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Proof of $\sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}$ The title pretty much summarizes my question. I am trying to prove the following:
$$\displaystyle \forall N \in \mathbb{N}: \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N}_{n=1} \frac{1}{N+n}.$$
I tried proving this using induction. Starting with the base case $N = 1$:
$$\displaystyle \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} = \frac{1}{1} + \frac{-1}{2} = \frac{1}{2} = \frac{1}{1+1} = \sum^{1}_{n=1} \frac{1}{N+n}.$$
My problem is the inductive step for $N+1$, starting with
$$\displaystyle \sum^{2(N+1)}_{n=1} \frac{(-1)^{n-1}}{n} = \sum^{N+1}_{n=1} \frac{1}{(N+1)+n}.$$
And now my problem:
\begin{align}
\sum^{2(N+1)}_{n=1} \frac{(-1)^{n-1}}{n} &\Leftrightarrow \sum^{2N}_{n=1} \frac{(-1)^{n-1}}{n} + \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n} \\[1em]
&\underset{\Leftrightarrow}{\text{ind. hyp.}} \sum^{N}_{n=1} \frac{1}{N + n} + \sum^{2}_{n=1} \frac{(-1)^{n-1}}{n}
\end{align}
Is this the correct start, and if so, how do I continue?
|
Instead of induction, you could start with
$\displaystyle \sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n}=\sum_{n=1}^{2N}\frac{1}{n}-2\sum_{k=1}^N\frac{1}{2k}=\sum_{n=1}^{2N}\frac{1}{n}-\sum_{n=1}^{N}\frac{1}{n}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find eigenvalues and eigenvectors of this matrix Problem: Let \begin{align*} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}. \end{align*} Compute all the eigenvalues and eigenvectors of $A$.
Attempt at solution: I found the eigenvalues by computing the characteristic polynomial. This gives me \begin{align*} \det(A - x \mathbb{I}_4) = \det \begin{pmatrix} 1-x & 1 & 1 & 1 \\ 1 & 1-x & 1 & 1 \\ 1 & 1 & 1-x & 1 \\ 1 & 1 & 1 & 1-x \end{pmatrix} = -x^3 (x-4) = 0 \end{align*} after many steps. So the eigenvalues are $\lambda_1 = 0$ with multiplicity $3$ and $ \lambda_2 = 4$ with multiplicity $1$.
Now I was trying to figure out what the eigenvectors are corresponding to these eigenvalues. Per definition we have $Av = \lambda v$, where $v$ is an eigenvector with the corresponding eigenvalue. So I did for $\lambda_2 = 4$: \begin{align*} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = 4 \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \end{align*} I think this is only possible when $x_1 = x_2 = x_3 = x_4 = 1$. So am I right in stating that all the eigenvectors corresponding to $\lambda_2$ are of the form $t \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}$ with $t$ some number $\neq 0$? For $\lambda_1 = 0$, I'm not sure how to find the eigenvectors. The zero vector is never an eigenvector. This means $x_1, x_2, x_3$ and $x_4$ can be anything aslong as they add to zero?
|
You are right, and to show how this applies more generally, look at the eigenspace corresponding to the other eigenvalue, $\lambda = 0$. This has multiplicity 3, so we expect exactly 3 linearly independent vectors in this space. The main equation looks like $A \vec{x} = 0 \vec{x} = \vec{0}$, in other words,
$$
\begin{pmatrix}
1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1
\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}
= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}
$$
It's not hard to see that we have 4 identical equations, so the only constraint is $x + y + z + w = 0$, so we define $w = -x-y-z$ and any vector in the eigenspace now looks like
$$
\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}
= \begin{pmatrix} x \\ y \\ z \\ -x-y-z \end{pmatrix}
= x \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix}
+ y \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}
+ z \begin{pmatrix} 0 \\ 0 \\ 1 \\ -1 \end{pmatrix}
$$
and the three vectors which form the basis of the eigenspace are now specified.
|
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|
How to find this limit : $x\sin{f(x)}$ How to find the limit
$$\lim_{x\to\infty}x\sin f(x)$$
where $$f(x)=\left(\sqrt[3]{x^3+4x^2}-\sqrt[3]{x^3+x^2}\right)\pi\ ?$$
Is it possible to solve without L'Hospital's rule ?
|
Setting $x=u^{-1}$, we have
\begin{eqnarray}
x\sin f(x)&=&\frac{1}{u}\sin\left[\pi\left(\sqrt[3]{u^{-3}+4u^{-2}}-\sqrt[3]{u^{-3}+u^{-2}}\right)\right]=\frac1u\sin\left[\pi\frac{\sqrt[3]{1+4u}-\sqrt[3]{1+u}}{u}\right]\\
&=&\frac1u\sin\left[\pi\frac{(1+\frac{4}{3}u-\frac{16}{9}u^2)-(1+\frac{u}{3}-\frac{u^2}{9})+o(u^2)}{u}\right]\\
&=&\frac1u\sin\left[\pi\frac{u-\frac{5}{3}u^2+o(u^2)}{u}\right]\\
&=&\frac1u\sin\left[\pi-\frac{5\pi}{3}u+o(u)\right]\\
&=&\frac1u\sin\left[\frac{5\pi}{3}u+o(u)\right]\\
&=&\frac1u\left[\frac{5\pi}{3}u+o(u)\right]\\
&=&\frac{5\pi}{3}+o(1).
\end{eqnarray}
It follows that
$$
\lim_{x\to\infty}x\sin f(x)=\lim_{u\to0}\frac1u\sin f\left(\frac1u\right)=\lim_{u\to0}\left[\frac{5\pi}{3}+o(1)\right]=\frac{5\pi}{3}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\iint_{R}(x+y)^2dxdy$ in $0\leq r\leq 1 \,\, ,\frac{\pi}{3}\leq \theta\leq\frac{2\pi}{3}$
$$\iint_{R}(x+y)^2dxdy$$
$$0\leq r\leq 1 \,\, ,\frac{\pi}{3}\leq \theta\leq\frac{2\pi}{3}$$
My attempt number 1:
$$=\iint_{R}(x^2+2xy+y^2)dxdy$$
$$x:=r\cos \theta \,\,\,,y:=r\cos \theta$$
$$\sqrt{x^2+y^2}=r$$
$$\int_{\pi/3}^{2\pi/3}\bigg[\int_{0}^{1}\bigg(r^2+(r\cos \theta)(r\sin \theta)\bigg)dr\bigg]d\theta$$
$$=\int_{\pi/3}^{2\pi/3}\bigg(\frac{1}{2}+\cos \theta \sin \theta\bigg)d \theta$$
Is it correct so far?
Attempt number 2:
$$\int_{\pi/3}^{2\pi/3}\bigg[\int_{0}^{1}r\bigg(r^2+\color{red}2(r\cos \theta)(r\sin \theta)\bigg)dr\bigg]d\theta$$
$$=\int_{\pi/3}^{2\pi/3}\bigg(\frac{1}{4}+ \frac{1}{4}\sin(2 \theta) \bigg)d \theta$$
$$=\boxed{\pi/12}$$
|
So you are trying to determine the surface area of the function $(x+y)^2$ in a part of a circle with radius 1. Let's define $R=\left( (r,\theta)\in \mathbb{R}^2 : 0 \leq r \leq 1 , \frac{\pi}{3} \leq \theta \leq \frac{2 \pi}{3}\right)$.
We have the co-ordinate transformation $X(x,y)=(r \cos \theta , r \sin \theta)$.
The Jacobian is:
$$J(r,\theta) = \left[ \begin{array}{cc} \frac{∂x}{∂r} &\frac{∂x}{∂\theta} \\ \frac{∂y}{∂r} & \frac{∂y}{∂\theta} \end{array} \right] = \left[ \begin{array}{cc} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{array} \right] $$ To determine the integral we have to calculate $|J(r,\theta)|=|r \cos^2\theta + r \sin^2 \theta| = r$.
$$ \int\int_R (x+y)^2 d(x,y) = \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \int_{0}^{1} (r \cos \theta + r \sin \theta)^2 \cdot r \ dr d\theta =$$
$$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \int_{0}^{1} r(r^2+2 r^2 \sin\theta \cos \theta)\ dr d\theta =$$
$$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \int_{0}^{1} r^3+2 r^3 \sin\theta \cos \theta\ dr d\theta =$$
$$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \left[\frac{r^4}{4}+\frac{r^4}{2}\sin\theta \cos\theta \right]^{1}_{r=0} d\theta=\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{1}{4}+\frac{1}{2}\sin\theta \cos\theta d\theta=$$
$$\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{1}{4} d\theta +\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{1}{4}\sin(2\theta) d\theta= \frac{1}{4}\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} 1 d\theta +\frac{1}{4}\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \sin(2\theta) d\theta=$$
$$\frac{1}{4}(\frac{2\pi}{3}-\frac{\pi}{3}) + \frac{1}{4} \left[-\frac{1}{2}\cos(2\theta) \right]_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} =$$
$$\frac{\pi}{12} -\frac{1}{8}(\frac{-1}{2}-\frac{-1}{2})= \frac{\pi}{12}$$
|
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|
Find numbers $\overline{abcd}$ so that $\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}$ Find the numbers $\overline{abcd}$, with digits not null that satisfy the equality
\begin{equation}\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}\end{equation}
where
\begin{equation}\overline{abcd}=1000a+100b+10c+d\end{equation}
I see that $4d+1=\overline{..a},\ d\neq0,\ a\neq 0,\ a<d$ but that gives too many pairs to start with $(d,a)\in\{(5,1),(6,5),(8,3),(9,7)\}$
|
\begin{array}{r}
abcd\\
bcd\\
cd\\
d\\
1\\
\hline
dcba
\end{array}
The leftmost $a$ in the top row cannot recieve more than $1$ as a carry. The worst cases are $(b,c)=(9,8)$ and $(b,c) = (8,9)$ as $(b,c) = (9,9)$ leads quickly to no solution. Hence $d = a + 1$ or $d = a$.
Case: $d = a + 1$
Summing the rightmost column, we get
\begin{align}
4d+1 &= a \pmod{10}\\
4a+5 &= a \pmod{10}\\
3a &= 5 \pmod{10}\\
a &= 5\\
d &= 6
\end{align}
Carrying the $2$ and summing the third column, we get
$3c + 2 = 10k + b$ where $k \in \{0, 1, 2\}$.
Finally, from the second column, we get
$2b + k = 10 + c$.
So $2b = 10 + c - k = 6c + 4 - 20k$
$5c = 19k + 6$
The only solution is $k = 1, c = 5$, and $ b = 7$.
So $\overline{abcd} = 5756$
Case: $d=a$.
Summing the rightmost column, we get
\begin{align}
4d+1 &= a \pmod{10}\\
4a+1 &= a \pmod{10}\\
3a &= 9 \pmod{10}\\
a &= 3\\
d &= 3
\end{align}
Carrying the $1$ and summing the third column, we get
$3c + 1 = 10k + b$ where $k \in \{0, 1, 2\}$.
Finally, from the second column, we get
$2b + k = c$.
So $2b = c - k = 6c + 2 - 20k$
$5c = 19k - 2$
which has no solution.
|
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|
Solve the trigonometric equation: $\sin {3x} = 4 \sin^2 x$ Solve the equation $\sin{3x} = 4 \sin^2 x$.
I tried to change the $\sin{3x}$ to $3\sin x\cos x$ then solve it, but I could not find the correct answer.
|
Hint: We have $$\sin 3x = 3\sin x - 4\sin^3 x\neq 3\sin x\cos x$$
So you get (let $\sin x = \alpha$) $$3\alpha - 4\alpha^3 = 4\alpha^2 \iff 4\alpha^3 + 4\alpha^2 - 3\alpha = 0$$
Which is a simple cubic in $\alpha$. Find the roots of the cubic, back-substitute and find the corresponding values of $x$.
|
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|
How to show that his series converges or diverges using LCT or CT? $$\sum_{n=1}^{\infty}\left (\sqrt{n^4+1}-n^2\right)$$
The question states that either the limit comparison or comparison test can be used to determine whether the series converge or diverge. I tried finding a $B_n$ in order to test $\frac{A_n}{B_n}$ for the limit comparison but having trouble coming up with $B_n$ that I know will converge or diverge. Maybe I'm going about this the wrong way. Any help would be appreciated.
|
\begin{align}
\sqrt{n^4+1}-n^2 &=\dfrac{\sqrt{n^4+1}-n^2}{1} \\
&=\dfrac{\left(\sqrt{n^4+1}-n^2\right)\left(\sqrt{n^4+1}+n^2\right)}{\left(\sqrt{n^4+1}+n^2\right)} \\
&=\dfrac{\left(\sqrt{n^4+1}\right)^2-n^4}{\sqrt{n^4\left(1-\dfrac{1}{n^4}\right)}+n^2} \\
&=\dfrac{n^4+1-n^4}{n^2\,\sqrt{1-\dfrac{1}{n^4}}+n^2} \\
&=\dfrac{1}{n^2\left(\sqrt{1-\dfrac{1}{n^4}}+1\right)}
\end{align}
As $\sqrt{1-\dfrac{1}{n^4}}\le 1,\ \forall n\in \mathbb{N}$ we have
$$\dfrac{1}{n^2\left(\sqrt{1-\dfrac{1}{n^4}}+1\right)}\le \dfrac{1}{2n^2}$$
therefore the series converges because the greater series converges.
|
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|
Prove by induction that $\frac{n^3}{3}+\frac{2n}{3}$ is an integer. The question that I am working on is:
Prove that $\dfrac{n^3}{3}+\dfrac{2n}{3} \in \mathbb Z \ \forall \ n \in \mathbb N$
The method that I think would be will work for this question is that I say that $3|(n^3+2n)$ and prove that.Would this be a good way to do this question?
So far I have done the following:
1) Base Case
n = 1
$3|n^3+2n = 3|3 \checkmark$
2) Assume, $n^3+2n$ is divisible by $3$ for $n =k, k \in \mathbb N$
$k^3+2k = 3m , m \in \mathbb N$
Let $n = k + 1$ ; Then:
$(k+1)^3 + 2(k+1)$
= $k^3+2k+3k^2+3k+3$
Since $k^3+2k = 3m$
$3m + 3k^2+3k+3$
Now, I'm stuck I don't know what else to do further.
|
$$( k + 1 )^3 = k^3 + 3 k^2 + 3k + 1$$
For $n = k + 1$ $$(k + 1)^3 + 2(k + 1) = k^3 + 3k^2 + 3k + 2k + 3$$
If $k^3 + 2k$ is divisible by $3,$ it is ovious that $$k^3 + 3k^2 + 3k + 2k +3=(k^2+2k)+3(k^2+k+1)$$ is dividable by $3.$
|
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|
Find the principal solutions of the trigonometric equation $\cos x-\sin x+\sin 2x+3\cos2x+1=0$ I am unable to simplify the expression. If I simplify the double angles, it leaves me with a nasty expression,
$\cos x-\sin x+2\sin x\cos x+6\cos^2 x-2=0$. What do I do next. Some hints, please. Also, is there some elegant solution? Thanks!
|
You have $\cos x-\sin x+\sin 2x+3\cos 2x+1=0$. Or $\sin x-\sin 2x=\cos x+3\cos 2x+1$. Or $\sin x(1-2\cos x)=\cos x+6\cos^2x-2$.
Either $\cos x =1/2$ or $\sin x=((6\cos^2x+\cos x-2)/(1-2\cos x))$. Or $\sin x=-3\cos x-2$. Squaring we get $10\cos^2x-12\cos x+3=0$. Thus $\cos x=((6\pm \sqrt{6})/(10))$.
|
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|
Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And then assume:
$$
\frac{6}{\sqrt{51}}= \cos \psi ; \frac{5}{\sqrt{51}}= \sin\psi ;
$$
$$
\cos \psi \cos x - \sin \psi \sin x = \cos (x+ \psi) = \cos(x + \arccos ( \frac{6}{\sqrt{51}}))
$$
$$
x + \arccos\left(\frac{6}{\sqrt{51}}\right) = \arcsin\left( \frac{8}{\sqrt{51}}\right)
$$
$$
x \approx 12^\circ
$$
But answer is:
$$
-\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n , n\in\Bbb Z
$$
|
By the Cauchy-Schwarz inequality:
$$ \left( 6\cos x-5\sin x\right)^2 \leq 36+25 = 61<64 $$
so it is not possible that $6\cos x-5\sin x$ equals $8$ for some real $x$.
|
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|
Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence of rearrangements.
Let $\{s_n\}$ be the sequence of partials sums of the series then for $n \ge 0$
$s_{3(n+1)} = \sum ^n _ {k=0} \frac{1}{4k+1} + \frac{1}{4k+3} - \frac{2}{4k+4}$
We can view it as the sequence(on $n$) of partials sums of
$\sum_0 a_n = \sum_0 \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{2}{4n+4}$
Where $|a_n| = a_n = \frac{1}{4n+4}\{\frac{3}{4n+1}+\frac{1}{4n+3}\} \le \frac{1}{4n^2}$.
By the comparison test $s_{3(n+1)}$ converges to some real $\alpha$.
But $s_{3(n+1)+1} = s_{3(n+1)}+ \frac{1}{4n+5}$ and $s_{3(n+1)+2} = s_{3(n+1)}+ \frac{1}{4n+5}+\frac{1}{4n+7} $hence we have a partition of $\{s_n\}$ into subsequences which tend to $\alpha$ and this implies $s_n \rightarrow \alpha$.
Is my proof correct? Any alternative solutions are appreciated.
|
By the Riemann-Dini theorem, we may take any series that is conditionally convergent but not absolutely convergent and rearrange it in order to get a series that converges to $\alpha$, for any $\alpha\in\mathbb{R}$.
In our case:
$$\begin{eqnarray*} \sum_{k\geq 0}\left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+2}\right)&=&\sum_{k\geq 0}\int_{0}^{1}\left(x^{4k}+x^{4k+2}-2 x^{4k+3}\right)\,dx\\&=&\int_{0}^{1}\frac{1+x^2-2x^3}{1-x^4}\,dx\\&=&\frac{3}{2}\log 2.\end{eqnarray*} $$
We may notice that we know in advance that the LHS is converging, since:
$$ \frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+2} = \frac{8k+5}{(4k+1)(4k+3)(2k+2)}=O\left(\frac{1}{k^2}\right).$$
Convergence also follows from Dirichlet's test, since the sequence $1,1,-2,1,1,-2,\ldots$ has bounded partial sums while the sequence $\frac{1}{1},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{7},\frac{1}{8},\ldots$ decreases to zero.
|
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|
$\lim_{n \to \infty} (\sqrt{9n^2 + 2n + 1} - 3n) = \frac{1}{3}$ Prove
$$\lim_{n \to \infty} (\sqrt{9n^2 + 2n + 1} - 3n) = \frac{1}{3}$$
I got this problem in Harro Heuser's "Lehrbuch der Analysis Teil 1". It is surely smaller than 1 because $\sqrt{9n^2 + 2n + 1} < \sqrt{9n^2 + 6n + 1} = 3n + 1$, but I cannot get closer than that, although it looks very simple...
|
$$\lim_{n\to\infty} (\sqrt{9n^2 + 2n+1} - 3n) = \lim_{n\to\infty} 3n\left(\sqrt{1 + \frac{2}{9n} + \frac{1}{9n^2}} - 1\right)$$
Now, using Taylor expansion for $\sqrt{1 + x} = 1 + \frac{x}{2} + o(x)$
$$\lim_{n\to\infty}3n\left(1 + \frac{1}{9n} + o\left(\frac{1}{n}\right) - 1\right) = \lim_{n\to\infty}\frac{3n}{9n} = \frac{1}{3}$$
|
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|
Differentiate the Function: $y=2x \log_{10}\sqrt{x}$ $y=2x\log_{10}\sqrt{x}$
Solve using: Product Rule $\left(f(x)\cdot g(x)\right)'= f(x)\cdot\frac{d}{dx}g(x)+g(x)\cdot \frac{d}{dx}f(x)$
and $\frac{d}{dx}(\log_ax)= \frac{1}{x\ \ln\ a}$
$(2x)\cdot [\log_{10}\sqrt{x}]'+(\log_{10}\sqrt{x})\cdot [2x]'$
$y'=2x\frac{1}{\sqrt{x}\ln 10}+\log_{10}\sqrt{x}\cdot 2$
Answer in book is $y'= \frac{1}{\ln10}+\log_{10}x$
|
Hint: You'll need to use the product rule with $2x$ and the logarithmic function.
Then, whilst applying the product rule, you use the chain rule on the logarithmic function.
Notice: differentiating $\log_a f(x)$ gives you (this is where you made your mistake) $$(\log_a f(x))' = \frac{f'(x)}{\ln a \cdot f(x)}$$
Use that with $f(x) = \sqrt{x}$. In fact, you should get $$\bbox[border: solid blue 1px, 10px]{(\log_{10} \sqrt{x})' = \frac{1}{2x \ln 10}}$$
Full solution:
Hence, using the product rule we have the derivative as $$2x \cdot \frac{1}{2x \ln 10} + 2 \cdot \frac{1}{2x \ln 10} = \frac{1}{\ln 10} + \frac{1}{x \ln 10}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Show that $\frac{1}{b-a_{1}} + \frac{1}{b-a_2}+ ...+ \frac{1}{b-a_n} \geq \frac{n}{b-\frac{1}{n}(a_1+a_2+...+a_n)}$ Let, $b> \max\{a_1,a_2,...,a_n\}.$ Show that $\frac{1}{b-a_{1}} + \frac{1}{b-a_2}+ ...+ \frac{1}{b-a_n} \geq \frac{n}{b-\frac{1}{n}(a_1+a_2+...+a_n)}$
f is convex if $f(t_1x_1+t_2x_2+...+t_nx_n)\leq t_1f(x_1)+t_2f(x_2)+...+t_nf(x_n)$
Let, $t_i = 1/n \Rightarrow \sum\limits_{i=1}^n t_i = 1$
Then, $f(\frac{a_1+a_2+...+a_n}{n}) \leq \frac{1}{n}[f(a_1)+f(a_2)+...+f(a_n)]$
NTS: $\frac{1}{b-\frac{1}{n}(a_1+a_2+...+a_n)} \leq \frac{1}{n}[\frac{1}{b-a_1}+...+\frac{1}{b-a_n}]$
Let$f(x) = \frac{1}{b-x}$ In order to prove the inequality, we have to show that $f''(x) \geq 0.$ So, $f''(x) = \frac{2}{(b-x)^3}$
Now, how can we show that $f''(x) \geq 0$? $b$ can be a negative number, can it?
|
Observe the inequality can be proven using the following inequality:
$(x_1+x_2+\cdots + x_n)\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\cdots + \dfrac{1}{x_n}\right)\geq n^2$ with $x_i = b-a_i > 0$. The answer follows.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding $[T_{|W_i}]_{C_i}$ Let $B=\{v_1,v_2,v_3\}$, a basis of $V$ above $\mathbb{R}$. Let $$ [T]_B = \left(\begin{array}{cccc} 6&-3&-2\\4&-1&-2\\10&-5&-3 \end{array}\right)$$
The characteristic polynomial is $f_T(x) = (x-2)(x^2+1)$. Hence, $m(x) = (x-2)(x^2+1)$
We have $W_1 = \ker (T-2I) = \text{span}\{v_1,2v_3\}$ and $W_2 = \ker (T^2+I) = \text{span}\{v_1+v_2, v_3\}$ and $V = W_1 \oplus W_2$.
Now, we denote $$C_1 = \{v_1 + 2v_3\} \\ C_2 = \{ v_1+v_2,v_3 \}$$
I don't quite understand why
$$[T_{|W_1}]_{C_1} = (2) \\ [T_{|W_2}]_{C_2} = \left(\begin{array}{cccc} 3&-2\\5&-3 \end{array}\right)$$
|
$ \{v_1 + 2v_3\}=\left(
\begin{array}{c}
2\\
0\\
4\\
\end{array}
\right)\\
\{v_1+v_2,v_3\} =\left(\begin{array}{c} 3\\
3\\
5\\
\end{array}\right),\left(\begin{array}{c} -2\\
-2\\
-3\\
\end{array}\right)$
and $T\left(
\begin{array}{c}
2\\
0\\
4\\
\end{array}
\right)\ = 2\left(
\begin{array}{c}
2\\
0\\
4\\
\end{array}
\right)\\$
thus your first matrix comes out as $(2)$
Now, $T\left(\begin{array}{c} 3\\
3\\
5\\
\end{array}\right)=3\left(\begin{array}{c} 3\\
3\\
5\\
\end{array}\right)+5\left(\begin{array}{c} -2\\
-2\\
-3\\
\end{array}\right)$
$T\left(\begin{array}{c} -2\\
-2\\
-3\\
\end{array}\right)=-2\left(\begin{array}{c} 3\\
3\\
5\\
\end{array}\right) -3\left(\begin{array}{c} -2\\
-2\\
-3\\
\end{array}\right)$
hence second matrix comes out as $\left(\begin{array}{cccc} 3&-2\\5&-3 \end{array}\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1371354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Point of intersection of $f(x)=\sin(2x)+\cos(2x)$ and the $x$-axis How can I algebraically (without looking at the graph) find the point of intersection of $f(x)=\sin(2x)+\cos(2x)$ and $x$-axis, in the interval $[0, \pi]$?
|
Notice, $$f(x)=\sin 2x+\cos 2x$$ $$\implies f(x)=\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos 2x+\frac{1}{\sqrt{2}}\sin 2x\right)$$ $$=\sqrt{2}\left(\cos 2x\cos\frac{\pi}{4}+\sin 2x\sin\frac{\pi}{4}\right)$$ $$=\sqrt{2}\cos \left(2x-\frac{\pi}{4}\right)$$ Now, for the intersection of $f(x)$ with the x-axis, we have $f(x)=0$ $$\implies \sqrt{2}\cos \left(2x-\frac{\pi}{4}\right)=0 $$ $$\implies \cos \left(2x-\frac{\pi}{4}\right)=0 $$ Writing the general solution as follows $$\implies 2x-\frac{\pi}{4}=\frac{(2n+1)\pi}{2}$$ $$\implies x=\frac{(2n+1)\pi}{4}+\frac{\pi}{8}$$ $$\implies x=\frac{(4n+3)\pi}{8}$$ Where, $n$ is any integer
For given interval $[0, \pi]$, put $n=0$ & $n=1$ in the general solution, we get $$\color{blue}{x\in \left\{\frac{3\pi}{8}, \frac{7\pi}{8}\right\}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1371434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Matrix Exponential and Logarithm Consider the following matrix $A$:
$A = \begin{bmatrix}
\cos^2(1) & -\sin(2) & \sin^2(1) \\
\cos(1)\sin(1) & \cos(2) & -\cos(1)\sin(1) \\
\sin^2(1) & \sin(2) & \cos^2(1)\\
\end{bmatrix}$
I want to find a matrix $B$ such that $\exp(B)=A$ (or essentially finding $\log(A))$. Is there a systematic way to approach these kinds of problems? I was thinking of using some properties involving diagonalization to get to the answer, which should be (obtained using Mathematica):
$B = \begin{bmatrix}
0 & -2 & 0 \\
1 & 0 & -1 \\
0 & 2 & 0\\
\end{bmatrix}$
However, I'm not sure how to get to this result. Thank you for your help.
|
First check that one can diagonalize $A$, which should give something like
$$\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & \frac{\cos ^2(1)-2 i \sin (1) \cos (1)-\sin ^2(1)}{\cos ^2(1)+\sin ^2(1)} & 0 \\
0 & 0 & \frac{\cos ^2(1)+2 i \sin (1) \cos (1)-\sin ^2(1)}{\cos ^2(1)+\sin ^2(1)} \\
\end{array}
\right)$$
Calculate the matrix of eigenvectors of $A$
$$V = \left(
\begin{array}{ccc}
-1 & i & 1 \\
-1 & -i & 1 \\
1 & 0 & 1 \\
\end{array}
\right)$$
Calculate the inverse
$$V^{-1} = \left(
\begin{array}{ccc}
-\frac{1}{4} & -\frac{1}{4} & \frac{1}{2} \\
-\frac{i}{2} & \frac{i}{2} & 0 \\
\frac{1}{4} & \frac{1}{4} & \frac{1}{2} \\
\end{array}
\right)$$
Calculate
$$A' = (V^T)^{-1} A V^T = \left(
\begin{array}{ccc}
-\frac{1}{4} & -\frac{1}{4} & \frac{1}{2} \\
-\frac{i}{2} & \frac{i}{2} & 0 \\
\frac{1}{4} & \frac{1}{4} & \frac{1}{2} \\
\end{array}
\right)^{\mathsf{T}}.\left(
\begin{array}{ccc}
\cos ^2(1) & -\sin (2) & \sin ^2(1) \\
\sin (1) \cos (1) & \cos (2) & \sin (1) (-\cos (1)) \\
\sin ^2(1) & \sin (2) & \cos ^2(1) \\
\end{array}
\right).\left(
\begin{array}{ccc}
-1 & i & 1 \\
-1 & -i & 1 \\
1 & 0 & 1 \\
\end{array}
\right)^{\mathsf{T}} = \left(
\begin{array}{ccc}
(\cos (1)+i \sin (1))^2 & 0 & 0 \\
0 & (\cos (1)-i \sin (1))^2 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)$$
then
$$\ln(A) = V^T \ln(A') (V^T)^{-1} = \left(
\begin{array}{ccc}
-1 & i & 1 \\
-1 & -i & 1 \\
1 & 0 & 1 \\
\end{array}
\right)^{\mathsf{T}}.\left(
\begin{array}{ccc}
2 i & 0 & 0 \\
0 & -2 i & 0 \\
0 & 0 & 0 \\
\end{array}
\right).\left(
\begin{array}{ccc}
-\frac{1}{4} & -\frac{1}{4} & \frac{1}{2} \\
-\frac{i}{2} & \frac{i}{2} & 0 \\
\frac{1}{4} & \frac{1}{4} & \frac{1}{2} \\
\end{array}
\right)^{\mathsf{T}} = \left(
\begin{array}{ccc}
0 & -2 & 0 \\
1 & 0 & -1 \\
0 & 2 & 0 \\
\end{array}
\right)$$
Here $\ln(A')$ means taking the natural logarithm for each diagonal entry.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1372370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
|
The effect of multiplication by $1/(1-x)$ to the sequence of coefficients is to calculate partial sums: if the original sequence is $c_0,c_1,\ldots$ then the new one is
$$ d_i = c_0 + \cdots + c_i. $$
The starting point is the sequence $1,0,0,\ldots$. Applying this operator twice, we get
$$
1,0,0,0,0,\ldots \\
1,1,1,1,1,\ldots \\
1,2,3,4,5,\ldots
$$
In this matrix, the first row is given, the first column is constant, and otherwise the value of a cell is the sum of the cell above it and the cell to its left.
I'll let you figure out the connection to Pascal's triangle on your own.
|
{
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"url": "https://math.stackexchange.com/questions/1372958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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|
Raising a number in Rectangular Form What is the value of $(-2 + 3i\sqrt3)^6$?
Answer is $4096$
Convert $(-2 + 3i\sqrt3)^6$ to Polar Form.
$${ (\sqrt{31} \angle 111.05)^6 }$$
I use something called De Moivre's Theorem
$${z^n = r^n( \cos(n\theta) + i\sin(n\theta) ) }$$
$${z^n = (\sqrt{31})^6( \cos(6\cdot 111.05) + i\sin(6\cdot 111.05) ) }$$
Even if I did continue this I know I wouldn't get a whole number sum thing...
$${z^n = 29791( 0.5920 - 8.0593i ) }$$
$${z^n = 17636 + 240009i }$$
What am I doing wrong?
|
First of all we know the following things:
$$a+bi=\left|a+bi\right|e^{\arg\left(a+bi\right)i}=\left|a+bi\right|\left(\cos\left(\arg\left(a+bi\right)\right)+\sin\left(\arg\left(a+bi\right)\right)i\right)$$
$$\left(-2+3i\sqrt{3}\right)^6=$$
$$\left(\left|-2+3i\sqrt{3}\right|e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$$
$$\left(\sqrt{\Re\left(-2+3i\sqrt{3}\right)^2+\Im\left(-2+3i\sqrt{3}\right)^2}e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$$
$$\left(\sqrt{(-2)^2+(3\sqrt{3})^2}e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$$
$$\left(\sqrt{4+27}e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$$
$$\left(\sqrt{31}e^{\arg\left(-2+3i\sqrt{3}\right)i}\right)^6=$$
$$\left(\sqrt{31}e^{\left(\pi-\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)i}\right)^6=$$
$$\sqrt{31}^6e^{6\left(\pi-\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)i}=$$
$$29791e^{\left(6\pi-6\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)i}$$
So in the three forms we get:
$$29791e^{\left(6\pi-6\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)i}=$$
$$29791\left(\cos\left(\left(6\pi-6\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)\right)+\sin\left(\left(6\pi-6\tan^{-1}\left(\frac{3\sqrt{3}}{2}\right)\right)\right)i\right)=$$
$$29791\left(\frac{17641}{29791}-\frac{13860\sqrt{3}}{29791}i\right)=17641-13860\sqrt{3}i$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1373195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find $x$ if $\frac {1} {x} + \frac {1} {y+z} = \frac {1} {2}$ I found this question from past year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Find $x$ if \begin{align}\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac{1}{3}\\ \frac {1} {z} + \frac {1} {x+y} &= \frac {1} {4}\end{align}
$(A)\;\frac 32$
$(B)\;\frac {17}{10}$
$(C)\;\frac {19}{10}$
$(D)\;\frac {21}{10}$
$(E)\;\frac {23}{10}$
EDIT: I'm very sorry guys, it should be $\frac {1} {x+y}$ not $xy$, I'm sorry for the typos (idk what is wrong with me)
|
Choosing
$p = \frac{1}{x} $
$q = \frac{1}{y} $
$r = \frac{1}{z} $
will certainly help writing the equations more neatly, I guess.
Then your equations can be written as:
$$2p + 2qr = 1 \\ 3q + 3pr = 1 \\ 4r + 4pq = 1$$
using first two equations you get:
$$ 2p + 2qr = 3q + 3pr \\ \implies r = \frac{3q - 2p}{3p - 2q}$$
Substitute in equations and solve as solved by Alan above.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$
Evaluate $$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$
I thought of rewriting this as $$\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}$$ and substituting $$u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}$$ and subsequently I got $$du=\dfrac{1}{4}\left(1+\frac{1}{x^4}\right)^{-3/4}\times (-4x^{-5})dx$$
However, I can not think of how to proceed further. Any help would be truly appreciated. Many thanks in advance!
|
$$\int\frac{1}{x^5}\frac{1}{(1+\frac{1}{x^4})^{3/4}}dx$$
$$u=1+\frac{1}{x^4}$$
$$-\frac{1}{4}du=\frac{1}{x^5}$$
The integral in the variable $u$ is then
$$-\frac{1}{4}\int u^{-3/4}du$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1373612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
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|
Simultaneous equations, $\frac{1}{x}+\frac{1}{y}=1$,$x+y=a$,$\frac{y}{x}=m$ By eliminating $x$ and $y$ from the following equations, I need to find the relation between $m$ and $a$.
\begin{align*}
\frac{1}{x}+\frac{1}{y}=1 \\
x+y=a \\
\frac{y}{x}=m
\end{align*}
I tried different ways, but cannot arrive at the answer. I end up with a quadratic.
This is what I have.
\begin{align*}
\frac{1}{x}=1-\frac{1}{y} \\
x(y-1)=y \\
x=\frac{y}{y-1}
\end{align*}
Now using second equation:
\begin{align*}
\frac{y}{y-1}+\frac{y(y-1)}{y-1}=a \\
\frac{y^2}{y-1}=a \\
y^2-ay+a=0
\end{align*}
Now roots are: $y_{1,2}=\displaystyle{\frac{a\pm\sqrt{a^2-4a}}{2}}$
Then it follows that: $x_{1,2}=\displaystyle{\frac{a\pm\sqrt{a^2-4a}}{2}}$
Hence there are three scenarios for the relation between $a$ and $m$.
First: $\frac{4a}{4}=m$, $a=m$
Second: $\displaystyle{\frac{(a+\sqrt{a^2-4a})^2}{4}=m}$
Third: $\displaystyle{\frac{(a-\sqrt{a^2-4a})^2}{4}=m}$
Is this correct? Thank you
|
Your system has three equations and only two unknowns, so it is probably incompatible. Nevertheless, it still can have solution for certain values of $a$ and $m$.
From the last equation, we get $y=mx$. With the second eq.:
$$(m+1)x=a$$
We see that the system is incompatible if $m=-1$ and $a\ne 0$. Also, if $m=-1$ and $a=0$ then $x=-y$ and the first equation is impossible.
So assume $m\neq-1$. Then
$$x=\frac{a}{m+1}$$
$$y=\frac{am}{m+1}$$
then
$$\frac1x+\frac1y=\frac{(m+1)^2}{am}$$
So the system has a solution only if
$$a=\frac{(m+1)^2}m$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1374380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem :
Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.
My approach:
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But this is quite time consuming by making six different equations and then solve them to get the values of $a,b,c,d,e,g,h$. Please suggest some alternate solution for this.
|
Since $f(x) - x$ is a polynomial of degree $6$ and has $6$ roots $1, 2, 3, 4, 5, 6$ by condition, we can factorize $f(x) - x$ as:
$$f(x) - x = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6).$$
Plug in $x = 0$ in the above expression, we have
$3 - 0 = C\times 6!$, hence $C = \dfrac{3}{6!}$.
Therefore,
$$f(7) = 7 + (f(7) - 7) = 7 + \frac{3}{6!}(7 - 1)(7 - 2)(7 - 3)(7 - 4)(7 -5 )(7 - 6) = 7 + \frac{3}{6!}\times 6! = 10.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1374551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
How to rotate a whole rectangle by an arbitrary angle around the origin using a transformation matrix? Suppose, I have a 2D rectangle ABCD like the following:
$A(0,0)$, $B(140,0)$, $C(140,100)$, $D(0,100)$.
I want to rotate the whole rectangle by $\theta = 50°$.
I want to rotate it around the Z-axis by an arbitrary angle using a rotation transformation matrix.
How to do that?
I know that, $$ A =
\begin{bmatrix}
\
0 & 0 & 1 \\
\end{bmatrix};
B =
\begin{bmatrix}
\
140 & 0 & 1 \\
\end{bmatrix};
C =
\begin{bmatrix}
\
140 & 100 & 1 \\
\end{bmatrix};
D =
\begin{bmatrix}
\
0 & 100 & 1 \\
\end{bmatrix}.
$$
And, I know that the rotation matrix is, $$R =
\begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$
Now, what is the calculation?
I have tried the following ways,
$$Rotation =
\begin{bmatrix}
\
0 & 0 & 1 \\
140 & 0 & 1 \\
140 & 100 & 1 \\
0 & 100 & 1 \\
\end{bmatrix}.\begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$
And,
$$
A' = \begin{bmatrix}
\
0 & 0 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\\
B' = \begin{bmatrix}
\
140 & 0 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\\
C' = \begin{bmatrix}
\
140 & 100 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\\
D' = \begin{bmatrix}
\
0 & 100 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
For example,
$$B' = \begin{bmatrix}
\
140 & 0 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos 50^\circ & -sin 50^\circ & 0 \\
sin 50^\circ & cos 50^\circ & 0 \\
0 & 0 & 1 \\
\end{bmatrix} = \begin{bmatrix}
\
89.99 & -107.24 & 1 \\
\end{bmatrix}\\
C' = \begin{bmatrix}
\
140 & 100 & 1 \\
\end{bmatrix} . \begin{bmatrix}
\
cos 50^\circ & -sin 50^\circ & 0 \\
sin 50^\circ & cos 50^\circ & 0 \\
0 & 0 & 1 \\
\end{bmatrix} = \begin{bmatrix}
\
166.59 & -42.96 & 1 \\
\end{bmatrix}$$
What is the right way to work with?
|
I have solved my problem.
$$Rotation =
\begin{bmatrix}
\
0 & 0 & 0 \\
140 & 0 & 0 \\
140 & 100 & 0 \\
0 & 100 & 0 \\
\end{bmatrix}.\begin{bmatrix}
\
cos \theta & -sin \theta & 0 \\
sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$
Alternatively,
$$
A' = \begin{bmatrix}
\
0 & 0 & 0 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & sin \theta & 0 \\
-sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
= \begin{bmatrix}0 & 0 & 0\end{bmatrix}\\
B' = \begin{bmatrix}
\
140 & 0 & 0 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & sin \theta & 0 \\
-sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
= \begin{bmatrix}98.9949 & 98.9949 & 0\end{bmatrix}\\
C' = \begin{bmatrix}
\
140 & 100 & 0 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & sin \theta & 0 \\
-sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
= \begin{bmatrix}28.2843 & 169.706 & 0\end{bmatrix}\\
D' = \begin{bmatrix}
\
0 & 100 & 0 \\
\end{bmatrix} . \begin{bmatrix}
\
cos \theta & sin \theta & 0 \\
-sin \theta & cos \theta & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
= \begin{bmatrix}-70.7107 & 70.7107 & 0\end{bmatrix}\\
$$
|
{
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"url": "https://math.stackexchange.com/questions/1375076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
unresolved partial fraction decomposition I'm having some trouble doing this partial fraction decomposition:
$$\frac{1}{t^3-2t+1}$$
using Ruffini rule i get:
$$\frac{1}{t^3-2t+1}= \frac{1}{(t-1)(t^2+t-1)}$$
i would like to decompose the previous result into partial fraction.
I did in this way:
$$\frac{1}{(t-1)(t^2+t-1)}=\frac{A}{t-1}+\frac{B}{t^2+t-1} \leftrightarrow$$
$$\leftrightarrow t^2A+t(A+B)+(-A-B)=1$$
comparing the coefficients i get the following system of equations:\begin{cases} A=0 \\ A+B=0 \\ -A-B=1 \end{cases}
that are not true..
what am i doing wrong?
|
Notice, In general $$\frac{1}{(ax+b)(px^2+qx+r)}=\frac{A}{ax+b}+\frac{Bx+C}{px^2+qx+r}$$ Now, factorizing the expression, we have $$\frac{1}{(t-1)(t^2+t-1)}=\frac{A}{t-1}+\frac{Bt+C}{t^2+t-1}$$ $$\implies \frac{1}{(t-1)(t^2+t-1)}=\frac{A(t^2+t-1)+(Bt+C)(t-1)}{(t-1)(t^2+t-1)}$$ $$\implies (A+B)t^2+(A-B+C)t-(A+C)=1$$ Now, comparing the corresponding coefficients of both the sides, we get $$\begin{cases}
A+B=0\\
A-B+C=0\\
A+C=-1
\end{cases}$$ On solving the above three equations, we get $A=1, B=-1$ & $C=-2$
Hence, the required partial fractions are as follows $$\frac{1}{(t-1)(t^2+t-1)}=\frac{(1)}{t-1}+\frac{(-1)t+(-2)}{t^2+t-1}$$$$=\color{blue}{\frac{1}{t-1}-\frac{t+2}{t^2+t-1}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
coefficient of $x^{17}$ in the expansion of $(1+x^5+x^7)^{20}$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
find the coefficient of $x^{17}$ in the expansion of $(1+x^5+x^7)^{20}$
(A)3400 (B)3410 (C)3420 (D)3430 (E)3440
so it would be $$x^{140} + ...... + 1$$
This requires binomial theorem and Multinomial theorem, but I'm not sure how to calculate it. Any tips or formula would be appreciate.
|
$(1+x^5+x^7)^{20}=\{(1+x^5)+x^7\}^{20}$
$=(1+x^5)^{20}+\binom{20}1(1+x^5)^{20-1}(x^7)^1+\binom{20}2(1+x^5)^{20-2}(x^7)^2+\cdots+(x^7)^{20}$
So the required sum will be
the coefficient of $x^{17}$ in $(1+x^5)^{20}$
$+\binom{20}1\cdot$ the coefficient of $x^{17-7}$ in $(1+x^5)^{20-1}$
$+\binom{20}2\cdot$ the coefficient of $x^{17-7\cdot2}$ in $(1+x^5)^{20-2}$
Clearly the exponent of $x$ in $(1+x^5)^n$ will be divisible by $5$
So, the first & the last summand must be zero
Now for the coefficient of $x^{17-7\cdot1}$ in $(1+x^5)^{20-1},$
the $r+1$th term $\binom{19}r(x^5)^r=\binom{19}rx^{5r}$ and we need $5r=10\iff r=?$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Simplification of rational expressions I have the following expression:
$${2\over x-2} + {2 \over{x^2} -5x +6}$$
So I can simplify this as:
$${2 \over x -2} + {2 \over (x -3) (x-2)}$$
I make the common denominator to be ${(x-3)(x-2)}$
So I then apply ${(x-3)}$ to the left hand side which gives me:
$${2(x-3) + 2 \over (x-3)(x-2)}$$
I have clearly taken a wrong step because the answer in the book to the original expression is ${2 \over x-3}$ so I'm not sure how that answer was arrived at.
|
Notice, the following steps $$\frac{2}{x-2}+\frac{2}{x^2-5x+6}$$ $$=2\left(\frac{1}{x-2}+\frac{1}{(x-2)(x-3)}\right)$$ $$=\frac{2}{x-2}\left(1+\frac{1}{x-3}\right)$$ $$=\frac{2}{x-2}\left(\frac{x-3+1}{x-3}\right)$$ $$=\frac{2}{x-2}\left(\frac{x-2}{x-3}\right)$$ $$=\frac{2}{x-3}$$
Your book has the correct expression.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove by induction that $\sum_{k=1}^{n} k^3 = \bigg( \sum_{k=1}^{n}k\bigg)^2$ Show the following for all positive integers using proof by induction:
$$\sum_{k=1}^{n} k^3 = \bigg( \sum_{k=1}^{n}k\bigg)^2$$
Base case (n = 1) passes: $1^3 = 1^2$
We assume the following: $$\sum_{k=1}^{p} k^3 = \bigg( \sum_{k=1}^{p}k\bigg)^2$$
This gives us:
$$\sum_{k=1}^{p+1} k^3 = \bigg( \sum_{k=1}^{p}k\bigg)^2 + (p+1)^3 = \bigg( \sum_{k=1}^{p+1}k\bigg)^2$$
This is as far as I have gotten. I have no Idea as to how I expand it so it becomes obvious that they are in fact equal.
|
We have $$\sum_{k=1}^{n} k^3 = \bigg( \sum_{k=1}^{n}k\bigg)^2$$
1. Putting $n=1$ in the above equality, we get $$\sum_{k=1}^{1} k^3 = \bigg( \sum_{k=1}^{1}k\bigg)^2$$ $$ (1)^3=(1)^2\iff 1=1$$
Hence, the equality holds for $n=1$
*Assuming that it holds for $n=p$ then we have
$$\sum_{k=1}^{p} k^3 = \bigg( \sum_{k=1}^{p}k\bigg)^2$$
$$\color{blue}{\sum_{k=1}^{p} k^3=\left(\frac{p(p+1)}{2}\right)^2}$$
*Putting $n=p+1$, we get
$$\sum_{k=1}^{p+1} k^3 = \bigg( \sum_{k=1}^{p+1}k\bigg)^2$$
$$\sum_{k=1}^{p} k^3+(p+1)^3=\left(\frac{(p+1)(p+2)}{2}\right)^2$$
$$\sum_{k=1}^{p} k^3=\left(\frac{(p+1)(p+2)}{2}\right)^2-(p+1)^3$$
$$=(p+1)^2\left(\frac{(p+2)^2}{4}-(p+1)\right)$$
$$=(p+1)^2\left(\frac{p^2+4p+4-4p-4}{4}\right)$$
$$=(p+1)^2\left(\frac{p^2}{4}\right)$$ $$=\left(\frac{p^2(p+1)^2}{4}\right)$$
$$=\left(\frac{p(p+1)}{2}\right)^2 $$ Which is true from (2)
Hence the equality holds for all positive integers $\color{blue}{n\geq 1}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$ If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$.
I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ... but I couldn't find anything.
|
Here's something I call the "miracle method" for this type of problem. Suspend your disbelief for a moment and suppose $A$ and $B$ were scalars, not matrices. Then, we would simply be looking for
$$ \frac{1}{B} = \frac{1}{A^2 - 2A - 2} = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots$$
where in the power series expansion, the coefficient of $A^n$ is
$$ c_n = \frac{1+i}{2^{n+2}} \left((1-i)^n-i (1+i)^n\right). $$
But we know that $A^3 = 2$, so this becomes
$$ \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A}{4}-\frac{A^2}{4} + \cdots $$
and by summing the resulting coefficients on $1$, $A$, and $A^2$, we find that
$$ \frac{1}{B} = \frac{2}{5} + \frac{3}{10}A + \frac{1}{10}A^2. $$
Now, what we've just done should be total nonsense if $A$ and $B$ are really matrices, not scalars. But try setting $B^{-1} = \frac{2}{5}I + \frac{3}{10}A + \frac{1}{10}A^2$, and you'll find that, miraculously, this answer works!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
Show that for any two positive real numbers $a$ and $b$, $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$
My attempt: $(\sqrt a-\sqrt b)^2\geq0\\\frac{a+b}{2}\geq \sqrt{ab}$
$(a-b)^2\geq0\\a^2+b^2\geq2ab \\\sqrt{\frac{a^2+b^2}{2}}\geq\sqrt{ab}$
I do not know how to link them together. Appreciate any tips.
|
By Bernoulli's Inequality,
$$\begin{align}
\sqrt{\frac{a^2+b^2}{2}}
&=\sqrt{\frac{(a+b)^2+(a-b)^2}{4}}=\frac{a+b}{2}\left(1+\left(\frac{a-b}{a+b}\right)^2\right)^{\frac12}
\\
&\leq \frac{a+b}{2}\left(1+\frac{1}{2}\left(\frac{a-b}{a+b}\right)^2\right)\,.
\end{align}$$
Thus,
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}\leq \frac{(a-b)^2}{4(a+b)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{4(a+b)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}\left(\frac{\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^2}{\frac{a+b}{2}}\right)\,.$$
Now, either by the AM-GM Inequality, by the Cauchy-Schwarz Inequality, or by the Power-Mean Inequality, we have
$\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^2\leq \frac{a+b}{2}\,.$ That is,
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}\leq \frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}\left(\frac{\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^2}{\frac{a+b}{2}}\right)\leq\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2}=\frac{a+b}{2}-\sqrt{ab}\,.$$
|
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|
Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means that we can set partial derivatives of $f$ equal to $0$ and try to solve for $x$ and $y$
$\nabla f = (1+2x, 1+2y) = (0,0) \implies (x,y) = (\frac{-1}{2}, \frac{-1}{2})$
So we get the minimal value $f(\frac{-1}{2}, \frac{-1}{2}) = \frac{-1}{2} + \frac{-1}{2} + (\frac{-1}{2})^2 \frac{-1}{2})^2 = -\frac{1}{2}$
But how about the maximal value? How does $x^2 + y^2 = 1$ restrict $f$?
|
Use the lagrange multiplier !
minimize the function:
$g(x,y,\lambda) = f(x,y)+\lambda(x^2+y^2-1)$
|
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|
If $\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)+1=0$,show that $\alpha-\beta$ or $\beta-\gamma$ or $\gamma-\alpha$ is multiple of $\pi$. This question is from SL Loney.
If $\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)+1=0$,
then show that $\alpha-\beta$ or $\beta-\gamma$ or $\gamma-\alpha$ is a multiple of $\pi$.
My try: Let $\alpha-\beta=A$, $\beta-\gamma=B$, $\gamma-\alpha=C$ so that $A+B+C=0$. So we have to prove that:
If $\cos A +\cos B+\cos C+1=0$, then show that $A,B$ or $C$ is a multiple of $\pi$.
$$2 \cos\frac{A+B}{2}\cos\frac{A-B}{2}+2\cos^2\frac{C}{2}=0\\
2 \cos\frac{C}{2}\cos\frac{A-B}{2}+2\cos^2\frac{C}{2}=0\\
4 \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}=0.$$
Either $\cos\frac{A}{2}=0$ or $\cos\frac{B}{2}=0$ or $\cos\frac{C}{2}$.
Either $\frac{A}{2}$=odd multiple of $\frac{\pi}{2}$ or $\frac{B}{2}$=odd multiple of $\frac{\pi}{2}$ or $\frac{C}{2}$=odd multiple of $\frac{\pi}{2}$.
Either $A$=odd multiple of $\pi$ or $B$=odd multiple of $\pi$ or $C$=odd multiple of $\pi$.
But the answer is:
either $A$=multiple of $\pi$ or $B$=multiple of $\pi$ or $C$=multiple of $\pi$
Is my approach correct? Or is there some other method to prove it.
|
using your abbreviations, since
$$
\cos A + \cos B + \cos (A+B) + 1 = 0
$$
we have
$$
(1+\cos A)(1+\cos B) = \sin A \sin B
$$
this leads directly to
$$
\cos \frac{A}2 \cos \frac{B}2 (\cos \frac{A}2 \cos \frac{B}2 -\sin \frac{A}2 \sin \frac{B}2)=0
$$
hence either $A=n\pi$ or $B=n\pi$ or
$$
\cos \frac{A}2 \cos \frac{B}2 -\sin \frac{A}2 \sin \frac{B}2=0
$$
which gives
$$
\tan \frac{A}2 = \cot \frac{B}2
$$
or
$$
A+B = (2n+1)\pi
$$
|
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|
Compare $e^2$ and $7$ without using calculator Which is bigger? $e^2$ or $7$? Any tricks?
Don't know quite how to approach those kind of things.
|
Here's an alternative. Try to compare $2$ and $\ln(7)$. $\ln(x)$ has a more natural geometric interpretation as area under $y=\frac{1}{x}$, compared to $e^x$, so some geometry might avail us.
Since $\ln(7)=\int_1^7\frac{1}{x}\,dx$, we consider approximating the integral with a trapezoid rule style Riemann sum. The trapezoids closer to $1$ will do a worse job since concavity is higher there, so we want these trapezoids to have a narrower base. Using a linear progression of base width would mean we need to cut up $[1,7]$ into $T_n$ equal sized pieces, where $T_n$ is a triangular number. Using $T_n=15$ (and therefore $5$ trapezoids with bases $\frac{2}{5}$, $\frac{4}{5}$, $\frac{6}{5}$, $\frac{8}{5}$, $\frac{10}{5}$) gives us:
And we can see/calculate that $$\ln(7)<\frac{1}{2}\left(\frac{2}{5}\left(1+\frac{5}{7}\right)+\frac{4}{5}\left(\frac{5}{7}+\frac{5}{11}\right)+\frac{6}{5}\left(\frac{5}{11}+\frac{5}{17}\right)+\frac{8}{5}\left(\frac{5}{17}+\frac{1}{5}\right)+\frac{10}{5}\left(\frac{1}{5}+\frac{1}{7}\right)\right)$$
$$\ln(7)<\frac{65376}{32725}<2$$
So $7<e^2$.
|
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|
$\frac{1}{x+1}+\frac{1}{y+1} +\frac{1}{z+1}$ minimum value if $xyz$ =k.$x,y,z$ are positive reals. $\frac{1}{x+1}+\frac{1}{y+1} +\frac{1}{z+1}$ minimum value if $xyz$ =k.$x,y,z$ are positive reals.I think the minimum should be when $x=y=z=k^{1/3}$. How do I show it? I tried to use AM-GM inequality but it doesnt seem to work.
|
First, note that it is not true that always the function takes minimum when $x=y=z$. For e.g., let $r = \sqrt[3]k, \; x = y = rt,\, z= \dfrac{r}{t^2}$ for $r, t>0$. Then our objective function is
$$f(t) = \frac1{rt+1}+\frac1{rt+1}+\frac{t^2}{r+t^2}$$
Now consider $t \to \infty$. Clearly $f(t) \to 1$. Hence for $\dfrac3{r+1} > 1 \implies r < 2$, the minimum cannot be when $x=y=z$.
Consider now the case $r \ge 2$. Further, let $x = r \dfrac{u}v, \, y = r \dfrac{v}w , \, z = r \dfrac{w}u $. Now we will show that
$$\sum_{cyc} \frac1{x+1} = \sum_{cyc} \frac{v}{ru+v} \ge \frac3{r+1}$$
so indeed in these cases we have the minimum when $x=y=z$.
By Cauchy Schwarz inequality,
$$\sum_{cyc} \frac{v^2}{ruv+v^2} \ge \frac{(u+v+w)^2}{r(uv+vw+wu)+(u^2+v^2+w^2)}$$
So it is enough to show that
$$(r+1)(u+v+w)^2 \ge 3\left(r(uv+vw+wu)+(u^2+v^2+w^2) \right)$$
$$\iff (r-2)(u^2+v^2+w^2-uv-vw-wu) \ge 0$$
which obviously holds for $r\ge 2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
how to solve $3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$ $$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$
My answer is:
$$\begin{align}
&A = 3 - \frac {2}{A}\\
\implies &\frac {A^2-3A+2}{A}=0\\
\implies &A^2-3A+2=0\\
\implies &(A-1)\cdot(A-2)=0\\
\implies &A=1\;\text{ or }\; A=2
\end{align}$$
I should note that I'm not sure if the above answer is true. Because I expected just one answer for A (A is a numeric expression), but I found two, $1$ and $2$. This seems to be a paradox.
|
Note that as we add more terms to the continued fraction, it oscillates between $1$ and slightly higher than $2$.
$$
\begin{align}
n&=1&
3&=3&
3-2&=1\\\\
n&=2&
3-\cfrac23&=\frac73&
3-\cfrac{2}{3-2}&=1\\\\
n&=3&
3-\cfrac{2}{3-\cfrac23}&=\frac{15}7&
3-\cfrac2{3-\cfrac2{3-2}}&=1\\\\
n&=4&
3-\cfrac2{3-\cfrac2{3-\cfrac23}}&=\frac{31}{15}&
3-\cfrac2{3-\cfrac2{3-\cfrac2{3-2}}}&=1\\\\
&&
3-\cfrac2{\cfrac{2^n-1}{2^{n-1}-1}}&=\frac{2^{n+1}-1}{2^n-1}&
3-\cfrac21&=1
\end{align}
$$
Therefore, the limit of the continued fractions with $2n-1$ twos and threes is
$$
\lim_{n\to\infty}\frac{2^{n+1}-1}{2^n-1}=2
$$
and the limit of the continued fractions with $2n$ twos and threes is
$$
\lim_{n\to\infty}1=1
$$
Therefore, one value represents the limit of an odd number of twos and threes and the other value represents the limit of an even number of twos and threes.
|
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"timestamp": "2023-03-29T00:00:00",
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Quadratic formula does not work If I put the equation: $5x^2-x-4 =0$ in the quadratic formula, than I get
$x = 1$ or $x = \frac{-4}{5}$
but the real zeros are: $x = -1$ or $x = \frac{4}{5}$
Can somebody explain me if the quadratic formula fails or me?
|
The quadratic formula comes from "completing the square". Let's do that to your examples:
\begin{eqnarray*}
5x^2-x-4 &=& 0 \\ \\
x^2 - \tfrac{1}{5}x - \tfrac{4}{5} &=& 0 \\ \\
x^2 - \tfrac{1}{5}x &=& \tfrac{4}{5} \\ \\
\left(x - \tfrac{1}{10}\right)^{2} - \tfrac{1}{100} &=& \tfrac{4}{5} \\ \\
\left(x - \tfrac{1}{10}\right)^{2} &=& \tfrac{4}{5} + \tfrac{1}{100} \\ \\
\left(x - \tfrac{1}{10}\right)^{2} &=& \tfrac{81}{100} \\ \\
x - \tfrac{1}{10} &=& \pm \tfrac{9}{10} \\ \\
x &=& \tfrac{1}{10} \pm \tfrac{9}{10} \\ \\
x&=& -\tfrac{4}{5} \ \ \text{or} \ \ 1
\end{eqnarray*}
|
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|
evaluate the integral Evaluate the integral:
$$\int_0^{\pi} \frac{\cos 2\theta}{1 -2a\cos \theta +a^2}d\theta$$
The way I approach this problem is:
since, $\cos \theta = \frac{e^{it} + e^{-it}}{2}$; and $cos 2\theta = Re(z^2)$. Then, the integral will be written as follow:
$$\frac{1}{2}\int_0^{2 \pi} \frac{Re(z^2)}{1 - 2a \left( \frac{z + z^{-1}}{2}\right) +a^2}dz$$
$$ = \frac{1}{2}\int_0^{2\pi} \frac{-Re(z^2)}{(z-z_1)(z-z_2)}dz$$
where:
$$z_1 = \frac{(2a^2 - 2) + \sqrt{4(a^4+1)}}{4a}; \; z_2 = \frac{(2a^2 - 2) - \sqrt{4(a^4+1)}}{4a}$$
until here, I don't know how to use the Residue theorem to evaluate the integral. Can someone show me ?
|
Another way is to use the following Fourier expansion: for $|a| < 1$,
$$ \frac{1-a^2}{1 - 2a\cos\theta + a^2} = \sum_{n=-\infty}^{\infty} a^{|n|} e^{in\theta} = 1 + 2 \sum_{n=1}^{\infty} a^n \cos(n\theta). $$
This series immediately gives us
$$ \int_{0}^{\pi} \frac{\cos n\theta}{1 - 2a\cos\theta + a^2} \, dx = \frac{\pi a^n}{1-a^2}, \quad n = 1, 2, \cdots. $$
|
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|
$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$ $\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=$
$(A)0\hspace{1cm}(B)\frac{-\pi}{2}\hspace{1cm}(C)\frac{\pi}{2}\hspace{1cm}(D)\frac{7\pi}{2}$
I tried and got the answer but my answer is not matching the options given.Is my method not correct?
$\int\limits_{-1/2}^{1/2}(\sin^{-1}(3x-4x^3)-\cos^{-1}(4x^3-3x))dx=\int\limits_{-1/2}^{1/2}(3\sin^{-1}(x)-3\cos^{-1}(x))dx$$=3\int\limits_{-1/2}^{1/2}(\sin^{-1}(x)-\cos^{-1}(x))dx=3\int\limits_{-1/2}^{1/2}(\frac{\pi}{2}-2\cos^{-1}(x))dx=\frac{3\pi}{2}-6\int\limits_{-1/2}^{1/2}\cos^{-1}(x)dx$
$=\frac{3\pi}{2}+6\int\limits_{2\pi/3}^{\pi/3}t \sin t dt=\frac{-3\pi}{2}$
|
Hint:
$$\arcsin{a} - \arccos{(-a)} = \arcsin{a}+\arcsin{(-a)} - \frac{\pi}{2} = - \frac{\pi}{2}$$
|
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|
Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understand?
$$\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$
How do I prove the above equation for all integers where $n\geq1$?
|
First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}\frac{1}{k(k+2)}=\frac34-\frac{2+3}{2(1+1)(1+2)}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}\frac{1}{k(k+2)}=\frac34-\frac{2n+3}{2(n+1)(n+2)}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}\frac{1}{k(k+2)}=$
$\color\red{\sum\limits_{k=1}^{n}\frac{1}{k(k+2)}}+\frac{1}{(n+1)(n+3)}=$
$\color\red{\frac34-\frac{2n+3}{2(n+1)(n+2)}}+\frac{1}{(n+1)(n+3)}=$
$\frac34-\frac{2(n+1)+3}{2(n+2)(n+3)}$
Please note that the assumption is used only in the part marked red.
|
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|
Find $\cos{A}+\cos{B}$
In $\Delta ABC$,if
$$\cos{C}\cdot(\sin{A}+\sin{B})=\sin{C}\cdot\cos{(A-B)}$$
Find $\cos{A}+\cos{B}$
Thus
$$\sin{A}+\sin{B}=\tan{C}\cdot\cos{(A-B)}=\dfrac{\sin{(A+B)}}{\cos{(A+B)}}\cos{(A-B)}$$
$$(\sin{A}+\sin{B})\cos{(A+B)}=\sin{(A+B)}\cos{(A-B)}$$
follow I can't figure it out.
|
Let me try. $\cos C (\sin A +\sin B)=\sin C \cos(A-B)$
$\implies2\left(1-2\sin^2\dfrac C2\right)\sin\dfrac{A+B}2\cos\dfrac{A-B}2 = 2\sin \dfrac C2\cos\dfrac C2\left(2\cos^2\dfrac{A-B}2-1\right)\ \ \ \ (1)$
Now as $\sin\dfrac{A+B}2=\sin\left(\dfrac\pi2-\dfrac C2\right)=\cos\dfrac C2$
and $\cos\dfrac C2=0\implies\dfrac C2=(2n+1)\dfrac\pi2\iff C=(2n+1)\pi$ where $n$ is any integer
But $0<C<\pi\implies\sin\dfrac{A+B}2=\cos\dfrac C2\ne0$
$(1)\implies$
$$\left(1-2\cos^2\dfrac{A+B}2\right)\cos\dfrac{A-B}2 = \cos\dfrac{A+B}2\left(2\cos^2\dfrac{A-B}2-1\right)$$
$$\implies\left(\cos\dfrac{A-B}2+\cos\dfrac{A+B}2\right)\left(1-2\cos\dfrac{A-B}2\cos\dfrac{A+B}2\right)=0$$
Now, $\cos\dfrac{A-B}2+\cos\dfrac{A+B}2=2\cos\dfrac A2\cos\dfrac B2$ which can not be zero due to the reason mentioned above.
$$\implies1-2\cos\dfrac{A-B}2\cos\dfrac{A+B}2=0$$
But $2\cos\dfrac{A-B}2\cos\dfrac{A+B}2=\cos A+\cos B$
|
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|
If $x^3+y^3=72$ and $xy=8$ then find the value of $x-y$. I recently came across a question,
If $x^3+y^3=72$ and $xy=8$ then find the value of $(x-y)$.
By trial and error I found that $x=4$ and $y=2$ satisfies both the conditions. But in general how can I solve it analytically? I tried using $a^3+b^3=(a+b)(a^2+b^2-ab)$ and also $a^3+b^3=(a+b)^3-3ab(a+b)$. But both ways aren't working.
Please explain how do I solve these types of questions analytically.
|
Hint:
$$x^3+y^3=72\text{ and }x^3y^3=512.$$
You know the sum, and you know the product of two cubes.
The cubes are $\dfrac{72\pm\sqrt{72^2-4\cdot 512}}2$, $8$ and $64$.
|
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|
Completing the square of $(x+a)(x+b)$ The problem is simple, to complete the square of $(x+a)(x+b)$. My calculations yield
$$\left(x+\frac{a+b}{2}\right)^2-\frac{(a+b)^2}{4}+ab,$$
But the textbook's answer is different ("problem 361", at the bottom of the page):
$$\left(x+\frac{a+b}{2}\right)^2-\frac{(a-b)^2}{4}$$
Did I do anything the wrong way?
$$(x+a)(x+b)=x^2+xb+ax+ab=x^2+x(a+b)+ab=$$
$$=\left(x^2+2*\frac{a+b}{2}*x+\left(\frac{a+b}{2}\right)^2\right)-\left(\frac{a+b}{2}\right)^2+ab=$$
$$=\left(x+\frac{a+b}{2}\right)^2-\frac{(a+b)^2}{4}+ab$$
|
You did nothing wrong.
Note that
$$-\frac{(a+b)^2}{4}+ab=\frac{-(a+b)^2+4ab}{4}=\frac{-a^2+2ab-b^2}{4}=-\frac{(a-b)^2}{4}$$
|
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|
Find $\int \frac{5x^4+4x^5}{(x^5+x+1)^2}$ $\displaystyle \int \frac{5x^4+4x^5}{(x^5+x+1)^2}$
Since in the denominator of the integrand,$(x^5+x+1)^2$ is there.So the answer must be in the form $\displaystyle \frac{f(x)}{(x^5+x+1)}$, but i could not figure out what $f(x)$ i should suppose.I could not imagine any other way out.Please help...
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$\displaystyle I = \int \frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \int\frac{5x^4+4x^5}{x^{10}\cdot \left(1+x^{-4}+x^{-5}\right)}dx = \int\frac{(5x^{-6}+4x^{-5})}{(1+x^{-4}+x^{-5})^2}dx$
Now Put $(1+x^{-4}+x^{-5}) = t\;,$ Then $\displaystyle \left(4x^{-5}+5x^{-6}\right)dx = -dt$
So $\displaystyle I = -\int\frac{1}{t^2} = \frac{1}{t}+\mathcal{C} = \frac{x^5}{x^5+x+1}+\mathcal{C}$
|
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|
Family of Lines Problem:
Consider a family of straight lines $(x+y)+\lambda(2x-y+1)=0$. Find the equation of the straight line belonging to this family which is farthest from $(1,-3)$.
$$$$
Any help with this problem would be really appreciated!
|
Notice, the equation of straight lines $$x+y+\lambda(2x-y+1)=0$$ $$(2\lambda +1)x+(1-\lambda)y+\lambda=0$$
Now, the distance say $D$ of the above line from the given point $(1, -3)$ is given as follows $$D=\left|\frac{(2\lambda +1)(1)+(1-\lambda)(-3)+\lambda}{\sqrt{(2\lambda +1)^2+(1-\lambda )^2}}\right|$$
$$=\left|\frac{(2\lambda +1)(1)+(1-\lambda)(-3)+\lambda}{\sqrt{(2\lambda +1)^2+(1-\lambda )^2}}\right|$$
$$=\left|\frac{6\lambda-2}{5\lambda^2+2\lambda+2}\right|$$
Now, considering the following function: $f(\lambda)=\frac{6\lambda-2}{5\lambda^2+2\lambda+2}$
$$\frac{df(\lambda)}{d\lambda}=\frac{d}{d\lambda}\left(\frac{6\lambda-2}{5\lambda^2+2\lambda+2}\right)$$ $$=\frac{(5\lambda^2+2\lambda+2)(6)-(6\lambda-2)(10\lambda+2)}{(5\lambda^2+2\lambda+2)^2}$$
$$=\frac{-30\lambda^2+20\lambda+16}{(5\lambda^2+2\lambda+2)^2}$$
Now, for maximum value of $D$ we have $$\frac{df(\lambda)}{d\lambda}=0$$
$$\frac{-30\lambda^2+20\lambda+16}{(5\lambda^2+2\lambda+2)^2}=0$$
$$15\lambda^2-10\lambda-8=0$$ $$\lambda=\frac{5\pm\sqrt{145}}{15}$$$$\iff \lambda =\frac{5+\sqrt{145}}{15}\ \vee\ \lambda=\frac{5-\sqrt{145}}{15}$$
Now, find $\frac{d^2f(\lambda)}{d\lambda^2}$ & substitute these values of $\lambda $ individually in $\frac{d^2f(\lambda)}{d\lambda^2}$ then you will find that for $\lambda=\frac{5+\sqrt{145}}{15}$, $\frac{d^2f(\lambda)}{d\lambda^2}<0$
Thus, the function $f(\lambda)$ will be maximum & hence the distance $D$ of the given point from the line will be maximum at $\lambda=\frac{5+\sqrt{145}}{15}$
Now, substituting this value of $\lambda $ in the given equation of family of straight lines we get the equation of the straight line farthest from the given point $(1, -3)$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{x+y+\left(\frac{5+\sqrt{145}}{15}\right)(2x-y-1)=0}}$$
|
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|
For what $p$ is $\frac{1}{(x(1+\ln(x)^2))^p}$ Lebesgue integrable? I'm trying to use the fact that given $f:[a,\infty)\to\mathbb{R}$ Riemann integrable for every closed interval $[c,d]\subset [a,\infty)$, then $f$ is Lebesgue integrable if, and only if, $\int_a^\infty|f(x)| \, dx$ exists.
In particular, $f(x)=\frac{1}{x(1+\ln(x)^2)}$ is $p$-Lebesgue integrable if $\int_0^\infty\frac{1}{x^p(1+\ln(x)^2)^p} \, dx<\infty $. Here using the fact that $f>0$.
But I can't solve this Riemann integral.
|
Using the inequality $\frac{x}{x+1} < \log (1+x) < x$ for ll $x>-1$ and $x \ne 0$. Let $x=u-1$ then the inequality becomes $\frac{u-1}{u} < \log (u) < u-1$ for all $u>0$. Thus, $\frac{(x-1)^2}{x^2}< (\ln x)^2<(x-1)^2$ so that $1+\frac{(x-1)^2}{x^2}<1+ (\ln x)^2<1+(x-1)^2$ which implies that $\frac{1}{1+\frac{(x-1)^2}{x^2}}>\frac{1}{1+ (\ln x)^2}>\frac{1}{1+(x-1)^2}$.
Thus, $\frac{x^{2p}}{(x^2+(x-1)^2)^p}>\frac{1}{(1+ (\ln x)^2)^p}>\frac{1}{(1+(x-1)^2)^p}$.
Alos, since $x>0$ then, $$\frac{1}{x^p(1+ (\ln x)^2)^p}<\frac{x^{p}}{(x^2+(x-1)^2)^p}.$$
Using Maple 12, I tried to integrate $\int_0^{\infty}{\frac{x^{p}}{(x^2+(x-1)^2)^p}dx}$ I get that the integral converges when $p\ge2$, diverges when $p<2$.
|
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How to solve the equation $x^3+y^3=0$ for real numbers $x$ and $y$? I'm finding stationary points of the function $f(x,y)=2(x-y)^2-x^4-y^4$, but stuck in the equation $x^3+y^3=0$ while solving the equations $f_x=0$ and $f_y=0$.
Please help me. Thanks in advance.
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HINT:
Notice, $$x^3+y^3=(x+y)(x^2+y^2-xy)$$ Then, we have $$(x+y)(x^2+y^2-xy)=0$$
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How to find the rightmost 25 digits in $100!$? The question is
Find the rightmost 25 digits in decimal expansion of $100!=1\times 2\times \dotsb \times100$
By counting the number of fives in the prime factorisation of $100!$, I know there are $\lfloor {100 \over 5} \rfloor + \lfloor {100 \over {5^2}} \rfloor =24$ trailing zeros.
But I am really stuck with finding the remaining 25th digit, could someone help me please?
|
Just figured out this myself.
It is easy to determine that there are 24 trailing zeros, so the only thing left is to determine $\frac{100!}{10^{24}}\pmod {10}$
Writing the fraction as $$\frac{(1 \cdot 3 \cdot 7 \cdot 9 \ \dotsc \cdot 99)(5 \cdot 10 \cdot 15 \cdot \dotsc \cdot 100)(2 \cdot 4 \cdot 6 \cdot 8 \cdot \dotsc \cdot 98) }{10^{24}} \equiv \frac{(1 \cdot 3 \cdot 7 \cdot 9)^{10} \cdot 5^{20} \cdot(1 \cdot 2 \cdot 3 \cdot \dotsc \cdot 20) \cdot 2^{40} \cdot (1 \cdot 2 \cdot 3 \cdot 4)^{10} }{5^{24} \cdot 2^{24}}\equiv \frac{(9)^{10} \cdot(1 \cdot 2 \cdot 3 \cdot \dotsc \cdot 20) \cdot 6 \cdot (4)^{10} }{5^{4}} \equiv 1 \cdot (1 \cdot 2 \cdot 3 \cdot 4)^4 \cdot 6 \cdot 6 \equiv 4$$
So the 25th digit is 4.
|
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|
Find inverse of 15 modulo 88. Here the question: Find an inverse $a$ for $15$ modulo $88$ so that $0 \le a \le 87$; that is, find an integer $a \in \{0, 1, ..., 87\}$ so that $15a \equiv1$ (mod 88).
Here is my attempt to answer:
Find using the Euclidean Algorithm, we need to find $\gcd(88, 15)$, that must equal to $1$ to be possible to find an inverse of $15 \pmod{88}$.
\begin{align*}
88 & = 5 \times 15 + 13\\
15 & = 1 \times 13 + 2\\
13 & = 6 \times 2 + 1\\
2 & = 2 \times 1 + 0
\end{align*}
So,
$$\gcd(88, 15) = 1$$
Now, we need to write this into the form:
$$\gcd(88, 15) = 88x + 15y.$$
And find $x$ and $y$.
\begin{align*}
1 & = 13(1) + 2(-6)\\
& = 13(7) + 15(-6)\\
& = 88(7) + 15(-41)
\end{align*}
So, $x = 7$ and $y = -41$.
So, an inverse of $15 \pmod{88} = -41$. Now, I need to find an inverse that is between $0$ and $87$. What is a good easy approach to find other inverses? Any ideas please?
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$15n=88k+1$. But $15n$ ends either in $0$ or in $5$. So we are looking for a multiple of $88$ which ends either in $9$ or in $4$. The former case is impossible, since $88$ is an even number. Now, what is the first multiple of $8$ to end in a $4$ ? So $k_0=3$ is our first suspect. Unfortunately, in this case, $n_0\not\in\mathbb N$. So let's check $k_1=3+5=8$, since the cycle of $8x\bmod10$ has legth $5$. Bingo ! We have $n_1=47$.
|
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$\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+.....+\cos(\alpha+(n-1)\beta)=0 $ If each side of a regular polygon of $n$ sides subtend an angle $\alpha$ at the center of the polygon and each exterior angle of the polygon is $\beta$,then prove that $\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+.....+\cos(\alpha+(n-1)\beta)=0 $
Since this is a regular polygon.Therefore,each $\alpha=\frac{2\pi}{n}$ and since each external angle is $\beta$.So by geometry,$\alpha=\beta.$
$\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+.....+\cos(\alpha+(n-1)\beta)=\frac{\cos\frac{n\beta}{2}}{\cos\frac{\beta}{2}}\cos\frac{2\alpha+(n-1)\beta}{2}$
Now putting $\alpha=\beta=\frac{2\pi}{n}$ does not give me answer.What mistake did i make?
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Notice, in a regular polygon with $n$ number of side,
the angle subtended by each side at the center of the polygon $$\alpha=\frac{2\pi}{n}$$
each exterior angle of the polygon $$\beta=\pi-\frac{(n-2)\pi}{n}=\frac{2\pi}{n}$$
$$\implies \alpha=\beta=\frac{2\pi}{n}$$ Now, we have
$$\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ldots +\cos(\alpha+(n-1)\beta)$$ setting $\beta=\alpha$
$$=\cos \alpha+\cos(\alpha+\alpha)+\cos(\alpha+2\alpha)+\ldots +\cos(\alpha+(n-1)\alpha)$$
$$=\cos \alpha+\cos(2\alpha)+\cos(3\alpha)+\ldots +\cos(n\alpha)$$ $$=\frac{1}{2}\left[\sin n\alpha\cot\frac{\alpha}{2}+\cos n\alpha-1\right]$$
Setting $\alpha=\frac{2\pi}{n}$ $$=\frac{1}{2}\left[\sin n\left(\frac{2\pi}{n}\right)\cot\frac{1}{2}\left(\frac{2\pi}{n}\right)+\cos n\left(\frac{2\pi}{n}\right)-1\right]$$
$$=\frac{1}{2}\left[\sin 2\pi\cot\frac{\pi}{n}+\cos 2\pi-1\right]$$
$$=\frac{1}{2}\left[0+1-1\right]=0$$ Hence, proved that
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\ldots +\cos(\alpha+(n-1)\beta)=\color{blue}{0}}}$$
|
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Evaluation of $\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$
Evaluation of $$\int\frac{x^7+2}{\left(x^2+x+1\right)^2}dx$$
$\bf{My\; Try::}$ Let $$\displaystyle \mathop{I = \int\frac{x^7+2}{(x^2+x+1)^2}}dx = \int\frac{(x^7-1)+3}{(x^2+x+1)^2}dx$$
$$\mathop{\displaystyle = \int\frac{x^7-1}{(x^2+x+1)^2}}+\displaystyle \int\frac{3}{(x^2+x+1)^2}dx$$
Now Using the formula $$\bullet x^7-1 = (x-1)\cdot \left[x^6+x^5+x^4+x^3+x^2+x+1\right]$$
So $$\bullet \; (x^7-1) = (x-1)\left[(x^4+x)\cdot (x^2+x+1)+1\right]$$
So we get $$\displaystyle I = \int\frac{(x-1)\cdot (x^4+x)\cdot (x^2+x+1)}{(x^2+x+1)^2}dx+\frac{1}{2}\int\frac{2x-2}{(x^2+x+1)^2}+3\int\frac{1}{(x^2+x+1)^2}dx$$
$$\displaystyle I = \underbrace{\int\frac{(x-1)(x^4+x)}{x^2+x+1}dx}_{J}+\underbrace{\frac{1}{2}\int\frac{(2x+1)}{(x^2+x+1)^2}dx}_{K}+\underbrace{\frac{3}{2}\int\frac{1}{(x^2+x+1)^2}dx}_{L}$$
Now $$\displaystyle J = \int\frac{(x-1)(x^4-x)+2x(x-1)}{(x^2+x+1)}dx = \int(x^3-2x^2+x)dx+\int\frac{2x^2-2x}{x^2+x+1}dx$$
Now we can solve the integral Using the formulae.
My question is can we solve it any other way, If yes then plz explain here
Thanks
|
Here is a sketch - not a full solution:
Say we know $\int\cos^2x\tan^jx dx$ for $j=0,1,...,7$. Then using $x+\frac12=\frac{\sqrt3}{2}\tan u$ we have \begin{align}
\int\frac{x^7+2}{(x^2+x+1)^2}dx&=\int\frac{x^7+2}{\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right)^2}dx\\
&=\frac{\sqrt3}{2}\int\frac{\Big(\frac{\sqrt3}{2}\tan u-\frac12\Big)^7+2}{\frac{9}{16}(1+\tan^2u)}du\\
&=\frac{\sqrt3}{2}\frac{32}{9}\int\cos^2udu+\frac{\sqrt3}{2}\sum_{j=0}^7\Big(-\frac12\Big)^j\Big(\frac{\sqrt3}{2}\Big)^{7-j}\frac{16}{9}\int\cos^2u\tan^{7-j}udu
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
$9 \mid 4n^2 + 15n - 1$ for $n \in \mathbb N$ How to prove by induction that $9 \mid 4n^2 + 15n - 1$ for every $n \in \mathbb N$?
For $n = 1$
$4 \cdot 1^2 + 15 \cdot 1 - 1 = 18$
For $n \ge 2$
If $4n^2 + 15n - 1 = 9k$ then $4(n+1)^2 + 15(n+1) - 1 = 4n^2 + 23n + 18 = 9k + 8n + 19$
|
If $4n^2+15n-1$ is divisible by $9$, then it's also divisible by $3$; however
$$
4n^2+15n-1\equiv n^2-1\pmod{3}
$$
but $n^2\equiv 1\pmod{3}$ if and only if $3\nmid n$. So, for $n=3k$, the number $4n^2+15n-1$ is not divisible by $3$ and, of course, not divisible by $9$ either.
|
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|
Linear transformation using non-standard basis Find a linear transformation $T:P_{2}({R})\longrightarrow P_{4}(R)$ so that
$T(1) = x^4$
$T(x+x^2) = 1$
$T(x-x^2) = x+x^3$
I have only solved problems using standard basis, and now I have no idea on how to deal with this.
|
You have that
$$x = \frac{(x+x^2) + (x-x^2)}{2}$$
So
$$T(x) = T\left( \frac{(x+x^2) + (x-x^2)}{2} \right) = \frac{T(x+x^2) + T(x-x^2)}{2}$$
$$ = \frac{1+x+x^3}{2}$$
And
$$T(x^2) = T(x+x^2-x) = T(x+x^2)-T(x)$$
$$=1 - \frac{1+x+x^3}{2} $$
|
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|
Solving a rational equation with multiple and nested fractions
This is the equation to solve: $\dfrac{\dfrac{x+\dfrac{1}{2}} {\dfrac{1}{2}+\dfrac{x}{3}}}{\dfrac{1}{4}+\dfrac{x}{5}}=3$
What I did:
$x+\dfrac{1}{2}=\dfrac{2x+1}{2}$
$\dfrac{x}{3}+\dfrac{1}{2}=\dfrac{2x+3}{6}$
$\dfrac{x}{5}+\dfrac{1}{4}=\dfrac{4x+5}{20}$
$\dfrac{2x+1}{2}\div \dfrac{2x+3}{6} = \dfrac{2x+1}{2}\times \dfrac{6}{2x+3}= \dfrac{6x+3}{2x+3}$
$\dfrac{6x+3}{2x+3}\div\dfrac{4x+5}{20}=\dfrac{6x+3}{2x+3}\times\dfrac{20}{4x+5}=3$
$\implies\dfrac{120x+60}{(2x+3)(4x+5)}=3$
I know how to solve the equation. But right now I tried several times and I got wrong answers.
So I appreciate your help. Thank you.
|
Notice, $$\frac{\frac{x+\frac{1}{2}}{\frac{1}{2}+\frac{x}{3}}}{\frac{1}{4}+\frac{x}{5}}=3$$
$$\frac{\frac{3(2x+1)}{(2x+3)}}{\frac{4x+5}{20}}=3$$
$$\frac{60(2x+1)}{(2x+3)(4x+5)}=3$$
$$8x^2-18x-5=0$$
$$x=\frac{18\pm\sqrt{(-18)^2-4(8)(-5)}}{2(8)}$$
$$x=\frac{18\pm22}{16}$$ $$x=\frac{5}{2}$$ or $$ x=-\frac{1}{4}$$
|
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|
Find the value of function at $\,x=5$ If $\,f(x)\,$ is a non-constant polynomial of $\,x\,$ such that $\,f\left(x^3\right)-f\left(x^3-2\right)=f^2\left(x\right)+12\,$ is true for all $\,x\,$ then find the value of $\,f\left(5\right).\,$
|
HINT: Substitute $\,f(x) = ax^3 + bx^2 + cx + d$ into your equation and collect coefficients of the different powers of $x$.
Solve obtained system of linear equations, thus getting an explicit expression for $f$.
Substitute $\,x = 5\,$ into obtained equation and get the answer.
How do we know that $f$ is the third-order polynomial?
On the left we have expression involving $\;f\left(x^3\right),\,$ and on the right $\;f^2\left(x\right).\,$
Assuming $\,f\,$ is a polynomial of degree $\,n\in\mathbb R,\,$ we conclude that
$\,3\left(n-1\right) = 2n.\,$
Therefore we conclude that polynomial $\,f\,$ is of order $3$, i.e.
$\;f(x) = ax^3 + bx^2 + cx + d.\,$
EDIT: Upon request of @AjaySharma I provide explicit solution:
$$
f(x) = a x^3 + b x^2 + c x + d
\implies
\begin{cases}
f\left(x^3\right) = a x^9 + b x^6 + c x^3 + d
\\
f\left(x^3 - 2\right) = a \left(x^3-2\right)^3 + b \left(x^3-2\right)^2 + c \left(x^3-2\right) + d
\end{cases}
$$
Thus the difference
$$
\begin{aligned}
f\left(x^3\right)-f\left(x^3-2\right)
& = a \left(\left(x^3\right)^3 - \left(x^3-2\right)^3 \right)
+ b \left( \left(x^3\right)^2 - \left(x^3-2\right)^2 \right)
+ c \left( \left(x^3\right) - \left(x^3-2\right) \right)
\\ & =
2a \left(\left(x^3\right)^2 + \left(x^3\right) \left(x^3-2\right)+ \left(x^3-2\right)^2 \right) +
2b \left( 2x^3-2 \right) + 2c
\\ & =
2a \left({x^6} + {x^6} - {2x^3} + {x^6} - {4x^3} + 4 \right) +
4b \left( x^3-1 \right) + 2c
\\ & =
6 a x^6 + \left(4 b - 12 a\right) x^3 + 8 a - 4 b + 2 c
\end{aligned}
$$
On the other hand,
$$
\begin{aligned}
f^2\left(x\right) & = \left(a x^3 + b x^2 + c x + d \right)^2 =
\left(a x^3 + b x^2 + c x + d \right)\cdot \left(a x^3 + b x^2 + c x + d \right)
= \\
& = a^2x^6 + \left(2ab\right)x^5 + \left(b^2 + 2ac\right)x^4 + \left(2ad + 2bc\right)x^3 + \left(c^2 + 2bd\right)x^2 + \left(2cd\right)x + d^2
\end{aligned}
$$
Therefore $\qquad f\left(x^3\right)-f\left(x^3-2\right) = f^2\left(x\right)+12
\implies $
$$
\begin{aligned}
\implies&
\ 6 a x^6 + \left(4 b - 12 a\right) x^3 + 8 a - 4 b + 2 c =
\\
=&\ a^2x^6 + \left(2ab\right)x^5 + \left(b^2 + 2ac\right)x^4 + \left(2ad + 2bc\right)x^3 + \left(c^2 + 2bd\right)x^2 + \left(2cd\right)x + d^2 + 12
\end{aligned}
$$
Let us gather coefficients by powers of $x$:
$$
\begin{aligned}
&x^6:& a^2 &= 6 a &\implies& & a &= 6 \\
&x^5:& 2ab &= 0 &\implies& & b &= 0 \\
&x^4:& b^2+2ac &= 0 &\implies& & c &= 0 \\
&x^3:& 2ad + 2bc &= 4 b - 12 a &\implies& & d &= -6 \\
&x^2:& c^2 + 2bd &= 0 \\
&x^1:& 2cd &= 0 \\
&x^0:& d^2 + 12 &= 8a - 4b + 2c
\end{aligned}
$$
Thus we get
$$
\bbox[5pt, border:2.5pt solid #FF0000]{f\left(x\right) = 6x^3-6}
$$
and so $\,f\left(5\right) = 744 .\,$
|
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|
Probability that the eventually a six on a dice will appear.
Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $ m$ and $ n$ be relatively prime positive integers such that $ \frac{m}{n}$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $ m+n$.
If you roll a die $r$ times, the probability of getting a six on the $r$th try is:
$\frac{1}{6} \cdot \frac{5^{r-1}}{6^{r-1}}$
Now I believe casework for Dave so:
1 roll: $ \frac{1}{6} $, 2 rolls: $\frac{1}{6} \frac{5}{6}$, 3 rolls: $\frac{1}{6} \frac{5^2}{6^2}$, .... , $r$ rolls, $\frac{1}{6} \frac{5^{r-1}}{6^{r-1}}$
I would add this up, but still, we never found $r$?
HINTS ONLY
EDIT
As pointed by 5xum, $L=1$ isn't possible with $D = L - 1$. So I got:
$P = 1 \cdot (1/6)\sum_{r=2}^{\infty} (5/6)^{r-2} + 1 + ...$
But $P > 1$ already, which is impossible?
|
Let $D, L$ be the obvious random variables. Then the desired probability is
\begin{align}
P & = P(D = 1, L = 1) + P(D = 1, L = 2) + P(D = 2, L = 1) \\
& + P(D = 2, L = 2) + P(D = 2, L = 3) + P(D = 3, L = 2) \\
& + P(D = 3, L = 3) + P(D = 3, L = 4) + P(D = 4, L = 3) \\
& + \cdots
\end{align}
Each row constitutes an element in a geometric series. Find the first row sum and the ratio, and the rest is trivial.
|
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|
Proving uniqueness of solutions to $\sin^2A + \sin^2B = \sin (A+B)$ without using multivariable calculus In the course of solving a trigonometric problem (see $a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle), in one approach the following equation needed to be solved:
$$\boxed{\sin^2A + \sin^2B = \sin (A+B)}$$
subject to $A\in(0,\frac{\pi}{2}),B\in(0,\frac{\pi}{2}),\pi-(A+B) > \max(A,B)$, i.e. $A$ and $B$ are the two angles of a triangle not opposite the longest side.
Clearly, any $A,B\:|\:A+B=\frac{\pi}{2}$ is a family of solutions. Since multivariable calculus is presumably beyond the level of the original problem:
How to prove that there are no other solutions $\underline{\text{without}}$ using multivariable calculus?
[I don't think a trig identity will suffice as there are other solutions if the restrictions on $A,B$ are relaxed.]
(For completeness - using multivariable calculus)
Part 1
Proof that $\sin^2A + \sin^2B < \sin (A+B)$ over region $R_1=\{0<A,B\land A+B<\frac{\pi}{2}\}$.
Consider $$f(x,y)=\sin^2x + \sin^2y - \sin (x+y)$$
Then
$$\begin{align}
f_x &= \sin 2x - \cos(x+y) \\
f_y &= \sin 2y - \cos(x+y) \\
f_{xx} &= 2\cos 2x + \sin(x+y) \\
f_{yy} &= 2\cos 2y + \sin(x+y) \\
f_{xy} &= \sin(x+y) &= f_{yx} \\
\end{align}$$
For local extrema we require $f_x=f_y=0$. But
$$f_x=f_y=0 \implies \sin2x=\sin2y \implies y=x \lor y=\frac{\pi}{2}-x$$
Exclude $y=\frac{\pi}{2}-x$ as it violates the restriction that $x+y<\frac{\pi}{2}$.
If $y=x$, $f_x=0 \implies \sin2x=\sin2y \implies x=y=\frac{\pi}{8}$.
The determinant of the Hessian is
$$D(x,y) = \begin{vmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{vmatrix} = f_{xx}f_{yy} - f_{xy}f_{yx} = f_{xx}f_{yy} - (f_{xy})^2$$
At $P(\frac{\pi}{8},\frac{\pi}{8})$, we have
$$\begin{align}
f_{xx}&=2\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{3}{\sqrt2} \\
f_{yy}&=2\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\frac{3}{\sqrt2} \\
f_{xy}&=\sin\frac{\pi}{4}=\frac{1}{\sqrt2} \\
D(x,y)&=\frac{9}{2}-\frac{1}{2}=4
\end{align}$$
Since both $f_{xx}$ and $D$ are positive at $P$, this is a local minimum (with $f(x,y)|_P=1-\sqrt2$).
On the boundaries:
*
*$f(0,y)=\sin^2y-\sin y<0$ on $x=0,y\in(0,\frac{\pi}{2})$
*$f(x,0)=\sin^2x-\sin x<0$ on $x\in(0,\frac{\pi}{2}),y=0$
*$f(0,0)=0$
*$f(x,y)=0$ for $x,y\geq0,x+y=\frac{\pi}{2}$
Since $(0,0)$ is not part of the domain, and there are no other local extrema, we have $f(x,y)<0$ over $R_1$.
Part 2
It will be sufficient to prove that $\sin^2A + \sin^2B > \sin (A+B)$ over region $R_2=\{0<A,B<\frac{\pi}{2} \land A+B>\frac{\pi}{2} \}$.
|
Well, fix $B$ and show that there's a unique solution for $A$. In other words, we want to show that for $0 < B < \pi/2$ the function
$$f_B(A) = \sin^2(A) + \sin^2(B) - \sin(A+B)$$
has a unique zero on the interval $(0, \pi/2)$.
First, note that
$$f_B(0) = \sin^2(B) - \sin(B) = \sin(B)[\sin(B) - 1] < 0$$
since $0 < \sin(B) < 1$. Now consider
\begin{align*}
f'_B(A) &= 2\sin(A) \cos(A) - \cos(A+B) \\
&= \sin(2A) - \cos(A+B)
\end{align*}
We have
$$f'(0) = - \cos(B) < 0$$
so $f_B$ starts out negative and decreasing. Now, where can $f'_B$ vanish? Well, that happens when
$$\sin(2A) = \cos(A+B).$$
Now $\sin(2A)$ increases from zero to one on $(0, \frac{\pi}{4})$ while $\cos(A+B)$ decreases from $\cos(B) > 0$ to $\cos(B + \frac{\pi}{4})$, which is between 0 and 1, on the same interval. Consequently there is some solution to this equation with $A \in (0, \pi/4)$. Moreover, there is only one solution on $(0, \pi/4)$, since to the right of this solution we have $\sin(2A)$ increasing and $\cos(A+B)$ decreasing.
What about solutions to
$$\sin(2A) = \cos(A+B).$$
when $\frac{\pi}{4} < A < \frac{\pi}{2}$? Well,
$$\cos(A+B) < \cos(A) < \sin(2A)$$
on this interval.
So we've shown that on the interval $(0, \pi/2)$ the function $f'_B(A)$ has a unique zero. Together with the facts that $f_B(0) < 0$ and $f'_B(0) < 0$ this means that on $(0, \pi/2)$ the function $f_B$ decreases from a negative value at 0 to a negative minimum, and then increases for the rest of the interval. Consequently, $f_B$ can have at most one zero on this interval.
I suspect if you also object to single-variable calculus you could eliminate it from the argument and do things entirely with trigonometry, although the argument would necessarily be longer.
|
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|
Polynomial with real roots Consider the polynomial: $$f=X^4+4X^3+6X^2+aX+b$$
We know that $f$ has four real roots. Let $x_1,x_2,x_3,x_4$ be the roots of this polynomial. How can one compute $$x_1^{2015}+x_2^{2015}+x_3^{2015}+x_4^{2015}?$$
If $a=4$ and $b=1$, we obtain a self-reciprocal (palindromic) polynomial. We can write $f=(X+1)^4$, thus $x_1=x_2=x_3=x_4=1$. Hence the sum computes to $-4$.
Are there any other cases to consider ($a,b$)? I thought using the formula for the quartic equation and paying attention to the cases where we have only real roots.
Any ideas? Thank you!
|
We know the following to be true:
$$
\tag1x_1+x_2+x_3+x_4 = -4
$$
$$
\tag2x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1x_3+x_2x_4 = 6
$$
$$
\tag3x_1 x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2 = -a
$$
$$
\tag4x_1x_2x_3x_4 = b
$$
Now we can square $(1)$ to get
$$
\begin{align}
(x_1+x_2+x_3+x_4)^2 = &x^2_1+x^2_2+x^2_3+x^2_4\\
&+2[x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1+x_3+x_2x_4]
\end{align}
$$
Substituting in values from $(1)$ gives
$$
\begin{align}
(-4)^2 &= x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}+2(6)\\
16&=x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}+12\\
4&=x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}
\end{align}
$$
Using the Cauchy-Schwartz Inequality, we get
$$
\left(x^2_{1}+x^2_{2}+x^2_{3}+x^2_{4}\right)(1^2+1^2+1^2+1^2)\geq (x_{1}+x_{2}+x_{3}+x_{4})^2
$$
where equality holds when
$$
\tag5x_{1} = x_{2} = x_{3} = x_{4}
$$
Thus, the equality condition holds, because $4\cdot 4= (-4)^2$. So from $(1)$ and $(5)$, we have $$x_{1} = x_{2} = x_{3} = x_{4}=-1$$
Setting the values of $x_{1} = x_{2} = x_{3} = x_{4}=-1$ in $(3)$ and $(4)$, we get $(a,b)=(4,1)$, which is thus the unique solution. So we can evaluate the sum as
$$
\begin{align}
x^{2015}_{1}+x^{2015}_{2}+x^{2015}_{3}+x^{2015}_{4} &= (-1)+(-1)+(-1)+(-1)\\
&=-4
\end{align}
$$
|
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|
Trying to solve the trig equation $\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$ The equation is
$$\sqrt{3+4\cos^2(x)}=\frac{\sin(x)}{\sqrt 3}+3\cos(x)$$
My solution goes like this
$$
\begin{cases}
3+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x) \\
\frac{\sin(x)}{\sqrt 3}+3\cos(x) \ge 0
\end{cases}
$$
$$3(\sin^2(x)+\cos^2(x))+4\cos^2(x)=\frac{\sin^2(x)}{3}+\frac{6}{\sqrt 3}\sin(x)\cos(x)+9\cos^2(x)$$
$$2\cos^2(x)+\frac{\sin^2(x)}{3}-3\sin^2(x)+\frac{6}{\sqrt 3}\sin(x)\cos(x)=0$$
I multiply by 3 and divide by $\cos^2(x)$:
$$8\tan^2(x)-6\sqrt{3}\tan(x)-6=0$$
Let $t=\tan(x)$, then
$$4t^2-3\sqrt{3}t-3=0$$
$$t_1=\frac{7\sqrt 3}{2}$$
$$t_2=\frac{11\sqrt 3}{4}$$
The solutions for $x$ would be arc-tangents of these values.
But the textbook's answer is
$$\color{green}{x_1=\frac{\pi}{3}+2\pi n; x_2=-\arctan\left(\frac{\sqrt 3}{4}\right)+2\pi n}$$
Where did I make a mistake?
P.S. From the textbook
|
$$4t^2-3\sqrt3t-3=0$$
$$\implies t=\dfrac{3\sqrt3\pm\sqrt{(3\sqrt3)^2-4\cdot4(-3)}}{2\cdot4}=\dfrac{3\sqrt3\pm5\sqrt3}8=?$$
|
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|
Integrating linear/trigonometric I have the following question-
$$\int \frac{x}{1+\cos x}\,\text{d}x$$
Do I do integration by parts or is there some other method?
Thanks for the help.
|
Notice, we have $$\int\frac{x}{1+\cos x}dx$$ $$=\int\frac{x}{1+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}dx$$
$$=\int\frac{x\left(1+\tan^2\frac{x}{2}\right)}{1+\tan^2\frac{x}{2}+1-\tan^2\frac{x}{2}}dx$$
$$=\int\frac{x\sec^2\frac{x}{2}}{2}dx$$
$$=\frac{1}{2}\int x\sec^2\frac{x}{2}dx$$
$$=\frac{1}{2}x \int \sec^2\frac{x}{2}-\frac{1}{2}\int\left( \int \sec^2\frac{x}{2} dx \right) dx$$
$$=\frac{1}{2}x 2\tan \frac{x}{2}-\frac{1}{2}\int 2\tan \frac{x}{2} dx$$
$$=x\tan \frac{x}{2}-2\ln\left|\sec \frac{x}{2}\right|+C$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
dice probability - same 2 dice in 6 dice rolls I have this simple probability problem that I am not sure I solved correctly. I am not interested in formulas, but rather the thought process of how to solve it.
Suppose we roll six 6-sided dice that are equal. I want to find the probability that at least two dice have the same face.
Let's consider the following roll notation:
$$
[x] [y] [a][b][c][d]
$$
I first started to calculate the probability of the first two dice being the same in an individual 2 die roll, which is $$\frac{1}{6^2}$$
Now considering the next 4 dice as static, the number of permutations of x and y in the 6 die roll would be: $$\frac{6!}{(6-2)!} = 6*5 = 30$$
And with the total number of possible outputs for a->d being 6^4 we would have the probability of rolling the same two dice in a 6 dice roll: $$\frac{\frac{1}{6^2} * 30 * 6^4}{6^6} = \frac{6^2 * 30}{6^6} = \frac{30}{6^4} < \frac{1}{6^2}$$ which doesn't make sense to me, since rolling 6 dice gives you a greater probability of having doubles as opposed to when rolling only 2.
Can anyone pinpoint a correct method of finding this probability?
|
When you say "two with the same dice face" do you mean "at least two"? If so then it is simply 1 minus the probability they are all different. The first one will come up with something. The probability the second is different is $\frac{5}{6}$. The probability the third is different from either of the first two is $\frac{4}{6}$, etc, until the probability the last is different from any of the first 5 is $\frac{1}{6}$. The probability all 6 dice are different is the product of those, $\frac{5!}{6^6}$ and the probability of "at least two different" is $1-\frac{5!}{6^6}$.
If you mean "exactly two", again the first can be anything. The probability the second is the same as the first is $\frac{1}{6}$. Then the probability the third, fourth, fifth, and sixth are different from the all previous rolls is is $\frac{5}{6}$, $\frac{4}{6}$, $\frac{3}{6}$, and $\frac{2}{6}$ respectively. The probability the first two are the same and the last four different is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{4}{6}\cdot\frac{3}{6}\cdot\frac{2}{6} = \frac{5!}{6^6}$.
There are $\binom{6}{2} = \frac{6!}{2!\ \cdot\ 4!} = 15$ different ways to order 6 things, two of which are the same so the probability of two rolls the same, the other 4 different in any order is $15\cdot \frac{5!}{6^6}$.
|
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|
A geometry question on finding the area of cyclic quadrilaterals The circumcircle of a cyclic quadrilateral $ABCD$ has radius $2$.
$AC, BD$ meet at $E$ such that $AE = EC$. If $AB^2 = 2\cdot AE^2$ and $BD^2 = 12$, what is the area of the quadrilateral?
|
Let $O$ be the circumcentre of $ABCD$ and $R$ the circumradius. Since $BD=R\sqrt{3}$, $\widehat{DAB}$ and $\widehat{DCB}$ are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, in some order. Let $AE=x$. We have $AC=2x$ and $AB=x\sqrt{2}$, so:
$$ \arcsin\frac{2x}{4}=\arcsin\frac{x\sqrt{2}}{4}+\arcsin\frac{BC}{4} $$
and:
$$ BC = 2x\sqrt{1-\frac{x^2}{8}}-x\sqrt{2-\frac{x^2}{2}}. $$
In the same way you may compute $AD$ and $CD$ in terms of $x$. By Ptolemy's theorem we have $AB\cdot CD+BC\cdot AD = AC\cdot BD = 4x\sqrt{3}$; moreover, $BE+EC=2\sqrt{3}$ and $BE\cdot ED=x^2$.
Along the previous lines, you may find every possible value for $x$, then the length of every segment in the configuration. Notice that the area of $ABCD$ is given by:
$$ \frac{1}{2}\left(BC\cdot CD\sin\widehat{BCD}+AB\cdot AD\sin\widehat{BAD}\right)=\frac{\sqrt{3}}{4}\left(BC\cdot CD+DA\cdot AB\right).\tag{1}$$
We may also use the following approach. We take $BD$ as the side of an equilateral triangle inscribed in a circle $\Gamma$, and $A\in\Gamma$. We take $A'$ as the symmetric of $A$ with respect to $BD$, then $C_1,C_2$ as the intersections between $\Gamma$ and the parallel to $BD$ through $A'$. In such a way, if we define $E_i=AC_i\cap BD$, we have $AE_i=E_i C_i$. Then we take $F$ on the segment $AC_1$ such that $AF=AB\sqrt{2}$.
It is not difficult to check that $F$ may lie on the $BD$ line only if $C_1=C_2$. That gives:
$$ AB=2R\sin\frac{\pi}{12},\quad BC=CD=2R\sin\frac{\pi}{6},\quad AD=2R\sin\frac{5\pi}{12} $$
hence by $(1)$:
$$ [ABCD]=R^2\sqrt{3}\left(\sin^2\frac{\pi}{6}+\sin\frac{\pi}{12}\sin\frac{5\pi}{12}\right)=R^2\frac{\sqrt{3}}{2}=\color{red}{2\sqrt{3}}.\tag{2}$$
|
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|
Minimum value of $\cos x+\cos y+\cos(x-y)$ What is the minimum value of $$ \cos x+\cos y+\cos(x-y). $$ Here $x,y$ are arbitrary real numbers. Mathematica gives (with NMinimize) $-3/2$. But I don't know if this is correct and if so, how to prove it.
|
Since cosine is even, the problem is equivalent to
Minimize $\cos(-x)+\cos(y)+\cos(x-y)$.
which is equivalent to
Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c=0$.
Since cosine is periodic with period $2\pi$, this is equivalent to
Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c\equiv 0\pmod{2\pi}$.
If two or more of $\cos a$, $\cos b$, $\cos c$ are positive, say $\cos a>0$ and $\cos b>0$, then replacing $a$ and $b$ with $a+\pi$ and $b+\pi$ will preserve the constraint $a+b+c\equiv 0\pmod{2\pi}$ and negate $\cos a$ and $\cos b$, reducing the sum. So we can assume that at most one of the three is positive. We thus reduce the problem to:
Minimize $\cos a + \cos b + \cos c$ subject to $a+b+c\equiv 0\pmod{2\pi}$, with $a,b\in[\frac\pi2,\frac{3\pi}2]$ and $c\in[0,2\pi)$.
And now we solve:
\begin{align*}
\cos a + \cos b + \cos c
&= \cos a + \cos b + \cos (2n\pi - a - b) \\
&= \cos a + \cos b + \cos(a+b) \\
&= \cos a + \cos b + 2\cos^2\big(\tfrac{a+b}2\big) - 1 \\
&\ge 2\cos\big(\tfrac{a+b}2\big) + 2\cos^2\big(\tfrac{a+b}2\big) - 1
&&\text{(cosine is convex on $[\tfrac\pi2,\tfrac{3\pi}2]$)} \\
&= 2\big(\cos\big(\tfrac{a+b}2\big) + \tfrac12\big)^2 - \tfrac32 \\
&\ge -\tfrac32
\end{align*}
with equality when $a=b$ and $\cos\big(\tfrac{a+b}2\big) = -\frac12$; for example, $a=b=c=\frac{2\pi}3$.
|
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|
Integrate $\tan^2(\frac{\pi}{12} \cdot y)$ Integrate $\tan^2(\frac{\pi}{12} \cdot y)$
Wolfram gives the answer:
$$\frac{12 \tan(\frac{\pi y}{12})}{\pi} -y + \text{constant}$$
But why is the value of $\tan^2$ not getting differentiated?
According to this rule, the answer should be:
$$\frac{\tan(\frac{\pi}{12} \cdot y)}{\frac{\pi}{12} \cdot (2-1)}$$
($\dfrac{\pi}{12}$ is a constant so it's not differentiated)
Also, what is meant by aX? In my case, what's my "aX"? $\frac{\pi}{12} \cdot y$? So does the "a" simply means what's before the variable y? $\frac{\pi}{12}$?
|
Recall that $\int \tan^2 x\ dx = \int {\sin^2 x \over 1 - \sin^2 x} dx = -x + \tan x$, we find:
$\int \tan^2 \left( {\pi y \over 12} \right) d y = \frac{12 \tan \left(\frac{\pi y}{12}\right)}{\pi }-y$
|
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"url": "https://math.stackexchange.com/questions/1422585",
"timestamp": "2023-03-29T00:00:00",
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|
There are 30 tokens numbers from 0 to 30. Find the number of ways of choosing 3 tickets such that the sum of the numbers on the tokens is 30 Then the number of solutions is divisible by
(A) 2 (B) 3 (C) 5 (D) 7
One way to solve its by finding the number of unequal integral solutions of $x+y+z=30$.
The possible cases are $x<y<z, y<x<z,..$ (3! Ways)
If $x<y<z$, let $x=a, y=a+b$ and $z=a+b+c$
Then $3a+2b+c=30$ where $0\le a,b,c\le 30$
The number of solutions is the coefficient of $x^{30}$ in $(1-x^3)^{-1}(1-x^2)^{-1}(1-x)^{-1}$ which is as difficult as counting the numbers manually.
Total number of ways would be $3!$ times the number obtained from above. Hence divisible by 2 and 3. Is this correct?
|
To determine three such tickets, we can pick two numbers $a<b$ in $\{1,\ldots,29\}$, which is possible in $29\choose 2$ ways; then let the tickets be $a,b-a,30-b$. Then do some inclusion and exclusion: Subtract the $14$ cases where $a=b-a$ ($1\le a\le 14$, $b=2a$); subtract the $14$ cases where $a=30-b$ ($1\le a\le 14$, $b=30-a$); subtract the $14$ cases where $b-a=30-b$ ($16\le b\le 29$, $a=2b-30$). Then add back (twice) the one case where $a=b-a=30-b$ ($a=10$, $b=20$). After this, we have counted each (unordered) tripel $\{x,y,z\}\subset\{1,\ldots,30\}$ of ticktes $3!$ times (namely as $x=a,y=b-a,z=30-b$ and all other permutations). Hence the number we look for is is $$\frac{{29\choose 2}-14-14-14+2}6=61. $$
If the tickets actually range from $0$ to $30$ (not $1$ to $30$) the calculation runs like this: There are ${31\choose 2}+31$ ways to pick two numbers $a\le b$ in $\{0,\ldots,30\}$; again let the tickets be $a,b-a,30-b$. Exclude the $16$ cases with $a=b-a$ ($0\le a\le 15$, $b=2a$), the $16$ cases with $a=30-b$ ($0\le a\le 15$, $b=30-a$), the $16$ cases with $b-a=30-b$ ($15\le b\le 30$, $a=2b-30$), and add back $a=10,b=20$ twice. As above divide by $3!$ to obtain
$$\frac{{31\choose 2}+31-3\cdot 16+2}6=75. $$
In the first interpretation, none of the answer suggestions is true, under the second, B and C are true.
|
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|
Which answer is correct for this product rule based probability problem The question reads like this:
A bag contains 5 black and 3 red balls. A ball is taken out of the bag and is not returned to it. If this process is repeated three times, then what is the probability of drawing a black ball in the next draw?
Solution 1
I think the three balls can be drawn in 4 ways
*
*All the three balls are black = 5/8 .4/7 .3/6 then the probability of next black ball can be drawn is 2/5.
*2 black and 1 red = 5/8 . 4/7 . 3/6 then the probability of next black ball can be drawn is 3/5.
*1 black and 2 red = 5/8 . 3/7 . 2/6 then the probability of next black ball can be drawn is 4/5.
*All red = 3/8 . 2/7 . 1/6 then the probability of next black ball can be drawn is 1.
And probability of each way is 1/4.
Hence desired probability $=\frac{1}{4}\times\frac{2}{5}+\frac{1}{4}\times\frac{3}{5}+\frac{1}{4}\times\frac{4}{5}+\frac{1}{4}\times1=\frac{7}{10} = 0.7$
Solution 2
Probability of 4th ball being black when
*
*All first first 3 draws are black balls $=\frac{5}{8}\times\frac{4}{7}\times\frac{3}{6}\times\frac{2}{5}=\frac{120}{1680}$
*First three draws contain all red balls $=\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}\times\frac{5}{5}=\frac{30}{1680}$
*First three draws contain 2 red and 1 black ball $=\frac{3}{8}\times\frac{2}{7}\times\frac{5}{6}\times\frac{4}{5}\times 3=\frac{360}{1680}$
*First three draws contain 1 red and 2 black ball $=\frac{3}{8}\times\frac{5}{7}\times\frac{4}{6}\times\frac{3}{5}\times 3=\frac{540}{1680}$
Summing all above probabilities gives $=\frac{5}{8}=0.625$
I cant decide which one is correct. I feel the second one is just far more precise. Is it?
|
You are overcomplicating things. There are eight balls, and any one of which could be the fourth ball drawn with equal probability. Five of these balls are black. The probability that a black ball is drawn on the fourth draw is thus: $$5/8$$
This agrees with your second solution, which correctly considers that there are three ways to arrange two balls of one colour and one of another.
To see this in action, lets examine ways to draw balls from a bag of two black and two red balls, and count those where the third ball is black. By my method we should expect $2/4$ of the possible arrangements to have a black ball drawn in third place. Let's see...
$$\begin{array}{cc}\bullet\bullet\circ\circ \\\bullet\circ\bullet\circ &\star\\ \bullet\circ\circ\bullet \\ \circ \bullet\bullet\circ & \star \\ \circ\bullet\circ\bullet \\ \circ\circ\bullet\bullet & \star\end{array}$$
|
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Trapped in Induction, how to get out? Example 1:
Prove by induction that $1+3+5+...+(2n-1)=n^2 \text{ for all } n \in \mathbb{N}....(*)$
Proof:
Step 1: For $n=1$, left-side we have $(2(1)-1) = 1$. Right-side we have $(1)^2 = 1$.
Step 2: Suppose (*) is true for some $n=k \in \mathbb{N}$ that is
$$1+3+5+...+(2k-1)=k^2$$
Step 3: Prove that (*) holds true for $n=k \in \mathbb{N}$ that is (adding $(2k+1)$ to both sides)
$$1+3+5+...+(2k-1)+(2k+1)=(k)^2+(2k+1)$$
we have
which shows both sides are equal?
but you can do this to several number of problems....
Example 2:
Prove by induction that $1^3+2^3+...+n^3=(1+2+...+n)^2 \text{ for all } n \in \mathbb{N}....(*)$
Proof:
Step 1: For $n=1$, left-side we have $1^3 = 1$. Right-side we have $(1)^2 = 1$. Which shows both sides are true.
Step 2: Suppose (*) is true for some $n=k \in \mathbb{N}$ that is
$$1^3+2^3+...+k^3=(1+2+...+k)^2$$
Step 3: Prove that (*) holds true for $n=k \in \mathbb{N}$ that is (adding $(k+1)^3$ to both sides)
$$1^3+2^3+...+k^3 + (k+1)^3=(1+2+...+k)^2 + (k+1)^3$$
we have
which shows both sides are equal again...
what am I fundamentally doing wrong?
|
In Step 3 of the second example, you write "Prove that (*) holds true for $n = k \in N$", but that should be $n = k + 1$.
I find it useful in proofs like this to write down something I call $P(n)$, the proposition that I want to prove, which is typically meant to be true for every integer $n$. In your example 3, $P(n)$ is the statement
$$
1^3+2^3+...+n^3=(1+2+...+n)^2
$$
That means that $P(1)$ is the statement
$$
1^3=(1)^2
$$
and $P(2)$ is the statement
$$
1^3 + 2^3=(1+2)^2
$$
and so on.
Now you can say this:
I'm going to assume that for some $k \in \mathbb N$, $P(k)$ is true, i.e., that
$$
1^3+2^3+...+k^3=(1+2+...+k)^2
$$
And using only algebraic manipulation, I'll use this to establish that $P(k+1)$ is true, i.e., that
$$
1^3+2^3+...+k^3 + (k+1)^3=(1+2+...+k + (k+1))^2.
$$
Now you have a starting point and a clear goal.
I'd say, at this point:
From the hypothesis, we have
$$
1^3+2^3+...+k^3=(1+2+...+k)^2
$$
Adding $(k+1)^3$ to each side, we get
$$
1^3+2^3+...+k^3+ (k+1)^3=(1+2+...+k)^2 + (k+1)^3
$$
To finish the proof, we have to show that the right hand side is the same as $(1 + 2 + \ldots + (k+1))^2$. To do so, let's look at the difference between these two,
\begin{align}
S &= (1+2+...+k)^2 + (k+1)^3 - ((1+2+...+k+(k+1))^2)
\end{align}
If we can show $S = 0$, we're done. Well,
\begin{align}
S
&= (1+2+...+k)^2 + (k+1)^3 - ((1+2+...+k+(k+1))^2)\\
&= (1+2+...+k)^2 - (1+2+...+k+(k+1))^2 + (k+1)^3
\end{align}
The first two terms look like $A^2 - B^2 = (A-B)(A+B)$, so
\begin{align}
S
&= (1 + 2 + \ldots + k)^2 - (1 + 2 + \ldots + k+(k+1))^2 + (k+1)^3 \\
&= ((1 + 2 + \ldots + k) - (1 + 2 + \ldots + k+(k+1)))\cdot((1 + 2 + \ldots + k) + (1 + 2 + \ldots + k + (k+1))) + (k+1)^3 \\
&= ((1 + 2 + \ldots + k) - (1 + 2 + \ldots + k +(k+1)))\cdot((1 + 2 + \ldots + k) + (1 + 2 + \ldots + k + (k+1))) + (k+1)^3 \\
&= -(k+1)\cdot(2 (1 + 2 + \ldots + k) + (k+1)) + (k+1)^3 \\
&= -(k+1)\cdot(2 \frac{k(k+1)}{2} + (k+1)) + (k+1)^3 \\
&= -(k+1)\cdot( k(k+1) + (k+1)) + (k+1)^3 \\
&= -(k+1)\cdot( k(k+1) + 1\cdot(k+1)) + (k+1)^3 \\
&= -(k+1)\cdot( (k+1)(k+1) ) + (k+1)^3 \\
&= -(k+1)^3 + (k+1)^3 \\
&= 0.
\end{align}
|
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Fourth degree polynomial in trajectory problem I have a problem where I'm trying to find the distance needed to lead a moving target with a projectile given variable parameters. I've had limited success so far and I think I may be close to a solution but I'm not sure how to go about the final steps involving a fourth degree polynomial. I've looked into Ferrari's method and tried using Cardano's formula as well as Lagrange's resolvents but I haven't been able to come to grips with them in terms of finding the real roots. Before showing the equation I'd also like to show the method of obtaining it in case I made an error in getting to where I am in the first place. This is a diagram of the problem:
The target is moving with velocity $v_1$ and the cannon firing the projectile is initially at an angle $a_1$ and distance $d_1$ from the target. The projectile is fired at velocity $v_2$ and angle $a_2$ over a distance $d_2$ where it meets the target after a distance of $x$. The unknowns are $a_2$, $d_2$, and $x$.
The relevant equations used are $$d_2 = \frac{v_2^2sin(2a_2)}{g}$$ $$t = \frac{d_2}{v_2cos(a_2)}$$$$x = v_1t$$
from which two equations in terms of $a_2$ are derived:
$$\frac{x}{v_1} = \frac{d_2}{v_2cos(a_2)}$$
$$\frac{x}{v_1} = \frac{2\,v_2^2\,sin(a_2)cos(a_2)}{g\,v_2\,cos(a_2)}$$
$$a_2 = arcsin\left(\frac{x\,g}{2\,v_1\,v_2}\right)$$
and then with the law of cosines:
$$d_2^2 = x^2 + d_1^2 - 2\,x\,d_1cos(a_1)$$
$$\frac{v_2^2sin(2a_2)}{g} = \sqrt{x^2 + d_1^2 - 2\,x\,d_1cos(a_1)}$$
$$a_2 = \frac12arcsin\left(\frac{g}{v_2^2}\sqrt{x^2 + d_1^2 - 2\,x\,d_1cos(a_1)}\right)$$
and finally setting these two equal to solve for x:
$$arcsin\left(\frac{x\,g}{2\,v_1\,v_2}\right) = \frac12arcsin\left(\frac{g}{v_2^2}\sqrt{x^2 + d_1^2 - 2\,x\,d_1cos(a_1)}\right)$$
$$2\left(\frac{x\,g}{2\,v_1\,v_2}\right)\sqrt{1-\left(\frac{x\,g}{2\,v_1\,v_2}\right)^2} = \frac{g}{v_2^2}\sqrt{x^2 + d_1^2 - 2\,x\,d_1cos(a_1)}$$$$\frac{x^2}{v_1^2}\left(1 - \frac{x^2\,g^2}{4\,v_1^2\,v_2^2}\right) = \frac{1}{v_2^2}(x^2 + d_1^2 - 2\,x\,d_1cos(a_1))$$$$\frac{x^2\,v_2^2}{v_1^2} - \frac{x^4\,g^2}{4\,v_1^4} = x^2 + d_1^2 - 2\,x\,d_1cos(a_1)$$$$\left(\frac{g^2}{4\,v_1^4}\right)\,x^4 + \left(1-\frac{v_2^2}{v_1^2}\right)\,x^2 + \left(-2\,d_1\,cos(a_1)\right)\,x + (d_1^2) = 0$$
and this is where I get stuck. I can recognize the fourth degree polynomial but I don't know how to go about getting the real roots. Any help would be appreciated, thanks.
|
I managed to find the equation for the solution of a depressed fourth degree polynomial on WolframAlpha and it yielded the expected results:
Let $$a = \frac{g^2}{4\,v_1^4}$$ $$c = 1-\frac{v_2^2}{v_1^2}$$ $$d = -2\,d_1\,cos(a_1)$$ $$e = d_1^2$$
$$y = \sqrt[3]{\sqrt{(-72\,e\,a\,c + 27\,a\,d^2 + 2\,c^3)^2 - 4(12\,e\,a + c^2)^3} - 72\,e\,a\,c + 27\,a\,d^2 + 2\,c^3}$$
Then the first root and the one that is real when the parameters are valid:
$$x = \frac{1}{2}\sqrt{\frac{y}{3\sqrt[3]{2}\,a} + \frac{\sqrt[3]{2}\,(12\,e\,a + c^2)}{3\,a\,y} - \frac{2\,c}{3\,a}} - \frac{1}{2}\sqrt{\frac{-1}{3\sqrt[3]{2}\,a}\,y - \frac{\sqrt[3]{2}\,(12\,e\,a + c^2)}{3\,a\,y} - \frac{2\,d}{a\,\sqrt{\frac{1}{\sqrt[3]{2}\,a}\,y + \frac{\sqrt[3]{2}\,(12\,e\,a + c^2)}{3\,a\,y} - \frac{2\,c}{3\,a}}} - \frac{4\,c}{3\,a}}$$
|
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Find 2 sums with the binomial newton Find the sum of:
i)$\displaystyle\sum_{k=0}^{n} k^2$ $\left(\begin{array}{c} n\\k\end{array}\right)$
ii) $\displaystyle\sum_{k=1}^{n} \frac{2k+5}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$
Thoughts:
i)(After the Edit)$\displaystyle\sum_{k=0}^{n} k^2$ $\left(\begin{array}{c} n\\k\end{array}\right)$ =
$\displaystyle\sum_{k=0}^{n} {k(k-1)}$ $\left(\begin{array}{c} n\\k\end{array}\right)$ + $\displaystyle\sum_{k=0}^{n}{k}$ $\left(\begin{array}{c} n\\k\end{array}\right)$=
$\displaystyle\sum_{k=0}^{n} {n(k-1)}$ $\left(\begin{array}{c} n-1\\k-1\end{array}\right)$ + $\displaystyle\sum_{k=0}^{n}{n}$ $\left(\begin{array}{c} n-1\\k-1\end{array}\right)$=
$\displaystyle\sum_{k=0}^{n} {n(n-1)}$ $\left(\begin{array}{c} n-2\\k-2\end{array}\right)$ + n$\displaystyle\sum_{k=0}^{n}$ $\left(\begin{array}{c} n-1\\k-1\end{array}\right)$=
$n(n-1)2^{n-2}+n2^{n-1}$=$n2^{n-2}(2+n-1)$=$n(n+1)2^{n-2}$
ii)(After the Edit)$\displaystyle\sum_{k=1}^{n} \frac{2k+5}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
$\displaystyle\sum_{k=1}^{n} \frac{(2k+2)+3}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
2$\displaystyle\sum_{k=1}^{n}$$\left(\begin{array}{c} n\\k\end{array}\right)$ + 3$\displaystyle\sum_{k=1}^{n} \frac{1}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
$2^{n+1}-2$+ $\frac{3}{n+1}$$\displaystyle\sum_{k=1}^{n} \frac{n+1}{k+1}$$\left(\begin{array}{c} n\\k\end{array}\right)$=
$2^{n+1}-2$+ $\frac{3}{n+1}$$\displaystyle\sum_{k=1}^{n}$$\left(\begin{array}{c} n+1\\k+1\end{array}\right)$=
$2^{n+1}-2+\frac{3}{n+1}(2^{n+1}-1)$
|
For the first one:
We know that:
$$
\sum_{k=0}^{n}{{n\choose k}x^k}={(1+x)}^{n}
$$
So let's differentiate it to get:
$$
\sum_{k=0}^{n}{k{n\choose k}{x}^{k-1}}=n{(1+x)}^{n-1}
$$
Multiply by $x$ and differntiate again:
$$
\sum_{k=0}^{n}{k^2{n\choose k}{x}^{k-1}}=(nx{(1+x)}^{n-1})'\\
\sum_{k=0}^{n}{k^2{n\choose k}{x}^{k-1}}=n({(1+x)}^{n-1}+(n-1)x{(1+x)}^{n-2})\\
$$
Multiply by $x$ one more time to get:
$$
\sum_{k=0}^{n}{k^2{n\choose k}{x}^{k}}=nx({(1+x)}^{n-1}+(n-1)x{(1+x)}^{n-2})\\
$$
Now just put $x=1$ and you'll get the answer.
Let's move on to the next one:
$$
\sum_{k=0}^{n}{\frac{2k+5}{k+1}{n\choose k}x^k} =2\sum_{k=0}^{n}{\frac{k}{k+1}{n\choose k}x^k}+5\sum_{k=0}^{n}{\frac{1}{k+1}{n\choose k}x^k}\\
=2\sum_{k=0}^{n}{(1-\frac{1}{k+1}){n\choose k}x^k}+5\sum_{k=0}^{n}{\frac{1}{k+1}{n\choose k}x^k}\\
=2\sum_{k=0}^{n}{{n\choose k}x^k}-2\sum_{k=0}^{n}{\frac{1}{k+1}{n\choose k}x^k}+5\sum_{k=0}^{n}{\frac{1}{k+1}{n\choose k}x^k}\\
=2{(1+x)}^{n}-2\sum_{k=0}^{n}{\frac{1}{k+1}{n\choose k}x^k}+5\sum_{k=0}^{n}{\frac{1}{k+1}{n\choose k}x^k}
$$
Now let's compute :
$$
\sum_{k=0}^{n}{\frac{1}{k+1}{n\choose k}x^k}
$$
We have:
$$
\sum_{k=0}^{n}{{n\choose k}x^k} = {(1+x)}^{n}
$$
Let's integrate to get:
$$
\sum_{k=0}^{n}{\frac{1}{k+1}{n\choose k}{x}^{k+1}}=\frac{{(1+x)}^{n+1}}{n+1}+C
$$
For $x=0$ the sum is $0$ so :
$$
C=-\frac{1}{n+1}
$$
We finally get:
$$
\sum_{k=0}^{n}{\frac{1}{k+1}{n\choose k}x^k}=\frac{1}{x}(\frac{{(1+x)}^{n+1}}{n+1}-\frac{1}{n+1})
$$
$$
\sum_{k=0}^{n}{\frac{2k+5}{k+1}{n\choose k}x^k} = 2{(1+x)}^{n}-2(\frac{1}{x}(\frac{{(1+x)}^{n+1}}{n+1}-\frac{1}{n+1}))+5(\frac{1}{x}(\frac{{(1+x)}^{n+1}}{n+1}-\frac{1}{n+1}))
$$
Now plug in $x=1$ and you'll get the answer
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $\sqrt{i}$ The results I get not the same as in the book, basically I need to get only this and the $x$'s of course:
$$y = \frac{\sqrt{2}}{2}$$
So this is what I did:
$$\sqrt{i} = x+yi$$
$$i = x^2 +2xyi -y^2$$
$$\begin{cases} x^2-y^2 = 0 \\
2xy = 1 \end{cases}$$
$$xy=0.5$$
$$x=\frac{0.5}{y}$$
$$(1)\space\space\space\space(\frac{0.5}{y})^2-y^2=0$$
$$0.25-y^4=0$$
$$y^4=0.25$$
$$y=\pm \sqrt[4]{0.25}$$
$$y=\pm \frac{\sqrt{2}}{2}$$
What is WRONG?
|
You have $y = \pm \dfrac{\sqrt 2} 2$.
If $y= \dfrac{\sqrt 2} 2$ then $x=\dfrac{0.5} y = \dfrac{0.5}{\sqrt 2/2}$, and this simplifies to $x=\dfrac{\sqrt 2} 2$.
So $\dfrac{\sqrt 2} 2 + i \dfrac{\sqrt 2} 2$ is a complex number whose square is $i$.
If you multiply that by $-1$ you get the other complex number whose square is $i$. That is the one that comes from $y=\dfrac{-\sqrt 2} 2\vphantom{\dfrac{\displaystyle\int}\int}$.
One can show that there cannot be more than two solutions: Recall from algebra that if $a$ is a solution to $x^2 -i=0$ then $x-a$ is a factor of $x^2-i$, so you get
$$
x^2 - i = (x-a)(\cdots\cdots).
$$
The other factor must be a first-degree polynomial, so it can't have more than one root.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Integration of an unusual trig function I am trying to solve the following definite integral:
$$ \int_{0}^{\frac{\pi}{4}} \sqrt{\tan^2t + \frac{1}{2}} \, dt $$
I have tried the usual trig substitutions with no avail.
|
Let $$\displaystyle I = \int \sqrt{\tan^2 t +\frac{1}{2}}dt\;,$$ Now Put $\displaystyle \tan t = \frac{1}{\sqrt{2}}\tan \phi\;,$
Then $$\displaystyle dt = \frac{1}{\sqrt{2}}\frac{\sec^2 \phi}{\sec^2 t} d\phi =\frac{1}{\sqrt{2}}\frac{\sec^2 \phi}{1+\tan^2 t}d\phi = \sqrt{2}\left(\frac{\sec^2 \phi}{2+\tan^2 \phi}\right)d\phi$$
So Integral $\displaystyle I = \int \frac{\sec^3 \phi}{2+\tan^2 \phi}d\phi = \int\frac{1}{\left(2\cos^2 \phi+\sin^2 \phi\right)}\cdot \frac{1}{\cos \phi}d\phi $
So Integral $$\displaystyle I = \int\frac{\cos \phi}{\left(1-\sin^2 \phi\right)(2-\sin^2 \phi)}d\phi$$
Now Put $\sin \phi = u\;,$ Then $\cos \phi d\phi = du$
So Integral $$\displaystyle I = \int\frac{1}{(1-u^2)(2-u^2)}du = \int\frac{1}{(u^2-1)(u^2-2)}du = \int \left[\frac{(u^2-1)-(u^2-2)}{(u^2-1)(u^2-2)}\right]du$$
So we get $$\displaystyle I = \int\frac{1}{u^2-2}du-\int \frac{1}{u^2-1}du$$
So we get $$\displaystyle I = \frac{1}{2\sqrt{2}}\ln \left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|-\frac{1}{2}\ln \left|\frac{u-1}{u+1}\right|+\mathcal{C}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$2(n-2)+1$ does not divide $(n-2)(n-3)/2$ for $n \ge 8$
For $n \ge 8$ the number $2(n-2)+1$ never divides $(n-2)(n-3)/2$.
Any ideas how to prove this? I see that $(n-2)(n-3)/2 = 1 + 2 + \ldots + (n-3)$. If I suppose that $2(n-2)+1$ divides $(n-2)(n-3)/2$ then it should also divide their difference
\begin{align*}
(n-2)(n-3)/2 - (2(n-2) + 1) & = \sum_{k=1}^{n-3} k - (2n - 3) \\
& = \sum_{k=1}^{n-4} k + (n-3) - 2n + 3 \\
& = \sum_{k=1}^{n-4} k - n \\
& = \frac{(n-3)(n-4)}{2} - n.
\end{align*}
That is all I have, hope you can give me some hints!?
|
Note that $2n-3$ is odd, so it is enough to show it does not divide $4(n-2)
(n-3)$. which is $4n^2-20n+24$. Do a polynomial division. We get $2n-7+\frac{3}{2n-3}$.
If $2n-3\gt 3$, then $\frac{3}{2n-3}$ is not an integer.
Another way: Let $n\gt 3$. Note that $2(n-2)+1$ and $n-2$ are relatively prime. So it is enough to show that $2n-3$ cannot divide $n-3$. This is obvious.
A Puzzle: We have given two straightforward arguments that we cannot have divisibility if $n\gt 3$. How did they come up with the weaker $\ge 8$?
|
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"timestamp": "2023-03-29T00:00:00",
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|
What are better approximations to $\pi$ as algebraic though irrational number? I know that $\pi \approx \sqrt{10}$, but that only gives one decimal place correct. I also found an algebraic number approximation that gives ten places but it's so cumbersome it's just much easier to just memorize those ten places.
What's a good approximation to $\pi$ as an irrational algebraic number (or algebraic integer if possible) that is easier to memorize than the number of places it gives correct?
EDIT: Algebraic number preferably of low degree, such as $2$ or $3$ (quadratic or cubic).
|
Some nice approximations can be produced by exploiting the ideas of Archimedes. The difference between a unit circle and an inscribed regular $n$-agon is made by $n$ circle segments. If we approximate them with parabolic segments and call
$$ A_n = \frac{n}{2}\sin\frac{2\pi}{n}=n\sin\frac{\pi}{n}\cos\frac{\pi}{n} $$
the area of the inscribed $n$-agon, we get that
$$ \pi \approx \frac{4 A_{2n}-A_n}{3} = \frac{n}{3}\sin\frac{\pi}{n}\left(4-\cos\frac{\pi}{n}\right)$$
where the absolute error behaves like $\frac{C}{n^5}$. Here some approximations derived through this geometric method:
$$\begin{array}{l|c|l}\hline n=12 & 4\sqrt{6}-4\sqrt{2}-1 & 3.141104722\\
\hline n=24 & \sqrt{2}-\sqrt{6}+8 \sqrt{8-4 \sqrt{2+\sqrt{3}}}&3.141561971\\ \hline\end{array}$$
This can be further improved. For instance, since the MacLaurin series of $\frac{x}{\sin x}$ and $\frac{1}{15}\left(68+11\cos(x)-64\cos(x/2)\right)$ agree up to the $x^6$ term (the same idea has been exploited here) we have
$$ \pi \approx \frac{n}{15}\sin\frac{\pi}{n}\left(68+11\cos\frac{\pi}{n}-15\cos\frac{\pi}{2n}\right) $$
and the following algebraic approximations:
$$\begin{array}{l|c|l}\hline n=6 & \frac{1}{10}\left(136+11\sqrt{3}-64\sqrt{2+\sqrt{3}}\right) & 3.141405312\\
\hline n=12 & \frac{\sqrt{3}-1}{5\sqrt{2}}\left(136+11\sqrt{2+\sqrt{3}}-64\sqrt{2+\sqrt{2+\sqrt{3}}}\right) &3.141589664\\ \hline \end{array}$$
Plenty of other approximations (both accurate and reasonably simple) can be derived by combining some version of the Shafer-Fink inequality and Machin formulas, for instance
$$\pi\approx \frac{180}{16 \sqrt{20+6 \sqrt{10}}+6 \sqrt{10}+21}+\frac{90}{8 \sqrt{10+4 \sqrt{5}}+3 \sqrt{5}+7}$$
whose error is $<10^{-6}$, or
$$ \pi \approx \frac{360}{7+7\sqrt{2}+6 \sqrt{2 \left(2+\sqrt{2}\right)}+16 \sqrt{2 \left(2+\sqrt{2}\right) \left(\sqrt{2+\sqrt{2}}+2\right)}}$$
whose error is $<4\cdot 10^{-7}$.
|
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|
Finding roots of a complex number I am asked to find $2^{1/5} $. Using the formula, and noticing $2 = 2e^{0 i} $, we have
$$ 2^{1/5} = 2^{1/5} \bigg( \cos( \frac{ 0}{5} + \frac{ 2 \pi k}{5} ) + i \sin( \frac{ 0}{n} + \frac{ 2 \pi k}{n} ) \bigg) $$
My confusing. I notice we can also write $2 = 2e^{2 \pi k} = 2e^{4 \pi k} = .... $ Do I need to consider every case, or would it be enough to just consider the case when $2 = 2e^{0i } $ as above?
|
One way to attack this baby is to remember that the number of $n$-th complex roots of a number is always $n$, and that those $n$ roots form the vertices of an $n$-sided regular polygon on the Complex plane. So there are $3$ cube roots, forming a triangle, $4$ cube roots, forming a square, $5$ fifth roots, forming a pentagon, and so forth.
Int the above diagram, the pentagon is superimposed on the Complex plane. It has been broken broken up into five segments. The angle between each segment is $1/5$ that of a full circle. That is, each angle is $2\pi/5$. Each of the points, $A$ to $E$ is located by two numbers: One number is the distance from the centre of the figure. The other number is the angle that around the circle that it's located away from point $A$.
The angles are marked on the diagram. $A$ is $0$, (but it's marked on the diagram as $10\pi/5=2\pi$ because travelling $2\pi$ radians from any point on a circle will take you back to where you started).
Your angles are as follows:
$A. 0$
$B. 4\pi/5$
$C. 2\pi/5$
$D. 6\pi/5$
$E. 8\pi/5$
The number you also want to know the argument which is the Real number value of the length of each of those spikes from the centre to a point. That number is the Real value of $2^{1/5}$.
How do you arrange all this information is your final stretch. If we designate $z$ as your Complex root, $r$ as your argument (which is $2^{1/5}$ in each case) and $\theta$ for your angles, your general forms are:
$$z = r(\cos\theta + i\sin\theta)$$ and $$z = e^{i\theta}$$
(Why these are equivalent is another story.)
So, your five roots are:
$$A.\; 2^{\frac{1}{5}}e^{0i}=2^{\frac{1}{5}}$$
$$B.\; 2^{\frac{1}{5}}e^{2\pi i/5}$$
$$C.\; 2^{\frac{1}{5}}e^{4\pi i/5}$$
$$D.\; 2^{\frac{1}{5}}e^{6\pi i/5}$$
$$E.\; 2^{\frac{1}{5}}e^{8\pi i/5}$$
Alternatively, these can be expressed as:
$$A.\; 2^{\frac{1}{5}}(\cos 0 + i\sin 0)= 2^{\frac{1}{5}}$$
$$B.\; 2^{\frac{1}{5}}(\cos \frac{2\pi}{5} + i\sin \frac{2\pi}{5})$$
$$C.\; 2^{\frac{1}{5}}(\cos \frac{4\pi}{5} + i\sin \frac{4\pi}{5})$$
$$D.\; 2^{\frac{1}{5}}(\cos \frac{6\pi}{5} + i\sin \frac{6\pi}{5})$$
$$E.\; 2^{\frac{1}{5}}(\cos \frac{8\pi}{5} + i\sin \frac{8\pi}{5})$$
These are your five roots, each expressed in two forms. The first form is polar and the second is rectangular.
Any of these numbers, multiplied by itself five times will yield $2$. Multiplying a rectangular form by itself five times is tedious. Multiplying a polar form is amazingly simple. I'll end off with one example.
$$[2^{\frac{1}{5}}e^{4\pi i/5}]^5=[2^{1/5}]^5[e^{4\pi i/5}]^5=
2e^{20\pi i/5}=2e^{4\pi i}$$
Since $4\pi$ is two full rotations about the circle, $e^{4\pi i}=e^{0 i}=1$
So the equation above finishes:
$$2e^{4\pi i}= 2(1) = 2$$
which is what we wanted to show.
EDIT:
To address the OP directly, you can always add $2\pi$ to any angle without changing the value of $z$ just as you can go around a circle exactly once and end up where you started.
So, you need only show the solution for $\theta=0$ and $\theta=2\pi$ takes care of itself.
|
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|
Find $\lim_\limits{x\to 0}{(\sqrt{f^2(x)+2f(x)+3}-f(x))}$ Let $f$ be a function such that $x^2f(x)\geq x^2+x+1,\forall x\in\mathbb{R^*}$. Find the value of: $$\lim_\limits{x\to 0}{\left(\sqrt{f^2(x)+2f(x)+3}-f(x)\right)}$$
I think that we may need to apply a sandwich theorem for the limit, so let $g(x)=\sqrt{f^2(x)+2f(x)+3}-f(x)$. It is easy to prove that $g(x)>1$ near/close to 0, since $f(x)>0$ from the original relation. However, I cannot find any function $h$, such that $g(x)<h(x)$ near/close to 0 with $\lim_\limits{x\to 0}{h(x)}=1$. Any hint?
|
If you want to use the sandwich theorem, observe that whenever $x$ is positive,
$f(x)$ also is positive and
\begin{align}
\left(f(x) + 1 + \frac{1}{f(x)}\right)^2
& = (f(x))^2 + 2f(x) + 3 + \frac{2}{f(x)} + \frac{1}{(f(x))^2} \\
& > (f(x))^2 + 2f(x) + 3 \\
& > \left(f(x) + 1\right)^2 \\
& > 0.
\end{align}
Therefore
$$f(x) + 1 < \sqrt{(f(x))^2 + 2f(x) + 3} < f(x) + 1 + \frac{1}{f(x)}$$
and
$$1 < \sqrt{(f(x))^2 + 2f(x) + 3} - f(x) < 1 + \frac{1}{f(x)}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int \frac{\sqrt{1+x^2}}{1-x^2}dx$ Problem :
$\int \frac{\sqrt{1+x^2}}{1-x^2}dx$
My approach :
Put $x = \tan\theta$
we get $$\int \frac{\sqrt{1+x^2}}{1-x^2}dx = \frac{\frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta \cos\theta}}{\frac{\cos^2\theta -\sin^2\theta}{\cos^2\theta}} d\theta $$
$$= \frac{1}{(\cos^2\theta -\sin^2\theta)\cos\theta}d\theta$$
But is it the right approach please guide will be of great help thanks.
|
Let $$\displaystyle I = \int\frac{\sqrt{1+x^2}}{1-x^2}dx = \int\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^2}}dx = -\int\frac{(x^2+1)}{(x^2-1)\sqrt{1+x^2}}dx$$
So $$\displaystyle I =-\int\frac{(x^2-1)+2}{(x^2-1)\sqrt{x^2+1}}dx = -\underbrace{\int\frac{1}{\sqrt{x^2+1}}dx}_{J}+2\underbrace{\int\frac{1}{(1-x^2)\sqrt{1+x^2}}dx}_{K}$$
So Here $$\displaystyle J = \int\frac{1}{\sqrt{x^2+1}}dx = \ln|x+\sqrt{x^2+1}|+\mathcal{C_{1}}$$
Now $$\displaystyle K = \int\frac{1}{(1-x^2)\sqrt{1+x^2}}dx\;,$$ Now Put $\displaystyle x=\frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt$
So we get $$\displaystyle K = -\int\frac{t}{(t^2-1)\sqrt{t^2+1}}dt\;,$$ Now Put $(t^2+1) = u^2\;,$ Then $tdt=udu$
So Integral $$\displaystyle K = -\int\frac{1}{u^2-\left(\sqrt{2}\right)^2}du = -\frac{1}{\sqrt{2}}\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|+\mathcal{C_{2}}$$
So we get $$\displaystyle K=-\frac{1}{\sqrt{2}}\ln\left|\frac{\sqrt{t^2+1}-\sqrt{2}}{\sqrt{t^2+1}+\sqrt{2}}\right|+\mathcal{C_{2}}$$
So we get $$\displaystyle I = -\ln|x+\sqrt{x^2+1}|-\sqrt{2}\ln\left|\frac{\sqrt{x^2+1}-\sqrt{2}x}{\sqrt{x^2+1}+\sqrt{2}x}\right|+\mathcal{C}$$
|
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|
In tetrahedron ABCD, prove that $r<\frac{AB\cdot CD}{2AB+2CD}$
Given tetrahedron $ABCD$, let $r$ be the radius of the sphere inscribed tetrahedron. Prove that $$r<\dfrac{AB\cdot CD}{2\,AB+2\,CD}$$
This is difficult question, my friends and I could not find any hint to solve it.
In the case that $ABCD$ is a regular tetrahedron, the inequality is easy to prove. Let $\ell$ be the side length of $ABCD$. Then, if $M$ is the centroid of $\triangle ABC$ and $I$ is the center of the insphere of the tetrahedron, then
we have
$$MA=\frac{1}{\sqrt{3}}\,\ell\,,$$
and
$$DM=\sqrt{\frac{2}{3}}\,\ell\,.$$
Since $r=DI$, we have
$$AI=DI=DM-DI=\sqrt{\frac{2}{3}}\,\ell-r\,.$$
Hence, by the Pythagorean Theorem,
$$r^2=IM^2=AI^2-MA^2=\left(\sqrt{\frac{2}{3}}\,\ell-r\right)^2-\frac{1}{3}\,\ell^2\,.$$
So
$$r=\frac{1}{2\sqrt{6}}\,\ell<\frac{1}{4}\,\ell=\frac{AB\cdot CD}{2\,AB+2\,CD}\,.$$
|
Let $MN$ be the common perpendicular line of $AB$ and $CD$ with $M$ a point in $AB$ and $N$ in $CD$. The volume $V$ of the tetrahedron $ABCD$ can be calculated in two ways:
$$\begin{align}
V &= \frac16 \times AB \times CD \times MN \times \sin(\alpha) \\
&= \frac13 \times r \times (S_{\triangle ABC} + S_{\triangle BCD} + S_{\triangle CDA} + S_{\triangle DAB})
\end{align}$$
where $\alpha$ is the angle between $AB$ and $CD$.
Notice that
$$ S_{\triangle ABC} = \frac12 \times AB \times MC > \frac12 \times AB \times MN, $$
and this is true for all other $3$ triangles. Thus
$$\begin{align}
V &= \frac13 \times r \times (S_{\triangle ABC} + S_{\triangle BCD} + S_{\triangle CDA} + S_{\triangle DAB}) \\
&> \frac16 \times r \times (AB \times MN + CD \times MN + CD \times MN + AB \times MN) \\
&= \frac13 \times r \times MN \times (AB + CD)
\end{align}$$
It follows that
$$\begin{align}
r \times MN \times (AB + CD) &< 3V \\
&= \frac12 \times AB \times CD \times MN \times \sin(\alpha) \\
&\leqslant \frac12 \times AB \times CD \times MN
\end{align}$$
So finally
$$ r < \frac{AB \times CD}{2(AB + CD)}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $n$, where its factorial is a product of factorials I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.
I have tried simplifying as follows:
$$\begin{array}{}
3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\
(3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^4 \cdot 5^2 \cdot 4^2 \cdot 7 &= n! \\
\end{array}$$
I really didn't see this helping me.
I then tried $6 \cdot 6 \cdot 6 \cdot 6 \cdot 25 \cdot 16 \cdot 7$, but $25$ only has $5$ as a double factor.
Any ideas?
|
Notice, $\color{red}{3}$, $\color{red}{5}$, $\color{red}{7}$ are prime numbers (can't be factorized), then the factorials can be successively reduced as follows $$3!5!7!=(3\cdot 2!)\cdot (5\cdot 4!)\cdot (7\cdot 6!)$$
$$=\color{red}{3}\cdot \color{red}{5}\cdot \color{red}{7}\cdot (2!)\cdot (4!)\cdot (6!)$$
$$=\color{red}{3}\cdot \color{red}{5}\cdot \color{red}{7}\cdot (2\cdot 1)\cdot (4\cdot 3!)\cdot (6\cdot 5!)$$
$$=1\cdot 2\cdot \color{red}{3}\cdot4\cdot \color{red}{5}\cdot 6\cdot \color{red}{7}\cdot (3! )\cdot (5!)$$
$$=1\cdot 2\cdot 3\cdot4\cdot 5\cdot 6\cdot 7\cdot (3\cdot 2\cdot1 )\cdot (5\cdot 4\cdot 3\cdot 2\cdot 1)$$
$$=1\cdot 2\cdot 3\cdot4\cdot 5\cdot 6\cdot 7\cdot ((2\cdot 4)\cdot(3\cdot 3) \cdot(2\cdot 5) )$$
$$=1\cdot 2\cdot 3\cdot4\cdot 5\cdot 6\cdot 7 \cdot8\cdot9\cdot 10$$$$=10!=n!$$
$$\color{red}{n=10}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate the limit of ratio of sums of sines (without L'Hopital): $\lim_{x\to0} \frac{\sin x+\sin3x+\sin5x}{\sin2x+\sin4x+\sin6x}$ Limit to evaluate:
$$\lim_{x \rightarrow 0} \cfrac{\sin{(x)}+\sin{(3x)}+\sin{(5x)}}{\sin{(2x)}+\sin{(4x)}+\sin{(6x)}}$$
Proposed solution:
$$
\cfrac{\sin(x)+\sin(3x)+\sin(5x)}{\sin(2x)+\sin(4x)+\sin(6x)}
\Bigg/ \cdot\ \cfrac{1/x}{1/x}\Bigg/=
\frac{\cfrac{\sin(x)}x + \cfrac{\sin(3x)}{3x} \cdot 3 + \cfrac{\sin(5x)}{5x} \cdot 5}
{\cfrac{\sin(2x)}{2x} \cdot 2 + \cfrac{\sin(4x)}{4x} \cdot 4 + \cfrac{\sin(6x)}{6x} \cdot 6}
$$
Using $\lim_{x \rightarrow 0} \frac{\sin x}x=1$, we get
$$\frac{1+1\cdot 3+1\cdot 5}{1\cdot 2+1\cdot 4+1\cdot 6} = \frac 9{12} = \frac 3 4$$
Please tell me if I am correct.
|
Use
$$
2·\cos x·\sin(kx) = \sin((k+1)x)+\sin((k-1)x)
$$
or
$$
2\sin(x/2)\sin(kx)=\cos((k-1/2)x)-\cos((k+1/2)x)
$$
or something similar.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1441163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Integral involving cube root and seventh root Find the value of $$\int_{0}^{1} (1-x^7)^{\frac{1}{3}}-(1-x^3)^{\frac{1}{7}}\:dx$$
My Approach:
Let $$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$ and
$$I_2=\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dx$$
For $I_1$ substitute $x^7=1-t^3$ so $dx=\frac{-3t^2}{7}(1-t^3)^{\frac{-6}{7}}\:dt$ Hence
$$I_1=\int_{1}^{0}\frac{-3t^3}{7}(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$
$$I_1=\frac{-3}{7}\int_{0}^{1}(1-t^3-1)(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$
$$I_1=\frac{-3I_2}{7}+\frac{3A}{7}$$
where $A=\int_{0}^{1}(1-t^3)^{\frac{-6}{7}}dt$
Similarly using substitution $x^3=1-t^7$ for $I_2$ and proceeding as above we get
$$I_2=\frac{-7I_1}{3}+\frac{7B}{3}$$ where $B=\int_{0}^{1}(1-t^7)^{\frac{-2}{3}}dt$
But we need to find $I_1-I_2$ so got stuck up here
|
Hint. A straightforward approach is to use the Euler beta function
$$
B(a,b) := \int _0^1 x^{a-1}(1-x)^{b-1}dx =\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},\quad a>0, b>0.
$$ Then you get
$$
\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dt-\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dt=0.
$$
|
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"source": "stackexchange",
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|
Find the roots of the equation $x^2-[x]=4$ where $[x]$ is the greatest integer less than or equal to $x$. Show that there is no rational number $x$ satisfying the equation $x^2-[x]=4$. Find the roots of the equation $x^2-[x]=4$ where $[x]$ is the greatest integer less than or equal to $x$.
I tried to solve this equation but stuck.
For the first part i supposed $x=\frac{p}{q}$ be the solution of the equation.
$$\frac{p^2}{q^2}-4=\left[\frac{p}{q}\right]$$
Now LHS is rational and RHS is an integer. Therefore our assumption is wrong. There is contradiction. So its roots are not rational.
$x^2-4=[x]\implies (x-2)(x+2)=[x]$
How should i move ahead? Please guide me. Its roots are $-\sqrt2,\sqrt6$.
Is my method of solving the first part correct?
|
Write $x = [x] + \varepsilon$. Then you are trying to solve
$$n^2 + 2\varepsilon n + \varepsilon^2 - n - 4= 0 $$
For an integer value of $n$ and some $\varepsilon \in [0,1)$. By the quadratic formula we can express $\varepsilon$ in terms of $n$,
$$\varepsilon^2 + (2n)\varepsilon +[n^2 - n - 4]= 0 $$
$$\varepsilon = \frac{-2n ± \sqrt{4n^2 - 4(n^2 - n - 4)}}{2} = -n ± \sqrt{n+4}$$
So now we want to find values of $n$ where
$$ -n + \sqrt{n+4} \in [0,1)$$
or $$-n -\sqrt{n+4} \in [0,1)$$
Lets try the first case, i.e. find $n$ such that
$-n + \sqrt{n+4} \in [0,1) $.
Both conditions below must hold for a solution:
$$ -n + \sqrt{n+4} \geq 0 \quad\text{ and } -n + \sqrt{n+4} < 1\quad$$
We try small values of $n\geq -4$,
$$\begin{array}{c|cccc}
n & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3\\ \hline
-n+\sqrt{n+4} & 4 & 4 & 2+\sqrt{2} & 1+\sqrt{3} & 2 & -1+\sqrt{5} & \color{red}{-2+\sqrt{6}} & -3+\sqrt{7}
\end{array}$$
There is no need to check $n<-4$ since $\sqrt{n+4}$ won't be defined, and no need to check $n>3$ since the gap only widens. ($-n + \sqrt{n+4}$ is decreasing for $n>3$).
so $n=2$ is the only solution for the first case, with $\varepsilon=-2+\sqrt{6}$ and $x=\sqrt{6}$.
Similarly for the second case, we need
$$ -n - \sqrt{n+4} \geq 0 \quad\text{ and } -n - \sqrt{n+4} < 1\quad$$
$$\begin{array}{c|cccccc}
n & -4 & -3 & -2 & -1 & 0 \\ \hline
-n-\sqrt{n+4} & 4 & 2 & \color{red}{2-\sqrt{2}} & 1-\sqrt{3}& -2 \\
\end{array}$$
There is no need to check $n\le-4$ or $n>0$, for similar reasons to the first case. Hence, the only two solutions to the whole problem are
$$ x=-\sqrt{2},\sqrt{6}.$$
Regarding the rest of the question as in your edit:
Your proof is incomplete. An integer is also rational! If $p/q$ is a solution, then we do indeed get $\frac{p^2}{q^2} - 4 = \left[\frac{p}{q}\right]$, but this itself is not a contradiction. It only means that $q^2=1$. Thus the equation reduces to
$$p^2 - p - 4 = 0, \quad p\in\Bbb Z$$
But of course the only real solutions to this are $\frac{1± \sqrt{1+16}}{2}$ which are not integers.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving this trig identity:$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
$$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$$
What I've tried,
$$\frac{((1+\cos\theta)+(\sin\theta))((1+\cos\theta)+(\sin\theta))}{(1+\cos\theta-\sin\theta) (1+\cos\theta+\sin\theta)}$$
$$=\frac{(1+\cos\theta+\sin\theta)^2}{(1+\cos\theta)^2-\sin^2\theta}$$
After simplifying,
$$=\frac{2(1+\sin\theta \cos\theta + \sin\theta + \cos\theta)}{\cos\theta(\cos\theta+2)}$$
I cant carry on further. Thus, any kind assistance would be much appreciated.
|
from Second last line, $$\displaystyle \frac{(1+\sin \theta+\cos \theta)^2}{(1+\cos \theta)^2-\sin^2 \theta}=\frac{2\left[1+\sin \theta +\sin \theta\cdot \cos \theta+\sin \theta\right]}{2\cos \theta \cdot (1+\cos \theta)} = \frac{2(1+\cos \theta)(1+\sin \theta)}{2\cos \theta\cdot (1+\cos \theta)}$$
So we get $$\displaystyle \frac{1+\sin \theta}{\cos \theta}$$
Here I have solved Using double angle formula.
Given $\displaystyle \bf{L.H.S}$ as $\displaystyle \frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}$
Let $\theta=2\phi\;,$ Then $$\displaystyle \frac{1+\cos 2\phi+\sin 2\phi }{1+\cos 2\phi-\sin 2\phi} = \frac{2\cos^2 \phi+2\sin \phi\cdot \cos \phi}{2\cos^2 \phi-2\sin \phi\cdot \cos \phi}$$
Above we use the formula $$\bullet \; 1+\cos 2\phi = 2\cos^2 \phi$$
and $$\bullet\; 1-\cos 2\phi = 2\sin^2 \phi$$
and $$\bullet \; \sin 2\phi = 2\sin \phi\cdot \cos \phi$$ and $$\bullet\; \cos^2\phi-\sin^2 \phi = \cos 2\phi$$
So we get $$\displaystyle \frac{\cos \phi+\sin \phi}{\cos \phi-\sin \phi} = \frac{\cos \phi+\sin \phi}{\cos \phi-\sin \phi}\times \frac{\cos \phi+\sin \phi}{\cos \phi+\sin \phi} = \frac{\sin^2 \phi +\cos^2 \phi+\sin 2\phi}{\cos^2 \phi-\sin^2 \phi}$$
So we get $$\displaystyle \frac{1+\sin 2\phi}{\cos 2\phi}$$
Now Put $2\phi = \theta\;,$ We get
$$\displaystyle = \frac{1+\sin \theta}{\cos \theta}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Real part of $\sqrt{ai-1}$ Is there a way to find the real and imaginary parts of
$$
z=\sqrt{ai-1},\qquad a>0
$$
where $i=\sqrt{-1}$. Thanks. I do not know what to to do.
Note
$$
i=e^{i\pi/2}=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}.
$$
|
Convert $ai-1$ into polar form. Let $ai-1 = r(\cos\theta + i\sin\theta)$. Then
$$\begin{align*}
a &= r\sin\theta\\-1 &= r\cos\theta\\
r^2 &= (r\sin\theta)^2 + (r\cos\theta)^2\\
&= a^2 + 1\\
r &= \sqrt{a^2 + 1}\\
\tan \theta &= \frac{r\sin\theta}{r\cos\theta}\\
&= -a\\
\theta &= \pi - \tan^{-1} a
\end{align*}$$
Let $r(\cos\theta + i\sin\theta) = [p(\cos\phi + i\sin\phi)]^2 = p^2(\cos2\phi + i\sin2\phi)$ for $p\ge 0$.
Then $p^2 = r$ and $p = \sqrt r$. Also, there are two arguments $\phi\in[0,2\pi)$ that will satisfy $2\phi\equiv \theta \pmod {2\pi}$:
$$\phi_1 = \frac\theta 2,\quad \phi_2 = \frac{\theta + 2\pi}2$$
So one of the roots is
$$\begin{align*}
p(\cos\phi_1 + i\sin\phi_1) &= \sqrt[4]{a^2+1}\left[\cos\frac{\pi-\tan^{-1}a}2 + i\sin\frac{\pi-\tan^{-1}a}2\right]\\
&= \sqrt[4]{a^2+1}\left[\sin\frac{\tan^{-1}a}2 + i\cos\frac{\tan^{-1}a}2\right]\\
\end{align*}$$
and the other root is
$$\begin{align*}
p(\cos\phi_2 + i\sin\phi_2) &= \sqrt[4]{a^2+1}\left[\cos\frac{3\pi-\tan^{-1}a}2 + i\sin\frac{3\pi-\tan^{-1}a}2\right]\\
&= \sqrt[4]{a^2+1}\left[-\sin\frac{\tan^{-1}a}2 - i\cos\frac{\tan^{-1}a}2\right]\\
\end{align*}$$
Simplify the nested inverse tangent in (co)sine using half-angle formulae.
$$\begin{align*}
\sin\frac{\tan^{-1} a}2 &= \operatorname{sgn} a \sqrt{\frac{1-\cos\tan^{-1}a}2}\\
&= \operatorname{sgn} a\sqrt{\frac{1-\frac1{\sqrt{a^2+1}}}2}\\
&= \operatorname{sgn} a\sqrt{\frac{\sqrt{a^2+1} - 1}{2\sqrt{a^2+1}}}\\
\cos\frac{\tan^{-1} a}2 &= \sqrt{\frac{1+\cos\tan^{-1}a}2}\\
&= \sqrt{\frac{1+\frac1{\sqrt{a^2+1}}}2}\\
&= \sqrt{\frac{\sqrt{a^2+1} + 1}{2\sqrt{a^2+1}}}
\end{align*}$$
For this question, $\operatorname{sgn} a$ is always $1$.
|
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|
Inequality involving an exponent I wish to prove the following inequality
$$x^{\frac{3}{x-1}} > 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}, \quad x > 1.$$
Graphically the above inequality appears to be true since if one plots
$$g(x) = x^{\frac{3}{x-1}} - \left (1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} \right )$$
it appears as though $g(x) > 0$ for all $x > 1$. I have however been unable to prove analytically this is true.
I know
$$1 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} < 4 \quad \mbox{for all} \,\, x> 1$$
and
$$1 < x^{\frac{3}{x - 1}} < \mathrm{e}^3 \quad \mbox{for all} \,\, x > 1,$$
but neither of these bounds seem to help me very much.
Any pointers in the right direction would be greatly appreciated.
|
Hint :
Consider the Taylor expansion at infinity of your exponent :
$$x^{\frac{3}{x-1}} = x ^{3\ \big(\frac{1}{x} +\frac{1}{x^2} + \frac{1}{x^3} +...\big)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Are there $a,b,c,d \in \mathbb N$ such that $\frac{a + b}{a + b + c + d} < \frac{a}{a + c} < \frac{b}{b + d}$? Consider the following $2 \times 2$ contingency table:
\begin{array}{c|cc|c}
& C & \overline C & \Sigma \\ \hline
V & 4000 & 3500 & 7500 \\
\overline V & 2000 & 500 & 2500 \\ \hline
\Sigma & 6000 & 4000 & 10000
\end{array}
In data mining, we informally say that the association rule $C \to V$ is misleading. We could justify this by observing that:
$$
\Pr[V \mid C] = \frac{4000}{6000} = 0.\overline{6} < 0.75 = \frac{7500}{10000} = \Pr[V]
$$
Alternatively, we could justify this by observing that:
$$
\Pr[V \mid C] = \frac{4000}{6000} = 0.\overline{6} < 0.875 = \frac{3500}{4000} = \Pr[V \mid \overline C]
$$
I'm trying to figure out which justification is more compelling. When I change the numbers in the contingency table, both inequalities always seem to be simultaneously satisfied. I can't seem to make one inequality true and the other false.
More generally, consider the contingency table:
\begin{array}{c|cc|c}
& C & \overline C & \Sigma \\ \hline
V & a & b & a+b \\
\overline V & c & d & c + d\\ \hline
\Sigma & a + c & b + d & a + b + c + d
\end{array}
Question: Do there exist $a,b,c,d \in \mathbb N$ such that either one of the following two inequalities are satisfied:
\begin{align*}
\frac{a + b}{a + b + c + d} &< \frac{a}{a + c} < \frac{b}{b + d} \tag{1} \\
\frac{b}{b + d} &< \frac{a}{a + c} < \frac{a + b}{a + b + c + d} \tag{2}
\end{align*}
|
No. In general, for $a,b,c,d\in \Bbb N$ with $\frac{a}{b} < \frac{c}{d}$, $$\frac{a}{b} <\frac{a+c}{b+d}<\frac{c}{d}.$$
Proof: The left inequality is equivalent to $$a(b+d) < b(a+c) \iff ad < bc \iff \frac{a}{b}<\frac{c}{d}.$$ Likewise for the right inequality.
|
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|
Evaluating the magnitude. Is this less than 1? Where $|x|<1$, I'm looking to determine if
$$\left|\frac{1}{x}(-1+\sqrt{1-x^2})\right|<1$$
I believe it is, since we can use a Taylor series to approximate
$$\sqrt{1-x^2} = 1 - \frac{1}{2}x^2-\frac{1}{4}(x^2)^2 + O(3) \approx 1 - \frac{1}{2}x^2$$
This then gives
$$\left|\frac{1}{x}(-1+\sqrt{1-x^2})\right| \approx \left|\frac{1}{x}(-1+1 - \frac{1}{2}x^2)\right| = \left|-\frac{x}{2}\right|<1$$
This, though, is not a proof. Any ideas?
|
HINT:
Let $2y=\arcsin x,0\le2y\le\dfrac\pi2$
$\implies x=\sin2y,\cos2y=+\sqrt{1-x^2}$
$$\dfrac{\sqrt{1-x^2}-1}x=\dfrac{\cos2y-1}{\sin2y}=\cdots=-\tan y$$
Now $$0\le y\le\dfrac\pi4\implies0\le\tan y\le1$$
|
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|
Is this a lower bound for this integral? Let $T$ be the triangular region in the $xy$ plane with vertices $(-2,0) (2,0) $ and $(0.2)$. Then is this inequality true? $$\displaystyle\int\int_T e^{-(x^2+y^2)}dA \geq \pi\left(1-\dfrac{1}{\sqrt{e}}\right)$$
I can inscribe the half disk $D=\{(x,y)\in\mathbb{R}^2$ | $x^2+y^2=2, y\geq 0\}$ inside $T$. Then it is clear that,
$$ \displaystyle\int\int_Te^{-(x^2+y^2)}dA \geq \displaystyle\int\int_De^{-(x^2+y^2)}dA$$
Changing to polar coordinates I get the RHS as $\dfrac{\pi}{2}\left(1-\dfrac{1}{e^2}\right)$
Clearly, $$\dfrac{\pi}{2}\left(1-\dfrac{1}{e^2}\right)\geq \dfrac{\pi}{2}\left(1-\dfrac{1}{\sqrt{e}}\right)$$
So, $$\displaystyle\int\int_T e^{-(x^2+y^2)}dA \geq\dfrac{\pi}{2}\left(1-\dfrac{1}{\sqrt{e}}\right)$$
Can I get the RHS without the $\dfrac{1}{2}$?
|
Twice your integral is the integral over the square with vertices at $(\pm 2,0)$ and $(0,\pm 2)$, hence:
$$2I = \int_{-\sqrt{2}}^{+\sqrt{2}}\int_{-\sqrt{2}}^{+\sqrt{2}} e^{-(x^2+y^2)}\,dx\,dy =\pi\cdot\text{Erf}(\sqrt{2})^2\tag{1}$$
but due to the continued fraction representation for the error function,
$$ \text{Erf}(\sqrt{2})\geq 1-\frac{1}{e^2\sqrt{2\pi}}\tag{2} $$
hence:
$$ I \geq \frac{\pi}{2}\left(1-\frac{2}{e^2\sqrt{2\pi}}\right)>\frac{7}{5}\tag{3}$$
but since:
$$ \text{Erf}(\sqrt{2}) \leq 1-\frac{4\sqrt{2}-2}{7e^2\sqrt{\pi}}\tag{4}$$
we also have:
$$ I \leq \frac{\left(2-4 \sqrt{2}+7 e^2 \sqrt{\pi }\right)^2}{98 e^4}<\frac{42}{29}\tag{5}$$
hence your initial inequality does not hold.
|
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|
Li Shanlan's combinatorial identities I am struggling to prove the following combinatorial identities:
$$(1)\quad\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\binom{p+r}{m+n} = \binom{p}{m}\binom{p}{n},\quad \forall n\in\mathbb N,p\ge m,n$$
$$(2)\quad\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\binom{p+m+n-r}{m+n} = \binom{p+m}{m}\binom{p+n}{n},\quad \forall p\in\mathbb N$$
The book I found them in says they were discovered and proved by Chinese mathematician Li Shanlan in the 19th century but I have failed to find any of his works translated into English in the Internet.
I am looking for an either combinatorial or algebraic solution.
|
Suppose we seek to verify that
$$\sum_{r=0}^{\min\{m,n,p\}}
{m\choose r} {n\choose r}
{p+m+n-r\choose m+n}
= {p+m\choose m} {p+n\choose n}.$$
Introduce
$${n\choose r} = {n\choose n-r} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-r+1}}
(1+z)^n \; dz$$
and
$${p+m+n-r\choose m+n} = {p+m+n-r\choose p-r} \\ =
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{p-r+1}}
(1+w)^{p+m+n-r} \; dw.$$
Observe carefully that the first of these is zero when $r\gt n$ and
the second one when $r\gt p$ so we may extend the range of $r$ to
infinity.
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p+m+n}}{w^{p+1}}
\sum_{r\ge 0} {m\choose r} z^r \frac{w^r}{(1+w)^r}
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p+m+n}}{w^{p+1}}
\left(1+\frac{zw}{1+w}\right)^m
\; dw \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p+n}}{w^{p+1}}
(1+w+zw)^m
\; dw \; dz.$$
The inner integral is
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p+n}}{w^{p+1}}
\sum_{q=0}^m {m\choose q} (1+z)^q w^q
\; dw$$
with residue
$$\sum_{q=0}^{\min(m,p)} {m\choose q} {p+n\choose p-q}
(1+z)^q$$
which in combination with the outer integral yields
$$\sum_{q=0}^{\min(m,p)} {m\choose q} {p+n\choose n+q}
{n+q\choose n}.$$
Now note that
$${p+n\choose n+q} {n+q\choose n}
= \frac{(p+n)!}{(p-q)! (n+q)!} \frac{(n+q)!}{q! n!}
\\ = \frac{(p+n)!}{(p-q)! p!} \frac{p!}{q! n!}
= {p+n\choose n} {p\choose q}.$$
Therefore we just need to verify that
$$\sum_{q=0}^{\min(m,p)} {m\choose q} {p\choose p-q}
= {p+m\choose m}$$
which follows by inspection.
It can also be done with the integral
$${p\choose p-q} =
\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p}}{w^{p-q+1}} \; dw$$
which is zero when $q\gt p$ so we can extend $q$ to infinity
to get for the sum
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p}}{w^{p+1}}
\sum_{q\ge 0} {m\choose q} w^q
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{(1+w)^{p+m}}{w^{p+1}}
\; dw
\\ = {p+m\choose m}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove using factor theorem. Using factor theorem, show that $a+b$,$b+c$ and $c+a$ are factors of
$(a + b + c)^3$ - $(a^3 + b^3 + c^3)$
How do we go about solving this ?
Thanks in advance !
|
If a+b is a Factor then -b is a root and vice versa. so just set a=-b and you get $((-b)+b+c)^3 - ((-b)^3+b^3+c^3) =c^3-c^3 = 0$. Therefor a+b is a Factor. Same for a+c and b+c.....
|
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"timestamp": "2023-03-29T00:00:00",
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|
A cone with diameter $12$cm and height $8$cm. Find the volume of the inscribed Sphere. A cone with diameter $12\ cm$ and height $8\ cm$. Find the volume of the inscribed Sphere. Can someone help me solve this maths problem?
|
Notice, let $r$ be the radius of the sphere inscribed in the cone having diameter of base $12\ cm$ (radius, $R=6\ cm$) & the vertical height $H=8\ cm$.
If $\alpha$ is the semi-apex angle of cone then using geometry in a right triangle we get
$$\tan \alpha=\frac{R}{H}=\frac{6}{8}=\frac{3}{4}\ \ (\forall\ \ 0<\alpha<\pi/2)$$
$$\implies \sin \alpha=\frac{\tan \alpha}{\sqrt{1+\tan^2\alpha}}$$
$$=\frac{\frac{3}{4}}{\sqrt{1+\left(\frac{3}{4}\right)^2}}$$$$\sin \alpha=\frac{3}{5}$$
Using geometry, we also have $$\sin\alpha=\frac{r}{h-r}$$
$$\frac{r}{8-r}=\frac{3}{5}$$ $$8r=24\implies r=3\ $$
Hence, the volume of sphere inscribed into the cone is
$$V=\frac{4\pi r^3}{3}=\frac{4\pi (3)^3}{3}=36 \pi$$
$$\color{red}{V=36\pi\ cm^3}$$
|
{
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"url": "https://math.stackexchange.com/questions/1461052",
"timestamp": "2023-03-29T00:00:00",
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|
A calculation problem about differential equation
Let $n>2$ and $n\in \mathbb{N}_+$, $\lambda \ge - \frac{{{{\left( {n - 1} \right)}^2}}}{4}$.
For $x>0$, we have $$f\left( x \right) = {\left( {\sinh x} \right)^{2 - n}}\int_0^\pi {{{\left( {\cosh x + \cos t} \right)}^\eta }{{\left( {\sin t} \right)}^{2\sqrt {\lambda + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} .$$
Solving the value of $\eta$ such that function $f(x)$ satisfies differential equation $$f''\left( x \right) + \left( {n - 1} \right)\frac{{\cosh x}}{{\sinh x}}f'\left( x \right) - \lambda f\left( x \right) = 0.$$
Adding: This is a difficult problem! I have get
\begin{align*}&f'\left( x \right) = \left( {2 - n} \right){\left( {\sinh x} \right)^{1 - n}}\cosh x\int_0^\pi {{{\left( {\cosh x + \cos t} \right)}^\eta }{{\left( {\sin t} \right)}^{2\sqrt {\lambda + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} \\+ &\eta {\left( {\sinh x} \right)^{3 - n}}\int_0^\pi {{{\left( {\cosh x + \cos t} \right)}^{\eta - 1}}{{\left( {\sin t} \right)}^{2\sqrt {\lambda + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} .\end{align*}
and
\begin{align*}&f''\left( x \right) = \left( {2 - n} \right)\left[ {\left( {1 - n} \right){{\left( {\sinh x} \right)}^{ - n}}{{\left( {\cosh x} \right)}^2} + {{\left( {\sinh x} \right)}^{2 - n}}} \right]\int_0^\pi {{{\left( {\cosh x + \cos t} \right)}^\eta }{{\left( {\sin t} \right)}^{2\sqrt {\lambda + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} \\+ &\left( {5 - 2n} \right)\eta {\left( {\sinh x} \right)^{2 - n}}\cosh x\int_0^\pi {{{\left( {\cosh x + \cos t} \right)}^{\eta - 1}}{{\left( {\sin t} \right)}^{2\sqrt {\lambda + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} \\+ &\eta \left( {\eta - 1} \right){\left( {\sinh x} \right)^{4 - n}}\int_0^\pi {{{\left( {\cosh x + \cos t} \right)}^{\eta - 2}}{{\left( {\sin t} \right)}^{2\sqrt {\lambda + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} .\end{align*}
Note that $$f''\left( x \right) + \left( {n - 1} \right)\frac{{\cosh x}}{{\sinh x}}f'\left( x \right) - \lambda f\left( x \right) = 0.$$
So we get \begin{align*}&\left( {2 - n-\lambda} \right)\int_0^\pi {{{\left( {\cosh x + \cos t} \right)}^\eta }{{\left( {\sin t} \right)}^{2\sqrt {\lambda + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} \\+& \left( {4 - n} \right)\eta \cosh x\int_0^\pi {{{\left( {\cosh x + \cos t} \right)}^{\eta - 1}}{{\left( {\sin t} \right)}^{2\sqrt {\lambda + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} \\+& \eta \left( {\eta - 1} \right){\left( {\sinh x} \right)^2}\int_0^\pi {{{\left( {\cosh x + \cos t} \right)}^{\eta - 2}}{{\left( {\sin t} \right)}^{2\sqrt {\lambda + \frac{{{{\left( {n - 1} \right)}^2}}}{4}} }}dt} = 0.\end{align*}
But I have no way to continue, thanks for your help!
|
This is not a complete answer, but rather another way of attacking the problem, that might be more fruitful. It is too long to put in a comment.
First, we write the differential equation as
$$
\bigl(f'(x)\sinh^{n-1}x\bigr)'=\lambda f(x)\sinh^{n-1}x.\tag{1}
$$
Differentiating the integral that should equal $f$, and multiplying with $\sinh^{n-1}x$, we find that
$$
f'(x)\sinh^{n-1}x=\int_0^\pi \bigl[(2-n)\cosh x+\frac{\eta\sinh x}{\cosh x+\cos t}\bigr](\cosh x+\cos t)^{\eta}(\sin t)^{2\sqrt{\lambda+(n-1)^2/4}}\,dt\tag{2}
$$
The right-hand side of $(1)$ equals
$$
\lambda f(x)\sinh^{n-1}x=\lambda\int_0^{\pi}\sinh x(\cosh x+\cos t)^{\eta}(\sin t)^{2\sqrt{\lambda+(n-1)^2/4}}\,dt\tag{3}
$$
Now, instead of differentiating $(2)$, it looks easier to integrate $(3)$,
$$
\int\lambda f(x)\sinh^{n-1}x\,dx
=\frac{\lambda}{1+\eta}\int_0^\pi(\cosh x+\cos t)^{\eta+1}(\sin t)^{2\sqrt{\lambda+(n-1)^2/4}}\,dt\tag{4}
$$
Maybe you can find $\eta$ so that $(2)$ and $(4)$ differs by a constant?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the improper integral: $\int _0^{+\infty }\:\frac{dx}{\left(x^2+9\right)\cdot \sqrt[3]{x^2+9}}$ $$\int _0^{+\infty }\:\frac{dx}{\left(x^2+9\right)\cdot \sqrt[3]{x^2+9}}$$
Wolfram says strange things. And you need to solve analytically. Help.
|
$$\begin{align*}I=\int_{0}^{+\infty}(x^2+9)^{-4/3}\,dx = 3^{-5/3}\int_{0}^{+\infty}(1+z^2)^{-4/3}\,dz=\frac{\sqrt{\pi}\cdot\Gamma\left(\frac{5}{6}\right)}{2\cdot 3^{5/3}\cdot\Gamma\left(\frac{4}{3}\right)}\end{align*}$$
by using the substitution $z=\tan\theta$ and Euler's beta function. That simplifies to:
$$ I =\frac{\pi^{3/2}}{3^{2/3}\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{6}\right)}. $$
|
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|
Extreme Values of $f(P)$ When $P$ Lies Inside a Triangle with Vectices $A$,$B$, $C$ I would appreciate if somebody could help me with the following problem
Q:$P(x,y)$ given point lie inside or on the boundary of triangle $\triangle ABC$.
Find maximum and minimum $f(a,b,c)$
$$f(P)=\vec{PA}\cdot \vec{PB}+\vec{PB}\cdot \vec{PC}+\vec{PC}\cdot \vec{PA}$$
|
I rewrite my friend H. R's answer and reach to solution by simpler math. First you should notice that f is a function of $\overrightarrow P$ not a,b,c ! you want to find appropriate vector $\overrightarrow P$ for fixed triangle.
$$\eqalign{
& f\left( {\overrightarrow P } \right) = \overrightarrow {PA} .\overrightarrow {PB} + \overrightarrow {PB} .\overrightarrow {PC} + \overrightarrow {PC} .\overrightarrow {PA} \cr
& \overrightarrow {PA} .\overrightarrow {PB} = \left( {\overrightarrow A - \overrightarrow P } \right).\left( {\overrightarrow B - \overrightarrow P } \right) = {\left| {\overrightarrow P } \right|^2} - \left( {\overrightarrow A + \overrightarrow B } \right).\overrightarrow P + \overrightarrow A .\overrightarrow B \cr
& \overrightarrow {PB} .\overrightarrow {PC} = \left( {\overrightarrow B - \overrightarrow P } \right).\left( {\overrightarrow C - \overrightarrow P } \right) = {\left| {\overrightarrow P } \right|^2} - \left( {\overrightarrow B + \overrightarrow C } \right).\overrightarrow P + \overrightarrow B .\overrightarrow C \cr
& \overrightarrow {PC} .\overrightarrow {PA} = \left( {\overrightarrow C - \overrightarrow P } \right).\left( {\overrightarrow A - \overrightarrow P } \right) = {\left| {\overrightarrow P } \right|^2} - \left( {\overrightarrow C + \overrightarrow A } \right).\overrightarrow P + \overrightarrow C .\overrightarrow A \cr} $$
Hence, you may have
$$f\left( {\overrightarrow P } \right) = 3\overrightarrow P .\overrightarrow P - 2\overrightarrow P .\left( {\overrightarrow A + \overrightarrow B + \overrightarrow C } \right) + \left( {\overrightarrow A .\overrightarrow B + \overrightarrow B .\overrightarrow C + \overrightarrow C .\overrightarrow A } \right)$$
or
$$f\left( {\overrightarrow P } \right) = 3\left( {\overrightarrow P - {1 \over 3}\left({\overrightarrow A + \overrightarrow B + \overrightarrow C}\right) } \right).\left( {\overrightarrow P - {1 \over 3}\left({\overrightarrow A + \overrightarrow B + \overrightarrow C}\right) } \right)-{1 \over 3}\left({\overrightarrow A + \overrightarrow B + \overrightarrow C}\right).\left({\overrightarrow A + \overrightarrow B + \overrightarrow C}\right) + \left( {\overrightarrow A .\overrightarrow B + \overrightarrow B .\overrightarrow C + \overrightarrow C .\overrightarrow A } \right)$$
or
$$f\left( {\overrightarrow P } \right) =3 {\left| {\overrightarrow P - {1 \over 3}\left({\overrightarrow A + \overrightarrow B + \overrightarrow C}\right) } \right|^2}-{1 \over 3}{\left| {\overrightarrow A + \overrightarrow B + \overrightarrow C } \right|^2}+\left( {\overrightarrow A .\overrightarrow B + \overrightarrow B .\overrightarrow C + \overrightarrow C .\overrightarrow A } \right)$$
Finally, you conclude easily that the minimum of $f$ happens when $\overrightarrow P $ is at the centroid of the triangle. (last two sentences are constant and first one is always equal or greater than zero.)
|
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|
Proof that $n+k+3$ divides $n(n+1)(n+2)(n+3) - k(k+1)(k+2)(k+3)$. I'm looking for proof that
$$
(n+k+3) \mid n(n+1)(n+2)(n+3) - k(k+1)(k+2)(k+3)\\
n,k \in \mathbb N^*, n>k
$$
I tried using induction, but i'm not sure how it would work with 2 parameters.
|
Since
$$n+3=n+k+3-k\qquad \text{and}\qquad k+3=n+k+3-n$$
we have
\begin{align}
n(n+1)(n+2)(n+3)&=n(n+1)(n+2)(n+k+3)-n(n+1)(n+2)k\\
k(k+1)(k+2)(k+3)&=k(k+1)(k+2)(n+k+3)-k(k+1)(k+2)n
\end{align}
Then it will be sufficient to show that $n+k+3$ divides $k(k+1)(k+2)n-n(n+1)(n+2)k$
But
\begin{align}
k(k+1)(k+2)n-n(n+1)(n+2)k&=nk(k^2+3k-n^2-3n)\\
&=nk\left[-(n-k)(n+k)-3(n-k)\right]\\
&=-nk(n-k)(n+k+3)
\end{align}
Which completes the proof.
|
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|
How do i show this equality without using reccurence method:$\prod_{k=0}^{k=n}{\cos\frac{\theta}{2^k}}={\frac {\sin\theta}{2^n\sin(2^{-n}\theta)}}$?? I would like to show this without using reccurence method for all $n$ $\in $ $\mathbb{N}$ and $\theta \in \mathbb{R}$ :
$${\cos\frac{\theta}{2}}\cos\frac{\theta}{2^2}\cos\frac{\theta}{2^3}\cdots \cos\frac{\theta}{2^n} ={\frac {\sin\theta}{2^n\sin(2^{-n}\theta)}} $$
Note: I have used many trigonomitrics transformations but i can't succed !!
$$\prod_{k=0}^{k=n}{\cos\frac{\theta}{2^k}}={\frac {\sin\theta}{2^n\sin(2^{-n}\theta)}}$$
Thank you for any help
|
First, notice that:
\begin{align}
2\cos x\sin x&=\sin 2x\\
\implies \cos x&=\frac{1}{2}\frac{\sin 2x}{\sin x}
\end{align}
Then, given an integer number $n>0$ we have
$$\cos \left(\frac{\theta}{2}\right)\cdot \cos \left(\frac{\theta}{2^2}\right)\cdot \ldots \cdot \cos \left(\frac{\theta}{2^n}\right)=\frac{1}{2^n}\frac{\sin\left(\theta\right)}{\sin \left(\frac{\theta}{2}\right)}\cdot \frac{\sin \left(\frac{\theta}{2}\right)}{\sin \left(\frac{\theta}{2^2}\right)}\cdot \ldots\cdot \frac{\sin \left(\frac{\theta}{2^{n-1}}\right)}{\sin \left(\frac{\theta}{2^n}\right)}$$
Which is a "telescopic product", and the inequality follows from here.
|
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|
How can we show polar coordinates r(theta) is an ellipse? $r(θ) = a(1 − β^2)/(1 + β \cos θ)$
and I want to show this $r(θ)$ is an ellipse described by
$\dfrac{(x+\sqrt{a^2 − b^2})^2}{a^2}+\dfrac{y^2}{b^2}= 1$, when $0<β<1$.
How can we show this?
|
With the relations
$$
r=\sqrt{x^2+y^2}, r\cos\theta=x
$$
we can rewrite the equation as
$$
r=\frac{ar(1-\beta^2)}{r+\beta x}
$$
or, equivalently (disregarding $r=0$ that's not a solution),
$$
r+\beta x=a(1-\beta^2)
$$
that becomes $r=a(1-\beta^2)-\beta x$; now square and get
$$
x^2+y^2=a^2(1-\beta^2)^2-2a(1-\beta^2)\beta x+\beta^2x^2
$$
Reorder:
$$
x^2(1-\beta^2)+2a(1-\beta^2)\beta x+y^2=a^2(1-\beta^2)^2
$$
Divide everything by $1-\beta^2$:
$$
x^2+2a\beta x+\frac{y^2}{1-\beta^2}=a^2(1-\beta^2)
$$
Complete the square:
$$
x^2+2a\beta x+a^2\beta^2+\frac{y^2}{1-\beta^2}=a^2
$$
Set $c=a\beta$ and $a^2(1-\beta^2)=b^2$:
$$
(x+c)^2+\frac{a^2}{b^2}y^2=a^2
$$
Divide by $a^2$:
$$
\frac{(x+c)^2}{a^2}+\frac{y^2}{b^2}=1
$$
Note that $c=\sqrt{a^2-b^2}$.
|
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|
If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex plane.Find the area of the square. For all real numbers $x$,let the mapping $f(x)=\frac{1}{x-i},\text{where} i=\sqrt{-1}$.If there exist real numbers $a,b,c,d$ for which $f(a),f(b),f(c),f(d)$ form a square on the complex plane.Find the area of the square.
$f(a)=\frac{1}{a-i}=\frac{a+i}{a^2+1}=(\frac{a}{a^2+1},\frac{1}{a^2+1})$
Similarly,$f(b)=\frac{1}{b-i}=\frac{b+i}{b^2+1}=(\frac{b}{b^2+1},\frac{1}{b^2+1})$
Similarly,$f(c)=\frac{1}{c-i}=\frac{c+i}{c^2+1}=(\frac{c}{c^2+1},\frac{1}{c^2+1})$
Similarly,$f(d)=\frac{1}{d-i}=\frac{d+i}{d^2+1}=(\frac{d}{d^2+1},\frac{1}{d^2+1})$
Now the area of the square$=(\frac{a}{a^2+1}-\frac{b}{b^2+1})^2+(\frac{1}{a^2+1}-\frac{1}{b^2+1})^2$
But this expression simplifies to $\frac{(a-b)^2}{(1+a^2)(1+b^2)}$.
How should i prove the area of the square to be $\frac{1}{2}$.Is my approach not correct?Please help me.Thanks.
|
Let $\Phi(z)=\dfrac{1}{\bar{z}}$, this is the inversion with respect to the unit circle. Now, for a real $x$, we have $f(x)=\Phi(x+i)$.
The image of the line $d=\{z:\Im z=1\}$ under the inversion $\Phi$ is the circle
of diameter $[0,i]$, So if $f(a)$, $f(b)$, $f(c)$ and $f(d)$ form a square, it must be inscribed in that circle of radius $\frac12$, and consequently it must have an area equal to $\frac12$.
|
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|
$Z^4 = -1$ How do I solve this without a calculator? Basically, the question is to solve $z^8= 1$. I have factored this down to $$(z+1)(z-1)(z^2+1)(z^4+1)=0$$
I have simplified $(z^4+1)$ to $z =\pm \sqrt{i}$ and online I know that this can be simplified to the following four solutions:
z= 0.7071 + 0.7071 i
z= -0.7071 + 0.7071 i
z= -0.7071 - 0.7071 i
z= 0.7071 - 0.7071 i
How would I get this without a calculator?
Thank you!
|
use the Euler's Identity
$$e^{\pi i}=-1$$
and
$$e^{ix}=\cos x+i\sin x$$
$$z^4=-1$$
$$z^4=e^{\pi i}$$
$$z=e^{\frac{\pi }{4}+\frac{n\pi}{2}}$$
$$z_1=\cos(\frac{\pi }{4}+\frac{\pi}{2})+i\sin(\frac{\pi }{4}+\frac{\pi}{2})$$
$$z_2=\cos(\frac{\pi }{4}+\frac{2\pi}{2})+i\sin(\frac{\pi }{4}+\frac{2\pi}{2})$$
$$z_3=\cos(\frac{\pi }{4}+\frac{3\pi}{2})+i\sin(\frac{\pi }{4}+\frac{3\pi}{2})$$
$$z_4=\cos(\frac{\pi }{4}+\frac{4\pi}{2})+i\sin(\frac{\pi }{4}+\frac{4\pi}{2})$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplify the fraction with radicals I want to simplify this fraction
$$ \frac{\sqrt{6} + \sqrt{10} + \sqrt{15} + 2}{\sqrt{6} - \sqrt{10} + \sqrt{15} - 2} $$
I've tried to group up the denominator members like $ (\sqrt{6} + \sqrt{15}) - (\sqrt{10} + 2) $ and then amplify with $ (\sqrt{6} + \sqrt{15}) + (\sqrt{10} + 2) $
|
For this system we implemented finding a split form for the denomiator:
$$a + \sqrt{p}\,b$$
Such that $\sqrt{p}$ is a new radical. For a quotient we then have:
$$\frac{c}{a + \sqrt{p}\,b} = \frac{c\,(a - \sqrt{p}\,b)}{a^2 - p\,b^2}$$
Lets give it a try:
$$\frac{2+\sqrt{6}+\sqrt{10}+\sqrt{15}}{-2+\sqrt{6}-\sqrt{10}+\sqrt{15}} = $$
$$\frac{2+\sqrt{15}+\sqrt{2}(\sqrt{3}+\sqrt{5})}{- 2+\sqrt{15}+\sqrt{2}(\sqrt{3}-\sqrt{5})} =$$
$$\frac{(2+\sqrt{15}+\sqrt{2}(\sqrt{3}+\sqrt{5}))\,(- 2+\sqrt{15}-\sqrt{2}(\sqrt{3}-\sqrt{5}))}{(- 2+\sqrt{15})^2-2(\sqrt{3}-\sqrt{5})^2} = $$
$$\frac{15+6\sqrt{6}}{3} = 5+2\sqrt{6}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the real values of a which satisfies both the equality $|z+\sqrt2|=a^2-3a+2$ and the inequality $|z+i\sqrt 2|
What are the real values of parameter 'a' for which at least one
complex number z satisfies both the equality $|z+\sqrt {2}|=a^2-3a+2$ and
the inequality $|z+i\sqrt 2|<a^2$ ?
Ok I tried to consider the equations as circles centred at $-\sqrt{2}$ and $-i\sqrt{2}$ respectively.Can't think of what to do after that.Suggestions please!
|
The equation
$$|z+\sqrt{2}|=a^2-3a+2 \tag{1}$$
defines a circle of radius $r_1=a^2-3a+2$ centered at $-\sqrt{2}$.
The inequality
$$|z+i\sqrt 2|<a^2 \tag{2}$$
defines the interior of a circle of radius $r_2=a^2$ centered at $-i\sqrt{2}$.
As others have pointed out, we must have
$$\begin{align}
a^2-3a+2>0 &\implies \\[2ex]
(a-1)(a-2)>0 &\implies \\[2ex]
a<1\text{ or }a>2
\tag{3}
\end{align}$$
and
$$a\ne0 \tag{4}$$
The distance between $-\sqrt{2}$ and $-i\sqrt{2}$ is $2$. So for the first circle to overlap with the interior of the second circle the following two inequalities need to hold:
$$\begin{align}
r_1+r_2>2 &\implies \\[2ex]
2a^2-3a+2>2 &\implies \\[2ex]
(a-\tfrac{3}{4})^2 > \tfrac{9}{16} &\implies \\[2ex]
a\in(-\infty,0)\cup(\tfrac{3}{2},\infty) \tag{5}
\end{align}$$
and
$$\begin{align}
r_1<r_2+2 &\implies \\[2ex]
a^2-3a+2<a^2+2 &\implies \\[2ex]
a>0 \tag{6}
\end{align}$$
Putting the restrictions from (3),(4),(5),(6) together, solutions for $z$ exist for all $a$ such that $a>2,\quad a\in\mathbb{R}$.
|
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}
|
A trigonometry equation: $3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$
What are the steps to solve this equation for $ \theta $?
Because, I am always unable to deal with the product $\sin \theta \cos \theta$.
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Following excellent advice by @raul, you write immediately
$$\tan(\theta)=\frac{-5\pm\sqrt{5^2+4\cdot3\cdot2}}{2\cdot3}=\frac13,-2.$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1481232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
}
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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.