems, Taylor series are very useful. Just start from the series of the most inner terms, cascade (replacing and simplifying) and be patient !$$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+\frac{17 x^7}{315}+O\left(x^9\right)$$ $$\sin(\tan(x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55 x^7}{1008}+O\left(x^9\right)$$ $$\tan(\sin(\tan(x)))=x+\frac{x^3}{2}+\frac{11 x^5}{40}+\frac{571 x^7}{5040}+O\left(x^9\right)$$ $$\sin(\tan(\sin(\tan(x))))=x+\frac{x^3}{3}+\frac{x^5}{30}-\frac{9 x^7}{70}+O\left(x^9\right)$$ Similarly $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^9\right)$$ $$\tan(\sin(x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+O\left(x^9\right)$$ $$\sin(\tan(\sin(x)))=x-\frac{x^5}{10}-\frac{x^7}{63}+O\left(x^9\right)$$ $$\tan(\sin(\tan(\sin(x))))=x+\frac{x^3}{3}+\frac{x^5}{30}-\frac{13 x^7}{210}+O\left(x^9\right)$$ So the numerator is $$\sin(\tan(\sin(\tan(x))))-\tan(\sin(\tan(\sin(x))))=-\frac{x^7}{15}+O\left(x^9\right)$$ and the denominator is $$\sin(\tan(x))-\tan(\sin(x))=-\frac{x^7}{30}+O\left(x^9\right)$$ from which $$\lim_{x\to 0}\dfrac{\sin{(\tan{(\sin{(\tan{x})})})}-\tan{(\sin{(\tan{(\sin{x})})})}}
{\sin{(\tan{x})}-\tan{(\sin{x})}}=2$$ Being much more patient and using an extra term for the expansions, you could show that
$$\dfrac{\sin{(\tan{(\sin{(\tan{x})})})}-\tan{(\sin{(\tan{(\sin{x})})})}}
{\sin{(\tan{x})}-\tan{(\sin{x})}}=2+\frac{5 x^2}{3}+O\left(x^3\right)$$ which shows the limit and how it is approached.
Edit
I forgot to precise that it is a good parctice to start investigating the simplest term; here, it is the denominator. Its expansion tells basically the degree of what you should do with the numerator.
Since the mechanism is in place, you could show for the next level
$$\frac{\sin (\tan (\sin (\tan (\sin (\tan (x))))))-\tan (\sin (\tan (\sin (\tan (\sin
(x))))))}{\sin (\tan (\sin (\tan (x))))-\tan (\sin (\tan (\sin (x))))}=\frac{3}{2}+\frac{5 x^2}{4}+O\left(x^3\right)$$
Added later
If we define $S(x)=\sin(\tan(x))$ and $T(x)=\tan(\sin(x))$, the expression given in the post is $$A_1=\frac{S(S(x))-T(T(x))}{S(x)-T(x)}$$ and the last given is $$A_2=\frac{S(S(S(x)))-T(T(T(x)))}{S(S(x))-T(T(x))}$$ What is interesting is that the asymptotic development of $A_n$ is simply given by $$A_n=\frac{n+1}{n}\big(1+\frac 56 x^2\big)$$ Amazing, isn't it ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Application of Implicit Function Theorem for Constrained Optimization Here's the problem:
Consider the subset $S \subset \mathbb R$ defined by
$$ x^4+2xy+y^4+yz+z^3 = 2 $$
Show that there exists a $C^1$ function $g: \mathbb R^2 \to \mathbb R$ defined near $(1,1)$ such that for $(x,y,z)$ near $(1,1,-1)$, then $(x,y,z) \in S $ if and only if $z=g(x,y)$. Compute $\nabla g $ near $(1,1)$.
Here's my attempt at a solution:
I set the equation equal to $0$ and computed $-\frac{F_x}{F_z}$ and got
$$\left. \frac{-4x^3+2y}{y+3z^2}\right|_{(1,1,-1)}=-\frac{1}{2}$$
So, can I say that, because this does not equal $0$, such a function exists? If so, how can I find that function so I can compute its gradient?
| From the equation:
$${x^4} + 2xy + {y^4} + yz + {z^3} = 2$$
We define a function:
$$F(x,y,z)= {x^4} + 2xy + {y^4} + yz + {z^3} - 2$$
Equation reads now:
$$F(x,y,z) = 0$$
Because of
$$dF(x,y,z) = (4{x^3} + 2y)dx + (2x + 4{y^3} + z)dy + (y + 3{z^2})dz$$
we have:
$$\nabla F(x,y,z) = \left( {\begin{array}{*{20}{c}}
{4{x^3} + 2y} \\
{2x + 4{y^3} + z} \\
{y + 3{z^2}}
\end{array}} \right)$$
So $$\frac{{\partial F}}{{\partial z}}(x,y,z) = y + 3{z^2}$$
and
$$\frac{{\partial F}}{{\partial z}}(1,1, - 1) = 1 + 3 = 4 \ne 0$$
Because of:
$$\frac{{\partial F}}{{\partial z}}(1,1, - 1) \ne 0$$
near $$(1,1, - 1)$$ exists a function $$z = g(x,y)$$ with
$$F(x,y,g(x,y)) \equiv 0$$
near $$(1,1)$$ and
$$\begin{gathered}
(x,y,z) = (x,y,g(x,y)) \hfill \\
(1,1, - 1) = (1,1,g(1,1)) \hfill \\
g(1,1) = - 1 \hfill \\
\end{gathered} $$
Setting
$$(4{x^3} + 2y)dx + (2x + 4{y^3} + z)dy + (y + 3{z^2})dz = 0$$
it follows:
$$dz(x,y) = - \frac{{4{x^3} + 2y}}{{y + 3{z^2}}}dx - \frac{{2x + 4{y^3} + z}}{{y + 3{z^2}}}dy$$
with:
$$\begin{gathered}
dz(1,1) = - \frac{6}{{1 + 3g{{(1,1)}^2}}}dx - \frac{{2x + 4{y^3} + g(1,1)}}{{1 + 3g{{(1,1)}^2}}}dy \hfill \\
dz(1,1) = - \frac{3}{2}dx - \frac{5}{4}dy \hfill \\
\end{gathered} $$
That means:
$$\nabla g(1,1) = - \left( {\begin{array}{*{20}{c}}
{\frac{3}{2}} \\
{\frac{5}{4}}
\end{array}} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve for $x$ in $2(x-5) + 4 (x-3) = -30$ In $2(x-5) + 4 (x-3) = -30$... I'm very confused as to how to solve for $x$, the correct response is $-2$, but I keep getting $4/3$.
| Assuming you transcribed the problem correctly, your answer is correct.
\begin{align*}
2(x - 5) + 4(x - 3) & = -30\\
x - 5 + 2(x - 3) & = -15 && \text{divide both sides of the equation by $2$}\\
x - 5 + 2x - 6 & = -15 && \text{distribute}\\
3x - 11 & = -15 && \text{combine like terms}\\
3x & = -4 && \text{add $11$ to each side of the equation}\\
x & = -\frac{4}{3} && \text{divide each side of the equation by $3$}
\end{align*}
Check: We can verify our answer by direct substitution. If $x = -\dfrac{4}{3}$, then
\begin{align*}
2(x - 5) + 4(x - 3) & = 2\left(-\frac{4}{3} - 5\right) + 4\left(-\frac{4}{3} - 3\right)\\
& = 2\left(-\frac{4}{3} - \frac{15}{3}\right) + 4\left(-\frac{4}{3} - \frac{9}{3}\right)\\
& = 2\left(-\frac{19}{3}\right) + 4\left(-\frac{13}{3}\right)\\
& = -\frac{38}{3} - \frac{52}{3}\\
& = -\frac{90}{3}\\
& = -30
\end{align*}
If you substitute $-2$ for $x$ in the expression $2(x - 5) + 4(x - 3)$, you should obtain $-34$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1208440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Closed form for an almost-elliptic integral Does
$$\int_0^{2 \pi} \log\left(\frac{1}{2}[1+\sqrt{1-(a \sin\phi)^2}]\right) d\phi $$ have a closed form ?
An approximation for small $a$ is $2E-\pi$, but it is the exact form that is needed for any $|a|<1$.
The integration standing to Jack is
$$ -\frac{\pi a^2}{4}\phantom{}_4 F_3\left(1,1,\frac{3}{2},\frac{3}{2};2,2,2;a^2\right) $$
The question now is: Can it be changed into a finite combination of elementary functions.
| For any $b\in(0,1)$ we have:
$$I(b)=\int_{0}^{2\pi}\log\left(1+\sqrt{1-b^2\sin^2 x}\right)\,dx=4\int_{0}^{1}\frac{\log\left(1+\sqrt{1-b^2 t^2}\right)}{\sqrt{1-t^2}}\,dt$$
hence:
$$ I(b)=\frac{1}{b}\int_{0}^{b}\frac{\log(1+\sqrt{1-u^2})}{\sqrt{1-\frac{u^2}{b^2}}}\,du=\frac{1}{b}\int_{0}^{\arcsin b}\frac{\cos\theta}{\sqrt{1-\frac{\sin^2\theta}{b^2}}}\,\log(1+\cos\theta)\,d\theta$$
and the last integral can be evaluated by exploiting the Fourier series of $\log(1+\cos\theta)$ that is pretty well-known. Another possible approach is differentiation under the integral sign.
We have:
$$ I'(b) = -4\int_{0}^{\pi/2}\frac{b\sin^2 t}{\sqrt{1-b^2\sin^2 t}\left(1+\sqrt{1-b^2\sin^2 t}\right)}\,dt = \frac{2\pi-4K(b^2)}{b}\tag{1}$$
where $K$ is the complete elliptic integral of the first kind. Since $I(0)=2\pi\log 2$, it follows that:
$$ I(b)=2\pi\log 2-\frac{\pi b^2}{4}\phantom{}_4 F_3\left(1,1,\frac{3}{2},\frac{3}{2};2,2,2;b^2\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability that minimum of two numbers is less than 4 Suppose I have to choose two numbers from set $$S=\{1,2,3,4,5,6 \}$$ without a replacement , then what is the probability that minimum of two is less than $4$?
I made two groups for this problem $A= \{1,2,3 \}$ and $B=\{4,5,6 \}$
.There are two possibilities , either both are from $A$ or one from $A$ and one from $B$ to satisfy our requirement. Hence
$$P(E)=\frac{3}{6}.\frac{2}{5} + \frac{3}{6}.\frac{3}{5}$$
but answer is incorrect. Help?
| It is easy as P.I.E.. Use the Principle of Inclusion and Exclusion.
$$\begin{align}
\mathsf P(\min(X,Y) < 4) & = \mathsf P(X < 4)+\mathsf P(Y < 4)- \mathsf P(X< 4 \cap Y< 4)
\\ &= \frac 3 6 + \frac 3 6 - \frac 3 6\cdot\frac 2 5
\\ & = \frac 4 5
\end{align}$$
Or you can use the Law of Complements
$$\begin{align}
\mathsf P(\min(X,Y) < 4) & = 1- \mathsf P(X \geq 4 \cap Y\geq 4)
\\ &= 1 - \frac 3 6\cdot\frac 2 5
\\ & = \frac 4 5
\end{align}$$
Or by your method (the Law of Total Probability), but remembering that there are two ways for one pick to be in A and the other in B.
$$\begin{align}
\mathsf P(\min(X,Y) < 4) & = \mathsf P(X < 4 \cap Y\geq 4)+\mathsf P(X \geq 4 \cap Y< 4) + \mathsf P(X< 4\cap Y< 4)
\\ & = \frac 3 6\cdot\frac 3 5 +\frac 3 6\cdot\frac 3 5 +\frac 3 6\cdot\frac 2 5
\\ & = \frac 4 5
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Eritrea's Theorem According to this newspaper, an Eritrean high school student named Saied Mohammed Ali has discovered a new geometric theorem. Another source seems to say that it's the following:
Say you have a triangle, with sides of length $a$, $b$, and $c$. Draw the medians (lines $\overline{AG}$, $\overline{BI}$, and $\overline{CH}$ in the diagram), and the altitudes (lines $\overline{AD}$, $\overline{BF}$, and $\overline{CE}$ in the digram).
Call the distance between where the median and altitude hit a given side the sma of that side. In the diagram, the smas are $\overline{GD}$, $\overline{IF}$, and $\overline{HE}$.
Call the length of the sma on side $a$, $\alpha$. Similarly, on sides $b$ and $c$ we have smas $\beta$ and $\gamma$.
The theorem is:
$$a\alpha+b\beta=c\gamma$$
In the picture, we have $5.19\times0.09+4.28\times0.9=4.39\times0.98$, which is true up to rounding error.
How would you prove this? I have almost no experience in geometry, so I wouldn't even know where to start on this. Thanks!
| Express α, β, γ in terms of a, b, c
According to the Encyclopedia of Triangle Centers, the orthocenter $X(4)$ has barycentric coordinates $[\tan A:\tan B:\tan C]$. From the cosine law you have $\cos C=\frac{a^2+b^2-c^2}{2ab}$ and likewise for the other angles. So you get
\begin{align*}
\tan C&=\frac{\sin C}{\cos C}=\frac{\sqrt{1-\cos C^2}}{\cos C}
=\frac{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}{a^2+b^2-c^2}
\\&=\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{a^2+b^2-c^2}
\\&=\frac{4V}{a^2+b^2-c^2}
\end{align*}
where $V$ denotes the area of the triangle, as obtained from Heron's formula. By canceling the $4V$ term and multiplying with all the denominators (which is allowed for homogeneous coordinates), you might write the barycentric coordinates of $X(4)$ also as
$$\begin{bmatrix}
(a^2-b^2+c^2)(a^2+b^2-c^2)\\
(b^2-c^2+a^2)(b^2+c^2-a^2)\\
(c^2-a^2+b^2)(c^2+a^2-b^2)
\end{bmatrix}$$
You get the footpoints of the heights by setting one of these coordinates to zero. So for example $D$ has barycentric coordinates
$$\begin{bmatrix}
0\\
(b^2-c^2+a^2)(b^2+c^2-a^2)\\
(c^2-a^2+b^2)(c^2+a^2-b^2)
\end{bmatrix}\sim\begin{bmatrix}
0\\
b^2-c^2+a^2\\
c^2+a^2-b^2
\end{bmatrix}$$
which means you can write its Euclidean coordinates as
\begin{align*}
D&=\frac{b^2-c^2+a^2}{(b^2-c^2+a^2)+(c^2+a^2-b^2)}B
+\frac{c^2+a^2-b^2}{(b^2-c^2+a^2)+(c^2+a^2-b^2)}C
\\&=\frac{a^2+b^2-c^2}{2a^2}B + \frac{a^2-b^2+c^2}{2a^2}C
\\&=B + \frac{a^2-b^2+c^2}{2a^2}(C-B)
\\&=B + \left(\frac12 + \frac{-b^2+c^2}{2a^2}\right)(C-B)
\\&=G + \frac{-b^2+c^2}{2a^2}(C-B)
\end{align*}
The distance between $D$ and $G$ is
$$\alpha=\lVert D-G\rVert
= \left\lvert\frac{-b^2+c^2}{2a^2}\right\rvert\cdot\lVert C-B\rVert
= \frac{\lvert b^2-c^2\rvert}{2a}\\
2a\alpha=\lvert b^2-c^2\rvert$$
Orientation and order
Up to here, the above is an alternative to the shorter deduction timon92 posted in his answer. The discussion which follows below applies no matter how one obtains that formula for $2a\alpha$.
Using this formula, the equation of the theorem (multiplied by $2$ to simplify things) would be
$$\lvert b^2-c^2\rvert + \lvert c^2-a^2\rvert = \lvert a^2-b^2\rvert$$
This is not always the case. But if you use signed distances, e.g. always measured in counter-clockwise direction, you can omit the absolute values. Then write the formula as “sum equals zero” as columbus8myhw suggests in his comment, and you obtain
$$(b^2-c^2) + (c^2-a^2) + (a^2-b^2) = 0$$
which is obviously true.
If you prefer unsigned distances, when does the equation with those hold? It holds if and only if the difference inside the absolute value function has equal sign for both terms on the left hand side of the equation but opposite sign on the right. So you have two cases to consider:
\begin{gather*}
b^2-c^2\ge0,\quad c^2-a^2\ge0,\quad a^2-b^2\le0
\quad\implies\quad b\ge c\ge a \\
b^2-c^2\le0,\quad c^2-a^2\le0,\quad a^2-b^2\ge0
\quad\implies\quad a\ge c\ge b
\end{gather*}
That's what the “middle side” g.kov quoted in his comment refers to: the $c$ on the right hand side of the equation must be the side of median length.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Reverse engineering a Taylor expansion 2 So there is the sum:
$$S(x) = \frac{x^3}{3(1!)} + \frac{x^6}{6(2!)} + \frac{x^9}{9(3!)} \text{ }...$$
and we are instructed to find the sum of the series in a small expression.
I took the derivative to cancel out the increasing terms on the bottom:
$$S'(x) = \frac{x^2}{(1!)} + \frac{x^5}{(2!)} + \frac{x^8}{(3!)} \text{ }...$$
And then I multiplied by X on both sides, and replaced X with X^(1/3) and added 1 to both sides:
$$xS'(x^{1/3})+1 = 1+ \frac{x}{(1!)} + \frac{x^2}{(2!)} + \frac{x^3}{(3!)} \text{ }...$$
which is
$$xS'(x^{1/3})+1 = e^x $$
But when working my way back, I get
$$\int{S'(x^{1/3})} = \int{\frac{e^x-1}{x}} $$
which is impossible to integrate. Is there any way to work around this?
| You properly identified that $$S'(x) = \frac{x^2}{1!} + \frac{x^5}{2!} + \frac{x^8}{3!}+ \cdots=\frac 1x\Big(\frac{x^3}{1!} + \frac{x^6}{2!} + \frac{x^9}{3!}+ \cdots\Big)=\frac{e^{x^3}-1}{x}$$ So, this write $$\frac{dS}{dx}=\frac{e^{x^3}-1}{x}$$ Now, changing variable $x=\sqrt[3]y$, this leads to $$\frac{dS}{dy}=\frac{e^y-1}{3 y}$$ and integration leads to $$S=\frac{1}{3} (\text{Ei}(y)-\log (y))+C$$ where appears the exponential integral $\text{Ei}$ which cannot be expressed in terms of any elementary functions and I suppose that this is the end of the story and $$S=\frac{1}{3}\text{Ei}\left(x^3\right)-\log (x)+C$$ The exponential integral can be expressed as incomplete gamma function but I do not think that this could help you in any manner.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to differentiate $y=(x+1)^3/x^{3/2}$ and $y=2x^4/(b^2-x^2)$ I need to solve a list of derivatives to help me on an exam; however, I'm in doubt when they use another variable (constant) or when I have a fraction with functions that use the power rule.
For example:
$$y = \frac{(x + 1)^3}{x^{3/2}}.$$
This one I tried having $u = x + 1$, deriving it and having $v = x^{3/2}$, but I couldn't get the right answer.
And:
$$y = \frac{2x^4}{b^2 - x^2}.$$
And this one I tried to separate the constant, like $y = \frac{1}{b^2} \frac{dx}{dy} \frac{2x^4}{-x^2}$, but also coudn't find the answer.
Can you give a tip on how to find which rule to apply to these kind of derivatives?
| You really only need the quotient rule and the power rule.
For the quotient rule, when $y=\frac{f(x)}{g(x)}$ (where $g(x)\neq 0$), we have $y'=\frac{f'g-fg'}{g^2}$.
Problem 1: Differentiate $y = \frac{(x + 1)^3}{x^{3/2}}$ with respect to $x$.
Solution. We have $y=\frac{f(x)}{g(x)}$ where $f(x)=(x+1)^3$ and $g(x)=x^{3/2}$. Thus, we also have $f'(x)=3(x+1)^2$ and $g'(x)=\frac{3}{2}x^{1/2}$. Using the quotient rule, we see that
$$
y' = \frac{3(x+1)^2\cdot x^{3/2}-(x+1)^3\cdot\frac{3}{2}x^{1/2}}{(x^{3/2})^2}.
$$
Problem 2: Differentiate $y = \frac{2x^4}{b^2 - x^2}$ with respect to $x$.
Solution. We have $y=\frac{f(x)}{g(x)}$ where $f(x)=2x^4$ and $g(x)=b^2-x^2$. Thus, we also have $f'(x)=8x^3$ and $g'(x)=-2x$. Using the quotient rule, we see that
$$
y'=\frac{8x^3\cdot(b^2-x^2)-2x^4\cdot(-2x)}{(b^2-x^2)^2}.
$$
If required, all that is left to do is for you to simplify.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The equality $\frac{3}{4}\zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} = \sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}$ I was reading an article and the author immediately states that the following identity is clear
$$\frac{3}{4}\zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} = \sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}.$$
I verified that $\frac{3}{4}\zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2}$, but I am having trouble seeing how it is obvious that $\sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}$ equates as well.
I tried expanding terms, but it did not get me anywhere. Am I missing something obvious?
| It's because
$$\sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} = \sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{\underset{n\,even}{n = 1}}^\infty \frac{1}{n^2} = \sum_{\underset{n\, odd}{n = 1}}^\infty \frac{1}{n^2} = \sum_{n = 0}^\infty \frac{1}{(2n+1)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the digits in $\frac{n}{2^k}$ (where $n$ is an integer divisible by 9 and $k$ is any integer) sum to a multiple of 9? I don't know where to start. Here are two examples of the statement:
$36/8=4.5$ and $(4+5)\mod9=0$
$9/4=2.25$ and $(2+2+5)\mod9=0$
| Hint $\ $ Multiply by a power of $10$ to get an integer multiple of $9$ with the same digits.
$\qquad x= \dfrac{9j}{2^k}\ \Rightarrow\ 10^k x = 5^k(9j)\ $ has same digits as $\,x\, $ (shifted left by $\,k$), except $0$'s at end
E.g. in your case $\ x = \dfrac{9}{2^2} = \color{#c00}{2.25}\ $ so $\,10^2 x = 5^2\cdot 9 = \color{#c00}{225}.\,$
and the other is $\,\ x = \dfrac{36}{2^3} = \color{#c00}{4.5}\ $ so $\,10^3 x = 5^3\cdot 36 = \color{#c00}{4500}$.
Being a multiple of $9$ we know by casting out nines that $9$ divides its digit sum (which is not altered by any trailing $0$'s).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$A$ is a sum of two postive integer squares? if $x,y,z,w$ be postive integer,and such $x^2+y^2$ is prime number,and $A=\dfrac{w^2+z^2}{x^2+y^2}\in N^{+}$
show that
$A$ is a sum of two postive integer squares?
maybe
$$A=\dfrac{(w^2+z^2)(x^2+y^2)}{(x^2+y^2)}=\dfrac{(wx+zy)^2}{(x^2+y^2)^2}+\dfrac{(wy-xz)^2}{(x^2+y^2)^2}$$
But I can't prove $\dfrac{wx+zy}{x^2+y^2},\dfrac{(wy-xz)}{x^2+y^2}\in Z$
| Approach 1
In this answer, it is shown that a number, $n$, can be written as the sum of two squares if and only if each prime in the prime factorization of $n$ which is $\equiv3\pmod4$ appears with even exponent. Thus, if
$$
\frac{w^2+z^2}{x^2+y^2}\in\mathbb{Z}
$$
each prime $\equiv3\pmod4$ in $w^2+z^2$ appears with even exponent, and each prime $\equiv3\pmod4$ in $x^2+y^2$ appears with even exponent. Therefore, each prime $\equiv3\pmod4$ in the quotient must appear with even exponent, and therefore, the quotient can be written as a sum of two squares.
Approach 2
Here is an alternate approach using the fact that $x^2+y^2$ is assumed to be a prime.
Since $x^2+y^2$ is a prime number, $x+iy$ and $x-iy$ are Gaussian primes. Thus, because
$$
x+iy\mid(w+iz)(w-iz)
$$
we can choose the sign of $z$ so that $x+iy\mid w+iz$, and then $x-iy\mid w-iz$, also.
Then
$$
\frac{w+iz}{x+iy}=a+ib\qquad\text{and}\qquad\frac{w-iz}{x-iy}=a-ib
$$
Thus,
$$
\frac{w^2+z^2}{x^2+y^2}=\frac{w+iz}{x+iy}\frac{w-iz}{x-iy}=(a+ib)(a-ib)=a^2+b^2
$$
Confirmation of the Hypothesis Regarding the Squares
Note in the last approach, we have that
$$
a+ib=\frac{w+iz}{x+iy}=\frac{(xw+yz)+i(xz-yw)}{x^2+y^2}
$$
and therefore,
$$
\frac{w^2+z^2}{x^2+y^2}
=\left(\frac{xw+yz}{x^2+y^2}\right)^2+\left(\frac{xz-yw}{x^2+y^2}\right)^2
$$
where
$$
\frac{xw+yz}{x^2+y^2},\frac{xz-yw}{x^2+y^2}\in\mathbb{Z}
$$
| {
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"url": "https://math.stackexchange.com/questions/1225309",
"timestamp": "2023-03-29T00:00:00",
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How to prove this limit using epsilon delta method? i know how to construct a proof using epsilon delta method but as much as i have done i have to just algebraically manipulate and i got the value but i cant get this one.
$$\lim_{x \to c} \frac{x}{1+x^2} = \frac{c}{1+c^2}$$
My try: $$\left|\frac{x^2}{1+x^2} - \frac{c^2}{1+c^2}\right|<\epsilon\implies\left|\frac{x^2-c^2}{1+c^2+x^2+x^2c^2}\right|<\epsilon\implies\left|\frac{(x-c)(x+c)}{1+c^2+x^2+x^2c^2}\right|<\epsilon$$
But i dont how to proceed after this?
| If $c=0,$ take $\delta = \epsilon.$ If $c\ne 0,$ take $\delta = \min (\epsilon/(1+2c^2),|c|).$ Why? We have
$$\left |\frac{(x-c)(1-xc)}{(1+x^2)(1+c^2)} \right | \le |x-c|\,|1-xc|.$$
If $c=0,$ then we're left with $|x-c|$ on the right, perfect. If $c\ne 0,$ $|x-c|<|c| \implies |x|\le |x-c|+|c|<2|c|\implies |1-xc| < 1+2c^2.$ As $|x-c|<\epsilon/(1+2c^2),$ we're done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Laurent series of $\frac1{\sin^2z}$ around 0 I tried to expand $\frac{z^2}{\sin(z)^{2}}$ using Taylor expansion, but the coefficient involved some limit of $\frac{0}{0}$ and was really difficult to calculate. (I tried to convince myself the constant term of the Taylor expansion is 1 because $\lim_{x\rightarrow 0}(\frac{z^2}{\sin(z)^{2}})$=1 and the coefficients of z with odd degree are $0$ because $\frac{z^2}{\sin(z)^{2}}$ is a even function). But how do I solve those coefficients explicitly...?
Thank you.
| You may just write, as $z \to 0$:
$$\begin{align}\frac{1}{\sin^2z}&=\frac{1}{\left(z-\dfrac{z^3}{3!}+\dfrac{z^5}{5!}+\mathcal{O}(z^7)\right)^2}\\\\&=\frac{1}{z^2\left(1-\dfrac{z^2}{3!}+\dfrac{z^4}{5!}+\mathcal{O}(z^6)\right)^2}
\\\\&=\frac{1}{z^2}\left( 1-2\times \left(-\frac{z^2}{3!}+\frac{z^4}{5!}\right)+3\times \left(-\frac{z^2}{3!}\right)^2+\mathcal{O}(z^6)\right)\\\\
&=\frac{1}{z^2}\left(1+\frac{z^2}{3}+\frac{z^4}{15}+\mathcal{O}(z^6)\right)\\\\
&=\frac{1}{z^2}+\frac{1}{3}+\frac{z^2}{15}+\mathcal{O}(z^4)
\end{align}
$$ and this gives the first terms of the Laurent series expansion of $\dfrac{1}{\sin^2z}$ near $z=0$.
To obtain the coefficient of order $n$, one may use the Bernoulli numbers, observing that
$$
\frac{1}{\sin^2z}=-2i\:\partial_z \left(\frac{1}{e^{2iz}-1}\right)
$$ then get
$$
\frac{1}{\sin^2z}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1} 2^{2n}(2n-1) }{(2n)!}B_{2n}z^{2n-2},\quad 0<|z|<\pi.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate: $\int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx$ Evaluate: $\int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx$
I solved this question using u-substitution.
But I got two answers:
1) When $x=u^2 \implies \int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx =
$2arcsinx $\left(\sqrt{x}\right)$ +c and
2) When $\sqrt{1-x}=u$ ===> $\int \frac{1}{\sqrt{x}\sqrt{1-x}}\:dx$ = -2arcsin $\sqrt{1-x}$ +c
So why did I get two answers? Are they equivalent?
| another way is :
$$\int \frac{dx}{\sqrt{x}\sqrt{1-x}}\\=\int \frac{dx}{\sqrt{x-x^2}}\\=\int \frac{dx}{\sqrt{\frac{1}{4}-\frac{1}{4}+x-x^2}}\\=\int \frac{dx}{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}\\=\int \frac{2dx}{\sqrt{1-(2x-1)^2}}\\$$now use this substittuion $u=2x-1$ so $du=2dx$ $$\int \frac{du}{\sqrt{1-u^2}}=arcsin(u)=arcsin(2x-1)$$ but answers are equal $$(arcsin(2x-1))'=\frac{2}{\sqrt{1-(2x-1)^2}}=\frac{2}{\sqrt{4x-4x^2}}=\frac{1}{\sqrt{x-x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1228641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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limit using taylor series I keep getting an error in the expansion
$$\lim_{x\to 0 }\frac{2\exp(\sin(x))-2-x-x^2-\arctan (x) }{x^3}$$
The numerator works out as
$$\approx 2 (1+\sin(x) + 1/2 \sin^2(x)) -2-x-x^2 - (x-x^3/3+x^5/5)$$
$$\approx 2(1+x-x^3/3!+1/2(x-x^3/3!)^2 -2-2x+x^3/3-x^2 -x^5/5 $$
$$ = o(x^4)$$
so that the limit is zero. But it is supposed to be $1/3$. Where is my mistake?
| $$\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\O}{{\mathbb O}}
\sin x=x-\frac16x^3+\O(x^5)\\
e^x=1+x+\frac12x^2+\frac16x^3+\O(x^5)\\
\arctan x=x-\frac13x^3+\O(x^5)$$
Now:
$$e^{\sin x}=1+\sin x+\frac12\sin^2 x+\O(x^3)\\
=1+\b{x-\frac16x^3+\O(x^5)}+\frac12\b{x-\frac16x^3+\O(x^5)}^2+\frac16\b{x-\frac16x^3+\O(x^5)}^3+O(x^5)\\
=1+\b{x-\frac16x^3+\O(x^5)}+\frac12\b{x^2-\frac13x^4+\O(x^5)}+\frac16\b{x^3+\O(x^5)}+\O(x^5)\\
=1+x+\frac12x^2+\O(x^5)$$
Now:
$$2e^{\sin x}-2-x-x^2=x+\O(x^5)$$
So:
$$2e^{\sin x}-2-x-x^2-\arctan x=\frac13x^3+\O(x^5)$$
Now:
$$\lim_{x\to0}\frac{2e^{\sin x}-2-x-x^2-\arctan x}{x^3}=\lim_{x\to0}\frac{\frac13x^3+\O(x^5)}{x^3}=\frac13$$
Anyways Why don't you use L'Hospital?
$$\lim_{x\to 0 }\frac{2\exp(\sin(x))-2-x-x^2-\arctan (x) }{x^3}
\\=\lim_{x\to 0 }\frac{2\cos x\exp(\sin(x))-1-2x-\frac1{x^2+1} }{3x^2}
\\=\lim_{x\to 0 }\frac{2(\cos^2x-\sin x)\exp(\sin(x))-2+\frac{2x}{(x^2+1)^2} }{6x}
\\=\lim_{x\to 0 }\frac{\frac{2-6x^2}{(x^2+1)^3}-2\sin xe^{\sin x}(\sin x+3)}{6}
\\=\frac{\frac{2-6\times0^2}{(0^2+1)^2}-2\times0\times e^0(0+3)}{6}
\\=\frac{2}{6}=\frac13$$
| {
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Convergence of the series: $\sum_{n\geq1}\frac{(-1)^n\arctan (n)}{n+n^{1/2}}$ $$\sum_{n=1}^{\infty} (-1)^n \tan^{-1}(n)/(n+(n)^{1/2}).$$
I know that the series is not absolutely converges.
I want to prove using Alternative test. I don't know how to prove that sequence $ tan^{-1}n/(n+(n)^{1/2})$ is decreasing sequence.
| Certainly using a comparison test is easier, but here's what the Alternating Series test requires.
1) Since the arctangent function has an upper bound of $ \ \frac{\pi}{2} \ $ , it is reasonably clear that $$ \ \lim_{n \rightarrow \infty} \ \ \frac{\arctan \ n}{n \ + \ n^{1/2}} \ \ = \ \ 0 \ \ , $$
so the series passes the "test for divergence".
2) Differentiating the real function $ \ \frac{\arctan \ x}{x \ + \ x^{1/2}} \ $ gives us
$$ \ \frac{(x \ + \ \sqrt{x}) \ \cdot \ \frac{1}{1 \ + \ x^2} \ - \ \arctan \ x \ \cdot \ ( 1 \ + \ \frac{1}{2 \ \sqrt{x}})}{( \ x \ + \ \sqrt{x} \ )^2} $$
$$ = \ \ \frac{(x^2 \ + \ x\sqrt{x}) \ - \ \arctan \ x \ \cdot \ ( x \ + \ \frac{1}{2} \ \sqrt{x}) \ (1 \ + \ x^2)}{x \ (1 \ + \ x^2) \ ( \ x \ + \ \sqrt{x} \ )^2} \ \ . $$
As $ \ x \ \rightarrow \ \infty\ $ , the denominator grows as $ \ x^5 \ $ , the first term in the numerator as $ \ x^2 \ $ , and the second term as $ \ x^3 \ $ . Since the arctangent factor only tends to a constant, this derivative will tend to roughly $ \ \frac{1 \ - \ \frac{\pi}{2} \ x}{x^3} \ $ . So at least beyond some finite value of $ \ n \ $ , the terms are always decreasing.
So the conditions for alternating-series convergence appear to be satisfied.
| {
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If the distance between two lines is $ \frac{1}{\sqrt{3}} $, then $ \alpha $ is...
If the shortest distance between the lines $\displaystyle \frac{x-1}{\alpha}=\frac{y+1}{-1}=\frac{z}{1}\;, (\alpha \neq - 1)$ and
$x+y+z+1=0 = 2x-y+z+3 = 0$ is $\displaystyle \frac{1}{\sqrt{3}}\;,$ Then $\alpha = $
$\bf{My\; Solution:: }$ Here $x+y+z+1=0$ and $2x-y+z+3=0$ represent $2$ plane and
we know that locus of point of Intersection of $2$ plane is also a Line.
Know Here we have to calculate the equation of Common line of these $2$ planes.
For Calculation of equation of line....
Add these $2$ equations... $\displaystyle 3x+2z+4=0\Rightarrow z= - \frac{\left(4+3x\right)}{2}$ and Put into $x+y+z+1=0$
$\displaystyle 2x+2y+2z+2=0\Rightarrow 2x+2y-4-3x+2=0\Rightarrow -x+2y=2$
So we get $\displaystyle x=2y-2\Rightarrow \frac{x-0}{1} = \frac{y-1}{\frac{1}{2}}$ and above we get $\displaystyle \frac{x-0}{1}=\frac{z+2}{-\frac{3}{2}}.$
So We get equation of line is $\displaystyle \frac{x-0}{1}=\frac{y-1}{\frac{1}{2}}=\frac{z+2}{-\frac{3}{2}}.$
Now we have to Calculate Distance b/w two lines $\displaystyle \frac{x-1}{\alpha}=\frac{y+1}{-1}=\frac{z}{1}\;, (\alpha \neq - 1)$
and $\displaystyle \frac{x-0}{1}=\frac{y-1}{\frac{1}{2}}=\frac{z+2}{-\frac{3}{2}}.$
Is my equation of line is Right or not, If not then how can i calculate common
equation of line and value of $\alpha.$
Help me , Thanks
| I would go about doing it like this: put your lines into vector notation, in this case, $$(x_1,y_1,z_1)=(0,1,0)+t_1(2,1,0)$$ $$(x_2,y_2,z_2)=(1,-1,0)+t_2(\alpha,-1,1)$$ Then solve the messy system of equations yielded by$$\frac{\partial}{\partial t_1 }||(x_1,y_1,z_2)-(x_2,y_2,z_2)||^2=0$$ $$\frac{\partial}{\partial t_2 }||(x_1,y_1,z_2)-(x_2,y_2,z_2)||^2=0$$ $$||(x_1,y_1,z_1)-(x_2,y_2,z_2)||^2=\frac{1}{3}$$
for $\alpha$.
Edit: Alternatively, you could use the fact that if you project the vector connecting a point on one line and the other onto the vector normal to both, you'll get the distance between them. That is, $$((0,1,0)-(1,-1,0))\cdot\left(\frac{(2,1,0)\times(\alpha,-1,1)}{||(2,1,0)\times(\alpha,-1,1)||}\right)=\alpha$$ where $||\textbf{v}||$ is the euclidean norm of $\textbf{v}$.
| {
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"source": "stackexchange",
"question_score": "1",
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Matrix addition/multiplication with different sizes I have the following two matrices:
$$A=\begin{pmatrix}1 & -2\\3 & 1\end{pmatrix}\text{ and }B=\begin{pmatrix}1 & 3 &
2\\-1 & 0 & 2\end{pmatrix}$$
So I have two matrixes with different sizes. Multiple sources tell me that I can't do multiplication or addition with matrix of different sizes. So I'm a bit confused. Can I do it with these? How?
|
So I have two matrixes with different sizes. Multiple sources tell me that I can't do multiplication or addition with matrix of different sizes.
You can't add matrixes of different sizes as stated by @meshal. Hence, A+B or B+A can't be performed. However, you can multiply them. Here's how:
$$A=\begin{pmatrix}1 & -2\\3 & 1\end{pmatrix}\text{ and }B=\begin{pmatrix}1 & 3 &
2\\-1 & 0 & 2\end{pmatrix}$$
$$AB=\begin{pmatrix}1 & -2\\3 & 1\end{pmatrix}\text{ * }\begin{pmatrix}1 & 3 &
2\\-1 & 0 & 2\end{pmatrix}$$
$$AB=\begin{pmatrix}1\cdot \:1+\left(-2\right)\left(-1\right)&1\cdot \:3+\left(-2\right)\cdot \:0&1\cdot \:2+\left(-2\right)\cdot \:2\\ 3\cdot \:1+1\cdot \left(-1\right)&3\cdot \:3+1\cdot \:0&3\cdot \:2+1\cdot \:2\end{pmatrix}$$
$$Finally, AB=\begin{pmatrix}3&3&-2\\ 2&9&8\end{pmatrix}$$
| {
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"url": "https://math.stackexchange.com/questions/1232835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trigonometric integrals How do I evaluate this indefinite integral ? Integral $$\int\frac{x^2+n(n-1)}{(x\sin(x)+n\cos(x))^2}dx$$ What type of integral is it ? Is there any intuition involved in the approach to solve it?
Edit: The complete term in denominator has 2 as exponent. There was a typo previously.
| $\bf{My\; Solution:}$ Using $$\displaystyle (x\cdot \sin x+n\cdot \cos x) = \sqrt{x^2+n^2}\left\{\frac{x}{\sqrt{x^2+n^2}}\cdot \sin x+\frac{n}{\sqrt{x^2+n^2}}\cdot \cos x\right\}$$
$$\displaystyle = \sqrt{x^2+n^2}\cdot \cos\left(x-\phi\right)\;,$$ where $$\displaystyle \sin \phi = \frac{x}{\sqrt{x^2+n^2}}$$ and $$\displaystyle \cos \phi = \frac{n}{\sqrt{x^2+n^2}}$$ and $$\displaystyle \tan \phi = \frac{x}{n}\Rightarrow \phi = \tan^{-1}\left(\frac{x}{n}\right)$$
So Integral is $$\displaystyle = \int \sec^2(x-\phi)\cdot \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx$$
Now Let $$\displaystyle (x-\phi) = y\Rightarrow \left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)=y.$$ Then $$\displaystyle \left(\frac{x^2+n(n-1)}{x^2+n^2}\right)dx = dy$$
So Integral is $$\displaystyle \int \sec^2(y)dy = \tan y +\mathbb{C} = \tan\left(x-\tan^{-1}\left(\frac{x}{n}\right)\right)+\mathcal{C}$$
So $$\displaystyle \int \frac{x^2+n(n-1)}{(x\cdot \sin x+n\cdot \cos x)^2}dx = \left(\frac{n\cdot \tan x-x}{n+x\cdot \tan x}\right)+\mathcal{C}.$$
| {
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Rotations of complex graphs Let $c_1 = -i$ and $c_2 = 3$. Let $z_0$ be an arbitrary complex number. We rotate $z_0$ around $c_1$ by $\pi/4$ counter-clockwise to get $z_1$. We then rotate $z_1$ around $c_2$ by $\pi/4$ counter-clockwise to get $z_2$.
There exists a complex number $c$ such that we can get $z_2$ from $z_0$ by rotating around $c$ by $\pi/2$ counter-clockwise. Find the sum of the real and imaginary parts of $c$.
I am having some trouble with this problem. I have tried thinking this problem as if it were to be on the cartesian plane, but I still could not solve it. Any help is appreciated.
| Geometrically, this is simply stating that we wish to define a circle with center $c$ such that the points $z_0$ and $z_2$ lie on this circle.
In the complex plane, a point can be describe in polar coordinates by $re^{i\theta}+c$ for $0 \leq \theta < 2 \pi$ where $c$ is the center and $r$ is the radius.
The first point can then be described as $z_1 = z_0e^{i\frac{\pi}{4}} + c_1$ and the second point $z_2 = z_1e^{i\frac{\pi}{4}} + c_2$. We can then describe $z_2$ in terms of a circle based on $z_0$ as:
$z_2 = \left (z_0 e^{i\frac{\pi}{4}} + c_1\right )e^{i\frac{\pi}{4}} + c_2= z_0e^{i\frac{\pi}{2}} + \underbrace{c_1e^{i\frac{\pi}{4}}+ c_2}_{\text{new center}, c} $
The center of this circle is $c = c_1e^{i\frac{\pi}{4}}+ c_2$.
From Euler's Formula: $re^{i\theta} = r(\cos(\theta) + i\sin(\theta))$. Using this, we can describe $c$ as:
$$ c = c_1\cos(\frac{\pi}{4}) + ic_1\sin(\frac{\pi}{4}) + c_2 = \frac{c_1}{\sqrt{2}} + i\frac{c_1}{\sqrt{2}} + c_2 $$
Note that in cartesian $c_1 = a_1+ b_1i$ and $c_2 = a_2+ b_2i$. Which implies $ic_1 = -b_1 + a_1 i$. Which leads to:
$$ c = \frac{a_1 + b_1i}{\sqrt{2}} + \frac{-b_1 + a_1 i}{\sqrt{2}} + \frac{\sqrt{2}a_2 + \sqrt{2}b_2i}{\sqrt{2}} = \frac{(a_1-b_1+\sqrt{2}a_2)}{\sqrt{2}} + \frac{(a_1+b_1+\sqrt{2}b_2)}{\sqrt{2}}i$$
Then we just need to find $\Re(c) + \Im(c)$:
$$ \Re(c) + \Im(c) = \frac{(a_1-b_1+\sqrt{2}a_2) + (a_1+b_1+\sqrt{2}b_2)}{\sqrt{2}}= \frac{a_1+ \sqrt{2}{a_2}+\sqrt{2}b_2}{\sqrt{2}}$$
or
$$ \Re(c) + \Im(c) = \frac{\Re(c_1) + \sqrt{2}\left ( \Re(c_2)+ \Im{c_2} \right ) }{\sqrt{2}}$$
| {
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Evaluate $\lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$
Evaluate
$$\lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$$
For this integral, I have tried using integration by parts and then evaluating the limit, but I don't think the integral inside converges. However, the limit does exist and the answer given in my book is $2$.
Any help will be appreciated.
Thanks in advance!
| First, consider the following two Lemmas,
Lemma $1$: $$\lim_{n \to \infty} \sum_{r=0}^n \left(\dfrac{1}{\displaystyle\binom{n}{r}}\right) =2$$
Proof : First of all, note that the limit exists, since, if we let
$$\text{S}(n)=\displaystyle \sum_{r=0}^n \left(\dfrac{1}{\dbinom{n}{r}} \right)$$
then $\text{S}(n+1)<\text{S}(n)$ for $n \geq 4$. Now,
$\text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}}$
$\implies \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}} \ \left[\text{since} \dbinom{n}{r}= \dfrac{n}{r} \times \dbinom{n-1}{r-1} \right]$
Also,
$$ \text{S}(n) = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}} = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{n-r+1}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{n-r}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} $$
$ \left[ \text{since} \ \displaystyle \sum_{r=a}^b f(r) = \displaystyle \sum_{r=a}^b f(a+b-r) \ \text{and} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]$
Thus, we have,
$$\begin{cases} \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}}\\ \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} \end{cases}$$
Adding the above two expressions, we get,
$ 2\text{S}(n) = 2 + \displaystyle \sum_{r=1}^n \left( \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}} + \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} \right) $
$= 2 + \dfrac{n+1}{n} \displaystyle \sum_{r=1}^n \dfrac{1}{\dbinom{n-1}{r-1}} $
$= 2 + \dfrac{n+1}{n} \times \text{S}(n-1)$
Since $ n \to \infty $, we have $\text{S}(n) = \text{S}(n-1) = \text{S}$ (say)
$ \implies 2\text{S} = \left(\dfrac{n+1}{n}\right) \times \text{S} +2 $
$ \implies \text{S} = \dfrac{2n}{n-1} = 2$ [since $n \to \infty $]
Lemma $2$ : $$\int_{0}^1 x^r (1-x)^{n-r} \mathrm{d}t = \dfrac{1}{(n+1)}\times \dfrac{1}{\dbinom{n}{r}}$$
Proof : Consider the $\text{R.H.S.}$,
$\text{I} = \displaystyle\int_{0}^1 x^r (1-x)^{n-r} \mathrm{d}t$
Let $x = \sin^2 \theta$
$\implies \text{I} = \displaystyle\int_{0}^{\frac{\pi}{2}} 2 \sin^{2r+1} \theta \cos^{2n-2r} \theta \ \mathrm{d}\theta $
Now, using Walli's Formula (or reduction formula) for the above integral, we have,
$\text{I} = \dfrac{1}{(n+1)}\times \dfrac{1}{\dbinom{n}{r}} $
This proves our Lemmas.
Now,
$$ \text{J} = \lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t $$
Since it is an even function in $t$, we have,
$$ \text{J} = \frac{1}{2} \times \lim_{n \to \infty} \int_{-1}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t $$
Let $t = 2x-1$
$\implies \text{J} = \displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) \left(\dfrac{x^{n+1}-(1-x)^{n+1}}{2x-1}\right) \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) (1-x)^n \left(\dfrac{\left(\frac{x}{1-x}\right)^{n+1}-1}{\frac{x}{1-x}-1}\right) \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) \sum_{r=0}^n (1-x)^n \left(\frac{x}{1-x}\right)^{r} \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n (n+1) \int_{0}^1 x^r(1-x)^{n-r} \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n \dfrac{1}{\dbinom{n}{r}}$ (Using Lemma 2)
$=\boxed{2}$ (Using Lemma 1).
Side Note : Another way to prove Lemma 1 is to use sandwich theorem.
| {
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Proving $\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}dx=\frac{π}{2}-1$ Proving $$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx= \frac{π}{2}-1$$
My attempt is:
I assumed the $1-x=u$
$du =-dx$
$$\int_0^1 \sqrt{\frac{u}{2+u}}\,(-du)$$
| Let $u = \sqrt{\dfrac{1-x}{1+x}} \to u^2 = \dfrac{1-x}{1+x} \to x = \dfrac{1-u^2}{1+u^2} \to dx = \dfrac{-4u}{(1+u^2)^2}du \to I = \displaystyle \int_{0}^1\dfrac{4u^2}{(1+u^2)^2}du$. Next let $u = \tan \theta \to du = \sec^2 \theta d\theta \to I = \displaystyle \int_{0}^{\pi/4} \dfrac{4\tan^2 \theta}{\sec^4 \theta}\cdot \sec^2 \theta\cdot d\theta = \displaystyle \int_{0}^{\pi/4} 4\sin^2 \theta d\theta = \displaystyle \int_{0}^{\pi/4} 2-2\cos (2\theta) d\theta = \pi/2-1$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Contour integration
Consider the real-valued function
$$u(t) = \frac{1}{13-12\cos(t)}$$
By converting it to a contour integral along the unit circle in $\mathbb{C}$, evaluate
$$\int_0^{2\pi} u(t)\;dt$$
I have attempted so solve this and have found the residue to be $-\frac{1}{5}$ and therefore the integral to be $$\frac{2\pi}{5}$$
could someone confirm if this is correct?
| It looks like your work is correct.
I get,
$$ \int_0^{2\pi} u(t)\; dt = \int_0^{2\pi} \frac{ie^{it}}{ie^{it} \left(13-12\frac{e^{it} + e^{-it}}{2}\right)}\; dt$$
$$ = \frac{1}{i} \oint_C \frac{dz}{-6z^2+13z-6},$$
where $C$ is the unit circle. From the quadratic equation, I find that $-6z^2+13z-6 = 0$ for $z = \frac{3}{2}, \frac{2}{3}$, so by the Residue Theorem,
$$ \int_0^{2\pi} u(t)\; dt = \frac{1}{i} 2\pi i \operatorname{Res}\left(\frac{1}{-6z^2+13z-6}, z=\frac{2}{3}\right).$$
Since $\frac{1}{-6z^2+13z-6}$ has a simple pole at $z=\frac{2}{3}$, we have,
$$\operatorname{Res}\left(\frac{1}{-6z^2+13z-6}, z=\frac{2}{3}\right) = \lim_{z\to\frac{2}{3}} \left(z-\frac{2}{3}\right) \frac{1}{-6z^2+13z-6} = \lim_{z\to\frac{2}{3}} \frac{1}{-6(z-\frac{3}{2})} = \frac{1}{5}$$
And so putting it all together,
$$ \int_0^{2\pi} u(t)\; dt = \frac{2\pi}{5}$$
as you also calculated.
| {
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Evaluate $\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$
I want to find a quick way of evaluating $$\int_0^{1/\sqrt{3}}\sqrt{x+\sqrt{x^2+1}}\,dx$$
This problem appeared on the qualifying round of MIT's 2014 Integration Bee, which leads me to think there should be a shortish way (no more than three minutes by hand) to solve it. Examining the indefinite integral (thanks to WolframAlpha) hasn't particularly helped me:
$$\int\sqrt{x+\sqrt{x^2+1}}\,dx=-\frac{2}{3} \left(\sqrt{x^2+1}-2x\right) \sqrt{x+\sqrt{x^2+1}}+C$$
The bounds on the integral hint at trigonometric substitution, but I got nowhere by trying $x=\tan u$.
I also noticed that we can transform the integral by multiplying $\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x}$ in the first square root, but there didn't seem to be anything to do after that either.
| Using formula:
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{1}{2} \left(a+\sqrt{a^2-b}\right)}+\sqrt{\frac{1}{2} \left(a-\sqrt{a^2-b}\right)}$$
so:
$$\sqrt{x+\sqrt{x^2+1}}=\frac{\sqrt{-i+x}}{\sqrt{2}}+\frac{\sqrt{i+x}}{\sqrt{2}}$$
and then:
$$\int \left(\frac{\sqrt{-i+x}}{\sqrt{2}}+\frac{\sqrt{i+x}}{\sqrt{2}}\right) \, dx=\frac{1}{3} \sqrt{2} (-i+x)^{3/2}+\frac{1}{3} \sqrt{2} (i+x)^{3/2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241346",
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"source": "stackexchange",
"question_score": "15",
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Find the values of $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$
Calculate $\sin 69^{\circ},\sin 18^{\circ} , \tan 23^{\circ}$. accurate upto two decimal places or in surds .
$\begin{align}\sin 69^{\circ}&=\sin (60+9)^{\circ}\\~\\
&=\sin (60^{\circ})\cos (9^{\circ})+\cos (60^{\circ})\sin (9^{\circ})\\~\\
&=\dfrac{\sqrt{3}}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\
&=\dfrac{1.73}{2}\cos (9^{\circ})+\dfrac{1}{2}\sin (9^{\circ})\\~\\
\end{align}$
$\begin{align}\sin 18^{\circ}&=\sin (30-12)^{\circ}\\~\\
&=\sin (30^{\circ})\cos (12^{\circ})-\cos (30^{\circ})\sin (12^{\circ})\\~\\
&=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{\sqrt3}{2}\sin (12^{\circ})\\~\\
&=\dfrac{1}{2}\cos (12^{\circ})-\dfrac{1.73}{2}\sin (12^{\circ})\\~\\
\end{align}$
$\begin{align}\tan 23^{\circ}&=\dfrac{\sin (30-7)^{\circ}}{\cos (30-7)^{\circ}}\\~\\
&=\dfrac{\sin (30)^{\circ}\cos 7^{\circ}-\cos (30)^{\circ}\sin 7^{\circ}}{\cos (30)^{\circ}\cos 7^{\circ}+\sin (30)^{\circ}\sin 7^{\circ}}\\~\\
\end{align}$
is their any simple way,do i have to rote all values of of $\sin,\cos $ from $1,2,3\cdots15$
I have studied maths upto $12$th grade.
| Please refer to : derivation of sin 18 on this page
Once you know sin 18, you can find sin 9, cos 9 etc. by using half angle formulas.
In brief:
sin 72° = 2 sin 36° cos 36° by the double angle relationship.
sin 72° = 4 sin 18° cos 18° (1 - 2sin^2 18°) by the double angle relationship, again.
cos 18° = 4 sin 18° cos 18° (1 - 2sin^2 18°)
sin 72° = cos 18°.
1 = 4 sin 18° (1 - 2sin^2 18°)
Let x = sin 18°, this is known as
1 = 4*x(1-2x^2) substitution
8*x^3-4*x+1 = 0 A product is zero only when one of its factors is zero.
8x^3-4x+1 = (2*x-1)(4*x^2+2*x-1)=0 (2*x-1)=0 implies x= ½=sin 30° > sin 18° ;
Since we know sin is increasing on [0°,90°].
x = (-2 ± \sqrt{(4 + 4•4•1))/8} So we must solve the other factor,
= (-2 ± \sqrt{20})/8 using the quadratic formula.
= (-2 ± \sqrt{4}\sqrt{5})/8
= (-1 ± \sqrt{5})/4 But the sin 18° > 0, so it cannot be negative.
sin 18° = (\sqrt{5}-1)/4 Hence the middle root is the one we want.
Here at the bottom of the page referred above you will see a comment about how to find sin 1 also.
From sin 1 you can find sin (1/2) and note that 23 = (22 +(1/2)) + (1/2). But 22 + ( 1/2 ) is 1/2 of ( 45 ).
| {
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Computing $\int_3^5 \frac{x^2\,dx}{\sqrt{(x-3)(5-x)}}$
$$
\int_3^5 \frac{x^2\,dx}{\sqrt{(x-3)(5-x)}}
$$
how? $x^2/\sqrt{8x-x^2-15}$ and what to do then?
| Set $x=3+2\sin^2(t)$. We then obtain
$$\int_3^5 \dfrac{x^2dx}{\sqrt{(x-3)(5-x)}} = \int_0^{\pi/2} \dfrac{2(3+2\sin^2(t))\sin(2t)dt}{\sqrt{(2\sin^2(t))\cdot(2\cos^2(t))}} = \int_0^{\pi/2}2(3+2\sin^2(t))dt$$
I trust you can finish from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve for $x:1 + \tan^2(x) = 8\sin^2(x)$ I have a tricky problem , I tried several methods and I can't seem to get a definite answer.
$1 + \tan^2(x) = 8\sin^2(x), x \in [\frac{\pi}{6} , \frac{\pi}{2}]$
I got to $8\cos^4(x)-8\cos^2(x)+1=0$ and found that $\cos^2(x) = \frac{1}{4}[2-\sqrt{2}]$ but that is not too useful.
I need to find the angle "x" which is:
a)$\frac{\pi}{8}\quad$ b)$\frac{\pi}{6}\quad$ c)$\frac{\pi}{4}\quad$ d)$\frac{5\pi}{6}\quad$ e)$\frac{3\pi}{4}\quad$ f)$\frac{3\pi}{8}$
| Here's another approach. Use the facts that $\tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)}$ and $\cos^2(x) = 1 - \sin^2(x)$ to rewrite the equation so that the only trigonometric expression in it is $\sin^2(x)$. Then let $u = \sin^2(x)$. Now your equation should read:
$$1 + \frac{u}{1-u} = 8u$$
Now solve for $u$, then back-substitute in $u = \sin^2(x)$ to your solution(s) for $u$ and solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Evaluation of $\int\frac{1}{x^2.(x^4+1)^{\frac{3}{4}}}dx$
Evaluate the integral
$$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx$$
My Attempt:
Let $x = \frac{1}{t}$. Then $dx = -\frac{1}{t^2}\,dt$. Then the integral converts to
$$
-\int \frac{t^3}{(1+t^4)^{3/4}}\,dt
$$
Now Let $(1+t^4) = u$. Then $t^3\,dt = \frac{1}{4}du$. This changes the integral to
$$
\begin{align}
-\frac{1}{4}\int t^{-3/4}\,dt &= -u^{1/4}+\mathcal{C}\\
&= -\left(1+t^4\right)^{1/4}+\mathcal{C}
\end{align}
$$
So we arrive at the solution
$$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx = - \left(\frac{1+x^4}{x^4}\right)^{1/4}+\mathcal{C.}$$
Question: Is there any other method for solving this problem?
| Let $$\displaystyle I = \int\frac{1}{x^2\cdot (x^4+1)^{\frac{3}{4}}}dx = \int \frac{(x^4+1)-x^4}{x^2\cdot (x^4+1)^{\frac{3}{4}}}dx$$
So we get $$\displaystyle I = \int\left[\frac{(x^4+1)^{\frac{1}{4}}-x^4\cdot (x^4+1)^{-\frac{3}{4}}}{x^2}\right]dx = \int \left[\frac{x(-x^3)\cdot (1+x^4)^{-\frac{3}{4}}+(x^4+1)^{\frac{1}{4}}}{x^2}\right]dx$$
So we get $$\displaystyle I = -\int \frac{d}{dx}\left[\frac{(1+x^4)^{\frac{1}{4}}}{x}\right]dx = -\frac{(1+x^4)^{\frac{1}{4}}}{x}+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding triplets $(a,b,c)$ such that $\sqrt{abc}\in\mathbb N$ divides $(a-1)(b-1)(c-1)$ When I was playing with numbers, I found that there are many triplets of three positive integers $(a,b,c)$ such that
*
*$\color{red}{2\le} a\le b\le c$
*$\sqrt{abc}\in\mathbb N$
*$\sqrt{abc}$ divides $(a-1)(b-1)(c-1)$
Examples : The followings are positive integers.
$$\frac{(2-1)(8-1)(49-1)}{\sqrt{2\cdot 8\cdot 49}},\ \frac{(6-1)(24-1)(529-1)}{\sqrt{6\cdot 24\cdot 529}},\frac{(7-1)(63-1)(3844-1)}{\sqrt{7\cdot 63\cdot 3844}}$$
Then, I began to try to find every such triplet. Then, I found
$$(a,b,c)=(k,km^2,(km^2-1)^2)$$
where $k,m$ are positive integers such that $k\ge 2$ and $km^2\ge 3$, so I knew that there are infinitely many such triplets. However, I can neither find the other triplets nor prove that there are no other triplets. So, here is my question.
Question : How can we find every such triplet $(a,b,c)$?
Added : There are other triplets : $(a,b,c)=(k,k,(k-1)^4)\ (k\ge 3)$ by a user user84413, $(6,24,25),(15,15,16)$ by a user Théophile. Also, from the first example by Théophile, I got $(2k,8k,(2k-1)^2)\ (k\ge 3)$.
Added : $(a,b,c)=(k^2,(k+1)^2,(k+2)^2)\ (k\ge 2)$ found by a user coffeemath. From this example, I got $(k^2,(k+1)^2,(k-1)^2(k+2)^2)\ (k\ge 2)$.
Added : I got $(a,b,c)=(2(2k-1),32(2k-1),(4k-3)^2)\ (k\ge 5)$.
Added : I got $(a,b,c)=(k,(k-1)^2,k(k-2)^2)\ (k\ge 4)$.
Added : A squarefree triplet $(6,10,15)$ and $(4,k^2,(k+1)^2)\ (k\ge 2)$ found by a user martin.
Added : user52733 shows that $(6,10,15)$ is the only squarefree solution.
| Too long for a comment:
In addition to the rather lengthy
\begin{align}
&(m^2,\\
&((-1)^{2 k} \left(2 (-1)^k k m+(-1)^{k+1} (m+2)+m-6\right)^2)/16,\\
&\left((-1)^k \left(2 (-1)^k k m+(-1)^{k+1} (m+2)+m-6\right)+1\right)^2/4)\\
\end{align}
we also have $(a,b,c):$
\begin{align}
&\left(k^3+k^2+k+1,k^3+k^2+k+1,k^4\right)\\
&\left(k^4+k^2+1,k^4+k^2+1,k^6\right)\\
&\left(k m^2,k m^2 \left(k m^2-2\right)^2,\left(k m^2 \left(k m^2-3\right)+1\right)^2\right)\\
\end{align}
and for $f(n)=(n-1)^2$ we also have
\begin{align}
&\left(k^2,f^{2 n-1} \left((k m+1)^2\right),f^{2 n} \left((k m+1)^2\right)\right)\\
\end{align}
where $f^n$ is $f$ iterated $n$ times for $n \geq 1.$
However, even for fixed $a,$ the above formulae don't catch all of the solutions (and they say nothing of non-square $a$ combinations), and yet for each $a$ there seem to be multiple (infinite?) solutions.
Examples: case $a=8:$
A straightforward brute-force search for $(8,b,c);\ (b,c)<1000$ gives triples
$(8,2,49),(8,8,49),(8,18,49),(8,18,289),(8,32,49),(8,32,961),(8,49,72),(8,49,288),(8,289,392),(8,392,529),$
where it is immediately apparent that the same numbers recur a number of times. Removing the $8$ and graphing shows the connectedness more clearly:
Searching for $c$ only, using the distinct elements from the initial search (eg $(8,49,c)$, etc.) up to $10^5$ reveals further connections:
$(8,49,c)$ for example turns up $6$ triplets: $(8,49,2),(8,49,8),(8,49,18),(8,49,32),(8,49,72),(8,49,288)$
It may be more pertinent to ask then, are there infinitely many triplets for fixed $a?$ Certainly where $a$ is square, this is the case, but it is less clear whether this is the case when it is not.
It may also be worthwhile pursuing the idea of primitive pairs $(a,b).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 3,
"answer_id": 1
} |
Does $\sum 3^{-\sqrt{n}}$ converge or diverge? I need to find out whether this series converges or diverges:
$$\sum_{n=1}^\infty \frac 1{3^{\sqrt{n}}}$$
The $n$th term, ratio, and root tests are inconclusive, Abel's test doesn't apply (or I can't think of how to separate out part of the sequence), and I can't figure out a series to compare this to that'll work.
WolframAlpha says it converges by the comparison test, BTW. It just doesn't tell me what it compared the series to.
| Here's something different, that doesn't require comparing $n^2$ with $3^{\sqrt{n}}$, or any similar comparison:
$$\frac{1}{3^{\sqrt{n}}}\leq\frac{1}{3^{\left\lfloor\sqrt{n}\right\rfloor}}$$
And the sequence $\{\left\lfloor\sqrt{1}\right\rfloor,\left\lfloor\sqrt{2}\right\rfloor,\left\lfloor\sqrt{3}\right\rfloor,\ldots\}$ is equal to $$\{\overbrace{1,1,1}^3,\overbrace{2,\ldots,2}^{5},\overbrace{3,\ldots,3}^{7},\ldots,\overbrace{k,\ldots,k}^{2k+1},\ldots\}\text{.}$$ This follows from understanding that consecutive perfect squares differ by increasing odd numbers. Or equivalently that the sum of consecutive odd integers $3+5+7+\cdots$ is always $1$ shy of a perfect square.
So
$$\begin{align}
\sum_{n=1}^{\infty}\frac{1}{3^{\sqrt{n}}}&<\sum_{n=1}^{\infty}\frac{1}{3^{\left\lfloor\sqrt{n}\right\rfloor}}\\
&=\sum_{k=1}^{\infty}\frac{2k+1}{3^{k}}\\
&=2\sum_{k=1}^{\infty}\frac{k}{3^{k}}+\sum_{k=1}^{\infty}\frac{1}{3^{k}}\\
&=2\cdot\frac{3}{4}+\frac{1}{2}\\
&=2
\end{align}$$
Not only is the sum convergent, it's less than $2$. You can get a better upper bound by leaving the initial terms alone instead of using the floor function. For instance, this same approach can be used with $$\begin{align}\frac{1}{3^{\sqrt{1}}}+\frac{1}{3^{\sqrt{2}}}+\frac{1}{3^{\sqrt{3}}}+\sum_{n=4}^{\infty}\frac{1}{3^{\left\lfloor\sqrt{n}\right\rfloor}}
&=\frac{1}{3^{\sqrt{1}}}+\frac{1}{3^{\sqrt{2}}}+\frac{1}{3^{\sqrt{3}}}+1\\
&\approx1.69\ldots
\end{align}$$
which is a better upper bound. (A CAS says the true value is approximately $1.34\ldots$)
For an even better approximation that you can't immediately tell is over or under, replace each term in $\sum_{n=N^2}^{(N+1)^2-1}\frac{1}{3^{\sqrt{n}}}$ (the portion of the series corresponding to one of the constant substrings in $\{\left\lfloor\sqrt{n}\right\rfloor\}$) with the average of the end terms: $\frac{1}{3^{\sqrt{N^2}}}$ and $\frac{1}{3^{\sqrt{(N+1)^2}}}$. So
$$\begin{align}
\sum_{n=1}^{\infty}\frac{1}{3^{\sqrt{n}}}
&\approx\sum_{N=1}^{\infty}\frac12\left(\frac{1}{3^N}+\frac{1}{3^{N+1}}\right)(2N+1)\\
&=\frac43
\end{align}$$
I'm not offering error analysis, but you can note that this does indeed get within $0.6\%$ of the exact value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Finding a matrix from equation we've got the following 4x4 Matrix
$$\begin{pmatrix}
4 & -2 & 3 & 2\\
3 & 5 & 1 & -4\\
-1 & 6 & -4 & -7\\
-2 & 0 & -2 & 4
\end{pmatrix}$$
and I need to find $B$ from the equation: $(A-3I)B=0$.
i started to solve it by finding first $A-3I$. and I got:
$$\begin{pmatrix}
1 & -2 & 3 & 2\\
3 & 2 & 1 & -4\\
-1 & 6 & -7 & -7\\
-2 & 0 & -2 & 1
\end{pmatrix}$$
now I know that every column $[AB]^j$ [$j$ represents column number] can be calculated by $A[B]^j$ [$j$ represents column number].
and I was trying to solve it by multiplying $A$ with a specific column in $B$ but I wasn't able to reach the zero matrix.
EDIT:
almost forgot to mention that B needs to be 4x4 matrix and different form 0!
| by row reducing $A-3I,$ using my ti-83, i get $$\pmatrix{1&-2&3&2\\3&2&1&-4\\-1&6&-7&-7\\-2&0&-2&1}\to\pmatrix{1&0&1&-0.5\\0&1&-1&-1.25\\0&0&0&0\\0&0&0&0} $$ we can see that two linearly independent vectors $a, b$ such that $(A-3I)a = 0, (A-3I)b = 0$ where $$a=\pmatrix{0.5\\1.25\\0\\1}, b = \pmatrix{-1\\1\\1\\0}.$$ you can make a matrix $B$ made up of columns $a, b, a, b.$ in fact each column can be any linear combination of $a, b$ would do as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $x$ and $y$ be real numbers such that $x^2 + y^2 = 1$. Find the maximum value of $2x - 5y$. Let $x$ and $y$ be real numbers such that $x^2 + y^2 = 1$. Find the maximum value of $2x - 5y$.
I do know how to solve this problem using trigonometry, however I need to solve it by using vectors. Except for defining vector $v=(x, y)$, such that $||v||=1$, I do not see any way in which I could apply vectors to solving this problem. Any ideas are appreciated. Thank you!
| The maximum value $k$ occurs when the line $2x-5y = k$ is tangent to the circle $x^2+y^2=1$.
Solving for $x$, we get $x = \frac{1}{2} (k+5y)$, so $\frac{1}{4}(k^2+10ky+25y^2) + y^2 = 1$. Now set up the quadratic and set $\Delta = 0$, as we only want there to be one value of $y$ for the intersection of the circle and line:
$$k^2+10ky + 25y^2 + 4y^2 = 4$$
$$\Rightarrow 29y^2+10ky+(k^2-4)=0$$
$$\Delta = b^2-4ac = 0: (10k)^2-4(29)(k^2-4)=0$$
$$\Rightarrow -16k^2+464=0, k = ±\sqrt{\frac{464}{16}} = ±\sqrt{29}$$
and we are looking for the maximum value of $k$. Thus $k = \boxed{\sqrt{29}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1254585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Show that $\sin\left(\frac\pi3(x-2)\right)$ is equal to $\cos\left(\frac\pi3(x-7/2)\right)$ Show that
$$\sin\left(\frac\pi3(x-2)\right)$$
is equal to
$$\cos\left(\frac\pi3(x-7/2)\right)$$
I know that $\cos(x + \frac\pi2) = −\sin(x)$ but i'm not sure how i can apply it to this question.
| Maybe you can try
\begin{align*}\sin\left(\frac{\pi}{3}(x-2)\right) &= \sin\left(\frac{\pi}{3}x-\frac{2\pi}{3}\right)\\
&=\sin\left( \pi - \frac{\pi}{3}x+\frac{2\pi}{3}\right) \\
&=\sin\left(\frac{5\pi}{3}-\frac{\pi}{3}x\right) \\
&=\sin\left(\frac{\pi}{2}+\frac{7}{2}\frac{\pi}{3} -\frac{\pi}{3}x\right)\\
&=\cos\left(\frac{\pi}{3}\left(x-\frac{7}{2}\right)\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Diophantine equation $x^2 + xy + y^2 = \left({{x+y}\over{3}} + 1\right)^3$. Solve in integers the equation$$x^2 + xy + y^2 = \left({{x+y}\over3} + 1\right)^3.$$
| Hint $1$ :
Let $t=\frac{x+y}{3}\in \mathbb Z$ the equation becomes:
$$y^2-3ty+(3t)^2-(t+1)^3=0 $$
a quadratic equation on $y$ which is soluble up to the condition $4t+1$ is a square.
Hint $2$: $\Delta_y=(t-2)^2(4t+1)$
Solutions $(t,y)=(a^2+a,-a^3+3a+1),(a^2+a,a^3+3a^2-1)$ for any parameter $a$
And it's your turn to do some work!
Edit
Because my answer is down-voted and I don't see the reasons, I will write a full solution based on my little hints below:
Claim If $(x,y)$ are integer solutions to the equation :
$$x^2 + xy + y^2 = \left({{x+y}\over3} + 1\right)^3\tag 1$$ then
$(x,y)\text{ or } (y,x) \in\left\{\left(-a^3+3a+1,a^3+3a^2-1\right)\big | a\in \Bbb Z \right\}$
Proof
Let $(x,y)$ be a solution to the given equation, and let $t=\frac{x+y}{3}\in \Bbb Z$ so that:
$$y^2-3ty+(3t)^2-(t+1)^3=0 \tag 2$$
and this is a quadratic equation on $y$ so let's compute its discriminant:
$$\Delta_y=(3t)^2-4\left((3t)^2-(t+1)^3\right)=4(t+1)^3-27t^2=(t-2)^2(4t+1)\tag 3$$
so this equation has an integer solution $y$ only if $\Delta_y$ is a square of an integer, and this is equivalent to $4t+1$ is a square of an integer which implies that $t=a^2+a$ for some integer $a$. and because the equation $(2)$ is quadratic and we know the value of its discriminant we can find the two solutions on $y$:
$$\begin{align}y&=&\frac12\left(3t-(t-2)\sqrt{4t+1}\right)&=-a^3+3a+1\tag 4\\ y&=&\frac12\left(3t+(t-2)\sqrt{4t+1}\right)&=a^3+3a^2-1 \tag 5\end{align}$$
now that we found the value of $t$ and $y$ we can find the value of $x=3t-y$ which gives:
$$\left\{\begin{matrix}
x=a^3+3a^2-1&\text{ and }& y=-a^3+3a+1 &\text{ or }\tag 6\\
x=-a^3+3a+1&\text{ and } & y=a^3+3a^2-1
\end{matrix}\right.$$
Finally:
$$(x,y)\text{ or } (y,x) \in\left\{\left(-a^3+3a+1,a^3+3a^2-1\right)\big | a\in \Bbb Z \right\}\tag 7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solving differential equation I want to solve the following differential equation with initial conditions:
$$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\frac{x \, y(x)}{\sqrt{1-x}}$$
But do not know how to actually solve it. Any suggestion?
| Hint:
Let $u=\sqrt{1-x}$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-\dfrac{1}{2\sqrt{1-x}}\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-\dfrac{1}{2\sqrt{1-x}}\dfrac{dy}{du}\right)=-\dfrac{1}{2\sqrt{1-x}}\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}=-\dfrac{1}{2\sqrt{1-x}}\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}=-\dfrac{1}{2\sqrt{1-x}}\dfrac{d^2y}{du^2}\left(-\dfrac{1}{2\sqrt{1-x}}\right)+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}=\dfrac{1}{4(1-x)}\dfrac{d^2y}{du^2}+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}$
$\therefore\dfrac{1}{4(1-x)}\dfrac{d^2y}{du^2}+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}=\dfrac{xy}{\sqrt{1-x}}$
$\dfrac{1}{4u^2}\dfrac{d^2y}{du^2}+\dfrac{1}{4u^3}\dfrac{dy}{du}-\dfrac{(1-u^2)y}{u}=0$
$u\dfrac{d^2y}{du^2}+\dfrac{dy}{du}+4u^2(u^2-1)y=0$
Consider generally change the abscissa and the ordinate:
Let $u=f(v)$ ,
Then $\dfrac{dy}{du}=\dfrac{\dfrac{dy}{dv}}{\dfrac{du}{dv}}=\dfrac{1}{f'(v)}\dfrac{dy}{dv}$
$\dfrac{d^2y}{du^2}=\dfrac{d}{du}\left(\dfrac{1}{f'(v)}\dfrac{dy}{dv}\right)=\dfrac{\dfrac{d}{dv}\left(\dfrac{1}{f'(v)}\dfrac{dy}{dv}\right)}{\dfrac{du}{dv}}=\dfrac{\dfrac{1}{f'(v)}\dfrac{d^2y}{dv^2}-\dfrac{f''(v)}{(f'(v))^2}\dfrac{dy}{dv}}{f'(v)}=\dfrac{1}{(f'(v))^2}\dfrac{d^2y}{dv^2}-\dfrac{f''(v)}{(f'(v))^3}\dfrac{dy}{dv}$
$\therefore\dfrac{f(v)}{(f'(v))^2}\dfrac{d^2y}{dv^2}-\dfrac{f(v)f''(v)}{(f'(v))^3}\dfrac{dy}{dv}+\dfrac{1}{f'(v)}\dfrac{dy}{dv}+4(f(v))^2((f(v))^2-1)y=0$
$\dfrac{d^2y}{dv^2}+\left(\dfrac{f'(v)}{f(v)}-\dfrac{f''(v)}{f'(v)}\right)\dfrac{dy}{dv}+4f(v)((f(v))^2-1)(f'(v))^2y=0$
For example when take $f(v)=\dfrac{pv+q}{rv+s}$ , the ODE becomes
$\dfrac{d^2y}{dv^2}+\left(\dfrac{ps-qr}{(pv+q)(rv+s)}+\dfrac{2r}{rv+s}\right)\dfrac{dy}{dv}+\dfrac{4(ps-qr)^2(pv+q)((pv+q)^2-(rv+s)^2)}{(rv+s)^7}y=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Evaluating $\sum_{n=1}^\infty \frac{n^2}{3^n} $ The following series converges to 3/2 but I do not know why.
$$ \sum_{n=1}^\infty \frac{n^2}{3^n} = \frac{3}{2} $$
Searching via Google did not yield anything useful. I'm wondering if there's some sort of algorithm that can be used to solve this, or better yet, a closed-form equation for evaluating infinite sums that are in a form similar to
$$ \sum_{n=1}^\infty \frac{n^a}{b^n} $$
Or maybe it's just a Taylor series that I don't recognize.
| A solution using only the fact that $\sum_{n=0}^\infty q^n = \frac{1}{1-q}$ when $|q|<1$:
Let $S_1 = \sum_{n=1}^\infty\dfrac{n^2}{3^n}$, then $3S_1 = \sum_{n=0}^\infty\dfrac{(n+1)^2}{3^n}$, so we have $$2S_1 = 3S_1 - S_1 = 1 + \sum_{n=1}^\infty\dfrac{2n+1}{3^n} = 1 + 2S_2 + \sum_{n=1}^\infty\dfrac{1}{3^n} = 1 + 2S_2 + \frac{1}{2}$$
where $S_2 = \sum_{n=1}^\infty\dfrac{n}{3^n}$
Similarly we have $$2S_2 = 3S_2 - S_2 = \sum_{n=0}^\infty\dfrac{n+1}{3^n} - \sum_{n=1}^\infty\dfrac{n}{3^n} = \sum_{n=0}^\infty\dfrac{1}{3^n} = \frac{3}{2}$$
So $S_1 = \dfrac{1+\frac{3}{2}+ \frac{1}{2}}{2} =\dfrac{3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Deriving a contradiction How can I derive a contradiction from the following nasty statement:
Assume $\sqrt{5} = a + b\sqrt[4]{2} + c\sqrt[4]{4} + d\sqrt[4]{8},$ with $a,b,c,d \in \mathbb{Q}$?
This is the last piece of an effort to prove that the polynomial $x^4-2$ is irreducible over $\mathbb{Q}(\sqrt{5}).$
| From your assumption
$$(\sqrt5-a)^2 =\sqrt2b^2+2c^2+2d^2\sqrt2 +2bc\sqrt[4]{8}+2cd\sqrt[4]{32}+4bd$$ which after rearrangement is $2c^2+4bd +(b^2+2d^2+4cd)\sqrt2+2\sqrt2bc\sqrt[4]2$.
So multiplying by $\sqrt2$ we get $$\sqrt2(\sqrt5-a)^2 = 2(b^2+2d^2+4cd)+(2c^2+4bd)\sqrt2 + 4bc\sqrt[4]2$$ Denoting the integers $2(b^2+2d^2+4cd),\ (2c^2+4bd)$ by $m$ and $n$ respectively bringing them to the right and squaring again we get
$$\bigg[\sqrt2[(\sqrt5-a)^2- n]-m\bigg]^2=16b^2c^2\sqrt2 $$
Now the problem is simpler as we have got rid of all 4th roots and everything is a square root in the above. Getting a contradiction from here should be manageable.
| {
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Limit and convergence of $\frac{1}{n^{n-1}}\sum_{p=2}^{n-1} \left[ {n \choose p} (n-p)^{n-2} p (-1)^p \right]$ This is a part of larger question, in which I was asked to show that a certain ratio has a limit of $e^{-1}$.
After much of algebraic manipulation, I've found this ratio to be
$$
\frac{1}{n^{n-1}}\sum_{p=2}^{n-1} \left[ {n \choose p} (n-p)^{n-2} p (-1)^p \right]
$$
and hopefully this is correct. Given that, this series must converge to $e^{-1}$ as $n\to\infty$, but I am stuck at showing this.
So knowing what this series should converge to, I can rewrite the summand as
\begin{align*}
\frac{1}{n^{n-1}}\left[ {n \choose p} (n-p)^{n-2} p (-1)^p \right]
&=\frac{n}{n} \frac{n-1}{n} \cdots \frac{n-p+1}{n} \cdot \frac{1}{(p-1)!} \frac{(n-p)^{n-2}}{n^{n-p-1}} (-1)^p
\end{align*}
and a slight hope of improvement seems to emerge with factorial terms. But at this point I'm rather hopelessly lost, having no idea how to proceed.
Obviously one approach would be bound the series by series that tend to $e^{-1}$, but I fail to spot any good candidate.
Any helps appreciated
| Suppose we seek to investigate
$$\frac{1}{n^{n-1}}
\sum_{p=2}^{n-1} {n\choose p} (-1)^p p (n-p)^{n-2}.$$
Taking $n\gt 2$ this becomes
$$\frac{1}{n^{n-1}}
\left(n\times (n-1)^{n-2}
+ \sum_{p=0}^{n} {n\choose p} (-1)^p p (n-p)^{n-2}\right).$$
The first term here is
$$\frac{(n-1)^{n-2}}{n^{n-2}}
= \left(1-\frac{1}{n}\right)^{n-2}
= \left(\frac{n}{n-1}\right)^2\left(1-\frac{1}{n}\right)^{n}.$$
This is $$\frac{1}{e}$$ in the limit by inspection.
The second term is the sum term which we now evaluate.
This is the sum:
$$\sum_{p=0}^{n} {n\choose p} (-1)^p p (n-p)^{n-2}.$$
Put $$(n-p)^{n-2}
= \frac{(n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}} \exp((n-p)z) \; dz.$$
This gives for the sum
$$\frac{(n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}}
\sum_{p=0}^{n} {n\choose p} (-1)^p p
\exp((n-p)z) \; dz
\\ = \frac{(n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}} \exp(nz)
\sum_{p=0}^{n} {n\choose p} (-1)^p p
\exp(-pz) \; dz.$$
We have
$$x((1+x)^n)' = x \sum_{q=1}^n {n\choose q} q x^{q-1}
= \sum_{q=0}^n {n\choose q} q x^{q}
= nx(1+x)^{n-1}.$$
Applying this to the sum
$$\frac{n \times (n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}} \exp(nz)
(-\exp(-z)) (1-\exp(-z))^{n-1} \; dz
\\ = -\frac{n \times (n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}} \exp((n-1)z)
(1-\exp(-z))^{n-1} \; dz
\\ = -\frac{n \times (n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}}
(\exp(z)-1)^{n-1} \; dz
\\ = -\frac{n! \times (n-2)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-1}}
\frac{(\exp(z)-1)^{n-1}}{(n-1)!} \; dz
\\ =
- n!\times (n-2)! \times {n-2\brace n-1} = 0.$$
The sum term evaluates to zero leaving only the contribution from
$$\frac{1}{e}$$ which was to be shown.
Remark. If the Stirling number troubles anyone just observe that $\exp(z)-1$ starts at $z$ and therefore the power $z^{n-2}$ does not appear in $(\exp(z)-1)^{n-1}.$
| {
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Choosing n objects from k types of objects, each of which is in limited supply Suppose I wanted to light my Christmas tree. In my basement, I find a cord that has $5$ sockets in which I can screw bulbs. I also locate $5$ red bulbs, $4$ green bulbs, and $3$ blue bulbs. How many unique strings of bulbs can I form? Let's say that reversed sequences do not count as unique sequences (so RRRGB and BGRRR would count as one unique sequence).
I do not know of an efficient approach to solving this type of problem. My initial thought is that the only way to compute the answer manually is through tedious case-by-case analysis. Thoughts?
| You can use dynamic programming. The first step is to find the number of unique strings without modding out by reversion. Consider the expression
$$ (r+g+b)^5 = b^5+5 b^4 g+5 b^4 r+10 b^3 g^2+20 b^3 g r+10 b^3 r^2+10 b^2 g^3+30 b^2 g^2 r+30 b^2 g r^2+10 b^2 r^3+5 b g^4+20 b g^3 r+30 b g^2 r^2+20 b g r^3+5 b r^4+g^5+5 g^4 r+10 g^3 r^2+10 g^2 r^3+5 g r^4+r^5. $$
What you want is the sum of coefficients of $r^ig^jb^k$ where $i \leq 5$, $j \leq 4$, $b \leq 3$. When you do the computation, there is no need to keep track of monomials which violate this constraint, and this leads to the dynamic programming solution, which is a slightly more efficient version of the generating series approach above. (I won't give the details of the dynamic programming approach, leaving them to you.)
In order to count the number of unique strings up to reversion, we will count the number of symmetric solutions (i.e. strings whose reverse is the same as the original string). Here is why. Suppose that $A$ is the number of strings, $B$ the number of symmetric strings, and $C$ the number of strings up to reversion. Then
$$ C = \frac{A+B}{2}. $$
(Figure out why on your own.)
How do we count the number of symmetric solutions? Using the same dynamic programming / generating function approach:
$$ (r^2+g^2+b^2)^2(r+g+b) = b^5+b^4 g+b^4 r+2 b^3 g^2+2 b^3 r^2+2 b^2 g^3+2 b^2 g^2 r+2 b^2 g r^2+2 b^2 r^3+b g^4+2 b g^2 r^2+b r^4+g^5+g^4 r+2 g^3 r^2+2 g^2 r^3+g r^4+r^5. $$
Again, we sum the coefficients of monomials $r^ig^jb^k$ with the same constraints as above, and using dynamic programming, we can get a slightly more efficient solution.
| {
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Find the change of basis matrix P from S to S'. Consider the following bases of $\mathbb{R}^2$:
$$S=\left\{\begin{pmatrix}
1\\
-2
\end{pmatrix},\begin{pmatrix}
3 \\
-4
\end{pmatrix}\right\}$$
$$S'=\left\{\begin{pmatrix}
1\\
3
\end{pmatrix},\begin{pmatrix}
3 \\
8
\end{pmatrix}\right\}$$
Find the change of basis matrix P from S to S'.
So I followed a guide online which said to express S in terms of S':
$$\begin{pmatrix}
1\\
-2
\end{pmatrix}= \begin{pmatrix}
1\\
3
\end{pmatrix}a+\begin{pmatrix}
3 \\
8
\end{pmatrix}b$$
$$\begin{pmatrix}
3\\
-4
\end{pmatrix}= \begin{pmatrix}
1\\
3
\end{pmatrix}c+\begin{pmatrix}
3 \\
8
\end{pmatrix}d$$
So I got $a=-14,b=5,c=-36,d=13$ so would the answer simply be:
$$\begin{pmatrix}
-14 & 5\\
-36 & 13
\end{pmatrix}$$
or am I missing a step?
| With your notation, I think your final matrix should be
$$\begin{pmatrix} a&c\\b&d \end{pmatrix}$$
| {
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} |
Solutions of the Congruence If $x^{10}\equiv 1\pmod{\!55^2}$, how do I know one must have $x^{10}\equiv 1\pmod{\!5^2}$ and $x^{10}\equiv 1\pmod{\!11^2}$?
| If $55$ divides a number $n$, then any divisor of $55$ must also divides the number $n$. $5$ and $11$ are divisors of $55$. Hence, $5$ and $11$ must both divide $x^{10}-1$. This mean we want $x$ such that
\begin{align}
x^{10} & \equiv 1\pmod5\\
x^{10} & \equiv 1\pmod{11}
\end{align}
From Fermat's little theorem, we have that $x^4 \equiv 1 \pmod5$ for $x \in \{1,2,3,4\}\pmod5$ and $x^{10} \equiv 1 \pmod{11}$, for $x \in \{1,2,3,4,5,6,7,8,9,10\}\pmod{11}$. This means we have $x^{10} \equiv x^2 \equiv 1\pmod{5}$.
Hence, our $x$ must be $\pm1 \pmod5$ and $x \equiv \pm1,\pm2,\pm3,\pm4,\pm5\pmod{11}$.
This gives us $20$ possible solutions.
| {
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Weights in the Dynkin Basis and Eigenvalues of the Cartan Generators for SU(3)? The Cartan Generators of $SU(3)$ in the three dimensional rep have eigenvalues $(1,-1,0)$ and $\frac{1}{\sqrt{3}} (1,1,-2)$. Therefore we have the weights:
$$ (1,\frac{1}{\sqrt{3}}) \quad (-1,\frac{1}{\sqrt{3}}) \quad (1,\frac{-2}{\sqrt{3}}) $$
In the Dynkin basis the weights of the 3 dimensional rep are
$$ [1,0] \quad [-1,1] \quad [0,1] $$
How are these connected to the eigenvalues of the Cartan generators quoted above? I thought that I get them by multiplying the weights in the Dynkin basis with the corresponding metric tensor (=the inverse of the Cartan matrix) of $SU(3)$:
$$ G=\frac{1}{3} \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} $$
For example,
$$ G [1,0] = \frac{1}{3} \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix}1 \\ 0 \end{pmatrix} = \begin{pmatrix}\frac{2}{3} \\ \frac{1}{3}\end{pmatrix} $$
Unfortunately this yields the wrong weights. What is wrong and how can I compute correctly the eigenvalues of the Cartan generators from the weights in the Dynkin basis?
| The thing is, by multiplying the weights in den Dynkin basis with the metric tensor, we get the weights in the simple root basis. If we want to know the weights in terms of the Cartan generator $H_1,H_2,...$ eigenvalues, we need to use the simple roots in this basis.
For $SU(3)$ the simple roots are $\alpha_1= (\frac{1}{2}, \frac{\sqrt{3}}{2})=\frac{1}{2} H_1 + \frac{\sqrt{3}}{2} H_2 $ (the last equality should be understood symbolically) and $\alpha_2= (\frac{1}{2}, \frac{-\sqrt{3}}{2})$.
Then we can rewrite the weight I computed in the OP $ \begin{pmatrix}\frac{2}{3} \\ \frac{1}{3}\end{pmatrix} = \frac{2}{3} \alpha_1 + \frac{1}{3} \alpha_2$. Using the explicit form of the simple roots then yields
$$\frac{2}{3} \alpha_1 + \frac{1}{3} \alpha_2 = \frac{2}{3} \begin{pmatrix}\frac{1}{2} \\ \frac{\sqrt{3}}{2}\end{pmatrix} + \frac{1}{3} \begin{pmatrix}\frac{1}{2} \\ \frac{-\sqrt{3}}{2}\end{pmatrix} =\begin{pmatrix}\frac{1}{2} \\ \frac{1}{2\sqrt{3}}\end{pmatrix}$$
Now, recalling that our generators are in fact $ \frac{1}{2} $ times the Gell-Mann matrices this is the correct weight in terms of eigenvalues of the Cartan generators.
| {
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Why do we invert then multiply when dividing fractions? I have looked up various sites online. They only explain it in a very basic arithmetic way. Would someone explain to me why is $\displaystyle\frac{x}{\frac{1}{2}}$ is $\displaystyle\frac{2x}{1}$?
| $\frac{x}{\frac{1}{2}}=\frac{x}{\frac{1}{2}}\cdot1=\frac{x}{\frac{1}{2}}\cdot\frac{2}{2}=\frac{x\cdot 2}{\frac{1}{2}\cdot2}=\frac{2x}{1}$
Generally,
$\frac{a}{\frac{b}{c}}=\frac{a}{\frac{b}{c}}\cdot\frac{c}{c}=\frac{ac}{b}=a\cdot\frac{c}{b}$
| {
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Finding $\int_{0}^{\infty }\frac{1}{1+x^4}dx$ finding $$\int_{0}^{\infty }\frac{1}{1+x^4}dx$$
My attempt is:
let $x=\sqrt{u}$
$dx=\frac{1}{2\sqrt{u}}$
$$\int_{0}^{\infty }\frac{1}{2\sqrt{u}(1+u^2)}du$$
here I stopped because I don't know how to complete this solution. Any help please.
| One way to deal with this is to write
$$
1+x^4 = (1 + 2x^2+x^4) - 2x^2 = (1+x^2)^2 - (x\sqrt 2)^2 = (1+x^2+x\sqrt2)(1+x^2-x\sqrt 2)
$$
Then use partial fractions:
$$
\frac {Ax+B} {1+x^2+x\sqrt2} + \frac {Cx+D} {1+x^2-x\sqrt2}
$$
Notice that you can tell that these quadratic polynomials are irreducible because their discriminants $b^2-4ac = (\sqrt 2)^2 - 4\cdot1\cdot1$ are negative. You have to do a bit of algebra to find $A$, $B$, $C$, and $D$.
If you let $u=x^2+x\sqrt2+1$ then $du=(2x+\sqrt2)\,dx$ and
$$
\frac{Ax+B}{x^2+x\sqrt2+1} \text{ becomes }\frac{\frac A 2 (2x+\sqrt 2)}{x^2+x\sqrt2+1} + \frac{\text{some consant}}{x^2+x\sqrt2+1}
$$
The substitution above handles the first fraction. The second one is done by the substitution below:
\begin{align}
\int \frac {dx} {1+x^2+x\sqrt2} & = \int \frac{dx}{\left(x^2+x\sqrt2+\frac 1 2 \right) + \frac 1 2} \qquad(\text{completing the square}) \\[8pt]
& = \int \frac {dx}{\left(x + \frac 1 {\sqrt2}\right)^2 + \frac 1 2} = \int \frac{2\,dx}{\left(x\sqrt2 + 1\right)^2 + 1}.
\end{align}
Then use the substitution
\begin{align}
x\sqrt2+1 & = \tan\theta \\[6pt]
\text{so that }\left(x\sqrt2+1\right)^2 + 1 & = \sec^2\theta \\[6pt]
\text{and }\sqrt 2\,dx & = \sec^2\theta\,d\theta
\end{align}
and the integral above becomes
$$
\int \sqrt2\,d\theta = \sqrt 2\ \theta = \sqrt 2\ \arctan(x\sqrt2 + 1) +C
$$
etc.
Perhaps I should add that the first time I ever saw this integral I did it by a more straightforward method and consequently saw how to do what I did above, but some people don't like that more straightforward method because it involves complex numbers. That method is this: to factor $x^4+1$ we solve $x^4+1=0$, getting
\begin{align}
x^4 & = -1 \\[6pt]
x^2 & = \pm i
\end{align}
To find square roots of $i$, notice that $i = \cos90^\circ+i\sin90^\circ$, so its square roots must be $\pm(\cos45^\circ+i\sin45^\circ)= \pm\left(\frac{1+i}{\sqrt2}\right)$ and similarly square roots of $-i$ are $\pm(\cos45^\circ-i\sin45^\circ)= \pm\left(\frac{1-i}{\sqrt2}\right)$ Now we have
$$
x^4+1=\underbrace{\left(x-\frac{1+i}{\sqrt2}\right)\left(x-\frac{1-i}{\sqrt2}\right)}_\text{conjugates}\ \underbrace{\left(x-\frac{-1+i}{\sqrt2}\right)\left(x-\frac{-1-i}{\sqrt2}\right)}_\text{conjugates}.
$$
When conjugates are multiplied, the imaginary parts cancel out:
$$
x^4+1 = (x^2-x\sqrt2+1)(x^2+x\sqrt2 + 1).
$$
| {
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Proving a little tough trigonometric identity Show that $$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}$$ How do I get the $A-B$ term in the denominator? Is RHS to LHS easier? Thanks.
| First, notice that
$$
\frac{1 + \sin A}{\cos A} = \frac{\cos A}{1 - \sin A}
$$
Next, regroup the denomitator as
$$
\sin(A - B) + \cos A - \cos B = \cos A(1 - \sin B) - \cos B(1 - \sin A)
$$
The right hand side becomes
$$
2\frac{\sin A - \sin B}{\cos A(1 - \sin B) - \cos B(1 - \sin A)} =
2\frac{\frac{\tan A}{\cos B} - \frac{\tan B}{\cos A}}{\frac{1 - \sin B}{\cos B} - \frac{1 - \sin A}{\cos A}}.
$$
Multiplying both sides with ${\frac{1 - \sin B}{\cos B} - \frac{1 - \sin A}{\cos A}}$ we get
$$
\left(\frac{1 + \sin A}{\cos A} +
\frac{1 + \sin B}{\cos B}\right)
\left(\frac{1 - \sin B}{\cos B} - \frac{1 - \sin A}{\cos A}\right)
=
2\left(\frac{\tan A}{\cos B} - \frac{\tan B}{\cos A}\right)
$$
Expanding the left side gives
$$
\left(\frac{1 + \sin A}{\cos A} +
\frac{1 + \sin B}{\cos B}\right)
\left(\frac{1 - \sin B}{\cos B} - \frac{1 - \sin A}{\cos A}\right) = \\ =
\frac{1-\sin^2 B}{\cos^2 B} - \frac{1-\sin^2 A}{\cos^2 A}
+ \frac{(1+\sin A)(1-\sin B)-(1-\sin A)(1+\sin B)}{\cos A\cos B} = \\ =
\frac{2\sin A - 2\sin B}{\cos A\cos B} = 2\left(\frac{\tan A}{\cos B} - \frac{\tan B}{\cos A}\right). \qquad \blacksquare
$$
| {
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Solve $\int_{0}^{1} \log(x)\log(1-x) dx$ without convolution Maybe it's too much to ask for, but is there a way to solve $\int \limits_{0}^{1} \log(x)\log(1-x) dx$ without convolution?
Note that $\log x =\log_e x$.
| Step 1: We prove that for each $n\in \mathbb{N}$, $$\int_{0}^{1} x^n \log(x) \, dx = -\frac{1}{(n+1)^2}$$ Starting with the identity $$\int_{0}^{1} x^n \, dx = \frac{1}{n+1}$$ differentiate both sides with respect to $n$ to get the desired result.
Step 2: Expanding $\log(1-x)$ as its Taylor series, we have $$\log(x)\log(1-x) = -x\log(x) - \frac{x^2 \log(x)}{2} - \frac{x^3 \log(x)}{3} - \cdots$$ Sweeping issues of convergence under the rug, integrating both sides over the interval $[0,1]$ gives $$\int_{0}^{1} \log(x) \log(1-x) \, dx = \frac{1}{2^2} + \frac{1}{2\cdot 3^2} + \frac{1}{3\cdot 4^2} + \cdots$$
Step 3: We must evaluate the sum $$\sum_{k=1}^{\infty} \frac{1}{k(k+1)^2} = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1} - \frac{1}{(k+1)^2} \right)$$ $$ = \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{k+1}\right) - \sum_{k=1}^{\infty} \frac{1}{(k+1)^2} = 1 - \left( \frac{\pi^2}{6} - 1\right) = 2-\frac{\pi^2}{6}$$
| {
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How to evaluate $\int_{-\infty}^{\infty} \frac{1}{x^6+1}\,dx$? I am trying to understand how to evaluate the following integral.
$$\int_{-\infty}^{\infty} \dfrac{1}{x^6+1}dx$$
We start by considering $f(z) = \dfrac{1}{z^6+1}dz.$ Then we find that $z_k = \exp \left (i \left (\frac{\pi}{6}+\frac{\pi k}{3}\right)\right )$, and simple poles occur when $k=0,1,2,3,4,5$.
Then
$$\text{Res}(f,z_k)= \dots =\dfrac{1}{6}\lim_{z \to z_k} z^{-5}= \dfrac{1}{6} \exp \left( -\frac{5 \pi i}{6} (1 +2k) \right)$$
I think that in the missing part, the following formula is used,
$$\text{Res}(f,c)=\frac{1}{(n-1)!}\lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}} ((z-c)^n f(z)) $$
that $n=4$ and that a cancellation occurs (otherwise it would be really messy), but I'm not sure how to do it. How do I proceed?
Thanks for your help.
| We first resolve the integrand into partial fractions.
$$
\frac{1}{x^{6}+1}=\frac{1}{3\left(x^{2}+1\right)}-\frac{x^{2}-2}{3\left(x^{4}-x^{2}+1\right)}
$$
Integrating both sides yields
$$
3 \int_{0}^{\infty} \frac{d x}{x^{6}+1}=\int_{0}^{\infty} \frac{d x}{x^{2}+1}-\int_{0}^{\infty} \frac{x^{2}-2}{x^{4}-x^{2}+1} d x
$$
Obviously,
$$
\int_{0}^{\infty} \frac{d x}{x^{2}+1}=\left[\tan ^{-1} x\right]_{0}^{\infty}=\frac{\pi}{2}
$$
Now let’s deal with the latter one.
$$
\begin{aligned}
\int_{0}^{\infty} \frac{x^{2}-2}{x^{4}-x^{2}+1} d x = &\int_{0}^{\infty} \frac{1-\frac{2}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x
\\=& \frac{3}{2} \int_{0}^{\infty} \frac{\left(1-\frac{1}{x^{2}}\right) d x}{x^{2}+\frac{1}{x^{2}}-1}-\frac{1}{2} \int_{0}^{\infty} \frac{\left(1+\frac{1}{x^{2}}\right) d x}{x^{2}+\frac{1}{x^{2}}-1} \\
=& \frac{3}{2} \int_{0}^{\infty} \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-3}-\frac{1}{2} \int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1} \\
=& \frac{\sqrt3}{4}\left[\ln \left|\frac{x+\frac{1}{x}-\sqrt3}{x+\frac{1}{x}+\sqrt3}\right|\right]_{0}^{\infty}-\frac{1}{2}\left[\tan ^{-1}\left(x-\frac{1}{x}\right)\right]_{0}^{\infty} \\
=&-\frac{\pi}{2}
\end{aligned}
$$
Hence
$$
3 \int_{0}^{\infty} \frac{d x}{x^{6}+1}=\frac{\pi}{2}+\frac{\pi}{2}=\pi
$$
Now we can conclude that
$$
\boxed{\int_{-\infty}^{\infty} \frac{d x}{x^{6}+1}=\frac{2 \pi}{3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove/disprove an inequality $\sqrt{1 + x} < 1 + \frac{x}{2} - \frac{x^2}{8}$ Is $\sqrt{1 + x} < 1 + \frac{x}{2} - \frac{x^2}{8}$ TRUE in $(0, \frac{π}{2})$ ?
I proceed by taking two functions $f(x) = \sqrt{1 + x}-1 $ and $g(x) = \frac{x}{2} - \frac{x^2}{8} $.
Then $f(0) = 0 = g(0),$ while $f(\frac{π}{2}) > g (\frac{π}{2})$.
Also both functions are strictly increasing in $(0, \frac{π}{2})$.
I found nothing else. Does it help?
| $$
\begin{align}
\sqrt{1+x}-(1+\frac x2-\frac{x^2}8)
&=\dfrac{1+x-(1+\frac x2-\frac{x^2}8)^2}{\sqrt{1+x}+(1+\frac x2-\frac{x^2}8)}\\
&=\dfrac{\frac{x^3}{64}(8-x)}{\sqrt{1+x}+(1+\frac x2-\frac{x^2}8)}\\
&=\dfrac{x^3}{8x+32-\frac{64}{3+\sqrt{1+x}}}
\end{align}
$$
The denominator is monotonically increasing, and at $x=0$, it is $16$. Thus, for $x\ge0$,
$$
\sqrt{1+x}\ge1+\frac x2-\frac{x^2}8
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving $9^n - 8n - 1$ is divisible by $8^2$ for $n\ge 0$? My textbook provided the following proof:
*Base case: When $n=0, 9^n-8n-1=0=64\cdot0$, so $64\mid\left(9^n-8n-1\right)$.
Induction step: Suppose that $n\in\mathbb N$ and $64\mid\left(9^n-8n-1\right)$. Then there is some integer $k$ such that $9^n-8n-1=0=64k$. Therefore:
$$\begin{align}
9^{n+1}-8(n+1)-1&=9^{n+1}-8n-9\\
&=9^{n+1}-72n-9+64n\\
&=9\left(9^n-8n-1\right)+64n\\
&=9(64k)+64n\\
&=64(9k+n)
\end{align}$$
so $64\mid\left(9^{n+1}-8(n+1)-1\right)$.
What I don't understand is how the equation goes from
$$= 9^{n+1} - 8n - 9$$
to
$$= 9^{n+1} - 72n - 9 + 64n$$
| Express $8n$ as $72n-64n$. We hence have
$$9^{n+1} - {\color{red}{8n}} - 9 = 9^{n+1} - \color{red}{(72n-64n)} - 9 = 9^{n+1} \color{red}{- 72n + 64n} - 9 = 9^{n+1} - \color{red}{72n} - 9 + \color{red}{64n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proof of inequality problem from Spivak, ch 1, 16 b) If $$4x^2+8xy+4y^2 \ge 0$$ it follows that $$4x^2+6xy+4y^2 \gt 0 $$ unless $x=0$ and $y=0$.
How can I prove that?
| The way that I think of it is as follows. First, we need to recognise that
$$ 4x^2+6xy+4y^2=3(x+y)^2 + x^2 + y^2$$
Then, based on the stated assumption we know that:
$$ 4x^2+8xy+y^2=4(x+y)^2 \ge 0 $$
$$ (x+y)^2\ge0$$
$(x+y)^2 = 0$ when $x=-y$ or when $x=y=0$
For all other values $(x+y)^2\gt0$
Similarly,
$$ x^2 + y^2 \ge 0$$
$x^2 + y^2 = 0$ where $x=y=0$
For all other values, $x^2+y^2\gt0$
Combining these two steps we find that where $x$ and $y\not=0$:
$$3(x+y)^2\ge0\; and\; x^2 + y^2 \gt0$$
$$\therefore\;3(x+y)^2+x^2+y^2\gt0$$
$$4x^2+6xy+4y^2\gt0$$
P.S. I'm just working through this chapter myself, so my understanding is pretty basic. I'm sure others have expressed this more elegantly, but hopefully my fumbling thought process will be helpful for other beginners.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Simplify the expression (combination and factorial) Simplify the following expression:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
My attempt:
$\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!} = \frac{(n+1)!}{3!(n-2)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$
and this is where I get stuck... How to continue?
When I put $\binom{n+1}{3} * \frac{(n-1)! + (n-2)!}{(n+1)!}$ into Wolfram Alpha it simplifies it into: $\frac{n}{6}$
When I put $\frac{(n+1)!}{3!(n+1-3)!} * \frac{(n-1)! + (n-2)!}{(n+1)!}$ into Wolfram Alpha it simplifies it into: $\frac{1}{6} * (n^{3} - n +1)$
| $$n/6$$
as ${n+1\choose 3}$ can be reduced to $(n+1)(n)(n-1)/3!$, now $(n+1)(n)(n-1)$ is canceled from the denominator and the $(n-2)!$ is canceled too so we are left with $(n-1) + 1$ numerator and $3!$ denominator. So the answer is $n/6 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove by induction that $1+4+7+...+(3n-2) = 2n(3n-1)$ I have an exercise where I, using induction, have to prove the following:
\begin{equation*}
1 + 4 + 7 + \ldots + (3n-2) = 2n(3n-1).
\end{equation*}
I immediately got stuck on the base case with $n=1$ because the following should be true: $1 = 2 \cdot 1 \cdot (3 \cdot 1 - 1) = 2 \cdot 2$ which is clearly not the case. However the exercise says to prove the given relation, not to check if it is correct, which is why I have elected to make this post to see if I am missing any obvious thing.
| First of all, the claim is wrong. A direct proof shows that.
The sequence $\{1,4,7,\ldots,3n-2\}$ is an arithmetic progression with common difference $3$ and there are $n$ number of terms in it. So, the sum of all the terms is:
$$\begin{align}S =1+4+7+\ldots+(3n-2)&=1+(1+3)+(1+6)+\ldots+(1+3(n-1))&\\ &=\sum_{i=1}^n(1)+\sum_{i=1}^{n-1}(3i)\\&=n+3\cdot\frac{(n-1)n}{2}=\frac{3n^2-n}{2}=\frac{n(3n-1)}{2}\neq 2n(3n-1)\end{align}$$
The "$\neq$" follows in general. Now, if you were to prove the corrected claim using induction, here's a hint for the inductive step:
$$\begin{align}S_{n+1} & =1+4+7+\ldots+(3n-2)+(3n+1)\\
&=S_n+(3n+1)\\
&=\frac{n(3n-1)}{2}+(3n+1)\tag{by I.H.}\\&=\frac{3n^2-n+6n+2}{2}\\&=\frac{3n^2+5n+2}{2}\\&=\frac{(n+1)(3n+2)}{2}\\
& =\ldots\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the eigenvectors corresponding to an eigenvalue I know how to find the eigenvectors corresponding to an eigenvalue of a matrix $A$: we basically need to find the vectors of the nullspace of $\lambda I - A$, but in my case, I have a matrix $A$ like this:
$$A = \left(\begin{matrix} \cos x & -\sin x \\ \sin x & \cos x\end{matrix}\right) \left(\begin{matrix} 3 & 0 \\ 0 & 7\end{matrix}\right) \left(\begin{matrix} \cos x & \sin x \\ -\sin x & \cos x\end{matrix}\right)$$
and I cannot multiply the individual component matrices.
I have found the eigen values $\lambda = 3$ and $\lambda = 7$ for the matrix $A$, now how do I found the eigen vectors, without multiplying $A$ first?
| Eigenvector for eigenvalue 3
Given that
$$
\left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right)
=
3
\left( \begin{array}{c} 1 \\ 0 \end{array} \right),
$$
we can also write
$$
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right)
=
3
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right),
$$
or
$$
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right)
\left( \begin{array}{cc} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right)
=
3
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right),
$$
or using $A$:
$$
A
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right)
=
3
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right),
$$
Meaning the the eigenvector for the eigenvalue $3$ is given by
$$
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 1 \\ 0 \end{array} \right)
=
\left( \begin{array}{c} \cos(x) \\ \sin(x) \end{array} \right)
$$
Eigenvector for eigenvalue 7Given that
$$
\left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right)
\left( \begin{array}{c} 0 \\ 1 \end{array} \right)
=
7
\left( \begin{array}{c} 0 \\ 1 \end{array} \right),
$$
we can also write
$$
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right)
\left( \begin{array}{c} 0 \\ 1 \end{array} \right)
=
7
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 0 \\ 1 \end{array} \right),
$$
or
$$
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right)
\left( \begin{array}{cc} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 0 \\ 1 \end{array} \right)
=
7
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 0 \\ 1 \end{array} \right),
$$
or using $A$:
$$
A
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 0 \\ 1 \end{array} \right)
=
3
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 0 \\ 1 \end{array} \right),
$$
Meaning the the eigenvector for the eigenvalue $7$ is given by
$$
\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x)
\end{array} \right)
\left( \begin{array}{c} 0 \\ 1 \end{array} \right)
=
\left( \begin{array}{c} -\sin(x) \\ \cos(x) \end{array} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
find an expression for $A^n$ for any positive integer N I have posted part of this question in another post which has gotten to long. I need help with the second part of the question which is finding an expression for $A^n$ for any positive integer $n$.
The previous question gave eigenvalues of $λ_1 =-\sqrt{2}, λ_2 =\sqrt{2} , λ_3 = 0$
and the corresponding eigenvectors are:
$$
v_1=\begin{pmatrix}
1 \\ 1 \\ -\sqrt 2
\end{pmatrix},\quad
v_2=\begin{pmatrix}
1 \\ 1 \\ \sqrt 2
\end{pmatrix},\quad
v_3=\begin{pmatrix}
1 \\ -1 \\ 0
\end{pmatrix}.
$$
The other question is here:
linear algebra eigenvalues/vectors and finding an expression for $A^n$
Which show the original question and previous working out. Any help would be much appreciated
| I see you have the eigenvectors and the eigenvalues of matrix $A$... So build a unitary matrix $U$ (in your case the matrix $U$ is orthogonal), with columns the eigenvectors of $A$. That is, $U=\begin{pmatrix}\ \frac{1}{2} &\frac{1}{2} &\frac{\sqrt{2}}{2}\\ \frac{1}{2} &\frac{1}{2} &\frac{-\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} &\frac{-\sqrt{2}}{2} &0 \end{pmatrix}.$ Then, $A^{n}=U^{*}\Lambda^{n}U=U^{T}\Lambda^{n}U$, as $U\in\mathbb{R^{n,n}}$. Matrix $\Lambda$ is a diagonal matrix with entries the eigenvalues of matrix $A$. So, $\Lambda^{n}=\begin{pmatrix} (-\sqrt{2})^n & 0 & 0\\ 0 & (\sqrt{2})^{n} & 0 \\ 0 & 0 & 0 \end{pmatrix}$. Note that the matrix $A$ is diagonalizable, but it is not nonsingular (invertible).
Right now i saw that $A=\begin{pmatrix} x & 0 & 1\\ 0 & x & 1 \\ 1 & 1 & x \end{pmatrix}$ and that you can write, $A=B+xI$. So $A^{n}=U^{T}\Lambda^{n}U$, but $\Lambda^{n}=\begin{pmatrix} (-\sqrt{2}+x)^n & 0 & 0\\ 0 & (\sqrt{2}+x)^{n} & 0 \\ 0 & 0 & x^{n} \end{pmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A limit with the harmonic series How can we prove the following (similar) limits?
$$\sum_{k=1}^n \frac{1}{k} (\ln 2 - \frac{1}{n+2} - \frac{1}{n+3} - \cdots -\frac{1}{2n + 2}) \to 0. $$
$$\sum_{k=1}^n \frac{1}{k} (\ln 3 - \frac{1}{n+2} - \frac{1}{n+3} - \cdots -\frac{1}{3n + 3}) \to 0. $$
| We have $$\sum_{k\leq n}\frac{1}{k}\left(\log\left(2\right)-\frac{1}{n+2}-\dots-\frac{1}{2n+2}\right)=H_{n}\left(\log\left(2\right)-\sum_{j=n+2}^{2n+2}\frac{1}{j}\right)=H_{n}\left(\log\left(2\right)-H_{2n+2}+H_{n+1}\right)
$$ where $H_{n}
$ is the $n
$-th armonic number. Recalling that $$H_{n}=\log\left(n\right)+\gamma+O\left(\frac{1}{n}\right)
$$ as $n\rightarrow\infty
$ we have $$H_{n}\left(\log\left(2\right)-H_{2n+2}+H_{n+1}\right)=\left(\log\left(n\right)+\gamma+O\left(\frac{1}{n}\right)\right)\left(\log\left(2\right)-\log\left(2n+2\right)+\log\left(n+1\right)+O\left(\frac{1}{n}\right)\right)=
$$ $$=\left(\log\left(n\right)+\gamma+O\left(\frac{1}{n}\right)\right)O\left(\frac{1}{n}\right)=O\left(\frac{\log\left(n\right)}{n}\right)
$$ then your claim. The other case is similar. In fact you will get $$\left(\log\left(3\right)-\log\left(3n+3\right)+\log\left(n+1\right)+O\left(\frac{1}{n}\right)\right)=\left(\log\left(\frac{3n+3}{3n+3}\right)+O\left(\frac{1}{n}\right)\right)=O\left(\frac{1}{n}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\alpha$ lies between $0$ and $4$.
Let $a,b,c$ be the length of the sides of the triangle $ABC$ .
Given $(a+b+c)(b+c-a)=\alpha bc$.Then Prove that the value of $\alpha$ lies in between $0$ and $4$.
$\begin{align}(a+b+c)(b+c-a)&=\alpha bc\\
\implies \alpha&=\dfrac{b^2+c^2-a^2}{bc}+2\\
\alpha&=2\cos A+2\\
\end{align}$
I also know a relation like $a+b>c\\b+c>a\\a+c>b$
I have studied maths up to $12th$ grade.
| Note that in the last line you should get $2\cos A+2$, not $\frac{\cos A}{2}+2$. Now you can easily use the trivial bound $-1\leq\cos A\leq 1$ (indeed, a strict inequalities hold, as $A$ is an angle of a triangle).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Need help with taylor series.
Evaluate the limit
$$\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}$$
The limit im trying to get is $-\frac{1}{π^2}$ as I've solved from l'Hopitals rule.
Now I need to solve the limit by using Taylor Series and this is what i did so far
$$\begin{align*}
f(x) &= 1-x + \ln x = 1 -x + (x-1) + \frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \frac{1}{4} (x-1)^4 + \ldots \\
g(x) &= 1+\cos πx = 1+\left[ 1+\frac{1}{2!} (πx)^2 + \frac{1}{4!} (πx)^4 - \frac{1}{6!} (πx)^6 +\ldots \right] \\
\frac {f(x)}{g(x)} & = \frac{\frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \frac{1}{4}(x-1)^4 + \ldots} {2-\frac{1}{2!} (πx)^2 + \frac{1}{4!} (πx)^4 - \frac{1}{6!} (πx)^6+\ldots}
\end{align*}$$
I have no idea where to go to solve for $-\frac{1}{π^2}$ now. Please help
| Hint. You have, near $x=1$,
$$1-x + \ln x = 1 -x + (x-1) -(x-1)^2/2+ O(x-1)^3$$
$$1-x + \ln x = -(x-1)^2/2+ O(x-1)^3$$
and
$$1+\cos πx = 1 -1+\frac{\pi^2}2 (x-1)^2 + O(x-1)^3$$
$$1+\cos πx = \frac{\pi^2}2 (x-1)^2 + O(x-1)^3$$ thus
$$\frac{1-x + \ln x}{1+ \cos πx} =\frac{-(x-1)^2/2+ O(x-1)^3}{\frac{\pi^2}2 (x-1)^2 + O(x-1)^3}=\frac{-1/2+ O(x-1)}{\pi^2/2 + O(x-1)}=-\frac{1}{\pi^2}+ O(x-1)$$
then
$$\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}=-\frac{1}{\pi^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Mean Value Theorem to show inequalities about numbers
Show $\sqrt{65}-8=\frac{1}{2\sqrt{c}}$ for some $c \in (64,65)$, and hence show:
$$8+\frac{1}{18}<\sqrt{65}<8+\frac{1}{16}$$
I managed to do the first part easily, but I don't know how to do the "hence show" part.
Working For the First Part:
Consider the function $f(x)=x^{\frac{1}{2}}$ on $[64,65]$. Then f is continuous on $[64,65]$ and differentiable on $(64,65)$. By the mean value theorem, there is a $c \in (64,65)$ such that:
$$f'(c)=\frac{f(65)-f(64)}{65-64}$$
$f'(x)=\frac{1}{2\sqrt{x}}$, so we have:
$$\frac{1}{2\sqrt{c}}=\sqrt{65}-8$$
| The first part looks good.
From the first part we know, that there exists a $c \in (64, 65)$, such that
$$\sqrt{65} = 8 + \frac{1}{2\sqrt{c}} \; .$$
Since $64 < c < 81$ and the square root is a monotone function, we see that $$8 < \sqrt{c} < 9 \; .$$
From this it follows that
$$ \frac{1}{18} = \frac{1}{2\cdot 9} < \frac{1}{2\sqrt{c}} < \frac{1}{2\cdot 8} = \frac{1}{16} \; ,$$
so by adding $8$ we get
$$ 8 + \frac{1}{18} < \underbrace{8 + \frac{1}{2\sqrt{c}}}_{= \sqrt{65}} < 8 + \frac{1}{16} \; .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Finding $\int\frac{1}{x^{11}+4x^6}dx$ I wanted to find out if there is an easy way to evaluate $$\displaystyle\int\frac{1}{x^{11}+4x^6}dx.$$
I substituted $u=x^5$ and then used partial fractions, but maybe there is a simpler way to find this.
| Let $t=\frac{1}{x}$, so then $x=\frac{1}{t}$ and $dx=-\frac{1}{t^2}dt$.
Then $\displaystyle\int\frac{1}{x^{11}+4x^6}dx=\int\frac{1}{\frac{1}{t^{11}}+\frac{4}{t^6}}\left(-\frac{1}{t^2}\right)dt=-\int\frac{t^9}{1+4t^5}dt$
$=-\int\big(\frac{1}{4}t^4-\frac{\frac{1}{4}t^4}{1+4t^5}\big)dt=-\frac{1}{20}t^5+\frac{1}{80}\ln\vert1+4t^5\vert+C=-\frac{1}{20x^5}+\frac{1}{80}\ln\vert1+\frac{4}{x^5}\vert+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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A not so hard basic calculus problem? But it appears to be very lengthy Find the coordinates of the two points on the curve $y=4-x^2$ whose tangents pass through the point $(-1,7)$.
My work:
Let the two points be $(a,b)$ and $(c,d)$. And $\frac{dy}{dx}=-2x$, so the gradients of the tangents are $-2a$ and $-2c$. We get $\frac{7-d}{-1-c}=-2c\implies 7-d=2c+2c^2$ and $\frac{7-b}{-1-a}=-2a\implies 7-b=2a+2a^2$. Now we have 2 equations and 4 unknowns. So we need two more equations.
We have the equations of the tangents are $y=-2cx+2c^2+d$ and $y=-2ax+2a^2+b$. The solutions of these two equations are $(-1,7)$, putting $(-1,7)$ in we get $7(a-c)=-2ac(a-c)+ad-bc$ and $2(a-c)=-2(a-c)(a+c)+d-b$.
We can also equate $4-x^2=-2cx+2c^2+d$ and $4-x^2=-2ax+2a^2+b$, then we get $a^2-c^2=d-b$. From here, we still have to substitute and do some algebra to get $(a,b)$ and $(c,d)$.
Is my working correct? But I feel like the working is too lengthy. Is there any faster way?
Many thanks!
| Your work is correct. Maybe you can shorten it by assuming that the points of tangency is $P=(t,4-t^2)$. Then slope at $P$ is $m_P=-2t$. Slope of line $LP=\dfrac{7-(4-t^2)}{-1-t}=\dfrac{3+t^2}{-(1+t)}$. Now this must be equal to $m_P$.
$$\dfrac{3+t^2}{-(1+t)}=-2t.$$
Solving this gives
$$t^2+2t-3=0.$$
Now solve for the two values of $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int \frac{\tan^4 \theta \,d \theta}{1-\tan^2 \theta}$? I have to evaluate this indefinite integral $$\int \frac{\tan^4 \theta\, d \theta}{1-\tan^2 \theta}$$
I tried it as follows $$I=\int\frac{(\sec^2 \theta-1)\tan^2 \theta\, d \theta}{1-\tan^2 \theta}=\int\frac{\sec^2 \theta \tan^2 \theta\, d \theta}{1-\tan^2 \theta}-\int\frac{\tan^2 \theta \,d \theta}{1-\tan^2 \theta}$$
First part of integration can be easily solved by substitution but how to solve the second part? Help to solve it by other method if you have. Thanks!
| Here is another way to proceed:
$\displaystyle\int\frac{\tan^4\theta}{1-\tan^2\theta}d\theta=\int\frac{\tan^4\theta-1}{1-\tan^2\theta}d\theta+\int\frac{1}{1-\tan^2\theta}d\theta=-\int(\tan^2\theta+1)d\theta+\int\frac{\cos^2\theta}{\cos^2\theta-\sin^2\theta}d\theta$
$=\displaystyle-\int\sec^2\theta \;d\theta+\int\frac{\frac{1}{2}(1+\cos 2\theta)}{\cos 2\theta}d\theta= -\tan\theta+\int\left(\frac{1}{2}\sec2\theta+\frac{1}{2}\right)d\theta$
$=\displaystyle-\tan\theta+\frac{1}{4}\ln\big|\sec2\theta+\tan2\theta\big|+\frac{1}{2}\theta+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Finding an angle which satisfies two equations I'd like to prove the following: Given any two real numbers $a$ and $b$, not both zero, there exists $c \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin c = \frac{a}{\sqrt{a^2 + b^2}}$ and $\cos c = \frac{b}{\sqrt{a^2 + b^2}}$. I'm trying to use as little geometry as possible, and appeal to the formal properties of sine and cosine.
So far what I have is this:
By hypothesis, $\sqrt{a^2 + b^2} \neq 0$. Thus, $\frac{a}{\sqrt{a^2 +b^2}} \in [-1,1]$,
and it follows from the intermediate value theorem
that there exists some $c \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ such
that $\sin c = \frac{a}{\sqrt{a^2 + b^2}}$. But $\sin^2 c + \cos^2 c = 1$
, which implies
$$ \cos^2 c = 1 - \frac{a^2}{a^2 + b^2} = \frac{b^2}{a^2 + b^2}. $$
It follows that $\cos c = \pm \frac{b}{\sqrt{a^2 + b^2}}$.
Of course, this is almost what I want, but is there a way for me to rule out the negative case?
Thanks in advance for any help!
| Your question is a trick question.
It is impossible to prove such a claim because it is not true. If you really want to prove the truthness of such a claim I'd start with the claim that for any real numbers $a$ and $b$ there exists an angle $\theta$ such that $\cos{\theta} = \frac{a}{\sqrt{a^2 + b^2}}$, $\sin{\theta} = \frac{b}{\sqrt{a^2 + b^2}}$, and $ \theta \in [\frac{\pi}{2}, \frac{3\pi}{2}]$
A case where the claim you're trying to prove does not hold is when $a=-1$ and $b=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to proceed with Euclidean algorithm for finding greatest common divisor of two polynomials. I am trying to find
\begin{equation*}
gcd(x^4-x^3-4x^2-x+5,x^2+x-2).
\end{equation*}
I have done the first step of long division and found.
\begin{equation*}
x^4-x^3-4x^2-x+5=(x^2-2x)(x^2+x-2)-5x+5
\end{equation*}
so we have
\begin{equation*}
gcd(x^4-x^3-4x^2-x+5,x^2+x-2)=gcd(x^2+x-2,-5x+5)
\end{equation*}
now is where I am stuck. For the next step do we need to divide $x^2+x-2$ by $-5x+5$ or can we simply divide $x^2+x-2$ by $-x+1$ since the $gcd$ needs to a monomial?
Also if I did divide by $-x+1$ instead of $-5x+5$ would this change my procedure at all when reversing the algorithm to find the polynomials that multiply $(x^4-x^3-4x^2-x+5,x^2+x-2)$ to give $gcd(x^4-x^3-4x^2-x+5,x^2+x-2)$. (Bezout's lemma)
| There is a theorem which states that if $f $ is a polynomial and $ f(a) =0 $ then $x-a$ is a factor of $f$, another theorem states that if $f$ is a polynomial the possible zeros of $ f$ are factors of $c/l$ where $l $is the leading coefficient and $ c$ is the constant. The two polynomials suppose $f(x)= x^4 - x^3 -4x^2-x +5 , g(x)=x^2+x-2$ you can see that $f(1)=0, g(1)=0$ hence $x-1$ is a factor for both. And no other factors remains between both the gcd is $x-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
Product of two integers of the form $x^2+my^2$ is of the same form. Let $x,y,a,b\in \mathbb Z$. Prove that there are integers $c$ and $d$
so that
\begin{equation*}
(x^2+y^2m)(a^2+b^2m)=c^2+d^2m.
\end{equation*}
I'm stuck, I took the product and got $x^2a^2+b^2y^2m^2+(a^2y^2+x^2b^2)m$ but the numbers in the parenthesis need not be squares. Thank you in advance.
| $$\begin{align}
(x^2+y^2m)(a^2+b^2m) &= x^2a^2 +mx^2b^2 +m^2y^2b^2+my^2a^2 \\
&= x^2a^2 +m^2y^2b^2 +m(x^2b^2 +y^2a^2) \\
&= x^2a^2 \color{red}{ +2mxyab } +m^2y^2b^2 +m(x^2b^2 \color{red}{ -2xyab }+y^2a^2)\\
&= (xa + myb)^2 + m(xb-ya)^2
\end{align}$$
Giving $c=(xa + myb)$ and $d=(xb-ya)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
taking the limit $\lim\limits_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$ I need help with a guide on how i will deal with this kind of problem.. This a part of my solution in series convergence. I find it hard taking the limit of this: $$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)} }$$
| There are a couple of properties of limits you can abuse here. The one that should pop out first is $\lim f(x)g(x) = \lim f(x) \cdot \lim g(x)$.
$\lim_{n\rightarrow \infty} {\frac{(3^{n+1} + 4)(7^n-47)}{(7^{n+1}-47)(3^n +4)}} = \lim_{n\rightarrow \infty} {\frac{3^{n+1}+4}{3^n+4}} \cdot \lim_{n\rightarrow \infty} {\frac{7^n-47}{7^{n+1}-47}} = 3 \cdot \frac{1}{7} = \frac{3}{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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if $4^{\alpha} \equiv k+1 \pmod{2k+1}$ prove there is no $\beta$ where $4^{\beta} \equiv k\pmod{2k+1}$. Suppose that $3 \nmid 2k+1$ and there is $\alpha$ with $4^{\alpha} \equiv k+1 \pmod{2k+1}$ where $0 \leq \alpha \leq k$. I want to prove that there is no $\beta$, $0\leq \beta \leq k$ such that $4^{\beta} \equiv k \pmod{2k+1}$. I have used the contradiction method but there was no result, it will be nice if you help me about it with your ideas, thanks.
| Let $m$ be the order of $2$ modulo $2k+1$, $m = \min \bigl\{ n\in \mathbb{N}\setminus \{0\} : 2^n \equiv 1 \pmod{2k+1}\bigr\}$.
If $2^{2\alpha} = 4^\alpha \equiv k+1 \pmod{2k+1}$, then $2^{2\alpha+1} \equiv 1 \pmod{2k+1}$, so $m\mid (2\alpha + 1)$, hence $m$ is odd.
If on the other hand $2^{2\beta} = 4^\beta \equiv k \pmod{2k+1}$, then $2^{2\beta + 1} \equiv -1 \pmod{2k+1}$ and $2^{2(2\beta+1)} \equiv 1 \pmod{2k+1}$. So then $m \mid 2(2\beta+1)$ but $m\nmid (2\beta+1)$, and that implies $2\mid m$, i.e. $m$ is even.
Thus at most one of the congruences $4^\alpha \equiv k+1 \pmod{2k+1}$ and $4^\beta \equiv k \pmod{2k+1}$ can be solvable for any $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show equality of a given function with a series in $ℝ$ Show that: $$2x\cos x-\sin x=4\sum_{n=2}^\infty \frac{(-1)^n}{n^2-1}\sin(nx)$$
Supposedly, this can be proved by using Fourier series, by choosing the right function but I have been thus far unable to do so. Any hints will be welcome, thanks in advance!
| By using the Fourier sine series
\begin{align}
f(x) &= \sum_{n=1}^{\infty} B_{n} \, \sin(nx) \hspace{15mm} x \in [0,\pi] \\
B_{n} &= \frac{2}{\pi} \, \int_{0}^{\pi} f(x) \, \sin(nx) \, dx
\end{align}
the function $f(x) = a x \cos(x) + b \sin(x)$ can be seen to be represented by
\begin{align}
a x \cos(x) + b \sin(x) = \frac{2b - a}{2} + 2a \, \sum_{n=2}^{\infty} \frac{(-1)^{n} \, n}{n^{2} -1} \, \sin(nx).
\end{align}
This is obtained by making use of
\begin{align}
B_{1} &= \frac{2}{\pi} \, \int_{0}^{\pi} (ax \cos(x) + b \sin(x)) \, \sin(x) \, dx = \frac{2b-a}{2}
\end{align}
and calculating the general case for $n \geq 2$. Now choosing $a$ and $b$ as
$a=2$ and $b=1$ then the reduction provides
\begin{align}
2 x \cos(x) + \sin(x) = 4 \, \sum_{n=1}^{\infty} \frac{(-1)^{n} \, n}{n^{2}-1} \, \sin(n x).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}$ $$\lim_{x\to 0}\frac{\sin x-x\cos x}{x^3}$$
How do I go about doing this? I can see no simple way of doing this. Application of l'Hopital's rule would be very laborious. A Taylor expansion seems feasible but is that the best way? It seems like it may be also very laborious.
| I may use classic limits (which can easily be computed by hospital's rule
(twice)
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}} &=&\frac{1}{2}
\end{eqnarray*}
The idea is to write the original expression using those above,
\begin{equation*}
\frac{\sin x-x\cos x}{x^{3}}=\frac{\sin x-x+x-x\cos x}{x^{3}}=\frac{\sin x-x%
}{x^{3}}+\frac{1-\cos x}{x^{2}}
\end{equation*}
It follows that
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x-x\cos x}{x^{3}}=\lim_{x\rightarrow 0}\frac{%
\sin x-x}{x^{3}}+\lim_{x\rightarrow 0}\frac{1-\cos x}{x^{2}}=-\frac{1}{6}+%
\frac{1}{2}=\frac{1}{3}.
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simple argument for $\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$ I would like to show that $\forall x,y\in\mathbb R^+:\frac{(x+y)^2}{x^2+xy+y^2}\le 4/3$.
The inequality is indeed true as the maximum of $\frac{(x+y)^2}{x^2+xy+y^2}$ is reached for $x=y$ and its value is $4/3$.
Except for the standard way of computing partial derivatives and finding the maximum, is there a simple argument that imply this inequality (perhaps using symmetry somehow?).
Thanks !
| $$\frac{(x+y)^2}{x^2+xy+y^2}=1+\frac{xy}{x^2+xy+y^2}\le 1+\frac{1}{3}$$
$$\iff 3xy\le x^2+xy+y^2\iff (x-y)^2\ge 0,$$
which is true, with equality iff $x=y$.
Note we could multiply by $x^2+xy+y^2$ without flipping ineq sign because it equals $\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
integer as sum of three binomials Prove that for any nonnegative integer $n$ $\exists x,y,z \in \mathbb{N}$ and $0\leq x<y<z$ so
$$n=\binom{x}{1}+\binom{y}{2}+\binom{z}{3}$$
Please give me a hint, I don't have any idea.
| Easy to see that
$$
1 = {x_1 \choose 1}+{y_1 \choose 2}+{z_1 \choose 3}
$$
where $x_1 = 0$, $y_1 = 1$, $z_1 = 3$.
Now suppose
$$
i = {x_i \choose 1}+{y_i \choose 2}+{z_i \choose 3}
$$
where $0 \leq x_i < y_i < z_i$. We want to decide $0 \leq x_{i+1} < y_{i+1}<z_{i+1}$ such that
$$
i + 1 = {x_{i+1} \choose 1}+{y_{i+1} \choose 2}+{z_{i+1} \choose 3}
$$
I provide a procedure to decide $x_{i+1}, y_{i+1}, z_{i+1}$ below.
*
*if $x_i < y_i - 1$
Let $x_{i+1} = x_i + 1, y_{i+1} = y_i, z_{i+1} = z_i$. Easy to see that the sum increment $1$.
*
*if $x_i = y_i - 1\textbf{ and }y_i < z_i - 1$
Let $x_{i+1} = 0, y_{i+1} = y_i + 1, z_{i+1} = z_i$. We have
\begin{align}
&{z_{i+1} \choose 3} + {y_{i + 1} \choose 2}+{x_{i+1} \choose 1} - {z_i \choose 3} - {y_i \choose 2}-{x_i \choose 1}\\
=& {z_i \choose 3} + {y_i + 1 \choose 2}+{0 \choose 1} - {z_i \choose 3} -{y_i \choose 2} - {y_i - 1 \choose 1} \\
=& \frac{(y_i + 1)y_i}{2} + 0 - \frac{y_i(y_i - 1)}{2} - (y_i - 1)\\
=& 1
\end{align}
*
*if $x_i = y_i - 1\textbf{ and }y_i = z_i - 1$
Let $x_{i+1} = 0, y_{i+1} = 1, z_{i+1} = z_i + 1$. We have
\begin{align}
&{z_{i+1} \choose 3} + {y_{i + 1} \choose 2}+{x_{i+1} \choose 1} - {z_i \choose 3} - {y_i \choose 2}-{x_i \choose 1}\\
=& {z_i + 1 \choose 3} + {1 \choose 2}+{0 \choose 1} - {z_i \choose 3} -{z_i - 1 \choose 2} - {z_i - 2 \choose 1} \\
=& \frac{(z_i + 1)z_i(z_i - 1)}{6} + 0 + 0 - \frac{z_i(z_i-1)(z_i-2)}{6} - \frac{(z_i - 1)(z_i - 2)}{2} - (z_i - 2)\\
=& \frac{z_i(z_i - 1)}{2} - \frac{(z_i-1)(z_i - 2)}{2} - (z_i - 2)\\
=& 1
\end{align}
Note that the procedure guarantees $0 \leq x_{i+1} < y_{i+1} < z_{i+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.