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Determine appropriate $c$ and $x_0$ for Big-O proofs. "Prove that $f(x)$ is $O(x^2)$:"
$$f(x) = \frac{x^4+2x-7}{2x^2-x-1}$$
Let $c=10$ (addition of coefficients of the numerator less the addition of coefficients of the denominator), and $x_0 = 1$ (the lowest coefficient among all elements in the function).
$$\frac{x^4+2x-7}{2x^2-x-1} \leq \frac{x^4+2x^4-7x^4}{2x^4-x^4-x^4} \leq 10x^2$$
Am I going on the right track? For any $x \geq 1$, the second fraction becomes troublesome, as you then get a division by zero. I'm not sure how to approach these questions if someone could assist me.
|
I will try to give you a more intuitive approach.
Theoretical Approach
The basic idea is the following: when $x$ get sufficiently large, the highest order term dominates over all the rest.
$x^4 + 2x - 7$ will eventually look like $x^4$.
$2x^2 -x - 1$ will eventually look like $2x^2$.
So
$\dfrac{x^4 + 2x - 7}{2x^2-x-1}$ will eventually look like $\dfrac{x^4}{2x^2} = \dfrac{1}{2}x^2$.
This is one way of seeing why your function should be $O(x^2)$.
Now, what should be $c$? You actually do NOT want to use $1/2$. This is because you want something that is definitely larger than $\frac{x^4 + 2x - 7}{2x^2-x-1}$. So you definitely want $c$ to be something bigger than $1/2$.
Theoretically speaking, you can choose $c$ to be any number bigger than $c$, and theoretically, $x_0$ exists. This is because when $x$ is very big, $\frac{x^4 + 2x - 7}{2x^2-x-1}$ is very close to $\dfrac{1}{2}x^2$, which will be smaller than $cx^2$.
More Practical Approach
You have a fraction. If you want an upper bound for a fraction, you need
*
*An upper bound for the numerator. (Bigger numerator means bigger value $\dfrac{2}{2} < \dfrac{4}{2}$)
*A lower bound for the denominator. (Smaller denominator means bigger value $\dfrac{2}{2} < \dfrac{2}{1}$)
Note from our theoretical approach that our top should be bounded by $x^4$, and the bottom should be bounded by $x^2$.
Easiest thing to do for the top: bound each term by $x^4$, the higher order term. It works precisely because higher order terms dominate eventually.
$x^4 \le x^4$, always.
$2x \le x^4$, when say $x > 2$. (To be more exact, this happens precisely when $2 \le x^3$, or $\sqrt[3]{2} \le x$. But for $O$ proofs this kind of precision is unnecessary.)
$-7 \le x^4$, always.
So, $x^4 + 2x - 7 \le x^4 + x^4 + x^4 = 3x^4$.
Now, what about the denominator?
I need something that will be eventually smaller than $2x^2 - x - 1$. Note from what I said earlier, that we want to use something like $x^2$. So $n x^2$ is good, for any $n$ less than 2. In fact, to make things easy, let's choose $n=1$.
The question now becomes: When will $2x^2 - x - 1 > x^2$?
$2x^2 - x - 1 > x^2$ becomes
$x^2 - x - 1 > 0 $.
$y = x^2 - x - 1$ is a parabola, and you just need to figure out some positive number $x$ where this parabola becomes positive.
How about $x = 10$? $(10^2) - 10 - 1 > 0$. And if you plug in a larger value for $x$, then it'll certainly be bigger.
Note: What Tim did in his answer was to find the precise point at which $2x^2 - x - 1 = x^2$. This gives you a much more precise $x_0$. But this isn't necessary for $O$ proofs.
So when $x>10$, we have both:
*
*$x^4 + 2x - 7 \le x^4 + x^4 + x^4 = 3x^4$.
*$2x^2 - x - 1 > x^2$.
And so
$\dfrac{x^4 + 2x - 7}{2x^2-x-1} < \dfrac{3x^4}{x^2} = 3x^2$
So my solution uses $c=3$ with $x_0 = 10$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Tips for solving linear equations Is there an easier way to solve linear equations than trial and error?
$$2(3b -1) - (4b -6) = 3$$
|
To solve $2(3b - 1) -(4b -6) =3$ for the unknown $ b $, it is best to simplify the equation's left side. To do that:
*
*Distribute: $2(3b - 1) -(4b -6) = (2 \cdot 3b - 2 \cdot 1) - 1 \cdot 4b - 1 \cdot {-6} = 6b - 2 - 4b + 6$
*Combine like terms: $ 6b - 2 - 4b + 6 = 6b - 4b - 2 + 6 = 2b + 4 $
Thus, $$ 2(3b - 1) -(4b -6) =3 $$ $$ <=> 2b + 4 = 3$$
Subtract $4$ from each side of the equation:
$$ 2b + 4 - 4 = 3 - 4$$ $$<=>2b = -1$$
Finally divide $2$ to each side to isolate the unknown $b$:
$$\frac{2b}{2} = \frac{-1}{2}$$ $$<=>b = -\frac{1}{2}$$
Thus, $ b = -\frac{1}{2} $.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Coefficient of $x^4$ in multinomial expansion What is the coefficient of $x^4$ in $(1 + x - 2x^2)^7$? What is a quick way to solve this problem using the binomial theorem (I have not learned multinomial theorem)?
|
By factorizing the inner expresion,
$$(1+x-2x^{2})^{7}=((1+2x)(1-x))^{7}=(1+2x)^{7}(1-x)^{7}$$
Using the Binomial Expansion on both, we get:
$$\begin{align}(1+x-2x^{2})^{7} &= \left(1 + \binom{7}{1}(2x) + \binom{7}{2}(2x)^2 + \binom{7}{3}(2x)^3 + \binom{7}{4}(2x)^4\right) \\
&\times\left(1 + \binom{7}{1}(-x) + \binom{7}{2}(-x)^2 + \binom{7}{3}(-x)^3 + \binom{7}{4}(-x)^4\right)\\
&= \left(1 + 14x + 84x^2 + 280x^3 + 560x^4 + ...\right)\\
&\times\left(1 - 7x + 21x^2 - 35x^3 + 35x^4 + ...\right)\\
&= (1\times35 + 14\times-35 + 84\times21 + 280\times-7 + 560\times1)x^4 + ...\\
&= -91x^4 + ... \end{align}$$
|
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|
How to solve for matrix $A$ in $AB = I$ Given $B$ = $\begin{bmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
1 & 1 & 1
\end{bmatrix}$
I know that $B$ is equal to inverse of $A$, how can I go backwards to solve for $A$ in $AB = I$?
|
Start by writing $[B|I]$ and apply row operations until you end up with $[I|A]$. This works because row operations are multiplications with suitable matrices from the left, and if this suitable matrix turns $B$ into $I$ it must be $B^{-1}=A$, which of course turns $I$ into $A$:
$$\begin{align}\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\
1 & 1 & 0 & | & 0 & 1 & 0 \\
1 & 1 & 1 & | & 0 & 0 & 1 \\
\end{bmatrix}&\text{subtract first from second}\\
\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\
0 & 1 & 0 & | & -1 & 1 & 0 \\
1 & 1 & 1 & | & 0 & 0 & 1 \\
\end{bmatrix}&\text{subtract first from third}\\
\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\
0 & 1 & 0 & | & -1 & 1 & 0 \\
0 & 1 & 1 & | & -1 & 0 & 1 \\
\end{bmatrix}&\text{subtract second from third}\\
\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\
0 & 1 & 0 & | & -1 & 1 & 0 \\
0 & 0 & 1 & | & 0 & -1 & 1 \\
\end{bmatrix}&\text{done}\end{align}$$
|
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|
Geometry problem about angles and triangles I've been working on this problem for a while. It doesn't seem to hard, but I cannot reach a satisfying solution.
The triangle $ABC$ is isosceles with base $\overline{AC}$. A point $O$ is also given.
Knowing: $\overline{OA}=R$, $\overline{AB}\equiv \overline{BC}=r$ and the angles $\widehat{OAB}=\varphi$, $\widehat{AOC}=\theta$, find the values of $\overline{OC}$ and $\widehat{OCB}$.
Blue objects are known, red are not.
Original (equivalent but messier) formulation:
On the plane $\pi$, two half lines: $s$, $t$ start from a point $O$ forming an angle $\theta$ between them.
A point $A$ is given on the line $s$ and a point $B$ is given on the plane $\pi$ such that $\overline{AB} >\text{dist}(B,t)$.
Find the biggest $\overline{OC}$ such that $\overline{AB}\equiv\overline{BC}$ with $C$ lying on the line $t$. What are the values of the segment $\overline{OC}$ and of the angle $\widehat{OCB}$?
So far I've been able to get:
$\widehat{CBA}=\widehat{AOC}+\widehat{OAB}+\widehat{OCB}$
which still contains two unknowns. I should have worked out a complete system of equations using triangles sine and cosine formulas, but then the substitutions become messy and I cannot reach anything as simple as the problem seems. Thanks for any hint!
|
I'm going to take as starting data the following, using your first formulation of things.
\begin{align}
\newcommand{\uvec}{{\bf u}}
\newcommand{\wvec}{{\bf w}}
P &= (x, y), \text{the coordinates of your point $O$, which I'll call $P$}\\
B &= (s, t), \text{the coordinates of $B$}\\
A &= (a, b), \text{the coordinates of $A$}\\
\uvec &= (h, k), \text{a vector pointing from $P$ towards $C$}
\end{align}
Thus points on the ray you called $t$ are all of the form
$$
P + c \uvec
$$
for some nonnegative number $c$.
Let
\begin{align}
r &= \sqrt{(s-a)^2 + (t-b)^2}, \text{the distance from $A$ to $B$}
\end{align}
For any value of $c$, we know that the vector from $P + c \uvec$ to $B$ is
\begin{align}
{\bf w} &= P + c\uvec - B \\
&= c \uvec + (P-B) \\
&= c (h, k) + (x-s, y-t) \\
&= (ch + x-s, ck + y - t).
\end{align}
We'd like to pick $c$ so that the length of $\wvec$ is exactly $r$, or, equivalently, so that its squared length is $r^2$. That means solving
\begin{align}
(ch + x-s)^2 + (ck + y - t)^2 = r^2
\end{align}
Expanding out the left-hand side, we get
\begin{align}
c^2h^2 + 2ch(x-s)+ (x-s)^2 + c^2k^2 + 2ck(y - t) + (y-t)^2 &= r^2\\
c^2(h^2 + k^2) + c\left(2h(x-s) + 2k(y - t) \right) + (x-s)^2 + (y-t)^2 - r^2 &= 0.
\end{align}
That's a quadratic expression in $c$, whose roots are
\begin{align}
c &= \frac{1}{2(h^2 + k^2)}\left(-\left(2h(x-s) + 2k(y - t) \right) \pm \sqrt{\left(2h(x-s) + 2k(y - t) \right)^2 - 4(h^2 + k^2)((x-s)^2 + (y-t)^2 - r^2)}\right)
\end{align}
The larger of these two roots finds the point farther along the ray from $P$; that larger root is
\begin{align}
c &= \frac{1}{2(h^2 + k^2)}\left(-\left(2h(x-s) + 2k(y - t) \right) + \sqrt{\left(2h(x-s) + 2k(y - t) \right)^2 - 4(h^2 + k^2)((x-s)^2 + (y-t)^2 - r^2)}\right)
\end{align}
and the resulting point, $C$, is at the location
$$
(x_c, y_c) = (x + ch, y + ck)
$$
where $c$ is the expression given above.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find possible minimal polynomial for $L^3 + 2L^2 + L + 3 \cdot I_v$ Given:
$L: V \rightarrow V ; L^2 + I_v = 0$
Find:
possibilities for the minimal polynom of $L^3+2L^2+L+3\cdot I_v$
|
$x - 1$. The one and only minimal polynomial of $L^3 + 2L^2 + L + 3I_V \;$ when $L^2 + I_V = 0$ is $x - 1$.
Observe that
$x^3 + 2x^2 + x + 3 = (x + 2)(x^2 + 1) + 1, \tag{1}$
which in turn implies, since $L^2 + I_V = 0$,
$L^3 + 2L^2 + L + 3I_V = (L + 2I_V)(L^2 + I_V) + I_V = I_V; \tag{2}$
the one and only minimal polynomial of $I_V$ is $x - 1$.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
|
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|
Compute $\int_0^\infty\frac{\cos(xt)}{1+t^2}dt$ Let $x \in \mathbb R$
Find a closed form of $\int_0^\infty\dfrac{\cos(xt)}{1+t^2}dt$ .
Let me give some context: this is an exercise from an improper integrals course for undergraduates. My teacher was not able to give a valid solution without resorting to Fourier series.
He actually wanted to prove first that the function is a solution of the ODE $$ y-y'' =0$$ without succeeding in proving it.
Anyway the result sould be $\dfrac{\pi}{2}e^{-x}$
Thanks for any help on this subject.
|
We can also find the differential equation by just differentiating under the integral sign, after rewriting a little:
$$F(x) = \int_0^\infty \frac{\cos (xt)}{1+t^2}\,dt$$
is continuous by the dominated convergence theorem, with $F(0) = \frac{\pi}{2}$ known and $\lvert F(x)\rvert \leqslant \frac{\pi}{2}$. For $x \neq 0$, we can integrate by parts,
$$\begin{align}
F(x) &= \left[\frac{1}{x}\cdot \frac{\sin (xt)}{1+t^2}\right]_0^\infty + \frac{2}{x} \underbrace{\int_0^\infty \frac{t\sin (xt)}{(1+t^2)^2}\,dt}_{G(x)}\\
&= \frac{2}{x}G(x).
\end{align}$$
$G(x)$ can be differentiated under the integral sign by the dominated convergence theorem, and
$$G'(x) = \int_0^\infty \frac{t^2\cos (xt)}{(1+t^2)^2}\,dt = F(x) - \underbrace{\int_0^\infty \frac{\cos (xt)}{(1+t^2)^2}\,dt}_{H(x)}.$$
$H(x)$ can also be differentiated under the integral sign by the dominated convergence theorem, and
$$H'(x) = -G(x).$$
Thus
$$\begin{align}
F'(x) &= -\frac{2}{x^2}G(x) + \frac{2}{x}G'(x)\\
&= -\frac{1}{x}F(x) + \frac{2}{x}\left(F(x) - H(x)\right)\\
&= \frac{1}{x}F(x) - \frac{2}{x}H(x),\quad\text{and}\\
F''(x) &= -\frac{1}{x^2}F(x) + \frac{1}{x} F'(x) + \frac{2}{x^2}H(x) - \frac{2}{x}H'(x)\\
&= -\frac{1}{x^2}F(x) + \frac{1}{x^2}F(x) - \frac{2}{x^2}H(x) + \frac{2}{x^2}H(x) + \frac{2}{x}G(x)\\
&= \frac{2}{x}G(x)\\
&= F(x).
\end{align}$$
From the boundedness of $F$ and the differential equation together with the initial value $F(0)$ we then obtain
$$F(x) = \frac{\pi}{2}e^{-\lvert x\rvert}.$$
|
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|
Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$ Prove by induction $6\cdot 7^n -2\cdot 3^n$ is divisible by $4$ for $n\ge 1$.
I am struggling on this problem very much.
So far I have Basis case = $6\cdot 7^1 - 2\cdot3^1 = 36$ which is divisible by $4$
Assume for a $n$ that $6\cdot 7^n-2\cdot3^n$ is divisible by $4$.
After that I am not sure what to do.
|
If $a$ is divisible by $4$, then $b$ is divisible by $4$ if and only if $b-a$ is divisible by $4$.
So, you can assume $6\cdot 7^n-2\cdot 3^n$ is divisible by $4$ and try for
\begin{align}
(6\cdot 7^{n+1}-2\cdot 3^{n+1})-(6\cdot 7^n-2\cdot 3^n)
&=
(6\cdot 7^{n+1}-6\cdot 7^{n})-(2\cdot 3^{n+1}-2\cdot 3^{n})
\\
&=
(6\cdot 7\cdot 7^{n}-6\cdot 7^{n})-(2\cdot 3\cdot 3^{n}-2\cdot 3^{n})\\
&=
\dots
\end{align}
Of course, using congruences modulo $4$ is easier:
$$
6\cdot 7^n-2\cdot 3^n\equiv 2\cdot 3^n-2\cdot 3^n\equiv 0 \pmod{4}
$$
|
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|
Why does this outcome change depending on infinity Why does the outcome of the limit as x approaches infinity of $$\sqrt{x^2+2x}- \sqrt{x^2-2x}$$
which simplifies to
$$\dfrac {4x}{x \left(\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}\right) }= \dfrac {4}{\sqrt{1 + \frac 2x} + \sqrt {1 - \frac 2x}}$$
change depending on whether we take the limit from minus infinity or positive infinity?
If positive, the answer is 2. If negative infinity, the answer is -2??
It seems to me that we get $$ \frac {4}{\sqrt{1} + \sqrt{1}}$$ no matter which side we approach.
|
The equation $$\sqrt{x^2+2x}=x\sqrt{1+\frac{2}{x}}$$
is true only for $x>0$. The correct equation is $$\sqrt{x^2+2x}=\sqrt{x^2}\sqrt{1+\frac{2}{x}}=|x|\sqrt{1+\frac{2}{x}}$$
The intuition is that for $x>0$ you are subtracting a smaller number from a larger one (i.e. the two square roots); however for $x<0$ you are subtracting a larger number from a smaller one.
|
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|
Formula help with this equation I don't know what the answer to this formula is, can someone please help me.
I've tried lots of things but getting no where.
If $x=\dfrac56+\dfrac{15}{18}-\dfrac{10}{12}$, then $(x-1)3=$ ?
|
The expression $x=5/6+15/18-10/12$ may be simplified considerably:
\begin{align*}
x &= \frac{5}{6}+\frac{15}{18}-\frac{10}{12}\\
&= \frac{5}{6}+\left(\frac{30}{36}-\frac{30}{36}\right)\\
&= \frac{5}{6}.
\end{align*}
Thus, $x-1=-\frac{1}{6}$ and $3(x-1)=-\frac{1}{2}$.
As @ParthKohli has pointed out in a comment, you may actually be interested in the expression $(x-1)^3$, rather than in $(x-1)3$. If that's the case, you can still use the simplification for $x$ provided above to obtain
$$(x-1)^3=-\frac{1^3}{6^3}=-\frac{1}{216}.$$
|
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|
How to evaluate $\sum_{n=1}^{\infty}\frac{H^3_{n}}{n+1}(-1)^{n+1}$. How Find this sum
$$I=\sum_{n=1}^{\infty}\dfrac{H^3_{n}}{n+1}(-1)^{n+1}$$
where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$
My idea: since
$$\dfrac{1}{n+1}(-1)^{n+1}=-\int_{-1}^{0}x^ndx$$
so
$$I=\sum_{n=1}^{\infty}H^3_{n}\int_{0}^{-1}x^ndx$$
then I can't.Thank you
This problem is not Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$
|
We will use the combinatorial identity, which can be proved through induction
$$\left(H_n^{(1)}\right)^3 - 3H^{(1)}_{n}H^{(2)}_{n} + 2H^{(3)}_{n} = \left [ n + 1 \atop 4\right] \frac{6}{(n-1)!}$$
Where the binomial-like notation of the right side is unsigned Stirling number. Multiplying by $x^n$ and summing both sides from $n =0$ to $\infty$, we get
$$\sum_{n=0}^\infty\left(H_n^{(1)}\right)^3 x^{n} - 3\sum_{n=0}^\infty H^{(1)}_{n-1}H^{(2)}_{n-1} x^{n} + 2\sum_{n=0}^\infty H^{(3)}_{n-1} x^{n} = 6\sum_{n=0}^\infty\left [ n+1 \atop 4\right]\frac{x^n}{(n-1)!} \tag{1}$$
Then note that we have the generating function
$$\sum_{n=1}^\infty (-1)^{n-k}\left [ n \atop k \right] \frac{z^n}{n!} = \frac{\log(1+z)^k}{k!}$$
Assuming $k = 4$, making the sub $z \mapsto -z$ gives
$$\sum_{n=1}^\infty \left [ n \atop 4 \right] \frac{z^n}{n!} = \frac{\log(1-z)^4}{24}$$
Diffing with respect to $z$ then gives
$$\sum_{n=1}^\infty \left [ n \atop 4\right ] \frac{z^{n-1}}{(n-1)!} = -\frac{1}{6}\frac{\log(1-z)^3}{1-z} \\ \!\!\!\!\!\!\!\!\implies \sum_{n=0}^\infty \left [ n + 1 \atop 4\right ] \frac{z^n}{(n-1)!} = -\frac{1}{6}\frac{\log(1-z)^3}{1-z} \tag{2}$$
Then subbing $(2)$ to the left side of $(1)$ gives us
$$\sum_{n=0}^{\infty}\left(H_n^{(1)}\right)^3x^n = \frac{\log^3(1-z)}{1-z} + 3\sum_{n=0}^\infty H^{(1)}_{n}H^{(2)}_{n} x^n - 2\sum_{n=0}^\infty H^{(3)}_{n} x^n \tag{3}$$
The rightmost sum is simply ${\text{Li}_3(x)}/(1-x)$, by summation interchange. The middle one is tricky.
$$\begin{align} \sum_{n=1}^\infty H_{n}H_{n}^{(2)} x^n &= -\sum_{n=1}^\infty x^n H_n \left( \psi_1(n+1)-\psi_1(1) \right) \\ &=-\frac{\psi_1(1)\log(1-x)}{1-x}-\sum_{n=1}^\infty x^n H_n \psi_1(n+1) \\ &= -\frac{\psi_1(1)\log(1-x)}{1-x}+\sum_{n=1}^\infty x^n H_n \int_0^1 \frac{z^n \log(z)}{1-z}dz \\ &= -\frac{\psi_1(1)\log(1-x)}{1-x}-\int_0^1 \frac{\log(z)\log(1-zx)}{(1-z)(1-xz)}dz \end{align}$$
Which is, through partial factorization, in turn
$$\!\!\!\!\!\!\!\!\!\!-\frac{\psi_1(1)\log(1-x)}{1-x}-\frac{1}{1-x}\int_0^1 \frac{\log(z)\log(1-zx)}{1-z}dz+\frac{x}{1-x}\int_0^1 \frac{\log(z)\log(1-zx)}{1-zx}dz \tag{4}$$
Evaluating the intermediate integral can be done, but it's quite a bit of tedious so I omit it. After some calculations, you can derive using some polylog identities that
$$\!\!\!\!\!\!\!\!\!\sum_{n=0}^\infty H^{(1)}_{n}H^{(2)}_{n} x^n = \frac{\text{Li}_3(1-x)+\text{Li}_3(x)+1/2\log^2(1-x)\log(x)-\zeta(2)\log(1-x)-\zeta(3)}{1-x} \tag{5}$$
Subbing $(5)$ and the polylog identity for the rightmost sum in $(3)$ gives
$$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\sum_{n=0}^{\infty}\left(H_n^{(1)}\right)^3x^n = \frac{-\pi^2/2\log(1-x)+3/2\log^{2}(1-x)\log(x)-\log^{3}(1-x)+\text{Li}_{3}(x)+3\text{Li}_{3}(1-x)-3\zeta(3)}{1-x}\tag{6}$$
Integrating with respect to $x$ and setting $x = -1$, carefully choosing the correct branch of logarithm, will give a closed form.
|
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|
$\triangle ABC$ has an angle $\angle A=60^{\circ}$. Also $AB=c$, $BC=a$, $AC=b$ and $2\cos B-1=\frac{a+b}{a+c}$. Find all the other angles.
$\triangle ABC$ has an angle $\angle A=60^{\circ}$. Also $AB=c$, $BC=a$, $AC=b$ and $2\cos B-1=\frac{a+b}{a+c}$. Find all the other angles.
And we can't use calculus, logarithms, limits or something else advanced. Of course, the $\cos$ and $\sin$ theorems are allowed.
The first thing I tried when I saw the problem was using the $\cos$ theorem to get that $a^2=b^2+c^2-bc$.
I don't think substituting $\cos B$ with an expression in terms of $a,b,c$ would help us find the angles.
There's probably a simple idea here that I can't see. So some observations would be great. Thanks.
|
Brute-force solution:
$$\frac{a+b}{a+c}=2\cos B-1=\frac{a^2+c^2-b^2-ac}{ac}=\frac{b^2+c^2-bc+c^2-b^2-ac}{ac}=\frac{2c-a-b}{a}$$
$$
0=a(a+b)+(a+c)(a+b-2c)=2a^2+2ab-ac+bc-2c^2=2(b^2+c^2-bc)+2ab-ac+bc-2c^2=(a+b)(2b-c)
$$
Therefore, $c=2b$, and consequently $a=\sqrt{3}b$. The angles follow easily.
|
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|
Two very difficult induction proofs; having trouble with the inductive step $$\sum_{k=0}^{n-2} \binom{n}{k}\binom{n}{k+1}\frac{n-2k-1}{k+1} = n-2 + \frac{1}{n+1}\binom{2n}{n}$$
$$\sum_{k=0}^{n-2} \binom{n}{k}\binom{n}{k+2}\frac{n-2k-1}{k+1} = -n + \frac{n}{(n+2)(n+1)}\binom{2n}{n}$$
The two are clearly related in some way, so proving one might yield the other, but I'm having a lot of difficulty knowing what to add to both sides to change all those n's to n+1's in the binomial coefficients of the sums. Do any of you have insight?
The induction will be on n, with base case n=2.
I've been using this:
$$\binom{n}{k} = \frac{n+1}{n+1-k}\binom{n+1}{k} $$
as a means of replacing n choose k with some factor of n+1 choose k, though current attempts are fairly circular. Substitute k+1 or k+1 for k to get a change to expand those binomials.
|
Let me present an algebraic proof of the first equality and a proof by
induction will perhaps appear.
Suppose we seek to verify that
$$\sum_{k=0}^{n-2} {n\choose k} {n\choose k+1}
\frac{n-2k-1}{k+1} = n-2+\frac{1}{n+1}{2n\choose n}.$$
The left has two pieces which are, first, piece $A$,
$$\sum_{k=0}^{n-2} {n\choose k} {n\choose k+1} \frac{n+1}{k+1}
= \sum_{k=0}^{n-2} {n+1\choose k+1} {n\choose k+1}$$
and second, piece $B$,
$$-2\sum_{k=0}^{n-2} {n\choose k} {n\choose k+1}.$$
Piece $A$ is
$$\sum_{k=1}^{n-1} {n+1\choose k} {n\choose k}
= -1 - (n+1) + \sum_{k=0}^{n} {n\choose k} {n+1\choose k}.$$
Introduce for the sum term
$${n+1\choose k} = {n+1\choose n+1-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+2-k}} (1+z)^{n+1} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+2}} (1+z)^{n+1}
\sum_{k=0}^{n} {n\choose k} z^k\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+2}} (1+z)^{n+1}
(1+z)^n \; dz
\\ = {2n+1\choose n+1}.$$
Therefore we have for piece $A$
$$-n-2 + {2n+1\choose n+1}.$$
Piece $B$ is
$$2n - 2\sum_{k=0}^{n-1} {n\choose k} {n\choose k+1}.$$
Introduce for the sum term
$${n\choose k+1} = {n\choose n-k-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-k}} (1+z)^{n} \; dz.$$
This is zero when $k=n$ so we may include this term in the sum which
yields
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n}} (1+z)^{n}
\sum_{k=0}^n {n\choose k} z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n}} (1+z)^{n}
(1+z)^n
\; dz
\\ = {2n\choose n-1}.$$
This yields for piece $B$
$$2n-2{2n\choose n+1}.$$
Collecting the two pieces we have
$$n-2 + {2n+1\choose n+1} - 2{2n\choose n+1}
\\ = n-2 + \frac{2n+1}{n+1} {2n\choose n}
- 2 \frac{n}{n+1} {2n\choose n}
\\ = n-2 + \frac{1}{n+1} {2n\choose n}.$$
|
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|
Did I solve these limits right? I have some limits that I tried solving:
$\lim_{x\rightarrow2}(4x^5+2x^3-1)=4(2)^5+2(2)^3-1=143$
$\lim_{x\rightarrow1}(\frac{5x}{x-1})=\frac{5}{0}=does~not~exist$
$\lim_{x\rightarrow(-3)}(\frac{x^2-5x-24}{4x+12})=\frac{(x+3)(x-8)}{4(x+3)}=\frac{x-8}{4}=-\frac{11}{4}$
$\lim_{x\rightarrow4}(\frac{x^2-16}{x-4})=\frac{(x-4)(x+4)}{x-4}=x+4=8$
$\lim_{x\rightarrow0}(\frac{\sqrt{x+1}-1}{x})=\frac{\sqrt{x+1}-1}{x}*\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}=\frac{1}{\sqrt{x+1}+1}=\frac{1}{2}$
$\lim_{x\rightarrow27}(\frac{\sqrt[3]{x}-3}{x-27})=\lim_{u\rightarrow3}\frac{u-3}{u^3-27}=\frac{u-3}{(u-3)(u^2+3u+9)}=\frac{1}{u^2+3u+9)}=\frac{1}{27}$
Did I do everything right? Was there any mistakes in my approach?
|
Your approach to these is spot on, in all cases. You've done very well! Nice work!
Note, that when canceling common factors in the numerator and denominator, for example, say, the factor $(x-a)$ in a limit for which $x \to a$, you should specify $x \neq a$.
|
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|
High degree polynom divisibility The problem is:
Show that $x^{44}+x^{33}+x^{22}+x^{11}+1$ is divisible by $x^4+x^3+x^2+x+1$
I am not sure how to approach this problem so any help would be very appreciated. Thank you in advance.
|
HINT:
Both are in Geometric Series
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$ and $$1+x^{11}+x^{22}+x^{33}+x^{44}=\frac{1-x^{55}}{1-x^{11}}$$
So, $$\frac{1+x^{11}+x^{22}+x^{33}+x^{44}}{1+x+x^2+x^3+x^4}=\frac{(1-x^{55})(1-x)}{(1-x^{11})(1-x^5)}$$
Now using Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$,
$1-x^{55}$ is divisible by $1-x^5,1-x^{11}$ and
$(1-x^{11},1-x^5)=1-x$
|
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|
Given, $2\le aGiven, $2\le a<b<c<d\le 16$ and $(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2$,$(d+1)^2+(a+1)^2=(b+1)^2+(c+1)^2$. Find all $a,b,c,d$
My work:
If $(a,b,c,d)\in \mathbb{N}$, for that I could show this
$$(d-1)^2=(a-1)^2+(b-1)^2+(c-1)^2\implies d^2-2d=(a^2+b^2+c^2)-2(a+b+c)+2$$
Now, From second equation we get,
$$d^2+2d+a^2+2b+2=b^2+c^2+2b+2c+2$$
Now, manipulating this equation to look like the above equation and substitution gives,
$$4d+2(a^2+1)=4(b+c)\implies a^2\equiv -1 \mod2$$
From this we can get that $a$ is odd and $b+c>d$.
Now what am I supposed to do? I could not approach this problem for $(a,b,c,d)\in \mathbb{R}\setminus \mathbb{N}$. Please help.
|
Start as you did (i.e. by expanding, simplifying, and “equating” the two given equalities), then solve for $a$ to obtain
$$a = \sqrt{2(b+c-d)-1}.$$
Note that $b+c \le 29$, so $b+c-d \le 24$, so $2(b+c-d)-1 \le 47$. Since $2(b+c-d)-1$ must be an odd square, and $a \ge 2$, we conclude $2(b+c-d)-1 \in \{9,25\}$. Hence $a=3$ or $a=5$. The latter is easily eliminated, which quickly leads to $(a,b,c,d)=(3,7,10,12)$ or $(a,b,c,d)=(3,6,15,16)$.
|
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|
Factor 9 terms with 3 variables into 4 expression I just got the determinant from a 4x4 matrix and the simplified version is below.
$$
det(M) = \begin{vmatrix}
2k-mw^2 & -k & 0 & 0 \\
-k & 2k-mw^2 & -k & 0 \\
0 & -k & 2k-mw^2 & -k \\
0 & 0 & -k & 2k-mw^2\\
\end{vmatrix}
$$
the polynomial I got after 1 hr is:
$$
5k^4 - 6k^2mw^2 + 6k^2m^2w^4 - 8k^3mw^2 + 2km^2w^4 - 4km^3w^6 - 3m^3w^6 + m^4w^8 - k^2m^2w^4
$$
I want to factor this out, I tried so many ways but just gave up.
Now I know, since it was defined in Physics that the system I am studying would produce 4 normal frequencies.
Usually, the form appear as:
$$
(k-mw^2)(3k-mw^2) \\
$$
(the example above is for 2x2 matrix, hence produces 2 normal frequencies)
which is very easy to solve for $ w $.
Please help, I just want to equate the whole equation to zero and get w.
|
First, let's make things a bit nicer to look at. Let's substitute $x = 2k - mw^2$:
$$
det(M) = \begin{vmatrix}
x & -k & 0 & 0 \\
-k & x & -k & 0 \\
0 & -k & x & -k \\
0 & 0 & -k & x\\
\end{vmatrix}
$$
I calculate the determinant to be $x^4 - 3k^2x^2 + k^4:$
$$
det(M) = x\begin{vmatrix}
x & -k & 0 \\
-k & x & -k \\
0 & -k & x\\
\end{vmatrix} + k\begin{vmatrix}
-k & -k & 0 \\
0 & x & -k \\
0 & -k & x\\
\end{vmatrix} \\
= x^2\begin{vmatrix}
x & -k \\
-k & x\\
\end{vmatrix} + xk\begin{vmatrix}
-k & -k \\
0 & x\\
\end{vmatrix} -k^2\begin{vmatrix}
x & -k \\
-k &x\\
\end{vmatrix} \\ = x^4 - x^2k^2 - x^2k^2 - x^2k^2+ k^4 \\
= x^4 - 3k^2x^2 + k^4.
$$
Using the quadratic formula on $x^2$ gives
$$x^2 = \frac{(3 \pm \sqrt{5})k^2}{2} = (2k - mw^2)^2.$$
Now take square roots of both sides, isolate $w$, and square root again, tossing the imaginary roots as you go along.
|
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|
Find the general formula for the partial sum? $\sum 2^{(n-1)} \cdot n$ from $n=1$ to $k$ I'm having a bit of trouble finding the general formula for the following partial sum
$$ \sum_{n=1}^k 2^{n-1} \cdot n $$
|
First note that $2^{n-1}+2^{n-1}=2^n$. By using this formula for different values of $k$ we can derive a formula for the sum without the $n$ factor:
\begin{eqnarray*}
2^{k} &=& 2^{k-1} + 2^{k-1}\\
&=& 2^{k-1} + 2^{k-2} + 2^{k-2}\\
&=& 2^{k-1} + 2^{k-2}+2^{k-3}+2^{k-3} \\
&=& \cdots\\
&=& (2^{k-1}+2^{k-2}+\cdots+1) + 1\\
&=& 1+\sum_{n=1}^k 2^{n-1}
\end{eqnarray*}
and therefore $\sum_{n=1}^k 2^{n-1} = 2^{n}-1$.
Now we can introduce the factor of $n$.
\begin{eqnarray*}
\sum_{n=1}^k n2^{n-1} &=& 1 + 2*2^1 + 3*2^2 + 4*2^3 + \cdots +n 2^{n-1}\\
&=& 1 \\
&& +2^1 + 2^1 \\
&& +2^2+2^2+2^2\\
&& + \vdots \\
&& +2^{k-1}+2^{k-1}+2^{k-1}+\cdots + 2^{k-1}
\end{eqnarray*}
Now sum each of the columns of the last line. The first column is
$1+2^1+2^2+\cdots+2^{k-1}=2^k-1$, the second column is $2^1+2^2+\cdots+2^{k-1}=2^k-2$
and the $i$'th column is $2^i+2^{i+1}+\cdots+2^{k-1}=2^k-2^i$. Therefore the whole sum can also be written
\begin{eqnarray*}
\sum_{n=1}^k n2^{n-1} &=& (2^k-1)+(2^k-2^1)+(2^k-2^2)+\cdots + (2^k-2^{k-1})\\
&=& k 2^k - \sum_{n=1}^{k}2^{n-1}\\
&=& k2^k-(2^k-1)\\
&=& 1+(k-1)2^k\\
\end{eqnarray*}
This is the most elementary derivation I can think of. There are faster and simpler methods as well. Hopefully this was helpful to you.
|
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|
Transform $\int \frac{x^2}{x - 2}$ to $ \int x +\frac{4}{x-2} + 2 $ I was trying to solve the following integral and I failed so I went to wolframalpha to see the step by step solution, but the following transformation is confusing me.
$\int \frac{x^2}{x - 2} = \int x +\frac{4}{x-2} + 2 $
I am unable to do the transformation by myself. Could you please explain the steps to get from $\int \frac{x^2}{x - 2} $ to $\int x +\frac{4}{x-2} + 2 $ ? The step is labeled as a "long division", but I have no idea what that means.
|
First you want to reduce the rational function by polynomial division to get
$$
\frac{x^2}{x - 2} = \frac{x^2 - 2x + 2x}{x - 2} = x + \frac{2x}{x - 2}
$$
Then you want to perform partial fractions to reduce the resulting rational function to get
$$
\frac{2x}{x - 2} = \frac{ 2x - 4 + 4 }{ x - 2 } = 2 + \frac{4}{x - 2}
$$
Now I did some tricks to avoid longer calculations, but these steps are the steps you want to follow.
|
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|
Given two determinants, find the det of this matrix? If det $ =\begin{bmatrix} a & 1 & d\\b & 1 & e\\c & 1 & f\\\end{bmatrix} = -2$
and If det $ =\begin{bmatrix} a & 1 & d\\b & 2 & e\\c & 3 & f\\\end{bmatrix} = -1$
what is the determinant of $\begin{bmatrix} a & -5 & d\\b & -7 & e\\c & -9 & f\\\end{bmatrix}$
I'm pretty lost here, I've tried multiplying matrices and subtracting them from each other but not getting the correct answer.
|
(This is essentially a restatement of the earlier answers, but expanding on how the multilinearity can be viewed.)
If we compute these determinants using an expansion along the second column, we find
$$\begin{vmatrix} a & 1 & d\\b & 1 & e\\c & 1 & f\\\end{vmatrix} = -1 \begin{vmatrix} b & e\\c & f\\\end{vmatrix}+1\begin{vmatrix} a & d\\c & f\\\end{vmatrix}-1\begin{vmatrix} a & d\\b & e\\\end{vmatrix} =-2 \tag{eq 1}$$
and
$$\begin{vmatrix} a & 1 & d\\b & 2 & e\\c & 3 & f\\\end{vmatrix} = -1 \begin{vmatrix} b & e\\c & f\\\end{vmatrix}+2\begin{vmatrix} a & d\\c & f\\\end{vmatrix}-3\begin{vmatrix} a & d\\b & e\\\end{vmatrix} =-1 \tag{eq 2}.$$
The determinant we want is
$$\begin{vmatrix} a & -5 & d\\b & -7 & e\\c & -9 & f\\\end{vmatrix} = -(-5) \begin{vmatrix} b & e\\c & f\\\end{vmatrix}+(-7)\begin{vmatrix} a & d\\c & f\\\end{vmatrix}-(-9)\begin{vmatrix} a & d\\b & e\\\end{vmatrix}$$
which is also given by $-3 \times (\text{eq } 1)-2 \times (\text{eq } 2).$ Hence the determinant is $8$.
|
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|
Nice Arctan Identity Prove that $ \text{arctan}\left(\frac{a+d}{c}\right)=2\text{arctan}\left(\frac{a}{c}\right) $ if $a, d, c$ are positive reals satisfying $$ a^4+a^2c^2+a^2d^2+2a^3d = c^2d^2 $$
(credit: bobthesmartypants)
|
Try to arrive at the equation that needs to be proved from the given equation.
$$ a^4+a^2c^2+a^2d^2+2a^3d = c^2d^2 $$
$$a^2(a^2+d^2+2ad)=c^2d^2-a^2c^2$$
$$(a^2)(a+d)^2+c^2(a^2-d^2)=0$$
$$(a+d)(a^2(a+d)+c^2(a-d))=0$$
$(a+d)$ cannot be equal to zero as $a,d$ are positive real numbers.So:
$$a^2(a+d)+c^2(a-d)=0$$
$$\frac{-a^2}{c^2}=\frac{a-d}{a+d}$$
Adding $1$ to both sides:
$$1-\frac{a^2}{c^2}=\frac{2a}{a+d}$$
Dividing both sides by $c$ and with a little rearrangement we get :
$$\frac{a+d}{c}=\frac{\frac{2a}{c}}{1-\frac{a^2}{c^2}}$$
Now apply $ \text{arctan}$ on both the sides and use the formula $2\text{arctan}\left(x\right)=\text{arctan}\left(\frac{2x}{1-x^2}\right)$:
$$ \text{arctan}\left(\frac{a+d}{c}\right)=2\text{arctan}\left(\frac{a}{c}\right) $$
|
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|
$a+b+c=3, a,b,c>0$, Prove that $a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$ $a+b+c=3, a,b,c>0$, Prove that $$a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$$
My work:
From the given inequality, we can have,
$a^2b^2c^2\ge 27-18(a+b+c)+12(ab+bc+ca)-8abc$
We can also have,$abc\le \bigg(\dfrac{a+b+c}{3}\bigg)^3=1$
So, $0\ge -36+12(ab+bc+ca)$
Again, we can have, $ab+bc+ca\le b(3-b)+\dfrac{1}{b}$
Now, I have to show that, $b(3-b)+\dfrac{1}{b}\le 3$
How can I prove this now? Please help.
|
We consider four cases for $(3- 2a),(3-2b),(3-2c)$ : The case when all of them are non-negative, the case when exactly one of them is negative, the case when exactly two of them are negative, and the case when all of them are negative.
For the first case when all of them are non-negative, notice that by the AM-GM inequality,
$$\begin{align}(3 - 2a) + (3 - 2b) &\ge 2\sqrt{(3-2a)(3-2b)}\\
6 - 2a - 2b &\ge 2\sqrt{(3-2a)(3 - 2b)}\\
2(a + b + c) - 2a - 2b &\ge 2\sqrt{(3-2a)(3 - 2b)}\\
c &\ge \sqrt{(3-2a)(3 - 2b)}\end{align}$$
Similarly for $(3 - 2a)(3-2c)$ and $(3 - 2b)(3 - 2c)$, we get $b \ge \sqrt{(3-2a)(3-2c)}$ and $a \ge \sqrt{(3-2b)(3-2c)}$ respectively.
Multiplying all $3$ together, we get :
$$\begin{align}abc &\ge\sqrt{(3-2a)^2(3-2b)^2(3-2c)^2}\\
\implies a^2b^2c^2 &\ge (3-2a)(3-2b)(3-2c)\end{align}$$
Hence the inequality holds for all $(3 - 2a), (3 - 2b), (3 - 2c) \ge 0$, with equality at $a = b = c = 1$.
For the second case when exactly one of them is negative, we assume without a loss of generality that $3 - 2a < 0$. Then,
$$\begin{align}&3 - 2a < 0\\
\implies &a > \frac{3}{2}\\
\implies &3 - b - c > \frac{3}{2}\\
\implies &b + c < \frac{3}{2}\\
\implies &(3 - 2b)(3 - 2c) > 0\\
\implies &(3 - 2a)(3 - 2b)(3 - 2c) < 0\end{align}$$
On the other hand, $a^2b^2c^2 \ge 0$, hence the inequality holds true for the case when exactly one of them is negative.
For the case when exactly two of them is negative, we see that (without a loss of generality) $(3 - 2a), (3 - 2b) < 0 \implies a, b > \frac{3}{2}\implies a + b > 3 \implies a + b + c > 3$. But this is impossible since $a + b + c = 0$, leading to a contradiction.
Similarly, we proof again by contradiction that not all of the three can be negative. If, on the contrary all the three are negative, then $a, b, c > \frac{3}{2} \implies a + b + c > \frac{9}{2}$. But this is a contradiction since $a + b + c = 3$.
Therefore, after considering all the cases, we conclude that $$a^2b^2c^2 \ge (3-2a)(3-2b)(3-2c)$$
My initial solution assumed that all of $(3-2a), (3-2b), (3-2c) \ge 0$. But clearly this is not necessarily true, so I took sometime to write out this new solution, though I'm pretty sure there are other shorter methods.
|
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|
Don't know how to continue modulo reduction Ok , so the question is to find the last three digits of $2013^{2012}$. After some reduction using Euler's Theorem I got $13^{12}$(mod 1000). I tried dividing it in 8 and 125 and later use the CRT , but this didn't help (or maybe I did it wrong) .
What is the shortest way to get the answer ? (ofc. without using very big numbers)
|
HINT:
$$13^{12}=(10+3)^{12}=3^{12}+12\cdot3^{11}10+\binom{12}2\cdot3^{10}10^2\pmod{1000}$$
Now, $\displaystyle3^{10}=9^5\equiv(-1)^5\pmod{10}\equiv-1$
$\displaystyle\implies\binom{12}2\cdot3^{10}10^2\equiv\binom{12}2\cdot(-1)10^2\pmod{100}$
and $\displaystyle3^{11}=3\cdot9^5=3(10-1)^5\equiv3(-1+\binom51\cdot10^1)\pmod{100}$
Finally, $3^{12}=(3^2)^6=9^6=(10-1)^6=1-\binom6110+\binom6210^2\pmod{1000}$
|
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|
determining order of $x+1$ given the $x$ has order three I was trying to expand $(x+1)^n$, then plug $x^3$ in to the expansion of the $(x+1)^n$, keep trying it until I get the order, are there any other ways?
So if $x^3\equiv 1\pmod y$, how would I determine order of $x+1$?
Can I brute force $x$ and $y$ for the condition then just find $x+1\equiv\pmod y$?
$y$ is prime
|
Clearly $x\neq1$ and hence
$$
x^3 \equiv 1 \Rightarrow x^2+x+1 \equiv 0 \tag1$$
Now consider
$(x+1)^6$ and show that
$$
(x+1)^6 - 1 \equiv x\,\left(x+2\right)\,\left(x^2+x+1\right)\,\left(x^2+3\,x+3\right)\tag2$$
and conclude that the order is $6$ by showing no smaller power works.
Added in response to comments
Using (1) in (2) we have
$$(x+1)^6 \equiv 1$$
Hence order of $x+1$ must divide $6$.
*
*If $x+1$ has order 1 then $x=0$ which is not possible. So the order cannot be 1.
*if $x+1$ has order 2, then from (1) $x+1 \equiv -x^2$, and hence $(x+1)^2 \equiv x^4$ and $x^4 \not \equiv 1$. So the order cannot be 2.
*if $x+1$ has order 3, then from (1) $x+1 \equiv -x^2$, and hence $(x+1)^3 \equiv -x^6 = -1$. This requires $1 \equiv -1$, which means there are only two elements in the multiplicative group but we know that $1$, $x$ and $x^2$ are distinct since order of $x$ is 3. So the order cannot be 3.
Hence the order has to be $6$
|
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Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
|
Using Cauchy-Schwarz Inequality twice:
$a^4 + b^4 +c^4 \geq a^2b^2 +b^2c^2 +c^2a^2 \geq ab^2c +ba^2c +ac^2b = abc(a+b+c)$
|
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|
Checking irreducibility I have the polynomial $f(X)=X^{2n}-2X^{n}+1-p$ where $p$ is a prime number and $n\in\mathbb{N}$. I want to check whether it is irreducible or not over $\mathbb{Q}[X]$.
If $2^{2}\nmid1-p$ then $f(X)$ is irreducible by Eisenstein's Criterion. However, I can't make any progress when I consider the polynomial $f(X)=X^{2n}-2X^{n}+4r, r\in\mathbb{Z}$.
Any hints?
|
You are correct, it is always irreducible.
Your polynomial factors as
$AB$ where $A=X^n-(1+\sqrt{p})$ and $B=X^n-(1-\sqrt{p})$. It will suffice
to show that $A$ (and therefore $B$ also) is irreducible over $K={\mathbb Q}[\sqrt{p}]$.
Thanks to Karpilovsky’s theorem (many thanks to Bill Dubeque for
quoting it here), it will suffice to show the following points :
(1) $c=1+\sqrt{p}$ is not a $m$-th power in $K$, for any $m\geq 2$.
(2) $c$ is not of the form $-4z^4$ with $z\in K$.
Proof of (1) : suppose $1+\sqrt{p}=(x+y\sqrt{p})^m$ with $x,y\in{\mathbb Q}$.
Then $1-p=d^m$ where $d$ is the rational number $d=x^2-py^2$. So $d$ is a rational root of the monic polynomial $X^m-(1-p)$, so $d$ is an integer. As $1-p<0$, $d$ must be a negative integer and
$m$ is odd. Then $p=1-d^m$ is divisible by $1-d>0$, so
$1-d$ can only be $1$ (clearly impossible) or $p$. So $1-d=p,d=1-p$ and hence
$(1-p)^{m-1}=1$, which occurs only when $p=2$.
We then have $1+\sqrt{2}=(x+y\sqrt{2})^m$, $m$ odd and $x^2-2y^2=-1$. Each real
number has a unique $m$-th real root, so $x+y\sqrt{2}=(1+\sqrt{2})^{\frac{1}{m}}$
and hence $x-y\sqrt{2}=\frac{x^2-2y^2}{x+y\sqrt{2}}=(1-\sqrt{2})^{\frac{1}{m}}$. Adding
those two last equalities, one obtains
$$
x=\frac{(1+\sqrt{2})^{\frac{1}{m}}+(1-\sqrt{2})^{\frac{1}{m}}}{2}
$$
Then, $r=2x$ is both rational and a sum of two algebraic integers, so it
must be an integer. Now,
$$
r=(\sqrt{2}-1)^{\frac{1}{m}}
\Bigg(\bigg(\frac{\sqrt{2}+1}{\sqrt{2}-1}\bigg)^m-1\Bigg) >0
$$
On the other hand, $\big(\frac{3}{2}\big)^3 > \sqrt{2}+1$ yields
$(\sqrt{2}+1)^{\frac{1}{m}} \leq \frac{3}{2}$, and
$\sqrt{2}-1 > \big(\frac{1}{2}\big)^3$ yields $1+(\sqrt{2}-1)^{\frac{1}{m}} \geq
\frac{3}{2}$. Combining the two, we obtain $r<1$. Finally $r$ is an integer
strictly between $0$ and $1$, which is impossible.
Proof of (2) : $1+\sqrt{p}=-4(x+y\sqrt{p})^4$ with $x,y\in{\mathbb Q}$ is clearly impossible as a fourth power cannot be negative.
|
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|
What is the maximum height of a snowman (consisting from 3 snowballs) when I have $\frac{4}{3}\pi$ cubic meters of snow Let balls $1,2,3$ have radia $r_1, r_2, r_3$. Then
$V_1=\frac{4}{3}\pi r_1^{3}$
,$V_2=\frac{4}{3}\pi r_2^{3}$
and $V_3=\frac{4}{3}\pi r_3^{3}$.
I know that $V_1 + V_2 + V_3 =\frac{4}{3}\pi$.
From this I get function $$f(V_1,V_3) = 2\left(\left(\frac{3V_1}{4\pi}\right)^{\frac{1}{3}} + \left(\frac{3V_3}{4\pi}\right)^{\frac{1}{3}} + \left(1 - \frac{3(V_1 + V_3)}{4\pi}\right)^\frac{1}{3}\right)$$
But I do not know how to derivate it since we have learnt only derivations of functions with $1$ variable; therefore, I cannot find maximum. Teacher gave us a tip that we should use Jensen inequality but I do not know how to apply it for this problem.
Thanks for your help.
|
Really, the problem is to maximize $h=2(r_1+r_2+r_3)$ given that $r_1^3+r_2^3+r_3^3=1$. You can find critical points by expressing
$$h(r_1,r_2) = 2 [ r_1+r_2 + (1-r_1^3-r_2^3)^{1/3})]$$
Consider
$$f(x,y) = x+y+(1-x^3-y^3)^{1/3}$$
$$f_x = 1 -\frac{x^2}{(1-x^3-y^3)^{2/3}}$$
$$f_y = 1 -\frac{y^2}{(1-x^3-y^3)^{2/3}}$$
Setting each of these to zero reveals that $x=y$, and thus, $r_1=r_2=r_3=r$, such that
$3 r^3=1 \implies r=3^{-1/3}$. The max height is thus $6 r = 2 \cdot 3^{2/3}$.
|
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|
Combinatorics on letters How many "words" of length n is it possible to create from {a,b,c,d} such that a and b are never next to each other?
|
Let the total number of such words on $n$ letters be $s_n$. Let the number of words which end with $a$ be $a_n$ and the number of words which end with $b$ be $b_n$.
Now consider a word on $n-1$ letters. We may add $c$ or $d$ to the end of the word to obtain a new word on $n$ letters. Conversely, all words on $n$ letters which end with $c$ or $d$ can be obtained in this way. Now to any word not ending with $a$ we may add $b$ to the end. To any word not ending in $b$ we may add $a$ to the end. All words ending with $a$ or $b$ are obtained in this fashion.
Therefore we have the recurrence for $a_n$ and $b_n$ as
$$s_n = 2s_{n-1} + (s_{n-1}-a_{n-1}) + (s_{n-1}-b_{n-1})=4s_{n-1} - a_{n-1} - b_{n-1}$$
Similarly, the number of words which end with $a$ or $b$ are given by
$$a_n = s_{n-1}-b_{n-1}$$
$$b_n = s_{n-1}-a_{n-1}$$
By symmetry, we must have $a_n = b_n$, so we will work solely with $a_n$ from now on. The above gives us two coupled recurrences
$$s_n = 4s_{n-1} - 2a_{n-1}$$
$$a_n = s_{n-1} - a_{n-1}$$
We can combine the two as a single recurrence for $a_n$
$$a_{n} = 3a_{n-1} + 2a_{n-2}$$
which has solution
$$a_n = \frac{(3+\sqrt{17})^n - (3-\sqrt{17})^n}{2^n\sqrt{17}}$$
Then we have
$$\begin{align}s_n &= a_{n+1} + a_n
\\&= \frac{(3+\sqrt{17})^{n+1} - (3-\sqrt{17})^{n+1}}{2^{n+1}\sqrt{17}} + \frac{(3+\sqrt{17})^n - (3-\sqrt{17})^n}{2^n\sqrt{17}}
\\&=\frac{(5+\sqrt{17})(3+\sqrt{17})^n - (5-\sqrt{17})(3-\sqrt{17})^n}{2^{n+1}\sqrt{17}}\end{align}$$
This sequence begins as
$$s_n = 4,\ 14,\ 50,\ 178,\ 636,\ 2258,\ 8042,\ 28642,\ 102010,\ \cdots$$
and is sequence A055099 in OEIS (with a shifted index).
|
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|
Finding $C$ if $ 3\sin A + 4\cos B = 6 $ and $ 3\cos A + 4\sin B = 1 $ in a triangle $ABC$
In a triangle $ABC$, it's given that the following two equations are satisfied:
$$ 3\sin A + 4\cos B = 6 $$
$$ 3\cos A + 4\sin B = 1 $$
Source: ISI B-math UGA 2017
We have to find the angle $ C$. Now, it's easy to see that $ \sin C = 0.5 $ (by squaring and adding both the equations). Now we have to decide whether $ C = \pi/6 $ or $5\pi/6$. The solution given in the book goes something like this:
Assume $ C = 5\pi/6$, then even if $B = 0$ and $A = \pi/6$, then the quantity $3\sin A + 4\cos B = 5.5 < 6$ and hence $ C \not= 5\pi/6 $.
But then if we do the same thing by setting $C = \pi/6$, again we face the same problem. So how to find the value of $C$?
EDIT:
Squaring and adding the two equations, we get:
$$ 9(\sin^2 A + \cos^2 A) + 16 (\sin^2 B + \cos^2 B) + 24\sin(A+B) = 37 $$
$$ 24\sin(A+B) = 24\sin C = 12 \implies \sin C = 0.5 $$
|
$ 3\sin A + 4\cos B = 6 $ ......(i)
$ 3\cos A + 4\sin B = 1 $.... (ii)
Squaring (i) we get :
$9 \sin^2 A +16\cos^2 B +24 \sin A \cos B = 36 $ ....(iii)
Now Squaring (ii) we get :
$9 \cos^2 A +16\sin^2 B +24 \sin B \cos A =1 $...(iv)
Now adding (iii) and (iv) we get :
$9(\sin^2 A +\cos^2 A) +16(\cos^2 B + \sin^2 B) +24 ( \sin A \cos B +\cos A \sin B ) = 37 $
$\Rightarrow 9 +16 + 24 \sin(A+B) = 37 $
$\Rightarrow 24 \sin C = 37 -25 = 12 $
$\Rightarrow \sin C =\frac{1}{2} = \sin \frac{\pi}{6}$
Now general solution of the equation $\sin\theta =\sin\alpha$ is given by :
$$\theta =n\pi +(-1)^n \alpha, n \in Z$$
$$\therefore C = n\pi +(-1)^n \frac{\pi}{6}$$ is the solution.
|
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Interesting and unexpected applications of $\pi$ $\text{What are some interesting cases of $\pi$ appearing in situations that do not seem geometric?}$
Ever since I saw the identity $$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$
and the generalization of $\zeta (2k)$, my perception of $\pi$ has changed. I used to think of it as rather obscure and purely geometric (applying to circles and such), but it seems that is not the case since it pops up in things like this which have no known geometric connection as far as I know. What are some other cases of $\pi$ popping up in unexpected places, and is there an underlying geometric explanation for its appearance?
In other words, what are some examples of $\pi$ popping up in places we wouldn't expect?
|
$\pi$ as a matrix eigenvalue
For a given positive integer $m$, let us form an $(m-1)\times(m-1)$ dimensional matrix $M$. The terms on the diagonal of $M$ will be $-2m^2$, the terms on the off diagonals of $M$ will be $m^2$ and all other terms will be zero. Thus,
$$M=m^2\left(
\begin{array}{cccccc}
-2 & 1 & 0 & \cdots & \cdots & 0 \\
1 & -2 & 1 & \cdots & \cdots & 0 \\
0 & 1 & -2 & 1 & \cdots & 0 \\
\vdots & \cdots & \vdots & \vdots & \cdots & \vdots \\
0 & \cdots & \cdots & 1 & -2 & 1 \\
0 & \cdots & \cdots & 0 & 1 & -2 \\
\end{array}
\right).
$$
For $m=10$, the matrix is
$$M = 100
\left(
\begin{array}{ccccccccc}
-2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & -2 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & -2 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & -2 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & -2 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & -2 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 \\
\end{array}
\right).
$$
If we numerically estimate the eigenvalues of this $M$ (for $m=10$), we get
$$
-390.211, -361.803, -317.557, -261.803, -200., -138.197, -82.4429,
-38.1966, -9.7887.
$$
Of note, they are all negative. If we take the square root of the absolute value of the largest eigenvalue (the last in the above list) we get $3.12869$.
If we perform this exact same process for $m=1000$, we find that the square root of the absolute value of largest eigenvalue is $3.14159$.
It can be proved that the limit as $m\rightarrow\infty$ of this process yields exactly $\pi$.
|
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Howto prove that $\sum\limits_{cyc}\cos\frac{A}{2}\cos\frac{B}{2}\le\frac{1+2\sqrt{2}}{2}+\frac{7-4\sqrt{2}}{R}r$ let $ABC$ is a triangle with inradius $r$ and circumradius $R$. Show that
$$\cos\frac{A}{2}\cos\frac{B}{2}+\cos\frac{C}{2}\cos\frac{B}{2}+\cos\frac{A}{2}\cos\frac{C}{2}\le\frac{1+2\sqrt{2}}{2}+\frac{7-4\sqrt{2}}{R}r$$
This problem is created by me. I believe that this is true, but I can't prove it.
This problem is motivated by a 1988 IMO Longlists problem
$$ \sin \left( \frac{A}{2} \right) \cdot \sin \left( \frac{B}{2} \right) + \sin \left( \frac{B}{2} \right) \cdot \sin \left( \frac{C}{2} \right) + \sin \left( \frac{C}{2} \right) \cdot \sin \left( \frac{A}{2} \right) \leq \frac{5}{8} + \frac{r}{4 \cdot R}. $$
I post the solution:
\begin{align*} & 4\sum \sin \frac A2 \sin \frac B2 \leq \frac 32 + 1 + \frac {r}{R} \\
\iff & 4\sum \sin \frac A2 \sin \frac B2 \leq \frac 32+ \sum \cos A \\
\iff & 4\sum \sin \frac A2 \sin \frac B2 \leq \frac 32 + \sum (1 - 2\sin^2 \frac A2) \\
\iff & 4\sum \sin \frac A2 \sin \frac B2 + 2\sum \sin^2 \frac A2 \leq \frac 92 \\
\iff & 2\left(\sum\sin\frac {A}{2}\right)^{2} \leq \frac 92 \\
\iff & \left(\sum\sin\frac {A}{2}\right)^{2}\leq \frac 94 \end{align*}
The last inequality is a well know inequality. We are done!
But My inequality is stronger this IMOLonglists problem, so I can't prove it. Thank you for you help!
|
your inequality isn't true, try $a=4,b=3,c=5$
|
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|
Difficulties evaluating the endpoints of the radius of convergence for a particular power series. I am having difficulties evaluating the endpoints of the radius of convergence for the following power series.
$$\sum_{k=0}^{\infty}\frac{(k!)^2 x^k}{(2k)!}$$
Using the ration test we get |x|<4. However, evaluating $x=4$ and $x=-4$ is proving immensely difficult. The solutions neglect this part, and my teacher is at loss. Any hints?
In other words, I am having difficulty evaluating
$$\sum_{k=0}^{\infty}\frac{(k!)^2 4^k}{(2k)!}.$$
|
$\begin{align*}
a_k = \frac{(k!)^2 4^k}{(2k)!} &= \frac{(k!)^2 4^k}{(2k)(2k-1)(2k-2) \cdots 3\cdot 2 \cdot 1} \\
&=\frac{(k!)^2 4^k}{((2k)(2k-2) \cdots 4 \cdot 2)((2k-1)(2k-3) \cdots 3 \cdot 1)} \\
&=\frac{(k!)^2 4^k}{(2^k k!)((2k-1)(2k-3) \cdots 3 \cdot 1)} \\
&=\frac{k! \cdot 2^k}{(2k-1)(2k-3) \cdots 3 \cdot 1} \\
&=\frac{k! \cdot 2^k}{2\cdot(k-\frac{1}{2})\cdot 2\cdot(k-\frac{3}{2}) \cdots 2\cdot(\frac{3}{2}) \cdot 2\cdot \frac{1}{2}} \\
&=\frac{k! \cdot 2^k}{2^k(k-\frac{1}{2})(k-\frac{3}{2}) \cdots \cdot(\frac{3}{2}) (\frac{1}{2})} \\
&= \left( \frac{k}{k-\frac{1}{2}}\right)\left( \frac{k-1}{k-\frac{3}{2}}\right) \cdots \left(\frac{2}{\frac{3}{2}}\right)\left(\frac{1}{\frac{1}{2}}\right)\\
&\geq 1 \cdot 1 \cdots 1 \cdot 1 \\
&=1
\end{align*}$
When $x = 4$, $a_k\geq 1$, so $\displaystyle \lim_{k \rightarrow \infty} a_k \neq 0$, and the series diverges. Similarly true when $x=-4$.
|
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Show that if $m,n$ are positive integers, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$. Show that if $m,n$ are positive integers and $m$ is odd, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$.
(Hint: Let $s=1^m+2^m+\cdots+(n-2)^m+(n-1)^m$. Obviously $s=(n-1)^m+(n-2)^m+\cdots+2^m+1^m$.
Consider these relations as equivalent ${}\bmod n$ and add them.)
$$1^m+2^m+\cdots+(n-2)^m+(n-1)^m=((n-1)^m+(n-2)^m+\cdots+2^m+1^m)\bmod n$$
$$(n-1)^m+(n-2)^m+\cdots+2^m+1^m=(1^m+2^m+\cdots+(n-2)^m+(n-1)^m)\bmod n$$
By adding them we get:
$$(1^m+(n-1)^m)+(2^m+(n-2)^m)+\cdots+((n-2)^m+2^m)+((n-1)^m+1^m)=((n-1)^m+1^m)+((n-2)^m+2^m)+\cdots+(2^m+(n-2)^m)+(1^m+(n-1)^m))\bmod n$$
That means that:
$$n|[(1^m+(n-1)^m)+(2^m+(n-2)^m)+\cdots+((n-2)^m+2^m)+((n-1)^m+1^m)]-[((n-1)^m+1^m)+((n-2)^m+2^m)+\cdots+(2^m+(n-2)^m)+(1^m+(n-1)^m))]$$
Or not??
And that is equal to $$n\mid 0$$
Is this correct? How can I continue??
|
Still not true. If $m = 1, n = 2,$ then $1$ isn't divisible by $2.$ You need $n$ to be odd as well.
|
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|
Finding the Absolute Maximum and Minimum of a 3D Function Find the absolute maximum and minimum values of the function:
$$f(x,y)=2x^3+2xy^2-x-y^2$$
on the unit disk $D=\{(x,y):x^2+y^2\leq 1\}$.
|
We first take the partial derivatives with respect to each variable and set them to zero:
$$\frac{\partial f}{\partial x}=6x^2 + 2y^2 - 1=0$$
$$\frac{\partial f}{\partial y}=4xy - 2y=0$$
From the second equation, we have either: $y=0$ or $x=\frac{1}{2}$. Now we substitute each value into the first equation to get the corresponding variable: for $y=0$, we get $x=\pm \frac{1}{\sqrt{6}}$, while for $x=\frac{1}{2}$ we have no real solutions, so we discard this.
Note that both $(0,\frac{1}{\sqrt{6}})$ and $(0,-\frac{1}{\sqrt{6}})$ satisfy our constraint. We evaluate $f(0,\frac{1}{\sqrt{6}})$ and $f(0,-\frac{1}{\sqrt{6}})$ which give us $f=-0.27$ and $f=0.27$, respectively.
We now need to check the perimeter of the unit disk. On this disk, $y^2 = 1-x^2$, so we substitute this into the original function to get:
$$ f = x^2 + x - 1 $$
We set $df/dx=0$ to get $x=-\frac{1}{2}$, and we evaluate $f$ at this point to get $-\frac{5}{4}$.
Finally we have to check the extreme values of $x$ on the unit disk: $x=-1$ and $x=1$, which give us $f=-1$ and $f=1$, respectively.
You now have your global minimum and maximum.
|
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|
How to show that this sequence is monotonic We are given that $(s_{n})$ is a bounded sequence. My goal is to show that $(s_{n})$ is monotonic, and thus it converges. It has the following property $$s_{n+1} \geq s_{n} - \frac{1}{2^n}$$
I have tried induction, because intuitively the sequence monotonically increasing, but it fails at the basis step but works on the induction step. Any ideas?
|
$s_n$ is not necessarily monotonic given the information provided.
Example.
Let $s_n = \frac{2}{3} \left(\frac{-1}{2}\right)^n $.
Then $s_n$ is bounded, and
\begin{align*}
s_{n+1} - s_n
&= \frac{2}{3}\left(\frac{-1}{2}\right)^{n+1} - \frac{2}{3}\left(\frac{-1}{2}\right)^n \\
&= \frac{2}{3}\left(\frac{-1}{2}\right)^n \left( \frac{-1}{2} - 1\right) \\
&= -\left(\frac{-1}{2}\right)^n \\
&= (-1)^{n+1} \left(\frac{1}{2}\right)^n \ge \frac{-1}{2^n}. \\
\end{align*}
However, $s_n$ is clearly not monotonic.
|
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|
PMF and CDF Calculating I've got the following homework problem that I'm stuck on.
Two fair six-sided dice are tossed independently. Let M= the maximum of the two tosses.
a. What is the PMF of M? [Hint: first determine P(1), then p(2), and so on.]
b. Determine the CDF of M.
I am just now learning how about PMF/CDF so this is new material for me. I was told by another person in my class that the following works for part a:
$\frac{2M - 1}{36}$
$P(1) = \frac{(2(1)-1)}{36}=1/36$
$P(2) = \frac{(2(2)-1)}{36}=3/36$
$P(3) = \frac{(2(3)-1)}{36}=5/36$
$P(4) = \frac{(2(4)-1)}{36}= 7/36$
$P(5) = \frac{(2(5)-1)}{36}=9/36$
$P(6) = \frac{(2(6)-1)}{36}=11/36$
I am not sure however if this is correct, and also do not know how to come to this conclusion by myself, so if someone could help me understand this it would be extremely useful.
As for part b, I don't know how to calculate the CDF.
|
First Question: Let us find the probability that the maximum is $3$, first a slightly tedious way, and then a less tedious way.
Imagine that the dice are Christmas dice, green and red. We record the result of the tossing as an ordered pair $(a,b)$, where $a$ is the number on the green, and $b$ is the number on the red. There are $36$ possibilities, all equally likely.
What is the probability that the maximum $M$ is exactly $3$. To do this, let us list all possibilities that give a maximum of $3$. These are $(1,3)$, $(2,3)$, $(3,3)$ and $(3,1)$, $(3,2)$. There are $5$ of them, so the probability that the maximum is $3$ is $\frac{5}{36}$.
Now we find the answer in the way that your friend used. The maximum is $3$ if (i) all numbers are $\le 3$ and it is not true that all numbers are $\le 2$.
The number of ways that all numbers are $\le 3$ is $3^2$, for $a$ can take on any one of $3$ values, and for each value taken on by $a$, the number $b$ can take on any one of $3$ values.
Similarly, the number of ways in which all the numbers are $\le 2$ is $2^2$.
So the number of ways in which they are all $\le 3$ but not all $\le 2$ is $3^2-2^2$.
It follows that our required probability is $\frac{3^2-2^2}{36}$.
We can use either the first type of argument or the second type to find $\Pr(M=k)$ for all values of $k$ between $1$ and $6$.
Second question: The cdf (cumulative distribution function) of $M$ is the function $F$ such that for all reals $x$ we have $F(x)=\Pr(M\le x)$.
First let us calculate $F(x)$ for $x\lt 1$. If $x\lt 1$, then $\Pr(M\le x)=0$. So $F(x)=0$ if $x\lt 1$.
Note that $F(1)=\frac{1}{6}$, since $\Pr(M\le 1)=\frac{1}{6}$. Similarly, $F(1.23)=\frac{1}{6}$. In fact, $F(x)=\frac{1}{6}$ for all $x$ with $1\le x\lt 2$.
But $F(2)$, the probability that $M$ is $\le 2$, is $\frac{2}{6}$. And it continues to be $\frac{2}{6}$ for all $x$ in the interval $2\le x\lt 3$.
At $3$, $F(x)$ jumps to $\frac{3}{6}$. And so on.
Continue. Finally, if $x\ge 1$, then $F(x)=1$.
You can think of $F(x)$ as the "weight" (total probability) up to and including $x$.
|
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|
Integral $\int_0^1\frac{\ln x}{x-1}\ln\left(1+\frac1{\ln^2x}\right)dx$ Is it possible to evaluate this integral in a closed form?
$$
I \equiv \int_{0}^{1}{\ln\left(x\right) \over x - 1}\,
\ln\left(1 + {1 \over \ln^{2}\left(x\right)}\right)\,{\rm d}x
$$
Numerically,
$$I\approx2.18083278090426462584033339029703713513\dots$$
|
Substitute $x=e^{-y}$:
$$
I=\int^\infty_0\frac{y}{e^y-1}\log(1+y^{-2})dy.
$$
We Start from Binet's Second Formula:
\begin{align}
\log\Gamma(z) &= (z-\frac12)\log z-z+\frac12\log(2\pi)+2\int^{\infty}_0\frac{\tan^{-1}(t/z)}{e^{2\pi t}-1}dt\\
&=(z-\frac12)\log z-z+\frac12\log(2\pi)+\frac1\pi\int^{\infty}_0\frac{\tan^{-1}(\frac{y}{2\pi z})}{e^{y}-1}dy
\end{align}
Integrate from $z=0$ to $\frac1{2\pi}$:
\begin{align}
\psi^{(-2)}\left(\frac1{2\pi}\right) &= -\frac{3}{16\pi^2}+\frac{1}{4\pi}-\frac{\log(2\pi)}{8\pi^2}+\frac{\log(2\pi)}{2\pi}
+\frac1\pi\int^{\infty}_0\frac1{e^{y}-1}\left(\frac{\tan^{-1}y}{2\pi}+\frac{y\log(1+y^{-2})}{4\pi}\right)dy\\
&=\frac{I}{4\pi^2}-\frac{3}{16\pi^2}+\frac{1}{4\pi}-\frac{\log(2\pi)}{8\pi^2}+\frac{\log(2\pi)}{2\pi}+\frac1{2\pi^2}\int^{\infty}_0\left(\frac{\tan^{-1}y}{e^{y}-1}\right)dy\\
&=\frac{I}{4\pi^2}-\frac{3}{16\pi^2}+\frac{1}{4\pi}-\frac{\log(2\pi)}{8\pi^2}+\frac{\log(2\pi)}{2\pi}+\frac1{\pi}\int^{\infty}_0\left(\frac{\tan^{-1}(2\pi t)}{e^{2\pi t}-1}\right)dt.
\end{align}
Therefore $$
I=4\pi^2\psi^{(-2)}\left(\frac1{2\pi}\right)+\frac{3}{4}-\pi+\frac{\log(2\pi)}{2}-2\pi\log(2\pi)-4\pi\int^{\infty}_0\left(\frac{\tan^{-1}(2\pi t)}{e^{2\pi t}-1}\right)dt.$$
We use Binet's second formula again:
$$
\log\Gamma(\frac{1}{2\pi})=(\frac{1}{2\pi}-\frac12)\log \frac{1}{2\pi}-\frac{1}{2\pi}+\frac12\log(2\pi)+2\int^{\infty}_0\frac{\tan^{-1}(2\pi t)}{e^{2\pi t}-1}dt\\
$$
Therefore $4\pi\int^{\infty}_0\frac{\tan^{-1}(2\pi t)}{e^{2\pi t}-1}dt=2\pi\log\Gamma(\frac{1}{2\pi})+1-(2\pi-1)\log(2\pi)$, and $$
I=4\pi^2\psi^{(-2)}\left(\frac1{2\pi}\right)-\frac{1}{4}-\pi-\frac{\log(2\pi)}{2}-2\pi\log\Gamma(\frac{1}{2\pi}).$$
Also, we have $$
\psi^{(-2)}(z)=\frac{z}{2}\log(2\pi)+(z-1)\log\Gamma(z)-z(z-1)/2-\log G(z),
$$
Therefore
$$
4\pi^2\psi^{(-2)}(\frac1{2\pi})=\pi\log(2\pi)-(4\pi^2-2\pi)\log\Gamma(\frac{1}{2\pi})-\frac{1}{2}+\pi-4\pi^2\log G(\frac{1}{2\pi})
$$
And we have the final answer
$$
I=\left(\pi-\frac12\right)\log(2\pi)-\frac{3}{4}-4\pi^2\left(\log\Gamma\left(\frac{1}{2\pi}\right)+\log G\left(\frac{1}{2\pi}\right)\right).$$
|
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|
Definition of matrix exponential Is there an alternative definition of a matrix exponential so I can use it to prove that $$e^{A}=\sum_{m=0}^{\infty} \frac{1}{m!}(A)^m \;?$$
Thanks a lot in advance!
|
$$\begin{align}&\lim_{n \to \infty} \left(I + \frac{1}{n} A \right)^n = \\ &\lim_{n \to \infty} \left(I^n+(n)I^{n-1}\frac{A}{n}+\frac{(n)(n-1)}{2}I^{n-2}\frac{A^2}{n^2} + \frac{(n)(n-1)(n-2)}{6}I^{n-3}\frac{A^3}{n^3} +\;...\right) = \\ &\lim_{n \to \infty}\left(I + A+\frac{(n)(n-1)}{2n^2}A^2 + \frac{(n)(n-1)(n-2)}{6n^3}A^3 ...\right)= \\ &\lim_{n \to \infty} I + \lim_{n \to \infty}A + \lim_{n \to \infty}\frac{(n)(n-1)}{2n^2}A^2 + \lim_{n \to \infty}\frac{(n)(n-1)(n-2)}{6n^3}A^3 +\; ...\; = \\ &I + \frac{A}{1!} + \frac{A^2}{2!} + \frac{A^3}{3!} +\; ... \;= \\ &\sum_{m=0}^\infty\frac{A^m}{m!} = e^A\end{align}$$
Pretty sure that binomial expansion for matrices is valid when the two matrices commute so this is valid because $IA = AI.$
|
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|
How to evaluate this definite integral $ \int_{-\pi/3}^{\pi/3} \frac{(\pi +4x^3)\,dx}{2-\cos(|x|+ \frac{\pi}{3})} $ $$ \int_{-\pi/3}^{\pi/3} \frac{(\pi +4x^3)\,dx}{2-\cos(|x|+ \frac{\pi}{3})} $$
I have separated the integral into two parts, then expanded using $\cos(a+b)$ formula, after that I am lost. Can someone provide me hint?
|
You don't need to separate the integrand. First of all, the $x^3$ piece vanishes; this follows immediately from the fact that this is an odd integrand over an even interval. That leaves the first piece, which I can reduce to
$$2 \pi \int_{\pi/3}^{2 \pi/3} \frac{dx}{2-\cos{x}}$$
This may be evaluated using a substitution of the form $t=\tan{x/2}$; $dx=2 dt/(1+t^2)$. Then the integral is equal to
$$\begin{align}4 \pi \int_{1/\sqrt{3}}^{\sqrt{3}} dt \frac{1}{1+ 3 t^2} &= \frac{4 \pi}{3 \sqrt{3}} \left [\arctan{\frac{t}{\sqrt{3}}}\right ]_{1/\sqrt{3}}^{\sqrt{3}}\\ &= \frac{4 \pi}{3 \sqrt{3}} \left (\arctan{1}-\arctan{\frac13} \right )\\ &= \frac{4 \pi}{3 \sqrt{3}} \arctan{\frac12}\end{align}$$
|
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|
Integrating polynomial fraction of same degree I'm to integrate:
$$ \int {\frac{x^2}{x^2+x-2} }dx$$
If there was a constant in the numerator then I could solve this but I'm not sure with $x^2$. Can I use long division? I tried and it didn't seem to work out (but my long division is very rusty).
|
$$\int {x^2\over x^2+x-2}dx=\int {x^2+x-2-x+2\over x^2+x-2}dx=\int1+{-x+2\over x^2+x-2}dx$$
$$=x-{1\over2}\int{2x-4\over x^2+x-2}dx=x-{1\over2}(\int{2x+1\over x^2+x-2}-{5\over x^2+x-2}dx)$$
$$=x-{1\over2}ln( x^2+x-2)+{5\over2}\int{1\over x^2+x-2}dx $$
After that you need to deal with $$\int{1\over x^2+x-2}dx=\int{1\over (x+2)(x-1)}dx $$
$$={1\over3} \int {1\over(x-1)}-{1\over(x+2)}dx={1\over3}(ln(x-1)-ln(x+2))+c$$
Total result is $$=x-{1\over2}ln( x^2+x-2)+{5\over6}(ln({x-1\over x+2}))+c $$
we are done.
|
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|
Calculus I: Tangent line at certain point, second derivative of the curve given. How to solve? I have never seen this problem in my homework.. I need help right before final!!
Let $\frac{d^2y}{dx^2}=y''=-x^3$ at every point on a curve. The equation of the tangent line at $(1,1)$ is $y = 3-2x$. Find the equation of the curve.
If someone can help me out with this problem, will greatly appreciate. Thank you.
|
Observe that via integration, it follows that
$$
y' = - \frac{x^4}{4} + C.
$$
Hence, at $(1,1)$, $y'(1) = C - \frac{1}{4}$, and the tangent line is of the form $y - 1 = ( C - \frac{1}{4} )(x - 1)$, that is ( using the knowledge of the tangent curve at (1,1) ) that
$$
3 - 2x = y = \left(C- \frac{1}{4} \right)x + \frac{5}{4} - C.
$$
Then, $C - \frac{1}{4} = - 2 \implies C = - \frac{7}{4}$. Also, we know that $\frac{5}{4} - C = 3 \implies C = - \frac{7}{4}$. Since these conditions agree, we have the solutions that
$$
y' = - \frac{1}{4} \left(x^4 + 7 \right)
$$
so that $y$ is any curve of the form $y = - \frac{1}{20} x^5 - \frac{7}{4} x + K$, for real numbers $K$. Further, we know that $(1,1)$ is a point on the curve, and hence that $K$ must be such that $1 = - \frac{1}{20} - \frac{7}{4} + K$. It follows that $K = \frac{56}{20} = \frac{14}{5}$. So,
$$
y = - \frac{1}{20} x^5 - \frac{7}{4} x + \frac{14}{5} .
$$
|
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|
Finding one sided limits algebraically I was wondering what the best method was for proving this limit algebraically:
$$\lim_{x \to 1}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}$$
I know the answer to this question is ;
$$\lim_{x \to 1^+}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={-\infty}$$
$$\lim_{x \to 1^-}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={+\infty}$$
The only way I can solve this is using the graph of the function as the limit becomes very apparent. What is the best way to answer this algebraically?
|
Since the numerator and denominator is zero at $1$, let's factor out $(x-1)$ from both of them to get an idea how the function behaves around $1$.
The fraction equals $\dfrac{(3x^3-5x^2-5x-5)(x-1)}{(x^2-1)(x-1)}=\dfrac{3x^3-5x^2-5x-5}{x^2-1}$.
At $x=1$, the numerator equals $-12$. So for values around and very close to $1$, the numerator stays near $-12$.
The denominator however, is negative for $x<1$ and is positive for $x>1$. Thus, as $x$ approaches $1$ from the left, $x^2-1$ takes on values like $-0.1,-0.01,-0.001,\ldots$ while the numerator remains close to $-12$. Hence, the fraction is positive and becomes arbitrarily large as $x\to 1^{-}$.
Similarly, as $x\to 1^{+}$, the denominator is positive and becomes small while the numerator remains near $-12$ so that your expression here approaches $-\infty$.
|
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|
How to prove for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$? I'm new to induction so please bear with me. How can I prove using induction that, for each positive integer $n$, the sum of the first $n$ odd positive integers is $n^2$?
I think $9$ can be an example since the sum of the first $9$ positive odd numbers is $1,3,5,7,9,11,13,15,17 = 81 = 9^2$, but where do I go from here.
|
Step 1.
For n = 1 it's true that 1 = 2*1-1
Step 2.
We suppose that 1+3+5+...+(2n-1) = n2
and want to prove that: 1+3+5+...+(2(n+1)-1) = (n+1)2
We add (2(n+1) -1) to this:
1+3+5+...+(2n-1) = n2
and get:
1+3+5+...+(2n-1) + (2(n+1) -1) = n2 + (2(n+1) -1)
so:
1+3+5+...+(2(n+1) -1) = n2 + 2n+2 -1
but n2+2n+1 = (n+1)2
so we finally have:
1+3+5+...+(2(n+1)-1) = (n+1)2
|
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|
One Diophantine equation I wonder now that the following Diophantine equation:
$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$
have only this formula describing his decision?
$a=-(k^2+2(p+s)k+p^2+ps+s^2)$
$b=2k^2+4(p+s)k+3p^2+3ps+2s^2$
$c=3k^2+4(p+s)k+2p^2+ps+2s^2$
$d=2k^2+4(p+s)k+2p^2+3ps+3s^2$
$k,p,s$ - what some integers.
By your question, I mean what that formula looks like this. Of course I know about the procedure of finding a solution, but I think that the formula would be better.
|
You can consider another equation:
$2(a^2+y^2+c^2+d^2+u^2)=(a+y+c+d+u)^2$
And write the formula to solve this equation.
$a=-(k^2+2(q+t+b)k+b^2+q^2+t^2+bq+bt+qt)$
$y=k^2+2(q+t+b)k+2b^2+q^2+t^2+2bq+2bt+qt$
$c=k^2+2(q+t+b)k+b^2+2q^2+t^2+2bq+bt+2qt$
$d=k^2+2(q+t+b)k+b^2+q^2+2t^2+bq+2bt+2qt$
$u=2k^2+2(q+t+b)k+b^2+q^2+t^2$
$k,q,t,b$ - what some integers.
|
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|
Finding Power Series Representations For f(x), find a power series representation centered at the given value of a and determine the radius of convergence.
$$ f(x) = \frac {4x} {x^2-2x-3} ; a=0.$$
How would i begin this?
And what does it mean at the given value of a?
Thanks!
|
Hint: Start with
$$
\frac {4x} {x^2-2x-3} = \frac {4x}{(x-1)^2-4}
=\frac {x-3+3(x+1)}{(x-3)(x+1)}.
$$
details:
then you get to
$$
\frac {4x} {x^2-2x-3} = \frac 1{x+1} + \frac 3{x-3}
= \sum_{n=0}^\infty (-1)^nx^n + \sum_{n=0}^\infty -3^{-n}x^n.
$$
|
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|
Integral $I=\int_0^\infty \frac{\ln(1+x)\ln(1+x^{-2})}{x} dx$ Hi I am stuck on showing that
$$
\int_0^\infty \frac{\ln(1+x)\ln(1+x^{-2})}{x} dx=\pi G-\frac{3\zeta(3)}{8}
$$
where G is the Catalan constant and $\zeta(3)$ is the Riemann zeta function. Explictly they are given by
$$
G=\beta(2)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}, \ \zeta(3)=\sum_{n=1}^\infty \frac{1}{n^3}.
$$
I have tried using
$$
\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n},
$$
but didn't get very far.
|
I have a simple solution for this problem. Let
$$ I(\alpha,\beta)=\int_0^\infty\frac{\ln(1+\alpha x)\ln(1+\beta x^{-2})}{x}dx. $$
Then $I(0,0)=I(\alpha,0)=I(0,\beta)=0$ and $I(1,1)=I$. It is easy to check
\begin{eqnarray*}
\frac{\partial^2 I}{\partial \alpha\partial\beta}&=&\int_0^\infty\frac{1}{(1+\alpha x)(x^2+\beta)}dx\\
&=&\frac{1}{2}\left(\frac{\pi}{\sqrt{\beta}}+2\alpha\ln\alpha+\alpha\ln\beta\right)\frac{1}{1+\alpha^2\beta}\\
&=&\frac{1}{2}\left(\frac{\pi}{\sqrt{\beta}}+2\alpha\ln\alpha+\alpha\ln\beta\right)\sum_{n=0}^\infty(-1)^n\alpha^{2n}\beta^n\\
&=&\frac{1}{2}\left(\pi\sum_{n=0}^\infty(-1)^n\alpha^{2n}\beta^{n-\frac{1}{2}}+2\sum_{n=0}^\infty(-1)^n(\alpha^{2n+1}\ln\alpha)\beta^n+\sum_{n=0}^\infty(-1)^n\alpha^{2n+1}\beta^n\ln\beta\right).
\end{eqnarray*}
Thus
\begin{eqnarray*}
I&=&\int_0^1\int_0^1\frac{\partial^2 I}{\partial \alpha\partial\beta}d\beta d\alpha\\
&=&\frac{1}{2}\int_0^1\left(\pi\sum_{n=0}^\infty\frac{(-1)^n}{n+\frac{1}{2}}\alpha^{2n}+2\sum_{n=0}^\infty\frac{(-1)^n}{n+1}\alpha^{2n+1}\ln\alpha-\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^2}\alpha^{2n+1}\right)d\alpha\\
&=&\pi\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}-\sum_{n=0}^\infty\frac{(-1)^n}{4(n+1)^3}-\frac{1}{2}\sum_{n=0}^\infty\frac{(-1)^n}{2(n+1)^3}\\
&=&\pi G-\frac{1}{2}\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^3}\\
&=&\pi G-\frac{3\zeta(3)}{8}.
\end{eqnarray*}
Here we use
$$\int_0^1 x^n\ln x dx=-\frac{1}{(n+1)^2}. $$
|
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|
Finding the coefficient of a generating function Given $f(x) = x^4\left(\frac{1-x^6}{1-x}\right)^4 = (x+x^2+x^3+x^4+x^5+x^6)^4$. This is the generating function $f(x)$ of $a_n$, which is the number of ways to get $n$ as the sum of the upper faces of four thrown dice.
How do I calculate a coefficient from said generating function, for example $a_{17}$?
I could of course write it all out, but that would take a lot of time. Is there a quicker way to do it?
I've rewritten the formula as $x^4(1-x^6)^4\left(\frac1{1-x}\right)^4$, giving me to find the coefficient of $x^{17-4}=x^{13}$ of $(1-x^6)^4\left(\frac1{1-x}\right)^4$. I'm, however, stuck there.
|
You want:
\begin{align}
[z^{17}] z^4 (1 - z^6)^4 (1 - z)^{-4}
&= [z^{13}] (1 - 4 z^6 + 6 z^{12} - 4 z^{18} + z^{24})
\cdot \sum_{k \ge 0} \binom{-4}{k} (-1)^k z^k \\
&= [z^{13}] (1 - 4 z^6 + 6 z^{12}) \cdot \sum_{k \ge 0} \binom{k + 3}{3} z^k \\
&= \binom{16}{3} - 4 \cdot \binom{10}{3} + 6 \cdot \binom{4}{3} \\
&= 104
\end{align}
|
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|
How do I integrate $\frac{1}{x^6+1}$ My technique so far was substitution with the intent of getting to a sum of three fractions with squares in their denominators.
$t = x^2 \\
\frac{1}{x^6 + 1} = \frac{1}{t^3+1} = \frac{1}{(t+1)(t^2-t+1)}$
Then I try to reduce this fraction into a sum of two fractions
$\frac{A}{t+1} + \frac{B}{t^2-t+1} = \frac{(At^2-(A-B)t) + A + B}{(t+1)(t^2-t+1)}$
And this is where I reach a dead-end
$\begin{cases}
At^2-(A-B)t &= 0 \\
A + B &= 1
\end{cases}$
Any techniques I'm overlooking?
|
You have to find the (complex) pairs of roots of the polynomial $x^6+1$, and then use the uncertain coefficients method as you did.
Notice that using the substitution $t=x^2$ leaves you with $x=\sqrt{t}$ and $dx=dt/\sqrt{t}$ which doesn't look quite nicely integrable.
When you have the polynomials $t^2+pt+q$ with complex roots, you can use a clever linear substitution to change that into $u^2+1$; then $\int du/(u^2+1) = \arctan u+C$.
--
As for finding the roots of $x^6+1=0$, we can either guess them or help ourselves. You already know that $x^2=t$ and $t^3+1=0$, which could be helpful. But we may even notice that $(x^6+1)(x^6-1)=x^{12}-1$, therefore $x$ are the 12th roots of unity which are not 6th roots of unity. These are
$$\xi, \xi^3, \xi^5, \xi^7=\overline{\xi^5}, \xi^9=\overline{\xi^3}\text{ and }\xi^{11}=\overline{\xi},$$
where $\xi=e^{2\pi i/12}=e^{i\pi/6}$ is the fundamental (or first) 12th root of unity.
|
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|
Proof, wheather a subset of a Group is a Subgroup I have to check, weather the following subset of a group is also a subgroup:
$$U = \left\{ \begin{pmatrix} a & -b \\ \overline{b} & \overline{a} \end{pmatrix} \in GL(2, \mathbb{C}) \bigg\vert |a|^2 + |b|^2 = 1, \ a,b \in \mathbb{C} \right\} \subset (GL(2,\mathbb{C}), \cdot)$$
I did the following:
1) $\forall x,y \in U: x\cdot y \in U$
$$\begin{pmatrix} a & -b \\ \overline{b} & \overline{a} \end{pmatrix} \cdot \begin{pmatrix} a' & -b' \\ \overline{b'} & \overline{a'} \end{pmatrix} = \begin{pmatrix} aa'-b\ \overline{b'} & -ab'-b\ \overline{a'} \\ \overline{b}a'+\ \overline{ab'} & -\overline{b}b' +\overline{a'a} \end{pmatrix}$$
For $|a|^2 + |b|^2 = 1$ and $|a'|^2 + |b'|^2 = 1$. To be honest - I am unable to check weather that is $\in U$ - how can this be done?
2) $e\in U$
Let $a = 1$ and $b = 0$ we have $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = e$ and $|1|^2 + |0|^2 = 1 \Rightarrow e \in U$.
3) $\forall x \in U: x^{-1} \in U$
What I did was just calculating the inverse matrix but I am stuck:
$$\left( \begin{array}{cc|cc} a & -b & 1 & 0 \\ \overline{b} & \overline{a} & 0 & 1 \end{array}\right)$$
$$a,\ \overline{a}\ne0,\ I \cdot \frac{1}{a}; II \cdot \frac{1}{\overline{a}} $$
$$\left( \begin{array}{cc|cc} 1 & -\frac{b}{a} & \frac{1}{a} & 0 \\ \frac{\overline{b}}{\overline{a}} & 1 & 0 & \frac{1}{\overline{a}} \end{array}\right)$$
And well - now what ever I add, multiply or subtract I am not able to get the unit matrix on the left side. Is there anything I've overlooked? Because since
$$\det\left| \begin{array}{cc} a & -b \\ \overline{b} & \overline{a} \end{array} \right| = a\overline{a} + b\overline{b} = |a|^2 + |b|^2 = 1 \ne 0$$
every matrix in $U$ should be invertible, right?
Thank you very much for your help!
FunkyPeanut
|
For part (1):
$$\begin{pmatrix} a & -b \\ \overline{b} & \overline{a} \end{pmatrix} \cdot \begin{pmatrix} a' & -b' \\ \overline{b'} & \overline{a'} \end{pmatrix} = \begin{pmatrix} aa'-b\ \overline{b'} & -ab'-b\ \overline{a'} \\ \overline{b}a'+\ \overline{ab'} & -\overline{b}b' +\overline{a'a} \end{pmatrix}$$
You want here that
$$
(aa'-b\ \overline{b'})(\overline{aa'-b\ \overline{b'}}) + (ab'+b\ \overline{a'})(\overline{ab'+b\ \overline{a'}}) =1
$$
but this is not hard to check. You "just" have to write it all out.
For part (3), note that the determinant of any matrix in your subset is $1$ (as you have shown). Recall that the inverse of a matrix in $SL(2,\mathbb{C})$
$$
\pmatrix{a & b \\ c & d}
$$
is
$$\pmatrix{d & -b \\ -c & a}
$$
Now that means that the inverse of one of the matrices
$$
\begin{pmatrix} a & -b \\ \overline{b} & \overline{a} \end{pmatrix}
$$
is
$$
\begin{pmatrix} \overline{a} & b \\ -\overline{b} & a \end{pmatrix}
$$
But this is clearly again in the subset.
|
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|
Hard Olympiad Inequality Let x,y,z be positive real numbers such that $xy+xz+yz=1$. Prove that $$\sqrt{x^3+x}+ \sqrt{y^3+y}+ \sqrt{z^3+z} \geq 2 \cdot \sqrt{x+y+z}$$.
I tried to square expand homogenize then majorize. But I couldn't make it work. Any help would be much appreciated.
|
problem: since $a,b,c>0$,and such $ab+bc+ac=1$, show that
$$\sqrt{a^3+a}+\sqrt{b^3+b}+\sqrt{c^3+c}\ge 2\sqrt{a+b+c}$$
Poof:
Using Holder inequality,we have
$$\left(\sum\sqrt{a^3+a}\right)^2\left(\sum\dfrac{a^2}{a^2+1}\right)\ge\left(\sum a\right)^3$$
it remains to prove that
$$\left(\sum a\right)^2\ge 4\sum\dfrac{a^2}{a^2+1}$$
which is true,because
\begin{align*}\left(\sum a\right)^2- 4\sum\dfrac{a^2}{a^2+1}&=\left(\sum a\right)^2-4\left(\sum bc\right)\left[\sum\dfrac{a^2}{(c+a)(a+b)}\right]\\
&=\sum\dfrac{a(b-c)^2(b+c-a)^2}{(b+a)(b+c)(a+c)}\\
&\ge 0
\end{align*}
|
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|
Why does $\prod_{n=1}^\infty \frac{n^3 + n^2 + n}{n^3 + 1}$ diverge? $\prod_{n=1}^\infty \frac{n^3 + n^2 + n}{n^3 + 1}$ diverges, and I have no idea why? It would seem using L'hop, $\frac{n^3 + n^2 + n}{n^3 + 1}$ goes to 1. So it should end up just being $1\cdot1\cdot1\cdots$, which makes me feel it converges. Is the problem that it always adds that $.000\ldots0001$?
|
Intuitively, as $\frac{n^3 + n^2 + n}{n^3 + 1} \sim 1$, if we erase $n$ from the top and $1$ from the bottom, the fraction should get smaller. Lets prove it:
$$\frac{n^3 + n^2 + n}{n^3 + 1}-\frac{n+1}{n}=\frac{n^2-n-1}{n(n^3+1)}$$
Thus, for $n \geq 2$ we have
$$\frac{n^3 + n^2 + n}{n^3 + 1} > \frac{n+1}{n}$$
Now, since it is telescopic:
$$\prod_{n=1}^N \frac{n+1}{n} =N+1$$
|
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|
Probability of eight dice showing sum of 9, 10 or 11 Suppose we roll eight fair dice. What is the probability that:
1) The sum of the faces is $9$
2) The sum of the faces is $10$
3) The sum of the faces is $11$
I'm thinking that we start with $8$ dice, each showing $1$. Then we think of the problem as assigning one $1$, two $1$'s or three $1$'s to the $8$ dice.
So that would give us:
1) $P(\sum = 9)$ = $8(\frac{1}{6})^8$
2) $P(\sum = 10)$ = $8^2(\frac{1}{6})^8$
3) $P(\sum = 11)$ = $8^3(\frac{1}{6})^8$
Is this right?
|
Sum of 9 : 1, 1, 1, 1, 1, 1, 1, 2 = $$\frac{8!}{7!} = 8$$
Sum of 10: 1, 1, 1, 1, 1, 1, 2, 2 =$$\frac{8!}{6!2!} = 28$$
1, 1, 1, 1, 1, 1, 1, 3 =$$\frac{8!}{7!} = 8$$
$$ 8+28 = 36$$
Sum of 11: 1, 1, 1, 1, 1, 1, 1, 4 :$$\frac{8!}{7!} = 8$$
1, 1, 1, 1, 1, 1, 2, 3 : $$\frac{8!}{6!} = 56$$
1, 1, 1, 1, 1, 2, 2, 2 : $$\frac{8!}{5!3!} = 56$$
$$ 8+56*2 = 120$$
Thus the probabilities for sums of 9, 10, 11 are $$8.\frac{1}{6^8}, 36.\frac{1}{6^8}, 120.\frac{1}{6^8}$$ respectively
|
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|
Inequality Exercise in Apostol's Calculus I Let p and n denote positive integers. Show that:
$$n^{p} \lt \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^{p}$$
Attempt at Solution
Using the identity $b^{p+1}-a^{p+1} = (b-a)\sum_{k=0}^{p}b^{p-k}a^{k}$, let $b = n+1$ and $a = n$. Then:
$$\frac{(n+1)^{p+1} - n^{p+1}}{p+1} = \frac{\sum_{k=0}^{p}(n+1)^{p-k}n^{k}}{p+1} = \frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1}$$
I'm not sure where to proceed from here. Am I supposed to use a geometric series in here somewhere?
EDIT: I may have figured it out if someone can confirm that $(n+1)^{p-k}n^{k} > n^{p}$, $1 \le k \le p$, $\forall n$
Assuming the above is true, we have:
$$\frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1} \gt \frac{n^{p} + n^{p} + .... + n^{p} + n^{p}}{p+1} = \frac{(p+1)n^{p}}{(p+1)} = n^{p}$$
Similarly:
$$\frac{(n+1)^{p} + (n+1)^{p-1}n + .... + (n+1)n^{p-1} + n^{p}}{p+1} \lt \frac{(n+1)^{p} + (n+1)^{p} + .... + (n+1)^{p} + (n+1)^{p}}{p+1} = \frac{(p+1)(n+1)^{p}}{(p+1)} = (n+1)^{p}$$
|
The middle term, $\dfrac{(n+1)^{p+1} - n^{p+1}}{p+1}$, equals $\int_n^{n+1} x^p dx$. On the interval $(n,n+1)$, which has length $1$, $n^p<x^p<(n+1)^p$, from which the inequality follows immediately.
|
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|
Consider the quadratic equation $ax^2-bx+c=0, a,b,c \in N. $ If the given equation has two distinct real root... Problem :
Consider the quadratic equation $~ax^2-bx+c=0, \quad a,b,c \in N. ~$ If the given equation has two distinct real roots belonging to the interval $~(1,2)~ $ then the minimum possible values of $~a~$ is
$(i) \quad -1 $
$(ii)\qquad 5 $
$(iii)~~\quad 2 $
$(iv)\quad -5 $
$(v) \qquad1 $
My approach :
We know the condition that two roots will between the two numbers viz. $(1,2)$
$(1) \quad f(1) >0 ; \qquad $
$(2)\quad f(2)>0\qquad$
$(3) \quad 1 < \frac{-b}{2a} <2\qquad $
$(4) \quad D \geq 0$
By using the above I got the following :
$(1) \quad a-b+c >0$
$(2)\quad 4a-2b+c >0$
$(3)\quad 1 < \frac{b}{2a} <2$
$(4) \quad b^2-4ac \geq 0$
Please guide further how to get the answer given in above five options. Thanks..
|
Firstly we can make some simple remarks about the problem:
*
*let $ax^2-bx+c = f(x)$
*$a,b,c \in \Bbb{N} \Rightarrow a,b,c \ge 0$
*there are $2$ distinct roots, so the parabola is not a degenerate one $\Rightarrow a \ge 1$
*$x_{1,2} \in ]1,2[ \Rightarrow x_1 \cdot x_2 > 1\cdot 1 = 1 \Rightarrow \frac{c}{a} > 1 \Rightarrow a < c \Rightarrow c \ge 2$
*by applying Rolle's Theorem in $]1,2[$ we get $e \in ]1,2[ | f'(e) = 0 = 2\cdot a\cdot e-b = 0$
*solutions $a=-1,-5$ can be excluded
*obviously $x_{1,2} \in \Bbb{R} \Rightarrow \Delta = b^2-4ac > 0$
By fact #5 we get $$\exists e \in ]1,2[ | f'(e) = 2\cdot a\cdot e-b = 0 \Rightarrow\\a=\frac{b}{2e}\Rightarrow\\max\{a\} = \frac{b}{2}, min\{a\} = \frac{b}{4}\Rightarrow\\\frac{b}{4} < a < \frac{b}{2}$$
Now let's plug in some values
*
*b=0 $\Rightarrow 0<a<0$ which is impossible
*b=1 $\Rightarrow \frac14<a<\frac12$, impossible too
*b=2 $\Rightarrow \frac14<a<\frac12$, another impossible case
*b=3 $\Rightarrow \frac34<a<\frac32 \Rightarrow a = 1$ but $x_{1,2} = \frac{3\pm\sqrt{9-4c}}{2} = \frac32 \pm \frac{\sqrt{9-4c}}{2} \notin ]1,2[$ if $c=2$; obviously $c>2 \Rightarrow \Delta <0$
*b=4 $\Rightarrow 1<a<2 \Rightarrow a \in \Bbb{N}$, impossible; so $a=1$ can be excluded
*b=5 $\Rightarrow \frac54<a<\frac52 \Rightarrow a =2$, yet $x_{1,2} = \frac{5\pm\sqrt{25-8c}}{4} = \frac54 \pm \frac{\sqrt{25-8c}}{4} \notin ]1,2[$ for $c\le3$, while $c>3 \Rightarrow \Delta <0$
*b=6 $\Rightarrow \frac32<a<3 \Rightarrow a =2$, but if you check, $x_{1,2}\notin]1,2[$ for any acceptable value of $c$
*b=7 $\Rightarrow \frac74<a<72 \Rightarrow a =2,3$, and you can check these values can fit
*b=8 $\Rightarrow 2<a<4 \Rightarrow a =2 $ can be discarded and the minimum value of $a$ is $5$; infact if $$a=5, b=15, c=11 \Rightarrow x_1 \approx 1.28, x_1 \approx 1.72$$
|
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If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $a^4c+b^4d\ge cd$.
If positive $a,b,c,d$ satisfy $(a^3+b^3)^4=c^3+d^3$, prove that $$a^4c+b^4d\ge cd$$
It kind of seems useful to begin with a division of both sides by $cd$:
$$\frac{a^4}{d}+\frac{b^4}{c}\ge1$$
It seems like a simple Cauchy-Schwarz would be of use in this case. However, after some attempts I can't come up with the right solution. It'd be great to hear some ideas. Thanks.
|
Holder's inequality for sums states that for $p, q \in (1, \infty)$ with $\frac{1}{p}+\frac{1}{q}=1$, $$\sum_{k=1}^{n}{|a_kb_k|} \leq \left(\sum_{k=1}^{n}{|a_k|^p} \right)^{\frac{1}{p}}\left(\sum_{k=1}^{n}{|b_k|^q} \right)^{\frac{1}{q}}$$
(For $p=q=2$ this reduces to Cauchy Schwarz)
Applying this for $p=\frac{4}{3}, q=4$ (Note $a, b, c, d>0$) gives
$$(a^3+b^3) \leq \left( (a^3c^{\frac{3}{4}})^{\frac{4}{3}}+(b^3d^{\frac{3}{4}})^{\frac{4}{3}} \right)^{\frac{3}{4}}\left( \left(\frac{1}{c^{\frac{3}{4}}}\right)^4+\left(\frac{1}{d^{\frac{3}{4}}}\right)^4 \right)^{\frac{1}{4}}$$
$$(a^3+b^3)^4 \leq (a^4c+b^4d)^3 \left( \frac{1}{c^3}+\frac{1}{d^3} \right)=(a^4c+b^4d)^3\frac{c^3+d^3}{c^3d^3}$$
Thus
$$a^4c+b^4d \geq \left(\frac{(a^3+b^3)^4c^3d^3}{c^3+d^3} \right)^{\frac{1}{3}}=cd$$
|
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|
Finding all Points on a Edwards curve I need to find all affine points on the Edwards curve:
$x^2 + y^2 = 1 - 5x^2y^2$ over $F_{13}$
I tackle this by transforming the equation to:
$y^2 = \frac{1-x^2}{1+5x^2}$
I then go from x = 0 to $\frac{p-1}{2}$ in this case x from 0 to 6.
If you can take the square root of $y^2$ you found a point. If you have a match you will find 4 points because of the symmetry of this equation. (x,y)(x,-y)(-x,y)(-x,-y).
I managed to find these points: (0,1)(0,12)(6,3)(6,10)(7,3)(7,10)(12,0)
However my computer algorithm says I'm missing the following points (3,6)(3,7)(10,6)(10,7).
Simply put I'm missing x = 3.
This is my calculation of x = 3:
$y^2 = \frac{1-3^2}{1+5*3^2} = \frac{-8}{46} = \frac{5}{7} \rightarrow 5 * 7^{-1} \mod{13} \equiv 10$
however the square root of 10 isn't an integer. Could anyone answer my question?
|
The congruence class of $10$ is a square in $\mathbb{F}_{13}$. Indeed,
$$6^2\equiv 36\equiv 10 \bmod 13$$
and similarly $(-6)^2\equiv 7^2\equiv 49\equiv 10\bmod 13$. Thus, for $x\equiv 3$ and $-3\equiv 10\bmod 13$, there are two possibilities for $y$, namely $6$ and $7\bmod 13$.
Notice that your algorithm could have run into trouble if $1+5x^2\equiv 0 \bmod 13$, for some $x\bmod 13$. Although $1+5x^2$ is never $0$ over $\mathbb{Q}$, it could have happened that $1+5x^2\equiv 0$ has a solution in $\mathbb{F}_{13}$. This is not the case, because a solution would imply that $x^2\equiv 5\bmod 13$, and $5$ is not a square in $\mathbb{F}_{13}$ (the squares are $1,3,4,9,10$, and $12$).
|
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let $a-b=6$ and $b\leq-1$ and $a\leq-2$. Find $A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}$ let $a-b=6$ and $b\leq-1$ and $a\leq-2$. Find $A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}$
My Try $$A=\sqrt{a^2+4a+4}+\sqrt{b^2+2b+1}=$$then I can't
Please just a hint.
|
Simply write $b$ in terms of $a$ (and mind the conditions on $a$ and $b$: $a \leq −2$ so $a+2\leq0$ and $b\leq-1$ so $b+1\leq0$) and you get
$$A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}=\sqrt{(a+2)^2}+\sqrt{(a-6+1)^2}=|a+2| + |a-5|=-a-2-a+5=-2a+3$$
since $a+2\leq0$ and $b+1\leq0$ ($\Rightarrow |a+2| = -a-2 \text{ and } |a-5| = -a-5$).
|
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|
Problem with LU decomposition I have this matrix:
$$
A =\begin{bmatrix}1 & -2 & 3\\ 2 & -4 & 5 \\ 1 & 1 & 2\end{bmatrix}
$$
After I decomposit it, I get:
$$
L = \begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\ 2 & 0 & 1 \end{bmatrix}\\
U = \begin{bmatrix} 1 & -2 & 3\\ 0 & 3 & -1\\ 0 & 0 & -1\end{bmatrix}\\
P = \begin{bmatrix}1 & 0 & 0\\0 & 0 & 1\\0 & 1 & 0\\\end{bmatrix}
$$
Now I need to solve $Ax=b$ using: $b_1 = \begin{bmatrix}2\\0\\1 \end{bmatrix}$ and $b_2 = \begin{bmatrix}-1\\1\\-1 \end{bmatrix}$
But I'm stuck and don't know how to permute the vectors and every time I try I get a different result...
As I checked in Octave, the solution of $Ax=b_1$ is: $x_1 = \begin{bmatrix}-8\\1\\4 \end{bmatrix}$, but I can't get that solution.
Can you please explain me how to get this done? Thank you!
|
All of your work is correct for $A = PLU$.
Now, we want to solve $Ax = b$ for $b_1$ and $b_2$, so we use forward and back substitution as:
$$Ly = Pb, Ux = y$$
Forward substitution yields:
$$Ly = \begin{bmatrix}1 & 0 & 0\\1 & 1 & 0\\ 2 & 0 & 1 \end{bmatrix}\begin{bmatrix}y_1\\y_2\\y_3 \end{bmatrix}=Pb = \begin{bmatrix}1 & 0 & 0\\0 & 0 & 1\\0 & 1 & 0\\\end{bmatrix}\begin{bmatrix}2\\0\\1 \end{bmatrix} = \begin{bmatrix}2\\1\\0 \end{bmatrix} \implies y = \begin{bmatrix}2\\-1\\-4 \end{bmatrix}$$
Next, we use back substitution to solve $Ux = y$, yielding:
$$Ux = \begin{bmatrix} 1 & -2 & 3\\ 0 & 3 & -1\\ 0 & 0 & -1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix}=y=\begin{bmatrix}2\\-1\\-4 \end{bmatrix} \implies \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}-8\\1\\4 \end{bmatrix}$$
Rinse and repeat for $b_2$, yielding:
$$x_2 = \begin{bmatrix}6\\-1\\-3 \end{bmatrix}$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
${2+6\over 4^{100}}+{2+2\cdot6\over 4^{99}}+{2+3\cdot6\over 4^{98}}+\cdots+{2+99\cdot6\over 4^2}+{2+100 \cdot6\over 4}$ Find the value of
$${2+6\over 4^{100}}+{2+2\cdot6\over 4^{99}}+{2+3\cdot6\over 4^{98}}+\cdots+{2+99\cdot6\over 4^2}+{2+100\cdot6\over 4}$$
My approach:
$${2\over 4^{100}}+{2\over 4^{99}}+\cdots+{2\over 4^{1}}$$
forms a G.P. But how to separate $\displaystyle{6\over 4^{100}}+{2\cdot6\over 4^{99}}+\cdots$?
|
The second piece is simply
$$6 \sum_{k=1}^{100} \frac{k}{4^{101-k}}= \frac{6}{4^{101}} \sum_{k=1}^{100} k \, 4^k$$
To evaluate, let $S = 4 + 2 \cdot 4^2 +3 \cdot 4^3 + 4 \cdot 4^4+\cdots+100\cdot 4^{100}$ Then
$$4 S = 4^2+ 2 \cdot 4^3 +3 \cdot 4^4 + 4 \cdot 4^5+\cdots+100 \cdot 4^{101}$$
$$\implies S-4S=-3 S = 4+4^2+4^3+4^4+\cdots+4^{100}-100\cdot 4^{101}$$
Now you have another geometric series, and use that fact to solve for $S$.
|
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|
How do I solve $(x^3-4x^2+5x-6)/(x^2-x-6)=4$ algebraically? How do I solve $\frac {(x^3-4x^2+5x-6)}{(x^2-x-6)=4}$ algebraically?
I tried:
$4(x^2-x-6)=x^3-4x^2+5x-6$
$4x^2-4x-24=x^3-4x^2+5x-6$
$x^3-8x^2+9x+18=0$
I don't know how to solve this algebraically.
|
Use the fact that:
$$\frac{ (x^3-4x^2+5x-6)}{(x^2-x-6)}=(x-3)+\frac{8}{(x+2)} $$
Therefore your solving:
$$ (x-3)+\frac{8}{x+2}=4 \\ x^2-5x-6+8=4x+8\\ x^2-5x-6=0$$
Can you solve this?
|
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|
Finding the Asymptotic Curves of a Given Surface I have to find the asymptotic curves of the surface given by $$z = a \left( \frac{x}{y} + \frac{y}{x} \right),$$ for constant $a \neq 0$.
I guess that what was meant by that statement is that surface $S$ can be locally parametrized by $$X(u,v) = \left( u, v, a \left( \frac{u}{v} + \frac{u}{v} \right) \right).$$ Do you think that my parametrization is correct (meaning that I read the description of the surface correctly), and do you know of a more convenient parametrization?
Assuming that parametrization, I derived the following ($E$, $F$, $G$, are the coefficients of the first fundamental form; $e$, $f$, $g$ are coefficients of the second fundamental form; $N$ is the normal vector to surface $S$ at a point; these quantities are all functions of local coordinates $(u,v)$):
$$E = 1 + a^2 \left( \frac{1}{v} - \frac{v}{u^2} \right)^2,$$
$$F = -\frac{a^2 (u^2 - v^2)^2}{u^3 v^3},$$
$$G = 1 + a^2 \left( \frac{1}{u} - \frac{u}{v^2} \right)^2.$$
$$N = \frac{1}{\sqrt{E G - F^2}} \left( a \left( \frac{v}{u^2}-\frac{1}{v} \right), a \left( \frac{u}{v^2}-\frac{1}{u} \right), 1 \right).$$
$$X_{u,u} = \left( 0,0, \frac{2 a v}{u^3} \right), X_{u,v} = \left( 0,0, -a \left( \frac{1}{u^2} + \frac{1}{v^2} \right) \right), X_{v,v} = \left( 0, 0, \frac{2 a u}{v^3} \right).$$
$$e = \frac{2 a v}{u^3 \sqrt{E G - F^2}},$$
$$f = - \frac{a (\frac{1}{u^2} + \frac{1}{v^2})}{\sqrt{E G - F^2}},$$
$$g = \frac{2 a u}{v3 \sqrt{E G - F^2}}.$$
Thus, the Gaussian curvature (from these calculations) is:
$$K = -\frac{a^2 u^4 v^4 (u^2 - v^2)^2}{(u^4 v^4 +
a^2 (u^2 - v^2)^2 (u^2 + v^2))^2}.$$
And the mean curvature would be:
$$H = \frac{a u^3 v^3 (u^4 + v^4)}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2
}}.$$
So, the principal curvatures are:
$$k_{\pm} = H \pm \sqrt{H^2 - K} = a u^2 v^2 \frac{u v (u^4 + v^4) \pm \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2}}.$$
In order to find the asymptotic curves, but trying to avoid the differential equation, I was hoping to find the angles $\theta (u,v)$ such that the normal curvature would always be $0$. In other words I was trying:
$0 = k_n = k_{+} \cos{(\theta)}^2 + k_{-} \sin{(\theta)}^2$, and solving for $\theta$.
Assuming sufficient niceness, this calculate would result in: $$(u v (u^4 + v^4) + \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \cos{(\theta)}^2 + (u v (u^4 + v^4) - \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \sin{(\theta)}^2$$
First of all, is this approach (solving for $\theta$ rather than solving the differential equation) valid? If it is, after I find that angle $\theta$, determined by location $(u,v)$ on $S$, what more work do I have to do? How do I find the equations for the asymptotic curves based on this angle?
If this whole method was for naught, how does one solve the differential equation. in this case, of: $$e (u')^2 + 2f u' v' + g (v')^2 = 2a v^4 (u')^2 - 2a u^3 v^3 \left( \frac{1}{u^2} + \frac{1}{v^2} \right)u' v' + 2a u^4 (v')^2 = 0?$$ (Again, assuming sufficient niceness.)
(See: https://math.stackexchange.com/questions/762195/differential-equation-for-the-asymptotic-directions-of-a-given-surface)
Thank you!
|
Proper sign to be considered for scalars after coming so far with normal curvature. For a negatively curved (Gauss curvature $K<0$ ) surface
$ k_n = -k_{+} \cos^2{\theta} + k_{-} \sin^2{\theta} = 0 $ and for real $\theta$,
$ \tan{\theta} = \sqrt {(k_{+} / k_{-})} $
For example, asymptotic lines on rotationally symmetric pseudospheres ($ K= -1/a^2 $ ) obey Sine-Gordon on Chebychev net with $2 \theta = 2 \psi,$ angle between asymptotic lines:
$ a\, k_1 = - \tan \psi, a\, k_2 = \cot \psi, \tan{\psi} = \sqrt {(k_{1} / k_{2})}. $
|
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|
Is $\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}}$ convergent? Is $\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}}$ convergent ?
I use it to compare with $1/n^2$, and then I used LHôpitals rule multiple times. Finally , I can solve it. However,I think we have other methods! somebody can help me?
|
By using the integral test:
The function $\frac{1}{3^{\sqrt{x}}}$ is continuous, positive and decreasing on $[1,\infty)$. We then evaluate the integral:
\begin{align}
\int_1^\infty\, \frac{1}{3^{\sqrt{x}}}\, dx = \frac{2}{3 \, \log\left(3\right)} + \frac{2}{3 \, \log\left(3\right)^{2}} \approx 1.15918311754471
\end{align}
and see that it has a finite value, which in turn proves that the sum is also finite.
It can also be used to get the lower and upper bounds:
\begin{align}
\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}} &\ge \int_1^\infty\, \frac{1}{3^{\sqrt{x}}}\, dx\\
&\ge \frac{2}{3 \, \log\left(3\right)} + \frac{2}{3 \, \log\left(3\right)^{2}} \approx 1.15918311754471
\end{align}
and
\begin{align}
\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}} &\le \frac{1}{3}+\int_1^\infty\, \frac{1}{3^{\sqrt{x}}}\, dx\\
&\le \frac{1}{3}+\frac{2}{3 \, \log\left(3\right)} + \frac{2}{3 \, \log\left(3\right)^{2}} \approx 1.49251645087804
\end{align}
The actual numerical result using a program:
\begin{align}
\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}} \approx 1.34070489742952420178654874099073605072839417287611074076358138561327
\end{align}
p.s. the integral can be easily solved by observing that $3^\sqrt{x} = e^{\left(\sqrt{x} \ln\left(3\right)\right)}$
|
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|
determining $n$ in a given sequence $\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $ Given that: $$\frac{1+3+5+...+(2n-1)}{2+4+6+...+(2n)} =\frac{2011}{2012} $$
Determine $n$.
The memorandum says the answer is 2011 but how is that so? Where did I go wrong?
|
Numerator :
\begin{eqnarray*}
T_n &=& 1+3+5+...+(2n-1)\\
&=& 2+4+6+...+2n - n\\
&=& 2(1+2+3+...+n) - n\\
&=& 2\frac{n(n+1)}{2} - n\\
&=& n^2
\end{eqnarray*}
Similarly for denominator (check) :
\begin{eqnarray*}
B_n &=& 2+4+6+...+2n\\
&=& n(n+1)
\end{eqnarray*}
But it back in the initial equation and you get :
$$\frac{n^2}{n(n+1)} = \frac{2011}{2012} \Longrightarrow 2012n = 2011(n+1)$$
Last line is easy enough to solve.
So in fact the fraction is written in the form $$\frac{4044121}{4046132} = \frac{2011}{2012}$$
I hope this explains well enough why assuming the sums are $2011$ and $2012$ is wrong.
|
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|
$2 \cos^2 x − 2 \cos x− 1 = 0$ Find the solutions if $0^\circ \le x < 360^\circ$
Find the solutions of $$2 \cos^2 x − 2 \cos x− 1 = 0$$ for all $0^\circ ≤ x < 360^\circ$.
For $0^\circ \le x < 360^\circ$, I'm getting $x=111.5^\circ$ and $x=248.5^\circ$.
Is this correct? Thanks!
|
Let $y = \cos(x)$. Then we have the following:
$$2y^2 - 2y - 1 = 0$$
Using the quadratic formula, we get the roots to be:
$$y = \cos(x) = \frac{1}{2} \pm \frac{\sqrt{3}}{2}$$
And so:
$$x = \arccos(\frac{1}{2} \pm \frac{\sqrt{3}}{2}) $$
Since $\frac{1}{2} + \frac{\sqrt{3}}{2}>1$, we can ignore this solution. Since both of your solutions satisfy $\cos(x) = \frac{1}{2} - \frac{\sqrt{3}}{2}$, you are indeed correct.
|
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|
What is the value of $\sin 47^{\circ}+\sin 61^{\circ}- \sin25^{\circ} -\sin11^{\circ}$? After simplification using sum to product transformation equations I keep ending up with
$$4\cos36^\circ\cdot\cos7^\circ\cdot\cos18^\circ$$
How do I simplify this to a single term?
|
Using Prosthaphaeresis Formulas
$$\sin47^\circ+\sin61^\circ=2\sin54^\circ\cos7^\circ$$
$$\sin25^\circ+\sin11^\circ=2\sin18^\circ\cos7^\circ$$
$$\implies\sin47^\circ+\sin61^\circ-(\sin25^\circ+\sin11^\circ)=2\cos7^\circ(\sin54^\circ-\sin18^\circ)$$
$$S=\sin54^\circ-\sin18^\circ$$
Method $\#1:$
$$S=\sin54^\circ-\sin18^\circ=2\sin18^\circ\cos36^\circ=2\cos72^\circ\cos36^\circ$$
$$=2\cos72^\circ\cdot\frac{2\sin36^\circ\cdot\cos36^\circ}{2\sin36^\circ}=2\cos72^\circ\cdot\frac{\sin72^\circ}{2\sin36^\circ}$$
$$=\frac{\sin144^\circ}{2\sin(180^\circ-36^\circ)}=\frac12$$
Method $\#2:$
$$S=\sin54^\circ-\sin18^\circ=\sin54^\circ+\sin(-18^\circ)$$
Multiplying either sides by $\displaystyle2\sin\left(\frac{54^\circ-(-18^\circ)}2\right)=2\sin36^\circ$
$$2\sin36^\circ\cdot S=2\sin36^\circ\sin54^\circ-2\sin36^\circ\sin18^\circ$$
Using Werner Formulas
$$2\sin36^\circ\cdot S=\cos18^\circ-\cos90^\circ-(\cos18^\circ-\cos54^\circ)=\cos54^\circ$$
$$\implies2S=\frac{\cos54^\circ}{\sin36^\circ}=1$$
Method $\#3:$
$$S=\sin54^\circ-\sin18^\circ=\cos36^\circ+\cos108^\circ\text{ as }\cos(90^\circ+y)=-\sin y$$
Multiplying either sides by $\displaystyle2\sin\left(\frac{108^\circ-36^\circ}2\right)=2\sin36^\circ$
The rest is like Method$\#2$
|
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|
Testing for convergence of a series Test the following series for convergence
$$\sum_{k=1}^\infty \sin k \sin \frac{1}{k}$$.
I am just wondering if the following method is ok:
$$\sum_{k=1}^\infty \sin k \sin \frac{1}{k} = \sum_{k=1}^\infty \sin (\frac{1}{k} - (\frac{1}{k} -k)) \sin \frac{1}{k} \\ \leq \sum_{k=1}^\infty (\sin \frac{1}{k} \cos(\frac{1}{k} -k) - \cos \frac{1}{k}\sin(\frac{1}{k}-1)) \sin \frac{1}{k} \\ \leq
\sum_{k=1}^\infty \sin \frac{1}{k} \cos(\frac{1}{k} -k)\sin\frac{1}{k} - \cos \frac{1}{k}\sin(\frac{1}{k}-1) \sin \frac{1}{k}
\\\sum_{k=1}^\infty \sin \frac{1}{k} \cos(\frac{1}{k} -k)\sin\frac{1}{k}
\\\sum_{k=1}^\infty \sin \frac{1}{k} \sin\frac{1}{k}
\\\sum_{k=1}^\infty \frac{1}{k^2}$$ which converges.
Is there any more elegant proof? Help appreciated thanks.
|
By Taylor's theorem:
$$\begin{align}
\sin k\sin \frac 1k&=\frac 12\left[\cos\left(k-\frac1k\right)-\cos\left(k+\frac1k\right)\right]=\\
&=\frac 12\left[\cos k-\frac1k\sin\xi_k-\cos k+\frac1k\sin\xi'_k\right]=\\
&=\frac1{k}\sin\left(\frac{\xi_k-\xi'_k}2\right)\cos\left(\frac{\xi_k+\xi'_k}2\right)
\end{align}
$$
where $\xi_k,\xi'_k\in[k-\frac1k,k+\frac1k]$. Note that $|\xi_k-\xi'_k|\leq 2/k$
Since $|\sin x|\leq|x|$ and $|\cos x|\leq 1$, the absolute value of the latter expresion is bounded by $1/k^2$.
Nice problem :)
|
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|
Finding a holomorphic function with a prescribed real part. I am to find a holomorphic function on $\mathbb{C}\setminus\{0\}$ with $u(x,y)=\dfrac{x+y}{x^2+y^2}$ and $f(1)=1$. Preceding this, I had to show we could write $f'(z) = \frac{\partial u}{\partial x} -i\frac{\partial u}{\partial y}$, which is clearly supposed to be a hint for this part.
The issue is I don't see how it helps at all. Other than taking $u(x,y)$ as I know it and doing lots of differentiation/integration along with the Cauchy Riemann equations (which would be a lot of work..) the only idea I had that used this as a hint was to compute $f'(z)$ with that formula, and see if it would arrange into an expression in $z$ which was readily integrable. I got to $f'(z(x,y)) = \dfrac{(y+ix)^2+i(y+ix)^2}{(x^2+y^2)^2}$, but given this all assumes $x$ is the real part, etc., you can't just swap $x$ and $y$ and there's no way to get to $x+iy$ nicely! so that approach seemed to fail..
This is a question from a past exam paper; one of the short 'more routine' questions in the first section. Typically these aren't so bad, but with this particular question is being more difficult to deal with.
|
Since $u(x,y) = \dfrac{x+y}{x^2+y^2}$,
$$\frac{\partial u}{\partial x} = \frac{1}{x^2+y^2}-\frac{2x(x+y)}{(x^2+y^2)^2} = \frac{x^2+y^2-2x^2-2xy}{(x^2+y^2)^2} = \frac{-x^2-2xy+y^2}{(x^2+y^2)^2}.$$
Likewise, by symmetry,
$$\frac{\partial u}{\partial y} = \frac{-y^2-2xy+x^2}{(x^2+y^2)^2}.$$
Putting these together shows that
$$\begin{align} f'(z) &= \frac{-x^2-2xy+y^2-i(-y^2-2xy+x^2)}{(x^2+y^2)^2} \\ &= \frac{-x^2-2xy+y^2+iy^2+2ixy-ix^2}{(x^2+y^2)^2}.\end{align}$$
The numerator can be rewritten a bit:
$$\begin{align}(-x^2+2ixy+y^2)+i(-x^2+2ixy+y^2) &= (1+i)(-x^2+2ixy+y^2) \\ &= -(1+i)(x^2-2ixy-y^2) \\ &= -(1+i)(x-iy)^2.\end{align}$$
Thus we have that
$$f'(z) = \frac{-(1+i)(x-iy)^2}{((x+iy)(x-iy))^2} = -\frac{1+i}{(x+iy)^2} = -\frac{1+i}{z^2}.$$
From here, it is a simple application of integration to get the final answer.
|
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A logic problem. No need for calculation Three people were told to go to a cave and each pick up a hat in pitch dark. The three then came out of the cave with the hat on their heads. There were five hats in the cave. Three of them are black, the rest is white. When the first one is asked to tell the color of his hat, he said that he doesn't know. The second person told then same. But the third person said he knows. Why does he know the color of his hat based on what he heard from the other two persons? And how?
|
Initially there are $7$ possibilities:
$$
\begin{array}{|c|c|c|c|}
\hline
1 & 2 & 3 \\ \hline
B & B & B \\ \hline
B & B & W \\ \hline
B & W & B \\ \hline
B & W & W \\ \hline
W & B & B \\ \hline
W & B & W \\ \hline
W & W & B \\ \hline
\hline
\end{array}
$$
First says that he doesn't know, so one of $2-3$ has a black, that leaves us with:
$$
\begin{array}{|c|c|c|c|}
\hline
1 & 2 & 3 \\ \hline
B & B & B \\ \hline
B & B & W \\ \hline
B & W & B \\ \hline
W & B & B \\ \hline
W & B & W \\ \hline
W & W & B \\ \hline
\hline
\end{array}
$$
Second says he doesn't know, that means one of the others has a black, which leaves us with:
$$
\begin{array}{|c|c|c|c|}
\hline
1 & 2 & 3 \\ \hline
B & B & B \\ \hline
B & B & W \\ \hline
B & W & B \\ \hline
W & B & B \\ \hline
W & W & B \\ \hline
\hline
\end{array}
$$
Moreover, the second knows that the first has seen a black among $2$ or $3$ and he sees that $3$ has white in row $2$, so if row two is the case then the second actually does know his color, so row $2$ gets eliminated:
$$
\begin{array}{|c|c|c|c|}
\hline
1 & 2 & 3 \\ \hline
B & B & B \\ \hline
B & W & B \\ \hline
W & B & B \\ \hline
W & W & B \\ \hline
\hline
\end{array}
$$
The third knows that his color is black, because in all remaining worlds that's the case.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral of $\int\frac{dx}{x^2+1}$ I know the formula$$\int\frac{dx}{x^2+1} = \tan^{-1}(x) + C$$
But, when integrating by parts:
$$u = x^2+1$$
$$u' = 2x$$
$$v' = 1$$
$$v = x$$
$$\int\frac{dx}{x^2+1} = uv - \int vu' dx = x(x^2+1) - \int 2x^2dx$$
$$ \int 2x^2dx = \frac{2x^3}{3} + C$$
So, finally: $$\int\frac{dx}{x^2+1} = x(x^2+1) - \frac{2x^3}{3} + C$$
Am I doing something wrong?
|
Look this is an easier approach: Put $x = \tan \alpha $ . then $dx = \sec^2 \alpha d \alpha $. Hence
$$ \int \frac{dx}{1 + x^2} = \int \frac{\sec^2 \alpha}{1 + \tan^2 \alpha} d \alpha = \int d \alpha = \alpha = \arctan x + C$$
|
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}
|
Solve $\cos x+8\sin x-7=0$
Solve $\cos x+8\sin x-7=0$
My attempt:
\begin{align}
&8\sin x=7-\cos x\\
&\implies 8\cdot \left(2\sin \frac{x}{2}\cos \frac{x}{2}\right)=7-\cos x\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=7-1+2\sin ^2\frac{x}{2}\\
&\implies 16\sin \frac{x}{2}\cos \frac{x}{2}=6+2\sin^2 \frac{x}{2}\\
&\implies 8\sin \frac{x}{2}\cos \frac{x}{2}=3+\sin^2 \frac{x}{2}\\
&\implies 0=\sin^2 \frac{x}{2}-8\sin \frac{x}{2}\cos \frac{x}{2}+3\\
&\implies 0=\sin \frac{x}{2}\left(\sin \frac{x}{2}-8\cos \frac{x}{2}\right)+3
\end{align}
I'm not sure how to proceed from here (if this process is even right at all?) . Any help would be appreciated. Thanks.
|
Another solution could use the tangent half-angle substitution. If you define $t=\tan \frac{x}{2}$, you have $\sin x=\frac{2t}{1+t^2}$ and $\cos x=\frac{1-t^2}{1+t^2}$. So, the equation becomes $$\cos x+8\sin x-7=\frac{16 t+2}{t^2+1}-8=0$$ that is to say $$-8 t^2+16 t-6=0$$ the roots of which being $t_1=\frac{1}{2}$ and $t_2=\frac{3}{2}$ and so $x_1=2 \tan ^{-1}\left(\frac{1}{2}\right)$ and $x_2=2 \tan ^{-1}\left(\frac{3}{2}\right)$.
I am sure that you can take from here.
|
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|
Simplifying $\frac{a}{(a-b)(a-c)(x-a)}+\frac{b}{(b-c)(b-a)(x-b)}+\frac{c}{(c-a)(c-b)(x-c)}$ We need to simplify $$\dfrac{a}{(a-b)(a-c)(x-a)}+\dfrac{b}{(b-c)(b-a)(x-b)}+\dfrac{c}{(c-a)(c-b)(x-c)}$$
The biggest problem is that the above expression has four variables.I transformed the expression into $$\dfrac{a}{-(a-b)(c-a)(x-a)}+\dfrac{b}{-(b-c)(a-b)(x-b)}+\dfrac{c}{-(c-a)(b-c)(x-c)}$$
and then tried to add them up,but as you can probably guess,it yielded something almost impossible to handle.I tried to factorize it using standard techniques,but that was fruitless too.A hint will be appreciated.
N.B: The chapter deals with Remainder-factor theorem,partial fractions,factoring cyclic homogenous polynomials and manipulating algebraic expressions.
|
$$
\frac{a}{(a-b)(a-c)(x-a)}=\frac{a((a-c)-(a-b))}{(b-c)(a-b)(a-c)(x-a)}=\frac{a}{(b-c)(a-b)(x-a)} - \frac{a}{(b-c)(a-c)(x-a)}
$$
Thus,
$$
\frac{a}{(a-b)(a-c)(x-a)} + \frac{b}{(b-a)(b-c)(x-b)} + \frac{c}{(c-a)(c-b)(x-c)}=\frac{1}{(b-c)(b-a)}(\frac{b}{x-b}-\frac{a}{x-a})+ \frac{1}{(c-a)(c-b)}(\frac{c}{x-c}-\frac{a}{x-a})
$$
$$
=\frac{x}{(b-c)(x-a)(x-b)}-\frac{x}{(b-c)(x-a)(x-c)}=\frac{x}{(x-a)(x-b)(x-c)}
$$
|
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|
Evaluate $\int\frac{1}{\sin(x-a)\sin(x-b)}\,dx$ I'm stuck in solving the integral of $\dfrac{1}{\sin(x-a)\sin(x-b)}$. I "developed" the sin at denominator and then I divided it by $\cos^2x$ obtaining $$\int\frac{1}{\cos(a)\cos(b)\operatorname{tan}^2x-\cos(a)\sin(b)\operatorname{tan}x-\sin(a)\cos(b)\operatorname{tan}x+\sin(a)\sin(b)}\frac{1}{\cos^2x}dx$$
Then I made a substitution by $t=\operatorname{tan}x$ arriving to this $$\int\frac{1}{\cos(a)\cos(b)t^2-(\cos(a)\sin(b)+\sin(a)\cos(b))t+\sin(a)\sin(b)}dt$$ How can I solve it now? (probably I forgot something, it easy to make mistakes here)
Thank you in advance!
|
Looks like I am late in the race. Let me present you a slight different way to integrate it.
let $\displaystyle x = y + \frac{a+b}{2}$, and also let $\displaystyle \frac{a-b}{2} = c$
$$\int \frac{1}{ \sin\left(y - \frac{a-b}{2}\right)\sin\left(y + \frac{a-b}{2}\right)} \, dx = \int \frac{1}{\sin(y-c) \sin(y+c)} \, dy $$
Expanding this we get,
$$\int \frac{1}{\sin^2(y) \cos^2(c) - \cos^2(y)\sin^2(c)}dy = \int \frac{\csc^2 (c) \sec^2(y) }{\tan^2(y) \cot^2(c) - 1} \, dy $$
Let, $\cot(c)\tan(y) = z$, then we get,
$$\frac{1}{2 \cos(c)\sin(c)} \int \frac{1}{z^2 - 1}dz=\frac{1}{2 \cos(c)\sin(c)} \log \left( \frac{z-1}{z+1} \right ) $$
Substituting values, we get
$$\frac{1}{\sin(a-b)} \log \left( \frac{ \sin \left( -\frac{a-b}{2} + \left(x- \frac{a+b}{2}\right ) \right )}{ \sin \left( \frac{a-b}{2} + \left( x- \frac{a+b}{2}\right ) \right )} \right ) = \color{red}{\frac{1}{\sin(a-b)} \log \left( \frac{\sin(x-a)}{\sin(x-b)} \right ) + \mathrm{constant}}$$
|
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|
Are there nice ways to solve $(2+x)^{0.25}-(2-x)^{0.25}=1$ Are there nice and elegant ways to solve this equation?
$(2+x)^{0.25}-(2-x)^{0.25}=1$
Thanks.
|
Let $t = 2 + x$, and $u = 2 - x$, then $u + t = 4$, and $t^{1/4} - u^{1/4} = 1$. Again let $p = t^{1/4}$, and $q = u^{1/4}$, then $p - q = 1$, and $p^4 + q^4 = 4$. So $(q+1)^4 + q^4 = 4$ and this gives: $2q^4 + 4q^3 + 6q^2 + 4q - 3 = 0$. Solving this we have:
$q = \dfrac{-1 + \sqrt{2\sqrt{10} - 3}}{2}$, and $u = q^4 = \dfrac{\left(-1 + \sqrt{2\sqrt{10} - 3}\right)^4}{16}$. So $x = 2 - u = 2 - \dfrac{\left(-1 + \sqrt{2\sqrt{10} - 3}\right)^4}{16}$
|
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|
How do you show the limit of a function? Using the epsilon-delta definition of a limit, how would you show that $\lim_{x \to 0} \frac{3+7x}{5+5x} = \frac{3}{5}$? The lecturer's definition and solution are nothing to shout about...
|
By definition, for a given $\epsilon > 0$, you want to $\delta > 0$ such that if $|x| < \delta$, then
$$ \left| \frac{ 3 + 7x}{5+5x} - \frac{3}{5} \right| < \epsilon $$
We want to find a bound for $\left| \frac{ 3 + 7x}{5+5x} - \frac{3}{5} \right|$. Lets see
$$ \left| \frac{ 3 + 7x}{5+5x} - \frac{3}{5} \right| = \left| \frac{ 3 + 7x - 3x -3}{5(x+1)} \right| = \left| \frac{3x}{5(x+1)} \right| = \frac{ 3|x|}{5|x+1|} $$
Notice we want to control the $|x+1|$ term in the denominator. To do so, let us assume a priori that $\delta < \frac{1}{2} $. Then
$$ |x| < \frac{1}{2} \iff - \frac{1}{2} < x < \frac{1}{2} \iff \frac{1}{2} < x+1 < \frac{3}{2} \iff 2 > \frac{1}{x+1} > \frac{2}{3}$$
Therefore,
$$ \frac{ 3|x|}{5|x+1|} < \frac{ 2\cdot3 |x|}{5} = \frac{6}{5} |x|$$
and the last quantity is less than $\epsilon$ precisely if we choose $\delta = \frac{5}{6} \epsilon $. Since we assumed a priori that $\delta < \frac{1}{2}$, then the choice
$$ \delta = \min\left\{ \frac{1}{2}, \frac{5}{6} \epsilon \right\} $$
will work, and the proof is complete.
|
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|
Diophantine equations problem/exercise 3 Find all the pythagorean tripples (x,y,z) with x=40. Well I started with the known formulas for the pythagorean tripples but got me nowhere. Or I was not able continue the thought process required. I do not have a lot of experience on Diophantine equations so I'm asking for a little bit detail help.Thank you.
|
The formula
$$
2mn, m^2 - n^2, m^2 + n^2
$$
where $m, n$ are relatively prime and of diferent parity,
and $m > n > 0$,
generates all primitive triples.
Thus, to get a triple including $x = 40$, we must solve for when
$2mn \; \mid 40$ or $m^2 - n^2 \; \mid \; 40$.
Case 1: $\boldsymbol{2mn \; \mid 40}$, i.e. $mn \; \mid \; 20$.
Since $m,n$ are relatively prime and not both odd and $m > n$, there are not too many cases
to consider.
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
m & n & 2mn & m^2 - n^2 & m^2 + n^2 & \text{Triple} \\
\hline
20 & 1 & 40 & 399 & 401 & (40, 399, 401) \\
10 & 1 & 20 & 99 & 101 & (40, 198, 202) \\
5 & 2 & 20 & 21 & 29 & (40, 42, 58) \\
5 & 4 & 40 & 9 & 41 & (40, 9, 41) \\
4 & 1 & 8 & 15 & 17 & (40, 75, 85) \\
2 & 1 & 4 & 3 & 5 & (40, 30, 50) \\
\hline
\end{array}
$$
Case 2: $\boldsymbol{m^2 - n^2 \; \mid \; 40}$
Here we factor $(m+n)(m-n) \; | \; 40$.
But since $m+n$ and $m-n$ are odd (since $m,n$ are different parity),
we must have $(m+n)(m-n) \; | \; 5$.
Thus there is just once case, $m = 3$ and $n = 2$.
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
m+n & m-n & m & n & 2mn & m^2 - n^2 & m^2 + n^2 & \text{Triple} \\
\hline
5 & 1 & 3 & 2 & 12 & 5 & 13 & (96, 40, 104) \\
\hline
\end{array}
In summary, all the triples are
$$
\begin{array}{|c|c|c|}
\hline
x & y & z \\
\hline
40 & 9 & 41 \\
40 & 30 & 50 \\
40 & 42 & 58 \\
40 & 75 & 85 \\
40 & 96 & 104 \\
40 & 198 & 202 \\
40 & 399 & 401 \\
\hline
\end{array}
$$
A quick search on the internet finds a couple
of sites
which show we didn't miss any.
|
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|
Evaluate $\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\cdots$ Evaluate
$$
\frac{2}{4}\frac{2+\sqrt{2}}{4}\frac{2+\sqrt{2+\sqrt{2}}}{4}\frac{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}{4}\cdots .
$$
First, it is clear that terms tend to $1$.
It seems that the infinity product is not 0. This is related to the post Sequence $x_{n+1}=\sqrt{x_n+a(a+1)}$.
|
Consider the corresponding finite product containing $n+1$ factors. Multiplying the last factor (with $n$ square roots) by
$$2-\underbrace{\sqrt{2+\sqrt{2+\ldots}}}_{x_n},$$
the product telescopes to $4^{-n}$.
On the other hand, since
$x_n^2-2=x_{n-1}$,
writing $x_n=2\cos\varphi_n$ we get $2\varphi_n=\varphi_{n-1}$, and therefore $\varphi_n=\frac{\varphi_1}{2^{n-1}}=\frac{\pi}{2^{n+1}}$. This allows to make an estimate
$$2-x_n=2-2\cos\varphi_n\approx \frac{\pi^2}{4}4^{-n},$$
which finally gives the answer:
$$\boxed{\displaystyle\lim =\frac{4}{\pi^2}}$$
|
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|
Find the maximum possible area of the rectangle ABZP Let ABZP be a rectangle. D and C lie on AP and BZ respectively such that ABCD is a square. X and Y lie on CD and ZP respectively such that CXYZ is a square.
Given: ar(ABCD) + ar(CXYZ) = 1
Find the maximum possible area of the rectangle ABZP.
|
Let $x=AD$ (that is, its length). Let $y=CX$.
Clearly $y<x$ and $x^2+y^2=1$. We are to maximize $f(x)=x(x+y)$. But
$$f(x)=x(x+y)=x^2+xy=x^2+x\sqrt{1-x^2}=x^2+\sqrt{x^2-x^4}$$
so
$$f'(x)=2x-\frac{x-2x^3}{\sqrt{x^2-x^4}}=2x-\frac{1-2x^2}{\sqrt{1-x^2}}$$
The value $x_0$ of $x$ that maximizes $f(x)$ satisfies $f'(x_0)=0$. Solve the equation and substitute.
|
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|
number of arrangements of $n$ zeroes and $n$ ones so that each zero has a neighbor zero and each one has a neighbor one this question is similar to a question in whitworth's choice and chance.
he gives the answer as $1 + (C(n-2,0) + C(n-3,1))^2 + (C(n-3,1) + C(n-4, 2))^2 + \ldots$
i found a combinatorial "proof" by distributing the zeroes and ones in boxes and arranged them in alternating sequence. i am not too happy with that arrangement.
i am wondering if there is a better combinatorial proof?
|
To get the generating function for the number of ways to arrange $m$ $0$s and $n$ $1$s, I generated all blocks of at least $2$ $x$s and at least $2$ $y$s, starting with an optional block of at least $2$ $y$s and ending with an optional block of at least $2$ $x$s
$$
\begin{align}
&\left(1+\frac{y^2}{1-y}\right)\left(1+\left(\frac{x^2}{1-x}\frac{y^2}{1-y}\right)+\left(\frac{x^2}{1-x}\frac{y^2}{1-y}\right)^2+\dots\right)\left(1+\frac{x^2}{1-x}\right)\\[6pt]
&=\frac{\left(1+\frac{y^2}{1-y}\right)\left(1+\frac{x^2}{1-x}\right)}{1-\frac{x^2}{1-x}\frac{y^2}{1-y}}\\[6pt]
&=\frac{(1-x+x^2)(1-y+y^2)}{1-x-y+xy-x^2y^2}\\[18pt]
&=\color{#C00000}{1}+x^2+\color{#C00000}{0xy}+y^2+x^3+y^3+x^4+\color{#C00000}{2x^2y^2}+y^4+x^5+2x^3y^2+2x^2y^3+y^5\\
&+x^6+3x^4y^2+\color{#C00000}{2x^3y^3}+3x^2y^4+y^6+\dots
\end{align}
$$
Picking out the coefficients of the $x^ny^n$ terms, I get, starting with $n=1$,
$$
0,2,2,6,14,34,84,208,518,1296,\dots
$$
The coefficient of $x^ny^m$ is
$$
\begin{align}
&\sum_{k\ge0}\overbrace{\binom{n-k-1}{n-2k}}^{\text{$k$ blocks of $x$s}}\overbrace{\binom{m-k-1}{m-2k}}^{\text{$k$ blocks of $y$s}}\overbrace{\left(2-\binom{k-1}{k}\right)}^{\text{$x$ or $y$ first if $k\ne0$}}\\
&+\sum_{k\ge0}\overbrace{\binom{n-k-2}{n-2k-2}}^{\text{$k+1$ blocks of $x$s}}\overbrace{\binom{m-k-1}{m-2k}}^{\text{$k$ blocks of $y$s}}\\
&+\sum_{k\ge0}\overbrace{\binom{n-k-1}{n-2k}}^{\text{$k$ blocks of $x$s}}\overbrace{\binom{m-k-2}{m-2k-2}}^{\text{$k+1$ blocks of $y$s}}
\end{align}
$$
Note that $\binom{n}{0}=1$ even if $n\lt0$ and $\binom{n}{k}=0$ if $k\lt0$.
Thus, the coefficient of $x^ny^n$ is
$$
\sum_{k\ge0}\binom{n-k-1}{n-2k}^2\left(2-\binom{k-1}{k}\right)+2\sum_{k\ge0}\binom{n-k-2}{n-2k-2}\binom{n-k-1}{n-2k}
$$
which, if we assume $n\gt0$, can be simplified to
$$
2\sum_{k\ge1}\binom{n-k-1}{n-2k}^2+2\sum_{k\ge1}\binom{n-k-2}{n-2k-2}\binom{n-k-1}{n-2k}
$$
|
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|
Find all solutions for $\cos(2x)\cos x-\sin(2x)\sin x=\frac{1}{\sqrt{2}}$ if $0\leq x<\pi$ Find all solutions for $\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{1}{\sqrt{2}}$ if $0\leq x< \pi$
Can you verify my work? Thanks!
$$\cos(2x)\cos(x)-\sin(2x)\sin(x)=\frac{1}{\sqrt{2}}$$
$$\cos(2x+x)=\frac{1}{\sqrt{2}}$$
$$\cos(3x)=\frac{1}{\sqrt{2}}$$
$$\cos^{-1}(\frac{1}{\sqrt{2}})$$
Reference angle: $\frac{\pi}{4}$
$$3x=\frac{\pi}{4}+2k\pi$$
$$x=\frac{\pi}{12}+\frac{2k\pi}{3}$$
$$3x=\frac{7\pi}{4}+2k\pi$$
$$x=\frac{7\pi}{12}+\frac{2k\pi}{3}$$
$$x=\frac{\pi}{12},\frac{3\pi}{4},\frac{17\pi}{12}, \frac{7\pi}{12},\frac{5\pi}{4},\frac{23\pi}{12} $$
|
Right idea, but after reducing, you'll get cos(3x) = 0 (reference circle), so you'll be left with x=π/6, x=π/2 and x=5π/6 after simplifying.
Realize that 5π/4 and 23π/12 are >π, all larger than π will not be included with your solution.
EDIT: SEE COMMENT BELOW.
|
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|
How to integrate $\sqrt{1-\sin 2x}$? I want to solve the following integral without substitution:
$$\int{\sqrt{1-\sin2x}} \space dx$$
I have:
$$\int{\sqrt{1-\sin2x}} \space dx = \int{\sqrt{1-2\sin x\cos x}} \space dx = \int{\sqrt{\sin^2x + \cos^2x -2\sin x\cos x}} \space dx$$
but this can be written in two ways: $\int{\sqrt{(\sin x - \cos x)^2}} \space dx$ or $\int{\sqrt{(\cos x - \sin x)^2} \space dx}$
and it seems to be pretty far from what the real result is: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ecvsatkj0te&mail=1
Can I get any hints?
EDIT:
Thank you for your answers! So as you all showed, we have:
$\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\sin x - \cos x)^2}} \space dx$
or
$\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\cos x - \sin x)^2} \space dx} = \int{\sqrt{(-(\sin x - \cos x))^2} \space dx}$
combined:
$\int{\sqrt{1-\sin2x}} \space dx = \cdots = \int{\sqrt{(\pm(\sin x - \cos x))^2} \space dx}$
so we actually have:
$$\int{|\sin x - \cos x|} \space dx$$
I drew myself a trigonometric circle and if I concluded correctly, we have:
1. $\int{\sin x - \cos x} \space dx$ for $x \in [ \frac{\pi}{4} + 2k\pi , \frac{5\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$
2. $\int{\cos x - \sin x} \space dx$ for $x \in [ -\frac{3\pi}{4} + 2k\pi , \frac{\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$
... which means:
1. $x \in [ \frac{\pi}{4} + 2k\pi , \frac{5\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$ :
$$\int{\sqrt{1-\sin2x}} \space dx = \int{\sin x - \cos x} \space dx = -\cos x - \sin x + C$$
2. $x \in [ -\frac{3\pi}{4} + 2k\pi , \frac{\pi}{4} + 2k\pi ] , \space k \in \mathbb{Z}$ :
$$\int{\sqrt{1-\sin2x}} \space dx = \int{\cos x - \sin x} \space dx = \sin x + \cos x + C$$
|
hint: $1- \sin (2x) = (\sin x - \cos x)^2$
|
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|
Solving $315 x \equiv 5 \pmod {11}$ I have to solve this: $$315 x \equiv 5 \pmod {11}$$
Isn't it like that?
$$315 \equiv (22+8) \cdot 10+15 \equiv 8 \cdot 3+4 \equiv 5+8+4 \equiv 6$$
Or have I done something wrong?
|
$315\equiv 7\pmod{11}\\7x\equiv5\pmod{11}\\-4x\equiv-6\pmod{11}\\2x\equiv3\pmod{11}\\2x\equiv-8\pmod{11}\\x\equiv-4\pmod{11}\\x\equiv7\pmod{11}$
|
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|
Solving a non-exact differential I started off solving the differential equation $$(xy^2 + 3e^{x-3})dx - x^2ydy = 0$$
It's a non-exact first order equation whose integrating factor is $1/x^4$.
Finally I got to the equation where I needed to integrate $$\frac{e^{x-3}}{x^4} dx$$ But I can't seem to proceed.
Is there any other way to solve this.
|
First, we rearrange the given equation (which is a Bernoulli equation):
$$
x^2y'y - xy^2 = 3e^{x-3} \Rightarrow 2y'y - \frac{2y^2}{x} = \frac{6e^{x-3}}{x^2}.
$$
Notice that we can simplify this result by letting $z(x) := y^2(x)$:
$$
z' - \frac{2z}{x} = \frac{6e^{x-3}}{x^2}.\tag{1}\label{eq:z}
$$
Let $\mu(x) := e^{\int -2/x\mathrm{d}x} = 1/x^2$ the integrating factor of \eqref{eq:z}. Multiplying both sides of this equation by $\mu(x)$,
$$
\frac{z'}{x^2} - \frac{2z}{x^3} = \frac{z'}{x^2} - z\frac{\mathrm{d}\phantom{x}}{\mathrm{d}x}\left(\frac{1}{x^2}\right) = \frac{\mathrm{d}\phantom{x}}{\mathrm{d}x}\left(\frac{z}{x^2}\right) = \frac{6e^{x-3}}{x^4}.
$$
Integrating both sides by $x$,
$$
\frac{z}{x^2} = \int \frac{6e^{x-3}}{x^4} \mathrm{d}x = \frac{\mathrm{Ei}(x)}{e^3} -\frac{e^{x-3}(x^2+x+2)}{x^3} + c_1,
$$
so
$$
z(x) = \frac{x^2\mathrm{Ei}(x)}{e^3} -\frac{e^{x-3}(x^2+x+2)}{x} + c_1x^2 \Rightarrow \boxed{y(x) = \pm \sqrt{\frac{x^2\mathrm{Ei}(x)}{e^3} -\frac{e^{x-3}(x^2+x+2)}{x} + c_1x^2}.}
$$
Here is a plot of the solution setting $10 \le z(1) \le 200$ for fixing a value of $c_1$.
|
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|
Roots of polynomial $x^3-3\sqrt 5x^2+13x-3\sqrt 5$ given the factor $x-\sqrt 5$
Given that $x-\sqrt 5 $ is a factor of the cubic polynomial
$x^3-3\sqrt 5x^2+13x-3\sqrt 5$, find all the values of
the polynomial
After the long division method I get $x^2-2\sqrt 5x+3$.
Now how to split the middle term to find the all zeros of the polynomial?
|
As an alternate method, you can use Vieta's formulas: if $a, b, c$ are the three roots, you have
$$\begin{eqnarray}
(i) & a+b+c&=&3\sqrt{5}\\
(ii) & ab+bc+ca&=&13\\
(iii) & abc&=&3\sqrt{5}\end{eqnarray}$$
One of the roots, say $c$, is $\sqrt{5}$, so
*
*from $(i)$, $a+b=2\sqrt{5}$,
*from $(iii)$, $ab=3$,
So
$$(a-b)^2=(a+b)^2-4ab=20-12=8$$
Then from $a+b=2\sqrt{5}$ and $a-b=2\sqrt{2}$, you have immadiately
$$a=\sqrt{5}+\sqrt{2}\\
b=\sqrt{5}-\sqrt{2}$$
|
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|
Another Inequality Let $a$, $b$ and $c$ be positive reals such that $$\dfrac{a^3+2a}{a^2+1}+\dfrac{b^3+2b}{b^2+1}+\dfrac{c^3+2c}{c^2+1}=\dfrac 92,$$
then is it true that $\dfrac 1a+\dfrac1b+\dfrac1c\ge3$ ?
|
$$\sum_{cyc}\frac{1}{a}-3=\sum_{cyc}\left(\frac{1}{a}-1\right)=\sum_{cyc}\left(\frac{1}{a}-1+\frac{a^3+2a}{a^2+1}-\frac{3}{2}\right)=$$
$$=\sum_{cyc}\frac{(a-1)^2(2a^2-a+2)}{2a(a^2+1)}\geq0.$$
Done!
|
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|
Maximum and Minimum value of an inverse function
Find the maximum and minimum value of
$\arcsin \left(x\right)^3+\arccos \left(x\right)^3$.
given that $-1\le x\le 1$
I have solved the problem but i am just curious to know if there are any other ways to solve this particular problem other than the method i used below.
By using the fact that $\arcsin \left(x\right)+\arccos \left(x\right)$ =$\frac{\pi }{2}$
i found that
$\arcsin \left(x\right)^3+\arccos \left(x\right)^3$=$3\left(\frac{\pi }{2}\right)^2\left(\left\{\arcsin \left(x\right)-\frac{\pi }{4}\right\}^2+\frac{\left(8\pi -3\pi ^2\right)}{48}\right)$
so it is minimum when $\left\{\arcsin \left(x\right)-\frac{\pi }{4}\right\}^2$$=0$
or $x=\sin \left(\frac{\pi }{4}\right)$
Therefore minimum value$=\frac{1}{32}\pi ^3$
and it is maximum when $\arcsin \left(x\right)=-\frac{\pi }{2}$
Therefore maximum value=$\frac{7}{8}\pi ^3$
|
Over the interval $[-1,1]$ the functions $\arcsin(x)$ and $\arccos(x)$ have a constant sum, $\frac{\pi}{2}$.
It follows that the problem is equivalent to finding the minimum and maximum value of
$$ g(t) = t^3+\left(\frac{\pi}{2}-t\right)^3 $$
over the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. By setting $t=\frac{\pi}{2}u$ this boils down to finding the minimum and maximum value of $h(u)=u^3+(1-u)^3 = 1-3u+3u^2$ over the interval $[-1,1]$. That is straightforward: the minimum lies at $u=\frac{1}{2}$ and the maximum lies at $u=-1$. By performing the inverse substitutions, the maximum of the original function is $\frac{7\pi^3}{8}$, attained at $x=-1$, and the minimum is $\frac{\pi^3}{32}$, attained at $x=\frac{1}{\sqrt{2}}$.
|
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|
Find the locus of $2/z$ given that $|z-(1+i)| = 2$
If complex numbers $z$ satisfy the equation $|z-(1+i)| = 2$ and $\displaystyle \omega = \frac{2}{z}$, then locus traced by $\omega$ in complex plane, is ...
My try
I want to solve it geometrically. Here $|z-(1+i)| = 2$ Represent a Circle whose center is at $(1,1)$ and Radius $=2$.
So $z$ lies on a given circle.
But I did not understand how can we find locus of $\displaystyle \omega = \frac{2}{z}$
|
Some ideas:
$$|z-(1+i)|=2\iff|z|\left|1-\frac{1+i}z\right|=2\iff|w|=\left|\frac2z\right|=\left|1-\frac{1+i}z\right|$$
If $\;z=x+iy\;$ , then
$$\frac{1+i}z=\frac{(1+i)\overline z}{|z|^2}=\frac{x+y}2+\frac{x-y}2i\implies 1-\frac{1+i}z=\frac{2-(x+y)}2-\frac{x-y}2i$$
But
$$2=|z-(1+i)|=|(x-1)+(y-1)i|\implies (x-1)^2+(y-1)^2=4$$
whereas
$$|w|^2=\frac14\left(4-4x-4y+x^2+2xy+y^2+x^2-2xy+y^2\right)=$$
$$=\frac12\left(x^2+y^2-2x-2y+2\right)=\frac12\left((x-1)^2+(y-1)^2\right)=2\implies$$
$$\color{red}{|w|=\sqrt2}$$
|
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|
Why does $\frac{4}{2} = \frac{2}{1}$? I take for granted that $\frac{4}{2} = \frac{2}{1}$.
Today, I thought about why it must be the case. My best answers amounted to $\frac{4}{2}=2$ and $\frac{2}{1}=2$; therefore $\frac{4}{2}=\frac{2}{1}$. However, that explanation seems circular:
*
*one can express $2$ as $\frac{2}{1}$.
*As such, to say $\frac{4}{2}$ equals $\frac{2}{1}$ because both equal $2$, is nearly saying $\frac{4}{2} = \frac{2}{1}$ (the question) and $\frac{2}{1}=\frac{2}{1}$ (trivial, at best).
So why does $\frac{4}{2} = \frac{2}{1}$?
|
If $c \neq 0$, do you agree that $\frac{c}{c} = 1$? Do you also agree that 1 is a unit and a multiplicative identity element?
Given any rational number $\frac{a}{b}$, if we multiply both the numerator and the denominator by the same number, we're computing $\frac{a}{b} \times \frac{c}{c} = \frac{a}{b} \times 1$. So, if we have $c = 2$, and we compute $\frac{2}{1} \times \frac{c}{c}$, we have $\frac{2}{1} \times \frac{2}{2} = \frac{4}{2} = \frac{2}{1} \times 1$.
|
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|
Finding a Jacobian determinant of transformation I want to find the Jacobian determinant of the transformation:
$$\begin{align}
\left(\begin{array}\\x\\y\\z\end{array}\right) = \left(\begin{array}\\\frac23 & \frac{-2}3 & \frac13\\\frac23 & \frac{-2}3 & \frac{-1}3\\\frac23 & \frac23 & \frac13 \end{array}\right) \left(\begin{array}\\u\\v\\w\end{array}\right)
\end{align}$$
What exactly does it want me to do? What does it mean by find the Jacobian determinant?
Note: Some of the values have been changed to not be cheating potentially, I hope that doesn't make the question impossible. I did change the values evenly though(take one off the top and give it to another for example).
|
$\def\d{\partial}$
To find the determinant of the transformation matrix, we need to complete the Jacobian matrix:
$$\begin{align}
\left(\begin{array}\\\frac{\d x}{\d u} & \frac{\d x}{\d v} & \frac{\d x}{\d w}\\\frac{\d y}{\d u} & \frac{\d y}{\d v} & \frac{\d y}{\d w}\\\frac{\d z}{\d u} & \frac{\d z}{\d v} & \frac{\d z}{\d w} \end{array}\right)
\end{align}$$
For:
$$ {\bf A}=\begin{align}
\left(\begin{array}\\x\\y\\z\end{array}\right) = \left(\begin{array}\\\frac23 & \frac{-2}3 & \frac13\\\frac23 & \frac{-2}3 & \frac{-1}3\\\frac23 & \frac23 & \frac13 \end{array}\right) \left(\begin{array}\\u\\v\\w\end{array}\right)
\end{align}$$
$$\begin{align}
\left(\begin{array}\\x\\y\\z \end{array}\right)=\left(\begin{array}\\\frac{2u}3- \frac{2v}3+ \frac{w}3\\\frac{2u}3 - \frac{2v}3 - \frac{w}3\\\frac{2u}3 +\frac{2v}3 +\frac{w}3 \end{array}\right)
\end{align}$$
Since the above is linear, and already constructed in $(u,v,w)$ order, the original matrix is indeed the Jacobian Determinant. Note that $(u,v,w)$ is transformed by this matrix.
$$\begin{align}
\left(\begin{array}\\\frac{\d x}{\d u} & \frac{\d x}{\d v} & \frac{\d x}{\d w}\\\frac{\d y}{\d u} & \frac{\d y}{\d v} & \frac{\d y}{\d w}\\\frac{\d z}{\d u} & \frac{\d z}{\d v} & \frac{\d z}{\d w} \end{array}\right) = \left(\begin{array}\\\frac23 & \frac{-2}3 & \frac13\\\frac23 & \frac{-2}3 & \frac{-1}3\\\frac23 & \frac23 & \frac13 \end{array}\right)
\end{align}$$
|
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|
How to find all $m,n$ such that $mn|m^2+n^2+1$? How to find all positive integers $m,n$ such that $m|n^2+1$ and $n|m^2+1$ ?
My work:- Let $n^2+1=mk , m^2+1=ng$ , then $mkg=gn^2+g=m^2n+n+g$ , hence
$m|n+g$ i.e. $m|n+\dfrac{m^2+1}n$ i.e. $\dfrac{n^2+m^2+1}{mn}$ is an integer i.e. $mn|m^2+n^2+1$ . Now conversely
assume $mn|m^2+n^2+1$ , then $m|m^2+n^2+1$ and $n|m^2+n^2+1$ i.e. $m|n^2+1$ and $n|m^2+1$
So the question equivalently asks to positive integers $m,n$ such that $mn|m^2+n^2+1$
|
Write a formula and although it is possible to draw some more.
Solutions of the equation: $\frac{X^2+aX+Y^2+bY+c}{XY}=j$
If a root. $t=\sqrt{(b+a)^2+4c(j-2)}$
Then using the equation Pell: $p^2-(j^2-4)s^2=1$
Solutions can be written:
$X=\frac{(b+a\pm{t})}{2(j-2)}p^2+(t\mp{(b-a)})ps-\frac{(b(3j-2)+a(6-j)\mp{(j+2)t})}{2}s^2$
$Y=\frac{(b+a\pm{t})}{2(j-2)}p^2+(-t\mp{(b-a)})ps-\frac{(b(6-j)+a(3j-2)\mp{(j+2)t})}{2}s^2$
Please note that the number $p,s$ Can be any characters.
|
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|
Integral $\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}$ I am trying to prove this interesting integral
$$
I:=\int_0^\infty \log \frac{1+x^3}{x^3} \frac{x \,dx}{1+x^3}=\frac{\pi}{\sqrt 3}\log 3-\frac{\pi^2}{9}.
$$
I tried using $y=1+x^3$ but that didn't help.
We can possibly try
$$
I=\int_0^\infty \frac{\log(1+x^3) x}{1+x^3} \,dx-\int_0^\infty \frac{\log(x^3) x}{1+x^3}\,dx.
$$ These integrals would be much easier had the bounds been from $0 $\ to $\infty$, however they are not. Perhaps partial integration will work but I didn't find the way if we try
$$
dv=\frac{x}{1+x^3}, \quad u= \log(1+x^3)
$$
but I ran into a divergent integral. Thanks how can we prove I?
|
Let us make the change of variables
$$v=\frac{x^3}{1+x^3}\iff x=\left(\frac{v}{1-v}\right)^{1/3}$$
This transforms the integral $I$ to the following form
$$
I=-\frac{1}{3}\int_0^1\log(v)\,v^{-1/3}(1-v)^{-2/3}dv
$$
Now, If
$$f(\alpha):=B(\alpha,\frac{1}{3})=\int_0^1v^{\alpha-1}(1-v)^{\frac{1}{3}-1}dv=\frac{\Gamma(\alpha)\Gamma(\frac{1}{3})}{\Gamma(\alpha+\frac{1}{3})}$$
then $I=-\frac{1}{3}f'(\frac{2}{3})$.
But, since $\Gamma(1)=1$, and $\Gamma'(1)=-\gamma$, we have
$$\eqalign{
f'\left(\frac{2}{3}\right)&=\Gamma'\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)-\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)\Gamma'(1)\cr
&=\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{1}{3}\right)\left(\psi\left(\frac{2}{3}\right)+\gamma\right)\cr
&\buildrel{\rm(1)}\over{=}\frac{\pi}{\sin(\pi/3)}\left(\psi\left(\frac{2}{3}\right)+\gamma\right)\cr
&\buildrel{\rm(2)}\over{=}\frac{2\pi}{\sqrt{3}}\left(
\frac{\pi}{2\sqrt{3}}-\frac{3}{2}\log 3\right)
}
$$
Where, for $(1)$ we used the Euler's reflection formula, and for $(2)$ we used Gauss' theorem for the digamma function. Combining our results we get
$$
I=-\frac{1}{3}\frac{2\pi}{\sqrt{3}}\left(
\frac{\pi}{2\sqrt{3}}-\frac{3}{2}\log 3\right)=\frac{\pi}{\sqrt{3}}\log 3-\frac{\pi^2}{9}.
$$
which is the desired result.$\qquad\square$
|
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|
How prove $((x-y)(y-z)(z-x))^2\le 2((x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2)$ let $x,y,z>0$, and such $$x^2+y^2+z^2=x^2y^2+y^2z^2+x^2z^2$$
show that
$$((x-y)(y-z)(z-x))^2\le 2((x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2)$$
My try: let
$$x-y=a,y-z=b,z-x=c\Longrightarrow a+b+c=0$$
then we only prove
$$2[a(a+b)^2+b(b+c)^2+c(c+a)^2]\ge a^2b^2c^2$$
or
$$2(ac^2+ba^2+cb^2)\ge a^2b^2c^2$$
then I can't
|
Indeed, we can prove the following sharper inequality holds
$$2\big[(x-y)(y-z)(z-x)\big]^{2}\leq (x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(z^{2}-x^{2})^{2}$$
Without loss of generality, assume that $x \geq y \geq z > 0,$ then
\begin{aligned} 2\big[(x-y)(y-z)(z-x)\big]^{2} \ & \leq 2(x-y)^2x^2y^2 \leq 2(x-y)^2(x^2y^2+y^2z^2+z^2x^2) \\ & =2(x-y)^2(x^2+y^2+z^2) \leq 2(x-y)^2(x^2+y^2+2xy) \\ & =2(x^2-y^2)^2 \leq (x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2.\end{aligned}
|
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|
Integral involving the spherical Bessel function of the first kind: $\int_{0}^{\infty} x^{-n}j_{n+1}(x) \, dx$ Let $j_{n}(x)$ be the spherical Bessel function of the first kind of nonnegative integer order $n$.
How can I prove the equation below using the spherical Bessel function recurrence relation?
$$ \int_0^\infty x^{-n} j_{n+1}(x) \, dx =
\frac1{(2n+1)(2n-1)(2n-3)\cdots3\cdot1} = \frac1{(2n+1)!!} \tag{1}$$
We can express $j_{n}(x)$ in terms of the Bessel function of the first kind:
$$ j_n(x)=\sqrt\frac{\pi}{2x} J_{n+\frac12}(x) $$
And Rayleigh's formula gives us a way to express $j_{n}(x)$ in terms of elementary functions:
$$j_n(x) = (-x)^n\left(\frac{1}{x}\frac{d}{dx}\right)^n\frac{\sin(x)}{x}$$
(See HERE for an explanation of the formula.)
But I don't know if these facts help us prove $(1)$.
EDIT:
The OP is not an active user, but I think he or she was referring to the recurrence relation $$\frac{d}{dx} \left(x^{-n}j_{n}(x) \right)=-x^{-n}j_{n+1}(x),\tag{1} $$ which reduces the evaluation to calculating a limit.
|
EDIT:
I think the OP was referring to the recurrence relation $$\frac{d}{dx} \left(x^{-n}j_{n}(x) \right)=-x^{-n}j_{n+1}(x). \tag{1} $$
From $(1)$ it follows immediately that $$\begin{align}\int_{0}^{\infty}x^{-n} j_{n+1}(x) \, dx &= -x^{-n} j_{n}(x) \Bigg|^{\infty}_{0} \\ &=-x^{n} \sqrt{\frac{\pi}{2x}} \, J_{n+\frac{1}{2}}(x) \Bigg|^{\infty}_{0} \\ &= \lim_{x \downarrow 0} \, x^{-n} \sqrt{\frac{\pi}{2x}} \, J_{n+\frac{1}{2}}(x) \tag{2} \\ &=\lim_{x \downarrow 0} \, x^{-n} \sqrt{\frac{\pi}{2x}}\left(\frac{1}{\Gamma\left(n + \frac{3}{2}\right)} \left(\frac{x}{2}\right)^{n+ \frac{1}{2}} +\mathcal{O}\left(x^{n+ \frac{5}{2}} \right)\right) \tag{3} \\ &= \sqrt{\pi} \, 2^{-n-1} \, \frac{1}{\Gamma \left(n+ \frac{3}{2}\right)} \\ &= \sqrt{\pi} \, 2^{-n-1} \, \frac{2^{n+1}}{\sqrt{\pi} (2n+1)!!} \tag{4}\\ &= \frac{1}{(2n+1)!!}. \end{align}$$
$(1)$: http://dlmf.nist.gov/10.51#E3
$(2)$: https://en.wikipedia.org/wiki/Bessel_function#Asymptotic_forms
$(3)$: https://en.wikipedia.org/wiki/Bessel_function#Bessel_functions_of_the_first_kind:_J.CE.B1
$(4)$: http://mathworld.wolfram.com/DoubleFactorial.html (2)
This might not be the particular approach you were seeking, but it can be evaluated using a property of the Mellin transform.
Ramanujan's master theorem states that if $f(x)$ has an expansion of the form $$ f(x) = \sum_{k=0}^{\infty} \frac{\phi(k)}{k!} (-x)^{k} ,$$
then
$$ \int_{0}^{\infty} x^{s-1} f(x) \, dx = \Gamma(s) \phi(-s) $$
for the values of $s$ for which the integral converges.
The hypergeometric representation of the Bessel function of the first kind of order $\alpha$ is $$ J_{\alpha}(x) = \frac{(\frac{x}{2})^{\alpha}}{\Gamma(\alpha +1)} \ {}_0F_{1} \left(\alpha +1; - \frac{x^{2}}{4} \right) = \frac{(\frac{x}{2})^{\alpha}}{\Gamma(\alpha +1)} \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 1)}{\Gamma(\alpha +1+k)} \frac{(-\frac{x^{2}}{4})^{k}}{k!} .$$
So for $s+n+1 >0$ and $s <2$, we have
$$ \begin{align} \int_{0}^{\infty} x^{s-1} j_{n+1}(x) \, dx &= \int_{0}^{\infty} x^{s-1} \sqrt{\frac{\pi}{2x}} \, J_{n+\frac{3}{2}} (x) \, dx \\ &= \int_{0}^{\infty} x^{s-1} \sqrt{\frac{\pi}{2x}} \frac{(\frac{x}{2})^{n + \frac{3}{2}}}{\Gamma(n +\frac{5}{2})} \, {}_0F_{1} \left(n + \frac{5}{2}; - \frac{x^{2}}{4} \right) \, dx \\ &= \frac{\sqrt{\pi}}{2^{n+2}} \frac{1}{\Gamma(n+\frac{5}{2})} \int_{0}^{\infty} x^{s+n} \, {}_0F_{1} \left(n + \frac{5}{2}; - \frac{x^{2}}{4} \right) \, dx \\ &= \frac{\sqrt{\pi}}{2^{n+2}} \frac{1}{\Gamma(n+\frac{5}{2})} \int_{0}^{\infty} (2 \sqrt{u})^{s+n} \, {}_0F_{1} \left(n + \frac{5}{2}; - u \right) \, \frac{du}{\sqrt{u}} \\ &= \frac{ \sqrt{\pi} \ 2^{s-2}}{\Gamma(n+\frac{5}{2})} \int_{0}^{\infty} u^{\frac{s+n+1}{2}-1} \, {}_0F_{1} \left(n + \frac{5}{2}; - u \right) \, du \\ &= \frac{ \sqrt{\pi} \ 2^{s-2}}{\Gamma(n+\frac{5}{2})} \, \Gamma \left(\frac{s+n+1}{2} \right) \frac{\Gamma(n+ \frac{5}{2})}{\Gamma(n + \frac{5}{2} - \frac{s+n+1}{2})} \\ &= \sqrt{\pi} \ 2^{s-2} \, \Gamma \left(\frac{s+n+1}{2} \right) \frac{1}{\Gamma(\frac{4-s+n}{2})} . \end{align}$$
Now let $s= -n+1$.
Then
$$ \begin{align} \int_{0}^{\infty} x^{-n} j_{n+1}(x) \ dx &= \sqrt{\pi} \ 2^{-n-1} \Gamma (1) \, \frac{1}{\Gamma \left(n + \frac{3}{2} \right)} \\ &= \sqrt{\pi} \ 2^{-n-1} \frac{2^{n+1}}{\sqrt{\pi} (2n+1)!!} \tag{1} \\ &= \frac{1}{(2n+1)!!} \end{align}$$
$(1)$: http://mathworld.wolfram.com/DoubleFactorial.html (2)
|
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|
A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$ How to prove the following
$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$
I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are not best ideas to approach the problem.
Any other ideas ?
|
Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{\ln\left( \frac{1+x}{1-x} \right)}{x \sqrt{1-x^{2}}} \ dx
\end{align}
when the transformation $x = \tanh(t)$ is made. The resulting integral is given by
\begin{align}
I &= \int_{0}^{\infty} \ln\left( \frac{1+\tanh(t)}{1-\tanh(t)}\right) \frac{\cosh^{2}(t)}{\sinh(t)} \ dt \\
&= \int_{0}^{\infty} \ln\left( \frac{e^{t}}{e^{-t}}\right) \frac{\cosh^{2}(t)}{\sinh(t)} \ dt \\
&= 2 \int_{0}^{\infty} \frac{t \cosh^{2}(t)}{\sinh(t)} \ dt \\
&= 2 \left( \frac{\pi^{2}}{4} \right)
\end{align}
which yields
\begin{align}
\int_{0}^{1} \frac{\ln\left( \frac{1+x}{1-x} \right)}{x \sqrt{1-x^{2}}} \ dx = \frac{\pi^{2}}{2}.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/805893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 10,
"answer_id": 0
}
|
How find this sum $ \frac{1}{1999}\binom{1999}{0}-\frac{1}{1998}\binom{1998}{1}+\cdots-\frac{1}{1000}\binom{1000}{999}$
prove or disprove :
$$S=\dfrac{1}{1999}\binom{1999}{0}-\dfrac{1}{1998}\binom{1998}{1}+\dfrac{1}{1997}\binom{1997}{2}-\dfrac{1}{1996}\binom{1996}{3}+\cdots-\dfrac{1}{1000}\binom{1000}{999}=\dfrac{1}{1999}\left(w^{1999}_{1}+w^{1999}_{2}\right)$$
where
$$w_{1}=\dfrac{1+\sqrt{3}i}{2},w_{2}=\dfrac{1-\sqrt{3}i}{2}$$
My idea: since
$$\dfrac{1}{1999-k}\binom{1999-k}{k}=\dfrac{1}{1999-k}\dfrac{(1999-k)!}{k!(1999-2k)!}=\dfrac{(1998-k)!}{k!(1999-2k)!}$$
Then I can't,maybe can use integral deal it
Thank you
|
This sum can also be done using Wilf / generatingfunctionology. Let
$$a_n = \sum_{k=0}^{\lfloor n/2 \rfloor}
\frac{(-1)^k}{n-k} {n-k \choose k}$$
and introduce the generating function
$$f(z) = \sum_{n\ge 1} a_n z^n
= \sum_{n\ge 1}
\sum_{k=0}^{\lfloor n/2 \rfloor}
\frac{(-1)^k}{n-k} {n-k \choose k} z^n
\\ =
\sum_{n\ge 1} \frac{z^n}{n} +
\sum_{n\ge 1}
\sum_{k=1}^{\lfloor n/2 \rfloor}
\frac{(-1)^k}{n-k} {n-k \choose k} z^n.$$
Switch the order of summation in the inner sum to obtain
$$\sum_{k\ge 1} \sum_{n\ge 2k}
\frac{(-1)^k}{n-k} {n-k \choose k} z^n
= \sum_{k\ge 1} \sum_{n\ge 0}
\frac{(-1)^k}{n+k} {n+k \choose k} z^{n+2k}
\\= \sum_{k\ge 1} z^{2k} \sum_{n\ge 0}
\frac{(-1)^k}{n+k} {n+k \choose k} z^n
= \sum_{k\ge 1} z^{2k} \sum_{n\ge 0}
\frac{(-1)^k}{k} {n+k-1 \choose k-1} z^n
\\ = \sum_{k\ge 1} \frac{(-1)^k}{k} z^{2k} \sum_{n\ge 0}
{n+k-1 \choose k-1} z^n
= \sum_{k\ge 1} \frac{(-1)^k}{k} z^{2k} \frac{1}{(1-z)^k}
\\ = \log\frac{1}{1 + \frac{z^2}{1-z}}.$$
This gives the following closed form for $f(z):$
$$f(z) = \log\frac{1}{1-z} +
\log\frac{1}{1 + \frac{z^2}{1-z}}
= \log\frac{1}{1-z+z^2}.$$
Finally put
$$w_{1,2} = \frac{1\pm i\sqrt{3}}{2}$$
so that $$1-z+z^2
= (1-w_1 z)(1-w_2 z).$$
This yields
$$f(z) = \log\frac{1}{1-z+z^2} =
\log\frac{1}{1-w_1 z} + \log\frac{1}{1-w_2 z}$$
so that
$$[z^n] f(z) = \frac{w_1^n}{n} + \frac{w_2^n}{n},$$
as obtained by @r9m in the comment.
Remark. Here we have made repeated use of the identity
$$\log\frac{1}{1-z} = \sum_{n\ge 1} \frac{z^n}{n}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/806590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
}
|
How to prove this inequality $\frac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\frac{\pi}{n+1}}$
Let $a_{1},a_{2},\cdots,a_{n},n\ge 2$ be real numbers,show that
$$\dfrac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\dfrac{\pi}{n+1}}$$
I think this result is interesting.
When $n=2$, clearly
$$\dfrac{a_{1}a_{2}}{a^2_{1}+a^2_{2}}\le\dfrac{1}{2}=\cos{\dfrac{\pi}{3}}=\dfrac{1}{2}$$
When $n=3$,
$$\dfrac{a_{1}a_{2}+a_{2}a_{3}}{a^2_{1}+a^2_{2}+a^2_{3}}\le\dfrac{\sqrt{2}}{2}.$$
This is true, because
$$a^2_{1}+\dfrac{1}{2}a^2_{2}\ge \sqrt{2}a_{1}a_{2},$$
$$\dfrac{1}{2}a^2_{2}+a^2_{3}\ge\sqrt{2}a_{2}a_{3}.$$
But for general $n$ I cannot prove it.
|
For $t \in (0,\frac{\pi}{n})$, we have $\sin kt > 0$, for $k \in 1,\cdots,n-1$.
Thus from AM-GM inequality, $\dfrac{\sin (k+1)t}{\sin kt}a_k^2+\dfrac{\sin kt}{\sin (k+1)t}a_{k+1}^2 \ge 2a_ka_{k+1}$,
with equality iff $a_{k+1}\sin kt = a_k\sin (k+1)t$.
Summing over we have, $\displaystyle \sum\limits_{k=1}^{n-1} \left(\dfrac{\sin (k+1)t}{\sin kt}a_k^2+\dfrac{\sin kt}{\sin (k+1)t}a_{k+1}^2\right) \ge 2\sum\limits_{k=1}^{n-1}a_ka_{k+1}$
$\displaystyle \implies 2a_1^2\cos t + \sum\limits_{k=2}^{n} \left(\dfrac{\sin (k-1)t}{\sin kt}+\dfrac{\sin (k+1)t}{\sin kt}\right)a_k^2 \ge \dfrac{\sin (n+1)t}{\sin nt}a_n^2 + 2\sum\limits_{k=1}^{n-1}a_ka_{k+1}$
$\displaystyle \implies 2\cos t\sum\limits_{k=1}^{n}a_k^2 \ge \dfrac{\sin (n+1)t}{\sin nt}a_n^2 + 2\sum\limits_{k=1}^{n-1}a_ka_{k+1}$
Putting, $t = \dfrac{\pi}{n+1}$, $\sin (n+1)t = 0$.
Therefore, $\displaystyle \cos \dfrac{\pi}{n+1}\sum\limits_{k=1}^{n}a_k^2 \ge \sum\limits_{k=1}^{n-1}a_ka_{k+1}$ as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/807326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 1
}
|
Measure of an Interior Angle Triangle $ABC$ has $AC = BC$ , $\angle ACB = 96^\circ$ . $D$ is a point in $ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$ . What is the measure (in degrees) of $\angle ACD$ ?
|
If we put $AB=1$ and apply the low of sines, we get
$$AC=\frac{\sin 42^\circ}{\sin 96^\circ}=\frac{\sin 42^\circ}{\cos 6^\circ},
\\ AD = \frac{\sin 30^\circ}{\sin 132^\circ}=\frac{1}{2\cos 42^\circ}.$$
But then it follows that $AC=AD$ from $2 \sin 42^\circ \cos 42^\circ=\sin 84^\circ = \cos 6^\circ$.
So $\triangle ADC$ is isosceles with the angle $24^\circ$ at $A$, which means the two equal angles are each $78^\circ$, and one of these is the desired $\angle ACD$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/808352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
how to calculate this double integral I have this double integral:
$$ \int_0^\frac{\pi}{4}\int_0^\frac{\pi}{2}(\cos x + \sin y) \, dy \, dx $$
Is this correct?
$$\begin{align*} \int_0^\frac{\pi}{2}(\cos x + \sin y) \, dy &= \int_0^\frac{\pi}{2}\cos x \, dy + \int_0^\frac{\pi}{2}\sin y \, dy \\ &= \cos x\int_0^\frac{\pi}{2} 1 \,dy + \int_0^\frac{\pi}{2} \sin y \, dy \\ &= (\cos x)\frac{\pi}{2} + \left(-\cos \frac{\pi}{2} + 1 \right) \\ &= (\cos x) \frac{\pi}{2} - 0 + 1 = \frac{\pi}{2} \cos x +1 \end{align*}$$
and now the second integral:
$$\begin{align*} \int_0^\frac{\pi}{4}\frac{\pi}{2} \cos x + 1 \, dx &= \int_0^\frac{\pi}{4}\frac{\pi}{2}\cos x + \int_0^\frac{\pi}{4} 1 \, dx \\ &= \frac{\pi}{2}\int_0^\frac{\pi}{4}(\cos x)dx + \left(\frac{\pi}{4}-0\right) \\ &= \frac{\pi}{2} \left(\sin \frac{\pi}{4} - \sin 0\right) + \frac{\pi}{4} \\ &= \frac{\pi}{2}\left(\frac{\sqrt{2}}{2} - 0\right) + \frac{\pi}{4} + = \frac{\pi \sqrt{2}}{4} + \frac{\pi}{4} = \frac{\pi (\sqrt{2}+1)}{4} \end{align*} $$
and now how it looks ?
|
You are doing fine just be careful of the antiderivative of $\sin x$ which is $-\cos x$. So the only mistake that I am seeing is in the first line: $$\int_{0}^{\pi/2}\sin ydy=[-\cos y]_{0}^{\pi/2}=1-0$$ not $-1$.
You have written $[-\cos \dfrac{\pi}{2}-1]$ instead of $[-\cos \dfrac{\pi}{2}+1]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/810370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to prove: prove $\frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta$ I need to prove that $\frac{1+\tan^2\theta}{1+\cot^2\theta}= \tan^2\theta.$
I know that $1+\tan^2\theta=\sec^2\theta$ and that $1+\cot^2\theta=\csc^2\theta$, making it now $$\frac{\sec^2\theta}{\csc^2\theta,}$$ but I don't know how to get it down to $ \tan^2\theta.$
HELP!!!!
I also need help proving that $\tan\theta + \cot\theta = \sec\theta\cdot\csc\theta.$
|
Note that $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$, and $\sin^2\theta+\cos^2\theta=1$. Hence
$$
\dfrac{1+\tan^2\theta}{1+\cot^2\theta}=\dfrac{1+\dfrac{\sin^2\theta}{\cos^2\theta}}{1+\dfrac{\cos^2\theta}{\sin^2\theta}}=\left(\dfrac{\cos^2\theta+\sin^2\theta}{\sin^2\theta+\cos^2\theta}\right)\cdot\dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/810453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 4
}
|
proving $\tan^{-1} \left( \frac{x}{1-x^2} \right) = \tan^{-1}x + \tan^{-1}(x^3)$ Two related questions, one easy, one just a bit harder:
1) Prove the identity
$$
\tan^{-1} \left( \frac{x}{1-x^2} \right) =
\tan^{-1}x + \tan^{-1}(x^3)
$$
2) Now try to find a geometric or trigonometric proof of that same geometry, without resorting to calculus.
I'll post an answer to both questions in a couple of days if nobody has one yet.
|
Part I: Consider $f(x) = tan^{-1}\left(\dfrac{x}{1-x^2}\right) - tan^{-1}x - tan^{-1}(x^3)$, taking derivative of $f$:
$f'(x) = \dfrac{1+x^2}{1-x^2+x^4} - \dfrac{1}{1+x^2} - \dfrac{3x^2}{1+x^6} = \dfrac{(1+x^2)^2 - (1-x^2+x^4) - 3x^2}{1+x^6} = \dfrac{1+ 2x^2 + x^4 - 1 + x^2 - x^4 - 3x^2}{1+x^6} = 0$. Thus $f$ is constant on $\mathbb{R}$, and $f(x) = f(0) = 0$, and the identity follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/812123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
How do I find the equation of a circle, given radius and centre coordinates? Say I am asked to find, in expanded form without brackets, the equation of a circle with radius 6 and centre 2,3 - how would I go on about doing this?
I know the equation of a circle is $x^2 + y^2 = r^2$, but what do i do with this information?
|
The general solution for the circle with centre $(a,b)$ and radius $r$ is
$$
(x-a)^2+(y-b)^2=r^2.
$$
Now, we have the centre $(2,3)$ and the radius $6$, therefore the equation of the circle is
\begin{align}
(x-2)^2+(y-3)^2&=6^2\\
x^2-4x+4+y^2-6x+9&=36\\
x^2+y^2-4x-6y+4+9-36&=0\\
\large\color{blue}{x^2+y^2-4x-6y-23}&\large\color{blue}{=0}.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/813715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 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.