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How to evaluate this definite integral? Here is the integral we have to evaluate:
$$\int_0^4\sqrt{x^2+4}\,dx+\int_2^{2\sqrt{5}}\sqrt{x^2-4}\,dx$$
After observing, i realise that i can't evaluate these integrals from area of circle, I say that $u=\sqrt{x^2+4}$. Then i can say $dx=\frac{udu}{\sqrt{u^2-4}}$. The first term would transform into:
$$\int_2^{2\sqrt{5}}\frac{u^2\,du}{\sqrt{u^2-4}}$$
Similarly i say $v=\sqrt{x^2-4}$. Then i can also say $dx=\frac{vdv}{\sqrt{v^2+4}}$ and the second term would transform into:
$$\int_0^4\frac{v^2\,dv}{\sqrt{v^2+4}}$$
But this integrals also doesn't seem easy to solve. It goes without saying that after substitution, the bounds of integrals interchanged. Maybe it can be helpful to solve the problem. Thank you for your effort!
| Let $f(x)=\sqrt{x^2+4}$ and $f^{-1}(x)=\sqrt{x^2-4}$. Note that
$$f(4)=2\sqrt{5} \\
f(0)=2
$$
We know that the integral of an inverse function can be calculated with
$$\int_{f(k)}^{f(l)}f^{-1}(x)\,dx=lf(l)-kf(k)-\int_k^lf(x)\,dx$$
So the integral
$$\int_0^4\sqrt{x^2+4}\,dx+\int_2^{2\sqrt{5}}\sqrt{x^2-4}\,dx$$
can be written as
$$\int_0^4\sqrt{x^2+4}\,dx+4\cdot2\sqrt{5}-2\cdot0-\int_0^4\sqrt{x^2+4}\,dx$$
Finally the result is
$$ \bbox[5px,border:2px solid red]
{
8\sqrt{5}
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
System of 5x5 non-linear equations from Lagrange multiplier problem I have a Lagrange multiplier problem, and I've got down to the system of equations, but solving these has proved to be a nightmare for me.
I have five equations:
$$x+y+z=0$$
$$x^2+y^2+z^2=24$$
$$y=\lambda_1+2x\lambda_2$$
$$x=\lambda_1+2y\lambda_2$$
$$2=\lambda_1+2z\lambda_2$$
I keep getting down to $x=y$ but I know that a local minimum occurs when $x=\frac{1-3\sqrt5}{2},y=\frac{1+3\sqrt5}{2},z=-1$ and I can't seem to get down to that one.
I can see the answers, I'm just struggling with the working.
For reference, the original problem is to find the extrema for:
$$f(x,y,z)=xy+2z$$
with the constraints of $x+y+z=0$ and $x^2+y^2+z^2=24$
Please help...
| One method is to eliminate the constraints. We have $x+y+z=0$ so $z=-x-y$. Also $x^2+y^2+z^2=1$ so
\begin{eqnarray*}
xy+yz+zx&=&-12. \\
f =xy+2z &=& -12+z(2-x-y)=12+z(2+z)=(z+1)^2-13
\end{eqnarray*}
So the extreme point occurs at $z=-1$. Now lets calulate $x$ & $y$
\begin{eqnarray*}
x+y&=&1 \\
xy &=& -11 \\
(x-y)^2 &=& (x+y)^2-4xy=45
\end{eqnarray*}
So $x=\frac{1+3 \sqrt{5}}{2}$ & $y=\frac{1-3 \sqrt{5}}{2}$ (or vice versa).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Abstract Algebra; Galois field extension and automorphisms I am really curious...
Can $\sqrt{4+\sqrt{7}}$ be written in the form $\sqrt a + \sqrt b $ with rational numbers $a$ and $b$ ?
I was thinking that we could try to brute force it by equating $\sqrt{4+\sqrt{7}} = x$ and then try to work out the $a$ and $b$ ? Any thoughts ? :)
| Wikipedia describes a simple method to write
$$
\sqrt{a+b \sqrt{c}\ } = \sqrt{d}+\sqrt{e}
$$
with
$$
d=\frac{a + \sqrt {a^2-b^2c}}{2},
\qquad
e=\frac{a - \sqrt {a^2-b^2c}}{2}
$$
This works iff $a^2 - b^2c$ is a square.
For $\sqrt{4+\sqrt{7}}$ we have $a^2 - b^2c=9$ and so
$$
\sqrt{4+\sqrt{7}} = \sqrt{\frac72}+\sqrt{\frac12}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216795",
"timestamp": "2023-03-29T00:00:00",
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Does the sum of digits squared of a number when we keep on doing it lead to a single digit number? When we add the digits of number for eg $2478125$
$$2+4+7+8+1+2+5=29;\\
2+9=11;\\
1+1=2$$
and when we square the digits and add them following this rule
$$2^2+4^2+7^2+8^2+1^2+2^2+5^2=163;\\
1^2+6^2+3^2=46; \\
4^2+6^2=52;\\
5^2+2^2=29;\\
2^2+9^2=85;$$and this goes on, so the question is
will it ever reach a single digit.And what is the proof of it.
| The answer is yes, it's the subject of this year's french olympiads : for one or two digits numbers, you prove by hand (or computer) verification that the resulting sequence is either stationnary at $1$, or loops indefinitely from $4$ to $16$, $37$, $58$, $89$, $145$, $42$, $20$, $4$ and so on.
Then you prove it for 3 digits numbers (again by hand verification), and finally, prove that if $x$ has $p\ge4$ digits, then $f(x)$ has at most $p-1$ digits.
There may be smarter solutions for $2$ and $3$ digits numbers, for example construct a graph, because the order of digits is irrelevant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2217832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Expand binomially to prove trigonometric identity Prompt: By expanding $\left(z+\frac{1}{z}\right)^4$ show that $\cos^4\theta = \frac{1}{8}(\cos4\theta + 4\cos2\theta + 3).$
I did the expansion using binomial equation as follows
$$\begin{align*}
\left(z+\frac{1}{z}\right)^4 &= z^4 + \binom{4}{1}z^3.\frac{1}{z} + \binom{4}{2}z^2.\frac{1}{z^2} + \binom{4}{3}z^3.\frac{1}{z}+\frac{1}{z^4}\\
&=z^4+4z^2+6+\frac{4}{z^2}+\frac{1}{z^4}\\
&=z^4+\frac{1}{z^4}+4\left(z^2+\frac{1}{z^2}\right) + 6. -(eqn 1)
\end{align*}
$$
I'm not sure how to go on about rest of the problem.
[update]
Reading comments, I tried assuming $z = e^{i\theta}$
$2\cos\theta = e^{i\theta} + e^{-i\theta}$
$(2\cos\theta)^4 = (e^{i\theta} + e^{-i\theta})^4$
$=e^{4i\theta} + e^{-4i\theta} + 8(e^{2i\theta}+e^{-2i\theta})+6$ (from eqn 1)
| HINT:
Set $z=e^{i\theta}$ and use Euler's formula.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Compute Flux (Divergence Theorem) Given that $\textbf{F} = \langle x\cos^2z, y\sin^2z, \sqrt{x^2+y^2}\:z \rangle $ and let E be the solid cone above the $xy$-plane and inside z = $1 -\sqrt{x^2+y^2}.$
I'm trying to use the divergence theorem to compute the flux.
$$\iint_{D} \textbf{F} \cdot \textbf{N} \: dS = \iiint_E \nabla \cdot \textbf{F}\:d\textbf{V}$$
Attempt:
$$\text{div}\textbf{F} \:=\: \nabla \cdot \textbf{F}
=\cos^2z + \sin^2z + \sqrt{x^2+y^2} \:
= z + r \:
$$
Since solid cone is above the $xy$-plane, $\text{z} \ge 0$ and $z = 1-\sqrt{x^2+y^2} = 1-r$. Hence $ 0 \le \text{z} \le 1-r$. Is this bound right for z?
\begin{align*}
\iiint_E \nabla \cdot \textbf{F}\:d\textbf{V} &= \int_0^{2\pi} \int_0^{1} \int_0^{1-r} (z+r) r\: dz\: dr\: d\theta \\
&= 2\pi \int_0^{1} \left(\frac{z^2r}{2} + r^2z\right)\bigg\rvert_0^{1-r} dr \\
&= 2\pi \cdot \frac{1}{2} \int_0^1 r-r^3\, dr \\
&= \pi \left(\frac{r^2}{2}-\frac{r^4}{4}\right)\bigg\rvert_0^1 \\
&= \frac{\pi}{4}.
\end{align*}
The answer given was $\frac{\pi}{2}.\:$
Perhaps I'm doing something wrong. One thing that I'm not sure of is picturing the surface that they're looking for. Also, not sure if my bounds are right. Where am I going wrong here? Would appreciate some help! Thank you.
| Since $\cos^2z + \sin^2z = 1$, you must have $\nabla \cdot \mathbf{F} = 1+r$. In this case, your integral becomes
$$
\iiint_E\nabla \cdot \mathbf{F} \, \mathrm{d}V = \int_0^{2\pi} \int_0^1 \int_0^{1-r} (1+r)r\, \mathrm{d}z\, \mathrm{d}r\, \mathrm{d}\theta = \frac{\pi}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219432",
"timestamp": "2023-03-29T00:00:00",
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Solve $a_n=7a_{n-1}-12a_{n-2}+3^n$ $a_0=1\:a_1=2$
Using generating functions I get:
$f\left(x\right)-1-2x=7xf\left(x\right)-7x-12x^2f\left(x\right)+\frac{1}{1+3x}-1-3x$
$f\left(x\right)=\frac{3}{14\left(1+3x\right)}-\frac{13}{2\left(3x-1\right)}\:+\:\frac{40}{7\left(4x-1\right)}$
Now how can I express $a_n$ from this?
| Hint. One may just recall that
$$
\frac{1}{1-u}=\sum_{n=0}^\infty u^n,\qquad |u|<1,
$$ then applying it to
$$
f\left(x\right)=\frac{3}{14\left(1+3x\right)}+\frac{13}{2\left(1-3x\right)}\:-\:\frac{40}{7\left(1-4x\right)}
$$ respectively with $u$ equals to $-3x$, $3x$ and $4x$.
Can you finish it?
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a$ , $b$, $c$ are positive integers satisfying the following $(a^2 +2)(b^2+3)(c^2+4)=2014$ What is the value of $ a+b+c $ If $a$ , $b$, $c$ are positive integers satisfying the following
$(a^2 +2)(b^2+3)(c^2+4)=2014$
What is the value of $ a+b+c $
I need details because i don't know how to solve simlar problems ?
Thank you for your help.
| $2014 = 2*19*53$
So each of the $a^2 + 2, b^2+3, c^2 + 4$ are some combination of $2, 19, 53$.
As none of them can equal 1, the must be that one of them equals $2$, another $19$ and the third $53$. What possible numbers work? (Note: just by looking only one can be small enough to equal $2$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Show the value of determinant is 0 I dont have any idea how to show the value of the following determinant is 0 without expanding the determinants
$$
\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab \\
\end{vmatrix}
$$
Any ideas?
| $\Delta=\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab \\
\end{vmatrix}=\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2\\
\end{vmatrix}-\begin{vmatrix}
1 & a & bc \\
1 & b & ca \\
1 & c & ab \\
\end{vmatrix}=\Delta_1-\Delta_2$, say.
Now, $\Delta_2=\begin{vmatrix}
1 & a & bc \\
1 & b & ca \\
1 & c & ab \\
\end{vmatrix}=\frac{1}{abc}\begin{vmatrix}
a & a^2 & abc \\
b & b^2 & abc \\
c & c^2 & abc \\
\end{vmatrix}$ $\quad(a,b,c\ne0)$
$\qquad\qquad=\begin{vmatrix}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1 \\
\end{vmatrix}=-\begin{vmatrix}
a & 1 & a^2 \\
b & 1 & b^2 \\
c & 1 & c^2 \\
\end{vmatrix}=\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2\\
\end{vmatrix}=\Delta_1$
The properties used are self-explanatory.
Alternatively, to show $\Delta_1=\Delta_2$ we can use the following argument:
Let $f(a,b,c)=\Delta_2-\Delta_1$
Then, $f(a,b,c)=k(a-b)(b-c)(a-c)$ for some constant $k$ as $f(b,b,c)=f(a,b,a)=f(a,b,b)=0$.
The coefficient of $a^2b$ in $f$ is $0\Rightarrow k=0.$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Compute $\prod_{k=5}^{\infty}\frac{k^4-17k^2+16}{k^4-8k^2+16}$ Compute the product: $$\prod_{k=5}^{\infty}\frac{k^4-17k^2+16}{k^4-8k^2+16}$$
I was able to factor in the following manner:
$$ \frac{k^4-17k^2+16}{k^4-8k^2+16}=\frac{(k-1)(k+1)(k-4)(k+4)}{(k-2)^2(k+2)^2}$$
but what should I do now?
| Take the partial products up to $N$ and use the fact that $$\prod_{k=5}^N \frac{k-1}{k-2} = \frac{4}{3} \cdot \frac{5}{4} \cdot ... \cdot \frac{N-1}{N-2} = \frac{N-1}{3},$$ and similarly $$\prod_{k=5}^N \frac{k+1}{k+2} = \frac{6}{N+2}, \quad \prod_{k=5}^N \frac{k-4}{k-2} = \frac{2}{(N-2)(N-3)}, \quad \prod_{k=5}^N \frac{k+4}{k+2} = \frac{(N+4)(N+3)}{7 \cdot 8}.$$ So $$\prod_{k=5}^N \frac{k^4 - 17k^2 + 16}{k^4 - 8k^2 + 16} = \frac{(N-1)(N+3)(N+4)}{14 (N-3)(N-2)(N+2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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use polar coordiantes and the Poincare-Bendixson Theorem. I am searching for some help with this question in review for a upcoming exam, I have been having some trouble understanding the workings that have been provided and am just looking for some clarification.
The question says to show that,
$$x'=x+y-x(x^2+2y^2)$$
$$y'=-x+y-y(x^2+2y^2)$$
has at least one periodic solution. The hint given was to use polar coordiantes and the Poincare-Bendixson Theorem.
So using polar coordiates,
$$r'=\frac{xx'+yy'}{r}=-r(F(\theta)r^2-1)$$
So how did they calculate $$-r(F(\theta)r^2-1)$$
and my final question is how does this next step come about? How did they calculate $F(\theta)$
$$F(\theta)=cos^4\theta+2sin^4\theta+3cos^2\theta sin^2\theta=1+sin^2\theta $$
| We calculate
$$r'=\dfrac{xx'+yy'}{r} = \dfrac{1}{2} r^3 \cos 2 \theta-\dfrac{3}{2}r^3 + r = -r\left(\left(\dfrac{3}{2} - \dfrac{1}{2} \cos 2 \theta \right)r^2 - 1 \right) = -r(F(\theta)r^2-1)$$
where
$$F(\theta) = \left(\dfrac{3}{2} - \dfrac{1}{2} \cos 2 \theta\right) = 1 + \sin^2 \theta$$
Update
From $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$, we have
$$\left(\dfrac{3}{2} - \dfrac{1}{2} \cos 2 \theta\right) = \dfrac{3}{2} -\dfrac{1}{2}\left(\cos^2 \theta - \sin^2 \theta\right) = \dfrac{3}{2} -\dfrac{1}{2}\left(1 - \sin^2 \theta\right) + \dfrac{1}{2} \sin^2 \theta = 1 + \sin^2 \theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Solving Recurrence Relation $a_n=6a_{n-1} - 9a_{n-2}$ for $n\geqslant2$ So the problem is to solve this recurrence relation with the initial conditions $a_0 = 2, a_1 = 21$.
$a_n=6a_{n-1} - 9a_{n-2}$ for $n\geqslant2$
And also find the value of $a_{2016}.$
Here is my solution but I'm not entirely sure if it's correct. Was wondering if anyone can confirm what I did is valid or perhaps I made a mistake somewhere? Thanks in advance.
| Generatingfunctionologically, define $A(z) = \sum_{n \ge 0} a_n z^n$, shift the recurrence by 2, multiply by $z^n$, sum over $n \ge 0$ and recognize resulting sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
&= 6 \sum_{n \ge 0} a_{n + 1} z^n - 9 \sum_{n \ge 0} a_n z^n \\
\frac{A(z) - a_0 - a_1 z}{z^2}
&= 6 \frac{A(z) - a_0}{z} - 9 A(z)
\end{align*}$
Solve for $A(z)$ with the given values of $a_0. a_1$, write as partial fractions:
$\begin{align*}
A(z)
&= \frac{2 + 9 z}{1 - 6 z + 9 z^2} \\
&= \frac{5}{(1 - 3 z)^2} - \frac{3}{1 - 3 z}
\end{align*}$
Use the generalized binomial theorem:
$\begin{align*}
(1 + x)^{-m}
&= \sum_{n \ge 0} (-1)^n \binom{-m}{n} x^n \\
&= \sum_{n \ge 0} \binom{n + m - 1}{m - 1} x^n
\end{align*}$
extract the desired coefficient:
$\begin{align*}
[z^n] A(z)
&= 5 \cdot \binom{n + 2 - 1}{2 - 1} \cdot 3^n - 3 \cdot 3^n \\
&= (5 n + 2) \cdot 3^n
\end{align*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the minimal polynomial of $\zeta_9+\zeta^{-1}_9$ over $\Bbb Q$ Knowing the degree of $\Bbb Q(\zeta_9+\zeta^{-1}_9)$ over $\Bbb Q$ is 3, now I want to find the minimal polynomial of $\zeta_9+\zeta^{-1}_9$ over $\Bbb Q$. I tried to use the relation $\zeta_9$ is the root of $x^6＋x^3＋1$ and $x^8＋x^7+x^6+x^5+x^4+x^3+x^2+x+1$. But it does not seems to work. Now I am stuck. Could someone please help？ Thanks so much！
| $\zeta_9$ is a root of $p(x)=\frac{x^9-1}{x^3-1}=x^6+x^3+1$ and
$$ \frac{p(x)}{x^3} = 1+x^{3}+x^{-3} = 1+\left(x+\frac{1}{x}\right)^3- 3\left(x+\frac{1}{x}\right) $$
hence $\zeta_9+\zeta_9^{-1}=2\cos\frac{2\pi}{9}$ is a root of $\color{red}{z^3-3z+1}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find the value of $(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a+1)$ if $\log_{b}a+\log_{c}b+\log_{a}c=13$ and $\log_{a}b+\log_{b}c+\log_{c}a=8$ Let $a,b$, and $c$ be positive real numbers such that
$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$
and
$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$
What is the value of
$$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1) ?$$
I tried to convert the entire thing to fractional logs and multiply the expression and add the two equations but it did not help.
| As lab bhattacharjee commented, convert to natural logarithms and get
$$\log_{a}b + \log_{b}c + \log_{c}a = 8\implies \frac{\log (b)}{\log (a)}+\frac{\log (a)}{\log (c)}+\frac{\log (c)}{\log (b)}=8$$ $$\log_{b}a + \log_{c}b + \log_{a}c = 13\implies \frac{\log (a)}{\log (b)}+\frac{\log (c)}{\log (a)}+\frac{\log (b)}{\log (c)}=13$$ Now expand
$$(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a + 1)$$ which is $$\frac{\log (a)}{\log (b)}+\frac{\log (b)}{\log (a)}+\frac{\log (a)}{\log
(c)}+\frac{\log (c)}{\log (a)}+\frac{\log (c)}{\log (b)}+\frac{\log (b)}{\log
(c)}+2$$ I am sure that you can take it from here and finish.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Finding the exact value of a radical How do I show that $\sqrt{97 +56\sqrt3}$ reduces to $7 +4\sqrt3?$. Without knowing intitially that it reduces to that value.
| If you suspect that $\sqrt{97+56\sqrt3}$ (or similar) equals something
like $a+\sqrt b$ with $a$, $b$ rational you must have
$$97+56\sqrt 3=a^2+b+2a\sqrt b.$$
To match up the surds you'll have to have $a^2+b=97$
and $2a\sqrt b=56\sqrt3$. Thus $a^2b=28^2\times 3=2352$. Therefore
$a^2$ and $b$ are the solutions of the quadratic
$$x^2-97x+2352=0.$$
The solutions are $x=48$ and $x=49$. Then $a^2=49$ as that's the one
that's a square, and $b^2=48$. So
$$\sqrt{97+56\sqrt3}=\sqrt{49}+\sqrt{48}=7+4\sqrt3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Triangles of the form $a^n+b^n=c^n$
Let $\Delta$ be a triangle with (real) sides length $a,b,c>0$. From this question, we know that $$a+b=c \qquad \iff \qquad \Delta \text{ is a straight line}$$ Now, we know from the Pythagorean theorem that
$$ a^2+b^2=c^2 \qquad \iff \qquad \Delta \text{ is a right triangle}$$
I'm wondering if it is possible, for a fixed $n>2$, to have a geometric characterization of the triangles which satisfy $a^n+b^n=c^n$, that is:
$$ a^n+b^n=c^n \qquad \iff \qquad \Delta \text{ is ?}.$$
Notes: For $n>1$, and any $a,b>0$ there exists always a non-degenerated triangle $\Delta$ with $c= (a^n+b^n)^{1/n}$. It seems that for the case $n=3$, things are already more complicated. All the following triangles satisfy
$$a^3+b^3=c^3.$$
| If $x = \frac{a}{c}$, $y = \frac{b}{c}$ (note: both are in $(0, 1)$)
then for $n \gt 2$,
$x + y \gt x^2 + y^2 > x^n + y^n = 1$
Thus $a + b \gt c$ and $a^2 + b^2 \gt c^2$ and so $a,b,c$ form the sides of a triangle and the triangle must be acute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $|\int_{x}^{x+1} \sin(t^2)dt| \leq \frac{1}{x}$? I have $F(x) = \int_{x}^{x+1} \sin(t^2)dt$, how can i proof that $|F(x)| \leq \frac{1}{x}$ for all $x > 0 $? I have no idea how to do it.
| Observe that
$$
\left|\int_x^{x+1}\sin(t^2)dt\right|=\left|\int_x^{x+1}\frac{t}{t}\sin(t^2)dt\right|.
$$
Using integration by parts, we have
\begin{align*}
u&=\frac{1}{t}&dv&=t\sin(t^2)dt\\
du=&-\frac{1}{t^2}&v&=-\frac{1}{2}\cos(t^2).
\end{align*}
Therefore,
\begin{align*}
\left|\int_x^{x+1}\frac{t}{t}\sin(t^2)dt\right|
&=\left|\left.-\frac{1}{2t}\cos(t^2)\right|_x^{x+1}-\int_x^{x+1}\frac{1}{2t^2}\cos(t^2)dt\right|\\
&\leq\left|\left.\frac{1}{2t}\cos(t^2)\right|_x^{x+1}\right|+\left|\int_x^{x+1}\frac{1}{2t^2}\cos(t^2)dt\right|\\
&\leq\frac{1}{2(x+1)}+\frac{1}{2x}+\int_x^{x+1}\frac{1}{2t^2}dt
\end{align*}
since $|\cos(t^2)|$ is at most $1$. Then, we can just integrate to get
$$=\frac{1}{2(x+1)}+\frac{1}{2x}+\left(\left.-\frac{1}{2t}\right|_{x}^{x+1}\right)
=\frac{1}{2(x+1)}+\frac{1}{2x}-\frac{1}{2(x+1)}+\frac{1}{2x}=\frac{1}{x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Probability that a passenger will catch the bus A bus and a passenger arrive at a bus stop at a uniformly distributed time over the interval $0$ to $1$ hour. Assume the arrival times of the bus and passenger are independent of one another and that the passenger will wait up to $15$ minutes for the bus to arrive. What is the probability that the passenger will catch the bus?
Hint: Let $Y1$ denote the bus arrival time and $Y2$ the passenger arrival time. Determine the joint density of $Y1$ and $Y2$ and determine $$P(Y2 \leq Y1 \leq Y2+\frac{1}{4}).$$
We already know that the answer to this question is $\frac{7}{32}$. Any help would be much appreciated.
| There is a $\frac{3}{4}$ chance the bus will be later than 10:15. Therefore there is a $\frac{1}{4}$ of an hour window of the person $Y2$, arriving on time. If $Y1 < \frac{1}{4}$ and $Y2 > \frac{1}{4}$ the probability of the person missing the bus is certainly $1$.
If $Y1 (\frac{1}{4})$ AND $Y2 (\frac{1}{4})$ arrives before 10:15, then $\frac{1}{4}+\frac{1}{4} = \frac{1}{2}$ chance of catching the bus.
So we have: $$\frac{3}{4} \cdot \frac{1}{4} + (\frac{1}{4}\cdot \frac{1}{4} \cdot \frac{1}{2}) = \dfrac{3}{16} + \dfrac{1}{32} = \dfrac{7}{32} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find $ \lim\limits_{{n \to \infty}} \frac1{2^n} \sum\limits_{k=1}^n \frac1{\sqrt{k}} \binom nk$ Find
$$\lim_{n \to \infty} \frac {1}{2^n} \sum_{k=1}^n \frac{1}{\sqrt{k}} \binom{n}{k}.$$
First time I thought about Stirling's approximation but didn't get anything by applying it. I would also think about a Riemann Sum, but no idea how to rewrite...
The answer is $0$.
| Let $a_k = k^{-1/2}$. Notice that $(a_k)$ decreases to $0$. Then for each fixed $m \geq 1$ and for all $n \geq m$,
$$ \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} a_k \leq \frac{1}{2^n} \underbrace{\sum_{k=1}^{m} \binom{n}{k} (a_k - a_m)}_{= \mathcal{O}(n^m)} + \frac{1}{2^n} \underbrace{\sum_{k=1}^{n} \binom{n}{k} a_m}_{=(2^n - 1)a_m}. $$
Taking limsup as $n\to\infty$, it follows that
$$ \limsup_{n\to\infty} \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} a_k \leq a_m $$
Since $a_m \to 0$ as $m\to\infty$, this proves
$$ \lim_{n\to\infty} \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} a_k = 0. $$
Addendum. I just saw that OP is a high-school student. Here is a little tweak of the argument above that does not use any fancy analysis stuffs.
Let $m_n = \lfloor \log n \rfloor$. Then for $n \geq 3$, we always have $1 \leq m_n \leq n$. Then
\begin{align*}
0
\leq \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} \frac{1}{\sqrt{k}}
&= \frac{1}{2^n} \sum_{k=1}^{m_n} \binom{n}{k} \frac{1}{\sqrt{k}}
+ \frac{1}{2^n} \sum_{k=m_n + 1}^{n} \binom{n}{k} \frac{1}{\sqrt{k}} \\
&\leq \frac{1}{2^n} \sum_{k=1}^{m_n} n^k
+ \frac{1}{2^n} \sum_{k=m_n + 1}^{n} \binom{n}{k} \frac{1}{\sqrt{m_n}} \tag{1} \\
&\leq \frac{n^{1+m_n}}{2^n}
+ \frac{1}{\sqrt{m_n}}. \tag{2}
\end{align*}
For $\text{(1)}$ I utilized the fact that $\binom{n}{k} \leq n^k$ and $\frac{1}{\sqrt{k}}$ is decreasing. For $\text{(2)}$ I utilized the geometric sum formula and the identity $\sum_{k=0}^{n} \binom{n}{k} = 2^n$.
Now taking $n\to\infty$ and applying the squeezing lemma proves the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 5
} |
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
| We can minimize $x^2 + y^2 + z^2$ subject to the constraint $x+y+z = 1$ using Lagrange multipliers, we then find that $x = y = z = \frac{1}{3}$, therefore $x^2 + y^2 + z^2\geq\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 15,
"answer_id": 5
} |
Using eigenvectors to find the general solution from a system of equations \begin{bmatrix}-13&40&-48\\-8&23&-24\\0&0&3\end{bmatrix}
Consider the matrix above. This corresponds to
\begin{align*}
x_{1}'&=-13x_{1}+40x_{2}-48x_{3}\\
x_{2}'&=-8x_{1}+23x_{2}-24x_{3}\\
x_{3}'&=3x_{3}
\end{align*}
What I want to do is use eigenvectors to find the general solution. First I computed $\det(A-\lambda I)=0$. From this I got my eigenvalues to be $\lambda = 7$ and $\lambda = 3$ (this one is multiplicity 2).
Next I would find $(A-\lambda I)x=0$ where $x$ is the eigenvector I am looking for. For $\lambda=7$, my eigenvector came out to be,
\begin{bmatrix} 2\\1\\0 \end{bmatrix}
So no problems here. On the other hand, I had some difficulties for when $\lambda = 3$. The matrix I would be solving for is
\begin{bmatrix} -16&40&-48&0 \\ -8&20&-24&0 \\ 0&0&0&0 \end{bmatrix}
When I do this I obtain,
\begin{bmatrix} 1&-2.5&3&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix}
so $$x_{1} -2.5x_{2}+3x_{3} = 0.$$
From here I am not sure how to get the two eigenvectors. The computer told me they are,
\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} and \begin{bmatrix} 5 \\ 2 \\ 0 \end{bmatrix}
but I have no idea how to get these from what I have remaining. Also how would I write the general solution?
| Solving the System of linear Equations
So it seems you want to solve the system of equations
$$
Ax =
\begin{pmatrix}-13&40&-48\\-8&23&-24\\0&0&3\end{pmatrix}
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}
=
\begin{pmatrix} x'_1 \\ x'_2 \\ x'_3 \end{pmatrix}
= x'
$$
Where the $x'$ is given and the $x_j$ are looked for.
You would normally solve this by calculating the inverse of $A$, since then
$$
x = A^{-1} x'
$$
But let's say you really want to solve this by using the eigenvalues and eigenvectors. You have already determined both of these, providing the change of base matrix $S$ which contains the eigenvectors
$$
S =
\begin{pmatrix}
2 & 5 & -3 \\
1 & 2 & 0 \\
0 & 0 & 1
\end{pmatrix}
$$
which accomplishes
$$
S^{-1} A S = D =
\begin{pmatrix}
7 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix}
$$
We can now solve for $x$. Using the above,
\begin{align*}
x &= A^{-1} x'
\\&= (SDS^{-1})^{-1} x'
\\&= SD^{-1}S^{-1} x'
\\&=
\begin{pmatrix}
2 & 5 & -3 \\
1 & 2 & 0 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
\frac{1}{7} & 0 & 0 \\
0 & \frac{1}{3} & 0 \\
0 & 0 & \frac{1}{7}
\end{pmatrix}
S^{-1} x'
\end{align*}
As you can see, we do not yet have the solution, since we don't know what $S^{-1}$ looks like. So what's left to do is: calculate the inverse of $S$, multiply everything and be done.
(So whatever method you use, you have to calculate the inverse of some matrix)
Getting the Eigenvectors
To determine the last two eigenvectors, you got the equation
$$
x_1 - 2.5 x_2 + 3 x_3 = 0
$$
whose solutions $(x_1,x_2,x_3)$ you want to find. What the equation tells us is, that if we are given two of the variables, the third one (for example $x_1$) will be determined.
$$
x_1 = 2.5 x_2 - 3 x_3
$$
Thus the solutions can be written as
$$
( 2.5 x_2 - 3 x_3 , x_2 , x_3 )
\equiv
\begin{pmatrix}
2.5 x_2 - 3 x_3 \\
x_2 \\
x_3
\end{pmatrix}
=
x_2
\begin{pmatrix}
2.5 \\
1 \\
0
\end{pmatrix}
+
x_3
\begin{pmatrix}
- 3 \\
0 \\
1
\end{pmatrix}
~~~~~~~ \text{with} ~~ x_2 , x_3 \in \mathbb{R}
$$
Any solution of this form is an eigenvector. Since you need two eigenvectors, you take two different solutions. For example the solutions with
$$
(x_2,x_3) = (2,0) ~~~\text{or}~~~ (0,1)
$$
Which yield the eigenvectors that the computer gave you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What are the solutions to: $x(x-3)=x^2 -4$? What are the solutions to: $x(x-3)=x^2 -4$ ?
Subtracting $x^2$ from both side results in the loss of the solution $(x=\infty)$.
My book says the solutions are $(x=\frac{4}{3},\infty)$
But is $(x=-\infty)$ another solution?
Is the answer $(x=\frac{4}{3},\pm\infty)$ or $(x=\frac{4}{3},\infty)$ ?
| $$\begin{align}
x(x-3)&=x^2 -4 \\ \\
\iff x^2-3x &= x^2 - 4\\ \\\iff 3x-4&=0\\ \\ \iff 3x&=4 \\ \\ \iff x &= \frac 43
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
How many zeros of $z^{17}+z-1$ lie in the region $|z|<1$? General problem:
Conjecture (B. Galo and Jack D'Aurizio) Let $p$ be a prime number then the number of zeros of $z^{p}+z-1$ that lie in the region $|z|<1$ is:
1) $⌊\frac{p}{3}⌋-2$ (where $⌊*⌋$ is the lower integer function). If $ p≡5 (mod6)$ but $p\neq 5$.
2) $⌊\frac{p}{3}⌋$. If this result is odd and $p\not\equiv 5 (mod6)$
3) $⌊\frac{p}{3}⌋+1$. If $⌊\frac{p}{3}⌋$ is even number.
I have to thanks Jack D'Aurizo for the "review" of my attempt and for sketching the proof.
| Visualiztion
Start with
$$
f(z) = z^{17}+z-1
$$
The Re and Im components are shown next.
The $0$ contours, the line where
$$
\color{blue}{\text{Re } f= 0}, \qquad \color{red}{\text{Im } f= 0}
$$
are shown below against the unit circle. The roots where $f(z)=0$ are where the blue and red contours intersect. We see six roots within the unit disk.
Observation
While using the argument principle, a pattern emerges.
Given
$$
g(k,z) = z^{k} - z + 1
$$
compute the series expansion for
$$
\frac{g'(z)}{g(z)}
$$
The expansion coefficients are
$$
\begin{array}{lrrrrrrrrrrrrrrrrrrrr}
k & const & w^1 & w^2 & w^3 & w^4 & w^5 & w^6 & w^7 & w^8 & w^9 & w^{10}\\\hline
2 & -1 & 1 & 2 & \dots \\
3 & -1 & -1 & 2 & 3 & 4 & \dots \\
4 & -1 & -1 & -1 & 3 & 4 & 5 & 6 & \dots \\
5 & -1 & -1 & -1 & -1 & 4 & 5 & 6 & 7 & 8 & \dots \\
6 & -1 & -1 & -1 & -1 & -1 & 5 & 6 & 7 & 8 & \dots \\
7 & -1 & -1 & -1 & -1 & -1 & -1 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \dots \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Solve the differential equation IVP
Solve $\displaystyle x \frac{dy}{dx} = 2y + x^2, y(1) = 2$
We can use the homogeneous form, $y = vx$, so $dy/dx = xdv/dx + v$
$x^2dv/dx + vx = 2x(v + x) \implies x^2 dv/dx = vx + x \implies x dv/dx = v + 1 \implies (v+1) dv = xdx \implies v^2/2 + v = x^2 + C$
Then we would just substitute $v = y/x$
But is there a simpler method?
| \begin{eqnarray}
xy'&=&2y+x^2\\
x^2y'-2xy&=&x^3\\
\dfrac{x^2y'-2xy}{x^4}&=&\dfrac{1}{x}\\
\left(\dfrac{y}{x^2}\right)'&=&\dfrac{1}{x}\\
\int\left(\dfrac{y}{x^2}\right)'&=&\int\dfrac{1}{x}\\
\dfrac{y}{x^2}&=&\ln x+C\\
y&=&x^2\ln x+2x^2
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Probability of drawing a spade on the first draw, a heart on the second draw, and an ace on the third draw My answer is
$(\frac{1}{52})(\frac{1}{51})(\frac{2}{50})$ + $2(\frac{1}{52})(\frac{12}{51})(\frac{3}{50})$ + $(\frac{12}{52})(\frac{12}{51})(\frac{4}{50})$
= $\frac{1}{204}$
But the answer in the textbook is
$(\frac{13}{52})(\frac{13}{51})(\frac{2}{50})$ + $2(\frac{13}{52})(\frac{13}{51})(\frac{3}{50})$ + $(\frac{13}{52})(\frac{13}{51})(\frac{4}{50})$
= $\frac{13}{850}$
Which answer is correct?
| $$P(A) = \frac{n(A)}{n(\Omega)}$$
Here, $$n(A) = n(\text{ace of spade}) \times n(\text{ace of heart}) \times n(\text{other aces}) + \\ n(\text{spade other than ace}) \times n(\text{ace of heart}) \times n(\text{other aces}) + \\ n(\text{ace of spade}) \times n(\text{ heart other than ace}) \times n(\text{other aces}) + \\ n(\text{spade other than ace}) \times n(\text{ heart other than ace}) \times n(\text{aces})$$
$$=1 \times 1 \times 2+ 2 \times 12 \times 1 \times 3 + 12 \times 12 \times 4$$
$$ = 2+72+576=650$$
$$n(\Omega)= 52 \times 51 \times 50 $$
So, $P(A) =1/204$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find $y'$ and $y''$ $\log(\sqrt{x^2 + y^2})=\arctan{\frac{y}{x}}$
Let $F(x,y)= \log(\sqrt{x^2 + y^2})-\arctan{\frac{y}{x}}$
I can't do it right, because I don't understand the formula $\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$
I think $z=y(x)$. So should I get something like that?
$$\frac{\partial F}{\partial x}=\frac{x}{x^2+y^2}$$ and
$$\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=\frac{y\cdot y'}{x^2+y^2}$$
| Frankly, I would just differentiate directly, using the chain rule, on both sides. And, before differentiating, reduce the left side: $\log(\sqrt{x^2+ y^2})= \frac{1}{2}\log(x^2+ y^2)$
The derivative of $\frac{1}{2}\log(x^2+ y^2)$ is
$$
\frac{1}{x^2+ y^2}(x+ yy')= \frac{x+ yy'}{x^2+ y^2}.
$$
The derivative of $\arctan(\frac{y}{x})$ is
$$
\frac{1}{1+\left(\frac{y}{x}\right)^2}
\left(-\frac{y}{x^2}+ \frac{1}{x}y'\right)= \frac{xy'- y}{y^2y'}.
$$
Now solve for $y'$ in
$$
\frac{x+ yy'}{x^2+ y^2}= \frac{xy'- y}{y^2y'}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which is greater $x_1$ or $x_2$? $$x_1=\arccos\left(\frac{3}{5}\right)+\arccos\left(\frac{2\sqrt{2}}{3}\right)$$
$$x_2=\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{2\sqrt{2}}{3}\right)$$ We have to find which is greater among $x_1$ and $x_2$
If we add both we get $$x_1+x_2=\pi$$ If we use formulas we get
$$x_1=\arccos\left(\frac{6\sqrt{2}-4}{15}\right)$$ and
$$x_2=\arcsin\left(\frac{3+8\sqrt{2}}{15}\right)$$
but how to compare now?
| (I'm going to do this in degrees rather than radians as this is very intuitive about right triangles).
Obviously if $\theta = \arccos \frac ah$ then $\theta$ represents an angle of a right triangle with adjacent side $a$ and hypotenuse $h$ (and opposite side $o = \sqrt {h^2 - a^2}$). So $\arcsin \frac ah = 90 - \theta$, the other angle of the same triangle where the side $a$ is now opposite rather than adjacent side.
[These are the triangles with sides $3,4,5$ and with sides $1, 2\sqrt{2}, 3$.]
So if $x <:> \frac{\sqrt{2}}{2}$ then $\arccos x = 45 \pm \phi$ for some positive angle $\phi$ and $\arcsin x = 45 \mp \phi$. (Draw a picture. It is obvious.)
So $\arccos \frac 35 + \arccos \frac {2\sqrt{2}}3 = 45 - \phi + 45 + \theta$
And $\arcsin \frac 35 + \arcsin \frac {2\sqrt{2}}3 = 45 + \phi + 45 -\theta$
So to solve this problem is simple and matter of figuring out which is larger $\phi$ or $\theta$. I.E. which triangle has the steeper slope the the $3$, $4$, $5$ triangle, or the $1, 2\sqrt{2}, 3$ triangle. Then answer is obviously the $1, 2\sqrt{2}, 3$ triangle
and so $\theta > \phi$ and $\arccos \frac 35 + \arccos \frac {2\sqrt{2}}3 = 45 - \phi + 45 + \theta > \arcsin \frac 35 + \arcsin \frac {2\sqrt{2}}3 = 45 + \phi + 45 -\theta$
===== or even more straightforward =====
$\sin x = \sqrt {1 - \cos^2 x} =y$
$\arcsin y = \arccos (\sqrt{1 - y^2})$
So $x_1 = \arccos \frac 35 + \arccos \frac{2\sqrt2}{3}= \arcsin \sqrt{ 1- \frac 35^2} + \arcsin \sqrt{1- \frac {2\sqrt{2}}3^2} = \arcsin \frac 45 + \arcsin \frac 13$
$x_2 - x_1 = (\arcsin \frac {2\sqrt 2}3 - \arcsin \frac 13)+(\arcsin \frac {3}{5} - \arcsin \frac 45)$
Now as $0 < 1/3 < 3/5 < 4/5 < \frac {2\sqrt 2} 2 < 1$ so $0 < \arcsin 1/3 <\arcsin 3/5 < \arcsin4/5 <\arcsin \frac {2\sqrt 2} 2 < 1$
So $x_2 - x_1 = (\arcsin \frac {2\sqrt 2}3 - \arcsin \frac 13)+(\arcsin \frac {3}{5} - \arcsin \frac 45) > 0$.
So $x_2 > x_1$.
| {
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"url": "https://math.stackexchange.com/questions/2259157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Finding a sum of $1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$ I need someone to find a mistake in my soliution or maybe to solf it much more easily... I have got a sum $$1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$$ and need to evaluate it. So here's my soliution:
$$S(x)=1+\frac{x^{4}}{4\cdot2^{4}}+\frac{x^{7}}{7\cdot2^{7}}+\frac{x^{10}}{10\cdot2^{10}}+\cdots=1+\sum_{n=1}^\infty \frac{x^{3n+1}}{(3n+1)\cdot2^{3n+1}}=1+S_1(x)$$
$$(S_1(x))_x'=\left(\sum_{n=1}^\infty \frac{x^{3n+1}}{(3n+1)\cdot2^{3n+1}}\right)_x'=\sum_{n=1}^\infty \frac{x^{3n}}{2^{3n+1}}=\frac{1}{x}\sum_{n=1}^\infty \left(\frac{x}{2}\right)^{3n+1}$$
Now let's take $\frac{x}{2}=y$, then
$$S_2(y)=\sum_{n=1}^\infty y^{3n+1}=y^4+y^7+y^{10}+\cdots=\frac{y^4}{1+y^3},|y|\le1$$
$$\left(S_1(y)\right)'=\frac{1}{2y}\cdot\frac{y^4}{1-y^3}=\frac{1}{2}\cdot\frac{y^3}{1-y^3}$$
$$S_1(y)=\frac{1}{2}\int\frac{y^3}{1-y^3}dy=\frac{1}{2}\int\left(-1+\frac{1}{1-y^3}\right)dy=-\frac{1}{2}y+\frac{\sqrt{3}}{6}\arctan\left(\frac{2\left(y+\frac{1}{2}\right)}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(y+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\lvert y-1\rvert+C$$
$$S_1(x)=-\frac{1}{4}x+\frac{\sqrt{3}}{6}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(\frac{x}{2}+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\left|\frac{x}{2}-1\right|+C$$
$$S_1(0)=0, C=-\frac{\sqrt{3}\pi}{36}$$
$$S(x)=-\frac{1}{4}x+\frac{\sqrt{3}}{6}\arctan\left(\frac{x+1}{\sqrt{3}}\right)+\frac{1}{12}\ln\left|\left(\frac{x}{2}+\frac{1}{2}\right)^2+\frac{3}{4}\right|-\frac{1}{6}\ln\left|\frac{x}{2}-1\right|-\frac{\sqrt{3}\pi}{36}+1$$
$$S(1)=\frac{3}{4}+\frac{\sqrt{3}}{3}\arctan\left(\frac{2\sqrt{3}}{3}\right)+\frac{1}{6}\ln(7)-\frac{\sqrt{3}\pi}{36}$$
By writing this for about 2 hours I deserve extra 50 points or at least good answers... Ha ha, thanks!
| It's worth cleaning things up a bit:
$$1+{1\over4\cdot2^4}+{1\over7\cdot2^7}+{1\over10\cdot2^{10}}+\cdots={1\over2}+{1\over2}\left(1+{1\over4\cdot2^3}+{1\over7\cdot2^6}+{1\over10\cdot2^9}+\cdots \right)\\={1\over2}+{1\over2}f(1)$$
where
$$f(x)=x+{x^4\over4\cdot2^3}+{x^7\over7\cdot2^6}+{x^{10}\over10\cdot2^9}+\cdots$$
Now
$$f'(x)=1+{x^3\over2^3}+{x^6\over2^6}+{x^9\over2^9}+\cdots={1\over1-\left(x\over2\right)^3}={8\over8-x^3}$$
Since $f(0)=0$, we have
$$1+{1\over4\cdot2^4}+{1\over7\cdot2^7}+{1\over10\cdot2^{10}}+\cdots={1\over2}+{1\over2}\int_0^1f'(x)dx={1\over2}+\int_0^1{4\over8-x^3}dx$$
Patient partial fractions gives
$$\int_0^1{4\over8-x^3}dx={1\over\sqrt3}\left(\arctan\left(2\over\sqrt3\right)-{\pi\over6}\right)+{1\over6}\ln7$$
and thus
$$1+{1\over4\cdot2^4}+{1\over7\cdot2^7}+{1\over10\cdot2^{10}}+\cdots={1\over2}+{\sqrt3\over3}\arctan\left(2\sqrt3\over3\right)+{1\over6}\ln7-{\pi\sqrt3\over18}\\
\approx1.016849$$
Remark: The OP's result has $3/4$ instead of $1/2$ and $\pi\sqrt3/36$ instead of $\pi\sqrt3/18$. Numerically it gives $1.417999...$, which is clearly too large.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Finding a function from a Maclaurin expansion Given the following Maclaurin expansion:
$$1+\frac{x}{2!}+\frac{x^2}{4!} + \frac{x^3}{6!} + \ldots + \frac{x^r}{(2r)!}+ \ldots$$
How would you go about finding the function? All I can see to do is use forms of trial and error.
| So you're looking for a closed form for the series:
$$\sum_{r=0}^{\infty} \frac{x^r}{(2r)!} = 1+\frac{x}{2!}+\frac{x^2}{4!} + \frac{x^3}{6!} + \ldots $$
You probably know that:
$$e^x = \sum_{r=0}^{\infty} \frac{x^r}{r!} = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots $$
And evaluating this at $\sqrt{x}$ makes the powers agree with the factorials:
$$e^\sqrt{x} = \sum_{r=0}^{\infty} \frac{\sqrt{x}^r}{r!} = 1\color{red}{+\sqrt{x}}+\frac{x}{2!} \color{red}{+ \frac{x\sqrt{x}}{3!}} + \ldots \tag{1}$$
You don't want the odd powers (in red), so add the following series to $(1)$:
$$e^{-\sqrt{x}} = \sum_{r=0}^{\infty} \frac{\left(-\sqrt{x}\right)^r}{r!} = 1\color{red}{-\sqrt{x}}+\frac{x}{2!} \color{red}{- \frac{x\sqrt{x}}{3!}} + \ldots$$
And divide by $2$ to get:
$$\frac{e^{\sqrt{x}}+e^{-\sqrt{x}}}{2}=1+\frac{x}{2!}+\frac{x^2}{4!} + \frac{x^3}{6!} + \ldots $$
Notice that this is also $\cosh(\sqrt{x})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the number of common, odd positive divisors of 27, 900 & 20, 700. Question as in title,
Find the number of common, odd positive divisors of $27, 900$ & $20, 700$.
explaining the steps in your method would be much appreciated.
I prime factorized and got that the common factor is $2^2\cdot3^2\cdot5^2$, then my notes from the relevant lecture simply go on to a solution implied from that step that the number of common odd positive divisors = $(2+1)(2+1)=9$. How the one leads to the other is where I've fallen down.
So how can you tell the number of common, odd positive divisors of $2^2\cdot3^2\cdot5^2$?
| Factorize $27{,}900$ and $20{,}700$ :
$$27{,}900=2^2 \times 3^2 \times 5^2 \times 31 ~;~ 20{,}700=2^2 \times 3^2 \times 5^2 \times 23$$
For their common odd divisors ; you need total number of divisors of $3^2 \times 5^2$.
For counting number of divisors of $3^2 \times 5^2$, you can have $0 , 1$ or $2$ power of $3$ from $3^2$. Hence three ways to chose exponent of $3$ in the common divisor. Similarly for exponent of $5$.
Therefore we have : $3 \times 3 = 9$ total number of common, odd positive divisors of $27{,}900$ & $20{,}700$.
In general, for exponent $\alpha$ of prime $p$ in a number, you'll have $\alpha+1$ ways to select an exponent of $p$ in the divisor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\lim_{n \to \infty} x_n$ Let $x$ be a positive real number. A sequence $x_n$ of real numbers is defined as follows:
$$x_1=\frac 12\left(x+\frac5x\right)$$ and $$x_{n+1}=\frac 12\left(x_n+\frac 5{x_n}\right) \quad\forall n\geq 1$$
The question is to show that $$\frac{x_n-\sqrt 5}{x_n+\sqrt 5}=\left(\frac{x-\sqrt 5}{x+\sqrt 5}\right)^{2^{n}}$$ and hence or otherwise evaluating $$\lim_{n \to \infty} x_n$$
To show first part I tried using induction. It is clear that $n=1$ is true. Let it be true for $n=k$. So $$\frac{x_n-\sqrt 5}{x_n+\sqrt 5}\left(\frac{x-\sqrt 5}{x+\sqrt 5}\right)^2=\left(\frac{x-\sqrt 5}{x+\sqrt 5}\right)^{2^{n+1}}$$
I tried manipulating this but could not proceed. Any ideas? Thanks.
| The limit, if it exists, should be a fixed point of $f(x) = \frac{1}{2} \left( x + \frac{5}{x} \right)$. Rearranging $x = \frac{1}{2} (x + \frac{5}{x})$ gives $x^2 - 5 = 0$ or $x = \pm \sqrt{5}$, and a negative limit is impossible as the series is manifestly positive. (And it's easy to show that the limit exists: note that if $x < \sqrt{5}$ then $x < f(x) < \sqrt{5}$, and vice versa if $x > \sqrt{5}$, so the sequence is monotonic and bounded and therefore has a limit.)
For the induction step of the proof of the closed form, you just have to prove that $$\frac{x_{n+1} - \sqrt{5}}{x_{n+1} + \sqrt{5}} = \left( \frac{x_n - \sqrt{5}}{x_n + \sqrt{5}} \right)^2.$$ Manipulating the RHS of this, using the fact that $2 x_n x_{n+1} = x_n^2 + 5$ (a rearrangement of the recursive relation that defines the sequence):
$$\left( \frac{x_n - \sqrt{5}}{x_n + \sqrt{5}} \right)^2 = \frac{x_n^2 - 2\sqrt{5} x_n + 5}{x_n^2 + 2\sqrt{5} x_n + 5} = \frac{2 x_n x_{n+1} - 2\sqrt{5}{x_n}}{2 x_n x_{n+1} + 2\sqrt{5} x_n}$$
which gives the desired result after cancelling $2 x_n$ from numerator and denominator.
Now to show the limit using the closed form: if $x$ is positive, then $$\left| \frac{x - \sqrt{5}}{x + \sqrt{5}}\right| < 1 $$ and so
$$0 = \lim_{n \to \infty} \left( \frac{x - \sqrt{5}}{x + \sqrt{5}} \right)^{2^n} = \lim_{n \to \infty} \frac{x_n - \sqrt{5}}{x_n + \sqrt{5}} = \frac{ \lim_{n \to \infty} x_n - \sqrt{5}}{\lim_{n \to \infty} x_n + \sqrt{5}}$$
which gives you $\lim_{n \to \infty} x_n = \sqrt{5}$.
| {
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"url": "https://math.stackexchange.com/questions/2265615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Infinite Sum Calculation: $\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$ Problem
Show the following equivalence: $$\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$$
Good afternoon, dear StackExchange community. I'm studying real analysis (the topic right now is exchanging limits) and I can't wrap my head around this problem. It might be a duplicate, but I couldn't find a thread that helped me to solve this problem.
Also, I know that both terms equals $\frac{\pi^2}{8}$ (courtesy of Wolframalpha), but I think I should show the original problems via exchanging limits of double series or with analyising the summands, because we didn't introduce the identity $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi}{6}$ yet.
My Attempts
If we consider the first summands of both sums, we have:
\begin{array}{|c|c|c|c|}
\hline
k & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\frac{1}{(2k+1)^2} & 1 & \frac{1}{9} & \frac{1}{25} & \frac{1}{49} & \frac{1}{81} & \frac{1}{121} & \frac{1}{169} \\ \hline
n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\frac{1}{n^2} & / & 1 & \frac{1}{4} & \frac{1}{9} & \frac{1}{16} & \frac{1}{25} & \frac{1}{36} \\ \hline
\end{array}
We immediately see that in $\frac{1}{n^2}$ is every summand of $\frac{1}{(2k+1)^2}$ is included. Additionally, we see that the extra summands in $\frac{1}{n^2}$ have the form $\frac{1}{(2x)^2}$.
Therefore, we could write,
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}$$
and the original equation would be
$$\Rightarrow 1 + \sum_{k=1}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \left( 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}\right)$$.
But I don't know how this could help me.
Anyways, series and infinite sums give me headaches and I would appreciate any help or hints. Thank you in advance.
Wow! Thank you for the really quick replies!
| You are almost done,:
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}= \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$
Notice that :
$$\sum_{k=1}^{\infty} \frac{1}{(2k)^2}=\sum_{k=1}^{\infty} \frac{1}{4k^2}=\frac14\sum_{k=1}^{\infty} \frac{1}{k^2}$$
Finally :
$$\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Differentiation - Power rule Find derivative of this function
$$g(t)=-3t(6t^4-1)^{4/3}$$
I have tried it till the answer:
$ -3t.\frac{4}{3} (6t^4 - 1)^{\frac{4}{3}-1} .\frac{d}{dt}(6t^4 -1) $
$-3t x \frac{4}{3} (6t^4-1)^\frac{1}{3} ( 6X4 t^3 -0 ) $
$ -4t (6t^4 -1)^\frac{4}{3} (24t^3) $
$-96t(6t^4-1)^\frac{1}{3} $
However , I checked and saw that the answer is
$$-3(6t^4-1)^{1/3}(38t^4-1)$$
I just started learning differentiation, and I don't understand my mistake on why I can't achieve the answer. Thanks!
| after the product and the chain rule we get
$$-3(6t^4-1)^{4/3}-3t\frac{4}{3}(6t^4-1)^{1/3}\cdot 24t^3$$
after simplifying this derivative we get
$$-3 \sqrt[3]{6 t^4-1} \left(38 t^4-1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Higher order interpolation Let we have some points $(x_i,y_i),~i=1,2,...n,$ that are scattered on a $2D$ domain like square. And $u_h$ is the solution of a numerical method at these points. Now, if I want to construct a higher order interpolation of $u_h$ at these points, how can I do that? Should I find $u_h^{new}$ with some more points and then use the value of $u_h^{new}$ at these points? Is there any higher order interpolation that can construct with the same points in MATLAB?
I could not use interp2.
| The griddata function is finding a surface. Here is a way to see how the fit works.
Problem statement
The data is a sequence of $m$ points of the form $$\left\{ x_{k}, y_{k}, z_{k}
\right\}_{k=1}^{m}$$
The model surface is an approximation of order $d$. Consider $d=2$:
$$
z(x,y) = a_{00} + a_{10}x + a_{01}y + a_{20}x^{2} + a_{11}xy + a_{02}y^{2}
$$
Linear system
$$
%
\begin{align}
%
\mathbf{A} a &= z \\
%
\left[ \begin{array}{cccccc}
1 & x_{1} & y_{1} & x_{1}^{2} & x_{1} y_{1} & y_{1}^{2} \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1 & x_{m} & y_{m} & x_{m}^{2} & x_{m} y_{m} & y_{m}^{2} \\
\end{array} \right]
%
\left[ \begin{array}{c}
a_{00} \\ a_{10} \\ a_{01} \\ a_{20} \\ a_{11} \\ a_{02}
\end{array} \right]
%
&=
\left[ \begin{array}{c}
z_{1} \\ \vdots \\ z_{m}
\end{array} \right]
%%
\end{align}
%
$$
Note: the matrix $\mathbf{A}$ is a Vandermonde matrix, the poster child for ill-conditioned matrices.
Normal equations
Forming the normal equations squares the condition number of the system. The solution here is for theoretical insight. If numerical problems manifest, investigate other methods like SVD or QR.
As $\mathbf{A}^{*}\mathbf{A}$ is symmetric, only the upper half is shown.
$$
%
\begin{align}
%
\mathbf{A}^{*}\mathbf{A} a &= \mathbf{A}^{*}z \\
%
\left[ \begin{array}{llllll}
m & \sum x & \sum y & \sum x^{2} & \sum xy & \sum y^{2} \\
& \sum x^{2} & \sum yx & \sum x^{3} & \sum x^{2}y & \sum xy^{2} \\
& & \sum y^{2} & \sum x^{2}y & \sum xy^{2} & \sum y^{3} \\
& & & \sum x^{3} & \sum x^{4}y & \sum x^{2}y^{2} \\
& & & & \sum x^{2}y^{2} & \sum xy^{3} \\
& & & & & \sum y^{4} \\
\end{array} \right]
%
\left[ \begin{array}{c}
a_{00} \\ a_{10} \\ a_{01} \\ a_{20} \\ a_{11} \\ a_{02}
\end{array} \right]
%
&=
\left[ \begin{array}{l}
\sum z \\ \sum xz \\ \sum yz \\ \sum x^{2}z \\ \sum xyz \\ \sum y^{2}z
\end{array} \right]
%%
\end{align}
%
$$
Normal equations solution
$$
a_{LS} = \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*}z
$$
Increase degree of fit
Increase $d$ from $2$ to $3$:
$$
z(x,y) = a_{00} + a_{10}x + a_{01}y + a_{20}x^{2} + a_{11}xy + a_{02}y^{2}
+ a_{30}x^{3} + a_{21}x^{2}y + a_{12}xy^{2} + a_{03} y^{3}
$$
The new system matrix is has new columns
$$
\mathbf{A} =
\left[ \begin{array}{ccccccc|ccc}
1 & x_{1} & y_{1} & x_{1}^{2} & x_{1} y_{1} & y_{1}^{2} & x_{1}^{3} & x_{1}^{2} y_{1} & x_{1}y_{1}^{2} & y_{1}^{3} \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
1 & x_{m} & y_{m} & x_{m}^{2} & x_{m} y_{m} & y_{m}^{2} & x_{m}^{3} & x_{m}^{2} y_{m} & x_{m}y_{m}^{2} & y_{m}^{3} \\
\end{array} \right]
$$
Algorithmically, increasing the fit order is a simple matter. Again, the numerics may present trouble.
Also, avoid the indeterminate form $0^{0}$. Force these terms to unity.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$ In MAO 1991,
Find $2x+5$ if $x$ satisfies $\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$
My attempt,
I squared the the equation then I got $144x^2+1648x+4480=144x^2-1632x+4624$, which results $x=-9$, and $2x+5=-13$.
I want to ask is there another way to solve this question as my method is very tedious. Thanks in advance.
| Substitution $3.5-x=t$ seems to help because $-x$ and $7-x$ are symmetric around the point $3.5$ you get
$$\sqrt{9t+8.5}=\sqrt{t-3.5}+2\sqrt{t+3.5}\\9t+8.5=t-3.5+4t+14+4\sqrt{t^2-3.5^2}\\4t-2=4\sqrt{t^2-3.5^2}\\(2t-1)^2=4(t^2-3.5^2)\\4t^2-4t+1=4t^2-4\cdot 3.5^2\\4t=4\cdot 3.5^2+1\\4t=(2\cdot 3.5)^2+1\\4t=50\\t=12.5\\x=-9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For integer m greather than 2, $\frac{1}{m} + \frac{1}{m+2}$, the numerators and denomitors are primitive pythagorean triples $a$ and $b$ For $m = 2$, the fraction is $\frac{3}{4}$. for $m=3$, the fraction is$\frac{8}{15}$. I was wondering why numerators and demoninators of $\frac{1}{m} + \frac{1}{m+2}$ show primitive pythagorean triples a and b.
| Notice that $$\frac{1}{m}+\frac{1}{m+2}=\frac{2m+2}{m^2+2m}.$$ We can check that $$(m^2+2m)^2+(2m+2)^2=m^4+4m^3+4m^2+4m^2+8m+4=m^4+4m^3+8m^2+8m+4=(m^2+2m+2)^2.$$ Note that this is always a primitive triple if you put the fraction in lowest form, but $2m+2$ and $m^2+2m$ are not coprime in general (and thus the triple $2m+2,m^2+2m, m^2+2m+2$ is not primitive in general).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all solutions to $|x^2-2|x||=1$ Firstly, we have that $$\left\{
\begin{array}{rcr}
|x| & = & x, \ \text{if} \ x\geq 0 \\
|x| & = & -x, \ \text{if} \ x<0 \\
\end{array}
\right.$$
So, this means that
$$\left\{
\begin{array}{rcr}
|x^2-2x| & = & 1, \ \text{if} \ x\geq 0 \\
|x^2+2x| & = & 1, \ \text{if} \ x<0 \\
\end{array}
\right.$$
For the first equation, we have
$$|x^2-2x|\Rightarrow\left\{\begin{array}{rcr}
x^2-2x & = & 1, \ \text{if} \ x^2\geq 2x \\
x^2-2x & = & -1, \ \text{if} \ x^2<2x \\
\end{array}
\right.$$
and for the second equation, we have
$$|x^2+2x|\Rightarrow\left\{\begin{array}{rcr}
x^2+2x & = & 1, \ \text{if} \ x^2+2x\geq 0 \\
x^2+2x & = & -1, \ \text{if} \ x^2+2x<0 \\
\end{array}
\right.$$
Solving for all of these equations, we get
$$\left\{\begin{array}{rcr}
x^2-2x & = 1 \Rightarrow& x_1=1+\sqrt{2} \ \ \text{and} \ \ x_2=1-\sqrt{2}\\
x^2-2x & =-1 \Rightarrow& x_3=1 \ \ \text{and} \ \ x_4=1\\
x^2+2x & = 1 \Rightarrow& x_5=-1-\sqrt{2} \ \ \text{and} \ \ x_6=-1+\sqrt{2}\\
x^2+2x & =-1 \Rightarrow& x_7=-1 \ \ \text{and} \ \ x_8=-1
\end{array}
\right.$$
So we have the roots $$\begin{array}{lcl}
x_1 = & 1+\sqrt{2} \\
x_2 = & 1-\sqrt{2} \\
x_3 = & -1+\sqrt{2} \\
x_4 = & -1-\sqrt{2} \\
x_5 = & 1 \\
x_6 = & -1
\end{array}$$
But according to the book, the answer is
\begin{array}{lcl}
x_1 & = & 1+\sqrt{2} \\
x_4 & = & -1-\sqrt{2} \\
x_5 & = & 1 \\
x_6 & = & -1
\end{array}
What happened to $x_2$ and $x_3$? Any other way to solve this equation quicker?
| As G.H.lee already pointed out, you did casework $x\geq 0$ and $x<0$ to simplify the expression but you completely disregarded it later. A correct way would be something like this:
$$|x^2-2|x||=1\implies
\begin{cases}|x^2-2x| = 1,& x\geq 0\\
|x^2+2x| = 1,& x < 0 \end{cases} \implies
\begin{cases} x^2-2x = 1,& x\geq 0,\ x^2-2x\geq 0\\
x^2-2x = -1,& x\geq 0,\ x^2-2x<0\\
x^2+2x = 1,& x < 0,\ x^2+2x\geq 0\\
x^2+2x = -1,& x < 0,\ x^2+2x<0\end{cases}$$
after which you solve the way you did, but remove extra solutions.
One way to simplify this is to notice that function $f(x) = |x^2-2|x||-1$ is even, i.e. $f(-x) = f(x)$, meaning that $x_0$ is root of $f$ if and only if $-x_0$ is root of $f$. Thus, we can assume that $x\geq 0$ while solving the equation, and we can just add "$\pm$" later to get all solutions.
Our equation now simplifies to $|x^2 -2x| = 1$, i.e. $x^2-2x = \pm 1$ or $(x-1)^2 = 1\pm 1$. This gives us solutions $x = 1$ and $x = 1\pm \sqrt 2$ and after we remove the negative $1-\sqrt 2$, we get $x =1$ and $x = 1+\sqrt 2$. To get all solutions, just add "$\pm$".
Personally, I like to draw graphs. Again, you can notice that $|x^2-2|x||$ is even, so we can assume that $x\geq 0$ and reflect the graph with respect to $y$-axis later.
To graph $|x^2-2x|$ (for $x\geq 0$), you can graph parabola $x^2-2x$ first and then reflect anything below $x$-axis. Afterwards, reflect with respect to $y$-axis to get $|x^2-2|x||$:
| {
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"url": "https://math.stackexchange.com/questions/2280212",
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"source": "stackexchange",
"question_score": "4",
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Triangular series perfect square formula 8n+1 derivation In triangular series
$$1 $$
$$1+2 = 3$$
$$1+2+3 = 6$$
$$1+2+3+4 =10$$
$$\ldots$$
Triangular number in 8n+1 always form perfect square .
Like $8\cdot 1+1 = 9 , 8\cdot 3+1 = 25$ .
How this formula is derived ?
| Let$$\begin{align}
A\cdot \overbrace{\frac {n(n+1)}2}^{T_n}+1&=(Bn+C)^2\\
\frac A2 n^2+\frac A2n+1&=B^2n^2+2BCn+C^2\\
\end{align}$$
Equating coefficients gives $C=1, A=4B, A=2B^2$, solving for which gives $B=2, A=8$.
Hence
$$8T_n+1=(2n+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
Let $x^3+ax^2+bx+c=0$ are $\alpha, \beta, \gamma$. Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
My attempt,
As I know from the original equation,
$\alpha+\beta+\gamma=-a$
$\alpha\beta+\beta\gamma+\alpha\gamma=b$
$\alpha\beta\gamma=-c$
I've tried to expand $(\alpha+\beta+\gamma)^3$ which is equal to $\alpha^3+\beta^3+\gamma^3+3\alpha^2\beta+3\alpha\beta^2+3\alpha^2\gamma+6\alpha\beta\gamma+3\beta^2\gamma+3\gamma^2\alpha+3\gamma\beta$
Basically, I know I've to find what's the value of $\alpha^3+\beta^3+\gamma^3$, $\alpha^3\beta^3+\alpha^3\gamma^3+\beta^3\gamma^3$ and $\alpha^3\beta^3\gamma^3$. But I
m stuck at it. I would appreciate can someone explain and guide me to it. Thanks a lot.
By the way, I would appreciate if someone provides another tactics to solve this kind of routine question. Thanks a lot.
| If we replace $x$ by $x^{1/3}$, we obtain $x + ax^{2/3} + bx^{1/3} + c = 0$. This equation does have the required roots, but it's not a polynomial.
To get the good ol' cubic form, write it as $x + c = -(ax^{2/3} + bx^{1/3})$ and cube both sides so that$$(x + c)^3 = -(ax^{2/3} + bx^{1/3})^3 = -(a^3 x^2 + b^3 x + 3abx(ax^{2/3} + bx^{1/3})) = -(a^3 x^2 + b^3 x - 3abx(x+c))$$
This method wouldn't generate extraneous roots because we're not squaring anything here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Laurent series expansion of a given function
Let $\sum_{-\infty}^{\infty} a_n z^n$ be the Laurent series expansion of $f(z)=\frac{1}{2z^2-13z+15}$ in an annulus $\frac{3}{2}< \vert z \vert < 5$.
What is the value of $\frac{a_1}{a_2}$?
My attempt:
First note that $f$ is analytic in the given annulus.
$f(z)=\frac{1}{2z^2-13z+15}=\frac{1}{(2z-3)(z-5)}$ ,
$\quad$ $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ $=\frac{\frac{-2}{7}}{2z-3}+\frac{\frac{1}{7}}{z-5}$ , by Partial Fraction.
How to goes further to find out the required function in the power series form and find $\frac{a_1}{a_2}$?
I'm Confused with which term can be taken outside of the expression.
Any help?
|
The function
\begin{align*}
f(z)&=\frac{1}{2z^2-13z+15}\\
&=-\frac{1}{7}\cdot\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\cdot\frac{1}{z-5}\\
\end{align*}
has two simple poles at $\frac{3}{2}$ and $5$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $\frac{3}{2}$ and $5$ and see they determine three regions.
\begin{align*}
|z|<\frac{3}{2},\qquad\quad
\frac{3}{2}<|z|<5,\qquad\quad
5<|z|
\end{align*}
*
*The first region $ |z|<\frac{3}{2}$ is a disc with center $0$, radius $\frac{3}{2}$ and the pole $\frac{3}{2}$ at the boundary of the disc. In the interior of this disc all two fractions with poles $\frac{3}{2}$ and $5$ admit a representation as power series at $z=0$.
*The second region $\frac{3}{2}<|z|<5$ is the annulus with center $0$, inner radius $\frac{3}{2}$ and outer radius $5$. Here we have a representation of the fraction with poles $\frac{3}{2}$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $5$ admits a representation as power series.
*The third region $|z|>5$ containing all points outside the disc with center $0$ and radius $5$ 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(z)$ at $z=0$ for all three regions. Here we have to consider
*
*Region 2: $\frac{3}{2}<|z|<5$
Since we only need to calculate $\frac{a_1}{a_2}$ it is sufficient to expand the power series part only. We obtain
\begin{align*}
f(z)&=-\frac{1}{7}\cdot\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\cdot\frac{1}{z-5}\\
&=-\frac{1}{7}\cdot\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\sum_{n=0}^\infty \frac{1}{(-5)^{n+1}}(-z)^n\\
&=-\frac{1}{7}\cdot\frac{1}{z-\frac{3}{2}}-\frac{1}{7}\sum_{n=0}^\infty \frac{1}{5^{n+1}}z^n\\
\end{align*}
We conclude since $a_1=-\frac{1}{7}\cdot\frac{1}{5^2}$ and $a_2=-\frac{1}{7}\cdot\frac{1}{5^3}$
\begin{align*}
\color{blue}{\frac{a_1}{a_2}=5}
\end{align*}
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
3-cocycle in $\mathbb{Z}/n\mathbb{Z}$ It is well known that $3$-cocycles in $H^3(\mathbb{Z}/n\mathbb{Z};U(1))$ are of the form
$$w(a,b,c)=\exp(\frac{2\pi ik}{n^2}a(b+c-[b+c])),$$
for $k\in\mathbb{Z}_n$, $a,b,c\in\mathbb{Z}_n$ and $[b+c]=b+c\mod{n}$.
(See https://mathoverflow.net/questions/121065/explicit-3-cocycle-of-a-cyclic-group)
How do I show that this is indeed a $3$-cocycle?
According to me, the identity that must be shown is $(\delta^3w)(a,b,c,d)=0$.
Here I use $\delta^nf(x_1,...,x_{n+1})=f(x_2,...,x_{n+1})(\Pi_{i=1}^nf(x_1,...,x_ix_{i+1},...,x_{n+1})^{(-1)^i})f(x_1,...,x_n)^{(-1)^{n+1}}$.
Since we have a product of exponentials, we can only regard the sum of the terms between brackets. Thus we will only look at $w'(a,b,c)=a(b+c-[b+c])$ and we want $(\delta^3w')(a,b,c,d)=0\mod{n^2}$.
We then obtain
$$\begin{align}
(\delta^3w')(a,b,c,d)=&w'(b,c,d)w'(a+b,c,d)^{-1}w'(a,b+c,d)w'(a,b,c+d)^{-1}w'(a,b,c)\\
\end{align}$$
$$\begin{align}
=b&(c+d-[c+d])+\\
-(a+b)&(c+d-[c+d])+\\
a&(b+c+d-[b+c+d])+\\
-a&(b+c+d-[b+c+d])+\\
a&(b+c-[b+c])\\
=a&(b-d+[c+d]-[b+c]).
\end{align}$$
How is $a(b-d+[c+d]-[b+c])$ equal to $0\mod{n^2}$?
| The mistake in the original question is the following: $w'(a+b, c, d) = [a + b]\cdot (c + d - [c + d])$ instead of $(a + b)\cdot (c + d - [c + d])$. So, you should make each argument modulo $n$ before using the formula of the 3-cocycle. With this changes the formula is correct. Indeed:
\begin{multline}
w'(b, c, d) - w'(a+b, c, d) + w'(a, b+c, d) - w'(a, b, c+d) + w'(a, b, c) = \\ = b\cdot(c + d - [c+d]) - [a+b]\cdot(c+d-[c+d]) + a\cdot([b+c] + d - [b+c+d]) - \\ - a\cdot(b + [c+d] - [b+c+d]) + a(b+c - [b+c]) = \\ = (a+b)\cdot (c+d) + [a+b]\cdot [c+d] - (a + b)\cdot [c + d] - [a + b]\cdot (c + d)
\end{multline}
This is quite symmetric expression. Let $a+b = \delta_{12} + [a+b]$ and $c+d = \delta_{34} + [c + d]$. Here $\delta_{12}, \delta_{34}\in\{0, n\}$, because the value $a + b$ can be less than $n$ (and in this case $\delta_{12} = 0$) or greater than $n$ (and in this case $\delta_{12} = n$). The situation with $c + d$ is the similar. Finally:
\begin{multline}
\delta^3(w')(a, b, c, d) = \delta_{12}\cdot \delta_{34} + \delta_{12}\cdot[c+d]+[a+b]\cdot \delta_{34} + [a+b]\cdot[c+d] - \\ - \delta_{12}\cdot[c+d] - [a+b]\cdot[c+d] - \delta_{34}\cdot[a+b] - [c+d]\cdot[a+b] + \\ + [a+b]\cdot[c+d] = \delta_{12}\cdot\delta_{34} = 0 (mod\ n^2).
\end{multline}
| {
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How many solutions does the equation $x^2-y^2 = 3^8\cdot 5^6\cdot 13^4$ have? Given that $x$ and $y$ are positive integers. Problem: How many solutions does the equation $x^2-y^2 = 3^8\cdot 5^6\cdot 13^4$ have? Given that $x$ and $y$ are positive integers.
I tried a similar approach to the ones described here, but to no avail. I can't split up the RHS so that all the factors have the same base.
| Factor
$$ (x+y)(x-y) = 3^8 \cdot 5^6 \cdot 13^4 $$
The number of ways to split up $3^8$ in a product is 9. The number of ways to split up $5^6$ in a product is 7. The number of ways to split up $13^4$ in a product is 5. Collectively this means $3^8 \cdot 5^6 \cdot 13^4$ can be split into a product in $9\cdot7\cdot5=315$ ways.
Since $2$ is not a factor, both factors in the product are odd. The value of $x$, that makes $(x+y)(x-y)$ be equal to the desired value, is the average of the the two factors in the product. Since the average of two positive odd integers is a positive integer, $x$ is always a positive integer.
Let $a = x+y$ and $b = x-y$. The 315 ways to split up the product can be divided into three groups: $a>b$, $a<b$ and $a=b$.
*
*The group $a=b$: There is one value in this group, and that's the one where $a=b=3^4\cdot5^3\cdot13^2$. In this case we have $y=0$, so it is not a solution.
*Every element in $a<b$ has a corresponding element in $a>b$: (the one where each power is swapped). Since only the ones in $a>b$ have $y>0$, only half of those not in $a=b$ are valid.
This means there are $(315-1)/2 = 157$ solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find sum and convergence of series $\sum_{n = 1}^{\infty}\frac{1}{(2n-1)^{2}\cdot(2n+1)^{2}}$ I laid this shot that:
$$\frac{1}{(2n-1)^2(2n+1)^2}=1/4\, \left( 2\,n+1 \right) ^{-2}-1/4\, \left( 2\,n-1 \right) ^{-1}+1/
4\, \left( 2\,n-1 \right) ^{-2}+1/4\, \left( 2\,n+1 \right) ^{-1}
$$
But I don't know what to do next. Help me please
| By partial fraction decomposition
$$ \frac{1}{(2x-1)^2 (2x+1)^2} = \frac{\frac{1}{16}}{\left(x-\frac{1}{2}\right)^2}+\frac{\frac{1}{8}}{x-\frac{1}{2}}-\frac{\frac{1}{8}}{x+\frac{1}{2}}+\frac{\frac{1}{16}}{\left(x+\frac{1}{2}\right)^2} \tag{1}$$
hence it follows that:
$$ \sum_{n\geq 1}\frac{1}{(2n-1)^2 (2n+1)^2}=\frac{1}{4}+\frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n-1)^2}+\frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n+1)^2} = \color{red}{\frac{\pi^2-8}{16}}\tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\lim_{(x,y)\to(0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^6+y^6}$
$$\lim_{(x,y)\to(0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^6+y^6}$$
How should I approach this?
| I would suggest using polar coordinates:
\begin{align}
x&=r\cos(\theta)\\
y&=r\sin(\theta)\\
r^2&=x^2+y^2.
\end{align}
Since $x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)$, we could also compute:
$$
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{r^6(\cos^4(\theta)-\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta))}.
$$
Since $\cos^6(\theta)+\sin^6(\theta)=\cos^4(\theta)-\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta))$ is always positive (on the unit circle), it is, therefore, bounded and bounded away from zero. Hence, there exist constants $C_1$ and $C_2$ so that
$$
0<C_1\leq \cos^4(\theta)-\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta))\leq C_2.
$$
Then, by the squeeze theorem,
$$
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{C_2r^6}\leq
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{r^6(\cos^4(\theta)-\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta))}\leq
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{C_1r^6}
$$
Therefore, it make sense to look at
$$
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{r^6}
$$
This is an indeterminate form of the form $\frac{0}{0}$. Using the substitution $t=\frac{1}{r}$, this limit becomes
$$
\lim_{t\rightarrow \infty}\frac{t^6}{e^{t^2}}.
$$
Using l'Hopital's rule a few times, you get that the limit is $0$. Hence, by the squeeze theorem, the desired limit is $0$.
| {
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"source": "stackexchange",
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$ \lim_{x \to \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $ fine limit :
$$ \lim_{x \to \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $$
such that : $$ a \in (0,2)$$
and :
$[x]: \ \ $ floor function
My Try :
$$f(x):=( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]}$$
$$\ln f(x)=(x-[x])\ln(\sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )$$
Now ?please help
| The expression under limit operation say $F(x) $ is of the form $\{f(x) \} ^{g(x)} $ and clearly $f(x) >0$ for all $x>0$ and $0\leq g(x) <1$ for all $x>0$. It is now clear that $F(x) $ lies between $1$ and $f(x) $ and it takes the value $1$ when $x$ is a positive integer. Moreover if $x$ is near a positive integer and less than it then $F(x) $ is near $f(x) $. Hence it is clear that if $F(x) $ tends to a limit as $x\to\infty$ then it must be $1$ and moreover this will happen if and only if $f(x) \to 1$ as $x\to\infty$.
Now it is easy to analyze $f(x) $ which is of the form $p-q$. Multiplying it by $(p^{2}+pq+q^{2})$ and dividing it by the same quantity we see that $f(x) $ is expressed as a fraction with numerator $p^{3}-q^{3}=3x^{a}$. Further if divide the numerator and denominator of $f(x) $ by $x^{4/3}$ then the denominator tends to $3$ and the numerator is $3x^{a-4/3}$. It follows that $f(x) \to 1$ if and only if $a=4/3$. Hence the desired limit is equal to $1$ if $a=4/3$ and the limit does not exist if $a\neq 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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how many $3\times 3$ matrices with entries from $\{0,1,2\}$.
How many $3 × 3$ matrices $M$ with entries from $\left\{0, 1, 2\right\}$ are there for which taken from
the sum of the main diagonal of $M^TM$ is $5$.
Attempt: Let $M = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$. where $a,b,c,d,e,f,g,h,i\in \{0,1,2\}$
$$M^{T}M= \begin{pmatrix}
a & d & g\\
b & e & h\\
c & f & i
\end{pmatrix}\begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix} $$.
sum of diagonal entries $$a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2 = 5$$
How can I form different cases?
| Either $5=1^2+1^2+1^2+1^2+1^2+0^2+0^2+0^2+0^2$ or $5=1^2+2^2+0^2+0^2+0^2+0^2+0^2+0^2+0^2$
thus there are two cases: 5 of them are 1 and the other 4 are 0, or 1 is 1, 1 is 2, and the other 7 are 0.
Case 1: there are $\binom{9}{5} = 126$ possibilities.
Case 2: there are $\binom{9}{1}\binom{8}{1}\binom{7}{7} = 72$ possibilities.
In total there are $126+72=198$ possibilities.
| {
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"url": "https://math.stackexchange.com/questions/2291491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
| Let $f (x)=x^5+x+1$
$$f'(x)=5x^4+1>0$$
$f $ is stricly increasing at $\mathbb R $, thus it has only one real root $\alpha \in (-1,0) $.
the factorisation will be of the form
$$(x-\alpha)(x^2+ax+b)(x^2+cx+d) $$
with $a^2-4b <0$ and $c^2-4d <0$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
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Calculate $\sum_{n=1}^{\infty}{c_n\over n}$ when $c_1=2, c_2=1, c_3=-3, c_{n+3}=c_n, n\in \mathbb{N}$ Let $(c_n)$ be defined as $c_1=2, c_2=1, c_3=-3, c_{n+3}=c_n, n\in \mathbb{N}$. I want to get $\sum_{n=1}^{\infty}{c_n\over n}$. I would appreciate any suggestions how to start on this. It is my suspicion that I would have to resort to integration.
| I'll expand on the (very clever!) method by Jack, giving some more details that may not have been obvious from his answer.
$$\begin{align}\sum_{n=1}^\infty\frac{c_n}{n}&=\sum_{k=0}^\infty\left(\frac{2}{3k+1} + \frac{1}{3k+2} - \frac{3}{3k+3}\right) \\
&=\sum_{k=0}^\infty\left(2 \int_0^1 x^{3k}dx + \int_0^1 x^{3k+1}dx - 3 \int_0^1 x^{3k+2}dx\right) \\
&=\sum_{k=0}^\infty \int_0^1 x^{3k}(2+x-3x^2) dx \\
&=\int_0^1 \sum_{k=0}^\infty x^{3k}(2+x-3x^2) dx \\
&=\int_0^1 \frac{2+x-3x^2}{1-x^3} dx \\
&=\int_0^1 \frac{2+3x}{1+x+x^2} dx \\
&=\int_0^1 \frac{\frac{3}{2}(2x+1) + \frac{1}{2}}{1+x+x^2} dx \\
&=\frac{3}{2}\int_0^1 \frac{2x+1}{1+x+x^2} dx +\frac{1}{2}\int_0^1 \frac{1}{1+x+x^2} dx\\
&=\frac{3}{2}\left[\log(1+x+x^2)\right]_0^1 +\frac{1}{2}\int_0^1 \frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}} dx\\
&=\frac{3}{2} \log 3 +\frac{1}{2} \left[\frac{2}{\sqrt{3}}\arctan\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]_0^1 \\
&=\frac{3}{2} \log 3 +\frac{1}{\sqrt{3}}\left(\arctan \sqrt{3}-\arctan \frac{1}{\sqrt{3}}\right) \\
&=\frac{3}{2} \log 3 +\frac{\pi}{6\sqrt{3}} \\
\end{align}$$
Switching the integral and sum is justified by convergence of both involved quantities, and the geometric series is justified as $0<x<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculation related to second partial derivative Problem: let $u(x,t) = \frac{1}{2\sqrt{\pi t}} e^{-\frac{x^2}{4t}}$, prove when $t \gt 0$, we have $|\frac{\partial^2u}{\partial x^2}| \le Ct^{-3/2}e^{-\lambda\frac{x^2}{t}}$, where $C \gt 0$ and $\lambda \le \frac{1}{4}$.
My Attempt:
\begin{align}
\frac{\partial u}{\partial x} &= \frac{-x}{4\sqrt{\pi}t^{3/2}}e^{-\frac{x^2}{4t}} \\
\frac{\partial^2 u}{\partial x^2} &= (\frac{x^2}{2t}-1)\frac{1}{4\sqrt{\pi}t^{3/2}}e^{-\frac{x^2}{4t}}
\end{align}
How to get the inequality from here? Can someone give me a hint? Thanks!
| We start from the question simplification by equivalent transformations, for all real $x$ and positive $t$ and $C$.
$$\left|\frac{\partial^2u}{\partial x^2}\right| \le Ct^{-3/2}e^{-\lambda\frac{x^2}{t}}$$
$$\left|\left(\frac{x^2}{2t}-1\right)\frac{1}{4\sqrt{\pi}t^{3/2}}e^{-\frac{x^2}{4t}}\right|\le Ct^{-3/2}e^{-\lambda\frac{x^2}{t}}$$
$$\left|\frac{x^2}{2t}-1\right|\le 4\sqrt{\pi}Ce^{(1-4\lambda)\frac{x^2}{4t}}$$
After a substitution $y=\frac{x^2}{4t}$ the last inequality becomes
$$\left|2y -1\right|\le 4\sqrt{\pi}Ce^{(1-4\lambda)y}.$$
Now we see that for each $\lambda<\frac 14$, there is $C=C(\lambda)$ such that the last inequality holds for all $y\le 0$. Since $e^{(1-4\lambda)y}\ge 1+(1-4\lambda)y $, it suffices to put $C(\lambda)=\max\left(\frac 1{4\sqrt{\pi}},\frac 2{1-4\lambda}\right)$. Then, for all real $x$ and positive $t$
$$\left|\frac{\partial^2u}{\partial x^2}\right| \le C(\lambda)t^{-3/2}e^{-\lambda\frac{x^2}{t}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find all prime $a, b, c$ such that $ab+bc+ac > abc$ Find all primes $a, b, c$ such that $ab + bc + ac > abc$
| Hint:
The above inequation is equivalent to
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} > 1$$
where $a,b,c$ are primes
Complete Solution:
WLOG we can assume $a \geq b \geq c$
This means that $c<3$ ie. $c=2$
Now, we have $$\frac{1}{a}+\frac{1}{b} > \frac{1}{2}$$
Since $a \geq b$, we have $b<4$ ie $b \in \{2,3\}$
If $b=2$, then $a$ can be any prime number.
If $b=3$, then $a<6$, so $a \in \{3,5\}$.
So, the required solutions are $(p,2,2),(3,3,2)$ and $(5,3,2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the correct way of expressing the general solution of $\sin x=0,1,-1$ and $\cos x=0,-1,1$? The general solution of $\sin x = \sin y$ is $x = n\pi + (-1)^ny$. Then if $\sin x = 1$ or $\sin x = \sin(\frac{\pi}{2})$ or $x = n\pi+(-1)^n\frac{\pi}{2}$. Is this correct?
Also, the general solution of $\cos x = \cos y$ is $x = 2n\pi \pm y$. Then if $\cos x = 0$ or $x = 2n\pi \pm \frac{\pi}{2}$. Is this correct?
Also for $\cos x = 1,-1$, I am confused.
| The clearest way of writing the solutions are
$$
\sin{x} = \sin{y} \Leftrightarrow x = y + n2\pi \text{ or } x = \pi - y + n2\pi$
$$
$$
\cos{x} = \cos{y} \Leftrightarrow x = \pm y + n2\pi
$$
where $n$ is a whole number.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integer values of $(\frac{19}{7}-\sum_{k=0}^N \frac{1}{k!})^{-1}$ The approximation for $e$ involved in An integral for $2\pi+e-9$
exactly matches a truncation of factorial reciprocal series
$$\sum_{k=0}^5 \frac{1}{k!} =\frac{163}{60}$$
In contrast, the difference
$$\frac{19}{7}-\sum_{k=0}^N \frac{1}{k!}$$
never cancels but it is a unit fraction for $N=3,4,5,7$.
$$\frac{19}{7}-\sum_{k=0}^3 \frac{1}{k!}=\frac{1}{21}=\frac{2}{6·7}$$
$$\frac{19}{7}-\sum_{k=0}^4 \frac{1}{k!}=\frac{1}{168}=\frac{3}{7·8·9}$$
$$\frac{19}{7}-\sum_{k=0}^5 \frac{1}{k!}=-\frac{1}{420}=-\frac{4}{5·6·7·8}$$
$$\frac{19}{7}-\sum_{k=0}^7 \frac{1}{k!}=-\frac{1}{252}=-\frac{2}{7·8·9}$$
Is this list complete?
| For ease of notation I'll denoting $\sum_{k=0}^N \frac{1}{k!}$ as $e_N$ for now.
Note that $(19/7 - e)^{-1} \approx -250.24...$, or in particular it's somewhere in-between $-250$ and $-251$. So if we can show that for $N \ge 8$ the sum $(\frac{19}{7}-e_N)^{-1}$ is in between $-250$ and $-251$ then from that point it can't have any more integer values.
Since $-1/250 < 19/7 - e$ it is enough to show that for $N \ge 8$ we have that $19/7 - e_{N} < -1/251$, and since $e_{N}$ is montonically increasing we only need to show this for $N=8$.
Now we know that $\frac{19}{7} - e_{7} = -1/252$, and $$1/252 - 1/251 = 1/(251 \cdot 252) = 1 / 63252.$$ Since $8! = 40320 < (251 \cdot 252)$ we find that $$\frac{19}{7} - e_{8} = \frac{19}{7} - e_{7} - 1/8! = -1/252 - 1/8! < -1/252 - (1/252 - 1/251) = -1/251.$$
Therefore for all $N \ge 8$ $$-1/250 < \frac{19}{7} - e_{N} < -1/251$$
So you indeed listed all options.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $(a-1)^{p+1}+a^p=(a+1)^{p-1}$
Suppose that $a,p$ are nonnegative integers such that $p$ is prime and $p\nmid (a-1)$. If $(a-1)^{p+1}+a^p=(a+1)^{p-1}$, find the sum of all possible values of $a$.
We can't have $p = 2$ since the equation $$(a-1)^3+a^2 = (a+1)$$ has no integer solutions. We now take two cases:
Case 1: $p \mid (a+1)$
In this case, $a \equiv -1 \pmod{p}$. Then $(a-1)^{p+1} \equiv (-2)^{p+1} \equiv 4 \equiv -a^p \equiv -a \pmod{p}$. Thus $a \equiv 4 \pmod{p}$ and therefore $p \mid 3$, so that $p = 3$. Thus $a = 2$ in this case.
Case 2: $p \nmid (a+1)$
We have $$(a-1)^{p+1}+a^p \equiv (a-1)^2+a^p \equiv 1 \pmod{p}.$$ If $a \not \equiv 0 \pmod{p}$, then we have $(a-1)^2+a \equiv 1 \pmod{p}$, which gives $a(a-1) \equiv 0 \pmod{p}$. Thus $p \mid (a-1)$, a contradiction. Thus $a \equiv 0 \pmod{p}$.
How do I continue from here?
| Hint
$$1 \geq \frac{(a-1)^{p+1}}{(a+1)^{p-1}}=(a-1)^2 (1-\frac{2}{a+1})^{p-1}\geq (a-1)^2 (1-\frac{2(p-1)}{a+1})$$
with the last inequality following by Bernoulli.
Thus
$$a+1 \geq (a-1)^2(a+3-2p) \\
2 =a+1-(a-1)\geq a+1-(a-1)^2\geq (a-1)^2(a+2-2p)$$
This implies that either $(a+2-2p) \leq 0$ or $a-1=1$ and $(a+2-2p) \in \{ 1,2 \}$.
Combine this with $p|a$ to complete the proof.
Added
If $a=p \geq 3$ then
$$(p-1)^{p+1}+p^p=(p+1)^{p-1}$$
and hence
$$(p-1)^{p+1}+p^p=(p+1)^{p-1} \pmod{p^3} \\
\binom{p+1}{2}p^2(-1)^{p-1}+\binom{p+1}{1}p(-1)^p+(-1)^{p+1}=\binom{p-1}{2}p^2+\binom{p-1}{1}p+1\pmod{p^3} \\
\frac{(p+1)p}{2}p^2-(p+1)p+1= \frac{(p-1)(p-2)}{2}p^2+(p-1)p+1 \pmod{p^3} \\
-p^2-p+1= p^2+p^2-p+1 \pmod{p^3} \\
3p^2=0 \pmod{p^3} \\
$$
This shows $p=3$. The case $p=2$ is easy to study separately.
| {
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"url": "https://math.stackexchange.com/questions/2307251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integer solutions to $y=(2^x-1)/3$ where $x$ is odd For the equation $y=(2^x-1)/3$ there will be integer solutions for every even $x$.
Proof: When $x$ is even the equation can be written as $y=(4^z-1)/3$ where $z=x/2$.
$$4^z =1 + (4-1)\sum_{k=0}^{z-1} 4^k$$
If you expand that out you get:
$$4^z=1+(4-1)4^0+(4-1)4^1+(4-1)4^2+\dots+(4-1)4^{z-2}+(4-1)4^{z-1}$$
Which becomes:
$$4^z=1+4^1-4^0+4^2-4^1+4^3-4^2+\dots+4^{z-1}-4^{z-2}+4^z-4^{z-1}$$
After canceling everything out you are left with:
$$4^z=4^z$$
More generally:
$$a^z =1 + (a-1)\sum_{k=0}^{z-1} a^k$$
Therefore: $(2^x-1)/3$ will always be an integer when $x$ is even.
My question is: will there ever be an integer solution to $(2^x-1)/3$ when $x$ is odd?
| If $x$ is odd, then $x=2k+1$ and then
$$2^{2k+1}-1=4^k\cdot 2-1\equiv1^k\cdot 2-1\equiv1\pmod 3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find integral solutions - rational points of $4p=x^2+27y^2$ for $p>3$ prime Let $p\geqslant 5$ be a prime. I want to solve $$4p=x^{2}+27y^{2} \tag{1}\quad x,y\in \Bbb N$$
This comes when considering the discriminant of a cubic polynomial $T^3 -pT-yp$ with cyclic group.
Are there only elementary ways to find solutions? I only achieved a bit of it.
Firstly, since the right hand side of (1) must be even, so x and y have same parity.
If $p$ is prime, then $\mathrm{gcd}(p,6)=1\Rightarrow p=1,5\mod6$.
Using the hint.
Case $p=5+6m$. If $x,y$ are even, let $x=2x',y=2y'$ so $5+6m=x'^{2}+27y'^{2}$. Then modulo 3, we get $2\equiv x'^{2}\mod3$, impossible. Thus $x,y$ are both odd. Let $x=2x'+1$ and $y=2y'+1$ so $6m=(x'^{2}+x')+23+27(y'^{2}+y')$. Modulo 3, this gives $x'^{2}+x'\equiv1\mod3$, without solutions in $\mathbb{F}_{3}$. Therefor, the equation (1) has no solution when $p=5+6m$.
So case $p=1+6m$. The equation (1) becomes $4+24m=x^{2}+27y^{2}$.
If $x,y$ are even, let $x=2x'$ and $y=2y'$. Then $1+6m=x'^{2}+27y'^{2}$. Since $x'^{2}-1=6m-27y'^{2}$, then $x$ and $y$ can be even only when $m\geqslant5\Leftrightarrow p\geqslant31$.
It follow that when $m\leqslant4$ that $x$ and $y$ are necessarily odd. Let $x=2x'+1$ and $y=2y'+1$. Then $6(m-1)=x'^{2}+x'+27(y'^{2}+y')$. Since $6(m-1)\leqslant18$, then $y'=0\Rightarrow y=1$. This gives $x'^{2}+x'=6(m-1)$. For example, $m=3\Leftrightarrow p=19$, then $12=x'+x'^{2}$ so $x'=3$ and $x=7$.
| All primes $1 \pmod 3$ are integrally represented by $x^2 + xy + 7 y^2,$ shorthand $\langle 1,1,7 \rangle.$ Discriminant $-27.$
The primitive quadratic forms of discriminant $-108$ are
$$ \langle 1,0,27 \rangle, \; \; \langle 4,2,7 \rangle, \; \; \langle 4,-2,7 \rangle $$
Among primes $p \equiv 1 \pmod 3,$ we have $p = x^2 + 27 y^2$ whenever $2$ is a cubic residue $\pmod p.$
When $p \equiv 1 \pmod 3$ but $2$ is not a cube, we have $p = 4 u^2 + 2uv + 7 v^2.$ Then $$4p = 16 u^2 + 8uv + 28 v^2 = (4u+v)^2 + 27 v^2$$
Ireland and Rosen do this by cubic reciprocity. However, there is a short version with less detail:
$$ p = u^2 + uv + 7 v^2, $$
$$ 4p = 4 u^2 + 4 uv + 28 v^2 = (2u+v)^2 + 27 v^2. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Finding the orthogonal projection of $(2,1,3)$. $V=\mathbb R^3,u=(2,1,3) $ and $W$={$(x,y,z):x+3y-2z=0$}.
Vectors ortogonal to $W$=Span{$(1,3,-2)$},Vectors on $W$=Span{$(-1,1,1)$}.
So,orthogonal basis is {$(1,3,-2),(-1,1,1)$}.Then,orthonormal basis is
{$\frac{1}{\sqrt {14}}(1,3,-2),\frac{1}{\sqrt {3}}(-1,1,1)$}.
So orthogonal projection of
(2,1,3)=$<\frac{1}{\sqrt {14}}(1,3,-2),(2,1,0)>\frac{1}{\sqrt {14}}(1,3,-2)$+$<\frac{1}{\sqrt {3}}(-1,1,1),(2,1,0)>\frac{1}{\sqrt {3}}(-1,1,1)>$
=$\frac{-1}{14}(1,3,-2)+\frac{2}{3}(-1,1,1)=(\frac{-25}{42},\frac{34}{42},\frac{22}{42})$
But it's correct answer is = $\frac{1}{
14}
(29,17,40)$
Where i'm missing?
| Let $(2,1,3)+k(1,3,-2)$ be the orthogonal projection.
Thus, $$(2+k,1+3k,3-2k)(1,3,-2)=0$$ or
$$2+k+3+9k-6+4k=0,$$
which gives $k=\frac{1}{14}$ and since $\left(2+\frac{1}{14},1+\frac{3}{14},3-\frac{1}{7}\right)=\frac{1}{14}(29,17,40)$,
we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximization of multivariable function $ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$ subject to constraint This is purely out of curiosity. I have this set of a high school entrance Math exam in Vietnam (a special high school for gifted kids, the exam was held 3 days ago, maybe 4, taking into account the time zone). Here's one question that I've been stuck with:
With $a, b, c$ $\in \mathbb{R}^{+}$ such that $ab + bc + ca + abc =2$, find the max of
$ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$
I'm just curious what kind(s) of techniques these junior high students can use, since as far as I remember when I was at that age, calculus and derivatives aren't taught until high school.
| Here Laplace multipliers, derivatives and gradients do not exist. Only multiplying and checking if something simplifies.
In this case, symmetry for simplifying the problem into:
$$
\mathcal{M}: 3{x+1 \over x^2+2x+2}\\
st. x^3+3x^2=2
$$
This holds because every term on the function is a sum of independent terms. In here symmetry is not a false friend.
Which reduces to finding the positive roots of $x^3+3x^2-2=0$. We do not know to solve a cubic equation, so we only can guess to break the polynomial. Our first guess is $x=-1$. Broken:
$$
(x^3+x^2+2x^2+2x-2x-2)/(x+1)=x^2+2x-2
$$
And the others two are given by, yes, we know how to solve that already:
$$
x=\frac 12 (-2\pm\sqrt{4+8})=-1\pm \sqrt{3}
$$
Hence the only positive solution is $x=\sqrt{3}-1=a=b=c$ and $M=3{\sqrt{3}\over 3+1-2\sqrt{3}+2\sqrt{3}-2+2}=3{\sqrt{3}\over 4}\sim 1.3$.
What if they are not equal?. Suppose $a=b=1,c=x$:
$$
\mathcal{M}: \frac45+{x+1 \over x^2+2x+2}\\
st. 3x=2
$$
then$x=2/3$ and $M=\frac45+{2/3+1 \over 4/9+4/3+2}=\frac45+{15 \over 34}\sim 1.2<1.3$
Yes, symmetry here helps, and the solution is a maximum.
| {
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"timestamp": "2023-03-29T00:00:00",
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$P[X\geq m/2]\geq\frac{1}{m+1}$ if $\mathbb{E}[X]=\frac{m}{2}$ Let $m\gt 0$ and let $X$ be a random variable on the set $\{0,\dots,m\}$.
Assuming that $\mathbb{E}[X]=\frac{m}{2}$, prove that:
$$P[X\geq m/2]\geq\frac{1}{m+1}$$
I tried induction but it didn't work, know I am trying to prove it by contradiction, someone got a hint?
| If $P(X \ge m/2) < \frac{1}{m+1}$, then
\begin{align}
E[X] &= \sum_{k < m/2} k P(X=k) + \sum_{k \ge m/2} kP(X=k)
\\
&\le \left(\left\lceil\frac{m}{2}\right\rceil - 1\right) P(X < m/2) + m P(X \ge m/2)
\\
&= \left(\left\lceil\frac{m}{2}\right\rceil - 1\right)(1-P(X \ge m/2)) + mP(X \ge m/2)
\\
&= \left(\left\lceil\frac{m}{2}\right\rceil - 1\right) +
\left(m+1 - \left\lceil\frac{m}{2}\right\rceil\right) P(X \ge m/2)
\\
&{\color{blue}<} \left(\left\lceil\frac{m}{2}\right\rceil - 1\right) +
\left(m+1 - \left\lceil\frac{m}{2}\right\rceil\right) \frac{1}{m+1}
\\
&= \left\lceil\frac{m}{2}\right\rceil \frac{m}{m+1}
\\
&= \begin{cases}
\frac{m}{2} \cdot \frac{m}{m+1} & \text{$m$ even}
\\
\frac{m+1}{2} \cdot \frac{m}{m+1} & \text{$m$ odd}
\\
\end{cases}
\\
&\le \frac{m}{2},
\end{align}
a contradiction.
Note that the statement is true even if $E[X]=m/2$ is replaced with $E[X] \ge m/2$.
Intuition: we want to see how small $P(X \ge m/2)$ can be while still keeping the expectation $E[X]$ large. Thus we should consider random variables that only take values $\lceil m/2 \rceil - 1$ and $m$. This is the first inequality in the work above.
| {
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Infinite zeros in infinite series The problem:
Given that
$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \ldots $$
Prove
$$\frac{\pi}{3} = 1 + \frac{1}{5} - \frac{1}{7} - \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \ldots$$
My solution:
We know
$$
\begin{align}
\frac{\pi}{4} & = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} -\frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \ldots \\
\\
\frac{\pi}{12} & = \frac{1}{3} - \frac{1}{9} + \frac{1}{15} - \frac{1}{21} + \frac{1}{27} -\frac{1}{33} + \frac{1}{39} - \frac{1}{45} + \ldots\\
\\
& = 0 + \frac{1}{3} + 0 + 0 - \frac{1}{9} + 0 + 0 + \frac{1}{15} + 0 + 0 - \frac{1}{21}
\end{align}
$$
now add them together:
$$
\begin{align}
\frac{\pi}{4} + \frac{\pi}{12} & = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \ldots \\
\\
& + 0 + \frac{1}{3} + 0 + 0 - \frac{1}{9} + 0 + 0 + \frac{1}{15} + \ldots \\
\end{align}
$$
and we will get:
$$
\begin{align}
\frac{\pi}{3} & = 1 + 0 + \frac{1}{5} - \frac{1}{7} + 0 -\frac{1}{11} + \frac{1}{13} + 0 + \ldots \\
& = 1 + \frac{1}{5} - \frac{1}{7} -\frac{1}{11} + \frac{1}{13} + \ldots
\end{align}
$$
My questions:
*
*I inserted/removed infinite zeros into/from the series, is that OK?
*My solution relies on the fact that $\Sigma a_n + \Sigma b_n = \Sigma (a_n + b_n)$ and $k \Sigma a_n = \Sigma k a_n$. Is this always true for convergent infinite series? If so, why is it? (yeah I know this is a stupid question, but since I'm adding infinite terms up, I'd better pay some attention.)
*Bouns question: Can I arbitrarily (arbitrariness isn't infinity, you know) insert/remove zeros into/from a convergent infinite series, without changing its convergence value?
| *
*Yes, that is ok.
*Yes it is. The partial sum of $\sum_k (a_k+b_k)$ is the partial sum of $\sum_ka_k$ plus the partial sum of $\sum_k b_k.$ The result follows from the sum property for limits.
*Yes. Adding zeros will only delay the inevitable convergence of the sequence of partial sums. Where you insert zeros, the sequence of partial sums will hold flat. For the $N$ you find in the proof of convergence of the original, simply replace with $N$ plus the number of zeros you inserted before the N-th and you'll have the same value that will be within $\epsilon$ of the number the sum converges to.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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} |
Solve $\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} = -7$ for $x$ I have been scratching my head for solving this equation but I am unable to do this. Even I am unable to get how to use the hint. A way to solve this would be of great help
Solve for $x$ :-
$$\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} = -7$$
(Hint: $-3-4=-7$, if $\frac 1a+\frac 1b +\frac 1c \ne 0$)
| Let $a^{-1}+b^{-1}+c^{-1}=k$. Your equation becomes
$$
\frac{a}{c}+\frac{b}{c}+
\frac{a}{b}+\frac{c}{b}+
\frac{b}{a}+\frac{c}{a}-
\frac{4abc}{a+b+c}+7
=kx
$$
The first six terms on the left-hand side can be rewritten as
$$
\frac{a^2b+ab^2+a^2c+ac^2+b^2c+cb^2}{abc}
$$
Now, set $S=a+b+c$, $Q=ab+bc+ca$ and $P=abc$. The numerator of the above fraction can be computed as
$$
a^2b+ab^2+a^2c+ac^2+b^2c+cb^2=SQ-3P
$$
using the theory of symmetric polynomials.
Basically, we must have
$$
a^2b+ab^2+a^2c+ac^2+b^2c+cb^2=\alpha S^3+\beta SQ+\gamma P
$$
for some constants $\alpha$, $\beta$ and $\gamma$; then
\begin{align}
a=1,b=0,c=0 &\implies 0=\alpha \\
a=1,b=1,c=0 &\implies 2=2\beta \\
a=1,b=1,c=1 &\implies 6=27\alpha+9\beta+\gamma
\end{align}
so $\alpha=0$, $\beta=1$ and $\gamma=-3$.
Since, easily, $k=Q/P$, the equation becomes
$$
\frac{Q}{P}x=\frac{SQ-3P}{P}-\frac{4P}{S}+7=
\frac{S^2Q-3PS-4P^2+7PS}{SP}=\frac{S^2Q+4PS-4P^2}{SP}
$$
and, finally,
$$
x=\frac{S^2Q+4PS-4P^2}{SQ}=S-\frac{4P(P-S)}{SQ}
$$
which agrees with the other solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proof that $\log_a b \cdot \log_b a = 1$
Prove that $\log_a b \cdot \log_b a = 1$
I could be totally off here but feel that I have at least a clue.
My proof is:
Suppose that $a = b$, then $a^{1} = b$ and $b^{1} = a$ and we are done.
Suppose now that $a \neq b$. We wish to show that: $\log_b a = \frac{1}{\log_a b}$.
\begin{align*}
1 &= \frac{1}{\log_a b} \cdot \frac{1}{\log_b a} \cdot \left(\log_a{b} \cdot \log_b a \right) \\
&= \left(\log_a{b} \cdot \frac{1}{\log_a b} \right) \cdot \left(\log_b a \cdot \frac{1}{\log_b a}\right) \\
&= \left(\log_b a \cdot \frac{1}{\log_b a}\right) \cdot 1 \\
&=1 \cdot 1 \\
&= 1
\end{align*}
| Your argument does not prove the identity. I can substitute any two values for $\log_a b$ and $\log_b a$ and get the "proof"
\begin{align}
1 &= \frac1x \cdot \frac1y \cdot \left( x \cdot y \right) \\
&= \left( x \cdot \frac1x \right)\left( y \cdot \frac1y \right) \\
&= \left( x \cdot \frac1x \right) \cdot 1 \\
&= 1 \cdot 1 \\
&= 1
\end{align}
But we certainly don't have $xy = 1$ for all values of $x$ and $y$. Your manipulations aren't wrong, but you've only shown that $1 = 1$ not that $\log_a b \log_b a = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find extremas of $f(x,y) = xy \ln(x^2+y^2), x>0, y>0$ As the title says I need to find extreme values(maximum and minimum) of
$$f(x,y) = xy \ln(x^2+y^2), x>0, y>0$$
I don't understand how to find critical points of this problem.
I start with finding partial derivative and set derivatives equal to zero. And that is where I am stuck currently. So any help would be appreciated.
So:
$
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (xy* ln(x^2+y^2)) = y*ln(x^2+y^2) + \frac{2xy^2}{x^2+y^2}
$
$
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (xy* ln(x^2+y^2)) = x*ln(x^2+y^2) + \frac{2x^2y}{x^2+y^2}
$
So we have now:
$
\nabla f(x,y) = (0,0)
$
$
y*ln(x^2+y^2) + \frac{2xy^2}{(x^2+y^2)^2} = 0
$
And
$
x*ln(x^2+y^2) + \frac{2x^2y}{(x^2+y^2)^2} = 0
$
After trying to solve these equations I get that $x=y$ Is that correct?. So I don't understand what are then critical points as It can't be (0,0)?
| We look for points for wich $\nabla f=0$
The method you describe is this:
$\dfrac{\partial f}{\partial x}=y\ln(x^2+y^2)+xy\dfrac{2x}{x^2+y^2}$
$\dfrac{\partial f}{\partial y}=x\ln(x^2+y^2)+xy\dfrac{2y}{x^2+y^2}$
(You need to practice the derivatives :)
Setting the partials to zero we can get the coordinates for those points.
$y\ln(x^2+y^2)+xy\dfrac{2x}{x^2+y^2}=0$
$x\ln(x^2+y^2)+xy\dfrac{2y}{x^2+y^2}=0$
$\dfrac{y}{x}\ln(x^2+y^2)=\dfrac{x}{y}\ln(x^2+y^2)\implies x^2=y^2\;,x^2+y^2\neq1$
Then, $y=x$ or $y=-x$, dropping the second as $x\gt0$ and $y\gt0$
$x\ln(2x^2)=-\dfrac{2x^3}{2x^2}\implies \ln(2x^2)=-1$ or $x=\pm\dfrac{e^{-1/2}}{\sqrt{2}}$
Dropping the negative solution:
$x=y=\dfrac{e^{-1/2}}{\sqrt{2}}$
(We need to consider the case $x^2+y^2=1$. It's easily seen that the partials doesn't vanish: for those points, $2x^2y=0$, that is not possible with $x\gt0\;,y\gt0$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solving equation of type $a\cos x+b\cos y-c=0$ and $a\sin x+b\sin y-d=0$ Here's the questions
There are two equations:
$a\cos x+b\cos y-c=0 $ and $ a\sin x+b\sin y-d=0$ .
For instance
What is the value of $x$ and $y$ in following question?
$$2\cos x+3\cos y-2=0$$
$$2\sin x+3\sin y-8=0$$
| I'll assume $a > 0$ and $b > 0$.
Let $R = \sqrt{c^2 + d^2}$, so that $c = R \cos(\theta)$ and $d = R \sin(\theta)$ for some $\theta \in [0,2\pi)$. The points $[0,0]$, $[a \cos(x), a \sin(x)]$ and $[c, d]$ form a triangle with sides of length $a$, $b$ and $R$.
We need $R \le a + b$, $a \le R + b$ and $b \le R + a$ for this to be possible.
By the law of cosines we get
$$\cos(x - \theta) = \frac{a^2 + R^2 - b^2}{2 a R} $$
Similarly,
$$\cos(y - \theta) = \frac{b^2 + R^2 - a^2}{2 b R} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle What I tried:
Let $A(2,0,4)$, $B(4,1,-1)$, $C(6,7,7)$ then
$$\vec{AB}=(2,1,-5), \vec{AC}=(4,7,3), \vec{BC}=(2,6,8)$$
Then I calculated the angle between vectors:
$$\begin{aligned}
\alpha_1 &= \cos^{-1}\left(\frac{(2,1,-5)(4,7,3)}{\sqrt{2^2+1^2+(-5)^2}\sqrt{4^2+7^2+3^2}}\right) \\
&= \cos^{-1}(0)=90° \\
\alpha_2 &= \cos^{-1}\left(\frac{(4,7,3)(2,6,8)}{\sqrt{4^2+7^2+3^2}\sqrt{2^2+6^2+8^2}}\right) \\
&= \cos^{-1}\left(\frac{74}{\sqrt{74}\sqrt{104}}\right)=32.49\\
\alpha_3 &= \cos^{-1}\left(\frac{(2,6,8)(2,1,-5)}{\sqrt{2^2+6^2+8^2}\sqrt{2^2+1^2+(-5)^2}}\right) \\
&= \cos^{-1}\left(\frac{-30}{\sqrt{104}\sqrt{30}}\right)=122.5°
\end {aligned}$$
As you can see, these angles don't even form a triangle, what am I doing wrong, any thoughts?
| The distances satisfy the Pythagorean theorem.
$d(A, B) = \sqrt{2^2 + 1^2 + 5^2} = \sqrt{30}$
$d(A, C) = \sqrt{4^2 + 7^2 + 3^2} = \sqrt{74}$
$d(B, C) = \sqrt{2^2 + 6^2 + 8^2} = \sqrt{104}$
And indeed:
$\sqrt{30}^2 + \sqrt{74}^2 = \sqrt{104}^2$
Therefore it is a right triangle (link).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 0
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Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 $ So far I've done this:
LHS $ =\cos^{2}3x-\sin^{2}3x$
$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$
$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$
I can tell I'm going in the right direction but how should I proceed further?
EDIT I used the identity $\cos{2x}=2\cos^{2}x-1$ to solve it in a simpler way. viz.
LHS $= 2\cos^{2}3x-1$
$=2{(4\cos^{3}x-3\cos{x})}^2-1$
$2(16\cos^{6}x+9\cos^{2}x-24\cos^{4}x)-1$
$=32\cos^{6}x+18\cos^{2}x-48\cos^{4}x-1$
Still thank you for the answers!
| Uain Prosthaphaeresis Formulas, $$\cos6x+\cos 2x=2\cos4x\cos2x\iff\cos6x=\cos2x(2\cos4x-1)$$
$\cos4x=2\cos^22x-1$ and $\cos2x=2\cos^2x-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Proof/Derivation of Closed form of Binomial Expression $\sum\limits_{k=0}^{2n}(-1)^k\binom{2n}{k}^2$ The binomial expression given as follows:
$$\sum_{k=0}^{2n}\left(-1\right)^{k}\binom{2n}{k}^{2}$$
results nicely into the following closed form:
$$(-1)^{n}\binom{2n}{n}$$
I wish to know how exactly is it being done? I haven't been able to make much progress in solving it.
My approach:
\begin{align}
\sum_{k=0}^{2n}(-1)^{k}\binom{2n}{k}^{2} = \binom{2n}{0}^{2} - \binom{2n}{1}^{2} + \binom{2n}{2}^{2} - ... -\binom{2n}{2n-1}^{2} + \binom{2n}{2n}^{2} \\
= \binom{2n}{0}.\binom{2n}{0} - \binom{2n}{1}.\binom{2n}{1} + \binom{2n}{2}.\binom{2n}{2} - ... -\binom{2n}{2n-1}.\binom{2n}{2n-1} + \binom{2n}{2n}.\binom{2n}{2n} \\
\text{By Symmetry of binomial coefficients} \\
= \binom{2n}{2n}.\binom{2n}{0} - \binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n-2}.\binom{2n}{2} - ... -\binom{2n}{1}.\binom{2n}{2n-1} + \binom{2n}{0}.\binom{2n}{2n} \\
= \binom{2n}{2n}.\binom{2n}{0} + \binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n-2}.\binom{2n}{2} + ... +\binom{2n}{1}.\binom{2n}{2n-1} + \binom{2n}{0}.\binom{2n}{2n} - 2.\left(\binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n -3}.\binom{2n}{3} + ... + \binom{2n}{1}.\binom{2n}{2n-1}\right) \\
\text{By Vandermond's identity, first component, i.e.not enclosed within -2.(...) evaluates to C(4n, 2n)} \\
\binom{4n}{2n} - 2.(\binom{2n}{2n -1}.\binom{2n}{1} + \binom{2n}{2n -3}.\binom{2n}{3} + ... + \binom{2n}{1}.\binom{2n}{2n-1})
\end{align}
I'm lost beyond this point. It will be extremely helpful if someone can direct me in the right direction or provide the answer to this perplexing and challenging problem. Thank you.
| Think generating functions. The sum is the $x^n$ coefficient of
$$\sum_{k=0}^n\binom nkx^k\sum_{l=0}^n(-1)^l\binom nlx^l.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$x^4 -ax^3 +2x^2 -bx +1$ has real root $\implies$ $a^2+b^2 \ge 8$ it is requested to show that if the quartic polynomial $f(x) \in \mathbb{R}[x]$, defined by:
$$
f(x) = x^4 -ax^3 +2x^2 -bx +1,
$$
has a real root, then
$$
a^2 +b^2 \ge 8
$$
this question was asked by @medo, then deleted a few minutes ago. however having spent a little time on it, i think the problem seems sufficiently instructive to be worth resuscitating. it is not deep or difficult, but to find the right way of rewriting the polynomial to demonstrate the result is an interesting coffee-break challenge.
| We have from Cauchy-Schwarz inequality: $((1+x^2)^2)^2=(x^4 + 2x^2 + 1)^2 = (ax^3+bx)^2 \le (a^2+b^2)(x^6+x^2) = (a^2+b^2)x^2(x^4+1)\implies a^2+b^2 \ge \dfrac{(1+x^2)^4}{x^2(x^4+1)}\ge 8 \iff (1+x^2)^4 \ge 8x^2(1+x^4) \iff (1+y)^4 \ge 8y(1+y^2), y = x^2 \ge 0$. Lastly, consider $f(y) = (1+y)^4 - 8y(1+y^2), y \ge 0\implies f'(y) = 4(1+y)^3 - 8-24y^2 = 4((1+y)^3 - 2 - 6y^2)= 4(y-1)^3$. Thus if $0 \le y \le 1 \implies f'(y) \le 0 \implies f(y) \ge f(1) = 0$. If $y \ge 1 \implies f'(y) \ge 0 \implies f(y) \ge f(1) = 0$. Either case $f(y) \ge 0\implies a^2+b^2 \ge 8$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Help to understand polynomial factoring I'm following some proof, but got stuck at how the factoring works. I can follow this part:
$$\begin{align*}
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\
&= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\
\end{align*}$$
The next two steps are not so clear to me anymore:
$$\begin{align*}
&= \frac{(k+1)^2(k^2 + 4k + 4)}{4}\\
&= \frac{(k+1)^2(k+2)^2}{4}.\\
\end{align*}$$
I understand that first $(k+1)^3$ was changed to have the same denominator as the main term (which is $4$). Can someone help me break down the steps how the polynomials are added then after that, the powers confuse me a bit.
| The first step that is troubling you is just factorization by $(k+1)^2$.
Consider the expression of the numerator:
$$k^2(k+1)^2+4(k+1)^3$$
which is equal to
$$(k+1)^2\times k^2+(k+1)^2\times4(k+1)=(k+1)^2(k^2+4(k+1))=(k+1)^2(k^2+4k+4).$$
The next step needs you to remark that
$$(k+2)^2=k^2+2\times k\times 2+2^2=k^2+4k+4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$ then find $B^{16}$.
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$, then find $B^{16}$.
My Method:
Given $$AB^2=BA \tag{1}$$ Post multiplying with $B^2$ we get
$$AB^4=BAB^2=B^2A$$ Hence
$$AB^4=B^2A$$ Pre Multiplying with $A$ and using $(1)$ we get
$$A^2B^4=(AB^2)A=BA^2$$ hence
$$A^2B^4=BA^2 \tag{2}$$ Now post multiplying with $B^4$ and using $(2)$we get
$$A^2B^8=B(A^2B^4)=B^2A^2$$ hence
$$A^2B^8=B^2A^2 \tag{3}$$
Now Pre Multiply with $B^2$ and use $(3)$ we get
$$B^2A^2B^8=B^4A^2$$ $\implies$
$$A^2B^8B^8=B^4A^2$$
$$A^2B^{16}=B^4A^2$$
Now pre multiply with $A^2$ and use $(2)$we get
$$A^4B^{16}=A^2B^4A^2$$ $\implies$
$$B^{16}=BA^4=B$$
is there any other approach to solve this?
| $$B^2=A^4B^2=A^3BA.$$
Thus,
$$B^4=A^3BAA^3BA=A^3B^2A=A^2BA^2.$$
Hence, $$B^8=A^2BA^2A^2BA^2=A^2B^2A^2=ABA^3$$
and from here
$$B^{16}=ABA^3ABA^3=AB^2A^3=BA^4=B$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 2
} |
Inequality involving the sum of square roots of distances from a fixed point in triangle to triangle sides Let $P$ be a point of a triangle $ABC$, and $d_1=dist(P,AB)$, $d_2=dist(P,BC)$, $d_3=dist(P,CA)$.
Prove that $\sqrt{d_1}+\sqrt{d_2}+\sqrt{d_3}\le \sqrt{\frac{9}{2}R}$
| I re-define $d_1$ as $d(P,BC)=d(P,a)$ and so on for the sake of symmetry.
By the Cauchy-Schwarz inequality
$$\begin{eqnarray*}\sqrt{d_1}+\sqrt{d_2}+\sqrt{d_3} &=& \frac{1}{\sqrt{a}}\sqrt{ad_1}+\frac{1}{\sqrt{b}}\sqrt{bd_2}+\frac{1}{\sqrt{c}}\sqrt{cd_3}\\ &\leq& \sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(ad_1+bd_2+cd_3)}\\&=&\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)2\Delta}\\&=&\sqrt{\left(\frac{1}{2R\sin A}+\frac{1}{2R\sin B}+\frac{1}{2R\sin C}\right)R^2(\sin(2A)+\sin(2B)+\sin(2C))}\\&\leq&\sqrt{3R\left(\cos A+\cos B+\cos C\right)} \end{eqnarray*}$$
where $R(\cos A+\cos B+\cos C)=R+r$ by Carnot's theorem and $R\geq 2r$ by Euler's theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Finding minima and maxima to $f(x,y) = x^2 + x(y^2 - 1)$ in the area $x^2 + y^2 \leq 1$ I'm asked to find minima and maxima on the function
$$f(x,y) = x^2 + x(y^2 - 1)$$
in the area
$$x^2 + y^2 \leq 1.$$
My solution:
$$\nabla (f) = (2x + y^2 - 1, 2xy)$$
Finding stationary points
$$2xy = 0$$
$$2x + y^2 - 1 = 0$$
gets me $(0,1),(0,-1),(\frac{1}{2},0)$.
Stuyding the boundary:
$$x^2 + y^2 = 1$$
$$y^2 = 1 - x^2 $$
$$f(x,y) = f(x, 1-x^2) = x^2 -x^3$$
Finding stationary points on the boundary:
$$f'(x,y) = 2x - 3x^2 = 0$$
gives $(0, 1), (0,-1), (\frac{2}{3}, \sqrt{\frac{5}{9}}),(\frac{2}{3}, -\sqrt{\frac{5}{9}})$.
So in total i've got the stationary points
$$(\frac{1}{2} , 0) | (0,1)| (0,-1)| (\frac{2}{3},\sqrt{\frac{5}{9}})|(\frac{2}{3}, -\sqrt{\frac{5}{9}}).$$
which give the function values of:
$$-\frac{1}{4}, 0,0,0.15,0.15$$.
which gives the minima: $-\frac{1}{4}$ and maxima $0.15$.
The minima is correct but the maxima should be 2. Why? What stationary point am I missing?
| You have a small problem. When you consider $f(x,1-x^2)$ it should be considered over the closed interval $x \in [-1,1]$, which means its endpoints must also be included into the possible list of points.
$f(-1)=2$...
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x\cos x}{2\sin x+ 2\sin x\cos x} & \text{since $\sin(2x) = 2\sin x\cos x$}&\\
& = \dfrac{2\sin x(1 - \cos x)}{2\sin x(1 + \cos x)} &&\\
& = \dfrac{(1- \cos x)}{(1+ \cos x)} &&\\
& = \dfrac{(1- \cos x)(1+ \cos x)}{(1+ \cos x)(1+ \cos x)}& \text{since $\dfrac{(1+ \cos x)}{(1+ \cos x)}=1$}&\\
& = \dfrac{(1)^2-(\cos x)^2}{(1+ \cos x)^2} & \text{since $a^2-b^2 = (a+b)(a-b)$}&\\
& = \dfrac{(\sin x)^2}{(1+ \cos x)^2} & \text{since $(\sin x)^2 + (\cos x)^2 =1$, so $(\sin x)^2 = 1- (\cos x)^2$.}&
\end{array}$$
Now, I understand that I have the $\sin x$ part on the numerator. What I have to do is get the denominator to be $\cos x$ somehow and also make the angles $\frac{x}{2}$ instead of $x$. How do I do that?
Please be through, and you can't use half-angle or triple angle or any of those formulas.
Also, we have to show left hand side is equal to right hand side, we can't do it the other way around. So please do not take $(\tan\frac{x}{2})^2$ and solve the equation.
Thank you for understanding and have a nice day :)
| Hint
If we cannot use the half angle, let us use the double angle formulae letting $x=2y$ $$\frac{1-\cos(x)}{1+\cos(x)}=\frac{1-\cos(2y)}{1+\cos(2y)}$$ Simplify and when finished reset $y=\frac x2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
| Use the same trick again to get
\begin{eqnarray*}
(z+\frac{1}{z})^2-2=7 \\
z+\frac{1}{z}=3
\end{eqnarray*}
\begin{eqnarray*}
(z+\frac{1}{z})(z^2+\frac{1}{z^2})=(z^3+\frac{1}{z^3})+(z+\frac{1}{z})=21 \\
z^3+\frac{1}{z^3}=\color{red}{18}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
calculating a limit using taylor series $$\lim_{x\rightarrow0}\frac{\cos(x) \sin(x)-x}{\sin^3 x}$$
I tried to use the identity $\cos(x) \sin(x)=\frac{1}{2}\sin(2x)$
and then use the taylor $\lim_{x\rightarrow0}\frac{x+\alpha(x)}{x^3+\beta(x)}$ where $\alpha(x), \beta(x)$ are the remainders which leads me to infinity but it's not the right answer because I did the same with l'hopital rule and got $-\frac{3}{2}$ (I used it three times!)
Can anyone help me how to solve it using taylor series?
Thanks in advance!
| $$\lim _{x\rightarrow 0} \frac {\cos x \sin x -x }{\sin^3 x} =\lim _{x\rightarrow 0} \frac { \sin 2x -2x }{2\sin^3 x} =\lim_{x\rightarrow 0} \frac { 2x-\frac {8x^3}{3!} +o(x^3) -2x }{ 2\sin^3 x} =\\
=\lim_{x\rightarrow 0} \frac {-\frac {8x^3}{3!} +o(x^3)}{2\sin^3 x} =\lim_{x\rightarrow 0} -\frac {8x^3}{12\sin^3 x} +\lim _{x\rightarrow 0} \frac { o\left(x^3 \right)}{2\sin^3 x} =-\frac 2 3$$
And by L'Hospital's rule $$\lim _{x\rightarrow 0} \frac {\cos x \sin x -x }{\sin^3 x} =\lim_{x\rightarrow 0} \frac {\sin 2x -2x}{2\sin^3 x} \overset {\text{L'Hospital}} = \lim_{x\rightarrow 0} \frac{2\cos 2x -2 }{6\sin^2 x\cos x} =\\
=4\lim_{x\rightarrow 0} \frac {-\sin^2 x}{6\sin^2 x\cos x} =-\frac 2 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians. Find the area of $ ABC $
Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians.
Find the area of $ ABC $
This problem is too difficult for me , the teacher said it is a challenge problem.
Any help will be appreciated , thank you.
| I tried to find an elegant geometric solution but couldn't. Here is the gung-ho algebraic version.
First by cosine rule about $A$ in $\Delta ADB$, and $\Delta AEC$, we get $$b^2 + c^2/4 -bc/2 = 36$$ and $$b^2/4 + c^2 -bc/2 = 81$$ From these we get by subtracting $$c^2=b^2 + 60 $$ Substituting this in the first equation we get
$$5b^2/4 -21 = b\sqrt{b^2+60} $$ Squaring both sides
$$25b^4/16 + 221 -105b^2/2 = b^4 + 60b^2 $$ $$b^4 - 200b^2 + 3536/9 = 0$$
You can solve this to get $b^2$. Since $c = \sqrt{b^2+60}$, and therfore the area is $$1/2 \cdot bc \cdot \sin(60)=\sqrt3/4 \cdot b \cdot \sqrt{b^2+60} $$
$$\Delta^2 = 3/16 \cdot b^2 \cdot (b^2 + 60) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Proof $13 \mid (k\cdot 2^n+1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$?
Proof $13 \mid (k \cdot 2^n + 1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$
Hint: for $k$ odd: $2^n \equiv-k' \pmod p$ and $kk' \equiv1 \pmod p$
My thoughts:
$13\mid(k-3) \Rightarrow k=13a+3$
and
$12|(n-2) \Rightarrow n=12b+2$
so
$\begin{align}k\cdot 2^n+1 &=(13a+3)2^{12b+2}+1 \\
&=4(13a+3)(2^{b})^{12}+1\\
\textrm{ or }
&=(k-3+3)\cdot 2^{n-2+2}+1\\
&=4\cdot 2^{n-2}(k-3+3)+1\\
&=4\cdot 2^{n-2}(k-3)+3\cdot2^n+1\end{align}$
I don't kow how to use the hint :(
| ${\rm mod}\ 13\!:\!\! \overbrace{\color{#c00}{ 2^{\large n}}\!\equiv 2^{\large 2+12j}}^{\large\quad\ \ \color{#c00}n\ \equiv\ 2\pmod{\!12}}\!\!\!\!\!\!\!\equiv 2^{\large 2}\color{#0a0}{(2^{\large 12})}^{\large j}\!\equiv 4\color{#0a0}{(1)}^{\large j}\!\equiv \color{#c00}4\,\ $ by $\ \rm\color{#0a0}{ Fermat}$
therefore $\ k\equiv 3\,\Rightarrow\, k\,\color{#c00}{2^{\large n}}\!+1\equiv 3\cdot \color{#c00}4+1\equiv 0\pmod{13}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Expand $\frac{1}{z-2}$ on $|z|>2$
Expand $\frac{1}{z-2}$ around $|z|>2$
So there is no singularity points
$$\frac{1}{(z-2)}=-\frac{1}{2-z}=-\frac{1}{2}\frac{1}{1-\frac{z}{2}}=-\frac{1}{2}\sum_{n=0}^{\infty} \frac{z^n}{2^n}=\sum_{n=0}^{\infty}-\frac{z^n}{2^{n+1}}$$
but we can also write:
$$\frac{1}{(z-2)}=\frac{1}{z}\frac{1}{1-\frac{2}{z}}=\frac{1}{z}\sum_{n=0}^{\infty} \frac{2^n}{z^n}=\sum_{n=0}^{\infty} \frac{2^n}{z^{n+1}}$$
So which one is true? the second one as for the first we get $|\frac{z}{2}|<1\Rightarrow |z|<2$ and we are looking at $|z|>2$?
| Yes, you are on the right track. The Laurent expansion of $\frac{1}{z-a}$ in $|z|>R\geq |a|$ is,
$$\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}}$$
Hence, for $|z|>R=2$,
$$\frac{1}{(z-1)(z-2)}=\frac{1}{z-2}-\frac{1}{z-1}=\sum_{n=0}^{\infty} \frac{2^n-1}{z^{n+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
If the roots of an equation are $a,b,c$ then find the equation having roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$. Actually I have come to know a technique of solving this kind of problem but it's not exactly producing when used in a certain problem.
Say we have an equation
$$2x^3+3x^2-x+1=0$$
the roots of this equation are $a,b,c$. If I were to find the equation having the roots $\frac{1}{2a}, \frac{1}{2b}$ and $\frac{1}{2c}$ then I can use the root co-efficient relations and use the rule of creating equations from roots.
This gives me the result : $$4x^3 + 2x^2 -3x -1 =0$$
A much much simpler way to solve this math is: if we pick that, $f(x)= 2x^3+3x^2-x+1$ and the values of $x$ are $a,b$ and $c$ then $f\left(\frac{1}{2x}\right)$ will denote an equation (if we just write $f\left(\frac{1}{2x}\right)=0$ ) which has the roots $\frac{1}{2a}, \frac{1}{2b}$ and $\frac{1}{2c}$. And this way the result also matches the result of my previous work.
But now I face an equation
$$x^3+3x+1=0$$ and if the roots are $a,b,c$ then I have to find the equation with the roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$.
If I use the root-coefficient relations then I have the result
$$x^3+6x^2+9x+5=0$$ (which is correct I think).
But here $f\left(\frac{1-x}{x}\right)= 3x^3-6x^2+3x-1$. So writing $f\left(\frac{1-x}{x}\right)=0$ won't give me the actual equation.
Where am I mistaking actually?
| First consider the polynomial with roots $a,b,c$:
$$g(x)=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$$
and let
$$r=a+b+c \quad\quad s=ab+ac+bc \quad\quad t=abc$$
And we accordingly write
$$g(x)=x^3-rx^2+sx-t$$
Now we multiply the linear factors of the expression having the desired roots:
$$f(x)=\Big(x-\frac{1-a}{a}\Big)\Big(x-\frac{1-b}{b}\Big)\Big(x-\frac{1-c}{c}\Big)$$
$$||$$
$$x^3-\frac{ab + ac+bc-3abc}{abc}x^2+\frac{a+b+c+-2(ab+ac+bc)+3abc}{abc}x-\frac{1-(a+b+c)+ab+ac+bc-abc}{abc}$$
and not that each of the coefficients can be expressed with $r,s,t$. Specifically we observe that
$$f(x)=x^3-\frac{s-3t}{t}x^2+\frac{3t-2s+r}{t}x-\frac{1-r+s-t}{t}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve an indefinite integral Find the value of
$$ \int\frac{\sqrt{1+x^2}}{1-x^2}dx $$
My Attempt: I tried to arrange the numerator as follows,
$$ \sqrt{1+x^2} = \sqrt{1-x^2 + 2x^2} $$
but that didn't help me.
Any help will be appreciated.
| Hint 1: Using $x=\tan(u)$, we get
$$
\begin{align}
\int\frac{\sqrt{1+x^2}}{1-x^2}\,\mathrm{d}x
&=\int\frac{\sec^3(u)}{1-\tan^2(u)}\,\mathrm{d}u\\
&=\int\frac{\sec(u)}{\cos^2(u)-\sin^2(u)}\,\mathrm{d}u\\
&=\int\frac{\sec^2(u)}{1-2\sin^2(u)}\,\mathrm{d}\sin(u)\\
&=\int\frac{\mathrm{d}\sin(u)}{\left(1-2\sin^2(u)\right)\left(1-\sin^2(u)\right)}\\
\end{align}
$$
Hint 2: Use partial fractions:
$$
\begin{align}
&\frac1{\left(1-2v^2\right)\left(1-v^2\right)}\\
&=\frac2{1-2v^2}-\frac1{1-v^2}\\
&=\frac1{1-\sqrt2v}+\frac1{1+\sqrt2v}-\frac{1/2}{1-v}-\frac{1/2}{1+v}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find $\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)$ I did it as follows: $$\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)=\tan\Bigg(\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)\Bigg)=\frac{\sqrt{2}-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}\sqrt{2}}=\frac{\sqrt{2}}{4}.$$ But there is no such an answer. What is wrong with it?
| $$\tan[\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)]=\frac{\sqrt{2}-1/\sqrt{2}}{1+1}=\frac{\sqrt{2}}{4},$$
on simplifying we see that
$$\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right) =\arctan(\frac{\sqrt{2}}{4})= \sin^{-1} \frac{1}{3}.$$
BTW, this identity suggests a Ruler & Compass construction to trisect a line segment $OA$ (Somos' trig simplification, Jack D'Aurizio's construction are put together):
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve$\int_0^a\frac{(a-x)^{n-1}}{(a+x)^{n+1}}\, dx$? How can I solve the following integral:
$$\int_0^a\frac{(a-x)^{n-1}}{(a+x)^{n+1}}\, dx.$$ I'm not posting my work because I don't even know where to start with this integral.
| Integrating by parts:
$$I_n=\int_0^a\frac{(a-x)^{n-1}}{(a+x)^{n+1}}\, dx = $$
$$(a-x)^{n-1}\cdot \left(-\frac{1}{n(a+x)^n}\right) \bigg |_0^a-\int_0^a \left(-\frac{1}{n(a+x)^n}\right)\cdot (-1)(n-1)(a-x)^{n-2}dx=$$
$$\frac{1}{na}-\frac{n-1}{n} \cdot I_{n-1}, \ \ \ I_1=\frac{1}{2a}.$$
$$I_2=\frac{1}{2a}-\frac{1}{2}I_1=\frac{1}{2a}-\frac{1}{4a}=\frac{1}{4a}.$$
$$I_3=\frac{1}{3a}-\frac{2}{3}I_2=\frac{1}{3a}-\frac{1}{6a}=\frac{1}{6a}.$$
$$\cdots$$
$$I_n=\frac{1}{2an}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
| If you expand you find its $a^2+b^2+c^2+2ab+2bc+2xa+a^2+b^2+c^2$. Now square first equation and we have $a^2+b^2+c^2+2ab+2bc+2ac=4. ..(3)$ thus now we want value of $4+a^2+b^2+c^2$ .rearranging $3$ and using your equation $2$ we have $a^2+b^2+c^2+2 (ab+bc+ac)=4$ thus $a^2+b^2+c^2+2=4$ thus value of expression we want is $4+a^2+b^2+c^2=4+2=6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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How to recognize a sum as a Riemann Sum Evaluate $$\frac{1}{1}+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\cdots+\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{2}{3n+3}+\cdots$$
answer choices:
a) $\ln 2$
b) $\ln 3$
c) $e^2$
d) $\dfrac 9 {25}$
Looking at the answer choices, I am almost certain that we need to rearrange the sum into a Riemann sum, but I am stuck on how to do so. Also are there any general techniques to recognize/rearrange summations such as these into Riemann sums? This problem is from a competition, so I would need to solve these problems quickly. Thanks!
| Since $-\frac{2}{3n+3} = \frac{1}{3n+3} - \frac{1}{n+1}$, the general term can be written
$$a_n = \left(\frac{1}{3n+1} + \frac{1}{3n+2} + \frac{1}{3n+3}\right) - \frac{1}{n+1}$$
so the sum (up to the $n$'th term) is the difference between two harmonic numbers
$$\left(1 + \frac{1}{2}+\ldots + \frac{1}{3n+3}\right) - \left(1 + \frac{1}{2} + \ldots + \frac{1}{n+1}\right) = \sum_{k=n+2}^{3n+3}\frac{1}{k} = \frac{1}{N}\sum_{k=N+1}^{3N}\frac{1}{\frac{k}{N}}\tag{1}$$
where $N = n+1$. This is one single term $\left(\frac{1}{N} = \frac{1}{n+1}~~\text{which vanishes as $n\to\infty$}\right)$ away from being an exact Riemann sum for the integral $\int_1^3\frac{{\rm d}x}{x}$. We therefore have
$$\lim_{n\to\infty }\sum_{k=n+2}^{3n+3}\frac{1}{k} = \lim_{N\to\infty} \frac{1}{N}\sum_{k=N}^{3N}\frac{1}{\frac{k}{N}} = \int_1^3\frac{{\rm d}x}{x} = \ln(3)$$
Alternatively: if you know that the harmonic numbers satisfy $H_n \sim \gamma + \log(n)$ then applying this to $(1)$ gives us the answer directly
$$H_{3n} - H_n \sim (\gamma + \ln(3n)) - (\gamma + \ln(n)) = \ln(3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2347522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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How many such three digit numbers are possible? Certain $3$-digit numbers have the following characteristics:
*
*all three digits are different
*the number is divisible by $7$
*the number on reversing the digits is also divisible by $7$
How many such numbers are there? I have tried using a brute force approach and found $168$ and $259$. Is there a better way to solve these questions?
| If the number is $10^2a+10b+c$ as digits then considering $(10^2a+10b+c) - (10^2c+10b+a)$ leads to realising $99(a-c)$ must be divisible by $7$ and cannot be zero and, since $99$ is coprime to $7$.
*
*$1b8$ but $108 \equiv 3 \pmod 7$ requiring $10b \equiv 4 \pmod 7$ making the only possible value of $b=6$, which is neither $1$ nor $8$ $\checkmark$
*$2b9$ but $209 \equiv 6 \pmod 7$ requiring $10b \equiv 1 \pmod 7$ making the only possible value of $b=5$, which is neither $2$ nor $9$ $\checkmark$
*$7b0$ but $700 \equiv 0 \pmod 7$ requiring $10b \equiv 0 \pmod 7$ making the only possible values of $b=0$ or $7$, which are either $7$ or $0$ $\times$
*$8b1$ but $801 \equiv 3 \pmod 7$ requiring $10b \equiv 4 \pmod 7$ making the only possible value of $b=6$, which is neither $8$ nor $1$ $\checkmark$
*$9b2$ but $902 \equiv 6 \pmod 7$ requiring $10b \equiv 1 \pmod 7$ making the only possible value of $b=5$, which is neither $9$ nor $2$ $\checkmark$
So the possible solutions are $168, 259, 861, 952$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to obtain the sum of the following series? $\sum_{n=1}^\infty{\frac{n^2}{2^n}}$ It seems that I'm missing something about this.
First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)
What I tried to do is to find a limit of a partial sum $\lim_{n\rightarrow\infty}S_n$ as follows: $S_n=\frac{\frac{1}{6} n (n+1) (2 n+1)}{\frac{1-\left(\frac{1}{2}\right)^n}{2 \left(1-\frac{1}{2}\right)}}$. Still, the limit is $\infty$ and I'm clearly doing something wrong.
| Note that
$$
n^2=\binom{n+1}{2}+\binom{n}{2}\tag{1}.
$$
Also
$$
\frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{k+n-1}{k-1}x^n\tag{2}.
$$
for $k\geq 1$ (by repeatedly differentiating the geometric series or by the extended binomial theorem). In particular
$$
\frac{1}{(1-x)^3}=\sum_{n=0}^\infty \binom{n+2}{2}x^n\tag{3}.
$$
Mutiply by $x^n$ and sum on $n$ in (1) to get that
$$
\sum_{n=0}^\infty n^2x^n
=\sum_{n=0}^\infty\binom{n+1}{2}x^n+\sum_{n=0}^\infty\binom{n}{2}x^n
=\frac{x}{(1-x)^3}+\frac{x^2}{(1-x)^3}
=\frac{x(1+x)}{(1-x)^3}.\tag{4}
$$
by (3).
Set $x=1/2$ in equation (4) to find that
$$
\sum_{n=0}^\infty \frac{n^2}{2^n}=\color{blue}{6}\tag{5}.
$$
| {
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"url": "https://math.stackexchange.com/questions/2348438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simplifying the Solution to the Cubic I am trying to solve the cubic. I currently have that, for $ax^3+bx^2+cx+d=0$, a substitution to make this monic. Dividing by $a$ gives
$$x^3+Bx^2+Cx+D=0$$
where $B=\frac{b}{a}, C=\frac{c}{a}, D=\frac{d}{a}$. Then, with the substitution $x=y-\frac{B}{3}$, I got
$$y^3+\left(C-\frac{B^2}{3}\right)y+\left(D-\frac{BC}{3}+\frac{2B^3}{27}\right)=0$$
Thus, to make things simpler, i made the substitution $p=C-\frac{B^2}{3}$ and $q=D-\frac{BC}{3}+\frac{2B^3}{27}$ we have the "depressed cubic"
$$y^3+py+q=0$$
Now, using the identity,
$$(m+n)^3=3mn(m+n)+(m^3+n^3)$$
we let $y=m+n$. This then translates to $p=-3mn,$ and $q=-(m^3+n^3)$ and gives us a system of equations in $m$ and $n$. Solving for $n$ gives $n=-\frac{p}{3m}$ and back substituting yields
$$q=-m^3+\frac{p}{3m}\qquad \Rightarrow \qquad m^6+qm^3-\frac{p^3}{27}=0$$ and now we can solve the quadratic for $m$;
$$m=\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}$$
and then that means, by back substitution
$$n=-\frac{p}{3\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}}$$
So, I think I am almost here, because now,
$$y=m+n=\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}-\frac{p}{3\sqrt[3]{\frac{-q\pm\sqrt{q^2+\frac{4p^2}{27}}}{2}}}$$
But how can I simplify this expression? I know I can back substitute for the original $a,b,c,d$ and solve for $x$. But this sum looks complicated and my attepts to simplify the sum have not worked.
| $m=\sqrt[3]{\frac{-q\pm\sqrt{q^2-\frac{4p^2}{27}}}{2}}$
Lets choose the positive root for m.
$m=\sqrt[3]{\frac{-q + \sqrt{q^2-\frac{4p^2}{27}}}{2}}$
and we know that this solves:
$q=-(m^3+n^3)$
So lets plug it into $m^3$
$q=\frac{q - \sqrt{q^2-\frac{4p^2}{27}}}{2} - n^3$
$n^3 = \frac{-q - \sqrt{q^2-\frac{4p^2}{27}}}{2}$
Which is the sign flipped other root.
$y = \sqrt[3]{\frac{-q + \sqrt{q^2-\frac{4p^2}{27}}}{2}} + \sqrt[3]{\frac{-q - \sqrt{q^2-\frac{4p^2}{27}}}{2}}$
One more note
$m^3 = \frac{-q\pm\sqrt{q^2-\frac{4p^2}{27}}}{2}$ has 2 complex roots that should not be forgotten.
$y = \omega \sqrt[3]{\frac{-q + \sqrt{q^2-\frac{4p^2}{27}}}{2}} + \omega\sqrt[3]{\frac{-q - \sqrt{q^2-\frac{4p^2}{27}}}{2}}$
where $\omega$ are the roots of $(z^3-1 = 0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + \cdots$ Attempt: I found the Fourier series for $f(x) = \begin{cases} 0,& -\pi < x < 0 \\ x/2,& 0 < x < \pi \end{cases}$
a) $a_0 = \frac{1}{2\pi}\int_0^{\pi} r\,dr = \pi/4$
$a_n = \frac{1}{2\pi}\int_0^r \frac{r\cos(nr)}{2}dr = \frac{(-1)^n - 1}{2\pi n^2}$
$b_n = \frac{1}{2\pi}\int_0^r r\sin(nr)\,dr = \frac{(-1)^n + 1}{2n}$
$f(x) = \frac{\pi}{8} - \sum_n [\frac{((-1)^n - 1)\cos(nx)}{2\pi n^2} + \frac{((-1)^n + 1)\sin(nx)}{2n}]$
The prof asked us to use this Fourier series to prove that $\pi^2/8 = 1+1/3^2+1/5^2+1/7^2+\cdots$. How do I do this?
| You can prove $$\sum \frac{1}{n^2}=\frac{\pi^2}{6}$$ using Fourier series.
Hence, $$\sum\frac{1}{(2n)^2}+\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}$$
Therefore,
$$\frac14 \sum \frac{1}{n^2}+\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}$$
This shows $$\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
What is the value of the expression $\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$? This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.
What is the value of
$\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$
Do I just use
$\cos(a-b)-\cos(a+b) = 2\sin(a)\sin(b)$
| As you observed,
$$\sin\alpha\sin\beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)]$$
Applying that observation here yields
\begin{align*}
\sin\left(\frac{2\pi}{7}\right)&\sin\left(\frac{4\pi}{7}\right) + \sin\left(\frac{4\pi}{7}\right)\sin\left(\frac{8\pi}{7}\right) + \sin\left(\frac{8\pi}{7}\right)\sin\left(\frac{2\pi}{7}\right)\\
& = \frac{1}{2}\left[\cos\left(-\frac{2\pi}{7}\right) - \cos\left(\frac{6\pi}{7}\right) + \cos\left(-\frac{4\pi}{7}\right) - \cos\left(\frac{12\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) - \cos\left(\frac{10\pi}{7}\right)\right]\\
& = \frac{1}{2}\left[\cos\left(-\frac{2\pi}{7}\right) + \cos\left(-\frac{4\pi}{7}\right) - \cos\left(\frac{12\pi}{7}\right) - \cos\left(\frac{10\pi}{7}\right)\right]\\
& = \frac{1}{2}\left[\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) - \cos\left(\frac{12\pi}{7}\right) - \cos\left(\frac{10\pi}{7}\right)\right]\\
& = \frac{1}{2}\left[\cos\left(\frac{2\pi}{7}\right) + \cos\left(\pi - \frac{3\pi}{7}\right) - \cos\left(2\pi - \frac{2\pi}{7}\right) - \cos\left(\pi + \frac{3\pi}{7}\right)\right]\\
& = \frac{1}{2}\left[\cos\left(\frac{2\pi}{7}\right) - \cos\left(\frac{3\pi}{7}\right) - \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{3\pi}{7}\right)\right]\\
& = 0
\end{align*}
where we have used the identities
\begin{align*}
\cos(-\theta) & = \cos\theta\\
\cos(\pi - \theta) & = -\cos\theta\\
\cos(\pi + \theta) & = -\cos\theta\\
\cos(2\pi - \theta) & = \cos\theta
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the probability of the sum of four dice being 22? Question
Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?
My Approach
I simplified it to the equation of the form:
$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $
Solving this equation results in:
$x_{1}+x_{2}+x_{3}+x_{4}=22$
I removed restriction of $x_{i} \geq 1$ first as follows-:
$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
$\Rightarrow \binom{18+4-1}{18}=1330$
Now i removed restriction for $x_{i} \leq 6$ , by calculating the number of bad cases and then subtracting it from $1330$:
calculating bad combination i.e $x_{i} \geq 7$
$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$
We can distribute $7$ to $2$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{2}$
We can distribute $7$ to $1$ of $x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$ i.e$\binom{4}{1}$ and then among all others .
i.e
$$\binom{4}{1} \binom{14}{11}$$
Therefore, the number of bad combinations equals $$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$
Therefore, the solution should be:
$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$
However, I am getting a negative value. What am I doing wrong?
EDIT
I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.
| The bad combinations criteria is atleast one $x_i \geq 7$.
The number of bad combinations when:
*
*One of $x_i$ is forced to be greater than or equal to $7$ is $\binom{4}{1}\binom{12 + 4 - 1}{12} = 1820$.
*Two of $x_i$'s are forced to be greater than or equal to $7$ is $\binom{4}{2}\binom{6+4-1}{6} = 504$.
*Three of $x_i$'s are forced to be greater than or equal to $7$ is $\binom{4}{3}\binom{0+4-1}{0} = 4$.
*Four of $x_i$'s are forced to be greater than or equal to $7$ is $0$.
So, total bad combinations $= 1820 - 504 + 4 - 0 = 1320$
I used $n(\cup_{i=1}^{4}A_i) = \sum_{i=1}^{4}n(A_i) - \sum_{i,j, i\neq j}n(A_i\cap A_j) + \ldots$
So, possible combinations $= 1330 - 1320 = 10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 11,
"answer_id": 8
} |
Roots of $abc^2x^2 + 3a^2 c x + b^2cx-6a^2 -ab +2b^2 = 0$ are rational We have to show that roots of $$abc^2x^2 + (3a^2 c + b^2c)x-6a^2 -ab +2b^2 = 0$$ are rational.
This can be possible if the discriminant is a perfect square. SO I tried converting it into perfect square but failed:
$$\text{Discriminant}=(3a^2c+b^2c)^2-4abc^2(2b^2-6a^2-ab)\\
c^2(9a^4+b^4+10a^2b^2+24a^3b-8ab^3)$$
I cannot proceed please help! Thanks!
| Starting with your discriminant:
$$
c^2(9a^4+b^4+10a^2b^2+24a^3b-8ab^3)
$$
we see that we can easily ignore the $c^2$ since that is already a square. Now, we look at the degrees of all the terms and notice that they are all degree $4$. Therefore, the factorization (if it exists) has to be of the form
$$
(9a^4+b^4+10a^2b^2+24a^3b-8ab^3)=(x_1a^2+x_2ab+x_3b^2)^2.
$$
Moreover, we can assume that $x_1>0$ since otherwise, we can multiply through by $-1$, which doesn't change anything since $(-1)^2=1$.
We can multiply out the RHS to get the following system of equations:
\begin{align}
9&=x_1^2&2x_1x_2&=24&x_3^2&=1\\
2x_2x_3&=-8&2x_1x_3+x_2^2&=10
\end{align}
Therefore, $x_1=3$ from the first equation. $x_2=4$ from the second equation, $x_3=-1$ from the fourth equation. We can check that all of the equations are satisfied with these choices, so
$$
(9a^4+b^4+10a^2b^2+24a^3b-8ab^3)=(3a^2+4ab-b^2)^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this question, but not able to proceed. How do I solve this?
| We need to prove that
$$\sum_{cyc}(a^8-a^3b^3c^2)\geq0$$ or
$$\sum_{cyc}(2a^8-2a^3b^3c^2)\geq0$$ or
$$\sum_{cyc}(a^8-a^6b^2-a^2b^6+b^6+a^6c^2-2a^3b^3c^2+b^6c^2)\geq0$$ or
$$\sum_{cyc}((a^2-b^2)^2(a^4+a^2b^2+b^4)+c^2(a^3-b^3)^2)\geq0,$$
which is obvious.
More way.
By AM-GM:
$$\sum_{cyc}a^8=\frac{1}{8}\sum_{cyc}(3a^8+3b^8+2c^8)\geq$$
$$\geq\frac{1}{8}\sum_{cyc}8\sqrt[8]{a^{24}b^{24}c^{16}}=\sum_{cyc}a^3b^3c^2.$$
We can prove it by Rearrangement, which is the Chebyshov's result.
Indeed, by Rearrangement:
$$\frac{a^8+b^8+c^8}{a^3b^3c^3}=\frac{1}{2}\sum_{cyc}\left(2\cdot\frac{a^5}{b^3c^3}\right)\geq\frac{1}{2}\sum_{cyc}\left(\frac{a^5}{a^3b^3}+\frac{a^5}{a^3c^3}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{b^3}+\frac{a^2}{c^3}\right)\geq\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{a^3}+\frac{a^2}{a^3}\right)=\sum_{cyc}\frac{1}{a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$
Evaluate the following limit.
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
My Attempt:
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-a} + \sqrt {bx}}$$
$$=\lim_{x\to \infty} \dfrac {x-a-bx}{\sqrt {x-a} + \sqrt {bx}}.$$
How do I proceed?
| Alternatively:
$$\lim_\limits{x\to\infty} \sqrt{x-a}-\sqrt{bx}=\lim_\limits{x\to\infty} \sqrt{x}-\sqrt{bx}=(1-\sqrt{b})\lim_\limits{x\to\infty} \sqrt{x}=\begin{cases} 0, \ if \ b=1 \\ \infty, \ if \ b\ne 1. \end{cases}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Simplifying $\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$ I'm struggling trying to simplify $$\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$$
Here's my attempt:
All of these coefficients can be converted to base $2$ exponentials.
$$\frac{2^{4^{(x+1)}}+20^\ (2^{2^{(2x)}})}{2^{x-3}2^{3^{(x+2)}}}$$
Doing some algebra:
$$\frac{2^{4x+4}+20^\ (2^{4x})}{2^{x-3}2^{3x+6}}$$
The denominator can be simplified.
$$\frac{2^{4x+4}+20^\ (2^{4x})}{2^{4x+3}}$$
Now, here's where I think I've made a mistake, but I'm gonna try to simplify it further by splitting this up into separate fractions.
$$\frac{2^{4x+4}}{2^{4x+3}} + \frac{5 \bullet 2^2 \ 2^{4x}}{2^{4x+3}}$$
$$2\ \frac{2^{4x+3}}{2^{4x+3}} + \frac{5\ \bullet 2^{4x+2}}{2^{4x+3}}$$
$$2 + 5\ \bullet 1/2\ \frac{2^{4x+2}}{2^{4x+2}}$$
$$2 + 5/2$$
$$4/2 + 5/2$$
$$9/2$$
Whelp, I just solved it properly while putting it in as a question. However, I still have two queries:
*
*Symbolab's answer is this: . Is there logic behind this answer or is outright not simplified as well?
*Does this algorithm to solving (changing all to the right base and trying to do algebra do them apply to ones even if they aren't connected by whole number bases? Like if I were trying to do one with base $5$ and base $7$ exponentials, could I solve it using logarithm laws just the same?
| why not use base 16? You have
$$
\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}} = \\
\frac{16 \cdot 16^{x}+20 \cdot 16^{x}}{16^{x} 2^{-3}8^{2}} = \\
\frac{36}{8} = \frac{9}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Double Series equals Gamma function I want to show $$\sum_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)}\cdot \frac{1}{3^{n+m+3x-1}} = \Gamma(x)$$ for $x>0$ and I don't know how.
| All credit goes to @Nemo. As he pointed out in mathoverflow.net, \begin{align*}
&\sum\limits_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)} \frac{1}{3^{n+m+3x-1}} \\
&= \int\limits_0^\infty t^{n+m+3x-1}e^{-t} \sum\limits_{n,m=0}^\infty \frac{1}{\Gamma(n+1+x)\Gamma(m+1+x)} \frac{1}{3^{n+m+3x-1}}\\
&\overset{t=3u}{=} 3\int\limits_{0}^{\infty}u^{3x-1} e^{-3u} \left(\sum\limits_{n=0}^\infty \frac{u^n}{\Gamma(n+1+x)}\right)^2 du .
\end{align*}
As written in the Wikipedia article to the Incomplete Gamma Function, $$\sum\limits_{n=0}^\infty \frac{u^n}{\Gamma(n+1+x)} = \frac{u^{-x}}{\Gamma(x)} e^u \int\limits_0^u v^{x-1} e^{-v} dv,$$ so we get
\begin{align*}
&\frac{1}{(\Gamma(x))^2} \cdot 3\int\limits_{0}^{\infty} u^{x-1}e^{-u} \left(\int\limits_0^u v^{x-1} e^{-v} dv\right)^2 du\\
&= \frac{1}{(\Gamma(x))^2} \cdot 3 \int\limits_0^{\Gamma(x)} s^2ds\\
&= \frac{1}{(\Gamma(x))^2} \cdot (\Gamma(x))^3\\
&= \Gamma(x),
\end{align*} where we used the substitution $s(u):= \int\limits_0^u w^{x-1}e^{-w} dw$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understand what the author tries to say here. Can't this problem be done in another manner?
| This is a brute force approach.
The square previous to $\left(n^2\right)^2=n^4$ is $\left(n^2-1\right)^2=n^4-2n^2+1$. Since it is impossible for
$$
n^4-n^2+64\le n^4-2n^2+1\implies n^2\le-63
$$
we are left with
$$
n^4-n^2+64=n^4\implies\color{#C00}{n^2=64}
$$
or
$$
n^4-n^2+64=n^4+2n^2+1\implies n^2=21
$$
or
$$
n^4-n^2+64=n^4+4n^2+4\implies n^2=12
$$
or
$$
n^4-n^2+64=n^4+6n^2+9\implies7n^2=55
$$
or
$$
n^4-n^2+64=n^4+8n^2+16\implies3n^2=16
$$
or
$$
n^4-n^2+64=n^4+10n^2+25\implies11n^2=39
$$
or
$$
n^4-n^2+64=n^4+12n^2+36\implies13n^2=28
$$
or
$$
n^4-n^2+64=n^4+14n^2+49\implies\color{#C00}{n^2=1}
$$
or
$$
n^4-n^2+64=n^4+16n^2+64\implies\color{#C00}{n^2=0}
$$
The next square is $\left(n^2+9\right)^2=n^4+18n^2+81$ and it is impossible for
$$
n^4-n^2+64\ge n^4+18n^2+81\implies19n^2\le-15
$$
Therefore, we have the solution set of
$$
n\in\{0,\pm1,\pm8\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
Solving the the Integral: $\int\frac{1}{u^2-2}du$ I am attempting to solve the following integral, in which I am not sure, how to proceed:
$\int\frac{1}{u^2-2}du$
I have tried the following, but I am not sure if it is correct:
$y = u^2-2$
$dy = 2u*du$
$du = \frac{dy}{2u}$
Therefore:
$\int\frac{1}{u^2-1}du = \int\frac{1}{y}*\frac{dy}{2u} = \int\frac{1}{y}*\frac{1}{2u}dy = \int\frac{1}{y2u}dy$
Solving for u:
$u^2 = y+2$
$u = \sqrt{y + 2}$
Substituting $u$ for $\sqrt{y + 2}$:
$\int\frac{1}{y2u}dy = \int\frac{1}{y2(\sqrt{y + 2})}dy$
I am not sure, where to go from here. The only thing I can remember is to use the logarithm rule where:
$\int\frac{1}{y2(\sqrt{y + 2})}dy = ln|{y2(\sqrt{y + 2})}| + C$
Would that be correct?
| Another approach: Let $u=\sqrt2\sec t$, so $du=\sqrt2\sec t\tan t\,dt$. Then we have:
$$\begin{align}
\int\frac1{u^2-2}du &= \int\frac{\sqrt2\sec t\tan t}{2\sec^2t-2}dt\\
&= \frac1{\sqrt2}\int\frac{\sec t \tan t}{\sec^2 t-1}dt\\
&= \frac1{\sqrt2}\int\frac{\sec t\tan t}{\tan^2 t}dt\\
&= \frac1{\sqrt2}\int\frac{\sec t}{\tan t}dt\\
&= \frac1{\sqrt2}\int\csc t\,dt\\
&= -\frac1{\sqrt2}\ln\left|\csc t + \cot t\right|+C
\end{align}$$
Since $\sec t=\frac{u}{\sqrt2}$, we have $\csc t=\frac{u}{\sqrt{u^2-2}}$ and $\cot t=\frac{\sqrt2}{\sqrt{u^2-2}}$. Thus, we end up with:
$$\begin{align}
-\frac{1}{\sqrt2}\ln\left|\frac{u+\sqrt2}{\sqrt{u^2-2}}\right|+C &= -\frac{1}{\sqrt2}\ln\left|\sqrt{\frac{(u+\sqrt2)^2}{u^2-2}}\right|+C\\
&= -\frac{1}{2\sqrt2}\ln\left|\frac{(u+\sqrt2)^2}{(u+\sqrt2)(u-\sqrt2)}\right|+C\\
&= -\frac{1}{2\sqrt2}\ln\left|\frac{u+\sqrt2}{u-\sqrt2}\right|+C\\
&= \frac{1}{2\sqrt2}\ln\left|\frac{u-\sqrt2}{u+\sqrt2}\right|+C
\end{align}$$
Messy, but possibly instructive. Partial fractions are cleaner than trig substitution, when you can use them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
generating function formula Suppose $\lambda$ is a young tableau, with one color boxes.
Prove the generating function,
$F(x) = \Sigma_{\lambda} x^{i} = \prod_{n \geq 1} \frac{1}{1-x^n} $, where $i$ is the number of boxes in the young tableau.
attempt:
$\Sigma_{\lambda} x^{i} = \prod_{n \geq 1} \frac{1}{1-x^n} = \frac{1}{1-x} \cdot \frac{1}{1-x^2} \cdot \frac{1}{1-x^3} \cdot \frac{1}{1-x^4}\cdots = (1 + x + x^2 + x^3 + \cdots)(1 + x^2 + x^4+ \cdots )(1 + x^3 + x^6 + x^9 + \cdots)( 1+ x^4 + x^8 + x^{12} \cdots ) = 1 + x + 2x^2 + 3x^3 + \cdots$ .
Here this one comes from when we have zero boxes so we would have $x^0$, and when there is only one box, then we have $x^1 $, similarly, there are $2x^2$ since there are two ways of having a young tableau with two boxes (horizontally, and vertically), and so on.
Can someone please help me?I am stuck after this. I would really appreciate it.
Thank you.
| Let $p(n,k)$ denote the number of partitions of $n$ with largest part at most $k$. Then, of the $p(n,k)$ such partitions, there are some that have a partition with part $k$ and some that don't. Those that don't have part $k$ are all precisely the partitions of $n$ with largest part at most $k-1$, so there are $p(n,k-1)$ of them. Those that do have part $k$ can be converted into a partition of largest part $k$ of $n-k$ by removing the $k$. Thus, we have the recurrence
$$
p(n,k) = p(n-k,k) + p(n,k-1).
$$
Defining the generating function $F_k(x) = \sum_{n\geq0}p(n,k)x^n$, the recurrence translates to
$$
F_k(x) = x^kF_k(x) + F_{k-1}(x)
$$
which rearranges to
$$
F_k(x) = \frac{F_{k-1}(x)}{1-x^k}.
$$
Can you proceed from here?
Edit:
Now, note that $F_1(x) = \sum_{n\geq0} x^n = \frac1{1-x}$ since there is always exactly one way to write $n$ as a partition with largest part $1$. Then, induction shows that $F_k(x) = \prod_{1\leq n\leq k}\frac1{1-x^n}$. It follows that for partitions of any large sized part, you take the desired infinite product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$?
$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$
I managed to get the answer as $1$ by standard methods of solving, learned from teachers, but my intuition says that the denominator of every term grows much faster than the numerator so limit must equal to zero.
Where is my mistake? Please explain very intuitively.
| Intuition should say:
the denominator grows with $n^2$, the numerator grows with $n$. However, the number of fractions also grows by $n$, so the total growth of the numerator is about $n^2$.
And that's where intuition stops. From here on, you go with logic and rigor, not intuition.
And it gets you to
$$\frac1{n^2} + \frac2{n^2}+\cdots + \frac{n}{n^2} = \frac{1+2+3+\cdots + n}{n^2} = \frac{\frac{n(n+1)}{2}}{n^2} = \frac{n^2+n}{2n^2}$$
and you find that the limit is $\frac12$ (not $1$!)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 11,
"answer_id": 0
} |
if $x+y+z=1$ then find the maximum of the form
Let $x>0$, $y>0$ and $z>0$ such that $x+y+z=1$. Find the maximal value of
$$x\sqrt{y}+y\sqrt{z}$$
I think $x=y=\dfrac{4}{9},z=\dfrac{1}{9}$ then the maximum $\dfrac{4}{9}$,but how to use AM-GM prove it?
| By AM-GM
$$x\sqrt{y}+y\sqrt{z}=\sqrt{y}\left(x+2\sqrt{\frac{y}{4}\cdot{z}}\right)\leq\sqrt{y}\left(x+\frac{y}{4}+z\right)=$$
$$=\sqrt{y}\left(1-y+\frac{y}{4}\right)=\frac{4}{9}-\frac{1}{36}\left(\left(3\sqrt{y}\right)^3+16-36\sqrt{y}\right)\leq$$
$$\leq\frac{4}{9}-\frac{1}{36}\left(3\sqrt[3]{\left(3\sqrt{y}\right)^3\cdot8^2}-36\sqrt{y}\right)=\frac{4}{9}.$$
The equality occurs for $x=y=\frac{4}{9}$ and $z=\frac{1}{9}$, which says that $\frac{4}{9}$ is the maximal value.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$x^2 + y^2 = 10 , \,\, \frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ I couldn't solve the above equation. Below, I describe my attempt at solving it.
$$x^2 + y^2 = 10 \tag{1}$$
$$\frac{1}{x} + \frac{1}{y} = \frac{4}{3} \tag{2}$$
Make the denominator common in the RHS of $(2)$.
$$\frac{y + x}{xy} = \frac{4}{3} \tag{2.1}$$
Multiply $(2.1)$ through by $3$:
$$\frac{3(y + x)}{xy} = 4 \tag{2.2}$$
Let $y = mx$
Substitute $y = mx$ into $(1)$ and $(2.2)$:
$$x^2 + m^2x^2 = 10 \tag{1.1}$$
$$\frac{3(mx + x)}{mx^2} = 4$$
Factorise:
$$\frac{3x(m + 1)}{x(mx)} = 4$$
$$\frac{3(m + 1)}{(mx)} = 4$$
Cross multiply:
$$3(m+1) = 4mx$$
Collect like terms:
$$4mx - 3m = 3$$
Factorise:
$$m(4x - 3) = 3$$
Divide through by $(4x - 3)$:
$$m = \frac{3}{4x - 3} \tag{3}$$
Substitute $(3)$ into $(1.1)$
$$x^2 + \left(\frac{3}{4x - 3}\right)^2x^2 = 10$$
$$x^2 + \frac{9x^2}{16x^2 - 24x + 9} = 10$$
Multiply through by ${16x^2 - 24x + 9}$:
$$16x^2 - 24x^3 + 9x^2 + 9x^2 = 10(16x^2 - 24x + 9)$$
Divide through by $2$:
$$8x^4 - 12x^3 + 9x^2 = 80x^2 - 120x + 45$$
$$8x^4 - 12x^3 - 71x^2 +120x - 45 = 0$$.
$x = 0$ is not a solution of the above equation, so divide through by $x^2$:
$$8x^2 - 12x - 71 + \frac{120}{x} - \frac{45}{x^2} = 0 \tag{1.2}$$
Let $v = x - \frac{k}{x}$.
$$v^2 = x^2 - 2k + \frac{k^2}{x^2}$$
Rewriting $(1.2)$:
$$\left(8x^2 - 71 - \frac{45}{x^2}\right) + \left(-12x + \frac{120}{x}\right)$$
$$\left(8x^2 - 71 - \frac{45}{x^2}\right) + \left(-12(x - \frac{10}{x}\right) \tag{1.3}$$
Putting $k = 10$, works for $v$, but not for $v^2$.
I don't know any other approach to solving the equation from $(1.3)$.
| from the second equation we get
$$x+y=\frac{4}{3}xy$$ and $$x,y\ne 0$$
solving this equation for $y$ we get $$y=\frac{x}{\frac{4}{3}x-1}$$
this can we insert in the first equation
$$x^2+\left(\frac{x}{\frac{4}{3}x-1}\right)^2=10$$
simplifying and factorizing we get
$$2\, \left( x-1 \right) \left( x-3 \right) \left( 8\,{x}^{2}+20\,x-15
\right)
=0$$
can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
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