Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to factor $r^6 -3r^4 +3r^2 - 1 = 0$ I tried two ways to factor $r^6 -3r^4 +3r^2 - 1 = 0$
When I factor $r^4$ out of $r^6 -3r^4$:
$r^4(r^2-3)+(3r^2-1) = 0$
When I factor $r^2$ out of $r^6 + 3r^2$:
$r^2(r^4+3)-(3r^4 + 1) = 0$
For both methods, I'm stuck at an equation that isn't factorable. But Wolfram Alpha says $r^6 -3r^4 +3r^2 - 1 = (r-1)^3(r+1)^3$. How do you get this?
| If you don't know the binomial theorem, you might rearrange the terms:
\begin{align*}
r^6-3r^4+3r^2-1 &= r^6-1-3(r^4-r^2) \\
&= (r^2-1)(r^4+r^2+1)-3r^2(r^2-1) \\
&= (r^2-1)(r^4+r^2+1-3r^2) \\
&= (r^2-1)(r^4-2r^2+1) \\
&= (r^2-1)(r^2-1)^2 \\
&= (r^2-1)^3 \\
&= (r+1)^3(r-1)^3
\end{align*}
Useful identities:
\begin{align*}
a^2 \pm 2ab+b^2&= (a \pm b)^2 \\
a^2-b^2 &= (a+b)(a-b) \\
a^3 \pm b^3 &= (a \pm b)(a^2 \mp ab + b^2)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$ Finding $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$
Attempt: Assume $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx =\int \frac{2x-x^{-2}}{\bigg[x^4-x^2+2x-\frac{2}{x}-\frac{1}{x}+\frac{1}{x^2}\bigg]}dx$
could some help me how to solve,thanks
| As Grant B. commented
$$x^6-x^4+2x^3-2x^2-x+1=(x-1)(x^2+x-1)(x^3+x+1)$$ Then, Using partial fraction decomposition,
$$\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}=\frac 13 \left(\frac{2 x+1}{ x^2+x-1}-\frac{3 x^2+1}{
x^3+x+1}+\frac{1}{ x-1}\right)$$ where you can notice that each numerator is the derivative of each denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing Matrix Representation of a Linear Transformation I'm having some trouble figuring out this question:
Let $V=P_3(\mathbb{R})$ and $W=P_4(\mathbb{R})$.
Let $D:W\to V$ be the derivative mapping $D(p)=p'$.
Let $Int:V\to W$ be the integration mapping $Int(p)=\int_0^x p(t)dt$.
Let $\alpha=\{1,x,x^2,x^3\}$ and $\beta=\{1,x,x^2,x^3,x^4\}$ be the standard bases in $V$ and $W$.
Compute : $[D]_{\alpha\beta}$ , $[Int]_{\beta\alpha}$ , $[DInt]_{\alpha\alpha}$ , $[IntD]_{\beta\beta}$.
I'm not sure how to go about finding the matrices that represent these transformations. Any help at all would be appreciated.
My reasoning is that $[D]_{\alpha\beta}$ would equal:
\begin{Bmatrix}
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0
\end{Bmatrix}
Since if you multiply that matrix by
$\beta=\{1,x,x^2,x^3,x^4\}$
You would end up with
$\alpha=\{1,x,x^2,x^3\}$
| Let's apply the transformation $D$ to each element of the given basis $\beta=\{1,x,x^2,x^3,x^4\}$ of $W=P_4(\mathbb{R})$ in the order they are listed, and write each result with respect to the given basis $\alpha=\{1,x,x^2,x^3\}$ of $V=P_3(\mathbb{R})$.
$D(1)=(1)'=0=0\cdot1+0\cdot x+0\cdot x^2+0\cdot x^3$, therefore the first column is $\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$.
$D(x)=(x)'=1=1\cdot1+0\cdot x+0\cdot x^2+0\cdot x^3$, therefore the second column is $\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$.
$D(x^2)=(x^2)'=2x=0\cdot1+2\cdot x+0\cdot x^2+0\cdot x^3$, therefore the third column is $\begin{bmatrix}0\\2\\0\\0\end{bmatrix}$.
$D(x^3)=(x^3)'=3x^2=0\cdot1+0\cdot x+3\cdot x^2+0\cdot x^3$, therefore the fourth column is $\begin{bmatrix}0\\0\\3\\0\end{bmatrix}$.
$D(x^4)=(x^4)'=4x^3=0\cdot1+0\cdot x+0\cdot x^2+4\cdot x^3$, therefore the fifth column is $\begin{bmatrix}0\\0\\0\\4\end{bmatrix}$.
Thus the desired matrix is $[D]_{\alpha\beta}=\begin{bmatrix}0&1&0&0&0\\0&0&2&0&0\\0&0&0&3&0\\0&0&0&0&4\end{bmatrix}$.
| {
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"url": "https://math.stackexchange.com/questions/2196147",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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$3\mid a^3+b^3+c^3$ if only if $3\mid a+b+c $
Prove the following equivalence: $3\mid (a^3 + b^3 + c^3) $ if and only if $3\mid (a + b + c) $.
My try:
I know $a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc$, but I can't seem to proceed from here.
Thanks all!
| *
*$1^3 = 1 \mod 3$
*$2^3 = 2 \mod 3$
*$3^3 = 3 \mod 3$
This means $x = x^3 \mod 3$ for any integer $x$. This also means that if $a^3 + b^3 + c^3 = 0 \mod 3$ then $a + b + c = 0 \mod 3$.
Thus it is divisible by three.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
Last three digits of $\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$ Find the last three digits of $$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$$
I don't know how to continue my work:
$$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}=\sqrt{2^{2(2016)}+2\cdot (3\cdot 2)^{2016}+3^{2(2016)}}=\sqrt{2^{4032}+2\cdot 3^{2016}\cdot 2^{2016}+3^{4032}}=\sqrt{2^{4032}+2^{2017}\cdot 3^{2016}+3^{4032}}=?$$
| HINT:
This can be factorised like this
$$(2^{2016})^2+2\cdot2^{2016}\cdot3^{2016}+({3^{2016}})^{2}=(2^{2016}+3^{2016})^2$$
So the question is to find last three digits of
$$2^{2016}+3^{2016}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the Laurent series expansion for: $\frac{1}{z^2-3z-4}$ Find the Laurent series expansion for:
$$\frac{1}{z^2-3z-4}$$
which converges for$\:$ $1 < |z| < 3.$
I understand the concept of Laurents expansion but I cant seem to make progress in this question so any help will be appreciated.
| $$\frac{1}{z^2-3z-4} = \frac{1}{(z+1)(z-4)}= \frac{1}{5}\left(\frac{1}{z-4}-\frac{1}{z+1}\right)$$
We look at the geometric series. For the first term we have
$$\frac{1}{z-4}=\frac{-1/4}{1-z/4}=\sum_{n=0}^{\infty }\left( -\frac{1}{4} \right) \left( \frac{z}{4}\right) ^{n}\qquad \left\vert z\right\vert <4\implies|z|<3 \\ =\sum_{n=0}^\infty \frac{-1}{4^{n+1}}z^n$$
For the second term we have the well known
$$\frac{1}{1+z}=\frac{1/z}{1-(-1/z)}=\sum_{n=0}^{\infty }\left(\frac{1}{z}\right)\left( \frac{-1}{z}\right) ^{n}\qquad \left\vert z\right\vert >1 \\ = \sum_{n=0}^\infty \frac{(-1)^n}{z^{n+1}}$$
We can combine these to get
$$\sum_{n=0}^\infty \left(\frac{(-1)^n}{z^{n+1}}-\frac{1}{4^{n+1}}z^n\right) \qquad 1<|z|<3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Constructive proof for $p|x^2+1$ It's easy to prove $x^2+1$ is never divisible by $4k+3$ primes. I know a non-constructive proof for existing $x$ so that $p|x^2+1$ for $4k+1$ primes. is there any constructive one?
| First note that
$(p-1)! = -1 \mod p$
by Wilson's Theorem. Since $p = 4k+ 1$, we can write this product as
$1\cdot 2 \cdots (\frac{p-1}{2}) (\frac{p+1}{2}) \cdots (p-1) \equiv -1 \mod p$
Note that this can be written as
$1 \cdot 2 \cdots (\frac{p-1}{2}) \cdot (p-\frac{p-1}{2}) \cdots (p-1) \equiv -1 \mod p$
so
$1 \cdot 2 \cdots (\frac{p-1}{2}) \cdot (-\frac{p-1}{2}) \cdots (-1) \equiv -1 \mod p$
so
$(\frac{p-1}{2})^2 \cdot (-1)^{\frac{p-1}{2}} \equiv -1 \mod p$
Since $p \equiv 1 \mod 4$, $\frac{p-1}{2}$ is even. Thus we are left with
$(\frac{p-1}{2})^2 \equiv -1 \mod 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x) = -3x^2 + 6x + 2$, prove that $f(x)$ is $O(x^2)$ I had a question on Big-O.
If $f(x) = -3x^2 + 6x + 2$, prove that $f(x)$ is $O(x^2)$
Generally my teacher gave us these strategies to follow:
1. Eliminate Negative Terms
2. Multiply to match highest order term.
So in this case Would I be getting rid of $-3x^2$ term and end up with $6x+2$?
So my $c = 8$ and $k = 1$.
Any help would be appreciated.
| Consider that by the triangle inequality we have,
$$|f(x)| = |-3x^2+6x+2| \leq |-3x^2|+|6x|+2$$
then for $x>0$ we have that
$$|f(x)| \leq 3x^2+6x+2$$
and for any $x>1$ (because $x^2>x>1$), we clearly than have that
$$|f(x)|\leq 3x^2+6x^2+2x^2 = 11x^2 = 11\cdot|x^2|.$$
Since this holds for all $x>1$, we have that $f(x)$ is $O(x^2)$ by definition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration with floor function I can't find the exact value of the following integral:
$$\int\limits_0^1 {[ax][bx]} dx$$
where $a$ and $b$ have as greatest common divisor 1.
(gcd(a,b)=1)
Thanks.
| As mentioned in the comments, we have
$$\int_0^1 \lfloor ax \rfloor \lfloor bx \rfloor \, dx = \frac{1}{ab} \sum_{n=0}^{ab-1} \lfloor n/a \rfloor \lfloor n/b \rfloor $$
We'll work with residues mod $a$ and $b$ instead of the floor function. We can write $\lfloor n/a \rfloor = n/a - \{n/a\}$, where $\{x\}$ denotes the fractional part of $x$, and this satisfies $\{n/a\} = (n \mbox{ mod } a)/a$ for $n, a \in \mathbb{Z}$, $a > 0$, so $\lfloor n/a \rfloor = (n - (n \mbox{ mod } a))/a$. Then our sum becomes
$$\frac{1}{a^2b^2} \sum_{n=0}^{ab-1} (n - (n \mbox{ mod } a)) (n - (n \mbox{ mod } b))$$
We can break this into four sums. First,
$$\sum_{n = 0}^{ab-1} n^2 = \frac{(ab-1)ab(2ab-1)}{6}$$
By the Chinese Remainder Theorem, each pair of a residue mod $a$ and a residue mod $b$ corresponds to a unique residue mod $ab$, so
$$\sum_{n=0}^{ab-1} (n \mbox{ mod } a)(n \mbox{ mod } b) = \left( \sum_{k=0}^{a-1} k \right) \left( \sum_{l=0}^{b-1} l \right) = \frac{ab(a-1)(b-1)}{4}$$
Now $n = a\lfloor n/a \rfloor + (n \mbox{ mod } a)$, so
\begin{align*}
\sum_{n=0}^{ab-1} n (n \mbox{ mod } a)
&= a \sum_{n=0}^{ab-1} \lfloor n/a \rfloor (n \mbox{ mod } a) + \sum_{n=0}^{ab-1} (n \mbox{ mod } a)^2 \\
&= a \left( \sum_{l=0}^{b-1} l \right) \left( \sum_{k=0}^{a-1} k \right) + b \sum_{k=0}^{a-1} k^2 \\
&= \frac{a^2 b (a-1)(b-1)}{4} + \frac{b(a-1)a(2a-1)}{6}
\end{align*}
and similarly $\sum_{n=0}^{ab-1} n (n \mbox{ mod } b) = \frac{ab^2 (a-1)(b-1)}{4} + \frac{a(b-1)b(2b-1)}{6}$, so our sum is
$$\frac{1}{a^2b^2} \left( \frac{(ab-1)ab(2ab-1)}{6} + \frac{ab(a-1)(b-1)}{4} - \frac{a^2 b (a-1)(b-1)}{4} - \frac{b(a-1)a(2a-1)}{6} - \frac{ab^2 (a-1)(b-1)}{4} - \frac{a(b-1)b(2b-1)}{6}\right)$$
which simplifies to $\frac{1}{12ab}(4a^2 b^2 - 3ab(a+b-1) - a^2 - b^2 + 1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $3(a^5b+b^5c+c^5a)\geq(a^2c+b^2a+c^2b)^2$
Let $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ be sides-lengths of a triangle. Prove that:
$$3(a^5b+b^5c+c^5a)\geq(a^2c+b^2a+c^2b)^2$$
I tried to apply the way like the proof of the following inequality, but without success.
Let $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ be sides-lengths of a triangle. Prove that:
$$2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq3+\frac{a}{c}+\frac{b}{a}+\frac{c}{b}.$$
Proof.
We need to prove that
$$\sum_{cyc}(2a^2c-a^2b-abc)\geq0$$ or
$$\sum_{cyc}(a^3-abc)-\sum_{cyc}(b^3-2ab^2+a^2b)\geq0$$ or
$$\frac{a+b+c}{2}\sum_{cyc}(a-b)^2-\sum_{cyc}b(a-b)^2\geq0$$ or
$$\sum_{cyc}(a-b)^2(a+c-b)\geq0.$$
Lemma.
Let $x+y+z\geq0$ and $xy+xz+yz\geq0$. Prove that:
$$(b-c)^2x+(a-c)^2y+(b-c)^2z\geq0.$$
Proof of the lemma.
Let $x+y\geq0$.
If $x+y=0$ then $xy+xz+yz=-x^2\geq0$,
which gives $x=y=0$ and $z\geq0$, which says that
$(b-c)^2x+(a-c)^2y+(a-b)^2z\geq0$ is true.
Thus, we can assume that $x+y>0$ and we need to prove that
$$(b-c)^2x+(a-b+b-c)^2y+(a-b)^2z\geq0$$ or
$$(x+y)(b-c)^2+2y(a-b)(b-c)+(y+z)(a-b)^2\geq0,$$
for which it's enough to prove that
$$y^2-(x+y)(y+z)\leq0$$ or
$$xy+xz+yz\geq0,$$
which ends a proof of the lemma.
Now, $\sum\limits_{cyc}(a+c-b)=a+b+c>0$ and
$$\sum_{cyc}(a+b-c)(a+c-b)=\sum_{cyc}(2ab-a^2)=$$
$$=(\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{b}+\sqrt{c}-\sqrt{a})>0,$$
which ends the proof by the lemma.
Thank you!
| $$3(a^5b + b^5c + c^5a) \ge (a^2b + b^2c + c^2a)^2$$
$$\iff 3(abc)^2\left[\left(\frac{a}{b}\right)^3 \cdot \left(\frac{b}{c}\right)^2 + \left(\frac{b}{c}\right)^3 \cdot \left(\frac{c}{a}\right)^2 + \left(\frac{c}{a}\right)^3 \cdot \left(\frac{a}{b}\right)^2\right] \ge (abc)^2\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right)^2$$
Let $\dfrac{a}{b} = x, \dfrac{b}{c} = y, \dfrac{c}{a} = z \implies xyz = 1$.
$$\iff 3\left(\frac{x^2y}{z} + \frac{y^2z}{x} + \frac{z^2x}{y}\right) \ge (x + y + z)^2$$
, which can be proven by using the Chebyshev inequality, the AM-GM inequality and the Cauchy inequality (which applications can be used in the order listed).
| {
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How can I show that $\sum_{k=0}^n [\frac{(n-2k)}{n}{n \choose k}]^2 = \frac{2}{n}{2n - 2 \choose n - 1}$? I want to show this
$\sum_{k=0}^n [\frac{(n-2k)}{n}{n \choose k}]^2 = \frac{2}{n}{2n - 2 \choose n - 1}$
I tried to solve it using this identities:
I) $\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}$
II) ${n \choose k} = {n-1 \choose k-1}\frac{n}{k}$
III) ${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$
Any hints?
Edit: I could solve it with a hint from TMM and the fact that $\sum_{k=0}^n {s \choose k}{r \choose n-k} = {s+r \choose n}$
| Solution
$\sum_{k=0}^{n}[\frac{n-2k}{n}\binom{n}{k}]^{2}$
$= \frac{1}{n^{2}}\sum_{k=0}^{n}(n-2k)^2\binom{n}{k}^2 $
$=\frac{1}{n^{2}}[\sum_{k=0}^{n}n^2\binom{n}{k}^2 + 4\sum_{k=0}^{n}k^2\binom{n}{k}^2 - 4\sum_{k=0}^{n}nk\binom{n}{k}^2] $
$= \sum_{k=0}^{n}\binom{n}{k}^2 + \frac{4}{n^{2}}\sum_{k=0}^{n}n^2\binom{n-1}{k-1}^2 - \frac{4}{n^{2}}\sum_{k=0}^{n}n^2\binom{n-1}{k-1}\binom{n}{k}$ , here we use the fact that ${n \choose k} = {n-1 \choose k-1}\frac{n}{k}$ so $k{n \choose k} = n{n-1 \choose k-1}$
$= \sum_{k=0}^{n}\binom{n}{k}^2 + 4\sum_{k=0}^{n}\binom{n-1}{k-1}^2 - 4\sum_{k=0}^{n}\binom{n-1}{n-k}\binom{n}{k}$ , we use in the last sum: $\binom{n}{k}=\binom{n}{n-k}$ so $\binom{n-1}{k-1}=\binom{n-1}{n-k}$
Finally, knowing that:
$\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}$ and $\sum_{k=0}^n {s \choose k}{r \choose n-k} = {s+r \choose n}$
$= \binom{2n}{n} + 4\binom{2n-2}{n-1} - 4\binom{2n-1}{n} = \binom{2n}{n} + 4(\binom{2n-2}{n-1} - \binom{2n-1}{n})$
$= \binom{2n}{n} - 4\binom{2n-2}{n}$ , because ${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$ then $ -{n-1 \choose k-1}={n-1 \choose k}-{n \choose k}$ in this case $\binom{2n-2}{n-1} - \binom{2n-1}{n} = -\binom{2n-2}{n}$
$= \frac{(2n)!}{(n!)^2}-2^2\frac{(2n-2)!}{n!(n-2)!}$
$= \frac{(2n)!-2^2(2n-2)!n(n-1)}{(n!)^2}$
$= \frac{(2n)!-(2n-2)!(2n)2(n-1)}{(n!)^2} = \frac{(2n)!-(2n-2)!(2n)(2n-1 - 1)}{(n!)^2}$
$= \frac{(2n)!-(2n-2)!(2n)(2n-1) -(2n-2)!(2n)(-1) }{(n!)^2}$
$= \frac{(2n)!-(2n)! +(2n-2)!(2n) }{(n!)^2}$
$= \frac{(2n-2)!}{((n-1)!)^2}\frac{2}{n}$
$= \frac{2}{n}\binom{2n-2}{n-1}$
| {
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How to calculate $f(x)$ in this question? I have a quation regarding composite functions.
if $f\left( \dfrac{x^2+1} x \right)=\dfrac{x^4+1}{x^2}$, then what is $f(x)$?
| HINT:
$$\left(\frac{x^2+1} x\right)^2=\frac{x^4+2x^2+1}{x^2}=\frac{x^4+1+2x^2}{x^2}$$
Alternatively you have $$\frac{x^2+1} x = x + \frac 1 x \\\frac{x^4+1}{x^2} = x^2 + \frac 1 {x^2}$$
| {
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"url": "https://math.stackexchange.com/questions/2216130",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Area inside outer curve of limacon $r=1+2\cos\theta$ I need to find the area inside the outer curve of limacon $r=1+2\cos\theta$
Here is what I tried:
$$
0=1+2\cos\theta
$$
$$
\theta=\cos^{-1}(-\frac{1}{2})
$$
$$
\theta=\frac{2}{3}\pi , \theta=\frac{8}{3}\pi
$$
$$
A = 2\int^{\frac{8}{2}\pi}_{\frac{2}{3}\pi}r^2d\theta=\frac{r^3}{3}|^{\frac{8}{3}\pi}_{\frac{2}{3}\pi}=2\frac{(1+2\cos\theta)^3}{3}|^{\frac{8}{3}\pi}_{\frac{2}{3}\pi}=0
$$
so, 0 area is clearly wrong. Where was my mistake?
| $\int\limits_{\theta = 0}^{2 \pi} (1 + 2 \cos (\theta )\ d \theta = 2 \pi$.
| {
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} |
Find the positive $n$ such $2^{n-2i}|\sum_{k=i}^{\lfloor\frac{n-1}{2}\rfloor}\binom{k}{i}\binom{n}{2k+1}$ Question:
if such $\forall i\in\{1,2,\cdots,\lfloor\dfrac{n-1}{2}\rfloor\}$,have
$$2^{n-2i}|\sum_{k=i}^{\lfloor\frac{n-1}{2}\rfloor}\binom{k}{i}\binom{n}{2k+1}$$ Find all the positive intger $n$
I want to find this sum $\sum_{k=i}^{\lfloor\frac{n-1}{2}\rfloor}\binom{k}{i}\binom{n}{2k+1}$?this step is right?
this problem is post my frend xi yong wang
| We have the sum
$$S_{n,q} = \sum_{k=q}^{\lfloor (n-1)/2\rfloor}
{k\choose q} {n\choose 2k+1}.$$
Introducing
$${n\choose 2k+1} = {n\choose n-2k-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k}} (1+z)^n
\; dz$$
this controls the range and we may extend $k$ to infinity, getting for
the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n}} (1+z)^n
\sum_{k\ge q} {k\choose q} z^{2k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2q}} (1+z)^n
\sum_{k\ge 0} {k+q\choose q} z^{2k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2q}} (1+z)^n
\frac{1}{(1-z^2)^{q+1}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2q}} (1+z)^{n-q-1}
\frac{1}{(1-z)^{q+1}}
\; dz.$$
This is
$$\sum_{p=0}^{n-2q-1} {n-q-1\choose p} {n-2q-1-p+q\choose q}
\\ = \sum_{p=0}^{n-2q-1} {n-q-1\choose n-q-p-1} {n-q-1-p\choose q}.$$
Observe that
$${n-q-1\choose n-q-p-1} {n-q-1-p\choose q}
= \frac{(n-q-1)!}{p! q! (n-2q-1-p)!}
\\ = {n-q-1\choose q} {n-2q-1\choose p}$$
so that the sum becomes
$${n-q-1\choose q} \sum_{p=0}^{n-2q-1} {n-2q-1\choose p}
= 2^{n-2q-1} {n-q-1\choose q}.$$
where we have $q\le \lfloor (n-1)/2\rfloor.$ If this is to be a
multiple of $2^{n-2q}$ then the binomial coefficient must be even. We
apply Lucas'
Theorem which says
that a binomial coefficient ${n\choose p}$ is odd iff all the bits of
$p$ are less than or equal to the corresponding bits of $n.$ Suppose
first that $n$ is not a power of two and take $q = n - 2^{\lfloor
\log_2 n\rfloor}.$ We have $q\ge 1$ as required. To verify the upper
range start from $\lceil (n-1)/2\rceil \le n/2$ which follows by
cases. This is $\lceil (n-1)/2\rceil \le 2^{-1 + \log_2 n}$ which is
less than $2^{\lfloor \log_2 n\rfloor}$ so that $n-1 - \lfloor
(n-1)/2\rfloor \lt 2^{\lfloor \log_2 n\rfloor}$ or $n - 2^{\lfloor
\log_2 n\rfloor} \lt \lfloor (n-1)/2\rfloor + 1$ or $n - 2^{\lfloor
\log_2 n\rfloor} \le \lfloor (n-1)/2\rfloor.$ This proves that the
choice of $q$ is within the range. We thus have $n-q-1 = 2^{\lfloor
\log_2 n\rfloor} - 1$, a string of one bits, which are larger than or
equal to the bits of $q$ by inspection. Hence we have found an odd
binomial coefficient and the claim does not hold when $n$ is not a
power of two. The remaining case is for $n$ a power of two. We thus
have $n-1$ a string of one bits. Hence the bits corresponding to $q$
in $n-1-q$ are the bits of $q$, but flipped. Now since all $q$ have at
least one bit that is set we get a correspondong zero bit in $n-1-q.$
Hence the binomial coefficient is even. This works for all $q.$
Therefore the answer to the query is that $S_{n,q}$ is divisible by
$2^{n-2q}$ with $q$ in the proposed range iff $n$ is a power of two.
Some of this material is sourced from the following MSE
link.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the degree and a basis for $\mathbb{Q}(\sqrt{3} + \sqrt{5})$ over $\mathbb{Q}(\sqrt{15})$ Find the degree and a basis for $\mathbb{Q}(\sqrt{3} + \sqrt{5})$ over $\mathbb{Q}(\sqrt{15})$.
My attempt:
I first proved that $\mathbb{Q}(\sqrt{3} + \sqrt{5}) = \mathbb{Q}(\sqrt{3}, \sqrt{5})$,
Then using tower law its fairly straight forward to establish that $[\mathbb{Q}(\sqrt{3} + \sqrt{5}) : \mathbb{Q}(\sqrt{15})] = 2$.
But I am having difficulty determining the basis for this extension.
I can only take a guess that the answer is $\{1, \sqrt{3}\}$ or $\{1, \sqrt{5}\}$.
Is there a better method than just trail and error to determine the basis in this case.
| One way to prove this is :
$a=(\sqrt{3}+\sqrt{5})$
$a^2=(\sqrt{3}+\sqrt{5})^2$
Thus $a^2-8+2\sqrt{15}=0$ and the polynomial $f(X)=X^2-8+2\sqrt{15}$ is irreducible over $\mathbb{Q}(\sqrt{15})$ because its roots do not lie in $\mathbb{Q}(\sqrt{15})$.
Thus a basis is the set $\{1,\sqrt{3}+\sqrt{5}\}$
Also $\{1,\sqrt{5}\}$ is indeed a basis for the extension because an element $x \in \mathbb{Q}(\sqrt{3},\sqrt{5})$ is of the form $x=a+b \sqrt{3} +c\sqrt{5}+d\sqrt{15}=a+\frac{b}{5} \sqrt{15} \sqrt{5} +c \sqrt{5}+d \sqrt{15}=(a+d \sqrt{15})+(c+ \frac{b}{5} \sqrt{15}) \sqrt{5}$.
In the same way you can prove that $\{1,\sqrt{3}\}$ is a basis of the extension you ask in your question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Singular value decomposition works only for certain orthonormal eigenvectors, not all? I'm trying to find the SVD of the following matrix:
$$A=
\begin{pmatrix}
1 & 1 \\
2 & -2 \\
2 & 2 \\
\end{pmatrix}
$$
I found the eigenvalues and vectors for $A'A$:
$$
\begin{array}{cc}
\lambda_1=10 & \lambda_2=8 \\
e_1'=(1,1) & e_2'=(-1,1). \\
\end{array}
$$
I find the eigenvalues and vectors for $AA'$:
$$
\begin{array}{ccc}
\lambda_1=10 & \lambda_2=8 & \lambda_3=0 \\
e_1=(1,0,2) & e_2=(0,1,0) & e_3=(-2,0,1), \\
\end{array}$$
and so my SVD should be:
$$\left(
\begin{array}{ccc}
\frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} \\
0 & 1 & 0 \\
\frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
\end{array}
\right).\left(
\begin{array}{cc}
\sqrt{10} & 0 \\
0 & \sqrt{8} \\
0 & 0 \\
\end{array}
\right).\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
However, this gives
$$\left(
\begin{array}{cc}
1 & 1 \\
-2 & 2 \\
2 & 2 \\
\end{array}
\right)$$ instead of $A$.
To get $A$ I need to decompose in the following way
$$\left(
\begin{array}{ccc}
\frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} \\
0 & 1 & 0 \\
\frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
\end{array}
\right).\left(
\begin{array}{cc}
\sqrt{10} & 0 \\
0 & \sqrt{8} \\
0 & 0 \\
\end{array}
\right).\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
This last decomposition is as if I had multiplied the first $e_2$ by $-1$.
These eigenvalues and vectors were derived from Mathematica, just to be sure I was using the correct elements.
Any help in explaining why my first decomposition doesn't work would be appreciated.
Edit: The book I'm using doesn't tell which of the orthonormal eigenvectors I have to use. For each eigenvalue, I have two orthonormal eigenvectors, $e_i$ and $-e_i$.
| I'm posting an answer just to complete the previous answers.
Let $A=UDV'$.
Then, if we have dim of $U$ greater than $V$ we can use the formula $v_i=\frac{1}{\sigma_i} A'u_i$, for given $U$.
However, if we have dim of $U$ smaller than $V$, we can use the formula $u_i=\frac{1}{\sigma_i}Av_i$, for given $V$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $f(x+1)+f(x-1)=4x^2-2x+10$ then what is $f(x)$
If $$f(x+1) +f(x-1)= 4x^2 -2x +10$$ then what is $f(x)$
What is strategy of solving this kind of problems ?
Thank you for help
| $$ 4x^2 -2x +10 = 2(x+1)^2-(x+1)+3+2(x-1)^2-(x-1)+3$$
which means $f(x)=2x^2-x+3$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How can I simplify $\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$? This problem got me thinking hard:
$$\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$$
Considering the fact that the common base is '$3$', I decided to do this:
$$\log_3\left(\frac{(n+1)(n-1)}{(n+2)(n)}\right)$$
$$\log_3\left(\frac{n^2-n+n+1}{n^2+2n}\right)$$
$$\log_3\left(\frac{n^2 + 1}{n^2 + 2n}\right)$$
$$\log_3\left(\frac{1}{2n}\right)$$
I am not 100% sure if have simplified the problem. I need some clarification.
| Factor out by $3^n $ to get
$$\frac{3^n (3+3^{-1})}{3^n (3^2+3^0)} $$
$$\frac {3+3^{-1}}{10}=\frac {1}{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove $8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c$
Let $a$, $b$ and $c$ be angles of an acute triangle. Prove that:
$$8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c$$
I tried am-gm, rm-gm rm-am and a couple of other inequalities but I didn't get anywhere.
| Let $u$, $v$ and $w$ be lengths-sides of the triangle such that
$\cos{a}=\frac{v^2+w^2-u^2}{2vw}$, $\cos{b}=\frac{u^2+w^2-v^2}{2uw}$, $\cos{c}=\frac{u^2+v^2-w^2}{2uv}$ and
let $u^2+v^2-w^2=r$, $u^2+w^2-v^2=q$, $v^2+w^2-u^2=p$.
Hence, we need to prove that
$$8\prod_{cyc}\left(1-\frac{r^2}{(p+r)(q+r)}\right)\geq\frac{27pqr}{(p+q)(p+r)(q+r)}$$ or
$$8(pq+pr+qr)^3\geq27pqr(p+q)(p+r)(q+r).$$
Let $pq=z$, $pr=y$ and $qr=x$.
Hence, we need to prove that
$$8(x+y+z)^3\geq27(x+y)(x+z)(y+z),$$
which is AM-GM:
$$\left(\frac{\sum\limits_{cyc}(x+y)}{3}\right)^3\geq\prod_{cyc}(x+y).$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Find the value of $3+7+12+18+25+\ldots=$ Now, this may be a very easy problem but I came across this in an examination and I could not solve it.
Find the value of
$$3+7+12+18+25+\ldots=$$
Now here is my try
$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$
After that, I could not proceed.
| This answer is unique in that it does not use trial and error to find out the expression for the $n$-th term.
Our sum is given by :
$$
S=3+7+12+18+...+T_n
\\
S=\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3+7+12+18+...+T_n-1+T_n
$$
Subtracting these two sums :
$$
0=3+4+...+(n+2)-T_n
\\
T_n=3+4+...+(n+2)
\\
T_n=\frac{(n+2)(n+3)}{2}-3
=\frac{n(n+5)}{2}
$$
Now, you can use Gregory's answer to figure the sum out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
definite integral- how to solve. I know the substitution. I know the substitution but how should I continue?
$$\int_0^1 x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$$
| An alternative way - by setting $x=u^2$, then $u=\frac{1-v}{1+v}$ and $v=z^2$, we are left with:
$$ \int_{0}^{1}2u^3\sqrt{\frac{1-u}{1+u}}\,du = 4\int_{0}^{1}\frac{(1-v)^3}{(1+v)^5}\sqrt{v}\,dv =8\int_{0}^{1}\frac{z^2(1-z^2)^3}{(1+z^2)^5}\,dz$$
Then by applying integration by parts we get
$$ \int_{0}^{1}\frac{1-9 z^2+15 z^4-7 z^6}{(1+z^2)^4}\,dz=-7\,I_1+36\,I_2-60\,I_3+32\,I_4 $$
with $I_k=\int_{0}^{1}\frac{dz}{(1+z^2)^k}.$ These integrals can be computed through differentiation under the integral sign, since for any $a>0$ we have
$$ \int_{0}^{1}\frac{a\,dz}{a+z^2}=\frac{\text{arccot}\sqrt{a}}{\sqrt{a}} $$
hence $I_1=\frac{\pi}{4}, I_2=\frac{1}{4}+\frac{\pi}{8}, I_3=\frac{1}{4}+\frac{3\pi}{32}, I_4=\frac{11}{48}+\frac{5\pi}{64}$. Summarizing,
$$ \int_{0}^{1}x\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\,dx = \color{red}{\frac{4}{3}-\frac{3\pi}{8}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
QUESTION on Logarithm This question was easy to solve, but when it came to matching options I failed. Hope you can help. Let,
*
*$a= \log_{24}(12)$
*$b= \log_{48}(36)$
*$c= \log_{36}(24)$
then $abc +1 = ?$
*
*a) $2ab$
*b) $2bc$
*c) $2ca$
*d) $ba + bc$
FYI :- I tried to solve and found $abc= log_{48}(12)$ and did several things but none were useful to get the right option I tried to take $24^a$ like that, but it didn't help me. Hope you can help. The answer was (b) but I didn't get it.
| It looks like you already know that $\log_a(b) = \frac{\log a}{\log b}$.
Since all of the numbers have $2$ and $3$ as their only prime factors, we can start by setting $x=\log 2$ and $y=\log 3$. We then have
$$ a = \log_{24}{12} = \frac{2x+y}{3x+y} \qquad
b = \log_{48}{36} = \frac{2x+2y}{4x+y} \qquad
c = \log_{36}{24} = \frac{3x+y}{2x+2y} $$
You can then compute $$abc+1 = \frac{2x+y}{4x+y} + 1 = \frac{6x+2y}{4x+y} $$
Do simiarly for each of the the answer options and see if one of them matches. For option (b) we get
$$ 2bc = 2\frac{2x+2y}{4x+y}\cdot\frac{3x+y}{2x+2y} = 2\frac{3x+y}{4x+y}=\frac{6x+2y}{4x+y} $$
which is the result we're searching for.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $p\mid (3^n+1)$, then $p\equiv 1\pmod{3}$
Show that if $p> 2$ is a prime, $n > 1$ is odd and $p\mid (3^n+1)$,
then $p\equiv 1\pmod{3}$.
Since $n$ is odd, we have $3^{n+1} \equiv -3 \pmod{p}$ is a quadratic residue. Then I thought about using Quadratic Reciprocity but didn't see how to apply it.
We have $x^2 \equiv -3 \pmod{p}$ for some integer $x$. By Fermat's Little Theorem we have $$x^{p-1} \equiv x^2 \cdot x^{p-3} \equiv -3 \cdot x^{p-3} \equiv 1 \pmod{p}.$$ Thus, $x^{p-3} \equiv -3^{-1} \pmod{p}$.
I didn't see how to get a contradiction if $p \equiv 2 \pmod{3}$.
| Lemma: for an odd prime $q,$ we always have
$$ (-3|q) = (q |3). $$
Proof: if $q \equiv 1 \pmod 4,$ then
$$ (-3|q) = (-1|q)(3|q) = (3|q) = (q|3). $$
If $q \equiv 3 \pmod 4,$ then
$$ (-3|q) = (-1|q)(3|q) = -(3|q) = (q|3). $$
In the lemma, $3$ can be replaced by any prime $r \equiv 3 \pmod 4.$
Let $n = 2k+1.$ We have $1 + 3^n = 1 + 3 (3^k)^2 = 1 + 3 w^2.$ If $p \equiv 2 \pmod 3$ and
$$ x^2 + 3 y^2 \equiv 0 \pmod p, $$ assume $y$ is not divisible by $p.$
Then $y$ has a multiplicative inverse $\pmod p.$
$$ x^2 \equiv -3 y^2 \pmod p, $$
$$ \frac{x^2}{y^2} \equiv -3 \pmod p, $$
$$ \left( \frac{x}{y} \right)^2 \equiv -3 \pmod p. $$
This contradicts $p \equiv 2 \pmod 3,$ so actually $y$ is divisible by $p.$ It follows that $x$ is also divisible by $p.$
Back to the original problem, we have $p \equiv 2 \pmod 3$ and
$$ 1 + 3 w^2 \equiv 0 \pmod p.$$
However, this implies $1$ is divisible by $p,$ which is false
====================================================================
The easy proposition is this: we are given a quadratic form
$$ a x^2 + b x y + c y^2 $$ with discriminant
$$ \Delta = b^2 - 4 a c,$$ finally an odd prime $q$ such that
$$ (\Delta | q) = -1. $$
IF
$$ a x^2 + b x y + c y^2 \equiv 0 \pmod q $$ THEN both
$$ x,y \equiv 0 \pmod q, $$ and we get the extra
$$ a x^2 + b x y + c y^2 \equiv 0 \pmod {q^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluation of $\sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$ Find the following sum
$$S= \sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$$
I am not able to make it telescopic series. Could someone help me with this?
| "Recall" this trigonometric identity:
$$
\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}
$$
Now think about applying that:
\begin{align}
& \arctan \frac{an+b}{cn+d} + \arctan \frac{en+f}{gn+h} \\[10pt]
= {} & \arctan \frac{\dfrac{an+b}{cn+d} + \dfrac{en+f}{gn+h}}{1 - \dfrac{an+b}{cn+d} \cdot \dfrac{en+f}{gn+h}} \\[10pt] = {} & \arctan \frac{(an+b)(gn+h) + (en+f)(cn+d)}{(cn+d)(gn+h) - (an+b)(en+f)} \\[10pt]
= {} & \arctan \frac{(ag+eh)n^2 + (bg+ah+ed+fe)n + (bh + fd)}{(cg-ae)n^2 + (ch+dg-af-be)n + (dh-bf)}
\end{align}
You'd want to choose $a,b,c,d,e,f,g,h$ in such a way that $ag+eh=0,$ $cg-ae=1,$ etc.
Then you've got two terms, $\dfrac{an+b}{cn+d} + \arctan \dfrac{en+f}{gn+h},$ where originally you had one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
The adjoint of a linear operator with respect to an inner product I have an operator $T (x_1, x_2) := (x_2, x_1)$ which is self-adjoint with respect to standard inner product, now I have a new inner product on $\Bbb R^2$
$$\langle x,y\rangle = x_1y_1 + \frac{1}{2} (x_1y_2 + x_2y_1) + \frac{1}{3} x_2y_2.$$
I find the matrix of this inner product to be $$A = \begin{pmatrix} 1 &\frac{1}{2}\\ \frac{1}{2} & \frac{1}{3}\end{pmatrix}.$$
How can I find the adjoint of $T$ - $(T^*)$ with respect to this new inner product? Is $T^* = A T A^{-1} $ ?
| By definition, the adjoint operator of $T$ corresponding to the inner $\langle \cdot, \cdot\rangle_A$ is given by
\begin{align}
\langle v, Tv\rangle_A = \langle T^\dagger v, v\rangle_A.
\end{align}
where $T^\dagger$ will be the symbol of the adjoint with respect to $A$. However, since
\begin{align}
\langle v, Tv\rangle_A = \langle v, ATv\rangle = \langle v, ATA^{-1}Av\rangle = \langle (ATA^{-1})^\ast v, Av\rangle= \langle (A^\ast)^{-1}T^\ast A^\ast v, v\rangle_A
\end{align}
then it follows
\begin{align}
T^\dagger = (A^\ast)^{-1}T^\ast A^\ast.
\end{align}
Edit: In your case, we have that
\begin{align}
T=
\begin{bmatrix}
0& 1\\
1 & 0
\end{bmatrix} \ \ \ \Rightarrow \ \ \
T^\ast = T
\end{align}
and
\begin{align}
A^\ast =
\begin{bmatrix} 1 &\frac{1}{2}\\
\frac{1}{2} & \frac{1}{3}\end{bmatrix} \ \ \ \
(A^{\ast})^{-1} =
\frac{1}{\frac{1}{3}-\frac{1}{4}}\begin{bmatrix} \frac{1}{3} &-\frac{1}{2}\\
-\frac{1}{2} & 1\end{bmatrix}
=
\begin{bmatrix}
4 & -6\\
-6 & 12
\end{bmatrix}
\end{align}
which means
\begin{align}
T^\dagger =
\begin{bmatrix}
4 & -6\\
-6 & 12
\end{bmatrix}
\begin{bmatrix}
0& 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix} 1 &\frac{1}{2}\\
\frac{1}{2} & \frac{1}{3}\end{bmatrix}
=
\begin{bmatrix}
4 & -\frac{5}{3}\\
9 & 4
\end{bmatrix}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Can the Laurent series be found in region where there is singularity? In class my teacher showed that we can use the expansion of $\frac{1}{1-z}$ to find an expansion for $\frac{1}{(z+1)(z+2)}$ in the regions $|z|<1$, $1<|z|<2$ and $|z|>2$
My question is can this method be used to find the expansion in the regions $|z|>1$ and $|z|<2$, if not then what method can be used?
|
The function
\begin{align*}
f(z)&=\frac{1}{(z+1)(z+2)}\\
&=\frac{1}{z+1}-\frac{1}{z+2}
\end{align*}
has two simple poles at $-1$ and $-2$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $-1$ and $-2$ and see they determine three regions.
\begin{align*}
|z|<1,\qquad\quad
1<|z|<2,\qquad\quad
2<|z|
\end{align*}
*
*The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the poles $-1$ at the boundary of the disc. In the interior of this disc all two fractions with poles $-1$ and $-2$ admit a representation as power series at $z=0$.
*The second region $1<|z|<2$ is the annulus with center $0$, inner radius $1$ and outer radius $2$. Here we have a representation of the fraction with poles $-1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $-2$ admits a representation as power series.
*The third region $|z|>2$ containing all points outside the disc with center $0$ and radius $2$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.
A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}
We can now obtain the Laurent expansion of $f(x)$ at $z=0$ for all three regions
*
*Region 2: $1<|z|<2$
\begin{align*}
f(z)&=\frac{1}{z+1}-\frac{1}{z+2}\\
&=-\sum_{n=1}^\infty\frac{1}{(-z)^n}-\sum_{n=0}^\infty \frac{1}{2^{n+1}}(-z)^n\\
&=\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{z^n}+\sum_{n=0}^\infty \left(-\frac{1}{2}\right)^{n+1}z^n\\
\end{align*}
The Laurent expansion for the other regions can be calculated similarly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequality by telescope It can be shown through less elementary methods that:
$\displaystyle \large \frac{2}{3}n\sqrt{n} < \sum_{k=1}^n \sqrt{k} < \frac{2}{3}n\sqrt{n} + \frac{\sqrt{n}}{2}$
It is easy, but tedious, through squaring, to show that:
Lemma:
$\displaystyle \large \frac{3}{2}\sqrt{k} > \left(k+\frac{1}{2}\right)^{\frac{3}{2}} - \left(k-\frac{1}{2}\right)^{\frac{3}{2}} $
Then telescope to obtain a slightly stronger lower bound than what we seek and acquire the lower bound with some basic inequalities
$\large 1. $ Does there exist an easier way to prove the lemma that doesn't use calculus or induction? Exploiting monotonicity and convexity does not count as calculus, for the purposes of this question.
$\large 2. $ Is there a telescope for the upper bound? I have tried with $\displaystyle \left(k+\frac{2}{3}\right)^{\frac{3}{2}} - \left(k-\frac{1}{3}\right)^{\frac{3}{2}}, \left(k+\frac{3}{4}\right)^{\frac{3}{2}} - \left(k-\frac{1}{4}\right)^{\frac{3}{2}} \text{ and } \left(k+\frac{3}{5}\right)^{\frac{3}{2}} - \left(k-\frac{2}{5}\right)^{\frac{3}{2}}$
to no success, as the upper bound is stronger than all three obtained from telescoping.
| Let $\frac{3}{2}\sqrt{k}=a$, $\sqrt{\left(k+\frac{1}{2}\right)^3}=b$ and $\sqrt{\left(k-\frac{1}{2}\right)^3}=c$.
Since $b>c$, we see that $a+b-c>0$ and $$b+c=\left(k+\frac{1}{2}\right)\sqrt{k+\frac{1}{2}}+\left(k-\frac{1}{2}\right)\sqrt{k-\frac{1}{2}}\geq\sqrt{k}+1\cdot\frac{1}{2}\sqrt{k}=\frac{3}{2}\sqrt{k}=a,$$
which says, that $b+c-a>0$.
But
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=$$
$$=2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4=\frac{1}{16}(3k^2-1)>0,$$
which says that $a+c>b$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$ If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$
My Attempt:
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2a.\dfrac {1}{a}$$
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2$$
How do I proceed further?
| This can be done by using Sridharacharya Formula.
The roots of a quadratic equation $ax^2+bx+c$ are $$\boxed{\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$
Thus,
$\Large a=\frac{6 \pm \sqrt{6^2+4}}{2}$
$\large a=3 \pm \sqrt{10}$
Now, $$a^2+a^{-2}=\bigg(a+\frac1a\bigg)^2-2$$
Now, one can put both the values of $a$, and then by rationalizing the denominator we get answer as $\boxed{38}$ in both the case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Transformation of $|z-1|=1$ and $\mathrm{Re}(z)^2 = \mathrm{Im }(z)^2-1$ under $f(z) = \sqrt{z}$ I have two regions in the complex plane, defined by
$|z-1|=1$ and $(\,\mathrm{Im}(z))^2 = (\,\mathrm{Re}(z))^2-1$, $\mathrm{Re}(z)>0$.
I am being asked to find and sketch the image of those regions under the mapping $f(z) = \sqrt{z}$ .
My attempt
We have $f(z) = \sqrt{r}e^{i\frac{\theta}{2}}$ and $|z-1|=1 \iff r = 2\cos(\theta)$
If $|z-1|=1 $, then:
$$f(z) = \sqrt{2|\cos(\theta)|}\cos\left(\frac{\theta}{2}\right) +
i\sqrt{2|\cos(\theta)|}\sin\left(\frac{\theta}{2}\right)$$
It means that $f(z)$ lies on the trace of the curve $\left(\sqrt{2|\cos(\theta)|}\cos\left(\frac{\theta}{2}\right);
\sqrt{2|\cos(\theta)|}\sin\left(\frac{\theta}{2}\right) \right)$ if $|z-1| = 1.$
Am I right so far?
I couldn't recognize this plane curve.
And I don't know how to proceed in the region $(\,\mathrm{Im}(z))^2 = (\,\mathrm{Re}(z))^2-1$.
How can I do it?
| Possible suggestion:
Let $z = x + yi$. We will emulate this problem in the Cartesian plane.
By transforming $Re(z)^2 = Im(z)^2 - 1$ under $f$, we are taking the root of each point. Conversely, $z^2 = (x^2 - y^2) + (2xy)i$ satisfies $Re(z^2)^2 = Im(z^2)^2 - 1$ instead for every point $z$ on the transformed curve.
This is the Cartesian graph $x^4 - 2x^2y^2 + y^4 = 4x^2y^2 - 1$
I.E. $x^4 - 6x^2y^2 + y^4 = - 1$
The first problem can be done similarly. The transformed curve is $|z^2 - 1| = 1$
This translates to the Cartesian plane as $4x^2y^2 + x^4 + y^4 + 1 - 2x^2y^2 + 2y^2 - 2x^2 = 1$
I.E. $0 = 2x^2y^2 + x^4 + y^4 + 2y^2 - 2x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Show this $|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\frac85$ Let $x\in R$ show that
$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{8}{5}$$
since
$$f(x)=f(x+\pi),$$it sufficient to show $x\in (0,\pi]$
| Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of $\sin$ on $[0, \pi]$ quickly reduces the problem to showing the inequality
$$2 \sin 1 > \frac{8}{5} .$$
Then, from Maclaurin expansion, we have
$$ \sin 1 = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \ldots $$
Observe that the absolute value of each of these terms is decreasing, hence in this alternating sum, we may conclude that
$$ \sin 1 = 1 - \frac{1}{3!} + \left( \frac{1}{5!} - \frac{1}{7!} \right) + \left( \ldots \right) + \ldots > 1 - \frac{1}{3!} = \frac{5}{6} > \frac{4}{5}. $$
In particular, we could have strengthened the original inequality to
$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{5}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
For $\Bbb K^{n \times n}$ matrices $A$ and $B$, $\exists p[p(A)=B] \implies \exists p[p(B)=A]$? For $\Bbb K^{n \times n}$ matrices $A$ and $B$, is it true that if there exists a matrix polynomial $p$ such that $p(A)=B$, then there exists another matrix polynomial $p$ (not necessarily the same) such that $p(B) = A$?
My attempt
Let $B = \displaystyle \sum_{k=0}^{n-1} a_k A^k$, where $a_k \in \Bbb K$.
From the Cayley-Hamilton theorem, there exists a polynomial $p$ of degree $n$ such that $p(A) = O$, so $A^m$ where $m \ge n$ can be expressed as linear combination of $\left\{ A^k \ | \ 0 \le k < n\right\}$.
So, we can express $B^m = \displaystyle \sum_{k=0}^{n-1} a_{mk} A^k$, where $a_{mk} \in \Bbb K$.
If $\{I, A, A^2, \cdots, A^{n-1}\}$ form a basis, then we can express a polynomial in $B$, $\displaystyle \sum_{k=0}^{n-1} b_k B^k$, as:
$$\begin{pmatrix}
a_{00} & a_{10} & a_{20} & \cdots & a_{(n-1)0} \\
a_{01} & a_{11} & a_{21} & \cdots & a_{(n-1)1} \\
a_{02} & a_{12} & a_{22} & \cdots & a_{(n-1)2} \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
a_{0(n-1)} & a_{1(n-1)} & a_{2(n-1)} & \cdots & a_{(n-1)(n-1)} \\
\end{pmatrix}
\begin{pmatrix}
b_0 \\ b_1 \\ b_2 \\ \vdots \\ b_{n-1}
\end{pmatrix}$$
Since we want the polynomial to equal $A$, we would need to solve:
$$\begin{pmatrix}
a_{00} & a_{10} & a_{20} & \cdots & a_{(n-1)0} \\
a_{01} & a_{11} & a_{21} & \cdots & a_{(n-1)1} \\
a_{02} & a_{12} & a_{22} & \cdots & a_{(n-1)2} \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
a_{0(n-1)} & a_{1(n-1)} & a_{2(n-1)} & \cdots & a_{(n-1)(n-1)} \\
\end{pmatrix}
\begin{pmatrix}
b_0 \\ b_1 \\ b_2 \\ \vdots \\ b_{n-1}
\end{pmatrix}
=
\begin{pmatrix}
0 \\ 1 \\ 0 \\ \vdots \\ 0
\end{pmatrix}$$
Until now I have transformed the problem into a problem in linear algebra, but is utterly useless. Firstly, $\{A^k \ | \ 0 \le k < n\}$ may not even be linearly independent.
| In a word, no. Let
$A=\pmatrix{1&0\\0&-1}$ and $p(X)=X^2$.
Then
$$B=p(A)=\pmatrix{1&0\\0&1}.$$
How can there be a polynomial with $q(B)=A$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $x$ and $y$ such that $15\sin(x+y)+7\sin x+7\sin y$ is maximized Question is: Find $x$ and $y$ such that
$$15\sin(x+y)+7\sin x+7\sin y$$
is maximized.
What I tried was
$$f(x,y)=15\sin(x+y)+7\sin x+7\sin y=15(\sin x\cos y+\cos x\sin y)+7(\sin x +\sin y)$$
$$\frac{\partial f}{\partial x}=15\cos(x+y)+7\cos x=0$$
$$\frac{\partial f}{\partial y}=15\cos(x+y)+7\cos y=0$$
$$\cos x=\cos y$$
$$y=-x$$
$$\cos x=-{15\over 7}\quad ??$$
| Since $f$ is periodic, we don't need to consider the other regions.
Hence, we can consider $x = \pm y$ without the $2n\pi$.
For the case $x=y$, we obtain:
$$\begin{array}{rcrcl}
\dfrac{\partial f}{\partial x} &=& 15\cos(x+y)+7\cos x &=& 0 \\
&& 15\cos(2x)+7\cos x &=& 0 \\
&& 15(2\cos^2x-1)+7\cos x &=& 0 \\
&& 30\cos^2x + 7\cos x - 15 &=& 0 \\
&& (6\cos x+5)(5\cos x - 3) &=& 0 \\
\end{array}$$
Hence, $\cos x=-\dfrac56$ or $\dfrac35$.
$$\begin{array}{rcl}
15\sin(x+y)+7\sin x+7\sin y
&=& 15\sin2x+14\sin x \\
&=& 30\sin x \cos x + 14\sin x \\
\end{array}$$
From this, we can obtain $4$ values.
For the case $x=-y$, we obtain $\cos x = -\dfrac{15}7$.
From this, we can obtain $2$ values.
Using the second partial derivative test:
$$H(x,y) = \begin{pmatrix}f_{xx}(x,y) &f_{xy}(x,y)\\f_{yx}(x,y) &f_{yy}(x,y)\end{pmatrix} = \begin{pmatrix}-15\sin(x+y)-7\sin x & -15\sin(x+y)\\-15\sin(x+y) & -15\sin(x+y)-7\sin y\end{pmatrix}$$
Then,
$$D(x,y) = \det(H(x,y)) = 105\sin(x+y)(\sin x + \sin y) + 7 \sin x \sin y$$
Find values from the $6$ values found, such that $D < 0$ and $f_{xx} < 0$.
For the case $x=-y$, $D$ must be negative, so they can be rejected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$2 + 2 = 5$ (Fake proof, or ?) Basically, there is no error in the following steps(as it seems), but there is some error due to which 2 + 2 = 5. What is it?
-20 = -20
16-36 = 25-45
16-36+(81/4) = 25-45+(81/4)
(4^2)-(2*4*9/2)+((9/2)^2) = (5^2)-(2*5*(9/2))+((9/2)^2)
(4-(9/2))^2 = (5-(9/2))^2
4-(9/2) = 5-(9/2)
4 = 5
2+2 = 5
| The problem is when you go from a quadratic to a linear.
$$(4-\frac{9}{2})^2 = (5-\frac{9}{2})^2$$
$$|4-\frac{9}{2}| = |5-\frac{9}{2}|$$
$$|\frac{-1}{2}| = |\frac{1}{2}|$$
$$\frac{1}{2} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Perturbation series for $x^5+\varepsilon x-1=0$ I want to find a closed form for the perturbation coefficients $a_n$ defined by the perturbative solution
$$
x(\varepsilon)=1+\sum_{n=1}^\infty a_n \varepsilon^n
$$
to the quintic equation
$$
x^5+\varepsilon x-1=0.
$$
By computing the determinant, as was suggested for this related question regarding the cubic case, we can argue that the radius of convergence of the above series must be
$$
\rho=\frac{5}{4^{4/5}}=1.64938\dots
$$
Furthermore, the Lagrange-Bürmann theorem allows us to formally write down a closed form for the coefficients, namely
$$
a_n=\frac{(-1)^n}{n!}\frac{d^{n-1}}{dx^{n-1}}\left(\frac{x}{1+x+x^2+x^3+x^4}\right)^n_{x=1}
$$
but this doesn't look very illuminating. (For the case of the cubic equation a slightly more explicit but still cumbersone rewriting was made possible by the simpler form of $a_n$).
What I would like to achieve is obtaining $\rho$ from the explicit closed expression for the $a_n$. Therefore I set out to compute some of them. Here they are:
$$
a_1=-\frac{1}{5},\
a_2=-\frac{1}{5^2},\
a_3=-\frac{1}{5^3},\
a_4=0,\\
a_5=\frac{21}{5^6},\
a_6=\frac{78}{5^7},\
a_7=\frac{187}{5^8},\
a_8=\frac{286}{5^9},\
a_{9}=0,\\
a_{10}=-\frac{9367}{5^{12}},\
a_{11}=-\frac{39767}{5^{13}},\
a_{12}=-\frac{105672}{5^{14}},\
a_{13}=-\frac{175398}{5^{15}},\
a_{14}=0.
$$
The behavior of the $a_n$ for $n=1,\ldots,30$ supports the following conjecture:
$$\boxed{
a_n = -(-1)^{\lfloor n/5\rfloor}\frac{c_n}{5^{\alpha_n}}
}
$$
where
$$
\alpha_n=\sum_{k=0}^\infty \left\lfloor \frac{n}{5^k} \right\rfloor
$$
and the $c_n$ are nonnegative integer coefficients which are not divisible by $5$ and vanish for $n=5m-1$.
I think that $c_n$ should be something of the form
$$
\frac{1}{4n+1}\binom{5n}{n}
$$
which for $n=4$ goes very near to reproducing $c_8=286$ and has a scaling similar to that exhibited by the $c_n$. This problem is also motivated by this video where it is suggested that the answer can be guessed with some effort by staring at the coefficients hard enough (27.08).
I think I need some help, however!
| For convenience let $\varepsilon = 5s$. Maple gives me the following:
$$\sum _{k=0}^\infty \frac{(-16 s^5)^k}{k!} \left(\frac{\left(\frac 2 5\right)_{2k} \left(-\frac 1 {10}\right)_{2k} }{\left(\frac 4 5\right)_k \left(\frac 3 5\right)_k \left(\frac 2 5\right)_k}
- \frac{\left(\frac 4 5\right)_{2k} \left(\frac 3 {10}\right)_{2k} s }{ \left(\frac 6 5\right)_k \left(\frac 4 5\right)_k \left(
\frac 3 5\right)_k }
- \frac{\left(\frac{6}{5}\right)_{2k} \left(\frac{7}{10}\right)_{2k} s^2 }{ \left(\frac{7}{5}\right)_k \left(\frac{6}{5}\right)_k \left(
\frac{4}{5}\right)_k }
- \frac{\left(\frac{8}{5}\right)_{2k} \left(\frac{11}{10}\right)_{2k} s^3 }{\left(\frac{8}{5}\right)_{k} \left(\frac{7}{5}\right)_k \left(
\frac{6}{5}\right)_k} \right)
$$
using the Pochhammer symbols $$(a)_m = \frac{\Gamma(a+m)}{\Gamma(a)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Sum of series = S
If $S=\frac35+\frac{10}{5^2}+\frac{21}{5^3}+\frac{36}{5^4}+\frac{55}{5^5}+\cdots$
In this we have to find value of S
Intianlly I thought it is geometric series .
After that I try to makes terms which can cancel other terms.
| We have
$$\frac{S}{5} = \frac{3}{5^2}+\frac{10}{5^3}+\frac{21}{5^4}+\frac{36}{5^5}+\frac{55}{5^6}+\cdots$$
Hence
$$S_1 = \frac{4S}{5} = S - \frac{S}{5} = \frac{3}{5}+\frac{7}{5^2}+\frac{11}{5^3}+\frac{15}{5^4}+\frac{19}{5^6}+\cdots$$
$$\frac{1}{5}S_1 = \frac{3}{5^2}+\frac{7}{5^3}+\frac{11}{5^4}+\frac{15}{5^5}+\frac{19}{5^7}+\cdots$$
Hence
\begin{align*}
\frac{4}{5}S_1 &= \frac{3}{5}+\frac{4}{5^2} + \frac{4}{5^3} + \cdots \\
&=\frac{3}{5} +\frac{4}{5^2}\frac{1}{1-\frac{1}{5}}\\
&= \frac{3}{5} + \frac{1}{5} = \frac{4}{5}
\end{align*}
Hence $S_1 = 1$ and $S = \frac{5}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve a simple equation: $x+x\sqrt{(2x+2)}=3$
$x+x\sqrt{(2x+2)}=3$
I must solve this, but I always get to a point where I don't know what to do. The answer is 1.
Here is what I did:
$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$
Then I got:
$-2x^{3}-x^{2}-6x+9=0$
| I usually check if the integer divisors of the independent term of the polynomial is a root in order to decompose it in factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Find Laurent Series for $f(z) = \frac{1}{z} \cdot \frac{1}{1+z^2}$ when $1 < |z| < \infty$ Q: Find the Laurent series that represents the following function when $1 < |z| < \infty$
\begin{align*}
f(z) &= \frac{1}{z} \cdot \frac{1}{1+z^2} \\
\end{align*}
The answer is given:
\begin{align*}
f(z) &= \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{z^{2n+1}} \\
\end{align*}
I don't see how to do the problem to arrive at the answer.
BTW, to solve the Laurent series of the same function when $|z| < 1$, it's very straightforward:
Start with standard Maclaurin Series
\begin{align*}
\frac{1}{1-z} = \sum\limits_{n=0}^\infty z^n \\
\end{align*}
Substitute $z$ with $-z^2$ to get:
\begin{align*}
\frac{1}{1+z^2} = \sum\limits_{n=0}^\infty (-1)^n z^{2n} \\
\end{align*}
Multiply both sides by $\frac{1}{z}$:
\begin{align*}
\frac{1}{z} \cdot \frac{1}{1+z^2} &= \sum\limits_{n=0}^\infty (-1)^n z^{2n-1} \\
\frac{1}{z} \cdot \frac{1}{1+z^2} &= \frac{1}{z} + \sum\limits_{n=0}^\infty (-1)^{n+1} z^{2n+1} \\
\end{align*}
How do I solve the $1 < |z| < \infty$ version?
| When $1 < |z| < \infty$ then $0 < \dfrac{1}{|z|} < 1$
$$\dfrac{1}{z}.\dfrac{1}{1+z^2}=\dfrac{1}{z^3}.\dfrac{1}{1+(\dfrac{1}{z})^2}=\\
\dfrac{1}{z^3}.\dfrac{1}{1-(-\dfrac{1}{z^2})}=\\
\dfrac{1}{z^3}(1+(-\dfrac{1}{z^2})+(-\dfrac{1}{z^2})^2+(-\dfrac{1}{z^2})^3+...)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos\left (\theta -\frac{\pi}{4}\right)$ If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos \left(\theta -\frac{\pi}{4}\right).$
I could not get any idea to solve. However I tried by using $\theta =0^\circ $. But could not get the answer.
| Since $$\tan(\pi\cos\theta)=\cot(\pi\sin\theta),$$
we can write $$\tan(\pi\cos\theta)=\cot(90-(\pi\cos\theta)).$$
Thus $$\cot(90-(\pi\cos\theta))=\cot(\pi\sin\theta)$$
$$\therefore\quad 90-(\pi\cos\theta)=\pi\sin\theta$$
$$90=\pi(\sin\theta+\cos\theta)$$
$$\frac{90}{\pi}=\sin\theta+\cos\theta$$
$$\frac{1}{2}=\sin\theta +\cos\theta$$
Now
\begin{align*}
\cos\left(\theta-\frac{\pi}{4}\right)&=\cos\theta \cos45^{\circ}+\sin\theta \sin45^{\circ}\\
&=\frac{\cos\theta}{\sqrt{2}}+\frac{\sin\theta}{\sqrt{2}}\\
&=\frac{\cos\theta +\sin\theta}{\sqrt{2}}
\end{align*}
Since $$\sin\theta + \cos\theta =\frac{1}{2},$$
$$\cos\left(\theta-\frac{\pi}{4}\right)=\frac{1}{2\sqrt{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int_0^{\pi/2}x e^x \sin(x)dx$. $$I=\int_0^{\pi/2}x e^x \sin(x)dx$$
I gave a lot of efforts in this question and replaced x with $\pi/2 -x $ but to no avail, this even became more complicated.
Kindly give me an idea on how to proceed in this very question.
| Define $$S(x) = \int e^x \sin x \, dx, \quad C(x) = \int e^x \cos x \, dx.$$ Then with the choice $$u = \sin x, \quad du = \cos x \, dx, \\ dv = e^x \, dx, \quad v = e^x,$$ we find that $$S(x) = e^x \sin x - \int e^x \cos x \, dx = e^x \sin x - C(x).$$ Similarly, we find $$C(x) = e^x \cos x + \int e^x \sin x \, dx = e^x \cos x + S(x).$$ It follows that $$S(x) = \frac{e^x}{2}(\sin x - \cos x) + K, \\ C(x) = \frac{e^x}{2}(\sin x + \cos x) + K$$ for any constant of integration $K$; moreover, $$\int S(x) \, dx = \int \frac{e^x}{2}(\sin x - \cos x) \, dx = \frac{1}{2}(S(x) - C(x)) = -\frac{e^x}{2} \cos x + K.$$
Thus prepared, we now easily compute with the choice $$u = x, \quad du = dx, \\ dv = e^x \sin x \, dx, \quad v = S(x),$$
$$\begin{align*} \int x e^x \sin x \, dx &= x S(x) - \int S(x) \, dx \\
&= \frac{x e^x}{2} (\sin x - \cos x) + \frac{e^x}{2} \cos x + K \\
&= \frac{e^x}{2} (x \sin x + (1-x) \cos x) + K. \end{align*}$$ The value of the definite integral is straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int \int \sqrt{(x^2 + y^2)}$ $dx dy$ over a triangle with corners (0, 0), (4, 4), and (4, 0) using $x = u$, $y = uv$. Help?
So far I have:
$$a = x = 4\\ b = y = 0\\ c = y = x$$
$$y=x \rightarrow uv = u\rightarrow v=u/u \rightarrow v=1$$
$$x = 4 \rightarrow u = 4$$
$$y = 0\rightarrow uv = 0 \rightarrow u = 0 \rightarrow v = 0$$
$$\sqrt {(x^2 + y^2)} = (u^2 +uv^2)^{(1/2)}$$
\begin{align} \|J\| & =det\left[ {\begin{array}{cc}
\frac {∂x}{∂u} & \frac{∂x}{∂v} \\
\frac{∂y}{∂u} & \frac{∂y}{∂v}\\
\end{array} } \right] = \left[ {\begin{array}{cc}
1 & v \\
0 & u\\
\end{array} } \right] = u = 4
\end{align}
\begin{align*}
\int_{0}^{4}\int_{0}^{1} \sqrt{(u^2 + uv^2)} + 4 ~dvdu = \int_{0}^{4}\int_{0}^{1} (u^2 + uv^2)^{1/2} + 4 ~dvdu
\end{align*}
I can already see this is incorrect due to the given answer.
This gives me J = 0, however J is supposed to = 1? and the answer to the problem is supposed to be:
$$\frac{32(\sqrt2+ln(1+\sqrt2))}3$$
according to the website: https://www.whitman.edu/mathematics/calculus_online/section15.07.html
| The Jacobian is not correct. It should be $J=u$. Then, we have
$$\begin{align}
\int_0^4\int_0^y \sqrt{x^2+y^2}\,dx\,dy&=\int_0^1\int_0^4 u^2\sqrt{v^2+1}\,du\,dv\\\\
&=\int_0^4 u^2 \,du\,\int_0^1\sqrt{v^2+1}\,dv\\\\
&=\frac{64}{3}\,\left(\frac12\left(\sqrt 2+\log(1+\sqrt 2)\right)\right)\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{32\left(\sqrt 2+\log(1+\sqrt 2)\right)}{3}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving an equation containing cube roots: $\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$ I'm trying to figure out a way to solve this equation:
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1.$$
I tried to cube both sides, but I ended up with an equation looking like this:
$$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1}-\sqrt[3]{5x+7})=-6.$$
At this point I'm out of stuff to do. Any help would be appreciated. Thanks in advance.
| Before cubing, separate your roots:
$$\sqrt[3]{5x+7}=\sqrt[3]{5x-12}+1$$
Now cube:
$$5x + 7 = 5x - 12 + 3\sqrt[3]{5x-12}^2 + 3\sqrt[3]{5x-12} + 1$$
Simplify:
$$ \sqrt[3]{5x-12}^2 + \sqrt[3]{5x-12} - 6 = 0$$
Solve for $\sqrt[3]{5x-12}$ (it is a quadratic equation). Finally solve to $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Series $(1+b^n/n)^n$ at infinity Is there a series expansion for
$$\left ( 1 + \frac{b^n}{n} \right )^n$$
at $n\rightarrow\infty$, assuming $0<b<1$?
Similar to
$$ \left ( 1 + \frac{b}{n} \right )^n \rightarrow e^b $$
I want to want to write this as
$$\left ( 1 + \frac{b^n}{n} \right )^n \rightarrow e^{b^n} $$
But I am not sure if this is correct. Thanks!
| For $0 < b < 1$, we have $b^n \to 0$ as $n \to \infty$ and it is true that
$$\lim_{n \to \infty} \left(1 + \frac{b^n}{n} \right)^n = e^0 = 1.$$
This follows because
$$\tag{*}\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n = e^x,$$
converging uniformly for $x$ in a compact set.
Justification for all of this requires showing, first, given sequences $f_n(x)$ and $a_n$, if $f_n(x) \to f(x)$ uniformly and $a_n \to a$, then $f_n(a_n) \to f(a)$. This is a straightforward application of the properties of uniform convergence.
Second, we have to verify the uniform convergence of the limit (*) for $x \in [0,1)$.
If $0 \leqslant x < n$, we have
$$\log(1+x/n) = \frac{x}{n} - \frac{x^2}{2n^2} + \frac{x^3}{3n^3} - \frac{x^4}{4n^4} + \ldots,$$
and, using the triangle inequality
$$\left|n\log(1+x/n) - x\right| \leqslant \frac{x^2}{2n} + \frac{x^3}{3n^2} + \frac{x^4}{4n^3} + \ldots.$$
Since, $0 \leqslant x < 1$, for all $n > 2 \geqslant 2x$ we have $x/n \leqslant 1/2$ and
$$\left|n\log(1+x/n) - x\right| \leqslant \frac{x^2}{n}\left[ \frac{1}{2} + \frac{x}{3n} + \frac{x^2}{4n^2} + \ldots\right] \\ \leqslant \frac{x^2}{n}\left[ \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 2^2} + \ldots\right] \\ \leqslant \frac{x^2}{n}\left[ \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots\right] \\ = \frac{x^2}{n} \\ \leqslant \frac{1}{n}.$$
Thus,
$$\lim_{n \to \infty} n\log(1+x/n) = \lim_{n \to \infty} \log(1+x/n)^n = x,$$
uniformly for $0 \leqslant x < 1$.
The function $x \mapsto \exp(x)$ is uniformly continuous on compact sets and it follows that
$$\lim_{n \to \infty} (1+x/n)^n = e^x, $$
uniformly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate this improper integral $\int_{0}^{\infty}\frac{1-x}{1-x^{n}}\,\mathrm dx$?
$\def\d{\mathrm{d}}$How to evaluate this improper integral?
$$\int_{0}^{\infty}\frac{1-x}{1-x^{n}}\,\d x.$$
What I tried is a substitution i.e $x^{n}=t$, but then things got complicated, and I'm stuck.
| We are going to evaluate the integral
$I:=\displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{} $
by the theorem
$ \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \textrm{ where } z\notin Z.\tag*{} $
We first split the integral into $2$ integrals
$ \displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{1-x}{1-x^{n}} d x+\int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{}$
Transforming the latter integral by the inverse substitution $ x\mapsto \frac{1}{x} $, we have
$$\displaystyle \int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{x^{n-3}-x^{n-2}}{1-x^{n}} d x \tag*{} $$
Putting back yields
$$ \begin{aligned}\displaystyle I&=\int_{0}^{1} \frac{1-x+x^{n-3}-x^{n-2}}{1-x^{n}} d x\\\displaystyle &=\int_{0}^{1}\left[\left(1-x+x^{n-3}-x^{n-2}\right) \sum_{k=0}^{\infty} x^{n k}\right] d x\\ \displaystyle & =\sum_{k=0}^{\infty} \int_{0}^{1}\left[x^{n k}-x^{n k+1}+x^{n(k+1)-3}-x^{n(k+1)-2}\right] d x\\\displaystyle & =\sum_{k=0}^{\infty}\left(\frac{1}{n k+1}-\frac{1}{n k+2}+\frac{1}{n(k+1)-2}-\frac{1}{n(k+1)-1}\right)\\\displaystyle & =\sum_{k=0}^{\infty}\left[\frac{1}{n k+1}-\frac{1}{n(k+1)-1}\right]+\sum_{k=0}^{\infty}\left[\frac{1}{n(k+1)-2}-\frac{1}{n k+2}\right]\\ \displaystyle & =\frac{1}{n}\left[\sum_{k=0}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-1}^{-\infty} \frac{1}{k+\frac{1}{n}}\right]+\frac{1}{n}\left(\sum_{k=1}^{\infty} \frac{1}{k-\frac{2}{n}}+\sum_{k=0}^{-\infty} \frac{1}{k-\frac{2}{n}}\right)\\\displaystyle & =\frac{1}{n}\left(\sum_{k=-\infty}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-\infty}^{\infty} \frac{1}{k-\frac{2}{n}}\right) \end{aligned}$$
By the Theorem
$\displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \tag*{} $
where $ z\notin Z.$
We can now conclude that
$\displaystyle \boxed{ I =\frac{1}{n}\left[\pi \cot \left(\frac{\pi}{n}\right)+\pi \cot \left(\frac{-2 \pi}{n}\right)\right]=\frac{\pi}{n}\left[\cot \left(\frac{\pi}{n}\right)-\cot \left(\frac{2 \pi}{n}\right)\right] =\frac{\pi}{n} \csc \frac{2 \pi}{n}} \tag*{} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Q6 polynomial complex Find the equation whose roots are the fourth powers of the roots of the equation $x^3 + x + 1 = 0$. Hence find the sum of the fourth powers of the roots of the equation $x^3 + x + 1 = 0$
I am really stuck. The answer at the back says cubic is $u^3 - 2u^2 + 5u -1 = 0$ and sum of roots is 2
| So we seek a cubic in $y = x^4, $ where $x^3+x+1=0$.
Hence we have $y = x^4 = x(x^3) = -x(x+1) = -x^2-x$, using which, $y^2 = (x^4)^2 = x^2(x+1)^2 = x^4+2x^3+x^2 = -3x-2$.
Similarly, $y^3 = (x^4)^3 = -x^3(x+1)^3 = 5x^2-x-3$.
Now it is easy to eliminate (the powers of) $x$ and get $y^3-2y^2 +5y = 1\dots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the angle between 2 intersection lines of four planes?
Given $L_1=\begin{cases}P_1: x-y+z=4\\ P_2: 2x+y-z=6 \end{cases}$
and $L_2=\begin{cases}P_1:x+y+z=4\\ P_2:2x+3y-z=6 \end{cases}$
$L_1$ and $L_2$ are the intersection lines of the given planes. Find the angle between the lines $L_1,L_2$
First in order to find the intersection lines between the respective planes it is enough to find to dots that belong to the intersection line, so for $L_1$:
$$
P_1+P_2 \Rightarrow 3x=10 \Rightarrow x={10 \over 3}
$$
Then:
$$
2 \cdot \frac{10}{3}+y-z=6 \Rightarrow y-z=-\frac{2}{3}
$$
For $z=0$ and $z=1$ we get $y=-\frac{2}{3}$ and $y=\frac{1}{3}$ respectively so now can represent $L_1$ parametrically: $(\frac{10}{3},\frac{1}{3}, 10)+t(0,-1,0)$ where $t(0,-1,0)$ is the direction vector.
After identical calculations we can find $L_2: (0,\frac{10}{4}, \frac{3}{2})+t(\frac{10}{3}, 0,-\frac{2}{3})$.
Now we can calculate the angle of $L_1, L_2$ via scalar product:
$$
\cos \theta=\frac{L_1 \cdot L_2}{\Vert{L_1}\Vert \cdot \Vert L_2 \Vert}=0 \Rightarrow \theta=90^{\circ}
$$
I'm not sure that my answer is correct and I feel I'm missing something.
| Let me answer the general case.
First finding intersection of planes:
In general planes in $\mathbb{R}^3$ are of the form:
$$ax+by+cz+d=0.$$
The intersection of two planes $ax+by+cz+d=0$ and $Ax+By+Cz+D=0$ exists if and only if $|aA+bB+cC|\neq ||(a,b,c)||.||(A,B,C)||$ or if $|aA+bB+cC|=||(a,b,c)||.||(A,B,C)||$, then $dC=Dc$. The latter case means planes coincide.
Note that $|aA+bB+cC|=||(a,b,c)||.||(A,B,C)||$ implies two planes are parallel.
If two planes have intersection then by letting one of the variables $x$, $y$ or $z$ equal to $t$ we will find other variables. Here let $z=t$ then solve:
$$\left\{\begin{array}{} ax+by=d-ct\\ Ax+By=D-Ct\end{array}\right.$$
we are now able to find values of $x$ and $y$. After calculation we have something like the following form:
$$\frac{x-a'}{a''}=\frac{y-b'}{b''}=\frac{z}{1}(=t).$$
Now, do the similar arguments for two other planes. We have:
$$\frac{x-A'}{A''}=\frac{y-B'}{B''}=\frac{z}{1}(=T).$$
Now we have two vectors $(A'',B'',1)$ and $(a'',b'',1)$ and by using the formula:
$a''A''+b''B''+1=\sqrt{a''^2+b''^2+1}.\sqrt{A''^2+B''^2+1}.cos(\alpha)$ the angle $\alpha$ is easy to obtain.
Now in your question: $$L_1: x=\frac{10}{3},y+\frac{2}{3}=z(=t)$$ here $v_1=(0,1,1)$ and $$L_2:\frac{x-6}{-4}=\frac{y+2}{3}=z(=t)$$ here $w=(-4,3,1)$. So, $$cos(\alpha)=\frac{4}{\sqrt{2}\sqrt{26}}=\frac{2\sqrt{13}}{13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Step-by-step derivative of $\left ( \frac{c_1 x}{c_2 x + c_3 + c_4 \sqrt{c_5 x}} \right)^{c_6x + c_7 + c_8 \sqrt{c_9 x}}$ Can someone please walk step by step on how to calculate the derivative
$\left ( \frac{c_1 x}{c_2 x + c_3 + c_4 \sqrt{c_5 x}} \right)^{c_6x + c_7 + c_8 \sqrt{c_9 x}}$
Where the $c_i$ are constants, which are positive.
| Define
$$
f(x) = g(x)^{h(x)} = e^{h(x)\ln g(x)}
$$
Compute
$$
f'(x) = e^{h(x)\ln g(x)} \left(
\frac{h(x)g'(x)}{g(x)} + h'(x) \ln g(x)
\right)
$$
Define
$$
\begin{align}
\alpha(x) &= c_{2} + c_{3} x + c_{4} \sqrt{c_{5}x} \\[2pt]
g(x) &= \frac{c_{1}x}{\alpha(x)} \\[2pt]
h(x) &= c_{6} + c_{7} x + c_{8} \sqrt{c_{9} x}
\end{align}
$$
Derivatives
$$
\begin{align}
\alpha'(x) &= c_{3} + \frac{c_{4}c_{5}}{2\sqrt{c_{5}x}} \\
g'(x) &= \frac{c_{1} \left(\alpha( x)- x \alpha'(x)\right)}{\alpha(x)^2} \\[3pt]
h'(x) &= c_{7}+\frac{c_{8} c_{9}}{2 \sqrt{c_{1} x}}
\end{align}
$$
Final answer
$$
f'(x) = \left(\frac{c_{1} x}{c_{2}+c_{3} x+c_{4} \sqrt{c_{5} x}}\right)^{c_{6}+c_{7} x+c_{8} \sqrt{c_{9} x}} \left(\left(c_{7}+\frac{c_{8} c_{9}}{2 \sqrt{c_{9} x}}\right) \log \left(\frac{c_{1} x}{c_{2}+c_{3} x+c_{4} \sqrt{c_{5} x}}\right)+\frac{c_{5} \left(2 c_{2} \sqrt{c_{5} x}+c_{4} c_{5} x\right) \left(c_{6}+c_{7} x+c_{8} \sqrt{c_{9} x}\right)}{2 (c_{5} x)^{3/2} \left(c_{2}+c_{3} x+c_{4} \sqrt{c_{5} x}\right)}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Chain rule differentiation
Given that $\overline{r} = \frac{r}{r_0}$ and $u = \frac{1}{\overline{r}}$, show that
$$r \frac{d^2r}{d\theta^2} - 2\left(\frac{dr}{d\theta} \right)^2 = r^2 - 3\frac{Gm}{c^2}r$$
is equivalent to the following;
$$\frac{d^2u}{d\theta^2} + u = \epsilon u^2 : \epsilon = \frac{3Gm}{r_0c^2}
$$
I understand that we can start with the working $\frac{dr}{d\theta} = r_0 \frac{d({\frac{1}{u})}}{d\theta}$ but then I cannot see how you would move further on.
Any help would be appreciated.
| \begin{equation}
r=\frac{r_{0}}{u}
\end{equation}
\begin{equation}
\frac{dr}{d\theta}=\frac{d}{d\theta}(\frac{r_{0}}{u})
\end{equation}
Applying chain rule of differentiation you'll get
\begin{equation}
\frac{dr}{d\theta}=\frac{-r_{0}}{u^{2}}\frac{du}{d\theta}
\end{equation}
\begin{equation}
\frac{d^{2}r}{d\theta^{2}}=\frac{d}{d\theta}(\frac{-r_{0}}{u^{2}}\frac{du}{d\theta})
\end{equation}
Here you'll have to use product rule of differentiation that is, if independent variable of differentiation is $t$ then,
\begin{equation}
\frac{d}{dt}(uv)=u\frac{dv}{dt}+v\frac{du}{dt}
\end{equation}
\begin{equation}
\therefore \frac{d^{2}r}{d\theta^{2}}=\frac{2r_{0}}{u^{3}}\left(\frac{du}{d\theta}\right)^{2}-\frac{r_{0}}{u^{2}}\frac{d}{d\theta}\left(\frac{du}{d\theta}\right)
\end{equation}
\begin{equation}
\therefore \frac{d^{2}r}{d\theta^{2}}=\frac{2r_{0}}{u^{3}}\left(\frac{du}{d\theta}\right)^{2}-\frac{r_{0}}{u^{2}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
Multiplying on both sides by r in above equation we get,
\begin{equation}
r\frac{d^{2}r}{d\theta^{2}}=\frac{2r_{0}r}{u^{3}}\left(\frac{du}{d\theta}\right)^{2}-\frac{r_{0}r}{u^{2}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
Substitue $r$ in terms of $u$
\begin{equation}
r\frac{d^{2}r}{d\theta^{2}}=\frac{2r_{0}^{2}}{u^{4}}\left(\frac{du}{d\theta}\right)^{2}-\frac{r_{0}^{2}}{u^{3}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
And
\begin{equation}
\left(\frac{dr}{d\theta}\right)^2=\frac{r_{0}^{2}}{u^{4}}\left(\frac{du}{d\theta}\right)^2
\end{equation}
Thus,
\begin{equation}
2\left(\frac{dr}{d\theta}\right)^2=\frac{2r_{0}^{2}}{u^{4}}\left(\frac{du}{d\theta}\right)^2
\end{equation}
Subtract appropriate equations and use original differential equation in terms of $r$
\begin{equation}
r\frac{d^{2}r}{d\theta^{2}}-2\left(\frac{dr}{d\theta}\right)^2=-\frac{r_{0}^{2}}{u^{3}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
\begin{equation}
r^{2}-\frac{3GMr}{c^{2}}=-\frac{r_{0}^{2}}{u^{3}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
Again on Left hand side convert $r$ to $u$
\begin{equation}
\frac{r_0^{2}}{u^{2}}-\frac{3GMr_0}{c^{2}u}=-\frac{r_{0}^{2}}{u^{3}}\left(\frac{d^{2}u}{d\theta^{2}}\right)
\end{equation}
Divide by $-r_0^{2}$ and multiply by $u^3$ on both sides
\begin{equation}
-u+\frac{3GMu^2}{c^{2}r_0}=\frac{d^{2}u}{d\theta^{2}}
\end{equation}
Rearranging and putting value of $\epsilon$ you'll have
\begin{equation}
\frac{d^{2}u}{d\theta^{2}}+u=\epsilon u^{2}
\end{equation}
(Note: This kind of manipulation is useful particularly when you're trying to find orbit equation in classical mechanics problem, I'm sure you're doing that only)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $x+y+z=1$ Find Maximum value of $x^5y+y^5z+z^5x$ $x,y,z$ are Non negative reals such that $x+y+z=1$ Find Maximum value of $$E=x^5y+y^5z+z^5x$$
The only idea i have is $E$ can be written as $f(x,y)$ and using partial differentiation for maxima...but too lengthy
| Let $x\geq y\geq z$. Hence, by AM-GM
$$x^5y+y^5z+z^5x\leq(x+z)^5y=5^5\left(\frac{x+z}{5}\right)^5y\leq5^5\left(\frac{5\cdot\frac{x+z}{5}+y}{6}\right)^6=\frac{5^5}{6^6}.$$
In the case $x\geq z\geq y$ we can use the same idea.
$$x^5y+y^5z+z^5x\leq x^5z+y^5x+z^5y\leq(x+y)^5z\leq5^5\left(\frac{5\cdot\frac{x+y}{5}+z}{6}\right)^6=\frac{5^5}{6^6}$$
The equality occurs for $z=0$ and $y=\frac{x}{5}$, which gives the answer: $\frac{5^5}{6^6}$.
| {
"language": "en",
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"source": "stackexchange",
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Closed Form for an Alternating Sum Involving Binomial Coefficients
Question. For integer $n \geq 0$, find the closed form for
$$
S_n = \sum_{k \leq 2^n} \binom{2^n - k}{k}(-1)^k
$$
My Attempt: I tried some small $n$ and got $S_0 = 1$, $S_1 = 1$, $S_2 = 0$, $S_3 = -1$ and $S_4 = -1$, but failed to come up with some patterns. Could you please provide some hints so that I can complete the remaining part by myself?
| The calculation is not correct in all aspects. We obtain
\begin{align*}
\color{blue}{S_0}&=\sum_{k=0}^1\binom{1-k}{k}(-1)^k=\binom{1}{0}=\color{blue}{1}\\
\color{blue}{S_1}&=\sum_{k=0}^2\binom{2-k}{k}(-1)^k=\binom{2}{0}-\binom{1}{1}
=1-1=\color{blue}{0}\\
\color{blue}{S_2}&=\sum_{k=0}^4\binom{4-k}{k}(-1)^k=\binom{4}{0}-\binom{3}{1}+\binom{2}{2}
=1-3+1=\color{blue}{-1}\\
\color{blue}{S_3}&=\sum_{k=0}^8\binom{8-k}{k}(-1)^k
=\binom{8}{0}-\binom{7}{1}+\binom{6}{2}-\binom{5}{3}+\binom{4}{4}\\
&=1-7+15-10+1=\color{blue}{0}\\
\color{blue}{S_4}&=\sum_{k=0}^{16}\binom{16-k}{k}(-1)^k\\
&=\binom{16}{0}-\binom{15}{1}+\binom{14}{2}-\binom{13}{3}+\binom{12}{4}-\binom{11}{5}\\
&\qquad+\binom{10}{6}-\binom{9}{7}+\binom{8}{8}\\
&=1-15+91-286+495-462+210-36+1=\color{blue}{-1}\\
\end{align*}
Hint: Since the index $2^n$ increases quickly it is not easy to manually derive the values of $S_n$. It looks plausible that we could more easily derive a pattern for
\begin{align*}
T_m&=\sum_{k \leq m} \binom{m - k}{k}(-1)^k\qquad\qquad m\geq 0
\end{align*}
Observe that $S_n=T_{2^m}$ for $m\geq 0$. So, if we can observe a pattern for the sequence $T_m$
we can derive from it a pattern for the sequence
\begin{align*}
S_n=T_{2^m}
\end{align*}
| {
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} |
How to solve $\frac{dx}{ds}=\frac{x}{x^2+y^2},\quad\frac{dy}{ds}=\frac{y}{x^2+y^2}$? I was wondering if my method to solve the following pair of equations:
$$\frac{dx}{ds}=\frac{x}{x^2+y^2},\quad\frac{dy}{ds}=\frac{y}{x^2+y^2}$$
is the right one.
My attempt: We have $\frac{x^2+y^2}{x}dx=ds$ and $\frac{x^2+y^2}{y}dy=ds$, so $$\frac{1}{x}dx=\frac{1}{y}dy.$$
The general solution of the first equation is: $y=Ax$, where $A$ is a constant.
Thank you in advance.
| You have $$\frac{\text{d}}{\text{d}s}\,\left(x^2+y^2\right)=2\,.$$
So, $x^2+y^2=2(s+c)$ for some constant $c$. Plugging this in, you will see that $$x=\pm\sqrt{2A(s+c)}\text{ and }y=\pm\sqrt{2(1-A)(s+c)}$$ for some constant $A$.
| {
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"source": "stackexchange",
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} |
Mysterious $w$ -- a function of triangle sides Suppose $a$, $b$, $c$ are the lengths of the three sides of a triangle. When studying rational Bézier curves inscribed in the triangle, a mysterious quantity $w$ emerges in the algebra:
$$
w = \frac{c}{\sqrt{2(a^2 + b^2)}}
$$
I'd like to get a better understanding of $w$. Specifically:
*
*Does it represent anything, geometrically? A ratio of lengths or areas, maybe, or some trig function of some angle??
*Can we say anything about its range of values? For example, is it true that $w \le 1$?
A couple of special cases I already figured out:
*
*If $a=b$ (the triangle is isosceles), then $w$ is the sine of the base angle (the angle between the sides of lengths $a$ and $c$).
*If the sides of length $a$ and $b$ form a right angle, then $w^2 = \tfrac12$.
| It looks as if $w^2$ is more tractable. So
$$2w^2=\frac{c^2}{a^2+b^2}$$
and
$$1-2w^2=\frac{a^2+b^2-c^2}{a^2+b^2}=\frac{2ab\cos C}{a^2+b^2}.$$
For fixed $a$ and $b$ as $C$ varies, this varies between $\pm2ab/(a^2+b^2)$. So $2w^2$ varies between $|a-b|^2/(a^2+b^2)$
and $|a+b|^2/(a^2+b^2)$. Thus
$$2w^2\le\frac{a^2+2ab+b^2}{a^2+b^2}\le2$$
by AM/GM so indeed $w\le1$.
I don't see any nice geometry so far.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Is this the correct way to find the Laurent Expansion of the complex function $f(z) = \frac{z}{z^2 - 1}$? The function
$$
f(z) = \frac{z}{z^2 - 1}
$$
has singularities at $z = \pm 1$. I will expand about the point $z = 1$. Then, with the substitution $w = z+1$,
\begin{align*}
f(x) =& \frac{z}{z^2 - 1} \\[3mm]
=& \frac{z}{(z+1)(z-1)} \\[3mm]
=& \frac{w-1}{w(w-2)} \\[3mm]
=& \frac{1-w}{2w} \cdot \frac{1}{1 - \frac{w}{2}} \\[3mm]
=& \frac{1-w}{2w} \sum_{n=0}^\infty \frac{w^n}{2^n} \\[3mm]
=& \sum_{n=0}^\infty \frac{w^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{w^n}{2^{n+1}} \\[3mm]
=& \sum_{n=0}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}
\end{align*}
This is almost in the form of a Laurent expansion, but not quite. How can I continue this to get the correct final result?
| From where you left off:
\begin{align}
&= \sum_{n=0}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px]
&=\frac{(z+1)^{-1}}{2}+
\sum_{n=1}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px]
&=\frac{(z+1)^{-1}}{2}+
\sum_{n=0}^\infty \frac{(z+1)^{n}}{2^{n+2}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px]
&=\frac{(z+1)^{-1}}{2}+\sum_{n=0}^{\infty}
\left(\frac{1}{2^{n+2}}-\frac{1}{2^{n+1}}\right)(z+1)^n\\[4px]
&=\frac{(z+1)^{-1}}{2}+\sum_{n=0}^{\infty}
-\frac{(z+1)^n}{2^{n+2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2262582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\sin^4(x)=A+B\cos(2x)+C\cos(4x)$, then find $A$, $B$, and $C$.
If $\sin^4(x)=A+B\cos(2x)+C\cos(4x)$, then find $A, B$, and $C$.
Would I use half-angle identities for this problem?
| By the double-angle formula for $\cos{2x}$,
$$ \cos{2x} = \cos^2{x} - \sin^2{x} = 2\cos^2{x}-1 = 1-2\sin^2{x}. $$
Rearranging,
$$ \sin^2{x} = \frac{1}{2}(1-\cos{2x}). $$
Squaring both sides an expanding,
$$ \sin^4{x} = \frac{1}{4}(1-2\cos{2x}+\cos^2{2x}), $$
and then using the double-angle formula $\cos{4x} = 2\cos^2{x}-1 $ gives
$$ \sin^4{x} = \frac{1}{4}-\frac{1}{2}\cos{2x}+\frac{1}{4}\frac{1}{2}(1+\cos{4x}) = \frac{3}{8} - \frac{1}{2}\cos{2x}+\frac{1}{8}\cos{4x}. $$
Alternatively,
$$ \sin^4{x} = \frac{(e^{ix}-e^{-ix})^4}{2^4i^4} = \frac{1}{16}(e^{4ix} - 4e^{2ix} + 6 - 4e^{-2ix} + e^{-4ix} ) = \frac{1}{8}(\cos{4x}-4\cos{2x}+3). $$
| {
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"source": "stackexchange",
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Show that $\sin(3x)$ is equivalent to $3\sin(x)\cos^2(x)-\sin^3(x)$
The expression $\sin(3x)$ is equivalent to:
A. ...
My book states the right answer is B which is $3\sin(x)\cos^2(x)-\sin^3(x)$.
I tried:
$$\sin(x)\cos(2x)+\cos(x)\sin(2x) = \\
\sin(x)(2\cos^2(x)-1)+\cos(x)\cdot2\sin(x)\cos(x) = \\
2\cdot\sin(x)\cos(x)\cos(x)-\sin(x)+\cos(x)2\sin(x)\cos(x) = \\
\sin(x)(2\cos^2(x)-1+2\cos^2(x)) = \\
\sin(x)(4\cos^2(x)-1) = \\
4\cos^2(x)\sin(x)-\sin(x) = ???$$
What do I do next?
| Almost there (from your 5th line):
$$
\sin x (4\cos^2 x - 1) = \sin x (3\cos^2 x + \cos^2 x - 1)
$$
$$
= \sin x (3\cos^2 x + (1-\sin^2 x) - 1) = 3\sin x\cos^2 x-\sin^3x.
$$
| {
"language": "en",
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"answer_id": 2
} |
Using Taylor's theorem show that $1 - \frac{x^2}{2} < \cos x < 1- \frac{x^2}{2} + \frac{x^4}{24}$ Using Taylor's theorem show that
$1-\frac{x^2}{2} < \cos x < 1-\frac{x^2}{2} + \frac{x^4}{24} \, \text{ for all } x \in \mathbb{R}$
I know this is true because of the Alternating series truncation error that alternates in sign, but is there a way to prove it?
| We keep integrating the inequality $\cos x\le x$, for $x\ge 0$, in the interval $[0,x]$, and we recursively obtain
$$
\cos x \le 1\quad\Longrightarrow\quad
\sin x \le x\quad\Longrightarrow\quad
1-\cos x \le \frac{x^2}{2!}\quad\Longrightarrow\quad
1- \frac{x^2}{2!}\le\cos x \quad\Longrightarrow\quad
x-\frac{x^3}{3!}\le \sin x\quad\Longrightarrow\quad \frac{x^2}{2!}-\frac{x^4}{4!}\le 1-\cos x\quad\Longrightarrow\quad
\cos x\le 1-\frac{x^2}{2!}+\frac{x^4}{4!}\quad\Longrightarrow\quad\cdots
$$
In general, continuing this we obtain
$$
\sum_{k=0}^{2n-1}\frac{(-1)^kx^{2k}}{(2k)!}\le\cos x\le \sum_{k=0}^{2n}\frac{(-1)^kx^{2k}}{(2k)!}
$$
Note that it holds, not only for $x\ge 0$, but for $x<0$, since all the functions above are even.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Factoring $x^5 - 5x^4 + 1$ The original problem was stated like this:
Prove that the polynomial $x^5 - 5x^4 + 1$ does not have roots of multiplicity 4.
So, factoring the polynomial would answer the question, but I don't know how to do it.
Using the root test, ${1, -1}$ should be roots, but If I divide by $x-1$ or $x+1$ $\frac {x^5 - 5x^4 + 1}{x-1}$, or $\frac {x^5 - 5x^4 + 1}{x-1}$ I get a remainder, so those are not roots?!
| extended gcd in $\mathbb Q[x].$ the original polynomial and its derivative are coprime; there are no multiple roots, not even double.
$$ \left( x^{5} - 5 x^{4} + 1 \right) $$
$$ \left( x^{4} - 4 x^{3} \right) $$
$$ \left( x^{5} - 5 x^{4} + 1 \right) = \left( x^{4} - 4 x^{3} \right) \cdot \color{magenta}{ \left( x - 1 \right)} + \left( - 4 x^{3} + 1 \right) $$
$$ \left( x^{4} - 4 x^{3} \right) = \left( - 4 x^{3} + 1 \right) \cdot \color{magenta}{ \left( \frac{ - x + 4 }{ 4 } \right)} + \left( \frac{ x - 4 }{ 4 } \right) $$
$$ \left( - 4 x^{3} + 1 \right) = \left( \frac{ x - 4 }{ 4 } \right) \cdot \color{magenta}{ \left( - 16 x^{2} - 64 x - 256 \right)} + \left( -255 \right) $$
$$ \left( \frac{ x - 4 }{ 4 } \right) = \left( -255 \right) \cdot \color{magenta}{ \left( \frac{ - x + 4 }{ 1020 } \right)} + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( x - 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( x - 1 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{\left( \frac{ - x + 4 }{ 4 } \right)} \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x^{2} + 5 x }{ 4 } \right) }{ \left( \frac{ - x + 4 }{ 4 } \right) } $$
$$ \color{magenta}{\left( - 16 x^{2} - 64 x - 256 \right)} \Longrightarrow \Longrightarrow \frac{ \left( 4 x^{4} - 4 x^{3} - 16 x^{2} - 319 x - 1 \right) }{ \left( 4 x^{3} - 255 \right) } $$
$$ \color{magenta}{ \left( \frac{ - x + 4 }{ 1020 } \right)} \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - x^{5} + 5 x^{4} - 1 }{ 255 } \right) }{ \left( \frac{ - x^{4} + 4 x^{3} }{ 255 } \right) } $$
$$ \left( x^{5} - 5 x^{4} + 1 \right) \left( \frac{ 4 x^{3} - 255 }{ 255 } \right) - \left( x^{4} - 4 x^{3} \right) \left( \frac{ 4 x^{4} - 4 x^{3} - 16 x^{2} - 319 x - 1 }{ 255 } \right) = \left( -1 \right) $$
| {
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Differentiation - Finding derivatives of this function -
$h(y)= (y^{-3}+2y)(2+3y^3)^{-1/2} $
I tried and weren't sure on how to simplify it to get the answer
$\frac{8-27y^{-1} - 12y^4-6y^3}{2(2+3y^3)^{3/2}} $
This is my try -
$ (y^{-3} + 2y) (\frac{-1}{2} (2+3y^3)^{-3/2} (0+9y^2)) + (2+3y^3)(-3y^{-4} + 2) $
I simplified it till -
$ (2+3y^3)^{-3/2}(-9/2 (y^{-3} + 2y) + (2+3y^3)^{5/2}(-3y^{-4} +2) $
$ \frac{(-9/2 (y^{-3} + 2y) + (2+3y^3)^{5/2}(-3y^{-4} +2)}{ (2+3y^3)^{3/2}}$
I'm not too sure how to simplify this further .. thanks !
| There are two typos in your calculation
We obtain applying the product rule
\begin{align*}
&(y^{-3}+2y)(2+3y^3)^{-1/2}\\
&\qquad= (y^{-3} + 2y) \left(-\frac{1}{2} (2+3y^3)^{-3/2} (0+9y^2)\right) + (2+3y^3)^{\color{blue}{-1/2}}(-3y^{-4} + 2)\tag{1}\\
&\qquad= (2+3y^3)^{-3/2} \left(-\frac{9}{2}\color{blue}{y^2} (y^{-3} + 2y)+ {{(2+3y^3)}}(-3y^{-4} + 2)\right)\tag{2}\\
&\qquad=(2+3y^3)^{-3/2}\left(-\frac{9}{2}y^{-1}-9y^{3}-6y^{-4}-9y^{-1}+4+6y^3\right)\\
&\qquad=2(2+3y^3)^{-3/2}\left(8-27y^{-1} - 12y^4-6y^3\right)
\end{align*}
and the claim follows.
Comment:
*
*In (1) a power $-\frac{1}{2}$ in $(2+3y^3)^{\color{blue}{-1/2}}$ was missing
*In (2) a factor $\color{blue}{y^2}$ was missing
| {
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How to find the sum of the series $(1) + (2 + 3) +(4 + 5 + 6) + \cdots$ to $n$ terms? How to find the sum of this series?
$$(1) + (2 + 3) +(4 + 5 + 6) + \cdots \text{ to }\, n \, \text{ terms }$$
Answer is given as:
$$\frac {1}{8} n (n + 1) (n^2 + n +2) $$
| The $k$th term is composed of $k$ integers, so the first $n$ terms, cumulatively, will be the composed of $\displaystyle F(n) = \sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$ consecutive integers overall.
And to actually sum up those consecutive integers, we apply the same idea again.
$$S = F(F(n)) = F\left(\frac{n(n+1)}{2}\right) = \sum_{k=1}^{\frac{n(n+1)}{2}} k= \dfrac{\dfrac{n(n+1)}{2}\left(\dfrac{n(n+1)}{2}+1\right)}{2} = \frac{n (n^3 + 2 n^2 + 3 n + 2)}{8}$$
Edit: All of this relies on the fact that the sum of the first $n$ integers can be expressed in the closed form $F(n) = \dfrac{n(n+1)}{2}$ in general.
So for this problem, we're first determining how many consecutive integers show up in the first $n$ terms, which is $F(n)$. But now we need to sum the first $F(n)$ terms, so it's like computing $F(F(n))$.
| {
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Rational numbers inequality proof - is this proof valid?
Knowing that $x^2+y^2 = 2 $ prove that $x+y \le 2$
I rewrite this as
$$x^2+y^2 \ge x+y$$
Now, multiply both sides by 2
$$x^2 -2x + 1 + y^2-2y + 1 -2 +x^2 + y^2 \ge 0$$
I substitue $2$ for $x^2+y^2$
$$(x-1)^2+(y-1)^2\ge0$$
Which is true
| Using Cauchy-Schwarz:
$$(1+1)(x^2+y^2)\ge(x+y)^2 \Rightarrow (x+y)^2\le4 \Rightarrow |x+y|\le2.$$
| {
"language": "en",
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"question_score": "4",
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Minimum value of $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}$ Let $x,y,z>0$ and $x+y+z=xyz$. What is the minimum value of $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}?$$
In the case when $x=y=z$, the equation $x+y+z=xyz$ translates to $3x=x^3$, or $x=\sqrt{3}$. If $A$ denotes the quantity that we want to minimize, then $A=\sqrt{3}$ as well.
If we use the inequality of arithmetic and geometric means, we get
$$A\geq \frac{3}{\sqrt[3]{xyz}}=\frac{3}{\sqrt[3]{x+y+z}}.$$
| The constraint allows the substitution $x=\tan A, y = \tan B, z = \tan C$ for some acute triangle $\triangle ABC$. Further from rearrangement inequality we have:
$$\sum_{cyc} \frac{x}{y^2} \geqslant \sum_{cyc} \frac1x = \sum_{cyc} \cot A$$
Now $x \mapsto \cot x$ is convex for $x \in (0, \frac{\pi}2)$, so we may use Jensen to conclude
$$\sum_{cyc} \cot A \geqslant 3 \cot \frac{\pi}3$$
Equality is when $x=y=z$.
| {
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Substitution Integral: $\int_0^3\frac {\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} dx$ $$\int_0^3\frac {\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} dx $$ I tried with substitution $t=3-x$ but I don't know what to do next.
Also how do I solve this limit: $$\lim_{x\to\infty}\frac{1}{x}\int_0^x\frac{dt}{2+\cos t} $$ I saw that most problems like this are solved by putting the period of the function in front,would it work here?
| Form the properties of definite integral we know that:
$$\int_b^af(x)\ dx=\int_b^af(a+b-x)\ dx\\
\implies \int_0^af(x)\ dx=\int_0^af(a-x)\ dx$$
Applying this property to your problem:
Let,
\begin{align*}I&=\int_0^3\dfrac{\sqrt x}{\sqrt{x}+\sqrt{3-x}}\ dx\\
&=\int_0^3\dfrac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt{3-(3-x)}}\ dx\\
I&=\int_0^3\dfrac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt x}\ dx
\end{align*}
Now, $$I+I=\int_0^3\dfrac{\sqrt x}{\sqrt{x}+\sqrt{3-x}}\ dx+\int_0^3\dfrac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt x}\ dx\\
\implies2I=\int_0^3\dfrac{\sqrt x+\sqrt{3-x}}{\sqrt{3-x}+\sqrt x}\ dx\\
\implies2I=\int_0^3dx=x\big|_0^3=3\\
\implies I=\dfrac{3}{2}\\
\implies\boxed{\int_0^3\dfrac{\sqrt x}{\sqrt{x}+\sqrt{3-x}}\ dx=\dfrac{3}{2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Limit $\lim_{n\to\infty}\int _{2n}^{3n}\frac{x+3}{x^2+1}\,dx?$ How to evaluate
$$\lim_{n\to\infty}\int _{2n}^{3n}\frac{x+3}{x^2+1}\,dx?$$
The answer in the book is $\ln\frac a2$ where $a$ is a constant.
Attempt:
$\lim_{n\to\infty }\int_{2n}^{3n}\:\frac{x+3}{x^2+1}dx =\lim_{n\to\infty}[\frac12\ln(x^2+1)+3\arctan x]^{3n}_{2n}=\lim_{n\to\infty}[\frac12\ln(9n^2+1)+3\arctan3n-\frac12\ln(4n^2+1)-3\arctan2n]=\infty
+\frac{3\pi}2-\infty-\frac{3\pi}2$
and no $\ln\frac a2$. Can someone explain to me how to evaluate this limit?
Thanks!!
| Well, for the integral:
$$\mathscr{I}_\text{n}:=\int_{2\text{n}}^{3\text{n}}\frac{3+x}{1+x^2}\space\text{d}x=\int_{2\text{n}}^{3\text{n}}\frac{3}{1+x^2}\space\text{d}x+\int_{2\text{n}}^{3\text{n}}\frac{x}{1+x^2}\space\text{d}x\tag1$$
*
*Substitute $\text{u}=1+x^2$:
$$\int_{2\text{n}}^{3\text{n}}\frac{x}{1+x^2}\space\text{d}x=\frac{1}{2}\int_{1+\left(2\text{n}\right)^2}^{1+\left(3\text{n}\right)^2}\frac{1}{\text{u}}\space\text{d}\text{u}=\ln\left|1+\left(3\text{n}\right)^2\right|-\ln\left|1+\left(2\text{n}\right)^2\right|\tag2$$
*$$\int_{2\text{n}}^{3\text{n}}\frac{3}{1+x^2}\space\text{d}x=3\int_{2\text{n}}^{3\text{n}}\frac{1}{1+x^2}\space\text{d}x=3\cdot\left(\arctan\left(3\text{n}\right)-\arctan\left(2\text{n}\right)\right)\tag3$$
So, we get:
$$\mathscr{I}_\text{n}=\ln\left|1+\left(3\text{n}\right)^2\right|-\ln\left|1+\left(2\text{n}\right)^2\right|+3\cdot\left(\arctan\left(3\text{n}\right)-\arctan\left(2\text{n}\right)\right)\tag4$$
Now, we know that $\text{n}>0$:
$$\mathscr{I}_\text{n}=\ln\left(\frac{1+\left(3\text{n}\right)^2}{1+\left(2\text{n}\right)^2}\right)+3\cdot\left(\arctan\left(3\text{n}\right)-\arctan\left(2\text{n}\right)\right)\tag5$$
So, for the limit:
$$\text{L}:=\lim_{\text{n}\to\infty}\space\mathscr{I}_\text{n}=\lim_{\text{n}\to\infty}\left\{\ln\left(\frac{1+\left(3\text{n}\right)^2}{1+\left(2\text{n}\right)^2}\right)+3\cdot\left(\arctan\left(3\text{n}\right)-\arctan\left(2\text{n}\right)\right)\right\}=$$
$$\lim_{\text{n}\to\infty}\ln\left(\frac{1+\left(3\text{n}\right)^2}{1+\left(2\text{n}\right)^2}\right)=\ln\left(\lim_{\text{n}\to\infty}\frac{18\text{n}}{8\text{n}}\right)=\frac{1}{2}\ln\left(\frac{9}{4}\right)=\ln\left(\frac{3}{4}\right)\tag6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that $\sum_{cyc}\frac{2a}{\sqrt{3a+b}}\le \sqrt{3(a+b+c)}$ Prove that if $a,b,c\in \mathbb{R}^+$ then $\sum_{cyc}\frac{2a}{\sqrt{3a+b}}\le \sqrt{3(a+b+c)}$
The equality case seems to hold if $a=b=c$ so I was thinking of using AM-GM but cant operate in any creative ways. Please help or provide alternate solutions.
| By C-S $$\left(\sum_{cyc}\frac{a}{\sqrt{3a+b}}\right)^2\leq(a+b+c)\sum_{cyc}\frac{a}{3a+b}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{a}{3a+b}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{a}{3a+b}-\frac{1}{3}\right)\leq\frac{3}{4}-1$$ or
$$\sum_{cyc}\frac{b}{3a+b}\geq\frac{3}{4},$$
which is true by C-S:
$$\sum_{cyc}\frac{b}{3a+b}=\sum_{cyc}\frac{b^2}{3ab+b^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(3ab+b^2)}\geq\frac{3}{4}$$
because the last inequality it's just
$$\sum_{cyc}(a-b)^2\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sqrt{3+\sqrt{3+x}}=x$
Solve
$$\sqrt{3+\sqrt{3+x}}=x$$
My try:
$$\sqrt{3+\sqrt{3+x}}=x \\
3+\sqrt{3+x}=x^2\\\sqrt{3+x}=x^2-3\\3+x=(x^2-3)^2$$
$$x^4-6x^2+9=x+3\\x^4-6x^2-x+6=0$$
Now ?
| There is no need to deal with any quartic equation in $x$. The key is the function
on LHS $x \mapsto \sqrt{ 3 + \sqrt{3+x}}$ is a composition of the map
$x \mapsto \sqrt{3+x}$ with itself.
The map $x \mapsto \sqrt{3+x}$ is strictly increasing whenever it is defined (i.e $x \ge -3$)
If $\sqrt{3+x} > x$, then $\sqrt{3 + \sqrt{3+x}} > \sqrt{3+x} > x$.
If $\sqrt{3+x} < x$, then $\sqrt{3 + \sqrt{3+x}} < \sqrt{3+x} < x$.
If we want $\sqrt{3 + \sqrt{3+x}} = x$, we need $\sqrt{3+x} = x$. This leads to
$$x^2 = 3 + x \iff x^2- x - 3 = 0 \implies x = \frac{1 \pm \sqrt{13}}{2}$$
Since $x = \sqrt{3 + \sqrt{3 + x}}$ is supposed to be non-negative,
$x = \frac{1 + \sqrt{13}}{2}$ is the only possible solution (and it does work).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Induction: $\frac1{1 \cdot 5} + \frac1{5 \cdot 9} + \cdots + \frac1{(4n-3)(4n+1)}$
For $n \ge 1$, let
$$a_n = \dfrac1{1 \cdot 5} + \dfrac1{5 \cdot 9} + \cdots + \dfrac1{(4n-3)(4n+1)}.$$
Guess a simple explicit formula for $a_n$ and prove it by induction.
Hi, I'm trying to answer this question. I was not provided with a solution.
| $$a_n=\frac{1}{4}\sum_{k=1}^n\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)=$$
$$=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{4n-3}-\frac{1}{4n+1}\right)=$$
$$=\frac{1}{4}\left(1-\frac{1}{4n+1}\right)=\frac{n}{4n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What are the roots of the equation $8x^3-6x+1=0$? Let the roots be, $a,b$ and $c$
We have $$a+b+c=0\tag1$$ $$ab +bc +ca=-\frac68\tag2$$ $$abc = -\frac18\tag3$$
| Note that
$$\cos 3\theta=4\cos^3\theta-3\cos\theta$$
Let $x=\cos\theta$. Then
\begin{align*}
8\cos^3\theta-6\cos\theta+1&=0\\
2\cos3\theta+1&=0\\
\cos3\theta&=\frac{-1}{2}\\
3\theta&=2n\pi\pm\frac{2\pi}{3}\\
\theta&=\frac{2n\pi}{3}\pm\frac{2\pi}{9}
\end{align*}
So $x=\cos\frac{2\pi}{9}\approx 0.7660$, $\cos\frac{4\pi}{9}\approx 0.1736$ or $\cos\frac{8\pi}{9}=-0.9397$.
By letting $x=\cos\theta$, we can only find roots with $|x|\le1$. But as we already have $3$ roots, we have found all the roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove an inequality involving powers of two Let $a,b,c$ be mutually distinct positive integers.
Prove
$$2^{a+b} + 2^{b+c} + 2^{a+c} < 2^{a+b+c} + 1$$
Any hints on how to start proving it?
| Note that the left-hand side is a sum of distinct powers of $2$, so (assuming WLOG that $a < b < c$)
$$2^{a+b}+2^{b+c}+2^{a+c}<\sum_{k=0}^{b+c}2^k=2^{b+c+1}-1<2^{a+b+c}+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find the "most upright" orthogonal vector to another vector? Given an arbitrary vector, I can find any orthogonal vector by solving $ax + by + cz = 0$.
I want to find the "most upright" orthogonal (unit) vector, where $z$ is maximized. There must be a straightforward closed form solution to this, right?
| Restating your problem (and adding a condition), you are given a vector $\vec{v}=\langle a,b,c\rangle$ and you want to find a vector $\vec{w}=\langle x,y,z\rangle$ so that
*
*$\vec{w}\cdot\vec{v}=0$,
*$\|\vec{w}\|=1$
*$\vec{w}\cdot\vec{e}_3$ is maximized
As you state, you want to maximize $z$ subject to
\begin{align*}
ax+by+cz&=0\\
x^2+y^2+z^2&=1
\end{align*}
Perhaps a Lagrange multiplier approach would be good here:
\begin{align*}
0&=\lambda a+2\mu x\\
0&=\lambda b+2\mu y\\
1&=\lambda c+2\mu z\\
0&=ax+by+cz\\
1&=x^2+y^2+z^2
\end{align*}
We should really assume that $a$ and $b$ are not both zero because otherwise $\vec{v}$ points vertically and there are no orthogonal, upward pointing vectors.
*
*If $\mu=0$, then we get that $\lambda=\frac{1}{c}$ (and $c\not=0$), but then the first two equations are $0=\frac{a}{c}$ and $0=\frac{b}{c}$, so $a=0=b$ This contradicts our assumption, so we're ok here.
*If $\mu\not=0$, then we can solve for $x$, $y$, and $z$:
\begin{align*}
x&=-\frac{\lambda a}{2\mu}\\
y&=-\frac{\lambda b}{2\mu}\\
z&=\frac{1-\lambda c}{2\mu}
\end{align*}
We can substitute these into the fourth equation (and cancel the $2\mu$'s) to get
$$
-\lambda a^2-\lambda b^2-\lambda c^2+c=0.
$$
Now, we make things simpler by adding the assumption that $\langle a,b,c\rangle $ is a unit vector, so we could rewrite this equation as $c=\lambda$. If not, we use $c=\lambda\|\vec{v}\|$ throughout the rest of this problem.
Therefore,
\begin{align*}
x&=-\frac{ac}{2\mu}\\
y&=-\frac{bc}{2\mu}\\
z&=\frac{1-c^2}{2\mu}
\end{align*}
Substituting all of this into the final equation gives
$$
4\mu^2=a^2c^2+b^2c^2+1-2c^2+c^4=(a^2+b^2+c^2)c^2+1-2c^2.
$$
Therefore
$$
4\mu^2=1-c^2
$$
or
$$
\mu=\pm\sqrt{\frac{1-c^2}{4}}
$$
Substituting these into the formulae for $x,y,z$ gives that
\begin{align*}
x&=\mp\frac{ac}{\sqrt{1-c^2}}\\
y&=\mp\frac{bc}{\sqrt{1-c^2}}\\
z&=\pm\frac{1-c^2}{\sqrt{1-c^2}}
\end{align*}
Depending on the signs, you should get the largest and smallest that $z$ could be (provided I made no errors).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2283842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Finding coefficient of $x^{15}$ $$(x+x^2+x^3+x^4+x^5)\cdot (x^2+x^3+x^4+…)^5$$
I have done a little :
$$x(1 + x+x^2+x^3+x^4)\cdot x^{10}(1 + x^2+x^3+…)^5$$
By generating functions:
$$\begin{align}&x^{11}\cdot\frac{1 - x^5}{1-x}\cdot\frac{1}{(1-x)^5}\\[1ex]
\implies &x^{11}(1 - x^5)\cdot\frac{1}{(1-x)^6}\\[1.5ex]
\implies &x^{11}(1 - x^5) \cdot\sum_{n=0}^{\infty}\binom{n+5}{5} x^n\\
\end{align}$$
How am I supposed to find $x^{15}$or any other one like $x^{18} , x^{19}$
Any clues? Thanks.
| It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{15}]}&\color{blue}{x^{11}(1-x^5)\sum_{n=0}^\infty \binom{n+5}{5}x^n}\\
&=[x^4](1-x^5)\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{1}\\
&=[x^4]\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{2}\\
&=\binom{9}{5}\tag{3}\\
&\color{blue}{=126}
\end{align*}
Comment:
*
*In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (2) we note the term $x^5$ does not contribute to the coefficient of $x^4$ and can be skipped.
*In (3) we select the coefficient of $x^4$.
In general we obtain for $k\geq 0$:
\begin{align*}
\color{blue}{[x^{k}]}&\color{blue}{x^{11}(1-x^5)\sum_{n=0}^\infty \binom{n+5}{5}x^n}\\
&=[x^{k-11}](1-x^5)\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{4}\\
&=\left([x^{k-11}]-[x^{k-16}]\right)\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{5}\\
&\color{blue}{=\binom{k-6}{5}-\binom{k-11}{5}}\tag{6}\\
\end{align*}
Comment:
*
*In (4) we apply the same rule as in (1).
*In (5) we use the linearity of the coefficient of operator and apply the same rule as in (1).
*In (6) we select the coefficients accordingly. Note that here we set $\binom{n}{k}=0$ if $k>n$ or $n<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Computing $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} $
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \space \text{ converges by the alternating series test, but what is its value?} $$
$1 - \sin(x) \space \text{ is an entire function} $, so by Weierstrass Factorization Theorem, it can be written as a product of its zeros, which occur at $x = \frac{(4n+1)\pi}{2} $ and $\frac{-(4n+3) \pi}{2}$
Equating this to its Taylor series, we get
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x)(1 + \frac{2}{3\pi}x)(1 - \frac{2}{5\pi}x)(1 + \frac{2}{7\pi}x)... $$
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x(1-\frac13)+O(x^2))(1 - \frac{2}{\pi}x(\frac15-\frac17)+O(x^2))... $$
$$1 - x + O(x^3) = 1 - \frac{2}{\pi}x(1-\frac13+\frac15-\frac17+...)+O(x^2)$$
$$x - O(x^3) = \frac{2}{\pi}x \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}-O(x^2)$$
Equating $x$ terms,
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = \frac{\pi}{2}$$
However, I seem to be off by a factor of $2$ (actual answer is $\frac{\pi}{4} $). Does anyone see where this went wrong?
|
I thought it might be instructive to present an approach that relies on elementary analysis only . To that end, we now proceed.
First, note that $\int_0^1 x^{2n}\,dx=\frac{1}{2n+1}$. Therefore, we can write
$$\begin{align}
\sum_{n=0}^N \frac{(-1)^n}{2n+1}&=\sum_{n=0}^N(-1)^n\int_0^1x^{2n}\,dx\\\\
&=\int_0^1 \sum_{n=0}^\infty (-x^2)^n\,dx\\\\
&=\int_0^1 \frac{1-(-x^2)^{N+1}}{1+x^2}\,dx\\\\
&=\frac{\pi}{4}+(-1)^N \int_0^1 \frac{x^{2N+2}}{1+x^2}\,dx\tag 1
\end{align}$$
Noting that by integrating by parts the integral on the right-hand side of $(1)$, with $u=\frac{1}{1+x^2}$ and $v=\frac{x^{2N+3}}{2N+3}$, we see that
$$\begin{align}
\left| \int_0^1 \frac{x^{2N+2}}{1+x^2}\,dx\right|&=\left|\frac{1}{2(2N+3)}+\frac{1}{2N+3}\int_0^1 \frac{2x}{(1+x^2)^2}x^{2N+3}\,dx\right|\\\\
&\le \frac{5/2}{2N+3}\\\\
&\to 0\,\,\text{as}\,\,N\to \infty\tag 2
\end{align}$$
Using the result from $(2)$, we see that
$$\lim_{N\to \infty}\sum_{n=0}^N \frac{(-1)^n}{2n+1}=\frac{\pi}{4}+\lim_{N\to \infty}\left((-1)^N \int_0^1 \frac{x^{2N+2}}{1+x^2}\,dx\right)=\frac{\pi}{4}$$
and hence we arrive at the coveted result
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}=\frac{\pi}{4}$$
as expected.
Alternatively, we can appeal to the Dominated Convergence Theorem. Since $-1\le -x^2\le 0$, we see that $\left|\frac{1-(-x^2)^{N+1}}{1+x^2}\right|\le \frac{2}{1+x^2}$. Since $\int_0^1 \frac{2}{1+x^2}\,dx<\infty$, then the Dominated Convergence Theorem guarantees that
$$\begin{align}
\lim_{N\to \infty}\int_0^1 \frac{1-(-x^2)^{N+1}}{1+x^2}\,dx
&=\int_0^1\lim_{N\to \infty}\left( \frac{1-(-x^2)^{N+1}}{1+x^2}\right)\,dx\\\\
&=\int_0^1\frac{1}{1+x^2}\,dx\\\\
&=\frac{\pi}{4}
\end{align}$$
as was to be shown!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
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How to find $\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}$ I am trying to find the limit as $x\to 8$ of the following function. What follows is the function and then the work I've done on it.
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}$$
\begin{align}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} &= \frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} \times \frac{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}} \\\\
& = \frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac{8-x}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac {-1}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}\end{align}
At this point I try direct substitution and get:
$$ = \frac{-1}{\frac{2}{3}}$$
This is not the answer. Could someone please help me figure out where I've gone wrong?
| $$
\begin{aligned}
\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}
& = \lim _{t\to 0}\left(\frac{\frac{1}{\sqrt{\left(t+8\right)\:+1}}\:-\:\frac{1}{3}}{\left(t+8\right)-8}\right)
\\& = \lim _{t\to 0}\left(\frac{\left(3-\sqrt{t+9}\right)\sqrt{t+9}}{3t^2+27t}\right)
\\& = \lim _{t\to \:0}\left(-\frac{1}{3\left(3+\sqrt{t+9}\right)\sqrt{t+9}}\right)
\\& = \color{red}{-\frac{1}{54}}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$ then $x+\dfrac{1}{x}=?$
let :
$$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$$
then :
$$x+\dfrac{1}{x}=?$$
My try :
$$\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18}=\sqrt3+\sqrt 3\times\sqrt2+4+3\sqrt2 \\ \sqrt3(1+\sqrt2+3)+4$$
Now what ?
| Suppose we say $u = \sqrt 2, v = \sqrt 3$
then
$\sqrt 2 + \sqrt 3 + \sqrt 4 = u+v + 2$
and $(\sqrt 3 +\sqrt 6 + \sqrt 16 + \sqrt 18) = (v+ uv + 3u + 4) = (uv + 2u+2)+(u+v+2)$
in the first term on the right hand side say $2 = u^2$
$(v + 2u+u^2)+(u+v+2) = u(u+v+2) + (u+v+2) = (u+1)(u+v+2)\\
x = \sqrt 2 +1$
$\frac 1x = \frac 1{1+\sqrt 2} = \sqrt 2-1$
$x + \frac 1x = 2\sqrt 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
I'm stuck at integrating an arc length over a circle + angle between curves Let $(\phi, \theta)$ be the usual spherical coordinates. And let $\gamma (t) = (\phi (t), \theta (t))$ be the curve given by
$$ \phi (t) = \ln\left(\cot\left(\frac{\pi}{4} - \frac{t}{2}\right)\right),$$ $$ \theta (t) = \frac{\pi}{2} - t.$$
$t \in (0, \frac{\pi}{2})$
a) Compute the arc length of $ \gamma (t). $
b) Show that the curve cuts each "parallel" in a constant angle, by computing $(\gamma (t))'$ and $\psi_{\phi}$
c) Make a diagram of the curves
So the formula for arc length would be $$ \begin{equation}
\ell_\gamma = \int_0^1 \sqrt{ \left ( \frac{dr}{dt} \right )^2 +
r^2\sin^2 \theta \left ( \frac{d \phi}{dt} \right)^2 +
r^2 \left ( \frac{d \theta}{dt} \right )^2} \, dt.
\end{equation}$$
In my case, as $$\phi'(t) = \frac 1 {2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right) \sin\left(\frac{\pi}{4} - \frac{t}{2}\right)}, $$ and $$ \theta ' (t) = -1 ,$$ then
\begin{align*}
\ell_\gamma &= \int_0^1 \sqrt{
\sin^2 \theta \left ( \frac{d \phi}{dt} \right)^2 +
\left ( \frac{d \theta}{dt} \right )^2} \, dt \\
&= \int_0^1 \sqrt{ \sin^2 \theta \left (\frac 1 {2\cos(\frac{\pi}{4} - \frac{t}{2})\sin(\frac{\pi}{4} - \frac{t}{2})} \right)^2 + 1} \, dt\\
&= \int_0^1 \sqrt{
\left(\frac{\sin \theta }{2\cos(\frac{\pi}{4} - \frac{t}{2})\sin(\frac{\pi}{4} - \frac{t}{2})} \right)^2 +
1} \, dt.
\end{align*}
I am stuck here. Before I go further, would this integral be correct? That $\sin^2(\theta)$ confuses me maybe, I don't know if I should change the argument to something else.
For exercise b) Could someone verify this reasoning?
I want the angle between two curves on the sphere, which is defined as the angle between the tangent vectors of this curves at the point of intersection. So I want to find $\alpha$ such that $\alpha = cos^{-1}(\frac{<\beta '(t), \gamma ' (t)>}{||\beta '(t) || ||\gamma ' (t)||}) $, because that is what the angle between two vectors in a vector space is.
So if $\beta (t) = (t, \theta_0)$ is a curve that describes the parallel curve in a sphere, with $\theta_0$ constant, Its tangents are given by $(1, 0),$. I already have the derivatives of $\gamma$ so $$ \alpha = cos^{-1}(\frac{<\beta '(t), \gamma ' (t)>}{||\beta '(t) || ||\gamma ' (t)||}) = cos^{-1}(\frac{<(1, 0) \dot (\frac{1}{sin{\frac{\pi}{2}- t}}, -1)>}{(1) \sqrt{csc^2(\frac{\pi}{2}-t) + 1}})$$
$$ = cos^{-1}(\frac{1}{sin{\frac{\pi}{2}- t}}{ \sqrt{csc^2(\frac{\pi}{2}-t) + 1}})$$.
Thanks in advance.
| \begin{align}
& \int_0^1 \sqrt{ \left(\frac{\sin \theta }{2\cos(\frac{\pi}{4} - \frac t 2) \sin(\frac \pi 4 - \frac t 2)} \right)^2 + 1} \, dt \\[10pt]
= {} & \int_0^1 \sqrt{\left(\frac{\sin\theta}{\cos t} \right)^2+1} \, dt = \int_0^1 \sqrt{ \left( \frac{\cos t}{\cos t} \right)^2 + 1} \, dt = \int_0^1 \sqrt 2 \, dt = \sqrt 2.
\end{align}
The first two equalities are trigonometric identities:
\begin{align}
& 2\cos\left( \frac \pi 4 - \frac t 2 \right) \sin \left( \frac \pi 4 - \frac t 2 \right) = \sin\left( \frac \pi 2 - t \right) = \cos t \\[10pt]
&
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Point of Tangency between Two Circles I made sure to look around this site a couple times before asking my question. A lot of people have already asked how to find the point of tangency or intersection between two circles, but none of them could really help me solve the problem.
I'm being asked to find the point of tangency between two circles and all I am given are just the two circles -- no equations of tangent lines, etc. Here's a picture of what I'm working with:
The equation of the green circle is $x^2$ + $y^2$ = 5.
The equation of the orange circle was given as $x^2$ - 12x + $y^2$ + 6y = -25, which I rewrote as ($x-6)^2$ + $(y+3)^2$ = 20.
How do I find the coordinates of the circles' point of tangency?
Thank you in advance!
| The equations are:
\begin{gather}
x^2 + y^2 = 5 \tag{1} \\
x^2 -12x + y^2 + 6y + 25 = 0 \tag{2}
\end{gather}
If you substitute (1) into (2), you get
\begin{align}
-12 x + 6y + 30 &= 0 \implies y=2x-5\tag{3}
\end{align}
Substitute this back into (1):
\begin{align}
x^2 + (2x-5)^2 &= 5 \\\implies
x^2 + 4x^2 - 20x + 25 &= 5 \\\implies
5x^2 - 20x + 20 &= 0 \\\implies
x^2 -4 x + 4 &= 0 \\\implies
(x-2)^2 &= 0
\end{align}
Therefore $x=2$. By (3), $y=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Number of ways to roll a 6-sided die and a 12-sided die (alternating) to sum to n There are two dice, a six-sided die and a twelve-sided die.
You roll the dice, alternating between the two, starting with the six-sided die. You keep track of the running total and stop when the total reaches $n$.
Show that the number of different ways to reach $n$ is:
$[x^n]$$ 1+x+x^2+\dots+x^{6}\over 1-(x+x^2+\dots+x^{6})(x+x^2+\dots+x^{12})$
Here is my attempt:
Let $S$ be the set of sets of ways to roll the dice any number of times (set of combinations).
Define the weight function, $w$, to be the sum of the set. $w((c_1,c_2,\dots,c_k))=c_1+c_2+\dots+c_k$
Let $\mathbb{N}_6 =\{1,2,3,\dots,6\}$
Let $\mathbb{N}_{12} =\{1,2,3,\dots,12\}$
Then
$\Phi_{\mathbb{N}_6} = x+x^2+\dots+x^{6} =$$ x-x^7\over 1-x$
$\Phi_{\mathbb{N}_12} = x+x^2+\dots+x^{12} =$$ x-x^{12}\over 1-x$
And
$ \begin{align}S
& = \mathbb{N}_6(\mathbb{N}_{12})^k(\mathbb{N}_6)^k\cup(\mathbb{N}_6)^k(\mathbb{N}_{12})^k \\
& = (\mathbb{N}_6)^{k+1}(\mathbb{N}_{12})^k\cup(\mathbb{N}_6)^k(\mathbb{N}_{12})^k \\
\end{align}$
This is because the six sided die can be rolled the same number of times as the twelve sided die or the six sided die can be rolled one more time than the twelve sided die.
Then
$\begin{align} \Phi_{\mathbb S}
& =(\Phi_{\mathbb {N}_6})^{k+1}(\Phi_{\mathbb{N}_{12}})^k+(\Phi_{\mathbb {N}_6})^k(\Phi_{\mathbb{N}_{12}})^k \\
& = (x+x^2+\dots+x^{6})^{k+1}(x+x^2+\dots+x^{12})^k+(x+x^2+\dots+x^{6})^k(x+x^2+\dots+x^{12})^k \\
\end{align}$
I am unsure how to proceed past this point. I tried using the geometric series formula but got stuck.
| It's actually very simple:
*
*$x+x^2+\dots+x^6$ gives the number of ways to roll $n$ on the d6, and similarly $x+x^2+\dots+x^{12}$ for the d12
*Their product $(x+x^2+\dots+x^6)(x+x^2+\dots+x^{12})$ gives the number of ways to roll $n$ as the sum of a d6-d12 pair
*$\frac1{1-(x+x^2+\dots+x^6)(x+x^2+\dots+x^{12})}$ gives the number of ways to roll $n$ on any number of d6-d12 pairs (i.e. an even number of individual die rolls), based on $\frac1{1-x}=1+x+x^2+x^3+\dots$ summing the number of ways to roll $n$ on any non-negative integer number of pairs
*Multiplying this last expression by $(x+x^2+\dots+x^6)$ gives the number of ways to roll $n$ on any number of pairs and one further d6 (an odd number of die rolls)
*Summing the last two expressions yields the desired result: the coefficient of $x^n$ in $\frac{1+x+x^2+\dots+x^6}{1-(x+x^2+\dots+x^6)(x+x^2+\dots+x^{12})}$ is the number of ways to roll $n$ on the sum of alternating d6s and d12s, the d6 being rolled first.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum product when the sum is constant and the variables are positive integers Today I came across an olympiad problem which after some time translated into a simple problem. I was supposed to prove that given $x+y+z = 73$ and $x,y,z ∈ I^+$. Show that the maximum value of $xyz = 14400$.
Now using the AM-GM inequality one can establish that the maximum value of the product for real values of the variable would be $≈ 14408.037037$. However we have the added clause that x,y,z must be positive integers.
Intuitively, it seems that whenever$$\sum_{i=1}^n |\frac{x}{n}-x_{i}|$$
is minimised for the general case, the product $x_{1}x_{2}....x_{n}$ is maximized. So 24,24,25 is the case in which the value of the above expression for positive integral values of the variables is minimized and hence the product is maximized. But I would like a proof of the above result.
| Let
$$x + y + z = c \tag{A}$$
and
$$f(x,y,z) = xyz$$
then using (A) to sub we have
\begin{align}
f(x,y) &= xy(c - x - y) \\
&=xyc - x^2y - xy^2
\end{align}
Take partial derivatives
\begin{align}
f_x(x,y) &= yc - 2xy - y^2 \tag{1}\\
f_y(x,y) &= xc - 2xy - x^2 \tag{2}
\end{align}
to find critical points of the 3d surface. So, set (1) and (2) equal to zero then subtract (2) from (1) to get
\begin{align}
(y-x)c + x^2 - y^2 &= 0 \\
(y - x)c - (y - x)(x+y) &= 0 \\
(y-x)(c - (x + y) &= 0\\
(y-x)(z) &= 0
\end{align}
So one stationary point when $z = 0$, which is clearly not a maximum for $xyz$, and a second stationary point for $y = x$. Taking second partials on (1) and (2), in order to verify this is an extremum in the plane $y = x$
\begin{align}
f_{xx}(x,y) &= -2x \\
f_{yy}(x,y) &= -2y \\
f_{xy}(x,y) &= c - 2(x + y)
\end{align}
we find that $f_{xx} < 0$ for all positive $x,y$ and
\begin{align}
D(x,y=x) &= f_{xx}(x,x)f_{yy}(x,x) - f_{xy}^2(x,x) \\
&= 4x^2 - (c^2 - 4cx + 4x^2) \\
&= c(4x - 1)
\end{align}
so $D > 0$ for all $c > 0$, and positive integers $x$. Therefore when $x,y \geq 1$, $x = y$, and $c > 0$ we have an extremum.
Now we can simplify the problem to $f(x,y = x,z) = f(x,z) = x^2z$ subject to
\begin{align}
x + x + z = 2x + z = c
\end{align}
So $z = c - 2x$, and subbing that we have $f(x,z = c - 2x) = f(x) = x^2(c-2x)$.
Take a good old fashion derivative to get
\begin{align}
f'(x) &= 2cx - 6x^2 \\
&= 2x(c - 3x)
\end{align}
So the max occurs when $x = c/3 = y$, which we can now use to say $z = c/3$ too.
We can see that our original function $f(x,y,z) = xyz$ is continuous. Secondly, we can see that the entire problem is symmetric in $x,y,z$ meaning our test for extrema would be the same for any subbing and pairing of $x,y,z$. That symmetry means that the identified extrema, $x=y=z=c/3$, is a local maximum along any approach in our plane $x + y + z = c$, meaning the shape of our function is concave along the part of plane $x + y + z = c$ of which we are concerned — and, importantly, that behavior was shown to not change for $x,y,z > 1$. Because of that concavity all we have to do is pick the closest integers to $c/3$ that satisfy x + y + z = c. In our case thats $24.33\dots$ so the closest numbers are is $24, 24, 25$.
In other words, if $x,y,z$ minimize their distance to $c/3$ in the plane $x + y + z = c$ then their product will necessarily be closer to the vertex of the paraboloid(s).
Similar to how $x = 6, 7$ maximizes the $2D$ version of this problem in the parabola defined by $xy$ in the plane $x + y = 13$. In this case $6,7$ are the closest satisfying integers to $6.5$. Visual:
https://www.desmos.com/calculator/z9jhl3fjpq
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2299901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim_{x\to 0} \left\lfloor \frac{\tan 2x}{\sin x} \right\rfloor $
Find the limit
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor $$
My try:
$$ \tan 2x =\dfrac{\sin 2x}{\cos 2x}$$
$$\sin 2x =2\cos x\sin x$$
So:
$$\dfrac{\tan 2x}{\sin x}=\dfrac{2\cos x}{\cos 2x}$$
So:
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor=
\lim_{x\to 0}\left\lfloor\dfrac{2\cos x}{\cos 2x}\right\rfloor $$
Now what?
| Note that
\begin{align}
\frac{2\cos x}{\cos 2x}-2&=\frac{2\cos x-2\cos 2x}{\cos 2x}\\
&=\frac{4\sin\frac{3x}{2}\sin\frac{x}{2}}{\cos 2x}\\
&>0
\end{align}
for $x\in(\frac{-\pi}{4},\frac{\pi}{4})$.
We also have
\begin{align}
3-\frac{2\cos x}{\cos 2x}&=\frac{3\cos 2x-2\cos x}{\cos 2x}\\
&=\frac{6\cos^2x-2\cos x-3}{\cos 2x}\\
&=\frac{36\cos^2x-12\cos x-18}{6\cos 2x}\\
&=\frac{(6\cos^2x-1)^2-19}{6\cos 2x}\\
&>0
\end{align}
for $x\in(-\alpha,\alpha)$, where $\alpha=\arccos(\frac{1+\sqrt{19}}{12})$
Note that $0<\alpha<\frac{\pi}{4}$.
For $x\in(-\alpha,\alpha)$, $\displaystyle\lfloor \frac{2\cos x}{\cos 2x}\rfloor=2$.
The limit is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
$\int_{\alpha}z^2 \log\left(\frac{z+1}{z-1}\right)\,dz$ where $\alpha$ is $|z-1|=1$ Calculate
$$\int_{\alpha}z^2 \log\left(\frac{z+1}{z-1}\right)\,dz$$
where $\alpha$ is $|z-1|=1$ and the initial point of integration $z_1=1+i$.
The function isn't analytic at $1$ and $-1$. Is there a theorem to solve it?
Transformation of the integral $z\to 1/z$
$$f(z)=\frac{\log\left(\frac{1+z}{1-z}\right)}{z^4}$$
How do I transform the circumference $1+e^{i\theta}$?
Is there any other way?
| $g(z)=z^2 (\log(1+z)-i\pi)$ is analytic on $|z-1| < 2$ so that $\int_{|z-1| = 1} g(z)dz = 0$.
Thus, it reduces to finding a primitive of $$h(z) = -z^2 \log(1-z) = \sum_{n=1}^\infty \frac{z^{n+2}}{n}$$ ie. $$H(z) = \sum_{n=1}^\infty \frac{z^{n+3}}{n(n+3)}=\frac13\sum_{n=1}^\infty z^{n+3} (\frac{1}{n+3}-\frac{1}{n}) \\ =-\frac13 z^3\log(1-z)+ \frac13 (\log(1-z)-z-\frac{z^2}{2})$$
Finally, being careful with the branches, and using that $H(1+ze^{2i\pi}) =H(1+z)- \frac{z^3-1}{3}2i\pi$
$$\int_{|z-1|=1} z^2 \log\left(\frac{z+1}{z-1}\right)\,dz=\int_{|z-1|=1} (g(z)+h(z))dz = \int_{|z-1|=1} h(z)dz = H(1+1)-H(1+ e^{2i\pi})\\= \frac{7}{3} 2i \pi$$
or something like that
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Does $\{f_n\}$ converge uniformly on $(0,1) $ and $(0,\infty)?$
Let $g_n(x)=\sin^2(x+\frac{1}{n}),x\in (0,\infty)$
and $f_n(x)=\int _0^x g_n(t) dt$.
Then show that $\{f_n\}$ converges uniformly on $(0,1)$.
Does $\{f_n\}$ converge uniformly on $(0,\infty)?$.
$f(x)=\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty}\int _0^x g_n(t) dt=\int _0^x\lim_{n\to \infty}\sin ^2(t+\frac {1}{n})=\int_0^x \sin^2 t$ [Since $\int$ and $\sin^2x$ are continuous functions].
Hence $f(x)=\int _0^x \sin ^2 t dt$.
Now $\sup_{(0,\infty)} |f_n(x)-f(x)|=\sup_{(0,\infty)}|\int _0^x \{\sin ^2(t+\frac{1}{n})-\sin^2 t\}|\le \sup_{(0,\infty)} \int _0^x |\{\sin ^2(t+\frac{1}{n})-\sin^2 t\}|$
Taking $|\{\sin ^2(t+\frac{1}{n})-\sin^2 t\}|=|\sin (t+\frac{1}{n})-\sin t||\sin (t+\frac{1}{n})+\sin t|\le 2\frac{1}{n}$
Hence $\sup_{(0,\infty)} |f_n(x)-f(x)|\le \sup_{(0,\infty)} 2\frac{x}{n}$
Hence if $x\in (0,1)\implies \sup_{(0,\infty)} |f_n(x)-f(x)|\le \frac{2}{n}\to 0$
Hence $f_n$ converges uniformly on $(0,1)$.
But $f_n$ does not converge uniformly on $(0,\infty)$ because I can choose $x$ quite large such that $\frac{x}{n}>1$.
Please validate.
| Note that
$$\int_0^x \sin^2(t +1/n) \, dt = \frac{x}{2} - \frac{1}{4}\sin(2x+2/n) + \frac{1}{4} \sin (2/n),$$
and
$$\int_0^x \sin^2(t) \, dt = \frac{x}{2} - \frac{1}{4}\sin(2x) .$$
Hence,
$$|f_n(x) - f(x)| \leqslant \frac{1}{4}\left|\sin(2x+2/n) - \sin(2x) \right|+ \frac{1}{4} |\sin(2/n)| \leqslant \frac{1}{2n} + \frac{1}{4}|\sin(2/n)|$$
We have uniform convergence on $(0,\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \sqrt{2x + 3} dx$ Integrate: $\int \sqrt{2x + 3} \ dx$
Doing some guesswork gives me: $\frac{1}{2}(2x + 3)^{3/2}$ to try. Differentiating this gives: $2\frac{3}{2}\frac{1}{2}(2x + 3)^{1/2}$ = $\frac{6}{4}(2x + 3)^{1/2}$
Too bad the answer should be: $\frac{1}{3}(2x + 3)^{1/2}$, so clearly I'm missing some factor of 2. Where am I going wrong?
| $$\int\sqrt{2x+3}dx=\frac{1}{2}\int(2x+3)^{\frac{1}{2}}d(2x+3)=$$
$$=\frac{1}{2}\frac{(2x+3)^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{3}\sqrt{(2x+3)^3}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
minimizing vectors Could please check whether my solution is right?
Q. $$A=\begin{pmatrix}
1 & 1 \\
1 & 1 \\
0 & 0 \\
\end{pmatrix}$$Find the set of vectors $x$ that minimize the value $$||
A
x-
\begin{pmatrix}
1 \\
2 \\
3 \\
\end{pmatrix}||$$
My solution.
$Ax-(1,2,3)^T=0$ (inconsistent) → least square method
$A^TAx=A^Tb$, $b=(1,2,3)^T$
$$
\begin{pmatrix}
2 & 2 \\
2 & 2 \
\end{pmatrix}\begin{pmatrix}
x_1 \\
x_2 \\
\end{pmatrix}=\begin{pmatrix}
3 \\
3 \\
\end{pmatrix}$$
$2x_1+2x_2=3$ → $x_2=-x_1+1.5$
Therefore the set of $x$ is {$
\begin{pmatrix}
k \\
-k+1.5 \\
\end{pmatrix}$| k∈ℝ }
| Problem statment
Given the matrix
$$
\mathbf{A} =
\left[
\begin{array}{cc}
1 & 1 \\
1 & 1 \\
0 & 0 \\
\end{array}
\right]
\in \mathbb{C}^{m\times n}_{\rho}
$$
and the data vector
$$
b =
\left[
\begin{array}{cc}
1 \\
2 \\
3 \\
\end{array}
\right]
$$
find the least squares solution
$$
x_{LS} = \left\{
x\in\mathbb{C}^{n} \colon
\lVert
\mathbf{A} x - b
\rVert_{2}^{2}
\text{ is minimized}
\right\}
$$
Solution
The general solution to the least squares problem is
$$
x_{LS} =
\color{blue}{\mathbf{A}^{+} b} +
\color{red}{
\left(
\mathbf{I}_{n} - \mathbf{A}^{+} \mathbf{A}
\right) y}, \quad y \in \mathbb{C}^{n}
\tag{1}
$$
where colors distinguish $\color{blue}{range}$ and $\color{red}{null}$ spaces.
The simple structure of the matrix $\mathbf{A}$ encourages a solution using the pseudoinverse built from the singular value decomposition (SVD).
Singular value decomposition
The SVD is straightforward:
$$
\mathbf{A} =
\mathbf{U} \, \Sigma \, \mathbf{V}^{*}
\tag{2}
$$
Resolve the eigensystem for
$$
\mathbf{W} = \mathbf{A}^{*} \mathbf{A} =
\left[
\begin{array}{cc}
2 & 2 \\
2 & 2 \\
\end{array}
\right]
$$
Singular values
The eigenvalues are
$$
\lambda \left( \mathbf{W} \right) =
\left\{ 4, 0 \right\}
$$
The matrix $\mathbf{A}$ has rank $\rho=1$, and the lone singular value is
$$
\sigma_{1} = \sqrt{\lambda_{1}} = 2
$$
The matrix of singular values is
$$
\mathbf{S} =
\left[
\begin{array}{c}
2
\end{array}
\right]
$$
and the sabot matrix is
$$
\Sigma =
\left[
\begin{array}{c}
\mathbf{S} & 0 \\
\mathbf{0} & \mathbf{0}
\end{array}
\right]
=
\left[
\begin{array}{c}
2 & 0 \\
0 & 0 \\
0 & 0 \\
\end{array}
\right]
$$
Domain matrix $\mathbf{V}$
The normalized eigenvectors of $\mathbf{W}$ are the column vectors of $\mathbf{V}$:
$$
\mathbf{V} = \frac{1}{\sqrt{2}}
\left[
\begin{array}{cr}
\color{blue}{1} & \color{red}{-1} \\
\color{blue}{1} & \color{red}{1} \\
\end{array}
\right]
$$
Domain matrix $\mathbf{U}$
Equation $(2)$ can be rewritten to provide the $\color{blue}{range}$ space vector for $\mathbf{U}$:
$$
\mathbf{U}_{1} =
\frac{1}{\sqrt{2}}
\color{blue}{\left[
\begin{array}{r}
1 \\ 0 \\ -1
\end{array}
\right]}
$$
For this simple problem, we can eyeball the $\color{red}{null}$ space vectors.
$$
\mathbf{U} =
\left[
\begin{array}{ccc}
%
\frac{1}{\sqrt{2}}
\color{blue}{\left[
\begin{array}{r}
1 \\ 0 \\ -1
\end{array}
\right]} &
%
\frac{1}{\sqrt{2}}
\color{red}{\left[
\begin{array}{r}
1 \\ 0 \\ 1
\end{array}
\right]} &
%
\color{red}{\left[
\begin{array}{c}
0 \\ 1 \\ 0
\end{array}
\right]}
%
\end{array}
\right]
$$
Pseudoinverse
The pseudoinverse matrix is
$$
\mathbf{A}^{+} = \mathbf{V} \, \Sigma^{+} \mathbf{U}^{*} =
\frac{1}{4}
\left[
\begin{array}{ccc}
1 & 1 & 0 \\
1 & 1 & 0 \\
\end{array}
\right]
$$
Least squares solution
Equation $(1)$ provides
$$
x_{LS} =
\color{blue}{\mathbf{A}^{+} b} +
\color{red}{
\left(
\mathbf{I}_{n} - \mathbf{A}^{+} \mathbf{A}
\right) y} =
\color{blue}{
\frac{3}{4}
\left[
\begin{array}{ccc}
1 \\
1 \\
\end{array}
\right]
}
+
\color{red}{
\left[
\begin{array}{rr}
1 & -1 \\
-1 & 1 \\
\end{array}
\right] y
}
, \quad y \in \mathbb{C}^{2}
$$
Plot
The following plot shows the least squares merit function $\lVert
\mathbf{A} x - b
\rVert_{2}^{2}
$ as a function of $x$. The white dot represents $\color{blue}{x_{LS}}$, the dashed, yellow line $\color{red}{x_{LS}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Giving a closed form combinatorial solution to finding the larger of two balls drawn from an urn I understand how to solve the following problem by listing out the sample space; however, I would like to know how to derive a combinatorial solution to the problem (that is, I would like to give a closed form solution to n(X=0), n(X=1), etc.)
An urn contains 4 balls numbered 0 to 3. If two balls are drawn with replacement, define the random variable X to be the larger of the two balls drawn. Find P(X=0), P(X=1), P(X=2), and P(X=3).
| Case 0: $X=0$
The only way that the largest ball drawn is $0$ is if we pick $0$ both times, so
$$n(X=0) = \boxed{1}$$
Case 1: $X=1$
We can either have both draws be $1$, or one draw be $0$ and the other $1$. There are two ways to draw $0$ and $1$, one for each order, so we get
$$n(X=1) = 1+2 = \boxed{3}$$
Case 2: $X=2$
Once again, we can have both draws be $2$, or one draw be $2$ and the other be $0$ or $1$. This time there are $4$ ways to have only one of the draws be $2$, which gives
$$n(X=2) = 1+2+2 = \boxed{5}$$
Case 3: $X=3$
Both draws could be $3$, or one draw could be $3$ and the other be $0$, $1$, or $2$. This time there are $6$ ways to have only one of the draws be $3$, so we get
$$n(X=3) = 1+2+2+2 = \boxed{7}$$
As a check, notice that $n(X=0)+n(X=1)+n(X=2)+n(X=3) = 16$, which is the size of the sample space.
You can also compactly write this (bof wrote something similar in the comments):
$$n(X=k) = 2k+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the different methods to determine the derivative of $f(x)=\sqrt{1+x}$? The methods that I can think of are:
1) Chain rule
2) Binomial series of $f(x)$
3) Through the formula $f\prime(g(x))=\frac{1}{g\prime (x)}$ at a particular point
What are the other methods to determine the derivative of $f(x)$?
| Using logarithmic differentiation:
\begin{align}
f'(x) &= f(x) \cdot \frac{d}{dx} \ln f(x)
= \sqrt{1+x} \cdot \frac{d}{dx} \ln \sqrt{1+x}\\
&= \sqrt{1+x} \cdot\frac{1}{2} \frac{d}{dx} \ln(1+x)
= \sqrt{1+x} \cdot \frac{1}{2} \frac{1}{1+x}\\
&= \frac{1}{2\sqrt{1+x}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate the definite integral $\int_0^{2016} x (x-1)(x-2)(x-3)... (x-2016)\,dx$
$$\int_0^{2016} x (x-1)(x-2)(x-3)... (x-2016)\,dx$$
I have tried integration by parts using calling $f (x)=(x-1)(x-2)... (x-2016)$ but it doesn't help. I can't think of anything else. Please help.
| Note that
$$\scriptsize\begin{align}
\int_0^2 x(x-1)(x-2)dx&=\int_{-1}^1 (x-1)x(x+1)dx&=0\\
\int_0^4 x(x-1)(x-2)(x-3)(x-4)dx&=\int_{-2}^2 (x-2)(x-1)x(x+1)(x+2)dx&=0\\
&\vdots\\
\int_0^{2016} x(x-1)(x-2)(x-3)\cdots (x-2016)dx&=\int_{-1008}^{1008} (x-1008)(x-1007)\cdots x\cdots (x+1007)(x+1008)dx&=\color{red}0\\
\end{align}$$
The result follows from the antisymmetric nature of the curve
$\scriptsize(x-a)(x-a+1)\cdots x\cdots (x+a-1)(x+a)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Find $S = \frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b}$ if values of $a+b+c$ and $\frac1{a+b}+\frac1{b+c}+\frac1{a+c}$ are given I just stumbled upon a contest question from last year's city olympiad math contest:
Question: For the real numbers $a,b,c$ such that: $a+b+c = 6, \dfrac{1}{a+b}+\dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{47}{60}$, find the value of $S = \dfrac{a}{b+c}+\dfrac{b}{c+a} + \dfrac{c}{a+b}$.
Since I just saw it from an online forum "elsewhere", I thought I'd want to hear from other more skilled and experienced MSE members about your tactics and approaches to the solution of this interesting question.
| Hint:$$\frac{a+b+c}{b+c}=1+\frac{a}{b+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 0
} |
Finding the geometric shape
Draw the geomatrix region of $|z^2-1|=1$
$$|z^2-1|=1$$
$$|(x+iy)^2-1|=1$$
$$|x^2-y^2-1+ixy|=1$$
$$\sqrt{(x^2-y^2-1)^2+(xy)^2}=1$$
$$\sqrt{(x^2-y^2-1)^2+(xy)^2}=1$$
Let $a=x^2, b=y^2$
$$\sqrt{a^2-ab+b^2-2a+2b+1}=1$$
$$a^2+b^2-ab-2(a+b)=0$$
How should I continue?
| Let us write $ z = r e^{i\varphi} $, then the equation says :
$$1=|z^2-1|^2=(z^2-1)\overline{ (z^2-1) } = \left(r^2e^{2i\varphi}-1\right)\cdot\left(r^2e^{-2i\varphi}-1\right)$$
Expanding the expression above rise in equation (using the Euler's formula in the form $e^{ix}+e^{-ix}=2\cos x$) :
$$1=1+r^4-2r^2 \cos(2\varphi)$$
Clearly if $r=0$ the equation is solved immediatelly so $z=0$ is a lies on the region. If $r\neq0$ then
$$r=\pm \sqrt{2\cos2\varphi}$$
This is the equation of a lemniscate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\sin(10^{\circ})$, if $\cos(40^{\circ}) = a$ I have this trig problem where it asks to find $ \sin(10^{\circ})$, given $\cos(40^{\circ})=a$.
I tried the write the $\sin(10^{\circ})$ as $\sin(40^{\circ}-30^{\circ})$ to use the sum formula but i didn't get anywhere near to the options given in the test:
a)$\sin(10^{\circ})=2a$
b)$\sin(10^{\circ})=2a^2$
c)$\sin(10^{\circ})=2a-1$
d)$\sin(10^{\circ})=2a^2+1$
| You are on the right lines
\begin{align*}
\sin(10) &= \sin(40-30) \\
&= \sin(40)\cos(30)-\cos(40)\sin(30) \\
&= \frac{\sqrt{3}}{2}\sin(40) - \frac{1}{2}\cos(40) \\
&= \frac{\sqrt{3}}{2}\sqrt{1-\cos^2(40)} - \frac{1}{2}\cos(40) \\
&= \frac{\sqrt{3}}{2}\sqrt{1-a^2} - \frac{1}{2}a \\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $p, q, r$ are the roots of $x^3 -x+1=0$, what is $p^5 + q^5 + r^5$? Can someone please help us solve this equation. It was on our college entry exam, but no one managed to solve it. The problem is:
Let $p,q,r$ be the roots of $x^3 -x+1=0$. Then $p^5+q^5+r^5 = ?$
Correct answer was $-5$, but no one managed to solve this problem. We tried different methods, but none came up with a solution. Any help is appreciated.
| Alternative approach: the characteristic polynomial of
$$ M = \begin{pmatrix}0 & 0 & -1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$$
is exactly $p(x)=x^3-x+1$, since $M$ is the companion matrix of $p(x)$. It follows that the wanted sum is
$$ p^5+q^5+r^5=\text{Tr }\begin{pmatrix}0 & 0 & -1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}^5 = \text{Tr }\begin{pmatrix}-1 & 1 & -1 \\ 1 & -2 & 2 \\ -1 & 1 & -2 \end{pmatrix}=\color{red}{-5}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving algebraic inequality I been struggling trying to solve this inequality by hand
$$x+\sqrt{x+3} < 0$$
how should I proceed?
| $$x+\sqrt { x+3 } <0\Rightarrow \begin{cases} x+3\ge 0 \\ x+3<{ x }^{ 2 } \end{cases}\Rightarrow \begin{cases} x\ge -3 \\ { x }^{ 2 }-x-3>0 \end{cases}\Rightarrow \cap \begin{cases} x\ge -3 \\ x\in \left( -\infty ;\frac { 1-\sqrt { 13 } }{ 2 } \right) \cup \left( \frac { 1+\sqrt { 13 } }{ 2 } ;+\infty \right) \end{cases}\Rightarrow \\ \Rightarrow -3\le x<\frac { 1-\sqrt { 13 } }{ 2 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Elements needed to derive the Riemann-Siegel Z function On this post I got a comment to motivate the present question:
To prove a zero $\zeta(s_0)=0, s_0 \approx 1/2+iy$ is exactly on the critical line, it is enough to prove the real function $Z(t) = \zeta(1/2+it) e^{i \,\text{arg}(\Gamma(1/4+it/2)}\, \pi^{-it/2}$ changes of sign around $t=y$. We know $Z(t)$ is real for $t$ real thanks to the functional equation.
So I would like to ask for some help understanding where the expression
$$\large Z(t) = \zeta\left(\frac{1}{2}+it\right)\quad e^{i \;\text{arg}\left(\Gamma(\frac{1}{4}+\frac{it}{2})\right)}\quad \pi^{\frac{-it}{2}}\tag 1$$
comes from.
The terser version $Z(t)=\zeta\left(\frac{1}{2}+it\right)\; e^{i\theta(t)}$ becomes Eq. $(1)$ when $\theta(t)$ is unfolded:
$$\theta(t)= \Im\left(\ln\Gamma\left(\frac{1}{4}+\frac{it}{2} \right) \right)-\frac{t}{2}\ln(\pi)$$
The closest I got digging out online material for the amateur math curious is the following, probably related, equation of the Riemann functional:
$$\large \pi^{\frac{-s}{2}}\Gamma\left(\frac{s}{2}\right)\,\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\,\zeta(1-s)$$
| The Riemann functional
$$\large \pi^{\frac{-s}{2}}\Gamma\left(\frac{s}{2}\right)\,\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\,\zeta(1-s)$$
relates $\zeta(s)$ to $\zeta(s-1)$, as a fraction expressed as
$$\chi(s) = \frac{\zeta(s)}{\zeta(s-1)}
=\frac{\Gamma\left(\frac{1}{2}(1-s)\right)}{\Gamma\left(\frac{1}{2} s \right)}\; \pi^{s-1/2}$$
But we are interested in $s = \frac{1}{2} + it$. Therefore,
$$\begin{align}
\chi\left(\frac{1}{2} + it \right)
&=\frac{\Gamma\left(\frac{1}{2}\,\left(1 - \left(\frac{1}{2}+it\right)\right) \right)}{\Gamma \left( \frac{1}{2}\,\left(\frac{1}{2}+it \right)\right)}
\,\pi^{\left(\frac{1}{2}+it\right)-\frac{1}{2}}\\[2ex]
&= \frac{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}
\,\pi^{it}
\end{align}$$
Now comes the definition of the Riemann-Siegel or Hardy Z-function:
$$Z(t) := \zeta\left(\frac{1}{2}+it\right)\;\left[\chi\left(\frac{1}{2} + it \right)\right]^{-1/2}$$
Substituting,
$$\begin{align}
Z(t) &= \zeta \left(\frac{1}{2}+it\right)\left(
\frac{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}
\right)^{-1/2}\; \,\pi^{\frac{-it}{2}}\\[2ex]
&=\zeta \left(\frac{1}{2}+it\right)\color{blue}{
\frac{\Gamma^{1/2} \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma^{1/2}\left(\frac{1}{4} - \frac{it}{2} \right)}}
\;\pi^{\frac{-it}{2}}\\[2ex]
&= \zeta \left(\frac{1}{2}+it\right)\color{blue}{\sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}}
\;\pi^{\frac{-it}{2}}\\[2ex]
&= \zeta \left(\frac{1}{2}+it\right)\color{blue}{e^{i\,\arg\left(\Gamma\left(\frac{1}{4}+\frac{it}{2}\right)\right)}}
\;\pi^{\frac{-it}{2}}
\end{align}$$
So, how to see that
$$\color{blue}{\sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}}=\color{blue}{e^{i\,\arg\left(\Gamma\left(\frac{1}{4}+\frac{it}{2}\right)\right)}}$$?
The argument of a complex number $z = x + i y=\rho\ e^{i\theta}$
$$\theta = \arg(z)= \frac{1}{i}\log\sqrt{\frac{z}{\bar z}}=\frac{\log z - \log \bar z}{2i}$$
so we can modify
$$\frac{\Gamma^{1/2} \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma^{1/2}\left(\frac{1}{4} - \frac{it}{2} \right)}= \sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}$$
Now $\Gamma\left(\frac{1}{4} - \frac{it}{2}\right)$ is the complex conjugate of $\Gamma\left(\frac{1}{4} + \frac{it}{2}\right)$, acknowledging the comment by Daniel Fischer on this post:
From whichever definition of $\Gamma$ you use, it should be easy to see that $\Gamma(x) \in \mathbb R$ for positive real $x$. So the two entire meromorphic functions $f \colon z \mapsto \Gamma(z)$ and $g \colon z \mapsto \overline{\Gamma(\overline{z})}$ coincide on a set that has an accumulation point in the domain where they are holomorphic. By the identity theorem, $f \equiv g$.
Hence,
$$\arg\left(\Gamma\left(\frac{1}{4} + \frac{it}{2}\right)\right)=\frac{1}{i}\;\log\left(\sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}\right)$$
and
$$\exp\left[i\;{\arg\left(\Gamma\left(\frac{1}{4} + \frac{it}{2}\right)\right)}\right]
=\exp\left[i\;{\frac{1}{i}\;\log\left(\sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}\right)}\right]$$
Finally,
$$\color{blue}{e^{i\,\arg\left(\Gamma\left(\frac{1}{4}+\frac{it}{2}\right)\right)}}=\color{blue}{\sqrt{\frac{\Gamma \left( \frac{1}{4}+\frac{it}{2}\right)}{\Gamma\left(\frac{1}{4} - \frac{it}{2} \right)}}}\quad\square$$
Now, of course, the most intriguing part, after all this algebra is the origin of the definition of the Hardy Z-function. So back to square one!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Spectral Decomposition calculating the Eigenvalue I currently have this problem i can't seem to figure out.
A symmetric matrix
.. 6 2/5 ..
S = [.. .. ..]
.. .. ..
It has the Eigenvalues 9, 18 and c is not equel to 9 and 18 with
eigenspace:
a
E9 = Span([ b ])
-2 1/2 ,
2
E18 = Span([ 1 ])
2 ,
1
Ec = Span([ -2 ])
0
Now i wonder why is a = 2 and b = 1
How do you proof these values?
and a second question, how do you calculate the Eigenvalue of c with Spectral Decomposition ?
Thank you so much in advantage
| The eigenspaces of a symmetric matrix associated with distinct eigenvalues are always orthogonal. Since all the eigenvalues of your matrix are distinct, this means we must have
$$ \left< \begin{pmatrix} a \\ b \\ -2.5 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} \right> = 2a + b - 5 = 0, \\
\left< \begin{pmatrix} a \\ b \\ -2.5 \end{pmatrix}, \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} \right> = a - 2b = 0. $$
Hence, $5 = 2a + b - 2(a - 2b) = 5b$ so $b=1$ and then $a = 2$. Since we know the eigenvectors of $S$, we must have
$$ \begin{pmatrix} \frac{4}{\sqrt{45}} & \frac{2}{3} & \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{45}} & \frac{1}{3} & -\frac{2}{\sqrt{5}} \\ -\frac{5}{\sqrt{45}} & \frac{2}{3} & 0 \end{pmatrix} \begin{pmatrix} 9 & 0 & 0 \\ 0 & 18 & 0 \\ 0 & 0 & c \end{pmatrix}
\begin{pmatrix} \frac{4}{\sqrt{45}} & \frac{2}{\sqrt{45}} & -\frac{5}{\sqrt{45}} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} & 0 \end{pmatrix} = \begin{pmatrix} * & \frac{32}{5} & * \\ * & * & * \\ * & * & * \end{pmatrix}. $$
Computing the $(1,2)$ entry of the product, we get
$$ \begin{pmatrix} \frac{4}{\sqrt{45}} & \frac{2}{3} & \frac{1}{\sqrt{5}} \end{pmatrix} \begin{pmatrix} \frac{18}{\sqrt{45}} \\ \frac{18}{3} \\ -\frac{2c}{\sqrt{5}} \end{pmatrix} = \frac{72}{45} + \frac{36}{9} - \frac{2c}{5} = \frac{32}{5}. $$
Hence, $c = -2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:
What is the blue area?
Hint: Think outside the box
My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as below:
This then gave me the following equation to solve:
\begin{align}y^2&=2+3+4+x\\
&\Downarrow\\
x&=y^2-9\end{align}
And the following equations to do so:
\begin{align}\frac {ya}2 &=4\\
\frac {bc}2 &=3\\
\frac {yd}2 &=2\\
a+b&=y\\
c+d&=y\end{align}
I solved these to obtain:
$$a=2,
b=2,
c=3,
d=1,
y=4$$
And thus $$x=4^2-9=7$$
My question:
Is there another way I could have solved this, using the hint to think outside the box?
| Here is an alternative solution. See the graph for labeling:
From the trapezium area formula we get:
$$\frac{a+x}{2}\cdot x=5+A \Rightarrow ax=10+2A-(9+A) \Rightarrow ax=A+1 \ \
(1)$$
$$\frac{b+x}{2}\cdot x=7+A \Rightarrow bx=14+2A-(9+A) \Rightarrow bx=A+5 \ \
(2)$$
$$ab=6 \ \ (3)$$
Now multiply $(1)$ and $(2)$ and substitute $(3)$:
$$abx^2=(A+1)(A+5) \Rightarrow 6(A+9)=(A+1)(A+5) \Rightarrow A^2=49 \Rightarrow A=7.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Can anyone explain this solution (contractive function, interval, first derivative related)
We are trying to find a positive null $x \geq 0$ of $$f(x) =
\sqrt{x+1}+\frac{\text{sin}(x)}{10}-2$$
Choose $X=\left[0,8\right]$. For $x \in X$ we have that
$$\frac{1}{15}= \frac{1}{6}-\frac{1}{10} \leq f'(x) =
\frac{1}{2\sqrt{x+1}}+\frac{\text{cos}(x)}{10}\leq
\frac{1}{2}+\frac{1}{10} = \frac{3}{5}$$
(YOU COULD STOP READING HERE BECAUSE I'M RATHER INTERESTING IN JUST UNDERSTANDING THE STEPS TILL HERE.)
Now try to find a $\beta$ such that $\text{sup} \left|1+\beta
f'(x)\right|<1$, i.e.
$-2 < \beta f'(x) <0 \text{ }\forall x \in M$. Because of the
inequality above, we have that $$-\frac{10}{3} < \beta <0$$
We can thus set $\beta = -3$ and $\phi(x) = x-3f(x) =
6+x-3\sqrt{x+1}-\frac{3}{10}\text{sin}(x)$. Thus
$\left|\phi'(x)\right| \leq \frac{4}{5}= 0.8$
So, we have a contractive mapping.
[... further things that need to be shown so we satisfy all conditions (that there even exists a null). I leave them out to not make this question longer.]
I'm mainly interesting in understanding the beginning where we have $$\frac{1}{15}= \frac{1}{6}-\frac{1}{10} \leq f'(x) =
\frac{1}{2\sqrt{x+1}}+\frac{\text{cos}(x)}{10}\leq
\frac{1}{2}+\frac{1}{10} = \frac{3}{5}$$
Can you tell me how you get that? I understand that in the middle we have the first derivative of $f$. On the right side, we took the beginning of interval zero and put it into the first fraction, so $\frac{1}{2\sqrt{0+1}}= \frac{1}{2}$. Then put zero in the second fraction: $\frac{\text{cos}(0)}{10}=\frac{1}{10}$. This explains why we have $\frac{3}{5}$ on the right side.
But how do you get the left side, especially how you get $-\frac{1}{10}?$ We insert the end of interval on the left, $8$, I think. But why we have $-\frac{1}{10}$ ?
| Consider $-1\leq \cos(x)\leq 1$ for all $x\in\mathbb{R}$ yields
$$
-\frac1{10}\leq \frac{\cos(x)}{10}\leq \frac1{10}.
$$
And consider that $x\mapsto\sqrt{x+1}$ is monoton increasing. Therefore you get $$
2\leq2\sqrt{x+1}\leq 6
$$
for $x\in[0,8]$ using the boundary points $x=0$ and $x=8$. This yields
$$
\frac16\leq \frac1{2\sqrt{x+1}}\leq \frac12
$$
for $x\in[0,8]$. Together we get
$$
\frac16-\frac1{10}\leq \frac1{2\sqrt{x+1}}+\frac{\cos(x)}{10}\leq \frac12+\frac1{10}
$$
for $x\in[0,8]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Area under the graph - integration
The region $P$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the line $x=a$ . The region $Q$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the lines $x=2a$ and $x=a$. Given that the area of $Q$ is twice the area of $P$, find the value of $a$ .
Firstly , on the first step , in already stuck ...
I used definite integral to find the area of $P$ -
$$\int^a_0\ (3x-x^2)dx=\frac{9a^2-2a^3}{6}$$
However when I calculate area of $Q$ , it's the same as Area of $P$ - $$\frac{9a^2-2a^3}{6}$$
Then since
$Q= 2P$
$9a^2 - 2a^3 = 18a^2 - 4a^3 $
From here, I definitely can't find the value of $a$ ... where have I gone wrong or misunderstood ?
| Since $a\neq 0$ (as this would be absurd), we can solve your equation as follows
\begin{align}9a^2−2a^3&=18a^2−4a^3\\
9-2a&=18-4a\tag{divide through by $a$}\\
2a&=9\\
a&=\frac 92\end{align}
However, as noted in other answers, this is incorrect.
Your error is somewhere in your calculation of the area of $Q$, you should get $$\int_a^{2a}(3x-x^2)\text dx =\frac{27a^2-14a^3}{6}$$
This then gives you \begin{align}Q&=2P\\\frac{27a^2-14a^3}{6}&=2\times \frac{9a^2-2a^3}{6}\\
27a^2-14a^3&=2(9a^2-2a^3)\\
27a^2-14a^3&=18a^2-4a^3\\
27-14a&=18-4a\tag{divide through by $a$}\\
9&=10a\\
a&=\frac 9{10}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Sitting in chairs with empty space There are n chairs in a row. In how many ways can a teacher sit k students on these chairs so that no 2 students sit next to each other (and obviously no 2 students sit on 1 chair)?
I was thinking about inclusion-exclusion, that each student will sit next to each other, however it isn't good enough since there could be more chairs than students.
Basically currently I have no idea how to solve it.
| Consider strings over the alphabet $\{0,1\}$ where $0$ denotes a chair and $1$ denotes a student. The strings that do not contain $11$ are captured by the regular expression $0^* (10^+)^* \{\varepsilon, 1\}$. If we give $0$ a weight of $x$ and $1$ a weight of $xy$ then we are looking for the coefficient of $[x^ny^k]$.
Using these weights, the generating function for $0^* (10^+)^* \{\varepsilon, 1\}$ is
$$ \frac{1}{1 - x} \cdot \frac{1}{1 - \frac{x^2y}{1 - x}} \cdot (1 + xy) = \frac{1 + xy}{1 - x - x^2y}. $$
Hence the answer is
$$
[x^ny^k]\frac{1 + xy}{1 - x - x^2y} = [x^ny^k]\frac{1}{1 - x - x^2y} + [x^{n - 1}y^{k-1}]\frac{1}{1 - x - x^2y}
$$
where
\begin{align}
[x^ny^k]\frac{1}{1 - x - x^2y} &= [x^ny^k] \sum_m (x + x^2y)^m \\
&= [x^ny^k] \sum_m \sum_r \binom{m}{r} x^{m-r}x^{2r}y^r \\
&= [x^ny^k] \sum_m \sum_r \binom{m}{r} x^{m + r}y^r \\
&= \binom{n - k}{k}.
\end{align}
Therefore
$$
[x^ny^k]\frac{1 + xy}{1 - x - x^2y} = \binom{n - k}{k} + \binom{(n - 1) - (k - 1)}{k - 1} = \binom{n - k + 1}{k}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
What is wrong with my integral calculation? I want to solve the integral $\int_1^\infty\frac{a}{a^2+x^2}dx$ where $a>0$ is a constant
I first tried simplifying so
\begin{align}
\frac{a}{a^2+x^2}
&=\frac{1}{a+\frac{x^2}{a}} \\[6px]
&=\frac{1}{a}\cdot\frac{1}{1+(\frac{x}{a})^2}
\end{align}
which got me
\begin{align}
\int_1^\infty\frac{a}{a^2+x^2}dx
&=\frac{1}{a}\int_1^\infty\frac{1}{1+(\frac{x}{a})^2}dx\\[6px]
&=\frac{1}{a}\left[\arctan \frac{x}{a}\right]_1^\infty
\end{align}
which is wrong, but I can't see my mistake, can anyone help me with this?
| \begin{align}
\int_1^\infty\frac{a}{a^2+x^2}dx
&=\frac{1}{a}\int_1^\infty\frac{1}{1+(\frac{x}{a})^2}dx\\[6px]
&=\int_1^\infty\frac{1}{1+(\frac{x}{a})^2}d\frac{x}{a}\\[6px]
&=\left[\arctan \frac{x}{a}\right]_1^\infty\\[6px]
&=\frac{\pi}{2}-\arctan\frac{1}{a}=\arctan a
\end{align}
If $a<0$, then the final step will give $\pi+\arctan a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Laplace Transform on Beam (w/ Loads) (Sorry about my english I'm a South American student)
The deflection of a fixed beam is zero at $x=0$ and $x=L$ on its boundaries. The beam resists a distributed load of $Wo$ per unit length. The equation is:
$$\frac{d^{4}y}{dx^4}=\frac{W_{0}}{EI}$$
$$0 < x < L$$
Where $Wo$, $E$, and $I$ are constants
$y(0) = y''(0) = y(L) = y''(L) = 0$
Solve using Laplace Transform
| Take Laplace Transform of the equation, but perform this with respect to x (where normally, it is performed with respect to time, t):
$$ \mathcal{L} \left[ \frac{d^4 y}{dx^4} \right] = \mathcal{L} \left[ \frac{W_0}{EI} \right] $$
$$ s^4 Y(s) - s^3 y(0) - s^2 y^\prime (0) - s y^{\prime \prime}(0) - y^{\prime \prime \prime}(0) = \frac{W_0}{EI} \frac{1}{s} $$
By applying the given boundary conditions, this equation reduces:
$$ s^4 Y(s) - s^2 y^\prime (0) - y^{\prime \prime \prime}(0) = \frac{W_0}{EI} \frac{1}{s} $$
$$ Y(s) = \frac{W_0}{EI} \frac{1}{s^5} + y^{\prime} (0) \frac{1}{s^2} + y^{\prime \prime \prime} (0) \frac{1}{s^4} $$
Now, take inverse Laplace Transform:
$$ \mathcal{L}^{-1} [Y(s)] = \mathcal{L}^{-1} \left[ \frac{W_0}{EI} \frac{1}{s^5} + y^{\prime} (0) \frac{1}{s^2} + y^{\prime \prime \prime} (0) \frac{1}{s^4} \right] $$
$$ y(x) = \frac{W_0}{EI} \frac{x^4}{24} + y^{\prime \prime \prime}(0) \frac{x^3}{6} + y^\prime(0)x $$
$$ y^{\prime \prime}(x) = \frac{W_0}{EI} \frac{x^2}{2} + y^{\prime \prime \prime}(0) x $$
Then, apply other boundary conditions:
$$ y(L) = 0 = \frac{W_0}{EI} \frac{L^4}{24} + y^{\prime \prime \prime}(0) \frac{L^3}{6} + y^\prime(0) L $$
$$ y^{\prime \prime}(L) = 0 = \frac{W_0}{EI} \frac{L^2}{2} + y^{\prime \prime \prime}(0) L $$
which turns out to be
$$ y^{\prime \prime \prime}(0) = - \frac{W_0 L}{2EI} $$
Substitute into first boundary condition:
$$ y(L) = 0 = \frac{W_0 L^4}{24EI} - \frac{W_0 L^4}{12EI} + y^\prime(0) L $$
$$ y^\prime (0) = \frac{W_0 L^3}{24EI} $$
Combining all the terms:
$$ y(x) = \frac{W_0}{24EI} x^4 - \frac{W_0}{12EI} x^3 L + \frac{W_0}{24EI}x L^3 $$
Finally,
$$ y(x) = \frac{W_0}{24EI} \left( x^4 - 2L x^3 + L^3 x \right) $$
which is the deflection of simply supported beam at both sides with distributed loading.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Non-Symmetric Algebraic inequality
Let $a,b,c$ be positive real numbers. Prove that : $$\frac{ab}{3a+b}+\frac{bc}{2c+b}+\frac{ac}{c+2a}\le \frac{2a+20b+27c}{49}$$
| Let $a=6x$, $b=3y$ and $c=2z$.
Hence, we need to prove that
$$\frac{xy}{6x+y}+\frac{yz}{3y+4z}+\frac{zx}{z+6x}\leq\frac{2x+10y+9z}{49},$$
which is true by AM-GM.
Indeed,
$$\frac{xy}{6x+y}+\frac{yz}{3y+4z}+\frac{zx}{z+6x}\leq\frac{x+6y}{49}+\frac{4y+3z}{49}+\frac{6z+x}{49}=\frac{2x+10y+9z}{49}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\frac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$, find the value of $P(n+1)$ Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\dfrac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$
I tried by making a $f(x) = (x^3+1) P(x) - x^2$. But this equation will have $n+3$ roots, out of which $n+1$ will be $0,1,2,...,n$. What about the other two roots? Can we simplify $x^2 \over x^3+1$ more to help?
| Using (Lagrange) polynomial interpolation:
\begin{align*}
P(x) &= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \prod_{j \neq k} \frac{x - j}{k-j} \right) \\
&= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \frac{x (x-1) \cdots (x - (k-1)) (x-(k+1)) \cdots (x-n) }{ k (k-1) \cdots (k-(k-1)) (k-(k+1)) \cdots (k-n)}\right) \\
&= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \frac{x (x-1) \cdots (x - (k-1)) (x-(k+1)) \cdots (x-n) }{ k! (n-k)!(-1)^{n-k} } \right) \\
\end{align*}
Substituting,
\begin{align*}
P(n+1) &= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \frac{(n+1) (n) \cdots (n+2-k) (n-k) \cdots (1) }{ k! (n-k)!(-1)^{n-k} } \right) \\
&= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \frac{(n+1)!}{ k! (n-k)!(-1)^{n-k} (n+1-k)} \right) \\
&= \sum_{k=0}^n (-1)^{k-n} \frac{k^2}{k^3+1} {n+1 \choose k} \\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How to solve these functional equations $f(x) = 1 + f(x) + f(x^2) + \ldots $ and $f(x) = 1 + f(x) + \big(f(x^2)\big)^2 + \ldots $? The identity $\frac{1}{1-f(x)}=1+f(x)+\big(f(x)\big)^2+\big(f(x)\big)^3 + \ldots$ is well known.
Is there somewhere in the literature a method allowing us to solve the following functional equations?
First equation:
$$f(x) = 1 + f(x) + f(x^2) + f(x^3) + \ldots $$
Second equation:
$$f(x) = 1 + f(x) + \big(f(x^2)\big)^2 + \big(f(x^3)\big)^3 + \ldots $$
| There is no solution for your first functional equation. Indeed :
$f\left( x\right) =1+f\left( x\right) +f\left( x^{2}\right) +f\left(
x^{3}\right) +f\left( x^{4}\right) +f\left( x^{5}\right) +f\left(
x^{6}\right) +\cdots $
so
$-1=f\left( x^{2}\right) +f\left( x^{3}\right) +f\left( x^{4}\right)
+f\left( x^{5}\right) +f\left( x^{6}\right) +\cdots $
so for $x^{2}$ we have
$-1=f\left( x^{4}\right) +f\left( x^{6}\right) +f\left( x^{8}\right)
+f\left( x^{10}\right) +f\left( x^{12}\right) +\cdots $
then
$0=f\left( x^{3}\right) +f\left( x^{5}\right) +f\left( x^{7}\right) +f\left(
x^{9}\right) +\cdots $
so
$-1=f\left( x^{2}\right) +f\left( x^{4}\right) +f\left( x^{6}\right)
+f\left( x^{8}\right) +\cdots $
then $f\left( x^{2}\right) =0$ then $f\left( x^{2n}\right) =0$ so
$-1=f\left( x^{3}\right) +f\left( x^{5}\right) +f\left( x^{7}\right)
+f\left( x^{9}\right) +\cdots $
so $-1=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.