Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that from the equalities, $\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z}$ follows $x^yy^x=y^zz^y=z^xx^z$. Problem :
Prove that from the equalities, $$\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z}$$ follows $$x^yy^x=y^zz^y=z^xx^z$$.
My approach :
$$\frac{x(y+z-x)}{\log x}=\frac{y(x+z-y)}{\log y}=\frac{z(y+x-z)}{\log z} = k $$
$$ \Rightarrow xy +zx -x^2 = k\log x .....(i)$$
$$\Rightarrow xy +zy -y^2 = k\log y ...(ii)$$
$$\Rightarrow zy +zx -z^2= k\log z ....(iii)$$
How to proceed further please guide thanks
| $$x^yy^x=y^zz^y=z^xx^z$$
$$\Rightarrow \left\{\begin{array}{ll}
x^{y-z}=(\frac{z}{y})^x \\
y^{x-z}=(\frac{z}{x})^y \\
z^{y-x}=(\frac{x}{y})^z
\end{array} \right. $$
Take the log of both sides
$$\Rightarrow \left\{\begin{array}{ll}
(y-z)\log{x}=x\log{\frac{z}{y}} ...(1) \\
(x-z)\log{y}=y\log{\frac{z}{x}} ...(2)\\
(y-x)\log{z}=z\log{\frac{x}{y}} ...(3)\\
\end{array} \right. $$
From (2)
$$\log{y}=\frac{y}{x-z}\log{\frac{z}{x}} ...(4)$$
subtitute (4) into (1)
$$(y-z)\log{x}=x\log{z}-x(\frac{y}{x-z}\log{\frac{z}{x}} )$$
$$\Rightarrow (z^2-xz-yz)\log{x}=(x^2 -xz-xy)\log{z}$$
$$\Rightarrow \frac{z(x+y-z)}{\log{z}}=\frac{x(y+z-x)}{\log{x}}$$
Similarly, we get the
$$\frac{z(x+y-z)}{\log{z}}=\frac{x(y+z-x)}{\log{x}}=\frac{y(x+z-y)}{\log{y}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Strange trigonometric proof. I was trying to find out how to prove
$$ \sin(A-\arcsin(0.3 \ \sin \ A)) \ \cdot \ \sin(A+\arcsin(0.3 \ \sin \ A)) \ = \ 0.91 \ \sin^2 \ A \ \ . $$
When I put this equation into my calculator both sides appear to be exactly the same, but I have no idea how to prove it.
| $$\sin { \left( A-\arcsin { \left( 0.3\sin { \left( A \right) } \right) } \right) } \cdot \sin { \left( A+\arcsin { \left( 0.3\sin { \left( A \right) } \right) } \right) =0.91\sin ^{ 2 }{ \left( A \right) } }$$
Solution
:
$$\left( \sin { A\cos { \left( \arcsin { \left( 0.3\sin { \left( A \right) } \right) } \right) } -\cos { A } \sin { \left( \arcsin { \left( 0.3\sin { \left( A \right) } \right) } \right) } } \right) \ast \left( \sin { A\cos { \left( \arcsin { \left( 0.3\sin { \left( A \right) } \right) } \right) } +\cos { A } \sin { \left( \arcsin { \left( 0.3\sin { \left( A \right) } \right) } \right) } } \right) =0.91\sin ^{ 2 }{ A } \\ \left( \sin { A\sqrt { 1-0.09\sin ^{ 2 }{ A } } -0.3\cos { A } \sin { A } } \right) \ast \left( \sin { A\sqrt { 1-0.09\sin ^{ 2 }{ A } } +0.3\cos { A } \sin { A } } \right) =\sin ^{ 2 }{ A\left( 1-0.09\sin ^{ 2 }{ A } \right) -0.09\cos ^{ 2 }{ A } \sin ^{ 2 }{ A } = } \\ =\sin ^{ 2 }{ A } \left( 1-0.09\sin ^{ 2 }{ A } -0.09\cos ^{ 2 }{ A } \right) =\sin ^{ 2 }{ A } \left( 1-0.09\left( \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } \right) \right) =\sin ^{ 2 }{ A\left( 1-0.09 \right) =091\sin ^{ 2 }{ A } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate $\lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)$ I've been studying limits on Rudin, Principles of Mathematical Analysis for a while, but the author doesn't exactly explain how to calculate limits...so, can you give me a hint on how to solve this? $$\lim_{x\to0}\frac{1}{x}\left(\sqrt[3]{\frac{1-\sqrt{1-x}}{\sqrt{1+x}-1}}-1\right)$$
| Using the basic limit $$\lim_{t \to a}\frac{t^{n} - a^{n}}{t - a} = na^{n - 1}\tag{1}$$ we can see by putting $n = 1/2, a = 1, t = 1 + x$ that $$\lim_{x \to 0}\frac{\sqrt{1 + x} - 1}{x} = \frac{1}{2}\tag{2}$$ Replacing $x$ by $-x$ we get $$\frac{1 - \sqrt{1 - x}}{x} = \frac{1}{2}\tag{3}$$ From the equation $(2), (3)$ we get (by division) $$\lim_{x \to 0}\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1} = 1\tag{4}$$ If $$u = \frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}$$ then $u \to 1$ as $x \to 0$ and hence by $(1)$ we get $$\lim_{u \to 1}\frac{\sqrt[3]{u} - 1}{u - 1} = \frac{1}{3}$$ Our desired limit is then given by
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt[3]{u} - 1}{x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt[3]{u} - 1}{u - 1}\cdot\frac{u - 1}{x}\notag\\
&= \lim_{u \to 1}\frac{\sqrt[3]{u} - 1}{u - 1}\cdot\lim_{x \to 0}\frac{u - 1}{x}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{2 - \sqrt{1 - x} - \sqrt{1 + x}}{x\{\sqrt{1 + x} - 1\}}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{2 - \sqrt{1 - x} - \sqrt{1 + x}}{x^{2}}\cdot\frac{x}{\sqrt{1 + x} - 1}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\frac{2 - \sqrt{1 - x} - \sqrt{1 + x}}{x^{2}}\notag\\
&= \frac{2}{3}\lim_{x \to 0}\frac{4 - \{\sqrt{1 - x} + \sqrt{1 + x}\}^{2}}{x^{2}\{2 + \sqrt{1 - x} + \sqrt{1 + x}\}}\notag\\
&= \frac{1}{6}\lim_{x \to 0}\frac{4 - \{2 + 2\sqrt{1 - x^{2}}\}}{x^{2}}\notag\\
&= \frac{1}{3}\lim_{x \to 0}\frac{1 - \sqrt{1 - x^{2}}}{x^{2}}\notag\\
&= \frac{1}{3}\cdot\frac{1}{2}\text{ (from equation }(3))\notag\\
&= \frac{1}{6}\notag\\
\end{align}
Note: The final limit calculation can be simplified greatly if we note that $u$ can be expressed as $$u = \frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1} = \frac{\sqrt{1 + x} + 1}{\sqrt{1 - x} + 1}$$ and then it is very easy to calculate the limit of $(u - 1)/x$ as $x \to 0$.
| {
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"url": "https://math.stackexchange.com/questions/1417409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$ It is asked to calculate $e^A$, where
$$A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$$
I begin evaluating some powers of A:
$A^0= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\; ; A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \; ; A^2 = \begin{pmatrix} -4 & 0 \\ 0 & -4\end{pmatrix} \; ; A^3 = \begin{pmatrix} 0 & -4 \\ 16 & 0\end{pmatrix}\; ; $
$ A^4=\begin{pmatrix} 16 & 0 \\ 0 & 16\end{pmatrix},\; \ldots$
I've noted that, since
$$e^A = \sum_{k=0}^\infty \frac{A^k}{k!}$$
we will have the cosine series at the principal diagonal for $\cos(2)$. But couldnt get what we will have in $(e^A)_{12}$ and $(e^A)_{21}$.
Also, we know that if $B=\begin{pmatrix} 0 & \alpha \\ -\alpha & 0 \end{pmatrix}$, then $e^B = \begin{pmatrix} \cos(\alpha) & \sin(\alpha) \\ -\sin(\alpha) & \cos(\alpha) \end{pmatrix} $. Is there a general formula for
$$B=\begin{pmatrix} 0& \alpha \\ \beta & 0 \end{pmatrix}$$?
Thanks!
| Note that
$$A=P\cdot\begin{bmatrix} 2i&0\\0&-2i\end{bmatrix}\cdot P^{-1}$$
With $P=\begin{bmatrix} -1&-1\\-2i&2i\end{bmatrix}$. We have
$$e^A=P\cdot e^{D}\cdot P^{-1}$$
With $D$ the diagonal matrix above
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that the group $G$ is abelian if $a^2 b^2 = b^2 a^2$ and $a^3 b^3 = b^3 a^3$
In a Group $G$, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds, $\forall a,b\in G$. Prove that the group $G$ is abelian.
My approach was the following:
Let $a,b\in G$
Then, $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ holds.
Now, $$\begin{align}
a^3b^3=b^3a^3 \\
\implies aaabbb=&bbbaaa\\
\implies a\cdot a^2\cdot b^2&\cdot b=b\cdot b^2\cdot a^2\cdot a\\
\implies a\cdot b^2\cdot a^2&\cdot b =b\cdot a^2\cdot b^2\cdot a\\
\implies ab \cdot ba \cdot a&b=ba\cdot ab\cdot ba
\end{align}$$
I am unable to proceed further.
What I need is a simple proof using simple theorems on groups, better if could be done using elementary properties.
| Hint: We can show that
$$a^{6}b=ba^{6}$$
for all $a, b\in G$. With assumptions, we have
$$a^{2}=b^{-2}a^{2}b^{2}\ \ \text{and}\ \ a^{3}=b^{-3}a^{3}b^{3},$$
and we get
$$a^{6}=b^{-2}a^{6}b^{2}\ \ \text{and}\ \ a^{6}=b^{-3}a^{6}b^{3},$$
thus
$$a^{6}=b^{-2}a^{6}b^{2}=b^{-3}a^{6}b^{3}$$
and the above relation implies that $a^{6}b=ba^{6}$ for all $a, b\in G$.
Proof. The center of a group $G$, denoted $Z(G)$ is the set of elements that commute with every element of $G$, which is subgroup and $Z(G)=G$ iff $G$ Abelian group.. With the above hint we get for any $a\in G$, $a^{6}\in Z(G)$, now $a^{8}b^{6}=b^{6}a^{8}$ and so
$$aa^{7}b^{6}=b^{6}a^{7}a;$$
therefore $a^{7}b^{6}\in Z(G)$ and since $Z(G)$ is a subgroup thus $a^{7}, a^{6}\in Z(G)$ (note that $a^{6}\in Z(G)$) and so $a^{7-6}=a\in Z(G)$ and this implies that $Z(G)=G$ i.e., $G$ is Abelian group.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve trigonometric inequality $ \sin x+2 \cos x<2$ $$ \sin x+2 \cos x<2$$
$$ \dfrac{2t}{1+t^2}+2\dfrac{1-t^2}{1+t^2}<2$$
$$ 4t^2-2t>0$$
$$ 2t(2t-1)>0$$
$$ t(2t-1)>0$$
$$ (t>0 \wedge t>\dfrac{1}{2}) \vee (t<0 \wedge t<\dfrac{1}{2})$$
From this, I can only find $x<2\pi+2k\pi$, and, $x<2k\pi$, these are good (I think), but I should find another two solutions.
| $\sin x+2\cos x=2(\frac{1}{2}\sin x + \cos x)$
Multiply and divide by $\frac{\sqrt{5}}{2}$
You will get: $$\sqrt{5}\big(\frac{1}{\sqrt{5}}\sin x + \frac{2}{\sqrt{5}}\cos x\big)$$
which can be expressed in the form of $a(\cos\theta\sin x+\sin\theta\cos x)=a\sin(x+\theta)$
You can express it as $$\sqrt{5}\sin \big(x+\arcsin( \frac{2}{\sqrt{5}})\big)<2$$
which is very easy to solve.
| {
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"url": "https://math.stackexchange.com/questions/1425452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ then $\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.$ Suppose that $\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ . Then , prove that $\displaystyle\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.$
Attempt :
From the given relation , $\displaystyle \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}=-2$.........(1)
Now I want to calculate , $\displaystyle \frac{a^5}{b^5}+\frac{b^5}{a^5}+\frac{b^5}{c^5}+\frac{c^5}{b^5}+\frac{a^5}{c^5}+\frac{c^5}{a^5}$. I tried by expanding pair of terms and putting the value of (1), but it can't help...
| Hint:
$$ (ab + bc + ca)(a + b + c) = (a + b)(b + c)(c + a) + abc.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the sum of the n terms of the series $2\cdot2^0+3\cdot2^1+4\cdot2^2+\dots$ Find the sum of the n terms of the series:
$2\cdot2^0+3\cdot2^1+4\cdot2^2+\dots$
I don't know how to proceed. Please explain the process and comment on technique to solve questions of similar type.
Source: Barnard and Child Higher Algebra.
Thanks in Advance!
| This lookslike a double sum. Try rewriting it the following way.
$$2\cdot 2^0+3\cdot 2^1+4\cdot 2^2+...+(n+2)\cdot 2^n =$$
$$=2\cdot 2^0+2\cdot 2^1+2\cdot 2^2+...+2\cdot 2^n$$
$$+1\cdot2^1+1\cdot 2^2+...+1\cdot 2^n$$
$$+1\cdot 2^2+...+1\cdot 2^n$$
$$ \cdots $$
The Terms are now a simple geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$ $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$
I tried to solve this question but no luck.
My try:
$$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx=\int x^4(x^8+x^4+1)(2x^8+3x^4+6)^{1/4}dx\\
\int x^4(x^8+x^4+1)x^2(2+3x^{-4}+6x^{-8})^{1/4}dx$$
Now i got stuck,please help me reach the answer.Answer is $$\frac{x^5}{30}(2x^8+3x^4+6)^{\frac54}+C$$
| Let $$\displaystyle I = \int (x^{12}+x^{8}+x^{4})\cdot (2x^8+3x^4+6)^{\frac{1}{4}}dx = \int (x^{11}+x^{7}+x^{3})\cdot (2x^{12}+3x^{8}+6x^{4})^{\frac{1}{4}}dx$$
Now Put $(2x^{12}+3x^{8}+6x^{4}) = t^4\;,$ Then $\displaystyle (x^{11}+x^{7}+x^{3})dx = \frac{t^3}{6}dt$
So Integral $$\displaystyle I = \frac{1}{6}\int t^{4}dt = \frac{t^5}{30}+\mathcal{C} = \frac{(2x^{12}+3x^{8}+6x^{4})^{\frac{5}{4}}}{30}+\mathcal{C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
How do I prove the above expression using mathematical induction? So far I have only been proving simpler stuff. The base case P(1) is easy enough but I am lost as to where I should even start with my inductive step. I really don't know what the steps for P(k + 1) should be, and so help would be greatly appreciated.
| $$P(k)=\frac{(-1)^{k-1} \cdot k \cdot (k + 1)}{2}$$
Therefore
$$P(k+1)=\frac{(-1)^{k} \cdot (k+1) \cdot (k + 2)}{2}$$
The way to go:
Write $P(k+1)$ as something containing $P(k)$.
Since you seem are asking only about some starting help, this should suffice. =)
After reading your comment:
The general way to go is starting from both direction and either go all the way are meet in the middle.
$$1^2 - 2^2 + 3^2 - 4^2 + ... + (-1)^{k-1}k^2+(-1)^k(k+1)^2 = P(k)+(-1)^k(k+1)^2\quad =\frac{(-1)^{k-1} \cdot k \cdot (k + 1)}{2} + (-1)^k(k+1)^2 =\frac{(-1)^{k-1}k\cdot (k+1) + (-1)^k2\cdot(k+1)^2}{2}\quad = \frac{(-1)^{k}(-k\cdot (k+1) + 2\cdot(k+1)^2)}{2}\quad = \frac{(-1)^{k}(-k\cdot (k+1) + 2\cdot(k+1)^2)}{2} \quad = \frac{(-1)^k(k^2+3k+2)}{2}=P(k+1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
Then, plugging the value $1$ for $x$, you get $\frac{1}{3}$.
I don't see how did they reach that conclusion. This is how I tried to tackle it:
$$\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} = \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} \cdot \frac{\sqrt[3]{x}+1}{\sqrt[3]{x}+1},$$
which then yields
$$\frac{x-1}{(2\sqrt{x}-2)(\sqrt[3]{x}+1)},$$
and that becomes
$$\frac{x-1}{2\cdot(\sqrt{x}-1)\cdot(\sqrt[3]x+1)}.$$
That's
$$\frac{x-1}{2\cdot(\sqrt[6]{x}+\sqrt{x}-\sqrt[3]{x}-1)},$$
and this will still evaluate to $\frac{0}{0}$.
How did they solve this, exactly?
| hint: Let $x = t^6$, and simplify to a nicer expression.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Taking n steps forward and m steps back: probability of returning where you started if n and m are determined by random dice rolls. You've all heard the phrase "Three steps forward, two steps back."
I am trying to figure out the probability of returning to my starting point if the number of steps forward is determined by rolling one six-sided die and the number of steps backward is determined by rolling another six-sided die.
Let's call rolling both dice a "trip." So if I roll a 3 on die #1, I go forward 3 steps, and then I roll a 2 on die #2, I go backward 2 steps. My total distance covered in the 'trip' is 1 step.
What is the probability that I return to the origin after 1 trip? What is the probability after 2 trips? What is the probability after n trips?
| The difficulty in this problem lies in the calculation of the number of different ways a sum of n throws of a die can be determined. For one throw the probability that the 'sum' of the second die equals the first is $\frac{1}{6}$, which is calculated by adding the probabilities of rolling the same number on two dice i.e. $6\times \frac{1}{6^2} = \frac{1}{6}$
For two throws you need to add together the probabilities of rolling each possible two dice total (i.e. 2, 3, 4 ... 12). The probability of a single throw is $$P_{total}=\frac{\text{Number of ways to roll the total}}{\text{Total number of possible rolls}}$$We square because we need the probability of it happening on the first die and the second (and = multiply). This is relatively simple:
$$\begin{matrix}
\text{Total for 2 dice}:&2&3&4&5&6&7&8&9&10&11&12\\
\text{Number of ways to roll}:&1&2&3&4&5&6&5&4&3&2&1\\
\text{Probability}:&\bigr(\frac{1}{36}\big)^2&\bigr(\frac{2}{36}\big)^2&\bigr(\frac{3}{36}\big)^2&\bigr(\frac{4}{36}\big)^2&\bigr(\frac{5}{36}\big)^2&\bigr(\frac{6}{36}\big)^2&\bigr(\frac{5}{36}\big)^2&\bigr(\frac{4}{36}\big)^2&\bigr(\frac{3}{36}\big)^2&\bigr(\frac{2}{36}\big)^2&\bigr(\frac{1}{36}\big)^2
\end{matrix}$$
Summing all the probabilities, which are all the different ways that the sum of the two rolls on the first dice equal the sum on the second, we get $$\frac{1}{36^2}(1^2+2^2+3^2+4^2+5^2+6^2+5^2+4^2+3^2+2^2+1^2) = \frac{73}{648}$$
Which is less than what we get for one trip, which makes sense when you think that there are many more ways to not be at the origin.
For three trips (this will be the last individual case I will work through) I wrote down the permutations of 3 dice rolling to achieve every outcome $k$ such that $3\le k\le6\times3$ which are all the possible totals for three dice.
I realised after a while that there was a pattern, and then after trying the first few for 4 trips I concluded that there was a Pascals triangular number thing going on. It arises because when writing down all the permutations you find that one number is fixed while the others permute through something you have done before i.e. for $n=2$. Then you move up the fixed number and repeat the process until the list is exhausted. This is hard to explain in words but makes sense if you write down the permutations orderly by hand.
For $n$ trips, the number of different ways of rolling a specific total $k$ where the allowed $k$ are given by $n\le k \le 6n$ are given by the $(k-n+1)^{th}$ column in the $n^{th}$ row of this modified Pascals triangle:
$$\begin{matrix}
&&&&&&&&&&1&&1&&1&&1&&1&&1\\
&&&&&1&&2&&3&&4&&5&&6&&5&&4&&3&&2&&1\\
1&&3&&6&&10&&15&&21&&25&&27&&27&&25&&21&&15&&10&&6&&3&&1
\end{matrix}$$
where each number is the sum of the 6 numbers above it i.e. 3 on either side. Only the first three rows are given because otherwise it would be much too wide.
So for 3 trips we have $$\left( \frac{1}{6^3}\ \right)^2(1^2+3^2+6^2+...+27^2+27^2+25^2+...+3^2+1^2)=\frac{361}{3888}$$
To generalise to $n$ trips using this method you would need to expand the modified 'Dicey' triangle out to row n, sum the square of every element in the row, and divide by $6^{2n}$.
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Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.
Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.
Here is what I have:
$a+b =b+a$ since $R$ is commutative.
Now,
$$(b+a) \cdot b^{-1} = 1+ab^{-1} \\
a^{-1} \cdot (1+ab^{-1}) = a^{-1} + b^{-1}
$$
Thus, $a^{-1} + b^{-1} =a^{-1} \cdot (a+b) \cdot b^{-1}$
Therefore, $(a^{-1} + b^{-1})^{-1} = b \cdot (a+b)^{-1} \cdot a$
And thus, $a^{-1} + b^{-1}$ is a unit in $R$ as well.
Does my answer sound logical. Or are there errors in it?
| Your proof is fine.
One way to discover it is by computing freely:
$$
\frac{1}{\dfrac{1}{a}+\dfrac{1}{b}}
=
\frac{1}{\dfrac{a+b}{ab}}
=
\frac{ab}{a+b}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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} |
Show using the definition of limit that $\lim_{ (x,y)\to(0,0)}\frac{ (1-\cos(xy))\sin y}{(x^2+y^2) }= 0$ can you help me with this excercise.
Show using the definition of limit that
$$\lim_{ (x,y)\to(0,0)}\frac{ (1-\cos(xy))\sin y}{(x^2+y^2) }= 0$$
Definition of limit:
$\lim_{(x,y)\to(a,b)} f(x,y) =L$ if and only if for every $\epsilon >0$ exist $\delta>0$ such that if $\sqrt{(x-a)^2+(y-b)^2}<\delta$ then $|f(x,y)-L|<\epsilon$.
Hi
I´ve tried this,
For taylor series.
Given $\epsilon>0$, find $\delta>0$ such that if
$$\sqrt{x^2+y^2}<\delta$$ then $$\bigg|\frac{(1-\cos xy)\sin y}{x^2+y^2}\bigg|<\epsilon$$
\begin{align*}
\bigg|\frac{(1-\cos xy)\sin y}{x^2+y^2}\bigg| & = \bigg|\frac{(1-1-\frac{x^2y^2}{2}+\frac{x^4y^4}{24})(y-\frac{y^3}{6}+\frac{y^5}{120})}{x^2+y^2}\bigg|\\
& =\bigg|\frac{(-\frac{x^2y^2}{2}+\frac{x^4y^4}{24})(y-\frac{y^3}{6}+\frac{y^5}{120})}{x^2+y^2}\bigg|
\end{align*}
Then what do I do?
| We need only use the inequality $\sin x<x$ for $x>0$ along with the trigonometric identity $\sin^2 x=\frac{1-\cos 2x}{2}$.
Then, we can write
$$\begin{align}
|1-\cos xy|&=|2\sin^2(xy/2)|\\\\
&\le\frac12(xy)^2\\\\
&\le \frac14(x^2+y^2)^2
\end{align}$$
along with
$$\begin{align}
|\sin y| &\le |y|\\\\
&\le(x^2+y^2)^{1/2}
\end{align}$$
Therefore, we have
$$\begin{align}
\left|\frac{(1-\cos xy)\sin y}{x^2+y^2}-0\right|&\le\frac{\frac14(x^2+y^2)^{5/2}}{x^2+y^2} \\\\
&=\frac14(x^2+y^2)^{3/2}\\\\
&<\epsilon
\end{align}$$
whenever $(x^2+y^2)^{1/2}<\delta = (4\epsilon)^{1/3}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solutions to the Laplace Equation $\Delta u =0$, where $u= \log p$ Find all real solutions to the two dimensional Laplace equation $U_{xx} + U_{yy} =0$ of the form $u=\log p(x,y)$, where $p$ is a quadratic polynomial.
Solution:
Let $p(x,y) = Ax^2 + By^2 +Cxy + D$ be a quadratic polynomial such that $A, B \not= 0$. Then
$$U_{x} = \frac{2Ax + Cy}{\ln(10)(Ax^2 + By^2 +Cxy + D},$$
$$U_{xx} = \frac{2A ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2Ax + Cy)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2},$$
$$U_{y} = \frac{2By + Cx}{\ln(10)(Ax^2 + By^2 +Cxy + D)},$$
$$U_{yy} = \frac{2B \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2By + Cx)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2}.$$
This implies
$$U_{xx} + U_{yy} = \frac{2A \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2Ax + Cy)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2} + \frac{2B \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2By + Cx)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2} = 0. $$
I feel like I am not doing this right. Is there a simpler way? Thanks. And also how do I find such solutions.
| This is an heavy method, but you can continue.
$$U_{xx} + U_{yy} = \frac{2A ln(10)(Ax^2 + By^2 +Cxy + D) - ln(10)(2Ax + Cy)^2}{ln(10) (Ax^2 + By^2 +Cxy + D)^2} + \frac{2B ln(10)(Ax^2 + By^2 +Cxy + D) - ln(10)(2By + Cx)^2}{ln(10) (Ax^2 + By^2 +Cxy + D)^2} = 0$$
After simplification :
$$2A(Ax^2 + By^2 +Cxy + D)-(2Ax + Cy)^2 + 2B(Ax^2 + By^2 +Cxy + D)-(2By + Cx)^2=0$$
It is easy to see that $A=B=1\:;\:C=D=0$ is solution. Hense : $p(x,y)=x^2+y^2$
$$U(x,y)=log(x^2+y^2)$$
A simpler method consists in solving first the PDE :
$$U(x,y)=f(x+iy)+g(x-iy)$$
any functions $f$ and $g$.
In case of $f=g=log$ then $U=log(x+iy)+log(x-iy)$
$$U(x,y)=log(x^2+y^2)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Plane of all points equidistant from two other points? Find the equation of the plane that contains all the points that are equidistant from the given points
$(-9, 3, 3), (6, -2, 4)$
I think the plane described lies in the midpoint of these points, and it is perpendicular to the line connecting the two points. This means that the point $\left(\dfrac{15}{2}, \dfrac{1}{2}, \dfrac{7}{2}\right)$ is on the plane, and vector line perpendicular to the plane is $<-9-6, 3-(-2), 3-4>=<-15, 5, -1>$ . So the equation of the plane is $$-15(x-\dfrac{15}{2})+5(y-\dfrac12)-(z-\dfrac72)=0$$ or $$15x-5y+z-\dfrac{227}{2}=0$$ However, this seems to be the wrong answer. What am I doing wrong?
| Notice, there is another easy method to find the equation of the plane
Let the parametric point be $(x, y)$ on the plane which is equidistant from the given points $(-9, 3, 3)$ & $(6, -2, 4)$ hence, we have $$\sqrt{(x-(-9))^2+(y-3)^2+(z-3)^2}=\sqrt{(x-6)^2+(y-(-2))^2+(z-4)^2}$$
$$(x+9)^2+(y-3)^2+(z-3)^2=(x-6)^2+(y+2)^2+(z-4)^2$$
$$((x+9)^2-(x-6)^2)+((y-3)^2-(y+2)^2)+((z-3)^2-(z-4)^2)=0$$
$$(x+9+x-6)(x+9-x+6)+(y-3+y+2)(y-3-y-2)+(z-3+z-4)(z-3-z+4)=0$$
$$15(2x+3)-5(2y-1)+(2z-7)=0$$
$$30x-10y+2z+43=0$$$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\text{Equation of the plane:}\ 30x-10y+2z+43=0}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a Quadratic Equation with roots of... I was given this problem at school to look at home as a challenge, after spending a good 2 hours on this I can't seem to get further than the last part of the equation. I'd love to see the way to get through 2) before tomorrow's lesson as a head start.
So the problem is as follows:
1) Quadratic Equation $$2x^2 + 8x + 1 = 0$$
i. Find roots $$\alpha + \beta$$
ii. Find roots $$\alpha\beta$$
2) Find an Equation with integer coefficients who's roots are:
$$2\alpha^4+\frac{1}{\beta^2}$$$$2\beta^4+\frac{1}{\alpha^2}$$
I'm completely puzzled on the second part of the question and I've tried following the method I was taught. Sorry if formatting is a bit off, first time posting here :)
Thanks in advance for any help!
| Given $\displaystyle \alpha,\beta$ are the roots of $2x^2+8x+4=0.$
So$\displaystyle \alpha+\beta = -\frac{8}{2}=-4$ and $\displaystyle \alpha\cdot \beta = \frac{1}{2}.$
Now for Second part, Using $\bullet\; \bf{x^2-(sum \; of \; roots)x+(product\; of \; roots) =0}$
So here $\displaystyle \bf{sum\; of \; roots } = 2\alpha^4+\frac{1}{\beta^2}+2\beta^4+\frac{1}{\alpha^2} = 2\left[\alpha^4+\beta^4\right]+\frac{1}{\alpha^2}+\frac{1}{\beta^2}$
So we get $\displaystyle = 2\left[(\alpha^2+\beta^2)^2-2(\alpha\cdot \beta)^2\right]+\frac{(\alpha+\beta)^2-2\alpha\cdot \beta}{(\alpha\cdot \beta)^2}$
$\displaystyle = 2\left[\left\{(\alpha+\beta)^2-2\alpha\cdot \beta\right\}^2-2(\alpha\cdot \beta)^2\right]+\frac{(\alpha+\beta)^2-2\alpha\cdot \beta}{(\alpha\cdot \beta)^2} = $
and $\displaystyle \bf{product\; of roots} = \left(2\alpha^4+\frac{1}{\beta^2}\right)\times \left(2\beta^4+\frac{1}{\alpha^2}\right)$
$\displaystyle = 4(\alpha\cdot \beta)^4+2\left[(\alpha+\beta)^2-2\alpha\cdot \beta\right]+\frac{1}{(\alpha\cdot \beta)^2}=$
| {
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"url": "https://math.stackexchange.com/questions/1438512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this Taylor series correct taken correctly? Confused reasoning I have $dx/dy=-ay, x(0)=1$ initial value problem.
Then $x(y)=\frac{x(0)}{0!}y^0+ \frac{x'(0)}{1!}y+\frac{x''(0)}{2!}y^2=1+(-a)y+a^2y^2...$
| Given
$$
x'(y) = \frac{dx}{dy} = -ay.
$$
The second derivative can be found with
$$
x''(y) = \frac{d^2x}{dy^2} = \frac{d}{dy} \frac{dx}{dy} = -a,
$$
and thus the third derivative will be equal to
$$
x^{(3)}(y) = \frac{d^3x}{dy^3} = \frac{d}{dy} \frac{d^2x}{dy^2} = 0,
$$
any higher derivatives will also be zero.
So the solution for $x(y)$ will just be a second order polynomial of the form
$$
x(y) = x(0) + x'(0) y + \frac{x''(0)}{2} y^2 = x(0) - 0 \cdot y - \frac{a}{2} y^2 = 1 - \frac{a}{2} y^2.
$$
The same solution can be found by integrating on both sides
$$
\int{dx} = \int{-aydy},
$$
$$
x = - \frac{a}{2} y^2 + c.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving equations with binomial coefficients
Let $n$ be a positive integer and let $x$ be a non zero real number. Prove the following.
*
*$\sum_{k=0}^n \dbinom{n}{k}2^{n-k} (1+x)(x^{-1}+x)^k = \frac{1}{x^n} (1+x)^{2n+1}$
*$\sum_{k=0}^n \dbinom{n}{k}2^{n-k} \dbinom{k}{\left\lfloor\frac{k}{2}\right\rfloor}= \dbinom{2n+1}{n}$
Sorry that I cannot make the title more clear. It is exactly my problem: I don't even know what the problem is asking, or say, what knowledge do I need to solve the problem.
It is a homework problem so I would appreciate some helps but not a complete solution. I have learnt to give combinatorial proof for some simple equalities, Sperner's Theorem, multinomial theorem and Newton's binomial theorem. But I just have no clue where to start and what to use. Thanks in advance.
| Here is a proof that leaves some work for you to do in completing the
details.
Suppose we seek to verify that
$$\sum_{k=0}^n {n\choose k} 2^{n-k} {k\choose \lfloor k/2 \rfloor}
= {2n+1\choose n}.$$
This is
$$\sum_{q=0}^n {n\choose 2q} 2^{n-2q} {2q\choose q}
+ \sum_{q=0}^n {n\choose 2q+1} 2^{n-2q-1} {2q+1\choose q}.$$
We treat these in turn.
First sum.
Observe that
$${n\choose 2q} {2q\choose q}
= {n\choose q} {n-q\choose q}.$$
This yields for the sum
$$2^n \sum_{q=0}^n {n\choose q} {n-q\choose q} 2^{-2q}.$$
Introduce
$${n-q\choose q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-q}}{z^{q+1}} \; dz$$
which yields for the sum
$$\frac{2^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z}
\sum_{q=0}^n {n\choose q} 2^{-2q} \frac{1}{z^q(1+z)^q}
\; dz
\\ = \frac{2^n}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z}
\left(1+\frac{1}{4z(1+z)}\right)^n
\; dz
\\ = \frac{2^{-n}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+2z)^{2n}}{z^{n+1}} \; dz
= 2^{-n} {2n\choose n} 2^n = {2n\choose n}.$$
Second sum.
Observe that
$${n\choose 2q+1} {2q+1\choose q}
= {n\choose q} {n-q\choose q+1}.$$
This yields for the sum
$$2^{n-1} \sum_{q=0}^n {n\choose q} {n-q\choose q+1} 2^{-2q}.$$
This time introduce
$${n-q\choose q+1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-q}}{z^{q+2}} \; dz$$
which yields for the sum
$$\frac{2^{n-1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^2}
\sum_{q=0}^n {n\choose q} 2^{-2q} \frac{1}{z^q(1+z)^q}
\; dz
\\ = \frac{2^{n-1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z^2}
\left(1+\frac{1}{4z(1+z)}\right)^n
\; dz
\\ = \frac{2^{-n-1}}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+2z)^{2n}}{z^{n+2}} \; dz
= 2^{-n-1} {2n\choose n+1} 2^{n+1} = {2n\choose n+1}.$$
Conclusion.
Collecting the two contributions we obtain
$${2n\choose n}+{2n\choose n+1} = {2n+1\choose n}$$
as claimed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $2^{2n} = \sum_{k=0}^{n}\binom{2n+1}{k}$ I'm trying to prove the following equation above. So far I have:
\begin{align}
2^{2n} &= (1+1)^{2n}\\
&= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{n-k} = \sum_{k=0}^{2n}\binom{2n}{k} & \text{(By the Binomial Theorem)}
\end{align}
I know I have to use the following identity somehow:
$$\binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}$$
How do I split my summation to get what I'm looking for? Thanks!
EDIT: HERE IS MY SOLUTION
\begin{align*}
2^{2n} &= (1+1)^{2n}\\
&= \sum_{k=0}^{2n}\binom{2n}{k}1^k1^{2n-k} & \text{(By the Binomial Theorem)}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{k}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=n+1}^{2n}\binom{2n}{2n-k} & \text{(Binomial Symmetry)}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=0}^{n-1}\binom{2n}{k}\\
&= \sum_{k=0}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\
&= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n}{k} + \sum_{k=1}^{n}\binom{2n}{k-1}\\
&= \binom{2n}{0} + \sum_{k=1}^{n}\binom{2n+1}{k} & \text{(By Identity listed above)}\\
&= \sum_{k=0}^{n}\binom{2n+1}{k}
\end{align*}
| $$\sum_{k=0}^{n}\binom{2n+1}{k}=\frac{1}{2}\sum_{k=0}^{2n+1}\binom{2n+1}{k}=\frac{2^{2n+1}}{2}=2^{2n}$$
Using the relation $$\binom{2n+1}{k}=\binom{2n+1}{2n+1-k}$$ for $0 \le k \le n$.
| {
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Prove $\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} + \frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} $ For postive integer $a, b, c $ prove the following inequality $$\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} + \frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} + \frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}} > \sqrt{2}$$
How we can prove this inequality?
| $$\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}} + \frac{1}{\sqrt{c+\frac{1}{b}+\frac{1}{2}}} + \frac{1}{\sqrt{a+\frac{1}{c}+\frac{1}{2}}} > \sqrt{2}$$
By$ AM-GM$ $$\sum_{cyc}\frac{1}{\sqrt{b+\frac{1}{a}+\frac{1}{2}}}=\sum_{cyc}\frac{\sqrt2}{2\cdot\frac{1}{\sqrt2}\sqrt{b+\frac{1}{a}+\frac{1}{2}}}\geq\sum_{cyc}\frac{\sqrt2}{b+\frac{1}{a}+1}=\sqrt2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int\frac{1}{\sin^2 x+\sin x+1}dx$
Evaluation of $\displaystyle \int\frac{1}{\sin^2 x+\sin x+1}dx$
$\bf{My\; Try::}$ Using $$\; \bullet\; x^2+x+1 = (x-\omega)\cdot (x-\omega^2)\;,$$ where $\omega,\omega^2$ are cube root of unity
So we can write Integal $$\displaystyle I = \int\frac{1}{(\sin x-\omega)\cdot (\sin x-\omega^2)}dx$$
So we get $$\displaystyle I = \frac{1}{\omega-\omega^2}\int\frac{(\sin x-\omega^2)-(\sin x-\omega)}{(\sin x-\omega)\cdot (\sin x-\omega^2)}dx$$
So $$\displaystyle I = \frac{1}{\omega-\omega^2}\int \left[\frac{1}{\sin x-\omega}-\frac{1}{\sin x-\omega^2}\right]dx$$
Now Substitute $$\displaystyle \sin x= \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$$
Can we solve it above method our we directly put $$\displaystyle \sin x= \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$$
in $$\displaystyle \int\frac{1}{\sin^2 x+\sin x+1}dx$$ and then solve it.
Or is there is any other process by which we can solve it.
Help me , Thanks
| By using Weierstrass substitution $x=2\arctan t$ the problem boils down to computing
$$ \int\frac{1+t^2}{1+2t+6t^2+2t^3+t^4}\,dt $$
through partial fraction decomposition. The roots of that palyndromic $4$th-degree polynomial are located at $t=-\frac{1}{2}\pm \frac{i \sqrt{3}}{2}-\sqrt{\frac{1}{2} \left(-3\pm i \sqrt{3}\right)}$. The remaining part is just tedious work.
| {
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Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{ c + 1} −\sqrt{c}$ for all real $c \geq 1$. Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{c + 1} −\sqrt{c}$ for all real $c \geq 1$.
Can anyone provide some form of guidance? So far all I have been able to think of is writing $c$ as $x^2$ for some $x$, or eliminating the radical on one side...
| We have:
$$
\sqrt{c}-\sqrt{c-1}≥\sqrt{c+1}-\sqrt{c}\iff\\
\\
\left(\sqrt{c}-\sqrt{c-1}\right)\frac{\sqrt{c}+\sqrt{c-1}}{\sqrt{c}+\sqrt{c-1}}≥\left(\sqrt{c+1}-\sqrt{c}\right)\frac{\sqrt{c+1}+\sqrt{c}}{\sqrt{c+1}+\sqrt{c}}\\
\\
\frac{1}{\sqrt{c}+\sqrt{c-1}}≥\frac{1}{\sqrt{c+1}+\sqrt{c}}\iff\\
\\
\sqrt{c+1}+\sqrt{c}≥\sqrt{c}+\sqrt{c-1}\iff\\
\\
\sqrt{c+1}≥\sqrt{c-1}
$$
and the last one is trivial.
| {
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Is there another way to solve this quadratic equation? $$\frac { 4 }{ x^{ 2 }-2x+1 } +\frac { 7 }{ x^{ 2 }-2x+4 } =2$$
Steps I took:
$$\frac { 4(x^{ 2 }-2x+4) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } +\frac { 7(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } =\frac { 2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } $$
$$4(x^{ 2 }-2x+4)+7(x^{ 2 }-2x+1)=2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1)$$
$$11x^{ 2 }-22x+23=2x^{ 4 }-8x^{ 3 }+18x^{ 2 }-20x+8$$
I can keep going with all the steps I took, but is there a more elegant way to arrive at the solution for this equation? It seems as if I keep going the way I am, I will hit a dead end. No actual solution, please. Hints are much better appreciated.
| Setting $t=x^2-2x+1$ gives
$$\frac{4}{t}+\frac{7}{t+3}=2$$
$$4(t+3)+7t=2t(t+3)$$
$$2t^2+6t-4t-7t-12=0$$
$$2t^2-5t-12=0$$
$$(2t+3)(t-4)=0$$
| {
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} |
How do I solve $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1})$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule as homework. So I tried rationalizing the denominator and numerator but it didn't work.
My best was: $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1}) = \lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x^2+x+1-2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{(x-1)(x+2)}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x+2}{(x^2+x+1)}) = 3 ???$
| Partial fractions save the day:
$$\frac{2}{x^3-1} = \frac{a}{x-1} + \frac{bx+c}{x^2+x+1}$$
You get $a=\frac{2}{3}$, so:
$$\frac{2}{x^3-1} =\frac{2}{3}\frac{1}{x-1} + \frac{bx+c}{x^2+x+1}$$
Now $$\lim_{x\to 1}\frac{bx+c}{x^2+x+1} = \frac{b+c}{3}.$$
So you only need to compute $$\lim_{x\to 1} \left(\frac{1}{x-1}-\frac{2}{3}\frac{1}{x-1}\right)=\lim_{x\to 1}\frac{1}{3}\frac{1}{x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solve the integral $\int \frac{dx}{\left(\sqrt{x^2+1}+x\right)^2}$ $$\int \frac{dx}{\left(\sqrt{x^2+1}+x\right)^2}$$
I think I need to use replace, but not sure:
$$x=\frac{u^2-1}{2\cdot u}$$
$$dx=\left(1-\frac{u^2-1}{2u^2}\right)$$
| Here, it might be a good idea to first simplify the integrand. We write
$$
\begin{aligned}
\frac{1}{\bigl(\sqrt{1+x^2}+x\bigr)^2}&=\frac{\bigl(\sqrt{1+x^2}-x\bigr)^2}{\bigl(\sqrt{1+x^2}+x\bigr)^2\bigl(\sqrt{1+x^2}-x\bigr)^2}\\
&=\bigl(\sqrt{1+x^2}-x\bigr)^2\\
&=1+2x^2-2x\sqrt{1+x^2}.
\end{aligned}
$$
Here, we have multiplied by the conjugate in the first step, simplified the denominator to 1 in the second step (conjugate rule twice, or squared if you want), and expanded the parenthesis in the last step. Thus,
$$
\int \frac{1}{\bigl(\sqrt{1+x^2}+x\bigr)^2}\,dx
=\int 1+2x^2-2x\sqrt{1+x^2}\,dx
=x+\frac{2}{3}x^3-\frac{2}{3}(1+x^2)^{3/2}+C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
How to get expected value from a probability mass function? A certain biased coin is flipped until it shows heads for the first time. If the probability of getting heads on a given flip is $5/11$ and $X$ is a random variable corresponding to the number of flips it will take to get heads for the first time, the expected value of $X$ is:
$$E[x] = \sum_{x=1}^\infty{x\frac{5}{11}\frac{6}{11}^{x-1}}$$
I'm not sure how to find an exact value for $E[x]$. I tried thinking about it in terms of a summation of an infinite geometric series but I don't see how that formula can be applied.
| The expectation is not a geometric series (at least not when you write
it directly), but its resemblance to a geometric series is a good observation.
First let's get that factor of $\frac{5}{11}$ out of the way,
because it will become annoying at some point if we keep it inside the
summation.
$$E[x] = \sum_{x=1}^\infty{x\frac{5}{11}\frac{6}{11}^{x-1}}
= \frac{5}{11} \sum_{x=1}^\infty{x \frac{6}{11}^{x-1}} = \frac{5}{11} S,$$
where
$$
S = \sum_{x=1}^\infty{x \frac{6}{11}^{x-1}}.
$$
Now write out $S$ and $\frac{6}{11}S$:
\begin{align}
\newcommand{x}{\left(\frac{6}{11}\right)}
S &= 1 \cdot \x^0 + 2 \cdot \x^1 + 3 \cdot \x^2 + 4 \cdot \x^3 + \cdots\\
\frac{6}{11}S &= \phantom{1 \cdot \x^0 + }\
1 \cdot \x^1 + 2 \cdot \x^2 + 3 \cdot \x^3 + \cdots
\end{align}
From here you should be able to work out what $S - \frac{6}{11}S$ is as a series, taking the difference of the right-hand sides of the two equations above, and then apply what you know about geometric series.
Notice how conveniently $S - \frac{6}{11}S = \frac{5}{11}S$,
which happens to be the value we need in the end.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Constructing Multiplication Table for Multiplication Modulo I have this question
Consider G={1,5,7,11,13,17} under Multiplication Modulo 18.Construct Multiplication Table for G.I have constructed the following
Im i correct ?
| When just multiplying in the reals I get:
$$\begin{array}{c|cccccc} \times & 1 & 5 & 7 & 11 & 13 & 17\\ \hline1 & 1 & 5 & 7 & 11 & 13 & 17\\ 5 & 5 & 25 & 35 & 55 & 65 & 85\\ 7 & 7 & 35 & 49 & 77 & 91 & 119\\ 11 & 11 & 55 & 77 & 121 & 143 & 187\\ 13 & 13 & 65 & 91 & 143 & 169 & 221\\ 17 & 17 & 85 & 119 & 187 & 221 & 289 \end{array}
$$
Then $\mod 18$ we get
$$\begin{array}{c|cccccc} \times & 1 & 5 & 7 & 11 & 13 & 17\\ \hline 1 & 1 & 5 & 7 & 11 & 13 & 17\\ 5 & 5 & 7 & 17 & 1 & 11 & 13\\ 7 & 7 & 17 & 13 & 5 & 1 & 11\\ 11 & 11 & 1 & 5 & 13 & 17 & 7\\ 13 & 13 & 11 & 1 & 17 & 7 & 5\\ 17 & 17 & 13 & 11 & 7 & 5 & 1 \end{array}$$
So only the $5$ on the bottom left is wrong, this should be $11$. As it is a commutative group, you should have a symmetry relative to the diagonal, this would provide a quick way that finds this error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Another beautiful arctan integral $\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx$ Do you think we can express the closed form of the integral below in a very nice and short way?
As you already know, your opinions weighs much to me, so I need them!
Calculate in closed-form
$$\int_{1/2}^1 \frac{\arctan\left(\frac{1-x^2}{7 x^2+10x+7}\right)}{1-x^2} \, dx.$$
I'm looking forward to your precious feedback!
Mathematica tells us the closed form is
$$\frac{1}{4} i \text{Li}_2\left(\frac{1}{5}+\frac{2 i}{5}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{2}{15}+\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{5}-\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{2}{15}-\frac{2 i}{5}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}+\frac{i}{8}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}-\frac{i}{8}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{10}+\frac{3 i}{10}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{10}-\frac{3 i}{10}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}+\frac{i}{12}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{1}{4}-\frac{i}{12}\right)-\frac{1}{4} i \text{Li}_2\left(\frac{4}{15}+\frac{8 i}{15}\right)+\frac{1}{4} i \text{Li}_2\left(\frac{4}{15}-\frac{8 i}{15}\right)+\frac{1}{4} \log (4) \tan ^{-1}(6)-\frac{1}{4} \log (4) \tan ^{-1}(9)+\frac{1}{4} \log (4) \tan ^{-1}\left(\frac{1}{6}\right)-\frac{1}{4} \log (4) \tan ^{-1}\left(\frac{1}{9}\right)-\frac{1}{4} \log (9) \tan ^{-1}(2)+\frac{1}{4} \log (9) \tan ^{-1}(3)-\frac{1}{4} \log (9) \tan ^{-1}(6)+\frac{1}{4} \log (9) \tan ^{-1}(9)-\frac{1}{4} \log (9) \tan ^{-1}\left(\frac{3}{55}\right)+\frac{1}{4} \log (16) \tan ^{-1}(2)-\frac{1}{4} \log (16) \tan ^{-1}(3).$$
| Exploiting integration by parts, the problem boils down to computing:
$$ \int_{1/2}^{1}\left(\log(1+x)-\log(1-x)\right)\left(\frac{3}{5x^2+8x+5}-\frac{4}{5x^2+6x+5}\right)\,dx $$
and by partial fraction decomposition that is equivalent to computing:
$$ I_{\pm}(\zeta)=\int_{1/2}^{1}\frac{\log(1\pm x)}{x-\zeta}\,dx $$
with $\zeta\in\left\{-\frac{3}{5}-\frac{4i}{5},-\frac{3}{5}+\frac{4i}{5},-\frac{4}{5}-\frac{3i}{5},-\frac{4}{5}+\frac{3i}{5}\right\} $. We may check that:
$$ \int \frac{\log(1+x)}{x-a} = \log(1+x)\log\left(1-\frac{1+x}{1+a}\right)+\text{Li}_2\left(\frac{1+x}{1-a}\right) $$
holds by differentiation to compute the closed form of the original integral in terms of dilogarithms and products of logarithms only.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ Solving without L'Hopital
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$
That's
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}\right)-\lim_{x\to-\infty}\left(\sqrt{4x^2+x}\right)$$
I have been taught to get the highest exponents, so...
$$\sqrt{4x^2}-\sqrt{4x^2}$$
It's the same for both sides, no?
$$2x-2x$$
$$-\infty+\infty$$
Which is wrong. The correct answer is
$$\frac{1}{4}$$
Why? I always just grab the highest exponent ($\sqrt{4x^2}$) and work with it. But this time it didn't go well.
| Set $-1/x=y\implies y\to0^+, |y|=+y$
and $4x^2-6=\dfrac{4-6y^2}{y^2},\sqrt{4x^2-6}=\dfrac{\sqrt{4-6y^2}}{|y|}=\dfrac{\sqrt{4-6y^2}}y$
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)=\lim_{y\to0^+}\dfrac{\sqrt{4-6y^2}-\sqrt{4-y}}y$$
$$=\lim_{y\to0^+}\dfrac{(4-6y^2)-(4-y)}{y(\sqrt{4-6y^2}+\sqrt{4-y})}$$
$$=\lim_{y\to0^+}\dfrac{y(1-6y)}{y(\sqrt{4-6y^2}+\sqrt{4-y})}$$
Cancel out $y$ as $y\ne0$ as $y\to0$
Then set $y=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Path Integral Problem Could anyone help understand why my answer disagrees with the given answer?
let the exact differential $dz = 3x^2(x^2+y^2)dx + 2y(x^3+y^4)dy$
Find $\int\limits_c dz$ from (1,2) to (2,1).
Here's my answer:
$\frac{\partial z}{\partial x}=3x^2(x^2+y^2)$
$\therefore$
$z = \int(3x^4+3y^2x^2) dx$
$\therefore$
$z = \frac{3}{5}x^5+y^2x^3+f(y)$
and
$\frac{\partial z}{\partial y} = 2y(x^3+y^4)$
$\therefore$
$z=\int(2x^3y+2y^5)dy$
$\therefore$
$z = x^3y^2+\frac{1}{3}y^6+g(x)$
so, in order to have $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial y}$ we need
$f(y)=\frac{1}{3}y^6$ and $g(x)=\frac{3}{5}x^5$
$\therefore$
$z = \frac{3}{5}x^5+y^2x^3+\frac{1}{3}y^6$
$\therefore$
$\int\limits_c dz$ from (1,2) to (2,1) =
$\left[\frac{3}{5}x^5+y^2x^3+\frac{1}{3}y^6\right]_{(1,2)}^{(2,1)}$
= $(\frac{96}{5}+8+\frac{1}{3})-(\frac{3}{5}+4+\frac{64}{3})$
= $\frac{8}{5}$
But the answer given in the book I'm using is $\frac{9}{2}$
Thanks,
Mitch.
| I write this only to confirm that I get the same $z$ as you (one could add an arbitrary constant, but that does not really change anything), and the same final answer. It looks like the book has a misprint.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof of the inequality $e^x\le e^{x^2} + x$ The question is to prove the inequality $e^x\le e^{x^2} + x$. I tried the Taylor expansion like ${e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ...$ and $x + {e^{{x^2}}} = 1 + x + \frac{{{x^4}}}{{2!}} + \frac{{{x^6}}}{{3!}} + ...$ but cannot see anything useful out of this. Anyone can provide some help? Thank you.
| If $x \geq 1$, then $x^2 \geq x$, we then have $e^x \leq e^{x^2}$, which in turn means that $e^x \leq e^{x^2}+x$.
For $0 \leq x \leq 1$, then $x^2 \leq x$, we then have $e^x$, we then have
\begin{align}
e^x & = \sum_{k=0}^{\infty} \dfrac{x^k}{k!} = 1 + x + \sum_{k=2}^{\infty} \dfrac{x^k}{k!} \leq 1 + x + \sum_{k=2}^{\infty} \dfrac{x^2}{2^{k-1}} \,\,\,\, (\because x \in [0,1] \text{ and }k! \geq 2^{k-1})\\
& = 1+x+x^2 \leq x + e^{x^2}
\end{align}
If $x \leq 0$, then $e^{x^2}+x \geq 1+x+\dfrac{x^2}2 = \dfrac{1+(1+x)^2}2 \geq e^x$.
Hence, $e^x \leq e^{x^2}+x$ for all $x \in \mathbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Calculus 1: limits using squeeze or sandwich theorem, verification. I have the next problem:
Suppose that a function $g$ satisfies that $-1\leq g(x) \leq 1$ for all $x\geq 0$.
Calculate $\lim_{x\to\infty}\dfrac{x+g(2x)}{g(3x)+4x}$
So: $$\lim_{x\to\infty}\dfrac{x+g(2x)}{g(3x)+4x}=\lim_{x\to\infty}\dfrac{\frac{x+g(2x)}{x}}{\frac{g(3x)+4x}{x}}=\lim_{x\to\infty}\dfrac{1+\frac{g(2x)}{x}}{\frac{g(3x)}{x}+4}$$
And because $-1\leq g(x) \leq 1$ then $\frac{g(2x)}{x}\to 0$ and $\frac{g(3x)}{x}\to 0$ as $x\to \infty$, therefore:
$$\lim_{x\to\infty}\dfrac{1+\frac{g(2x)}{x}}{\frac{g(3x)}{x}+4}=\frac{1}{4}$$
But I don't know in which part I am using the squeeze/sandwich theorem, this is the second bullet of an exercise and the first bullet is just to enunciate the theorem.
| Get rid of $g(x)$ entirely. $-1\le g(x) \le 1$. So $x - 1 \le x + g(2x) \le x + 1$.
Likewise. $4x - 1 \le g(3x) + 4x \le 4x + 1$.
So $\dfrac{x - 1}{4x + 1} \le \dfrac{x + g(2x)}{g(3x) + 4x} \le \dfrac{x + 1}{4x - 1}$
So $1/4 = \lim\dfrac{x-1}{4x+1} \le \lim \text{wholeness} \le \lim \dfrac{x + 1}{4x - 1} = \dfrac14$.
Your method worked too. You used $\frac{-1}x \le \frac{g(2x)}x < \frac1x$ and $\frac{-1}x \le \frac{g(3x)}x < \frac1x$ for the squeeze theorem twice. It also works.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$ Determine the value of $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$. I've tried forcing in a difference of squares, looked for various difference of $n$s or sum of odd powers that I could equate this to, but have yet to find a solution.
| Clearly, $x\ne0$ in $x^2-3x+1=0$
Dividing both sides by $x,$
$$\dfrac{x^2-3x+1}x=0\implies x+\dfrac1x=3$$
$$x^{2n}+\dfrac1{x^{2n}}=\left(x^n+\dfrac1{x^n}\right)^2-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the value of the following determinant? \begin{vmatrix}
1 & n & n & \dots & n \\
n & 2 & n & \dots & n \\
n & n & 3 & \dots & n \\
\vdots & \vdots& \vdots & \ddots & \vdots\\
n & n & n & & n\\
\end{vmatrix}
I've the feeling that I should to eliminate with the last row the others ove and after that should I multiply with the elements of the main diagonal.
| By Gaussian elimination,
$$\begin{vmatrix}
1 & n & n & \dots & n \\
n & 2 & n & \dots & n \\
n & n & 3 & \dots & n \\
\vdots & \vdots& \vdots & \ddots & \vdots\\
n & n & n & \dots & n\\
\end{vmatrix}=\begin{vmatrix}
1 & n & n & \dots & n \\
n-1 & 2-n & 0 & \dots & 0 \\
n-1 & 0 & 3-n & \dots & 0 \\
\vdots & \vdots& \vdots & \ddots & \vdots\\
n-1 & 0 & 0 & \dots & 0\\
\end{vmatrix}$$
so if we expand along the last column, then along the last row, we get that the determinant equals $\color{red}{n!(-1)^{n+1}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Showing $\pi/(2\sqrt3)=1-1/5+1/7-1/11+1/13-1/17+1/19-\cdots$ I am struggling to show that $$\dfrac \pi{2\sqrt3}=1-\dfrac 15+\dfrac 17-\dfrac 1{11}+\dfrac 1{13}-\dfrac 1{17}+\dfrac 1{19}-\cdots$$ by using the Fourier series $$\frac \pi2-\frac x2=\sum_1^\infty \dfrac {\sin(nx)}{n}.$$
Can somebody give me any hint?
| In the figure below, in which the trigonometric circle has been divided into six equal parts, we can see clearly that for $x= \frac {\pi}{3}$ the following equalities are verified:
$\sin( \frac{n\pi}{3})=\frac{\sqrt3}{2}$ for $n=1,2,7,8,……, 1+6n,2+6n,…..$
$\sin( \frac{n\pi}{3})=\frac{-\sqrt3}{2}$ for $n=4,5,10,11,......,4+6n,5+6n$
$\sin\left( \frac{n\pi}{3}\right)=0$ for $n=3,6,9,12,……,3n$
Hence, for $x= \frac {\pi}{3}$
$$\frac {\pi}{2} - \frac {\pi}{6}=\frac{\pi}{3}$$
$$\frac{\pi}{3}=\frac{\sqrt 3}{2}\sum_{n=1}^\infty \left[\frac{1}{1+6n} +\frac{1}{2+6n}-\frac{1}{4+6n}-\frac{1}{5+6n}\right]$$
We have
$$\frac{\pi}{3}=\frac{\sqrt 3}{2}A\iff \frac{\pi}{2\sqrt3}=\frac{3}{4}A=(1-\frac14)A$$
From which we see the odd-denominator terms $$\frac{1}{1+6n} -\frac{1}{5+6n}$$ remain unchanged and the even-denominator terms $$\frac{1}{2+6n} -\frac{1}{4+6n}-\frac{1}{4+24n}-\frac{1}{8+24n}+\frac{1}{16+24n}+\frac{1}{20+24n}$$
give a telescopic series of total sum equal to zero.
(Note that $4+24n=4+6(4n)$;$8+24n=2+6(4n+1)$; $16+24n=4+6(4n+2)$;$20+24n=2+6(4n+3)$; we obtain $$0=\frac12-\left(\frac14+\frac14\right) = \frac18-\frac18=\frac{1}{10}-\left(\frac{1}{20}+\frac{1}{20}\right)=\frac{1}{14}-\left(\frac{1}{28}+\frac{1}{28}\right)$$ an so on.
Thus we end with the addition of the terms $$\frac{1}{1+6n} -\frac{1}{5+6n}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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If $2x+3y \propto \sqrt{xy}$ ,then prove that $9x^2+4y^2 \propto xy$. I am stuck on the following problem which one of my friends gave me:
If $2x+3y \propto \sqrt{xy}$ ,then prove that $9x^2+4y^2 \propto xy$.
The problem could have been easier if I had to prove $4x^2+9y^2 \propto xy$ but this problem has stumped me.
Can someone please help ? Thanks in advance for your time.
| $$\text{A) }2x+3y \propto \sqrt{xy}\qquad \text{B) }9x^2+4y^2 \propto xy$$
From A we get
$$4 x^2+12 x y+9 y^2 \propto xy$$
Combining A and B we get
$$4 x^2+12 x y+9 y^2 = k(9x^2+4y^2)$$
$$\Rightarrow 4 x^2+12 x y+9 y^2 = 9kx^2+4ky^2$$
From this we see that no $k$ will give an $xy$ term, so the statement is false. If we continue, we get
$$ 12xy = x^2(9k-4)+y^2(4k-9)$$
$$ xy = c_1x^2+c_2y^2$$
This is even more nonsensical. Further, if we substitute in the definition of $k$ from A we get
$$4 x^2+12 x y+9 y^2 = \frac{2x+3y}{\sqrt{xy}}(9x^2+4y^2)$$
Solving this for $y$, we get
$$y = \frac{-2x}{3}$$
Since the signs of $x$ and $y$ differ for all non-zero $x$,$y$ we conclude that the original statement only holds for $x$,$y = 0$, and does not hold in general
| {
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"timestamp": "2023-03-29T00:00:00",
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How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What surprises me is that the result implies
$$y=\frac{1}{2}\tan^{-1}x+C$$
Can anyone tell me how to see that directly?
| By virtue of solving for $x$, the original equation $$y = \tan^{-1} (x - \sqrt{1+x^2})$$ implies $$x = \frac{1}{2}(\tan y - \cot y) = \frac{\sin^2 y - \cos^2 y}{2 \sin y \cos y} = -\cot 2y,$$ hence $$y = \frac{1}{2}\cot^{-1} (-x) = -\frac{1}{2} \cot^{-1} x.$$ Now recalling that $$\cot^{-1} x + \tan^{-1} x = \frac{\pi}{2},$$ we readily obtain $$y = \frac{1}{2} \tan^{-1} x - \frac{\pi}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 1
} |
separating equation $$
\begin{align}
\frac {dy}{dx} &= \frac{(y-1)(x+2)}{(y+1)(x-3)}\\
\frac{y+1}{y-1}dy &= \frac{x+2}{x-3}dx
\end{align}
$$
Integrate both sides :
$$
\begin{align}
\int\frac{y+1}{y-1}dy & =\int\frac{x+2}{x-3}dx\\
\int 1+\frac{2}{y-1}dy &=\int 1+\frac{5}{x-3}dx\\
y+2\ln|y-1|&=x+5\ln|x-3|+C\\
\end{align}
$$
trapped here because I cannot get rid of $2\ln|y-1|$, any techniques I should use to solve this?
| You are not trapped and the solution cannot be expressed in terms of elementary functions.
If fact, there is a solution in terms of Lambert function you will learn at a time and the solution would be $$y=1+2 W\left(\pm\frac{c \sqrt{(x-3)^5 e^{x}}}{2 \sqrt{e}}\right)$$ where $c$ is the integration constant you missed.
You could be interested knowing that any equation which can write $$A+B x+C \log(D+Ex)=0$$ has solutions in terms of Lambert function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Maximisation and minimisation of sum of squares, if sum is equal to 15 Find the numbers whose sum is $15$ and sum of squares is minimum
My answer:
Let the numbers be $x$ and $15-x$. Then
$$y=x^2+625-30x+x^2$$
$$=2x^2-30x+625$$
$$dy/dx=0$$
$$4x-30=0$$
$$X=7.5$$
Did I go wrong?
| Approach without using calculus:
Geometric viewpoint. This is the same as finding the point on the line $x+y=15$ which is closest to the origin/farthest from the origin. If you draw the picture, you can see that the point with coordinates $x=y=15/2$ is closest to the origin. (It lies on the perpendicular line $x=y$.) The distance from the origin can be arbitrarily large, so there is no maximum. If you have restriction that $x,y\ge0$, then the maximum is attained for $(x,y)=(0,15)$ and $(x,y)=(15,0)$.
Inequalities. We know from the inequality between quadratic mean and arithmetic mean that for $x,y\ge0$ we have
$$\sqrt{\frac{x^2+y^2}2}\ge\frac{x+y}2$$
which can be simplified to
$$x^2+y^2\ge 2\left(\frac{x+y}2\right)^2.$$
For two variables this can be derived even easier:
$$
\begin{align*}
(x-y)^2&\ge0\\
x^2-2xy+y^2\ge0\\
x^2+y^2&\ge2xy\\
2x^2+2y^2 &\ge x^2+2xy+y^2\\
2x^2+2y^2 &\ge (x+y)^2\\
x^2+y^2 &\ge \frac12(x+y)^2\\
\end{align*}
$$
| {
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"url": "https://math.stackexchange.com/questions/1474314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the inequation for $x$ Solve for $x$ :
$ (x-1)^{2005} x^{2006} (x+1)^{2007} \le 0 $
I tried cases like : $ x-1 \le 0 $ , $ x \le 0 $ , $ x+1 \le 0 $
| Well, let's see if we can simplify your expression first:
$$ (x-1)^{2005} x^{2006} (x+1)^{2007} \le 0 $$
$$(x^2 - 1)^{2005}x^{2006}(x+1)^{2} \le 0$$
Think, first, what qualifies $x$ to be $\le 0$? If $x$ is $0$ or a negative number, correct?
Well, our expression can be $0$ if $x = -1, 1$
Thus, now lets look when our expression yields a negative number.
Therefore, look at the powers and see which ones could be negative:
$$(x^2 - 1)^{2005}x^{2006}(x+1)^{2} \le 0$$
We can see that $(x^2 - 1)^{2005}$ can only be negative, since the other $2$ terms have even powers, meaning they will only be positive.
Therefore, when is $(x^2 - 1)^{2005} \le 0$?
Well,
$$x^2 - 1 \le 0$$
$$x^2 \le 1$$
$$-1 \le x \le 1$$
And thus we arrive at our answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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module of sum is less than... Somehow need to prove:
$$|x_1 + x_2 + ... + x_n| \le \sqrt{n(x_1^2 + x_2 ^ 2 + ... +x_n^2)}$$
$x_i$ is a real number; $i = 1,...,n$
Here's mentioned that mathematical induction should help. So I tried to use it but right after squaring, cause I had no idea how to do it without squaring first.
So it looks like:
$$(x_1 + x_2 + ... + x_n + x_{n+1})^2 \le (n+1)(x_1^2 + x_2^2 + ... +x_n^2 + x_{n+1}^2) $$
And some transformation based on the induction hypothesis(part which is more should be transformed into less:
$$(x_1 + x_2 + ... + x_n + x_{n+1})^2 \le (x_1 + x_2 + ... + x_n )^2 + (n+1)x_{n+1}^2$$
Community I need a hint for the next step or even another idea for making a proof.
| Your induction hypothesis is $|x_1 + \cdots + x_n| \le \sqrt{n(x_1^2 + \cdots + x_n^2)}$, which becomes $(x_1 + \cdots + x_n)^2 \le n(x_1^2 + \cdots + x_n^2)$ by squaring. Write
\begin{align}(x_1 + \cdots + x_n + x_{n+1})^2 &= (x_1 + \cdots + x_n)^2 + x_{n+1}^2 + 2(x_1 + \cdots + x_n)x_{n+1}\\
&= (x_1 + \cdots + x_n)^2 + x_{n+1}^2 + 2x_1x_{n+1} + \cdots + 2x_nx_{n+1}.\tag{*}
\end{align}
Since $2xy \le x^2 + y^2$ for all $x,y\in \Bbb R$, then
$$2x_1x_{n+1} + \cdots + 2x_nx_{n+1} \le (x_1^2 + x_{n+1}^2) + \cdots + (x_n^2 + x_{n+1}^2) = (x_1^2 + \cdots + x_n^2) + nx_{n+1}^2.$$
Using the induction hypothesis, we find that the expression (*) is less than or equal to
$$n(x_1^2 + \cdots + x_n^2) + x_{n+1}^2 + (x_1^2 + \cdots + x_n^2) + nx_{n+1}^2,$$
which is
$$(n+1)(x_1^2 + \cdots + x_n^2) + (n+1)x_{n+1}^2 = (n+1)(x_1^2 + \cdots + x_n^2 + x_{n+1}^2)$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Help showing $\lim\limits _{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right) = \frac{5}{4}$ I am stuck solving the following limit. I know the answer is 5/4, I just can't get it. This is the steps I have done so far.
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right)$
Multiply by Conjugate
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right)\cdot \frac{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Multiply Out
$\lim\limits_{x\to -\infty \:}\cdot \frac{\left(4\cdot \:\:\:x^2-5\cdot \:\:\:x-4\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Combine Like Terms
$\lim\limits_{x\to -\infty \:}\cdot \frac{\left(4\cdot \:\:\:x^2-9\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Factor out x
$\lim\limits_{x\to -\infty \:}\cdot \frac{x\left(4\cdot \:x-9\right)}{\left(\sqrt{x^2\left(4-\frac{5}{x}\right)}-2\cdot \:\:x\right)}$
Pull out x of sqrt and factor again
$\lim \limits_{x\to -\infty \:}\cdot \frac{x\left(4\cdot \:x-9\right)}{x\left(\sqrt{\left(4-\frac{5}{x}\right)}-2\right)}$
Now cancel x terms
$\lim \limits_{x\to -\infty \:}\frac{4\cdot \:\:x-9}{\sqrt{\left(4-\frac{5}{x}\right)}-2}$
Now I don't know what to do next.
If I plug in I get
$\frac{4\cdot \:\:\:-\infty \:-9}{\sqrt{\left(4-0\right)}-2}=\:\frac{-\infty \:}{0}$ Which doesn't equal $\frac{5}{4}$?
| Hint: You can use the fact that
$$
\sqrt{4-5t} = 2\sqrt{1-\frac{5}{4}t} = 2\left(1-\frac{5}{8}t + o(t)\right)
$$
when $t\to 0$. (This is the Taylor approximation of $\sqrt{1+t}$ around $0$.) Note that when $x\to-\infty$, $t=\frac{1}{x}\to 0$.
Additionally, you have a few issues in your derivation. For instance, $x\to -\infty$, so in particular is negative:
$$
\sqrt{x^2} = \lvert x\rvert = -x
$$
(when you factor). Also, even before, when you multiply by the conjugate you should have obtained a $-4x^2$, not $-4x$, in the numerator.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by induction: $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$ $\sum\limits_{k=1}^{n}sin(kx)=\frac{sin(\frac{n+1}{2}x)sin\frac{nx}{2}}{sin\frac{x}{2}}$
Base case: For $n=1$
$sinx=\frac{sinx\cdot sin\frac{x}{2}}{sin\frac{x}{2}}=sinx$
Induction hypothesis: For $n=m$
$\sum\limits_{k=1}^{m}sin(kx)=\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}$
Induction step: $n=m+1$
$\sum\limits_{k=1}^{m+1}sin(kx)=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$
Prove: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}}{sin\frac{x}{2}}+sin(m+1)x=\frac{sin(\frac{m+2}{2}x)sin\frac{(m+1)x}{2}}{sin\frac{x}{2}}$
Left side: $\frac{sin(\frac{m+1}{2}x)sin\frac{mx}{2}+sin\frac{x}{2}sin(m+1)x}
{sin\frac{x}{2}}$
How to prove this equality? I used $sin(u)sin(v)$ identity but that didn't help.
| HINT:
Double angle formula:
$$\sin(m+1)x=2\sin\dfrac{(m+1)x}2\cos\dfrac{(m+1)x}2$$
Werner Formula: $$2\sin\dfrac x2\cos\dfrac{(m+1)x}2=\sin\dfrac{(m+2)x}2-\sin\dfrac{mx}2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
Integral $\int \sqrt{x+\sqrt{x^2+2}}dx$ $$\int \sqrt{x+\sqrt{x^2+2}}\ dx$$
I tried various solving methods but I am not coming forward. I unformed the term also to ${x+\sqrt{x^2+2} \over \sqrt{x+\sqrt{x^2+2}}}$ and even multiplied with ${\sqrt{x-\sqrt{x^2+2}} \over \sqrt{x-\sqrt{x^2+2}}}$.
Trigonometric and hyperbolic substitution didn't help either.
| Notice, let $$x+\sqrt{x^2+2}=t^2\implies x=\frac{t^4-2}{2t^2}$$$$ \left(1+\frac{x}{\sqrt{x^2+2}}\right)dx=2tdt\iff \left(\frac{x+\sqrt{x^2+2}}{\sqrt{x^2+2}}\right)dx=2tdt$$$$\implies dx=\frac{t^4+2}{t^3}dt$$
Now, we get
$$\int\sqrt{x+\sqrt{x^2+2}}dx=\int t\frac{t^4+2}{t^3}dt$$
$$=\int \frac{t^4+2}{t^2}dt=\int\left(t^2+\frac{2}{t^2}\right)dt$$
$$=\frac{t^3}{3}-\frac{2}{t}+C$$
$$=\color{}{\frac{(x+\sqrt{x^2+2})\sqrt{x+\sqrt{x^2+2}}}{3}-\frac{2}{\sqrt{x+\sqrt{x^2+2}}}+C}$$
$$=\color{blue}{\frac{(x+\sqrt{x^2+2})^2-6}{3\sqrt{x+\sqrt{x^2+2}}}+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How is $\frac{ds}{dt}$ related to $\frac{dx}{dt}$? The problem states: Let $x$ and $y$ be differentiable functions of $t$, and let $s = \sqrt{4x^2+6y^2}$ be a function of $x$ and $y$. How is $\frac{ds}{dt}$ related to $\frac{dx}{dt}$ if $y$ is constant?
My attempt:
\begin{align}
\frac{ds}{dt} & = \frac{d}{dt} \left( \sqrt{4x^2+6y^2} \right) \\
& = \frac{1}{2\sqrt{4x^2+6y^2}}*\left(8x\frac{dx}{dt}+12y\frac{dy}{dt}\right)
\end{align}
If $y$ is constant, then $\frac{dy}{dt}=0$
\begin{align}
\frac{ds}{dt} & = \frac{1}{2\sqrt{4x^2+6y^2}}*\left(8x\frac{dx}{dt}+12y*0\right)\\
& =\left(\frac{8x}{2\sqrt{4x^2+6y^2}}\right)*\frac{dx}{dt}
\end{align}
This is the relation I have found, but it is not correct. What is the relation between $\frac{ds}{dt}$ and $\frac{dx}{dt}$ ?
| This is an example of the usefulness of Leibniz notation in dealing with these sorts of problems.
$$\frac{ds}{dt} = \frac{ds}{dx} \cdot \frac{dx}{dt}$$
This, intuitively, is because the $dx$'s "cancel out".
All to do here is to find $\frac{ds}{dx}$.
Since $y$ is not a function of $x$, we can treat it as constant. Therefore:
$$\frac{ds}{dt}=\frac{dx}{dt} \cdot \frac{d}{dx} \left[\sqrt{4x^2+6y^2} \right]$$
And since $$\frac{d}{dx} \left[\sqrt{4x^2+6y^2} \right] = \frac{4x}{\sqrt{4x^2+6y^2}}$$
The only difference I can find is that you did not simplify your answer. To find out more, how exactly did you know your answer is incorrect?
| {
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"url": "https://math.stackexchange.com/questions/1486747",
"timestamp": "2023-03-29T00:00:00",
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How to solve $x^6-x^5+x^4-x^3+x^2-x+1=0$? Can anyone tell me how to solve this?
$x^6-x^5+x^4-x^3+x^2-x+1=0$
What I got to was $x^7+1=0$.
Thanks in advance.
| $$x^6-x^5+x^4-x^3+x^2-x+1=0 \Longleftrightarrow$$
$$(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)=0 \Longleftrightarrow$$
$$x^7+1=0 \Longleftrightarrow$$
This introduces the extraneous root of $x=-1$, so from now on we assume that $x\ne -1$:
$$x^7+1=0 \Longleftrightarrow$$
$$x^7=-1 \Longleftrightarrow$$
$$x^7=e^{\pi i} \Longleftrightarrow$$
$$x=\left(e^{(\pi+2\pi k) i}\right)^{\frac{1}{7}} \Longleftrightarrow$$
$$x=e^{\frac{1}{7}(\pi+2\pi k) i}$$
With $k\in\mathbb{Z}$ and $k:0-6$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate equilibrium points of a 2nd order system The following 2nd order system is given:
$$
\frac{d\textbf{x}}{dt}=\begin{bmatrix} -6 & -\frac{2}{\pi} \\ 0 & \frac{1}{2} \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}+\begin{bmatrix}\sin(\frac{\pi}{2}-x_2) \\ x_1x_2\end{bmatrix}
$$
The task is to calculate all equilibrium points, which I started as follows. I wrote the system down as two equations:
$$
I: -6x_1-\frac{2}{\pi}x_2+\sin(\frac{\pi}{2}-x_2)=0 \\
II: \frac{1}{2}x_2+x_1x_2=0
$$
We can see that the second equation is satisfied if either $x_2=0$ or $x_1=-\frac{1}{2}$. When I plug that into equation one I get:
$$
I, x_2=0: \\ -6x_1+\sin (\frac{\pi}{2}) = 0\\
-6x_1+1=0\\
x1=\frac{1}{6}
$$
$$
I, x_1=-\frac{1}{2}:\\ 3-\frac{2}{\pi}x_2+\sin (\frac{\pi}{2}-x_2)=0 \\
x_2=3\frac{\pi}{2}
$$
So as a result there are two equilibrium points:
$$
\textbf{x}_{R1}=\begin{pmatrix} \frac{1}{6} \\ 0 \end{pmatrix} \\
\textbf{x}_{R2}=\begin{pmatrix} -\frac{1}{2} \\ \frac{3\pi}{2} \end{pmatrix}
$$
But my problem description states that there should be four equilibrium points! How do I calculate the rest? Are the ones I already have correct?
| What you have found is correct. But $x_2 = \pi$ and $x_2 = 2 \pi$ are also solutions in the second equations.
| {
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Ordinary differential equation with polynomial terms The original equation I had was:
$$-y''+(x^2+2x^4-2\alpha)y=0$$
Where $\alpha$ is a real parameter $\geq 0$ and we require the solutions to go to $0$ at infinity. With the substitution $y=p(x)e^{-\frac{x^2}{2}}$ I got the equation:
$$-p''+2xp'+p(1+2x^4-2\alpha)=0$$
I tried the power series method, getting:
$$a_2=(\frac{1}{2}-\alpha)a_0$$
$$a_3=\frac{(\frac{3}{2}-\alpha)a_1}{3}$$
$$a_4=\frac{(\frac{5}{2}-\alpha)(\frac{1}{2}-\alpha)a_0}{6}$$
$$a_5=\frac{(\frac{7}{2}-\alpha)(\frac{3}{2}-\alpha)a_1}{30}$$
And after $n=4$:
$$a_{n+2}=\frac{2(n+\frac{1}{2}-\alpha)a_n+2a_{n-4}}{(n+1)(n+2)}$$
Are the solutions to this equation known? What would you do?
| Hint:
Let $u=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=2x\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{dy}{du}\right)=2x\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)+2\dfrac{dy}{du}=2x\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}+2\dfrac{dy}{du}=2x\dfrac{d^2y}{du^2}2x+2\dfrac{dy}{du}=4x^2\dfrac{d^2y}{du^2}+2\dfrac{dy}{du}=4u\dfrac{d^2y}{du^2}+2\dfrac{dy}{du}$
$\therefore4u\dfrac{d^2y}{du^2}+2\dfrac{dy}{du}-(2u^2+u-2\alpha)y=0$
| {
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How to Calculate $x^6+x^3y^3+y^6$ Given that $x,y$ real numbers such that :
$x^2+xy+y^2=4$
And
$x^4+x^2y^2+y^4=8$
How can one calculate :
$x^6+x^3y^3+y^6$
Can someone give me hint .
| Note that
$$(x^2+xy+y^2)^2-2xy(x^2+xy+y^2)=x^4+x^2y^2+y^4 $$
so that
$$ xy = 1.$$
Then
$$\begin{align}x^6+x^3y^3+y^6&=(x^4+x^2y^2+y^4)(x+xy+y^2)-xy^5-2x^2y^4-2x^4y^2-yx^5\\
&=8\cdot 4-xy(x^4-2x^3y-2xy^3-y^4)\\
&=32-1\cdot ((x^4+x^2y^2+y^4)-xy(2x^2-xy-2y^2))\\
&=24+1\cdot(2(x^2+xy+y^2)-3xy)\\
&=29
\end{align}
$$
(I suppose) Just keep subtracting simple products and powers of the given polynomials to get rid of moniomials not divisible by $xy$, then divide out $xy$ from the rest and continue.
Alternatively to the above, you might also start by subtracting $(x^2+xy+y^2)^3$, which would also "kill" the $x^6$ and $y^6$.
| {
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Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ I had an example in the book given as follows:
Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .
Solution: $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$
$(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$
Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$
Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$.
Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$
So degree of $p(x)$ is atleast $4$.
But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .
So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$
But I can't get the step how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$
Kindly help with this.
| Good question: that is a big leap in the argument in my opinion.
If we let $K = \mathbb{Q}(\sqrt{2},\sqrt{3})$, then $\sqrt{2} + \sqrt{3}$, $\sqrt{2} - \sqrt{3}$, $-\sqrt{2} + \sqrt{3}$, and $-\sqrt{2} - \sqrt{3}$ all lie in $K$.
In order to prove that they are all roots of $p(x)$, the minimal polynomial of $\sqrt{2} + \sqrt{3}$, we follow An Hoa's idea: find automorphisms $\sigma, \tau$ of $K / F$ such that $\sigma(\sqrt{2}) = -\sqrt{2}$, $\sigma(\sqrt{3}) = \sqrt{3}$, $\tau(\sqrt{2}) = \sqrt{2}$, and $\tau(\sqrt{3}) = -\sqrt{3}$.
The argument is then completed by noting that $\sigma, \tau$ both preserve $p(x)$ (since its coefficients are in $K$), so they preserve the roots of $p(x)$. (If $\alpha$ is a root of $p(x)$, then $\sigma(\alpha)$ is a root of $\sigma(p(x)) = p(x)$ and $\tau(\alpha)$ is a root of $\tau(p(x)) = p(x)$.)
Therefore what we need to show is that these automorphisms $\sigma$ and $\tau$ exist.
$K = \mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$ is the splitting field of $(x^2 - 2)(x^2 - 3)$, so it is a Galois extension.
To argue the extension has degree $4$, further notice that it contains both $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$. These two fields are distinct because one contains a square root of two and one doesn't (all elements of $\mathbb{Q}(\sqrt{3})$ can be written $a + b\sqrt{3}$ and you can show that such a thing squared cannot be $2$). So $K$ has degree divisible by $2$ and strictly greater than $2$ over $\mathbb{Q}$, and it also has degree at most $4$ since it is the splitting field of a degree-4 polynomial, so it has degree exactly $4$.
Now that we know $K$ has degree $4$ over $\mathbb{Q}$ and is Galois, it must have four automorphisms (this is the definition of Galois, that the number of automorphisms equals the degree). These automorphisms permute the roots of $(x^2 - 2)(x^2 - 3)$ and are defined by where they send these roots. They must send roots of $(x^2 - 2)$ to roots of $(x^2 - 2)$ and roots of $(x^2 - 3)$ to roots of $(x^2 - 3)$. But there are only $4$ different ways to do this. Therefore, every possible way of sending $\sqrt{2} \mapsto \pm \sqrt{2}$ and $\sqrt{3} \mapsto \pm \sqrt{3}$ is an automorphism. In particular, $\sigma$ and $\tau$ are automorphisms.
$\square$
| {
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"timestamp": "2023-03-29T00:00:00",
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What line is determined by the following complex equation? $$\left | z+1-i \right |= \frac{\Re z-\Im z}{\sqrt{2}}$$
This leads to the following equation, if we write $z=x+iy$ :
$$\sqrt{\left ( x+1 \right )^{2}+\left( y-1 \right)^2}=\frac{x-y}{\sqrt{2}}$$
which according to desmos, defines a set in $\mathbb{C}$ which is empty.
| Suppose that there exist $(x,y)$ such that
$$\sqrt{(x+1)^2+(y-1)^2}=\frac{x-y}{\sqrt 2}$$
(note here that $x-y\ge 0$)
Then we have
$$\begin{align}(x+1)^2+(y-1)^2=\frac{(x-y)^2}{2}&\Rightarrow x^2+y^2+2x-2y+2=\frac{x^2}{2}+\frac{y^2}{2}-xy\\&\Rightarrow \frac{x^2}{2}+\frac{y^2}{2}+xy+2x-2y+2=0\\&\Rightarrow x^2+y^2+2xy+4x-4y+4=0\\&\Rightarrow (x+y)^2+4=4(y-x)\end{align}$$
The LHS is positive and the RHS is non-positive. This is a contradiction.
| {
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Is it possible to solve this limit without Hopital / Taylor / derivatives: $\lim\limits_{x \to 0} \frac{x-\sin(x)}{x^3} = \frac{1}{6}$? It's simple to prove with Hopital that
$$ \lim_{x \to 0} \frac{x-\sin(x)}{x^3} = \frac{1}{6}$$
Is it possible to solve this limit without Hopital or Taylor (without derivatives)?
| Let us assume $$\displaystyle \lim_{x\rightarrow 0}\frac{x-\sin x}{x^3} = L$$ (A finite quantity).
Now replace $x\rightarrow 3y$, then we get $$\displaystyle \lim_{y\rightarrow 0}\frac{3y-\sin 3y}{27y^3} = L$$
Now, using the formula $$\sin 3y = 3\sin y-4\sin^3 y$$
we get $$\displaystyle \lim_{y\rightarrow 0}\frac{3y-3\sin y+4\sin^3 y}{27y^3} = L$$
So $$\displaystyle \frac{1}{9}\lim_{y\rightarrow 0}\frac{y-\sin y}{y^3}+\frac{4}{27}\displaystyle \lim_{y\rightarrow 0}\left(\frac{\sin y}{y}\right)^3=L$$
So $$\frac{1}{9}L+\frac{4}{27} = L\Rightarrow \frac{8}{9}L = \frac{4}{27}\Rightarrow L=\frac{1}{6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the value of $\sum_{i = 1}^{\infty } (\frac{1}{i} - \frac{1}{2i + 1} - \frac{1}{2i - 1})$ Is there any way to solve this summation?
$$\sum_{i = 1}^{\infty } (\frac{1}{i} - \frac{1}{2i + 1} - \frac{1}{2i - 1})$$
The value is 1 - log4... but I'm not able to proove it.
| @Olivier Oloa, obviously, there is something wrong in your solution.
For i>0, $\frac{1}{i}-\frac{1}{2i+1}-\frac{1}{2i-1}=\frac{1}{2i}-\frac{1}{2i+1}+\frac{1}{2i}-\frac{1}{2i-1}<\frac{1}{2i}-\frac{1}{2i+1}$, but using your method, $\sum_{i=1}^{\infty}\frac{1}{2i}-\frac{1}{2i+1}=\sum\int dx{x^{2i-1}-x^{2i}}=\int dx\frac{x}{1-x^2}-\frac{x^2}{1-x^2}=\int dx \frac{x}{1+x}=1-log2 $. But I cannot see where lies the wrong. And mathematica tells the answer is 1-log4.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $\arctan n$ always equal to $\arccos\sqrt{\frac{1}{n^2+1}}$? $$\arccos\sqrt\frac{1}{2}=\arctan 1$$
$$\arccos\sqrt\frac{1}{5}=\arctan 2$$
$$\arccos\sqrt\frac{1}{10}=\arctan 3$$
$$\arccos\sqrt\frac{1}{17}=\arctan 4$$
$$\arccos\sqrt\frac{1}{26}=\arctan 5$$
$$\arccos\sqrt\frac{1}{37}=\arctan 6$$
$$\arccos\sqrt\frac{1}{50}=\arctan 7$$
The answer is a sequence $n^2+1$ for the slope which is in the inverse of tangent. Digits that are whole numbers. Is there any explanation as to why this is true? Is it a well-known problem?
| Let $\arctan n = a$, then $\tan(a)=n$ and...
$$1 + \tan^2 (a) = 1/(\cos^2 (a) )$$
$$1 + n^2 = 1/ (\cos^2 (a))$$
$$\cos^2 (a) = 1/(1+n^2)$$
$$\cos(a) = (1/(1+n^2) )^{1/2}$$
$$\arccos (1/(1+n^2) )^{1/2} = a = \arctan(n)$$
| {
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For the partial differential equation $u_{xx}-u_{yy}=-1$ find $u(1/2,1/2)$
Let , $u(x,y)$ be a solution of $u_{xx}-u_{yy}=-1$ , $x\in \mathbb R$ , $y>0$ with $u(x,0)=0$ , $u_y(x,0)=0$ , $x\in \mathbb R$. Then , find $u\left(\frac{1}{2},\frac{1}{2}\right)$.
Attempt :
Solution is $u(x,y)=f_1(y+x)+f_2(y-x)-\frac{x^2}{2}$. With the help of initial conditions I can't eliminate the arbitrary functions $f_1$ and $f_2$. How I can eliminate these two functions ?
| $$
u(x, y) = f_1(y + x) + f_2(y-x) - \frac {x^2}2 \implies u_y = f_1' + f_2'
$$
Now, substitute ICs
\begin{align}
u(x, 0) &= f_1(x) + f_2(-x) - \frac {x^2}2= 0 \tag 1 \\
u_y(x, 0) &= f_1'(x) + f_2'(-x) = 0 \tag 2
\end{align}
Differentiate $(1)$
$$
f_1'(x) - f_2'(-x) -x = 0 \tag 3
$$
And solve it in conjunction with the $(2)$
$$
2f_1'(x) = x \implies f_1 = \frac {x^2}4 + C \tag 4
$$
and
$$
f_2(-x) = \frac {x^2}2 - \frac {x^2}4 - C = \frac {x^2}4 - C
$$
or
$$
f_2(x) = \frac {x^2}4 - C \tag 5
$$
So, finally you can write
$$
u(x, y) = \frac {(y+x)^2}4 + \frac {(y-x)^2}4 - \frac {x^2}2
$$
| {
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Finding the quaternion that performs a rotation I managed to find this answer here where Christian Rau says "axis/angle rotation (a,x,y,z) is equal to quaternion (cos(a/2),xsin(a/2),ysin(a/2),z*sin(a/2))"
Assuming I know what rotation I need to perform, how would I represent it?
eg, finding the quaternion that rotates 30 degrees around the z axis.
Any help would be greatly appreciated.
edit** I got as far as writing out "(cos(15),0,0," and then got confused on z * sin.
| To rotate about the $z$ axis (yaw) by $\alpha$ you need the following quaternion
$\begin{aligned}q = \begin{bmatrix}\cos(\tfrac{\alpha}{2})\\0\\0\\\sin(\tfrac{\alpha}{2})\end{bmatrix}\end{aligned}\tag{1},$
to rotate about the $x$ (pitch) axis you need
$\begin{aligned}q = \begin{bmatrix}\cos(\tfrac{\alpha}{2})\\0\\ \sin(\tfrac{\alpha}{2})\\0\end{bmatrix}\end{aligned}\tag{2},$
and to rotate about $y$ by $\alpha$ you need
$\begin{aligned}q = \begin{bmatrix}\cos(\tfrac{\alpha}{2})\\\sin(\tfrac{\alpha}{2})\\0\\0\end{bmatrix}\end{aligned}\tag{3}.$
If you have a rotation described by the Euler angles $(\phi, \theta, \psi)$ (in the standard order), then, the corresponding quaternion is
$\begin{aligned}
q = \begin{bmatrix}
\cos \tfrac{\phi}{2} \cos \tfrac{\theta}{2} \cos \tfrac{\psi}{2} + \sin \tfrac{\phi}{2} \sin \tfrac{\theta}{2} \sin \tfrac{\psi}{2} \\
\sin \tfrac{\phi}{2} \cos \tfrac{\theta}{2} \cos \tfrac{\psi}{2} - \cos \tfrac{\phi}{2} \sin \tfrac{\theta}{2} \sin \tfrac{\psi}{2} \\
\cos \tfrac{\phi}{2} \sin \tfrac{\theta}{2} \cos \tfrac{\psi}{2} + \sin \tfrac{\phi}{2} \cos \tfrac{\theta}{2} \sin \tfrac{\psi}{2} \\
\cos \tfrac{\phi}{2} \cos \tfrac{\theta}{2} \sin \tfrac{\psi}{2} - \sin \tfrac{\phi}{2} \sin \tfrac{\theta}{2} \cos \tfrac{\psi}{2}
\end{bmatrix}
\end{aligned}\tag{4}.$
If you are rotating your object about an axis described by the vector $u=(u_x, u_y, u_z)\in\mathbb{R}^3$ and by an angle $\alpha$ about that axis, then
$\begin{aligned}q = \begin{bmatrix}\cos(\tfrac{\alpha}{2})\\\sin(\tfrac{\alpha}{2})u\end{bmatrix}=
\begin{bmatrix}\cos(\tfrac{\alpha}{2})\\\sin(\tfrac{\alpha}{2})u_x\\\sin(\tfrac{\alpha}{2})u_y\\\sin(\tfrac{\alpha}{2})u_z\end{bmatrix}
\end{aligned}
\tag{5}.$
| {
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Legendre symbol $(-3/p)$ where $p = 1 \mod 3$ Suppose $p = 1 \bmod 3$, prove the following statements:
*
*prove that $x^2 + x + 1 = 0 \mod p$ has a solution
*Prove that $\left(\frac{-3}{p}\right) = 1$ if $p = 1\mod 3$
*Determine the discriminant of $x^2 + x + 1$
*Prove using 2,3 that $\left(\frac{-3}{p}\right) = -1$ if $p = -1\mod 3$
This is what I've tried by each question:
*
*prove $x^2 + x = -1 \mod p$ has a solution, i tried to find an x such that: $x^2 + x = a^{\frac{p-1}{2}} = -1 \mod p$, where we use that a is equal to a quadratic non-residue and use Euler. I don't seem to see why this is true though.
*Note the following:
$(2x+1)^2 = 4x^2 + 4x + 4 = 4(x^2 + x + 1) - 3 = -3 \mod p$. So this solution exist and thus $-3$ must be a quadratic residue mod p.
*-3
*?
| *
*You will need to use Cauchy's theorem (or similar) as mentioned in the answer that Jack D'Aurizio provided to the question mentioned in his comment. So if $p \equiv 1 \mod 3$ then there is an $x \ne 1$ for which $x^3 = 1$, and then $(x-1)(x^2+ x+1) =0$ so we have our required root of $x^2 + x + 1 = 0$. By the way, any quadratic equation that has a root will automatically also have a second root, which can be found by noting that $x^2[(x^{-1})^2+x^{-1}+1]=x^2+x+1=0$ and thus $x^{-1}$ is the second root. Alternatively, note that $(-(x+1))^2+-(x+1)+1=x^2+x+1=0$, so the second root is $-(x+1)$, and we find as a bonus that $x^{-1}=-(x+1)$. This of course also follows easily by multiplying $x^2+x+1=0$ by $x^{-1}$.
*Agree with your own answer except for the typo:
$$(2x+1)^2=4x^2+4x+1=4(x^2+x+1)-3=-3$$ so $(\frac{-3}{p})=1$
*Indeed $D=1^2-4\cdot 1\cdot 1 = -3$. Effectively 2. says that $\sqrt{D}$ exists, hence we can find the roots of the equation $x^2+x+1=0$ from $x_{1,2}=\frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm (2x+1)}{2}$, thus $x_1=x$, $x_2=-(x+1)$, as mentioned under 1.
*If $p \equiv -1 \mod 3$ then there will be no non-trivial root of $x^3=1$ (as 3 does not divide the group order $p-1$), thus $\sqrt{-3}$ does not exist (otherwise we would have solutions of $x^2+x+1=0$), which is the same as $(\frac{-3}{p})=-1$
| {
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Show $\big\lvert\frac{1}{2} - \frac{1}{2}e^{-i2\pi f}\big\vert^{2} = \sin^2(\pi f)$? $\big\lvert\frac{1}{2} - \frac{1}{2}e^{-i2\pi f}\big\vert^{2} = \sin^2(\pi f)$?
Any help?
| HINT:
Assuming $f\in\mathbb{R}$:
$$\left|\frac{1}{2} - \frac{1}{2}e^{-2i\pi f}\right|^{2} = \sin^2(\pi f)\Longleftrightarrow$$
$$\left|\frac{1}{2}\left(1 - \frac{1}{2}e^{-2i\pi f}\right)\right|^{2} = \sin^2(\pi f)\Longleftrightarrow$$
$$\frac{1}{4}\left|1 - \frac{1}{2}e^{-2i\pi f}\right|^{2} = \sin^2(\pi f)\Longleftrightarrow$$
$$\frac{1}{4}\left(\sqrt{\frac{1}{4}\sin^2(2\pi f)+\left(-\frac{1}{2}\cos(2\pi f)-1\right)^2}\right)^{2} = \sin^2(\pi f)\Longleftrightarrow$$
$$\frac{1}{4}\left(\frac{1}{4}\sin^2(2\pi f)+\left(-\frac{1}{2}\cos(2\pi f)-1\right)^2\right)= \sin^2(\pi f)$$
| {
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Proving that the sum of fractions has an odd numerator and even denominator. I'm struggling to show that, for all $n>1$
$$
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \frac{k}{m}
$$
where $k$ is an odd number and $m$ is an even number.
Proof: The proof is by induction on $n$.
Base Case: $1 + \frac{1}{2} = \frac{3}{2}$
Assume the theorem is true for $n$ and consider $n+1$.
$$
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + \frac{1}{n+1}
$$
We know by the induction hypothesis that the first $n$ terms take the form $\frac{k}{m}$, where $k$ is odd and $m$ is even.
$$
\frac{k}{m} + \frac{1}{n+1}
$$
To combine the terms, we find the least common multiple of $m$ and $n+1$. Since
$m$ is even, the least common multiple must be even.
Now What? I'm not sure how to show that the numerator remains odd. I don't think there's enough information to do just a case analysis on whether $n+1$ is odd or even.
This question is from Manber's Introduction to Algorithms: A Creative Approach, which I'm using for personal development. He marked this question as "substantially more difficult."
| To complete your proof, first observe that
$$
\frac{k}{m} + \frac{1}{n+1} = \frac{k(n+1)+m}{m(n+1)}
$$
now $m$ is even so let $m=2^\alpha a$,where $\alpha$ is the biggest power of $2$ in $m$, so $a$ is odd.
if $n+1$ is odd obviously $k(n+1)+m$ is odd and ${m(n+1)}$ is even, so we are done.
So consider the case when $n+1$ is even. Then $n+1 = 2^\beta b$, where $\beta$ is the biggest power of 2 in $n+1$, so b is odd.
so $ \frac{k(n+1)+m}{m(n+1)}= \frac{k(2^\beta b)+2^\alpha a}{(2^\beta b)(2^\alpha a)}= \frac{k(2^\beta b)+2^\alpha a}{2^{\alpha + \beta}ab} $
now look at whether $\alpha$ or $\beta$ is smaller. Say $\alpha$ is smaller, then
$\frac{k(2^\beta b)+2^\alpha a}{2^{\alpha + \beta}ab} = \frac{k(2^{\beta-\alpha} b)+ a}{2^{ \beta}ab} $
now we have an even denominator, and on the numerator $k(2^{\beta-\alpha} b)$ is even and $a$ is odd, so the numerator is odd.
Now if $\beta$ is smaller, then
$\frac{k(2^\beta b)+2^\alpha a}{2^{\alpha + \beta}ab} = \frac{k b+ 2^{\alpha-\beta} a}{2^{ \alpha}ab} $
now $kb$ is odd so the numerator is odd, and our denominator is obviously even.
By induction, you have proved statement as required
| {
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Proving that $\lim \limits_{x \to \infty}\frac{x^3}{x^2 +10}=\infty$ using $\epsilon-\delta$ definition $\lim \limits_{x \to \infty}\frac{x^3}{x^2 +10}=\infty$
Therefore $\forall M>0$ we need to find a $\delta=\delta(M)$ such that $f(x)>M$ or $x<\delta(M)$
so $\frac{x^2}{x^2+10}>M$
$x^2>M(x^2+10)$
$x^2-Mx^2>10M$
$x^2(1-M)>10M$
$x>\sqrt{\frac{10}{1-M}}$
But this doesn't help me because x is greater than the function $\delta(M)$
| The limit
$$
\lim_{x\to \infty} \frac{x^2}{x^2+10} = 1
$$
and not equal to $\infty$. You can see this from
$$
\lim_{x\to \infty} \frac{x^2 / x^2}{(x^2+10)/x^2} = \lim_{x\to \infty} \frac{1}{1 + 10/x^2} = 1.
$$
If you are considering the limit
$$
\lim_{x\to \infty} \frac{x^3}{x^2+10} = \lim_{x\to \infty} \frac{x}{1 + 1/x^2}
$$
then let $M$ be given. You want to find a $\delta$ such that if $x > \delta$, then $\frac{x}{1 + 1/x^2} > M$. Now let $\delta_1>0$ be be such that $\frac{1}{x^2} < 1$ for $\delta > 0$. You probably know this exists because $\lim_{x\to \infty} \frac{1}{x^2} = 1$. So you have $1+\frac{1}{x^2} < 2$ for $x > \delta_1$.
Let $\delta_2 > 2M$. Then let $\delta = \max\{\delta_1, \delta_2\}$. This will work.
| {
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roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$ are real
If $a,b,c,d\in \mathbb{R}$ and roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$
are real.Then prove that roots are equal.
$\bf{My\; Try::}$ Given $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0\;$ Then we can write it as
$$\left[(a^2x)^2+c^4-2a^2c^2x+(b^2x)^2+d^4-2b^2d^2x+2a^2c^2x+2b^2d^2x+4abcdx\right]=0$$
So $$(a^2x-c^2)^2+(b^2x-d^2)^2+2x(ac+bd)^2=0$$
Now I did not understand How can I proceed further.
Although I have a knowledge of $\bf{Discriminant\; Method.}$
So plz explain me above method which i am trying above.
Thanks.
| You want to see when the discriminant is non negative, that is,
$$
(4abcd)^2-4(a^4+b^4)(c^4+d^4)\ge0
$$
that can be simplified into
$$
4a^2b^2c^2d^2-(a^4+b^4)(c^4+d^4)\ge0
$$
Consider the left-hand side:
\begin{align}
4a^2b^2c^2d^2-(a^4+b^4)(c^4+d^4)
&=4a^2b^2c^2d^2-a^4c^4-b^4c^4-a^4d^4-b^4d^4\\
&=(-a^4c^4+2a^2b^2c^2d^2-b^4d^4)+(-b^4c^4+2a^2b^2c^2d^2-a^4d^4)\\
&=-(a^2c^2-b^2d^2)^2-(b^2c^2-a^2d^2)^2\\
&=-\bigl((a^2c^2-b^2d^2)^2+(b^2c^2-a^2d^2)^2\bigr)
\end{align}
Since a sum of squares is non negative, the discriminant is non negative if and only if it is zero; this implies the root are coincident, when real.
| {
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Proving the irrationality of $\sqrt{5} = x \sqrt{7} + y$ I need to prove that there are no rational numbers $x, y$ that
$$\sqrt{5} = x \sqrt{7} + y$$
We know that square root of prime is irrational so $y = 5 - 7x$ so the only number for it to be rational is if $y = 0$ so $x = \frac{\sqrt{7}} {\sqrt {5}}$ but that is irrational or when $x = 0$, $y = \sqrt{5}$ which is irrational.
$x = \sqrt{5} - y \sqrt {7}$, so I'm stuck here.
Or I need to try to prove each at turn, like
1 case : let $x =$ blah blah and do it as irrational prime proof
2 case : let $y =$ ???
Help would be appreciated.
| Assume we have rationals $x,y$ such that $\sqrt{5} = x\sqrt{7} + y$.
Square both sides, we get $5 = 7x^2 + y^2 + 2xy\sqrt{7}$.
So $\sqrt{7} = {{5 - 7x^2 - y^2} \over {2xy}}$ which is rational, contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 (n^2 +1)}\right)^n
}
$$
I think I got the correct limit by using fast converging limits to $e$.
In particular I used truncated Taylor series for the sqrt and 4th root.
Or squares and bisquares.
Example
$(1+1/2n)^{2n}$
Becomes
$(1 + 1/n + 1/4n^2)^n.$
In combination with l'hopital it gives me the answer.
But I guess that is not a very good (fast) method.
| Perhaps a way to 'see the answer' is to use the MVT: for the numerator, write $f( x) = x^n$, $a = 1 + 1/n -1/n^2$, $b = 1 + 1/n + 1/n^2$. Then
$$ f( b) - f(a) = f'(c)(b-a),$$
for some $c \in (a,b)$.
Therefore the numerator is
$$ n\left(1 + 1/n + o(1/n^2)\right)^{n-1} 2/n^2.$$
Similarly, in the denominator, using the same trick twice, one gets
$$ n\left(1 +1/n + o(1/n^2)\right)^{n-1} ( 2/n^2 + 1/(n^2+1) + 1/n^2(n^2+1) ).$$
Taking the limit (of the ratio!), one gets Hamed's answer of $2/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Greatest Common Divisor Proof: $\gcd(m^2-n^2, m^2+n^2) = 1$ Prove that if $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$ then $\gcd(m^2-n^2, m^2+n^2) = 1$
Workings:
Suppose that $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$
$\gcd(m^2-n^2, m^2+n^2)$
$= gcd((m-n)(m+n), (m-n)(m+n)+2n^2)$
Now I know that $m+n=1 (\text{mod} ~2)$ means that one of $m$ or $n$ is odd or even
But now I'm not too sure on what to do.
Any help will be appreciated.
| We can make use of the fact that $\gcd(a, b) = \gcd(a, b-ka)$ for any integer $k$. Note that
$\gcd(m^2 - n^2, m^2 + n^2) = \gcd(m^2 - n^2, 2n^2).$ You noted that one of $m, n$ is odd and one is even, so $2 \nmid m^2 -n ^2$. So $$\gcd(m^2 - n^2, 2n^2) = \gcd(m^2 - n^2, n^2) (\text{why?}) = \gcd(m^2, n^2).$$
From there, you can argue that this gcd must be 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding maximum $b$ in $x^5-20x^4+bx^3+cx^2+dx+e=0$ Let $b, c, d, e$ be real numbers such that the following equation
$$x^5-20x^4+bx^3+cx^2+dx+e=0$$
has real roots only. Find the largest possibe value of $b$.
What I have done is:
Let $x_1, x_2, x_3, x_4, x_5$ be the 5 real roots of the equation. Then we have
$$x_1+x_2+x_3+x_4+x_5=20$$ and
$$b=\sum_{0<i\le j \le 5} x_ix_j=\frac{1}{2}[(x_1+x_2+x_3+x_4+x_5)^2-(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)]$$
To find maximum $b$, we can find minimum $x_1^2+x^2_2+x_3^2+x_4^2+x_5^2$. Cauchy-Schwartz Inequality yields,
$$(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)(1+1+1+1+1)\ge(x_1+x_2+x_3+x_4+x_5)=20$$
Thus,
$$x_1^2+x^2_2+x_3^2+x_4^2+x_5^2\ge\frac{20}{5}=4$$
So,
$$b_{max}=\frac{1}{2}[20^2-4]=198$$
However, the answer was 160, yet I am pretty sure I am correct. Where did I go wrong?
Thanks in advance!
| The answer is correct and you are wong.
The mistake in your solution happens when you apply the Cauchy Inequality where you missed to square the right side.
$(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)(1+1+1+1+1)\ge(x_1+x_2+x_3+x_4+x_5)^2=400$
So $x_1^2+x^2_2+x_3^2+x_4^2+x_5^2\ge\frac{400}{5}=80$
and $b_{max}=\frac{1}{2}[20^2-80]=160$.
A simple check would see your answer is wrong, the third coefficient in $(x-4)^5=160$ and not $198$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 }$ So I'm trying to show that:
$\dfrac{ -1 }{ 2 } \le \dfrac{ x }{ 1+x^2 } \le \dfrac{ 1 }{ 2 }$
for every value of x.
I know I have to use mean value theorem so I tried to show it with cases. First I tried showing that $\dfrac{-1}{2} \le \dfrac{x}{1+x^2}$ and then $\dfrac{1}{2} \ge \dfrac{x}{1+x^2}$ using MVT. Is that correct? Can someone guide me?
Thanks!
| Let $u=|x|$, then $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 } \iff \frac{ u }{ 1+u^2 } \le \frac{ 1 }{ 2 }$.
You are using $AM-GM$ here and not Mean Value Theorem since $1+u^2\geq2u$ by $AM-GM$.
To do the calculus way, just find the derivative of $x\over {1+x^2}$.
$df\over dx$$={1+x^2-2x^2\over{(1+x^2)^2}}=0\implies x=\pm1$.
Sub in $x=1$ you get $f={1\over2}$ and sub in $x=-1$ you get $f={-{1\over2}}$.
And $lim_{x\to \infty}f=0$ shows it is a global maximum. Minimum similarly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Equation of locus Point P$(x, y)$ moves in such a way that its distance from the point $(3, 5)$ is proportional to its distance from the point $(-2, 4)$. Find the locus of P if the origin is a point on the locus.
Answer:
$$(x-3)^2 + (y-5)^2 = (x+2)^2 + (y-4)^2$$
or, $$10x+2y-14=0$$
or, $$5x+y-7=0$$
but answer given is $$7x^2+7y^2+128x-36y=0$$
| $$(x-3)^2 + (y-5)^2 = k\left[ (x+2)^2 + (y-4)^2 \right]$$ where $k$ is a constant
Now $(0,0)$ lies on the locus.
Therefore $$9+25=k(4+16) \Rightarrow k=\frac{34}{20} = \frac{17}{10}$$
Using this value of $k$ in the equation, we get
$$(x-3)^2 + (y-5)^2 = \frac{17}{10}\left[ (x+2)^2 + (y-4)^2 \right]$$
$$10\left[(x-3)^2 + (y-5)^2 \right]= 17\left[ (x+2)^2 + (y-4)^2 \right]$$
$$10(x^2-6x+9+y^2-10y+25)=17(x^2+4x+4+y^2-8y+16)$$
$$7x^2+7y^2+128x-36y=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Sum of the rows of Pascal's Triangle. I've discovered that the sum of each row in Pascal's triangle is $2^n$, where $n$ number of rows. I'm interested why this is so. Rewriting the triangle in terms of C would give us
$0C0$ in first row.
$1C0$ and $1C1$ in the second, and so on and so forth. However, I still cannot grasp why summing, say, 4C0+4C1+4C2+4c3+4C4=2^4.
| There are various different ways to look at this. Here's one:
Two adjacent numbers in a row get added to get the number in the row below it:
$$
\begin{array}{cccccccccc}
& & 1 & & & & & 8 & & & & 28 & & & & 56 & & & & 70 & & \cdots \\
& & & & & & & & \searrow & & \swarrow \\
1 & & & & & 9 & & & & 36 & & & & 84 & & & & 126 & & \cdots & & & & \cdots
\end{array}
$$
That means every number in a row is added into the next row twice:
$$
\begin{array}{cccccccccc}
& & 1 & & & & & 8 & & & & 28 & & & & 56 & & & & 70 & & \cdots \\
& & & & & & \swarrow & & \searrow \\
1 & & & & & 9 & & & & 36 & & & & 84 & & & & 126 & & \cdots & & & & \cdots
\end{array}
$$
Since every number is added into the next row twice, the sum of the numbers in the next row is twice as big.
Here's another: In row $9$ (which is the tenth row, since the first row is "row $0$), the entries are.
$$
\binom 9 0 = 1,\ \binom 9 1 = 9,\ \binom 9 2 = 36,\ \binom 9 3 = 84,\ \binom 9 4 = 126,\ \ldots
$$
These are
*
*the number of subsets of size $0$ of a set of size $9$, and
*the number of subsets of size $1$ of a set of size $9$, and
*the number of subsets of size $2$ of a set of size $9$, and
*the number of subsets of size $3$ of a set of size $9$, and
*and so on.
Their sum is therefore the total number of subsets of a set of size $9$. If you know that that is $2^9$, you've got it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Use integration by parts to find the integral $\int\frac{\sqrt {4x^2-9}}{x^2}dx$ $$\int\frac{\sqrt {4x^2-9}}{x^2}dx$$
I tried to solve this using integration by parts, but I come up with something that is much more difficult to solve. How can this be solved?
| Trigonometric substitution There is another simpler method to solve the problem
Let $2x=3\sec\theta \implies dx=\frac{3}{2}\sec\theta \tan\theta \ d\theta$
$$\int \frac{\sqrt{4x^2-9}}{x^2}\ dx=\int \frac{\sqrt{9\sec^2\theta-9}}{\frac{9}{4}\sec^2\theta}\ \frac{3}{2}\sec\theta \tan\theta \ d\theta$$
taking positive value,
$$=\frac{2}{3}\int \frac{3\tan\theta}{\sec\theta}\tan\theta \ d\theta$$
$$=2\int \frac{\tan^2 \theta}{\sec\theta}\ d\theta$$
$$=2\int \frac{\sec^2 \theta-1}{\sec\theta}\ d\theta$$
$$=2\int (\sec\theta-\cos\theta)\ d\theta$$
$$=2\left(\int \sec\theta\ d\theta-\int \cos\theta\ d\theta\right)$$
$$=2\left(\ln\left|\sec\theta+\tan\theta\right| -\sin\theta\right)+c$$
substituting $\sec\theta=\frac{2x}{3}$, one should get $$=2\ln|2x+\sqrt{4x^2-9}|-\frac{\sqrt{4x^2-9}}{x}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Finding the shortest distance between an arbitrary point and a parabola I'm attempting to find the shortest distance between a point and a parabola. The point in question is (0,b), for any b, and the parabola that we are given is$\ y = x^2 $.
How would you approach the problem and find the shortest distance for any given b?
What about if the point was (0,0,b), and the equation was $\ z = x^2 + y^2$?
| 2D
We consider the distance between a query point $Q=(0,b)$ and some point $P(x) = (x, x^2)$ on the graph of the function.
This leads to
$$
d(x) = d(Q, P(x)) = \sqrt{x^2 + (x^2 - b)^2}
$$
The distance to the graph is the minimum of those distances:
\begin{align}
d &= \min_{x \in \mathbb{R}} d(x) \\
&= \min_{x \in \mathbb{R}} \lVert (x, x^2)) - (0,b)\rVert
\end{align}
So we look for local extrema via
$$
0
= d'(x)
= \frac{2x + 2(x^2 -b)\,2x}{2\sqrt{x^2 + (x^2 - b)^2}}
= \frac{2x^3+(1-2b)x}{\sqrt{x^2 + (x^2 - b)^2}}
$$
which happens for $x = 0$ or
$$
2x^2 = 2b - 1 \iff \\
x = \pm\sqrt{b - 1/2}
$$
which has real solutions for $b \ge 1/2$. The resulting distance for both of these solutions is
$$
d
= \sqrt{b - 1/2 + (b - 1/2 - b)^2}
= \sqrt{b-1/4}
= \sqrt{4b - 1 }/2
$$
while the distance for $x=0$ is between $(0,0)$ and $(0,b)$ thus $\lvert b \rvert$.
So the minimum distance is
$$
d =
\begin{cases}
\min(\lvert b \rvert, \sqrt{4b - 1}/2 & \text{for } b > 1/2 \\
\lvert b \rvert & \text{for } b \le 1/2
\end{cases}
$$
(Large version)
3D
The above generalizes to three dimensions. Query point is $(0,0,b)$ and the points on the graph are $P(x,y) = (x, y, x^2 + y^2)$.
This leads to
$$
d(x,y) = d(Q, P(x,y)) = \sqrt{x^2 + y^2 + (x^2 + y^2 - b)^2}
$$
For local minima we look where the gradient vanishes:
\begin{align}
0 = \text{grad } d(x, y)
&= \frac{1}{2\sqrt{x^2 + y^2 + (x^2 + y^2 - b)^2}}
(2x + 2(x^2 + y^2 - b)2x, 2y + 2(x^2 + y^2 - b)2y) \\
&= \frac{1}{\sqrt{x^2 + y^2 + (x^2 + y^2 - b)^2}}
(2x(x^2 + y^2) + (1-2b)x, 2y(x^2+y^2) + (1-2b)y)
\end{align}
This is the case for $(x,y) = (0,0)$ otherwise if
$$
2(x^2 + y^2) = 2b - 1 \iff \\
x^2 + y^2 = r^2 \wedge r = \sqrt{b - 1/2}
$$
which has real solutions for $b \ge 1/2$. The resulting distance for solutions on this circle in the $x$-$y$-plane with radius $r$ is:
$$
d = \sqrt{b - 1/2 + (b - 1/2 - b)^2} = \sqrt{4b-1}/2
$$
while the distance for $(x,y)=(0,0)$ is between $(0,0,0)$ and $(0,0,b)$ thus $\lvert b \rvert$.
So the minimum distance is again
$$
d =
\begin{cases}
\min(\lvert b \rvert, \sqrt{4b - 1}/2 & \text{for } b > 1/2 \\
\lvert b \rvert & \text{for } b \le 1/2
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Proving a complex equality. How to prove the following equality:
$$2i\tan^{-1} \left(\frac{ia}{b}\right) = \log \left|\frac{a-b}{a+b}\right|$$
where $i = \sqrt{-1}$ and $a,b \in \mathbb{R},a \neq \pm b$
| Hint:
$$\int_0^x \frac{1}{x^2-a^2} \mathrm{d}x = \frac{1}{2a}\int_0^x \frac{1}{x-a} - \frac{1}{x+a} \mathrm{d}x = \frac{1}{2a} \log \left|\frac{x-a}{x+a}\right|$$
$$\int_0^x \frac{1}{x^2-a^2} \mathrm{d}x = \int_0^x \frac{1}{x^2+(ia)^2} \mathrm{d}x = \frac{i}{a}\tan^{-1} \left(\frac{ix}{a}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Rewrite an expression as a sum of even powers - proving injectivity without calculus I teach algebra to undergraduates - nothing too fancy. But we recently covered injective functions, and I was a little disappointed that I couldn't give rigorous justifications why certain functions were injective. For the handful of functions that they are expected to know, the only justifications I've seen that these functions are indeed injective proceed by showing that they are strictly increasing or decreasing on certain intervals by means of calculus - much more machinery than is covered in this class. However, there was one function which I - excitedly - got to prove to them was injective. Namely $f(x)=x^3$. For if, $x^3=y^3$, this implies that
$$(x-y)(x^2+xy+y^2)=0$$
Now, we 'note' that
$$x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2$$
Thus if the product above is zero, we must have that either
$$x-y=0\qquad\text{or}\qquad\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2=0$$
and this latter can only happen if $x=y=0$. Thus $x=y$ in either case, and $f(x)$ is injective.
I wondered if this approach could tackle $g(x)=x^5$. And, indeed, $x^5=y^5$ implies
$$(x-y)\left(\frac{3}{4}\left[x^2-\frac{1}{2}y^2\right]^2+\frac{1}{4}\left[x+y\right]^4+\frac{2}{3}y^4\right)=0$$
from which we see $g(x)$ is injective. Clearly, this method is going to become extremely cumbersome as the power increases, and we'd like to have elementary calculus to show that $h(x)=x^{2n+1}$ is injective on all of $\mathbb{R}$. But, just for entertainment, I'd like to see
$$x^6+x^5y+x^4y^2+x^3y^3+x^2y^4+xy^5+y^6$$
written as a sum of even-degree terms (if it can) to give an elementary proof - if retractable proof - that $k(x)=x^7$ is indeed injective. How one approached the problem would also be insightful. My attempts haven't been successful.
Any help is appreciated.
| There is a simple recursive reduction, at least for $7$.
Write:
$$\begin{align}
x^6+x^5+x^4+x^3+x^2+x+1 &= \left(x^3+\frac{1}{2}x^2\right)^2 + \frac{3}{4}x^4+x^3+x^2+x+1\\
&=\left(x^3+\frac{1}{2}x^2\right)^2 + \frac{3}{4}\left(x^2+\frac{2}{3}x\right)^2 + \frac{2}{3}x^2+x+1\\
&=\left(x^3+\frac{1}{2}x^2\right)^2 + \frac{3}{4}\left(x^2+\frac{2}{3}x\right)^2+\frac{2}{3}\left(x+\frac{3}{4}\right)^2 + \frac{5}{8}
\end{align}
$$
Basically, the most obvious induction works here. If $p_n(x)=\frac{x^{2n+1}-1}{x-1}$ then
Claim: $p_n(x)$ can be written as a sum of squares with one constant term $\alpha_n$ in the range $(1/2,1]$.
Then prove inductively that since:
$$p_{n+1}(x) = x^2p_n(x) + x + 1$$
And we can just write:
$$\alpha_n x^2 + x+1 = \alpha_n\left(x+\frac{1}{2\alpha_n}\right)^2 + \left(1-\frac{1}{4\alpha_n}\right)$$
Since $\alpha_n\in(1/2,1]$ implies $1-\frac{1}{4\alpha_n}\in (1/2,1]$, we are done. (Well, technically, we need to show true for $n=1$.)
There is a closed form for $\alpha_n$:
$$\alpha_n = \frac{n+2}{2(n+1)} = \frac{1}{2} + \frac{1}{2(n+1)}$$
Since $\alpha_n$ is also used to determine the terms, that is $$\frac{1}{2\alpha_n} = \frac{n+1}{n+2}$, we can get a closed form:
$$1+x+\dots x^{2n} = \frac{1}{2}+\frac{1}{2(n+1)} + \sum_{k=0}^{n-1} \frac{k+2}{2(k+1)}\left(x^{n-k}+ \frac{k+1}{k+2}x^{n-k-1}\right)^2$$
If you must get the result back for two variables, we write $y^{2n}p_n(x/y)$, you get:
$$\frac{x^{2n+1}-y^{2n+1}}{x-y} =\left( \frac{1}{2}+\frac{1}{2(n+1)}\right)y^{2n} + \sum_{k=0}^{n-1} \frac{k+2}{2(k+1)}\left(x^{n-k}y^k+ \frac{k+1}{k+2}x^{n-k-1}y^{k+1}\right)^2$$
An alternative question is whether you can always write it as as a symmetric sum in $x,y$. For example, solving: $y^2+xy+x^2=(ax+by)^2+(bx+ay)^2$ yields $a,b=\frac{\sqrt{2\pm\sqrt{3}}}2$.
We can fudge with $2n$ terms by taking $\frac{f(x,y)+f(y,x)}{2}$ using our original formula, but is that the best we can do, or can we find $n+1$ terms that give us this result so that the terms are "obviously" symmetric, as we see with $n=1$ even, we get ugly results constants.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int_0^{2\pi}\frac{1}{a\cos \theta+b\sin\theta+d}d\theta$ where $a,b,d\in\mathbb{R}$ and $a^2+b^2$\int_0^{2\pi}\frac{1}{a\cos \theta+b\sin\theta+d}d\theta$ where $a,b,d\in\mathbb{R}$ and $a^2+b^2<d^2$
Here, I solve it by Residue Theory. By substituting $d\theta=dz/iz,\cos \theta=(1/2)(z+1/z),\sin\theta=(1/2i)(z-1/z),$ I get
$$\frac{2}{i}\int_{C^+_1(0)}\frac{1}{(a-ib)z^2+2dz+(a+ib)}dz$$
So there are two singularities which are
$$z=\frac{-d\pm\surd{d^2-a^2-b^2}}{a-ib}$$
In order to use the Cauchy's Residue Theorem, I have to identify the singularities that lie inside the unit circle. But, if I take $d=3,a=2,b=2$
$$|\frac{-d-\surd{d^2-a^2-b^2}}{a-ib}|=\frac{4}{2\surd2}>1$$
While if I take $d=-3,a=2,b=2$
$$|\frac{-d-\surd{d^2-a^2-b^2}}{a-ib}|=\frac{2}{2\surd2}<1$$
So I can't sure that this term will be in the unit circle or not. Or should I assume that $d>0$?
| $I
=\int\frac{1}{a\cos \theta+b\sin\theta+d}d\theta
$
Letting
$\tan(\theta/2)
=t
$,
$d\theta
=\frac{2dt}{1+t^2}
$,
$\sin \theta
=\frac{2t}{1+t^2}
$,
and
$\cos \theta
=\frac{1-t^2}{1+t^2}
$.
Therefore
$\begin{array}\\
I
&=\int\frac{1}{a\cos \theta+b\sin\theta+d}d\theta\\
&=\int\frac{1}{a\frac{1-t^2}{1+t^2}+b\frac{2t}{1+t^2}+d}\frac{2dt}{1+t^2}\\
&=2\int\frac{1}{a(1-t^2)+b(2t)+d(1+t^2)}dt\\
&=2\int\frac{1}{a+d+2bt+(d-a)t^2}dt\\
&=2\left( -\frac{\tan^{-1}\left(\frac{(t (a-d)-b)}{\sqrt{-a^2-b^2+d^2}}\right)}{\sqrt{-a^2-b^2+d^2}}\right)
\qquad\text{(according to Wolfram)}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1524329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series Represent $f(x)=\frac{1}{(1-x^2)^4}$ as a power series
$$\sum\limits_{n=0}^{\infty}x^{2n}=\frac{1}{1-x^2}$$
Second derivative is
$$\left(\frac{1}{1-x^2}\right)^{''}=\frac{1}{(1-x^2)^4}\cdot x(1+8x-2x^2-8x^3+x^4)$$
This gives
$$\frac{1}{(1-x^2)^4}=\sum\limits_{n=0}^{\infty}\frac{2n(2n-1)x^{2n-2}}{x(1+8x-2x^2-8x^3+x^4)}$$
I have a proof in combinatorics which involves this series.
How to represent this series using binomial coefficients?
| $$\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$
$$\frac{d^3}{dx^3}(\frac{1}{1-x})=\frac{6}{(1-x)^4}=\sum_{n=3}^{\infty }n(n-1)(n-2)x^{n-3}$$
$$\frac{1}{(1-x)^4}=\frac{1}{6}\sum_{n=3}^{\infty }n(n-1)(n-2)x^{n-3}$$
now replace $x$ by $x^2$
$$\frac{1}{(1-x^2)^4}=\frac{1}{6}\sum_{n=3}^{\infty }n(n-1)(n-2)x^{2n-6}=1+4x^2+10x^4+20x^6+35x^8+..$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the best way to solve an equation of the form $(f(x))^2-a(f(x))+b=x$? On a math contest I was told to solve the equation $$(x^2-3x+1)^2-3(x^2-3x+1)+1=x$$
For this particular problem I simplified by letting $$a\equiv x^2-3x+1$$
Then I continued with $$a^2-3a+1-x=0$$
$$a=\frac{3\pm\sqrt{9-4\left(1-x\right)}}{2}$$
$$3\pm\sqrt{5+4x}=2a=2x^2-6x+2$$
$$\pm\sqrt{5+4x}=2x^2-6x-1$$
I am not sure how to finish off this problem. Also, I have seen a bunch of problems like this in the past. What is the best way to approach a problem like this and also how could I finish solving this problem? Could someone also explain why f(f(x))=x has the same solutions as f(x)=x.
| Let $f(x) = x^2-3x+1$.
Idea
The important observation here is following: If you have a solution $a$ that satisfies $f(a) = a$ then obviously
$a$ also solves the original equation because $f(f(a)) = f(a) = a$.
Finding the $a$ that solve $f(a) = a$ is easy, that is just a quadratic equation. This means you can divide $f(f(x))-x$ by $(x-a)$ for both solutions $a$ and you get a quadratic equation in $x$ which again can be solved easily.
Calculations
So lets first solve $f(x) = x$ This is aequivalent to $0 = x^2-4x+1$ which has the solutions $x_{1,2} = 2 \pm \sqrt{3}$
Now we can divide $f(f(x))-x$ by $(x-x_1)$ and $(x-x_2)$ or combined by $(x-x_1)(x-x_2) = x^2-4x+1 $.
Via polynomial division we get $$(f(f(x))-1)/(x^2-4x+1) = x^2-2x+1$$
Remember taht $f(f(x))-x = 0$ is equivalent now to
$0=f(f(x))-x=(x-x_1)(x-x_2)(x^2-2x+1) = f(f(x))-x$.
As $x^2-2x+1$ has the zeros $x_{3,4} = 1 \pm \sqrt{2}$ we can further factorize
$0=f(f(x))-x=(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ and we have all four solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
integrations by parts. Show $\int_0^\infty 2y^2e^{-y^2} dy = .886$
$2\int_0^\infty y^2e^{-y^2} dy$
Then by parts;
$f=y^2$ $dg=e^{-y^2}$
$df=2y$ $g=-2ye^{-y^2}$
$2\int_0^\infty y^2e^{-y^2} dy$ = $-2y^3e^{-y^2} -\int_0^\infty -4y^2e^{-y^2}$=$-2y^3e^{-y^2} +4\int_0^\infty y^2e^{-y^2}$
equivalently; $-2\int_0^\infty y^2e^{-y^2} dy = -2y^3e^{-y^2}$
or $2\int_0^\infty y^2e^{-y^2} dy = 2y^3e^{-y^2}$
$\left(2(\infty)^3e^{-\infty^2}\right)$-$\left(2(0)^3e^{-0^2}\right)=0-0=0$
Where have I gone wrong?
| $$\int y^2e^{-y^2}dy=\int \underbrace{y}_{=u}\cdot\underbrace{ ye^{-y^2}}_{=v'}dy=[-\frac{1}{2}ye^{-y^2}]+\frac{1}{2}\int_0^\infty e^{-y^2}dy$$
The integral $\int_0^\infty e^{-y^2}dy$ is not that obvious. If you set $I=\int_0^\infty e^{-y^2}dy$, you can compute $I^2$ using polar coordinate. You can also know that $$\int_{\mathbb R}e^{-x^2}dx=\sqrt\pi$$
or make the substitution $u=y^2$ to get$$\int_0^\infty e^{-y^2}dy=\int_0^\infty e^{-y^2}dy=\frac{1}{2}\int_0^\infty u^{\frac{1}{2}-1}e^{-u}du=\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\int_{0}^\infty \mathrm{d}y\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}x=\int_{0}^\infty \mathrm{d}x\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}y=\pi/4$ How can we prove that
\begin{aligned}
&\int_{0}^\infty \mathrm{d}y\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}x\\
=&\int_{0}^\infty \mathrm{d}x\int_{0}^\infty \sin(x^2+y^2)\mathrm{d}y\\=&\cfrac{\pi}{4}
\end{aligned}
I can prove these two are integrable but how can we calculate the exact value?
| I do not know if you are supposed to know this. So, if I am off-topic, please forgive me.
All the problem is around Fresnel integrals. So, using the basic definitions,$$\int_{0}^t \sin(x^2+y^2)dx=\sqrt{\frac{\pi }{2}} \left(C\left(\sqrt{\frac{2}{\pi }} t\right) \sin
\left(y^2\right)+S\left(\sqrt{\frac{2}{\pi }} t\right) \cos
\left(y^2\right)\right)$$ where appear sine and cosine Fresnel integrals. $$\int_{0}^\infty \sin(x^2+y^2)dx=\frac{1}{2} \sqrt{\frac{\pi }{2}} \left(\sin \left(y^2\right)+\cos
\left(y^2\right)\right)$$ Integrating a second time,$$\frac{1}{2} \sqrt{\frac{\pi }{2}}\int_0^t \left(\sin \left(y^2\right)+\cos
\left(y^2\right)\right)dy=\frac{\pi}{4} \left(C\left(\sqrt{\frac{2}{\pi }}
t\right)+S\left(\sqrt{\frac{2}{\pi }} t\right)\right)$$ $$\frac{1}{2} \sqrt{\frac{\pi }{2}}\int_0^\infty \left(\sin \left(y^2\right)+\cos
\left(y^2\right)\right)dy=\frac{\pi}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $E((X+Y)^p)\leq 2^p (E(X^p)+E(Y^p))$ for nonnegative random variables $X,Y$ and $p\ge0$ Suppose $X \geq 0$ and $Y \geq 0$ are random variables and that $p\geq 0$
*
*Prove
$$E((X+Y)^p)\leq 2^p (E(X^p)+E(Y^p))$$
Proof
Since $(X+Y)^p \leq (2 \> \max\{X,Y\})^p=2^p \> \max \{X^p,Y^p\}\leq 2^p(X^p+Y^p)$ $ \implies E((X+Y)^p)\leq 2^p (E(X^p)+E(Y^p))$
*If $p>1$ the factor $2^p$ may be replaced by $2^{p-1}$
*If $0 \leq p \leq 1$ the factor $2^p$ can be replaced by $1$
Need help with part 2 and 3 any suggestions
| If $p>1$, then by Holder inequality,
\begin{align*}
X+Y &\le (X^p+Y^p)^{\frac{1}{p}} 2^{1-\frac{1}{p}}.
\end{align*}
That is,
\begin{align*}
(X+Y)^p \le 2^{p-1} (X^p+Y^p).
\end{align*}
For $0 \le p \le 1$, we note that
\begin{align*}
\left(\frac{X}{X+Y} \right)^p + \left(\frac{Y}{X+Y} \right)^p \ge \frac{X}{X+Y}+ \frac{Y}{X+Y}=1,
\end{align*}
and then
\begin{align*}
(X+Y)^p \le X^p+Y^p.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
An Algebraic Proof that $|y^3 - x^3| \ge |(y - x)|^3/4 $ I can prove this using calculus, but not by simple algebra: can anyone help ?
Calculus Proof:
Fix the separation of $x$ and $y$ so that $y = x + d$ with $d>0$ ($ \implies y > x \implies y^3 > x^3$) and now consider $ f(x) = y^3 - x^3 = (x+d)^3 - x^3$.
$ f(x) = (x+d)^3 - x^3$ = $3x^2d + 3d^2x +d^3$ which is a quadratic in $x$ with a minimum given by $f`(x) = 0 = 6xd + 3d^2$ giving $x = - d/2$ and therefore $y = d/2$. This is not surprising considering the geometry: it says that for a fixed separation of $x$ and $y$, $y^3 - x^3$ is minimised when $x$ and $y$ are symetrically placed around the inflexion point of the cubic.
So, $y^3 - x^3 = d^3/8 - (-d^3)/8 = d^3/4$ and since this is the minimum value it follows that for $x \in (-\infty, +\infty)$ and $y > x$ then $y^3 - x^3 \ge (y - x)^3/4 $, so that $|y^3 - x^3| \ge |(y - x)|^3/4 $ whether $y > x$ or $y < x$ (and clearly this is true for $y = x$.)
(The original interest in the inequality comes from this question: cauchy sequence on $\mathbb{R}$)
| $ |y^3-x^3| \geq |y-x|^3/4 \\
\iff |y-x|(x^2+xy+y^2) \geq |y-x|(y-x)^2/4 \\
\iff 4(x^2+xy+y^2) \geq (x^2-2xy+y^2)$
(Assuming that $y\neq x$)
$\iff 3x^2+6xy+3y^2 \geq 0\\
\iff (x+y)^2 \geq 0$
EDIT
$x^2+xy+y^2 = (x+y/2)^2 + 3y^2/4 \geq 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Minimizing $f(x)=A^{\frac{tx-1}{x-1}} \left( c^x \frac{\Gamma(0.5+x)}{\sqrt{\pi}} \right)^{\frac{1-t}{x-1}}$ subject to the constraint Let $f(r)$ be a function defined as follows
\begin{align}
f(x)=A^{\frac{tx-1}{x-1}} \left( c^x \frac{\Gamma(0.5+x)}{\sqrt{\pi}} \right)^{\frac{1-t}{x-1}}
\end{align}
where $0 < A,c$ and $ t\in (0,1)$.
I want to solve
\begin{align}
\min_{\frac{1}{t}<x} f(x)
\end{align}
My approach:
Let $g(x)=\ln(f(x))$ then both $g(x)$ and $f(x)$ have the same minimum where
\begin{align}
g(x)=\frac{tx-1}{x-1} \ln(A)+\frac{(1-t)x}{x-1} \ln(c)+\frac{1-t}{x-1} \ln \left(\Gamma(0.5+x) \right)-\frac{1-t}{2(x-1)}\ln(\pi)
\end{align}
and
\begin{align}
&g'(x)=\\
&=\frac{1-t}{(x-1)^2} \ln(A)-\frac{1-t}{(x-1)^2} \ln(c)+(1-t)\frac{(x-1)\psi^{(0)}(x+0.5)-\log(\Gamma(x+0.5))}{(x-1)^2}+\frac{1-t}{2(x-1)^2}\ln(\pi)\\
&=\frac{1-t}{(x-1)^2} \left(\ln \left( \frac{ \sqrt{\pi} A}{c\Gamma(x+0.5)} \right)+ (x-1)\psi^{(0)}(x+0.5)\right)
\end{align}
where $\psi(x)$ is the so called digamma function.
So, this means we have to focus on
\begin{align}
h(x)=\ln \left( \frac{ \sqrt{\pi} A}{c\Gamma(x+0.5)} \right)+ (x-1)\psi^{(0)}(x+0.5)
\end{align}
for $\frac{1}{t} <x$.
But how to solve $h(x)=0$ or say for what $x$ is $h(x)>0$ ???
If this impossible to do then the approximated solution is also fine?
I also feel that there might be a simpler approach with out using derivative.
Thank you for any help.
| Hopefully the following can somehow help or give some idea.
Rewrite
\begin{align}
h(x)=\ln \left( \frac{ \sqrt{\pi} A}{c\Gamma(x+0.5)} \right)+ (x-1)\psi^{(0)}(x+0.5)
\end{align}
as
\begin{align}
h(x)= (x-1)\psi^{(0)}(x+0.5)-
\ln \left( \Gamma(x+0.5) \right) +\alpha
\end{align}
where $\alpha =\ln \left( \frac{ \sqrt{\pi} A}{c} \right)$.
Plot using Mathematica
*
*graph of $h(x)$ intersection with $0$-plane
Plot3D[{[Alpha] + (x - 1) PolyGamma[0, 1/2 + x] -
Log[Gamma[1/2 + x]], 0}, {x, 0, 3}, {[Alpha], -2, 2}]
*Contour plot of $h(x)=0$
ContourPlot[{[Alpha] + (x - 1) PolyGamma[0, 1/2 + x] -
Log[Gamma[1/2 + x]], 0}, {x, 0, 5}, {[Alpha], -3, 4}]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the following limit 0/0 $$\lim_{t\to 0} \frac{\left(\sqrt{t+9}-3 \sqrt[3]{2 t+1}\right)}{\sqrt[3]{t+8}-2 \sqrt[3]{3 t+1}}$$
I tried to multiply the nominator by conjugation but got even bigger expression.
| Let $a=\sqrt{t+9}$, $b=3\sqrt[3]{2t+1}$, $c=\sqrt[3]{t+8}$ and $d=2\sqrt[3]{3t+1}$.
We have
$$ \lim_{t\to 0} \frac{a-b}{c-d} = \lim_{t\to 0}\left(\frac{a^6 -b^6}{c^3-d^3}\right)\left( \frac{c^2+cd + d^2}{a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5}\right)
= \lim_{t\to 0}\left(\frac{(t+9)^3 -(3^6(2t+1)^2)}{t+8 - 8(3t+1)}\right)\left( \frac{c^2+cd + d^2}{a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5}\right)=\left(\frac{3(9^2) -4(3^6)}{-23}\right)\left(\frac{3(4)}{6(3^5)}\right)=\frac{22}{23}.
$$
Note that on the last line we used: $$\lim_{t\to 0}\frac{(t+9)^3 -(3^6(2t+1)^2)}{t+8 - 8(3t+1)} = \lim_{t\to 0}\frac{(t^3+3(9)t^2+3(9)^2t + 9^3) - (3^6(4t^2+4t+1))}{-23t} = \lim_{t\to 0}\frac{(t^3+3(9)t^2+3(9)^2t) - (3^6(4t^2+4t))}{-23t} = \lim_{t\to 0}\frac{(t^2+3(9)t^1+3(9)^2) - (3^6(4t+4))}{-23} = \frac{3(9^2) -4(3^6)}{-23}$$
and $\lim_{t\to 0} a = \lim_{t\to 0} b=3$ and $\lim_{t\to 0} c = \lim_{t\to 0} d=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Compute the limit of the sequence of functions of $\lim_{n\to \infty} f_n(x) = \frac{x^2}{x^2+(1-nx)^2}$ Compute the limit of the sequence of functions of
$$
\lim_{n\to \infty} f_n(x) = \frac{x^2}{x^2+(1-nx)^2}.
$$
Attempt.
$$\lim_{n\to \infty} f_n(x) = \lim_{n\to \infty} \frac{x^2}{x^2+(1-nx)^2} = \lim_{n\to \infty} \frac{x^2}{x^2+n^2x^2-2nx + 1} $$
$$= \lim_{n\to \infty} \frac{x^2}{x^2+\frac{x^2}{n^2}-\frac{2x}{n} + \frac{1}{n}} = \lim_{n\to \infty} \frac{x^2}{x^2} = 1$$
| For $x=0$ and all $n\in \mathbf{N}$, we see that $$f_{n}(0)=\frac{0}{0+1}=0$$ For $x\neq 0$, then we note that $$\lim_{n\rightarrow \infty}(1-nx)$$ is $+\infty$ for $x<0$ and is $-\infty$ for $x>0$. Thus, $$\lim_{n\rightarrow \infty}(1-nx)^{2}=\left(\lim_{n\rightarrow \infty}(1-nx)\right)^{2}=+\infty$$ and $$\lim_{n\rightarrow \infty} (x^{2}+(1-nx)^{2})=x^{2}+\lim_{n\rightarrow \infty}(1-nx)^{2}=x^{2}+\infty=\infty$$ (remember that while we take the sequential limit, we are holding $x$ fixed). It follows that $$\lim_{n\rightarrow \infty}\frac{x^{2}}{x^{2}+(1-nx)^{2}}=\frac{\lim_{n\rightarrow \infty}x^{2}}{\lim_{n\rightarrow \infty}(x^{2}+(1-nx)^{2})}=\frac{x^{2}}{\infty}=0$$ since we have restricted our attentions to $x\neq 0$. Therefore, $$\lim_{n\rightarrow \infty}f_{n}(x)=0$$ for all real $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove by induction $ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $
Prove by induction
$$ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $$
What I have for now:
$$ \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} + \sin(n + 1)x = \frac {\sin (\frac {n + 2} {2} x)} {\sin \frac{x}{2}} \sin \frac{(n + 1)x}{2}$$
Letting $y = \frac {(n + 1)x} {2} $
$$\frac {\sin y} {\sin \frac{x}{2}} \sin (y - \frac{x}{2}) + \sin2y =
\frac {\sin (y + \frac {x} {2} )} {\sin \frac{x}{2}} \sin y $$
Then I used $\sin (\alpha + \beta)$ and $\sin (\alpha - \beta)$ formulas, bit they didn't help.
| You can also see that
\begin{align}
\sum_{k=0}^{n}\sin kx & =\Im \sum_{k=0}^{n}\mathrm{e}^{\mathrm{i}kx}\\
& = \Im \sum_{k=0}^{n}\left(\mathrm{e}^{\mathrm{i}x}\right)^k\\
& = \Im\frac{\left(\mathrm{e}^{\mathrm{i}x}\right)^{n+1}-1}{\mathrm{e}^{\mathrm{i}x}-1}\\
& = \Im\frac{\mathrm{e}^{\mathrm{i}(n+1)x}-1}{\mathrm{e}^{\mathrm{i}x}-1}\\
& = \Im\frac{\mathrm{e}^{\mathrm{i}\frac{n+1}{2}x}}{\mathrm{e}^{\mathrm{i}\frac{x}{2}}}\frac{\mathrm{e}^{\mathrm{i}\frac{n+1}{2}x}-\mathrm{e}^{-\mathrm{i}\frac{n+1}{2}x}}{\mathrm{e}^{\mathrm{i}\frac{x}{2}}-\mathrm{e}^{-\mathrm{i}\frac{x}{2}}}\\
& = \Im\mathrm{e}^{\mathrm{i}\frac{nx}{2}}\frac{\sin\left(\frac{n+1}{2}x\right)}{\sin\left(\frac{x}{2}\right)}\\
& = \sin\left(\frac{nx}{2}\right)\frac{\sin\left(\frac{n+1}{2}x\right)}{\sin\left(\frac{x}{2}\right)}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$ The line $\dfrac{x+6}{5}=\dfrac{y+10}{3}=\dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the remaining sides.
My Attempt:
Let the right angled triangle be $ABC$,right angled at $B(7,2,4).$The remaining two sides $AB$ and $BC$ of the right angled triangle will pass through the vertex $B(7,2,4)$.
Let the equation of $AB$ be $\dfrac{x-7}{a_1}=\dfrac{y-2}{b_1}=\dfrac{z-4}{c_1}$
and the equation of $BC$ be $\dfrac{x-7}{a_2}=\dfrac{y-2}{b_2}=\dfrac{z-4}{c_2}$
As $AB$ and $BC$ are perpendicular to each other. So $a_1a_2+b_1b_2+c_1c_2=0.....(1)$
Also as the triangle is a isosceles right triangle, so angle between $AC$ and
$BC$ will be same as the angle between $CA$ and $BA$. Let it be $\theta$.
So $\cos\theta=\dfrac{5a_1+3b_1+8c_1}{\sqrt{5^2+3^2+8^2}\sqrt{a_1^2+b_1^2+c_1^2}}=\dfrac{5a_2+3b_2+8c_2}{\sqrt{5^2+3^2+8^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
Squaring both sides we get
$\dfrac{(5a_1+3b_1+8c_1)^2}{a_1^2+b_1^2+c_1^2}=\dfrac{(5a_2+3b_2+8c_2)^2}{a_2^2+b_2^2+c_2^2}.............(2)$
But I am stuck here.I don't know how to find $a_1,b_1,c_1,a_2,b_2$ and $c_2$. Please help me. Thanks.
| There is a much simpler and efficient way to solve this problem.
First, find the foot of the perpendicular from the point B to the hypotenuse, and let's call it $D$. Now calculate the length of the perpendicular $BD$ and let's call it $d$.
Now since the triangle is isosceles and right-angled it should be obvious that the points $A$ and $C$ will be obtained by traversing a distance $d$ along the given line.
So if the direction cosines of the line are $l,m,n$ and the point $D$ is at $(x_0,y_0,z_0)$ then clearly we have the coordinates of $A$ and $C$ as $(x_0\pm ld,y_0\pm md,z_0\pm nd)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $\frac{1+4n^2}{2+2n^2}$ is a cauchy sequence? I know that the following sequence is a
$$\frac{1 + 4n^2}{2+2n^2}$$
How can I show that it is a cauchy sequence - well it has to cause its convergent but I want to understand it with the cauchy definition.
I know that i have to get the $|a_m - a_n | < \varepsilon$ meaning that the distance between two elements has to be smaller than epsilon. But how big is epsilon? I only know that there is an element called $N_0$ when this will become true.
This is where my problem starts - what will be the next step?
So I have $ \frac{1+4n^2+1}{2+2n^2+1} - \frac{1+4n^2}{2+2n^2} < \varepsilon $ but I don't know how much $\varepsilon$ is nor do I know where $N_0$ is.
| For any $p > 1$, we have
\begin{align}
& \left|\frac{1 + 4(n + p)^2}{2 + 2(n + p)^2} - \frac{1 + 4n^2}{2 + 2n^2} \right|\\
= & \frac{2 + 2n^2 + 8(n + p)^2 + 8n^2(n + p)^2 - 2 - 2(n + p)^2- 8n^2 - 8n^2(n + p)^2}{4(1 + n^2)(1 + (n + p)^2)} \\
= & \frac{8p(2n + p) - 2p(2n + p)}{4(1 + n^2)(1 + (n + p)^2)} \\
= & \frac{3p(2n + p)}{2(1 + n^2)(1 + (n + p)^2)} \\
< & \frac{3p(2n + 2p)}{2n^2(n + p)(n + p)} \\
= & \frac{3}{n^2} \times \frac{p}{n + p} \times \frac{n + p}{n + p}\\
< & \frac{3}{n^2}.
\end{align}
From which you can see that the sequence is Cauchy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Matrix multiplication of $4 \times 1$ matrix with $4 \times 4$ matrix I'm new to matrix multiplication and just wondered how I would evaluate the following:
$$
\begin{pmatrix}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & -1& 0 & 0 \\
-1& 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
a \\
b \\
c \\
d \\
\end{pmatrix}
$$
and even:
$$
\begin{pmatrix}
a & b & c & d
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & -1& 0 & 0 \\
-1& 0 & 0 & 0
\end{pmatrix}
$$
| There is a formula to compute the $(i, j)$th entry of $C = AB$ (provided the sizes of $A$ and $B$ match): assume $A$ is $n \times p$ and $B$ is $p \times m$, then $C$ is $n \times m$, and
$$c_{i, j} = \sum_{k = 1}^p a_{\color{red}{i}k}b_{k\color{red}{j}}.$$
Take your first question for example, the result should be a $4 \times 1$ matrix, it is quite straightforward to compute as follows:
$$
\begin{pmatrix}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & -1& 0 & 0 \\
-1& 0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
a \\
b \\
c \\
d \\
\end{pmatrix}
= \begin{pmatrix}
0 \times a + 0 \times b + 0 \times c + 1 \times d \\
0 \times a + 0 \times b + 1 \times c + 0 \times d \\
0 \times a + (-1) \times b + 0 \times c + 0 \times d \\
(-1) \times a + 0 \times b + 0 \times c + 0 \times d
\end{pmatrix}
= \begin{pmatrix}
d \\
c \\
-b \\
-a
\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Upper bound of $a_{n+1}=a_n + \frac{1}{a_n}$ How can I prove:
If $$a_0=\alpha>0\quad and\quad a_{n+1}=a_n + \frac{1}{a_n}$$, then $$a_n^2<\alpha^2+2n+\frac{1}{\alpha^2}+\frac{1}{2}ln\left ( \frac{2n}{\alpha^2}+1 \right )$$
?
I'll really appreciate your help. Thanks.
| We have $\displaystyle a_{n+1}^2=a_n^2+\frac{1}{a_n^2}+2$. Hence for $n\geq 1$
$$a_n^2=a_0^2+2n+\sum_{k=0}^{n-1}\frac{1}{a_k^2}$$
This imply that $\displaystyle a_k^2\geq a_0^2+2k$ for $k\geq 1$. Hence
$$a_n^2\leq a_0^2+2n+\frac{1}{a_0^2}+\sum_{k=1}^{n-1}\frac{1}{a_0^2+2k}$$
Now $$\sum_{k=1}^{n-1}\frac{1}{a_0^2+2k}\leq \int_0^{n}\frac{dt}{a_0^2+2t}=\frac{1}{2}\log (\frac{2}{a_0^2}n+1)$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Taking the PMF to find the CDF, probability. Question:We toss a fair coin three times. Let X be the number of heads minus the number of tails obtained.
Write the cumulative distribution function of X.
What I already know: I know how to get the pmf from this problem.
$$
\ P(X=x) = \left\{
\begin{array}{l l}
\ {3 \choose 0} (1/2)^3 & \quad \text{for $k = -3$}\\
{3 \choose 1} (1/2)^3 & \quad \text{for $k = -1$}\\
{3 \choose 2} (1/2)^3 & \quad \text{for $k = 1$}\\
{3 \choose 3} (1/2)^3 & \quad \text{for $k = 3$}\\
\end{array} \right.
$$
I'm not sure how my professor got this cdf:
$$
\ P(X \le x) = \left\{
\begin{array}{l l}
\ 0 & \quad \text{for $x \lt -3$}\\
(1/2)^3 & \quad \text{for $-3 \le x \lt -1$}\\
(1+3) (1/2)^3 & \quad \text{for $-1 \le x \lt 1$}\\
(1+3+3) (1/2)^3 & \quad \text{for $1 \le x \lt 3$}\\
(1+3+3+1)(1/2)^3 & \quad \text{for $3 \le x$}\\
\end{array} \right.
$$
| The c.d.f. is not $x\mapsto \Pr(X=x)$; it is $x\mapsto\Pr(X\le x)$. So, for example, its value at $x=2$ is $\Pr(X\le 2) = \Pr(X=-3) + \Pr(X=-1) + \Pr(X=1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do I find the maximum perimeter of a rectangle inscribed in an ellipse? The problem I've been stuck on is this:
A rectangle is inscribed in the ellipse $$\frac{x^2}{20} + \frac{y^2}{12} = 1$$ What is the maximum perimeter of the rectangle?
I don't even know if I'm taking the right approach. So far, I've been trying to solve for $y$, giving me $y = \sqrt{12-(3/5)x}$, and plugging that into the equation $P = 4x + 4y$, which should be the equation for the perimeter of an inscribed rectangle. I then took the derivative of $P$ after plugging in the equation for $y$, giving me $$P' = 4 - \frac{12x}{5\sqrt{12-(3/5)x}}.$$ To find a maximum, I'd set the equation to zero right? Well, I don't know where to go from this step, since simplifying from here only seems to make it harder.
Any help would be much appreciated, even a nudge in the right direction. I have no idea where to go from here, or even if I got to the right place. Thanks for your time
| One simple way of solving this problem is by Lagrange multipliers method. Note that if $(x,y)$ is in the first quadrant on the ellipse $x^2/a^2+y^2/b^2 = 1$, then the perimeter of the inscribed rectangle represented by $(x,y)$ is simply $4(x+y)$. Therefore you want to maximize $x+y$ given the constraint that $x^2/a^2+y^2/b^2 = 1$. Define
$$
f(x,y,\lambda) = x+y -\lambda\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)
$$
Hence by maximizing $f$
$$
1 = \frac{2x\lambda}{a^2}=\frac{2y\lambda}{b^2}\Longrightarrow \frac{x}{a} = \frac{y}{b}\left(\frac{a}{b}\right)
$$
but then
$$1=\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{y^2}{b^2}\left(1+\frac{a^2}{b^2}\right)\Longrightarrow y=\frac{b^2}{\sqrt{a^2+b^2}}, \quad x=\frac{a^2}{\sqrt{a^2+b^2}}$$
The maximum perimeter is therefore $4(x+y) = 4\sqrt{a^2+b^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
If $z_1$ and $z_2$ are two complex numbers,and if $z_1^3-3z_1^2z_2=2,3z_1z_2^2-z_2^3=11$, If $z_1$ and $z_2$ are two complex numbers,and if $z_1^3-3z_1^2z_2=2,3z_1z_2^2-z_2^3=11$,then find the value of $|z_1^2+z_2^2|$.
$z_1^3-3z_1^2z_2=2$
$3z_1z_2^2-z_2^3=11$
Adding them,we get
$z_1^3-3z_1^2z_2+3z_1z_2^2-z_2^3=13$
$(z_1-z_2)^3=13$
We need to find $|z_1^2+z_2^2|$,
I could not solve from here,I am stuck.Please help me.Thanks.
| dividing both the equations we get
$$\frac{z_1^2}{z_2^2} \times \frac{z_1-3z_2}{3z_1-z_2}=\frac{2}{11}$$ $\implies$ assuming $\frac{z_1}{z_2}=p$ we get
$$\frac{p^2(p-3)}{3p-1}=\frac{2}{11}$$ i.e.,
$$11p^3-33p^2-6p+2=0$$ and this cubic equation has three real roots which means
$\frac{z_1}{z_2}$ is Real.
Now $$|z_1^2+z_2^2|=|p^2z_2^2+z_2^2|=|z_2^2|(p^2+1) \tag{1}$$
also $$(z_1-z_2)^3=13$$ $\implies$
$$(pz_2-z_2)^3=13$$, so
$$z_2^3=\frac{13}{(p-1)^3}$$ so
$$|z_2|^2=\frac{13^{\frac{2}{3}}}{(p-1)^2}$$ substituting this in $(1)$ we get
$$|z_1^2+z_2^2|=\frac{13^{\frac{2}{3}}(p^2+1)}{(p-1)^2}$$ so we have three answers for each $p \in \mathbb{R}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Can this approximation be made more formal? When considering oscillating systems in physics, we end up with some response function like $$F(\omega) = \frac{\omega^2}{(\omega_0^2 - \omega^2)^2 + (\omega/\tau)^2},$$ where $\omega_0$ and $\tau$ are characteristic properties of the system, and $\omega$ is the driving frequency. We are generally interested in the behavior of $F(\omega)$ close to the maximum, $\omega = \omega_0$, which is the resonant frequency. However, the usual approach to approximate $F(\omega)$ is the following $$F(\omega) = \frac{\omega^2}{(\omega_0 - \omega)^2 (\omega_0 + \omega)^2 + (\omega/\tau)^2} \approx \frac{\omega_0^2}{4 \omega_0^2 (\omega_0 - \omega)^2 + (\omega_0/\tau)^2} = \frac{\frac{1}{4}}{(\omega_0 - \omega)^2 + (1/2\tau)^2}.$$
Is there a way to formalize and justify the above approximation? Could this be a form of Pade approximation?
| Let $F(\omega)$ be given by
$$\begin{align}
F(\omega)&=\frac{\omega^2}{(\omega^2-\omega_0^2)^2+(\omega/\tau)^2}\\\\
&=\frac{1/4}{(\omega_0/2)^2(1-\omega/\omega_0)^2(1+\omega/\omega_0)^2+(1/2\tau)^2}
\end{align}$$
Now, let's denote $\omega/\omega_0=1+x$. Then, we have
$$\begin{align}
F(\omega)&=F(\omega_0(1+x))\\\\
&=\frac{1/4}{(\omega_0x)^2\,(1+x/2)^2+(1/2\tau)^2}\\\\
&=\frac{1/4}{(\omega_0x)^2+(1/2\tau)^2}\times\left(\frac{(\omega_0x)^2+(1/2\tau)^2}{(\omega_0x)^2\,(1+x/2)^2+(1/2\tau)^2}\right)\\\\
&=\frac{1/4}{(\omega_0x)^2+(1/2\tau)^2}\times\left(1+O\left(x^3\right)\right)\\\\
&=\frac{1/4}{(\omega-\omega_0)^2+(1/2\tau)^2}\times\left(1+O\left((\omega-\omega_0)^3\right)\right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Finding the $n^{\text{th}}$ term of $-1,-1,-1,-1,1,1,1,1,...$ as a repeating 8-block In my work I came across that sequence
$-1,-1,-1,-1,1,1,1,1,\dots$ and repeating this 8-block so on forever
Now I cant find an ( e.g. trigonometric/complex ) expression
$f(n)$ ( e.g. $f(n) =(-1)^g(n)$ ) which gives me the sequence starting with
$n=2,3,4,5,6,7,8,9,…$
and so on forever.
| By symmetry, we can assume
$$
f(n) =
a_1 \sin\frac{\pi n}{4}
+a_3 \sin\frac{3\pi n}{4}
+b_1 \cos\frac{\pi n}{4}
+b_3 \cos\frac{3\pi n}{4}.
$$
Then
$$
\begin{aligned}
f(1) &= \frac{a_1}{\sqrt{2}} +\frac{a_3}{\sqrt{2}} + \frac{b_1}{\sqrt{2}} -\frac{b_3}{\sqrt{2}} = -1\\
f(2) &= a_1 - a_3 = -1\\
f(3) &= \frac{a_1}{\sqrt{2}} +\frac{a_3}{\sqrt{2}} - \frac{b_1}{\sqrt{2}} +\frac{b_3}{\sqrt{2}} = -1 \\
f(4) &= -b_1 - b_3 = -1.
\end{aligned}
$$
So
$$
\begin{aligned}
a_1 &= -\frac{1}{2} -\frac{1}{\sqrt 2}, \\
a_3 &= +\frac{1}{2} -\frac{1}{\sqrt 2}, \\
b_1 &= b_3 = \frac{1}{2}.
\end{aligned}
$$
So
$$\begin{aligned}
f(n)
&=
\left( -\frac{1}{2} -\frac{1}{\sqrt 2} \right) \sin\frac{\pi n}{4}
+
\left( \frac{1}{2} -\frac{1}{\sqrt 2} \right) \sin\frac{3 \pi n}{4}
+
\frac{1}{2} \left( \cos\frac{\pi n}{4} + \cos\frac{3 \pi n}{4} \right) \\
&=
\sin\frac{\pi n}{4} \cos\frac{\pi n}{2}
-\sqrt 2 \cos\frac{\pi n}{4} \sin\frac{\pi n}{2}
+
\cos\frac{\pi n}{4} \cos\frac{\pi n}{2} \\
&=
\sqrt 2 \left[ \sin\frac{\pi (n+1)}{4} \cos\frac{\pi n}{2}
- \cos\frac{\pi n}{4} \sin\frac{\pi n}{2} \right].
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
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Why is there at least a $50$% chance that greatest common divisor $(n,x-y)$ is a factor of $n$ not equal to $0$, $1$, or $n$? Given integers $x$, $y$, and $n$ such that $x^2\equiv y^2 ($mod$ n)$, there is at least a $50$% chance that greatest common divisor $(n,x-y)$ is a factor of $n$ not equal to $1$, $0$, or $n$. Why is it at least $50$%?
I think it is because if $x\not\equiv±y($mod$ n)$, then gcd$(n,x-y)$ is a factor of $n$ not equal to $1$, $0$, or $n$, but if $x\equiv±y($mod$ n)$, then gcd$(n,x-y)$ must equal $n$, $1$ or $0$.
My Guess:
Let’s say we find an $x$ and $y$ such that $x^2 ≡ y^2($mod$ n)$.
Then $(x – y)(x + y) = 0 ($mod$ n)$
First, if either $x$ or $y$ is $0$, then gcd$(n, x - y) = 0$
If $x ≡ y($mod$ $n)$, then gcd$(n, x - y) = 0$ or $n$.
If $x ≡ -y($mod$ $n)$, then gcd$(n, x - y) = 1$ or $n$
If $x ≡ y(mod$ $n)$, then gcd$(n, x + y) = 1$ or $n$
If $x ≡ -y(mod$ $n)$, then gcd$(n, x + y) = 0$ or $n$
But if $x$ is not congruent to $± y($mod $n)$, then gcd$(n, x + y)$ does not equal $0$, $1$, or $n$.
So when $x^2 ≡ y^2(mod$ $n)$, there is at least a $50$% chance that gcd$(n, x + y)$ is a non-trivial factor of $n$ (a factor not equal $0$, $1$, or $n$).
Is this correct?
| The given statement is correct for numbers $n$ with at least two distinct odd prime factors. (to be more precise, if there is no primitive root modulo $n$ , but the case of $n$ odd and $n$ has two distinct prime factors is the most important in practice)
Suppose, $n$ is divisible by the odd distinct primes $p$ and $q$ and $x^2\equiv y^2\ ($mod $n$).
If $gcd(x,y)\ne1$, then there exists a prime $r$ such that $r|x$ and $r|y$.
In such a case case, $r$ also divides $x-y$ and therefore $x^2-y^2$. Hence $r|gcd(n,x-y)$.
If $x\equiv y\ ($mod $n\ )$, then $gcd(x-y,n)=$$n$. Because $n > 2r$,
this occurs with probability at most $\frac{1}{2}$.
If $gcd(x,y)=1$, then we have that $gcd(n,y)=1$. If it didn't, there would be a prime $r$ dividing $n$ and $y$, so we would have that $x^2\equiv 0\ ($mod $r)$, implying $x\equiv 0\ ($mod $r )$, which is a contradiction to $gcd(x,y)=1$.
Hence, $x^2\equiv y^2\ (\ mod\ n)$ can be transformed into $z^2\equiv 1\ (\ mod\ n)$.
The solution of the system $p\equiv 1\ (\ mod\ n)$ $q\equiv -1\ (\ mod\ n)$ and the solution of the system $p\equiv-1\ (\ mod\ n)$ $q\equiv 1\ (\ mod\ n)$ are two non-trivial solutions. So, the case $gcd(x,y)=1$ gives a chance of at least $\frac{1}{2}$ to find a factor because the equation $z^2\equiv 1\ (\ mod\ n)$ has at least $4$ solutions, two of which are trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Volume between cone and sphere - First octant Find the volume between $z=\sqrt{x^2+y^2}$ and the sphere $x^2+y^2+z^2=1$ that lies in the first octant using cylindrical coordinates.
So I found the intersection and got $r=\frac{\sqrt2}{2}$.
I know theta has to be between $0$ and $\frac{\pi}{2}$ but not sure about z
| Using Cylindrical coordinates:
$\displaystyle\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r}^{\sqrt{1-r^2}}rdzdrd\theta=\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sqrt{2}}}{r(\sqrt{1-r^2}-r)}drd\theta=\int_{0}^{\frac{\pi}{2}}[-\frac{1}{3}(1-r^2)^{\frac{3}{2}}-\frac{r^3}{3}]_{0}^{\frac{1}{\sqrt{2}}}=\frac{\pi}{2}[(-\frac{1}{3(2\sqrt{2})}-\frac{1}{6\sqrt{2}})-(\frac{1}{3})]=\frac{\pi}{2}[\frac{-2\sqrt{2}}{12}+\frac{1}{3}]=\frac{\pi(2-\sqrt{2})}{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Calculate Point Coordinates As you can see, In the image a rectangle gets translated to another position in the coordinates System.
The origin Coordinates are A1(8,2) B1(9,3) from the length 7 and the height 3 you can also guess the vertices of the rectangle.
Now the Rectangle gets moved.
Now A1 is at A2(16,9) and B1 is located at B2(16,11).
It means that the rectangle got translated, rotated and stretched.
How can I calculate the Coordinates, of the left-upper corner?
I first tried to calculate the stretching-factor but then I got stuck when I trying to calculate the angle and translation
Thanks for your help
| Your transformation contains translation (2 parameter), rotation (1 parameter) and stretching, which I hope means scaling (1 parameter).
This in general is no linear but an affine transform, except for the case that the origin gets mapped to the origin, which I doubt here.
The transform would be like this, using homogeneous coordinates:
$$
T = \left(
\begin{matrix}
t_{11} & t_{12} & t_{13} \\
t_{21} & t_{22} & t_{23} \\
0 & 0 & 1
\end{matrix}
\right)
\quad (*)
$$
Assuming we first do the translation, then the roation, then the scaling we get a matrix:
\begin{align}
T
&=
T_\text{scale} \, T_\text{rot} \, T_\text{trans} \\
&=
\left(
\begin{matrix}
s & 0 & 0 \\
0 & s & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 0 & t_x \\
0 & 1 & t_y \\
0 & 0 & 1
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
s \cos \theta & -s \sin \theta & 0 \\
s \sin \theta & s \cos \theta & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 0 & t_x \\
0 & 1 & t_y \\
0 & 0 & 1
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
s \cos \theta & -s \sin \theta & s(t_x \cos \theta - t_y \sin \theta) \\
s \sin \theta & s \cos \theta & s(t_x \sin \theta + t_y \cos \theta) \\
0 & 0 & 1
\end{matrix}
\right)
\end{align}
Your two points give four equations, it might be enough to determine the four parameters $t_x, t_y, \theta, s$.
Inserting the points and their images
$$
\left(
\begin{matrix}
s \cos \theta & -s \sin \theta & s(t_x \cos \theta - t_y \sin \theta) \\
s \sin \theta & s \cos \theta & s(t_x \sin \theta + t_y \cos \theta) \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
x_1 \\
y_1 \\
1
\end{matrix}
\right)
=
\left(
\begin{matrix}
x_2 \\
y_2 \\
1
\end{matrix}
\right)
$$
leads to the equations
$$
\cos \theta - \sin \theta = 0 \\
\sin \theta + \cos \theta = 2
$$
where the second one has no real solution. So this is not working.
Looking at the rectangle, the ratio of the sides seems to have changed.
This means we have
\begin{align}
T
&=
T_\text{scale} \, T_\text{rot} \, T_\text{trans} \\
&=
\left(
\begin{matrix}
s & 0 & 0 \\
0 & t & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 0 & t_x \\
0 & 1 & t_y \\
0 & 0 & 1
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
s \cos \theta & -s \sin \theta & 0 \\
t \sin \theta & t \cos \theta & 0 \\
0 & 0 & 1
\end{matrix}
\right)
\left(
\begin{matrix}
1 & 0 & t_x \\
0 & 1 & t_y \\
0 & 0 & 1
\end{matrix}
\right)
\\
&=
\left(
\begin{matrix}
s \cos \theta & -s \sin \theta & s(t_x \cos \theta - t_y \sin \theta) \\
t \sin \theta & t \cos \theta & t(t_x \sin \theta + t_y \cos \theta) \\
0 & 0 & 1
\end{matrix}
\right)
\end{align}
And we end up with five parameters $t_x, t_y, \theta, s, t$ and only four equations.
This agrees with the six unknowns of equation $(*)$ and making use of the property of the rotation that $1 = \text{det}(R) = \cos^2 \theta + \sin^2 \theta$. I fail to spot a fifth equation so far.
Update: It seems the given data $A1, B1, A2, B2$ does not match the other part of the drawing, especially the original rectangle boundary and its transformed image.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate determinant with induction I need to prove the following, with induction to every $1 \leq n$:
$$D(a_1,...,a_n) = \left| \begin{array}{ccc}
a_1+x& a_2 & a_3 & \cdots & a_n \\
a_1& a_2+x & a_3 & \cdots & a_n \\
a_1& a_2 & a_3+x & \cdots & a_n \\
\vdots & \vdots & \vdots & & \vdots \\
a_1& a_2 & a_3 & \cdots & a_n + x
\end{array} \right| = x^n + (a_1 + \cdots + a_n)x^{n-1}$$
I played with it a bit and couldn't find a way to prove it.
This is what I did: I assumed that it is correct for $n$, and tried to solve it for $n+1$
$$D(a_1, \ldots , a_n, a_{n+1}) = \left| \begin{array}{ccc}
a_1+x& a_2 & a_3 & \cdots & a_{n+1} \\
a_1& a_2+x & a_3 & \cdots & a_{n+1} \\
a_1& a_2 & a_3+x & \cdots & a_{n+1} \\
\vdots & \vdots & \vdots & & \vdots \\
a_1& a_2 & a_3 & \cdots & a_{n+1} + x
\end{array} \right| $$
and I did the following operation on the determinant ($R_{n+1} \to R_{n+1} - R_1$) and got:
$$ \left| \begin{array}{ccc}
a_1+x& a_2 & a_3 & \cdots & a_{n+1} \\
a_1& a_2+x & a_3 & \cdots & a_{n+1} \\
a_1& a_2 & a_3+x & \cdots & a_{n+1} \\
\vdots & \vdots & \vdots & & \vdots \\
-x& 0 & \cdots & 0 & x
\end{array} \right| $$
And I wasn't sure on how to proceed from here, or even if I'm on the right path.
| Developing with respect to the last row, after performing those elementary row operations (that don't change the determinant), you get
$$
D(a_1,\dots,a_n,a_{n+1})=\\
xD(a_1,\dots,a_n)+(-1)^{(n+1)+1}(-x)\det\begin{bmatrix}
a_2 & a_3 & \dots & a_n & a_{n+1} \\
a_2+x & a_3 & \dots & a_n & a_{n+1} \\
a_2 & a_3+x & \dots & a_n & a_{n+1} \\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_2 & a_3 & \dots & a_n+x & a_{n+1}
\end{bmatrix}
$$
Doing $n-1$ row swaps, the determinant we need is
\begin{multline}
\det\begin{bmatrix}
a_2+x & a_3 & \dots & a_n & a_{n+1} \\
a_2 & a_3+x & \dots & a_n & a_{n+1} \\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_2 & a_3 & \dots & a_n+x & a_{n+1} \\
a_2 & a_3 & \dots & a_n & a_{n+1}
\end{bmatrix}=\\
\det\begin{bmatrix}
a_2+x & a_3 & \dots & a_n & a_{n+1} \\
a_2 & a_3+x & \dots & a_n & a_{n+1} \\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_2 & a_3 & \dots & a_n+x & a_{n+1} \\
a_2 & a_3 & \dots & a_n & a_{n+1}+x
\end{bmatrix}-\\
\det\begin{bmatrix}
a_2+x & a_3 & \dots & a_n & 0 \\
a_2 & a_3+x & \dots & a_n & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_2 & a_3 & \dots & a_n+x & 0 \\
a_2 & a_3 & \dots & a_n & -x
\end{bmatrix}=\\[6px]
D(a_2,\dots,a_{n+1})-xD(a_2,\dots,a_n)
\end{multline}
Therefore
$$
D(a_1,\dots,a_n,a_{n+1})=
xD(a_1,\dots,a_n)+
x(D(a_2,\dots,a_{n+1})-xD(a_2,\dots,a_n))
$$
By the induction hypothesis,
\begin{multline}
xD(a_1,\dots,a_n)+
x(D(a_2,\dots,a_{n+1})-xD(a_2,\dots,a_n))=\\
x(x^n+(a_1+\dots+a_n)x^{n-1})+\\
\qquad x(x^n+(a_2+\dots+a_{n+1})x^{n-1}-x^n-(a_2+\dots+a_n)x^{n-1})=\\
x^{n+1}+(a_1+\dots+a_n+a_{n+1})x^n
\end{multline}
Note that you have to check the induction basis for $n=1$ and $n=2$, which is easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If an integer $n$ is chosen at random from $1$ to $96$ inclusive ,what is the probability that $n(n+1)(n+2)$ is divisible by 8? In this one If I consider that n is even then probability that the number $n(n+1)(n+2)$ will be divisible by 8 will be 1/2 ,Now if n is odd then for n(n+1)(n+2) to be divisible by 8 ,n+1 should be a multiple of 8 ,Now how to find the probability among 96 numbers such that it is a multiple of 8 ?
| the total numbers are $96$. for an odd number to be multiple of 8 using expression $n(n+1)(n+2)$ the number $n$ should be a number preceding to a multiple of $8$ starting from $8$ itself so first number is $7$ and such is an AP whose last term is $95$. So total terms which are $1$ less than a multiple of $8$ ie odd are $12$ thus the probability is $\frac{12}{96}=\frac{1}{8}$. Numbers with $4m-2=\frac{1}{4}$ and then $4m=\frac{1}{4}$ so addition is $\frac{5}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Convergence rate of the sequence $a_{n+1} = a_n-a_n^2, a_0=1/2$. The sequence converges to zero at a rate that seems to be slightly faster than $1/n$.
What are the best known results on the convergence rate of this sequence?
| Here is a slightly better bound:
first, it is more convenient to look at $b_n = 1/a_n$. We have $$\frac{1}{b_{n+1}}=\frac{1}{{b_n}}-\frac{1}{{b_n^2}}=\frac{b_n-1}{b_n^2}$$ or, $$b_{n+1}=\frac{b_n^2}{b_n-1}=\frac{(b_n+1)(b_n-1)+1}{b_n-1}=b_n+1+\frac{1}{b_n-1}.$$
It is easy to see that $b_k\geq 2$ for all $k$, which implies
$$b_{n}\leq b_{n-1}+1+\frac{1}{2-1}\leq\dots\leq 2n+2.$$
Now,
$$
b_{n+1}\geq b_n+1+\frac{1}{1+2n}\geq b_{n-1}+2+\frac{1}{2n+1}+\frac{1}{2n-1}\geq \dots\geq b_0+n+1+\sum_{k=0}^n \frac{1}{2k+1},
$$
and since
$$
\sum_{k=0}^n \frac{1}{2k+1} > \frac{1}{2}\sum_{k=0}^n \frac{1}{k+1} \geq \frac{1}{2}\ln(n+2)
$$
we get
$$
b_{n+1}\geq n+\frac{1}{2}\ln(n+2)+3.
$$
Using the same reasoning, from $b_n\geq n+2$ (and using $\sum_{k=1}^n \frac{1}{k} \leq \ln n+1$) we get
$$
b_{n+1}\leq n+\ln(n+1)+4,
$$
hence
$$
\frac{1}{n+\ln(n+1)+4}\leq a_{n+1} \leq \frac{1}{n+\frac{1}{2}\ln(n+2)+3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Define $a$ and $b$ rational numbers so they satisfy equation Define $a, b \in \mathbb{Q}$ so that
$$\frac{a}{\sqrt{7 + 4\sqrt{3}}} + \frac{b}{\sqrt{7 - 4\sqrt{3}}} = \sqrt{4 + 2\sqrt{3}}$$
Using $\sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} \pm \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}$ I got $\frac{a}{2 + \sqrt{3}} + \frac{b}{2 - \sqrt{3}} = \sqrt{3}+1$ which results in $2(a+b)+\sqrt{3}(b-a) = \sqrt{3}+1$, I'm not sure how to proceed from that.
| Let's use the completion of square as @Kay K: did.
$\sqrt{ 4 + 2 \sqrt{3}} = \sqrt{ \sqrt {3} ^ 2 + 2 \sqrt{3} + 1} = \sqrt{ (\sqrt{3} + 1)^2} = \sqrt{3} + 1$
$\sqrt{ 7 + 4 \sqrt{3}}= \sqrt{ 2^2 + 2\cdot 2 \sqrt{3} + \sqrt{3}^2} = \sqrt{ (2+ \sqrt{3})^2}= 2+ \sqrt{3}$
$\sqrt{ 7 - 4 \sqrt{3}}= \sqrt{ 2^2 - 2\cdot 2 \sqrt{3} + \sqrt{3}^2} = \sqrt{ (2- \sqrt{3})^2}= 2- \sqrt{3}$ since $2- \sqrt{3}>0$.
Note now that $(2+ \sqrt{3})(2- \sqrt{3} ) = 1$. Thus we need
$$a(2- \sqrt{3})+ b (2+ \sqrt{3}) = \sqrt{3} + 1$$ that is
$$2 a + 2 b = 1\\
b-a =1$$
with solution $a = -1/4, b = 3/4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.