Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove for a,b,c,d > 0 $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \ge \frac{1}{3}$ where $ab + bc + cd + da = 1$ The task requires for Cauchy-Schwarz inequality to be used.
My attempt: Using Titu's Lemma (direct consequence of the inequality), I got:
$\frac{a^4}{a(b+c+d)} + \frac{b^4}{b(a+c+d)} + \frac{c^4}{c(a+b+d)} + \frac{d^4}{d(a+b+c)} \ge \frac{(a^2 + b^2 + c^2 + d^2)^2}{a(b+c+d) \ + \ b(a+c+d) \ + \ c(a+b+d) \ + \ d(a+b+c)}$
Using the condition that $ab + bc + cd + da = 1$, the RHS results in $\frac{(a^2+b^2+c^2+d^2)^2}{2+2(ac+bd)}$.
Nothing more seems to help. Opening up the numerator or using AM-GM on it doesn't work and I have no idea on how to turn this result in a $\frac{1}{3}$ fraction.
| \begin{align*}
\sum \frac{a^3}{b+c+d} &\geq \frac{4(a^3+b^3+c^3+d^3)}{3(a+b+c+d)}\\
&\geq \frac{(a+c)^3+(b+d)^3}{3(a+b+c+d)}\\
&=\frac{1}{3}((a+c)^2+(b+d)^2-(a+c)(b+d))\\
&\geq \frac{(a+c)(b+d)}{3}\\
&=\frac{1}{3}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $a+b+c \geq \frac{3+\sqrt{7}}{2}$ for $a,b,c \in [0, \frac32]$ with $a^2+b^2+c^2+abc=4$
Let $a,b,c \in \left[0;\frac{3}{2}\right]$ and $a^2+b^2+c^2+abc=4$. Prove that
$$a+b+c \geq \dfrac{3+\sqrt{7}}{2}.$$
Source: This is a math problem that my teacher gave me $3$ months ago (the submission deadline has expired). My teacher wrote a book and sent me to test the difficulty of the problem.
My attempt: I have converted $a$ to $b,c$, used to trigonometric conversion, but all failed.
Related problem (the same source, with the same conditions): https://artofproblemsolving.com/community/c6h2975620p26673082
Please give me a suggestion! Thank you!
| It suffices to prove that
$$2a + 2b + 2c - 3 \ge \sqrt 7.$$
Using $(a + b + c)^2 \ge a^2 + b^2 + c^2$ and $(a + b + c)^3/27 \ge abc$, we have
$(a + b + c)^2 + (a + b + c)^3/27 \ge 4$
which results in $a + b + c > 3/2$ or
$2a + 2b + 2c - 3 > 0$.
Thus, it suffices to prove that
$$(2a + 2b + 2c - 3)^2 - 7 \ge 0.$$
We have
\begin{align*}
&(2a + 2b + 2c - 3)^2 - 7 - 4(a^2 + b^2 + c^2 + abc - 4)\\
={}& (-4ab + 8a + 8b - 12)c + 2(3-2a)(3-2b)\\
\ge{}& 0. \tag{1}
\end{align*}
(Note: If $-4ab + 8a + 8b - 12 \ge 0 $, clearly (1) is true.
If $-4ab + 8a + 8b - 12 < 0$, we have
$(-4ab + 8a + 8b - 12)c + 2(3-2a)(3-2b)$
$\ge (-4ab + 8a + 8b - 12)\cdot \frac32 + 2(3-2a)(3-2b)$
$ = 2ab \ge 0$.)
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factor $1-64m^6$ under $\mathbb{Q}$ Factor $1-64m^6$ under $\mathbb{Q}$. We can employ the well known formula: $a^3-b^3 = (a-b)\cdot(a^2+ab+b^2)$.
$1-64m^6 = (1)^3-(4m^2)^3 = (1-4m^2)(1+4m^2+16m^4)=(1+2m)(1-2m)(1+4m^2+16m^4)$.
However, the answer is provided as $(1+2m)(1-2m)(1-2m+4m^2)\cdot(1+2m+4m^2)$.
I cannot understand how it is possible to simplify $(1+4m^2+16m^4)$ to $(1-2m+4m^2)(1+2m+4m^2)$?
| There is a simpler approach. We could first factor $1 - 64m^6$ as a difference of squares, then factoring the resulting sum of cubes and difference of cubes.
\begin{align*}
1 - 64m^6 & = (1 + 8m^3)(1 - 8m^3)\\
& = (1 + 2m)(1 - 2m + 4m^2)(1 - 2m)(1 + 2m + 4m^2)
\end{align*}
As for your method of first factoring $1 - 64m^3$ as a difference of cubes, we can add and subtract $4m^2$ to the term $1 + 4m^2 + 16m^4$ to form a difference of squares.
\begin{align*}
1 - 64m^6 & = (1 - 4m^2)(1 + 4m^2 + 16m^4)\\
& = (1 - 4m^2)(1 + 8m^2 + 16m^4 - 4m^2)\\
& = (1 - 4m^2)[(1 + 4m^2)^2 - 4m^2]\\
& = (1 + 2m)(1 - 2m)(1 + 4m^2 + 2m)(1 + 4m^2 - 2m)\\
& = (1 + 2m)(1 - 2m)(1 + 2m + 4m^2)(1 - 2m + 4m^2)
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+c\sqrt{b}+a\sqrt{c})^2+18}{a+b+c} \geq 9$.
I don't know what to do next, any help is appreciated.
Thanks!
| Write given expression as $A+B$ where $$A = \frac{ab^2}{a+c} +\frac{bc^2}{b+a} +\frac{ca^2}{c+b}$$
and
$$B= \frac{2}{a+c} + \frac{2}{b+a} + \frac{2}{c+b}$$
By AM-GM inequality we have
$$ A\geq \frac{3abc}{\sqrt[3]{(a+c)(c+b)(b+a)}}$$
and $$ B\geq \frac{6}{\sqrt[3]{(a+c)(c+b)(b+a)}}$$
Further again by AM-GM we have $$\sqrt[3]{(a+c)(c+b)(b+a)}\leq {2(a+b+c)\over 3}$$
So $$A\geq {9abc\over 2(a+b+c)}\;\;\;{\rm and}\;\;\; B\geq {9\over a+b+c}$$
We are left to check if $${9abc\over 2(a+b+c)} +{9\over a+b+c} \geq {9\over 2}$$ i.e. $$\boxed{abc+2\geq a+b+c}$$ is true.
Rewrite it like $$(ab-1)c+2\geq a+b$$
and notice that linear function $f(c) = (ab-1)c+2$ is increasing so $f(c)\geq f(1)$ and we are left to check if $f(1)\geq a+b$ i.e. if $ab+1\geq a+b$ which is clearly true since $$(a-1)(b-1)\geq 0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$ I was only able to observe that:
$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$
This hints at telescoping, but I would need an $r$ term in the numerator.
The original question was
Evaluate $\sum_{r=1}^{\infty} \dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)}$
I was able to simplify it to the following:
$\dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)} = \dfrac{(r^4 + r^3 - r^2 - r)}{(r^4 + r^2 + 1)(r^2 + r)} + \dfrac{3r^2+r+1}{\{(r+1)(r^2 + r + 1)\}\{r(r^2 - r + 1)\}} = \dfrac{r^2 - 1}{r^4 + r^2 + 1} + \left[\dfrac{1}{r(r^2 - r + 1)} - \dfrac{1}{(r+1)(r^2 + r + 1)}\right]$
The second term can be evaluated using telescoping, and the first term is what this post is asking for.
Any other ways of solving the original question are also welcome.
| $$r^4+r^2+1=(r^2+1)^2-r^2=(r^2+r+1)(r^2-r+1)$$
Observe that if $f(n)=n^2+n+1, f(r-1)=(r-1)^2-(r-1)+1=r^2-r+1$
If $g(n)=an+b, g(n-1)=a(n-1)+b=an+b-a$
$$\text{Let }\dfrac{r^2-1}{r^4+r^2+1}=\dfrac{g(r)}{f(r)}-\dfrac{g(r-1)}{f(r-1)}$$ $$\text{so that the partial sum becomes }\sum_{r=1}^n\dfrac{r^2-1}{r^4+r^2+1}=\dfrac{g(n)}{f(n)}-\dfrac{g(0)}{f(0)}$$
$$\text{Now, }\lim_{n\to\infty}\dfrac{g(n)}{f(n)}=\lim_{n\to\infty}\dfrac{an+b}{n^2+n+1}=\lim_{n\to\infty}\dfrac{\dfrac an+\dfrac b{n^2}}{1+\dfrac1n+\dfrac1{n^2}}=0\text{ for finite a,b}$$
$$\text{ and } \dfrac{g(0)}{f(0)}=?\text{ so we don't need the actual value of }a$$
$$\begin{align}\implies r^2-1=(ar+b)(r^2-r+1)-(ar+b-a)(r^2+r+1)=-ar^2+(a-2b)r+a\end{align}$$
$$\implies -a=1\iff a=-1, a-2b=0\iff b=\dfrac a2=-\dfrac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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sum of 100 terms of logarithmic expression
Calculate value of
$\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}\bigg)$
My try :: $\displaystyle x^4+4y^4$
$=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$
So sum $\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{1+4(2k+1)^4}{1+4(2k)^4}\bigg)$
$\displaystyle =\sum^{100}_{k=1}\ln\bigg[\frac{(1+2(2k+1)+2(2k+1)^2)(1-2(2k+1)+2(2k+1)^2)}{(1+2(2k)+2(2k)^2)(1-2(2k)+2(2k)^2)}\bigg]$
How can I decompose that complex expression into partial fractions?
| $\ln (16241)$. Work out the first few cases, use $\ln a + \ln b = \ln ab$ and then prove by induction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simplifying the surd $\sqrt{3 - \sqrt{8}}$ Recently I was solving a math problem and I came across the following ( Only part of the problem ):
$$\sqrt{3 - \sqrt{8}}$$
Here's is what I did to simplify the above:
$$(a-b)^2=a^2+b^2-2ab$$
$$\sqrt{1+2-2\sqrt{2}}$$
$$\sqrt{(1-\sqrt{2})^2}$$
$$=1-\sqrt{2}$$
When you expand $(1-\sqrt{2})^2$ you do get $3 - \sqrt{8}$. However, I found that upon the expansion of $(\sqrt{2}-1)^2$ I also arrive at the same answer as before $3 - \sqrt{8}$.
Yet the in the solution of the problem I was doing it required me to use the latter $(\sqrt{2}-1)^2$ to find the correct answer. $(1-\sqrt{2})^2$ did not help to solve the problem. Why is this the case? Why does the latter only hold true?
I suspect this might have something to do with the square root function only returning a positive value. Might this be the case? If so, could someone explain why a square root function can only return a positive value. My working above seems to give a negative answer ( As $\sqrt{2}>1 )$ in the square root function, but I can't seem to find what I'm doing wrong.
| We have $\sqrt {3-\sqrt 8 }=\sqrt {3-2\sqrt 2 }.$ We ask whether there exist rational $a,b$ such that that $$3-2\sqrt 2=(a+b\sqrt 2)^2=(a^2+2b^2)+2ab\sqrt 2.$$ Since $a,b\in\Bbb Q$, this requires $$ (\bullet) \quad a^2+2/a^2=3 \land -2ab=2.$$ So $b=-1/a$ from the 2nd equation of $(\bullet).$ So $3=a^2+2b^2=a^2+2/a^2.$
Let $a^2=x.$ Then $3=x+2/x.$ So $x^2-3x+2=0.$ Take the solution $x=1.$ Now if $a=-1$ then $a^2=1=x,$ and with $b=-1/a=1,$ we see that $(\bullet)$ must hold.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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How do I solve this, do I have to continue calculating with rational fractions? $$\int \frac{\sin^2(x)dx}{\sin(x)+2\cos(x)}$$
I tried to use different substitutions such as $t=\cos(x)$, $t=\sin(x)$, $t=\tan(x)$, and after expressing $\sin$ and $\cos$ through $\tan(\frac{x}{2})$, I've got $ \int -4 \frac{t^2dt}{(1+t^2)^2(t^2-t-1)}.$
Rational fractions didn’t work.
| We have
$$(\sin x+2\cos x)(2\cos x-\sin x)=4-5\sin^2x=5\cos^2x-1$$
Therefore
$${\sin^2x\over \sin x+2\cos x} ={\sin^2x(2\cos x-\sin x)\over (\sin x+2\cos x)(2\cos x-\sin x)}\\ =
2{\sin^2x\over 4-5\sin^2x}\cos x-{1-\cos^2x\over 5\cos^2x -1}\sin x$$
Hence substituting $s=\sin x$ and $t=\cos x,$ respectively, gives
$$\int {\sin^2x\over \sin x+2\cos x}\,dx=2\int{s^2\over 4-5s^2}\,ds +\int{1-t^2\over 5t^2-1}\,dt \\
=-{2\over 5}s+{8\over 5}\int {1\over 4-5s^2}\,ds-{1\over 5}t+{4\over 5}\int{1\over 5t^2-1}\,dt $$
The remaining integrals can be evaluated by easy partial fraction decompositions. At the end we have to get back to $x=\arcsin s$ and $x=\arccos t,$ respectively.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^\pi \left(\frac{\sin{2x}\sin{3x}\sin{5x}\sin{30x}}{\sin{x}\sin{6x}\sin{10x}\sin{15x}}\right)^2 dx$ (from MIT Integration Bee 2023) This is from the final round of MIT Integration Bee 2023.
$$\int_0^\pi \left(\frac{\sin{2x}\sin{3x}\sin{5x}\sin{30x}}{\sin{x}\sin{6x}\sin{10x}\sin{15x}}\right)^2 dx$$
The given answer is $7\pi$.
I found a way to do it using contour integrals (see answer below), but this is not a calculation I can finish within 4 minutes (the time limit in the competition). I am still looking for other elegant methods, possibly without using contour integrals.
| From @Archisman Panigrahi,
\begin{eqnarray}
I&=&\frac{1}{2} \int_0^{2\pi} \left(\frac{\cos{x}\cos{15x}}{\cos{3x}\cos{5x}}\right)^2 dx= \frac{1}{2} \int_0^{2\pi} \left(\frac{\cos{14x}+\cos{16x}}{\cos{2x}+\cos{8x}}\right)^2 dx\\
&=&\frac{1}{4} \int_0^{4\pi} \left(\frac{\cos{7x}+\cos{8x}}{\cos{x}+\cos{4x}}\right)^2 dx=\frac{1}{2} \int_0^{2\pi} \left(\frac{\cos{7x}+\cos{8x}}{\cos{x}+\cos{4x}}\right)^2 dx.
\end{eqnarray}
Using
$$ \cos\theta=\frac12(z+z^{-1}),z=e^{i\theta} $$
one has
\begin{eqnarray}
&&\frac{\cos{7x}+\cos{8x}}{\cos{x}+\cos{4x}}=\frac{z^7+z^{-7}+z^8+z^{-8}}{z+z^{-1}+z^4+z^{-4}}\\
&=&\frac{1+z+z^{15}+z^{16}}{z^4(1+z^3+z^5+z^8)}=-1-(z+z^{-1})+(z^3+z^{-3})+(z^4+z^{-4})\\
&=&-1-2\cos x+2\cos3x+2\cos4x
\end{eqnarray}
and hence
$$I=\frac{1}{2} \int_0^{2\pi} \left(\frac{\cos{7x}+\cos{8x}}{\cos{x}+\cos{4x}}\right)^2 dx=\frac12\int_0^{2\pi}(-1-2\cos x+2\cos3x+2\cos4x)^2=7\pi.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using Maclaurin series to solve $y'=xe^x$. I am trying to solve a differential equation using the Maclaurin series. The differential equation is
$$y'=xe^x$$
My solution 1:
$xe^x$ expressed in sigma notation is: $\sum_{k=0}^\infty\frac{x^{k+1}}{k!}$ and $y'=\sum_{k=0}^\infty(k+1)a_{k+1}x^k$. Equating them will lead to
$$\sum_{k=0}^\infty(k+1)a_{k+1}x^k=\sum_{k=0}^\infty\frac{x^{k+1}}{k!}$$
I was trying to comparing the coefficients, so as to obtain a recurrence relation. But I couldn't do it because the $x$ on both sides of the equation has different exponents ($k$ and $k+1$) respectively.
My solution 2:
I integrate the differential equation to obtain $y=xe^x-e^x+c$ which in sigma notation corresponds to $$y=c+\sum_{k=0}^\infty\frac{x^{k+1}}{k!}-\sum_{k=0}^\infty\frac{x^k}{k!}$$ which can be samplified as
$$y=c+\sum_{k=0}^\infty\frac{x^k(x-1)}{k!}$$
Question:
Am I going in the right direction? In both solutions, I am stuck and am unable to arrive at the textbook solution of
$$y=c+\sum_{k=0}^\infty\frac{x^{k+2}}{(k+2)k!}$$
Thank you in advance.
| Method 1:
LHS of the series is given by
$$\begin{aligned} \sum_{k=0}^\infty (k+1)a_{k+1}x^k &= a_1 + \sum_{k=1}^{\infty}(k+1)a_{k+1}x^{k} \\
&= a_1 + \sum_{k=0}^{\infty}(k+2)a_{k+2}x^{k+1}\end{aligned}$$
Note that we have replaced $k$ by $k+1$ in the second equality.
Equality of power series implies that $a_1 = 0$ and
$$(k+2)a_{k+2} = \frac{1}{k!}$$
for all integers $k \geq 0$.
This implies that $a_{k} = \frac{1}{k(k-2)!}$
for $k \geq 2$, and the series solution is given by
$$\begin{aligned} y(x) &= \sum_{k=0}^\infty a_k x^k \\ &= a_0 + \sum_{k=2}^\infty \frac{x^k}{(k-2)!k} \\
&= a_0 + \sum_{k=0}^\infty \frac{x^{k+2}}{k!(k+2)}. \end{aligned}$$
The sum starts from $k = 2$ since we have the general formula for $k \geq 2$. Furthermore, we have deduced that $a_1 = 0$ and we have no information on $a_0$ (which thus serves as an arbitrary constant to be determined).
Method 2:
Starting from your derived solution, we see that
$$y = C + \sum_{k=0}^\infty \frac{x^{k+1}}{k!} - \sum_{k=0}^\infty \frac{x^{k}}{k!}.$$
In the second summation term, the $k = 0$ term is yet another constant, so we can group it with the $C$ to obtain another arbitrary constant $C'$ and simplify it as follows:
$$\begin{aligned} y &= C' + \sum_{k=0}^\infty \frac{x^{k+1}}{k!} - \sum_{k=1}^\infty \frac{x^{k}}{k!} \\ &=C' + \sum_{k=0}^\infty \left(\frac{1}{k!}- \frac{1}{(k+1)!}\right) x^{k+1}. \end{aligned}$$
At $k = 0$, note that $k! = 0! = 1 = 1! = (k+1)!$, so the sum starts from 1 instead. We thus have
$$\begin{aligned} y &= C' + \sum_{k=1}^\infty \left(\frac{1}{k!}- \frac{1}{(k+1)!}\right) x^{k+1} \\ &= C' + \sum_{k=0}^\infty \left(\frac{1}{(k+1)!}- \frac{1}{(k+2)!}\right) x^{k+2} \\ &= C' + \sum_{k=0}^\infty \left(\frac{k+2}{(k+2)!}- \frac{1}{(k+2)!}\right) x^{k+2} \\ &= C' + \sum_{k=0}^\infty \left(\frac{k+1}{(k+2)!}\right) x^{k+2} \\
&= C' + \sum_{k=0}^\infty \left(\frac{k+1}{k!(k+1)(k+2)}\right) x^{k+2} \\
&= C' + \sum_{k=0}^\infty \left(\frac{1}{k!(k+2)}\right) x^{k+2} \end{aligned}$$
and thus you have the same solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac a{a+2b}+\frac bc+\frac{c^2}{\left(c+2a\right)^2 }\ge1$
(sqing) Let $a$, $b$, $c>0$. Prove that $\dfrac a{a+2b}+\dfrac bc+\dfrac{c^2}{\left(c+2a\right)^2 }\ge1$. Source (unsolved)
In this inequality, $b$ appears the least frequent, so I want to deal with it. I took $a$, $c$ as parameters and derived $b$, we have
\[\frac{\text d}{\text db}~\left(\dfrac a{a+2b}+\dfrac bc\right)=\frac1c-\frac{2a}{(a+2b)^2}.\tag1\label1\]
When $2c>a$, $\eqref1=0$ gives $b=\dfrac{\sqrt{2ac}-a}2$. When $2c\le a$, $\eqref1>0$ is true for all $b$, so $b\to0$ shall minimize the left side.
But plugging back makes the expressions complicated. What should I do?
| Using Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
&\frac{a}{a + 2b} + \frac{b}{c} + \frac{c^2}{(c + 2a)^2} - 1\\
\ge{}& \frac{(a + b)^2}{a(a + 2b) + bc} + \frac{c^2}{(c + 2a)^2} - 1 \\
={}& \frac{(2ab - ac + bc)^2}{(a^2 + 2ab + bc)(c + 2a)^2}\\
\ge{}& 0
\end{align*}
with equality if $a:b:c = 2:1:4$.
We are done.
| {
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Find out limit of the following question let, $f(x) =x^\frac{1}{3}$ be a diffrentiable function on $ (0, \infty).$
Given that
$$\frac{f(3+h) -f(3)}{h}=f'(3+\theta(h)h)$$
Then find out
$\lim_{h\to 0+} \theta(h) =? $
Since, $f$ is diffrentiable at $3$ , I think limit must be $0$ as it tends to $f'(3)$
Please help me
| You can obtain explicitly that
$$\theta(h)=\frac{1}{h}\left(\left(3\frac{\sqrt[3]{3+h}-\sqrt[3]{3}}{h}\right)^{-3/2}-3\right)$$
Now, by noticing that $h=(\sqrt[3]{3+h}-\sqrt[3]{3})(\sqrt[3]{(3+h)^2}+\sqrt[3]{3(3+h)}+\sqrt[3]{9})$, compute the desired limit as
$$\lim_{h\to0^+}\theta(h)=\lim_{h\to0^+}\frac{1}{h}\left(\left(3\frac{\sqrt[3]{3+h}-\sqrt[3]{3}}{h}\right)^{-3/2}-3\right) = \lim_{h\to0^+}\frac{1}{h}\left(\left(\frac{\sqrt[3]{(3+h)^2}+\sqrt[3]{3(3+h)}+\sqrt[3]{9}}{3}\right)^{3/2}-3\right)$$
and now treat this limit as a derivative to finally get
$$\lim_{h\to0^+}\theta(h)=\left(\dfrac{\left(\frac{2}{3\sqrt[3]{3+h}}+\frac{1}{
\sqrt[3]{9(3+h)^2}}\right)\sqrt{\sqrt[3]{(3+h)^2}+\sqrt[3]{3(3+h)}+\sqrt[3]{9}}}{2\sqrt{3}}
\right)_{h=0} = \frac{\left(\frac{2}{3\sqrt[3]{3}}+\frac{1}{
\sqrt[3]{9\cdot 9}}\right)\sqrt{3\sqrt[3]{9}}}{2\sqrt{3}}=\frac{1}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate: $\int\frac{3e^{2x}-2e^x}{e^{2x}+2e^x-8}dx$ Given $\int\frac{3e^{2x}-2e^x}{e^{2x}+2e^x-8}dx$
My attempt- Substitute $u=e^x \Rightarrow du=dx.e^x$
The integral becomes
$$\int \frac{3u-2}{u^2+2u-8}du $$
Now, the numerator is a linear function and the denominator is a quadratic function. We can rewrite the numerator as follows
$$\frac{g(x)}{f(x)}=\frac{f^{'}(x)}{f(x)}+\frac{C}{f(x)}$$
Thus the integral may be rewritten as
$$\int \frac{\frac{3}{2}(2u+2)-5}{u^2+2u-8}du$$
Now, break the integral into $2$ sub-integrals
$\Rightarrow \int \frac{\frac{3}{2}(2u+2)}{u^2+2u-8}du-\int \frac{5}{u^2+2u-8}du$
$\Rightarrow \frac{3}{2}\int \frac{(2u+2)}{u^2+2u-8}du-\int \frac{5}{u^2+2u-8}du$
$\Rightarrow \frac{3}{2}\log(u^2+2u-8)-5\int \frac{1}{u^2+2u-8}du$
$\Rightarrow \frac{3}{2}\log(u^2+2u-8)-5\int \frac{1}{(u+1)^2-3^2}du$
$\Rightarrow \frac{3}{2}\log(u^2+2u-8)-\frac{5}{9} \log(\frac{u-2}{u+4})+C$
$\Rightarrow \frac{3}{2}\log(e^{2x}+2e^x-8)-\frac{5}{9} \log(\frac{e^x-2}{e^x+4})+C$
$\Rightarrow \log(\frac{[{e^{2x}+2e^x-8}]^{\frac{3}{2}} } {{[\frac{e^x-2}{e^x+4}}]^{\frac{5}{9}}})+C$
Is my solution correct? Are there other ways to solve such integrals?
Edit: Correction $\Rightarrow \log(\frac{[{e^{2x}+2e^x-8}]^{\frac{3}{2}} } {{[\frac{e^x-2}{e^x+4}}]^{\frac{5}{6}}})+C$
| No, it is not correct, since$$\int\frac1{(u+1)^2-3^2}=\frac1{\color{red}6}\log\left(\frac{u-2}{u+4}\right).$$So, your approach should lead to$$\log\left(\frac{\left(e^{2x}+2e^x-8\right)^{\frac32}}{\left(\frac{e^x-2}{e^x+4}\right)^{\frac5{\color{red}6}}}\right)+C.$$But there is a simpler approach: using the fact that$$\frac{3u-2}{u^2+2u-8}=\frac{2}{3(u-2)}+\frac{7}{3(u+4)},$$you get that$$\int\frac{3u-2}{u^2+2u-8}\,\mathrm du=\frac23\log\left(\left|u-2\right|\right)+\frac73\log\left(\left|u+4\right|\right).$$
| {
"language": "en",
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Formula for $\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ b} \ln \left(a^2+b^2+a b \sqrt{2}\right) $ In my post, I had proved that $$
\int_0^{\infty} \frac{\ln \left(x^2+a^2\right)}{b^2+x^2} d x=\frac{ \pi}{b} \ln (a+b) \tag*{(*)}
$$
To go further, I guess that
$$\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ |b|} \ln \left(a^2+b^2+|a|| b| \sqrt{2}\right) $$
Proof:
For $a,b>0$,
Using $\ln \left(a^2+b^2\right)=2 Re(\ln (a+b i))$, we can reduce the power $4$ to $2$.
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x
& = 2\int_{0}^{\infty} \frac {Re\left[\ln \left(x^2+a ^2i\right)\right]}{b^2+x^2} d x \\
& =2 Re\left(\int_0^{\infty} \frac{\ln \left(x^2+\left[\left(\frac{1+i}{\sqrt{2}}\right) a\right]^2\right)}{b^2+x^2} d x\right)
\end{aligned}
$$
Using (*), we have $$
\begin{aligned}\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x&=2 Re\left[\frac{\pi}{b} \ln \left(\frac{1+i}{\sqrt{2}} a+b\right)\right] \\&=\frac{2 \pi}{b} R e\left[\ln \left(\frac{a}{\sqrt{2}}+b+\frac{a}{\sqrt{2}}i\right)\right] \\&= \boxed{\frac{\pi}{b} \ln \left(a^2+b^2+a b \sqrt{2}\right)}\end{aligned}
$$
In general, for any $a, b \in \mathbb{R} \backslash\{0\}$, replacing $a$ and $b$ by $|a|$ and $|b|$ yields
$$\boxed{\int_0^{\infty} \frac{\ln \left(x^4+a^4\right)}{b^2+x^2} d x =\frac{\pi}{ |b|} \ln \left(a^2+b^2+|a|| b| \sqrt{2}\right) }$$
For example, $$
\int_0^{\infty} \frac{\ln \left(x^4+16\right)}{9+x^2} d x= \frac{\pi}{3} \ln (13+6 \sqrt{2})
$$
Comments and alternative methods are highly appreciated.
| An option using a semicircular contour in the upper half-plane: due to its even integrand, if $a,\,b>0$ your integral is$$\begin{align}\Re\int_{\Bbb R}\frac{\ln(x^2-a^2+aix\sqrt{2})}{b^2+x^2}dx=\Re\left[2\pi i\lim_{z\to bi}\frac{\ln(z^2-a^2+aiz\sqrt{2})}{z+bi}\right]=\frac{\pi}{b}\ln(a^2+b^2+ab\sqrt{2}).\end{align}$$The trick is to multiply the linear factors of $x^4+a^4$ whose roots have negative imaginary parts, so the integrand is meromorphic inside the contour. In the above calculation, the pole makes the logarithm's argument negative, so it's an especially simple $\Re\ln z=\ln|z|$ calculation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve the equation $\cos\frac{2x}{3}+\sqrt3\sin\frac{2x}{3}=\frac{8}{3+\cos{4x}}$ Solve the equation $$\cos\dfrac{2x}{3}+\sqrt3\sin\dfrac{2x}{3}=\dfrac{8}{3+\cos{4x}}$$
Let's divide both sides of the equation by $2$ to get $$\dfrac12\cos\dfrac{2x}{3}+\dfrac{\sqrt3}{2}\sin\dfrac{2x}{3}=\dfrac{4}{3+\cos{4x}}\\\iff \sin\left(\dfrac{\pi}{6}+\dfrac{2x}{3}\right)=\dfrac{4}{3+\cos4x}$$ The RHS appears to be $\ge1$. Then isn't then the equation equivalent to $$\sin\left(\dfrac{\pi}{6}+\dfrac{2x}{3}\right)=1$$ which gives $x=\dfrac{\pi}{2}+3k\pi,k\in\mathbb{Z}$?
Why in the authors' solution is the equation said to be equivalent to the system $$\begin{cases}\sin\left(\dfrac{\pi}{6}+\dfrac{2x}{3}\right)=1\\\cos{4x}=1\end{cases}$$ and then it's said that all of the solutions of the first equation satisfy the second. Why is the second necessary at all?
| To begin with, I would recommend you to notice that
\begin{align*}
-1 \leq \cos(4x) \leq 1 & \Longleftrightarrow 2 \leq \cos(4x) + 3 \leq 4\\\\
& \Longleftrightarrow \frac{1}{4} \leq \frac{1}{\cos(4x) + 3} \leq \frac{1}{2}\\\\
& \Longleftrightarrow 1 \leq \frac{4}{\cos(4x) + 3} \leq 2
\end{align*}
Hence the LHS equals the RHS iff one has that
\begin{align*}
1 \leq \frac{4}{\cos(4x) + 3} = \sin\left(\frac{2x}{3} + \frac{\pi}{6}\right) \leq 1
\end{align*}
Hopefully this helps!
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $\csc x = a+bi$ defined? A question in my assignment asks to find which equation can have its roots $\sec^2 x, \csc^2 x$ (no restrictions in $x$ were given).
The equations were quadratic ($x^2+bx+c$) and the one which satisfies $-b=c$ can be the required equations.
There were two which satisfies this,
$x^2-3x+3=0, x^2-9x+9=0$
The second one is absolutely correct but the first in the first one $\csc^2x$ is in the form of $a+bi$. And so first one wasn't the correct answer.
But I am having doubt in the answer, as by Euler's formula,
$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$
And here putting a suitable complex value of $x$, $\sin x$ can have a value in form of $a+bi$ and so $\csc^2 x$.
So, am I correct? Can we put a complex value in trigonometric functions or can they have a complex value? And also tell what if $\sin x=2$?
| If you want to solve the equation $$\csc (x+i y)=a+i\,b$$ expand first the cosecant, go to hyperbolic functions, use the conjugate, separate the real and imaginary parts to end with
$$\frac{\sin (x) \cosh (y)}{\cos ^2(x) \sinh ^2(y)+\sin
^2(x) \cosh ^2(y)}=a \tag 1$$
$$\frac{\cos (x) \sinh (y)}{\cos ^2(x) \sinh ^2(y)+\sin
^2(x) \cosh ^2(y)}=-b \tag 2$$ Computing the ratio
$$\tan (x) \coth (y)=-\frac ab\quad \implies \quad \color{blue}{y=-\coth ^{-1}\left(\frac{a }{b}\cot (x)\right)} \tag 3$$ Replacing $y$ in $(1)$ give the simple
$$\csc (x) \sqrt{1-\frac{b^2 }{a^2}\tan
^2(x)}=\frac{a^2+b^2}{a}$$ Squaring
$$\csc ^2(x)-\frac{b^2 }{a^2}\sec ^2(x)=\left(\frac{a^2+b^2}{a}\right)^2$$ Cross multiply by $\sin^2(x)\,\cos^2(x)$
$$\cos ^2(x)-\frac{b^2 }{a^2}\sin ^2(x)=\left(\frac{a^2+b^2}{a}\right)^2\sin^2(x)\,\cos^2(x)$$ Let $x=\sin^{-1}(t)$ to obtain
$$t^4-\frac{a^2+b^2+1}{a^2+b^2} t^2+\frac{a^2}{\left(a^2+b^2\right)^2}=0$$ which is just a quadratic in $t^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the coordinates in an isosceles triangle if the triangle it self is in positive axis A at $(45,10)$, B at $(10,20)$, $AB=AC$ and angle $C=20$ degree find the coordinates of $C$.suggest the formula so i can write code in Perl.
| We have $A(45,10),B(10,20),C(x_c,y_c)$.
$AB=AC=\sqrt{(45-10)^{2}+(10-20)^{2}}=5\sqrt{53}$
$C=B=20^{%
%TCIMACRO{\U{ba}}%
%BeginExpansion
{{}^o}%
%EndExpansion
}=\pi /9$ rad
$A=180^{%
%TCIMACRO{\U{ba}}%
%BeginExpansion
{{}^o}%
%EndExpansion
}-40^{%
%TCIMACRO{\U{ba}}%
%BeginExpansion
{{}^o}%
%EndExpansion
}=140^{%
%TCIMACRO{\U{ba}}%
%BeginExpansion
{{}^o}%
%EndExpansion
}=7\pi /9$ rad
Let's make the following change of variables: $x=X+45,y=Y+10$ (translation of axes). Then $A$ becomes the origin of the $XY$ referential.
The vector $\overrightarrow{AB}$ can be written in this $XY$ referential as
$\overrightarrow{AB}=(5\sqrt{53}\cos \left( \pi -\arctan \frac{2}{7}\right)
,5\sqrt{53}\sin \left( \pi -\arctan \frac{2}{7}\right) )=(-35,10)$
and the vector $\overrightarrow{AC}$ as
$\overrightarrow{AC}=(5\sqrt{53}\cos \left( \pi -\frac{7\pi }{9}-\arctan
\frac{2}{7}\right) ,5\sqrt{53}\sin \left( \pi -\frac{7\pi }{9}-\arctan \frac{%
2}{7}\right) )$
Therefore in the original referencial $xy$, we have
$x_{C}=5\sqrt{53}\cos \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi
\right) +45\approx 78.239$
$y_{C}=5\sqrt{53}\sin \left( -\arctan \frac{2}{7}-\frac{7\pi }{9}+\pi
\right) +10\approx 24.837$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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} |
How closely can we estimate $\sum_{i=0}^n \sqrt{i}$ By looking at an integral and bounding the error?
| This was an interesting challenge, to try and come up with an estimate for this sum in an elementary way.
The estimate I got was
$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$
for some constant $C$ which appears to be close to $-0.207$ (which I guess will be $\zeta(-1/2)$).
(By $a_{n} \sim b_{n}$ I mean $\lim_{n \rightarrow \infty} (a_{n}-b_{n}) = 0$)
I believe here is a completely elementary proof of that fact:
First consider the inequality for $x > 0$ and $k > 0$.
$$ \sqrt{x} \le \frac{x}{2\sqrt{k}} + \frac{\sqrt{k}}{2}$$
This follows easily by the arithmetic mean $\ge$ geometric mean inequality.
Thus we have that
$$\int_{k}^{k+1} \sqrt{x} \ dx \le \int_{k}^{k+1} (\frac{x}{2\sqrt{k}} + \frac{\sqrt{k}}{2}) \ dx$$
$$ = \frac{(k+1)^2 - k^2}{4\sqrt{k}} + \frac{\sqrt{k}}{2} = \sqrt{k} + \frac{1}{4\sqrt{k}}$$
Thus $$\sum_{k=1}^{n-1} \int_{k}^{k+1} \sqrt{x} \ dx \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$
i.e.
$$\int_{1}^{n} \sqrt{x} \ dx \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$
and so
$$ \frac{2}{3} n^{3/2} - \frac{2}{3} \le \sum_{k=1}^{n-1} (\sqrt{k} + \frac{1}{4\sqrt{k}})$$
Now we have inequality
$$\frac{1}{2\sqrt{k}} < \frac{1}{\sqrt{k} + \sqrt{k-1}} = \sqrt{k} -\sqrt{k-1}$$
And so, we have that
$$ \sum_{k=1}^{n-1} \frac{1}{4\sqrt{k}} < \frac{\sqrt{n-1}}{2}$$
Thus,
$$ \frac{2}{3} n^{3/2} - \frac{2}{3} \le \frac{\sqrt{n-1}}{2} + \sum_{k=1}^{n-1} \sqrt{k} $$
So if $$S_{n} = \sum_{k=1}^{n} \sqrt{k}$$ we have that
$$S_{n} \ge \frac{2}{3} n^{3/2} - \frac{2}{3} + \sqrt{n} - \frac{\sqrt{n-1}}{2} \ge \frac{2}{3} n^{3/2} - \frac{2}{3} + \frac{\sqrt{n}}{2}$$
Now let $$G_{n} = S_{n} - \frac{2}{3} n^{3/2} - \frac{\sqrt{n}}{2}$$
We have that $$G_{n} \ge -\frac{2}{3}$$
We can easily show that (using tedious but not too complicated algebra*) $$G_{n+1} < G_{n}$$ and so $G_{n}$ is a convergent sequence, as it is bounded below and monotonically decreasing.
Thus there exists a constant $C$ (the limit of $G_{n}$) such that
$$1 + \sqrt{2} + \dots + \sqrt{n} \sim \frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + C$$
* For the sake of completeness, we show that $G_{n+1} < G_{n}$.
Consider $$6(G_{n}-G_{n+1}) = 4(n+1)\sqrt{n+1} + 3\sqrt{n+1} - 6\sqrt{n+1} - 4n\sqrt{n} - 3\sqrt{n}$$
$$ = \sqrt{n+1} + 4n(\sqrt{n+1} -\sqrt{n}) - 3\sqrt{n}$$
Multiplying by $\sqrt{n+1} + \sqrt{n}$ does not change the sign, so we look at
$$ \sqrt{n+1}(\sqrt{n+1} + \sqrt{n}) + 4n - 3\sqrt{n}(\sqrt{n+1} + \sqrt{n})$$
$$ = 2n+1 - 2\sqrt{n^2 + n}$$
Now $$ (2n+1)^2 = 4n^2 + 4n + 1 > 4n^2 + 4n = (2\sqrt{n^2+n}) ^2$$
Hence
$$ 2n+1 > 2\sqrt{n^2+n}$$
and so
$$G_{n} > G_{n+1}$$
| {
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"question_score": "21",
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Sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ Are there known sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$ for $k > 1$ subject to the constraint $\sum_{p \mid k} \frac{1}{p+1} < 1$? (The factor of +1 in the denominator and the constraint of the sum of inverse "shifted prime divisors" is bounded by 1 are both necessary.)
A related question: Suppose $(p_i)$ is a set of $n$ consecutive primes which minimizes $1 - \sum_{i = 1}^{n} \frac{1}{p_i+1} > 0$ for a given $n > 1$. Are there known bounds for $1 - \sum_{i = 1}^{n} \frac{1}{p_i + 1}$ from below in terms of $n$, e.g., $n^{-\delta n}$ for some fixed $\delta > 0$?
Thanks!
| My answer on MO:
If we are allowed to consider somewhat weaker bounds, both answers depend only on the number of unitary divisors of $k$, which is $2^{\omega(k)}$. By the quoted result of O. Izhboldin and L. Kurliandchik (see Fedor's and Myerson's comments here), for any set of $n$ positive integers {$a_{1}, \dots, a_{n}$} such that $\sum_{i = 1}^{n} \frac{1}{a_{i}} <1$, we have
\begin{eqnarray}
\sum_{i = 1}^{n} \frac{1}{a_{i}} \leq \sum_{i = 1}^{n} \frac{1}{d_{i}} = \frac{d_{n+1} - 2}{d_{n+1} -1} < 1,
\end{eqnarray}
where $d_{i}$ is the $i^{\text{th}}$-Euler number, which satisfies the quadratic recurrence $d_{i} = d_{1} \cdots d_{i-1} + 1$ with $d_{1} = 2$. The first few terms of the sequence are 2, 3, 7, 43, 1807.... (A000058). It is relatively straightforward to show that the aforementioned recurrence is equivalent to $d_{i} = d_{i-1}(d_{i-1}-1) + 1$, and it is known from the recurrence that $d_{i} = \lfloor \theta^{2^{i}} + \frac{1}{2} \rfloor$, where $\theta \approx 1.2640$.... Thus,
\begin{eqnarray}
\sum_{p \mid k} \frac{1}{p + 1} \leq \sum_{i = 1}^{\omega(k)} \frac{1}{d_{i}} = \frac{d_{\omega(k) + 1} - 2}{d_{\omega(k)+1} - 1} = \frac{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 2}{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 1} < 1.
\end{eqnarray}
One can also show that $d_{i} - 1 = \lfloor \vartheta^{2^{i-1}} - \tfrac{1}{2} \rfloor$, where where $\vartheta \approx 1.5979$...., so we have
\begin{eqnarray}
1 - \sum_{p \mid k} \frac{1}{p + 1} \geq 1 - \frac{d_{\omega(k)+1} - 2}{d_{\omega(k)+1} - 1} = \frac{1}{d_{\omega(k)+1} - 1} = \lfloor \vartheta^{2^{\omega(k)}} - \tfrac{1}{2} \rfloor^{-1} > 0.
\end{eqnarray}
Remark E. Deutsche points out that the sequence {$d_{i} - 1$} (A007018) counts the number of ordered rooted trees with out-going degree up to 2 with all leaves at top level.
| {
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What type of triangle satisfies: $\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $ If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ?
| Begin with the Law of Cosines to compute $\cot^2\frac{A}{2}$ in terms of the lengths of the sides of the triangle:
$$\cos{A} = \frac{b^2+c^2-a^2}{2bc}$$
$$\Rightarrow \begin{cases}
\cos^2{\frac{A}{2}}=\frac{1+\cos{A}}{2}=\frac{2bc+b^2+c^2-a^2}{4bc} = \frac{(b+c)^2-a^2}{4bc} \\
\sin^2{\frac{A}{2}}=\frac{1-\cos{A}}{2}=\frac{2bc-(b^2+c^2-a^2)}{4bc} = \frac{a^2-(b-c)^2}{4bc}
\end{cases}$$
$$\Rightarrow \cot^2\frac{A}{2}=\frac{\cos^2{\frac{A}{2}}}{\sin^2{\frac{A}{2}}}=\frac{(b+c)^2-a^2}{a^2-(b-c)^2}$$
Now, work in the condition from the problem, and let Algebra bring you home ...
$$\begin{eqnarray}
\frac{(b+c)^2}{a^2} &=& \frac{(b+c)^2-a^2}{a^2-(b-c)^2} \\
(b+c)^2\left(a^2-(b-c)^2\right)&=& a^2\left((b+c)^2-a^2\right) \\
a^2(b+c)^2-(b+c)^2(b-c)^2 &=& a^2 (b+c)^2-a^4 \\
a^4-(b+c)^2(b-c)^2 &=& 0 \\
a^4-(b^2-c^2)^2 &=& 0 \\
\left(a^2+(b^2-c^2)\right)\left(a^2-(b^2-c^2)\right) &=& 0 \\
\end{eqnarray}$$
$$\Rightarrow \hspace{0.25in} a^2+b^2=c^2 \hspace{0.25in}\text{or}\hspace{0.25in} a^2+c^2=b^2$$
Nice little problem. I suspect there's a more-direct path to the solution, though. (Edit I see a couple such paths had been posted when I was working on my answer! :)
| {
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Baffling identity: $\prod \log_{10} \tan = \sum \log_{10} \tan$ I am quite a baffled now, I am not getting by how it can be written that :
\begin{align*}
&\log_{10} \tan 40^\circ \cdot \log_{10} \tan 41^\circ \cdot \log_{10} \tan 42^\circ \cdot\log_{10} \tan 43^\circ \cdots \log_{10} \tan 50^\circ \\
&= \log_{10} \tan 40^\circ + \log_{10} \tan 41^\circ + \log_{10} \tan 42^\circ + \log_{10} \tan 43^\circ + \cdots + \log_{10} \tan 50^\circ
\end{align*}
Is it even valid ? If yes,how ?
| Yes it is valid.
$\log_{10} \tan 40^{\circ} + \log_{10} \tan 50^{\circ} = \log_{10} \tan 40^{\circ} + \log_{10} \cot 40^{\circ} = \log_{10} 1 = 0$
Similarly combine 41 and 49, 42 and 48 etc.
The product on the left side is $0$ as $\log_{10} \tan 45^{\circ} = 0$.
| {
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How would I create a rotation matrix that rotates X by a, Y by b, and Z by c? How would I create a rotation matrix that rotates X by a, Y by b, and Z by c?
I need to formulas, unless you're using the ardor3d api's functions/methods.
Matrix is set up like this
xx, xy, xz,
yx, yy, yz,
zx, zy, zz
A Quaternion is fine too.
| The rotation matrices around the x, y, and z axes, respectively, are
$$R_x(\theta) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{pmatrix}$$
$$R_y(\phi) = \begin{pmatrix} \cos \phi & 0 & \sin \phi \\ 0 & 1 & 0 \\ - \sin \phi & 0 & \cos \phi \end{pmatrix}$$
$$R_z(\psi) = \begin{pmatrix} \cos \psi & - \sin \psi & 0 \\ \sin \psi & \cos \psi & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
If you want to rotate in the order specified in your comment on mathcast's answer, then you want
$$R_z(\psi) \cdot R_y(\phi) \cdot R_x(\theta) = $$
$$\begin{pmatrix} \cos \phi \cos \psi & \cos \psi \sin \theta \sin \phi - \cos \theta \sin \psi & \cos \theta \cos \psi \sin \phi + \sin \theta \sin \psi \\ \cos \phi \sin \psi & \cos \theta \cos \psi + \sin \theta \sin \phi \sin \psi & \cos \theta \sin \phi \sin \psi - \cos \psi \sin \theta \\ - \sin \phi & \cos \phi \sin \theta & \cos \theta \cos \phi \end{pmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$ $$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$
This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it?
| It might not be elliptic after all... (unless I have made some mistake)
$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$
Let $\displaystyle u = x -\frac{1}{x}$.
Then $\displaystyle du = (1 + \frac{1}{x^2})dx$.
Now $$\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}} = -\frac{x^2(1 + 1/x^2)}{x(x-1/x)\sqrt{x^2(x^2 + 1/x^2)}} = -\frac{1 + 1/x^2}{(x-1/x)\sqrt{(x - 1/x)^2 + 2}}$$
Thus
$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$
$$= -\int \ \frac{\mathrm{du}}{u \sqrt{u^2 + 2}}$$
| {
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"source": "stackexchange",
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How to factorize $4x^4+12x^{10/3} y^{2/3}+ \dots $? Does anyone know how to factorize the following expression:
$$4x^4+12x^{10/3} y^{2/3}+33x^{8/3} y^{4/3}+46x^2 y^2+33x^{4/3} y^{8/3}+12x^{2/3} y^{10/3}+4 y^4$$
?
| It is possible, but likely messy.
First set $\displaystyle z = \left(\frac{x}{y}\right)^{2/3}$ and divide your expression by $\displaystyle y^4$.
We get
$$4z^6 + 12 z^5 + 33 z^4 + 46 z^3 + 33 z^2 + 12 z + 4$$
Now divide this by $\displaystyle z^3$ and set $\displaystyle t = z + 1/z$.
We get
$$4(z^3 + 1/z^3) + 12(z^2 + 1/z^2) + 33(z+1/z) + 46$$
$$ = 4(t^3 - 3t) + 12(t^2 - 2) + 33t + 46$$
$$ = 4t^3 + 12t^2 + 21t + 22$$
Which is a messy cubic, according to Wolfram Alpha.
(Note it is always possible to factorize a cubic in "closed form", because of Cardano's method of finding the roots).
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial fraction integration So I want to find all antiderivaties of
$\frac{x}{x^3-1}$
Since the denominator is of a lesser degree than the numerator, partial fraction is to be used instead of long division.
I've started by doing:
$\frac{x}{x^3-1} = \frac{A}{x^2+x+1} + \frac{B}{x-1}$
Hence,
$x = A(x-1)+B(x^2+x+1)$
Then setting $x = 1$ to get $3B = 1 => B = \frac{1}{3}$
So I have $\int({\frac{A}{x^2+x+1}}+{\frac{1}{3(x-1)}})$
However, I'm not sure how to continue from here.
| Like most techniques of integration, the ideal of partial fractions is to reduce a difficult integral to integrals that are, if not easy, at least doable. The three types of integrals that show up when doing partial fractions are:
*
*$\displaystyle \int\frac{1}{(ax+b)^n}\,dx$, with $a,b$ constants, $a\neq 0$, and $n\geq 1$.
*$\displaystyle \int\frac{x}{(ax^2+bx+c)^n}\,dx$, with $n\geq 1$, $ax^2+bx+c$ irreducible quadratic.
*$\displaystyle \int\frac{1}{(ax^2+bx+c)^n}\,dx$ with $n\geq 1$, $ax^2+bx+c$ irreducible quadratic.
The first type is easy to solve: do a substitution $u=ax+b$ and go at it.
Second type is a bit more involved. By doing a substitution $u=ax^2+bx+c$, we get
\begin{align*}
\int\frac{x}{(ax^2+bx+c)^n}\,dx &= \frac{1}{2a}\int \frac{2ax\,dx}{(ax^2+bx+c)^n}\\
&= \frac{1}{2a}\left(\int\frac{2ax+b}{(ax^2+bx+c)^n}\,dx - \int\frac{b}{(ax^2+bx+c)^n}\,dx\right)\\
&= \frac{1}{2a}\int\frac{du}{u^n} - \frac{b}{2a}\int\frac{dx}{(ax^2+bx+c)^n}.
\end{align*}
The first integral can be done; the second reduces to the third type mentioned above.
And so we come down to the third type. When $n=1$, the simplest thing to do is to complete the square; factoring out $a$ we may assume we have $x^2+Bx+C$. Completing the square, you get $(x+\frac{B}{2})^2 + (C - \frac{B^2}{4})$. Because we are assuming that the original quadratic is irreducible, that means that $B^2 - 4C\lt 0$, so that $C-\frac{B^2}{4}\gt 0$. Substituting $u=X+\frac{B}{2}$ turns this into a fraction of the form $\frac{1}{u^2+r^2}$; factor out $r^2$, do another substitution, and you can turn it into a fraction of the form $\frac{1}{w^2+1}$. But $\int\frac{dw}{w^2+1}$ is an easy integral: you have an immediate antiderivative for it.
So, modulo a bunch of algebra and some substitution, you can solve $\int\frac{1}{ax^2+bx+c}\,dx$ with $ax^2+bx+c$ irreducible quadratic: complete the square, do some substituttions, and turn it into $\int\frac{1}{w^2+1}\,dx$.
What if you have $\int\frac{dx}{(ax^2+bx+c)^n}$ with $n\gt 1$, $ax^2+bx+c$ irreducible quadratic? Those are more complicated, but not too bad; you can still complete the square and do a bit of algebra, so that you bring it to the form
$$\int \frac{du}{(u^2+r^2)^n}$$
for some positive $r$. Then one can use the reduction formula (obtained by doing integration by parts):
$$\int\frac{du}{(u^2+r^2)^n} = \frac{1}{2r^2(n-1)}\left(\frac{u}{(u^2+r^2)^{n-1}} + (2n-3)\int\frac{du}{(u^2+r^2)^{n-1}}\right)$$
and continuing this way you will eventually end in the integral with denominator $u^2+r^2$, which we know how to do.
| {
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"timestamp": "2023-03-29T00:00:00",
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Singular matrix Suppose I have a singular matrix given by
$$A = \begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14}\\
a_{12} & a_{11} & a_{14} & a_{13}\\
a_{31} & a_{32} & a_{33} & a_{34}\\
a_{32} & a_{31} & a_{34} & a_{33}
\end{pmatrix}$$
Which is a homogeneous system in $(x_1, x_2, x_3, x_4)^T$ where there is a variable $\Lambda$ in the coefficients that makes the matrix singular if chosen such that the determinant is zero.
Now I have the matrices:
$$B = \begin{pmatrix}
a_{11} + a_{12} & a_{13} + a_{14}\\
a_{31} + a_{32} & a_{33} + a_{34}
\end{pmatrix}$$
$$C = \begin{pmatrix}
a_{11} - a_{12} & a_{13} - a_{14}\\
a_{31} - a_{32} & a_{33} - a_{34}
\end{pmatrix}$$
which are homogeneous systems in $(x_1 + x_2, x_3 + x_4)^T$ and $(x_1 - x_2, x_3 - x_4)^T$ respectively.
If I make the matrices $B$ and $C$ out of $A$, and I determine the $\Lambda$ such that those determinants are zero, can I say anything about the determinant of $A$ with those values of $\Lambda$?
The idea is that I won't have a fourth degree polynomial to solve (which Maple or Mathematica can seem to do in this case) and then I can obtain the original result by summing or subtracting and such.
| Yes. We have $\det A=\det B\cdot \det C$. There are some different ways to see this; here is one:
Your matrix $A$ can be written as the block matrix $\left(\begin{array}{cc} X&Y\\ U&W\end{array}\right)$, where $X$, $Y$, $U$, $W$ are the following $2\times 2$ matrices:
$X=\left(\begin{array}{cc} a_{11}&a_{12}\\ a_{12}&a_{11}\end{array}\right)$;
$Y=\left(\begin{array}{cc} a_{13}&a_{14}\\ a_{14}&a_{13}\end{array}\right)$;
$Z=\left(\begin{array}{cc} a_{31}&a_{32}\\ a_{32}&a_{33}\end{array}\right)$;
$W=\left(\begin{array}{cc} a_{33}&a_{34}\\ a_{34}&a_{33}\end{array}\right)$.
Now, these matrices $X$, $Y$, $U$, $W$ are circulant matrices, and thus can be diagonalized by the unitary discrete Fourier transform matrix
$F_2=\frac{1}{\sqrt 2}\left(\begin{array}{cc} 1&1\\ 1&-1\end{array}\right)$.
So we have
$X=F_2\mathrm{diag}\left(a_{11}+a_{12},a_{11}-a_{12}\right)F_2^{-1}$;
$Y=F_2\mathrm{diag}\left(a_{13}+a_{14},a_{13}-a_{14}\right)F_2^{-1}$;
$Z=F_2\mathrm{diag}\left(a_{31}+a_{32},a_{31}-a_{32}\right)F_2^{-1}$;
$W=F_2\mathrm{diag}\left(a_{33}+a_{34},a_{33}-a_{34}\right)F_2^{-1}$.
As a consequence, the block matrix $A=\left(\begin{array}{cc} X&Y\\ U&W\end{array}\right)$ can be written as
$A=\left(\begin{array}{cc} F_2&0\\ 0&F_2\end{array}\right)\left(\begin{array}{cc} \mathrm{diag}\left(a_{11}+a_{12},a_{11}-a_{12}\right) & \mathrm{diag}\left(a_{13}+a_{14},a_{13}-a_{14}\right) \\ \mathrm{diag}\left(a_{31}+a_{32},a_{31}-a_{32}\right) & \mathrm{diag}\left(a_{33}+a_{34},a_{33}-a_{34}\right) \end{array}\right) \left(\begin{array}{cc} F_2&0\\ 0&F_2\end{array}\right)^{-1}$
(check this!), so that
$\det A = \det \left(\begin{array}{cc} \mathrm{diag}\left(a_{11}+a_{12},a_{11}-a_{12}\right) & \mathrm{diag}\left(a_{13}+a_{14},a_{13}-a_{14}\right) \\ \mathrm{diag}\left(a_{31}+a_{32},a_{31}-a_{32}\right) & \mathrm{diag}\left(a_{33}+a_{34},a_{33}-a_{34}\right) \end{array}\right) $.
Now, the determinant on the right hand side can be even simplified by transposing the second row with the third row and transposing the second column with the third column:
$\det A = \det \left(\begin{array}{cccc} a_{11}+a_{12} & a_{13}+a_{14} & 0 & 0 \\ a_{31}+a_{32} & a_{33}+a_{34} & 0 & 0 \\ 0 & 0 & a_{11}-a_{12} & a_{13}-a_{14} \\ 0 & 0 & a_{31}-a_{32} & a_{33}-a_{34} \end{array}\right)$.
Now the matrix on the right hand side is obviously just the block matrix $\left(\begin{array}{cc} B&0\\ 0&C\end{array}\right)$, so its determinant is $\det B\cdot \det C$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How $m^4+4 = 0 \Rightarrow m = 1 \pm i,-1\pm i$? This is given in my module as a part of a problem's solution:
$$m^4 + 4 = 0 $$ $$\Rightarrow m = 1 \pm i,-1\pm i$$
I am not getting how this conversion is taking place,could somebody explain?
| Here $m = (-4)^{1/4}$.
Now $-4 = 4(\cos(\theta) + i\sin(\theta))$ in polar form where $\theta = (2n-1)\pi$.
By de Moivre's theorem
$m = (-4)^{1/4} = 4^{1/4}(\cos(\frac{\theta}{4}) + i\sin(\frac{\theta}{4}))$
For $n = 0$, $\displaystyle m = \sqrt{2}\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = 1 - i$,
For $n = 1$, $\displaystyle m = \sqrt{2}\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = 1 + i$,
For $n = 2$, $\displaystyle m = \sqrt{2}\left(-\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = -1 + i$,
For $n = 3$, $\displaystyle m = \sqrt{2}\left(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = -1 - i$
| {
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"timestamp": "2023-03-29T00:00:00",
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Help solving $\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}dx}$ $\displaystyle\int {\frac{8x^4+15x^3+16x^2+22x+4}{x(x+1)^2(x^2+2)}\,\mathrm{d}x}$
I used partial fractions, solved $A = 2, C = 3$.
$$\frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} +\frac{(Dx+E)}{(x^2+2)}$$
\begin{align*}
&8x^4+15x^3+16x^2+22x+4\\
&\quad = A(x+1)^2(x^2+2)+B(x)(x+1)(x^2+2)+C(x)(x^2+2)+(Dx+E)(x)(x+1)^2
\end{align*}
Substitute in $x=0$ to get $4=A(1)(2)$, so $A = 2$
$$6x^4+11x^3+10x^2+14x = B(x)(x+1)(x^2+2)+C(x)(x^2+2)+(Dx+E)(x)(x+1)^2$$
Substitute in $x=-1$ to get
$$6-11+10-14 = C(-1)(1+2)$$
so $-9=-3C$, thus $C=3$.
Leaving me what I have below:
Which brings me to where I am currently stuck.
$$6x^4 +8x^3 +10x^2+8x = B(x)(x+1)(x^2+2) + (Dx + E) (x) (x+1)^2$$
Is the next best move to use substitution to solve for $B$?
| Given what you have, what happens when you set $x = -1$?
| {
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"timestamp": "2023-03-29T00:00:00",
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The inverse of a certain tricky function What is the explicit form of the inverse of the function $f:\mathbb{Z}^+\times\mathbb{Z}^+\rightarrow\mathbb{Z}^+$ where $$f(i,j)=\frac{(i+j-2)(i+j-1)}{2}+i?$$
| Since your function seems to be Cantor's pairing function $p(x,y) = \frac{(x+y)(x+y+1)}{2} + y$ applied to $x= j-2, y = i$, and since the inverse of the pairing function is $p^{-1}(z) = (\frac{\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor^2 + 3\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor}{2}-z,z-\frac{\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor^2 + \lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor}{2})$, the inverse of your function is: $f^{-1}(z)=(z-\frac{\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor^2 + \lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor}{2},2+ \frac{\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor^2 + 3\lfloor \frac{\sqrt{8z+1}-1}{2} \rfloor}{2}-z)$, which can be a bit ugly. What is your motivation for inverting this function?
| {
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"timestamp": "2023-03-29T00:00:00",
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Question regarding solving polynomial of congruence? In my textbook, they said:
$$2x^{3} + 7x - 4 \equiv 0 \pmod{5}$$
The solution of this equation are the integers with $x \equiv 1 \pmod{5}$, as can be seen by testing $x = 0, 1, 2, 3,$ and $4.$
And I have no clue how do they had $x \equiv 1 \pmod{5}$. I tested as they suggested:
Let $y = 2x^{3} + 7x - 4$, we have:
$$x = 0: y = -4, \implies y \equiv 1 \pmod{5}$$
$$x = 1: y = 5, \implies y \equiv 0 \pmod{5}$$
$$x = 2: y = 26, \implies y \equiv 1 \pmod{5}$$
$$x = 3: y = 71, \implies y \equiv 1 \pmod{5}$$
$$x = 4: y = 152, \implies y \equiv 2 \pmod{5}$$
What I did not understand is how these five modulo equations become $x \equiv 1 \pmod{5}$? Can anyone help me explain this?
Thanks,
| What your textbook means is $$\left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right) \iff \left(x \equiv 1 \pmod{5} \right)$$
This is what you checked by plugging in the different cases for $x$.
By taking $x \equiv 1 \pmod{5}$, you proved that $2x^{3} + 7x - 4 \equiv 0 \pmod{5}$ and hence $$ \left(x \equiv 1 \pmod{5} \right) \Rightarrow \left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right)$$
Now by taking $x \neq 1 \pmod{5}$, you proved that $2x^{3} + 7x - 4 \neq 0 \pmod{5}$ and hence $$\left(2x^{3} + 7x - 4 \equiv 0 \pmod{5} \right) \Rightarrow \left(x \equiv 1 \pmod{5} \right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given a matrix $A$ find a matrix $C$ such that $C^3$=$A$ This is a question I had on a test, we were told not to use brute-force and figure out a smart way to solve the problem.
We have a matrix $A =$
$\displaystyle\begin{bmatrix}
2 & 3\\
3 & 2
\end{bmatrix}$. Find a matrix $C$ such that $C^{3}$=$A$.
What is the 'smart not brute-force' way to solve this, without picking numbers, looking for patterns and so on?
it was in eigenvalues section" in the end?
| One thing that jumps out at me is that this matrix has two eigenvalues. Find them: det
$
(\begin{bmatrix}
2-\lambda & 3\\
3 & 2-\lambda
\end{bmatrix})$=
$(2-\lambda)^{2}-9$=$-5-4\lambda+\lambda^{2}$. Factoring this you get $(\lambda+1)(\lambda-5)=0$, so your eigenvalues are -1 and 5. Now, find bases for their respective eigenspaces.
Basis $\xi_{-1}$ = ker$(A+I)$=ker $(\begin{bmatrix}
3 & 3\\\
3 & 3
\end{bmatrix})$. Row-reducing, you get ker $(\begin{bmatrix}
1 & 1\\\
0 & 0
\end{bmatrix})$, which is equal to the span of $\begin{bmatrix}
-1\\
1
\end{bmatrix}$. So a basis of $\xi_{-1}$ = $\begin{bmatrix}
-1\\
1
\end{bmatrix}$.
Same for $\xi_{5}$ -- ker $(\begin{bmatrix}
-3 & 3\\\
3 & -3
\end{bmatrix})$, row-reducing we get ker $(\begin{bmatrix}
1 & -1\\\
0 & 0
\end{bmatrix})$, so ker = span $\begin{bmatrix}
1\\
1
\end{bmatrix}$. So a basis of $\xi_{5}$ = $\begin{bmatrix}
1\\
1
\end{bmatrix}$.
Now you know that $A$=$C^{3}$=$CCC$. So if $\vec{x}$ is an eigenvector of $A$, it is clearly an eigenvector of $C$, but perhaps with a different eigenvalue. You know that $A$$
\begin{bmatrix}
1\\
1
\end{bmatrix}$=$CCC$$
\begin{bmatrix}
1\\
1
\end{bmatrix}$=$\begin{bmatrix}
5\\
5
\end{bmatrix}$. So, $\sqrt[3]{5}$ must be an eigenvalue of $C$.
Same goes for $A$$\begin{bmatrix}
-1\\
1
\end{bmatrix}$=$CCC$$\begin{bmatrix}
-1\\
1
\end{bmatrix}$=$\begin{bmatrix}
1\\
-1
\end{bmatrix}$. This would be possible only if -1 was an eigenvalue of $C$, since $(-1)^{3}$=$-1$.
So now you can construct a system of equations -- you know that $C$ is of some form $(\begin{bmatrix}
a & b\\\
c & d
\end{bmatrix})$, and based on the eigenvectors and eigenvalues of $C$ that we just found out, you can devise the following system:
$\left\{\begin{matrix}
a & + & b & = & \sqrt[3]{5}\\
c & + & d & = & \sqrt[3]{5}\\
-a & + & b & = & 1\\
-c & + & d & = & -1
\end{matrix}\right.$
Now solve for the variables -- $b=a+1$, substitute into the other $a, b$ equation to get $2a+1=\sqrt[3]{5}$, so $a=\frac{\sqrt[3]{5}-1}{2}$. $b$, then, is equal to $\frac{\sqrt[3]{5}-1}{2}+1$, as obvious from the third equation. Same for $c$ and $d$ -- $c=d+1$, so substituting, we get $2d+1=\sqrt[3]{5}$, so $d=\frac{\sqrt[3]{5}-1}{2}$, and hence $c=\frac{\sqrt[3]{5}-1}{2}+1$.
This results into $C$=$
\begin{bmatrix}
\frac{\sqrt[3]{5}-1}{2} & \frac{\sqrt[3]{5}-1}{2}+1 \\
\frac{\sqrt[3]{5}-1}{2}+1 & \frac{\sqrt[3]{5}-1}{2}
\end{bmatrix}$.
| {
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Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$
I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question). I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.
I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.
| By induction we can easily prove that for any nonnegative real numbers $a_k$
$$1-\sum_{k=1}^na_k+\sum_{1\le i<j\le n}a_ia_j-\sum_{1\le i<j<k\le n}a_ia_ja_k\le\prod_{k=1}^n(1-a_k)\le1-\sum_{k=1}^na_k+\sum_{1\le i<j\le n}a_ia_j$$
Taking $a_k=\frac{x^2}{k^2\pi^2}$,we get
$$1-\frac{x^2}{\pi^2}\zeta_n(2)+\frac{x^4}{\pi^4}\frac{\zeta_n(2)^2-\zeta_n(4)}2-\frac{x^6}{\pi^6}\frac{\zeta_n(2)^3-\zeta_n(6)}{6}\le\prod_{k=1}^n(1-\frac{x^2}{k^2\pi^2})\le1-\frac{x^2}{\pi^2}\zeta_n(2)+\frac{x^4}{\pi^4}\frac{\zeta_n(2)^2-\zeta_n(4)}2$$
Since $\prod\limits_{k=1}^{\infty}(1-\frac{x^2}{k^2\pi^2})=\frac{\sin(x)}x$, by taking $n\to\infty$
$$1-\frac{x^2}{\pi^2}\zeta(2)+\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2-\frac{x^6}{\pi^6}\frac{\zeta(2)^3-\zeta(6)}{6}\le1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\le1-\frac{x^2}{\pi^2}\zeta(2)+\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2$$
subtraciting $1-\frac{x^2}{\pi^2}\zeta_n(2)$
$$\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2-\frac{x^6}{\pi^6}\frac{\zeta(2)^3-\zeta(6)}{6}\le\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\le\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2$$
dividing by $x^4$ and putting $x=0$ we get
$$\frac1{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2=\frac1{5!}$$
and this follows that $\zeta(4)=\frac{\pi^4}{90}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Summing the power series $\sum\limits_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}\prod\limits_{k=1}^n\frac{2k-1}{2k} $ I'd like to determine the function corresponding to the following power series:
$$x + \sum_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} \frac{x^{2n+1}}{2n+1},
$$
where $|x|<1$.
| Consider the following series,
\begin{align}
\arcsin x = \sum_{n \geq 0} \frac{(2n)!}{2^{2n} (n!)^{2}} \frac{x^{2n+1}}{2n+1} = \sum_{n \geq 0} \frac{(2n-1)!!}{(2n)!!} \frac{x^{2n+1}}{2n+1},
\end{align}
which converges for $|x| \leq 1$ and where $\frac{(2n-1)!!}{(2n)!!} = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots 2n}$. This can be derived by integrating the series expansion of $(1 -x^2)^{-1/2}$ termwise and paying careful attention to questions of convergence. Observe that
\begin{align}
-i \arcsin i x = x + \sum_{n \geq 1} (-1)^n \frac{(2n-1)!!}{(2n)!!} \frac{ x^{2n + 1}}{2n+1},
\end{align}
which is the series in question.
| {
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"source": "stackexchange",
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} |
$x^3+48=y^4$ does not have integer (?) solutions How does one find all positive integer solutions to the equation $x^3+48=y^4$?
| I realize this is a massive revive, but here is a solution without any powerful theorems about elliptic curves.
We have $x^3 + 64 = y^4 + 2^4$. It is easy to show all (odd) primes dividing the RHS are $1 \pmod{8}$. Now, looking modulo $16$ we deduce either $x \equiv 1 \pmod{16}$ and $y$ is odd or $x,y$ are both even. For the second case if $v_2(x) = 1$ we derive a contradiction because $v_2(LHS) = 3$ while $v_2(RHS) \ge 4$. If $v_2(x) \ge 2$ we have $v_2(LHS) \ge 6$. Thus we must have $v_2(y) = 1$ for $v_2(y^4 + 16) > 4$. Writing $y = 2k$ we need $v_2(k^4 + 1) \ge 2$. But this is absurd since $v_2(k^4 + 1) \le 1$ by checking modulo $4$. Thus it follows $x \equiv 1 \pmod{16}$.
But now utilize $x^3 + 64 = (x+4)(x^2 - 4x + 16)$. $x+4$ is $5 \pmod{8}$, which is absurd so we are done.
| {
"language": "en",
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Calculating prime numbers I have a program which determines if a number is prime or not. The basic algorithm simply checks for the number being divisible by 1 and itself.
My question is, is there an upper limit to checking numbers for divisibility without a remainder? If yes, can you explain why?
My "gut" feeling is that any number above 1/2 the number being checked should be ignored, but I can't prove it.
| I will try to explain by using an example.
Let's check if $37$ is a prime number.
*
*It's obviously not divisible by $2$,
*it's not divisible by $3$ (because $3+7=10$ which is not divisible by $3$),
*and it's not divisible by five (because the last digit is not $0$ or $5$).
These are all the possible candidates for factors of $37$, and I will try to explain why. If $7$ is a factor of $37$ (which would mean that $37$ is divisible by $7$, and hence not a prime) then the other factor must be greater or equal to $7$, since we have already shown that there is no factor less than $7$. But $7 \cdot 7 = 49 > 37$. Then there is no point in trying with greater factors, because then the product would have to be even larger.
So we have shown that for $37$, we only have to try to divide by $2$, $3$ and $5$. As both Raeder and Theo has pointed out we don't have to go higher than the square root of the number we want to check, and I hope that my example will help explaining why.
Edit: As Picakhu suggested, I will explain why these methods for checking divisibility by $2$, $3$, and $5$ works.
Checking divisibility by $2$ and $5$
When checking for divisibility by $2$ or $5$, it is sufficient to check that only the last digit is divisible $2$ or $5$ respectively. This has to do with the fact that we are using $10$, which is the product of $2$ and $5$, as our base. I will explain using three-digit numbers. The explanation would be very similar if we had used more digits. All numbers $abc$ (where $a$ is the first digit, $b$ the second, and $c$ the third) can be written on the form
$$abc = a\cdot 10^2 + b\cdot 10^1 + c\cdot 10^0 = a \cdot 100 + b \cdot 10 + c.$$
The notation I'm using can be confusing. To clear up: If, for example, $a=b=c=4$, then $abc = 444 \neq 4\cdot 4\cdot 4$. Let's see what happens if we divide by 2.
$$\frac{abc}{2} = \frac{a \cdot 100}{2} + \frac{b \cdot 10}{2} + \frac{c}{2} = a \cdot 50 + b \cdot 5 + \frac{c}{2}$$
We notice that this is a whole number if and only if $\frac{c}{2}$ is a whole number, no matter what $abc$ is. We get a similar result if we divide by $5$.
$$\frac{abc}{5} = a \cdot 20 + b \cdot 2 + \frac{c}{5}$$
Hence it is sufficient to check the last digit.
Checking divisibility by $3$
When checking for divisibility by $3$, it is sufficient to check that the sum of the digits is divisible by $3$. This is easiest to show using modular arithmetics. Keep in mind that $10 \equiv 1 \pmod 3$. If we use our three-digit number $abc$ again we see that
$$a \cdot 10^2 + b \cdot 10 + c \equiv a \cdot 1 + b \cdot 1 + c = a + b + c \pmod 3.$$
Hence $abc$ is divisible by $3$ if and only if $a + b + c$ is divisible by $3$.
Wikipedia has a list of more divisibility tests like these.
| {
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"timestamp": "2023-03-29T00:00:00",
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Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$ Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.
Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$.
Therefore $c$ should be integers, but I don't know how to do it at next step ?
| HINT $\rm\:\ \ mod\ 9:\ (x+3)^3\equiv\:x^3\ \Rightarrow\ f(n+1)\: \equiv\ f(n)\ $ so, by induction, $\rm\ f(n)\equiv\: f(1)\equiv\: 0 $
E.g. a specific inductive step: $\rm\: mod\ 9\::\ \ 3^3\:\equiv\: 0^3\ \Rightarrow\ 1^3+ 2^3+3^3\ \equiv\ 0^3+1^3+2^3\ \equiv\ 0\:.$
NOTE $\ $ The period $3$ repeating behavior is illustrated vividly in quanta's calculations, where one can explicitly see the repeating $3$-cycle $\ -1+0+1,\ \ 0+1-1,\ \ 1-1+0,\ \cdots$
| {
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Proof of inequality I have problems with proving inequality :
$${a^{2}}+b^2+c^2+\frac{2}{5}abc<50$$
where $a,b,c$ are the lengths of triangle's sides, and the circumference of the triangle is $10$.
Thanks.
| Consider the polynomial $(x-a)(x-b)(x-c)$. Multiplied out, this is
$$
\begin{eqnarray}
(x-a)(x-b)(x-c)
&=&
x^3-(a+b+c)x^2+(ab+ac+bc)x-abc
\\
&=&
x^3-10x^2+(ab+ac+bc)x-abc\;.
\end{eqnarray}
$$
We also have
$$10^2=(a+b+c)^2=(a^2+b^2+c^2) + 2(ab+ac+bc)\;,$$
$$ab+ac+bc=\frac{100-(a^2+b^2+c^2)}{2}\;,$$
and thus
$$
(x-a)(x-b)(x-c)
=
x^3-10x^2+\frac{100-(a^2+b^2+c^2)}{2}x-abc\;.
$$
Now since $a$, $b$ and $c$ form a triangle with perimeter $10$, they must all be less than $5$. Thus the value of the polynomial for $x=5$ is positive, that is,
$$
5^3-10\cdot5^2+\frac{100-(a^2+b^2+c^2)}{2}\cdot5-abc>0\;,
$$
which upon rearrangement becomes your inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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To sum $1+2+3+\cdots$ to $-\frac1{12}$ $$\sum_{n=1}^\infty\frac1{n^s}$$
only converges to $\zeta(s)$ if $\text{Re}(s)>1$.
Why should analytically continuing to $\zeta(-1)$ give the right answer?
| Today I stumbled on a totally elementary derivation of this that does not rely on regulators or any reference to the Riemann Zeta Function. We start by looking at the geometric series
$$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + ... $$
Now instead of computing the taylor series of this function we can to compute its $\log$-taylor series that is the coefficients in terms of $\log(x)^n $
First we recall that $x^n = e^{n \log(x)} $ then it follows that:
$$ \frac{1}{1-x} = 1 + e^{2 \log(x)} + e^{3 \log(x)} +e^{4 \log(x)} + ... $$
Therefore:
$$ \frac{1}{1-x} = 1 + \left(1 + 2 \log(x) + \frac{1}{2!} \left( 2 \log(x) \right)^2+ ... \right) + \left( 1 + 3 \log(x) + \frac{1}{3!} (3 \log(x))^2 \right) + ... $$
Which we can re-arrange as:
$$ \frac{1}{1-x} = 1 + \left(1+1+1... \right) + (1+2+3+4....) \log(x) + \frac{1}{2!} (1^2+2^2 +3^2 ...) \log(x)^2 +... $$
Which obviously don't converge.
But we have ONE more observation. We can get the log taylor series one other way. First we find the laurent series (since theres a negative power term) of
$$ \frac{1}{1-e^x} = -\frac{1}{x} - \frac{1}{2} - \frac{1}{12}x + \frac{1}{720}x^3 + ... $$
So we conclude then that
$$ \frac{1}{1-x} = -\frac{1}{\ln(x)} + \frac{1}{2} - \frac{1}{12}\ln(x) + \frac{1}{720}\ln(x)^3 + ... $$
So then in order for this to be consistent with our earlier result it must be the case that
$$ 1 + \left( 1 + 1 + 1 ... \right) = 0! *\left( \frac{1}{2} \right) \rightarrow \left( 1 + 1 + 1 ... \right) = -\frac{1}{2} $$
$$ 1 + 2 + 3 + 4 ... = (1!) *\left( - \frac{1}{12} \right) = - \frac{1}{12} $$
$$ 1^2 + 2^2 + 3^2 + 4^2 ... = 2! * 0 = 0 $$
$$ 1^3 + 2^3 +3^3 + 4^3 ... = 3!* \frac{1}{720} = \frac{1}{120} $$
etc..
CAVEAT:
There is that mysterious $- \frac{1}{\ln(x)}$ that is not accounted for. I'm not really sure what to do about that, although I have some undeveloped ideas brewing which ill update this with once i have refined them.
Another Elementary Approach:
A function $G(x)$ can be written as a series of the form $a_n f(x)^n$ for your choice of $f(x)$ if we define the operator $ D = \frac{1}{f'(x)} \frac{d}{dx}$ and $c$ is a constant such that $\frac{1}{f'(c)}$ is well defined and $f(c)=a$ then
$$ G(x) = G(c) + D[f](c)(f(x)-a) + \frac{1}{2!}D^2[f](c)(f(x)-a)^2 + \frac{1}{3!}D^3[f](c)(f(x)-a)^3 ... $$
Now if $G(x)$ has a singularity $x=c$ as is the case for $G(x)=\frac{1}{1-x}$ $f(x) = \ln(x)$, $c=1$, then you basically need an analogue of the residue theorem to extract its laurent series w.r.t $f$, i'm working on building that right now, but assuming such a thing can be constructed it gives us a different elementary framework which explains the $-\frac{1}{\ln(x)}$ term, but the mystery of why the first method doesn't catch it remains...
| {
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"timestamp": "2023-03-29T00:00:00",
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Understanding algebraic manipulation If $a+b+c \neq 0 $ where $a,b$ and $c$ are three non-zero distinct integers, then find the value of:
$$\frac{ab+ca}{a^2+ab+ca} + \frac{ab+cb}{b^2+ab+bc} + \frac{ac+cb}{c^2+ac+bc}$$
What confusing me here, is the not so obvious hint which is given with the problem,which says that that form could be written as:
$$3- \frac{a^2}{a^2+ab+ca} - \frac{b^2}{b^2+ab+bc} - \frac{c^2}{c^2+ac+bc}$$
But how is this possible?
| Or just add and subtract $a^2$, $b^2$, $c^2$ to the numerators of the three fractions respectively.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sqrt[x]{y} = y^{1 / x}$? Need help with algebra of exponentiation I need help with algebra of exponentiation.
$$\begin{align*}
\sqrt{x^2} &= (x^2)^{1/2} &\qquad &\text{(since }\sqrt[x]{y}=y^{1/x}\text{)}\\
&= x^{2(1/2)} &&\text{(since }(x^y)^z = x^{yz}\text{)}\\
&= x^{(1/2)2} &&\text{(since }xy=yx\text{)}\\
&= (x^{1/2})^2 &&\text{(since }(x^y)^z = x^{yz}\text{)}\\
&= \left(\sqrt{x}\right)^2 &&\text{(since }\sqrt[x]{y} = y^{1/x}\text{)}
\end{align*}$$
$\sqrt{x^2}$ is a real number.
$(\sqrt{x})^2$ is a real number, if $x\geq 0$.
$\sqrt{x^2} = (\sqrt{x})^2$.
| The equality
$$
(x^2 )^{1/2} = x^{2(1/2)}
$$
holds only for $x \geq 0$. If $x < 0$, then
$$
(x^2 )^{1/2} = |x| > 0,
$$
but
$$
x^{2(1/2)} = x < 0.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$a,b,c>0 \text{ s.t. }a+b+c=1 \implies \sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}$ How can we show that the assumption $a,b,c>0$ and $a+b+c=1$ implies
$$\sqrt{ab+c}+\sqrt{bc+a}+\sqrt{ca+b} \ge 1+ \sqrt{ab}+\sqrt{bc}+\sqrt{ca}~?$$
| First we claim that $\sqrt{ab+c} \geq \sqrt{ab}+c$.
Proof of the claim:
$$\begin{eqnarray*}
\sqrt{ab}+c
&=& \sqrt{ab + 2c\sqrt{ab} + c^2}
\\ &\leq& \sqrt{ab + (a+b) c + c^2}
\\ &=& \sqrt{ab + c(a+b+c)} = \sqrt{ab+c}.
\end{eqnarray*}
$$
The only inequality used here is the AM-GM inequality: $\sqrt{ab} \leq \frac{a+b}{2}$. $\Box$
Now, the full inequality follows easily from the claim. By symmetry, we also have $\sqrt{bc+a} \geq \sqrt{bc}+a$ and $\sqrt{ac+b}\geq \sqrt{ac}+b$. Adding all three inequalities, we get the result.
| {
"language": "en",
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On calculating the minor determinant of adjoint matrix Question:
$A$ is an $n$ order square matrix, and $B=A^{(*)}$ is the adjoint matrix of $A$ (i.e. $AB=BA=\det A\cdot I_n$).
Show that
$$\det B
\begin{pmatrix}
1 & 2 & \cdots & k \\
1 & 2 & \cdots & k
\end{pmatrix}
=
(\det A)^{k-1}\cdot
\det A
\begin{pmatrix}
k+1 & k+2 & \cdots & n \\
k+1 & k+2 & \cdots & n
\end{pmatrix}.
$$
Here, $B
\begin{pmatrix}
1 & 2 & \cdots & k \\
1 & 2 & \cdots & k
\end{pmatrix}$ denotes the $k\times k$ submatrix of $B$ on the top left corner,
$A
\begin{pmatrix}
k+1 & k+2 & \cdots & n \\
k+1 & k+2 & \cdots & n
\end{pmatrix}$ denotes the the $(n-k)\times (n-k)$ submatrix of $A$ on the bottom right corner.
I have tried to use the Laplace expansion theorem but failed.
| It is actually more related to this. Maybe it is better to express it by partitioning.
Given an "invertible" square matrix $A$,
$$
A = \left(\begin{array}{c|c}A_1 &A_2\\\hline A_3 &A_4\end{array}\right), A^{-1} = \left(\begin{array}{c|c}\hat A_1 &\hat A_2\\\hline \hat A_3 &\hat A_4\end{array}\right)
$$
and by stealing the latex code from wikipedia
$$
\begin{multline}
\det{(A^{-1})}=\det\left[ \begin{matrix} A & B \\ C & D \end{matrix}\right]^{-1} = \underbrace{\det\left[ \begin{matrix} I & 0 \\ -D^{-1}C & I \end{matrix}\right]}_1\det\left[ \begin{matrix} (A-BD^{-1}C)^{-1} & 0 \\ 0 & D^{-1} \end{matrix}\right]\\ \underbrace{\det\left[ \begin{matrix} I & -BD^{-1} \\ 0 & I \end{matrix}\right]}_{1}
\end{multline}
$$
another detail is that $(A-BD^{-1}C)^{-1} = \hat{A}_1$. Therefore $\det{(A^{-1})}=\det{(\hat{A}_1)}\det{(A_4^{-1})} $also
$B$ is defined as
$$
B = \det{(A)} A^{-1} = \left(\begin{array}{c|c}B_1 &B_2\\\hline B_3 &B_4\end{array}\right)
$$
partitioned accordingly. From the Schur Complement formula and from $\det{(\alpha A) = \alpha^n\det{(A)}}$, we know that
$$
\begin{align}
\det{(A^{-1})} &= \det{(A_4^{-1})}\det{(\hat A_1)}\\
&= \det{(A_4^{-1})}\left[ (\det(A))^{-k}\det{(B_1)}\right] \\
&= (\det(A))^{-k}\frac{1}{\det{(A_4)}}\det{(B_1)}\\
&=\frac{1}{\det{(A)}}
\end{align}
$$
Then, we have the desired result,
$$
\det{(A_4)} = (\det{(A)})^{k-1}\det{(B_1)}
$$
| {
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If $N= (323232 \cdots 50 \text{ digits})_9$ (i.e in base $9$) then how to find the remainder when this $N$ is divided by $8$?
If $N= (\underbrace{323232 \cdots}_{50 \text{ digits}})_9$ (i.e in base $9$) then how to
find the remainder when this $N$ is divided by $8$?
I am looking for a "fast" approach that could be used to solve this in less than $2$ minutes.
| The same way we find the remainder of dividing a number by $9$ when the number is written in base 10: add the digits.
Why?
Because remember that writing, say, $(38571)_9$ "really" means
$$1 + 7\times 9 + 5\times 9^2 + 8\times 9^3 + 3\times 9^4.$$
When you divide $9$ by $8$, the remainder is $1$; when you divide $9^2$ by $8$, the remainder is $1^2 = 1$; then you divide $9^3$ by $8$, the remainder is $1^3=1$, etc. So the remainder of this number when divided by $8$ is the same as the remainder of
$$1 + 7\times 1 + 5\times 9^2 + 8\times 1^3 + 3\times 1^4$$
which is the same as the remainder of $1+7+5+8+3 = 24$, which is $0$.
In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.
(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$).
(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).
| {
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Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction I recently proved that
$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$
using mathematical induction. I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.
| We begin by writing $k^3$ in a more clever fashion: $k^3 = k(k-1)(k-2) + 3k^2 - 2k$ :
$$\sum_{k=0}^n k^3 = \sum_{k=0}^n k(k-1)(k-2) + 3k^2 -2k$$
Distributing the summation and adjusting our indices we obtain: $$ \sum_{k=3}^n k(k-1)(k-2) + \sum_{k=0}^n 3k^2 - \sum_{k=0}^n 2k$$
Notice, $$k(k-1)(k-2) = \frac {k!}{(k-3)!}$$
Now we have
$$\sum_{k=3}^n \frac {k!}{(k-3)!} + \sum_{k=0}^n 3k^2 - \sum_{k=0}^n 2k$$
Notice that we have nearly obtained the binomial expansion of K choose 3, all we need to do is divide by 3! So we offset this by also taking the product of 3!
$$\sum_{k=3}^n \frac {k!}{(k-3)!} = 3!\sum_{k=3}^n \frac {k!}{(k-3)!3!} = 3!\sum_{k=3}^n\binom{k}{3} = 3!\binom{n+1}{4}$$
We have now obtained
$$\sum_{k=0}^n k^3 = 3!\binom{n+1}{4} + 3\sum_{k=0}^n k^2 - 2\sum_{k=0}^n k$$
Focusing solely on the right-hand side we have
$$6\biggl(\frac {(n+1)!}{(n-3)!4!}\biggr) + 3\sum_{k=0}^n k^2 - 2\sum_{k=0}^n k$$
Assuming we already know the sum of the sequence of integers and squared integers (the 2 sums we have left) we have
$$ \frac {(n+1)(n)(n-1)(n-2)}{4} + 3\frac{n(n+1)(2n+1)}{6} - 2\frac {n(n+1)}{2}$$
Generating common denominators and with a bit of algebra we now have
$$ \frac {n^4-2n^3-n^2+2n+4n^3+6n^2+2n-4n^2-4n}{4}$$
Combining like-terms we have reached our solution:
$$ \frac {n^4+2n^3+n^2}{4} = \biggl(\sum_{k=0}^n k\biggr)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "184",
"answer_count": 28,
"answer_id": 10
} |
Proving a series for the Watson Triple Integrals? It's long known that the first Watson triple integral evaluates to,
$$I_1 = \frac{1}{\pi^3} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \frac{dx \, dy \, dz}{1-\cos x \cos y \cos z} = \frac{\Gamma^4\left(\tfrac{1}{4}\right)}{4\pi^3} = 1.3932039\dots$$
Apparently, it is also equivalent to the simple infinite series,
$$I_1= \frac{4}{3} \sum_{n = 0}^{\infty} \frac{(4n)!}{n!^4} \frac{1}{(2 \cdot 18^2)^n} $$
which is connected to the Ramanujan-type pi formula,
$$\frac{1}{\pi}= \frac{2}{9} \sum_{n = 0}^{\infty} \frac{(4n)!}{n!^4} \frac{7n+1}{(2 \cdot 18^2)^n} $$
I found such series for $I_1$, $I_2$, and $I_3$. Using Mathematica, it is easy to see that they agree to an arbitrary number of decimal digits. But if you can prove them rigorously, it would be interesting to know how. More details are at,
http://sites.google.com/site/tpiezas/0025
| The sum
$$
\frac{4}{3} \sum_{n \ge 0} \frac{(4 n)!}{(n!)^4} z^4 = \frac{4}{3} \sum_{n \ge 0} \frac{ (\frac{1}{4})_n (\frac{3}{4})_n (\frac{1}{2})_n }{ (1)_n (1)_n} \frac{(256 z)^n}{n!} = \frac{4}{3} {}_3F_2\left( \left. \begin{array}{c} \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \\ 1, 1 \end{array} \right| 256 z \right)
$$
This particular hypergeometric function can be expressed in term of complete elliptic integral:
$$
{}_3F_2\left( \left. \begin{array}{c} \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \\ 1, 1 \end{array} \right| w \right) = \frac{4 \sqrt{2} }{ \pi^2 } \frac{ K\left( \frac{1}{2}- { \frac{1}{\sqrt{2(1+ \sqrt{1-w} )} } } \right)^2}{ \sqrt{(1 + \sqrt{1 - w}) } }
$$
For $z =\frac{1}{2 \cdot 18^2}$, $w =\frac{32}{81}$. Then, it results that
$$
I_1 = \frac{4 \sqrt{2}}{\pi^2} \left(K\left(\frac{1}{2} - \frac{3 \sqrt{2}}{8}\right)\right)^2
$$
Now elliptic integrals for certain moduli are known to related to gamma functions.
These are known as singular values of elliptic integrals.
| {
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Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
| We can try by induction, that is
*
*base case: $1^3=1^2$
*induction step: assume $1^3 + 2^3 +...+ n^3 = (1 + 2 +...+ n)^2=\frac{n^2(n+1)^2}{4}$ then
$$1^3 + 2^3 +...+ n^3+(n+1)^3=\frac{n^2(n+1)^2}{4}+(n+1)^3=\\=(n+1)^2\left(\frac{n^2}{4}+(n+1)\right)=(n+1)^2\frac{n^2+4n+4}{4}=\frac{(n+1)^2(n+2)^2}{4}$$
| {
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The square of an integer has form $3n$ or $3n+1$ Prove that if $n\in\mathbb Z$, then $n^2$ is of the form $3q$ or $3q+1$ for some $q\in\mathbb Z$
I would like to show that 3q+2 is = 3q+1 thus $n^2$ can be of the form of 3q or 3q+1.
Case one
$(3k)^2=(3k)(3k)=9k^2=3(3k^2)$ and is still of the form $3q$
When $q=3k^2$
$3q$
Case two
$(3k+1)^2= (3k+1)(3k+1)= 9k^2+6k+1$
$(9k^2+6k)+1 =3(3k^2+2k)+1$
this is of the form $3q+1$ when $q=3k^2+2k$
$3q+1$
Case three
$(3k+2)^2= (3k+2)(3k+2) =9k^2+12k+4$
$(9k^2+12k+4) = 3(3k^2+4k+1)+1$
This is of the form $3q+1$ when $q=3k^2+4k+1$
$3q+1$
using a direct proof with cases we see that when $n$ is of the form $3k$ it's in the form $3q$ after squaring. Also when n is in the form $3k+1$, $n$ squared is still in the form $3q+1$ after squaring. Lastly we saw that when $n$ was in the form $3k+2$ we could simplify to the form $3q+1$.
| When working by cases, you need to go through all three of them. So you should square $3q$ and find $(3q)^2=9q^2=3k$ for $k=3q$, then square $3q+1$ and $3q+2$ and find they equal $3k+1$ for some $k$. Under case two, you need an equal sign between the two lines. You skipped squaring $3q+2$. You also need parentheses around $3q+1$ when you square it-$3q+1^2\ne(3q+1)^2$. But the basic idea is fine.
| {
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Properties of Fermat primes Fermat primes 17 and 257 appear a lot in the prime composition of numbers of the form $a^{2^n}+1$. For example, $11^8+1$ is divisible by 17 and $11^{32}+1$ is divisible by 257.
I have verified the following statement for 3, 5, 17 and 257 but can't prove it. I would be grateful if someone could provide a hint.
Let $F = 2^{2^k} + 1$ be a Fermat prime.
Then for $n=0, 1, ..., 2^k-1, x^{2^n} + 1 \equiv 0 \mod F$ has exactly $2^n$ solutions.
For example, when k=2 we have:
$(n=0) \hspace{10pt} x^1 + 1 \equiv 0 \mod 17$ has one solution (x=16).
$(n=1) \hspace{10pt} x^2 + 1 \equiv 0 \mod 17$ has two solutions (x=4 and 13).
$(n=2) \hspace{10pt} x^4 + 1 \equiv 0 \mod 17$ has four solutions (x=2, 8, 9 and 15).
$(n=3) \hspace{10pt} x^8 + 1 \equiv 0 \mod 17$ has eight solutions (x=3, 5, 6, 7, 10, 11, 12 and 14).
| This sort of rephrases Henning's answer. (That one is bottom up, this one is top down, which I like more personally)
Fermat's little theorem says that $x^{2^{2^k}}-1$ has $2^{2^k}$ solution mod $p$. Now $x^{2^{2^k}}-1 = \left (x^{2^{2^{k-1}}} - 1 \right)\left(x^{2^{2^{k-1}}} + 1 \right)$, each factor having at most $2^{2^{k-1}}$ solutions because the degree of each factor is $2^{2^{k-1}}$. As their product has $2^{2^k}$ solutions we are forced to conclude that each factor has exactly $2^{2^{k-1}}$ solutions.
Now take the factor $x^{2^{2^{k-1}}} - 1$ and repeat the same argument.
| {
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"question_score": "6",
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Trouble with elementary number theory I'm trying to solve the following two problems:
Show $n^{12}\equiv 1\pmod{72}$ when $(72,n)=1$
and
Compute ${7^8}^9\pmod{100}$
For the first, I saw that $n^{24}=n^{\varphi(72)}\equiv1\pmod{72}$ by Euler's theorem, but this leaves us with $n^{12}n^{12}\equiv 1\pmod{72}$. This could mean $n^{12}\equiv 1\pmod{72}$ or $n^{12}\equiv 71\pmod{72}$. I'm not sure how to argue correctly to conclude it's only $n^{12}\equiv 1\pmod{72}$.
For the second one, I practically did brute force. ${7^8}^9\pmod{100}$ can be solved by considering $8^{9}\pmod{\varphi(100)=40}$. Since $8$ is cyclic with order $4: 8^5\equiv 8\pmod{40}$, we know $8^9=8^5\cdot8^4\equiv 8^5\equiv 8$. So look at $7^8\pmod{100}$. 7 is also cyclic with $7^4\equiv 1\pmod{100}$. So $7^8=(7^4)^2\equiv 1\pmod{100}$. I knew these "cyclic" things by manually computing and reducing, though.
Thank you for any help... (note this is not homework)
| $(1)$ Using Carmichael function, $$\lambda(72)=\text{lcm}(\lambda(8),\lambda(9))=\text{lcm}(2,6)=6$$
$$\implies n^6\equiv1\pmod{72}\text{ if }(n,72)=1 $$
$(2)$ Observe that $\displaystyle7^2=49=50-1\implies 7^4=(50-1)^2=50^2-2\cdot50\cdot1+1\equiv1\pmod{100}$
$\displaystyle\implies7^n\equiv1\pmod{100}$ if $4|n$
Now, $8^9=(2^3)^9=2^{27}$
| {
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} |
Matrix multiplication: is C(AB) the same as (CA)B? I would like to show that $(\mathbf{A} \mathbf{B})^{-1} = \mathbf{B}^{-1} \mathbf{A}^{-1}$, where $\mathbf{A}$ and $\mathbf{B}$ are $N \times N$ square matrices.
I think that this can be done as follows:
First, note that $(\mathbf{A}\mathbf{B})^{-1} (\mathbf{A} \mathbf{B}) = \mathbf{1}$ and also that $\mathbf{B}^{-1} \mathbf{A}^{-1} \mathbf{A} \mathbf{B} = \mathbf{B}^{-1} \mathbf{1} \mathbf{B} = \mathbf{B}^{-1} \mathbf{B} = \mathbf{1}$ (where $\mathbf{1}$ is the unit matrix). Thus
$(\mathbf{A} \mathbf{B})^{-1} \mathbf{A} \mathbf{B} = \mathbf{B}^{-1} \mathbf{A}^{-1} \mathbf{A} \mathbf{B}$
which implies that
$(\mathbf{A} \mathbf{B})^{-1} = \mathbf{B}^{-1} \mathbf{A}^{-1}$.
I am not sure that this is correct. It seems almost too easy! The correctness of my working is based on my assumption that $\mathbf{C} (\mathbf{A}\mathbf{B}) = (\mathbf{C} \mathbf{A}) \mathbf{B}$. Is this correct?
In normal linear algebra, this is the case. But is it the case in matrix multiplication? I am not sure whether the brackets result in a different order of multiplication and thus a different result.
| As mentioned in the comments on my question, it is possible to demonstrate to yourself the property of associativity by switching to the index representation of matrices. I did this to verify to myself what was said above.
To show that $\mathbf{C}(\mathbf{A}\mathbf{B}) = (\mathbf{C} \mathbf{A}) \mathbf{B}$, note that
$ (\mathbf{C}(\mathbf{A}\mathbf{B}))_{ij} = \sum_{l} C_{il}(\mathbf{A}\mathbf{B})_{lj} = \sum_{l}\sum_{k} C_{il}A_{lk}B_{kj}$
and similarly
$ ((\mathbf{C}\mathbf{A})\mathbf{B})_{ij} = \sum_{l} (\mathbf{C}\mathbf{A})_{il} B_{lj} = \sum_{l}\sum_{k} C_{ik}A_{kl}B_{lj}$.
Both double summations can be seen to run over the same range of terms, and so we can see that the matrix elements $(\mathbf{C}(\mathbf{A}\mathbf{B}))_{ij}$ and $((\mathbf{C}\mathbf{A})\mathbf{B})_{ij}$ are the same and thus $\mathbf{C}(\mathbf{A}\mathbf{B}) = (\mathbf{C} \mathbf{A}) \mathbf{B}$.
| {
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} |
Beta Function -- finding a lower bound based on parameters I would like to show that
$$ 1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) \geq \frac{1}{c+1}.$$
for all $c \geq 2$.
I have plotted it out for $c$ up through 200, and it seems to hold.
Does anyone have any tips on if this can be formally shown? I imagine an expert on the beta function might know...
| The definition of the beta function is
$$Beta\left(c+1,\frac{1}{c}\right) = \int_0^1 t^c (1-t)^{1/c-1} dt.$$
Since $c \geq 2$ and $t \in [0,1]$, $t^c \leq t$. Thus (applying integration by parts to the first integral), we have
$$Beta\left(c+1,\frac{1}{c}\right) \leq \int_0^1 t (1-t)^{1/c-1} dt = \left. -c t (1-t)^{1/c}\right|_0^1 + c\int_0^1 (1-t)^{1/c} dt$$ $$= \left. \frac{-c (1-t)^{1/c+1}}{1+1/c} \right|_0^1 = \frac{c}{1+1/c} = \frac{c^2}{c+1}.$$
Therefore,
$$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) \geq 1- \frac{c}{c+1} = \frac{1}{c+1}.$$
The bound isn't very tight as $c$ increases, though, as we can see from the asymptotics:
$$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) = \frac{\log c + \gamma}{c} + O\left(\frac{(\log c)^2}{c^2}\right).$$
Derivation:
$$Beta\left(c+1,\frac{1}{c}\right) = \frac{\Gamma(c+1) \Gamma \left(\frac{1}{c}\right)}{\Gamma\left(c+1+\frac{1}{c}\right)} = \frac{c \Gamma(c+1) \Gamma \left(1+\frac{1}{c}\right)}{\Gamma\left(c+1+\frac{1}{c}\right)}.$$
Now, $\Gamma \left(1+\frac{1}{c}\right) = 1 - \frac{\gamma}{c} + O\left(\frac{1}{c^2}\right)$. (This is via the Maclaurin series for $\Gamma(x+1)$; see Expression 8.321 in Gradshteyn and Ryzhik.)
Also (see the DLMF), $$\frac{\Gamma(c+1)}{\Gamma\left(c+1+\frac{1}{c}\right)} = c^{-1/c} \left(1 + O\left(\frac{1}{c^2}\right)\right) = \exp\left(\frac{-\log c}{c}\right)\left(1 + O\left(\frac{1}{c^2}\right)\right)$$
$$= \left(1 - \frac{\log c}{c} + O\left(\frac{(\log c)^2}{c^2}\right)\right)\left(1 + O\left(\frac{1}{c^2}\right)\right) = \left(1 - \frac{\log c}{c} + O\left(\frac{(\log c)^2}{c^2}\right)\right).$$
Putting this all together, we have
$$1-\frac{1}{c}Beta\left(c+1,\frac{1}{c}\right) = 1 - \frac{c}{c}\left(1 - \frac{\gamma}{c} + O\left(\frac{1}{c^2}\right)\right)\left(1 - \frac{\log c}{c} + O\left(\frac{(\log c)^2}{c^2}\right)\right)$$
$$= \frac{\log c + \gamma}{c} + O\left(\frac{(\log c)^2}{c^2}\right).$$
| {
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Number of solutions - please check my solution?
Determine, in accordance to $k$, how many solutions does the given
system of equations have:
$$ \begin{cases}kx+(k+1)y=k-1\\4x+(k+4)y=k\end{cases} $$
And check, for which values of $k$ this system has exactly one solution lying within the boundaries of a triangle of vertexes: $A=(0, 0), B=(\frac{2}{3},
0), C=(0, 2)$.
Can I just solve it using determinants? I mean:
$$D=k(k+4)-4(k+1)=k^2-4=(k+2)(k-2)$$
$$D_x=(k-1)(k+4)-(k+1)k=2k-4$$
$$D_y=k^2-4(k-1)=k^2-4k+4$$
Our system has one solution if $D$ is not $0$. This happens when $k$ is neither $2$ nor $-2$.
It has no solutions when $D=0$ and ($D_x \neq 0$ or $D_y \neq 0$). $D=0$ for $k=2$ or $k=-2$, $D_x \neq 0$ when $k \neq 2$ and $D_y \neq 0$ when $k \neq 2$. Though for $k=2$ our $D$ equals $0$, neither $D_x$ nor $D_y$ is $\neq 0$ so it has no solutions only when $k=-2$.
It has infinitely many solutions for $D=0$ (which means $k=2$ or $k=-2$), $D_x = D_y = 0$. This happens when $k=2$.
Summing up:
$$
\begin{cases}
\text{no solution}, &\text{for } k=-2,
\\
\text{infinitely many solutions}, &\text{for } k=2,
\\
\text{one solution}, &\text{for every other } k.
\end{cases}
$$
Is this all right? I need to know that before proceeding to the triangle-thing.
| First, let's get back to the mainstream terminology: We have a linear system $Ax=b$ with the data
$$
\pmatrix{k &k+1\\4 &k+4}\pmatrix{x\\y}=\pmatrix{k-1\\k}
$$
Now if $A$ is invertible then there exists a unique solution which is given by
$$
\pmatrix{x\\y} = \pmatrix{k &k+1\\4 &k+4}^{-1}\pmatrix{k-1\\k} = \frac{1}{k^2-4}\pmatrix{k+4 &-(k+1)\\-4&k}\pmatrix{k-1\\k} = \pmatrix{\frac{2}{k+2}\\ 1-\frac{4}{k+2}}
$$
As you have computed, the easiest way to conclude invertibility is having the determinant nonzero. Computing the determinant gives: $\det A = k^2-4 = (k-2)(k+2)$, Thus, we have a unique solution for all $|k|\neq 2$.
Regarding the zero determinant cases, let's look at the system at those points:
($k=2$)
$$
\pmatrix{2&3\\4&6}\pmatrix{x\\y} = \pmatrix{1\\2}
$$
We can clearly see that the second equation is the 2-multiple of the first one. Hence we actually only have $2x+3y = 1$
($k=-2$)
$$
\pmatrix{-2&-1\\4&2}\pmatrix{x\\y} = \pmatrix{-3\\-2}
$$
We can see that the right hand side is not in the range of $A$, hence there is no solution. Another way to look at it is rewriting as
$$
\pmatrix{-2\\4}x+\pmatrix{-1\\2}y = \pmatrix{-1\\2}(2x+y) = \pmatrix{-1\\2}\alpha = \pmatrix{-3\\-2}
$$
It is obvious that no alpha can satisfy that equality hence no solution.
Regarding the triangle question, you can rewrite the constraints as
$$
\pmatrix{-3 &-1\\1 &0\\0&1}\pmatrix{x\\y}\geq \pmatrix{-2\\0\\0}
$$
Now plug in the $k$ dependent solution into this and we get
$$
\pmatrix{1-\frac{2}{(k + 2)}\\
\frac{2}{(k + 2)}\\
1 - \frac{4}{(k + 2)}}\geq 0
$$
This gives us the condition on $k$:
$$
0\leq \frac{2}{k+2} \leq \frac{1}{2} \implies k>2
$$
Note that we ruled out $k=2$ case since there are infinitely many solutions in that case violating the requirement given in the question.
| {
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$n! \leq \left( \frac{n+1}{2} \right)^n$ via induction I have to show $n! \leq \left( \frac{n+1}{2} \right)^n$ via induction.
This is where I am stuck:
$$\left( \frac{n+2}{2} \right)^{n+1}
\geq \dots \geq
=2 \left( \frac{n+1}{2} \right)^{n+1}
= \left( \frac{n+1}{2} \right)^n(n+1)
\geq n!(n+1)
= (n+1)! $$
I approached this from both sides and this is the closest I can get. I realize that $n+2$ on the left has to be bigger than $n+1$ on the right, but I do not know who to show that it overpowers the factor two I have from the right.
What could I do to fill the dots? Currently, I just have it without the dots, but I would be happier if I could back it up.
| Assuming $n! \le \left( \frac{n+1}{2} \right)^n$ is true, carry the induction step
$$
(n+1) n!\leq (n+1) \left(\frac{n+1}{2}\right)^n =2
\left(\frac{n+1}{2}\right)^{n+1} \stackrel{?}{\leq} \left(\frac{n+2}{2}\right)^{n+1}
$$
But the last inequality is just
$$
2 \le \left( \frac{n+2}{n+1} \right)^{n+1} = \left( 1 + \frac{1}{n+1} \right)^{n+1}
$$
It follows because:
$$
\left( 1 + \frac{1}{n+1} \right)^{n+1} =
\sum_{k=0}^{n+1} \binom{n+1}{k} \frac{1}{(n+1)^k} \ge
\sum_{k=0}^{1} \binom{n+1}{k} \frac{1}{(n+1)^k} = 1 + (n+1) \frac{1}{n+1} = 2
$$
| {
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Which base of numerical system have $\frac 15 = 0.33333\ldots$? Which base of numerical system have $\frac{1}{5} = 0.33333\ldots$?
I need assistance in solving this one.
| If we are working in base $b$ (we must have $b\gt3$), then $0.3333\ldots$ is
$$0.3333\ldots = \frac{3}{b} + \frac{3}{b^2} + \frac{3}{b^3}+\cdots$$
Since
$$\sum_{n=1}^{\infty}\frac{3}{b^n} = \frac{3}{b}\sum_{n=0}^{\infty}\frac{1}{b^n} = \frac{3}{b}\left(\frac{1}{1-\frac{1}{b}}\right) =\frac{3}{b-1},$$
then...
| {
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Midpoint-Convex and Continuous Implies Convex Given that $$f\left(\frac{x+y}{2}\right)\leqslant \frac{f(x)+f(y)}{2}~,$$
how can I show that $f$ is convex.
Thanks.
Edit: I'm sorry for all the confusion. $f$ is assumed to be continuous on an interval $(a,b)$.
|
I would like to properly show by induction that if $ m\in\{0,1,2,\ldots,2^{n}-1\} $ then
$$
f\left(\frac{m}{2^n}x+\Big(1-\frac{m}{2^n}\Big)y\right)\le \frac{m}{2^n}f(x)
+\Big(1-\frac{m}{2^n}\Big)f(y).
$$
and then the result will follows by contuinity since $$ m=\lfloor2^nt\rfloor \in\{0,1,2,\ldots,2^{n}\} ~~~and~~\frac{\lfloor2^nt\rfloor}{2^n}\to t$$
The initial state $n = 1$ is trivial by hypothesis.
Now we assume that for every $k<n$, whenever $m\in\{0,1,2,\ldots,2^{k-1}-1\}$,we have
$$
f\left(\frac{m}{2^k}x+\Big(1-\frac{m}{2^k}\Big)y\right)\le \frac{m}{2^k}f(x)
+\Big(1-\frac{m}{2^k}\Big)f(y). \tag{I}\label{I}
$$
we want to prove \eqref{I} for $k=n$.
Let $m \in\{0,1,2,\ldots,2^{n}-1\}$ then the division by 2 yields $m =2p +r$ with $r\in \{0,1\}$
\begin{align}X&:=\left(\frac{m}{2^n}x+\Big(1-\frac{m}{2^n}\Big)y\right)= \left(\frac{2p +r}{2^n}x+\Big(1-\frac{2p +r}{2^n}\Big)y\right)
\\&= \left(\frac{p }{2^n}x+\frac{p +r}{2^n}x+\Big(1-\frac{p }{2^n}y-\frac{p +r}{2^n}y\Big)y\right)
\\&= \frac12 \left(\frac{p }{2^{n-1}}x+\frac{p +r}{2^{n-1}}x+\Big(2-\frac{p }{2^{n-1}}y-\frac{p +r}{2^{n-1}}y\Big)y\right)
\\&=\frac12\left(\frac{p }{2^{n-1}}x+ \Big(1-\frac{p }{2^{n-1}}y\Big)\right)+\frac12\left(\frac{p+r }{2^{n-1}}x+ \Big(1-\frac{p +r}{2^{n-1}}y\Big)\right) \\&:=
\color{red}{\frac{1}{2}\left(X_1+X_2\right)}\end{align}
On the other hand, since $r\in \{0,1\}$ and $m\in\{0,1,2,\ldots,2^{n}-1\}$ it is easy to check using parity that $p,p+1 \in \{0,1,2,\ldots,2^{n-1}-1\}$
that is $p+r \in \{0,1,2,\ldots,2^{n-1}-1\}$
By hypothesis of induction we obtain
\begin{align}f(X) &= f\left(\frac{1}{2}\left(X_1+X_2\right)\right)\le \frac12f(X_1) +\frac12f(X_2) \\&=
\frac12f\left(\frac{p }{2^{n-1}}x+ \Big(1-\frac{p }{2^{n-1}}y\Big)\right)+\frac12f\left(\frac{p+r }{2^{n-1}}x+ \Big(1-\frac{p +r}{2^{n-1}}y\Big)\right)
\\&\le\frac12\left(\frac{p }{2^{n-1}}f(x)+ \Big(1-\frac{p }{2^{n-1}}f(y)\Big)\right)+\frac12\left(\frac{p+r }{2^{n-1}}f(x)+ \Big(1-\frac{p +r}{2^{n-1}}f(y)\Big)\right) \\&= \frac{2p+r}{2^n}f(x)
+\Big(1-\frac{2p+r}{2^n}\Big)f(y)\\&= \frac{m}{2^n}f(x)
+\Big(1-\frac{m}{2^n}\Big)f(y).\end{align}
| {
"language": "en",
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"answer_count": 6,
"answer_id": 1
} |
A question about series $$\sum\frac{1}{2^n} = 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$
$$\sum \frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$$
Why is the first one convergent and the second divergent? First has $$\lim_ {n\to \infty}\ \frac{1}{2^n}$$ and the second $$\lim_{n\to \infty}\ \frac{1}{n}$$ Both limis are $0$. Right? Now, why does the first series converge and the second diverge? The first one looks like "the half of the other".
| The fact that the terms go to zero is a necessary condition for convergence, but it is not sufficient. These two series show that it is not enough for the terms to get small in order for the series to converge, you need the partial sums to converge.
Remember that the series is really the limit of the partial sums:
$$\sum_{i=1}^{\infty}a_i = \lim_{n\to\infty}\left(\sum_{i=1}^n a_i\right).$$
In order for the sequence on the right to converge, the terms must get closer to one another; that's why we need (require) $a_n\to 0$ as $n\to\infty$. But this is not sufficient in order to get a sequence that converges.
The first series,
$$\sum_{i=0}^{\infty}\frac{1}{2^i}$$
converges because the sequence of partial sums converges. The $n$th partial sum is equal to
$$\sum_{i=0}^{n}\frac{1}{2^i} = \frac{1 - \frac{1}{2^{n+1}}}{\frac{1}{2}} = 2 - \frac{1}{2^n},$$
so the sequence converges to $2$ as $n\to\infty$. See also this answer.
The second series,
$$\sum_{i=1}^{\infty}\frac{1}{i}$$
does not converge because the sequence of partial sums does not converge; even though the terms differ from the preceding one by very little, they eventually get very far away from any particular term. For instance,
$$\begin{align*}
\sum_{i=1}^2\frac{1}{i} &= 1+\frac{1}{2} =\frac {3}{2}\\
\sum_{i=1}^4\frac{1}{i} &= 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\
&\gt 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4} = 2\\
\sum_{i=1}^8 \frac{1}{i} &= 1+ \frac{1}{2}+\frac{1}{3}+\frac{1}{4} + \frac{1}{5}+\frac{1}{6} + \frac{1}{7}+\frac{1}{8}\\
&\gt \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) + \left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)\\
&\gt 2 + \frac{1}{2}\\
\sum_{i=1}^{16}\frac{1}{i} &= \sum_{i=1}^{8}\frac{1}{i} + \sum_{i=9}^{16}\frac{1}{i}\\
&\gt \frac{5}{2} + \sum_{i=9}^{16}\frac{1}{16} = \frac{5}{2}+\frac{8}{16}\\
&= 3\\
&\vdots
\end{align*}$$
And so on. The sum up to $\frac{1}{2^n}$ will be greater than $\frac{n}{2}+1$, which gets infinitely large.
Note that the first series is not "half the other"; half the second series would be
$$\frac{1}{2}+\frac{1}{4}+\frac{1}{6} + \frac {1}{8} + \frac{1}{10} + \frac{1}{12} + \frac{1}{14} + \frac{1}{16}+\cdots$$
you keep losing more and more terms for larger $n$ (first you are missing $\frac{1}{6}$, then $\frac{1}{10}$ and $\frac{1}{12}$ and $\frac{1}{14}$, etc.). If you did in fact have the series of "half the second", that is, if you had
$$\sum_{i=1}^{\infty}\frac{1}{2i}\quad\text{instead of}\quad \sum_{i=1}^{\infty}\frac{1}{2^i}$$
then the series would not converge, precisely because it is half a divergent series.
| {
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If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then $a + b = c + d$ I came across this problem today. I would be interested to see if anyone knows a proof for it:
If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then show that $a + b = c + d$.
| The statement has a natural geometric interpretation when $a, b, c, d \geq 0$. A circle $x^2 + y^2 = r$ and a curve $x^3 + y^3 = s$ for $x, y > 0$ are symmetric about the line $x = y$ and if they intersect, they either intersect at a point $(x,x)$ or two symmetric points $(x,y)$ and $(y,x)$ for $x \neq y$. Thus if $a^2 + b^2 = c^2 + d^2$ and $a^3 + b^3 = c^3 + d^3$ for $a, b, c, d \geq 0$, then $(a,b) = (c,d)$ or $(a,b) = (d,c)$. In either case, $a + b = c + d$.
Why the curves intersect the way they do: If $p > q$ then $x^p + y^p = 1$ is "fatter" than $x^q + y^q = 1$, and so scaled versions of these curves will intersect as above. Maybe to prove this might require algebra like in Sivaram's answer, but the picture seems clear (to me anyway).
| {
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Limit of geometric mean of N radii of an ellipse Is this equation correct?
$$\lim_{N \to \infty} \prod_{n=1}^N (a^2\cos^2 (2\pi n/N)+b^2\sin^2(2\pi n/N))^{1/N}=ab$$
If so, why?
| This is not an answer, but simply a verification of the integral in Srivatsan's answer.
Using $\cos^2(y)=\frac{1+\cos(2y)}{2}$ and $\sin^2(y)=\frac{1-\cos(2y)}{2}$,
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(a^2\cos^2(y)+b^2\sin^2(y)\right)\mathrm{d}y\tag{1}
$$
becomes
$$
\log\left(\frac{a^2+b^2}{2}\right)+\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+\frac{a^2-b^2}{a^2+b^2}\cos(2y)\right)\;\mathrm{d}y\tag{2}
$$
Letting $x=\frac{a^2-b^2}{a^2+b^2}$ and substituting $y\mapsto y/2$, the integral in $(2)$ becomes
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+x\cos(y)\right)\;\mathrm{d}y\tag{3}
$$
Taking the derivative of $(3)$ with respect to $x$ yields
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+x\cos(y)\right)\;\mathrm{d}y
&=\frac{1}{2\pi}\int_0^{2\pi}\frac{\cos(y)}{1+x\cos(y)}\mathrm{d}y\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{\frac{1-z^2}{1+z^2}}{1+x\frac{1-z^2}{1+z^2}}\frac{2\mathrm{d}z}{1+z^2}\\
&=\frac{1}{\pi(1+x)}\int_{-\infty}^\infty\frac{(1-z^2)\;\mathrm{d}z}{\left(1+\frac{1-x}{1+x}z^2\right)(1+z^2)}\\
&=\frac{1}{\pi}\int_{-\infty}^\infty\left(\frac{\frac{1}{x}}{1+z^2}-\frac{\frac{1}{x(1+x)}}{1+\frac{1-x}{1+x}z^2}\right)\mathrm{d}z\\
&=\frac{1}{x}-\frac{1}{x(1+x)}\sqrt{\frac{1+x}{1-x}}\\
&=\frac{1}{x}-\frac{1}{x\sqrt{1-x^2}}\tag{4}
\end{align}
$$
where $z=\tan(y/2)$.
Integrating $(4)$ gives
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+x\cos(y)\right)\;\mathrm{d}y=\log\left(\frac{1+\sqrt{1-x^2}}{2}\right)\tag{5}
$$
Substituting $x=\frac{a^2-b^2}{a^2+b^2}$ into $(5)$ and $(5)$ into $(2)$ yields
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(a^2\cos^2(y)+b^2\sin^2(y)\right)\mathrm{d}y=2\log\left(\frac{a+b}{2}\right)\tag{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/84987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Count the number of ways of four distinct numbers showing up when six dice are rolled Suppose we roll six fair dice, how many ways can four distinct numbers show up?
| I'll assume that you mean six-sided dice.
There are $\binom64$ ways of choosing the $4$ distinct numbers. They can either appear $3,1,1,1$ times or $2,2,1,1$ times. In the first case, there are $4$ choices of the number appearing thrice and $6\cdot5\cdot4$ choices for the positions. In the second case, there are $6$ choices for the two numbers appearing twice and $6\cdot5\cdot\binom42$ choices for the positions. Thus, the total is
$$\binom64\left(4\cdot6\cdot5\cdot4+6\cdot6\cdot5\cdot\binom42\right)=15\cdot6\cdot5\cdot(16+36)=23400\;.$$
Thus, the probability of this happening is $23400/6^6=23400/46656\approx50\%$.
The corresponding probabilities for the other numbers of distinct numbers are:
$$
\begin{eqnarray}
p(1)&=&\binom61/6^6\approx0.01\%\;,\\
p(2)&=&\binom62(2^6-2)/6^6\approx2\%\;,\\
p(3)&=&\binom63\left(3\cdot6\cdot5+3!\cdot6\cdot\binom52+\binom62\binom42\right)/6^6\approx23\%\;,\\
p(5)&=&\binom65\cdot5\cdot6\cdot5\cdot4\cdot3/6^6\approx23\%\;,\\
p(6)&=&\binom666!/6^6\approx2\%\;.
\end{eqnarray}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there any way to evaluate this limit without applying de l'Hôpital rule nor series expansion? Is there a way to evaluate this limit:
$$\lim_{x \to 0} \frac{\sin(e^{\tan^2 x} - 1)}{\cos^{\frac35}(x) - \cos(x)}$$
without using de l'Hôpital rule and series expansion?
Thank you,
| This is similar to the accepted answer by user Pacciu, but avoids the use of "asymptotics" and $\approx$ symbols as these are totally unnecessary and perhaps can be confusing to a beginner. Rules of limits combined with the following standard limits is sufficient: $$\lim_{x \to 0}\frac{\sin x}{x} = 1,\,\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\,\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ where $n$ is rational.
We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\sin(e^{\tan^{2}x} - 1)}{\cos^{3/5}x - \cos x}\notag\\
&= \lim_{x \to 0}\frac{\sin(e^{\tan^{2}x} - 1)}{e^{\tan^{2}x} - 1}\cdot\frac{e^{\tan^{2}x} - 1}{\cos^{3/5}x - \cos x}\notag\\
&= \lim_{t \to 0}\frac{\sin t}{t}\cdot\lim_{x \to 0}\frac{e^{\tan^{2}x} - 1}{\cos^{3/5}x - \cos x}\text{ (putting }t = e^{\tan^{2}x} - 1)\notag\\
&= 1\cdot\lim_{x \to 0}\frac{e^{\tan^{2}x} - 1}{\tan^{2}x}\cdot\frac{\tan^{2}x}{\cos^{3/5}x - \cos x}\notag\\
&= \lim_{t \to 0}\frac{e^{t} - 1}{t}\cdot\lim_{x \to 0}\frac{\tan^{2}x}{\cos^{3/5}x - \cos x}\text{ (putting }t = \tan^{2}x)\notag\\
&= 1\cdot\lim_{x \to 0}\frac{\sin^{2}x}{\cos^{2}x}\cdot\frac{1}{\cos^{3/5}x - \cos x}\notag\\
&= \lim_{x \to 0}\frac{\sin^{2}x}{1\cdot\{\cos^{3/5}x - \cos x\}}\notag\\
&= \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{x^{2}}{\cos^{3/5}x - \cos x}\notag\\
&= 1\cdot\lim_{x \to 0}\frac{x^{2}}{(1 - \cos x) - (1 - \cos^{3/5}x)}\notag\\
&= \lim_{x \to 0}\frac{x^{2}}{(1 - \cos x)}\cdot\dfrac{1}{1 - \dfrac{1 - \cos^{3/5}x}{1 - \cos x}}\notag\\
&= 2\lim_{x \to 0}\dfrac{1}{1 - \dfrac{\cos^{3/5}x - 1}{\cos x - 1}}\notag\\
&= 2\lim_{t \to 1}\dfrac{1}{1 - \dfrac{t^{3/5} - 1}{t - 1}}\text{ (putting }t = \cos x)\notag\\
&= 2\cdot\dfrac{1}{1 - \dfrac{3}{5}} \text{ (here }n = 3/5, a = 1)\notag\\
&= 5\notag
\end{align}
| {
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"source": "stackexchange",
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A version of Hardy's inequality involving reciprocals. How can one prove for any sequence of positive numbers $a_n, n\ge1,$ we have
$$\sum_{n=1}^\infty \frac{n}{a_1+a_2+a_3+\cdots+a_n}\le 2\sum_{n=1}^\infty \frac{1}{a_n}$$
Added later:
Apparently, this is a version of Hardy's inequality. The above is the case $p=-1$. (See the wiki for what $p$ is).
The case $p=2$ appears here: Proving $A: l_2 \to l_2$ is a bounded operator
| There is also the following N.Sedrakyan's proof.
By C-S
$$2\sum_{k=1}^n\frac{1}{a_k}>2\sum_{k=1}^n\frac{k^2}{a_k}\left(\frac{1}{k^2}-\frac{1}{(n+1)^2}\right)=2\sum_{k=1}^n\frac{k^2}{a_k}\sum_{m=k}^n\left(\frac{1}{m^2}-\frac{1}{(m+1)^2}\right)=$$
$$=2\sum_{k=1}^n\left(\frac{1}{k^2}-\frac{1}{(k+1)^2}\right)\sum_{m=1}^k\frac{m^2}{a_m}=2\sum_{k=1}^n\left(\frac{2k+1}{k^2(k+1)^2\sum\limits_{m=1}^ka_m}\sum\limits_{m=1}^ka_m\sum_{m=1}^k\frac{m^2}{a_m}\right)\geq$$
$$\geq2\sum_{k=1}^n\left(\frac{2k+1}{k^2(k+1)^2\sum\limits_{m=1}^ka_m}\left(\sum\limits_{m=1}^km\right)^2\right)=2\sum_{k=1}^n\frac{(2k+1)k^2(k+1)^2}{4k^2(k+1)^2\sum\limits_{m=1}^ka_m}>\sum_{k=1}^n\frac{k}{\sum\limits_{m=1}^ka_m}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "12",
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Is there a formula for $(1+i)^n+(1-i)^n$? I'm wondering if there is a formula for the value of $(1+i)^n+(1-i)^n$?
I calculated the first terms starting with $n=1$ to be, in order, $2$, $0$, $-4$, $-8$, $-8$, $0$, $16$, $\dots$
So it seems to be some sequence of positive and negative powers of $2$ with $0$s thrown in. Is there a more explicit formulation of what $(1+i)^n+(1-i)^n$ is, based on $n$?
With the binomial theorem, I get it equal to
$$
\sum_{k=0}^n\binom{n}{k}i^{n-k}(1+(-1)^{n-k}).
$$
Can this be made nicer?
Thanks.
| Calculate!
$(1+i)^0=1$, so $(1-i)^0=1$, so $(1+i)^0+(1-i)^0=2$.
$(1+i)^1=1+i$, so $(1-i)^1=1-i$, so $(1+i)^1+(1-i)^1=2$ .
$(1+i)^2=2i$, so $(1-i)^2=-2i$, so $(1+i)^2+(1-i)^2=0$.
$(1+i)^3=-2+2i$, so $(1-i)^3=-2-2i$, so $(1+i)^3+(1-i)^3=-4$.
$(1+i)^4=-4$, so $(1-i)^4=-4$, so $(1+i)^4+(1-i)^4=-8$.
Now the game starts all over again. The pattern of the first four entries continues forever, except that every time $n$ is incremented by $4$, we multiply by $-4$, for the simple reason that $(1+i)^4=(1-i)^4=-4$.
Let $n=4k+r$, where $r=0$, $1$, $2$, or $3$.
Then
$(1+i)^n+(1-i)^n=2(-4)^k=(-1)^k 2^{2k+1}$ if $\;r=0\;$ or $\;r=1$.
$(1+i)^n+(1-i)^n=0$ if $\;r=2$.
$(1+i)^n+(1-i)^n=(-4)^{k+1}=(-1)^{k+1}2^{2k+2}$ if $\;r=3$.
Comment: More briefly, since $(1+i)^4=(1-i)^4=-4$, we have
$$(1+i)^{4k+r}=(-4)^k (1+i)^r\qquad\text{and}\qquad (1-i)^{4k+r}=(-4)^k (1-i)^r,$$
and therefore
$$(1+i)^{4k+r}+(1-i)^{4k+r}=(-4)^k\left((1+i)^r+ (1-i)^r \right).$$
Note that the "cases" expression for $(1+i)^n+(1-i)^n$ can be made into a single expression in various ways.
| {
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"question_score": "15",
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Write the equations of the system for which $Ax=b$ is compatible $A =\left[ {\begin{array}{cc}
1&-2&1 \\ -2&1&1 \\ 1&1&-2 \\
\end{array} } \right]
\ \ \ \
\left[ {\begin{array}{cc}
1&-2&1&a \\ -2&1&1&b \\ 1&1&-2&c \\
\end{array} } \right]
\ \ \ \
\left[ {\begin{array}{cc}
1&-2&1&a \\ 0&-3&3&b+2a \\ 0&3&-3&c-a \\
\end{array} } \right]
\ \ \ \
\left[ {\begin{array}{cc}
1&-2&1&a \\ 0&-3&3&b+2a \\ 0&0&0&a+b+c \\
\end{array} } \right]
$
$A$ is the matrix. I concatenated $[A|b]$ and tried to reduce it. I end up with the last matrix, does it mean that the system is not compatible? Is there another way to solve it?
Or should I concatenate $[A|0]$ to make a homogeneous system?
UPDATE:
$A =\left[ {\begin{array}{cc}
1&-2&1 \\ -2&1&1 \\ 1&1&-2 \\
\end{array} } \right]
\ \ \ \
\left[ {\begin{array}{cc}
1&-2&1&0 \\ -2&1&1&0 \\ 1&1&-2&0 \\
\end{array} } \right]
\ \ \ \
\left[ {\begin{array}{cc}
1&-2&1&0 \\ 0&-3&3&0 \\ 0&0&0&0 \\
\end{array} } \right]
\ \ \
\left[ {\begin{array}{cc}
1&0&-1&0 \\ 0&1&-1&0 \\ 0&0&0&0 \\
\end{array} } \right]
$
After solving $[A|0]$ I get $a-c=0$ and $b-c=0$, the equations of the system seem to be: $a=c,b=c$, so $a=b=c$. Is this the correct solution?
| What you did in the Update just shows that $A{\bf x}=\bf0$ has one (independent) solution. That is the dimension of the null space of $A$ is 1.
So $A{\bf x}=\bf b$ has two independent solutions. That is, the dimension of the column space of $A$ is 2. You have an echelon form of $A$. What you need to do is find a basis for the column space of $A$. Then the set of linear combinations of elements of this basis will be the answer to your problem (that is $\bf y$ is in the column space of $A$ if and only if $A{\bf x}=\bf y$ has a solution).
That is, if I'm interpreting the problem correctly: for what vectors $\bf b$ does $A{\bf x}=\bf b$ have a solution?
You could also keep going with what you did at the start: find conditions on $a$, $b$, and $c$ so that the system represented by $\left[ {\begin{array}{cc}
1&-2&1&a \\ 0&-3&3&b+2a \\ 0&0&0&a+b+c \\
\end{array} } \right]$ has a solution.
| {
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Circle locus, how to satisfy the equation.
$A(-3,1), B(0,-5), P(X,Y)$
If $|AP| = 2|BP|$ prove that $x$ and $y$ satisfy the equation:
\begin{aligned}
\ x^2+y^2-2x+14y+30 =0
\end{aligned}
I get as far as determining the co-ordinates like so
\begin{aligned}
\ \sqrt{(x+3)^2+(y-1)^2}= 2\sqrt{(x)^2+(y+5)^2}
\end{aligned}
To
\begin{aligned}
\ x^2+y^2+2y+40-6x =0
\end{aligned}
Which gives me $(3, -1)$, this won't satisfy, is my method correct?
| If we start from the equation
$$
\sqrt{(x+3)^2+(y-1)^2}= 2\sqrt{(x)^2+(y+5)^2},
$$
and square both sides, we obtain the equivalent equation
$$x^2+6x+9+y^2-2y+1=4x^2+4y^2+40y+100,$$
which simplifies to
$$3x^2+3y^2-6x+42y+90=0,$$
which is equivalent to
$$x^2+y^2-2x+14y+30=0.$$
Comment: Draw the line segment that joins $A$ to $B$. It is fairly easy to see that $(-1,-3)$ is the point on this line segment that is twice as far from $A$ as it is from $B$. So $(-1,-3)$ ought to be on our circle, and indeed it is. That provides a partial check on the correctness of our computations.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$ Q. Prove
$$\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$$
My try: expand and got:
$$\frac{5^{2x-2}+2(5^{x^2-1})+5^{2x+2}}{5^{2x-2}+5^{2x+2}}$$
Now what? I find my pre-calculus skills esp with Indices, Logarithms & Trigo lacking ... need to know how to apply the formulas more
| Factor as:
$$\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{(5^{x-1})^2(1+5^2)^2}{(5^{x-1})^2(1+25^2)}$$
Then simplify.
| {
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"timestamp": "2023-03-29T00:00:00",
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A question about finding the time difference Two people $A,~B$ are participating a running race.Initially of course they are both at rest.They then proceed with constant acceleration. $A$ covers the last $\dfrac{1}{4}$ of the distance in $3$ second while $B$ covers the last $\dfrac{1}{3}$ of his distance in $4$ second.
How long after the winner finishing the race will the other complete the race, assuming he completes the race?
| Let $a$ and $b$ be the accelerations of $A$ and $B$, respectively, and let $d$ be the length of the race. Let $x_A(t)$ and $x_B(t)$ be the distances covered by $A$ and $B$, respectively, at time $t$ after the start of the race; $x_A(t)=\frac12at^2$, and $x_B(t)=\frac12bt^2$. $A$ reaches the $3/4$ mark at time $t_1$ given by the equation $$\frac34d=\frac12at_1^2\tag{1}$$ and finishes the race at time $t_2$ given by the equation $$d=\frac12at_2^2\;,\tag{2}$$ and we’re told that $t_2-t_1=3$, so we can rewrite $(2)$ as $$d=\frac12a(t_1+3)^2\tag{3}$$ and solve $(1)$ and $(3)$ for $t_1$. Multiplying $(1)$ by $4/3$, we get $$d=\frac23at_1^2\;,$$ so
$$\begin{align*}
&\frac23at_1^2=\frac12a(t_1+3)^2\;,\\
&4t_1^2=3(t_1+3)^2=3t_1^2+18t_1+27\;,\\
&t_1^2-18t_1-27=0\;,\\
&t_1=\frac12(18\pm\sqrt{432})=9\pm 6\sqrt3\;,
\end{align*}$$
and since $3-6\sqrt3<0$, we must have $t_1=9+6\sqrt3$ and $t_2=12+6\sqrt3$.
Following the same procedure for $B$, starting with the system
$$\left\{\begin{align*}
&\frac23d=\frac12bt_3^2\\
&d=\frac12b(t_3+4)^2\;,
\end{align*}\right.$$
we get $3t_3^2=2(t_3+4)^2=2t_3^2+16t_3+32$, $t_3^2-16t_3-32=0$, $t_3=8+4\sqrt6$, and $t_3+4=12+4\sqrt6$.
Thus, $A$ finishes at time $12+6\sqrt3\approx 22.392$, and $B$ finishes at time $12+4\sqrt6\approx 21.798$. $B$ is the winner by $6\sqrt3-4\sqrt6=(6-4\sqrt2)\sqrt3\approx 0.343$ seconds.
| {
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Writing a Polar Equation for the Graph of an Implicit Cartesian Equation
If $(x^2+y^2)^3=4x^2y^2,$ then $r=\sin 2\theta$ for some $\theta$.
Using $r^2=x^2+y^2, x=r\cos\theta,y=r\sin\theta$, it's easy to get $r^2=\sin^22\theta$.
But I don't know what to do next, since $r$ could be negative in $r=\sin2\theta.$
Actually the original problem is to show that the affine variety $V((x^2+y^2)^3-4x^2y^2)$ is contained in the four-leaved rose, whose polar equation is certainly $r=\sin 2\theta$. (Exercise 7(b), section 1.2, Ideals, Varieties and Algorithms, 3rd edition, David Cox etc.)
| Avoiding the issue entirely:
$$r = \sin(2\theta) = 2\sin\theta\cdot \cos\theta$$
$$r^3 = 2(r\sin\theta)(r\cos\theta)$$
$$x = r\cos\theta$$
$$y = r\sin\theta$$
$$r^3 =2xy$$
$$r = (x^2 + y^2)^{\frac 12}$$
$$(x^2+y^2)^{\frac 32} =2xy$$
$$(x^2+y^2)^3=4x^2y^2$$
| {
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Simplify _Elementary Calculus_ section 1.6 problem 25 Let's try this again. We're still on problem 25 in section 1.6 of Elementary Calculus.
$$\frac{3-\sqrt{c+2}}{c-7}$$
My first thought is (again) to multiply by $3+\sqrt{c+2}$:
$$=\frac{(3-\sqrt{c+2})(3+\sqrt{c+2})}{(c-7)(3+\sqrt{c+2})}$$
$$=\frac{9-(c+2)}{(c-7)(3+\sqrt{c+2})}$$
$$=\frac{9-(c+2)}{3c+c\sqrt{c+2}-7\sqrt{c+2}-21}$$
This looks "simplified" to me, so I proceed to substitute $c=7+\epsilon, \epsilon \in \mathbb{R}^*, \epsilon \approx 0$:
$$=\frac{9-(9+\epsilon)}{3(7+\epsilon)+(7+\epsilon)\sqrt{7+\epsilon+2}-7\sqrt{9+\epsilon}-21}$$
$$=\frac{-\epsilon}{21+3\epsilon+7\sqrt{9+\epsilon}+\epsilon\sqrt{9+\epsilon}-7\sqrt{9+\epsilon}-21}$$
$$=-\frac{\epsilon}{3\epsilon + \epsilon\sqrt{9+\epsilon}}$$
Knowing the answer is $-\frac{1}{6}$, it seems likely that this somehow reduces to $-\frac{\epsilon}{6\epsilon}$ (apart from some error), but I don't see how to get from $3\epsilon+\epsilon\sqrt{9+\epsilon}$ to $3\epsilon+3\epsilon=6\epsilon$.
Thanks for your help again!
| Everything is fine, apart from the fact that you don't think it is fine.
First, a small piece of advice. Don't multiply unless you have to. At a certain stage you had
$$\frac{9-(c+2)}{(c-7)(3+\sqrt{c+2})},$$
which you expanded (but certainly did not simplify) to
$$=\frac{9-(c+2)}{3c+c\sqrt{c+2}-7\sqrt{c+2}-21}.$$
This is correct but looks worse to me. Instead, simplify the top to $7-c$, which cancels with the $c-7$ at the bottom to give $-1$. Or if you really want to, note that $7-c=-\epsilon$, and $c-7=\epsilon$, and cancel the $\epsilon$ (or, to be fancy, divide top and bottom by $\epsilon$).
You managed, however, to get through the mess to get the right expression
$$-\frac{\epsilon}{3\epsilon + \epsilon\sqrt{9+\epsilon}},$$
but there you seemed to be stuck.
Note then that the two bottom terms have a common factor of $\epsilon$. "Take it out" to get $\epsilon(3+\sqrt{9+\epsilon})$, and cancel with the $\epsilon$ on top.
However we do these things, we end up with
$$-\frac{1}{3+\sqrt{9+\epsilon}}.$$
Find the standard part of this. The standard part of $9+\epsilon$ is $9$, so the standard part of $\sqrt{9+\epsilon}$ is $3$, so the standard part of the whole thing is $-\dfrac{1}{6}$.
| {
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Solving integral equation with Laplace's Transform. I'm trying to prove the following
$$\int\limits_0^\infty {\frac{{\cos tu}}{{{u^2} + 1}}\log udu} = - \frac{\pi }{2}\int\limits_0^\infty {\frac{{\sin tu}}{{{u^2} + 1}}du} $$
The original problem suggested the use of the Laplace Transform, thus I have
$$\int\limits_0^\infty {\frac{s}{{{u^2} + {s^2}}}\frac{{\log u}}{{{u^2} + 1}}du = } - \frac{\pi }{2}\int\limits_0^\infty {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$$
ADD The RHS transform can be evaluated as follows:
$$- \frac{\pi }{2}\int\limits_0^\infty {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$$
$$ - \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} $$
Now by partial fractions you can separate and get:
$$ - \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} = - \frac{\pi }{4}\frac{1}{{{s^2} - 1}}\left[ {\log \left( {\frac{{m + 1}}{{m + {s^2}}}} \right)} \right]_0^\infty $$
$$ - \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} = -\frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$
For the second one I make a similar manipulation:
$$\frac{1}{{{s^2} + {u^2}}}\frac{1}{{{u^2} + 1}} = \frac{1}{{{s^2} - 1}}\left( {\frac{1}{{{u^2} + 1}} - \frac{1}{{{u^2} + {s^2}}}} \right)$$
$$\frac{s}{{{s^2} - 1}}\int\limits_0^\infty {\left( {\frac{{\log u}}{{{u^2} + 1}} - \frac{{\log u}}{{{u^2} + {s^2}}}} \right)du} $$
Substitution to show integral over $\mathbb{R}$ of an odd function is zero so,
$$\frac{s}{{{s^2} - 1}}\left( {\int\limits_{ - \infty }^\infty {\frac{{x{e^x}}}{{{e^{2x}} + 1}}dx} - \int\limits_0^\infty {\frac{{\log u}}{{{u^2} + {s^2}}}} }du \right)$$
$$ - \frac{s}{{{s^2} - 1}}\int\limits_0^\infty {\frac{{\log u}}{{{u^2} + {s^2}}}}du $$
And again a suitable $u = m s$ produces
$$ - \frac{1}{{{s^2} - 1}}\int\limits_0^\infty {\frac{{\log m + \log s}}{{{m^2} + 1}}} dm = - \frac{{\log s}}{{{s^2} - 1}}\int\limits_0^\infty {\frac{1}{{{m^2} + 1}}} dm = - \frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$
| $$- \frac{\pi }{2}\int\limits_0^\infty {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$$
$$ - \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} $$
Now by partial fractions you can separate and get:
$$ - \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} = - \frac{\pi }{4}\frac{1}{{{s^2} - 1}}\left[ {\log \left( {\frac{{m + 1}}{{m + {s^2}}}} \right)} \right]_0^\infty $$
$$ - \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} = -\frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$
For the second one I make a similar manipulation:
$$\frac{1}{{{s^2} + {u^2}}}\frac{1}{{{u^2} + 1}} = \frac{1}{{{s^2} - 1}}\left( {\frac{1}{{{u^2} + 1}} - \frac{1}{{{u^2} + {s^2}}}} \right)$$
$$\frac{s}{{{s^2} - 1}}\int\limits_0^\infty {\left( {\frac{{\log u}}{{{u^2} + 1}} - \frac{{\log u}}{{{u^2} + {s^2}}}} \right)du} $$
Substitution to show integral over $\mathbb{R}$ of an odd function is zero so,
$$\frac{s}{{{s^2} - 1}}\left( {\int\limits_{ - \infty }^\infty {\frac{{x{e^x}}}{{{e^{2x}} + 1}}dx} - \int\limits_0^\infty {\frac{{\log u}}{{{u^2} + {s^2}}}} }du \right)$$
$$ - \frac{s}{{{s^2} - 1}}\int\limits_0^\infty {\frac{{\log u}}{{{u^2} + {s^2}}}}du $$
And again a suitable $u = m s$ produces
$$ - \frac{1}{{{s^2} - 1}}\int\limits_0^\infty {\frac{{\log m + \log s}}{{{m^2} + 1}}} dm = - \frac{{\log s}}{{{s^2} - 1}}\int\limits_0^\infty {\frac{1}{{{m^2} + 1}}} dm = - \frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$
| {
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$\lim _{(x,y) \to (0,0)}$ $\frac{(x^3-y^2)^2}{x^4+y^4}$- two methods different conclusion. I tried to use couple of inequalities in this way:
$0\leq\left|\frac{(x^3-y^2)^2}{x^4+y^4}\right| \leq \frac{(x^3)^2}{x^4}=x^2$, so now I can use the squeeze theorem and conclude that this limit is 0, but when I choose a route such as $y=Kx$ I get that the limit is $\frac {K^4}{1+K^4}$, which depends on K, so apparently there's no limit when $(x,y) \to (0,0)$.
Where's my mistake?
Thanks a lot.
| This can be done simply:
Take a path with $x=0$ and $y\rightarrow0$; then ${(x^3-y^2)^2\over x^4+y^4}={y^4\over y^4}=1$.
Take a path with $y=0$ and $x\rightarrow0$; then ${(x^3-y^2)^2\over x^4+y^4}={x^6\over x^4}=x^2\rightarrow0$.
So $ \lim\limits_{(x,y)\rightarrow(0,0)} { (x^3-y^2)^2 \over x^4+y^4}$ does not exist.
As mentioned in the comments, your initial inequality is incorrect.
Your second argument is correct. For $y=Kx$, you have:
$$
{(x^3-y^2)^2\over x^4+y^4} ={(x^3-K^2x^2)^2\over x^4+K^4 x^4}
={(x-K^2)^2\over 1+K^4}\quad\buildrel{x\rightarrow0}\over{\longrightarrow}\quad{K^4\over 1+K^4} .
$$
From this, you can see that the limit does not exist (take, e.g., $K=0$ and $K=1$).
| {
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reducing $x^4+1$ in $\mathbb{Z}_p[x]$
Possible Duplicate:
reducible polynomial modulo every prime
Ok so we have to prove the following: If $R=\mathbb{Z}_p$ for $p$ a prime, then, $x^4+1$ is reducible over $R[x]$. This is are my ideas (please let me know if there is an easier way to tackle this problem):
First if $p=2$ then we know that $x+1$ is a factor since $1^4+1=0$ and it follows that $x-1=x+1$ is a factor.
Hence from now on let us assume $p$ is odd. First if $x^4+1=(x^2+ax+b)(x^2+cx+d)$, I have that the leading coefficients in the factors is $1$ since I can always force this, and now I obtain the following:
$bd=1$
$ad+bc=0$
$d+ac+b=0$
$a+c=0$
Where the above are obtained by multiplying out the factors and setting the coefficients for the value they ought to be.
However, we can use the last equation and plug in the second equation, to obtain $a(b-d)=0$ and since $\mathbb{Z}_p$ is a field, then we know it has no zero divisors.
If $a=0$ then we have that $b=-d$ and hence $-d^2=1$, which means that $d^2=-1$. By using quadratic residues (here is where I worry since my class has not cover that) we have that if $p\equiv 1\pmod{4}$ then we know that (The Legendre Symbol) $(\frac{a}{p})=\bar{a}^{(p-1)/2}$, using $a=-1$, we have that $(\frac{a}{p})=1$, so $-1$ is not a quadratic residue, i.e., it is a square., hence
$x^4+1=(x^4-(-1))=(x^2-d)(x^2+d)$
I am stuck in the case when $p\equiv 3\pmod{4}$. Any help, comments, hints, or thoughts would be appreciated!
| If $p \equiv 1 \pmod{4}$, we have the factorization $$x^4 + 1 = (x^2 + i)(x^2 - i),$$ where $i \in \mathbb{F}_p$ is the element such that $i^2 = -1$.
When $p \equiv 3 \pmod{4}$, if $2$ is a square $\bmod{p}$ (say $a^2 = 2$), then
$$
x^4 + 1 = (x^2 + ax + 1)(x^2 - ax + 1).
$$
If $2$ is not a square, then $-2$ is a square $\bmod{p}$ (see my answer here for a proof). If, say $b^2 = -2$, then
$$
x^4 + 1 = (x^2 +bx -1)(x^2 -bx -1).
$$
| {
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Area of a circle I've tried to find as a personnal exercise where the formula $A=\pi R^2$ comes from.
After drawing the problem, I've found that $A = 2\int\limits_{-R}^{R}\sqrt{R^2-t^2}dt$. How can I calculate this ?
| This is a classic case of trigonometric substitution. Set $t=R\sin\theta$, $-\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}$.
Then $R^2 - t^2 = R^2(1-\sin^2\theta) = R^2\cos^2\theta$, hence $\sqrt{R^2-t^2} = \sqrt{R^2\cos^2\theta} = |R\cos\theta| = R\cos\theta$, because $R\geq 0$ and $\cos\theta\geq 0$ for $\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$.
Since $t=R\sin\theta$, $dt = R\cos\theta\,d\theta$. When $t=-R$, we have $\theta=-\frac{\pi}{2}$; when $t=R$, we have $\theta=\frac{\pi}{2}$. So we get
$$\int_{-R}^R\sqrt{R^2-t^2}\,dt = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}R\cos^2\theta\,d\theta.$$
Now you can use the formula
$$\int \cos^2\theta\,d\theta = \frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta+C$$
(which can be found by using integration by parts, then replacing $\sin^2\theta$ with $1-\cos^2\theta$, and "solving" for the integral of the cosine squared) to get the desired result.
| {
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Find the value of a? Let $a$ and $b$ be two positive numbers such that $a\gt b$. Let $G$ be the geometric mean of $a$ and $b$ (that is, $G=\sqrt{ab}$), and $H$ be the Harmonic mean of $a$ and $b$, that is, $$H = \frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$
If $4G = 5H$, what is the value of $a$?
| The Harmonic Mean of $a$ and $b$ is
$$\frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$
The Geometric Mean of $a$ and $b$ is
$$\sqrt{ab}.$$
So, to state the problem you have in a way that would be actually intelligible would be:
Let $a$ and $b$ be positive numbers such that $a\gt b$; assume that
$$4\times\text{geometric-mean(a,b)} = 4\sqrt{ab} = 5\left(\frac{2ab}{a+b}\right) = 5\times\text{harmonic-mean}(a,b).$$
What is the value of $a$?
We have
$$\begin{align*}
4\sqrt{ab} &= \frac{10ab}{a+b}\\
4(a+b) &= \frac{10ab}{\sqrt{ab}}\\
2(a+b) &= 5\sqrt{ab}\\
4(a+b)^2 &= 25ab\\
4a^2 + 8ab + 4b^2 &= 25ab\\
4a^2 -17ab + 4b^2 &=0.
\end{align*}$$
You can view this as a quadratic equation in $a$; the solutions are given by
$$\frac{17b - \sqrt{(17b)^2 - 64b^2}}{8} = \frac{17b-\sqrt{225b^2}}{8} = \frac{17b-15b}{8} = \frac{b}{4}$$
(which is impossible since $a\gt b$) and
$$\frac{17b + \sqrt{(17b)^2 - 64b^2}}{8} = \frac{17b + \sqrt{225b^2}}{8} = \frac{32b}{8} = 4b.$$
So the answer is that $a$ must be $4b$.
You can verify this works: the Geometric Mean of $b$ and $4b$ is $\sqrt{4b^2} = 2b$; the Harmonic mean is
$$\frac{2(4b)b}{4b+b} = \frac{8b^2}{5b} = \frac{8b}{5}.$$
And
$$4(2b) = 5\left(\frac{8b}{5}\right).$$
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Finding derivative of $\sqrt[3]{x}$ using only limits I need to finding derivative of $\sqrt[3]{x}$ using only limits
So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator
$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \lim_{h \to 0} \frac{x+h-x}{h(\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2})}$$
$$= \lim_{h \to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x^2}}$$
$$= \frac{1}{2 \sqrt[3]{x^2}}$$
But I think it should be $\frac{1}{3 \sqrt[3]{x^2}}$ (3 instead of 2 in denominator?)
UPDATE
I found that I am using the wrong conjugate in step 1. But this (wrong) conjugate gives the same result when I multiply the numerator by it. So whats wrong with it? (I know its wrong, but why?)
| You can use a similar 'trick' to find the derivative of $ y=\sqrt[n] x $. The limit will involve multiplying the numerator $(x+h)^{1/n} - x^{1/n} $ by an appropriate expression to get $ (x + h) - x $ .
| {
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Find the value of $A/B$? $$A = x^4 ( 1 + y^4 ) + y^4 ( 1 + z^4 ) + z^4 ( 1 + x^4 )$$
$$B = x^2 y^2 z^2$$
If $x$, $y$ and $z$ are real, which of the following can be the value of $A/B$?
Options
i) 0
ii) 2
iii) 4
iv) 8
| Hint:
Expand $A$ to be a sum of $6$ terms and use $\text{AM} \ge \text{GM}$.
For the sake of completeness, we add more detail.
Since we talk of the ratio $\frac{A}{B}$, we can assume that $B \neq 0$, and thus none of $x,y,z$ are non zero.
Now $A = x^4 + x^4y^4 + y^4 +y^4z^4 + z^4 + z^4x^4 \ge 6 x^2y^2z^2 = 6B$, using $\text{AM} \ge \text{GM}$.
Thus the ratio $\frac{A}{B}$ is atleast $6$. So if there is an answer, it must be iv) 8. Note, we can eliminate $0$, as we need $B \neq 0$.
Extra Credit:
We now show that for any $r \ge 6$, there are $x,y,z$, such that we have $\frac{A}{B} = r$.
We arbitrarily choose $y = z = 1$.
This gives us the equation
$3x^4 + 3 = r x^2$.
This is a quadratic in $x^2$ and has the positive root $\frac{r + \sqrt{r^2 - 36}}{6}$.
Thus we have that $\frac{A}{B}$ can only take values $\ge 6$, and for each such value, there are $x,y,z$ for which we attain that ratio.
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"language": "en",
"url": "https://math.stackexchange.com/questions/113885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does $n^2+(n+k)^2+(n+2k)^2+ \ldots +(n+mk)^2$ have a general equation? Does $n^2+(n+k)^2+(n+2k)^2+\ldots+(n+mk)^2$ have a general formula?
e.g.
$$1^2+2^2+3^2+\ldots+n^2=n(n+1)(2n+1)/6$$
| The sum can we written as $\sum_{j=0}^m(n+jk)^2=\sum_{j=0}^mn^2+2jkn+j^2k^2$ and using the formulas $\sum_{j=0}^mj=\frac{m(m+1)}2$, $\sum_{j=0}^mj^2=\frac{m(m+1)(2m+1)}6$, we get
\begin{align*}\sum_{j=0}^m(n+jk)^2&=(m+1)n^2+knm(m+1)+k^2\frac{m(m+1)(2m+1)}6\\
&=(m+1)\left(n^2+knm+\frac{k^2m(2m+1)}6\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/114814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do you find the sine and cosine from the tangent? I'm given the problem:
If $\cot(\theta) = 1.5$ and $\theta$ is in quadrant 3, what is the value of $\sin(\theta)$?
I looked at all the related answers I could find on here, but I haven't been able to piece together the answer I need from them.
I know that $\sin^2\theta + \cos^2\theta = 1$, $ \cot^2\theta + 1 = \csc^2\theta $, and $\csc^2\theta = \frac{1}{\sin^2\theta}$
Substituting 3.25 for $\cot^2\theta + 1$ and $\frac{1}{\sin^2\theta}$ for $\csc^2\theta$ I get:
$3.25 = \frac{1}{\sin^2\theta}$
then
$\sin\theta = -\sqrt{\frac{1}{3.25}}$
This doesn't seem correct though. Can anyone help please?
edit: Sorry, meant to make that answer negative.
| Indeed, we know that
$$1.5 = \cot\theta = \frac{\cos\theta}{\sin\theta}$$
hence
$$1.5\sin\theta = \cos\theta.$$
Squaring both sides we have
$$2.25\sin^2\theta = \cos^2\theta$$
and since $\cos^2\theta = 1-\sin^2\theta$ we have
$$\begin{align*}
2.25\sin^2\theta &= 1-\sin^2\theta\\
2.25\sin^2\theta + \sin^2\theta &= 1\\
3.25\sin^2\theta &= 1.
\end{align*}$$
From this, you can figure out the value of $\sin^2\theta$. Taking square roots will tell you something about the absolute value of $\sin\theta$.
Now... why did they tell you $\theta$ was in the third quadrant?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/115865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$ How can I evaluate the integral
$$\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$$
I tried manipulating the known integral
$$\int_0^1 \frac{\ln(1 - x)}{x}dx = -\frac{\pi^2}{6}$$
but couldn't do anything with it.
| You can use double integration:
$$\int\limits_0^1 {\frac{{\log \left( {1 - x} \right)}}{{1 + x}}dx} = \int\limits_0^1 {\int\limits_0^{ - x} {\frac{{du \cdot dx}}{{\left( {1 + u} \right)\left( {1 + x} \right)}}} } $$
$$\int\limits_0^1 {\int\limits_0^x {\frac{{dm \cdot dx}}{{\left( {m - 1} \right)\left( {1 + x} \right)}}} } $$
Now make
$$m = ux $$
$$\int\limits_0^1 {\int\limits_0^1 {\frac{{x \cdot du \cdot dx}}{{\left( {ux - 1} \right)\left( {1 + x} \right)}}} } = \int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)}}} } - \int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)\left( {1 + x} \right)}}} } $$
We have that (partial fraction decomposition)
$$\frac{1}{ \left( ux - 1 \right)\left( x + 1 \right) } = \frac{u}{ \left( u + 1 \right)\left( ux - 1 \right) } - \frac{1}{ \left( x + 1 \right)\left( u + 1 \right) }$$
So we get
$$\int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)}}} } - \int\limits_0^1 {\int\limits_0^1 {\frac{{u \cdot du \cdot dx}}{{\left( {ux - 1} \right)\left( {u + 1} \right)}}} } + \int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {x + 1} \right)\left( {u + 1} \right)}}} } $$
Now:
$$\int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)}}} } = \int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{u}} du = - \frac{{{\pi ^2}}}{6}$$
$$\int\limits_0^1 {\int\limits_0^1 {\frac{{du\cdot dx}}{{\left( {x + 1} \right)\left( {u + 1} \right)}}} } = {\log ^2}2$$
For our last one,note it is the integral we're looking for
$$\int\limits_0^1 {\int\limits_0^1 {\frac{{u\cdot du\cdot dx}}{{\left( {ux - 1} \right)\left( {u + 1} \right)}}} \mathop = \limits^{ux = m} } \int\limits_0^1 {\int\limits_0^u {\frac{{dm\cdot du}}{{\left( {m - 1} \right)\left( {u + 1} \right)}}} } \mathop = \limits^{m = - x} \int\limits_0^1 {\int\limits_0^{ - u} {\frac{{dx\cdot du}}{{\left( {x + 1} \right)\left( {u + 1} \right)}}} } = \int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{ {u + 1} }}} du$$
We get
$$\int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{ {u + 1} }}} du = {\log ^2}2 - \frac{{{\pi ^2}}}{6} - \int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{ {u + 1} }}} du$$
or
$$\int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{{u + 1} }}} du = \frac{{{{\log }^2}2}}{2} - \frac{{{\pi ^2}}}{{12}}$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 7,
"answer_id": 1
} |
Why Doesn't This Integral $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$ Work? I am trying to solve this integral, which is incorrect compared to Wolfram|Alpha. Why doesn't my method work?
Find $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$
Side work:
$\sin{\theta} = \frac{3}{x}$
$x = \frac{3}{\sin{\theta}} = 3 \csc{\theta}$
$dx = -3 \csc{\theta}\cot{\theta} \ d\theta$
$-\int \frac{\sqrt{9\csc^2{\theta} - 9}}{27 \csc^3{\theta}} \cdot 3 \csc{\theta}\tan{\theta} \ d\theta$
$-\int \frac{3 \cdot \sqrt{\csc^2{\theta} - 1}}{27 \csc^3{\theta}} \cdot 3 \csc{\theta}\cot{\theta} \ d\theta$
$-\int \frac{\cot{\theta}\csc{\theta}\cot{\theta}}{3 \csc^3{\theta}} \ d\theta$
$-\frac{1}{3} \int \cos^2{\theta} \ d\theta$
$-\frac{1}{3} \int \frac{1}{2} \left(1 + \cos{\left(2\theta\right)}\right) \ d\theta$
$-\frac{1}{6} \int 1 + \cos{\left(2\theta\right)} \ d\theta$
$-\frac{1}{6} \int \ d\theta + \frac{1}{6} \int \cos{\left(2\theta\right)} \ d\theta$
Sidework:
$u = 2\theta$
$du = 2 \ d\theta$
$-\frac{1}{6} \int \ d\theta + \frac{1}{3} \int \cos{u} \ du$
$-\frac{\theta}{6} + \frac{1}{3} \cdot \sin{u} + C$
$-\frac{\theta}{6} + \frac{1}{3} \cdot \sin{2\theta} + C$
Sidework:
$\sin{\left(2\theta\right)} = 2 \sin{\theta} \cos{\theta}$
$\sin{\theta} = \frac{3}{x}$
$\cos{\theta} = \frac{\sqrt{x^2 - 9}}{x}$
$\theta = \sin^{-1}{\left(\frac{3}{x}\right)}$
$-\frac{\sin^{-1}{\left(\frac{3}{x}\right)}}{6} + \frac{2}{3} \cdot \frac{3}{x} \cdot \frac{\sqrt{x^2 - 9}}{x} + C$
$-\frac{\sin^{-1}{\left(\frac{3}{x}\right)}}{6} + \frac{2\sqrt{x^2 - 9}}{x^2} + C$
Thank you for your time.
| The derivative of $\csc \theta$ is not $\csc \theta \tan \theta$ it is $-\csc \theta \cot \theta$.
After your edit:
Try writing $\sin^{-1} (3/x)$ in the form of $\tan^{-1}$.
Also you have the integral of $\cos 2\theta$ wrong.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\int_0^{\frac{\pi}{2}}t(\frac{\sin nt}{\sin t})^4dt<\frac{\pi^2n^2}{4}$ I have a question about integral. Prove:
$$\int_0^{\frac{\pi}{2}}t\left(\dfrac{\sin(nt)}{\sin(t)}\right)^4dt<\dfrac{\pi^2n^2}{4}$$
I have tried several methods including $\sin(t)\geq\frac{2t}{\pi}$, but I can't work it out.
| First note that
$$I:=\int_0^{\frac{\pi}{2}}x\left(\frac{\sin nx}{\sin x}\right)^4\mathrm{d}x=\int_0^{\frac{\pi}{2}}\frac{x}{\sin x}\frac{|\sin nx|}{\sin x}|\sin nx|\left(\frac{\sin nx}{\sin x}\right)^2\mathrm{d}x.$$
We can obtain the following inequalities with little effort :
$$1<\frac{x}{\sin x}<\frac{\pi}{2}(0<x<\pi),\ |\sin nx|\leqslant n|\sin x|(x\in \mathbb{R}).$$
Thus with them we get
$$I<\frac{n\pi}{2}\int_0^{\frac{\pi}{2}}\left(\frac{\sin nx}{\sin x}\right)^2\mathrm{d}x=:\frac{n\pi}{2}J_n.\tag{1}$$
To finish the proof, it's sufficient to evaluate that :
\begin{align*}
J_{n+1}-J_n& =\int_0^{\frac{\pi}{2}}\frac{\sin^2 (n+1)x-\sin^2 nx}{\sin^2 x}\mathrm{d}x\\
& =\int_0^{\frac{\pi}{2}}\frac{\sin(2n+1)x}{\sin x}\mathrm{d}x=:K_n.
\end{align*}
Since $K_{n+1}-K_n=\int_0^{\frac{\pi}{2}}2\cos 2nx\ \mathrm{d}x=0$, we claim that $J_n=\frac{n\pi}{2}$.
With (1), we get $I_n<\frac{n^2\pi^2}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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Is there a way to create a summation function for this? I have an algorithm that generates terms as follows:
$$\begin{align*}
\text{Term 1}: &2\times 3\times\left(\frac{1}{2}+\frac{1}{3}\right) + 4\\
\text{Term 2}: &2\times 3\times 4\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) + 3\times 5\\
\text{Term 3}: &2\times 3\times 4\times 5\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right) + 3\times4\times 6\\
\text{Term 4}: &2\times 3\times 4\times 5\times 6\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right) + 3\times4\times 5\times 7
\end{align*}$$
and so on.
The pattern in the first half is obvious; the pattern in the second half is a little trickier. It "splits" the last multiplier into (multiplier-1)*(multiplier+1) so 4 changes into 3*5 and then the 5 on that one changes into 4*6 and so on for the next term.
I am trying to find a clever way to say "sum terms 1-N" given this pattern
| Term $k$ is $$(k+2)!(H_{k+2}-1)+\frac12(k+1)!(k+3)\;,$$ where $H_k$ is as usual the $k$-th harmonic number, $\sum_{i=1}^k\frac1k$, so the sum of the first $n$ terms is $$\sum_{k=1}^n\Big((k+2)!(H_{k+2}-1)+\frac12(k+1)!(k+3)\Big)\;.$$ You may of course replace $\frac12(k+1)!(k+3)$ by $\dfrac{(k+3)!}{2(k+2)}$ if you wish.
Added: Come to think of it, $n!H_n=\left[{n+1}\atop 2\right]$, a Stirling number of the first kind, so this can also be written $$\begin{align*}\sum_{k=1}^n\left(\left[{k+3}\atop 2\right]-(k+2)!+\frac{(k+3)!}{2(k+2)}\right)&=\sum_{k=1}^n\left(\left[{k+3}\atop 2\right]-\frac{(k+1)(k+1)!}2\right)\;.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2$? I have tried something to solve the series
$$\binom{n}{1}^2+2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2+\cdots + n\binom{n}{n}^2.$$
My approach is :
$$(1+x)^n=\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots + \binom{n}{n}x^n.$$
Differentiating the above equation
$$n(1+x)^{n-1} = \binom{n}{1} + \binom{n}{2}x + \cdots + n\binom{n}{n}x^{n-1}$$
Also,
$$
\left(1+\frac{1}{x}\right)^n =\binom{n}{0} + \binom{n}{1}\frac{1}{x} + \binom{n}{2}\left(\frac{1}{x}\right)^2 + \cdots + \binom{n}{n}\left(\frac{1}{x}\right)^n$$
Multiplying above two equation I get,
$$\begin{align*}
&{n(1+x)^{n-1}\left(1 + \frac{1}{x}\right)^n}\\
&\quad= \left(
\binom{n}{1}^2 + 2\binom{n}{2}^2 + 3\binom{n}{3}^2 + 4\binom{n}{4}^2 + \cdots + n\binom{n}{n}^2\right)\left(\frac{1}{x}\right) + \text{other terms}
\end{align*}$$
So I can say that coefficient of $\frac{1}{x}$ in expansion of $n(1+x)^{n-1}(1+\frac{1}{x})^n$ will give me the required answer.
Am I doing it correct,please correct me if I'm wrong ?
If I'm right,please tell me how to calculate the coefficient of $\frac{1}{x}$ ?
Based on the answers,I tried to implement the things in a C++ code.
I tried implementing the code using extended euclidean algorithm so that the problem of truncated division can be eliminated but still not abled to figure out why am I getting wrong answer for n>=3. This is my updated code : http://pastebin.com/imS6rdWs I'll be thankful if anyone can help me to figure out what's wrong with this code.
Thanks.
Solution:
Finally abled to solve the problem.Thanks to all those people who spent their precious time for my problem.Thanks a lot.This is my updated code :
http://pastebin.com/WQ9LRy6F
| This kind of looks like you want to appeal to Vandermonde's convolution, or at least the method you'd use to prove it. It can be applied directly as follows:
Let $S = \sum\limits_{k=0}^n k\binom{n}{k}^2$ be the sum we want to compute. Note that $S = \sum\limits_{k=0}^n (n-k) \binom{n}{n-k}^2 = \sum\limits_{k=0}^n (n-k) \binom{n}{k}^2$. Therefore $2S = n\sum\limits_{k=0}^n \binom{n}{k}^2 = n \binom{2n}{n}$. Then
$$S = \frac{n}{2}\binom{2n}{n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/122147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 0
} |
If $f(x) = \frac{4^x}{4^x+2},$ find the value of $\sum \limits_{i=1}^{1999} f\left(\frac{i}{1999}\right) $
If
$$f(x) = \frac{4^x}{4^x+2} $$
then find the value of
$$f\left(\frac{1}{1999}\right) + f\left(\frac{2}{1999}\right) + f\left(\frac{3}{1999}\right) +\cdots+f\left(\frac{1999}{1999}\right).$$
I tried it by changing expression to
$$f(x) =1 - \frac{2}{4^x+2}$$
but I am not able to cancel any term.
Is there is any other trick to do it?
Thanks in advance.
| Hint:
Observe
$$f(x)+f(1-x)=1$$
So, $$\begin{align}&f\left(\dfrac 1 {1999}\right)+f\left(\dfrac 2 {1999}\right)+ \cdots+f\left(\dfrac{1998}{1999}\right)+f\left(\dfrac{1999}{1999}\right)\\&=f\left(\dfrac 1 {1999}\right)+f\left(\dfrac{1998}{1999}\right)+f\left(\dfrac 2 {1999}\right)+f\left(\dfrac {1997} {1999}\right)+\cdots +f\left(\dfrac{999}{1999}\right)+f\left(\dfrac{1000}{1999}\right)+f(1)\\&=999+f(1)\\&=999+\dfrac 2 3\\&=\dfrac {2999} 3\end{align}$$
$$\begin{align}f(x)+f(1-x)&=\dfrac{4^x}{4^x+2}+\dfrac{4^{1-x}}{4^{1-x}+2}\\&=\dfrac{4^x}{4^x+2}+\dfrac{\dfrac{4}{4^x}}{\dfrac{4}{4^x}+2}\\&=\dfrac{4^x}{4^x+2}+\dfrac 4 {2 \cdot4^x+4}\\&=\dfrac{4^x}{4^x+2}+\dfrac{\not{4}~~2}{\not2(4^x+2)}\\&=\dfrac{4^x+2}{4^x+2}\\&=1\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/123327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a,b \in {\mathbb{Z_+}}$ such that $a|b^2, b^3|a^4, a^5|b^6, b^7|a^8 \cdots$, Prove $a=b$ Let $a,b \in {\mathbb{Z_+}}$ such that $$a|b^2, b^3|a^4, a^5|b^6, b^7|a^8 \cdots$$
Prove $a=b$
| consider the prime factorisations of $a = \prod_p p^{\nu_p(a)}$ and $b = \prod_p p^{\nu_p(b)}$. Your assumptions yield
*
*$\nu_p(a) \le \nu_p(b^2) = 2\nu_p(b)$ for each $p$
*$3\nu_p(b) \le 4\nu_p(a)$
*$\ldots$
*$(4n+1)\nu_p(a) \le (4n+2)\nu_p(b)$ for each $p$, $n$
*$(4n+3)\nu_p(b) \le (4n + 4)\nu_p(a)$, each $p$, $n$
So we have for each $p$, $n$
$$
\frac{4n + 3}{4n+4} \cdot \nu_p(b) \le \nu_p(a) \le \frac{4n+2}{4n+1}\nu_p(b)
$$
letting $n\to \infty$ yields $\nu_p(a) = \nu_p(b)$ for each $p$, so $a = b$.
AB,
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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How does $(x_n - \sqrt{2})^2/(2x_n ) \geq 0$ mean $x_n \geq \sqrt{2}$ when $x_1=1$? I don't understand one step of proving of the Babylonian Method for $a=2,x_1=1$, and
$$x_{n+1} =\frac{1}{2}( x_n + \frac{a}{x_n}). \quad (n=1,2,\ldots)$$
$$x_{n+1} - \sqrt{2} = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right) - \sqrt{2} = \frac{(x_n^2-2\sqrt{2} x_n+2)}{2x_n} = \frac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$$
For all values of $x_n>0$. Conclude $x_n \geq \sqrt{2} \Rightarrow x_{n+1} \geq \sqrt{2}$.
After using $x_n \geq \sqrt{2}$ we find
$$x_{n+1} - x_n = \frac{1}{2} \left(x_n + \frac{2}{x_n}\right) - x_n = \frac{x_n^2+2-2x_n^2}{2x_n} =\frac{2-x_n^2}{2x_n} \leq 0,$$
which means the sequence $(x_n)$ is monotonically decreasing. Now we know $(x_n)$ is bounded below by $\sqrt{2}$ and $(x_n)$ is monotonically decreasing so $(x_n)$ has a limit.
My problem is I don't understand how could we say $x_n \ge \sqrt{2}$ by using $\dfrac{(x_n-\sqrt{2})^2}{2x_n} \geq 0$?
| There seems to be confusion on what you think they claim, and what they actually claim. The step you are talking about is as follows.
Conclude that $x_n \geq \sqrt{2} \Rightarrow x_{n+1} \geq \sqrt{2}$.
What is claimed here is that if $x_n \geq \sqrt{2}$, then $x_{n+1} \geq \sqrt{2}$. We know this holds, since, as shown in the previous step,
$$x_{n+1} - \sqrt{2} = \frac{(x_n - \sqrt{2})^2}{2 x_n} \geq 0. \quad \quad (\text{if } x_n > 0)$$
So a stronger statement actually holds: $x_n > 0 \Rightarrow x_{n+1} \geq \sqrt{2}$. Since $x_n \geq \sqrt{2}$ implies $x_n \geq 0$, this then also implies $x_{n+1} \geq \sqrt{2}$.
However, this does not mean that $x_n \geq \sqrt{2}$ for any $n$! This only means that if $x_{n_0} \geq \sqrt{2}$ for some $n_0 \in \mathbb{N}$, then we can use induction to show that $x_n \geq \sqrt{2}$ for all $n \geq n_0$.
So probably you would now also like to find such an $n_0$. Since $x_n > 0 \Rightarrow x_{n+1} \geq \sqrt{2}$ as shown above, and $x_1 = 1 > 0$, it follows that $x_2 \geq \sqrt{2}$. And indeed, $x_2 = 2 \geq \sqrt{2}$. So then, by induction we get $x_n \geq \sqrt{2}$ for all $n \geq 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of : $\lim_{x \to \infty} \sqrt{4x^2 + 4} - (2x + 2)$
Possible Duplicate:
Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$
$$\lim_{x \to \infty} \sqrt{4x^2 + 4} - (2x + 2)$$
So, I have an intermediate form of $\infty - \infty$ and I tried multiplying by the conjugate; however, I seem to be left with another intermediate form of $\frac{\infty}{\infty}$ and wasn't sure what else to to do. Is there anything else I can do other than L'Hopital's rule?
| Multiplying and dividing by $\sqrt{4x^2 + 4} + (2x + 2)$ you obtain:
$$\begin{split}
\sqrt{4x^2 + 4} - (2x + 2) &= \frac{(\sqrt{4x^2 + 4} - (2x + 2))(\sqrt{4x^2 + 4} + (2x + 2))}{\sqrt{4x^2 + 4} + (2x + 2)} \\
&= \frac{(4x^2 + 4) - (2x+2)^2}{\sqrt{4x^2 + 4} + (2x + 2)} \\
&= \frac{-8x}{\sqrt{4x^2 + 4} + (2x + 2)} \\
&= \frac{-8x}{2x (\sqrt{1+\frac{1}{x^2}} + 1 + \frac{1}{x})} \\
&= - \frac{4}{\sqrt{1+\frac{1}{x^2}} + 1 + \frac{1}{x}}
\end{split}$$
hence the limit equals $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Different approaches to evaluate this determinant How to evaluate this determinant $$\det\begin{bmatrix}
a& b&b &\cdots&b\\ c &d &0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots& 0\\c&0&\cdots&0&d
\end{bmatrix}?$$
I am looking for the different approaches.
| Your (upper) arrowhead matrix can be decomposed as follows:
$$\begin{pmatrix}a&b&b&\cdots&b\\c&d&0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots&0\\c&0&\cdots&0&d\end{pmatrix}=\color{red}{\begin{pmatrix}a-b-c&&&&\\&d&&&\\&&d&&\\&&&\ddots&\\&&&&d\end{pmatrix}}+\color{blue}{\begin{pmatrix}1&c\\&c\\&c\\&\vdots\\&c\end{pmatrix}}\cdot\color{magenta}{\begin{pmatrix}b&b&b&\cdots&b\\1&&&&\end{pmatrix}}$$
Now, one can then use the Sherman-Morrison-Woodbury formula for determinants:
$$\det(\color{red}{\mathbf A}+\mathbf{\color{blue}{U}\color{magenta}{V^\top}}) = \det(\mathbf I + \color{magenta}{\mathbf V^\top}\color{red}{\mathbf A}^{-1}\color{blue}{\mathbf U})\det(\color{red}{\mathbf A})$$
to yield
$$\begin{align*}
&\begin{vmatrix}a&b&b&\cdots&b\\c&d&0&\cdots&0\\c&0&d&\ddots&\vdots\\\vdots &\vdots&\ddots&\ddots&0\\c&0&\cdots&0&d\end{vmatrix}\\
&=\det\left(\mathbf I+\color{magenta}{\begin{pmatrix}b&b&\cdots&b\\1&&&\end{pmatrix}}\color{red}{\begin{pmatrix}\frac1{a-b-c}&&&\\&\frac1d&&\\&&\ddots&\\&&&\frac1d\end{pmatrix}}\color{blue}{\begin{pmatrix}1&c\\&c\\&\vdots\\&c\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\
&=\det\left(\mathbf I+\color{magenta}{\begin{pmatrix}b&b&\cdots&b\\1&&&\end{pmatrix}}\color{orange}{\begin{pmatrix}\frac1{a-b-c}&\frac{c}{a-b-c}\\&\frac{c}{d}\\&\vdots\\&\frac{c}{d}\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\
&=\det\left(\mathbf I+\color{green}{\begin{pmatrix}\frac{b}{a-b-c}&bc\left(\frac1{a-b-c}+\frac{n-1}{d}\right)\\\frac1{a-b-c}&\frac{c}{a-b-c}\end{pmatrix}}\right)\color{red}{(a-b-c)d^{n-1}}\\
&=\frac{bc(1-n)+ad}{d(a-b-c)}(a-b-c)d^{n-1}\\
&=(bc(1-n)+ad)d^{n-2}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/126901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to derive a differential equation of an ellipse I am quite new to differential equations and derivatives. I want to derive an differential form for equation of an ellipse. If i start with an ordinary ellipse equation
\begin{equation}
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
\end{equation}
How do i derive it then to get this form
$$
-\frac{dx}{dy} = \frac{a^2}{b^2} \frac{y}{x}
$$
I would need an equation and some brief explanation on the procedure.
| The equation $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \tag{1}$$
has two variables: $\{ x, y \}.$ By "derive," it seems that you mean
$ \frac{dx}{dy}.$
Well, differentiating equation $(1)$ w.r.t $y,$ we get:
$$
\frac{d}{dy} \frac{x^2}{a^2} + \frac{d}{dy} \frac{y^2}{b^2} = \frac{d}{dy} 1
\tag{2}$$
First, note that $\dfrac{d}{dy} 1 = 0,$ $\dfrac{d}{dy} y = 1,$ and $\dfrac{d}{dy} f^2 = 2 f \dfrac{df}{dy}.$
So
*
*$ \dfrac{d}{dy} \dfrac{x^2}{a^2} = 2 \dfrac{x^{2-1}}{a^2} \dfrac{dx}{dy}$
*$ \dfrac{d}{dy} \dfrac{y^2}{b^2} = 2 \dfrac{y^{2-1}}{b^2} \dfrac{dy}{dy} $
In other words, equation $(2)$ becomes:
$$
2\frac{x}{a^2} \frac{dx}{dy} + 2 \frac{y}{b^2} = 0.
$$
The rest is simple algebra, you can isolate $\dfrac{dx}{dy}$ one side, and get:
$$ -\frac{dx}{dy} = \frac{a^2}{b^2} \frac{y}{x} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Showing $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$ for $x,y,z \ge 0$ How do I show that $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$ for $x,y,z \ge 0$?
The answer is to:
Let $A = x+y+z$, then $z=A-x-y$. Maximize $f(x,y)=xy(A-x-y)$ on region enclosed by $x=y=0$ and $x+y=A$
What I don't understand is the idea behind this method.
*
*Why $A=x+y+z$ not $A=\frac{x+y+z}{3}$?
*I guess $f(x,y)=xy(A-x-y)$ is to see whats the maximum of the equation
*But why regions $x=y=0$ and $x+y=A$?
Sorry this seems like an easy question but somehow I don't get it ...
| For your first question, the answer is that it doesn't matter. We could just as well let $A=\frac{x+y+z}{3}$. Then $z=3A-x-y$, and so on. The calculations end up very similar. Indeed things are a little prettier, I think, if we let $\frac{x+y+z}{3}=A$. But it makes no real difference.
The point is that we fix $x+y+z$, and show using tools from the calculus that for the fixed value $x+y+z=A$, we must have $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$. Since our argument works for any non-negative $A$, that shows that the desired inequality always holds.
In order to show that $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$, we find the maximum value of $xyz$, given that $x+y+z=A$ and $x,y,z\ge 0$. Since $z=A-x-y$, we want to find the maximum value of $f(x,y)=xy(A-x-y)$. It will turn out when you maximize that this maximum value is $A^3/27$, which gives exactly the right result. For it implies that the maximum of $\sqrt[3]{xyz}$ given that $x+y+z=A$ is $\sqrt[3]{A^3/27}$, which is $A/3$.
Of course we will only consider $x,y\ge 0$. The condition $z\ge 0$ becomes, in terms of $x$ and $y$, that $A-x-y \ge 0$. This can be rewritten as $x+y \le A$.
So we are maximizing the function (not equation) $xy(A-x-y)$ in the region that satisfies $x\ge 0$, $y\ge 0$, $x+y \le A$.
This is the same problem as maximizing $xy(A-x-y)$ over the triangle bounded by the lines $x=0$, $y=0$, and $x+y=A$. (The condition $x+y \le A$ just says that $(x,y)$ lies below the line $x+y=A$.)
I am sure you can handle the partial derivatives stuff.
Remark: Actually, because the inequality we are interested in is homogeneous (it holds at $(x,y,z)$ iff it holds at $(tx,ty,tz)$), we could even assume that $x+y+z$ is some specific number, like $3$.
| {
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"url": "https://math.stackexchange.com/questions/128983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Solving $\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}}$ How to get from
$$\lim_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}} = \lim_{n\to\infty} \frac{\frac{\frac{2}{n}}{\tan{\frac{2}{n}}}\cdot\frac{1}{2n^2}+\frac{\frac{3}{n^2}}{\sin{\frac{3}{n^2}}}\cdot\frac{1}{3}}{\frac{\frac{3}{n}}{\sin{\frac{3}{n}}}\cdot \frac{1}{3n^2}+\frac{\frac{2}{n^2}}{\tan{\frac{2}{n^2}}}\cdot\frac{1}{2}}=...=\frac{2}{3}$$
? It doesn't appear like l'Hôpital's Rule was applied here?
| $$\Large{\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}} = 2}$$
There should be a typo in the book. ( the second term in the bottom should have $n^3$ )
$$\Large{\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^3}}} = \frac{2}{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/133814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the rational form of a $2\times 2$ rotation matrix Given the matrix
$$A = \begin{pmatrix}
\cos x & \sin x\\ -\sin x & \ \cos x \end{pmatrix}$$
I have to express it in rational form
The characteristic polynomial is $y^2 - 2y\cos x+1 $
Accordingly, the rational form $R$ if $x$ is neither $0$ or $2 \pi$ is
$$ R = \begin{pmatrix}
0 & -1\\ 1 & \ 2 \cos x\end{pmatrix}$$
Please suggest if it is correct.
If $x = 0$ or $2 \pi$, then Min poly of $A = x-1$.
Then, it seems the rational form is
$$ R = [1]$$
Please help.
| The characteristic polynomial of
$$\left(\begin{array}{rr}
\cos\theta & \sin\theta\\
-\sin\theta & \cos\theta
\end{array}\right)$$
is, as you note, $t^2 -2(\cos\theta)t + 1$.
The discriminant of this polynomial is
$$4(\cos^2\theta) - 4 = 4(\cos^2\theta - 1).$$
Therefore, the discriminant is always nonpositive, and is equal to zero if and only if $\cos^2\theta = 1$, if and only if $\theta = n\pi$ for some $n\in\mathbb{Z}$.
In particular, if $\theta\neq n\pi$ then the characteristic polynomial is irreducible over $\mathbb{Q}$, so the minimal polynomial agrees with the characteristic polynomial, and so the rational canonical form is just the companion matrix of the characteristic polynomial (just as you wrote):
$$\left(\begin{array}{rr}
0 & -1\\
1 & 2\cos\theta
\end{array}\right).$$
If $\theta=n\pi$, then $A$ is one of the following matrices:
$$\begin{align*}
A &= \left(\begin{array}{rr}
1 & 0\\
0 & 1
\end{array}\right) &\text{if }n\text{ is even,}\\
A &=\left(\begin{array}{rr}
-1 & 0\\
0 & -1
\end{array}\right) &\text{if }n\text{ is odd.}
\end{align*}$$
Both of these matrices are already in rational canonical form.
(Your error: the minimal polynomial is indeed $t-\cos\theta$ in this situation; since the characteristic polynomial is then the square of the minimal polynomial, the rational canonical form will have two blocks, each associated to a degree 1 polynomial; not just a single block. Remember that the exponent of the irreducible factor on the minimal polynomial gives you the size of the largest block, in this case $1\times 1$; but the sum of the sizes of the blocks has to add up to the exponent of the irreducible factor in the characteristic polynomial, in this case $2$. So you need two $1\times 1$ blocks here. )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx = 0$
Evaluate the integral: $$ \int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx$$
The answer is $0,$ but I am unable to get it. There is some symmetry I can not see.
| Observing the inverse of the function $f(x) = \sqrt[3]{1 - x^7}$ on the interval $[0,1]$ is $f^{-1} (x) = \sqrt[7]{1 - x^3}$, using the result
$$\int^b_a f(x) \, dx + \int^{f(b)}_{f(a)} f^{-1} (x) \, dx = b f(b) - a f(a),$$
since $f(a) = f(0) = 1$ and $f(b) = f(1) = 0$ it immediately follows that
$$\int^1_0 \sqrt[3]{1 - x^7} \, dx + \int^0_1 \sqrt[7]{1 - x^3} \, dx = 0,$$
or
$$\int^1_0 \left \{\sqrt[3]{1 - x^7} - \sqrt[7]{1 - x^3} \right \}\, dx = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 3,
"answer_id": 2
} |
Finding derivative of $\sqrt{9-x}$ I am trying to find the derivative of $\sqrt{9-x}$ using the definition of a derivative
$$\lim_{h\to 0} \frac {f(a+h)-f(a)}{h} $$
$$\lim_{h\to 0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h} $$
So to simplify I multiply by the conjugate
$$\lim_{h\to0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}\cdot \frac{ \sqrt{9-(a+h)}+ \sqrt{9-a}}{\sqrt{9-(a+h)}+\sqrt{9-a}}$$
which gives me
$$\frac {-2a-h}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}$$
I have no idea what to do from here, obviously I can easily get the derivative using other methods but with this one I have no idea how to proceed.
| Everything you have done is right except for the last step.
$$\begin{align}
&\lim_{h\to0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}\cdot \frac{ \sqrt{9-(a+h)}+ \sqrt{9-a}}{\sqrt{9-(a+h)}+\sqrt{9-a}}=\\
&\lim_{h\to0} \frac{9-(a+h)-(9-a)}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\
&\lim_{h\to0} \frac{9-a-h-9+a}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\
&\lim_{h\to0} \frac{h}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\
&\lim_{h\to0} \frac{1}{\sqrt{9-(a+h)}+\sqrt{9-a}}
\end{align}$$
The limit is then easy to evaluate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/143274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Derivative of $\ln(x\sqrt{x^2-1})$ I am trying to find the derivative of $\ln(x\sqrt{x^2-1})$ but I can not get what the book gets.
I get $$\frac{1}{x \sqrt{x^2-1}} \cdot \sqrt{x^2-1} + x\cdot\frac{1}{2}(x^2-1)^\frac{-1}{2}\cdot2x$$
which I reduce to
$$\begin{align}
&\frac{1}{x\sqrt{x^2-1}}\sqrt{x^2-1} + x^2(x^2-1)^\frac{-1}{2}=\\
&\frac{\sqrt{x^2-1}}{x\sqrt{x^2-1}} + \frac{x^2(x^2-1)^\frac{-1}{2}}{\sqrt{x^2-1}}=\\
&\frac{1}{x} + \frac{x^2}{x^2-1}= \frac{x^2 - 1 +x^3}{x^3 - x}
\end{align}$$
From here I am not sure what to do. This is not the right answer and I do not know what to do.
| You obtained the correct derivative, but you need parentheses as such:
$$\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + x \frac{1}{2}(x^2-1)^\frac{-1}{2}2x\Bigr)$$
Clean this up a bit to get
$$\tag{1}
\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + x^2 (x^2-1)^\frac{-1}{2} \Bigr).
$$
You were ok up to this point. The rest of your work contains an algebraic error: in the second line of the displayed equations after you say "which I reduce to", the second term is off, it needs an "$x$" downstairs.
But other than that, you did fine. For what it's worth here is the derivation with the correction:
Equation ${1}$ can be written as
$$
\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + {x^2\over\sqrt{x^2-1}}\Bigr)
$$
Now multiply through
$$\eqalign{
\frac{1}{x \sqrt{x^2-1}} \Bigl( \sqrt{x^2-1} + {x^2\over\sqrt{x^2-1}}\Bigr)
&=
\frac{1}{x \color{maroon}{\sqrt{x^2-1}}} \cdot
\color{maroon}{\sqrt{x^2-1} }+
\frac{1}{\color{darkblue}x\color{darkgreen}{ \sqrt{x^2-1}}} \cdot {\color{darkblue}{x^2}\over\color{darkgreen}{\sqrt{x^2-1}}} \cr
&={1\cdot\color{maroon}1\over x} +\frac{\color{darkblue}x}{ \color{darkgreen}{{x^2-1}}} \cr
&={{(x^2-1)}\cdot1+x\cdot x\over x(x^2-1)}\cr
&={2x^2-1\over x(x^2-1)}.\cr
}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Evaluating $\int_0^a \frac{x^{b}+x^{c}}{(1+x)^{b+c+2}}$ Let $0 <a < 1$ and let $b,c \in \mathbb{N}$, evaluate $$\int_0^a \frac{x^{b}+x^{c}}{(1+x)^{b+c+2}}$$
How to evaluate in terms of $a$,$b$ and $c$?
| Since the question only cares about natural numbers of $b$ and $c$ , it should be no problem about any binomial series expansions.
$\int_0^a\dfrac{x^b+x^c}{(1+x)^{b+c+2}}dx$
$=\int_0^a\dfrac{x^b}{(1+x)^{b+c+2}}dx+\int_0^a\dfrac{x^c}{(1+x)^{b+c+2}}dx$
$=\int_0^a\dfrac{(x+1-1)^b}{(x+1)^{b+c+2}}dx+\int_0^a\dfrac{(x+1-1)^c}{(x+1)^{b+c+2}}dx$
$=\int_0^a\dfrac{\sum\limits_{n=0}^bC_n^b(-1)^n(x+1)^{b-n}}{(x+1)^{b+c+2}}dx+\int_0^a\dfrac{\sum\limits_{n=0}^cC_n^c(-1)^n(x+1)^{c-n}}{(x+1)^{b+c+2}}dx$
$=\int_0^a\sum\limits_{n=0}^bC_n^b(-1)^n(x+1)^{-n-c-2}~dx+\int_0^a\sum\limits_{n=0}^cC_n^c(-1)^n(x+1)^{-n-b-2}~dx$
$=\left[\sum\limits_{n=0}^b\dfrac{C_n^b(-1)^n(x+1)^{-n-c-1}}{-n-c-1}\right]_0^a+\left[\sum\limits_{n=0}^c\dfrac{C_n^c(-1)^n(x+1)^{-n-b-1}}{-n-b-1}\right]_0^a$
$=\sum\limits_{n=0}^b\dfrac{C_n^b(-1)^n}{n+c+1}-\sum\limits_{n=0}^b\dfrac{C_n^b(-1)^n}{(n+c+1)(a+1)^{n+c+1}}+\sum\limits_{n=0}^c\dfrac{C_n^c(-1)^n}{n+b+1}-\sum\limits_{n=0}^c\dfrac{C_n^c(-1)^n}{(n+b+1)(a+1)^{n+b+1}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Splitting polynomials I have a polynomial ${\frac{{{{({z^2})}^p} \pm {p^p}}}{{{z^2} \pm p}}}$ where $p$ is an odd prime number, and I know it splits into two factors $$ \sum_{i = 0}^{p - 1} a_i z^i \text{ and } \sum_{i = 0}^{p - 1} ( - 1)^i a_i z^i $$
For example, when $p=5$
$$
\begin{eqnarray*}
\frac{x^{10}-5^5}{x^2-5} &=& x^8+5x^6+25x^4+125x^2+625\\
&=& (x^4 + 5x^3+15x^2+25x+25)(x^4-5x^3+15x^2-25x+25)
\end{eqnarray*}
$$
Does anyone know a nice method for determining these two factoring polynomials?
| Here is the solution I found, which to me is unsatisfactory.
This solution uses Table 24 from Riesel’s “Prime Numbers and Computer Methods for Factorization”. Look up the number you are interested in in the n column, and note down the two sets of coefficients given there: they are $U_n(x)$ and $V_n(x)$. Multiply each coefficient in $U_n(x)$ by $p^0$, $p^1$, $p^2$, …, $p^{p-1}$, and those in $V_n(x)$ by $p^1$, $p^2$, … $p^{p-1}$. Then take a coefficient from each list alternately and you have your factor. Do the same for the other factor after multiplying $V_n(x)$ by $-1$.
Example for $p=11$:
$U_n(x)=1, 5, -1, -1, 5, 1$
$V_n(x)=1, 1, -1, 1, 1$
Multiplying by powers of $11$:
$U_n = 1, 55, -121, -1331, 73205, 161051\\
V_n = 11, 121, -1331, 14641, 161051$
Taking the coefficients one at a time from each list one factor is
$$x^{10} + 11x^9 +55x^8 +121x^7 -121x^6 -1331x^5 -1331x^4 +14641x^3 +73205x^2 +161051x +161051$$
and the other factor is
$$x^{10} - 11x^9 +55x^8 -121x^7 -121x^6 +1331x^5 -1331x^4 -14641x^3 +73205x^2 -161051x +161051$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How did they get this result? Please, explain these computations:
1) $-\left(\frac{1}{2}\right)^2 +1 = \cos^2x$
$\frac{\sqrt{3}}{2} = \cos x$
How did we get $\frac{\sqrt{3}}{2}$ from $-\left(\frac{1}{2}\right)^2 +1$?
2) $-\left(\frac{\sqrt{2}}{2}\right)^2 +1 = \cos^2x$
$\frac{\sqrt{2}}{2} = \cos x$
How did we get $ \frac{\sqrt{2}}{2}$ from $-\left(\frac{\sqrt{2}}{2}\right)^2 +1 $?
| Here's another way:
$$ \cos^2(x) = 1 - (\frac{1}{2})^2 $$ and
$$ \cos^2(x) = 1 - \sin^2(x) $$ so immediately we have
$$ \sin(x) = \pm\frac{1}{2} $$ then, since $\sin(x) = \frac{opp}{hyp}$ we have from the reference triangle, $$ \cos(x) = \frac{adj}{hyp} = \pm\frac{\sqrt{3}}{2} $$ and also
$$ \tan(x) = \frac{opp}{adj} =\pm \frac{1}{\sqrt{3}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/146785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve an equation involving the floor of a number? I'm looking for a solving procedure for this type of exercises.
If $[x]$ represents the floor of $x$, solve the equation:
$$\left[\frac{6x+5}8\right]=\frac{15x-7}5$$
Choose the correct answer:
a) {$\frac{4}5$}
b) {$\frac{3}4$}
c) {$\frac{7}{15}$,$\frac{4}5$}
d) {$\frac{6}{15}$}
e) {$\frac{1}{2}$,$\frac{3}4$}
f) {$\frac{1}{2}$,$\frac{4}5$}
Can someone please explain how I can solve this equation? Thank you very much!
ps. I asked a similar question, unfortunately, the same method doesn't seem to work.
| Let $\mathbb{Z} \ni k = \lfloor \frac{6 x+ 5}{8} \rfloor$. Then the equation says $x = \frac{k}{3} + \frac{7}{15}$. Now we write the equation for $k$:
$$
k = \left\lfloor \frac{k}{4} + \frac{39}{40} \right\rfloor
$$
It now remains to test few small values of $k$. Values $k=0$ and $k=1$ work out.
These correspond to $x = \frac{7}{15}$ and $x=\frac{12}{15} = \frac{4}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/149237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Taylor expansion around infinity of a fraction I want to Taylor expand the following function:
$f(x)=\frac{1}{a-x^2}$ when $x \rightarrow \infty$.
I know the result (from Wolfram Alpha) to be $-\frac{1}{x^2} - \frac{a}{x^4} + O(\frac{1}{x^6})$
but I have no idea how to calculate it. Any help is greatly appreciated!
| One way is as follows. Set $x= \dfrac1y$ and expand it around $y=0$. $$g(y) = \dfrac{1}{a - \dfrac1{y^2}} = \dfrac{y^2}{ay^2-1} = - y^2 \dfrac1{1-ay^2} = -y^2 \left( 1 + ay^2 +a^2y^4 + a^3 y^6 + \cdots\right)$$
Now replace $y$ by $1/x$ to get that
$$f(x) = -\dfrac1{x^2} \left( 1 + \dfrac{a}{x^2} + \dfrac{a^2}{x^4} + \dfrac{a^3}{x^6} + \cdots\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/152570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Multiple roots of a polynomial over a field of characteristic $p$ I have to show for what value of the prime $p$ does the polynomial $ x ^4 + x + 6$
have a root of multiplicity $>1$ over the field of characteristic $p$.
$ p=2, 3, 5, 7 $
Please help.
For $F$ a field of characteristic $3$, $f(x)= x^4 + x = x(x^3+1)$ and $f'(x) = x^3+1$. Hence, $f^′(x)= 0$ for $x=2$. Therefore in an algebraically closed field of characteristic $3$, $f(x)$ has multiple roots.
| The discriminant of $f(x)$ is $55269 = (3^3)(23)(89)$. So the characteristics in which $f(x)$ has a multiple root are $3$, $23$ and $89$. In fact, $f(x) = x(x+1)^3$ in characteristic $3$, $(x^2+7x+8)(x+8)^2$ in characteristic $23$ and $(x+8)^2(x+28)(x+45)$ in characteristic $89$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.