Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Integral $\int \frac{2x^2-1}{\pm x(2x^2-1)+2x+ \sqrt{1+2x^2}} d x$ Consider the indefinite integral
$$F_{\pm}(x):=\int \frac{2x^2-1}{\pm x(2x^2-1)+2x+ \sqrt{1+2x^2}} \mathbb{d} x.$$
For all $x\in \mathbb{R}$ it is
$$ F_{+}(x)= \log(-x + \sqrt{1 + 2 x^2}).$$
But what is $F_{-}$?
Note that numerical studies suggest that it does exist in a certain range. Math software already struggles to prove the formula for $F_{+}$.
| The problem is $$F_\pm(x)=\int\frac{2x^2-1}{\pm x(2x^2-1)+2x+\sqrt{1+2x^2}}\,\mathrm{d}x.$$ Let $x=2t/(2-t^2)$, then $$\begin{aligned}F_+(t)&=2\int\bigg(\frac{t}{2-t^2}+\frac{1+t}{t^2+2t+2}\bigg)\,\mathrm{d}t,\\F_-(t)&=2\int\bigg(\frac{t}{t^2-2}-\frac{t^3-t^2-6t-2}{t^4-2t^3-12t^2-4t+4}\bigg)\,\mathrm{d}t.\end{aligned}$$ You can integrate $F_+$ directly: $$F_+(t)=-\log(2-t^2)+\log(t^2+2t+2)+C.$$
For $F_-$ you need only solve a quadratic equation to factor the denominator and then integrate directly: $$\begin{aligned}F_-(t)=\log(2-t^2)&+\frac{-17+\sqrt{17}}{34}\log(t^2+(\sqrt{17}-1)t+2)\\&+\frac{-17-\sqrt{17}}{34}\log(-t^2+(\sqrt{17}+1)t-2)+C.\end{aligned}$$
After this, solve a quadratic equation to get a formula for $F_\pm(x)$.
| {
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Values of $a$ such that $x^5-x-a$ has quadratic factor I would like to find all integers $a$ such that $x^5-x-a$ has a quadratic factor in $\mathbb{Z}[x]$.
My Attempt
Let $x^5-x-a=(x^2+bx+c)(x^3+dx^2+ex+f)$, so that we have the following:
$$\begin{array}{rcl}
b+d&=&0\\
e+bd+c&=&0\\
f+be+cd&=&0\\
bf+ce&=&-1\\
cf&=&-a
\end{array}$$
Hence:
$$\begin{array}{rcccl}
d&=&-b\\
e&=&-bd-c&=&b^2-c\\
f&=&-be-cd&=&-b^3+2bc
\end{array}$$
and we have:
$$1=-bf-ce=b^4-3b^2c+c^2,$$
so that:
$$(2c-3b^2)^2=5b^4+4.$$
Question
How can I find all values of $n$ such that $5n^4+4$ is a perfect square?
My Attempt
If $m^2=5n^4+4$, then $m^2-5n^4=4$.
If $m=2m_*$, then $n$ is even, so that $n=2n_*$, and we have the equation $m_*^2-20n_*^4=1$. By Pell equation, since $(a,b)=(9,2)$ is the least non-trivial solution of $a^2-20b^2=1$, then the general solution has the form $(a_n,b_n)$ where $a_n+b_n\sqrt{20}=(9+2\sqrt{20})^n$, but I do not know how to find out what values of $n$ make $b_n$ a square.
| Another way is to do a long division of $x ^ 5-x-a$ by the arbitrary trinomial $x ^ 2 + bx + c$ and set the remainder to zero. This gives the remainder
$$(c ^ 2-3b ^ 2c + b ^ 4-1) x + (cb ^ 3-2bc ^ 2-a) = 0$$ from where we have
$a = cb (b ^ 2-2c)$ and $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$.
This means that for every solution of $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$ we have a corresponding value $a = cb (b ^ 2-2c)$.
Some solutions of $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$ are $(b,c)=(1,3),(0,1),(12,55),(12,377)$.
EXAMPLES.-$(b,c)=(1,3)$ gives $a=-15$ and we have
$$x^5-x+15=(x^2+x+3)(x^3-x^2-2x+5)$$
$(b,c)=(12,377)$ gives $a=-2759640$ and we have
$$x^5-x+2759640=(x^2+12x+377)(x^3-12x^2-233x+7320)$$
| {
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We know that $x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$. Prove ${x_n}$ is convergent Suppose $a>0, x_0>0$ and $x_{n+1}=\frac{1}{3}\left(2x_n+\frac{a}{x_n^2}\right)$. How to prove ${x_n}$ is convergent? Can someone helps me?
I tried to prove that $x_n$ is monotonically decreasing and have a lower bound: $a^\frac{1}{3}$.
| Easy to see that all terms of $(x_n)$ are positive.
Now by AM-GM, we have the lower bound:
$$x_{n+1} = \frac 13\left(x_n + x_n + \frac{a}{x_n^2}\right)\ge \sqrt[3] a, \quad (\forall n \in \mathbb N) \tag{1}$$
Also, from the given formula we have
$$x_{n+1}-x_n = -\frac{1}{3}x_n + \frac {a}{3x_n^2} = \frac{-x^3_n + a}{3x_n} \tag{2}$$
But from $(1)$ we have $$-x_n^3 \le -a$$
Plugging this into $(2)$, we have
$$x_{n+1}-x_n \le \frac{- a + a}{3x_n^2} = 0 \\ x_n \ge x_{n+1}$$
So, the sequence is decreasing and lower-bounded and, therefore, convergent.
Finally, let the limit be $x$. Then, we have
$$3x = 2x + \frac{a}{x^2} \\ x = \frac{a}{x^2} \implies x = \sqrt[3]a \\ \bbox[5px,border:2px solid #C0AA00]{\lim_{n\to\infty}x_n = \sqrt[3]a}$$
| {
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Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$? How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.
| \begin{align}
5^{6n+6}-2^{2n+2} &\equiv 2^{6n+6}-2^{2n+2} \pmod{3} \\
&\equiv 2^{2n+2}(2^{(2n+2)2}-1) \pmod{3}\\
&\equiv2^{2n+2}(1-1) \pmod{3}\\
&\equiv 0 \pmod{3}
\end{align}
Notice that I used $\gcd(3, 2^{2n+2})=1$ and Fermat's theorem to conclude that $2^{(2n+2)2}\equiv1 \pmod{3}$.
| {
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△ with altitude from and angle bisector of intersecting at $P$. Find $BP$ Let $\triangle ABC$ be a triangle with side lengths $AB = 17, BC = 28, AC = 25$. Let the altitude from $A$ to $BC$ and the angle bisector of angle $B$ meet at $P$. Given the length of $BP$ can be expressed as $\frac{a\sqrt{b}}{c}$ for positive integers $a, b, c$ where $gcd(a, c) = 1$ and $b$ is not divisible by the square of any prime, find $a + b + c$.
I drew my diagram out, and assumed that the intersection point is the midpoint of the altitude. Then i used pythag to find the length of the segment where the altitude intersects BC. I got a huge answer which is 99% incorrect so i need help on solving it.
| Let $\angle B=2\beta$ and $D$ be the foot of the altitude from $A$. Then
$$\cos\beta = \frac{BD}{BP}, \>\>\> \cos 2\beta = \frac{BD}{17}
$$
leading to
$$BP= \frac{17}{\cos\beta}(2\cos^2\beta-1)\tag1$$
From the cosine rule
$$\cos2\beta= \frac{ 17^2+28^2-25^2}{2\cdot 17\cdot 28}=2\cos\beta^2-1
$$
which yields $\cos^2\beta=\frac{25}{34}$. Plug into (1) to obtain
$$BP= \frac{8\sqrt{34}}5$$
Thus, $a+b+c= 47$.
| {
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Vertex $E$ of equilateral $\Delta ABE$ is in interior of $\square ABCD.$ $F$ is the midpoint of intersection of $AE$ and diagonal $BD$.
Vertex $E$ of equilateral $\Delta ABE$ is in interior of $\square ABCD.$ $F$ is the midpoint of intersection of $AE$ and diagonal $BD$. If $AB = \sqrt{1 + \sqrt{3}}$ and $[\Delta ABF] = A$, find $(4A)^2$.
What I Tried: Here is a picture :-
From a bit of angle chasing, we have the angles of $\Delta ABF$ .
Also $AB = \sqrt{1 + \sqrt{3}}$, let $AF = y$. We have :-
$$\frac{y}{\sin B} = \frac{AB}{\sin C}$$
We have $\sin 75^\circ = \frac{1 + \sqrt{3}}{2\sqrt{2}}$ , so :-
$$\rightarrow y = \frac{\sqrt{1 + \sqrt{3}} * \sqrt{2}}{\frac{1 + \sqrt{3}}{2\sqrt{2}}*2}$$
Let $\sqrt{1 + \sqrt{3}} = k$. After continuous solving we will have :-
$$y = \frac{2\sqrt{k}}{k}$$
From here I can get $$[\Delta ABF] = \frac{1}{2}(AB)(AF)\sin 60^\circ$$
$$\rightarrow \sqrt{k} * \frac{2\sqrt{k}}{k} * \frac{\sqrt{3}}{2}$$
$$\rightarrow [\Delta ABF] = \sqrt{3}$$
Hence $(4A)^2 = 48$ .
The question is, am I correct? Was it a coincidence that the $k$'s got cancelled out?
I am probably not correct, because right now I found out that $\sqrt{1 + \sqrt{3}} < \sqrt{3}$ , and this is impossible if my answer would have correct.
Can anyone let me know where I went wrong? Thank You.
| Let's use $AB=k=\sqrt{1+\sqrt 3}$. Then $$\sin 75^\circ=\frac{1+\sqrt 3}{2\sqrt 2}=\frac {k^2}{2\sqrt 2}$$
Area of triangle $ABF$ is then:
$$[\triangle ABF]=\frac12 AB\cdot AF\sin60^\circ=\frac12 k\frac {k\sin 45^\circ}{\sin 75^\circ}\sin 60^\circ\\=\frac 12 k^2\frac{\frac{\sqrt 2}2}{\frac{k^2}{2\sqrt 2}}\frac{\sqrt 3}2=\frac{\sqrt 3}2$$
| {
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Combinatorics with repetition - is my textbook right? I am suspicious that my textbook is incorrect.
Here's the question:
In how many ways can we distribute 20 candies among 6 children so that the youngest gets at most 2 candies?
My solution:
There are 20 candies (identical) and 6 children.
This is a combinations with repetition question so it can be represented by $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 20$
And we can consider the cases where $x_1 = 0, 1, \text{or } 2$ where $x_1$ is the youngest child.
$x_1 = 0$: No candies for the youngest one. Then $r=20$ and $n = 5$ and so there are $\binom{20+5-1}{20}$ combinations.
$x_1 = 1$: We give one candy to the youngest one. Then $r=19$ and $n = 5$ and so there are $\binom{19+5-1}{19}$ combinations.
$x_1 = 2$: We give two candies to the youngest one. Then $r=18$ and $n = 5$ and so there are $\binom{18+5-1}{18}$ combinations.
The number of ways we can distribute the candies is $\binom{24}{20} + \binom{23}{19} + \binom{22}{18} = 26,796$.
Textbook solution:
Who is right?
| You are correct. The flaw in the textbook's solution is that if the youngest child gets $0$, $1$, or $2$ candies, only the number of candies the other children may receive can vary. In the textbook's solution, $y_1$ should be $0$ in each case.
That said, there is another way to do this problem.
If we ignore the restriction that the youngest child can receive at most two candies, the number of ways to distribute the indistinguishable candies to the children is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 20 \tag{1}$$
in the nonnegative integers. A particular solution corresponds to the placement of $6 - 1 = 5$ addition signs in a row of $20$ ones. The number of such solutions is
$$\binom{20 + 6 - 1}{6 - 1} = \binom{25}{5} = \binom{25}{20}$$
since we must choose which five of the $25$ positions required for twenty ones and five addition signs will be filled with addition signs or, equivalently, which $20$ of those $25$ positions will be filled with ones.
From these, we must subtract those cases in which the youngest child receives more than two candies. Suppose the youngest child does receive at least three candies. Let $x_1' = x_1 - 3$. Then $x_1$ is a nonnegative integer. Substituting $x_1' + 3$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 3 + x_2 + x_3 + x_4 + x_5 + x_6 & = 20\\
x_1' + x_2 + x_3 + x_4 + x_5 + x_6 & = 17 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{17 + 6 - 1}{6 - 1} = \binom{22}{5} = \binom{22}{17}$$
solutions.
Hence, the number of ways the candies can be distributed to the six children so that the youngest child receives at most two candies is
$$\binom{25}{5} - \binom{22}{5} = 26,796$$
as you found.
| {
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Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$? Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &= 1 \\
a^2 + 2b^2 &= 1
\end{align*}
We write $b = \frac{1}{2a}$ and substitute this in the second equation.
\begin{align*}
a^2 + 2\left(\frac{1}{2a}\right)^2 &= 1 \\
a^2 + \frac{1}{2a^2} &= 1 \\
2a^4 - 2a^2 + 1 &= 0
\end{align*}
Let $z = a^2$, so $z^2 = a^4$. The equation is then
\begin{equation}
2z^2 - 2z + 1 = 0
\end{equation}
Using the quadratic formula we find $z = \frac{1 \pm i}{2}$. This worked out when checked. Thus $a = \sqrt{\frac{1 \pm i}{2}}$.
We then find $b$ using $a$ in our original system of equations.
\begin{align*}
\frac{1 \pm i}{2} + 2b^2 &= 1 \\
1 \pm i + 4b^2 &= 2 \\
\pm i + 4b^2 &= 1 \\
4b^2 &= 1 \pm i \\
2b &= \sqrt{1 \pm i} \\
b &= \frac{\sqrt{1 \pm i}}{2} \\
\end{align*}
Substituting $a$ and $b$ into the equation $2ab = 1$, leads to inconsistent solutions.
What do I need to reconsider? How can I improve my answer?
| The truth is that $\sqrt{1+\sqrt2}$ cannot be expressed as $a+b\sqrt2$ where $a,b$ are integers or rationals. If they were reals then the problem becomes trivial.
(Another way to see the first fact above is that the minimal polynomial of $\sqrt{1+\sqrt2}$ is degree-$4$, whereas if it were expressible as $a+b\sqrt2$ with $a,b\in\mathbb Q$ the minimal polynomial would only be quadratic.)
| {
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Show that $4x^2-yz$ is a perfect square Here is my problem.
$A=xy+yz+zx$, where $x,y,z\in\mathbb{Z}$. It is known that if we add $1$ to $x$, and subtract $2$ from both $y$ and $z$, the value $A$ won't change. Prove that $-A$ is a square of whole number.
My attempt:
\begin{align}
A=xy+yz+zx
\end{align}
and
\begin{align}
A=(x+1)(y-2)+(y-2)(z-2)+(z-2)(x+1)
\end{align}
comparing them we get
\begin{align}
4x+y+z=0.
\end{align}
Now we insert last equation to first equation:
\begin{align}
-A&=-xy-yz-zx\\
&=-x(y+z)-yz\\
&=4x^2-yz
\end{align}
From here I don't know how to show that $4x^2-yz$ is a perfect square.
| If we expand, we get
$$A=(x+1)(y-2)+(y-2)(z-2)+(z-2)(x+1)\quad =\quad x y + x z - 4 x + y z - y - z$$
Subtracting the original from this equation should be zero
$$(x y + x z - 4 x + y z - y - z)-(xy+yz+xz)=0\\
\implies 4x+y+z=0\implies z = -4 x - y$$
Substituting $z$ now allows the equation to be viewed more simply
$$xy+y(-4 x - y)+x(-4 x - y)= -4 x^2 - 4 x y - y^2=A\\
\implies A=-(4x+4xy+y^2)$$
$$\therefore -A=(2x+y)^2$$
\begin{align*}
\end{align*}
| {
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Evaluate $\lim_{n\to \infty} (n+1)^{\frac 23} -(n-1)^{\frac 23}$ Clearly, $\infty -\infty \not =0$
I have a feeling the squeeze theorem can be applied here, but I am not sure how to write the required terms separately. Can I get a hint?
| Let $f:[n-1,n+1]\to\mathbb{R}$ and $f(x)=x^{\frac{2}{3}}$. Then from Mean value theorem (Lagrange’a) exist $\xi\in (n-1,n+1)$ such that:
$$\frac{f(n+1)-f(n-1)}{n+1-(n-1)}=f'(\xi)$$
but (check it out)
$$\frac{2}{3 \sqrt[3]{n+1} }\le f'(\xi)\le \frac{2}{3 \sqrt[3]{n-1} } $$
so
$$\frac{4}{3 \sqrt[3]{n+1} }\le f(n+1)-f(n-1 )\le \frac{4}{3 \sqrt[3]{n-1} } $$
$$\frac{4}{3 \sqrt[3]{n+1} }\le (n+1)^{\frac{2}{3}}-(n-1 )^{\frac{2}{3}}\le \frac{4}{3 \sqrt[3]{n-1} } $$
now You can aplay Squeeze theorem.
| {
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Determining the average area as the point $X$ varies We consider a fixed triangle $ABC$ with side lengths $a = BC$, $b = CA$, $c = AB$, and
a variable point $X$ in the interior. The lines through $X$ parallel to $AB$ and $AC$,
together with line $BC$, determine a triangle $T_a$. The triangles $T_b$ and $T_c$ are defined
in a similarly way .
Let $S$ and $p$ denote the average area and perimeter, respectively, of the three triangles $T_a, T_b, T_c$.
(a) Determine all possible values of $S$ as $X$ varies, in terms of $a, b, c.$
(b) Determine all possible values of $p$ as $X$ varies, in terms of $a, b, c$.
Can someone give me hints for this problem ? I don't know how to start (a).
I got that (b) is $\frac {1}{3}(a+b+c)$, because of the parallel property , sum of perimeters of $T_a, T_b, T_c$ is $a+b+c$.
| Let $[ABC]=S_0$ , its perimeter $= P_0$
Let bases (parallel to $a$) of ${T_a, T_b, T_c}$ be ${x,y,z}$ respectively.
Note each $T_i \sim \triangle ABC$ with similarity ratios ${x/a, y/a, z/a}$ respectively.
These three triangles leave three parallelograms out of $\triangle ABC$. Equating base lengths, we get $x+y+z=a$
Thus $$P=\frac{1}{3} \cdot \big( \dfrac{x}{a}+\dfrac{y}{a}+\dfrac{z}{a} \big)P_0$$
$$S=\frac{1}{3} \cdot \big( \dfrac{x^2}{a^2}+\dfrac{y^2}{a^2}+\dfrac{z^2}{a^2} \big)S_0$$
Now find the range of $$\dfrac{x^2+y^2+z^2}{a^2}$$
| {
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$\frac{b^{2n}+b^{n+1}+3b-5}{b-1}$ is square Find all $b>5$ so that $x_n = \frac{b^{2n}+b^{n+1}+3b-5}{b-1}$ is square for all sufficiently so large integers n.
I think the only value of $b$ is 10.
If there is $p \in \mathbb{P}$ (prime), $p \neq 3, p \mid b-1$, then $v_p(b-1)=1$. If $v_p(b-1) > 1$, we can choose n.
If $b=3^m+1$, then m must be even.(If m is odd, $r_8(x_n)=3$)
What should I do next?
| COMMENT: May be using these ideas help:
1-Suppose $\frac{b^{2n}+b^{n+1}+3b-5}{b-1}= (b+a)^2$
If polynomial $f(b)=b^{2n}+b^{n+1}+3b-5$ must have a double root like $-a$ then we must have:
$f(-a)=f'(-a)=0$
This gives following system of equations:
$\begin{cases}\frac{(-a)^{2n}+(-a)^{n+1}-3a-5}{-a-1}=0\\2n(-a)^{2n-1}+(n+1)(-a)^n+3=0 \end {cases}$
Which gives a and n.Plugging these in fraction gives an equation in terms of b which gives b.
2- We must have:
$\frac{b^{2n}+b^{n+1}+3b-5}{b-1}=(b+a)^2$
Which gives:
$b^{2n}+b^{n+1}+3b-5=(b-1)(b+a)^2$
Or:
$b^{2n}+b^{n+1}-b^3-(2a-1)b^2-(a^2-2a)b+a^2-5$
Which is a Diophantine equation to be solved.
| {
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Prove $1^3 + 2^3 + \ldots n^3 = \frac{n^2(n + 1)^2}{4}$ with mathematical induction Hello guys I'm trying to prove this equation with mathematical induction method.
So for n = 1 I know that :
1 = 1
Now I know that it works for some k
I wanna prove it for k + 1:
Since :
I can rewrite it as :
So I'm stuck here.. I don't know what should I do.. Should I first square (k + 1)^2 or rewrite (k + 1)^3 ??'
Either way I get weird result and don't know how to proceed with this next step to prove mathematical
induction. I mean I can multiple k^2 with the first parenthesis but I think I should leave it that way and do smth other than that. I'll appreciate any help. Thanks :)
| Note: It's enough to prove $\frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4 = n^3$.
That is to say, if we assume Equation 1: $1^3 + 2^3 + ...... + (n-1)^3= \frac {(n-1)^2n^2}4$
And we want to prove Equation 2: $1^3 + 2^3 +.... + (n-1)^3 + n^3 = \frac {n^2(n+1)^2}4$
That is a matter of proving by subtraction Equation 2 for Equation 1, that
$n^3= \frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4$
So... Lets do that.
$\frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4=$
$\frac {n^2}4[ (n+1)^2 - (n-1)^2] =$
$\frac {n^2}4[\{(n+1)+(n-1)\}\{(n+1)-(n-1)\}]=$
$\frac {n^2}4[\{2n\}\{2\}]=$
$\frac {n^2}4\cdot 4n = n^3$.
And ... that's that.
....
If somehow that doesn't feel right..... you can expand...
$\frac {n^2(n+1)^2}{4} - \frac {(n-1)^2n^2}4=$
$\frac {n^2 (n^2 + 2n + 1)}4 -\frac {n^2(n^2 -2n+1)}4 = $
... and, yeah, yeah, we get it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Probability Q on rolling two 8-sided dice Let's say I roll two 8-sided dice. I win if the sums '7' and '11' show up before we see the sum '9' TWICE. What is my probability of winning?
So this is my answer and please correct me if I am wrong:
Prob of getting either 7 or 9 —> 6/64 + 6/64 = 12/64 —> 18.75%
Prob of getting 9 twice —> 8/64 x 8/64 = 1/64 —> 1.6%
But now to find the probability of winning, would it be 18.75 - 1.6 = 17.15% ? Am I right or wrong?
Thanks!
EDIT
From the answers provided below, we have three different results:
84%
49%
60%
Which one would be the correct answer?
| This answer assumes the person wins if there is a sum of $\, 7\,$ or $\,11\,$ in any toss before a sum of $\,9\,$ appears twice.
Probability of winning in first toss $ = \frac{12}{64}$
a) If the person gets a sum of $9$ once, the chance of winning from there
$\displaystyle P(W1) = \frac{12}{64} + \frac{44}{64} \times (\frac{12}{64} + \frac{44}{64} \times ...)... = \frac{12}{64} (1 + \frac{44}{64} + (\frac{44}{64})^2 + ...)$
$\displaystyle = \frac{3}{5}$
[Using sum of infinite geometric series $a + ar + ar^2 + ... = \frac{a}{1-r}$]
b) If the first toss is any sum other than $7, 9, 11, \,$starting 2nd toss
$P(W2) = \frac{12}{64} + \frac{8}{64} \times P(W1) + \frac{44}{64} \times (\frac{12}{64} + \frac{8}{64} \times P(W1) + ...)... $
$$= (\frac{12}{64} + \frac{8}{64} \times P(W1))(1 + \frac{44}{64} + (\frac{44}{64})^2 + ...) = \frac{21}{25}$$
$\,$
So person's chance of winning $ \displaystyle = \frac{12}{64} + \frac{8}{64} \times P(W1) + \frac{44}{64} \times P(W2) = \frac{21}{25}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3941894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
How to put $(-\sqrt{3}-i)^{\frac{5}{7}}$ into polar form How to put $(-\sqrt{3}-i)^{\frac{5}{7}}$ into polar form and find all roots.
What I tried:
$$w = -\sqrt{3}-i$$
$$\arg(w)=\arctan(\frac{1}{\sqrt 3})-\pi = \frac{\pi}{6} - \pi = \frac{-5\pi}{6}$$
$$w = 2(\cos(\frac{-5\pi}{6})+ i\sin(\frac{-5\pi}{6})) $$
$$w^5 = 2^5(\cos(-\frac{5\pi}{6}*5)+i\sin(-\frac{5\pi}{6}*5)) $$
$$z^7 = w^5$$
$$z = w^{\frac{5}{7}} $$
$$z = 32^{\frac{1}{7}}(\cos(\frac{-5\pi*5}{6*7}+\frac{2k\pi}{7})+i\sin(\frac{-5\pi*5}{6*7}+\frac{2k\pi}{7})) $$
Here I get stuck and I don't know how to continue...
| you have :
$$(-\sqrt{3}-i)^{\frac{5}{7}}=\frac{1}{2^{5/7}}\left(\frac{-\sqrt{3}}{2}-\frac{1}{2}i\right)^{5/7}=\frac{1}{2^{5/7}} \left(\exp{\frac{i7\pi}{6}}\right)^{5/7}=\frac{1}{2^{5/7}}\exp{\frac{i5\pi}{6}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate of $\int_{0}^{\infty}\frac{e^{x\pi}-1}{e^{x\pi}+1}\cdot \frac{x}{(1+x^2)(9+x^2)} dx$ Would like to know how would you evaluate $(1)$:
$$\int_{0}^{\infty}\frac{e^{x\pi}-1}{e^{x\pi}+1}\cdot \frac{x}{(1+x^2)(9+x^2)} dx=\frac{1}{8}?\tag1$$
Notices that, $$\tanh\left(\frac{x\pi}{2}\right)=\frac{e^{x\pi}-1}{e^{x\pi}+1}$$
$$\int_{0}^{\infty}\tanh\left(\frac{x\pi}{2}\right)\cdot \frac{x}{(1+x^2)(9+x^2)} dx\tag 2$$
$$8\int_{0}^{\infty}\tanh\left(\frac{x\pi}{2}\right)\cdot \frac{x}{1+x^2} dx-8\int_{0}^{\infty}\tanh\left(\frac{x\pi}{2}\right)\cdot \frac{x}{9+x^2} dx \tag3$$
| Basically, we note
$$
\int_{0}^{\infty} e^{-at}\sin(xt) \,\mathrm{d}t = \frac{x}{a^2+x^2}
$$
$$
\int_{0}^{\infty} {\frac{\cos(\alpha x)}{\cosh(\beta x)} \mathrm{d}x} = \frac{\pi}{2\beta} \operatorname{sech}\left(\frac{\pi\alpha}{2\beta}\right)
$$
then denote desired integral with $I(a,b)$ that is
$$
\begin{aligned}
I(a,b)
& = \int_{0}^{\infty} \frac{x}{(a^2+x^2)(b^2+x^2)}\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x\\
& = \frac1{b^2-a^2} \int_{0}^{\infty} \left(\frac{x}{a^2+x^2}-\frac{x}{b^2+x^2}\right)\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} \left(\int_{0}^{\infty} (e^{-at}-e^{-bt})\sin(xt) \,\mathrm{d}t\right)\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} (e^{-at}-e^{-bt}) \left( \int_{0}^{\infty} \sin(xt)\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x\right)\mathrm{d}t\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} (e^{-at}-e^{-bt}) \left( -\frac{i}{2}\int_{0}^{\infty} \frac{\cos\left(t-\frac{i\pi}{2}\right)x-\cos\left(t+\frac{i\pi}{2}\right)x}{\cosh\left(\frac{\pi x}{2}\right)}\mathrm{d}x\right)\mathrm{d}t\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} (e^{-at}-e^{-bt}) \left( -\frac{i}{2}\left(\operatorname{sech}\left(t-\frac{i\pi}{2}\right)-\operatorname{sech}\left(t+\frac{i\pi}{2}\right)\right)\right)\mathrm{d}t\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} \frac{e^{-at}-e^{-bt}}{\sinh t} \mathrm{d}t = \frac1{b^2-a^2}\left(\psi\left(\frac{1+b}{2}\right) - \psi\left(\frac{1+a}{2}\right)\right)
\end{aligned}
$$
which easily reveals $I(1,3)=\frac1{8}$
Supplement for some identites
$$
\sinh\left(\frac{\pi x}{2}\right)=-i\sin\left(\frac{i\pi x}{2}\right), \quad\cosh\left(t\pm\frac{i\pi}{2}\right)=\pm i\sinh t
$$
and
Gauss's representation of digamma function for $\Re(z)>0$, namely
$$
\psi(z) = \int_{0}^{\infty} {\left( \frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}} \right) \mathrm{d}t}
$$
Edit for some necessary explanation, actually this integral
$$
\int_{0}^{\infty} \sin(xt)\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x
$$
does not converge under common meaning, there are some abuse of notation need to be justified, a quick insight is to recall
$$
\int_{0}^{\infty} {\frac{\sin(\alpha x)}{\sinh(\beta x)} \mathrm{d}x} = \frac{\pi}{2\beta} \tanh\left(\frac{\pi\alpha}{2\beta}\right)
$$
and we have
$$
\begin{aligned}
\int_{0}^{\infty} \sin(xt)\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x
& = \int_{0}^{\infty} \frac{\mathrm{d}w}{\sinh w} \left(\frac2{\pi}\int_{0}^{\infty} \sin(xt)\sin(xw)\mathrm{d}x\right)\\
& = \int_{0}^{\infty} \frac{\delta(t-w)-\delta(t+w)}{\sinh w} \mathrm{d}w = \frac1{\sinh t}
\end{aligned}
$$
by utilizing the Dirac's delta in such integral.
Or more directly, recalling $(w>0)$
$$
\int_{0}^{\infty} \frac{x\sin(wx)}{a^2+x^2}\mathrm{d}x = \frac{\pi}{2}e^{-aw}
$$
hence
$$
\begin{aligned}
I(a,b)
& = \int_{0}^{\infty} \frac{x}{(a^2+x^2)(b^2+x^2)}\tanh\left(\frac{\pi x}{2}\right)\mathrm{d}x\\
& = \frac1{b^2-a^2} \int_{0}^{\infty} \left(\frac{x}{a^2+x^2}-\frac{x}{b^2+x^2}\right) \left( \frac2{\pi}\int_{0}^{\infty} {\frac{\sin(wx)}{\sinh(w)} \mathrm{d}w} \right) \mathrm{d}x\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} \frac{\mathrm{d}w}{\sinh w} \left(\frac{2}{\pi} \int_{0}^{\infty} \left(\frac{x\sin(wx)}{a^2+x^2} - \frac{x\sin(wx)}{b^2+x^2}\right) \mathrm{d}x\right)\\
& = \frac1{b^2-a^2}\int_{0}^{\infty} \frac{e^{-aw}-e^{-bw}}{\sinh w} \mathrm{d}w = \frac1{b^2-a^2}\left(\psi\left(\frac{1+b}{2}\right) - \psi\left(\frac{1+a}{2}\right)\right)
\end{aligned}
$$
which appears to be a better solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to write this rational sequence : $(u_n=\frac{2^{n-1}+1}{2^n})_{n \ge 1}$ using only the rational numbers : $1$; $\dfrac{1}{2}$; $\dfrac{1}{3}$; $\dfrac{1}{4}$ and the operation $+,-,\times$ ?
I get : $u_1 = 1 = \dfrac{1}{2} +\dfrac{1}{2}= \dfrac{1}{2}\left(1+1\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\right)$
$u_2 = \dfrac{3}{4} = \dfrac{1}{2} +\dfrac{1}{4}= \dfrac{1}{2}\left(1+\dfrac{1}{2}\right)= \dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}+\dfrac{1}{4}\right)$
$u_3 = \dfrac{5}{8}= \dfrac{1}{2} +\dfrac{1}{8}= \dfrac{1}{2}\left(1+\dfrac{1}{4}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)$
$u_4=\dfrac{9}{16}=\dfrac{1}{2} +\dfrac{1}{16}=\dfrac{1}{2}\left(1+\dfrac{1}{8}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{8}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\times \dfrac{1}{4}\right)$
etc...
$u_n = \dfrac{2^{n-1}+1}{2^n}= \dfrac{1}{2} +\dfrac{1}{2^{n}}=\dfrac{1}{2}\left(1+\dfrac{1}{2^{n-1}}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2^{n-1}}\right)$ and following the value of $n$ we can make appear several products of $\dfrac{1}{2}$ and $\dfrac{1}{4}$.
Can we find a better form for $u_n$ ?
Moreover it seems possible to write the terms of this sequence (for $n>1$) using only once the rational numbers $1$; $\dfrac{1}{2}$; $\dfrac{1}{3}$; $\dfrac{1}{2^n}$ :
$u_2 = \dfrac{1}{3}\left(1+\dfrac{1}{2}\right)+\dfrac{1}{4}$
$u_3 = \dfrac{1}{3}\left(1+\dfrac{1}{2}\right)+\dfrac{1}{8}$
etc...
$u_n = \dfrac{2^{n-1}+1}{2^n}= \dfrac{1}{2} +\dfrac{1}{2^{n}}=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)+\dfrac{1}{2^{n}}$
Thanks in advance !
| As mentioned in the comments, you can use a recurrence, which needs only $\frac{1}{2},+$ and $\times$:
Since $u_n=\frac{1}{2}+\frac{1}{2^n} \Rightarrow \frac{1}{2^n}=u_n-\frac{1}{2}$, we have $u_{n+1}=\frac{1}{2}+\frac{1}{2^{n+1}}=\frac{1}{2}(1+\frac{1}{2^n})$ and the recurrence relation $u_{n+1}=\frac{1}{2}(\frac{1}{2}+u_n)$. By $u_1=\frac{1}{2}+\frac{1}{2}$ we are finished.
The first n are:
$u_1=\frac{1}{2}+\frac{1}{2}$,
$u_2=\frac{1}{2}(\frac{1}{2}+u_1)=\frac{1}{2}(\frac{1}{2}+\frac{1}{2}+\frac{1}{2})$,
$u_3=\frac{1}{2}(\frac{1}{2}+u_2)=\frac{1}{2}(\frac{1}{2}+\frac{1}{2}(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}))$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Differential equation system solution: do I get the right solution? I am very stuck with differential equation systems.
For example:
$
Y'(x) =
\begin{pmatrix}
2 & 0 & 1 \\
0 & 2 & 0 \\
0 & 1 & 3 \\
\end{pmatrix}
Y(x)
$
I get the eigenvalues and eigenvectors:
$ \lambda = 2 (double) \rightarrow \vec v_{\lambda2} = \begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
$
and
$ \lambda = 3 \rightarrow \vec v_{\lambda3} = \begin{pmatrix}
1 \\
0 \\
1 \\
\end{pmatrix}
$
I obtain just one vector asociated with $ \lambda = 2 $ value, so that I suppose I've to get another vector.
For this eigenvalue:
$ (A-\lambda I) = \begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 1 & 1 \\
\end{pmatrix}
$
So that the vector I'm looking must meet:
$
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 1 & 1 \\
\end{pmatrix}
\begin{pmatrix}
v_{x} \\
v_{y} \\
v_{z} \\
\end{pmatrix}
=
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
$
From doing this I get:
$
\begin{pmatrix}
v_{x} \\
v_{y} \\
v_{z} \\
\end{pmatrix}
\sim
\begin{pmatrix}
\alpha \\
-1 \\
1 \\
\end{pmatrix}
$, and I choose the vector $
\begin{pmatrix}
0 \\
-1 \\
1 \\
\end{pmatrix}
$
The problems come from this point. I think the general solution should be:
$
y(x) = C_{1}
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
e^{2x}
+
C_{1}x
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
e^{2x}
+
C_{2}
\begin{pmatrix}
0 \\
-1 \\
1 \\
\end{pmatrix}
e^{2x}
+
C_{3}
\begin{pmatrix}
1 \\
0 \\
1 \\
\end{pmatrix}
e^{3x}
$
But this result is wrong. I've tried to proove it and it doesn't work.
I've read in some books about it but I think I'm following the mathematic method fine...
Does anyone know what I'm doing wrong?
Thanks in advance.
| The eignevalues and eigenvectors looks correect to me.
$$y(x) = C_{1}
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
e^{2x}
+
C_{1}x
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
e^{2x}
+
C_{2}
\begin{pmatrix}
0 \\
-1 \\
1 \\
\end{pmatrix}
e^{2x}
+
C_{3}
\begin{pmatrix}
1 \\
0 \\
1 \\
\end{pmatrix}
e^{3x}$$
It should be:
$$y(x) = C_{1}
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
e^{2x}
+
\color {red}{C_{2}x}
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}
e^{2x}
+
C_{2}
\begin{pmatrix}
0 \\
-1 \\
1 \\
\end{pmatrix}
e^{2x}
+
C_{3}
\begin{pmatrix}
1 \\
0 \\
1 \\
\end{pmatrix}
e^{3x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$
For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$
This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}=\frac{\pi}{2}$$
This is where I'm unsure I made a 'legal' move. I reasoned that if I take the tangent of both sides of the equation, I would end up with a fraction that must be undefined as $\tan\frac{\pi}{2}$ is undefined; hence the denominator of the fraction I obtain must be $0$:
$$\tan\left(\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}\right)=\frac{\frac{y}{x-8}-\frac{y}{x-2}}{1+\frac{y^2}{(x-8)(x-2)}}=\frac{y(x-2)-y(x-8)}{(x-8)(x-2)+y^2}$$
But this is undefined; hence
$$(x-8)(x-2)+y^2=0\implies (x-5)^2+y^2=9$$
so it appears that the required locus is a circle with centre at $(5,0)$ on the Argand diagram with radius $3$.
Is my reasoning fully acceptable throughout my solution?
| If $A(z_A),B(z_B),M(z_M)$ are distinct points in Argand plane, then $\arg \left(\frac{z_M-z_B}{z_M-z_A}\right)$ is an oriented angle between vectors $(\vec{AM},\vec{BM}).$
In the present case $z_M=z,z_A=2,z_B=8.$
$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ says that the vectors are orthogonal, and $M$ lies on an open semi-circle with diameter $AB.$
(It is the upper half of the circle with diameter $AB,$ without $A,B.$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_{x\to+∞}\frac{(\sum_{n=0}^∞{(\frac{x^n}{n!})^2})^2}{(\sum_{n=0}^∞{(\frac{x^n}{n!})^1}) (\sum_{n=0}^∞{(\frac{x^n}{n!})^3})}$ Prove the following limit:
$$
\lim_{x\rightarrow +\infty} \frac{\left( \sum_{n=0}^{\infty}{\left( \frac{x^n}{n!} \right) ^2} \right) ^2}{\left( \sum_{n=0}^{\infty}{\left( \frac{x^n}{n!} \right) ^1} \right) \left( \sum_{n=0}^{\infty}{\left( \frac{x^n}{n!} \right) ^3} \right)}=\frac{\sqrt{3}}{2} \tag{1}
$$
Mathematica tells me that
$$
\sum _{n=0}^{\infty } \left(\frac{x^n}{n!}\right)^2=I_0(2 x)\\
\sum _{n=0}^{\infty } \left(\frac{x^n}{n!}\right)^3 = \, _0F_2\left(;1,1;x^3\right)
$$
but they don't make sense to me for calculating the limits, since I don't have any knowledge about Special Functions.
How can I prove $(1)$ in an elementary way(not involving special functions)?
| We investigate a heuristic idea that lead the correct answer. Consider a Poisson random variable $N$ with rate $x$, so that
$$ \mathbb{P}(N = n) = \frac{x^n}{n!} e^{-x} $$
for $n \geq 0$. Then by the (local) central limit theorem, we know that $Z = \frac{N-x}{\sqrt{x}}$ approximates the standard normal distribution, loosely in the sense that
\begin{align*}
\mathbb{P}(N = n)
&= \mathbb{P}\left( \left| N - n \right| < \frac{1}{2}\right) \\
&\approx \mathbb{P}\biggl( \left| Z - \frac{n-x}{\sqrt{x}} \right| < \frac{1}{2\sqrt{x}} \biggr) \\
&\approx \phi\left(\frac{n-x}{\sqrt{x}}\right) \frac{1}{\sqrt{x}}
\end{align*}
where $\phi(t) = \frac{1}{\sqrt{2\pi}} e^{-t^2/2}$ is the p.d.f. of the standard normal distribution. So we expect that, for each given $\alpha > 0$, as $x\to\infty$,
\begin{align*}
\sum_{n=0}^{\infty} \left(\frac{x^n}{n!}\right)^{\alpha}
&\sim e^{\alpha x} \sum_{n=0}^{\infty} \left[ \phi\left(\frac{n-x}{\sqrt{x}}\right) \frac{1}{\sqrt{x}} \right]^{\alpha} \\
&\sim \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\alpha t^2/2} \, \mathrm{d}t \\
&= \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \cdot \frac{1}{\sqrt{\alpha}}. \tag{*}
\end{align*}
Assuming that $\text{(*)}$ indeed holds, then the limit easily follows.
Indeed, some hard analysis using the idea of Laplace's method shows that, for each fixed $\alpha > 0$ and sufficiently small $\epsilon > 0$,
$$ \sum_{n=0}^{\infty} \left(\frac{x^n}{n!}\right)^{\alpha}
= \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \cdot \frac{1}{\sqrt{\alpha}} \bigl( 1 + \mathcal{O}(x^{-\frac{1}{2}+3\epsilon}) \bigr) \tag{$\diamond$} $$
This can be established by splitting the sum into two parts, one over the range $\left|n - x\right| \leq x^{\frac{1}{2}+\epsilon}$ that realizes the leading term of $(\diamond)$, and the other over the range $\left|n - x\right| \geq x^{\frac{1}{2}+\epsilon}$ that only contribute to the relative error $(\diamond)$.
As the computation is a bit lengthy, let me estimate the first sum only:
*
*Fix a sufficiently small $\epsilon > 0$ and consider the case $n \in \mathcal{I} = \{ n \in \mathbb{N}_0 : \left| n - x \right| \leq x^{\frac{1}{2}+\epsilon} \}$. Also, introduce a new parameter $u$ which is related to $n$ by
$$n = x + u\sqrt{x}.$$
In particular, $\left| u \right| \leq x^{\epsilon}$ and $ \frac{n}{x} = 1 + \frac{u}{\sqrt{x}} $. Now by the Stirling's approximation,
\begin{align*}
\frac{x^n}{n!}
&= \exp\left( n \log x - \frac{1}{2}\log(2\pi n) + n - n \log n + \mathcal{O}\bigl(n^{-1}\bigr) \right) \\
&= \exp\left( - \frac{1}{2}\log(2\pi x) - \frac{1}{2}\log\frac{n}{x} + x \left( \frac{n}{x} - \frac{n}{x} \log \frac{n}{x} \right) + \mathcal{O}\bigl(x^{-1}\bigr) \right) \\
&= \exp\left( - \frac{1}{2}\log(2\pi x) + \mathcal{O}\bigl(x^{-\frac{1}{2}+\epsilon}\bigr) + x \left( 1 - \frac{u^2}{2x} + \mathcal{O}\bigl(x^{-\frac{3}{2}+3\epsilon}\bigr) \right) + \mathcal{O}\bigl(x^{-1}\bigr) \right) \\
&= \exp\left( - \frac{1}{2}\log(2\pi x) + x - \frac{u^2}{2} + + \mathcal{O}\bigl(x^{-\frac{1}{2}+3\epsilon}\bigr) \right) \\
&= \bigl( 1 + \mathcal{O}\bigl(x^{-\frac{1}{2}+3\epsilon}\bigr) \bigr) e^{x} \phi(u) \frac{1}{\sqrt{x}}.
\end{align*}
So it follows that, for each given $\alpha > 0$,
\begin{align*}
\sum_{n \in \mathcal{I}} \left( \frac{x^n}{n!} \right)^{\alpha}
&= \bigl( 1 + \mathcal{O}(x^{-\frac{1}{2}+3\epsilon}) \bigr) \frac{e^{\alpha x}}{(2\pi x)^{(\alpha-1)/2}} \sum_{n \in \mathcal{I}} \frac{1}{\sqrt{2\pi x}} e^{-\alpha u^2/2}.
\end{align*}
Now the summation part in the right-hand side can be regarded as the Riemann sum. Indeed, using the fact that
$$ \left| \sum_{a \leq n \leq b} f(n) - \int_{a}^{b} f(x) \, \mathrm{d}x \right| \leq 3\sup_{a \leq x \leq b} \left| f(x) \right| $$
holds for any continuous unimodal function $f : [a, b] \to \mathbb{R}$,
$$ \sum_{n \in \mathcal{I}} \frac{1}{\sqrt{2\pi x}} e^{-\alpha u^2/2}
= \int_{-x^{\epsilon}}^{x^{\epsilon}} \frac{1}{\sqrt{2\pi}} e^{-\alpha u^2/2} \, \mathrm{d}u + \mathcal{O}(x^{-1/2})
= \frac{1}{\sqrt{\alpha}} + \mathcal{O}(x^{-1/2}). $$
Combining all the estimates, we obtain the desired leading term in $(\diamond)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Can we write $\sin(\frac{\pi}{14})$ as an finite expression using only basic operations? First, let us recall some trigonometric values
$\sin(0)=0$
$\sin(\pi/6)=\frac{1}{2}$
$\sin(\pi/3)=\frac{\sqrt{3}}{2}$
$\sin(\pi/2)=1$
$\sin(\pi/10)=\frac{\sqrt{5}-1}{4}$
$\sin(\pi/12)=\frac{\sqrt{3}-1}{2\sqrt{2}}$
Here, we can observe that for some values of $\theta$, $\sin(\theta)$ can be expressed as a sum of finite radical terms. It is also easy to see that, there exists infinite such $\theta$'s. Say, I have given $\sin(\pi/14)$, can we determine whether it can also be expressed as sum of finite radical terms? What about the general case? Are there any criteria that $\theta$ must obey in order to do that?
| $8x^3−4x^2−4x+1=0$
Where does this polynomial come from?
Suppose you take a regular polynomial and plot the coordinates.
Consider a n-gon, with one vertex on $(1,0)$ all the vertices one unit from the center, and the center at $(0,0).$
$(1,0)\\
(\cos \frac {2\pi}{n},\sin \frac {2\pi}{n})\\
(\cos \frac {4\pi}{n},\sin \frac {4\pi}{n})\\
\cdots\\
(\cos 2(n-1)\pi,\sin 2(n-1)\pi)$
If we average, or sum, all of these coordinates we get the center of the circle.
$1 + \cos\frac {2\pi}{n} + \cdots + \cos \frac {2(n-1)\pi}{n} = 0$
and similarly
$1 + \sin\frac {2\pi}{n} + \cdots + \sin {2(n-1)\pi}{n} = 0$
We could rotate the polygon to make one vertex associate with the angle $\frac {\pi}{14},$ but that isn't what I did. Instead, I kept the same orientation and said:
$\sin \frac {\pi}{14} = \cos (\frac {\pi}{2}-\frac {\pi}{14}) = -\cos (\frac {\pi}{2}+\frac {\pi}{14}) = -\cos \frac {4\pi}{7}$
Working with:
$1 + \cos\frac {2\pi}{7} + \cos\frac {4\pi}{7} + \cdots + \cos \frac {12\pi}{7} = 0$
This polygon has symmetry about the x-axis.
$\cos \frac {2\pi}{7} = \cos \frac {12\pi}{7}\\
\cos \frac {4\pi}{7} = \cos \frac {10\pi}{7}\\
\cos \frac {6\pi}{7} = \cos \frac {8\pi}{7}$
$1 + 2\cos\frac {2\pi}{7} + 2\cos\frac {4\pi}{7} + 2\cos\frac {6\pi}{7}=0$
Now we use multiple-angle identities.
$1 + 2\cos\frac {2\pi}{7} + 2(2\cos^2\frac {2\pi}{7}-1) + 2(4\cos^3\frac {2\pi}{7} - 3\cos\frac {2\pi}{7})$
Say $\cos\frac {2\pi}{7} = x$
$8x^3 +4x^2 - 4x - 1 = 0$
This polynomial has 3 roots. One equals $\cos\frac {2\pi}{7}$, another equals $\cos\frac {4\pi}{7}$ and the third equals $\cos\frac {6\pi}{7}$
But we wanted the negative of one of those roots. Then replace $x$ with $-x$ in the equation above, remembering that $(-x)^n = x^n$ when $n$ is even and $-x^n$ when $n$ is odd.
$-8x^3 +4x^2 + 4x - 1 = 0\\
8x^3 -4x^2 - 4x + 1 = 0$
$\sin \frac {\pi}{14}$ is a root of $8x^3 -4x^2 - 4x + 1 = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The limit of a series I need help with this exercise!
$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}.$
I saw somewhere online that $1^4+2^4+...+n^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.
But I dont understand why!
So following that then,
$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}=$
$=\lim\limits_{n \to \infty} \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}=$
$=\lim\limits_{n \to \infty} \dfrac{(n+1)(2n+1)(3n^2+3n-1)}{30n^4}=\dfrac15.$
And after that what!? Do I multiply and then use L'Hopital?
| Note $\sum_{k=1}^nk^4$ must be divisible by $n(n+1)$, so that the obvious extension to negative $n$ achieves $0-0^4=0$ at $n=-1$. The large-$n$ behaviour is asymptotic to $\int_0^nx^4dx=\tfrac15n^5$. We can determine coefficients in $\sum_{k=1}^nk^4=\tfrac15n(n+1)(n^3+An^2+Bn+C)$ from the sum at three values of $n$, say $n\in\{1,\,2,\,3\}$, thereby acquiring simultaneous equations. This gives $A=\tfrac32,\,B=\tfrac16,\,C=-\tfrac16$, after which the rational root theorem finds the factor $n+\tfrac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$
Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)
My approach:
Solving the quadratic, I get:
$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$
Case 1: If the $b^2 - 4ac$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.
Case 2: If $b^2 - 4ac$ is not a perfect square, $b = \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$
Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)
So the equation is not quadratic and $c = 0$.
$a + bp^\frac{1}{3} = 0$
$-\dfrac{a}{b} = p^\frac{1}{3}$
This is a contradiction and hence $b = 0$ and $a = 0$
Is there any other way to solve this?
| I have somewhat weird way to see this. Consider the system
\begin{align*}
a + b p^{1/3} + c p^{2/3} & = 0\\
cp + a p^{1/3} + b p^{2/3} & = 0\\
bp + cp p^{1/3} + a p^{2/3} & = 0
\end{align*}
or
$$\begin{pmatrix} a & b & c \\
cp & a & b \\
bp & cp & a \end{pmatrix} \begin{pmatrix} 1 \\ p^{1/3} \\ p^{2/3} \end{pmatrix} =0.
$$
So the coefficient matrix has zero determinant, i.e.
$$a^3 + b(b^2-3ac)p + c^3 p^2=0.$$
Now we can proceed with infinite descent.
(The essence is until here, and below is just some calculations.)
Note that we can assume that $p$ is an integer; If $a + b (n/d)^{1/3} + c (n/d)^{2/3}=0$ then
$$ad^2 + bd\cdot d^{2/3} n^{1/3} + c d^{4/3}n^{2/3}=0,$$
i.e.
$$ad^2 + bd\cdot (d^2n)^{1/3} + c (d^2n)^{2/3}=0$$ so we are reduced to the integer $p$ case.
Let $q$ be a prime factor or $p$. One can assume that $q^3 \not\mid p$; in this case $q$ factor is absorbed into coefficients $b$ and $c$. Assume $(a, b, c)$ is a nontrivial integer solution. We have two cases;
*
*Case 1: Let $p = qN$ with $q\not\mid N$. Then
$$a^3 + b(b^2-3ac)qN + c^3 q^2N^2=0,$$
i.e. $a = qA$. Then
$$q^2A^3 + b(b^2-3qAc)N + c^3 qN^2=0,$$
i.e. $b = qB$. Then again
$$qA^3 + B(q B^2-3Ac)qN + c^3 N^2=0,$$
i.e. $q|c$, i.e. $c = qC$, and
$$A^3 + B(B^2-3AC)qN + C^3 q^2 N^2 = A^3 + B(B^2-3AC)p + C^3 p^2 =0.$$
Thus, if $(a, b, c)$ is an integer solution then $(a/q, b/q, c/q)$ also is an integer solution; this descent cannot be done infinitely since $a, b, c$ are finite, i.e. a contradiction.
*Case 2 : Let $p = q^2 N $ with $q\not\mid N$. Then
$$a^3 + b(b^2-3ac)q^2 N + c^3 q^4 N^2=0.$$
One can assert $a = q A$, then
$$q A^3 + b(b^2-3qAc) N + q^2c^3 N^2=0,\quad \mathbf{(**)}$$
i.e. $b = qB$. Thus
$$ A^3 + B(q B^2-3Ac)q N + q c^3 N^2=0,$$
i.e. $A$ can be divided by $q$ once again. Let $A = qA'$ to have
$$ q^2 A'^3 + B( B^2-3A'c)q N + c^3 N^2=0,$$
i.e. $c = qC$,
$$ q A'^3 + B( B^2-3qA'C) N + q^2 C^3 N^2=0 \quad \mathbf{(**)}$$
Compare two equations marked by (**);
If $(A, b, c)$ satisfies
$$q A^3 + b(b^2-3qAc) N + q^2c^3 N^2=0 $$ then we have another integer solution $(A/q, b/q, c/q)$. So, again by infinite descent, there is no such $(a, b, c)$.
This method also works for $p^{1/4}$ case.
I think I have never seen the matrix of the form
$$\begin{pmatrix} a & b & c \\
cp & a & b \\
bp & cp & a \end{pmatrix} $$
or its variants. Are there any reference?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Stuck on a step for finding the closed formula for catalan numbers from generating function! I am looking at the Frazer Jarvis paper, entitled Catalan Numbers.
http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec01.pdf
In this paper, he derives the closed formula from the generating function.
The $nth$ Catalan number, denoted $C_n$, is the number of ways
of multiplying together $n$ symbols.
I am stuck on page 7, particularly the following steps:
$\begin{equation}C_n=-\frac{1}{2}\bigg\{\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-3}{2})}{n!}(-4)^n\bigg\}\end{equation}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(1 \cdot 2 \cdot 3 \cdot 4...(2n-4)(2n-3)(2n-2)}{n!(n-1)!}2^{2}\bigg\}$
Particularly, from the 2nd to the 4th line, I am struggling to understand the steps taken here.
I am confused with what he is doing with the $\frac{1}{2}$. Is he pulling it out in the 3rd step? And most of all, I am very confused with how the $2^n$ went away in the 4th step. If anyone could please help me understand some of the steps that he skipped here, I would really appreciate it! Thank you so much!!!
| Start from the second line,
$$
\begin{aligned}
C_n=&\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}\\
\text{line 3 } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
=&\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}\\
=&\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot \frac 22 \cdot\frac{3}{2} \cdot \frac 42...(\frac{2n-4}{2})(\frac{2n-3}{2})(\frac{2n-2}{2})]}{n!(n-1)!}2^{2n}\bigg\}\\
=&\frac{1}{2}\bigg\{\frac{1}{2^{2n-1}}\frac{(2n-2)!}{n!(n-1)!}2^{2n}\bigg\}\\
=& \frac{(2n-2)!}{n!(n-1)!}=\frac 1n \binom{2n-2}{n-1}.
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve three-variable system $x^2 - yz = 1,\>y^2 - xz = 2,\>z^2 - xy = 3$ I am going around in circles on this system
$$x^2 - yz = 1\\
y^2 - xz = 2\\
z^2 - xy = 3$$
I have tried a few things (below) but keep hitting a wall. I know that by adding the three equations we get:
$x^2 + y^2 + z^2 - yz - xz - xy =6$. And also that:
$(x-y)^2 + (x-z)^2 + (y-z)^2 = 2x^2+2y^2+2z^2-2xy-2xz-2yz=12$
I have tried the following:
$(x-y)(x+y+z)=(x^2-yz)-(y^2-xz)=-1$
$(x-z)(x+y+z)=(x^2-yz)-(z^2-xy)=-2$
$(y-z)(x+y+z)=(y^2-xz)-(z^2-xy)=-1$
So I think it follows that $x-y=y-z$ and therefore $x-z=2(x-y)$
Substituting in:
$(x-y)^2+[2(x-y)]^2+(x-y)^2=12$
$6(x-y)^2=12$
$x-y=\sqrt{2}$
Therefore: $y-z=\sqrt{2}$ and $x-z=2\sqrt{2}$
But when I rearrange and substitute into one of the original equations, the solutions I get for x, y and z don't actually work.
$y=x-\sqrt{2}$ and $z=x-2\sqrt{2}$
Substituting into the first equation:
$x^2-(x-\sqrt{2})(x-2\sqrt{2})=1$
$x=\frac{3\sqrt{2}}{2}$
Giving $y=\frac{\sqrt{2}}{2}, z=-\frac{\sqrt{2}}{2}$
Unfortunately, substituting these values into the three original equations only satisfies the second one. Can anyone please help?
| This type of the system is well known. Its general form reads
$$x^2 - yz = a,\>\>\>y^2 - xz = b,\>\>\>z^2 - xy = c$$
and has the solutions given in the symmetric expression below
$$(x,y,z)=\pm \frac{(a^2-bc, b^2-ca, c^2 -ab)}{\sqrt{a^3+b^3+c^3-3abc}}
$$
So, substitute $a=1,\>b=2,\>c=3$ to obtain
$\pm(\frac{-5}{3 \sqrt{2}},\frac{1}{3 \sqrt{2}},\frac{7}{3
\sqrt{2}})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If a, b and c are positive integers such that $ab = 432$, $bc = 96$ and $c < 9$, then the smallest possible value of $a + b + c$ is Below are the steps I have done so far.
Find $a$ in terms of $c$
$$
ab = 432 \\
bc = 96 \\
\frac{ab}{bc} = \frac ac = \frac 92 \implies a = 4.5c.
$$
Find $b$ in terms of $c$
$$
ab = 432 \\
\implies 4.5c \cdot b = 432 \\
\implies b = \frac{96}c.
$$
Substituting, we get
$$
a + b + c = 4.5c + \frac{96}c + c = 5.5c + \frac{96}c.
$$
How should I proceed from here?
| The factors of $96$ that are less than $9$ are $\{1,2,3,4,6,8\}$
Plug these values of $c$ into your equation above to find the set of $a+b+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving $9x \equiv 21 \bmod 30$
Solving $9x \equiv 21 \bmod 30$
My approach:
Finding $gcd(9,30)$:
\begin{align}
30 &= 9 \cdot 3 + 3 \\ 9 &= 3 \cdot 3 + 0
\end{align}
So $gcd(9,30)=3$ and since $3 \mid 21$ so the congruence has $3$ incongruent solutions. Express $3$ as a linear combination of $9$ and $30$, we get: $$3 = 30-9 \cdot 3 \ \ \ \ \ \ (1)$$
We have to solve this equation to find $x_0$:
$$9x-30y=21 \ \ \ \ \ \ (2)$$
Multiplying $(1)$ by $7$ we get:
$$21=30 \cdot7-9 \cdot 21$$
Rewritten in form of $(2)$:
$$-9 \cdot (21) + 30 \cdot (7) = 21 $$
So $x_0=21$, the incongruent solutions are:
$$21 + 10n, \ \ \ \ \ n = 0,1,2$$
Then $x=21,31,41$. But my answer is wrong, the answer is: $x=9,19,29$. Can you tell me which step is wrong. Thanks for your help!
| if $9x = 21 + 30 t,$ we see that $3x = 7 + 10 t,$ the reverse also holding. Solve
$$ 3x \equiv 7 \pmod {10} $$
for which you need a reciprocal of $3.$ Since $21 \equiv 1 \pmod {10} $ we see that
$$ \frac{1}{3} \equiv 7 \pmod {10} , $$
and.......
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Identity with the floor function I'm struggling to complete proof of the following identity:
$$
\Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m - n +1}{2} \Bigr\rfloor =m,
$$
where $m$ and $n$ are both integer.
By definition of floor function, $x-1 < \lfloor x \rfloor \leq x$.
Then
\begin{align}
\frac{m+n}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor \leq & \frac{m+n}{2} \\
\frac{m-n + 1}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor \leq & \frac{m-n +1}{2}.
\end{align}
By adding member to member, we obtain the following result:
$$
m +\frac{1}{2} -2 < \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \leq m +\frac{1}{2}.
$$
Which implies
$$
-1.5 < \Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2} \Bigr\rfloor -m \leq .5
$$
This results in
$$
\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \in \left\{0,1\right\}.
$$
How to decide that the result is $0$?
Any help is welcome.
| Notice that $\forall x \in \mathbb Z, \left\lfloor \frac x2 \right\rfloor = \frac x2 - \left\lfloor \frac{x \mod 2}{2}\right\rfloor$, therefore
$$\left\lfloor \frac{m+n}{2}\right\rfloor + \left\lfloor \frac{m-n+1}{2}\right\rfloor\\
=\frac{m+n}{2} - \left\lfloor \frac{m+n \mod 2}{2}\right\rfloor + \frac{m-n+1}{2}- \left\lfloor \frac{m-n+1 \mod 2}{2} \right\rfloor$$
$$= m + \frac 12 - \left\lfloor \frac{m+n \mod 2}{2}\right\rfloor - \left\lfloor \frac{m-n+1 \mod 2}{2} \right\rfloor\tag1$$
Since $m+n$ and $m-n$ have the same parity, the last two terms in $(1)$ sum to $-\frac 12$, and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.
Is there a shorter/simpler solution than the one presented below? It feels there is some ‘trick’ to it. The solution presented below is more a ‘straight forward’ one.
Solution
We have
\begin{gather*}
\left\{
\begin{aligned}
x+y&=a\\
x^3+y^3&=10a^3
\end{aligned}
\right.
\quad\Leftrightarrow\quad
x^3+(a-x)^3=10a^3
\quad\Leftrightarrow\quad
x^2-ax-3a^2=0
\end{gather*}
which has the solutions
$$
x_{1,2}=\tfrac{1}{2}(1\pm\sqrt{13}\,)a
\qquad \Rightarrow \qquad
y_{1,2}=\tfrac{1}{2}(1\mp\sqrt{13}\,)a.
$$
Since
$$
(1+z)^4+(1-z)^4=2(1+6z^2+z^4)
$$
we have
\begin{align*}
x_1^4+y_1^4
&
=\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}
=\tfrac{a^4}{16}\cdot2\bigl(1+6z^2+z^4\bigr)\big|_{z=\sqrt{13}}
\\&=\tfrac{a^4}{8}(1+6\cdot13+13^2)
=\tfrac{a^4}{8}\cdot248
=31a^4
\end{align*}
and, as above,
$$
x_2^4+y_2^4
=\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}
=31a^4.
$$
Hence, the answer is $31a^4$.
The original exam
| Since $x+y=a$,
$$10a^3=x^3+y^3=(x+y)^3-3xy(x+y)=a^3-3axy\,,$$
and $xy=-3a^2$.
$$\begin{align*}
x^4+y^4&=(x+y)^4-2xy\left(2x^2+3xy+2y^2\right)\\
&=(x+y)^4-2xy\left(2(x+y)^2-xy\right)\\
&=a^4+6a^2\left(2a^2+3a^2\right)\\
&=a^4+30a^4\\
&=31a^4\,.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Remainder when $\frac{f(x)}{x^3+4x^2+x-6}$ and given two other remainders Determine the remainder $r$ when $$\frac{f(x)}{x^3+4x^2+x-6}\rightarrow remainder \ r$$.
The following is known $$\frac{f(x)}{x^2+2x-3}\rightarrow remainder\ (x+2)\\
\frac{f(x)}{x+2}\rightarrow remainder\ (1)\\$$
My work:
$$(x^3+4x^2+x-6)=(x^2+2x-3)(x+2)\\(x^2+2x-3)=(x-1)(x+3)\\
(x+2)\\
f(x)=(x^3+4x^2+x-6)g(x)+(Ax+B)\\
f(1)=A+B=x+2\\
f(-3)=-3A+B=x+2\\
f(-2)=-2A+B=1\\
\left\{\begin{matrix}
A+B=x+2& x=-1\\
-3A+B=x+2& B=1\\
-2A+B=1& A=0
\end{matrix}\right.\\
r=(-1)(0)+1=1$$
Can the remainder really be $r=1$? My intuition says it's wrong.
EDIT:
$$f(x)=(x^3+4x^2+x-6)g(x)+(Ax^2+Bx+C)\\
f(1)=A+B+C=x+2\\
f(-3)=9A-3B+C=x+2\\
f(-2)=4A-2B+C=1\\
\left\{\begin{matrix}
A+B+C=x+2& A=\frac{x+3}{6}\\
9A-3B+C=x+2& B=\frac{3x+5}{6}\\
4A-2B+C=1 &C=\frac{4-2x}{6}
\end{matrix}\right.\\
r=\frac{x+3}{6}x^2+ \frac{3x+5}{6}x+\frac{4-2x}{6}=\frac{x^3+6x^2+3x+9}{6}$$
| Since:
$f(x) \equiv x+2 \ (mod \ x^2+2x-3)$
$f(x) \equiv 1 \ (mod \ x+2)$
$x^2+2x-3 \equiv -3 \ (mod \ x+2)$
We have that:
$f(x)=(-1/3x^2-2/3x+1)+(x+2)+k \ (x^3+4^2x+x-6)$
so the remainder is: $(-1/3x^2-2/3x+1)+(x+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct? I've found this limit by this way. Am I correct?
Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$
Let's see that:
\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\frac{n^2-n^2-2n-1}{n^2+n}\\&=\frac{-2n-1}{n^2+n}\\&=\frac{-\frac{2}{n}-1}{1+\frac{1}{n}}
\end{align}
Así,
\begin{align}
\lim_{n \rightarrow \infty}\left ( \frac{n}{n+1}-\frac{n+1}{n} \right ) &=\lim_{n \rightarrow \infty} \frac{-\frac{2}{n}-1}{1+\frac{1}{n}}=\frac{-1}{1}=-1
\end{align}
Am I correct? Is there another way to find it? I would really be very grateful if you can help me with this. Thank you very much!
| $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n} \implies \lim_{n\rightarrow \infty} \dfrac{1}{1+\frac{1}{n}}-1-\dfrac{1}{n} $ equals?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
$\lim_{n \rightarrow \infty} \frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}$. Am I correct? I have to find this limit:
\begin{align} \lim_{n \rightarrow \infty}
\frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1} \end{align}
My attempt:
Since we know that $-1\leq \sin (x) \leq 1$ for all $x \in \mathbb{R}$ we have:
\begin{align}
\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot(-1)}{n+1}\underbrace{\leq}_{\text{Is this fine?}}&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{(1)^{n}\cdot\sqrt{n}\cdot(1)}{n+1}\\ \\ \lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{n}}{n+1}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n+1}
\end{align}
My doubt in the inequeality is because of the $(-1)$ terms. If everything is correct, then we have
\begin{align}
\lim_{n \rightarrow \infty} \frac{(-1)^{n+1}\cdot\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\leq&\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq \lim_{n \rightarrow \infty} \frac{\sqrt{\frac{1}{n}}}{1+\frac{1}{n}}\\ \\ \Rightarrow \ \ \ 0 \leq &\lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1} \leq 0 \\ \\ \therefore& \lim_{n \rightarrow \infty} \frac{(-1)^{n}\cdot\sqrt{n}\cdot\sin{(n^{n})}}{n+1}=0
\end{align}
Am I correct? If I am not, how can I solve it? There are other ways to find this limit? I really appreciate your help!
| The first inequality is not true in general: if $a\leqslant b$ and $c<0$, then $ac\color{red}{\geqslant}bc$.
But note that$$\left|\frac{(-1)^{n}\sqrt{n}\sin(n^{n})}{n+1}\right|\leqslant\frac{\sqrt n}{n+1}\to0,$$and therefore your limit is $0$ indeed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3974238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Is it possible for $x_{n+1}-x_{n}$ to converge to $0$, but the limit of $x_n$ to not exist? I was reading about subsequences today and I came across an example which said the if $x_{n+1}-x_{n}$ converges to $0$, it's possible for $x_n$ to not have a finite limit (for example if we take $x_n$ to be the harmonic series). But then I thought "is it possible for the limit of $x_n$ to not exist at all?" and I think the answer is no,because I couldn't find any examples, but I am not sure. Could you please tell me what your opinion is on this?
Thanks in advance!
| The answer is yes!
Not only you can find examples were the limit is not finite but is for example $\infty$. You can also have examples were a full interval is a set of limit points of the sequence.
Look at a sequence that will slide back and forth from $0$ to $1$ with steps decreasing to zero for example. Let's for example consider $\{x_n\}$ whose initial terms are
$$
\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \dots$$
Formal definition of $\{x_n\}$ is
$$
x_n=\begin{cases}
\frac{1}{2} &\text{ for } n= 1\\
\frac{n}{k+2} - \frac{k(k+1)}{2(k+2)} &\text{ for } \frac{k(k+1)}{2} \lt n \le \frac{(k+1)(k+2)}{2}
\end{cases}$$
The values of the sequence $\{x_n\}$ are in $(0,1) \cap \mathbb Q$. Moreover, one can notice that $\{x_n\}$ takes each rational number of $(0,1)$ as value an infinite number of times. Indeed for $\frac{p}{q} \in (0,1)$ with $1 \le p \lt q$ and $m \ge 1$ we have $$
\frac{(mq-2)(mq-1)}{2} \lt \frac{(mq-2)(mq-1)}{2} + mp$$ and $$\begin{aligned}
\frac{(mq-2)(mq-1)}{2} + mp &\le \frac{(mq-2)(mq-1)}{2} + m(q-1)\\
&\le \frac{(mq-2)(mq-1)}{2} + mq-1\\
&= \frac{(mq-1)mq}{2}
\end{aligned}$$ Hence $$\begin{aligned}
r_{\frac{(mq-2)(mq-1)}{2} + mp} &= \frac{(mq-2)(mq-1)}{2mq} + \frac{mp}{mq} - \frac{(mq-2)(mq-1)}{2mq}\\
&= \frac{mp}{mq}\\
&= \frac{p}{q}
\end{aligned}$$ proving the desired result. As the rational numbers of the segment $(0,1)$ are dense in $[0,1]$, we can conclude that the set of limit points of $\{x_n\}$ is exactly the interval $[0,1]$.
See here for additional examples.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Eliminate $\theta$ and prove $x^2+y^2=1$ We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$
And want to prove $x^2+y^2=1$
My works:
I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
By extracting $\sin\theta\cos\theta$ we get: $$2x\cos\theta-y\sin\theta=2y\sin\theta-x\cos\theta$$
$$x\cos\theta=y\sin\theta$$
But I don't know whether this helps or not.
| You got $x=y\tan\theta$. Now substitute back into the original equations to get $y=\cos\theta$ and $x=y\tan\theta=\sin\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Question about proving differentiability I'm asked to prove $f\left(x\right)=\left(x\cdot\ln x\right)^{\ln x}$ is differentiable for every x > 1
and calculate it's deriviate.
I have no idea how to approach this question.. Do I need to define a function g that will be the inverse of f?
i.e. $g\left(x\right)=\left(\frac{e^{x}}{x}\right)^{e^{x}}$, prove it's continuous, differential and that it's derivate is not 0 and then use the chain rule?
My attempt:
$f\left(x\right)=\left(x\cdot\ln x\right)^{\ln x}=e^{\ln\left(x\cdot\ln x\right)^{\ln x}}=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}$
Note that by the algebra of differential functions $e^x,ln(x)$ are differential for every $x > 1$. Therefore $g(x)={\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}$ is also differential for every $x > 1$, and $f$ as a whole is differential for every $x > 1$. Therefore by the chain rule we have
$f'\left(x\right)=\left(e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\right)^{'}=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{1}{x}\cdot\ln\left(x\cdot\ln x\right)\right)+\left(\frac{1}{x\cdot\ln x}\cdot\left(\ln x+1\right)\right)\cdot\ln x\right)$
$=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x\cdot\ln x\right)}{x}\right)+\left(\frac{\ln^{2}x+\ln x}{x\cdot\ln x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x\right)+\ln\left(\ln x\right)}{x}\right)+\left(\frac{\ln x+1}{x}\right)\right)$
$=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x\right)+\ln\left(\ln x\right)+\ln x+1}{x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{2\ln\left(x\right)+\ln\left(\ln x\right)+1}{x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x^{2}\right)+\ln\left(\ln x\right)+1}{x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x^{2}\cdot\ln x\right)+1}{x}\right)\right)$
| if $x \ge e$ then $\ln x \ge 1$ so $x\cdot \ln x > 0$ and $\ln (x\cdot \ln x)$ exists and $x\cdot \ln x = e^{\ln (x\cdot \ln x)}$.
And then $f(x) = (x\cdot \ln x)^{\ln x}= [e^{\ln (x\cdot \ln x)}]^{\ln x} = e^{\ln x\dot \ln(x\cdot \ln x)}$.
And that can be differentiated by the using the chain rule and the product rule many times.
If we use the notation $exp(x)= e^x$ and $exp'(x) = $ derivative of $exp(x) =$ derivative of $e^x = e^x=exp(x)$ then
First use of chain rule: $[exp(\ln x\dot \ln(x\cdot \ln x))]' = exp'(\ln x\dot \ln(x\cdot \ln x))\times (\ln x\dot \ln(x\cdot \ln x))' =$
$exp(\ln x\dot \ln(x\cdot \ln x))\times (\ln x\dot \ln(x\cdot \ln x))'=$
$(x\cdot \ln x)^{\ln x} \times (\ln x\dot \ln(x\cdot \ln x))'=$
Now apply the product rule;
$(x\cdot \ln x)^{\ln x} \times [\ln' x\cdot \ln(x\cdot \ln x) + \ln x (\ln(x\cdot \ln x))'=$
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\ln(x\cdot \ln x))'=$
Now apply the chain rule again:
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\ln'(x\cdot \ln x)\times (x\cdot \ln x)')=$
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (x\cdot \ln x)')=$
And use the product rule again
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (x\cdot (\ln x)'+ (x)'\ln x))=$
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (x\cdot\frac 1x+ 1\cdot\ln x))=$
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (1+ \ln x))=$
$(x\cdot \ln x)^{\ln x}[\frac {\ln(x\cdot\ln x)}x + \ln x(\frac 1{x\ln x} + \frac 1x)]$.
Sof $1 < x < e$ we have $x\ln x < 0$ so $\ln (-x\cdot \ln x)$ exists and $x\cdot \ln x= e^{-\ln(-x\ln x)}$.
And do what we did above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$
This solution isn't true. Where am I wrong?
EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.
| $$\int x\sqrt{1-x^2}sin^{-1}x dx$$
$$x = sin\theta$$
$$cos\theta= \frac{dx}{d\theta}$$
$$dx= cos\theta d\theta$$
$$\int x\sqrt{1-x^2}sin^{-1}x dx= \int sin\theta(\sqrt{cos^2\theta})\theta (cos\theta d\theta)$$
$$=\int \theta sin\theta cos^2\theta d\theta$$
$$=\int \theta sin\theta (1-sin^2\theta) d\theta$$
$$=\int \theta sin\theta -\int \theta sin^3\theta d\theta$$
$$=\int \theta sin\theta -\int \theta\Bigl(\frac{3sin\theta}{4}-\frac{sin3\theta}{4}\Bigl) d\theta$$
$$=\int \theta sin\theta -\int\frac{3\theta sin\theta}{4}+ \int\frac{\theta sin3\theta}{4}d\theta$$
Applying product rule,
$$\int \theta sin\theta= \theta \int sin\theta - \int(\theta)' \int sin\theta d\theta= sin\theta - \theta cos\theta$$
$$-\int\frac{3\theta sin\theta}{4}= \frac{3\theta cos\theta}{4} -\frac{3 \sin\theta}{4}$$
$$\int\frac{\theta sin3\theta}{4}d\theta=\frac{1}{4}[\frac{sin3\theta}{9}-\frac{\theta cos3\theta}{3}]$$
$$\int x\sqrt{1-x^2}sin^{-1}x dx=sin\theta - \theta cos\theta\ + \frac{3\theta cos\theta}{4} -\frac{3 \sin\theta}{4}+\frac{sin3\theta}{36}-\frac{\theta cos3\theta}{12}$$
$$= \frac{1}{4}(sin \theta - \theta cos\theta)+ \frac{sin3\theta}{36}-\frac{\theta cos3\theta}{12}+ c$$
Substitute theta with corresponding expression in x
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Time period of a nonuniform oscillator I am given the equation of a nonuniform oscillator as
$$\dot\theta=\omega-a\sin(\theta)\tag{1}\label{eq1},$$
where $a<\omega$, and I'm told that the period, $T$, of this is given by
$$T=\frac{2\pi}{\sqrt{\omega^2-a^2}}\tag{2}\label{eq2}.$$
To work this out, I first separate the variables for $\eqref{eq1}$ to get
$$\frac{1}{\omega-a\sin(\theta)}d\theta=dt.$$
I know that for the time to increase from $0$ to $T$, then the oscillator has gone from $0$ to $2\pi$. Therefore
$$\int_0^{2\pi}\frac{1}{\omega-a\sin(\theta)}d\theta=\int_0^Tdt$$
and so
$$T=\int_0^{2\pi}\frac{1}{\omega-a\sin(\theta)}d\theta.$$
This is where the fun begins. I use the Weierstrass substitution of $u=\tan\left(\frac{\theta}{2}\right)$ to evaluate this which changes my integral to
$$T=\int_{\theta=0}^{\theta=2\pi}\frac{2}{\omega u^2+\omega-2ua}du.$$
I take the factor of $2$ outside of the integral and complete the square of the quadratic in the denominator to arrive at
$$T=2\int_{\theta=0}^{\theta=2\pi}\frac{1}{\omega\left(\left(u-\frac{a}{\omega}\right)^2+1-\frac{a^2}{\omega^2}\right)}du.$$
I factor out $\frac{1}{\omega}$ from the integral, and use the substitutions $x=u-\frac{a}{\omega}$ and $y=1-\frac{a^2}{\omega^2}$ to make this
$$T=\frac{2}{\omega}\int_{\theta=0}^{\theta=2\pi}\frac{1}{x^2+y}dx.$$
I then use the substitution $x=\sqrt{y}\tan(v)$, meaning $x^2=y\tan^2(v)$ and $dx=\sqrt{y}\sec^2(v)dv$. This makes my integral
$$T=\frac{2}{\omega}\int_{\theta=0}^{\theta=2\pi}\frac{\sqrt{y}\sec^2(v)}{y\tan^2(v)+y}dv.$$
I take out a factor of $\frac{\sqrt{y}}{y}$ and find the following:
$$T=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}\frac{\sec^2(v)}{\tan^2(v)+1}dv=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}\frac{\sec^2(v)}{\sec^2(v)}dv=\frac{2\sqrt{y}}{\omega y}\int_{\theta=0}^{\theta=2\pi}1 dv=\frac{2\sqrt{y}}{\omega y}\left[v\right]^{\theta=2\pi}_{\theta=0}.$$
Then I start the process of re-inserting my many substitutions. First $v=\tan^{-1}\left(\frac{x}{\sqrt{y}}\right)$, then $x=u-\frac{a}{\omega}$, then $u=\tan\left(\frac{\theta}{2}\right)$, and finally $y=1-\frac{a^2}{\omega^"}$. This gets me
$$T=\frac{2\sqrt{1-\frac{a^2}{\omega^2}}}{\omega\left(1-\frac{a^2}{\omega^2}\right)}\left[\tan^{-1}\left(\frac{\tan\left(\frac{\theta}{2}\right)-\frac{a}{\omega}}{\sqrt{1-\frac{a^2}{\omega^2}}}\right)\right]^{2\pi}_{0}.$$
I can simplify this slightly. Firstly,
$$\frac{2\sqrt{1-\frac{a^2}{\omega^2}}}{\omega\left(1-\frac{a^2}{\omega^2}\right)}=\frac{2}{\omega\sqrt{1-\frac{a^2}{\omega^2}}}=\frac{2}{\omega\sqrt{\frac{\omega^2-a^2}{\omega^2}}}=\frac{2}{\sqrt{\omega^2-a^2}},$$
and secondly
$$\tan^{-1}\left(\frac{\tan\left(\frac{\theta}{2}\right)-\frac{a}{\omega}}{\sqrt{1-\frac{a^2}{\omega^2}}}\right)=\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right).$$
So now I have
$$T=\frac{2}{\sqrt{\omega^2-a^2}}\left[\tan^{-1}\left(\frac{\omega\tan\left(\frac{\theta}{2}\right)-a}{\sqrt{\omega^2-a^2}}\right)\right]^{2\pi}_{0}.$$
I feel like I'm so very close to $\eqref{eq2}$ now, but this is where I become stuck. I know that $\tan\left(\frac{2\pi}{2}\right)=\tan\left(\frac{0}{2}\right)=0$, which means when I evaluate the square brackets between $0$ and $2\pi$ they just cancel out and I'm left with
$$T=0.$$
Could someone please tell me where I have gone wrong, or what I am missing here? Been stuck on this one all day to no avail!
| Note that
$$
T=
\int_0^{2\pi}\frac{2e^{-i\theta}}{2i\omega-a(e^{i\theta}-e^{-i\theta}) }ie^{i\theta}\,d\theta=
\int_0^{2\pi}\frac{2}{2i\omega e^{i\theta}-a(e^{2i\theta}-1) }ie^{i\theta}\,d\theta=\int_\gamma f,
$$
with $\gamma\colon[0,2\pi]\to\mathbb C$ given by $\gamma(\theta)=e^{i\theta}$ and $f(z)=2/(-az^2+2i\omega z+a)$. By the Residues' theorem:
$$
T=2\pi i \,{\rm Res}\left(f,\frac{i(\omega+\sqrt{\omega^2-a^2})}{a}\right)=2\pi i\frac{2}{-a}\cdot\frac1{-\frac{2i\sqrt{\omega^2-a^2}}{a}}= \frac{2\pi}{{\sqrt{\omega^2-a^2}}}
$$
since the poles satisfy $|i(\omega+\sqrt{\omega^2-a^2})/a|<1$ and $|i(\omega-\sqrt{\omega^2-a^2})/a|>1$.
| {
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"url": "https://math.stackexchange.com/questions/3985408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial fractions of $\frac{x^4}{(x^2-1)^3}$ (is it possible to find coefficients without expanding?) I want to decompose the fraction $\frac{x^4}{(x^2-1)^3}$ :
$$\frac{x^4}{(x^2-1)^3}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{(x-1)^3}+\frac D{x+1}+\frac E{(x+1)^2}+\frac F{(x+1)^3}$$
Because $f(x)=\frac{x^4}{(x^2-1)^3}$ is an even function we have $f(x)=f(-x)$. Hence we have $A=-D\;,\;B=E\;,\;C=-F$ :
$$\frac{x^4}{(x^2-1)^3}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{(x-1)^3}+\frac {-A}{x+1}+\frac B{(x+1)^2}+\frac {-C}{(x+1)^3}$$
Hence:
$$A(x-1)^2(x+1)^3+B(x-1)(x+1)^3+C(x+1)^3-A(x+1)^2(x-1)^3+B(x+1)(x-1)^3-C(x-1)^3=x^4$$
If we plug in $x=1$ we get $C=\frac18$. but I can't find $A$ and $B$ without expanding $(x-1)^2(x+1)^3\ldots$ is it possible to find them without expanding whole expression?
| Following my comments, we may try two simple values of $x$ to reduce computations. Here I believe $x=0$ and $x=2$ are simple enough.
Plugging in $x=0$ gives $$A-B+C-(-A)+(-B)-(-C)=0\implies A-B=-\frac 1 8$$
and plugging in $x=2$ gives $$A\cdot 3^3 - B\cdot 3^3 + C\cdot 3^3 - A\cdot 3^2 + B\cdot 3 - C = 2^4\implies A + \frac 5 3 B =\frac {17} {24}$$
which gives us $A=\frac 3 {16}, B=\frac 5 {16}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$.
Integrate the following integral:
$$\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$$
I did this question a while ago and the answer is correct. However, when I started revising this topic again, I can't seem to figure out how I went from the first step $\left( \text{for this integral}\int \log _{3x}\left(x^2\right)dx\right)$ to the second step. Here's my attempt:
\begin{align}
\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx &= {\int{\frac{2}{\sin{\left(2 x \right)}} d x}} \\
&= {\int{\frac{1}{\sin{\left(x \right)} \cos{\left(x \right)}} d x}} \\
&= {\int{\frac{\sec^{2}{\left(x \right)}}{\tan{\left(x \right)}} d x}} \\
&\stackrel{{u = \tan{\left(x\right)}}}{=} {\int{\frac{1}{u} d u}} \\
&= {\ln{\left(u \right)}} + c \\
&= \ln{\left(\left|{\tan{\left(x \right)}}\right| \right)}+
\end{align}
\begin{align}
\int \log _{3x}\left(x^2\right)dx &= x\log _{3x}\left(x^2\right)-\int \frac{2\ln \left(3\right)}{\ln ^2\left(3x\right)}dx \\
&= x\log _{3x}\left(x^2\right)-\left(-\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}+\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right)\right) \\
&= 2x\log _{3x}\left(x\right)+\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}-\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right) + c
\end{align}
Therefore:
$$\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx = \left(\ln \left|\tan \left(x\right)\right|+c\right)\left(2x\log _{3x}\left(x\right)+\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}-\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right)+c\right)$$
| $ \sin{(2x)} = 2 \cos{(x)} \sin{(x)}$ if that's what you are asking for...
| {
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"question_score": "1",
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prove or disprove $(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}$ The following question I read in a book, but the book does not give proof. I doubt the correctness of the result
let $p>3$ be prime number. prove or disprove
$$(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}\tag{1}$$
I think use binomial theorem
it maybe show $$\binom{2p^2}{k}\equiv 0,\pmod {p^2},k=1,2,\cdots,p^2-1,p^2+1,\cdots,2p^2-1\tag{2}$$
but other hand I think (2) is not right.becuase let $p=5,k=5$
we have $$\binom{2p^2}{k}=\binom{50}{5}=\dfrac{50\cdot 49\cdot 48\cdot 47\cdot 46}{5\cdot 4\cdot 3\cdot 2\cdot 1}\ne 0\pmod {25}$$,so I think $(1)$ is not right?
| You got it right. By Lucas correspondence the congruence (1) would be correct if it were modulo $p$ only. If we did this modulo $p$ we would be doing arithmetic in the ring $\Bbb{Z}_p[x]$ — a commutative ring of characteristic $p$. This means that we can apply Freshman's Dream twice, and get
$$
\begin{aligned}
(x+1)^{2p^2}&=\left((x+1)^2\right)^{p^2}\\
&=\left((x^2+2x+1)^p\right)^p\\
&\equiv\left(x^{2p}+2^px^p+1\right)^p\pmod{p}\\
&\equiv\left(x^{2p}+2x^p+1\right)^p\pmod{p}\\
&\equiv x^{2p^2}+2x^{p^2}+1\pmod{p}.
\end{aligned}
$$
But, this is only modulo $p$.
Your example is the first failure modulo $p^2$. Well, the same way you would see that $\binom{18}3$ is not divisible by $9$, but the question specified $p>3$. Well done.
Kummer's theorem (see Qiaochu's link) says the same thing. When you do grade school addition of $5+45$ in base five there will only be a single carry, from the $5^1$-digit to the $5^2$-digit. Hence $\binom{50}5$ is divisible by $5^1$ but not divisible by $5^2$. For the same reason $\binom{2p^2}p$ is not divisible by $p^2$ for any prime $p>2$.
As the last observation, when $p=2$ the congruence (1) will be correct modulo $4$. For example the binomial coefficient $\binom 82$ is divisible by $2^2$ as $8$ has a non-zero bit at position $2^3$, and we get the required two carries when calculating $2+6$ in base two.
| {
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Given $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$, find $\tan x\cot 2x$ I can't solve this problem. I tried to find $\tan x$ directly by solving cubic equations but I failed.
The problem is to find $\tan x\cot 2x$ given that
$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$
How am I supposed to solve this problem?
| If $t:=\tan x$ then $\frac{t(3-t^2)}{1-t^2}=\tfrac{2}{\sqrt{3}}$ so $t^3-\tfrac{2}{\sqrt{3}}t^2+3t+\tfrac{2}{\sqrt{3}}=0$. This has one real root, $t=\tfrac{2+33/a-a}{3\sqrt{3}}$ with $a:=\sqrt[3]{154+9\sqrt{443}}$. Now just calculate $\frac{1-t^2}{2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain. $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$
I know I have to switch the $f(x)$ and the $y$:
$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\frac{2f^{-1}(x)+1}{3} \to f^{-1}=\frac{3\arcsin^2(x)-1}{2}$
To find the domain:
$$-\frac{3\pi+1}{2}\le \frac{2x+1}{3}< -\frac{3\pi+2}{4}$$
$$-\frac{9\pi+3}{2}\le 2x+1< -\frac{9\pi+6}{4}$$
$$-\frac{9\pi+5}{2}\le 2x< -\frac{9\pi+10}{4}$$
$$-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$$
The inverse function is $f^{-1}=\frac{3\arcsin^2(x)-1}{2}$ and its domain is $-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$
But I am slightly confused since Desmos will not map it correctly. I'm asking if this is the correct solution or I went wrong somewhere.
Thanks
| $$\text{Let $y^2 = f(x)=\sin^2\left(\frac{2x+1}{3}\right)$}$$
So $y \in [-1,1]$ and
$y=\sin\left(\dfrac{2x+1}{3}\right)$
where $\dfrac{2x+1}{3} = 2n\pi + \theta $ and $-\pi \le \theta \le \pi$
for some $n \in \mathbb Z$.
We compute
\begin{align}
\theta &= \arcsin y \\
\dfrac{2x+1}{3} &= 2n \pi + \arcsin y \\
2x &= 6n \pi - 1 + 3 \arcsin y \\
x &= 3n \pi - \dfrac 12 + \dfrac 32 \arcsin y \\
\end{align}
| {
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"source": "stackexchange",
"question_score": "1",
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Probability of finding a non-repeating combination This is a real life problem, which I'm turning into balls just to make it easier to understand.
Suppose balls can be numbered from 1 to 12.
Suppose I have 10 boxes. Each box has 3 distinct balls, distributed at random.
A valid combination is if I can find, from those 10 boxes and taking 1 ball from each, a combination where all balls have distinct numbers (i.e. there are (12,10) valid combinations).
Here are some examples, to make it more concrete:
*
*Box 1 has (1,2,3), Box 2 has (2,3,4), Box 3 has (3,4,5)... Box 10 has (10,11,12). This is the best case scenario, I can easily make valid combinations such as (1,2,3,4,5,6,7,8,9,10); (1,2,3,..,9,11); and so on
*Box 1 has (1,2,3), Box 2 has (1,2,3), Box 3 has (1,2,3), Box 4 has (1,2,3). This is the worst case scenario, as I can't find any valid combination. I will always have to repeat 1, 2 or 3.
I would like to know what is the probability of finding scenarios that yield valid combinations from such a setup. In other words - given a scenario of 10 box with 3 distinct balls in each, what is the probability that from those 10 boxes I can find at least one valid combination (it doesn't matter if there is more than 1 valid combination).
Here's a potential approach:
The universe of potential combinations: A = C(12,3)^10
The cases where numbers repeat and I get an invalid scenario:
B = C(12,9) * C(9,3)^10
Probability is then 1 - B/A.
| Consider a simpler case: 4 boxes and balls numbered from 1 to 6. We still put 3 balls in each box. I have chosen this case, since it is possible to write out all combinations and check corresponding probabilities.
For each box, we now have $6\choose3$ possible combinations.
In the first box, we can select any ball, yielding a probability of 1.
For the second box, we have two options to select a valid ball (i.e. a ball not selected previously):
*
*A box without the ball that we already selected. The total number of possibilities for this case are ${1\choose0}{5\choose3}$. ${1\choose0}$ specifies the number of balls we select from the previously selected balls. ${5\choose3}$ specifies the number of combinations of choosing three balls from the remaining five that have not been selected previously. The probability of selecting a valid ball is 1, since the previously selected ball is not in the box.
*A box with the ball that we already have selected. Out of the $6\choose3$ possibilities, ${1\choose1}{5\choose2}$ have the already selected ball. Selecting a valid ball has a probability of $\frac{2}{3}$.
Thus, selecting a valid ball from the second box is given by $\frac{1\times {1\choose0}{5\choose3} + \frac{2}{3}\times{1\choose1}{5\choose2}}{6\choose3} = \frac{5}{6}$.
For the third box, we have the following cases:
*
*A box without the previously selected balls: ${2\choose0}{4\choose3}$ possibilities. The probability of selecting a valid ball is 1.
*A box with one previously selected ball: ${2\choose1} {4\choose2}$. The probability of selecting a valid ball is $\frac{2}{3}$.
*A box with two previously selected balls: ${2\choose2}{4\choose1}$ possibilities, with a probability of $\frac{1}{3}$ of not selecting a previously selected ball.
For the third box, the probability of selecting a valid ball is consequently:
$\frac{1 \times {2\choose0}{4\choose3} + \frac{2}{3} \times {2\choose1} {4\choose2} + \frac{1}{3}\times {2\choose2}{4\choose1}}{6\choose3} = \frac{4}{6}$.
For the fourth box, we have the following cases:
*
*No previously selected balls: ${3\choose0}{3\choose3}$ possibilities. We have a probability of 1 of selecting a valid ball.
*One previously selected ball: ${3\choose1}{3\choose2}$. Corresponding probability is $\frac{2}{3}$.
*Two previously selected balls: ${3\choose2}{3\choose1}$. Corresponding probability is $\frac{1}{3}$.
*Three previously selected balls: ${3\choose3}{3\choose0}$. Corresponding probability is 0.
For the fourth box, the probability of selecting a valid ball is thus given by:
$\frac{1 \times {3\choose0}{3\choose3} + \frac{2}{3}\times {3\choose1}{3\choose2} + \frac{1}{3} \times {3\choose2}{3\choose1} + 0 \times {3\choose3}{3\choose0}}{6\choose3} = \frac{3}{6}$.
We thus find the probability of valid combinations for this scenario as follows:
$\text{P(selected balls are distinct)} = \frac{6}{6}\times\frac{5}{6}\times\frac{4}{6}\times\frac{3}{6} = \frac{\frac{6!}{(6-4)!}}{6^4}$.
This pattern also arises in your scenario. In your scenario, we find a probability of:
$\text{P(Selected balls are all distinct)} = \frac{\frac{12!}{(12-10)!}}{12^{10}}\approx 0.003868$.
| {
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"url": "https://math.stackexchange.com/questions/3993947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given a sequence $\{u_{n}\}$ so that $\left\{\begin{matrix} u_0=a>1\\u_{n+1}=\sqrt{1+\frac{1}{1+u_n}}\end{matrix}\right..$ Find $\lim u_n.$
Given a sequence $\left \{ u_{n} \right \}$ so that $\left\{\begin{matrix} u_{0}= a> 1\\ u_{n+ 1}= \sqrt{1+ \frac{1}{1+ u_{n}}} \end{matrix}\right..$ Find $\lim u_{n}.$
If there exists a $\lim u_{n},$ I think with $n\rightarrow \infty\Rightarrow u_{n+ 1}= u_{n}= \sqrt{1+ \frac{1}{1+ u_{n}}}\Rightarrow 1< u_{n}< 2.$ What does that help for my problem here ? I need your thought on it ? Thank you.
| Hint: Let $f(x) = \sqrt {1 + \frac{1}{{1 + x}}}$ for $x>0$. Then, for all $x>0$, $y>0$,
\begin{align*}
&\left| {f(x) - f(y)} \right| = \left| {\sqrt {1 + \frac{1}{{1 + x}}} - \sqrt {1 + \frac{1}{{1 + y}}} } \right| \\ &= \frac{{\left| {x - y} \right|}}{{(1 + x)(1 + y)\left( {\sqrt {1 + \frac{1}{{1 + x}}} + \sqrt {1 + \frac{1}{{1 + y}}} } \right)}} \le \frac{1}{2}\left| {x - y} \right|,
\end{align*}
i.e., $f$ is a contraction.
| {
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Proving $\sum_{n=0}^\infty\frac{x^n}{n!}>0$ for all $x\in\Bbb R$ without invoking any Taylor series knowledge base. While studying basic Taylor expansions, I came across this expansion of $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}.$$We already know this function is positive for all $x$. But let us assume we don't know about the Taylor expansion , and try to prove this in the expanded form only, how to will we prove it ?
It's already clearly visible for $ x > 0$ as well as for $ -1<x<0 $ . But how to prove it for $x<-1$ ? I have tried much but not able to build up a strong convincing proof.
Any help would be highly appreciated. thanks!
P.S : prove it 'WITHOUT' using $e^{x}$ function.
| To elaborate on some of the hints in the comments, let $x,y \in \mathbb{R}$. Then:
\begin{align*}
e^{x+y} &= \sum_{n=0}^{\infty}\frac{(x+y)^n}{n!} \\
&= \sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}\frac{x^ky^{n-k}}{n!} \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!} \\
&= 1 + \left(x+y\right) + \left(\frac{x^2}{2!}+xy + \frac{y^2}{2!}\right) + \left(\frac{x^3}{3!} + \frac{x^2y}{2!} + \frac{xy^2}{2!} + \frac{y^3}{3!}\right) + \ldots \\
&= \left(1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right) + y\left(1+x+\frac{x^2}{2!} + \ldots\right) + \frac{y^2}{2!}\left(1+x + \ldots\right) + \ldots \\
&= \left(1+y+\frac{y^2}{2!} + \frac{y^3}{3!} + \ldots\right)\left(1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right) \\
&= e^{x}e^{y}
\end{align*}
This factoring trick is what people in the comments are calling the Cauchy Product. I thought that seeing the algebra done by hand (sort of) might convince you it's not a crazy sophisticated trick. It follows easily from this identity that $e^{-x} = 1/e^{x} > 0$.
| {
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Probability of a certain roll with six dice, each with a single marked side There are six regular (6-sided) dice being rolled. However, each dice has one side colored gold. The 1st has a gold "1", the 2nd has a gold "2"... and the 6th has a gold "6".
I calculate the probability of rolling three sets of two with 6 dice:
$$\frac{{6 \choose 3} {6 \choose 2} {4 \choose 2} {2 \choose 2}}{6^6} = \frac{1800}{6^6}$$
As there are ${6 \choose 3}$ ways to assign the numbers, ${6 \choose 2}$ ways to arrange the 1st pair of dice into the available positions, then ${4 \choose 2}$ for the 2nd pair and then ${2 \choose 2}$ for the 3rd pair.
An example of this kind of roll would be "112233" or "242554".
How can I determine how many of these rolls have 0 gold sides and how many have 1 gold side?
Finding how many have 2 gold sides is:
$${3 \choose 2} \times \frac{2}{6} \times \frac {2}{5} \times \frac {2}{4} \times 1800 = 360$$
As there are ${3 \choose 2}$ ways to assign the gold sides, $\frac{2}{6}$ chance the dice with the gold A is an A, then $\frac{2}{5}$ chance the dice with the gold B is a B, and then $\frac{2}{4}$ chance the dice with the gold C is not a C (in the context "AABBCC", where A, B, and C are different numbers).
And similarly 3 gold sides:
$${3 \choose 3} \times \frac{2}{6} \times \frac {2}{5} \times \frac {2}{4} \times 1800 = 120$$
But for 0-1 gold sides it's more complicated as the chance for the dice with the gold B depends on the position of the dice with the gold A, and so on for the gold C.
| Actually I didn't get how you calculated the number of possibilities for $2,3$ gold sides (though I have the same answers) but anyways this is my method (we'll solve for $2,3$ gold sides and then go for $1,0$ gold sides):
First of all, let's partition our $6$ dices into $3$ pairs of dices. Clearly there are $\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}=15$ ways to do that. Throughout this solution, the number of the two dices in each pair shall be equal and the number of each dice means the number of its golden side. Also we'll name pairs with a dice with gold side as a golden pair. (Trivially no golden pair has more than one dice with gold side)
Now for the case with $3$ gold sides, exactly one dice in each pair shall be golden. Hence, there are $2^3=8$ ways to choose the number of the dices of each pair. So there are $8*15=120$ ways for that.
For the case with $2$ gold sides, we'll first choose the pairs that shall have a golden side. There are $\binom{3}{2}=3$ choices for this. Then there are two choices for each of these pairs to choose which one of the dices of each of these pairs shall have golden side. So there are $3*2^2=12$ choices for this and finally for the third pair, there are $2$ possibilities for the number of the dices of this pair. (Excluding two numbers because it shouldn't be a golden pair and excluding another couple of numbers because the dices of this pair shall have different numbers than the number of first and second pairs) So finally there are $15\cdot12\cdot2=360$ choices in this case.
For the case for only one golden side, there are $3$ ways to choose the golden pair and $2$ choices to choose the dice with golden side. Now for the two other pairs, if the number of the second pair is the number of one of the dices in the third pair, then there are two choices for the number of the second pair and $3$ choices for the number of the third pair. Otherwise there is only $1$ choice for the number of the second pair, and $2$ choices for the number of the third pair. Therefore there are $8$ choices for the numbers of the two other pairs. So at last there are $15\cdot3\cdot2\cdot8=720$ possibilities for this case.
Finally for no golden sides, there are $4$ choices to choose the number on the dices of the first pair and WLOG assumer that this number is the number of one of the dices in the third pair. Then like the previous case, if the number of the second pair is in the third pair, then there is only $1$ possibility for the number of the second pair and $4$ choices for the number of the third pair. Otherwise there are $2$ possibilities for the number of the second pair and $3$ choices for the number of the third pair. Therefore there are $10$ different possibilities for choosing the number of the two other pairs. Hence there are $15\cdot4\cdot10=600$ different possibilities for this case.
P.S. You can verify the answer by noting that $120+360+720+600=1800$.
| {
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If $x^5=2$, find $\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$
If $x^5=2$, find
$$\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$
My attempt-
Since $x^5=2,x^6=2x,x^7=2x^2..$and so on
The equation is equivalent to
$$x^5\left(\frac{1}{2x+x^4}+\frac{1}{2x^2+x^3}+\frac{1}{x^3+x^2}+\frac{1}{x^4+x}\right)$$
Which simplifies to
$$2\left[\frac{3x^4+7x^3+5x^2+8x+12}{2x^4+6x^3+4x^2+6x+12}\right]$$
Now I am struck here.
Any help will be appreciated.
Note: Please don't use root tables and complex numbers too!
| Working mod $x^5-2$,
$$
\scriptsize\frac{3x^3}{1+x}=x^3\frac{1+x^5}{1+x}=x^3-x^4+x^5-x^6+x^7=2-2x+2x^2+x^3-x^4\tag1
$$
$$
\scriptsize\frac{5x}{1+x^2}=x\frac{1+x^{10}}{1+x^2}=x-x^3+x^5-x^7+x^9=2+x-2x^2-x^3+2x^4\tag2
$$
$$
\scriptsize\frac{9x^4}{1+x^3}=x^4\frac{1+x^{15}}{1+x^3}=x^4-x^7+x^{10}-x^{13}+x^{16}=4+8x-2x^2-4x^3+x^4\tag3
$$
$$
\scriptsize\frac{17x^2}{1+x^4}=x^2\frac{1+x^{20}}{1+x^4}=x^2-x^6+x^{10}-x^{14}+x^{18}=4-2x+x^2+8x^3-4x^4\tag4
$$
Adding $\frac13\text{(1)}+\frac15\text{(2)}+\frac19\text{(3)}+\frac1{17}\text{(4)}$, we get
$$
\scriptsize\frac{x^3}{1+x}+\frac{x}{1+x^2}+\frac{x^4}{1+x^3}+\frac{x^2}{1+x^4}=\frac{1336+233x+79x^2+122x^3-44x^4}{765}\tag5
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Puzzling solution for $1/x = 3 - 2\sqrt{x}$ What is the solution for
$$\frac{1}{x} = 3 - 2\sqrt{x}$$
When I plot $1/x$ and $3-2\sqrt{x}$ separately, it meets at $x = 1$.
However when I solve the equation $1/x = 3 - 2\sqrt{x}$ algebraically, I get two solutions $-1/4$ and $1$.
What am I missing?
| If we let $y = \sqrt{x}$, we get
\begin{align*}
\frac{1}{y^{2}} = 3 - 2y & \Longleftrightarrow 1 = 3y^{2} - 2y^{3}\\
& \Longleftrightarrow 2y^{3} - 3y^{2} + 1 = 0\\\\
& \Longleftrightarrow (2y^{3} - 2y^{2}) - (y^{2} - 1) = 0\\\\
& \Longleftrightarrow 2y^{2}(y-1) - (y-1)(y+1) = 0\\\\
& \Longleftrightarrow (2y^{2} - y - 1)(y-1) = 0\\\\
& \Longleftrightarrow \left(y + \frac{1}{2}\right)(y-1)^{2} = 0\\\\
& \Longleftrightarrow \left(y = -\frac{1}{2}\right)\vee(y = 1)
\end{align*}
Since $y = \sqrt{x} > 0$, there is only one possible solution, which is $x = 1$.
Hopefully this helps!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\angle AEF =90^\circ$ given a square $ABCD$ Let $ABCD$ be a square and $E\in CD$ such that $DE=EC$. Let $F\in BC$ such that $BC=4FC$. Prove that $\angle AEF =90^\circ$.
My attempt:
Proving that $\angle AEF =90^\circ$ is the same as proving that $\triangle AEF$ is a right triangle. In other words, we wish to prove that $AE^2 +EF^2 = AF^2$. I’ve tried a lot of methods to reach this point but none of them worked.
| The idea in your attempt definitely works here (although it is not the shortest proof, see the answer by Narasimham): Let $a$ be the sidelength of the square. Then we have \begin{align*}
AE^2&=AD^2+DE^2=a^2+\left(\frac{1}{2}a\right)^2\\
EF^2&=EC^2+CF^2=\left(\frac{1}{2}a\right)^2+\left(\frac{1}{4}a\right)^2\\
AF^2&=AB^2+BF^2=a^2+\left(\frac{3}{4}a\right)^2
\end{align*}
And thus: $$AE^2+EF^2=\frac{25}{16}a^2=AF^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum and minimum value of $\frac{8x - 3y}{\sqrt{4x^2+y^2}}$? For real numbers $\,x, y\neq 0\,$ consider
$$\frac{8x - 3y}{\sqrt{4x^2+y^2}}\,.$$
How to find the maximum and minimum value?
I've already got the maximum by using the Cauchy–Schwarz inequality
$$\big[(2x)^2 + y^2\big]\big[4^2 + (-3)^2\big] \geq (8x - 3y)^2\\[3ex]
25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\\[4ex]
5 \geq \frac{8x - 3y}{\sqrt{(2x)^2 + y^2}}$$
But I cannot get the minimum value.
| You are almost there: Note that by Cauchy-Schwarz you did
$$25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\,.$$
Thus
$$\left|\frac{8x-3y}{\sqrt{4x^2+y^2}}\right|\le 5\tag {*}\\[5ex]
\implies -5\le \frac{8x-3y}{\sqrt{4x^2+y^2}}\le 5$$
The important part was in step (*). The modulus was crucial as $8x-3y$ could be negative too.
| {
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Radius of convergence of $f(z)=\frac{z^4e^z}{(6-z)(2i-z)} $ The Question is:
Compute the radius of convergence for the expansion at the point
$z=4+3i$ of the function $$f(z)=\frac{z^4e^z}{(6-z)(2i-z)}. $$
In my text book the solution is written like this:
$\sqrt{(4-6)^2+3^2}$, but I don't understand how they calculate it. Can someone please explain it?
Thanks!
| The function $f$ is holomorphic on $\mathbb{C} \setminus \{6, 2i\}$. The Radius of Convergence is given by the minimum of the distances from $4+3i$ to $z = 6$ or $z = 2i$. That's due to the fact that we can write $f$ as a power series on $D_R(4+3i)$ (meaning the open disc of radius $R$ around $4+3i$). You also might fancy a look at Radius of Convergence.
The distance from $4 + 3i$ to $z = 2i$ is given by $\sqrt{4^2 + 1^2}$. The distance from $4 +3i$ to $z = 6$ is given by $\sqrt{2^2 + 3^2}$. This will immediately become clear if you draw a figure and use Pythagoras.
Since
\begin{align}
\sqrt{17} = \sqrt{4^2 + 1^2} > \sqrt{2^2 + 3^2} = \sqrt{13}
\end{align}
and $R = \operatorname{min}\{\sqrt{4^2 + 1^2}, \sqrt{2^2 + 3^2}\}$ we find $R = \sqrt{2^2 + 3^2}$ or as the text book has written it: $R = \sqrt{(4-6)^2 + 3^2}$.
| {
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Show that if $r$ and $s$ are positive roots to the equation $x^3-cx+d=0$, then $r+s$ is a root of $x^3-cx-d=0$ If $r$ and $s$ are roots of $x^3-cx+d=0$, then
$$r^3-cr+d=s^3-cs+d , $$
for $r \neq s$. Solving the equation for $c$ gives me
$$ c=r^2+rs+s^2$$
Now to find an expression for $d$
$$ r^3-cr+d=0 $$
$$ d=r(c-r^2) $$
Using the expression for $c$ gives me
$$d=sr^2+rs^2$$
Assume $t$ is a root of the second equation
$$t^3-ct+d=0$$
We can represent $c$ and $d$ in terms of $r$ and $s$ which gives us
$$s*r^2+r*s^2=t(t^2-r^2-rs-s^2)$$
Plugging this into wolfram alpha gives us $t=r+s$, which is what we wanted to show. But how can I solve for $t$ algebraically? Or perhaps there is another way to solve this question?
| Let $p(x) = x^3-xc-d$ then $p(s)=p(r)=0$.
This is the part you already made
$$s^3-cs =d = r^3-rc$$ so
$$ (s-r)(r^2+rs+s^2)= (s-r)c$$ so
$$r^2+rs+s^2=c$$
Let $q(x)=x^3-cx+d$.
Now we have \begin{align}q(s+r)&=(s+r)^3-(r+s)c+d\\
&= p(s)+p(r)+3d +3sr(r+s)\\
&= 0+0+3d-3s(c-s^2)\\
&= 3p(s)\\
&=0
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Leading term asymptotic integral containing logarithm I want to get the leading term in the asymptotic of the integral $$\int_{a}^{+\infty}\frac{x^3}{(1+x^2)^{\alpha}}\ln\frac{x+a}{x-a} \,dx$$ for $a\to +\infty$. Here $a>0$ and $\alpha>\frac{3}{2}$ to ensure integrability at infinity.
My main concern is the integrable singularity in $a$. Usually I do an appropriate change of variables to remove it, but I can't seem to find an appropriate one.
| First, we make a change $x=ta$. The integral becomes $I(a, \alpha)=a^4\int_{1}^{\infty}\frac{t^3}{(1+a^2t^2)^{\alpha}}\ln\frac{t+1}{t-1} \,dt=a^{4-2\alpha}\int_{1}^{\infty}\frac{t^{3-2\alpha}}{(1+\frac{1}{a^2t^2})^{\alpha}}\ln\frac{t+1}{t-1} \,dt$.
Integral converges; the only (integrable) singularity is at $t=1$.
We can expand the integrand into the series
$$I(a,\alpha)=a^{4-2\alpha}\int_{1}^{\infty}t^{3-2\alpha}\ln\frac{t+1}{t-1}{(1-\frac{\alpha}{a^21!}\frac{1}{t^2}+\frac{\alpha(\alpha+1)}{a^42!}\frac{1}{t^4}+...)} \,dt$$ where all integrals converge ($3-2\alpha<1$ and $\ln\frac{t+1}{t-1}\sim\frac{1}{t}$ at $t\to\infty$)
The mian asymptotics term is $a^{4-2\alpha}\int_{1}^{\infty}t^{3-2\alpha}\ln\frac{t+1}{t-1}dt=a^{4-2\alpha}\int_{1}^{\infty}t^{b}\ln\frac{t+1}{t-1}dt$
where $b=3-2\alpha$.
It seems that the easiest way to evaluate the integral $J(b)=\int_{1}^{\infty}t^{b}\ln\frac{t+1}{t-1}dt=\int_{0}^{1}x^{-b-2}\ln\frac{1+x}{1-x}dx$ ($t=\frac{1}{x}$) is to expand the integrand into the series:
$J(b)=2\int_{0}^{1}x^{-b-2}\left(x+\frac{x^3}{3}+\frac{x^5}{5}+...\right)dx=2\left(\frac{1}{-b}+\frac{1}{3(-b+2)}+\frac{1}{5(-b+4)}+...\right)=$$=\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{k+\frac{1}{2}}\frac{1}{k-\frac{b}{2}}=\frac{1}{-b-1}\left(\Psi(-\frac{b}{2})-\Psi(\frac{1}{2})\right)$,
where $\Psi$ is digamma function: $\Psi(1+s)=-\gamma-\sum_{k=1}^{\infty}\left(\frac{1}{k+s}-\frac{1}{k}\right)$
Finally, we get the main asymptotics term$$I(a, \alpha)\sim{a}^{4-2\alpha}\frac{1}{2(\alpha-2)}\left(\Psi(\alpha-\frac{3}{2})-\Psi(\frac{1}{2})\right)$$
At $\alpha\to2$ we have to disclose the uncertainty
$I(a,\alpha\to2)\sim\frac{{a}}{2}^{0}\lim_{\epsilon\to0}\frac{1}{\epsilon}\left(\Psi(\frac{1}{2}+\epsilon)-\Psi(\frac{1}{2})\right)=\frac{1}{2}\Psi'(\frac{1}{2})=\sum_{k=1}^{\infty}\frac{2}{(2k-1)^2}$
$$I(a,\alpha=2)\sim\int_{1}^{\infty}t^{-1}\ln\frac{t+1}{t-1}dt=\frac{3}{2}\zeta(2)=\frac{\pi^2}{4}$$
Other asymptotics terms can be evaluated in the same way.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate $\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$ Calculate $$\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$$
I think that firstly I should use Taylor's theorem, so I have:$$\int_0^\infty \frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots}{(x^2+1)^2}dx$$
However I don't know what I can do the next.
| Note that $I(a) = \int_0^\infty \frac{\sin at} {t(t^2+1)}dt
= \frac\pi2 (1-e^{-a}) $, which can be obtain by solving
$$I’’(a)-I(a) = -\int_0^\infty \frac{\sin at}t dt= -\frac\pi2$$
Then
\begin{align}
\int_{0}^{\infty}\frac{\cos 3x}{x^4+x^2+1} dx
= &\frac14\int_{-\infty}^{\infty}
\overset{x=\frac{\sqrt3}2 t-\frac12}{\frac{(1+x)\cos 3x}{x^2+x+1}} +\overset{x= \frac{\sqrt3}2 t +\frac12}{\frac{(1-x)\cos 3x}{x^2-x+1}} d x\\
=& \frac1{\sqrt3}\cos\frac3{2} \int_{0}^{\infty} \frac{\cos \frac {3\sqrt3}{2}t}{t^2+1} dt + \sin\frac3{2} \int_{0}^{\infty} \frac{t\sin\frac {3\sqrt3}{2}t}{t^2+1} dt \\
=& \frac1{\sqrt3}\cos\frac3{2}\cdot I’(\frac {3\sqrt3}{2})
-\sin\frac3{2}\cdot I’’(\frac {3\sqrt3}{2})\\
=&\frac\pi2 e^{-\frac {3\sqrt3}{2} }\left( \frac1{\sqrt3}\cos\frac32+ \sin\frac32\right)
=\frac\pi{\sqrt3}e^{-\frac {3\sqrt3}{2} }\sin\left(\frac\pi6+\frac32\right)
\end{align}
| {
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"question_score": "7",
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Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$ Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$
| Writing
$$\frac 1{x^3-1}=\frac{1}{(a-1) (a-b) (x-a)}+\frac{1}{(b-1) (b-a) (x-b)}+\frac{1}{(a-1) (b-1) (x-1)}$$ where
$$a=-\frac{1}{2}-\frac{i \sqrt{3}}{2} \qquad \text{and} \qquad b=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$$ we face three integrals
$$I(c)=\int \frac{\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{x}{c}\right)+\log (x) \log \left(1-\frac{x}{c}\right)$$
$$J(c)=\int_0^t \frac{\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{t}{c}\right)+\log (t) \log \left(1-\frac{t}{c}\right)$$
Recombining the three terms, computing at the bounds and using the values of the polylogarithms leads to
$$\int_0^\infty\frac{\ln(x)}{x^3-1}dx=\frac{5 \pi ^2}{54}-\left(-\frac{\pi ^2}{18} \right)=\frac{4 \pi ^2}{27}$$
| {
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Factorisation of a given polynomial into another with integer roots Source: Challenge and Thrill of Pre-College Mathematics
"Find all integers $a$ such that
$$(x-a)(x-12)+2$$
can be factored into $(x-b)(x-c)$, such that $b$ and $c$ are integers."
My attempt: Simplifying the given polynomial and by the given condition, we get
$$x^2+(-12-a)x+(12a+2)=x^2-(b+c)x+ bc,$$ from where we obtain the system
$$b+c= (12+a)$$
$$bc= 12a+2$$
I can't proceed with the same process. A hint would be appreciated.
| We could also look at the properties of the family of "upward-opening" parabolas described by $ \ y \ = \ (x-a)·(x-12) + 2 \ \ . $ Each member of this set must pass through the point $ \ (12 \ , \ 2 ) \ \ , $ which we shall see imposes an important restriction on its zeroes $ \ b \ $ and $ \ c \ \ . $ This forces the family of parabolas to be arranged symmetrically about the line $ \ x = 12 \ \ . $
The vertex of a particular parabola is then located at $ \ x = \frac{12 + a}{2} \ = \ \frac{b + c}{2} \ \ $ (which is the geometric interpretation of the relevant Viete relation). The additional requirement that the zeroes be integers will prove to be a very stringent constraint as we then must have for these zeroes $ \ b \ , \ c \ \le \ 11 \ $ or $ \ b \ , \ c \ \ge \ 13 \ \ . $
The difference of the zeroes is given by $ \ b - c \ = \ \sqrt{\Delta} \ \ , $ which must therefore be an integer. We can then say that either the larger of the two zeroes must satisfy $ \ \frac{12 + a}{2} + \frac{\sqrt{\Delta}}{2} \ \le \ 11 \ \Rightarrow \ a + \sqrt{\Delta} \ \le \ 10 \ \ $ or the smaller of the zeroes must satisfy $ \ \frac{12 + a}{2} - \frac{\sqrt{\Delta}}{2} \ \ge \ 13 \ \Rightarrow \ a - \sqrt{\Delta} \ \ge \ 14 \ \ . $
As for the discriminant of $ \ x^2 - (12 + a) + (12a + 2) \ $ itself, we have $$ \ \Delta \ \ = \ \ (12 + a)^2 - 4·1·(12a + 2) \ \ = \ \ a^2 + 24a + 144 - 48a - 8 \ \ = \ \ (a - 12)^2 - 8 \ \ . $$ Our earlier stated specification means that this number must be a perfect square. What we also learn from this is that since we must have $ \ \Delta \ \ge \ 0 \ $ in order for a particular parabola to have real zeroes, then either $ \ a - 12 \ \ge \ \sqrt8 \ \Rightarrow \ a \ \ge \ 12 + \sqrt8 \ \approx \ 14.83 \ \ $ or $ \ -(a - 12) \ \le \ \sqrt8 \ \Rightarrow \ a \ \le \ 12 - \sqrt8 \ \approx \ 9.17 \ \ . $ The integral condition calls for these inequalities to be interpreted as $ \ a \ \ge \ 15 \ \ $ or $ \ a \ \le \ 9 \ \ . $
While we could devote effort to further analysis from here, we quickly see that our inequalities have very narrowly "pinned-down" the permissible values of $ \ a \ \ : $
$$ \mathbf{a = 9} \ \ \Rightarrow \ \Delta \ = \ (9 - 12)^2 - 8 \ = \ 1 \ \ \Rightarrow \ \ a + \sqrt{\Delta} \ = \ 10 \ , \ 12 + a \ = \ 21 \ \ \Rightarrow \ \ b \ , \ c \ \ = \ \ 10 \ , \ 11 $$
and the associated polynomial is $ \ x^2 \ - \ 21x \ + \ 110 \ \ ; $
$$ a = 8 \ \ \Rightarrow \ \Delta \ = \ (8 - 12)^2 - 8 \ = \ 8 \ \ \Rightarrow \ \ a + \sqrt{\Delta} \ = \ 8 + \sqrt8 \ > \ 10 \ \ \text{[inadmissible]} \ \ ; $$
$$ \mathbf{a = 15} \ \ \Rightarrow \ \Delta \ = \ (15 - 12)^2 - 8 \ = \ 1 \ \ \Rightarrow \ \ a - \sqrt{\Delta} \ = \ 14 \ , \ 12 + a \ = \ 27 $$ $$ \Rightarrow \ \ b \ , \ c \ \ = \ \ 13 \ , \ 14 $$
and the associated polynomial is $ \ x^2 \ - \ 27x \ + \ 182 \ \ ; $
$$a = 16 \ \ \Rightarrow \ \Delta \ = \ (16 - 12)^2 - 8 \ = \ 8 \ \ \Rightarrow \ \ a - \sqrt{\Delta} \ = \ 16 - \sqrt8 \ < \ 14 \ \ \text{[inadmissible]} \ \ . $$
Consequently, there are only two solutions, represented in the graph below by the two parabolas located symmetrically about $ \ x = 12 \ \ . $
| {
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Prove that $\frac{x^2+y^2}{z}+\frac{y^2+z^2}{x}+\frac{z^2+x^2}{y}\geq 2(x+y+z)$ $x,y,z \in \mathbb R^+$ different than $0$, Prove that :
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
My attempt:
First, we should prove that :
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$
I will consider it as proven, we gonna start by :
$$(x-y)^2 \geq 0 \iff x^2+y^2 \geq 2xy $$
by the same method we have :
$x^2+z^2 \geq 2xz$ and $y^2+z^2 \geq 2yz$
so we can replace $2xy$ by $x^2+y^2$ in this inequality
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$
so :
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
I'm new to AM-GM inequality, if it exists a better proof using it please post it.
| $$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
$$\Leftrightarrow (x^2 + y^2 + z^2) \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq 3(x+y+z)$$
Note that
$$(x^2 + y^2 + z^2) \geq \frac{1}{3} (x+y+z)^2 $$ (Cauchy-Schwarz ineq., equality: x = y = z)
and
$$(x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \geq (1+1+1)^2 = 9$$ (Cauchy-Schwarz ineq., equality: $x=y=z$).
Applying them implies the original inequality (equality holds when $x=y=z$).
| {
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Question on solving one algebra relation to obtain another If $a,b,c\in \mathbb R$ and,
$$\left(\frac{a-b}{a+b}\right)\left(\frac{b-c}{b+c}\right)\left(\frac{c-a}{c+a}\right)=-27$$
Evaluate $$\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$
I noticed that you can reduce the question to the following:
If $(1-2x)(1-2y)(1-2z)=-27$, then evaluate $x+y+z$ where $x=\frac{b}{a+b}, \ y=\frac{c}{b+c}, \ z=\frac{a}{c+a}$
I'm not sure how to solve this. I think it may use inequalities or pure algebra.
Please help me out here. Thanks a lot :)
| let me continue from your work. Note that $$\frac{1}{x}-1=\frac{a}{b} $$ so $$\frac{(1-x)(1-y)(1-z)}{xyz}=1$$
$$\Rightarrow 1-(x+y+z)+xy+yz+xz-2xyz=0$$
$$\Rightarrow xy+xz+yz-2xyz=(x+y+z)-1$$
Also $$(1-2x)(1-2y)(1-2z)=-27$$
$$\Rightarrow 1-2(x+y+z)+4(xy+yz+xz)-8xyz=-27$$
$$1-2(x+y+z)+4(x+y+z)-4=-27$$
$$\Rightarrow x+y+z=-12$$
| {
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Binomial identity arising from Catalan recurrence. The following identity features right at the end of the answer to another question I asked: https://math.stackexchange.com/a/4019598/155881.
$$F(x) = \frac{1}{\sqrt{1-4x}} \left(\frac{1- \sqrt{1-4x}}{2x}\right)^n = \sum\limits_{k=0}^{\infty}{n+2k \choose k}x^k$$
Haven't been able to prove this. We know the expansions of both terms on the left side, making the identity:
$$\left(\sum\limits_{j=0}^{\infty} \left(\binom{n+2j-1}{j} - \binom{n+2j-1}{j-1}\right)x^j\right)
\left(\sum\limits_{l=0}^{\infty}{2l \choose l} x^l\right) = \sum\limits_{k=0}^{\infty}{2n+k \choose k}x^k$$
My attempt:
An obvious thing to do is to try a convolution of the left side and equate coefficients of $x^k$ (also using an alternate expression for the first summation in the product of the left side):
$$\sum\limits_{l=0}^k {n+2l-1\choose l}\frac{l}{n+l} {2k-2l\choose k-l} = {2n+k\choose k}$$
Not sure how to make progress on this. Expanding the binomial terms doesn't lead anywhere.
| We seek to show that with
$$Q(z) = \frac{1}{\sqrt{1-4z}}
\left(\frac{1-\sqrt{1-4z}}{2z}\right)^n$$
we have
$$[z^k] Q(z) = {n+2k\choose k}.$$
Now with the branch cut on $[1/4, \infty)$ for $\sqrt{1-4z}$ we have
analyticity of $Q(z)$ in a neighborhood of the origin (note that the
exponentiated term does not in fact have a pole at $z=0$) and the Cauchy
Coefficient Formula applies. We obtain
$$[z^k] Q(z) = \frac{1}{2\pi i}
\int_{|z|=\varepsilon} \frac{1}{z^{k+1}}
\frac{1}{\sqrt{1-4z}}
\left(\frac{1-\sqrt{1-4z}}{2z}\right)^n \; dz.$$
We put $\sqrt{1-4z} = w$ so that $\frac{1}{\sqrt{1-4z}} \; dz =
-\frac{1}{2} \; dw$ and $z=(1-w^2)/4.$ With $w = 1 - 2z - \cdots$ we get
as the image of $|z|=\varepsilon$ a contour that winds around $w=1$
counterclockwise once and may be deformed to a circle, so that we obtain
$$[z^k] Q(z) =
-\frac{1}{2}
\frac{1}{2\pi i}
\int_{|w-1|=\gamma}
\frac{4^{k+1}}{(1-w^2)^{k+1}}
(1-w)^n
\frac{1}{2^n} \frac{4^n}{(1-w^2)^n}
\; dw
\\ = \frac{(-1)^k \times 2^{n+2k+1}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{(w-1)^n}{(w^2-1)^{n+k+1}} \; dw
\\ = \frac{(-1)^k \times 2^{n+2k+1}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(w-1)^{k+1}} \frac{1}{(w+1)^{n+k+1}} \; dw
\\ = \frac{(-1)^k \times 2^{k}}{2\pi i}
\int_{|w-1|=\gamma}
\frac{1}{(w-1)^{k+1}} \frac{1}{(1+(w-1)/2)^{n+k+1}} \; dw$$
Apply the Cauchy Residue Theorem to get
$$(-1)^k \times 2^k \times (-1)^k \frac{1}{2^k}
{n+k+k\choose n+k} = \bbox[5px,border:2px solid #00A000]{
{n+2k\choose k}}$$
as claimed.
| {
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Prove equality using sum of squares and determinants Let $a$, $b$, $c$, $d$ be four positive real numbers satisfying that $a^2+b^2 = S = c^2+d^2$. I want to show that, under that hypotheses, it also holds:
$$ (ab+cd)S = (ac+bd)(ad+bc) $$
It seems to me that this equality can be deduced using suitable determinants of matrices. At a certain point one should use that $a^2+b^2$ and $c^2+d^2$ coincide.
| You can simply calculate this algebraically, no need to consider matrices:
$$\begin{align*}(ac+bd)(ad+bc) &= a^2cd + abc^2 + abd^2 + b^2cd \\
&= abc^2 + abd^2 +a^2cd + b^2cd \\ &= ab(c^2+d^2) + cd(a^2+b^2)\\
&= abS+cdS \\
&= (ab+cd)S\end{align*}$$
| {
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Find all complex roots of $(z+1+i)^4 - 1 + i =0$. Find all complex roots of $(z+1+i)^4 - 1 + i =0$.
Attempt:
I got
\begin{align*}
(z+1+i)^4 &= 1 - i \\
(z + \sqrt{2} e^{i(\frac{\pi}{4})})^4 &= \sqrt{2} e^{i(-\frac{\pi}{4})} \\
z + \sqrt{2} e^{i(\frac{\pi}{4})} &= \sqrt[8]{2} \exp\left(i\frac{-\frac{\pi}{4} + 2k\pi}{4} \right) \\
z &= \sqrt[8]{2} \exp\left(i\frac{-\frac{\pi}{4} + 2k\pi}{4} \right) - \sqrt{2}e^{i(\frac{\pi}{4})}
\end{align*}
for $k=0,1,2,3$.
Am I true? If not yet, how to find it?
| I got
\begin{align*}
(z+1+i)^4 &= 1 - i \\
(z + \sqrt{2} e^{i(\frac{\pi}{4})})^4 &= \sqrt{2} e^{i(-\frac{\pi}{4})} \\
z + \sqrt{2} e^{i(\frac{\pi}{4})} &= \sqrt[8]{2} \exp\left(i\frac{-\frac{\pi}{4} + 2k\pi}{4} \right) \\
z &= \sqrt[8]{2} \exp\left(i\frac{-\frac{\pi}{4} + 2k\pi}{4} \right) - \sqrt{2}e^{i(\frac{\pi}{4})}
\end{align*}
for $k=0,1,2,3$.
| {
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Help with inverse trigonometric substitutions $ \int x^2\sqrt{a^2+x^2}\,dx $. how would I go about integrating this? It is a lecture exercise.
$$
\int x^2\sqrt{a^2+x^2}\,dx
$$
I used a substitution of $x=a\tan\theta$ and ended up with $$\int a^4\tan^2\theta \sec^3\theta \,d\theta.$$ I was thinking by parts but then I would have to integrate a $\ln|\sec\theta+\tan\theta|$. Can't seem to think of the appropriate identity to collapse it. I am given the answer to verify but I don't know any mental algorithms to even begin doing inverse trigonometric substitution. The answer given is:
$$
\frac{x}{8}(a^2+2x^2)\sqrt{a^2+x^2}-\frac{a^2}{8}\ln\left(x+\sqrt{a^2+x^2}\right)
$$
Also, what are some things I should be thinking about first when looking at such questions when I need to substitute trigonometric functions into the x-variable? Should I attempt to make odd powers into even? How do I do so?
Thank you very much for your help.
| Integrate by parts directly without substitutions
\begin{align}
\int x^2\sqrt{a^2+x^2}dx
= &\frac14 \int \frac x{\sqrt{a^2+x^2}}d[(a^2+x^2)^2]\\
= &\frac14 x{(a^2+x^2)^{3/2}}- \frac{a^2}4\int \frac{\sqrt{a^2+x^2}}{2x}d(x^2)\\
=& \frac14 x{(a^2+x^2)^{3/2}}- \frac{a^2}8x{\sqrt{a^2+x^2}}-\frac{a^4}8\int \frac{dx}{\sqrt{a^2+x^2}}\\
\end{align}
where $\int \frac{dx}{\sqrt{a^2+x^2}}= \sinh^{-1}\frac xa$.
| {
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Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$ Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$
and by the same method we have :
$$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$
Adding them together will give us the desired inequality.
and please don't use any $\sum_{cyc}$ because I get confused with it.
| Here's the inevitable 'expand and see if everything works out' solution.
$$
\frac{a^2}{a^2 + bc} + \frac{b^2}{b^2 + ac}+\frac{c^2}{c^2 + ab} < 2 \\
\iff a^2 (b^2 + ac)(c^2 + ab) + b^2 (a^2 + bc)(c^2 + ab) + c^2(a^2 + bc)(b^2 + ac) < 2 (a^2 + bc)(b^2 + ac)(c^2 + ab) \\
\iff 3a^2b^2c^2 + 2(a^3b^3 + b^3c^3 + c^3a^3) + a^4 bc + b^4 ac + c^4ab < 2(2a^2b^2c^2 + a^3 b^3 + a^3c^3 + b^3 c^3 + a^4 bc + b^4ac + c^4 ab)
$$
It's important when doing these sort of calculations to make sure you didn't miss a term. Here the number of terms on the left (before summing them up) is $12 = 3 \times 2^2$ and the number of terms on the right is $8 = 2^3$ as expected.
This inequality simplifies to
$$
a^2b^2c^2 +a^4 bc + b^4ac + c^4 ab > 0
$$
Which is evidently true.
| {
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Difference between the maximum and minimum values of $a+b$ that satisfy $a+b+\frac{1}{a}+\frac{9}{b}=10, (a,b\in\mathbb{R}^+)$
Find the difference between the maximum and minimum values of $a+b$
that satisfy
$$a+b+\frac{1}{a}+\frac{9}{b}=10,\quad(a,b\in\mathbb{R}^+)$$
I'm trying to use Cauchy–Schwarz inequality, but I can't get a sense of how to transform and solve the given equation. Can anyone help me?
| Using the CS-inequality "Engels" form we have: $10 = a+b+\dfrac{1^2}{a}+\dfrac{3^2}{b} \ge a+b+\dfrac{(1+3)^2}{a+b}=a+b+\dfrac{16}{a+b}=x+\dfrac{16}{x}\implies 10x\ge x^2+16\implies x^2-10x+16 \le 0\implies (x-2)(x-8) \le 0\implies 2 \le x = a+b \le 8 \implies \text{max - min} = 8 - 2 = 6$. The minimum value of $a+b$ is $2$ which occurs when $a+b=2, \dfrac{1}{a} = \dfrac{3}{b} \implies a = \dfrac{1}{2}, b = \dfrac{3}{2}$. The maximum of $a+b$ is $8$ which occurs when $a+b = 8, b = 3a \implies a=2,b=6$.
| {
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Proof that a certain fraction is always an integer Prove that $$\frac{(n+1)(n)^2(n-1)^2...(n-k+2)^2(n-k+1)}{(k+1)(k)^2(k-1)^2...(2)^2(1)}$$ is or is not an integer for $0\leq k \leq n$, where $k$ and $n$ are integer values.
This looks like $\frac{(n+1)!(n)!}{(n-k+1)(n-k)(k+1)!(k)!}$. The above statement is true for k=1 and k=2 as seen by observing modular residues. Not sure how to proceed.
| I’m going to give the answer more or less as I worked through the problem, since these numbers turn out to be rather interesting; if you just want a quick and easy computational proof, skip to the end.
I rewrote it as
$$\frac{n!}{k!(n-k)!}\cdot\frac{(n+1)!}{(k+1)!(n-k+1)!}=\frac1{k+1}\binom{n}k\binom{n+1}k\,,\tag{1}$$
and calculated some values:
$$\begin{array}{c|cc}
n\backslash k&0&1&2&3&4&5\\\hline
0&1\\
1&1&1\\
2&1&3&1\\
3&1&6&6&1\\
4&1&10&20&10&1\\
5&1&15&50&50&15&1
\end{array}$$
The row sums are familiar: $1,2,5,14,42,132$ are the Catalan numbers $C_{n+1}$. This does suggest that these numbers may be counting something. And if we look up the triangle in OEIS by searching on the sequence
$$1,1,1,1,3,1,1,6,6,1,1,10,20,10,1\,,$$
the very first return is A001263, the sequence of Narayana numbers, whose first FORMULA entry is an offset version of the righthand side of $(1)$. If the entry in row $n$, column $k$ of my table is $t(n,k)$, then $t(n,k)$ is the Narayana number $N(n+1,k+1)$ and is the number of Dyck paths of length $2(n+1)$ having exactly $k+1$ peaks. Thus, it must be an integer.
That $N(n,k)$ is the number of Dyck paths having exactly $k$ humps is not obvious; one proof can be found in this PDF. However, the FORMULA section of the OEIS entry also notes that
$$N(n,k)=\binom{n-1}{k-1}\binom{n+1}k-\binom{n}{k-1}\binom{n}k\,,$$
which suggests the following purely computational proof that the original expression is an integer:
$$\begin{align*}
&\binom{n}k\binom{n+2}{k+1}-\binom{n+1}k\binom{n+1}{k+1}\\\\
&\qquad=\frac{n!(n+2)!}{k!(k+1)!(n-k)!(n-k+1)!}-\frac{(n+1)!^2}{k!(k+1)!(n-k+1)!(n-k)!}\\\\
&\qquad=\frac{n!(n+2)!-(n+1)!^2}{k!(k+1)!(n-k+1)!(n-k)!}\\\\
&\qquad=\frac{(n+1)!^2\left(\frac{n+2}{n+1}-1\right)}{{k!(k+1)!(n-k+1)!(n-k)!}}\\\\
&\qquad=\frac{n!(n+1)!}{{k!(k+1)!(n-k+1)!(n-k)!}}\,.
\end{align*}$$
Here the first expression is clearly an integer, and the last is easily seen to be equivalent to the original fraction.
| {
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Using $z = x + yi$ does not reveal all solutions The question is to find all solutions in $\mathbb{C}$ of the equation: $$z^2 - (i+1)z + i = 0$$
After expanding with $z = x + yi$, I get $(x^2 - y^2+y-x) + (2xy -x-y+1)i = 0$.
Solving $(x^2 - y^2+y-x)=0$ gives $x = y$ or $x = y-1$.
Now, we plug that into $(2xy -x-y+1) = 0$ .
Since $x,y \in \mathbb{R}$, this only gives the solution $x= 0, y = 1$.
However, there is another solution which is $z=1$. Why this solution does not appear with this method?
|
Solving $(x^2 - y^2+y-x)=0$ gives $x = y$ or $x = y-1$.
That is your mistake here. $0=x^2-y^2+y-x=(x-y)(x+y-1)$ so $x=y$ or $x=1-y$, not $y-1$.
Then substituting into the second equation $2xy-x-y+1=0$ gives $(x,y)=(0,1)$ or $(1,0)$.
| {
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If $a^3b-ab^2$ is divisible by $3$ then either $a$ is divisible by $3$ or $b$ is divisible by $3$ or $(b-1)$ is divisible by $3$ Question:
Prove for all integers $a,b$ that if $a^3b-ab^2$ is divisible by $3$ then either $a$ is divisible by $3$ or $b$ is divisible by $3$ or $(b-1)$ is divisible by $3$
Attempt:
Suppose by contradiction that $3 \nmid a$ and $3 \nmid b$ and $3 \nmid (b-1)$,
Then $a^3b-ab^2=(a)(b)(a^2-b)$, and so since $3$ is prime by Euclid's lemma we have
$3|a$ or $3|b$ or $3|(a^2-b)$, but by assumption $3 \nmid a$ and $3 \nmid b$ so then
$3|(a^2-b)$
Now since $3 \nmid a$ and $3 \nmid b$ and $3 \nmid b-1$ then there exists some integers $k,j$ with
$a = 3k+1$ or $a=3k+2$ and $b=3j+1$
Here is where I am not sure how to continue, Consider
$a^2-b \mod 3 \equiv 9k^2+6k+1 - (3j+1) \equiv 0 \mod 3$ or
$a^2-b \mod 3 \equiv 9k^2+12k+4 - (3j+1) \equiv 0 \mod 3$
where now I can't seem to find a contradiction. I think I have made a small mistake somewhere along the way but I can't seem to find it.
Any insights appreciated.
| Your error is that if $3\not \mid b$ (So $b\not \equiv 0\pmod 3$) and $3\not \mid b-1$ (So $b-1\not \equiv 0 \pmod 3$ and $b\not \equiv 1 \pmod 3$ then $b \equiv 2 \pmod 3$ (and not $b \equiv 1 \pmod 3$). So $b = 3j + 2\ne 3j + 1$.
(If $b = 3k$ then $3\mid b$. If $b = 3j+ 1$ then $3\mid b-1$. So to have $3\not \mid b, b-1$ we must have $b= 3j + 2$.)
And then you contradiction is right in front of our eyes.
$a^2 - b = 9k^2 + \begin{cases}6k+1\\12k + 4\end{cases} - (3j+2)\equiv -1\pmod 3$.
(Minor error: $(3k+2)^2 = 9k^2 + 12k + 4$ and not $9k^2 +6k + 4$. But it's better to use $a\equiv -1\pmod 3; a =3k -1; (3k-1)^2 = 9k^2 -6k + 1$.)
=======
But I'd do it directly. With Euclids lemma we have if $3|a^3b-ab^2 = ab(a^2-b)$ we must have either $3|a$ or $3|b$ or $3|a^2-b$.
If $3|a$ or $3|b$ we are done. SO it suffices to show: Assume $3\not\mid a$ and $3\not \mid b$ then $3|a^2 -b\implies 3|b-1$.
And if $3\not \mid a$ then $a\equiv 1,2\pmod 3$ and $a^2 \equiv 1 \pmod 3$ so
So $b-1 \equiv b - a^2 \equiv -(a^2 -b)\equiv 0 \pmod 3$.
And we are done.
| {
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Use Newton's Binomial Theorem to prove series Verify that for all integer $n\ge 2$,
\begin{eqnarray*}
\sum_{k=0}^n\binom{\frac{1}{2}}{k}\binom{\frac{1}{2}}{n-k}=0.
\end{eqnarray*}
(Hint: Consider Newton's Binomial Theorem.)
=================================
EDIT:
From Newton's Binomial Theorem:
\begin{eqnarray*}
(1+x)^p &=& \sum_{n=0}^\infty \begin{pmatrix}
p \\n
\end{pmatrix} x^n.
\end{eqnarray*}
So,
\begin{eqnarray*} | (1+x)^\frac{1}{2} &=& \sum_{k=0}^\infty \begin{pmatrix}
\frac{1}{2} \\k
\end{pmatrix} x^k\\
B: (1+x)^\frac{1}{2} &=& \sum_{k=n}^\infty \begin{pmatrix}
\frac{1}{2} \\{n-k}
\end{pmatrix} x^{n-k}
\end{eqnarray*}
I assume finding $A-B$ will lead to the answer $0$ but I'm not sure how to continue from here
| {
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Solve in the set of integers $2^x +5^x = 3^x + 4^x$
Find the number of integer solutions(both positive and negative) of the equation:
$$2^x +5^x = 3^x + 4^x$$
With induction we see that for $x\geq 2$ we have $$5^x\geq 3^x+4^x$$
It is trivially true for $x=2$ and $x=3$. Now say it is true for $x$ and prove it for $x+2$: $$5^{x+2} \geq 5^2(3^x+4^x) = (3^2+4^2)(3^x+4^x) > 3^{x+2}+4^{x+2}$$
So equation has no solution for $x\geq 2$. Clearly $x=0$ and $x= 1$ are solutions.
If $x<0$ you can write $x=-n$ where $n$ is positive integer and we get $$3^n\cdot 4^n(2^n+5^n) = (3^n+4^n)2^n\cdot 5^n$$
so $ 3^n\mid 40^n$ which is clearly nonsense and thus no solution (for negative $x$).
Any easier solution?
| Re write the Eq. as $$3^x-2^x=5^x-4^x....(1)$$
Use LMVT on $f(t)=t^x$ in $(2,3)$ and $(4,5)$
to write $$\frac{3^x-2^x}{3-2}=xt_1^{x-1}, t_1\in(2,3)~~~~(2)$$
and
$$\frac{5^x-4^x}{5-4}=xt_2^{x-1}, t_2\in (4,5)~~~~(3)$$
so $t_1 \ne t_2$, By (1), we can equate $xt_1^{x-1}=xt_2^{x-1} \implies x=0~or~ x=1.$
So (1) has only two solutions: $x=0,1$.
| {
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Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Here's what I have done so far:
$$\begin{align*}
S &=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\[5pt]
&= \left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{n}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{n+1}\right)\\[5pt]
&= \sum_{k=1}^n \frac{1}{2k-1} - \sum_{k=1}^{n+1} \frac{1}{2k} \\[5pt]
\end{align*}$$
How do I continue the problem from here? Is it even possible?
| So , we need to solve for : $$\sum_{k=1}^{\frac{n+1}{2}}\frac{1}{2k-1}-\sum_{k=1}^{\frac{n+1}{2}}\frac{1}{2k}=\psi$$ Which is also , $$\psi=\left(1+\frac{1}{3}+\frac{1}{5}+........+\frac{1}{n}\right)-\frac{1}{2}S_{n+1}$$ $$\psi=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{n}+\frac{1}{n+1}\right)-S_{n+1}$$ Hence , $$\psi=0$$
| {
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Supremum of $x^2y^2(x^2+y^2)$ Let $S$ be the set of all tuples $(x, y) $ with $x, y$ non-negative real numbers satisfying $x+y=2n$,for a fixed $n\in \Bbb{N} $. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is
*
*$3n^2$
*$2n^6$
*$4n^6$
*$n^6$
My thinking : take $f(x, y) =x^2y^2(x^2+y^2)$ then express $f(x, y) $ as a function of $x$ by putting $y=2n-x$. Then applying calculus we can find the maximum for some value of $x$.
But calculation will be cumbersome. Is there any smart way?
Thanks in advance.
| This is a good opportunity to use Lagrange's multipliers method:
Let $g(x,y)=x+y-2n$. We have
$$\nabla g=\langle 1,1\rangle \textrm{ and }\nabla f = \langle 2xy^2(y^2+2x^2), 2yx^2(x^2+2y^2)\rangle$$
At a maximum, $\nabla f$ and $\nabla g$ are parallel, so $xy^2(y^2+2x^2)=yx^2(x^2+2y^2)$. If $x=0$ or $y=0$, we have $f(x,y)=0$, which his the minimum value, so we can assume that $x\neq 0$ and $y\neq 0$. Therefore,
$$ y(y^2+2x^2)=x(x^2+2y^2) \iff y^3+2yx^2=x^3+ 2xy^2 \iff y^3-x^3=2xy^2-2yx^2$$
so $$ (y-x)(y^2+xy+y^2)=2xy(y-x)$$
This means that $x=y$ or $y^2+x^2-xy=0 \iff (y-\frac{x}{2})^2+\frac{3}{4}x^2=0 \iff x=y=0$, which we excluded.
Therefore, $x=y$ at the maximum, so $(x,y)=(n,n)$. Substituting, the maximum value is $f(n,n)=2n^6$.
Note: to be totally rigorous, Lagrange's method only says that, if the maximum value exists, then $\nabla f$ and $\nabla g$ are parallel at the maximum. The maximum actually exists here because $\{ (x,y) | x\ge 0, y\ge 0, x+y=2n \}$ is compact.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve determinants with $|AB| = |A||B|$ How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.
(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & \dots & 1+x_ny_n
\end{vmatrix}$
(2) $D_2 = \begin{vmatrix}
1 & \cos(a_1-a_2) & \cos(a_1-a_3) & \dots & \cos(a_1-a_n) \\
\cos(a_1-a_2) & 1 & \cos(a_2-a_3) & \dots & \cos(a_2-a_n) \\
\cos(a_1-a_3) & \cos(a_2-a_3) & 1 & \dots & \cos(a_3-a_n) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\cos(a_1-a_n) & \cos(a_2-a_n) & \cos(a_3-a_n) & \dots & 1
\end{vmatrix}$
(3) $D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}$
(4) Let $s_k=a_1^k+a_2^k+a_3^k+a_4^k \quad (k=1,2,3,4,5,6)$,
$$D_4 = \begin{vmatrix}
4 & s_1 & s_2 & s_3 \\
s_1 & s_2 & s_3 & s_4 \\
s_2 & s_3 & s_4 & s_5 \\
s_3 & s_4 & s_5 & s_6
\end{vmatrix}$$
My Attempt:
(1) I noticed for any $i,j$,
$1+x_iy_j = \begin{bmatrix} 1 & x_i \end{bmatrix}
\begin{bmatrix} 1 \\ y_j \end{bmatrix}$.
So, the matrix corresponds $D_1$ equals $\begin{bmatrix}
1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n
\end{bmatrix}_{n \times 2}
\begin{bmatrix}
1 & 1 & \dots & 1 \\
y_1 & y_2 & \dots & y_n
\end{bmatrix}_{2 \times n}$.
But it's not helpful at all. :(
(2) It's obvious that
$a_{ij} = \cos(a_i - a_j) = \cos a_i \cos a_j + \sin a_i \sin a_j$.
(3) I've no idea about this problem at all. All I came up with, is
$$D_3 = a^4 \begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
1 & -1 & -1 & 1
\end{vmatrix}$$
(4) I noticed that $a_{ij} = s_{i+j-2}$. But still, not helpful.
Plz give me some hints. Thx in advance.
| for the case $\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
x_2*y_1+1 & x_2*y_2+1 & x_2*y_3+1 \\
x_3*y_1+1 & x_3*y_2+1 & x_3*y_3+1
\end{matrix}\right)$
also LU decopmostion
$$\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
x_2*y_1+1 & x_2*y_2+1 & x_2*y_3+1 \\
x_3*y_1+1 & x_3*y_2+1 & x_3*y_3+1
\end{matrix}\right)=\\
\left(\begin{matrix}
1 & 0 & 0 \\
\frac{x_2*y_1+1}{x_1*y_1+1} & 1 & 0 \\
\frac{x_3*y_1+1}{x_1*y_1+1} & \frac{x_1-x_3}{x_1-x_2} & 1
\end{matrix}\right)\left(\begin{matrix}
x_1*y_1+1 & x_1*y_2+1 & x_1*y_3+1 \\
0 & \frac{x_1*y_1-x_2*y_1-x_1*y_2+x_2*y_2}{x_1*y_1+1} & \frac{x_1*y_1-x_2*y_1-x_1*y_3+x_2*y_3}{x_1*y_1+1} \\
0 & 0 & 0
\end{matrix}\right)$$so $$\left|\begin{matrix}
1+x_1*y_1 & 1+x_1*y_2 & 1+x_1*y_3 \\
1+x_2*y_1 & 1+x_2*y_2 & 1+x_2*y_3 \\
1+x_3*y_1 & 1+x_3*y_2 & 1+x_3*y_3
\end{matrix}\right|=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Squeeze theorem 2 bounded functions $$\lim_{(x,y) \to (0,0)} \frac{x^2 \sin (y)}{x^2+y^2} $$
Squeeze Theorem:
$$g(x) \leq f(x) \leq h(x)$$
$$-1 \leq \sin(y) \leq 1$$
$$- \frac{x^2}{x^2+y^2}\leq \frac{x^2 \sin (y)}{x^2+y^2} \leq \frac{x^2}{x^2+y^2} $$
Also
$$ 0 \leq \frac{x^2}{x^2+y^2} \leq 1 $$
What should be my next step
How does this comes out to be zero
| I would use polar coordinates and equivalence of functions:
$$\frac{x^2 \sin (y)}{x^2+y^2}=\frac{r^2\cos^2\theta\sin(r\sin\theta)}{r^2}\sim_{r\to 0}r\sin\theta\cos^2\theta.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
any physical interpretation for these two matrices? I know that the physical interpretation of $\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin\theta & \cos \theta \end{pmatrix}$ is the rotation matrix.
But what are the physical interpretations of the matrices $\begin{pmatrix} \sin \theta & \cos \theta \\ -\cos\theta & \sin \theta \end{pmatrix}$ and $\begin{pmatrix} \sec \theta & \tan \theta \\ \tan\theta & \sec \theta \end{pmatrix}$ ?
I have tried with a couple of values but was unable to arrive at a conclusion.
| $$\begin{pmatrix} \sin \theta & \cos \theta \\ -\cos\theta & \sin \theta \end{pmatrix} = \begin{pmatrix} \cos \left(\theta-\frac{\pi}{2}\right) & -\sin \left(\theta-\frac{\pi}{2}\right) \\ \sin \left(\theta-\frac{\pi}{2}\right) & \cos \left(\theta-\frac{\pi}{2}\right) \end{pmatrix}$$
So it is also a rotation matrix.
A physical interpretation for the other is not so clear to me. But the diagonalization may help.
$$\begin{pmatrix} \sec \theta & \tan \theta \\ \tan\theta & \sec \theta \end{pmatrix}=\begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \sec \theta - \tan \theta & 0 \\ 0 & \sec \theta + \tan \theta \end{pmatrix}\begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4048106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the class of a function Let $$f(x) = \begin{cases}
x^3\sin(\frac{1}{x}); & \text{ if, } x \neq 0 \\
0; & \text{ $x=0$ }
\end{cases}$$
How do I find the greatest value of n such that $f \in C^n ([-1,1])$?
I know to find n, I must find the greatest n such that $f^{(n)}(x)$ is differentiable in the interval $[-1,1]$. But how do I prove its differentiability?
| You have to check if $f^{(n)}$ is continuous $n$ by $n$.
Start with $n=0$.
The function is continuous because
$$\lim_{x \to 0^+} x^3 \sin{\left(\tfrac{1}{x}\right)} = \lim_{x \to 0^-} x^3 \sin{\left(\tfrac{1}{x}\right)}=0.$$
For $n=1$,
$f'(x) = \begin{cases} -x \cos{\left(\tfrac{1}{x}\right)} + 3x^2 \sin{\left(\tfrac{1}{x}\right)} , & \text{ if } x \neq 0\\
0, & \text{ if } x = 0
\end{cases}$
is continuous because
$$\lim_{x \to 0^+} -x \cos{\left(\tfrac{1}{x}\right)} + 3x^2 \sin{\left(\tfrac{1}{x}\right)} = \lim_{ x \to 0^-} -x \cos{\left(\tfrac{1}{x}\right)} + 3x^2 \sin{\left(\tfrac{1}{x}\right)} = 0.$$
For $n = 2$,
$f''(x) = \begin{cases} -4 \cos{\left(\tfrac{1}{x}\right)} - \tfrac{1}{x}\sin{\left(\tfrac{1}{x}\right)}+6x\sin{\left(\tfrac{1}{x}\right)} , & \text{ if } x \neq 0\\
0, & \text{ if } x = 0
\end{cases}$ is not continuous because
$$\lim_{ x \to 0}-4 \cos{\left(\tfrac{1}{x}\right)} - \tfrac{1}{x}\sin{\left(\tfrac{1}{x}\right)}+6x\sin{\left(\tfrac{1}{x}\right)}$$ does not exist.
Thus the greatest $n$ for which $f \in C^n([-1,1])$ is $n=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2? Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?
I started off by noticing that the base of the triangle is 3, so the height must be greater than $\frac{4}{3}$ if the area is greater than 2 because $\frac{1}{2}(3)(h)>2 \rightarrow h >\frac{4}{3}.$
Since $y>\frac{4}{3}, x > \frac{1}{3}$ because the line is $y=x+1$.
So for the area of the triangle to be greater than 2, $P$ can be any point from $(\frac{1}{3}, \frac{4}{3})$ to $(3,4)$ on the line $y=x+1$.
Therefore the probability of the triangle area being greater than 2 is $\dfrac{\text{length of segment (1/3, 4/3) to (3,4)}}{\text{length of segment (0,1) to (3,4)}}=\dfrac{2/3\sqrt{17}}{3\sqrt{2}}=\dfrac{\sqrt{34}}{9}.$
However the solution says the answer is 8/9, so what am I doing wrong here? I'm also pretty sure I overcomplicated a lot of things, and I was also wondering if there's a simpler solution. Thanks!
| If the upper vertex of the triangle is $(x,y)$, the area of the triangle formed by $(0,0)$, $(3,0)$, and $(x,y)$ is ${1 \over 2}bh$ or ${3h \over 2}$.
So the probability the area is at least $2$ is the probability that ${3h \over 2} > 2$, or equivalently the probability that $h > {4 \over 3}$. Since $h$ is uniformly distributed in $[1,4]$, the probability of this event is ${4 - {4 \over 3} \over 4 - 1} = {8 \over 9}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the sequence $ \frac{ (2^n-1)(2^{n-1}-1)...(2^{n-k+1}-1) } { (2^1-1)(2^2-1)...(2^k-1) }$ is an integer For any given integer $n\ge 1$ and $k \in\{1, 2, \dots, n\},$
$$ F_n(k) = \frac{ (2^n-1)(2^{n-1}-1)...(2^{n-k+1}-1) } { (2^1-1)(2^2-1)...(2^k-1) } $$
For example, if $n = 1,$ then $k \in \{1\}$ and $$F_1(1) = \frac{2^1-1}{2^1-1} $$
For $n=3,$ then $k\in\{1, 2, 3\}$ and
$$ F_3(1) = \frac{2^3-1}{2^1-1} = 7 $$
$$ F_3(2) = \frac{(2^3-1)(2^2-1)}{(2^1-1)(2^2-1)} =7 $$
$$ F_3(3) = \frac{(2^3-1)(2^2-1)(2^1-1)}{(2^1-1)(2^2-1)(2^3-1)} =1 $$
| Cobsider $(n)_q:=\frac{q^n-1}{q-1} $, the so called q-analogue of $n$. Observe that this is a polynomial of degree n-1 in the variable q.
Define now the q-factorial and the q-binomial as follow:
$$(n)_q!:= \prod_{i=1}^n (i)_q \qquad \binom{n}{k}_q:=\frac{(n)_q!}{(k)_q! \,(n-k)_q!}$$
It is easy to prove that both are polynomials, for example using the identity
$$
\binom{n}{ k}_q = q^k \binom{n-1}{k}_q + \binom{n-1}{k-1}_q
$$
What you need follows from the evaluation in 2 on the q-binomial.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\Delta ABC$ is an equilateral triangle if and only if the complex coordinates $a$,$b$, and $c$ satisfy a relation.
Let $\triangle ABC$ be a triangle in the complex plane and let $a$,$b$ and $c$, respectively, be the complex coordinates of its vertices. Suppose that the tiangle is inscribed in the circle $C(0,1)$. Prove that $\triangle ABC$ is equilateral if and only if $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$ and $$(a+b)(b+c)(c+a)\neq0\,.$$
I tried to convert $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$ into a simpler equation by multipling it by $(a+b)(b+c)(c+a)$
In the end I obtained $$a^2+b^2+c^2+3ab+3bc+3ca=0$$ but I don't know how to continue from here. Can you help me, please?
| Given equilateral triangle, i.e.
$$a= e^{i\alpha}, \>\>\>\>\>b = e^{i(\alpha+\frac{2\pi} 3)}, \>\>\>\>\>c= e^{i(\alpha-\frac{2\pi} 3)}
$$
it is straightforward to verify $a^2+b^2+c^2+3ab+3bc+3ca=0$.
Conversely, given
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=0$$
rewrite the equation in term of $b =a e^{i x}$, $c =a e^{ i y} $
$$\frac{1}{1+ e^{i x}}+\frac{1}{1+ e^{i y}}+\frac{1}{e^{i x} + e^{i y}}=0\tag1$$
Suppose $(x,y)$ is a solution, then $(y,x)$ is also a solution due to party, as well as $(-x,-y)$ via conjugate. which implies $x = -y$. Substitute into (1) to obtain $\cos x=-\frac12$, or $x= -y= \pm \frac {2\pi}3$. Thus
$$b = a e^{\pm i\frac{2\pi}3}, \>\>\>\>\>c=a e^{\mp i \frac{2\pi} 3}
$$
hence, equilateral triangles.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4055070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
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Proving that $a \cos A+b\cos B+c \cos C$ is the semi-perimeter of $\triangle ABC$
With each symbol having its usual meaning, prove that in $\triangle ABC$,
$$a \cos A+b\cos B+c \cos C=s$$
where $s$ is the semi-perimeter.
I started by applying the rule of sines:
*
*$a=k\sin A$
*$b=k\sin B$
*$c=k \sin C$
where $k$ is a constant
so, we have:
$k(\sin A\cos A+\sin B\cos B+\sin C\cos C)$
manipulating this, we get:
$\frac{k}{2}(\sin 2A+\sin 2B+\sin 2C)$
$\frac{k}{2}(2\sin(A+B)\cos(A-B) + \sin 2C)$
$\frac{k}{2}(2\sin C\cos(A-B)+\sin 2C)$
$\frac{k}{2}2\sin C(\cos(A-B)+\cos C)$
But this doesn't seem to be yielding desired results. Could someone help me out with this? Or maybe tell me a better approach to this problem?
| The semi-perimeter formula you're trying to use here is not true
$$a\cos{A}+b\cos{B}+c\cos{C} \ne s $$
If each symbol has its usual meaning in $\Delta ABC$
$$\cos{A} = \frac{b^2+c^2-a^2}{2bc}$$
$$a \frac{b^2+c^2-a^2}{2bc} + b \frac{a^2+c^2-b^2}{2ac} + c \frac{a^2+b^2-c^2}{2ab} $$
$$\frac{ a^2(b^2+c^2-a^2)+b^2(a^2+c^2-b^2)+c^2(a^2+b^2-c^2) }{2abc}$$
$$\frac{ 2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}{2abc}$$
Simplifying this and checking it out doesn't yield the semi-perimeter $s$
| {
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Find a power series representation for $\frac{1}{2+x}$
Find a power series representation for $\frac{1}{2+x}$
The question did not specify a real number $c$ and an expansion on the form
$$f(x)=\sum_{n\geq 0} a_n(x-c)^n.$$
I did this problem as follows, but I seem to be off by a factor of $1/3$.
$$\frac{1}{2+x} = \frac{1}{1+(x+1)}=\frac{1}{1-[-(x+1)]}.$$
Putting this into the geometric form of:
$$\sum_{n=0}^{\infty} \frac{a}{1-r},$$ I get $a=1$ and $r=-(x-1).$ So, this could be written as $$\sum_{n=0}^{\infty} [(-1)(x-1)]^n = \sum_{n=0}^{\infty} (-1)^n(x-1)^n.$$
Wolfram's widget gets the following:
$$\frac{1}{2+x}=\sum^\infty_{n=0}(-1+x)^n(-1)^n3^{-1-n}\quad \text{for}|-1+x|\lt3$$
which looks like my answer except for the $3^\left(-1-n\right)$ term.
Can you assist me in finding my error? I don's see where the $1/3$ term gets introduced.
| Here is another answer: For this rational function it is possible to give an elementary construction of a powerseries and an "equality"
$$ \frac{1}{2+x}=\sum_{n \geq 0} a_n(x-a)^n$$
for any $a\neq -2$:
$$\frac{1}{2+x}=\frac{1}{2(1-(-x/2))} =\frac{1}{2}\sum_{n\geq 0}(-\frac{x}{2})^n=$$
$$ \sum_{n\geq 0}(-1)^n\frac{x^n}{2^{n+1}}.$$
I used
$$\frac{1}{1-t}=\sum_{n\geq 0} t^n$$
with $t:=-\frac{x}{2}$.
We can generalize as follows (let $a\neq -2$):
$$ \frac{1}{2+x}=\frac{1}{2+a-(-(x-a))} =$$
$$ \frac{1}{2+a}\frac{1}{1-(-(\frac{x-a}{2+a}))}=$$
(here we let $t:=\frac{x-a}{2+a}$)
$$ \frac{1}{2+a}\sum_{n\geq 0}(-(\frac{x-a}{2+a}))^n.$$
We get a power series
$$\frac{1}{2+x}= \frac{1}{2+a}\sum_{n\geq 0}(-1)^n(\frac{x-a}{2+a})^n.$$
With $a:=1$ we get
$$\frac{1}{2+x}=\frac{1}{3}\sum_{n \geq 0}(-1)^n(\frac{x-1}{3})^n.$$
When I write $=$ in the above equation this means for all $x$ where the powerseries converge. You should be able to calculate this radius of convergence yourself.
In general: If $\frac{f(x)}{g(x)}$ is a rational function over the reals, there is a product decomposition
$$P1.\text{ }g(x)=\prod_{i} (x-c_i)^{l_i}\prod_j p_j(x)^{k_j}$$
where $c_i$ are real numbers and $p_j(x)=b_j^2+(x-a_j)^2$ where $a_j,b_j\neq 0$ are real numbers. For a rational function on the form $\frac{1}{p_j(x)}$
you get a similar calculation:
$$ \frac{1}{p_j(x)}=\frac{1}{b_j^2+(x-a_j)^2}=$$
$$\frac{1}{p_j(x)}"="\frac{1}{b_j^2}\sum_{n\geq 0}(-1)^n(\frac{x-a_j}{b_j})^{2n} = \sum_{n \geq 0} \frac{(-1)^n}{b_j^{2(n+1)}}(x-a_j)^{2n}.$$
Hence you can "in principle" construct an explicit formal power series expansion for any rational function $\frac{f(x)}{g(x)}$ using the geometric series and partial fraction decomposition of the product in P1.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Logic behind base to number conversion I am trying to understand and visualize the logic behind the base conversion method.
By "base" I mean how many numbers in a number system:
*
*The decimal number system we use every day has 10 digits
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and so it is Base 10
*A binary digit can only be 0 or 1, so is Base 2
*A hexadecimal digit can be {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C,
D, E, F}, so is Base 16
*And we can use letters {A, B, C, ..., X, Y, Z}, and we get Base 26
Like in base 26 for AAA its 1* 26*26 + 1*26 + 1
Why formula is:- digit * ( base )^position index
| Because it works.
We need a way to express every possible whole numbers.
By having a base $b$ (assuming $b$ is a whole number larger than one) we can simply use the $b$ different digits to list the first $b$ whole numbers from $0$ to $(b-1)$.
(You might ask why $0$ to $b-1$ rather than $1$ to $b$. Technically it is arbitrary. But the indexing works out better if we start at "nothing")
That's the first group of $b$ numbers. We can do a second batch of $b$ numbers by putting a $1$ in front and relisting the $b$ digits after the $1$ for $1\color{blue}0$ to $1\color{blue}{(b-1)}$. This next group of number will go from $1\color{blue}0= b$ to $1\color{blue}{(b-1)} = b + (b-1) =2*b - 1$.
So those are the first $2b$ numbers from $0$ to $2b -1$.
We can do a third, fourth and fifth group of $b$ numbers (assuming our $b$ is that large by placing a $2,3,4$ befor and repeating the $b$ digits. We can do this for $b$ groups of $b$ from:
$\underbrace{\underbrace{0, 1,2 ,...,(b-1)}_{b} + \underbrace{10, 11,12 ,...,1(b-1)}_{b}+ \underbrace{20, 21,22 ,...,2(b-1)}_{b}+...... \underbrace{(b-1)0, (b-1)1,(b-1)2 ,...,(b-1)(b-1)}_{b}}_{b\text{ groups of }b\text{ numbers each}}$
This $b$ groups of $b$ numbers so this is $b^2$ numbers from $0$ to $b^2 -1$. The the number $ka$ would the int $k+1$th group which has the numbers from $k*b$ to $k*b + (b-1)$ and this is the $a+1$ number in the group and is the number $k*b + a$. (Okay, I admit it. The indexing from $0$ is confusing sometimes... but trust me... on the whole it is easier.
Now we have run out of two digit combos after $b^2$ numbers. (Which makes sense. We can have $b$ options for the first digit and $b$ for the second of $b^2$ total.)
But we can do another set of $b^2$ numbers but putting a $1$ before the two digits.
And if we put the $b$ digits before the two digits we can get $b$ groups of $b^2$ number making a total of $b^3$ numbers ranging from $0$ to $b^3-1$ but having $0...(b-1)$ being the numbers $0$ to $b-1$. $10$ to $(b-1)(b-1)$ being so that the number $an = a*b + n$ being the numbers from $b$ to $(b-1)*b + (b-1) = b^2-1$. And then numbers $100$ to $(b-1)(b-1)(b-1)$ where $amn$ would be the $a+1$th group of $b^2$ numbers which go from $ab^2$ to $a*b^2 + (b^2-1)$ and within that group it is the $m+1$th group of $b$ numbers that go from $a*b^2 + m*b$ to $a*b^2 + m*b + (b-1)$ and it is the $n+th$ number in the group so is $n$ more than $a*b^2 + m*b$ or is $a*b^2 + m*b + n$.
And we bootstrap up. We can get $(b-1)$ more groups of $b^3$ numbers but putting $1$ to $(b-1)$ before them to represent the first $b^4$ numbers with $4$ digits and so on.
It works.
And that's really all there is that can be said.
We do it because it will work.
| {
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Evaluating $\int_0^1 (1-x^2)^n dx$ In an exercise I'm asked the following:
a) Find a formula for $\int (1-x^2)^n dx$, for any $n \in \mathbb N$.
b) Prove that, for all $n \in \mathbb N$: $$\int_0^1(1-x^2)^n dx = \frac{2^{2n}(n!)^2}{(2n + 1)!}$$
I used the binomial theorem in $a$ and got:
$$\int (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) (-1)^k \ \frac{x^{2k + 1}}{2k+1} \ \ \ + \ \ C$$
and so in part (b) i got:
$$\int_0^1 (1-x^2)^n dx = \sum_{k=0}^n \left( \begin{matrix} n \\ k \end{matrix} \right) \ \frac{(-1)^k}{2k+1}$$
I have no clue on how to arrive at the expression that I'm supposed to arrive. How can I solve this?
| Part b) can be obtained just by induction. The result is immediate for $n=1$ and
\begin{align*}
\int_0^1 (1-x^2)^{n+1}dx = & \int_0^1 (1-x^2)(1-x^2)^ndx = \int_0^1(1-x^2)^n dx-\int_0^1 x^2 (1-x^2)^n dx\\
= & \frac{2^{2n}(n!)^2}{(2n+1)!} + \frac{1}{2} \int_0^1 x (-2x) (1-x^2)^n dx\\
= & \frac{2^{2n}(n!)^2}{(2n+1)!}-\frac{1}{2n+2}\int_0^1(1-x^2)^{n+1} dx
\end{align*}
This implies that
$$
\int_0^1 (1-x^2)^{n+1}dx = \frac{2n+2}{2n+3} \frac{2^{2n}(n!)^2}{(2n+1)!}=\frac{(2n+2)^2 2^{2n} (n!)^2}{(2n+1)!(2n+2)(2n+3)}=\dfrac{2^{2n+2}((n+1)!)^2}{(2n+3)!}.
$$
\begin{align*}
\int_0^1 x (-2x) (1-x^2)^n dx = &\left[x \dfrac{(1-x^2)^{n+1}}{n+1} \right]_0^1 - \frac{1}{n+1}\int_0^1 (1-x^2)^{n+1}
\end{align*}
| {
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Simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ Any simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ that does not involve Fermat's last theorem?
| $$1 = a^3+b^3 = (a+b)(a^2-ab+b^2) = (a+b)((a+b)^2-3ab)$$
implies $a+b=\pm 1$ and so $(a+b)^2 - 3ab = 1 - 3ab \not\in \{-1,1\}$ if $ab \ne 0$
| {
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Show that $a^2 - 2021b^2 = 13$ has no integer solutions
Show that $a^2 - 2021b^2 = 13$ has no integer solutions.
I'm no number theory expert, but here's what I've thought of so far. If there exist $a,b$ that satisfy the above equation, then there must also exist $a,b$ that satisfy $$a^2 \bmod 4 = b^2\bmod 4 + 1$$
A square is $0$ or $1 \bmod 4$, so we get $a^2\equiv 1\bmod 4$ and $b^2 \equiv 0 \bmod 4$. Thus, $2|b$ and $a^2 = 4k+1$. I do not see a contradiction yet. What else should I try?
Also tried to use the fact that squares have units digit only $0,1,4,5,6,9$, so if $a,b$ exist that satisfy the equation, we must have $a^2$ ending in $\{0,1,4,5,6,9\} \cap \{3,4,7,8,9,2\} = \{4,9\}$. Moreover, we must have $b^2$ ending in $\{0,1,4,5,6,9\} \cap \{7, 8, 1,2,3,6\} = \{1,6\}$.
If $b^2$ ends in $1$, then $a^2$ ends in $4$ - but we just saw that $a^2 = 4k+1$, so this is not a possibility.
If $b^2$ ends in $6$, then $a^2$ ends in $9$. How do I contradict this? Thank you!
| $a^2=13+2021b^2 \Rightarrow a^2 \equiv 6b^2 \mod 13 \iff \Big(\frac{a}{b}\Big)^2 \equiv 6 \mod 13 $ for $b\neq0$. But as $\genfrac(){}{}{6}{13}=-1$ then $6$ is not a quadratic residue so the equation does not have solutions unless $a \equiv b \mod 13$ and $a \equiv 0 \mod 13$. But even in this case, $13^2$ divides the LHS but does not divide the RHS.
| {
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Determine all $f:\Bbb Z \to \Bbb Z$ st $f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$. (Vietnamese TST 2005) Determine all $f:\Bbb Z \to \Bbb Z$ st $$f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$$ for all $x,y,z\in\Bbb Z$.
Source : Vietnamese TST 2005
I'm sure that the only solutions are the zero function, the identity, and the negative of the identity. First, when $x=y=z=0$ we have $f(0)=3f(0)^3$ so $f(0)=0$ is the only integer solution. Next if $y=-x$ and $z=0$ then $f(-x)=-f(x)$ so $f$ is odd. Then $x=y=1$ and $z=-1$ give $f(1)^3=f(1)$ so $f(1)=0,1,-1$.
Since $f$ is a soln iff $-f$ is, we can take $f(1)=0,1$. Then $f(2)=f(1^3+1^3+0^3)=0,2$. Then $f(3)=f(1^3+1^3+1^3)=0,3$. Also $f(4^3-3^3-3^3)=f(10)=f(1^3+1^3+2^3)$, so $f(4)=0,4$. Then $f(5^3-4^3-4^3)=f(-3)$ so $f(5)=0,5$. Then $f(6)=f(2^3-1^3-1^3)=0,6$. Then, $f(7)=f(2^3-1^3+0^3)=0,7$. Then $f(8)=f(2^3+0^3+0^3)=0,8$. Then $f(9)=f(2^3+1^3+0^3)=0,9$ and $f(10)=f(2^3+1^3+1^3)=0,10$. Etc...
There should be a smart way to make induction, but I haven't found that. May be a nice identity. This is the closest to the claim that $n^3$ is the sum of five cubes of integers with absolute values $<|n|$:
$$n^3=6^3+(n-2)^3+(n-3)^3-(n-6)^3-(n-11)^3+(n-12)^3.$$
At least the identity above gives me $f(11)=0,11$ and $f(12)=0,12$.
| As requested I'm posting the above comment as an answer here.
In order to show that every $n\geq 8$ can be written as sum of five cubes with absolute value less than $n$:
*
*we can write for odd numbers $n=2k+1$ with $k\geq 4$
$$(2k+1)^3 = (2k-1)^3 + (k+4)^3 - (k-4)^3 - 5^3 - 1^3$$
*for even numbers $n\geq 8$, we can write $n=2^a m$ where $m$ is $4$, $6$, or an odd number $\geq 5$. Then $m$ can be decomposed as $m=m_1^3+m_2^3+m_3^3+m_4^3+m_5^3$ so that
$$(2^a m)^3 = (2^a m_1)^3 + (2^a m_2)^3 + (2^a m_3)^3 + (2^a m_4)^3 + (2^a m_5)^3$$
| {
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(intergation) area of region bounded by a curve and a line with unknown a Let $S1$ be the area of the region bounded by the curve $y = x^2$ and the lines $y = ax, (0 < a < 1)$, and let $S2$ be the area of the region bounded by the curve $y = x^2$, the lines $y = ax, (0 < a < 1)$ and the line $x = 1$, where $a$ is the same constant.
(a) Find the value of $a$ that minimizes the value of $S1 + S2$. What is the minimum value?
(b) Find the volume of the solid generated by revolving the region, that takes the minimum value of $S1 + S2$, about the x-axis.
I find $S1 = \displaystyle \frac{a^3}{6}$ and $S2 = \displaystyle \frac{a^3}{3} - \frac{a}{2} + \frac{1}{3}$, but how can I calculate the value of $a$?
| You have calculated value of $\small S_1$ and $\small S_2$ correctly. Now to find value of $a$ that minimizes $ \small S = S_1+S_2$, take the derivative of $S$ with respect to $a$ and equate to zero.
$\small S = S_1 + S_2 = \displaystyle \frac{a^3}{3} - \frac{a}{2} + \frac{1}{3}$
$\small S' = \displaystyle a^2 - \frac{1}{2} = 0 \implies a = \frac{1}{\sqrt2}$
That leads to minimum area of $\small S_1 + S_2 = \displaystyle \frac{\sqrt2 - 1}{3 \sqrt2}$
Now you need to find volume of solid generated by revolution of the region about x-axis $\big($for $a = \displaystyle \small \frac{1}{\sqrt2}\big)$. For $\small S_1$, both shell and washer methods are fine. For $\small S_2$, it is easier to apply washer method as the shell method will require you to split the integral into two.
| {
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Where did I go wrong with my differentiation here? This question involves the product rule and power of a function rule for differentiation. I often make silly algebra errors, but here I cannot find them. The answer is supposed to be y'=22. Thanks!
When $x=-1$, $y=(2x+1)^5(3x+2)^4$.
Hence, the derivative,
$$y'=5(2x+1)^4 \cdot (2)(3x+2)^4 + (2x+1)^5 \cdot 4(3x+2)^3 \cdot 3$$
Plugging in $x=-1$,
$$y'= 5(2(-1)+1)^4 \cdot (2)(3(-1)+2)^4 + (2(-1)+1)^5 \cdot 4(3x+2)^3 \cdot 3\\
y' = 10-12\\
y' = -2$$
| We are computing the derivative at $x=-1$ when $y=(2x+1)^5 (3x+2)^4$. To find the derivative, we will be using chain rule and the product rule in correlation.
The product rule: $f'g+fg'$
In our case, Let $f=(2x+1)^5$ and $g=(3x+2)^4$
Hence, $f'=5(2x+1)^{5-1}\cdot(2x+1)^{'}\\ \;\;\;=5(2x+1)^4\cdot 2\\ \;\;\;=10(2x+1)^4$ and $g'=4(3x+2)^{4-1}\cdot (3x+2)^{'}\\ \;\;\;=4(3x+2)^3\cdot3\\ \;\;\;=12(3x+2)^3$
Completing the product rule ($f'g+fg'$): $$y'=10(2x+1)^4*(3x+2)^4+(2x+1)^5*12(3x+2)^3$$
Evaluating at $x=-1$, $$y'(-1)=10(2(-1)+1)^4\cdot (3(-1)+2)^4 +(2(-1)+1)^5\cdot 12(3(-1)+2)^3\\y'(-1) =10+12
\\y'(-1)=22$$
| {
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How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.
| Your sum can be put as
$$(1+\frac 23)\Bigl(\frac 13+\frac{1}{3^3}+\frac{1}{3^5}...\Bigr)=$$
$$\frac 53\frac 13\Bigl(1+\frac 19+\frac{1}{9^2}+...\Bigr)=$$
$$\frac 53\frac 13 \frac{1}{1-\frac 19}=\frac 58$$
| {
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Calculate the limit $\lim_{(x,y)\to (0,0)} \frac{x-\cos(x^2+y^2)-\arctan(x)+1}{x^2+y^2}$ Find the limit, $$\lim_{(x,y)\to (0,0)} \frac{x-\cos(x^2+y^2)-\arctan(x)+1}{x^2+y^2}.$$
My attempt:
I have tried several paths for evaluating the limit. For instance, $y=0$, $x=0$, $y=x$, $x=\tan y$, etc. I always obtain $0$. But I haven't been able to prove that the limit is equal to $0$.
| You may use the following limits which you can easily show using Taylor or L'Hosp.:
*
*$\lim_{t\to 0}\frac{1-\cos t}{t^2}=\frac 12$, hence, $\lim_{t\to 0}\frac{1-\cos t}{t}=0$
*$\lim_{x\to 0}\frac{x-\arctan x}{x^2}=0$
Now, just rewrite the given expression as follows:
\begin{eqnarray*}\frac{x-\cos(x^2+y^2)-\arctan(x)+1}{x^2+y^2}
& = & \underbrace{\frac{1-\cos(x^2+y^2)}{x^2+y^2}}_{\stackrel{(x,y)\to (0,0)}{\longrightarrow}0} + \underbrace{\frac{x-\arctan x}{x^2+y^2}}_{\left|\frac{x-\arctan x}{x^2+y^2}\right|\leq\left|\frac{x-\arctan x}{x^2}\right|\stackrel{(x,y)\to (0,0)}{\longrightarrow}0} \\
\end{eqnarray*}
| {
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Messed up on the inverse of $\cosh(x)$ but I don't know where I went wrong If someone can tell me where I went wrong, that would be extremely appreciated :)
The function $f:[0,\infty)\mapsto[1,\infty)$ defined by $f(x)=\cosh x$ is bijective. Find a formula for $f^{-1}.$
If $y=\cosh x$, then
\begin{align}
& y = \frac{1}{2}\left(e^x+e^{-x}\right) \\[5pt]
\implies & 2y = e^x+e^{-x} \\[5pt]
\implies & 2ye^x = e^x(e^x+e^{-x}) \\[5pt]
\implies & 2ye^x = e^{2x} + e^0 \\[5pt]
\implies & 2ye^x = e^{2x}+1 \\[5pt]
\implies & -e^{2x}+2ye^x - 1 = 0 \\[5pt]
\implies & e^x = \frac{-2y \pm \sqrt{4y^2-4(-1 \times -1)}}{-2} \\[5pt]
\implies & e^x = y + \sqrt{-2y+2} \text{ as $e^x>0$} \\[5pt]
\implies & x = \ln(y+\sqrt{-2y+2}) \\[5pt]
\implies & \cosh^{-1}x = \ln(y+\sqrt{-2y+2})
\end{align}
| We have
\begin{align}
e^x &= \frac{-2y\pm\sqrt{(2y)^2-4}}{-2} \\[5pt]
&= \color{red}{y\pm\frac{\sqrt{4y^2-4}}{-2}} \\[5pt]
&= y\pm\frac{\sqrt{4}\sqrt{y^2-1}}{-2} \\[5pt]
&= y\pm \frac{2\sqrt{y^2-1}}{-2} \\[5pt]
&= y \mp\sqrt{y^2-1} \, .
\end{align}
Since $x$ is nonnegative, we know that $e^x \geq 1$. Hence, $y\mp\sqrt{y^2-1} \geq 1$. Notice that it has to be greater than or equal to $1$, not $0$ as you wrote in your question. Since $0<y-\sqrt{y^2-1}\leq 1$ for $y\geq1$, the negative branch doesn't work. Hence, $e^x=y\color{red}{+}\sqrt{y^2-1}$, and the result follows.
| {
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Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My solution is as follow
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$
$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$
The solution is correct but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.
Is there any generalized method
| Problem with the Approach in the Question
One must take extra care when taking the limit of the sum of an increasing number of terms, each of which is tending to $0$. For example,
$$
\overbrace{\frac1{n+1}+\frac1{n+2}+\frac1{n+3}+\cdots+\frac1{n+n}}^{\text{$n$ terms, each of which tends to $0$}}
$$
If we sum the limits of each term, we get $0$. However, each of the $n$ terms is $\ge\frac1{2n}$, therefore, the $\liminf\limits_{n\to\infty}$ of the sum is $\ge\frac12$.
Inequalities and the Limit
First, we have the bound
$$
\begin{align}
\sum_{k=1}^n\frac1k
&\le1+\sum_{k=2}^n\int_{k-1}^k\frac{\mathrm{d}x}x\tag{1a}\\
&=1+\int_1^n\frac{\mathrm{d}x}x\tag{1b}\\[6pt]
&=1+\log(n)\tag{1c}
\end{align}
$$
Explanation:
$\text{(1a)}$: $\frac1k\le\int_{k-1}^k\frac{\mathrm{d}x}x$
$\text{(1b)}$: the sum of an integral over disjoint intervals
$\phantom{\text{(1b):}}$ is the integral over the union of those intervals
$\text{(1c)}$: integrate
Furthermore, for $x\gt-1$ and $n\ge1$ and $nx\lt1$,
$$
\begin{align}
(1+x)^n
&\le\left(\frac1{1-x}\right)^n\tag{2a}\\
&\le\frac1{1-nx}\tag{2b}
\end{align}
$$
Explanation:
$\text{(2a)}$: $(1+x)(1-x)=1-x^2\le1$
$\text{(2b)}$: Bernoulli's Inequality applied to the reciprocal
Thus,
$$
\begin{align}
\lim_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n
&\le\lim_{n\to\infty}\left(1+\frac{1+\log(n)}{n^2}\right)^n\tag{3a}\\
&\le\lim_{n\to\infty}\frac1{1-\frac{1+\log(n)}n}\tag{3b}\\[3pt]
&=1\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: apply $(1)$
$\text{(3b)}$: apply $(2)$
$\text{(3c)}$: $\lim\limits_{n\to\infty}\frac{1+\log(n)}n=0$
Therefore, since $\lim\limits_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n\ge1$,
$$
\lim_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n=1\tag4
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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solve $|\sin(a)|=\cos(3a)$ $|\sin(a)|=\cos(3a)$ is an alternative version of an equation $\sqrt{1-x^2}=4x^3-3x$, where I made a substitution $x=\cos(a)$ for $x \in [-1, 1]$. Unfortunately, I have no idea how to solve trigonometric equations.
| Consider
Case 1: $0 \le a \le \frac \pi 6$ then $0 \le 3a \le \frac \pi 2$. Than $\sin a \ge 0$ and $\cos 3a \ge 0$.
Then $a$ and $3a$ are both in the first quandrant. And the identity that will apply is $\sin a = \cos (\frac \pi 2 - a)$.
So we must have $\frac\pi 2 - a= 3a$ or $a =\frac {\pi} 8$.
Case 2: $\frac \pi 6 < a < \frac \pi 2$. Then $\frac \pi 2 < 3a < \frac {3\pi}2$. But then $\cos 3a < 0$ and that's not the case.
Case 3: $\frac \pi 2 \le a \le \frac {2\pi}3$. And $\frac {3\pi}2 \le 3a \le 2\pi$. Then $\sin a \ge 0$ and $\cos 3a \ge 0$.
Now $a$ is in the second quadrant and $a$ is in the fourth and the indentity that is relevent is.
$\sin (\frac \pi 2 + \alpha) = \cos (2\pi - \alpha)$.
So $a = \frac \pi 2 + \alpha$ and $3a = 2\pi - \alpha$ so $a+ 3a = 2\pi + \frac \pi 2$ so $4a = \frac {5\pi }2$ and $a = \frac {5\pi} 8$.
Case 4: $\frac {2\pi}3 < a \le \frac {5\pi}6$ and $2\pi < 3a \le 2\pi + \frac {\pi}2$.
Then $a$ is in second quadrant but $3a$ is in the first but over $2\pi$.
The identity that applies here is $\cos (2\pi + \alpha) = \sin (\frac \pi 2 + \alpha)$.
So $a = \frac \pi 2 + \alpha$ and $3a = 2\pi + \alpha$ ans so $3a - a = \frac {3\pi }2$ and $a= \frac {3\pi}4$.
Case 5: $\frac {5\pi}6 < a < \frac {7\pi}6$ and $\frac {5\pi}2 < 3a < \frac {7\pi}2$. But then $\cos 3a < 0$ which isn't possible.
Case 6: $\frac {7\pi}6 \le a \le \frac {4\pi}3$ so $\frac {7\pi}2 \le 3a \le 4\pi$ then $\sin a \le 0$ but $\cos 3a = -\sin a \ge 0$.
The identity here is $-\sin (\pi + \alpha) = \sin (\pi - \alpha)=\cos(\frac {7\pi}2 + \alpha)$.
So $a = \pi + \alpha$ and $3a = \frac {7\pi}2 + \alpha $ so $2a = \frac {5\pi}2$ and $a =\frac {5\pi}4$.
Case 7: $\frac {4\pi}3 \le a \le \frac {3\pi}2$ so $4\pi \le 3a \le \frac {9\pi}2$.
Then identity to use here is $-\sin (\frac {3\pi} 2 - \alpha)= \cos (4\pi + \alpha)$
So $a = \frac {3\pi}2 -\alpha$ and $3a =4\pi +\alpha$ so $4a = 4\pi + \frac {3\pi}2$ so $a =\pi +\frac {3\pi}8 = \frac {11\pi}8$.
Case 8: $\frac {3\pi}2 < a < \frac {11\pi } 6$ so $\frac {9\pi}2< 3a < \frac {11\pi}2$ so $\cos 3a < 0$.
And final final case
9: $\frac {11\pi}6 \le a < 2\pi$ then $\frac {11\pi}2 \le 3a < 6\pi$.
The identity we need is $-\sin (\frac {3\pi}2 + \alpha) = \cos(6\pi - \alpha)$.
So $a + 3a = 6\pi + \frac {3\pi }2$ and so $a = \frac {15\pi}8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Help with arcsin integral I have the following integral:
$$ I =\int x^2\sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) dx$$
Where $a$ and $b$ are non-zero positive integers, and $x<a$. I have started by integration by parts:
$$ I = uv - \int vu'$$
$$u = \sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right )$$
$$u' = \left [ \sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) \right ]'$$
We substitute the inside content of the arcsin by $z=\frac{\sqrt{a^2-x^2}}{b}$:
$$u' = \left [ \sin^{-1}(z) \right ]' = \frac{1}{\sqrt{1-z^2}} \cdot z'$$
$$z' = - \frac{x}{b} (a^2-x^2)^{- \frac{1}{2}}$$
Hence we obtain:
$$u' = -\frac{x}{\sqrt{(a^2-x^2)(b^2-a^2+x^2)}} $$
On the other hand of the integration by parts we have $v'$:
$$v' = x^2$$
$$v = \frac{1}{3}x^3$$
Hence we end up with:
$$I = \frac{1}{3}x^3\sin^{-1}\left ( \frac{\sqrt{a^2-x^2}}{b} \right ) + \frac{1}{3}\int \frac{x^4}{\sqrt{(a^2-x^2)(b^2-a^2+x^2)}} $$
I cant really figure out how to solve this final integral. Any help would be appreciated
| Make the substitution
$$u = \frac{\sqrt{a^2-x^2}}b.$$
Then $x = \sqrt{a^2 - u^2 b^2},$ and so $d x = - \frac{b^2 du}{2 \sqrt{a^2-u^2}},$
and so the integral becomes
$$\frac{b^2}2\int \sqrt{a^2 - u^2 b^2} \arcsin(u) du= \frac{b^2 a}{2} \int \sqrt{1- (\frac{ub}a)^2} \arcsin u du$$ Now integrate by parts.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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Find the matrix of the linear transformation $T$ Let $A=\begin{pmatrix}0\\0\end{pmatrix},B=\begin{pmatrix}1\\0\end{pmatrix},C=\begin{pmatrix}0\\1\end{pmatrix}$ and $D=\begin{pmatrix}1\\1\end{pmatrix}$. The images of these vectors under the linear transformation is $A^*=\begin{pmatrix}0\\0\end{pmatrix},B^*=\begin{pmatrix}2\\2\end{pmatrix},C^*=\begin{pmatrix}0\\1\end{pmatrix}$ and $D^*=\begin{pmatrix}2\\3\end{pmatrix}$ respectively. Find the matrix of the linear transformation $T$.
Let $T=\begin{pmatrix}a & b\\c & d\end{pmatrix}$
$$
T(A)=A^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}
$$
$$
T(B)=B^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}a\\c\end{pmatrix}=
\begin{pmatrix}2\\2\end{pmatrix}
$$
$$
T(C)=C^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}b\\d\end{pmatrix}=
\begin{pmatrix}0\\1\end{pmatrix}
$$
$$
T(D)=D^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}a+b\\c+d\end{pmatrix}=
\begin{pmatrix}1\\1\end{pmatrix}
$$
$(2)$ and $(3)$ implies $T=\begin{pmatrix}2&0\\2&1\end{pmatrix}$ but it's not satisfy $(4)$. Am I doing it in wrong way?
| It's correct! Note $D^*=\begin{pmatrix}2\\3\end{pmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
There is a shortcut for finding the equation of a tangent to a conic. To what other curves can this shortcut be applied? The conics can be written in Cartesian and parametric form:
Conic
Cartesian equation
Parametric equation
Ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$x=a \cos t, y=b \sin t$
Parabola
$y^2=4ax$
$x=at^2,y=2at$
Hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$x=a \sec t, y=b \tan t$
One task that students are frequently asked to perform is to find the equation of a tangent to a curve at a given point.
Here is the usual method being applied for an ellipse:
$x=a \cos t$
$\frac {dx}{dt}=-a\sin t$
$y = b\sin t$
$\frac {dy}{dt}=b\cos t$
$\frac{dy}{dx}=\frac{b \cos t}{-a\sin t}$
Tangent passes through point $(a\cos t,b\sin t)$
$y-b\sin t=-\frac{b \cos t}{a\sin t}(x-a\cos t)$
$ay\sin t-ab\sin^2 t=-bx\cos t+ab\cos^2 t$
$ay\sin t+bx\cos t=ab\cos^2 t+ab\sin^2 t$
$ay\sin t+bx\cos t=ab$
In a similar way we can find the equations of the tangents for the other conics:
Conic
Equation of tangent
Ellipse
$ay\sin t+bx\cos t=ab$
Parabola
$ty-x=at^2$
Hyperbola
$bx \sec t-ay \tan t=ab$
I have noticed an interesting shortcut to finding the equation of a tangent.
I'm applying this shortcut to a hyperbola.
Start with the Cartesian equation: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Perform a partial substitution using the parametric forms: $x=a \sec t, y=b \tan t$. By a partial substitution I mean that I am replacing just one of the $x$ and $y$ with the parametric equivalent.
$\frac{xa \sec t}{a^2}-\frac{yb \tan t}{b^2}=1$
$\frac{x \sec t}{a}-\frac{y \tan t}{b}=1$
$bx \sec t-ay \tan t=ab$
This shortcut works for both the ellipse and the hyperbola.
The parabola needs a slightly different approach.
Rewrite the Cartesian equation $y^2=4ax$ as $y^2=2ax+ 2ax$
Then perform the partial substitution:
$y \times 2at=2ax+ 2a \times at^2$
$2aty=2ax+ 2a^2t^2$
$ty=x+ at^2$
$ty-x=at^2$
Neat as the shortcut is, it seems unreasonable that it should work so effectively.
I see that there is another similar question: The Instant Tangent , but I would like to know if there are any other curves where this shortcut can be applied successfully. Or is there something special about the conics that make this shortcut work?
| For unified expressions, let $(x_1, y_1)$ be the point on the ellipse $\frac{x^2}{a^2} +\frac{y^2}{b^2}=1$, the hyperbola $\frac{x^2}{a^2} -\frac{y^2}{b^2}=1$, and the parabola $y=ax^2$. Then, their tangent lines are respectively
\begin{align}
\text{Ellipse:} &\>\>\>\>\> \frac{xx_1}{a^2}+ \frac{yy_1}{b^2}=1\\
\text{Hyperbola:} & \>\>\>\>\> \frac{xx_1}{a^2}- \frac{yy_1}{b^2}=1\\
\text{Parabola:}& \>\>\>\>\> \frac{y+y_1}2={axx_1}
\end{align}
The ‘short-cuts’ also work in cases where the point $(x_1, y_1)$ is not on the curve. Then, there are a pair of tangent lines drawn to the curve from the point, which are given by
\begin{align}
\text{Ellipse:} &\>\>\>\>\>
(\frac{x^2}{a^2} +\frac{y^2}{b^2}-1)(\frac{x_1^2}{a^2} +\frac{y_1^2}{b^2}-1) =(\frac{x_1x}{a^2} +\frac{y_1y}{b^2}-1)^2\\
\text{Hyperbola:} &\>\>\>\>\>
(\frac{x^2}{a^2} -\frac{y^2}{b^2}-1)(\frac{x_1^2}{a^2} -\frac{y_1^2}{b^2}-1) =(\frac{x_1x}{a^2} -\frac{y_1y}{b^2}-1)^2\\
\text{Parabola:} &\>\>\>\>\>
(y-a x^2)(y_1 -ax_1^2)= \left( \frac{y+y_1}2-axx_1\right)^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty$ as $x\rightarrow +\infty$ I am trying to show that $$\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty \ \ \text{as} \ \ x\rightarrow +\infty.$$
My attempt is as follows:
\begin{align}
\frac{x^{2x}}{(x+1)^{x+1}}&=\frac{x^{x}}{x+1}\left(\frac{x^x}{(x+1)^x}\right) \\
&=\frac{x^{x}}{x+1}\left(\frac{x}{x+1}\right)^x \\
&=\frac{x^{x}}{x+1}\left(\frac{1}{(1+1/x)^x}\right).
\end{align}
I can see that the second fraction will converge to $1/e$, but I am unsure of how to approach the first fraction.
| Taking the $\log$ of both sides for big enough $x$
$$\begin{align*}\log \frac{x^{2x}}{(x+1)^{x+1}}&=2x\log x-(x+1)\log(x+1)\\&=x\log \frac{x^2}{x+1}-\log(x+1)\\&\geq x\log \frac{x^2}{2x}-\log(2x)\\&=\log \frac{x^{x-1}}{2^{x+1}}.\end{align*}$$
Now take $e$ to both sides to get, for large enough $x$
$$\frac{x^{x}}{(x+1)^{x+1}}\geq \frac{x^{x-1}}{2^{x+1}}\geq\frac{2^{2x-2}}{2^{x+1}}\to\infty .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
find the roots $(2z+3)^3=\frac{1}{64}$ There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$ so $z=\frac{-11}{8}$ however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. thank you
| For any $z = a + ib \in \mathbb{C}$,
$$
z = \rho\left(\cos{\varphi} + i\sin{\varphi}\right),
$$
where
$$
\rho = \sqrt{a^2 + b^2}, \text{ and }\varphi = \arctan\left(\frac{b}{a}\right).
$$
Then,
$$
\sqrt[n]{z} = \sqrt[n]{\rho}\left(\cos{\left(\frac{\varphi + 2\pi k}{n}\right)} + i\sin{\left(\frac{\varphi + 2\pi k}{n}\right)}\right), \quad k = 0, 1, \ldots, n-1.
$$
In our example,
$$
\frac{1}{64} = \frac{1}{64} + i0, \text{ i. e., }a = \frac{1}{64}\text{ and }b = 0 \Rightarrow
$$
$$
\frac{1}{64} = \frac{1}{64}\left(\cos(0) + i\sin(0)\right) \Rightarrow
$$
$$
\sqrt[3]{\frac{1}{64}} = \sqrt[3]{\frac{1}{64}}\left(\cos\left(\frac{2\pi k}{3}\right) + i\sin\left(\frac{2\pi k}{3}\right)\right), \quad k = 0, 1, 2.
$$
$$
\begin{array}{rl}
k = 0 \Rightarrow & z_1 = \frac{1}{4} \\
k = 1 \Rightarrow & z_2 = \frac{1}{4}\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -\frac{1 - i\sqrt{3}}{8}\\
k = 2 \Rightarrow & z_3 = \frac{1}{4}\left(\cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -\frac{1 + i\sqrt{3}}{8}
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral of Euler. Here is an integration problem I found in an old book:
Integrate
$$\int\frac{1+x^2}{1-x^2}\frac{dx}{\sqrt{1+x^4}}$$
The integral is attributed to Euler.
My solution is
let $$y=\frac{1+x^2}{1-x^2}$$ then get
$$\frac{1}{\sqrt{2}} \int \frac{ydy}{\sqrt{y^4-1}}$$
After another substitution $z=y^2$ get
$$\frac{1}{2\sqrt{2}}\int\frac{dz}{\sqrt{z^2-1}}=\frac{1}{2\sqrt{2}}\cosh^{-1}z$$ so my answer is
$$\frac{1}{2\sqrt{2}}\cosh^{-1}\left(\frac{1+x^2}{1-x^2}\right)^2$$
However the answer in the back is
$$\frac{1}{\sqrt{2}} \sinh^{-1}\frac{\sqrt{2}x}{1-x^2}$$
After much work (!!!) I have shown that the two forms are equal to a constant.
My question is: How can one solve the original integral to get the alternative answer directly. What substitution to use ?
| Another shortcut: Let $I=\int\frac{1+x^2}{1-x^2}\frac{dx}{\sqrt{1+x^4}}=\int\frac{1+\frac 1{x^2}}{\frac 1x-x}\frac{dx}{\sqrt{x^2+\frac 1{x^2}}}$ (By dividing $N^r$ and $D^r$ by $x^2$)
$I= -\int\frac{1+\frac 1{x^2}}{-x+\frac 1x}\frac{dx}{\sqrt{(-x+\frac 1x)^2+2}}$
Now substitute $-x+\frac 1x=t$ so that $(1+\frac 1{x^2})dx=-dt$ and it follows that:
$I=-\int\frac{dt}{t\sqrt{(t^2+2)}}=\frac 1 {\sqrt 2}\ln|\frac {\sqrt 2} t+\sqrt {\frac {2}{t^2}+1}|=\frac 1{\sqrt 2}\sin h^{-1}(\frac {\sqrt 2}t) +C$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving $\frac{3^x+2^x}{3^x-2^x}=7$: More than one answer? How to solve for all $x$?
I am trying to solve this problem (math for fun):
$$\frac{3^x+2^x}{3^x-2^x}=7$$
Step 1. Let$\:a=3^x\:and\:b=2^x$
Step 2. $\frac{a+b}{a-b}=7$
Step 3. $a+b=7\left(a-b\right)=7a-7b$
Step 4. $6a-8b=0$
Step 5. $6a=8b=3\cdot 2\cdot a=4\cdot 2\cdot b$
Step 6. $3a=4b$
Step 7. $3\cdot 3^x=2\cdot 2\cdot 2^x=3^{x+1}=2^{x+2}$
Step 8. Log both sides
Step 9. $\ln \left(3\right)+x\ln 3=x\ln2+\ln 4$
Step 10. $\ln \left(3\right)-\ln 4=x\:\left(\ln 2-\ln 3\right)$
Step 11. $\frac{\:\left(\ln 3-\ln 4\right)}{\left(\ln 2-\ln 3\right)\:}=x≈0.71$
SOLVED^
| To answer you question how do you know how many solutions there are and how to you find them all: By keeping track of your steps and assuring you never add extraneous solution and that you steps are all one to one reversable so you don't lose or add solutions.
So $\frac {3^x + 2^x}{3^x -2^x} = 7$ gives us the condition that $3^x \ne 2^x$.
Substitute $a =3^x$ and $b=2^x$. This is just substitution. But we temporarily lose all information we may have about $3^x, 2^x$. But we don't have to worry we will get the information back.
$\frac {a+b}{a-b} = 7\implies a+b = 7(a-b)$. This adds an extraneous solution where $a = b$ which we know can not happens. (This is actually some of the information we lost when substitute. We know $2^x, 3^x > 0$ so $2^x + 3^x > 0$ so $a-b =0$ is not possible... okay lets not worry. We'll have to keep track $a \ne b$.
$8b = 6a$ just adding to each side. Completely reversible. This doesn't affect our solutions.
$4b = 3a$ ditto. This is a restriction.
$4\cdot 2^x = 3\cdot 3^x$
$2^{x+2} = 3^{x+1}$. We know longer are dealing with $a,b$ so the caviet $a\ne b$ no longer is a concern.
$\ln 2^{x+2} = \ln 3^{x+1}$. As $\ln$ is one to one this keeps the number of solutions the same.
$(x+2)\ln 2 = (x+1)\ln 3$ Linear shifting doesn't affect anything so
$x(\ln 2 - \ln 3) = (\ln 3 -2\ln 2)=(\ln 3-\ln 4)$. As $\ln 2-\ln 3$ is a constant and $\ln 2 \ne\ln 3$ we can divide both sides and that doesn't affect the number of solutions.
And $x -\frac {\ln 3 - \ln 4}{\ln 2-\ln 3}$.
Nothing we did lost any solutions and the one thing that added extraneous solution took care of itself.
So we know this is the only solution.
BTW $\frac {\ln 3-ln 4}{\ln 3- \ln 2} = \frac {\ln \frac 34}{\ln \frac 32}$ or $\frac {\ln 3 -2\ln 2}{\ln 3 - \ln 2} = 1 -\frac {\ln 2}{\ln 3 - \ln 2}= 1-\frac {\ln 2}{\ln \frac 32}$
If is helps (probably doesn't) $\frac {\ln \frac 34}{\ln \frac 32} =\log_{\frac 32} \frac 34=\log_{\frac 32} \frac 32\cdot \frac 12 = 1 - \log_{\frac 32} 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Limit $\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x}$ equals to $\frac{5}{12}$ A friend of mine asked this question to me. It seems it's from Stewart.
Find the values of a and b such that
$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12}$
This is what I tried with better results.
For $b$:
$$\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12} $$
$$\sqrt[3]{a x+b} - 2 = x\frac{5}{12} $$
$$\underset{x\to 0}{\text{lim}} \sqrt[3]{a x+b} - 2 = \underset{x\to 0}{\text{lim}} x\frac{5}{12} $$
$$\sqrt[3]{b} - 2 = 0 $$
$$\sqrt[3]{b} = 2 $$
$$ b = 8$$
For $a$:
$$a x+8 = (x\frac{5}{12} + 2)^3$$
$$a x+8 = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x+8$$
$$a x = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x$$
$$a = \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$\underset{x\to 0}{\text{lim}} a =\underset{x\to 0}{\text{lim}} \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$a = 5 $$
But the limit $$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{5x+8} - 2}{x}$$ doens't go to $\frac {5}{12}$.
May someone help.
| You alos could use the binomial expansion or Taylor series
$$\sqrt[3]{a x+b}=\sqrt[3]{b}+\frac{a x}{3 b^{2/3}}-\frac{a^2 x^2}{9 b^{5/3}}+O\left(x^3\right)$$
$$\frac{\sqrt[3]{a x+b} - 2}{x}=\frac{\sqrt[3]{b}-2}{x}+\frac{a}{3 b^{2/3}}-\frac{a^2 x}{9
b^{5/3}}+O\left(x^2\right)$$ So, $$\sqrt[3]{b}-2=0 \implies b=8$$
$$\frac{a}{3 b^{2/3}}=\frac{a}{12}=\frac{5}{12}\implies a=5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086122",
"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.