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How to prove the Vandermonde's determinant for a $3\times 3$ matrix when rows and columns have been swapped? The problem is as follows:
The following determinant is named after french mathematician Alexandre-Théophile Vandermonde who lived in the late 18th century.
Prove this determinant
$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=(b-a)(c-b)(c-a)$
What I've attempted to do was to use this well known identity on matrices:
$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=1\left|\begin{matrix}b&c\\b^2&c^2\end{matrix}\right|-1\left|\begin{matrix}a&c\\a^2&c^2\end{matrix}\right|+1\left|\begin{matrix}a&b\\a^2&b^2\end{matrix}\right|$
Then working in the right side of this equation I'm getting:
(Here I'm using the notation of capital $V$ for Vandermonde's determinant)
$V=(bc^2-cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$
Factoring the common terms I'm getting
$V=bc(c-b)-ac(c-a)+ab(b-a)$
But now what? I'm stuck, what else should it be done here? Can someone help me here on how to prove this determinant? Could it be that my method wasn't that right? Please help.
|
$$\begin{align}
V &= bc^2 - cb^2 - ac^2 + ca^2 + ab^2 - ba^2\\
&=bc^2 \color{fuchsia}{- bac} - cb^2 + ab^2 - ac^2 + ca^2 \color{fuchsia}{+ abc} - ba^2\\
&=b(c^2-ac-bc+ab)-a(c^2-ac+bc+ab)\\
&=(b-a)(c^2-ac-bc+ab)\\
&=(b-a)(c-b)(c-a)
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
False proof that $\frac{13}{6}=0$
At start, the length of a line segment is $a_0=0$. When $3$ hours have elapsed since start, its length is $a_3$. When $1$ hour has elapsed since start, its length increased by $\frac{a_3}{2}$ with respect to $a_0$ (call the new length $a_1$). When $2$ hours have elapsed since start, its length increased by $\frac{a_3}{3}$ (call the new length $a_2$) with respect to $a_1$. When $3$ hours have elapsed since start, its length increased by $\frac{a_3}{4}$ with respect to $a_2$. What is the value of $\frac{a_3}{a_1}$?
At $t=0$, the length is $a_0=0$. At $t=1$ hour, the length is $a_1=\frac{a_3}{2}$. At $t=2$ hours, the length is $a_2=\frac{a_3}{2}+\frac{a_3}{3}$. At $t=3$ hours, the length is $\color{red}{a_3=\frac{a_3}{2}+\frac{a_3}{3}+\frac{a_3}{4}=\frac{13}{12}a_3}$. So
$$\frac{a_3}{a_1}=\frac{\frac{13}{12}a_3}{\frac{a_3}{2}}=\frac{13}{6}$$
is the answer. But notice that the red equation enables solving for $a_3$ which gives $a_3=0$. Therefore $\frac{a_3}{a_1}=0$ and, by transitivity, we have:
Conclusion: $\frac{13}{6}=0$
Where is the mistake?
|
I wrote my answer in spoiler blocks so you can decide when to see it.
It is defined that for $t = 3$, $l = a_{3}$.
For $t = 0$, $l = 0$. The remaining length should be $a_{3}$.
For $t = 1$, $l = \frac{a_{3}}{2}$. The remaining length should be $\frac{a_{3}}{2}$.
For $t = 2$, $l = \frac{a_{3}}{2} + \frac{a_{3}}{3}$. The remaining length should be $\frac{a_{3}}{6}$.
For $t = 3$, $l = \frac{a_{3}}{2} + \frac{a_{3}}{3} + \frac{a_{3}}{4}$. This is where the problem arises, as the added length is greater than the remaining $(\frac{a_{3}}{4} > \frac{a_{3}}{6})$. This is the mistake.
|
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|
Multiplication of matrices of the given form is a group. \begin{equation*}
A_{2,2} =
\begin{pmatrix}
a & b \\
0 & 1 \\
\end{pmatrix}
\end{equation*} for $a,b \in $ $\mathbf{Q}$. Find $Z(A)$ and all elements and subgroups of finite order in $A$, provided that $A$ is a group under multiplication of matrices.
My attempt so far is that I just multiplied two matrices to check when they commute. Namely, \begin{equation*}
A_1.A_2=
\begin{pmatrix}
a_1 & b_1 \\
0 & 1 \\
\end{pmatrix} . \begin{pmatrix}
a_2 & b_2 \\
0 & 1 \\
\end{pmatrix} =A_2.A_1 \leftrightarrow a.b_1+b-b_1=a_1.b;
\end{equation*}
for $b\neq 0$ and then we can form the matrices of $Z(A)$ to be of the form \begin{pmatrix}
a_1 & b_1 \\
0 & 1 \\
\end{pmatrix} for $ a_1 =(a.b_1+b-b_1)/b $ and $b_1 \neq b/(1-a) $ for $ a\neq {0,1} $ and b $\neq 0$.
But what does mean elements of finite order? I mean that the set of rational numbers is only countable and so how I can count all the element of a given subgroup of A?
Any help is welcome!
|
Note that the identity matrix is the identity element in this group. An element is of finite order if multiplying it finitely many times gives us the identity. Now note that
\begin{equation}
\begin{pmatrix}a&b\\0&1\end{pmatrix}\cdot\begin{pmatrix}a&b\\0&1\end{pmatrix} = \begin{pmatrix}a^2&b(a+1)\\0&1\end{pmatrix}
\end{equation}
By induction you get $n$th power of a matrix to be
\begin{align}
\begin{pmatrix}a&b\\0&1\end{pmatrix}^n = \begin{pmatrix}a^n&b(1+a+a^2+\dots+a^{n-1})\\0&1\end{pmatrix}
\end{align}
For this to be identity for some finite $n$, we have $a^n=1$ and $b(1+a+\dots+a^{n-1})=0$. We make two cases:
Case 1: If $a=1$ then $b(1+a+\dots+a^{n-1})=nb$, which is zero for some natural $n$ iff $b=0$. So we get the matrix $\begin{pmatrix}a&b\\0&1\end{pmatrix}$ to be identity itself!
Case 2: If $a=-1$, then $b(1+a+a^2+\dots+a^{n-1}) = 0$ for $n=2$. So every element of the form $\begin{pmatrix}-1&b\\0&1\end{pmatrix}$ has order $2$ which is finite.
No other element is of finite order.
|
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|
Simplifying $\frac{2\sqrt n + \frac{1}{\sqrt n}-3}{2\sqrt n -1}$ Consider the term $$\dfrac{2\sqrt n + \dfrac{1}{\sqrt n}-3}{2\sqrt n -1} \tag1$$
I simplified it to
$$\dfrac{4 n-4 \sqrt{n}+\dfrac{1}{\sqrt{n}}-1}{4 n-1} \tag2$$
or also
$$\dfrac{2 n+1-3 \sqrt n}{2 n-\sqrt{n}} \tag3$$
but apparently one can simplify this to
$$\dfrac{n-\sqrt n}{n} \tag4$$
but I don't really get to this expression.
How can one arrive at $(4)$?
|
You were almost there:
I simplified it to $\dfrac{4 n-4 \sqrt{n}+\dfrac{1}{\sqrt{n}}-1}{4 n-1}$
From there, the key is to spot the $4n-1$ in the numerator and pull it out:
$$\begin{align}
\ldots &= \frac{4n-1}{4n-1} + \frac{-4\sqrt n + \frac1{\sqrt n}}{4n-1}\\
&= 1 + \frac1{\sqrt n}\frac{-4n+1}{4n-1}\\
&= 1 - \frac1{\sqrt n}
\end{align}$$
|
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|
Find the number of solutions in nonnegative integers of the equation $x_1+x_2+x_3+3x_4=7$. Find the number of solutions in nonnegative integers of the equation $x_1+x_2+x_3+3x_4=7$.
The term $3x_4$ will only be equal to $0,3$, and $6$. So $x_4\in\{0,1,2\}$. By letting $3x_4=0,3$, and $6$ we will have three cases and then by the addition principle we will get the result. So the number of solutions is ${9 \choose 2}+{6 \choose 2}+{3 \choose 2}$. Is this the correct method? Are there other or more slick ways of doing this?
|
Lets use generating functions for it such that :
$x_1 = \frac {1}{1-x}$ ,$x_2 = \frac {1}{1-x}$, $x_3 = \frac {1}{1-x}$ , $x_4 =\sum_{0}^{\infty} {x}^{3m}$
So , the result of multiplication is ${\frac {1}{1-x}}^3 \times x^{3m} = C(3+k-1,k)\times x^k \times x^{3m}$ ,because the coefficient of $x^{3m}$ is always $1$
We want to calculate the coefficient of $x^7$ , so we should find the values for $k,m$ , when $k+3m=7$
It is possible when $m=0,k=7$ or $m=1,k=4$ or $m=2,k=1$
So , $C(3+7-1,7)+C(3+4-1,4)+C(3+1-1,1)=54$
|
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|
Number of possible values of $4x-z$ if $x+y+z=20$ Given that $x, y,z$ are non negative integers such that $x+y+z=20$. If $S$ is the set of all possible values of $4x-z$.Find $n(S)$.
My try:
By stars and bars number of non negative integer solutions of $x+y+z=20$ is $\binom{22}{2}=210$
Among these $210$ ordered triplets of $(x,y,z)$ we need to find number of possible values taken by $4x-z$.
Obviously the least value taken by $4x-z$ is $-20$ when $x=0,z=20$.
The next value taken by $4x-z$ is $-19$ when $x=0,z=19$ and so on.
So once we fix $x=0$ ,since $z$ varies from $[0,20]$ number of values taken by $4x-z$ is $21$.
But when we fix $x=1$, we get some overlaps. For example when $x=1,z=18$ we have $4x-z=-14$ and this value has already been counted in the previous case when $x=0,z=14$.
So any way to count total number of values taken by $4x-z$ excluding overlaps?
|
We can count the total possible unique values of $4x-z$ by summing the size of the following 5 mutually disjoint sets of values:
*
*Non-positive values: corresponds to $x=0$, there are $21$ such values of $4x-z$ from $-20$ to $0$, corresponding to $z=0,1,2,\ldots , 20$.
*Positive values of form $\equiv 0 \pmod 4$: corresponds to $z=0$ and $x=1,2,\ldots, 20$, there are 20 such values.
*Positive values of form $\equiv 3 \pmod 4$: corresponds to $z=1$ and $x=1,2,\ldots, 19$, there are 19 such values.
*Positive values of form $\equiv 2 \pmod 4$: corresponds to $z=2$ and $x=1,2,\ldots, 18$, there are 18 such values.
*Positive values of form $\equiv 1 \pmod 4$: corresponds to $z=3$ and $x=1,2,\ldots, 17$, there are 17 such values.
Generalizing the above mod 4 cases, we have the following mutually disjoint sets of values:
*
*Non-positive values: corresponds to $x=0$, there are $21$ such values of $4x-z$ from $-20$ to $0$, corresponding to $z=0,1,2,\ldots, 20$.
*Positive values of form $\equiv -k \pmod 4$: corresponds to $z=k$ and $x=1,2,\ldots, (20-k)$, there are $20-k$ such values of $4x-z$, for $k \in \{0,1,2,3\}$.
Hence, the total number of distinct values of $4x-z$ will be $=21+\sum\limits_{k=0}^{3}(20-k)=21+80-6=95$
|
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|
Finding the solutions of an equation in positive intgers We are asked to find all positive integer solutions of the equation
$$x^7+7=y^2$$
or to show that it does not have any solutions. It is clear that $x$ can not be even. So $x$ is odd and $y$ is even (In fact $x$ is of the form $4k+1$). Adding $121$ to both sides of the equation, we get $x^7+128=y^2+121$. Hence
$$(x+2)(x^6-2x^5+4x^4-8x^3+16x^2-32x+64)=y^2+121.$$ Since the second expression on LHS is of the form $4k+3$, it does have a prime divisor $p$ with the same form. So
$$p|y^2+121$$, and from a previous lemma, we can conclude that $p|11$ and $p|y$, from which we obtain $p=11$ and $484| x^7+7$. If we proceed further, for $x^7+7$ to be divisible by 11, $x$ has to be of the form $11k+9$, where $k$ itself is shown to be of the form $11t+10$ giving us $x$ is of the form $121t+119$. How to continue in this way to show that such an equation does not have any solutions? Any help will be appreciated. Thank you.
|
I’m turning my comment into an answer.
Let $x$ be an integer such that $11|x^7+128$. Then clearly $11$ and $x$ are coprime and $11|(x^7)^3+128^3=x^{21}+2^{21}$. By (Pierre) Fermat’s little theorem, $11|x^{21}-x$, $11|2^{21}-2$ and thus $11|x+2$. Then write $x=11y-2$, then $\frac{x^7+128}{x+2}=\frac{(11y-2)^7+2^7}{11y}=\frac{11^2Ny+11\times 14y}{11y}=11N+14$ isn’t divisible by $11$.
In other words, for any integer $x$, $\frac{x^7+128}{x+2}$ is coprime to $11$. But we know from the OP that if $x^7+7=y^2$, then $11$ divides $\frac{x^7+128}{x+2}$, and we get a contradiction.
|
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|
Prove that $a^{b} +1\geq b(a+1)$ $a, b \in \mathbb{N} $ and $a>1$,$b>2$
Prove that $a^{b} +1\geq b(a+1)$
My attempt :
suppose $a=p+1$, ($ p \in \mathbb{N}$,$ p\geq 2$)
Remarks :(if $ p=1$ hence $ a=2$,in this case isn't difficult)
So let's show $(p+1)^b +1\geq bp+2b$
We know that $(p+1)^b=\sum_{k=0}^{b}\left( {\begin{array}{c} b \\ k \\ \end{array}} \right) ×p^b\geq \left( {\begin{array}{c} b \\ 1 \\ \end{array}} \right)
×p+\left( {\begin{array}{c} b \\ b \\ \end{array}} \right) ×p^b$
So $(p+1)^b+1\geq bp+p^b+1\geq bp+2b$
Because $p\geq 2$$\Rightarrow$$ p^b\geq 2^b\geq 2b$
Does my attempt is true?
|
We can prove this by induction on $b$ when $b\gt2 ,a\gt 1$.
Step 1:
$b=3$, $a^3+1\geq 3a+3$ is easy to prove, note that $a(a^2-3)\geq2$ since $a \geq2$.
Step2:
Now suppose we have $a^b+1\geq ab+b$.
$a^{b+1}+1=a(a^b+1)-a+1$
$ \geq a(ab+b)-a+1=
ab(a+1)-a+1\geq2b(a+1)-a+1$,
and since $b(a+1)-a+1 \gt 2(a+1)-a+1 \gt a+1$,we have $a^{b+1}+1\gt b(a+1)+a+1=(b+1)(a+1)$and we are done.
|
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|
Get $f(x)=u_x\frac{x}{u}$ from ODE for $u$ Consider the Cauchy-Euler ODE
\begin{align*}
\frac{1}{2}x^2u_{xx}+xu_x-u=0.
\end{align*}
Guessing $u(x)=Cx^n$ gives
\begin{align*}
\frac{1}{2}n(n-1)Cx^n+nCx^n-Cx^n=0,
\end{align*}
which we can solve to get
\begin{align*}
n_1&=-2,\\
n_2&=1.
\end{align*}
Given initial conditions and boundary behavior, we can pin down a unique solution.
I'm really interested in the function $$f(x)=u_x\frac{x}{u}.$$ Given the solution for $u$, we can compute $f$ to be $f(x)=n$, which is constant!
Q: I wonder whether I can find $f$, without solving the ODE first?
Set $f=u_x\frac{x}{u}$. Then, $u_x=\frac{fu}{x}$ and $u_{xx}=\frac{f_xu}{x}+\frac{fu_x}{x}-\frac{fu}{x^2}$. The ODE then turns into
\begin{align*}
\frac{1}{2}\left(xf_xu+xfu_x-fu\right)+fu-u=0.
\end{align*}
Dividing this ODE by $u$ gives
\begin{align*}
\frac{1}{2}\left(xf_x+f^2-f\right)+f-1=0.
\end{align*}
If I assume $f_x=0$, then I get
\begin{align*}
\frac{1}{2}f^2+\frac{1}{2}f-1=0,
\end{align*}
which is a normal quadratic equation with solutions
\begin{align*}
f_1&=-2,\\
f_2&=1.
\end{align*}
These are precisely the solutions I expected from the previous calculations.
However, I had to assume $f_x=0$. I only knew this because I already knew the solution.
Why doesn't the second approach work? Is there a way to compute $f(x)$ without first solving the ODE for $u(x)$?
|
Your claim that $f$ is a constant for any $u$ which satisfies the given equation is false. Indeed,
$$u(x) = \frac{x^3 + 1}{x^2}$$
certainly satisfies the equation, but using that $u$ we have
$$ f(x) = \frac{x^3-2}{x^3+1} $$
Now, it's certainly true that some $u$ which satisfy the original equation make $f$ constant. In particular, it seems you were considering the $x^n$ solution. In general,
$$ u(x) = ax + bx^{-2} $$
so if we force one of $a,b$ to be zero, then we have the desired identity. We can get those by imposing the initial conditions $ u(0) = 0 $ for $b$ to vanish, and $ u(\infty) = 0 $ for $a$ to vanish.
In fact, setting $f'$ to zero gives an equation for $u$ with solution
$$u(x) = \alpha x^\beta$$
showing that this result is independent of the original equation for $u$.
|
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|
$f_{n}$ uniformly converges to $f$ and $|f'_{n} (x)| ≤ C $, is $f$ necessarily differentiable? Assume $f(x)$ is the uniform limit of real differentiable functions $f_{n}(x)$ on $[−1,1]$. Assume that $|f'_{n} (x)| ≤ C $ for some $C$ independent of $n$ and $x ∈ [−1, 1]$. Is the function $f(x)$ necessarily differentiable?
|
No. Choose $f_n(x) := \sqrt{x^2 + \frac{1}{n}}$. Then, for all $x \in [-1, 1]$:
$$
\left \lvert \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \right \rvert = \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \leq \lvert x \rvert + \frac{1}{\sqrt{n}} - \lvert x \rvert = \frac{1}{\sqrt{n}} \overset{n \rightarrow \infty}{\longrightarrow}0
$$
Therefore
$$
\sup_{x \in [-1, 1]}\left \lvert \sqrt{x^2 + \frac{1}{n}} - \lvert x \rvert \right \rvert \leq \frac{1}{\sqrt{n}} \overset{n \rightarrow \infty}{\longrightarrow}0
$$
and thus uniform convergence. The limit, i.e. the absolute value is not differentiable in $0$. Furthermore, for all $n \in \mathbb{N}$, $x \in [-1, 1]$:
$$
\lvert f_n'(x) \rvert = \left \lvert \frac{x}{\sqrt{x^2 + \frac{1}{n}}} \right \rvert = \frac{\sqrt{x^2}}{\sqrt{x^2 + \frac{1}{n}}} \leq \frac{\sqrt{x^2 + \frac{1}{n}}}{\sqrt{x^2 + \frac{1}{n}}} = 1
$$
|
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|
Find the limit of following sequence $\lim\limits_{n \to inf}\frac{n×1^r+(n-1)×2^r+...+1×n^r}{n^{r+2}}$
Ok so the only idea I got was to use Stholz theorem and then use $1^r+2^r+...+n^r$ ~ $
\frac{1}{(r+1)} (n^{r+1})$ somehow but that didn't lead anywhere (or at least I didn't see it)
|
Igor Rivin's idea is great, but the limit is not $0$. We need to assume that $r\neq -1$ and $r\neq -2$. The general term of the given sequence is
\begin{align*}
S_n(r) & = \frac{1}{n^{r+2}} \sum_{k=1}^{n} (n-k+1)k^r \\
& = \frac{n+1}{n^{r+2}} \sum_{k=1}^{n} k^r - \frac{1}{n^{r+2}} \sum_{k=1}^{n} k^{r+1} \\
& = \frac{n+1}{n} \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^{r} - \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^{r+1}.
\end{align*}
We have that
\begin{equation*}
\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^{r} = \frac{1}{r+1},
\end{equation*}
since the terms of this seqence are Riemann sums of $\int_{0}^{1} x^r dx = \frac{1}{r+1}$ and
\begin{equation*}
\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^{r+1} = \frac{1}{r+2},
\end{equation*}
since the terms of this sequence are Riemann sums of $\int_{0}^{1} x^{r+1} dx = \frac{1}{r+2}$.
Therefore
\begin{equation*}
\lim_{n\to\infty} S_n(r) = \frac{1}{r+1} - \frac{1}{r+2} = \frac{1}{(r+1)(r+2)}.
\end{equation*}
For $r=-1$ we have
\begin{equation*}
S_n(-1) = \frac{n+1}{n} H_n - 1.
\end{equation*}
Therefore,
\begin{equation*}
\lim_{n\to\infty} S_n(-1) = \infty.
\end{equation*}
For $r=-2$ we have
\begin{equation*}
S_n(-2) = (n+1) H_{n,2} - H_n.
\end{equation*}
Since,
\begin{equation*}
\lim_{n\to\infty} H_{n,2} =\frac{\pi^2}{6},
\end{equation*}
we have
\begin{equation*}
\lim_{n\to\infty} S_n(-2) = \infty.
\end{equation*}
|
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Find all pair of primes $(p,q)$ such that both $p^2+q^3$ and $p^3+q^2$ are perfect squares. Let
$p^2+q^3=a^2$ and $p^3+q^2=b^2$. Let's suppose $ p \neq q$. When one of $p,q$ equals $2$, it yields system of equations with no solution, so $p,q \geq 3$.
Since any two primes numbers are coprime, then all $a,b,p,q$ are coprime.
$$(a-p)(a+p)=q^3$$
$$(b-q)(b+q)=p^3$$
$a \pm p$ does not divide $q$, and $b \pm q$ doesn't divide $p$ (if any of those are not equal to $1$). WLOG let's assume that $b-q\neq1$, then:
$$
\begin{cases}
b+q=p^2\\
b-q=p
\end{cases}$$
From this, it concludes that $$b^2=(p+q)^2=p^3+q^2\Rightarrow p^2+2pq+q^2=p^3+q^2 \Rightarrow p\frac{p-1}{2}=q $$ It's a contradiction with primality of $q$. When $\frac{p-1}{2}=1$, then $p=q$ - also contradiction.
The case when both $a-p$ and $b-q$ are equal $1$:
$$a-b=p-q=q^3-p-p^3+q$$
$$2(p-q)=(q-p)(q^2+qp+p^2)$$
$$-2=q^2+qp+p^2<0$$
Contradiction!
Now let's check the problem under condition $p=q$.
$$(a-p)(a+p)=p^3$$
$a+p$ is bigger than $p$, and $a-p$ is bigger than $1$, so
$$ \begin{cases}
a+p=p^2\\
a-p=p
\end{cases}
\Rightarrow p^2-p=2p \Rightarrow p(p-3)=0$$
The only solution pair is $(3,3)$
Is my solution correct?
|
Your proof is way more complex than it needs to be. No need to first treat $2$ as a special case. No need to assume $p\ne q$ and arrive at contradictions.
$p^3+q^2=a^2 \Rightarrow (a-q)(a+q)=p^3$. Plainly $(a-q)\text{ and }(a+q)$ are factors of $p^3$ and multiply to $p^3$. Since $p$ is prime, the only factors of $p^3$ are $\{1,p,p^2,p^3\}$.
Case 1: $a-q=1,\ a+q=p^3$. Then $a=q+1 \Rightarrow a^2=q^2+2q+1=p^3+q^2$ or $p^3=2q+1$. This leads to $q=\frac{p^3-1}{2}=\frac{(p-1)(p^2+p+1)}{2}$. Since $q$ is prime, this requires either $p-1=1$ or $\frac{p-1}{2}=1$. This forces $(p,q)=(2,\frac{7}{2})\text{ or }(3,13)$. The first pair does not consist only of integers. The second pair consists of primes and satisfies $p^3+q^2=27+169=196=14^2$. However, $13^3+3^2=2197+9=2206$ and is not the square of an integer. There is no Case 1 answer.
Case 2: $a-q=p,\ a+q=p^2$. Your reasoning here was good, but ran off track because you assumed unnecessarily that $p\ne q$. $$a^2=p^2+2pq+q^2=p^3+q^2 \Rightarrow p^3=p^2+2pq \\ \Rightarrow q=\frac{p^2-p}{2}=p\cdot \frac{p-1}{2}$$
Since $q$ is prime, this last result requires $q=p$ and $\frac{p-1}{2}=1 \Rightarrow p=3$. This turns out to be a valid solution: $3^3+3^2=27+9=36=6^2$
If you choose to analyze the other equation, $q^3+p^2=b^2$, the analysis is completely parallel and arrives at the same results.
|
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|
Prove that $(n+1)^n < n^{n+1}$ for all $n>3$
Prove that $(n+1)^n < n^{n+1}$ for all $n>3$
At $n=4$, $$5^4<4^5$$ which is indeed true.
By mathematical induction, we need to prove that $$(n+2)^{n+1} < (n+1)^{n+2}$$
$$\implies (n+2)^n\times (n+2) < (n+1)^n\times(n+1)^2 $$
I am not getting how to proceed further than this. Any hints or help would be highly appreciated! Thanks.
|
Using binomial expansion, we can write
\begin{align*}
\text{LHS} &= (n+1)^n\\
&= (n+1)^{n-1}\times (n+1)\\
&= \left[n^{n-1} + {n-1 \choose 1}n^{n-2} + {n-1 \choose 2}n^{n-3} + \cdots + {n-1 \choose n-2}n + {n-1 \choose n-1}\right] \times (n+1)\\
&= \left[n^{n-1} + {n-1 \choose 1}n^{n-2} + {n-1 \choose 2}n^{n-3} + \cdots + {n-1 \choose n-2}n\right] \times (n+1) + (n+1)\\
\end{align*}
For $n > 3$, it is straightforward to show that
$${n-1 \choose k} < n^k$$
for all $k = 1,2,\dots, n-2$. Thus, we have
\begin{align*}
\text{LHS} &< \left[n^{n-1} + n^1n^{n-2} + n^2n^{n-3} + \cdots + n^{n-2}n\right] \times (n+1) + (n+1)\\
&= \left[(n-1)n^{n-1}\right] \times (n+1) + (n+1)\\
&= (n^2 - 1)n^{n-1} + (n+1)\\
&= n^{n+1} - n^{n-1} + (n+1)\\
&< n^{n+1} = \text{RHS},
\end{align*}
where the last step follows from the fact that
$$n^{n-1} > n+1$$
for all $n > 3$.
Remark: In contrast to other answers, this gives a proof without using induction.
|
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|
Prove that $\sum_{k=0}^{n-1}\left \lfloor x+\dfrac{k}{n}\right \rfloor=\lfloor nx\rfloor$ Let $x,y\in\Bbb R$ and $n\in\Bbb N$. Prove that, $$\sum_{k=0}^{n-1}\left \lfloor x+\dfrac{k}{n}\right \rfloor=\lfloor nx\rfloor.$$
Try:
We have to $x<\lfloor x\rfloor+1$. By adding $k/n$ to $x$, could be that $x+(k/n)$ keep being less than $\lfloor x\rfloor+1$, or they can be equal to or greater than it. We are going to have the following list of inequalities $$\begin{align*}\lfloor x\rfloor&\leq x+\dfrac{0}{n}<\lfloor x\rfloor +1\\ \lfloor x\rfloor &\leq x+\dfrac{1}{n}<\lfloor x\rfloor +1\\ & \vdots\\ \lfloor x\rfloor &\leq x+\dfrac i n<\lfloor x\rfloor +1\\ \lfloor x\rfloor +1&\leq x+\dfrac{i+1}{n}< \lfloor x\rfloor +2\\ &\vdots \\ \lfloor x\rfloor +1&\leq x+\dfrac {n-1} {n}<\lfloor x\rfloor +2 \end{align*} $$ where $ i $ is the maximum value of $ k $ so that the amount $ x + (k / n) $ does not reach $\lfloor x\rfloor+1$. So,
$$\begin{align*}\sum_{k=0}^{n-1}\left \lfloor x+\dfrac{k}{n}\right \rfloor&=\sum_{k=0}^i \left \lfloor x+\dfrac{k}{n}\right \rfloor+\sum_{k=i+1}^{n-1}\left \lfloor x+\dfrac{k}{n}\right \rfloor\\
&=\sum_{k=0}^i \lfloor x\rfloor+\sum_{k=i+1}^{n-1} (\lfloor x\rfloor+1)\\
&= (i+1)\lfloor x\rfloor+(n-1-i)(\lfloor x\rfloor+1)\\ &=n\lfloor x\rfloor+n-i-1\\ &= \lfloor nx\rfloor +(n\lfloor x\rfloor-\lfloor nx\rfloor+n-i-1). \end{align*}$$ I got to that last part. I would only need to prove that $n\lfloor x\rfloor-\lfloor nx\rfloor+n-i-1=0$. Someone help me with that.
|
Let's write $x = \lfloor x \rfloor + \{x \}$, there exists $u \in \Bbb N, u \in \{ 0,...,(n-1)\}$ such that $$ \frac{u}{n}\le \{x \} < \frac{u+1}{n}$$
We have
\begin{align}
\sum_{k=0}^{n-1}\left \lfloor x+\dfrac{k}{n}\right \rfloor &= \sum_{k=0}^{n-1}\left \lfloor \lfloor x \rfloor +\{x \}+\dfrac{k}{n}\right \rfloor \\
&=n\lfloor x \rfloor+ \sum_{k=0}^{n-1}\left \lfloor \{x \}+\dfrac{k}{n}\right \rfloor \\
&=n\lfloor x \rfloor+ \sum_{k=0}^{n-1-u}\left \lfloor \{x \}+\dfrac{k}{n}\right \rfloor +\sum_{k=n-u}^{n-1}\left \lfloor \{x \}+\dfrac{k}{n}\right \rfloor \tag{1}\\
\end{align}
*
*For $k = 0,...,(n-1-u)$:
$$\{x \}+\dfrac{k}{n}< \frac{u+1+k}{n}<1 \implies \sum_{k=0}^{n-1-u}\left \lfloor \{x \}+\dfrac{k}{n}\right \rfloor =0$$
*
*For $k = (n-u),...,(n-1)$:
$$2>\{x \}+\dfrac{k}{n} \ge \frac{u+1+k}{n} \ge 1 \implies \sum_{k=n-u}^{n-1}\left \lfloor \{x \}+\dfrac{k}{n}\right \rfloor =(n-1)-(n-u)+1 = u$$
From $(1)$, we deduce then
$$\sum_{k=0}^{n-1}\left \lfloor x+\dfrac{k}{n}\right \rfloor = n\lfloor x \rfloor+ u =n\lfloor x \rfloor +n\times \frac{u}{n} = n(\lfloor x \rfloor +\{x \}) = \lfloor nx \rfloor $$
Q.E.D
|
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|
what is $a^2+9=b^2+16=1+(a+b)^2$ solve for $a,b$ This is for a geometry question, and through a construction arrived at this equation. I could not solve it and after plugging it into wolfram got the correct answer but can anyone show a method for finding a,b
they are $\frac{5}{\sqrt3}$ and $\frac{2}{\sqrt3}$ respectively (note since this is a geometry question the lengths of a side cannot be negative so the negative solution sets do not matter)
|
You can start from
$$
1+(a+b)^2=b^2+16 \\
\iff a^2+2ab+b^2+1=b^2+16 \\
\iff a^2+2ab-15=0 \\
\iff b= \frac{15-a^2}{2a}
$$
Now substitute in another equality
$$
a^2+9=b^2+16 \\
\iff a^2+9=\left(\frac{15-a^2}{2a}\right)^2+16 \\
\iff a^2+9=\frac{15^2-30a^2+a^4}{4a^2}+16 \\
\iff 3a^4+2a^2-225 = 0 \\
$$
Solve the quadratic for $a^2$, you get
$$
\implies a^2=\frac{25}{3} \implies a=\frac{5}{\sqrt{3}}
$$
Can you finish from here?
|
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|
How am I supposed to expand $\sin^2 A + \sin^4 A = 1$ into $1 + \sin^2A = \tan^2A$? My question is how can i expand
$$\sin^2 A + \sin^4 A = 1$$
into:
$$1 + \sin^2A = \tan^2A$$
I tried quite a few ways I know but all of them kinda felt random. i am not sure how to share my trials here. I am quite beginner in trigonometry. it is one of the extra test question from my textbook. I don't need it but cant control curiosity. so pls help me.
Thanks in advance!
EDIT:
found the solution, dropping it here,
\begin{align}
\sin^2 A + \sin^4 A & = 1 \\
\sin^4 A & = 1 - sin^2 A \\
\sin^2 A . \sin^2 A & = cos^2 A \\
\sin^2 A . (1 - \cos^2 A) & = cos^2 A \\
\sin^2 A - \sin^2 A.\cos^2 A & = cos^2 A \\
\sin^2 A & = cos^2 A + \sin^2 A.\cos^2 A \\
\sin^2 A & = \cos^2 A(1 + \sin^2 A) \\
1 + \sin^2 A & = \cfrac{\sin^2 A}{\cos^2 A} \\
1 + \sin^2 A & = \tan^2 A \\
\end{align}
|
$$\sin^2 A = 1 - \sin^4 A$$ $$ \sin^2A= (1 - \sin^2 A)(1 + \sin^2 A) $$ $$\sin^2 A= (\cos^2 A) (1 + \sin^2 A)$$ $$\therefore \tan^2 A = 1 + \sin^2 A$$
|
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|
Finding minima from simultaneous equations We are given that a point $(x,y,z)$ in $\mathbb{R}^3$ satisfies the following equations
$x\cos\alpha-y\sin\alpha+z =1+\cos\beta$
$x\sin\alpha+y\cos\alpha+z =1-\sin\beta$
$x\cos(\alpha+\beta)-y\sin(\alpha+\beta)+z=2$
Where $\alpha,\beta\in\mathbb(0,2\pi)$
We need to find the Minimum value , $M$ of $x^2+y^2+z^2$.
\begin{align}M&=2\end{align}
My attempt:
By using Cramers rule, we find that the equations have an unique solution in the angle range mentioned. Solving for $x,y,z$ and then squaring and adding them and then finding the minimum value would take so much time. Is there any easier alternative way to do this?
|
When $\alpha = \beta = \pi$ and $x = 1, y = 0, z = 1$, the system of equations is met. So there exists $(x, y, z)$, which is a solution of the system of equations for some $\alpha, \beta \in (0, 2\pi)$, such that $x^2 + y^2 + z^2 = 2$.
On the other hand, by Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
&(x\cos (\alpha + \beta) - y\sin(\alpha + \beta) + z)^2 \\
\le\ & (x^2 + y^2 + z^2)(\cos^2 (\alpha + \beta) + \sin^2(\alpha + \beta) + 1^2 )\\
=\ & 2(x^2 + y^2 + z^2)
\end{align*}
which results in
$$x^2 + y^2 + z^2 \ge 2.$$
Thus, the minimum of $x^2 + y^2 + z^2$ is $2$.
|
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|
Maximizing $x^2y$ given $x^2+y^2=100$, without using the AM-GM inequality and calculus tools Problem says:
Let $x^2+y^2=100$, where $x,y>0$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?
I know these possible tools:
*
*AM-GM inequality
*Calculus tools
Here, I want to escape from all of the tools I mentioned above.
I will try to explain my attempts in the simplest sentences. (my english is not enough, unfortunately). I will not prove any strong theorem and also I'm not sure what I'm doing exactly matches the math, rigorously.
Solution I made:
First, it is not necessary to make these substitutions. I'm just doing this to work with smaller numbers.
Let, $x=10m, ~y=10n$,
where $0<m<1,~ 0<n<1$, then we have
$$ x^2+y^2=100 \iff m^2+n^2=1$$
$$x^2y=1000m^2n$$
This means,
$$\max\left\{x^2y\right\}=10^3\max\left\{m^2n\right\}$$
$$m^2n=n(1-n^2)=n-n^3$$
Then suppose that,
$$\begin{align}\max\left\{n-n^3 \mid 0<n<1\right\}&=a, a>0&\end{align}$$
This implies
$$n-n^3-a≤0,~ \forall n\in\mathbb (0,1)$$
$$n^3-n+a≥0,~\forall n\in\mathbb (0,1)$$
Then, we observe that
$$\begin{align}n^3-n+a≥0, \forall n\in (0,1) ~ \text{and} ~ \forall n≥1\end{align}$$
This follows
$$ n^3-n+a≥0, ~ \forall n>0.$$
Using the last conclusion, I assume that there exist $u,v>0$, such that
$$n^3-n+a=(n-u)^2(n+v)≥0.$$
If $n>0$, then the equality occurs, if and only if
$$n=u>0$$
Based on these, we have:
$$\begin{align}n^3-n+a= (n-u)^2(n+v)≥0 \end{align}$$
$$\begin{align}n^3-n+a = & n^3 - n^2(2u-v)+ n(u^2 - 2 u v ) + u^2v & \end{align}$$
$$\begin{align} \begin{cases} 2u-v=0 \\ u^2-2uv=-1 \\u^2v=a \\u,v>0 \end{cases} &\implies \begin{cases} v=2u \\ u^2-4u^2=-1 \\ 2u^3=a \\ u,v>0 \end{cases}\\
&\implies \begin{cases} u=\frac{\sqrt 3}{3} \\ v=\frac{2\sqrt 3}{3}\\ a=2\left(\frac{\sqrt 3}{3} \right)^3=\frac{2\sqrt 3}{9} \end{cases} \end{align}$$
$$\begin{align}n^3-n+\frac{2\sqrt 3}{9} &=\left(n-\frac{\sqrt 3}{3} \right)^2\left(n+\frac{2\sqrt 3}{3}\right)≥0.&\end{align}$$
As a result, we deduce that
$$\begin{align}n-n^3-\frac{2\sqrt 3}{9} &=-\left(n-\frac{\sqrt 3}{3} \right)^2\left(n+\frac{2\sqrt 3}{3}\right)≤0, &\forall n\in (0,1).&\end{align}$$
$$\begin{align}\max\left\{n-n^3 \mid 0<n<1\right\}&=\frac{2\sqrt 3}{9}, ~ \text{at }~ n=\frac{\sqrt 3}{3}&\end{align}$$
Finally, we obtain
$$m=\sqrt{1-n^2}=\sqrt{1-\frac 13}=\frac{\sqrt 6}{3}$$
$$\frac xy=\frac mn=\sqrt 2.$$
Question:
*
*How much of the things I've done here are correct?
|
Your approach is good and the easiest way without calculus.
An Approach that Exposes the Core Ideas
The following approach is a slight modification of yours that simplifies the algebra using Vieta's formulas.
If
$$
x^2+y^2=100\tag1
$$
then
$$
\begin{align}
x^2y
&=100\cos^2(\theta)\cdot10\sin(\theta)\tag{2a}\\
&=1000\left(\sin(\theta)-\sin^3(\theta)\right)\tag{2b}
\end{align}
$$
So we wish to maximize $\sin(\theta)-\sin^3(\theta)$ for $0\lt\theta\lt\frac\pi2$ (since $x,y\gt0$). Suppose the maximum is $m$. Then your assumption/lemma says that $s-s^3-m=0$ has a double root. That is,
$$
s^3-s+m=(s-r)^2(s-q)\tag3
$$
Vieta's formulas say that for a monic cubic polynomial,
$$
\begin{align}
r+r+q&=0\tag{4a}\\
r^2+rq+rq&=-1\tag{4b}\\
r^2q&=-m\tag{4c}
\end{align}
$$
Explanation:
$\text{(4a)}$: the sum of the roots is the negative of the coefficient of $s^2$
$\text{(4b)}$: the sum of the pairwise products of the roots is the coefficient of $s$
$\text{(4c)}$: the product of the roots is the negative of the constant coefficient
Now things just fall into place:
$\text{(4a)}$ says that $q=-2r$
$\text{(4b)}$ says that $-3r^2=-1$, that is $r=\frac1{\sqrt3}$
$\text{(4c)}$ says that $-2r^3=-m$, that is $m=\frac2{3\sqrt3}$
Thus, the maximum of $s-s^3$ is $\frac2{3\sqrt3}$ which happens at $s=\frac1{\sqrt3}$, which is a value that $\sin(\theta)$ attains for $0\lt\theta\lt\frac\pi2$.
Thus, the maximum of $x^2y=1000\cdot\frac2{3\sqrt3}=\frac{2000}{3\sqrt3}$.
|
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|
Prove that $\operatorname{Re}\left(\frac{1-z^{n+1}}{1-z}\right)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\frac{\theta}{2})}$. My objective is to prove that:
$$\operatorname{Re}\left(\frac{1-z^{n+1}}{1-z}\right)=\frac{1}{2}+\frac{\sin((n+\frac{1}{2})\theta)}{2\sin(\frac{\theta}{2})}\text{ , where $$z is a complex number }.$$
I have developed a good reasoning, but I cannot conclude. Let's go:
$$\operatorname{Re}\left(\frac{1-z^{n+1}}{1-z}\right)=\frac{[1-\cos((n+1)\theta)](1-\cos\theta)+\sin((n+1)\theta)\sin\theta}{[1-2\cos\theta + cos^2\theta +sen^2\theta]}=$$
$$=\frac{1-\cos\theta-\cos((n+1)\theta)+cos\theta\cos((n+1)\theta)+\sin\theta\sin((n+1)\theta)}{2-2\cos\theta}=$$
$$=\frac{1-\cos\theta}{2-2\cos\theta}+\frac{\cos((n+1)\theta)(\cos\theta-1)+\sin\theta\sin((n+1)\theta)}{2-2\cos\theta}=$$
$$=\frac{1}{2}-\frac{\cos((n+1)\theta)}{2}+\frac{\sin\theta\sin((n+1)\theta)}{2-2\cos\theta}.$$
After that, i was unable to continue. I tried to go the other way, that is, try to develop the right side of equality. However, I was not successful. Does anyone have any idea how I can make progress?
Note: I need to do it using only trigonometric relations. I cannot use exponential rules.
Note: The previous steps I did not put in, because I am sure that it is right and it is not necessary for the continuation. I just need to know how to continue to develop to get to the right side of the requested equality.
|
If you are unable to continue, you can look at the following steps:
$\dfrac{1}{2}-\dfrac{\cos\big((n+1)\theta\big)}{2}+\dfrac{\sin\theta\sin\big((n+1)\theta\big)}{2-2\cos\theta}=$
$=\dfrac{1}{2}+\dfrac{-\cos\big((n+1)\theta\big)(1-\cos\theta)+\sin\theta\sin\big((n+1)\theta\big)}{2(1-\cos\theta)}=$
$=\!\dfrac{1}{2}\!+\!\dfrac{-\!\cos((n\!+\!1)\theta)\!+\!\cos\theta\cos((n\!+\!1)\theta)\!+\!\sin\theta\sin((n\!+\!1)\theta)}{2(1-\cos\theta)}=$
$=\dfrac{1}{2}+\dfrac{-\cos\big((n+1)\theta\big)+\cos\big((n+1)\theta-\theta\big)}{2(1-\cos\theta)}=$
$=\dfrac{1}{2}+\dfrac{-\cos\big((n+1)\theta\big)+\cos(n\theta)}{2(1-\cos\theta)}=$
$=\dfrac{1}{2}+\dfrac{-2\sin\big((n+\frac12)\theta\big)\sin\left(-\frac12\theta\right)}{2(1-\cos\theta)}=$
$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)\sin\left(\frac12\theta\right)}{1-\cos\theta}=$
$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)\sin\left(\frac12\theta\right)}{2\left(\sqrt{\frac{1-\cos\theta}2}\right)^2}=$
$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)\sin\left(\frac{\theta}2\right)}{2\sin^2\left(\frac{\theta}2\right)}=$
$=\dfrac{1}{2}+\dfrac{\sin\big((n+\frac12)\theta\big)}{2\sin\left(\frac{\theta}2\right)}\;.$
|
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|
How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involving the natural logarithm, that is similar to the identity that:
$$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{\psi(a) - \psi(b)}{a-b}$$
Also potentially related, the Lüroth analogue of Khintchine’s constant can be defined as the following:
$$\sum_{n=1}^{\infty} \frac{\ln (n)}{n(n+1)}$$
as mentioned here.
After some work, the following can be shown:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{5\ln(2) + 4\ln(3)}{16} + \frac{1}{2} \sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right)$$
and furthermore:
$$\sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right) = \int_{0}^{2} \left( \frac{\left(1-\pi x \cot(\pi x) \right)}{2x^2} + \frac{1}{x^2 - 1} + \frac{1}{x^2 -4} \right) \, dx$$
EDIT
I have derived yet another form for my sum of interest, however, I found this one interesting as it seems like it could potentially be solvable?
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \int_{0}^{\infty} \left( \frac{\psi^{(0)} (s+3) + \gamma}{(s+2)(s-2)} - \frac{25}{16 (s-2)(s+1)} \right) \, ds$$
From this, it is possible to obtain the following:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4} i + \frac{25}{48} (\ln (2) - i \pi) - \frac{1}{8} + \frac{1}{16} i \pi + \frac{1}{4} \int_{0}^{i \pi} \psi^{(0)} \left( \frac{4}{1+ e^{u}} \right) \, du$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4}i+\frac{25}{48} (\ln (2)-i \pi )+\frac{7 i \pi }{48}-\frac{1}{8}-\frac{\ln (2)}{3} -2 \int_0^{\infty } \frac{t \ln (\Gamma (1-i t))}{\left(t^2+4\right)^2} \, dt$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = -\frac{1}{8}-\frac{i \pi }{4}+\frac{i \gamma \pi }{4}-\frac{\ln (2)}{16} - 2 \int_{0}^{\infty} \frac{t \ln (\Gamma (-i t)) }{(4+t^2)^2} \, dt$$
$$\implies \sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} =\frac{25}{48} \ln (2) -\frac{1}{8} + \int_{1}^{\infty} \frac{\ln (v-1) \text{li} (v^2)}{v^5} \, dv$$
Where $\text{li}$ is the logarithmic integral function.
$$\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} = \frac{3 \ln (2)}{16} - \frac{\pi^2+1}{8} - \frac{\pi}{2} \int_{0}^{\infty} \sin(4\pi x) (\psi (x) - \ln (x)) \, dx$$
|
There is a connection to $\zeta'$ at positive even values
Using partial fractions, we have
$$\frac{1}{k^{2n}(k^2-a)}=\frac{1}{a^n(k^2-a)}-\sum_{j=1}^{n}\frac{1}{a^{n-j+1}k^{2j}},$$
so that
$$S_r(n,a)=\sum_{k\ge r}\frac{\ln k}{k^{2n}(k^2-a)}=\frac{1}{a^n}\left(\sum_{k\ge r}\frac{\ln k}{k^2-a}-\sum_{j=1}^{n}a^{j-1}\sum_{k\ge r}\frac{\ln k}{k^{2j}}\right),$$
provided that $r>\sqrt a$. It is then easy to show that
$$\sum_{k\ge r}\frac{\ln k}{k^{2j}}=\zeta'(2j)+\sum_{k=2}^{r-1}\frac{\ln k}{k^{2j}},$$
with the $\zeta'(2j)$ terms having no simpler form that I'm aware of. As of now, I don't know how to deal with the sum $\sum_{k}\ln(k)/(k^2-a)$, but I will update once I do.
|
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"url": "https://math.stackexchange.com/questions/4123446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
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|
The ODE $y''(x)=\sinh(x)-3y'(x)-2y(x)$ I am trying to solve the differential equation that is in the title as a System of first order ode.
My Approach:
$\frac{d}{dx} \left(\begin{array}{c} y \\ y' \end{array}\right)=\left(\begin{array}{c} y' \\ \sinh(x)-3y'-2y \end{array}\right)=$
$\left( \begin{array}{rrr}
0 & 1 \\
-2 & -3 \\
\end{array}\right)\left(\begin{array}{c} y \\ y' \end{array}\right)+\left(\begin{array}{c} 0 \\ \sinh(x) \end{array}\right)$
Then I calculate the characteristic polynomial of the coefficient matrix, which leads to the eigenvalues $\lambda_1=-2,\lambda_2=-1$. Let A denote the matrix above.
Calculating the matrix exponential, I get $e^{Ax}$=$\frac{1}{e^{2x}} \left( \begin{array}{rrr}
2e^x-1 & e^x-1 \\
-2e^x+2 & -e^x+2 \\
\end{array}\right)$
Now I am variating the parameters and get
$y(x)=\frac{1}{e^{2x}} \left( \begin{array}{rrr}
2e^x-1 & e^x-1 \\
-2e^x+2 & -e^x+2 \\
\end{array}\right)y_0+\frac{1}{e^{2x}} \left( \begin{array}{rrr}
2e^x-1 & e^x-1 \\
-2e^x+2 & -e^x+2 \\
\end{array}\right) \int_0^s \left(\begin{array}{c} \sinh(x)(e^{-s}-1) \\ \sinh(x)(-e^{-s}+2) \end{array}\right)ds$ =
$y(x)=\frac{1}{e^{2x}} \left( \begin{array}{rrr}
2e^x-1 & e^x-1 \\
-2e^x+2 & -e^x+2 \\
\end{array}\right)y_0+\frac{1}{e^{2x}} \left( \begin{array}{rrr}
2e^x-1 & e^x-1 \\
-2e^x+2 & -e^x+2 \\
\end{array}\right) \left(\begin{array}{c} \frac{1}{4}e^{-2x}(e^{2x}(2x+3)-2e^x-2e^{3x}+1+C_1) \\ \frac{-x}{2} -\frac{e^{-2x}}{4}+e^{-x}+e^{x} +C_2\end{array}\right)$
My Questions are: In the exercise description there was no value for $y_0$, is there a way to find the value for it?
Is my calculation correct (does it seem correct) or are there any mistakes?
|
Rewrite equation as
$$y''+3y'+2y=\frac{e^x}{2}-\frac{e^{-x}}{2}$$
Characteristic equation is
$$\lambda^2+3\lambda + 2=0$$
Then $\lambda_1=-2,\lambda_2=-1$. Solution of $y''+3y'+2y=0$ is
$$y_h=c_1e^{-2x}+c_2e^{-x}$$
For particular solution use method of undetermined coefficients.
$$y_p=Ae^x+Bxe^{-x}$$
We get
$$6Ae^x+Be^{-x}=\frac{e^x}{2}-\frac{e^{-x}}{2}$$
Then $A=\frac{1}{12}, B=-\frac{1}{2}$.
General solution is
$$y=y_h+y_p=c_1e^{-2x}+c_2e^{-x}+\frac{e^x}{12}-\frac{xe^{-x}}{2}.$$
|
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"url": "https://math.stackexchange.com/questions/4126196",
"timestamp": "2023-03-29T00:00:00",
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|
Minimizing $a_1x_1^2 + a_2x_2^2$ for positive $a_i$, where $a_1x_1+a_2x_2=B$
Find $$\min\{a_1x_1^2 + a_2x_2^2\}$$
Where $ a_1x_1 + a_2x_2 = B$, and $a_1>0$ and $a_2>0 $. Find $x_1$ and $x_2$.
Can we do it usig AG mean inequality?
Let's say we have the problem to find the minimum value of $ x_1^2 + x_2^2 $.
From: $ (x_1 - x_2)^2 ≥0 $,
$ x_1^2 + x_2^2 ≥ 2x_1x_2 $
So the minimum value is: $2x_1x_2 $ for $x_1=x_2$.
Can be this done in a simillar manner for the starting problem.
why cannot we put $ x^2 =a_1 x_1^2, y^2 =a_2 x_2^2$ and solve it like root mean square inequality (without generalization?) We get:
$x^2 = y^2 $
.
$ a_1 x_1^2 = a_2 x_2^2 $
but not $x_1 = x_2$?
|
Going along your lines, we can write
\begin{align}
&a_1a_2(x_1-x_2)^2\ge0\\
\iff&a_1a_2(x_1^2+x_2^2)\ge 2a_1a_2x_1x_2\\
\iff&a_1^2x_1^2+a_2^2x_2^2+a_1a_2(x_1^2+x_2^2)\ge a_1^2x_1^2+a_2^2x_2^2+2a_1x_1a_2x_2\\
\iff&(a_1x_1^2+a_2x_2^2)(a_1+a_2)\ge(a_1x_1+a_2x_2)^2\\
\iff&a_1x_1^2+a_2x_2^2\ge\frac{B^2}{a_1+a_2}
\end{align}
This is the required lower bound.
|
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|
Approximate $\log_{10}$ values without a calculator I've got this problem:
$1,000,000^{{1,000,000}^{1,000,000}} < n^{n^{n^n}}$
What is the first positive integer value of n for which this inequality holds?
I managed to reduce it to this:
$6+\log_{10}(6) < n\log_{10}(n)$
by using $\log_{10}$ three times (and discarding some of the insignificant values – I can explain why these are insignificant at the end).
The only problem is I don't know how to approximate this (using an upper bound on the left and lower on the right as an upper bound for n, and then the reverse for a lower bound of n) accurately enough.
Does anyone have any ideas?
Also, if anyone has another way of doing this problem without the method I used, avoiding my issue altogether, that would be helpful.
|
Here is a solution with limited hand multiplications/additions. Starting from your condition that $6+\log_{10} 6< n\log_{10} n$, notice that
$$
10<6^2<100
\rightarrow 1< 2\log_{10} 6 < 2
\rightarrow 1/2 < \log_{10} 6 < 1
\longrightarrow \fbox{$6.5 < 6+\log_{10} 6 < 7$}.
$$
Also, $6^6=46,656$ (by hand!) therefore
$$
10^4<6^6<10^5
\rightarrow 4 < 6\log_{10} 6 < 5
$$
therefore $n=6$ is too small. On the other hand
$$
\text{with } n=10\quad 10\log_{10} 10 = 10 > 7
$$
therefore $n=10$ is large enough. Between 6 (too small) and 10 (large enough), try 8 :
\begin{align}
& 8^8 = (2^3)^8=2^{24}=(2^{10})^2\times 2^4 > 1000^2\times 16\\
\longrightarrow\quad
&
8\log_{10} 8 > 2\log_{10} 1000 + \log_{10} 16 > 6+1=7
\end{align}
therefore 8 is also enough. The only remaining possibility is 7:
\begin{align}
& 7^2=49<50
\rightarrow
7^6<50^3=125,000
\rightarrow
7^7<7\times 125,000=875,000 < 10^6\\
\longrightarrow\quad &
7\log_{10} 7 < 6
\end{align}
and thus $n=7$ is not enough. The answer is 8.
PS. Using Knuth's up-arrow notation, you get $7\uparrow\uparrow4<(10^6)\uparrow\uparrow3<8\uparrow\uparrow4$.
|
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|
Differentiate $x^{a^x}$ without logarithmic differentiation? Problem. Compute $\frac{d}{dx} x^{a^x}$.
Method 1: Logarithmic Differentiation (Correct):
\begin{align*}
y&= x^{a^x} \\
\ln(y)&=a^x \cdot \ln(x) \\
\frac{1}{y} \frac{dy}{dx} &=a^x \frac{1}{x} + \ln(x) \ln(a) a^x \\
\frac{dy}{dx} &= y\left(\frac{1}{x} a^x + \ln(a) \ln(x) a^x\right) \\
\frac{dy}{dx} &= x^{a^x}\left(\frac{1}{x} a^x + \ln(a) \ln(x) a^x\right) \\
\frac{dy}{dx} &= a^x x^{a^x}\left(\frac{1}{x} + \ln(a) \ln(x)\right) \\
\frac{dy}{dx} &= a^x x^{a^x-1}\left(1 + x\ln(a) \ln(x) \right) \\
\end{align*}
Method 2: Chain Rule (Incorrect):
\begin{align*}
y&= x^{a^x} \\
\frac{dy}{dx} &= a^x x^{a^x-1} \cdot \left[\ln(a) \cdot a^x\right]
\end{align*}
Method 2 is incorrect because it $x^{a^x}$ is not a power function, so we cannot apply power rule (thanks @Alann_Rosas and @Parcly_Taxel).
My Question: Can Ninad Munshi's answer here be adopted to correct method 2 without the use of logarithmic differentiation? What is the name (and/or proof) of this generalized version of chain rule?
Thank you!
|
As suggested, one can use the answer in Why the chain rule does not work for this question? to solve this problem.
\begin{align*}
\frac{d}{dx} x^{a^x}
&= a^x \cdot x^{a^x-1} + x^{a^x} \cdot \ln(x) \cdot \frac{d}{dx} a^x \\
&= a^x \cdot x^{a^x-1} + x^{a^x} \cdot \ln(x) \cdot \ln(a) \cdot a^x
\end{align*}
This is a special case of the multivariable chain rule.
|
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|
Prove $\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8}$ As the title says, prove $$\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8},$$ for $n>1$. This inequality is from Erdős, "Problems and results on the theory of interpolation". I, Lemma 3.
My attempt: since $H_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k} > \int_{1}^n\frac{1}{t}dt = \log(n),$ we then have
$$
\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} = \sum_{k=1}^{n-1}\frac{n^2 - 2kn + k^2}{2k} = \frac{n^2}{2}H_{n-1} - n(n-1) + \frac{n(n-1)}{4} > \frac{n^2\log(n)}{2} - \frac{3n(n-1)}{4}.
$$
Am I missing something here?
|
Your bound is much tighter for sufficiently large $n$. Define $$f(n) = \frac{n^2 \log n}{8}, \\ g(n) = \frac{n^2 \log n}{2} - \frac{3n(n-1)}{4}.$$ Then the ratio $$\frac{g(n)}{f(n)} = 4 - \frac{6(n-1)}{n \log n}.$$ For $n \ge 5$, $\log n > 1.6$, hence $\frac{6(n-1)}{n \log n} < 3$, thus $g > f$.
In the case where $n \le 4$, we would directly compare the bound $\frac{n^2 \log n}{8}$ with the sum.
|
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|
How to manipulate the Algebra in the parenthesis of this question? Consider the series $1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + ....$
Since the pattern of signs repeats every 3 terms, it is natural to consider only the partial sums that include entire blocks of 3 terms, i.e.,
$s_{3n} = \sum_{k=1}^{n} (\frac{1}{3k - 2} + \frac{1}{3k-1} - \frac{1}{3k})$
a) Use a comparison test against the Harmonic Series to show that $(s_{3n})$ diverges. To this end, rewrite the expression in parenthesis above appropriately.
Now from my understanding, we have to show that $(\frac{1}{3k - 2} + \frac{1}{3k-1} - \frac{1}{3k}) = \frac{C}{k} + a_{k}$ where $a_{k}$ is a positive sequence and C is a positive constant or sequence of positive constants, therefore, our original sequence bounds the sequence of partials sums of the harmonic series and we conclude by the Comparison test that the original diverges. If I am right, how would one go about doing this?
|
$$s_{3n} = \sum_{k=1}^{n} \left(\frac{1}{3k - 2} + \frac{1}{3k-1} - \frac{1}{3k}\right)\geqslant \sum_{k=1}^{n}\frac{1}{3k - 2}$$
as last diverged, then it have limit $+\infty$, so have $s_{3n}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can we use Vieta's formula in solving Trigonometric equations? **The value of $$\sec\frac{\pi}{11}-\sec\frac{2\pi}{11}+\sec\frac{3\pi}{11}-\sec\frac{4\pi}{11}+\sec\frac{5\pi}{11}$$
is ...
My Approach
I used the fact that $$\sec (\pi-x)=-\sec x$$ to simplify the equation to $$\sec\frac{\pi}{11}+\sec\frac{3\pi}{11}+\sec\frac{5\pi}{11}+\sec\frac{7\pi}{11}+\sec\frac{9\pi}{11}$$
Now I tried to devise an equation whose roots are $$\sec\frac{\pi}{11}, \sec\frac{3\pi}{11}, \sec\frac{5\pi}{11}, \sec\frac{7\pi}{11}, \sec\frac{9\pi}{11}$$
Afterwards, I found that the equation $$\cos \frac{11x}{2}=0 $$ satisfy the condition. But the equation has infinite number of roots, so my plan to use Vieta's formula to calculate the required sum did not work.
Please suggest how to proceed in this problem or share any other method.
|
Let $\alpha = \dfrac{(2n+1)\pi}{11}$, which is the form of angles we are interested. Then $6\alpha = (2n+1)\pi - 5\alpha$ which implies $\cos 6\alpha +\cos 5\alpha = 0$. Now, expressing this in terms of $\cos \alpha$, we observe
$$32\cos^6\alpha + 16\cos^5\alpha- 48\cos^4\alpha - 20\cos^3\alpha + 18\cos^2\alpha + 5\cos\alpha - 1 = 0$$
From this, we conclude that the equation
$$32x^6 + 16x^5-48x^4-20x^3+18x^2+5x-1 = 0$$
has the roots of exactly same angles that you want (with $\cos$) and $1$, so divide by $(x-1)$ and substitute $1/x$ to get $\sec$ and use Vieta's formula.
Note: We partitioned $11\alpha$ to $5\alpha$ and $6\alpha$ since we needed a polynomial with (possibly exactly) $5$ roots (and $\lfloor 11 / 2 \rfloor = 5$).
|
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|
The area of $\triangle AMN$ where $MN$ is midsegment $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$. If the area of $\triangle ABC$ is $24$ $cm^2,$ what is the area of $\triangle AMN$?
We can say for sure that $MN\parallel BC$ and $MN=\dfrac12 BC$. I don't know what to do next. Any help would be appreciated. Thank you in advance!
|
We know that $$S_\triangle = \frac{1}{2}ab\sin\alpha$$
where $a$ and $b$ are sides and $\alpha$ the angle between them. Using this, we have
$$\frac{S_{ABC}}{S_{AMN}} = \frac{\frac{1}{2}\cdot \color{red}{AB}\cdot \color{blue}{AC}\cdot \sin\alpha}{\frac{1}{2}\cdot \color{red}{AM}\cdot \color{blue}{AN} \cdot \sin \alpha} = \color{red}{2}\cdot \color{blue}{2}=4
\Rightarrow S_{AMN} = \frac{S_{ABC}}{4} = \frac{24}{4} = 6$$
after some cancellations.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$
Prove that $$\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$$
for $a,b \in \mathbb{R}^+$. Establish when the equality holds.
My approach was to use AM-GM however trying both sides individually assert to $\geqslant 2ab.$ RHS is baby AM-GM if considered.
Edit:
From L.H.S -
$\dfrac{\frac{a^3}{b} + \frac{b^3}{a}}{2} \geqslant ab \implies \dfrac{a^3}{b} + \dfrac{b^3}{a} \geqslant 2ab$
|
Bring all terms to LHS$$\frac{a^3}b-\frac{a^3}a+\frac{b^3}a-\frac{b^3}b=(a^3-b^3)(1/b-1/a)$$
Now if $a\ge b$, both factors are $\ge0$. When $a<b$, both are negative. Can you see when the equality holds?
|
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|
solve gcd $n^3+3n^2-5$ and $n+2$ with bezout Hello I have to solve $\gcd (n^3+3n^2-5 , n+2)$ with bezout
Here’s how I did it:
bezout says $a$ and $b$ are co prime if and only if $au+bv = 1$.
Then I did :
$(n+2)(n^2+n-2) -(n^3+3n^2-5)$
I found one so their $\gcd$ is equal to $1$.
Did I get the right method ?
Sorry for my bad english !!!
|
$$A(n^3+3n^2-5) + B(n+2) = 1$$
This isn't a blueprint for a general method of finding a solution. I am going to take advantage of the fact that the second polynomial is $n+2$, a first degree polynomial.
Let $m = n+2$. Then
$$n^3+3n^2-5 = (m-2)^3+3(m-2)^2-5 = m^3 - 3m^2 -1$$
So we need to solve
$$A(m^3 - 3m^2 -1) + B(m) = 1$$
If we let $A=-1$, we get
\begin{align}
A(m^3 - 3m^2 -1) + B(m) &= 1 \\
-1(m^3 - 3m^2 -1) + B(m) &= 1 \\
Bm &= m^3 -3m^2 \\
B &= m^2 - 3m \\
B &= (n+2)^2-3(n+2) \\
B &= n^2 + n - 2
\end{align}
It is easy to check that
$$A(n^2 + n - 2) + B(n+2) = (-1)(n^3+3n^2-5) + (n^2 + n - 2)(n+2) = 1 $$
|
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|
Filling in details for calculation of the limit $\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right)$ I want to evaluate the following limit using asymptotics
\begin{equation}
\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) \tag{1}
\end{equation}
This is an example problem and this was the solution:
\begin{gather}
\frac{x^{3} +x}{1+x^{3}} =\left( 1+\frac{1}{x^{2}}\right)\left( 1+\frac{1}{x^{3}}\right)^{-1} =\left( 1+\frac{1}{x^{2}}\right)\left( 1-\frac{1}{x^{3}} +\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) \tag{2}\\
=1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{3}
\end{gather}
And then
\begin{gather}
\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} =\left( 1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\right)^{\frac{1}{7}} \tag{4}\\
=1+\frac{1}{7} .\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{5}
\end{gather}
\begin{equation}
\cos\frac{1}{x} =1-\frac{1}{2x^{2}} +\mathcal{O}\left(\frac{1}{x^{4}}\right) \tag{6}
\end{equation}
From above, we obtain:
\begin{equation}
\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{7}
\end{equation}
Hence the required limit is:
\begin{equation}
\lim _{x\rightarrow \infty }\left(\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\right) =\frac{9}{14} \tag{8}
\end{equation}
I tried to fill in the details for the steps involved in the above steps.
I supplied the details for $\displaystyle ( 3)$ as below:
\begin{gather}
\left( 1+\frac{1}{x^{2}}\right)\left( 1-\frac{1}{x^{3}} +\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) =1+\frac{1}{x^{2}} +\frac{1}{x^{3}}\left( -1-\frac{1}{x^{2}} +\left( x^{3} +x\right)\mathcal{O}\left(\frac{1}{x^{6}}\right)\right) \tag{9}\\
=1+\frac{1}{x^{2}} +\frac{1}{x^{3}}\left( -1-\frac{1}{x^{2}} +\frac{\left( x^{3} +x\right)}{x^{6}}\mathcal{O}( 1)\right) =1+\frac{1}{x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) \tag{10}
\end{gather}
where in (9) and (10), I have used the standard result: $\displaystyle \frac{\mathcal{O}( f( x))}{g( x)} =\mathcal{O}\left(\frac{f( x)}{g( x)}\right)$, if $\displaystyle g( x) \neq 0$.
I tried to get (5) from (4) but failed.
Then I supplied details for $\displaystyle ( 7)$ as below:
\begin{gather*}
\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right) -\mathcal{O}\left(\frac{1}{x^{4}}\right) =\frac{9}{14x^{2}} +\frac{1}{x^{3}}\left(\mathcal{O}( 1) -\mathcal{O}\left(\frac{1}{x}\right)\right)\\
=\frac{9}{14x^{2}} +\mathcal{O}\left(\frac{1}{x^{3}}\right)\\
\Longrightarrow \lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right) =\lim _{x\rightarrow \infty }\left(\frac{9}{14} +\mathcal{O}\left(\frac{1}{x}\right)\right) =\frac{9}{14}
\end{gather*}
Any help in getting (5) from (4) is much appreciated. Thanks.
\begin{equation*}
\end{equation*}
\begin{equation*}
\end{equation*}
|
For positive integers, we have the well known binomial formula
$$(1 + x)^n = \sum_{k\geq 0} \binom{n}{k} x^k = \sum_{k=0}^n \binom{n}{k} x^k$$
where
$$\binom{n}{k} = \frac{(n)_k}{k!} = \frac{n(n - 1)\cdots(n - k + 1)}{k!}$$
$(n)_k$ is the Pochhammer symbol. It is not difficult to show that if we replace $n$ by an arbitrary number $\alpha$, and define analogously
$$\binom{\alpha}{k} = \frac{\alpha(\alpha - 1)\cdots(\alpha - k + 1)}{k!}$$
we get the Taylor series
$$ (1 + x)^\alpha = \sum_{k\geq 0} \binom{\alpha}{k}x^k = 1 + \alpha x + \mathcal{O}(x^2)$$
Substituting $\frac{1}{x^2} + \mathcal{O}\left(\frac{1}{x^3}\right)$ into the Taylor series for $(1+x)^{\frac{1}{7}}$ yields the desired result.
Edit: Note that
$$\left[\frac{1}{x^2} + \mathcal{O}\left(\frac{1}{x^3}\right)\right]^2 = \mathcal{O}\left(\frac{1}{x^4}\right)$$
so we have
\begin{align*}
\left(1+\frac{1}{x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)\right)^\frac{1}{7} &= 1 + \frac{1}{7}\left[\frac{1}{x^2} + \mathcal{O}\left(\frac{1}{x^3}\right)\right] + \mathcal{O}\left(\left[\frac{1}{x^2} + \mathcal{O}\left(\frac{1}{x^3}\right)\right]^2\right)\\
&= 1 + \frac{1}{7x^2} + \mathcal{O}\left(\frac{1}{x^3}\right) + \mathcal{O}\left(\frac{1}{x^4}\right)\\
&= 1 + \frac{1}{7x^2} + \mathcal{O}\left(\frac{1}{x^3}\right)
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Find steps to solve : $\int_0^t \frac{x \cos u - x^2}{1 - 2x \cos u + x^2} \, du $ $$ \int_0^t \frac{x \cos u - x^2}{1 - 2x \cos u + x^2} \, du\quad\text{with}\quad x \in ]-1,1[ $$
$$ \int_0^t \frac{x \cos u - x^2}{1 - 2x \cos u + x^2} \, du = \arctan \left( \frac{x \sin t}{1- x \cos t} \right) $$
I have this integral with a known result. The "proof" the book gives is :
"We recognize that the fraction is $\dfrac{h'(u)}{1+h(u)^2}$ with $h(u) = \dfrac{x \sin u}{1- x \cos u } $"
But it's not that easy to spot, and I would like to know how to find this result without guessing that out of nowhere and without knowing it beforehand.
I've tried substitutions but I can't find a way to make it work.
Any idea on how to solve this ? Thanks !
|
I'm not sure this helps but ...
\begin{align}\mathcal{I}&=\displaystyle\int \frac{x\cos u-x^2}{(-2x)\cos u+(x^2+1)}\mathrm du\\&=\underbrace{\displaystyle\int \frac{x\cos u}{-2x\cos u+(x^2+1)}\mathrm du}_{\mathcal{I_1}}-x^2\overbrace{\displaystyle\int \frac{\mathrm du}{-2x\cos u+(x^2+1)}}^{\mathcal{I_2}}\end{align}
\begin{align}\mathcal{I_2}&=\displaystyle\int \frac{\mathrm du}{-2x\cos u+(x^2+1)}\\&=\displaystyle\int \frac{1}{\frac{1-t^2}{1+t^2}(-2x)+x^2+1}\left(\frac{2\mathrm dt}{1+t^2}\right)\, \text{ ,via substituting $t=\tan(u/2)$}\\&=2\displaystyle\int \frac{\mathrm dt}{(t(x+1))^2+(x-1)^2}\\&=\frac{2}{x^2-1}\arctan\left(\frac{(x+1)t}{x-1}\right)\, \text{ ,using the formula $\displaystyle\int \frac{dx}{a^2x^2+b^2}=\frac{1}{ab}\arctan(ax/b)$}\\&=\frac{2}{x^2-1}\arctan\left(\frac{(x+1)\tan(u/2)}{x-1}\right)\end{align}
\begin{align}\mathcal{I_1}&=\displaystyle\int \frac{x\cos u}{-2x\cos u+x^2+1}\, \mathrm du\\&=-x\displaystyle\int\left(\frac{1}{2x}+\frac{x^2+1}{2x(2x\cos u-x^2-1)}\right)\, \mathrm du\\&=\frac{-u}{2}-\frac{x^2+1}{2}\displaystyle\int\frac{\mathrm du}{2x\cos u-x^2-1}\\&=\frac{-u}{2}+\frac{(x^2+1)}{(x^2-1)}\arctan\left(\frac{(x+1)\tan(u/2)}{x-1}\right)\, \text{ ,using $\mathcal{I_2}$}\end{align}
Therefore,
\begin{align}\mathcal{I}&=-\frac{u}{2}+\frac{x^2+1}{x^2-1}\arctan\left(\frac{(x+1)\tan(u/2)}{x-1}\right)-\frac{2x^2}{x^2-1}\arctan\left(\frac{(x+1)\tan(u/2)}{x-1}\right)\\&=\frac{-u}{2}-\arctan\left(\frac{(x+1)\tan(u/2)}{x-1}\right)\end{align}
Now, you can put the limits.
Footnotes
Weierstrass Substitution
|
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|
$f(a)=b,f(b)=c,f(c)=a$, find $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Let $f(x)=x^2-x-2$. It is given that $a, b,c \in \mathbb{R}$ such that
$$f(a)=b,f(b)=c,f(c)=a$$ Then find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.
My attempt:
We have:
$$\begin{array}{l}
a^{2}-a-2=b \\
b^{2}-b-2=c \\
c^{2}-c-2=a
\end{array}$$
Subtracting pair wise we get:
$$(a-b)(a+b)=a-c$$
$$(b-c)(b+c)=b-a$$
$$(c-a)(c+a)=c-b$$
Trivially $a=b=c$ satisfies. So we get $a=b=c=\sqrt{3}+1, 1-\sqrt{3}$
Thus we get
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3\sqrt{3}+3}{2}, \frac{-3-3\sqrt{3}}{2}$$
Now if $a\ne b \ne c$
Then we have:
$$(a+b)(b+c)(c+a)=-1$$
Letting $p=a+b+c$ we have
$$(p-a)(p-b)(p-c)=-1$$ $\implies$
$$p(ab+bc+ca)-abc+1=0$$
Any hint here?
|
The following uses the symmetry of the problem and "lucky" cancellations of terms to get the result with somewhat less brute work.
The case $a=b=c$ is obvious and will be left out.
Otherwise, with $u=a+b+c, v=ab+bc+ca, w=abc$, the OP showed that:
$$
uv - w + 1 = 0 \tag{1}
$$
A second relation can be derived by summing the three equalities $a^2-a-2-b=0$ and using the Newton identity $a^2+b^2+c^2=u^2-2v\,$:
$$
0 = u^2 - 2u - 2v - 6 \tag{2}
$$
Then, multiplying $a^2-a-2-b=0$ by $a$ and eliminating the $a^2=a+2+b$ term gives:
$$
\begin{align}
0 &= a^3 - a^2 - 2a - ab
\\ &= a^3 - (a + 2 + b) - 2 a - a b
\\ &= a^3 - 3a - b - a b - 2 \tag{3}
\end{align}
$$
Summing the three equalities and using another Newton identity $a^3+b^3+c^3=u^3 - 3uv + 3 w$:
$$
\begin{align}
0 &= (u^3-3uv+3w) - 4 u - v - 6
\\ &= u^3 - 4 u - v - 3 uv + 3 w - 6\tag{4}
\end{align}
$$
At this point $(1),(2),(4)$ are three equations that can be resolved for $u,v,w$.
Substituting $uv = w - 1$ from $(1)$ into $(4)$:
$$
\require{cancel}
\begin{align}
0 &= u^3 - 4 u - v - 3 (\cancel{w} - 1) + \cancel{3 w} - 6
\\ &= u^3 - 4u - v - 3\tag{5}
\end{align}
$$
Eliminating $v$ between $(2)$ and $(5)$ by subtracting $2 \times (5)-(2)\,$:
$$
\begin{align}
0 &= 2 u^3 - 8 u - \bcancel{2 v} - \cancel{6} - (u^2 - 2u - \bcancel{2v} - \cancel{6})
\\ &= 2 u^3 - u^2 - 6 u
\\ &= u(2u+3)(u-2)\tag{6}
\end{align}
$$
For each $u \in \{-\frac{3}{2}, 0, 2\}$, the other variables $v,w$ follow from $(2)$ then $(4)$:
$$
(u,v,w) \in \left\{\; \left(-\frac{3}{2}, -\frac{3}{8}, \frac{25}{16}\right), \left(0, -3, 1\right), \left(2, -3, -5\right) \;\right\}
$$
The first triplet corresponds to the cubic $16 z^3 + 24 z^2 - 6 z - 25$ which has two complex conjugate non-real roots, but the other two triplets qualify, so in the end $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}= \dfrac{v}{w} \in \left\{ -3, \dfrac{3}{5}\right\}\,$ (again, not counting the two cases where $a=b=c$ are all equal to either root of the original quadratic).
[ EDIT ] The steps above are not reversible, so the possibility exists that extraneous solutions were introduced along the way. Therefore the two solutions $(u,v,w) \in \{ \left(0, -3, 1\right), \left(2, -3, -5\right) \}$ still need to be verified to work, as was done towards the end of @Macavity's answer for the same two cubics, where both sets of roots turned out to be eligible solutions.
|
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|
Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations and worked out examples convince me that this inequality indeed holds true. I have tried to solve this problem by induction. Clearly, for $N=1$ we have,
$$\mathcal{P}(1) = \left(\frac{2}{3}\right)^{\tfrac{3}{4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{4}} \approx 0.8774 < 1.$$
Now assume the inequality holds for $N$, then for $N+1$ we have,
\begin{align}
\mathcal{P}(N+1) &=\left(\frac{2N+2}{2N+3}\right)^{\tfrac{2N+3}{2N+4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{2N+4}} \\
&= \left(\left(\frac{2N}{2N+1}\right)\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)\right)^{\left(\frac{2N+1}{2N+2}\right)\left(\frac{2N+2}{2N+1}\cdot\frac{2N+3}{2N+4}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\frac{2N+2}{2N+4}}\\[1em]
&= \left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)\left(1 + \frac{1}{2(N+1/2)(N+2)}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\left(1 - \frac{1}{N+2}\right)}\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \small\left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}}\cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \mathcal{P}(N) \cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \cdot\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}
\end{align}
Now from here we know that the first three terms are all smaller than 1 ($\mathcal{P}(N) < 1$ by induction hypothesis). However the last term is larger than one. For the proof by induction to work out, we need that this last term cancels against,
$$\left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}}.$$ But I do not see how it does. Any help is greatly appreciated.
|
We present a proof that utilises a special case of the well known Bernoulli's Inequality.
Lemma:
$\forall \ x \geq 0, (1+x)^n \geq 1+nx, n\in \mathbb{N}.$
(This can easily be proven by applying the Binomial Theorem)
Now,
\begin{align}
&\left(\dfrac{2N}{2N+1}\right)^\dfrac{2N+1}{2N+2} \cdot \left(\dfrac{2}{1}\right)^\dfrac{1}{2N+2} < 1 \\
& \iff \left(\dfrac{2N}{2N+1}\right)^{2N+1} \cdot 2 < 1 \\
& \iff \left(\dfrac{2N+1}{2N}\right)^{2N+1} \cdot \dfrac{1}{2} > 1 \\
& \iff \left(1+\dfrac{1}{2N}\right)^{2N+1} > 2
\end{align}
The last inequality follows immediately from our lemma - since $\dfrac{1}{2N} > 0 \ \forall \ N \in \mathbb{N}$, $\left(1+\dfrac{1}{2N}\right)^{2N+1} \geq 1+\dfrac{2N+1}{2N} >2.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all ordered pairs of integers$(x,y)$ which satisfy the equation $2(x^2+y^2)+x+y=5xy$
Find all ordered pairs of integers$(x,y)$ which satisfy the equation $2(x^2+y^2)+x+y=5xy$
We can transform this equation as
$$\begin{equation}
2(x^2+y^2-2xy)+x+y-xy=0\\
\implies2(x-y)^2+x+y-xy=0\\
\implies(xy-x-y)\ge0\end{equation}$$
How to proceed now?
|
First, let's expand $$2(x^2+y^2)+x+y=5xy$$ We get : $$2x^2+2y^2+x+y=5xy$$ Then, subtract $5xy$ from both sides and we get $$2x^2+2y^2+x+y-5xy=0$$ On factoring it we get : $$2x^2+(1-5y)x+2y^2+y=0$$ Now this is a quadratic equation with $a=2, b=1-5y, c=2y^2+y$ So we get :$$x=\frac{-(1-5y)\pm \sqrt{(1-5y)^2-4 \times 2(2y^2+y)}}{2 \times 2}$$ On simplifying "the inside of the radical" we get :
$$\sqrt{9y^2-18y+1}$$ Replacing it we get : $$x=\frac{-(1-5y)\pm \sqrt{9y^2-18y+1}}{4}$$Now just separate the solutions that is separate the positive and the negative solutions :
$$x=\frac{-(1-5y)+ \sqrt{9y^2-18y+1}}{4}$$ and $$x=\frac{-(1-5y)- \sqrt{9y^2-18y+1}}{4}$$ With all this we get our solutions. Hope it helps
|
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|
How do you find the shortest distance from the point $\left(-1,7\right)$ to a given $\mbox{curve}\ ?$ I find this difficult to solve. It involves maxima and minima of differential calculus.
Since it's requiring the shortest distance, I have to find $y'$ and set it to zero.
Here's how I did it, but I still can't seem to find the correct answer.
$$
9x^2+25y^2-18x+100y-116=0\quad \mbox{is the given curve.}
$$
I resolved it into an equation of an ellipse,
*
*Completing the square
$$
9x^2-18x + 25y^2+100y = 116
$$
$$
9(x^2-2x+1) + 25(y^2+4y+4) = 116+9+100
$$
*Equation of ellipse:
$$\tag{1}
\frac{(x-1)^2}{25} + \frac{(y+2)^2}{9} = 1
$$
*I used the distance formula between two points
$$\tag{2}
d = \sqrt{(x+1)^2 + (y-7)^2}
$$
Let $x$ be $\sqrt{(225-25(y+2)^2) / 9)} + 1 $ from the curve equation
*I input $x$ into the distance formula, and this is the result.
$$
d = \sqrt{(-\frac{16}{9}y^2 -\frac{226}{9}y + \frac{602}{9} + \frac{20}{3} \sqrt{-y^2 -4y+5} }
$$
I find its derivative so it will be equal to zero to be the shortest distance to the curve.
The derivative is $\frac {-32}{9}y -\frac {226}{9} + [\frac {\frac {20}{3}(\frac 12)(-2y-4)}{ \sqrt(-y^2-4y+5)}]$
Equating the derivative of the curve equation to zero,
$$y = 0.921$$
$$x = 2.14$$
Input the values into distance formula,
$d = 6.84$ units
The correct answer is $6.14$ units.
Where could the error be? Please guide me.
|
The ellipse equation is
$ Q_1(x,y) = 9 x^2 + 25 y^2 - 18 x + 100 y -116 = 0 $
The objective function (to be minimized) is the square of the distance between $(x,y)$ and $(-1, 7)$
$ f(x,y) = Q_2(x,y) = (x + 1)^2 + (y - 7)^2 $
So this is a quadratic optimization subject to a quadratic constraint problem.
Using Lagrange multiplier method, the objective function is modified to include the constraint, as follows
$ g(x,y) = Q_2(x,y) + \lambda Q_1(x,y) $
Differentiating $g(x,y)$ with respect to $x, y, \lambda$ gives us
$ g_x = Q_{2x} + \lambda Q_{1x} = 0 $
$ g_y = Q_{2y} + \lambda Q_{1y} = 0 $
$ g_{\lambda} = Q_1 $
Eliminating $\lambda$ from the first two equations results in
$ Q_{2x} Q_{1y} - Q_{2y} Q_{1x} = 0 \hspace{30pt}(1)$
and, in addition, from the third equation, we have,
$ Q_1 = 0 \hspace{30pt} (2)$
Equations $(1), (2)$ are as follows for this particular problem
$ 2 (x + 1) (50 y + 100) - 2 (y - 7) (18 x - 18) = 0 $
which simplifies to
$ (x + 1)(25 y + 50) - (y - 7) (9 x - 9) = 0 \hspace{30pt}(3) $, and
$ 9 x^2 + 25 y^2 - 18 x + 100 y -116 = 0 \hspace{30pt}(4)$
Feeding these equations to wolframalpha.com gives the following two solutions
$ (x,y) = (-0.143256, 0.920525) $
and
$ (x,y) = (3.90826, -4.44032) $
The distances associated with these two points are
$ d_1 = \sqrt{ (-0.143256 + 1 )^2 + (0.920525 - 7)^2 } = 6.139546 $
$ d_2 = \sqrt{ (3.90826 + 1 )^2 + (-4.44032 - 7)^2 } = 12.44877$
So the first pair is the point on the ellipse that is closest to $(-1,7)$ while the second point is the farthest.
|
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|
Find a linear transformation $T\begin{pmatrix}-1\\-2\end{pmatrix}$ $T:\mathbb{R}^2 \Rightarrow\mathbb{R}^3, T\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}-1\\4\\3\end{pmatrix}$ and $T\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}4\\-1\\1\end{pmatrix}$. Find $T\begin{pmatrix}-1\\-2\end{pmatrix}$ and $T\begin{pmatrix}-3\\-1\end{pmatrix}$.
For $T\begin{pmatrix}-1\\-2\end{pmatrix}=-1T\begin{pmatrix}1\\0\end{pmatrix}-2T\begin{pmatrix}0\\1\end{pmatrix}=-\begin{pmatrix}-1\\4\\3\end{pmatrix}-2\begin{pmatrix}4\\-1\\1\end{pmatrix}=\begin{pmatrix}1\\-4\\-3\end{pmatrix}+\begin{pmatrix}-8\\2\\-2\end{pmatrix}=\begin{pmatrix}-7\\-2\\-5\end{pmatrix}$.
Is this correct? I would like to know if my procedure is correct by finding T.
|
Note also that $$T \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}= \begin{pmatrix} -1 & 4 \\ 4 & -1\\ 3 & 1\end{pmatrix} \implies T= \begin{pmatrix} -1 & 4 \\ 4 & -1\\ 3 & 1\end{pmatrix} I^{-1}= \begin{pmatrix} -1 & 4 \\ 4 & -1\\ 3 & 1\end{pmatrix}.$$
Next $$T\begin{pmatrix} -1 \\ -2 \end{pmatrix}=\begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix}$$
|
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|
integration by integration by parts: $\cos^2(\frac{\pi}{L}\,x)$ (inner argument) Really simple integrale that can by solved much easier, but I wanna try straight integration by parts:
$\begin{array}{ccc}
&D&I \\
+&\cos(\frac{\pi}{L}\,x)&\cos\frac{\pi}{L}\,x \\ \\
-&-\dfrac{L}{\pi}\,\sin(\frac{\pi}{L}\,x)& \dfrac{L}{\pi}\,\sin(\frac{\pi}{L}\,x)
\\\\\\
&\Rightarrow \int\cos^2(\frac{\pi}{L}\,x)
=& \dfrac{L}{\pi}\cos(\frac{\pi}{L}\,x)\,\sin(\frac{\pi}{L}\,x)+\int\left(\dfrac{L}{\pi}\right)^2\sin\left((\frac{\pi}{L})\right)^2 \\\\
&& \dfrac{L}{\pi}\cos(\frac{\pi}{L}\,x)\,\sin(\frac{\pi}{L}\,x)+\int\left(\dfrac{L}{\pi}\right)^2\left(1-\cos\left((\frac{\pi}{L})\right)^2\right)\\\\
&\Rightarrow 2\,\int\left(\dfrac{L}{\pi}\right)^2\cos^2(\frac{\pi}{L}\,x)
=&\dfrac{L}{\pi}\cos(\frac{\pi}{L}\,x)\,\sin(\frac{\pi}{L}\,x)-\left(\dfrac{L}{\pi}\right)^2\,x \\\\
&\Rightarrow \int\cos^2(\frac{\pi}{L}\,x) = &\dfrac{\pi\,(\cos(\frac{\pi}{L}\,x)\,\sin(\frac{\pi}{L}\,x))}{2\,L}-\dfrac{x}{2} \\\\
&\text{However, checking the integral gives:} \\\\
&\int\cos^2(\frac{\pi}{L}\,x) = &\dfrac{L\,(\cos(\frac{\pi}{L}\,x)\,\sin(\frac{\pi}{L}\,x))}{2\,\pi}-\dfrac{x}{2}\\\\
&\text{Where does the difference come from? }
\end{array}$
|
We have
\begin{align*}
\int \cos^{2}\left(\frac{\pi x}{L}\right) \ dx &= \frac{L}{\pi}\cos\left(\frac{\pi x}{L}\right)\sin\left(\frac{\pi x}{L}\right) + \int \sin^{2}\left(\frac{\pi x}{L}\right) \ dx \\
&= \frac{L}{\pi}\cos\left(\frac{\pi x}{L}\right)\sin\left(\frac{\pi x}{L}\right) + \int\left(1-\cos^{2}\left(\frac{\pi x}{L}\right)\right)\end{align*}
Therefore
$$2\int \cos^{2}\left(\frac{\pi x}{L}\right) \ dx =\frac{L}{\pi}\cos\left(\frac{\pi x}{L}\right)\sin\left(\frac{\pi x}{L}\right)+x$$
|
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|
Find the range of possible values of $\sqrt{a^2+a+1}-\sqrt{a^2-a+1}$
Let $x=\sqrt{a^2+a+1}-\sqrt{a^2-a+1},x\in \mathbb R$. Find range of possible values of $x$.
I tried drawing the graph and obtained this:
Through which the answer came out to be $(-1,1)$.
What should be the procedure through algebra?
|
Square both sides to get
$$
x^2 = 2(a^2 + 1 - \sqrt{a^4 + a^2 + 1})
$$
We can then use $a^4 + a^2 + 1 > (a^2 + 1/2)^2$ to see that $x^2 < 1$. Now rearrange and square again to get
$$
(x^2 - 2a^2 - 2)^2 = 4(a^4 + a^2 + 1).
$$
Assuming $|x| < 1$, we can solve for $a$ to get
$$
a =x\sqrt{\frac{1- (x/2)^2}{1-x^2}}.
$$
This is defined for every $x \in (-1,1)$, so $x(a)$ covers the entire range $(-1,1)$.
|
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|
On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part II The topic of odd perfect numbers likely needs no introduction.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
If $n$ is odd and $\sigma(n)=2n$, then we call $n$ an odd perfect number. Euler proved that a hypothetical odd perfect number must necessarily have the form $n = p^k m^2$ where $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds. Dris conjectured that the inequality $p^k < m$ is true in his M. Sc. thesis, and Brown (2016) eventually produced a proof for the weaker inequality $p < m$.
Now, recent evidence suggests that $p^k < m$ may in fact be false.
THE ARGUMENT
Let $n = p^k m^2$ be an odd perfect number with special prime $p$.
Since $p \equiv k \equiv 1 \pmod 4$ and $m$ is odd, then $m^2 - p^k \equiv 0 \pmod 4$. Moreover, $m^2 - p^k$ is not a square (Dris and San Diego (2020)).
This implies that we may write
$$m^2 - p^k = 2^r t$$
where $2^r \neq t$, $r \geq 2$, and $\gcd(2,t)=1$.
It is trivial to prove that $m \neq 2^r$ and $m \neq t$, so that we consider the following cases:
$$\text{Case (1): } m > t > 2^r$$
$$\text{Case (2): } m > 2^r > t$$
$$\text{Case (3): } t > m > 2^r$$
$$\text{Case (4): } 2^r > m > t$$
$$\text{Case (5): } t > 2^r > m$$
$$\text{Case (6): } 2^r > t > m$$
We can rule out Case (5) and Case (6), and under Case (1) and Case (2), we can prove that the inequality $m < p^k$ holds.
So we are now left with Case (3) and Case (4):
Under both cases left under consideration, we have
$$(m - 2^r)(m - t) < 0$$
$$m^2 + 2^r t < m(2^r + t)$$
$$m^2 + (m^2 - p^k) < m(2^r + t)$$
$$2m^2 < m(2^r + t) + p^k.$$
Since we want to prove $m < p^k$, assume to the contrary that $p^k < m$. We get
$$2m^2 < m(2^r + t) + p^k < m(2^r + t) + m < m(2^r + t + 1)$$
which implies, since $m > 0$, that
$$2m < 2^r + t + 1.$$
Here then is our question:
Will it be possible to derive a contradiction from the inequality
$$2m < 2^r + t + 1,$$
under Case (3) and Case (4) above, considering that $2m$ is large? (In fact, it is known that $m > {10}^{375}$.)
|
On OP's request, I am converting my comment into an answer.
*
*$p^k\lt m$ is equivalent to $$m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\tag7$$ since we have$$\begin{align}p^k\lt m&\iff m^2-2^rt\lt m
\\\\&\iff m^2-m-2^rt\lt 0
\\\\&\iff m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\end{align}$$
*
*$(7)$ is better than $2m\lt 2^r+t+1$ since
$$\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \frac{2^r+t+1}{2}\tag8$$
holds. To see that $(8)$ holds, note that
$$\begin{align}(2)&\iff \sqrt{1+2^{r+2}t}\lt 2^r+t
\\\\&\iff 1+2^{r+2}t\lt 2^{r+1}+2^{r+1}t+t^2
\\\\&\iff (2^r-t)^2\gt 1
\\\\&\iff |2^r-t|\gt 1\end{align}$$
which does hold.
*
*We can say that $$\bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}-t\bigg)\bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}-2^r\bigg)\lt 0\tag9$$
since
$$\begin{align}(9)&\iff \bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\bigg)^2-\dfrac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0
\\\\&\iff \frac{1+\sqrt{1+2^{r+2}t}+2^{r+1}t}{2}-\dfrac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0
\\\\&\iff 1+\sqrt{1+2^{r+2}t}+2^{r+1}t-(1+\sqrt{1+2^{r+2}t})(t+2^r)+2^{r+1}t\lt 0
\\\\&\iff 2^{r+2}t-2^r-t+1\lt (t+2^r-1)\sqrt{1+2^{r+2}t}
\\\\&\iff (2^{r+2}t-2^r-t+1)^2\lt (t+2^r-1)^2(1+2^{r+2}t)
\\\\&\iff 2^{r + 2} t (2^r - t - 1) (2^r - t + 1)\gt 0
\\\\&\iff (2^r-t)^2\gt 1
\\\\&\iff |2^r-t|\gt 1\end{align}$$
which does hold.
*
*It follows from $(7)(9)$ that if $p^k\lt m$ with $(m-t)(m-2^r)\lt 0$, then $$\min(t,2^r)\lt m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \max(t,2^r)$$
|
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|
Induction Squares Proof Prove that for every positive integer $n$ there exist positive integers $$a_{11}, a_{21}, a_{22}, a_{31}, a_{32}, a_{33}, \dots ,a_{n1}, a_{n2},\dots,a_{nn}$$
such that
$$
a_{11}^2 = a_{21}^2 + a_{22}^2 = a_{31}^2 + a_{32}^2 + a_{33}^2 = a_{n1}^2 + a_{n2}^2 + \cdots + a_{nn}^2.
$$
We're doing a chapter on proofs by induction so I'm pretty sure that would be the way to go. My general thought is to somehow prove that a square exists that can be the sum of any number of squares but I'm not too sure. Thank you for the help!
|
We proof by induction. You have already found the initial case.
$$5^2 = 3^2 + 4^2$$
We now use the formula
$$\left(\frac{x^2+1}2\right)^2=x^2+\left(\frac{x^2-1}2\right)^2$$ to find a sequence
$$z^2_n=x^2_n+y^2_n$$
such that $x_n=z_{n-1}$.
$$\begin{array}{r,r,r}
5^2&=&4^2&+&3^2\\
13^2&=&12^2&+&5^2\\
85^2&=&84^2&+&13^2\\
3613^2&=&3612^2&+&85^2\\
6 526 885^2&=&6 526 884^2&+&3613^2\\
&\cdots&&&&&
\end{array}$$
From this we get
$$\begin{eqnarray}
6 526 885^2\\
=6 526 884^2&+&3613^2\\
=6 526 884^2&+&3612^2+85^2\\
=6 526 884^2&+&3612^2+84^2+13^2\\
=6 526 884^2&+&3612^2+84^2+12^2+5^2\\
=6 526 884^2&+&3612^2+84^2+12^2+4^2+3^2
\end{eqnarray}
$$
So we have found a number that can be represented as the sum of 1, 2, 3, 4, 5 or 6 squares. I think it is clear to continue this process and how to proof it by induction.
|
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For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$. For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$.
My attempt: We have
$$2(a^2+b^2)\geq (a+b)^2$$
so $$-2\leq a+b \leq 2$$
In other hand $$ab=\frac{(a+b)^2-2}{2}=(a+b)^2-1$$
|
let $a=x-y$ and $b=x+y$
$a^2+b^2=2 \implies x^2+y^2=1 $
substituting $y$ given that $x^2+y^2=1 \implies $
the problem now is to prove that $6x+x^2-y^2 \ge -5 \iff 2x^2+6x-1 \ge -5 $ (1)
$y^2=1-x^2 \ge 0 \implies -1 \ge x \ge 1$
it's enough to see that (1) $\iff (x-1)(2x-4) $
which is easy to check that it's positive when $-1 \ge x \ge 1$
|
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|
Finding $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$ Find $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$.
I wrote this proof:
Let $f(x)=e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$
$|\sqrt{e}-P_N(\frac{1}{2})|=|R_N(\frac{1}{2})| < \frac{1}{100}$
From the Taylor theorem we get that there exists $0<c<\frac{1}{2}$ such that:
$|R_N(\frac{1}{2})| = |\frac{f^{(N+1)}(c)}{(N+1)!2^N}| = \frac{e^c}{(N+1)!2^N} \leq \frac{e}{(N+1)!2^N}\leq \frac{3}{(N+1)!2^N}\leq \frac{3}{(N+1)!}<10^{-2}$
hence for $N=5$ we get $720>300$ and the inequality holds.
Therefore:
$P_5(\frac{1}{2})=1+\frac{1}{2}+\frac{1}{8}+\frac{1}{6\cdot8}+\frac{1}{16\cdot 24}+\frac{1}{32\cdot 120} = \frac{32\cdot120+16\cdot 120+4\cdot 120+4\cdot 20+2\cdot 5}{32\cdot 120} = \frac{n}{m}$
Did I get it right? Was there a simpler method?
|
Your solution is correct, but unnecessarily complicated. Also, there seems to be an error in your expression for $\left\lvert{R_N\left(\tfrac12\right)}\right\rvert,$ which doesn't invalidate your numerical answer, but adds to the complication of finding it.
You're using Taylor's theorem with the Lagrange form of the remainder. Surely the expression should be:
$$
\left\lvert{R_N\left(\frac12\right)}\right\rvert = \left\lvert{\frac{f^{(N+1)}(c)}{(N+1)!\cdot2^{N+1}}}\right\rvert
$$
Having $2^{N+1}$ rather than $2^N$ in the denominator makes things slightly easier. Also, you have gone rather too far in your simplifying approximations; the approach is valid, but the calculation is harder than it needs to be.
Continuing in a simpler way, using the fact that $e < 4,$ so $\sqrt{e} < 2$:
$$
\frac{e^c}{(N+1)!\cdot2^{N+1}} < \frac{\sqrt{e}}{(N+1)!\cdot2^{N+1}} < \frac{1}{(N+1)!\cdot2^N}.
$$
Therefore, all we need is to find $N$ large enough that:
$$
(N+1)!\cdot2^N > 100.
$$
So we don't need $N = 5$; it is good enough to take $N = 3.$ Then we find:
$$
\sqrt{e} - \frac{79}{48} < \frac{1}{192}.
$$
We could get the same value for $\tfrac{n}{m}$ even more simply, without using the remainder expression for Taylor's theorem - it's quite easy to get it wrong, and I'm not sure I haven't! - instead just using the infinite series for $\sqrt{e}$:
\begin{align*}
\sqrt{e} & = 1 + \frac12 + \frac18 + \frac1{48} + \frac1{384}\left(1 + \frac1{5\cdot2} + \frac1{5\cdot6\cdot2^2} + \cdots\right) \\
& < \frac{79}{48} + \frac1{384}\left(1 + \frac1{2} + \frac1{2^2} + \cdots\right) \\
& = \frac{79}{48} + \frac1{192},
\end{align*}
an approximation which is still more than good enough $\ldots$ indeed, the accuracy is just the same! (I miscalculated at first.) I presume this is just a coincidence.
(Continued fraction methods, as used in the two previous answers, give more accurate approximations than these - in fact, the most accurate approximations possible - but we don't need that level of precision for this question.)
|
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|
Compute $\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2$ How would you evaluate
$$\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2$$
Wolfram|Alpha says it equals $\;\pi/2-1/2.\;$ If the solution is complicated, ... I can handle complicated.
|
$\displaystyle\sum_{n=1}^∞ \dfrac{\sin(nx)}{n} =\frac{1}{2i} \sum_{n=1}^∞ \dfrac{(e^{ix})^n}{n}-\dfrac{(e^{-ix})^n}{n}=-\frac{1}{2i}({\log(1-e^{ix})-\log(1-e^{-ix})})$
$= \displaystyle \dfrac{1}{2i} \log\left(\frac{(1-e^{-ix})}{(1-e^{ix})}\right)= \dfrac{1}{2i} \log(-e^{-ix})= \dfrac{\log(e^{iπ})-\log(e^{ix})}{2i} =\dfrac{π-x}{2}. $
Now $\displaystyle\int_0^x \sum_{n=1}^∞\dfrac{ \sin(nx)}{n}dx = \int_0^x \dfrac{π-x}{2}dx $
$\implies \displaystyle -\sum_{n=1}^∞ \left(\dfrac{\cos(nx)}{n^2} -\dfrac{\cos(0)}{n^2}\right)= \dfrac{2πx-x^2}{4}$
$\implies \displaystyle\sum_{n=1}^∞ \dfrac{2\sin^2(\frac{nx}{2})}{n^2} = \dfrac{2πx-x^2}{4} .$
Put $x=2$ the answer will be $\displaystyle\sum_{n=1}^∞ \dfrac{\sin^2(n)}{n^2} =\dfrac{π-1}{2} $.
Edit
In this way we can also solve Basel problem.
$\displaystyle\sum_{n=1}^∞ \dfrac{\sin^2(\frac{nπ}{2})}{n^2} = \dfrac{\frac{π}{2}(π-\frac{π}{2})}{2} = \dfrac{π^2}{8} $
Which means $\displaystyle\dfrac{1}{1^2} +\dfrac{1}{3^2} +\dfrac{1}{5^2} +...= \dfrac{π^2}{8} \implies \dfrac{3\zeta(2)}{4} =\dfrac{π^2}{8} \implies \zeta(2)=\dfrac{π^2}{6}$
|
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Does the maxima of $\sin x + \sin (\sin x ) + \sin(\sin (\sin x )) + \sin(\sin(\sin (\sin x ))) + ...$ converge? Does the maxima of $\sin x + \sin (\sin x ) + \sin(\sin (\sin x )) + \sin(\sin(\sin (\sin x ))) + ...$ converge?
Given that $f^{n+1}(x) = f(f^n(x))$, where $f(x) = \sin x$, each "next" term grows by $f^{n+1}(x)$. Thus, if there's a convergence point, it would mean that $f^{\infty}(\pi/2) = 0$. Is there any way to prove this?
Another form of this expression would be the repeated integral bounds of $\cos x$, something like $\int_0^{\int^{pi/2}_0\cos xdx}\cos x dx$ repeating.
|
Given that, when $0<x<1$ you have $$x>\sin x >x\left(1-\frac{x^2}6\right)$$ you get, when $x=1/n$ for $n\geq 2,$ $$\begin{align}\sin\frac1n&>\frac{n^2-1/6}{n^3}\\&>\frac{n^2-1}{n^3+1}\\&=\frac{n-1}{n^2-n+1}\\&=\dfrac{1}{n+\frac{1}{n-1}}\\&\geq \frac1{n+1}
\end{align}$$
So $\sin(1/n)\geq \frac{1}{n+1}.$
So let $x_n=f^n(x)$ then when $\frac{1}{n}<x_k<1$ then $\frac{1}{n+1}<x_{k+1}<1$ and thus by induction, $\frac{1}{n+m}<x_{k+m}.$
So $\sum_{i=1}^{\infty} x_i$ diverges because $$\sum_{i=0}^N x_{k+i}>\sum_{i=0}^N \frac1{n+i}$$
The same happens when $-1<x_k\leq -\frac1{n}.$ We know $|\sin(\sin x))|<1.$ If it is non-zero, then $|\sin(\sin x))|>\frac{1}{n}$ for some $n,$ and thus the sum diverges.
So the only time you can get convergence is when $\sin(\sin x)=0.$ but that is only possible if $\sin(x)=0.$
|
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Prove $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $ without expansion It's easy to prove that if $a,b,c \neq 0 $: $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $$
as $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff \frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\iff (ab+bc+ca)(a+b+c)=abc \iff (a+b)(b+c)(c+a)=0 $
I'm curious to prove this some other ways. I tried to use function and inequality but still no progress.
|
You could define:
$$F(a,b,c) = (a+b+c)(ab+bc+ca) - abc$$
and check that $F(a, -a,c) = F(a,b,-b) = F(-c,b,c) = 0$ easily. This means that $F$ as a polynomial must contain the linear factors $a+b, b+c$ and $c+a.$ Then, you argue that $F$ must be divisible by all of them at the same time, which generates a cubic polynomial where $F$ itself is cubic. Therefore,
$$F(a,b,c) = (a+b)(b+c)(c+a)$$
by matching a leading coefficient.
But this argument, while seems clever, requires unnecessary sophisticated methods from Number Theory/Algebra than just fully expanding $F.$
|
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|
Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$.
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.
My attempt was to write $f(x)=(x^4+x^2+1)q(x)+Ax^3+Bx^2+Cx+D=(x^2+x+1)(x^2-x+1)q(x)+Ax^3+Bx^2+Cx+D$ and then factor out $x^2+x+1$ to get $f(x)=(x^2+x+1)[(x^2-x+1)q(x)+B]+Ax^3+(C-B)x+(D-B)$, then do the same for $x^2-x+1$ and then use that along with the known remainder of $f$ divided by $x^2+x+1$ and $x^2-x+1$ to obtain $A$, $B$, $C$, $D$. However, $Ax^3$ is in the way so I don't know how to proceed nor do I have any other ideas to start with.
|
This is where things go wrong:
My attempt was to write $f(x)=(x^4+x^2+1)q(x)+Ax^3+Bx^2+Cx+D=(x^2+x+1)(x^2-x+1)q(x)+Ax^3+Bx^2+Cx+D$ and then factor out $x^2+x+1$ to get $f(x)=(x^2+x+1)[(x^2-x+1)q(x)+B]+Ax^3+(C-B)x+(D-B)$
The remainder inside the brackets can have a linear term as well, so you should have written it as
$$
f(x) = (x^2 + x + 1)[(x^2 - x + 1)q(x) + Ax + (B-A)] + (C-B)x + (D-B+A).
$$
And similarly,
$$
f(x) = (x^2 - x + 1)[(x^2 - x + 1)q(x) + Ax + (B+A)] + (C+B)x + (D-B-A),
$$
at which point you can solve for all four coefficients.
|
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Solution set to non-linear equation of $3$ variables I have the following trigonometric equation of $3$ variables:
$$f(\theta,\lambda,\phi)=3 \cos (\theta ) \cos (\lambda ) \cos (\phi
)-(\cos (\theta )+3) \sin (\lambda ) \sin
(\phi )$$
$$-\sin (\theta ) \cos (\lambda )+\sin
(\theta ) \cos (\phi )+3 \cos (\theta )+\cos
(\lambda ) \cos (\phi )-7$$
I want to prove that the solution set to the equation $f(\theta,\lambda,\phi)=0$ is
$$S_f=\{(2\pi k,\lambda+2\pi m,-\lambda+2\pi n):k,m,n\in\mathbb{Z},\lambda\in\mathbb{R}\}$$
Graphically, it is easy to see that this is the case but everything I have tried so far has failed to prove the conjecture. Perhaps my best attempt was extrapolating this equation into an unwieldy polynomial of $3$ variables
$$P(x,y,z)=9 x^4 y^2+16 x^4-18 x^3 y^2-192 x^3+9 x^2
y^4+z^4 \left(\left(x^2-1\right)
y^2+1\right) \left(16 \left(x^2-1\right)
y^2+(3 x+5)^2\right)-64 x^2 y^2+z^2
\left((x-1) (x+1) \left(9 x^2+30 x+41\right)
y^4+2 x (x (48 (x-5) x+187)+234) y^2+x (x (9
(x-2) x-64)-78)-74 y^2+151\right)+736 x^2+2
y z^3 \left(9 x^4+12 x^3-20 x^2+(x-1) (x+1)
(x (31 x-78)-33) y^2-96 x-33\right)+2 y z
\left(31 x^4-270 x^3+560 x^2+(x (x (3 x (3
x+4)-20)-96)-33) y^2-210 x-111\right)+30 x
y^4-78 x y^2-960 x+25 y^4+151 y^2+400$$
over the domain $(x,y,z)\in [-1,1]^3$. If I can prove that the solution set to $P(x,y,z)=0$ over this domain is
$$S_P=\{(1,y,y):y\in[-1,1]\}$$
then the original conjecture would be solved. The motivation behind this is actually proving a certain type of quantum error detection encoding exists. It's a little difficult to explain (although if anyone wants details I am more than happy to provide them) but suffice to say that after a lot of work I managed to whittle my existence proof down to the conjecture above.
|
Let $\;x=\tan\dfrac\theta2,\;y=\dfrac{\lambda+\varphi}2,\;z=\dfrac{\lambda-\varphi}2,\;$ then
$$g(x,y,z) = (1-x^2)(2\cos2y+\cos 2z+3)+4x\sin y\sin z\\
+(1+x^2)(-\cos2y+2\cos2z-7)\\
= (-3\cos2y+\cos2z-10)x^2+4x\sin y\sin z+\cos2y+3\cos2z-4\\
= (-6\cos^2y-2\sin^2z-6)x^2+4x\sin y\sin z - 2\sin^2y-6\sin^2z=0,$$
$$-6(1+\cos^2y)x^2-2(x\sin z-\sin y)^2-6\sin^2z=0.\tag1$$
From $(1)$ should
$$x=0,\quad y=\pi j,\quad z=\pi l,\tag2$$
and finally
$$\color{green}{\mathbf{\theta=2\pi k,\quad \lambda=\pi m,\quad \varphi=2\pi n-\lambda,\quad k,m,n\in\mathbb Z.}}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Alternative approaches to maximize $y=x\sqrt{100-x^2}$ I could find three good approaches to find maximum of the function $y=x\sqrt{100-x^2}$. I will explain them briefly :
First: Finding $x$ satisfies $y'=0$ then plugging it in the function.
Second: Using the substitution $x=10\sin\theta$ (or $x=10\cos\theta)$ for
$\theta\in(0,\frac{\pi}2)$ to get $y=100\sin\theta\cos\theta=50\sin(2\theta)$ hence the maximum is $50$.
Third: Using AM-GM inequality: It is obvious that maximum occurs for $x>0$ So we can rewrite $y$ as $y=\sqrt{x^2(100-x^2)}$ . Now the sum of $x^2$ and $100-x^2$ is $100$ so the maximum of product happens when $x^2=100-x^2$ or $x^2=50$ Hence $y_{\text{max}}=50$.
Just for fun, can you maximize $y=x\sqrt{100-x^2}$ with other approaches?
|
$$x\sqrt{100-x^2}=A$$
$$\begin{align}&\implies \left(\frac Ax\right)^2=100-x^2,~x≠0 \\
&\implies \frac{A^2}{x^2}+x^2=100 \\
&\implies\left(\frac Ax-x\right)^2+2A=100 \\
&\implies \left(\frac Ax-x\right)^2=100-2A\\
&\implies 100-2A≥0\\
&\implies A≤50.\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $abc\geq ab+2c\geq ab+c+c\geq a+b+c+c\geq a+b+c+2$.
The main problem I face is to justify that $$abc=a+b+c+2\implies a=b=c=2.$$
My idea is to assume that $a>2$ and try to get a contradiction.
|
Another way,
The condition gives:
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1.$$
Thus, $$1=\sum_{cyc}\frac{1}{1+a}\leq\sum_{cyc}\frac{1}{1+2}=1,$$ which gives $a=b=c=2.$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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|
Does $ a_{\lfloor n + \sqrt{n/2}\rfloor} \le a_n\left(1 + \frac{b}{n^{\frac{1}{2}+\varepsilon}}\right) $ imply that $a_n$ is bounded? Let $(a_n)_{n\ge 1}$ be an increasing sequence of positive real numbers and suppose there is a constant $b > 0$ and $\varepsilon > 0$ such that
$$
a_{\lfloor n + \sqrt{n/2}\rfloor} \le a_n\left(1 + \frac{b}{n^{\frac{1}{2}+\varepsilon}}\right)
$$
for all $n$ sufficiently large. Does it follow that $a_n$ is bounded? i.e., is $\limsup a_n < \infty$?
If the left hand side of the inequality was $a_{n+1}$, then we could use induction to derive
$$
a_{n + 1} \le a_{n_0}\prod_{k=n_0}^{n}\left(1 + \frac{b}{k^{\frac{1}{2}+\varepsilon}}\right)
$$
and even then, it is not clear whether $a_n$ is bounded. Moreover, the form of the index on the left hand side of the inequality makes it difficult to apply this method. Is there a way to do this?
Any help or comments are welcome.
|
Write $f(n) = \bigl\lfloor n + \sqrt{n/2} \bigr\rfloor$ and consider the sequence $(n_k)_{k\geq 0}$ defined by
$$ n_0 \geq 2 \qquad\text{and}\qquad n_{k+1} = f(n_k). $$
Since $f(n) \geq n+1$ for $n \geq 2$, we find that $(n_k)_{k\geq 0}$ is strictly increasing. Moreover, for any sufficiently large $n$, we have
$$
\sqrt{f(n)}
\geq \biggl( n + \frac{\sqrt{n}}{2} \biggr)^{1/2}
= \sqrt{n} \biggl( 1 + \frac{1}{2\sqrt{n}} \biggr)^{1/2}
\geq \sqrt{n} \biggl( 1 + \frac{1}{5\sqrt{n}} \biggr)
= \sqrt{n} + \frac{1}{5}. \tag{*}
$$
So by choosing $n_0$ to be sufficiently large, it follows that
$$ \sqrt{n_k} \geq \sqrt{n_0} + \tfrac{1}{5}k \quad \text{for all} \quad k \geq 0 $$
Then
$$ a_{n_k}
\leq a_{n_0} \prod_{j=0}^{k-1} \biggl( 1 + \frac{b}{n_j^{0.5+\epsilon}} \biggr)
\leq a_{n_0} \prod_{j=0}^{k-1} \biggl( 1 + \frac{b}{(\sqrt{n_0} + \frac{1}{5}j)^{1+2\epsilon}} \biggr) $$
and this upper bound converges as $k\to\infty$. Therefore $(a_n)$ is bounded above.
Addendum. In fact we can give a precise range of $n$ for which the inequality $\text{(*)}$ holds.
*
*To begin with, the first step of $\text{(*)}$ holds precise when $ \left\lfloor \sqrt{\frac{n}{2}} \right\rfloor \geq \frac{\sqrt{n}}{2} $, which turns out to be true if and only if $n \geq 18$.
*Next, it follows that $\sqrt{1+x} \geq 1 + \frac{2}{5}x$ if and only if $0 \leq x \leq \frac{5}{4}$. So the third step of $\text{(*)}$ holds for any $n$.
Combining altogether, $\text{(*)}$ holds for any $n \geq 18$ and we may take $n_0 = 18$.
|
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|
Solving matrix equation with element-wise products I am wondering if there is a way to solve this equation for a:
$$(as^T ⊙ b)n = t.$$
where:
⊙ is element-wise multiplication
a is an unknown v x 1 vector
s is an i x 1 $\vec{1}$ vector
b is a known v x i matrix
n is a known i x 1 vector
t is a v x 1 $\vec{1}$ vector
Thank you in advance
|
Yes. Multiply it out as elements. You will have a system of $v$ uncoupled linear equations in the components of $a$. Then solve each, by itself.
Example with $v=3$, $i=2$:
\begin{align*}
\left( \begin{pmatrix}a_1 \\ a_2 \\ a_3\end{pmatrix} (s_1 \, s_2) \odot \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{pmatrix}\right)\begin{pmatrix}n_1 \\n_2\end{pmatrix} &= \begin{pmatrix}t_1 \\ t_2 \\ t_3 \end{pmatrix} \\
\left( \begin{pmatrix}a_1s_1 & a_1s_2 \\ a_2s_1 & a_2s_2 \\ a_3s_1 & a_3s_2\end{pmatrix} \odot \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ b_{31} & b_{32} \end{pmatrix}\right)\begin{pmatrix}n_1 \\n_2\end{pmatrix} &= \begin{pmatrix}t_1 \\ t_2 \\ t_3 \end{pmatrix} \\
\begin{pmatrix}a_1b_{11}s_1 & a_1b_{12}s_2 \\ a_2b_{21}s_1 & a_2b_{22}s_2 \\ a_3b_{31}s_1 & a_3b_{32}s_2\end{pmatrix} \begin{pmatrix}n_1 \\n_2\end{pmatrix} &= \begin{pmatrix}t_1 \\ t_2 \\ t_3 \end{pmatrix} \\
\begin{pmatrix}a_1b_{11}n_1s_1 + a_1b_{12}n_2s_2 \\ a_2b_{21}n_1s_1 + a_2b_{22}n_2s_2 \\ a_3b_{31}n_1s_1 + a_3b_{32}n_2s_2\end{pmatrix} &= \begin{pmatrix}t_1 \\ t_2 \\ t_3 \end{pmatrix} \\
\begin{cases}
a_1 = \frac{t_1}{b_{11}n_1s_1 + b_{12}n_2s_2} \\
a_2 = \frac{t_2}{b_{21}n_1s_1 + b_{22}n_2s_2} \\
a_3 = \frac{t_3}{b_{31}n_1s_1 + b_{32}n_2s_2}
\end{cases}\text{,}
\end{align*}
assuming those $v$ divisions are defined.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve for variable inside and outside of a log I came across this problem:
$$
0 = \ln(c_1 + c_2 \cdot x) + (c_3 + c_4 \cdot x)^{1/2} + c_5
$$
I simplified it down to:
$$
0 = \ln(c_1 + c_2 \cdot x) + c_3 \cdot x + c_4
$$
Is there a way to solve this analytically?
|
You can isolate $x$ only if you make use of special functions (specifically Lambert's product log function $W$):
That may or may not be acceptable to you, depending of what you're trying to do.
For example:
$$
\begin{align}
\ln(5+6x)&=2+3x\\
\ln(5+6x)&=\frac{5+6x}{2}-\frac{1}{2}\\
5+6x&=e^{\frac{5+6x}{2}}\cdot\frac{1}{\sqrt{e}}\\
\frac{5+6x}{2}&=e^{\frac{5+6x}{2}}\cdot\frac{1}{2\sqrt{e}}\\
-\frac{5+6x}{2}\cdot e^{-\frac{5+6x}{2}}&=-\frac{1}{2\sqrt{e}}\\
-\frac{5+6x}{2}&=W\left(-\frac{1}{2\sqrt{e}}\right)\\
x&=-\frac{1}{3}W\left(-\frac{1}{2\sqrt{e}}\right)-\frac{5}{6}
\end{align}
$$
If solving on real numbers (as opposed to complex), one gets two solutions corresponding to branches $W_0$ and $W_{-1}$:
$$x_1=-\frac{1}{3}W_0\left(-\frac{1}{2\sqrt{e}}\right)-\frac{5}{6}=-\frac{2}{3}$$
and:
$$x_2=-\frac{1}{3}W_{-1}\left(-\frac{1}{2\sqrt{e}}\right)-\frac{5}{6}\approx -0.247856$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
For any $ a \in \mathbb{N}$ ; $f(x)=x^4+(4a+2)x^2+1$ is irreducible in $\mathbb{Z}[X]$ & reducible mod $p$ for all $p$. A.1 First we show $f(x)$ is reducible mod $p$ for all $p$.
Taking $y=x^2$ ; $f(x)$ becomes $y^2+(4a+2)y+1$. Solving the equation $y^2+(4a+2)y+1=0$ we get
$$
y=-(2a+1)\pm2\sqrt{a(a+1)}
$$
Now replacing $y$ by $x^2$ we have
$$
x=\pm \sqrt{-(2a+1)+2\sqrt{a(a+1)}} = \pm \ (i\sqrt{a} + i\sqrt{a+1})
$$
&
$$
x=\pm \sqrt{-(2a+1)-2\sqrt{a(a+1)}} = \pm \ (i\sqrt{a} - i\sqrt{a+1})
$$
Hence we have obtained all the roots of the polynomial $f(x)$ say $x_1, x_2, x_3, x_4$ where
$$
x_1= (i\sqrt{a} + i\sqrt{a+1}) \ ; \ x_3= -(i\sqrt{a} + i\sqrt{a+1})
$$
$$
x_2= (i\sqrt{a} - i\sqrt{a+1}) \ ; \ x_4= -(i\sqrt{a} - i\sqrt{a+1})
$$
A.2 The three ways of expanding this quartic are as follows:
*
*$\{(x-x_1)(x-x_2)\}\{(x-x_3)(x-x_4)\}=(x^2+1-2\sqrt{a}ix)(x^2+1+2\sqrt{a}ix)\\
= (x^2+1)^2-(2\sqrt{a}ix)^2 \\
= (x^2+1)^2-2^2\mathbf{(-a)}x^2$
*$\{(x-x_1)(x-x_4)\}\{(x-x_2)(x-x_3)\}=(x^2-1-2\sqrt{a+1} \ ix)(x^2-1+2\sqrt{a+1} \ ix)\\
= (x^2-1)^2-(2\sqrt{a+1} \ ix)^2\\
=(x^2-1)^2-2^2 \ \mathbf{(-(a+1))} \ x^2$
*$\{(x-x_1)(x-x_3)\}\{(x-x_2)(x-x_4)\}\\
=(x^2+(2a+1)+2\sqrt{a(a+1)})(x^2+(2a+1)-2\sqrt{a(a+1)})\\
=(x^2+(2a+1))^2-(2\sqrt{a(a+1)})^2\\
=(x^2+(2a+1))^2-2^2\mathbf{(a(a+1))}$
A.3 Now the final piece.
$f(x)$ can be factored in any of the above 3 ways. Consider any prime $p$.
*
*If $\mathbf{-a}$ is a square element in $\mathbb{F}_p$; there exist $b \in \mathbb{F}_p$ such that $-a=b^2$. So factoring $f(x)$ as in form A.2.1. Hence considering $f(x)$ in $\mathbb{F}_p[x]$ we have
$$
f(x)=(x^2+1)^2-2^2(-a)x^2=(x^2+1)^2-(2bx)^2=(x^2+1-2bx)(x^2+1+2bx)
$$
Hence $f(x)$ is reducible modulo $p$ if $-a$ is a square element.
*If $\mathbf{-(a+1)}$ is square element in $\mathbb{F}_p$, there exist $c \in \mathbb{F}_p$ such that $-(a+1)=c^2$. So we factor $f(x)$ as in form A.2.2. Now considering $f(x)$ in $\mathbb{F}_p[x]$ we have
$$
f(x) =(x^2-1)^2-2^2 \ \mathbf{(-(a+1))} \ x^2=(x^2-1)^2-(2cx)^2=(x^2-1+2cx)(x^2-1+2cx)
$$
Hence we get $f(x)$ is reducible modulo $p$ if $2$ is a square element.
*Now if both $\mathbf{-a}$ and $\mathbf{-(a+1)}$ are non square elements in $\mathbb{F}_p$ then $\mathbf{a(a+1)}$ is a square element in $\mathbb{F}_p$. Hence factoring $f(x)$ as in the form A.2.3 we get that $f(x)$ is reducible modulo $p$.
Hence we have $f(x)$ is reducible modulo $p$ for all $p$.
How to show $f(x)$ is irreducible in $\mathbb{Z}[X]$?
|
The following gives a mild generalization, but the idea is essentially what has been discussed.
Proposition 1. Let $f(x)=x^4+ax^2+b,~a,b\in {\mathbb N},$ where $a^2-4b$ is a non-square, and either (1) $b$ is a non-square, or (2) $b$ is a square and $\pm2\sqrt{b}-a$ is a non-square. Then $f(x)$ is irreducible over ${\mathbb Z}.$
Proof. Assume that $f$ satisfies the conditions. Clearly $f$ has no rational roots (in fact, no real roots). By comparison of coefficients, if $f$ were reducible, it must be of the form $$f(x)=(x^2+cx+d)(x^2-cx+e), c,d,e\in {\mathbb Z},$$ where $c\neq 0$ implies $e=d$. Necessarily $c\neq 0$ and $e=d$, since $a^2-4b$ is a non-square. Now by comparison of coefficients again, one has $$2d-c^2=a,d^2=b.$$ If $b$ is not a square, one immediately gets a contradiction. If $b=d^2$, then $d=\pm \sqrt{b},$ and one would have $\pm 2\sqrt{b}-a=c^2,$ contradicting the assumption that $\pm2\sqrt{b}-a$ is not a square. It follows that $f$ is irreducible. $\Box$
Proposition 2. Let $f(x)=x^4+ax^2+1$ with $a\in {\mathbb N}$ such that $a>2$ and $a^2-4$ is a non-square. Then $f$ is irreducible over ${\mathbb Z}$, but $f$ is reducible over ${\mathbb F}_p$ for any prime $p$.
Proof.
Part 1. $f(x)$ is irreducible over ${\mathbb Z}$.
This follows from Proposition 1, since $a^2-4$ is not a square, and $\pm 2(1)-a<0$ is not a square.
Part 2. $f(x)$ is reducible over ${\mathbb F}_p$ for any prime $p$.
Case 0. The case $p=2$ is immediate, since $$x^4+1=(x^2+1)^2$$ and $$x^4+x^2+1=(x^2+x+1)^2.$$ Assume from now on that $p$ is odd. Using the Legendre symbol, it suffices to consider the following cases.
Case 1. $p\mid D$ or $\left(\frac{D}p\right)=1,$ where $D=a^2-4=(2-a)(-2-a).$ Writing $D=u^2$ over ${\mathbb F}_p,$ one has $$x^4+ax^2+1=\left(x^2-\frac{-a+u}2\right)\left(x^2-\frac{-a-u}2\right),$$ which might be factored further.
Case 2. Either $\left(\frac {2-a}p\right)=1$ and $\left(\frac {-2-a}p\right)=-1,$ or $\left(\frac {-2-a}p\right)=1$ and $\left(\frac {2-a}p\right)=-1.$
Over ${\mathbb F}_p,$ one seeks to factor $$x^4+ax^2+1=(x^2+cx+d)(x^2-cx+d),$$ which is equivalent to $$2d-c^2=a,d^2=1$$
$$\Leftrightarrow (d=1~{\rm and~}2-a=c^2)~{\rm or~}(d=-1~{\rm and~}-2-a=c^2).$$ This means that when $\left(\frac {2-a}p\right)=1,$ one can set $d=1$, and when $\left(\frac {-2-a}p\right)=1,$ one can set $d=-1.$ The $c$ can be solved accordingly. This completes the proof of the reducibility of Case 2, and Proposition 2. $\Box$
To answer the question when $f(x)=x^4+(4a+2)x^2+1,~a\in {\mathbb N},$ just apply Proposition 2. One has $$(4a+2)^2-4=16a(a+1),$$ which is not a square, since $\gcd(a,a+1)=1$ and $a,a+1$ cannot be simultaneously squares. The other condition is obviously satisfied. Hence $f(x)$ is irreducible over $\mathbb Z$, and reducible over ${\mathbb F}_p$ for any prime $p.$
Remark. It may be tempting to assert the reducibility of $f(x)$ over ${\mathbb F}_p$ for any prime $p$ under the conditions in Proposition 1. But this is not true. For example, let $f(x)=x^4+5x^2+2.$ Then $f(x)$ is irreducible both over ${\mathbb Z}$ and ${\mathbb F}_5.$
|
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|
Infinite series of matrix product
If $t$ is real and positive and matrix $A$ and $B$ are such that $A = \begin{pmatrix} \dfrac{t^2+1}{t} & 0 & 0 \\ 0 & t & 0 \\ 0 & 0 & 25 \end{pmatrix}$ and $B=\begin{pmatrix} \dfrac{2t}{t^2+1} & 0 & 0 \\ 0 & \dfrac{3}{t} & 0 \\ 0 & 0 & \dfrac{1}{5} \end{pmatrix}$ and $X=(AB)^{-1} + (AB)^{-2} + (AB)^{-3} + \cdots\infty, Y=X^{-1}$ then which option is correct?
$1.$ $\text{det}(Y)=10$
$2.$ $X\cdot \text{adj(adj}(Y)=8I$
$3.$ $\text{det}(Y)=20$
$4.$ $X\cdot \text{adj(adj}(Y)=5I$
My attempt:
I managed to calculate some of the important matrices, but am stuck in a part where I have to find the determinant of a matrix that has some elements tending to infinity.
As $t>0$ $$AB= \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{pmatrix}$$
The inverse of a diagonal matrix is another diagonal matrix that's diagonal elements are reciprocals of the diagonal elements of the original matrix, so
$$(AB)^{-1}= \begin{pmatrix} \dfrac{1}{2} & 0 & 0 \\ 0 & \dfrac{1}{3} & 0 \\ 0 & 0 & \dfrac{1}{5} \end{pmatrix}, (AB)^{-2}=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{pmatrix}, (AB)^{-3}=\begin{pmatrix} \dfrac{1}{2} & 0 & 0 \\ 0 & \dfrac{1}{3} & 0 \\ 0 & 0 & \dfrac{1}{5} \end{pmatrix}, \cdots$$
My problem is, adding these matrices results in a matrix that's diagonal elements tend to infinity, as $$X=\begin{pmatrix} 2+1/2+2+1/2+\cdots \infty & 0 & 0 \\ 0 & 3+1/3+3+1/3+\cdots \infty \\ 0 & 0 & 5+1/5+5+1/5+\cdots \infty\end{pmatrix} = \begin{pmatrix} S_2 & 0 & 0 \\ 0 & S_3 & 0\\ 0 & 0 & S_5 \end{pmatrix}$$
And $$\text{adj}(X) = \begin{pmatrix} S_3S_5 & 0 & 0 \\ 0 & S_2S_5 & 0\\ 0 & 0 & S_2S_3 \end{pmatrix}$$
I know that $Y=\dfrac{\text{adj}(X)}{|X|}$ but I can't figure out how to calculate $|X|$. The determinant of a diagonal matrix is simply the product of the diagonal elements, but how to calculate this value when the diagonal elements tend to infinity?
|
$$AB = diag(2,3,5)$$
$$(AB)^{-1}=diag\left(\frac12, \frac13, \frac15\right)$$
$$(AB)^{-n}=diag\left(\frac1{2^n}, \frac1{3^n}, \frac1{5^n}\right)$$
$$X=diag\left(\frac{1/2}{1-1/2}, \frac{1/3}{1-1/3},\frac{1/5}{1-1/5} \right)=diag(1, 1/2, 1/4)$$
Hence now, you can compute quantity about $X$ and $Y$.
|
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|
How do you write $4(x +1)^2 + 1$ in the form $(ax +b)^2 + c\;$? We can write $4x^2+8x+5$ in the form $a(x+b)^2+c$ as $4(x+1)^2+1$. However, the question I am doing asks me to write it in the form $(ax+b)^2+c$. How do I change it to that form?
|
You can also solve the equation $(ax+b)^2+c\equiv 4x^2+8x+15$ (the $\equiv$ sign means "true for all values of $x$"—the coefficients on both sides of the equation have to be the same). If we expand the left-hand side, we get
$$
(ax+b)^2+c\equiv a^2x^2+2abx+b^2+c\equiv 4x^2+8x+5 \, .
$$
This means that $a^2=4$ and $2ab=8$ and $b^2+c=15$. If $a^2=4$ then $a=2$ or $a=-2$. Suppose that $a=2$. Then, $4b=8$, and so $b=2$. Finally, $4+c=5$, and so $c=1$. This gives us the solution
$$
(2x+2)^2+1\equiv 4x^2+8x+5 \, .
$$
If $a=-2$, then $-4b=8$, and so $b=-2$. Finally, $4+c=5$, and so $c=1$. This gives us the solution
$$
(-2x-2)^2+1 \equiv 4x^2+8x+5 \, .
$$
However, the first solution is arguably simpler because it doesn't have any minus signs in it.
|
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|
How we solve $0=y\sin\frac{3t}{2}+x\cos\frac{3t}{2}-y\sin(t)-x\cos(t)$ for real $t$ ($\frac{4\pi}{3}I have a system of equation:
*
*$x^2+y^2=y\sin\frac{3t}{2}+x\cos\frac{3t}{2}$
*$x^2+y^2=y\sin(t)+x\cos(t)$
I want to get an algebraic curve depending only from $x$ and $y$. If it is too complex, it would be already sufficient to determine, which degree the resulting curve has.
Can we at least determine the degree (the largest power $n$ that appears in $x^n$) of the curve equation in $x$ and $y$?
I am interested only in a (real) range $\frac{4\pi}{3}<t<2\pi$, which hopefully gives a condition that reduces the effort in solving the equation.
|
You already isolated a linear equation in $x$ and $y$ for each value $t$:
$$ y \sin \frac{3t}{2} + x \cos \frac{3t}{2} - y \sin t - x \cos t = 0 $$
$$ y = \frac{\cos t - \cos \frac{3t}{2}}{\sin \frac{3t}{2} - \sin t} x $$
$$ y = m(t) \, x \quad\mathrm{\ where\ } m(t) = \frac{\cos t - \cos \frac{3t}{2}}{\sin \frac{3t}{2} - \sin t} $$
unless $\sin \frac{3t}{2} - \sin t = 0$, which occurs at $t=2\pi m$ and at $t=\frac{2\pi}{5} (2n+1)$ for any integer $m$ or $n$. This never happens in the given domain $\frac{4\pi}{3} < t < 2\pi$.
$$ m(t) = \frac{2 \sin \frac{5t}{4}\, \sin \frac{t}{4}}{2 \cos \frac{5t}{4}\, \sin \frac{t}{4}} $$
$$ m(t) = \tan \frac{5t}{4} $$
Based on this ratio, let's define $x(t)$ and $y(t)$ in terms of the unknown magnitude $r(t)$:
$$ x = r \cos \frac{5t}{4} $$
$$ y = r \sin \frac{5t}{4} $$
Then from the second original equation,
$$ r^2 = r \sin \frac{5t}{4}\, \sin t + r \cos \frac{5t}{4}\, \cos t $$
The solution $r=0$ is just the extra point $(0,0)$ which always satisfies the equations for any $t$. Ignoring that, we can divide out $r$.
$$ r = \cos \left(\frac{5t}{4} - t\right) = \cos \frac{t}{4} $$
So
$$ x = \cos \frac{5t}{4}\, \cos \frac{t}{4} $$
$$ x^2+y^2 = \cos^2 \frac{t}{4} $$
The fifth Chebyshev polynomial of the first kind is $T_5(z) = 16z^5 - 20z^3+5z$, so
$$ \cos \frac{5t}{4} = T_5\left(\cos \frac{t}{4}\right) = 16 \cos^5 \frac{t}{4} - 20 \cos^3 \frac{t}{4} + 5 \cos \frac{t}{4} $$
$$ x = 16 \cos^6 \frac{t}{4} - 20 \cos^4 \frac{t}{4} + 5 \cos^2 \frac{t}{4} $$
$$ x = 16 (x^2+y^2)^3 - 20 (x^2+y^2)^2 + 5 (x^2+y^2) $$
This is a polynomial equation in just $x$ and $y$ of degree 6.
|
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|
let $a_{n}=n(a_{1}+a_{2}+\dots+a_{n-1})$, $n\geqslant 2$, $a_{1}=1$. Find $a_{n}$ let $a_{n}=n(a_{1}+a_{2}+\dots+a_{n-1})$, $n\geqslant 2$, $a_{1}=1$. Find $a_{n}$
My Approach:
so I thought, since this is a homogeneous relation, I expanded the expression and wrote the characteristic equation of the same, since that was $n(x^1+x^2+\dots+x^{n-1})$. So I substituted the values of $n=2$, and my answer was $n(n-1)$ which is wrong.
EDIT: I cannot use the characteristic equation since the coeff is not constant
|
We have $$\frac{a_n}{n} = a_1+a_2+\cdots +a_{n-1}= (a_1+a_2+\cdots +a_{n-2}) + a_{n-1} = \frac{a_{n-1}}{n-1}+a_{n-1} = n\cdot \frac{a_{n-1}}{n-1}$$
which is valid for $n\ge 3$ (notice that it's not valid for $n=2$ because we used $a_{n-2}$).
Then $$\frac{a_n}{n}= n\cdot \frac{a_{n-1}}{n-1}=n(n-1)\cdot \frac{a_{n-2}}{n-2}=\cdots = n(n-1)\cdots 3\cdot \frac{a_{2}}{2} = \frac{n!}{2}$$
|
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|
Solutions to $2^a3^b+1=2^c+3^d$
Find all $a,b,c,d$ positive integer such that:
$2^a3^b+1=2^c+3^d$
My progress:
One solution satisfying is $$\boxed{a=1,b=1,c=2,d=1} $$
We first take $\mod 3$ which gives $$ L.H.S\equiv 1\mod 3,~~R.H.S\equiv 2^c\mod 3$$
Hence we get $c$ even. So let $c=2k.$
We get $$2^a3^b-3^d=2^{2k}-1=(2^k-1)(2^k+1). $$
Now note that $2^k-1,2^k+1$ are relatively prime. Because, if not then let $d$ be the common divisor.
Then $$d|2^k-1,~~d|2^k+1\implies d|(2^k+1)-(2^k-1)=2\implies d|2^k\implies d|2^k+1-2^k=1.$$
Now there are two cases. So using the fact that $2^k-1,2^k+1$ are relatively prime and then for odd k $3|2^k+1$ and for even $k$ $3|2^k-1$
Case 1: When $d<b$ then $$2^a3^b-3^d=3^d(2^a3^{b-d}-1)=(2^k-1)(2^k+1)$$
*
*$K$ is odd $$\implies v_3(2^k+1)=d,~~3\nmid 2^k-1. $$
*$K$ is even $$\implies v_3(2^k-1)=d,~~3\nmid 2^k+1. $$
Case 2: When $d>b$ then $$2^a3^b-3^d=3^b(2^a -3^{d-b})=(2^k-1)(2^k+1)$$
*
*$K$ is odd $$\implies v_3(2^k+1)=b,~~3\nmid 2^k-1. $$
*$K$ is even $$\implies v_3(2^k-1)=b,~~~~3\nmid 2^k+1. $$
P.S. This is my 100th post in MSE. The other solutions are there in the chat. Any elementary method?
|
This is not an answer, it is a comment for discussion:
We can construct similar equation as follows:
$$\begin{cases}3^m-2^n=1\\3^r-2^s=1\end{cases}$$
Multiplying both sides we get:
$$3^{m+r}-3^m\cdot 2^s-3^r\cdot 2^n +2^{n+s}=1$$
$$\big(\frac{3^m}{3^r}-\frac{2^n}{2^s}\big)2^s\cdot3^r+1=2^{n+s}+3^{m+r}$$
$$\Rightarrow \big(3^{m-r}-2^{n-s}\big)2^s\cdot 3^r +1=2^{n+s}+3^{m+r}$$
If $m-r=1$ and $n-s=1$ then we have:
$$2^s\cdot3^r +1=2^{n+s}+3^{m+r}$$
Substituting $s=a, r=b, n+s=c$ and $m+r=d$ we get the equation. Each of equations in system have similar solutions in $\mathbb Z$ which may give solutions of final equation, i.e we may proceed reversely and find the solutions under constrains $m-r=1$ and $ n-s=1$; $m, n, r, s $ $\in \mathbb Z$.
|
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|
What is the order of magnitude of $\sum\limits_{n=1}^kn\binom{k}{n}\frac{(2k-2n-1)!!}{(2k-1)!!}$ as $k\to\infty$? What is the order of magnitude of the function $f(k)$ below in the limit as $k \rightarrow \infty$? Does it diverge, converge to a positive limit, or converge to zero?
$$
f(k) = \sum\limits_{n=1}^{k} n \binom{k}{n} \frac{(2k - 2n - 1)!! }{(2k-1)!! }
$$
Here $!!$ represents the double factorial. Does anyone have an hint?
|
We can write your $f(k)$ in the form
$$
\sum\limits_{n = 1}^k {\frac{1}{{2^n (n - 1)!}}\frac{{\Gamma (k + 1)\Gamma \!\left( {k - n + \frac{1}{2}} \right)}}{{\Gamma \!\left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}}} .
$$
Suppose that $k$ is large and fixed and consider
$$\tag{$*$}
\frac{1}{{(n - 1)!}}\frac{{\Gamma (k + 1)\Gamma \!\left( {k - n + \frac{1}{2}} \right)}}{{\Gamma \!\left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}}
$$
The logarithmic derivative for $2\leq n\leq k$ satisfies
$$
\psi (k - n + 1) - \psi \!\left( {k - n + \tfrac{1}{2}} \right) - \psi (n) < \frac{2}{{k - n + 1}} - \log \left( {n - \tfrac{1}{2}} \right)<0,
$$
i.e., the sequence starts to decrease after $n=2$. For $n=1,2$ the terms of $(*)$ are
$$
\frac{{2k}}{{2k - 1}},\quad \frac{{2k(2k - 2)}}{{(2k - 1)(2k - 3)}}
$$
which are at most $2$, say, for large $k$. Thus,
$$
0 \leq \frac{1}{{2^n (n - 1)!}}\frac{{\Gamma (k + 1)\Gamma\! \left( {k - n + \frac{1}{2}} \right)}}{{\Gamma \!\left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}} \le 2 \times \frac{1}{{2^n }}
$$
for all large $k$ and any $n\geq 1$. Consequently, we can apply Tannery's theorem and Stirling's formula, to deduce
\begin{align*}
& \mathop {\lim }\limits_{k \to + \infty } \sum\limits_{n = 1}^k {\frac{1}{{2^n (n - 1)!}}\frac{{\Gamma (k + 1)\Gamma \!\left( {k - n + \frac{1}{2}} \right)}}{{\Gamma\! \left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}}}
\\ &
= \sum\limits_{n = 1}^\infty {\frac{1}{{2^n (n - 1)!}}\mathop {\lim }\limits_{k \to + \infty } \frac{{\Gamma (k + 1)\Gamma \!\left( {k - n + \frac{1}{2}} \right)}}{{\Gamma \!\left( {k + \frac{1}{2}} \right)\Gamma (k - n + 1)}}} = \sum\limits_{n = 1}^\infty {\frac{1}{{2^n (n - 1)!}}} = \frac{{\sqrt e }}{2}.
\end{align*}
|
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|
How to prove that problem $\frac{\partial ^m f_n(x)}{\partial x^m}$? Let $n,N\in\mathbb{N}$ and $x\in\mathbb{R}$.
Let $f_{n}(x)=(1-\frac{x}{n})^n-(1-\frac{x}{N})^{N}$.
Prove the following:
If ${N}≤n$ then
$$\frac{\partial ^{M}f_{n}}{\partial x^M}(x)=(-1)^{M}\displaystyle\Pi_{k=0}^{M-1}\left(1-\frac{k}{n}\right)\left(1-\frac{x}{n}\right)^{n-M} \\+(-1)^{M-1}\Pi_{k=0}^{M-1}\left(1-\frac{k}{N}\right)\left(1-\frac{x}{N}\right)^{N-M}.$$
I tried to calculate $\frac{\partial^m f_n(x)}{\partial x^m}$ where $f_n(x)=(1-\frac{x}{n})^n-(1-\frac{x}{b})^b$ if $b≤n$. I found
$$\frac{\partial f_{n}}{\partial x}(x)=-(1-\frac{x}{n})^{n-1}+(1-\frac{x}{b})^{b-1},$$
$$\frac{\partial^{2} f_{n}}{\partial x^{2}}=(\frac{n-1}{n})(1-\frac{x}{n})^{n-2}-(\frac{b-1}{b})(1-\frac{x}{b})^{b-2},$$
$$\frac{\partial^{3}f_{n}}{\partial x^3}(x)=(-1)^{3} (1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{x}{n})^{n-3} +\\
(-1)^{2=3-1} (1-\frac{1}{b})(1-\frac{2}{b})(1-\frac{x}{b})^{b-3}.$$
|
We put $f_n(x)=\sigma_1(x)-\sigma_2(x)$ where $\sigma_1(x)=(1-\frac{x}{n})^n$ and $ \sigma_2(x)=(1-\frac{x}{N})^N$
then we calculate $\frac{\partial ^M\sigma_1(x)}{\partial x^M}, \frac{\partial^M\sigma_2(x)}{\partial x^M }$
$\frac{\partial \sigma_1(x)}{\partial x}=(-1)^1(1-\frac{x}{n})^{n-1}$
$\frac{\partial^2\sigma_1(x)}{\partial x^2}=(-1)^2\frac{n-1}{n}(1-\frac{x}{n})^{n-2}$
$\frac{\partial^3\sigma_1(x)}{\partial x^3}=(-1)^3(\frac{1}{n})^2(n-1)(n-2)(1-\frac{x}{n})^{n-3}$
$\frac{\partial ^4\sigma_1(x)}{\partial x^4} = (-1)^4(\frac{1}{n^3}(n-1)(n-2)(n-3)(1-\frac{x}{n})^{n-4}$
So, we can now easily deduce the $m$-order derivative, starting from the first derivatives and we find:
$\frac{\partial ^M \sigma_1(x)}{\partial x^M }=(-1)^{M}(\frac{1}{n})^{M-1}(n-1)(n-2).......(n-(M-1))(1-\frac{x}{n})^{n-M}=(-1)^M\prod_{k=0}^{k=M-1}(1-\frac{k}{n})(1-\frac{x}{n})^{n-M}$
In the same way, we find:
$\frac{\partial ^M \sigma_2(x)}{\partial x^M}=(-1)^M\prod _{k=0}^{k=M-1}(1-\frac{k}{N})(1-\frac{x}{N})^{n-N}$
So :
$\frac{\partial ^M f_n(x)}{\partial x^M }=(-1)^M\prod_{k=0}^{k=M-1}(1-\frac{k}{n})\left((1-\frac{x}{n})^{n-M}-(1-\frac{x}{N})^{n-N}\right)$ .
|
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|
Show that $\int_0^{\frac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta=\pi$
Show that $$\int_0^{\frac{\pi}{2}} \sqrt{\sin\theta}\mathrm d\theta \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin \theta}} \mathrm d \theta=\pi$$
My book wrote that
$$=\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{5}{4})} \cdot \frac{\Gamma(\frac{1}{4}) \Gamma(\frac{1}{2})}{2\Gamma(3/4)}$$
My work says that there wasn't $\Gamma(5/4)$ in the denominator of the left one.
$$\beta(\frac{\frac{1}{2}+1}{2},\frac{1}{2})\beta(\frac{-\frac{1}{2}+1}{2},\frac{1}{2})$$
$$\frac{\Gamma(\frac{3}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{3}{4}+1)}\cdot \frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{1}{4}+\frac{1}{2})}$$
$$\frac{\Gamma(\frac{3}{4})\pi}{2\Gamma(\frac{7}{4})} \cdot \frac{\Gamma(\frac{1}{4})}{2\Gamma(\frac{3}{4})}$$
For that reason, my work doesn't match with their even answer.
|
You made a small calculation error. In the second expression you wrote $1$ instead of $\frac12$:
$$\beta\left(\frac{\color{red}{\frac{1}{2}+1}}{2},\color{green}{\frac{1}{2}}\right)\beta\left(\frac{-\frac{1}{2}+1}{2},\frac{1}{2}\right)$$
$$\frac{\Gamma\left(\color{red}{\frac{3}{4}}\right)\Gamma\left(\color{green}{\frac{1}{2}}\right)}{2\Gamma(\color{red}{\frac{3}{4}}+\color{green}{\frac{1}{2})}}\cdot \frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{2\Gamma(\frac{1}{4}+\frac{1}{2})}$$
$$\frac{\Gamma(\color{red}{\frac{3}{4}})\pi}{2\Gamma(\color{green}{\frac{5}{4}})} \cdot \frac{\Gamma(\frac{1}{4})}{2\Gamma(\frac{3}{4})}$$
|
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|
Why is AM-GM giving 2 contradictory results for the same problem? The problem is the following:
given x and y such that $x+y=1$ and $x,y \geq 0$, find the maximum value of $x^2y$.
The point is, I tried to solve it in 2 different ways, which are shown below, but the second method I used gives a wrong result, and I can't understand where the mistake is.
First Method
Using AM-GM inequality you can say
$ \displaystyle{\left[{\frac{x}{2}\cdot \frac{x}{2} \cdot y}\right]^{\frac{1}{3}} \leq \frac{\frac{x}{2}+\frac{x}{2}+y}{3}}$
But we also have
$\displaystyle{\frac{x}{2}+\frac{x}{2}+y = x+y=1}$
Hence
$\displaystyle{\frac{x^2y}{4} \leq \left(\frac{1}{3}\right)^3 \quad {\LARGE\Rightarrow} \quad x^2y \leq \frac{4}{27}}$
So the solution is
$ max=\frac{4}{27} $, where $x= \frac {2}{3}$ and $y=\frac{1}{3}$
Second Method
Using AM-GM inequality again, we have
$ \displaystyle{\left[x\cdot x \cdot y\right]^{\frac{1}{3}} } \leq \frac{x+x+y}{3}$
Now if you substitute $x+y=1$, you get
$\displaystyle{\left[x\cdot x \cdot y\right]^{\frac{1}{3}} } \leq \frac{x+1}{3}$
Cubing both sides of the inequality, it becomes
$\displaystyle {x^2y\leq \frac{(x+1)^3}{27} }$
It follows that $x^2y$ is maximum when it is equal to the other side of the inequality, so we have:
$\displaystyle {x^2y - \frac{(x+1)^3}{27}= 0 }$
As it is given that $x+y =1$, then $ y=1-x$ and we have a cubic equation which is:
$\displaystyle {x^2(1-x) - \frac{(x+1)^3}{27}= 0 }$
I expected that its solution would be $ x= \frac{2}{3} $ as above, but instead, you can check that its only real solution is $x=y=\frac{1}{2}$
Clearly the correct solution is given by the first method that I showed, as it can also be confirmed by looking for the maximum using the derivative. I just don't know why this happens, apparently I don't see any mistake in the second method.
If anyone is able to come up with an explanation, I would appreciate it, thanks.
|
By your second way the equality occurs for $x=y=\frac{1}{2}$, which gives $x^2y\leq\frac{1}{8},$ which is indeed the maximal value for $x=y$.
But why does $x=y$?
Maybe for $x\neq y$ we can get a greater value?
And indeed, your first solution does not depend on the assuming $x=y$ and it gives a right answer.
|
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|
Does there exist $\ n,m\in\mathbb{N}\ $ such that $\ \left\lvert \left(\frac{3}{2}\right)^n - 2^m \right\rvert < \frac{1}{4}\ $? Does there exist $\ n,m\in\mathbb{N}\ $ such that $\ \left\lvert \left(\frac{3}{2}\right)^n - 2^m \right\rvert < \frac{1}{4}\ $ ?
I have tried for the first few integers $\ n,m\ $ up until $\ m\approx30\ $ with no $\ n,m\ $ satisfying the inequality. However, I can't think of techniques for trying to prove it False. So I'm stuck.
Edit: To be honest, I'm not even sure, for example, how to try to find $\ p,q\in\mathbb{N}_{\geq 2}\ $ such that $\ \lvert 5^p - 7^q \rvert < 10,\ $ which might be an easier type of problem (or harder? I'm not sure...).
Edit:
$$\left(\frac{3}{2}\right)^n - 2^m = \left(\left(\frac{3}{2}\right)^{n/m}\right)^m - 2^m = \left(\left(\frac{3}{2}\right)^{n/m} - 2 \right)\left( \left(\left(\frac{3}{2}\right)^{n/m} \right)^{m-1} + \left(\left(\frac{3}{2}\right)^{n/m} \right)^{m-2} \cdot 2 + \left(\left(\frac{3}{2}\right)^{n/m} \right)^{m-3} \cdot 2^2 + \ldots + \left(\left(\frac{3}{2}\right)^{n/m} \right)^{2} \cdot 2^{m-3} + \left(\left(\frac{3}{2}\right)^{n/m} \right) \cdot 2^{m-2} + \left(\left(\frac{3}{2}\right)^{n/m} \right) \cdot 2^{m-1} \right). $$
Since $\left(\frac{3}{2}\right)^{n/m}\ $ is close to $\ 2,\ $ we therefore have:
$$\left(\frac{3}{2}\right)^n - 2^m \approx \left(\left(\frac{3}{2}\right)^{n/m} - 2 \right)\left( 2^{m-1} +2^{m-2} \cdot 2 + 2^{m-3} \cdot 2^2 + \ldots +2^2 \cdot 2^{m-3} +2 \cdot 2^{m-2} + 2 \cdot 2^{m-1} \right) = \left(\left(\frac{3}{2}\right)^{n/m} - 2 \right)\cdot m \cdot 2^{m-1}.$$
I'm not sure if this helps, but maybe it relates to mjqxxxx's answer. Maybe this is what he/she means by "where "very close" means exponentially close as a function of that rational's denominator".
Edit: This is an open problem in number theory, so perhaps this means the question here is also an open problem?
|
You can also use this (relies on Baker/Rhin): $2^n<2^l-3^n<3^n-2^n$, where the left inequality holds except for $n$ in $\{1,3,5\}$ and the right inequality holds except for $n$ in $\{2\}$ and $l={\lceil n \log_23\rceil}$ is the smallest exponent of $2$ making $2^l-3^n$ positive.
As stated by blamethelag, we can write your inequality like this:
$$ \lvert 3^n - 2^{m+n} \rvert < 2^{n-2}$$
What is next is similar to what Gottfried exposed in his answer (use of transcendence theory). There are 2 cases:
*
*case 1: $2^{m+n}>3^n$ and since $l$ is the smallest possible exponent "$m+n$" for this case we have
$$2^{m+n}-3^n\geq 2^l-3^n>0$$
and using the inequality from first line
$$2^{m+n}-3^n> 2^n>2^{n-2}$$
Note: that taking $2^{n-2}$ as reference removes the exception list mentioned in the introduction.
*case 2: $3^n>2^{m+n}$ and since $l-1$ is the largest possible exponent "$m+n$" for this case we have
$$3^n-2^{m+n}\geq 3^n-2^{l-1}>0$$
and using the inequality from first line
$$2^l-3^n<3^n-2^n$$
$$2\cdot 2^{l-1}-2\cdot 3^n<-2^n$$
$$3^n-2^{l-1}>2^{n-1}$$
you end up with
$$3^n-2^{m+n}>2^{n-1}>2^{n-2}$$
except for $n=2$ from the exception list where we can have equality
which leads to
$$ \lvert 3^n - 2^{m+n} \rvert \ge 2^{n-2}$$
or $\ \lvert \left(\frac{3}{2}\right)^n - 2^m \rvert \ge \frac{1}{4}\ $
|
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|
Solve $xy''+y'-xy=0$ using Frobenius method
Using Frobenius method solve $$xy''+y'-xy=0$$
Comparing the given equation with $y''+P(x)y'+Q(x)y=0$ we have,
$$P(x)=\frac{1}{x}\qquad Q(x)=-1$$
Here $x=0$ is a singular point of the given differential equation. Now
$$
\begin{aligned}
\lim_{x\rightarrow 0}(x-0)P(x)&=1\\
\lim_{x\rightarrow 0}(x-0)^2Q(x)&=0
\end{aligned}
$$
Hence $x=0$ is a regular singular point. The indicial equation is,
$$
\begin{aligned}
r(r-1)+pr+q&=0\\
r^2-r+r&=0\\
r=0
\end{aligned}
$$
$$
\begin{aligned}
y'&=\sum_{k=0}^\infty a_k(k+r)x^{k+r-1}\\
y''&=\sum_{k=0}^\infty a_k(k+r)(k+r-1)x^{k+r-2}
\end{aligned}
$$
Substituting these on the DE,
$$
\begin{aligned}
\sum_{k=0}^\infty(k+r)(k+r-1)a_kx^{k+r-1}+\sum_{k=0}^\infty(k+r)x^{k+r-1}-\sum_{k=0}^\infty a_kx^{k+r+1}&=0\\
\sum_{k=0}^\infty ((k+r)(k+r-1)+(k+r))a_k x^{(k+r-1)} +\sum_{k=2}^\infty a_{k-2}x^{(k+r-1)} &=0\\
r^2a_0x^{r-1}+(r+1)^2a_1x^r+\sum_{k=2}^\infty [(n+r)^2a_n-a_{n-2}]x^{n+r-1} &=0
\end{aligned}
$$
$$r^2a_0=0\qquad (r+1)^2a_1=0\implies a_1=0$$
$$
\begin{aligned}
(n+r)^2a_n-a_{n-2} &=0\\
an&=\frac{a_{n-2}}{(n+r)^2}\\
{a_{n+2}}&=\frac{a_n}{(n+2+r)^2}\\
&=\frac{a_n}{(n+2)^2}\quad\text{ Using }r=0
\end{aligned}
$$
$a_3=a_5=\cdots=a_{2n+1}=0$ because $a_1=0$. Now,
$$
\begin{aligned}
a_2&=\frac{a_0}{2^2}\\
a_4&=\frac{a_2}{4^2}=\frac{a_0}{4^22^2}=\frac{a_0}{2^{2.2}(2.1)^2}\\
a_6&=\frac{a_4}{6^2}=\frac{a_0}{6^24^22^2}=\frac{a_0}{2^{2.3}(3.2.1)^2}\\
&\vdots\\
a_{2n}&=\frac{a_0}{2^{2n}(n!)^2}
\end{aligned}
$$
One solution of the DE is,
$$
y(x)=x^0(\sum_{k=0}^\infty a_kx^k)=a_0(1+\frac{1}{2^2}x^2+\cdots+\frac{1}{2^{2n}n!}x^{2n}+\cdots)
$$
what about another solution? How to get that one?
|
The indicial equation has double root $r=0$. So the second solution will involve $\log x$.
$$
y_1(x)=\, \left(1+{\frac{1}{4}}{x}^{2}+{\frac
{1}{64}}{x}^{4}+\dots\right),
\\
y_2(x) =
\log(x) \left(1+{\frac{1}{4}}{x}^{2}+{\frac{1
}{64}}{x}^{4}+\dots \right) +\left(-{\frac{1}{4}}{x}^
{2}-{\frac{3}{128}}{x}^{4}+\dots\right)
$$
Maple code
|
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"url": "https://math.stackexchange.com/questions/4216722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Exercise on limit of indeterminate form: $\lim_{x \rightarrow 1} \frac{\sqrt{x+3} - \sqrt{5-x}}{\sqrt{1+x} - \sqrt{2}} = \sqrt2$ I'm not able to show that
$\lim_{x \rightarrow 1} \frac{\sqrt{x+3} - \sqrt{5-x}}{\sqrt{1+x} - \sqrt{2}} = \sqrt{2}$
I proceeded as follows:
Substituting 1 to $x$ gives the indeterminate form $\frac{0}{0}$.
I rationalized by multiplying the numerator and the denominator by $\sqrt{1+x} + \sqrt{2}$ but I was not able to prosecute, since I obtained again an indetermiante form $\frac{0}{0}$.
I used
$y = x -1$
$x = y + 1$
and computed the limit for $y \rightarrow 0$ and then I rationalized the denominator, but I was again stuck with the indeterminate form $\frac{0}{0}$.
The exercise is taken from a high school book and the chapter comes before the one about de l'Hopital rule, so it must be solved without using it.
|
If you rationalize the denominator of your expression, the limit becomes$$\lim_{x\to1}\frac{\left(\sqrt{x+3}-\sqrt{5-x}\right)\left(\sqrt{1+x}+\sqrt2\right)}{x-1}.\tag1$$Now define $f(x)=\left(\sqrt{x+3}-\sqrt{5-x}\right)\left(\sqrt{1+x}+\sqrt2\right)$, and then $(1)=f'(1)=\sqrt2$.
|
{
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"url": "https://math.stackexchange.com/questions/4217254",
"timestamp": "2023-03-29T00:00:00",
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|
Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$. I want to find limit of $a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}$ as $n\to \infty$.
$\displaystyle a_{n} =\frac{\left( 2n^{\frac{1}{n}} -1\right)^{n}}{n^{2}} =\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right)^{n}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\log a_{n} =n\log\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right) =\frac{\log\left( 1+\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)\right)}{\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)} .\left(\frac{2n}{n^{\frac{1}{n}}} -\frac{n}{n^{\frac{2}{n}}} -n\right)\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\log\left( 1+\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)\right)}{\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}} -1\right)} .\left(\frac{n^{\frac{1}{n}} -1}{n^{\frac{1}{n}}}\right)^{2} .( -n)
\end{array}$
The first term on RHS has limit equal to $\displaystyle 1\ $but the second term is giving me problem.
Please help. Thanks.
|
Here is yet another estimation of the $\lim_n\frac{(2n^{\frac 1n}-1)^n}{n^2}$.
Since
$$a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}=\Big(\frac{2n^{1/n}-1}{n^{2/n}} \Big)^n=\Big(\frac{-(1-2n^{1/n}+n^{2/n})+n^{2/n}}{n^{2/n}}\Big)^n=\Big(1-\Big(\frac{1-n^{1/n}}{n^{1/n}}\Big)^2\Big)^n$$
we obtain that
$$1\leq a_n\leq \exp\left(n\Big(\frac{1-n^{1/n}}{n^{1/n}}\Big)^2\right)$$
(Here we use the inequality $1+v\leq e^v$ for all $v$). The expression in the exponent can be rewritten as
$$n\Big(\frac{1-n^{1/n}}{n^{1/n}}\Big)^2=\frac{1}{n^{2/n}}\Big(\frac{n^{1/n}-1}{n^{-1/2}}\Big)^2$$
Since $n^{2/n}\xrightarrow{n\rightarrow0}1$, it is enough to check that $\frac{n^{1/n}-1}{n^{-1/2}}$ converges and find the limit. Here we may try L'Hospital rule, for we have a indeterminacy of the type $\frac{0}{0}$ as $n\rightarrow\infty$. Set $f(x)=x^{1/x}-1$ and $g(x)=x^{-1/2}$.
$$\frac{f(x)}{g(x)}=\frac{x^{1/x}-1}{x^{-1/2}}\sim \frac{f'(x)}{g'(x)}=2x^{1/x}\frac{\log x -1}{x^{1/2}}$$
Now $x^{1/x}\xrightarrow{x\rightarrow\infty}1$, while $\frac{\log x-1}{x^{1/2}}\xrightarrow{x\rightarrow\infty}0$ (this follows from a the well known limit $\lim_{u\rightarrow\infty}\frac{u^\alpha}{(1+p)^u}=0$ for any $p>0$ and $\alpha\in\mathbb{R}$) Consequently
(by L'Hospital theorem) $\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=L$ exists and equals $L=0$. Putting things together, we obtain that
$$
\exp\left(n\Big(\frac{1-n^{1/n}}{n^{1/n}}\Big)^2\right)\xrightarrow{n\rightarrow\infty}1$$
Therefore, $\lim_{n\rightarrow\infty}a_n=1$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Find the volume of the region which lies inside both $x^2+y^2=r^2$ and $y^2+z^2=r^2$ Find the region inside two cylinders $x^2+z^2 \le r^2$ and $y^2+z^2 \le r^2$.
I attempted this question by combining the two inequalities $x^2+2z^2+y^2 \le 2r^2$ and I got the bounds for $z$ which is $-\sqrt{{(2r^2-x^2-y^2)}/{2}} \le z \le \sqrt{{(2r^2-x^2-y^2)}/{2}}$.
And I found the bound on $y$ to be $-\sqrt{(2r^2-x^2)} \le y \le \sqrt{(2r^2-x^2)}$, the bound on x is $0 \le x \le r$.
I am not sure if the above attempt is correct, if not, can anyone provide the correct way to find the bounds on z and y?
|
If you visualize it carefully, you might be able to see that for fixed $y$ the cross-section is a square. Then:
$$|x| \le \sqrt{r^2 - y^2},$$
$$|z| \le \sqrt{r^2 - y^2}.$$
Then, the cross-section is a square of area $4 (r^2 - y^2)$ and the volume should be:
$$\int_{-r}^r dy 4 (r^2 - y^2) = 16 r^3/3.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4220114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Evaluate $\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2\sin^2(\theta) + b^2\cos^2(\theta)} \ d\theta$ Evaluate
$$\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2\sin^2(\theta) + b^2\cos^2(\theta)} \ d\theta$$
I transformed it into:
$$\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2(1-\cos^2(\theta)) + b^2\cos^2(\theta)} \ d\theta\\ \rightarrow \frac{b}{\sqrt{b^2-a^2}}\int_{0}^{2\pi} |\sin(\theta)| \sqrt{\frac{a^2}{b^2-a^2} + \cos^2(\theta)} \ d\theta$$
Using $u = \cos(\theta)$ I can make the substitution:
$$\frac{-b}{\sqrt{b^2-a^2}}\int_{1}^{0} \sqrt{\frac{a^2}{b^2-a^2} + u^2} \ du$$
Now I want to use the formula for $\int \sqrt{x^2 + a^2} dx$ but that will give me an expression of $\ln(x)$ I somehow need to end up with :
$$b^2 +\frac{a^2b}{\sqrt{a^2-b^2}}\arcsin(\frac{\sqrt{a^2-b^2}}{a})$$
Does anyone see a route to get this?
|
Assuming $b^2-a^2>0$,
$$
\begin{align}
&\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2(1-\cos^2(\theta)) + b^2\cos^2(\theta)} \ d\theta\\
& =b\sqrt{b^2-a^2}\left[\int_{0}^{\pi} \sin(\theta) \sqrt{\frac{a^2}{b^2-a^2} + \cos^2(\theta)} \ d\theta+\int_{\pi}^{2\pi} \sin(\theta)\sqrt{\frac{a^2}{b^2-a^2} + \cos^2(\theta)} \ d\theta\right]\\
&=b\sqrt{b^2-a^2}\left[-\int_1^{-1}\mathrm{d}x\sqrt{\frac{a^2}{b^2-a^2}+x^2}-\int_{-1}^{0}\mathrm{d}x\sqrt{\frac{a^2}{b^2-a^2}+x^2} \right]\\
&=b\sqrt{b^2-a^2}\int_0^1\mathrm{d}x\sqrt{\frac{a^2}{b^2-a^2}+x^2}\\
&=b\sqrt{b^2-a^2}\frac{1}{\sqrt{b^2-a^2}}\left[b+a\log\left(\frac{b+\sqrt{b^2-a^2}}{a}\right)\right].
\end{align}
$$
The last equality can be safely left to the reader. Using the identity $i \arcsin x = \ln(i x + \sqrt {1 - x^2} )$ we get the expression you wrote in the bottom.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given a polynomial $W(x)$. Find all pairs of integers $a,b$ that satisfy $W(a)=W(b)$ Given a polynomial $W(x) = x^4-3x^3+5x^2-9x$. Find all pairs of distinct integers $a,b$ that satisfy $W(a)=W(b)$. My approach was to factor the polynomial.
$$\begin{align*}
x^4-3x^3+5x^2-9x &= (x^4-3x^3+2x^2)+(3x^2-9x+6)-6\\
&= x^2(x^2-3x+2)+3(x^2-3x+2)-6\\
&=(x-1)(x-2)(x^2+3)-6
\end{align*}$$
As $-6$ is a constant we can now consider the following expression $$\\(x-1)(x-2)(x^2+3)$$ form this form we see that $(a,b)=(1,2)$ and $(a,b)=(2,1)$ satisfy the condition. We can also graph this function and see that $(1)$WLOG $a\ge 2$ and $b\le 1$ but as $1$ and $2$ have been used we get $a\ge 3$ and $b\le 0$. Now, if $|b|>|a|$ the following holds $$\\(b-1)(b-2)(b^2+3)>(a-1)(a-2)(a^2+3)$$
So we get that $|a|>|b|$ and $(1)$. How do I proceed? Any help appreciated.
|
I would proceed from there as follows :
One can say that if $a\geqslant 3,b\leqslant 0$ and $|b|\color{red}{\geqslant}|a|$, then $$\\(b-1)(b-2)(b^2+3)>(a-1)(a-2)(a^2+3)$$
So one gets that $a\geqslant 3,b\leqslant 0$ and $|a|>|b|$.
Note that if $a\geqslant 3$ and $b\leqslant 0$, then $|a|\gt |b|$ is equivalent to $a+b\gt 0$.
One has
$$\begin{align}0&=(a-1)(a-2)(a^2+3)-(b-1)(b-2)(b^2+3)
\\\\&=(a - b) (a^3+ b^3 + a^2 b + a b^2 - 3 a^2 - 3 b^2- 3 a b + 5 a + 5 b - 9)
\\\\&=(a-b)\bigg((a+b)^2(a+b-3)-ab(2(a+b)-3)+5(a+b)-9\bigg)\end{align}$$
So, if $a\geqslant 3,b\leqslant 0$ and $a+b\geqslant 3$, then
$$0=(\underbrace{a-b}_{\text{positive}})\bigg(\underbrace{(a+b)^2(a+b-3)-ab(2(a+b)-3)}_{\text{non-negative}}+\underbrace{5(a+b)-9}_{\text{positive}}\bigg)$$
where LHS is zero while RHS is positive, which is impossible.
So, if $a\geqslant 3,b\leqslant 0$ and $a+b\gt 0$, then $a+b=1$ or $2$.
*
*If $b=1-a$, then $0 = (2 a - 1) (a^2 - a + 6)$ has no integer solutions.
*If $b=2-a$, then $0 = (a - 3) (a - 1) (a + 1)$ implies $a=3$.
In conclusion, the answer is
$$(a,b)=(2,1),(1,2),(3,-1),(-1,3)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sum_{i,j=1}^N \frac{1}{(i^2 + j^2)}$ (i..e. $1/r^2$ over the square lattice $i,j\in\{1,2,\ldots,N\}$) Does anyone have any insight as to how I might find a closed-form expression for
$\sum_{i=1}^N \sum_{j=1}^N\frac{1}{(i^2 + j^2)}$ ?
It feels like it should be straightforward because it's easy to do as an integral in polar coordinates if you include the axes. But here the summation is over a square lattice, and excludes the axis. Wolfram Alpha can't do it, and neither can I!
|
An asymptotic formula for large $N$ may be derived as follows. First,
$$
\sum\limits_{m = 1}^N {\frac{1}{{m^2 + n^2 }}} = - \frac{1}{{2n^2 }} + \frac{\pi }{{2n}}\coth (\pi n) - \frac{{\operatorname{Im} \psi (1 + N + \mathrm{i}n)}}{n}
$$
where $\psi$ is the digamma function. From the known asymptotics of the digamma function,
\begin{align*}
\operatorname{Im} \psi (1 + N + \mathrm{i}n) &= \arg (N + \mathrm{i}n) + \frac{n}{{2(n^2 + N^2 )}} + \mathcal{O}\!\left( {\frac{1}{{N^2 }}} \right) \\ & = \arctan \left( {\frac{n}{N}} \right) + \frac{n}{{2(n^2 + N^2 )}} + \mathcal{O}\!\left( {\frac{1}{{N^2 }}} \right).
\end{align*}
Now
\begin{align*}
\sum\limits_{n = 1}^N {\frac{1}{n}\arctan \left( {\frac{n}{N}} \right)} & = \frac{1}{N}\sum\limits_{n = 1}^N {\frac{N}{n}\arctan \left( {\frac{n}{N}} \right)} \\ & = \int_0^1 {\frac{{\arctan x}}{x}\mathrm{d}x} + \mathcal{O}\!\left( {\frac{1}{N}} \right) = G + \mathcal{O}\!\left( {\frac{1}{N}} \right)
\end{align*}
where $G$ is Catalan's constant. Also
$$
\sum\limits_{n = 1}^N {\frac{1}{{2(n^2 + N^2 )}}} = \mathcal{O}\!\left( {\frac{1}{N}} \right),\;\, \sum\limits_{n = 1}^N {\mathcal{O}\!\left( {\frac{1}{{N^2 }}} \right)} = \mathcal{O}\!\left( {\frac{1}{N}} \right),\;\, \sum\limits_{n = 1}^N { - \frac{1}{{2n^2 }}} = - \frac{{\pi ^2 }}{{12}} + \mathcal{O}\!\left( {\frac{1}{N}} \right)
$$
and
\begin{align*}
\sum\limits_{n = 1}^N {\frac{\pi }{{2n}}\coth (\pi n)} & = \frac{\pi }{2}\sum\limits_{n = 1}^N {\frac{1}{n}} + \frac{\pi }{2}\sum\limits_{n = 1}^N {\frac{{\coth (\pi n) - 1}}{n}} \\ & = \frac{\pi }{2}\log N + \frac{\pi }{2}\gamma + \frac{\pi }{2}\sum\limits_{n = 1}^\infty {\frac{{\coth (\pi n) - 1}}{n}} + \mathcal{O}\!\left( {\frac{1}{N}} \right),
\end{align*}
using the asymptotics of the harmonic numbers ($\gamma$ is the Euler–Mascheroni constant). In summary
$$
\sum\limits_{n = 1}^N {\sum\limits_{m = 1}^N {\frac{1}{{m^2 + n^2 }}} } = \frac{\pi }{2}\log N + C + \mathcal{O}\!\left( {\frac{1}{N}} \right)
$$
with
$$
C = \frac{\pi }{2}\gamma - G - \frac{{\pi ^2 }}{{12}} + \frac{\pi }{2}\sum\limits_{n = 1}^\infty {\frac{{\coth (\pi n) - 1}}{n}} =-0.825861175988396\ldots .
$$
Addendum. Note that, using the Dedekind $\eta$-function,
$$
\sum\limits_{n = 1}^\infty {\frac{{\coth (\pi n) - 1}}{n}} = - 2\log ({\rm e}^{\pi /12} \eta ({\rm i})) = 2\log \left( {\frac{{2\pi ^{3/4} }}{{\Gamma (1/4)}}} \right) - \frac{\pi }{6}
$$
and hence
$$
C = \frac{\pi }{2}\gamma - G - \frac{{\pi ^2 }}{6} + \pi \log \left( {\frac{{2\pi ^{3/4} }}{{\Gamma (1/4)}}} \right).
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Continuous function satisfying $f\left( {2{x^2} - 1} \right) = \left( {{x^3} + x} \right)f\left( x \right)$ If $f\colon\left[ { - 1,1} \right] \to \mathbb R$ be continuous function satisfying $f\left( {2{x^2} - 1} \right) = \left( {{x^3} + x} \right)f\left( x \right)$, then $\mathop {\lim }\limits_{x \to 0} \frac{{f\left( {\cos x} \right)}}{{\sin x}}$ is _______.
My solution is as follow
$x = \cos \left( {\frac{\theta }{2}} \right)$
$f\left( {2{{\cos }^2}\left( {\frac{\theta }{2}} \right) - 1} \right) = \left( {{{\cos }^3}\left( {\frac{\theta }{2}} \right) + \cos \left( {\frac{\theta }{2}} \right)} \right)f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)$
$\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} = \frac{{\left( {{{\cos }^3}\left( {\frac{\theta }{2}} \right) + \cos \left( {\frac{\theta }{2}} \right)} \right)}}{{2\sin \left( {\frac{\theta }{2}} \right)\cos \left( {\frac{\theta }{2}} \right)}}f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)$
$\frac{{f\left( {\cos \theta } \right)}}{{\sin \theta }} = \frac{{\left( {{{\cos }^2}\left( {\frac{\theta }{2}} \right) + 1} \right)}}{{2\sin \left( {\frac{\theta }{2}} \right)}}f\left( {\cos \left( {\frac{\theta }{2}} \right)} \right)$
How do I proceed from here
|
First we note that $f$ is odd. The left hand side $f(2x^2-1)$ is even, and the factor $x^3+x$ in the right hand side is odd, so the other factor in the right hand side, $f(x),$ must also be odd. This implies that $f(0)=0.$
Taking $x=\sin\theta$ gives
$$
f(2\sin^2\theta - 1) = \sin\theta(\sin^2\theta+1) f(\sin\theta)
$$
so
$$
\frac{f(\cos 2\theta)}{\sin 2\theta}
= \frac{f(-(2\sin^2\theta - 1))}{2 \sin\theta \cos\theta}
= \frac{-\sin\theta(\sin^2\theta+1) f(\sin\theta)}{2 \sin\theta \cos\theta} \\
= -\frac{(\sin^2\theta+1) f(\sin\theta)}{2 \cos\theta}
\to -\frac{(0^2+1) f(0)}{2\cdot 1}
= 0
$$
as $\theta\to 0.$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Show that the following non-linear group of equations only have zero solution
If
$$\left\{\begin{array}{c}
(\lambda_1 + 1)(\lambda_2 + 1) \cdots (\lambda_n + 1) = 1 \\
(\lambda_1^2 + 1)(\lambda_2^2 + 1) \cdots (\lambda_n^2 + 1) = 1 \\
\vdots \\
(\lambda_1^n + 1)(\lambda_2^n + 1) \cdots (\lambda_n^n + 1) = 1
\end{array}\right.$$
with $\lambda_1, \lambda_2, \ldots, \lambda_n \in \Bbb{C}$, then
$$\lambda_1 = \lambda_2 = \cdots = \lambda_n = 0.$$
Sorry I misunderstood the question before, actually it is stated as follows:
If
$$
\
(\lambda_1^k + 1)(\lambda_2^k + 1) \cdots (\lambda_n^k + 1) = 1\\
.$$
hold for all positive integers $k$, with $\lambda_1, \lambda_2, \ldots, \lambda_n \in \Bbb{C}$, then
$$\lambda_1 = \lambda_2 = \cdots = \lambda_n = 0.$$
|
In the case of $n=2$ you have
$$
a+b+ab=0\\
a^2+b^2+a^2b^2=0
$$
The second equation transforms to
$$
0=(a+b)^2-2ab+a^2b^2=2(a^2b^2-ab).
$$
The solution $ab=0$ leads to $a=b=0$. The other solution $ab=1$ gives $a+b=1$, so that $a,b$ are the solutions of the quadratic $x^2-x+1=0$ which has non-zero solutions $\frac{1\pm i\sqrt3}2$.
The claim needs some work.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $(x^2 - y^2) = (x + y)(x-y)$ I am trying to prove from the field axioms that $(x^2 - y^2) = (x+y)(x-y)$. I am going to take for granted the definition of subtraction as the addition of a negation and that $x(-y) = (-x)y = -(xy)$. Here is my attempt.
We have:
\begin{align*}
(x+y)(x-y) & = x(x-y) + y(x-y) & & \text{distributive law} \\
& = x(x + (-y)) + y(x + (-y)) & & \text{definition of subtraction} \\
& = (xx + x(-y)) + (yx + y(-y)) & & \text{distributive law} \\
& = (x^2 + (-(xy))) + (yx + (-y^2)) & & \text{uniqueness of additive inverse; definition of square} \\
& = (x^2 + (-(yx))) + (yx + (-y^2)) & & \text{commutativity of multiplication} \\
& = (x^2 + \left(-(yx) + yx \right) + (-y^2) & & \text{associativity of addition} \\
& = (x^2 + 0) + (-y^2) & & \text{additive inverse axiom} \\
& = x^2 + (-y^2) & & \text{additive identity axiom} \\
& = x^2 - y^2 & & \text{definition of subtraction}
\end{align*}
How does this look?
|
Another way to approach it for the sake of curiosity:
\begin{align*}
x^{2} - y^{2} & = (x^{2} - xy) + (xy - y^{2})\\\\
& = x(x-y) + y(x - y)\\\\
& = (x+y)(x-y)
\end{align*}
Hopefully this helps!
|
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"url": "https://math.stackexchange.com/questions/4231411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
How to show $\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \Sigma_{k=1}^n \sqrt{k-1} = \frac{2}{3}$? How to show $$\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \sum_{k=1}^n \sqrt{k-1} = \frac{2}{3}$$
In fact, this is the lower sum of the integral of $\sqrt{x}$ from 0 to 1. So the value of the above must be $\frac{2}{3}$. But how to show?
|
Using right Riemann sums, \begin{align*}
\frac{1}{{\sqrt n }}\frac{1}{n}\sum\limits_{k = 1}^n {\sqrt {k - 1} } & = \frac{1}{{\sqrt n }}\frac{1}{n}\sum\limits_{k = 1}^{n - 1} {\sqrt k } = \frac{1}{n}\sum\limits_{k = 1}^{n - 1} {\sqrt {\frac{k}{n}} } \\ & = \left( {\frac{1}{n}\sum\limits_{k = 1}^n {\sqrt {\frac{k}{n}} } } \right) - \frac{1}{n} \to \int_0^1 {\sqrt x dx} - 0 = \frac{2}{3}.
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4232317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
What should be the winning strategy for Bob? Alice and Bob are playing a calculator game in which the calculator can only display positive integers and is used like this: starting with the integer $x$ one player types an integer $1<=y<=99$ in and if $y\%$ of $x$ (meaning $\frac{xy}{100}$) is again an integer the calculator shows that result, otherwise the job fails and the one whose turn it was loses. How many starting numbers $1<=x<=2017$ guarantee a winning strategy for Bob who plays second?
|
Let $x = 2^a \cdot 5^b$ and $y = 2^c \cdot 5^d$, each round the calculator does $\frac{xy}{100} = 2^{a+c-2} \cdot 5^{b+d-2}$. The goal is to either reduce both exponents to zero so the opponent can't reach another integer or to reduce them just enough so the opponent can't win in the next move. Essentially there are $14$ moves (or $y$-values):
$$\begin{array}{c|c|c} y & c-2 & d-2 \\ \hline 1 & -2 & -2 \\ 2 & -1 & -2 \\ 5 & -2 & -1 \\ 10 & -1 & -1 \\ \hline 4 & \pm 0 & -2 \\ 20 & \pm 0 & -1 \\ 25 & -2 & \pm 0 \\ 50 & -1 & \pm 0 \end{array} \qquad \begin{array}{c|c|c} y & c-2 & d-2 \\ \hline 8 & +1 & -2 \\ 16 & +2 & -2 \\ 32 & +3 & -2 \\ 64 & +4 & -2 \\ 40 & +1 & -1 \\ 80 & +2 & -1 \end{array}$$
If the starting number is not divisible by $2$ or $5$, Alice would lose on the first move. There are only $33$ tuples $(a,b)$ for which $x<2017$ and the only ones where Bob has a guaranteed winning strategy are
$$(3,0) \to x = 8 \quad (6,0) \to 64 \quad (9,0) \to 512 \quad (0,3) \to 125 \quad (3,3) \to 1000$$
In the words of gnasher729's answer, call them "losers" for Alice. All other tuples can be reduced in one move to these and become therefore "losers" for Bob. So the starting numbers from the tuples above for which Bob can win are $n \cdot 2^a \cdot 5^b$ with $n$ not divisible by $2$ or $5$, and according to my calculations there are $123$ of them
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find min $ P=\frac{x}{x+2}+\frac{y}{y+2}+\frac{z}{z+2}$ under $x,y,z >0$ and $xyz=1$ Let $x,y,z >0$ and $xyz=1$
Find min: $P=\frac{x}{x+2}+\frac{y}{y+2}+\frac{z}{z+2}$
My trying but it false :<<
We have: $VT=\frac{1}{\frac{x+2}{x}}+\frac{1}{\frac{y+2}{y}}+\frac{1}{\frac{z+2}{z}}$
$=\frac{1}{1+\frac{2}{x}}+\frac{1}{1+\frac{2}{y}}+\frac{1}{1+\frac{2}{z}} $
-> $Bunhiacopxki$:
$⇒P \geq \frac{9}{3+2.(\frac{xy+yz+xz}{xyz})}=\frac{9}{3+2.(xy+yz+xz)}$
We have $\frac{(x+y+z)^2}{3} \geq xy+yz+xz $ and $x+y+z \geq 3\sqrt[]{xyz}=3$
(it flase, please don't talking about that :<<)
$⇒P \geq \frac{9}{3+2.(xy+yz+xz)} \geq \frac{9}{6}=\frac{3}{2}$
$=>P \geq \frac{3}{2}$
$x=y=z=1$
|
Let $x=a/b$ and $y=b/c,$ where $a,$ $b$ and $c$ are positives.
Thus, $z=c/a$ and by C-S we obtain: $$\sum_{cyc}\frac{x}{x+2}=\sum_{cyc}\frac{a}{a+2b}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+2ab)}=1$$
Can you end it now?
|
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"timestamp": "2023-03-29T00:00:00",
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|
$0~$ division error occurs as I use $~ t= \tan^{}\left( \frac{\theta_{}}{ 2 } \right) ~$ of the integration. I have to integral the below formula of integration .
$$ a, d \in\mathbb{R}_{>0} ~~\wedge~~ d > a $$
$$ \alpha := \int_{0 }^{\frac{\pi}{2} } \frac{ d }{ d- a \cdot \cos^{}\left(\theta_{} \right) } \,d\theta $$
$$ = d\int_{0 }^{\frac{\pi}{2} } \frac{ 1 }{ d- a \cdot \cos^{}\left(\theta_{} \right) } \,d\theta $$
My tries are as below so far . By the way , I wonder if such system exists which enable 2-column-display to be used against the below equations .
$$ t= \tan^{}\left( \frac{\theta_{}}{ 2 } \right) $$
$$ \theta_{} : 0 ~\rightarrow~ \frac{\pi}{2} $$
$$ t : 0 ~\rightarrow~1= \tan^{}\left( \frac{ 1 }{ 2 } \cdot \frac{\pi}{2} \right) $$
$$ \frac{ dt }{ d\theta } = \frac{1}{2} \sec^2\left( \frac{\theta}{ 2 } \right) $$
$$ \frac{ d\theta }{ dt } = \frac{ 2 }{ \sec^2\left( \frac{\theta}{2} \right) } $$
$$ 1+ t^2 = \sec^2\left( \frac{\theta}{2} \right) $$
$$ \therefore ~~ d\theta = \frac{ 2 \,dt }{ 1+ t^2 } = 2 \, dt \left( 1 + t^2 \right)^{-1} $$
$$ \cos(\theta) = 2 \cos^2\left(\frac{\theta}{ 2 } \right) -1 $$
$$ = 2 \left( \cos\left(\frac{\theta}{ 2 } \right) \right)^2 -1 $$
$$ = 2 \left( \sec^{-1}\left( \frac{\theta}{2} \right) \right)^2 -1 $$
$$ = 2 \left( \sec^2\left( \frac{\theta}{2} \right) \right)^{-1} -1 $$
$$ \therefore ~~ \cos(\theta) = 2(1+ t^2)^{-1} -1 $$
$$ \cos(\theta) = \frac 2 {1+ t^2} -1 $$
$$ = \frac 2 {1 + t^2} - \frac{ (1+ t^2) }{1+t^2} $$
$$ = \frac{ 2 - (1 + t^2) }{ 1+ t^2} $$
$$ = \frac{ 2-1-t^2}{1+ t^2} $$
$$ = \frac{ 1- t^2}{ 1 + t^2} = \cos(\theta) = (1-t^2) (1+t^2)^{-1} $$
$$ \alpha = \int_0^{\frac{\pi}{2} } \frac d { d- a \cos(\theta)} \,d\theta $$
$$ = d\int_{0 }^{\frac{\pi}{2} } \frac{ d\theta }{ d- a \cdot \cos^{}\left(\theta_{} \right) } $$
$$ = d\int_{0 }^{1 } \left( \frac{ 2 \, dt \left( 1 + t ^{2} \right) ^{-1} }{ d- a \cdot \left\{ \left( 1- t ^{2} \right) \left( 1+t ^{2} \right) ^{-1} \right\} } \right) $$
$$ = d \int_{0 }^{1 } \frac{ 2 \,dt (1+ t^2)^{-1} }{ d - a(1- t^2) (1+ t^2)^{-1}} $$
$$ = 2d \int_0^1 \frac{ (1+ t^2)^{-1} \,dt }{ d - a(1-t^2) (1+t^2)^{-1}} $$
$$ = 2d \int_0^1 \frac{(1+t^2)^{-1} \, dt }{ \left\{d - a(1- t^2) (1+ t^2)^{-1} \right\} } \cdot \frac{ (1 + t^2)^1 }{ (1 + t^2)^1} $$
$$ = 2 d \int_{0 }^{1 } \frac{ dt }{ d \left( 1+ t ^{2} \right) -a \cdot \left( 1-t ^{2} \right) } $$
$$ = 2 d \int_{0 }^{1 } \frac{ 1 }{ d + d \cdot t ^{2} - a + a \cdot t ^{2} } dt $$
$$ = 2 d \int_{0 }^{1 } \frac{ 1 }{ \left( d-a \right) + \left( d+a \right)t ^{2} } dt $$
$$ = 2d \left[ \frac{ \ln\left( \left( d-a \right) + \left( d+ a \right) t ^{2} \right) }{ 2 \left( d + a \right)t } \right]_{0}^{1} ~~ \leftarrow~~ \text{As } t= 0 ~~\text{is set , the zero division error occurs .} $$
I need hints to solve $~ \alpha ~$
|
Use the below general equation to proceed the given integration .
$$ a \in \mathbb R_{> 0} $$
$$ \frac{ a }{ a ^{2} + x ^{2} } = \frac{ d }{ dx } \left( \tan^{-1} \left( \frac{ x }{a } \right) \right) $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Wrong answer in Thomas Calculus 14th Edition textbook There is this question on derivatives to which the answer is given as $\frac{43}{75}$rad/sec in the answers section. This answer appears to be wrong.
My Solution
$$
\theta+\tan^{-1}\left(\frac{6}{4-x}\right)+\tan^{-1}\left(\frac{3}{x}\right)=\pi
$$
which gets reduced to
$$
\theta=\pi-\tan^{-1}\left(\frac{3x+12}{4x-x^2-18}\right)
$$
Therefore,
$$
\frac{d\theta}{dt}=\left(\frac{3x^2+24x-102}{(4x-x^2+18)^2+(3x+12)^2}\right)\frac{dx}{dt}
$$
Given $x=4$ and $\frac{dx}{dt}=2\text{ cm/sec}$,
$$
\frac{d\theta}{dt}=-\frac{7}{75}\text{rad/sec}
$$
I've got the graph of $\theta$ as a function of $x$ here, which also indicates my answer is correct. Or am I? Kindly help.
|
Differentiating your formula
$$\theta+\arctan\frac{6}{4-x}+\arctan\frac{3}{x}=\pi$$
with respect to time gives
$$\dot\theta+\frac1{1+\left(\dfrac6{4-x}\right)^2}\cdot\frac6{({4-x})^2}\dot x+\frac1{1+\left(\dfrac3x\right)^2}\cdot\frac{-3}{x^2}\dot x=0.$$
After substituting $\dot x=2$, this may be written
$$\dot\theta=6\left(\frac1{x^2+9}-\frac2{(4-x)^2+36}\right).$$
Now substituting $x=4$ yields the answer $\dot\theta=-\frac7{75}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $a,b,c$ be non-negative real numbers .Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $ Let $a,b,c$ be non-negative real numbers
Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
My idea is to use the $(p,q,r)$ method:
$p=a+b+c$
$q=ab+bc+ca$
$r = abc $
$\Rightarrow a^2+b^2+c^2 = p^2-2q $
$ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
or
$ p^2-2q +\sqrt{2} r + 2\sqrt{2} +3 \geq (2+\sqrt {2} )p $
or
$ (p - (1+\sqrt{2} ))^2 -2q +\sqrt{2} r \geq 0 $
or
The problem I am facing is exactly what I want to prove : $\sqrt{2} r \geq 2q$ is completely wrong .
I hope to get help from everyone. Thanks very much !
|
Another way.
Since $$\prod_{cyc}(a-1)^2=\prod_{cyc}((b-1)(c-1))\geq0,$$ we can assume $(b-1)(c-1)\geq0$, which gives $$a(b-1)(c-1)\geq0$$ or$$abc\geq ab+ac-a$$ and since $$b^2+c^2\geq\frac{1}{2}(b+c)^2,$$ it's enough to prove that
$$a^2+\frac{1}{2}(b+c)^2+\sqrt2(ab+ac-a)+2\sqrt2+3\geq(2+\sqrt2)(a+b+c)$$ or $$(\sqrt2a+b+c-2-\sqrt2)^2\geq0$$ and we are done!
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4237369",
"timestamp": "2023-03-29T00:00:00",
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|
Integral Involving Bessel function and trigonometric function Consider the following integral
\begin{equation}
\int_0^{2 \pi} \int_0^{\pi} e^{A \cos \phi \cos \theta} \, I_0(B \sin \phi \sin \theta) \sin \theta \, d\theta \, d\phi
\label{1}
\end{equation}
where $A$ and $B$ are two constants and $I_0(.)$ is the modified Bessel function of the first kind.
To evaluate the integral, I tried to write the expansion of $I_0$ which (by Methemtica) results in
$$
\sum_{n=0}^{\infty} \frac{1}{4^n} \frac{1}{(n!)^2} \int_0^{2 \pi} \int_0^{\pi} e^{A \cos \phi \cos \theta} \, (B \sin \phi \sin \theta)^{2n} \sin \theta \, d\theta \, d\phi
$$
$$
= \sum_{n=0}^{\infty} \frac{1}{4^n} \frac{1}{(n!)^2} \sqrt{\pi} \, \Gamma(n+1) \int_0^{2 \pi} \operatorname{HGR}_{01}\left(n+\frac{3}{2}, \frac{1}{4} A^2 \cos^2 \phi\right) \, (B^2 \sin^2 \phi)^{n} \sin \theta \, d\phi
$$
where $\operatorname{HGR}_{01}$ is the regularized confluent hypergeometric function.
Again, by expanding this function I found (up to some constants)
$$
\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} A'^m B'^n \frac{\Gamma(n+\frac{1}{2}) \Gamma(m+\frac{1}{2})\Gamma(m + n + \frac{3}{2})}{\Gamma(n+1) \Gamma(m+1)\Gamma(m+n+1)}
$$
where $A',B'$ are (properly) rescaled version of $A,B$.
I couldn't go further, and stucked with this double sum.
I appreciate if anyone can provide me some insights on this sum, or any hints on the original integral .
|
In this paper by VNP Anghel, a result obtained by Glasser is generalized (eq. 32):
\begin{equation}
I_{\frac{m+1}{2}}(b)=\left( \frac{b}{2\pi \sin^m\alpha}\right)^{1/2} \int_0^\pi \exp(b\cos\alpha\cos\theta)I_{\frac m2}(b\sin\alpha\sin\theta)(\sin\theta)^{\frac{m+2}{2}}\,d\theta \tag{1}\label{eq1}
\end{equation}
Then, in the case $A=B=b$, by choosing $m=0$, we have directly
\begin{align}
J(b,b)&=\int_0^{2 \pi} \int_0^{\pi} e^{b \cos \phi \cos \theta} \, I_0(b \sin \phi \sin \theta) \sin \theta \, d\theta \, d\phi\\
&=\int_0^{2 \pi} \, d\phi \left( \frac{2\pi}{b} \right)^{1/2}I_{1/2}(b)\\
&=4\pi\frac{\sinh b}{b}
\end{align}
The result \eqref{eq1} may also be used in the case $A\ne B$ to express the integral as a series. We remark first that $A$ can be chosen to be positive (parity of the integral wrt $A$ is clear from the substitution $\phi\to \pi-\phi$ in the integral). By the multiplication theorem,
\begin{equation}
I_{\nu}\left(\lambda z\right)=\lambda^{\nu}\sum_{k=0}^{\infty}
\frac{(\lambda^{2}-1)^{k}(\frac{1}{2}z)^{k}}{k!}I_{\nu+ k}\left(z
\right)
\end{equation}
with $\nu=0,\lambda=B/A,z=A\sin\phi\sin\theta$, one can express
\begin{equation}
I_0\left(B\sin\phi\sin\theta\right)=\sum_{k=0}^{\infty}
\frac{\left( B^2-A^2 \right)^k}{2^kk!A^k}\sin^k\phi\sin^k\theta \,I_{ k}\left(A\sin\phi\sin\theta\right)
\end{equation}
and thus, by interverting integral and summation,
\begin{align}
J(A,B)&=\int_0^{2 \pi} \int_0^{\pi} e^{A \cos \phi \cos \theta} \, I_0(B \sin \phi \sin \theta) \sin \theta \, d\theta \, d\phi\\
&=\sum_{k=0}^{\infty}\frac{\left( B^2-A^2 \right)^k}{2^kk!A^k}\int_0^{2 \pi}\sin^k\phi\,d\phi
\int_0^{\pi} e^{A \cos \phi \cos \theta}I_{ k}\left(A\sin\phi\sin\theta\right) \sin^{k+1}\theta \, d\theta
\end{align}
With $m=2k$ in eq. \eqref{eq1}, it may be expressed as
\begin{align}
J(A,B)&=\sum_{k=0}^{\infty}\frac{\left( B^2-A^2 \right)^k}{2^kk!A^k}\int_0^{2 \pi}\left( \frac{2\pi\sin^{2k}\phi}{A} \right)^{1/2}\sin^k\phi \,I_{k+1/2}(A)\,d\phi\\
&=\sqrt{\frac{2\pi}{A}}\sum_{k=0}^{\infty}\frac{\left( B^2-A^2 \right)^k}{2^kk!A^k}I_{k+1/2}(A)\int_0^{2 \pi}\sin^{2k}\phi\,d\phi\\
&=\frac{(2\pi)^{3/2}}{\sqrt{A}}\sum_{k=0}^{\infty}\frac{(2k)!}{2^{3k}(k!)^3}\frac{\left( B^2-A^2 \right)^k}{A^k}I_{k+1/2}(A)
\end{align}
Numerical experiments show that this series converges very quickly, which is related to the asymptotic expansion of the modified Bessel function for large orders (DLMF).
|
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|
Nonlinear ODE and $_2\text{F}_1$ Consider the differential equation $y'(x) = B \, y(x)^{-b} - A \, y(x)^a$ to which WA returns the following:
$$x - c_1 = \frac{y(x)^{b+1}}{B\,(b+1)} \; _2\text{F}_1 \left( 1, \frac{b+1}{a+b}; 1 + \frac{b+1}{a+b}; \frac{A}{B} \, y(x)^{a+b} \right)$$
Is there a way to turn this into an explicit expression for $y(x)$ or to approximate the solution? Thanks for your help.
|
Firstly, the question should have your ideas on how to solve the problem.
The answer is yes, but it needs a limit since your hypergeometric function is a Lerch Transcendent. Here is the “closed form” solution using the Wolfram Language’s Incomplete Beta function $\text B_z(a,b)$ and Inverse Beta Regularized $\text I^{-1}_z(a,b)$:
$$x - c= \frac{y^{b+1}}{B\,(b+1)} \; _2\text{F}_1 \left( 1, \frac{b+1}{a+b}; 1 + \frac{b+1}{a+b}; \frac{A}{B} \, y^{a+b} \right)= \frac{y^{b+1}}{B\,(b+1)} \frac{b+1}{a+b}\left( \frac{A}{B} \, y^{a+b} \right)^{-\frac{b+1}{a+b}}\text B_{ \frac{A}{B} \, y^{a+b} }\left(\frac{b+1}{a+b},0\right)\mathop=^{a,b\in\Bbb R^+} \frac{\text B_{ \frac{A}{B} \, y^{a+b} }\left(\frac{b+1}{a+b},0\right)}{\sqrt[a+b]{A^{b+1}B^{a-1}}(a+b)}$$
Now invert:
$$\boxed{y’=By^{-b}-Ay^a\implies y=\lim_{t\to0}\sqrt[a+b]{\frac BA\text I^{-1}_{\left(\frac AB\right)^\frac{b+1}{a+b}(a+b)Bt(x-c)}\left(\frac{b+1}{a+b},t\right)}}$$
which works when you use the decimal answer from the left bolded link and compare use it like in the right bolded link.
Take $t\to0$ to adjust the accuracy
|
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|
Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I think that in the case $A=0$ one of the results of the two formulas should match $-\frac{C}{B}$. Yet, if $A=0$, $\frac{-B + \sqrt{B^2 - 4AC}}{2\times0}$, we will get $\frac{0}{0}$, and with $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$ we will get $-\frac{B}{0}$, so none of the two possible solutions will match $-\frac{C}{B}$. Shouldn't there be at least one solution that has the same value as $-\frac{C}{B}$?
|
When $A=0$, you do not have $A$ in the denominator.
You reached the quadratic formula assuming that $A\neq 0$. Didn't you?
|
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|
What is the value of side $AC$ in the triangle below? For reference:The angle B of a triangle ABC measures 60°. The AN and CM medians are plotted. The radius of the circle inscribed in the MBNG quadrilateral
(G is centroid(barycenter) of ABC) measures $\sqrt3$ . Calculate AC.
My progress
$\triangle BED: \\sen30 = \frac{\sqrt3}{BD}\therefore BD = 2\sqrt3\\
cos 30 = \frac{BE}{BD}\rightarrow \frac{\sqrt3}{2}=\frac{BE}{2\sqrt3}\therefore BE = 3\implies BN = 3+\sqrt3\\
\triangle BNG:cos 30 = \frac{BN}{BG}\rightarrow \frac{\sqrt3}{2}=\frac{3+\sqrt3}{BG}\rightarrow BG = \frac{6+2\sqrt3}{\sqrt3}=2\sqrt3+2\\
BG = \frac{2BP}{3}\rightarrow BP = 3\sqrt3 + 3\\
\triangle BPC: tg30 = \frac{PC}{BP}\rightarrow \frac{\sqrt3}{3} = \frac{PC}{3\sqrt3+3} \implies \boxed{ PC = 3+\sqrt3}\\
\therefore \boxed{\color{red}AC = 2(3+\sqrt3) = 6+2\sqrt3}$
My question...only the equilateral triangle meets the conditions? Why if the quadrilateral is indescribable $\measuredangle MGN=120^o$
|
Since $BNGM$ has an inscribed circle, it must follow pitot theorem i.e. the opposite sides must sum to the same value. Let $x,y,d$ be the lengths of $BM, BN, BG$ respectively.
By law of cosines, we have
$$MG=\sqrt{x^2+d^2-xd\sqrt{3}}$$
$$NG=\sqrt{y^2+d^2-yd\sqrt{3}}$$
By pitot theorem,
$$x+\sqrt{y^2+d^2-yd\sqrt{3}}=y+\sqrt{x^2+d^2-xd\sqrt{3}}$$
$$x-\sqrt{x^2+d^2-xd\sqrt{3}}=y-\sqrt{y^2+d^2-yd\sqrt{3}}$$
Consider the function $f(x)=x-\sqrt{x^2+a^2-xa\sqrt{3}}$ for some real parameter $a$. We have that
$$f'(x)=1-\frac{2x-a\sqrt{3}}{2\sqrt{x^2+a^2-xa\sqrt{3}}}$$
If $u=\frac{x}{a}$, this is equivalent to
$$1-\frac{2u-\sqrt{3}}{2\sqrt{u^2-u\sqrt{3}+1}}$$
$$=1-\sqrt{\frac{4u^2-4u\sqrt{3}u+3}{4u^2-4u\sqrt{3}+4}}$$
$$=1-\sqrt{1-\frac{1}{4u^2-4u\sqrt{3}+4}}$$
Since $u\in\mathbb{R}$, we have $4u^2-4u\sqrt{3}+4\in [1,\infty)$, which implies
$$\frac{1}{4u^2-4u\sqrt{3}+4}\in (0,1]$$
$$\implies 1-\frac{1}{4u^2-4u\sqrt{3}+4}\in [0,1)$$
$$\implies \sqrt{1-\frac{1}{4u^2-4u\sqrt{3}+4}}\in [0,1)$$
$$\implies 1-\sqrt{1-\frac{1}{4u^2-4u\sqrt{3}+4}}\in (0,1]$$
Hence, $f'(x)$ is positive$~\forall x\in\mathbb{R}$, which means $f(x)$ is monotonically increasing.
This means that the only solution to
$$f(x)=f(y)$$
is when $x=y$. So $BM=BN$. From there it follows that the original triangle is equilateral. The rest of the answer follows.
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|
Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$
With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
3 y z^2 + 6 x y z$$
taking advantage of the fact that there is an $xyz=-1$ in the equation, but I'm not getting anywhere, someone less myopic than me.how to solve it?
Thanks
Edit : Will there be any university way to solve this problem , they posed it to a high school friend and told him it was just manipulations of remarkable products.
His answers I understand to some extent but I don't think my friend understands all of it.
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You got already that $$x^2+y^2+z^2=2.$$
Also, we have:
$$\sum_{cyc}x^2y^2=\left(\sum_{cyc}xy\right)^2-2xyz(x+y+z)=1$$ and $$x^2y^2z^2=1.$$
Thus, $$\sum_{cyc}x^8=\left(\sum_{cyc}x^4\right)^2-2\sum_{cyc}x^4y^4=$$
$$=\left(\left(\sum_{cyc}x^2\right)^2-2\sum_{cyc}x^2y^2\right)^2-2\left(\left(\sum_{cyc}x^2y^2\right)^2-2x^2y^2z^2\sum_{cyc}x^2\right)=$$
$$=(4-2)^2-2(1-2\cdot2)=10.$$
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Foci of Ellipse lies on Hyperbola and vice-versa Let the foci of the hyperbola $\frac{{{x^2}}}{{{A^2}}} - \frac{{{y^2}}}{{{B^2}}} = 1$ , (A,B > 0) be vertices of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , (a, b > 0) and foci of ellipse be vertices of hyperbola. Let eccentricities of the ellipse and hyperbola be $e_1$ & $e_2$ , respectively $L_1$ and $L_2$ are length of Latus rectum of ellipse and hyperbola respectively. Then find the minimum value of $[e_1+e_2]$ (where [.] denotes greatest integer function)
My approach is as follow
Let $ {e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{B^2}}}{{{A^2}}}} $
Foci of the ellipse are the vertices of the hyperbola and vice-versa then we get the following
$\frac{{{A^2} + {B^2}}}{{{a^2}}} = 1;\frac{{{a^2} - {b^2}}}{{{A^2}}} = 1 \Rightarrow {a^2} - {A^2} = {b^2}$ and ${B^2} = {a^2} - {A^2}$ hence $b=B$
${e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{B^2}}}{{{A^2}}}} \Rightarrow {e_1} = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} ;{e_2} = \sqrt {1 + \frac{{{b^2}}}{{{a^2} - {b^2}}}} $
Not able to proceed further
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Further you can write,
$e_1 = \sqrt{\dfrac{a^2-b^2}{a^2}} , e_2 = \sqrt{\dfrac{a^2}{a^2-b^2}}=\dfrac1{e_1} $
Now we have, $e_1+\dfrac1{e_1}$ where $0<e_1<1 $. For the Floor function to achieve minimum value, $e_1$ must be as close as possible to $1$.
For $0<e_1<1$, $~~2<e_1+\dfrac{1}{e_1}<\infty$
So, $\min\left(\Big{\lfloor}e_1+\frac1{e_1}\Big{\rfloor}\right) =2$
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Integer solutions to $2^{^{11}} a + 2^{^{11}} b + ab = 1$ $2^{^{11}} a + 2^{^{11}} b + ab = 1$
By guessing that $a+b = 0$, I was able to find the solutions (a, b) = (-63, 65), (65, -63). Is there any practical way of finding other solutions to the equation?
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Your equation is equivalent to the following
\begin{align*}
2^{22} + 2^{11}a + 2^{11}b +ab = 2^{22}+1 \\
(2^{11} + a)(2^{11} + b) = 2^{22} +1
\end{align*}
Now lets factor $2^{22} +1$ using Sophie Germain identitity
\begin{align*}
2^{22} + 1 =4(2^5)^4 +1^4 = (1 + 2^{11} - 2^6)(1 + 2^{11} + 2^6) \\
2^{22} + 1 = 1985 \cdot 2113 = 5 \cdot 397 \cdot2113
\end{align*}
These numbers ($5, 397, 2113$)are prime factors, so they divide either $2^{11} + a$ or $2^{11} + b$, but not both of them. There are very few posibilities (be careful and consider that both factors can be negative)
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Find $\int x^2\arcsin(2x)dx$ Find $\int x^2\arcsin(2x)dx$
My work.
$\frac{1}{3}\int \arcsin(2x)dx^3=\frac{1}{3}(x^3\arcsin(2x)dx-\int x^3d(\arcsin(2x))$
This yields to finding $\int \frac{2x^3}{\sqrt{(1-4x^2)}}dx$ with which I have problem finding.
$Edit$
$x=\frac{1}{2}sin\theta$
$\frac{2}{8}\int\frac{sin^3\theta cos\theta d\theta}{\sqrt{1-sin^2\theta}} = \frac{2}{8}\int\frac{sin^3\theta cos\theta d\theta}{|cos\theta|}$
Now what should I do with $|cosx|?$It is $cosx$ or $-cosx$.
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Using integration by parts:
$\int u\mathrm dv= uv-\int v\mathrm du$
$u=\arcsin {2x} \implies \mathrm du=\frac{2}{\sqrt{1-4x^2}}$
$\,dv=x^2 \mathrm dx \implies v=\frac{x^3}{3}$
$\int x^2 \arcsin{2x}\mathrm dx=\frac{x^3}{3}\arcsin {2x}-\int { \frac{2x^3}{3\sqrt{1-4x^2}}\mathrm dx}=\frac{x^3}{3}\arcsin {2x}+\frac{1}{72}(3\sqrt {1-4x^2} - (1-4x^2)\sqrt {1-4x^2})+c$
$\int { \frac{2x^3}{3\sqrt{1-4x^2}}\mathrm dx}=-\frac{1}{48}\int \frac {1-t}{\sqrt t}\mathrm dt=-\frac{1}{72}(3\sqrt t - t\sqrt t)+c=-\frac{1}{72}(3\sqrt {1-4x^2} - (1-4x^2)\sqrt {1-4x^2})+c$
$t=1-4x^2\implies \mathrm dt= -8x \mathrm dx$
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Inequality with directional derivative
Suppose that $f$ is differentiable on $\mathbb{R}^2$, $\boldsymbol{l}_1$, $\boldsymbol{l}_2$ are given directions, and the intersection angle between them is $\varphi$($0<\varphi<\pi$). Prove that:
$$
\left( \frac{\partial f}{\partial x} \right) ^2+\left( \frac{\partial f}{\partial y} \right) ^2\leqslant \frac{2}{\sin ^2\varphi}\left[ \left( \frac{\partial f}{\partial \boldsymbol{l}_1} \right) ^2+\left( \frac{\partial f}{\partial \boldsymbol{l}_2} \right) ^2 \right] .
$$
Let $\boldsymbol{l}_1=(\cos \alpha,\sin \alpha)$, $\boldsymbol{l}_2=(\cos(\varphi+\alpha),\sin(\varphi+\alpha))$
$$
RHS\cdot \frac{\sin ^2\varphi}{2}=\left( \frac{\partial f}{\partial x} \right) ^2\left[ \cos ^2\alpha +\cos ^2\left( \varphi +\alpha \right) \right] +\left( \frac{\partial f}{\partial y} \right) ^2\left[ \sin ^2\alpha +\cos ^2\left( \varphi +\alpha \right) \right] +\frac{\partial f}{\partial x}\cdot \frac{\partial f}{\partial y}\left[ \sin 2\alpha +\sin \left( 2\alpha +2\varphi \right) \right]
$$
I want to use Cauchy inequality,but $\left( \frac{\partial f}{\partial x} \right) ^2\left[ \cos ^2\alpha +\cos ^2\left( \varphi +\alpha \right) \right] +\left( \frac{\partial f}{\partial y} \right) ^2\left[ \sin ^2\alpha +\cos ^2\left( \varphi +\alpha \right) \right] \leqslant \left( \frac{\partial f}{\partial x} \right) ^4+\left( \frac{\partial f}{\partial y} \right) ^4$ is useless.
Or let $t=\frac{f_x}{f_y}$,$f(t)=\frac{\sin ^2\varphi}{2}\left( t^2+1 \right) -t\left( \cos ^2\alpha +\cos ^2\left( \varphi +\alpha \right) \right) -\frac{1}{t}\left( \sin ^2\alpha +\sin ^2\left( \varphi +\alpha \right) \right) +\sin 2\alpha +\sin \left( 2\alpha +2\varphi \right)$
$f'\left( t \right) =t\sin ^2\varphi +\frac{1}{t^2}\left( \sin ^2\alpha +\sin ^2\left( \varphi +\alpha \right) \right) -2+\sin ^2\alpha +\sin ^2\left( \varphi +\alpha \right) $,the maximum of $f(t)$ seems too hard to find.
How can I prove this inequality?
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We have
$$\mathrm{RHS} - \mathrm{LHS}
= \frac{2}{\sin^2 \varphi}
\left[A\left( \frac{\partial f}{\partial x} \right) ^2 + B \left( \frac{\partial f}{\partial y} \right) ^2
+ C \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \right]
$$
where
\begin{align*}
A &= \cos ^2\alpha +\cos ^2\left( \varphi +\alpha \right) - \frac{\sin^2 \varphi}{2}, \\
B &= \sin ^2\alpha +\sin ^2\left( \varphi +\alpha \right) - \frac{\sin^2 \varphi}{2}, \\
C &= \sin 2\alpha +\sin \left( 2\alpha +2\varphi \right).
\end{align*}
It suffices to prove that
$$A \ge 0, \quad B \ge 0, \quad 4AB - C^2 \ge 0.$$
For convenience, denote $c = \cos \alpha,
~ s = \sin \alpha, ~ r = \cos \varphi, ~ t = \sin \varphi$.
We have
\begin{align*}
A &= \frac12 t^2 + 2r^2c^2 - 2csrt ,\\
B &= \frac12 t^2 + 2r^2s^2 + 2csrt, \\
C &= 4c^2rt + 4cr^2s - 2rt.
\end{align*}
It is easy to prove that $A \ge 0$ and $B \ge 0$. Also, we have $4AB - C^2 = t^4 \ge 0$.
We are done.
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|
Proving $\left\lfloor \sqrt{x^2+3x+3}\right\rfloor = x+1$
Let $$f(x)=\sqrt{x^2+3x+3}$$
Claim: $$\lfloor f(x)\rfloor = x+1$$
I came across this expression in a question and made this claim based on observations
$$⌊f(1)⌋=\lfloor\sqrt{7}\rfloor=2$$
$$⌊f(2)⌋=\lfloor\sqrt{13}\rfloor=3$$
$$⌊f(3)⌋=\lfloor\sqrt{21}\rfloor=4$$
$$⌊f(4)⌋=\lfloor\sqrt{31}\rfloor=5$$
and so on ...
However, I can't think of a formal proof for my claim.
It would be appreciated if someone could help me with the proof or give some kind of hint.
Thank you.
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I am assuming that $x\in\Bbb Z_+$.
Note that $x^2+3x+3=x^2+2x+1+x+2=(x+1)^2+x+2$. So, $x^2+3x+3>(x+1)^2$. Could it be that $x^2+3x+3\geqslant(x+2)^2$? No, because\begin{align}x^2+3x+3\geqslant(x+2)^2&\iff x^2+3x+3\geqslant x^2+4x+4\\&\iff3x+3\geqslant4x+4.\end{align}So$$(x+1)^2<x^2+3x+3<(x+2)^2$$and therefore$$x+1<\sqrt{x^2+3x+3}<x+2.$$
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|
Find the minimum value of: $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$ Let $a,b,c>0$, find the minimum value of: $$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$$
I have tried:
$\bullet$ The minimum value is $\dfrac{3}{4}$ occur when $a=b=c$
$\bullet \dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}= \dfrac{a^2}{3a^2+ab}+\dfrac{b^2}{3b^2+bc}+\dfrac{c^2}{3c^2+ca}\ge\dfrac{(a+b+c)^2}{3(a^2+b^2+c^2)+ab+bc+ca}=\dfrac{(a+b+c)^2}{3(a+b+c)^2-5(ab+bc+ca)}$
$\bullet$ $ab+bc+ca\le\dfrac{(a+b+c)^2}{3}$ so we have $\dfrac{(a+b+c)^2}{3(a+b+c)^2-5(ab+bc+ca)}\le \dfrac{3}{4}$ but we need to prove $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c} \ge \dfrac{3}{4}$
Can you give me some hint ?
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For $c\rightarrow0^+$ and $b\rightarrow+\infty$ we see that our expression is closed to $\frac{1}{3}$.
But $\sum\limits_{cyc}\frac{a}{b+3a}\geq\frac{1}{3}$ it's just $$\sum_{cyc}\left(9a^2b+\frac{53}{3}abc\right)\geq0,$$ which is obvious.
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|
Solving Inequality $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$ If someone could help with solving the inequality above, that would be awsome!
Here is my thinking of using AGM
*
*$\sqrt{\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)}\le \frac{\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}}{2}$
*$\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)\le \frac{\left(x\sqrt{x}+y\sqrt{y}\right)^2}{4xy}$ (Square both sides)
*$\sqrt{x}\sqrt{y}\le \:\frac{\left(x\sqrt{x}+y\sqrt{y}\right)^2}{4xy}$ (Simplify Left Side)
*$4xy\sqrt{x}\sqrt{y}\le x^3+2xy\sqrt{x}\sqrt{y}+y^3$ (Move 4xy to the other side and Square remaining)
*But now I see that this will not end up in $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$
Please help????
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Here's an elementary way to handle it. First put on the same denimator
$$\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}=\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{xy}}$$
then factor the numerator using the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$
$$\sqrt{x^3}+\sqrt{y^3}=(\sqrt{x}+\sqrt{y})(\sqrt{x^2}-\sqrt{xy}+\sqrt{y^2})$$
but
$$\frac{\sqrt{x^2}-\sqrt{xy}+\sqrt{y^2}}{\sqrt{xy}}=
\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}-1\geqslant 1$$
by the inequality $a+\frac{1}{a}\geqslant2$ for any $a>0$.
Alternative: use Cauchy-Schwarz inequality to prove that
$$\left(\frac{a^2}{b}+\frac{b^2}{a}\right)(b+a)\geqslant (a+b)^2$$
for all positive $a,b$, thus
$$\frac{a^2}{b}+\frac{b^2}{a}\geqslant a+b.$$
Now just take $x=a^2$ and $y=b^2$.
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Error in my approach to prove that $(m-1)^{m-1} \equiv m - 1 \pmod m$ using a concrete example I read somewhere that if $m$ is composite then $(m - 1)^{m - 1} \equiv m - 1 \pmod m$ and I was curious to try to prove it myself.
So I took as $m = 6$.
Now I can see that $5^5 \equiv 5 \pmod 6$
I was thinking along the following lines:
Since $\gcd(6,5) = 1$ this means that if we multiply $5$ with all the numbers from $1$ to $5$ we have to get all the numbers in different order.
I.e.
The numbers less than $6$: $1,2,3,4,5$.
Multiply the above by $5$: $5,4,3,2,1$ i.e. we get the same set of numbers but in different order.
Since: $1\cdot 2\cdot 3\cdot 4\cdot 5 = 5\cdot 4\cdot 3\cdot 2\cdot 1$ we also have:
$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow 1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 1 \equiv 5^5$
Which is actually wrong.
What is the problem in my approach and thought process?
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your statement $$1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv (1\cdot 5) \cdot (2\cdot 5) \cdot (3\cdot 5) \cdot (4\cdot 5) \cdot (5\cdot 5) \Leftrightarrow 1\cdot 2\cdot 3\cdot 4\cdot 5 \equiv 5^5 (1\cdot 2\cdot 3\cdot 4\cdot 5) \Leftrightarrow 1 \equiv 5^5$$
is not corrrect .
Note that $$1.2.3.4.5\equiv 0 \pmod 6 $$
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|
What is the solution to the differential equation $(1+x^2) \frac{dy}{dx} -2xy =2x$? I am using integrating factor method to solve $(1+x^2) \frac{dy}{dx} -2xy =2x$ but having some issues.
When I divide through by $(1+x^2)$ I get
$ \frac{dy}{dx}-\frac{2x}{(1+x^2)}y=\frac{2x}{1+x^2}$
Then I integrate $\frac{2x}{(1+x^2)}$ to give $\ln(1+x^2)$. Thus the integrating factor is $(1+x^2)$.
When you multiply through by the i.f then you are simply left with the same equation as you start with.
Can anybody kindly point out where I may be going wrong?
Thanks
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As an alternative method, you can rearrange the given ODE to form a separable equation. We are given
$$(1+x^2) y' -2xy =2x$$
Rewrite as
$$y' = \frac{2x}{1+x^2} + \frac{2x}{1+x^2}y=\frac{2x}{1+x^2}(1+y)$$
which is a separable ODE.
$$\frac{dy}{1+y}=\frac{2x\:dx}{1+x^2}$$
Therefore we may integrate to find
$$\ln\left|y+1\right|=\ln\left|x^2+1\right|+\text{constant}$$
$$y+1=C(x^2+1)$$
Thus
$$y(x)=C(x^2+1)-1$$
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FInding the exact value of a series I would like to see if anyone can find the EXACT value of
$$5(\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+...)+7(\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+...)+9(\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}+...)+...$$
I have tried to regroup the terms and obtain
$$5(\zeta(4)-\frac{1}{1^4})+7(\zeta(4)-\frac{1}{1^4}-\frac{1}{2^4})+9(\zeta(4)-\frac{1}{1^4}-\frac{1}{2^4}-\frac{1}{3^4})+11(\zeta(4)-\frac{1}{1^4}-\frac{1}{2^4}-\frac{1}{3^4}-\frac{1}{4^4})+...$$
Seems like this is a diverging series. Any idea how to resolve this?
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Since all summands are positive, you can rearrange and get
$$
\sum_{n=2}^\infty (2n+1)\sum_{k=n}^\infty \frac1{k^4}=\sum_{n=2}^\infty\frac1{n^4}\sum_{k=2}^n (2k+1)
$$
The coefficient of $\frac1{n^4}$ is $5+7+\dots+(2n+1)=n^2+2n-3$, so by absolute convergence,
\begin{align*}
\sum_{n=2}^\infty\frac{n^2+2n-3}{n^4}
&=\sum_{n=1}^\infty\frac{n^2+2n-3}{n^4}\\
&=\sum_{n=1}^\infty\frac1{n^2}+2\sum_{n=1}^\infty\frac1{n^3}+(-3)\sum_{n=1}^\infty\frac1{n^4}\\
&=\zeta(2)+2\zeta(3)-3\zeta(4).
\end{align*}
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How to solve $\lim _{x\to \infty}\dfrac{x^5}{2^x} $ without L'Hospital's Rule Considering that asymptotically, $2^x$ grows faster than $x^5$ (in the beginning, $x^5$ grows faster than $2^x$, but there will be a point where $2^x$ outgrows $x^5$) then $\dfrac{x^5}{2^x} \rightarrow 0$ as $x \rightarrow \infty$. Therefore,
$$\lim _{x\to \infty}\dfrac{x^5}{2^x} = 0$$
But in order to solve the limit, I applied L'Hospital's Rule five times
\begin{align}
\lim _{x\to \infty}\dfrac{x^5}{2^x} & =\lim _{x\to \infty}\dfrac{5x^4}{2^x\ln 2}\\
& = \lim _{x\to \infty}\frac{20x^3}{\ln^2(2)\cdot 2^x} \\
& = \lim _{x\to \infty}\frac{60x^2}{\ln^3(2)\cdot 2^x} \\
& = \lim _{x\to \infty}\frac{120x}{\ln^4(2)\cdot 2^x} \\
& = \lim _{x\to \infty}\frac{120}{\ln^5(2)\cdot 2^x} \\
& = \frac{120}{\ln^5(2)}\cdot\lim _{x\to \infty}\frac{1}{2^x} \\
& = 0
\end{align}
What would be a more elegant way solve it without using L'Hospital's Rule?
Edit
Even though, the Limit: $\lim_{n\to \infty} \frac{n^5}{3^n}$ is similar, I found the link provided by Axion004, How to prove that exponential grows faster than polynomial? more interesting. Also, the answer provided by user trancelocation was very interesting and is what I was expecting.
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As an alternative way, by ratio test
$$\frac{\dfrac{(n+1)^5}{2^{n+1}}}{\dfrac{n^5}{2^n}}=\frac12\left(1+\frac1n\right)^5 \to \frac12 \implies \dfrac{n^5}{2^n} \to 0$$
and since $\forall x>0\quad \exists n$ such that $n\le x\le n+1$ we have
$$\dfrac{x^5}{2^x}\le \dfrac{(n+1)^5}{2^{n}}=2 \dfrac{(n+1)^5}{2^{n+1}} \to 0$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4265184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
}
|
Find two complex numbers which satisfy the equation $\frac{3+2i}{z^2}=1$ There is one really obvious solution to the equation which I ignored because it's lazy $$z=\pm \sqrt{3+2i}$$
When trying to answer the question I tried many different approaches but never got an answer. I tried thinking of a general case where $z=x+iy$ and expanding to see what $z^2$ would be. I then compared the expansion to $3+2i$ and got $x^2-y^2=3$ and $2xy=2$. This yields two quartic equations with ugly roots : $y^4-3y^2+1=0$ and $x^4-3x^2-1=0$. This method does tie into polar coordinates because if $x=cos(\theta)$ and $y=sin(\theta)$ then $2xy=2cos(\theta)sin(\theta)=sin(2\theta)=2$ and $x^2-y^2=cos^2(\theta)-sin^2(\theta)=cos(2\theta)=3$. You could then setup $tan(2\theta)$ and find $r$. I used a different approach though.
To approach this using polar coordinates I did the following: Let $u=z^2$ so that $u=3+2i$. I assumed $u$ is a complex number so that $u=re^{i\theta}$ where $r=\sqrt{3^2+2^2}$ and $\theta=tan^{-1}(\frac{2}{3})$. Thus $u=\sqrt{13}e^{i{tan}^{-1}(\frac{2}{3})}$. Thus $z^2=u=\sqrt{13}(cos(tan^{-1}(\frac{2}{3}))+isin(tan^{-1}(\frac{2}{3})))$. And indeed $$\frac{3+2i}{\sqrt{13}(cos(tan^{-1}(\frac{2}{3}))+isin(tan^{-1}(\frac{2}{3})))}=1$$Therefore, $z=\sqrt{u}$. This means that$$z=\pm\biggr(\sqrt{13}(cos(tan^{-1}(\frac{2}{3}))+isin(tan^{-1}(\frac{2}{3})))\biggr)^{\frac{1}{2}}$$or$$z=\pm\Bigr(\sqrt{13}e^{itan^{-1}(\frac{2}{3})}\Bigr)^\frac{1}{2}$$I didn't apply de Moivre's theorem to the sqrt to simplify it because I learned that it doesn't always work as intended for fractional powers. I really can't see how this could be the answer that they are looking for. Because in the end its just the polar coordinate version of $\pm\sqrt{3+2i}$.
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Your quartic equations are not so ugly, because they are really just quadratic equations in $x^2$ and $y^2$ respectively. So for instance $x^4-3x^2-1$ has solutions
$$x^2=\frac32\pm\frac12\sqrt{13}$$
And $x$ is real, so we can discard the solution $x^2=\frac32-\frac12\sqrt{13}$. Therefore
$$x=\pm\sqrt{\frac32+\frac12\sqrt{13}}$$
and similarly for $y$.
Having said that, I'm not really sure what your question is, so this might or might not be helpful.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4266028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Convergent limit $\lim_{n\to+\infty} \left(\frac{(3^n-1)^2-9^n}{\sqrt{9^n+n^9}}\right)$ which seems divergent by graph. Question $$\lim_{n\to+\infty} \left(\frac{(3^n-1)^2-9^n}{\sqrt{9^n+n^9}}\right)$$
My solution $$=\lim_{n\to+\infty} \left(\frac{9^n-2\cdot3^n+1-9^n}{\sqrt{9^n\cdot(1+\frac{n^9}{9^n})}}\right)$$
$$=\lim_{n\to+\infty} \left(\frac{-2+\frac{1}{3^n}}{\sqrt{(1+\frac{n^9}{9^n})}}\right)$$
$$= \frac{0-2}{\sqrt{1+0}}$$
$$=-2$$
But the graph of this sequence looks divergent.
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Your work is fine, as an alternative way to check the result we can proceed as follows
*
*$(3^n-1)^2-9^n=((3^n-1)+3^n)((3^n-1)-3^n)\sim -2\cdot 3^n$
*$\sqrt{9^n+n^9}\sim \sqrt{9^n}=3^n$
that is
$$\frac{(3^n-1)^2-9^n}{\sqrt{9^n+n^9}} \sim \frac{ -2\cdot 3^n}{3^n} \to -2$$
As noticed, the problem with the graph is a numerical issue since exponential terms grow very fast (here a better graph obtained by WA). As a suggestion, we should be very careful when evaluating or check limits by graphs, better to try with alternative ways to obtain the result.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4266844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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|
Shifted Gaussian I am looking for a closed form expression for the following Gaussian integral, shifted by $N$ parameters $b_i$:
$$
\frac{1}{2\pi}\int_{-\infty}^\infty e^{-\frac{1}{2}\sum_{i=1}^N (x-b_i)^2}dx.
$$
The first few $i=1,2,3,\dots$ are easy to compute but I haven’t spotted the pattern yet. Even better than answering the above question would be to give an expression for the more general integral
$$
\frac{1}{2\pi}\int_{-\infty}^\infty e^{-\frac{1}{2}\sum_{i=1}^N a_i(x-b_i)^2}dx
$$
in terms of the parameters $a_i$ $>0$ and $b_i$. Thanks for the help.
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$$\sum_{i=1}^N a_i(x-b_i)^2=\left(\sum_{i=1}^N a_i\right)x^2 -2\left(\sum_{i=1}^N a_ib_i\right)x+\left(\sum_{i=1}^N a_i(b_1)^2\right)$$
$$\begin{cases}
A=\sum_{i=1}^N a_i\\
B=\sum_{i=1}^N a_ib_i\\
C=\sum_{i=1}^N a_i(b_1)^2
\end{cases} $$
$$\sum_{i=1}^N a_i(x-b_i)^2=Ax^2-2Bx+C=A\left(x-\frac{B}{A}\right)^2+C-\frac{B^2}{A}
$$
$$
\frac{1}{2\pi}\int_{-\infty}^\infty e^{-\frac{1}{2}\sum_{i=1}^N a_i(x-b_i)^2}dx=\frac{1}{2\pi} e^{-\frac{1}{2}\left(C-\frac{B^2}{A}\right) }\int_{-\infty}^\infty e^{-\frac{1}{2}A\left(x-\frac{B}{A}\right)^2 } dx$$
$\int_{-\infty}^\infty e^{-\frac{1}{2}A\left(x-\frac{B}{A}\right)^2 } dx = \sqrt{\frac{2\pi}{A}}$
$$\frac{1}{2\pi}\int_{-\infty}^\infty e^{-\frac{1}{2}\sum_{i=1}^N a_i(x-b_i)^2}dx=\frac{1}{\sqrt{2\pi A}} e^{-\frac{1}{2}\left(C-\frac{B^2}{A}\right)}$$
$$\boxed{\frac{1}{2\pi}\int_{-\infty}^\infty e^{-\frac{1}{2}\sum_{i=1}^N a_i(x-b_i)^2}dx=\frac{1}{\sqrt{2\pi \sum_{i=1}^N a_i}} e^{-\frac{1}{2}\left(\sum_{i=1}^N a_i(b_1)^2-\frac{\left(\sum_{i=1}^N a_ib_i\right)^2}{\sum_{i=1}^N a_i}\right)}}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4271142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Solve below somewhat symmetric equations Solve $x,y,z$ subject to
$$x^2+y^2 - xy = 3$$
$$(x-z)^2+(y-z)^2 - (y-z)(x-z) = 4$$
$$(x-z)^2+y^2 - y(x-z) = 1$$
$$x,y,z \in R^{+}$$
My attempts:
$(x-z)^2+(y-z)^2 - (y-z)(x-z) - ((x-z)^2+y^2 - y(x-z)) = xz - 2yz = (x-2y)z = 3$
$x^2+y^2 - xy - ((x-z)^2+y^2 - y(x-z)) = 2xz - yz - z^2 = (2x-y-z)z = 2$
divide above, we have $\frac{2x-y-z}{x-2y} = \frac{2}{3}$, or $3z = 4x+y$
Then the rest is easy. But I start to get interested in if there are some geometric solution. Like the comment says, they are the elliptical cylinders in the positive octant
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We multiply the thrid equation by four, subtract it from two, and then substitute with equation one solved for x.
\begin{align*}
&[ (x-z)^2+(y-z)^2 - (y-z)(x-z) ]\\
- 4&[ (x-z)^2+y^2 - y(x-z) ]=0 \\
,\space &x = \dfrac{\sqrt{3}\sqrt{4 - y^2} + y}{2}\\
\\
\implies &y = \dfrac{14 - 3 \sqrt{3}}{13},
\space z = \pm\sqrt{3}
\end{align*}
It should be easy to figure out $x$ from here.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4271316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 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.