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Finding solutions to $2^x+17=y^2$
Find all positive integer solutions $(x,y)$ of the following equation:
$$2^x+17=y^2.$$
If $x = 2k$, then we can rewrite the equation as $(y - 2^k)(y + 2^k) = 17$, so the factors must be $1$ and $17$, and we must have $x = 6, y = 9$.
However, this approach doesn't work when $x$ is odd.
| A largely complete answer:
Starting from $2^x+17=y^2$, we obtain $2^x+2^4=(y-1)(y+1)$. By observation, $x=4$ fails as $33\ne y^2$. Note that $(y-1),(y+1)$ are two consecutive even numbers, so their difference is $2$, and one of them contains a single factor of $2$ while the other contains more than one factor of $2$.
Case 1, $x<4 \Rightarrow x=1,2,3$: Hence, $2^x+2^4=18,20,24$. Of these, only $24=4\times 6$ is the product of two consecutive even numbers. This corresponds to $x=3$.
Case 2, $x>4$: $16(2^{x-4}+1)=(y-1)(y+1)$. We see $(2^{x-4}+1)$ is an odd number, which we can factor as $ab$, where both $a,b$ are odd integers $\ge 1$. It must be the case that $(y-1)(y+1)=8a\cdot 2b$ where $8a-2b=\pm 2$, or $4a-b=\pm 1 \Rightarrow b=4a\pm 1$ and $(2^{x-4}+1)=ab=4a^2\pm a$
For $a=1$ we get $4a^2\pm a=3,5$ corresponding to $2^1+1$ and $2^2+1$, whence $x-4=1,2 \Rightarrow x=5,6$
For $a=3$ we get $4a^2\pm a=33,39$. Of these, $33=2^5+1 \Rightarrow x-4=5 \Rightarrow x=9$
I note that for $x\ge 9 \Rightarrow x-4 \ge 5$ it is the case that $(2^{x-4}+1) \equiv 1 \bmod 32$ and $a$ being odd means $a^2 \equiv 1 \bmod 8 \Rightarrow 4a^2 \equiv 4 \bmod 32$. Thus $(2^{x-4}+1)=4a^2\pm a \Rightarrow 4 \pm a \equiv 1 \bmod 32$, or $a\equiv \pm 3 \bmod 32$. Looking at the first $200$ members of $\{3,29,35,61,67,\dots\}$ I find no further suitable values of $a$
For larger $a$, I have not found a proof that none exist. If I find such a proof, I will update this answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
} |
Proving Trigonometric Equality I have this trigonometric equality to prove:
$$\frac{\cos x}{1-\tan x}-\frac{\sin x}{1+\tan x}=\frac{\cos x}{2\cos^2x-1}$$
I started with the left hand side, reducing the fractions to common denominator and got this:
$$\frac{\cos x+\cos x\tan x-\sin x+\sin x\tan x}{1-\tan^2x}\\=\frac{\cos x+\cos x(\frac{\sin x}{\cos x})-\sin x+\sin x(\frac{\sin x}{\cos x})}{1-\left(\frac{\sin^2x}{\cos^2x}\right)}\\=\frac{\cos x+\left(\frac{\sin^2x}{\cos x}\right)}{1-\frac{\sin^2}{\cos^2x}}$$and by finding common denominator top and bottom and then multiplying the fractions i got: $$\frac{\cos^2x}{\cos^3x-\cos x\sin^2x}$$ which is far from the right hand side and I don't know what am I doing wrong. What is the correct way to prove this equality?
| $$\frac { cosx }{ 1-tanx } -\frac { sinx }{ 1+tanx } =\frac { cosx }{ 2cos^{ 2 }x-1 } \\ \frac { \cos { x } \left( 1+tanx \right) }{ \left( 1-tanx \right) \left( 1+tanx \right) } -\frac { \sin { x } \left( 1-tanx \right) }{ \left( 1-tanx \right) \left( 1-tanx \right) } =\\ =\frac { \cos { x } +\sin { x } -\sin { x } +\frac { \sin ^{ 2 }{ x } }{ \cos { x } } }{ 1-\tan ^{ 2 }{ x } } =\frac { \frac { \cos ^{ 2 }{ x+\sin ^{ 2 }{ x } } }{ \cos { x } } }{ \frac { \cos ^{ 2 }{ x-\sin ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } } =\\ =\frac { \cos { x } }{ \cos ^{ 2 }{ x-\sin ^{ 2 }{ x } } } =\frac { \cos { x } }{ 2\cos ^{ 2 }{ x-1 } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2299015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find limit $a_{n + 1} = \int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt,$ Find the limit of the sequence:
$$a_{n + 1} = \int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt,$$
such that $a_0 \in (0, 2 \pi)$
That was one of the tasks in the Olympiad.
Here is my approach.
First, I wanted to simplify the integral:
$\int_{0}^{a_n}(1 + \frac{1}{4} \cos^{2n + 1} t)dt = \int_{0}^{a_n}(dt) + \frac{1}{4} \int_{0}^{a_n} \cos^{2n + 1} (t) dt$
That leads to the following relation:
$$a_{n + 1} = a_n + \frac{1}{4} \int_{0}^{a_n} \cos^{2n + 1} (t) dt$$
Now, there is a $\cos t$ with some power which reminded me of the standart integral $\int \cos^n(x) dx$. We can find a recursive formula for it in the following way:
$I_n = \int \cos^n(x) dx = \int \cos(x) \cos^{n - 1}(x) dx = \sin x \cos^{n - 1}x - (n - 1)\int \sin(x) \cos^{n - 2}(x) (- \sin (x)) dx.$
This leads to
$I_n = \sin x \cos^{n - 1}x + (n - 1) I_{n - 2} - (n - 1) I_n$
And final recurrence relation is
$$I_n = \frac{1}{n} \sin x \cos^{n - 1}x + \frac{n - 1}{n} I_{n - 2}$$
For a long time I am trying to make a connection between the original integral $\int_{0}^{a_n} \cos^{2n + 1} (t) dt$ and this recurrence relation, but I have failed to come up with anything meaningful at the moment.
Well, I guess we can just plug in $2n + 1$ instead of $n$ and we get
$$I_{2n + 1} = \frac{1}{2n + 1} \sin x \cos^{2n}x + \frac{2n}{2n + 1} I_{2n - 1}$$
Ok, now if we try to evaluate this as definite integral we should get
$I_{2n + 1}(a_n) - I_{2n + 1}(0) = (\frac{1}{2n + 1} \sin a_n \cos^{2n}a_n + \frac{2n}{2n + 1} I_{2n - 1}(a_n)) - (0 + \frac{2n}{2n + 1} I_{2n - 1}(0))$
$I_{2n + 1}(a_n) - I_{2n + 1}(0) = \frac{1}{2n + 1} \sin a_n \cos^{2n}a_n + \frac{2n}{2n + 1} I_{2n - 1}(a_n) - \frac{2n}{2n + 1} I_{2n - 1}(0).$
So,
$$\frac{1}{4} \int_{0}^{a_n} \cos^{2n + 1} (t) dt = \frac{1}{4(2n + 1)} \sin a_n \cos^{2n}a_n + \frac{2n}{4(2n + 1)} \big[ I_{2n - 1}(a_n) - I_{2n - 1}(0) \big] $$
$$\frac{1}{4} \int_{0}^{a_n} \cos^{2n + 1} (t) dt = \frac{1}{4(2n + 1)} \sin a_n \cos^{2n}a_n + \frac{2n}{4(2n + 1)} \big[ I_{2n - 1}(a_n) - \cos a_0 \big] $$
I would appreciate any help if you provide me with some insights or clues on how to proceed.
| You can easily verify that $$I_(2n+1) \to 0$$,$$ as n\to \infty$$
And $$\lim_{n\to\infty} a_(n+1)-a_n =0$$
And a_1=\int_{0}{a_0} (1+cost)dt=2π
| {
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"timestamp": "2023-03-29T00:00:00",
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Using De Moivre's Theorem for summation
I am able to get the required denominator but I keep ending with 4 terms in numerator. Can somebody help me out in this question?( I am using geometric sum identity)
| $$\Im\sum_{n=1}^{10} \left(\frac{1}{2}e^{i\pi/10}\right)^n=\Im\left(\frac{1}{2}e^{i\pi/10}\frac{1-(\frac{1}{2}e^{i\pi/10})^{10}}{1-\frac{1}{2}e^{i\pi/10}}\right)$$
$$=\Im\left(\frac{1}{2}e^{i\pi/10}\frac{1-\frac{1}{1024}e^{i\pi}}{1-\frac{1}{2}e^{i\pi/10}}\right)$$
$$=\Im\left(\frac{1}{2}e^{i\pi/10}\frac{1+\frac{1}{1024}}{1-\frac{1}{2}e^{i\pi/10}}\right)$$
$$=\frac{1025}{1024}\Im\left(\frac{\frac{1}{2}e^{i\pi/10}}{1-\frac{1}{2}e^{i\pi/10}}\right)$$
$$=\frac{1025}{1024}\Im\left(\frac{\frac{1}{2}e^{i\pi/10}}{1-\frac{1}{2}e^{i\pi/10}}\frac{1-\frac{1}{2}e^{-i\pi/10}}{1-\frac{1}{2}e^{-i\pi/10}}\right)$$
$$=\frac{1025}{1024}\Im\left(\frac{\frac{1}{2}e^{i\pi/10}-\frac{1}{4}}{\frac{5}{4}-\frac{1}{2}(2\cos\frac{\pi}{10})}\right)$$
$$=\frac{1025}{1024}\frac{\frac{1}{2}\sin\frac{\pi}{10}}{\frac{5}{4}-\frac{1}{2}(2\cos\frac{\pi}{10})}$$
$$=\frac{1025\sin\frac{\pi}{10}}{2560-2048\cos\frac{\pi}{10}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find power series of $f(z)=\frac{e^z}{1-z}$ at $z_0=0$? I tried to calculate few derivatives, but I cant get $f^{(n)}(z)$ from them. Any other way?
$$f(z)=\frac{e^z}{1-z}\text{ at }z_0=0$$
| $$
g(z) = a_0+a_1z+a_2z^2+...,\\
\frac{g(z)}{1-z}=a_0\frac{1}{1-z}+a_1\frac{z}{1-z}+a_2\frac{z^2}{1-z}+...
$$
Using the power series for $\frac{1}{1-z}$ gives
$$
g(z)=a_0(1+z+z^2+...)+a_1z\cdot(1+z+z^2+...)+a_2z^2\cdot(1+z+z^2+...),\\
g(z)=a_0(1+z+z^2+...)+a_1\cdot(z+z^2+z^3+...)+a_2\cdot(z^2+z^3+z^4...),\\
g(z)=a_0+(a_0+a_1)z+(a_0+a_1+a_2)z^2+...
$$
For the specific case $g(z)=e^z$
$$
\frac{e^z}{1-z}=\frac{1}{0!}+\left(\frac{1}{0!}+\frac{1}{1!}\right)z+\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}\right)z^2+...+\left(\sum_{r=0}^n\frac{1}{r!}\right)z^n+...
$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Calculating $\sum_{k=1}^\infty \frac{k^2}{2^k}=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots+\frac{k^2}{2^k}+\cdots$ I want to know the value of $$\sum_{k=1}^\infty \frac{k^2}{2^k}=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots+\frac{k^2}{2^k}+\cdots$$
I added up to $k=50$ and got the value $5.999999999997597$, so it seems that it converges to $6.$ But, I don't know how to get the exact value. Is there any other simple method to calculate it?
| Alternatively, note that if $T = \sum_{k \geq 0}k^2 2^{-k}$ then $$T = 2T-T = \sum_{k\geq 0}\frac{2k+1}{2^k} = 2\sum_{k \geq 0}\frac{k}{2^k} + \sum_{k \geq 0}2^{-k} $$
But we can let $S= \sum_{k \geq 0}k2^{-k}$ and note that $$S = 2S-S = \sum_{k\geq 0}2^{-k} = 2.$$
Hence $T = 2\cdot 2 + 2 = 6.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Solve the Inequality $x+\frac{x}{\sqrt{x^2-1}} \gt \frac{35}{12}$ Solve the Inequality $$x+\frac{x}{\sqrt{x^2-1}} \gt \frac{35}{12}$$
First of all the Domain of LHS is $(-\infty \:\: -1) \cup (1 \:\: \infty)$
So i assumed $x=\sec y$ since Range of $\sec y$ is $(-\infty \:\: -1) \cup (1 \:\: \infty)$
So
$$\sec y+ |\csc y| \gt \frac{35}{12}$$
Any help here to proceed?
| We need only concerntrate on the interval $(1,\infty)$ ( the expression is clearly negative in the other interval) ... now change the inequality into an equation, in order to find the points where the expression will change sign
\begin{eqnarray*}
x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12} \\
\frac{x^2}{(x^2-1)}= \left( \frac{35}{12} -x \right)^2 \\
144 x^2 =(35-12x)^2(x^2-1) \\
144x^4-840x^3+937x^2+840x-1225=0 \\
(12x^2-35x-49)(4x-5)(3x-5)=0
\end{eqnarray*}
So we need to consider the intervals $(1,\frac{5}{4}),(\frac{5}{4},\frac{5}{3}),(\frac{5}{3},\infty)$
So the intervals where the inequality is satisfied is $\color{red}{(1,\frac{5}{4})}$ &$\color{red}{(\frac{5}{3},\infty)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2301181",
"timestamp": "2023-03-29T00:00:00",
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How to find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$?
Find the area of the surface enclosed between a sphere $x^2+y^2+z^2=4a^2$ and a cylinder $x^2+(y-a)^2=a^2$. The correct answer should be $(8\pi-16)a^2$.
This is an illustration:
First, there's the following formula to calculate surface area:
$$
\int\int_R\sqrt{Z_x^2+Z_y^2+1}\cdot dA
$$
In our case it's:
$$
\int\int_R \sqrt{\frac{x^2+y^2}{4a^2-x^2-y^2}+1}dA
$$
We can move into polar coordinates (and multiply the original operand by $r$ as the Jacobian in polar), then:
$$
\int\int_R\ r\sqrt{\frac{r^2}{4a^2-r^2}+1}drd\theta
$$
Now the only thing left is to find the bounds of $r$ and $\theta$.
The projection of the surface to $xy$ plane will be a circle with the center in $(0,a)$ with the radius $a$ so it's symmetric around $y$ axis. Then $0\le\theta\le\pi$.
Because $x^2+y^2=2ay$ then $r=2a\sin\theta$, therefore we now have the integral:
$$
\int_0^{\pi}\int_0^{2a\sin\theta}r\sqrt{\frac{r^2}{4a^2-r^2}+1}drd\theta
$$
And this is where I'm stuck. How can I integrate $dr$?
| After some chat and subsequent corrections, all is clear now. Continuing from the end of the question,
$$S/2=\int_0^{\pi}\int_0^{2a\sin\theta}r\sqrt{\dfrac{r^2}{4a^2-r^2}+1}drd\theta=$$
$$=\int_0^{\pi}\int_0^{2a\sin\theta}\frac {2ar}{\sqrt{4a^2-r^2}}drd\theta=2a\int_0^{\pi}\left(-\sqrt{4a^2-4a^2\sin^2\theta}+2a\right)d\theta=$$
$$=4a^2\int_0^\pi(-\vert\cos\theta\vert+1)d\theta=4a^2(\pi-2)$$
This is the area of the upper part, but upper and lower are symmetric, so,
$S=a^2(8\pi-16)$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the number of solution pairs for this equation? Question: Consider the equation $(1+a+b)^2=3(1+a^2+b^2)$ where $a$ and $b$ are real numbers. How many solution pair(s), ($a$,$b$), are possible for this equation?
My attempt:
$(1+a+b)^2=3(1+a^2+b^2)$
=>$1+a^2+b^2+2(a+ab+b)=3+3a^2+3b^2$
=>$1+a^2+b^2+2a+2ab+2b=3+3a^2+3b^2$
=>$2a+2ab+2b=2+2a^2+2b^2$
=>$2(a+ab+b)=2(1+a^2+b^2)$
=>$a+ab+b=1+a^2+b^2$
=>$a-ab+b-1=a^2+b^2-2ab$
=>$-a(b-1)+1(b-1)=(a-b)^2$
=>$(b-1)(1-a)=(a-b)^2$
I have simplified the given equation to this point. I cannot understand how to proceed. I would be grateful for any help i receive.
| Hint: after expanding and collecting, the equation can be written as:
$$a^2 - a b - a + b^2 - b + 1 = 0 \;\;\iff\;\; a^2 - (b+1)\,a + b^2-b+1=0$$
Considering it as a quadratic in $\,a\,$, its discriminant is:
$$
\Delta=(b+1)^2-4(b^2-b+1)=-3b^2+6b-3=-3(b-1)^2
$$
For the quadratic to have real roots the discriminant must be non-negative, so $\;\cdots$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sketch the graph of $F_4^n(x)$ on the unit interval, where $F_4(x) = 4x(1-x)$. Conclude that $F_4$ has at least $2^n$ periodic points of period $n$. Sketching the graph I found that $F_4$ has exactly $2^{n-1}$ point wich prime period $n$. My $F_4^n$ graph look like a $|sin(x)|$ graph with $2^n$ intersections on $Id$, but I removed the $2^{n-1}$ to remove points that have period less then n. What is wrong?
| Incomplete and with typos ... but contains a few hints and references.
Given the following functions:
$$\varphi(x)=-4x+2,\varphi^{-1}(x)=\frac{x-2}{-4}, g(x)=x^2-2$$
we have:
$$F_4(x)=(\varphi^{-1} \circ g\circ \varphi)(x)$$
or
$$F_4^{\circ n}(x)=(\varphi^{-1} \circ g^{\circ n} \circ \varphi)(x)$$
Further to this, function $\gamma(x)=x+\frac{1}{x}$ has the property that
$$(g \circ \gamma)(x)=x^2+\frac{1}{x^2}=\gamma(x^2)\Rightarrow (\gamma^{-1} \circ g \circ \gamma)(x)=x^2\Rightarrow \\g(x)=(\gamma \circ t \circ \gamma^{-1})(x), t(x)=x^2$$
As a result
$$F_4(x)=(\varphi^{-1} \circ \gamma \circ t \circ \gamma^{-1} \circ \varphi)(x)$$
or
$$F_4^{\circ n}(x)=(\varphi^{-1} \circ \gamma \circ t^{\circ n} \circ \gamma^{-1} \circ \varphi)(x) \tag{1}$$
Effectively, by introducing $\gamma$ we have to exclude $x=0$ perioding point, which is not of period $n$ though.
Periodic points of $t(x)=x^{2}$ are periodic for $F_4(x)$ "through" $\gamma(x)$ and $\varphi(x)$. Complex roots of $x^{2^n-1}=1$ are periodic points for $t(x)$ (and $F_4(x)$ "through" $\gamma(x)$ and $\varphi(x)$) and, except $x=1$, all the other roots have period $n$ (there is an entire chapter in this book, starting with page 6, dedicated to this function). To see all these working, let's take $x=\sin^2{\frac{\alpha}{2}} \in (0,1)$ then (I will omit the details as those are easy to check)
$$\varphi(x)=2\cos{\alpha}$$
$$\gamma^{-1}\left(2\cos{\alpha}\right)=e^{i\alpha}$$
$$t^{\circ n}\left(e^{i\alpha}\right)=e^{i2^{n}\alpha}$$
$$\gamma\left(e^{i2^{n}\alpha}\right)=2\cos{\left(2^{n}\alpha\right)}$$
$$\varphi^{-1}\left(2\cos{\left(2^{n}\alpha\right)}\right)=\sin^2{\left(2^{n-1}\alpha\right)}$$
or
$$F_4^{\circ n}\left(\sin^2{\frac{\alpha}{2}}\right)=\sin^2{\left(2^{n-1}\alpha\right)} \tag{2}$$
Complex roots of $z^{2^n-1}=1$ are of the form
$$z_k=\exp{\left(\frac{2i\pi k}{2^n-1}\right)},k=\overline{0..2^n-1}$$
thus taking $$\alpha=\frac{2\pi k}{2^n-1}$$
we have
$$F_4^{\circ n}\left(\sin^2{\frac{\pi k}{2^n-1}}\right)=\sin^2{\left(2^{n-1}\frac{2\pi k}{2^n-1}\right)}=\sin^2{\left(\frac{(2^{n}-1+1)\pi k}{2^n-1}\right)}=\\
\sin^2{\left(k\pi + \frac{\pi k}{2^n-1}\right)}=\sin^2{\frac{\pi k}{2^n-1}},k=\overline{0..2^n-1} \tag{3}$$
Note: we could simply apply (from $(2)$) $\sin^2{\frac{\alpha}{2}}=\sin^2{\left(2^{n-1}\alpha\right)} \Rightarrow \frac{\alpha}{2}+k\pi=2^{n-1}\alpha$ and arrive to the same conclusion, but without conjugacy functions reaching this point would require having a good intuition.
However
$$F_4^{\circ n}\left(\sin^2{\frac{\pi (2^n-k)}{2^n-1}}\right)=\sin^2{\left(\frac{\pi (2^n-k)}{2^n-1}\right)}=\sin^2{\left(\frac{\pi (2^n-1-k+1)}{2^n-1}\right)}=\\
\sin^2{\left(\pi - \frac{\pi (k-1)}{2^n-1}\right)}=\sin^2{\left(\frac{\pi (k-1)}{2^n-1}\right)}=F_4^{\circ n}\left(\sin^2{\left(\frac{\pi (k-1)}{2^n-1}\right)}\right) \tag{4}$$
this last statement reduces the number of periodic points with period $n$ to $2^{n-1}$.
Note: $(4)$ can also simply be obtained from $\frac{2\pi k}{2^n-1}=\frac{2\pi m}{2^n-1} + p\pi$, where $0\leq k,m \leq 2^n-1$ and concluding that $p=1$ and $k-m=2^n-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trouble understanding this Equality of Complex number question. I have this question here
$$(3-2j)(x+yj)=4+j^9$$
I know that $a + bj$ and $x + yj$ are equal when $a = x$ and $b = y$.
This question is confusing me because I have had three different answers but they are all wrong.
What feels like my closest attempt was:
$$(3-2j)(x+yj)=4+j^9.$$
$x + yj = (-3+2j)(4-j)$
$= -12 + 3j + 8j -2j^2$
$= -12 + 11j -2(-1)$
$= -12 + 11j + 2$
$= -10 + 11j \:\: \Rightarrow$
$x = -10$
$y = 11$
But the answer is supposed to be
$x = 10/13$
$y = 11/13$
I am just looking for some guidance on how to properly solve this and figure out where I went wrong.
Thank you!
| \begin{align*}(3-2i)(x+yi)=4+i^9&\Longleftrightarrow x+yi=\frac{4+i}{3-2i}\\&\Longleftrightarrow x+yi=\frac{(4+i)(3+2i)}{(3-2i)(3+2i)}\\&\Longleftrightarrow x+yi=\frac{10+11i}{13}=\frac{10}{13}+\frac{11}{13}i\\&\Longleftrightarrow x=\frac{10}{13}\text{ and }y=\frac{11}{13}\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to calculate $\int{\frac {x^2}{ ( x\cos x -\sin x ) ^2}}\,{\rm d}x$? When I want to calculate
$$\int{\frac {x^2}{ ( x\cos x -\sin x ) ^2}}\,{\rm d}x$$
I have tested with software and get
$${\frac {x\sin \left( x \right) +\cos \left( x \right) }{x\cos \left( x
\right) -\sin \left( x \right) }}$$
But I can not come to this conclusion, neither using integration by parts, nor using trigonometric identities, nor multiplying by their conjugate, Even by rational trigonometric substitution. I do not know what else to try. Could you give me any suggestions?
| We can rewrite $$\begin{align}\int \frac{x^2}{(\sin x - x\cos x)^2} \, \mathrm{d}x &= \int \frac{x\sin x(x\sin x + \cos x)}{(\sin x - x\cos x)^2} \, \mathrm{d}x - \int \frac{x\cos x}{\sin x - x\cos x} \, \mathrm{d}x \\ & = -\frac{x\sin x + \cos x}{\sin x - x\cos x} \color{green}{+} \int \frac{x\cos x}{\sin x- x\cos x} \, \mathrm{d}x \color{red}{-} \int \frac{x\cos x}{\sin x - x\cos x} \, \mathrm{d}x \\ & = -\frac{x\sin x + \cos x}{\sin x - x\cos x}\end{align}$$
via IBP with $u = x\sin x + \cos x \implies u' = x\cos x$ and $$\mathrm{d}v = \frac{x\sin x}{(\sin x - x\cos x)^2} \implies v = -\frac{1}{\sin x -x\cos x}$$ via straightforward $f = \sin x - x\cos x \implies f' = x\sin x$ sub.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Ellipse in a Rectangle What is the equation for an ellipse (or rather, family of ellipses) which has as its tangents the lines forming the rectangle $$x=\pm a, y=\pm b\;\; (a,b>0)$$?
This question is a modification/extension of this other question here posted recently.
| Here's a purely algebraic approach without any use of trigonometric functions.
By symmetry, the center of the ellipse is at the origin, so its
equation is just
$$
Ax^2 + Bxy + Cy^2 = 1
$$
with $A>0,$ $C>0,$ and $B^2 < 4AC.$
The tangents at $y = \pm b$ will occur at values of $y$ for which the
quadratic equation in $x,$ $Ax^2 + Bxy + Cy^2 - 1 = 0,$
has a double root.
This occurs when
$$(By)^2 - 4A(Cy^2 - 1) = 0,$$
that is, at
$$
y^2 = \frac{4A}{4AC - B^2}.
$$
For similar reasons, the tangents at $x=\pm a$ occur when
$$
x^2 = \frac{4C}{4AC - B^2}.
$$
(Note that the conditions on $A,$ $B,$ and $C$ guarantee that
$\frac{4A}{4AC - B^2} > 0$ and $\frac{4C}{4AC - B^2} > 0.$)
Set
$$
\frac{4C}{4AC - B^2} = a^2, \quad
\frac{4A}{4AC - B^2} = b^2.
$$
It follows that $C=A a^2/b^2.$ Using this to substitute for $C$
in one of the equations above and solving for $B,$ we get
$$B = \pm\sqrt{ \frac{ 4A(a^2A - 1)}{b^2}}.$$
It follows that $a^2A - 1\geq 0,$ that is, $A \geq \frac1{a^2}.$
If $A = \frac1{a^2}$ then $B = 0$ and $C = \frac1{b^2},$
that is, we get the equation of the inscribed ellipse whose axes are aligned with the coordinate axes.
For any other value of $A$ we get an ellipse whose axes are inclined;
the direction of inclination depends on whether we choose the positive
or negative.
For very large values of $A$ the major axis of the ellipse is very
nearly the diagonal of the rectangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculate the value of the series $\,\sum_{n=1}^\infty\frac{1}{2n(2n+1)(2n+2)}$ Calculate the infinite sum
$$\dfrac{1}{2\cdot 3\cdot 4}+ \dfrac{1}{4\cdot 5\cdot 6}+\dfrac{1}{6\cdot 7\cdot 8}+\cdots$$
I know this series is convergent by Comparison Test, but I can't understand how can I get the value of the sum.
Is there any easy way to calculate this?
Please someone help.
| In the style suggested by Michael Rozenberg's comment (solution 2), let us take one and two terms out of Mercator series
$$\begin{align}
S_0=\log(2)&=\sum_{k=0}^\infty\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)\\
&=1-\sum_{k=0}^\infty \left(\frac{1}{2k+2}-\frac{1}{2k+3}\right)\\
&=1-\frac{1}{2}+\sum_{k=0}^\infty \left(\frac{1}{2k+3}-\frac{1}{2k+4}\right)\\
&=\frac{1}{2}+\sum_{k=0}^\infty \left(\frac{1}{2k+3}-\frac{1}{2k+4}\right)\\
\end{align}$$
to obtain
$$S_1 = \sum_{k=0}^\infty \left(\frac{1}{2k+2}-\frac{1}{2k+3}\right)=1-\log(2)$$
and
$$S_2=\sum_{k=0}^\infty \left(\frac{1}{2k+3}-\frac{1}{2k+4}\right)=\log(2)-\frac{1}{2}$$
This reduces the evaluation to
$$\begin{align}S&=\sum_{k=0}^\infty \frac{1}{(2k+2)(2k+3)(2k+4)}\\
\\
&=\frac{1}{2}\sum_{k=0}^\infty\left(\frac{1}{2k+2}-\frac{2}{2k+3}+\frac{1}{2k+4}\right)\\
\\
&=\frac{1}{2}\sum_{k=0}^\infty\left(\frac{1}{2k+2}-\frac{2}{2k+3}+\frac{1}{2k+4}\right)\\
\\
&=\frac{1}{2}\sum_{k=0}^\infty\left(\frac{1}{2k+2}-\frac{1}{2k+3}-\frac{1}{2k+3}+\frac{1}{2k+4}\right)\\
\\
&=\frac{S_1-S_2}{2}=\dfrac{1-\log(2)-\left(\log(2)-\dfrac{1}{2}\right)}{2}=\dfrac{\dfrac{3}{2}-2\log(2)}{2}\\
\\
&=\frac{3}{4}-\log(2)\\
\end{align}$$
the semidifference of the precomputed series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Proof of limit using epsilon-L proof We have $\lim \frac{n+6}{n^2-6}=0$. I need to prove this limit using the epsilon-L proof method.
I wrote down the following proof:
$\dfrac{n+6}{n^2-6}<\dfrac{2n}{\frac{1}{2}n^2}=\dfrac{4}{n}<\epsilon$ (why are we allowed to say this?) so $n>\dfrac{4}{\epsilon}$.
Proof. Let $\epsilon>0$ and let $N=\max\{4,\dfrac{4}{\epsilon}\}$ (since the inequality is only valid for $n>6$ right? Since $n+6<2n$ gives $n>6$ and $n^2-6>\frac{1}{2}n^2$ gives $n>4$)?? I don't understand this too.
Can anyone clarify? I can't seem to understand anything about this *** proof
| For $n>6$, we have
$$n^2-6 >n^2-36>0$$
$$n^2-36=(n+6)(n-6) $$
thus
$$\frac {n+6}{n^2-6}<\frac {1}{n-6} $$
To satisfy $\frac {n+6}{n^2-6}<\epsilon $, it sufficient to have
$n>6$ and $\frac {1}{n-6}<\epsilon $
or $n>6$ and $n>6+\frac {1}{\epsilon} $
so we will take $N=6+\frac {1}{\epsilon} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Linear combination of others Write one of the vectors as a linear combination of the others.
$$ \left[
\begin{array}{cc|c}
0\\
1\\
1
\end{array}
\right],\left[
\begin{array}{cc|c}
1\\
0\\
-1
\end{array}
\right],\left[
\begin{array}{cc|c}
4\\
5\\
1
\end{array}
\right] $$
So far, this is what I have done:
I took the 3 vectors and made them into a matrix.
$$ \left[
\begin{array}{ccc}
0&1&4\\
1&0&5\\
1&-1&1
\end{array}
\right] $$
Then I put it into RREF and got the matrix below.
$$ \left[
\begin{array}{ccc}
1&0&5\\
0&1&4\\
0&0&0
\end{array}
\right] $$
I don't know how to continue this problem.
| If one can be expressed as a linear combination of the others, they are linearly dependent. This means that we can find constants $x,y,z $ s.t.
$$ \left[
\begin{array}{cc|c}
0\\
1\\
1
\end{array}
\right]x+
\left[
\begin{array}{cc|c}
1\\
0\\
-1
\end{array}
\right]y + \left[
\begin{array}{cc|c}
4\\
5\\
1
\end{array}
\right]z =
\left[
\begin{array}{cc|c}
0\\
0\\
0
\end{array}
\right]$$
$$ \left[
\begin{array}{ccc|c}
0&1&4\\
1&0&5\\
1&-1&1\\
\end{array}
\right] \left[
\begin{array}{cc|c}
x\\
y\\
z\\
\end{array}
\right] = \left[
\begin{array}{cc|c}
0\\
0\\
0\\
\end{array}
\right]$$
As you found, rref tells us:
$$ \left[
\begin{array}{ccc|c}
1&0&5\\
0&1&4\\
0&0&0\\
\end{array}
\right] \left[
\begin{array}{cc|c}
x\\
y\\
z\\
\end{array}
\right] = \left[
\begin{array}{cc|c}
0\\
0\\
0\\
\end{array}
\right]$$
This leads to $2$ equation:
$$x + 5z = 0$$
$$y + 4z = 0$$
If we let $z = -1$
$$x = 5 \space\text { and } \space y = 4$$
So:
$$ 5\left[
\begin{array}{cc|c}
0\\
1\\
1
\end{array}
\right]+
4\left[
\begin{array}{cc|c}
1\\
0\\
-1
\end{array}
\right] - \left[
\begin{array}{cc|c}
4\\
5\\
1
\end{array}
\right] =
\left[
\begin{array}{cc|c}
0\\
0\\
0
\end{array}
\right]$$
Or, equivalently,
$$ 5\left[
\begin{array}{cc|c}
0\\
1\\
1
\end{array}
\right]+
4\left[
\begin{array}{cc|c}
1\\
0\\
-1
\end{array}
\right] = \left[
\begin{array}{cc|c}
4\\
5\\
1
\end{array}
\right] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that if $\frac1a+\frac1b+\frac1c = a+b+c$, then $\frac1{3+a}+\frac1{3+c}+\frac1{3+c} \leq\frac34$
Show that if $a,b,c$ are positive reals, and
$\frac1a+\frac1b+\frac1c = a+b+c$, then
$$\frac1{3+a}+\frac1{3+b}+\frac1{3+c} \leq\frac34$$
The corresponding problem replacing the $3$s with $2$ is shown here:
How to prove $\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$?
The proof is not staggeringly difficult: The idea is to show (the tough part) that if
$\frac1{2+a}+\frac1{2+c}+\frac1{2+c}=1$ then
$\frac1a+\frac1b+\frac1c \ge a+b+c$. The result then easily follows.
A pair of observations, both under the constraint of $\frac1a+\frac1b+\frac1c = a+b+c$:
*
*If $k\geq 2$ then $\frac1{k+a}+\frac1{k+c}+\frac1{k+c}\leq\frac3{k+1}$.
This is proven for $k=2$ but I don't see how for $k>2$
*If $0<k<2$ then there are positive $(a,b,c)$ satisfying the constraint such that $\frac1{k+a}+\frac1{k+c}+\frac1{k+c}>\frac3{k+1}$.
So one would hope that the $k=3$ case, lying farther within the "valid" region, might be easier than $k=2$ but I have not been able to prove it.
Note that it is not always true, under our constraint, that
$\frac1{2+a}+\frac1{2+c}+\frac1{2+c}\geq \frac1{3+a}+\frac1{3+c}+\frac1{3+c}+\frac14$, which if true would prove the $k=3$ case immediately.
| This is a follow-on to Michael Rozenberg's answer, generalizing to all real $k>2$, given only the theorem that under the given constraint,
$$
\sum_{\mbox{cyc}}\frac1{2+a}\leq 1
$$
Use (C-S)
$$ \sum \left(\frac{\alpha_i^2}{x_i}\right) \geq \frac{\left(\sum \alpha_i\right)^2}{\sum x_i}
$$
for the $n=2$ case with
$$
\matrix{ \alpha_1 = 3 & \alpha_2 = k-2\\x_1 = a+2 & x_2 = k-2
}
$$giving
$$
\frac{3^2}{2+a}+\frac{(k-2)^2}{k-2}\geq \frac{(k+1)^2}{k+a}
$$
Similarly for $b$ in place of $a$, and for $c$ in place of $a$.
Adding the $a$, $b$, and $c$ inequalities we get
$$
\sum_{\mbox{cyc}}\frac1{k+a}\leq \frac1{(k+1)^2}
\sum_{\mbox{cyc}}\left( \frac{3^2}{2+a}+(k-2) \right)
= 3\frac{(k-2)}{(k+1)^2}+\frac{9}{(k+1)^2}\sum_{\mbox{cyc}} \frac{3^2}{2+a}
$$
Now we use our given theorem to write
$$
\sum_{\mbox{cyc}}\frac1{k+a}\leq 3\frac{(k-2)}{(k+1)^2}+\frac{9}{(k+1)^2}=\frac{3k+3}{(k+1)^2}=\frac3{k+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
For a polynomial with integer coefficients, is it true that if constant term is prime then it cannot be the root of the polynomial.
For a polynomial with integer coefficients, is it true that if constant term is prime then it cannot be the root of the polynomial.
Let $p$ be a polynomial with constant term $a_0$ and if $a_0$ is prime then $p(a_0) \ne 0$
I just thought of this while working on some other problem.
Is this true ?
My attempt :-
$$|a_0/a_n| = |r_1 ... r_n|$$
If, $r_1 = a_0$
$$1 = |a_n||r_2 ... r_n|$$
Also $$|a_1/a_n| = |r_2...r_n + r_1r_3...r_n + ... + r_1...r_{n-1}|$$
Or $$|a_1| = |r_1|\left| \dfrac{1}{r_1} + \dfrac{1}{r_2} + \cdots + \dfrac1{r_n}\right|$$
I am having difficulty in proving that $\left| \dfrac{1}{r_1} + \dfrac{1}{r_2} + \cdots + \dfrac1{r_n}\right|$ not a integer if $|r_1 ... r_n|^{-1}$ is a integer.
Any hints ?
Ok this is false but can somebody prove/disprove this if $p(1) \ne p(-1) \ne 0$ ?
| Consider the polynomial $$f(x)=x^2 - 6 x + 5$$
The constant term here is $5$, which is prime.
However, \begin{align}f(5)&=5^2-6\times 5+5\\
&=25-30+5\\
&=0\end{align}
And thus $5$ is a root of the equation.
Therefore, I have disproved your hypothesis through contradiction
For higher degree polynomials, anything of the form $$f(x)=(x\pm p)(x+1)^a(x-1)^b$$ for $p$ prime, and $a,b$ integers will form a contradiction.
Examples include $$x^4 + 8 x^3 + 6 x^2 - 8 x - 7 = (x+7)(x+1)^2(x-1)$$ which has $p=-7$, $a=2$, $b=1$
and $$x^6 - 14 x^5 + 11 x^4 + 28 x^3 - 25 x^2 - 14 x + 13 = (x-13)(x+1)^2(x-1)^3$$ which has $p=13$, $a=2$, $b=3$
To add the constraint that $f(1)\neq f(-1)\neq 0$, then we can choose $$x^3 - 11 x^2 + x - 11=(x-11)(x^2+1)$$
Here we have $f(1)=-20$, $f(-1)=-24$ but $f(11)=0$
As long as the second polynomial is an irreducible one, with a constant of $1$, then this will form a contradiction
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
How can I find maximum and minimum modulus of a complex number? I have this problem. Let be given complex number $z$ such that
$$|z+1|+ 4 |z-1|=25.$$
Find the greastest and the least of the modulus of $z$.
I tried with minimum.
Put $A(-1,0)$, $B(1,0)$ and $M(x,y)$ present of $z$.
We have $O(0,0)$ is the midpoint of the segment $AB$. Therefore
$$OM^2 = \dfrac{AM^2 + BM^2}{2}-\dfrac{AB^2}{4}.$$
Another way
$$25=AM+4BM \leqslant \sqrt{(1^2 + 4^2)(AM^2 + BM^2)},$$
Therefore
$$AM^2 + BM^2 \geqslant \dfrac{625}{17}.$$
$$OM^2 \geqslant \dfrac{625}{17} -1 = \dfrac{591}{17}.$$
Thus, minimum of $z$ is $\sqrt{\dfrac{591}{17}}$.
This answer is not true with Mathematica. Mathematica give $\dfrac{22}{5}$.
Where is wrong in my solution and how can I find the maximum?
| Put $A(-1,0)$, $B(1,0)$, $M(x,y)$ is the point present for $z$. Note that $O(0,0)$ is midpoint of the segment $AB$ and $AB=2$. We have $AM+4BM=25$. We need to find the least and the greastest of the segment $OM$.
Put $a=AM$, $b=BM$ $(a,\, b >0)$, we have $a+4b=25$ or $a=25-4b$.
Because of $|MA-MB|\leqslant AB$ or
$$|a-b|\leqslant 2 \Leftrightarrow |25-4b-b| \leqslant 2 \Leftrightarrow \dfrac{23}{5}\leq b\leq
\dfrac{27}{5}.$$
Anotherway,
$$OM^2 = \dfrac{AM^2 + BM^2}{2}-\dfrac{AB^2}{4}=\dfrac{(25-4b)^2+b^2}{2}-1=\frac{17 b^2}{2}-100
b+\frac{623}{2}.$$
Consider the function $f(b)=\dfrac{17 b^2}{2}-100
b+\dfrac{623}{2}$ here $\dfrac{23}{5}\leq b\leq
\dfrac{27}{5}$, we get the maximum of $f(b)$ is $\dfrac{784}{25}$ and the minimum of $f(b)$ is $\dfrac{484}{25}$. Then, the maximum of $|z|$ is $\dfrac{28}{5}$ and the minimum of $|z|$ is $\dfrac{22}{25}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integration of $1/(1+a \csc^2(x))$
Integration of $$\int_{0}^{\frac{(M-1)\pi}{M} }\frac{1}{1+\alpha \csc^2(x)} dx,$$ where $ \alpha $ is a constant.
I tried taking $\cot(x) = t$, then differentiating it w.r.t $dx$ we get, $-\csc^2(x)dx = dt$. And as we know that, $\csc^2(x)= \cot^2(x) +1$, so tried substituting these values, but did not got any outcome.
| $$\int^{ }_{} \frac{1}{1+\alpha \csc^2(x)}dx=\int \frac{\csc^2(x)}{\csc^2(x)+\alpha \csc^4(x)}dx$$
and $$\csc^2(x)=\cot^2(x)+1$$
$\int \frac{\cot^2 (x) +1}{\cot^2 (x)+1+\alpha (\cot^2 (x) +1)^2}dx$
$\int \frac{\csc^2x}{\cot^2 (x)+1+\alpha (\cot^2 (x) +1)^2}dx$
Then substitute $t=\cot(x)=> dt = -\csc^2 x \,dx.$
$\int \frac{-dt}{t^2 +1+\alpha (t^2 +1)^2}$
$\int \frac{-dt}{(t^2 +1)(1+\alpha (t^2 +1))}$
$\int (\frac {\alpha}{(\alpha t^2 + \alpha + 1)} -\frac{ 1}{(t^2 + 1)})dt$
$\int \frac {\alpha}{(\alpha t^2 + \alpha + 1)}dt - \int \frac{ 1}{(t^2 + 1)}dt$
$ \int \frac {\alpha}{(\alpha t^2 + \alpha + 1)}dt=\frac{\alpha}{\alpha+1}\int \frac {1}{1+\frac{\alpha t^2}{\alpha+1}}dt\,\,\,\,\,$
[substitute $u=t\sqrt{\frac{\alpha}{\alpha+1}}] $
$=\frac{\alpha}{\alpha+1}[\sqrt{\frac{\alpha}{\alpha+1}}+\frac{\sqrt{\frac{\alpha}{\alpha+1}}}{a}]\int \frac {1}{u^2+1}du=\frac{\alpha}{\alpha+1}[\sqrt{\frac{\alpha}{\alpha+1}}+\frac{\sqrt{\frac{\alpha}{\alpha+1}}}{a}]\tan^{-1}(t\sqrt{\frac{\alpha}{\alpha+1}})$
$\int \frac {\alpha}{(\alpha t^2 + \alpha + 1)}dt - \int \frac{ 1}{(t^2 + 1)}dt=\frac{\alpha}{\alpha+1}[\sqrt{\frac{\alpha}{\alpha+1}}+\frac{\sqrt{\frac{\alpha}{\alpha+1}}}{a}]\tan^{-1}(t\sqrt{\frac{\alpha}{\alpha+1}})-\tan^{-1}t$
$$=(\frac{\sqrt {a}}{\sqrt{\alpha+1}})\tan^{-1}(\cot (x)\sqrt{\frac{\alpha}{\alpha+1}})-\tan^{-1} (\cot x)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$
Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$
I get the answer $\frac{-1}{ \sqrt2}$
By solving like this
\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\
2\sin^{-1}(- x) &= \frac\pi2\\
\sin^{-1}(- x) &= \frac\pi4\\
-x &= \sin\left(\frac\pi4\right)\end{align}
Thus $x =\frac{-1}{\sqrt2}$
But correct answer is $\pm\frac{1}{\sqrt2}$
Where am I going wrong?
| Let $\sin^{-1}(-x)=u\implies\sin u=-x$
$$\cos(2\sin^{-1}(-x))=\cos2u=1-2\sin^2u=1-2(-x)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Very hard hyperbolic inequality I would like to solve this
Your inequality is equivalent to :
$$\sum_{cyc}\frac{a}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2\geq \frac{a+b+c}{18}$$
Each side is divided by $b$
We get :
$$\frac{a}{13b} \sin(\arctan(\sqrt{\frac{13}{5}}\frac{a}{b}))^2+\frac{1}{13}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{b}{c}))^2+\frac{c}{13b}\sin(\arctan(\sqrt{\frac{13}{5}}\frac{c}{a}))^2\geq \frac{1+\frac{a}{b}+\frac{c}{b}}{18}$$
Now we put :
$\sqrt{\frac{13}{5}}\frac{a}{b}=\frac{A}{B}$
$\sqrt{\frac{13}{5}}\frac{b}{c}=\frac{B}{C}$
$\sqrt{\frac{13}{5}}\frac{c}{a}=\frac{C}{A}$
We find :
$$\sqrt{\frac{5}{13}}\frac{A}{13B}\sin(\arctan(\frac{A}{B}))^2+\frac{1}{13}\sin(\arctan(\frac{B}{C}))^2+\sqrt{\frac{13}{5}}\frac{C}{13B}\sin(\arctan(\frac{C}{A}))^2\geq \frac{1+\sqrt{\frac{5}{13}}\frac{A}{B}+\sqrt{\frac{13}{5}}\frac{C}{B}}{18}$$
After that we put :
$\frac{A}{B}=\frac{x+y}{1-xy}$
$\frac{B}{C}=\frac{z+y}{1-zy}$
$\frac{C}{A}=\frac{x+z}{1-xz}$
With $xy<1$ , $yz<1$ et $zx<1$
Now we use the following identity :
$\arctan(x)+\arctan(y)=\arctan(\frac{x+y}{1-xy})$
We find :
$\sqrt{\frac{5}{13}}\frac{x+y}{13(1-xy)}\sin(\arctan(x)+\arctan(y))^2+\frac{1}{13}\sin(\arctan(z)+\arctan(y))^2+\sqrt{\frac{13}{5}}\frac{1-zy}{13(z+y)}\sin(\arctan(x)+\arctan(z))^2$
$\geq \frac{1+\sqrt{\frac{5}{13}}\frac{x+y}{1-xy}+\sqrt{\frac{13}{5}}\frac{1-zy}{y+z}}{18}$
Moreover $\sin(\arctan(u)+\arctan(v))^2=\frac{(u+v)^2}{(u^2+1)(v^2+1)}$
So we get the following inequality :
$\sqrt{\frac{5}{13}}\frac{x+y}{13(1-xy)}\frac{(x+y)^2}{(x^2+1)(y^2+1)}+\frac{1}{13}\frac{(z+y)^2}{(z^2+1)(y^2+1)}+\sqrt{\frac{13}{5}}\frac{1-zy}{13(z+y)}\frac{(x+z)^2}{(x^2+1)(z^2+1)}\geq \frac{1+\sqrt{\frac{5}{13}}\frac{x+y}{1-xy}+\sqrt{\frac{13}{5}}\frac{1-zy}{y+z}}{18}$
After that I think we can use trigonometric or hyperbolic properties but I don't know how .
| So, my attempt is based on putting: $\tfrac{b}{a}=x$ and $\tfrac{c}{b}=y$, so $x,y>0$. Then our OP inequality:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
$$\frac{a}{13+5x^2}+\frac{b}{13+5y^2}+\frac{c}{13+5(\tfrac{1}{xy})^2}\geq\frac{a+b+c}{18}$$
$$\frac{\tfrac{b}{x}}{13+5x^2}+\frac{b}{13+5y^2}+\frac{yb}{13+5(\tfrac{1}{xy})^2}\geq\frac{\tfrac{b}{x}+b+yb}{18}$$
$$\frac{\tfrac{1}{x}}{13+5x^2}+\frac{1}{13+5y^2}+\frac{y}{13+5(\tfrac{1}{xy})^2}\geq\frac{\tfrac{1}{x}+1+y}{18}$$
$$\frac{1}{13+5x^2}+\frac{x}{13+5y^2}+\frac{(xy)^3}{13(xy)^2+5}\geq\frac{1+x+xy}{18}$$
$$\frac{1}{13+5x^2}+\frac{x}{13+5y^2}+\frac{(xy)^3}{13(xy)^2+5}-\frac{1+x+xy}{18}\geq0$$
So, it's enough to study 2-variables inequality, but I can't continue it in terms of elementary inequalities, by the way wolf says
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int \frac {dx}{\cos x+C}$? I want to evaluate -
$$\int \frac {dx}{\cos x+C}$$
Where $C$ is an arbitrary constant.I tried substitution and parts but could not do it.
Note that for $C=1$ one can simply do this by using compound angle formulas.But what about other values of $C $?
Thanks for any help!!
| Hint:
Make the substitution $ \theta = \frac{x}{2}$.
\begin{align}
\frac{1}{C + \cos 2 \theta} &= \frac{1}{C -1 + 2\cos ^2 \theta} \\
&= \frac{\sec^2 \theta}{(C-1)\sec^2 \theta + 2} \\
&= \frac{\sec^2 \theta}{C + 1 + (C-1)\tan^2 \theta}
\end{align}
Hence
\begin{align}
\int \frac{1}{C + \cos x} \text{d}x &= \int \frac{2}{C + \cos 2 \theta} d \theta \\
&= \int \frac{ 2\sec^2 \theta}{ C + 1 + (C-1)\tan^2 \theta} d \theta
\end{align}
Now make the second substitution $t = \tan \theta$ and continue as above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrating rational functions. $\int\frac{x^2-3x+1}{x^4-x^2+1}\mathrm{d}x$ Is there really another way of solving this problem $$\int\frac{x^2-3x+1}{x^4-x^2+1}\mathrm{d}x$$
avoiding partial fractions decomposition? I have tried the partial fractions decomposition, but it is not serving me right.
| The integral can be split into two pieces:
$$\begin{aligned}
\int {\frac{x}{{{x^4} - {x^2} + 1}}dx} &= \frac{1}{2}\int {\frac{1}{{{u^2} - u + 1}}du} \\
&= \frac{1}{{\sqrt 3 }}\arctan \left( {\frac{{2u - 1}}{{\sqrt 3 }}} \right) + C \\&= \frac{1}{{\sqrt 3 }}\arctan \left( {\frac{{2{x^2} - 1}}{{\sqrt 3 }}} \right) + C
\end{aligned}$$
where $u=x^2$, and
$$\int {\frac{{{x^2} + 1}}{{{x^4} - {x^2} + 1}}dx} = \int {\frac{{1 + \frac{1}{{{x^2}}}}}{{{x^2} + \frac{1}{{{x^2}}} - 1}}dx} = \int {\frac{{d(x - \frac{1}{x})}}{{{{(x - \frac{1}{x})}^2} + 1}}dx} = \arctan (x - \frac{1}{x}) + C$$
Therefore
$$\begin{aligned}
\int {\frac{{{x^2} - 3x + 1}}{{{x^4} - {x^2} + 1}}dx} &= \arctan (x - \frac{1}{x}) - 3\left[ {\frac{1}{{\sqrt 3 }}\arctan \left( {\frac{{2{x^2} - 1}}{{\sqrt 3 }}} \right)} \right] + C \\
&=
\arctan (x - \frac{1}{x}) - \sqrt 3 \arctan \left( {\frac{{2{x^2} - 1}}{{\sqrt 3 }}} \right) + C
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\sum_{n=1}^{\infty}\frac{n}{3^n-1}.$ Evaluate $$\sum_{n=1}^{\infty}\frac{n}{3^n-1}.$$
Solution(Partial):
$|x|<1$
$$\sum_{n=1}^{\infty}\frac{nx^{n-1}}{3^n-1}=\frac{d}{dt}\sum_{n=1}^{\infty}\frac{t^n}{3^n-1}\big{|}_{t=x}$$
$$\sum_{n=1}^{\infty}\frac{t^n}{3^n-1}=\sum_{n=1}^{\infty} t^n\sum_{k=1}^{\infty}\frac{1}{3^{kn}}=\\ \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{t^n}{3^{kn}}=\sum_{k=1}^{\infty}\frac{\frac{t}{3^k}}{1-\frac{t}{3^k}}=\sum_{k=1}^{\infty}\frac{t}{3^k-t}$$
$$\frac{d}{dt}\sum_{k=1}^{\infty}\frac{t}{3^k-t}\big{|}_{t=x}=\sum_{k=1}^{\infty}\frac{1}{3^k-x}+\frac{x}{(3^k-x)^2}$$
as $x\to 1-$
$$\sum_{n=1}^{\infty}\frac{n}{3^n-1}=\sum_{k=1}^{\infty}\frac{1}{3^k-1}+\frac{1}{(3^k-1)^2}=\sum_{k=1}^{\infty}\frac{1}{3^k-1}+\sum_{k=1}^{\infty}\frac{1}{(3^k-1)^2}$$
$$\Rightarrow \sum_{n=1}^{\infty}\frac{n-1}{3^n-1}=\sum_{k=1}^{\infty}\frac{1}{(3^k-1)^2}$$
But $\frac{n-1}{3^n-1}>\frac{1}{(3^n-1)^2}$ for all $n\geq 2$
Next part of the solution after the answer of Professor Vector:
$$\sum_{k=1}^{\infty}\frac{1}{3^k-1}+\frac{1}{(3^k-1)^2}=\sum_{k=1}^{\infty} \frac{3^k}{(3^k-1)^2}=\sum_{k=1}^{\infty} \frac{1}{3^k-2+\frac{1}{3^k}}=\sum_{k=1}^{\infty} \frac{1}{(3^{k/2}-3^{-k/2})^2}\\=\frac{1}{4}\sum_{k=1}^{\infty} \frac{1}{\big{(}\frac{e^{\frac{\log 3}{2}k}-e^{-\frac{\log 3}{2}k}}{2}\big{)}^2}=\sum_{k=1}^{\infty}\frac{1}{4\sinh^2\big{(}\frac{\log 3}{2}k\big{)}}$$
I can not go more further after the last expression. I do not have much experience about Hyperbolic Trigonometric series.
The main post is here. The main poster is inactive for 8 months, so I had to post it here. I could not do it 5 years ago. I could not do it now sadly!
| Even though $\frac{n-1}{3^n-1}>\frac{1}{(3^n-1)^2}$ for all $n\geq 2$, your sum
$$\sum_{n=1}^{\infty}\frac{n-1}{3^n-1}=\sum_{k=1}^{\infty}\frac{1}{(3^k-1)^2}$$
starts with $n=1,$ so that first term may compensate the others. In fact, a quick numerical check shows that both sides essentially coincide (well, 0.26727973753486445 and 0.2672797375348645, let's not be too picky). Your derivation is absolutely correct, as far as I can see.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$ Find the equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$
Let the required line is tangent to the parabola at the point $(x_1,y_1)$.It passes through $(\frac{1}{2},2)$.Its equation is $y-2=-x_1(x-\frac{1}{2})$.
This line is also secant to the the curve $y=\sqrt{4-x^2}$.
I solved $y=\sqrt{4-x^2}$ and the line $y-2=-x_1(x-\frac{1}{2})$.
I am stuck here.
| Problem
$$
\color{red}{y = 2 - \frac{x^{2}}{2}}
\tag{1}
$$
$$
\color{blue}{y = \sqrt{4-x^2}}
\tag{2}
$$
Equation of the line
Find the equation of the tangent line
$$
y = m x + b
\tag{3}
$$
tangent to $\color{red}{y(x)}$
We are given the a point
$$
p = \left( \frac{1}{2}, 2 \right)
$$
Find the slope, $m$, and the intercept, $b$.
Slope
To be tangent to the red curve, the slope of the line must match the slope of red curve. The slope is of the red curve is
$$
\color{red}{y'} = -1
$$
Intercept
The intercept is computed from $(3)$ using the point $p$:
$$
b = y - m x \qquad \Rightarrow \qquad b = 1 - (-1) \frac{1}{2} = \frac{5}{4}
$$
Solution
The equation for the tangent line (the dashed line below) is
$$
\boxed{ y = -x + \frac{5}{4}}
\tag{4}
$$
Tangent point
Where does the dashed line, $(4)$, touch the red curve, $(1)$? Solve
$$
\begin{align}
y &= \color{red}{y} \\
%
-x + \frac{5}{4} &= \color{red}{2 - \frac{x^{2}}{2}} \\
%
x &= 1
\end{align}
$$
Using $(4)$, we have the tangent point is
$$
q = \left( 1, \frac{3}{2} \right),
$$
where the dashed line touches the red curve.
Secant points
Where are the two points where the dashed line, $(4)$ intercepts the blue curve, $(2)$? Solve
$$
\begin{align}
y &= \color{blue}{y} \\
-x + \frac{5}{4} &= \color{blue}{\sqrt{4-x^2}} \\
\end{align}
$$
The solution is $x = \frac{1}{4} \left(5 \pm \sqrt{7} \right)$. Therefore the two points define the secant chord are
$$
%
\frac{1}{4}
\left( \left(5 - \sqrt{7}\right),
\left(5 + \sqrt{7}\right) + 10 \right), \qquad
%
\frac{1}{4}
\left( \left(5 + \sqrt{7}\right),
\left(-5 - \sqrt{7}\right) + 10 \right)
%
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplification of Trigo expression
Simplify $$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}$$
My attempt,
$$=\frac{(\cos^2x+\frac{\sin^2x}{\cos^2x})}{\frac{\sin x \cos x+1}{\cos x}} $$
$$=\frac{\cos^4 x+\sin^2 x}{\cos^2 x}\cdot \frac{\cos x}{\sin x \cos x+1}$$
$$=\frac{\cos^4 x+\sin^2 x}{\cos x(\sin x \cos x+1)}$$
I'm stuck at here. The given answer is $\sec x -\sin x$
| $$\scriptsize\frac {\tan^2x+\color{blue}{\cos^2x}}{\sin x+\sec x}=\frac {\overbrace{\tan^2x+\color{blue}1}^{\sec^2 x}\color{blue}{-\sin^2x}}{\sin x+\sec x}=\frac {\sec^2x-\sin^2x}{\sec x+\sin x}=\frac{(\sec x -\sin x)(\sec x+\sin x)}{\sec x+\sin x}=\sec x-\sin x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the minimal value of $\sum\limits_{n=1}^8|x-n|$
Find the minimum value of $( |x-1|+|x-2|...+|x-8|) $
My attempt
Using triangle inequality i.e.
$|x-y|≤|x|+|y|$ $|x-1|=|x-1-0|≤|x-1|+0$ $|x-2|=|x-1-1|≤|x-1|+1$ $|x-3| =|x-1-2| ≤|x-1|+2$
...
...
...
$|x-8| =|x-1-7| ≤|x-1|+7$
Adding all these inequalities, $( |x-1|+|x-2|...+|x-8|) $$≤$ $8|x-1| + 28$ Clearly the minimum value is 28 at x=1. The answer is wrong. Also it is intuitive that the x must lie somewhere in between 1 to 8 so that the sum of the distances of x from each number is the least.
What am I doing wrong? How else to approach this question. Are there any other ways?
| Use the triangle inequality: $|x+y|\le |x|+|y|$.
$|x-1|+|8-x|\ge |x-1+8-x|=7$,
$|x-2|+|7-x|\ge |x-2+7-x|=5$,
$|x-3|+|6-x|\ge |x-3+6-x|=3$,
$|x-4|+|5-x|\ge |x-4+5-x|=1$,
Summing we get min $16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$
And,
$$x^2-5x+1=0$$
??
| If $y=\sqrt x + 1/\sqrt x$ then your value of $y$ is given as a solution of $y$ for the system
\begin{align}
x^2&-5x+1=0\\
t^2&=x\\
t^2&=ty-1\\
\end{align}
Putting the second and third equation together $x=ty-1$ so
$$(ty-1)^2-5(ty-1)+1=t^2y^2-2ty+1-5ty+5+1=(ty-1)y^2-7(ty-1)=0$$
If $ty-1\neq 0$ then $y^2=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
how to compute $E(X^4)$ when $X$ follow the normal law $N(0,1)$ Suppose $X$ follow the normal law $N(0,1)$
We have the density $f= \frac{1}{\sqrt{\sigma^2 2\pi}} e^{- \frac{(x-m)^2}{2\sigma^2}}$
We want to compute $E(X^4)$
We have by definition that $\displaystyle E(X^4) = \frac{1}{\sqrt{2\pi}} \int_\infty^\infty{e^{-x^2/2}} x^4\,dx $
But i dont know how to continue
Thank you for helping me
| \begin{align}
& \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty x^4 e^{-x^2/2}\,dx \\[10pt]
= {} & 2\cdot\frac 1 {\sqrt{2\pi}} \int_0^\infty x^4 e^{-x^2/2} \, dx & & \text{since the integral is of an even} \\ & & & \text{function over an interval that} \\[8pt]
& & & \text{is symmetric about 0} \\[10pt]
= {} & \sqrt{\frac 2 \pi} \int_0^\infty x^3 e^{-x^2/2} (x\,dx) \\[10pt]
= {} & \sqrt{\frac 2 \pi} \int_0^\infty (2u)^{3/2} e^{-u} \, du \\[10pt]
= {} & \frac 4 {\sqrt\pi} \int_0^\infty u^{3/2} e^{-u} \, du \\[10pt]
= {} & \frac 4 {\sqrt\pi}\,\, \Gamma\left(\frac 5 2 \right) \\[10pt]
= {} & \frac 4 {\sqrt\pi} \cdot\frac 3 2 \Gamma\left( \frac 3 2 \right) \\[10pt]
= {} & \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \cdot \Gamma\left( \frac 1 2 \right) \\[10pt]
= {} & \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \cdot \sqrt\pi \\[10pt]
= {} & 3.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to integrate $\int \frac {\cos^4 x}{\sqrt {1- \sin x}} $ I frankly don't know where to start. Looked for some identities to get rid of the square root below.
$$\int \frac {\cos^4 x}{\sqrt {1- \sin x}} \, dx $$
| Write $\cos(x)^4 = \cos(x)(1-\sin^2(x))^{3/2} = \cos(x)(1-\sin(x))^{3/2}(1+\sin(x))^{3/2}.$ Thus $$\int \frac{\cos(x)^4}{\sqrt{1-\sin(x)}}dx = \int (1-\sin(x))(1+\sin(x))^{3/2}\cos(x) dx$$ Put $u = \sin(x)$. Then $$\int\frac{\cos(x)^4}{\sqrt{1-\sin(x)}}dx = \int(1-u)(1+u)^{3/2}du.$$ Now put $z = 1+u$ so $$\int\frac{\cos(x)^4}{\sqrt{1-\sin(x)}}dx = \int(2-z)z^{3/2}dz = \int (2z^{3/2} - z^{5/2}) dz = \frac{4}{5} z^{5/2} - \frac{2}{7} z^{7/2} + C.$$ Un-doing the substitutions gives $$\int\frac{\cos(x)^4}{\sqrt{1-\sin(x)}}dx =\frac{4}{5}(1+\sin(x))^{5/2} - \frac{2}{7}(1+\sin(x))^{7/2} + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Given $u_n=au_{n-1}+b$ show that $u_n=A\cdot a^n + d$ where $d$ is a fixed point of $u_n$ Given $u_n=au_{n-1}+b$ show that $u_n=A\cdot a^n + d$ where:
*
*$d$ is a fixed point of the recurrence relation $u_n=au_{n-1}+b$;
*$A$ is a constant.
My attempt:
I have tried evaluating the recurrence relation and I've come to the following equation
$u_{n+k}=a^{k+1}u_{n-1}+b(a_k...(a_2(a_1+1)+1)...+1)$
| Just for illustration, a brute force approach yields
$$
\begin{split}
u_n &= au_{n-1} + b \\
&= a(au_{n-2} + b) + b
= a^2 u_{n-2} + b(a+1)\\
&= a^2 (au_{n-3} + b) + b(a+1)
= a^3 u_{n-3} + b(a^2+a+1)\\
&= a^nu_0 + b \sum_{k=0}^{n-1}a^k\\
&= a^nu_0 + b \frac{1-a^n}{1-a} \\
&= a^n \left( u_0-\frac{b}{1-a}\right) + \frac{b}{1-a}
\end{split}
$$
It remains to show that $$\frac{b}{1-a}$$ is a fixed point of the relation, to do that note that
$$
a\left(\frac{b}{1-a}\right) + b
= \frac{b}{1-a} \left[a + (1-a)\right]
= \frac{b}{1-a}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$ where $\mathcal{C}$ is the unit circle On the generalization of a recent question, I have shown, by analytic and numerical means, that
$$\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2+\cdots+z^{2n}|^2~dz =2n$$
where $\mathcal{C}$ is the unit circle. Thus, $z=e^{i\theta}$ and $dz=iz~d\theta$. There remains to prove it, however.
What I have done: consider the absolute value part of the integrand,
$$
\begin{align}
|1+z+z^2+\cdots+z^{2n}|^2
&=(1+z+z^2+\cdots+z^{2n})(1+z+z^2+\cdots+z^{2n})^*\\
&=(1+z+z^2+\cdots+z^{2n})(1+z^{-1}+z^{-2}+\cdots+z^{-2n})\\
&=(1+z+z^2+\cdots+z^{2n})(1+z^{-1}+z^{-2}+\cdots+z^{-2n})\frac{z^{2n}}{z^{2n}}\\
&=\left(\frac{1+z+z^2+\cdots+z^{2n}}{z^n} \right)^2\\
&=\left(\frac{1}{z^n}\cdots+\frac{1}{z}+1+z+\cdots z^n \right)^2\\
&=(1+2\cos\theta+2\cos 2\theta+\cdots+2\cos n\theta)^2\\
\end{align}
$$
We now return to the integral,
$$
\begin{align}\frac{1}{2\pi i}\int_C |1+z+z^2+\cdots z^n|^2dz
&=\frac{1}{2\pi}\int_0^{2\pi}(1+2\cos\theta+2\cos 2\theta+\cdots+2\cos n\theta)^2 (\cos\theta+i\sin\theta)~d\theta\\
&=\frac{1}{2\pi}\int_0^{2\pi}(1+2\cos\theta+2\cos 2\theta+\cdots+2\cos n\theta)^2 \cos\theta~d\theta
\end{align}$$
where we note that the sine terms integrate to zero by virtue of symmetry. This is where my trouble begins. Clearly, expanding the square becomes horrendous as $n$ increases, and even though most of the terms will integrate to zero, I haven't been able to selectively find the ones that won't.
The other thing I tried was to simplify the integrand by expressing it in terms of $\cos\theta$ only using the identity
$$\cos n\theta=2\cos (n-1)\theta\cos\theta-\cos(n-2)\theta$$
but this too unfolds as an algebraic jungle very quickly. There are various other expressions for $\cos n\theta$, but they seem equally unsuited to the task. I'll present them here insofar as you may find them more helpful than I did.
$$
\cos(nx)=\cos^n(x)\sum_{j=0,2,4}^{n\text{ or }n-1} (-1)^{n/2}\begin{pmatrix}n\\j\end{pmatrix}\cot^j(x)=\text{T}_n\{\cos(x)\}\\
\cos(nx)=2^{n-1}\prod_{j=0}^{n-1}\cos\left(x+\frac{(1-n+2j)\pi}{2n} \right)\quad n=1,2,3,\dots
$$
where $\text{T}_n$ are the Chebyshev polynomials. Any suggestions will be appreciated.
| For $z\in\mathcal C$, we have $$|1+...+z^{2n}|^2 = (1+z+...+z^{2n})(1+z^{-1}+...+z^{-2n}) = \sum_{j=0}^{2n}\sum_{k=0}^{2n}z^jz^{-k} = \sum_{j=0}^{2n}\sum_{k=0}^{2n}z^{j-k}.$$
Therefore $$\frac{1}{2\pi i}\int_{\mathcal C}|1+z+...+z^{2n}|^2 dz = \sum_{j=0}^{2n}\sum_{k=0}^{2n}\frac{1}{2\pi i}\int_{\mathcal C}z^{j-k}dz.$$
If $j-k\neq -1$, this integral vanishes, so you just have to count the summands for which $j-k=-1$ and calculate $\frac{1}{2\pi i}\int_{\mathcal C}z^{-1}dz$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove $5^n +5 <5^{n+1}$ $∀n∈N$ using induction Prove $5^n +5 <5^{n+1}$ $∀n∈N$
Base Case: $n=1$
$\implies 5^1 +5 <25$
$\implies 10<25$ ; holds true
Induction hypothesis: Suppose $5^k +5 < 5^{k+1}$ is true for k∈N
Then;
$\implies 5^{k+1} +5 < 5^{k+2}$
$\implies 5\cdot 5^k +5 < 25*5^k$
I don't know how to proceed after this step.
| $$5^{k+1}+5=5\times 5^k+5<5(5^{k+1}-5)+5=5^{k+2}-25+5<5^{k+2}$$
Notice that, by the induction step, we have $$5^k+5<5^{k+1}\implies 5^k<5^{k+1}-5$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\tan(\frac{1}{2}\arcsin(\frac{5}{13}))$ I wanted to solve it by using this formula: $\tan(\frac{x}{2})=\pm\sqrt{\frac{1-x^2}{1+x^2}}.$ I thought it wouldn't work (because there are $\pm$). Then used the right triangle method: $$\frac{1}{2}\arcsin(\frac{5}{13})=\alpha\Rightarrow\frac{5}{26}=\sin\alpha$$ $$a^2+b^2=c^2\Rightarrow a^2+25=676\Rightarrow a=\sqrt{651}\Rightarrow\tan(\frac{5}{\sqrt{651}}).$$ It turned out to be wrong. How to get the right answer?
| Let $\arcsin(\frac{5}{13})=x$, thus $\sin(x)=\frac{5}{13}$
So, this problem reduced to find value of $\tan(\frac{x}{2})=u$,
According to $\sin(x)=\frac{2u}{u^2+1}$ identity for $u=\tan(\frac{x}{2})$;
$\frac{2u}{u^2+1}=\frac{5}{13}$
$5*(u^2+1)=13*2u$
$5u^2+5=26u$
$5u^2-26u+5=0$
$(u-5)*(5u-1)=0$
Due to $\tan(x)>0$ constraint, $\tan(\frac{x}{2})$ must be equal to $\frac{1}{5}$.
| {
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"url": "https://math.stackexchange.com/questions/2334402",
"timestamp": "2023-03-29T00:00:00",
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Calculating integral $\oint\limits_{|z|=4} \frac{\sin^2(z)}{(z-\frac{\pi}{6})^2(z+\frac{\pi}{6})}dz$
Calculating $$\oint\limits_{|z|=4} \frac{\sin^2(z)}{(z-\frac{\pi}{6})^2(z+\frac{\pi}{6})}dz$$
Due to the fact of the denominator I've looked for the partial fractions:
$$1=A\left(z-\frac{\pi}{6}\right)^3+B\left(z+\frac{\pi}{6}\right)\left(z-\frac{\pi}{6}\right)+C\left(z^2-\left(\frac{\pi}{6}\right)^2\right)$$
$$\Leftrightarrow 1=z^3(A+B)+z^2\left(-\frac{\pi}{2}A-\frac{\pi}{6}B+C\right)+z\left(\frac{\pi^2}{12}A-\frac{\pi^2}{36}B\right)-\frac{\pi^3}{216}A+\frac{\pi^2}{36}C$$
$$\begin{align}\Rightarrow 0&=A+B\\
0&=\frac{\pi^2}{12}A-\frac{\pi^2}{36}B\\
&...
\end{align}$$
Unfortunately there I made a mistake. It's a quite heavy equation and maybe there is a more efficient way to solve this integral. Any hints? Thank you!
| It happens that$$\frac1{\left(z-\frac\pi6\right)^2\left(z+\frac\pi6\right)}=\frac9{\pi ^2\left(z+\frac\pi6\right)}-\frac9{\pi ^2\left(z-\frac\pi6\right)}+\frac3{\pi\left(z-\frac\pi6\right)^2}$$Therefore\begin{multline*}\oint_{|z|=4}\frac{\sin^2(z)}{\left(z-\frac\pi6\right)^2\left(z+\frac\pi6\right)}dz=\\=\frac9{\pi^2}\oint_{|z|=4}\frac{\sin^2(z)}{z+\frac\pi6}dz-\frac9{\pi^2}\oint_{|z|=4}\frac{\sin^2(z)}{z-\frac\pi6}dz+\frac3\pi\oint_{|z|=4}\frac{\sin^2(z)}{\left(z-\frac\pi6\right)^2}dz.\end{multline*}
Added note: Here's how to obtain the partial fraction decomposition. You want numbers $A$, $B$, and $C$ such that$$\frac1{\left(z-\frac\pi6\right)^2\left(z+\frac\pi6\right)}=\frac A{z+\frac\pi6}+\frac B{z-\frac\pi6}+\frac C{\left(z-\frac\pi6\right)^2}.$$Putting together these three fractions on the right hand side, you get:$$\frac{A z^2-\frac\pi3 A z+\frac{\pi^2}{36}A+B z^2-\frac{\pi ^2}{36} B+C z+\frac\pi6C}{\left(z+\frac\pi6\right)\left(z-\frac\pi6\right)^2.}$$This is equal to $\frac1{\left(z-\frac\pi6\right)^2\left(z+\frac\pi6\right)}$ if and only if$$\left\{\begin{array}{l}\frac{\pi^2}{36}A-\frac{\pi^2}{36}B+\frac\pi6C=1\\-\frac\pi3A+C=0\\A+B=0.\end{array}\right.$$The solution of this system is $A=\frac9{\pi^2}$, $B=-\frac9{\pi^2}$ and $C=\frac3\pi$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $\tan{\left(\frac{2\pi}{5}\right)}$ to find that the answer is $10$. However, I want to know whether there is a faster way that does not involve calculating those values.
So far, I have this: let $a=\tan{\left(\frac{\pi}{5}\right)}$ and $b=\tan{\left(\frac{2\pi}{5}\right)}$; then, $b=\frac{2a}{1-a^2}$ and $a=-\frac{2b}{1-b^2}$, so multiplying the two and simplifying gives:
$$a^2+b^2={\left(ab\right)}^2+5$$
Any ideas? Thanks!
| $$\tan^2\frac{\pi}{5}+\tan^2\frac{2\pi}{5}=\frac{1-\cos\frac{2\pi}{5}}{1+\cos\frac{2\pi}{5}}+\frac{1-\cos\frac{4\pi}{5}}{1+\cos\frac{4\pi}{5}}=$$
$$=\frac{2-2\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}{1+\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}+\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}=$$
$$=\frac{2-\frac{4\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}{2\sin\frac{2\pi}{5}}}{1+\frac{2\sin\frac{\pi}{5}\cos\frac{2\pi}{5}+2\sin\frac{\pi}{5}\cos\frac{4\pi}{5}}{2\sin\frac{\pi}{5}}+\frac{4\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}}{4\sin\frac{2\pi}{5}}}=$$
$$=\frac{2-\frac{\sin\frac{8\pi}{5}}{2\sin\frac{2\pi}{5}}}{1+\frac{\sin\frac{3\pi}{5}-\sin\frac{\pi}{5}+\sin\frac{5\pi}{5}-\sin\frac{3\pi}{5}}{2\sin\frac{\pi}{5}}+\frac{\sin\frac{8\pi}{5}}{4\sin\frac{2\pi}{5}}}=\frac{2+\frac{1}{2}}{1-\frac{1}{2}-\frac{1}{4}}=10.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding number of asymptotes in $[ 0 , 2\pi]$ interval Consider $f(x) = \frac{2\tan 2x + 1}{2\cos x + 1}$ . Now find number of asymptotes in $[ 0 , 2\pi]$ interval . I know we can simplify it to $\frac{2\sin 2x + 1}{\cos 2x (2\cos x + 1)}$ and then solve $\cos 2x (2\cos x + 1) = 0$ but is there any way for doing it straightly without simplifying ? Also I'm looking for an instruction in order to finding number of asymptotes .
| $\cos2x\neq0$ gives asymptotes $x=\frac{\pi}{4}$, $x=\frac{3\pi}{4}$, $x=\frac{5\pi}{4}$ and $x=\frac{7\pi}{4}$.
$\cos{x}\neq-\frac{1}{2}$ gives asymptotes $x=\frac{2\pi}{3}$ and $x=\frac{4\pi}{3}$.
Id est, we have six asymptotes.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving trigonometric identity $\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$ Show that
$$\cos^6A+\sin^6A=1-3 \sin^2 A\cos^2A$$
Starting from the left hand side (LHS)
\begin{align}
\text{LHS} &=(\cos^2A)^3+(\sin^2A)^3 \\
&=(\cos^2A+\sin^2A)(\cos^4A-\cos^2A\sin^2A+\sin^4A)\\
&=\cos^4A-\cos^2A\sin^2A+\sin^4A
\end{align}
Can anyone help me to continue from here
| Let $a=\cos^2 (A),b=\sin^2 (A) $ . Now use $a^3+b^3=(a+b)^3-3ab (a+b) $ also note that $a+b=1$. Hence the proof.
| {
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Proving trigonometric identity $\frac{\sin(A)}{1+ \cos(A )}+\frac{1+ \cos(A )}{\sin(A)}=2 \csc(A)$
$$
\frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=2\csc(A)
$$
\begin{align}
\mathrm{L.H.S}&= \frac{\sin^2A+(1+\cos^2(A))}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{\sin^2A+2\sin(A)\cos(A)+\cos^2(A)+1}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{2+2\sin(A)\cos(A)}{\sin(A)(1+\cos(A))}
\end{align}
What should be done from here?
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\i}{\mathrm{i}} \newcommand{\text}[1]{\mathrm{#1}} \newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}$
$$\frac{\sin(a)}{1+ cos(a
)}+\frac{1+ \cos(a
)}{\sin(a)}=2 cosec(a)$$
$$\frac{\sin^2a + (1+\cos a)^2}{\sin a (1+ \cos a)} = \frac{2}{\sin a}$$
$$\frac{\sin^2a + (1+\cos^2 a + 2\cos a)}{\sin a (1+ \cos a)} = \frac{2}{\sin a}$$
$$\frac{2+2\cos a}{\sin a (1+ \cos a)} = \frac{2}{\sin a}$$
$$\bbx{\frac{1+1\cos a}{1 (1+ \cos a)} = \frac{1}{1}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve three equations with three unknowns Solve the system:
$$\begin{cases}a+b+c=6\\ab+ac+bc=11\\abc=6\end{cases}$$
The solution is:
$a=1,b=2,c=3$
How can I solve it?
| By method of substitution:
$$\begin{cases} b+c=6-a \\ a(6-a)+\frac{6}{a}=11 \\ bc=\frac{6}{a}\end{cases} \Rightarrow$$
$$a^3-6a^2+11a-6=0 \Rightarrow (a-1)(a-2)(a-3)=0 \Rightarrow a=1; 2; 3.$$
Substituting these and solving $$\begin{cases} b+c=6-a \\ bc=\frac{6}{a}\end{cases}$$
six solutions will be found:
$$(a,b,c)=(1,2,3); (1,3,2); (2,1,3); (2,3,1); (3,1,2); (3,2,1).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving inequality $\ 1+\frac14+\frac19+\cdots+\frac1{n^2}\le 2-\frac1n$ using induction Question:
Prove $$\ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\le 2-\frac{1}{n},
\text{ for all natural } n$$
My attempt:
Base Case: $n=1$ is true:
I.H: Suppose $1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{k^2}\le 2-\frac{1}{k},$ for some natural $k.$
Now we prove true for $n = k+1$
$$ 1+\frac{1}{4}+\cdots+\frac{1}{k^2}+\frac{1}{\left(k+1\right)^2}\le 2-\frac{1}{k}+\frac{1}{\left(k+1\right)^2},\text{ by induction hypothesis} $$
Now how do I show that $2-\frac{1}{k}+\frac{1}{\left(k+1\right)^2}\le 2-\frac{1}{\left(k+1\right)}\text{ ?}$
Have I done everything correctly up until here?
If yes, how do I show this inequality is true?
Any help would be appreciated.
| You are right!
We need to prove that:
$$\frac{1}{(k+1)^2}<\frac{1}{k}-\frac{1}{k+1}$$ or
$$\frac{1}{(k+1)^2}<\frac{1}{k(k+1)},$$
which is obvious.
| {
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"timestamp": "2023-03-29T00:00:00",
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Binomial vs Probability
A shipment of 12 microwave ovens contains 3 defective units. A vending company has ordered 4 of these units, and because all are packaged identically, the selection will be at random. What is the probability that at least 2 units are good?
There are two ways to solve this problem:
*
*Using the binomial distribution:
\begin{align*}
P(\text{defective}) & = \frac{3}{12} = \frac{1}{4}\\
P(\text{good}) & = 1 - \frac{1}{4} = \frac{3}{4}\\
P(2~\text {good}) & = \binom{4}{2} \times 0.75^{2} \times 0.25^{2} = 0.211\\
P(3~\text{good}) & = \binom{4}{3}\times 0.75^{3} \times 0.25 = 0.422\\
P(4~\text{good}) & = 0.75^{4} = 0.316
\end{align*}
The probability that at least 2 units are good is given by:
$$P(2~\text{good}) + P(3~\text{good}) + P(4~\text{good}) = 0.211 + 0.422 + 0.316 = 0.949$$
*Using combinations:
$$\frac{\dbinom{9}{2}\dbinom{3}{2} + \dbinom{9}{3}\dbinom{3}{1} + \dbinom{9}{4}}{\dbinom{12}{4}} = \frac{486}{495} = \frac{54}{55} = 0.9818$$
Why are these two answers different?
| You cannot use the binomial distribution, as this is an experiment without replacement: once you pick a defective oven, there are only two defective ovens left. Using combinations, the probability of selecting at least two good units indeed equals:
$$\frac{{9 \choose 4}{3 \choose 0} + {9 \choose 3}{3 \choose 1} + {9 \choose 2}{3 \choose 2}}{12 \choose 4} \approx 0.982$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the value of this logarithmic/trigonometric expression? I need the value of this expression, when simplified.
$$ \log_{10}(\cot(1)°) + \log_{10}(\cot(2°)) + \cdots + \log_{10}(\cot(89°))$$
All the $\log$ have base $10$.
| The sum is essentially
\begin{align}
S &= \sum_{k=1}^{89} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\right) \\
&= \sum_{k=1}^{44} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\right) + \log_{10}\left(\cot\left(\frac{\pi}{4}\right)\right) + \sum_{k=46}^{89} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\right) \\
&= \sum_{k=1}^{44} \left[\log_{10}\left(\cot\left(\frac{k \pi}{180}\right)\right) + \log_{10}\left(\cot\left(\frac{\pi}{2} - \frac{k \pi}{180} \right)\right) \right] + \log_{10}\left(\cot\left(\frac{\pi}{4}\right)\right) \\
&= \sum_{k=1}^{44} \log_{10}\left(\cot\left(\frac{k \pi}{180}\right) \, \tan\left(\frac{k \pi}{180}\right)\right) + \log_{10}\left(\cot\left(\frac{\pi}{4}\right)\right) \\
&= \sum_{k=1}^{44} \log_{10}(1) + \log_{10}(1) \\
&= \sum_{k=1}^{45} \log_{10}(1) = 0.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Hyperbolas: Deriving $\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$ My textbook's section on Hyperbolas states the following:
If the foci are $F_1(-c, 0)$ and $F_2(c, 0)$ and the constant difference is $2a$, then a point $(x, y)$ lies on the hyperbola if and only if $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$.
To simplify this equation, we move the second radical to the right-hand side, square, isolate the remaining radical, and square again, obtaining $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$.
I've attempted to "move the second radical to the right-hand side, square, isolate the remaining radical, and square again", but I cannot derive $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$. I could be misunderstanding the instructions, but my attempts to derive the textbook's solution by precisely following the instructions have not been successful.
I would greatly appreciate it if people could please take the time to demonstrate the derivation of $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$, as mentioned in the textbook. Are the textbook's instructions incorrect/insufficient or am I simply misunderstanding them?
| Let's do the case
$$
\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = 2a
$$
which holds provided $x>0$.
Move the second radical to the right-hand side and square:
$$
(x + c)^2 + y^2=4a^2+(x - c)^2 + y^2 + 4a\sqrt{(x - c)^2 + y^2}
$$
Lots of terms cancel:
$$
2cx=4a^2-2cx+4a\sqrt{(x - c)^2 + y^2}
$$
which becomes
$$
cx-a^2=a\sqrt{(x - c)^2 + y^2}
$$
Note that $cx\ge a^2$, so we can square again and arrive easily to the result.
The case $x<0$ is identical.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\frac{BF}{FC}$ in terms of $b,c$ If $AD$ is the bisector of angle $BAC$ and $E$ is the reflection of $D$ to point $M$(midpoint of $BC$).We construct point $F$ such that angles $BAF=EAC$.Then find $\frac{BF}{FC}$ in terms of $b,c$.
My attempt:We have:
$\frac{BF}{FC}=\frac{AB}{AC}*\frac{\sin BAF}{\sin CAF}$
But then what?
| By the bisector theorem $BD=\frac{c}{b+c}a$ and $CD=\frac{b}{b+c}a$.
Since $BM=CM=\frac{1}{2}a$, $BE=\frac{b}{b+c}a$ and $CE=\frac{c}{b+c}a.$
If $\widehat{BAF}=\widehat{EAC}$, the lines $AE$ and $AF$ have to be symmetric with respect to $AD$.
The trilinear coordinates of $E$ are $[0,c^2,b^2]$, hence the trilinear coordinates of $F$ are $[0,b^2,c^2]$ and
$$\frac{BF}{FC}=\frac{c^2/b}{b^2/c}=\color{red}{\frac{c^3}{b^3}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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An identity form: $x_1^3+x_2^3+x_3^3+x_4^3=y_1^3+y_2^3+y_3^3+y_4^3$ I found an identity form: $x_1^3+x_2^3+x_3^3+x_4^3=y_1^3+y_2^3+y_3^3+y_4^3$ as follows:
$$(a+b+c)^3+a^3+b^3+c^3=(a+b)^3+(b+c)^3+(c+a)^3+(6y)^3$$
Where $abc=36y^3$
Poof of this identity is very simple. But the identity nice.
Which reference of the identity? can generalized of the identity?
| Equation (shown below) has parametrization:
$(a+b+c)^3+a^3+b^3+c^3=(a+b)^3+(b+c)^3+(c+a)^3+(6y)^3$
$(y,a,b,c)=((8)(3k-4),(256),(24)(3k-4),(3)(3k-4)^2)$
For $k=2$, we get: $(3,12,64,79)^3=(15,24,67,76)^3$
| {
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Riccati D.E., vertical asymptotes
For the D.E.
$$y'=x^2+y^2$$
show that the solution with $y(0) = 0$ has a vertical asymptote at some point $x_0$. Try to find upper and lower bounds for $x_0$:
$$y'=x^2+y^2$$
$$x\in \left [ a,b \right ]$$
$$b> a> 0$$
$$a^2+y^2\leq x^2+y^2\leq b^2+y^2$$
$$a^2+y^2\leq y'\leq b^2+y^2$$
$$y'\geq a^2+y^2$$
$$\frac{y}{a^2+y^2}\geq 1$$
$$\int \frac{dy}{a^2+y^2}\geq \int dx=x+c$$
$$\frac{1}{a}\arctan \frac{y}{a}\geq x+c$$
$$\arctan \frac{y}{a}\geq a(x+c)$$
$$\frac{y}{a}\geq\tan a(x+c)$$
$$y\geq a\tan a(x+c)$$
$$a(x+c)\simeq \frac{\pi}{2}$$
But where to from here?
| From numerical solution comes out that $x=2$ and $x=-2$ are vertical asymptotes. Trying to solve as a Bernoulli equation gives a mess and the substitution $w=\frac{1}{y}$ gives problems as
$$w=\frac{1}{y};\;w'=-\frac{y'}{y^2}$$
Divide the original equation by $y^2$
$$\frac{y'}{y^2}=\frac{x^2}{y^2}+1\rightarrow -w'=x^2w^2+1$$
$$w'+x^2w^2=-1\rightarrow w(x)=c\;e^{-\frac{x^3}{3}}+\frac{e^{-\frac{x^3}{3}} x \Gamma \left(\frac{1}{3},-\frac{x^3}{3}\right)}{3^{2/3} \sqrt[3]{-x^3}}$$
The problem is now with the initial value, as $w\to\infty$ as $x\to 0$
Anyway the general solution is
$$y=\frac{1}{w(x)}=\frac{3 e^{\frac{x^3}{3}}}{3 c+x E_{\frac{2}{3}}\left(-\frac{x^3}{3}\right)}$$
where $E_k(x)$ is the integral exponential function defined by
$$E_k(x)=\int_1^{\infty }\frac{e^{-tx}}{t^k}\,dt$$
and has a vertical asymptote for any $c\in\mathbb{R}$
Hope this helps
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$ Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$
I need to specify the value of $f(x,y)$ which makes the given function continuous.
I had tried to give a arbitral relationship between $y$ and $x$, but as $(x, y)$ gets closer to $(0, 0)$ the arbitral movement cannot always be represented as a functional representation between $y$ and $x$.
How could I solve this problem rigorously?
| Recalling that $|x|\,|\cos(x)|\le |\sin(x)|\le |x|$ for $0\le |x|\le \pi/2$, we assert that for $x^2+y^2\le \pi/2$
$$|\cos(x^2+y^2)|\le \left|\frac{\sin(x^2+y^2)}{x^2+y^2}\right|\le 1$$
whence application of the squeeze theorem reveals
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Common complex roots
If the equations $ax^2+bx+c=0$ and $x^3+3x^2+3x+2=0$ have two common roots then show that $a=b=c$.
My attempts:
Observing $-2$ is a root of $x^3+3x^2+3x+2=0\implies x^3+3x^2+3x+2=(x+2)(x^2+x+1)=0$
Hence $ax^2+bx+c=0$ can have complex roots in common, comming from $(x^2+x+1)=0$
Both the roots of $(x^2+x+1)=0$ and $ax^2+bx+c=0$ are common should imply $a=b=c$ not only this but $a=b=c=1$.
Is this solution correct?
| We have that
$$
\eqalign{
& x^{\,3} + 3x^{\,2} + 3x + 2 = 0\quad \Rightarrow \quad x^{\,3} + 3x^{\,2} + 3x + 1 = - 1\quad \Rightarrow \cr
& \Rightarrow \quad \left( {x + 1} \right)^{\,3} = - 1\quad \Rightarrow \quad \left( {x + 1} \right) = e^{\,i\;\,\left( {1 + 2k} \right)\pi /3} = e^{\, \pm \,i\;\,\pi /3} ,e^{\,i\;\pi } = {1 \over 2} \pm i{{\sqrt 3 } \over 2},\;\; - 1 \cr}
$$
and
$$
\eqalign{
& 0 = ax^2 + bx + c\quad \Rightarrow \quad a\left( {x + 1} \right)^2 + \left( {b - 2a} \right)\left( {x + 1} \right) + \left( {a + c - b} \right) = 0\quad \Rightarrow \cr
& \Rightarrow \quad \left( {x + 1} \right) = {{ - \left( {b - 2a} \right) \pm \sqrt {b^2 - 4ac} } \over {2a}} \cr}
$$
Comparing the two results, the condition to have two identical roots imposes that it shall be
$$
\left\{ \matrix{
{{2a - b} \over {2a}} = {1 \over 2} \hfill \cr
{{\sqrt {4ac - b^2 } } \over {2a}} = {{\sqrt 3 } \over 2} \hfill \cr} \right.\;\quad \Rightarrow \quad \left\{ \matrix{
b = a \hfill \cr
c = a \hfill \cr} \right.
$$
either in the real and in the complex field.
| {
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"url": "https://math.stackexchange.com/questions/2353367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
If $a + \frac{1}{a} = -1$, then the value of $(1-a+a^2)(1+a-a^2)$ is?
If $a + \frac{1}{a} = -1$ then the value of $(1-a+a^2)(1+a-a^2)$ is?
Ans. 4
What I have tried:
\begin{align}
a + \frac{1}{a} &= -1 \\
\implies a^2 + 1 &= -a \tag 1 \\
\end{align}
which means
\begin{align}
(1-a+a^2)(1+a-a^2) &=(-2a)(-2a^2) \\
&=4a^{3}
\end{align}
as $1 + a^{2} = -a$ and $1 + a = -a^{2}$ from $(1)$.
| $a^3=1$.
Thus, $$(1-a+a^2)(1+a-a^2)=-2a\cdot(-2a^2)=4$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
If a is a non real root of $x^7 = 1$, find the equation whose roots are $ a + a^6 , a^2 + a^5, a^3 + a^4$ If a is a non real root of $ x^7 = 1$, find the equation whose roots are $a + a^6 , a^2 + a^5, a^3 + a^4$. This is one of the questions I have encountered while preparing for pre rmo. I feel the question requires the concept of the nth roots of unity and de moivre's theorem. But i actually couldnt work it out. Any help will be appreciated. Thanks in advance.
| Think of the roots as $a+a^{-1}$, $a^2+a^{-2}$ and $a^3+a^{-3}$. Then $a$ satisfies $a^6+a^5+a^4+a^3+a^2+a+1=0$ or equivalently,
$a^3+a^2+a+1+a^{-1}+a^{-2}+a^{-3}=0$. Can you write the expression
$x^3+x^2+x+1+x^{-1}+x^{-2}+x^{-3}$ as a linear combination of $(x+x^{-1})^3$, $(x+x^{-1})^2$, $x+x^{-1}$ and $1$? Putting in $a$ for $x$
would then give you an equation satisfied by $a+a^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Another limit involving an integral How to find the following limit (if exists ) : $\lim _{x \to \infty} \dfrac 1 x \int_0^x \dfrac {t}{1+x^2 \cos^2 t}dt$ ?
This looks quite similar to the previous To find a limit involving integral , but the huge difference is , there is apparently no closed form formula for the indefinite integral $\int \dfrac {t}{1+x^2 \cos^2 t}dt$ , where there was a closed form formula for the previous one , which made its evaluation slightly easy . Please help .
| I am using the same method as in this answer you linked. Let
$$
F(x)=\frac{1}{x}\int_0^x \frac{t}{1+x^2\cos^2{t}}\mathrm{d}t.
$$
If $n$ is a positive integer, we have
$$
F(n\pi)=\frac{1}{n\pi}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi} \frac{t}{1+(n\pi)^2\cos^2{t}}\mathrm{d}t,
$$
so we can control the $t$ in the integrand :
$$\frac{1}{n}\sum_{k=0}^{n-1}k\int_{k\pi}^{(k+1)\pi} \frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}} \leq F(n\pi)\leq \frac{1}{n}\sum_{k=0}^{n-1}(k+1)\int_{k\pi}^{(k+1)\pi} \frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}}.
$$
Using the identity $\frac{1}{\pi}\int_0^\pi\frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}}=\frac{1}{\sqrt{1+n^2\pi^2}}$, that yields
$$\frac{1}{n}\frac{\pi}{\sqrt{1+n^2\pi^2}}\sum_{k=0}^{n-1}k\leq F(n\pi)\leq \frac{1}{n}\frac{\pi}{\sqrt{1+n^2\pi^2}}\sum_{k=0}^{n-1}(k+1),
$$
hence
$$F(n\pi) \to \frac{1}{2}.
$$
Now when $n\pi \leq x < (n+1)\pi$,
$$
\begin{align}
F(x) &\leq \frac{1}{n\pi}\int_0^{(n+1)\pi} \frac{t}{1+(n\pi)^2\cos^2{t}}\mathrm{d}t \\
&=F(n\pi)+\frac{1}{n\pi}\int_{n\pi}^{(n+1)\pi} \frac{t}{1+(n\pi)^2\cos^2{t}}\mathrm{d}t \\
&\leq F(n\pi)+\frac{n+1}{n}\int_0^\pi \frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}} \\
&= F(n\pi)+\frac{(n+1)\pi}{n\sqrt{1+n^2\pi^2}}.
\end{align}
$$
Using the same kind of inequality on the other side it shouldn't be difficult to show that
$$
F(x)\to \frac{1}{2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Quadratic polynomial with integer roots. The quadratic polynomial $ax^2+bx+c$ has positive coefficients $a,b,c$ in A.P. in the given order. If it has integer roots $\alpha,\beta,$ find $\alpha+\beta+\alpha \beta.$
I tried with Vieta's theorem and putting $b=\frac{a+c}{2}$ to get $\alpha+\beta+\alpha \beta=\frac{b}{a}-1=\frac{c-a}{2a}$ but couldn't arrive at a solution.
P.S. The question had the following options given of which one and only one is the correct answer (if they are of any help)-$3,5,7,14$.
| Unpacking J.G.'s answer for the masses:
$ax^2+bx+c$
$$\text{if} \quad (a,b,c) \quad \text{are in "AP", then} \quad $$
$$\begin{align}\begin{cases}
a&=a \\
b&=a+d \\
c&=a+2d
\end{cases}
\ \ &\underbrace{\implies}_{\text{b is the average of a and c}} \ \
\left[\frac{a+c}{2} =\frac{2a+2d}{2}=a+d=b\right]
\\
&\qquad \quad \implies \qquad \quad \ \left[c=2b-a\right]\end{align}$$
Vieta: if $\alpha$ and $\beta$ are the roots $\implies \frac{-b}{a}=\alpha + \beta$ and $\frac{c}{a}=\frac{2b-a}{a}=\alpha \beta $
$$\begin{align}
\implies \alpha+\beta+\alpha \beta&=\frac{c-b}{a} \\
&=\frac{b-a}{a} \\
&=\frac{b}{a}-1 \\
&=-(\alpha+\beta)-1 \\
\alpha \beta +2(\alpha+\beta)&=-1 \\
\alpha \beta +2(\alpha +\beta)+4&=3 \\
(\alpha+2)(\beta+2)&=3=d_1 \cdot d_2 \\
\text{such that}\quad (d_1,d_2) &\in \{(1,3),(3,1),(-1,-3),(-3,-1) \\
&\begin{cases}
\alpha=d_1-2 \\
\beta=d_2-2
\end{cases} \\
\text{so} \quad (\alpha,\beta)&\in\{(-1,1),(1,-1),(-3,-5),(-5,-3) \\
\implies \alpha + \beta + \alpha \beta&=-1 \quad \lor \quad 7
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{x^2-4)^2}=2x^2-4+2x\sqrt{x^2-4}$
Now,
$a^2-ax=2x^2-4+2x\sqrt{x^2-4}-x^2-x\sqrt{x^2-4}=x^2+x\sqrt{x^2-4}-4=xa-4$
Why am I not getting the intended value?
| You should check your rationalization of $a$ again. I believe you are missing a factor of $\frac{1}{2}$.
Additionally, you could find the answer by choosing some easily computable value of $x$, say $x = 2$, so $a = 1$ and $a^2 - ax = -1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How does the trigonometric identity $1 + \cot^2\theta = \csc^2\theta$ derive from the identity $\sin^2\theta + \cos^2\theta = 1$? I would like to understand how would the original identity of $$ \sin^2 \theta + \cos^2 \theta = 1$$ derives into
$$ 1 + \cot^2 \theta = \csc^2 \theta $$
This is my working:
a) $$ \frac{\sin^2 \theta}{ \sin^2 \theta } + \frac{\cos^2 \theta }{ \sin^2 \theta }= \frac 1 { \sin^2 \theta } $$
b) $$1 + \cot^2 \theta = \csc^2 \theta
$$
How did the $ \tan^2 \theta + 1 = \sec^2 \theta$ comes into the picture?
| Just as you divide $\sin^2(\theta)+\cos^2(\theta)=1$ by $\sin^2(\theta)$ to get $ 1 + \cot^2 (\theta) = \csc^2 (\theta)$,
you can divide $\sin^2(\theta)+\cos^2(\theta)=1$ by $\cos^2(\theta)$ to get $\tan^2 (\theta) + 1 = \sec^2(\theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does $\frac{\cos\Delta x - 1 }{\Delta x} \to 0$ I'm watching Lecture 3 in MIT single variable calculus.
https://www.youtube.com/watch?v=kCPVBl953eY&list=PL590CCC2BC5AF3BC1&index=3
And at one point the instructor does the following:
I was under the impression that when evaluating limits we need to avoid having $0/0$ in the denominator. However, in the notes here, it says that
$\frac{\cos\Delta x - 1 }{\Delta x} \to 0$
How does this work?
| Since you know that
$$\lim_{x \rightarrow 0}\frac{\cos x -1}{x}=\lim_{x \rightarrow 0}\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}-1}{x} $$
$$\lim_{x \rightarrow 0}\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}-1}{x} =\lim_{x \rightarrow 0}\frac{-\sin^2\frac{x}{2}-\sin^2\frac{x}{2}}{x}$$
$$\lim_{x \rightarrow 0}\frac{-\sin^2\frac{x}{2}-\sin^2\frac{x}{2}}{x}=\lim_{x \rightarrow 0}\frac{-2\sin^2\frac{x}{2}}{x}$$
$$ \lim_{x \rightarrow 0}\frac{-2\sin^2\frac{x}{2}}{x}=\lim_{x \rightarrow 0}-\sin\frac{x}{2}\frac{\sin \frac{x}{2}}{\frac{x}{2}}$$
$$\lim_{x \rightarrow 0}-\sin\frac{x}{2}\frac{\sin \frac{x}{2}}{\frac{x}{2}}=\lim_{x \rightarrow 0}-\sin\frac{x}{2}$$
$$\lim_{x \rightarrow 0}-\sin\frac{x}{2}=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sequence $a_n = a_{n - 1} + a_{\lfloor n/2 \rfloor}$ Given a sequence with $a_1 = 1$. And $a_n = a_{n - 1} + a_{\lfloor n/2 \rfloor}$ for $n \geq 2$, prove that $a_n$ is not divisible by 4 for any $n \in \mathbb{N}$.
Don't have any conjectures.
| A few hints
Notice that $a_{2n+1}=a_{2n-1}+2a_n$. Since $a_1$ is odd, all $a_n$ are odd for odd $n$.
Then $a_{4n+2}=a_{4n+1}+a_{2n+1}$ even.
And $$a_{4n+2}=a_{4n+1}+a_{2n+1}=a_{4n-1}+2a_{2n}+a_{2n-1}+2a_n=a_{4n-2}+a_{2n-1}+3a_{2n-1}+4a_n=a_{4n-2}+4a_{2n-1}+4a_n$$ which shows by recurrence that $a_{4n+2}$ is not divisible by 4.
$a_{4n+4}=a_{4n+3}+a_{2n+2}$.
In the case where $n$ is even, this is the sum of an even and an odd term. If $n$ is odd, $$a_{4n+4}=a_{4n+3}+a_{2n+2}=a_{4n+2}+a_{2n+1}+a_{2n+1}+a_{n+1}=a_{4n+2}+2a_{2n-1}+4a_n+a_{n+1}=2+2+a_{n+1} \ (\textrm{mod} \ 4) = a_{n+1}\ (\textrm{mod} \ 4)$$ Again by recurrence and contradiction (saying that $a_{4n+4}$ is the smallest element in the list to be divisible by 4) we show that $a_{4n+4}$ is not divisible by 4.
Normally this covers all the cases.
| {
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Find number of real solutions of $3x^5+2x^4+x^3+2x^2-x-2=0$
Find a number of real roots of $$f(x)=3x^5+2x^4+x^3+2x^2-x-2$$
I tried using differentiation:
$$f'(x)=15x^4+8x^3+3x^2+4x-1=0$$ and I found number of real roots of $f'(x)=0$ by drawing graphs of $g(x)=-15x^4$ and $h(x)=8x^3+3x^2+4x-1$ and obviously from graph there are two real solutions as $h(x)$ is an increasing function.
But now how to proceed?
| $LHD=x^3(3x^2+2x+1)=x^3(3(x+\frac13)^2+\frac23)$
$RHD=-2x^2+x+2=-2(x-\frac14)^2+\frac{17}8$
if $x>\frac54$, obviously this equality has no solution.
if $\frac14<x<\frac54$, since LHD is monotoniously increasing, RHD is its opposite,
so there is one solution.
if $0<x<\frac14$ ,$LHD<1<RHD$ there is no solution.
if $-1<x<0$ ,$-2<LHD<0, 1<RHD<2$ there is no solution.
if $x<-1$ at $x=-1$, $LHD=-2$, $RHD=-1$ and since $3x^2+2x+1-(-2x^2+x+2)=5x^2+x-1$$=5(x+\frac1{10})^2-\frac1{20}>0,$ then $LHD<RHD$
Finally there is only one solution at $\frac14<x<\frac54$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Need help calculating a simple inverse function Quick question.
$$y=\sqrt{2x-x^2}$$
I need the inverse function for some other problem, but I just can't find it.
Could you please point me the steps to solve this?
Thanks
| The domain of $f(x)=\sqrt{2x-x^2}$ is $$\mbox{dom}(f)=\{x\in\mathbb{R} \ | \ 0\le x\le 2\}$$ moreover its image is $$\mbox{Im}(f)=\{y\in\mathbb{R} \ | \ 0\le y\le 1 \}$$ Its graph is a semicircumpherence with centre $C(1,0)$ and radius $r=1$ infact
$\\ y=\sqrt{2x-x^2}\implies y^2=2x-x^2\implies \\ \\ \\ x^2+y^2-2x=0\implies \implies x^2+y^2-2x+1-1=0$
so graph of $f$ is a part of the circumpherence of equantion $(x-1)^2+y^2=1$ that lies in the first quadrant. This implies that $f(x)=\sqrt{2x-x^2}$ is not globally injective hence it can't be invertible on the domain.
but... consider the equation
$y=\sqrt{2x-x^2}$
square both side
$y^2=2x-x^2\implies y^2-2x+x^2=0\to (x-1)^2+y^2=1$
solve with respect of $x-1$
$(x-1)^2=1-y^2\to |x-1|=\sqrt{1-y^2}$
If $1\le x\le 2$ then $|x-1|=x-1$ so
$|x-1|=\sqrt{1-y^2}\implies x-1=\sqrt{1-y^2}\implies x=\sqrt{1-y^2}+1$
swap the variables
$y=\sqrt{1-x^2}+1$ is the inverse of $f(x)$ for $x\in [0,1]\wedge y\in [1,2]$
If $0\le x\le 1$ then $|x-1|=1-x$ so $|x-1|=\sqrt{1-y^2}$ becomes $$1-x=\sqrt{1-y^2}\implies x=1-\sqrt{1-y^2}$$
Swap the variables $y=1-\sqrt{1-y^2}$
Hence $y=1-\sqrt{1-x^2}$ is the inverse of $f(x)$ when $x\in [0, 1]\wedge y\in [0,1]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}\right)}{y}$$
using L'Hopital's rule we get
$$L=\lim_{y \to 0} \frac{1}{1+\left(\frac{1+y}{1+4y}\right)^2} \times \frac{-3}{(1+4y)^2}$$
$$L=\lim_{y \to 0}\frac{-3}{(1+y)^2+(1+4y)^2}=\frac{-3}{2}$$
is this possible to do without Lhopita's rule
| It would be much simpler to proceed directly without any change of variables. Note that the expression under limit can be written as $$x\arctan\dfrac{\dfrac{1+x}{4+x}-1}{\dfrac{1+x}{4+x}+1}$$ which is same as $$-\frac{3x}{5+2x}\cdot\dfrac{\arctan\dfrac{3}{5+2x}}{\dfrac{3}{5+2x}}$$ and first factor tends to $-3/2$ and second factor tends to $1$ so that the desired limit is $-3/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $-\arctan(\frac{s}{c}) + \pi\Theta(c)$? I know that
$$-\arctan\left(\frac{1}{c}\right)+\pi\Theta(c)=\arctan(c)+\frac{\pi}{2}$$
where $\Theta(x)$ is the Heaviside step function.
I was wondering if it is possible to find a similar expression for
$$-\arctan\left(\frac{s}{c}\right) + \pi\Theta(c)$$
| Consider a right angled triangle:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
The sum of the angles $\measuredangle CAB$ and $\measuredangle ABC$ is $\frac{\pi}{2}$. We know that $\tan(\measuredangle CAB) = \frac{a}{b}$ and $\tan(\measuredangle ABC) = \frac{b}{a}$ so for $a,b>0$ we have
$$\frac{\pi}{2} = \arctan\left(\frac{b}{a}\right) + \arctan\left(\frac{a}{b}\right)$$
One can extend this to $\frac{a}{b} < 0$ by using that $\tan$ is an odd function so $\arctan(-x) = -\arctan(x)$ giving us
$$\arctan\left(\frac{b}{a}\right) + \arctan\left(\frac{a}{b}\right) = \left\{\matrix{\frac{\pi}{2}, & ab > 0\\ -\frac{\pi}{2} & ab < 0}\right. = -\frac{\pi}{2} + \pi\Theta\left(ab\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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integral $\int_{}^{}\frac{dx}{1+x^4+x^8} $ looking for help for the following integral -
$$
\int_{}^{}\frac{dx}{1+x^4+x^8}
$$
what I tried to do:
$$\int_{}^{}\frac{dx}{1+x^4+x^8} = \int_{}^{}\frac{dx}{\frac14+x^4+x^8 + \frac{3}{4}}= \int_{}^{}\frac{dx}{\left(x^4+\frac{1}{2}\right)^2 + \frac{3}{4}} $$
and now I am stuck :-(
| Factorise the denominator $x^8+x^4+1=(x^4+x^2+1)(x^4-x^2+1)=(x^2+x+1)(x^2-x+1)(x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1)$. Now do partial fractions
\begin{eqnarray*}
\frac{1}{1+x^4+x^8}=\frac{1}{4(1+x+x^2)}+\frac{1}{4(1-x+x^2)}+\frac{2x+\sqrt{3}}{4\sqrt{3}(1+\sqrt{3}x+x^2)}-\frac{2x-\sqrt{3}}{4\sqrt{3}(1-\sqrt{3}x+x^2)}
\end{eqnarray*}
Each of these terms can be integrated using standard logs and inverse tangents.
| {
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Inclusion-Exclusion Principle; why is this wrong? I'm unsure where I'm going wrong with this.
In a class of 40 people studying music: 2 play violin, piano and recorder, 7 play at least violina nd piano, 6 play at least piano and recorder, 5 play at least recorder and violin, 17 play at least violin, 19 play at least piano, and 14 play at least recorder. How many play none of these instruments?
This is what I did:
$|A| = $ plays violin
$|B| = $ plays piano
$|C| = $ plays recorder.
From the information:
$|A\cap B\cap C| = 2 \\ |A\cup B| = 7 \\ |A\cup C| =5 \\ |B \cup C| = 6 \\ |A| = 17 \\ |B| = 19 \\ |C| = 14.$
I want $|A^c \cap B^c \cap C^c|$.
By considering
$$|A^c \cup B^c \cup C^c| = |A^c| + |B^c| + |C^c| - |A^c\cap B^c| - |A^c \cap C^c| - |B^c \cap C^c| + |A^c \cap B^c \cap C^c|\\ = (40-17)+(40-19) + (40-14) - (40-7) - (40-5) + (40-6) + |A^c \cap B^c \cap C^c|\\ = -32 + |A^c \cap B^c \cap C^c| = 40 - |A \cap B \cap C| = 38 \\ \implies |A^c \cap B^c \cap C^c| = 70...?$$
| As previously stated, take the information you know. Your events for more than 1 instrument should be intersections, not unions. That is $\cap$ not $\cup$
$|A\cap B\cap C| = 2 \\ |A\cap B| = 7 \\ |A\cap C| =5 \\ |B \cap C| =
6 \\ |A| = 17 \\ |B|
= 19 \\ |C| = 14.$
You want $|A^c \cap B^c \cap C^c|$ which is simply $40-|A \cup B \cup C|$
Then for $|A \cup B \cup C|$ we have $|A| + |B| + |C| - |A \cap B| - |A\cap C| - |B \cap C| + |A \cup B \cup C|$
$=17+19+14-7-5-6+2=34$
Don't forget final step of $40-34=6$
| {
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solution set of the inequality $\frac{x^2-1}{(x+2)(x+3)}>2$
Question: Find the solution set of the inequality $$\frac{x^2-1}{(x+2)(x+3)}>2$$
From the answer given in the previous problem I got this:
First $x\neq -2,-3$. solving the equation I get $-(2\sqrt{3}+5)<x<(2\sqrt{3}-5)$.
Is this ok?
| $$\frac { x^{ 2 }-1 }{ \left( x+2 \right) \left( x+3 \right) } >2\\ \frac { x^{ 2 }-1 }{ \left( x+2 \right) \left( x+3 \right) } -2>0\\ \frac { -{ x }^{ 2 }-10x-13 }{ \left( x+2 \right) \left( x+3 \right) } >0\\ \frac { { x }^{ 2 }+10x+13 }{ \left( x+2 \right) \left( x+3 \right) } <0\\ \frac { \left( x+2\sqrt { 3 } +5 \right) \left( x+5-2\sqrt { 3 } \right) \left( x+2 \right) \left( x+3 \right) }{ { \left( x+2 \right) }^{ 2 }{ \left( x+3 \right) }^{ 2 } } <0\\ \left( x+2\sqrt { 3 } +5 \right) \left( x+5-2\sqrt { 3 } \right) \left( x+2 \right) \left( x+3 \right) <0\\ \\ \\ \\ $$can you take from here? (your answer is wrong by the way)
| {
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What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$?
$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$
I managed to get the answer as $1$ by standard methods of solving, learned from teachers, but my intuition says that the denominator of every term grows much faster than the numerator so limit must equal to zero.
Where is my mistake? Please explain very intuitively.
| Notice that:
$$\lim_{n\rightarrow\infty}\dfrac{1}{n^2} + \dfrac{2}{n^2} + \dfrac{3}{n^2}+\cdots+\dfrac{n}{n^2}=\lim_{n\rightarrow\infty} \frac{1}{n^2}\left(1+2+\dots n\right)$$
and this last sum can be replaced by the Gauss formula, so it becomes:
$$\lim_{n\rightarrow\infty}\frac{1}{n^2}\left(\frac{n(n+1)}{2}\right)$$
$$=\lim_{n\rightarrow\infty}\frac{n+1}{2n}=\frac{1}{2}\left(\lim_{n\rightarrow\infty} \frac{n+1}{n}\right)=\frac{1}{2}\left(\lim_{n\rightarrow\infty} 1+\frac{1}{n}\right)=\frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum of the given expression
For all real numbers $a$ and $b$ find the minimum of the following expression.
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2$$
I tried expressing the entire expression in terms of a single function of $a$ and $b$. For example, if the entire expression reduces to $(a-2b)^2+(a-2b)+5$ then its minimum can be easily found. But nothing seems to get this expression in such a form, because of the third unsymmetric square.
Since there are two variables here we can also not use differentiation.
Can you please provide hints on how to solve this?
| \begin{eqnarray*}
(a-b)^2+(2-a-b)^2+(2a-3b)^2=6a^2-12ab+11b^2-4(a+b)+4 \\
=6\left(a-b-\frac{1}{3}\right)^2+5\left(b-\frac{4}{5}\right)^2+\color{red}{\frac{2}{15}}.
\end{eqnarray*}
| {
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How can I achieve this transformation? I calculate a integration which results in the below.
$$
\log{\left|\frac{x+\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}+1}\right|}
$$
I've checked the integration with Wolfram Alpha and it give me this result.
$$
\log{\left|\frac{\sqrt{x^2+1}-1}{x}\right|}
$$
These expressions seems to have the equal value, but I cannot find the way to transform one to the other.
Is it possible to perform the transformation?
If possible, how can I do that?
| $$
\begin{align}
\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}\cdot\frac{x+1-\sqrt{x^2+1}}{x+1-\sqrt{x^2+1}}
&=\frac{(x^2-1)+2\sqrt{x^2+1}-(x^2+1)}{(x+1)^2-(x^2+1)} \\
&= \frac{2\sqrt{x^2+1}-2}{2x} \\
&= \frac{\sqrt{x^2+1}-1}{x} \\
\end{align}
$$
| {
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Recurrence relation $a_n = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$ I didn't do a lot of maths in my career, and we asked me to solve the following recurrence relation:
$$a_{n} = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$$
with
$a_0 = 2$, $a_1 = 3$ and $a_2 = 1$
What is the procedure to solve such relation? So far, I just know that $a_n$ could be split into two difference recurrence relations (homogeneous and non-homogeneous) as $a_n = b_n + c_n$, where
$$b_n = 11b_{n-1} - 40b_{n-2} + 48b_{n-3}$$
and
$$c_n = n2^n$$
| The general way of solving this kind of problem is to define relation $G(n)$ as
$$G(n) = (a_n - 11a_{n - 1} + 40a_{n - 2} - 48a_{n - 3} = n2^n)$$
Then write out:
$$\begin{array} {rrl}
G(n) = &(a_n + \dots &= n2^n) \\
G(n + 1) = &(a_{n + 1} + \dots &= 2n2^n + 2 \cdot 2^n) \\
G(n + 2) = &(a_{n + 2} + \dots &= 4 n 2^n + 8 \cdot 2^n) \\
\end{array}$$
until you have more equations than nonlinear terms per equation. Then let variables replace the nonlinear terms, $u = n2^n$ and $v = 2^n$ to get:
$$\begin{align}
a_n + \dots &= u \\
a_{n + 1} + \dots &= 2u + 2v \\
a_{n + 2} + \dots &= 4u + 8v \\
\end{align}$$
2 variables, 3 equations, so you can eliminate $u$ and $v$ and you get:
$$a_{n+2} - 15 a_{n+1} + 88 a_{n} - 252 a_{n-1} + 352 a_{n-2} - 192 a_{n-3}=0 \tag{T}$$
And that is just a regular linear recursion. Note that if the characteristic polynomial of the linear part of $G(n)$ is $g'$, and the characteristic polynomial of (T) is $t'$, then $g'$ is polynomial factor of $t'$. So 3 roots of $t'$ are known and you only need to find the remaining $2$ if you wish to continue in the standard fashion.
In this case, the roots of $g'$ are $\{4, 4, 3\}$ so the roots of $t'$ are $\{4, 4, 3, 2, 2\}$, so the final equation is $a_n = (An + B)4^n + (Cn + D)2^n + E3^n$ based on whatever the initial conditions are.
| {
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Computing the composition of a piecewise function Question:
$$ \text{Define } f:\mathbb{Z}\rightarrow\mathbb{Z}\text{ by }f(x)=\begin{cases}x+3\text{ if }x\text{ is ODD}\\ x-5\text{ if }x\text{ is EVEN}\end{cases} $$
Compute $ \ f \circ f$
My attempt:
$ \ (f \circ f)(x) = f(f(x))$
From here do I have to create another piecewise function and consider the cases when $ f(x) = x+3$ and when $ f(x) = x-5$?
| Yes, you could regard the composition as a piecewise function.
$$(f\circ f)(n) = f( f(n) ) = \begin{cases}
f(n) + 3 & \text{if $f(n)$ is odd, and} \\
f(n)-5 & \text{if $f(n)$ is even}. \\
\end{cases}$$
But notice that $f(n)$ is $n$, plus some odd integer (either $+3$ or $-5$). Thus $f(n)$ is even when $n$ is odd, and vice versa. Thus we have
$$f( f(n) ) = \begin{cases}
f(n) + 3 & \text{if $n$ is even, and} \\
f(n)-5 & \text{if $n$ is odd}. \\
\end{cases}$$
But then we know what to do with $f(n)$. When $n$ is even, $f(n) = n-5$, so $$f(n) + 3 = (n-5) + 3 = n-2.$$ Thus the first part of the piecewise definition can be rewritten to give
$$f( f(n) ) = \begin{cases}
n-2 & \text{if $n$ is even, and} \\
f(n)-5 & \text{if $n$ is odd}. \\
\end{cases}$$
Similarly, of $n$ is odd, then $f(n) = n+3$, and so
$$f(n)-5 = (n+3)-5 = n-2.$$
Hence the second part of the piecewise defined composition can be rewritten, giving
$$f( f(n) ) = \begin{cases}
n-2 & \text{if $n$ is even, and} \\
n-2 & \text{if $n$ is odd}. \\
\end{cases}$$
Since we get the same thing either way, we conclude that
$$(f\circ f)(n) = n-2.$$
That begin said, it is perhaps more direct to do the following: there are two cases to consider, either $n$ is even, or $n$ is odd (since all integers are either odd or even). If $n$ is even, then
$$(f\circ f)(n) = f(f(n)) = f(n-5).$$
But $n-5$ is odd (even - odd = odd), so
$$f(n-5) = (n-5) + 3 = n-2.$$
Similarly, if $n$ is odd, then
$$(f\circ f)(n) = f(n+3) = (n+3) - 5 = n-2.$$
In either case, we conclude that
$$(f\circ f)(n) = n-2.$$
| {
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Prove this inequality $2(a+b+c)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$ For $a,b,c$ are positive real numbers satisfy $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that $$2\left(a+b+c\right)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$$
We have:$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}\Leftrightarrow \:\left(a+b+c\right)^2\ge \:9\Leftrightarrow \:a+b+c\ge \:3$
By Cauchy-Schwarz: $R.H.S^2\le \left(1+1+1\right)\left(a^2+b^2+c^2+27\right)$
$=3\left(a^2+b^2+c^2+9\right)\Rightarrow R.H.S\le \sqrt{3\left(a^2+b^2+c^2\right)+27}$
Need to prove $4(a+b+2)^2\ge 3(a^2+b^2+c^2)+27$
$\Leftrightarrow \left(a+b+c\right)^2+3\left(a+b+c\right)^2\ge 3\left(a^2+b^2+c^2\right)+27$
$\Leftrightarrow \left(a+b+c\right)^2\ge 3\left(a^2+b^2+c^2\right)$ It's wrong. Help me
| We need to prove that
$$\sum_{cyc}\left(2a-\sqrt{a^2+3}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-1}{2a+\sqrt{a^2+3}}\geq0$$ or
$$\sum_{cyc}\left(\frac{a^2-1}{2a+\sqrt{a^2+3}}+\frac{1}{4}\left(\frac{1}{a}-a\right)\right)\geq0$$ or
$$\sum_{cyc}(a^2-1)\left(\frac{1}{2a+\sqrt{a^2+3}}-\frac{1}{4a}\right)\geq0$$ 0r
$$\sum_{cyc}\frac{(a^2-1)(2a-\sqrt{a^2+3})}{a(2a+\sqrt{a^2+3})}\geq0$$ or
$$\sum_{cyc}\frac{(a^2-1)^2}{a(2a+\sqrt{a^2+3})^2}\geq0.$$
Done!
| {
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Rationalizing the denominator having square roots and cube roots In middle-school mathematics, the teachers always tell you that if you have radicals on the denominator of a fraction, then it isn't fit to be a final answer - you have to rationalize the denominator, or get rid of all of the radicals in the denominator by moving them to numerator.
Rationalizing the denominator is usually very easy, and can be done quickly using its conjugate. For example, consider
$$\frac{1}{2+\sqrt 2}$$
This denominator can be easily rationalized by using its conjugate:
$$\frac{2-\sqrt 2}{(2+\sqrt 2)(2-\sqrt 2)}$$
$$\frac{2-\sqrt 2}{4-2}$$
$$\frac{2-\sqrt 2}{2}$$
However, I have stumbled upon a new class of denominator-rationalization problems that I can't figure out how to solve. I was thoroughly stumped when I tried to rationalize the denominator of this fraction:
$$\frac{1}{2+\sqrt 2+\sqrt[3]{2}}$$
Can anybody figure out how to rationalize this? Is it even possible?
Or, more interestingly, if anyone suspects that it is not possible, how might one prove something like this?
| It's always possible.
You want to multiply top and bottom by $M$ to get that $denominator*M$ has not radical.
As you have figured out: If the denominator is $a + b\sqrt{c}$ you mulitply by the conjugate to get $(a + b\sqrt{c})(a - b\sqrt{c}) = a^2 - b^2*c$.
This will also work with $(\sqrt a + \sqrt b)(\sqrt a - \sqrt b) = a - b$.
So it's the same idea for $a + \sqrt[k] b$. The trick is to realize that $(a + \sqrt[k]b)(a^{k-1} - a^{k-2}\sqrt[k]b + a^{k-3}(\sqrt[k]b)^2-..... \pm a(\sqrt[k]b)^{k-2} \mp (\sqrt[k]b)^{k-1} = a^k \pm b$.
Example: To deradicalize $5 + \sqrt[3]7$ multiply by $5^2 - 5*\sqrt[3]7 + (\sqrt[3]7)^2$ to get $(5 + \sqrt[3]7)(5^2 - 5*\sqrt[3]7 + (\sqrt[3]7)^2) = 5^3 + 5^2\sqrt[3]7 -5^2\sqrt[3]7 - 5*(\sqrt[3]7)^2 + 5*(\sqrt[3]7)^2 + (\sqrt[3]7)^3 = 125 + 7$.
So to deradicalize $(2 + \sqrt 2 + \sqrt[3] 2)$ just deradicalize it term by term.
First let's get rid of the $\sqrt[3]2$ term. So we multiply top and bottom by $(2+\sqrt 2)^2 - (2 +\sqrt 2)*\sqrt[3]2 + (\sqrt[3]2)^2$ to get $(2 + \sqrt 2 + \sqrt[3] 2)*[(2+\sqrt 2)^2 - (2 +\sqrt 2)*\sqrt[3]2 + (\sqrt[3]2)^2] = (2 + \sqrt 2)^3 + 2= 8 + 12 \sqrt 2 + 12\sqrt 2 + 2\sqrt 2 + 2 = 10 + 26\sqrt 2$. Then we multiply that by $10 - 26 \sqrt 2$ to get $(10 + 26\sqrt 2)(10 - 26\sqrt 2) = 100 - 2*26^2$.
So example:
\begin{align} &\frac 1 {2 + \sqrt 2 + \sqrt[3] 2} \\&=
\frac {(2 + \sqrt 2)^2 - (2+\sqrt2)\sqrt[3]2 + \sqrt[3]2^2}{(2+\sqrt 2)^3 + 2}\\&=
\frac {(4 + 4\sqrt 2 + 2) -2\sqrt[3] 2 - \sqrt 2\sqrt[3]2 + \sqrt[3]2^2}{10 + 26\sqrt 2}\\&=
\frac {[(4 + 4\sqrt 2 + 2) -2\sqrt[3] 2 - \sqrt 2\sqrt[3]2 + \sqrt[3]2^2](10 - 26\sqrt{2})}{100 - 2*26^2} \end{align}
Okay... admittedly that is a bear... but it is doable.
| {
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Pythagorean Triple: $\text{Area} = 2 \cdot \text{perimeter}$ Find the unique primitive Pythagorean triple whose area is equal to twice the perimeter.
So far I set the sides of the triangle to be $a, b,~\text{and}~c$ where $a$ and $b$ are the legs of the triangle and c is the hypotenuse.
I came up with 2 equations which are:
$\dfrac{ab}2 = 2(a+b+c)\;\;$ and $\;\;a^2+b^2=c^2$
but I'm not sure how to proceed and solve for $a, b, c$.
| Hint: formula for primitive triples is
$a = mn; b = \frac {m^2 -n^2}2; c = \frac {m^2 + n^2}2$ for $\gcd(m,n) = 1$. And as $m^2 - n^2 $ is even $m$ and $n$ must both be odd.
So we want $\frac {mn(m^2 -n^2)}4 = 2(mn + \frac {m^2 - n^2}2 + \frac {m^2 - n^2} 2) = 2(mn + m^2)$
So $mn(m-n)(m+n) = 8m(m+n)$
So $n(m-n)=8$
For $n=1,2,4,8$ we have $m = 9,6,6,9$. Only $n =1; m= 9$ have $\gcd(m,n) =1$ and $m^2 -n^2$ is even.
So solutions are $a = 9; b = 40; c=41$
====earlier answer with and an easier but not exhaustive formula for a triple ====
==== it worked out nicely, but it doesn't rule out that there aren't any others; nor did it guarentee I'd find a solution ===
Hint: Formula for primitive triples is $k, \frac {k^2 - 1}2, \frac {k^2+1}2$ where $k$ is odd. Ex. $3,4,5; 5,12,13; 7,24,25$ etc.
So we need to solve $\frac {ab}2 = \frac {k (k^2 -1)}4 = 2(a+b+c) = 2(k + \frac {k^2 - 1}2 + \frac {k^2 + 1}2) = 2(k + k^2)$.
Has a very nice solution!.
| {
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Formula for consecutive residue of primitive modulo n. \begin{align*}
3^0 \equiv 1\mod 7\\
3^1 \equiv 3\mod 7\\
3^2 \equiv 2\mod 7\\
3^3 \equiv 6\mod 7\\
3^4 \equiv 4\mod 7\\
3^5 \equiv 5\mod 7\\
3^6 \equiv 1\mod 7\\
3^7 \equiv 3\mod 7\\
\end{align*}
Now just focusing on 1, 3, 2, 6, 4, 5, 1....
How to devise a formula to find the next number.
Like if 2 is given how to find 6 or if 4 is given how to find 5?
I am looking for an explicit function.
| If you don't like modulus calculations, you could use:
For odd $a_k:\quad a_{k+1} = (7-a_k)/2$
For even $a_k:\quad a_{k+1} = 7-a_k/2$
| {
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} |
Is there a simple proof of this inequality? I want to prove that
$$ (a-1)^2+(b-1)^2 \ge 2(\sqrt{ ab}-1)^2 $$ for any positive real numbers $a,b$, and that equality holds iff $a=b$.
Edit: My "proof" for this inequality was wrong! It turns out the inequality holds iff $a=b$ or $\sqrt{a}+\sqrt{b} \ge \sqrt{2}$. (See the answers for details).
| $$ (a-1)^2+(b-1)^2 - 2(\sqrt{ab}-1)^2 = a^2+b^2-2ab-2(a+b)+4\sqrt{ab} = (\sqrt{a}-\sqrt{b})^2((\sqrt{a}+\sqrt{b})^2-2), $$
which is nonnegative if and only if $\sqrt{a}+\sqrt{b} \geq \sqrt{2}$ or $a=b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Logically, can you not subtract $\cos^2(x)$ to the other side from this Pythagorean identity $\sin^2(x)+\cos^2(x)=1?$
When I look up trig identities, however, it says $\sin^2(x) = \frac{1-\cos(2x)}{2}$.
Why is this?
| Let's address two things, the question in the post and the confusion others have pointed out.
Part 1: Proof
Here's a proof that $\cos^2(\theta)+\sin^2(\theta)=1$, from which you can show that $\sin^2(\theta)=1-\cos^2(\theta)$. It requires a bit of set up using the Euler identity and that $i=\sqrt{-1}$. If you are not familiar with complex numbers, find yourself a video on YouTube, you'll get the hang of them in next to no time.
$$e^{i\theta}=\cos(\theta)+i\sin(\theta)\tag{1}$$
$$e^{-i\theta}=\cos(\theta)-i\sin(\theta)\tag{2}$$
Now, add (1) and (2).
$$e^{i\theta}+e^{-i\theta}=\cos(\theta)+i\sin(\theta)+\cos(\theta)-i\sin(\theta)=2\cos(\theta)\tag{3}$$
Subtract (2) from (1)
$$e^{i\theta}-e^{-i\theta}=\cos(\theta)+i\sin(\theta)-\cos(\theta)+i\sin(\theta)=2i\sin(\theta)\tag{4}$$
From (3) and (4) we get (5) and (6)
$$\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}\tag{5}$$
$$\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}\tag{6}$$
now, add the squares of (5) and (6)
$$\cos^2(\theta)=\frac{e^{2i\theta}+2e^{i\theta-i\theta}+e^{-2i\theta}}{4}=\frac{e^{2i\theta}+2+e^{-2i\theta}}{4}\tag{7}$$
$$\sin^2(\theta)=\frac{e^{2i\theta}-2e^{i\theta-i\theta}+e^{-2i\theta}}{-4}=\frac{e^{2i\theta}-2+e^{-2i\theta}}{-4}\tag{8}$$
$$\cos^2(\theta)+sin^2(\theta)=\frac{e^{2i\theta}+2+e^{-2i\theta}}{4}+\frac{e^{2i\theta}-2+e^{-2i\theta}}{-4}=\frac{e^{2i\theta}+2+e^{-2i\theta}-e^{2i\theta}+2-e^{-2i\theta}}{4}\tag{9}$$
$$\cos^2(\theta)+sin^2(\theta)=\frac{e^{2i\theta}+2+e^{-2i\theta}-e^{2i\theta}+2-e^{-2i\theta}}{4}=\frac{4}{4}=1\tag{10}$$
Part 2: Addressing the confusion of $\cos(2x)$ and $\cos^2(x)$
$$\cos(2\theta)=\frac{e^{2i\theta}+e^{-2i\theta}}{2}= \frac{\left(e^{i\theta}\right)^2+\left(e^{-i\theta}\right)^2}{2} = \frac{(\cos(\theta)+i\sin(\theta))^2+(\cos(\theta)-i\sin(\theta))^2}{2}\tag{11}$$
Processing this result further, we obtain
$$\cos(2\theta)=\frac{\cos^2(\theta)+2i\cos(\theta)\sin(\theta) - \sin^2(\theta) + \cos^2(\theta) -2i\cos(\theta)\sin(\theta)-\sin^2(\theta) }{2}=\frac{2\cos^2(\theta)-2\sin^2(\theta)}{2}$$
Therefore
$$\cos(2\theta)=\cos^2(\theta)-\sin^2(2\theta)\tag{12}$$.
Ok, you stated that $$\sin^2(x)=\frac{1-\cos(2x)}{2}\tag{13}$$
well, if you substitute the result from (12) into (13), you find that
$$\sin^2(\theta)=\frac{1-\cos^2(\theta)+\sin^2(2\theta)}{2}\tag{14}$$
Rearrange this (multiply both sides by 2 and then subtract the $sin^2(\theta)$ from the right side to the left) to get
$$\sin^2(\theta)=1-\cos^2(\theta)\tag{15}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Is there a mistake in my evaluation of $\int\frac{3x+1}{\sqrt{5x^2+1}}dx$? I want to evaluate the following integral:
$$\int\frac{3x+1}{\sqrt{5x^2+1}}dx$$
The answer given in my textbook is $\frac{3}{5}\sqrt{5x^2+1}+\frac{1}{\sqrt{5}}\ln(x\sqrt{5}+\sqrt{5x^2+1})$.
I think the author has excess factor $\sqrt{5}$ in logarithm.
My solution is
$$\begin{align}
\int\frac{3x+1}{\sqrt{5x^2+1}}dx &= \frac{3}{2}\int\frac{d(x^2)}{\sqrt{5x^2+1}}+\int\frac{1}{\sqrt{5x^2+1}}dx \\[6pt]
&=\frac{3}{\sqrt{5}}\sqrt{x^2+\frac{1}{5}}+\frac{1}{\sqrt{5}}\ln\left(x+\sqrt{x^2+\frac{1}{5}}\right)
\end{align}$$
since
$$\int\frac{dx}{\sqrt{x^2+a}}=\ln\left(x+\sqrt{x^2+a}\right)$$
Am I right?
N.B. Just to be short I skipped additive constant in antiderivatives, so don't mention it.
| Thanks to all comments, the difference between two forms was exactly in additive constant. My problem is solved and answered now.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Partial fractions and linear vs quadratic factors I was watching some videos on partial fraction decompistion and I got confused on one of the examples:
Say for example you have $$\frac{x+4}{x^2(x^2 +3)^2}.$$
The partial fraction equation of this is apparently:
$$\frac{A}{x} + \frac{B}{x^2} + \frac{Cx+E}{x^2 +3} + \frac{Dx+F}{(x^2 +3)^2}$$
My question is why $A/x+B/x^2$ do not have numerators with an $ax+b$ form, cause $x^2$ is a quadratic not a linear right? Is it because the $x^2$ is in brackets, so you can perceive it as $(x+0)^2$?
| hint
if
$$\frac {x+4}{x^2(x^2+3)^2}=\frac {ax+b}{x}+\frac {cx+d}{x^2}+$$
$$\frac {Ax+B}{x^2+3}+\frac {Cx+D}{(x^2+3)^2} $$
$$=a+\frac {b+c}{x}+\frac {d}{x^2}+... $$
then $x\to +\infty $ gives $a=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Definition of convergence of a series My text book gives the following definition of convergence of a series:
A sequence of real numbers is said to be convergent if it has a finite real number $L$ as its limit. We then say that the sequence $x$ converges to $L$.
But I find it to be confusing, as from the above definition of convergence, $1/n$ turns out to be a Cauchy sequence.
But wikipedia seems to be providing a different definition of convergence Definition of Convergent Series
PS: I am just a newbie in real analysis, so I do request you to be a little more elaborative.
Edit: Oh, that question was stupid enough. I really did mix up the two terms "sequence" and "series". Thanks for pointing this out.
| There are two different kinds of objects that you are studying: sequences, and series.
A sequence $(a_n)$ of real numbers converges if there is some finite real number $L$ such that $\lim_{n\to\infty} a_n = L$. What this means is that we can make the difference between $a_n$ and $L$ as small as we like by choosing a number $n$ that is large enough. More formally,
A sequence $(a_n)$ converges to $L$ if for any $\varepsilon > 0$ there exists some $N$ so large that $n \ge N$ implies that $|a_n - L| < \varepsilon$.
A series $\sum_{n=1}^{\infty} a_n$ converges if the sequence of partial sums $S_N$ converges, where
$$ S_N := \sum_{n=1}^{N} a_n. $$
That is, in order to discuss the convergence of a series, we first turn the series into a sequence, then seek to understand the properties of that sequence. Thus a series is said to converge to a limit $S$ if the sequence $(S_N)$ (as defined above) converges to $S$ as a sequence. In notation, we might write
$$ \sum_{n=1}^{\infty} a_n = S \iff \lim_{N\to\infty} S_n = \lim_{N\to\infty} \left( \sum_{n=1}^{N} a_n \right) = S.$$
The classic example (cited by other responses to your question) is the harmonic series,
$$ \sum_{n=1}^{\infty} \frac{1}{n}. $$
The individual terms $\frac{1}{n} \to 0$ as $n\to \infty$, but the series does not converge. To see this, write
$$ \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \dotsb.$$
Notice that $\frac{1}{3} > \frac{1}{4}$, thus
$$ \frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}. $$
Similarly, each term in the next group is bigger than $\frac{1}{8}$, and so
$$ \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}. $$
By a similar process, we can continue to lump terms together to add up to $\frac{1}{2}$, which gives
$$ \sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{2} + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \dotsb \ge \sum_{k=1}^{\infty} \frac{1}{2} = \infty, $$
where the inequality can be justified via the Squeeze Theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) = \frac{(1 + \sin\theta + i\cos\theta)^2}{(1 + \sin\theta)^2 + \cos^2\theta}$$
$$=\frac{(1 + \sin\theta)^2 + \cos^2\theta + 2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
thus
$$x = \frac{(1 + \sin\theta)^2 + \cos^2\theta }{(1 + \sin\theta)^2 + \cos^2\theta} $$
$$y= \frac{2i(1 + \sin\theta)\cos\theta }{(1 + \sin\theta)^2 + \cos^2\theta}$$
According to the binomial theorem,
$$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k$$
we get
$$z = \frac{1}{(1 + \sin\theta)^2 + \cos^2\theta}\sum_{k=0}^n \binom{n}{k} ((1 + \sin\theta)^2 + \cos^2\theta)^{n-k}\cdot(2i(1 + \sin\theta)\cos\theta)^k$$
...and that is where I'm stuck. What do you think? Thanks for the attention.
| It is convenient to use $e^{i\theta}=\cos\theta + i\sin\theta$.
We obtain
\begin{align*}
\color{blue}{z}&\color{blue}{=\left(\frac{1+\sin \theta +i\cos \theta}{1+\sin\theta-i\cos \theta}\right)^n}\\
&=\left(\frac{1+ie^{-i\theta}}{1-ie^{i\theta}}\right)^n\\
&=\left(\frac{1+ie^{-i\theta}}{-ie^{i\theta}(1+e^{-i\theta})}\right)^n\\
&=\left(ie^{-i\theta}\right)^n\\
&=i^ne^{-in\theta}\\
&=\color{blue}{i^n\left(\cos (n\theta) - i\sin(n\theta)\right)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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rotation about $x$ and $y$ axis on the Bloch sphere $$R_x(\theta)= \begin{bmatrix} \cos \left( \frac{\theta}{2} \right) & -i\sin \left( \frac{\theta}{2} \right) \\ -i\sin \left( \frac{\theta}{2} \right) & \cos \left( \frac{\theta}{2} \right)\end{bmatrix}$$
$$R_y(\theta)= \begin{bmatrix} \cos \left( \frac{\theta}{2} \right) & \sin \left( \frac{\theta}{2} \right) \\ \sin \left( \frac{\theta}{2} \right) & \cos \left( \frac{\theta}{2} \right)\end{bmatrix}$$
$$R_z(\theta)=\begin{bmatrix} \exp\left(-i\frac{\theta}{2}\right) & 0 \\ 0 & \exp\left(-i\frac{\theta}{2}\right)\end{bmatrix}$$
This is the rotation gates as matrices on the Bloch sphere. It was easy to show that $R_z$ to be the rotation about $z$ axis on the Bloch sphere. However, I can't find a way to show that $R_x$ and $R_y$ are rotations about $x$ and $y$ axis respectively and I can't find any solutions on the google.... Could anyone show me why $R_x$ and $R_y$ are rotations about $x$ and $y$ axis.
| I will leave $R_y(\theta)$ as an exercise for you.
Below is a verification of $R_x(\theta)$ is a rotation about the $x$-axis.
Denote the Pauli matrices as
$$X=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, Y=\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}, Z=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$
We have the following expression for $R_x(\theta)$
$$R_x(\theta) = \cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X$$
Consider $$\rho=\frac12(I+r_xX+r_yY+r_zZ)$$
$\begin{align}\rho' &= R_x(\theta)\rho R_x(\theta)^\dagger \\
&=\frac12(I+r_xR_x(\theta)XR_x(\theta)^\dagger+r_yR_x(\theta)YR_x(\theta)^\dagger+r_zR_x(\theta)ZR_x(\theta)^\dagger)\end{align}\tag{1}$
Now, let's analyze each separate term:
\begin{align}R_x(\theta)XR_x(\theta)^\dagger &= \left(\cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X\right)X\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\left(\cos \left( \frac{\theta}{2} \right)X-i\sin\left( \frac{\theta}{2}\right)I\right)\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\cos^2 \left( \frac{\theta}{2}\right)X-i\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)I+i\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)I+\sin^2 \left( \frac{\theta}{2}\right)X\\
&= X\end{align}
\begin{align}R_x(\theta)YR_x(\theta)^\dagger &= \left(\cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X\right)Y\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\left(\cos \left( \frac{\theta}{2} \right)Y+\sin\left( \frac{\theta}{2}\right)Z\right)\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\cos^2 \left( \frac{\theta}{2}\right)Y+\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Z+\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Z-\sin^2 \left( \frac{\theta}{2}\right)Y\\
&= \cos (\theta) Y + \sin(\theta)Z\end{align}
\begin{align}R_x(\theta)ZR_x(\theta)^\dagger &= \left(\cos \left( \frac{\theta}{2} \right)I-i\sin\left( \frac{\theta}{2}\right)X\right)Z\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\left(\cos \left( \frac{\theta}{2} \right)Z-\sin\left( \frac{\theta}{2}\right)Y\right)\left(\cos \left( \frac{\theta}{2} \right)I+i\sin\left( \frac{\theta}{2}\right)X\right)\\
&=\cos^2 \left( \frac{\theta}{2}\right)Z-\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Y-\sin \left(\frac{\theta}{2} \right)\cos \left(\frac{\theta}{2} \right)Y-\sin^2 \left( \frac{\theta}{2}\right)Z\\
&= -\sin (\theta) Y + \cos(\theta)Z\end{align}
\begin{align}\rho'&=\frac12 \left(I+r_xX+r_y(\cos (\theta) Y + \sin(\theta)Z)+r_z(-\sin (\theta) Y + \cos(\theta)Z)\right)\\
&= \frac12 \left(I+r_xX+(r_y(\cos (\theta) - r_z \sin (\theta)) Y + (r_y\sin(\theta) + r_z\cos(\theta))Z\right)\\\end{align}
Hence if we write
$$\rho'=\frac12 \left(I+r_x'X+r_y'Y+r_z'Z\right)$$
$$\begin{bmatrix}r_x' \\ r_y' \\ r_z'\end{bmatrix}=\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos \theta & - \sin \theta\\ 0 & \sin \theta & \cos \theta\end{bmatrix}\begin{bmatrix}r_x \\ r_y \\ r_z\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387431",
"timestamp": "2023-03-29T00:00:00",
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Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}\tag1$$
Moving the things in RHS to LHS:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} - \frac{1}{(x-3)} + \frac{1}{(x-4)} = 0\tag2$$
Writing everything above a common denominator:
$$\frac{1}{(x-4)(x-1)(x-2)(x-3)}\bigg[(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3)\bigg] = 0\tag3$$
Multiplying both sides with the denominator to cancel the denominator:
$$(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3) = 0\tag4$$
Multiplying the first two factors in every term:
$$(x^2-3x-2x+6)(x-4) - (x^2-3x-x+3)(x-4) - (x^2-x-2x+2)(x-4) + (x^2-2x-x+2)(x-3) = 0\tag5$$
Simplifying the first factors in every term:
$$(x^2-5x+6)(x-4) - (x^2-4x+3)(x-4) - (x^2-3x+2)(x-4) + (x^2-3x+2)(x-3) = 0\tag6$$
Multiplying factors again:
$$(x^3-4x^2-5x^2+20x+6x-24) - (x^3-4x^2-4x^2+16x+3x-12) - (x^3-4x^2-3x^2-12x+2x-8) + (x^3-3x^2-3x^2+9x+2x-6) = 0\tag7$$
Removing the parenthesis yields:
$$x^3-4x^2-5x^2+20x+6x-24 - x^3+4x^2+4x^2-16x-3x+12 - x^3+4x^2+3x^2+12x-2x+8 + x^3-3x^2-3x^2+9x+2x-6 = 0\tag8$$
Which results in:
$$28x - 10 = 0 \Rightarrow 28x = 10 \Rightarrow x = \frac{5}{14}\tag9$$
which is not correct. The correct answer is $x = \frac{5}{2}$.
| The error is in the third group of parentheses in the "multiplying factors again" step. What you have as $-12x$ should be positive.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers such that: $n^2-3|3n^3-5n^2+7$
Find all positive integers such that: $$n^2-3|3n^3-5n^2+7.$$
I did the following:
$$n^2-3|3n^3-5n^2+7,n^2-3\Rightarrow n^2-3|3n^3-5n^2+7,3n^3-9n$$
$$\Rightarrow n^2-3|5n^2-9n-7,5n^2-15\Rightarrow n^2-3|9n-8$$
Now we must have: $9n-8\geq n^2-3$.We should determine the sign of $n^2-9n+5$, but the equation: $n^2-9n+5=0$ does not have integer roots.How can we proceed?
| Continuing from $\,n^2\!-\!3\mid 9n\!-\!8\ $ we can avoid quadratic inequalities and checking $n = 1,\ldots,8$
$\!\bmod n^2\!-3\!:\ \color{#c00}{n^2\equiv 3}\,\Rightarrow\, 0 \equiv \color{#c00}{n}(9\color{#c00}{n}\!-\!8)\equiv {27}\!-\!8n\,\Rightarrow\,0\equiv 9n\!-\!8+27\!-\!8n \equiv n\!+\!19\,$ hence $\,n\equiv -19\,$ so $\,0\equiv 8\!-9n\equiv 179\,$ is prime, so its divisor $\,n^2\!-\!3 = \pm1,\pm 179,\,$ so $\,n = 2.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve a system of non-linear equations: $2(a-b)=29+4ab$, $2(c-b)= 11+4bc$, $2(c+a) = 9-4ac$
So I've solved this system of equations: $$\begin{array}{lcl}2(a-b) & =& 29+4ab \\2(c-b)& = & 11+4bc \\2(c+a) & = & 9-4ac\end{array}$$
by simply solving for each variable in terms of the others, separately and plugging them into the other equations. It does get tedious and I don't want to sit with this for $40$ min in an exam. Is there any efficient way to solve a system like this?
| It's $$(2a+1)(2c+1)=10,$$ $$(2b-1)(2c+1)=-12$$ and $$(2a+1)(2b-1)=-30.$$
The rest is smooth:
$$(2a+1)^2(2c+1)^2(2b-1)^2=3600,$$
which gives
$$(2a+1)(2c+1)(2b-1)=60$$ or $$(2a+1)(2c+1)(2b-1)=-60.$$
If $(2a+1)(2c+1)(2b-1)=60$ then we obtain:
$2a+1=\frac{60}{-12}$, $2c+1=\frac{60}{-30}$ and $2b-1=\frac{60}{10}$, which is
$a=-3$, $b=\frac{7}{2}$, $c=-\frac{3}{2}$.
If $(2a+1)(2c+1)(2b-1)=-60$ then the work is similar
and we get the answer:
$$\left\{\left(-3,\frac{7}{2},-\frac{3}{2}\right),\left(2,-\frac{5}{2},\frac{1}{2}\right)\right\}$$
| {
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how to show limit of $\frac{x^2-y^2}{x^3-y^3}$ as (x,y) goes to $(0,0)$ does not exist I know that $\lim\limits_{(x,y)\to (0,0)} \frac{x^2-y^2}{x^3-y^3}$ does not exist, but i'm not sure how to show it.
What I have done is let $y=kx$
and then,
$\lim\limits_{x\to 0} \frac{x^2-(k x)^2}{x^3-(kx)^3} = \lim\limits_{x\to 0} \frac{1-k^2}{x(1-k^3)}$
and k is just a constant,
is this enough to show that the limit does not exist?
| Using polar coordinates, $x=r\cos\theta, y=r\sin\theta,$ we have $$\frac{r^2(\cos^2\theta - \sin^2\theta)}{r^3(\cos^3\theta-\sin^3\theta)}=\frac{\cos\theta + \sin\theta}{r(\cos^2\theta+\cos\theta\sin\theta+\sin^2\theta)}=\frac{\cos\theta+\sin\theta}{r(1+\cos\theta\sin\theta)}$$
Choose any $\theta$ such that the numerator is not $0$. Then the limit is undefined since the denominator tends to $0$ as $r\to 0.$
On the other hand, if you choose $\theta$ such that the numerator is zero, then the entire limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
If $x$, $y$ and $z$ are distinct positive integers and $x+y+z=11$ then what is the maximum value of $(xyz+xy+yz+zx)$?
We know that product is maximum when difference between $x$, $y$ and $z$ is minimum.
So, I assumed $x=3$, $y=4$ and $z=4$.
Now putting this value in $xyz+xy+yz+zx$ I got my answer $88$. But actual answer is $78$. Where am I doing it wrong?
| Note that $(x+1)(y+1)(z+1)=1+(x+y+z)+(xy+yx+zx)+xyz$ so the sum you want is $$(x+1)(y+1)(z+1)-12$$ which justifies your comment about the product.
If $a\gt a-1\ge b+1\gt b$ we have $$(a-1)(b+1)=ab+(a-b)-1\gt ab$$ since $a-b\ge 2$.
The best selection of distinct integers is (as others have noted) $5+4+2=11$. And the product formula gives $90-12=78$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the integral $\int\frac{dx}{x^2\sqrt{4x^2-1}}$ I solved the question, but Wolfram Alpha and Symbolab both give me two completely different answers.
Here's my work:
Let $u = 2x$ and $a = 1$
Then $du = 2dx$ and $dx = \frac{du}{2}$
Then $\int\frac{dx}{x^2\sqrt{4x^2-1}}$ = $\int\frac{1}{u\sqrt{u^2-a^2}}$
$\therefore \space\int\frac{dx}{x^2\sqrt{4x^2-1}} = sec^{-1}(2x) + C$
Can anyone verify this solution?
| $\int\frac{dx}{x^2\sqrt{4x^2-1}}$
take $x=\dfrac{\sec \theta}{2}$, then $dx=\dfrac{\sec \theta\tan \theta}{2}d\theta$
Now $\sin\theta=\pm\dfrac{\sqrt{4x^2-1}}{2x}$
When $x>\dfrac{1}{2}$, then $\sqrt{4x^2-1}=\tan \theta$
$$\int\dfrac{dx}{x^2\sqrt{4x^2-1}}=2\int \dfrac{\sec \theta\tan \theta}{\sec^2\theta\sqrt{\sec^2\theta-1}}d\theta=2\int \dfrac{\sin\theta}{\tan \theta}d\theta\\=2\int \cos\theta d\theta=2\sin\theta+C=2\sin\Big(\cos^{-1}\frac{1}{2x}\Big)+C$$
Similarly, When $x<-\dfrac{1}{2}$, then $\sqrt{4x^2-1}=-\tan \theta$
$$\int\dfrac{dx}{x^2\sqrt{4x^2-1}}=-2\sin\Big(\cos^{-1}\frac{1}{2x}\Big)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$
Evaluate $$\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$$
My attempt. Taking $\sin x \sin y=t$, $\iint\sin x \ dx \ dy$.
I am not sure but I think the limits would change to $x=0$ to $\pi/2$ and $t=0$ to $1$.
I do not know how to change the elemental area i.e $dx\, dy$. Can I apply Jacobian where I am only changing one variable (i.e $y$ to $t$)?
Is this the correct way. Is there any easier way to go about it?
| Here is an approach using power series expansions.
It is known that for $t\in (-1,1)$
$$\frac{1}{\sqrt{1-t}} =(1-t)^{-1/2}=\sum_{n=0}^\infty
\binom{-1/2}{n} (-t)^{n}=
\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} t^{n},$$
and for $t\in [-1,1]$,
$$\arcsin(t) =\int_0^t\frac{ds}{\sqrt{1-s^2}}= \sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{ t^{2n+1}}{2n+1}.$$
Hence, the given integral is equal to
\begin{align*}I&=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}\int_0^{\pi/2}\int_0^{\pi/2} \sin x (\sin x \sin y)^{2n+1} \ dx \ dy\\
&=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}W_{2n+2} W_{2n+1} \\
&=\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}\cdot \frac{\pi}{4(n+1)}\\
&=\frac{\pi}{4}\left(2\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{2n+1}-\sum_{n=0}^\infty \frac{1 }{2^{2n}}\binom{2n}{n} \frac{1}{n+1}\right)\\
&=\frac{\pi}{4}\left(2\arcsin(1)-\int_0^1\frac{dt}{\sqrt{1-t}}\right)\\
&=\frac{\pi(\pi-2)}{4}.
\end{align*}
where $W_r=\int_{0}^{\pi/2}\sin^r(t)dt $ denotes the Wallis integral of order $r$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result that simplifies to an integer; (b) tried to find $a$ and $b$ such that $(a+\sqrt{b})^3=2+\sqrt{5}$ without success.
The answer stated for the problem in the original source (a local Math Olympiad Constest) is $x=1$.
| Let $a=\sqrt[3]{2+\sqrt{5}}, b=\sqrt[3]{2-\sqrt{5}}\,$, then:
$$\require{cancel}
a^3+b^3 = 2+\cancel{\sqrt{5}} + 2-\cancel{\sqrt{5}} = 4 \\
ab = \sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})} = \sqrt[3]{2^2 - 5} = -1
$$
From $\,(a+b)^3 = a^3+b^3+3ab(a+b)\,$ and given that $\,x=a+b\,$ it follows that $\,x^3=4-3x\,$ $\iff 0 = x^3+3x-4=(x-1)(x^2+x+4)\,$, where only the first factor has a real root.
| {
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"url": "https://math.stackexchange.com/questions/2395060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculate $\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}$ I need to calculate:
$$\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}$$
I get $0/0$ and can then use l'hopital's rule to find the limit, I can do this but someone asked me how I can do this without using l'hopital's rule. I guess I have to seperate $(x-a)$ in the nominator and denominator. The nominator can be written as $(x-a)(x+a)$ but I don't see how to seperate $(x-a)$.
| You want
$\lim_{x \rightarrow a} \frac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}
$.
Let
$x = y+a$.
Then
$\begin{array}\\
\dfrac{x^2 + ax - 2a^2}{\sqrt{2x^2 - ax} -a}
&=\dfrac{(y+a)^2 + a(y+a) - 2a^2}{\sqrt{2(y+a)^2 - a(y+a)} -a}\\
&=\dfrac{y^2+2ay+a^2 + ay+a^2 - 2a^2}{\sqrt{2y^2+4ya+2a^2 - ay-a^2} -a}\\
&=\dfrac{y^2+3ay}{\sqrt{2y^2+3ya+a^2} -a}\\
&=\dfrac{y(y+3a)}{\sqrt{2y^2+3ya+a^2} -a}
\dfrac{\sqrt{2y^2+3ya+a^2} +a}{\sqrt{2y^2+3ya+a^2} +a}\\
&=\dfrac{y(y+3a)(\sqrt{2y^2+3ya+a^2} +a)}{2y^2+3ya+a^2 -a^2}\\
&=\dfrac{y(y+3a)(\sqrt{2y^2+3ya+a^2} +a)}{2y^2+3ya}\\
&=\dfrac{y(y+3a)(\sqrt{2y^2+3ya+a^2} +a)}{y(2y+3a)}\\
&=\dfrac{(y+3a)(\sqrt{2y^2+3ya+a^2} +a)}{(2y+3a)}\\
&\to \dfrac{(3a)(\sqrt{a^2} +a)}{(3a)}
\qquad\text{as } y \to 0\\
&= a+|a|
\qquad\text{since } \sqrt{z^2} = |z|\\
\end{array}
$
| {
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"url": "https://math.stackexchange.com/questions/2395782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial fractions - integration $$\int \frac{4}{(x)(x^2+4)} $$
By comparing coefficients,
$ 4A = 4 $,
$A = 1$
$1 + B = 0 $,
$B= -1 $
$xC= 0 $,
$C= 0 $
where $\int \frac{4}{x(x^2+4)}dx =\int \left(\frac{A}{x} + \frac{Bx+C}{x^2 + 4}\right)dx$.
So we obtain $\int \frac{1}{x} - \frac{x}{x^2+4} dx$.
And my final answer is
$\ln|x| - x \ln |x^2 + 4| + C$.
However my answer is wrong , the answer is - $\ln|x| - \frac{1}{2} \ln |x^2 + 4| + C$.
Where have I gone wrong?
| Yes, $4 = A(x^2+4)~+~(Bx+C)x \implies A=1, B=-1, C=0$
So
$$\require{enclose}\int \dfrac{4}{x(x^2+4)}~\mathrm d x ~{= \int \dfrac{1}{x}+\dfrac{\enclose{circle}{~-x~}}{(x^2+4)}~\mathrm d x \\ = \ln\lvert x\rvert -\int\dfrac{\tfrac 12\mathrm d (x^2+4)}{(x^2+4)} \\ = \ln\lvert x\rvert -\tfrac 12\ln\lvert x^2+4\rvert+D}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Difference of two complex polynomials Prove that for any integer $m>1$, $(z+a)^{2m}-(z-a)^{2m}=4maz\prod_{k=1}^{m-1}z^2+a^2\cot^2{\frac{k\pi}{2m}}$
I started off by expanding the left hand side using the binomial therem and noticed that some terms cancelled out and I ended up with the expression below for the LHS
$2\sum_{k=1}^{m-1}$$2m\choose{2k-1}$$z^{2m-2k+1}$$a^{2k-1}$
I am stuck as to where to go past this. Some help would be appreciated. Thanks
| We follow the comments and consider the polynomial
\begin{align*}
u^{2m}=1\qquad\text{ with roots }\qquad \exp\left(\frac{k\pi i}{m}\right), 1\leq k\leq 2m
\end{align*}
The function
\begin{align*}
u(z)=\frac{z-a}{z+a}
\end{align*}
has the inverse function
\begin{align*}
z(u)=-a\frac{u+1}{u-1}
\end{align*}
We conclude the roots of the equation
\begin{align*}
\left(\frac{z-a}{z+a}\right)^{2m}=1\tag{1}
\end{align*}
are if $-m+1\leq k\leq m-1$
\begin{align*}
-a\cdot\frac{\exp\left(\frac{k\pi i}{m}\right)+1}{\exp\left(\frac{k\pi i}{m}\right)-1}
=-ai\cdot\frac{\frac{1}{2}\left(\exp\left(\frac{k\pi i}{2m}\right)+\exp\left(-\frac{k \pi i}{2m}\right)\right)}{\frac{1}{2i}\left(\exp\left(\frac{k\pi i}{2m}\right)-\exp\left(-\frac{k \pi i}{2m}\right)\right)}
=ia\cot \frac{k\pi}{2m}
\end{align*}
together with the root $z=0$.
We can now write (1) in polynomial form and as product of factors
\begin{align*}
(z+a)^{2m}-(z-a)^{2m}&=Az\prod_{k=1}^{m-1}\left[\left(z-ia\cot \frac{k\pi}{2m}\right)\left(z+ia\cot \frac{k\pi}{2m}\right)\right]\\
&=Az\prod_{k=1}^{m-1}\left(z^2+a^2\cot ^2\frac{k\pi}{2m}\right)
\end{align*}
with $A$ constant.
In order to determine $A$ it is convenient to calculate the coefficient of $z^{2m-1}$, denoted with $[z^{2m-1}]$. We obtain
\begin{align*}
[z^{2m-1}]&\left((z+a)^{2m}-(z-a)^{2m}\right)
=\binom{2m}{2m-1}a-\binom{2m}{2m-1}(-a)=4ma\\
[z^{2m-1}]&Az\prod_{k=1}^{m-1}\left(z^2+a^2\cot ^2\frac{k\pi}{2m}\right)=A
\end{align*}
We finally obtain with $A=4ma$
\begin{align*}
\color{blue}{(z+a)^{2m}-(z-a)^{2m}=4maz\prod_{k=1}^{m-1}\left(z^2+a^2\cot ^2\frac{k\pi}{2m}\right)}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove or disprove : If $a\equiv b$ mod $m$, when $a,b,m\in \mathbb{Z}$ , then $a^3\equiv b^3$ mod $m^2$. Prove or disprove : If $a\equiv b$ mod $m$, when $a,b,m\in \mathbb{Z}$ , then
$a^3\equiv b^3$ mod $m^2$
My attempt:
since $a\equiv b$ mod $m$ then $a=b+mk$ for some $k$
Now $a^3=m^3k^3+b^3+3m^2k^2b+3mkb^2$
$a^3-b^3=m^3k^3+3m^2k^2b+3mkb^2$
from here I stuck
| $$1\equiv 5\pmod 4$$
but $$5^3\equiv 13\not\equiv 1\pmod{16}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Orthogonal projection of point onto line not through origin
What is the orthogonal projection of the point $(8,3)$ onto the line $y = -2x - 3$?
Using basic analytical methods, I get the point $(-4/5,-7/5)$, but I want to obtain this point by seeing what happens to the standard basis vectors. But I always end up with the matrix
$$
\begin{bmatrix}-1&-\frac{8}{5}\\-1&\frac{1}{5}\end{bmatrix},
$$
and
$$
\begin{bmatrix}-1&-\frac{8}{5}\\-1&\frac{1}{5}\end{bmatrix}\begin{bmatrix}8\\3\end{bmatrix}
=\begin{bmatrix}-\frac{64}{5}\\-\frac{37}{5}\end{bmatrix}
\neq\begin{bmatrix}-\frac{4}{5}\\-\frac{7}{5}\end{bmatrix}.
$$
Can anyone tell me where I'm going wrong?
| $y = -2x-3$ is denoted by $<(1,-2)>$ named A
The orthogonal projection of point $<(8,3)>$ named $\bar x$ subtract (-3) from the y vector to obtain a new $\bar x$ to give you $<(8,6)>$ is given by
$ A(A^{T}A)^{-1}A^{T}\bar x$
$(1,-2)\left((1,-2).(1,-2)\right)^{-1}\times\left((1,-2).(8,6)\right) = (1,-2)(\frac{1}{5}\times -4)= \frac{-4}{5}<(1,-2)> = <(\frac{-4}{5},\frac{8}{5})>$ and now add -3 to y vector to get the projected point which will give you
$<(\frac{-4}{5},\frac{-7}{5})>$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397750",
"timestamp": "2023-03-29T00:00:00",
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3-D geometry: line intersecting 2 lines and parallel to plane Find a surface generated by line intersecting lines $$y=a=z$$ and $$x+3z=a=y+z$$ and parallel to plane $$x+y=0$$
I tried to form a line equation which intersects the given two lines i.e.
$(y-a)+k1(z-a)=0$ and $(x+3z-a)+k2(y+z-a)=0$. But don't know how to use the other (plane) condition.
| Reordering the equation of the line
$$\alpha: y+k_1 z-a k_1-a=0;\;\beta:x+k_2 y+(k_2+3) z -ak_2-a=0$$
The normal vector of plane $\alpha$ is $n_1=(0,1,k_1)$ and the normal vector of $\beta$ is $n_2=(1,k_2,k_2+3)$
the direction vector of the line is the cross product $m=n_1\times n_2=(3+k_2-k_1 k_2,k_1,-1)$
The line $(y-a)+k_1(z-a)=0;\;(x+3z-a)+k_2(y+z-a)=0$ is parallel to the plane $x+y=0$ if its direction vector is perpendicular to the normal vector of the plane $x+y=0$ which is $n=(1,1,0)$
$m\cdot n=0\to (3+k_2-k_1 k_2,k_1,-1)\cdot (1,1,0)=3 + k_1 + k_2 - k_1 k_2$
the condition is that $3 + k_1 + k_2 - k_1 k_2=0$
$k_2=\dfrac{k_1+3}{k_1-1}$
$\alpha:y+k_1 z-a-a k_1=0$
$\beta:(-1+k_1) x+(3+k_1) y+(3+3 (-1+k_1)+k_1) z-a (-1+k_1)-a (3+k_1)=0$
The direction of the line is $m=n_1\times n_2=(3+k_2-k_1 k_2,k_1,-1)$ where $k_2=\dfrac{k_1+3}{k_1-1}$
$m=(k_1^2-k_1,k_1-k_1^2,k_1-1)$
a point on the line is $P\left(-6 a,0,\dfrac{3 a}{2}\right)$
a parametric equation of the line is
$(x,y,x)=P+tm$ that is
$x=-k_1 t+k_1^2 t-6a,y=k_1 t-k_1^2 t,z=\dfrac{3 a}{2}-t+k_1 t$
Note that from the first two equations $x=-y-6a$
Furthermore from the three equation we get
$k_1= -\dfrac{4 y}{x+y+4 z},\;t= -\dfrac{(x+y+4 z)^2}{4 (x+5 y+4 z)}$
substitute in the third equation $z=\dfrac{3 a}{2}-t+k_1 t$
$z=\dfrac{3 a}{2}+\dfrac{(x+y+4 z)^2}{4 (x+5 y+4 z)}+\dfrac{y (x+y+4 z)}{x+5 y+4 z}$
simplify and get the result.
The requested surface is the plane $x + y+6a=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that $\frac{\tan x}{x}>\frac{x}{\sin x}, x\in(0,\pi/2)$
Prove that $$\frac{\tan x}{x}>\frac{x}{\sin x},\;\;\; x\in(0,\pi/2).$$
My work
I formulated $$f(x)=\tan x \sin x - x^2$$ in hope that if $f'(x)>0$ i.e. monotonic then I can conclude for $x>0, f(x)>f(0)$ and hence, prove the statement.
However, I got $$f'(x)=\sin x + \sec x \tan x -2x, $$ where I am unable to conclude if $f'(x)>0.$
I also found $$f''(x)=\cos x + 2\sec^3x-\sec x-2,$$
$$f'''(x)=-\sin x (1-6\sec^4x+\sec^2x).$$
But I am not able conclude the sign of any of the higher derivatives either. Am I doing something wrong? Or is there some other way?
| We can prove a stronger inequality
$$\frac{\tan x}{x} > \left(\frac{x}{\sin x}\right)^2$$ for $x \in (0, \pi/2)$. Indeed the function
$\tan x \sin^2 x - x^3$ has derivative $\tan^2 x + 2 \sin^2 x - 3 x^2$. Now, from this answer we see that $\frac{\tan x + 2 \sin x }{3} > x$ for $x \in (0, \frac{\pi}{2})$ so
$\frac{\tan^2 x + 2 \sin^2 x}{3}> \left(\frac{ \tan x + 2 \sin x }{3}\right)^2 > x^2$ and we are done.
Note that the Taylor series of $1-\frac{x^3}{\tan x \sin^2 x}$ has all coefficients positive
$$1-\frac{x^3}{\tan x \sin^2 x}=\frac{x^4}{15} + \frac{4 x^6}{189} + \frac{x^8}{225}+\cdots$$
We will prove a weaker stamement than the above (which was only checked for some coefficients) by proving that the function
$$f(x) = \frac{\sin x}{\cos^{\frac{1}{3}} x}$$ has the Taylor series at $0$ with all coefficients positive. For this, we notice that
$$f^{(2)}(x) =\frac{4}{9} \frac{\sin^3 x}{\cos^{\frac{7}{3}}x} =\frac{4}{9} \sec^{\frac{4}{3}}x \cdot f^3(x)$$
Now, $\sec x$ at any positive power has all Taylor coefficients positive. This follows from the fact that $\sec x$ has a product expansion with factors $\frac{1}{1- a_k x^2}$ and the series $(1-t)^{-\alpha}$ has all Taylor coefficients positive. Now, the fact that $f$ has all coefficients positive follows by induction, by using the recurrence given by the differential equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x}$ I came across this problem a few days ago and I have not been able to solve it. Wolfram Alpha says the answer is 1/2 but the answer I came up with is 0. Can anyone see what is wrong with my work and/or provide the correct way of solving this problem?
$$\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x} $$
$$\lim_{x \to \infty} \frac{ex^{x+1}-x[(x)(1+\frac{1}{x})]^x}{[(x)(1+\frac{1}{x})]^x} $$
$$\lim_{x \to \infty} \frac{ex^{x+1}-x^{x+1}(1+\frac{1}{x})^x}{x^x(1+\frac{1}{x})^x} $$
$$\lim_{x \to \infty} \frac{ex^{x+1}-x^{x+1}e}{x^xe} $$
$$\lim_{x \to \infty} \frac{x-x}{1} $$
$$\lim_{x \to \infty} \frac{0}{1} $$
$$0$$
I understand my mistakes may be simple and trivial, but I'm trying to learn. Thank you for your help!
| This solution is not as elegant as @farruhotas, but it's more direct. This solution relies on Taylor approximations and handwaving. I realize it's not rigorous; the purpose of this is providing intuition as to why the limit is $\frac{1}{2}$. (I'll be assuming that $x$ is an integer. Assuming the limit exists, since $x \to \infty$, this isn't an issue.)
Consider the numerator. The highest power terms are $x^{x+1}$ terms. We will first show that the $x^{x+1}$ terms cancel as $x \to \infty$. Consider the expansion of $(x+1)^x$:
$$ (x+1)^x = x^x + \binom{x}{1} x^{x-1} + \binom{x}{2} x^{x-2} + \cdots \binom{x}{\lfloor x/2 \rfloor} x^{x-\lfloor \frac{x}{2} \rfloor}+ \cdots + \binom{x}{x-1} x^1 + 1 $$
Its highest power terms are of order $x^x$. Since the $x^k$ term of $\binom{x}{k}$ is approximately $\frac{1}{k!} x^x$, the $x^x$ term of $(x+1)^x$ is approximately
$$ x^x (1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{x!} ) \approx x^x e$$
as $x \to \infty$. Thus, the $x^{x+1}$ term of $x(x+1)^x$ is $ex^{x+1}$. So the $x^{x+1}$ powers in the numerator cancel.
Now we need to find the $x^x$ powers in the numerator. The first summand $ex^{x+1}$ doesn't provide any, so let's look at the second summand $x(x+1)^x$. Go back to the expansion of $(x+1)^x$:
$$ (x+1)^x = x^x + \binom{x}{1} x^{x-1} + \binom{x}{2} x^{x-2} + \cdots \binom{x}{\lfloor x/2 \rfloor} x^{x-\lfloor \frac{x}{2} \rfloor}+ \cdots + \binom{x}{x-1} x^1 + 1 $$
Now we are looking for all terms with $x^{x-1}$ powers. Since the $x^k$ term of $\binom{x}{k}$ is approximately
$$ \frac{-1-2-\cdots -k}{k!} = -\frac{k(k-1)}{2 k!} = -\frac{1}{2} \frac{1}{(k-2)!}$$
we have that the $x^{x-1}$ term of $(x+1)^x$ is approximately
$$ x^{x-1} \cdot (-\frac{1}{2}) (0 + 0 + 1 + 1 + \frac{1}{2} + \frac{1}{6} + \cdots) \approx x^{x-1} (-\frac{e}{2}) $$
as $x \to \infty$. Thus, the $x^{x}$ term in the numerator is
$$ 0 + x \cdot (x^{x-1} (-\frac{e}{2} )) = \boxed{\frac{e}{2} x^{x}} .$$
Now consider the denominator. We already know that the $x^x$ term of $(x+1)^x$ tends to $e x^x$ as $x \to \infty$.
So, the limit becomes
\begin{align*}
\lim_{x \to \infty} \frac{ex^{x+1}-x(x+1)^x}{(x+1)^x} &= \lim_{x \to \infty} \frac{\frac{e}{2} x^x }{e x^x} \\
&= \frac{1}{2}.
\end{align*}
No elegant tricks. Just Taylor expansions and handwaving.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Show that $a^4+b^4\geq \frac18$ given that $a+b=1$
Show that $a^4+b^4\geq \frac18$ given that $a+b=1$
$$b=1-a\Rightarrow a^4+b^4=2a^4-4a^3+6a^2-4a+1$$
If we try to find the minimum of a one-variable function,we must solve a 3rd degree equation,on the other hand making perfect squares seems somewhat difficult!
Please help.
| Hint: by the generalized mean inequality: $\;\displaystyle \sqrt[4]{\frac{a^4+b^4}{2}} \ge \frac{a+b}{2}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Equation of locus of points satisfied by $\frac{\left|z+3i\right|}{\left|z-6i\right|}=1$ Equation of locus of points satisfied by $\frac{\left|z+3i\right|}{\left|z-6i\right|}=1$
The answer I got is $y=\frac{3}{2}$, but the answer given in my book is $y=-0.5x+2.25$
Can anyone please confirm which is the right answer
Edit: Sorry, the modulus was on each numerator and denominator, not for the whole thing, though I don't think this makes a difference.
My working
subbing $z=x+yi$ gives
$\frac{\left|x+yi+3i\right|}{\left|x+yi-6i\right|}=1$
$\frac{x^2+(y+3)^2}{x^2+(y-6)^2}=1^2$
Expanding brackets and then solving gives
$y=3/2$
Thank You
| Given two points,
the set of points
equidistant from them
is the perpendicular bisector
of the line joining them.
If the points are
$(a, b)$ and
$(c,d)$,
the midpoint is
$((a+c)/2, (b+d)/2)$.
The slope of the line is
$\dfrac{d-b}{c-a}$,
so the slope of the normal
is the negative reciprocal
of this or
$-\dfrac{c-a}{d-b}
=\dfrac{a-c}{d-b}
$.
Therefore the equation
of the perpendicular bisector is
$\dfrac{y-(b+d)/2)}{x-(a+c)/2}
=\dfrac{a-c}{d-b}
$.
If $a=c=0$,
as in this case,
the equation is
$\dfrac{y-(b+d)/2)}{x}
=0
$
or
$y=(b+d)/2$.
In this case,
$b=-3$ and $d=6$,
so the equation is
$y
=(6-3)/2
=3/2
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Approximating a series through successive telescopic series I am trying to approximate $\sum \frac{1}{n^2}$ with telescopic series. So far, I've understood up to this:
$$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq
m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq
m}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\\&+&\frac{1}{6}\sum_{n\geq
m}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)-\frac{1}{6}\sum_{n\geq
m}\frac{1}{n^3(n+1)^3}\end{eqnarray*} $$
I am told that the last term, $-\displaystyle\frac{1}{6}\sum_{n\geq
m}\frac{1}{n^3(n+1)^3}$, can be represented as $C\sum(\frac{C}{n^5}-\frac{C}{(n+1)^5})$ plus sum other higher order term, where $C$ is a just a number, different in each appearance.
Questions
How do we make this cubic into a telescopic sum of quintics?
And after we do that, how do we keep going and reating more and more telescopic terms, like $C\sum(\frac{C}{n^7}-\frac{C}{(n+1)^7})$ or something?
| $$\frac{1}{n^3(n+1)^3}=\frac{1}{n^6}-\frac{3}{n^7}+\frac{6}{n^8}+\ldots \tag{1}$$
$$\frac{1}{n^5}-\frac{1}{(n+1)^5} = \frac{5}{n^6}-\frac{15}{n^7}+\frac{35}{n^8}\ldots\tag{2} $$
hence
$$ \frac{1}{n^3(n+1)^3}-\color{blue}{\frac{1}{5}}\left(\frac{1}{n^5}-\frac{1}{(n+1)^5}\right) = -\frac{1}{n^8}+\ldots \tag{3} $$
where the LHS equals
$$ -\frac{1+5 n+5 n^2}{5 n^5 (1+n)^5} \tag{4}$$
which can be approximated by something of the form $\frac{D}{n^7}-\frac{D}{(n+1)^7}$ by just repeating the steps $(1),(2),(3)$.
I am quite fond of the celebrity my proof of Stirling's approximation is acquiring. I got that idea while writing the first section of these notes, before discovering such approach was already well-known in the literature.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2403098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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