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Multivariable limit exists? Does the limit
$$\lim_{(x,y)\rightarrow (0,0)} \frac{y^4}{x^\beta(x^2+y^4)}$$ exists for $\beta>0$? I don't think it exists but how do you prove it rigorously. Thanks
| $$\lim\limits_{(x,y)\to (0,0)} \frac{y^4}{x^\beta\left(x^2+y^4\right)}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+} \frac{(r\sin\phi)^4}{(r\cos\phi)^\beta\left((r\cos\phi)^2+(r\sin\phi)^4\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{r^4\sin^4\phi}{r^\beta\left(\cos^\beta\phi\right)\left(r^2\cos^2\phi+r^4\sin^4\phi\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{r^4\sin^4\phi}{r^\beta r^2\left(\cos^\beta\phi\right)\left(\cos^2\phi+r^2\sin^4\phi\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{r^2\sin^4\phi}{r^\beta\left(\cos^\beta\phi\right)\left(\cos^2\phi+r^2\sin^4\phi\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{\sin^4\phi}{r^{\beta-2}\left(\cos^\beta\phi\right)\left(\cos^2\phi+r^2\sin^4\phi\right)}$$
Now let's attempt to find a bound that is independent of $\phi$
$$\left|\frac{\sin^4\phi}{\left(\cos^\beta\phi\right)\left(\cos^2\phi+r^2\sin^4\phi\right)}\right|\leq \left|\frac{1}{\left(\cos^\beta\phi\right)\left(\cos^2\phi+r^2\sin^4\phi\right)}\right| $$
The right hand side is unbounded, which implies that the limit is dependent on $\phi$. Therefore
$$\lim\limits_{(x,y)\to (0,0)} \frac{y^4}{x^\beta\left(x^2+y^4\right)}\Rightarrow \mbox{does not exist}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the derivative using the chain rule and the quotient rule
$$f(x) = \left(\frac{x}{x+1}\right)^4$$ Find $f'(x)$.
Here is my work:
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4\left(x+1\right)^3x^4}{\left(x+1\right)^8}$$
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4x^4\left(x+1\right)^3}{\left(x+1\right)^8}$$
I know the final simplified answer to be:
$${4x^3\over (x+1)^5}$$
How do I get to the final answer from my last step? Or have I done something wrong in my own work?
| Apparently you complicated things too much.
Starting from $$f(x) = \left(\frac{x}{x+1}\right)^4$$ with the chain rule we get first
\begin{align}
f'(x) &= 4\left(\frac{x}{x+1}\right)^3\cdot \left(\frac{x}{x+1}\right)^\prime\notag\\
&= 4\left(\frac{x}{x+1}\right)^3\cdot \frac{x'(x+1)-x(x+1)'}{(x+1)^2}\notag\\
&= 4\frac{x^3}{(x+1)^3}\cdot \frac{(x+1)-x}{(x+1)^2}\notag\\
&= \frac{4x^3}{(x+1)^3}\cdot \frac 1{(x+1)^2}\notag\\
&= \frac{4x^3}{(x+1)^5}\notag
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Generating Functions - Extracting Coefficients In many counting problems, we find an appropriate generating function which allows us to extract a given coefficient as our answer.
In cases where the generating function is not one that is easily used as an infinite sum, how does one alter the generating function for simpler coefficient extraction? The question I'm looking at involves the following generating function:
f(x) = ($x^3$ + $x^4$ + . . . + $x^8$)$^4$, and we seek the coefficient of $x^2$$^4$ in f(x). To start, we convert to a simpler form. With ($x^3$ + $x^4$ + . . . + $x^8$)$^4$ = $x^1$$^2$(1 + x + $x^2$ + . . . + $x^5$)$^4$ = $x^1$$^2$((1 - $x^6$)/(1 - x))$^4$, the answer is the coefficient of $x^1$$^2$ in (1 - $x^6$)$^4$(1 - x)$^-$$^4$
Why is this true? How does $x^1$$^2$ come about, allowing for the common (1 + x + $x^2$ + . . . ) sum ?
| You are looking for the coefficient of $x^{24}$ in:$$x^{12}\left(\frac{1-x^6}{1-x}\right)^4$$
Which is the same as the coefficient of $x^{24-12}=x^{12}$ in $$\left(\frac{1-x^6}{1-x}\right)^4\tag{1}$$
Now, $$(1-x^6)^4 = 1-4x^6+6x^{12}-4x^{18}-x^{24}$$
And $$\frac{1}{(1-x)^4} = \sum_{n=0}^{\infty} \binom{n+3}{3}x^n$$
So what is the coefficient of $x^{12}$ in the product? It is $$\binom{12+3}{3}\cdot 1 + \binom{6+3}{3}\cdot (-4) + \binom{3}{3}\cdot 6$$
More generally, the coefficient of $x^k$ in (1) is:
$$\binom{k+3}{3}\cdot 1 + \binom{k-3}{3}\cdot (-4) + \binom{k-9}{3}\cdot 6 +\binom{k-15}{3}\cdot (-4) + \binom{k-21}{3} \cdot 1$$
And yes, if $k>20$ this will evaluate as zero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does every $9 \times 9$ Latin square contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?
Q: Does every $9 \times 9$ Latin square on the symbol set $\{1,2,\ldots,9\}$ contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?
This one has $1728$ such submatrices, which is as low as I've gotten:
$$
\begin{bmatrix}
6 & 7 & 8 & 9 & 1 & 4 & 2 & 3 & 5 \\
5 & 6 & 1 & 7 & 2 & 8 & 3 & 4 & 9 \\
9 & 1 & 6 & 2 & 4 & 3 & 7 & 5 & 8 \\
4 & 5 & 3 & 6 & 8 & 7 & 1 & 9 & 2 \\
1 & 2 & 4 & 8 & 3 & 5 & 9 & 6 & 7 \\
2 & 3 & 7 & 4 & 9 & 6 & 5 & 8 & 1 \\
8 & 9 & 2 & 3 & 5 & 1 & 6 & 7 & 4 \\
7 & 8 & 5 & 1 & 6 & 9 & 4 & 2 & 3 \\
3 & 4 & 9 & 5 & 7 & 2 & 8 & 1 & 6 \\
\end{bmatrix}$$
It doesn't seem likely that random Latin squares will help much; they average in the thousands of such submatrices. The one above is the best random Latin square I've found so far (although, I haven't busted a gut doing this; it seems like it won't work anyway).
The groups of order $9$ have lots ($C_9$ has $5832$ and $C_3 \times C_3$ has $19440$).
This question was motivated by answering this math.SE question which asks if any $9 \times 9$ Latin square can have its rows and columns permuted to give a sudoku square.
One way to find an explicit counterexample would be to find a $9 \times 9$ Latin square with no $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$. But this attempt didn't work since I couldn't find one. Hence my question.
| We might consider the general concept of a saturated sub-matrix, or $SSM$, being an $M \times K$ sub-matrix within a Latin square of of rank $N = M \times K$, that contains all $N$ values.
We are looking for Latin squares that are completely void of $SSM$s, so they might well be called poly-unsaturated!
For example $6 = 3 \times 2$. Of the 9408 reduced forms, just 6 are poly-unsaturated. Here is one of them:
$$\begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\
2 & 3 & 4 & 1 & 6 & 5 \\
3 & 1 & 5 & 6 & 2 & 4 \\
4 & 5 & 6 & 2 & 1 & 3 \\
5 & 6 & 1 & 3 & 4 & 2 \\
6 & 4 & 2 & 5 & 3 & 1
\end{bmatrix}$$
The counter is still running for $9 = 3 \times 3$, but these are definitely scarce objects.
I have also confirmed existence for $8 = 4 \times 2$. One is tempted to conjecture that for all $M,K \gt 1$, whenever $M \neq N$ they will be found to exist.
For $M = K$, well we now know that $N=9$ has poly-unsaturated forms, and $N=4$ does not. I don't know enough about cyclic groups to predict the result for $N=16$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Posed in regional mathematics Olympiad 1995 Call a positive integer $n$ good if there are $n$ integers, positive or negative, and not necessarily distinct, such that their sum and product are both equal to $n$
For example, $8$ is good since
$$\begin{align}
8 = 4\times2\times1\times1&\times1\times1\times(-1)\times(-1) \\
&\textrm{and}\\
8 = 4+2+1+1&+1+1+(-1)+(-1)
\end{align}$$
Show that integers of the form $4k+1$ ($k\ge0$) and $4l$ ($l\ge2$) are good
| As “Your Ad Here” wrote, if $n=4k+1$, take the following numbers:
$$4k+1,\underbrace{1,\dots,1}_{2k\,\text{times}},\underbrace{-1,\dots,-1}_{2k\,\text{times}}.$$
If $n=4k$, where $k>1$, write $n=2^j q$, where $q$ is odd and $j\ge2$, and proceed by induction on $j$.
If $j=2$, $n=4k$ with $k$ odd. Then take these numbers:
$$-2, 2k,\underbrace{1,\dots,1}_{3k\,\text{times}},\underbrace{-1,\dots,-1}_{k-2\,\text{times}}.$$
Their product is positive, because there are $1+3k-2$ (evenly many, when $k$ is odd) negative numbers, and its absolute value is $4k$, so the product is $4k$.
Their sum is $-2+2k+3k-(k-2)=4k$.
If $j>2$, $n=2N$, where $N=2k$ is “good.” Let $a_i$ be the $N$ numbers with sum and product equal to $N$. Now consider these $2N$ numbers:
$$2, a_1,\dots,a_N,\underbrace{1,\dots,1}_{\frac{N}{2}-2\,\text{times}},\underbrace{-1,\dots,-1}_{\frac{N}{2}\,\text{times}}.$$
Their product is $2\left(\prod{a_i}\right)(-1)^{\frac{n}{2}}=2N=n$, because $N$ is even.
Their sum is $2+\left(\sum a_i\right)+\frac{N}{2}-2-\frac{N}{2}=2+N+N-2=2N=n$.
(Note that $\frac{N}{2}-2=\frac{2k}{2}-2$ is at least zero, because $k>1$ was given.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor series for the function $f(z) = \frac{1}{(z-5)(z-7)}$ on a disc centered at point $z_0=3$ I started by expressing the function as sum of two fractions using partial fraction decomposition to get $\frac{-1}{2(z-5)} + \frac{1}{2(z-7)}$
However I could only then end up writing that as the sum of two power series :
$$\frac{1}{10} \sum_{n=0}^\infty \frac{z^n}{3^n} - \frac{1}{14} \sum_{n=0}^\infty \frac{z^n}{7^n}$$ and I'm unsure how to write this in the form of a Taylor series.
| If we are looking to construct the series around $z_0 = 3$, we could have $\lvert z - 3\rvert < 1$ or $\lvert z - 3\rvert > 1$ which would be the series inside and outside a disc of radius one. In the first case,
\begin{align}
\frac{1/2}{z - 3 - 4} - \frac{1/2}{z - 3 - 2}&= \frac{-1/8}{1 - (z-3)/4} + \frac{1/4}{1-(z-3)/2}\\
&= -1/8\sum_{n=0}^{\infty}\Bigl(\frac{z-3}{4}\Bigr)^n+1/4\sum_{n=0}^{\infty}\Bigl(\frac{z-3}{2}\Bigr)^n\\
&= \frac{1}{2}\sum_{n=0}^{\infty}\biggl[\frac{1}{2^{n+1}}-\frac{1}{4^{n+1}}\biggr](z-3)^n
\end{align}
This was due to the definition of a geometric series
$$
\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}
$$
for $\lvert r\rvert < 1$.
If we want a power series convergent outside the disc, $\lvert z - 3\rvert > 1\iff 1/\lvert z -3\rvert < 1$.
$$
\frac{1}{2(z - 3)}\frac{1}{1 - 4/(z-3)} - \frac{1}{2(z-3)}\frac{1}{1 - 2/(z-3)}
$$
which leads to a Laurent series. What do you get for the Laurent series convergent outside of the disc?
| {
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"url": "https://math.stackexchange.com/questions/1195674",
"timestamp": "2023-03-29T00:00:00",
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Length and breadth of a rectangle enclosed between two semi-circles of given radii
34. It is required to take a rectangular frame in a horizontal position along a corridor bounded by vertical walls of which a horizontal cross-section is two concentric semicircles of radii $r$ and $r\ \sqrt{3}$; the frame is of length $2x$ and breadth $y$. One side of length $2x$ is tangential to the inner wall, and the two ends of the opposite side are in contact with the outer wall, as shown in Fig. Misc. Ex. 1-34. Prove that
$$x^{2} = 2r^{2} - 2ry - y^{2}.$$
Prove that, if $x$ and $y$ may vary, the greatest possible area enclosed by the frame is $\frac{1}{2}r^{2}\ \sqrt{3}.$
I noticed a few things that I believe are relevant. The first is that one of the sides of the rectangle is a chord of the larger circle and the second is that the angle between the radius of the smaller circle and the opposite side of rectangle is 90 degrees.
It's been a long time since I had fun with circles, but using Wolfram MathWorld I was able to piece together this much:
$2x = 2r\sqrt{3}\sin\left(\dfrac{\theta}{2}\right) \Leftrightarrow x = r\sqrt{3}\sin\left(\dfrac{\theta}{2}\right)$
$h = r\sqrt{3} - (y + r) \Leftrightarrow y = r\sqrt{3} - r - h$
$y + r = r\sqrt{3}\cos\left(\dfrac{\theta}{2}\right) \Leftrightarrow y = r\sqrt{3}\cos\left(\dfrac{\theta}{2}\right) - r$
$x^{2} = 3r^{2}\sin^{2}\left(\dfrac{\theta}{2}\right)$
$y^{2} = 3r^{2}\cos^{2}\left(\dfrac{\theta}{2}\right) - 2r^{2}\sqrt{3}\cos\left(\dfrac{\theta}{2}\right) + r^{2}$
But that's about everything of use - does anybody have any hints?
Edit to incorporate the suggestion of André Nicolas:
May we form a triangle as shown in dashed pink and so derive $y = \sqrt{3r^{2} - x^{2}} - r$
| Place the circle centers at the origin of a Cartesian grid. Without loss of generality, we can assume optimal rectangle is oriented with its base parallel to the $X$ axis, touching the smaller circle at the base midpoint at $(r,0)$.
Then the coordinates of the upper right hand corner can be taken to be $(rp,rq)$ where, since it must be on the outer circle,
$$
q = \sqrt{3-p^2}$$
Then the width of the rectagle is $2rp$ and the height is $r(q-1) = r(\sqrt{3-p^2}-1)$.
The area is
$$A=wh = 2r^2p(\sqrt{3-p^2}-1)$$
$$\frac{dA}{dp} = 2r^2\left[\sqrt{3-p^2}-1-\frac{p^2}{\sqrt{3-p^2}} \right]$$
We can set this to zero and solve for $p$ but since this is not easy I will show the steps. The expression is simpler if we solve for $q$, noting that $p^2 = 3-q^2$.
$$ q-1- \frac{3-q^2}{q}=0\\
q^2 - q - 3 + q^2 = 0 \\
2q^2 -q+3 = 0 \\
q = \frac{1+\sqrt{1+4\cdot 3\cdot 2}}{4}= \frac{3}{2}
\\p = \frac{\sqrt{3}}{2}
\\A_{\max} = 2r^2p(q-1) = 2r^2\frac{\sqrt{3}}{2}\frac{1}{2}=\frac{\sqrt{3}r^2}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Aproximation of $a_n$ where $a_{n+1}=a_n+\sqrt {a_n}$ Let $a_1=2$ and we define $a_{n+1}=a_n+\sqrt {a_n},n\geq 1$.
Is it possible to get a good aproximation of the $n$th term $a_n$?
The first terms are $2,2+\sqrt{2}$, $2+\sqrt{2}+\sqrt{2+\sqrt{2}}$ ...
Thanks in advance!
| Here is a method to develop the asymptotic behaviour of $a_n$ with a precision of two terms.
First, simply remark that $a_n$ grows to infinity since $a_n \geq n$ (by induction).
Then remark that $\sqrt{a_{n+1}} - \sqrt{a_n} = \frac{a_{n+1}-a_{n}}{\sqrt{a_{n+1}} + \sqrt{a_n}} = \frac{1}{1+\sqrt{1+\sqrt{\frac{1}{a_n}}}} \rightarrow \frac{1}{2}$ since $a_n\rightarrow\infty$.
It follows that $\sqrt{a_n} - 2=\sum_{i=0}^{n-1}\sqrt{a_{i+1}} - \sqrt{a_i}=\frac{1}{2}n+\mathcal{o}(n)$, then $a_n=(\frac{n}{2}+\mathcal{o}(n))^2 = \frac{n^2}{4}+\mathcal{o}(n^2)$
For the next term, we plug the asymptotic development we have just obtained into $\frac{1}{1+\sqrt{1+\sqrt{\frac{1}{a_n}}}}$ and look for an equivalent of $\frac{1}{1+\sqrt{1+\frac{1}{\sqrt{\frac{n^2}{4}+\mathcal{o}(n^2)}}}}-\frac{1}{2} \sim \frac{1-\sqrt{1+\frac{1}{\sqrt{\frac{n^2}{4}+\mathcal{o}(n^2)}}}}{4} \sim \frac{1}{4}\frac{1}{2}\frac{1}{\sqrt{\frac{n^2}{4}+\mathcal{o}(n^2)}} \sim \frac{1}{4n}$
From this, we deduce $\sqrt{a_{n+1}} - \sqrt{a_n} = \frac{1}{2}-\frac{1}{4n}+\mathcal{o}(\frac{1}{n})$, then $\sqrt{a_n} = \frac{n}{2}-\frac{1}{4}S_n + \mathcal{o}(Sn)$ where $Sn=\sum_0^n\frac{1}{k}=ln(n)+\mathcal{o}(ln(n))$.
It follows that $a_n=(\frac{n}{2}-\frac{1}{4}ln(n)+\mathcal{o}(ln(n)))^2=\frac{n^2}{4}-\frac{nln(n)}{4}+ \mathcal{o}(nln(n))$.
I believe that plugging again this new asymptotic development into the first formula would produce subsequent terms.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integrating reciprocals of functions with known antiderivatives If
$$\int_{}^{} f(x)\,dx$$
is known, is there a way to directly find
$$\int_{}^{} \frac{1}{f(x)}\,dx$$
| Consider the function $f(x)$ in a power series. Let
\begin{align}
f(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n}
\end{align}
then it can be seen that
\begin{align}
\frac{1}{f(x)} = \frac{1}{a_{0}} - \frac{a_{1}}{a_{0}^{2}} \, x + \frac{a_{1}^{2} - a_{2}}{a_{0}^{3}} \, x^{2} + \cdots
\end{align}
Integration of each shows
\begin{align}
\int f(x) \, dx &= a_{0} x + \frac{a_{1}}{2} \, x^{2} + \frac{a_{2}}{3} \, x^{3} + \cdots \\
\int \frac{dx}{f(x)} &= \frac{x}{a_{0}} - \frac{a_{1}}{2 \, a_{0}^{2}} \, x^{2} + \frac{a_{1}^{2} - a_{2}}{3 \, a_{0}^{3}} \, x^{3} + \cdots
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Radius of convergence of the power series $\sum_{n=1}^{\infty}a_nz^{n^2}$
Find the radius of convergence of the power series $$\sum_{n=1}^{\infty}a_nz^{n^2}$$ where , $a_0=1$ and $a_n=\frac{a_{n-1}}{3^n}$.
My Work :
We, have, $$\frac{a_1}{a_0}.\frac{a_2}{a_1}...\frac{a_n}{a_{n-1}}=\frac{1}{3}.\frac{1}{3^2}...\frac{1}{3^n}$$
$$\implies a_n=3^{-\frac{n(n+1)}{2}}$$
Now put $n^2=p$. Then the series becomes , $$\sum_{p=1}^{\infty}a_pz^p$$ where, $a_p=3^{\frac{p+\sqrt {p}}{2}}$.
Now $$\left|\frac{a_{p+1}}{a_p}\right|\to \frac{1}{\sqrt 3} \text{ as } p\to \infty$$
So , radius of convergence is $\sqrt 3$.
I think my answer is correct..But I want to know does there any other trick so that we can find the radius of convergence without finding $a_n$ explicitly ?
Please help...
| Just to add, if $\sum_n b_nz^n$ is a power series, then there is an explicit formula for the radius of convergence: $$R = \frac{1}{\limsup_n |b_n|^{\frac{1}{n}}}$$ In this case we have that $b_{n^2}=a_n=3^{-\frac{n(n+1)}{2}}$. Hence $b_{n^2}^{\frac{1}{n^2}}=3^{-\frac{1}{2}(1+\frac{1}{n})}$. Thus, substituting $k$ for $n^2$, we have that
$|b_k|^{-\frac{1}{k}} = \left\{
\begin{array}{}
3^{-\frac{1}{2}(1+\frac{1}{\sqrt{k}})} & : k \text{ is a perfect square}\\
0 & : \text{ otherwise}
\end{array} \right.$
It follows that $\limsup_k |b_k|^{\frac{1}{k}}=3^{-\frac{1}{2}}$, which gives $R=\sqrt{3}$.
| {
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Let $∑_{n=0}^∞c_n z^n $ be a representation for the function $\frac{1}{1-z-z^2 }$. Find the coefficient $c_n$ Let $∑_{n=0}^∞c_n z^n $ be a power series representation for the function $\frac{1}{1-z-z^2 }$. Find the coefficient $c_n$ and radius of convergence of the series.
Clearly this is a power series with center $z_0=0$, and $f(z)=\frac{1}{1-z-z^2 }$ is analytic, because it's represented by a power series. I also know that
$$c_n =\frac{1}{n!} f^{(n)}(0)$$
but this doesn't get me anywhere, I also try the special case of Taylor series, but nothing look like this. I wonder if any one would give me a hint please.
| Using the factorization $1 - z - z^2 = -(z + \varphi)(z + \bar{\varphi})$, where $\varphi = (1 + \sqrt{5})/2$ and $\bar{\varphi} = (1 - \sqrt{5})/2$, we write
\begin{align}\frac{1}{1 - z - z^2} &= \frac{1}{\sqrt{5}}\left(\frac{1}{z + \varphi} - \frac{1}{z + \bar{\varphi}}\right) \\
& = \frac{1}{\sqrt{5}}\left(\frac{1/\varphi}{(z/\varphi) + 1} - \frac{1/\bar{\varphi}}{(z/\bar{\varphi}) + 1}\right)\\
& = \frac{1}{\sqrt{5}}\left(\frac{-\bar{\varphi}}{1 - \bar{\varphi}z} + \frac{\varphi}{1 - \varphi z}\right)\\
& = \frac{1}{\sqrt{5}}\sum_{n = 0}^\infty [-\bar{\varphi}(\bar{\varphi} z)^n + \varphi(\varphi z)^n]\\
& = \sum_{n = 0}^\infty \left(\frac{\varphi^{n+1} - \bar{\varphi}^{n+1}}{\sqrt{5}}\right)z^n.
\end{align}
So $c_n = (\varphi^{n+1} - \bar{\varphi}^{n+1})/\sqrt{5}$ (which is equal to $F_{n+1}$, the $(n+1)$st Fibonacci number) and the radius of convergence is $1/\varphi$.
| {
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Sum $\pmod{1000}$
Let $$N= \sum_{k=1}^{1000}k(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor).$$ Find $N \pmod{1000}$.
Let $\lceil x \rceil$ be represented by $(x)$ and $\lfloor x \rfloor$ be represented by $[x]$.
Consider $0 < x < 1$ then:
$$(x) - [x] = 1 - 0 = 1$$
Consider $x=0$ then:
$$(x) - [x] = 0$$
Consider $x=1$ then:
$$(x) - [x] = 1-1 = 0$$
So we must find $k$ such that:
$$\log_{\sqrt{2}}k \in \mathbb{Z}$$
$$\log_{\sqrt{2}}k = \frac{2\log k}{\log 2} = y \in \mathbb{Z}$$
$$\log k = \log(\sqrt{2}^y)\implies k = \sqrt{2}^y$$
So for a few integer $y$ values,
$$k = \sqrt{2}, 2^1, \sqrt{2}^3, 2^2,..., 2^3$$...
$k$ is an integer for powers of $2$.
$$\log_{\sqrt{2}} 2^x = x\log_{\sqrt{2}} 2 = x \frac{2\log_2 2}{\log_2 2} = 2x \in \mathbb{Z}$$
The sum is $1$ for $k=1$ but then $0$ for $k=2$.
$2^9 = 512 < 1000$ but $2^{10} = 1024 > 1000$.
but how can I compute the sum?
| Note that $(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor)=1$ unless $k$ is an integer power of $\sqrt 2$, which is when $k$ is a power of $2$. In that case it is zero. So your sum is $$N= \sum_{k=1}^{1000}k-\sum_{m=0}^92^m$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A problem I didn't know since high school algebra Determine all positive integers which can be written as a sum of two squares of integers.
This is a problem I saw when I was in high school...
*
*sum of two squares of integers can be (4k, 4k+1, 4k+2, but no 4k+3?)
| Every integer $n$ can be expressed as $n=4k+r$ for $r=0,1,2,3$. $r$ is called the remainder after division by 4. Now $n^2 = (4k+r)^2 = 4^2k^2 + 8kr + r^2$.
If $r = 0$ then $r^2=0$. So $n^2=4k'$ for some $k'$.
If $r=1$ then $r^2=1$. So $n^2=4k'+1$ for some $k'$.
If $r=2$ then $r^2=4$. So $n^2=4k'$ for some $k'$.
If $r=3$ then $r^2 = 9 = 2\cdot 4 + 1$. So $n^2 = 4k'+1$ for some $k'$.
Thus the square of every integer can be expressed as $n^2 = 4k + r$ where $r=0$ or $r=1$.
Thus the sum of two squares can have the following remainders after division by 4:
$$1+1=2$$
$$1+0=1$$
$$0+0=0$$
Note that we cannot have a remainder of 3, since this has exhausted all the possibilities.
A neat thing that you can show with a bit of imaginary numbers is that the product of two numbers that are a sum of squares is again a sum of squares.
$$(n^2+m^2)(a^2+b^2)$$ $$= (n+im)(n-im)(a+ib)(a-ib)$$ $$= (an-bm+i(nb+ma))(an-bm-i(nb+ma))$$
$$=(an-bm)^2+(nb+ma)^2$$
Thus if we take $10=3^2+1$ and $8=2^2+2^2$ we can write $80$ as the sum of two squares:
$$80 = (6-2)^2+(6+2)^2 = 4^2 + 8^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Use logarithmic differentiation to find $\frac{dy}{dx}$ Here is the problem as well as my work so far. Any advice or hints regarding where I should go from here would be appreciated.The arrow indicates where I got stuck.
Where do I go from here?
EDIT
I realize my mistake in not using logarithms. Here is my second attempt. What should I try from here?
| here is a way to do check your i will do it little differently. i will start with
$$\ln y = 5 \ln (x^2 + 3) - \frac 12 \ln (x+1) $$ differencing this we get $$\frac{dy}{y} = 5 \frac{d(x^2 + 3)}{x^2 + 3} - \frac{d(x+1)}{2(x+1)}=\frac{10x\, dx}{x^2 + 3} - \frac{dx}{2(x+1)} $$ therefore
$$\frac{dy}{dx} = y\left(\frac{10x}{x^2 + 3} - \frac{1}{2(x+1)}\right)= \frac{(x^2+3)^3}{\sqrt{x+1}}\left(\frac{10x}{x^2 + 3} - \frac{1}{2(x+1)}\right)=\frac{(x^2+3)^2(19x^2+20x - 3)}{2(x+1)^{3/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Showing $1+2+\cdots+n=\frac{n(n+1)}{2}$ by induction (stuck on inductive step) This is from this website:
Use mathematical induction to prove that
$$1 + 2 + 3 +\cdots+ n = \frac{n (n + 1)}{2}$$
for all positive integers $n$.
Solution to Problem 1: Let the statement $P(n)$ be
$$1 + 2 + 3 + \cdots + n = \frac{n (n + 1)}{2}.$$
STEP 1: We first show that $P(1)$ is true.
Left Side $= 1$
Right Side $= \frac{1 (1 + 1)}{2} = 1$
Both sides of the statement are equal hence $P(1)$ is true.
STEP 2: We now assume that $P(k)$ is true
$$1 + 2 + 3 + \cdots + k = \frac{k (k + 1)}{2}$$
and show that $P(k + 1)$ is true by adding $k + 1$ to both sides of the above statement
$$
\begin{align}
1 + 2 + 3 + \cdots + k + (k + 1) &= \frac{k (k + 1)}{2} + (k + 1) \\
&= (k + 1)\left(\frac{k}{2} + 1\right) \\
&= \frac{(k + 1)(k + 2)}{2}
\end{align}
$$
The last statement may be written as
$$1 + 2 + 3 + \cdots + k + (k + 1) = \frac{(k + 1)(k + 2)}{2}$$
Which is the statement $P(k + 1)$.
My question is how in the very last line is the statement $P(k + 1)$ equal to $\frac{(k + 1)(k + 2)}{2}$. I don't get the last step.
| We know that
$P(k) = 1 + 2 + 3 + ... + k$
Therefore:
$P(k+1) = 1 + 2 + 3 + ... + k + (k+1)$
By induction hypothesis we have:
$1 + 2 + 3 + ... + k + (k+1) = \frac{(k+1)(k+2)}{2}$
so
$P(k+1) = 1 + 2 + 3+...+k+(k+1) = \frac{(k+1)(k+2)}{2}$
so
$P(k+1) = \frac{(k+1)(k+2)}{2}$
By induction we now know that since this is true for one integer $k$, it is true for all integers greater than or equal to $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $A,B>0$ and $A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$. If $A,B>0$ and $\displaystyle A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$.
$\bf{My\; Try::}$ Given $$\displaystyle A+ B = \frac{\pi}{3}$$ and $A,B>0$.
So we can say $$\displaystyle 0< A,B<\frac{\pi}{3}$$. Now taking $\tan $ on both side, we get
$$\displaystyle \tan(A+B) = \tan \left(\frac{\pi}{3}\right).$$ So $\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B} = \sqrt{3}$.
Now Let $\displaystyle \tan A\cdot \tan B=y\;,$ Then $\displaystyle \tan B = \frac{y}{\tan A}.$
So $$\displaystyle \frac{\tan A+\frac{y}{\tan A}}{1-y}=\sqrt{3}\Rightarrow \tan^2 A+y=\sqrt{3}\tan A-y\sqrt{3}\tan A$$
So equation $$\tan^2 A+\sqrt{3}\left(y-1\right)\tan A+y=0$$
Now for real values of $y\;,$ Given equation has real roots. So its $\bf{Discrimnant>0}$
So $$\displaystyle 3\left(y-1\right)^2-4y\geq 0\Rightarrow 3y^2+3-6y-4y\geq 0$$
So we get $$3y^2-10y+3\geq 0\Rightarrow \displaystyle 3y^2-9y-y+3\geq 0$$
So we get $$\displaystyle y\leq \frac{1}{3}\cup y\geq 3$$, But above we get $\displaystyle 0<A,B<\frac{\pi}{3}$
So We Get $$\bf{\displaystyle y_{Max.} = \left(\tan A \cdot \tan B\right)_{Max} = \frac{1}{3}}.$$
My Question is can we solve above question using $\bf{A.M\geq G.M}$ Inequality or Power Mean equality.
Thanks
| $$(\tan A+\tan B)^2-4\tan A\tan B=(\tan A-\tan B)^2\ge0$$
$$\iff4\tan A\tan B\le(\tan A+\tan B)^2$$
Also by $\bf{A.M\geq G.M},$ $\dfrac{\tan A+\tan B}2\ge\sqrt{\tan A\tan B}$
By Jensen's inequality, $$\frac{\tan A+\tan B}2\le\tan\frac{A+B}2=\frac1{\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Apostol (3.8.22): Finding $f \circ g (x)$ $$
f(x)=
\begin{cases}
1 &\text{if} & |x|\leq 1,\\
0 &\text{if} & |x|>1,\\
\end{cases}
\qquad
g(x)=
\begin{cases}
2-x^2 &\text{if} & |x|\leq 2,\\
2 &\text{if} & |x|>2.
\end{cases}
$$
I need to find the composition $h(x)$ of $f(x)$ and $g(x)$ such that $h(x)=f(g(x))$. My answer to this question is $h(x)=f(x)$ since the values of $f$ are constants. But the answer of Apostol is
$$
h(x)=
\begin{cases}
1 &\text{if} & 1\leq|x|\leq\sqrt{3},\\
0 &{} &\text{otherwise}.
\end{cases}
$$
How can this be?
| $$
f(g(x)) = \begin{cases} 1 & \left|g(x)\right| \leq 1 \\
0 & \left|g(x)\right| > 1
\end{cases}
$$
So you need to find those cases:
First, it's clear that when $|x| > 2$, $g(x) = 2 > 1$ and thus when $|x| > 2$, $f(g(x)) = 0$. So now we need to investigate when $|x| \leq 2$ and $g(x) = 2 - x^2$. It might help to right down $\left|g(x)\right|$:
$$
\left|2 - x^2\right| = \begin{cases}
2 - x^2 & 2 - x^2 \geq 0 \\
x^2 - 2 & 2 - x^2 < 0
\end{cases}
$$
So lets solve the following:
\begin{align}
2 - x^2 \geq 0 \\
(\sqrt{2} - x)(\sqrt{2} + x) \geq 0 \\
\sqrt{2} - x \geq 0 \wedge \sqrt{2} + x \geq 0 \rightarrow x \leq \sqrt{2} \wedge x \geq -\sqrt{2} \rightarrow -\sqrt{2} \leq x \leq \sqrt{2} \\
\textit{or}\\
\sqrt{2} - x < 0 \wedge \sqrt{2} + x < 0 \rightarrow x > \sqrt{2} \wedge x < -\sqrt{2} \rightarrow False
\end{align}
Therefore we can now write down the absolute value of $g(x)$ as follows:
$$
\left|g(x)\right| = \begin{cases}
2 - x^2 & |x| \leq \sqrt{2} \\
x^2 - 2 & |x| > \sqrt{2}
\end{cases}
$$
(which should be obvious even without the above rigmarole).
Now let's look at the two cases:
1) $|x| \leq \sqrt{2}$: $\left|g(x)\right| = 2 - x^2$
Find when $\left|g(x)\right| \leq 1$:
\begin{align}
2 - x^2 \leq 1 \rightarrow 1 - x^2 \leq 0 \rightarrow (1 - x)(1 + x) \leq 0 \\
1 - x \leq 0 \wedge 1 + x \geq 0 \rightarrow x \geq 1 \wedge x \geq -1 \rightarrow x \geq 1 \\
1 - x \geq 0 \wedge 1 + x \leq 0 \rightarrow x \leq 1 \wedge x \leq -1 \rightarrow x \leq -1
\end{align}
Combined this is the constraint that $|x| \geq 1$. This along with the original constraint that $\left|x\right| \leq \sqrt{2}$ gives $1 \leq |x| \leq \sqrt{2}$.
2) $|x| > \sqrt{2}$: $\left|g(x)\right| = x^2 - 2$.
Again, find when $\left|g(x)\right| \leq 1$:
\begin{align}
x^2 - 2 \leq 1 \rightarrow x^2 - 3 \leq 0 \rightarrow (x - \sqrt{3})(x + \sqrt{3} \leq 0 \\
x - \sqrt{3} \leq 0 \wedge x + \sqrt{3} \geq 0 \rightarrow x \leq \sqrt{3} \wedge x \geq -\sqrt{3} \rightarrow -\sqrt{3} \leq x \leq \sqrt{3} \\
x - \sqrt{3} \geq 0 \wedge x + \sqrt{3} \leq 0 \rightarrow x \geq \sqrt{3} \wedge x \leq -\sqrt{3} \rightarrow False
\end{align}
Combined with the original assumption that $|x| > \sqrt{2}$ then $\left|g(x)\right| \leq 1$ when $\sqrt{2} < |x| \leq \sqrt{3}$ and $\left|g(x)\right| > 1$ otherwise.
We can naively combine all of this:
$$
f(g(x)) = \begin{cases}
\begin{cases}
1 & 1 \leq |x| \leq \sqrt{2} \\
0 & |x| <1
\end{cases} & |x| \leq \sqrt{2} \\
\begin{cases}
1 & \sqrt{2} < |x| \leq \sqrt{3} \\
0 & |x| > \sqrt{3}
\end{cases} & |x| > \sqrt{2}
\end{cases}
$$
You should be able to see that the two cases when $f(g(x)) = 1$ combine to create the interval $1 \leq |x| \leq \sqrt{3}$ such that:
$$
f(g(x)) = \begin{cases}
1 & 1 \leq |x| \leq \sqrt{3} \\
0 & |x| < 1 \vee |x| > \sqrt{3}
\end{cases}
$$
or simply:
$$
f(g(x)) = \begin{cases}
1 & 1 \leq |x| \leq \sqrt{3} \\
0 & \text{otherwise}
\end{cases}
$$
Note: I was sloppy and started to write $g(x)$ as if it was always $g(x) = 2 - x^2$. I never explicitly combined the fact that $g(x) = 2$ when $|x| > 2$. However, since $2 > \sqrt{3}$ and $f(2) = 0$, the above is indeed correct, but we should be careful to note that $|x| > 2$ gives $f(x) = 0$ (by the definition of $f(x)$) and that this agrees with the above formulation since we say that $f(x) = 0$ when $|x| > \sqrt{3}$ which includes when $|x| > 2$.
| {
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"url": "https://math.stackexchange.com/questions/1209828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Does this series $\sum_{i=0}^n \frac{4}{3^n}$ diverge or converge? I a newbie to series, and I have not done too much yet. I have an exercise where I have basically to say if some series are convergent or divergent. If convergent, determine (and prove) the sum of the series.
This is the first series:
$$\sum_{i=0}^n \frac{4}{3^i}$$
I have heard about the ratio test, so I decided to try to apply it in this case to see if the series is or not convergent.
The ratio test is basically defined like this:
$$L = \lim{\left|\frac{a_{n+1}}{a_n}\right|}$$
Where $a_n$ is in this case $\frac{4}{3^n}$ and $a_{n+1}$ is consequently $\frac{4}{3^{n+1}}$. Thus, we have:
$$L = \lim{\left|\frac{\frac{4}{3^{n+1}}}{\frac{4}{3^n}}\right|} = \lim{\left|\frac{4 \cdot 3^n}{3^{n+1} \cdot 4}\right|} = \lim{\left|\frac{3^n}{3^{n+1}}\right|} = \lim{\left|\frac{3^n}{3^n\cdot 3}\right|} = \lim{\left|\frac{1}{3}\right|} = \frac{1}{3} < 1$$
From the ratio test, we know if $L< 1$, then the series converges.
Now, I need to find the sum. I have followed a video which explain how to find the sum using the ratio, which sincerely I have not understand well what it is. Can you explain a general rule to find the ratio, and what exactly is it?
Ok. So, following the same process done during the video, I have:
$$\sum_{i=0}^n \frac{4}{3^n} = 4 \sum_{i=0}^n \frac{1}{3^n} = 4 \sum_{i=0}^n \left(\frac{1}{3}\right)^n$$
Apparently, our ratio $r$ is $\frac{1}{3}$.
Now, to find the sum of the series, he uses a formula:
$$\frac{\text{first term}}{1 - r}$$
Where first term is the first term of the series. Can you explain where this formula comes from?
Applying that formula, we obtain:
$$\frac{\left(\frac{1}{3}\right)^0}{1 - \frac{1}{3}} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$$
I checked the sum using a online calculator, and apparently it is not $\frac{3}{2}$, so I decided to manipulate, and I thought I left a $4$ outside the series, so I multiply that $4$ by $\frac{3}{2}$, which produces $\frac{12}{2} = 6$, which is exactly the sum I found using 2 online calculators.
Now, my 3rd question is: why what I did is correct (or not correct)?
|
A proof that, if the left-hand-side converges:
$$a+ar+ar^2+\dotsb=\frac a{1-r}$$
Call the sum $S$. Thus, we have $S=a+ar+ar^2+\dotsb$.
$\phantom rS=a+ar+ar^2+ar^3+\dotsb$
$rS=\phantom{a+{}}ar+ar^2+ar^3+\dotsb$ (using the distributive property)
Thus:
\begin{align}
S&=a+rS\\
S-rS&=a\\
S(1-r)&=a\\
S&=\frac a{1-r}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
$(\delta,\varepsilon)$ Proof of Limit I wish to prove that $\lim_{x\to 2} {x+1 \over x+2} = {3 \over 4} $.
The $(\delta,\varepsilon)$ limit definition in this case is:
$\forall \epsilon >0, \exists \delta >0$ such that $0<|x-2|<\delta \Rightarrow |{x+1 \over x+2} - {3 \over 4}| < \epsilon.$
Thus, I need to provide a $\delta$, which is a function of $\epsilon$ in order to satisfy the above definition.
I am having a bit of difficulty finding an inequality to continue my work below.
$|{x+1 \over x+2} - {3 \over 4}| = {1 \over 4}|{x-2 \over x+2}|$
| You want to prove
$$\lim_{x\to 2}\frac{x+1}{x+2}=\frac{3}{4}.$$
By the definition of limit you need to find, for all $\epsilon>0$, a $\delta$ such that $0<|x-2|<\delta$ implies $$\left|\frac{x+1}{x+2}-\frac{3}{4}\right|<\epsilon.$$ Your best bet is to work back from the complicated expression. We can simplify,
$$\left|\frac{4(x+1)-3(x+2)}{4(x+2)}\right|=\left|\frac{x-2}{4(x+2)}\right|=\frac{1}{4}\left|\frac{x-2}{x+2}\right|=\frac{1}{4}\frac{|x-2|}{|x+2|}<\epsilon.$$ Look what just appeared in the numerator - it's the $|x-2|$ you needed to get a handle on.
Rearranging we have $$|x-2|<4|x+2|\epsilon.$$
Suppose $|x-2|<1$ (which we suppose because we want to be "close" to 2, here within a distance of 1), then $$-1<x-2<1,$$ or adding 4 to both sides, $$3<x+2<5,$$ from which we can write $|x+2|<5$.
Substituting, we get
$$|x-2|<(4\cdot 5)\epsilon,$$ or $$|x-2|<20\epsilon.$$ So let $\delta = \min\{1,20\epsilon\}$.
Now let's check this works: For all $\epsilon>0$, let $|x-2|<\delta=\min\{1,20\epsilon\}$. We can see from this inequality that $\delta\leq 20\epsilon$ holds true regardless of what the minimum turns out to be. Hence, we have
$$|x-2|<20\epsilon\iff \frac{1}{4}|x-2|<5\epsilon.$$
Rewriting this as $$\frac{1}{4}\frac{|x-2|}{|x+2|}|x+2|<5\epsilon,$$ and then using the fact that
$$|x-2|<1\iff -1<x-2<1\iff 4<x+2<5\implies|x+2|<5,$$
then we deduce that $$\frac{1}{4}\left|\frac{x-2}{x+2}\right|<\epsilon,$$ as required.
| {
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"url": "https://math.stackexchange.com/questions/1212672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the derivative of the function $F(x) = \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt$. $$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+t^4}}2x - \frac{1}{\sqrt{2+t^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+t^4}} - \frac{\sec^2{x}}{\sqrt{2+t^4}} \\
&= \frac{2x-\sec^2{x}}{\sqrt{2+t^4}} \\
\end{align}$$
Update:
Is this looking better??
$$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+(x^2)^4}}2x - \frac{1}{\sqrt{2+(\tan{x})^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+x^8}} - \frac{\sec^2{x}}{\sqrt{2+\tan^4{x}}} \\
\end{align}$$
Is this correct?
| No. $t$ is a mute variable. The answer must have $x$ as the only variable. Yor computations are similar to
$$
\int_1^2t\,dt=2\cdot t-1\cdot t=t,
$$
which is obviously wrong.
| {
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"url": "https://math.stackexchange.com/questions/1213093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Limit as $x$ tend to zero of: $x/[\ln (x^2+2x+4) - \ln(x+4)]$ Without making use of LHôpital's Rule solve:
$$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)}$$
$ x^2+2x+4=0$ has no real roots which seems to be the gist of the issue.
I have attempted several variable changes but none seemed to work.
| In the eternal words of Claude Leibovici, love Taylor Series.
We have
\begin{align*}
\ln(x^{2}+2x+4) &= \ln(4) + \frac{x}{2} + \frac{x^{2}}{8} + \mathcal{O}(x^{3})\\
\ln(x+4) &=\ln(4) + \frac{x}{4} -\frac{x^{2}}{32} + \mathcal{O}(x^{3}).
\end{align*}
So
$$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)} = \lim_{x\to 0} \frac{x}{\ln(4) + \frac{x}{2} + \frac{x^{2}}{8} - \left( \ln(4) + \frac{x}{4} -\frac{x^{2}}{32}\right) + \mathcal{O}(x^{3})} = \lim_{x\to 0} \frac{x}{\frac{x}{4} + \frac{5x^{2}}{32} + \mathcal{O}(x^{3})}.$$
Divide the numerator and denominator by the lowest power of $x$ in the denominator to get
$$\lim_{x\to 0} \frac{x}{\frac{x}{4} + \frac{5x^{2}}{32} + \mathcal{O}(x^{3})} = \lim_{x\to 0} \frac{\frac{1}{x} x}{\frac{1}{x}\left( \frac{x}{4} + \frac{5x^{2}}{32} + \mathcal{O}(x^{3})\right)} = \lim_{x\to 0}\frac{1}{\frac{1}{4} + \frac{5x}{32} + \mathcal{O}(x^{2})} = \frac{1}{\frac{1}{4}} = 4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $z$ s.t. $\frac{1+z}{z-1}$ is real I must find all $z$ s.t. $\dfrac{1+z}{z-1}$ is real.
So, $\dfrac{1+z}{1-z}$ is real when the Imaginary part is $0$.
I simplified the fraction to $$-1 - \dfrac{2}{a+ib-1}$$
but for what $a,b$ is the RHS $0$?
| Hint:
$$\begin{align}\frac{1+z}{z-1} \,\,\, \text{is real} \ &\iff \frac{1+z}{z-1} = \overline{\frac{1+z}{z-1}} = \frac{\overline{1+z}}{\overline {z-1}} \\&\iff (1+z)(\overline {z}-1) = (1+\overline {z})(z-1)\\&\iff \mathfrak {Im} (z) = 0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A sequence is defined by $f(x) = 4^{x-1}$, find the sum of the first $8$ terms. A sequence is defined by $f(x) = 4^{x-1}$, find the sum of the first $8$ terms.
$\dfrac{a(1-r^n)}{1-r}$
$\dfrac{1(1-4^7)}{1-4} = 5461$.
The answer in the book is $21845$. How is this so?
Thank you
| For $a_n = 4^{n-1}$,
$$
a\frac{r^n-1}{r-1} = \frac{4^8-1}{4-1} = \frac{65536 -1}{3} = 21845.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that equation $x^6+x^5-x^4-x^3+x^2+x-1=0$ has two real roots Prove that equation
$$x^6+x^5-x^4-x^3+x^2+x-1=0$$ has two real roots
and $$x^6-x^5+x^4+x^3-x^2-x+1=0$$
has two real roots
I think that:
$$x^{4k+2}+x^{4k+1}-x^{4k}-x^{4k-1}+x^{4k-2}+x^{4k-3}-..+x^2+x-1=0$$
and
$$x^{4k+2}-x^{4k+1}+x^{4k}+x^{4k-1}-x^{4k-2}-x^{4k-3}-..+x^2+x-1=0$$
has two real roots but i don't have solution
| I'm not sure of the general case; but for the first polynomial divide through by $x^3$, since $x\neq0$, and substitute $y=x-1/x$. You then obtain $y^3+y^2+2y+1=0$, which must have just one real root, in the interval $(-1,0).$ Call this root $y=a=x-1/x$. Then $x^2-ax-1=0$, which has a positive discriminant ($a^2+4$), implying two distinct real roots for $x$. I'm unsure if you can do something similar for the next polynomial...
| {
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"url": "https://math.stackexchange.com/questions/1217316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to factorize $(x-2)^5+x-1$? This is a difficult problem.
How to factorize this?
$$(x-2)^5+x-1$$
we can't do any thing now and we should expand it first:
$$x^5-10x^4+40x^3-80x^2+81x-33$$
but I can't factorize it.
| finally found how looking in my previous questions.
first substitute $n=x-2$ then
$$(x-2)^5+x-1=n^5+n+1$$
$$n^5+n+1=n^5-n^2+n^2+n+1=n^2(n^3-1)+n^2+n+1$$
$$=n^2(n-1)(n^2+n+1)+n^2+n+1$$
$$=(n^2+n+1)(n^3-n^2+1)$$
| {
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"url": "https://math.stackexchange.com/questions/1222420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof of sum results I was going through some of my notes when I found both these sums with their results
$$
x^0+x^1+x^2+x^3+... = \frac{1}{1-x}, |x|<1
$$
$$
0+1+2x+3x^2+4x^3+... = \frac{1}{(1-x)^2}
$$
I tried but I was unable to prove or confirm that these results are actually correct, could anyone please help me confirm whether these work or not?
| Others have answered on how to justify that $S=1+x+x^2+\cdots+x^n+\cdots$ equals $\frac{1}{1-x}$ for $|x|<1$. To get the second identity without differentiation, note that
\begin{align*}
1+2x+3x^2+4x^3+\cdots&=1+(x+x)+(x^2+x^2+x^2)+(x^3+x^3+x^3+x^3)+\cdots\\
&=(1+x+x^2+x^3+\cdots)+(x+x^2+x^3+\cdots)+(x^2+x^3+\cdots)+\cdots\\
&=S+xS+x^2S+\cdots=S(1+x+x^2+\cdots)\\
&=S^2\\
&=\frac{1}{(1-x)^2}\cdot
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Minimising the distance covered I am trying to solve the question: You are trying to get to go from A to B 10 times. At each journey, a coin is flipped and if its heads, a wall appears in the middle as in scenario 2. If tails, no wall appears. Construct a method to minimize the distance covered. What is the expected distance covered over those 10 walks?
Attempt: we can treat the 10 trials independently, and the expected distance of any one walk is 0.5 ( 2m) + 0.5 (4m) = 3m, so the expected total distance covered is 30 metres. But what strategy can we construct to minimize this distance? My only thought is to walk along the diagonals so that we are guareenteed to walk $2\sqrt{2}$ each walk, this minimizes the distance to approximately 28 metres.
| I assume that the coin is tossed each trip only upon reaching the wall position, and also that the diagram is a top view so we are walking around, not climbing over, the wall.
Firstly, the default expected distance should probably assume, in the case of tails, that you walk diagonally on the second part of the trip, so we would have $E(D) = \frac{1}{2}2 + \frac{1}{2}(2+\sqrt{2}) = 2+1/\sqrt{2}$.
The idea to walk the diagonal on the first half assumes the worst case (tails). There might be a middle ground that's optimal. Say you aim for a spot on the wall at distance $0\leq x\leq 1$ from its centre. Then the expected distance of one trip is
\begin{eqnarray*}
E(D) &=& \frac{1}{2} (2\sqrt{x^2+1}) + \frac{1}{2}(\sqrt{x^2+1} + (1-x) + \sqrt{2}) \\
&& \\
&=& \frac{3}{2} \sqrt{x^2+1} - \frac{x}{2} + \frac{1}{2} + \frac{1}{\sqrt{2}} \\
&& \\
\dfrac{dE(D)}{dx} &=& \frac{3x}{2\sqrt{x^2+1}} - \frac{1}{2} = 0 \implies 3x = \sqrt{x^2+1}\implies x = \dfrac{1}{2\sqrt{2}}. \\
\end{eqnarray*}
With this value of $x$ we get $E(D)=\dfrac{6+\sqrt{2}}{2\sqrt{2}}.\;$ And $10E(D) = \dfrac{30+5\sqrt{2}}{\sqrt{2}}\approx 26.21$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate $\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$ I came across this strange limit whilst showing convergence of a series:
$$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$$
How can I calculate this limit?
| Use the sandwich/squeeze theorem. Since
$$\frac{3^n}{2\cdot 4^n}<\frac{2^n + 3^n}{3^n + 4^n} < \frac{2\cdot 3^n}{4^n}$$
for all $n$,
$$\frac{1}{2^{1/n}} \cdot \frac{3}{4} < \sqrt[n]{\frac{2^n + 3^n}{3^n + 4^n}} < 2^{1/n}\cdot \frac{3}{4}$$
for all $n$. Since $2^{1/n} \to 1$ as $n\to \infty$, by the squeeze theorem, your limit is $3/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What will be the sum of the series of binomial co-efficients? What will be the sum of the following binomial co-efficent series
$$\binom{z+1}{z} + \binom{z+2}{z} + \binom{z+3}{z} + \dots + \binom{z+r}{z} = \sum\limits_{i=1}^r \binom{z+i}{z}$$
Thank you
| $\dbinom{z+1}{z} +\dbinom{z+2}{z} +\cdots+\dbinom{z+r}{z}$
= Coeff of $x^{z}$ in $(1+x)^{z+1}$+ Coeff of $x^{z}$ in $(1+x)^{z+2}$+$\cdots$+ Coeff of $x^{z}$ in $(1+x)^{z+r}$
= Coeff of $x^{z}$ in $((1+x)^{z+1}+ (1+x)^{z+2}+\cdots+ (1+x)^{z+r})$
= Coeff of $x^{z}$ in $((1+x)^{z+1}\frac{(1+x)^{r}-1}{1+x-1})$
= Coeff of $x^{z+1}$ in $((1+x)^{z+1}.((1+x)^{r}-1))$
= Coeff of $x^{z+1}$ in $((1+x)^{z+r+1}-(1+x)^{z+1})$
=$\dbinom{z+r+1}{z+1}-\dbinom{z+1}{z+1}$
=$\dbinom{z+r+1}{r}-1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can you derive $\sin(x) = \sin(x+2\pi)$ from the Taylor series for $\sin(x)$? \begin{eqnarray*}
\sin(x) & = & x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\\
\sin(x+2π) & = & x + 2\pi - \frac{(x+2π)^3}{3!} + \frac{(x+2π)^5}{5!} - \ldots \\
\end{eqnarray*}
Those two series must be equal, but how can you show that by only manipulating the series?
| With series manipulation, we can get
$$
\begin{align}
\sin(x+y)
&=\sum_{k=0}^\infty(-1)^k\frac{(x+y)^{2k+1}}{(2k+1)!}\tag{1a}\\
&=\sum_{k=0}^\infty\sum_{j=0}^{2k+1}\frac{(-1)^k}{(2k+1)!}\binom{2k+1}{j}x^jy^{2k+1-j}\tag{1b}\\
&=\sum_{k=0}^\infty\sum_{j=0}^{2k+1}(-1)^k\frac{x^j}{j!}\frac{y^{2k+1-j}}{(2k+1-j)!}\tag{1c}\\
&=\sum_{k=0}^\infty\sum_{j=0}^k(-1)^j\frac{x^{2j+1}}{(2j+1)!}(-1)^{k-j}\frac{y^{2k-2j}}{(2k-2j)!}\tag{1d}\\
&+\sum_{k=0}^\infty\sum_{j=0}^k(-1)^j\frac{x^{2j}}{(2j)!}(-1)^{k-j}\frac{y^{2k-2j+1}}{(2k-2j+1)!}\tag{1e}\\
&=\sum_{j=0}^\infty(-1)^j\frac{x^{2j+1}}{(2j+1)!}\sum_{k=0}^\infty(-1)^{k}\frac{y^{2k}}{(2k)!}\tag{1f}\\
&+\sum_{j=0}^\infty(-1)^j\frac{x^{2j}}{(2j)!}\sum_{k=0}^\infty(-1)^k\frac{y^{2k+1}}{(2k+1)!}\tag{1g}\\[6pt]
&=\sin(x)\cos(y)+\cos(x)\sin(y)\tag{1h}
\end{align}
$$
Explanation:
$\text{(1a)}$: series definition
$\text{(1b)}$: binomial theorem
$\text{(1c)}$: break binomial coefficient into factorials
$\text{(1d)}$: add the odd powers of $x$ from $\text{(1c)}$
$\text{(1e)}$: to the even powers of $x$ from $\text{(1c)}$
$\text{(1f)}$: change the order of summation and substitute $k\mapsto k+j$
$\text{(1g)}$: change the order of summation and substitute $k\mapsto k+j$
$\text{(1g)}$: series definitions
We can prove that $\cos(2\pi)=1$ and $\sin(2\pi)=0$ using a couple of derivatives and some geometry. First,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)
&=\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}\\
&=\sum_{k=0}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\\[6pt]
&=\cos(x)\tag{2}
\end{align}
$$
and
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\cos(x)
&=\frac{\mathrm{d}}{\mathrm{d}x}\left(1+\sum_{k=1}^\infty(-1)^k\frac{x^{2k}}{(2k)!}\right)\\
&=\sum_{k=1}^\infty(-1)^k\frac{x^{2k-1}}{(2k-1)!}\\
&=\sum_{k=0}^\infty(-1)^{k+1}\frac{x^{2k+1}}{(2k+1)!}\\[6pt]
&=-\sin(x)\tag{3}
\end{align}
$$
Using $(2)$ and $(3)$, we get
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin^2(x)+\cos^2(x)\right)
&=2\sin(x)\cos(x)-2\cos(x)\sin(x)\\
&=0\tag{4}
\end{align}
$$
Equation $(4)$ says that $\sin^2(x)+\cos^2(x)$ is a constant, and since $\sin(0)=0$ and $\cos(0)=1$, we have
$$
\sin^2(x)+\cos^2(x)=1\tag{5}
$$
Thus, $(\cos(x),\sin(x))$ is on the unit circle and moves with speed
$$
\begin{align}
\left|\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x),\sin(x))\right|
&=\left|\vphantom{\frac12}(-\sin(x),\cos(x))\right|\\[6pt]
&=1\tag{6}
\end{align}
$$
To go around the unit circle once at speed $1$ takes $2\pi$. Thus,
$$
\begin{align}
(\cos(2\pi),\sin(2\pi))
&=(\cos(0),\sin(0))\\[6pt]
&=(1,0)\tag{7}
\end{align}
$$
Combining $(1)$ and $(7)$ yields
$$
\begin{align}
\sin(x+2\pi)
&=\sin(x)\cos(2\pi)+\cos(x)\sin(2\pi)\\[6pt]
&=\sin(x)\tag{8}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of m. Problem :
If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of $m$.
My Approach:
$|x^2+4x+3|-mx+2m=0$
Case I : $x^2+4x+3-mx+2m=0$
$\Rightarrow x^2+ x (4-m) + 3+2m=0 $
Discriminant of above qudratic is
$D = (4-m)^2 -4(3+2m) \geq 0$
$D = 16+m^2-8m-12-8m$
Solving for $m$ we get the values $-8 \pm 2\sqrt{15}$
Case II :
Similarly solving for the given equation taking negative sign of modulus we get the solution
for $m =$$8 \pm 2\sqrt{15}$
Can we take all the values of m to satisfy the given condition of the problem , please suggest which value of m should be neglected in this. Thanks.
| The parabola $x^2 + 4x + 3 = (x+3)(x+1)$ is upward facing and has roots at $x = -3, x = -1$, so it is negative in $(-3, -1)$. Break it into cases. One of these cases has two solutions, the other has exactly one (i.e. discriminant = 0).
Case 1: $-3 \leq x \leq -1$. The equation is $-x^2 - 4x - 3 - mx + 2m = 0$ so $x^2 + (4+m)x + (3-2m) = 0$. Discriminant is $\sqrt{(m+4)^2 - 4(3-2m)}$.
Case 2: The other case. The equation is $x^2 + (4-m) x + (3+2m) = 0$, with discriminant $\sqrt{(4-m)^2 - 4(3+2m)}$.
Suppose Case 1 has one solution and Case 2 has two. Then $m^2 + 8m + 16 - 12 + 8m = 0 \to m^2 + 16m + 4 = 0$ so $m = -8 \pm 2 \sqrt{15}$. The root of the quadratic is $\frac{-(m+4)}{2}$, and so $m = -8 + 2 \sqrt{15}$ works for the domain. This also gives valid roots in the other quadratic, so this is our solution.
Remark. If we instead suppose that Case 2 has the one solution we end up finding two values of $m$ which give invalid solutions for the Case 1 quadratic.
| {
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"timestamp": "2023-03-29T00:00:00",
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substitution integration question I want to integrate $\int \sqrt{1 - x^2} dx $.
When I substitute $x = \sin θ$ , I get the right answer. ( $ \cos^2\theta$ before integration)
But when I substitute $x = \cos θ$ , I don't get the right answer. ( $ -\sin^2\theta$ before integration).
What step is wrong here? If I proceed like this, don't I end up with a wrong answer?
| The short answer is they become the same answer once you back-substitute $\theta$ for $x$.
It makes more sense if you actually do the integrals.
$$\begin{align} \int \cos^2 \theta \,d\theta &= \frac{1}{2}\int (1 +\cos 2\theta)\,d\theta \\&= \frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C \\&= \frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta + C \\&= \frac{1}{2}\arcsin x + \frac{1}{2}x\sqrt{1-x^2} + C \end{align}$$
$$\begin{align} \int -\sin^2 \theta \,d\theta &= -\frac{1}{2}\int (1 -\cos 2\theta)\,d\theta \\&= -\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C \\&= -\frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta + C \\&= -\frac{1}{2}\arccos x + \frac{1}{2}x\sqrt{1-x^2} + C \end{align}$$
The key here is that $\arcsin x$ and $-\arccos x$ differ by a constant
$$\arcsin x + C_1 = -\arccos x + \frac{\pi}{2} + C_1 = -\arccos x + C_2$$
What this means is when you take the derivative, the constant goes away, leaving the same function, so the 2 answers you get are equivalent.
| {
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Logarithm multivariable limit $\frac{\ln(1+x^3+y^3)}{\sqrt{x^2+y^2}}$ Find multivariable limit $$\lim_{\left( x,y \right) \rightarrow (0,0)}\frac{\ln(1+x^3+y^3)}{\sqrt{x^2+y^2}}$$
I was trying to find and inequality i've found out that:
$$\frac{\ln(1+x^3+y^3)}{\sqrt{x^2+y^2}} \le \frac{\sqrt{x^2+y^2} \ln(1+x^3+y^3)}{2xy} $$
and after that i am stuck. From iterated limits i've calculated before i knwo that the limit exists for certain. I have also another idea to do that precisely to $y=x$ and $y=-x$ but i do not know whether this is working properly.
| Using Polar coordinates.
Let $x = r\cos(\theta)\ and \ y=r\sin(\theta)$ hence the function becomes $\rightarrow$ $\frac{\ln( 1 +r^3\cos^3(\theta)sin^3(\theta))}{r}$
.
As $r^3\cos^3(\theta)\sin^3(\theta)$ $\rightarrow$ $0$ as $r$ $\rightarrow$ $0$ hence $\ln(1 + r^3\cos^3(\theta)\sin^3(\theta))$ $\sim$ $r^3\cos^3(\theta)\sin^3(\theta)$ and so the function becomes $\frac{r^3\cos^3(\theta)\sin^3(\theta)}{r}$ that tends to $0$ as r$\rightarrow$ $0$.
| {
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Prove the set $\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$ is a ring.
Prove that if $d$ is a non-square integer with $d \equiv 1 \mod 4$ then the set $$\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$$ is a ring, and in particular a integral domain.
Little bit stuck here.
For $x,y \in \mathcal{O}_d,$ we define $+: \mathcal{O}_d \times \mathcal{O}_d \to \mathcal{O}_d$ by $(x + y) = \dfrac{a + b \sqrt{d}}{2} + \dfrac{a' + b'\sqrt{d}}{2} = \dfrac{(a+a') + (b+b')\sqrt{d}}{2} = \dfrac{a'+b'\sqrt{d}}{2} + \dfrac{a + b\sqrt{d}}{2} = (y + x) \in \mathcal{O}_d.$
We have that $0 = \dfrac{0 + 0\sqrt{d}}{2} \in \mathcal{O}_d$ and $\dfrac{a + b \sqrt{d}}{2} + \dfrac{-a + -b\sqrt{d}}{2} = \dfrac{(a-a) + (b-b)\sqrt{d}}{2} = 0 .$ Hence $-a \in \mathcal{O}_d.$ Also $$\begin{align} (x + y) + z &= \left(\dfrac{a + b\sqrt{d}}{2} + \dfrac{a' + b'\sqrt{d}}{2}\right) + \dfrac{a'' + b''\sqrt{d}}{2} \\&= \dfrac{(a + a') + (b + b')\sqrt{d}}{2} + \dfrac{a'' + b''\sqrt{d}}{2} \\&= \dfrac{(a + a' + a'') + (b + b' + b'')\sqrt{d}}{2} \\&= \dfrac{a + b\sqrt{d}}{2} + \dfrac{(a'+a'')+(b+b'')\sqrt{d}}{2} \\&= \dfrac{a + b\sqrt{d}}{2} + \left(\dfrac{a'+b'\sqrt{d}}{2} + \dfrac{a''+b''\sqrt{d}}{2}\right) \\&= x + (y+z) \end{align}$$
for all $x,y,z \in \mathcal{O}_d.$
Hence $\mathcal{O}_d$ is a group under addition.
I've had difficulty showing the multiplicative operation $\circ: \mathcal{O}_d \times \mathcal{O}_d \to \mathcal{O}_d$ stays in the set. How do we define the operation so that $x \circ y \in \mathcal{O}_d?$ When I multiply two elements from $\mathcal{O}_d$ I get $x \circ y = \left(\dfrac{a + b \sqrt{d}}{2} \right)\left(\dfrac{a' + b' \sqrt{d}}{2} \right) = \dfrac{(a + b \sqrt{d})(a' + b'\sqrt{d})}{4} = \dfrac{aa' + ab'\sqrt{d} +ba'\sqrt{d} + bb'd}{4}= \dfrac{aa' + bb'd + (ab' +ba')\sqrt{d}}{4}$ which doesn't seem to be an element of $\mathcal{O}_d.$
| Hint: $\mathcal O_d$ is a subset of real numbers, so it is enough to show that it is closed under $+,-,\cdot$ and that it contains $0$ and $1$. I'm pretty sure the operations are (implicitly) the standard addition and multiplication.
For the closure under multiplication, just consider the four cases (depending on the remainders of $a,a'$ modulo $2$).
| {
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Find generating functions for the Perrin and Padovan sequences
The Perrin sequence is defined by $a_0 = 3, a_1 = 0, a_2 = 2$ and $a_k = a_{k-2}+a_{k-3}$ for $k \ge 3$. The Padovan sequence is defined by $b_0 = 0, b_1=1, b_2=1$ and $b_k=b_{k-2}+b_{k-3}$ for $k\ge 3$.
Find generating functions in the form of rational functions for the Perrin sequence and the Padovan sequence.
I am a little bit confused about this question , any hint?
| Denote the generating function for the Perrin sequence as $A(x) = \sum_{n=0}^\infty a_n x^n$, where $\{a_k\}_{k=0}^\infty$ is the Perrin sequence.
We get
$$\begin{align}
A(x) &= \sum_{n=0}^\infty a_n x^n \\
&= \sum_{n=3}^\infty a_n x^n + a_2x^2+a_1x^1+a_0x^0\\
&= \sum_{n=3}^\infty \left(a_{n-2} + a_{n-3}\right)x^n+ a_2x^2+a_1x^1+a_0x^0\\
&=x^2\sum_{n=3}^\infty a_{n-2} x^{n-2}+x^3\sum_{n=3}^\infty a_{n-3} x^{n-3}+ a_2x^2+a_1x^1+a_0x^0\\
&=x^2(A(x)-a_0x^0)+x^3A(x)+ a_2x^2+a_1x^1+a_0x^0\\
\end{align}$$
Now, substitute the known initial conditions and solve for $A(x)$.
(The same procedure can be used also for the Padovan sequence, since they differ only in the initial conditions.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $6^n - 1$ is always divisible by $5$ by induction I'm trying to prove the following, but can't seem to understand it. Can somebody help?
Prove $6^n - 1$ is always divisible by $5$ for $n \geq 1$.
What I've done:
Base Case:
$n = 1$: $6^1 - 1 = 5$, which is divisible by $5$ so TRUE.
Assume true for $n = k$, where $k \geq 1$:
$6^k - 1 = 5P$.
Should be true for $n = k + 1$
$6^{k + 1} - 1 = 5Q$
$= 6 \cdot 6^k - 1$
However, I am unsure on where to go from here.
| We can show by induction that $6^k$ has remainder $1$ after division by $5$.
The base case $k=1$ (or $k=0$) is straightforward, since $6=5\cdot 1+1$.
Now suppose that $6^k$ has remainder $1$ after division by $5$ for $k\ge 1$. Thus $6^k = 5\cdot m+1$ for some $m \in \mathbb{N}$. We can then see that $$6^{k+1}=6\cdot 6^{k} = (5+1)(5\cdot m +1) = 5^2 \cdot m + 5 + 5\cdot m + 1$$
$$=5(5\cdot m + m + 1) + 1.$$
Thus $6^{k+1}$ has remainder $1$ after division by $5$.
Therefore for every $k$, we can write $6^k = 5\cdot m +1$ for some $m$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Use the generating function to solve a recurrence relation We have the recurrence relation $\displaystyle a_n = a_{n-1} + 2(n-1)$ for $n \geq 2$, with $a_1 = 2$.
Now I have to show that $\displaystyle a_n = n^2 - n +2$, with $n \geq 1$ using the generating function.
The theory in my book is scanty, so with the help of the internet I have the following:
$\displaystyle \sum_{n = 2} ^\infty (a_n - a_{n-1}) x^n = \sum_{n = 2} ^\infty a_n x^n - \sum_{n = 2} ^\infty a_{n-1} x^n = \sum_{n = 2} ^\infty a_n x^n - x \sum_{n = 2} ^\infty a_{n-1} x^{n-1} = (a(x) - a_1 x) - x(a(x)) = (a(x) - 2 x) - x(a(x)) = a(x) (1-x) - 2x$
But how I have to work out $\displaystyle \sum_{n = 2} ^\infty 2(n-1) x^n$ ?
If I have this, an expression for $a(x)$ can be found. How should $a_n$ be found from $a(x)$?
| Note: Here is another approach using generating functions.
According to the recurrence relation
\begin{align*}
a_1&=2\\
a_n&=a_{n-1}+2(n-1)\qquad\qquad n\geq 2
\end{align*}
we set
\begin{align*}
A(x):=\sum_{n=1}^{\infty}a_nx^n
\end{align*}
and use the Ansatz
\begin{align*}
\sum_{n=2}^{\infty}a_nx^n&=\sum_{n=2}^{\infty}a_{n-1}x^n+2\sum_{n=2}^{\infty}(n-1)x^n\tag{1}
\end{align*}
We observe
\begin{align*}
A(x)-2&=\sum_{n=1}^{\infty}a_nx^{n+1}+2\sum_{n=1}^{\infty}nx^{n+1}\tag{2}\\
&=xA(x)+2x^2\sum_{n=1}^{\infty}nx^{n-1}\\
&=xA(x)+2x^2\frac{d}{dx}\sum_{n=1}^{\infty}x^{n}\tag{3}\\
&=xA(x)+2x^2\frac{d}{dx}\left(\frac{1}{1-x}-1\right)\\
&=xA(x)+\frac{2x^2}{(1-x)^2}\tag{4}
\end{align*}
Comment:
*
*In (2) we use the generating function $A(x)-a_1=A(x)-2$ at the LHS of (1) and do some index shifting at the RHS.
*In (3) we use the (formal) differential operator $\frac{d}{dx}$ applied to the geometric series $\frac{1}{1-x}$
We obtain from (4) by collecting terms with $A(x)$ at the LHS
\begin{align*}
A(x)(1-x)&=2\left(1+\frac{x^2}{(1-x)^2}\right)\\
A(x)&=2\left(\frac{1}{1-x}+\frac{x^2}{(1-x)^3}\right)\tag{5}
\end{align*}
Note: We have now derived a closed expression for $A(x)$ and it's time to harvest. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. So, we can write
$$a_n=[x^n]A(x)$$
and we get from (5)
\begin{align*}
a_n&=[x^n]A(x)\\
&=2[x^n]\frac{1}{1-x}+2[x^n]\frac{x^2}{(1-x)^3}\tag{6}\\
&=2[x^n]\sum_{n=0}^{\infty}x^n-2[x^{n}]x^2\sum_{n=0}^{\infty}\binom{-3}{n}(-x)^n\tag{7}\\
&=2+2[x^{n-2}]\sum_{n=0}^{\infty}\binom{n+2}{2}x^n\\
&=2+2\binom{n}{2}\\
&=n^2-n+2
\end{align*}
Comment:
*
*In (6) we use the formula for the binomial series
$\frac{1}{(1-x)^\alpha}=\sum_{n=0}^{\infty}\binom{-\alpha}{n}x^n$
*In (7) we use $\binom{-\alpha}{n}(-1)^n=\binom{\alpha+n-1}{n}=\binom{\alpha+n-1}{\alpha-1}$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1)^2 = 0$ are real and distinct, Then set of all values of $a$
If the two roots of the equation $(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1)^2 = 0$ are
real and distinct, Then the set of all values of $a$ is.
$\bf{Options::}$ $(a)\;\; \displaystyle \left(0,\frac{1}{2}\right)\;\;\;\; (b)\;\; \displaystyle \left(-\frac{1}{2},0\right)\cup \left(0,\frac{1}{2}\right)\;\;\;\; (c)\;\; \left(-\infty, -2 \right)\cup (2,\infty)\;\;\;\; (d)\;\; \displaystyle \left(-\frac{1}{2},0\right)$
$\bf{My\; Try::}$ We can write above equation as $$\displaystyle \frac{a-1}{a+1} = -\frac{(x^2+x+1)^2}{x^4+x^2+1} = -\left[\frac{x^2+x+1}{x^2-x+1}\right]$$
Now We calculate the range of $$\displaystyle \frac{1}{3}\leq \left[\frac{x^2+x+1}{x^2-x+1}\right]\leq 3$$
So We Get $$\displaystyle -3 \leq -\left[\frac{x^2+x+1}{x^2-x+1}\right]\leq - \frac{1}{3}$$
So We get $$\displaystyle -3 \leq \frac{a-1}{a+1}\leq -\frac{1}{3}\Rightarrow -3 -1 \leq 1-\frac{2}{a+1}-1 \leq -\frac{1}{3}-1$$
So we get $$\displaystyle -4\leq -\frac{2}{a+1}\leq -\frac{4}{3}\Rightarrow \frac{2}{3}\leq \frac{1}{a+1} \leq 2$$
So We get $$\displaystyle a\in \left(-\frac{1}{2}\;,\frac{1}{2}\right)$$.
By from above no answer is matching, plz explain me where I am Wrong.
Thanks
| A slightly different method. Seemed easier.
$$(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1)^2=0$$
$$(x^2+x+1)\left[(a-1)(x^2-x+1)+(a+1)(x^2+x+1)\right]=0$$
$x^2+x+1=0$ has no real roots. So
$$(a-1)(x^2-x+1)+(a+1)(x^2+x+1)=0$$
$$2ax^2+2x+2a=0$$
$$ax^2+x+a=0$$
For this to have real and distinct roots, we have $\mathrm{Discriminant}>0$.
$$1-4a^2>0$$
$$a\in\left(-\frac{1}{2},0\right)\cup\left(0,\frac{1}{2}\right)$$
$0$ gets excluded because the above quadratic wouldn't be a quadratic for $a=0$, it would only be a linear with one root, and the question asks for distinct roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $p_n$ denote the $n$th prime and $c_n$ the $n$th composite number. Determine all positive integers $n$ such that $|p_n − c_n| = 1$. The only ones with the property in title I could find are $n = 5$ and $n = 6$: $|11 − 10| = 1$ and $|13 − 12| = 1$. Past the $20$th prime, the list of primes grows too fast for composite numbers. Is there more general way to show this?
| If $|p_n-c_n| = 1$ then there must exist a $N$ such that the set $\{2,3,\ldots,N\}$ contains an equal number of primes and composite numbers.
The number of composite numbers in the $N-1$ element set $\{2,\ldots, N\}$ is larger than (numbers divisible by $2$ + numbers divisible by $3$ - numbers divisible by $2\cdot 3$ since these have been counted twice)
$$M = \left\lfloor\frac{N-1}{2}\right\rfloor + \left\lfloor\frac{N-1}{3}\right\rfloor - \left\lfloor\frac{N-1}{6}\right\rfloor - 2$$
where we have subtracted $2$ for the two primes $2$ and $3$.
When $N>25$ we have $M \geq \frac{4N-21}{6} > \frac{N-1}{2} + 1$ so there will always be at least two more composite numbers than prime numbers in the set $\{2,3,\ldots,N\}$ and since $p_{10} = 29 > 25$ it follows that $|p_n-c_n|=1$ has no solutions for $n\geq 10$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this proof for 1/4 mod 9 = x, correct? Find an integer x so (1/4) mod 9 = x
Proof:
> 1/4 mod 9 = x
> 1 mod 9 = 4 * x
> - using x = 7 -
> 1 mod 9 = 28
> 28 mod 9 = 1 (to validate)
> - using Euler division theorem m = nq + r -
> 28 = 9(3) + 1
> r = 1, so 1 = 1
I appreciate the feedback.
| Here is a systematic approach to finding the inverse.
Since $\gcd(4, 9) = 1$, $4$ has a multiplicative inverse modulo $9$. To find it, we must solve the equivalence
$$4x \equiv 1 \pmod{9}$$
To find the solution, we can use the Euclidean Algorithm.
$$9 = 2 \cdot 4 + 1$$
Solving for $1$ yields
$$9 - 2 \cdot 4 = 1$$
Hence, $1 \equiv -2 \cdot 4 \pmod{9}$, so
$$x \equiv -2 \equiv 7 \pmod{9}$$
Therefore,
$$4^{-1} \equiv \frac{1}{4} \equiv 7 \pmod{9}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to Find the Function of a Given Power Series? (Please see edit below; I originally asked how to find a power series expansion of a given function, but I now wanted to know how to do the reverse case.)
Can someone please explain how to find the power series expansion of the function in the title? Also, how would you do it in the reverse case? That is, given the power series expansion, how can you deduce the function $\frac{x}{1-x-x^3}$?
I've also rewritten the function as $$\frac{x}{1-x-x^3} = x \cdot \frac{1}{1-x(1-x^2)}$$ which is of the form of the Maclaurin series $\frac{1}{1-x} = 1+x+x^2+x^3+\ldots$
So the series expansion would then be $$x \cdot (1+(x-x^3)+(x-x^3)^2+(x-x^3)^3+\ldots)$$
However, expanding this to find the simplified power series expansion becomes complicated. According to WolframAlpha (link: http://www.wolframalpha.com/input/?i=power+series+of+x%2F%281-x-x%5E3%29) it eventually works out to $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots$$ which is a function defined recursively by $f_n = f_{n-1} + f_{n-3}$ for all $n\gt 3$ with the initial condition that $f_1=f_2=f_3=1$
Appreciate any and all help!
Thanks for reading,
A
Edit: I actually want to find the reverse of the original question. Given the recursively defined function (see above) $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots ,$$ how can I show that the function of this power series expansion is $\frac{x}{1-x-x^3}$?
| $x = (1-x-x^3) x + (x^2 + x^4)$
$x^2 + x^4 = (1-x-x^3) * x^2 + (x^3 + x^4 + x^5)$
$(x^3 + x^4 + x^5) = (1-x-x^3) * x^3 + (2*x^2+x^5 + x^6)$
$(2*x^2+x^5 + x^6) = (1-x-x^3) * 2x^4 + (x^5+3*x^6+x^7)$ then
$\frac{x}{1-x-x^3} = 1 + x + x^2 + x^3 + 2x^4 +\frac{(x^5+3*x^6+x^7)}{1-x-x^3}$
if you continue you all the coefficients of power series
| {
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How does $\frac{-1}{x^2}+2x=0$ become $2x^3-1=0$? Below is part of a solution to a critical points question. I'm just not sure how the equation on the left becomes the equation on the right. Could someone please show me the steps in-between? Thanks.
$$\frac{-1}{x^2}+2x=0 \implies 2x^3-1=0$$
| For $x\neq 0$, we have $0= - \frac{1}{x^2} + 2x =- \frac{1}{x^2} + \frac{x^2}{x^2} \times 2x = - \frac{1}{x^2} + \frac{2x^3}{x^2} = \frac{2x^3 -1}{x^2}$.
So we have $\frac{2x^3 -1}{x^2} = 0$ which will give us $2x^3 -1 = 0.$
And we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $2^2=4$ imply $2=\pm \sqrt{4}$? I read the square root property from the book, College Algebra by Raymond A Barnett and Micheal R Ziegler that,
The square root property says,
If $A^2=C$ then $A=\pm \sqrt{C}$
I took the equality,
$2^2=4$
$\implies 2=\pm \sqrt{4}$
$\implies 2=\pm {2}$
but, How $2$ can be equal to $-2$?
| Let's see how "If $A^2=C$ then $A=\pm \sqrt{C}$" is obtained:
$A^2=C \Rightarrow A^2-C=0 \Rightarrow (A-\sqrt{C})(A+\sqrt{C})=0 \Rightarrow$
$A-\sqrt{C}=0$ or $A+\sqrt{C}=0 \Rightarrow A_1=\sqrt{C}$ or $A_2=-\sqrt{C} \Rightarrow A=\pm \sqrt{C}$.
Now let's see what happens to "$2^2=4$":
$2^2=4 \Rightarrow 2^2-4=0 \Rightarrow (2-\sqrt{4})(2+\sqrt{4})=0 \Rightarrow$
$2-\sqrt{4}=0$ or $2+\sqrt{4}=0 \Rightarrow 2=\sqrt{4}$ or $2\ne-\sqrt{4} \Rightarrow 2=\sqrt{4}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How Deficient a Number is? (Finding numbers having a certain deficiency) This question was edited, in particular equations were corrected:
A number N is said to be deficient by an integer $d$ if:
$\sigma(N)=2N-d$
Note that powers of 2 are deficient by 1.
While a prime $p$ is deficient by $p-1$
My question will be this:
Is there an integer $N$ that is deficient by an integer $d$ that lies between $(2/3)N$ and $N-\sqrt{N}$, that is
$(2/3)N<d<N-\sqrt{N}$ and can we characterize them?
Thanks a lot for your help.
| Let us try using the following criterion:
Let $1 < d = 2N - \sigma(N)$. Then
$$\frac{2N}{N + d} < I(N) < \frac{2N + d}{N + d}$$
where $I(N)=\sigma(N)/N$ is the abundancy index of $N$.
Since you require
$$\frac{2}{3}N < d < N - \sqrt{N},$$
then we have
$$\frac{2}{3} < \frac{d}{N} = 2 - I(N) < 1 - \sqrt{\frac{1}{N}}$$
so that we obtain
$$\frac{2}{2 - \sqrt{\frac{1}{N}}}< \frac{2}{1 + \frac{d}{N}} = \frac{2N}{N+d} < I(N) < \frac{2N+d}{N+d} = \frac{2+\frac{d}{N}}{1+\frac{d}{N}} < \frac{3 - \sqrt{\frac{1}{N}}}{\frac{5}{3}} = \frac{3}{5}\left(3 - \sqrt{\frac{1}{N}}\right).$$
However, from
$$\frac{2}{3} < \frac{d}{N} = 2 - I(N) < 1 - \sqrt{\frac{1}{N}}$$
you have the bounds
$$1 + \sqrt{\frac{1}{N}} < I(N) < \frac{4}{3},$$
so that
$$\frac{2}{2 - \sqrt{\frac{1}{N}}}<I(N)<\frac{4}{3}$$
and
$$1 + \sqrt{\frac{1}{N}}<I(N)<\frac{3}{5}\left(3 - \sqrt{\frac{1}{N}}\right).$$
The first inequality yields:
$$6 < 8 - 4\sqrt{\frac{1}{N}} \implies 4\sqrt{\frac{1}{N}} < 2 \implies 2 < \sqrt{N} \implies 4 < N.$$
The second inequality yields:
$$5\sqrt{N} + 5 < 9\sqrt{N} - 3 \implies 8 < 4\sqrt{N} \implies 2 < \sqrt{N} \implies 4 < N.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate this multivariable limit? $$
\lim_{(x,y,z)\to (0,0,0) } \frac{\sin(x^2+y^2+z^2) + \tan(x+y+z) }{|x|+|y|+|z|}
$$
I know the entire limit should not exist. In addition, the limit:
$$
\lim_{(x,y,z)\to (0,0,0) } \frac{\tan(x+y+z) }{|x|+|y|+|z|}
$$
does not exist and it seems like the limit:
$$
\lim_{(x,y,z)\to (0,0,0) } \frac{\sin(x^2+y^2+z^2) }{|x|+|y|+|z|}
$$
is zero. But, how can I calculate this limit (the last one) ?
Will you please help me?
Thanks a lot in advance
| Write $\frac{\sin(x^2+y^2+z^2)}{|x|+|y|+|z|}=\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}\frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|}\sqrt{x^2+y^2+z^2}$. The first factor tends to $1$, the second is bounded and the last tends to zero.
To answer your question in your comment:
$$x^2+y^2+z^2\leq x^2+y^2+z^2+2|x||y|+2|x||z|+2|y||z|=(|x|+|y|+|z|)^2$$
Therefore $$\frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|}\leq1$$
A good-to-know fact is that the norms in $\mathbb{R}^n$ are all equivalent. $\|(x,y,z)\|_2=\sqrt{x^2+y^2+z^2}$ and $\|(x,y,z)\|_1=|x|+|y|+|z|$ are to norms of $\mathbb{R}^3$ and equivalence means that there are non-zero constants $c,C$ such that
$$c\|\cdot\|_1\leq \|\cdot\|_2\leq C\|\cdot\|_1$$
In our case $C=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving convergence/divergence via the ratio test Consider the series
$$\sum\limits_{k=1}^\infty \frac{-3^k\cdot k!}{k^k}$$
Using the ratio test, the expression $\frac{|a_{k+1}|}{|a_k|}$ is calculated as:
$$\frac{3^{k+1}\cdot (k+1)!}{(k+1)^{k+1}}\cdot \frac{k^k}{3^k\cdot k!}=\frac{3}{(k+1)^{k}}\cdot {k^k}=3\cdot \frac{k^k}{(k+1)^{k}}=3\cdot \left(\frac{k}{k+1}\right)^k$$
How to continue?
| $$\lim_{k\to\infty} 3\cdot \bigg(\frac{k}{k+1}\bigg)^k = \lim_{k\to\infty} 3\cdot \bigg(1 - \frac{1}{k+1}\bigg)^k = 3\cdot \lim_{k\to\infty}\bigg(1-\frac{1}{k}\bigg)^k = 3\cdot e^{-1}$$
| {
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Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should be:
If we pick 2 numbers $a$ and $b$ from the domain of a function $f$, where $a < b$, then $f(a) < f(b)$.
Now (if that is a correct definition), I have tried to apply it to my case:
Let $a$, $b \in [-1, 1]$ and $a < b$. We want to show that $f(a) < f(b)$ or that $f(b) - f(a) > 0$.
We know that $x^2 + 3x + 2 = (x + 2)(x + 1)$, thus we have that $f(a) = (a + 2)(a + 1)$ and $f(b) = (b + 2)(b + 1)$ , therefore we need to show that:
$$(b + 2)(b + 1) - (a + 2)(a + 1) > 0$$
We can see that $(b + 2)$ and $(a + 2)$ will always be positive, and that $(b + 2) > (a + 2)$ , since $b > a$ (by assumption).
Since $b > a$, we know that $b > -1$ (otherwise $a \geq b$), thus $(b + 1) > 0$ (so we know that $(b + 2)(b + 1) > 0$ . We also have at most $(a + 1) = 0$, and thus we have that $(a + 2)(a + 1) \ge 0$.
So, here is the proof that $f(b) - f(a) > 0$ .
Am I correct? If yes, what can I improve it? If not, where are the erros and possible solutions?
| Seems good to me. I would be lazier, though. Pick $a,b\in[-1,1]$, with $a<b$. Since $a \geq -1$, then $a+1 \geq 0$. Since $b > a$, we have also that $b+2,b+1,a+2 > 0$. So: $$\begin{cases} b+2>a+2 >0 \\ b+1 > a+1 \geq 0 \end{cases} \implies (b+2)(b+1) > (a+2)(a+1) \implies f(b) > f(a).$$
| {
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Finding equation of hyperbola with only foci and asymptote This is a concept we learned in class today, which I still can't seem to grasp. I have no specific question that necessarily has to be done, so I will use one of the examples my book gives me:
Given the foci are: (6√5, 10) and (-6√5,10)
The asymptote is y = 1/2x + 10
Find the equation of the hyperbola
I know that...
(y-k)²/a² - (x-h)²/b² = 1 (Hyperbola equation)
Center (h,k)
foci (h, k±c), c² = a²+b²
Asmyptote y = -(a/b)x + k+(a/b)h (The -/+ are interchangeable)
I am just confused on how to find the equation with the givens above. Thanks!
| It looks like you know all of the equations you need to solve this problem. I also see that you know that the slope of the asymptote line of a hyperbola is the ratio $\dfrac{b}{a}$ for a simple hyperbola of the form $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$
So, given $y=\dfrac{1}{2}x+10$, we can see that the ratio $\dfrac{b}{a}=\dfrac{1}{2}$. Furthermore, we can say that $b=n$ and $a=2n$ since the $n's$ would cancel in the ratio.
Now that we have our $a$ and $b$ depending on only one variable $n$, we can use the relationship you mentioned, $c^2=a^2+b^2$, to determine the value of $n$.
\begin{align*}
a^2+b^2&=c^2\\
(2n)^2+n^2&=(6\sqrt{5})^2\\
n^2&=36\\
n&=6\\
\end{align*}
Therefore, we have $a=12$ and $b=6$, so then our equation for this hyperbola can be expressed with the following:
\begin{align*}
\dfrac{x^2}{144}-\dfrac{(y-10)^2}{36}=1
\end{align*}
Where the 10 came from shifting the hyperbola up 10 units to match the $y$ value of our foci.
| {
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Find $\int \limits_0^1 \int \limits_x^1 \arctan \bigg(\frac yx \bigg) \, \, \, dx \, \, dy$ Find $$\int \limits_0^1 \int \limits_x^1 \arctan \bigg(\frac yx \bigg)dx \, \, dy$$
So obviously using cylindrical is the way to go to give $\theta r$ inside the integral (after considering the jacobian term).
But how can the limits of $x$ be $(x,1)$. I am guessing it is meant to say $dy \, \, dx$ right so those limits are for $y$? Please tell me if you disagree.
But even if this is the case, it is not possible to compute the integral.
After sketching out $x<y<1$ and $0<x<1$, I realised that the region is the triangle bounded by origin, $(0,1)$ and $(1,0)$. So the limits of $r$ and $\theta$ are $(0,1/\cos\theta)$ and $(\pi/4 , \pi /2)$. So we have $$\int \limits_{\pi/4}^{\pi/2} \int \limits_0^{1/\cos\theta} r\theta \, \, \, dr \, \, d\theta$$ which gives $$\int \limits_{\pi/4}^{\pi/2} \frac 12 \theta \sec^2 \theta \, \, \, d\theta$$
The integral of $\sec^2 \theta $ is $ln |\cos \theta |$ and one of the limits is $\pi/2$ which is undefined. Please help.
| Let $I=\int_{0}^{1} \int_{x}^{1} arctan(\frac{y}{x}) dy dx $, then considering polar coordinates we have the foolowing region :
so as we see from the picture, the region is bounded by $\theta$ between $\frac{\pi}{4} $ and $\frac{\pi}{2} $. Now taking a ray from the origin crossing the region, we recignize that it starts from origin and finaly left the region meeting the line $y=1$, so that $r\sin \theta= 1$, i.e. $ r $ is bounded between $0$ and $\frac{1}{\sin \theta}$. so
$$I = \int_{\frac{\pi}{4} }^{\frac{\pi}{2}} \int_{ 0}^{\frac{1}{\sin \theta} } r \theta dr d\theta = \int_{\frac{\pi}{4} }^{\frac{\pi}{2}} \frac{\theta}{ 2 sin^2 \theta } d\theta= - \frac{\theta}{2\tan \theta } \mid_{\frac{\pi}{4}}^{\frac{\pi}{2}} + \int_{\frac{\pi}{4} }^{\frac{\pi}{2}} \frac{d\theta}{2\tan \theta } = -\lim_{\theta \rightarrow \frac{\pi}{2} } \frac{\theta}{2\tan \theta } + \frac{\pi}{8 } + \frac{1}{2} \ln (\sin \theta)\mid_{\frac{\pi}{4}}^{\frac{\pi}{2}}= \frac{\pi}{8 } -\frac{1}{2} \ln (\sin \frac{\pi}{4}). $$
| {
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Positivity of power function. Prove that
$6^a-7^a+2\cdot 4^a-3^a-5^a\ge0$ for $-\frac{1}{2}\le a\le0$.
I tried to do it by first derivative test but derivative become more complicated (same with 2nd derivative for convexity).
Here is it's plotting.
| We want to show that
$$\left((3^x+7^x+5^x)/3\right)^{1/x} > \left((6^x+4^x+4^x)/3\right)^{1/x}\hbox{ for }-0.5\le x<0$$ "as it may be seen from the plot".
It's not too hard to test that
$$\hbox{(1) }4.57 < \left((3^{-\frac{1}{2}}+7^{-\frac{1}{2}}+5^{-\frac{1}{2}})/3\right)^{-2} $$
$$\hbox{(2) }\left((6^{-\frac{1}{4}}+4^{-\frac{1}{4}}+4^{-\frac{1}{4}})/3\right)^{-4} < 4.57$$
$$\hbox{(3) }\sqrt[3]{6\cdot 4\cdot 4} < \left((3^{-\frac{1}{4}}+7^{-\frac{1}{4}}+5^{-\frac{1}{4}})/3\right)^{-4}$$
So, using Generalized mean inequality we have
$$M_x(6,4,4)< M_{-\frac{1}{4}}(6,4,4)<M_{-\frac{1}{4}}(4.57,4.57,4.57)=$$
$$M_{-\frac{1}{2}}(4.57,4.57,4.57)<M_{-\frac{1}{2}}(3,5,7)\le M_x(3,5,7)\ \forall x:-\frac{1}{2}\le x<-\frac{1}{4},$$
$$M_x(6,4,4)\le M_0(6,4,4)<M_{-\frac{1}{4}}(3,5,7)<M_x(3,5,7)\ \forall x:-\frac{1}{4}\le x<0.$$
| {
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} |
Show that only one prime can be expressed as $n^3-1$ for some positive integer $n$
Show that only one prime can be expressed as $n^3-1$ for some positive integer $n$
$n^3-1=(n-1)(n^2+n+1)$. If $n^3-1$ is prime then $n-1=1$ or $n=2$ so $n^3-1=7$.
But I need a concrete proof of it, not from intuition.
| $$n^3-1=(n-1)(n^2+n+1)$$ Suppose that $n^2+n+1\le n-1\implies n^2\le -2$ which leads to contradiction.
Therefore $n^2+n+1\gt n-1$ and since $n^3-1$ is prime only possibility is $$n-1=1\implies \color{Green}{n=2}\,\,\,\,\,\text{and} \,\,\,\,\,\,\color{Red}{n^3-1=7}$$
| {
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Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$
Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$
I must use Vieta's formulas in my solution since that is the lesson we are covering with our teacher.
My solution:
Let p and q be the roots of the quadratic.
$$\begin{align} p + q = & -8 \\ pq = & -1 \end{align}$$
If the roots are each six more than the roots of the quadratic, then we will have:
$$\begin{align} p + q + 12 = & -8 \\
p + q = & -20 \\
pq = & - 3 \end{align}$$
Also,
$$ \begin{align} (p+6)(q+6) = & pq + 6p + 6q + 36\\
=&pq + 6(p+q) + 36 \\
=&-3 + 6(-8) + 36 \\
=&-3 - 48 + 36 \\
&36 - 51 = -15 \end{align} $$
Hence, -15 = constant.
Thus, the quadratic equation is $x^2 + 20x - 15$
My worksheet gives an answer of $x^2 - 4x - 13$, how am I wrong?
Thanks!
| The original quadratic equation is given to you as $x^2 + 8x - 1 = 0$ . Let it's roots be $(x_1 , x_2)$ . The roots of required equation be $(\alpha ,\beta)$ (say).
Now, $$x_1 + x_2 = -8 \\ x_1 x_2 = -1$$
And $$\alpha = x_1 + 6 \\ \beta = x_2 + 6 $$
So, $$\alpha + \beta = x_1 + x_2 + 12 \\ \alpha \beta = (x_1 + 6)(x_2 + 6) = x_1 x_2 + 6(x_1 + x_2) + 36 $$
You already know the values of $\color{red}{x_1 x_2 = -1} ; \ \color{blue}{x_1 + x_2 = -8}$ . So, just plug-in the values in the above formula.
$$\begin{align}
&\alpha + \beta = \color{blue}{x_1 + x_2} + 12 = \color{blue}{-8} + 12 = 4 \\
&\alpha \times \beta = \color{red}{x_1 x_2} + 6\color{Blue}{(x_1 + x_2)} + 36 = \color{red}{-1} + 6\color{blue}{(-8)} + 36 = -13\end{align}$$
So, you can just finalize it and write the quadratic equation as :
$$\boxed{\color{brown}{x^2 - 4x - 13}}$$
NOTE : Any quadratic equation can be written as : $x^2 - \color{blue}{S} x + \color{red}{P} $ .
Where:
$\color{blue}{S}$ = Sum of Zeroes(or roots) of the equation.
$\color{red}{P}$ = Product of Zeroes(or roots) of the equation.
Hope it helps!
| {
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What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the center
$$-\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
This means that $\frac{\sqrt{b^2 - 4ac}}{2a}$ is the (horizontal) distance from the vertex to the roots. If I squint, the two squared-ish quantities being subtracted under a square root sign reminds me of the Euclidean distance formula
$$\sqrt{(x_0 - x_1)^2 + (y_0 - y_1)^2}$$
Is there a connection? If not, is there any intuitive or geometric reason why $\frac{\sqrt{b^2 - 4ac}}{2a}$ should be the horizontal distance from the vertex to the roots?
| Without loss of generality, we can find the horizontal distance between $x=\dfrac{-b}{2a}$ and $x=\dfrac{-b+\sqrt{b^2-4ac}}{2a}.$ So inputting these coordinates into the distance formula, with both $y-$coordinates equal to zero, we have
\begin{align}
d&=\sqrt{\left(\dfrac{-b+\sqrt{b^2-4ac}}{2a}-\dfrac{-b}{2a}\right)^2+(0-0)^2}\\
&=\dfrac{\sqrt{b^2-4ac}}{2a}
\end{align}
This is the horizontal distance from the positive root of $f(x)=ax^2+bx+c$ "to the right" of the point $\left(\dfrac{-b}{2a},0\right).$ So both roots are located this distance left and right of the vertex's $x-$coordinate, giving the roots $x-$coordinates
$$\left(\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},0\right).$$
As a side note, the quadratic formula is derived from completing the square as follows:
\begin{align}
y&=ax^2+bx+c\\
&=a\left(x^2+\frac ba x+\frac ca\right)\\
&=a\left(x^2+\frac ba x+\frac{b^2}{4a^2}+\frac ca-\frac{b^2}{4a^2}\right)\\
&=a\left(x+\frac b{2a}\right)^2+c-\frac{b^2}{4a}.
\end{align}
Now letting $y=0,$ we get
\begin{align}
a\left(x+\frac b{2a}\right)^2&=\frac{b^2}{4a}-c\\
\left(x+\frac b{2a}\right)^2&=\frac{b^2}{4a^2}-\frac ca\\
\left(x+\frac b{2a}\right)&=\pm\sqrt{\frac{b^2}{4a^2}-\frac ca}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm\sqrt{b^2-4ac}}{2a}\\
x&=\dfrac{-b+\sqrt{b^2-4ac}}{2a}.
\end{align}
| {
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I cannot solve this limit $$
\lim_{n\to\infty}\frac{(\frac{1}{n}+1)^{bn+c+n^2}}{e^n}=e^{b-\frac{1}{2}}
$$
I am doing it like this, and I cannot find the mistake:
$$
\lim_{n\to\infty}\frac{1}{e^n}e^{n+b+c/n}=
\lim_{n\to\infty}e^{n+b-n+c/n}=e^b
$$
| You could use L'Hospital's rule and $x=\frac{1}{n}$ to get
$\displaystyle\lim_{n\to\infty}\ln\bigg[\frac{(1+\frac{1}{n})^{bn+n^2+c}}{e^n}\bigg]=\lim_{n\to\infty}(bn+n^2+c)\ln\big(1+\frac{1}{n}\big)-n$
$=\displaystyle\lim_{x\to0}\bigg(\frac{b}{x}+c+\frac{1}{x^2}\bigg)\ln(1+x)-\frac{1}{x}=\lim_{x\to0}\frac{(bx+cx^2+1)\ln(1+x)-x}{x^2}$
$=\displaystyle\lim_{x\to0}\frac{\ln(1+x)-\frac{x}{bx+cx^2+1}}{\frac{x^2}{bx+cx^2+1}}=\displaystyle\lim_{x\to0}\frac{\frac{1}{1+x}-\frac{1-cx^2}{(bx+cx^2+1)^2}}{\frac{bx^2+2x}{(bx+cx^2+1)^2}}=\lim_{x\to0}\frac{(bx+cx^2+1)^2-(1+x)(1-cx^2)}{(1+x)(bx^2+2x)}$
$\displaystyle=\lim_{x\to0}\frac{b^2x+2b+2bcx^2+3cx+c^2x^3+cx^2-1}{(1+x)(bx+2)}=\frac{2b-1}{2}=b-\frac{1}{2}$,
so $\displaystyle\lim_{n\to\infty}\bigg(\frac{(1+\frac{1}{n})^{bn+n^2+c}}{e^n}\bigg)=e^{b-\frac{1}{2}}$.
| {
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Finding the inverse of a number under a certain modulus How does one get the inverse of 7 modulo 11?
I know the answer is supposed to be 8, but have no idea how to reach or calculate that figure.
Likewise, I have the same problem finding the inverse of 3 modulo 13, which is 9.
| To find the inverse of $7$ modulo $11$, we must solve the equivalence $7x \equiv 1 \pmod{11}$. To do this, we use the Extended Euclidean Algorithm to express $1$ as a linear combination of $7$ and $11$. The coefficient of $7$ will be the inverse modulo $11$. By the Euclidean Algorithm,
\begin{align*}
11 & = 1 \cdot 7 + 4\\
7 & = 1 \cdot 4 + 3\\
4 & = 1 \cdot 3 + 1\\
3 & = 3 \cdot 1
\end{align*}
We now take the equation $4 = 1 \cdot 3 + 1 = 3 + 1$, solve for $1$, then work backwards until we obtain $1$ as a linear combination of $7$ and $11$.
\begin{align*}
1 & = 4 - 3\\
& = 4 - (7 - 4)\\
& = 2 \cdot 4 - 7\\
& = 2(11 - 7) - 7\\
& = 2 \cdot 11 - 3 \cdot 7
\end{align*}
Since $2 \cdot 11 - 3 \cdot 7 = 1$,
$$-3 \cdot 7 = 1 - 2 \cdot 11 \Longrightarrow -3 \cdot 7 \equiv 1 \pmod{11}$$
Hence, $-3$ is the inverse of $7 \pmod{11}$. To express the inverse as one of the residues $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, we add $11$ to $-3$ to obtain $-3 + 11 \equiv 8 \pmod{11}$. Hence, $7^{-1} \equiv 8 \pmod{11}$.
Check: $7 \cdot 8 \equiv 56 \equiv 1 + 5 \cdot 11 \equiv 1 \pmod{11}$.
To verify you understand the algorithm, try to find the inverse of $3$ modulo $13$.
| {
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For $n>2, n\in\mathbb{Z}$, why is this true: $\left\lfloor 1/\left(\frac{1}{n^2}+\frac{1}{(n+1)^2}+\cdots+\frac{1}{(n+n)^2}\right)\right\rfloor=2n-3$ Let $n>2$ be a positive integer, prove that
$$\left\lfloor \dfrac{1}{\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}+\cdots+\dfrac{1}{(n+n)^2}}\right\rfloor=2n-3?$$
before I use hand Calculation $n=2,3,4$,maybe I calculation some wrong,I can't use computer it!
so maybe this result is right. But how to prove the statement?
| Let $f(n) = \dfrac1{n^2} + \dfrac1{(n+1)^2} + \cdots + \dfrac1{(2n)^2}$.
We then have
$$\int_{n-1}^{2n} \dfrac{dx}{x^2} > \dfrac1{n^2} + \dfrac1{(n+1)^2} + \cdots + \dfrac1{(2n)^2} \geq \int_n^{2n+1} \dfrac{dx}{x^2}$$
This gives us
$$\dfrac1n-\dfrac1{2n+1} < f(n) < \dfrac1{n-1} - \dfrac1{2n} \implies \dfrac{n+1}{n(2n+1)} < f(n) < \dfrac{n+1}{2n(n-1)}$$
Hence, we have
$$\dfrac{2n^2-2n}{n+1}< \dfrac1{f(n)} < \dfrac{2n^2+n}{n+1} \implies 2n - \dfrac{4n}{n+1} < \dfrac1{f(n)} < 2n - \dfrac{n}{n+1}$$
A tighter bound using this integral approach (ala Euler–Maclaurin) should provide the answer.
EDIT For a more precise answer, we have
\begin{align}
f(n) & = \int_{n^-}^{2n^+} \dfrac{d\lfloor x \rfloor}{x^2} = \left.\dfrac{\lfloor x \rfloor}{x^2} \right \vert_{n^-}^{2n^+} + \int_{n^-}^{2n^+} \dfrac{2\lfloor x \rfloor}{x^3}dx =\dfrac1{2n} - \dfrac{n-1}{n^2} + \int_{n^-}^{2n^+}\dfrac{2x-2\{x\}}{x^3}dx\\
& = \dfrac{n-2n+2}{2n^2} + \dfrac1n - 2\int_{n^-}^{2n^+} \dfrac{\{x\}}{x^3}dx = \dfrac{n+2}{2n^2} - 2\int_{n^-}^{2n^+} \dfrac{\{x\}-1/2}{x^3}dx -\int_{n^-}^{2n^+} \dfrac{dx}{x^3}\\
& = \dfrac{n+2}{2n^2} - 2\int_{n^-}^{2n^+} \dfrac{\{x\}-1/2}{x^3}dx -\dfrac3{8n^2} = \dfrac1{2n} + \dfrac5{8n^2} + \mathcal{O}(1/n^3) = \dfrac1{2n}\left(1+\dfrac5{4n} + \mathcal{O}(1/n^2)\right)
\end{align}
This means
$$\dfrac1{f(n)} = \dfrac{2n}{1+\dfrac5{4n} + \mathcal{O}(1/n^2)} = 2n\left(1-\dfrac5{4n} + \mathcal{O}(1/n^2)\right) = 2n-\dfrac52 + \mathcal{O}(1/n)$$
Hence, we have
$$\left\lfloor \dfrac1{f(n)} \right\rfloor = 2n-3$$ eventually (in fact for $n > 4$).
| {
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Finding the evolute of a parabola I previously tried to find the evolute of a parabola by using parameterisation by arc length. It didn't work. While I was hoping for an answer I kept working on it and came up with the following method (unfortunately, something is not quite right as it leads to a different result than the answer to my previous question).
I would appreciate it if someone could tell me where I went wrong:
Given $\gamma (t) = (t,t^2)$ we find $\gamma'(t) = (1,2t)$. Normalising we find the unit tangent vector to be
$$ v(t) = {1 \over \sqrt{1 + 4 t^2}} \left (\begin{matrix} 1 \\ 2t \end{matrix}\right)$$
Since this vector is unit lenght, its derivative will be orthogonal to it so that $v'(t)$ should be the normal vector to the plane curve:
$$ n(t) := v'(t) = {1\over (1+4 t^2)^{3/2}} \left (\begin{matrix} -4t \\ 2(1 + 4t^2)^{2} +8t^2 \end{matrix}\right)$$
The curvature is $\kappa (t) = \|n(t)\|$ hence the radius of curvature is $r = {1\over \kappa (t)}$.
Hence the equation for the evolute should be
$$ e(t) = \gamma (t) + r n(t) = \left (\begin{matrix} t \\ t^2 \end{matrix}\right) + {1\over \sqrt{16t^2 + (2(1+4t^2)^2 +8t^2)^2}} \left (\begin{matrix} -4t \\ 2(1+4t^2) + 8t^2 \end{matrix}\right)$$
(But the equation of the evolute (according to the answer by Chappers) should be
$$x = -4t^3, \qquad y = \frac{1}{2}+3t^2$$)
| Right, your method is not inherently flawed, but you have calculational mistakes: they first appear in the normal vector: differentiating the tangent vector should give
$$ -\frac{4t}{(1+4t^2)^{3/2}} \begin{pmatrix} 1 \\ 2t \end{pmatrix} + \frac{1}{(1+4t^2)^{1/2}}\begin{pmatrix} 0 \\ 2 \end{pmatrix} = \frac{1}{(1+4t^2)^{3/2}} \begin{pmatrix} -4t \\ -8t^2+2(1+4t^2) \end{pmatrix} \\
=\frac{2}{(1+4t^2)^{3/2}} \begin{pmatrix} -2t \\ 1 \end{pmatrix}
, $$
Now you do as you suggested, (and here is your second mistake, in that you should have $\hat{n}$ instead of $n$, for obvious reasons):
$$ e(t) = \gamma(t) + r\hat{n}(t) = \gamma(t) + \frac{1}{\lVert n(t) \rVert} \hat{n}(t), $$
and here, obviously
$$ \hat{n}(t) = \frac{1}{(1+4t^2)^{1/2}}\begin{pmatrix} -2t \\ 1 \end{pmatrix}, $$
so
$$ e(t) = \begin{pmatrix} t \\ t^2 \end{pmatrix} + \frac{(1+4t^2)^{3/2}}{2} \frac{1}{(1+4t^2)^{1/2}}\begin{pmatrix} -2t \\ 1 \end{pmatrix} \\
= \begin{pmatrix} t \\ t^2 \end{pmatrix} + \frac{(1+4t^2)}{2} \begin{pmatrix} -2t \\ 1 \end{pmatrix} \\
= \begin{pmatrix} -4t^3 \\ 1/2 + 3t^3 \end{pmatrix} $$
| {
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If $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$, prove that $\frac{a}{d}=\sqrt{\frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}}$ What I've done so far;
$$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\\
a=bk, b=ck, c=dk\\
a=ck^2, b=dk^2\\
a=dk^3$$
I tried substituting above values in the right hand side of the equation to get $\frac{a}{d}$ but couldn't.
| Having $a=dk^3,b=dk^2,c=dk$ gives you$$\begin{align}\sqrt{\frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}}&=\sqrt{\frac{d^5k^{15}+d^2k^4\cdot d^2k^2+d^3k^9\cdot d^2k^2}{d^4k^8\cdot dk+d^4+d^2k^4\cdot dk\cdot d^2}}\\&=\sqrt{\frac{dk^{15}+k^6+dk^{11}}{dk^9+1+dk^5}}\\&=\sqrt{\frac{k^6(dk^9+1+dk^5)}{dk^9+1+dk^5}}\\&=\sqrt{k^6}\\&=|k^3|\end{align}$$
| {
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Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Attempt:
I want to use the following theorem:
The largest exponent of $e$ of a prime $p$ such that $p^e$ is a divisor of $n!$ is given by $$e=[\frac{n}{p}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]+\cdots$$.
The number of times the prime divisor $5$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $5$ contained in $(5^{25}-1)!$, which is $e_1=[\frac{(5^{25}-1)}{5}]+[\frac{(5^{25}-1)}{5^2}]+[\frac{(5^{25}-1)}{5^3}]+[\frac{(5^{25}-1)}{5^4}]+\cdots$
The number of times the prime divisor $2$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $2$ contained in $(5^{25}-1)!$, which is $e_2=[\frac{(5^{25}-1)}{2}]+[\frac{(5^{25}-1)}{2^2}]+[\frac{(5^{25}-1)}{2^3}+[\frac{(5^{25}-1)}{2^4}]+\cdots$
Therefore, the number of zeros at the right end equals the greatest exponent of $10$ contained in $(5^{25}-1)!$$=\min{(e_1, e_2)}$.
I am unable to simplify $e_1, e_2$. Please help me.
| Since $e_1\le e_2$, all you need is $e_1$.
$$\begin{align}e_1&=\sum_{k=1}^{25}\left\lfloor\frac{5^{25}-1}{5^k}\right\rfloor\\&=\sum_{k=1}^{25}\left\lfloor 5^{25-k}-\frac{1}{5^k}\right\rfloor\\&=\sum_{k=1}^{25}5^{25-k}+\sum_{k=1}^{25}\left\lfloor-\frac{1}{5^k}\right\rfloor\\&=\frac{5^{25}-1}{4}+25\times(-1)\\&=\frac{5^{25}-101}{4}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic manipulation is done between the first step and the second step, and I can't figure out what this manipulation could be.
The full breakdown comes from this solution
$$
\small\begin{align}
\frac1{x^2-5x+6}
&=\frac1{(x-2)(x-3)}
=\frac1{-3-(-2)}\left(\frac1{x-2}-\frac1{x-3}\right)
=\bbox[4px,border:4px solid #F00000]{-\frac1{x-2}+\frac1{x-3}}\\
&=\bbox[4px,border:4px solid #F00000]{\frac1{2-x}-\frac1{3-x}}
=\sum_{n=0}^\infty\frac1{2^{n+1}}x^n-\sum_{n=0}^\infty\frac1{3^{n+1}}x^n
=\bbox[4px,border:1px solid #000000]{\sum_{n=0}^\infty\left(\frac1{2^{n+1}}-\frac1{3^{n+1}}\right)x^n}
\end{align}
$$
Original image
| $$
\frac{1}{x-a} = \frac{1}{-(a - x)} = - \frac{1}{a - x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
How to compute $\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$? I have to compute the series $\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$.
$$\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}} = \sum_{n=0}^{\infty}{\frac{3^n\frac{1}{2}}{n!}} + \sum_{n=0}^{\infty}{\frac{3^nn}{n!}} = \frac{e^3}{2} + \sum_{n=0}^{\infty}{\frac{3^nn}{n!}},$$ but how to compute the $\displaystyle\sum_{n=0}^{\infty}{\frac{3^nn}{n!}}$?
| We have
$$
x(e^x)'=xe^x=\sum_{n=0}^\infty \frac{nx^n}{n!}
$$
Now take $x=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Find the domain of $x$ in $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$
Solve this equation for $x$: $4\sqrt{x+1}+2\sqrt{2x+3}\leq(x-1)(x^2-2)$
I have no idea to solve that, but I know solutions are $x=-1$ or $x\ge 3$.
| After more than two years I managed to find a more simple and creative solution of this homework-looking inequality. :-)
Define the function from the left hand side of the inequality by $f(x)$ and the function from the right hand side of the inequality by $g(x)$. Draw the graphs of these functions
at a segment containing the proposed key points $-1$ and $3$. Since $\sqrt{x+1}$ exists, admissible $x$ should be not less than $-1$, so as the left endpoint of the segment we pick $-1$. As its right endpoint we pick $3+1=4$.
At the picture the graph of the function $f(x)$ is drawn in red and the graph of the function $g(x)$ is drawn in blue. Note the equality of the values of $f$ and $g$
at the key points $x=-1$ and $x=3$. The picture suggests to take a linear function $h(x)$ which graph is drawn in green, separating the functions $f$ and $g$. That is
$f(x)\ge h(x)\ge g(x)$ for $-1\le x\le 3$ and $f(x)\le h(x)\le g(x)$ for $x\ge 3$.
The values $f(-1)=g(-1)=h(-1)=2$ and $f(3)=g(3)=h(3)=14$ uniquely determine the linear function $h(x)=3x+5$.
It remains to check that the function $h$ satisfies the required inequalities. For this purpose we use the following equivalent transformations.
Case 1. $4\sqrt{x+1}+2\sqrt{2x+3}\le 3x+5$
$16(x+1)+4(2x+3)+16\sqrt{(x+1)(2x+3)}\le 9x^2+30x+25$
$16\sqrt{(x+1)(2x+3)}\le 9x^2+6x-3$
If $9x^2+6x-3<0$ that is when $-1<x<1/3$, the inequality is false, in the opposite cases we can square both its sides.
$256(x+1)(2x+3)\le (9x^2+6x-3)^2$
$0\le 81x^4+108x^3-530x^2-1316x-759$
The right hand side looks frustrating, but recall that we have the equality when $x=-1$ and $x=3$, so these values are its roots. Dividing we obtain
$0\le (x+1)(x-3)(81x^2+270x+253)=(x+1)(x-3)((9x+15)^2+28)$.
Thus (for $x\ge -1$) the inequality
$16(x+1)+4(2x+3)+16\sqrt{(x+1)(2x+3)} \le (x-1)^2(x^2-2)^2$
holds iff $x=-1$ or $x\ge 3$.
Case 2. $3x+5\le (x-1)(x^2-2)$
$0\le x^3-x^2-5x-3$
Again dividing by $(x+1)(x-3)$ we obtain
$0\le (x+1)^2(x-3)$.
Thus (for $x\ge -1$) the inequality
$3x+5\le (x-1)(x^2-2)$
holds iff $x\ge 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1276510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculating all possible sums of the numbers $2^0, 2^1, \ldots, 2^{(n-1)}$ Using the simple equation $2^{n-1}$ you get values such as:
$1,2,4,8,16,32,64,128,256,$ etc.
How can I find all possible number combinations within this range? For example, the numbers $1,2,4,8$ give:
\begin{align}
1 &= 1 \\
2 &= 2 \\
1 + 2 &= 3 \\
4 &= 4 \\
1 + 4 &= 5 \\
2 + 4 &= 6 \\
1 + 2 + 4 &= 7 \\
8 &= 8 \\
1 + 8 &= 9 \\
2 + 8 &= 10 \\
1 + 2 + 8 &= 11 \\
4 + 8 &= 12 \\
1 + 4 + 8 &= 13 \\
2 + 4 + 8 &= 14 \\
1 + 2 + 4 + 8 &= 15
\end{align}
The numbers calculable are: $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$.
Background I'm a computer programmer trying to get a unique number which represents a series of other numbers without the unique IDs being lost.
If I assign $5$ people with the id's, $1,2,4,8,16$, then in order to pass the list of them I don't need to list them all, just the mathematical equation to work it out. For example, if I wanted people $1,2,4$ then I could just pass the number $7$ and therefore it would HAVE to be people $1,2,4$ as there is no other possible combination to make this.
Please let me know if my logic is flawed somewhere!
Thanks,
Nick
| Every integer from $1$ to $2^n-1$ is expressible using the sums of the numbers $1,2,4,8, ... 2^{n-2}, 2^{n-1}$, and no other numbers. This can be proven using induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\int \frac{1}{x^4+x^2+1} \,\, dx$ Find $$\int \frac{1}{x^4+x^2+1} \,\, dx$$
I tried to find like that:
$\int \frac{1}{x^4+x^2+1} = \int \frac{\frac{1}{2}x + \frac{1}{2}}{x^2+x+1} \,\, dx + \int \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2-x+1} \,\, dx = \frac{1}{2} \Big(\int \frac{2x + 1}{x^2+x+1} - \int \frac{x}{x^2+x+1} \Big) - \frac{1}{2} \Big( \int \frac{2x - 1}{x^2-x+1} \,\, dx - \int \frac{x}{x^2-x+1} \,\, dx \Big)$
but then I don't know how to find integrals:
$\, \int \frac{x}{x^2-x+1} \,\, dx \,$ and $\, \int \frac{x}{x^2+x+1} \,\, dx \,$
Is there another way to integrate this function or way to end my calculations?
| $$\int\frac{xdx}{x^2-x+1}=\int\frac{(x-\frac12)dx}{x^2-x+1}+\int\frac{\frac12dx}{(x-\frac12)^2+\frac34}$$. I'm guessing you know how to solve that. The other half is very similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$ Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$
To show a function is monotonically increasing, I started by saying that:
A function $f$ is monotonically increasing in an interval $(a, b)$ if for all $x, y \in (a, b)$, with $x$ < $y$, we have that $f(x) \leq f(y)$.
So I started by tryin to show:
Let $x$, $y$ be arbitrary fixed numbers in $(-1, 1)$, with $x < y$, we want to show that $f(x) \leq f(y)$:
$$f(x) = \frac{1 - 2x}{x^2 - 1} \leq \frac{1 - 2y}{y^2 - 1} = f(y)$$
We multiply the left side by $\frac{y^2 - 1}{y^2 - 1}$, which is positive because $y^2$ is always less than $1$, so we don't need to change the sign:
$$\frac{1 - 2x}{x^2 - 1}\cdot \frac{y^2 - 1}{y^2 - 1} \leq \frac{1 - 2y}{y^2 - 1}$$
We know multiply by $\frac{x^2 - 1}{x^2 - 1}$ the right side:
$$\frac{1 - 2x}{x^2 - 1}\cdot \frac{y^2 - 1}{y^2 - 1} \leq \frac{1 - 2y}{y^2 - 1} \cdot \frac{x^2 - 1}{x^2 - 1}$$
So we can simplify:
$$(1 - 2x)\cdot (y^2 - 1) \leq (1 - 2y) \cdot (x^2 - 1)$$
Now, how do we proceed? Is this correct?
| A differentiable function in some interval is monotonically aascending if its derivative is positive:
$$f'(x)=\frac{-2(x^2-1)-2x(1-2x)}{(x^2-1)^2}=\frac{-2x^2+2-2x+4x^2}{(x^2-1)^2}=$$
$$=2\frac{x^2-x+1}{(x^2-1)^2}$$
Now just prove the above indeed is positive for all $\;x\in\Bbb R\setminus\{-1,1\}\;$ , in particular in $\;(-1,1)\;$ .
Another way: Suppose $\;x<y\;$ , then ( observe that $\;t^2-1<0\;$ for $\;t\in(-1,1)\;$ )
$$f(x)<f(y)\iff\frac{1-2x}{x^2-1}<\frac{1-2y}{y^2-1}\iff y^2-1-2xy^2+2x>x^2-1-2x^2y+2y$$
$$\iff(y-x)\left[(y+x)-2xy-2\right]>0$$
Since $\;y-x>0\;$ , it is enough to show the other factor is positive: this part is for you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the value of this sum. $$\sum_{n=5}^\infty \frac {2^n-4}{3^n}$$
I tried treating it as a geometric sum but I get 2 as the value, when it should be $\frac {10}{27}$
| $\sum=\sum_{n=0}^{\infty} (\frac{2}{3})^n-4\sum_{n=0}^\infty (\frac{1}{3})^n-\sum_{n=0}^4 \frac{2^n-4}{3^n}$
$=\frac{1}{1-\frac{2}{3}}-\frac{4}{1-\frac{1}{3}}+\frac{91}{27}$
$=3-6+\frac{91}{27}$
$=\frac{10}{27}$
QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
complex nos in ellipse. I was practising some ques on ellipses when I came a criss this question:
If normal at four points $(x_1,y_1)$..... on the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ are concurrent then find the value of $$(x_1+x_2+x_3+x_4)\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\right)$$
I know how to solve this question by co ordinate geometry formulas, but I want to do it with complex Nos. I let $z=\cos(\theta)$ and replaced $\sin$ and $\cos$ in the equation of normal by $z$ but I cannot simplify the second bracket. Can anybody help me to show how its done ?
| Equation of tangent at $(x_k,y_k)$,
$$\frac{x_kx}{a^2}+\frac{y_ky}{b^2}=1$$
Equation of normal at $(x_k,y_k)$,
$$\frac{y_k}{b^2}(x-x_k)-\frac{x_k}{a^2}(y-y_k)=0$$
If $(X,Y)$ is the common point of the four normals, then
$$\frac{y_k}{b^2}(X-x_k)-\frac{x_k}{a^2}(Y-y_k)=0$$
Hence, all the points $(x_k,y_k)$ lie on another conic:
$$\frac{y}{b^2}(X-x)-\frac{x}{a^2}(Y-y)=0$$
Re-arrange, $$y=\frac{b^2 Y x}{a^2(x-X)+b^2x}$$
Put back into the ellipse:
\begin{align}
0 &=(a^2+b^2)^2 x^4-2a^2X(a^2+b^2)x^3-a^2[(a^2+b^2)^2-a^2 X^2+b^2 Y^2]x^2 \\
& \qquad +2 a^4X(a^2+b^2)x-a^6 X^2
\end{align}
By Vieta's formulae,
\begin{align}
(x_1+x_2+x_3+x_4)
\left(
\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}
\right)
&= \frac{\displaystyle
\left( \sum x_i \right)
\left(\sum x_i x_j x_k \right)}{\displaystyle \prod x_i} \\[5pt]
&=
\frac{
\left( \dfrac{2a^2X}{a^2+b^2} \right)
\left( -\dfrac{2a^4X}{a^2+b^2} \right)}
{-\dfrac{a^6 X^2}{(a^2+b^2)^2}} \\[5pt]
&= 4
\end{align}
providing $X\ne 0$.
It won't be the case for oblique central conics. See another answer here.
| {
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"url": "https://math.stackexchange.com/questions/1281470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Roots of a polynomial whose coefficients are ratios of binomial coefficients
Prove that $\left\{\cot^2\left(\dfrac{k\pi}{2n+1}\right)\right\}_{k=1}^{n}$ are the roots of the equation
$$x^n-\dfrac{\dbinom{2n+1}{3}}{\dbinom{2n+1}{1}}x^{n-1} + \dfrac{\dbinom{2n+1}{5}}{\dbinom{2n+1}{1}}x^{n-2} - \ldots \ldots \ldots + \dfrac{(-1)^{n}}{\dbinom{2n+1}{1}} =0 $$
Hence prove that
$$\sum_{r=1}^{\infty} \dfrac{1}{r^2}=\dfrac{\pi^{2}}{6}$$
I was stumped on the first sight. Then I tried using complex numbers but it was in vein. I further tried simplifying the equation, but since it contains only half of the binomial coefficients, I wasn't able to get a simpler equation.
Any help will be appreciated.
Thanks.
| Like Sum of tangent functions where arguments are in specific arithmetic series,
$$\displaystyle\tan(2n+1)x=\dfrac{\binom{2n+1}1\tan x-\binom{2n+1}3\tan^3x+\cdots}{1-\binom{2n+1}2\tan^2x+\cdots}$$
Multiplying the denominator & the numerator by $\cot^{2n+1}x$ we get,
$$\displaystyle\tan(2n+1)x=\dfrac{\binom{2n+1}1\cot^{2n}x-\binom{2n+1}3\cot^{2n-2}x+\cdots}{\cot^{2n+1}x-\binom{2n+1}2\cot^{2n-1}x+\cdots}$$
If $\tan(2n+1)x=0,(2n+1)x=r\pi$ where $r$ is any integer
$\implies x=\dfrac{r\pi}{2n+1}$ where $r=-n,-(n-1),\cdots,0,1,\cdots,n\pmod{2n+1}$
So, the finite roots of $$\binom{2n+1}1y^{2n}-\binom{2n+1}3y^{2(n-1)}+\cdots=0$$
are $y=\cot x,$ where $x=\dfrac{r\pi}{2n+1}$ where $r=\pm1,\pm2\cdots,\pm n\pmod{2n+1}$
But $\cot(-A)=-\cot A$
So, the finite roots of $$\binom{2n+1}1z^n-\binom{2n+1}3z^{(n-1)}+\cdots=0$$
are $z=\cot^2x,$ where $x=\dfrac{r\pi}{2n+1}$ where $r=1,2\cdots,n\pmod{2n+1}$
The rest has already been taken care of in the other answer
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving simple trigonometric identity I need help with this one
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha = \sin \alpha
$$
I tried moving sin a on the other side of the eqation
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha - \sin \alpha = 0
$$
This are the operations I was able to do
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \frac{\sin^2\alpha}{\cos^2\alpha}} - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{\frac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha} } - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha*\cos^2\alpha + \cos^3 \alpha}{\cos^2\alpha- \sin^2\alpha} - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\cos^3 \alpha}{\sin\alpha} - \cos\alpha - \sin \alpha = 0
$$
I don't see what else I can do with this, so I tried to solve the left part of the equation.
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \tan^2\alpha} - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \frac{\sin^2\alpha}{\cos^2\alpha}} - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{\frac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha} } - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - cos\alpha} + \frac{\sin\alpha\cdot\cos^2\alpha + \cos^3 \alpha}{\cos^2\alpha- \sin^2\alpha} - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\cos^3 \alpha}{\sin\alpha} - \cos\alpha = \sin \alpha
$$
And I get to nowhere again. I have no other ideas, I didn't see some formula or something. Any help is appreciated.
| $$LHS=\frac{\sin^2 \alpha}{\sin \alpha-\cos \alpha}+\frac{\sin\alpha+\cos\alpha}{1-\tan^2\alpha}-\cos\alpha\\=\frac{\sin^2 \alpha}{\sin \alpha-\cos \alpha}-\frac{\cos^2\alpha}{\sin\alpha-\cos\alpha}-\cos\alpha\\=\sin\alpha+\cos\alpha-\cos\alpha=\sin\alpha$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Understanding density of irrational numbers and Archemedian property From
Density of irrationals
I know this much of the proof of the density of irrational numbers
"We know that $y-x>0$.
By the Archimedean property, there exists a positive integer $n$ such that $n(y-x)>1$ or $1/n < y-x$. There exists an integer $m$ such that $m \leq nx < m+1$ or $\displaystyle \frac{m}{n} \leq x \leq \frac{m+1}{n} < y$. "
Why does the following flow from the Archimedean property? As I know it the property states only what's above in the proof.
"Pick your favorite positive irrational, which is $\sqrt{2}$. By the Archimedean property, there exists $n$ such that $\frac{\sqrt{2}}{n}\lt \frac{y-x}{2}$. Again by the Archimedean property, we know there exists an integer $m$ such that $m\left(\frac{\sqrt{2}}{n}\right)\gt x$. Pick $M$ to be the least such $m$. Can you show that $M\left(\frac{\sqrt{2}}{n}\right)$ is strictly between $x$ and $y$?
(Above quote By Arturo Magidin)
I've seen the Archimedean property used similarly in this post Proof that the set of irrational numbers is dense in reals
"By the density of rational numbers, there exists a rational number $r \in (x, y)$.
Since $\frac{y - r}{2} > 0$, by the Archimedian Property there exists $n \in \mathbb{N}$ such that $\frac{y - r}{2} > \frac{1}{n}$. Then we have $x < r + \frac{\sqrt{2}}{n} < r + \frac{\sqrt{4}}{n} < y$. Now check that $s = r + \frac{\sqrt{2}}{n}$ is an irrational number sitting in $(x, y)$."(above proof written by Akech)
Thanks for explaining
| If $M$ is the least $m$
such that
$m\left(\frac{\sqrt{2}}{n}\right)\gt x
$,
then
$(M-1)\left(\frac{\sqrt{2}}{n}\right)
\le x
\lt M\left(\frac{\sqrt{2}}{n}\right)
$.
Since
$\frac{\sqrt{2}}{n}\lt \frac{y-x}{2}
$,
$\begin{array}\\
y
&\gt 2\frac{\sqrt{2}}{n}+x\\
&\ge 2\frac{\sqrt{2}}{n}+(M-1)\left(\frac{\sqrt{2}}{n}\right)\\
&= 2\frac{\sqrt{2}}{n}+M\left(\frac{\sqrt{2}}{n}\right)-\left(\frac{\sqrt{2}}{n}\right)\\
&= \frac{\sqrt{2}}{n}+M\left(\frac{\sqrt{2}}{n}\right)\\
&\gt M\left(\frac{\sqrt{2}}{n}\right)\\
\end{array}
$
Actually,
it looks like
all that is needed is
$\frac{\sqrt{2}}{n}\lt y-x
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving the following number is real Let $z_i$ be complex numbers such that $|z_i| = 1$ .
Prove that :
$$ z\, :=\, \frac{z_1+z_2+z_3 +z_1z_2+z_2z_3+z_1z_3}{1+z_1z_2z_3} \in \mathbb{R} $$
This problem was featured on my son's final exam today, I tried helping him with this, but I guess I got a bit rusty myself.
Any ideas ?
| If $|z|=1$, then $z+\frac1z=z+\bar{z}=2\mathrm{Re}(z)$. Therefore, the following are both real:
$$
\begin{align}
\small\left(z_1+\frac1{z_1}\right)\left(z_2+\frac1{z_2}\right)\left(z_3+\frac1{z_3}\right)
&\small=\frac{z_1^2z_2^2z_3^2+z_1^2z_2^2+z_2^2z_3^2+z_3^2z_1^2+z_1^2+z_2^2+z_3^2+1}{z_1z_2z_3}\tag{1}\\
z_1z_2z_3+\frac1{z_1z_2z_3}
&=\frac{z_1^2z_2^2z_3^2+1}{z_1z_2z_3}\tag{2}
\end{align}
$$
The difference of $(1)$ and $(2)$ is also real:
$$
\frac{z_1^2z_2^2+z_2^2z_3^2+z_3^2z_1^2+z_1^2+z_2^2+z_3^2}{z_1z_2z_3}\tag{3}
$$
The ratio of $(3)$ and $(2)$ is also real:
$$
\frac{z_1^2z_2^2+z_2^2z_3^2+z_3^2z_1^2+z_1^2+z_2^2+z_3^2}{z_1^2z_2^2z_3^2+1}\tag{4}
$$
For any $w_1$, $w_2$, and $w_3$ whose absolute values are $1$, we can find $z_1$, $z_2$, and $z_3$ whose absolute values are $1$ so that $w_1=z_1^2$, $w_2=z_2^2$, and $w_3=z_3^2$. Plugging into $(4)$, we get that for any $w_1$, $w_2$, and $w_3$ whose absolute values are $1$, the following is real:
$$
\frac{w_1+w_2+w_3+w_1w_2+w_2w_3+w_3w_1}{1+w_1w_2w_3}\tag{5}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 4
} |
Matrix Problem of form Ax=B The matrix $A$ is given by
$$\left(\begin{array}{ccc}
1 & 2 & 3 & 4\\
3 & 8 & 11 & 8\\
1 & 3 & 4 & \lambda\\
\lambda & 5 & 7 & 6\end{array} \right)$$
Given that $\lambda$=$2$, $B$=$\left(\begin{array}{ccc}
2 \\
4 \\
\mu \\
3 \end{array} \right)$ and $X$=$\left(\begin{array}{ccc}
x \\
y \\
z \\
t \end{array} \right)$
Find the value of $\mu$ for which the equations defined by $AX=B$ are consistent and solve the equations in this case. State the rank of A.
So I began by reducing matrix $A$ to reduced row echelon form (kind of like taking the null space, except I'm dealing with $Ax=B$ instead of $Ax=0$) but since I have 5 variables and only 4 equations, I'm not sure how to continue onward.
| First, note that
$$
\det A=2\,(\lambda-2)^2
$$
so $A$ is invertible if and only if $\lambda\neq 2$. In this case we can use Cramer's rule to solve for $X$. For example, we have
$$
x=
\frac{
\begin{vmatrix}
2 & 2 & 3 & 4 \\
4 & 8 & 11 & 8 \\
\mu & 3 & 4 & \lambda \\
3 & 5 & 7 & 6
\end{vmatrix}
}
{
\begin{vmatrix}
1 & 2 & 3 & 4\\
3 & 8 & 11 & 8\\
1 & 3 & 4 & \lambda\\
\lambda & 5 & 7 & 6
\end{vmatrix}
}
=
\frac{0}{2\,(\lambda-2)^2}
=
0
$$
and solving for $y$, $z$, and $t$ is similar.
Now, if $\lambda=2$, then row-reducing shows that
$$
\DeclareMathOperator{rref}{rref}\rref
\begin{bmatrix}
2 & 2 & 3 & 4 & 2 \\
3 & 8 & 11 & 8 & 4 \\
1 & 3 & 4 & 2 & \mu \\
2 & 5 & 7 & 6 & 3
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 1 & 8 & 0 \\
0 & 1 & 1 & -2 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
which implies the system $AX=B$ is inconsistent (do you see why?).
Finally, the above also shows that
$$
\rref A
=
\begin{bmatrix}
1 & 0 & 1 & 8 \\
0 & 1 & 1 & -2 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
What does this say about the rank of $A$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to compute $\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$? $$\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$$
I have difficulty to evaluating above integrals.
First I try the substitution $x^4 =t$ or $x^4 +x^2+1 =t$ but it makes integral worse.
Using Mathematica I found the result $\dfrac{\pi}{48\sqrt{3}}$ I want to know the procedure of evaluating this integral.
| For the antiderivative, you could also "simplify" the problem using partial fraction decomposition since $$\frac{x^4}{\left(x^4+x^2+1\right)^3}=-\frac{3 (x-1)}{16 \left(x^2-x+1\right)}+\frac{3 (x+1)}{16
\left(x^2+x+1\right)}-\frac{3}{16 \left(x^2-x+1\right)^2}-\frac{3}{16
\left(x^2+x+1\right)^2}+\frac{x}{8 \left(x^2-x+1\right)^3}-\frac{x}{8
\left(x^2+x+1\right)^3}$$ which leads to the result. But, I suspect that using residues will make the problem easier.
Could you prove that $$\int_0^1 \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{1}{288} \left(-28+\sqrt{3} \pi +27 \log (3)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Derivative of $f(x) = \frac{\cos{(x^2 - 1)}}{2x}$ Find the derivative of the function $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}$$
This is my step-by-step solution: $$f'(x) = \frac{-\sin{(x^2 - 1)}2x - 2\cos{(x^2 -1)}}{4x^2} = \frac{2x\sin{(1 - x^2)} - 2\cos{(1 - x^2)}}{4x^2} = \frac{x\sin{(1 - x^2)} - \cos{(1 - x^2)}}{2x^2} = \frac{\sin{(1 - x^2)}}{2x} - \frac{\cos{(1-x^2)}}{2x^2}$$
and this is the output of WolphramAlpha: http://www.wolframalpha.com/input/?i=derivative+cos(x^2+-+1)%2F(2x)
Where is the mistakes?
| Differentiating $\cos(x^2-1)$ gives you $-2x\sin(x^2-1)$, using the Chain Rule
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Do the spaces spanned by the columns of the given matrices coincide? Reviewing linear algebra here.
Let $$A = \begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 1 & 1
\end{pmatrix} \qquad \text{and} \qquad
B = \begin{pmatrix}
1 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}.$$
Is the space spanned by the columns of $A$ the same as the space spanned by the columns of $B$?
My answer: Yes. For $A$, we have
$$C(A) =
\left\{\left.a_1\begin{pmatrix}
1 \\
1 \\
0 \\
0
\end{pmatrix}+ a_2
\begin{pmatrix}
1 \\
1 \\
0 \\
0
\end{pmatrix}+ a_3
\begin{pmatrix}
0 \\
0 \\
1 \\
1
\end{pmatrix}+ a_4
\begin{pmatrix}
0 \\
0 \\
0 \\
1
\end{pmatrix} =
\begin{pmatrix}
a_1 + a_2 \\
a_1 + a_2 \\
a_3 \\
a_3 + a_4
\end{pmatrix} \,
\right\vert a_i \in \mathbf{R}\right\}\text{.}
$$
Denote $a_1 + a_2 = c_1$, $a_3 = c_2$, and $a_3 + a_4 = c_3$. Then since $a_4$ is arbitrary, any vector in $C(A)$ can be described as $(c_1, c_1, c_2, c_3)^{\prime}$.
For $B$, we have
$$
C(B) =
\left\{\left.b_1\begin{pmatrix}
1 \\
1 \\
0 \\
0
\end{pmatrix}+ b_2
\begin{pmatrix}
0 \\
0 \\
1 \\
0
\end{pmatrix}+ b_3
\begin{pmatrix}
0 \\
0 \\
0 \\
1
\end{pmatrix} =
\begin{pmatrix}
b_1 \\
b_1 \\
b_2 \\
b_3
\end{pmatrix} \,
\right\vert b_i \in \mathbf{R}\right\}\text{.}
$$
Thus, any vector in $C(B)$ can be described as $(b_1, b_1, b_2, b_3)^{\prime}$, the same form as a vector in $C(A)$.
| You are correct. One can also verify this by checking that $A, B$ are column-equivalent, or equivalently, that is, that $A^T, B^T$ are row-equivalent. One can check the latter algorithmically by checking that the reduced row echelon form of the $A^T, B^T$ coincide (at least after padding with an appropriate number of zero rows to the smaller matrix).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $3x^3\equiv 7\pmod{925}$ I am trying to solve $3x^3\equiv 7\pmod{925}$.
I thought of using brute force, but $925$ is too big for that.
I also tried raising both sides of the equation to the power of $3$, but it didn't help.
How can I solve it?
| With these, it's often easier to consider the problem after breaking the moduli down and then recombining the results using the Chinese Remainder Theorem.
So we consider the pair of problems
$$ \begin{align}
3x^3 &\equiv 7 \pmod {37}\\
3x^3 &\equiv 7 \pmod {25}.
\end{align}$$
In the first, it's easy to check that the inverse of $3$ mod $37$ is $25$ (as $3 \cdot 25 = 75 = 1 + 2\cdot 37$), so multiplying through by $25$, we get
$$x^3 \equiv 27 \pmod {37}.$$
This yields the obvious solution $x \equiv 3 \pmod {37}$. But we can find two more. In particular, since $3^3 \equiv 27 \pmod {37}$, Fermat's Little Theorem gives us that $3^{3 + 36} = 3^{13 \cdot 3} \equiv 27 \pmod {37}$ and $3^{3 + 72} = 3^{25 \cdot 3} \equiv 27 \pmod{37}$. So we have the three solutions
$$3, 3^{13} \equiv 30, 3^{25} \equiv 4 \pmod{37}.$$
In the second, it's easy to check that the inverse of $3 \pmod {25}$ is $17$ (as $3\cdot 17 = 1 + 2\cdot 25$), so multiplying by $17$ we get
$$ x^3 \equiv 19 \pmod {25}.$$
There is no obvious solution here. If we note this also means that
$$ x^3 \equiv 4 \pmod 5,$$
then we can quickly check by hand that the only solution mod $5$ is $4$. This at least indicates that mod $25$, we only need to check $4,9,14,19,24$. A few quick pen strokes later, we can conclude that
$$x \equiv 14 \pmod {25}$$
is the only solution here.
Now we patch these together using the Chinese Remainder Theorem. We have three choices mod $37$ and one choice mod $25$, so we will get three answers overall. Executing the Chinese Remainder Theorem reveals the three answers to be
$$ 114, 189, 289,$$
as claimed above. $\diamondsuit$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding a nullspace of a matrix - what should I do after finding equations? I am given the following matrix $A$ and I need to find a nullspace of this matrix.
$$A =
\begin{pmatrix}
2&4&12&-6&7 \\
0&0&2&-3&-4 \\
3&6&17&-10&7
\end{pmatrix}$$
I have found a row reduced form of this matrix, which is:
$$A' =
\begin{pmatrix}
1&2&0&0&\frac{23}{10} \\
0&0&1&0&\frac{13}{10} \\
0&0&0&1&\frac{22}{10} \end{pmatrix}$$
And then I used the formula $A'x=0$, which gave me:
$$A' =
\begin{pmatrix}
1&2&0&0&\frac{23}{10} \\
0&0&1&0&\frac{13}{10} \\
0&0&0&1&\frac{22}{10}
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{pmatrix}=
\begin{pmatrix}
0 \\
0 \\
0
\end{pmatrix}$$
Hence I obtained the following system of linear equations:
$$\begin{cases} x_1+2x_2+\frac{23}{10}x_5=0 \\ x_3+\frac{13}{10}x_5=0 \\ x_4+\frac{22}{10}x_5=0 \end{cases}$$
How should I proceed from this point?
Thanks!
| \begin{cases} x_1+2x_2+\frac{23}{10}x_5=0 \\ x_3+\frac{13}{10}x_5=0 \\ x_4+\frac{22}{10}x_5=0 \end{cases}
$$
x_1=-2x_2-\dfrac{23}{10}x_5
$$
$$
x_3=-\dfrac{13}{10}x_5
$$
$$
x_4=-\dfrac{11}{5}x_5
$$
Therefore,basis of null space=
$$
\begin{pmatrix}
-2 \\
1 \\
0 \\
0 \\
0
\end{pmatrix},
\begin{pmatrix}
-\dfrac{23}{10} \\
0 \\
-\dfrac{13}{10} \\
-\dfrac{11}{5} \\
1
\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solutions of $\sqrt{x+4+2\sqrt{x+3}}-(x^2+4x+3)^{1/3}=1$ $\sqrt{x+4+2\sqrt{x+3}}-(x^2+4x+3)^{1/3}=1$
I get that $-3$ as a solution, but apparently 1 is as well a solution, and I don't see a mechanism through which I could find it. Any help would be appreciated.
| we have $$\sqrt{x+4+2\sqrt{x+3}}-(x^2+4x+3)^{1/3}=1 $$ let us make a change of variable $u = x + 3 \ge 0, x = u - 3.$ with that we have
$$\sqrt{u+1+2\sqrt u}-(u(u-2))^{1/3}=1 \to 1+\sqrt u=(u(u-2))^{1/3}+1$$ this gives us $$\sqrt u = (u(u-2))^{1/3}\tag 1 $$
now, exponentiating $(1)$ implies $$ 0=u^3 - u^2(u-2)^2 \to 0=u^2(u^2 -5u + 4)=u^2(u-4)(u-1) $$
subbing $u = 1,$ in $(1),$ we have $1 = -1$ therefore is an extraneous solution and needs to be rejected.
the roots are $u = 0, u = 4$ are solutions of $1.$ this translates to $$x = -3, x = 1. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the solutions of $x^2\equiv 9 \pmod {256}$. Find the solutions of $x^2\equiv 9 \pmod {256}$. I try to follow an algorithm shown us in class, but I am having troubles doing so. First I have to check how many solutions there are. Since $9\equiv 1 \pmod {2^{\min\{3,k\}}}$ where k fulfills $256=2^k$, then since $k\ge 3$, there are 4 solutions.
As I understood, I have to take a smaller power of 2 and check $x^2\equiv 1 \pmod{2^m}$. I took m=4 and found that x=5 is a solution. According to the algorithm, I should take $x=5+Ax$ where A is a power of 2 and raise it by 2 and check congruence modulo, 256(I guess?), but the examples I were given didn't tell me accurately what I am to pick. I could really use your help on this.
| This is equivalent to finding all $0\leq x \leq 255$ such that $256\mid x^2-9 $. But $256 = 2^8$ and $x^2-9 = (x-3)(x+3)$. But the greatest common divisor of $x-3$ and $x+3$ must divide $(x+3)-(x-3) = 6$. Hence, if $x+3$ is divisible by $4$, then $x-3$ is only divisible by $2$, and vice versa.
The statement is thus equivalent to finding $x$ such that $2^7 \mid x+3$ or $2^7\mid x-3$. Because $0\leq x \leq 255$ and $2^7 = 128$ there are four solutions:
$$\{3,128-3,128+3,256-3\} = \{3,125,131,253\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Having problem in last step on proving by induction $\sum^{2n}_{i=n+1}\frac{1}{i}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i} $ for $n\ge 1$ The question I am asked is to prove by induction $\sum^{2n}_{i=n+1}\frac{1}{i}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i} $ for $n\ge 1$
its easy to prove this holds for $n =1$ that gives $\frac{1}{2}=\frac{1}{2}$
Now assuming $n$ its true I want to say it is true $n+1$. So,
$\sum^{2n}_{i=n+1}\frac{1}{i}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i} $
$\sum^{2n}_{i=n+1}\frac{1}{i}+\frac{(-1)^{1+(2n+1)}}{2n+1}+\frac{(-1)^{1+(2n+2)}}{2n+2}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i}+\frac{(-1)^{1+(2n+1)}}{2n+1}+\frac{(-1)^{1+(2n+2)}}{2n+2} $
$\sum^{2n}_{i=n+1}\frac{1}{i}+(-1)^{2n+2}[ \frac{1}{2n+1}+\frac{(-1)}{2n+2}]=\sum^{2n+2}_{i=1}\frac{(-1)^{1+i}}{i} $
$\sum^{2n}_{i=n+1}\frac{1}{i}+ \frac{1}{2n+1}+\frac{(-1)}{2n+2}=\sum^{2(n+1)}_{i=1}\frac{(-1)^{1+i}}{i} $
$\sum^{2n+1}_{i=n+1}\frac{1}{i}+\frac{(-1)}{2n+2}=\sum^{2(n+1)}_{i=1}\frac{(-1)^{1+i}}{i} $
i don't know what can i do next if the numerator of $\frac{1}{i}+\frac{(-1)}{2n+2}$ was positive i knew. is there a way i can turn it positive?or my approach is wrong ?
| $$\sum_{i=n+2}^{2n+2} \frac{1}{i}=\sum_{i=n+1}^{2n}\frac{1}{i}+ \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i}+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$$
It remains to prove that:
$$ \sum_{i=1}^{2n}\frac{(-1)^{i+1}}{i}+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\sum_{i=1}^{2n+2}\frac{(-1)^{i+1}}{i}$$
that is
$$ \frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}=\frac{1}{2n+1}+\frac{-1}{2n+2} \Leftrightarrow \frac{2}{2n+2}=\frac{1}{n+1} (A). $$
EDIT: If you want to continue in your way, you can do like this:
$$ \sum_{i=n+1}^{2n+1} \frac{1}{i} + \frac{-1}{2n+2}=\sum_{i=n+1}^{2n+1} \frac{1}{i}+ \frac{1}{2n+2}-\frac{1}{n+1}=\sum_{i=n+2}^{2n+2} \frac{1}{i}.$$
So,
$$ \sum_{i=n+2}^{2n+2} \frac{1}{i} =\sum_{i=1}^{2n+2} \frac{(-1)^{i+1}}{i}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Reducing the form of $2\sum\limits_{j=0}^{n-2}\sum\limits_{k=1}^n {{k+j}\choose{k}}{{2n-j-k-1}\choose{n-k+1}}$. I've been toying around with simplifying the expression $2\sum\limits_{j=0}^{n-2}\sum\limits_{k=1}^n {{k+j}\choose{k}}{{2n-j-k-1}\choose{n-k+1}}$ (for integer only $n$) for a while, as I was hoping it would have some sort of simpler closed form representation, and so I plugged it into Mathematica's FullSimplify function and got back the expression: $$-\dfrac{2\pi n \csc (2 \pi n)}{\Gamma (n-1) \Gamma (-2 n) \Gamma (n+3)}$$ Which can be rewritten as $$-\dfrac{4\pi n^2 \csc (2 \pi n)}{2n\Gamma (n-1) \Gamma (-2 n) \Gamma (n+3)}$$ and then simplified via the property that $\Gamma (x) \Gamma (-x) = -\dfrac{\pi \csc (\pi x)}{x}$, as $$-\dfrac{4\pi n^2 \csc (2 \pi n)}{2n\Gamma (n-1) \Gamma (-2 n) \Gamma (n+3)}$$ $$=-\dfrac{\pi \csc (\pi 2n)}{2n} \dfrac{4n^2}{\Gamma (n-1) \Gamma (-2 n) \Gamma (n+3)}$$ $$=\Gamma (2n) \Gamma (-2n) \dfrac{4n^2}{\Gamma (n-1) \Gamma (-2 n) \Gamma (n+3)}$$ $$= \dfrac{4n^2\Gamma (2n)}{\Gamma (n-1) \Gamma (n+3)}$$ and as $n$ is strictly an integer, the expression can be rewritten as $$\dfrac{4n^2(2n-1)!}{(n-2)!(n+2)!}$$ $$= \dfrac{2n(2n)!}{(n-2)!(n+2)!}$$ $$ = 2n {{2n}\choose{n-2}}$$ My problem here is that I have no idea how to get to that first step without the use of Mathematica. I wouldn't even know where to start with this thing if I didn't have the computer's help to break it down first for me. I would very much appreciate advice about how I could solve this by hand, or a hint about how to start reducing it from its double sum form, if that's possible. I would also appreciate feedback about the correctness of my simplification. Thank you in advance.
| Alternate solution.
As before we start trying to evaluate
$$S(n) =
\sum_{q=0}^{n-2}
\sum_{k=1}^n {k+q\choose k} {2n-q-k-1\choose n-k+1}$$
which we re-write as
$$-\sum_{q=0}^{n-2} {2n-q-1\choose n+1}
-\sum_{q=0}^{n-2} {n+1+q\choose n+1}
+ \sum_{q=0}^{n-2}
\sum_{k=0}^{n+1} {k+q\choose k} {2n-q-k-1\choose n-k+1}.$$
Call these pieces up to sign from left to right $S_1, S_2$ and $S_3.$
The two pieces in front cancel the quantities introduced by extending
$k$ to include the values zero and $n+1.$
Evaluation of $S_1.$
Introduce
$${2n-q-1\choose n+1} = {2n-q-1\choose n-q-2} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-q-1}}{z^{n-q-1}} \; dz.$$
This vanishes when $q\gt n-2$ so we may extend the sum to infinity to
get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-1}}{z^{n-1}}
\sum_{q\ge 0} \frac{z^q}{(1+z)^q}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-1}}{z^{n-1}}
\frac{1}{1-z/(1+z)}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n-1}}
\; dz
\\ = {2n\choose n-2}.$$
Evaluation of $S_2.$
Introduce
$${n+1+q\choose n+1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1+q}}{z^{n+2}} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}}
\sum_{q=0}^{n-2} (1+z)^q
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+2}}
\frac{(1+z)^{n-1}-1}{1+z-1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n+1}}{z^{n+3}}
((1+z)^{n-1}-1) \; dz
\\ = {2n\choose n+2}.$$
A more efficient evaluation is to notice that when we re-index
$q$ as $n-2-q$ in $S_2$ we obtain
$$\sum_{q=0}^{n-2} {n+1+n-2-q\choose n+1}
= \sum_{q=0}^{n-2} {2n-q-1\choose n+1}$$
which is $S_1.$
Evaluation of $S_3.$
Introduce
$${2n-q-k-1\choose n-k+1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-q-k-1}}{z^{n-k+2}} \; dz.$$
This effectively controls the range so we can let $k$ go to infinity
to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-1}}{z^{n+2}}
\sum_{q=0}^{n-2} \sum_{k\ge 0} {k+q\choose q} \frac{z^k}{(1+z)^{q+k}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n-1}}{z^{n+2}}
\sum_{q=0}^{n-2} \frac{1}{(1+z)^q} \frac{1}{(1-z/(1+z))^{q+1}} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+2}}
\sum_{q=0}^{n-2} \frac{1}{(1+z)^{q+1}} \frac{1}{(1-z/(1+z))^{q+1}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+2}}
\times (n-1) \times \; dz
\\ = (n-1) \times {2n\choose n+1}.$$
Finally collecting the three contributions we obtain
$$(n-1) \times {2n\choose n+1}
- 2{2n\choose n+2}
= (n+2) {2n\choose n+2}
- 2{2n\choose n+2}
\\ = n\times {2n\choose n+2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Finding two solutions to $x^2 - 6y^2 = 1$ using continued fractions Can anyone show me how to find the solutions to $x^2-6y^2=1$ by using continued fractions? I know how to find the fractions for $\sqrt6$ but do not know how to proceed. THANK YOU!!!
| Suppose you are able to find a solution to $x^2-6y^2=1$ like $u_1=5$ and $v_1=2$ as indicated in the answer given by user17762. Then ALL other positive integer solutions $(u_n,v_n)$ can be found by
$$u_n+\sqrt{6}v_n=(u_1+\sqrt{6}v_1)^n.$$
Thus another solution will be
$$u_2+\sqrt{6}v_2=(u_1+\sqrt{6}v_1)^2=(5+2\sqrt{6})^2=49+20\sqrt{6}.$$
In general, you can get a recurrence relation as follows:
$$u_n+\sqrt{6}v_n=(u_{n-1}+\sqrt{6}v_{n-1})(u_1+\sqrt{6}v_1)=(u_{n-1}+\sqrt{6}v_{n-1})(5+2\sqrt{6}).$$
Upon comparison you get:
\begin{align*}
u_n & = 5u_{n-1}+12v_{n-1}\\
v_n & = 2u_{n-1}+5v_{n-1}
\end{align*}
This is exactly the recurence relation satisfied by the convergents of the continued fraction of $\sqrt{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help with Lagrange multipliers on an intresting function Hi guys I am trying to do Lagrange multipliers to figure out $\lambda$
$$F=a \log(x^2-y)+b\log(x^3-z)-\lambda (x^2-y+x^3-z -1)$$
Where a and b are constants and we have the constraint $x^2-y+x^3-z =1$
What I did was take the partials and added them together and got $a(\frac{2x-1}{x^2-y})+b(\frac{3x-1}{x^3-z})- \lambda (2x+3x^2)+2 \lambda$
I am not sure how to find this lambda
| Suppose your $\log$ means natural logarithm. Then set the partial derivatives to $0$:
$$\frac{2ax}{x^2-y}+\frac{3bx^2}{x^3-z}-2x\lambda+3x^2\lambda=0 \\
\frac{-a}{x^2-y}+\lambda=0 \\
\frac{-b}{x^3-z}+\lambda=0\\
x^2-y+x^3-z =1$$
From equation 2 and 3:
$$\lambda=\frac{a}{x^2-y}=\frac{b}{x^3-z}$$
Plug these information into 1:
$$2x\lambda+3x^2\lambda-2x\lambda+3x^2\lambda=0$$
Can you go on from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving differential equation $x'=\frac{x+2t}{x-t}$ I am trying to solve the following differential equation:
$$x'=\frac{x+2t}{x-t}$$
with initial value condition: $x(1)=2$
This is what I have so far:
Substitution: $u=\frac{x}{t}$
$$\implies u't+u=\frac{2t+tu}{-t+tu}$$
Separation of variables:
$$\implies \frac{u'(u-1)}{-u^2+2u+2}=\frac{1}{t}$$
Integrate both sides:
$$\implies \int \frac{u'(u-1)}{-u^2+2u+2} dt=\int \frac{1}{t} dt \iff -\frac{1}{2}\ln(-u^2+2u+2)=\ln(t)+C$$
Solve for "$u$":
$$\implies u=1 \pm\sqrt{\frac{-1+3e^{2C}t^2}{e^{2C}t^2}}$$
Resubstitution: $u=\frac{x}{t}$
$$\implies x=t \pm t\sqrt{\frac{-1+3e^{2C}t^2}{e^{2C}t^2}}$$
Solving for "$C$" using $x(1)=2$:
$$x(1)=2=1\pm 1\sqrt{\frac{-1+3e^{2C}1^2}{e^{2C}1^2}} \implies C=-\frac{\ln(2)}{2}$$
Substituting "$C$" into the solution:
$$x=t \pm t\sqrt{\frac{-1+\frac{3}{2}t^2}{\frac{1}{2}t^2}}=t \pm t\sqrt{3t^2-2}$$
Plugging "$x$" into the differential equation:
$$1 \pm (\sqrt{3t^2-2}+\frac{6t^2}{\sqrt{3t^2-2}}) \not = \frac{3t \pm t\sqrt{3t^2-2}}{\pm t\sqrt{3t^2-2}}$$
Where did I go wrong?
| it must be $$\frac{u-1}{-u^2+2u+2}du=\frac{dt}{t}$$
we get
$$-\frac{1}{2}\ln|-u^2+2u+2|=\ln|t|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit as $x$ tends to zero of $\frac{\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}$ How would I find the $$\lim_{x\to0}\left(\frac {\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}\right)$$ The only way I know how to do this is with l'hopitals rule but I don't see it helping here as we have x's in our denominator.
| you have $$ \left(\frac {\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}\right)= \frac{2x^2 - (e^x - e^{-x})\sin x}{2x^5\sin x}$$ we will look at maclaurin expansion of $$\begin{align} (e^x - e^{-x})\sin x - 2x^2 &=2\left(x+ \frac16 x^3+\frac1{120}x^5 + \cdots\right)\left(x - \frac16x^3 + \frac1{120}x^5+\cdots\right)-2x^2 \\
&=2\left(x^2+\left(\frac1{60} - \frac1{36}\right)x^6+ \cdots \right)-2x^2 \\
&= -\frac1{45}x^6 + \cdots\end{align} $$
therefore $$ \lim_{x \to 0}\left(\frac {\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}\right) = \frac1{90}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1304433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Matrices $\begin{pmatrix} a &b \\ c &d \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}=k\begin{pmatrix}x\\y \end{pmatrix}$ If $\begin{pmatrix}
a &b \\
c &d
\end{pmatrix}\begin{pmatrix}
x\\y \end{pmatrix}=k\begin{pmatrix}x\\y \end{pmatrix}$, prove that $k$ satisfies the equation $k^2-(a+d)k+(ad-bc)=0$.
If the roots of this quadractic equation are $\alpha$ and $\beta$, find the value of $\alpha+\beta$ and $\alpha \beta$ in terms of $a,b,c$ and $d$. Hence, or otherwise, prove that $\begin{pmatrix}
a &b \\
c &d
\end{pmatrix}\begin{pmatrix}
b &b \\
\alpha+a &\beta-a
\end{pmatrix}=\begin{pmatrix}
b &b \\
\alpha+a &\beta-a
\end{pmatrix}\begin{pmatrix}
\alpha &0 \\
0&\beta
\end{pmatrix}$
Can anyone give me some hints for solving this question? Thanks
| we can rewrite the matrix equation as s system of two linear equations. they are $$\begin{align} (a-k)x + by &= 0\\cx + (d-k)y &=0\end{align}$$
multiply the first equation by $(d-k)$, the second one by $b$, and subtracting gives $$\left((a-k)(d-k)-bc\right)x = 0.$$
we have two choices: (a) $x = 0$ or (b) $$(a-k)(d-k)-bc = 0.\tag 1$$
the first choice will lead to $y \neq 0, b = 0, d = k$ which also satisfies $(1).$
therefore it is necessary for $k$ to satisfy $(1)$ called the characteristic equation of the matrix $\pmatrix{a&b\\c&d}.$
rewriting $(1)$ as a quadratic in $k,$ we have $$k^2 - (a+d)k + ad - bc = (k - \alpha)(k-\beta) = 0 $$ equating coefficients you get $$\alpha + \beta = a + d, \quad \alpha \beta = ad - bc. $$
you can verify the last matrix equation by just multiplying it out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Showing that $x^3+2y^3+4z^3=2xyz$ has no integer solutions except $(0,0,0)$. Let $x,y,z\in \mathbb{Z}$ satisfy the equation:
$$
x^3+2y^3+4z^3=2xyz
$$
How do I prove that the only solution is $x=y=z=0$?
| If following equation has a non-trivial solution
$$x^3 + 2y^3 + 4z^3 = 2xyz\tag{*1}$$
there must be one whose $x^2 + y^2 + z^2$ is minimal. Let $(x_1,y_1,z_1)$ be such a minimal non-trivial solution. We have
$$x_1^3 + 2y_1^3 + 4z_1^3 = 2x_1y_1z_1 \implies x_1^3 \equiv 0 \pmod 2 \implies 2|x_1$$
Let $(x_2,y_2,z_2) = (y_1,z_1,\frac{x_1}{2})$, we have
$$8z_2^3 + 2x_2^3 + 4y_2^3 = 4z_2x_2y_2\quad\iff\quad
x_2^3 + 2y_2^3 + 4z_2^3 = 2x_2y_2z_2$$
This means $(x_2,y_2,z_2)$ is again a solution for $(*1)$. Since $(x_1,y_1,z_1)$ is minimal, we have
$$x_1^2 + y_1^2 + z_1^2 \le x_2^2 + y_2^2 + z_2^2 \implies x_1^2 \le z_2^2 = \frac14 x_1^2 \implies x_1 = 0$$
This leads to $2y_1^3 + 4z_1^3 = 0$. Since $(x_1,y_1,z_1)$ is assumed to be non-trivial, $z_1 \ne 0$ and this implies $\sqrt[3]{2} = \left|\frac{y_1}{z_1}\right|$ is a rational number!
This is absurd and hence the original assumption on existence of non-trivial solution for $(*1)$ is simply wrong!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Remainder of $(1+x)^{2015}$ after division with $x^2+x+1$
Remainder of $(1+x)^{2015}$ after division with $x^2+x+1$
Is it correct if I consider the polynomial modulo $5$
$$(1+x)^{2015}=\sum\binom{2015}{n}x^n=1+2015x+2015\cdot1007x^2+\cdots+x^{2015}$$
RHS stays the same and then The remainder must be of the form $Ax+B$
$$x^{2015}+1\equiv Ax+B\pmod{1+x+x^2}$$
plug in $x=0\implies B=0$
plug in $x=1\implies A=-1\implies $ the remainder is $-x$
Is this a good way to solve the problem or were we lucky ?
| Recall that $x^2+x+1=\Phi_3(x)$. If we set $\omega=\exp\left(\frac{2\pi i}{3}\right),\bar\omega=\left(\frac{2\pi i}{3}\right)$, we must have:
$$ (1+x)^{2015} = q(x)\cdot \Phi_3(x)+(Ax+B), \tag{1}$$
and evaluating the previous identity for $x\in\{\omega,\bar{\omega}\}$ we get:
$$ (1+\omega)^{2015} = A\omega+B,\quad (1+\bar\omega)^{2015} = A\bar\omega+B\tag{2}$$
but since $1+\omega=\exp\left(\frac{2\pi i}{6}\right)$ and $2015\equiv 5\pmod 6$, by setting $\zeta=\exp\left(\frac{2\pi i}{6}\right)$ we have that $(2)$ translates into:
$$ A\zeta^2 + B = \zeta^5,\quad A\zeta^4 + B = \zeta,\tag{3} $$
hence by Cramer's rule:
$$ A = \frac{\zeta^5-\zeta}{\zeta^2-\zeta^4}=-1, \qquad B=\frac{\zeta^3-\zeta^9}{\zeta^2-\zeta^4}=0.\tag{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Determine whether or not the limit exists: $\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$
Determine whether or not the limit $$\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$$
exists. If it does, then calculate its value.
My attempt:
$$\begin{align}\lim \frac{(x+y)^2}{x^2+y^2} &= \lim \frac{x^2+y^2}{x^2+y^2} + \lim \frac {2xy}{x^2+y^2} =\\&= 1 + \lim \frac 2{xy^{-1}+yx^{-1}} = 1+ 2\cdot\lim \frac 1{xy^{-1}+yx^{-1}}\end{align}$$
But $\lim_{x\to 0^+} x^{-1} = +\infty$ and $\lim_{x\to 0^-} x^{-1} = -\infty$
Likewise, $\lim_{y\to 0^+} y^{-1} = +\infty$ and $\lim_{y\to 0^-} y^{-1} = -\infty$
So the left hand and right hand limits cannot be equal, and therefore the limit does not exist.
| Consider $$f(x,y)=\frac{(x+y)^2}{x^2+y^2} .$$
If you take the path $(0,y)$, then:
$$\displaystyle\lim_{y\to0} f(0,y) =\lim_{y\to0} \frac{y^2}{y^2}=1$$
If you take the path $(x,x)$, then:
$$ \lim_{x\to0}f(x,x) =\lim_{x\to0} \frac{(2x)^2}{2x^2}=2$$
So, the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 2
} |
find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, after that. I don't know how to do next in order to find the minimum value. Please help!
| $$\begin{align} f'(x)=0 &\iff \frac{x}{\sqrt{x^2+4}} = \frac{12-x}{\sqrt{x^2-24x+153}}\\
&\implies \frac{x^2}{{x^2+4}} = \frac{x^2 - 24x + 144}{{x^2-24x+153}}\\
&\iff 1 - \frac{4}{{x^2+4}} = 1 - \frac{9}{{x^2-24x+153}}\\
&\iff 4(x^2 -24x + 153) = 9(x^2+4) \end{align}$$
from which you find the roots (if I'm not mistaken, only one of them is valid because of the squaring in the second line).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
What's the relation between different antiderivatives? If a function $f(x)$ has different forms of antiderivatives:
$\frac { d }{ dx } { F }_{ 1 }(x)=f(x)$
$\frac { d }{ dx } { F }_{ 2 }(x)=f(x)$
What's the relationship between $F_1$ and $F_2$, is that ${F}_{1}(x)-{F}_{2}(x)=constant$ correct?
For example, question find $\int { \frac { dx }{ { x }^{ 4 }-1 } = } $ ?
Method 1: $\int { \frac { dx }{ { x }^{ 4 }-1 } =\int { \frac { dx }{ \left( { x }^{ 2 }-1 \right) \left( { x }^{ 2 }+1 \right) } =\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }-1 } -\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }+1 } } =\frac { 1 }{ 4 } ln\left| \frac { x-1 }{ x+1 } \right| -\frac { 1 }{ 2 } arctan(x) } +c } } $
Method 2:$\int { \frac { dx }{ { x }^{ 4 }-1 } =\frac { 1 }{ 2 } \int { \frac { d{ x }^{ 2 } }{ { \left( { x }^{ 2 } \right) }^{ 2 }-1 } =\frac { 1 }{ 2 } ln\left| \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right| +c } } $
Ok, now the question is: what's the relation between $ln\left| \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right| $ and $\frac { 1 }{ 2 } ln\left| \frac { { x }-1 }{ { x }+1 } \right| -arctan(x)$ ?
Does the equation below is correct and how to prove it?
$ln\left| \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right| =\frac { 1 }{ 2 } ln\left| \frac { { x }-1 }{ { x }+1 } \right| -arctan(x) +constant$
| Your method 1 is correct.
Your method 2 is not:
$$\int\frac{dx}{x^4-1} \neq \frac12\int\frac{d(x^2)}{(x^2)^2-1}$$
If you're going to substitute for $x^2$, $d(x^2) = 2x dx$, so the equality would be
$$\int\frac{dx}{x^4-1} = \frac12\int\frac{du}{\sqrt{u}(u^2-1)}$$
which gets you nowhere.
But had both methods been correct, yes, the two anti-derivatives would differ by a constant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Evaluate $\lim\limits_{x\to\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})$ When using Maclaurin series, the limit is
$$\lim\limits_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}=0$$
If we expand the expression with two limits
$$\lim\limits_{x\to\infty}\sin\sqrt{x+1}-\lim\limits_{x\to\infty}\sin\sqrt{x}$$
it diverges.
Which solution is right?
| Using the prosthaphaeresis formulae,
$$ \sin{A}-\sin{B} = 2\sin{\tfrac{A+B}{2}} \cos{\tfrac{A-B}{2}}, $$
which gives you
$$ 2\sin{\left( \frac{\sqrt{x+1}-\sqrt{x}}{2} \right)} \cos{\left( \frac{\sqrt{x+1}+\sqrt{x}}{2} \right)} $$
Then you have
$$ (\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x}) = 1+x-x=1, $$
so we have
$$ 2\sin{\left( \frac{1}{2(\sqrt{x+1}+\sqrt{x})} \right)} \cos{\left( \frac{\sqrt{x+1}+\sqrt{x}}{2} \right)} $$
Then the limit is the same as
$$ \lim_{y \to \infty} 2(\cos{\tfrac{1}{2}y} )\sin{\left(\frac{1}{2y}\right)}, $$
and the bracket is bounded, the last term tends to zero since $\sin{z} \to 0$ as $z \to 0$. Hence the whole expression tends to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
| Consider $\mathbb Z[\sqrt{2}]= \{a + b\sqrt 2 ; a,b \in \mathbb Z\}$. Take $\alpha = \sqrt 2 - 1 \in \mathbb Z [\sqrt 2]$, then $$0 < \alpha < 1 \implies \alpha ^k \to 0 \,\,\text{as} \,\, k \to \infty \tag {*}$$
Say $\sqrt 2 = \frac{p}{q}$, since $\mathbb Z[\sqrt 2]$ is closed under multiplication and addition we have
$$\alpha^k = e + f \sqrt 2 = \frac{eq + fp}{q} \geq \frac{1}{q}$$
which contradicts $(*)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 12
} |
solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3 + xy^2$$
imaginary component
$$-2yx^2 i +yi + x^2 yi - y^3 i=0i$$
$$-2yx^2 +y + x^2 y - y^3 =0$$
...
can't solve this question
| HINT:
Let $z=r(\cos A+i\sin A)$ where $r\ge0,A$ are real
As $y\ne0,\sin A\ne0$
$1+z^2=1+r^2\cos2A+i(r^2\sin2A)$
$\dfrac1{1+z^2}=\dfrac{1+r^2\cos2A-ir^2\sin2A}{(1+r^2\cos2A)^2+(r^2\sin2A)^2}$
$r(\cos A+i\sin A)\cdot(1+r^2\cos2A-ir^2\sin2A)$
$=\cdots+ir[\sin A(1+r^2\cos2A)-r^2\cos A\sin2A]=\cdots+ir[\sin A-r^2\sin(2A-A)]$
$=\cdots+ir\sin A(1-r^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Laurent Series of $\frac{z+1}{z(z-4)^3}$ in $0<|z-4|<4$ Find the Laurent Series of $\displaystyle \frac{z+1}{z(z-4)^3}$ in $0<|z-4|<4$
I thought about doing partial fraction decomposition first, so I'd have
$\displaystyle \frac{A}{z}+\frac{B}{z-4}+\frac{C}{(z-4)^2}+\frac{D}{(z-4)^3}$, but that would take awhile...so I'm not sure if this is the right method..?
| Setting $\xi=\frac{z-4}{4}$, i.e. $z=4\xi+4$ we have $0<|\xi|<1$ and therefore
\begin{eqnarray}
\frac{z+1}{z(z-4)^3}&=&\frac{4\xi+5}{(4\xi+4)(4\xi)^3}=\frac{4\xi+5}{4^4\xi^3}\cdot\frac{1}{\xi+1}=\left(\frac{1}{4^3\xi^2}+\frac{5}{4^4\xi^3}\right)\cdot\frac{1}{1+\xi}\\
&=&\left(\frac{1}{4^3\xi^2}+\frac{5}{4^4\xi^3}\right)\sum_{k=0}^\infty(-1)^k\xi^k=\frac{1}{4^3}\sum_{k=0}^\infty(-1)^k\xi^{k-2}+\frac{5}{4^4}\sum_{k=0}^\infty(-1)^k\xi^{k-3}\\
&=&\frac{1}{4^3}\sum_{k=-2}^\infty(-1)^k\xi^k-\frac{5}{4^4}\sum_{k=-3}^\infty(-1)^k\xi^k\\
&=&\frac{5}{4^4\xi^3}-\frac{1}{4^4}\sum_{k=-2}^\infty(-1)^k\xi^k\\
&=&\frac{5}{4(z-4)^3}+\sum_{k=-2}^\infty\frac{(-1)^{k+1}}{4^{k+4}}(z-4)^k.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $p_{n+1} + \frac 16 p_n = \frac 1 2 (\frac 5 6 ) ^{n-1}$ I'm trying to solve:
$$p_{n+1} + \frac 16 p_n = \frac 1 2 \left(\frac 5 6 \right) ^{n-1}$$
with initial condition: $p_1 = 1$.
First, I search particular solution of the form $p_n^* = \lambda (\frac 5 6 ) ^n$. I found $\lambda = \frac 3 5$.
Next, I know that $p_n = \alpha (-\frac 1 6)^n + \frac 3 5 (\frac 5 6 )^n$, and using $p_1 = 1$, I found: $\alpha = -3$.
Which leads to: $p_n = -3(-\frac 1 6)^n + \frac 3 5 (\frac 5 6 )^n$.
Well... But thanks to the recurrence formula, I have $p_2 = \frac 1 4$ and with this new formula, I have $p_2 = \frac 1 3$.
Where is my mistake?
| An alternate method of finding the solution of the difference equation is through generating functions. The following illustrates this method.
For the difference equation
\begin{align}
p_{n+1} + \frac{1}{6} \, p_n = \frac{1}{2} \, \left(\frac{5}{6} \right) ^{n-1}
\end{align}
it is seen that
\begin{align}
\sum_{n=0}^{\infty} p_{n+1} \, t^{n} + \frac{1}{6} \, \sum_{n=0}^{\infty} p_{n} \, t^{n} &= \frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{5}{6} \right) ^{n-1} \, t^{n} \\
\sum_{n=1}^{\infty} p_{n} \, t^{n-1} + \frac{1}{6} \, P(t) &= \frac{3}{5} \, \frac{1}{1 - \frac{5 t}{6}} \\
\frac{1}{t} \left( - p_{0} + P(t) \right) + \frac{P(t)}{6} &= \frac{18}{5} \, \frac{1}{6 - 5 t} \\
- 6 p_{0} + (t+6) P(t) &= \frac{108 t}{5(6- 5t)}
\end{align}
or
$$P(t) = \frac{3}{5} \frac{1}{1- \frac{5t}{6}} + \frac{p_{0} - 3}{1 + \frac{t}{6}}$$
where $P(t) = \sum_{n=0}^{\infty} p_{n} t^{n}$. From this the result becomes
\begin{align}
p_{n} = \frac{1}{2} \, \left(\frac{5}{6}\right)^{n-1} + \frac{(-1)^{n}(p_{0} - 3)}{6^{n}}.
\end{align}
Since $p_{1} = 1$ then $p_{0} = 0$ and
\begin{align}
p_{n} = \frac{3}{6^{n}} \, \left(5^{n-1} - (-1)^{n} \right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding a polynomial by divisibility
Let $f(x)$ be a polynomial with integer coefficients. If $f(x)$ is divisible by $x^2+1$ and $f(x)+1$ by $x^3+x^2+1$, what is $f(x)$?
My guess is that the only answer is $f(x)=-x^4-x^3-x-1$, but how can I prove it?
| In fact, there are infinitely many solutions. If $f(x)=-x^4-x^3-x-1$ and $g(x)$ is any polynomial, then
$$f(x)+g(x)(x^2+1)(x^3+x^2+1)$$
is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all the natural numbers which are coprimes to $n$ and are not a fermat witness to compositeness of $n$. The number $n=35$ is given. Find all the natural numbers $1 \leq a \leq n-1$ which are coprimes to $n$ and are not a fermat witness to compositeness of $n$.
Is it enough to say that we are looking for these numbers that satisfy the following conditions?
$n=35=5 \cdot 7$
$(a,n)=1$
$$a^{5-1} \equiv 1 \pmod 5 , \forall a \in \{1, \dots , 4\} \\ a^{7-1} \equiv 1 \pmod 7, \forall a \in \{1, \dots , 6\}$$
Can we say also the following for the inverse of the wanted number?
$(a,35)=1 \Leftrightarrow (a,5)=1 \text{ and } (a,7)=1$
$(a,5)=1 \Rightarrow \exists x_1, y_1 : ax_1+5y_1=1 \\ \Rightarrow ax_1 \equiv 1 \pmod 5 \Rightarrow a^4x_1^4 \equiv 1 \pmod 5 \\ \Rightarrow x_1^4 \equiv 1 \pmod 5 \Rightarrow 5 \mid x_1^4 -1=(x_1^2-1)(x_1^2+1)=(x_1-1)(x_1+1)(x_1^2+1) \\ \Rightarrow x_1\equiv 1 \pmod 5 \text{ or } x_1 \equiv -1\pmod 5 \text{ or } x_1^2 \equiv -1\pmod 5$
Similar for $7$.
| We want to find the $a$ in the interval $1\le a\le 34$ such that $a^{34}\equiv 1\pmod{35}$.
Note that since $a$ is relatively prime to $5$ and $7$, we have $a^4\equiv 1\pmod{5}$ and $a^{6}\equiv 1\pmod{7}$. It follows that $a^{12}\equiv 1$ modulo each of $5$ and $7$, and hence modulo $35$.
We have $a^{12}\equiv 1\pmod{35}$ and $a^{34}\equiv 1\pmod{35}$ if and only if $a^d\equiv 1\pmod{35}$, where $d=\gcd(12,34)=2$.
Thus the non-witnesses to primality are the solutions of $x^2\equiv 1\pmod{35}$. This congruence has $4$ solutions, obtained by splicing together the $2$ solutions of $x^2\equiv 1\pmod{5}$ with the $2$ solutions of $x^2\equiv 1\pmod{7}$ using the Chinese Remainder Theorem.
There are the two obvious solutions $x\equiv \pm 1\pmod{35}$. This gives $a=1$ and $a=34$. Now solve the system $x\equiv 1\pmod{5}$, $x\equiv -1\pmod{7}$. That gives the solution $a=6$. The final solution is obtained by taking the negative of $6$ modulo $35$, which gives $a=29$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
System of equations with complex numbers-circles The system of equations
\begin{align*}
|z - 2 - 2i| &= \sqrt{23}, \\
|z - 8 - 5i| &= \sqrt{38}
\end{align*}
has two solutions $z_1$ and $z_2$ in complex numbers. Find $(z_1 + z_2)/2$.
So far I have gotten the two original equations to equations of circles,
$(a-2)^2 +(b-2)^2=23$ and $(a-8)^2+(b-5)^2=38$.
From here how do I find the solutions?
Thanks.
| We can also treat this in a way that has little to do with complex numbers. The line connecting the centers of the circles at $ \ (2, \ 2) \ $ and $ \ (8, \ 5 ) \ $ has a slope of $ \ \frac{5 - 2}{8 - 2} \ = \ \frac{1}{2} \ $ and has the equation
$$ y \ - \ 2 \ = \ \frac{1}{2} \ (x - 2) \ \ \Rightarrow \ \ y \ = \ \frac{1}{2} \ x \ + \ 1 \ \ . $$
Using a theorem from classical geometry, the two intersection points of the two circles lie on a mutual chord, of which the line connecting the centers of the circles is its perpendicular bisector (a fact also used in some way by M. Wind and John). Thus, we expect the chord to have a slope of $ \ -2 \ $ .
A relation for the coordinates of the intersection points of the circles can be found by subtracting one circle equation from the other; thus,
$$ \ x^2 \ - \ 16x \ + \ 64 \ + \ y^2 \ - \ 10y \ + \ 25 \ = \ 38 $$
$$ - \ ( \ x^2 \ - \ 4x \ + \ 4 \ + \ y^2 \ - \ 4y \ + \ 4 \ = \ 23 \ ) $$
$$ -------------------- $$
$$ -12x \ + \ 60 \ - \ 6y \ + \ 21 \ = \ 15 $$
$$ \Rightarrow \ \ y \ = \ -2x \ + \ 11 \ \ . $$
The chord and its perpendicular bisector meet at
$$ \frac{1}{2} \ x \ + \ 1 \ = \ -2x \ + \ 11 \ \ \Rightarrow \ \ \frac{5}{2} \ x \ = \ 10 \ \ \Rightarrow \ \ x \ = \ 4 \ \ \Rightarrow \ \ y \ = \ 3 \ \ , $$
found by inserting the value for $ \ x \ $ into either of the linear equations we've determined. Hence, the midpoint of the chord containing the intersections of the circles lies at $ \ z \ = \ 4 \ + \ 3i \ $ in the complex plane.
Though we aren't asked to find them, we can also locate $ \ z_1 \ \ \text{and} \ \ z_2 \ $ . Inserting $ \ y \ = \ -2x \ + \ 11 \ $ into the first circle equation, we produce
$$ (x - 2)^2 \ + \ (-2x + 11 - 2)^2 \ = \ 23 \ \ \Rightarrow \ \ 5x^2 \ - \ 40x \ + \ 62 \ = \ 0 \ \ , $$
for which the roots are $ \ x \ = \ 4 \ \pm \ \frac{3}{5} \ \sqrt{10} \ $ . [Using the other circle equation instead yields exactly the same quadratic equation.]
We may then calculate $ \ y \ = \ -2 \ (4 \ \pm \ \frac{3}{5} \ \sqrt{10}) \ + \ 11 \ = \ 3 \ \mp \ \frac{6}{5} \ \sqrt{10} \ $ . Thus, we find
$$ z_1 \ = \ (4 \ - \ \frac{3}{5} \ \sqrt{10}, \ 3 \ + \ \frac{6}{5} \ \sqrt{10}) \ \ , \ \ z_2 \ = \ (4 \ + \ \frac{3}{5} \ \sqrt{10}, \ 3 \ - \ \frac{6}{5} \ \sqrt{10}) \ \ . $$
(Some brief work with right triangles then tells us that the length of the chord between $ \ z_1 \ \ \text{and} \ \ z_2 \ $ is $ \ \sqrt{\frac{36 + 144}{25} \ \cdot \ 10} \ = \ \sqrt{72} \ = \ 6 \ \sqrt{2} \ $ .)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Quick way to solve the system $\displaystyle \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$, $xy-x+y=118$. Consider the system
$$\begin{aligned} \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & = \frac{65}{36}, \\ xy -x +y & = 118. \end{aligned}$$
I have solved it by performing the substitutions $x-y=u$ and $xy=v$. Then I multiplied the first equation by $6^u$ and used $a^2-b^2=(a+b)(a-b)$ to find
$$(3^u+2^u)(3^u-2^u) = 65 \cdot 6^{u-2}.$$
By inspection I found $u=2$ and $v=120$. I solved the original system in $x,y$ and got the answers. Is there another quicker way to solve this without resorting to this sort of ninja inspection? I have found a second solution by solving $a^u +1/a^u = 65/36$, which assures $u=2$ but takes much more time.
Could there be a third way faster than these?
| If $\left(\dfrac32\right)^u=a$
If $a+\dfrac1a=\dfrac{65}{36}<2$
and for $a>0, a+\dfrac1a\ge2\sqrt{a\cdot\dfrac1a}=2,$
I assume $a-\dfrac1a=\dfrac{65}{36}$
$\iff36a^2-65a-36=0$
$a=\dfrac{65\pm\sqrt{65^2+4\cdot36^2}}{2\cdot36}=\dfrac94,-\dfrac49$
For real $u,$
$\left(\dfrac32\right)^u=\dfrac94=\left(\dfrac32\right)^2\implies u=2$
Consequently, $x-y=u=2,xy=118+x-y=120$
Replacing $x$ with $y+2,$ we have $y(y+2)=120\iff y^2+2y-120=0$
$y^2+2y-120=(y+12)(y-10)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
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