Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How many distinct real roots does $ (x^2 + x – 2)^3 + (8–2x^2 )^3 = (x^2 + 3x + 2)^3 $have?
How many distinct real roots does $ (x^2 + x – 2)^3 + (8–2x^2 )^3 =
(x^2 + 3x + 2)^3 $have?
If I didn't make any mistake that equation could be reduced to the form:$$(2+6x^2)(x+2)^3=(8–2x^2 )^3$$
However,I don't know how to approach next as more simplification and then forming up the factors again would be more time consuming,I suppose this problem could be solved in a more easy way,any ideas?
| We are trying to solve the equation
$$(x^2 + x – 2)^3 + (8–2x^2 )^3 -(x^2 + 3x + 2)^3=0.$$
Let $P(x)$ be the polynomial on the left-hand side. Note that $x^2+x-2=(x+2)(x-1)$, and $8-2x^2=-2(x-2)(x+2)$, and $x^2+3x+2=(x+2)(x+1).$
Thus
$$P(x)=(x+2)^3[(x-1)^3-8(x-2)^3-(x+1)^3]=(x-2)^3Q(x).$$
Expand the remaining cubes. We get
$$Q(x)=-8x^3+42x^2-96x+62.$$
We get very lucky: $x=1$ is a root of $Q(x)$. Divide.
$$Q(x)=(x-1)(-8x^2+34x-62).$$
It is easy to verify, say by the Quadratic Formula, that $8x^2-34x+62=0$ has no real roots.
So there are altogether two distinct real roots, $x=-2$ and $x=1$.
Your calculation: Things would have worked out fine, if after the (correct) simplification of the difference of cubes you had used the fact that $8-2x^2=-2(x+2)(x-2)$. But except in special circumstances, of which this happens to be one, it is easier to deal with a polynomial equation of shape $P(x)=0$ than with one of shape $P_1(x)=P_2(x)$.
Comment: The question has a quite artificial character, because of the occurrence of an obvious root $x=-2$ of multiplicity at least $3$. Given an "arbitrary" degree $6$ polynomial, the techniques for finding the number of real roots would be quite different. Nowadays, an almost automatic first step is to see what graphing software, or a tool like Wolfram Alpha, has to say.
| {
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"timestamp": "2023-03-29T00:00:00",
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Trig substitution integration of $\int1/(x^2\sqrt{x^2 - 9}) dx$ - stuck on a problem I am getting stuck on this trig substitution problem.
$$\int\frac1{x^2\sqrt{x^2 - 9}}~\mathrm dx.$$
$$x = 3 \sec\theta,\qquad\theta = \sec^{-1} \sqrt{\frac{x^2}{9}},\qquad\mathrm dx = \sec\theta\tan\theta\ \mathrm d\theta$$
I can get to here, but I don't know how to finish it (perhaps I have made a mistake before this point?)
$$\int\frac{3\sec\theta\tan\theta}{9\sec^2\theta(3\sec\theta -3)}~\mathrm d\theta.$$
If anyone could help from here, I'd appreciate it.
Thanks.
| Using your substitution $x=3\sec \theta $ and cancelling $\sec \theta$ in the numerator and denominator, I got
$$I=\int \frac{1}{x^{2}\sqrt{x^{2}-9}}\ \textrm{d}x=\int \frac{ \tan \theta }{3\ \sec
\theta \ \sqrt{9\sec ^{2}\theta -9}}\ \textrm{d}\theta=\int \frac{\tan \theta }{9\ \sec
\theta \ \sqrt{\sec ^{2}\theta -1}}\ \textrm{d}\theta.$$
It is easy to see that
$$\frac{\tan \theta }{ \sec \theta \sqrt{\sec ^{2}\theta -1}}=
\cos \theta .$$
So
$$I=\int \frac{1}{9}\cos \theta \ \textrm{d}\theta = \dots .$$
Added. Just to confirm Christian Blatter's evaluation.
$$\begin{eqnarray*}
I &=&\int \frac{1}{9}\cos \theta \,d\theta =\frac{1}{9}\sin \theta +C \\
&=&\frac{1}{9}\sqrt{1-\cos ^{2}\theta }+C=\frac{1}{9}\sqrt{1-\frac{1}{\sec
^{2}\theta }} +C\\
&=&\frac{1}{9}\sqrt{1-\frac{9}{x^{2}}}+C=\frac{\sqrt{x^{2}-9}}{9x}+C.
\end{eqnarray*}$$
| {
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If $a+\frac1b=b+\frac1c=c+\frac1a$ for distinct $a$, $b$, $c$, how to find the value of $abc$?
If $a, b, c$ be distinct reals such that $$a + \frac1b = b + \frac1c = c + \frac1a$$ how do I find the value of $abc$?
The answer says $1$, but I am not sure how to derive it.
| From the fact that Expressions $1$ and $2$ are equal, we obtain
$$a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc}.$$
From the fact that Expressions $2$ and $3$ are equal, we obtain
$$b-c=\frac{1}{a}-\frac{1}{c}=\frac{c-a}{ca}.$$
From the fact that Expressions $3$ and $1$ are equal, we obtain
$$c-a=\frac{1}{b}-\frac{1}{a}=\frac{a-b}{ab}.$$
Multiply the left-hand sides, the right-hand sides. We get
$$(a-b)(b-c)(c-a)=\frac{(b-c)(c-a)(a-b)}{(abc)^2}.$$
Since $a$, $b$, and $c$ are distinct, $(a-b)(b-c)(c-a)\ne 0$. We conclude that
$(abc)^2=1$. This yields the two possibilities $abc=1$ and $abc=-1$.
In a logical sense we are finished: We have shown that if $(a,b,c)$ is a solution of the system with $a$, $b$, and $c$ distinct, then $abc=\pm 1$.
But it would be nice to show that there are solutions of the desired type. So let's find such a solution, with $abc=1$.
Look for a solution with $a=1$. Then we need $c=1/b$. In order to satisfy our equations, we need $1+1/b=2b$. Beside the useless solution $b=1$, this has the solution $b=-1/2$. We conclude that
$$a=1,\quad b=-\frac{1}{2},\quad c=-2$$
is a solution of the desired type, with $abc=1$. By changing all the signs, we find that
$$a=-1,\quad b=\frac{1}{2},\quad c=2$$
is a solution with $abc=-1$.
Added: It is easy to see that if two 0f $a$, $b$, $c$ are equal, then they are all equal, giving the parametric family $(t,t,t)$, where $t\ne 0$. Now assume that $abc=\pm 1$. We find all solutions with $abc=1$. The solutions with $abc=-1$ are then obtained by changing all the signs.
Let $a=t$. Our equations will be satisfied if $t+1/b=b+1/c$, where $c=1/bt$. We therefore obtain the equation $t+1/b=b+bt$, which simplifies to $(1+t)b^2 -tb-1=0$. Now we can solve this quadratic equation for $b$, and get $c$ from $tbc=1$. There is undoubtedly a more symmetric way to obtain the complete parametric description of the solutions!
| {
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A quadratic equation $ax^2+bx+c=0$ has equal roots at $a=2c$. How could we find the sum of reciprocals of the roots of this equation?
A quadratic equation $ax^2+bx+c=0$ has equal roots at $a=2c$. How
could we find the sum of reciprocals of the roots of this equation?
I need some hints for solving this problem.
| This shall answer your question.
A polynomial is of the form,
$$P(x) = (x-r_1)(x-r_2)........(x-r_n)$$
where $r_1,r_2,.....r_n $ are the roots of the polynomial.
Then the first derivative of the polynomial will be,
$$P^{'}(x) = \frac{(x-r_1)^{'}}{(x-r_1)}P(x) + \frac{(x-r_2)^{'}}{(x-r_2)}P(x) + \frac{(x-r_1)^{'}}{(x-r_1)}P(x)........\frac{(x-r_n)^{'}}{(x-r_n)}P(x)$$
since $(x-k)^{'} = 1$, we have,
$$\frac{P^{'}(x)}{P(x)} = \frac{1}{(x-r_1)} + \frac{1}{(x-r_2)} + \frac{1}{(x-r_3)} +........ \frac{1}{(x-r_n)}$$
for $x = 0$
$$\frac{P^{'}(0)}{P(0)} = \frac{1}{(-r_1)} + \frac{1}{(-r_2)} + \frac{1}{(-r_3)} +........ \frac{1}{(-r_n)}$$
$$-\frac{P^{'}(0)}{k} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} +........ \frac{1}{r_n} $$
where $k$ is the constant term of the polynomial.
Therefore the sum of the reciprocals of the roots of any polynomial is equal to minus of the first derivative of the polynomial at $x = 0$ divided by the constant term of the polynomial.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to create 2x2 matrix to rotate vector to right side? I have vector u=(x,y) and i need to create matrix M: M*u=(1,0).
But that matrix has to rotate vector, instead of keep and scale the x unit. So when i apply it on different vectors, the angle between them won't change.
Btw, this isn't homework! We haven't learned any matrices at school yet. ;)
| Your problem is equivalent to find the transformation between the $x,y$
coordinates of a point and the $x^{\prime },y^{\prime }$ coordinates of the
same point in a rotated system of coordinates, followed by a multiplication
by the factor $k=1/\sqrt{x^{2}+y^{2}}$, so that $x^{\prime \prime
}=kx^{\prime }=1$ and $y^{\prime \prime }=kx^{\prime }=0$. The rotation
angle should be $\theta =\arctan \frac{y}{x}$.
From trigonometry, we know that
$$
\begin{eqnarray*}
&&\left\{
\begin{array}{c}
x^{\prime }=x\cos \theta +y\sin \theta =\sqrt{x^{2}+y^{2}} \\
y^{\prime }=-x\sin \theta +y\cos \theta =0
\end{array}
\right.
\end{eqnarray*}
$$
and since
$$
\begin{eqnarray*}
\cos \left( \arctan \frac{y}{x}\right) &=&\frac{x}{\sqrt{x^{2}+y^{2}}} \\
\sin \left( \arctan \frac{y}{x}\right) &=&\frac{y}{\sqrt{x^{2}+y^{2}}}, \\
\end{eqnarray*}
$$
we have
$$\begin{eqnarray*}
\left\{
\begin{array}{c}
x^{\prime \prime }=\frac{1}{\sqrt{x^{2}+y^{2}}}x^{\prime }=\frac{x^{2}}{
x^{2}+y^{2}}+\frac{y^{2}}{x^{2}+y^{2}}=1 \\
y^{\prime \prime }=\frac{1}{\sqrt{x^{2}+y^{2}}}y^{\prime }=-\frac{xy}{
x^{2}+y^{2}}+\frac{xy}{x^{2}+y^{2}}=0.
\end{array}
\right.
\end{eqnarray*}
$$
We haven't learned any matrices at school yet.
In matrix notation$^1$
$$
\begin{eqnarray*}
\begin{pmatrix}
x^{\prime \prime } \\
y^{\prime \prime }
\end{pmatrix}
&=&\frac{1}{\sqrt{x^{2}+y^{2}}}
\begin{pmatrix}
x^{\prime } \\
y^{\prime }
\end{pmatrix}
=
\begin{pmatrix}
\frac{x}{x^{2}+y^{2}} & \frac{y}{x^{2}+y^{2}} \\
-\frac{y}{x^{2}+y^{2}} & \frac{x}{x^{2}+y^{2}}
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
1 \\
0
\end{pmatrix}.
\end{eqnarray*}
$$
So
$$
M=
\begin{pmatrix}
\frac{x}{x^{2}+y^{2}} & \frac{y}{x^{2}+y^{2}} \\
-\frac{y}{x^{2}+y^{2}} & \frac{x}{x^{2}+y^{2}}
\end{pmatrix}.
$$
--
$^1$ Product of a $2\times 2$ matrix by a $2\times 1$ matrix
$$
\begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{pmatrix}
\begin{pmatrix}
b_{1} \\
b_{2}
\end{pmatrix}
=
\begin{pmatrix}
a_{11}b_{1}+a_{12}b_{2} \\
a_{21}b_{1}+a_{22}b_{2}
\end{pmatrix}
$$
and product between a scalar $\alpha$ and a $2\times 1$ matrix
$$\alpha
\begin{pmatrix}
c_{1} \\
c_{2}
\end{pmatrix}
=
\begin{pmatrix}
\alpha c_{1} \\
\alpha c_{2}
\end{pmatrix}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}\mathrm dx$ I need help to evaluate the integral: $$ \int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}\mathrm dx.$$
The procedure on Wolfram Alpha is very long and complicated. Is there any easier way ?
Thanks.
| As suggested in the comments, you can substitute to get rid of the $-1$ in the third root. Let $y = x - \frac{1}{4}$. To write $p(x) = x^2 - 2x + 7$ as a polynomial in $y$, note that $p(x) = p(y + \frac{1}{4})$ so substituting $x = y + \frac{1}{4}$ in $p(x)$ gives $x^2 - 2x + 7 = y^2 - \frac{3}{2} y + \frac{105}{16}$. So we get
$$\int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}dx = \int \frac{y^{2}-\frac{3}{2}y+\frac{105}{16}}{\sqrt[3]{4y}}dy = \frac{1}{\sqrt[3]{4}} \left( \int y^{5/3} dy - \frac{3}{2} \int y^{2/3} dy + \frac{105}{16} \int y^{-1/3} dy \right)$$
Solving each integral individually gives
$$\frac{1}{\sqrt[3]{4}} \left(\frac{3}{8} y^{8/3} - \frac{9}{10} y^{5/3} + \frac{315}{32} y^{2/3} + C\right)$$
Plugging in $y = x - \frac{1}{4}$ gives
$$\int \frac{x^{2}-2x+7}{\sqrt[3]{4x-1}}dx = \frac{3 (4x - 1)^{2/3} \left(80 x^2 - 232 x + 2153\right)}{2560} + C$$
| {
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Difficult Integral: $\int\frac{x^n}{\sqrt{1+x^2}}dx$ How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?
The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.
And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?
| Here's one way to go about deriving a recursion relation for integrals of the form
$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx$$
Split the integral like so:
$$\int x^{n-1}\frac{x}{\sqrt{1+x^2}}\mathrm dx$$
and integrate by parts:
$$\int x^{n-1}\frac{x}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int\sqrt{1+x^2} x^{n-2}\mathrm dx$$
Noting that $1+x^2$ is always positive for real $x$, we then complicate things a little:
$$\int \frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int(1+x^2)\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx$$
Perform another split:
$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\left(\int \frac{x^n}{\sqrt{1+x^2}}\mathrm dx+\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx\right)$$
and we see something we can isolate:
$$n\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx$$
and then we finally divide both sides by $n$:
$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=\frac1{n}\left(x^{n-1}\sqrt{1+x^2}-(n-1)\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx\right)$$
We can use the starting values $\int\frac{\mathrm dx}{\sqrt{1+x^2}}=\mathrm{arsinh}\,x$ and $\int\frac{x \mathrm dx}{\sqrt{1+x^2}}=\sqrt{1+x^2}$ for the recursion.
(This is a response to Srivatsan's comment, which got too long for the comment box.)
| {
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How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$? How can the following inequation be proven?
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
| This inequality can be solved by simple algebra
we have the equation
$$a^2+b^2+c^2 >= ab+ac+cb$$
multiply and divide 2 on both sides
$$\frac{2}{2}(a^2+b^2+c^2) >= \frac{2}{2}(ab+ac+cb)$$
putting the left side equation on the right side
$$\frac{2a^2+2b^2+2c^2 - 2(ab+ac+cb)}{2} >= 0$$
then factorize and multiply by 2 on both the sides
$$\frac{2a^2+2b^2+2c^2 - 2(ab+ac+cb)}{2} >= 0$$
$$\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} >= 0$$
$$\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} >= 0$$
$$2*\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} >= 2*0$$
$$(a-b)^2+(b-c)^2+(c-a)^2 >= 0$$
So the above equation is zero when $a=b=c$
So the above inequality is proved
| {
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How can I compute $\sum\limits_{k = 1}^n \frac{1} {k + 1}\binom{n}{k} $? This sum is difficult. How can I compute it, without using calculus?
$$\sum_{k = 1}^n \frac1{k + 1}\binom{n}{k}$$
If someone can explain some technique to do it, I'd appreciate it.
Or advice using a telescopic sum, I think with a telescopic could go, but do not know how to assemble it.
| \begin{align*}\sum_{k=1}^n \binom{n}{k} \frac{1}{k+1}
= \frac{1}{n+1} \sum_{k=1}^n \binom{n+1}{k+1}
= \frac{2^{n+1} - 1 - (n+1)}{n+1} = \frac{2^{n+1} - n-2}{n+1}.
\end{align*}
The first step follows from the identity $\binom{n}{k} \frac{n+1}{k+1} = \frac{n!}{k! (n-k)!} \frac{n+1}{k+1} = \frac{(n+1)!}{(k+1)! (n-k)!} = \binom{n+1}{k+1}$. The second step uses the fact that $\sum_{k=0}^{n+1} \binom{n+1}{k} = 2^{n+1}$, while noting that $\binom{n+1}{0}$ and $\binom{n+1}{1}$ are not included in the sum.
| {
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The $3 = 2$ trick on Google+ I found out this on Google+ yesterday and I was thinking about what's the trick. Can you tell?
How can you prove $3=2$?
This seems to be an anomaly or whatever you call in mathematics. Or maybe I'm just plain dense.
See this illustration:
$$ -6 = -6 $$
$$ 9-15 = 4-10 $$
Adding $\frac{25}{4}$ to both sides:
$$ 9-15+ \frac{25}{4} = 4-10+ \frac{25}{4} $$
Changing the order
$$ 9+\frac{25}{4}-15 = 4+\frac{25}{4}-10 $$
This is just like $a^2 + b^2 - 2a b = (a-b)^2$. Here $a_1 = 3, b_1=\frac{5}{2}$ for L.H.S, and $a_2 =2, b_2=\frac{5}{2}$ for R.H.S. So it can be expressed as follows:
$$ \left(3-\frac{5}{2} \right) \left(3-\frac{5}{2} \right) =
\left(2-\frac{5}{2} \right) \left( 2-\frac{5}{2} \right) $$
Taking positive square root on both sides:
$$ 3 - \frac{5}{2} = 2 - \frac{5}{2} $$
$$ 3 = 2 .$$
I think it's something near the root.
| On the right side, when you say to take the positive square root of $(2-5/2)(2-5/2)$, you're taking a $-.5 [(2-5/2)]$ instead of $.5$
It's easy to see if you multiply out all the numbers in each step.
| {
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Evaluating $\sum_{k=0}^n \binom{n}{k} 2^{k^2}$ Can someone please help me simplifying this sum
$$\sum_{k=0}^n \binom{n}{k} 2^{k^2}$$
Wolframalpha fails (see here).
Thanks in advance.
The sum counts the number of (labelled) digraphs (with loops) with vertex set $V \subseteq [n]$.
| I agree with J.M.'s comment above that this is not going to resolve into something simple. Asymptotically, though, the last term will dominate, and maybe that will be useful. We have
$$2^{n^2} \leq \sum_{k=0}^n \binom{n}{k} 2^{k^2} \leq 2^{n^2}\left(1 + \frac{2n^2}{4^n}\right).$$
The relative remainder term $R(n) = \frac{2n^2}{4^n}$ goes to $0$ fairly quickly as $n \to \infty$, and so $\sum_{k=0}^n \binom{n}{k} 2^{k^2} \approx 2^{n^2}$.
Derivation: The terms in the sum are strictly increasing. Since $n \geq k$, for $k \geq 1$ we have $$\frac{\binom{n}{k} 2^{k^2}}{\binom{n}{k-1}2^{(k-1)^2}} = 2^{2k-1}\left(\frac{n}{k}-1+\frac{1}{k}\right) \geq 2^{2k-1}\left(1-1+\frac{1}{k}\right) = \frac{2^{2k}}{2k}> 1.$$
Therefore,
$$2^{n^2} \leq \sum_{k=0}^n \binom{n}{k} 2^{k^2} = 2^{n^2} + \sum_{k=0}^{n-1} \binom{n}{k} 2^{k^2} \leq 2^{n^2} + \sum_{k=0}^{n-1} \binom{n}{n-1} 2^{(n-1)^2} = 2^{n^2} + n^2 2^{n^2-2n+1}$$
$$= 2^{n^2}\left(1 + \frac{2n^2}{4^n}\right).$$
| {
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Problem with generating functions and binary recurrences I am considering the following recurrence:
$a_0 = 1$;
$a_1 = 2$
$a_{n} = 2 (a_{n - 1} + a_{n - 2})$
Then I proceeded with the generating function:
$F(x) = \displaystyle\sum_{n = 0}^\infty a_n x^n = 1 + 2x + \displaystyle\sum_{n = 2}^\infty a_{n} x^{n} = 1 + 2x + \displaystyle\sum_{n = 2}^\infty 2x^n(a_{n - 1} + a_{n - 2})$
$F(x) = 1 + 2x + \displaystyle\sum_{n = 2}^{\infty} 2x^{n} a_{n - 1} + \displaystyle\sum_{n = 2}^{\infty} 2x^{n} a_{n - 2}$
$F(x) = 1 + 2x + (2x \displaystyle\sum_{n = 2}^{\infty} x^{n - 1} a_{n - 1}) + (2x^{2} \displaystyle\sum_{n = 2}^{\infty} x^{n - 2} a_{n - 2})$
$F(x) = 1 + 2x + 2x(F(x) - 1) + 2x^{2}F(x)$
$F(x) = \frac{1}{1 - 2x - 2x^{2}}$ Let a, b be the roots of the quadratic.
$F(x) = \frac{1}{(x - a)(x - b)} = \displaystyle\sum_{n = 0}^{\infty} \frac{x^{n}(b^{-1 - n} - a^{-1 - n})}{\sqrt{3}}$
We should then have $a_{n} = \frac{b^{-1 - n} - a^{-1 - n}}{\sqrt{3}}$, but I know that this is false. Where have I gone wrong?
| OK, using the recurrence equation $a_0 = 1$, $a_1=2$, $a_2 = 6$, $a_3 = 16$, $a_4 = 44$ and $a_5=120$.
Verifying this with the generating function directly:
$$
\frac{1}{1-2x - 2x^2} \sim \sum_{k=0}^5 (2 x+2 x^2)^k \sim 1+ 2x + 6 x^2 + 16 x^3 + 44 x^4 + 120 x^5 + o(x^5)
$$
Now using roots of the denominator $1-2x-2x^2 = -2( x - a)(x-b)$, where $a= -\frac{1}{2} - \frac{\sqrt{3}}{2}$ and $b= -\frac{1}{2} + \frac{\sqrt{3}}{2}$. Therefore
$$
\frac{1}{1-2x-2x^2 } = -\frac{1}{2}\frac{1}{x-a}\frac{1}{x-b}=-\frac{1}{2(a-b)} \left( \frac{1}{x-a} - \frac{1}{x-b} \right)
$$
It is readily seen that $-\frac{1}{2(a-b)} = \frac{1}{2 \sqrt{3}}$ we thus get
$$
\frac{1}{1-2x-2x^2 } = \sum_{n=0}^\infty x^n \frac{1}{2 \sqrt{3}} \left( -a^{-n-1} + b^{-n-1} \right)
$$
Now check with WolframAlpha again, and scroll to the alternative forms.
| {
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Formula for completing the square? My math teacher said that this was the formula for completing the square.
Original function: $$ax^2 + bx + c$$
Completed square: $$a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c$$
However, using this formula I'm not getting the same answers that I would get just by determining the stuff myself. Is this correct?
| Let me derive it for you,
$$ax^2+bx+c= a \left( x^2+\frac{b}{a} x +\frac ca \right) = a\left(x^2+2\frac{b}{2a} x + \left( \frac b{2a} \right) ^2 - \left( \frac b{2a} \right) ^2+\frac ca \right)$$
$$ = a \left\{ \left(x+\frac{b}{a}\right)^2 - \frac{(b^2-4ac)}{4a^2} \right\} = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c $$
Btw how are you applying this?
| {
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Summation Identity: $\sum_{i=1}^ni^3 = \left( \frac{n(n+1)}{2} \right)^2$ I have to prove:
$$\sum\limits_{i = 1}^n i^3 = \Bigg( \frac{n(n+1)}{2}\Bigg)^2$$
Using the following:
$$n^3 = 6 {n \choose 3} + 6 {n \choose 2} + n \quad \forall n \in \mathbb{N}$$
My work is that first I substitute $n^3$ for $6 {n \choose 3} + 6 {n \choose 2} + n$. Then I go and invoke the sum over that (I am assuming this is how it works). That is,
$$\sum\limits^n_{i=1} \bigg(6 {i \choose 3} + 6 {i \choose 2} + i \bigg)$$
$$6 \sum\limits^n_{i=1} {i \choose 3} + 6 \sum\limits^n_{i=1} {i \choose 2} + \sum\limits^n_{i=1} {i \choose 1}$$
The summation identity is:
$$\sum\limits^n_{i=0} {i \choose k} = {n+1 \choose k+1}$$
invoking it over the sums yields:
$$6{n+1 \choose 4} + 6{n+1 \choose 3} + {n+1 \choose 2}$$
I think these are the right steps, but simplification seems a bit difficult to get the ending result.
| Don't stop, just do! =)
$$
\begin{eqnarray*}
&& 6 \binom{n+1}{4} + 6\binom{n+1}{3} + \binom{n+1}{2}
\\ &=& 6 \frac{(n+1) n (n-1)(n-2)}{24} + 6 \frac{(n+1) n (n-1)}{6} + \frac{(n+1) n}{2}
\\ &=& \frac{n(n+1)}{4} \left( (n-1)(n-2) + 4 (n-1) + 2 \right)
\\ &=& \cdots
\end{eqnarray*}
$$
I presume you can simplify the expression further to get the right hand side.
| {
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Need help finding smallest value of $x^2 + y^2$ I need to find the smallest value of $x^2 + y^2$ with the restriction $2x + 3y = 6$. This chapter focuses on the vertex formula.
| Let's solve for $y$ in the equation $2x + 3y = 6$. This gives $y = 2 - \frac{2x}{3}$.
After rewriting $x^2 + y^2$ in terms of $x$, we have
$x^2 + y^2 = x^2 + (2 - \frac{2x}{3})^2 = x^2 + (4 - 2 \cdot \frac{4x}{3} +\frac{4x^2}{9})$
This is a quadratic "in $x$" that we would like to minimize. You can write it in standard form $ax^2 + bx + c$ and then use the methods you have learned.
| {
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Difficult integral: $\sin^3\theta / (\sin^3\theta - \cos^3\theta)$ The question is, find the integral to the function:
$\sin^3\theta / (\sin^3\theta - \cos^3\theta)$
The only thing I could think of was to factor the denominator. But then I couldn't make any further progress.
| We take advantage of the symmetry, indeed expand on it. Let
$$I=\int \frac{\sin^3\theta\,d\theta}{\sin^3\theta-\cos^3\theta} \qquad\text{and}\qquad J=\int \frac{\cos^3\theta\,d\theta}{\sin^3\theta-\cos^3\theta}.$$
Note that
$$\frac{\sin^3\theta}{\sin^3\theta-\cos^3\theta}=1+ \frac{\cos^3\theta}{\sin^3\theta-\cos^3\theta},$$
and therefore
$$I-J=\theta.$$
If we can find $I+J$ we will be finished.
So we want to find
$$\int\frac{\sin^3\theta+\cos^3\theta}{\sin^3\theta-\cos^3\theta}\,d\theta=
\int\frac{(\sin\theta+\cos\theta)(\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta)}{(\sin\theta-\cos\theta)(\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta) }\,d\theta.$$
Let $u=\sin\theta-\cos\theta$. Then $du=(\cos\theta+\sin\theta)\,d\theta$. Also,
$u^2=1-2\sin\theta\cos\theta$. From this we find that $\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta=\frac{1+u^2}{2}$ and $\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta=\frac{3-u^2}{2}$. Thus
$$I+J=\int\frac{1+u^2}{u(3-u^2)}\,du.$$
We do a partial partial fraction decomposition:
$$\frac{1+u^2}{u(3-u^2)}=\frac{1}{3}\left(\frac{1}{u}+\frac{4u}{3-u^2}\right).$$
Integrate: $I+J=(1/3)\ln\left(\dfrac{|u|}{(3-u^2)^2}\right).$
| {
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Proving sequence equality using the binomial theorem The problem: Prove that for $n \in \mathbb N$:
$$ \left(1 + \frac{1}{n} \right)^n = 1 + \sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right). $$
The hint is to use the binomial theorem. So the left side can become:
$$ \sum_{m=0}^{n} \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m $$
I don't really know where to go from here, I've tried manipulating the expressions to make them look similar but I'm not really getting anywhere.
| Take your second sum
$$
\sum_{m=0}^{n} \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m
$$
and write it as
$$
1+\sum_{m=1}^n \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m
$$
to get the indices to match.
In your first sum
$$
\sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right)
$$
ignoring the $\frac{1}{m!}$ for now, notice
$$
\left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right)=\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots\left(\frac{n-m+1}{n}\right).
$$
Multiplying by $1=\frac{n}{n}$ gives
$$
\frac{n}{n}\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots\left(\frac{n-m+1}{n}\right)=\dots
$$
Can you take it from there?
| {
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alternative proof to induction, binomial sum identity
$\displaystyle \sum_{q=0}^{k} \begin{pmatrix}n-1+q\\ n-1 \end{pmatrix} = \begin{pmatrix}n+k\\n \end{pmatrix} $
induction
$\begin{pmatrix}n \\ k \end{pmatrix} := \frac{n!}{(n-k)!k!}$
Beginning of induction : $k=0 \rightarrow \begin{pmatrix} n-1 \\ n-1 \end{pmatrix} = \begin{pmatrix} n \\ n \end{pmatrix} = 1 $
Induction step : $k\rightarrow k+1: \sum_{q=0}^{k+1} \begin{pmatrix} n-1+q\\n-1 \end{pmatrix} = (\sum_{q=0}^{k} \begin{pmatrix} n-1+q\\n-1 \end{pmatrix}) + \pmatrix{n-1+k+1\\n-1} $
$\displaystyle = \begin{pmatrix}n+k \\ n \end{pmatrix} + \begin{pmatrix} n+k \\ n-1 \end{pmatrix} = \frac{(n+k)!}{k!(n)!}+ \frac{(n+k)!}{(k+1)!(n-1)!} =$
$\displaystyle =\frac{(n+k)!(k!n!+(k+1)k!(n-1)!}{k!n!(k+1)!(n-1)!} = \frac{(n+k)!(n+k+1)}{n!(k+1)!} = \begin{pmatrix}n+k+1 \\ n \end{pmatrix}$
Do you know any other way to show this? Please do tell.
| Another way is to use snake oil:
$\begin{align}
\sum_{n \ge 1} z^n \sum_{0 \le q \le k} \binom{n + q - 1}{n - 1}
&= \sum_{0 \le q \le k}
z^{1 - q} \sum_{n \ge 1} \binom{n + q - 1}{q} z^{n + q - 1} \\
&= \sum_{0 \le q \le k}
z^{1 - q} \sum_{s \le 0} \binom{s}{q} z^s \\
&= \sum_{0 \le q \le k}
z^{1 - q} \frac{z^q}{(1 - z)^{q + 1}} \\
&= \frac{z}{1 - z} \sum_{0 \le q \le k} (1 - z)^{-q} \\
&= \frac{z}{1 - z} \frac{1 - (1 - z)^{-k - 1}}{1 - (1 - z)^{-1}} \\
&= \frac{1}{(1 - z)^{k + 1}} - 1
\end{align}$
Now we want the coefficient of $z^n$:
$\begin{align}
[z^n] \frac{1}{(1 - z)^{k + 1}}
&= (-1)^n \binom{-k - 1}{n} \\
&= \binom{n + k + 1 - 1}{k + 1 - 1} \\
&= \binom{n + k}{k} \\
&= \binom{n + k}{n}
\end{align}$
| {
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Solve $8^{x+1} = 32 \cdot\sqrt2$ without $\log$ I need help solving the equation $8^{x+1} = 32 \cdot\sqrt2$. The obvious answer is to use log, but that is reserved for the next section. The example given for this section of questions is:
$4^x = 8$
$(2^2)^x = 2^3$
$2x = 3$
$x = \dfrac{3}{2}$
The example looks obvious and easy to solve, but I do not understand where you'd use this for my question.
| $$
8^{x+1} = 32 \sqrt{2}
$$
$$
8\times 8^x = 32 \sqrt{2}
$$
$$
8^x =2^{3x}= 4\sqrt{2}=2^2\times 2^{1/2}= 2^{2,5}
$$
$$
3x=2,5 \to x=\frac{5}{6}
$$
| {
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The last digit of $n^5-n$ What will be the last digit of
$$n^5 - n \bmod 1000,$$
where $n$ is a natural number and $m \bmod 1000$ is the remainder when $m$ is divided by $1000$.
| The unit digit of $n^{4k+1}$ is same as the unit digit of $n$
This is because the unit digit is periodic with period $4$,or $2$, or $1$, For instance
$2^1=2 _{U.D=2}$
$2^2=4_{U.D=4}$
$2^3=8_{U.D=8}$
$2^4=16_{U.D=6}$
$2^5=32_{U.D=2}$
$2^6=64_{U.D=4}$
Hence we observe that the unit digit is $2,4,8,6$ and it oscillates with period $4$, in case of $2^n$,
For numbers like $5^n$,$11^n$,$6^n$ this period is $1$,
and for $9^n$ the period is $2$.
Hence the last digit of $n^5-n$ is $0$.
| {
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Is this proof about Mersenne numbers acceptable? I want to prove following property of Mersenne numbers :
If $p > 3$ then $M_p\equiv 1 \pmod {6\cdot p}$
So, according to Fermat's Little Theorem we may write :
$2^p\equiv 2 \pmod p \Rightarrow 2^p-2=a\cdot p \Rightarrow 2^p-1=a\cdot p +1 \Rightarrow$
$\Rightarrow 2^p-1\equiv 1 \pmod p \Rightarrow M_p\equiv 1 \pmod p$
Now, since $M_p\equiv 7 \pmod {24} $ it follows that :
$M_p=24\cdot a +7 =24\cdot a+6+1=6\cdot(4a+1)+1 \Rightarrow M_p\equiv 1 \pmod 6$
Therefore we may conclude :
$(6 \mid (M_p-1) \land p \mid (M_p-1))\Rightarrow 6\cdot p \mid(M_p-1) \Rightarrow M_p \equiv 1 \pmod{6\cdot p}$
Am I correct ?
| There is nothing incorrect, but there are a few things that could be changed. We only need $p>2$.
From $2^p \equiv 2 \pmod {p}$ one should conclude $M_p=2^p -1\equiv 1 \pmod{p}$ immediately, without the detour through $a\cdot p +1$.
A similar needless detour is made when from $M_p\equiv 7\pmod{24}$ it is argued that $M_p \equiv 1 \pmod{p}$. Instead, if we are going to go that route, we could note that since $6$ is a factor of $24$, we have $M_p\equiv 7\equiv 1\pmod{6}$. But the assertion $M_p \equiv 7 \pmod{24}$ may not be familiar to people, and it is easy to prove. We won't bother, and will only prove what you need.
To prove your result, all you need is to show that $M_p \equiv 1 \pmod{6}$. It is clear that $M_p$ is odd, that is, $M_p\equiv 1 \pmod{2}$. So all we need is to show that $M_p \equiv 1 \pmod 3$. We have $2\equiv -1\pmod 3$. So if $n$ is odd, then $2^n\equiv (-1)^n \pmod{3}$, and therefore $M_p=2^p-1\equiv (-1)-1\equiv 1\pmod {3}$.
It is a standard fact about congruences that if $x\equiv a \pmod{m}$ and $x\equiv a \pmod{n}$, then $x\equiv a \pmod{\text{lcm}(m,n)}$. You chose to prove it in the special case $m=6$, $n=p$. That's fine, perhaps not needed.
Comment: The congruence machinery is very powerful, and it is useful to learn to exploit it efficiently.
| {
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Evaluate the limit of $\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$ $$\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$$
Can't find a means to resolve. The answer is $10$ by graphing.
| Divide through by $x, \lim \limits_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}=\lim \limits_{x\to\infty}\frac{1 - \sqrt{1+5/x+2/x^2}}{1-\sqrt{1+0.5/x + 1/x^2}}$ then use the expansion around $x=\infty \lim \limits_{x\to\infty}\frac{1 - \sqrt{1+5/x+2/x^2}}{1-\sqrt{1+0.5/x + 1/x^2}} \approx \frac{1-[1+5/(2x)]}{1-[1+1/(4x)]}=10$ where the $\approx$ means here that the first orders are the same.
| {
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Congruence and diagonalizations How does one find matrix $M\in M_3(\mathbb R)$ such that $M^TAM=I$ where $$A=\left(\begin{array}{rrr}2&0&0\\ 0&3&-1\\ 0&-1&3\end{array}\right)$$
and simultaneously, $M^TBM$ is some diagonal matrix where $$B=\left(\begin{array}{rrr}1&3&-3\\ 3&3&1\\ -3&1&3\end{array}\right)$$?
Thanks.
| As noted in my answer to your related question, there is a theorem regarding simultaneous diagonalization by congruence: if $A,B$ are real symmetric and $A$ is nonsingular, then they are simultaneously congruent to diagonal matrices if and only if $C=A^{−1}B$ is diagonalizable by similarity transform (Horn and Johnson, Matrix Analysis, Theorem 4.5.15.).
When all eigenvalues of $C$ are distinct, the matrix for diagonalizing $C$ via similarity transform is by itself the matrix for simultaneous diagonalizing $A$ and $B$ by congruence. Unfortunately, in your problem, $C$ has repeated eigenvalues and hence some additional manual work is needed. Specifically, we first try to diagonalize $C$ and obtain
$$
C = A^{-1}B =
\begin{pmatrix}
1/2&3/2&-3/2\\
3/4&5/4&3/4\\
-3/4&3/4&5/4
\end{pmatrix}
=M_1\ D\ M_1^{-1},
$$
where
$$
M_1 =
\begin{pmatrix}
2&1&0\\
-1&1&1\\
1&0&1
\end{pmatrix},\quad D=\textrm{diag}(-1,2,2).
$$
So, our first candidate for $M$ is $M_1$. However, observe that $D$ has repeated eigenvalues. So there is a chance that $M_1^\top AM_1$ or $M_1^\top BM_1$ are not diagonal. Indeed,
$$
M_1^\top AM_1 =
\begin{pmatrix}
16&0&0\\
0&5&2\\
0&2&4
\end{pmatrix},
\quad M_1^\top BM_1 =
\begin{pmatrix}
-16&0&0\\
0&10&4\\
0&4&8
\end{pmatrix}.
$$
So we need to further diagonalize $\begin{pmatrix}5&2\\2&4\end{pmatrix}$ and $\begin{pmatrix}10&4\\4&8\end{pmatrix}$ simultaneously via congruence. Since the two matrices are multiples of each other and they are symmetric, it suffices to find a matrix $M_2$ that orthogonally diagonalizes either one of them. It is not difficult to work out such a matrix: for
$$
\lambda=\frac{9+\sqrt{17}}{2},\quad
M_2=\frac{1}{\sqrt{\lambda+4}}\begin{pmatrix}\lambda-4&-2\\2&\lambda-4\end{pmatrix},
$$
we have
$$
\begin{pmatrix}5&2\\2&4\end{pmatrix} = M_2\begin{pmatrix}\lambda&0\\0&9-\lambda\end{pmatrix}M_2^\top.
$$
Now we may set
$$
M = M_1\begin{pmatrix}1\\&M_2\end{pmatrix}.
$$
For this particular problem, since $A$ is positive definite, there is a simpler solution -- find the square root of $A$, then orthogonally diagonalize $(\sqrt{A}^\top)^{-1}B\sqrt{A}^{-1}$. In other words, if $\sqrt{A}$ is a matrix such that $\sqrt{A}^\top\sqrt{A}=A$, and $Q$ is an orthogonal matrix such that $Q^\top(\sqrt{A}^\top)^{-1}B\sqrt{A}^{-1}Q$ is diagonal, then you may set $M=\sqrt{A}^{-1}Q$. In this case you will get $M^\top AM=I$.
| {
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Compositeness of number $k\cdot 2^n+1$? Every odd prime number can be expressed in the form $k \cdot 2^n+1$ ,where $k$ is an odd number .
For $n>2$ number $k \cdot 2^n+1$ is composite if :
$1.$ $k\equiv 1 \pmod {30} \land (n\equiv 2 \pmod 4 \lor n \equiv 1 \pmod 2 ) $
$2.$ $k\equiv 3 \pmod {30} \land n\equiv 3 \pmod 4$
$3.$ $k\equiv 5 \pmod {30} \land n\equiv 0 \pmod 2$
$4.$ $k\equiv 7 \pmod {30} \land n\equiv 1 \pmod 2$
$5.$ $k\equiv 9 \pmod {30} \land n\equiv 0 \pmod 4$
$6.$ $k\equiv 11 \pmod {30} \land n\equiv 0 \pmod 2$
$7.$ $k\equiv 13 \pmod {30} \land n\equiv 1 \pmod 2$
$8.$ $k\equiv 17 \pmod {30} \land (n\equiv 1 \pmod 4 \lor n\equiv 0 \pmod 2)$
$9.$ $k\equiv 19 \pmod {30} \land (n\equiv 0 \pmod 4 \lor n\equiv 1 \pmod 2)$
$10.$ $k\equiv 21 \pmod {30} \land n\equiv 2 \pmod 4$
$11.$ $k\equiv 23 \pmod {30} \land (n\equiv 3 \pmod 4 \lor n\equiv 0 \pmod 2)$
$12.$ $k\equiv 25 \pmod {30} \land n\equiv 1 \pmod 2$
$13.$ $k\equiv 27 \pmod {30} \land n\equiv 1 \pmod 4$
$14.$ $k\equiv 29 \pmod {30} \land n\equiv 0 \pmod 2$
Are there some other similar relations between coefficient $k$ and exponent $n$ that ensure compositeness of number $k \cdot 2^n+1$ ?
| Your findings have been generalized in this paper. Here, a covering system of integers is constructed and that allows for infinitely many numbers of the form $k\times2^n+1$ to be composite in the sequence of Lucas nubers.
So, for you, I guess the important part is that there are infinitely many Sierpinski numbers i.e. numbers of the form $k\times2^n+1$ are composite.
http://home.wlu.edu/~finchc/Research/RieselLucasArticle11.pdf
Just in case you are wondering what is a covering system, see this wiki entry
http://en.wikipedia.org/wiki/Covering_system
And this thing by Carl Pomerance
http://www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf
| {
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recurrence relation, linear, second order, homogeneous, constant coefficients, generating functions How to solve this by using the generating functions? What is the possible solution for this?
recurrence relation $$ a_n = 5a_{n-1} – 6a_{n-2}, n \ge 2,\text{ given }a_0 = 1, a_1 = 4.$$
Thanks.
| $$
\begin{align}
a_n & = 5 a_{n-1} - 6 a_{n-2}\\
a_n - 3a_{n-1} & = 2 (a_{n-1} - 3 a_{n-2})
\end{align}
$$
Letting $T_n = a_n - 3a_{n-1}$ we get that $T_n = 2T_{n-1}$ and $T_1 = a_1 - 3a_0 = 1$. Hence, $$T_n = 2T_{n-1} = 2^2 T_{n-2} = \cdots = 2^{n-1} T_1 = 2^{n-1}$$
Hence,
$$
\begin{align}
a_n - 3a_{n-1} & = 2^{n-1}\\
a_{n-1} - 3a_{n-2} & = 2^{n-2}\\
a_{n-2} - 3a_{n-3} & = 2^{n-3}\\
\vdots \\
a_2 - 3a_1 & = 2
\end{align}
$$
Multiply the second equation by $3$, the third equation by $3^2$, and in general the $k^{th}$ equation by $3^{k-1}$ and add them up to get,
$$
\begin{align}
a_n - 3^{n-2} \times 3a_1 & = 2^{n-1} + 3 \times 2^{n-2} + 3^2 \times 2^{n-3} \cdots + 3^{n-2} \times 2\\
& = 2^{n-1} \times \left( 1 + \frac32 + \left(\frac32 \right)^2 + \cdots + \left( \frac32 \right)^{n-2} \right)\\
& = 2^{n-1} \left( \frac{\left(\frac32\right)^{n-1}-1}{\frac32-1} \right)\\
& = 2 \left( 3^{n-1} - 2^{n-1} \right)\\
a_n & = 2 \left( 3^{n-1} - 2^{n-1} \right) + 3^{n-1} a_1\\
& = 6 \times 3^{n-1} - 2^n
\end{align}
$$
Hence, $$a_n = 2 \times 3^n - 2^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/91418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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} |
How to find $\sqrt[3]{8i}$ How do I find the following cube root?
$$\sqrt[3]{8i} = ?$$
I tried by adding $\sqrt[3]{i^3 + 8i + i}$ but that is where my imagination quits.
| If you want a general method for finding roots of complex numbers, you may try my answer here, for instance.
In your case, it would work as follows: write your complex number in exponential form,
$$
8i = 8 e^{i\pi/2} \ .
$$
Then,
$$
\sqrt[3]{8e^{i\pi/2}} = \sqrt[3]{8} e^{i(\pi/2 + 2k\pi)/3} \qquad \text{for} \quad k=0,1,2 \ .
$$
That is,
$$
\sqrt[3]{8e^{i\pi/2}} = 2 e^{i(\pi/6 + 2k\pi/3)} \qquad \text{for} \quad k=0,1,2
$$
And this gives:
$$
2e^{i\pi/6} = 2\left(\cos\frac{\pi}{6} + i \sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2} \right) \ ,
$$
$$
2e^{i5\pi/6} = 2\left(\cos\frac{5\pi}{6} + i \sin\frac{5\pi}{6}\right) = 2\left(\frac{\sqrt{3}}{2} - i\frac{1}{2} \right) \ ,
$$
and
$$
2e^{i3\pi/2} = 2\left(\cos\frac{3\pi}{2} + i \sin\frac{3\pi}{2}\right) = -2i \ .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/93130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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} |
Can a regular heptagon be constructed using a compass, straightedge, and angle trisector? Euclid has a magical compass with which he can trisect any angle. Together with a regular compass and a straightedge, can he construct a regular heptagon?
| Gleason's article "Angle Trisection, the Heptagon, and the Triskaidecagon" (also available here) mentions a construction due to Plemelj:
Draw the circle with center $O$ passing through $A$ and on it find $M$ so that $AM=OA$. Bisect $OM$ at $N$, and trisect at $P$, and find $T$ on $NP$ so that $\angle NAT=\frac13\angle NAP$. $AT$ is the needed side of the heptagon.
To validate Plemelj's construction, we must prove that $AT=OA\left(2\sin\dfrac{\pi}{7}\right)$. Since $2\cos\dfrac{2\pi}{7}=2-\left(2\sin\dfrac{\pi}{7}\right)^2$, it follows from $x=2\cos\dfrac{2\pi}{7}$ being a root of $x^3+x^2-2x-1=0$ that $2\sin\dfrac{\pi}{7}$ is a root of
$$(2-x^2)^3+(2-x^2)^2-2(2-x^2)-1=0$$
the other roots being $-2\sin\dfrac{\pi}{7}$, $\pm 2\sin\dfrac{2\pi}{7}$, and $\pm 2\sin\dfrac{3\pi}{7}$. The equation can be factored as
$$\left(x^3+\sqrt 7\left(x^2-1\right)\right)\left(x^3-\sqrt 7\left(x^2-1\right)\right)=0$$
The roots corresponding to the first cubic factor are $2\sin\dfrac{\pi}{7}$, $-2\sin\dfrac{2\pi}{7}$, and $-2\sin\dfrac{3\pi}{7}$. Writing the first cubic factor in the form
$$\left(\frac1{x}\right)^3-\frac1{x}=\frac1{\sqrt 7}$$
and making the substitution $\dfrac1{x}=\dfrac2{\sqrt 3}\cos\,\psi$ yields the equation $\cos\,3\psi=\sqrt{\dfrac{27}{28}}$. The desired root corresponds to the choice
$$\psi=\frac13\arccos\sqrt{\frac{27}{28}}=\frac13\arctan\frac1{3\sqrt 3}$$
which yields
$$2\sin\frac{\pi}{7}\cos\,\psi=\frac{\sqrt 3}{2}$$
From the figure, we have $\angle NAP=\arctan\dfrac1{3\sqrt 3}$, so $\angle NAT=\psi$. We thus have $AT\cos\,\psi=AN=\dfrac{\sqrt 3}{2}OA$, and from this we also have $AT=OA\left(2\sin\dfrac{\pi}{7}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/93476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Dice Question Probability A fair die is rolled until a $6$ appears. What is the probability that is must be cast more than 5 times?
So this is $1- P(\text{dice has to be cast less than or equal to}\ 5 \ \text{times})$. So this probability is equal to $$ P =\frac{1}{6}+ \frac{5}{6} \frac{1}{6}+\left(\frac{5}{6} \right)^{2} \frac{1}{6} + \cdots + \left(\frac{5}{6} \right)^{4} \frac{1}{6}$$
So just add $-1$ to this?
| The probability of getting a non-$6$ the first five times is $(5/6)^5$.
$$
\begin{align}
& \Pr(\text{a non-}6\text{ on the 1st }5\text{ trials}) \\ \\
& = \Pr(\text{a non-}6\text{ on the first trial and a non-}6\text{ on the 2nd trial and a non-}6\text{ on the 3rd trial and }\ldots) \\ \\
& = \Pr(\text{a non-}6\text{ on the 1st trial})\cdot\Pr(\text{a non-}6\text{ on the 2nd trial})\cdot\Pr(\text{a non-}6\text{ on the 3rd trial})\cdots\cdots \\ \\
& = \frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac56.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/94079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why do we assume $a+b=b+a$ in a Ring with 1. Also, is it true with Rings without 1? I was wondering why do some people use redundant axioms in definitions?
If you just expand $(a+1)(b+1)=(a+1)b+a+1=ab+b+a+1$ $(a+1)(b+1)=a(b+1)+b+1=ab+a+b+1$. Hence, $ab+a+b+1=ab+b+a+1$, then cancel ab and 1. Then, you get it's commutative for free.
Why do we then assume it?
Also, is it even stronger. If we drop the condition that 1 is in R, then can we still deduce a+b=b+a? As clearly the argument just done falls apart if you haven't got 1 in R.
| Without a $1$, you cannot deduce the condition. Take $(R,\cdot,\times)$ where $(R,\cdot)$ is any nonabelian group with identity element $e$, and let $a\times b = e$ for all $a$ and $b$ (the "zero multiplication ring"). This satisfies all the axioms of a ring without $1$, except for commutativity of the first operation. So $a+b=b+a$ cannot be deduced from the rest of the axioms of a ring without a $1$.
Another method to deduce commutativity when there is a $1$ is to consider $(a+b)(1+1)$ and distribute both ways:
$$\begin{align*}
(a+b)(1+1) &= (a+b)+(a+b) = a+b+a+b,\\
(a+b)(1+1) &= a(1+1) + b(1+1) = a+a+b+b.
\end{align*}$$
From $a+b+a+b = a+a+b+b$ we deduce $b+a=a+b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/94350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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} |
solve complex equation $x^8 = \frac{1+i}{\sqrt{3} - i} = \frac{\sqrt[8]{\frac{2}{\sqrt{2}}}(\cos \frac{\pi}{4} + i \sin{\frac{\pi}{4}})}{2 \cos \frac{\pi}{6} + i \sin \frac{3\pi}{2}}$
What's the way to solve this kind of equation? I think there must be 8 solutions.
I tried to solve the following two equations
$a^6 = 1+i$
$b^6 = \sqrt{3} - i$
I tried the following for the first equation
$-1 = i^2 = (a^6 - 1)^2 = a^{12} - 2a^6 +1$
Then I would need to solve this:
$a^{12} - 2a^6 +2 = 0$
Is this the right approach?
| Any equation of the form $z^n=\alpha$, where $\alpha$ is a fixed complex number can be solved by switching $\alpha$ to polar form
$$\alpha= r (\cos(\theta)+i\sin(\theta) ) \,.$$
Now, can you find the polar forms of $z$?
For second equation, since the equation only has powers of $a^6$, a substitution $t=a^6$ should work ;)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/95226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Geometric inequality: $2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$ Suppose $a$, $b$, and $c$ are the lengths of the sides of a triangle, and $R$ and $r$ are its circumradius and inradius respectively. How can one prove the following inequality? $$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$
| We need to prove that
$$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$ or
$$2\cdot\frac{4S^2}{(a+b+c)^2}+8\cdot\frac{abc}{4S}\cdot\frac{2S}{a+b+c}\leq\frac{a^2+b^2+c^2}{2}$$ or
$$\frac{\prod\limits_{cyc}(a+b-c)}{2(a+b+c)}+\frac{4abc}{a+b+c}\leq\frac{a^2+b^2+c^2}{2}$$ or
$$(a^2+b^2+c^2)(a+b+c)-8abc+\sum_{cyc}\left(a^3-a^2b-a^2c+\frac{2}{3}abc\right)\geq0$$ or
$$a^3+b^3+c^3-3abc\geq0,$$
which is AM-GM.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
} |
Showing that $(2^a - 1)\bmod (2^b - 1) = 2^{a \; \bmod \; b} - 1 $ I've been thinking on this proof for two days. I'm stuck.
Show that,
$$ (2^a - 1)\bmod (2^b - 1) = 2^{a \! \! \mod b} - 1 $$
where $a,b \in \mathbb{Z}^+$.
I would be happy if someone can help me.
Thanks.
| I have another approach which I think it should be easier.
We must prove that $2^a-1 \equiv 2^{a mod b}-1 \pmod{2^b-1}$
Thus, We should prove
$$2^a \equiv 2^{a mod b} \pmod{2^b-1}$$
Proof: Using dividing algorithm, we have:
$a = bq + r$ then $r = a - bq$
Thus we have: $2^a \equiv 2^r \pmod{2^b-1}$ then $2^a \equiv 2^{a - bq} \pmod{2^b-1}$
Now if multiply both sides of the congruence to $2^{br}$ we have:
$$2^a * 2^{bq} \equiv 2^a \pmod{2^b-1}$$
We know that $gcd(2^a, 2^b-1)=1$ (because $2^a$ is even and $2^b-1$ is odd), so we can divide both sides of the congruence by $2^a$, so we have:
$$2^{bq} \equiv 1 \pmod{2^b-1}$$
From the definition of congruence we have:
$$2^b-1 | 2^{bq}-1$$
We know that $2^{bq}-1 = (2^b-1)(2^{b(q-1)} + 2^{b(q-2)} + ... + 2^b + 1)$
Thus we have:
$$2^b-1 | (2^b-1)(2^{b(q-1)} + 2^{b(q-2)} + ... + 2^b + 1)$$
Which is trivial.
So we reached a trivial and thus all of the conclusions are reversible, then the proof is done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find a sum of appropriate values of $\cos$ and $\sin$ to determine the value of a series The task is to find a sum of multiple values $\cos$ and $\sin$ to determine the value of
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}$$
Since I had no clue how to approach this I consulted Wolfram|Alpha which returned this result:
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \frac{\sin(1)}{2} + \cos(1) - \cos(0)$$
So I wrote down the partial sums of the given series and $\sin(1)$ and $\cos(1)$:
$$
\qquad\qquad\quad\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \quad - \frac{1}{4!} + \frac{2}{6!} - \frac{3}{8!} \cdots
$$
$$
\;\;\sin(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n+1}}{(2n+1)!} = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} \cdots
$$
$$
\cos(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n}}{(2n \qquad)!} = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} \cdots
$$
Looking at the numbers I can see that Wolfram|Alpha's result is correct: $\frac{1}{2}1 - \frac{1}{2!} = 0$ and $\frac{1}{2}\frac{-1}{3!} + \frac{1}{4!} = \frac{1}{4!}$, so the $\cos(1)$-series is shifted by $1$ since there is no $1$ at the beginning of the given series, so it needs to be subtracted from $\cos(1)$: $-cos(0)=-1$. But how do I get here without Wolfram|Alpha?
| let $m=n+1$, so your series reduces to $$\sum_{m=2}^\infty(-1)^{m-1}\frac{m-1}{(2m)!}=-\sum_{m=2}^\infty(-1)^{m}\frac{m}{(2m)!}+\sum_{m=2}^\infty(-1)^{m} \frac{1}{(2m)!}$$
$$=-\sum_{m=2}^\infty(-1)^{m}\frac{1}{2(2m-1)!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1+\frac{1}{2!}$$
Note that $\frac{m}{(2m)!}=\frac{m}{(2m)(2m-1)!}=\frac{1}{(2)(2m-1)!}$, and also we evaluated the second series at $m=0$ and $m=1$
Replacing $m$ by $k+1$ in the first series and simplifying, we get,
$$\sum_{k=1}^\infty(-1)^{k}\frac{1}{2(2k+1)!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1+\frac{1}{2!}$$
$$\sum_{k=0}^\infty(-1)^{k}\frac{1}{2(2k+1)!}-\frac{1}{2!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1+\frac{1}{2!}$$
$$\sum_{k=0}^\infty(-1)^{k}\frac{1}{2(2k+1)!}+\sum_{m=0}^\infty(-1)^{m} \frac{1}{(2m)!}-1$$
$$=\frac{\sin (1)}{2}+\cos(1)-\cos(0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/96571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Let $f(x)=x^2+2x$ and $g(x)=x^3$, finding the roots of $f\circ g(x)=g\circ f(x)$ Let $f(x)=x^2+2x$ and $g(x)=x^3$, how many different roots does $f\circ g(x)=g\circ f(x)$ have?
I solved it and found $x=-1$ but it says it has two different roots, could someone clarify?
| Other than the nice answer from Alex, you can also do it concretely:
$$f\circ g(x)=f(x^3)=x^6+2x^3=x^3(x^3+2)$$ and
$$g\circ f(x)=g(x^2+2x)=(x^2+2x)^3=x^3(x+2)^3.$$
Therefore, $f\circ g(x)=g\circ f(x)$ can be written as
$x^3(x^3+2)-x^3(x+2)^3=0$, or equivalently,
$$x^3\Big[(x+2)^3-(x^3+2)\Big]=x^3\Big[(x^3+6x^2+12x+8)-(x^3+2)\Big]$$
$$=x^3(6x^2+12x+6)=6x^3(x+1)^2=0,$$
which gives $x=0$ or $x=-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Perfect Squares ending in 576 I want to find out perfect squares ending in 576, after the number 576.
Here is my derivation to arrive at such a number. Let the perfect square ending in $576$ be $1000k+576$. Every perfect square can be expressed as a the sum of a certain number of consecutive odd numbers. For eg: $2^2 = 1+3$, $3^2 = 1+3+5$, $4^2 = 1+3+5+7$, and so on..
Hence I can write my required perfect square ending in 576 as -
$$(1+3+5+7+\cdots+49) + \underbrace{(51+53+55+57+\cdots)}_{(n\text{ summands})}$$
Therefore,
$$(1+3+5+7+\cdots+49) + \underbrace{(51+53+55+57+\cdots)}_{
(n \text{ summands})} = 1000k +576.$$
Since $(1+3+5+7+ ....49) = 576$, the equation reduces to
$$\underbrace{(51+53+55+57+\cdots)}_{n\text{ summands})} = 1000k$$
Using formula for Arithmetic Progression starting with 51 and a common difference of 2,
$$\begin{align*}
\frac{n}{2}\left(2(51) + (n-1)2\right) &= 1000k\\
n(n+50) &= 1000k
\end{align*}$$
Put $n = 100$, $100\times 150 = 1000k$, hence $k = 15$.
Put $k = 15$ in the perfect square term $1000k+576$ we get the number $15576$.
But $15576$ is not a perfect square.
What is flawed in my derivation? Kindly help.
| If $x^2=24^2\pmod{1000}$, then $(x-24)(x+24)=0\pmod{1000}$. Thus, exactly one of $x-24$ or $x+24$ must be divisible by $125$ because their product is and if one is divisible by 5, the other is not. Furthermore, both $x-24=x+24=0\pmod{4}$ because they are equal $\!\!\pmod{4}$ and none of $1^2$, $2^2$, or $3^2$ are $0\pmod{8}$.
We need to solve one of
$$
\begin{align}
x=24&\pmod{125}\\
x=0&\pmod{4}
\end{align}\tag{1}
$$
or
$$
\begin{align}
x=101&\pmod{125}\\
x=0&\pmod{4}
\end{align}\tag{2}
$$
and these will give all the solutions $\pmod{500}$.
The solution to $(1)$ is $24\pmod{500}$ and the solution to $(2)$ is $476\pmod{500}$. Thus, numbers ending in $024$, $476$, $524$, and $976$ are the only ones whose squares end in $576$.
| {
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Three maximal ideals lying over $3\mathbb{Z}$? A few weeks ago I asked a question about finding the number of maximal ideals lying above $3\mathbb{Z}$ in $B$, where $B$ is the integral closure of $\mathbb{Z}$ in a splitting extension $E\supset\mathbb{Q}$ for the polynomial $f(x)=x^3+x+1$.
I'm reading over the following solution, but would like to ask a few follow up questions.
Since the discriminant of $f(x)$ is negative, two of the roots are complex conjugates, and since conjugation has order 2, the degree of the extension must be 6. Hence we may think of $E$ as adjoining two roots $a$ and $b$ of $f(x)$ to $\mathbb{Q}$, where $a$ satisfies $a^3+a+1$ and $b$ satisfies the polynomial $(x^3+x+1)/(x-a)=x^2+ax+a^2+1$.
So factor $(3)$ in $\mathbb{Z}[a]$ by factoring $x^3+x+1\pmod{3}$. Since there is one root modulo 3, it follows that
$$
x^3+x+1=(x-1)(x^2+x-1)\pmod{3}
$$
and so as we have seen, $(3)$ factors in $\mathbb{Z}[a]$ as $(3)=(3,a-1)(3,a^2+a-1)$.
Now since $a\equiv 1\mod (3,a-1)$, the polynomial $x^2+ax+a^2+1$ becomes $x^2+x+2$, which is irreducible modulo 3, and thus does not reduce further. Also, since $a^2=-a+1$ modulo this ideal, the polynomial can be further simplified to $x^2+ax-a+2$, so any root has for $ra+s$ for $r$ and $s$ integers modulo 3. This gives the set of equations
$$
\begin{align*}
(ra + s)^2 + a(ra + s) - a + 2 = 0 &\iff r^2a^2 + 2ras + s^2 + ra^2 + sa - a + 2 = 0,\\
&\iff r^2(-a + 1) + 2ras + s^2 + r(-a + 1) + sa - a + 2 = 0.
\end{align*}
$$
and so $-r^2 + r + s - 1 = 0$, and $r^2 + s^2 + r + 2 = 0 \pmod{3}$. This former equation implies $s\equiv (r+1)^2\pmod{3}$, and substituting into the second gives $r^2+(r+1)^4+r+2\equiv 0\pmod{3}$. This is true when $r=0$ and $s=1$, so $x=1$ is a root of $x^2+ax-a+2\mod(3,a^2+a-1)$. This in turn implies it factors as $x^2+ax-a+2=(x-1)(x+a+1)$.
Hence $(3, a^2 + a - 1) = (3, a^2 + a - 1, b - 1)(3, a^2 + a -1, b + a + 1)$, and hence there are three prime factors, and hence maximal ideals since $3\mathbb{Z}$ is maximal in $\mathbb{Z}$, lying above $3\mathbb{Z}$ in $\mathbb{Z}$.
I see that the $3\mathbb{Z}$ is the product of the $3$ factors found, but is there a more detailed explanation about why this immediately implies that there are $3$ maximal ideals lying above $3\mathbb{Z}$? I don't see how the conclusion follows so quickly. Would there need to be some uniqueness about this factoring for it to work as it does? Thank you kindly for your responses.
| Yes. The integral closure of $\mathbb{Z}$ in a finite extension of $\mathbb{Q}$ is a Dedekind domain, where each (nonzero) ideal factors uniquely as a product of (maximal) prime ideals. For proofs of these facts (and much more), see Robert Ash's notes here.
| {
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Using the complex logarithm to find the sum of angles in a triangle. Suppose you have a triangle with vertices $a$, $b$, and $c$. I asked earlier how you can define the angles in a triangle based on the $\log$ function. I received the answer that, for instance, the angle at $a$ is found as $\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|$.
Can this be used to show that the sum of angles in a triangle is $\pi$? I summed the angles as
$$
\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|+\left|\Im\log\left(\frac{a-b}{c-b}\right)\right|+\left|\Im\log\left(\frac{a-c}{b-c}\right)\right|.
$$
I noticed that $\left|\Im\log\left(\frac{c-a}{b-a}\frac{a-b}{c-b}\frac{b-c}{a-c}\right)\right|=\left|\Im\log(-1)\right|=\pi$, when evaluating on the principal branch.
I had to cheat a bit and flip the $\frac{a-c}{b-c}$. Is there a more systematic way to prove this somehow?
| Flipping the $\frac{a-c}{b-c}$ (or any of the fractions in your angle expressions) actually isn't cheating. First, perhaps more intuitively, when you said that the measure of angle $a$ is $\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|$, swapping $b$ and $c$ should not change the measure of the angle, so you should expect $\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|=\left|\Im\log\left(\frac{b-a}{c-a}\right)\right|$. More formally, $\frac{b-a}{c-a}=(\frac{c-a}{b-a})^{-1}$, so $$\begin{align}
\left|\Im\log\left(\frac{b-a}{c-a}\right)\right|&=\left|\Im\log\left(\left(\frac{c-a}{b-a}\right)^{-1}\right)\right|
\\
&=\left|-\Im\log\left(\frac{c-a}{b-a}\right)\right|
\\
&=\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|
\end{align}$$ (since the factor of $-1$ only changes the sign of the imaginary part, and that sign change is wiped out by the absolute value).
What might be cheating, though, is combining the absolute values.
I'd go about this in a slightly different way. Let's start by backing up to $$\begin{align}
m\angle a&=\left|\Im\log\left(\frac{c-a}{b-a}\right)\right|
\\
&=\left|\Im\left(\log(c-a)-\log(b-a)\right)\right|
\\
&=\left|\Im\log(c-a)-\Im\log(b-a)\right|.
\end{align}$$ $\Im\log(c-a)$ and $\Im\log(b-a)$ are the directed angles from the positive real axis to the ray from $0$ to $c-a$ and $b-a$, respectively, so $\Im\log(c-a)-\Im\log(b-a)$ is the directed angle from $b-a$ to $c-a$. When I say "directed" angle, I mean that a positive angle is a counterclockwise rotation.
Now, without loss of generality, let the vertices be labeled $a$, $b$, and $c$ in a counterclockwise direction around the triangle:
Working carefully, we can ensure that we measure each angle in the positive direction, and thus avoid the absolute values:
$$\begin{align}
m\angle a&=\Im\log(c-a)-\Im\log(b-a)=\Im\log\frac{c-a}{b-a}
\\
m\angle b&=\Im\log(a-b)-\Im\log(c-b)=\Im\log\frac{a-b}{c-b}
\\
m\angle c&=\Im\log(b-c)-\Im\log(a-c)=\Im\log\frac{b-c}{a-c}
\\
\\
m\angle a+m\angle b+m\angle c&=\Im\log\frac{c-a}{b-a}+\Im\log\frac{a-b}{c-b}+\Im\log\frac{b-c}{a-c}
\\
&=\Im\log\left(\frac{c-a}{b-a}\cdot\frac{a-b}{c-b}\cdot\frac{b-c}{a-c}\right)
\\
&=\Im\log\left((-1)^3\right)
\\
&=\pi.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
How to find$\int x^2 \arctan(x) \;dx $ I was solving $\int x^2 \arctan(x) \;dx $
I set $u=x^2$, $dv= \arctan(x)$, so I could get $du=2x$, $v=x\arctan(x)-\frac12\ln(1+x^{2})$.
From $\int x^2 \arctan(x)\;dx = uv - \int v \; du$
I got
$$\begin{align*}&\int x^2 \arctan(x) \; dx =\\
&x^3\arctan(x) -\frac12\ln(1+x^{2})-\int 2x\left[x^{2}\arctan x -\frac12\ln(1+x^2)\right]\;dx \end{align*}$$
and simplified if; then I got
$$3\int x^2\arctan(x) \;dx = x^3 \arctan(x)-\frac12\ln(1+x^2)+\int x\ln(1+x^2) \;dx$$
after that I use $w=\ln(1+x^2) \; dv =dx$ to find $\int x\ln(1+ x^2)$
but I got
$$x^2 \arctan(x)-\int 2x\arctan(x) \; dx$$
If I got $$x^2 \arctan(x)-\int 2x^2 \arctan(x)\;dx$$ instead, it would be easy to solve the question....
How can I solve this question and if you find any my mistake could you post this wall ??
Thank you !
| $$\int{ x^2 \cdot \tan^{-1} x} dx = $$
$$ \tan^{-1} x dx = du $$
$$ x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) = u $$
$$ x^2 = v $$
$$ 2xdx = dv $$
$$\int{ x^2 \cdot \tan^{-1} x} dx = x^2\left(x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) \right)-\int 2x\left(x\cdot\tan^{-1} x - \frac{1}{2}\log(x^2+1) \right)dx$$
$$I = {x^2}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) - 2I + \int {x\log \left( {{x^2} + 1} \right)} dx$$
$$3I = {x^2}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) + \frac{1}{2}\int {\log u} du$$
$$3I = {x^2}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) + \frac{{{x^2} + 1}}{2}\left[ {\log \left( {{x^2} + 1} \right) - 1} \right]$$
$$I = \frac{{{x^2}}}{3}\left( {x\cdot{{\tan }^{ - 1}}x - \frac{1}{2}\log ({x^2} + 1)} \right) + \frac{{{x^2} + 1}}{6}\left[ {\log \left( {{x^2} + 1} \right) - 1} \right]$$
This simplifies to $$I = \frac{{{x^3}}}{3}\cdot{\tan ^{ - 1}}x + \frac{1}{6}\log \left( {{x^2} + 1} \right) - \frac{{{x^2}}}{6} + C $$
which coincides with WA's solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/106827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Finding the value of $\sum k/3^k$
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$
please help me with this
$$\sum\limits_{k=0}^{\infty}\frac{k}{3^k}$$
I need just a hint, not a full answer. Thanks!
| There are several ways to do this. One involves the fact that
$$
\frac{k}{3^k} = kx^k = x\cdot kx^{k-1} = x\cdot \frac{d}{dx} x^k
$$
where $x=1/3$, and $\displaystyle\sum_{k=0}^\infty x\cdot \frac{d}{dx} x^k$ can be found.
Another looks like this:
$$
\sum_{k=0}^\infty \frac{k}{3^k} = \sum_{k=1}^\infty \frac{k}{3^k}
= \left\{\begin{array}{cccccccccccccccc}
& & 1/3 \\
& + & 1/9 & + & 1/9 \\
& + & 1/27 & + & 1/27 & + & 1/27 \\
& + & 1/81 & + & 1/81 & + & 1/81 & + & 1/81 \\
& + & \cdots
\end{array}\right.
$$
Then you can sum the first column, then the second column, etc., and finally find the sum of all of the column sums.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/107468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I make the following 2 fractions integers? Let $m,n$ be integers. I want to find the possible values of $m,n$ such that $4(m+n)\over (2m+n)^2+3n^2$ and $4n\over (2m+n)^2+3n^2$ are both integers too. Would someone please help? Of course letting $(2m+n)^2+3n^2=4$ gives some good values, but is this all the $m,n$ I can get?
Added: I can see that the problem can be reduced to asking for $4k\over (2m+n)^2+3n^2$ to be an integer for both $k=m,n$
| We can't have $m=n=0$. Let's suppose for a moment that $n=0$ then (using Graphth's answer) $n/(m^2+mn+n^2)=0$ and $m/(m^2+mn+n^2)=1/m$ and $1/m$ is integer only if $m=-1$ or $m=1$. Since the problem is symmetric in $m$ and $n$ we found the solutions $(m,n)=(0,-1),(0,1),(1,0),(-1,0)$.
We know that if $a|b$ and $a,b\neq0$ then $|a|\le|b|$. Since $m^2+mn+n^2$ is always positive, we can say $m^2+mn+n^2\le |n|$ and since $|n|\le n^2$ for $n\in\mathbb{Z}$ we have $m^2+mn+n^2\le n^2$ then $m^2+mn\le 0$. Analogously $mn+n^2\le 0$. Summing up we get $m^2+2mn+n^2\le0$ or $(m+n)^2\le 0$. This inequality is only true if $(m+n)=0$. Now suppose $m=-n$.
We get that $m^2=m^2+mn+n^2|m$ then $m=1$ or $m=-1$. Using the symmetry we get two more solutions $(m,n)=(1,-1),(-1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
limit $\lim\limits_{n\to\infty}\left(\sum\limits_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$ Calculate below limit
$$\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{\sqrt{i}} - 2\sqrt{n}\right)$$
| The following is an elementary consideration, which shows how to compute the limit in terms of an infinite series. It's evaluation requires usage of Euler's summation formula, already covered by Dane.
Consider the following transformation
$$
\sum_{k=1}^n \frac{1}{\sqrt{k}} = \sum_{k=1}^n \left(\frac{1}{\sqrt{k}} - \frac{2}{\sqrt{k} + \sqrt{k+1}} \right) + \sum_{k=1}^n \frac{2}{\sqrt{k} + \sqrt{k+1}}
$$
Then use $\sqrt{k+1}-\sqrt{k} = \frac{\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}+\sqrt{k}} = \frac{(k+1)-k}{\sqrt{k+1}+\sqrt{k}} = \frac{1}{\sqrt{k+1}+\sqrt{k}}$:
$$
\sum_{k=1}^n \frac{1}{\sqrt{k}} = \sum_{k=1}^n \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} + 2 \sum_{k=1}^n \left( \sqrt{k+1}-\sqrt{k} \right)
$$
The latter sum telescopes:
$$
\sum_{k=1}^n \left( \sqrt{k+1}-\sqrt{k} \right) = \left( \sqrt{2}-\sqrt{1} \right) + \left( \sqrt{3}-\sqrt{2} \right) + \cdots + \left( \sqrt{n+1}-\sqrt{n} \right) = \sqrt{n+1}-1
$$
From here:
$$ \begin{eqnarray}
\left(\sum_{k=1}^n \frac{1}{\sqrt{k}} \right)- 2 \sqrt{n} &=&
\sum_{k=1}^n \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} + 2 \left( \sqrt{n+1}-\sqrt{n}-1\right) \\
&=& \sum_{k=1}^n \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2} + 2 \left( \frac{1}{\sqrt{n+1}+\sqrt{n}}-1\right)
\end{eqnarray}
$$
In the limit:
$$
\lim_{n\to \infty} \left(\sum_{k=1}^n \frac{1}{\sqrt{k}} \right)- 2 \sqrt{n} = -2 +
\sum_{k=1}^\infty \frac{1}{\sqrt{k} \left( \sqrt{k} + \sqrt{k+1} \right)^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Finding $\frac{d}{dx} \frac{x^2}{y}$ $$\frac{d}{dx} \frac{x^2}{y}$$
According to Wolframalpha
I "factor out constants"
$$\frac{\frac{d}{dx} x^2}{y}$$
Then I will get $\frac{2x}{y}$. Is that right? But $y$ is not a constant? What I did actually (quotient rule got me stuck)
The actual question is "Find $\frac{d^2y}{dx^2}$ of $2x^3 - 3y^2 = 8$"
I got
$$\frac{dy}{dx} = \frac{x^2}{y}$$
Then
$$\frac{d^2y}{dx^2} = \frac{y \cdot 2x - x^2 \cdot \frac{dy}{dx}}{y^2}$$
$$ = \frac{2xy - x^2 \cdot \frac{x^2}{y}}{y^2}$$
$$ = \frac{2xy^2 - x^4}{y^3}$$
Is this correct? It doesn't look like a "simple" answer (or whats in wolfram)?
| $$
2x^3-3y^2 = 8.
$$
Differentiate both sides with respect to $x$:
$$
6x^2 - 6yy' = 0.
$$
Solve for $y'$:
$$
y' = \frac{x^2}{y}.
$$
Differentiate again with respect to $x$:
$$
y'' = \frac{y(2x)- x^2y'}{y^2}.
$$
Put $x^2/y$ in place of $y'$ and simplify:
$$
y'' = \frac{2xy - x^2\frac{x^2}{y}}{y^2} = \frac{2xy^2 - x^4}{y^3}.
$$
In a sense, you're done now, but notice that in place of $y^2$ you can put $(2x^3-8)/3$:
$$
y'' = \frac{2x(2x^3-8)/3 - x^4 }{y(2x^3-8)/3} = \frac{3x^4 - 16x}{y(2x^3-8)}.
$$
Maybe depending on the purpose, you might prefer to put $(8+3y^2)/2$ in place of $x^3$:
$$
y'' = \frac{2xy^2-x(8+3y^2)/2}{y^3} = \frac{4xy^2-8x-3xy^2}{2y^3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/113433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove this property of congruences? Prove that for each $n \in \mathbb{N}, s \in \mathbb{N}$ the following is true
(i) $n \equiv Q_s(n) \left(\bmod\ 10^s - 1\right)$
(ii) $n \equiv Q'_s(n)\left(\bmod\ 10^s + 1\right)$
where
$$Q_s(n) = \sum_{i=0}^{\infty}(a_{is+s-1}\dots a_{is+1}a_{is})$$
for example
$$Q_3 (6154328103) = 103 + 328 + 154 + 006 = 591$$
and
$$Q'_s(n) = \sum_{i=0}^{\infty}(-1^i)(a_{is+s-1}\dots a_{is+1}a_{is})$$
for example
$$Q'_3 (6154328103) = 103 - 328 + 154 - 006 = -77$$
Also $n$ can be expressed as
$$n = \sum_{i=0}^{\infty}(a_{is+s-1}\dots a_{is+1}a_{is}) \cdot 10^{is}$$
for example
$$6154328103 = 103 \cdot 10^{0 \cdot 3} + 328 \cdot 10^{1 \cdot 3} + 154 \cdot 10^{2 \cdot 3} + 6 \cdot 10^{3 \cdot 3}$$
Any ideas?
| Hint: mod $10^s -1$ we have $10^s\equiv 1$, and mod $10^s + 1$ we have $10^s\equiv -1$. Use these congruences as rewrite rules to replace powers of $10^s$ by powers of $\pm 1$ in radix $10^s$ representation.
This leads to simple universal divisibility tests: evaluate a radix polynomial in nested Horner form, using modular arithmetic. For example, consider evaluating a $3$ digit radix $10$ number modulo $7$. In Horner form $\rm\ d_2\: d_1\:d_0 \ $ is $\rm\: (d_2\cdot 10 + d_1)\ 10 + d_0\ \equiv\ (d_2\cdot 3 + d_1)\ 3 + d_0\ (mod\ 7)\ $ since $\rm\ 10\equiv 3\ (mod\ 7)\:.\:$ So we compute the remainder $\rm\ (mod\ 7)\ $ as follows. Start with the leading digit then repeatedly apply the operation: multiply by $3$ then add the next digit, doing all of the arithmetic $\rm\:(mod\ 7)\:.\:$
For example, let's use this algorithm to reduce $\rm\ 43211\ \:(mod\ 7)\:.\:$ The algorithm consists of repeatedly replacing the first two leading digits $\rm\ d_n\: d_{n-1}\ $ by $\rm\ d_n\cdot 3 + d_{n-1}\:\ (mod\ 7),\:$ namely
$\rm\qquad\phantom{\equiv} \color{red}{4\ 3}\ 2\ 1\ 1$
$\rm\qquad\equiv\phantom{4} \color{green}{1\ 2}\ 1\ 1\quad $ by $\rm\quad \color{red}4\cdot 3 + \color{red}3\ \equiv\ \color{green}1 $
$\rm\qquad\equiv\phantom{4\ 3} \color{royalblue}{5\ 1}\ 1\quad $ by $\rm\quad \color{green}1\cdot 3 + \color{green}2\ \equiv\ \color{royalblue}5 $
$\rm\qquad\equiv\phantom{4\ 3\ 5} \color{brown}{2\ 1}\quad $ by $\rm\quad \color{royalblue}5\cdot 3 + \color{royalblue}1\ \equiv\ \color{brown}2 $
$\rm\qquad\equiv\phantom{4\ 3\ 5\ 2} 0\quad $ by $\rm\quad \color{brown}2\cdot 3 + \color{brown}1\ \equiv\ 0 $
Hence $\rm\ 43211\equiv 0\:\ (mod\ 7)\:,\:$ indeed $\rm\ 43211 = 7\cdot 6173\:.\:$ Notice that for modulus $11$ or $9\:$ the above method reduces to the well-known divisibility tests by $11$ or $9\:$ (a.k.a. "casting out nines" for modulus $9$). Your examples are analogous: casting out $11\cdots1$ or $99\cdots9$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $f(x)f(y)=f(\sqrt{x^2+y^2})$ how to find $f(x)$ As we know, for the $$f(x)f(y)=f(x+y)$$ $f(x)=\mathrm e^{\alpha x}$ is a solution.
What about
$f(x)f(y)=f(\sqrt{x^2+y^2})$?
Does anybody know about the solution of the function equation?
I tried to find $f(x)$.
See my attempts below to find $f(x)$.
$$f(x)=a_0+a_1x+\frac{a_2x^2}{2!}+\frac{a_3x^3}{3!}+\cdots$$
$$f(y)=a_0+a_1y+\frac{a_2y^2}{2!}+\frac{a_3y^3}{3!}+\cdots$$
$$f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$
$$f(\sqrt{x^2+y^2})=a_0+a_1\sqrt{x^2+y^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3(x^2+y^2)^{3/2}}{3!}+\cdots=$$
$$f(\sqrt{x^2+y^2})=a_0+a_1y\sqrt{1+(x/y)^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3y^2(1+(x/y)^2)^{3/2}}{3!}+\cdots=f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$
if we use binom expansion for $(1+(x/y)^2)^{m}$
$$(1+(x/y)^2)^{m}=1+\frac{mx^2}{y^2}+\frac{m(m-1)x^4}{2!y^4}+\frac{m(m-1)(m-2)x^6}{3!y^6}+\cdots$$
Let's put the expansion to the equation $f(\sqrt{x^2+y^2})$
$$
\begin{align}
& f(\sqrt{x^2+y^2}) =a_0 + a_1 y \left( 1 + \frac{(1/2)x^2}{y^2} + \frac{(1/2)((1/2)-1)x^4}{2!y^4} \right. \\ \\
& \left. {} + \frac{(1/2)((1/2)-1)((1/2)-2)x^6}{3!y^6} + \cdots\right) + \frac{ a_2 (x^2+y^2)}{2!} \\ \\
& + \frac{a_3y^2 \left(1+\frac{(3/2)x^2}{y^2}+\frac{(3/2)((3/2)-1)x^4}{2!y^4}+\frac{(3/2)((3/2)-1)((3/2)-2)x^6}{3!y^6}+\cdots\right)}{3!} +\cdots \\ \\
& = a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots
\end{align}
$$
If we equal for all $x^n$ terms in both sides
we can see $a_{2n-1}=0$, but to find $a_{2n}$ seems hard for me.
Any idea to find $a_{2n}$
Thanks in advice.
| Change variable, $g(u) = f(\sqrt{u})$. You need to decide what you want for negative $u$. Then this functional equation becomes $g(u+v)=g(u)g(v)$.
| {
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Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
&= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
&= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2} - \cdots- x y^{n-1} - y^n \\
&= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n \\
&\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:
*
*Associate law for addition
*Existence of an additive identity
*Existence of additive inverses
*Commutative law for additions
*Associative law for multiplication
*Existence of a multiplicative identity
*Existence of multiplicative inverses
*Commutative law for multiplication
*Distibutive law
| Sorry but I don't speak English well but I will try to write as correctly as possible.
Your demonstration is correct, but your idea of cancellation is incorrect.
$x^n−y^n=(x−y)(x^\left(n−1\right)+x^\left(n−2\right)y+⋯+xy^\left(n−2\right)+y^\left(n−1\right))$
When we multiply first for 'x' to our terms of the second parenthesis let's notice that all terms that contains 'x' go up one grade, this by exponent laws.
When we multiplies for 'y' to our terms of the second parenthesis let's notice that all terms that contains y go up one grade but the terms that contains x remain the same.
Therefore at the moment to cancel our terms, "ellipsis" of the formula cancel the terms left over in your demostration.
This is your result $x^n+x^2y^\left(n−2\right)−x^\left(n−2\right)y^2−y^n$ the extra term $+x^2y^\left(n−2\right)$is "cancel" by the "ellipsis" because the term that is not visible but is in the "ellipsis" in the second parenthesis is
$(x^\left(n−1\right)+x^\left(n−2\right)y+⋯ (x^2y^\left(n-3\right)) ⋯+xy^\left(n−2\right)+y^\left(n−1\right)$ because on the left one degree goes up the 'x' and one degree goes down the y and cancel your term. Analogously the other extra term $-x^\left(n−2\right)y^2$ is cancel by the other "ellipsis" that came out of the first multiplication by x.
$x^n−y^n=(x−y)(x^\left(n−1\right)+x^\left(n−2\right)y+⋯+xy^\left(n−2\right)+y^\left(n−1\right))$
Dem.
By distributive law we have
$x(x^\left(n-1\right)+x^\left(n-2\right)y+...+xy^\left(n-2\right)+y^\left(n-1\right))$
$-y(x^\left(n-1\right)+x^\left(n-2\right)y+...+xy^\left(n-2\right)+y^\left(n-1\right))$ from where
$\quad x^n+x^\left(n-1\right)y+.........+x^2y^\left(n-2\right)+xy^\left(n-1\right)$
$\qquad -(x^\left(n-1\right)y+x^\left(n-2\right)y^2+.........+xy^\left(n-1\right)+y^n)$ therefore
$x^n-y^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 9,
"answer_id": 4
} |
Limit of a Sequence involving $\frac {n^2} {\sqrt{n^{6}+k}}$ I am stuck on this problem:
Compute the limit of the sequence $(a_{n})_{n=1}^{\infty}$ defined by
$$a_{n}:=\frac {n^2} {\sqrt{n^{6}+1}}+\frac {n^2} {\sqrt{n^{6}+2}}+\cdot \cdot \cdot + \frac {n^2} {\sqrt{n^{6}+n}}=\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}$$
So I am trying to find:
$$\lim_{n \to \infty}\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}$$
In a situation like this should I be noting that the denominator is increasing in value faster than the numerator?
My first thought was to do some manipulation. I may have done something incorrectly. I began with the following.
$$\frac {n^2} {\sqrt{n^6+k}}=\frac {n^2} {\sqrt{n^6(1+k/n^6)}}=\frac {n^2} {\sqrt{n^6}\sqrt{1+k/n^6}}=\frac {1} {n} \cdot \frac {1} {\sqrt {1+k/n^6}}$$
So I now have
$$\lim_{n \to \infty} \frac {1} {n} \sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}$$
Now, I am unsure about the following. It looks to me as if $k/n^6$ goes to zero as $n \to \infty $. That would result in $\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}=n$. So I would be left with
$$\lim_{n \to \infty} \frac {n} {n}=1$$
| I think this should work, $$\frac{n\cdot n^2}{\sqrt{n^6+k}}\leq \sum_{k=1}^n\frac{n^2}{\sqrt{n^6+k}}\leq \frac{n\cdot n^2}{\sqrt{n^6+1}}$$Taking limits at both sides says the limit is $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Divide with remainder $\frac{x^2}{x^2 + x + 2}$ I am having a hard time long dividing:
$$\frac{x^2}{x^2 + x + 2}.$$
Could someone please show a step by step way to divide this, as I can only get it down to : $1 + \frac{x^2}{x + 2}$.
Thank you for your time!
| Since $x^2 - 1 \cdot (x^2 + x + 2) = -x - 2$ (one step of polynomial division), we get $x^2 : (x^2 + x + 2) = 1$, remainder $-x-2$, or equivalently, $$\frac{x^2}{x^2 + x + 2} = 1 + \frac{-x-2}{x^2 + x + 2}.$$ This is, in fact, the final result, since the degree of $-x-2$ is less than that of $x^2 + x + 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Exponent Law of Addition? Is there a rule for adding exponential terms of like bases just like there are rules for multiplying and dividing such terms?
For example we know that:
$x^1 \cdot x^2 = x^{1+2} = x^3$
But what about for addition (or subtract for that matter)?
$x^1 + x^2 = x^?$
If no such pattern exists, why is that?
| Observe that
$ \log(A+B) = \log(A) + \log \left(\frac{B}{A} + 1 \right) $. Using this fact, one can show that
$
\begin{align}
\log (x^a + x^b) &= \log(x^a) + \log ( x^{b-a} + 1 ) \\
&= a\log x + \log ( x^{b-a} + 1 ) \\
&= \log x \left( a + \frac{\log ( x^{b-a} + 1 )}{\log x} \right) \\
\Rightarrow x^a + x^b &= x^{\left( a + \frac{\log ( x^{b-a} + 1 )}{\log x} \right)} \\
\text{In your case, with }& a=1, b=2, \text{ we get} \\
x^1 + x^2 &= x^{\left( 1 + \frac{\log ( x + 1 )}{\log x} \right)}
\end{align}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Given that $a+b\sqrt[3]{2} +c\sqrt[3]{4} =0$, where $a,b,c$ are integers. Show $a=b=c=0$ Given that $\displaystyle{a+b\sqrt[3]{2} +c\sqrt[3]{4} =0}$, where $a,b,c$ are integers. Show $a=b=c=0$
Do I use modular arithmetic?
| Let $\alpha=\sqrt[3]{2}$
Then $m_{\alpha}({x})=x^3-2$ is the minimal polynomial of $\alpha$ over $\Bbb{Q}$.
$f(x)=a+bx+cx^2$ is an anhilating polynomial of $\sqrt[3]{2}$ as $f(\alpha) =0$
Hence $m_{\alpha}(x) \mid f(x) $
$x^3-2\mid a+bx+cx^2$ implies $a=b=c=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
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} |
Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ d}x$ I'm learning integral. Here is my homework:
$$\int_0^1 \sqrt{1+x^2}\;dx$$
I think this problem solve by change $x$ to other variable. Can you tell me how please. (just direction how to solve)
thanks :)
| Since the integrand is a function of $x$ and $\sqrt{ax^{2}+bx+c}$ another
option is to use the Euler substitution $\sqrt{ax^{2}+bx+c}=\pm
\sqrt{a}x\pm t$, with $a>0$. Choosing $\sqrt{1+x^{2}}=t-x$, squaring both
sides and solving for $x$, we obtain $x=\frac{t^{2}-1}{2t}$ and $dx=\frac{
t^{2}+1}{2t^{2}}dt$. The integrand becomes an easily integrable rational fraction of $t$
$$\begin{equation*}
\sqrt{1+\left( \frac{t^{2}-1}{2t}\right) ^{2}}\frac{t^{2}+1}{2t^{2}}=\frac{1
}{4}\frac{\left( t^{2}+1\right) ^{2}}{t^{3}}=\frac{1}{2t}+\frac{1}{4t^{3}}+\frac{1}{4}t.
\end{equation*}$$
So
$$\begin{eqnarray*}
\int_{0}^{1}\sqrt{1+x^{2}}dx &=&\int_{1}^{\sqrt{2}+1}\left( \frac{1}{2t}+\frac{1}{4t^{3}}+\frac{t}{4}\right) dt \\
&=&\left. \frac{1}{2}\ln t-\frac{1}{8t^{2}}+\frac{1}{8}t^{2}\right\vert
_{1}^{\sqrt{2}+1}.
\end{eqnarray*}$$
Added: I've checked the final result:
$$\begin{eqnarray*}
\left. \frac{1}{2}\ln t-\frac{1}{8t^{2}}+\frac{1}{8}t^{2}\right\vert _{1}^{\sqrt{2}+1} &=&\frac{\ln \left( \sqrt{2}+1\right) }{2}-\frac{1}{8\left( \sqrt{2}
+1\right) ^{2}}+\frac{1}{8}\left( \sqrt{2}+1\right) ^{2}-0 \\
&=&\frac{\ln \left( \sqrt{2}+1\right) }{2}+\frac{\sqrt{2}}{2}.
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 1
} |
$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$ is a $C^1$-function I would like to show that the function:
$$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$$
is a $C^1$-function.
$$ \frac{\partial f}{\partial x}(x,y)=\frac{\sin(y)-y\cos(x)}{x^2+y^2}+\frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2}$$
$$ \frac{\partial f}{\partial y}(x,y)=-\frac{\partial f}{\partial y}(y,x)=... $$
So I just have to show that:
$$ \frac{\partial f}{\partial x}(x,y)\rightarrow_{(0,0)}0$$
When $y\geq0$ :
$$ -\frac{y^3}{6(x^2+y^2)}+\frac{x^2y}{x^2+y^2}-\frac{x^4y}{4!(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}$$
When $y<0$ :
$$ -\frac{y^3}{6(x^2+y^2)}+\frac{y^5}{5!(x^2+y^2)}+\frac{x^2y}{2(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}-\frac{yx^4}{4!(x^2+y^2)}$$
So $$ \frac{\sin(y)-y\cos(x)}{x^2+y^2}\rightarrow_{(0,0)}0 $$
How can I directly find an upper bound of $$ \left| \frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2} \right|$$ that tends to 0 ?
| A partial answer to at least show that $f$ is differentiable at $(0,0)$. Then it remains to show that its derivative is continuous. I didn't check if that also follows from this bound directly.
Combine $\tan(x)\geq x$ and $\sin(x) \leq x$ for $x \in [0,\pi/2)$ to get
$$
\cos(x) \leq \frac{\sin(x)}{x} \leq 1
$$
for $x \in (-\pi/2,\pi,2)$. Then for $x,y \in (-\pi/2,\pi/2)$
$$
-\frac{y^2}{2} \leq \cos(y)-1 \leq \frac{\sin(y)}{y} - \frac{\sin(x)}{x} \leq 1 - \cos(x) \leq \frac{x^2}{2}.
$$
Taking the absolute value:
$$
\left| \frac{\sin(y)}{y} - \frac{\sin(x)}{x} \right| \leq \frac{\max(x^2,y^2)}{2}\leq \frac{x^2+y^2}{2}.
$$
This results in the following estimate of $f$:
$$
\left|\frac{x\sin(y) - y\sin(x)}{x^2+y^2}\right| \leq \left| \frac{xy}{x^2+y^2}\right| \cdot \left| \frac{\sin(y)}{y} - \frac{\sin(x)}{x}\right| \leq \frac{x^2+y^2}{4}
$$
This is sufficiently sharp to conclude that $f$ is differentiable in $(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding inverse of polynomial in a field I'm having trouble with the procedure to find an inverse of a polynomial in a field. For example, take:
In $\frac{\mathbb{Z}_3[x]}{m(x)}$, where $m(x) = x^3 + 2x +1$, find the inverse of $x^2 + 1$.
My understanding is that one needs to use the (Extended?) Euclidean Algorithm and Bezout's Identity. Here's what I currently have:
Proceeding with Euclid's algorithm:
$$ x^3 + 2x + 1 =(x^2 + 1)(x) + (x + 1) \\\\
x^2 + 1 = (x + 1)(2 + x) + 2$$
We stop here because 2 is invertible in $\mathbb{Z}_3[x]$. We rewrite it using a congruence:
$$(x+1)(2+x) \equiv 2 \mod{(x^2+1)}$$
I don't understand the high level concepts sufficiently well and I'm lost from here. Thoughts?
Wikipedia has a page on this we a decent explanation, but it's still not clear in my mind.
Note that this question has almost the same title, but it's a level of abstraction higher. It doesn't help me, as I don't understand the basic concepts.
Thanks.
| Write $f := x^3+2x+1$ and $g := x^2+1$. We want to find the inverse of $g$ in the field $\mathbb F_3[x]/(f)$ (I prefer to write $\mathbb F_3$ instead of $\mathbb Z_3$ to avoid confusion with the $3$-adic integers), i.e. we are looking for a polynomial $h$ such that $gh \equiv 1 \pmod f$, or equivalently $gh+kf=1$ for some $k\in \mathbb F_3[x]$. The Euclidean algorithm can be used to find $h$ and $k$:
\begin{align}
f &= x\cdot g+(x+1)\\
g &= (x+2)\cdot(x+1) + 2\\
(x+1) &= (2x)\cdot2 + 1
\end{align}
Working backwards, we find
\begin{align}
1 &= (x+1)-(2x)\cdot 2\\
&= (x+1)-(2x)(g-(x+2)(x+1))\\
&= (2x^2+x+1)(x+1)-(2x)g\\
&= (2x^2+x+1)(f-xg)-(2x)g\\
&= (2x^2+x+1)f- (x^3+2x^2)g\\
&= (2x^2+x+1)f - (2x^3+x^2)g\\
&= (2x^2+x+1)f + (x^3+2x^2)g.
\end{align}
So, the inverse of $g$ modulo $f$ is $h = x^3+2x^2 \pmod f = 2x^2+x+2 \pmod f$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 4
} |
Finding $x^2+y^2+z^2$ given that $x+y+z=0$, $x^3+y^3+z^3=3$ and $x^4+y^4+z^4=15$ I just ran into this:
$$\begin{align}
x^1+y^1+z^1&=0\\
x^3+y^3+z^3&=3\\
x^4+y^4+z^4&=15\\
x^2+y^2+z^2&=\text{?}
\end{align}$$
I know the answer, but can this be solved without guessing?
| Note that $0 = (x+y+z)^2 = (x^2+y^2+z^2) + 2(xy+xz+yz)$, so $x^2+y^2+z^2 = -2(xy+xz+yz)$.
And $$\begin{align*}
15 &= x^4+y^4+z^4\\
& = (x^2+y^2+z^2)^2 - 2(x^2y^2 + x^2z^2 + y^2z^2).\end{align*}$$
And
$$\begin{align*}
(xy+xz+yz)^2 &= (x^2y^2 + x^2z^2 + y^2z^2) + 2(x^2yz+xy^2z+xyz^2)\\
&= (x^2y^2+x^2z^2 + y^2z^2) + 2xyz(x+y+z)\\
&= x^2y^2+x^2z^2+y^2z^2 + 2xyz(0)\\
&= x^2y^2+x^2z^2+y^2z^2
\end{align*}$$
hence
$$x^2y^2+x^2z^2+y^2z^2 = (xy+xz+yz)^2.$$
So
$$\begin{align*}
15 &= x^4+y^4+z^4\\
&= (x^2+y^2+z^2)^2 - 2(x^2y^2 + x^2z^2+y^2z^2)\\
&= (x^2+y^2+z^2)^2 - 2(xy+xz+yz)^2\\
&= (x^2+y^2+z^2)^2 -2(xy+xz+yz)(xy+xz+yz)\\
&= (x^2+y^2+z^2)^2 +(x^2+y^2+z^2)(xy+xz+yz)\\
&= (x^2+y^2+z^2)^2 + (x^2+y^2+z^2)\left(-\frac{1}{2}(x^2+y^2+z^2)\right)\\
&= (x^2+y^2+z^2)^2 -\frac{1}{2}(x^2+y^2+z^2)^2\\
&=\frac{1}{2}(x^2+y^2+z^2)^2
\end{align*}$$
so $(x^2+y^2+z^2)^2 = 30$, hence $x^2+y^2+z^2=\sqrt{30}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/126069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Plot $|z - i| + |z + i| = 16$ on the complex plane
Plot $|z - i| + |z + i| = 16$ on the complex plane
Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellipse algebraically.
I get to the point:
$x^2 + (y - 1) ^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} + x^2 + (y+ 1)^2 = 256$
$2x^2 + 2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$
It seems like I'm and doing something the hard way.
| Just keep going:
$$2x^2+2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$$
$$\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 127 - (x^2+y^2)$$
$$(x^2 + (y - 1)^2)(x^2 + (y + 1)^2) = (127 - (x^2+y^2))^2$$
$$x^4 + x^2[(y - 1)^2 + (y + 1)^2] + [(y+1)(y-1)]^2 = 127^2 -2\cdot127\cdot(x^2+y^2)+ (x^2+y^2)^2$$
$$x^4 + 2x^2y^2 +2x^2 + (y^2-1)^2 = 127^2 -2\cdot127\cdot(x^2+y^2)+ x^4+y^4+2x^2y^2$$
$$x^4 + 2x^2y^2 +2x^2 + y^4-2y^2+1 = 127^2 -2\cdot127\cdot(x^2+y^2)+ x^4+y^4+2x^2y^2$$
$$2x^2 -2y^2+1 = 127^2 -2\cdot127\cdot(x^2+y^2)$$
$$x^2(2+2\cdot127) +y^2(-2+2\cdot127) = 127^2 -1$$
$$256x^2 +252y^2 = 16128$$
$$\frac {x^2} {63} +\frac {y^2} {64} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/126518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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Recurrence relation for number of ternary strings that contain two consecutive zeros The question is: Find a recurrence relation for number of ternary strings of length n that contain two consecutive zeros.
I know for ternary strings with length one, there are 0. For a length of 2, there is just 1 (00), and for a length of 3, there are 5 (000,001,002,100,200).
I did a similar problem, finding a relation for the number of bit strings of length n with two consecutive zeros: $$a_n = a_{n-1} + a_{n-2} + 2^{n-2}$$
Since you can add "1" to the end of all the $a_{n-1}$ strings, "10" to all the $a_{n-2}$ strings, and "00" any string of size $n-2$.
For the ternary string problem, I'm pretty sure you would replace the $2^{n-2}$ with $3^{n-2}$, but confused about the other terms of the relation. My guess is that it would have the coefficient $2$ in front of the other terms, since you can add either $1$ or $2$ to the end of $a_{n-1}$ and either $01$ or $02$ at the end of $a_{n-1}$.
So I believe the answer for the relation is: $$a_n = 2a_{n-1} + 2a_{n-2} + 3^{n-2}$$
How does that look?
| Let us count the number of strings $a_n$ that do not contain $00$ as a substring. The required number of strings that contain $00$ is $3^n - a_n$.
Let us call a string that does not contain $00$ as good. Let $x_n, y_n, z_n$ be the number of good strings of length $n$ that end with $0, 1, 2$ respectively. Clearly,
\begin{align*}
x_{n+1} &= y_n + z_n \\
y_{n+1} &= x_n + y_n + z_n \\
z_{n+1} &= x_n + y_n + z_n
\end{align*}
Thus
\begin{align*}
y_n = z_n, \qquad x_{n+1} = 2y_n, \qquad y_{n+1} = x_n + 2y_n
\end{align*}
and $y_n$ satisfies the recurrence
$y_{n+1} - 2y_n - 2y_{n-1} = 0$. The characteristic roots are $1 \pm \sqrt{3}$ and
$$y_n = A(1+\sqrt{3})^n + B(1-\sqrt{3})^n$$
where $A, B$ are constants. Since $y_2 = 3$ and $y_3 = 8$ we get
$$A = \frac{1+\sqrt{3}}{4\sqrt{3}}, \qquad B=-\frac{1-\sqrt{3}}{4\sqrt{3}}$$
Hence
$$y_n = \frac{1}{4\sqrt{3}}\left(\alpha^{n+1} -\beta^{n+1}\right) $$
Since $a_n = x_n + y_n + z_n = 2y_{n-1} + 2y_n$, we get
\begin{align*}
a_n &= \frac{1}{2\sqrt{3}}\left(\alpha^n(1+\alpha) - \beta^n(1+\beta)\right)\\
&= \left(\frac{2+\sqrt{3}}{2\sqrt{3}}\right)\alpha^n - \left(\frac{2-\sqrt{3}}{2\sqrt{3}}\right)\beta^n
\end{align*}
Hence the number of sequences of length $n$ that contain $00$ is
$$3^n - \left(\frac{2+\sqrt{3}}{2\sqrt{3}}\right)(1+\sqrt{3})^n + \left(\frac{2-\sqrt{3}}{2\sqrt{3}}\right)(1-\sqrt{3})^n$$
| {
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"url": "https://math.stackexchange.com/questions/127818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
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Proving a simple inequality. I am trying to prove that
$\frac{n}{4n^2+1} > \frac{n+1}{4(n+1)^2+1}, \forall n\in\mathbb{N}$.
What I did so far was
$n < n+1\\
\Rightarrow \frac{n^2}{n} < \frac{(n+1)^2}{n+1}\\
\Rightarrow \frac{n}{n^2} > \frac{n+1}{(n+1)^2}\\
\Rightarrow \frac{n}{4n^2} > \frac{n+1}{4(n+1)^2}\\
\Rightarrow \frac{n}{4n^2+1} > \frac{n+1}{4(n+1)^2+1}\\
$
I'm not so sure about the last step though... basically, is
$\frac{a}{b} < \frac{c}{d} \Rightarrow \frac{a}{b+1} < \frac{c}{d+1} (b\neq -1 \wedge d \neq -1)$ a correct assumption? It was just a gut feeling for me, and I can't really justify it.
And yes, this is a homework question. I just didn't know what exactly I would have to look for, so I had to ask.
| Inequality is equivalent to the :
$4n^2+4n-1>0$
which is equivalent to the :
$(2n+1)^2>2$
You can use induction to prove this inequality :
*
*for $n=1$ it follows $9>2$
*suppose $(2n+1)^2>2$
*we want to to prove that $(2n+3)^2>2$
Since $(2n+1)^2>2$ it follows :
$4n^2+4n+1 > 2$
$4n^2+4n+1+8n+8>2+8n+8>2$
Hence :
$(2n+3)^2>2$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/129290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to calculate the improper integral $\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx$ How to Prove: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$
| Rewrite it as
$$
I =\int_0^\infty \frac{\log\left(x+\frac{1}{x}\right)}{x+\frac{1}{x}} \frac{\mathrm{d} x}{x} = \left.\frac{\mathrm{d}}{\mathrm{d}s} \mathcal{I}(s)\right|_{s=-1}
$$
where
$$
\mathcal{I}(s) = \int_0^\infty \left( x+\frac{1}{x}\right)^s \frac{\mathrm{d} x}{x} \stackrel{x=\sqrt{\frac{u}{1-u}}}{=} \frac{1}{2} \int_0^{1} (u(1-u))^{-s/2-1} \mathrm{d} x
$$
Thus
$$
\mathcal{I}(s) = \frac{1}{2} \operatorname{B}\left(-\frac{s}{2},-\frac{s}{2} \right) = \frac{1}{2} \frac{\Gamma^2\left(-\frac{s}{2}\right)}{\Gamma(-s)}
$$
It remains to evaluate the derivative at $s=-1$:
$$
I = \left.\frac{\mathrm{d}}{\mathrm{d}s} \mathcal{I}(s)\right|_{s=-1} = \mathcal{I}(-1)\left( \psi(1) - \psi\left(\frac{1}{2}\right) \right) = \pi \log(2)
$$
The result follows from $\mathcal{I}(-1) = \frac{1}{2} B(1/2,1/2) = \frac{\pi}{2}$, and from duplication identity for digamma, evaluated as $s=1$:
$$
\psi(s) = \log(2) + \frac{1}{2} \left( \psi\left( \frac{s}{2} \right) + \psi\left( \frac{s+1}{2} \right) \right) \stackrel{s=1}{\implies } \psi(1) - \psi(1/2) = 2 \log(2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How do you integrate $\int \frac{1}{a + \cos x} dx$? How do you integrate $\int \frac{1}{a + \cos x} dx$? Is it solvable by elementary methods? I was trying to do it while incorrectly solving a homework problem but I couldn't find the answer.
Thanks!
| Also a generalised solution, borrowing from and expanding upon user1357113's answer,
I. For the case $|a| > |b|$, note that the substitution $t=\tan \left( \frac{x}{2} \right)$ is not injective. So to retrieve a continuous antiderivative, we have as a complete answer
$$
\int \frac{1}{a+b\cos(x)} {\rm d}x
= \frac{2}{\sqrt{a^2-b^2}} \arctan \left( \sqrt{\frac{a-b}{a+b}}\tan \left( \frac{x}{2} \right) \right) + \frac{2\pi}{\sqrt{a^2-b^2}} \left\lfloor \frac{x+\pi}{2\pi} \right\rfloor +C
$$
However, we can simplify the ugly floor function in terms of arctangents and tangents. First, we have
$$
\pi \left\lfloor \frac{x+\pi}{2\pi} \right\rfloor
= \frac{x}{2} - \arctan \left( \tan \left( \frac{x}{2} \right) \right)
$$
So, implementing this and then asking What is $\arctan(x) + \arctan(y)$? We get
$$
\begin{align}
\int \frac{1}{a+b\cos(x)} {\rm d}x
& = \frac{2}{\sqrt{a^2-b^2}} \arctan \left( \sqrt{\frac{a-b}{a+b}}\tan \left( \frac{x}{2} \right) \right) + \frac{2}{\sqrt{a^2-b^2}} \left( \frac{x}{2} - \arctan \left( \tan \left( \frac{x}{2} \right) \right) \right) +C \\
& = \frac{1}{\sqrt{a^2-b^2}} \left( x + 2\arctan \left( \sqrt{\frac{a-b}{a+b}}\tan \left( \frac{x}{2} \right) \right) - 2\arctan \left( \tan \left( \frac{x}{2} \right) \right) \right) + C \\
& = \frac{1}{\sqrt{a^2-b^2}} \left( x + 2\arctan \left( \frac{ \sqrt{\frac{a-b}{a+b}}\tan \left(\frac{x}{2}\right) - \tan \left(\frac{x}{2}\right) }{1 + \sqrt{\frac{a-b}{a+b}} \tan^2 \left(\frac{x}{2}\right) } \right) \right) + C \\
\end{align}
$$
Which ultimately simplifies to a satisfying compact
$$ \int \frac{1}{a+b\cos(x)} {\rm d}x
= \frac{1}{\sqrt{a^2-b^2}} \left( x - 2\arctan \left( \frac{ (\sqrt{a+b}-\sqrt{a-b})\tan \left(\frac{x}{2}\right) }{\sqrt{a+b} + \sqrt{a-b} \tan^2 \left(\frac{x}{2}\right) } \right) \right) + C
$$
II. And for the case $|a|<|b|$, we will have, from user1357113's partial fraction decomposition,
$$ \int \frac{1}{a+b\cos(x)} {\rm d}x
= \frac{1}{\sqrt{b^2-a^2}}\ln\left(\left|\frac{b+a\cos\left(x\right)+\sqrt{b^2-a^2}\sin\left(x\right)}{a+b\cos\left(x\right)}\right|\right) +C
$$
III a. And for the case $a = b$, we will have
$$
\begin{align}
\int \frac{1}{a+b\cos(x)} {\rm d}x
& = \frac1a \int \frac{1}{1+\cos(x)} {\rm d}x \\
& = \frac1a \int \frac{1}{1+2\cos^2(\frac{x}{2})-1}{\rm d}x \\
& = \frac1{2a} \int \sec^2 \left(\frac{x}{2}\right) {\rm d}x \\
& = \frac1a \tan\left(\frac{x}{2}\right) + C
\end{align}
$$
III b. Finally, for the case $a = -b$, we will have
$$
\begin{align}
\int \frac{1}{a+b\cos(x)} {\rm d}x
& = \frac1a \int \frac{1}{1-\cos(x)} {\rm d}x \\
& = \frac1a \int \frac{1}{1 - 1 + 2\sin^2(\frac{x}{2})}{\rm d}x \\
& = \frac1{2a} \int \csc^2 \left(\frac{x}{2}\right) {\rm d}x \\
& = \frac1b \cot\left(\frac{x}{2}\right) + C
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 0
} |
Using linear algebra, how is the Binet formula (for finding the nth Fibonacci number) derived? If possible, please refrain from any type of proof besides linear algebra. So, using the recursion formula $F_{n+1} = F_{n-1} + F_n$, for $n\gt 1$, and where $F_0 = 0$ and $F_1 = 1$, and the Fibonacci matrix, derive the golden ratio and ultimately the Binet formula.
| The Fibonacci numbers are defined by a second-order linear recurrence equation:
$$F_{n+2} = F_{n+1} + F_n$$
This means we can treat the solution of $F_n$ in terms of $n$ as a problem in linear algebra involving only $2$-dimensional vectors. In some sense, what we are doing is modelling this as a dynamical process on a $2$-dimensional state space.
Let $V = \mathbb{R}^2$. We define a linear operator $T : V \to V$ by
$$T(x, y) = (y, x + y)$$
Notice that $T(F_{n}, F_{n+1}) = (F_{n+1}, F_{n+2})$, so you can think of $V$ as being a sliding $2$-entry window on the Fibonacci sequence and $T$ as the operator which advances the window along the Fibonacci sequence. The initial conditions $F_0 = 0, F_1 = 1$ then imply that
$$T^n(0, 1) = (F_n, F_{n+1})$$
so all we need to do to find $F_n$ in terms of $n$ is to find an effective way to compute iterates of the operator $T$!
Now, we get our hands dirty and represent $T$ as a matrix:
$$T = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$$
Imagine if we could somehow find an invertible matrix $P$ and a diagonal matrix $D$ such that $T = P D P^{-1}$; then, by matrix algebra, we would have $T^n = P D^n P^{-1}$, and it is easy to compute powers of diagonal matrices. The theory of eigenvectors and eigenvalues gives us one way to find such a factorisation of $T$. Notice that
$$\det (T - x I) = \det (P(D - x I)P^{-1}) = (\det P)(\det (D - x I))(\det P^{-1}) = \det (D - x I)$$
but a simple calculation shows
$$\det (D - x I) = \det \begin{pmatrix} \lambda_1 - x & 0 \\ 0 & \lambda_2 - x \end{pmatrix} = (\lambda_1 - x)(\lambda_2 - x)$$
so whatever $\lambda_1$ and $\lambda_2$ are, they must be the zeros of the polynomial
$$\det (T - x I) = \det \begin{pmatrix} -x & 1 \\ 1 & 1 - x \end{pmatrix} = x^2 - x - 1$$
which, surprise surprise, is the minimal polynomial of the golden ratio. So let $\lambda_1 = \frac{1}{2} (1 + \sqrt{5})$ and $\lambda_2 = \frac{1}{2} (1 - \sqrt{5})$. These are the eigenvalues of $T$. By construction, $\det (T - \lambda_1 I) = \det (T - \lambda_2 I) = 0$, so there must be non-zero vectors $\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ and $\begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$ such that
$$
T
\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} =
\lambda_1 \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}
$$
$$
T
\begin{pmatrix} x_2 \\ y_2 \end{pmatrix} =
\lambda_2 \begin{pmatrix} x_2 \\ y_2 \end{pmatrix}
$$
These vectors are called the eigenvectors of $T$. I leave you to verify that $x_1 = \lambda_1 - 1$, $y_1 = 1$, $x_2 = \lambda_2 - 1$, $y_2 = 1$ works, but there are other solutions.
Define the matrix $P$ by
$$P = \begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix}$$
Notice that
$$\begin{pmatrix} x \\ y \end{pmatrix} =
\alpha_1 \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} +
\alpha_2 \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} =
P \begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix}$$
so by linearity we have
$$T \begin{pmatrix} x \\ y \end{pmatrix} =
\lambda_1 \alpha_1 \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} +
\lambda_2 \alpha_2 \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} =
P \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix}$$
but we can invert $P$ to find $\alpha_1$ and $\alpha_2$ in terms of $x$ and $y$, so we have obtained the desired factorisation of $T$ as $P D P^{-1}$ with $D$ diagonal. Putting everything together, we get the formula
$$T^n = P \begin{pmatrix} {\lambda_1}^n & 0 \\ 0 & {\lambda_2}^n \end{pmatrix} P^{-1}$$
and applying this to the vector $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ gives Binet's formula.
| {
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"url": "https://math.stackexchange.com/questions/135478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
A 'should be simple' derivative Help! I can't seem to prove that $$\Bigl[\log(x+\sqrt{x^2 + 1})\Bigr]' = \frac{1}{\sqrt{x^2 + 1}}$$ I keep getting some horrible answer namely $$\frac{x + \sqrt{x^2 + 1}}{x\sqrt{x^2 + 1} + 1 + x^2}$$ does this cancel down at all?
| we have in your denominator
$$ x\sqrt{x^2 + 1} + (1 + x^2) = \bigl(x + \sqrt{x^2 + 1}\bigr)\cdot \sqrt{x^2 + 1}$$
so by canceling $x + \sqrt{x^2 + 1}$ we get $\frac 1{\sqrt{x^2 + 1}}$ as wished.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/135920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
How to $\int \sqrt{10x-x^2}dx$ Theres a hint to use $x=5+5\sin{t}$. Ok, but how do I know what substitution to use if a hint wasn't given? Is it "trivial" or perhaps, its very unlikely that that will appear?
Anyways, I did:
$\int \sqrt{10(5+5\sin{t}) - (5+2\sin{t})^2} dx \\
= \int \sqrt{50+50\sin{t} - (25+50\sin{t} + 25\sin^2{t})} dx\\
= \int \sqrt{ 25-25\sin^2{t} } dx \\
= 5 \int \sqrt{1-\sin^2{t}} dx \\
= 5\sin^{-1}{\sin{t}} \\
= 5t \\
= 5 \sin^{-1}{\frac{x-5}{5}}$
But the answer was:
$$\frac{25}{2}\sin^{-1}{\frac{x-5}{5}}+\frac{x-5}{2}\sqrt{10x-x^2}+c$$
What did I do wrong? Or is the answer wrong perhaps?
| put dx = 5cost dt
from your fourth step,
5∫cost * 5cost dt = 25∫(cos^2)(t) dt
= 25∫(1 + cos2t)/2 dt
= (25/2)∫(1 + cos2t)dt
= (25/2)(t + (sin2t)/2) + c
=(25/2)(t + (2 sint cost)/2) + c
and substitute the value of t and sint and cost in above. Then you will get your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/136175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find tangent line of curve that intersects point. How do I find the tangent line of the curve $y=x^2$ that intersects the point $(8,2)$?
| Starting with Andre's equation of
$y-a^2=2a(x-a)$,
$y=2ax-2a^2+a^2 = 2ax-a^2$,
or $a^2-2ax = -y$
or $a^2-2ax + x^2 = x^2-y$
or $(a-x)^2 = x^2-y$
or $a = x \pm \sqrt{x^2-y}$.
Note that this requires $x^2 \ge y$,
which means that the point that the tangent goes through
has to be below (or on) the parabola
and that, if the point is below, there are two tangents
going through it - see if you can visualize why.
Also note that the two values of $a$ are equally spaced around $x$.
Also$^2$ note that if $y < 0$
then there are always two values of $a$,
one of which is negative.
If $x=8$ and $y=2$,
$x^2-y = 62$,
so $ a = 8 \pm \sqrt{62}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/137065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\int |x|^3 \; dx $ $$\int |x|^3 \; dx $$
In my module it is suggest to use integration by parts,
$$ \text{ Set }I = \int (|x|^3 \cdot 1) \; dx = |x|^3 \cdot x - \int \color{red}{\frac {x^3}{|x|^3}3x^2}\cdot x \; dx$$
$$ \implies I = |x|^3 \cdot x - \color{red}{3\int |x|^3\;} dx$$
$$ \implies I = \frac 1 4 |x|^3 \cdot x +C $$
I am having trouble understanding the red parts, to be precise the differentiation of $|x|^3 $.
I will also appreciate if somebody wants to share a different approach (if any).
Thanks,
| Let $\displaystyle I= \int|x|^3 dx$
\begin{align*}I =& \int x^2|x| dx \\
=& |x|\int x^2 dx - \int\frac{d}{dx}(|x|)\left(\int x^2 dx \right)dx \\
=&\frac{|x|x^3}{3} - \int\left(\frac{|x|}{x}\cdot\frac{x^3}{3}\right)dx \text{ since } \frac{d}{dx}(|x|) = \frac{|x|}{x} \\
=&\frac{|x|x^3}{3} - \frac{1}{3}\int|x|x^2dx \\
=&\frac{|x|x^3}{3} - \frac{I}{3}\end{align*}
That implies,
\begin{align*}I + \frac{I}{3} =& \frac{|x|x^3}{3} \\
\frac{4I}{3} =& \frac{|x|x^3}{3} \\
4I =& |x|(x^3)\end{align*}
Therefore $\displaystyle I = \frac{|x|x^3}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/137943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Show that $2\tan^{-1}(2) = \pi - \cos^{-1}(\frac{3}{5})$
Show that $2\tan^{-1}(2) = \pi - \cos^{-1}(\frac{3}{5})$
So, taking $\tan$ of both sides I get:
LHS $=\frac{2\tan(\tan^{-1}(2))}{1 - \tan^2(\tan^{-1}(2))} = -\frac{4}{3}$
and
RHS $= \tan(\pi - \cos^{-1}(\frac{3}{5})) = ~...$
I wasn't sure how to treat the RHS given the $\pi$ term. If it were just $\cos^{-1}(\frac{3}{5})$ I guess I would try to manipulate $\sin^2(\theta) + \cos^2(\theta) = 1$ to try and find something appropriate for substitution; however, I am not sure where to even start with the given $\theta$.
| From this or Ex$\#5$ of Page $\#276$ of this,
$$\tan^{-1}x+\tan^{-1}y=\begin{cases} \tan^{-1}\frac{x+y}{1-xy} &\mbox{if } xy<1 \\\pi+ \tan^{-1}\frac{x+y}{1-xy} & \mbox{if } xy> 1. \end{cases} $$
$$\implies2\tan^{-1}x=\begin{cases} \tan^{-1}\frac{2x}{1-x^2} &\mbox{if } x^2<1\iff -1\le x\le1 \\\pi+ \tan^{-1}\frac{2x}{1-x^2} & \mbox{ else where } \end{cases} $$
$$\implies2\tan^{-1}2=\tan^{-1}\left(\frac{2\cdot2}{1-2^2}\right)=\pi+\tan^{-1}\left(-\frac43\right) $$
Let $\theta=\tan^{-1}\left(-\frac43\right)\implies \tan\theta=-\frac43$
Now, as the principal value of $\displaystyle\tan$ lies $\displaystyle\in\left[-\frac\pi2,\frac\pi2\right],\cos\theta=+\frac1{\sqrt{1+\tan^2\theta}}$
Now, $\cos(\pi+y)=-\cos y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/138310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How do I show that the sum $(a+\frac12)^n+(b+\frac12)^n$ is an integer for only finitely many $n$?
Show that if $a$ and $b$ are positive integers, then $$\left(a +\frac12\right)^n + \left(b+\frac{1}{2}\right)^n$$is an integer for only finitely many positive integers $n$.
I tried hard but nothing seems to work. :(
| Note that
$$\left(a+\frac{1}{2}\right)^n + \left(b+\frac{1}{2}\right)^n=\frac{1}{2^n}\left(c^n+d^n\right),$$
where $c=2a+1$ and $d=2b+1$.
Let $2^e$ be the highest power of $2$ that divides $c+d$.
If $n$ is even, the highest power of $2$ that divides $c^n+d^n$ is $2$.
For odd $n$, note that $u^n+v^n=(u+v)(u^{n-1}+u^{n-2}v+\cdots +v^{n-1})$. The second term in this product is odd. It follows that the highest power of $2$ that divides $c^n+d^n$ is $2^e$.
Thus if $n\gt e$, then $\left(a+\frac{1}{2}\right)^n + \left(b+\frac{1}{2}\right)^n$ is not an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
A trigonometric identity: $(\sin x)^{-2}+(\cos x)^{-2}=(\tan x+\cot x)^2$ I've been trying to prove it for a while, but can't seem to get anywhere.
$$\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta} = (\tan \theta + \cot \theta)^2$$
Could someone please provide a valid proof?
I am not allowed to work on both sides of the equation.
Work so far:
RS:
$$
\begin{align}
& \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} + 2 \\[10pt]
& = \frac{\sin^4\theta}{(\cos^2\theta)(\sin^2\theta)} + \frac{\cos^4\theta}{(\sin^2\theta) (\cos^2\theta)} + \frac{(\sin^4\theta)(\cos^2\theta)}{(\sin^2\theta)(\cos^2\theta)} + \frac{(\sin^2\theta)(\cos^4\theta)}{(\sin^2\theta)(\cos^2\theta)} \\[10pt]
& = \frac{\sin^4\theta + \cos^4\theta + (\sin^4\theta)(\cos^2\theta) + (\sin^2\theta)(\cos^4\theta)}{(\cos^2\theta)(\sin^2\theta)}
\end{align}
$$
I am completely lost after this.
| Hint: $$(\tan\theta+\cot \theta)^2=\left(\frac{\sin\theta}{\cos \theta} +\frac{\cos \theta}{\sin \theta}\right)^2$$ $$= \left(\frac{\cos^2 \theta+\sin^2\theta}{\cos \theta \sin \theta}\right)^2.$$ Now try using the fact that $\cos^2\theta+\sin^2\theta=1$, twice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/142252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Inverse of $f^{-1}(x)=x^5+2x^3+3x+1$ question Let $f$ be a one-to-one function whose inverse function is $f^{-1}(x)=x^5+2x^3+3x+1$.
Compute the value of $x_0$ such that $f(x_0)=1$.
I am confused as to what this question is asking me, particularly since I don't understand the subscript under the $x$ variable.
| $f^{-1}(x)=x^5+2x^3+3x+1$.
$f(f^{-1}(x))=f(x^5+2x^3+3x+1)$.
$f(f^{-1}(x))=x$ as function property
$x=f(x^5+2x^3+3x+1)$.
$f(x^5+2x^3+3x+1)=x$
$f(x_0)=1$
You can see that $x=1$ so $x_0=x^5+2x^3+3x+1$
$x_0=1^5+2.1^3+3.1+1=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/142509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Derivative of $\ln\sqrt{\frac{e^{x^2}}{e^x+2}}$?
Being: $\ln f(x)=\log_ef(x)$
I started derivating $$\ln\sqrt{\dfrac{e^{x^2}}{e^x+2}}$$ but I get to a point that I don't know how to follow.
I try to get it by derivating the logarithm directly and by using the logarithmic properties such us: $$\ln f(x)^n=n\ln f(x)$$ and $$\ln \frac{f(x)}{Q(x)}=\ln f(x)-\ln Q(x)$$ but I don't get it.
By the time I have done this:
\begin{align}
f(x)&=\ln\sqrt{\dfrac{e^{x^2}}{e^x+2}}\\
&=\frac{\ln e^{x^2}-\ln (e^x+2)}{2}\\
&=\frac{x^2-\ln (e^x+2)}{2}\\
\text{And derivating:}\\
f'(x)&=\frac{1}{2}\left(\frac{d}{dx}x^2-\frac{d}{dx}\ln (e^x+2)\right)\\
&=x-\frac{1}{2}\left(\frac{e^x}{e^x+2}\right)
\end{align}
So I finally get this:
$f'(x)=x-\dfrac{e^x}{2e^x+4}$
But WolframAlpha says that $$f'(x)=\frac{x(x^2-5)}{x^2-4}$$
If I'm wrong, what do I do wrong? And if I'm right, how can I get from $x-\dfrac{e^x}{2e^x+4}$ to $\dfrac{x(x^2-5)}{x^2-4}$
| A check, if you trusting your integration skills more than your derivating ones:
Integrate your $f'(x)$, and if you compute your derivative correctly, you will then get your original function, $f(x)$ with some constant. (For more reading: Fundamental theorem of calculus)
$$\int\frac{x(x^2-5)}{x^2-4}\ dx=\int (x-\frac{1}{2(x-2)}+\frac{1}{2(x+2)})\ dx = $$
$$=\frac{x^2}{2}-\frac{1}{2}\ln|x-2|-\frac{1}{2}\ln|x+2|+C=\frac{1}{2}(x^2-\ln|x^2-4|)+C=\frac{1}{2}{\ln{e^{x^2}}-\frac{1}{2}\ln|x^2-4|}+C=$$
$$=\frac{1}{2}\ln\frac{e^{x^2}}{x^2-4}+C=\ln\sqrt{\frac{e^{x^2}}{x^2-4}}+C$$
Now, the question, is there $C\in\mathbb{R}$, such that:
$$\ln\sqrt{\frac{e^{x^2}}{x^2-4}}+C=\ln\sqrt{\frac{e^{x^2}}{e^x+2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/143135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$ I met an inequality, I ask, do not mathematical induction to prove that:
Prove \[ \left(\frac n2\right)^n > n! > \left(\frac n3\right)^n \] without using induction
| Let $a_n =\displaystyle \frac{2^n n!}{n^n}.$ Note that $a_6 = 80/81 < 1.$ We also have $$ \frac{a_{n+1}}{a_n} = \frac{2^{n+1} (n+1)! }{(n+1)^{n+1}} \cdot \frac{n^n}{2^n n!} = 2 \left(\frac{n}{n+1}\right)^n < 1.$$ The sequence $$x_n = \left(1- \frac{1}{n+1} \right)^n$$ is monotonically decreasing to $1/e.$ Since $e>2$, $a_{n+1}/a_n < 1$ so $(a_n)$ is a monotonically decreasing sequence. Thus the first inequality holds.
By considering Taylor series, $\displaystyle e^x \geq \frac{x^n}{n!}$ for all $x\geq 0,$ and $n\in \mathbb{N}.$ In particular, for $x=n$ this yields $$ n! \geq \left( \frac{n}{e} \right)^n $$ and this is stronger than the second inequality.
We could have used the same proof method for the second inequality as we did for the first: Let $b_n= \displaystyle \frac{3^n n!}{n^n}.$ Then $b_6 = 45/4 > 1.$ Also, $$ \frac{b_{n+1}}{b_n} = \frac{3^{n+1} (n+1)! }{(n+1)^{n+1}} \cdot \frac{n^n}{3^n n!} = 3 \left(\frac{n}{n+1}\right)^n $$ and this is greater than $1$ since $e<3.$
What we have just done suggests we can prove the following: If $a,b$ are positive real numbers such that $a<e<b$ then $$ \left(\frac{n}{a} \right)^n > n! > \left( \frac{n}{b} \right)^n $$ holds for sufficiently large $n$. This is turn suggests something stronger about the growth of the factorial function: $n!$ is asymptotically equal to $(n/e)^n$ to within at most a sub-exponential factor.
| {
"language": "en",
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"source": "stackexchange",
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Showing this sequence converges. Given a sequence $x_n=\left(\dfrac{2n^3+n}{n^3} \right)+ i\left(\dfrac{3n}{n+1}\right)$, how would I show it converges? How would I choose $N$?
I did the following.
Given $\epsilon >0,$ choose $N>[?]$. Then for $n>N$
\begin{align}
\left \lvert x_n-(2+3i) \right \rvert & = \left \lvert \frac{2n^3+n}{n^3}-2 \right \rvert+ \lvert \frac{3n}{n+1}-3 \rvert\\
& = \left \lvert \frac{1}{n^2} \right \rvert+ \left \lvert -\frac{3}{n+1} \right \rvert\\
& =\frac{1}{n^2}+\frac{3}{n+1} \\
& = \frac{n+1+3n^2}{n^2+n^3} <?<\epsilon
\end{align}
Hence $x_n \rightarrow 2+3i$.
| For sequences of complex numbers it occurs that a given sequence of complex numbers $a_{n}=x_{n}+y_{n}i$ converges to some complex number $x+yi$ iff sequence $x_{n}$ converges to $x$ and as one may expect $y_{n}$ converges to $y$. Sequence diverges if at least one of the mentioned conditions does not hold.
| {
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"source": "stackexchange",
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General expression for determinant of a block-diagonal matrix Consider having a matrix whose structure is the following:
$$
A =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & a_{2,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{3,1} & a_{3,2} & a_{3,3} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & a_{4,4} & a_{4,5} & a_{4,6} & 0 & 0 & 0\\
0 & 0 & 0 & a_{5,4} & a_{5,5} & a_{5,6} & 0 & 0 & 0\\
0 & 0 & 0 & a_{6,4} & a_{6,5} & a_{6,6} & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & a_{7,7} & a_{7,8} & a_{7,9}\\
0 & 0 & 0 & 0 & 0 & 0 & a_{8,7} & a_{8,8} & a_{8,9}\\
0 & 0 & 0 & 0 & 0 & 0 & a_{9,7} & a_{9,8} & a_{9,9}\\
\end{pmatrix}
$$
Question.
What about its determinant $|A|$?.
Another question
I was wondering that maybe matrix $A$ can be expressed as a product of particular matrices to have such a structure... maybe using these matrices:
$$
A_1 =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}\\
\end{pmatrix}
$$
$$
A_2 =
\begin{pmatrix}
a_{4,4} & a_{4,5} & a_{4,6}\\
a_{5,4} & a_{5,5} & a_{5,6}\\
a_{6,4} & a_{6,5} & a_{6,6}\\
\end{pmatrix}
$$
$$
A_2 =
\begin{pmatrix}
a_{7,7} & a_{7,8} & a_{7,9}\\
a_{8,7} & a_{8,8} & a_{8,9}\\
a_{9,7} & a_{9,8} & a_{9,9}\\
\end{pmatrix}
$$
I can arrange $A$ as a compination of those: $A = f(A_1,A_2,A_3)$
Kronecker product
One possibility can be the Kronecker product:
$$
A=
\begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{pmatrix} \otimes A_1 +
\begin{pmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\\
\end{pmatrix} \otimes A_2 +
\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1\\
\end{pmatrix} \cdot A_3
$$
But what about the determinant??? There are sums in this case which is not good...
| First write
$$\left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] = \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& \text{I}_{n_k} \end{array} \right] \dots \left[ \begin{array}{cccc} \text{I}_{n_1} \hspace{-5pt} &&& \\ & \text{I}_{n_2} \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] $$
Also,
$$\det \left( \left[ \begin{array}{ccccc} \text{I}_{n_1} \hspace{-5pt} &&&& \\[-3pt] & \ddots \hspace{-5pt} &&& \\ && A_j \hspace{-5pt} && \\[-3pt] &&& \ddots \hspace{-5pt} & \\ &&&& \text{I}_{n_k} \end{array} \right] \right) = \det (A_j)$$
which can be seen by using the cofactor formula and repeatedly expanding along a row or column with all 0's and one 1
$$ \implies \det \left( \left[ \begin{array}{cccc} A_1 \hspace{-5pt} &&& \\ & A_2 \hspace{-5pt} && \\[-3pt] && \ddots \hspace{-5pt} & \\ &&& A_k \end{array} \right] \right) = \det (A_1) \det (A_2) \cdots \det (A_k)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Find multiplicative inverse existense. I cannot find whether multiplicative inverse of $x^3+x^2+x+1 \pmod{x^5+x^4+x^3-x^2-x+1}$ over $\mathrm{GF}(3)$ exists. This problem must be solved with Extended Euclidean algorithm.
I tried to divide $x^5+x^4+x^3-x^2-x+1$ by $x^3+x^2+x+1$.
I think I divided it wrong. I got $x^2-2x-2$.
Thanks for any help
| The problem you are trying to solve is not clearly stated, but what is clear is that you are having some difficulty with polynomial division, or (more likely) just think that you are. What probably happened is something that happens to everybody, a slip with signs.
We will do ordinary polynomial division. To start the division process, imitate school division. The polynomial $x^3+x^2+x+1$ "goes into" $x^5+x^4+x^3-x^2-x+1$ how many tines? Clearly $x^2$ times. So multiply $x^3+x^2+x+1$ by $x^2$, subtract from
$x^5+x^4+x^3-x^2-x+1$. The raw remainder we get is $-2x^2-x+1$. Since we are working over the $3$ element field, this can be rewritten in various ways. It is sensible to replace the $-2$ by $1$, obtaining remainder $x^2-x+1$, or $x^2+2x+1$.
If negatives give you trouble, you could start by replacing $x^5+x^4+x^3-x^2-x+1$ by $x^5+x^4+x^3+2x^2+2x+1$.
Anyway, we have quotient $x^2$, remainder $x^2-x+1$. Continue the Euclidean algorithm process by dividing $x^3+x^2+x+1$ by $x^2-x+1$. You should get quotient $x+2$ (or $x-1$), remainder some version of $2x-1$. Continue.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find an equation for the plane that contains the following line and passes through point P How do you determine the plane which contains the line
\begin{align}
x & = -1 + 3t \\
y & = 5 + 2t \\
z & = 2 + t
\end{align}
and passes through the point $P = (2,4,-1)$?
| the line is
\begin{equation}
(-1,5,2) + t(3,2,1)
\end{equation}
Then the plane contains the segment
$\overline{(-1,5,2)(2,4,-1)}$ $\quad((-3,1,3) = (-1,5,2) -((2,4,-1))$ and the plane $\Pi$ has normal $(3,2,1) \times (-3,1,3) = (5,-12,9)$. Hence the plane is $5x -12y + 9z + d = 0$. As $(2,4,-1) \in \Pi$, we have $d = 47$. Hence, the plane is
$5x -12y +9z +47=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\sin x + \sin x \cdot \cot^2 x = \csc x $ The exercise is to prove the trig identity by rewriting each side of the equation into the same form. However only the following identities can be used in the process:
$$\begin{align*}
\tan \theta &= \frac{\sin \theta}{\cos \theta}\\
(\sin \theta)^2 + (\cos \theta)^2 &= 1\\
\csc \theta &= \frac{1}{\sin \theta}\\
\sec \theta &= \frac{1}{\cos \theta}\\
\cot \theta &= \frac{1}{\tan \theta}.
\end{align*}$$
The trig identity given is:
$$\sin x + \sin x \cdot \cot^2 x = \csc x $$
I simplified it to:
$$\sin x + \sin x \cdot \left(\frac{\cos x}{\sin x}\right)^2 = \frac{1}{\sin x} $$
After which I get lost in a mess of term rewriting that never seems to lead anywhere fruitful. Any hints would be greatly appreciated.
| Start with $\sin x + \sin x\cot^2 x$. Factoring out $\sin x$ we get
$$\sin x + \sin x\cot^2 x = \sin x (1 + \cot^2 x).$$
Now use the definition of $\cot x$ and rewrite the expression in the parenthesis as a single fraction:
$$\begin{align*}
\sin x + \sin x\cot^2x &= \sin x\Bigl( 1 + \cot^2 x\Bigr)\\
&= \sin x \left( 1 + \frac{\cos^2 x}{\sin^2 x}\right)\\
&= \sin x \left(\frac{\sin^2 x}{\sin ^2x} + \frac{\cos^2 x}{\sin^2x}\right)\\
&= \sin x\left(\frac{\sin^2 x + \cos^2x}{\sin^2x}\right)\\
&= \sin x\left(\frac{1}{\sin^2 x}\right).
\end{align*}$$
Note that I am not manipulating the identity as a whole. One has to be careful when doing that, because you can start with a false equality and end with a true one if not all your steps are reversible. Instead, we start with one side, and apply known identities to it until it becomes the other side.
(I am guessing you can take it home from where I left it, of course...)
| {
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"timestamp": "2023-03-29T00:00:00",
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Summing the series $ \sum_{n=1}^{\infty} \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}$ How could I find the sum of the series
$$ \sum_{n=1}^{\infty} \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}$$
With: $$ \int_0^{\infty} \frac{\mathrm dx}{n(1+x^3)^n}=\frac{2\pi\Gamma(n-1/3)}{\Gamma(2/3)3^{3/2}n!}$$
(Previous post)
?
| Since integrals are taken over positive measurable functions we can interchange integration and summation
$$
\sum_{n=1}^{\infty} \int_0^{\infty} \frac{dx}{n(1+x^3)^n}=
\int_0^{\infty} \sum_{n=1}^{\infty}\left(\frac{1}{n(1+x^3)^n}\right)dx
$$
Consider the following Taylor expansion
$$
\log(1-q)=-\sum\limits_{n=1}^\infty\frac{q^n}{n}\qquad-1<q<1
$$
With $q=(1+x^3)^{-1}$ we get
$$
\sum\limits_{n=1}^\infty\frac{1}{n(1+x^3)^n}=
-\log\left(1-\frac{1}{1+x^3}\right)=
\log\left(\frac{1+x^3}{x^3}\right)
$$
Hence
$$
\sum_{n=1}^{\infty} \int_0^{\infty} \frac{dx}{n(1+x^3)^n}=
\int_0^{\infty}\log\left(1+\frac{1}{x^3}\right)dx
$$
Let's proceed to calculation of the last integral. For the first we use integration by parts
$$
\int_0^{\infty}\log\left(1+\frac{1}{x^3}\right)dx=
x\log\left(1+\frac{1}{x^3}\right)\biggr|_0^\infty-
\int_0^{\infty}x\frac{d}{dx}\log\left(1+\frac{1}{x^3}\right)dx=
3\int_0^{\infty}\frac{1}{1+x^3}dx
$$
Now lets make substitution $u=x^{-1}$, then we get
$$
\int_0^{\infty}\frac{1}{1+x^3}dx=\int_0^{\infty}\frac{u}{1+u^3}dx
$$
Hence,
$$
\int_0^{\infty}\frac{1}{1+x^3}dx=
\frac{1}{2}\left(\int_0^{\infty}\frac{1}{1+x^3}dx+\int_0^{\infty}\frac{x}{1+x^3}dx\right)=
\frac{1}{2}\int_0^{\infty}\frac{1}{1-x+x^2}dx=
$$
$$
\frac{1}{2}\int_0^{\infty}\frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}d x=
\frac{1}{2}\frac{2}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)\Biggl|_0^\infty=\frac{2\pi}{3\sqrt{3}}
$$
Finally,
$$
\int_0^{\infty}\log\left(1+\frac{1}{x^3}\right)dx=3\int_0^{\infty}\frac{1}{1+x^3}dx=\frac{2\pi}{\sqrt{3}}
$$
| {
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Integral of $\int \frac {dx}{x^2 \sqrt{4-x^2}}$ I am trying to find $$\int \frac {dx}{x^2 \sqrt{4-x^2}}$$
I make $t=2\sin x$
$$\int \frac {dx}{x^2 \sqrt{4-4\sin^2 t}}$$
$$\int \frac {dx}{x^2 \sqrt{4(1-\sin^2 t)}}$$
$$\int \frac {dx}{x^2 \sqrt{4(\cos^2 t)}}$$
$$\int \frac {dx}{x^2 \cdot 2\cos t}$$
I do not really know where to go from here. I have two variables and that is really bad but I do not know how to write $x$ in terms of $t$.
| You must replace every occurrence of $x$. Let $x=2\sin t$. This is so that the square root of $4-x^2$ will be nice. Then $dx=2\cos t\,dt$, and $\sqrt{4-x^2}=2\cos t$. So we end up needing to find
$$\int \frac{2\cos t}{(4\sin^2 t)(2\cos t)}dt.$$
There is cancellation, and we end up needing to find $\int\frac{1}{4}\csc^2 t\,dt$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Parametric representation of rectangular form in terms of parameters $\rho$ & $\theta$
I need to represent the cone $z=\sqrt{3x^2+3y^2}$ parametrically in terms of $\rho$ and $\theta$ where $(\rho,\theta,\phi)$ are spherical coordinates.
Attempt. I tried using: $$x=\rho\sin\phi\cos\theta \\y=\rho\sin\phi\sin\theta\\z=\rho\cos\phi$$
and $$\rho^2=x^2+y^2+z^2\\\cos\phi=\frac{z}{\sqrt{x^2+y^2+z^2}}\\\rho^2\sin^2\phi=x^2+y^2$$
I cannot find a way to get rid of $\phi$. Hints please.
This is the graph of it. It is a cone.
| If you have $z=\sqrt{3x^2+3y^2}$ and $z^2=3x^2+3y^2$, making the appropriate replacements in $\cos\phi=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$ gives
$$\cos\phi=\dfrac{z}{\sqrt{\frac{z^2}{3}+z^2}}$$
Can you see what to do next? (Hint: your Cartesian equation guarantees that $z$ is always positive, so $\sqrt{z^2}=?$)
| {
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Plotting a quadratic equation in the $\,xy\,$- plane My question is:
Represent the following set of points in the $\,xy\,$- plane:
$$\left\{ (x,y)\,\, |\,\, x^2 + y^2 - 2x - 2y + 1 = 0 \right\}$$
What i got: $\,\,(x-2)^2 + (y-2)^2 = 1\,\,$
I am not getting what to do next.
Any help to solve this question would be greatly appreciated.
Thank you,
| When given an equation of the form $x^2-2x+y^2-2y+1=0$ the first step is to complete the square for $x$ and for $y$.
The idea is that if we have $x^2-2x$ we can write it as $(x+C)^2+D$ instead. Since know those that the coefficient of $x$ is $2C$, we know that $C=-1$, so we have: $$(x-1)^2=x^2-2x+1\implies x^2-2x = (x-1)^2-1$$
Therefore we can write it as $(x-1)^2-1$.
Similarly we can replace $y^2-2y$ by a similar term, so we have now:
$$\begin{align}
&\underbrace{x^2-2x}+\underline{y^2-2y}+1 = 0 &&\text{complete the squares}\\
&\underbrace{(x-1)^2-1}+\underline{(y-1)^2-1}+1 =0 &&\text{sum the }1\text{'s}\\
&(x-1)^2+(y-1)^2-1=0 &&+1\\
&(x-1)^2+(y-1)^2=1 &&\text{ circle!}
\end{align}$$
Therefore we have a circle of radius $1$ whose center is at $(1,1)$.
| {
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Multiplication in the field $F = \mathbb{Z}_2[x]/f(x)$ Let $f(x) = x^6 + x + 1$ and define the field $F = \mathbb{Z}_2[x]/f(x)$
Compute the following in this field:
1. $(x^5 + x + 1)(x^3 + x^2 +1)$
I start by multiplying (in $\mathbb{Z}_2[x]$):
$(x^5 + x + 1)(x^3 + x^2 +1)$ = $(x^8 + x^7 + x^5 +x^4 + x^2 + x + 1)$
Then dividing the result with $f(x)$:
$(x^2 + x)$ and the remainder $(x^5 + x^4 + x^3 + x^2 + 1)$
Is this the right approach for solving this problem? Do I understand it correct that I want the result of my multiplication mod $f(x)$? Can I think of it as a simple modulus calculation:
$11$ mod $7 = a$
$7*1 + 4$ mod $7 = 4$
In my case I have
$(x^6 + x +1)*(x^2 + x) + \mbox{remainder}$ mod $(x^6 +x + 1) = \mbox{remainder}$
So my answer to the question would be the remainder, $(x^5 + x^4 + x^3 + x^2 + 1)$?
2. $(x + 1)^{-1}$
I read (wiki) that the inverse to $(x + 1)$ could be found by using the extended euclidean alg. for $a = (x^6 + x +1)$, $b = (x+1)$ but I don't really get it since $a$ is irreducible.
Any hints in the right direction would be appreciated!
| Your idea for (1) is correct.
As for (2), it is the same idea as for the ring of integers modulo $n$. The $gcd$ of $x+1$ and $x^6+x+1$ will be $1$ since $x+1$ and $x^6+x+1$ are coprime. Using the extended Euclidean algorithm you'll get
$$
A(x+1) + B(x^6+x+1) = 1
$$
where $A, B \in \Bbb{F}_2[x]/f(x)$. So as you can see, modulo $x^6+x+1$ we have that $$
A(x+1) \equiv 1 \pmod{x^6+x+1}
$$
and hence $A$ is the inverse of $x+1$.
NOTE: remember that in $\Bbb{Z}_2$ (which I hope you don't mean the $p$-adic integers), $-1 \equiv 1 \pmod{2}$.
| {
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Sum of Cosine/Exponential I try to simplify to get rid of sum
$$ \sum_{k=0}^{n-1}\cos(2 \pi fk)$$
I discover I shall use euler equation to form:
$$ \sum_{k=0}^{n-1}\frac{1}{2}(e^{2 \pi fki}+e^{-2 \pi fki})$$
but how to sum exponentials?
| You could proceed using the fact that $\cos(k \theta) = \dfrac{\exp(i k \theta) + \exp(-i k \theta)}{2}$, where $\theta = 2 \pi f$, as others have suggested.
Here is another method. Let $I_n = \displaystyle \sum_{k=0}^{n-1} \cos(k \theta)$. Now multiply by $\sin \left( \dfrac{\theta}{2}\right)$. Hence,
\begin{align}
\sin \left( \dfrac{\theta}{2}\right) I_n & = \sin \left( \dfrac{\theta}{2}\right) \displaystyle \sum_{k=0}^{n-1} \cos(k \theta) = \displaystyle \sum_{k=0}^{n-1} \cos(k \theta) \sin \left( \dfrac{\theta}{2}\right)\\
& = \dfrac12 \left(\displaystyle \sum_{k=0}^{n-1} \left( \sin \left( \left(k+\dfrac12 \right) \theta \right) - \sin \left( \left(k- \dfrac12 \right) \theta \right) \right) \right)
\end{align}
Now telescopic cancellation gives us
$$\sin \left( \dfrac{\theta}{2}\right) I_n = \dfrac12 \left( \sin \left( \left( n - \dfrac12\right) \theta \right) + \sin \left( \dfrac{\theta}{2}\right)\right) = \sin \left( \left(\dfrac{n}{2} \right) \theta \right) \cos \left( \left(\dfrac{n-1}{2} \right)\theta\right)$$
Hence, $$I_n = \dfrac{\sin \left( \left(\dfrac{n}{2} \right) \theta \right) \cos \left( \left(\dfrac{n-1}{2} \right)\theta\right)}{\sin \left( \dfrac{\theta}{2}\right)}$$
The same idea works for $J_n = \displaystyle \sum_{k=0}^{n-1} \sin(k \theta)$ as well. Multiply by $\sin \left( \dfrac{\theta}{2}\right)$ and write each term as a difference of cosines to get
$$J_n = \dfrac{\sin \left( \left(\dfrac{n}{2} \right) \theta \right) \sin \left( \left(\dfrac{n-1}{2} \right)\theta\right)}{\sin \left( \dfrac{\theta}{2}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/157143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Hypergeometric functions inequality Let $_2F_1(a,b;c,z)$ be the (Gauss) hypergeometric function, and $m$ and $n$ positive integers.
From a simple plot it looks like
$_2F_1(m+n,1,m+1,\frac{m}{m+n})>\frac{m}{n} \,_2F_1(m+n,1,n+1,\frac{n}{m+n})$
when $m<n$, but how do I prove this?
Hope this is not too trivial, I am not very familiar with such functions.
| Using Euler's transformation of the Gauss's hypergeometric function:
$$
{}_2 F_1\left(a,b; c; z\right) = (1-z)^{c-a-b} \cdot {}_2F_1\left(c-a,c-b; c; z\right)
$$
for $b=1$, $a=m+n$ and $c=m+1$ we have a "closed form" expression:
$$ \begin{eqnarray}
{}_2 F_1\left(m+n,1; m+1; \frac{m}{m+n}\right) &=& \left(\frac{n}{n+m}\right)^{-n} {}_2F_1\left( 1-n, m; m+1; \frac{m}{m+n} \right) \\ &=& m \left(\frac{n+m}{n}\right)^{n} \cdot \sum_{k=0}^{n-1} \frac{1}{k+m} \frac{(1-n)_k}{k!} \left(\frac{m}{n+m}\right)^k \\
&=& m \left(\frac{n+m}{n}\right)^{n} \cdot \sum_{k=0}^{n-1} \frac{(-1)^k}{k+m} \binom{n-1}{k} \left(\frac{m}{n+m}\right)^k \\
&=& m \left(\frac{n+m}{n}\right)^{n} \cdot \left(\frac{n+m}{m}\right)^{m} \int_0^{\frac{m}{n+m}} \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x
\end{eqnarray}
$$
Thus:
$$
\frac{1}{m} {}_2 F_1\left(m+n,1; m+1; \frac{m}{m+n}\right) -
\frac{1}{n} {}_2 F_1\left(m+n,1; n+1; \frac{n}{m+n}\right) =
\frac{(n+m)^{n+m}}{n^n m^m} \left( \int_0^{\frac{m}{n+m}} \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x - \int_0^{\frac{n}{n+m}} \left(1-x\right)^{m-1} x^{n-1} \mathrm{d} x \right) =
\frac{(n+m)^{n+m}}{n^n m^m} \left( \int_0^{\frac{m}{n+m}} \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x - \int_{\frac{m}{n+m}}^1 \left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x \right) = \\
\frac{(n+m)^{n+m}}{n^n m^m} \int_0^1 \operatorname{sgn}\left({\frac{m}{n+m}}-x\right)\left(1-x\right)^{n-1} x^{m-1} \mathrm{d} x
$$
Interpreting $x$ as a beta random variable the last expression becomes:
$$
\frac{(n+m)^{n+m}}{\Gamma(n+m)} \frac{n^n}{\Gamma(n)} \frac{m^m}{\Gamma(m)} \mathbb{E}\left( \operatorname{sgn}\left(\mu-X\right) \right) =
\frac{(n+m)^{n+m}}{\Gamma(n+m)} \frac{n^n}{\Gamma(n)} \frac{m^m}{\Gamma(m)} \left( \mathbb{P}\left( X < \mu \right) - \mathbb{P}\left( X > \mu \right)\right)
$$
where $\mu = \mathbb{E}(X) = \frac{m}{m+n}$, where $X$ follows $\mathrm{Be}(m,n)$. Given that $m<n$, random variable $X$ is positively skewed, and the expression is positive.
| {
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"url": "https://math.stackexchange.com/questions/159297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| My hint:
$$\int \frac{1}{1+x^4}dx=\frac{1}{2}\left[\int \frac{1+x^2}{1+x^4}dx+\int\frac{1-x^2}{1+x^4}dx\right]=\frac{1}{2}\left[\int \frac{1}{\left(x-\frac{1}{x}\right)^2+2}d\left(x-\frac{1}{x}\right)+\int\frac{1}{\left(x+\frac{1}{x}\right)^2-2}d\left(x-\frac{1}{x}\right)\right]=\frac{1}{2}\left[\frac{1}{\sqrt{2}}\arctan\frac{x^2-1}{\sqrt{2}x}+\frac{}{2\sqrt{2}}\ln\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right]+Constant$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 20,
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To simplify $f_a(x)= \int_{-a}^{+a} e^ {-\frac{x}{t^2-a^2}}\;dt$ Let $x\leq0$, then $$ f_a(x)= \int_{-a}^{+a} e^ {-\frac{x}{t^2-a^2}}\;dt$$
$$ f'(x)= -\int_{-a}^{+a} \frac{1}{t^2-a^2} e^ {-\frac{x}{t^2-a^2}}dt$$
$$ f'(x)= -\int_{-a}^{+a} \frac{t^2-(t^2-a^2)}{a^2(t^2-a^2)} e^ {-\frac{x}{t^2-a^2}}dt$$
$$ f'(x)= -\int_{-a}^{+a} \frac{t^2}{a^2(t^2-a^2)} e^ {-\frac{x}{t^2-a^2}}dt + \int_{-a}^{+a} \frac{1}{a^2} e^ {-\frac{x}{t^2-a^2}}dt$$
$$ f'(x)= -\int_{-a}^{+a} \frac{t^2}{a^2(t^2-a^2)} e^ {-\frac{x}{t^2-a^2}}dt + \frac{f(x)}{a^2} $$
$$ f''(x)= \int_{-a}^{+a} \frac{t^2}{a^2(t^2-a^2)^2} e^ {-\frac{x}{t^2-a^2}}dt + \frac{f'(x)}{a^2} $$
$$ f''(x)= \frac{1 }{2xa^2}\int_{-a}^{+a} t\frac{2xt }{(t^2-a^2)^2} e^ {-\frac{x}{t^2-a^2}}dt + \frac{f'(x)}{a^2} $$
$$ f''(x)= \frac{1 }{2xa^2}(te^ {-\frac{x}{t^2-a^2}}|{_{-a}^{+a}})-\frac{1 }{2xa^2}\int_{-a}^{+a} e^ {-\frac{x}{t^2-a^2}}dt + \frac{f'(x)}{a^2} $$
$$ f''(x)= -\frac{f(x) }{2xa^2} + \frac{f'(x)}{a^2} $$
$$ a^2f''(x) - f'(x)+\frac{f(x) }{2x}=0 $$
wolframalpha gave me that long result :
Can you please help me with some technics to simplify $f_a(x)$ ?
Thanks a lot for any help.
| How about
$$
f_a(x) = \frac{\operatorname{e} ^{x/(2a^2)} x}{a} \biggl(\mathrm K_0 \biggl(\frac{-x}{2 a^{2}}\biggr) - \mathrm K_1 \biggl(\frac{-x}{2 a^{2}}\biggr)\biggr),\qquad x<0
$$
in terms of Bessel functions.
added
It should work like this: The function $f$ listed on the right-hand-side satisfies the differential equation. And the "initial conditions"
$$
f(0) = 2a,\qquad
f'(x) \sim \frac{-1}{a}\,\log\left(-x\right)\qquad\text{as } x \uparrow 0
$$
match $f_a$, therefore this one is equal to $f_a$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all integer solutions for $x^2 - 2y^2 =2 $ I'd love your help with finding all the integer solutions to the following equation:
$x^2 - 2y^2 =2 $. I want to use Pell's theorem so I changed the equation to $-\frac{1}{2}x^2+ y^2 =-1$, Can I use Pell's Theorem now? I got a private solution for $-\frac{1}{2}x^2+ y^2 =1$ $y=3, x=4$, so form Pell I get that $\alpha= (4+3\sqrt{2})^n$ for every integer $n$, and a private solution for $-\frac{1}{2}x^2+ y^2 =-1$ is $y=1, x=2$, so the total solution is $\alpha= (1+\sqrt{2}) \cdot (+/- (4+3\sqrt{2})^n)$. Are all these steps correct? and if not- how should I solve this one?
Thank you!
| Another way is to note that $x$ has to be even. Hence if $x=2x_1$, we get that
$$2x_1^2-y^2 = 1 \implies y^2 - 2x_1^2 = -1 \implies \left(y + \sqrt2 x_1\right)\left(y - \sqrt2 x_1\right) = -1$$
Now clearly, one solution is $(x_1,y) = (1,1)$.
Raise both sides to any odd power, i.e.,
$$\left(y + \sqrt2 x_1\right)^{2n-1}\left(y - \sqrt2 x_1\right)^{2n-1} = (-1)^{2n-1} = -1$$
and note that
$$\left(y + \sqrt2 x_1\right)^{2n-1} = Y_n(y,x_1) + \sqrt2 X_n(y,x_1)$$
$$\left(y - \sqrt2 x_1\right)^{2n-1} = Y_n(y,x_1) - \sqrt2 X_n(y,x_1)$$
which in turn gives us infinite other solutions since
$$Y_n^2 -2 X_n^2 = -1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\lim \limits_{y\rightarrow\infty}\left (\ln^2y\,-2\int_{0}^y\frac{\ln x}{\sqrt{x^2+1}}dx\right)$ I have difficulty with this limit. Where to start?
$$\lim_{y\rightarrow\infty}\left (\ln^2y\,-2\int_{0}^y\frac{\ln x}{\sqrt{x^2+1}}dx\right)$$
| By simple integration by parts, we have
$$ \int_{0}^{y} \frac{\log x}{\sqrt{1+x^2}} \; dx = \log y \, \sinh^{-1}y - \int_{0}^{y} \frac{\sinh^{-1}x}{x} \; dx. $$
Now by the substitution
$$x = \frac{u^2-1}{2u} \quad \Longleftrightarrow \quad u = x + \sqrt{x^2+1},$$
and the easy equality $ \sinh^{-1} y = \log \left( y + \sqrt{y^2+1} \right)$, we have
$$ \int_{0}^{y} \frac{\sinh^{-1}x}{x} \; dx = -\int_{1}^{y + \sqrt{y^2+1}} \left( \frac{2u}{1-u^2} + \frac{1}{u} \right) \log u \; du = -F\bigg(y+\sqrt{y^2+1}\bigg),$$
where
$$F(s) := \int_{1}^{s} \left( \frac{2u}{1-u^2} + \frac{1}{u} \right) \log u \; du.$$
Now simple observation shows that for $s > 0$ we have
$$F\left( \frac{1}{s} \right) = -F(s).$$
Since $y + \sqrt{y^2+1} \gg 1$ whenever $y \gg 1$, in view of the identity above, we may calculate $F(s)$ for $s = \left( y + \sqrt{y^2+1} \right)^{-1} \in (0, 1)$ instead since
$$ F\left(y+\sqrt{y^2+1}\right) = -F\left(\frac{1}{y+\sqrt{y^2+1}}\right) = -F(s) .$$
Now we introduce the dilogarithm function, defined by
$$ \mathrm{Li}_{2} (x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2} = -\int_{0}^{x} \frac{\log(1-t)}{t} \; dt.$$
Then
$$ \begin{align*}
F(s)
&= \frac{1}{2} \int_{1}^{s} \frac{\log(u^2)}{1-u^2} \; (2udu) + \int_{1}^{s} \frac{\log u}{u} \; du \\
&= \frac{1}{2} \int_{1}^{s^2} \frac{\log v}{1-v} \; dv + \frac{1}{2} \log^2 s \qquad (v = u^2) \\
&= - \frac{1}{2} \int_{0}^{1-s^2} \frac{\log (1-w)}{w} \; dw + \frac{1}{2} \log^2 s \qquad (w = 1-v) \\
&= \frac{1}{2} \mathrm{Li}_{2}(1-s^2) + \frac{1}{2} \log^2 s. \qquad (w = 1-v)
\end{align*}$$
Thus plugging back, we have
$$ \int_{0}^{y} \frac{\sinh^{-1}x}{x} \; dx = \frac{1}{2} \left[ \mathrm{Li}_{2} \left( \frac{2y}{y+\sqrt{y^2+1}}\right) + \log^2 \left(y+\sqrt{y^2+1}\right) \right].$$
This shows that
$$ \begin{align*}
\log^2 y - 2\int_{0}^{y} \frac{\log x}{\sqrt{x^2+1}}\;dx
&= \log^2 y - 2 \log y \, \log \left( y + \sqrt{y^2+1} \right) \\
& \qquad + \mathrm{Li}_{2} \left( \frac{2y}{y+\sqrt{y^2+1}}\right) + \log^2 \left(y+\sqrt{y^2+1}\right) \\
&= \mathrm{Li}_{2} \left( \frac{2y}{y+\sqrt{y^2+1}}\right) + \log^2 \left(1+\sqrt{1+y^{-2}}\right),
\end{align*}$$
which clearly converges to
$$ \mathrm{Li}_{2}(1) + \log^2 2 = \zeta(2) + \log^2 2 = \frac{\pi^2}{6} + \log^2 2.$$
Numerical experiment shows that this converges to its limit relatively fast.
In fact, reflection formula for dilagorithm gives the following estimate.
$$ \log^2 y - 2\int_{0}^{y} \frac{\log x}{\sqrt{x^2+1}}\;dx = \frac{\pi^2}{6} + \log^2 2 - \frac{\log y}{2y^2} + O\left(\frac{1}{y^2}\right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Count the number of solutions of the inequality $x + y + z \leq N$
Problem
Given $A, B, C $ and $N$. How many integer solutions are there of the following inequality:
$$x + y + z \leq N$$
where $0 \leq x \leq A, 0 \leq y \leq B, 0 \leq z \leq C$?
When $A + B + C \leq N$, the solution is obvious $(A + 1) \cdot (B + 1) \cdot (C + 1)$
The general formula for the number of solutions of the form $x + y + z = N$, for $x, y , z \geq 0$ is $\dbinom{N + r -1}{r - 1}$. Unfortunately, I couldn't find a way to bring this idea into place. Any suggestion?
| The main trick is to add a fourth variable to take up the slack between $x+y+z$ and $N$. You know that there are $\binom{N+3}3$ solutions to $w+x+y+z=N$ in non-negative integers, so there are also $\binom{N+3}3$ solutions to $x+y+z\le N$ in non-negative integers. Now we have to throw out the solutions that exceed one of the caps $A,B$, and $C$.
Counting non-negative solutions to $w+x+y+z=N$ in which $x>A$ is the same as counting non-negative solutions to $w+x+y+z=N-A-1$, of which there are $$\binom{N-A-1+3}3=\binom{N-A+2}3\;.$$ Similarly, there are $\binom{N-B+2}3$ non-negative solutions to $w+x+y+z=N$ in which $y>B$ and $\binom{N-C+2}3$ in which $z>C$. Thus, a better approximation to the desired number is
$$\binom{N+3}3-\binom{N-A+2}3-\binom{N-B+2}3-\binom{N-C+2}3\;.\tag{1}$$
However, it’s possible that a non-negative solution to $w+x+y+z=N$ exceeds two caps, and $(1)$ overcorrects for those unwanted solutions by subtracting each of them twice. The same reasoning used to calculate the correction terms above shows that there are $$\binom{N-A-B+1}3$$ solutions exceeding the $A$ and $B$ caps, $$\binom{N-A-C+1}3$$ exceeding the $A$ and $C$ caps, and $$\binom{N-B-C+1}3$$ exceeding the $B$ and $C$ caps. Thus, we must add to the expression in $(1)$ the sum
$$\binom{N-A-B+1}3+\binom{N-A-C+1}3+\binom{N-B-C+1}3\;.\tag{2}$$
Finally, we have to subtract the number of solutions that exceed all three caps, which is $$\binom{N-A-B-C}3\;.\tag{3}$$
The final count is $(1)+(2)-(3)$, where the parenthesized numbers refer to the numbered expression above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of a succession of additions $$1^2+2^2+3^2+...+10000$$
How do you find the exact value of that?
I'm studying induction, and I'm still not sure how to get that value.
| The sum of squares formula is
$\sum_{i= 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$
Clearly it holds for $n = 1$.
Suppose it held for $n$. That is
$\sum_{i= 1}^n i^2 = \frac{n(n + 1)(2n + 1)}{6}$
Then $\sum_{i = 1}^{n + 1} i^2 = \sum_{i = 1}^n i^2 + (n + 1)^2 = \frac{n(n + 1)(2n + 1)}{6} + (n + 1)^2 = \frac{2n^3 + 2n + n}{6} + \frac{6(n^2 + 2n + 1)}{6}$
$=\frac{2n^3 + 9n^2 + 13n + 6}{6} = \frac{(n + 1)((n + 1) + 1)(2(n + 1) + 1)}{6}$
Thus by induction the formula holds.
Your series above is $\sum_{i = 1}^{100} i^2 = \frac{(100)(101)(201)}{6}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing
Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.
I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$
I am trying to go ahead of this step.
| We have that
$$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n>0 \iff \ln \left(1+\frac{1}{n+1}\right)^{(n+1)} - \ln \left(1+\frac{1}{n}\right)^n>0$$
and
$$(n+1) \ln \left(1+\frac{1}{n+1}\right) - n \ln \left(1+\frac{1}{n}\right)
\ge(n+1)\left[ \ln \left(1+\frac{1}{n+1}\right) - \ln \left(1+\frac{1}{n}\right)\right]=$$
$$=(n+1)\left[ \ln \left(\frac{n+2}{n+1}\right) - \ln \left(\frac{n+1}{n}\right)\right]=$$
$$=(n+1) \ln \left(\frac{(n+2)(n+1)}{(n+1)^2}\right)=(n+1) \ln \left(\frac{n^2+2n+2}{n^2+2n+1}\right) >0$$
since $\log x>0$ for $x>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Pythagorian quadruples From my work on hyperelliptic equations I found how to get infinitely many solutions of the equation $a^4+b^4+c^2=d^4$. I call these solutions harmonic:
$$\begin{array}{rcccccl}
1^4 &+& 2^4 &+& 8^2 &=& 3^4\\
2^4 &+& 3^4 &+& 48^2 &=& 7^4\\
3^4 &+& 4^4 &+& 168^2 &=& 13^4\\
4^4 &+& 5^4 &+& 440^2 &=& 21^4
\end{array}$$
and so on. All numbers natural.
Does anyone know if there are infinite non harmonic solutions of this equation?
| EDIT
There seems to be a lot of non-harmonic solutions.
Here are couple of one parameter family of non-harmonic solutions.
$b = a(a-1)$, $d = a^2 - a + 1$ and $c = (a-1)(2a^2 - a +1)$ which relies on the identity $$a^4 + \left( a(a-1)\right)^4 + \left((a-1)(2a^2-a+1) \right)^2 = (a^2 - a + 1)^2$$ You could scale these up appropriately i.e. $$\left(ka,ka(a-1),k^2(a-1)(2a^2 - a + 1),k \left(a^2-a+1 \right) \right),$$ to get other solutions.
$b = a(a+1), d = a^2 + a + 1$ and $c = (a+1)(2a^2+a+1)$ which relies on the identity $$a^4 + \left( a(a+1)\right)^4 + \left((a+1)(2a^2+a+1) \right)^2 = (a^2 + a + 1)^2$$ You could again scale these up appropriately i.e. $$\left(ka,ka(a+1),k^2(a+1)(2a^2 + a + 1),k \left(a^2+a+1 \right) \right),$$ to get other solutions.
You could also take your harmonic solution $(a,a+1,a(a+1)(a^2 + a + 2),a^2+a+1)$ and scale appropriately, i.e. $$\left(ka,k(a+1),k^2a(a+1)(a^2 + a + 2),k \left(a^2+a+1 \right) \right),$$ to get other solutions.
Yes. Below is a one parameter family of infinite solutions.
$b = a+1$, $d = a^2+a+1$, $c = (a^2+a+1)^2-1$.
$$b^4 = (a+1)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1$$
$$d^4 = (a^2 + a +1)^4 = 1+4 a+10 a^2+16 a^3+19 a^4+16 a^5+10 a^6+4 a^7+a^8$$
Hence,
\begin{align}
d^4 - b^4 - a^4 & = 4 a^2+12 a^3+17 a^4+16 a^5+10 a^6+4 a^7+a^8\\
& = a^2 \left(4 +12 a+17 a^2+16 a^3+10 a^4+4 a^5+a^6 \right)\\
& = a^2 (a+1)^2 (a^2 + a + 2)^2
\end{align}
Hence, choose $c = a(a+1)(a^2+a+2) = (a^2 + a + 1 -1)(a^2 + a + 1 +1) = \left( \left(a^2 + a + 1 \right)^2 -1 \right)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the area enclosed by the curve $r=2+3\cos \theta$. the question is
Find the area enclosed by the curve:
$r=2+3\cos \theta$
Here's my steps:
since when $r=0$, $\cos \theta=0$ or $\cos\theta =\arccos(-2/3)$.
so the area of enclosed by the curve is 2*(the area bounded by $\theta=\arccos(-2/3)$ and $\theta=0$)
the answer on my book is $5\sqrt{5}+（17/2）*\arccos(-2/3)$
I have no idea why there is a $5\sqrt{5}$ since $\arccos(-2/3)=2.300523984$ on my calculator.
| Area of curve, $$A=2\int_0^{\arccos(-2/3)}\frac{r^2}{2}d\phi$$ $$\implies A=2\int_0^{\arccos(-2/3)} \frac{(2+3\cos\phi)^2}{2}d\phi$$ $$=2\int_0^{\arccos(-2/3)} \frac{4+9\cos^2\phi+12\cos\phi }{2}d\phi$$ $$=\int_0^{\arccos(-2/3)} (4+9\cos^2\phi+12\cos\phi) d\phi$$ $$=3\sqrt{5}+\frac{17}{2}\cos^{-1}(\frac{-2}{3}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Integral of $e^{(x-x^3)/3n}$ from $0$ to $\infty$ How can you compute the following integral assuming $n>0$?
$$\int_{x=0}^{\infty}e^{\frac{x -x^3}{3n}}dx $$
Mathematica etc. fail to produce anything useful.
EDIT: I would be happy with an asymptotic result in $n$ if it is too hard to compute exactly. I don't know if it helps but
$$\int_{x=0}^{\infty}e^{\frac{-x^3}{3n}}dx \approx 1.29 n^{1/3}.$$
| I guess Maple has one of the useless answers:
$$\int_{0}^{\infty} \operatorname{e} ^{\frac{-x (-1 + x^{2})}{3 n}} d x = \frac{i}{9} \pi \mathrm{BesselY} \biggl(\frac{1}{3},\frac{2 i \sqrt{3}}{27 n}\biggr) \sqrt{3} - \\
\quad{}\quad{}\frac{i}{9} \pi \mathrm{AngerJ} \biggl(\frac{1}{3},\frac{2 i \sqrt{3}}{27 n}\biggr) + \frac{2 i}{3} \mathrm{BesselK} \biggl(\frac{1}{3},\frac{-2\sqrt{3}}{27 n}\biggr) + \\
\quad{}\quad{}\frac{i}{9} \pi \mathrm{BesselJ} \biggl(\frac{1}{3},\frac{2 i \sqrt{3}}{27 n}\biggr) + \frac{i}{9} \sqrt{3} \pi \mathrm{WeberE} \biggl(\frac{1}{3},\frac{2 i \sqrt{3}}{27 n}\biggr)
$$
Maple reports asymptotics for this as
$$
\frac{2 \sqrt[3]{n} \pi 3^{\frac{5}{6}}}{9 \Gamma \Bigl(\frac{2}{3}\Bigr)}
$$
which differs by factor $3^{5/6}$ from Raymond's answer. Perhaps this is more accurate??
More terms
$$
\frac{2 \sqrt[3]{n} \pi 3^{\frac{5}{6}}}{9 \Gamma \Bigl(\frac{2}{3}\Bigr)}
+ \frac{\Gamma \Bigl(\frac{2}{3}\Bigr) 3^{\frac{2}{3}}}{9 \sqrt[3]{n}}
- \frac{1}{36 n}
+ O(n^{-5/3})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Determine $a$ values allowing $x^2+ax+2$ to be divided by $x-3$ in $\mathbb Z_5$ Determine for which $a$ values $f = x^2+ax+2$ can be divided by $g= x-3$ in $\mathbb Z_5$.
I don't know if there are more effective (and certainly right) ways to solve this problem, I assume there definitely are, but as I am not aware of them, I thought I could proceed like this: I have divided $f$ by $g$, pretending $a$ to be a constant in $\mathbb Q$, the resulting quotient is $x+(a+3)$, the reminder is $2+3(a+3)$. In order to have an exact division it needs to happen:
$$\begin{aligned} 2+3(a+3) = 0 \end{aligned}$$
$$\begin{aligned} 2+3a + 4 = 0 \end{aligned}$$
$$\begin{aligned} 3a + 1 = 0 \Rightarrow 3a=-1 \Rightarrow 3a = 4 \Rightarrow a= \frac{4}{3}=3 \end{aligned}$$
now I would expect $x^2+3x+2 = (x+1)(x-3)$, but it isn't the case because $(x+1)(x-3) = x^2-2x-3$. Is my way to solve this exercise totally wrong and it would be better if I'd set my notebook on fire (and in this case please feel free to jump in) or I am just doing some calculation wrong?
| $x-a$ is the factor of polynimial $f(x)$ iff $f(a)=0\implies (x-3)$ divides $f(x)=x^2+ax+2$ iff $f(3)=11+3a\equiv 0\pmod 5\implies 3a\equiv-11\pmod 5\implies 9\pmod 5\implies a\equiv3\pmod 5$. In your solution you are getting $(x+1)(x-3)=x^2-2x-3\equiv x^2+3x+2\pmod 5$ as $-2\equiv 3\pmod 5$ and $-3\equiv 2\pmod 5$. so your solution is fine.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Write each expression in the form $ca^pb^q$ Write each expression in the form $ca^pb^q$
c) $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$
\begin{align*}
&= \frac{a\left(\frac{2}{b}\right)}{1}*\frac{\left(\frac{a}{3}\right)}{3}=\dfrac{a^2\left(\frac{2}{b}\right)}{3}=\frac{a^2}{1}*\frac{2}{b}*\frac{1}{3}=\frac{2a^2}{3b}*\frac{b}{1}=\frac{2a^2b}{3}=\frac{2}{3}a^2b^1
\end{align*}
e) $\dfrac{a^{-1}}{(b^{-1})\sqrt{a}}$
\begin{align*}
&= \frac{1}{(b^{-1})a\sqrt{a}}=\frac{1b}{1a^1a^{\frac{1}{2}}}=\frac{1b^1}{1a^{\frac{2}{3}}}=1a^{\frac{-2}{3}}b^1
\end{align*}
These are my steps. Any corrections help.
| If you remember that dividing by a fraction is the same as multiplying by its inverse, we get at once:
$$c)\,\,\frac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}=\frac{2a}{b}\frac{a}{3}=\frac{2a^2}{3b}=\frac{2}{3}a^2b^{-1}$$
$$(e)\,\,\frac{a^{-1}}{b^{-1}a^{1/2}}=a^{-1-1/2}\,b=a^{-3/2}\,b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\int^1_0 \frac{dx}{x^2+x+1}$ Here is an integral which I don't know the answer:
$$\int^1_0 \frac{dx}{x^2+x+1}$$
I tried to use complex number to solve i.e. the root of $x^2+x+1$ is $(-1/2 + \sqrt {3}i/2)$.
Let w=$(-1/2 + \sqrt {3}i/2)$ , then it becomes $ \int_{0}^{1} 1/(x-w)^2\,\mathrm{d} (x-w)$ , the answer is in terms of complex number. I wanna ask whether my method is correct and seek for another method. Thank you.
| \begin{eqnarray}
\int_{0}^{1} \dfrac{1}{x^2+x+1} dx &=& \int_{0}^{1} \dfrac{1}{(x+1/2)^2 +3/4} dx \\
&=& \int_{1/2}^{3/2} \dfrac{1}{y^2 +3/4}dy \\
&=& \dfrac{4}{3}\int_{1/2}^{3/2} \dfrac{1}{(\dfrac{2}{\sqrt{3}}y)^2 + 1} dy \\
&=& \dfrac{\sqrt{3}}{2} \dfrac{4}{3} \int_{\sqrt{3}/3}^{\sqrt{3}} \dfrac{1}{z^2+1} dz\\
&=& 2 \dfrac{\sqrt{3}}{3} (\arctan(\sqrt{3}) - \arctan(\sqrt{3}/3))\\
&=&2 \dfrac{\sqrt{3}}{3}( \dfrac{\pi}{3} - \dfrac{\pi}{6})\\
&=& \dfrac{\sqrt{3}\pi}{9}\\
&=& 0,6045997880781
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
How to solve this equation involving $()^x$? I have the equation:
$\left (\sqrt{3+2\sqrt{2}} \right )^x- \left (\sqrt{3-2\sqrt{2}} \right )^x=\frac{3}{2}$
I wrote the left side of the equation as square roots.
$(1+\sqrt{2})^x-(1-\sqrt{2})^x=\frac{3}{2}$
How do I found out the final solution? Thank you very much!
P.S. The answers I can choose from are:
a) $x=1$
b) $x=2$
c) $x=\frac{2\lg2}{\lg(3+2\sqrt2)}$
d) $x=\frac{2\lg2}{\lg(3-2\sqrt2)}$
e) no solution
f) $x=2\lg2$
| $$(\sqrt{3+2\sqrt{2}})^x-(\sqrt{3-2\sqrt{2}})^x=\frac{3}{2}$$
Multiply both side with $(\sqrt{3+2\sqrt{2}})^x$, then you get
$$(\sqrt{3+2\sqrt{2}})^{2x}-(\sqrt{(3-2\sqrt{2})(3+2\sqrt{2})})^x=\frac{3}{2}(\sqrt{3+2\sqrt{2}})^x$$
$$(\sqrt{3+2\sqrt{2}})^{2x}-(1)^x=\frac{3}{2}(\sqrt{3+2\sqrt{2}})^x$$
Let $y =(\sqrt{3+2\sqrt{2}})^x$, then
$y^2 -1 = \frac{3}{2}y$
Solve above equation, $y =2$ or $y = \frac{-1}{2}$(its never true)
Therefore, $y =2$ that implies $(\sqrt{3+2\sqrt{2}})^x =2$
Apply $log$ both side,$$x\log(\sqrt{3+2\sqrt{2}}) =log(2)$$
Therefore, $$x=\frac {2\log(2)}{\log({3+2\sqrt{2}})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Find center, radius and a tangent to $x^2+y^2+6x-4y+3=0$
For the circle $x^2+y^2+6x-4y+3=0$ find
a) The center and radius
b) The equation of the tangent line at the point $(-2,5)$
Now, I solved a) and got the equation
$$(x+3)^2+(y-2)^2=10$$ with center $=(-3,2)$ and radius $=\sqrt{10}$
Now, I've never learned about the tangent of a circle, but I think that it's a line that touches the outer end of a circle. But I'm not 100% on that. So if anyone can help me out with this, that would be very beneficial. And please do not solve this for me.
| Let us complete the square. We get $x^2+6*x+9+y^2-4*y+4-9-4+3=0$ or $(x+3)^2+(y-2)^2=10$. Your circle is with center $O(-3,2)$ and radius $r=\sqrt{10}$
For the second question, let us calculate slope, which would be $\text{rise}/\text{run}$ or in our case center is $(-3,2)$ so slope=$(5-2)/(-2+3)=3$ slope of tangent line would be $-1/3$ put this conditions into form of tangent line $y=mx+b$ we get $5=-1/3\cdot(-2)+b$ or $b=5-2/3=13/3$
We get equation of tangent line
$y=-1/3\cdot x+13/3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Inequality problems I have Maths test tomorrow and was just doing my revision when I came across these two questions. Would anyone please give me a nudge in the right direction?
$1)$ If $x$ is real and $$y=\frac{x^2+4x-17}{2(x-3)},$$ show that $|y-5|\geq2$
$2)$ If $a>0$, $b>0$, prove that $$\left(a+\frac1b\right)\left(2b+\frac1{2a}\right)\ge\frac92$$
| For the first problem:
Write it as
$$
\begin{eqnarray}
\left(y-5\right)^2-4&=&\left(\frac{x^2+4x-17}{2(x-3)}-5\right)^2-4\\
&=&\left(\frac{x^2+4x-17-10x+30}{2(x-3)}\right)^2-4\\
&=&\left(\frac{x^2-6x+13}{2(x-3)}\right)^2-4\\
&=&\frac{(x^2-6x+13)^2 - 16(x-3)^2}{4(x-3)^2}\\
&=&\frac{169-156 x+62 x^2-12 x^3+x^4 - 16x^2+96x-144}{4(x-3)^2}\\
&=&\frac{x^4 -12x^3+46x^2 -60x+25}{4(x-3)^2}\\
&=&\frac{(x^2 -6x+5)^2}{4(x-3)^2}\\
&=&\frac{(x-5)^2(x-1)^2}{4(x-3)^2}\ge 0
\end{eqnarray}
$$
So only squares show up, hence it's positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding a pair of elements to satisfy an inequation Let $F$ be a field of characteristic 2 with more than 2 elements. Show that there are elements $a$ and $b$ in $F$ such that $(a+b)^3 \not= a^3 + b^3$.
$F$ couldn't possibly have less than 2 elements, and if it had exactly 2 — that is, $F = \mathbb Z_2$ —, $(a+b)^3$ would actually always be $a^3+b^3$. With $\#F>2$, how do I find $a$ and $b$ to violate that?
All I could do so far is simplify the inequation, using the characteristic and the commutative prroperty:
$$\begin{align}
(a+b)^3 &\not= a^3 + b^3\\
a^3 + 3a^2b + 3ab^2 + b^3 &\not= a^3 + b^3\\
a^3 + a^2b + ab^2 + b^3 &\not= a^3 + b^3\\
a^2b + ab^2 &\not= 0\\
ab(a + b) &\not= 0\quad.
\end{align}$$
| In a field, product is $0$ iff atleast one of the element in product is $0$. Here, if $a\neq0,b\neq 0$ and $a\neq -b$ then , $ab(a+b)\neq 0$, so choose non-zero elements of field $\Bbb F$ such that one is not the additive inverse of other one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ How would I verify the following double angle identity.
$$
\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B
$$
So far I have done this.
$$
(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)
$$But I am not sure how to proceed.
| Your question involves the basic algebra identity which says, $(a + b)(a - b) = a^2 - b^2 $. For targeting your question, it is easy to assume $ a = \sin A\cos B $ and $b = \cos A \sin B$. The process becomes easy now.
$$\begin{align}(a + b)(a - b)& =& a^2 - b^2\\
& = & (\sin A \cos B)^2 - (\cos A \sin B)^2\\ & = & \sin^2A\cos^2 B - \cos^2A\sin^2B
\\ & = & \sin^2A(1 - \sin^2 B) - \cos^2 A\sin^2B
\end{align} $$Proceed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 5
} |
Confusing double angle identity How would I solve the following double angle identity.
$$\cos^4x=\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$$
So far my work is
$$\frac{3}{8}+\frac{2\cos^x-1}{2}+\frac{1}{8}(2\cos^2x-1)$$
But how would I proceed.
| Notice that
\begin{eqnarray}
\cos(2x)&=& \cos^2 x - \sin^2 x \\
&=& 2 \cos^2 x - 1.\\
\end{eqnarray}
Then
\begin{equation}
\cos^2 x = \dfrac{1}{2}(1+\cos(2x)).
\end{equation}
Hence,
\begin{eqnarray}
\cos^4 x &=& (\cos^2 x)^2\\
&=& \left[\dfrac{1}{2}(1 + \cos(2x))\right]^2\\
&=& \dfrac{1}{4}(1 +2 \cos(2x)+ \cos^2(2x))\\
&=& \dfrac{1}{4} +\dfrac{1}{2} \cos(2x) + \dfrac{1}{4}\dfrac{1}{2}(1+\cos(4x))\\
&=& 3/8 + 1/2 \cos(2x) +1/8 \cos(4x)
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.