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Stuck with solving inequality to find the product of highest and lowest integer solutions The inequality in question is
$$\sqrt{(5+2\sqrt{6})^{2x}}+\sqrt{(5-2\sqrt{6})^{2x}}\leq98$$
This time our job is to find the product of highest and lowest integer solutions.
My attempt
$$\sqrt{(5+2\sqrt{6})^{2x}}+\sqrt{(5-2\sqrt{6})^{2x}}\leq98$$
$$\sqrt{((5+2\sqrt{6})^{x})^2}+\sqrt{((5-2\sqrt{6})^{x})^2}\leq98$$
$$(5+2\sqrt{6})^x+(5-2\sqrt{6})^x\leq98$$
What next?
The solution is $-4$.
| As $(5+2\sqrt6)(5-2\sqrt6)=1$
if $(5+2\sqrt6)^x=a>0$
$$a+\dfrac1a\le98\iff a^2-98a+1\le0$$
$a^2-98a+1=0\implies a=\dfrac{98\pm40\sqrt6}2=49\pm20\sqrt6=(5\pm2\sqrt6)^2$
$\implies(5+2\sqrt6)^{-2}=(5-2\sqrt6)^2\le a\le(5+2\sqrt6)^2 $
$\implies-2\le x\le2$
| {
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"url": "https://math.stackexchange.com/questions/2795546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a_0=0$, $a_{n+1}=a_n-\frac{1}{2}(a_n ^2-a)\rightarrow \sqrt{a}$ using banach's theorem Let's have a sequence $a_0=0$, $a_{n+1}=a_n-\frac{a_n ^2-a}{2}$. This sequence converges for $0<a<1$ to $\sqrt{a}$.
One way of proving this is to prove that $T(x)=x-\frac{1}{2}(x ^2-a)$ is a contraction mapping.
$$x-\frac{x ^2-a}{2}=-\frac{1}{2}(x-1)^2+\frac{a+1}{2}$$ so for $0< a<1$ T maps $[0,\frac{a+1}{2}]\subset [1-\sqrt{1+a^2},1+\sqrt{1+a^2}]$ into $[0,\frac{a+1}{2}]$
If we define metrics on $[0,\frac{a+1}{2}]$ as $|x-y|$, then, since $\frac{\mathrm{d}}{\mathrm{d}x}(x-\frac{x ^2-a}{2})=1-x$, using mean value theorem
$$|Tx-Ty|=|x-\frac{x ^2-a}{2}-(y-\frac{y ^2-a}{2})|=|1-\xi||x-y|$$ where $\xi=\xi(x,y) $ is between x and y.
Since $x,y\in [0,\frac{1+a}{2}]$, $0<a<1$ $|1-\xi| $ has to be less than 1 and at least 0 so we have proven T to be a contraction mapping. Is this the correct way?
| $T$ is not a contraction on $\left[0,\frac{a+1}2\right]$.
We have
\begin{align}
|Tx - Ty| &= \left|x - \frac12(x^2 - a) - y + \frac12(y^2 -a )\right| \\
&= \left|x-y - \frac12(x^2 - y^2)\right| \\
&= \left|x-y - \frac12(x-y)(x+y)\right| \\
&= \left|(x-y)\left(1 - \frac{x+y}2\right)\right| \\
&= \left|1-\frac{x+y}2\right||x-y|\\
\end{align}
If there exists $C > 0$ such that $|Tx - Ty | \le C|x-y|, \forall x,y \in \left[0,\frac{a+1}2\right]$ then in particular for $x = 0$ and $y_n = \frac1n$ we have
$$ \left(1-\frac1{2n}\right)\underbrace{|x-y_n|}_{=\frac1n} = \left|1-\frac{x+y_n}2\right||x-y_n|= |Tx -Ty_n| \le C\underbrace{|x - y_n|}_{=\frac1n} \implies 1-\frac1{2n} \le C$$
Letting $n\to\infty$ we get $1 \le C$ so $T$ cannot be a contraction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the Jordan canonical form of A and Choose the correct option Let $$ A = \begin{pmatrix}
0&0&0&-4 \\ 1&0&0&0 \\ 0&1&0&5 \\ 0&0&1&0
\end{pmatrix}$$
Then a Jordan canonical form of A is
Choose the correct option
$a) \begin{pmatrix}
-1&0&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$b) \begin{pmatrix}
-1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$c) \begin{pmatrix}
1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$d) \begin{pmatrix}
-1&1&0&0 \\ 0&-1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
My attempt : I know that Determinant of A = product of eigenvalues of A, as option c and d is not correct because Here Determinant of A = 4
that is $ \det A = -(-4) \begin{pmatrix}1 & 0 &0\\0& 1 & 0\\ 0&0&1\end{pmatrix}$
I'm in confusion about option a) and b).......how can I find the Jordan canonical form of A ?
PLiz help me.
Any hints/solution will be appreciated.
Thanks in advance
| HINT
Since all options are compatible with the check on det(A)=4, we need to determine the eigenvalues by $det(A-\lambda I)=0$ and the evaluate again the given options.
Note also that b is not a Jordan normal form.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of coefficients of even powers of x in $(1+x)^5 (1+x^2)^5$? the exact question is asking the sum of coefficients of even powers of x in the expansion of $(1+x+x^2+x^3)^5$ and in the solution this expression is simplified to $(1+x)^5(1+x^2)^5$. The solution to this question says this expression is equal to $(1+5x+10x^2+10x^3+5x^4+x^5)(1+x^2)^5$, this I understood, then it says that required sum of coefficients is equal to $(1+10+5)2^5$. How did they get this?
| The sum of coefficients of even power in $P(x)$ equals $$\frac{P(1)+P(-1)}{2}$$ in your case it's $$\frac{0+2^5\cdot 2^5}{2}=2^9=512$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to use the rearrangement inequality and conclude that $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x}$ is more than or equal to $\dfrac{\sin^3x}{\sin x} +\dfrac{\cos^3x}{\cos x}$ which is equal to $1$? Thanks.
| Correct me if wrong:
Another option.
Let $x \in (0,π/2)$.
$f(x):=$
$2 \dfrac{\sin^4 x}{2\sin x \cos x} +2 \dfrac{\cos^4 x}{2 \sin x \cos} =$
$\dfrac{2}{\sin 2x}×$
$((\sin^2x+\cos^2x)^2 -(1/2)\sin^2 2x)$
$=\dfrac{2}{\sin 2x}(1-(1/2)\sin^2 2x)$;
Let $y:= \sin 2x$, then $0 <y \le 1,$
and consider $g(y): = 2(1/y - (1/2)y)$.
Need to find the minimum of $g(y)$.
$g'(y) = 2(-1/y^2 -1/2) <0.$
$g$ is strictly decreasing
$\min_{0<y \le1} g(y)= 1.$
The minimum of the given expression is at $y=\sin 2x =1$,
i.e. at $x=π/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$
Solve $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$.
I am able to reduce the LHS to $\sqrt{x}=3^{\log_4(x)} \cdot \dfrac{4}{3}$. Squaring both sides do not seem to lead to a result. Do you know how to proceed?
| Easy step by step
$$\begin{align}
3^{\log_4x+\frac{1}{2}}+3^{\log_4x-\frac{1}{2}}&=\sqrt{x} \\
\sqrt{3}\cdot 3^{\log_4x} + (\sqrt 3)^{-1} \cdot 3^{\log_4x}&=\sqrt x \\
3 \cdot 3^{\log_4x}+3^{\log_4x}&=\sqrt{3x} \\
4\cdot3^{\log_4x}&=\sqrt{3x} \\
3^{\log_4x}&=\frac{\sqrt{3x}}{4} \\
4^{(\log_4 3) \cdot (\log_4 x)} &=\frac{\sqrt{3x}}{4} \\
(\color{red}{4^{\log_4 x}})^{\log_4 3} &= \frac{\sqrt{3x}}{4} \quad \text{a logarithm in an exponent is an inverse operation}\\
x^{\log_4 3}&=\frac{\sqrt 3}{4} \cdot \sqrt{x} \\
x^{2\log_4 3}&=\frac{3}{16} \cdot x \\
&\color{red}{x \neq 0} \quad \text{by the original equation} \\
x^{2\log_4 3 -1}&=\frac{3}{16} \\
x&=\left(\frac{3}{16}\right)^{\frac{1}{{2\log_4 3 -1}}} \approx 0.0571725372071
\end{align}$$
The online Desmos Graphing Calculator
| {
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Prediction of index in Lexicographical ordering of binary sequences? Suppose all $n$ sequences of 0's and 1's were displayed in Lexicographical order. Can the index of sequences that sum to $k$ be predicted?
| $\textbf{Better stated question}$: given all possible sequences of $n$ coin flips (a total of $2^n$ many), arrange them lexicographically. Can we find the index of those sequences that have $k$ heads?
$\textbf{Solution}:$ First let us consider a small case example where $n = 10$ and $k = 4$:
\begin{cases}
0000001111 & = 2^{3} + 2^{2} + 2^1 + 2^0 \quad (\text{all $k$ 1's are at the end})\\
0000010111 & = 2^{4} + 2^{2} + 2^1 + 2^0 \quad (k-1\text{ 1's are at the end; one 1 popped to the left})\\
0000011011 & = 2^{4} + 2^{3} + 2^1 + 2^0 \quad (k-2\text{ 1's are at the end; one 1 popped to the left})\\
0000011101 & = 2^{4} + 2^{3} + 2^2 + 2^0\\
0000011110 & = 2^{4} + 2^{3} + 2^2 + 2^1\\
0000100111 & = 2^{5} + 2^{2} + 2^1 + 2^1\\
0000101011 & = 2^{5} + 2^{3} + 2^1 + 2^0\\
& \vdots\\
1111100000 &= 2^{10} + 2^{9} + 2^8 + 2^7 \quad (\text{all} \ k \ \text{1's are to the left})
\end{cases}
Note for the equality I'm just translating binaries into decimals. For the general case, if $k > n$ then no such sequence exist, so assume $k \leq n$. Then the sequence we desire is
\begin{cases}
0....01...1 & = 2^{k-1} + 2^{k-2} + 2^{k-3} +... 2^0 \quad (\text{all $k$ 1's are at the end})\\
0...01011...1 & = 2^{k} + 2^{k-2} + 2^{k-3} + ... + 2^0 \quad (k-1\text{ 1's are at the end; one 1 popped to the left})\\
& \vdots\\
1...10....0 &= 2^n + 2^{n-1} + ... + 2^{n-k+1}
\end{cases}
so the answer is yes.
| {
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Show differential equation solution is arc of great circle I have to show that the solution of a differential equation is an arc of a great circle. The differential equation is as follows $($in spherical coordinates$)$:
$$\frac{\sin ^2\theta\phi '}{(1+\sin ^2\theta (\phi ')^2)^{\frac{1}{2}}}=C$$ where $C$ is an arbitrary constant and $\phi '$ denotes the derivative of $\phi$ with respect to $\theta$.
My reasoning:
By setting $\phi (0)=0$, any arc of a great circle will have no change in $\phi$ with respect to $\theta$, so with this initial condition the answer follows by proving that $\phi '=0$. My issue is that upon working this round I end up with $$(\phi ')^2=\frac{C^2}{\sin ^4\theta -C^2\sin ^2\theta}$$ From this I can see no way forward.
Where do i go from here/ what should I do instead?
| Let $u=\cot \theta$, then $1+u^2=\csc^2 \theta$ and $du=-\csc^2 \theta \, d\theta$.
\begin{align}
\frac{d\phi}{d\theta} &= \frac{d\theta}{\sin \theta \sqrt{\sin^2 \theta-C^2}} \\
&= \frac{C\csc^2 \theta}{\sqrt{1-C^2\csc^2 \theta}} \\
d\phi &= -\frac{C\,du}{\sqrt{1-C^2(1+u^2)}} \\
&= -\frac{C\,du}{\sqrt{(1-C^2)-C^2u^2}} \\
&= -\frac{du}{\sqrt{\tan \alpha^2-u^2}} \tag{$C=\cos \alpha$} \\
\phi &=\cos^{-1} \left( \frac{u}{\tan \alpha} \right)+\beta \\
\cos (\phi-\beta) &= \frac{\cot \theta}{\tan \alpha} \\
\cot \theta &= \tan \alpha \cos (\phi-\beta) \\
\end{align}
Rearrange,
$$(\sin \theta \cos \phi)(\sin \alpha \cos \beta)+
(\sin \theta \sin \phi)(\sin \alpha \sin \beta)=
(\cos \theta)(\cos \alpha)$$
which lies on the plane
$$x\sin \alpha \cos \beta+y\sin \alpha \sin \beta-z\cos \alpha=0$$
| {
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Let $f(x)=(2+x+ax^2)\ln (1+x)-2x$ with $f(x)$ achieving its maximum value at $x=0$. Find $a$. Problem
Let $f(x)=(2+x+ax^2)\ln (1+x)-2x$ with $f(x)$ achieving its maximum value at $x=0$. Find $a$.
Solution
At first, let's review a theorem, which states as follows:
Let $f^{(n)}(x)$ denote the $n-$order derivative of $f(x)$,and $k$ be
a positive integer such that $f^{(n)}(x_0) \neq 0$ holds for $n=k$ and
$f^{(n)}(x_0)=0$ holds for all $n=1,2,\cdots,k-1$. Then, if $k$ is
odd, $f(x_0)$ is neither a maximum value nor a minimum value, and if
$k$ is even, $f(x_0)$ is either a maximum value or a minimum value,
which depends on the sign of $f^{(k)}(x_0)$.
Now, let's take up to solve the problem. Notice that, for $f(x)=(2+x+ax^2)\ln (1+x)-2x$, we have $$f'(x)=\frac{a x^2+x+2}{x+1}+(2 a x+1) \ln (x+1)-2,$$$$f''(x)=-\frac{a x^2+x+2}{(x+1)^2}+\frac{2 (2 a x+1)}{x+1}+2 a \ln(x+1),$$$$f'''(x)=\frac{2 \left(a x^2+x+2\right)}{(x+1)^3}+\frac{6 a}{x+1}-\frac{3 (2 a x+1)}{(x+1)^2}.$$ According to these, it's clear that $f'(0)=0, f''(0)=0,f'''(0)=6a+1.$If $f'''(0)=6a+1 \neq 1$. Then, by the theorem above, $f(0)$ is neither a maximum value nor a minimum value, which is a contradiction. Hence, $f'''(0)=6a+1=0$. As a result, $a=-\dfrac{1}{6}.$ Moreover, in this case, $f''''(0)=-2$. This shows that $f(0)$ is a maximum value, which indeed satisfies the requirement.
P.S.
In fact, we can further prove that, in this case $a=-\dfrac{1}{6}$, $f(0)$ is also the global maximum value. Just notice $$f''(x)=\frac{2x-3x^2}{6(x+1)^2}-\frac{1}{3}\ln(x+1)\leq \frac{2x-3x^2}{6(x+1)^2}-\frac{1}{3}\cdot\frac{x}{x+1}=-\frac{5x^2}{6(1+x)^2}\leq 0$$ holds for all $x>-1$,and $f''(x)=0$ holds for $x=0$ only.
Please correct me if I'm wrong. And I'm always waiting for other nicer solutions. Thanks.
| Here's a nicer solution: using the Maclaurin series of $\log(1+x)$,
$$
-2x+\left(a x^2+x+2\right) \log (x+1) = -2x+\left(a x^2+x+2\right) \left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O\left(x^5\right)\right)
$$
$$
=\left(a+\frac{1}{6}\right) x^3+\left(-\frac{a}{2}-\frac{1}{6}\right)
x^4+O\left(x^5\right)
$$By your theorem, we must have $a=-1/6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5$
Let $P(x)$ be a polynomial of degree $4$ such that $P(n) = \frac{120}{n}$ for $n=1,2,3,4,5.$ Determine the value of $P(6)$.
Let $P(x) = ax^4 + bx^3 + cx^2 + dx + e$. For $n=1,2,3,4,5$ I have plugged it into this polynomial and got the following —
$$P(1) = a+b+c+d+e = 120$$
$$P(2) = 16a + 8b + 4c + 2d + e = 60$$
$$P(3) = 81a + 27b + 9c + 3d + e = 40$$
$$...$$
And what the problem asks for is $$P(6) = 1296a + 216b + 36c + 6d + e .$$
However, I'm not sure if all this is helping me very much. So noticing that $2P(2) = P(1) = 3P(3)$ (which is also equal to $4P(4), 5P(5)...$) From solving simultaneous equations I got that $31a + 15b + 7c + 3d + e=0$ and similarly $211a + 65b + 19c + 5d + e=0$, but they seem rather useless at this point.
| Hint: We can write $P(x)$ in the form of
$$P(x) =c_1(x-2)(x-3)(x-4)(x-5)\ +\ c_2(x-1)(x-3)(x-4)(x-5)\ +\dots+\ c_5(x-1)(x-2)(x-3)(x-4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816980",
"timestamp": "2023-03-29T00:00:00",
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Finding out the last digit of number $3a/4$ knowing the last two digits of $a$ Natural number $a$ ends with digits $16$. If $a$ is not divisible by $8$, what is the last digit of $3a/4$?
Solution:
$$7$$
How do we arrive at that solution?
| Note that $a = 100k + 16$ where $k$ is a positive integer.
Recall the divisibility criteria for $ 8$ : a number is divisible by $8$ if and only if its last three digits are.
Let $b$ be the last digit of $k$. Then, the last three digits of $a$ are $\overline{b16}$.
When is $\overline{b16}$ divisible by $8$? It is not difficult to see that this is a multiple of $8$ if and only if $b$ is a multiple of $2$, since $\overline{b16}$ is a multiple of $8$, if and only if $\overline{b00}$ is a multiple of $8$, if and only if $b $ is even.
In conclusion, $a = 100k + 16$ where the last digit of $k$ is odd. In other words, where $k$ is odd, since $a$ is not a multiple of $8$.
Now, $\frac a4 = 25k + 4$, and $\frac {3a}4 = 75k + 12$.
Note that $k $ is odd, so $75k$ is an odd multiple of $5$. Odd multiples of $5$ always end with $5$, so $75k + 12$ ends with $5+2 = 7$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $\lim_{x\to1}=\frac{x^2+x-2}{1-\sqrt{x}}$? let $f(x)=\dfrac{x^2+x-2}{1-\sqrt{x}}$
How do I solve this limit?
$$\lim_{x\to1}f(x)$$
I can replace the function with its content
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}$$
Then rationalizing the denominator
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}\cdot\dfrac{1+\sqrt{x}}{1+\sqrt{x}}$$
With $(a + b)(a - b) = a^2 - b^2$, I can remove the irrational denominator.
$$\lim_{x\to1}\dfrac{(x^2+x-2)(1+\sqrt{x})}{1-x}$$
I the multiply the two parenthesis
$$\lim_{x\to1}\dfrac{\sqrt{x} \cdot x^2+\sqrt{x}\cdot x- \sqrt{x} \cdot 2 + x^2 + x - 2}{1-x}$$
I'm not sure where to continue to solve this limit.
| Observe that
$$ x^2 + x - 2 = (x-1)(x+2). $$
Therefore
\begin{align} \lim_{x\to 1}\frac{(x^2+x-2)(1+\sqrt{x})}{1-x}
&= \lim_{x\to 1} \frac{(x-1)(x+2)(1+\sqrt{x})}{1-x} \\
&= \lim_{x\to 1} \frac{-(1-x)(x+2)(1+\sqrt{x})}{1-x} \\
&= -\lim_{x\to 1} (x+2)(1+\sqrt{x}) \\
&= -(1+2)(1+\sqrt{1}) && \text{($\ast$)}\\
&= -6. \end{align}
At step ($\ast$), we are using the fact that both factors are continuous at $x=1$ and that the product of continuous functions is continuous.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}} $ $$\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}} $$
I tried writing it as $\lim\limits_{x \to 0} \dfrac{\dfrac{\tan^{12}x}{x^{12}}-1}{x^{2}} $ thinking that if I applied L'hospital two times I would reach my result but with no luck. Any suggestions or tricks I could use?
| $\lim_{x \to 0} \frac{\tan^{12}x-x^{12}}{x^{14}}
$
$\begin{array}\\
\lim_{x \to 0} \dfrac{\tan^{12}x-x^{12}}{x^{14}}
&=\lim_{x \to 0} \dfrac{(\tan(x)/x)^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\sin(x)/(x\cos(x))^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\frac{\sin(x)}{x\cos(x)})^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\frac{x-x^3/6+O(x^5)}{x(1-x^2/2+O(x^4))})^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\frac{x-x^3/6+O(x^5)}{x-x^3/2+O(x^5))})^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(\frac{1-x^2/6+O(x^4)}{1-x^2/2+O(x^4)})^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{(1+x^2/3+O(x^4))^{12}-1}{x^{2}}\\
&=\lim_{x \to 0} \dfrac{1+4x^2+O(x^4)-1}{x^{2}}\\
&=\lim_{x \to 0} 4+O(x^2)\\
&=4\\
\end{array}
$
| {
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Without relating the problem to polynomials, how can we solve this inequality? $a,b,c$ are real;
$a<b<c$ ;
$a+b+c=6$ ;
$ab+bc+ca = 9$.
Prove that $0<a<1<b<3<c<4$.
I solved this by treating $a,b,c$ as roots of $x^3 - 6x^2 + 9x + d$ for some $d$, but I have not been able to solve it in a more direct way.
| I want to note that solving this problem using polynomials may be the most natural and, in fact, direct way to do. You can for example prove each inequality individually. To show $a>0$, you have $$0 < (b-c)^2=(b+c)^2-4bc=(6-a)^2-4\big(9-a(b+c)\big)=(6-a)^2-4\big(9-a(6-a)\big)\,,$$
which gives $3a(4-a)>0$, whence $a>0$ (and similarly, $c<4$).
Now, since $0<a<b<c$, we have
$$
\begin{align}0<(b-a)(c-a)&=bc-(b+c)a+a^2=\big(9-(b+c)a\big)-(b+c)a+a^2
\\&=9-2(b+c)a+a^2=9-2(6-a)a+a^2=3(a-1)(a-3)\,.
\end{align}$$
That is, either $a<1$ or $a>3$. Clearly, $a<\frac{a+b+c}{3}=2$, so $a<1$ must hold. In the same way, a similar proof shows that $c>3$.
Finally, note that
$$\begin{align}
0<(b-a)(c-b)=-ac+(a+c)b-b^2=-9+2(a+c)b-b^2=-9+2(6-b)b-b^2\,.
\end{align}$$
This shows that $0<3(b-1)(3-b)$, or $1<b<3$. To be honest, I find this proof rather cumbersome and inelegant.
In general, suppose that $p(x)=(x-a)(x-b)(x-c)=x^3-ux^2+vx-w$, where $a,b,c,u,v,w\in\mathbb{R}$ with $a\leq b\leq c$. By observing that $$p'(x)=3\left(x^2-\frac{2}3ux+\frac13v\right)=3\left(x-\frac{1}{3}u+\frac{1}{3}\Delta\right)\left(x-\frac{1}{3}u-\frac{1}{3}\Delta\right)\,,$$
where $\Delta:=\sqrt{u^2-3v}$ (i.e., $u^2\geq 3v$ must hold), we conclude that $$a\leq \frac{u-\Delta}{3}\leq b\leq \frac{u+\Delta}{3} \leq c\,.$$
The minimum possible value of $a$ occurs when $w$ is minimized, which happens when $b=c=\frac{u+\Delta}{3}$, and so $a=u-b-c=\frac{u-2\Delta}{3}$. Consequently,
$$a\geq \frac{u-2\Delta}{3}$$
must hold in general. Similarly, the maximum possible value of $c$ occurs when $w$ is maximized, so that $a=b=\frac{u-2\Delta}{3}$ and $c=u-b-c=\frac{u+2\Delta}{3}$. This means that
$$c\leq \frac{u+2\Delta}{3}$$ in general. In other words, if real numbers $a,b,c$ satisfy $a\leq b\leq c$, $a+b+c=u$, and $bc+ca+ab=v$, then
$$\frac{u-2\sqrt{u^2-3v}}{3}\leq a \leq \frac{u-\sqrt{u^2-3v}}{3}\leq b \leq \frac{u+\sqrt{u^2-3v}}{3} \leq c \leq \frac{u+2\sqrt{u^2-3v}}{3}\,.$$
We also have $u^2\geq 3v$ and
$$\frac{-2u^3+9uv-2\left(u^2-3v\right)^{3/2}}{27}\leq w \leq \frac{-2u^3+9uv+2\left(u^2-3v\right)^{3/2}}{27}\,.$$
(If $a<b<c$ is assumed, then all inequalities above are strict.)
When $u=6$ and $v=9$, we get
$$0\leq a\leq 1 \leq b\leq 3 \leq c\leq 4\,.$$
Furthermore, $0\leq w \leq 4$ ($w=-d$ in OP's notation).
P.S. The general case (with $u,v,w$) can also be verified, using the inequalities $(b-c)^2\geq 0$, $(a-b)^2\geq 0$, $(b-a)(c-a)\geq 0$, $(b-a)(c-b)\geq 0$, and $(c-a)(c-b)\geq 0$, as in the first part of this answer. However, the work is much more tedious than the proof using polynomials as illustrated in the second part of this answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that $\cosh(x)\cosh(y) \geq \sqrt{x^2 + y^2}$ I'm trying to prove the inequality :
$\cosh(x)\cosh(y) \geq \sqrt{x^2 + y^2}$
I played a bit with Geogebra and it looks true. I've tried proving it via convexity but so far I'm unsuccesful.
Thank you for your help.
| By Taylor's theorem with the Lagrange form of the remainder,
$$ \cosh{x} = 1 + \frac{x^2}{2}\cosh{\xi} $$
for some $\xi$ with $0 \leq \lvert\xi\rvert \leq \lvert x\rvert$. Hence $\cosh{x} \geq 1+\frac{1}{2}x^2$. Thus
$$ \cosh{x}\cosh{y} \geq (1+\tfrac{1}{2}x^2)(1+\tfrac{1}{2}y^2) \geq 1 + \frac{1}{2}(x^2+y^2), $$
since $x^2y^2>0$.
Now by the AM–GM inequality $\frac{1}{2}(a+b) \geq \sqrt{ab}$, taking $a=2$ and $b=x^2+y^2$,
$$1+\tfrac{1}{2}(x^2+y^2) \geq \sqrt{2(x^2+y^2)} > \sqrt{x^2+y^2}, $$
as required.
| {
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"url": "https://math.stackexchange.com/questions/2822687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Connection Formula for Hypergeometric Function 2F1 Suppose I have the function $_2F_1\left(a,b;c;x^2\right)$ with $a=\frac{3}{4}+\frac{k}{4}$, $b=\frac{3}{4}-\frac{k}{4}$ and $c=\frac{1}{2}$.
I want to know the behaviour about $x=1.\,$
I go to DLMF equation 15.10.21 and choose
$$
w_1\left(x^2\right) = {\frac {\Gamma \left( c \right) \Gamma \left( c-a-b \right)}{
\Gamma \left( c-a \right) \Gamma \left( c-b \right) }} \, w_3\left(x^2\right) +{\frac {\Gamma
\left( c \right) \Gamma \left( a+b-c \right)}{\Gamma \left(
a \right) \Gamma \left( b \right) }} \, w_4\left(x^2\right).
$$
Since $w_4$ is singular at $x=1$ (and it should be finite) I expected this typical constraint that $a$ or $b$ must be some $n\leq 0 \in \mathbb{Z}$ for the second term to vanish. Now the first term has $\Gamma(-1)$. Is that a problem or can I absorb this into a constant? However if I do so, then the original function $w_1$ is not really defined. Does this mean the solution is not valid unless $k=3$?
PS: Actually if $k=4n+3$ $(n\geq 0)$ then the second term vanishes and in the first term the poles cancel?
| Let
$$F(x) = {_2F_1}\left( \frac {3+k} 4, \frac {3-k} 4; \frac 1 2; x \right),
\quad k \geq 0.$$
If $(3-k)/4$ is an integer, $F(x)$ becomes a polynomial and we have
$$F(x) = F(1) + O(|1-x|) =
\frac {(-1)^{(k-3)/4} \sqrt \pi \,\Gamma \left( \frac {k+5} 4 \right)}
{\Gamma \left( \frac {k-1} 4 \right)} + O(|1-x|).$$
Otherwise this is the logarithmic case, the formulas for which can be found here. The leading term is
$$F(x) = \frac {\sqrt \pi}
{\Gamma\left( \frac {3+k} 4 \right)
\Gamma\left( \frac {3-k} 4 \right)
(1-x)} +
O(|\!\ln(1-x)|).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...
The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1)$
The Book's Answer: $4x^3(3x+1)(x+1)^7$
| Note that $(x^3+2x^2+x)=x(x^2+2x+1)=x(x+1)^2$.
Also, $3x^2+4x+1=(x+1)(3x+1)$
Apply those factorizations to your answer, and collect like factors together; you should get the book's answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the inverse of $x^3+x^2+1$ in $\Bbb F_2[x]/(x^4+x^2)$ with the Euclidean algorithm My first thought was successful: $x^4+x^2=x^2(x^2+1)$ and $x^3+x^2+1=x^2(x+1)+1$ so it is its own inverse because $(x^2(x+1)+1)^2\equiv x^4(x+1)^2+1\equiv x^4(x^2+1)+1\equiv1.$
The given solution claims to use the Euclidean algorithm, succintly, but I don't quite get it:
$x^4+x^2=(x^3+x^2+1)(x+1)+(x+1); \\ x^3+x^2+1=x^2(x+1)+1$
hence $1=(x^3+x^2+1)(x^3+x^2+1)+x^2(x^4+x^2)$
What is the logic in the above lines? How does it work in general?
| A possible way to "visualize" what happens, in analogy with the same algorithm for (the ring of) integers, is to (formally) use the formalism of continued fractions,
so let us write:
$$
\begin{aligned}
\frac{x^4+x^2}{x^3+x^2+1}
&=
(x+1)+\frac{x+1}{x^3+x^2+1}
\\
&=
(x+1)+\frac1{\frac{x^3+x^2+1}{x+1}}
\\
&=
(x+1)+\frac1{x^2+\frac1{x+1}}
\\
&=[\ x+1,\ x^2,\ x+1\ ]\text{ (in notation)}\ .
\end{aligned}
$$
Now we consider the "best approximation", which is $[\ x+1,\ x^2\ ]=x+1+\frac 1{x^2}$, and compute the difference:
$$
\begin{aligned}
&
\frac{x^4+x^2}{x^3+x^2+1}
-
\frac{x^3+x^2+1}{x^2}
\\
&\qquad
=
\frac
{x^2(x^4+x^2)-(x^3+x^2+1)^2}
{x^2(x^3+x^2+1)}
\\
&\qquad
=
\frac
{1}
{x^2(x^3+x^2+1)}
\\
\ .
\end{aligned}
$$
From the above computation we need only the last numerator computation,
$$
x^2(x^4+x^2)-(x^3+x^2+1)^2=1\ ,
$$
and on the R.H.S we have a constant polynomial... Exactly what we need!
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\sin (x/2)+2\sin (x/4)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$
Given that $$f (x)=\sin (x/2)+2\sin (x/4)$$
Show that $f (x)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$.
My try
$f (x)=\sin (x/2)+2\sin (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}(\cos(x/2)+\cos (x/4))$
$f' (x)=\cos(3x/4)\cos (x/4))$
$f'(x)=0$ $\implies$ $\cos(3x/4)\cos (x/4))=0$
$\cos(3x/4)=0\:$ or $\:\cos (x/4)=0$
$3x/4= \pi/2\:$ and so $\:x= 2\pi/3$ or $ x=2\pi$
$x/4= \pi/2\:$ and so $\:x= 2\pi $
Is my work ok ? What should I do now?
| Put $\theta = \dfrac{x}{4} \implies \text{ We prove :} f(\theta) = \sin (2\theta)+2\sin(\theta) \le \dfrac{3\sqrt{3}}{2}, \theta \in [0,\pi/2]$ . Put $t = \sin(\theta)\implies t \in [0,1], f(t) = 2t\sqrt{1-t^2}+ 2t\implies f'(t)=2\sqrt{1-t^2}-\dfrac{2t^2}{\sqrt{1-t^2}}+2= 0 \iff t = 0, \dfrac{\sqrt{3}}{2}$ . We have $f(0) = 0, f(1) = 2, f(\frac{\sqrt{3}}{2}) = \dfrac{3\sqrt{3}}{2}$, and this is the largest value and it means $f(t) \le \dfrac{3\sqrt{3}}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Approximate integral: $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx.$
Approximate integral: $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx.$
My attempt:
Let I = $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx$
$u=7-x\implies I=\frac 72 \int_3^4 \frac 1{\sqrt{(7x-10-x^2)^3}}dx$ Now I have to approximate this...
I got to this point too:
$u=3-x \implies I= \int_0^1 \frac {1}{\sqrt{(-x^2-x+2)^3}}dx.$
and: $\frac 1{-x^2-x+2}\leq1\implies \int_0^1 \frac{1}{\sqrt{(-x^2-x+2)^3}}dx\leq \int_0^1-\frac{1}{(x+2)(x-1)}dx$ and the right one does not converge...
I have these values:
$a) (\frac 9{10};\frac {19}{20})$
$b) (\frac {19}{20},1)$
$c) (1;\frac {21}{20})$
$d) (\frac {21}{20};\frac {11}{10})$
$e) (\frac {11}{10},\frac {23}{20})$
$f) (\frac {23}{20}; \frac 65)$.
| I will try to solve the problem.
The integrand has the form $P_1 \cdot Q_2 ^ {- 3/2} $.
It is logical to assume that the antiderivative has the form $ R_1 \cdot Q_2 ^ {- 1/2} $.
Differentiate and equate.
$\bigg((ax+b)(7x-10-x^2)^{-\frac{1}{2}}\bigg)'=x\cdot (7x-10-x^2)^{-\frac{3}{2}}$
So $\displaystyle a(7x-10-x^2)^{-\frac{1}{2}}+(ax+b)(7-2x)\cdot -\frac{1}{2}\cdot (7x-10-x^2)^{-\frac{3}{2}}=x\cdot (7x-10-x^2)^{-\frac{3}{2}}$
So $\displaystyle a(7x-10-x^2)+(ax+b)(7-2x)\cdot -\frac{1}{2}=x$
So we get $\displaystyle a=\frac{14}{9}$ and $\displaystyle b=-\frac{40}{9}$
So $\displaystyle \int^{4}_{3}\frac{x}{\sqrt{(7x-10x-x^2)^3}}dx=-\frac{1}{9}\bigg(\frac{14x-40}{\sqrt{7x-10-x^2}}\bigg)\bigg|^{4}_{3}=\frac{7\sqrt{2}}{9}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Bounds for an integral I am trying to show that $$\frac{1}{5} < \int_5^8 \frac{2x-7}{2x+5} dx <1$$ since for the integral $$5\le x \le 8 \rightarrow 15\le 2x+5 \le 21$$$$-\frac{12}{15}\le-\frac{12}{2x+5}\le -\frac{12}{21}\rightarrow \frac{3}{15}\le 1- \frac{12}{2x+5}\le \frac{3}{7}$$ By taking integral $$\frac{3}{5}\le \int_5^8 \left(1-\frac{12}{2x+5}\right)dx\le\frac{9}{7}$$ Whilee the first one could be reduced to what is needed, how can I deal with the right bound?
| We wish to show that $$\int_5^8{\frac{2x-7}{2x+5}dx}<1.$$ Let $g:(0,\infty)\to\mathbf R$ be given by $$g(x)=1-\int_5^x{1-\frac{12}{2w+5}dw}.$$ Then $$g'(x)={12\over{2x+5}}-1$$ and $$g''(x)={-24\over{(2x+5)^2}}<0\,\, \forall x.$$ Thus, $g$ is concave. Also, $g'=0$ at $x=3.5$, so that $g$ has its turning point $\left( g'(3)>0\,\text{and}\,g'(4)<0 \right)$ at $x=3.5$, where $g=2.5-6\log \frac 54>0$ since $4e^{5/12}>5.$ Thus, $g$ has just two roots.
Clearly, one of the roots must be $<3.5$, so that we need only consider the other root $>3.5$ (since we're interested only in how this function behaves in $[5,8]$). I claim that the other root is also $>8.$ Now we have that $$g(x)=6-x-6\log(2x+5),$$ so that $$g(8)=6\log \frac75 -2>0.$$ That last inequality holds because clearly $7>5e^{1/3}.$ Also, $$g(20)=-14+6\log 3<-14+12=-2<0,$$ since $3<e^2=(2+\varepsilon)^2=4+\delta$, where $1/2<\varepsilon<1$ and $\delta=4\varepsilon+\varepsilon^2.$
Consequently, the other root of $g$ must be in $(8,20)$, so that $g>0$ in $[5,8]$ (since $g(5)=1>0$). The result follows by setting $x=8$ in $g(x)>0.\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Showing that $4b^2+4b = a^2+a$ has no non-zero integer solutions? The problem is
Show that $4b^2+4b = a^2+a$ has no integer solutions where none of $a, b$ are zero.
I have a solution but I think there must be some better ways:
My Solution:
$$4b^2+4b = a^2+a$$
$$(2b+a)(2b-a)+4b-a= 0$$
Now letting $x = 2b + a$ and $y = 2b-a$, we see that $x+y = 4b$. Substituting,
$$xy+\dfrac {x+y}{2}+y=0$$
$$2xy+x+3y=0$$
From this we see that $y|x$, so we can substitute $x = ky$ for some integer $k$
$2ky^2+ky+3y = 0$
$$k = \dfrac {3}{2y+1}$$
From here we get that $y \in \{-2, -1 , 1 \}$, and each of the cases can be checked individually.
| Multiply by $4$ and complete the square on both sides. This gives
$(4b+2)^2-4=(2a+1)^2-1$
$(4b+2)^2-(2a+1)^2=3$
What are the only two squares differing by exactly $3$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$ If the circle $(x-a)^2+(y-b)^2=r^2$ and the line $y=mx+c$ do not meet:
Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$
These are the steps I have thus taken (although they may be wrong/useless):
*
*Rearranged the circle equation for y:
$y=\sqrt{r^2-x^2+2ax-a^2}+b$
*Set the circle and line equations equal to each other as in simultaneous equations.
I was then planning on finding some discriminant and as we know the line and circle do not meet I could set the discriminant < 0 and show it is the same as the long inequality.
Any solutions/pointers appreciated as I cannot seem to work this problem out.
| The line is outide the circle hance putting the line equation inside the formula for the circle is larger then $r^2$ hence
$$
(x - a)^2 + (mx + c - b)^2 > r^2
$$ l.h.s is minimized for $2(x-a) + 2m(mx + c - b) = 0$ or for $x = (m(b - c) + a)/(m^2+1))$. let $f=m^2+1$, then the expression is minimized by
$$
(m(b - c) + a - af)^2 + (m(m(b - c) + a) + cf - bf)^2 > f^2r^2
$$ or
$$
(m(b - c) - m^2a)^2 + (ma + c - b)^2 > f^2r^2.
$$ Which is
$$
f(ma + c - b)^2 > f^2r^2
$$
$$
(ma + c - b)^2 > fr^2 = (m^2 + 1)r^2
$$
And rearranged gives the formula above.
| {
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"timestamp": "2023-03-29T00:00:00",
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Q: Proving Existence I'm currently stuck on a problem right now for my Intro to Proofs Class. The problem says:
Let $a,b ∈ ℕ$. Prove that if $a+b$ is even, then there exists nonnegative integers $x$ and $y$ such that $x^2-y^2= ab$.
So far I've tried it directly, and by contrapositive and came to a similar road block.
Direct:
Assume $a,b ∈ ℕ$, and that $a+b$ is even.
$a+b$ being even $\implies$ $a+b=k_1$, such that $k_1 ∈ \mathbb{Z}$.
$a+b=k_1$ $\implies$ $a=k_1-b$.
Multiplying both sides of $a$ by $b$ we get: $ab= b(k_1-b)=bk_1-b^2$.
observing the conclusion $x^2-y^2=(x+y)(x-y)=ab$
It's at this point I'm at a road block. I'm not sure if direct is the way to go. I've also tried contrapositive, and contradiction, but i've also hit a roadblock for both of those as well.
| Well, brainstorm first.
$x^2 - y^2 = ab$
$(x - y)(x + y) = ab$. Can I say $x - y = a$ and $x+y = b$? Why or why not? Well, first off $x - y$ would have to equal the smaller of $a$ or $b$ and $x + y$ would have to be the larger. But we can assume without loss of generality that $a \le b$. So can we say that $x, y$ exist where $x-y = a$ or $x + y = b$?
That would mean $x$ is the midpoint of $a, b$ or that $x = \frac {a + b}2$. And that $y =$ the distance from the midpoint to either of the extreme $a$ or $b$ $= |\frac {a+b}2 - a| =|b - \frac {a+b}2|=|\frac {b-a}2|$.
And that is possible because $a+b$ is even!
So:
Pf: Let $x = \frac {a+b}2; y =|\frac {b-a}2|=$
Then ... (well, let's just push the throttle down and see what happens... we know it must work so ... let's go with it....)
then $x^2 - y^2 = (\frac {a+b}2)^2 - (\frac {b-a}2)^2 =$
$\frac {a^2 + 2ab + b^2}4 - (\frac {b^2 -2ab + a^2}4)=$
$\frac {4ab}4 = ab$.
======
Note: we did get lucky. There are probably other answers where $x-y \ne \min (a,b)$ and $x+y \ne \max (a,b)$ but those worked.
===
Alternatively... once we see that $x-y$ and $x + y$ can be extreme points of a segment we can say: Let $m = x+y$ and $d= x-y$. Now $d= m -2y$ so $m$ and $d$ are both even or both odd.
So we can solve for $m,d$ and set $x = \frac {m+d}2$ and $y = \frac {m-d}2$.
$x^2 - y^2 = md = ab$. So $m,d$ can be any complimentary factors of $ab$ BUT with the only stipulation that $m \ge d$ and that $m$ and $d$ are both the same parity.
As $a + b$ is even $a$ and $b$ are the same parity two such factors will always exists.
(We can always chose $m = \max (a,b)$ and $d = \min (a,b)$ if we want. But if we want another solution and if $a$ nor $b$ are primes. We can choose any complimentary factors provided they are both even or both odd.)
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"language": "en",
"url": "https://math.stackexchange.com/questions/2831864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding critical points of $f(x,y)= \sin x+\sin y + \cos(x+y)$
Find the critical points of function$$
f(x,y)=\sin x + \sin y + \cos(x+y),$$
where $0<x<\dfrac{\pi}{2}$, $0<y<\dfrac{\pi}{2}$.
What I have done:
$$f_{x}=\cos(x)-\sin(x+y),\\
f_{y}=\cos(y)-\sin(x+y).$$
From $f_{x}=0$, $\cos(x)=\sin(x+y)$. From $f_{x}=0$, $\cos(y)=\sin(x+y)$. I do not know where to go from here.
My attemps:
$$\sin\left(\frac{\pi}{2}-x\right)=\sin(x+y)=\sin\left(\frac{\pi}{2}-y\right).$$
| Hint:
$$\frac{df}{dx}=\cos x - \sin(x+y) = 0$$
$$\frac{df}{dy}=\cos y - \sin(x+y) = 0$$
Exploding the $\sin$s
$$\cos x - \sin x \cos y - \cos x \sin y = 0$$
$$\cos y - \sin x \cos y - \cos x \sin y = 0$$
Dividing the first with $\cos x$, the second with $\cos y$:
$$1 - \tan x \cos y - \sin y = 0$$
$$1 - \sin x - \tan x \cos y = 0$$
Expressing $\tan x$ from the first:
$$\tan x = \frac{1- \sin y}{\cos y}$$
...and now substitute it into the second equation. You have to express $\sin x$ with $\tan x$, there is a trigonometrical identity for that, but unfortunately this small margin is too narrow here to contain. ;-)
But the result will be a second degree equation and you will simply solve it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Then general values of $\theta$ in inverse Trigo sum
If $\displaystyle \theta = \tan^{-1}\bigg(2\tan^2 \theta\bigg)-\frac{1}{2}\sin^{-1}\bigg(\frac{3\sin 2 \theta}{5+4\cos 2 \theta}\bigg).$ Then general values of $\theta$ is
Try: Let $\alpha =\tan^{-1}\bigg(2\tan^2 \theta\bigg)\Rightarrow \tan \alpha =2 \tan^2 \theta$
and $\displaystyle \beta = \frac{1}{2}\sin^{-1}\bigg(\frac{3\sin 2 \theta}{5+4\cos 2 \theta}\bigg)\Rightarrow \sin (2\beta) = \frac{3\sin 2 \theta}{5+4\cos 2 \theta} = 3\frac{2\tan \theta}{1+\tan^2 \theta}\cdot \frac{1}{5+4\bigg(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\bigg)}$
So $\displaystyle \sin (2\beta) = \frac{6\tan \theta}{9+\tan^2 \theta}.$
could some help me how to find range of $\theta$. Thanks
| Hint:
$$\dfrac{3\sin2t}{5+4\cos2t}=\dfrac{6\tan t}{9+\tan^2t}=\dfrac{2\cdot\dfrac{\tan t}3}{1+\left(\dfrac{\tan t}3\right)^2}=\sin2y$$ where $3\tan y=\tan t$
$$\implies\sin^{-1}\dfrac{3\sin2t}{5+4\cos2t}=\begin{cases}2\tan^{-1}\dfrac{\tan t}3 &\mbox{if }-1\le\dfrac{\tan t}3\le1\\
\pi-2\tan^{-1}\dfrac{\tan t}3 & \mbox{if }\dfrac{\tan t}3>1\\
-\pi-2\tan^{-1}\dfrac{\tan t}3 & \mbox{if }\dfrac{\tan t}3<-1 \end{cases}$$
For $-1\le\dfrac{\tan t}3\le1,$
$t=\tan^{-1}(2\tan^2t)-\tan^{-1}\left(\dfrac{\tan t}3\right)$
$\implies\tan t=\dfrac{2\tan^2t-\dfrac{\tan t}3}{1+2\tan^2t\cdot\dfrac{\tan t}3}$
Clearly, $\tan t=0$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\cos{x} \cdot \cos{\sqrt{3}x} = 1$ has exactly one solution at $x=0$
Prove that $\cos{x} \cdot \cos{\sqrt{3}x} = 1$ has exactly on solution at $x=0$
I've recently asked a question on periodicity of the $f(x) = \cos{x}\cdot\cos{\sqrt{3}x}$
One of my statements was:
But this equation has only one solution at $x=0$
No prove of that was given in the OP since that statement was based on W|A numerical solution of the equation. A comment by @fleehood was added with a suggestion to prove that. And that is actually something I'm trying to do.
I've given it a try with analyzing Taylor series expansion for $\cos{x}$ and $\cos{\sqrt{3}x}$:
$$
f(x) = \cos{x} \cdot \cos{\sqrt{3}x} = \left( 1-\frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}+\ldots \right) \times \\ \times \left(1-\frac{(\sqrt{3}x)^2}{2!} + \frac{(\sqrt{3}x)^4}{4!} - \frac{(\sqrt{3}x)^6}{6!}+\ldots\right) = 1
$$
In order to find some pattern I've expanded several terms and eventually got a monstrous polynomial. I have to somehow prove that the polynomial has only one solution, but I'm not sure how to do that given infinite terms.
There may be other ways to prove that, so any ideas are appreciated.
| $$\cos \sqrt 3x \cdot \cos x
= \frac 12 \cos(\sqrt 3 + 1)x + \frac 12 \cos(\sqrt 3-1)x$$
So $\cos \sqrt 3x \cdot \cos x = 1 \iff
\cos(\sqrt 3 + 1)x + \cos(\sqrt 3-1)x = 2$
This can only happen when $\cos(\sqrt 3 + 1)x = \cos(\sqrt 3-1)x = 1$
So, for some integers $m$ and $n$,
\begin{align}
(\sqrt 3 + 1)x &= 2 \pi m \\
(\sqrt 3 - 1)x &= 2 \pi n \\
\end{align}
Adding we find $\sqrt 3x = \pi(m+n)$
and subtacting we find $x = \pi(m-n)$.
So
\begin{align}
\pi(m-n) &= \dfrac{\pi}{\sqrt 3}(m+n) \\
m-n &= \dfrac{1}{\sqrt 3}(m+n) \\
\sqrt 3(m-n) &= m+n \\
\end{align}
The only integer solution is $m=n=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Confusion about roots Solving $$z^4 + 1 = 0$$ where $z = x+iy$, I tried by factorizing the expressiong into $$z^4 + 1 = (z^2 - 2z + 2)(z^2 + 2z + 2).$$ But then solving $$z^2 - 2z + 2 = 0$$ for $z$ gives the solutions $z = 1 \pm i$, and solving $$z^2 + 2z + 2 = 0$$ for $z$ gives the solutions $z = -1 \pm i$. The actual roots should be $z = \frac{1 \pm i}{\sqrt{2}} $ and $ z= \frac{-1 \pm i}{\sqrt{2}}$. I'm possibly missing something that should be obvious to me here, but what about $1/\sqrt{2}$?
| Your starting factorization is not correct. It should be:
$$z^4+1 = z^4+2z^2+1-2z^2 = (z^2+1)^2-(\sqrt{2}z)^2=...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is
Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$
$$=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})+2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$
$$=2-2(a_{1}+a_{3})(a_{2}+a_{4})=2+2(a_{1}+a_{3})^2\geq 2$$
and equality hold when $\displaystyle a_{1}=-a_{3}$ and $a_{2}=-a_{4}$
Could some help me how to find its upper bound, Thanks
| Since $$a_{1}+a_{2}+a_{3}+a_{4}=0,$$ we can denote $$a_1+a_3=-(a_2+a_4)=t.$$Thus,
\begin{align}
&(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2\\=&2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})-2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})\\
=&2-2(a_{1}+a_{3})(a_{2}+a_{4})\\
=&2+2t^2.
\end{align}
Notice that
$$a_1a_3 \leq \frac{1}{4}(a_1+a_3)^2,$$ and
$$ a_2a_4 \leq \frac{1}{4}(a_2+a_4)^2$$
Hence,
$$2(a_1a_3+a_2a_4) \leq t^2.$$
But
$$2(a_1a_3+a_2a_4)=(a_1+a_3)^2+(a_2+a_4)^2-(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})=2t^2-1.$$
Therefore, $$2t^2-1 \leq t^2.$$Thus,
$$t^2 \leq 1.$$
As a result,$$2+2t^2 \leq 2+2 \cdot 1=4,$$with equality holding if and only if $a_1=a_3,a_2=a_4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Evaluate the Determinants A? Evaluate the Determinants
$$A=\left(\begin{matrix}
1 & 1 & 0 & 0 & 0\\
-1 & 1 & 1 & 0 & 0\\
0 & -1& 1 & 1 & 0\\
0& 0 & -1 & 1 & 1\\
0 & 0 & 0 & -1 & 1\\
\end{matrix}\right)$$
My attempst : I was thinking abouts the
Schur complement https://en.wikipedia.org/wiki/Schur_complement
$\det \begin{pmatrix}
A&B\\
C&D
\end{pmatrix}= \det (A-BD^{-1}C)\det D
$
As i not getting How to applied this formula and finding the Determinant of the Given question,,
Pliz help me or is There another way to find the determinant of the matrix.
Thanks u
| Given $A=\left(\begin{matrix}
1 & 1 & 0 & 0 & 0\\
-1 & 1 & 1 & 0 & 0\\
0 & -1& 1 & 1 & 0\\
0& 0 & -1 & 1 & 1\\
0 & 0 & 0 & -1 & 1\\
\end{matrix}\right)$
To find the determinant you need to find the upper triangular matrix and then multiply the diagonal elements of the matrix.
$$=\begin{vmatrix}
1 & 1 & 0 & 0 & 0\\
-1 & 1 & 1 & 0 & 0\\
0 & -1& 1 & 1 & 0\\
0& 0 & -1 & 1 & 1\\
0 & 0 & 0 & -1 & 1\\
\end{vmatrix}_{R_2->R_2+R_1}$$
$$=\begin{vmatrix}
1 & 1 & 0 & 0 & 0\\
0 & 2 & 1 & 0 & 0\\
0 & -1& 1 & 1 & 0\\
0& 0 & -1 & 1 & 1\\
0 & 0 & 0 & -1 & 1\\
\end{vmatrix}_{R_3->R_3+\dfrac12R_1}$$
$$=\begin{vmatrix}
1 & 1 & 0 & 0 & 0\\
0 & 2 & 1 & 0 & 0\\
0 & 0& \dfrac32 & 1 & 0\\
0& 0 & -1 & 1 & 1\\
0 & 0 & 0 & -1 & 1\\
\end{vmatrix}_{R_4->R_4+\dfrac23R_3}$$
$$=\begin{vmatrix}
1 & 1 & 0 & 0 & 0\\
0 & 2 & 1 & 0 & 0\\
0 & 0& \dfrac32 & 1 & 0\\
0& 0 & 0 & \dfrac53 & 1\\
0 & 0 & 0 & -1 & 1\\
\end{vmatrix}_{R_5->R_5+\dfrac35R_4}$$
$$=\begin{vmatrix}
1 & 1 & 0 & 0 & 0\\
0 & 2 & 1 & 0 & 0\\
0 & 0& \dfrac32 & 1 & 0\\
0& 0 & 0 & \dfrac53 & 1\\
0 & 0 & 0 & 0 & \dfrac85\\
\end{vmatrix}_{R_5->R_5+\dfrac35R_4}$$
Now the determinant $=1\times 2\times \dfrac32\times \dfrac53 \times \dfrac85=8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$a+b+c=1.$ Show that $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Assume $a,b,c>0$ and $a+b+c=1.$ Show that
$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq
\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Here's what I tried:
$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+b}+\frac{a+b+c}{a+b} \geq
\frac{1+a+b+c}{1+a}+\frac{1+a+b+c}{1+b}+\frac{1+a+b+c}{1+c}$
$\rightarrow \frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+b} \geq
\frac{b+c}{1+a}+\frac{a+c}{1+b}+\frac{a+b}{1+c}$
Let $A=b+c, B=a+c,$ and $ C=a+b,$ so $A+B+C=2.$
$\rightarrow \frac{1-A}{A}+\frac{1-B}{B}+\frac{1-C}{C} \geq \frac{A}{2-A}+\frac{B}{2-B}+\frac{C}{2-C}.$
Or $\frac{1}{A}+\frac{1}{B}+\frac{1}{C} - 3 \geq \frac{A}{2-A}+\frac{B}{2-B}+\frac{C}{2-C}.$
I am stuck here, thought about using AM-GM-HM to get rid of the reciprocal on the RHS but it doesn't work if applied directly.
| From Jensen's Inequality with the convex function $f(t):= \frac{1}{1-t}$ for $t\in (0,1)$, we have
$$\frac{a}{1-c}f(a)+\frac{b}{1-c}f(b)\geq f\left(\frac{a^2}{1-c}+\frac{b^2}{1-c}\right)\,.$$
Since $f$ is increasing, the Power-Mean Inequality $\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2$ implies that
$$f\left(\frac{a^2+b^2}{1-c}\right)\geq f\left(\frac{(a+b)^2}{2(1-c)}\right)=f\left(\frac{1-c}{2}\right)\,.$$
Thus,
$$\frac{a}{1-a}+\frac{b}{1-b}\geq (1-c)\,f\left(\frac{a^2+b^2}{1-c}\right)\geq (1-c)\,f\left(\frac{1-c}{2}\right)=\frac{2(1-c)}{1+c}\,.$$
We have two more similar inequalities, and by summing all three of them (and dividing the result by $2$), we have
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{1-a}{1+a}+\frac{1-b}{1+b}+\frac{1-c}{1+c}\,.$$
Adding $3$ to both sides of the inequality above, the required result is yielded. The equality case is, of course, $(a,b,c)=\left(\frac13,\frac13,\frac13\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculation of the modulus of elasticity of a stretched string I posted concerning this question a little while ago, not asking for the answer but for an understanding of the setup. Well, I thought I would solve it but I seem to be unable to obtain the required answer. If anyone could help I'd be very grateful.
Here's the question again:
Here's my attempt:
We are told that for the weight attached to the unstretched string that AC = $\frac{4a}{3}$ and CB = $\frac{4a}{7}$ so that means that AB = $\frac{4a}{3}+\frac{4a}{7}=\frac{40a}{21}$ = natural length of string
Below is my diagram of the final situation:
My method will be by consideration of energy.
Once the string is stretched over the bowl, but before the weight falls to touch the inner surface, it will have elastic potential energy due to having been stretched. I'll calculate this.
Then I'll calculate the energy stored in the string due to both it's stretching over the diameter AND the weight having fallen.
The difference in these two energies will be equal to the loss in potential energy of the weight.
So, natural length = $\frac{40a}{21}$
Length once stretched over the diameter is $2a =\frac{42a}{21}$
and so extension due to this is $\frac{2a}{21}$ and so the elastic potential energy in the string due to only the stretching over the diameter is $\frac{\lambda(\frac{2}{21})^2a^2}{2*\frac{40}{21}a}=\frac{4}{441}\lambda a*\frac{21}{80}$
I'll leave it in this form for convenience later on.
Ok. from the diagram:
$x\,cos\,30+y\,cos\,60=2a$ therefore
$\frac{\sqrt{3}}{2}x+\frac{y}{2}=2a$ therefore
$\sqrt{3}x+y=4a$ call this equation 1
also
$h=x\,cos\,60=\frac{x}{2}$ and $h=y\,cos\,30=\frac{\sqrt{3}}{2}y$
and so $x=2h$ and $y=\frac{2h}{\sqrt{3}}$
And so, from equation 1, we have
$2\sqrt{3}h+\frac{2h}{\sqrt{3}}=4a$ therefore
h = $\frac{\sqrt{3}a}{2}$
and so
$x=\sqrt{3}a$ and $y=\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{2}a=a$
So the new length, due to stretching over diameter AND falling of weight = $a(1+\sqrt{3})$
So the extension is now:
$a(1+\sqrt{3})-\frac{40a}{21} = (\sqrt{3}-\frac{19}{21})a$
So energy now stored in string is:
$\frac{\lambda(\sqrt{3}-\frac{19}{21})^2a^2}{2\frac{40a}{21}}$ = $\frac{21}{80}\lambda(\sqrt{3}-\frac{19}{21})^2a$
So, the change in elastic potential energy in going from just stretched across the diameter to streched across the diameter AND having the weight fallen is:
$\frac{21}{80}\lambda a[(\sqrt{3}-\frac{19}{21})^2-\frac{4}{441}]$
and this can be equated to the loss in gravitational potential energy of the weight W giving:
$\frac{21}{80}\lambda a[(\sqrt{3}-\frac{19}{21})^2-\frac{4}{441}]=wh=\frac{\sqrt{3}}{2}wa$
So I have $\lambda$ in terms of w BUT I do not have $\lambda=w$
Is my physical reasoning incorrect?
If not, have I made a mathematical mistake(s)?
Thanks for any help,
Mitch.
| Assuming the Hooke's law
$$
F = \lambda\left(\frac{l-l_0}{l_0}\right)
$$
Calling
$$
|AC|_0 = a\frac 43\\
|CB|_0 = a\frac 47\\
|AC| = 2a\sin(\frac{\pi}{3})\\
|CB| = 2a\sin(\frac{\pi}{6})\\
F_{AC} = \lambda\left(\frac{|AC|-|AC|_0}{|AC|_0}\right)(-\cos(\frac{\pi}{6}),\sin(\frac{\pi}{6}))\\
F_{CB} = \lambda \left(\frac{|CB|-|CB|_0}{|CB|_0}\right)(\cos(\frac{\pi}{3}),\sin(\frac{\pi}{3}))\\
R = r(-\cos(\frac{\pi}{3}),\sin(\frac{\pi}{3}))\\
W = w(0,-1)
$$
we have in equilibrium
$$
F_{AC}+F_{CB}+R+W=0
$$
or
$$
\left\{
\begin{array}{c}
-\frac{3 \sqrt{3} \left(\sqrt{3} a-\frac{4 a}{3}\right) \lambda }{8 a}+\frac{3 \lambda }{8}-\frac{r}{2}=0 \\
\frac{3 \left(\sqrt{3} a-\frac{4 a}{3}\right) \lambda }{8 a}+\frac{3 \sqrt{3} \lambda }{8}+\frac{\sqrt{3} r}{2}-w=0 \\
\end{array}
\right.
$$
now solving for $\lambda, r$ we have
$$
\begin{array}{c}
r=\left(\sqrt{3}-\frac{3}{2}\right) w \\
\lambda =w \\
\end{array}
$$
NOTE
$R = $ normal surface reaction force
$\lambda = $ string elastic modulus.
$|\cdot| = $ stretched lenght
$|\cdot|_0 = $ unstrectched length.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ a. at $\frac{1}{4}<|z|<1$ b. $|z|>1$
I know that $\mathbb{C}$ is a closed algebraic field so we can write the polynomial has a product of first degree polynomials, so we will have to guess one root and divide and find the others. but it is hard to guess here.
So I set $z=x+iy$ and got
$$(x^3-y^2x+2xy^2-2x^2+2y^2+\frac{1}{4})+i(-2x^2y+yx^2-y^3+4xy)=0$$
Looking at the imaginary part we get
$y(-x^2+4x-y^2)=0$ so or $y=0$ or $-x^2+4x-y^2=0\iff (x-2)^2+y^2=4$
But how I continue from here? and what does
I read and now I am given it a try:
for a. we look at $|z|<1$ and $\frac{1}{4}<|z|$ we look at the boundary
so fo $|z|=1$ $|z^3+\frac{1}{4}|<|-2z^2|$ so we choose $g(z)=z^3+\frac{1}{4}$ and $f(z)=-2z^2$
So $|z^3+\frac{1}{4}|\leq 1+\frac{1}{4}\leq 2=|-2z^2|$
So we can conclude that there are $2$ zeros in $|z|<1$?
| Using Rouche's theorem, in $|z|\leq\dfrac14$ we consider $f(z)=z^3-2z^2$ and $g(z)=\dfrac14$ then
$$|f(z)|=|z^3-2z^2|\leq|z|^3+2|z|^2=\dfrac{9}{64}<|g(z)|$$
then $z^3-2z^2+\frac14=0$ hasn't zero in $|z|\leq\dfrac14$.
In $|z|\leq1$ with $f(z)=z^3+\dfrac14$ and $g(z)=-2z^2$ then
$$|f(z)|=|z^3+\dfrac14|\leq\dfrac54<2=|g(z)|$$
and $g(z)$ has two zeros in $\dfrac14<|z|<1$, then $z^3-2z^2+\frac14=0$ has two zeros there.
The third root will be in $|z|>1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Trying to evaluate $\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{dx}{1+x^3}$ I would like to work this out:
$$I=\large\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{\mathrm dx}{1+x^3}$$
Making a sub: $u=x^3$, $dx=\frac{du}{3x^2}$
$$I=\frac{1}{3}\int_{0}^{\infty}\frac{\ln(1+u)}{u^{2/3}(1+u)^2}\mathrm du$$
Making a sub: $u=\tan^2(y)$, $du=\sec^2(y)dy$
$$I=\frac{2}{3}\int_{0}^{\pi/2}\frac{\ln\sec(y)}{\sec^2(y)\sqrt[3]{\tan^4(y)}}\mathrm dy$$
$$I=\frac{2}{3}\int_{0}^{\pi/2}\frac{\cot(y)\cos^2(y)\ln\sec(y)}{\sqrt[3]{\tan(y)}}\mathrm dy$$
I can't continue.
Maybe there is another alternative way to simplify $I$
| The substitution $t = (1 + x^3)^{-1}$ yields
$$ I = \frac{1}{3} \int \limits_0^1 - \ln(t) \left(\frac{t}{1-t}\right)^{2/3} \, \mathrm{d} t = f'\left(\frac{2}{3}\right) \, , $$
where
$$ f(\alpha) \equiv - \frac{1}{3} \int \limits_0^1 \frac{t^\alpha}{(1-t)^{2/3}} \, \mathrm{d} t = - \frac{1}{3} \operatorname{B} \left(\frac{1}{3}, \alpha +1 \right) = - \frac{\operatorname{\Gamma} \left(\frac{1}{3}\right)}{3} \frac{\operatorname{\Gamma}(\alpha + 1)}{\operatorname{\Gamma}\left(\alpha + \frac{4}{3} \right)} ~~~ , \, \alpha > -1 \, .$$
The differentiation under the integral sign can be justified using the dominated convergence theorem.
In terms of the digamma function $\psi$ we now have
$$ I = f'\left(\frac{2}{3}\right) = \frac{2}{9} \operatorname{\Gamma} \left(\frac{1}{3}\right) \operatorname{\Gamma} \left(\frac{2}{3}\right) \left[\operatorname{\psi} (2) - \operatorname{\psi} \left(\frac{5}{3}\right)\right] \, . $$
We can use the reflection formulas for $\Gamma$ and $\psi$ and the recurrence relation $\operatorname{\psi}(x + 1) = \operatorname{\psi}(x) + \frac{1}{x}$ to find
$$ I = \frac{2}{9} \frac{\pi}{\sin\left(\frac{\pi}{3}\right)} \left[\operatorname{\psi}(1) - \operatorname{\psi} \left(\frac{1}{3}\right) - \pi \cot\left(\frac{\pi}{3}\right) - \frac{1}{2} \right] \, .$$
The special values $\operatorname{\psi}(1) = - \gamma$ and $\operatorname{\psi} \left(\frac{1}{3}\right) = - \frac{\pi}{2 \sqrt{3}} - \frac{3}{2} \ln (3) - \gamma$ then lead to the final result
$$ I = \frac{2 \pi}{27} [\sqrt{3} (3 \ln(3) - 1) - \pi] \, .$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the system has a limit cycle. The full question reads:
Show that the system with $\dot{x} = x-y-(x^2 + \frac{3}{2} y^2)x$ and $\dot{y} = x+y-(x^2 + \frac{1}{2} y^2)y$ has a limit cycle.
I wanted to check the correctness of my method. My thought was to find the equilibrium points (If I am correct, the only one is $(0,0)$, then find the stability of that point. Then I attempted to show that if we go far enough out in the domain, that the flow of the direction field is going in the opposite direction of the direction field directly surrounding the equilibrium point. Then we can use Poincare-Bendixon and be done. Is this correct, or should I handle it differently? Thanks!
| Yes, your approach looks good to me. If $r^2 = x^2 + y^2$ then we have
$r\dot{r} = x\dot{x} + y\dot{y} = x^2(1-x^2-\frac{3}{2}y^2) + y^2(1-x^2-\frac{1}{2}y^2)$
$\Rightarrow r(1-x^2-\frac{3}{2}y^2) \le \dot{r} \le r(1-x^2-\frac{1}{2}y^2)$
and since $1-\frac{3}{2}r^2 \le 1-x^2 - \frac{3}{2}y^2 \le 1-x^2-\frac{1}{2}y^2 \le 1 - \frac{1}{2}r^2$
which suggests that the circles $r=\sqrt{\frac{2}{3}}$ and $r=\sqrt{2}$ may be significant ...
| {
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Find number of integers satisfying $x^{y^z} \times y^{z^x} \times z^{x^y}=5xyz$ Find number of integers satisfying $$x^{y^z} \times y^{z^x} \times z^{x^y}=5xyz$$
My try:
we can rewrite the equation as
$$x^{y^z-1} \times y^{z^x-1} \times z^{x^y-1}=5$$
Then since all are integers we get
$x=5$, $y=1$ ,$z=1$ and like wise we get other two triplets.
is there a way to do formally?
| The factor of $5$ must come from somewhere, so $x, y$ or $z$ must have a factor of $5$ (let's just assume it's $x$, as the expression is symmetric). The power of $x$, i.e. $y^z - 1$, must be strictly positive. If there were any other prime factor of $x$, because $x$ is raised to a positive integer power, it would have to divide $5$, hence $5$ is the only prime factor of $x$. Moreover, we can similarly see that $25$ does not divide $x$, so $x$ is $5$ (or $y$ or $z$, when you account for symmetry).
Since $x = 5$, we must have $y^z - 1 = 1$. We already knew it had to be strictly larger than $0$, but if it were at least $2$, then $25$ would divide the left side but not the right side. Thus we have $y^z = 2$.
Similarly, this means $y = 2$. We know via the same arguments that $2$ must divide $y$, that $2$ is the only prime factor of $y$, and if $y$ were a power of $2$ greater than $2$ itself, then $y^z$ would be divisible by $4$. Similarly, $z = 1$.
So, our only possible solutions are $(x, y, z) = (5, 2, 1), (2, 1, 5), (1, 5, 2)$, by symmetry. Are they solutions?
No. There are no solutions.
| {
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Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
| Just for a variation, using Lagrange’s method:
$$
f(a,b,t)=\frac{b+1}{a+b-2}-t(a^2+b^2-1)
$$
Then
\begin{align}
\frac{\partial f}{\partial a}&=-\frac{b+1}{(a+b-2)^2}-2at \\[6px]
\frac{\partial f}{\partial b}&=\frac{a-3}{(a+b-2)^2}-2bt
\end{align}
If these equal $0$, then
$$
-\frac{b+1}{a(a+b-2)^2}=\frac{a-3}{b(a+b-2)^2}
$$
so that $-b^2-b=a^2-3a$, that gives $3a-b=1$ and from $a^2+b^2=1$ we derive $a=0$ or $a=3/5$.
The critical points are thus $(0,-1)$ and $(3/5,4/5)$. We have
$$
f(0,-1,0)=\frac{2}{3},\quad
f(3/5,4/5,0)=-3,\quad
$$
This also shows the maximum.
| {
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Can we conclude that two of the variables must be $0$?
Assuming $$a^2+b^2+c^2=1$$ and $$a^3+b^3+c^3=1$$ for real numbers $a,b,c$, can we conclude that two of the numbers $a,b,c$ must be $0$ ?
I wonder whether mathworld's result that only the triples $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ satisfy the given equation-system , is actually true.
Looking at $(a+b+c)^3$ and $(a+b+c)^2$ , using
\begin{align}
&(a+b+c)^3= (a+b+c)^2(a+b+c)=\\
&(1+2(ab+ac+bc))(a+b+c)= \\
&2(a+b)(a+c)(b+c)+a+b+c+2abc
\end{align}
and eliminating $(a+b)(a+c)(b+c)$, with $S:=a+b+c$ , I finally got $$(S-1)^2(S+2)=6abc$$
I guess this is not enough to show the above result (if it is true at all).
This question is inspired by an exercise to determine the possible values of $a+b+c$ assuming the above equations, so this question could be a duplicate, but I am not sure whether it actually is.
| Since $|a|,|b|,|c|\in[0,1]$ (because $a^2+b^2+c^2=1$), we have
$$
a^2(1-a)+b^2(1-b)+c^2(1-c)=0
$$
The LHS is a sum of non-negative terms; to be zero, all the terms must be zero. This means that
$$
a^2(1-a)=b^2(1-b)=c^2(1-c)=0
$$
and therefore $|a|,|b|,|c|\in\{0,1\}$. Along with $a^2+b^2+c^2=1$, we get that exactly one of them is $1$, and two must be zero: this implies the result.
| {
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Minimizing in 3 variables
Find the least possible value of the fraction $\dfrac{a^2+b^2+c^2}{ab+bc}$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3\sqrt[3]{-2ac/b(a+c)}$
And I cant somehow move on.
| Yet another approach: The expression does not change if $(a, b, c)$
are multiplied by a common factor, so we can assume that $a+c=2$. Then
$$
\frac{a^2+b^2+c^2}{ab+bc} = \frac 12 \left (\frac{a^2+(2-a)^2}{b} + b
\right) \ge \sqrt{a^2 + (2-a)^2} = \sqrt{2(a-1)^2 + 2} \, ,
$$
using $AM \ge GM$. It follows that
$$
\frac{a^2+b^2+c^2}{ab+bc} \ge \sqrt 2 \, ,
$$
with equality if and only if $(a, b, c)$ is a multiple of $(1, \sqrt 2, 1)$.
| {
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prove this inequality $(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$ Let $x,y,z>0$,show that
$$(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$$
I have prove this inequality
$$(x+y-z)(y+z-x)(x+y-z)\le xyz$$
because it is three schur inequality
$$\Longleftrightarrow x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x)$$
how to solve this inequiality $xyz ⩾ (x+y-z)(y+z-x)(z+x-y)$
But I can't prove this inequality to prove $(1)$
| It's enough to prove our inequality for $\prod\limits_{cyc}(x+y-z)\geq0.$
Now, if $x+y-z\leq0$ and $x+z-y\leq0$ then $x+y-z+x+z-y\leq0$ or $2x\leq0$,
which is a contradiction.
Thus, we can assume that $x+y-z>0$, $x+z-y>0$ and $y+z-x>0.$
Now, let $x+y-z=c$, $x+z-y=b$ and $y+z-x=a$.
Thus, we need to prove prove that
$$27(a+b)^2(a+c)^2(b+c)^2\geq64abc(a+b+c)^3$$ and since
$$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ it's just $$\sum_{cyc}c(a-b)^2\geq0,$$ it's enough to prove that
$$3(ab+ac+bc)(a+b)(a+c)(b+c)\geq8abc(a+b+c)^2$$ or
$$\sum_{cyc}(3a^3b^2+3a^3c^2-2a^3bc-4a^2b^2c)\geq0,$$
which is true by Muirhead.
A bit of easier way it's to use $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ for the all term $\prod\limits_{cyc}(a+b)^2.$
If so, it's enough to prove that $$(ab+ac+bc)^2\geq3abc(a+b+c),$$ which is obvious.
The following much more stronger inequality is also true.
Let $x$, $y$ and $z$ be real numbers. Prove that:
$$4x^2y^2z^2\geq(x+y-z)(x+z-y)(y+z-x)(x^3+y^3+z^3+xyz).$$
| {
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Solve Recurrence Relation $a_k=(a_{k-1})^2-2$ $$a_k=\left(a_{k-1}\right)^2-2$$
$a_0=\frac{5}{2}$
Then find $$P=\prod_{k=0}^{\infty} \left(1-\frac{1}{a_k}\right)$$
My try:
I rewrote the Recurrence equation as
$$a_k+1=(a_{k-1}-1)(a_{k-1}+1)$$ $\implies$
$$\frac{1}{a_{k-1}-1}=\frac{a_{k-1}+1}{a_k+1}$$ $\implies$
$$\frac{a_{k-1}}{a_{k-1}-1}=\frac{(a_{k-1})^2+a_{k-1}}{a_k+1}=\frac{a_k+a_{k-1}+2}{a_k+1}$$
any hint here?
| Hint: There are three steps. First, for each $k=0,1,2,\ldots$, show that $$a_k=2^{2^k}+\frac{1}{2^{2^k}}\,.$$
Second, write $$1-\frac{1}{a_k}=\frac{a_k-1}{a_k}=\left(\frac{a_{k+1}+1}{a_k+1}\right)\frac{1}{a_k}\,,$$
for all $k=0,1,2,\ldots$.
Finally, show that
$$\prod_{k=0}^n\,a_k=\frac{2}{3}\left(2^{2^{n+1}}-\frac{1}{2^{2^{n+1}}}\right)\,,$$ using the identity
$$(x-y)\,\prod_{k=0}^n\,\left(x^{2^k}+y^{2^k}\right)=x^{2^{n+1}}-y^{2^{n+1}}\,,$$
for all $n=0,1,2,\ldots$.
You should in the end obtain $$\prod_{k=0}^\infty\,\left(1-\frac{1}{a_k}\right)=\lim_{n\to\infty}\,\frac{3}{7}\left(\frac{2^{2^{n+1}}+1+\frac{1}{2^{2^{n+1}}}} { 2^{2^{n+1}}-\frac{1}{2^{2^{n+1}}} }\right)=\frac{3}{7}\,.$$
| {
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Determining $\sin(2x)$
Given that
$$\sin (y-x)\cos(x+y)=\dfrac 1 2$$
$$\sin (x+y)\cos (x-y) = \dfrac 1 3 $$
Determine $\sin (2x)$.
As stated in my perspective, the question does not make any sense. We know that the double angle identity for $\sin(2x)$ is given by
$$\sin(2x) = 2\sin\cos$$
Let us try simpiflying the second equation
$$\sin(x+y)-\cos(x-y)=\sin x \cos y+\cos x \sin y-\cos x \cos y-\sin x\sin y=
\cos x(\sin y-\cos y)+\sin x(\cos y-\sin y)=\color{blue}{(\cos x-\sin x)(\sin y-\cos y})$$
However, there seems to be nothing useful.
| You were on the right track but the equations contain the product, not the difference.
\begin{align}
\frac12 &= \sin (y-x)\cos(x+y)\\
&=(\sin y\cos x - \sin x \cos y)(\cos x\cos y - \sin x\sin y) \\
&= \sin y\cos y - \cos x \sin x
\end{align}
\begin{align}
\frac13 &= \sin (x+y)\cos(x-y)\\
&=(\sin x\cos y + \sin y \cos x)(\cos x\cos y + \sin x\sin y) \\
&= \sin y\cos y + \cos x \sin x
\end{align}
So subtracting them gives
$$\sin 2x = 2\cos x\sin x = ( \sin y\cos y + \cos x \sin x) - ( \sin y\cos y - \cos x \sin x) = \frac13 - \frac12 = -\frac16$$
| {
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knowing: $\cos x+\sin x=\frac{5}{4}$, obtain: $\cos(4x)$
knowing: $\cos x+\sin x=\frac{5}{4}$,
obtain: $\cos(4x)$
$$\cos x+\sin x=\frac{5}{4}$$
$$\sin^2x+\cos^2x+2\sin x\cos x=\frac{25}{16}$$
$$\sin2x=\frac{25}{16}-\frac{16}{16}=\frac{9}{16}$$
$$\cos4x=1-2\sin^22x=1-2\Bigl(\frac{9}{16}\Bigr)^2=\frac{47}{128}$$
Taken out of one of the TAU entry tests but unfortunately, they don't give solutions to most of the exercises, so… Am I correct?
| Another way to do this is as follows-
$\cos x+\sin x=\frac{5}{4}$, obtain: $\cos(4x)$
$$\Rightarrow 4\cos x + 4\sin x = 5$$
$$\Rightarrow 16 \cos^2x +25-40 \cos x = 16 \sin x$$
$$\Rightarrow 15 \cos^2x -56 \sin x+25=0$$
$$\Rightarrow \cos x = \frac{\sqrt{409}}{15}$$
Now, by knowing the value of x, the cosine value of $4x$ can be calculated in the similar manner.
| {
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Give an asymptotic developement of $I_n=\int_0^1 (x^{n}-x^{n-2})\ln(1+x^n)dx.$ Let $$I_n=\int_0^1 (x^n+x^{n-2})\ln(1+x^n)dx.$$
Give an asymptotic developement of $I_n$ at order $O(\frac{1}{n^3})$ when $n\to \infty $.
I wanted to use the fact that $$\sum_{k=1}^\infty \frac{(-1)^{k+1} x^{kn}}{k},$$
and thus $$I_n=\int_0^1(x^n+x^{n-2})\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^{kn}dx,$$
but since the convergence of this series is a priori not uniform and the coefficient are not positive on $[0,1]$ I can't permute the sum and the integral.
Any other idea ?
| Assuming that
you can reverse the order,
$\begin{array}\\
I_n
&=\int_0^1(x^n+x^{n-2})\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^{kn}dx\\
&=\sum_{k=1}^\infty \int_0^1(x^n+x^{n-2})\frac{(-1)^{k+1}}{k}x^{kn}dx\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k} \int_0^1(x^n+x^{n-2})x^{kn}dx\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k} \int_0^1(x^{n(k+1)}+x^{n(k+1)-2})dx\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}(\frac1{n(k+1)+1}+\frac1{n(k+1)-1})\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}(\frac{2n(k+1)}{n^2(k+1)^2-1})\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{kn}(\frac{2(k+1)}{(k+1)^2-1/n^2})\\
&=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k(k+1)n}(\frac{2}{1-1/(n^2(k+1)^2)})\\
&=\frac{2}{n}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k(k+1)}\sum_{m=0}^{\infty}\frac{1}{n^{2m}(k+1)^{2m}}\\
&=\frac{2}{n}\sum_{m=0}^{\infty}\frac{1}{n^{2m}}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k(k+1)}\frac{1}{(k+1)^{2m}}\\
&=2\sum_{m=0}^{\infty}\frac{1}{n^{2m+1}}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k(k+1)^{2m+1}}\\
&=2(\frac{\ln(4/e)}{n}-\frac{0.110304071913...}{n^3}+O(\frac1{n^5}))
\qquad\text{(according to Wolfy)}\\
\end{array}
$
Note:
The Inverse Symbolic Calculator
did not find anything for
0.110304071913.
| {
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Calculate boundary of two domains $D_1 =\{ -1 \le x \le 1-\sqrt{y^2+z^2}\}$
$D_2 =\{ x^2+y^2+z^2\le 8,z\ge 2\}$
Is it correct to say that the boundary of $D_1$ is :
$B(D_1) =\{ (x-1)^2=y^2+z^2,(x+1)^2=y^2+z^2\}$
and the boundary of $D_2$ :
$B(D_2) =\{z\ge 2 ,x^2+y^2+z^2=8 \} \cup \{x^2+y^2=4\}$
?
Especially the last one. I know for sure that the boundary is $x^2+y^2=4$, but i'm not sure that the boundary is also the 'cover' of the shpere($x^2+y^2+z^2=8$)
| $C= \{x \in \mathbb R^3 ; x^2+y^2=4\}$ is not a subset of $B(D_2)$. $C$ is the right circular cylinder of axis $Oz$ and with radius $2$. $C$ is unbounded while $D_2$ is bounded. What is true is that
$$B(D_2) = \{z\ge 2 ,x^2+y^2+z^2=8 \} \cup \{z=2, x^2+y^2 \le 4\}.$$
Your $B(D_1)$ isn't correct either. To "compute" the boundary, use the fact that one of the inequality has to be an equality and look at what it implies for the second inequality.
| {
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Find the positive value of $x$ satisfying the given equation $${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = \sec(y)$ but couldn't even come close to the solution.
| Let's see how insurmountable this sextic equation is.
Start by deriving the equation. First multiply by $\sqrt{x}$, square both sides, and isolate the remaining radical:
$(x^2-1)+2\sqrt{(x^2-1)(x-1)}+(x-1)=x^3$
$2\sqrt{(x^2-1)(x-1)}=x^2-x^2-x+2$
Square again, expand and collect to get our monster:
$x^6-2x^5-x^4+2x^3+x^2=0$
Clearly since $x$ must be positive (certainly not zero, which the fractional radicands in the original equations would forbid), the factor $x^2$ can be divided out. We are down to degree $4$:
$x^4-2x^3-x^2+2x+1=0$
Next observe that this quartic equation has the following property:
(Linear coefficient)/(Cubic coefficient) $=c$
(Constant)/(Quartic coefficient)$=c^2$
When this occurs, the roots of the quartic equation occur in pairs giving the product $c$ and thus we must have a factorization:
$(x^2+ax+c)(x^2+bx+c)$
Here, $c=-1$ and so:
$x^4-2x^3-x^4+2x+1=0=(x^2+ax-1)(x^2+bx-1)$
Expanding the right side and matching like terms leads to two independent equations, thus:
$a+b=-2; b=-2-a$
$ab=1$
Then $a(-2-a)=1, a^2+2a+1=0$ and we have the root $a=-1$. Thus $b=-1$ and our polynomial is now reduced to a single, squared factor:
$(x^2-x-1)^2=0$
And so, from the quadratic formula,
$x=\frac{1+\sqrt{5}}{2}$
Which the reader can verify, checks out! Note that in the checking process we should find $(x^2-1)/x=1$.
| {
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Reduction of Order via Substitution
Suppose $u_1=\sin{x^2}$ is a solution of
$$xu''-u'+4x^3u=0\Rightarrow u''-x^{-1}u'+4x^2u=0 \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
I am trying to find a second, linearly independent, solution (say $u_2$) to the above equation using reduction of order.
Now, using the formula
$$u_2=u_1\int\frac{dx}{u^2_1\text{exp}\left(\int p(x)\ dx\right)} \ \ \ \ ,\ p(x)=-x^{-1}$$
I have found that $$u_2=\frac{\sin{x^2}}{2}\int \text{cosec}^2x \ dx=-\frac{\cos{x^2}}{2}$$
Now I tried to replicate this using the substitution $u_2=u_1v(x)$.
After finding the first and second derivatives of this equation and substituting into $(1)$, I get
$$v''\sin{x^2}+v'\left(4x\cos{x^2}-x^{-1}\sin{x^2}\right)=0$$
Letting $w=v'$,
$$\frac{dw}{dx}\sin{x^2}+w\left(4x\cos{x^2}-x^{-1}\sin{x^2}\right)=0$$
I tried to simplify this using the integrating factor
$$\text{exp}\left(\int 4x\cos{x^2}-x^{-1}\sin{x^2} \ dx\right)$$ but could not compute it.
How can I solve this problem using my suggested substitution?
| The ODE for $w$ is actually separable:
\begin{align*}
\sin(x^2)\frac{dw}{dx} + \left(4x\cos(x^2) - \frac{\sin(x^2)}{x}\right)w & = 0 \\
\int \frac{1}{w}\, dw & = \int \frac{1}{\sin(x^2)}\left(\frac{\sin(x^2)}{x} - 4x\cos(x^2)\right)\, dx \\
\int\frac{1}{w}\, dw & = \int\frac{1}{x} - 4x\cot(x^2)\, dx \\
\end{align*}
Making a change of variable $s = x^2$ on the second integrand, we obtain
\begin{align*}
\int\frac{1}{w}\, dw & = \int \frac{1}{x}\, dx - 2\int\cot(s)\, ds \\
\ln|w| & = \ln|x| - 2\ln|\sin(x^2)| + C
\end{align*}
You may then exponentiate each side and get
$$ |w| = e^C\frac{|x|}{\sin^2(x^2)} = \frac{A|x|}{\sin^2(x^2)}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{-1}^{1} \cot^{-1} \left(\frac{1}{\sqrt{1-x^2}}\right) \cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-(x^2)^{|x|}}}\right)$
$$\int_{-1}^{1} \left(\cot^{-1} \dfrac{1}{\sqrt{1-x^2}}\right) \left(\cot^{-1}\dfrac{x}{\sqrt{1-(x^2)^{|x|}}}\right)= \dfrac{\pi^2(\sqrt a-\sqrt b )}{\sqrt c}$$
, where a,b, c are natural numbers and are there in their lowest form, then find the value of a+b+c.
Using $\int_a^b f(x)dx= \int_a^bf(a+b-x{) dx}$, I got:
$$2I =2\pi\int_{0}^1 \cot^{-1}\left(\dfrac{1}{\sqrt{1-x^2}}\right) dx$$
Then, letting $x = \sin \theta$
$I = \pi\displaystyle\int_{0}^{\pi/2}\arctan (\cos\theta) \cos \theta d\theta$
After this I tried integration by parts but it gets really complicated with that? How do I continue?
EDIT:
Please note that arccot(x) + arccot(-x)= $\pi$ $\ne 0$
Principal range of $\cot^{-1}x$ considered in the question is $(0,\pi)$
| I usually see $\cot^{-1}(-x)=-\cot^{-1}(x)$, but working with what is stated in the question,
$$
\begin{align}
&\int_{-1}^1\cot^{-1}\left(\frac1{\sqrt{1-x^2}}\right)\cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-\left(x^2\right)^{|x|}}}\right)\,\mathrm{d}x\tag1\\
&=\int_0^1\cot^{-1}\left(\frac1{\sqrt{1-x^2}}\right)\cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-\left(x^2\right)^{|x|}}}\right)\,\mathrm{d}x\\
&+\int_0^1\cot^{-1}\left(\frac1{\sqrt{1-x^2}}\right)\left[\pi-\cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-\left(x^2\right)^{|x|}}}\right)\right]\,\mathrm{d}x\tag2\\
&=\pi\int_0^1\tan^{-1}\left(\sqrt{1-x^2}\right)\,\mathrm{d}x\tag3\\
&=\pi\int_0^1\frac{\sqrt{1-x^2}}{1+x^2}\,\mathrm{d}x\tag4\\
&=\pi\int_0^{\pi/2}\frac{\cos^2(x)}{1+\sin^2(x)}\,\mathrm{d}x\tag5\\
&=\pi\int_0^{\pi/2}\frac2{1+\sin^2(x)}\,\mathrm{d}x-\frac{\pi^2}2\tag6\\
&=\pi\int_0^{\pi/2}\frac{2\sec^2(x)}{1+2\tan^2(x)}\,\mathrm{d}x-\frac{\pi^2}2\tag7\\
&=\pi\sqrt2\int_0^\infty\frac1{1+u^2}\,\mathrm{d}u-\frac{\pi^2}2\tag8\\
&=\pi^2\frac{\sqrt2-1}2\tag9
\end{align}
$$
Explanation:
$(2)$: use $\cot^{-1}(-x)=\pi-\cot^{-1}(x)$ and substitute $x\mapsto-x$ on $[-1,0]$
$(3)$: $\cot^{-1}(x)=\tan^{-1}\left(\frac1x\right)$
$(4)$: substitute $x\mapsto\sqrt{1-x^2}$ then integrate by parts
$(5)$: substitute $x\mapsto\sin(x)$
$(6)$: add $1$ to the integrand and subtract $\frac{\pi^2}2$
$(7)$: multiply numerator and denominator of the integrand by $\sec^2(x)$
$(8)$: substitute $u=\sqrt2\tan(x)$
$(9)$: substitute $u=\tan(\theta)$ and integrate
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
| $$
\cos x-\cos(3x) = \cos x\left(\sin^2x+3\sin^2 x\right)
$$
then
$$
\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} = \frac{x^2}{x^2}\cos x\left(\frac{\sin^2x+3\sin^2 x}{\sin (3x^2)-\sin (x^2)}\right)=\cos x\left(\frac{\left(\frac{\sin x}{x}\right)^2+3\left(\frac{\sin x}{x}\right)^2}{\frac{3\sin(3x^2)}{3x^2}-\frac{\sin (x^2)}{x^2}}\right)
$$
hence
$$
\lim_{x\to 0}=\cos x\left(\frac{\left(\frac{\sin x}{x}\right)^2+3\left(\frac{\sin x}{x}\right)^2}{\frac{3\sin(3x^2)}{3x^2}-\frac{\sin (x^2)}{x^2}}\right) = 1\cdot\left(\frac{1+3}{3-1}\right) = 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 7
} |
Evaluate the intgral $\int{\frac{dx}{x^2(1-x^2)}}$ (solution verification) I have to solve the following integral
$$\int{\frac{dx}{x^2(1-x^2)}}$$
What I've got:
\begin{split}
\int{\frac{dx}{x^2(1-x^2)}} &=\int{\frac{(1-x^2+x^2)dx}{x^2(1-x^2)}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{1-x^2}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{(1+(xi)^2)}}\\
&=-x^{-1}+\arctan{xi}+C
\end{split}
Is this correct?
Thanks in advance!
| Don't forget $\dfrac1i$ before $\arctan$
$$\int{\frac{dx}{x^2(1-x^2)}}=-x^{-1}+\dfrac{1}{i}\arctan{xi}+C$$
Also better to write
$$\int\dfrac{1}{1-x^2}dx=\dfrac12\ln\dfrac{1+x}{1-x}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find the values of $x$ and $y$ that satisfies $\sin(x+y)=\sin x+\sin y$. I know that in general the following equality does not hold: $\sin(x+y)=\sin x + \sin y$. However, I have been looking for specific values of $x$ and $y$ that satisfies the given equation. This is what I have done so far:
$\sin(x+y)=\sin x\cos y + \sin y \cos x = \sin x + \sin y$. Then from this equation, I got
$\sin x(1-\cos y)+\sin y(1-\cos x)=0$.
There are two possibilities:
Case 1: $\sin x(1-\cos y)=0$ and $\sin y(1-\cos x)=0$. Solving equations for $x$ and $y$, we get that $x=y=2\pi k$ for some integer $k$.
Case 2: $\sin x(1-\cos y)=n$ and $\sin y(1-\cos x)=-n$ where $n$ is a nonzero real number. Then we have
$\sin x(1-\cos y)=\sin y(\cos x-1)$ Since $n\neq0$ then we can divide both sides of this equation by $\sin x \sin y)$ to obtain $\csc y - \cot y = \cot x - \csc x$. I'm stuck in here. Any suggestions and hints (not answers) will be welcomed...
| We know that
$$\sin(x+y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)$$
and by sum to product formula we have
$$\sin(x)+\sin(y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
therefore
$$\sin(x+y)=\sin(x)+\sin(y) $$
$$\iff 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
then we need to consider two cases
$$2\sin\left(\frac{x+y}{2}\right)=0 $$
or otherwise we can cancel out $2\sin\left(\frac{x+y}{2}\right)$ and obtain
$$\cos\left(\frac{x+y}{2}\right)=\cos\left(\frac{x-y}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
compute $\frac{1}{2}(\frac{1}{3.5........(2n-1)} -\frac{1}{3.5........(2n+1)})$ compute the summation
$\sum_ {n=1}^{\infty} \frac{n}{3.5........(2n+1)}= ?$
My attempts : i take $a_n =\frac{n}{3.5........(2n+1)}$
Now =$\frac{1}{2}$ .$\frac{(2n+ 1) - 1}{3.5........(2n+1)}= \frac{1}{2}(\frac{1}{3.5........(2n-1)} -\frac{1}{3.5........(2n+1)})$
after that i can not able to proceed further,,,
pliz help me..
thanks u
| You don't need to compute $a_n = \frac{1}{2}(\frac{1}{3.5…(2n-1)} -\frac{1}{3.5…(2n+1)})$
You need to compute $\sum_{n=1}^K a_n$.
Which is $\frac 12(\frac 11 - \frac 13) + \frac 12(\frac 13-\frac 1{3\cdot 5})+.....+ \frac{1}{2}(\frac{1}{3.5…(2K-3)} -\frac{1}{3.5…(2K-1)}) + \frac{1}{2}(\frac{1}{3.5…(2K-1)} -\frac{1}{3.5…(2K+1)})$
And the compute $\sum_{n=1}^\infty a_n =\lim\limits_{K\to \infty}\sum_{n=1}^K a_n$
====
Note: If $a_n = \frac{1}{2}(\frac{1}{3.5…(2n-1)} -\frac{1}{3.5…(2n+1)})$ we can let $b_n =\frac{1}{3.5…(2n-1)}$
and then we have
$$\sum_{n=1}^{\infty}\frac 12 (b_n - b_{n+1})$$.
And you should be able so solve that even if you don't have any frigging idea what $b_n$ was defined as.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Coin change with fixed denominations with duplicate values Suppose I have $\{1, 1\} \{2, 2, 2\} \{5\} \{10\}$ coins.
How many ways I can get 15 by adding coins?
If I use DP then it counts 7, because $\{1+2+2+10\} \{1+2+2+10\} \{1+2+2+10\} \{1+2+2+10\} \{1+2+2+10\} \{1+2+2+10\}$ and $\{10 + 5\}$. But actual result is $\{1+2+2+10\}$ and $\{10 + 5\}$. How do I solve efficiently?
Now I able to figure out the solution. I have to find integral solutions of equation $x+2y+5z+10m=15$ with constraints $x <= 2$, $y <= 3$, $z <= 1$ and $m <= 1$. Now how to do that?
| This is a supplement to the already given answer showing that calculating the coefficient is not that cumbersome. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a polynomial.
We obtain
\begin{align*}
\color{blue}{[x^{15}]}&\color{blue}{(1+x+x^2)(1+x^2+x^4+x^6)(1+x^5)(1+x^{10})}\\
&=[x^{15}](1+x^5+x^{10}+x^{15})(1+x+x^2)(1+x^2+x^4+x^6)\tag{1}\\
&=\left([x^{15}]+[x^{10}]+[x^5]+[x^0]\right)(1+x+x^2)(1+x^2+x^4+x^6)\tag{2}\\
&=\left([x^5]+[x^0]\right)(\color{blue}{1}+x+x^2)(\color{blue}{1}+x^2+x^4+x^6)\tag{3}\\
&=1+[x^5](1+\color{blue}{x}+x^2)(1+x^2+\color{blue}{x^4}+x^6)\tag{4}\\
&=1+1\tag{5}\\
&\,\,\color{blue}{=2}
\end{align*}
Comment:
*
*In (1) we multiply out the terms with the highest powers which is convenient as we will see in step (3).
*In (2) we use the linearity of the coefficient of operator and apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
*In (3) we can skip $[x^{15}]$ and $[x^{10}]$ since they do not contribute as the highest power of $x$ is $8$.
*In (4) we select the coefficient of $x^0$.
*In (5) we select the coefficient of $x^5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $4\cos^4x-2\cos2x-1/2\cos4x$ is independent of x $4\cos^4x-2\cos2x-1/2\cos4x$
I don't know how to proceed. $\cos4x=2\cos^2(2x)-1$ Is this useful at all?
| Hint: using $\cos2x = 1-2\sin^2(x)$, $\sin(2x) = 2\sin(x)\cos(x)$ and $\sin^2(x) + \cos^2(x) = 1$ to get the following and simplify it:
$$S = 4\cos^4(x)-2(1-2sin^2(x))+1/2(1-2sin^2(2x))$$
$$S = 4\cos^4(x)-2+4\sin^2(x)+1/2 - 4sin^2(x)cos^2(x)$$
$$S = 4\cos^4(x)-2+1/2 + 4\sin^2(x)(1-\cos^2(x)) = 4\cos^4(x)-2+1/2 + 4\sin^4(x) = 4 - 2 + 1/2 = 2.5$$
I should notice that I think one of the $-$ is $+$ (in my case the second minus).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to solve $\frac{|x+2|}{x-1}>\frac{x+1}{2x+1}$? I am working on the following problem:
$$\frac{|x+2|}{x-1}>\frac{x+1}{2x+1}$$
Here's what I have done so far:
$$|x+2|>\frac{x+1}{2x+1}\times(x-1)$$
$$-\left(\frac{(x+1)(x-1)}{(2x+1)}\right)<x+2<\frac{(x+1)(x-1)}{(2x+1)}$$
This is where I stopped. I am not entirely sure how to go about solving this type of inequality. Am I using the correct approach?
| We need to consider 2 cases
*
*for $x+2\ge 0 \implies x\ge -2$ we need to solve
$$\frac{x+2}{x-1}>\frac{x+1}{2x+1}$$
*for $x+2< 0 \implies x< -2$ we need to solve
$$\frac{-x-2}{x-1}>\frac{x+1}{2x+1}$$
then the final solution is given by the union of the solution obtained for each case.
For case 1 we can proceed as follow
$$\frac{x+2}{x-1}>\frac{x+1}{2x+1}\iff \frac{x+2}{x-1}-\frac{x+1}{2x+1}>0\iff \frac{(x+2)(2x+1)-(x+1)(x-1)}{(x-1)(2x+1)}>0$$
$$\iff \frac{x^2+5x+3}{(x-1)(2x+1)}>0\iff \frac{\left(x-\frac{5+\sqrt 13}{2}\right)\left(x-\frac{5-\sqrt 13}{2}\right)}{(x-1)(2x+1)}>0$$
then we ca easily find the solutions under the condition $x\ge -2$.
In a similar way we can study case 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the quotient and the remainder of $(n^6-7)/(n^2+1)$
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.
So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
| Yet another way:
$$
\begin{align}
n^6-7 &= \big(\left(n^2\color{red}{+1}\right)\color{red}{-1}\big)^3-7 \\
&= \left(n^2+1\right)^3-3\left(n^2+1\right)^2+3\left(n^2+1\right)-1 -7 \\
&= \left(n^2+1\right)\big(\left(n^2+1\right)^2-3\left(n^2+1\right)+3\big) - 8 \\
&= \left(n^2+1\right)\left(n^4-n^2+1\right) - 8
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2+b^2)}$$
After partial fraction decomposition and simplifying I get:
$u = \dfrac{\pi}{2(a+b)}$
But answer is $\frac \pi 6$.
Where have I gone wrong?
| I'm not sure you have done anything wrong. You have $a+b$ in your answer, but $a$ is specifically $\sqrt{\frac{7-\sqrt{45}}{2}}=\frac{3-\sqrt{5}}{2}$. And $b=\frac{3+\sqrt{5}}{2}$. So $a+b=3$, making your answer agree with what you are expecting.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating L'hospital] and $\underset{x \to \infty}{\lim}2. \tan^{-1}(x+2) = \pi$
so $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]= \pi$
Can anyone please correct me If I have gone wrong anywhere?
| Here is another straightforward way to obtain the limit.
Using the integral relation
$$\arctan u=\int_0^u{dt\over1+t^2}$$
we have
$$\begin{align}
(x+2)\arctan(x+2)-x\arctan x
&=2\arctan(x+2)+x\int_0^{x+2}{dt\over1+t^2}-x\int_0^x{dt\over1+t^2}\\
&=2\arctan(x+2)+x\int_x^{x+2}{dt\over1+t^2}
\end{align}$$
Now $2\arctan(x+2)\to\pi$ as $x\to\infty$, while
$$0\le x\int_x^{x+2}{dt\over1+t^2}\le x\int_x^{x+2}{dt\over x^2}={2\over x}\to0$$
Thus
$$\lim_{x\to\infty}((x+2)\arctan(x+2)-x\arctan x)=\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Find the roots of $3x^3-4x-8$ It is given that $\alpha$, $\beta$ and $\gamma$ are the roots of the polynomial $3x^3-4x-8$.
I have been asked to calculate the value of $\alpha^2 + \beta^2 + \gamma^2$.
However I am unsure how to find these roots, seeing as though I haven't been given a root to start with.
I began by identifying that
$\alpha + \beta + \gamma = 0$
$\alpha\beta + \alpha\gamma + \beta\gamma = -4/3$
$\alpha\beta\gamma = 8/3$
However I am unsure how to continue to find $\alpha^2 + \beta^2 + \gamma^2$.
| Yet another approach: $\,\dfrac{1}{\alpha}, \dfrac{1}{\beta}, \dfrac{1}{\gamma}\,$ are the roots of $\,8x^3+4x^2-3\,$, so $\,\dfrac{1}{\alpha}+ \dfrac{1}{\beta}+ \dfrac{1}{\gamma}=-\dfrac{1}{2}\,$.
But $\,3\alpha^3=4\alpha+8 \iff \alpha^2=\dfrac{4}{3} + \dfrac{8}{3\alpha}\,$, and therefore: $$\alpha^2+\beta^2+\gamma^2=3 \cdot \frac{4}{3}+ \frac{8}{3}\left(\frac{1}{\alpha}+ \frac{1}{\beta}+ \frac{1}{\gamma}\right)=4-\frac{4}{3} = \frac{8}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Intersection of 3 planes along a line I have three planes:
\begin{align*}
\pi_1: x+y+z&=2\\
\pi_2: x+ay+2z&=3\\
\pi_3: x+a^2y+4z&=3+a
\end{align*}
I want to determine a such that the three planes intersect along a line. I do this by setting up the system of equations:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
x+ay+2z&=3\\
x+a^2y+4z&=3+a
\end{align*}
\end{cases}
$$
and solve for x, y and z. I subtract the first row from the second and third
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
(a-1)y+z&=1\\
(a^2-1)y+3z&=1+a.
\end{align*}
\end{cases}
$$
I subtract $(a+1)*\textrm{row}_1$ from the third row:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
(a-1)y+z&=1\\
(2-a)z&=1+a.
\end{align*}
\end{cases}
$$
I think I am supposed to find an $a$ such that $0=0$ in the third row, but obviously there isn't any such $a$. Do I have the right idea about how to solve this? If so, where is my mistake?
| Hint
See, that from your system of equations you can obtain two lines:
$$l_1:\begin{cases}x=(a-2)t+1\\y=t\\z=(1-a)t+1\end{cases}$$
$$l_2:\begin{cases}x=\frac{5-a}{3}+\frac{a^2-4}{3}t\\y=t\\z=\frac{a+1}{3}+\frac{1-a^2}{3}t\end{cases}$$
| {
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} |
Express $z = \dfrac{3i}{\sqrt{2-i}} +1$ in the form $a + bi$, where $a, b \in\Bbb R$.
Express $$z = \frac{3i}{\sqrt{2-i}} +1$$ in the form $a + bi$, where $a, b \in\Bbb R$.
I figure for this one I multiply by the conjugate of $\sqrt{2 +1}$? But I’m still struggle to achieve the form $a+bi$.
| Given $$\dfrac{3i}{\sqrt{2-i}}+1=\dfrac{3i+\sqrt{2-i}}{\sqrt{2-i}}$$
Now $$=\dfrac{3i+\sqrt{2-i}}{\sqrt{2-i}}\times \dfrac{\sqrt{2+i}}{\sqrt{2+i}}$$$$=\dfrac{3i\sqrt{2+i}+\sqrt{5}}{\sqrt{5}}$$
$$=\dfrac{3i\sqrt{2+i}}{\sqrt{5}}+\dfrac{\sqrt{5}}{\sqrt{5}}$$
| {
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"answer_count": 4,
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} |
Probability of at least two being grey Say we have $11$ grey and $15$ white mice, so $26$ in total in a container we can't see. We want to take $5$ of them home. What is the probability of at least two of them being grey?
*
*2 grey: The probability of the first being grey is $\frac{11}{26}$ the second grey is $\frac{10}{25}$, the last three white $\frac{15\cdot 14 \cdot 13}{24\cdot 23\cdot 22}$.
*3 grey: $\frac{11\cdot 10 \cdot 9 \cdot 15\cdot 14\cdot 13}{26\cdot 25\cdot 24\cdot 23\cdot 22}$
*4 grey: $\frac{11\cdot 10 \cdot 9 \cdot 8 \cdot 15\cdot 14}{26\cdot 25\cdot 24\cdot 23\cdot 22}$
*5 grey: $\frac{11\cdot 10 \cdot 9 \cdot 8 \cdot 7\cdot 15}{26\cdot 25\cdot 24\cdot 23\cdot 22}$
And the probability of at least $2$ grey is the sum of these?
| The probability that at least two are grey is one minus the probability that at most one is grey.
The probability that none are grey is : $\dfrac{15}{26}\dfrac{14}{25}\dfrac{13}{24}\dfrac{12}{23}\dfrac{11}{22}$ or in short $\left.\dbinom {15}5\middle/\dbinom{26}5\right.$
The probability that one is grey is : $\dfrac{11}{26}\dfrac{15}{25}\dfrac{14}{24}\dfrac{13}{23}\dfrac{12}{22}\times 5$ Since the order does not matter we count all the orders. Or in short $\left.\dbinom 51\dbinom{11}1\dbinom {15}4\middle/\dbinom{26}5\right.$
Put it together
| {
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} |
Find the roots of $x^3 -6x^2 +13x -12$ I am trying to find the roots of $$\tag{1} x^3 -6x^2 +13x -12$$ by applying the method outlined here (I think this Cardan’s method).
Letting $y= x-2$, we can transform $(1)$ into $$\tag{2} y^3 + y -2=0.$$
It is clear that $1$ is a root of $(2)$, and, from the formula $$u - \frac{h}{u}=1,$$ we get $$\tag{3} u= \frac{3+ \sqrt{21}}{6}.$$
Given that the formula for the other two roots of $(2)$ are $$u\omega -\frac{h\omega^2}{u}, \quad u\omega^2 -\frac{h\omega}{u},$$ we can use $(3)$ and the fact that $\omega=\frac{1}{2}\Big(-1+\sqrt{3}i \Big)$ in order to find them, but this seems an incredibly tedious method.
Question: Is there a more expedient way to evaluate the last two roots of $(2)$?
Note: I am aware that you can use the fact that $x=3$ is a root of $(1)$ in order to factorise the equation, but I am trying here to use the Cubic Formula.
| $$x^3+x-2= 0$$
We can eliminate the $x^1$ by using a quadratic tschirnhausen transformation $y= x^2+mx+n$, to get the cubic in a binomial form
$$(y-(x_{1}^{2}+mx_{1}+n))(y-(x_{2}^{2}+mx_{2}+n))(y-(x_{3}^{2}+mx_{3}+n)) = 0$$
Collect the new variable $y$
$$y^3+(\dots \dots)y^2+(\dots \dots)y+(\dots \dots) = 0$$
Solving for the unknowns $m$, $n$ to eliminate the $y^2$ and $y$ term
$$y^3+(2-3n)y^2+(1-6m+m^2-4n+3n^2)y+(-4-2m-2m^3-n+6mn-m^2n+2n^2-n^3) = 0$$
Therefore
$n= \frac{2}{3}$, $m = \frac{9\pm 2\sqrt{21}}{3}$
Using the negative case, the binomial cubic becomes
$$y^3-\frac{112 \cdot {112-6\sqrt{3\cdot 112}}}{54} = 0$$
$$y^3 = (\frac{2}{3})^3\cdot 7\cdot (112-24\sqrt{21})$$
$$y = \frac{2}{3}\sqrt[3]{7(112-24\sqrt{21})}$$
Don't forget $x^2+mx+n-y = 0$ and check for sign
$$x = -\frac{3}{2}+\frac{\sqrt{21}}{3} \pm \frac{1}{2}\sqrt{\frac{47}{3}-4\sqrt{21}+\frac{8}{3}\sqrt[3]{7(112-24\sqrt{21})}}$$
The trigonometric solution is
$$x = 2\sqrt{\frac{-1}{3}} \cos(\frac{1}{3}\arccos(-\sqrt{-27}))$$
EDITED to adjust the type settings
| {
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} |
Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x - \sqrt{x},$ then we are trying to solve $f(a)=f(b).$ Using some simple calculus I found that the turning point of $f(x)$ is $(\frac{1}{4}, -\frac{1}{4})$. Hence $0 \le b \le \frac{1}{4}$ and $\frac{1}{4} \le a \le 1$. From here, I have no idea how to proceed.
I used trial and error to find that when $a$ increases, the value of $a+b$ increases as well. Hence I hypothesise that $a+b$ is at a maximum when $a=1$ and $b=0$, which implies that $a+b=1$ is a maximum. Can anyone confirm this?
| Calling
$$
a = x^2\\
b = y^2
$$
the problem reads now
$$
\max (x^2+y^2)\;\;\mbox{s.t.}\;\; x^2-y^2=x-y\Rightarrow x+y=1
$$
now the problem is reduced to:
Find $r$ in $C\to x^2+y^2-r^2=0$ such that $C$ is tangent to $x+y = 1$
| {
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"answer_id": 5
} |
Sum of a repeating pattern of numbers Consider a pattern, in which 5 repeats 3 times & 4 repeats 1 time.
$$5,5,5,4,5,5,5,4,5,5,5,4,...$$
Help me find terms count, when sum of the pattern is at least S.
For example,
if S=29, pattern would be $5+5+5+4+5+5=29$, terms count is 6.
if S=20, pattern would be $5+5+5+4+5=24$, terms count is 5.
| Note that you have cycles of $$5+5+5+4$$ and may be a few extra terms.
We have $$5+5+5+4=19$$ so the first step is to divide your S by 19 and then fit the remainder into your cycle.
For example if you have $S=123$, then $123= 6(19)+9$ so you have 6 cycles and the remainder of 9 indicates that you can fit two $5$ into your string.
Therefore the length of your string is $6(4)+2 = 26$ and the string is $$5+5+5+4 + 5+5+5+4 +5+5+5+4+5+5+5+4+5+5+5+4+5+5+5+4+5+5$$
| {
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How can I factorize this: "$X^3 + X^2 + X - 3$" I am going to elementary school & I am living in one of those deprived areas of Africa.
I can solve mathematical questions like this:
$$X^3 + X^2 + X +1 = X^2(X+1)+(X+1) = (X+1)(X^2+1)$$
Or even
\begin{align}X^2 − 2X + X^2 - X + 1 &= (X^2 - 2X + 1) + (X^2 - X) \\
&= (X - 1)^2 + X(X - 1) \\
&= (X-1)(X-1+X) \\
&= (X - 1)(2X - 1)
\end{align}
But for a few months I have not been able to find a teacher around here who can factorize this:
$$X^3 + X^2 + X - 3$$
Do we have to solve it in this way?
$$X^3 + X^2 + X - 3 = X^2(X + 1) + X - 3$$
Or something else?
I'd appreciate your help with this.
| For this kind of problem it is worth knowing about the factor theorem
This says if $a$ is a root of your polynomial $f(x)$ i.e. $f(a)=0$ then $x-a$ is a factor of, i.e. divides, the polynomial.
In these kinds of problems it is worth trying a few values such as $\pm1, \pm2$.
In your example $f(1)=0$ so $x-1$ divides your polynomial, allowing you to factorize it as $(x-1)(Ax^2+Bx+C)$ where you need to find $A,B,C$.
| {
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} |
Find two $2\times2$ real matrix $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$ Find two $2\times2$ real matrices $A$ and $B$ such that $A$, $B$ , $A+B$ are all invertible with $(A+B)^{-1}=A^{-1}+B^{-1}$
Tried to write the matrices as
$$A=\pmatrix {a&b\\c&d},B=\pmatrix {e&f\\g&h}$$
and solve $(A+B)^{-1}=A^{-1}+B^{-1}$, But make it too complex. Any more convenient ways?
| You get $I = (A^{-1}+B^{-1})(A+B) = I + A^{-1}B + B^{-1}A + I$. Thus we get:
$$(A^{-1}B) + (A^{-1}B)^{-1} = -I$$
$$(A^{-1}B)^2 + I = -(A^{-1}B)$$
So $A^{-1}B$ satisfies the polynomial $x^2 + x + 1 = 0$. Take any matrix satisfying this polynomial; for example you can take
$$A^{-1}B = \begin{bmatrix} -1 &1 \\ -1&0 \end{bmatrix}$$
$$B = A\begin{bmatrix} -1 &1 \\ -1&0 \end{bmatrix}$$
Hence you can take any invertible $A$ and produce $B$ of the wanted form.
| {
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matrix identity extends to real entries? Let a matrix be $(m+n)$ by $(m+n)$ be all reals with determinant $1.$ Call the matrix $S.$ Name the top $m$ rows $A$ and the next $n$ rows $V.$ Now, $S$ has an inverse $T,$ $ST = I.$ Name the left $m$ columns $W^T$ and the right $n$ columns $B^T.$
We get two Gram matrices, first $G = A A^T$ is $m$ by $m.$ Next, $H = B B^T$ is $n$ by $n.$
All the examples I have done so far have
$$\det(G) = \det(H)$$
Theorems:
(I) true when integer entries
(II) true when $S \in SO_n$
(III) true for 2 by 2, that would be $m=n=1$
The question is, is this thing always true with real entries? i have a limited ability to look for counterexamples in low dimension, none so far. I have tried a few examples up to $m+n \leq 4,$ however rational entries with small denominators. I also cannot prove it, it is in the area of Schur complement but I do not see a proof either. I should add that, should the thing be true over the reals, i would expect a proof to have appeared in the (few) places where there is a proof with integer entries.
I did one by hand last night, irrational entries. Let $w = 2 \cos \frac{2 \pi}{9},$ so that $w > 1$ and $w^3 - 3w + 1 = 0.$ The initial matrix with determinant $1$ is
$$
S =
\left(
\begin{array}{ccc}
w & 1 & 1 \\
1 & w & 1 \\
1 & 1 & w \\
\end{array}
\right)
$$
with
$$
T =
\left(
\begin{array}{ccc}
w^2-1 & 1-w & 1-w \\
1 -w & w^2-1 & 1-w \\
1-w & 1-w & w^2-1 \\
\end{array}
\right)
$$
From $$ A = ( w,1,1)$$
we get $G = \left( w^2 + 2 \right)$ and
$$ \det G = w^2 + 2. $$
Then
$$
B =
\left(
\begin{array}{ccc}
1 -w & w^2-1 & 1-w \\
1-w & 1-w & w^2-1 \\
\end{array}
\right)
$$
gives, using relations $w^3 = 3 w - 1$ and $w^4 = 3 w^2 - w,$
$$
H =
\left(
\begin{array}{cc}
3 w^2 - 5 w + 3 & 3w^2-6w+1 \\
3w^2 - 6 w + 1 & 3 w^2 - 5 w + 3 \\
\end{array}
\right)
$$
and, after using the same relations,
$$ \det H = w^2 + 2 = \det G $$
| Example with integer coefficients I had partly typed up. Detail that could be important, could be irrelevant: here, i did not work out a complete matrix $S$ that contained $A$ as the top two rows. i went directly to putting $AR$ in Hermite form by column operations. Hmmm: with the square matrix $R$ far below, the top two rows of $R^{-1}$ really do turn out to be $A.$ maybe this is not a problem.
The whole point of the exercise is that the determinant of the Gram matrix for $M$ is the same as
the determinant of the Gram matrix for $M^\perp \; .$
We begin with a lattice $M$ with basis given by the two rows of
$$
A =
\left(
\begin{array}{ccccc}
1&1&0&0&2 \\
1&1&1&0&3
\end{array}
\right)
$$
The inner products give the Gram matrix $G = A A^T$ as
$$
G =
\left(
\begin{array}{cc}
6&8 \\
8&12
\end{array}
\right)
$$
This gives
$$
G^{-1} =
\left(
\begin{array}{cc}
\frac{3}{2}&-1 \\
-1&\frac{3}{2}
\end{array}
\right)
$$
This allows us to define a basis with rational elements, for the dual lattice $M^\ast,$ given by $A^\ast = G^{-1}A.$ The dual lattice is integer valued linear functionals on $M,$ but we can expres it using rational vectors.
$$
A^\ast =
\left(
\begin{array}{ccccc}
\frac{1}{2}&\frac{1}{2}&-1&0&0 \\
-\frac{1}{4}&-\frac{1}{4}&\frac{3}{4}&0&\frac{1}{4}
\end{array}
\right)
$$
We have, automatically, $A^\ast A^T = I$
We also need the Smith Normal Form of $G,$ in symbols $USV = G.$ this one came out with $U = I$ and
$$
S =
\left(
\begin{array}{cc}
2&0 \\
0&4
\end{array}
\right) ,
$$
$$
V =
\left(
\begin{array}{cc}
3&4 \\
2&3
\end{array}
\right) .
$$
This allows us to give an improved basis for $A^\ast,$ while keeping $A$ because $U^{-1} = I$ and $U^{-1}A=A.$ Let vectors $x_1, x_2$ be the rows of
$$
VA^\ast =
\left(
\begin{array}{ccccc}
\frac{1}{2}&\frac{1}{2}&0&0&1 \\
\frac{1}{4}&\frac{1}{4}&\frac{1}{4}&0&\frac{3}{4}
\end{array}
\right)
$$
as in the proof of Lemma (2.3) in\cite{looijenga},
while vectors $a_1, a_2$ are the rows of
$$
U^{-1}A =
\left(
\begin{array}{ccccc}
1&1&0&0&2 \\
1&1&1&0&3
\end{array}
\right) .
$$
Note that the basis vectors are aligned. We have $a_1 = 2 x_1$ and $a_2 = 4 x_2.$ The quotient group $M^\ast / M $ is therefore $Z/2Z \oplus Z/4Z.$ This follows the method on page 36 of Newman~\cite{newman}. We also need vectors $z_1, z_2$ in the ambient lattice such that $z_i \cdot a_j = x_i \cdot a_j. $ This was easy enough, I took $z_1, z_2$ as the rows of
$$
Z =
\left(
\begin{array}{ccccc}
0&1&0&0&1 \\
1&1&1&0&0
\end{array}
\right)
$$
We are finally ready to solve the linear Diophantine system $A B^T = 0,$ where $B$ will be a basis matrix for the full rank lattice $M^\perp.$
Alright, our lattice $M$ is primitively embedded. We take the definition to be that the matrix $A$ can be completed to become a square matrix of integers with determinant $1.$ What we want is to follow\cite{gilbert} and use column operations to force $A$ into Hermite Normal Form. We find integer matrix $R$ such that $\det R = 1,$
$$
R =
\left(
\begin{array}{ccccc}
1&0&1&0&-2 \\
0&0&-1&0&0 \\
-1&1&0&0&-1 \\
0&0&0&1&0 \\
0&0&0&0&1 \\
\end{array}
\right).
$$
we get
$$
AR =
\left(
\begin{array}{ccccc}
1&0&0&0&0 \\
0&1&0&0&0
\end{array}
\right)
$$
So, what is a basis for $M^\perp \; ?$ We know that an integer column vector $X$ that solves $ARX =0$ can be any
$$
X =
\left(
\begin{array}{c}
0 \\
0 \\
p \\
q \\
r \\
\end{array}
\right)
$$
Thus $RX$ can be any integer linear combination of the three right columns of $R.$ Writing as rows, we find a basis of
$M^\perp \;$ given by
$$
B =
\left(
\begin{array}{ccccc}
1&-1&0&0&0 \\
0&0&0&1&0 \\
-2&0&-1&0&1
\end{array}
\right)
$$
The Gram matrix for $B$ is
$$
H =
\left(
\begin{array}{ccc}
2&0&-2 \\
0&1&0 \\
-2&0&6
\end{array}
\right)
$$
which has determinant $8,$ same as $\det G.$ Who Knew?
| {
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Summation of Binomial Distribution 2 So we have
$\sum _{k=0}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k} = (0.3+0.7)^n=1^n=1$
Now I want to solve
$\sum _{k=\frac{n}{2}+2}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}$
This is what I tried doing
$\sum _{k=0}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}=\sum _{k=\frac{n}{2}+2}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{\frac{n}{2}+2-k} +\sum _{k=0}^{\frac{n}{2}+1}\binom{\frac{n}{2}+1\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{\frac{n}{2}+1-k}=1 $
$\sum _{k=\frac{n}{2}+2}^n\binom{\frac{n}{2}+2\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{\frac{n}{2}+2-k}= 1-\sum _{k=0}^j\binom{j\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{j-k}, j=\frac{n}{2}+1$
$\sum _{k=\frac{n}{2}+2}^n\binom{\frac{n}{2}+2\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{\frac{n}{2}+2-k}= 1-1^{\frac{n}{2}+1}=0$
Now this is obviously wrong so I am not sure how to make it right.
| Hint: $$\sum _{k=0}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}=1$$
if $n$ is even
$$S=\sum _{k=0}^{\frac{n}{2}-1}\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}=\sum _{k=\frac{n}{2}+1}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}$$
leaving one term in between $$\binom{n\:}{\frac{n}{2}}\:\left(0.3\right)^k\left(0.7\right)^{n-\frac{n}{2}}$$
$$S+S+\binom{n\:}{\frac{n}{2}}\:\left(0.3\right)^k\left(0.7\right)^{n-\frac{n}{2}}=1$$
if $n$ is odd
$$\sum _{k=0}^{\frac{2n-1}{2}}\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}=\sum _{k=\frac{2n+1}{2}}^n\binom{n\:}{k}\:\left(0.3\right)^k\left(0.7\right)^{n-k}$$
| {
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Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \cdot \frac{a+1-bi}{a+1-bi} = \frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a} = \frac{bi}{1+a} $$
and therefore $\ \frac{z-1}{z+1}$ is imaginary
now let me assume $\ \frac{z-1}{z+1} $ is imaginary number, how could I conclude that $\ |z| =1 $ I really can't think of any direction..
Thanks
| Another approach
$${\bf Re}\dfrac{z-1}{z+1}={\bf Re}\dfrac{(z-1)(\bar{z}+1)}{|z+1|^2}=\dfrac{|z|^2-1}{|z+1|^2}=0 \iff |z|=1$$
| {
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Converting complex numbers into Cartesian Form 3 When calculating the real and imaginary parts of the complex number, do we take the angle as shown or the magnitude of it? I thought that we would just take the angle as shown, but apparently not, according to my textbook (unless its a typo):
$$z=2\sqrt{3}\operatorname{cis}\left(\frac{\pi }{3}\right)\:w=4\operatorname{cis}\left(\frac{-\pi }{6}\right)$$
Find $z + w$ in:
*
*Cartesian form
*Modulus-argument form
I worked out the complex number to be $3\sqrt{3}+i$:
$$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:=\:2\sqrt{3}-2i\\
2\sqrt{3}\cos\left(\frac{\pi }{3}\right)+2\sqrt{3}i\:\sin\left(\frac{\pi }{3}\right)=\:\sqrt{3}\:+3i\\
\sqrt{3}\:+3i\:+\:2\sqrt{3}-2i\:=\:3\sqrt{3}+i$$
(Following from this I got the mod-arg form to be $[2\sqrt{7},0.19]$)
The textbook says pretty much the same thing, except
$$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:$$ is taken to be: $4\cos\left(\frac{\pi }{6}\right)+4i\sin\left(\frac{\pi }{6}\right)$, resulting in the complex number $=3\sqrt{3}+5i$.
| You are correct and the text book is wrong.
$$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:=\:2\sqrt{3}-2i\\
2\sqrt{3}\cos\left(\frac{\pi }{3}\right)+2\sqrt{3}i\:\sin\left(\frac{\pi }{3}\right)=\:\sqrt{3}\:+3i\\
\sqrt{3}\:+3i\:+\:2\sqrt{3}-2i\:=\:3\sqrt{3}+i$$
is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $10^n \gt 6n^2+n$ Show for all $n \in \mathbb{N}$ $(n\geq1):$ $10^n \gt 6n^2+n$
My solution:
Base case: For $n=1$
$10^1 \gt 6 \cdot 1^2+1$
Inductive hypothesis:
$10^n \gt 6n^2+n \Rightarrow 10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
Inductive step:
$10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
$\Rightarrow$ $10^{n+1} \gt 6(n^2+2n+1)+n+1$
$\Rightarrow$ $10^{n+1} \gt 6n^2+12n+6+n+1$
$\Rightarrow$ $10^{n+1} \gt 6n^2+13n+7$
$\Rightarrow$ $10^n \cdot 10^1 \gt 6n^2+13n+7$
$\Rightarrow$ $(6n^2+n)\cdot10 \gt 6n^2+13n+7$
$\Rightarrow$ $60n^2+10n \gt 6n^2+13n+7$
I am stuck at this point. What techniques or tricks are there to solve the rest?
It is of interest to me, because I am currently practicing a lot of exercises related to convergences, inequalities and mathematical induction.
Any hints guiding me to the right direction I much appreciate.
| For the inductive step we can proceed as follow
$$10^{n+1}=10\cdot 10^n \stackrel{Ind.Hyp.}>10\cdot(6n^2+n) \stackrel{?}>6\cdot(n+1)^2+(n+1)$$
and the latter requires
$$10\cdot(6n^2+n) \stackrel{?}>6\cdot(n+1)^2+(n+1)$$
$$60n^2+10n \stackrel{?}>6n^2+12n+6+n+1$$
$$54n^2-3n-7 \stackrel{?}>0$$
which can be shown to be true by quadratic formula or again by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Unable to prove that $\sqrt{i} + \sqrt{-i}$ is a real number. I did this :-
$$
\sqrt{i} = \sqrt{\frac{1}{2}.2i} = \sqrt{\frac{1}{2}.(1 + 2i - 1)} = \sqrt{\frac{1}{2}.(1 + 2i + i^2)} = \sqrt{\frac{1}{2}.(1+i)^2} = \frac{1}{\sqrt{2}}(1+i) \\= \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ ---------
1} \\ \mbox{Also, }
\sqrt{-i} = \sqrt{-1.i} = \sqrt{-1}.\sqrt{i} = i\sqrt{i} = i(\frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}}) \mbox{ --------from 1} \\ = \frac{1}{\sqrt{2}}i - \frac{1}{\sqrt{2}} \\ = -\frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ --------- 2} \\
\mbox{Adding 1 and 2, we get} \\ \sqrt{i} + \sqrt{-i} \\ = \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \\= i.\frac{2}{\sqrt{2}} \\= i.\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}} \\= i\sqrt{2}
$$
This is a complex number, not a real number.
What am I doing wrong here? Is there another way to prove this ?
| Note that $i=e^{(\pi/2) i}$ and $-i=e^{(-\pi/2) i}$
Thus $\sqrt i = e^{(\pi/4) i}$ and $\sqrt -i = e^{(-\pi/4) i}$
Adding the two we get $$\sqrt i+\sqrt -i =e^{(\pi/4) i}+e^{(-\pi/4) i}=2 \cos (\pi/4)=\sqrt 2$$ Which is real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Extremum Problem
How big is the circle?
My first steps:
*
*$ x^2 + y^2 =r^2$
*$ f(x)=y=e^{-x^2}$
Substitute $y^2$ in $x^2 + y^2 =r^2.$
So, $x^2 + e^{-2x^2} =r^2$
Is this way correct? Because after calculating the first derivative $2x -4xe^{-2x^2}$ and so on my solution at the end is $x^2 = \dfrac {\ln(2)}{2}, $ and this is not correct.
| Your solution is right. Clearly the circle and the curve meet at $(x,y)$ in first quadrant where $$x^2+y^2=r^2\\y=e^{-x^2}\\-2xe^{-x^2}=\dfrac{-x}{\sqrt{r^2-x^2}}$$which means that $$2y=\dfrac{1}{\sqrt {r^2-x^2}}=\dfrac{1}{y}$$therefore $$y=\dfrac{\sqrt 2}{2}$$and $$x=\sqrt{\dfrac{\ln 2}{2}}$$which means that $$r=\sqrt{\dfrac{1+\ln 2}{2}}\approx 0.9200943377$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $$\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$$
Here's how I tried it
$$x+y+z=xyz $$
So, by De Moivre's Theorem,
$$(\cos a + \cos b + \cos c) + i(\sin a + \sin b + \sin c) = \cos(a+b+c) + i \sin(a+b+c)
$$
Equating real and imaginary parts,
$$\cos a + \cos b + \cos c= \cos(a+b+c)$$
and similarly for sine. Now,
$$(a-b) + (b-c) + (c-a) =0$$
What to do now? Please help. And please use De Moivre Theorem!
| Squaring & adding we get
$$\cos^2(a+b+c)+\sin^2(a+b+c)=(\cos a+\cos b+\cos c)^2+(\sin a+\sin b+\sin c)^2$$
$$\implies1=1+1+1+2\sum_{\text{cyc}(a,b,c)}\cos(a-b)$$
Observe that we actually don't need $x+y+z=xyz$
The sufficient condition is $$ x+y+z=\cos p+i\sin p\iff|x+y+z|=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$
Calculate
$$\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$$
My Attempt:
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{\sin{x}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\frac{-2\sin^2{\frac{x}{2}}}{-2\sin^2{\frac{x}{2}}}}\cdot\frac{\frac{-2\sin^2{\frac{x}{4}}}{-2\sin^2{\frac{x}{4}}}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{\sin{\frac{x}{2}}}+\ln{2}}$$
$$\frac{1}{4}\cdot 16\cdot 1$$
$$=4$$
Am I solving this correct?
| You can rewrite the function as
$$
\frac{\log\cos(x)}{\log\cos(x/2)}\frac{\log\sin(x/2)}{\log\sin(x)}
$$
The first fraction is easy to deal with:
$$
\log\cos x=\log(1-x^2/2+o(x^2))=-\frac{x^2}{2}+o(x^2)
$$
and similarly
$$
\log\cos\frac{x}{2}=\log(1-x^2/8+o(x^2))=-\frac{x^2}{8}+o(x^2)
$$
Therefore
$$
\lim_{x\to0}\frac{\log\cos(x)}{\log\cos(x/2)}=4
$$
For the second fraction, it's simpler to consider $x=2t$, so the limit becomes
$$
\lim_{t\to0^+}\frac{\log\sin(t)}{\log\sin(2t)}=
\lim_{t\to0^+}\frac{\dfrac{\cos(t)}{\sin(t)}}{\dfrac{2\cos(2t)}{\sin(2t)}}=
\lim_{t\to0^+}\frac{\cos(t)}{\cos(2t)}\frac{\sin(2t)}{2\sin(t)}=1
$$
with a simple application of l’Hôpital.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Induction. Am I missing something or is there a mistake in the question? $$\sum_{k=1}^n k*3^k=\frac {3(3^n(2n-1)+1)} 4 $$
So let f(n)= $\sum_{k=1}^n k*3^k $
and g(n)=$\frac {3(3^n(2n-1)+1)} 4$
By induction hypothesis, $f(n+1) = f(n) + (n+1)3^{n+1} \overset{\text{i.h.}}{=} g(n) + (n+1) 3^{n+1} = g(n+1).$
$$\frac{3(3^{n+1}(2n+1)+1)}4=(n+1)(3^{n+1})+\frac{3(3^n(2n-1)+1))}4 $$
I am stuck afterward, please help, thanks.
| You can calculate $$\frac{3^{n+2}(2n+1)-3-3^{n+1}(2n-1)-3}{4}=\frac{3^{n+1}(3(2n+1)-2n+1)}{4}=\frac{3^{n+1}(6n+3-2n+1)}{4}=\frac{3^{n+1}(6n+3-2n+1)}{4}=…$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that $\lim _{x\to 1}f(x)=3$ if $f(x)=\frac{x^3-1}{x-1}$ Prove that $\lim _{x\to 1}f(x)=3$ where $f:(0,\infty)\to \mathbb{R}$ is given by $f(x)=\frac{x^3-1}{x-1}$.
I proved by definition of the limit
$$|f(x)-3|=\left|\frac{x^3-1}{x-1}-3\right|=|x^2+x-2|=|x-1||x+2|\leq |x-1|(|x|+2)$$
how to processed from this
| Let $\varepsilon > 0$. We have
$$\left|\frac{x^3-1}{x-1} - 3\right| = \left|\frac{x^3-3x+2}{x-1}\right| = \left|x^2+x-2\right| = |x-1||x+2| \le |x-1|(|x-1| + 3)$$
Solving the quadratic inequation $y^2+3y - \varepsilon < 0$ gives $y \in \left\langle \frac{-3-\sqrt{9+4\varepsilon}}2, \frac{-3+\sqrt{9+4\varepsilon}}2\right\rangle$.
Therefore if we pick $\delta = \frac{-3+\sqrt{9+4\varepsilon}}2$, then $|x-1| < \delta$ implies
$$\left|\frac{x^3-1}{x-1} - 3\right| \le |x-1|^2 + 3|x-1| < \varepsilon$$
We conclude $\lim_{x\to 1} \frac{x^3-1}{x-1} = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How we derive $y \geq (b+1)$ from $(b+1)b^n < (b+1)^n b$? In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page 5 that-
$y \geq (b+1)$ since, $(b+1)b^n < (b+1)^n b$ and it was given that $(b+1)x^n - by^n =1$.
but $(b+1)x^n - by^n =1 \implies (b+1)x^n > by^n $, and together with $(b+1)b^n < (b+1)^n b$, I don't see how we reach at $y \geq (b+1)$
Please explain, Thanks in advance.
| The given equation is-
$(b+1)x^n - by^n =1 \cdots (1)$.
Case $1$: If $x=y$,
then equation (1) becomes $(b+1)x^n - bx^n =1 \implies bx^n+x^n - bx^n =1 \implies x^n =1 \implies x=1 $, but it is given that $(x, y) \neq (1, 1)$. So, $x \neq y$.
Case $2$: If $x>y$,
then, let $x=y+c$ (where $c>0$, $c$ is an integer since x,y are integers) we substitute the value of $x$ in equation (1) -
$(b+1)(y+c)^n - by^n =1\implies b(y+c)^n+(y+c)^n - by^n =1 $. By inspection, we find that the difference between $b(y+c)^n+(y+c)^n $ and $by^n$ is larger than $1$ (reader will see it more clearly if he expands $(y+c)^n $ (binomial expansion), note that $c$ is an integer).
Case $3$: So, $x<y$ (the only possibility). Let, $y=x+d$. Then equation (1) becomes-
$(b+1)x^n - b(x+d)^n =1 $ [substituting the value of $y=x+d$ in $(1)$]
$\implies \frac{b+1}{b} - (\frac{x+d}{x})^n =\frac{1}{bx^n} $ [dividing both sides of the equitation by $bx^n$ ]
$\implies \frac{b+1}{b} - (\frac{x+d}{x})^n >0 $ [Since, $\frac{1}{bx^n}>0$]
$\implies \frac{b+1}{b} > (\frac{x+d}{x})^n $
$\implies \frac{b+1}{b} > \frac{x+d}{x}$ [Since, $(\frac{x+d}{x})^n > (\frac{x+d}{x}) $ ]
$\implies 1+ \frac{1}{b} > 1 + \frac{d}{x} $
$\implies \frac{1}{b} >\frac{d}{x} $
$\implies x > bd $. Here, $d \geq 1$, we see that for $d=1, x>b$, as we have found earlier $y>x$, we deduce, $y>b \implies y\geq b+1$.
| {
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"source": "stackexchange",
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Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)\right]=f(x)\implies \\ \left({x \over x^2+x}+{x \over x^2+x}+\cdots +{x\over x^2+x}\right)\lt f(x)\lt \left({x \over x^2+1}+{x \over x^2+1}+\cdots +{x\over x^2+1}\right) \\ {x^2 \over x+x^2}\lt f(x) \lt {x^2\over 1+x^2}\\ \text{applying limit on both sides }\\ \implies\lim_{x\to \infty}{x^2 \over x+x^2}= \lim_{x\to \infty}{x^2\over 1+x^2}=1\\ \implies \lim_{x\to \infty}f(x)=1$$
Can we do this without squeeze theorem?
| Just using that $\frac{1}{1+\epsilon}=1-\epsilon+O(\epsilon^2)$ and $\sum_{i=1}^n i=\frac{n(n+1)}{2}$
So $$\frac{x}{x^2+c}=\frac{1}{x}\frac{x^2}{x^2+c}=\frac{1}{x}\frac{1}{1+\frac{c}{x^2}}=\frac{1}{x}\left(1-\frac{c}{x^2}+O(\frac{c}{x^4})\right)$$
$$\sum_{c=1}^x\frac{x}{x^2+c}=\frac{1}{x}\sum_{c=1}^x1-\frac{c}{x^2}+O(\frac{c}{x^4})=1-\frac{x (x+1)}{2x^3}+O(\frac{1}{x^2})=1-\frac{1}{2x}+O(\frac{1}{x^2})$$
Note that using $\frac{1}{1+\epsilon}=1+O(\epsilon)$ is in fact enough and simpler here, but you don't get the $\frac{-1}{2x}$, only $O(\frac{1}{x})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of p when the equation has 3 real roots( Derived from a symmetric line question)
Question statement: Two points M and N are on a parabola $y^2=2px$ ($p\neq0$) such that they are symmetric points about a straight line $x+y=1$. Determine the range of $p$.
What I did was I reflected that parabola over $x+y=1$, obtaining the equation:
$$(1-x)^2=2p(1-y)$$
So now we got 2 equations.
$$(1-x)^2=2p(1-y)\tag{1}$$
$$y^2=2px \tag{2}$$
I solved for $y^2$ from the first equation then substitute it into the second equation. After simplification I end up with the following:
$$0=x^4 - 4 x^3 +(6-4p)x^2 +(8p-4-8p^3) x +1$$
According to the question I have to find the range where the equation has 4 roots, but then I realized I can just find the two points where the function has only 3 roots, and the range between them would be the answer.
That's how far I've gone through. Thank you.
| Let $M(a,b),N(c,d)$ where $a\not=c$ or $b\not=d$.
These points are on the parabola $y^2=2px$, so
$$b^2=2pa\iff a=\frac{b^2}{2p}\tag1$$
$$d^2=2pc\iff c=\frac{d^2}{2p}\tag2$$
Also, the two points are symmetric points about a straight line $x+y=1$, so the midpoint of the line segemnt $MN$ is on the line $x+y=1$, and the line $MN$ is perpendicular to $y=-x+1$, so
$$\frac{a+c}{2}+\frac{b+d}{2}=1\iff a+b+c+d=2\tag3$$
$$\frac{b-d}{a-c}\times (-1)=-1\iff b-d=a-c\tag4$$
From $(1),(2),(3),(4)$, we have
$$\frac{b^2}{2p}+b+\frac{d^2}{2p}+d=2\tag5$$
$$b-d=\frac{b^2}{2p}-\frac{d^2}{2p}\tag6$$
From $(6)$, we get
$$(b-d)(2p-b-d)=0$$
Supposing that $b=d$ gives $a=c$, so $b\not=d$.
So, we have $2p-b-d=0$,
i.e.$$d=2p-b\tag7$$
From $(5),(7)$, we get
$$\frac{b^2}{2p}+b+\frac{(2p-b)^2}{2p}+2p-b=2,$$
i.e.
$$b^2-2pb+4p^2-2p=0\tag8$$
What we are seeking is the condition on $p$ that $(8)$, a quadratic equation on $b$, has at least one real solution and $b=p$ is not a solution of $(8)$ (because of $b\not=d$) .
So, the answer is
$$(-2p)^2-4\cdot 1\cdot (4p^2-2p)\ge 0\quad\text{and}\quad p^2-2p^2+4p^2-2p\not=0,$$
i.e.
$$\color{red}{0\lt p\lt\frac 23}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral $\int_0^\infty dp \, \frac{p^5 \sin(p x) e^{-b p^2}}{p^4 + a^2}$: any clever ideas? I am trying to solve the following integral, with $a>0,$ $b>0$:
$I \equiv \int_0^\infty dp \, \frac{p^5 \sin(p x) e^{-b p^2}}{p^4 + a^2} $
By expanding the $\sin$, I get
$I = \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!} \int_0^\infty dp \, \frac{p^{4+2n} e^{-b p^2}}{p^4 + a^2} \\
= \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!}\Bigg\{
\frac{1}{2} b^{\frac{3}{2}-n} \Gamma (n-{3}/{2}) \, _1F_2\left(1;\frac{5}{4}-\frac{n}{2},\frac{7}{4}-\frac{n}{2};-\frac{1}{4} a^2 b^2\right)
+\frac{1}{4} \pi a^{n-\frac{3}{2}}
\left[\csc \left((2 \pi n+\pi )/4\right] \cos (a b)
-\sec \left[ (2 \pi n+\pi )/4\right] \sin (a b)\right)
\Bigg\}.$
Here $_1F_2$ is a hypergeometric function. The first summation can be done, so that we are left with
$I= \frac{\pi}{2a} \left[\cos (a b) \cos (x\sqrt{a/2}) \sinh (x\sqrt{a/2})+\sin (a b) \sin (x\sqrt{a/2}) \cosh (x\sqrt{a/2})\right]
+
\sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!}
\frac{1}{2} b^{\frac{3}{2}-n} \Gamma (n-{3}/{2}) \, _1F_2\left(1;\frac{5}{4}-\frac{n}{2},\frac{7}{4}-\frac{n}{2};-\frac{1}{4} a^2 b^2\right).$
I am unable to find a closed form for the second summation.
Is there an easier way to solve $I$? Presumably one can apply the residue theorem, but I did not find a quick way. Any ideas would be appreciated!
| We have
$$\frac {p^5 e^{-b p^2} \sin x p} {p^4 + a^2} =
p e^{-b p^2} \sin x p \left( 1 +
\frac {i a} {2 (p^2 - i a)} - \frac {i a } {2 (p^2 + i a)} \right), \\
\frac d {db} \left( e^{i a b} \int_0^\infty
\frac {p e^{-b p^2} \sin x p} {p^2 - i a} dp \right) =
-e^{i a b} \int_0^\infty p e^{-b p^2} \sin x p \,dp,$$
and, after some calculations,
$$\int_0^\infty \frac {p^5 e^{-b p^2} \sin x p} {p^4 + a^2} dp =
F(a) + F(-a) + \frac {\sqrt \pi x e^{-x^2/(4 b)}} {4 b^{3/2}}, \\
F(a) = \frac {i \pi a e^{i a b}} 8 \left(
e^{x \sqrt {i a}} \operatorname{erfc} \frac { 2 b \sqrt{i a} + x} {2 \sqrt b} -
e^{-x \sqrt {i a}} \operatorname{erfc} \frac { 2 b \sqrt{i a} - x} {2 \sqrt b}
\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Volume of a tetrahedron whose 4 faces are congruent. Suppose that I have a tetrahedron such that all four faces consist of congruent triangles, says with the lengths $a,b$ and $c$ for each side. Is there a beautiful method to compute its volume?
PS. The reason for me tagging calculus and linear algebra is that I figured that the technique used to calculate such a problem may come from this areas.
| If the given triangle is acute-angled, define positive numbers $x, y, z$ by:
\begin{align*}
x^2 & = \tfrac{1}{2}(b^2 + c^2 - a^2), \\
y^2 & = \tfrac{1}{2}(c^2 + a^2 - b^2), \\
z^2 & = \tfrac{1}{2}(a^2 + b^2 - c^2).
\end{align*}
Let $\mathbf{u}, \mathbf{v}, \mathbf{w}$ be mutually orthogonal vectors of unit length, and define:
$$
\mathbf{t}_{ijk} = ix\mathbf{u} + jy\mathbf{v} + kz\mathbf{w} \quad (i, j, k = 0, 1).
$$
The cuboid with these 8 vertices has volume $xyz$. It contains a tetrahedron $T$ with vertices $\mathbf{t}_{100}\mathbf{t}_{010}\mathbf{t}_{001}\mathbf{t}_{111}$, and faces which are triangles with the same sides:
\begin{align*}
a & = \sqrt{y^2 + z^2}, \\
b & = \sqrt{z^2 + x^2}, \\
c & = \sqrt{x^2 + y^2}.
\end{align*}
The remainder of the cuboid consists of 4 tetrahedra, each with sides $x, y, z$ meeting at right angles at one of the vertices $\mathbf{t}_{000}, \mathbf{t}_{011}, \mathbf{t}_{101}, \mathbf{t}_{110}$, and each therefore having volume $\tfrac{1}{6}xyz$, implying that $T$ has volume $\tfrac{1}{3}xyz$.
Wikipedia illustration (with different notation):
Another Wikipedia illustration, to accompany the comment:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\sum_{k=2}^\infty \frac{1}{k^2 - 1}$ I'm wondering if someone could check on this working:
$$\sum_{k=2}^\infty \frac{1}{k^2 - 1} = \sum_{k=2}^\infty \frac{1}{(k - 1)(k + 1)} = \frac{1}{2}\sum_{k=2}^\infty \frac{1}{k-1} - \frac{1}{k+1}$$
This is a nice telescoping thing that has a $k$-th partial sum that looks like this:
$$\frac{1}{1} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + ... + \frac{1}{k-1} - \frac{1}{k+1} = 1 + \frac{1}{2} - \frac{1}{k+1} - \frac{1}{k+1}$$
Skipping some formalities we find:
$$\sum_{k=2}^\infty \frac{1}{k^2 - 1} = \frac{1}{2} . \lim_{k \rightarrow \infty}\frac{3}{2}-\frac{2}{k+1} = \frac{1}{2} . \frac{3}{2} = \frac{3}{4}$$
| You have
$$
...= 1 + \frac{1}{2} - \frac{1}{k+1} - \frac{1}{k+1}
$$
and there should be
$$
= 1 + \frac{1}{2} - \frac{1}{\color{cyan}{k}} - \frac{1}{k+1}
$$
but it is easy to correct and even with this mistake does not change the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is this proof correct (Rationality of a number)? Is $\sqrt[3] {3}+\sqrt[3]{9} $ a rational number? My answer is no, and there is my proof. I would like to know if this is correct:
Suppose this is rational. So there are positive integers $m,n$ such that $$\sqrt[3]{3}+\sqrt[3]{9}=\sqrt[3]{3}(1+\sqrt[3]{3})=\frac{m}{n}$$
Let $x=\sqrt[3]{3}$. We get $x^2+x-\frac{m}{n}=0 \rightarrow x=\frac{-1+\sqrt{1+\frac{4m}{n}}}{2}$.
We know that $x$ is irrational and that implies $\sqrt{1+\frac{4m}{n}}$ is irrational as well (Otherwise $x$ is rational). Write $x=\sqrt[3]{3}$, multiply both sides by $2$ and then raise both sides to the power of $6$ to get:
$$24^2=\left(\left(\sqrt{1+\frac{4m}{n}}-1\right)^3\right)^2\rightarrow 24=\left(\sqrt{1+\frac{4m}{n}}-1\right)^3=\left(1+\frac{4m}{n}\right)\cdot \sqrt{1+\frac{4m}{n}}-3\left(1+\frac{4m}{n}\right)+3\sqrt{1+\frac{4m}{n}}-1$$
Let $\sqrt{1+\frac{4m}{n}}=y,1+\frac{4m}{n}=k $.
We get: $25=ky-3k+3y\rightarrow y=\frac{25+3k}{k+3}$, So $y$ is rational. But we know $y$ is irrational (again, otherwise $\sqrt[3]{3}$ is rational) which leads to a contradiction. So the answer is No, $\sqrt[3]{3}+\sqrt[3]{9}$ is irrational.
Is this proof correct? Is there another way to prove this? Thanks!
| Alternatively, denote $x=\sqrt[3] {3}+\sqrt[3]{9}$ and cube it to get a cubic equation with integer coefficients:
$$x^3=3+3^2(\sqrt[3]3+\sqrt[3]{9})+9 \Rightarrow \\
x^3-9x-12=0.$$
According to the rational root theorem, the possible rational roots are: $\pm (1,2,3,4,6,12)$. However, none of them satisfies the equation. Hence, $x$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Calculate $A^5 - 27A^3 + 65A^2$, where $A$ is the matrix defined below. If $A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$, find $A^5 - 27A^3 + 65A^2$
$$A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$$
Let $\lambda$ be its eigenvalue, then
$$(A-\lambda I) = \begin{bmatrix} 0-\lambda & 0 & 1 \\
3 & 1-\lambda & 0 \\
-2&1&4-\lambda\end{bmatrix}$$
$$|A-\lambda I| = -(\lambda)^3 + 5(\lambda)^2 - 6(\lambda) +5$$
Using Cayley-Hamilton theorem
$$A^3-5^2+6A-5=0$$
How do I use this find $A^5 - 27A^3 + 65A^2$?
| Let
$p(x) =x^5 -27x^3 + 65x^2; \tag 1$
one can always compute $p(A)$ directly, but perhaps the following requires less arithmetical work:
Let
$\chi(x) = x^3 - 5x^2 +6x - 5 \tag 2$
be the characteristic polynomial of $A$.
By the division algorithm for polynomials, there exist unique
$q(x), r(x) \in \Bbb Q[x] \tag 3$
such that
$p(x) = q(x)\chi(x) + r(x), \tag 4$
with either $r(x) = 0$ or $\deg r(x) < 3 = \deg \chi(x)$; thus, since
$\chi(A) = 0, \tag 5$
we have from (4)
$r(A) = p(A) - \chi(A)q(A) = p(A) - 0 \cdot q(A) = p(A); \tag 6$
thus, we compute $r(x)$ by synthetic (long) division of polynomials; it's not too much work; we find
$q(x) = x^2 + 5x - 8, \tag 7$
and
$r(x) = 73x - 40; \tag 8$
the reader may easily check that
$p(x) = x^5 - 27x^3 + 65x^2$
$= (x^2 + 5x - 8)(x^3 - 5x^2 +6x - 5) + 73 x - 40 = q(x)\chi(x) + r(x); \tag 9$
thus
$p(A) = r(A) = 73A - 40I = 73 \begin{bmatrix} 0 & 0 & 1 \\ 3 & 1 & 0 \\ -2 & 1 & 4 \end{bmatrix} - 40I = \begin{bmatrix} -40 & 0 & 73 \\ 219 & 33 & 0 \\ -146 & 73 & 252 \end{bmatrix}. \tag{10}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find the sum of the power series $\sum_{n=1}^\infty n*(n+1)*x^n$ $\sum_{n=1}^\infty n*(n+1)*x^n$
Hello everyone, I need help in solving the question above.
I started with $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$, if differentiated it once so it became $\frac{1}{(1-x)^2} =\sum_{n=1}^\infty nx^{n-1} $ but from here I don't know how to continue.
Thanks.
| Using
$$\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$$
then
$$\sum_{n} n \, x^{n+1} = x^2 \, \frac{d}{dx} \, \frac{1}{1-x}$$
and
$$\sum_{n} n \, (n+1) \, x^n = \frac{d}{dx} \left( x^{2} \, \frac{d}{dx} \frac{1}{1-x} \right).$$
After some work it is found that:
\begin{align}
\sum_{n=0}^{\infty} x^{n} &= \frac{1}{1-x} \\
\sum_{n=0}^{\infty} n \, x^{n} &= \frac{x}{1-x} \\
\sum_{n=0}^{\infty} n(n+1) \, x^{n} &= \frac{2 x \, (x^2 - x + 1)}{(1-x)^{3}} = \frac{2 x (1+x^3)}{(1-x)^2 \, (1-x^2)}.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
positive integer solutions to $x^3+y^3=3^z$ I am seeking all positive integer solutions to the equation $x^3+y^3=3^z$.
After doing number crunching, I think there are no solutions. But I am unable to prove it.
Attempt
If $x$ and $y$ have common divisor $d$, we have $d^3(m^3+n^3)=3^z$. So $d$ must be a power of $3$, and we are back to where we started. So we assume $x$ and $y$ are coprime.
Testing the parity, we have sum of 2 cubes to be odd. WLOG, we can assume $x$ is even and $y$ is odd.
Trying mod $3$, we have $x+y=0 \pmod 3$. Since $x$ and $y$ are coprime, $x$ and $y$ must be congruent to $1$ and $-1$ or vice-versa.
If I assume $x=3m+1$ and $y=3n-1$, expand out and simplify, I get $27(m^3+n^3)+27(m^2-n^2)+9(m+n)=3^z$. If I assume $z \geq 3$, this gives $(m^3+n^3)+(m^2-n^2)+\frac{m+n}{3}=3^{z-3}$. But I don't see how to proceed.
I also tried mod $9$ but didn't get anywhere, it didn't cut down the possibilities by much.
I also tried letting $y=x+r$. Then
\begin{align*}
x^3+y^3 &= x^3+(x+r)^3 \\
&= x^3 + (x^3+3x^2r+3xr^2+r^3) \\
&= 2x^3+3x^2r+3xr^2+r^3 \\
&= 3^z
\end{align*}
Then $3\mid 2x^3+3x^2r+3xr^2+r^3$, and $3\mid 3x^2r+3xr^2$, so this implies $3 \mid 2x^3+r^3$. But this doesn't yield any contradiction.
Can anyone supply a proof? Or if my hypothesis is wrong, how to derive all the integer solutions?
Thank you.
| You already proved that we can assume that $x,y$ are relatively prime.
We can find easly solution for $n\leq 2$. Let $n\geq 3$.
So $x+y = 3^a$ and $x^2-xy+y^2=3^b$ for some nonegative integers $a+b=n$.
If $b\geq 2$ then $9\mid x^2-xy+y^2$ and $3\mid x+y$, so $$9\mid (x+y)^2-(x^2-xy+y^2)=3xy\implies 3\mid xy$$
A contradiction. So $b\leq 2$ and this should be easly done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Find $\lim\limits_{x \to\infty} (\frac {2+3x}{2x+1})^{x+1}$ without L'Hopital $$
\lim_{x\to\infty}\left(\frac {2+3x}{2x+1}\right)^{x+1}
$$
Not sure how to deal with this, I've tried doing the following
$$
\lim_{x\to\infty}\left(\frac {2+3x}{2x+1}\right)^{x}\cdot \lim_{x\to\infty}\left(\frac{2+3x}{2x+1}\right)
$$
Then I tried dividing by $x,$ but my teacher told me it was wrong.
| We have
\begin{align}
\lim_{x\to\infty}\left(\frac {2+3x}{1+2x}\right)^{1+x}
&=
\lim_{x\to\infty}\left(\frac {(1+2x)+(1+x)}{1+2x}\right)^{1+x}
\\
&=
\lim_{x\to\infty}\left(1+\frac {1+x}{1+2x}\right)^{1+x}
\\
&=
\lim_{x\to\infty}\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{1+x}
\\
&=
\lim_{x\to\infty}\left[\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{\frac{1+x}{1+2x}}\right]^{1+2x}
\\
\end{align}
For $ x $ large enough we have
$$
1.3<\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)<3
$$
and
$$
1.3^{1+2x}
<\left[\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{\frac{1+x}{1+2x}}\right]^{1+2x}
<3^{1+2x}
$$
By the sandwich theorem we have to
$$
\lim_{x\to \infty}\left[\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{\frac{1+x}{1+2x}}\right]^{1+2x}=\infty
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 5
} |
The limit of $\frac{n^3-3}{2n^2+n-1}$ I have to find the limit of the sequence above.
Firstly, I tried to multiply out $n^3$, as it has the largest exponent.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} =
\lim_{n\to\infty}\frac{n^3(1-\frac{3}{n^3})}{n^3(\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3})} =
\lim_{n\to\infty}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\[1ex]
\lim_{n\to\infty}\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3} = 0 \\[1ex]
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{0}
\end{align}
$$
Then, after realizing $\frac{1}{0}$ might not be a plausible limit, I tried to multiply out the variable with the largest exponent in both the dividend and the divisor.
$$\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \lim_{n\to\infty}\frac{n^3(1 - \frac{3}{n^3})}{n^2(2 + \frac{1}{n} - \frac{1}{n^2})} =
\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
$$
\begin{align}
\lim_{n\to\infty}1-\frac{3}{n^3} = 1 \\
\lim_{n\to\infty}2 + \frac{1}{n} - \frac{1}{n^2} = 2 \\
\lim_{n\to\infty}\frac{n^3-3}{2n^2+n-1} = \frac{1}{2} \\
\lim_{n\to\infty}n = \infty
\end{align}
$$
So, my questions about this problem:
*
*Could $\frac{1}{0}$ be a valid limit?
*Does $\infty\cdot\frac{1}{2}$
equal to $\infty$?
*In conclusion, what is the limit of the sequence
above? $\infty?$
Thank you!
| The second way is preferable and since the product "$\infty \cdot \frac12$" is not an indeterminate form from here
$$\lim_{n\to\infty}n\cdot\frac{1 - \frac{3}{n^3}}{2 + \frac{1}{n} - \frac{1}{n^2}}$$
we can conclude that the sequence diverges.
Note that also with the first method from here
$$\lim_{n\to\infty}\frac{1-\frac{3}{n^3}}{\frac{2}{n} + \frac{1}{n^2} - \frac{1}{n^3}}$$
since also "$\frac1 0$" is not an indeterminate form and the denominator is $>0$ we can conclude that the sequence diverges at $\infty$.
In both cases, what we cannot do is to calculate separetely the limit of each part and then make a direct calculation (that's allowed only is the final operation in well defined).
As an alternative since eventually
*
*$n^3-3\ge n^3-6n^2+9n \iff6n^2-9n-3\ge 0$
*$2n^2+n-1 \le 4n^2-24n+36\iff2n^2-25n+35 \ge 0$
by squeeze theorem we have
$$\frac{n^3-3}{2n^2+n-1} \ge\frac{n^3-6n^2+9n}{4n^2-24n+36}=\frac{n(n-3)^2}{4(n-3)^2}=\frac n 4 \to \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 1
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Proof by induction, $1/2 + ... + n/2^n < 2$ So I'm having trouble with proving this homework question by induction.
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2
$$
I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is where I get confused.
Base case is easy, n=1.
$$
\frac{1}{2^1}<2
$$
Induction case we assume that
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} <2
$$
Then we get to fun old induction. How do I show that
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} <2 ?
$$
| Hint:
$$
\begin{align}
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{k-1}{2^{k-1}} + \frac{k}{2^k} + \frac{k+1}{2^{k+1}}
&=\qquad\;\;\frac12\Big(\frac{1}{2^1} + \frac{2}{2^2} + \dots +\frac{k-1}{2^{k-1}} + \frac{k}{2^k}\Big) \\
&\quad+\Big(\frac{1}{2^1} + \frac{1}{2^2}+\frac1{2^3} + \dots + \;\;\frac{1}{2^k}\;\;+\frac1{2^{k+1}}\Big)
\end{align}
$$
On the right hand side, the first summand corresponds to the induction hypothesis, and you can bound the second summand by...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Fourier series of $f(x)=\sin^2x\cos^2x$ at $(-\pi,\pi)$ Find Fourier series of $f(x)=\sin^2 x \cos^2 x$ at $(-\pi,\pi)$
$f(x)$ is even so we only have to evaluate $a_0,a_n$
$$
a_0 = \frac{1}{\pi}
\int_{-\pi}^{\pi} \sin^2 x \cos^2 x dx
= \frac{1}{4\pi}
\int_{-\pi}^{\pi}\sin^2(2x)
= \frac{1}{4\pi}
\int_{-\pi}^{\pi}\frac{1-\cos 2x}{2} dx
= \frac{1}{4}
$$
and
$$
\begin{split}
a_n &= \frac{1}{\pi}
\int_{-\pi}^{\pi}\sin^2 x \cos^2 x \cos(nx)dx\\
&= \frac{1}{4\pi}
\int_{-\pi}^{\pi}\frac{\cos(nx)-\cos(2x)\cos(nx)}{2}dx\\
&=\frac{-1}{16\pi}\int_{-\pi}^{\pi}(\cos((n+2)x)+\cos((n-2)x))dx\\
&=0
\end{split}
$$
So $f(x)\approx\frac{1}{8}$
Is it correct?
| As a more effective way we have that
$$
\begin{split}
f(x) &= \sin^2 x \cos^2 x=\frac12(1-\cos 2x)\cdot \frac12(1+\cos 2x) \\
&=\frac14(1-\cos^2 2x)=\frac14\left[1-\frac12\left(1+\cos 4x\right)\right]=\frac18-\frac18\cos 4x
\end{split}
$$
Using the same expression in the integral obviously we obtain
$$a_n= \frac{1}{\pi}
\int_{-\pi}^{\pi}\sin^2 x \cos^2 x \cos(nx)dx= \frac{1}{\pi}
\int_{-\pi}^{\pi} \cos(nx)\left(\frac18-\frac18\cos 4x\right)dx=$$
$$= \frac{1}{8\pi}\int_{-\pi}^{\pi} \cos(nx)dx-\frac{1}{8\pi}\int_{-\pi}^{\pi} \cos(nx)\cos 4xdx$$
which is equal to $-\frac18$ for $n=4$ and equal to $0$ otherwise.
| {
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"url": "https://math.stackexchange.com/questions/2916053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}$ How can I prove this equality:
$$
\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}= \frac{-1}{n(n+1)(n+1)!}
$$
| Given $$\frac{1}{(n+1)!} + \frac{1}{(n+1)(n+1)!} - \frac{1}{n(n!)}$$
$$=\frac{1}{(n+1)n!} + \frac{1}{(n+1)(n+1)n!} - \frac{1}{n(n!)}$$
$$=\dfrac{1}{n!}\left(\frac{1}{(n+1)} + \frac{1}{(n+1)(n+1)} - \frac{1}{n}\right)$$
$$=\dfrac{1}{n!}\left(-\dfrac{1}{n(n+1)}+ \frac{1}{(n+1)(n+1)}\right)$$
$$=\dfrac{-1}{n(n+1)(n+1)!}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{\left(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\right)^2}dx$ The goal is to show (preferably without contour integration, as my knowledge is pretty limited there, but if you can do it that way there is no problem to share it) that: $$I= \int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{\left(\pi^2+\ln^2\left(\frac{1+x}{1-x}\right)\right)^2}\mathrm dx=\frac{1}{240}$$
I am still trying to find more relevant approaches, but so far I thought of: Considering $$J(a)= \int_0^1 \frac{x\ln\left(\frac{1+x}{1-x}\right)}{a^2+\ln^2\left(\frac{1+x}{1-x}\right)}dx$$ We have that $J'(\pi)=-2\pi I$, also $$J(a)=\int_0^\infty e^{-at}\int_0^1 x\sin\left(t\ln\left(\frac{1+x}{1-x}\right)\right)dx dt $$ Which is nice, however I couldnt solve the inner integral.
Another approach would be to let $\ln\left(\frac{1+x}{1-x}\right)=t \rightarrow x=\frac{e^t-1}{e^t+1}\rightarrow dx=\frac{2e^t}{(e^t+1)^2}$
$$I=2\int_0^\infty \frac{e^t(e^t-1)}{(e^t+1)^3}\frac{t}{(\pi^2+t^2)^2}dt$$ This doesnt look nice, but notice that $$\frac{d^2}{dx^2} \left(\frac{1}{1+e^x}\right)=\frac{e^x(e^x-1)}{(e^x+1)^3}$$ Integrating by parts two times I get $$I=\frac{1}{\pi^2} +24 \int_0^\infty \frac{x^2-\pi^2}{(x^2+\pi^2)^4}\frac{x}{1+e^x}dx$$ Now I am thinking to use series for $\frac{x}{1+e^x}$, but I dont know a series that satisfies it, similary we have for $\frac{x}{e^x-1}$ in terms of Bernoulli numbers. Could you land some help for this integral?
| We will use $3$ instances of Schroder's formula which evaluate Gregory coefficients https://en.wikipedia.org/wiki/Gregory_coefficients#Computation_and_representations
\begin{eqnarray*}
-G_2=\int_0^{\infty} \frac{dx}{(1+x)^2(\pi^2+(\ln x)^2)} =\frac{1}{12} \\
G_3=\int_0^{\infty} \frac{dx}{(1+x)^3(\pi^2+(\ln x)^2)} =\frac{1}{24} \\
-G_4=\int_0^{\infty} \frac{dx}{(1+x)^4(\pi^2+(\ln x)^2)} =\frac{19}{720} \\
\end{eqnarray*}
I am not sure how you would show these without using residue calculus ... but try & ignore this $\ddot \smile$
We want to show
\begin{eqnarray*}
I=\int_0^1 \frac{x \ln (\frac{1+x}{1-x}) } {\left(\pi^2+\left(\ln (\frac{1+x}{1-x}) \right)^2\right)^2 } dx = \frac{1}{240}.
\end{eqnarray*}
Substitute $u=\frac{1-x}{1+x}$ ( so $dx = \frac{-2 du}{(1+u)^2}$) and the intgeral becomes
\begin{eqnarray*}
I=-2\int_0^1 \frac{(1-u) \ln (u) } {(1+u)^3(\pi^2+\left(\ln (u) \right)^2 } du.
\end{eqnarray*}
Now the substitution $v=\frac{1}{u}$ gives exactly the same integrand (but with different limits)
\begin{eqnarray*}
I=-2\int_1^{\infty} \frac{(1-v) \ln (v) } {(1+v)^3(\pi^2+\left(\ln (v) \right)^2 } dv
\end{eqnarray*}
and the integral becomes
\begin{eqnarray*}
I=-\int_0^{\infty} \frac{(1-x) \ln (x) } {(1+x)^3(\pi^2+\left(\ln (x) \right)^2 } dx.
\end{eqnarray*}
Now observe that
\begin{eqnarray*}
\frac{d}{dx} \frac{1}{\pi^2+(\ln(x))^2} = \frac{-2 \ln(x)}{x(\pi^2+(\ln(x))^2)^2} \\
\end{eqnarray*}
\begin{eqnarray*}
\frac{d}{dx} \frac{x(1-x)} {(1+x)^3} =\frac{x^2-4x+1}{(1+x)^4}
\end{eqnarray*}
so we can integrate by parts to get
\begin{eqnarray*}
I=- \frac{1}{2}\int_0^{\infty} \frac{x^2-4x+1}{(1+x)^4(\pi^2+(\ln(x))^2)} dx.
\end{eqnarray*}
Rewrite $x^2-4x+1=(x+1)^2-6(x+1)+6$ and use the three results stated at the begining of this answer and we have
\begin{eqnarray*}
I= \frac{1}{2} \left(-6 \frac{19}{720}+6 \frac{1}{24} -\frac{1}{12} \right) = \color{red}{\frac{1}{240}}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Solving the equation $\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$ for $x$ or $y$ One day, while making and doing math problems, I came across this equation:
$$\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$$
After some simple steps, I found $g$, but I couldn't find $x$ or $y$.
Here's a picture, since I'm not good with MathJax.
| Multiply $$\frac{gy}{x} + \frac{2y^2}{x} = \frac{gy}{y^2} + \frac{2x}{y^2}$$ by $x y^2$ to obtain
$$g y^3 + 2 y^4 = g x y + 2 x^2.$$
By grouping the terms together this becomes $g y (y^2 - x) + 2 (y^4 - x^2) = 0$ and factoring the last term leads to
$$(y^2 - x)(2 y^2 + g y + 2 x) = 0.$$
Solving to $x$ yields two possible values:
$$x \in \left\{y^2, - \frac{y (2 y + g)}{2} \right\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$ The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$.
I tried the following concept, how to sum 9 using 1-9 only without repetition.
1:9, 2:1+8,3:7+2, 4: 6+3, 5: 5+4,
6:1+2+6 ,7:1+3+5, 8:2+3+4
The answer is 8.
How will it be solved using Binomial theorem.
| As Theo Bendit has already commented this is not solved by Newton's binomial theorem. The solution is the same way that is used to prove Newton's bionomian theorem that is the proper use of combinatorial principles. Let
$$
p(x)=(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)(1+x^6)(1+x^7)(1+x^8)(1+x^9)
$$
We have
\begin{align}
p(x)
=&
1
+\sum_{1\leq i_1\leq 9}x^{i_1}
+\sum_{1\leq i_1<i_2\leq 9}x^{i_1}\cdot x^{i_2}
+\sum_{1\leq i_1<i_2<i_3\leq 9}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}
+
\\
\\
&\qquad
+\sum_{1\leq i_1<i_2<i_3<i_4\leq 9}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}\cdot x^{i_4}
+\ldots +\sum_{1\leq i_1<\cdots <i_9\leq 9}x^{i_1}\cdot \dots \cdot x^{i_9}
\end{align}
or for $m_1=\max\{i_1: 1\leq i_1\leq 9\}$, $m_2=\max\{i_1+i_2: : 1\leq i_1<i_2\leq 9\}$,$\ldots$, $m_9=\max\{i_1+\ldots+i_9: 1\leq i_1<\ldots <i_9\leq 9\}$
\begin{align}
p(x)
=&
1
+\sum_{k=1}^{m_1}\sum_{i_1= k}x^{i_1}
+\sum_{k=1+2}^{m_2}\sum_{\substack{ i_1<i_2 \\i_1+i_2=k}}x^{i_1}\cdot x^{i_2}
+\sum_{k=1+2+3}^{m_3}\sum_{\substack{ i_1<i_2<i_3 \\i_1+i_2+i_3=k}}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}
+\ldots
\\
\\
&\qquad
+\sum_{k=1+2+3+4}^{m_4}\;\;\sum_{i_1<i_2<i_3<i_4}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}\cdot x^{i_4}
+
\qquad\ldots+\sum_{k=1+\ldots+9}^{m_9}\sum_{\substack{ i_1<\cdots <i_9\\ i_1+\ldots+i_9=k}}x^{i_1}\cdot \dots \cdot x^{i_9}
\end{align}
All possible powers of $ x $ equal to $ x ^ 9 $ are generated by the first three summations.
Let $ c_9 $ be the coefficient of $ x ^ 9 $ in the polynomial $ p (x) $. So
\begin{align}
c_9\cdot x^9
=&
\sum_{\substack{1\leq i_1\leq 9\\ i_1=9}}x^{i_1}
+\sum_{\substack{1\leq i_1<i_2\leq 9\\ i_1+i_2=9}}x^{i_1}\cdot x^{i_2}
+\sum_{\substack{1\leq i_1<i_2<i_3\leq 9\\ i_1+i_2+i_3=9}}x^{i_1}\cdot x^{i_2}\cdot x^{i_3}
\\
=& 1\cdot x^9+1\cdot x^1\cdot x^8+1\cdot x^2\cdot x^7+1\cdot x^3\cdot x^6+1\cdot x^4\cdot x^5
\\
&+1\cdot x^1\cdot x^2\cdot x^6+1\cdot x^1\cdot x^3\cdot x^5+1\cdot x^2\cdot x^3\cdot x^4
\\
&=8x^9
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is a fast way to evaluate $\cos(\pi/4 + 2\pi/3)$ and $\cos(\pi/4 + 4\pi/3)$ (maybe using inspection) Given
$\cos(\pi/4 + 2\pi/3)$
and
$\cos(\pi/4 + 4\pi/3)$
I wish to show that these values are given by $(\pm(\sqrt{6}) - \sqrt{2})/4$.
What is a quick way to arrive at these exact solutions (perhaps even by inspection)?
I am currently using the addition of cosine formula, but it is too slow. I'm thinking perhaps a graphic method can be used?
| Is it really that slow?
$\cos (\frac{\pi}{4} + \frac{2\pi}{3}) =\cos \frac{\pi}{4}\cos \frac{2\pi}{3} - \sin \frac{\pi}{4}\sin\frac{2\pi}{3} = (\frac {\sqrt{2}}{2})(\frac {-1}2) - (\frac {\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) = \frac {-\sqrt 2 - \sqrt 6}{4}$
That seems pretty straight forward to me.
$\cos (\frac{\pi}{4} + \frac{4\pi}{3})$ isn't any more difficult.
If you want to use symmetry arguments...
$\cos (\frac{\pi}{12}) = \cos (\frac{\pi}{3}-\frac{\pi}{4}) = \frac {\sqrt 6 + \sqrt 2}{4}\\
\sin (\frac{\pi}{12}) = \sin (\frac{\pi}{3}-\frac{\pi}{4}) = \frac {\sqrt 6 - \sqrt 2}{4}$
(which is every bit as much work as done in the first line.)
Then you can say:
$\cos (\frac{\pi}{4} + \frac{2\pi}{3}) = \cos (\frac{11\pi}{12}) = \cos(\pi - \frac {\pi}{12}) = -\cos \frac {\pi}{12}$
$\cos (\frac{\pi}{4} + \frac{4\pi}{3}) = \cos (\frac{19\pi}{12}) = \cos(2\pi - \frac {5\pi}{12}) = \cos \frac {5\pi}{12} = \cos(\frac {\pi}{2} - \frac {\pi}{12}) =\sin \frac{\pi}{12} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding $\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x$ results in loss of information Let $f(x) = \sqrt{x^2-5x+1}-x$
Find $\lim_{x\to\infty}f(x)$
$$\lim_{x\to\infty} \sqrt{x^2-5x+1}-x$$
$$\lim_{x\to\infty} \dfrac{x^2-5x+1-x^2}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{-5x+1}{\sqrt{x^2-5x+1}+x}$$
$$\lim_{x\to\infty} \dfrac{\dfrac{-5x+1}{x}}{\dfrac{\sqrt{x^2-5x+1}+x}{x}}$$
$$\lim_{x\to\infty} \dfrac{-5+\dfrac{1}{x}}{\sqrt{\dfrac{x^2}{x^2}-\dfrac{5}{x}+\dfrac{1}{x^2}}+1}$$
From here, I know that $\lim_{x\to\infty} \dfrac{1}{x} = 0$, $\lim_{x\to\infty} \dfrac{5}{x} = 0$ and $\lim_{x\to\infty} \dfrac{1}{x^2} = 0$
$$\lim_{x\to\infty} \dfrac{-5}{\sqrt{1}+1}$$
$$\lim_{x\to\infty} \dfrac{-5}{2}$$
$$\lim_{x\to\infty} f(x) = \dfrac{-5}{2}$$
Everything up to here seems fine. The issue is when I try to find $\lim_{x\to-\infty} f(x)$
I also know that $\lim_{x\to-\infty} \dfrac{1}{x} = 0$, $\lim_{x\to-\infty} \dfrac{5}{x} = 0$ and $\lim_{x\to-\infty} \dfrac{1}{x^2} = 0$
This would make me conclude that $\lim_{x\to-\infty}f(x) = \dfrac{-5}{2}$.
However, this is not the case because $\lim_{x\to-\infty}f(x) = \infty$
Desmos view of $f(x)$
Why am I arriving to the wrong answer and how can I algebraically prove that the answer is $\infty$?
| Note that for $x\to -\infty$
*
*$\sqrt{x^2-5x+1} \to +\infty$
*$-x \to +\infty$
and therefore
$$\sqrt{x^2-5x+1}-x \to +\infty$$
As an alternative by $x=-y\to -\infty$ with $y\to +\infty$ we have
$$\lim_{x\to-\infty}\sqrt{x^2-5x+1}-x=\lim_{y\to +\infty}\sqrt{y^2+5y+1}+y \to +\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Application of Fermat's little theorem to check divisibility Using Fermat's little theorem to prove: $(i)19\mid 2^{2^{6k+2}}+3$, where $k=0,1,2.....$$(ii)13\mid 2^{70}+3^{70}$My Approach: I couldn't think of how to go with $(i)$ but i tried $(ii)$ to show $2^{70} \equiv 0\pmod {13}$ and $3^{70} \equiv 0\pmod {13}$.Since,
$$2^{12} \equiv 1 \pmod {13}\Rightarrow (2^{12})^5\equiv 1 \pmod {13}\Rightarrow2^{60} \equiv 1 \pmod {13}$$Again, $$2^4 \equiv 3 \pmod {13}\Rightarrow2^8.2^2 \equiv 10\pmod {13}$$Using both result: $2^{70} \equiv 10\pmod {13}$I failed again to show that. Any hints or solution will be appreciated.Thanks in advance.
| $n|a + b$ does not mean that $n|a$ and $n|b$. It means that if $a \equiv k \mod n$ then $b \equiv -k \mod n$.
So if $2^{70}\equiv 10 \mod 13$ (which it is) then $13|2^{70} + 3^{70}$ if $3^{70} \equiv -10 \mod 13$.
Does $3^{70}\equiv -10 \mod 13$?
Well If $0 < a < 13$ then $a^{12} \equiv 1 \mod 13$ and $a^{60} = (a^{12})^5\equiv 1 \mod 13$ so $a^{70} \equiv a^{10}\mod 13$.
And $2^{10} \equiv 10 \mod 13$.
And $3^3 \equiv 27 \equiv 1 \mod 13$ so $3^9 \equiv 1^3 \mod 13$ and $3^{10} \equiv 3 \equiv -10 \mod 13$.
So $2^{70} + 3^{70}\equiv 2^{10} + 3^{10} \equiv 10 + (-10)\equiv 0 \mod 13$.
i) is a lot harder but the idea is that as $2^{18} \equiv 1 \mod 19$ then if $2^{6k + 2}\equiv m \mod 18$ then $2^{2^{6k+2}}\equiv 2^m \mod 19$.
$2^{6k+2} = 64^k*4 = (7*9 + 1)^k*4$. Note $(7*9 + 1)^k$ will be $1$ more than a multiple of $9$. So $(7*9 + 1)^k*4$ will be $4$ more than a multiple of $4*9 = 2*18$. So $2^{6k+2} \equiv 4 \mod 18$.
And $2^{2^{6k + 2}}\equiv 2^4 \equiv 16 \mod 19$.
So $2^{2^{6k+2}}+3 \equiv 16 + 3 \equiv 0 \mod 19$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve a system of equation with absolute value and rotation Solve
$\begin{cases} \left|x_1-x_2\right|=\left|x_2-x_3\right|=...=\left|x_{2018}-x_1\right|,\\ x_1+x_2+...+x_{2018}=2018. \end{cases}$
I think there must be such a way to solve systems of equation with the form of rotation and absolute value like this.
I have difficulty in solving the first equation. Since to me, there are quite a lot of cases to consider, for example, $\left|x_1-x_2\right|=\left|x_2-x_3\right|$ leads to $x_1-x_2=x_2-x_3$ and $x_1-x_2=x_3-x_2$ and so on.
| Here is the answer to verify my comment.
Let $|x_{1} - x_{2}| = C \geq 0$ and note that
\begin{align}
|x_{1} - x_{2}| = C \implies x_{1} = \pm C + x_{2}
\end{align}
Similarly, we have
\begin{align}
x_{1} &= \pm C + x_{2} \\
x_{2} &= \pm C + x_{3} \\
& \; \; \vdots \\
x_{2017} &= \pm C + x_{2018} \\
x_{2018} &= \pm C + x_{1} \\
\end{align}
Keep substituting, we obtain
\begin{align}
x_{1} &= \pm C + x_{2} \\
&= \pm C \pm C + x_{3} \\
& \; \; \vdots \\
&= \pm C \pm C \pm \cdots \pm C + x_{1}
\end{align}
Thus, we have
\begin{align}
\pm 2018C = 0 \implies C = 0
\end{align}
It means that $x_{1} = x_{2} \cdots = x_{2018}$ and let them be $A$.
So now, we use the second equation
\begin{align}
x_{1} + x_{2} + \cdots + x_{2018} &= 2018 \\
2018A &= 2018 \\
A &= 1
\end{align}
Conclusion: $x_{1} = x_{2} \cdots = x_{2018} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2932479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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