Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculating limits with Strange Rearrangement How did they get to this third step?
$\lim\limits_{x \to 3} \frac{\sqrt{x+1}-2}{\sqrt{x-2}-1}$ =
$\lim\limits_{x \to 3} \frac{(\sqrt{x+1}-2)(\sqrt{x-2}+1)}{x-3}$ =
$\lim\limits_{x \to 3} \frac{(x-3)(\sqrt{x+1}-2)}{(x-3)(\sqrt{x+1}-2)}$
| $\frac{\sqrt{x+1}-2}{\sqrt{x-2}-1} \cdot \frac{\sqrt{x+1}+2}{\sqrt{x-2}+1} \cdot \frac{\sqrt{x-2}+1}{\sqrt{x+1}+2} \\ \text{ multiply the first two fractions } \\ \text{ Also I multiplied the initial fraction you have by one } $
| {
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Proof by induction that $\frac1{n+1} + \frac1{n+2} + \ldots + \frac1{2n} \geq \frac7{12}$ The question is prove that for every integer greater than or equal to 2
$$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} \geq \frac{7}{12}$$
So far I have
Base case let $p(2)$
\begin{align*}
\frac{1}{2}+ \frac{1}{2+2} + \frac{1}{4} & \geq \frac{7}{12}\\
\frac{5}{6} & \geq \frac{7}{12}
\end{align*}
therefore base case is true.
Let's assume it is a true statement for $p(n+1)$ for all $n$ greater than or equal to $2.$
\begin{align*}
\frac{1}{(n+1)+1} + \frac{1}{(n+1)+2} + \frac{1}{2(n+1)} \geq \frac{7}{12}\\
\frac{1}{n+2} + \frac{1}{n+3} + \frac{1}{2n+2} \geq \frac{7}{12}
\end{align*}
then I wrote that if $n$ is greater than or equal to $2$ is a true statement then the inequality is true. This is a proof by induction and I'm not sure if I concluded properly
| The three dots $\ldots$ are an ellipsis, indicating continuation of the preceding trend of terms up to the final term.
There are only two terms in the $n=2$ case. The ellipsis form of the formula really only definitely indicates the start point, the end point and the general nature of interim points; it doesn't set a minimum number of terms or indeed the maximum number of terms - you are left to infer the exact quantity.
In this case, there are $n$ terms in each sum, one each for all the numbers from $n+1$ to $2n$. So for the sake of clarity, here are a few terms:
$$\begin{align}
\frac{1}{3} + \frac{1}{4} &= \frac{7}{12} \tag{n=2} \\[0.5em]
\frac{1}{4} + \frac{1}{5} + \frac{1}{6} &= \frac{37}{60} \tag{n=3} \\[0.5em]
\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} &= \frac{533}{840} \tag{n=4} \\[0.5em]
\end{align}$$
You can see that that $\frac{1}{3}$ has been replaced by $\frac{1}{5} + \frac{1}{6}$ in going from $n=2$ to $n=3$, and $\frac{1}{4}$ has been replaced by $\frac{1}{7} + \frac{1}{8}$ in going from $n=3$ to $n=4$. This should indicate the correct line of the induction argument.
| {
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Probability Distribution sum The sum regarding probability distribution is:
There is a group of 50 people who are patriotic out of which 20 believe in non-violence. Two persons are selected at random out of them. Write the probability distribution for the selected persons who are non-violent. Also find the mean of the distribution.
My solution:
$$\begin{array}{r|lll}
x & 0 & 1 & 2 \\
P(X) & {n \choose 0}\frac25^0\frac35^2 & {n \choose 1}\frac25^1\frac35^1 & {n \choose 2}\frac25^2\frac35^0 \\
P(X) & \frac9{25} & \frac{12}{25} & \frac4{25}
\end{array}$$
$$\text{Mean}=\dfrac{9}{25} + 2 \cdot \dfrac{4}{25} = \dfrac{17}{25}$$
However, my book is suggesting $\dfrac{196}{245}$.
Which one is correct?
| Among the $\binom{50}{2}$ possible couples we can select, $\binom{20}{2}$ of them are made of two non-violent people, $\binom{30}{2}$ of them are made of violent people and $\binom{50}{2}-\binom{30}{2}-\binom{20}{2}=20\cdot 30$ of them are mixed, so:
$$\mathbb{P}[X=0]=\frac{\binom{30}{2}}{\binom{50}{2}}=\frac{87}{245},$$
$$\quad \mathbb{P}[X=1]=\frac{20\cdot 30}{\binom{50}{2}}=\frac{120}{245},$$
$$ \mathbb{P}[X=2]=\frac{\binom{20}{2}}{\binom{50}{2}}=\frac{38}{245}$$
and:
$$\mathbb{E}[X]=\frac{87\cdot 0+120\cdot 1+38\cdot 2}{245}=\color{red}{\frac{196}{245}}=\frac{4}{5}.$$
| {
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Sum $\sum _{k=0}^{\infty } \frac{(-1)^k \psi (k+1)}{\left(k-\frac{3}{2}\right)^2}$ Is it possible to get a closed form for:
$$\sum _{k=0}^{\infty } \frac{(-1)^k \psi (k+1)}{\left(k-\frac{3}{2}\right)^2}$$ where $\psi$ is the polygamma function?
| Since $\psi(k+1)=H_k-\gamma$ our series is given by:
$$ -4+\frac{32}{9}\gamma+4\sum_{k\geq 2}\frac{(-1)^k\left(H_k-\gamma\right)}{(2k-3)^2}=-4+\frac{32}{9}\gamma-4\gamma K+4\sum_{n\geq 0}\frac{(-1)^n H_{n+2}}{(2n+1)^2}$$
where $K$ is the Catalan constant. The last series can be computed by evaluating:
$$S_1 = \sum_{n\geq 0}\frac{(-1)^n}{(n+1)(2n+1)^2},\qquad S_2 = \sum_{n\geq 0}\frac{(-1)^n}{(n+2)(2n+1)^2},\qquad S_3 = \sum_{n\geq 0}\frac{(-1)^n\, H_n}{(2n+1)^2}$$
through the integral identities:
$$ S_1 = -\int_{0}^{1}\frac{\log(1+x^2)\log x}{x^2}\,dx, \qquad S_2 = -\int_{0}^{1}\frac{\left(x^2-\log(1+x^2)\right)\log x}{x^4}\,dx,$$
$$ S_3 = \int_{0}^{1}\frac{\log(1-x^2)\log x}{1-x^2}\,dx.$$
In particular, we have:
$$ S_1 = 2K-\frac{\pi}{2}+\log 2, $$
$$ S_2 = \frac{2}{3}K-\frac{\pi}{18}+\frac{1-\log 2}{9}, $$
$$ S_3 = \frac{7}{4}\zeta(3)-\frac{\pi^2}{4}\log 2.$$
| {
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Find all integers $x$, $y$, and $z$ such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ Characterize all positive integers $x$, $y$, and $z$ such that:
$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$$
For example, $\dfrac{1}{x+1} + \dfrac{1}{x(x+1)} = \dfrac{1}{x}$.
| $\frac{1}{x} + \frac{1}{y}=\frac{x+y}{xy}$ then we have to found $(x,y)\in \mathbb N^{2}$ such that $x+y|xy$ then $k(x+y)=xy$ if and only if $x=\frac{ky}{y-k}$ then $k<y$ and $y-k|ky$ , then fixed y you can find a solution choosing $y-k$ as a divisor of y.(Note that if y is prime we not have solution)
| {
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the total differential equation $(y^2z - y^3 +x^2y)dx - (x^2z +x^3 -xy^2)dy +(x^2y - xy^2)dz=0$ In solving the total differential equation $$(y^2z - y^3 +x^2y)dx - (x^2z +x^3 -xy^2)dy +(x^2y - xy^2)dz=0$$
Substituting $x =uz$ and $y=vz$ and reached at a position where, i face the following
$$(v^2 - v^3 +u^2v)du - (u^2+u^3-uv^2)dv=0$$
I am stuck here please help!!
| It's all equation devided by x^2 y^2.
Now we get
(z-y/x^2 +1/y)dx-(z+x/y^2 - 1/x)dy+(1/y-1/x)dz
d(x/y) +d(y/x) +z(dx-dy) +(1/y-1/x)dz
d(x/y) +d(y/x) +d(z/y) - d(z/x) = 0
x+y + log y - log x = c
x+y+ log y/ x = c
x + y + log y / x = c.
Is a required solution.
| {
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Find the maximum and minimum of $x^2+2y^2$ if $x^2-xy+2y^2=1$.
Find the maximum and minimum of $x^2+2y^2$ if $x,y\in\mathbb R$ and $$x^2-xy+2y^2=1$$
My attempt:
Clearly, since $x^2-x(y)+(2y^2-1)=0$ and $2y^2-y(x)+(x^2-1)=0$, we have that
$$\Delta_1=y^2-8y^2+4=4-7y^2\ge 0$$
and
$$\Delta_2=x^2-8x^2+8=8-7x^2\ge 0$$
so that $y^2\le \frac{4}{7}$ and $x^2\le \frac{8}{7}$.
Thus $x^2+2y^2\le \frac{8}{7}+\frac{8}{7}=\frac{16}{7}$.
But clearly $x^2+2y^2\neq \frac{16}{7}$, since if $\begin{cases}x^2=\frac{8}{7}\\y^2=\frac{4}{7}\end{cases}$, then $x^2-xy+2y^2\neq 1$
so that equality can't be achieved and we only have that $x^2+2y^2<\frac{16}{7}$.
| Since your answer is tagged precalculus, I'll stick to precalculus methods. Note that one way to solve this would be using Lagrange multipliers from calculus. That calculation is straight-forward, but the numbers get messy. (In some sense, the solution below is a Lagrange multiplier solution, but without the calculus).
In your problem, the first step that I will take is to change variables to $z=\sqrt{2}y$. Then, your problem is to maximize or minimize
$x^2+z^2$ subject to
$x^2-\frac{1}{\sqrt{2}}xz+z^2=1$.
Note that the function that you're maximizing is the square of the distance function from the origin and the constraint equality is an ellipse centered at the origin. The closest point to the origin is on the minor axis and the furthest point from the origin is on the major axis. The problem is that these axes are not axis aligned.
In this case, the major axis is along the line $x=z$ and the minor axis is along the line $x=-z$. This can be seen because the equation is symmetric in $x$ and $z$. Now, when $x=z$, we can substitute $z=x$ into $x^2-\frac{1}{\sqrt{2}}xz+z^2=1$ to get
$x^2-\frac{1}{\sqrt{2}}x^2+x^2=1$.
We can isolate $x^2$ to get
$\left(2-\frac{1}{\sqrt{2}}\right)x^2=1$ or that
$x^2=\frac{\sqrt{2}}{2\sqrt{2}-1}$.
Since $x=z$, the maximum value is then
$\frac{2\sqrt{2}}{2\sqrt{2}-1}$.
On the other hand, along the line $x=-z$, we can substitute into $x^2-\frac{1}{\sqrt{2}}xz+z^2=1$ to get
$x^2+\frac{1}{\sqrt{2}}x^2+x^2=1$.
Once again, we can isolate for $x^2$ to get
$x^2=\frac{\sqrt{2}}{2\sqrt{2}+1}$.
Since $x^2=z^2$, the minimum value is
$\frac{2\sqrt{2}}{2\sqrt{2}+1}$.
| {
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Maximum of a trigonometric expression Maximize $f(x)=4\sin x+48\sin x\cos x+3\cos x+14\sin^2x$ I broke it in 2 parts but then realized they don't have their maxima at the same points. So I am stuck. This is what I did. I wrote it as $$3\cos x+4\sin x+2\sin x(24\cos x+7\sin x)$$ Now we can maximize the first two terms and the terms inside the bracket.
| $f(x)=4\sin x+48\sin x\cos x+3\cos x+14\sin^2x$
Step 1: Take derivative and equate it to zero
$f'(x)=4\cos x+48(\cos^x-\sin^2 x)-3\sin x+14(2\sin x\cos x)$
$\implies 4\cos x + 48\cos 2x-3\sin x+14\sin 2x=0$
Step 2: Find the solution to above equation.
Step 3: It will attain a maxima if $f''(x)<0$, that means second derivative of the function.
Follow this wolfram Alpha link for solutions.
| {
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Algebra Splitting fields Take the irreducible polynomial $x^3 + x^2 + 1$ over $F_2$ (field of order $2$). Find the splitting field and its roots in that field.
Where I am:
I understand what splitting fields are, and I also know that the splitting field is $F_8$ because $8=2^3$ and we can work in the field $F_2[\theta]/(\theta^3 + \theta^2 + 1)$ in which the polynomial $x^3 + x^2 + 1$ will have linear factors.
What I want to know is: is there are a quick method for finding such roots? Should I simply plug in each element of the splitting field every time given such a question in order to find a linear factor, and work from there? However, I understand that this only works for polynomials up to deg $3$.
| in $F_2[Y]/(Y^3+Y^2+1)$, an obvious root of $P(X)=X^3+X^2+1$ is $Y$.
Another one will be $Y^2$ because in $F_2[Y]$, for any polynomial $Q$, $Q(Y^2)=Q(Y)^2$
Hence $$X^3+X^2+1=(X-Y)(X-Y^2)(X-z)$$
But looking at the coefficient of $X^2$, you get $z=Y^2+Y+1$
Or as $Y^2$ is a root $(Y^2)^2=Y^4$ is too. But $Y^4=Y^2+Y+1$.
$$X^3+X^2+1=(X-Y)(X-Y^2)(X-(Y^2+Y+1))$$
Note that $(Y^2+Y+1)^2=Y^4+Y^2+1=Y$, so there are only 3 different roots.
| {
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AIME 2013 Solutions (divisiblity) Problem 2
Find the number of five-digit positive integers, $n$, that satisfy the following conditions:
(a) the number $n$ is divisible by $5,$
(b) the first and last digits of $n$ are equal, and
(c) the sum of the digits of $n$ is divisible by $5.$
Obviously, if $n$ satisfies divisiblity by five and takes the form,
$$n = abcda$$ then $a=5$, deductively,
$$n = 5bcd5$$ And:
$$10 + b + c + d \equiv 0 \pmod{5} \implies b + c + d \equiv -10 \equiv 0 \pmod{5}$$
What do I do next? HINTS are appreciated!
| Big hint: Note that the digits $b$ and $c$ can be chosen freely, ($100$ choices total); and then, whatever the choices for $b$ and $c$, there are $2$ choices for $d$. For instance if $b$ and $c$ are chosen to be $7$ and $6$ respectively, then $d$ could be $2$ or $7$ to make $b+c+d$ congruent to $0 \pmod{5}$.
| {
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Pell Fermat equation If $3n + 1$ and $4n + 1$ are perfect squares then $56$ divides $n$.
I somehow proved $n$ is even so proceeding further proved that $8$ divides $n$ but don't know how to manage for $7$.
| Let $3k+1=x^2,4k+1=y^2$ As you proved $8|k$.
Notice that $z^2-3y^2=1\ \ (*)$ when $z=2a$. All solutions of this Pell equation are obtained by substitution of $(a_n,b_n)$ in the equation: $(2+\sqrt{3})^n=a_n+\sqrt{3}b_n$. Because $z=2a$, w're looking for solutions with even $a_n$.
Let's prove that $a_n$ is even if and only if $n$ is odd (you can use induction) but here we will use the binomial formula :
$$(2+\sqrt{3})^n=\sum_{k=0}^n 2^{n-k}(\sqrt 3)^{k}=\sum_{k=0,k\, even}^n 2^{n-k}(\sqrt 3)^{k}+(\sum_{k=0,k\, odd}^n 2^{n-k}(\sqrt 3)^{\frac{k-1}{2}})\sqrt 3 $$.
We observe that all terms in $a_n$ are even except when $k$ is even and $n-k=0$ we will have the term $3^{\frac{k}{2}}$ which makes $a_n$ odd. So finally $a_n$ is even if and only is $n$ is odd, hence all solutions of our equation $(*)$ are given by
$(2+\sqrt{3})^{2n+1}=z_n+\sqrt{3}y_n$.
We can easily get the recursions:
$z_{n+1}=2x_{n+1}=14x_n+12y_n, y_{n+1}=8x_n+7y_n.$
This means that if $3k+1, 4k+1$ are squares then there exist $n$ such that $3k+1=x_n^2, 4k+1=y_n^2$ or $k=y_n^2-x_n^2$, the question now is equivalent to prove for all $n$ we have $7|y_n^2-x_n^2$.
By induction on $n$ , for $n=0$ note that $ y_0^2-x_0^2\equiv 1^2-1^2 \equiv 0\pmod{7}$
Suppose that $y_n^2-x_n^2 \equiv 0\pmod{7}$ , So $x_{n+1}=7x_n+6y_n \equiv -y_n\pmod{7}$ and $y_{n+1}=8x_n+7y_n\equiv x_n\pmod{7}$,we get $y_{n+1}^2-x_{n+1}^2 \equiv y_n^2-x_n^2 \equiv 0\pmod{7}$.
| {
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Integral: $\int_0^{1/2}\frac{\ln(1+2x)}{1+4x^2}{\rm d}x$ [Other method type question.]
Integral: $$I=\int_0^{1/2}\frac{\ln(1+2x)}{1+4x^2}{\rm d}x$$
One thing to quickly do it take $y=2x$:
$$I=\frac12\int_0^1\frac{\ln(1+y)}{1+y^2}{\rm d}y$$
I took $y=\tan z$:
$$I=\frac12\int_0^{\pi/4}\ln(1+\tan z){\rm d}z$$
Now substitute $u=\pi/4-z$:
$$I=\frac12\int_0^{\pi/4}[\ln2-\ln(1+\tan z)]{\rm d}z=\frac\pi8\ln2-I\\I=\frac\pi{16}\ln 2$$
Any other methods?
| We have
\begin{align}I &= \frac{1}{2} \int_0^1 \frac{\ln(1 + y)}{1 + y^2}\, dy\\
& = \frac{1}{2}\int_0^1\int_0^1 \frac{y}{(1 + ry)(1 + y^2)}\, dr\, dy\\
& = \frac{1}{2}\int_0^1 \int_0^1 \frac{y}{(1 + ry)(1 + y^2)}\, dy\, dr\\
&= \frac{1}{2}\int_0^1 \int_0^1 \left(\frac{r}{(1 + r^2)(1 + y^2)} + \frac{y}{(1 + r^2)(1 + y^2)}- \frac{r}{(1 + r^2)(1 + ry)}\right)\, dy\, dr\\
&= \frac{1}{2}\int_0^1 \left(\frac{r}{1 + r^2}\cdot \frac{\pi}{4} + \frac{1}{2(1 + r^2)}\ln(2) - \frac{\ln(1 + r)}{1 + r^2}\right)\, dr\\
&= \frac{1}{2}\left(\frac{\pi}{8}\ln(2) + \frac{\pi}{8}\ln(2)\right) - I\\
&= \frac{\pi}{8}\ln(2) - I
\end{align}
So $2I = \frac{\pi}{8}\ln(2)$, or $$I = \frac{\pi}{16}\ln(2).$$
| {
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Integer solution to $19x^3-84y^2=1984$ Show that there exist no integer values $x,y$ such that $19x^3-84y^2=1984$.
Please help me in understanding no solution problems.
I tried to check the modulo $7$ of both sides but couldn't reject some cases.
| Working modulo 7,
$$
19x^3 − 84y^2 \equiv 1984 \\
5x^3 \equiv 3 \\
15x^3 \equiv 9 \\
x^3 \equiv 2 \\
$$
By inspection, $x^3 \equiv 2 \mod 7 \,$ has no solutions:
$$
\begin{array}{c|r|c}
x & x^3 & x^3\mod 7 \\
\hline
0 & 0 & 0 \\
1 & 1 & 1 \\
2 & 8 & 1 \\
3 & 27 & 6 \\
4 & 64 & 1 \\
5 & 125 & 6 \\
6 & 216 & 6 \\
\end{array}
$$
Or we could invoke Fermat's little theorem and note that for all $x: x\mod 7 \neq 0,\, x^6 = 1$, hence $x^3 = \pm 1 \mod 7$
Alternatively, working mod 19
$$
19x^3 − 84y^2 \equiv 1984 \\
− 8y^2 \equiv 8 \\
$$
Note that $7 \times 8 = 56 \equiv -1 \mod 19$
So
$$y^2 \equiv -1 \mod 19$$
But according to the the first supplement to the law of quadratic reciprocity
$$y^2 \equiv -1 \mod p$$ for prime $p$ only has solutions if $p \equiv 1 \mod 4 $ . And since $19 \mod 4 \equiv 3\,$ there are no solutions for $y$.
| {
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How to integrate $\frac{x^n}{1+x}$ How to integrate the function
$ \frac{x^n}{1+x} $
I tried to do it by parts but it was of no use.
I did it by taking x as $-\theta$ and did the series expansion.
I need a better method.
| Write $x^n$ as $(x^n\pm 1)\mp 1$ depending on $n$ being odd or even. By integrating you will get $\pm \log(x+1)$ plus the integral of a polynomial in $x$. For instance:
$$\int\frac{x^3}{1+x}\,dx = -\log(1+x)+\int\frac{x^3+1}{x+1}\,dx = \log(1+x)+\int(x^2-x+1)\,dx\\ =-\log(1+x)+\frac{x^3}{3}-\frac{x^2}{2}+x, $$
$$\int\frac{x^4}{1+x}\,dx = \log(1+x)+\int\frac{x^4-1}{x+1}\,dx = \log(1+x)+\int(x^3-x^2+x-1)\,dx\\ =\log(1+x)+\frac{x^4}{4}-\frac{x^3}{3}+\frac{x^2}{2}-x. $$
| {
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Taking the cube root of a sum of radicals I am wondering how to derive the following simplification without knowing it beforehand:
$$^3\sqrt{10 + 6\sqrt{3}} = 1 + \sqrt{3}$$
After the fact, it is easy to verify algebraically. The problem arose when applying Cardano's method to solve
$$y^3 + 6y = 20$$
I was able to derive a similar but less complicated simplification,
$$\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}$$
by assuming that $3 + 2\sqrt{2} = (a + b\sqrt{2})^2$, expanding, equating coefficients of $\sqrt{2}$, and solving for $a$ and $b$ using the quadratic formula. However, using the same method, i.e. assuming that $10 + 6\sqrt{3} = (a + b\sqrt{3})^3$, expanding, and equating coefficients of $\sqrt{3}$ yields cubic equations in $a$ and $b$:
$$a^3 + 9ab^2 = 10, a^2b + b^3 = 2$$
Guess-and-check yields $a = 1$ and $b = 1$, but I would prefer a more systematic method of solution. I did not use the cubic formula again, figuring that this would probably yield another nested radical.
According to Wolfram|Alpha, $10 + 6\sqrt{3} = 1 + 3\sqrt{3} + 9 + 3\sqrt{3} = 1 + 3\sqrt{3} + 3(\sqrt{3})^2 + (\sqrt{3})^3 = (1 + \sqrt{3})^3$. However, I'm not sure how I would arrive at that chain of reasoning except by chance or by using Wolfram|Alpha.
| $N(10+6\sqrt{3})=10^2-3\cdot6^2=-8$. So in $\mathbb Z[\sqrt3]$, it is possible that $10+6\sqrt{3}$ is a perfect cube. The only possible solution to $\sqrt[3]{10+6\sqrt{3}} $ will be one with norm $-2$. That yields possible cube roots $(1\pm \sqrt{3})(2\pm \sqrt{3})^k$ for some $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1178996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Algebra of trigonometry and some limits I am singlehandedly trying to get through a book on divergent series, however one sequence is really strange and I cant seem to understand what the authors use to come to these conclusions!
Imagine the following is our sum. ($0 < x < 2{\pi}$)
As the authors attempt to determine the mean of the partial sums, they go on to do the following:
Is there any way to understand any of the following steps, including the determination of the limit as $n$ approaches positive infinity?
Thanks a lot for your help!
| Rewrite
$$
S_n+\frac{1}{2}=\frac{\sin\left(n+\dfrac{1}{2}\right)x}{2\sin\dfrac{x}{2}}
$$
Then
$$
\sum_{m=1}^n \left(S_m+\frac{1}{2}\right)=
\frac{1}{2\sin(x/2)}\sum_{m=1}^n\sin\left(m+\frac{1}{2}\right)x
$$
or
$$
\frac{n}{2}+\sum_{m=1}^n S_m=
\frac{1}{2\sin(x/2)}\sum_{m=1}^n\sin\left(m+\frac{1}{2}\right)x
$$
that can also be rewritten as
$$
4\sin^2\frac{x}{2}\left(\frac{n}{2}+\sum_{m=1}^n S_m\right)=
\sum_{m=1}^n 2\sin\frac{x}{2}\sin\left(m+\frac{1}{2}\right)x
$$
Now we can recall that
$$
2\sin\frac{x}{2}\sin\left(m+\frac{1}{2}\right)x=\cos mx-\cos(m+1)x
$$
and therefore
$$
\sum_{m=1}^n2\sin\frac{x}{2}\sin\left(m+\frac{1}{2}\right)x=
\sum_{m=1}^n(\cos mx-\cos(m+1)x)=\cos x-\cos(n+1)x
$$
(by telescoping).
Hence
$$
4\sin^2\frac{x}{2}\left(\frac{n}{2}+\sum_{m=1}^n S_m\right)=
\cos x-\cos(n+1)x
$$
Divide both sides by $n$:
$$
4\sin^2\frac{x}{2}\left(\frac{1}{2}+\frac{1}{n}\sum_{m=1}^n S_m\right)=
\frac{\cos x-\cos(n+1)x}{n}
$$
Since $0<x<2\pi$, we know that $\sin(x/2)\ne0$ and so
$$
\frac{1}{2}+\lim_{n\to\infty}\frac{1}{n}\sum_{m=1}^n S_m=0
$$
because
$$
\lim_{n\to\infty}\frac{\cos x-\cos(n+1)x}{n}=0
$$
as the numerator is bounded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the difference between $\ln(\sin\frac{x}{2}+\cos\frac{x}{2} )- \ln(\sin\frac{x}{2}-\cos\frac{x}{2})$ and $\ln(\tan x+\sec x)$ I've noticed if you ask wolfram what is the integral of $ \sec x$ they will answer
$$\ln\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right) - \ln\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)$$
I've been taught that the $\int\sec x = \ln|\tan x + \sec x | + C$ which is in my opinion way simpler to read and write.
However, even in the alternatives form of Wolfram, I could not find it there which let me believe that what I've been taught is probably not exact (but near) and they teach us that so it is simpler for us to learn.
Am I right ? If yes, what is the difference ?
| there is no difference. here is a reason
$$\begin{align}\tan x + \sec x &= \frac{\sin x}{\cos x} + \frac 1 {\cos x}\\
& = \frac{1+\sin x}{\cos x} \\
&= \frac{\sin^2x/2 + \cos^2x/2 + 2\sin x/2 \cos x/2}{\cos^2 x/2 - \sin^2 x/2} \\
&= \frac{(\sin x/2 + \cos x/2)^2}{(\cos x/2 - \sin x/2)(\cos x/2 + \sin x/2)}\\
&=\frac{\sin x/2 + \cos x/2}{\cos x/2 - \sin x/2}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How do I calculate the number of permutations of the list $(6, 6 ,5, 4)$? I have the list $l = (6, 6, 5, 4)$ and want to how to calculate the possible number of permutations.
By using brute force I know that there are 12 possible permutations:
$$\{(6, 5, 6, 4),
(6, 6, 5, 4),
(5, 6, 6, 4),
(6, 4, 5, 6),
(6, 5, 4, 6),
(4, 6, 6, 5),
(4, 5, 6, 6),
(4, 6, 5, 6),
(6, 4, 6, 5),
(6, 6, 4, 5),
(5, 4, 6, 6),
(5, 6, 4, 6)\}$$
But how would I calculate this?
I know we talk about permutation as order matters also I know that repetition is allowed (at least for $x = 6$).
| Imagine there are two different $6$'s, say $6_a$ and $6_b$. Then there would be $4!=24$ permutations. Now let the two $6$'s be the same, so $(6_a,7,5,6_b)=(6_b,7,5,6_a).$ This halves the number of permutations, giving the answer of $24/2=12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the roots of $(1 + i)^{\frac{1}{4}}$ The professor says that the $n = 4$ roots of this are in the form: $\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})$, where $k = 0, 1, 2, 3$.
So to find $\theta$, we find the $r = \sqrt{(1)^2 + (1)^2} = \sqrt{2}$ since $Re(1+i) = 1$ and $Im(1+i) = 1$. So $\sqrt{2}\cos\theta = 1$ and $\sqrt{2}\sin\theta = 1$, so the angle $\theta$ is $\frac{\pi}{4}$.
However, if we do $k=0$, then we get that one of the roots is $1$, which is obviously not true since $1^4 \neq 1 + i$. The professor says that the solutions are: $k=1: \cos(\frac{9\pi}{16}) + i\sin(\frac{9\pi}{16})$, $k=2: \cos(\frac{17\pi}{16}) + i\sin(\frac{17\pi}{16})$, and $k=3: \cos(\frac{25\pi}{16}) + i\sin(\frac{25\pi}{16})$. I plugged these into WolfRamAlpha and rose them to the $4$th power, but none of them return the form $1+i$.
What is incorrect about these steps?
| $1+i = \sqrt{2}\cdot {e^{i\pi/4}}= \sqrt{2}\cdot e^{i\pi/4+i2k\pi}\to \left(1+i\right)^{1/4}= 2^{1/8}\cdot e^{i(\pi/16+k\pi/2)}, k=0,1,2,3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Problem with definite integral $\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$ $$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$$
The question says 'evaluate the integral using the suggested substitution. It gives $u=\cos x$. But I think Let $u=1-2\sin x$ is better.
$$\int_{0}^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x} dx$$
$$u=1-2\sin x$$
$$du=-2\cos x dx$$
$$=-\frac{1}{2}\int_{1}^{0}\sqrt{u}du$$
$$=\frac{1}{2}\int_{0}^{1}\sqrt{u}du$$
$$=\left | \frac{u^{\frac{3}{2}}}{3} \right |_{0}^{1}$$
$$=\frac{1}{3}$$
My question is how to solve it by using $u=\cos x$. Can anyone show the solution for it? Thanks a lot!
| Well, @Nilan has the better way to go. But, here is another "Brute Force," double substitution method that works.
Write, $\sin x = \sqrt{1-\cos^2 x}$ (the positive square root is appropriate here since $0\le x\le \frac{\pi}{6}$. Then, with $u = \cos x$, $du = -\sin x dx=-\sqrt{1-u^2}du$, and the limits of integration extend from $u=1$ to $u=\frac{\sqrt{3}}{2}$. Thus,
$$\int_0^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x}dx= \int_{\frac{\sqrt{3}}{2}}^1 \frac{u\sqrt{1-2\sqrt{1-u^2}}}{\sqrt{1-u^2}}du$$
Now, we make a second substitution. Let $y=\sqrt{1-u^2}$. Then, $dy=\frac{-2u}{\sqrt{1-u^2}}du$ and the limits of integration go from $y=\frac12$ to $y=0$. (Note: This second substitution is identical to making the original substitution $u=\sin x$). Thus,
$$\int_0^{\frac{\pi}{6}}\cos x\sqrt{1-2\sin x}dx=\int_0^{\frac12} \sqrt{1-2y}dy=\frac13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\int\frac{1}{5\cos x+\sin x+7}~dx$ Evaluating $$\int\frac{1}{5\cos x+\sin x+7}~dx.$$
This can be done by substituting $$\sin x = \frac{2t}{1 + t^2}$$ and $$\cos x = \frac{1 - t^2}{1 + t^2}.$$
However after I substitute it I cannot simplify it to get anything easier to integrate.
After substituting I got: integral $$\frac{1 + t^2}{2(t + 2)(t + 3)}$$ or $$\frac{1 + t^2}{12 + 2t^2 + 10t}.$$
Could someone give me a hint?
Many thanks.
| Generally, consider the real function
$$f(x)=\frac{1}{a+b\cos x+c\sin x}$$
With $a^2>b^2+c^2$ so that $f$ is defined on $\mathbb{R}$.
It is not hard to check that the derivative of
$$F(x)=\frac{x}{d}+\frac{2}{d}\arctan\left(\frac{c\cos x-b\sin x}{d+a+b\cos x+c\sin x}\right)$$
with $d=\sqrt{a^2-b^2-c^2}$, is $f(x)$. So $\int f(x)dx=F(x)+k$. The advantage
of this expression of $F$ is that it is also defined on $\mathbb{R}$. In particular,
$$\int \frac{1}{7+5\cos x+\sin x}=\frac{x}{\sqrt{23}}+\frac{2}{\sqrt{23}}\arctan\left(\frac{\cos x-5\sin x}{\sqrt{23}+7+5\cos x+\sin x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find a segment value of Isosceles Triangle $M$ and $N$ are points from equal sides $CA$ and $CB$ from isosceles triangle $ABC$ as what $CM = CN$. Determine the segment $CM$, knowing that $AB = 2k$ and that $2P$ and $2p$ are the perimeters respectively from triangle $ABC$ and the trapeze AMNB.
Answer: $\Large CM = \frac{(P-k)(P-p)}{P-2k}$
|
Given isosceles $\triangle ABC$,
\begin{align}
AB+BC+AC&=2P, \tag{1}\label{1}\\
AB+BN+MN+AM&=2p. \tag{2}\label{2}
\end{align}
Subtraction $\eqref{1}-\eqref{2}$ gives
\begin{align}
2x-y&=2P-2p. \tag{3}\label{3}
\end{align}
$\triangle MNC\propto\triangle ABC$, hence
\begin{align}
\frac{y}{2k}&=
\frac{x}{AC}=\frac{x}{BC},\\
\frac{y}{2k}&=
\frac{y+x+x}{2k+AC+BC}
=\frac{y+2x}{2P},\\
y&=\frac{2x k}{P-k}.
\tag{4}\label{4}
\end{align}
Finally, a combination of $\eqref{3}$ and $\eqref{4}$ gives the result
\begin{align}
x&=
\frac{(P-p)(P-k)}{P-2k}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An identity satisfying the divisors of a positive integer I saw a hard competition problem with long and ugly proof in http://solmu.math.helsinki.fi/olympia/valmennus/2013/vt2013_12var.pdf ? The question is from Australian mathematical olympiad 1985. Is there a nice way to solve the following:
A positive integer $n$ has factors $1=d_1<d_2<\ldots < d_k=n$. Determine those $n$ that satisfies $n=d_6^2+d_7^2-1$.
| From the equality $n = d_6^2 + d_7^2 - 1$, we see
*
*$d_6$ and $d_7$ are mutually prime,
*$d_7 \mid d_6^2 - 1 = (d_6 - 1)(d_6 + 1)$, and
*$d_6 \mid d_7^2 - 1 = (d_7 - 1)(d_7 + 1)$.
Suppose $d_6 = ab$ and $d_7 = cd$ with $1 < a < b$ and $1 < c < d$.
Then $n$ has $6$ divisors smaller than $d_6$, namely $1$, $a$, $b$, $c$, $d$, $ac$. Hence one of $d_6$ and $d_7$ is either a prime $P$ or the square of a prime $p^2$. But $p$ is not $2$, therefore $d_7 = d_6 + 1$.
Write $d_6 = x$, $d_7 = x + 1$ to give$$n = x^2 + (x + 1)^2 - 1 = 2x(x + 1).$$
*
*Assume either $x$ or $x + 1$ is a prime $p$. So the other one has at most $6$ divisors and possible candidates are $2^3$, $2^4$, $2^5$, $2s$, $2s^2$, $4s$ where $s$ is an odd prime. $2^3$ and $2^4$ are not solutions. For $2^5$ we get the solution $x = 31$, $x + 1 = 32$, $n = 1984$. $2s$ is no solution because $n = 4ps$ has at most $4$ divisors less than $x$, namely $1$, $2$, $4$, $s$. $2s^2$ is no solution because $n = 4ps^2$ has $6$ divisors less than $x$, namely $1$, $2$, $4$, $s$, $2s$, $s^2$. $4s$ is no solution because $n = 8ps$ has $6$ divisors less than $x$, namely $1$, $2$, $4$, $8$, $s$, $2s$.
*Assume either $x$ or $x + 1$ is $p^2$. So the other one has at most $5$ divisors and possible candidates are $2^3$, $2^4$, $2s$ where $s$ is an odd prime. For $2^3$ we get the solution $x = 8$, $x + 1 = 9$, $n = 144$. $2^4$ is no solution. Also $2s$ is no solution because $n = 4p^2s$ has $6$ divisors less than $x$, namely $1$, $2$, $4$, $p$, $2p$, $s$.
Thus there are two solutions in all: $144$ and $1984$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding x in matrix algebra The question asked me to calculate $|A|$ for the given matrix below and state whether or not it is invertible.
$$
\left(
\begin{matrix}
1 & 0 & 2 & 1\\
0 & x & x & x\\
x & x & 3x & 1\\
0 & x & x & 0\\
\end{matrix}
\right)
$$
I got my answer to be $-2x^3 + 2x^3 = 0$, indicating that the matrix is not invertible because the determinant is a zero, regardless of the value of $x$.
However, I'm not certain whether I'm right after doing all the calculations. Could you help me verify?
| Okay, here's every step:
$$\left|\begin{matrix} x & x & x \\ x & 3x & 1 \\ x & x & 0\end{matrix}\right|+2\left|\begin{matrix} 0 & x & x \\ x & x & 1 \\ 0 & x & 0\end{matrix}\right| - \left|\begin{matrix} 0 & x & x \\ x & x & 3x \\ 0 & x & x\end{matrix}\right| \\ = x^2\left|\begin{matrix} 1 & 1 & x \\ 1 & 3 & 1 \\ 1 & 1 & 0\end{matrix}\right|+2x^2\left|\begin{matrix} 0 & 1 & x \\ 1 & 1 & 1 \\ 0 & 1 & 0\end{matrix}\right| - x^3\left|\begin{matrix} 0 & 1 & 1 \\ 1 & 1 & 3 \\ 0 & 1 & 1\end{matrix}\right| \\ = x^2\left(\left|\begin{matrix} 3 & 1 \\ 1 & 0\end{matrix}\right| - \left|\begin{matrix} 1 & 1 \\ 1 & 0\end{matrix}\right| + x\left|\begin{matrix} 1 & 3 \\ 1 & 1\end{matrix}\right|\right) + 2x^2\left(- \left|\begin{matrix} 1 & 1 \\ 0 & 0\end{matrix}\right| + x\left|\begin{matrix} 1 & 1 \\ 0 & 1\end{matrix}\right|\right) -x^3\left(- \left|\begin{matrix} 1 & 3 \\ 0 & 1\end{matrix}\right| + \left|\begin{matrix} 1 & 1 \\ 0 & 1\end{matrix}\right|\right) \\ = -2x^3+2x^3-0 \\ =0$$
So you did it correctly.
Or, or course, for the easier way, just recognize that twice the first column plus the second column equals the third column of your matrix so the determinant must be $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194117",
"timestamp": "2023-03-29T00:00:00",
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Solving an equation $\pmod {13}$ Suppose:
$$1 + \frac12 +\frac13 + \dots + \frac1{23} = \frac{a}{23!}$$
I would like to find $a \pmod {13}$.
My attempt:
I'm attempting to use Wilson's theorem which states:
$$(n-1)!= -1 \pmod n$$
consider:
$$\begin{align}
\frac{23!a}{13} & = 23\times22\times21\dots\times14\times a\times(12!) &\\
&= -23\times22\times21\dots\times14\times a &\pmod {13} && \text{By Wilson's theorem}\\
&= -10\times9\times8\times\dots\times1\times a &\pmod{13} \\
&= -10!a &\pmod{13} \\
\end{align}$$
I have no idea where to go from here or even if I'm going in the right direction.
Any and all help is appreciated.
| We have $\frac{23!}{k}\equiv 0\,\text{(mod $13$)}$ for $k=1,2,\dots,23$ except when $k=13$.
Sou you are left with $$
\begin{align}
a&\equiv \frac{23!}{13}=12!\cdot 14\cdot 15\cdots 23\\
&\equiv (-1)\cdot 1\cdot 2\cdots 10\equiv (-1)\cdot 12! \cdot 11^{-1}\cdot 12^{-1}\\
&\equiv (-1)\cdot (-1)\cdot (-2)^{-1}\cdot (-1)^{-1}\equiv -6\,\text{(mod $13$)}
\end{align}
$$
because $-2\cdot 6=-12\equiv 1\,\text{(mod $13$)}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Bernoulli Number analog using Cosine I know that Bernoulli Numbers can be found with the generating function
$$\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n$$
I was wondering if any work has been done using a similar equation
$$\frac{x^2}{\cos{x}-1}=\sum_{n=0}^{\infty}\frac{C_n}{n!}x^{2n}$$
I'm particularly interested in the $C_n$. Can anyone help me with a reference to work with this particular generating function or to the coefficients $C_n$?
| Your sequence (negated) starts
$$
2, \frac{1}{6}, \frac{1}{60}, \frac{1}{504}, \frac{1}{3600}, \frac{1}{22176}, \frac{691}{82555200}, \frac{1}{570240}, \frac{3617}{8821612800}, \frac{43867}{414096883200}, \frac{174611}{5822264448000}, \ldots
$$
We know that
$$
\frac{1-\cos x}{x^2} = \frac{1}{2!} - \frac{1}{4!}x^2 + \frac{1}{6!}x^4 - \frac{1}{8!}x^6 + \cdots.
$$
Let $c_n = C_n/n!$. The formula for the inverse of a power series gives
$$
c_0 = 2, \quad c_n = 2\frac{c_{n-1}}{4!} - 2\frac{c_{n-2}}{6!} + 2\frac{c_{n-3}}{8!} - 2\frac{c_{n-4}}{10!} + \cdots.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Bound on the Beta function For positive integers x and y, we have that
$$
B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} = \frac{1}{x} \left( \begin{array}{c} x+y-1 \\ x \end{array} \right)^{-1} .
$$
However,
$$
\left( \begin{array}{c} a+b \\ a \end{array} \right) = \left(\frac{b+1}{1}\right)\left(\frac{b+2}{2}\right) ... \left(\frac{b+a}{a}\right) \leq (b+1)^a .
$$
Therefore, using $a=x$ and $b=y-1$, we have that
$$
B(x,y) \geq \frac{y^{-x}}{x}.
$$
Is the final inequality true for real values $x \geq 1$ and $y\geq 1$? If so, how is this proved?
| OK, I finally figured it out. This is for integer $x\geq 1$ and real valued $y \geq 1$, which is actually all that I needed. Note that you can swap x and y due to the symmetry of the beta function. I believe that the inequality should hold when neither $x$ nor $y$ are integers (but still at least 1), but I did not actually need this result.
We have that
\begin{equation*}
B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.
\end{equation*}
However, note that $\Gamma(x) = (x-1)! = \frac{x!}{x}$ for integers $x \geq 1$. We also have that
$$
\begin{split}
\Gamma(x+y) & = (x+y-1) \Gamma(x+y-1) \\
&= (x+y-1)(x+y-2) \Gamma(x+y-2) \\
&= \cdots \\
&= \left({\prod_{j=1}^x (x+y-j)}\right) \Gamma(x+y-x) \\
&= \Gamma(y) \left({\prod_{j=1}^x (y-1+j)}\right) .
\end{split}
$$
Putting all of this together, we have that (Equation (1))
$$
\begin{split}
B(x,y) &= \frac{x! \Gamma(y)}{x \Gamma(y) \left({\prod_{j=1}^x (y-1+j)}\right) } \\
&= \frac{1}{x} \frac{x!}{\left({\prod_{j=1}^x (y-1+j)}\right) } \\
&= \frac{1}{x\left({\prod_{j=1}^x \frac{y-1+j}{j} }\right) } .
\end{split}
$$
However, for $j\geq 1$, note that
$$
\begin{split}
y \geq 1 &\implies y-1 \geq 0 \\
&\implies (y-1)(j-1) \geq 0 \\
&\implies 1 - y - j + jy \geq 0 \\
&\implies y-1+j \leq jy \\
&\implies \frac{y-1+j}{j} \leq y .
\end{split}
$$
We also have that $0 < \frac{y-1+j}{j}$, and therefore we have that
$$
{\prod_{j=1}^x \frac{y-1+j}{j} } \leq \prod_{j=1}^x y = y^x .
$$
Combining this inequality with Equation (1), we have that
$$
B(x,y) \geq \frac{y^{-x}}{x} . \phantom{mmmm}\blacksquare
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196130",
"timestamp": "2023-03-29T00:00:00",
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3 urns, each with 4 balls. select one ball from each Three urns are labeled $1,2,3$. Each urn contains $4$ balls labeled $1,2,3,4$. A ball is drawn from each urn such that any ball is equally likely to be drawn. The number on the ball is compared to the number of the urn.
Let $X$ be the number of balls with label higher than the urn it was drawn from. For example if balls $3,3,1$ where drawn from urns $1,2,3$ respectively, $X$ would have a value of $2$.
Calculate the discrete probability mass function for $X$.
| Examine each urn individually, and for urn $i$ for $i\in\{1,2,3\}$ let $X_i=1$ if the the number of the drawn ball is greater than the urn label, or else $X_i=0$.
For urn $1$, the we have $P(X_1=0)=\frac{1}{4}$ and $P(X_1=1)=\frac{3}{4}$
For urn $2$, the we have $P(X_2=0)=\frac{1}{2}$ and $P(X_2=1)=\frac{1}{2}$
For urn $3$, the we have $P(X_3=0)=\frac{3}{4}$ and $P(X_3=1)=\frac{1}{4}$
The task is to calculate the probability mass function of $X_1+X_2+X_3$.
As we have only three urns, we can exhaustively list all eight possible outcomes:
$$P(X_1=0, X_2=0,X_3=0)=\frac{1}{4}\frac{1}{2}\frac{3}{4}=\frac{3}{32}\\
P(X_1=0, X_2=0,X_3=1)=\frac{1}{4}\frac{1}{2}\frac{1}{4}=\frac{1}{32}
\\
P(X_1=0, X_2=1,X_3=0)=\frac{1}{4}\frac{1}{2}\frac{3}{4}=\frac{3}{32}
\\
P(X_1=0, X_2=1,X_3=1)=\frac{1}{4}\frac{1}{2}\frac{1}{4}=\frac{1}{32}
\\
P(X_1=1, X_2=0,X_3=0)=\frac{3}{4}\frac{1}{2}\frac{3}{4}=\frac{9}{32}
\\
P(X_1=1, X_2=0,X_3=1)=\frac{3}{4}\frac{1}{2}\frac{1}{4}=\frac{3}{32}
\\
P(X_1=1, X_2=1,X_3=0)=\frac{3}{4}\frac{1}{2}\frac{3}{4}=\frac{9}{32}
\\
P(X_1=1, X_2=1,X_3=1)=\frac{3}{4}\frac{1}{2}\frac{1}{4}=\frac{3}{32}$$
Thus, we have
$$P(X_1+X_2+X_3=0)=\frac{3}{32}\\
P(X_1+X_2+X_3=1)=\frac{13}{32}
\\
P(X_1+X_2+X_3=2)=\frac{13}{32}
\\
P(X_1+X_2+X_3=3)=\frac{3}{32}$$
| {
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Taylor Series for $\frac{1}{1+e^z}$ and radius of convergence I have done some manipulation and got that $$\frac{1}{1+e^z} = \sum_{n=0}^\infty \frac{n!}{n!+z^n}$$ by the fact that:
$$\frac{1}{1+e^z}= \frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}=\frac{1}{2}+\frac{1}{1+z}+\frac{1}{1+\frac{z^2}{2}}+\ldots = \frac{1}{2}+\frac{1}{1+z}+\frac{2!}{2!+z^2}+\frac{3!}{3!+z^3}+\ldots$$ $$= \sum_{n=0}^\infty\frac{n!}{n!+z^n}$$
Assuming I did the above right, I am having trouble finding the radius of convergence of the Taylor series given above. I tried the ratio test but got stuck.
Edit: I just realized I did this completely wrong, thinking that I was taking $$\sum\frac{1}{1+\sum_{n=0}^\infty\frac{z^n}{n!}}$$.
Could anyone help me? My question wants me to compute the first four terms of the taylor series for $\frac{1}{1+e^z}$ and find the radius of convergence. Perhaps they do not want me to actually find the explicit form?
| One option is: Write
$$\frac{1}{1 + e^z} \cdot (1 + e^z) = 1$$
and expand both of the factors on the l.h.s. in Taylor series" Since $e^z \sim \sum_{k = 0}^{\infty} \frac{1}{k!} z^k$, we have
$$1 + e^z = 1 + \left(1 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + O(z^5)\right) = 2 + z + \frac{z^2}{2} + \frac{z^3}{6} + \frac{z^4}{24} + O(z^5).$$
Then, expand the (t.b.d.) Taylor series of $\frac{1}{1 + e^z}$ as $$a_0 + a_1 z + a_2 z^2 + a_3 z^3 + a_4 z^4 + O(z^5).$$
The product of these two series is
$$2 a_0 + (a_0 + 2 a_1) z + \left(\frac{1}{2}a_0 + a_1 + 2 a_2\right) + (\text{third- and fourth-order terms}) + O(z^5),$$
and the polynomial part here must agree with $1$ to fourth order. (Note that we only need consider the involved Taylor expansions here to order $4$ to determine $a_0, \ldots, a_4$, that is, to determine the desired Taylor series to that order.)
So, we get a system of five linear equations in the coefficients $a_0, \ldots, a_4$:
\begin{align}
1 &= 2 a_0 \\
0 &= a_0 + 2 a_1 \\
0 &= \frac{1}{2} a_0 + a_1 + 2 a_2 \\
\vdots &= \vdots .
\end{align}
Solving and substituting gives the desired Taylor series.
| {
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Evaluating $ \sum\limits_{n=1}^\infty \frac{1}{n^2 2^n} $
Evaluate
$$ \sum_{n=1}^\infty \dfrac{1}{n^2 2^n}. $$
I have tried using the Maclaurin series of $2^{-n}$ but it further complicated the question. Moreover, I have also tried taking help from another question when the $n^2$ is in the numerator, but no significant progress so far.
Any help will be appreciated.
Thanks!
| Given: $$ \text{S}= \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2.2^n} $$
Consider the following Lemma,
Lemma:
$$ \displaystyle \int_{0}^1 \dfrac{ \ln x }{x+1} \mathrm{d}x = -\dfrac{\pi^2}{12}$$
Proof:
$ \displaystyle \int_{0}^1 \dfrac{ \ln x }{x+1} \mathrm{d}x = \displaystyle \int_{0}^1 \dfrac{ -\ln x }{1-x}\mathrm{d}x + \displaystyle \int_{0}^1 \dfrac{ -2\ln x }{x^2-1}\mathrm{d}x$
Also,
$\displaystyle \int_{0}^1 \dfrac{ -\ln x }{1-x}\mathrm{d}x = \displaystyle \int_{0}^1 \dfrac{ -\ln (1-x) }{x}\mathrm{d}x = \displaystyle \int_{0}^1 \dfrac{x+\frac{x^2}{2}+\frac{x^3}{3}+...}{x}\mathrm{d}x = \sum_{k=1}^\infty \dfrac{1}{k^2} = \frac{\pi^2}{6}$
and,
$\displaystyle \int_{0}^1 \dfrac{ \ln x }{x^2-1}\mathrm{d}x = \displaystyle \sum_{n=0}^{\infty} \int_{0}^1 x^{2n}.{\ln{x}} \ \mathrm{d}x $
$=- \displaystyle \sum_{n=0}^{\infty} {\dfrac{1}{(2n+1)^2}}$ (Using Integration By Parts)
$= \displaystyle \sum_{n=0}^{\infty} {\dfrac{1}{(2n)^2}} - \sum\limits_{n=0}^{\infty} {\dfrac{1}{n^2}}$
$=\dfrac{-3}{4}\zeta(2)= -\dfrac{\pi^2}{8}$
$\therefore \displaystyle \int_{0}^1 \dfrac{ \ln x }{x+1} \mathrm{d}x = \dfrac{\pi^2}{6} - \dfrac{\pi^2}{4} = -\dfrac{\pi^2}{12}$
This completes the proof of our Lemma.
Now,
$ \text{S}= \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2.2^n} $
$= - \displaystyle \sum_{n=1}^{\infty} \left( \dfrac{2^n-1}{n^2.2^n} - \dfrac{1}{n^2} \right)$
$ = - \displaystyle {\ln 2} \sum_{n=1}^{\infty} \left(\dfrac{1}{n} \int_{0}^1 2^{-nx} \mathrm{d}x \right) + \dfrac{\pi^2}{6}$
$ = - \displaystyle {\ln 2} \int_{0}^1 \left(\sum_{n=1}^{\infty} \dfrac{1}{n} \times {2^{-nx}} \right)\mathrm{d}x + \dfrac{\pi^2}{6}$
$ = \displaystyle {\ln 2} \int_{0}^1 \ln \left( \dfrac{2^x - 1}{2^x} \right)\mathrm{d}x + \dfrac{\pi^2}{6}$
$ = \displaystyle {\ln 2} \int_{0}^1 \ln (2^x - 1)\ \mathrm{d}x + \dfrac{\pi^2}{6} -\dfrac{\ln^2 2}{2}$
Now, substituting $(2^x - 1) = t$, we have,
$\text{S} = {\ln 2} \times \dfrac{1}{\ln 2} \displaystyle \int_{0}^1 \dfrac{ \ln t }{t+1} \mathrm{d}t + \dfrac{\pi^2}{6} -\dfrac{\ln^2 2}{2}$
$ = -\dfrac{\pi^2}{12} + \dfrac{\pi^2}{6} -\dfrac{\ln^2 2}{2}$ (Using the Lemma)
$=\boxed{\dfrac{\pi^2}{12} - \dfrac{\ln^2 2}{2}}$
| {
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How does this line came? :)
Today I started to learn algebra with my own. And the first chapter which I'm learning is Ratio. I'm kind of confuse in an example, in all the lines starting with (*). Until first three lines of the solution I understand but I didn't get how
$$\frac{ny + mz}a = \frac{lz + nx}b = \frac{mx + ly}c$$ came.
Please help and thank you in advance.
And sorry for my bad English.
Example 2. If $\frac{x}{l(mb+nc-la)}=\frac{y}{m(nc+la-mb)}=\frac{z}{n(la+mb-nc)}$, prove that $\frac{l}{x(by+cz-ax)}=\frac{m}{y(cz+ax-by)}=\frac{n}{z(ax+by-cz)}$.
We have
\begin{align} \frac{\frac xl}{mb+nc-la} &= \frac{\frac ym}{nc+la-mb} = \frac{\frac zn}{la+mb-nc} \\ &= \frac{\frac ym + \frac zn}{2la} \\ &= \text{two similar expressions}; \end{align}
$$\therefore \quad \frac{ny+mz}a = \frac{lz+nx}b = \frac{mx+ly}c. \tag{*}$$
Multiply the first of these fractions above and below by $x$, the second by $y$, and the third by $z$; then
\begin{align} \frac{nxy+mxz}{ax} &= \frac{lyz+nxy}{by} = \frac{mxz+lyz}{cz} \\ &= \frac{2lyz}{by+cz-ax} \\ &= \text{two similar expressions}; \end{align}
$$\therefore \quad \frac{l}{x(by+cz-ax)}=\frac{m}{y(cz+ax-by)}=\frac{n}{z(ax+by-cz)}$$
| If we have:$$\frac{a}{b}=\frac{c}{d}$$then this implies:$$\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\tag{1}$$You have:$$\frac{\frac{x}{l}}{mb+nc-la}=\frac{\frac{y}{m}}{nc+la-mb}=\frac{\frac{z}{n}}{la+mb-nc}$$Since these 3 terms are all equal, we can apply (1) to any two of them. So, lets start by applying (1) to the first 2 terms to get:$$\frac{\frac{x}{l}+\frac{y}{m}}{mb+nc-la+nc+la-mb}=\frac{\frac{x}{l}+\frac{y}{m}}{2nc}$$Similarly, applying (1) to the 2nd the 3rd terms gives:$$\frac{\frac{y}{m}+\frac{z}{n}}{nc+la-mb+la+mb-nc}=\frac{\frac{y}{m}+\frac{z}{n}}{2la}$$And finally, applying (1) to the 1st the 3rd terms gives:$$\frac{\frac{x}{l}+\frac{z}{n}}{mb+nc-la+la+mb-nc}=\frac{\frac{x}{l}+\frac{z}{n}}{2mb}$$All 3 of these must be equal to one another, therefore:$$\frac{\frac{x}{l}+\frac{y}{m}}{2nc}=\frac{\frac{y}{m}+\frac{z}{n}}{2la}=\frac{\frac{x}{l}+\frac{z}{n}}{2mb}$$We can multiply everything by 2 to get:$$\frac{\frac{x}{l}+\frac{y}{m}}{nc}=\frac{\frac{y}{m}+\frac{z}{n}}{la}=\frac{\frac{x}{l}+\frac{z}{n}}{mb}$$$$\therefore\frac{mx+ly}{lmnc}=\frac{ny+mz}{lmna}=\frac{nx+lz}{lmnb}$$Finally we multiply everything by $lmn$ to get:$$\frac{mx+ly}{c}=\frac{ny+mz}{a}=\frac{nx+lz}{b}$$
| {
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"url": "https://math.stackexchange.com/questions/1211782",
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"source": "stackexchange",
"question_score": "3",
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Interpretation of a combinatorial identity involving iterated binomial coefficients I am trying to find an combinatorial interpretation for the following combinatorial identity involving iterated binomial coefficients, which appeared in the November 1980 edition of The American Math Monthly:
$\dbinom{\binom{n}{b}}{2}=\displaystyle\sum_{j=1}^b\dbinom{\binom{b}{j}+e_j}{2}\binom{n+b-j}{2b},$
where $e_j=\frac{1+(-1)^{j+1}}{2}$, that is, $e_j=1$ if $j$ is odd, and $0$ otherwise.
Essentially, the left hand side of the identity can be interpreted as the number of ways to choose 2 $b$-element subsets of $\{1,2,\cdots,n\}$. What I am interested in is the derivation of the right hand side of the identity; I have read the proof in The American Math Monthly, and the author mentioned that the expression $\binom{n+b-j}{2b}$ refers to the selection of $2b$ objects from the original set $\{1,2,\cdots,n\}$, augmented by the adjunction of $b-j$ "jokers", where $0\leq j\leq b-1$, to allow for the fact that the intersection of two distinct $b$-element subsets of $\{1,2,\cdots,n\}$ can have a minimum and a maximum cardinality of $0$ and $b-1$ respectively.
What I am struggling to understand here is the expression $\dbinom{\binom{b}{j}+e_j}{2}$; how does one interpret this expression? Alternatively, is there any other way for which one can prove the identity? Personally, I have proven via a combination of combinatorial and algebraic methods that the identity does hold for small values of $b$, but the expression is not that tractable for large values of $b$.
| Here is an answer using formal power series that is somewhat simpler
than the first one. We start as before:
$$\frac{1}{2}
\sum_{j=0}^b \left({b\choose j} + e_j\right)
\left({b\choose j} + e_j-1\right)
{n+b-j\choose 2b}$$
where we have $e_j = [[j\;\text{is odd}]].$
Expanding we get three pieces, the first is
$$A = \frac{1}{2}
\sum_{j=0}^b {b\choose j}^2
{n+b-j\choose 2b},$$
the second is
$$B = \frac{1}{2}
\sum_{j=0}^b
{b\choose j} (2e_j-1)
{n+b-j\choose 2b}
= \frac{1}{2}
\sum_{j=0}^b
{b\choose j} (-1)^{j+1}
{n+b-j\choose 2b}
$$
and the third
$$C = \sum_{j=0}^b
e_j (e_j-1)
{n+b-j\choose 2b} =0,$$
which leaves $A$ and $B.$ Starting with $A$ we observe that we must
have $n\ge b$ or else the third binomial coefficient vanishes for all
$j$ in the range, making for a zero contribution. We find
$$\frac{1}{2}
\sum_{j=0}^b {b\choose j} [z^{b-j}] (1+z)^b
[w^{n-b-j}] (1+w)^{n+b-j}
\\ = \frac{1}{2} [z^b] (1+z)^b [w^{n-b}] (1+w)^{n+b}
\sum_{j=0}^b {b\choose j} z^j w^j (1+w)^{-j}
\\ = \frac{1}{2} [z^b] (1+z)^b [w^{n-b}] (1+w)^{n+b}
\left(1+\frac{wz}{1+w}\right)^b
\\ = \frac{1}{2} [z^b] (1+z)^b [w^{n-b}] (1+w)^{n}
(1+w+wz)^b.$$
Recalling that $n\ge b$ this becomes
$$\frac{1}{2} [z^b] (1+z)^b [w^{n-b}] (1+w)^{n}
\sum_{q=0}^b {b\choose q} w^q (1+z)^q
\\ = \frac{1}{2} [z^b] (1+z)^b
\sum_{q=0}^b {b\choose q} {n\choose n-b-q} (1+z)^q
\\ = \frac{1}{2}
\sum_{q=0}^b {b\choose q} {n\choose n-b-q} {b+q\choose b}.$$
Factoring the two binomials on the right we get
$$\frac{n!}{(n-b-q)! \times b! \times q!}
= {n\choose b} {n-b\choose q}.$$
We continue with
$$\frac{1}{2} {n\choose b}
\sum_{q=0}^b {b\choose q} {n-b\choose q}
= \frac{1}{2} {n\choose b}
\sum_{q=0}^b {b\choose q} [z^{n-b-q}] (1+z)^{n-b}
\\ = \frac{1}{2} {n\choose b} [z^{n-b}] (1+z)^{n-b}
\sum_{q=0}^b {b\choose q} z^q
= \frac{1}{2} {n\choose b} [z^{n-b}] (1+z)^{n-b} (1+z)^b
\\ = \frac{1}{2} {n\choose b}^2.$$
The piece labeled $B$ is next. We get
$$\frac{1}{2} \sum_{j=0}^b
{b\choose j} (-1)^{j+1}
[w^{n-b-j}] (1+w)^{n+b-j}
\\ = - \frac{1}{2} [w^{n-b}] (1+w)^{n+b}
\sum_{j=0}^b
{b\choose j} (-1)^{j} w^j (1+w)^{-j}
\\ = - \frac{1}{2} [w^{n-b}] (1+w)^{n+b}
\left(1-\frac{w}{1+w}\right)^b
\\ = - \frac{1}{2} [w^{n-b}] (1+w)^{n}
= - \frac{1}{2} {n\choose b}.$$
We have shown that
$$A+B = \frac{1}{2} {n\choose b}^2 - \frac{1}{2} {n\choose b}
= \frac{1}{2} {n\choose b}
\left({n\choose b} - 1\right)$$
as claimed.
Remark. The piece $B$ was also evaluated in the first answer.
| {
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Prove that $n(r) < 2\pi \sqrt[3]{r^{2}}$
Suppose that $n(r)$ denotes the numbers of points with integer coordinates on a circle of radius $r > 1$. Prove that
$$
n(r) < 2\pi \sqrt[3]{r^{2}}
$$
What process would you use to resolve the last one?
| Label the $n(r)$ points $P_1, P_2, \ldots, P_n$ counterclockwise. A general strategy with problems regarding lattice points is to somehow use the area (using the shoelace formula with integer coordinates gives you good bounds). In this case, looking at any triangle the area will be at least $1/2.$ Now we can in particular pick a triangle with "small" area. Restricting our attention to triangles $P_i P_{i+1} P_{i+2}$ and realizing that this area effectively depends on the angle $P_i P_{i+1} P_{i+2},$ and hence the arc $P_i P_{i+2},$ we relabel so that $P_1OP_3$ has angle at most $\frac{4\pi}{n}.$
Now $P_1 P_2 P_3$ has area at most $\frac{1}{2} P_1P_2\cdot P_2P_3\sin \frac{2\pi}{n} \le \frac{\pi}{n} P_1P_2 \cdot P_2P_3.$ But $P_1 P_2 = 2r\sin \alpha,$ where $\alpha$ is half the measure of the arc $P_1 P_2$ (divide by $r$ of course) and $P_2 P_3 = 2r \sin \beta,$ with $\beta$ defined similarly. Notice that $\alpha + \beta$ is at most $\frac{2\pi}{n}$ and thus $\sin \alpha \cdot \sin\beta = \frac{1}{2} \cos(\alpha-\beta) - \frac{1}{2}\cos(\alpha + \beta)\le \frac{1}{2} - \cos\frac{2\pi}{n},$ with equality when $\alpha = \beta = \frac{\pi}{n}.$
Consequently $\frac{1}{2}\le [\triangle P_1 P_2 P_3] \le \frac{4\pi r^2}{n}\sin \alpha \sin\beta \le \frac{4\pi^3 r^2}{n^3},$ whence $n \le 2\pi \sqrt[3]{r^2}.$
Strict inequality isn't difficult, if you really want it. Question for you: where was the assumption $r>1$ used?
| {
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Using factorisation, $A=PJP^{-1}$ to compute $A^k$ Using factorisation, $A=PJP^{-1}$ to compute $A^k$, where $k$ represents an arbitrary positive integer.
$$ \begin{bmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{-1} & \mathbf{2} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{1} \\ \mathbf{0} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{-1} & \mathbf{1} \end{bmatrix}. $$
Not sure how to finish this problem as the matrix $J$ is not diagonal, but $J^k$ can still be found somehow. How exactly is $J^k$ obtained in this case?
| You can easily prove (for example with induction) that
$${\begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix}}^k=\begin{pmatrix} 1 & k \\ 0 & 1 \\ \end{pmatrix}$$
Now $A^k$ can be computed as follows:
$A^k=(S^{-1}BS)^{k}=\underbrace{S^{-1}BS\cdot S^{-1}BS\dots S^{-1}BS}_{k \text{ times }}=S^{-1}B^{k}S=S^{-1}\begin{pmatrix} 1 & k \\ 0 & 1 \\ \end{pmatrix}S$
| {
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"question_score": "2",
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How to solve this limit: $\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$ Compute the limit $\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}$
my attempt:
I tried to multiply top and bottom by the conjugate
$$\begin{align}
\lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-1}\right)\frac{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{\left(\sqrt{x+\sqrt{x}}\right)^2-\left(\sqrt{x-1}\right)^2}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{(x+\sqrt{x})-(x-1)}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\
&=\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}
\end{align}$$
But I don't know what I can do after this.
| We have $$\frac{1}{2}=\lim_{x\rightarrow\infty}\frac{\sqrt{x}}{2\sqrt{x}\left(\sqrt{\left(1+\frac{1}{\sqrt{x}}\right)}\right)}\leq\lim_{x\rightarrow\infty}\frac{\sqrt{x}}{\sqrt{x\left(1+\frac{1}{\sqrt{x}}\right)}+\sqrt{x}}\leq\lim_{x\rightarrow\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\leq\lim_{x\rightarrow\infty}\frac{1+\sqrt{x}}{2\sqrt{x}}=\frac{1}{2}$$
| {
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"source": "stackexchange",
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Indefinite Integral with "sin" and "cos": $\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $ Indefinite Integral with sin/cos
I can't find a good way to integrate: $$\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx $$
| We can split this into two integrals:
$$3 \int \frac{\sin(x)}{2\sin(x)+3\cos(x)}dx + 2 \int \frac{\cos(x)}{2\sin(x)+3\cos(x)}dx.$$
Focusing on the second integral we find:
$$\int \frac{\cos(x)}{2\sin(x)+3\cos(x)}dx = \int \frac{1}{2\tan(x)+3}dx$$
Make the substitution $u=2\tan(x)+3$ which makes $$du = 2\sec^2(x)dx = 2(\tan^2(x)+1)dx = 2(u^2+1)dx.$$
Thus we have
$$\int \frac{\cos(x)}{2\sin(x)+3\cos(x)}dx = \frac12 \int \frac{1}{u(u^2+1)} du = \frac12 \int \left(\frac{1}{u} - \frac{u}{u^2+1}\right) du.$$
This integral can be computed by another substitution:
$$\int \frac{\cos(x)}{2\sin(x)+3\cos(x)}dx=\frac12 \left(\ln|2\tan(x)+3| - \frac12 \ln|(2\tan(x)+3)^2+1|\right) + C.$$
That completes the second integral. The first can be handled in a similar manner.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 1
} |
What is a closed form for $\sum_{r=1}^k \binom nr$, where $k\leqslant n$. What will be the closed form of the following equation
$$\sum_{r=1}^k C(n,r), $$ where $n$ and $k$ are positive integers with $k\leqslant n$?
| $$\begin{align}
S=\sum_{\begin{array}{c}0\le k\le n\\1\le r\le k\end{array}}\binom{n}{r}&=\sum_{k=0}^n\sum_{r=1}^k\binom{n}{r}\\
&=\left[\binom{n}{1}\right]+\left[\binom{n}{1}+\binom{n}{2}\right]\dotsm+\left[\binom{n}{1}+\dotsm+\binom{n}{n}\right]\\
&=n\binom{n}{1}+(n-1)\binom{n}{2}+(n-2)\binom{n}{3}+\dotsm+\binom{n}{n}\\
&=\sum_{i=1}^n(n+1-i)\binom{n}{i}\\
&=-(n+1)\binom{n}{0}+\sum_{k=0}^n(n+1-i)\binom{n}{i}
\end{align}$$
By the binomial theorem,
$$\begin{align}
(1+x)^n&=\sum_{i=0}^n\binom{n}{i}x^{n-i}&\text{multiply by $x$ ...}\\
x(1+x)^n&=\sum_{i=0}^n\binom{n}{i}x^{n+1-i}&\text{differentiate ...}\\
(1+x)^n+nx(1+x)^{n-1}&=\sum_{i=0}^n(n+1-i)\binom{n}{i}x^{n-i}&\text{set $x=1$ ...}\\
2^n+n2^{n-1}&=\sum_{i=0}^n(n+1-i)\binom{n}{i}
\end{align}$$
and combining these two result gives us
$$
\sum_{\begin{array}{c}0\le k\le n\\1\le r\le k\end{array}}\binom{n}{r}=2^n+n2^{n-1}-1=(n+2)2^{n-1}-(n+1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find ways to pick 3 numbers such that their product is a multiple of 8 $Let$ $A=\{1,2,3,4,5,6,7,8,9\}$
How many ways are there to pick 3 non-repeated numbers from the set $A$ such that their product is a multiple of 8?
My attempt:
I need to fit three $2$s in the multiplication and after that I can pick whatever to get 3 numbers.
I tried separating this by cases:
I pick $8$ then I have $2$ slots left, and so I pick $2$ out of the $8$ numbers remaining.
Case 2: I pick $2$ first, then $4$, then whatever number is left.
Case 3: I pick $6$, etc...
But then I have repeated scenarios between my cases, how can I approach this?
| To obtain a multiple of $8$, we must select numbers containing at least three factors of $2$. We can do this by selecting an $8$, by selecting a $2$ and a $4$, or by selecting a $4$ and a $6$.
Let $E$ be the event that you select $8$ and two other numbers from set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Let $F$ be the event that you select $2$, $4$, and a third number from set $A$. Let $G$ be the event that you select $4$, $6$, and a third number from set $A$. The number of ways to pick three elements from set $A$ such that their product is a multiple of $8$ is $|E \cup F \cup G|$, where $|S|$ denotes the number of elements in set $S$.
The Inclusion-Exclusion Principle states that
$$|E \cup F \cup G| = |E| + |F| + |G| - |E \cap F| - |E \cap G| - |F \cap G| + |E \cap F \cap G|$$
As you observed, the number of elements in event $E$ is $\binom{8}{2}$ since we must select $8$ and two of the other eight numbers in set $A$.
The number of elements in event $F$ is $7$ since we must select $2$, $4$, and one of the other seven numbers in set $A$.
The number of elements in event $G$ is also $7$ since we must select $4$, $6$, and one of the other seven numbers in set $A$.
Since $E \cap F = \{2, 4, 8\}$, $|E \cap F| = 1$.
Since $E \cap G = \{4, 6, 8\}$, $|E \cap G| = 1$.
Since $F \cap G = \{2, 4, 6\}$, $|F \cap G| = 1$.
Observe that $E \cap F \cap G = \emptyset$ since no three element set contains $2$, $4$, $6$, and $8$.
Hence, the number of three elements of set $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ in which the product is a multiple of $8$ is
\begin{align*}
|E \cup F \cup G| & = |E| + |F| + |G| - |E \cap F| - |E \cap G| - |F \cap G| + |E \cap F \cap G|\\
& = \binom{8}{2} + 7 + 7 - 1 - 1 - 1 + 0\\
& = 28 + 7 + 7 - 1 - 1 - 1 + 0\\
& = 39
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Sum of square patterns Can anyone give the name of this pattern
$$136^2+137^2+138^2+139^2+140^2+141^2+142^2+143^2+144^2 =\\
145^2+146^2+147^2+148^2+149^2+150^2+151^2+152^2$$
| If we consider,
$3^2+4^2=5^2$ as triangle,
$10^2+11^2+12^2=13^2+14^2$ as pentagon,
$21^2+22^2+23^2+24^2=25^2+26^2+27^2$ as heptagon.
Then, interestingly, every structure has a right angled triangle as below.
Triangle: $3^2+4^2=5^2,$
Pentagon: $5^2+12^2=13^2,$
Heptagon: $7^2+24^2=25^2.$
The least length leg gives the count of the series as explained above in every right angled triangle.
Trigonometric Structure of the above series gives a common pattern that looks like a Olympic Torch or similar ones.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Prove this form $2^{n+2}-7a^2_{n}$ always is square numbers. Let sequence $\{a_{n}\}$ such $a_{1}=1,a_{2}=-1$,and
$$a_{n}+a_{n-1}+2a_{n-2}=0,n\ge 3$$
show that
$2^{n+2}-7a^2_{n}$ is square numbers.
such as
$$2^3-7=1,n=1$$
$$2^4-7=9=3^2,n=2$$
$a_{3}=-1$
$$2^{5}-7=25=5^2,n=3$$
$a_{4}=3$
$$2^6-7\cdot 9=1^2,n=4$$
| One finds
$$a_n=\frac{1}{\sqrt 7i}\left(A^n-B^n\right)\ \ (n\ge 1)$$ where $A=-\frac 12-\frac{\sqrt 7}{2i},B=-\frac 12+\frac{\sqrt 7}{2i}$.
Noting that $AB=2$, one finds
$$\begin{align}2^{n+2}-7{a_n}^2&=2^{n+2}-7\cdot\frac{1}{-7}\left(A^{2n}+B^{2n}-2\cdot 2^n\right)\\&=2^{n+2}+A^{2n}+B^{2n}-2^{n+1}\\&=2^{n+1}(2-1)+A^{2n}+B^{2n}\\&=\left(A^n+B^n\right)^2.\end{align}$$
Now it should be easy to prove that $A^n+B^n$ is an integer for every $n\ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating limit at infinity (algebric issue) Q: evaluate $\lim_{x \to \infty}$ $ (x-1)\over \sqrt {2x^2-1}$
What I did:
when $\lim_ {x \to \infty}$ you must put the argument in the form of $1/x$ so in that way you know that is equal to $0$
but in this ex. the farest that I went was
$\lim_{x \to \infty}$ $x \over x \sqrt{2}$ $1-(1/x) \over (1/x) - ??$
| $$\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-2}}$$
$$\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2}x}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2(x^2-1)}}$$
$$\lim_\limits{x\to\infty}\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}x}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\sqrt{\dfrac{x-1}{2(x+1)}}$$
$$\dfrac{1}{\sqrt{2}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\sqrt{\dfrac{x+1-2}{2(x+1)}}$$
$$\dfrac{1}{\sqrt{2}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\sqrt{\dfrac{1}{2}-\dfrac{1}{x+1}}$$
$$\dfrac{1}{\sqrt{2}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\dfrac{1}{\sqrt{2}}$$
In fact we can say $\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\to
\dfrac{1}{\sqrt{2}}$ from beneath.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1224180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\int^2_0 x(8-x^3)^\frac{1}{3}dx=\frac{16\pi}{9\sqrt{3}}$ I have to prove that:
$$\int^2_0 x(8-x^3)^\frac{1}{3}dx=\frac{16\,\pi}{9\sqrt{3}}.$$
I tried substituting $x^3=8u$ but I just got stuck.
Any help would be appreciated.
| By setting for first $x=2z$ we have:
$$I=\int_{0}^{2}x(8-x^3)^{1/3}\,dx = 8\int_{0}^{1}z(1-z^3)^{1/3}\,dz \tag{1}$$
and by setting $z=t^{1/3}$ we have:
$$I=\frac{8}{3}\int_{0}^{1}t^{-1/3}(1-t)^{1/3}\,dt = \frac{8}{3}\cdot\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma(2)}\tag{2}$$
through Euler's Beta function. Exploiting the reflection formula for the $\Gamma$ function:
$$ I = \frac{8}{9}\,\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)=\frac{8\pi}{9\sin\frac{\pi}{3}}=\color{red}{\frac{16\,\pi}{9\sqrt{3}}}\tag{3}$$ as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\frac{t^2}{t+1}$ to the form $\frac{1}{t+1} +t -1$ One of my attempts would look like below.
$\frac{t^2}{t+1}$ = $\frac{t \times t+1-1}{t+1}$ = 1+ $\frac{t-1}{t+1}$ = $\frac{t-1+1-1}{t+1} + 1$ = $\frac{-2}{t+1} +2$
Also, I put into t an arbitrary number in both equations and have gotten the same answer. The first number was 18785 to which both contained the answer $\frac{352876225}{18786}$.
| we have $$\frac{t^2}{t+1}=\frac{t^2-1+1}{t+1}=\frac{(t-1)(t+1)+1}{t+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Possible solutions of a diophantine equation: $p^2+pq+275p+10q=2008$ What are couples of prime integers that verify this diophantine equation: $$p^2+pq+275p+10q=2008?$$
I tried to solve this equation trough the rules of modular-arithmetic. I rewrite the equation as: $$p^2+pq\equiv 2008 \pmod5.$$ The equation can be rewrite also as: $p^2+pq\equiv 3 \pmod5$.
Now I analyzed five possible cases: $p^2\equiv 0$, $p^2\equiv 1$, $p^2\equiv 2$, $p^2\equiv 3$ and $p^2\equiv 4$.
*
*We can not that if $p^2\equiv 0$ also $pq\equiv 0$ therefore the equation is impossible
*$p^2\equiv 2$ is impossible because $2$ is a non-quadratic residue $\pmod5$
*also $p^2\equiv 3$ is impossible because $3$ is a non-quadratic residue $\pmod5$.
*If $p^2\equiv 1 \pmod5$, $p\equiv \pm 1$, therefore $p\equiv -1\equiv 4 \pmod5$ and $p2\equiv 1 \pmod5$. These values are not acceptable because if $p=9$ ($9\equiv 4 \pmod5$) $275\cdot 9>2008$, if $p=11$ ($11\equiv 1\pmod5$) $275\cdot 9>2008$ therefore two values aren't acceptable.
*If $p^2\equiv 4 \pmod5$, $p\equiv \pm2$ and if $p\equiv -2\equiv 3 \pmod5$ there not exist solutions for the same reason of previous example. If $p\equiv 2 \pmod5$ possible values of $p$ are $2$ or $7$ because $275p<2008$. Replacing $p=2$ we obtain a valor not integer of $q$ while replacing $p=7$ we obtain $q=2$ therefore this is the only solution of equation.
Is correct my process or are there other solutions?
Thanks:)
| This isn't the most clever approach, but it is a universal approach to quadratic diophantine equations to start by removing linear terms (completing the square).
$$
\begin{align}
p^2+pq+275p+10q
&=2008
\end{align}
$$
Shift to remove linear terms: $p=u+a,q=v+b$
$$
\begin{align}
(u+a)^2+(u+a)(v+b)+275(u+a)+10(v+b)
&=2008\\
u^2+uv+(2a+b+275)u+(a+10)v
&=2008-a^2-ab-275a-10b
\end{align}
$$
Now we can set $a=-10$ and $b=-255$ to eliminate the linear terms:
$$
\begin{align}
u^2+uv
&=2008-100-2550+2750+2550\\
u(u+v)&=4658=2\cdot17\cdot137
\end{align}
$$
There are only $16$ possible integers solutions for $(u,v)$, which then are shifted to corresponding values for $(p,q)$. You can inspect which pairs $(p,q)$ are positive primes. Actually since you know you are looking for positive $p$, and $p=u-10$, then $u$ is at least $12$, so there are only six solutions for $u$: $17,34,137,274,2329$, and $4658$. This is just a shortcut to save yourself some checking.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to determine if an equation represents a cubic spline? Given the equation
$$
f(x) = \left\{
\begin{array}{lr}
2x^3+x^2+4x+5 & : 0 \le x \le 1\\
(x-1)^3 + 7(x-1)^2 + 12(x-1)+12 & : 1 \le x \le 2
\end{array}
\right.
$$
What is the process used to determine if this represents a cubic spline?
It's obviously a piecewise interpolation, and the second function simplifies, so perhaps it's better to write like:
$$
f(x) = \left\{
\begin{array}{lr}
2x^3+x^2+4x+5 & : 0 \le x \le 1\\
x^3+4x^2+x+6 & : 1 \le x \le 2
\end{array}
\right.
$$
| Both expression are obviously cubic. If they both have the same value and the same first and second derivatives at $x=1$, your set of expressions fits the definition of a cubic spline. A pretty simple spline, but there it is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve these equations simultaneously (trig) Solve for $ x,y: $
\begin{equation}\cos x -\cos(x+y) = 0
\end{equation}
\begin{equation}\cos y -\cos(x+y) = 0
\end{equation}
The answers are $(0, 0), (\frac{2\pi}{3}, \frac{2\pi}{3})$.
I get $(0, 0)$, but how do you get the latter?
Sorry if this is a bit basic - I don't remember trig well...
tfa
| Given
$$\cos x-\cos(x+y)=0\\
\cos y-\cos(x+y)=0$$
Subtracting the second equation from the first yields
$$\cos x=\cos y$$
Recall that $\cos x=\pm\sqrt{1-\sin^2x}$ and $\cos y=\pm\sqrt{1-\sin^2x}$.
This now becomes $\sqrt{1-\sin^2x}=\pm\sqrt{1-\sin^2y}$ yielding
$$\sin x=\pm\sin y$$
Recall that $\cos(x+y)=\cos x\cos y-\sin x\sin y$.
Thus, $\cos x-\cos(x+y)=0$ now becomes $\cos x-\cos^2x\pm\sin^2x=0$.
Solving $\cos x-\cos^2x-\sin^2x=0$ yields $\cos x=1$.
Solving $\cos x-\cos^2x+\sin^2x=0$ yields $\cos x-\cos^2x+1-\cos^2x=0$ which is equivalent to
$$(2\cos x+1)(-\cos x+1)=0$$
This yields two possibilities: $\cos x=1$ or $\cos x=-\frac12$. Can you do the rest?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Working out $\tan x$ using sin and cos expansion
Using only the series expansions $\sin x = x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...$ and $\cos x = 1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ...$
Find the series expansions of the $\tan x$ function up to the $x^5$ term.
So it is:
$$
\frac {x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...} {1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ...}
$$
However many times I retry, trying long division results :
$$
\require{enclose}
\begin{array}{rll}
Q(x) \\[-3pt]
1 - \dfrac{x^2} {2!} + \dfrac{x^4}{4!} + ... \enclose{longdiv}{x- \dfrac{x^3} {3!} + \dfrac{x^5} {5!} + ...}\kern-.2ex \\[-3pt]
\underline{x-\dfrac{x^3} {2!} + \dfrac{x^5} {4!}\phantom{0000}} \\[-3pt]
\dfrac{-x^3} {3!} + \dfrac{x^3} {2!}+ \dfrac{x^5} {5!}-\dfrac{x^5} {4!} \phantom{00}&& \\[-3pt]
\underline{\phantom{0}\dfrac{-x^3} {3!} + \dfrac{x^3} {2!}- \dfrac{x^5} {2!2!}+ \dfrac{x^5} {3!2!}} && \\[-3pt]
\dfrac{x^5} {5!} - \dfrac{x^5} {4!} + \dfrac{x^5} {2!2!}-\dfrac{x^5} {3!2!}&& \\[-3pt]
\end{array}
$$
Sorry for the horrible format. but this is the best i can do..
$$
Q(x) = x + (\frac {x^3}{2!} - \frac {x^3}{3!}) + (\dfrac {x^5}{5!} - \dfrac {x^5}{4!} + \dfrac {x^5}{2!2!} - \dfrac {x^5}{3!2!}) + ...
$$
Which results the coefficients of the $x^5$ term $\dfrac 1 {20}$.
This is clearly a wrong value.
I can't seem to get $\dfrac 2 {15}$ instead.
Where did I get wrong?
Many thanks in advance.
| You get that the coefficient for $x^5$ is $\dfrac {1}{5!} - \dfrac {1}{4!} + \dfrac {1}{2!2!} - \dfrac {1}{3!2!}$, which is indeed $\dfrac2{15}$. So, no need to change method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Area of triangle formed by angle bisector, altitude and median Question:-
Given a triangle ABC with side length a, b and c. Calculate the area of a triangle in terms of a, b and c formed by angle bisector from vertex A, altitude from vertex B and median from vertex C.
looking for solution using complex number, vectors and co-ordinate geometry
Regards,
vishal
|
Given $\triangle ABC$, $BC=a,\ AC=b,\ AB=c$;
$\angle BAD=\angle CAD$,
$BE\perp AC$, $AF=BF$,
$\angle BAC=\alpha$,
$\angle BCA=\gamma$.
First, define coordinates of the points $A,B,C,E$ and $F$:
\begin{align}
A &=(0,0),\\
B &=(c\cos\alpha,c\sin\alpha), \\
C &=(b,0),\\
E &=(c\cos(\alpha),0),\\
F &=(c/2\cos\alpha,c/2\sin\alpha) \\
\end{align}
Coordinates of the point $D$ can be defined as follows:
\begin{align}
D_y &= x\tan\gamma = (b-x)\tan\frac{\alpha}{2},
\\
x&=\frac{b}{\tan\gamma+\tan\frac{\alpha}{2}},
\\
D&=\left(
\frac{b}{\tan\frac{\alpha}{2}(1+\tan\frac{\alpha}{2}/\tan\gamma)}
,\frac{b}{1+\tan\frac{\alpha}{2}/\tan\gamma}
\right).
\end{align}
Next, find the intersection points
$U=AD \cap CF$,
$V=AD \cap BE$,
$W=BE \cap CF$:
\begin{align}
U&=\left(
c b \frac{1+\cos\alpha}{c+2 b}
,
c b \frac{\sin\alpha}{c+2 b}
\right)
\\
V &= \left(
c\cos\alpha, c\cos\alpha\frac{(1-\cos\alpha)}{\sin\alpha}
\right)
\\
W &= \left(c \cos(\alpha),c\sin(\alpha)
\frac{b-c\cos(\alpha)}{2b-c\cos(\alpha)}
\right)
\end{align}
\begin{align}
S_{\triangle UVW}
&=
|(V_x-U_x)(W_y-U_y)-(V_y-U_y)(W_x-U_x)|/2.
\end{align}
Finally, after substitution $\sin\alpha=\sqrt{1-\cos^2\alpha}$,
$\cos\alpha=\frac{b^2+c^2-a^2}{2bc}$
and simplification,
the area of $\triangle UVW$ is given
in terms of $a,b$ and $c$
by quite a nice construction:
\begin{align}
S_{\triangle UVW}
&=
\frac{1}{4}
\sqrt{\frac{(a-b+c)(a+b-c)}{(a+b+c)(b+c-a)}}
\frac{\left(b^3-b^2 c+b c^2+c^3-a^2 (b+c)\right)^2}
{b|3 b^2-c^2+a^2| (c+2 b)}
.\quad(*)
\end{align}
Test:
the first triangle: $a=8$, $b=10$, $c=12$,
$S_{\triangle UVW}=3.266386$,
the second triangle: $a=8$, $b=10$, $c=15$,
$S_{\triangle UVW}=16.835510$.
And as expected, when $a=b=c$, $S_{\triangle UVW}=0.$
Edit:
Equation (*) also reveals another extreme case,
when the altitude $BE$ is parallel to the median $CF$ and
$S_{\triangle UVW}=\infty$ due to the condition $3 b^2-c^2+a^2=0$
in the denominator:
Indeed in this case, $c^2-(2b)^2=a^2-b^2$.
| {
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"url": "https://math.stackexchange.com/questions/1241428",
"timestamp": "2023-03-29T00:00:00",
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Complex Numbers Question, IIT JEE [2006]. Please tell me whether I solved it properly? $Q.$The value of $\sum\limits_{k=1}^{10}(\sin{\frac{2k\pi}{11}-i\cos\frac{2k\pi}{11}})$ is-?
I solved it like this-
$\frac{\sum\limits_{k=1}^{10}(\cos{\frac{2k\pi}{11}+i\sin\frac{2k\pi}{11}})}{i}$
If we observe these are the roots of the equation $z^{11}=1$
So $1+z_1+z_2+...+z_{10}=1$ (De Moivre
s Theorem)
So $z_1+z_2+...+z_{10}=-1$
$-1=i^2$
So $\frac{i^2}{i}=i$
| $\begin{array}{l}
\\
\sum \limits^{10}_{k=1}(\sin \ \frac{2k\pi }{11} -i\cos \frac{2k\pi }{11} )\\
=(-i)\sum \limits^{10}_{k=1}(\cos \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11} )\\
=(-i)(\sum \limits^{10}_{k=0}(\cos \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11} )-1)\\
=(-i)(-1)\\
=i\\
Note:
\sum \limits^{10}_{k=0}(\cos \frac{2k\pi }{11} +i\sin \frac{2k\pi }{11} )=0\\
\end{array}$
Sum of $11^{th}$ roots of unity =$0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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this sum inequality $\sum_{i=1}^{n}\frac{1}{4i(i+1)-1}<\frac{2}{7}$ show that
$$\sum_{i=1}^{n}\dfrac{1}{4i(i+1)-1}<\dfrac{2}{7}\tag{1}$$
we have
$$4i(i+1)-1>4i^2$$
But
$$\sum_{i=1}^{n}\dfrac{1}{i^2}<\dfrac{8}{7}$$ it is clear not hold, because
$$\sum_{i=1}^{n}\dfrac{1}{i^2}>1+\dfrac{1}{4}>1+\dfrac{1}{7}=\dfrac{8}{7}$$
I can't prove that this (1) is true. Can anyone help?
| Your estimation of denominator wasn't enough to bound the sum. I have noticed that using $4k^2 + 4k- 8$ your method will work.
\begin{align*}
\sum_{k=1}^n \frac{1}{4k^2+4k-1} & \le \frac 19+\sum_{k=2}^n \frac{1}{4k^2+4k-8} = \frac 19+\sum_{k=2}^n \left(\frac{1}{12(k-1)} - \frac{1}{12(k+2)}\right) \\
& = \frac 19 + \frac{11}{72}- \frac{1}{12n}- \frac{1}{12n+12}- \frac{1}{12n+24} < \frac{19}{72} < \frac 27,
\end{align*}
because the sum is telescopic.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Unilateral Z-Transform of sen($\frac{\pi}{4}n$) I was calculating the Unilateral Z-Transform of $sen(\frac{\pi}{4}n)$
I calculated it this way, but then the calculation become very difficult and i think i have made some mistakes.
It is : $\sum_{n=0}^{+\infty} sen(\frac{\pi}{4}n)z^{-n}$
$sen(\frac{\pi}{4}n) $ is periodic of a period 8.
And if $a_n=sen(\frac{\pi}{4}n)$
$a_0=0, \, a_1=\frac{\sqrt{2}}{2}, \, a_2= 1, \, a_3 = \frac{\sqrt{2}}{2},
\, a_4= 0, \, a_5 = -\frac{\sqrt{2}}{2}, \, a_6= -1, \, a_7 = -\frac{\sqrt{2}}{2}$
Then $a_{n+8k}=a_n, k \in Z$
Then:
\begin{align}
\sum_{n=0}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} &= (\sum_{n=4k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} + \sum_{n=1+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} + \sum_{n=2+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} + \sum_{n=3+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} \\
& \hspace{5mm} + \sum_{n=5+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} + \sum_{n=6+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n}+ \sum_{n=7+8k}^{+\infty} sen(\frac{\pi}{4}n)z^{-n} )
\\
&= 0 + \frac{\sqrt{2}}{2} \sum_{n=1+8k}^{+\infty} z^{-n}+ \sum_{n=2+8k}^{+\infty} z^{-n}+ \frac{\sqrt{2}}{2} \sum_{n=3+8k}^{+\infty} z^{-n} \\
& \hspace{5mm} -\frac{\sqrt{2}}{2} \sum_{n=5+8k}^{+\infty} z^{-n} - \sum_{n=6+8k}^{+\infty} z^{-n} - \frac{\sqrt{2}}{2} \sum_{n=7+8k}^{+\infty} z^{-n}
\end{align}
Those Z-Transforms are not difficult (as $\sum_{n=0}^{+\infty} z^{-n}= \frac{z}{z-1} $ but very long ones.
Wolframs shows a simplier Z-Transform of this function.
Is there any more efficient way to calculate it?
| For the series
\begin{align}
f(t) = \sum_{n=0}^{\infty} \sin\left(\frac{n \pi}{4} \right) \, t^{n}
\end{align}
then it is seen that
\begin{align}
f(t) &= \sin\left(\frac{\pi}{4} \right) \, t + \sin\left(\frac{2 \pi}{4} \right) \, t^{2} + \sin\left(\frac{3 \pi}{4} \right) \, t^{3} + \cdots \\
&= \sum_{n=0}^{\infty} \left[ \sin\left(\frac{(4n+1) \pi}{4} \right) \, t^{4n+1} + \sin\left(\frac{(4n+2) \pi}{4} \right) \, t^{4n+2} + \sin\left(\frac{(4n+3) \pi}{4} \right) \, t^{4n+3}\right] \\
&= \left[\sin\left(\frac{\pi}{4} \right) \, t + \sin\left(\frac{2 \pi}{4} \right) \, t^{2} + \sin\left(\frac{3 \pi}{4} \right) \, t^{3} \right] \, \frac{1}{1+t^{4}} \\
&= \frac{t(1 + \sqrt{2} t + t^2)}{\sqrt{2} \, (1+t^4)}.
\end{align}
By making the change $t \to z^{-1}$ the desired result is obtained.
| {
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"url": "https://math.stackexchange.com/questions/1243717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral by using substitution (How to proceed?) Using the substitution $x=a\sin\theta$, or otherwise, find $\int\frac{1}{x^2\sqrt{a^2-x^2}}dx$.
My attempt,
$x=a\sin\theta$
$dx=a\cos (\theta)d\theta$. Then $\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin ^2(\theta)}$
The given answer is $-\frac{\sqrt{a^2-x^2}}{a^2x}+c$
How to proceed then?
| Other possibility:
Use $x=1/y$, the integral becomes:
$$
I(a)=\int -\frac{1}{y^2}\frac{1}{\frac{1}{y^2} \sqrt{a^2-\frac{1}{y^2}}}dy=-\int \frac{y}{\sqrt{a^2y^2-1}}dy=-\frac{1}{a^2}\int \partial_y\left(\sqrt{a^2y^2-1}\right)dy=-\frac{\sqrt{a^2y^2-1}}{a^2}+C
$$
Resubstitute:
$$
I(a)=-\frac{\sqrt{a^2-x^2}}{ a^2 x}+C
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this inequality $25ab+25a+10b\le38$ let $a,b>0$,and such $a^2+b^2=1$,show that
$$25ab+25a+10b\le38$$
Now I have found this inequality $"="$,if and only if $a=\dfrac{4}{5},b=\dfrac{3}{5}$
then How to prove this inequality by AM-GM or other ?
| Given the solution $a=4/5,b=3/5$ gives equality, try $$a=\frac45\cos\theta-\frac35\sin\theta,b=\frac35\cos\theta+\frac45\sin\theta$$
(sorry, my comment was wrong at first) I think it becomes
$$12(\cos^2\theta-\sin^2\theta)+26\cos\theta-7\sin\theta+7\cos\theta\sin\theta)\leq38\\
24(\cos^2\theta-1)+26(\cos\theta-1)+7\sin\theta(\cos\theta-1)\leq0\\
(\cos\theta-1)(50+24\cos\theta+7\sin\theta)\leq0 $$
which of course is true, equality if $\theta=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Derivative of $\frac{1}{2} x + x^2 sin{\frac{1}{x}}$ How do you evaluate the derivative of:
$h(x) = \frac{1}{2} x + x^2 \sin{\frac{1}{x}}$ at $x=0$? It seems that the answer is $\frac{1}{2}$ but I don't know how to take care of the $x^2 \sin{\frac{1}{x}}$.
When you differentiate $h$, you get $h'(x) = \frac{1}{2} + 2x\sin{\frac{1}{x}} - \cos{\frac{1}{x}}$ and at $x=0$, $\cos{\frac{1}{x}}$ is not well defined.
| You're right that $x^2 \sin\left(\frac 1 x\right)$ is not defined at $x=0$. But we can get over this discontinuity.
Think about it like we're approaching $x=0$ from the left and the right. That's like taking:
$$\lim_{x\to 0} \left|x^2 \sin\left(\frac 1 x\right)\right|$$
Here's the big trick: we know that $\sin$ can only take values on the interval $[-1,1]$. So the absolute value of the sine function is always less than or equal to $1$. So $|\sin\left( \frac 1 x \right)| \leq 1$, regardless of the value of $x$. This means that:
$$\left|x^2 \sin\left(\frac 1 x\right)\right| \leq \left|x^2\right|$$
So:
$$\lim_{x \to 0} \left|x^2 \sin\left(\frac 1 x\right)\right| \leq \left|x^2\right|$$
And you know that $\lim_{x \to 0} |x^2| = 0$. So
$$\lim_{x \to 0} \left|x^2 \sin\left(\frac 1 x\right)\right| \leq 0$$
But the smallest value the absolute value of anything can take is $0$. So that inequality holds only if the limit of the left side is $0$. So we conclude:
$$\lim_{x \to 0} \left|x^2 \sin\left(\frac 1 x\right)\right| = 0$$
Now we use a definition of the derivative at $0$:
$$h'(0) = \lim_{t \to 0} \frac{h(t) - h(0)}{t-0}$$
So what's $h(0)$? From the argument we developed above, we know that $\lim_{x \to 0} \left|x^2 \sin\left(\frac 1 x\right)\right| = 0$, and the other part of $h(0)$ is $\frac 1 2 \cdot 0 = 0$. So $h(0) = 0$.
So
$$h'(0) = \lim_{t \to 0} \frac{h(t)}{t} = \lim_{t \to 0} \frac{\frac 1 2t + t^2 \sin \left( \frac 1 t \right)}{t}
=
\lim_{t \to 0}
\frac{\frac 1 2t}{t} + \frac{t^2 \sin \left( \frac 1 t \right)}{t}
=
\lim_{t \to 0}
\frac 1 2 + t \sin \left( \frac 1 t \right)
$$
And we established previously that $\lim_{t \to 0} t \sin \left( \frac 1 t \right) = 0$. And $\lim_{t \to 0} \frac 1 2 = \frac 1 2$, clearly.
So to conclude, $h'(0) = \frac 1 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Simplifying Square Roots Frustration Okay, I'm really frustrated with this.
So, when you have $3 \sqrt 5 + 5 \sqrt 5$,
you get $8\sqrt5$, right?
Okay, so what do I do for here:
$\sqrt{11} - 3 \sqrt{11}$
Is it just $-3 \sqrt{11}$ ?
What about for $6 \sqrt 2 + 4 \sqrt{50}$ ?
Do I multiply the $6 \sqrt{2}$ so that I can make what's inside of the square root equal to $50$?
| Compare:
\begin{align}
x - 3 x
&= 1x - 3 x \\
&= (1 - 3)x \\
&= -2 x
\end{align}
to
\begin{align}
\sqrt{11} - 3 \sqrt{11}
&= 1\sqrt{11} - 3 \sqrt{11} \\
&= (1 - 3)\sqrt{11} \\
&= -2 \sqrt{11}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove that $\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)$ How prove that sum
$$\frac{1}{\sin^2\frac{\pi}{2n}}+\frac{1}{\sin^2\frac{2\pi}{2n}}+\cdots+\frac{1}{\sin^2\frac{(n-1)\pi}{2n}} =\frac{2}{3}(n-1)(n+1)$$
| To develop further on Jean-Claude Arbaut's comment, you'll find here the proof that:
$$\sum_{k=1}^{n-1}{\frac{1}{\sin^2\left(\frac{k\pi}{n}\right)}}=\dfrac{(n-1)(n+1)}{3}$$
All you have to do now, is show that:
$$\sum_{k=1}^{n-1}{\frac{1}{\sin^2\left(\frac{k\pi}{\color{red}{2n}}\right)}}=\color{red}{2}\sum_{k=1}^{n-1}{\frac{1}{\sin^2\left(\frac{k\pi}{\color{red}{n}}\right)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute the values of two infinite products whose factors are the same I have the following question:
How to prove that
$(1-\frac{1}{2})\cdot (1+\frac{1}{3})\cdot (1-\frac{1}{4})\cdot (1+\frac{1}{5})\cdot (1-\frac{1}{6})\cdot (1+\frac{1}{7})\cdot (1-\frac{1}{8})\cdot\ ...\ = \frac{1}{2}\ $, but
$(1-\frac{1}{2})\cdot (1-\frac{1}{4})\cdot (1+\frac{1}{3})\cdot (1-\frac{1}{6})\cdot (1-\frac{1}{8})\cdot (1+\frac{1}{5})\cdot (1-\frac{1}{10})\cdot (1-\frac{1}{12})\cdot (1+\frac{1}{7})\cdot\ ...\ = \frac{1}{2 \sqrt{2}}\ ?$
Thanks for the help!
| For the first product note that, taking $k\geq3
$ $$\left(1+\frac{1}{k}\right)\left(1-\frac{1}{k+1}\right)=1
$$ then $$\prod_{k\geq2}\left(1+\frac{\left(-1\right)^{k+1}}{k}\right)=1-\frac{1}{2}=\frac{1}{2}.
$$ For the second product, which can be written as kennytm suggest, we have $$\prod_{k\geq1}\left(1-\frac{1}{4k-2}\right)\left(1-\frac{1}{4k}\right)\left(1+\frac{1}{2k+1}\right)=\lim_{n\rightarrow\infty}\prod_{k\leq n}\left(1-\frac{1}{4k-2}\right)\left(1-\frac{1}{4k}\right)\left(1+\frac{1}{2k+1}\right)
$$ and using the identity, for a positive integer $a
$ $$\Gamma\left(a+\frac{1}{2}\right)=\frac{\sqrt{\pi}\left(2a-1\right)!!}{2^{a}}
$$ and the functional equation $$\Gamma\left(1+z\right)=z\Gamma\left(z\right)
$$ we get $$\prod_{k\leq n}\left(1-\frac{1}{4k-2}\right)\left(1-\frac{1}{4k}\right)\left(1+\frac{1}{2k+1}\right)=\frac{\sqrt{\pi}\left(n+1\right)\Gamma\left(2n+\frac{1}{2}\right)}{2^{2n+1}\Gamma\left(n+\frac{1}{2}\right)\Gamma\left(n+\frac{3}{2}\right)}
$$ hence $$\prod_{k\geq1}\left(1-\frac{1}{4k-2}\right)\left(1-\frac{1}{4k}\right)\left(1+\frac{1}{2k+1}\right)=\lim_{n\rightarrow\infty}\frac{\sqrt{\pi}\left(n+1\right)\Gamma\left(2n+\frac{1}{2}\right)}{2^{2n+1}\Gamma\left(n+\frac{1}{2}\right)\Gamma\left(n+\frac{3}{2}\right)}
$$ which can be calculated using the Stirling's approximation. Finally $$\prod_{k\geq1}\left(1-\frac{1}{4k-2}\right)\left(1-\frac{1}{4k}\right)\left(1+\frac{1}{2k+1}\right)=\frac{1}{2\sqrt{2}}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a^2+b^2+c^2\geq [2(a-b)^2(b-c)^2(a-c)^2]^{1/3}$ Mathematica seems to know that this statement is true, yet I am struggling to prove it. Possible useful inequalities are Minkowski and the geometric mean. Using the geometric mean inequality I can prove a less strict bound:
$$(a-b)^2+(b-c)^2+(a-c)^2 \geq 0 \implies 4 a^2+4 b^2+4 c^2 \geq (a-b)^2+(b-c)^2+(a-c)^2,$$
then the geometric mean inequality gives
$$ (a-b)^2+(b-c)^2+(a-c)^2 \geq 3[(a-b)^2(b-c)^2(a-c)^2]^{1/3} .$$
Combining the two we get
$$ a^2+b^2+c^2 \geq \frac{3}{4}[(a-b)^2(b-c)^2(a-c)^2]^{1/3},$$
which is not strong enough.
Can anyone prove this for $2^{1/3}$ in place of $3/4$?
| Hint:
WOLOG, suppose $a>b>c$. Let $b=c+y$ and $a=c+y+x$. One needs to show
$$
(c+x+y)^2+(c+y)^2+c^2 \ge (2x^2y^2(x+y)^2)^{\frac{1}{3}}.
$$
Notice that $(c+x+y)^2+(c+y)^2+c^2$ attains minimal when $c=-\frac{x+2y}{3}$. It remains to show
$$
(x+2y)^2+(x-y)^2+(2x+y)^2 \ge 9 (2x^2y^2(x+y)^2)^{\frac{1}{3}},
$$
or
$$
x^2+y^2+xy\ge 3 (xy\frac{x+y}{2})^{\frac{2}{3}}.
$$
for positive $x, y$. Finally, by rearranging we reach
$$
x^2 + \frac{y(x+y)}{2}+ \frac{y(x+y)}{2} \ge 3 (xy\frac{x+y}{2})^{\frac{2}{3}},
$$
which is true due to the geometric mean inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $2^{12^{7}+3}\equiv x \pmod{36}$, then what is the value of $x$?
If $2^{12^{7} + 3} \equiv x \pmod{36}$, then what is the value of $x$?
We have:
$$
\begin{align}
2^5 & \equiv - 4 \pmod{36} \\
2^{10} & \equiv 16 \pmod{36} \\
2^{12} & \equiv - 8 \pmod{36} \\
(2^{12})^{12} & \equiv 8^{12} \pmod{36} \\
2^{12^2} & \equiv - 8 \pmod{36} \text{, as } 8^{12} \equiv - 8 \pmod{36} \\
(2^{12^2})^{12^5} & \equiv - 8 \pmod{36} \\
2^{12^7} & \equiv - 8 \pmod{36} \\
2^{{12^7} + 3} & \equiv - 64 \pmod{36} \\
2^{{12^7} + 3} & \equiv 8 \pmod{36} \\
\end{align}
$$
Am I right or wrong? Is there any simpler way to find it?
| We need $x \in \{0,1,2,\ldots,35\}$ such that $2^{12^7+3} = 36M + x$. Note that $4$ divides $2^{12^7+3}$ and $36M$. Hence, $x$ must be a multiple of $4$, say $4y$. We now need $y$ such that
$$2^{12^7+1} = 9M + y$$
where $y \in \{0,1,2,\ldots,8\}$. Since $\gcd(2,9) = 1$, we have $$2^{\phi(9)} \equiv 1 \pmod9 \implies 2^6 \equiv 1\pmod9$$
We have $12^7+1 = 6t+1$. Hence,
$$2^{12^7} \equiv 1 \pmod9 \implies 2^{12^7+1} \equiv 2 \pmod9$$
Hence, $y=2$, which gives us that $x=8$.
| {
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Let $x$ be an integer and $n$ be a positive integer. Find the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$. I need help proving the following: Let $x$ be an integer and $n$ be a positive integer. Find the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$. I know that the smallest $n$ is 8 by testing $n=1,2,3,...$. I need help proving that $n=8$ is the smallest $n$ such that $x^4+n^2$ is not a prime for any $x$.
| It is well-known that $x^4+4=\Big(x^2\Big)^2+2^2+\underbrace{2\cdot2x^2-2\cdot2x^2}_0=\Big(x^2+2\Big)^2-4x^2=$
$=\big(x^2-2x+2\big)\big(x^2+2x+2\big)$. However, since $x^2\pm2x+2=\pm1$ has as viable integer
solutions $x=\pm1$, we must extend the search to polynomials of the form $x^2+4a^4$, which
has no such solutions for $a=2\iff n=8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Domain of the function $f(x) = \sqrt{\frac{3^x-4^x}{x^2-4x-4}}$ will be? I tried solving this question by
$1.$ $-1$ and $4$ will not be in domain because denominator can not be zero .
$2.$ Either both denominator and numerator will be positive or negative so that whole term in root becomes positive.
But I am not able to solve the upper part , can it be done by taking $\log$ ?
| To obtain the domain of $f(x) = \sqrt{\frac{3^x-4^x}{x^2-4x-4}}$, the expression inside the radical must be positive and the denominator of the expression must be nonzero.
The numerator $3^ x - 4^x$ is positive when $x < 0$. The denominator $x^2-4x-4$ is positive when $x > 2(1+ \sqrt{2})$ or when $x < 2 (1 - \sqrt{2})$. Hence, the expression inside the radical sign is positive when $x < 2 (1 - \sqrt{2})$ or when $0 <x < 2(2 + \sqrt{2})$.
The values where the denominator are zero are $2(1 \pm \sqrt{2})$. They are not in the stated intervals above.
Thus the domain (in interval notation) is $(- \infty,2 (1 - \sqrt{2})) \cup (0,2 (1 + \sqrt{2}))$.
| {
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"answer_id": 2
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Find a projective transformation that sends the locus $x^2-y^2=z^2 $ to $yz=x^2$ So I'm trying to find a function that sends $x^2-y^2=z^2$ to $yz=x^2$. I know it can be done because all conics are projectively equivalent. I think I have to use a matrix of some kind but I don't know how.
Thanks for any help.
| In general you have three degrees of freedom when mapping one conic to another. One approach is the following: choose three distinct points $A,B,C$ resp. $A',B',C'$ on each conic. Then construct a point $D$ as follows: take the tangents to the first conic in $A$ and $B$, intersecting these in a point $P$ and join that point of intersection with $C$. That last line will intersect the conic in a second point, which is that point $D$. Do the same to construct $D'$ on the second conic. Now you can compute the projective transformation which maps these four preimages to their corresponding images.
Applied to your situation, where you can use any matching transformation, you could choose simple coordinates on each conic. E.g.
\begin{align*}
&& M &= \begin{pmatrix}1&0&0\\0&-1&0\\0&0&-1\end{pmatrix} &
&& M' &= \begin{pmatrix}-2&0&0\\0&0&1\\0&1&0\end{pmatrix}
\end{align*}
\begin{align*}
A &= \begin{pmatrix}1\\0\\1\end{pmatrix} &
MA &= \begin{pmatrix}1\\0\\-1\end{pmatrix} &
A' &= \begin{pmatrix}0\\1\\0\end{pmatrix} &
M'A' &= \begin{pmatrix}0\\0\\1\end{pmatrix} \\
B &= \begin{pmatrix}1\\0\\-1\end{pmatrix} &
MB &= \begin{pmatrix}1\\0\\1\end{pmatrix} &
B' &= \begin{pmatrix}0\\0\\1\end{pmatrix} &
M'B' &= \begin{pmatrix}0\\1\\0\end{pmatrix} \\
C &= \begin{pmatrix}1\\1\\0\end{pmatrix} &
P &= \begin{pmatrix}0\\1\\0\end{pmatrix} &
C' &= \begin{pmatrix}1\\1\\1\end{pmatrix} &
P' &= \begin{pmatrix}1\\0\\0\end{pmatrix}
\end{align*}
\begin{align*}
D &= (P^TMP)C-(2C^TMP)P &
D' &= (P'^TM'P')C'-(2C'^TM'P')P'
\\
D &= -C+2P = \begin{pmatrix}-1\\1\\0\end{pmatrix} &
D' &= -2C'+4P' \sim \begin{pmatrix}-1\\1\\1\end{pmatrix}
\end{align*}
If you now follow the steps from my other post you will find that the projective transformation which maps $A,B,C,D$ to $A',B',C',D'$ can be described by the matrix
$$\begin{pmatrix}0&1&0\\1&0&1\\1&0&-1\end{pmatrix}$$
Of course, this is ony one suitable transformation from the three-parameter family of possible transformations.
| {
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"url": "https://math.stackexchange.com/questions/1266319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Show that $\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ is an algebraic integer. Let $m$ be an integer such that $m \equiv 2 \pmod 3$. Show that the number $$\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$$ is an algebraic integer.
The usual technique, doing $x = \dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ and trying to find an algebraic expression in terms of $x$ seems not to work in this case (at least I couldn't do it). Can anyone help me? This is a question from an old exam.
| Let $x=\frac {m-\sqrt [3] 2}{\sqrt [3] 3}$
Let $m=3k+2$
Then $x=\frac {3k+2-\sqrt [3] 2}{\sqrt [3] 3}$
$x\sqrt [3] 3=3k+2-\sqrt [3] 2$
$x\sqrt [3] 3+\sqrt [3] 2 =3k+2$
$\left( x\sqrt [3] 3+\sqrt [3] 2 \right)^3 =\left(3k+2 \right)^3$
$3x^3+3\sqrt [3] {18}x^2+3\sqrt [3] {12}x+2=27k^3+54k^2+36k+8$
... no not getting anywhere helpful!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that: $2(ab+bc+ca)-a^2-b^2-c^2\le6$. Let $a,b,c>0$ such that: $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}=1$.
Prove that: $2(ab+bc+ca)-a^2-b^2-c^2\le6$.
I have no idea for solve this problem.
| $2=\dfrac{a^2}{1+a^2}+\dfrac{b^2}{1+b^2}+\dfrac{c^2}{1+c^2}\ge \dfrac{(a+b+c)^2}{a^2+b^2+c^2+3} \iff 2(a^2+b^2+c^2+3)\ge (a+b+c)^2 \iff 2(ab+bc+ac)-a^2-b^2-c^2\le6 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Question regarding proving by induction I am struggling with a math problem I have been assigned. The problem is as follows:
Let $X_1 = -3$ and $X_2 = 0$. Given that for every natural number $n \geq 2, X_{n+1} = 7X_n - 10X_{n-1}$, prove by induction that for every $n$ belonging to $\mathbb{N}$, $X_n = 2 \cdot 5^{n-1} - 5 \cdot 2^{n-1}$
Right now all I have proven so far is the two base cases, I am not sure how to take this proof any further. Any assistance you can offer is greatly appreciated, thank you.
| Start with your recurrence relation and substitute your inductive assumption:
$$X_{n+1} = 7(2 \cdot 5^{n-1} - 5 \cdot 2^{n-1}) - 10(2 \cdot 5^{n-2} - 5 \cdot 2^{n-2}) \\
= 14 \cdot 5^{n-1} - 35 \cdot 2^{n-1} - 20 \cdot 5^{n-2} + 50 \cdot 2^{n-2} \\
= 10 \cdot 5^{n-1} - 10 \cdot 2^{n-1} \\
= 2 \cdot 5^n - 5 \cdot 2^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solvability of the equation $2a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$ when $n \geq 5$ is prime? As a natural extension of the question titled
Solvability of $a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$ when $n \geq 5$ is prime?,
I wonder if the equation
$$2a_{1}^{2} = a_{2}^{n} + a_{3}^{n} + a_{4}^{n}$$
is solvable for co-prime $a_i$ when $n \geq 5$ is prime?
The Darmon-Merel theorem asserts that $a_{1}^{2} \neq a_{2}^{n} + a_{3}^{n}$ for all $n \geq 3$ and all integers $a_{1}, a_{2}, a_{3}$ not all
$0$, but I am not sure about the result when $2a_{1}^{2}$ is considered instead.
If there is already any known result in existence, bringing it in is more than welcome.
| I do not know why co-prime solutions are more interesting and must these be mutual co-prime or co-prime altogether, but there are few mutual co-prime ones $a_1,a_2,a_3<1000$ (not sure all of these found due to precision trunks):
$$\begin{align}4^5+239^5+659^5=2\cdot 7907819^2\\
40^5+ 617^5+ 633^5= 2\cdot 9773625^2\\
73^5+ 121^5+ 144^5= 2\cdot 212047^2\\
96^5+ 709^5+ 877^5= 2\cdot 18681029^2\\
161^5+ 380^5+ 479^5= 2\cdot 4077240^2\\
211^5+335^5+704^5=2\cdot 9422435^2\\
224^5+387^5+541^5=2\cdot 5271996^2\\
360^5+493^5+517^5=2\cdot 6004405^2\\
369^5+640^5+881^5=2\cdot 17957625^2\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$17!=3556xy428096000,$then $(x+y)$ equals?(without using a calculator) $17!=3556xy428096000,$then $(x+y)$ equals?
a)$15$ b)$6$ c)$12$ d)$13$
With help of calculator $(x+y)$ can be easily calculated as $15$.But without a calculator,I can only conclude that the sum of digits $(3+5+5+6+x+y+4+2+8+9+6)$ is a multiple of $3$, & excluding $(x+y)$ the sum of the digits $(3+5+5+6+4+2+8+9+6) =48$.Thus excluding option d) the ans can be any of a),b) or c).
Please help me to find the exact answer.
Thank you
| Actually, the number was copied incorrectly (somehow) and it was driving me crazy trying to figure out what I was doing wrong.
Since $17 > 13$, $17!$ must be divisible by $7 \cdot 11 \cdot 13 = 1001$. That means if we take the digits preceding the trailing zeros, and place them in groups of three, like so—$355, 6\text{XY}, 428, 096$—the alternating sums must be the same modulo $1001$. We have, then,
$$
355+428 = 6\text{XY}+96
$$
and then $\text{XY} = 87$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
prove an identity involving beta function and gamma function We know that $B(p,q)=\Gamma(p)\Gamma(q)/\Gamma(p+q)$ where $p, q>0$,
and $B(p,q)$ is related to binomial coefficients if one of $p,q$ is an integer. I want to prove the following identity.
$$\frac{\Gamma(b)}{\Gamma(b+2n+3)}\sum_{k=0}^n \frac{(-1)^{k+1}\Gamma(b+n+k+2)}{(n-k+1)!(k+1)!\Gamma(b+k+1)}=\frac{1}{(n+1)!(n+2)!}\left[(-1)^{n+1}B(b,n+2)-B(b+n+1,n+2)\right],$$
where all variables except $b$ are nonnegative integers, and $b=p/2+q$ for nonnegative integers $p,q.$ In fact, the value of $b$ is not essential here, as long as $b\geq 0$.
I tried to convert the above expression into an identity about binomial coefficients but still could not figure it out.
Note that for the special case where $n=0$, it can be verified that the above identity holds.
| Suppose we are interested in the value of
$$S_b(n) =
\sum_{k=0}^n
\frac{(-1)^{k+1}}{(n-k+1)!(k+1)!}
\frac{\Gamma(b+n+k+2)}{\Gamma(b+k+1)}.$$
This is
$$\sum_{k=0}^n
\frac{(-1)^{k+1}}{(n-k+1)!(k+1)!}
(n+1)! {b+n+k+1\choose n+1}
\\ = \frac{1}{n+2} \sum_{k=0}^n
\frac{(-1)^{k+1} (n+2)!}{(n-k+1)!(k+1)!}
{b+n+k+1\choose n+1}
\\ = \frac{1}{n+2} \sum_{k=0}^n (-1)^{k+1}
{n+2\choose k+1} {b+n+k+1\choose n+1}
\\ = \frac{1}{n+2} \sum_{k=1}^{n+1} (-1)^k
{n+2\choose k} {b+n+k\choose n+1}.$$
This becomes
$$-\frac{1}{n+2} {n+b\choose n+1}
- \frac{1}{n+2} (-1)^n {2n+b+2\choose n+1}
\\ + \frac{1}{n+2} \sum_{k=0}^{n+2} (-1)^k
{n+2\choose k} {b+n+k\choose n+1}.$$
To evaluate the sum we introduce
$${b+n+k\choose n+1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{b+n+k}}{z^{n+2}} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{b+n}}{z^{n+2}}
\sum_{k=0}^{n+2} (-1)^k {n+2\choose k} (1+z)^{k} \; dz
\\ = \frac{(-1)^{n+2}}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{b+n}}{z^{n+2}}
(-1+1+z)^{n+2} \; dz
\\ = \frac{(-1)^{n}}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{b+n} \; dz = 0.$$
It follows that
$$S_b(n) =
-\frac{1}{n+2} {n+b\choose n+1}
- \frac{1}{n+2} (-1)^n {2n+b+2\choose n+1}.$$
Converting this into the gamma function format is pure algebra and
left as an exercise to the reader.
Remark. We use $$(1+z)^{b+n} = \exp((b+n)\log(1+z))$$ when $b$ is
not an integer where the branch cut of the logarithm is on the
negative real axis so that the branch point is at $z=-1.$ This ensures
that we have analyticity in a disk enclosing the origin.
| {
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"url": "https://math.stackexchange.com/questions/1275743",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How do I find $\int_{0}^{\infty} \frac{\sin^4 x}{x^2}\,dx$? I need help evaluating the following integral
$$\int_{0}^{\infty} \frac{\sin^4 x}{x^2}\,dx$$
which should probably be equal to $\frac{\pi}{4}$
Using some trigonometric manipulations I got $\frac{3}{8} - \frac{\cos{2x}}{2} + \frac{\cos{4x}}{8}$ which using integration by parts doesn't lead me to anything pretty.
Update: Not sure if I should post this as a separate question but getting explanation why $\int_{0}^{\infty} \frac{\sin{ax}}{x} = \frac{\pi}{2}$ for a positive integer $a$ could help me solve this question.
| Another way to solve this problem is to use parametrized integral: Let
$$ I(s)=\int_0^\infty e^{-sx}\frac{\sin^4x}{x^2}dx. $$
Then
\begin{eqnarray}
I''(s)&=&\int_0^\infty e^{-sx}\sin^4xdx\\
&=&\int_0^\infty e^{-sx}(\frac{3}{8} - \frac{\cos{2x}}{2} + \frac{\cos{4x}}{8})dx\\
&=&\frac{3}{8s}-\frac{s}{2(s^2+4)}+\frac{s}{8(s^2+16)}
\end{eqnarray}
so
\begin{eqnarray}
I(0)&=&\int_0^\infty\int_{t}^\infty(\frac{3}{8s}-\frac{s}{2(s^2+4)}+\frac{s}{8(s^2+16)})dsdt\\
&=&\frac{1}{16}\int_0^\infty(-6\ln t+4\ln(t^2+4)-\ln(t^2+16))dt\\
&=&\left.-\frac{3}{8} t \ln t+\frac{1}{4} t \ln \left(t^2+4\right)-\frac{1}{16} t \ln
\left(t^2+16\right)-\frac{1}{2} \arctan
^{-1}\left(\frac{t}{4}\right)+\arctan ^{-1}\left(\frac{t}{2}\right)\right|_0^\infty\\
&=&\frac{\pi}{4}.
\end{eqnarray}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
System of equations involving sin and cos I'm trying to solve the following system:
$$
\sin(x) + \cos(y) = 0.6\\
\cos(x) - \sin(y) = 0.2\\
$$
Solving for y in terms of x:
$$
y=\arccos(0.6-\sin(x))=\arcsin(\cos(x) -0.2)
$$
Therefore:
$$
\frac{\pi}{2} - \arcsin(0.6-\sin(x)) = \arcsin(\cos(x) - 0.2)
$$
Forgot to add:
$$
\frac{-\pi}{2}\le x,y \le \frac{\pi}{2}
$$
Not too sure how to proceed from here, any hints?
| here is one way to do this.
$$\begin{align}\sin x + \cos y &= 0.6\\
\sin (x) + \sin(\pi/2- y) &= 0.6 \\
\sin(\pi/4+x/2-y/2)\cos(x/2+y/2-\pi/4) &= 0.3\tag 1\end{align}$$
in the same way
$$\begin{align}\cos x - \sin y &= 0.2\\
\cos(x) - \cos(\pi/2- y) &= 0.2 \\
\sin(\pi/4+x/2-y/2)\sin(x/2+y/2-\pi/4) &= -0.1\tag 2\end{align}$$
dividing $(1)$ by $2$, we have $$\tan(x/2+y/2-\pi/4) = -\frac13\tag 3 $$ which gives you $$x+y=\pi/2 -2\tan^{-1}(1/3)+ 2k\pi \tag 4 $$
using $(3)$ in $(2)$ we find that $$\sin(\pi/4+x/2-y/2)\sin(x/2+y/2-\pi/4) = \sin(\pi/4+x/2-y/2)\left(\mp\frac1{\sqrt{10}}\right)=-0.1 $$
so that $$x-y = -\pi/2 \pm 2\sin^{-1}(1/\sqrt{10}), -\pi/2 \mp 2\sin^{-1}(1/\sqrt{10}) \tag 5$$
solving $(4)$ and $5$ you get four pairs of solutions. i will do one pair.
$$x = 0, y = \pi/2-2\tan^{-1}(1/3) $$ you can verify that $$\sin x + \cos y = 0.6, \cos x - \sin y = 0.2 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the inequality $2\left|x+\frac{1}{4}\right| < 9$
I keep trying to figure this out and I can't.
I tried to split up the absolute value first.
$2 (x+\frac{1}{4}) < 9$
$2+x+\frac{1}{4} < 9$
subtract $2$ from both sides? $x+\frac{1}{4} < 7$
and
$2 -(x+\frac{1}{4}) < 9$
$2-x-\frac{1}{4} < 9$
subtract $2$ from both sides? $-x-\frac{1}{4} < 7$
well that's not right. I suck at this. :c
Thanks for any helpers.
| Write $2|\frac{x+1}{4}|<9$ as $\frac{-9}{2}<\frac{x+1}{4}<\frac{9}{2}.$ This gives $-18<x+1<18$ which is $-19<x<17$
| {
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"timestamp": "2023-03-29T00:00:00",
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What's wrong?: Find the infinite sum $S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$ The answer I got by hand is not the same to the one I found using a spreadsheet.
$\displaystyle S = 1 + \frac{3}{4} + \frac{7}{16} + \frac{15}{64} + \frac{31}{256} + \ldots$
$\displaystyle \frac{1}{4}S = \hspace{8.5pt} \frac{1}{4} + \frac{3}{16} + \frac{7}{64} + \frac{15}{256} + \ldots$
$\displaystyle \frac{3}{4}S = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \qquad \leftarrow S- \frac{1}{4}S$
For the Infinite Sum on the RHS $\displaystyle \left(S = \frac{a}{1-r}\right)$:
$\displaystyle a = 1$
$\displaystyle r = \frac{1}{2}$
Then
$\displaystyle \frac{3}{4}S = \frac{1}{1-\frac{1}{2}}$
$\displaystyle \frac{3}{4}S = 2$
$\displaystyle S = \frac{8}{3}$
Using Excel the answer is $\displaystyle \frac{5}{3}$, but I don't know where is the issue.
Thanks!!
| At first, note that $S=\dfrac{2^1-1}{4^{1-1}} + \dfrac{2^2-1}{4^{2-1}}+ \dfrac{2^3-1}{4^{3-1}}+\dfrac{2^4-1}{4^{4-1}} + \dfrac{2^5-1}{4^{5-1}}+\ldots $
So: $$S=\displaystyle\sum_{k=1}^{\infty}\dfrac{2^k-1}{4^{k-1}}?$$
$$S=\displaystyle\sum_{k=1}^{\infty}\dfrac{2^k}{4^{k-1}}-\dfrac{1}{4^{k-1}}$$
$$S=\displaystyle\sum_{k=1}^{\infty}\dfrac{4}{2^k}-\dfrac{1}{4^{k-1}}$$
And you can go on since here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1280948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integrate $\int_{-\infty}^\infty \frac{x}{\sinh x}~dx$ From a Problem Set on residues:
Evaluate $$\int_0^\infty \frac{\log x}{(x-1) \sqrt{x}}~dx.$$
After the substitution $x = e^u$ and easy computations, the integral becomes $$\int_{-\infty}^\infty \frac{x}{\sinh x}~dx.$$
| We have
$$\int_1^{\infty} \dfrac{\log(x)}{(x-1)\sqrt{x}}dx = \int_1^0 \dfrac{\log(1/x)}{(1/x-1)1/\sqrt{x}} \left(-\dfrac{dx}{x^2}\right) = \int_0^1 \dfrac{-\log(x)}{(1-x)\sqrt{x}}dx = \int_0^1 \dfrac{\log(x)}{(x-1)\sqrt{x}}dx$$
Hence, the integral is
\begin{align}
I & = -2\int_0^1 \dfrac{\log(x)dx}{(1-x)\sqrt{x}} = -2 \sum_{k=0}^{\infty}\int_0^1 x^{k-1/2}\log(x)dx = 2\sum_{k=0}^{\infty} \dfrac1{(k+1/2)^2} =8 \sum_{k=0}^{\infty} \dfrac1{(2k+1)^2} = \pi^2
\end{align}
A very similar method to evaluate $$I=\int_{-\infty}^{\infty} \dfrac{x}{\sinh(x)}dx = 2\int_0^{\infty} \dfrac{x}{\sinh(x)}dx$$
We have
\begin{align}
I & = 2 \cdot \int_0^{\infty} \dfrac{2x}{e^x-e^{-x}}dx = \int_0^{\infty} \dfrac{4xe^{-x}}{1-e^{-2x}}dx = 4 \sum_{k=0}^{\infty} \int_0^{\infty}xe^{-(2k+1)x}dx = 4 \sum_{k=0}^{\infty} \dfrac1{(2k+1)^2} = \dfrac{\pi^2}2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $ The Question reads -
$$ \frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x $$
I tried using the obvious approach by converting $5x , 4x $ and $ 3x$ to either $2x$ or $x$ but all that seemed to do was to further complicate the fraction. Any hints would be much appreciated.
I also tried applying various identities but to no avail.
| Main idea: Stick to the $\cos x$ function and use brute force.
LHS:
The numerator
$=\cos5x+\cos4x$
$=\cos3x\cos2x-\sin3x\sin2x+2\cos^22x-1$
$=(\cos2x\cos x-\sin2x\sin x)(2\cos^2x-1)-(\sin2x\cos x+\sin x\cos2x)(2\sin x\cos x)+2(2\cos^2x-1)^2-1$
$=((2\cos^2x-1)\cos x-2\sin^2 x\cos x)(2\cos^2x-1)-(2\sin x\cos^2x+\sin x(2\cos^2x-1))(2\sin x\cos x)+2(4\cos^4x-4\cos^2x+1)-1$
$=(2\cos^3 x-\cos x-2\sin^2x\cos x)(2\cos^2x-1)-(4\sin x\cos^2x-\sin x)(2\sin x\cos x)+8\cos^4x-8\cos^2x+1$
$=(2\cos^3 x-\cos x-2(1-\cos^2x)\cos x)(2\cos^2x-1)-(4(1-\cos^2x)\cos^2x-(1-\cos^2 x))2\cos x+8\cos^4x-8\cos^2x+1$
$=(4\cos^3x-3\cos x)(2\cos^2x-1)-(5\cos^2x-4\cos^4x-1)2\cos x+8\cos^4x-8\cos^2x+1$
$=8\cos^5x-6\cos^3x-4\cos^3x+3\cos x-10\cos^3x+8\cos^5x+2\cos x+8\cos^4x-8\cos^2{x}+1$
$=16\cos^5x+8\cos^4x-20\cos^3x-8\cos^2x+5\cos x+1$
$=(1+6\cos x - 8\cos^3 x)(1-\cos x - 2\cos^2 x)$
The denominator $=1-2\cos3x$
$=1-2(\cos2x\cos x - \sin2x\sin x)$
$=1-2((2\cos^2x-1)\cos x-2\sin^2x\cos x)$
$=1-2(2\cos^3x-\cos x-2(1-\cos^2x)\cos x)$
$=1-2(4\cos^3x-3\cos x)$
$=1+6\cos x-8\cos^3x$
The fraction $=1-\cos x - 2\cos^2 x$
$= - (2\cos^2x-1)-\cos x$
$=-\cos2x-\cos x$
$=$ RHS
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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} |
Integration of $\frac{1}{\sin x+\cos x}$ I'm given this $\int\frac{1}{\sin x+\cos x}dx$.
My attempt,
$\sin x+\cos x=R\cos (x-\alpha)$
$R\cos \alpha=1$ and $R\sin \alpha=1$
$R=\sqrt{1^2+1^2}=\sqrt{2}$,
$\tan\alpha=1$
$\alpha=\frac{\pi}{4}$
So, $\sin x+\cos x=\sqrt{2}\cos (x-\frac{\pi}{4})$
$\int\frac{1}{\sin x+\cos x}dx=\int \frac{1}{\sqrt{2}\cos (x-\frac{\pi}{4})}dx$
$=\frac{1}{\sqrt{2}}\int \sec (x-\frac{\pi}{4})dx$
$=\frac{1}{\sqrt{2}} \ln \left | \sec (x-\frac{\pi}{4})+\tan (x-\frac{\pi}{4}) \right |+c$
Am I correct? Is there another way to solve this integral? Thanks in advance.
| Your computations are correct. Now, if you make the substitution $t = \tan\left(\frac{x}{2}\right)$, you get
\begin{align}
\cos x &= \frac{1-t^{2}}{1+t^{2}} \\
\sin x &= \frac{2t}{1+t^{2}} \\
dx &= \frac{2 \, dt}{1+t^{2}}
\end{align} then
\begin{align}
\int \frac{dx}{\cos x + \sin x }
&= -\int \frac{dt}{t^{2}-2t-1 } = \frac{1}{\sqrt{2}}\rm arctanh \left(\frac{t-1}{\sqrt{2}}\right)+C.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Is it possible to find the sum of all integer values that $x$ can take? Is it possible to find the sum of all integer values that $x$ can take? In:
$$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$$
| Make the substitution $x=t^2+1$
$$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$$
$$\Rightarrow\sqrt{(t^2+1)+3-4\sqrt{(t^2+1)-1}}+\sqrt{(t^2+1)+8-6\sqrt{(t^2+1)-1}}=1$$
$$\Rightarrow\sqrt{t^2+1+3-4|t|}+\sqrt{t^2+1+8-6|t|}=1$$
$$\Rightarrow\sqrt{|t|^2-4|t|+4}+\sqrt{|t|^2-6|t|+9}=1$$
Make the substituion $|t|=a$
$$\Rightarrow\sqrt{(a-2)^2}+\sqrt{(a-3)^2}=1$$
$$\Rightarrow |a-2|+|a-3|=1$$
$$\Rightarrow |3-a|+|a-2|=1$$
We can see $(3-a)+(a-2)=1$ And we know that
$$|a|+|b|\geq|a+b|$$
And equality holds when $a$ and $b$ have like sign. So we have,
$$(3-a)(a-2)\geq 0$$
$$\Rightarrow (a-3)(a-2)\leq 0$$
$$\Rightarrow a\in [2,3] $$
Reverting back to $t$
$$\Rightarrow t\in [2,3]\cup [-3,-2] $$
Since $t=\sqrt{a-1}$, we have
$$x\in [5,10] $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A supposed to be easy calculus problem Find the values of $m$ if the line $y=mx+2$ is a tangent to the curve $x^2-2y^2=1$.
My working:
First we differentiate $x^2-2y^2=1$ with respect to $y$ to get the gradient. We get $y^2=\frac{1}{2}x^2-\frac{1}{2}\implies y=\pm\sqrt{\frac{1}{2}x^2-\frac{1}{2}}$.
We take the positive one for demonstration
$\frac{dy}{dx}=\frac{1}{2}x(\frac{1}{2}x^2-\frac{1}{2})^{-\frac{1}{2}}=\frac{x}{2\sqrt{\frac{1}{2}x^2-\frac{1}{2}}}$
$\implies(1-2m^2)x^2=-2m^2$
Since the tangent touches the curve, we can make $x^2-2(mx+2)^2=1$, we then get $(1-2m^2)x^2=9+8mx$
$\implies(1-2m^2)x^2=-2m^2$ and $(1-2m^2)x^2=9+8mx$ are two equations with two unknowns, then we should be able to find the values of $m$, but I couldn't find any easy way to solve those 2 simultaneous equations. Is there any easier method?
I tried solving $9+8mx=-2m^2$ but we still have two unknowns in one equation?
Also, if we don't use those two simultaneous equations, can we solve this question with a different method?
I am trying to solve WITHOUT implicit differentiation.
Many thanks for the help!
| Let the given line be tangent to the curve at the point $(x_0, y_0).$ Then we have $$y_0 = m x_0 + 2, \\ x_0^2 = 1 + 2y_0^2.$$ Using the fist equation in the second, we have $$x_0^2 = 1 + 2 (m x_0^2 + 4 m x_0 + 4),$$ which is $$m = \frac {x_0^2 - 9} {2 x_0^2 + 8 x_0}.$$ We also know from the second equation that $$y' = \frac{x} {\sqrt{2x^2-2}}.$$ Hence, $$m = \frac{x_0} {\sqrt{2x_0^2-2}}.$$ Solving the equation $$\frac{x_0} {\sqrt{2x_0^2-2}} = \frac {x_0^2 - 9} {2 x_0^2 + 8 x_0},$$ we get $x_0 = -3.44122\cdots$ and $m = -.73899\cdots$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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prove using Lagrange multipliers that for $x,y>0,\space n\in \mathbb N,\space (\frac{x+y}2)^n \leq \frac{x^n+y^n}2 $ I have been asked to prove using Lagrange multipliers that for
\begin{equation*}
\space (\frac{x+y}2)^n \leq \frac{x^n+y^n}2,~x,y>0,~n\in \mathbb {N}
\end{equation*}
I am familiar with the proof by induction, but I can't think of a way to prove it using Lagrange multipliers. I tried working with the function
\begin{equation*}
f(x,y)=\frac{x^n+y^n}2-(\frac{x+y}2)^n
\end{equation*}
but couldn't figure how to constrain it to $\{(x,y):x,y>0\}$. Maybe the constrain should be $g(x,y)=\sqrt x +\sqrt y -r$ for a given $r>0$ and then prove it for all $r$?
| Let $a=\frac{x}{x+y},b=\frac{y}{x+y}$. The inequality
$$(\frac{x+y}2)^n \leq \frac{x^n+y^n}2$$
is equivalent to
$$\frac{a^n+b^n}2\ge\frac{1}{2^n} $$
for $a,b>0$ and $a+b=1$. Let
$$ f(a,b,\lambda)=\frac{a^n+b^n}2-\lambda[(\frac{a+b}2)^n-1]. $$
Then solving
$$ \frac{\partial f}{\partial a}=\frac n2a^{n-1}-\frac{\lambda n}{2^n}(a+b)^{n-1}=0,\frac{\partial f}{\partial b}=\frac n2b^{n-1}-\frac{\lambda n}{2^n}(a+b)^{n-1}=0,a+b=1$$
gives
$$ a=b=\frac12,\lambda=1. $$
Thus $\frac{a^n+b^n}2$ reaches the minimum $\frac{1}{2^n}$ when $a=b=\frac12$ or
$$ \frac{a^n+b^n}2\ge \frac{1}{2^n}. $$
So
$$ \frac{x^n+y^n}2\ge (\frac{x+y}{2})^n. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding value of a quadratic The polynomial
\begin{equation*}
p(x)= ax^2+bx+c
\end{equation*}
has $1+\sqrt{3}$ as one of it's roots and also $p(2)=-2$. Is there any way to know the value of $a$, $b$ and $c$? I tried but I can form only $2$ equations how can I figure out value of $3$ unknown with $2$ equations so something must be missing.
Also $a$, $b$ and $c$ are rationals.
| Let $x = 1+\sqrt{3} \Rightarrow (x-1)^2 = 3 \Rightarrow x^2-2x-2 = 0 \Rightarrow a = k, b = -2k,c = -2k$, and $p(2) = -2 \Rightarrow 4a+2b+c=-2\Rightarrow 4k - 4k - 2k = -2 \Rightarrow k = 1 \Rightarrow (a,b,c) = (1,-2,-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the following equation: $\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x= 1$ A past examination paper had the following question that I found interesting. I tried having a go at it but haven't come around with any solutions. How would one go about tackling it?
$$\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x = 1$$
I'm seeing a relation between $4x$, $16x$ and $64x$ so maybe the larger can be simplified to the smaller?
I do encourage you to working within an examination environment (thus not making use of anything other than pen and paper and possibly a calculator).
EDIT: My question had a missing $-\sqrt x$ at the end, sorry!
| \begin{align*}
\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} - \sqrt x & = 1\\
\sqrt {x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}}} & = 1+\sqrt{x}\\
x + \sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+x+2\sqrt{x}\\
\sqrt {4x + \sqrt {16x + \sqrt {64x + 5}}} & = 1+2\sqrt{x}\\
4x + \sqrt {16x + \sqrt {64x + 5}} & = 1+4x+4\sqrt{x}\\
\sqrt {16x + \sqrt {64x + 5}} & = 1+4\sqrt{x}\\
16x + \sqrt {64x + 5} & = 1+16x+8\sqrt{x}\\
\sqrt {64x + 5} & = 1+8\sqrt{x}\\
64x + 5 & = 1+64x+16\sqrt{x}\\
16\sqrt{x} & = 4\\
x & =\frac{1}{16}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
Taylor series $\ln(1+e^x)$ about $x=0$ What's the best way to determine this up $x^3$ terms?
I thought it would be to take the series for $\ln(1+x)$ and the series for $e^x$ up to $x^3$ and sub the second series into the first.
$$\ln(1+x) = x - \frac {x^2}{2} +\frac {x^3}{3} + \cdots$$
$$e^x = 1+ x + \frac {x^2}{2} +\frac {x^3}{6} + \cdots$$
This gave me $$\frac 5{6} + x + \frac {3x^2}{2} + \frac {2x^3}{3}$$
but this is not the answer in my book. help?
| A standard way to obtain the Taylor series about $0$ is
$$\sum_{k=0}^{\infty} \dfrac{f^{(k)}(0)}{k!}\cdot x^k$$
Since $f(x) = \log(1+e^x)$, we have
$$f(0) = \log(1+e^0) = \log(2)$$
$$f'(0) = \left.\dfrac{e^x}{1+e^x} \right \vert_{x=0} = \dfrac12$$
$$f''(0) = \left.\dfrac{e^x}{(1+e^x)^2} \right \vert_{x=0} = \dfrac14$$
$$f'''(0) = \left.\dfrac{e^x(1-e^x)}{(1+e^x)^3} \right \vert_{x=0} = 0$$
Hence, we have the Taylor series about $0$ to be
$$\log(2) + \dfrac{x}2 + \dfrac{x^2}8 + \mathcal{O}(x^4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the largest triangle inscribed in the unit circle Among all triangles inscribed in the unit circle, how can the one with the largest area be found?
| For the maximum area the normal height of the triangle must be maximum for a given base Hence, the triangle ABC inscribed in unit circle should be an isosceles triangle having, $\angle B=\angle C$, base BC & angle between equal sides AB & AC be $\theta$ ($=\angle A$). Now, from right triangle, we get $$\sin\theta=\frac{\frac{BC}{2}}{1}\implies BC=2\sin\theta$$ Length of equal sides of triangle is determined as $$AB=AC=\frac{BC}{2\sin\frac{\theta}{2}}=\frac{2\sin\theta}{2\sin\frac{\theta}{2}}=2\cos\frac{\theta}{2}$$ Now, the area of isosceles $\Delta ABC$ is given as
$$A=\frac{1}{2}(AB)(AC)\sin \theta=\frac{1}{2}\left(2\cos\frac{\theta}{2}\right)^2\sin \theta=2\sin\theta\cos^2\frac{\theta}{2}=\sin\theta(1+\cos\theta)$$$$\implies A=\sin\theta+\frac{1}{2}\sin2\theta$$ Now differentiating the area (A) w.r.t. $\theta$ & equating it to zero, we get $$\frac{dA}{d\theta}=\cos\theta+\cos2\theta=0$$$$ \implies\cos\theta=-\cos2\theta=\cos(\pi-2\theta)$$ $$\implies \theta=\pi-2\theta \quad \text{or} \quad 3\theta=\pi \quad \text{or} \quad \theta=\frac{\pi}{3}=60^o=\angle A$$ $$\implies \angle B=\angle C=\frac{180^o-\angle A}{2}=60^o$$
Hence, the triangle having maximum area in a unit circle must be an equilateral triangle
($\color{#0ae}{\angle A=\angle B=\angle C=60^o}$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 8,
"answer_id": 5
} |
Integral does not 'converge' despite describing a well-defined area.... I have almost evaluated (where all variables are real including the variable $i$)
$$
C_1\int_{a + bt^2}^{i} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r \\ = \left[ \frac{1}{2} r^2 (i-r)^{\frac{3 (i-1)}{2}} \left(1-\frac{r}{i}\right)^{\frac{1}{2} (-3) (i-1)} \left(1-\frac{r}{i+j-R}\right)^{\frac{3}{2} (i+j-1)} \\(i+j-r-R)^{\frac{1}{2} (-3) (i+j-1)} F_1\left(2;\frac{1}{2} (-3) (i-1),\frac{3}{2} (i+j-1);3;\frac{r}{i},\frac{r}{i+j-R}\right) \right] _{a+bt^2}^{i}
$$
where $a+bt^2$ lies somewhere between $i-1$ and $i$. Also assume $i,j,r,R>0$; $F_1$ is the Appell hypergeometric series $F_1$.
There is a singularity at the upper limit (i.e. at $r=i$), despite the integral describing a well defined area (plotted here for $i=2,\hspace{1mm} j=2,\hspace{1mm}$ and $R=j-\frac{1}{5}$), so I can't continue...
Can anyone explain how I'm meant to perform this definite integral if there is a singularity at $r=i$?
| To answer my own question, after looking at Convergence of the improper integral $\int_{0}^{\pi/2}\tan^{p}(x) \; dx$ I tried the substitution $u = i-r$ and got
$$
\int_{a + bt^2}^{i} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r =\\\frac{2\left(-a-b t^2+i\right)^{\frac{3 i}{2}-\frac{1}{2}}}{\left(9 i^2-1\right) \left(-\frac{a+b t^2-i-j+R}{j-R}\right)^{3/2}} \\ \left(\left(\frac{1}{-a-b t^2+i+j-R}\right)^{\frac{3}{2} (i+j-1)} \left(1-\frac{a+b t^2-i}{j-R}\right)^{\frac{3 (i+j)}{2}} \\ \left(i (3 i+1) \, _2F_1\left(\frac{1}{2} (3 i-1),\frac{3}{2} (i+j-1);\frac{1}{2} (3 i+1);\frac{b t^2+a-i}{j-R}\right) \\ +(3 i-1) \left(a+b t^2-i\right) \, _2F_1\left(\frac{1}{2} (3 i+1),\frac{3}{2} (i+j-1);\frac{3 (i+1)}{2};\frac{b t^2+a-i}{j-R}\right)\right)\right)
$$
since the upper limit is now $0$ and makes the integral proper. Using a limit
$$
\int_{a + bt^2}^{i} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r =\lim_{W \to i}\int_{a + bt^2}^{W} \frac{r \left(i-r\right)^{\frac{3}{2}\left(i-1\right)}}{\left(j+i-R-r\right)^{\frac{3}{2}\left(j+i-1\right)}}\mathrm{d} r
$$
may also work (see http://en.wikipedia.org/wiki/Improper_integral), though would be tricky due to the complexity of the evaluated definite integral between $a+bt^2$ and $W$ (if it even exists without using a substitution).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"Rationalizing the denominator" of $1/(a + b\sqrt[3]{2} + c\sqrt[3]{4})$? If $(a, b, c) \in \mathbb{Q}^3 \setminus \{(0, 0, 0)\}$, so that $a + b\sqrt[3]{2} + c\sqrt[3]{4}$ is a nonzero element of $\mathbb{Q}(\sqrt[3]{2})$, is there a formula for $${1\over{a + b\sqrt[3]{2} + c\sqrt[3]{4}}}$$ as a rational linear combination of $1$, $\sqrt[3]{2}$, and $\sqrt[3]{4}$?
| Hint:
The product of conjugates elements is an integer (the norm of that element), which is equal to $2$ (see Viète's formulae). The norm of this element, setting $\omega=\mathrm e^{\tfrac{2\mathrm i\pi}3}$, is:
$$(a + b\sqrt[3]{2} + c\sqrt[3]{4})(a + b\omega\sqrt[3]{2} + c\omega^2\sqrt[3]{4})(a + b\omega^2\sqrt[3]{2} + c\omega\sqrt[3]{4})=2.$$
Thus \begin{align*}(a + b\sqrt[3]{2} + c\sqrt[3]{4})^{-1}&=\frac12(a + b\omega\sqrt[3]{2} + c\omega^2\sqrt[3]{4})(a + b\omega^2\sqrt[3]{2} + c\omega\sqrt[3]{4})\\
&=a^2-2bc+(2c^2-ab)\sqrt[3]{2}+(b^2-ac)\sqrt[3]{4}.\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$ Solve this equation: $(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$
This is my try
Let $t=\sqrt{2x+3}-2\sqrt{x+1}$ or $t^2=6x+7-4\sqrt{2x^2+5x+3}=1$
The equation is equivalent to: $t^2-4(x+2)t-6x-3=0\qquad(*)$
I have no idea how to solve the equation $(*)$. Who can help me?
| $$(x+2)(\sqrt{2x+3}-2\sqrt{x+1})+\sqrt{2x^2+5x+3}=1$$
$$(x+2)(\sqrt{2x+3}-2\sqrt{x+1})=1-\sqrt{2x^2+5x+3}$$
$$(x+2)^2(6x+7-4\sqrt{(2x+3)(x+1)})=2x^2+5x+4-2\sqrt{2x^2+5x+3}$$
$$(x+2)^2(6x+7)-(2x^2+5x+4)=2(2(x+2)^2-1)\sqrt{(2x+3)(x+1)})$$
$$(6 x^3+29 x^2+47 x+24)^2=4(2x^2+8x+7)^2(2x+3)(x+1)$$
Latter can be factored as roots $-1,\,-\frac{1}{2},3$ are known.
$\frac{4 x^6+12 x^5-19 x^4-106 x^3-135 x^2-68 x-12 }{(x+\frac{1}{2})(x+1)(x-3)}$$=2 (2 x+1) (x+2)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Coordinates of the center of the circle I am stuck on this problem:
If the lines $y=x+\sqrt{2}$ and $y=x-2\sqrt{2}$ are two tangents of a
circle and $(0,\sqrt{2})$ lies on this circle then what is the equation of the circle?
I found out the distance between the two tangents $y=x+\sqrt{2}$ and $y=x-2\sqrt{2}$ is $3$. The radius is $3/2$ but I don't know how to find the center. I tried forming the equations but could not succeed. Please tell me the easiest way. Thank you
| Hint: A sketch of the situation should be helpful:
What can we say about the position of the center of a circle with those two tangents?
What is its radius?
The equation of a circle with radious $r$ and center $(x_0,y_0)$ is:
$$
(x - x_0)^2 + (y - y_0)^2 = r^2
$$
Solution:
a) Center line: The center points lie on a line in the middle of the lines, thus on
\begin{align}
y_c(x)
&= \frac{1}{2}(y_1(x) + y_2(x)) \\
&= \frac{1}{2}((x + \sqrt{2}) + (x - 2 \sqrt{2})) \\
&= x - \frac{1}{\sqrt{2}}
\end{align}
b) Radius: The radius of the circle is two times the distance of the tangent lines.
An orthogonal line to both tangents is $y = - x$ it intersects $y_1$ if
$$
x + \sqrt{2} = - x \iff 2x = -\sqrt{2} \iff x = - 1/\sqrt{2}
$$
thus at $P = (-1/\sqrt{2}, 1/\sqrt{2})$.
The distance to the origin is $\lVert OP \rVert = \sqrt{1/2 + 1/2} = 1$
It intersects $y_2$ if
$$
x - 2 \sqrt{2} = -x \iff 2x = 2 \sqrt{2} \iff x = \sqrt{2}
$$
thus at $Q = (\sqrt{2}, -\sqrt{2})$.
The distance to the origin is $\lVert OQ \rVert = \sqrt{2 + 2} = 2$.
So both tangents are at a distance $3$ and the radius is $3/2$.
c) Equation of any circle with those tangents:
So all circles with those two tangents have the equation
$$
(x - x_0)^2 + \left(y - x_0 + \frac{1}{\sqrt{2}}\right)^2 = 9/4
$$
d) The circle with those tangents that contains $(0, \sqrt{2})$:
We want the one that contains $(0, \sqrt{2})$ thus
$$
9/4 = x_0^2 + (\sqrt{2} - x_0 + 1/\sqrt{2})^2
= x_0^2 + (3/\sqrt{2} - x_0)^2
= x_0^2 + 9/2 + x_0^2 - 3\sqrt{2} x_0 \Rightarrow \\
-9/4 = 2 x_0^2 - 3\sqrt{2} x_0 \Rightarrow \\
-9/8 = x_0^2 - (3/\sqrt{2}) x_0
= (x_0-3/(2\sqrt{2}))^2 - 9/8 \Rightarrow \\
x_0 = \frac{3}{2\sqrt{2}}
$$
This gives the center
$C=(3/(2\sqrt{2}), 3/(2\sqrt{2}) - 1/\sqrt{2}) = (3/(2\sqrt{2}), 1/(2\sqrt{2}))$.
And we have the equation
$$
\left(x - \frac{3}{2\sqrt{2}} \right)^2 +
\left(y - \frac{1}{2\sqrt{2}} \right)^2 = 9/4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Given $\csc\theta=-\frac53$ and $\pi<\theta<\frac32\pi$, evaluate sine ,cosine, and tangent of $2\theta$ If $\csc\theta=\frac{-5}{3}$, what is the exact value of $\tan(2\theta)$, $\sin(2\theta)$, and $\cos(2\theta)$ on the interval of $\left(\pi, \frac{3\pi}{2}\right)$?
I think I'm getting the fraction negatives wrong. I've used the sine formula($2\sin{\theta}\cos{\theta}$), the cosine formula ($2\cos^2{\theta}-1$) and the tangent formula, $\left(\frac{\sin(2\theta)}{\cos(2\theta)}\right)$
originally answering the problem I got $\sin(2\theta)=\frac{24}{25}$,
$\cos(2\theta)=\frac{7}{25}$, and $\tan(2\theta)=\frac{24}{7}$
| Notice that $$\csc{x} = \frac{1}{\sin{x}}$$
and so $$\csc {x} = \frac{-5}{3} = \frac{1}{\sin{x}}$$ and so $\sin{x} = \frac{-3}{5}$
now since $\color{blue}{\sin^2(x) + \cos^2(x) = 1}$ then we have that $$\big(\frac{-3}{5}\big)^2 + \cos^2(x) = 1$$ and so we get that $$\frac{9}{25} + \cos^2(x) = 1$$ and so $\cos^2{x} = 1 - \frac{9}{25} = \frac{16}{25}$ and so $$\cos{x} = \pm \frac{4}{5}$$
Now you want to find $\sin{2x}$ then you should use the identity $\color{red}{\sin{2x} = 2\sin{x}\cos{x}}$ and so you get $$\sin{2x} = 2\times {\frac{-3}{5}} \times \pm \frac{4}{5}$$.
Now you should the use other identity $\color{purple}{\cos{2x} = \cos^2{x} - \sin^2{x}}$ to find $\cos{2x}$ and finally you should use the identity that $$\color{green}{\tan{2x} = \frac{\sin{2x}}{\cos{2x}}}$$
To finish up your question.
I advise you to take a look at that link to study your identities
Double angle indentities
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sin(a)$ + $\cos(a)\leq\sqrt{2}$ $$\begin{align*}
\sin (a) + \cos(a) &\leq \sqrt{2}\\
(\sin(a)+ \cos(a))^2 &\leq (\sqrt{2})^2\\
\sin^2(a) + 2\sin(a)\cos(a) + \cos^2(a) &\leq \text{2}
\end{align*}$$
Am I doing it right? I need help.
| If @Zev Chonoles guessed right, you can prove this by computing
$$\max_{x\in [0,2\pi]} \sin x + \cos x$$
The critical points are precisely those where $\cos x - \sin x = 0$ (derivative). These are $x = \frac\pi4$ and $x = \frac{5\pi}4$. Evaluating at those and the boundary points gives
$$\sin \frac\pi4 + \cos \frac\pi4 = \frac1{\sqrt2} + \frac1{\sqrt2} = \sqrt2 \\
\sin \frac{4\pi}4 + \cos \frac{5\pi}4 = -\frac1{\sqrt2} - \frac1{\sqrt 2} = -\sqrt2 \\
\sin 0 + \cos 0 = 0 + 1 = 1\\
\sin 2\pi + \cos 2\pi = 0 + 1 = 1$$
Thus the maximum is $\max(\sqrt 2, -\sqrt 2, 1, 1) = \sqrt2$, wich proves
$$\sin x + \cos x \le \sqrt2 \qquad\forall\ x\in[0,2\pi]$$
Now since the function is $2\pi$-periodic, we can conclude
$$\sin x + \cos x \le \sqrt2 \qquad \forall\ x\in\mathbb R$$
| {
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"url": "https://math.stackexchange.com/questions/1304646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+c}$ Let $a,b,c\in \mathbb{R^+}$.
Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+c}$
PS: I don't have ay ideas about this problem :(
Thanks
| $\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}\ge 3 \sqrt[3]{\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}} \\ \iff \sqrt[3]{\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}} \ge 3\dfrac{\sqrt[3]{abc}}{a+b+c} $
now we prove a stronger one :
$\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}\ge \dfrac{3\sqrt[3]{abc}}{a+b+c} \iff \dfrac{(a^{2}+bc)(b^{2}+ac)(c^{2}+ab)}{(b^2+c^2)(a^2+c^2)(a^2+b^2)} \ge \dfrac{3^3abc}{(a+b+c)^3} \iff \dfrac{(a+b+c)^3}{3^3abc} \ge \dfrac{(b^2+c^2)(a^2+c^2)(a^2+b^2)}{(a^{2}+bc)(b^{2}+ac)(c^{2}+ab)}$
there is a lemma:
$\dfrac{(a+b)(b+c)(c+a)}{8abc} \ge \dfrac{(b^2+c^2)(a^2+c^2)(a^2+b^2)}{(a^{2}+bc)(b^{2}+ac)(c^{2}+ab)}$
Michael Rozenberg give this lemma a nice proof:
$(c^{2}+ab)(a+b)=a(b^2+c^2)+b(a^2+c^2) \ge 2 \sqrt{ab(b^2+c^2)(a^2+c^2)}$
with AM-GM, it is easy to prove:
$\dfrac{(a+b+c)^3}{3^3abc} \ge \dfrac{(a+b)(b+c)(c+a)}{8abc}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Zeroes of polynomials and their sum Let $a, b$ are zeroes of the polynomial $x^2-10cx-11d$ and $c,d$ are the zeroes of the polynomial $x^2-10a x-11b $ where $a,b,c, d$ are distinct reals then $a+b+c+d=?$
| K.Dutta gave you the good track to follow.
From the first equation, we have $$a+b=10c\ \ \ \ (1)$$ $$a b=-11d\ \ \ \ (2)$$ From the second equation, we have $$c+d=10a\ \ \ \ (3)$$ $$c d=-11b\ \ \ \ (4)$$ Use the first and the third of these to eliminate $c$ and $d$. This gives $$c=\frac{a+b}{10}$$ $$d=\frac{99 a-b}{10}$$ Now $$0=ab+11d=a \left(b+\frac{1089}{10}\right)-\frac{11 b}{10}$$ from which we can extract $b$ $$b=\frac{1089 a}{11-10 a}$$ Now, replacing in $cd+11b=0$ we get $$\frac{99 a^4-10890 a^3-119790 a^2+131769 a}{(11-10 a)^2}=0$$ By inspection, there are two simple roots for the numerator, namely $a_1=0$ and $a_2=-11$ and so the solution of $$99 a^2-11979 a+11979=0$$ provides the other two $a_3,a_4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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I think I can complete the square of any quadratic, is it true? (Any reason to ever use Quad. Formula?) I was taught that you could only complete the square of a quadratic if the coefficient on the $x^2$ term is 1.
However, playing a little bit with other quadratics, I've found that it's just not true. Based on the CTS algorithm, you just need to divide the coefficient of the $x$ term by twice the square root of the coefficient of the $x^2$ term.
So, if you have $ax^2 + bx + c$, your perfect square would be $(\sqrt{a}x + \frac{b}{2 \sqrt{a}})^2$
If $a$ is not a perfect square it could get nasty, but then you can just square the whole quadratic and go from there.
For example:
In the equation $5x^2 + 6x + 5 = 0$, we could do:
$25x^2 + 30x + 25 = 0$
$(5x+3)^2 = -16$
$5x+3 = \pm4i$
$x = \pm \frac{4i}{5} - \frac{3}{5}$
My questions are:
-Is this correct?
-Is there ever an advantage to using the quadratic formula?
-Are there quadratics that are unsolvable this way?
| The proposed way of completing the square is correct. The identity is:
$$
ax^2 + bx + c = \left(\sqrt{a}x + \frac{b}{2 \sqrt{a}}\right)^2 + \left(c-\frac{b^2}{2a}\right).
$$
One can of course also do what is more usually done and write
$$
ax^2+bx+c = a\left( x + \frac{b}{2a} \right)^2 + \left(c-\frac{b^2}{2a}\right).
$$
The latter way is done for the sake of convenience, not because it is the only way possible.
PS: Someone points out in comments that this works only if $a>0$. If $a<0$, one can write:
$$
ax^2 + bx + c = -\left(\sqrt{|a|}x + \frac{b}{2 \sqrt{|a|}}\right)^2 + \left(c-\frac{b^2}{2a}\right).
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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} |
Prove factorial problem $\forall n\in\mathbb N$ $C_n=\frac{1}{n+1}\left(\begin{matrix}2n\\n\end{matrix}\right)$.
Prove that $C_{n+1}=\left(\begin{matrix}2n\\n\end{matrix}\right)-\left(\begin{matrix}2n\\n-1\end{matrix}\right)$
I tried many ways but got nothing. Appreciate any tips.
My attempt: By defining, $C_{n+1}= \dfrac{1}{n+2}\left(\begin{matrix}2n+2\\n+1\end{matrix}\right)=\dfrac{1}{n+2}\dfrac{(2n+2)!}{(n+1)!(n+1)!}=\dfrac{1}{n+2}*\dfrac{(2n+2)(2n+1)n!}{(n+1)^2n!n!}$
$=\dfrac{1}{n+2}\dfrac{2(2n+1)n!}{(n+1)n!}$
| You’ve misstated the result: you want to show that
$$C_n=\binom{2n}n-\binom{2n}{n-1}\;,$$
not that
$$C_{n+1}=\binom{2n}n-\binom{2n}{n-1}\;.$$
Note that $\binom{2n-1}{n-1}=\binom{2n-1}n$, so
$$\binom{2n}{n-1}=\frac{2n}{n+1}\binom{2n-1}{n-1}=\frac{n}{n+1}\left(\binom{2n-1}{n-1}+\binom{2n-1}n\right)\;.$$
Now use Pascal’s identity and a little algebra.
| {
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"timestamp": "2023-03-29T00:00:00",
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Easier ways to prove $\int_0^1 \frac{\log^2 x-2}{x^x}dx<0$ Prove that
$$\int_0^1 \frac{\log^2 x-2}{x^x}dx<0$$
One way to do this is use the idea in the proof of Sophomore's dream. We have
$$x^{-x}=\exp(-x\log x)=\sum_{n=0}^\infty\frac{(-1)^nx^n\log^n x}{n!}$$
Therefore, using the change of variable $x=\exp(-t/(n+1))$ we have
$$\begin{aligned}\int_0^1 \frac{\log^2 x}{x^x}dx=&\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1x^n\log^{n+2}xdx\\
=&\sum_{n=0}^\infty\frac{(n+1)^{-(n+3)}}{n!}\int_0^\infty t^{n+2}e^{-t}dt\\
=&\sum_{n=0}^\infty\frac{(n+1)^{-(n+3)}(n+2)!}{n!}\\
=&\sum_{n=0}^\infty(n+1)^{-(n+2)}(n+2)\\
=&\sum_{n=1}^\infty n^{-(n+1)}(n+1)\\
=&\sum_{n=1}^\infty n^{-n}+n^{-(n+1)}\\
<&2\sum_{n=1}^\infty n^{-n}\\
=&\int_0^1\frac{2}{x^x}dx
\end{aligned}$$
Hence the result follows.
I am curious if there are any other methods to prove this, especially I am interested in easier approaches.
P.S. This was a bonus problem in an assignment from a multivariable calculus class.
| Using the inequality $\log(x)<x-1$ for all $x>0$ we have
\begin{align}
\int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx} < \int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{x^x }}dx}
\end{align}
Also we use the inequality $e^{x} \ge 1+x$ for all $x>0$ to express the denumeantor
$$x^x=\exp(x\ln(x))\ge 1+x\ln(x).$$
Therefore,
\begin{align}
\int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx} < \int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{x^x }}dx} <\int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{1+x\ln(x)}}dx}.
\end{align}
Now, using the inequality $\log(x) > \frac{x-1}{x}$ for all $x > 0$, we egt
\begin{align}
\int_0^1 {\frac{{\log ^2 \left( x \right) - 2}}{{x^x }}dx} < \int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{x^x }}dx} &<\int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{1+x\ln(x)}}dx}
\\
&<\int_0^1 {\frac{{\left( {x - 1} \right)^2 - 2}}{{1+x\cdot \frac{x-1}{x}}}dx}
\\
&<\int_0^1 {e\cdot x\cdot \frac{{\left( {x - 1} \right)^2 - 2}}{{1+x\cdot \frac{x-1}{x}}}dx} \,\,\,\,\,\,\,\,\,\,\,\,(\text{since $0<x<1$})
\\
&<\int_0^1 {e\cdot \left({\left( {x - 1} \right)^2 - 2}\right)dx} = -\frac{5}{3} e <0
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$. Hint: $x^2-y^2 = (x+y)(x-y)$. This is the exercise verbatim:
An integer n is a square modulo p if there exists another integer x
such that $n \equiv x^2 \pmod p$. Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$.
Hint: $x^2-y^2 = (x+y)(x-y)$.
This is my attempt to solve it:
If $x \equiv y \pmod p$ then $\displaystyle \frac{x-y}{p} = q_1 \Rightarrow (x-y)=pq_1$.
If $x \equiv -y$ then $\displaystyle \frac {x+y}{p}=q_2 \Rightarrow (x +y)=pq_2$.
Then $x^2-y^2 = (x+y)(x-y)=(q_1p)\cdot(q_2p)=q_1q_2p^2$
If $p$ is to divide $ x^2-y^2 $ evenly, then it must also divide $q_1q_2p^2$, therefore $\displaystyle \frac {x^2-y^2}{p}=q_1q_2p$.
To satisfy the condition of divisibility by $p$, $q_1q_2p$ must be an integer. Since the assertion to be proven is "Prove that $x^2 \equiv y^2 \pmod p$ if and only if $x \equiv y \pmod p$ or $x \equiv -y \pmod p$.", we must show that the premise is true if either $q_1$ or $q_2$ is an integer for sure.
Is this line of reasoning correct up until this point?
How to guarantee that $q_1q_2p$ is an integer?
| Well, use your hint:
$$x^2 - y^2 \equiv 0 \pmod p$$
$$(x+y)(x-y) \equiv 0 \pmod p$$
Clearly, ths is satisfied if $x=y$ or $x=-y$. Because $\mathbb{Z}_p$ has no zero divisors, these are the only solutions.
| {
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Part of an induction question I might have done or not realize something stupid, but I can't seem to prove the following...
Inductive hypothesis
Assume $\exists$k$\in$N such that P(k) is true.
P(k): $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot (2k - 1)}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot (2k)}$ $\leq$ $\frac{1}{\sqrt{k + 1}}$
Inductive step
Prove P(k+1) is true.
P(k+1): $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot [2(k + 1)]}$ $\leq$ $\frac{1}{\sqrt{(k + 1) + 1}}$
$\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot [2(k + 1)]}$
= $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot (2k - 1) \cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot (2k) \cdot [2(k + 1)]}$
$\leq$ $\frac{1}{\sqrt{k + 1}}$ $\cdot$ $\frac{2(k + 1) - 1}{2(k + 1)}$ #by IH
= $\frac{2(k + 1) - 1}{2(k + 1) \cdot \sqrt{k + 1}}$
$\leq$ $\frac{2(k + 1)}{2(k + 1) \cdot \sqrt{k + 1}}$
= $\frac{1}{\sqrt{k + 1}}$
So what I seem to have done is proven that $\frac{1 \cdot 3 \cdot 5 \cdot\cdot\cdot [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdot\cdot\cdot [2(k + 1)]}$ $\leq$ $\frac{1}{\sqrt{k + 1}}$ and not $\frac{1}{\sqrt{(k + 1) + 1}}$. I seem to need a minor adjustment somewhere.
| We first (directly) prove:
$$\frac{1}{\sqrt{k+1}}\cdot\frac{2k+1}{2k+2} < \frac{1}{\sqrt{k+2}}$$
for $k>1$. Starting with:
$$9k+2<12k+4$$
which is obviously true for $k>1$. Then, we add $4k^3+12k^2$ to both sides:
$$4k^3+12k^2+9k+2<4k^3+12k^2+12k+4$$
Splitting and factoring:
$$4k^3+4k^2+k+8k^2+8k+2<4k^3+8k^2+4k+4k^2+8k+4$$
$$(4k^2+4k+1)(k+2)<(4k^2+8k+4)(k+1)$$
$$(2k+1)^2(k+2)<(2k+2)^2(k+1)$$
Taking square roots:
$$(2k+1)\sqrt{k+2}<(2k+2)\sqrt{k+1}$$
Dividing by $(2k+2)\sqrt{k+1}\sqrt{k+2}$ on both sides:
$$\frac{1}{\sqrt{k+1}}\cdot\frac{2k+1}{2k+2} < \frac{1}{\sqrt{k+2}}$$
Now we start the induction step:
$$\frac{1 \cdot 3 \cdot 5 \cdots[2k-1] [2(k + 1) - 1]}{2 \cdot 4 \cdot 6 \cdots [2k][2(k + 1)]} \leq \frac{1}{\sqrt{k + 1}}\cdot \frac{2(k+1)-1}{2(k+1)}=\frac{1}{\sqrt{k + 1}}\cdot \frac{2k+1}{2k+2}<\frac{1}{\sqrt{k + 2}}$$
There is our desired result! Q.E.D.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
Question: Is it possible to have an $n\times n$ real matrix $A$ such that $A^TA$ has an eigenvalue of $-1$?
I can prove that it is not possible for $n=1,2$, but I am not sure for the general case.
Case $n=1$: $a^2v=-v$ $\implies$ $(a^2+1)v=0$ $\implies$ $v=0$.
Case $n=2$: Write $A=\begin{pmatrix}a&b \\ c&d\end{pmatrix}$. Then, $A^T A=\begin{pmatrix}a^2+c^2&ab+cd\\ ab+cd&b^2+d^2\end{pmatrix}$, so
$$
\begin{align}
\det(A^TA+1)
&= \begin{vmatrix}a^2+c^2+1&ab+cd\\ ab+cd&b^2+d^2+1\end{vmatrix}\\
&= 1+a^2+b^2+c^2+d^2+(ad-bc)^2\\
&\neq 0.
\end{align}
$$
Now, can we generalize to all $n$?
| suppose $\lambda,$ necessarily real, is an eigenvalue of the symmetric matrix $A^\top A$ and $x \neq 0$ an eigenvector. then $$A^\top A x = \lambda x \tag 1$$ multiplying by $x^\top$ on the left, we get $$\lambda x^\top x =x^\top A^\top Ax = (Ax)^\top Ax \implies\lambda = \frac{|Ax|^2}{|x|^2} \ge 0. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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The maximum of $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}$
Let $a,b,c$ be positive reals, such that $\frac1a+\frac1b+\frac1c=a+b+c\ (\star)$, find the maximum of $\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}$
This should be an application of Jensen's Inequality, so I have to find a concave function to maximize the sum, I thought to define;
$f(\frac1x):=\frac{1}{x+a+b+c},\quad\frac{\partial^2f(x)}{\partial x^2}<0$
so $f$ is concave and then the sum above is;
$f(\frac1a)+f(\frac1b)+f(\frac1c)$
using Jensen it should be less or equal to
$3\cdot f(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}3)=3\frac{1}{\left(\frac3{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+a+b+c\right)^2}\overset{(\star)}=3\frac{1}{\left(\frac3{a+b+c}+a+b+c\right)^2}$
and the denominator can be minimized with AM-GM;
$\left(\frac3{a+b+c}+a+b+c\right)^2\ge\left(2\sqrt{3\frac{a+b+c}{a+b+c}}\right)^2=12$
so the result is $\frac14$, but it is wrong, where does it fail, maybe AM-GM value cannot be attained ?
| I would rather show the maximum is $\frac3{16}$ by getting rid of the cumbersome constraint by homogenizing to:
$$\frac{abc(a+b+c)}{ab+bc+ca}\sum_{cyc} \frac1{(2a+b+c)^2} \le \frac3{16}$$
So if $a+b+c=3$, this is equivalent to
$$16\sum_{cyc} \frac1{(3+a)^2} \le \frac1a+\frac1b+\frac1c$$
which follows from $f(a)+f(b)+f(c)\ge 0$ with
$$f(x) = \frac1x-\frac{16}{(3+x)^2}+\frac12(x-1) = \frac{(x-1)^2(x^2+7x+18)}{2x(3+x)^2} \ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Combining probability of dependent events?
Two dice are thrown simultaneously. What is the probability of getting a multiple of $2$ on one die and a multiple of $3$ on the other?
According to me, the answer should be $\frac16$, as the probability of getting a multiple of $2$ is $\frac12$, and the probability of getting a multiple of $3$ is $\frac13$. So the combined probability is $\left(\frac12\right)\left(\frac13\right) = \frac16$.
But the answer in the book says that the probability is $\frac{11}{36}$.
What is the error in my method? How does one solve this problem properly?
| The error you've made is that you haven't considered the possibility that either dice might have the multiple of 2, or the multiple of 3.
For a problem with these kinds of small numbers we can use fairly weak combinatorial methods.
The total number of outcomes is 36, we count the pairs of numbers such that one is a multiple of 2, and one is a multiple of 3.
The multiples of 2 are 2, 4, 6 the multiples of 3 are 3, 6. So the pairs satisfying the question are
$(2,3)(4,3)(6,3)(3,2)(3,4)
(3,6)(2,6)(6,2)(4,6)(6,4)
(6,6)$
This is eleven pairs of 36 so
$$\frac{11}{36}$$
We could also consider that each multiple of 2 gives 4 pairs (2,3)(3,2)(2,6)(6,2)
with three multiples of 2 we have 12 pairs, but we have to discount (6,6) as it is not distinct when we swap the numbers. So we have
$$\frac{3 \times 4 -1}{36} = \frac{11}{36}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ To find the above minimal polynomial, let
$$x=\sqrt{2}+\sqrt{3}+\sqrt{5}$$
$$x^2=10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}$$
Subtracting 10 and squaring gives
$$x^4-20x^2+100=4(31+2\sqrt{60}+2\sqrt{90}+2\sqrt{150})$$
$$x^4-20x^2+100=4(31+4\sqrt{15}+6\sqrt{10}+10\sqrt{6})$$
$$x^4-20x^2-24=40\sqrt{6}+24\sqrt{10}+16\sqrt{15}$$
$$x^4-20x^2-24=8(2\sqrt{6}+2\sqrt{10}+2\sqrt{15})+24\sqrt{6}+8\sqrt{10}$$
$$x^4-20x^2-24=8(x^2-10)+24\sqrt{6}+8\sqrt{10}$$
$$x^4-28x^2-104=24\sqrt{6}+8\sqrt{10}$$
Again, squaring both sides
$$x^8-56x^6+576x^4+5428x^2+10816=4096+765\sqrt{6}$$
But if I square again, I will get a degree 16 polynomial. Mathematica says the minimal polynomial is degree 8, which would make sense since elements of $\mathbb{Q}[\sqrt{2},\sqrt{3},\sqrt{5}]$ look like
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5}+e\sqrt{6}+f\sqrt{10}+g\sqrt{15}+h\sqrt{30}$$
Where am I making mistakes?
| I thought I'd get rid of the surds one-by-one: first get rid of $\sqrt2$, then $\sqrt3$, and finally $\sqrt5$. To get rid of a surd, I would get everything with that surd in it to one side, and then square. (Note that $\sqrt{15}$ has a $\sqrt3$ in it.)
\begin{align}
x&=\sqrt2+\sqrt3+\sqrt5\\
x-\sqrt3-\sqrt5&=\sqrt2\\
(x-\sqrt3-\sqrt5)^2&=2\\
x^2+(-2\sqrt5-2\sqrt3)x+(2\sqrt{15}+8)&=2\\
x^2+(-2\sqrt5-2\sqrt3)x+(2\sqrt{15}+6)&=0\\
x^2-2\sqrt5x+6&=2\sqrt3x-2\sqrt{15}\\
(x^2-2\sqrt5x+6)^2&=(2\sqrt3x-2\sqrt{15})^2\\
x^4-4\sqrt5x^3+32x^2-24\sqrt5x+36&=12x^2-24\sqrt5x+60\\
x^4-4\sqrt5x^3+20x^2-24&=0\\
x^4+20x^2-24&=4\sqrt5x^3\\
(x^4+20x^2-24)^2&=80x^6\\
x^8+40x^6+352x^4-960x^2+576&=80x^6\\
x^8-40x^6+352x^4-960x^2+576&=0
\end{align}
This, of course, is equal to the Galois theory answer of $\displaystyle\prod(x\pm\sqrt2\pm\sqrt3\pm\sqrt5)$, where you multiply every possible choice of signs together.
| {
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"url": "https://math.stackexchange.com/questions/1313897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 2
} |
Expansion of Generating Functions If you roll $10$ dice, how many ways can you get a total sum of top faces of $25$?
I understand how to write the generating function of $(x+x^2+ \dots +x^6)^{10}$ and the fact that you need to find the coefficient of $x^{25}$, but how do you do this? The conventional binomial theorem only works for binomial expansion....
| Write $x+x^2+\cdots+x^6$ as $(1-x^7)(1-x)^{-1}$. Then
$$\begin{align}
[x^{25}]&(x+x^2+\cdots+x^6)^{10}\\
&=[x^{25}](x-x^7)^{10}(1-x)^{-10}\\
&=[x^{15}](1-x^6)^{10}(1-x)^{-10}\\
&=[x^{15}]\sum_{r\ge0}\binom{10}r(-x^6)^r
\sum_{s\ge0}\binom{-10}s(-x)^s\\
&=\sum_{r\ge0}(-1)^r\binom{10}r\binom{10+s-1}{10-1}\quad\textrm{with $s=15-6r$}\\
&=\sum_{r=0}^2(-1)^r\binom{10}r\binom{24-6r}{9}\quad\textrm{noting $15-6r\ge0$}\\
&=\binom{10}0\binom{24}9 -\binom{10}1\binom{18}9 +\binom{10}2\binom{12}9\\
&=831204
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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