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$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
|
\begin{align}
\sqrt[5]{2^4 \sqrt[3]{16}}
&=\sqrt[5]{2^4 \sqrt[3]{2^4}}
=\sqrt[5]{2^4 \sqrt[3]{2^3 \cdot 2}}
=\sqrt[5]{2^4 \cdot 2 \sqrt[3]{2}}\\
&=\sqrt[5]{2^5 \sqrt[3]{2}}
=2 \sqrt[5]{\sqrt[3]{2}}
=2 \cdot \sqrt[15]{2}
=2 \cdot 2^{1/15}\\
&=2^{16/15}.
\end{align}
|
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|
Compute $\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$
Compute $\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$
This question came after an exercise involving finding the $7$th roots of $-1$. The roots were $\operatorname{cis}\frac{\pi}{7},\operatorname{cis}\frac{3\pi}{7},\dots$
This made me wonder if I could somehow use those roots, along with the geometry of complex numbers, to compute the expression. Any insight would be helpful. Thanks!
|
As $\cos(\pi-x)=-\cos(x)$ this is the same as to compute:
$$\cos \left( \frac{\pi}{7} \right)+\cos \left( \frac{3 \pi}{7} \right)+\cos \left(5 \frac{\pi}{7} \right)$$
But denoting by $\omega_i$ $i\in{0,\ldots 6}$ the $7$th roots of $-1$.
You have:
$$\sum_{i} \omega_i=0$$
and by taking the real part:
$$-1+2\left( \cos \left( \frac{\pi}{7} \right)+\cos \left( \frac{3 \pi}{7}\right)+ \cos \left(5 \frac{\pi}{7} \right) \right)=0$$
The geometric interpretation is that the center of mass of a regular heptagon (the $7$th roots of $-1$) is $0$ so it must be the same for the center of mass of the projections on the $x$-axis of it vertices.
|
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Estimation of $\int_0^1 \frac{1}{1+x^4}dx$
Prove that $\dfrac34<\displaystyle\int_0^1 \frac{1}{1+x^4}\,\mathrm dx<\dfrac {9}{10}$.
My working:
We can easily prove that
$$\begin{align}
\frac{1}{1+x^2}&<\frac{1}{1+x^4}<1-x^4+x^8,\forall x\in (0,1) \\
\implies\int_0^1\frac{1}{1+x^2}\,\mathrm dx&<\int_0^1\frac{1}{1+x^4}\,\mathrm dx<\int_0^1(1-x^4+x^8)\,\mathrm dx \\
\implies\frac34&<\frac{\pi}{4}<\int_0^1\frac{1}{1+x^4}\,\mathrm dx<(1-\frac 15+\frac 19)=\frac{41}{45}
\end{align}$$
But unfortunately $\dfrac {9}{10}<\dfrac{41}{45}$.
|
We can easily prove that
$$\begin{align}
1-x^3&<\frac{1}{1+x^4}<1-\frac{x^4}{2},\forall x\in (0,1) \\
\implies\int_0^1 1-x^3\,\mathrm dx&<\int_0^1\frac{1}{1+x^4}\,\mathrm dx<\int_0^11-\frac{x^4}{2}\,\mathrm dx \\
\implies\frac34&<\int_0^1\frac{1}{1+x^4}\,\mathrm dx<1-\frac {1}{10}=\frac{9}{10}
\end{align}$$
|
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A Problem Dealing with putting balls in bin and Expected Value - possible wrong answer Please consider the problem below. Is my answer correct. If is not correct then where did I go wrong?
Problem:
You keep tossing balls into $n$ bins until one of the bins has two balls. For each toss there is a $\frac{1}{n}$ probability that the ball you toss lands in any one of the bins. What is the expected number of tosses?
Answer:
Let $p_i$ be the probability that after $i$ tosses we have at least one bin
with two balls.
\begin{eqnarray*}
p_1 &=& 0 \\
p_2 &=& 1 - ( 1 - \frac{1}{n} ) = \frac{1}{n} \\
p_3 &=& 1 - (1 - \frac{1}{n})(1 - \frac{1}{n-1}) \\
p_3 &=& 1 - (\frac{n-1}{n})(\frac{n-1-1}{n-1}) \\
p_3 &=& 1 - (\frac{n-2}{n}) = \frac{2}{n} \\
p_4 &=& 1 - (1 - \frac{1}{n})(1 - \frac{1}{n-1})(1 - \frac{1}{n-2}) \\
p_4 &=& 1 - ( \frac{n-1}{n} )( \frac{n-2}{n-1} )( \frac{n - 2 -1}{n - 2} ) \\
p_4 &=& 1 - \frac{n-3}{n} = \frac{3}{n} \\
\end{eqnarray*}
Now for $1 <= i <= n$ we have: $p_i = \frac{i-1}{n}$.
\begin{eqnarray*}
E &=& 2p_2 + 3p_3 + 4p_4 + ... (n+1)p_{n+1} \\
E &=& \sum_{i = 1}^{n} \frac{i(i+1)}{n} = \frac{1}{2n} \sum_{i=1}^{n} i^2 + i \\
E &=& \frac{1}{2n}(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} ) \\
E &=& \frac{n+1}{4n} ( \frac{2n+1}{3} + 1 ) \\
\end{eqnarray*}
Here is an update to my answer:
Let $p_i$ be the probability that after $i$ tosses we have at least one bin
with two balls.
\newline
\begin{eqnarray*}
p_1 &=& 0 \\
p_2 &=& 1 - ( 1 - \frac{1}{n} ) = \frac{1}{n} \\
p_3 &=& 1 - (\frac{n-1}{n})( \frac{n-2}{n}) \\
p_3 &=& 1 - \frac{(n-1)(n-2)}{n^2} = \frac{n^2 - (n^2 - 3n + 2)}{n^2} \\
p_3 &=& \frac{3n-2}{n^2} \\
p_4 &=& 1 - (\frac{n-1}{n})(\frac{n-2}{n})(\frac{n-3}{n}) \\
p_4 &=& 1 - \frac{(n^2-3n+2)(n-3)}{n^3}\\
p_4 &=& 1 - \frac{n^3-3n^2+2n - 3n^2 +9n - 6}{n^3}\\
p_4 &=& \frac{3n^2-2n + 3n^2 - 9n + 6}{n^3}\\
p_4 &=& \frac{3n^2 + 3n^2 - 11n + 6}{n^3}\\
\end{eqnarray*}
\begin{eqnarray*}
E &=& 2p_2 + 3p_3 + 4p_4 + ... (n+1)p_{n+1} \\
\end{eqnarray*}
Now, am on the right track? That is, is what I have so far correct?
Thanks,
Bob
|
Use the pigeonhole principle.
If the number of tosses ($n$) $\geq b+1$ , then at least one bin will contain $2$ balls. If the number of tosses is $0 < n < b$, the probability that each ball will go into a different bin is $$\frac{1 \times 2 \times ...(b-n+1)}{b^n} = \frac{b!}{n!\times b^n}$$
So the probability that a bin contains more than one ball after $n$ tosses is $$1-\frac{b!}{n!\times b^n}$$
So the expected number of tosses can be found with $$E(b) = 1+ \sum_{k=1}^{b} \frac{b!}{\left(b-k\right)!b^k}$$
Visualized, this looks like:
|
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|
$30\mid p_1^4+p_2^4+\cdots+p_{31}^4$. Prove that $p_1=2,p_2=3,p_3=5$ Let $p_1<p_2<\cdots<p_{31}$ be prime numbers such that $30\mid p_1^4+p_2^4+\cdots+p_{31}^4$. Prove that $p_1=2,p_2=3,p_3=5$
I have no clue whatsoever about how to even approach this problem. Any hint is welcome.
|
There are $30$ terms in $p_2^4+...+p_{31}^4$ and each of them is $\equiv 1\pmod 2.$ So $2$ divides $p_1^4+30$ so $2$ divides $p_1^4$ so $p_1=2$.
There are $29$ terms in $p_3^4+...+p_{31}^4 $ and each of them is $\equiv 1 \pmod 3 .$ So $3$ divides $2^4+p_2^4 +29$ so $3$ divides $p_2^4$ so $p_2=3.$
There are $28$ terms in $p_4^4+...+p_{31}^4$ and each of them is $\equiv 1 \pmod 5 .$ So $5$ divides $2^4+3^4+p_3^4+28,$ so $5$ divides $p_3^4$ so $p_3=5.$
|
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A question about a palindromic polynomial of even degree In Wikipedia it is stated,
If $p(x)$ is a palindromic polynomial of even degree $2d$, then there is a polynomial $q$ of degree $d$ such that $p(x) = x^d q(x + \frac{1}{x})$.
My question is:
How does one find the polynomial $q$ given $p$.
The specific example I have in mind is:
$p(t) = t^{12} - t^{11} - t^{10} + t^8 + t^7 - 2t^6 + t^5 + t^4 - t^2 - t + 1 =(t^2 + t + 1)^2(t^2 - t + 1)(t + 1)^2(t - 1)^4 $
Any help is highly appreciated.
|
Write $q(y) = a_6 y^6 + \cdots + a_1 y + a_0$. Using your example, we have $$\begin{align*}q(x + x^{-1}) =& x^6 - x^5 - x^4 + x^2 + x -2 + x^{-1} + x^{-2} - x^{-4} - x^{-5} + x^{-6} \\
=& a_6 x^6 + a_5 x^5 + (6 a_6 + a_4) x^4 + (5 a_5 + a_3) x^3 + \\&(15 a_6 + 4 a_4 +a_2) x^2 + (10 a_5 + 3 a_3 + a_1) x + \\
&(20a_6 + 6 a_4 + 2 a_2 + a_0) + \text{symmetrical part}\end{align*}.$$
This gives two linear systems: $$\begin{align*} a_6 &=1 \\ 6 a_6 + a_4 &= -1 \\ 15a_6 + 4 a_4 + a_1 &=1 \\ 20 a_6 + 6 a_4 + 2 a_2 + a_0 &= -2 \\ \end{align*}$$
which has solution $(a_6, a_4, a_2, a_0) = (1, -7, 14, -8)$, and $$\begin{align*} a_5 &= -1 \\ 5a_5 + a_3 &= 0 \\ 10 a_5 + 3a_3 + a_1 &= 1 \end{align*}$$
which has solution $(a_5, a_3, a_1) = (-1, 5, -4)$, so $$q(y) = y^6 - y^5 - 7y^4 + 5y^3 + 14y^2 - 4y - 8.$$
You can confirm this in WolframAlpha.
To solve the problem in more generality, let's construct polynomials $S_n$ with the property that $S_n(x + x^{-1}) = x^n + x^{-n}$. Obviously $S_0(y) = 2$ and $S_1(y) = y$, and from then on we have a recurrence relation $$S_{n+1}(y) = y S_n(y) - S_{n-1}(y)$$ which you can confirm by substituting $y = x + x^{-1}$, $S_n(y) = x^n + x^{-n}$, $S_{n-1}(y) = x^{n-1} + x^{1-n}$ and simplifying. The first few such polynomials are $$\begin{align*}S_0(y) &= 2 \\ S_1(y) &= y \\ S_2(y) &= y^2 - 2 \\ S_3(y) &= y^3 - 3y \\ S_4(y) &= y^4 - 4y + 2 \\ S_5(y) &= y^5 - 5y^3 + 5y \\ S_6(y) &= y^6 - 6y^4 + 9y - 2. \end{align*}$$
It's not hard to see that in general, $S_n(y) = 2 T_n(y/2)$, where $T_n$ is the Chebyshev polynomial of the first kind defined by $T_0(y) = 1, T_1(y) = y, T_{n+1}(y) = 2y T_n(y) - T_{n-1}(y).$ Then given a palindromic polynomial $p(x) = a_d x^{2d} + \cdots + a_1 x^{d+1} + a_0 x^d + a_{1} x^{d-1} + \cdots + a_d$, the polynomial $q$ such that $p(x) = x^d q(x + x^{-1})$ is simply $$q(y) = 2 \sum_{i=0}^d a_i T_i\left(\frac{y}{2}\right).$$
|
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How to solve this complex limits at infinity with trig? Please consider this limit question
$$\lim_{x\rightarrow\infty}\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{x\cdot \sin\left(\frac{a}{2x}\right)}$$
How should I solve this? I have no idea where to start please help.
|
Trivial with equivalents:
as $x\to\infty$,
*
*$\dfrac{a(x+1)}{2x}\to \dfrac a2$, so $\sin\dfrac{a(x+1)}{2x}\sim_\infty \sin \dfrac a2$,
*$\sin\dfrac{a}{2x}\sim_\infty\dfrac{a}{2x} $,
so that
$$\frac{a\sin \dfrac{a(x+1)}{2x}\,\sin\dfrac{a}{2}}{x\cdot \sin \dfrac{a}{2x}}\sim_\infty\frac{a\sin^2\dfrac a2}{x\,\dfrac a{2x}}=2\sin^2\dfrac a2=1-\cos a.$$
|
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Probability that if three balls numbered from 1-20 are selected without replacement that at least one will be numbered at least 17 I'm a new user here. I am currently self-learning probability theory (long journey ahead). I was wondering as to why my answer is not correct. I have listed the question below.
Three balls are to be randomly selected without replacement from an urn containing 20 balls numbered 1 through 20. If we bet that at least one of the balls that are drawn has a number as large as or larger than 17, what is the probability that we win the bet?
My solution: $ {{ 4 \choose 1}{ 19 \choose 2} \over { 20 \choose 3}} $
Reason: ${ 4 \choose 1}$ is of choosing either 17, 18, 19, or 20. ${ 19 \choose 2}$ once a urn is chosen which is either 17,18, 19 ,20. There are 19 balls left and the value does not matter.
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Find the probability that at least one of three balls selected from an urn containing $20$ balls, numbered from $1$ to $20$, displays a number that is at least $17$.
Method 1: We subtract the probability that none of the three selected balls have numbers larger than $16$ from $1$.
There are
$$\binom{20}{3}$$
ways to select three of the $20$ balls.
There are
$$\binom{16}{3}$$
ways to select three of the $16$ balls with numbers less than $17$.
Hence, the probability that none of the three balls have a number that is at least $17$ is
$$\frac{\dbinom{16}{3}}{\dbinom{20}{3}}$$
Therefore, the probability that at least one ball has a number that is at least $17$ is
$$1 - \frac{\dbinom{16}{3}}{\dbinom{20}{3}}$$
Method 2: We count directly.
There are four balls with numbers at least $17$ in the urn and $20 - 4 = 16$ balls with smaller numbers. The number of ways of selecting exactly $k$ balls with numbers that are at least $17$ and $3 - k$ balls with numbers less than $17$ is
$$\binom{4}{k}\binom{16}{3 - k}$$
If at least one ball with a number at least $17$ is selected, then $k \geq 1$, so the number of favorable cases is
$$\binom{4}{1}\binom{16}{2} + \binom{4}{2}\binom{16}{1} + \binom{4}{3}\binom{16}{0}$$
The probability that at least one ball that displays a number at least $17$ is thus
$$\frac{\dbinom{4}{1}\dbinom{16}{2} + \dbinom{4}{2}\dbinom{16}{1} + \dbinom{4}{3}\dbinom{16}{0}}{\dbinom{20}{3}}$$
What was your mistake?
In designating one of the balls to be the one that displays a number that is at least $17$, you counted selections in which two balls display a number that is at least $17$ twice and selections in which all three balls display a number that is at least $17$ three times, once for each way you could designate one of those balls as the one that displays a number that is at least $17$. Notice that
$$\binom{1}{1}\binom{4}{1}\binom{16}{2} + \binom{2}{1}\binom{4}{2}\binom{16}{1} + \binom{3}{1}\binom{4}{3}\binom{16}{0} = \binom{4}{1}\binom{19}{2}$$
|
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Evaluate$\int_0^\infty \frac{\log(x)^2}{(1+x)^2}dx$ Trying to evaluate the integral
$$\int_0^\infty \frac{\log(x)^2}{(1+x)^2}dx$$
or equivalently $$\int_{-\infty}^\infty e^x\frac{x^2}{(1+e^x)^2}dx$$
The answer should be $\frac{\pi^2}{3}$
Looks like we can use residue theorem, but the integration path does not seem to be easy to choose
|
We can compute directly.
$$\int_0^{\infty}\frac{\log^2x}{(1+x)^2}dx = \int_{0}^1
\frac{\log^2x}{(1+x)^2}dx + \int_1^{\infty}\frac{\log^2x}{(1+x)^2}dx. $$
With a change of variable $t=\frac{1}{x}$ in the second integral we have
$$\int_{1}^{\infty} \frac{\log^2x}{(1+x)^2} dx = \int_0^1\frac{\log^2t}{(1+t)^2}dt.$$
Then we have
$$\int_0^\infty\frac{\log^2x}{(1+x)^2} dx = 2 \int_0^1 \frac{\log^2x}{(1+x)^2} dx$$
and we only need to compute the last integral.
However,
$$\int_0^1 \frac{\log^2x}{(1+x)^2} dx = \int_0^1 \frac{\log^2x}{(1+x)} dx
- \int_0^1 \frac{x\log^2x}{(1+x)^2}dx$$
and a integration by parts gives
$$\int_0^1\frac{x\log^2x}{(1+x)^2} dx= -\int_0^1 x\log^2x d\left(\frac{1}{1+x}\right) = \int_0^1 \frac{\log^2x}{(1+x)} dx + 2\int_0^1\frac{\log x}{1+x}dx.$$
As a result, we have
$$\int_0^1\frac{\log^2x}{(1+x)^2} dx = -2 \int_0^1 \frac{\log x}{1+x}dx
= -2\int_0^1 \log x \sum_{n=0}^\infty (-x)^n dx
=2 \sum_{n=0}^\infty (-1)^n \frac{1}{(n+1)^2} = \frac{\pi^2}{6},$$
where the term by term integration can be justified by the Fubini's theorem.
After all, the result is $\frac{\pi^2}{3}$.
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Evaluate $\int \frac{e^{2x}-1}{e^{2x}+1}dx$ using partial fractions
Integrate the function $\frac{e^{2x}-1}{e^{2x}+1}$ using partial fractions.
My Attempt
$$
\int\frac{e^{2x}-1}{e^{2x}+1}dx=\frac{1}{2}\int\frac{2.e^{2x}}{e^{2x}+1}dx-\int\frac{dx}{e^{2x}+1}=\frac{1}{2}\log|e^{2x}+1|-\int\frac{dx}{e^{2x}+1}
$$
Put $t=e^x\implies dt=e^xdx=tdx\implies dx=\frac{dt}{t}$
$$
\int\frac{dx}{e^{2x}+1}=\int\frac{dt}{t(t^2+1)}
$$
Using partial fractions,
\begin{equation}
\frac{1}{t(t^2+1)}=\frac{A}{t}+\frac{Bt+C}{t^2+1}\implies 1=A(t^2+1)+t(Bt+C)
\end{equation}
$$
\begin{multline}
\begin{aligned}
&t=0\implies \boxed{A=1}\\
&1=t^2+1+t(Bt+C)\implies -t^2=t(Bt+C)\\
&\implies-t=Bt+C\text{ if } t\neq 0\\
&t=0\implies \boxed{C=0}\implies \boxed{B=-1}
\end{aligned}
\end{multline}
$$
$$
\begin{multline}
\begin{aligned}
\int\frac{dt}{t(t^2+1)}&=\int\frac{dt}{t}-\frac{1}{2}\int\frac{2t.dt}{t^2+1}=\log|t|-\frac{1}{2}\log|t^2+1|\\
&=\log|e^x|-\frac{1}{2}\log|e^{2x}+1|+C_1
\end{aligned}
\end{multline}
$$
$$
\int\frac{e^{2x}-1}{e^{2x}+1}dx=\frac{1}{2}\log|e^{2x}+1|-\log|e^x|+\frac{1}{2}\log|e^{2x}+1|+C=\log|e^x+e^{-x}|+C
$$
Doubt
While doing partial fractions, first i have assumed $t=0$ but $t=e^x\neq{0}$ and even if I accept that, in the second step the equation is transformed into a new form assuming $t\neq{0}$. But, in the next step I need to again assume $t=0$ to get $B$ and $C$. How can I jusify this and why am I getting the right answer after all ?
|
$$\begin{equation}
\frac{1}{t(t^2+1)}=\frac{A}{t}+\frac{Bt+C}{t^2+1}\implies 1=A(t^2+1)+t(Bt+C)
\end{equation}$$
This is for all t you get that
$$1=t^2(A+B)+Ct+A \quad \color{red}{\forall t}$$
$$A+B=0, A=1, C=0$$
$$(A,B,C)=(1,-1,0)$$
Note that you just integrate $\tanh(x)$
$$\int \frac{e^{2x}-1}{e^{2x}+1} dx= \int \tanh(x)dx= \ln|\cosh(x)|+K$$
|
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British Maths Olympiad (BMO) 2004 Round 1 Question 1 alternative approaches? The questions states:
Solve the simultaneous equations (which I respectively label as $
> \ref{1}, \ref{2}, \ref{3}, \ref{4}$)
$$\begin{align} ab + c + d &= 3 \tag{1} \label{1} \\ bc + d + a &= 5
\tag{2} \label{2} \\ cd + a + b &= 2 \tag{3} \label{3} \\ da + b + c
&= 6 \tag{4} \label{4} \end{align}$$
where $a,b,c,d$ are real numbers.
I solved this system after quite a while by taking
$eqns$ 1 - 3 = $eqns$ 4 - 2
which yields $a + c = 2$
You can then substitute that in and find the other variables
I also noticed that $(a+1)(b+1) + (a+1)(d+1) + (c+1)(b+1) + (c+1)(d+1) = 20$ but that line didnt really help me.
I'm interested in seeing the other approaches people can take with this system.
Additionally, is there a sufficient enough hint to take another route? Did I miss an easy solution?
|
Step 1: Obtain $a+c=2$.
Step 2:
Note that
$$ab+bc+cd+ad=(a+c)(b+d)$$
Adding four equations gives
$$(a+c)(b+d)+2(a+c)+2(b+d)=16$$
$$(a+c)(b+d+2)+2(b+d+2)=20$$
$$(a+c+2)(b+d+2)=20$$
With $a+c=2$, we have $b+d=3$.
Step 3:
Further manipulating, 1-2+3-4 gives
$$(a-c)(b+d)=6$$
Thus, $a-c=2$.
Therefore, we have $a=2, c=0$.
Step 4: Put $a,c$ into 1,
$$2b+d=3$$
With $b+d=3$, $b=0, d=3$.
|
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|
Given that exactly 2 Jacks appear, what is the expected number of Aces that appear? Pick 5 cards from standard 52 cards without replacement. Given that exactly 2 jack cards appear, find the expected number of ace cards that appear.
ATTEMPT
Let $X$ be number of aces chosen and $Y$ number of jacks chosen so we want find $E(X|Y=2)$. In the definition we have
$$ E(X|Y=2) = \sum_{x=0}^5 x \frac{ p_{XY}(x,2) }{p_Y(2)} $$
First we find
$$ p_{XY}(1,2) = P(X=1 \cap Y=2) = \frac{ {4 \choose 1 } {4 \choose 2} {13 \choose 2} }{ {52 \choose 5} } $$
$$ p_{XY}(2,2) = P(X=2 \cap Y=2) = \frac{ {4 \choose 2} {4 \choose 2} {13 \choose 2} }{ {52 \choose 5} } $$
$$ p_{XY}(3,2) = P(X=3 \cap Y=2) = \frac{ {4 \choose 3 } {4 \choose 2} {13 \choose 2} }{ {52 \choose 5} } $$
after $x=3$ we have $0$ since we cant have more that 5 cards. finally we find
$$ p_Y(2) = P(Y=2) = \frac{ {4 \choose 2}{13 \choose 3} }{ {52 \choose 5} } $$
Now, pluggin in into the first equation should give the answer. Is this a correct approach to tackle this problem?
|
Now, pluggin in into the first equation should give the answer. Is this a correct approach to tackle this problem?
Yes, that would work, although your evaluations are off.
$p_{\small Y}(2) =\mathsf P(Y{=}2) = \left.\binom 4 2\binom {48}{3}\middle/\binom{52}{5}\right.$ is the probability for selecting $2$ from $4$ Jacks and $3$ from $48$ non-Jacks when selecting $5$ from $52$ cards.
$p_{\small X,Y}(x,2) =\mathsf P(X{=}x, Y{=}2) =\left.\binom 42\binom 4x\binom {44}{3-x}\middle/\binom{52}{5}\right. $ is the probability for selecting $2$ from $4$ Jacks, $x$ from $4$ aces, and $3-x$ from $44$ non-Jack-or-Aces when selecting $5$ from $52$ cards.
Thus $p_{\small X\mid Y}(x\mid 2) =\mathsf P(X{=}x\mid Y{=}2) =\left.\binom 4x\binom {44}{3-x}\middle/\binom{48}{3}\right. $ is the probability for selecting $x$ from $4$ aces, and $3-x$ from $44$ non-Jack-or-Aces when selecting $3$ from $48$ non-Jack cards.
Alternatively, I place the hand with the non-Jack cards face down on the table, and the two Jacks face up, asserting they are the only Jacks in the hand. I point to one of the face down cards and ask: What is the expectation that this card is an Ace? I do this for each from the three face-down cards in the hand, and the answer is the same.
So, by using the Linearity of Expectation, the expected count for Aces in the hand when given exactly two Jacks are in the hand, will be...
|
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Induction principle: $n^2-7n+12≥0$ for every $n≥3$ How can I prove that $n^2-7n+12≥0$ for every $n≥3$?
I know that for $n=3$ I have $0≥0$ so the inductive Hypothesis is true.
Now for $n+1$ I have $(n+1)^2-7(n+1)+12=n^2-5n+6$ and now I don't know how to go on...
|
Use completing the square:
$$n^2-7n+12=\bigg(n-\frac{7}{2}\bigg)^2-\bigg(\frac{7}{2}\bigg)^2+12$$
$$\to\bigg(n-\frac{7}{2}\bigg)^2-\frac{49}{4}+\frac{48}{4}=\bigg(n-\frac{7}{2}\bigg)^2-\frac{1}{4}$$
Note that $k^2 \ge 0$ $\forall k \in \Bbb R$, and for $n\ge 3, \bigg(n-\frac{7}{2}\bigg)^2\ge\frac{1}{4}$, hence $\bigg(n-\frac{7}{2}\bigg)^2-\frac{1}{4}\ge 0$
|
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|
Deriving a recurrence equation I've really been stuck on this problem for a while. We have the equation: $s_t = (s_{t-1}/2)+3$. I need to show the steps it would take to show this can be written as: $s_t = 2^{-t}(s_0-6) +6$. I figure it has something to do with telescoping but I'm not sure how this would be done.
|
This is just a direct approach. First apply the recursion a couple of times to find a pattern. Then suppose you apply it $i$ times. Then substitute $i=t$ to show what it would look like if you applied it all the way. Bam you're done.
$$\begin{align}
s_t &= \frac{1}{2}s_{t-1}+3 \\
s_t &= \frac{1}{2}\left(\frac{1}{2}s_{t-2}+3\right)+3 \\
&= 2^{-2}s_{t-2}+3\left(\frac{1}{2}+1\right)\\
s_t &= \frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2}s_{t-3}+3\right)+3\right) + 3 \\
&= 2^{-3} s_{t-3}+3\left(\frac{1}{2^2}+\frac{1}{2}+1\right)\\
&\vdots \\
&\vdots \\
s_t&=2^{-i} s_{t-i}+3\left(\left(\frac{1}{2}\right)^{i-1}+\left(\frac{1}{2}\right)^{i-2}+ \cdots + \left(\frac{1}{2}\right)^{1} + \left(\frac{1}{2}\right)^{0}\right)\\
&=2^{-i} s_{t-i}+3\left(\frac{1-(\frac{1}{2})^i}{1-\frac{1}{2}}\right)\\
&=2^{-i} s_{t-i}+6\left(1-2^{-i}\right)\\
&=2^{-i} (s_{t-i}-6)+6\\
&\vdots \\
s_t&=2^{-t}(s_0-6)+6\\
\end{align}$$
Now just verify with math induction
|
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|
If $z^{23}=1$ then evaluate $\sum^{22}_{z=0}\frac{1}{1+z^r+z^{2r}}$
If $z$ is any complex number and $z^{23}=1$ then evaluate $\displaystyle \sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}$
Try: From $$z^{23}-1=(z-1)(1+z+z^2+\cdots +z^{22})$$
And our sum $$\sum^{22}_{r=0}\frac{1}{1+z^r+z^{2r}}=\frac{1}{3}+\frac{1}{1+z+z^2}+\frac{1}{1+z^2+z^4}+\cdots +\frac{1}{1+z^{22}+z^{44}}$$
Now i did not understand how can i simplify it.
Could some help me. Thanks.
|
Define $f(x)=\frac{1}{1+x+x^2}$. Then we have to calculate
$$
\sum_{k=0}^{22}f(z^k)=\frac{1}{3}+\sum_{k=1}^{22}f(z^k).
$$
Let's work out the polynomial expression for $f$ using $(a+b)(a-b)=a^2-b^2$.
$$
f(s)=\frac{s-1}{s^3-1}=\frac{(s-1)(s^3+1)}{s^6-1}=\frac{(s-1)(s^3+1)(s^6+1)}{s^{12}-1}=\frac{(s-1)(s^3+1)(s^6+1)(s^{12}+1)}{s^{24}-1}.
$$
Since $s^{24}=s$ for the $23$d roots of unity, we get
$$
f(z^k)=(z^{3k}+1)(z^{6k}+1)(z^{12k}+1)=1+z^{3k}+z^{6k}+z^{9k}+z^{12k}+z^{15k}+z^{18k}+z^{21k}.\tag{1}
$$
Now as the OP has noticed
$$
s^{23}-1=(s-1)(1+\underbrace{s+s^2+\ldots+s^{22}}_{\Phi(s)}).
$$
For the $z^k$, $k=1,2,\ldots,22$, it makes (as all those are primitive roots)
$$
0=\underbrace{(z^k-1)}_{\ne 0}(1+\Phi(z^k))\quad\Leftrightarrow\quad \Phi(z^k)=-1.
$$
Finally, summing up (1) we get
$$
\sum_{k=1}^{22}f(z^k)=22+\Phi(z^3)+\Phi(z^6)+\Phi(z^9)+\Phi(z^{12})+\Phi(z^{15})+\Phi(z^{18})+\Phi(z^{21})=22-7=15.
$$
Plus $\frac13$.
|
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|
A common tangent line
The graph of $f(x)=x^4+4x^3-16x^2+6x-5$ has a common tangent line at $x=p$ and $x=q$. Compute the product $pq$.
So what I did is I took the derivative and found out that $p^2+3p+q^2+3q+pq=0$. However when I tried to factorize it I didn't find out an obvious solution. Can someone hint me what to do next? Thanks in advance
|
The tangent line equation at $x=p$ and $x=q$ is:
$$y=p^4+4p^3-16p^2+6p-5+(4p^3+12p^2-32p+6)(x-p);\\
y=q^4+4q^3-16q^2+6q-5+(4q^3+12q^2-32q+6)(x-q).\\$$
By equating the coefficients, simplifying and denoting $a=p+q, b=pq$ we get:
$$\begin{cases}b=a^2+3a-8\\
3a^3-(6a+8)b+8a^2-16a=0\end{cases}$$
Solving the system we get:
$$a=-2,b=-10=pq.$$
|
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|
Inequality $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1$ Show that $$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}>1,\:\forall n\in\mathbb{N}$$
This is a 9th grade problem.
I was trying to take the greatest numerator, which is the last numerator of the last fraction. But there are only $2n+1$ terms. Right?
After that I have no idea. Thx!
|
Use the method given in this answer by Jack D'Aurizio.
Note: $H_n=1+\frac12+\frac13+\cdots+\frac1n=\sum_{k=1}^n \frac1n$ is called $n$-th harmonic number.
Then:
$$\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}=\sum_{k=1}^{2n+1}\frac{1}{n+k}=H_{3n+1}-H_{n}.$$
Consider the sequence: $a_n=H_{3n+1}-H_n$. We will show that it is an increasing sequence:
$$a_{n+1}-a_n=(H_{3(n+1)+1}-H_{n+1})-(H_{3n+1}-H_n)=\\
(H_{3n+4}-H_{3n+1})-(H_{n+1}-H_n)=\\
\frac{1}{3n+4}+\frac{1}{3n+3}+\frac{1}{3n+2}+\frac{1}{3n+1}-\frac{1}{n+1}>\\
\frac{1}{3n+4}+\frac{1}{3n+3}+\frac{1}{3n+\color{red}{3}}+\frac{1}{3n+\color{red}{3}}-\frac{1}{n+1}=\\
\frac{1}{3n+4}>0 \Rightarrow a_{n+1}>a_n.$$
Hence:
$$a_1=H_{3+1}-H_1=\sum_{k=1}^{2+1} \frac{1}{1+k}=\frac12+\frac13+\frac14=\frac {13}{12}>1.$$
|
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|
How many ways can dominoes cover a $2 \times n$ rectangle? Justify proposed solution. I was able to get that $d_n = d_{n-1}+d_{n-2}$
It isn't finished, because I have to solve this recursive equation.
I read about Binet's formula, but I don't know the steps between this
$$d_n = d_{n-1}+d_{n-2}$$
and this
$$f_n=\frac{1}{\sqrt{5}} \left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$$
Which is exactly what I'm looking for - function with $n$ as parameter.
|
Your recurrence relation is correct. The sequence either begins with one vertical tile followed by one of $d_{n - 1}$ admissible sequences of length $n - 1$ or two horizontal tiles followed by one of the $d_{n - 2}$ admissible sequences of length $n - 2$. Hence,
$$d_n = d_{n - 1} + d_{n - 2}$$
The recursion $d_n = d_{n - 1} + d_{n - 2}$ has characteristic equation $r^2 = r + 1$.
\begin{align*}
r^2 & = r + 1\\
r^2 - r & = 1\\
r^2 - r + \frac{1}{4} & = 1 + \frac{1}{4}\\
\left(r - \frac{1}{2}\right)^2 & = \frac{5}{4}\\
r - \frac{1}{2} & = \pm \frac{\sqrt{5}}{2}\\
r & = \frac{1 \pm \sqrt{5}}{2}
\end{align*}
so
$$d_n = A\left(\frac{1 + \sqrt{5}}{2}\right)^n + B\left(\frac{1 - \sqrt{5}}{2}\right)^n$$
where $A$ and $B$ are constants, which can be determined from the initial conditions $d_1 = 1$ (one vertical tile), $d_2 = 2$ (both tiles are horizontal or both are vertical). Hence, $d_n = F_{n + 1}$, where $F_n$ is the $n$th Fibonacci number.
Since Binet's formula gives an explicit formula for $F_n$, the $n$th Fibonacci number,
$$F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n\right]$$
the formula for $d_n$ can be found using that formula and the observation that $d_n = F_{n + 1}$.
|
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Integration of $ \int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2} \mathrm dx$ Evaluate $$\int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\mathrm dx$$I believe I should use the partial fractions technique to evaluate this integral, but I am not getting anywhere when I try it.
|
\begin{align} \frac {3x^3 + 3x^2 - 5x + 4}{3x^3 -2x^2 + 3x - 2}
& = \frac {(3x^3 - 2x^2 + 3x - 2) + 5x^2 - 8x + 6}{3x^3 -2x^2 + 3x - 2}\\ \\
&\ = 1 + \frac{5x^2 - 8x + 6}{3x^3 -2x^2 + 3x - 2} \end{align}
Factor the denominator
$$3x^3 -2x^2 + 3x - 2 = (3x-2)(x^2 + 1)$$
$$\frac {5x^3 -8x^2 +6}{(3x-2)(x^2+1)} = \frac {A}{3x-2}+ \frac {Bx+C}{x^2 + 1}$$
The "obvious" way to do this..
$$5x^2 -8x +6 = Ax^2 + A + 3Bx^2 + (-2B+3C)x - 2C$$
Giving a system of equations:
\begin{align} A+3B &= 5\\
-2B+3C &= -8\\
A - 2C &= 6 \end{align}
Here is a trick which you may, or may not find easier.
$$\frac {5x^3 -8x^2 +6}{(3x-2)(x^2+1)} = \frac {A}{3x-2}+ \frac {Bx+C}{x^2 + 1}$$
Multiply through by $(3x-2)$
$$\frac {5x^3 -8x^2 +6}{(x^2+1)} = A+ \frac {Bx+C}{x^2 + 1}(3x-2)$$
and evaluate at $x = 2/3$
$$\frac{\frac{26}{9}}{\frac{13}{9}} = 2 = A \implies A = 2 $$
$$\frac {3x^3 + 3x^2 - 5x + 4}{3x^3 -2x^2 + 3x - 2} = 1 + \frac {2}{3x- 2} + \frac {x-2}{x^2+1}$$
|
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Show that a sequence does not have a limit How can I show that the sequence $x_n=\frac{(-1)^nn}{n+1}$ does not have a limit by using only the definition of limits?
Attempts:
Let's assume that our sequence has a limit $x$. Then there exists such a number $N$ so that for any $n>N$ $|a_n-a| < \epsilon$. This is also true for $\epsilon = \frac{1}{2}$. Now I take the numbers $N_1$, $2N_1$ and $2N_1+1$. But when I start subtracting (just like it is done with the sequence $x_n=(-1)^n$), I get a result which has $N_1$ in it, and thus I can not get a contradiction. What do I do now?
|
If
$x_n=\frac{(-1)^nn}{n+1}
$
has a limit $v$,
then,
for any $c > 0$
there is a $n(c)$ such that
$|x_n-v| < c$
for all $n > n(c)$.
In this case,
$x_n$ has two subsequences
that have limits:
$x_{2n} \to 1$
and
$x_{2n+1} \to -1$.
So let's choose a $c$
that is less than
half the distance
between the limits of
these two subsequences,
for example, $c = \frac12$.
Then for $n > n(c)$
$|x_n-v| < c$.
Suppose also that
$n > 10$.
Then
$x_{2n}
= \dfrac{2n}{2n+1}
=1- \dfrac{1}{2n+1}
$
so
$c
\gt |x_{2n}-v|
= |1- \dfrac{1}{2n+1}-v|
\ge |1-v|- \dfrac{1}{2n+1}
$.
Therefore
$|1-v|
\le c+\dfrac{1}{2n+1}
$
or
$-c-\dfrac{1}{2n+1}
\le 1-v
\le c+\dfrac{1}{2n+1}
$
or
$1-c-\dfrac{1}{2n+1}
\le v
\le 1+ c+\dfrac{1}{2n+1}
$.
Similarly,
$x_{2n-1}
= -\dfrac{2n-1}{2n}
=-1+\dfrac{1}{2n}
$
so
$c
\gt |x_{2n}-v|
= |-1+ \dfrac{1}{2n}-v|
\ge |-1-v|- \dfrac{1}{2n}
$.
Therefore
$|-1-v|
\le c+\dfrac{1}{2n}
$
or
$-c-\dfrac{1}{2n}
\le 1-v
\le c+\dfrac{1}{2n}
$
or
$-1-c-\dfrac{1}{2n}
\le v
\le -1+ c+\dfrac{1}{2n}
$.
Therefore
$1-c-\dfrac{1}{2n+1}
\le v
-1+ c+\dfrac{1}{2n}
$
so
$2
\le 2c+\dfrac{1}{2n}+\dfrac{1}{2n+1}
$
which is false for
$c \le \frac12$
and
$n > 10$.
This is readily modified
to show that
any sequence
with two subsequences
that have different limits
can not iself
have a limit.
Even more general results
can be similarly proved.
|
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${1 \over \sqrt{3}} \sum\limits _{r=0}^4 \tan \left( \frac{\pi}{15}+\frac{r\pi}{5} \right)=$? $$\sum _{r=0}^4 \tan \left( \frac{\pi}{15}+\frac{r\pi}{5} \right)=k \sqrt{3}$$. Then evaluation of $k$
solution i try
$$\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(120^\circ)+\tan(156^\circ)$$
$$=\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(156^\circ)+\sqrt{3}$$
$$=\frac{\sin(60^\circ)}{\cos(12^\circ)\cos(48^\circ)}+\frac{\sin(240^\circ)}{\cos(84^\circ)\cos(156^\circ)}-\sqrt{3}.$$
plz help me how to simplify denominator.
|
$$S=\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(120^\circ)+\tan(156^\circ)$$
$$S=\tan(12^\circ)+\tan(48^\circ)+\tan(84^\circ)+\tan(156^\circ)+\sqrt{3}$$
$$S=\frac{\sin(60^\circ)}{\cos(12^\circ)\cos(48^\circ)}+\frac{\sin(240^\circ)}{\cos(84^\circ)\cos(156^\circ)}-\sqrt{3}.$$
$$S=\frac{\sqrt{3}}{\cos (60^\circ)+\cos(36^\circ)}-\frac{\sqrt{3}}{\cos(240^\circ)+\cos(72\circ)}-\sqrt{3}$$
use $\displaystyle \cos (36^\circ)=\frac{\sqrt{5}+1}{4}$ and $\displaystyle \cos(72^\circ)=\sin(18^\circ)=\frac{\sqrt{5}-1}{4}$
$$=4\sqrt{3}\bigg[\frac{6}{4}\bigg]-\sqrt{3}=5\sqrt{3}$$
|
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Evaluating $\int\sqrt{1+\sin x}dx$. The integral is:$$\int\sqrt{1+\sin x} dx$$
My first attempt was to multiply by:
$$\frac{\sqrt{1-\sin x}}{\sqrt{1-\sin x}}$$
giving:
$$\int\frac{\sqrt{1-\sin^2(x)}}{\sqrt{1-\sin x}}dx$$
which is:
$$\int\frac{\cos x}{\sqrt{1-\sin x}}dx$$
then make the substitution:
$$u=1-\sin x$$
which gives
$$\int\frac{-1}{\sqrt{u}}du$$
which gives:
$$-2\sqrt{u}+c$$
so:
$$I=-2\sqrt{1-\sin x}+c$$
but I am not sure if this is correct.
Also sorry if the formatting is a bit rubbish
|
$$I = \int \sqrt{1+\sin x}dx$$
Apply substitution $ u=\tan \left(\frac{x}{2}\right)$
(Tangent half-angle substitution)
$$\int \frac{2\sqrt{\left(u+1\right)^2}}{\left(1+u^2\right)\sqrt{u^2+1}}du$$
$$2\int \frac{u+1}{\left(1+u^2\right)\sqrt{u^2+1}}du$$
$$2\left(\underbrace{\int \frac{u}{\left(1+u^2\right)\sqrt{u^2+1}}du}_{I_1}+\underbrace{\int \frac{1}{\left(1+u^2\right)\sqrt{u^2+1}}du}_{I_2}\right)$$
Set $v=1+u^2$
$$I_1 = \int \frac{1}{2v\sqrt{v}}dv = \frac{1}{2}\int v^{-\frac{3}{2}}dv =\frac{1}{2}\frac{v^{\left(-\frac{3}{2}+1\right)}}{\left(-\frac{3}{2}+1\right)} =-\frac{1}{\sqrt{v}}= -\frac{1}{\sqrt{1+u^2}}$$
By Trigonometric substitution $u=\tan \left(v\right)$
$$I_2 = \int \frac{1}{\left(u^2+1\right)^{\frac{3}{2}}}du= \int \frac{\sec ^2\left(v\right)}{\left(\tan ^2\left(v\right)+1\right)^{\frac{3}{2}}}dv = \int \frac{\sec ^2\left(v\right)}{\sec ^3\left(v\right)}dv =\int \frac{1}{\sec \left(v\right)}dv =\int \cos \left(v\right)dv =\sin \left(v\right) +C$$
$$I_2 = \sin \left(v\right) = \sin \left(\arctan \left(u\right)\right) = \frac{u}{\sqrt{1+u^2}}$$
So
$$ I = 2\left(-\frac{1}{\sqrt{1+u^2}}+\frac{u}{\sqrt{1+u^2}}\right) +C$$
$$I = 2\left(-\frac{1}{\sqrt{1+\tan ^2\left(\frac{x}{2}\right)}}+\frac{\tan \left(\frac{x}{2}\right)}{\sqrt{1+\tan ^2\left(\frac{x}{2}\right)}}\right) +C$$
|
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|
How to explain this point transformation? Yesterday we were having a lecture on point coordinates after rotation. The prof. explained that the position of a point after a counterclockwise rotation is obtained from the following formula,
$x=x_0 \operatorname{Cos}(\theta)-y_0 \operatorname{Sin}(\theta), \qquad y=y_0 \operatorname{Cos}(\theta)+x_0\operatorname{Sin}(\theta)$,
where $\theta$ is the angle of rotation.
And then, out of nowhere, for an exercise he used the following formula for the displacements of the point after rotation,
$u_x=(-x_0 \operatorname{Cos}(\frac{\pi-\theta}{2})-y_0\operatorname{Sin}(\frac{\pi-\theta}{2}))\times 2\operatorname{Sin}(\frac{\theta}{2}), \qquad \\ u_y=(x_0 \operatorname{Sin}(\frac{\pi-\theta}{2})-y_0\operatorname{Cos}(\frac{\pi-\theta}{2}))\times 2\operatorname{Sin}(\frac{\theta}{2}).$
Why the signs are different? why instead of just $\theta$ he has used $\frac{\pi-\theta}{2}$? and most importantly what is the role of $2\operatorname{Sin}(\frac{\theta}{2})$?
Thanks in advance.
|
You must use some trig equalities:
$$
\cos\frac{\pi-\theta}{2}=\sin\frac{\theta}{2},\quad
\sin\frac{\pi-\theta}{2}=\cos\frac{\theta}{2},\\
\cos\theta=1-2\sin^2\frac{\theta}{2}=2\cos^2\frac{\theta}{2}-1,\quad
\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}
$$
to obtain:
$$
\begin{align}
&\left(-x_0 \cos\frac{\pi-\theta}{2}-y_0\sin\frac{\pi-\theta}{2}\right)\cdot 2\sin\frac{\theta}{2}\\
=&\left(-x_0 \sin\frac{\theta}{2}-y_0\cos\frac{\theta}{2}\right)\cdot 2\sin\frac{\theta}{2}\\
=&-2x_0 \sin^2\frac{\theta}{2}-2y_0\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\
=&x_0 (1-\cos\theta)-y_0\sin\theta\\
=&x_0-x_0\cos\theta-y_0\sin\theta\\
=&x_0-x,
\end{align}
$$
and the analogous for the second equation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2786510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Interpolating $f$ and approximating its derivative We are given $x_0 = 1, x_1 = \frac{4}{3}$ and $x_2 = 2$.
Find a parabola which agrees with function $f(x) = (x + 1)\sin(x)$ in the given points. Afterwards derive a formula for the
approximation of the $f'$ i.e. the derivative in $x_1$. What is the approximation of $f'(x_1)$ for function f?
I did the first step using interpolation and got the polynomial $p(x) = -1.0647x^2 + 4.2390x - 1.4914.$ Now I do not know how to use this to approximate the derivative. Should I just compute the derivative of the polynomial $p$? What about the approximation then?
|
$$
f\left(x \right) = \left(x + 1\right)\sin x
\implies \left\lbrace
\begin{alignedat}{4}
f\left(x_0\right)&= 2\sin \left(1\right) &&\approx 1.68,\qquad &\text{where} \quad x_0&=1 \\
f\left(x_1\right)&= \tfrac{7}{3}\sin \left(\tfrac{4}{3}\right)&\, &\approx 2.27, &\text{where} \quad x_1&=4/3 \\
f\left(x_2\right)&= 3\sin \left(2\right) &&\approx 2.73,& \text{where} \quad x_2&=2& \\
\end{alignedat} \right.
$$
The general form of equation of parabola is
$$
y-\widehat{y} = k\left(x-\widehat{x}\right)^2\qquad
\text{ or }\qquad
y\left(x\right) = ax^2+bx+c
$$
Substituting known values we get system of equations
$$
\left\lbrace\begin{aligned}
y\left(x_0\right) =f\left(x_0\right)\\ y\left(x_1\right) =f\left(x_1\right) \\ y\left(x_2\right) =f\left(x_2\right)
\end{aligned}\right. \implies
\left\lbrace\begin{aligned}
ax_0^2+bx_0+c &= 2\sin \left(1\right) \\
ax_1^2+bx_1+c &= \tfrac{7}{3}\sin \left(\tfrac{4}{3}\right) \\
ax_2^2+bx_2+c &= 3\sin \left(2\right) \\
\end{aligned}\right.
\iff
\left\lbrace\begin{aligned}
a+b+c & \approx 1.68 \\
\tfrac{16}{9}a+\tfrac{4}{3}b+c &\approx 2.27 \\
4a +2b+c &\approx 2.73\\
\end{aligned}\right.
$$
Solving the last system of linear equations for coefficients $a,b$, and $c$ will give you general form of parabola.
As for approximating derivatives, I recommend following advise of @caverac and using Divided differences, or use straightforward Finite difference formula
$$
f'\left(x_1\right)\approx\frac{f\left(x_2\right)-f\left(x_0\right)}{x_2-x_0}
= \frac{2\sin 1 -3\sin 2}{1}\approx -1.045
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Area of square under a curve.
A square having sides parallel to the coordinate axes is inscribed in the region.
{$(x,y):x,y>0:y\le -x^3+3x$}.
If the area of the square is written as $A^{1/3}+B^{1/3}$ square units where $A,B\in \Bbb Z$ and $A>B$, then find
(i)$\sqrt{A-B}$
(ii)Slope of line with x and y intercepts as $A,B$ respectively
(iii)$A+B\over A-B$
(iv)Circumradius of $\triangle OPQ$ where $O$ is origin, $P(A^{1/3},0)$ and $Q(0,B^{1/3})$
My approach:
Let the square be $KLMN$ where $K,L$ lie on x-axis with coordinates $(a,0);(b,0)$ respectively.
This will give coordinates of other points i.e. $M,N$ in terms of $a,b$.
Now taking out the length of all sides and equating them will give $a,b$.
Problem with my approach:
I end up in a cumbersome equation involving $a,b$
Tips: I know nothing more than simple differentiation and limits. So please avoid solutions with integration and other stuffs.
PS: No need to downvote as I'm on the verge of getting banned. Please comment if you have any problem with the question.
|
Define $f(x):=3 x - x^3$. Obviously, the square must have a side length of $f(q)$, and we need to find $q$ that gives the largest square inside $f(x)$.
The line $y=f(q)$ intersects $y=f(x)$ at three points:
$$\left(q,f\left(q\right)\right)\\
\left(\frac{1}{2} \left(-\sqrt{3} \sqrt{4-q^2}-q\right),f\left(\frac{1}{2} \left(-\sqrt{3} \sqrt{4-q^2}-q\right)\right)\right)\\
\color{red}{\left(\frac{1}{2} \left(\sqrt{3} \sqrt{4-q^2}-q\right),f\left(\frac{1}{2} \left(\sqrt{3} \sqrt{4-q^2}-q\right)\right)\right)}\\
$$
We are only interested at the last point. The top left vertex of the square is $(q,f(q))$, and we need to ensure that $\sqrt{(f(q)-f(z))^2+(q-z)^2}=f(q)$, where $z=\frac{1}{2} \left(\sqrt{3} \sqrt{4-q^2}-q\right)$. So we need to find $q$ for the equation below:
$$\sqrt{(f(q)-f(z))^2+(q-z)^2}-f(q)=0\tag{1}$$
You can use this implementation to get the intuition for the solution.
This gives us two solutions:
$$\left\{\left\{q\to \sqrt{1+\sqrt[3]{2}}\right\},\left\{q\to \sqrt{3-\sqrt[3]{2}-2^{2/3}}\right\}\right\}$$
Which means the area of the maximum square is:
$$\bbox[5px,border:2px solid black]{f(q)^2 \iff f\left(\sqrt{3-\sqrt[3]{2}-2^{2/3}}\right)^2=6-3\times 2^{2/3}}$$
To get $A$ and $B$ set:
$$A^{1/3}=6\iff A=6^3=216\\
B^{1/3}=3\times2^{2/3} \iff B=(3\times2^{2/3})^3=108$$
Since $\sqrt[3]A+\sqrt[3]{-B}=\sqrt[3]A-\sqrt[3]{B}$, for reals, we have the solution:
$$\therefore{A=216\\B=-108}$$
The questions can now be easily answered:
*
*$\sqrt{216+108}=\sqrt{324}=18$
*Line passes through $(216,0)$ and $(0,-108)$, therefore: $y=\frac{-108}{216}(x-216)\iff y=\frac12x+108$
*$\frac{108}{324}=\frac13$
*The circumcircle is centered at:
$$\left(\frac62,-\frac{3\times2^{2/3}}{2}\right)$$ therefore the radius is $$r=\sqrt{3^2+\left(-\frac{3\times2^{2/3}}{2}\right)^2}=\sqrt{9+\frac{9}{2^{2/3}}}=\frac{3 \sqrt{1+2^{2/3}}}{\sqrt[3]{2}}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluation of $\int\frac{1}{(\sin x+a\sec x)^2}dx$
Evaluation of $$\int\frac{1}{(\sin x+a\sec x)^2}dx$$
Try: Let $$I=\int\frac{1}{(\sin x+a\sec x)^2}dx=\int\frac{\sec^2 x}{(\tan x+a\sec^2 x)^2}dx$$
put $\tan x=t$ and $dx=\sec^2 tdt$
So $$I=\int\frac{1}{(a+at^2+t)^2}dt$$
Could some help me how to solve above Integral , thanks in advance.
|
HINT
Let
$$b=\dfrac1{2a},\quad y=t+b,\tag1$$
then
$$I=4b^2\int\dfrac{dt}{(t^2+2bt+1)^2} = 4b^2\int\dfrac{dy}{(y^2+1-b^2)^2} = \dfrac{4b^2}{1-b^2}\int\dfrac{(y^2+1-b^2)-y^2}{(y^2+1-b^2)^2}\,dy,$$
$$I=\dfrac{4b^2}{1-b^2}(I_1+I_2),\tag2$$
where
$$I_1=\int\dfrac{dy}{y^2+1-b^2} = const +
\begin{cases}\dfrac1{\sqrt{1-b^2}}\arctan\dfrac{y}{\sqrt{1-b^2}},\text{ if } b<1\\[4pt]
-\dfrac1y,\text{ if }b=1\\[4pt]
\dfrac1{2\sqrt{b^2-1}}\ln\left|\dfrac{y-\sqrt{b^2-1}}{y+\sqrt{b^2-1}}\right|, \text{ if }b>1,
\end{cases}\tag3$$
$$I_2 = -\int\dfrac{y^2dy}{(y^2+1-b^2)^2} = \dfrac12\int y\cdot d\left(\dfrac{1}{y^2+1-b^2}\right) = \dfrac12\dfrac{y}{y^2+1-b^2} - \dfrac {I_1}2.\tag4$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Partial fraction decomposition trouble I'm trying to do a partial fraction expansion on
$$Y = n \cdot \frac{e^{-pt_{0}}}{(p+n)(p^2+\omega^2)}$$
which gives $A(p^2+\omega^2)+(Bp+C)(p+n) = 1$
which implies
$$A = -B$$
$$Bpn=0\rightarrow B=0\rightarrow A=0$$
which is incorrect. Any help with what I'm missing would be greatly appreciated
|
The condition
$$
A(p^2 + \omega^2) + (B p + C)(p + n ) = 1
$$
leads to the equations
\begin{eqnarray}
A + B &=& 0 \\
B n + C &=& 0 \\
A \omega^2 + C n &=& 1
\end{eqnarray}
whose solution is
\begin{eqnarray}
A &=& \frac{1}{\omega^2 + n^2} = -B \\
C &=& \frac{n}{\omega^2 + n^2}
\end{eqnarray}
So the fraction is
$$
Y = \frac{n e^{-pt_0}}{\omega^2 + n^2}\left[ \frac{1}{p + n} + \frac{n-p}{p^2 + \omega^2} \right]
$$
|
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"url": "https://math.stackexchange.com/questions/2794585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why is there a pattern to the last digits of square numbers? I was programming and I realized that the last digit of all the integer numbers squared end in $ 0, 1, 4, 5, 6,$ or $ 9 $.
And in addition, the numbers that end in $ 1, 4, 9, 6 $ are repeated twice as many times as the numbers that end in $ 0, 5$
I checked the numbers from $1$ to $1000$, and the results are:
$1.$ The numbers on the left are the last digit of each digit squared.
$2.$ The numbers on the right are the number of times that the last digit is repeated.
$$
\begin{array}{cc}
0: &100, \\
1: &200, \\
4: &200, \\
5: &100, \\
6: &200, \\
9: &200
\end{array}
$$
So, why does this happen? What is the property that all integers have?
|
What you are looking at is the residues of squares modulo $10$.
$0^2=\color{red}0\bmod 10\\1^2=\color{blue}1\bmod 10\\2^2=\color{orange}4\bmod 10\\3^2=9\bmod 10\\4^2=\color{green}6\bmod 10\\5^2=\color{brown}5\bmod 10\\6^2=\color{green}6\bmod 10\\7^2=9\bmod 10\\8^2=\color{orange}4\bmod 10\\9^2=\color{blue}1\bmod 10$
As you can see, $0$ and $5$ are half as frequent as the other residues which are indeed $1,4,6,9$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Matrix exponential Let $A,B,C \in \operatorname{Mat}_2(\mathbb{R})$ define the real matrices:
$$A = \begin{pmatrix}
0 & 1 \\
0 & 0 \\
\end{pmatrix},\ B =\begin{pmatrix}
0 & 0 \\
0 & 1 \\
\end{pmatrix},\ C = A+B=\begin{pmatrix}
0 & 1 \\
0 & 1 \\
\end{pmatrix}
$$
Show that
$$\exp(A;t) = \begin{pmatrix}
1 & t \\
0 & 1 \\
\end{pmatrix}\text{ and } \exp(B;t)=\begin{pmatrix}
1 & 0 \\
0 & e^t \\
\end{pmatrix}
\text{ for } t\in \mathbb{R}$$ and show that $$\exp(C;t)=\begin{pmatrix}
1 & e^t-1 \\
0 & e^t \\
\end{pmatrix}\text{ for } t\in \mathbb{R}$$
I am a bit unsure with the notation used here for $\exp(A;t)=\begin{pmatrix}
1 & t \\
0 & 1 \\
\end{pmatrix}$ and how am I supposed to show this based on the information for the matrices $A,B,C$ ?
|
I assume that $e^A = I + A + {1 \over 2!} A^2+ \cdots$.
Note that $A^2 = 0$, hence $e^{At} = I + tA$.
Note that $B^k = B$ for all $k \ge 1$, hence $e^{Bt} = I + B(t + {t^2 \over 2} + \cdots) )= \begin{bmatrix} 1 & 0 \\ 0 & e^t \end{bmatrix}$.
Finally, as Jean-Claude noted, $C^k = C$ for $k \ge 1$.
|
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|
Trigamma identity $\psi_1\left(\frac{11}{12}\right)-\psi_1\left(\frac{5}{12}\right)=4\sqrt 3 \pi^2-80G$ Regarding this integral: Integral $\int_0^1 \frac{\sqrt x \ln x} {x^2 - x+1}dx$ The following conjecture comes: $$\psi_1\left(\frac{11}{12}\right)-\psi_1\left(\frac{5}{12}\right)=4\sqrt 3 \pi^2-80G$$ Where $G$ is Catalan's constant. How can we show this? Wolfram-alpha agrees on this: https://www.wolframalpha.com/input/?i=trigamma(11%2F12)-trigamma(5%2F12)%3D4sqrt3pi%5E2-80Catalan
|
\begin{align}
\psi_1(\tfrac{11}{12})
-\psi_1(\tfrac{5}{12})
&=4\sqrt 3 \pi^2-80G
\tag{1}\label{1}
.
\end{align}
The constant $G$ - Catalan's constant
has known to appear in relations
\begin{align}
\psi_1(\tfrac14)&=\pi^2+8G
,\\
\psi_1(\tfrac34)&=\pi^2-8G
,
\end{align}
so we can try to start from $\psi_1(\tfrac14)$.
Applying triplication identity
\begin{align}
9\psi_1(3x) &=
\psi_1(x)
+\psi_1(x+\tfrac13)
+\psi_1(x+\tfrac23)
\tag{2}\label{2}
\end{align}
to $\psi_1(\tfrac14)=\psi_1(3\cdot\tfrac1{12})$,
we get
\begin{align}
9\psi_1(\tfrac14)
&=
\psi_1(\tfrac1{12})
+\psi_1(\tfrac5{12})
+\psi_1(\tfrac34)
,\\
\psi_1(\tfrac1{12})
+\psi_1(\tfrac5{12})
&=9\psi_1(\tfrac14)-\psi_1(\tfrac34)
\\
&=10\psi_1(\tfrac14)-(\psi_1(\tfrac14)+\psi_1(\tfrac34))
\\
&=10\psi_1(\tfrac14)-2\pi^2
\\
&=10(\pi^2+8G)-2\pi^2
\\
\psi_1(\tfrac1{12})
+\psi_1(\tfrac5{12})
&=8\pi^2+80G
,
\end{align}
and we are almost there.
Now, applying the identity
\begin{align}
\psi_1(1-x)&=
\frac{\pi^2}{\sin(\pi x)^2}
-\psi_1(x)
\end{align}
to $\psi_1(\tfrac1{12})=\psi_1(1-\tfrac{11}{12})$,
we get
\begin{align}
\psi_1(\tfrac1{12})
&=\frac{\pi^2}{\sin(\tfrac\pi{12})^2}
-\psi_1(\tfrac{11}{12})
,\\
&=8\pi^2+4\sqrt3\pi^2
-\psi_1(\tfrac{11}{12})
,
\end{align}
and \eqref{1} follows.
|
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|
${3^n\choose k}$ is divisible by $3$? How can I prove that ${3^n\choose k}$ is divisible by $3$ for all positive integer values of $n$? (where $k$ is any positive integer smaller than $3^n$) Can you use induction? Thanks.
|
I just want to demonstrate the relevance of if $p$ is prime and $0<t<p$ then $p$ divides $\displaystyle \binom p t$
Consider $$\begin{split}(x+y)^p
& =\sum_{k=0}^p \binom p k x^{p-l} y^{k} \\
& =x^p+\binom p 1x^{p-1}y+\binom p 2x^{p-2}y^2+\cdots+\binom{p}{p-1}xy^{p-1}+y^p\\
& = x^p + y^p \mod p
\end{split}$$
Hence
$$\begin{split}
(x+y)^{p^n} &= \left((x+y)^p\right)^{p^{n-1}} \\
&= \left(x^p+y^p \mod p\right)^{p^{n-1}}\\
&= (\left(x^p+y^p \mod p\right)^p)^{p^{n-2}}\\
&= \left(x^{p^2}+y^{p^2} \mod p\right)^{p^{n-2}}\\
& \vdots\\
& = x^{p^n}+y^{p^n} \mod p
\end{split}$$
But also
$$\begin{split}
(x+y)^{p^n}
& =\sum_{k=0}^{p^n} \binom{p^n}{k} x^{p^n-l} y^{k} \\
& =x^{p^n}+\binom{p^n}{1}x^{p^n-1}y+\binom{p^n}{2}x^{p^n-2}y^2+\cdots+\binom{p^n}{p^n-1}xy^{p^n-1}+y^{p^n}\\
& = x^{p^n}+y^{p^n} \mod p
\end{split}$$
Therefore $p$ divides $\displaystyle\binom{p^n}{k}$ for all $k$ with $0<k<p^n$
|
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|
Find $x$ if $\cot^{-1}\left(\frac{1}{x}\right)+\cos ^{-1}(-x)+\tan^{-1}(x)=\pi$ if $x \lt 0$ Then Find value of $$\frac{(1-x^2)^{\frac{3}{2}}}{x^2}$$ if
$$ \cot^{-1} \left(\frac{1}{x}\right)+\cos^{-1}(-x)+\tan^{-1}(x)=\pi$$
My try:
Since $x \lt 0$ we have $$ \cot^{-1}\left(\frac{1}{x}\right)=\pi +\tan ^{-1}x$$
Also $$\cos^{-1}(-x)=\pi -\cos ^{-1}x$$
Hence the given equation becomes
$$\pi +2 \tan^{-1} x+\pi -\cos^{-1}{x}=\pi$$
$$\pi+2 \tan^{-1}x=\cos^{-1}x$$
taking $\cos$ both sides we get
$$\frac{x^2-1}{x^2+1}=x$$
Now how to proceed further?
|
A further step to a solution is along the following lines:
Note that if $$cot^{-1}\left(\frac{1}{x}\right)=\theta \Rightarrow{cot(\theta)=\frac{1}{x}=\frac{1}{tan(\theta)}}\Rightarrow{x=tan(\theta)}$$
Therefore
$$cot^{-1}\left(\frac{1}{x}\right)=tan^{-1}\left(x\right)$$
The original problem becomes
$$2tan^{-1}\left(x\right)+cos^{-1}\left(-x\right)=\pi$$
Taking the $cosine$ of both sides gives:
$$cos\left[2tan^{-1}\left(x\right)+cos^{-1}\left(-x\right)\right]=-1$$
Using the formula $cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$:
$$cos\left[2tan^{-1}\left(x\right)\right]cos\left[cos^{-1}\left(-x\right)\right]-sin\left[2tan^{-1}\left(x\right)\right]sin\left[cos^{-1}\left(-x\right)\right]=-1 \tag{1}$$
Now, recall from Calculus II that if we let $\theta=tan^{-1}\left(x\right)\Rightarrow {x=tan(\theta)}$, which means (by using a right angle triangle) that:
$$sin(\theta)=\frac{x}{\sqrt[]{1+x^2}}$$
and
$$cos(\theta)=\frac{1}{\sqrt[]{1+x^2}}$$
Therefore
$$cos\left[2tan^{-1}\left(x\right)\right]=cos\left[2\theta\right]=2cos^2{\theta}-1=2\left(\frac{1}{\sqrt[]{1+x^2}}\right)^2-1 \tag{2}$$
$$cos\left[cos^{-1}\left(-x\right)\right]=-x \tag{3}$$
$$sin\left[2tan^{-1}\left(x\right)\right]=sin\left[2\theta\right]=2sin(\theta)cos(\theta)=2\left({\frac{x}{\sqrt[]{1+x^2}}}\right)\left({\frac{1}{\sqrt[]{1+x^2}}}\right) \tag{4}$$
$$sin\left[cos^{-1}\left(-x\right)\right]=\sqrt[]{1-x^2} \tag{5}$$
Substituting equations $(2), (3)$ and $(4)$ and $(5)$ into $(1)$ gives:
$$\left[2\left(\frac{1}{\sqrt[]{1+x^2}}\right)^2-1\right]{\left(-x\right)}-2\left({\frac{x}{\sqrt[]{1+x^2}}}\right)\left({\frac{1}{\sqrt[]{1+x^2}}}\right)\sqrt[]{1-x^2}=-1$$
Simplifying and collecting like terms gives:
$$x^3 + x^2 - x + 1 -2x\sqrt[]{1-x^2}=0$$
Now, to try and solve for $x$, first let:
$$x^3 + x^2 - x + 1=2x\sqrt[]{1-x^2}$$
$$\Rightarrow {\left(x^3 + x^2 - x^2 + 1\right)^2=\left(2x\sqrt[]{1-x^2}\right)^2}$$
Again, simplifying and collecting like terms, gives:
$$x^6+2x^5-x^4+3x^2-2x+1=0$$
But (after much work!) $$x^6+2x^5-x^4+3x^2-2x+1=\left(x^3+x^2+x-1\right)^2=0$$
Finally, the value of $x$ can be found by solving the cubic equation:
$$x^3+x^2+x-1=0 \tag{6}$$
When $x=0$, the equation is negative, and when $x=1$ the equations is positive. Hence, since the equation is a polynomial, by the Intermediate Value Theorem (IVT) the equation $(5)$ has a real root between $0$ and $1$.
Unfortunately, you may have to use the (very complicated) formula for finding the roots of a cubic equation or a numerical solver to determine an accurate value of $x$. Using a numerical solver at this point kind of nullifies all the previous work, since you could have just used a numerical solver for the original problem.
|
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"timestamp": "2023-03-29T00:00:00",
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Prove using induction that $n^3 − n$ is divisible by 6 whenever $n > 0$. Prove using induction that $n^3 − n$ is divisible by $6$ whenever $n > 0$
My attempt:
Base step: For $n=1$
$1^3 - 1 = 0$.
$0$ which is divisible by $6$.
Thus, $n= 1$ is true.
Assumption step: Let $n=k$
$k^3-k$
Inductive step: $f(k+1)-f(k)$
$(k+1)^3-(k+1)-k^3-k$
This is where I am stuck. How do I go forward to prove using induction that $n^3 − n$ is divisible by $6$?
|
In your assumption step, you need to assume the statement is true for $n=k$, i.e. $k^3-k$ is divisible by $6$.
In the induction step, expand and simplify $(k+1)^3-(k+1)$:
$$\begin{aligned}(k+1)^3-(k+1)&=k^3+3k^2+3k+1-k-1\\&=k^3-k+3k^2+3k\\&=(k^3-k)+3k^2+3k\\&=(k^3-k)+3k(k+1)\end{aligned}$$
Note that $k^3-k$ is divisible by $6$ by the inductive hypothesis (the assumption). $k$ and $k+1$ are consecutive integers, so their product must be even. Hence $3k(k+1)$ is also divisible by $6$. Thus, whenever $k^3-k$ is divisible by $6$, we also have that $(k+1)^3-(k+1)$ is divisible by $6$.
|
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|
Using the limit definition to find a derivative for $\frac{-5x}{2+\sqrt{x+3}}$ I am trying to find the derivative of $\frac{-5x}{2+\sqrt{x+3}}$ using the limit definition of a derivative.
$$\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}$$
What I did is
$$\lim_{h \to 0} \frac{\dfrac{-5(x+h)}{2+\sqrt{x+h+3}}\dfrac{2-\sqrt{x+h+3}}{2-\sqrt{x+h+3}}-\dfrac{-5x}{2+\sqrt{x+3}}\dfrac{2-\sqrt{x+3}}{2-\sqrt{x+3}}}{h}$$
But I am having problem finding the correct answer.
|
The difference quotient to which we are to apply the operation $\;\lim\limits_{h \rightarrow 0}\;$ is
$$ \frac{f(x+h) - f(x)}{h} \;\; = \;\; \left[\frac{-5(x+h)}{2 + \sqrt{(x+h) + 3\;}\;} \; - \; \frac{-5x}{2 + \sqrt{x+3\;}\;} \right] \; \div \; h $$
$$ = \;\; \left[\frac{-5(x+h)\left(2+\sqrt{x+3\;}\right) \;\; - \;\; (-5x)\left(2 + \sqrt{x+h+3\;}\right)\;}{\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)}\right] \; \div \; h $$
$$ = \;\; \frac{-10x - 5x\sqrt{x+3\;} - 10h - 5h\sqrt{x+3\;} + 10x + 5x\sqrt{x+h+3\;}\;}{h\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} $$
$$ = \;\; \frac{-5x\sqrt{x+3\;} - 10h - 5h\sqrt{x+3\;} + 5x\sqrt{x+h+3\;}\;}{h\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} $$
$$ = \;\; \frac{-5x\sqrt{x+3\;} \; + \; 5x\sqrt{x+h+3\;}\;}{h\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} \;\; + \;\; \frac{-10h \; - \; 5h\sqrt{x+3\;}\;}{h\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} $$
$$ = \;\; \frac{5x}{\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} \; \cdot \; \frac{\sqrt{x+h+3\;} \; - \; \sqrt{x+3\;}\;}{h} $$
$$ + \;\; \frac{-10 \; - \; 5\sqrt{x+3\;}\;}{\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} $$
This last expression has the form $\;A \cdot B \; + \; C,\;$ where for $\;h \rightarrow 0\;$ we have
$$ A \;\; = \;\; \frac{5x}{\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} \;\; \longrightarrow \;\; \frac{5x}{\left(2 + \sqrt{x+3\;}\right)^2} $$
and
$$ B \;\; = \;\; \frac{\sqrt{x+h+3\;} \; - \; \sqrt{x+3\;}}{h} \;\; = \;\; \frac{\sqrt{x+h+3\;} \; - \; \sqrt{x+3\;}}{h} \cdot \frac{\sqrt{x+h+3\;} \; + \; \sqrt{x+3\;}}{\sqrt{x+h+3\;} \; + \; \sqrt{x+3\;}} $$
$$ = \;\; \frac{(x+h+3) \; - \; (x+3)\;}{h\left(\sqrt{x+h+3\;} \; + \; \sqrt{x+3\;}\right)} \;\; = \;\; \frac{h}{h\left(\sqrt{x+h+3\;} \; + \; \sqrt{x+3\;}\right)} $$
$$ = \;\; \frac{1}{\sqrt{x+h+3\;} \; + \; \sqrt{x+3\;}\;} \;\; \longrightarrow \; \frac{1}{2\sqrt{x+3\;}\;} $$
and
$$ C \;\; = \;\; \frac{-10 \; - \; 5\sqrt{x+3\;}\;}{\left(2 + \sqrt{x+h+3\;}\right)\left(2+\sqrt{x+3\;}\right)} \;\; \longrightarrow \;\; \frac{-10 \; - \; 5\sqrt{x+3\;}}{\left(2 + \sqrt{x+3\;}\right)^2} $$
Therefore, the derivative is
$$ A \cdot B \; + \; C \;\; = \;\; \frac{5x}{\left(2 + \sqrt{x+3\;}\right)^2} \; \cdot \; \frac{1}{2\sqrt{x+3\;}\;} \;\; + \;\; \frac{-10 \; - \; 5\sqrt{x+3\;}}{\left(2 + \sqrt{x+3\;}\right)^2} $$
I'll leave to you the verification that this expression is equal to the expression you get by differentiating using short-cut rules.
|
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solve $L=\int_0^4 \sqrt{1+\frac{9}{4}y^4+\frac{9}{2}y^2}dy$ I need help with this excercise.
Find the arc length of the function $$x=\frac{1}{2}(y^2+2)^{3/2}$$
from $y=0$ to $y=4$.
$$L=\int_a^b \sqrt{1+(\frac{dx}{dy})^2}dy$$
Now, $$x=\frac{1}{2}(y^2+2)^{3/2}$$
$$\frac{dx}{dy}=\frac{3}{2}y\sqrt{y^2+2}$$
Then,
$$L=\int_0^4 \sqrt{1+(\frac{3}{2}y\sqrt{y^2+2})^2}dy$$
$$L=\int_0^4 \sqrt{1+\frac{9}{4}y^4+\frac{9}{2}y^2}dy$$
How to solve this integral?
|
We can instead take the parametrization
$y=\sqrt{2}\tan(a), x=\sqrt{2}\sec^3({a})$
$y'^2=2\sec^4(a), x'^2=18\sec^6(a)\tan^2(a)$
We then need to compute
$\int_{0}^{\arctan(2\sqrt{2})} \sqrt{18\sec^6(a)\tan^2(a)+2\sec^4(a)}da$
Making the substiution $u=\tan{a}$ transfomrs the integral to
$\int_{0}^{2\sqrt{2}} \sqrt{18u^2+20}du$
Now take $u=\sqrt{\frac{20}{18}}\sinh{v}$
then we're done hope you can continue from here
|
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|
Find real part of $\frac{1}{1-e^{i\pi/7}}$ How can you find
$$\operatorname{Re}\left(\frac{1}{1-e^{i\pi/7}}\right).$$
I put it into wolframalpha and got $\frac{1}{2}$, but I have no idea where to begin. I though maybe we could use the fact that $$\frac{1}{z}=\frac{\bar{z}}{|z|^2},$$ where $\bar{z}$ is the conjugate of $z$. Unfortunately, the magnitude doesn't seem to be a nice number. I feel like this might be a trigonometry question in disguise, but converting $e^{i\pi/7}=\cos\left(\frac{\pi}{7}\right)+i\sin\left(\frac{\pi}{7}\right)$ hasn't been very fruitful.
|
Well, firstly, since $z\bar{z} = |z|^2$ we know $\frac{1}{z} = \frac{\bar{z}}{|z|^2}$ (1). So let $z = 1 - e^{\frac{i \pi}{7}}$. Then from (1), $$\frac{1}{1 - e^{\frac{i \pi}{7}}} = \frac{\bar{z}}{|z|^2}.$$
Now, by Euler's formula, $e^{\frac{i \pi}{7}} = \cos(\frac{\pi}{7})+i\sin(\frac{\pi}{7}),$ so $z = (1- (\cos(\frac{\pi}{7})+i\sin(\frac{\pi}{7}))) = (1 - \cos(\frac{\pi}{7})) - i\sin(\frac{\pi}{7})$.
Now this means that $\bar{z} = (1 - \cos(\frac{\pi}{7})) + i\sin(\frac{\pi}{7})$. Moreover, $|z|^2 = (1 - \cos(\frac{\pi}{7}))^2 + \sin(\frac{\pi}{7})^2$, therefore, $|z|^2 = 1 + (\cos^2(\frac{\pi}{7}) + \sin^2(\frac{\pi}{7})) - 2\cos(\frac{\pi}{7}) = 2 - 2\cos(\frac{\pi}{7})$, so; $$\frac{1}{z} = \frac{1 - \cos(\frac{\pi}{7})}{2 - 2\cos(\frac{\pi}{7})} + i\frac{\sin(\frac{\pi}{7})}{2-2\cos(\frac{\pi}{7})}.$$
So, $Re(\frac{1}{z}) = \frac{1 - \cos(\frac{\pi}{7})}{2 - 2\cos(\frac{\pi}{7})} = \frac{1 - \cos(\frac{\pi}{7})}{2(1 - \cos(\frac{\pi}{7}))} = \frac{1}{2}$. I hope that helps :P
|
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$1^n-3^n-6^n+8^n$ is divisible by $10$ Prove that $1^n-3^n-6^n+8^n$ is divisible by $10$ for all $n\in\mathbb{N}$
It is divisible by $2$ and $5$ if we rearrange it will it be enough
$(1^n -3^n)$ and $(6^n -8^n)$ is divisible by $2$.
And
$(1^n-6^n)$ and $(8^n-3^n)$ is divisible by $5$.
Hence $\gcd(2,5)$ is $1$ and it is divisible by $2\cdot5=10$.
Is it correct?
|
Note: $p,q,r,s,t, a ,b, n \in \mathbb{N},$
$f(n):= (8^n-3^n) - (6^n-1^n)= (8-3)p-(6-1)q=5(p-q).$
On the other hand:
$f(n)= (8^n-6^n) -(3^n-1^n)=(8-6)r -(3-1)s=2(r-s)$
Hence:
$f(n)=5(p-q)=2(r-s).$
Euclid's Lemma:
$2$ divides $(p-q)$, i.e.
$p-q=2t$.
Combining:
$f(n)= 5\cdot 2 t.$
Used:
$(a^n-b^n)=$
$(a-b)(a^{n-1}+a^{n-2}b+.....+b^{n-1}).$
$p,q,r$ and $s$ were used for the above second factor.
Euclid's Lemma:
If a prime $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$.
|
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Proof by induction: is it enough to show 0=0? Let $P_n$ be a proposition, so that $$P_n: 3+11+...+(8n-5)=4n^2-n$$
$P_1:3=4\cdot1^2-1$, $P_2:3+11=4\cdot2^2-2$ etc.
When proving $P_n\to P_{n+1}$, is it enough to show in the second induction step that, by subtracting from both sides we have $$4(n+1)^2-(n+1)=4n^2-n+(8(n+1)-5)\leftrightarrow0=0$$ and therefore $P_n\to P_{n+1}$. Or do I must rewrite $4n^2-n+(8(n+1)-5)$ to $4(n+1)^2-(n+1)$?
|
I would rather
make the use of induction
extremely clear.
First of all,
I would write
$P_n: 3+11+...+(8n-5)=4n^2-n
$
as
$P_n:
\sum_{k=1}^n (8n-5)
=4n^2-n
$.
This makes the start and end terms
of the sum clear
and reduces the chance of error.
Then
I would show the step
from $P_n$
to $P_{n+1}$
like this:
$\begin{array}\\
\sum_{k=1}^{n+1} (8n-5)
&=\sum_{k=1}^{n} (8n-5)+8(n+1)-5
\qquad\text{split off the last term}\\
&=4n^2-n+8(n+1)-5
\qquad\text{use the induction hypothesis}\\
&=4n^2+7n+3
\qquad\text{algebra}\\
&=4(n+1)^2-(n+1)
\qquad\text{more algebra to get the right side of }P_{n+1}\\
\end{array}
$
which is
$P_{n+1}$,
and we are done.
More generally,
suppose we are given
$a$ and $b$,
and we want to find
$u, v, $ and $w$ such that
$P_n:
\sum_{k=1}^n (an+b)
=un^2+vn+w
$
is true.
Arguing exactly as above,
$\begin{array}\\
\sum_{k=1}^{n+1} (an+b)
&=\sum_{k=1}^{n} (an+b)+a(n+1)+b
\qquad\text{split off the last term}\\
&=un^2+vn+w+a(n+1)+b
\qquad\text{use the induction hypothesis}\\
&=un^2+(v+a)n+w+a+b
\qquad\text{algebra}\\
\end{array}
$
$P_{n+1}$ is
$\sum_{k=1}^{n+1} (an+b)
=u(n+1)^2+v(n+1)+w
$,
so we want
$un^2+(v+a)n+w+a+b\\
=u(n+1)^2+v(n+1)+w\\
=un^2+(2u+v)n+u+v+w.
$
Equating coefficients,
$v+a = 2u+v$
and
$w+a+b = u+v+w$.
From the first,
$u = a/2$.
From the second,
$a+b = u+v
=a/2+v$
so
$v = a/2+b$.
For your problem,
$a=8, b=-5$,
so
$u = 4,
v=8/2-5
=-1
$.
Note that
$w$ does not seem to be
determined.
For this,
we need the initial value;
either $n=0$ or $n=1$
will work.
Using $n=0$,
the sum is empty,
so we want
$0 = w$.
Using $k=1$,
$P_1$ is
$a+b
= u+v+w$
or
$w
=a+b-u-v
=a+b-a/2-(a/2+b)
= 0
$.
It is comforting that
both cases lead
to the same result.
Therefore
we have shown that
$\sum_{k=1}^n (ak+b)
=(a/2)n^2+(a/2+b)n
$.
This type of argument
enables us to
both discover a result
and prove it.
In some problems,
this technique
allows us to show that
a particular form
of a summation does not exist,
because the assumption
that the form exists
leads to a contradiction.
|
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|
Is this proof of the following integrals fine $\int_{0}^{1} \frac{\ln(1+x)}{x} dx$? $$\int_{0}^{1} \frac{\ln(1+x)}{x} dx = \int_{0}^{1} \frac1x\cdot (x-x^2/2 + x^3/3 -x^4/4 \ldots)dx = \int_{0}^{1}(1 - x/2 + x^2/3 - x^4/4 \dots)dx = 1-\frac1{2^2} + \frac1{3^3} - \frac{1}{4^4} \dots = \frac{\pi^2}{12}$$
Also,
$$\int_{0}^{1} \frac{\ln(1-x)}{x}dx = -\int_{0}^{1}\frac1x \cdot (x+x^2/2 +x^3/3 \ldots)dx =- \int_{0}^{1}(1+x/2+x^2/3 \dots )dx = -( 1 + \frac1{2^2} + \frac1{3^3} \dots) = -\frac{\pi^2}{6}$$
Using these integral, I evaluate $$ \int_{0}^{1} \frac{\ln(x)}{1-x} $$
by letting $1-x=t$
$$ \int_{0}^{1} \frac{\ln(x)}{1-x} =\int_{0}^{1} \frac{\ln(1-t)}{t} dx = -\frac{\pi^2}{6}$$
However, I'm facing problem while integrating
$$\int_{0}^{1} \frac{\ln(x)}{1+x}$$ using the same above way.
|
Let,
$\begin{align}I&=\int_0^1 \frac{\ln x}{1-x}\,dx\\
J&=\int_0^1 \frac{\ln x}{1+x}\,dx\\
\end{align}$
$\begin{align}I-J&=\int_0^1 \frac{2x\ln x}{1-x^2}\,dx\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}I-J&=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}\,dx\\
&=\frac{1}{2}I\end{align}$
Therefore,
$J=\dfrac{1}{2}I$
If $I=-\dfrac{\pi^2}{6}$ then $J=-\dfrac{\pi^2}{12}$
|
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|
Inequalities for $\operatorname{rad}(n\text{th pentagonal number})$ and the diagonal of the regular pentagon that represents this polygonal number We denote for integers $n\geq 1$ the square-free kernel, or radical of the natural $n>1$, as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing our integer $n>1$ with the definition $\operatorname{rad}(1)=1$.
Motivation studying the case of the squares. Then it is obvious that $$\operatorname{rad}(n^2)=\operatorname{rad}(n)\leq n,\tag{1}$$
and from this inequality we deduce a worse inequality $$\operatorname{rad}(n^2)<\sqrt{2}(n-1)\tag{2}$$
that holds for integers $n>3$. This inequality $(2)$ tell us (the obvious claim) that the square-free kernel of the perfect square $n^2$ is strictly lesser than the diagonal of the corresponding polygonal number, the square number $$\text{Square Number}(n)=n^2.$$
In this post we explore the similar inequality than $(2)$ for pentagonal numbers: seems that the most of the naturals $n> 1$ satisfy the inequality
$$\operatorname{rad}\left(\frac{3n^2-n}{2}\right)>\text{diagonal of regular pentagon that represents the }n\text{th pentagonal number},$$
that is
$$\operatorname{rad}\left(\frac{3n^2-n}{2}\right)>\left(\frac{1+\sqrt{5}}{2}\right)(n-1).\tag{3}$$
Question. But also seems that there exist many integers $m>1$ satisfying $$\operatorname{rad}\left(\frac{3m^2-m}{2}\right)<\left(\frac{1+\sqrt{5}}{2}\right)(m-1).\tag{4}$$
Prove or refute that there exist infinitely many integers $m>1$ for which the inequality $(4)$ holds.
Many thanks.
I hope that there aren't mistakes in my presentation of previous problem and that is well motivated. I've calculated with the help of a Pari/GP program the first few terms of the sequence corresponding to $(4)$, but I don't see an easy pattern to hypothesize a family of infinitely many solutions of $(4)$.
As references I add here, for example, the Wikipedia's article dedicated to Pentagonal numbers and the MathWorld's article dedicated to Pentagon.
|
If the $m^{\text{th}}$ pentagonal number is a square, then we have $\DeclareMathOperator{\rad}{rad}$
$$\rad \biggl(\frac{3m^2-m}{2}\biggr) \leqslant \sqrt{\frac{3m^2-m}{2}} \leqslant \sqrt{\frac{3}{2}}\cdot m\,.$$
Since $\sqrt{\frac{3}{2}} < \frac{3}{2} < \frac{1 + \sqrt{5}}{2}$, in this case we will have $(4)$ except for a few very small $m$. So if there are infinitely many pentagonal numbers that are also squares, $(4)$ holds for infinitely many $m$.
Hence let's look when a pentagonal number is a square.
\begin{align}
&& \frac{3m^2-m}{2} &= k^2 \\
&\iff& 3m^2 - m &= 2k^2 \\
&\iff& 36m^2 - 12m &= 24k^2 \\
&\iff& (6m - 1)^2 &= 24k^2 + 1 \\
&\iff& (6m-1)^2 - 24k^2 &= 1
\end{align}
One knows that Pell's equation $x^2 - Dy^2 = 1$ always has infinitely many solutions (where $x$ and $y$ are positive integers) if $D$ is not a perfect square. Since $24$ is not a perfect square, it remains to see that of the infinitely many solutions of $x^2 - 24 y^2 = 1$, infinitely many have $x \equiv 5 \pmod{6}$. The solutions are given by
$$x_r + y_r\sqrt{24} = (5 + \sqrt{24})^r\,,$$
so we have the recurrence
$$x_{r+1} + y_{r+1}\sqrt{24} = (x_r + y_r\sqrt{24}) (5 + \sqrt{24}) = (5x_r + 24y_r) + (x_r + 5y_r)\sqrt{24}$$
and in particular
$$x_{r+1} \equiv 5x_r \pmod{6}\,.$$
Since $x_1 = 5 \equiv 5 \pmod{6}$, the solutions
$$x_{2r+1} + y_{2r+1}\sqrt{24} = (5 + \sqrt{24})^{2r+1}$$
yield pentagonal square numbers for
$$m_r = \frac{x_{2r+1} + 1}{6}\,.$$
|
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|
What is the probability of a symmetric coin fall
The prompt: We have 10 coins and 1 of them is non-symmetric (with probability of head equal $\frac{1}{3}$). We toss a randomly selected coin 6 times, and obtain 3 tails. What is the probability that we tossed a symmetrical coin?
The way I went on about solving the problem was first assuming it to be a win or success if we obtain a Tails, assuming that
NS = Non symmetric coin, S = Symmetric coin, H = heads, T = tails
$$P(S) = \frac{9}{10}; \text{ } P(NS) = \frac{1}{10}; \text{ }
P(T|S) = \frac{1}{2}; \text{ }P(T|NS) = \frac{1}{3}$$
Now Probability of getting a Tail can be obtained using total probability theorem,
$$P(T) = P(T|S)\cdot P(S) + P(T|NS) \cdot P(NS)$$
$$P(T) = \frac{1}{2}\cdot \frac{9}{10} + \frac{1}{3} \cdot \frac{1}{10}
= \frac{29}{60}$$
Using Bernoulli's trials, where winning/success(p) is defined as getting a tails, given by $P(T) = \frac{29}{60}$ and failure(q) is defined by getting a head, $P(H) = \frac{31}{60}$ and n = number of tosses.
$$P(X = x) = {n \choose x} (p)^n \cdot (q)^{n - x}$$
$$P(X = 3) = {6 \choose 3} (\frac{29}{60})^6 \cdot (\frac{31}{60})^3$$
Assuming this to be an event E, we now have $P(E) = {6 \choose 3} (\frac{29}{60})^6 \cdot (\frac{31}{60})^3$,
The desired result is $P(S|E)$, I'm not sure how to go on about finding that, any hint would be much appreciated.
|
Borrowing your notation, the value we want is: $P(S \mid E) = \frac{P(E \mid S)~P(S)}{P(E)}$, which is Bayes' rule.
Note that:
$$P(E \mid S) = {6 \choose 3} \left( \frac{1}{2} \right)^6 = 0.3125$$
$$P(E \mid NS) = {6 \choose 3} \left( \frac{1}{3} \right)^3 \left(\frac{2}{3} \right)^3=0.219$$
Thus, $P(E) = \sum_i P(E \mid C_i)P(C_i)=\frac{1}{10} \cdot 0.3125 + \frac{9}{10} \cdot 0.219 = 0.0566$. Plug these numbers into the quotient above to get the answer.
|
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|
Expected length of longest stick The problem is the same as here.
A stick of 1m is divided into three pieces by two random points. Find the average length of the largest segment.
I tried solving it in a different way, and the logic seems fine, however I get a different result to $\frac{11}{18}$.
Here is my solution. Please let me know what I did wrong.
Let $X$ be the length of the stick from the beginning to the first cut. $Y$ be the length of the stick between the first and second cut and $1-X-Y$ the length between the second cut and the end of the stick.
We want to find the CDF of the following random variable: $Z=\max(X,Y,1-X-Y)$. (I believe that if anything is wrong, this might be it).
$$\begin{split}
F_Z(z) = P(Z\leq z) & = P(\max(X,Y,1-X-Y) \leq z)\\ & = P(X\leq z, Y\leq z, 1-X-Y\leq z)\\ &= P(1-Y-z\leq X \leq z, Y\leq z)
\end{split}
$$
Since we have $1-Y-z\leq z$ we deduce that $Y\geq 1-2z$. Hence:
$$\begin{split}
F_Z(z) &= \int_{1-2z}^z\int_{1-y-z}^z 1 dx dy = \int_{1-2z}^z (z-1+y+z) dy\\ &= (2z-1)(z-1+2z) + \left. \frac{y^2}{2}\right|_{y=1-2z}^{y=z} \\ &=(2z-1)(3z-1) + \frac{1}{2}(z^2- (2z-1)^2) \\ & = (2z-1)(3z-1) +\frac{1}{2}(-3z^2 + 4z -1) \\ & = \frac{1}{2}(3z-1)^2
\end{split}
$$
Now, the pdf of $Z$ is :
$$f_Z(z) = \frac{d}{dz}F_Z(z) = 9z-3
$$
And now, in order to find the expected value of the largest length, we need to integrate over $(\frac{1}{3},1)$ as the largest piece needs to be greater than $\frac{1}{3}$. Hence
$$\begin{split}
E[Z] = \int_{\frac{1}{3}}^{1} z f_Z(z) dz = \int_{\frac{1}{3}}^{1} z (9z-3) dz = \frac{14}{9}
\end{split}
$$
The result is obviously wrong as it needs to be something between $0$ and $1$, however after going over the solution multiple times, and checking the calculations with Wolfram, I cannot seem to figure out what went wrong.
|
Here is how I would do it.
Lets define $x$ to be the short stick, $y$ to be the medium stick and $z$ to be the long stick.
$x\le y\le z\\
z = 1-x-y\\
x\le y \le \frac {1-x}{2}\\
x\le \frac 13$
$$ \bar z = \frac {\displaystyle\int_0^\frac 13\int_x^{\frac {1-x}{2}} 1-x-y\ dy\ dx}{\displaystyle\int_0^\frac 13\int_x^{\frac {1-x}{2}} 1\ dy\ dx}$$
|
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|
Maximize $f(x,y)=xy$ subject to $x^2-yx+y^2 = 1$
Use Lagrange multipliers method to find the maximum and minimum values of the function
$$f(x,y)=xy$$
on the curve
$$x^2-yx+y^2=1$$
Attempt:
First I set let $g(x,y)=x^2-xy+y^2-1$ and set $$\nabla f=\lambda\nabla g$$
so
$$(y,x)=\lambda(2x-y,2y-x)$$
then
$$\begin{cases}
\lambda=\frac{y}{2x-y} & (1) \\
\lambda=\frac{x}{2y-x} & (2)\\
x^2-yx+y^2=1
\end{cases}
$$
Solving $(1)$ and $(2)$ simultaneously, I get that $$y^2=x^2$$
Substitutiting into $(3)$ and following through with the arithmetic, I get four candidates for max and min, namely $$(1,1),(-1,-1),\big(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\big),\big(\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\big)$$
Evaluating these points on $f$, I get that the maximum value is $$1 \ \text{at} \ (\pm1,\pm1)$$
and the minimum value is $$-\frac{1}{3} \ \text{at} \ \big(\pm\frac{1}{\sqrt{3}},\mp\frac{1}{\sqrt{3}}\big)$$
Am I correct? I am unsure if there are indeed four critical points.
|
Geometrical ways:
\begin{align}
x^2-xy+y^2 &= 1 \\
\frac{(x+y)^2}{4}+\frac{3(x-y)^2}{4} &= 1 \tag{1}
\end{align}
With the transformation $(X,Y)=
\left( \dfrac{x+y}{\sqrt{2}}, \dfrac{x-y}{\sqrt{2}} \right)$,
$$\frac{X^2}{2}+\frac{3Y^2}{2}=1 \tag{2} $$
which is an ellipse with semi-major and minor axes $\sqrt{2}$ and $\sqrt{\dfrac{2}{3}}$ respectively.
Also, $$xy=\frac{X^2-Y^2}{2}$$
Now $(2)$ touches
*
*$X^2-Y^2=2$ at $\left( \pm \sqrt{2},0 \right)$ which gives maximum $xy$ of $1$
*$X^2-Y^2=-\dfrac{2}{3}$ at $\left( 0, \pm \sqrt{\dfrac{2}{3}} \right)$ which gives minimum $xy$ of $-\dfrac{2}{3}$
|
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|
Why are the solutions of the $3\times 3$ system like that? Consider the problem:$$
\min \quad -x_1^2-4x_1x_2-x_2^2\\ \text{s.t.} \quad x_1^2 + x_2^2 = 1$$
The KKT system is given by\begin{align*}
x_1 (-1 + v) + 2 x_2 &= 0 \tag{1} ,\\ x_2 (-1 + v) + 2 x_1 &= 0 \tag{2},\\ x_1^2 + x_2^2 &= 1 \tag{3}
\end{align*}
The solutions according to the book are$$
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\ v=3;\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\ v=3;\\
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right),\ v=-1,\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),\ v=-1,\\
$$
I tried to solved by hand: Substracting (1) and (2) I get $(x_1-x_2)(1+v)=0$. As $v$ cannot be zero (by the KKT condition) we have that $x_1=x_2$ if $v\neq -1$.
Thus I get the solutions:$$
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right);\\
(x_1,x_2)=\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right);\\
$$
only asking $v\neq -1$.
Why the book ask $v=-1$ and $v=3$? Could someone explain please?
|
$(x_1-x_2)^2\geq0$ gives $$x_1x_2\leq\frac{x_1^2+x_2^2}{2}=\frac{1}{2}.$$
The equality occurs for $x_1=x_2$.
Thus,
$$-x_1^2-4x_1x_2-x_2^2=-1-4x_1x_2\geq-1-4\cdot\frac{1}{2}=-3.$$
The equality occurs for
$$(x_1,x_2)\in\left\{\left(\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right),\left(-\frac{1}{\sqrt2},-\frac{1}{\sqrt2}\right)\right\}$$ only.
|
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|
Interesting facts about the numbers 1 - 31 I'm a maths teacher and want to create a calendar for my classroom.
I'm looking to compile some interesting facts about each of the numbers 1 through to 31.
The hard part is they must be at a level where 11-18 year-olds can understand them.
Along the lines of: 2 is the smallest prime number, only even prime number.
$ $
Any help would be greatly appreciated.
|
Very interesting idea- I like it. How about these:
$1$ is the only number with one factor, or the only number which is the factorial of two numbers ($0!=1!=1)$
$2$ is the only even prime
$3$ is the best integer approximation of $\pi$ and $e$.
Every even square is a multiple of $4$.
$5$ is the only odd number for which every number ending in it is composite.
$6$ is the smallest perfect number
$7$ is the smallest number that cannot be represented as the sum of three integer squares (e.g. $6=2^2+1^2+1^2$). Or $\frac n7$ always features the same digits in the same order with the start varying.
$8$ is the only Fibonacci Number besides $1$ that is a perfect cube.
$9$ is roughly $2\pi+e$
$10$ is the sum of the first three primes, the first four factorials, and the first four positive integers.
The powers of $11$ can be found by Pascal's Triangle.
$12$ is the smallest sublime number, of which there are only two known.
$13$ is the smallest emirp.
$14$ is the smallest satisfier of the Shapiro Inequality.
$15$ is $1|5$ and $1+2+3+4+5$
$16$ is the smallest perfect fourth power besides $0$ and $1$
$17\approx12\sqrt2$ and so we can use it to approximate $\sqrt2$
$18$ is the first non-square number expressible as $p\cdot q^2$, with $p,q$ integers and $q>p$
Any natural number can be made by summing $19$ $4$th powers.
The product of the divisors $(1,2,4,5,10)$ with the proper divisors $(2,4,5,10)$ of $20$ is $20$.
$21$ is a triangular number.
$22\approx7\pi$, thus $\pi\approx\frac{22}{7}$ is used where a fractional approximation is needed.
$23$ is the smallest number irrepresentable as the sum of nine or fewer positive cubes.
$a=24$ is the largest and only non-trivial solution to $1^2+2^2+...+a^2=b^2$
If the last two digits of a number are a multiple of $25$, the number is a multiple of $25$.
$26$ is the only number of both the form $a^2+1$ and the form $a^3-1$.
$27$ is the only integer that is three times the sum of its digits. Or the first significant input of the Collatz Conjecture.
$28$ is the only known number which is the sum of the first $a$ primes, the first $b$ non-negative integers, and the sum of the first $c$ non-primes. (Here, $a=7, b=5, c=5$)
$29$ cannot be made using addition, subtraction, multiplication and division with $1,2,3,4$, suing each only once.
$30=(2220422932)^3+(-2218888517)^3+(-283059965)^3$, and was the first number that was significantly challenging to represent as the sum of three cubes.
$31,331,3331,33331,333331,3333331,33333331$ are all prime.
|
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|
How to find the minimum of $f(x)=\frac{4x^2}{\sqrt{x^2-16}}$ without using the derivative?
Find the minimum of function $$f(x)=\frac{4x^2}{\sqrt{x^2-16}}$$ without using the derivative.
In math class we haven't learnt how to solve this kind of problems (optimization) yet. I already know that is solvable using derivatives, but there should be another way. Thanks in advance!
|
Render
$\frac{4x^2}{\sqrt{x^2-16}}=a$
Here, $a$ is the minimum function value.
Square both sides and convert the resulting polynomial relation to the form $P(x)=0$:
$\frac{16x^4}{x^2-16}=a^2$
$x^4-a^2x^2+16a^2=0$
Now if $a$ is to be the minimum function for $f(x)$ value then the polynomial will have a squared factor for that value of $a$; the value of $x$ where that occurs (called $x_0$) is the resulting double root:
$x^4-a^2x^2+16a^2=(x-x_0)^2(x^2+px+q)=0$
Then
$x^4-a^2x^2+16a^2=(x-x_0)^2(x^2+px+q)=x^4+(p-2x_0)x^3+(q-2px_0+x_0^2)x^2+(-2qx_0+px_0^2)x+qx_0^2$
For this factorization to hold The cubic and linear terms must satisfy $p-2x_0=0$ and $-2q+px_0=0$ (the function would not be defined at $x_0=0$). Therefore $p=2x_0, q=x_0^2$. Put these into the quadratic and constant terms:
$q-2px_0+x_0^2=-2x_0^2=-a^2$
$qx_0^2=x_0^4=16a^2$
Dividing the second of these two equations by the first leads to $x_0^2=32$, thus $x_0=\pm 4\sqrt{2}$ for the minimizing $x$ value. The minimum function value, which is clearly positive, is then $a=\sqrt(2x_0^2)=8$.
|
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|
$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$ $a, b, c ∈ \mathbb{R}+.$
WLOG assume $a \leq b \leq c.$ I tried substitution: $x=\frac{1} {1+a^2}, y=\frac{1} {1+b^2}, z=\frac{1} {1+c^2},$ so $x \geq y \geq z$ and $(1-x)+(1-y)+(1-z)=2 \to x+y+z=1.$
We want to prove $ax+by+cz \leq \sqrt{2}.$ This somewhat looks like Cauchy-Schwarz so I tried that: $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq (ax+by+cz)^2.$ The problem becomes $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq 2,$ since $a,b,c,x,y,z>0.$
Expressing $a,b,c$ in terms of $x,y,z$: $(\frac {1}{x} + \frac {1}{y} + \frac {1}{z} - 3)(x^2+y^2+z^2)$
$= x+y+z+\frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 2.$
$\to \frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 1.$ Stuck here. Thinking about using AM-GM but not sure how. Help would be greatly appreciated.
|
Let $a,b,c>0$
$$\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}=2\rightarrow \dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\leq \sqrt{2}$$
$$\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}=\left( 1-\dfrac{1}{1+a}\right) +\left( 1-\dfrac{1}{1+b}\right) +\left( 1-\dfrac{1}{1+c}\right) =$$
$$=3-\left( \dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)$$
$$\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\leq 2\leftrightarrow 3-\left( \dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right) \leq \sqrt{2}\leftrightarrow$$
$$\leftrightarrow \dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\geq 3-\sqrt{2}\leftrightarrow \dfrac{a}{a+a^2}+\dfrac{b}{b+b^2}+\dfrac{c}{c+c^2}\geq 3-\sqrt{2}\leftrightarrow$$
$$\leftrightarrow \dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\geq \dfrac{a}{a+a^2}+\dfrac{b}{b+b^2}+\dfrac{c}{c+c^2}\geq 3-\sqrt{2}\leftrightarrow$$
$$\leftrightarrow 2\geq 3-\sqrt{2}\quad \text{(true)}$$
|
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|
In $\triangle ABC$, $(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = $?
In $\triangle ABC$,
$$(b^2-c^2) \cot A+(c^2-a^2) \cot B +(a^2-b^2) \cot C = \text{?}$$
I tried solving this by cosine rule but it is becoming too long. Any short step solution for this?
|
Using Proof of the sine rule
and Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$$(b^2-c^2)\cot A$$
$$=4R^2(\sin^2B-\sin^2C)\cot A$$
$$=4R^2\sin(B+C)\sin(B-C)\cdot\dfrac{-\cos(B+C)}{\sin(B+C)}$$
$$=-4R^2\sin(B-C)\cos(B+C)=2R^2(\sin2C-\sin2B)$$
|
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|
Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^2)+4n^3-4n-3
\\ &=(n^4-4n^2) + (4n^3+6n^2-4n)-3
\end{align}$$
Now $(n^4-4n^2)$ is divisible by 3, and $-3$ is divisible by 3. Now I am stuck on what to do to the remaining expression.
So, how to show that $4n^3+6n^2-4n$ should be divisible by 3? Or is there a better way to prove the statement in the title? Thank you!
|
Given that $6n^2$ is divisible by $3$, it's all a mater of showing that $4n^3-4n$ is divisible by $3$, and we can show that by showing that $n^3-n$ is divisible by $3$
And we know the later is true as follows:
if $n$ itself is divisble by $3$, then obviously $n^3-n$ is divisible by $3$. So, the only other options are that $n=3k+1$ or that $n=3k+2$ for some $k$.
If $n=3k+1$, then $n^3-n$ works out to all multiples of $3$ except for the $+1$ at the end, and so subtracting $n$ will get rid of that $1$
And if $n=3k+2$, then $n^3-n$works out to all multiples of $3$ except for the $+8$ at the end, and so subtracting $n$ will leave $6$, so that's all divisible by $3$ as well.
|
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|
Distance of point $P$ from an ellipse If $ \frac {x^2}{a^2} + \frac {y^2}{b^2} = r^2$ is an ellipse, with the parameterization $x(θ)≔r(a \cos θ,b \sin θ ),$ I have to find the value of $θ$ giving the minimum distance from $P(p,q)$ (not on the ellipse) to the ellipse is given by a quartic in $t= \tan( \frac {θ}{2}).$
A necessary condition for $x$ to be the closest point to $P$ is that $P-x$ is perpendicular to the tangent vector in $x ,$ i.e. $(P-x(θ) ). x' (θ)=0$
I can't handle the above condition to make an fuction (e.g. $f(θ)$) to find the minimum value by calculating the rerivative $f'(θ)=0$ for example. Then I have to prove that rational, non-zero values of $a, b, p, q$ can be found such that the quartic factorises as the product of two quadratics with rational coefficients. Any help? Thank you
|
One can normalize the length to $r=1$. The distance from $P$ to a point of the ellipse can be written as
\begin{equation}
d(\theta)=\sqrt{(p-a\cos \theta)^2+(q-b\sin\theta)^2}
\end{equation}
It is minimal if
\begin{equation}
a(p-a\cos\theta)\sin\theta-b(q-b\sin\theta)\cos\theta=0
\end{equation}
with $u=\tan\theta/2$, the condition reads
\begin{equation}
u^{4}bq+ \left( 2a^{2}+2ap-2b^{2} \right) u^{3}+
\left( -2a^{2}+2ap+2b^{2} \right) u-bq=0
\end{equation}
To find rational values for the parameters $a,b,p,q$ which allow a factorization of the quartic, one may choose them to verify
\begin{equation}
\frac{bq}{ -2a^{2}+2ap+2b^{2}}=\frac{ 2a^{2}+2ap-2b^{2}}{-bq}
\end{equation}
or
\begin{equation}
b^2q^2=4\left[ \left( a^2-b^2\right)^2-a^2p^2 \right]
\end{equation}
which can be written as
\begin{equation}
b^2q^2+4a^2p^2=4\left( a^2-b^2\right)^2
\end{equation}
By comparison to the pythagorean triples
\begin{equation}
(3n)^2+(4n)^2=(5n)^2
\end{equation}
$a^2-b^2=5n$ is verified by $a=4,b=1,n=6$, for example, and thus $q=18,p=3$. The condition reads
\begin{equation}
6 (u+3)(3u^3-1)=0
\end{equation}
which has 2 real roots $u=-3$ and $u=3^{-1/3}$.
|
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|
The number of incongruent solutions of $x^4-14x^2+36\equiv 0\pmod{2019}$ I was asked to find the number of the incongruent solutions of $x^4-14x^2+36\equiv 0\pmod{2019}$, and the fact $2019=673\times 3$ is given.
My attempt: Given congruence is equivalent to
$$
\begin{cases}
x^4-14x^2+36\equiv 0\pmod{673}\\
x^4-14x^2+36\equiv 0\pmod{3}
\end{cases}
$$
The second one is easily reduced to $1-14\equiv 0\pmod{3}$ if $3\nmid x$, so $x\equiv 0\pmod{3}$. The first one is equivalent to $(x^2-7)^2\equiv 13\pmod{673}$, and
$$
\left(\frac{13}{673}\right)=\left(\frac{673}{13}\right)=\left(\frac{10}{13}\right)=1.
$$
Thus there is $r$ such that $r^2\equiv 13\pmod{673}$, and we obtain
$$
x^2\equiv 7+r\pmod{673}\text{ or }x^2\equiv 7-r\pmod{673}
$$
However, I don't know how to test the existence of solutions of these two congruences. How to do it, or are there other ways?
|
I found a solution without finding the square root of 13 modulo 673 directly. $x^4-14x^2+36\equiv 0\pmod{673}$ is equivalent to
$$
x^4-12x^2+36\equiv 2x^2\pmod{673}.
$$
Also, there is an integer $r$ such that $r^2\equiv 2\pmod{673}$, since $\left(\dfrac{2}{673}\right)=1$. Thus given congruence is factored to
$$
(x^2-rx-6)(x^2+rx-6)\equiv 0\pmod{673}.
$$
Since 673 is prime, $x^2+rx-6\equiv 0\pmod{673}$ or $x^2-rx-6\equiv 0\pmod{673}$. Let $y=2x\pm r$, then each congruence is equivalent to $y^2\equiv r^2+24\equiv 26\pmod{673}$. Since
$$
\left(\dfrac{26}{673}\right)=\left(\dfrac{2}{673}\right)\left(\dfrac{13}{673}\right)=1,
$$
each congruence has two solutions. Therefore $x^4-14x^2+36\equiv 0\pmod{2019}$ has four incongruent solutions.
|
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|
Finding the shaded area in a triangle Here is the diagram:
I only know that the middle segment is a median of the big triangle. But nothing else.
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Let $h$ be the height of the triangles. Then the area of the large triangle is $$\Delta_{\text{ large}} = \frac{1}{2}\times 6\times \sqrt{h^2+36} \times \sin a$$ and the area of the white triangle is $$\Delta_{\text{ white}} =\frac{1}{2} \times\sqrt{h^2+1}\times\sqrt{h^2+9}\times\sin a$$
But since the large triangle has the same height as the white triangle, but three times its base, we have $\Delta_{\text{ large}} = 3 \Delta_{\text{white}}$. So
$$ 2\sqrt{h^2+36} = \sqrt{h^2+1}\sqrt{h^2+9}$$
Squaring both sides and simplifying gives
$$h^4+10h^2+9=4h^2+144$$
$$\Rightarrow (h^2+15)(h^2-9) = 0$$
So $h=3$, and the shaded area is $6$.
|
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|
What tools one should use for inequalities? If $a,b,c>0$ prove that:
$$\frac{1}{a+4b+4c}+\frac{1}{4a+b+4c}+\frac{1}{4a+4b+c}\leq \frac{1}{3\sqrt[3]{abc}}.$$
My first try was the following:
$$\sum_{cyc}\frac{1}{a+4b+4c}\leq\sum_{cyc}\frac{1}{\sqrt[3]{16abc}}=\frac{1}{\sqrt[3]{16abc}}$$
But $\frac{1}{\sqrt[3]{16abc}}\geq \frac{1}{3\sqrt[3]{abc}}$
The I have tried your method from another post:
$$\sum_{cyc}\frac{1}{a+4b+4c}=\sum_{cyc}\frac{1}{a+2b+2(b+2c)}$$
$$\sum_{cyc}\frac{1}{a+2b+2(b+2c)}\leq\sum_{cyc}\frac{1}{9}\left (\frac{1^2}{a+2b}+\frac{2^2}{b+2c} \right )$$
Where I got $$\sum_{cyc}\frac{1}{3}\left ( \frac{1}{a+2b} \right )\leq \frac{1}{3\sqrt[3]{abc}}$$ wich is false.
$$$$
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This is sort of an ugly proof which uses Maclaurin inequalities.
Define constants $A,B,C, \alpha,\beta,\gamma$ through following polynomial:
$$P(\lambda) = (\lambda-a)(\lambda-b)(\lambda-c)
= \lambda^3 - A\lambda^2 + B\lambda - C
= \lambda^3 - 3\alpha\lambda^2 + 3\beta^2\lambda - \gamma^3$$
By Vieta's formulas, we have
$$a + b + c = A = 3\alpha\quad\text{ and }\quad abc = C = \gamma^3$$
The LHS of the inequality at hand can be rewritten as
$$\begin{align}{\rm LHS} &= \sum_{cyc} \frac{1}{4A - 3a}
= \frac13 \sum_{cyc}\frac{1}{4\alpha-a}
= \frac13 \frac{P'(4\alpha)}{P(4\alpha)}
= \left.\frac{\lambda^2-2\alpha\lambda+\beta^2}{\lambda^3-3\alpha\lambda^2+3
\beta^2\lambda - \gamma^3}\right|_{\lambda=4\alpha}\\
&= \frac{8\alpha^2+\beta^2}{16\alpha^3 + 12\alpha\beta^2 - \gamma^3}
\end{align}
$$
while the RHS equals to $\displaystyle\;\frac{1}{3\gamma}$. The inequality we want to prove
is equivalent to
$$\begin{align}
{\rm LHS} \stackrel{?}{\le} {\rm RHS}
\iff & \frac{8\alpha^2+\beta^2}{16\alpha^3 + 12\alpha\beta^2 - \gamma^3} \stackrel{?}{ \le} \frac{1}{3\gamma}\\
\iff &
16\alpha^3 + 12\alpha\beta^2 - \gamma^3 - 3\gamma(8\alpha^2 + \beta^2) \stackrel{?}{\ge} 0
\end{align}\tag{*1}
$$
By Maclaurin's inequality, we have $\alpha \ge \beta \ge \gamma$.
This implies
$$\begin{align}
&\; 16\alpha^3 + 12\alpha\beta^2 - \gamma^3 - 3\gamma(8\alpha^2 + \beta^2)\\
\ge &\; 16\alpha^3 + 12\alpha\beta^2 - \beta^3 - 3\beta(8\alpha^2 + \beta^2)\\
= &\; 16\alpha^3 - 24\alpha^2\beta + 12\alpha\beta^2 - 4\beta^3\\
= &\; 4(\alpha^3 - \beta^3) + 12\alpha(\alpha-\beta)^2\\
\ge &\; 0
\end{align}$$
This establish the last line in $(*1)$. As a result, the inequality at hand is valid.
|
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|
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Is there a way to solve this rather than just declaring a matrix $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ and then trying to solve a system of cubic equations? My attempt:
$$A^3 -I_2 = \begin{bmatrix} 3 & 3\\ -3 & -3\end{bmatrix} = 9 \begin{bmatrix} 1 & 1\\ -1 & -1\end{bmatrix}$$
and
$$B=\frac {A^3-I_2}{A-I_2}-I_2$$
$$(A-I_2)(A^2+A+I_2)=9\begin{bmatrix}1&1\\-1&-1\end{bmatrix}$$
But I get stuck here.
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Presumably $A$ is real. Observe that $A^3-I$ is nonzero but nilpotent (because $(A^3-I)^2=0$). Therefore $A^3$ and in turn $A$ are not diagonalisable. Hence $A$ has repeated eigenvalues. So, if $A$ is real, its eigenvalues must be real (or else the trace of $A$ would become non-real) and equal to $1$ (because $A^3-I$ is nilpotent). Hence the Jordan form of $A$ is $\pmatrix{1&1\\0&1}$. Since
$$
\pmatrix{1&1\\0&1}^2+\pmatrix{1&1\\0&1}=\pmatrix{2&3\\0&2}
=\pmatrix{1&1\\0&1}^3+I,
$$
we conclude that $A^2+A=A^3+I$.
Remark. Note that the above conclusion does not hold when $A$ can be non-real. (Thus the other answers here are either wrong or incomplete.) E.g. suppose $w=\exp(2\pi i/3)$ and
$$
A=\pmatrix{2w&w\\ -w&0}.
$$
Then $A^3$ is indeed equal to $\pmatrix{4&3\\ -3&-2}$ but $(A^2+A)-(A^3+I)$ is non-real.
|
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|
Solution of polynomial Find the set of values of $k$ for which the equation $x^4+kx^3+11x^2+kx+1=0$ has four distinct positive root.
Attempt:
$x^4+kx^3+11x^2+kx+1=0$
$x^2+kx+11+{k\over x}+{1\over x^2}=0$
$x^2 + {1\over x^2} +k(x+{1\over x})+11=0$
$(x + {1\over x})^2 +k(x+{1\over x})+13=0$
I don't know how to proceed after this......
|
By your work let $f(u)=u^2+ku+13,$ where $u=x+\frac{1}{x}.$
Hence, $|u|>2$ and we need to solve the following system:
$$f(2)>0,$$
$$\frac{-k}{2}>2$$ and
$$k^2-52>0$$ or
$$f(-2)>0,$$
$$\frac{-k}{2}<-2$$ and
$$k^2-52>0,$$ which gives
$$-8.5<k<-\sqrt{52}$$ or
$$\sqrt{52}<k<8.5.$$
|
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|
Evaluating indefinite integrals.
If $$\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\dfrac{d}{b}{\sin^{-1}} \bigg(\dfrac{x^\frac{3}{2}}{a^\frac{3}{2}}\bigg)+C$$ where $b,d$ are relatively prime find $b+d$.
My solution:
$$\displaystyle\int\sqrt{\dfrac{x}{a^3-x^3}}dx=\int\dfrac{1}{x}\sqrt{\dfrac{x^3}{a^3-x^3}}dx$$
Then let $x^3=t\implies 3x^2\cdot dx=dt$
$$\implies\dfrac{1}{3}\int\dfrac{1}{x}\cdot\dfrac{3x^2}{x^2}\sqrt{\dfrac{x^3}{a^3-x^3}}dx\\=\dfrac{1}{3}\int\dfrac{1}{t}\sqrt{\dfrac{t}{a^3-t}}dt\\ =\dfrac{1}{3}\int{\dfrac{1}{\sqrt{ta^3-t^2}}}dt\\=\dfrac{1}{3}\int{\dfrac{1}{\sqrt{\frac{a^6}{4}-\bigg(t-\frac{a^3}{2}\bigg)^2}}}dt\\=\dfrac{1}{3}\sin^{-1}\bigg(\dfrac{2t-a^3}{a^3}\bigg)+C$$ $$=\dfrac{1}{3}\sin^{-1}\bigg[2\bigg(\dfrac{x^{3/2}}{a^{3/2}}\bigg)^2-1\bigg]+C\tag{1}$$
This is not of desired form but when I drew graph of: $\color{grey}{1.57+\sin^{-1}\left(2x^2-1\right)}$ and $\color{green}{2\sin^{-1}x}$, they coincides for $\color{red}{x>0}$. Green graph is for $\ 2\sin^{-1}x\ $.
So from $(1)$ $$\dfrac{1}{3}\sin^{-1}\bigg[2\bigg(\dfrac{x^{3/2}}{a^{3/2}}\bigg)^2-1\bigg]+C=\dfrac{2}{3}{\sin^{-1}} \bigg(\dfrac{x^\frac{3}{2}}{a^\frac{3}{2}}\bigg)+C$$
If I didn't made any mistake then please help me getting $1.57+\sin^{-1}\left(2x^2-1\right)\equiv\ 2\sin^{-1}x\ $ for $x>0$ without any graphs.
Also, I found that if I substitute $x^3=a^3\sin^{2}x$ then we reach the desired form of the result directly. But I'm an enthusiast to continue my previous solution. Please help.
Thanks!
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Hint: As suggested in the comments, differentiate both sides of the equation.
Edit: If you do this, you obtain $$\left(\frac{x}{a^3-x^3}\right)^{1/2}=\frac db \frac {1}{\sqrt{1-[\left(\frac xa\right)^{3/2}]^2}}\cdot \left(\frac1 a\right)^{3/2}\frac32 x^{1/2}.$$ Hopefully, you can rearrange this to obtain $$\frac{x^{1/2}}{(a^3-x^3)^{1/2}}=\frac{3d}{2b}\cdot \frac{x^{1/2}}{(a^3-x^3)^{1/2}},$$ which implies $$\frac db=\frac23,$$ so that $d=2$ and $b=3$ since $d$ and $b$ are coprime.
|
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|
The logic in radical symplification I'm having troubles while studying radicals, namely with converting expressions with the form $$\sqrt{a+b\sqrt{c}}$$ to $$a+b\sqrt{c}$$ and vice versa. When I'm dealing with these kind of problem, I usually add and subtract $\sqrt{c}^2$. In example: $$\sqrt{27+10\sqrt{2}}$$ I factor ${27+10\sqrt{2}} = 27+10\sqrt{2}+\sqrt{2}^2-2=\left(\sqrt{2}^2+5\sqrt{2}\right)+\left(5\sqrt{2}+25\right)$ ... until I get to $\sqrt{\left(\sqrt{2}+5\right)^2}$ and finally $\sqrt{2}+5$. However, either I'm missing something, or this does not always work. For instance, I cannot solve $\sqrt{113+8\sqrt{7}}$ using this technique.I would like to fully comprehend the logic behind these conversions so that I can apply it indiscriminately.
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For a different approach, let $\alpha = \sqrt{113+8\sqrt{7}}$.
Then $\alpha$ is a root of $$(x^2-113)^2-8^27=(x^2 - 2 x - 111) (x^2 + 2 x - 111)$$ whose roots are $\pm 1 \pm 4 \sqrt 7$. We just have find the correct signs. Since $\alpha$ is positive, $\alpha= \pm 1 + 4 \sqrt 7$. Now $(1 + 4 \sqrt 7)^2=113+8\sqrt{7}$ and so $\alpha= 1 + 4 \sqrt 7$.
In the general case, let $\alpha = \sqrt{a+b\sqrt{c}}$.
Then $\alpha$ is a root of $x^4- 2 a x^2 +a^2 - b^2 c$. The key in this approach is that sometimes this quartic polynomial factors as $(x^2+ux-v)(x^2-ux-v)$. This happens exactly when $u^2+2v=2a$ and $v^2=a^2 - b^2 c$. We want $u,v$ integers and this places restrictions on $a,b,c$.
This approach works for $\alpha = \sqrt{27+10\sqrt{2}}$ because $\alpha$ is a root of $$(x-27)^2-10^2 2 = x^4 - 54 x^2 + 529 = (x^2 - 10 x + 23) (x^2 + 10 x + 23)$$ whose roots are $\pm 5 \pm \sqrt{2}$. As above, this yields $\alpha = 5+\sqrt{2}$.
|
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|
Range of $f(x)= \frac{\tan{x}}{\tan{3x}} $
Prove that for the values of $x$ where the following $f(x)$ is defined, $f(x)$ does not lie between $\frac{1}{3}$ and $3$. $$f(x)=\frac{\tan{x}}{\tan{3x}}$$
My Attempt:
I wrote down, $$\tan{3x}=\frac{3\tan{x}-\tan^3{x}}{1-3\tan^2{x}}$$
This reduced $f(x)$ to,
$$f(x)=\frac{1-3\tan^2{x}}{3-\tan^2{x}}$$
I don't know how to solve any further. I thought of using derivative, but the function is dicontinuous at times. How do I solve it? Any hints would be helpful.
Thanks.
|
\begin{align*}
f(x)&=\frac{1-3\tan^2{x}}{3-\tan^2{x}}\\
&=\frac{1-3\frac{\sin^2{x}}{\cos^2{x}}}{3-\frac{\sin^2{x}}{\cos^2{x}}}\\
&=\frac{\cos^2{x}-3\sin^2{x}}{3\cos^2{x}-\sin^2{x}}\\
&=\frac{\cos^2{x}-3(1-\cos^2{x})}{3\cos^2{x}-(1-\cos^2{x})}\\
&=\frac{4\cos^2{x}-3}{4\cos^2{x}-1}\\
&=\frac{4\cos^2{x}-1}{4\cos^2{x}-1}-\frac{2}{4\cos^2{x}-1}\\
&=1-\frac{2}{4\cos^2{x}-1}
\end{align*}
Now $0\le \cos^2{x}\le1$, so at $\cos^2{x}=0$, $1-\frac{2}{4\cdot0-1}=1-\frac{2}{-1}=3$, and at at $\cos^2{x}=1$ $1-\frac{2}{4\cdot1-1}=1-\frac{2}{3}=\frac{1}{3}$.
To investigate further take the derivative of $f$:
$$f'(x)=-\frac{16 \cos x \sin x}{(4 \cos x^2-1)^2}$$
Now at $\cos^2x=0$, we have $\cos x=0$ and so $f'(x)=0$. Similarly at $\cos^2x=1$, we have $\sin x=0$ and so $f'(x)=0$. Hence these are both stationary points for $f(x)$.
To investigate these stationary points take the scond derivative of $f$:
$$f''(x)=-\frac{256 \cos^2x \sin^2x}{(4 \cos x^2-1)^3}+
\frac{16 (\sin^2x -\cos^2x)}{(4 \cos x^2-1)^2}$$
Now at $\cos^2x=0$, we have $\sin^2x=1$ and so $f''(x)=16$, implying $f(x)=3$ is a local minimum. Similarly at $\cos^2x=1$, we have $\sin^2x=0$ and so $f''(x)=-16$ implying $f(x)=\frac{1}{3}$ is a local maximum. Putting these two results together shows $f(x)$ cannot have any value between $\frac{1}{3}$ or $3$ as required. (See plot of $f(x)$ below for illustration of result.)
|
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|
Find the minimum value of $\frac{a+b+c}{b-a}$ Let $f(x)=ax^2+bx+c$ where $(a<b)$ and $f(x)\geq 0$ $\forall x\in R$.
Find the minimum value of $$\frac{a+b+c}{b-a}$$
If $f(x)\geq 0$ $\forall x\in R$ then $b>a>0$ and $b^2-4ac\leq 0$ implying that $c>0$. After this not able to find way out.
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As you noted, we have $b > a > 0$, and $b^2-4ac\le 0$, hence $c\ge {\large{\frac{b^2}{4a}}}$.
Letting $t=b-a$, we have $t > 0$, and $b=a+t$.
\begin{align*}
\text{Then}\;\;&\frac{a+b+c}{b-a}\\[4pt]
&\ge \frac{a+b+\frac{b^2}{4a}}{b-a}\\[4pt]
&=\frac{(2a+b)^2}{4a(b-a)}\\[4pt]
&=\frac{(3a+t)^2}{4at}\\[4pt]
&=\frac{9a^2+6at+t^2}{4at}\\[4pt]
&=\frac{9a}{4t}+\frac{3}{2}+\frac{t}{4a}\\[4pt]
&=\frac{3}{2}+\left(\frac{9a}{4t}+\frac{t}{4a}\right)\\[4pt]
&\ge \frac{3}{2}+2\sqrt{\frac{9}{16}}\qquad\text{[by $\text{AM-GM}$]}\\[4pt]
&=3\\[4pt]
\end{align*}
so $3$ is a lower bound.
To show that $3$ is realizable, we can use $a=1$, and $t=3$ (which makes the $\text{AM-GM}$ inequality an equality), so $b=a+t=4$, and finally, letting $c={\large{\frac{b^2}{4a}}}=4$, we get
$$\frac{a+b+c}{b-a}=3$$
which gives the minimum possible value.
|
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|
What is the number of connected components of solutions to $y^3 +3xy^2 - x^3 = 1$
What is the number of connected components of solutions to $y^3 +3xy^2 - x^3 = 1$
Attempt
$x^3 - 3y^2x + 1 - y^3 = 0$
$p := -3y^2$, $q := 1 - y^3$
The discriminant is
$Q = (p/3)^3 + (q/2)^2 = (-y^2)^3 + (1 - y^3)^2 / 8 = -y^6 + 1/8 - 2y^3 / 8 + y^6 / 8 = (-7y^6 - 2y^3 + 1/8) / 8$
Hence there exist $y$ such that $Q < 0$. They yield 3 real solutions and therefore 3 connected components.
But we need to prove the components do not intersect.
|
There are three disjoint components to the solution curves.
Consider the disjoint regions formed by the following half lines:
$1)$ $$ \{ (x,0): x\ge 0\}$$
$2)$ $$ \{(x,-x): x\le 0\}$$
$3)$ $$ \{(0,y): y\le 0 \}$$
Each component is contained in one and only one of the rgions.
Thus the three branches of the solution curve are disjoint.
|
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|
Prove by induction $\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}$ In this problem I had to use the method of mathematical induction in order to solve $$\sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}{4}$$
After proving that $P_1$ is true, and writing out $P_k$ I proceeded to solve for $P_{k+1}$. I wrote it out, using $P_k$ to substitute, as $$\frac{k^2(k+1)^2}{4}+\frac{4}{4}(k+1)^3=\frac{(k+1)^2(k+2)^2}{4}$$
I continued expanding on the LHS:=
$$\frac{k^2(k^2+2k+1)+(4k+4)^3}{4}=$$
$$\frac{k^4+2k^3+k^2+(4k+4)(16k^2+32k+16}{4}=$$
$$\frac{k^4+68k^3+192k^2+192k+64}{4}=$$
And on the RHS I did the same:
$$=\frac{(k+1)^2(k+2)^2}{4}$$
$$=\frac{(k^2+2k+1)(k^2+4k+4)}{4}$$
When I make the LHS=RHS I get equations that arent equal to each other.
$$\frac{k^4+68k^3+192k^2+192k+64}{4}=\frac{(k^2+2k+1)(k^2+4k+4)}{4}$$
Any ideas?
EDIT: As pointed out by LordSharktheUnknown, $4(k+1)^3≠(4k+4)^3$, currently working on it to see if this will resolve the problem.
|
We have that
$$\sum_{i=1}^{n+1} i^3=\frac{n^2(n+1)^2}{4}+\frac{4}{4}(n+1)^3=\frac{(n+1)^2(n^2+4n+4)}{4}=\frac{(n+1)^2(n+2)^2}{4}$$
|
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|
Prove the equation $\left(2x^2+1\right)\left(2y^2+1\right)=4z^2+1$ has no solution in the positive integers Prove the equation
$$\left(2x^2+1\right)\left(2y^2+1\right)=4z^2+1$$
has no solution in the positive integers
My work:
1) I have the usually problem
$$\left(nx^2+1\right)\left(my^2+1\right)=(m+n)z^2+1$$
in the positive integers. Initially I use case $\gcd(m,n)=1$
2) Let $m=n=2$. It is this case. I need to prove that $(x^2-y^2)^2+(x^2+y^2) $ is not perfect square for any $x,y$
|
The question is concerned only with $x,y,z\ge 1$. $x=y$ does not give a solution (see comments). WLOG assume $x>y$. Look at the equation $\mod x$.
$1\cdot (2y^2+1)\equiv 4z^2+1\mod x$.
Let $z\equiv a\mod x$. Then $2y^2+1=4(nx+a)^2+1$.
Subtracting $1$ from each side and dividing by $2$ we get $y^2=2(nx+a)^2\Rightarrow 2=\frac{y^2}{(nx+a)^2}\ $ which would make $\sqrt{2}$ rational.
Hence there are no positive integers that satisfy the equation.
|
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|
Find min of $P = \dfrac{1}{(a-b)^2} + \dfrac{1}{(b-c)^2} + \dfrac{1}{(c-a)^2}$ Let $a, b, c \in \mathbb{R}^+$ such that $a^2 + b^2 + c^2 = 3$. Find the minimum value of $P = \dfrac{1}{(a-b)^2} + \dfrac{1}{(b-c)^2} + \dfrac{1}{(c-a)^2}$?
In fact, we have that $P \geq \dfrac{4}{ab +bc +ca}$. However, the equality can not occur in this exercise because $a >0, b > 0$ and $c> 0$. Would you please send me a hint for this problem? Thank you so much!
|
Let $D = \{ (a,b,c) \in (0,\infty)^3 : a^2 + b^2 + c^2 = 3 \}$ and $\bar{D}$ be its closure. Instead of looking for a minimum of $P = \frac{1}{(a-b)^2} + \frac{1}{(b-c)^2} + \frac{1}{(c-a)^2}$ over $D$, we need to look at the minumum over the larger $\bar{D}$.
This is because $P$ doesn't achieve any local minimum over $D$.
Assume the contrary, let's say $P$ achieve a local minimum at $(a,b,c) \in D$. WOLOG, we can assume $0 < a < b < c$. For this to happen, the methods of Lagrange multipliers tell us there is a $\lambda \in \mathbb{R}$ such that
$$
\begin{align}
\frac{1}{(a-b)^3} + \frac{1}{(a-c)^3} &= \lambda a\\
\frac{1}{(b-a)^3} + \frac{1}{(b-c)^3} &= \lambda b\\
\frac{1}{(c-a)^3} + \frac{1}{(c-b)^3} &= \lambda c\\
\end{align}
$$
Summing these 3 equations together give us $0 = \lambda (a + b + c)$. Since $a + b + c \ne 0$, this forces $\lambda = 0$. It is easy to see it is impossible
for the first equation to work.
Next, let us switch to the problem of finding the minimum over $\bar{D}$.
Since $\bar{D}$ is compact and $P : \bar{D} \to \mathbb{R} \cup \{ +\infty \}$ is continuous and bounded from below, $P$ reaches an absolute minimum somewhere in $\bar{D}$. This minimum cannot lie in $D$. At least one of $a,b,c$ need to vanish. However, we can't have more than one to vanish or $P$ will blow up. This means for the minimum, one and only one of $a,b,c$ is $0$. WOLOG, we will assume
$0 = a < b < c$.
Let $t = \frac{b}{c} \in (0,1)$ and $u = t+t^{-1} \in (2,\infty)$, the target function $P$ becomes
$$\begin{align}P(u) = \frac{1}{b^2} + \frac{1}{(b-c)^2} + \frac{1}{c^2}
&= \frac13\frac{b^2+c^2}{c^2}\left(\frac{1}{t^2} + \frac{1}{(1-t)^2} + 1\right)\\
&= \frac13(t + t^{-1})\left(t^{-1} + \frac{1}{t+t^{-1}-2} + t\right)\\
&= \frac{u}{3}\left(u + \frac{1}{u-2}\right)
\end{align}
\tag{*1}
$$
Let $u_{min}$ be the $u$ corresponds to the absolute minimum.
When $u = u_{min}$, we have
$$P'(u) = \frac23\frac{(u-1)(u^2-3u+1)}{(u-2)^2} = 0
\quad\iff\quad u = 1, \frac{3\pm \sqrt{5}}{2}$$
Since $u_{min} > 2$, the last root is the only root relevant to us.
This means $u_{min} = \frac{3 + \sqrt{5}}{2}$ and for any admissible $u$, we have the inequality:
$$P = P(u) \ge P(u_{min}) = \frac{11+5\sqrt{5}}{6} \approx 3.696723314583158$$
The minimum value of $P$ over $\bar{D}$ (and the infimum over $D$) is $\frac{11+5\sqrt{5}}{6}$.
|
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"url": "https://math.stackexchange.com/questions/2848219",
"timestamp": "2023-03-29T00:00:00",
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|
Is $\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=1$?
Evaluate $$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}$$
My attempt:
$$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{-\sqrt{4x^2}}=\frac{-5}{2}$$ According to the answer key, it actually equals $1$.
Thanks in advance.
|
Note that $\sqrt {4x^2} = 2|x|$ and in this case $|x|=-x$, thus $$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{3x-\sqrt{4x^2}}= \lim_{x\to-\infty} \frac{5x}{3x+2x }=1$$
|
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"url": "https://math.stackexchange.com/questions/2849212",
"timestamp": "2023-03-29T00:00:00",
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|
Let $V_m(r)=\operatorname{vol}B(0,r)$ and prove $V_{n+1}(r) = 2V_n(1)\int_0^r(r^2-t^2)^{n/2}\,\mathrm dt$
Let $V_{m}(r)$ the volume of the ball with center in origin and radius $r$ in $\mathbb{R}^{m}$. Prove the inductive relation $V_{n+1}(r) = 2V_{n}(1)\int_{0}^{r}(r^{2}-t^{2})^{n/2}\,\mathrm dt$ and conclude that
$$V_{m}(r) = \frac{r^{m}\pi^{m/2}}{\left(\frac{m}{2}\right)!}$$
when $m$ is even and
$$V_{m}(r) = r^{m}\pi^{(m-1)/2}\frac{2^{m}\left(\frac{m-1}{2}\right)!}{m!}$$
if $m$ is odd. Note that of these formulas results a curious fact:
$$\lim_{m \to \infty}V_{m}(r) = 0.$$
For $V_{n+1}(r) = 2V_{n}(1)\int_{0}^{r}(r^{2}-t^{2})^{n/2}\,\mathrm dt$, I checked the case $n=1$. For general case, I think I should apply a change of variables, but I could not do it correctly. Can anybody help me? Please, only with this equality because I want to try to do the other parts on my own.
Edit. I try to use the Oliver hint, but I could not complete the proof.
|
$\def\peq{\mathrel{\phantom{=}}{}}\def\d{\mathrm{d}}$For any $n \geqslant 1$ and $(x_1, \cdots, x_{n + 1}) \in \mathbb{R}^{n + 1}$, note that\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}} (x_1, \cdots, x_{n + 1}) \in B_n(0, r)\\
&\Longleftrightarrow \sum_{k = 1}^{n + 1} x_k^2 < r^2 \Longleftrightarrow -r < x_{n + 1} < r \text{ and } \sum_{k = 1}^n x_k^2 < r^2 - x_{n + 1}^2\\
&\Longleftrightarrow -r < x_{n + 1} < r \text{ and } (x_1, \cdots, x_n) \in B_n(0, \sqrt{\smash[b]{r^2 - x_{n + 1}^2}})
\end{align*}
and $V_n(r) = r^n V_n(1)$, thus\begin{align*}
V_{n + 1}(r) &= \int\limits_{B_{n + 1}(0, r)} \d x = \int_{-r}^r \d x_{n + 1} \int\limits_{B_n(0, \sqrt{\smash[b]{r^2 - x_{n + 1}^2}})} \d x\\
&= \int_{-r}^r V_n(\sqrt{\smash[b]{r^2 - x_{n + 1}^2}}) \,\d x_{n + 1} = \int_{-r}^r (r^2 - x_{n + 1}^2)^{\frac{n}{2}} V_n(1) \,\d x_{n + 1}\\
&= V_n(1) \int_{-r}^r (r^2 - x_{n + 1}^2)^{\frac{n}{2}} \,\d x_{n + 1} = 2V_n(1) \int_0^r (r^2 - t^2)^{\frac{n}{2}} \,\d t.
\end{align*}
Now for any $n \geqslant 1$,$$
I_n := \int_0^1 (1 - t^2)^{\frac{n}{2}} \,\d t = \int_0^{\tfrac{π}{2}} (1 - \sin^2 θ)^{\frac{n}{2}} \,\d(\sin θ) = \int_0^{\tfrac{π}{2}} \cos^{n + 1} θ \,\d θ,
$$
and $I_1 = \dfrac{π}{4}$, $I_2 = \dfrac{2}{3}$. For $n \geqslant 3$, because\begin{align*}
I_n &= \int_0^{\tfrac{π}{2}} \cos^{n + 1} θ \,\d θ = \int_0^{\tfrac{π}{2}} \cos^n θ \,\d(\sin θ)\\
&= \cos^n θ \sin θ \biggr|_0^{\tfrac{π}{2}} - \int_0^{\tfrac{π}{2}} \sin θ \,\d(\cos^n θ)\\
&= n \int_0^{\tfrac{π}{2}} \sin^2 θ\cos^{n - 1} θ \,\d θ = n \int_0^{\tfrac{π}{2}} (1 - \cos^2 θ) \cos^{n - 1} θ \,\d θ\\
&= n(I_{n - 2} - I_n),
\end{align*}
then $I_n = \dfrac{n}{n + 1} I_{n - 2}$. Thus for $k \geqslant 1$,\begin{gather*}
I_{2k - 1} = I_1 \prod_{j = 2}^k \frac{2j - 1}{2j} = \frac{π}{2} \cdot \frac{(2k - 1)!!}{(2k)!!},\quad I_{2k} = I_2 \prod_{j = 2}^k \frac{2j}{2j + 1} = \frac{(2k)!!}{(2k + 1)!!}.
\end{gather*}
Therefore for $k \geqslant 1$,\begin{align*}
V_{2k - 1}(1) &= 2V_{2k - 2}(1) I_{2k - 2} = \cdots = 2^{2k - 2} V_1(1) \prod_{j = 1}^{2k - 2} I_j\\
&= 2^{2k - 1} \cdot \left( \frac{π}{2} \right)^{k - 1} \frac{1}{(2k - 1)!!} = π^k \cdot \frac{2^{2k - 1} (k - 1)!}{(2k - 1)!},
\end{align*}\begin{align*}
V_{2k}(1) &= 2V_{2k - 1}(1) I_{2k - 1} = \cdots = 2^{2k - 1} V_1(1) \prod_{j = 1}^{2k - 1} I_j\\
&= 2^{2k} \cdot \left( \frac{π}{2} \right)^k \frac{1}{(2k)!!} = \frac{π^k}{k!},
\end{align*}
and$$
V_{2k - 1}(r) = r^{2k - 1} V_{2k - 1}(1) = π^k r^{2k - 1} \cdot \frac{2^{2k - 1} (k - 1)!}{(2k - 1)!},\\
V_{2k}(r) = r^{2k} V_{2k}(1) = \frac{π^k r^{2k}}{k!}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?
|
Notice
$$\begin{align}
a + b + \frac1a + \frac1b
= &\; \left(a + \frac{1}{2a} + \frac{1}{2a}\right)
+ \left(b + \frac{1}{2b} + \frac{1}{2b}\right)\\
\\
\color{blue}{\rm AM \ge \rm GM \rightarrow\quad}
\ge &\; 3\left[\left(\frac{1}{4a}\right)^{1/3} + \left(\frac{1}{4b}\right)^{1/3}\right]\\
\color{blue}{\rm AM \ge \rm GM \rightarrow\quad}
\ge &\; 6 \left(\frac{1}{16ab}\right)^{1/6}\\
\color{blue}{a^2 + b^2 \ge 2 ab \rightarrow\quad}
\ge &\; 6 \left(\frac{1}{8(a^2+b^2)}\right)^{1/6}\\
= &\; 6 \left(\frac18\right)^{1/6} = \frac{6}{\sqrt{2}} = 3\sqrt{2}
\end{align}
$$
Since the value $3\sqrt{2}$ is achieved at $a = b = \frac{1}{\sqrt{2}}$, we have
$$\min \left\{ a + b + \frac1a + \frac1b : a, b > 0, a^2+b^2 = 1 \right\} = 3\sqrt{2}$$
Notes
About the question how do I come up with this. I basically know the minimum is
achieved at $a = b = \frac{1}{\sqrt{2}}$. Since the bound $3\sqrt{2}$ on RHS of is optimal, if we ever want to prove the inequality, we need to use something that is tight when $a = b$. If we want to use AM $\ge$ GM, we need to arrange the pieces so that all terms are equal. That's why I split $\frac1a$ into $\frac{1}{2a} + \frac{1}{2a}$ and $\frac1b$ into $\frac{1}{2b} + \frac{1}{2b}$ and see what happens. It just turns out that works.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\tan(a) = 3/4$ and $\tan (b) = 5/12$, what is $\cos(a+b)$ It is known that
$$\tan(a) = \frac{3}{4}, \:\:\: \tan(b) = \frac{5}{12} $$
with $a,b < \frac{\pi}{2}$.
What is $\cos(a+b)$?
Attempt :
$$ \cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b) $$
And we can write $\tan(a) = \sin(a)/\cos(a) = 0.3/0.4 $
and $ \sin(b)/\cos(b) = 0.05/0.12 $
so
$$ \cos(a+b) = (0.4)(0.12) - (0.3) (0.05) = 33/1000$$
Is this correct? Thanks.
|
$$ \tan(a+b)= \frac {\tan a +\tan b}{1-\tan a \tan b} = \frac {3/4 +5/12}{1-(3/4)(5/12)}= \frac {56}{33}$$
$$ \sec^2(a+b)=1+\tan^2(a+b)=\frac {4225}{1089}=(\frac {65}{33})^2$$
$$\cos(a+b)= 33/65$$
|
{
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"url": "https://math.stackexchange.com/questions/2854021",
"timestamp": "2023-03-29T00:00:00",
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|
Positive integer solutions to $a^3 + b^3 = c^4$ Let $n$ and $m$ be positive integers. We know that
$n^3 + m^3 = n^3 + m^3$
Multiply both sides by $(n^3 + m^3)^3$; on the LHS you distribute, and on the RHS you use power addition rule;
$(n^4 + nm^3)^3 + (m^4 + mn^3)^3 = (n^3 + m^3)^4$
So we get an infinite array of solutions with
$a = n^4 + nm^3$
$b = m^4 + mn^3$
$c = n^3 + m^3$
Are there any solutions to the equation (in the title) that cannot be expressed in this form?
|
$$793^3+854^3=183^4$$ and $$183$$ is not the sum of two cubes.
|
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|
Verifying the limit of the expressions I have calculated the limit of the following expressions as follows:
$$\lim_{x \rightarrow 0} \frac{x \times \frac{2}{x}}{5x + 3 + e^{-2x}} = \lim_{x \rightarrow 0} \frac{2x}{5x^2 + 3x + xe^{-2x}} = \lim_{x \rightarrow 0} \frac{2}{10x + 3 + e^{-2x} -2xe^{-2x}} (\text{using Le Chatelier's principle}) = \frac{2}{3+1}$$
$$\lim_{x \rightarrow 0} \frac{x \times \frac{2}{x}}{4 + \frac{1-x}{1+x}} = \lim_{x \rightarrow 0} \frac{2x(1+x)}{4x(1+x) + x(1-x)} = \lim_{x \rightarrow 0} \frac{2x(1+x)}{5x-3x^2} = \lim_{x \rightarrow 0} \frac{4x+2}{5 - 6x } (\text{using Le Chatelier's principle}) = \frac{2}{5}$$
Are my calculations correct?
|
As suggested by Bernard in the above comment, more simply we have
$$\lim_{x \rightarrow 0} \frac{\color{red}x \times \frac{2}{\color{red}x}}{5x + 3 + e^{-2x}} = \lim_{x \rightarrow 0} \frac{2}{5x + 3 + e^{-2x}} \stackrel{\text{by continuity}}= \frac2{3+1}=\frac12$$
and similarly for the second we have
$$\lim_{x \rightarrow 0} \frac{\color{red}x \times \frac{2}{\color{red}x}}{4 + \frac{1-x}{1+x}} =\lim_{x \rightarrow 0} \frac{2}{4 + \frac{1-x}{1+x}} \stackrel{\text{by continuity}}= \frac2{4+1}=\frac25$$
Refer also to: Why are we allowed to cancel fractions in limits?
|
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|
Prove that $\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac{\pi}{3}$ I was trying to solve the integral $$\int_{0}^{\infty}\frac{x^4}{1+x^6}dx$$
using series. Now I'm stuck at the series below.
How to prove that
$$\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac{\pi}{3}?$$
|
Denote
$$
S = \sum_{j=0}^{\infty}(-1)^j \left( \frac{1}{6j+1} + \dfrac{1}{6j+5}\right).
$$
And denote
$$
T= \sum_{j=0}^{\infty} (-1)^j \dfrac{1}{2j+1}.
$$
According to Leibniz formula for $\pi$, $\;T=\dfrac{\pi}{4}$.
Then
$$S - \dfrac{1}{3}T = \\
S - \dfrac{1}{3}\sum_{j=0}^{\infty} (-1)^j \dfrac{1}{2j+1} = \\ S-\sum_{j=0}^{\infty}(-1)^{j}\left(\dfrac{1}{6j+3}\right) = \\
\sum_{j=0}^{\infty}(-1)^j \left( \frac{1}{6j+1} - \frac{1}{6j+3} + \dfrac{1}{6j+5}\right) = \\
\sum_{k=0}^{\infty} (-1)^k \dfrac{1}{2k+1} = \\ T.
$$
Therefore $$S-\dfrac{1}{3}T = T,$$
$$S=\dfrac{4}{3}T,$$
$$S=\dfrac{4}{3}\cdot \dfrac{\pi}{4}=\dfrac{\pi}{3}.$$
|
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|
Double fractional part integral Let $\{\}$ denote the fractional part, does the following integral have a closed form ?
$$\int_{0}^{1}\int_{0}^{1}\bigg\{\frac{1}{x\,y}\bigg\}^2dx\,dy$$
|
This is an incomplete answer that only addresses the 1-dimensional case.
We split the integral into continuous pieces:
$$\begin{align}
\int_0^1 \left\{ \frac{1}{x} \right\}^2 \mathrm{d}x
&= \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} \left\{ \frac{1}{x} \right\}^2 \mathrm{d}x \\\\
&= \sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac{1}{n}} \left( \frac{1}{x} - n\right)^2 \mathrm{d}x \\\\
&= \sum_{n=1}^{\infty} \left( n^2x - 2n \ln |x| - \frac{1}{x} \right)\biggr\rvert_{\frac{1}{n+1}}^{\frac{1}{n}} \\\\
&= \sum_{n=1}^{\infty} \left( n - 2n \ln \frac{1}{n} - n - \frac{n^2}{n+1} + 2n \ln \frac{1}{n+1} + n + 1 \right) \\\\
&= \sum_{n=1}^{\infty} \left( 2n \ln \frac{n}{n+1} + \frac{2n + 1}{n+1} \right)
\end{align}$$
With the aid of computer algebra, we obtain that this series converges to $$\ln (2\pi) - \gamma - 1$$ where $\gamma$ is the Euler-Mascheroni constant.
|
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|
Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi}{20}$$
and that,
$$\cos\frac{\pi}{20}-\cos\frac{3\pi}{20}=-2\sin\frac{2\pi}{20}\sin\frac{\pi}{20}$$
So,
$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}-\sin\frac{\pi}{20})=\frac{\sqrt2}{2}=\sin\frac{5\pi}{20}$$
Unfortunately, I can't find a way to continue this, any ideas or different ways of proof?
*Taken out of the TAU entry exams (no solutions are offered)
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Since $\surd2\sin\frac14\pi=\surd2\cos\frac14\pi=1$, we may write the difference as$$\surd2\left(\cos\frac\pi4\cos\frac\pi{20}+\sin\frac\pi4\sin\frac\pi{20}\right)-\surd2\left(\cos\frac\pi4\cos\frac{3\pi}{20}-\sin\frac\pi4\sin\frac{3\pi}{20}\right)-\frac{\surd2}2$$$$=\frac{\surd2}2\left(2\cos\frac\pi5-2\cos\frac{2\pi}5-1\right)\qquad\qquad\qquad\qquad\qquad$$$$=\frac{\surd2}2\left(2\cos\frac\pi5\sin\frac\pi5-2\cos\frac{2\pi}5\sin\frac\pi5-\sin\frac\pi5\right)/\sin\frac\pi5$$$$=\frac{\surd2}2\left(\sin\frac{2\pi}5+\sin\frac\pi5-\sin\frac{3\pi}5-\sin\frac\pi5\right)/\sin\frac\pi5\qquad$$$$=0,$$since $\sin\frac35\pi=\sin(\pi-\frac35\pi)=\sin\frac25\pi$.
|
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|
Prove the limit below by using epsilon-delta definition Prove $$\lim_{z\rightarrow4+3i}\overline{z}^2 =(4-3i)^2$$ by using epsilon-delta definition.
My working:
Assume $|z-(4-3i)|<1 \implies |z|<6$
\begin{align}
|\overline{z}^2-(4-3i)^2| &=|[\overline{z}-(4-3i)][\overline{z}+(4-3i)]|\\
&=|\overline{z}-(4-3i)||\overline{z}+(4-3i)|\\
&=|\overline{z}-(4+3i)+6i||\overline{z}-(4+3i)+8|\\
&=|\overline{z}-(4+3i)|\left[\left|1+\frac{6i}{\overline{z}-(4+3i)}\right|\left|1+\frac{8}{\overline{z}-(4+3i)}\right|\right]\\
&\le \delta \left[\left(1+\frac{|6i|}{|\overline{z}-(4+3i)|}\right)\left(1+\frac{8}{|\overline{z}-(4+3i)|}\right)\right]\\
&< 63\delta
\end{align}
Is this correct?
|
Careful, you can assume $|z - (4+3i)| < \delta$, not $|\overline{z} - (4+3i)| < \delta$.
If we pick $0 < \delta < -5+\sqrt{25+\varepsilon}$ we have
\begin{align}
\left|\overline{z}^2 - (4-3i)^2\right| &= |\overline{z} - (4-3i)|\cdot |\overline{z} + (4-3i)|\\
&= \left|\overline{z - (4+3i)}\right|\cdot \left|\overline{z + (4+3i)}\right|\\
&= |z - (4+3i)|\cdot |z + (4+3i)|\\
&= |z - (4+3i)|\cdot |z - (4+3i) + (8+6i)|\\
&\le |z - (4+3i)|\cdot \big(|z - (4+3i)| + |8+6i|\big)\\
&< \delta(\delta + 10)\\
&< \varepsilon
\end{align}
|
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|
Verifying that $\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx$ without Cauchy-Schwarz
For real numbers $a,b,p$, verify that $$\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx\tag{1}$$ without using the Cauchy-Schwarz inequality.
Note that this is obviously an immediate consequence of C-S, and I assumed that it had to be shown by integrating directly.
Working backwards, we get $$\left[\frac{b^2-a^2}2-p(b-a)\right]^2\le(b-a)\frac{(b-p)^3-(a-p)^3}3$$ and since $m^3-n^3=(m-n)(m^2+mn+n^2)$, the RHS can be rewritten as $$\frac{(b-a)^2}4(b+a-2p)^2\le\frac{(b-a)^2}3(b^2+ab+b^2+3p^2-3p(b+a))$$ or $$3(b+a-2p)^2\le4((b+a)^2+3p^2-3p(b+a))-4ab$$ which gives $$12p^2-12p(b+a)\le(b-a)^2+12p^2-12p(b+a)$$ or that $$(b-a)^2\ge0$$ which is true.
Any alternative methods?
|
Do the substitution $t=x-p$, so the inequality becomes
$$
\left(\int_{a-p}^{b-p} t\,dt\right)^{\!2}\le (b-a)\int_{a-p}^{b-p} t^2\,dt
$$
Setting $a-p=A$ and $b-p=B$, this is the same as proving that
$$
\left(\int_{A}^{B} t\,dt\right)^{\!2}\le (B-A)\int_{A}^{B} t^2\,dt
$$
that is,
$$
\frac{(B^2-A^2)^2}{4}\le (B-A)\frac{B^3-A^3}{3}
$$
that becomes
$$
3(B+A)^2\le 4(B^2+AB+A^2)
$$
This becomes
$$
A^2-2AB+B^2\ge0
$$
which is true.
|
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"url": "https://math.stackexchange.com/questions/2862491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
The probability density function of a random variable is given by $f_X(x) = \frac{3}{208}x^2$ Find $E(X)$ The probability density function of a random variable is given by $$f_X(x) = \frac{3}{208}x^2$$ whenever $2 < x < 6$. $0$ otherwise
$(a)$ Compute $E(X)$
$(b)$ Compute $E[X(X+1)]$
My attempt
$(a)$
$$E(X) = \int_{2}^{6} x\frac{3}{208}x^2dx = \frac{3}{208}\cdot\frac{1}{4}(x^4)\bigg|_{2}^{6} = 60/13$$
$(b)$
$$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$
$$E(X^2) = \int_{2}^{6}x^2 \frac{3}{208}x^2 dx = \frac{3}{208}\frac{1}{5}(x^5)\bigg|_{2}^{6} = 1452/65$$
Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 \approx 26.9538 $$
Would this be right?
|
The probability density function of a random variable is given by $$f_X(x) = \frac{3}{208}x^2$$ whenever $2 < x < 6$. $0$ otherwise
$(a)$ Compute $E(X)$
$(b)$ Compute $E[X(X+1)]$
$(a)$
$$E(X) = \int_{2}^{6} x\frac{3}{208}x^2dx = \frac{3}{208}\cdot\frac{1}{4}(x^4)\bigg|_{2}^{6} = 60/13$$
$(b)$
$$E(X(X+1)) = E(X^2 + X) = E(X^2) + E(X) $$
$$E(X^2) = \int_{2}^{6}x^2 \frac{3}{208}x^2 dx = \frac{3}{208}\frac{1}{5}(x^5)\bigg|_{2}^{6} = 1452/65$$
Thus, $$E(X^2) + E(X) = 1452/65 + 60/13 \approx 26.9538 $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\int _0^\infty\,\frac{\ln x}{(1+x^2)^2}\,\text{d}x$ using the Residue Theorem To evaluate this integral
$$\int\limits_0^\infty\frac{\ln x}{(1+x^2)^2} \text{d}x\,,$$
Let us consider the following function.
$$f(z)=\frac{\ln^2 z}{(1+z^2)^2}=\frac{\ln^2 z}{(z+i)^2(z-i)^2}\,.$$
The integral over $C_R$ and $C_\epsilon$ vanish as $R$ approaches $\infty$ and $\epsilon$ approaches $0$. So we only need to evaluate the residues
$$\oint\limits_C f(z)dz = 2\pi i \sum \text{Res}\big(f(z)\big)\,.$$
The poles are $\pm i$ and of second order. Here is our chosen contour.
Let us recall the residue of an $m$-th order of pole is
$$\text{Res}[f(z);z_0]=\frac{1}{(m-1)!}\lim_{z\to z_0} \frac{\text{d}^{m-1}}{\text{d}z^{m-1}}\,(z-z_0)^m\,f(z)\,.$$
Consequently the residues are
$$\text{Res}[f(z);i]=\lim_{z \to i}\Bigg(\frac{2\ln z}{z(z+i)^2}-\frac{2\ln^2 z}{(z+i)^3}\Bigg)=\frac{2\cdot \left(\frac{i\pi}{2}\right)}{-4i}-\frac{2\cdot (-\frac{\pi^2}{4})}{-8i}=-\frac{\pi}{4}+i\frac{\pi^2}{16}$$
As for the other pole
$$\text{Res}[f(z);-i]=\lim_{z \to -i}\Bigg(\frac{2\ln z}{z(z-i)^2}-\frac{2\ln^2 z}{(z-i)^3}\Bigg)=\frac{2\cdot \left(\frac{i3\pi}{2}\right)}{4i}-\frac{2\cdot \left(-\frac{9\pi^2}{4}\right)}{8i}=\frac{3\pi}{4}-i\frac{9\pi^2}{16}\,.$$
Adding these two together and remembering the residue theorem, we have
$$\oint\limits_C f(z)\,\text{d}z = 2\pi i \cdot \left(\frac{\pi}{2}-i\frac{\pi^2}{2}\right)=i\pi^2+\pi^3$$
On the other hand, for the integrals over $C_+$ and $C_-$, $z=x$ and $z=xe^{2\pi i}$ respectively. Hence, we can write
$$\int\limits_\epsilon^R\frac{\ln^2 x}{(1+x^2)^2} \,\text{d}x + \int\limits_R^\epsilon \frac{(\ln x +2\pi i)^2}{(1+x^2)^2} \,\text{d}x=i\pi^2+\pi^3\,.$$
If we let $R \to \infty$ and $\epsilon \to 0$ and change the direction of the second integral, we get
$$-4\pi i\int\limits_0^\infty\frac{\ln x}{(1+x^2)^2}\,\text{d}x + 4\pi^2\int\limits_0^\infty\frac{\text{d}x}{(1+x^2)^2}=i\pi^2+\pi^3\,.$$
Comparing the real and the imaginary parts
$$\int\limits_0^\infty\frac{\ln x}{(1+x^2)^2}\,\text{d}x = -\frac{\pi}{4}$$
$$\int\limits_0^\infty\frac{\text{d}x}{(1+x^2)^2} = \frac{\pi}{4}$$
|
Let $f(z):=\dfrac{\big(\ln(z)\big)^2}{\left(1+z^2\right)^2}$ for $z\in\mathbb{C}\setminus\mathbb{R}_{\leq 0}$ (i.e., the chosen branch cut of $\ln(z)$ is the negative real line). For a real number $\epsilon$ such that $0<\epsilon<1$, define $\gamma_\epsilon$ to be the contour
$$\left[\epsilon,\frac{1}{\epsilon}\right]\cup A_\epsilon \cup \left[\frac{1}{\epsilon}\,\exp\big(\text{i}(\pi-\epsilon)\big),\epsilon\,\exp\big(\text{i}(\pi-\epsilon)\big)\right]\cup A'_\epsilon\,,$$
where $A_\epsilon$ is the arc in the upper half-plane (namely, the set of complex numbers with nonnegative imaginary parts) of the circle centered at $0$ with radius $\frac{1}{\epsilon}$ starting from $\frac{1}{\epsilon}$ to $\frac{1}{\epsilon}\,\exp\big(\text{i}(\pi-\epsilon)\big)$ (i.e., in the counterclockwise direction), and $A'_\epsilon$ is the arc in the upper half-plane of the circle centered at $0$ with radius $\epsilon$ starting from $\epsilon\,\exp\big(\text{i}(\pi-\epsilon)\big)$ to $\epsilon$ (i.e., in the clockwise direction).
We note that
$$\lim_{\epsilon\to0^+}\,\oint_{\gamma_\epsilon}\,f(z)\,\text{d}z=2\pi\text{i}\,\text{Res}_{z=\text{i}}\big(f(z)\big)=2\pi\text{i}\,\left(-\frac{\pi}{4}+\frac{\pi^2\text{i}}{16}\right)=-\frac{\pi^3}{8}-\frac{\pi^2\text{i}}{2}\,.$$
Furthermore,
$$\lim_{\epsilon\to0^+}\,\oint_{\gamma_\epsilon}\,f(z)\,\text{d}z=2\pi\text{i}\,\int_0^\infty\,\frac{\ln(x)}{\left(1+x^2\right)^2}\,\text{d}x+\,\int_0^\infty\,\frac{2\big(\ln(x)\big)^2-\pi^2}{\left(1+x^2\right)^2}\,\text{d}x\,.$$
Consequently,
$$\int_0^\infty\,\frac{\ln(x)}{\left(1+x^2\right)^2}\,\text{d}x=\frac{1}{2\pi\text{i}}\left(-\frac{\pi^2\text{i}}{2}\right)=-\frac{\pi}{4}\,.$$
In fact, with the same contour, we can see that
$$\int_0^\infty\,\frac{1}{\left(1+x^2\right)^2}\,\text{d}x=\pi\text{i}\,\text{Res}_{z=\text{i}}\left(\frac{1}{\left(1+z^2\right)^2}\right)=\pi\text{i}\,\left(-\frac{\text{i}}{4}\right)=\frac{\pi}{4}\,.$$
Therefore,
$$\int_0^\infty\,\frac{\big(\ln(x)\big)^2}{\left(1+x^2\right)^2}\,\text{d}x=\frac{\pi^2}{2}\,\int_0^\infty\,\frac{1}{\left(1+x^2\right)^2}\,\text{d}x-\frac{\pi^3}{16}=\frac{\pi^3}{8}-\frac{\pi^3}{16}=\frac{\pi^3}{16}\,.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given a matrix $A$ find $A^n$. $A=$$
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} $
Find $A^n$.
My input:
$A^2= \begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix} = \begin{bmatrix}
1 & 4\\
0 & 1
\end{bmatrix} $
$A^3 = \begin{bmatrix}
1 & 6\\
0 & 1
\end{bmatrix} $
......
$A^n = \begin{bmatrix}
1 & 2n\\
0 & 1
\end{bmatrix} $
This was very basic approach. I want to know if there is any other way a smart trick or something to solve this problem ?
|
You can use this too
$$A=\begin{pmatrix}1&2\\0&1\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}0&2\\0&0\end{pmatrix}$$
$$A=I_2+B$$
And B is a nilpotent matrix $\implies B^2=0$
$$A^n=(I_2+B)^n$$
Use binomial theorem
|
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|
Evaluate $\int_0^{\frac{\pi}{2}}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx$
Evaluate
$$
\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx
$$
I used the substitution $\sin x =t$, then I got the integral as $$\int_0^1 \frac{t}{2t^4-2t^2+1}dt $$
After that I don't know how to proceed. Please help me with this.
|
Perform the change of variable $y=\sin^2 x$,
$\begin{align}J&=\int_0^{\frac{\pi}{2}}\frac{\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)}dx\\
&=\frac{1}{2}\int_0^1 \frac{1}{x^2+(1-x)^2}\,dx\\
&=\frac{1}{2}\int_0^1 \frac{1}{2x^2-2x+1}\,dx\\
&=\int_0^1 \frac{1}{(2x-1)^2+1}\,dx\\
\end{align}$
Perform the change of variable $y=2x-1$,
$\begin{align}
J&=\frac{1}{2}\int_{-1}^1 \frac{1}{x^2+1}\,dx\\
&=\frac{1}{2}\Big[\arctan x\Big]_{-1}^1\\
&=\frac{1}{2}\times\frac{\pi}{2}\\
&=\boxed{\frac{\pi}{4}}
\end{align}$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
The product of three consecutive integers is ...? Odd? Divisible by $4$? by $5$? by $6$? by $12$? If i have the product of three consecutive integers:
$n(n+1)(n+2)$, so the result is:
$A)$ Odd
$B)$ Divisible by $4$
$C)$ Divisible by $5$
$D)$ Divisible by $6$
$E)$ Divisible by $12$
My thought was:
$i)$ If we have three consecutive numbers, $a, (a + 1), (a + 2)$, one of these three numbers must be divisible by $3$.
$ii)$ If we have two consecutive numbers, $a, (a + 1)$, one of these two numbers must be divisible by $2$, also one of these numbers will be even and the other will be odd.
$iii)$ Any number is divisible by $6$, when is divisible by $3$ and $2$ at the same time.
So, the correct answer must be $D)$
Well, I would like to know if:
*
*My answer is correct
*What is the formal proof of what I said in $ i) $
|
To prove your statement $i)$ you can use the technique of induction.
First consider $a=1$ therefore the product becomes
$$1\cdot(1+1)\cdot(1+2)=1\cdot2\cdot3$$
which is clearly divisible by $3$ since one of the factors is $3$ itself. Next, by setting $a=n$ and $a=n+1$, we get
$$\begin{align}
n(n+1)(n+2)&\equiv0\mod(3)\\
(n+1)(n+2)(n+3)&\equiv0\mod(3)
\end{align}$$
To derive the $n+1$-statement from the truth of the $n$-statement just split up the last factor which yields to
$$(n+1)(n+2)(n+3)=n(n+1)(n+2)+3(n+1)(n+2)$$
where the first term is the given hypotesis and the second one has a $3$ in it and so your $i)$ is right.
For the overall answer for the divisibilty by $6$ it gets a little bit tricky and so I am not quite sure about my own approach. Howsoever:
Again lets assume the cases $a=n$ and $a=n+1$ which yields to the equations
$$\begin{align}
n(n+1)(n+2)&\equiv0\mod(6)\\
(n+1)(n+2)(n+3)&\equiv0\mod(6)
\end{align}$$
This time the rewrite part is a little bit different
$$\begin{align}
(n+1)(n+2)(n+3)&=(n+1)[n^2+5n+6]\\
&=(6)(n+1)+n(n+1)(n+5)\\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)\\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)+n(n+1)(3)-n(n+1)(3)\\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(6)-n(n+1)(3)\\
&=(6)(n+1)+(n-1)n(n+1)+n(n+1)(6)\\
\end{align}$$
Two of these terms contain a $6$ and are so by defintion divisible by $6$. The remaining term is again the sum of three consecutive integers but in a different form than the given statement. I guess this should be valid nevertheless but I am not sure at all.
Besides the trivial $1\cdot2\cdot3=6$ the divisibilty by $12$ and $4$ fails with $5\cdot6\cdot7=210$. The divisibility by $5$ fails for example with $2\cdot3\cdot4=24$.
|
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"url": "https://math.stackexchange.com/questions/2870587",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Evaluating the integral $\int \frac1{(2+3\cos x)^2}\mathrm dx$ Please someone give me an idea to evaluate
this: $$\int \frac1{(2+3\cos x)^2}\mathrm dx$$
I don't even know how to start cause even multiplying an dividing by $\cos^2x$ does not work, so help me here.
|
Integrating by parts,
$$\int\dfrac{dx}{(a+b\cos x)^2}=\int\dfrac{\sin x}{(a+b\cos x)^2}\cdot\dfrac1{\sin x}dx$$
$$=\dfrac1{\sin x}\int\dfrac{\sin x}{(a+b\cos x)^2}dx-\left(\dfrac{d(1/\sin x)}{dx}\cdot\int\dfrac{\sin x}{(a+b\cos x)^2}dx\right)dx$$
$$=\dfrac1{b\sin x(a+b\cos x)}+\int\dfrac{\cos x}{b(1-\cos^2x)(a+b\cos x)}\,dx$$
Use Partial Fraction Decomposition,
$$\dfrac{\cos x}{(1-\cos^2x)(a+b\cos x)}=\dfrac A{1+\cos x}+\dfrac B{1-\cos x}+\dfrac C{a+b\cos x}$$
The first two integral can be managed easily, for the last use Weierstrass substitution
|
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|
Rows of a Matrix is divisible by 19, show that its Determinant is also divisible by 19 I came across the following problem while self studying:
Let
\begin{equation}
A =
\begin{bmatrix}
2 & 1 & 3 & 7 & 5\\
3 & 8 & 7 & 9 & 8\\
3 & 4 & 1 & 6 & 2\\
4 & 0 & 2 & 2 & 3\\
7 & 9 & 1 & 5 & 4\\
\end{bmatrix}
\end{equation}
Use the fact that 21375, 38798, 34162, 40223, and 79154 are divisible by 19 to show, without evaluating, that $\det[A]$ is divisible by 19.
I noticed that each of these numbers are the entries in the rows of A, but I don't see how that helps me.
|
Note that $$10^{10} \det(A) = 10^{4+3+2+1+0} \det(A) = \det \begin{pmatrix} 2 \cdot 10^{4} & 1\cdot 10^3 & 3\cdot 10^2 & 7\cdot 10 & 5 \\
3\cdot 10^{4} & 8 \cdot 10^3 & 7\cdot 10^2 & 9\cdot 10 & 8 \\
3\cdot 10^{4} & 4\cdot 10^3 & 1\cdot 10^2 & 6\cdot 10 & 2 \\
4\cdot 10^{4} & 0 \cdot 10^3 & 2 \cdot 10^2 & 2 \cdot 10 & 3 \\
7 \cdot 10^{4} & 9\cdot 10^3 & 1\cdot 10^2 & 5\cdot 10 & 4 \end{pmatrix}$$ $$ = \det \begin{pmatrix} 21375 & 1\cdot 10^3 & 3\cdot 10^2 & 7\cdot 10 & 5 \\
38798 & 8 \cdot 10^3 & 7\cdot 10^2 & 9\cdot 10 & 8 \\
34162 & 4\cdot 10^3 & 1\cdot 10^2 & 6\cdot 10 & 2 \\
40223 & 0 \cdot 10^3 & 2 \cdot 10^2 & 2 \cdot 10 & 3 \\
79154 & 9\cdot 10^3 & 1\cdot 10^2 & 5\cdot 10 & 4 \end{pmatrix}$$ which is evidently divisibly by $19$ when calculated via expansion by minors along the first column.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "12",
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|
How to show that $\cos(\sin^{-1}(x))$ is $\sqrt{1-x^2}$?
How to show that $\cos(\sin^{-1}(x))$ is $\sqrt{1-x^2}$?
I remember having to draw a triangle, but I'm not sure anymore.
|
Given $\cos(\sin^{-1}x)$
Let $\sin^{-1}x=\theta$
$$\implies\sin\theta=x\\\sin^2\theta=x^2\\1-\cos^2\theta=x^2\\\cos^2\theta=1-x^2\\\cos\theta=\pm\sqrt{1-x^2}\\\theta=\cos^{-1}\pm\sqrt{1-x^2}$$
Now, plug in $\theta=\cos^{-1}\pm\sqrt{1-x^2}$ in $\cos(\sin^{-1}x)$ and we get$$\pm\sqrt{1-x^2}$$But note that $\sin^{-1}x$ is in $\left[\dfrac{-\pi}{2},\dfrac{\pi}{2}\right]$ so $\cos(\sin^{-1}x)\ge0$
Therefore, $\cos(\sin^{-1}x)=\sqrt{1-x^2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2871629",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Show that:$\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}} \geq n^2 $ Show that:$$\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}} \geq n^2 $$
The following hints are also given: $$\left(\frac{x}{y} + \frac{y}{x} \geq 2 \right) \land x,y \gt 0$$
Base Case: For n = 2
$$\left(1+2\right) \cdot \left(\frac{1}{1}+\frac{1}{2}\right) \geq 2^2$$
Inductive hypothesis:
$$\sum_{i=1}^{n+1} x_{i} \cdot \sum_{i=1}^{n+1} \frac{1}{x_{i}} \geq \left(n+1\right)^2 = n^2+2n+1$$
Inductive step:
$$\sum_{i=1}^{n+1} x_{i} \cdot \sum_{i=1}^{n+1} \frac{1}{x_{i}} = \left(\sum_{i=1}^n x_{i}+(n+1)\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_{i}}+\frac{1}{n+1}\right) = \left(\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) + \left((n+1) \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) + \left(\sum_{i=1}^n x_{i} \cdot \frac{1}{n+1}\right) + (n+1) \cdot \frac{1}{n+1}$$
Final words:
I came to the conclusion that:
$$\left(\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) = n^2$$
I inserted for n = 1 so that $$\left((n+1) \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) = \frac{2}{1} \land \left(\sum_{i=1}^n x_{i} \cdot \frac{1}{n+1}\right) = \frac{1}{2}$$ Since $$\left(\frac{x}{y} + \frac{y}{x} \geq 2 \right) \land x,y$$ was given as a hint in the beginning I thougt I can say that $$\left((n+1) \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) + \left(\sum_{i=1}^n x_{i} \cdot \frac{1}{n+1}\right) = 2n$$
Furthermore it's obvious that: $$(n+1) \cdot \frac{1}{n+1} = 1$$
It's pretty standard proof by induction and I hope you can maybe give me some advices on what I could have done differently and verify the legitimacy of this proof.
|
Alternative way :
It's a direct consequence of the inequality of Chebyshev .
|
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|
Integral with inverse trigonometric function How do I integrate $$\int \cot^{-1}\sqrt{x^2+x+1}\ dx$$
I don't understand how to proceed?
I did try the question using integration by parts, but it didn’t help.
|
A solution without using hyperbolic functions
Integration by parts gives
$$x\cot^{-1}\left( \sqrt{x^2+x+1} \right)+\frac{1}{2}\int \frac{x(2x+1)}{\sqrt{x^2+x+1}(x^2+x+2)}\,dx$$
i can split the integrated on the right as
$$
\begin{align}
& =2\int{\frac{1}{\sqrt{{{x}^{2}}+x+1}}dx}-\int{\frac{1+2x}{2\sqrt{{{x}^{2}}+x+1}\left( {{x}^{2}}+x+2 \right)}dx}-\frac{7}{2}\int{\frac{1}{\sqrt{1+x+{{x}^{2}}}\left( {{x}^{2}}+x+2 \right)}dx} \\
& =2A-B-\frac{7}{2}C \\
\end{align}
$$
to calculate $A$ notice that ${{x}^{2}}+x+1={{\left( x+\frac{1}{2} \right)}^{2}}+{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}$ and using $x+\frac{1}{2}=\frac{\sqrt{3}}{2}u$
$$A=\int{\frac{1}{\sqrt{{{u}^{2}}+1}}du}=\ln \left( u+\sqrt{{{u}^{2}}+1} \right)=\ln \left( \frac{2x+1}{\sqrt{3}}+\sqrt{{{\left( \frac{2x+1}{\sqrt{3}} \right)}^{2}}+1} \right)$$
that’s because i accept the fact that ${{\left( \ln \left( u+\sqrt{{{u}^{2}}+1} \right) \right)}^{\prime }}=\frac{1}{\sqrt{{{u}^{2}}+1}}$
for $B$ we have
$$\frac{1+2x}{2\sqrt{1+x+{{x}^{2}}}\left( 2+x+{{x}^{2}} \right)}=\frac{{{\left( \sqrt{1+x+{{x}^{2}}} \right)}^{\prime }}}{\left( 1+{{\left( \sqrt{1+x+{{x}^{2}}} \right)}^{2}} \right)}\Rightarrow B={{\tan }^{-1}}\left( \sqrt{1+x+{{x}^{2}}} \right)$$
for $C$ use $u=\frac{2x+1}{\sqrt{{{x}^{2}}+x+1}}$ hence $du=\frac{3}{2{{\left( \sqrt{{{x}^{2}}+x+1} \right)}^{3}}}dx$ also
$${{u}^{2}}-7=\frac{4{{x}^{2}}+4x+1}{{{x}^{2}}+x+1}-\frac{7{{x}^{2}}+7x+7}{{{x}^{2}}+x+1}=\frac{-3{{x}^{2}}-3x-6}{{{x}^{2}}+x+1}=-3\left( \frac{{{x}^{2}}+x+2}{{{x}^{2}}+x+1} \right)$$
$$\begin{align}
& C=\frac{2}{3}\int{\frac{1}{\sqrt{{{x}^{2}}+x+1}\left( {{x}^{2}}+x+2 \right)}{{\left( \sqrt{{{x}^{2}}+x+1} \right)}^{3}}du} \\
& \quad =\frac{2}{3}\int{\frac{{{x}^{2}}+x+1}{{{x}^{2}}+x+2}du} \\
& \quad =-2\int{\frac{1}{{{u}^{2}}-7}du}=\frac{1}{\sqrt{7}}\ln \left( \frac{\sqrt{7}+u}{\sqrt{7}-u} \right)=\frac{1}{\sqrt{7}}\ln \left( \frac{\sqrt{7}+\frac{2x+1}{\sqrt{{{x}^{2}}+x+1}}}{\sqrt{7}-\frac{2x+1}{\sqrt{{{x}^{2}}+x+1}}} \right) \\
\end{align}$$
finally,
$$
\int \cot^{-1}\left( \sqrt{x^2+x+1} \right) \, dx=x\cot^{-1}\left( \sqrt{{{x}^{2}}+x+1} \right)+\ln \left( \frac{2x+1}{\sqrt{3}}+\sqrt{{{\left( \frac{2x+1}{\sqrt{3}} \right)}^{2}}+1} \right)-\frac{1}{2}{{\tan }^{-1}}\left( \sqrt{1+x+{{x}^{2}}} \right)-\frac{\sqrt{7}}{4}\ln \left( \frac{\sqrt{7}+\frac{2x+1}{\sqrt{{{x}^{2}}+x+1}}}{\sqrt{7}-\frac{2x+1}{\sqrt{{{x}^{2}}+x+1}}} \right)
$$
|
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Let $m$ be the largest real root of the equation $\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4$ find $m$
Let $m$ be the largest real root of the equation $$\frac3{x-3} + \frac5{x-5}+\frac{17}{x-17}+\frac{19}{x-19} =x^2 - 11x -4.$$ Find $m$.
do we literally add all the fractions or do we do something else
I have no clue how to solve this
|
We need to solve
$$\frac3{x-3}+1+ \frac5{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1 =x^2 - 11x $$ or
$$x\left(\frac{1}{x-3} + \frac{1}{x-5}+\frac{1}{x-17}+\frac{1}{x-19}\right) =x^2 - 11x$$ or
$$2x(x-11)\left(\frac{1}{(x-3)(x-19)}+\frac{1}{(x-5)(x-17)}\right)=x(x-11),$$ which gives $x_1=0$, $x_2=11$ or
$$\frac{1}{x^2-22x+57}+\frac{1}{x^2-22x+85}=\frac{1}{2}.$$
Let $x^2-22x+57=a$.
Thus, $$\frac{1}{a}+\frac{1}{a+28}=\frac{1}{2}$$ or
$$a^2+24a-56=0$$ or
$$a^2+24a+144=200,$$ which gives
$$a=-12+10\sqrt2$$ or
$$a=-12-10\sqrt2,$$
which gives
$$x^2-22x+69\pm10\sqrt2=0$$ or
$$x^2-22x+121=52\pm10\sqrt2$$ or
$$x_{3,4,5,6}=11\pm\sqrt{52\pm10\sqrt2}$$ and we got a maximal root:
$$11+\sqrt{52+10\sqrt2}.$$
|
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|
How to prove $1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\left(2n-1\right)^{1/4} $
Prove that $$1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\Bigl(2n-1\Bigr)^{\frac{1}{4}} $$
My Approach :
I tried by applying Tchebychev's Inequality for two same sets of numbers;
$$1 , \frac{1}{\sqrt{2}} ,... ,\frac{1}{\sqrt{n}}$$
And got , $$\Bigl(1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}}\Bigr)^2 <n\Bigl(1 + \frac{1}{2} +... +\frac{1}{n}\Bigr) $$
Again I tried by applying Tchebychev's Inequality for another two same sets of numbers;
$$1,\frac{1}{2},...,\frac{1}{n}$$ And got, $$\Bigl(1 + \frac{1}{2} +... +\frac{1}{n}\Bigr)^2 <n\Bigl(1 + \frac{1}{2^2} +... +\frac{1}{n^2}\Bigr)$$
With these two inequities i tried solving further more, but i couldn't. So can you please help me solving this further. And if there is some other approach for this question then please answer that way too.
Thank you.
|
Your inequality should be not strong.
We can prove that for all natural $n\geq1$ the following inequality holds:
$$\sum_{k=1}^n\frac{1}{\sqrt{k}}\leq\sqrt{n}\sqrt[4]{2n-1}.$$
Indeed,
$$\sum_{k=1}^n\frac{1}{\sqrt{k}}\leq1+\int\limits_1^{n}\frac{1}{\sqrt x}dx=2\sqrt{n}-1.$$
Thus, it's enough to prove that
$$2\sqrt{n}-1\leq\sqrt{n}\sqrt[4]{2n-1}.$$
Let $n=(x+1)^2,$ where $x\geq0$.
Thus, we need to prove that
$$2x+1\leq (x+1)\sqrt[4]{2x^2+4x+1}$$ or
$$(x+1)^4(2x^2+4x+1)\geq(2x+1)^4$$ or
$$x^3(2x^3+12x^2+13x+4)\geq0,$$ which is obvious.
Done!
|
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|
Prove $\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0$. I want to show that
$$\lim_{(x,y) \to (0,0)} \frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2} =0.$$
Is it valid to do it like this:
$$\lim_{(x,y) \to (0,0)} \left|\frac{\exp(xy)\cdot xy\cdot(x^2-y^2)}{x^2+y^2}\right| \leq \lim_{(x,y) \to (0,0)} |\exp(xy)\cdot xy\cdot (x^2-y^2)|=0$$
|
We have that $e^{xy}\to 1$ and by polar coordinates
$$\left|\frac{xy\cdot(x^2-y^2)}{x^2+y^2}\right|=r^2|(\cos \theta\sin \theta)(\cos^2\theta-\sin^2\theta)|\le r^2\to 0$$
|
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|
Compare $\arcsin (1)$ and $\tan (1)$
Which one is greater: $\arcsin (1)$ or $\tan (1)$?
How to find without using graph or calculator?
I tried using $\sin(\tan1)\leq1$, but how do I eliminate the possibility of an equality without using the calculator?
|
$\arcsin(1)=\frac{\pi}{2}$ while from the Weierstrass product for the cosine function we have
$$ \tan(1) = \sum_{n\geq 0}\frac{8}{(2n+1)^2 \pi^2-4 }=2\sum_{n\geq 0}\left[\frac{1}{(2n+1)\pi-2}-\frac{1}{(2n+1)\pi+2}\right]$$
such that an effective integral representation of $\tan(1)$ through an almost-Gaussian integral is
$$\tan(1)=\int_{\mathbb{R}}\frac{\sinh(2x)}{\sinh(\pi x)}\,dx.$$
As an alternative, by expanding $\frac{8}{(2n+1)^2\pi^2-4}$ as a geometric series we get
$$ \tan(1) = 2 \sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)}{\pi^{2m}} $$
which is equivalent to the previous integral representation via $\zeta(2m)=\frac{1}{(2m-1)!}\int_{0}^{+\infty}\frac{z^{2m-1}}{e^z-1}\,dz$.
In order to prove that $\tan(1)<\frac{\pi}{2}$ it is enough to show that $\tan\left(\frac{1}{2}\right)<\frac{\pi}{2+\sqrt{\pi^2+4}}$, since $\tan(2z)=\frac{2\tan z}{1-\tan^2 z}$. $\tan\left(\frac{1}{2}\right)$ has a fast-convergent series representation
$$ \tan\left(\tfrac12\right)=4\sum_{m\geq 1}\frac{\zeta(2m)}{\pi^{2m}}\left(1-\frac{1}{4^m}\right) $$
which allows to state
$$ \tan\left(\tfrac12\right) < 4\sum_{m= 1}^{3}\frac{\zeta(2m)}{\pi^{2m}}\left(1-\frac{1}{4^m}\right)+4\,\zeta(8)\sum_{m \geq 4}\frac{1}{\pi^{2m}}\left(1-\frac{1}{4^m}\right)$$
or
$$ \tan\left(\tfrac12\right)<\frac{131}{240}+\frac{\pi ^2 \left(85 \pi^2-21\right)}{50400 \left(\pi^2-1\right) \left(4 \pi^2-1\right)}$$
(this is extremely accurate). The proof is finished by exploiting $\frac{227}{23}<\pi^2<\frac{79}{8}$ which follows from the study of the Beuker-like integrals
$$ \iint_{(0,1)^2}\frac{x^m(1-x)^m y^n (1-y)^n}{1-xy}\,dx\,dy.$$
An alternative approach. The Shafer-Fink inequality gives that for any $x\in(0,3/2)$
$$ \tan(x) < \frac{3x+2x\sqrt{9-3x^2}}{9-4x^2} $$
holds, so by the duplication formula for the tangent function the sharper inequality
$$ \tan(x) < \frac{4 x\left(9-x^2\right) \left(3+\sqrt{36-3 x^2}\right)}{324-117 x^2+7 x^4-6 x^2 \sqrt{36-3 x^2}} $$
holds too. This gives
$$ \tan(1) < \frac{32 \left(3+\sqrt{33}\right)}{214-6 \sqrt{33}}=\frac{4}{697} \left(105+29 \sqrt{33}\right) $$
so it is enough to show that $\pi>\frac{32(3+\sqrt{33})}{107-3\sqrt{33}}$, which is a pretty loose inequality.
|
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What does unique factorization exactly mean here? Given $a$ and $b$ are integers:
$$ 3\cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b $$
The answer is as follows:
Since both sides of this equation are integers and have unique factorizations, it follows that $a = 1$ and $b = 2$ is the only solution.
What does unique factorization exactly mean here? I looked up the word unique factorization on google but I don't really understand it. Can you explain it at a pre-calculus level? Also, what aspect of 'integers' makes the solution $a = 1$, and $b = 2$ ??
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Every positive integer is either prime or composite or the number one.
If the number is composite then it can be factored as a product of smaller numbers. Ex: $120 = 12\cdot10$ or we could say $120 = 30\cdot4$ or we could say $120 = 6\cdot10\cdot2$.
If we factor a number into smaller composites and then factored those number we'd eventually factor it do to a factorization containing only primes. Ex: $120 = 12\cdot10 = (3\cdot4)\cdot(2\cdot5) = 3\cdot2\cdot2\cdot2\cdot5$ or we could say $120 = 30\cdot4 =(6\cdot5)\cdot(2\cdot2) = 2\cdot3\cdot5\cdot2\cdot2$ or $120=6\cdot10\cdot2 = 2\cdot3\cdot2\cdot5\cdot2$.
And if we have multiple occurrences of the same prime factor we can list them as a power. Ex: $120 = 3\cdot2^3\cdot5$ or $120 = 2^3\cdot3\cdot5$ or $120 =3\cdot5\cdot2^3$.
Now perhaps the most anti-climatic conclusion you'll ever see.
No matter how you break a number down into primes you will always and up with the same primes and the same number of each of them. Each number has only one unique way of being broken down into primes.
And that is what unique prime factorization means.
So if you have $3^2 \cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b$ then you have a number. By the Left-Hand side we know it factors into: two $3$s, one $2$, two $5$s, and some $7$ but we don't know how many. On the RHS we see that it factors down into: one $2$, two $3$s, one $7$, and some $5$s but we don't know how many.
But by the Unique prime factorizations we know the two lists of components must be exactly the same. So how many $7$s are there? Well, by the RHS we know there is one $7$. And how many $5$s are there? Well, by the LHS we know there are two $5$s.
So that's that. If $3^2 \cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b$ then $a =1$ and $b = 2$ and then number is $3^2 \cdot2\cdot5^2\cdot7 = 9\cdot2\cdot5^2\cdot7 = 3150$.
Note: The really hard way to do it:
$3^2 \cdot 2 \cdot 5^2 \cdot 7^a = 2 \cdot 3^2 \cdot 7 \cdot 5^b$ Divide both sides by $3^2\cdot2$
$5^2\cdot7^a = 7\cdot5^b$. Divide both sides by $7$.
$5^2\cdot7^{a-1} = 5^b$. Divide both sides by $5^2$.
$7^{a-1} = 5^{b-2}$.
So.... the question is: what (integer) power can you raise $5$ to get a (integer) power of $7$ (or vice versa). The answer is: $5$ and $7$ are prime; there isn't any such power. The only way this makes sense is if $7^0 = 5^0 = 1$. So $a-1 = b-2 =0$.
The point being: if you have ever told that $3^a\cdot5^b = 3^c\cdot5^d$ and you are told $a$ and $b$ are integers, then it should be immediately clear $a = c$ and $b = d$.
|
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Number of Non negative integer solutions of $3a+2b+c+d=19$ Find Number of Non negative integer solutions of $3a+2b+c+d=19$
My attempt:
we have $$2b+c+d=19-3a$$ Required solutions is coefficient of $t^{19-3a}$ in
$$( 1-t^2)^{-1}(1-t)^{-1}(1-t)^{-1}=\frac{1}{(1-t)^3(1+t)}$$
By partial fractions we get
$$( 1-t^2)^{-1}(1-t)^{-1}(1-t)^{-1}=\frac{1}{2}\times (1-t)^{-3}+\frac{1}{4} \times (1-t)^{-2}+\frac{1}{4}\times (1-t^2)^{-1}$$
Required coefficient is
$$\frac{1}{2} \sum_{a=0}^{6} \binom{21-3a}{2}+\frac{1}{4} \sum_{a=0}^{6}\binom{20-3a}{1}+\frac{1}{4}(4)=294+\frac{77}{4}+1$$
why i am not getting integer answer?
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HINT.-Checking directly is not hard.
$$3a+2b=19$$ has only three solutions $(1,8),(3,5),(5,2)$.
When $a=6$,$$2b+c+d=1$$ has two solutions $(0,0,1),(0,1,0)$.
When $a=5$, $$2b+c+d=4$$ has five solutions from which one has to be discarded $(1,1,1)(2,0,0),(0,1,3),(0,2,2),(0,3,1)$ we have till now
$$(1,8,0,0),(3,5,0,0),(5,2,0,0)\\(6,0,0,1),(6,0,1,0)\\(5,1,1,1),(5,0,1,3),(5,0,2,2),(5,0,3,1)$$ There are nine solutions as far and we can continue with the remaining values of $a$.
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if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$ This may be trivial but I am not able to prove that if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ for $a,b,c\in\mathbb{R}$ then $\rho\ge-1/2$. Can anybody help me please?
Thanks!
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You can show that the inequality is true for all $a,b,c\in\mathbb{R}$ if and only if $-\dfrac{1}{2}\leq \rho \leq 1$. Indeed, setting $a$, $b$, and $c$ to be $1$ gives
$$3+6\rho\geq 0\,,\text{ or equivalently }\rho\geq -\frac12\,.$$
Taking $(a,b,c)$ to be $(1,0,-1)$ leads to
$$2-2\rho\geq 0\,,\text{ whence }\rho\leq 1\,.$$
We shall prove that the inequality
$$a^2+b^2+c^2+2\rho\,(bc+ca+ab)\geq 0$$
holds for all $\rho\in\left[-\dfrac12,1\right]$. This is simply because
$$\begin{align}a^2+b^2+c^2+&2\rho\,(bc+ca+ab)\\&=\frac{1+2\rho}{3}\,(a+b+c)^2+\left(\frac{1-\rho}{3}\right)\,\left((b-c)^2+(c-a)^2+(a-b)^2\right)\geq0\,.\end{align}$$
The inequality $a^2+b^2+c^2+2\rho\,(bc+ca+ab)\geq 0$ for $\rho\in\left[-\dfrac12,1\right]$ becomes an equality if and only if
*
*$\rho=-\frac12$ and $a=b=c$,
*$-\frac12<\rho<1$ and $a=b=c=0$, or
*$\rho=1$ and $a+b+c=0$.
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order of the splitting field of $x^5 +x^4 +1 $? what is the order of the splitting field of
$x^5 +x^4 +1 = (x^2 +x +1)( x^3 +x+1)$ over $\mathbb{Z_2}$
i thinks it will $6$ because $2.3 = 6$
Pliz help me...
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Yes, it is everything ok. This is an answer, not a comment, in order to insert the following confirmation using computer assistance, here sage:
sage: F = GF(2)
sage: R.<x> = PolynomialRing(F)
sage: f = x^5+x^4+1
sage: f.factor()
(x^2 + x + 1) * (x^3 + x + 1)
sage: K.<a> = f.splitting_field()
sage: K
Finite Field in a of size 2^6
sage: a.minpoly()
x^6 + x^4 + x^3 + x + 1
sage: f.base_extend(K).factor()
(x + a^3 + a^2 + a)
* (x + a^3 + a^2 + a + 1)
* (x + a^4 + a^2 + a + 1)
* (x + a^5 + a)
* (x + a^5 + a^4 + a^2 + 1)
The code was minimally rearranged for the last result to fit in the window.
For the first polynomial in the factorization $(x^2 + x + 1) (x^3 + x + 1)$ we need an extension from $\Bbb F_2$ to $\Bbb F_{2^2}$, then from this one a new degree three extension to $\Bbb F_{(2^2)^3}$ to split also the second factor $x^3+x+1$.
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|
Double series convergent to $2\zeta(4)$? Using a computer I found the double sum
$$S(n)= \sum_{j=1}^n\sum_{k=1}^n \frac{j^2 + jk + k^2}{j^2(j+k)^2k^2}$$
has values
$$S(10) \quad\quad= 1.881427206538142 \\ S(1000) \quad= 2.161366028875634 \\S(100000) = 2.164613524212465\\$$
As a guess I compared with fractions $\pi^p/q$ where $p,q$ are positive integers and it appears
$$\lim_{n \to \infty} S(n) = \frac{\pi^4}{45} = 2\zeta(4) \approx 2.164646467422276 $$
I'd be interested in seeing a proof if true.
|
So before I start, I've never even attempted to evaluate a double sum before, so there could very well have been an easier way.
$$\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{j^2+jk+k^2}{j^2k^2(j+k)^2} = \sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{k^2(j+k)^2} +\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{jk(j+k)^2} + \sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2}= $$
$$ 2\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{k^2(j+k)^2} +\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{jk(j+k)^2}$$
Through partial fraction decomposition the above equals:
$$\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^3k} -\frac{1}{j^3(j+k)}-\frac{1}{j^2(j+k)}+2\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{2}{j^3(j+k)}-\frac{2}{j^3k}+\frac{1}{j^2k^2}+\frac{1}{j^2(j+k)^2} $$
Collecting like-terms:
$$3\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^3(j+k)}-3\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^3k} +\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2} + 2\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2k^2} = $$
The final sum clearly equals $2\zeta(2)^2$ or $\frac{\pi^4}{18}$. I then evaluate the first two sums by combining them to get:
$$3\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^3}(\frac{1}{j+k}-\frac{1}{k}) $$
Interchanging j and k to and averaging the two sums to get:
$$\frac{3}{2}\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^3}(\frac{1}{j+k}-\frac{1}{k})+\frac{1}{k^3}(\frac{1}{j+k}-\frac{1}{j}) $$
This can be rewritten as:
$$-\frac{3}{2}\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2k(j+k)}+\frac{1}{k^2j(j+k)}= -\frac{3}{2}\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2k^2} = -\frac{3}{2}\zeta{(2)}^2 = -\frac{\pi^4}{24}$$
So putting that back into the original problem:
$$\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2} + \frac{\pi^4}{18} -\frac{\pi^4}{48} = \sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2} + \frac{\pi^4}{72} $$
This is all I got to. I couldn't evaluate the other sum the way I did before. Using a calculator there is a very good chance it equals $\frac{\pi^4}{120}$.
Just for fun I was able to write the remaining sum as:
$$\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2} = \sum_{n=1}^\infty \frac{\zeta(2,n+1)}{n^2}$$
Where $\zeta(x,y)$ is the Hurwitz Zeta Function. Wolfram Alpha was able to calculate the sum as $\frac{\pi^4}{120}$ as desired.
|
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|
Find the maximum of the $| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|$ with $|w|=1$ Let $w \in \mathbb{C}$, and $\left | w \right | = 1$. Find the maximum of the function $| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|$
Since $$|(w+2)^3(w-3)^2|=|w^5-15w^3-10w^2+60w+72|$$
Let $w=\cos x+i \sin x$. Then we have an ugly form
|
Calling
$$
z_1=\rho_1 e^{i\phi_1} = \sqrt{(\cos\phi+2)^2+\sin^2\phi} = \sqrt{5+4\cos\phi}\\
z_2=\rho_2 e^{i\phi_2} = \sqrt{(\cos\phi-2)^2+\sin^2\phi} = \sqrt{10-6\cos\phi}
$$
so
$$
|z_1^3z_2^2| = \rho_1^3\rho_2^2 = \left(5+4\cos\phi\right)^{\frac 32}|10-6\cos\phi|
$$
now
$$
\frac{d}{d\phi}\left( \left(5+4\cos\phi\right)^{\frac 32}(10-6\cos\phi)\right) = 6 \sin\phi (4 \cos\phi+5)^{3/2}-6 \sin\phi (10-6 \cos\phi) \sqrt{4 \cos\phi+5}=0
$$
for $\phi \in \{0,\pm\frac{\pi}{3}\}$ so the maximum is
$$
108=\max_{w}| \left( w + 2 \right) ^3 \left( w - 3 \right)^2|\;\;\mbox{s. t.}\;\;|w| = 1
$$
for $\phi = 0$
|
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For positive number $a,b$, when $a,b$ satisfies $2a^2 +7ab+3b^2=7$, what is maximum value of $a+{\sqrt{3ab}}+2b$ Question is
For positive numbers $a,b$ such that $2a^2 +7ab+3b^2 = 7$, what is the maximum value of $a+{\sqrt{3ab}}+2b$?
I use AM-GM Inequality to do this
$$(2a+b)(a+3b)=7\text{ and }2a+b=\frac{7}{a+3b}.$$
So, maximum value $m$ is $\frac{7}{a+3b}+{\sqrt{3ab}}$ so $m=\frac{7}{2{\sqrt{3ab}}}+{\sqrt{3ab}}$ is $2{\sqrt{\frac{7}{2}}}$ so $2m^2=28$.
But it didn't satisfy equal condition so how do I get it?
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Let $a＝x^2,b＝3y^2$ and $t＝\frac{x}{y}$,only need to find the maxium of this
$$\frac{7(t^2+3t+6)^2}{2t^4+21t^2+2}$$
and ...
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2882430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Differential equation: $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$
The solution of $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ is given by:
a) $(x+2)^4 (1+\frac{2y}{x})= ke^{\frac{2y}{x}}$
b) $(x+2)^4 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
c) $(x+2)^3 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
d) None of these
Attempt:
I have expanded and checked but couldn't spot any exact differentials.
Secondly, it's not a homogeneous equation, so couldn't use $y = vx$.
How do I go about solving this problem?
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Let $X=x+2$ and $Y=y-2$, so the given DE is equivalent to
$$\dfrac{\mathrm d Y}{\mathrm d X}=\frac{(X+Y)^2}{X Y}$$
last DE is homogeneous, so can be transformed into a separable DE by making $Y =uX$ as follows
$$\dfrac{\mathrm d Y}{\mathrm d X}=\frac{(X+Y)^2}{X Y}\qquad\iff\qquad u+X\frac{\mathrm d u}{\mathrm dX}=\frac{X^2(1+u)^2}{X^2u}$$
Then we get
$$X\frac{\mathrm d u}{\mathrm d X}=\frac{1+2u}{u}\quad\implies\quad \frac{u}{2u+1}\dfrac{\mathrm d u}{\mathrm d X}=\frac1X$$
Integrating both sides respect to $X$ (supposing that $u$ depends on $X$) we get
\begin{align*}
\int\frac{u}{2u+1}\dfrac{\mathrm d u}{\mathrm d X}\mathrm d X&=\int\frac1X\mathrm dX\\[4pt]
\int\left(\frac12-\frac{1/2}{2u+1}\right)\mathrm du&=\int\frac1X\mathrm dX\\[4pt]
\frac12u-\frac14\ln|2u+1|&=\ln|X|+c_1
\end{align*}
Last equality is equivalent to
$$\frac{y-2}{x+2}-\frac12\ln\left|\frac{2(y-2)}{x+2}+1\right|=2\ln|x+2|+2c_1$$
Notice that this solution can be carried to the form b):
\begin{align*}
\frac{2(y-2)}{x+2}-4c_2&=4\ln|x+2|+\ln\left|\frac{2(y-2)}{x+2}+1\right|\\[4pt]
k e^{\frac{2(y-2)}{x+2}}&=(x+4)^4\left[\frac{2(y-2)}{x+2}+1\right]
\end{align*}
where $k=\pm e^{-4c_1}$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2883253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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|
Why isn't this true for $x<0$?
Prove that
$$\lfloor{x}\rfloor=\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+
\ldots$$
$x\geq{0}$
I asked this question here yesterday. Although I was able to come up with a proof, the second part was left unanswered.
I proved it using,
Subtract $\lfloor{x}\rfloor$ from both sides.
Then we need to prove that,
$$-\lfloor{x}\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+
\ldots=0$$
We know that,
$\lfloor{x}\rfloor=\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor$
Therefore,
$-\lfloor{x}\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor=-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor$
Subsituting this,
$$-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+
\ldots$$
$$-\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{\frac{x}{2}+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+
\ldots$$
Similarly, I reach,
$$\lim_{n\to\infty}-\bigg\lfloor{\frac{x}{2^n}}\bigg\rfloor+\bigg\lfloor{\frac{\frac{x}{2^n}+1}{2}}\bigg\rfloor$$
i.e. $$=0$$
My question is,as $\lfloor{x}\rfloor=\bigg\lfloor{\frac{x}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor$ is true $\forall x\in{\mathbb{R}}$.
Why isn't
$$\lfloor{x}\rfloor=\bigg\lfloor{\frac{x+1}{2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2}{2^2}}\bigg\rfloor+\bigg\lfloor{\frac{x+2^2}{2^3}}\bigg\rfloor+
\ldots$$ true $\forall x\in{\mathbb{R}}$.
Also how would the expression change for $x\lt{0}$.
Thank you.
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For $x=-1$ the lest side is $-1$ and the right side is $0$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2883364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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|
How to find the auxiliary circle of a non-standard ellipse?
Given the equation of conic C is $5x^2 + 6xy + 5y^2 = 8$, find the equation S of its auxiliary circle?
Now, I know that the equation C is an ellipse.
Since $\Delta = 5*5(-8) - (-8)(3^2) \neq 0$
And,
$ab - h^2 = 5*5 - 3^2 > 0$
Which is the condition for an ellipse.
But this isn't a standard one!. In my school, we have only worked with ellipses of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
So, how do get the equation of the auxiliary circle for the given ellipse?
Any help would be appreciated.
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Since the conic is centre origin, we may use polar coordinates:
\begin{align}
8 &= 5r^2\cos^2 \theta+6r^2\cos \theta \sin \theta+5r^2\sin^2 \theta \\
r^2 &= \frac{8}{5+6\cos \theta \sin \theta} \\
&= \frac{8}{5+3\sin 2\theta}
\end{align}
Now,
$$\frac{8}{5+3} \le r^2 \le \frac{8}{5-3} \implies 1 \le r^2 \le 4 \\$$
Hence, the auxiliary circle is $r=2$ or equivalently,
$$x^2+y^2=4$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2886063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
How to solve $\sin^2(x)+\sin2x+2\cos^2(x)=0$ How do you solve $\sin^2(x)+\sin2x+2\cos^2(x)=0$?
I have been able to rewrite it as $(\sin(x)+\cos(x))^2+\cos^2(x)=0$. Not obviously useful, I think
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$\sin^2(x) + \cos^2(x) = 1$, so we have:
$$\sin 2x + \cos^2 x + 1 = 0$$
Now, $\cos^2 x ≥ 0$, and $\sin 2x ≥ -1$, so the only solutions are when $\cos^2 x = 0$ and $\sin 2x = -1$.
When $\cos^2 x = 0$, $x = -\frac{\pi}{2} + 2\pi n$ or $x = \frac{\pi}{2} + 2\pi n$. However, $\sin \left( 2(-\frac{\pi}{2}) \right) = 0$ and $\sin \left( 2(\frac{\pi}{2}) \right) = 0$, so there are no solutions.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2887265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
How to find all the $z$ that satisfy $(1+i)z^4=(1-i)|z|^2$? Would you please help me solve this? I need all the $z$ that satisfy the equality
$$(1+i)z^4=(1-i)|z|^2.$$
I tried doing this:
$$
\begin{aligned}
(1+i)z^4&= (1-i)z\overline z\\
(1+i)z^4 -(1-i)z\overline z &= 0\\
z[(1+i)z^3-(1-i)\overline z]&= 0
\end{aligned}
$$
Then $z= 0$ and
$$(1+i)z^3-(1-i)\overline z= 0.$$
I don't know what to do with $\overline z$.
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$$(1+i)z^4=(1-i)|z|^2\iff iz^4=z\bar z\iff\begin{cases}z=0\\iz^3=\bar z\end{cases}$$ $$iz^3=\bar z\iff i(a+ib)^3=(a-ib)\iff {ia^3 - 3 a^2 b -i 3 a b^2 + b^3}=(a-ib)\\\iff b^3- 3 a^2 b +i(a^3 - 3 a b^2)=a-ib\iff\begin{cases}b^3- 3 a^2 b=a\\a^3 - 3 a b^2=-b\end{cases}$$
We can also notice that $|b^3- 3 a^2 b +i(a^3 - 3 a b^2)|=\sqrt{(a^2+b^2)^3}=\sqrt{a^2+b^2}=|a-ib|\implies
|z|=\begin{cases}1\\0\end{cases}$, so it comes with terms with @mfl's answer
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{
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"url": "https://math.stackexchange.com/questions/2888218",
"timestamp": "2023-03-29T00:00:00",
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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.