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Question involving some roots of the equation $31z^{15} - z^{10} + 32 = 0 $ Consider the equation $$31z^{15} - z^{10} + 32 = 0.$$ What would be the sum of all those roots of the equation whose real part is positive? Only trivially trying to solve the equation I find not helpful. Even factorizing we get , by putting $z^5=x $ ,
$31x^3 - x^2 + 32 = (x+1)(31x^2 - 32x + 32) =0 $ which is of no help .
|
The equation $31x^3-x^2+32=0$, has the roots,
$$
-1\qquad \frac{4}{31(4 + i\sqrt{46})}\qquad \frac{4}{31(4 - i\sqrt{46})}
$$
so take all them to the $1/5$-th power and sum there real parts.
|
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|
Formula for A119016 : Numerators of "Farey fraction" approximations to $\sqrt{2}$ Can someone please verify and maybe simplify the closed formula for A119016 :
$$ a(n)=\frac{1}{4} \left(1+\sqrt{2}\right)^{n/2} \left((-1)^n+1\right)+\frac{1}{2}
\left(\sqrt{2}-1\right)^{n/2} \cos \left(\frac{\pi n}{2}\right)
+\frac{1}{4 \sqrt{8119+5741\sqrt{2}}} \left(58+41 \sqrt{2}\right) \left(1+\sqrt{2}\right)^{n/2} \left(1-(-1)^n\right)-2 \left(140+99
\sqrt{2}\right) \left(\sqrt{2}-1\right)^{n/2} \sin \left(\frac{\pi n}{2}\right) $$
Considering even and odd functions I could simplify the formula to :
$$
a(n)=\frac{1}{4} \left(\sqrt{2}-2\right) \left(1+\sqrt{2}\right) \left(1-(-1)^n\right)
\left(\left(1-\sqrt{2}\right)^{\frac{n-1}{2}}-\left(1+\sqrt{2}\right)^{\frac{n-1}{2}}\right)+\frac{1
}{4} \left((-1)^n+1\right) \left(\left(1-\sqrt{2}\right)^{n/2}+\left(1+\sqrt{2}\right)^{n/2}\right)
$$
Thanks
|
One observation is that
$$\cos\left(\frac{\pi n}2\right)=\left\{\begin{array}{cc}(-1)^{n/2}& n\equiv0,2(\mod4)\\0&n\equiv1,3(\mod4) \end{array}\right.$$
and
$$\sin\left(\frac{\pi n}2\right)=\left\{\begin{array}{cc}(-1)^{n/2}& n\equiv1,3(\mod4)\\0&n\equiv0,2(\mod4) \end{array}\right.$$
|
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|
"reverse" diophantine equation Suppose we define $A= (8+\sqrt{x})^{1/3} + (8-\sqrt{x})^{1/3}$. How can we find, algebraically, all values of x for which $A$ is an integer?
I was not able this problem save for with Mathematica. How can we solve this using the tool of our brains?
|
$A^3 = 16+3A\sqrt[3]{64-x}$, or $A^3 = 16+3Ay$, where $y=\sqrt[3]{64-x}$.
For each $A \ne 0$ we have $y=\frac{A^3-16}{3A}$, and $x=64-y^3$.
If $A\in (0,4]$, then $x\ge 0$.
If $A\in (4,\infty)$, then $x<0$, so $\sqrt{x}$ will be imaginary.
(if $A<0$, then we have no solutions, because RHS of your formula has $Re > 0$. (?))
Writing by one row, we have
$$x=64-\left(\frac{A^3-16}{3A}\right)^3.$$
\begin{array}{|l|l|}
A & x \\
--- & --- \\
1 & 189 \\
2 & 1792/3^3 \\
3 & 45325/3^6 \\
4 & 0 \\
5 & -1079029/(3\cdot 5)^3 \\
6 & -7626752/(3\cdot 6)^3 \\
7 & -34373079/(3\cdot 7)^3 \\
8 & -236600/3^3 \\
9 & -361207385/(3\cdot 9)^3 \\
\cdots & \cdots
\end{array}
|
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|
How to find the roots of $x^4 +1$ I'm trying to find the roots of $x^4+1$. I've already found in this site solutions for polynomials like this $x^n+a$, where $a$ is a negative term. I don't remember how to solve an equation when $a$ is a positive term as the equation above.
Thanks
|
Using this, $x^4=-1=e^{i\pi}=e^{(2n+1)\pi i}$ as $e^{2m\pi i}=1$ where $m,n$ are integers.
So, $x=e^{\frac{(2n+1)\pi i}4}=\cos\frac{(2n+1)\pi}4 +i\sin \frac{(2n+1)\pi}4 $ where $n$ has any $4$ in-congruent values $\pmod 4$, the most simple set of values of $n$ can be $\{0,1,2,3\}$.
If $x_m=e^{\frac{(2m+1)\pi i}4},x_{m+2}=e^{\frac{(2m+3)\pi i}4}=e^{\frac{(2m+1)\pi i}4}\cdot e^{\frac {i\pi}2}=-x_m$
Also, observe that if $y$ is a solution of $x^4=-1$, so is $-y$
$x_0=\cos\frac{\pi}4 +i\sin \frac{\pi}4=\frac{1+i}{\sqrt 2}$
$x_1=\cos\frac{3\pi}4 +i\sin \frac{3\pi}4=\frac{-1+i}{\sqrt 2}$
$x_2=-x_0$
$x_3=-x_1$
So, the values of $x$ are $\pm\left(\frac{1+i}{\sqrt 2}\right),\pm\left(\frac{-1+i}{\sqrt 2}\right)$
|
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|
Evaluation of $\lim_{n\rightarrow\infty} \int_{-\pi}^\pi x^2 \frac{\sin(2nx)}{\sin x} dx$ I need an approach to analytically evaluating this limit:
$$\lim_{n\rightarrow\infty} \int_{-\pi}^\pi x^2 \frac{\sin(2nx)}{\sin x} dx$$
Numerically, I see that the answer is $-\pi^3$. Similarly, if I replace $x^2$ with $x^4$, I get $-\pi^5$. I vaguely recall seeing this result obtained analytically and not necessarily using advanced ideas, but I just can't remember any details. I know the fraction in the integrand relates to Chebyshev polynomials of the second kind. Thoughts anyone? Thanks!
|
Here is an (admittedly somewhat horrible) solution using both the Riemann-Lebesgue lemma and the residue theorem. Let
$$ f(x) = \frac{x^2}{\sin x} \quad \text{ and } \quad g(x) = \frac{\pi^2}{2} \frac{1-\cos x}{\sin x} $$
on $(-\pi,\pi)$. Then
$$ h(x) = f(x)-g(x) = \frac{x^2 - \frac{\pi^2}2(1-\cos x)}{\sin x}$$
is actually continuous on the closed interval $[-\pi,\pi]$, since both the numerator and denominator have simple zeros at $\pm \pi$ and at $0$, so that the quotient has removable singularities there. Then the Riemann-Lebesgue lemma implies that
$$ \lim_{n\to\infty} \int_{-\pi}^\pi h(x) \sin (2nx) \, dx = 0, $$
so that
$$ \lim_{n\to\infty} \int_{-\pi}^\pi f(x) \sin (2nx) \, dx = \lim_{n\to\infty} \int_{-\pi}^\pi g(x) \sin (2nx) \, dx $$
if either of those limits exists. The integral on the right-hand side for fixed $n$ can now be evaluated by standard residue theorem techniques, using the substitutions $z=e^{ix}$, $\cos x = \frac12(z+z^{-1})$, $dx = \frac{dz}{iz}$, $\sin x = \frac{1}{2i}(z-z^{-1})$, $\sin (2nx) = \frac{1}{2i} (z^{2n}-z^{-2n})$, which yields
$$
\begin{align*}
\frac{\pi^2}{2} \int_{|z|=1|} & \frac{1-\frac12 (z+z^{-1})}{\frac1{2i}(z-z^{-1})}\frac{z^{2n}-z^{-2n}}{2i} \frac{dz}{iz}
= \frac{\pi^2}{4i} \int_{|z|=1} \frac{2-z-z^{-1}}{z^{2n}} \frac{z^{4n}-1}{z^2-1}\, dz \\
&=\frac{\pi^2}{4i} \int_{|z|=1} \frac{2-z-z^{-1}}{z^{2n}} (1+z^2+z^4\ldots + z^{4n-2}) \, dz \\
&=\frac{\pi^2}{4i} \int_{|z|=1} (2-z-z^{-1}) (z^{-2n}+z^{-2n+2}+z^{-2n+4}\ldots + z^{2n-2}) \, dz \\
&= \frac{\pi^2}{4i} 2\pi i a_{-1} = \frac{\pi^3}{2} a_{-1} = \frac{\pi^3}{2}(-2) = -\pi^3
\end{align*}
$$
where $a_{-1}=-2$ is the coefficient of $z^{-1}$ in the integrand which is easily found by multiplying the terms in parentheses.
|
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How to calculate $3^{45357} \mod 5$? I wrote some code, here is what it gives:
\begin{align*}
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
3^7 \mod 5 = 2 \\\\
3^8 \mod 5 = 1 \\
3^9 \mod 5 = 3 \\
3^{10} \mod 5 = 4 \\
3^{11} \mod 5 = 2 \\\\
3^{12} \mod 5 = 1 \\
3^{13} \mod 5 = 3 \\
...
\end{align*}
As I see, there's a period $1, 3, 4, 2$. And intuitively, it's easy to predict that $3^{45357} \mod 5 = 3$. Unfortunately, I'm almost unfamilar with modular math. Could you give me strict analytic explaination of this thing?
|
Since you say you are unfamiliar with modular arithmetic, here is an alternative proof. Your exponent $45357$ has form $\rm\:4n+1\:$ and your small table of values all satisfy $\rm\: mod\ 5\!:\ 3^{4n+1}\! \equiv 3,\:$ i.e. $\rm\:5\mid 3^{4n+1}\!-3 = 3(3^{4n}-1).\:$ This is true since $\rm\:5\cdot 16 = 3^4-1\:|\:3^{4n}\!-1\ $ by $\rm\:x-1\:|\:x^n-1$.
|
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How to prove $\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$ With $a + b + c = 3$ and $a, b, c>0$ prove these inequality:
1)$$\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$$
2)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+abc\geq 4$$
3)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+\frac{9}{4}abc\geq \frac{21}{4}$$
|
(Note: The notation $\sum$ refers to a cyclic sum here.)
For the first one:
We can assume WLOG that $a\ge b\ge c$. We will prove a lemma:
Let $x, y$ be positive reals with $xy\ge 1$. Then: $$\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge \frac{2}{1+xy}$$
Proof: Upon direct expanding, the inequality is equivalent to $(x-y)^2(xy-1)\ge 0$, which is indeed true.
Now applying the lemma with $x=\sqrt{\frac{ab}{c}}, y=\sqrt{\frac{ac}{b}}$, we have $xy=a\ge 1$ and thus:
$$\sum\frac{a}{a+bc}=\sum\frac{1}{1+\frac{bc}{a}}\ge \frac{2}{1+a}+\frac{a}{a+bc}$$
Note that $bc\le\frac{(b+c)^2}{4}=\frac{(3-a)^2}{4}=\frac{a^2-6a+9}{4}$. Thus:
$$\frac{2}{1+a}+\frac{a}{a+bc}\ge \frac{2}{1+a}+\frac{4a}{a^2-2a+9}$$ It suffices to show that the above is at least $\frac{3}{2}$. Indeed, direct calculation gives:
$$2(1+a)(a^2-2a+9)(\frac{2}{1+a}+\frac{4a}{a^2-2a+9}-\frac{3}{2})=3(3-a)(a-1)^2\ge 0$$
because $a\le 3$.
|
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For any odd integer $x,y$, $(x^2+2) \nmid (y^2+4)$
Possible Duplicate:
Quadratics and divisibility
Prove that for any odd integers $x$ and $y$, we have $(x^2+2) \nmid (y^2+4)$
|
If $x$ is odd, then $x^2+2\equiv 3\pmod{4}$, so $x^2+2$ must have a prime divisor $p$ of the form $4k+3$.
If $x^2+2$ divides $y^2+4$, then $y^2+4\equiv 0 \pmod{p}$, or equivalently $y^2\equiv -4\pmod{p}$. Let $w$ be the inverse of $2$ modulo $p$. Then $w^2y^2\equiv -4w^2\equiv -1\pmod{p}$, and therefore
$$(wy)^2\equiv -1\pmod{p}.$$
This is impossible, since that would mean that $-1$ is a quadratic residue of $p$. But it is a standard theorem that if $p$ is of the shape $4k+3$, then $-1$ is not a quadratic residue of $p$.
|
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Prove via induction $\frac{(n+1)^2}{2^n}\le\frac{9}{4}$ How do you prove in induction that:
$$\frac{(n+1)^2}{2^n}\le\frac{9}{4}$$
This is what I keep getting:
Checking for $n=1$ we get $2\le\frac{9}{4}$.
Assuming it's true for $n$ and checking for $n+1$ I get this:
$$\frac{(n+2)^2}{2^{n+1}}=\frac{2(n+1)^2-n^2+2}{2\times2^n}\le\frac{9}{4}-\frac{n^2-2}{2\times2^n}\le\frac{9}{4}$$
Which is true only for $n>1$.
|
For another proof, you can see that $\displaystyle \frac{u_{n+1}}{u_n}= \frac{1}{2} \left( 1+ \frac{1}{n+1} \right)^2 \leq 1$ with $n\geq 2$ and $\displaystyle u_n= \frac{(n+1)^2}{2^n}$. So $(u_n)$ decreases for $n\geq 2$ whence $\displaystyle u_n \leq u_2=\frac{9}{4}$ for $n \geq 2$. Finally, you only have to check the case $n=1$.
|
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Finding a value of $a$ to satisfy an expression of the form $a*(1-\frac{1}{b})^{(a-1)} = r$ Consider the following expression:
$a*(1-\frac{1}{b})^{(a-1)} = r$
Provided some real number value for $b$, I need to find a positive real number $0 < a \leq b$ to satisfy the above equation, where $0 < r < 1$.
Must we appeal to an approximation for the above expression to solve for $a \leq b$? If so, what is a good approximation that becomes better as $a \to Inf$?
|
We have
$$
\begin{align*}
a\left(1-\frac{1}{b}\right)^{a-1} &= r \\
a\left(1-\frac{1}{b}\right)^a &= \left(1-\frac{1}{b}\right)r \\
a e^{a \log\left(1-\frac{1}{b}\right)} &= \left(1-\frac{1}{b}\right)r \\
a \log\left(1-\frac{1}{b}\right) e^{a \log\left(1-\frac{1}{b}\right)} &= \left(1-\frac{1}{b}\right)r\log\left(1-\frac{1}{b}\right),
\end{align*}
$$
so that
$$
\begin{align*}
a \log\left(1-\frac{1}{b}\right) &= W\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right) \\
a &= \frac{W\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)},
\end{align*}
$$
where $W$ is the the Lambert W function. Note that $W(x)$ is double-valued when $x \in (-1/e,0)$, and the solution you want is given by the principal branch:
$$
a_0 = \frac{W_0\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)}
$$
If you'd like, you can expand this in a series which converges for $b$ large:
$$
a_0 = r + \frac{r(r-1)}{b} - \frac{3r^2(r-1)}{2b^2} + O(b^{-3}).
$$
Let us denote the other solution, given by the other branch of $W$, by
$$
a_{-1} = \frac{W_{-1}\left(\left(1-\frac{1}{b}\right) r \log\left(1-\frac{1}{b}\right)\right)}{\log\left(1-\frac{1}{b}\right)}.
$$
One can use the asymptotic series derived in this paper this paper (pp. 19-23) to calculate an expression for this solution as $b \to \infty$:
$$
a_{-1} = -\frac{1}{\log\!\left(1-\frac{1}{b}\right)}\left\{\log b + \log \log b - \log r + \frac{\log \log b}{\log b} - \frac{\log r}{\log b} + O\!\left(\frac{\log \log b}{\log b}\right)^2\right\}.
$$
This is an okay approximation but note that the absolute error does not decrease to $0$ so it won't be very helpful for numerics.
|
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Showing that such sequence is monotone. I have trouble figuring out how to show that
$\frac{\sqrt{n+1}}{(n+1)^{2}+1}<\frac{\sqrt{n}}{n^{2}+1},\quad\forall n\geq1$.
I have solved several inequalities before, but I can't prove this one.
Thanks!
|
This is equivalent to
$\sqrt{1+1/n} < ((n+1)^2+1)/(n^2+1)$.
Since $\sqrt{1+1/n} < 1+1/(2n)$
and $((n+1)^2+1)/(n^2+1) = 1+(2n+1)/(n^2+1)$,
this is true if
$1/(2n) < (2n+1)/(n^2+1)$
or $4n^2+2n > n^2+1$,
which is true for $n \ge 1$.
Note that more than this is true: Let's try to find the largest value of $c > 1$
such that
$\sqrt{1+c/n} < ((n+1)^2+1)/(n^2+1)$.
As before,
$\sqrt{1+c/n} < 1+c/(2n)$
so we want
$c/(2n) < (2n+1)/(n^2+1)$
or $n^2+1 < (2/c)n(2n+1)
=(4/c)n^2+2/c$.
For $c = 4$ this is false,
but it is true for large enough $n$
for any $c < 4$.
To see this, let $4/c = 1+d$ where $d > 0$.
Then we want
$n^2+1 < (1+d)n^2 + (1+d)/2$
or $d n^2 > 1-(1+d)/2 = (1-d)/2$
or, assuming $d < 1$ (this is true for all $n$ if $d \ge 1$),
$n^2 > (1-d)/(2d) = 1/(2d) -1/2$.
Thus $n > \sqrt{1/2d}$ works.
So, there is no largest value of $c$
such that
$\sqrt{1+c/n} < ((n+1)^2+1)/(n^2+1)$
(at least with this elementary method),
but the inequality is true for any $c < 4$ for all large enough $n$.
|
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|
Unusual pair of dice Suppose we roll two ordinary, 6-sided dice, $f_i=s_i=\{1,2,3,4,5,6\} $.
All $36$ possible rolls with a pair of dice (first dice - $f_i$, second dice $s_i$):
$$
\begin{array}{ccccccc}
f_i/s_i & 1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
2 & 3 & 4 & 5 & 6 & 7 & 8 \\
3 & 4 & 5 & 6 & 7 & 8 & 9 \\
4 & 5 & 6 & 7 & 8 & 9 & 10 \\
5 & 6 & 7 & 8 & 9 & 10 & 11 \\
6 & 7 & 8 & 9 & 10 & 11 & 12
\end{array}
$$
So we have $1$ variant to roll the sum $2, 2$ variants to roll $3, 3$ variants to roll $4, ..., 2$ variants to roll $11, 1$ variant to roll $12.$
Is there some another pair $(f_i, s_i)$ of $6$-sided dice with different numbers on its sides, but with the same set of variants to roll the same sums $(2,\ldots,12)$? And all $f_i>0, s_i>0$. And how much?
$f_i$ and $s_i$ in such pair can be different.
Special upd $f_i \in \mathbb{N}$, $s_i \in \mathbb{N}$.
|
Yes.
If we factor the generating function $(x+x^2+x^3+x^4+x^5+x^6)^2$, we obtain
$$(x+x^2+x^3+x^4+x^5+x^6)^2= x^2(x+1)^2(x^2-x+1)^2(x^2+x+1)^2$$
and have the option of distributing the factors unevenly to the dice. But (apart from the standard dice) the only possibility to have nonnegative coefficients, no constant term and coefficient sum six is given by
$$(x^2 -x+1)^2(x^2+x+1)(x+1)=x^8 + x^6 + x^5 + x^4 + x^3 + x$$
and
$$(x^2+x+1)(x+1)x=x^4 + 2x^3 + 2x^2 + x$$
corresponding to dice with labels $(1, 3, 4, 5, 6, 8)$ and $(1, 2, 2, 3, 3, 4)$.
There is no other way because each die must obtain one factor of $x$ (to have positive labels), one factor of $(x+1)$ to have even face count and one factor of $1+x+x^2$ to have face count divisible by three. Only the factors $x^2-x+1$ are "expendible".
|
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Determining the number $N$ Let $1 = d_1 < d_2 <\cdots< d_k = N$ be all the divisors of $N$ arranged in increasing order. Given that $N=d_1^2+d_2^2+d_3^2+d_4^2$, determine $N$. The divisors include $N$. It seems that $130$ is an answer. Is there another possible answer for $N$?
|
The equation $$N = 1 + d_1^2 + d_2^2 + d_3^2$$ implies $N$ even by reducing $\mod 2$. Suppose $2^2|N$ then there are three possible cases:
*
*$N = 1 + 2^2 + 4^2 + 8^2$ - but this is impossible by arithmetic.
*$N = 1 + 2^2 + 4^2 + p^2$ (with $p>3$) - but reducing $\mod p$ we find $0 \equiv 13 \pmod p$ so $p = 13$ but this too is impossible by arithmetic.
*$N = 1 + 2^2 + 3^2 + 4^2$ - impossible by arithmetic.
So $2$ is the highest power of $2$ dividing $N$, thus we have the cases:
*
*$N = 1 + 2^2 + p^2 + q^2$ (with $q < 2p$) but this is impossible by reducing $\mod 2$
*$N = 1 + 2^2 + p^2 + (2p)^2$ so $N = 5(1+p^2)$ and $5|N$ so $p$ must be $3$ (impossible by arithmetic) or $5$.
This shows the only possible solution is $130$.
|
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Variance of binomial distribution Why for $X\sim B(n,p)$ is $Var(X)=np(1-p)$?
$Var(X)=\sum x_i^2 p_i -(\sum x_i p_i)^2=\sum_{r=0}^n r^2 \binom{n}{r}p^r(1-p)^{n-r}+( \sum_{r=0}^n r \binom{n}{r}p^r(1-p)^{n-r} )^2$
In my short-sightedness, I don't see any viable ways to derive the variance from this.
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An easier way is to recognize that $X = Y_1 + Y_2 + \cdots Y_n$ where $Y_k$ are independent Bernoulli random variables with parameter $p$. For a Bernoulli random variable $Y_k$, we have $$\text{Var}(Y_k) = p(1-p)$$ Since $Y_k$ are independent, we have that
$$\text{Var}(X) = \text{Var}(Y_1) + \text{Var}(Y_2) + \cdots + \text{Var}(Y_n) = np(1-p)$$
To go the direct way, we need to first evaluate couple of summations.
We will evaluate the sums $$\sum_{k = 0}^n k \mathbb{P}(X=k) \text{ and }\sum_{k = 0}^n k^2 \mathbb{P}(X=k)$$
First note that $\mathbb{P}(X=k) = \dbinom{n}k p^k (1-p)^{n-k}$. Hence, $$\sum_{k = 0}^n k \mathbb{P}(X=k) = \sum_{k = 0}^n k \dbinom{n}k p^k (1-p)^{n-k}$$
Note that $$k \dbinom{n}k = \dfrac{n!}{(n-k)! (k-1)!} = n \dbinom{n-1}{k-1}$$ Hence,
\begin{align}
\sum_{k = 0}^n k \mathbb{P}(X=k) & = \sum_{k = 1}^n n \dbinom{n-1}{k-1} p^k (1-p)^{n-k} = np \sum_{k = 1}^n \dbinom{n-1}{k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\
& = np \left( p + (1-p)\right)^{n-1} = np
\end{align}
Similarly,
\begin{align}
k^2 \dbinom{n}k & = k \dfrac{n!}{(n-k)! (k-1)!} = n k \dbinom{n-1}{k-1}\\
& = n (k-1) \dbinom{n-1}{k-1} + n \dbinom{n-1}{k-1}\\
& = n(n-1) \dbinom{n-2}{k-2} + n \dbinom{n-1}{k-1}
\end{align}
Hence,
\begin{align}
\sum_{k = 0}^n k^2 \mathbb{P}(X=k) & = \sum_{k = 2}^n n(n-1) \dbinom{n-2}{k-2} p^k (1-p)^{n-k} + \sum_{k = 1}^n n \dbinom{n-1}{k-1} p^k (1-p)^{n-k}\\
& = n(n-1)p^2 + n p
\end{align}
Hence,
\begin{align}
\text{Var}(X) & = \sum_{k = 0}^n k^2 \mathbb{P}(X=k) - \left(\sum_{k = 0}^n k \mathbb{P}(X=k) \right)^2\\
& = n(n-1)p^2 + n p - (np)^2 = n^2p^2 - np^2 + np - n^2 p^2\\
& = np(1-p)
\end{align}
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|
Find the area of largest rectangle that can be inscribed in an ellipse The actual problem reads:
Find the area of the largest rectangle that can be inscribed in the ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
I got as far as coming up with the equation for the area to be $A=4xy$ but then when trying to find the derivative I don't think I'm doing it right.
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let L and H be the length and breadth of the required rectangle respectively
$$\frac{(L/2)^2}{a^2}+\frac{(H/2)^2}{b^2}=1$$
$$\frac{(L)^2}{4a^2}+\frac{(H)^2}{4b^2}=1$$
$$H=\frac{b}{a}\sqrt{4a^2-L^2}$$
Area=L*H
$$A= L*\frac{b}{a}\sqrt{4a^2-L^2}$$
$$\frac{dA}{dL} =\frac{b}{a}\sqrt{4a^2-L^2}-\frac{L^2b}{a\sqrt{4a^2-L^2}}=0$$
$$\frac{b*(4a^2-2L^2)}{a\sqrt{4a^2-L^2}}$$
$$=> 4a^2-2L^2=0$$
$$=2a^2=L^2$$
$$L=a\sqrt{2}$$
$$H=b\sqrt{2}$$
$$Area=L*H=2ab$$
$$\frac{d^2A}{dL^2}=\frac{\sqrt{4a^2-L^2}*(-4L)-\frac{4a^2-2L^2}{2\sqrt{4a^2-L^2}}}{4a^2-L^2}$$
Putting L=$a\sqrt{2}$
$$\frac{d^2A}{dL^2}=\frac{-a\sqrt{2}(4a\sqrt{2})-\frac{0}{2\sqrt{4a^2-2a^2}}}{4a^2-2a^2}$$
$$=\frac{-8a^2}{2a^2}$$
-4<0.
Therefore the area is max
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|
Integral $\int \sqrt{4+4x^2+1/x^2} dx$ I'm totally lost on how I would integrate the following function:
$f(x) = \sqrt{4+4x^2+1/x^2}$
If anyone could even just point me to the method of integration that would be grand.
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$$4 + 4x^2 + \dfrac1{x^2} = \left(2x + \dfrac1x \right)^2$$
Hence,
\begin{align}
\sqrt{4 + 4x^2 + \dfrac1{x^2}} & = \begin{cases} 2x + \dfrac1x & x > 0\\ -\left( 2x + \dfrac1x\right) & x < 0\end{cases}\\
& = 2 \vert x \vert + \dfrac1{\vert x \vert}
\end{align}
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|
(Wolframalpha) this shouldnt give x=50? This is my problem:
I need x or y for the triangle area that forms between the vertical axis(y) and the function y=100+2x where the area is equal to 2500.
so I used for condition to the linear function:
knowing that the triangle area in this case should be like:
x*y/2=area, so:
x-100*y/2=2500
x-100*y=5000
y=5100/x
and then:
5100/x=100+2x
5100=100+2x*x
5000=2x^2
sqrt(2500)=x
50=x
the weird thing is that works for any area, and gives me the correct result for what I'm looking for, wich is x=50 and y=f(50)=200, if the area is calculated as is shown in the condition: 200-100*50/2=2500 !
5100/x=100+2x [http://www.wolframalpha.com/input/?i=5100%2Fx%3D100%2B2x][1]
it outputs x=-5 (5+sqrt(127)) and x=5 (-5+sqrt(127))
how can I get the same results on wolframalhpa ?
thank you ! (:
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You start with the equation $\frac{5100}{x} = 100 + 2x.$ If the left side and the right side are equal, then I can do the same to both sides and they'll still be equal. Let's multiply both sides by $x$. I get $5100 = 100x + 2x^2.$ Bringing all of the terms over to one side, we get $2x^2 + 100x - 5100 = 0.$ Next, notice that there is a common factor: $2(x^2 + 50x - 2550) = 0.$ Finally, we use the quadratic formula where $a = 1,$ $b = 50$ and $c = -2550$. We have:
$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \, ,$$
$$x = \frac{-50 \pm \sqrt{50^2-4\times 1 \times (-2550)}}{2 \times 1} \, , $$
$$x = \frac{-50 \pm \sqrt{12700}}{2} \, , $$
$$x = \frac{-50 \pm 10\sqrt{127}}{2} \, , $$
$$x = -25 \pm 5\sqrt{127} \, .$$
It seems that the website was correct. Notice that $-5(5\pm \sqrt{127}) = -25 \mp 5\sqrt{127}.$ If you don't see why $\sqrt{12700} = 10\sqrt{127}$ then notice that $12700 = 2^2 \times 5^2 \times 127$, where $127$ is prime and so:
$$\sqrt{12700} = \sqrt{2^2 \times 5^2 \times 127} = \sqrt{2^2} \times \sqrt{5^2} \times \sqrt{127} = 2 \times 5 \times \sqrt{127} = 10\sqrt{127} \, . $$
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Does the equation have infinite number of solutions in integers? Does either of the equations ${a^2} - 10{b^2} = \pm 1$ have infinite number of solutions in integers?
If the answers is yes, a hint about how to reduce this problem to the problem of Pythagorean triples in the Gaussian integers $\mathbb{Z}\left[ i \right]$ would be enough for solution.
If not, how would one proceed to prove that?
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Clearly $$3^2 - 10 \cdot 1^2$$ is a solution.
Now observe that:
*
*$(a + b \sqrt{10})(a - b \sqrt{10}) = a^2 - 10 b^2$ for any integers $a,b$.
*$(a + b \sqrt{10})(c + d \sqrt{10}) = (ac+10bd) + (ad + bc)\sqrt{10}$
therefore every power $r>1$, $(3+\sqrt{10})^r$ will give a new solution of the Diophantine equation (clearly the components increase in size).
*
*$(3+\sqrt{10})^2 = (19 + 6 \sqrt{10})$ and $19^2 - 10\cdot 6^2 = 1$
*$(3+\sqrt{10})^3 = (117 + 37 \sqrt{10})$ and $117^2 - 10\cdot 37^2 = -1$
*...
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|
Infinitely many odd and even integers in a sequence. Consider the sequence $\{a_n\}_{n\in\Bbb N}$ where $a_n = \lfloor n \sqrt2 \rfloor + \lfloor n \sqrt3 \rfloor $. Show that there exists infinitely many odd and even integers in this sequence.
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$1 \lt (n+1) \sqrt2 - n\sqrt2 \lt 1.5 $ so $1 \le \lfloor (n+1) \sqrt2 \rfloor- \lfloor n \sqrt2 \rfloor \le 2$
$1.5 \lt (n+1) \sqrt3 - n\sqrt3 \lt 2 $ so $1 \le \lfloor (n+1) \sqrt3 \rfloor- \lfloor n\sqrt3 \rfloor \le 2$
So $a_{n+1}-a_{n}$ must be $2$, $3$ or $4$.
Similarly $2 \lt (n+2) \sqrt2 - n \sqrt2 \lt 3 $ so $2 \le \lfloor (n+2) \sqrt2 \rfloor- \lfloor n \sqrt2 \rfloor \le 3$
and $3 \lt (n+2) \sqrt3 - n \sqrt3 \lt 4 $ so $3 \le \lfloor (n+2) \sqrt3 \rfloor- \lfloor n \sqrt3 \rfloor \le 4$
So $a_{n+2}-a_{n}$ must be $5$, $6$ or $7$.
To avoid infinitely many odd and even integers in this sequence we would need $a_{n+2k}-a_{n}=6k$ (i.e. even) for all $n$ greater than some $N$ and all positive $k$. But that is not possible, as $\sqrt2 +\sqrt3 \gt 3.1$ and so for large enough $k$ (e.g $10$ or more) we have $a_{n+2k}-a_{n} \gt 6k$.
|
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|
Finding Where A Parametric Curve Intersects Itself The problem I am working on is to find the where the curve intersects itself, using the parametric equations.
These are: $x=t^2-t$ and $y=t^3-3t-1$
For the graph to intersect itself, there must be two distinct t-values, $a$ and $b$, that when plugged into the parametric equations, produce the same output. These two t-values create two ordered-pairs that are the same.
My system of equations: $a^2-a=b^2-b$ and $a^3-3a-1=b^3-3b-1$.
I solved for $a$, but am not sure if I did it correctly: $a(a-1)=b^2-b$ then either $a=b^2-b$ or $a-1=b^2-b$. I would then have two values that I have to test. When I plugged in $a$, I ended up with a 6-degree polynomial, did I do something wrong?
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Well, $\,xy=\alpha\beta\,$ doesn't necessarily means $\,x=\alpha\,\,\vee\,\,x= \beta\,$ , of course. Form a quadratic in $\,a\,$ and solve:
$$a^2-a=b^2-b\Longrightarrow a^2-a+b-b^2=0\Longrightarrow \Delta=1-4(b-b^2)=(2b-1)^2\Longrightarrow$$
$$a_{1,2}=\frac{1\pm\sqrt\Delta}{2}=\begin{cases}b\\{}\\1-b\end{cases}$$
If $\,a=b\,$ we're done, otherwise take $\,a=1-b\,$ and substitute into the second equation:
$$a^3-3a-1=b^3-3b-1\Longrightarrow 1-3b+3b^2-b^3-3+3b=b^3-3b\Longrightarrow$$
$$0=2b^3-3b^2-3b+2=(b+1)(2b^2-5b+2)\Longrightarrow b=-1\,,\,2\,,\,\frac{1}{2}$$
Thus, we have the solution:
$$t=-1\,,\,t=2\Longrightarrow (x,y)=(2,1)$$
Note that the other two possibilities give us the same $\,t\,$!
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Defining/constructing an ellipse Years ago I was confronted with a (self imposed) problem, which unexpectedly resurfaced just recently... I don't know whether it makes sense to explain the background or not, so I'll be brief.
If I was given two points in a plane, and two lines passing through them would I be able to construct (or get a equation of) an ellipse to which those lines are tangents in given points? At a first glance, four equations (each point belonging to the ellipse, and each of the lines being a tangent in respective point) would be insufficient to solve for five unknowns (major and minor axis, coordinates of the ellipse centre and the angle for coordinate system transform)- so there would be infinite number of ellipses that could be drawn with those constraints. But it was intuitive to me that only finite number would indeed be possible... So, what is the correct answer to this?
Hopefully, I wont be considered rude- but there is a twist to this... Firstly, I believe to have solved this problem, but recently I was working in a top of the line CAD program, where those constraints seemed to yield an infinite number of solutions. This bit puzzled me greatly- because it doesn't seem correct (so I'm assuming the ellipses they use are not proper ellipses). Secondly, I hope it won't be taken against me because I'm trying to pick your brains and having a theory of my own... which I will disclose a bit later.
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OK, if it's of any interest, here it goes...
Assuming the equation of the ellipse $ b^2x^2+a^2y^2=a^2b^2 $, we get:
$$ y=\pm\frac{b}{a}\sqrt{a^2-x^2}$$
$$y´=-\frac{b^2x}{a^2y}=\mp\frac{b}{a}\frac{x}{\sqrt{a^2-x^2}} $$
For two points $ i=1,2 $ equation of tangent is:
$$ t_i ... y=y´(x-x_i)+y_i=\pm\frac{b}{a}\frac{a^2-x \centerdot x_i}{\sqrt{a^2-x_i^2}} $$
Intersection of those tangents will be ( $ x_I $ is calculated first by equating right sides of tangent equations):
$$ \pm\frac{b}{a}\frac{a^2-x_Ix_1}{\sqrt{a^2-x_1^2}}=\pm\frac{b}{a}\frac{a^2-x_Ix_2}{\sqrt{a^2-x_2^2}} $$
$$ \Downarrow $$
$$ \left( a^2-x_Ix_1\right)\sqrt{a^2-x_2^2}=\left( a^2-x_Ix_2\right)\sqrt{a^2-x_1^2} $$
$$ \Downarrow $$
$$ x_I=a^2\frac{\sqrt{a^2-x_2^2}-\sqrt{a^2-x_1^2}}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}} $$
$$ y_I=\pm\frac{b}{a}\frac{a^2-x_Ix_1}{\sqrt{a^2-x_1^2}}=\text{ ... }=\pm ab\frac{x_1-x_2}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}} $$
Coordinates of midpoint of a secant:
$$ x_S=\frac{x_1+x_2}{2}\text{ , }y_S=\frac{y_1+y_2}{2}=\pm \frac{b}{a}\frac{\sqrt{a^2-x_1^2}+\sqrt{a^2-x_2^2}}{2} $$
If we assume the equation of the line connecting intersection of tangents and midpoint of secant to be: $ y=\frac{y_I-y_S}{x_I-x_S}x+l $.
In order to prove my claim that that line must pass through the centre of the ellipse, it will be: $ l=0 $- so let's calculate l:
$$ l=y_S-\frac{y_I-y_S}{x_I-x_S}x_S=\frac{x_Iy_S-x_Sy_I}{x_I-x_S} $$
Concentrating on the numerator:
\begin{eqnarray*}
x_Iy_S-x_Sy_I&=&a^2\frac{\sqrt{a^2-x_2^2}-\sqrt{a^2-x_1^2}}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}}\left(\pm \frac{b}{a}\frac{\sqrt{a^2-x_1^2}+\sqrt{a^2-x_2^2}}{2}\right)-\frac{x_1+x_2}{2}\left(\pm ab\frac{x_1-x_2}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}}\right)\\
&=&\pm\frac{1}{2} ab \frac{\left(\sqrt{a^2-x_2^2}-\sqrt{a^2-x_1^2}\right) \left( \sqrt{a^2-x_1^2}+\sqrt{a^2-x_2^2}\right)-\left(x_1+x_2 \right)\left(x_1-x_2 \right)}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}}\\
&=&\pm\frac{1}{2} ab \frac{\left[ \left(a^2-x_2^2 \right)-\left(a^2-x_1^2 \right) \right]-\left(x_1^2-x_2^2 \right)}{x_1\sqrt{a^2-x_2^2}-x_2\sqrt{a^2-x_1^2}}=0
\end{eqnarray*}
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Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I want to try to prove that for example $$a+\frac{bc}{a} \geq a+b$$
but I don't know if is is a good idea.
Thanks:)
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By C-S $$\sum_{cyc}(a^2+bc)=\prod_{cyc}\sqrt{(a^2+bc)(ac+b^2)}\geq\prod_{cyc}\left(\sqrt{a^3c}+\sqrt{b^3c}\right)=abc\prod_{cyc}(a+b).$$
Done!
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Prove that $5/2 < e < 3$?
Prove that $$\dfrac{5}{2} < e < 3$$
By the definition of $\log$ and $\exp$,
$$1 = \log(e) = \int_1^e \dfrac{1}{t} dt$$
where $e = \exp(1)$.
So given that $e$ is unknown, how could I prove this problem? I think I'm missing some important facts that could somehow help me rewrite $e$ in some form of $3$ and $5/2$. Any idea would be greatly appreciated.
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Let $$e_n=\left(1+\frac{1}{n}\right)^n,\quad s_n=\sum_{k=0}^{n}\frac{1}{k!}\ (0!=1),$$
then $e_1=2=s_1,e_2=\frac{9}{4}<\frac{5}{2}=s_2$, when $n>2$,
\begin{align*}
\left(1+\frac{1}{n}\right)^n
&=1+1+\sum_{k=2}^{n}\frac{1}{k!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\\
&<1+1+\sum_{k=2}^{n}\frac{1}{k!}\\
&=1+1+\frac{1}{2}+\sum_{k=3}^{n}\frac{1}{k!}\\
&\leq\frac{5}{2}+\sum_{k=3}^{n}\frac{1}{k(k-1)(k-2)}\\
&=\frac{5}{2}+\frac{1}{2}\sum_{k=3}^{n}\frac{1}{k-1}\left(\frac1{k-2}-\frac1k\right)\\
&=\frac{5}{2}+\frac{1}{2}\sum_{k=3}^{n}\frac{1}{(k-1)(k-2)}-\frac{1}{2}\sum_{k=3}^{n}\frac{1}{k(k-1)}\\
&=2.75+\frac{1}{2}\left(\frac1n-\frac{1}{n-1}\right)\\
&<2.75.
\end{align*}
Let $n\to \infty$, we have $$e\leq 2.75<3.$$
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Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$ I've tried following this way, but I haven't succeeded.
Thank you!
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Let $\displaystyle\arctan x= y$
$\implies(i) \tan y =x$
and $(ii)\displaystyle-\frac\pi2\le y\le\frac\pi2$ (using the definition of principal value)
$\implies \cos y\ge0$
We have $$\frac{\sin y}x=\frac {\cos y }1=\pm\sqrt{\frac{\sin^2y+\cos^2y}{x^2+1^2}}=\pm\frac1{\sqrt{x^2+1}}$$
$\displaystyle\implies \cos y=+\frac1{\sqrt{x^2+1}}$ and $\displaystyle\sin y=\frac x{\sqrt{x^2+1}}$
So, $\displaystyle\arctan x= y=\arcsin\frac x{\sqrt{x^2+1}}=\arccos\frac1{\sqrt{x^2+1}}$
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Determine that improper integral converge or diverge
Possible Duplicate:
Convergence/divergence of $\int_0^{\infty}\frac{x-\sin x}{x^{7/2}}\ dx$
Determine the improper integral $\int_0^{\infty} \frac{x-\sin x}{x^{7/2}}dx$ converge or diverge. Prove that please.
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Write for $0< \varepsilon<1$
$$\int\limits_{0}^{+\infty}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}=\int\limits_{0}^{\varepsilon}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}+\int\limits_{\varepsilon}^{1}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}+\int\limits_{1}^{+\infty}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}.$$
Since for small $x, \;\; 0<x<\varepsilon$ $$x-\sin{x}=\dfrac{x^3}{3!}+O(x^5), $$ then
$$\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\sim {\dfrac{1}{3!}}x^{3-\frac{7}{2}}={\dfrac{1}{3!}}x^{-\frac{1}{2}},$$
therefore the first integral in RHS converges.
The second integral is proper integral, therefore it is finite.
The third integral converges, since $\left|\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\right| \leqslant \dfrac{x+1}{x^{\frac{7}{2}}} \leqslant \dfrac{2x}{x^{\frac{7}{2}}}=2x^{-\frac{5}{2}}.$
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|
If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
|
A solution and some generalization:
First of all, we can replace $b$ by $-b$, since $\gcd(a,b)=\gcd(a,-b)$. So we need to show $\gcd(a^2+ab+b^2,a-b)$.
If $p$ is a prime dividing the gcd, then it also divides $a-b$ which means $a \equiv b \mod p$, hence:
$$0 = a^2+ab+b^2=a^2+a^2+a^2=3a^2 \mod p$$
So either $p|3$ or $p|a$. $p$ can't divide $a$, because then it also divides $b$ and so $\gcd(a,b) \neq 1$. Thus $p=3$. So $\gcd(a^2+ab+b^2,a-b)$ is a power of 3 (this includes 1).
Can it be divisible by 9? No, since $a^2+ab+b^2=(a-b)^2+3ab$, so if $9$ divides the gcd, it divides $(a-b)^2$ and $a^2+ab+b^2$, and so it divides $3ab$, i.e. $3|ab$, so one of $a,b$ is divisible by 3, a contradiction ($3|a-b,a\implies 3|b \implies (a,b)!= 1$).
The proof resembles Hagen von Eitzen's proof, but it can be generalized. What is very noticeable is that $a^2+ab+b^2,a-b$ are both homogenizations of Cyclotomic polynomials, $\phi_1(x)=x-1$ and $\phi_3(x)=x^2+x+1$.
The following more general statement can be proved in a similar way: let $\phi_n(x), \phi_m(x)$ be 2 Cyclotomic polynomials. Assume $m=1$ and $n=p$, a prime, so $\phi_m(x)=x-1,\phi_n(x)=x^{p-1}+x^{p-2} + \cdots + 1$.
Define their homogenization as $h_n(x,y)=\phi_n(\frac{x}{y})y^{\phi(n)}=\sum_{i=0}^{p-1}x^i y^{p-1-i},h_m(x,y)=\phi_m(\frac{x}{y})y^{\phi(m)}=x-y$.
I claim that $\gcd(x,y)=1 \implies \gcd(\phi_m(x,y),\phi_m(x,y)) \in \{1,p \}$.
Proof (for $p \neq 2$):
*
*If a prime $q$ divides the gcd, then $\sum_{i=0}^{p-1}x^i y^{p-1-i} = px^{p-1} \equiv 0 \mod q \implies p=q$, as before.
*$p^2$ can't divide the gcd: assume $x \equiv y \mod p$, i.e. $x=y+pk$ for some integer $k$. Note that $x^i = y^i + y^{i-1}pki \mod p^2$. Thus:
$$\sum_{i=0}^{p-1}x^i y^{p-1-i} = \sum_{i=0}^{p-1} ( y^i + y^{i-1}pki) y^{p-1-i} =$$
$$p y^{p-1} + pk y^{p-2} \sum_{i=1}^{p-1} i = py^{p-1} + pk y^{p-2} \binom{p}{2} =$$
$$ y^{p-2}p (y+k\frac{p-1}{2}) \mod {p^{2}}$$
Since $p$ doesn't divide $y$ (as $\gcd(x,y)=1$), if this is zero modulo $p^2$ it means $y+k\frac{p-1}{2} \equiv 0 \mod p$, and so $p \nmid k$ (else $p|y$, contradiction), so $p^2 \nmid x-y$.
A new question arises - what are the possible values of $gcd(\phi_n(x), \phi_m(x))$, for general $n,m$? From this problem I can probably deduce a similar result for homogeneous polynomials.
EDIT: I think that the fact $\phi_n(1) = p$ if $n=p^{k}$ and $1$ otherwise should help, it gives $\gcd(h_n(x,y),h_m(x,y))=1$ if $(n,m)=1$ and one of $n,m$ is not a prime power or 1, and $\gcd(h_n(x,y),h_m(x,y))$ is a power of $p$ if $n=1, m=p^k, k>1$.
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"url": "https://math.stackexchange.com/questions/257392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Prove that $\int_0^{+\infty} \frac{\ln x}{a^2+x^2} dx = \frac{\pi\ln a}{2a}$ Is that true $$\int_0^{+\infty} \cfrac{\ln x}{a^2+x^2} dx = \cfrac{\pi\ln a}{2a},$$ where $a>0$ ? And how to compute it?
|
Another approach :
Consider
$$
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx.\tag1
$$
Rewrite $(1)$ as
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac1{a^2}\int_0^\infty\frac{x^{b-1}}{1+\left(\frac{x}{a}\right)^2}\ dx.\tag2
\end{align}
Putting $x=ay\;\color{blue}{\Rightarrow}\;dx=a\ dy$ yields
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=a^{b-2}\int_0^\infty\frac{y^{b-1}}{1+y^2}\ dy,
\end{align}
where
\begin{align}
\int_0^\infty\frac{y^{b-1}}{1+y^2}\ dy=\frac{\pi}{2\sin\left(\frac{b\pi}{2}\right)}.
\end{align}
Hence
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac\pi2\cdot\frac{a^{b-2}}{\sin\left(\frac{b\pi}{2}\right)}.\tag3
\end{align}
Differentiating $(3)$ with respect to $b$ and setting $b=1$ yields
\begin{align}
\int_0^\infty\frac{\partial}{\partial b}\left[\frac{x^{b-1}}{a^2+x^2}\right]_{b=1}\ dx&=\frac\pi2\frac{\partial}{\partial b}\left[\frac{a^{b-2}}{\sin\left(\frac{b\pi}{2}\right)}\right]_{b=1}\\
\int_0^\infty\frac{\ln x}{x^2+a^2}\ dx&=\large\color{blue}{\frac{\pi\ln a}{2a}}.
\end{align}
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"url": "https://math.stackexchange.com/questions/260621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Plotting $(x^2 + y^2 - 1)^3 - x^2 y^3 = 0$ I have no idea how this equation:
\begin{equation}
(x^2 + y^2 - 1)^3 - x^2 y^3 = 0
\end{equation}
Produces this picture:
Can someone provide a general explanation of plotting this function?
|
It's easier to be understood in polar coordinates.
(x^2 + y^2 - 1)^3 - x^2*y^3 == 0 /. {x -> r*Cos[\[Theta]],
y -> r*Sin[\[Theta]]} // FullSimplify
% // ApplySides[#/r^5 &, #] &
$$
\left(r^2-1\right)^3=r^5 \sin ^3(\theta ) \cos ^2(\theta ) \\
\frac{\left(r^2-1\right)^3}{r^5}=\sin ^3(\theta ) \cos ^2(\theta )
$$
Plot[Sin[\[Theta]]^3*Cos[\[Theta]]^2, {\[Theta], 0, 2 Pi}]
Plot[(r^2 - 1)^3/r^5, {r, 0, 5}]
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/263737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $m,n\in\mathbb{N}$ and $f(m) \mid f(n)$ for the $f$ given below, must $m=n$? If for two natural number $m$ and $n$, $(2^{2m+1}-1)2^{4m-2}(2^{2m+1}+2^{m+1}+1)\mid(2^{2n+1}-1)2^{4n-2}(2^{2n+1}+2^{n+1}+1)$, then $m=n$?
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Another answer. Let $p(z)=(2z^2-1)(2z^2+2z+1)$. Then you want $m<n$ and $p(2^m) \mid p(2^n)$.
Given $m$, let $D_m=p(2^m)$. Since $D_m$ is odd, if $n\equiv m \pmod {\phi(D_m)}$, then $2^n \equiv 2^m \pmod {D_m}$ and thus $D_n=p(2^n)\equiv P(2^m)=D_m \equiv 0\pmod {D_m}$
So for any $m$, there are infinitely many $n$.
Actually, you can show that $$2^{8m+4}\equiv 1\pmod {2^{2m+1}+2^{m+1}+1}$$
and
$$2^{8m+4}\equiv 1\pmod {2^{2m+1}-1}$$
So $2^{8m+4}=1\pmod {D_m}$. So, if $m\equiv n \pmod {8m+4}$, then $p(2^m)\mid p(2^n)$.
E.g., if $m=2$, $8m+4=20$, and we get $D_2\mid D_{22}$.
More generally, if $p(z)$ is an integer polynomial with $p(0)$ odd, then for any $m$, there are infinitely many $n>m$ such that $p(2^m)\mid p(2^n)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/264649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Help me solve this olympiad challenge? Given:
$$p(x) = x^4 - 5773x^3 - 46464x^2 - 5773x + 46$$
What is the sum of all arctan of all the roots of $p(x)$?
|
Nothing additional, but for the OP a more elementary approach.
$(x-a_1)(x-a_2)(x-a_3)(x-a_4) = x^4 - c_1x^3 + c_2x^2 - c_3x + a_1a_2a_3a_4.$ Where
$c_1 = \Sigma a_i$, $c_2 = \Sigma_{i<j}a_ia_j$ and $c_3 = \Sigma_{i <j < k}a_ia_ja_k$. Let $t_i=\tan^{-1}a_i.$ You want to compute $\Sigma t_i.$ Here, the trick is knowing the $\tan(x+y)$ expansion
$$\tan(x+y) = \frac{\tan x + \tan y}{1- \tan x \tan y}.$$
Apply twice for $x=t_1+t_2$ and $y=t_3+t_4$.
$$\tan(t_1+t_2+t_3+t_4) = \frac{\tan(t_1+t_2) + \tan(t_3+t_4)}{1+\tan(t_1+t_2)\tan(t_3+t_4)}=\frac{\frac{a_1 + a_2}{1- a_1a_2}+\frac{a_3 + a_4}{1- a_3a_4}}{1-\frac{a_1 + a_2}{1- a_1a_2}\frac{a_3 + a_4}{1- a_3a_4}} = \frac{(a_1+a_2)(1-a_3a_4)+(a_3+a_4)(1-a_1a_2)}{(1-a_1a_2)(1-a_3a_4)-(a_1+a_2)(a_3+a_4)}=\frac{c_1-c_3}{1+c_4-c_2}.$$ For this polynomial $c_1=c_3$, thefore $\tan\Sigma t_i =0,$ so is $\Sigma t_i =0$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
A relationship between matrices, bernoulli polynomials, and binomial coefficients We define the following polynomials, for $n≥0$:
$$P_n(x)=(x+1)^{n+1}-x^{n+1}=\sum_{k=0}^{n}{\binom{n+1}{k}x^k}$$
For $n=0,1,2,3$ this gives us,
$$P_0(x)=1\enspace P_1(x)=2x+1\enspace P_2(x)=3x^2+3x+1\enspace P_3(x)=4x^3+6x^2+4x+1$$
We then define the set $P_{(3)}=\{P_0,P_1,P_2,P_3\}$. It can be easily shown that this set is a basis over the vector space of polynomials of degree $3$ and lower. We take $3$ for the sake of brevity.
Taking the coefficients of these polynomials and turning them into column vectors, we can construct the matrix (coefficients from the lowest term to the highest term)
$$\large{M_{P_{(3)}}}=\begin{pmatrix}
1 & 1 & 1 & 1 \\
0 & 2 & 3 & 4 \\
0 & 0 & 3 & 6 \\
0 & 0 & 0 & 4
\end{pmatrix}$$
We'll call this matrix the pascal in the context of this post, and the above polynomials as pascal polynomials.
The inverse of this matrix is the matrix,
$$M_{P_{(3)}}^{-1}=\begin{pmatrix}
1 & -\frac{1}{2} & \frac{1}{6} & 0 \\
0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{4} \\
0 & 0 & \frac{1}{3} & -\frac{1}{2} \\
0 & 0 & 0 & \frac{1}{4}
\end{pmatrix}$$
We'll factor this matrix into two matrices as follows:
$$M_{P_{(3)}}^{-1}=\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & \frac{1}{3} & 0 \\
0 & 0 & 0 & \frac{1}{4}
\end{pmatrix}×\begin{pmatrix}
1 & -\frac{1}{2} & \frac{1}{6} & 0 \\
0 & 1 & -1 & \frac{1}{2} \\
0 & 0 & 1 & -\frac{3}{2} \\
0 & 0 & 0 & 1
\end{pmatrix}$$
We can see the Bernoulli numbers in the first row of the matrix. Every column is a coefficient vector of a Bernoulli polynomial
The following are extended versions of these matrices:
*
*Pascal Matrix
*Inverse Pascal Matrix
*Last, Bernoulli Matrix
Are there accepted names for these matrices and polynomials? What is the meaning of these relationships?
In particular, is there some treatment of using these matrices as change of basis transformations between representations of polynomials? E.g. from a linear combination of pascal polynomials to a linear combination of monomial terms.
|
This is a very fascinating problem, but in the $P_{(4)}$ case, we have
$$\tag{1}
M_{P_{(4)}}^{-1}=\begin{pmatrix}
1 & -1/2 & 1/6 & 0 & -1/30\\
& 1/2 & -1/2 & 1/4 & 0\\
& & 1/3 & -1/2 & 1/3\\
& & & 1/4 & -1/2\\
& & & & 1/5
\end{pmatrix}
$$
This factors out as
$$
M_{P_{(4)}}^{-1}=\begin{pmatrix}
1 & -1 & 1 & 0 & -1/6\\
& 1 & -3/2 & 1 & 0\\
& & 1 & -2 & 5/3\\
& & & 1 & -5/2\\
& & & & 1
\end{pmatrix}
\begin{pmatrix}
1 & & & & \\
& 1/2 & & & \\
& & 1/3 & & \\
& & & 1/4 & \\
& & & & 1/5
\end{pmatrix}
$$
Unfortunately, as we can read off, the columns in the matrix on the left do not give us Bernoulli polynomials :(
Although, please note the first row of the matrix in (1) gives us Bernoulli numbers. Perhaps the first row of the inverse matrix does give you the Bernoulli numbers, but you do not obtain the Bernoulli polynomials.
Again, read this with caution and suspicion: the good reader should compute this and verify it for him or herself!
Addendum
Write out
$$
M_{P_{(n)}} = DU
$$
where $D$ is a diagonal matrix, and $U=I+X$ is a unit upper triangular matrix (here $X$ is the strictly upper triangular part of $U$). The claim is that
$$(DU)^{-1}=(I+X)^{-1}D^{-1}=(I-X+X^{2}-X^{3}+\dots+(-1)^{n}X^{n})D^{-1}$$
produces the coefficients of Bernoulli polynomials in the columns, and the first row is the Bernoulli numbers. I claim the first (Bernoulli polynomial data) implies the latter (the first row consists of Bernoulli numbers), and claim this is obvious.
Conjecture
It seems that the OP found the matrix version for the Bernoulli polynomials described on Wikipedia using forward differences.
I would have urged the OP to change notation, and use lower triangular matrices. Why? Because then you could write
$$
M_{P_{(n)}}\begin{bmatrix}0\\ 0\\\vdots\\0\\ x^{n}\end{bmatrix} \cong (1+x)^{n+1}-x^{n+1}
$$
in vector form. Current notation demands we use row vectors for polynomials.
I'm about to go to bed, so I do not believe I have time to prove my conjecture. If no one has proven it by tomorrow, I'll try writing up a proof.
|
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Smallest inradius in a triangle inradius http://sphotos-a.ak.fbcdn.net/hphotos-ak-ash4/603253_4699189150138_1902686882_n.jpg
Inside triangle ABC there are three circles with radius $r_1$, $r_2$, and $r_3$ each of which is tangent to two sides of the triangle and to its incircle with radius r. All of $r$, $r_1$, $r_2$, and $r_3$ are distinct perfect square integers. Find the smallest value of inradius $r$.
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For typographic convenience (and reduced visual clutter) in the following, I write "$A_2$", "$B_2$", "$C_2$" for the half-angles "$A/2$", "$B/2$", "$C/2$".
Let $P$ be the incenter of the triangle with radius $r =: s^2$; and let $P_1$, $P_2$, $P_3$ be the centers of the circles with respective radii $r_1 =: s_1^2$, $r_2 =: s_2^2$, $r_3 =: s_3^2$.
If $Q$ is the point of tangency of the incircle with $AB$, then $\triangle APQ$ has a right angle at $Q$, and we have $|AP| = \frac{r}{\sin A_2}$. With $Q_1$ the corresponding point of tangency creating $\triangle AP_1Q_1$, we have
$$\frac{r_1}{r} = \frac{|AP_1|}{|AP|} = \frac{|AP|-r-r_1}{|AP|} = \frac{r-(r+r_1)\sin A_2}{r}$$
so that
$$\sin A_2 = \frac{r-r_1}{r+r_1} = \frac{s^2 - s_1^2}{s^2+s_1^2}$$
Likewise,
$$\sin B_2 = \frac{s^2-s_2^2}{s+s_2^2} \qquad \sin C_2 = \frac{s^2-s_3^2}{s+s_3^2}$$
whence
$$\cos B_2 = \frac{2 s s_2}{s^2+s_2^2} \qquad \cos C_2 = \frac{2 s s_3}{s^2+s_3^2}$$
(Note: We know the cosines of the half-angles must be non-negative.)
Now, $A_2 + B_2 + C_2 = \pi_2$, so that
$$\begin{align}
\sin A_2 &= \cos(B_2+C_2) \\
\sin A_2 &= \cos B_2 \cos C_2 - \sin B_2 \sin C_2 \\
\frac{s^2-s_1^2}{s^2+s_1^2} &= \frac{4 s^2 s_2 s_3 - ( s^2 - s_2^2)( s^2 - s_3^2)}{(s^2+s_2^2)(s^2+s_3^2)}
\end{align}$$
Thus,
$$2 s^2 \left( s^2 - s_1 s_2 - s_2 s_3 - s_3 s_1 \right) \left( s^2 + s_1 s_2 - s_2 s_3 + s_3 s_1 \right) = 0$$
and we have
$$s^2 = s_1 s_2 + s_2 s_3 + s_3 s_1 \qquad \text{or} \qquad - s_1 s_2 + s_2 s_3 - s_3 s_1$$
As the latter option is clearly less than $\max\{s_1^2, s_2^2, s_3^2\}$, whereas $s^2$ must exceed that value (the incircle is bigger than the other three), we are left to solve the Diophantine equation
$$s^2 = s_1 s_2 + s_2 s_3 + s_3 s_1$$
in distinct positive integers dominated by $s$. One solution (found via Mathematica's FindInstance function) is
$$(s,s_1,s_2,s_3) = (9, 7, 6, 3)$$
which (approximately) corresponds to an $106.26^\circ$-$45.24^\circ$-$28.5^\circ$ triangle. No guarantees that this minimizes $s$ (and thus $r$).
|
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|
If $\sum\limits_{n=1}^{\infty}a_{n}$ diverges, does $\sum\limits_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ diverge? Suppose $\displaystyle\sum_{n=1}^{\infty}a_{n}$ diverges. Does $\displaystyle\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ diverge?
|
Eric has proven that the divergence of
$$
\sum_{n=1}^\infty a_n\tag{1}
$$
does not imply the divergence of
$$
\sum\limits_{n=1}^\infty\frac{a_n}{1+na_n}\tag{2}
$$
even if $a_n\ge0$.
However, if $a_n$ decreases monotonically to $0$, then $(2)$ also diverges.
First, note that $\displaystyle\frac{a_n}{1+na_n}$ also decreases monotonically to $0$ since
$$
\frac{a_{n+1}}{1+(n+1)a_{n+1}}=\frac1{1/a_{n+1}+(n+1)}\le\frac1{1/a_n+n}=\frac{a_n}{1+na_n}\tag{3}
$$
Next, note that for $x\ge0$,
$$
\frac{x}{1+x}\ge\frac12\min\left(x,1\right)\tag{4}
$$
Setting $x=na_n$ and dividing by $n$, $(4)$ becomes
$$
\frac{a_n}{1+na_n}\ge\frac12\min\left(a_n,\frac1n\right)\tag{5}
$$
Now, there are two cases:
*
*$a_n\ge\frac1n$ infinitely often
*there is an $N$ so that for $n\ge N\Rightarrow a_n\lt\frac1n$.
In case 1, we can generate a sequence $n_k$ so that $n_{k+1}\ge2n_k$ and $a_{n_k}\ge\frac1{n_k}$. Then
$$
\begin{align}
\sum_{n=n_k+1}^{n_{k+1}}\frac{a_n}{1+na_n}
&\ge(n_{k+1}-n_k)\frac{a_{n_{k+1}}}{1+n_{k+1}a_{n_{k+1}}}\\
&\ge(n_{k+1}-n_k)\frac1{2n_{k+1}}\\
&\ge\frac14\tag{6}
\end{align}
$$
Since we can find infinitely many such $k_n$, we have that $(2)$ diverges.
In case 2, for $n\ge N$, we have $\frac{a_n}{1+na_n}\ge\frac{a_n}{2}$. Therefore, since $(1)$ diverges, $(2)$ also diverges.
|
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|
Prove that $\sum_{k=1}^{\infty} \large\frac{k}{\text{e}^{2\pi k}-1}=\frac{1}{24}-\frac{1}{8\pi}$ Prove that
$$\sum_{k=1}^{\infty} \frac{k}{\text{e}^{2\pi k}-1}=\frac{1}{24}-\frac{1}{8\pi}$$
|
Rewrite
$$\frac{1}{e^{2\pi k} -1} = \sum_{n=1}^\infty e^{-2\pi k n}.$$
So we need to evaluate
$$\sum_{n,k=1}^\infty k e^{-2\pi k n}.$$
Summing first over $k$, we have
$$ \sum_{k=1}^\infty k e^{-2\pi k n} = \frac1{2\pi}\frac{\partial}{\partial n} \sum_{k=1}^\infty e^{-2\pi k n} = \frac1{2\pi}\frac{\partial}{\partial n} \frac{1}{e^{2\pi n} -1} =\frac{e^{2\pi n}}{(e^{2\pi n}-1)^2} = \frac{1}{4 \sinh^2(\pi n)} .$$
The sum
$$\sum_{n=1}^\infty \frac{1}{\sinh^2(\pi n)} =\frac{1}{6} - \frac{1}{2\pi} $$ is evaluated here, see also page 3 here, and the quoted result follows.
|
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|
Cauchy comparison test - if the series limit tends to 1 from bottom I've a following task
Check convergence of:
$$\sum_{n=2}^\infty (-1)^{\lceil 1+\sin^{2} n^{5} \rceil}\left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n^{2}(\sqrt{n+1}-\sqrt{n-1})}$$
My solution is:
$\lceil 1+\sin^{2} n^{5} \rceil$ is always 2, because $\forall {k\in \mathbb{Z}}$, $\sin^{2} n^{5}\neq k\Pi$, where $\Pi$ is irrational.
$$\sum_{n=2}^\infty \left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n^{2}(\sqrt{n+1}-\sqrt{n-1})}$$
Checking Cauchy test:
$$\lim_{n\rightarrow\infty} \sqrt[n]{\left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n^{2}(\sqrt{n+1}-\sqrt{n-1})}}=\lim_{n\rightarrow\infty}\left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n(\sqrt{n+1}-\sqrt{n-1})}$$
$$=\lim_{n\rightarrow\infty} \left(1+\frac{-2n-7}{n^{2}+5n+17}\right)^{n(\sqrt{n+1}-\sqrt{n-1})}=\lim_{n\rightarrow\infty}e^{\frac{-2n-7}{n^{2}+5n+17} n(\sqrt{n+1}-\sqrt{n-1})}=$$
$$=\lim_{n\rightarrow\infty}e^{\frac{(-2n-7)2n}{(n^{2}+5n+17)(\sqrt{n+1}+\sqrt{n-1})}}=-1^{-}$$
Because the series limit tends to 1 from bottom, the series converge. Is that a true assumption?
|
According to your calculation, there is no indication of whether the series converges or diverges. ($|a_n^{1/n}|$ converges or diverges as $n \rightarrow \infty$ according to the limit being > or < 1, and is inconclusive if = 1.)
You made a mistake above:
$$\frac{n^2+3 n+10}{n^2+5 n+17} \approx 1 - \frac{2}{n} $$
so that
$$\left ( \frac{n^2+3 n+10}{n^2+5 n+17} \right )^{n^2 [\sqrt{n+1} - \sqrt{n-1}]} \approx \left ( 1 - \frac{2}{n} \right )^{n^{3/2}} \approx \exp{(-2\sqrt{n})} $$
as $n \rightarrow \infty$. The above convergence test would be inconclusive here, but since the behavior at infinity is a decaying exponential, we can say that the series converges.
|
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|
What do cycle decompositions mean? I am reading about cycle decompositions where a cycle $(a_1 a_2 a_3\cdots a_m)$ is defined as the permutation which sends $a_i$ to $a_{i+1}$, $1\leq i\leq m-1$ and sends $a_m$ to $a_1$.
But I am unable to understand this particular notation:
$(1 2 3)\circ ( 1 2)(3 4)$. What is this supposed to mean and how do we compute it?
|
You want to think of your permutations as functions. In your example, let $f:\{1,2,3,4\}\rightarrow\{1,2,3,4\}$ be the function represented by $(1,2,3)$. This means that
\begin{align*}
f(1)=2,\quad f(2)=3,\quad f(3)=1,\quad \text{and} \quad f(4)=4.
\end{align*}
In a similar way, let $g:\{1,2,3,4\}\rightarrow\{1,2,3,4\}$ be the function represented by $(1,2)(3,4)$. So
\begin{align*}
g(1)=2,\quad g(2)=1,\quad g(3)=4\quad\text{and}\quad g(4)=3.
\end{align*}
Then if we do $(1,2,3)$ and then do $(1,2)(3,4)$, then this is the same as doing $f$ and then $g$. So we get that
\begin{align*}
1\stackrel{f}{\longmapsto}2\stackrel{g}{\longmapsto}1\\
2\stackrel{f}{\longmapsto}3\stackrel{g}{\longmapsto}4\\
3\stackrel{f}{\longmapsto}1\stackrel{g}{\longmapsto}2\\
4\stackrel{f}{\longmapsto}4\stackrel{g}{\longmapsto}3.\\
\end{align*}
So we may represent the effect of doing $f$ and then doing $g$ by $(2,4,3)$. Since this is the effect of doing $(1,2,3)$ and then doing $(1,2)(3,4)$, you have that
\begin{align*}
(1,2)(3,4)(1,2,3)=(2,4,3)
\end{align*}
if you are composing permutations from right to left.
Note: If you are composing permutations from left to right then you would get
\begin{align*}
(1,2)(3,4)(1,2,3)=(1,3,4).
\end{align*}
|
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|
Minimum value of $|z-w|$ where $z,w \in \mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$? I was thinking about the problem:
What is the minimum value of $|z-w|$ where $z,w \in \mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$?
My attempts: I notice that $|z-w| \geq |z|-|w|=11-|w|$. Also if i take $w=u+iv$ then $|w+4+3i|=5$ gives $|w|^2+8u+6v=0$. Now i can not proceed. Can someone point me in the right direction? Thanks in advance for your time.
|
let $z=11(\cos A+i\sin A), w=r(\cos B+i\sin B)$ so $|w+4+3i|=\sqrt{(r\cos B+4)^2+(r\sin B+3)^2}=\sqrt{r^2+25+2r(4\cos B+3\sin B)}$
So, $r^2+25+2r(4\cos B+3\sin B)=25\implies r=-2(4\cos B+3\sin B)$ as $r\ne0$
$|z-w|^2=(11\cos A-r\cos B)^2+(11\sin A-r\sin B)^2=11^2+r^2-22r\cos(A-B)\ge (11-r)^2$
Now, let $R\sin C=4,R\cos C=3$ where $R>0$ so that $R^2=4^2+3^2=5^2\implies R=5$ and $\sin C=\frac45,\cos C=\frac35$
so $4\cos B+3\sin B=R\sin(B+C)=5\sin(B+C)$
So, $-5\le 4\cos B+3\sin B\le 5\implies -10\le r\le 10$
$\implies|z-w|^2\ge (11-10)^2=1$ which occurs if $A=B$ and if $B+C=-\frac\pi2\implies \sin B=\sin(-\frac\pi2-C)=-\sin(\frac\pi2+C)=-\cos C=-\frac34$ and $\cos B=\cos(-\frac\pi2-C)=\cos(\frac\pi2+C)=-\sin C=-\frac45$
So, $z=-\frac{11}5(4+3i),w=-\frac{10}5(4+3i)=-(8+6i)$
|
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|
Proof of tangent half identity
Prove the following: $$\tan \left(\frac{x}{2}\right) = \frac{1 + \sin (x) - \cos (x)}{1 + \sin (x) + \cos (x)}$$
I was unable to find any proofs of the above formula online. Thanks!
|
Hints:
*
*$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$
*$1 - \cos x = 2 \sin^2 \frac{x}{2}$
*$1 + \cos x = 2 \cos^2 \frac{x}{2}$
|
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|
Tips on proving this convergence. We have an inductively defined sequence $x_n=x_{n-1}+2y_{n-1}$ and $y_n=x_{n-1}+y_{n-1}$ where $x_n^2-2y_n^2=\pm 1$, where $x_0=1$ and $y_0=0$.
I need to prove that the sequence $\left(\frac{x_n}{y_n}\right)_{n=1}^\infty$ converges to $\sqrt2$.
Now I can see that $\left(\frac{x_n^2}{y_n^2}\right)_{n=1}^\infty$ converges to $2$, so that may be easier to prove.
It follows now that in order to prove this, $\forall\epsilon >0, \exists n>N$ s.t. $|\frac{x_n^2}{y_n^2}-2|<\epsilon$.
Can anybody point me in the right direction for solving this?
|
The recursion can be written as
$$
\begin{bmatrix}x_n\\y_n\end{bmatrix}=\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}x_{n-1}\\y_{n-1}\end{bmatrix}
$$
The matrix has two eigenvalues: the roots of $\lambda^2-2\lambda-1=0$; that is, $\lambda=1\pm\sqrt{2}$.
The eigenvector with the eigenvalue of $1-\sqrt{2}$ is
$$
\begin{bmatrix}1&2\\1&1\end{bmatrix}\color{#C00000}{\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}=\begin{bmatrix}-2+\sqrt{2}\\-1+\sqrt{2}\end{bmatrix}=(1-\sqrt{2})\color{#C00000}{\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}
$$
The eigenvector with the eigenvalue of $1+\sqrt{2}$ is
$$
\begin{bmatrix}1&2\\1&1\end{bmatrix}\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}}=\begin{bmatrix}2+\sqrt{2}\\1+\sqrt{2}\end{bmatrix}=(1+\sqrt{2})\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}}
$$
Therefore,
$$
\begin{align}
\begin{bmatrix}x_n\\y_n\end{bmatrix}
&=\begin{bmatrix}x_0\\y_0\end{bmatrix}\begin{bmatrix}1&2\\1&1\end{bmatrix}^n\\
&=\left(a\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+b\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}\right)\begin{bmatrix}1&2\\1&1\end{bmatrix}^n\\
&=a(1+\sqrt2)^n\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+b(1-\sqrt2)^n\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}\\
&=a(1+\sqrt2)^n\left(\color{#C00000}{\begin{bmatrix}\sqrt{2}\\1\end{bmatrix}+\frac{b}{a}(2\sqrt{2}-3)^n\begin{bmatrix}\sqrt{2}\\-1\end{bmatrix}}\right)
\end{align}
$$
Assuming $a\ne0$, what is the limit of $x_n/y_n$ in red?
|
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|
Help figuring out all the alternative solutions to the integrals of sine and cosine I always worry a lot when doing integrals with trigonometric functions because there's always many ways to write the final answer. I am trying to figure out the general pattern for the various different solutions.
The integral
$$ \int \sin^3\left(x\right)\cos^2\left(x\right)\, \mathrm{d}x $$
My answer (which never seems to be the same with wolfram alpha's):
$$\frac{\cos^5\left(x\right)}{5} + \frac{-\cos^3\left(x\right)}{3} + C$$
wolfram alpha lists many more:
$$\cos^3\left(x\right)\left(\frac{1}{10}\cos\left(2x\right)-\frac{7}{30}\right) + C$$
$$\frac{1}{240}\left(-30\cos\left(x\right) - 5\cos\left(3x\right)+3\cos\left(5x\right)\right)+C$$
$$\frac{\cos^5\left(x\right)}{80} -\frac{\cos^3\left(x\right)}{48}-\frac{\cos\left(x\right)}{8}-\frac{1}{8}\sin^2\left(x\right)\cos^3\left(x\right) + \frac{1}{16}\sin^4\left(x\right)\cos\left(x\right)+\frac{1}{16}\sin^2\left(x\right)\cos\left(x\right) + C$$
Could anyone help me understand what integration procedure would result in those answers?
|
Since
$$\sin^2x\cos^2x=\frac{1}{4}\left(2\sin x\cos x\right)^2=\frac{1}{4}\sin 2x$$
we get
$$\int\sin^2x\cos^2x=\frac{1}{4}\int\sin 2x\,dx=-\frac{1}{8}\cos 2x\;\;\;(**)$$
and then, integrating by parts:
$$u=\sin x\;\;,\;\;u'=\cos x\\v'=\sin^2x\cos^2 x\;;\,\;\;v=-\frac{1}{8}\cos 2x\,\Longrightarrow$$
$$\int \sin^3x\cos^2 x\,dx=\int\sin x\left(\sin^2x\cos^2 x\right)dx=$$
$$-\frac{1}{8}\cos 2x\sin x+\frac{1}{8}\int\cos x\cos 2x\,dx=-\frac{1}{8}\cos 2x\sin x+\frac{1}{8}\int (1-\sin^2x)\cos x\,dx=$$
$$=-\frac{1}{8}\left(\cos 2x\sin x+\sin x-\frac{1}{3}\sin^3x\right)+C$$
|
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|
Find the maximum and minimum values of $A=\frac{y}{x}-\frac{x}{y}+\frac{z}{y}-\frac{y}{z}+\frac{x}{z}-\frac{z}{x}$ Find the maximum and minimum values of
$$A=\frac{y}{x}-\frac{x}{y}+\frac{z}{y}-\frac{y}{z}+\frac{x}{z}-\frac{z}{x}$$
$x,y,z$ are possitive real numbers satisfying $M\le4m$
where $M=max{(x,y,z)}$, $m=min{(x,y,z)}$
My try: Let $\frac{y}{x}=a,\frac{z}{y}=b,\frac{x}{z}=c$ so $abc=1$
$A=a+b+c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=a+b+c-ab-bc-ca=(a-1)(b-1)(c-1)$
$minA$ (may be) $=0$
And I don't know what to do more.
Thanks
|
From $M \leq 4m$, you get $1/4 \leq a,b,c \leq 4$. And $abc=1$ implies atleast one of $a-1$, $b-1$, $c-1$ is non-positive and atleast one is non-negative. The third one can be either positive or negative (or zero).
For minimum value, without loss of generality, assume $a-1$ is negative and $b-1$, $c-1$ are positive. For any fixed $a<1$, $A$ is minimum when $(b-1)(c-1) = bc + 1 -(b+c)$ is maximum. As $bc$ is fixed, maximum is achieved when $b=c$. So for a fixed $a$, minimum value of $A$ is $(a-1)(\frac{1}{\sqrt{a}}-1)^2$. This can be seen to be strictly increasing for $a \in (0,1]$. So minimum value is at $a=1/4$.
Maximum value of $A$ can be proved similarly, or can be obtained directly by noticing that maximum will exactly be negative of minimum (just replace $a,b,c$ by $\frac{1}{a},\frac{1}{b},\frac{1}{c}$).
|
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|
Evaluate $\int\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}\,dx$ Question:
Solve the integral
$$
\int\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}\,dx
$$
My solution:
Multiply both the numerator and denominator by $\sqrt{x+1}-\sqrt{x}$. This changes the integral to
$$
\begin{align}
\int&\left(1+x+\sqrt{1+x^2}\right)\left(\sqrt{x+1}-\sqrt{x}\right)\,dx\\
&= \int{(1+x)^{3/2}}\,dx-\int{\sqrt{(1+x)(1+x^2)}}\,dx-\int{\sqrt{x}(1+x)}\,dx-\int{\sqrt{x}\sqrt{1+x^2}}\,dx\\
& = \frac{2}{5}(1+x)^{5/2}-I-\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}-J
\end{align}
$$
where $I = \int{(1+x)^{1/2}(1+x^2)^{1/2}}\,dx$ and $J = \int{x^{1/2}(1+x^2)^{1/2}}\,dx$.
How can I solve the integrals $I$ and $J$?
|
One can be brought to the form $\int u^a (1-u)^b \mathrm{d}u$ which is discussed in this answer of mine:
$$\begin{eqnarray}
J &=& \int \sqrt{x} \sqrt{1+x^2} \mathrm{d}x \stackrel{u=x^2}{=} \frac{1}{2} \int u^{-1/4} (1+u)^{1/2} \mathrm{d} u \\ &=& \frac{2}{3} u^{3/4} \cdot {}_2F_1\left(-\frac{1}{2}, \frac{3}{4}; \frac{7}{4}; -u\right) +\text{const.} = \frac{2}{3} x^{3/2} \cdot {}_2F_1\left(-\frac{1}{2}, \frac{3}{4}; \frac{7}{4}; -x^2\right) +\text{const.}
\end{eqnarray}
$$
The other integral is an elliptic integral:
$$\begin{eqnarray}
I &=& \frac{4}{15} \sqrt{2\alpha} \left(\alpha \operatorname{F}\left( \arcsin \left( \frac{x-\frac{1}{\alpha}}{x+\alpha}\right) ,-\alpha^2\right) -\frac{1}{\alpha} \operatorname{E}\left(\arcsin \left( \frac{x-\frac{1}{\alpha}}{x+\alpha}\right) ,-\alpha^2\right)\right) \\ && + \frac{2}{15} \sqrt{1+x}\sqrt{1+x^2} \left(3x+1 + \frac{4}{x+\alpha}\right) + \text{const.}
\end{eqnarray}
$$
where $\alpha = \sqrt{2}+1$ and
$$
\operatorname{E}\left(\phi, m\right) = \int_0^\phi \sqrt{1-m \sin^2\varphi}\, \mathrm{d}\varphi, \quad \operatorname{F}\left(\phi, m\right) = \int_0^\phi \frac{\mathrm{d}\varphi}{\sqrt{1-m \sin^2\varphi}}
$$
It can be evaluated using the Jacobi elliptic functions substitution:
$$
\operatorname{sn}\left(t, -\alpha^2\right) = \frac{x-\frac{1}{\alpha}}{x+\alpha}
$$
as described in Byrd and Friedman.
|
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|
Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
|
Final solution :
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=527572&p=2997801#p2997801
$$ (a-1)^4\geq0\Rightarrow a^3+a+1\leq\frac{(a^2+1)(a^2+5)}{4} $$
$$ \prod(a^3+a+1)\leq\frac{1}{64}\prod(a^2+1)\prod(a^2+5)$$
$$ \leq\frac{1}{64}\frac{(a^2+b^2+c^2+3)^3}{27}\frac{(a^2+b^2+c^2+15)^3}{27}=27 $$
|
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|
Inequality. $\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $ Let $x,y,z$ be real positive numbers such that $x^2+y^2+z^2=3$. Prove that :
$$\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $$
I try to write this expression as:
$$\frac{x^4}{x(y^2+z^2)}+\frac{y^4}{y(z^2+x^2)}+\frac{z^4}{z(x^2+y^2)}$$ and then I try to apply Cauchy-Buniakowsky but still nothing.
I need a proof/idea without derivatives.
thanks for your help.
|
The function $f(w)=w^{3/2}/(3-w)$ is convex on $(0,3)$ so
$${1\over 3}\left[f(x^2)+f(y^2)+f(z^2)\right]\geq f\left([(x^2+y^2+z^2)/3]\right).$$
|
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|
Relative error and roots of significant digits
a.) The maximal relative error of the volume of a ball is allowed
within $1\%$. What is the maximal relative error measuring the
radius of the ball?
b.) Given the equation $x^2-40x+1=0$, find its roots to five
significant digits using $\sqrt{399} = 19.975$, correctly rounded to
five digits.
What I did and know for a is that the volume of a sphere is $V = 4/3\pi * R^3$ and the only term that can have a relative error is the R term. So, I let $dV$ be the relative error of the volume and $dR$ the relative error of the radius then
$dV = dR + dR + dR = 3dR$
$dV/3 = dR$
$1\%/3 = dR$
and I am completely lost in b.
|
Solve for $x$ over the real numbers:
$x^2-40 x+1 = 0$
Solve the quadratic equation by completing the square.
Subtract $1$ from both sides:
$x^2-40 x = -1$
Take one half of the coefficient of $x$ and square it, then add it to both sides.
Add $400$ to both sides:
$x^2-40 x+400 = 399$
Factor the left hand side.
Write the left hand side as a square:
$(x-20)^2 = 399$
Eliminate the exponent on the left hand side.
Take the square root of both sides:
$x-20 = \sqrt{(399)} \text{ or } x-20 = -\sqrt{(399)}$
Look at the first equation: Solve for $x$.
Add $20$ to both sides:
$x = 20+\sqrt{(399)} \text{ or } x-20 = -\sqrt{(399)}$
Look at the second equation: Solve for x.
Add $20$ to both sides:
Answer:
$$
x = 20+\sqrt{(399)} \text{ or } x = 20-\sqrt{(399)}
$$
For question a):
$$
V = \frac{4}{3} \pi R^3 \Rightarrow dV = 4\pi R^2 dR \Rightarrow \frac{dV}{V} = \frac{4\pi R^2dR}{\frac{4}{3} \pi R^3} \le .01 \Rightarrow \frac{3dR}{R} \le .01 \Rightarrow \frac{dR}{R} \le \frac{1}{300} \Rightarrow \max{\left(\frac{dR}{R}\right)} = \frac{1}{300}
$$
|
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|
Problem on improper integrals I'm having some trouble evaluating the following integrals:
$$
I=\int_{-\infty}^0\frac{dx}{\sqrt{1-x^3}},\quad
J=\int_0^1\frac{dx}{\sqrt{1-x^3}}.
$$
Any help will be appreciated.
|
The $J$ integral is reduced to the Euler integral of the first kind using $u$-substitution $u=x^3$:
$$
J = \frac{1}{3} \int_0^1 \left(1-u\right)^{1/2-1} u^{1/3-1} \mathrm{d}u = \frac{1}{3} \operatorname{Beta}\left(\frac{1}{3}, \frac{1}{2} \right)
$$
The $I$ integral can be reduced to the beta integral using $x^3 = \frac{u}{1-u}$, i.e. $u=\frac{x^3}{1+x^3}$:
$$
I = \int_0^\infty \frac{\mathrm{d}x}{\sqrt{1+x^3}} =\frac{1}{3} \int_0^1 u^{1/3-1} \left(1-u\right)^{1/6-1} \mathrm{d}u = \frac{1}{3} \operatorname{Beta}\left(\frac{1}{3}, \frac{1}{6} \right)
$$
|
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|
even and odd perfect numbers existence Thank for my previous post. Also, thank you so much for this site (m.s.e)
1) If odd perfect numbers there, those numbers can be expressible $12k + 1$ or $324k + 81$
or $468k + 117$. If yes, please discuss, how far I am correct.
2) If $K$ = $(4^n - 2^n)$/2 is perfect, when $k = 1^3 + 3^3 + ...$
|
For part 2):
The formula
$$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}$$
is well-known and leads to
$$ 1^3+3^3+\ldots+(2m-1)^3=\sum_{k=1}^{2m}k^3-\sum_{k=1}^{m}(2k)^3\\=\frac{(2m)^2(2m+1)^2}{4}-8\cdot\frac{m^2(m+1)^2}{4}\\=m^2(2m^2-1).$$
Since it is well-known that even perfect numbers $N$ are of the form $N=2^{p-1}(2^p-1)$ with $p=2n+1$ an odd prime and $2^p-1$ a Mersenne prime, letting $m=2^n$, you find that indeed $$N=\frac{4^n-2^n}{2}=m^2(2m^2-1)=1^3+3^3+\ldots+(2m-1)^3.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $x$ such that $12+13^x$ be a perfect square Find $x \in N$ such that $12+13^x$ be a perfect square
I am going to limit $k < 12 + 13^x < k+i$ so that I can have $t<x<t+u$, I don't know how to do it, if $x=2k$, it pretty easy but x can also equal $2k +1$ too. So... Stuck here
Update 2:
I can prove that $x$ can't be $2k$, if so, x = 2k $(k \in \mathbb{N})$ then
$13^{2k}<12+13^x = 12 + 13^{2k}<(13^k+1)^2$ => $12+13^x$ can't be a perfect square.
~# if $x=2k+1$
=> $12+13^x = 12+13^{2k+1}$.
Now we need prove that $k$ can not greater than $1$ (how to do that ?, stuck again)
|
I think algebraic approach is appropriate to this problem. With some calculation, we get
$$(y+2\sqrt{3})(y-2\sqrt{3})=(4+\sqrt{3})^x (4-\sqrt{3})^x.$$
and $4\pm\sqrt{3}$ is prime on $\mathbb{Z}[\sqrt{3}]$. And
$$\frac{5-2\sqrt{3}}{4+\sqrt{3}}=2-\sqrt{3}$$
$$\frac{47+2\sqrt{3}}{(4+\sqrt{3})^3}=2-\sqrt{3}$$
So I conjectured following propositions:
*
*If $(x,y)$ is solution of this equation, then $y+2\sqrt{3}$ associates $(4+\sqrt{3})^x$ or $(4-\sqrt{3})^x$.
*And each case ($(4+\sqrt{3})^x$ associates $y+2\sqrt{3}$ or $(4-\sqrt{3})^x$ associates $y+2\sqrt{3}$) gives only one solution.
But I can't get more.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I write a vector as a linear combination of other vectors. Write $\begin{pmatrix} 5 \\ 3 \\15 \end{pmatrix}$ as a linerar combination of the following vectors: $u=\begin{pmatrix} 1 \\ 2 \\5 \end{pmatrix}$, $v=\begin{pmatrix} 3 \\ -4 \\-1 \end{pmatrix}$, $w=\begin{pmatrix} -1 \\ 1 \\1 \end{pmatrix}$.
My attempt:
$$\begin{bmatrix}
1 & 3& -1 & 5\\
2 & -4 & 1& 3\\
5&-1&1&15\\
\end{bmatrix}\sim\to\begin{bmatrix}
1 & 0& 0 & 3\\
0 & 1 & 0 & 1\\
0&0&1&1\\
\end{bmatrix}$$
Obviously I skipped a lot of reduction steps, because it's a pain to type matrices here, but I was wondering if I even did the right thing.
|
You can set this problem up as linear system of 3 equations
$$ \begin{pmatrix} 5 \\ 3 \\ 15 \end{pmatrix}
= \begin{bmatrix} 1 & 3 &-1 \\ 2 & -4 & 1 \\ 5 & -1 & 1 \end{bmatrix}
\begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix} $$
You do this by arranging the basis vectors as columns to the coefficient matrix.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that
$$
f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}}
$$
We can see that
$$\begin{align}
f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\
&= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
\end{align}$$
And we begin solving by
$$\begin{align}
f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\
f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\
&= 2^x + f(x + 1) \\
f(x + 1) &= f(x)^2 - 2^x
\end{align}$$
At this point I find myself stuck, as I have little experience with recurrence relations.
How would this recurrence relation be solved? Would the method extend easily to
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$
|
Let $x_0 = \sqrt{1}$, $x_1 = \sqrt{1 + \sqrt{2}}$, $x_3 = \sqrt{1 + \sqrt{2 + \sqrt{4}}}$, and so on. Then we have:
$$\sqrt{1 + \sqrt{2 + \sqrt{4 + \cdots}}} = \lim_{n \to \infty} x_n$$
Clearly, this sequence is monotonically increasing. It also converges, since we can see each new term appears under $n$ square roots, and hence $| x_n - x_{n - 1} | \propto 2^{-n}$ which should be enough.
From observation, the value of $x_n - x_{n - 1}$ is a root of a polynomial of order $2^{2n - 3}$. In this sense, a closed-form solution is very unlikely to exist. Not a full answer, but it looks complicated.
|
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|
Another limit from a math contest $\lim_{n\to\infty}\frac{x_n^2y_n}{3x_n^2-2x_ny_n+y_n^2}$ Let $(x_{n})_{n\ge1}$, $(y_{n})_{n\ge1}$ be real number sequences and both converge to $0$.
Evaluate $$\lim_{n\to\infty}\frac{x_n^2y_n}{3x_n^2-2x_ny_n+y_n^2}$$
|
I don't know how many of your alters (see http://en.wikipedia.org/wiki/United_States_of_Tara ) know Lagrange multipliers, so I will give a bit extra. $3 x^2 - 2 x y + y^2$ is a positive definite quadratic form. Setting it equal to a positive constant gives and ellipse, a bounded figure. As a result, the function $x^2$ is bounded on such an ellipse, and Lagrange Multipliers tells us that $$ x^2 \leq \frac{1}{2} \left( 3 x^2 - 2 x y + y^2 \right) $$ with the optima occurring where $x=y.$ As a result, except at the origin itself,
$$ \frac{x^2}{3 x^2 - 2 x y + y^2} \leq \frac{1}{2}, $$ and
$$ \left| \frac{x^2 y}{3 x^2 - 2 x y + y^2} \right| \leq \left| \frac{y}{2} \right| . $$
A little late now. I see that Andre had the better presentation, then deleted it in anguish over not having quite been first. So, as seen elsewhere,
$$ 3 x^2 - 2 x y + y^2 = 2 x^2 + (x-y)^2 \geq 2 x^2. $$ Completing the square the other way gives
$$ 3 x^2 - 2 x y + y^2 = \frac{1}{3} \left( 9 x^2 - 6 x y + 3 y^2 \right) = \frac{1}{3} \left( (3x-y)^2 + 2 y^2 \right) \geq \frac{1}{3} \cdot 2 y^2, $$ and
$$ 3 x^2 - 2 x y + y^2 \geq \frac{2}{3} y^2. $$
So, if either of $x,y$ is nonzero, the denominator is also nonzero and positive. That's called positive definite.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$ Prove by Mathematical Induction . . .
$1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$
I tried solving it, but I got stuck near the end . . .
a. Basis Step:
$(1)(1!) = (1+1)!-1$
$1 = (2\cdot1)-1$
$1 = 1 \checkmark$
b. Inductive Hypothesis
$1(1!) + 2(2!) + \cdot \cdot \cdot +k(k!) = (k+1)!-1$
Prove k+1 is true.
$1(1!) + 2(2!) + \cdot \cdot \cdot +(k+1)(k+1)! = (k+2)!-1$
$\big[RHS\big]$
$(k+2)!-1 = (k+2)(k+1)k!-1$
$\big[LHS\big]$
$=\underbrace{1(1!) + 2(2!) + \cdot \cdot \cdot + k(k+1)!} + (k+1)(k+1)!$ (Explicit Last Step)
$= \underbrace{(k+1)!-1}+(k+1)(k+1)!$ (Inductive Hypothesis Substitution)
$= (k+1)!-1 + (k+1)(k+1)k!$
$= (k+1)k!-1 + (k+1)^{2}k!$
My [LHS] looks nothing like my [RHS] did I do something wrong?
EDIT:
$ = (k+1)k! + (k+1)^2k! -1 $
$ = (k+1)(k!)(1 + (k+1))-1$
$ = (k+1)(k!)(k+2)-1 = (k+2)(k+1)k!-1$
|
There is a mistake in the [LHS], it should look like this:
$$
\underbrace{1(1!) + \ldots + k(k!)}_{=(k+1)! - 1} + (k+1)(k+1)! = \ldots
$$
|
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|
Find the value of "a," so that the lines intersect, Given: $$(l_1) =(x,y,z)=(2,1,2)+t(a,1,0)$$ $$(l_2) =(x,y,z)=(5,2,3)+t(2,3,1)$$
Find the value of "$a$", if any, so that $(l_1)$ and $(l_2)$ intersect
|
For the two lines to intersect at a point we need all of their coordinates to be the same. For that reason we want to solve the following:
$$\begin{pmatrix}2\\1\\2\end{pmatrix} + t_1 \begin{pmatrix}a\\1\\0\end{pmatrix} = \begin{pmatrix}5\\2\\3\end{pmatrix} + t_2 \begin{pmatrix}2\\3\\1\end{pmatrix}$$
Which, if we split them up into the parametric equations for each axis, are the following three equations:
\begin{cases}
2 + a t_1 = 5 + 2 t_2 \\
1 + t_1 = 2 + 3 t_2 \\
2 = 3 + t_2
\end{cases}
Rearranging them somewhat we have the following:
\begin{cases}
-3 = - a t_1 + 2 t_2 \\
-1 = - t_1 + 3 t_2 \\
-1 = t_2
\end{cases}
Substituting the third expression, $t_2 = -1$ into the second and we find that
\begin{align*}
-1 &= - t_1 + 3 \cdot - 1 \\
-1 &= - t_1 - 3 \\
2 &= -t_1 \\
\therefore t_1 &= -2
\end{align*}
Putting both $t_1$ and $t_2$ back into the first of the simultaneous equations above and solve for $a$:
\begin{align*}
-3 &= - a \cdot -2 + 2 \cdot -1 \\
-3 &= 2 a - 2 \\
-1 &= 2a \\
\therefore a &= - \frac{1}{2}
\end{align*}
Finally, to confirm our findings (and to double check for mistakes), we plug it all into the original equation:
\begin{align*}
\begin{pmatrix}2\\1\\2\end{pmatrix} - 2 \begin{pmatrix}-1/2\\1\\0\end{pmatrix} &= \begin{pmatrix}5\\2\\3\end{pmatrix} - \begin{pmatrix}2\\3\\1\end{pmatrix} \\
\begin{pmatrix}2\\1\\2\end{pmatrix} - \begin{pmatrix}-1\\2\\0\end{pmatrix} &= \begin{pmatrix}3\\-1\\2\end{pmatrix} \\
\begin{pmatrix}3\\-1\\2\end{pmatrix} &= \begin{pmatrix}3\\-1\\2\end{pmatrix}
\end{align*}
And thus it was shown!
Edit: Notice how I have used two different parameters for the two lines. This is because two lines in space aren't dependent upon each other; using the same parameter causes problems. Take for example two expressions of the very same line:
\begin{align*}
l_1: (x,y,z)=(1,1,1) + t_1 (1,1,1) \\
l_2: (x,y,z)=(1,1,1) + t_2 (2,2,2)
\end{align*}
If we use the same parameter, we have this:
\begin{align*}
l_1: (x,y,z)=(1,1,1) + t (1,1,1) \\
l_2: (x,y,z)=(1,1,1) + t (2,2,2)
\end{align*}
which is quite problematic if we are to solve for points of intersection (of which there are an infinite amount, of course, since the two lines are the same). Each parametric equation will look like $1 + t = 1 + 2 t$, which leads to $t = 2t$ and we certainly aren't happy about that.
If, on the other hand, we use two different parameters, we simply find that $t_1 = 2 t_2$, which does indeed give us our desired result (that is, the two lines being the same).
|
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|
The remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$ How to find the remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$
|
Hint $\rm\ mod\ 8\!:\,\ odd^2 = \{\pm 1,\, \pm3\}^2 \equiv \color{#C00}1,\:$ so $\rm\,1^2\!+3^2\!+\,\cdots + (2n\!-\!1)^2\equiv \color{#C00}1+\,\cdots\, + \color{#C00}1\equiv\, n\, $
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $4^{47}\equiv 4\pmod{12}$ I know this is a simple exercise, but I was wondering if I can make the following logical jump in my proof:
We see that $4\equiv 4\pmod{12}$ and $4^2\equiv 4\pmod{12}$. Then we can recursively multiply by $4$ to get $4^{47}\equiv 4\pmod{12}$.
|
$$4^{47}\equiv 4\pmod{12}$$
$$4^{47}-4=4 \cdot \left(4^{46}-1\right)=4 \cdot \left(4^{23}-1\right)\cdot \left(4^{23}+1\right)=$$
$$=4 \cdot \left(4-1\right)(4^{22}+4^{21}\cdot1+\ldots+1)\cdot \left(4^{23}+1\right)=4 \cdot3 \cdot \left(4^{22}+4^{21}\cdot1+\ldots+1\right)\cdot \left(4^{23}+1\right)=$$ $$=12 \cdot \left(4^{22}+4^{21}\cdot1+\ldots+1\right)\cdot \left(4^{23}+1\right)$$
which is divisible with $12$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$X=(1 + \tan 1^{\circ})(1 + \tan 2^{\circ})(1 + \tan 3^{\circ})\ldots(1 + \tan {45}^{\circ})$. what is the value of X? $$X=(1 + \tan 1^{\circ})(1 + \tan 2^{\circ})(1 + \tan 3^{\circ})\ldots(1 + \tan {45}^{\circ})$$
$$\tan(90-\theta)=\cot\theta=\frac{1}{\tan\theta}$$
|
Let $\theta = 1^\circ = \frac{\pi}{180} \operatorname{rad}$. Then$$
X = \prod_{k=1}^{45} \left(1 + \tan(k \theta)\right) = \prod_{k=1}^{45} \frac{\sin(k \theta) + \cos(k \theta)}{\cos(k \theta)} = \prod_{k=1}^{45} \sqrt{2} \frac{\sin((k+45) \theta)}{\cos(k \theta)}
$$
Furthermore
$$
\prod_{k=1}^{45} \sqrt{2} \frac{\sin((k+45) \theta)}{\cos(k \theta)} = 2^{45/2} \frac{\prod_{k=1}^{45} \sin((k+45) \theta)}{\prod_{k=1}^{45} \cos(k \theta)} \stackrel{45+k = 90-n}{=} 2^{45/2} \frac{\prod_{n=0}^{44} \sin(90^\circ-n \theta)}{\prod_{k=1}^{45} \cos(k \theta)}
$$
Now, using $\sin(90^\circ - \alpha) = \cos(\alpha)$:
$$
2^{45/2} \frac{\prod_{n=0}^{44} \sin(90^\circ-n \theta)}{\prod_{k=1}^{45} \cos(k \theta)} = 2^{45/2} \frac{\prod_{n=0}^{44} \cos(n \theta)}{\prod_{k=1}^{45} \cos(k \theta)} = 2^{45/2} \frac{\cos( 0 )}{\cos(45^\circ)} = 2^{23}
$$
|
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|
Real Analysis Inequality Questions I am working on a few questions, and I was hoping for some feedback regarding whether or not my answers make sense.
(a) Let $\mathbf{x}, \mathbf{y}, \mathbf{z} \in \mathbb{R}^d$. If $||\mathbf{x}||<2$, $||\mathbf{y}||<3$ and $||\mathbf{z}||<4$, show that
$$| \mathbf{x} \cdot \mathbf{y} - \mathbf{x} \cdot \mathbf{z}| < 14.$$
Observe that $\mathbf{x} \cdot \mathbf{y}=|| \mathbf{x} |||| \mathbf{y}|| cos (\theta)$, where $\theta $ is the angle between $\mathbf{x}$ and $\mathbf{y}$, and $\mathbf{x} \cdot \mathbf{z}=|| \mathbf{x} |||| \mathbf{z}|| cos (\alpha)$, where $\alpha $ is the angle between $\mathbf{x}$ and $\mathbf{z}$. Then we have
$$\mathbf{x} \cdot \mathbf{y} - \mathbf{x} \cdot \mathbf{z} = || \mathbf{x} |||| \mathbf{y}|| cos (\theta) - || \mathbf{x} |||| \mathbf{z}|| cos (\alpha).$$
The function $cos(x)$ ranges between $-1$ and $1$. Now, I believe that the extreme values of the expression above are $-14$ and $14$. Note that $|| \mathbf{x} |||| \mathbf{y}||<6$ and $|| \mathbf{x} |||| \mathbf{z}|| < 8$. Furthermore, $ -1 \leq cos(\theta) \leq 1$ and $ -1 \leq cos(\alpha) \leq 1$. If $cos(\theta)=1$, and $cos(\alpha)=-1$, then
$$|| \mathbf{x} |||| \mathbf{y}|| cos (\theta) - || \mathbf{x} |||| \mathbf{z}|| cos (\alpha) = || \mathbf{x} |||| \mathbf{y}|| + || \mathbf{x} |||| \mathbf{z}|| < 6 +8=14.$$
If $cos(\theta)=-1$, and $cos(\alpha)=1$, then we multiply the previous inequality on each sides by $(-1)$, and change the direction of the inequality accordingly, to arrive at
$$- || \mathbf{x} |||| \mathbf{y}|| - || \mathbf{x} |||| \mathbf{z}|| > -6 -8=-14.$$
This gives,
$$-14 <|| \mathbf{x} |||| \mathbf{y}|| cos (\theta) - || \mathbf{x} |||| \mathbf{z}|| cos (\alpha) < 14$$
or,
$$\left | || \mathbf{x} |||| \mathbf{y}|| cos (\theta) - || \mathbf{x} |||| \mathbf{z}|| cos (\alpha) \right | < 14,$$
$$| \mathbf{x} \cdot \mathbf{y} - \mathbf{x} \cdot \mathbf{z}| < 14.$$
Is this correct? If it is, is my explanation of the extreme values rigorous enough?
(b) Set $B:= \left \{ \mathbf{x} \in \mathbb{R}^n : ||\mathbf{x}|| <1 \right \}.$ If $\mathbf{x}, \mathbf{y}, \mathbf{z} \in B$, show that
$$\mathbf{w} = \frac{(\mathbf{x} \cdot \mathbf{y})\mathbf{z} + (\mathbf{x} \cdot \mathbf{z})\mathbf{y} + (\mathbf{z} \cdot \mathbf{y})\mathbf{x}}{3}$$
belongs to B.
Using the same reasoning as before, and defining the dot products as the product of the lengths of the vectors and the cosine of the angle between the vectors, I get
$$|| \mathbf{w} || = \left \| \frac{(\mathbf{x} \cdot \mathbf{y})\mathbf{z} + (\mathbf{x} \cdot \mathbf{z})\mathbf{y} + (\mathbf{z} \cdot \mathbf{y})\mathbf{x}}{3} \right \| < \left \| \frac{\mathbf{z} + \mathbf{y} + \mathbf{x}}{3} \right \| < \frac{||\mathbf{z}||}{3} + \frac{||\mathbf{y}||}{3} + \frac{||\mathbf{x}||}{3} < 1$$
So, $\mathbf{w} \in B$.
(c) Let B be as above, and let $ \mathbf{z}$, $\mathbf{w} \in B$. Prove that
$$|\mathbf{x} \cdot \mathbf{z} - \mathbf{y} \cdot \mathbf{w} | \leq \left \| \mathbf{y}-\mathbf{z} \right \| + \left \| \mathbf{x}-\mathbf{w} \right \|$$
$$|\mathbf{x} \cdot \mathbf{z} - \mathbf{y} \cdot \mathbf{w} | \leq |\mathbf{x} \cdot \mathbf{z} - \mathbf{z} \cdot \mathbf{w} + \mathbf{z} \cdot \mathbf{w} - \mathbf{y} \cdot \mathbf{w} | $$
$$\leq |\mathbf{z}| \cdot |\mathbf{x} - \mathbf{w}| + |\mathbf{w}| \cdot |\mathbf{z} - \mathbf{y}|$$
$$\leq ||\mathbf{z}|| \cdot ||\mathbf{x} - \mathbf{w}|| + ||\mathbf{w}|| \cdot ||\mathbf{z} - \mathbf{y}||$$
$$\leq ||\mathbf{x} - \mathbf{w}|| + ||\mathbf{z} - \mathbf{y}||$$
Any input would be appreciated.
|
a) follows from $|x \cdot y - x \cdot z| = |x \cdot (y -z)| \leq \|x\| \|y-z\| \leq \|x\| (\|y\|+\|z\|) = 2 (3+4)$.
The first inequality in your answer for b) is incorrect. First do the triangle inequality, then upper bound using the fact that $|x \cdot y| < 1$ when $x,y \in B$.
|
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|
Showing that $ |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$ For every nonnegative integer $n$ and every real number $ x$ prove the inequality:
$$\sum_{k=0}^n|\cos(2^kx)|= |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$$
|
Actually, the statement is true even when one restrict the sum to start at $k=1$.
i.e. For all $x \in \mathbb{R}$,
$$\sum_{k=1}^n |\cos(2^k x)| \ge \frac{n}{2\sqrt{2}} \Leftrightarrow \varphi_n(2x) \ge 0$$
where $\varphi_n(x) = \sum_{k=0}^{n-1}|\cos(2^kx)| - \frac{n}{2\sqrt{2}}$.
Notice:
$$\begin{align}
\varphi_4(x) &= \varphi_2(x) + \varphi_2(4x)\\
\varphi_5(x) &= \varphi_2(x) + \varphi_3(4x)\\
&\,\,\vdots\\
\varphi_n(x) &= \varphi_2(x) + \varphi_{n-2}(4x)
\end{align}$$
One only need to show $\varphi_2(x), \varphi_3(x) \ge 0$ for all $x$.
As other answer suggested, $|\cos(2^k x)|$ are piecewise concave functions. We have:
$$\frac{d^2}{dx^2}|\cos(2^k x)| = -4^k |\cos(2^k x)| < 0 \,\,\,\text{ for }
x \ne \pm 2^{-k}(l+\frac12)\pi \text{ where } l \in \mathbb{N}
$$
This in turn implies
$$\frac{d^2}{dx^2}\varphi_n(x) < 0 \,\,\,\text{ for } x \ne \pm2^{-n}(l+1)\pi \text{ where } l \in \mathbb{N}$$
The absolute minimum of $\varphi(x)$ must be located at a $x_{min}$ of the form $\pm 2^{-n}(l+1)\pi$.
Since $\varphi(x)$ is even and has period $\pi$, we only need to look for $x_{min}$ over the interval $[0, \frac{\pi}{2}]$.
For all x, we have:
$$\begin{align}
\varphi_2(x) &\ge \min(\varphi_2(\frac{\pi}{4}),\varphi_2(\frac{\pi}{2})) = \varphi_2(\frac{\pi}{4}) = 0\\
\varphi_3(x) &\ge \min(\varphi_3(\frac{\pi}{8}), \varphi_3(\frac{\pi}{4}), \varphi_3(\frac{3\pi}{8}), \varphi_3(\frac{\pi}{2})) = \varphi_3(\frac{3\pi}{8}) = \cos(\frac{3\pi}{8}) - \frac{1}{2\sqrt{2}} > 0
\end{align}$$
|
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How to evaluate this rational limit Evaluate:
$$\lim_{x\to0^+}\frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln (1+x)+x^4-x^6}$$
|
$$\lim_{x\to0^+}\frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln (1+x)+x^4-x^6} \\= \lim_{x\to0^+} \frac{-3\sqrt{x}+x^3+\sin(x^6)}{\sqrt x} \times \frac{e^{4\sqrt{x}}-1)}{\sqrt x} \times \frac{1}{4\frac{\ln (1+x)}{x} + x^3 - x^5 }$$
|
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|
Evaluate $\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$ How can one evaluate $\displaystyle\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$?
My attempt:
$$\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2} = \frac{1}{2}\int_{0}^{2\pi} \frac{d\theta}{(2+\cos\theta)^2}$$
To find the singularity, I solve:
$ (2+\cos\theta)^2 = 0 $ and therefore, $\cos\theta = -2$.
Substituting: $\cos z = \frac{e^{iz} + e^{-iz}}{2} = \frac{z + \frac{1}{z}}{2}$,
I find that $z = -2 + \sqrt{3} $ is the singular point that lies in the unit circle $|z| = 1$.
From this point, I have little idea how to go about solving this problem. I know I have to find the residue and then just sum them but to get the expression that would cancel out the pole is where I am currently stuck.
|
I will start from the point where you left off. The integral may be written as, upon substitution of $z=e^{i \theta}$:
$$\begin{align}\frac{1}{2} \int_0^{2 \pi} \frac{d\theta}{(2+\cos{\theta})^2} &= -i 2 \oint_{|z|=1} dz \frac{z}{(z^2+4 z+1)^2}\\ &= -i 2 \oint_{|z|=1} dz \frac{z}{(z+2-\sqrt{3})^2 (z+2+\sqrt{3})^2}\end{align}$$
after a bit of algebra. This integral is $i 2 \pi$ times the sum of the residues of the poles within the integration contour. The only pole inside this contour, as you point out, is the pole at $z=-2+\sqrt{3}$. The other pole at $z=-2-\sqrt{3}$ is outside this contour and is not counted.
To compute the residue at this pole, note that we have double roots. For such roots, we have to take a derivative:
$$\mathrm{Res}_{z=-2+\sqrt{3}} \frac{-i 2 z}{(z+2-\sqrt{3})^2 (z+2+\sqrt{3})^2} = \lim_{z \rightarrow -2+\sqrt{3}} \left [\frac{d}{dz} \frac{-i 2 z}{(z+2+\sqrt{3})^2} \right ]$$
I will leave the algebra to the reader; the result is $-i \sqrt{3}/9$. The integral is then
$$\int_0^{\pi} \frac{d\theta}{(2+\cos{\theta})^2} = i 2 \pi \frac{-i \sqrt{3}}{9} = \frac{2 \sqrt{3}}{9} \pi$$
|
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|
How can we prove the Inequality : $ \frac {n!}{ 2^{n-1}((\frac {n-1}{2})!)^2} \leq \sqrt{n}$ How can we prove the following inequality? For every odd positive integer $n$,
$$ \frac {n!}{ 2^{n-1}((\frac {n-1}{2})!)^2} \leq \sqrt{n}$$
Thank You.
|
For completeness, here is the proof by induction:
Base Case: The case $ n = 1$ is easy to check.
Inductive Step: Assume that $$\frac{k!}{2^{k-1} \left( \frac{k-1}{2} \right)!^2} \leq \sqrt{k}.$$ We aim to show that this implies $$\frac{(k+2)!}{2^{k+1} \left( \frac{k+1}{2} \right)!^2} \leq \sqrt{k+2}.$$
We have
\begin{align}
\frac{(k+2)!}{2^{k+1} \left( \frac{k+1}{2} \right)!^2} &= \frac{(k+2)(k+1)}{4(\frac{k+1}{2})^2} \cdot \frac{k!}{2^{k-1} \left( \frac{k-1}{2} \right)!^2}
\\
&\leq \frac{(k+2)(k+1)}{4(\frac{k+1}{2})^2} \cdot \sqrt{k}
\\
&= \frac{k+2}{k+1} \sqrt{k}
\\
&\leq \sqrt{k+2}
\end{align}
which occurs if and only if $\left( \frac{k+2}{k+1}\right)^2 k \leq k+2$. I will leave the last inequality for you to check.
|
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|
limit of $\sqrt[n]{\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}}$ using geometric mean How to find this question using Geometric Mean?
$$\sqrt[n]{\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}}$$
Thanks!
|
Your product is itself a geometric mean.
\[ \sqrt[n]{\frac{1\cdot 3\cdot \cdot (2n-1)}{2\cdot 4\cdot \cdot (2n)}}
= \left(\frac{1}{2}\right)^{1/n} \cdot \left(\frac{3}{4}\right)^{1/n} \dots \left(\frac{2n-1}{2n}\right)^{1/n}\]
The factor is approaching the same number:
\[ \frac{2n-1}{2n} = 1 - \frac{1}{2n} \to 1\]
If you take geometric mean 1 , 1 and ... 1, what should the answer be?
|
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|
$3^{2n+3}+2^{n+3}$ is divisible with 7 Prove that $$3^{2n+3}+2^{n+3}$$ is divisible with $7$ for every $n \in \Bbb N$
$$3^{2n}\cdot27+2^{n}\cdot8=3^{2n}(28-1)+2^{n}(7+1)=\left(3^{2n}\cdot28+2^{n}\cdot7\right)-\left(9^{n}-2^{n}\right)=$$
$$\left(3^{2n}\cdot28+2^{n}\cdot7\right)-(9-2)\left(9^{n-1}+\ldots+2^{n-1}\right)$$ if $n=2k+1$. If $n=2k$ is also true. So the number is divisible with $7$.
Is there any proof without using $a^{n}-b^{n}$
|
Hint $\rm\ mod\ 7\!:\,\ 3^{2n+3}\!+2^{n+3}\equiv\, 27\cdot 9^n\! + 8\cdot 2^n \equiv\, 35\cdot 2^n\equiv 0\ \ $ by $\rm\ 9\equiv 2;\ 35\equiv 0$
Remark $\ $ Per your comment, for a proof without mod, subtract and add $\rm\,27\cdot 2^n\,$ above, i.e.
$\rm\quad 27\cdot 9^n\! + 8\cdot 2^n =\, 27\, (9^n\!-2^n)\! + (\color{#C00}{27\!+\!8})\cdot 2^n $ is divisible by $\,7\,$ by $\rm\ 9\!-\!2\,|\,9^n\!-2^n,\,$ and $\rm\, 7\,|\, \color{#C00}{35}$
|
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|
Calculus Midterm Question. I'm studying for my calc midterm right now, and I was finding the limit of the question:
$$\lim_{x \rightarrow 2}{\frac{\sqrt{x+2} - \sqrt{2x}}{x-2}}\tag{1}$$
When I was trying to solve it I came up with this:
$$\lim_{x \rightarrow 2}{\frac{\sqrt{x+2}^{2} - \sqrt{2x}^{2}}{x-2(\sqrt{x-2}+ \sqrt{2x})}}\tag{2}$$
$$\lim_{x \rightarrow 2}{\frac{x + 2 - 2x}{x-2(\sqrt{x+2}+ \sqrt{2x})}}\tag{3}$$
$$\lim_{x \rightarrow 2}{\frac{-2x}{\sqrt{x+2} + \sqrt{2x}}}\tag{4}$$
However, the solution is this: Solution
I'm not understanding what happens to that $-2x.$
|
Your error is your move from line $(3)$ to line $(4)$. Those two expressions are not equal.
Multiplying the numerator and denominator by the conjugate of $\;\displaystyle \frac{\sqrt{x+2} - \sqrt{2x}}{ x-2}\;$ gives us:
$$
\begin{align}
\frac{\sqrt{x+2} - \sqrt{2x}}{x-2}\cdot \frac{\sqrt{x+2} + \sqrt{2x}}{\sqrt{x+2} + \sqrt{2x}}
& = \frac{x+2 - 2x}{(x- 2)(\sqrt{x+2} + \sqrt{2x)}} \tag{a}\\ \\
& = \frac{-(x-2)}{(x- 2)(\sqrt{x+2} + \sqrt{2x)}} \tag{b} \\ \\
& = \frac{-1}{(\sqrt{x+2} + \sqrt{2x)}}\tag{c}\\ \\
\end{align}
$$
Now evaluate your $\displaystyle \;\;\lim_{x\to 2} \; \frac{-1}{\sqrt{x+2} + \sqrt{2x}}$.
Note that the $x + 2 - 2x = -x + 2 = -(x-2)$, and so we are able to cancel the factor $(x - 2)$ from both numerator and denominator.
$$\lim_{x\to 2} \frac{\sqrt{x+2} - \sqrt{2x}}{x-2} = \lim_{x\to 2}\;\frac{-1}{(\sqrt{x+2} + \sqrt{2x)}} = \frac{-1}{4}$$
|
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|
Simple AM-GM inequality Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that
$$
a+b+c \geq ab+bc+ca
$$
I was able to prove that
$$
\begin{align}
a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\
&\ge \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\frac{2\sqrt{a^2c^2}}{2} \\
&= ab+bc+ca
\end{align}
$$
but now I am stuck. I don't know how to use the fact that $a+b+c=3$ to prove the inequality. Anybody can give me a hint?
|
$$9=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 3(ab+bc+ca)$$
|
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|
Inequality: $x^2+y^2+z^2+t(xy+yz+zx) \geq 0$ Prove that $x^2+y^2+z^2+t(xy+yz+zx) \geq 0$ for any $x,y,z \in \mathbb{R}$ and any $t \in [-1,2].$
One try:
for $t=-1$: $x^2+y^2+z^2-xy-yz-zx \geq 0$ is true .
for $t=2$: $x^2+y^2+z^2+2(xy+yz+zx) \geq 0$ is true
also for $t=0$.
But how can prove for $t \in (-1,0)$ and $t \in (0,2)$.
Also I don't know if what I did is the right step.
Thanks :)
|
Note that
\begin{eqnarray}
x^2+y^2+z^2+t(xy+yz+zx)&=&\frac{2-t}{3}(x^2+y^2+z^2-xy-yz-zx)\\
&&+\frac{1+t}{3}(x^2+y^2+z^2+2xy+2yz+2yz) \, .
\end{eqnarray}
Your previous work shows that both terms on the right-hand side are always non-negative for $t \in [-1, 2]$; thus the left-hand side is as well.
|
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|
Proving the inequality $\arctan\frac{\pi}{2}\ge1$ Do you see any nice way to prove that
$$\arctan\frac{\pi}{2}\ge1 ?$$
Thanks!
Sis.
|
By Taylor's theorem with remainder in the integral form for the function $f=\arctan$ at the point $1$, we have:
\begin{equation}\arctan x=\arctan 1+(x-1)f'(1)+\int_1^x\frac{f''(t)}{2!}(x-t)dt \end{equation}
then we substitute $\displaystyle f'(1)=\frac{1}{2}$, $\displaystyle f''(t)=\frac{-2t}{(1+t^2)^2}$ and $\displaystyle x=\frac{\pi}{2}$ we find
\begin{equation}\arctan \frac{\pi}{2}=\frac{\pi}{4}+(\frac{\pi}{2}-1)\frac{1}{2}-\int_1^{\frac{\pi}{2}}\frac{t}{(1+t^2)^2}(\frac{\pi}{2}-t)dt\\=1+\frac{\pi-3}{2} -\int_1^{\frac{\pi}{2}}\frac{t}{(1+t^2)^2}(\frac{\pi}{2}-t)dt.\end{equation}
Now, we have
\begin{align}D=\frac{\pi-3}{2} -\int_1^{\frac{\pi}{2}}\frac{t}{(1+t^2)^2}(\frac{\pi}{2}-t)dt&\geq \frac{\pi-3}{2} -(\frac{\pi}{2}-1)\int_1^{\frac{\pi}{2}}\frac{t}{(1+t^2)^2}dt\\&=\frac{\pi-3}{2}-(\frac{\pi}{2}-1)\frac{1}{4}\frac{\pi^2-4}{\pi^2+4}=S.\end{align}
The approximation $\pi\approx 3.1416$ gives $D\geq S\approx 4.27\ 10^{-3}$ so we conclude:
$$\arctan\frac{\pi}{2}\geq 1.$$
|
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|
How to solve this irrational equation? How to solve this equation in the set real numbers $$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$ Using Mathematica, I know, this equation has two solutions $x = 1$ and $x = 3.$
|
Write:
$$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$
As:
$$(\sqrt{8x + 1} - \sqrt{6x - 2}) = (2x^2 - 8x + 7).$$
Squaring both sides yields:
$$14 x-2 \sqrt{6 x-2}\sqrt{8 x+1}-1 = 4x^4-32x^3+92x^2-112x+49$$
Simplifying:
$$-2 \sqrt{6 x-2}\sqrt{8 x+1} = 4x^4-32x^3+92x^2-112x+49$$
Squaring both sides again yields:
$$ 4(6x-2)(8x+1) = 16 x^8-256 x^7+1760 x^6-6896 x^5+16928 x^4-26384 x^3+25076 x^2-12600 x+2500$$
Simplifying:
$$16 x^8-256 x^7+1760 x^6-6896 x^5+16928 x^4-26384 x^3+24884 x^2-12560 x+2508 = 0$$
This gives us more roots, so care must be taken, but indeed, we get back two of the $8$ roots as $x = 1$ and $x = 3$.
Numerical methods would also have worked earlier and guessing too.
Regards
|
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|
Writing a matrix as multiplication of orthogonal and diagonal matrices Given is the matrix A = \begin{pmatrix}
5 & 3 \\
0 & -4 \\
\end{pmatrix}
I would like to write this matrix as the product A = $U S V^T$
whereas U and V both are orthogonal matrices and S is a diagonal matrix. What is the approach to find suitable matrices?
|
Given:
$$A = \begin{pmatrix}
5 & 3 \\
0 & -4 \\
\end{pmatrix}$$
We want to find the Singular Value Decomposition (SVD) where $U$ and $V$ both are orthogonal matrices and $\Sigma$ is a diagonal matrix, such that:
$$A = U \cdot \Sigma \cdot V^T$$
We arrive at:
$$A = U \cdot \Sigma \cdot V^T = \begin{pmatrix} 5 & 3 \\ 0 & -4 \\ \end{pmatrix} = \begin{pmatrix} \frac{2}{\sqrt{5}} & -\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & -\frac{2}{\sqrt{5}} \\ \end{pmatrix} \cdot \begin{pmatrix} 2\sqrt{10} & 0 \\ 0 & \sqrt{10} \\ \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{pmatrix} $$
|
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Critical point of a function? Can anyone show me how to simplify get the critical point of this function.
$f(x)=x^2\sqrt[3]{2+x}$
I did the product rule and got
$$2x\sqrt[3]{2+x}+\frac{x^2}{3\sqrt[3]{(2+x)^2}}$$
but I am having touble simplifying such this how would I simplify it can anyone show me how this would be done.
|
You need only find the points at which $f'(x) = 0$, and where $f'(x)$ is undefined.
To simplify, find a common denominator and add terms, then set equal to zero: The common denominator is $3\sqrt[\large 3]{(2 + x)^2}$. Do you recall how to bring all terms over this common denominator?
We have $$2x\sqrt[\large3]{2+x}+\frac{x^2}{3\sqrt[\large3]{(2+x)^2}}$$
And want
$$
\begin{align} f'(x)
& = \dfrac{2x \sqrt[\large 3]{2+x}\times 3\sqrt[\large 3]{(2+x)^2} + x^2}{3\sqrt[\large 3]{(2+x)^2}} \\ \\
& = \frac{6x \sqrt[\large 3]{(2+x)(2+x)^2} + x^2}{3 \sqrt[\large 3]{(2+x)^2}} \\ \\
& = \frac{6x \sqrt[\large 3]{(2+x)^3} + x^2}{3 \sqrt[\large 3]{(2+x)^2}} \\ \\
& = \dfrac{6x(2 + x) + x^2}{3\sqrt[\large 3]{(2+x)^2}} \\ \\
& = \dfrac{12x + 6x^2 + x^2}{3 \sqrt[\large 3]{(2+x)^2}} \\ \\
& = \dfrac{x(12 + 7x)}{3\sqrt[\large 3]{(2 + x)^2}} = 0 \\ \\
\end{align}
$$
$f'(x)$ is undefined when $x = -2$.
$f'(x) = 0$ when the numerator equals zero: one point at which this occurs is when $x = 0$.
$f'(x) = 0$ when $(12 + 7x) =0 \iff 7x = -12 \iff x = -\large\frac{12}{7}$
Three critical points in all. The blue line below is your function of interest. Note the sharp corner at $x = -2$. It happens to be a local minimum. Also note the local maximum at $x = -\large\frac{12}{7}$, and the minimum at $x = 0$. Graphs are really helpful to confirm the work you're doing, and better understand the behavior of the function.
ASIDE: Personally I think using fractional exponents to express roots makes this sort of problem a bit clearer, in terms of algebraic manipulation, particularly when we're talking about roots other than the square root, and especially when they appear in fractions.
|
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|
How to show $f(x,y)=|xy|^\alpha\log(x^2+y^2)$ is differentiable at $(0,0)$? Show that if $\alpha > 1/2$, then
$$f(x, y)=\begin{cases}
|xy|^\alpha\log(x^2+y^2), ~(x, y) \ne (0,0)\\\\
~~~0, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{otherwise}\;.
\end{cases}$$
is differentiable at $(0,0)$.
|
Note that $|x| = \sqrt{x^2} \leq \sqrt{x^2+y^2}$ and similarly $|y| = \sqrt{y^2} \leq \sqrt{x^2+y^2}$.
Thus $|xy|^\alpha \leq (x^2+y^2)^\alpha$.
Then
$$
\left|\frac{f(x,y)}{\sqrt{x^2+y^2}}\right| \leq (x^2+y^2)^{\alpha-\frac12} \log(x^2+y^2) = (r^2)^{\alpha-\frac12} \log(r^2)
$$
Of course $(x,y) \to 0$ if and only if $r^2 = x^2+y^2 \to 0$. Since $\alpha>1/2$ the limit is zero by an elementary argument using L'Hospital's rule. It follows that $f$ is differentiable at $(0,0)$.
|
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|
Find reduction formula using integration by parts
Let $\displaystyle I(n,\alpha)=\int_0^\pi \sin(n\theta)e^{\alpha \theta} d\theta$. Show that $\displaystyle I(n,\alpha)=\frac{n}{n^2+\alpha^2}(1-(-1)^ne^{\alpha \pi})$.
This is done by parts, but I can't get the expression to equal the RHS.
|
Note: I adapted this from Arturo's answer at: Integration by parts: $\int e^{ax}\cos(bx)\,dx$
Also, please forgive me for not using your variable names as I found them confusing (easy to modify the entire result by just changing them to your names).
Note: we are going to save the integration limits to the end for ease.
So, doing IBP twice yileds:
$$\int e^{ax}\sin(bx)dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\left(\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).$$
Multiplying out you get
$$\int e^{ax}\sin(bx)\,dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a^2}e^{ax}\cos(bx) - \frac{b^2}{a^2}\int e^{ax}\cos(bx)\,dx.$$
At this point, you should move that last integral on the right hand side to the left hand side and "leave oout" the constant of integration on the right (because we'll use limits at the end as noted earlier).
Moving the last integral to the left hand side, you get
$$\left(1 + \frac{b^2}{a^2}\right)\int e^{ax}\sin(bx)\,dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a^2}e^{ax}\cos(bx).$$
Now, $\displaystyle 1 + \frac{b^2}{a^2} = \frac{a^2+b^2}{a^2}$, and we want to clear this from the left side, so multiply both sides by the reciprocal, that is by, $\displaystyle \frac{a^2}{a^2+b^2}$.
If you do that, from
$$\frac{a^2+b^2}{a^2}\int e^{ax}\sin(bx)\,dx = \frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a^2}e^{ax}\cos(bx),$$
multiplying both sides by $\frac{a^2}{a^2+b^2}$, we get:
$$\tag 1 \int e^{ax}\sin(bx)\,dx = \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx).$$
Now, we just need to deal with the RHS of $(1)$ with the limits from $\Pi$ to $0$.
So we have:
$\displaystyle \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx)$ evaluated at $\pi$, yields: $\displaystyle \frac{a}{a^2+b^2} e^{a \pi}(a \sin b \pi - b \cos b \pi)$
Of course, $\sin b \pi= 0$ for all $b$ (odd or even) and $\cos b \pi = \pm 1$ (depending on parity of b, that is $-1$ if $b$ is odd and is $+1$ if b is even, so we will account for this by using $(-1)^b$).
$\displaystyle \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx)$ evaluated at $0$, yields: $\frac{a}{a^2+b^2} (0 - b).$
Subtracting these expressions (because we are doing over limits $\pi$ and $0$), yields:
$\displaystyle \frac{a}{a^2+b^2}e^{ax}\sin(bx) - \frac{b}{a^2+b^2}e^{ax}\cos(bx)|{_0^\pi} = \frac{b}{a^2+b^2}\left(1 - (-1)^b e^{a \pi}\right)$.
The desired result.
Regards
|
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|
Proof: Two non identical circles have at most 2 same points I'm struggeling with an analytic proof for the fact, that two different circles have at most 2 same points. (I try to solve it analytical, because geometrical I already prooved it).
I tried to start with the equations
$r_1^2=(x-a_1)^2+(y-b_1)^2$ and $r_2^2=(x-a_2)^2+(y-b_2)^2$, further
$r_1^2=x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2$ and the same for the 2nd equation.
Then I get
\begin{eqnarray*}
r_1^2&=&x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2\\
r_2^2&=&x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2\\
0&=&x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2-r_1^2\\
0&=&x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2-r_2^2\\
&\Rightarrow& x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2-r_1^2=x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2-r_2^2\\
&\Leftrightarrow& 0=-2xa_1+a_1^2-2xb_1+b_1^2-r_1^2+2xa_2-a_2^2+2yb_2-b_2^2+r_2^2
\end{eqnarray*}
But now I don't know how to move on. Can someone give a hint?
|
You have worked too hard. The first circle has equation of the shape $x^2+y^2+ax+by+c=0$. The second circle has equation of the shape $x^2+y^2+dx+ey+f=0$. The two equations are not identical, since the circles are not.
If $(x,y)$ lies on both circles, then both equations are satisfied, and therefore (subtract) we have
$$(a-d)x+(b-e)y+(c-f)=0.\tag{$1$}$$
If $b-e=0$, then $x$ is determined, and substituting in the equation of one of the circles, we get a quadratic equation in $y$, which has at most $2$ solutions.
If $b-e\ne 0$, then we can use the linear equation to solve for $y$ in terms of $x$. Substitute in the equation of one of the circles. We get a quadratic equation in $x$, which has at most $2$ solutions. And from $(1)$, each corresponds to a unique value for $y$.
Note that we do not need to carry out the substitutions, all that is needed is to imagine carrying them out.
Remark: The line with Equation $(1)$ is called the radical axis of the two circles. If the two circles do indeed meet in $2$ distinct points, the radical axis is the line through these $2$ points.
|
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|
$2\frac{d^2y}{dx^2} + 7\frac{dy}{dx} -y =0$ where y(0)=1 and y'(0)=3 I need help with this slightly tedious question
the got the solution of the auxiliary equation to be $s=\frac{-7+\sqrt{57} }{2}$ and $t=\frac{-7-\sqrt{57} }{2}$ thus the general solution $y=Ae^{sx}+Be^{tx}$.
Given the initial conditions I obtained the two simultaneous equations $1=A+B$ and $\frac{20}{\sqrt{57} }=A-B$ which gave me $A=1.824532357$ and $B=-0.824532357$
so when x=2 my y value was 3.16 (2dp) hich was incorrect but I don't see where I went wrong.
Can anyone PLEASE confirm what answer they get to 2dp (Note: you must be as accurate as possible)
Many thanks
|
As L.F. stated, you have an issue with your denominator from your auxiliary equation.
You should have, $2m^2 + 7m -1 = 0$, yielding:
$~m = \frac{1}{4} (-\sqrt{57}-7) $ and $m = \frac{1}{4} (\sqrt{57}-7)$
From this and the IC's, we arrive at:
$$\displaystyle \large y(x) = \frac{1}{6} e^{-\frac{1}{4} (7+\sqrt{57}) x} \left((3+\sqrt{57}) e^{\frac{\sqrt{57} x}{2}}+3-\sqrt{57}\right)$$
At $x = 2$, we get:
$$y(2) = \displaystyle \frac{1}{6} e^{\frac{1}{2} (-7-\sqrt{57})} \left(3-\sqrt{57}+(3+\sqrt{57}) e^{\sqrt{57}}\right) \approx 2.3141465430$$
|
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|
A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.
|
If we are allowed to use Coordinate Geometry,
WLOG we can assume the coordinates of $A,B,C$ to be $(x,y),(-a,0),(a,0)$ respectively and $D$ to be $(h,k).$
As $\triangle ABC$ is equilateral $(x-a)^2+(y-0)^2=(x+a)^2+(y-0)^2=(-a-a)^2+(0-0)^2$
From the first relation we find $4ax=0\implies x=0$ as $a\ne0$
From the second relation we find $y^2=3a^2\implies y=\sqrt3a$ assuming $a>0$
So, $\overline{AB}=2a$
So, $\overline{AD}^2=(h-0)^2+(k-\sqrt3a)^2$ But $\overline{AD}=13$
$\implies (h-0)^2+(k-\sqrt3a)^2=13^2--->(1)$
Similarly, for $BD,CD$ respectively,
$(h+a)^2+k^2=12^2--->(2)$ and $(h-a)^2+k^2=5^2--->(3)$
So here we have three equations with three unknowns $a,h,k$
One way to solve this could be as follows :
From $(2)-(3),4ah=119\implies h=\frac{119}{4a}$
From $(2)-(1),2ah+2\sqrt3ak=2a^2-25\implies k=\frac{4a^2-169}{4\sqrt3a} $ (putting $ah=\frac{119}4$)
Putting the values of $h,k$ in terms of $a$ in $(1),$
$$\left(\frac{119}{4a}\right)^2+\left(\frac{4a^2-169}{4\sqrt3a}-\sqrt3a\right)^2=13^2$$
$$\implies 64a^4-32\cdot169a^2+169^2+3(119^2)=0$$
which is a Quadratic Equation in $a^2$
|
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|
How do I find the maximum and minimum values of $1−4\cos(2\theta)+3\sin(2\theta)$? To find the maximum of $$1 - 4\cos(2\theta) + 3\sin(2\theta) $$I tried:
$$1-4(1)+3(1)=0.$$
To find the minimum I tried to substitute with the minimum values of sin and cos:
$$1-4(-1)+3(-1)=2$$
I know I'm wrong, could someone explain why? And how I should go around obtaining the correct answer?
Thanks
|
Following is an algebraic way of finding the range of $
As $\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$ and $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$
So, $1-4\cos2\theta+3\sin2\theta=\frac{(1+t^2)-4(1-t^2)+3(2t)}{1+t^2}=y$(say) putting $t=\tan\theta$
So, $t^2(y-5)-6t+y+3=0$
This is a Quadratic Equation in $t=\tan\theta$
As $\theta$ is real, so will be $t,$ hence the discriminant $(-6)^2-4(y-5)(y+3)\ge0$
$\implies y^2-2y-24\le0$
(i)$\implies (y-6)(y+4)\le 0\implies$ either $(y\le 6, y\ge -4\implies -4\le y\le 6)$ or $(y\le -4,y\ge 6)$ which is impossible
(ii)$\implies (y-1)^2\le 25=5^2\implies -5\le y-1\le 5\implies -4\le y\le 6$
|
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|
Show $ \frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}=1 $ given $xyz = 1$ Please help me prove the equality: If $xyz=1$, prove that
$$
\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}=1
$$
|
Hint: Rewrite the $1$ in the first denominator as $xyz,$ so that the left hand side becomes $$\begin{align}\frac1{y+1+yz}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1} &= \frac{1+y}{yz+y+1}+\frac{z}{xz+z+1}\\ &= 1-\frac{yz}{yz+y+1}+\frac{z}{xz+z+1}.\end{align}$$ Can you go from here?
|
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|
Finding $u(x)$ using Green's Function Let $(Lu)(x)= -\frac{d}{dx}\big(\frac{1}{x} \frac{du}{dx} )$ where $u(x)$ is twice differentiable function defined on $[1,2]$.
A) I need to find Green's function $G(x,t)$ such that for any $h(x)$ defined on $[1,2]$ the solution to the boundary value problem
$Lu=h$ for $1\le x \le2$
$u(1)=u^{'}(2)=0 $
is given by $u(x)=\int_1^2G(x,t)h(t)dt= \int_1^xG(x,t)h(t)dt+\int_x^2G(x,t)h(t)$.
B) Let $h(x)=x$. Use the Green's function $G(x,t)$ to find the solution $u(x)$.
So here's what I have:
$\frac{1}{x}\frac{du}{dx}=A$
$\frac{du}{dx}=Ax$
$du=Axdx$
$u(x)=\frac{Ax^2}{2}+B$
$u(1)=\frac{A}2+B=0 \Rightarrow B=-\frac{A}2$
So $u(x)=\frac{Ax^2}{2}-\frac{A}2$
$u_1(x)=\frac{x^2}2$
$u^{'}(x)=2Ax$
$u^{'}(2)=4A=0 \Rightarrow A=0$ so $U_2(x)=B=1$
Now calculating $ c=p(x) \begin{vmatrix}
u_1 & u_2 \\
u_1^{'} & u_2^{'} \\
\end{vmatrix}$ where $p(x)=-\frac1{x}$.
$c= -\frac1{x} \begin{vmatrix}
\frac{x^2}2 & 1 \\
2x & 0 \\
\end{vmatrix}=-\frac1{x}(-x)=1$
So green's function yields
$G(x,t)=\begin{cases}
\frac1{c}u_1(t)u_2(x) & \text 1 \le t \le x\\
\frac1{c}u_1(x)u_2(t) & \text x \le t \le 2\\
\end{cases}$
$G(x,t)=\begin{cases}
(\frac1{1})\frac{t^2}2 (1) & \text 1 \le t \le x\\
(\frac1{1})\frac{x^2}2 (1) & \text x \le t \le 2\\
\end{cases}$
Therefore, with $h(t)=t$
$u(x)=\int_1^2G(x,t)h(t)dt= \int_1^x\frac{t^2}2tdt+\int_x^2\frac{x^2}2t dt$.
After integrating, I obtain
$u(x)=\frac{x^4}8-\frac{x^2}8+x^2-\frac{x^4}4$
But then the boundary conditions do not hold. Where did I go wrong?
|
Hint: The correct formula for the Green's function is
$$G(x,t) =
\begin{cases}
\frac{1}{2}(x^2-1), & x\le t \\
\frac{1}{2}(t^2-1), & t\le x.
\end{cases}$$
|
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|
How to prove that $\frac{(5m)!(5n)!}{(m!)(n!)(3m+n)!(3n+m)!}$ is a natural number?
How to prove that $$\frac{(5m)! \cdot (5n)!}{m! \cdot n! \cdot (3m+n)! \cdot (3n+m)!}$$ is a natural number $\forall m,n\in\mathbb N$ , $m\geqslant 1$ and $n\geqslant 1$?
If $p$ is a prime, then the number of times $p$ divides $N!$ is $e_p(N)=\sum_{k=1}^\infty\left\lfloor\frac{N}{p^k}\right\rfloor$. So I need
$$e_p(5m)+e_p(5n)\geq e_p(m)+e_p(n)+e_p(3m+n)+e_p(3n+m).$$
What to do next? Thanks in advance.
|
Let $p$ be a prime number. You should show that
$$\left\lfloor \frac{5m}{p^k}\right\rfloor+\left\lfloor \frac{5n}{p^k}\right\rfloor \geq \left\lfloor \frac{m}{p^k}\right\rfloor+\left\lfloor \frac{n}{p^k}\right\rfloor+\left\lfloor \frac{3m+n}{p^k}\right\rfloor+\left\lfloor \frac{3n+m}{p^k}\right\rfloor$$
for $k \geq 1.$ By Hermite's identity you will get
$$\left\lfloor \frac{5m}{p^k}\right\rfloor=\left\lfloor \frac{m}{p^k}\right\rfloor+\left\lfloor \frac{m}{p^k}+\frac{1}{5}\right\rfloor+\left\lfloor \frac{m}{p^k}+\frac{2}{5}\right\rfloor+\left\lfloor \frac{m}{p^k}+\frac{3}{5}\right\rfloor+\left\lfloor \frac{m}{p^k}+\frac{4}{5}\right\rfloor$$
Then you should show that
$$\sum_{i=1}^{4} \left\lfloor \frac{m}{p^k}+\frac{i}{5}\right\rfloor+\sum_{j=1}^{4}\left\lfloor \frac{n}{p^k}+\frac{j}{5}\right\rfloor \geq \left\lfloor \frac{3m+n}{p^k}\right\rfloor+\left\lfloor \frac{3n+m}{p^k}\right\rfloor$$
which can be done by considering different cases for the values $\frac{m}{p^k}, \frac{n}{p^k}$ i.e. each can be in the following intervals $(0,\frac15), [\frac15, \frac25), [\frac25, \frac35), [\frac35, \frac45), [\frac45, 1),$ since $\lfloor x+n \rfloor=\lfloor x\rfloor+n,$ for and real number $x$ and an integer $n,$ you can just ignore the case when $\frac{m}{p^k}, \frac{n}{p^k}$ are bigger that $1.$
Earlier thought: The expression can be rewritten as
$$\frac{\binom{5m}{3m}(3m)!\binom{5n}{n}n!}{(3m+n)!} \frac {\binom{4n}{3n}(3n)!\binom{2m}{m}m!}{(3n+m)!}$$
I was hoping to find a combinatorial interpretation for this expression but I haven't found yet!
|
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|
Calculate: $\int_{2}^{7}\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx $ How to calculate following integration?
$$\int_{2}^{7}\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx $$
|
Replace $x$ by $9-x$. We then get
$$I = \underbrace{\int_2^7 \dfrac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} dx = \int_2^7 \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{9-x}} dx}_{x \to 9-x}$$
Adding both we get that
$$2I = \int_2^7 dx = 5 \implies I = \dfrac52$$
In general, if we have
$$I = \int_a^b \dfrac{f(a+b-x)}{f(x) + f(a+b-x)} dx$$ where $f(y) + f(a+b-y)$ doesn't vanish in $[a,b]$ then replacing $x$ by $a+b-x$, we get
$$I = \int_a^b \dfrac{f(a+b-x)}{f(x) + f(a+b-x)} dx = \int_a^b \dfrac{f(x)}{f(x) + f(a+b-x)} dx$$
Adding both gives us
$$2I = \int_a^b dx \implies I = \dfrac{b-a}2$$
|
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|
Finding polynomial given the remainders Question: Find a polynomial $f(x) \in \mathbb{Q} (x)$ of minimal degree that has both the following properties:
When $f(x)$ is divided by $(x-1)^2$, the remainder is $2x$; and when $f(x)$ is divided by $(x-2)^3$, the remainder is $3x$.
Answer provided: $f(x)=(x-2)^3 \cdot (4x-3)+3x$
Work so far: Okay, so I know this problem shouldn't be difficult, but I've been stumped. I know that this is probably a simple application of the division algorithm, where $f(x)=q(x) \cdot d(x)+r(x)$ but I can't seem to get an answer. The polynomial's minimum degree should be $4$, intuitively.
So I have that $f(x)=(x-1)^2 \cdot q_1(x) + 2x$ and $f(x)=(x-2)^3 \cdot q_2(x)+3x$. Then I know that for the second equation, $q_2(x)$ should be of form $(ax+b)$ for some $a,b \in \mathbb{Q}$, also I also applied the remainder theorem to find that $f(1)=4$ and that $f(2)=6$, and so
$f(1)=4=(-1)^3 \cdot (a+b) +3(1) \rightarrow a+b=1$
so it makes sense that the answer is $q_2(x)=4x-3$, but I don't know how to get there after finding $a+b=1$.
|
Rather than attempting to match two solutions, construct the solution using one remainder, and then solve using the other one, like this:
$$
f(x) = (x-2)^3g(x)+3x
$$
Therefore, mod $(x-1)^2$,
$$
f(x) = (3x-4) g(x) + 3x = 2x
$$
Therefore, $(3x-4)g(x) = -x$ under mod $(x-1)^2$.
Now, we just need to find the polynomial that is the inverse of $(3x-4)$.
$$
(ax+b)(3x-4) = 3ax^2+(3b-4a)x-4b = (2a+3b)x-(3a+4b) = 1
$$
Now, $2a+3b=0$ and $-3a-4b=1$, so $b=2$ and $a=-3$. Therefore,
$$
g(x) = -x(-3x+2) = 3x^2-2x = 3(2x-1)-2x = 4x-3
$$
So $g(x)=4x-3$, and the answer is
$$
f(x) = (x-2)^3(4x-3)+3x
$$
|
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|
Twice a triangle is triangle The question is to prove that there are infinitely many triangular numbers $T_n$ where $2 \times T_n$ is also a triangular number, and give the first few as an example.
My attempt:
$$2 \cdot {x(x+1) \over 2} = {y(y+1) \over 2} \\
\Leftrightarrow {x(x+1)} = {y(y+1) \over 2} \\
\Leftrightarrow 2x^2 + 2x = y^2 + y \\
\Leftrightarrow 4x^2 + 4x = 2y^2 + 2y \\
\Leftrightarrow 4x^2 + 4x +1= 2y^2 + 2y +1\\
\Leftrightarrow (2x +1)^2= 2y^2 + 2y +1\\
\Leftrightarrow 2(2x +1)^2= 4y^2 + 4y +2\\
\Leftrightarrow 2(2x +1)^2= (2y +1)^2 + 1\\
\Leftrightarrow Y^2 - 2X^2 = -1, \quad X=(2x+1) \text{ and } Y=(2y + 1)\\
$$
This is Pell's equation in 2 variables, and I obtained $(1, 1)$ and $(7, 5)$ as 2 solutions.
However, I'm unable to prove infinite of them. I know that this is a very famous Pell equation, but I haven't been able to find good answers as to how the recurrence is established, firstly between $X$ and $Y$, and then between $x$ and $y$. I know that there exists one, which I obtained from this diophantine equation solver, but I don't understand the solution given there.
Can anyone please help me derive the recurrence relation?
|
Note that
$$(a^2-2b^2)(c^2-2d^2)=(ac+2bd)^2 -2(ad+bc)^2.$$
This is a special case of the Brahmagupta Identity. It can be readily proved by expanding the two sides.
A nice way of looking at things is to let $(a,b)$ be a particular solution of your Pellian $x^2-2y^2=1$, say $(1,1)$. Then you can generate infinitely many solutions $(x_n,y_n)$ by letting
$$x_n+y_n\sqrt{2}= (1+\sqrt{2})(3+2\sqrt{2})^n.$$
You can simplify calculations by noting that $x_{n+1}+y_{n+1}\sqrt{2}=(x_n+y_n\sqrt{2})(3+2\sqrt{2})$. This yields the recurrences
$$x_{n+1}=3x_n+2y_n,\qquad y_{n+1}=2x_n+3y_n.$$
|
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|
Sum of the series $\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} $ for $n>3$, The sum of the series $\displaystyle \sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} = $
where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$
My try:: I have expand the expression
$\displaystyle 1.2.\binom{n}{r}-2.3.\binom{n}{r-1}+3.4.\binom{n}{r-2}+........+(-1)^r.(r+1).(r+2).\binom{n}{0}$
Now after that how can i calculate it
Thanks
|
Suppose we seek to evaluate
$$\sum_{k=0}^r (-1)^k (k+2)(k+1) {n\choose r-k}.$$
Start from
$${n\choose r-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r-k+1}} (1+z)^n \; dz.$$
This yields the following expression for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{k=0}^r (-1)^k (k+2)(k+1)
\frac{1}{z^{r-k+1}} (1+z)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}}
\sum_{k=0}^r (-1)^k (k+2)(k+1) z^k \; dz.$$
Observe that the defining integral of the binomial coefficient is zero
when $k > r.$ This condition is precisely $r-k+1 < 1$ which makes the
integrand into an entire function. Therefore we may extend the sum to
infinity.
We thus have
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}}
\sum_{k=0}^\infty (-1)^k (k+2)(k+1) z^k \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}}
\left(\frac{1}{1+z}\right)'' \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{r+1}}
\frac{2}{(1+z)^3} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{2(1+z)^{n-3}}{z^{r+1}} \; dz$$
Therefore the value of the sum is given by
$$[z^r] 2 (1+z)^{n-3} = 2{n-3\choose r}.$$
This may be re-written as
$$2{n\choose r} \frac{(n-r)(n-r-1)(n-r-2)}{n(n-1)(n-2)}$$
which is indeed the same as the earlier results, namely
$${n\choose r} \frac{(n-r)(2r^2 + (6-4n)r + 2n^2-6n+4}{n(n-1)(n-2)}.$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
|
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"timestamp": "2023-03-29T00:00:00",
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|
The quadratic diophantine $ k^2 - 1 = 5(m^2 - 1)$ Here's the problem.
Find the solutions of the following equation:
$$ k^2 - 1 = 5(m^2 - 1).$$
Here's my idea:
The original equation can be written as:
$$ k^2 = 5m^2 - 4 \Longleftrightarrow k^2 - 5m^2 = -4$$
I know this is Quadratic Diophantine Equation and i've done some searching on the internet and I couldn't find a particular way for solving equations of this type. Also I know that this is a varition of Pell's equation, because instead of 1 we have -4.
I actually found the fundamental solution to this equation (by guessing) and it's (1,1). Then using this algorithm (that works for Pell's equation) I tried to generate another solution and I get:
$$ X_k+_1 = aX_k + nbY_k$$
$$ X_2 = aX_1 + nbY_1$$
$$ X_2 = 1 \times 1 + 5 \times 1 \times 1 = 6$$
$$ Y_k+_1 = bX_k + aY_k$$
$$ Y_2 = bX_1 + aY_1$$
$$ Y_2 = 1 \times 1 + 1 \times 1 = 2$$
We can easily check that (6,2) isn't a solution.
So how can I transform a Quadratic Diophantine Equation into a Pell's equatuon and how can I generate more solution for Pell's equation if the constant isn't 1 (in this case it's -4)?
|
$$ A =
\left( \begin{array}{cc}
9 & 20 \\
4 & 9
\end{array}
\right) ,
$$ and
$$ A^{-1} =
\left( \begin{array}{cc}
9 & -20 \\
-4 & 9
\end{array}
\right).
$$
$$
\left( \begin{array}{cc}
9 & 20 \\
4 & 9
\end{array}
\right)
\left( \begin{array}{c}
1 \\
1
\end{array}
\right) =
\left( \begin{array}{c}
29 \\
13
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
9 & 20 \\
4 & 9
\end{array}
\right)
\left( \begin{array}{c}
29 \\
13
\end{array}
\right) =
\left( \begin{array}{c}
521 \\
233
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
9 & 20 \\
4 & 9
\end{array}
\right)
\left( \begin{array}{c}
521 \\
233
\end{array}
\right) =
\left( \begin{array}{c}
9349 \\
4181
\end{array}
\right),
$$
Switching to $-A^{-1},$ we get
$$
\left( \begin{array}{cc}
-9 & 20 \\
4 & -9
\end{array}
\right)
\left( \begin{array}{c}
1 \\
1
\end{array}
\right) =
\left( \begin{array}{c}
11 \\
-5
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
-9 & 20 \\
4 & -9
\end{array}
\right)
\left( \begin{array}{c}
11 \\
-5
\end{array}
\right) =
\left( \begin{array}{c}
-199 \\
89
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
-9 & 20 \\
4 & -9
\end{array}
\right)
\left( \begin{array}{c}
-199 \\
89
\end{array}
\right) =
\left( \begin{array}{c}
3571 \\
-1597
\end{array}
\right),
$$
If you want to allow common factors,
$$
\left( \begin{array}{cc}
9 & 20 \\
4 & 9
\end{array}
\right)
\left( \begin{array}{c}
4 \\
2
\end{array}
\right) =
\left( \begin{array}{c}
76 \\
34
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
9 & 20 \\
4 & 9
\end{array}
\right)
\left( \begin{array}{c}
76 \\
34
\end{array}
\right) =
\left( \begin{array}{c}
1364 \\
610
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
9 & 20 \\
4 & 9
\end{array}
\right)
\left( \begin{array}{c}
1364 \\
610
\end{array}
\right) =
\left( \begin{array}{c}
24476 \\
10946
\end{array}
\right).
$$
Switching to $-A^{-1},$ we get
$$
\left( \begin{array}{cc}
-9 & 20 \\
4 & -9
\end{array}
\right)
\left( \begin{array}{c}
4 \\
2
\end{array}
\right) =
\left( \begin{array}{c}
4 \\
-2
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
-9 & 20 \\
4 & -9
\end{array}
\right)
\left( \begin{array}{c}
4 \\
-2
\end{array}
\right) =
\left( \begin{array}{c}
-76 \\
34
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
-9 & 20 \\
4 & -9
\end{array}
\right)
\left( \begin{array}{c}
-76 \\
34
\end{array}
\right) =
\left( \begin{array}{c}
1364 \\
-610
\end{array}
\right),
$$
so you see nothing new happens this time.
|
{
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"url": "https://math.stackexchange.com/questions/340142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Factorize $3m^4-6m^3+14m^2-6m+11$ I have this expression:
$3m^4-6m^3+14m^2-6m+11=0$ and I want to factorize it in $(m^2+1)(3m^2-6m+11)$.
How can I do it? Thanks for any help!
|
You already factorized the $3m^4 −6m^3 +14m^2 −6m+11$ with $(m^2 +1)(3m^2 −6m+11)$.
First factor equals to $(m^2 +1)$ and second $(3m^2 −6m+11)$.
$(m^2 +1)(3m^2 −6m+11) = 3m^4-6m^3+11m^2+3m^2-6m+11 = 3m^4-6m^3+14m^2+11$
Note:
But if you need solve $3m^4 −6m^3 +14m^2 −6m+11 = 0$ using factorization $(m^2 +1)(3m^2 −6m+11)$ it's a little different problem. $(m^2 +1)(3m^2 −6m+11) = 0$ is true if 1.:$m^2 +1 = 0$ or 2.:$3m^2 −6m+11=0$.
1. $m^2 +1 = 0$ => $m_{1,2} = \pm\sqrt{-1} = \pm{i}$
2. $3m^2 −6m+11=0$ => $m_{3,4} = 1\pm{2}i\sqrt{\frac{2}{3}}$
Here are four solutions:
1. $m = i$ : $3i^4-6i^3+14i^2-6i+11 = 3+6i+14-6i+11 = 0$
2. $m = -i$ : $3(-i)^4-6(-i)^3+14(-i)^2-6(-i)+11$$ = 3(-1)^4i^4-6(-1)^3i^3+14(-1)^2i^2-6(-i)+13 = 3 -6i-14+6i+13 = 0$
3. $m = 1+{2}i\sqrt{\frac{2}{3}}$: substitute to equation as in 1. or 2.
4. $m = 1-{2}i\sqrt{\frac{2}{3}}$: substitute to equation as in 1. or 2.
|
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"url": "https://math.stackexchange.com/questions/340822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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|
$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1$, or $p$ Let $p$ be prime number ($p\gt2$) and $a,b\in\mathbb Z$ ,$a+b\neq0$ ,$\gcd(a,b)=1$ how to prove that $$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1~~\text{or}~~ p$$
Thanks in advance .
|
We already now that $\frac{a^p+b^p}{a+b}=a^{p-1}-a^{p-2}b+\ldots+b^{p-1}$
Using polynomial division, we find that
$$a+b\,|\,a^{p-1}-a^{p-2}b+\ldots+b^{p-1}=(a^{p-2}-2a^{p-2}b+3a^{p-3}b^2-\ldots-pb^{p-2})+pb^{p-1}$$
Hence $\gcd(\frac{a^p+b^p}{a+b},a+b)=\gcd(pb^{p-1},a+b)=\gcd(p,a+b)=1\text{ or } p$
as desired.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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|
Combinatorial interpretation of a sum identity: $\sum_{k=1}^n(k-1)(n-k)=\binom{n}{3}$ I solved $\sum_{k=1}^n(k-1)(n-k)$ algebraically
\begin{eqnarray*}
\sum_{k=1}^n(k-1)(n-k)&=&\sum_{k=1}^n(nk-n-k^2+k)\\
&=&\sum_{k=1}^nnk-\sum_{k=1}^nn-\sum_{k=1}^nk^2+\sum_{k=1}^nk\\
&=&\frac{n(n^2+n)}{2}-n^2-\frac{n(2n^2+3n+1)}{6}+\frac{n^2+n}{2}\\
&=&n\left(\frac{3n^2+3n-6n-2n^2-3n-1+3n+3}{6}\right)\\
&=&n\left(\frac{n^2-3n+2}{6}\right)=\frac{n(n-1)(n-2)}{6}\\
&=&\frac{n!}{(n-3)!3!}=\binom{n}{3}
\end{eqnarray*}
But now I am interested in a combinatorial interpretation of it. For the right hand side $\binom{n}{3}$ is the number of ways to choose 3 from a total of $n$, and for the left hand side, it looks like dividing $n$ into 2 groups, one of size $k$, and other of size $(n-k)$, then sum over all possible $k$, but I do not see any clues that how I can choose 3 from these 2 groups.
|
The left hand side represents picking a "middle" element in a set of $3.\ $ Then you have $k-1$ choices for picking the smallest element and $n-k$ choices for picking the largest element.
For example if $n=5$ and $k=3$, then you have $2 \times 2$ ways of $3$ being the middle element:
$$(1 2) 3 (4 5)$$
If $n = 7$ and $k=5$, you have $4 \times 2$ ways for $5$ to be the middle element:
$$(1 2 3 4) 5 (6 7)$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Prove ${a^2+ac-c^2=b^2+bd-d^2}$ and $a > b > c > d \implies ab + cd$ is not prime Let $a>b>c>d$ be positive integers and suppose that
$${a^2+ac-c^2=b^2+bd-d^2}$$
Prove that $ab+cd$ is not prime? I don't know if this problem is true.
I found that this same problem has also been posted on AOPS.
But I can't prove this problem. Can anyone help me?
|
Rewrite as:
$$a^2-b^2+ac-bc=bd-bc+c^2-d^2$$
$$(a-b)(a+b+c)=(c-d)(c+d-b)$$
Since $a>b>c>d$, each of $a-b, a+b+c, c-d, c+d-b$ is positive. By factoring lemma (excerpted below) there exists $w, x, y, z \in \mathbb{Z}^+$ s.t.
$$a-b=wx, a+b+c=yz, c-d=wy, c+d-b=xz$$
Solving for $a, b, c, d$, we get:
\begin{align}
5a=3wx+2yz-wy-xz \\
5b=-2wx+2yz-wy-xz \\
5c=-wx+yz+2wy+2xz \\
5d=-wx+yz-3wy+2xz
\end{align}
Thus:
\begin{align}
& 25(ab+cd) \\
& =(3wx+2yz-wy-xz)(-2wx+2yz-wy-xz) \\
& +(-wx+yz+2wy+2xz)(-wx+yz-3wy+2xz) \\
& =5(z^2-wz-w^2)(x^2+y^2)
\end{align}
$$5(ab+cd)=(z^2-wz-w^2)(x^2+y^2)$$
Since $b>c$,
$$-2wx+2yz-wy-xz=5b>5c=-wx+yz+2wy+2xz$$
$$yz>wx+3wy+3xz$$
In particular, $yz>wx+3wy+3xz>3xz$ implies $y>3x$ and $yz>wx+3wy+3xz>3wy$ implies $z>3w$.
Thus
$$x^2+y^2>x^2+9x^2>5$$
$$z^2-wz-w^2=(z-\frac{w}{2})^2-\frac{5w^2}{4}>(3w-\frac{w}{2})^2-\frac{5w^2}{4}=5w^2 \geq 5$$
If $ab+cd$ is a prime, then $ab+cd \geq 4(3)+2(1)>5$, then $5(ab+cd)=(z^2-wz-w^2)(x^2+y^2)$ implies that $ab+cd$ divides exactly 1 of $z^2-wz-w^2$ and $x^2+y^2$. However, the term not divisible by $ab+cd$ must necessarily divide $5$, and thus be $\leq 5$. Since both $z^2-wz-w^2>5$ and $x^2+y^2>5$, we obtain a contradiction.
Therefore $ab+cd$ is not prime.
Below is the linked "factoring lemma".
|
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|
Proving that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$
Prove that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$.
I know that for proving the $\gcd(a,b) = c$ you need to prove
*
*$c|a$ and $c|b$
*$c$ is the greatest number that divides $a$ and $b$
Number 2 is what I'm struggling with. Does anybody have any ideas?
|
By euclid's algorithm $$\begin{align}gcd(5^{98}+3,5^{99}+1) &= gcd(5^{98}+3,5^{99}-1-5*(5^{98}-3))
\\ & =gcd(5^{98}+3,14)\end{align}$$Now we can see that 2 divides $5^{98}+3$. Also $5^3$ is -$1 (mod 7)$. Therefore $5^{98}=5^2*5^{96}=4(mod 7)$. So $5^{98}+3=0(mod 7)$. Hence 14 divides $5^{98}+3$. So $gcd(5^{98}+3,14)=14.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/346524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Showing that $T(n)=2T([n/2]+17)+n$ has a solution in $O(n \log n)$ How can we prove that $T(n)=2T([n/2]+17)+n$ has a solution in $O(n \log n)$? What is the resulting equation I get after the substitution?
$$ T(n) = 2c \cdot \frac n2 \cdot \log \frac n2 + 17 + n $$
or
$$ T(n) = 2c \cdot \left(\frac n2 + 17\right) \cdot \log \left(\frac n2 + 17\right) + n $$
To me the first equation looks the correct, but the professor came to the second equation after the substitution. Which one of the two is correct, and how do I proceed in solving it (because that $17$ can be quite tricky)?
|
After the inductive step we have:
$T(n) \le 2c(\lfloor \frac{n}{2} \rfloor + 17) \log (\lfloor \frac{n}{2} \rfloor + 17) + n$
Note that $2 \lfloor \frac{n}{2} \rfloor \le n$ and we can get
$T(n) \le c(n + 34) \log (\lfloor \frac{n}{2} \rfloor + 17) + n$
Expand to
$ T(n) \le cn\log (\lfloor \frac{n}{2} \rfloor + 17) + 34c\log (\lfloor \frac{n}{2} \rfloor + 17) + n$
Now add and subtract $cn \log n$
$ T(n) \le cn \log n - cn \log n + cn\log (\lfloor \frac{n}{2} \rfloor + 17) + 34c\log (\lfloor \frac{n}{2} \rfloor + 17) + n$
$ T(n) \le cn \log n + cn\log (\frac{\lfloor \frac{n}{2} \rfloor + 17}{n}) + 34c\log (\lfloor \frac{n}{2} \rfloor + 17) + n$
Now show the expression $cn\log (\frac{\lfloor \frac{n}{2} \rfloor + 17}{n}) + 34c\log (\lfloor \frac{n}{2} \rfloor + 17) + n$ is negative for $n$ large enough and a correct choice of $c$ so that we can replace with
$ T(n) \le cn \log n$
Showing negativity
$cn\log (\frac{\lfloor \frac{n}{2} \rfloor + 17}{n}) + 34c\log (\lfloor \frac{n}{2} \rfloor + 17) + n \le 0$
For large $n$, the first term $cn\log (\frac{\lfloor \frac{n}{2} \rfloor + 17}{n})$ will be close to $cn\log(\frac{1}{2})$ which is negative. The other two terms are $O(n)$ so we hope we can choose a $c$ so that the first term can overtake the other two.
$cn\log (\frac{\lfloor \frac{n}{2} \rfloor + 17}{n}) + 34c\log (\lfloor \frac{n}{2} \rfloor + 17) + n \lt $
$cn\log (\frac{ \frac{n}{2} + 17}{n}) + 34c\log ( \frac{n}{2} + 17) + n = $
$cn\log (\frac{1}{2} + \frac{17}{n}) + 34c\log ( n (\frac{1}{2} + \frac{17}{n})) + n = $
$cn\log (\frac{1}{2} + \frac{17}{n}) + 34c\log n + 34c\log(\frac{1}{2} + \frac{17}{n}) + n = $
$cn\log (\frac{1}{2} ( 1 + \frac{34}{n})) + 34c\log n + 34c\log(\frac{1}{2}(1 + \frac{34}{n})) + n = $
$cn\log \frac{1}{2} + cn \log( 1 + \frac{34}{n}) + 34c\log n + 34c\log\frac{1}{2}
+ 34c\log(1 + \frac{34}{n}) + n \lt$
... now choosing any $ 0 < \epsilon \ll 1$ and $n$ large enough ...
$cn\log \frac{1}{2} + cn \epsilon + 34c\log n + 34c\log\frac{1}{2}
+ 34c\epsilon + n =$
$c(n\log \frac{1}{2} + n \epsilon + 34\log n + 34\epsilon) + 34c\log\frac{1}{2}
+ n \lt$
$c(n\log \frac{1}{2} + n \epsilon + 34\log n + 34\epsilon) + n $
Now note that $n\log \frac{1}{2} + n \epsilon + 34\log n + 34 \epsilon= n(\log \frac{1}{2} + \epsilon ) + 34 \log n + 34\epsilon \equiv g(n)$ is negative for large enough $n$ since we have a negative linear function competing with a postive $\log$ function.
It only remains to show we can choose $c$ such that for large enough $n$,
$cg(n) + n \lt 0$
But we can do so, since $|g(n)| \in O(n)$, a $c$ exists such that it can outcompete the postive term $ n$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
What is the remainder when $25^{889}$ is divided by 99? What is the remainder when $25^{889}$ is divided by 99 ?
$25^3$ divided by $99$ gives $26$ as a remainder.
$25*(25^3)$ divided by $99$ gives (remainder when $25*26$ is divided by $99$) as a remainder.
i.e. $25*(25^3)$ divided by $99$ gives $56$ as a remainder.
$(25^3)*(25^3)$ divided by $99$ gives (remainder when $26*26$ is divided by $99$) as a remainder.
i.e. $(25^3)*(25^3)$ divided by $99$ gives $82$ as a remainder.
|
Using Carmichael Function,
$\lambda(99)=$lcm $(\lambda(9),\lambda(11))=$lcm$(3(3-1),10)=30$
So, $5^{30}\equiv1\pmod {99}$
Now, $25^{889}=(5^2)^{889}=5^{1788}$
Also, $1778\equiv 8\pmod {30}\implies 25^{889}=5^{1780}\equiv5^8\pmod {99}$
$5^2=25,5^3=125\equiv26\pmod{99},5^4\equiv26\cdot5\equiv31\pmod{99},$
$5^8=(5^4)^2\equiv(31)^2\pmod{99}\equiv961\equiv-29\pmod{99}$ as $990=99\cdot10$
So, $5^8\equiv-29\pmod{99}\equiv70$
|
{
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"url": "https://math.stackexchange.com/questions/346900",
"timestamp": "2023-03-29T00:00:00",
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|
How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$ Is there an easy way to prove the identity?
$$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3} \right )$$
While solving one question, I am stuck, which looks obvious but without any feasible way to approach.
Few observations, not sure if it would help
$$
\begin{align}
\dfrac{\dfrac{\pi}{7}+\dfrac{3\pi}{7}}{2} &= \dfrac{2\pi}{7}\\\\
\dfrac{\pi}{7} + \dfrac{3\pi}{7} + \dfrac{2\pi}{7} &= \pi - \dfrac{\pi}{7}
\end{align}
$$
|
Let $w = \cos \left ( \frac{2\pi}{7} \right ) + i\sin \left ( \frac{2\pi}{7} \right )$ so that $w^7 = 1$. Thus
$$\begin{align*}
w^7 - 1 &= 0\\
(w-1)(w^6 + w^5 + w^4 + w^3 + w^2 + w + 1) &= 0\\
w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 &= 0 &&\text{since } w \ne 1\\
\left ( w^3 + w^{-3} \right ) + \left ( w^2 + w^{-2} \right ) + \left ( w + w^{-1} \right ) &= -1 &&\text{since } w \ne 0\\
\end{align*}$$
Since $w + w^{-1} = \cos \left ( \frac{2\pi}{7} \right ) + i\sin \left ( \frac{2\pi}{7} \right ) + \cos \left ( - \frac{2\pi}{7} \right ) + i\sin \left ( - \frac{2\pi}{7} \right ) = 2\cos \left ( \frac{2\pi}{7} \right )$, using de Moivre's theroem:
$$\begin{align*}
2\cos \left ( 3\times \frac{2\pi}{7} \right ) + 2\cos \left ( 2\times \frac{2\pi}{7} \right ) + 2\cos \left ( \frac{2\pi}{7} \right ) &= -1\\
\cos \left ( \frac{6\pi}{7} \right ) + \cos \left ( \frac{4\pi}{7} \right ) + \cos \left ( \frac{2\pi}{7} \right ) &= -\frac{1}{2}= -\cos \left (\frac{\pi}{3} \right )
\end{align*}$$
Using $\cos(\theta) = -\cos \left (\pi - \theta \right )$:
$$-\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{3\pi}{7} \right ) + \cos \left ( \frac{2\pi}{7} \right ) = -\cos \left (\frac{\pi}{3} \right )$$
And hence
$$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3} \right )$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving a system of n linear equations has only one solution I have been given the system:
$-x_1 + 2x_2 + ... + 2x_{n-1} + 2x_n = 1$
$2x_1 - x_2 + ... 2x_{n-1} + 2x_n = 2$
$.$
$.$
$.$
$2x_1 + 2x_2 + ... + 2x_{n-1} - x_n = n$
And the assignment to prove that it has only one solution. I am aware of the existence of discriminants and their role of solving system of linear equations, although I have no idea on how to use that to prove this system has only one solution. (And I currently lack materials on infinite determinants and recurrent relations, any links would be appreciated).
Could anyone provide suggestions or hints on how to prove this system has exactly one solution?
|
Your equation is of the form $Ax = b$ where $$A = \begin{bmatrix}-1 & 2 & 2 & 2 & 2 & \cdots & 2 & 2\\ 2 & - 1 & 2 & 2 & 2 &\cdots & 2 & 2\\ 2 & 2 & -1 & 2 & 2 &\cdots & 2 & 2\\ 2 & 2 & 2 & -1 & 2 &\cdots & 2 & 2\\ 2 & 2 & 2 & 2 & -1 &\cdots & 2 & 2\\ \vdots & \vdots & \vdots & \vdots & \vdots &\ddots & \vdots & \vdots\\ 2 & 2 & 2 & 2 & 2 &\cdots & -1 & 2\\ 2 & 2 & 2 & 2 & 2 &\cdots & 2 & -1 \end{bmatrix}$$
We then have
$$A = -3I_{n \times n} + 2 \begin{bmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \vdots\\ 1\\ 1\end{bmatrix} \begin{bmatrix} 1& 1& 1& 1& 1& \cdots& 1& 1\end{bmatrix}$$
By Sherman-Morrison, we have
$$A^{-1} = -\dfrac13 I_{n \times n}-\dfrac2{3(3-2n)}\begin{bmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \vdots\\ 1\\ 1\end{bmatrix}\begin{bmatrix} 1& 1& 1& 1& 1& \cdots& 1& 1\end{bmatrix}$$
Clearly, $A^{-1}$ exists for all $n$.
|
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|
Equation $(8\cos^3x+1)^3=162\cos x-27$
Solve equation $$(8\cos^3x+1)^3=162\cos x-27$$
I saw this equation before 5 month, and I couldn't solve it. This isn't homework, etc. (I don't do stuff like this anymore). I am just curious.
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This approach is based off knowing that the answers are $ \frac {2 \pi}{9} , \frac {4 \pi }{9} , \frac { 8 \pi }{9} $. Another equation whose solutions are exactly those values, is $ 2 \cos 3 x + 1 = 0$, so we'd look to factorize that out.
We know that $ 2 \cos 3x + 1 = 8\cos^3 - 6 \cos x + 1$, so $ (8 \cos^3 x + 1) = (2 \cos 3x + 1 + 6 \cos x)$. Henceforth, let $w = 2 \cos 3x + 1$.
We have $ ( w + 6 \cos x) ^3 = 162 \cos x - 27$, or that $w^3 + 3 w^2 6\cos x + 3 w (6 \cos x)^2 + 216 \cos^3 x = 162 \cos x - 27$.
Continue to focus on factorizing out $w$, we get
$w^3 + 18 w^2 \cos x + 108 w \cos^2x = - 216 \cos^3 x + 162 \cos x - 27 = - 27 w$,
hence
$w^3 + 18w^2 \cos x + 108 w \cos^2x + 27 w = 0$.
This factorizes into $0 = w ( w^2 + 18 w \cos x + 108 \cos^2 x + 27) = w [ (w + 9 \cos x)^2 + 27 \cos^2 x + 27 ]. $
It is clear that the other term is strictly positive, so has no solution for $x$ (regardless of what $w$ is), so we must have $w = 0$. Hence, the solutions are $x = \frac {2\pi}{9}, \frac { 4\pi}{9}, \frac {8\pi}{9}$.
|
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|
Prove $(n^5-n)$ is divisible by 5 by induction. So I started with a base case $n = 1$. This yields $5|0$, which is true since zero is divisible by any non zero number. I let $n = k >= 1$ and let $5|A = (k^5-k)$. Now I want to show $5|B = [(k+1)^5-(k+1)]$ is true....
After that I get lost.
I was given a supplement that provides a similar example, but that confuses me as well.
Here it is if anyone wants to take a look at it:
Prove that for all n elements of N, $27|(10n + 18n - 1)$.
Proof:
We use the method of mathematical induction. For $n = 1$, $10^1+18*1-1 = 27$.
Since $27|27$, the statement is correct in this case.
Let $n = k = 1$ and let $27|A = 10k + 18k - 1$.
We wish to show that $27|B = 10k+1 + 18(k + 1) - 1 = 10k+1 + 18k + 17$.
Consider $C = B - 10A$ ***I don't understand why A is multiplied by 10.
$= (10k+1 + 18k + 17) - (10k+1 + 180k - 10)$
$= -162k + 27
= 27(-6k + 1)$.
Then $27|C$, and $B = 10A+C$. Since $27|A$ (inductive hypothesis) and $27|C$, then
$B$ is the sum of two addends each divisible by $27$. By Theorem 1 (iii), $27|B$, and
the proof is complete.
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Your induction hypothesis is that $5\mid k^5-k$, which means that $k^5-k=5n$ for some integer $n$. Now
$$\begin{align*}
(k+1)^5-(k+1)&=\left(k^5+5k^4+10k^3+10k^2+5k+1\right)-(k+1)\\
&=k^5+5k^4+10k^3+10k^2+5k-k\\
&=(k^5-k)+5k^4+10k^3+10k^2+5k
\end{align*}$$
can you see why that must be a multiple of $5$?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof of $\sqrt{2^{2^k}} = 2^{2^{k-1}}$? It's quite easy to observe that for $k \ge 0$:
$$
\begin{align}
2^{2^k} &= 4, 16, 256, 65536, \dots\\
\sqrt{2^{2^k}} &= 2, 4, 16, 256,\dots
\end{align}
$$
More in general:
$$
\sqrt{2^{2^k}} = 2^{2^{k-1}}
$$
How can I prove this identity?
|
$$\begin{align}
\left(2^{2^{k-1}}\right)^2 & = \left(2^{2^{k-1}}\right)\left(2^{2^{k-1}}\right)\tag{1}\\
& = \left(2^{2^{k-1}+2^{k-1}}\right)\tag{2}\\
& = \left(2^{2(2^{k-1})}\right)\tag{3}\\
& = 2^{2^{k}}\tag{4}\\
\end{align}$$
To go from $(1)$ to $(2)$, we apply the rule that $a^ba^c=a^{b+c}$.
Going from $(2)$ to $(3)$, we note that $a+a = 2a$.
Going from $(3)$ to $(4)$, we again apply the rule that $a^ba^c=a^{b+c}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Sin inverse of a complex number Is it possible to calculate the value of $\delta$ from the relation
$\delta=\sin^{-1}(5.4i)$ ? where $i=\sqrt{-1}$
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Let $\delta=x+iy$
So, $\sin(x+iy)=5.4i$
Now, $\sin(x+iy)=\sin x\cos(iy)+\cos x\sin(iy)=\sin x\cosh y+i\cos x\sinh y$
Equating the real parts, $\sin x\cosh y=0\implies \sin x=0$ as $\cosh y\ge 1$ for real $y$
So, $\cos x=\pm1$
If $\cos x=1, x=2m\pi$ where $m$ is any integer
and $\sinh y=5.4\implies \frac{e^y-e^{-y}}2=\frac{27}5\implies 5(e^y)^2-54e^y-5=0$
Solve for $e^y$ which is $>0$ for real $y$
If $\cos x=-1, x=2(n+1)\pi$ where $n$ is any integer
and $\sinh y=-5.4\implies \frac{e^y-e^{-y}}2=-\frac{27}5\implies 5(e^y)^2+54e^y-5=0$
Solve for $e^y$ which is $>0$ for real $y$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find $\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$? Integration by parts is of no success. What else to try?
$$\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$$
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If $a$ is real-valued and $|a| >1$, $$\sum_{n=1}^{\infty} \frac{\cos (n \theta)}{n} \left(\frac{1}{a} \right)^{n} = - \frac{1}{2} \log \left(1- \frac{2}{a} \cos \theta+ \frac{1}{a^{2}} \right) = \log|a| - \frac{1}{2} \log \left(1-2 a \cos \theta +a^{2} \right). $$
This identity can be derived from the Maclaurin series $$\sum_{n=1}^{\infty} \frac{z^{n}}{n} = - \log(1-z) \ , \ |z| <1$$ by replacing $z$ with $\displaystyle \frac{e^{i \theta}}{a}$ and equating the real parts on both sides.
Now rearrange the identity, square both sides, and integrate.
$$ \begin{align} \int_{0}^{\pi} \log^{2}(1-2a \cos \theta+a^{2}) \ d \theta &= 4 \log^{2}|a| \int_{0}^{\pi} d \theta -4 \log|a| \int_{0}^{\pi} \sum_{n=1}^{\infty} \frac{\cos (n \theta)}{n} \left(\frac{1}{a} \right)^{n} \ d \theta \\ &+ 4 \int_{0}^{\pi} \sum_{k=1}^{\infty} \frac{\cos(k \theta)}{k} \left(\frac{1}{a} \right)^{k} \sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n} \left(\frac{1}{a} \right)^{n} \ d \theta \\ &= 4 \pi \log^{2}|a| - 4 \log|a| \sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{1}{a} \right)^{n} \int_{0}^{\pi} \cos (n \theta) \ d \theta \\ &+ 4 \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{kn} \left(\frac{1}{a} \right)^{k+n} \int_{0}^{\pi} \cos(k \theta) \cos(n \theta) \ d \theta \\ &= 4 \pi \log^{2}|a| - 0 + 4 \sum_{n=1}^{\infty} \frac{1}{n^{2}} \left(\frac{1}{a} \right)^{2n} \int_{0}^{\pi} \cos^{2}(n \theta) \ d \theta \qquad (1) \\ &= 4 \pi \log^{2}|a| + 2 \pi \sum_{n=1}^{\infty} \frac{1}{n^{2}} \left( \frac{1}{a^{2}}\right)^{n} \\ &= 4 \pi \log^{2}|a| + 2 \pi \ \text{Li}_{2} \left(\frac{1}{a^{2}} \right) \end{align}$$
$(1)$ $\displaystyle \int_{0}^{\pi} \cos(k\theta) \cos(n \theta) \ d \theta = 0$ unless $k=n$
|
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|
Finding all orthogonal matrices of a given form I am given this problem. I need to find all $a,b,c,d \in \mathbb{R}$ for which the matrix $A$ is orthogonal.
$$A:=\frac{1}{3} \begin{pmatrix} -1 & 2 & a \\ 2 & 2 & b \\ 2 & c & d \end{pmatrix}$$
I know that $A$ is orthogonal if $A^T=A^{-1}$. But I am stuck how to find all unknowns and prove its orthogonality.
Can you guys please help me out?
|
Let $u,v,w$ be the first, second and third collumns of $3A$, respectively.
You wish to have $u\cdot v=0\wedge u\cdot w=0 \wedge v\cdot w=0$ (among other things).
This yields $-2+4+2c=0 \wedge -a+2b+2d= 0\wedge 2a+2b+cd=0$. Note that this is a linear system in (a very poor) disguise. So:
$$\begin{align} \begin{cases} -2+4+2c=0 \\ -a+2b+2d= 0 \\ 2a+2b+cd=0 \end{cases}&\iff \begin{cases}c=-1\\-a+2b+2d= 0 \\2a+2b-d=0 \end{cases}\\
&\iff \begin{cases}c=-1\\ -a+2b+2d= 0\\ 3a+0b-3d=0 \end{cases}\\
&\iff \begin{cases}c=-1 \\a+2b=0\\a=d\end{cases}\\
&\iff\begin{cases}c=-1\\ b=-a/2\\ a=d\end{cases} \end{align}$$
and you get $3A=\begin{pmatrix} -1 & 2 & a \\ 2 & 2 & -a/2 \\ 2 & -1 & a \end{pmatrix}$
Now you want $||(a,-a/2,a)||=3$, that is $\displaystyle \sqrt {a^2+\frac{a^2}{4} +a^2}=3$, i.e., $\displaystyle \sqrt \frac{9a^2}{4}=3$ which gives $a=\pm 2.$
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|
Strategies to prove inequalities with interval notation How to prove a inequalities with interval notation, for example:
Find minimum of $a^3+b^3+c^3$ with $a,b,c \in [-1;\infty), a^2+b^2+c^2=9$
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By Holder: if $a,b,c>0$
$$(a^3+b^3+c^3)(a^3+b^3+c^3)(1+1+1)\ge (a^2+b^2+c^2)^3$$
then
$$a^3+b^3+c^3\ge 9\sqrt{3}$$
and when $a=b=c=\sqrt{3}$
case 2:let $a\ge b\ge 0>c\ge -1$
then $a^2+b^2>8$,then $a>2sqrt{2}$,so $a^3+b^3+c^3>a^3+c^3>a^3-1>\sqrt{512}-1>9\sqrt{3}$
case 3:let $a>0>b>c\ge -1$,and we have $a^2>7$,so $a>\sqrt{7}$,then
$a^3+b^3+c^3>\sqrt{343}-2>9\sqrt{3}$
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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.