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Finding a function with certain properties I ran into a problem, and I'm not sure how to continue.
Problem: Let $f$ be a function such that $\sqrt {x - \sqrt { x + f(x) } } = f(x)$, for $x > 1$. In that domain, $f(x)$ has the form $\dfrac{a+\sqrt{cx+d}}{b}$, where $a$, $b$, $c$, $d$ are integers and $a$, $b$ are relatively prime. Find $a+b+c+d$.
So, I tried to cancel out the radicals, and got $(f(x))^4-2x(f(x))^2-f(x)+x^2-x=0$. Setting $y=f(x)$, I tried to apply the quadratic formula to find $x$ in terms of $y$. I got
$$
x=\frac{x+2xy^2 \pm \sqrt{(x+2xy^2)^2-4(y^4-y)}}{2}.
$$
From here, I tried simplifying the radical, but got
$$
x=\frac{x+2xy^2 \pm \sqrt{(x^2)(2y^2+1)^2-y(4y^3+1)}}{2}.
$$
I don't know if I factored it wrong, or if I'm missing something painfully obvious. Can I have a hint as to how to continue? It would be greatly appreciated.
| Putting $y=f(x)$, your equation becomes:
$y^4 - 2xy^2 + x^2 - x - y = 0$
This is hard to solve for $y$, but we have the suggestion that $y$ has the form $y=\frac{a+\sqrt{cx+d}}{b}$. This can be substituted into our polynomial and then simplified:
$\left(\frac{a+\sqrt{cx+d}}{b}\right)^4 - 2x\left(\frac{a+\sqrt{cx+d}}{b}\right)^2 + x^2 - x - \frac{a+\sqrt{cx+d}}{b} = 0$
$\implies\left(a+\sqrt{cx+d}\right)^4 - 2b^2x\left(a+\sqrt{cx+d}\right)^2 +b^4x^2 - b^4x - b^3\left(a+\sqrt{cx+d}\right) = 0$
$\implies a^4+4a^3\sqrt{cx+d}+6a^2(cx+d)+4a(cx+d)^{3/2}+(cx+d)^2-2a^2b^2x-4ab^2x\sqrt{cx+d}-2b^2x(cx+d)+b^4x^2-b^4x-ab^3-b^3\sqrt{cx+d}=0$
$\implies a^4+6a^2(cx+d)+(cx+d)^2-2a^2b^2x-2b^2x(cx+d)+b^4x^2-b^4x-ab^3=b^3\sqrt{cx+d}+4ab^2x\sqrt{cx+d}-4a^3\sqrt{cx+d}-4a(cx+d)^{3/2}$
$\implies a^4+6a^2(cx+d)+(cx+d)^2-2a^2b^2x-2b^2x(cx+d)+b^4x^2-b^4x-ab^3 = (b^3+4ab^2x-4a^3-4a(cx+d))\sqrt{cx+d}$
$\implies (a^4+6a^2(cx+d)+(cx+d)^2-2a^2b^2x-2b^2x(cx+d)+b^4x^2-b^4x-ab^3)^2 = (b^3+4ab^2x-4a^3-4a(cx+d))^2(cx+d)$
If you multiply this all out, and move everything to the left, you'll have a polynomial of the form $g(x)=M_4x^4 + M_3x^3 + M_2x^2 + M_1x + M_0 = 0$, where each of the $M_i$'s is a polynomial in $a$, $b$, $c$ and $d$. Since $g(x)=0$ for all $x>1$, we can say that $M_4=M_3=M_2=M_1=M_0=0$. If there is a solution to this system of five polynomials in four variables, then that's your answer.
In general, systems of polynomials can be difficult, but certain software packages allow for Groebner basis calculations, which can work, depending on the computer's capabilities and the user's patience.
One hopes there's an easier way to solve this, but the above is a sort of brute-force approach.
EDIT The solution posted here by Hagen von Eitzen using limits is considerably nicer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that $A+A^2+A^3$ has eigenvalue $\lambda+\lambda^2+\lambda^3$ where $\lambda$ is an eigenvalue of $A$
I know that $$P= \begin{pmatrix} 4 & 5 & 1 \\ -9 & -6 & 0 \\ 8 & 4 & 2 \end{pmatrix}$$
I also know that
$$D= \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix}$$
Where $a,b,c$ are eigenvalues of $A+A^2+A^3$
In this case the eigenvalues of :
$$A+A^2+A^3= \lambda + \lambda^2 + \lambda^3$$
Where $\lambda$ is an eigenvalue of A. Can someone prove this statement for me.
| We know that a matrix $M^n = PD^nP^{-1}$. Then $A = PDP^{-1}$ so
$$
A + A^2 + A^3 = PDP^{-1} + PD^2P^{-1} + PD^3P^{-1} = P(D+D^2+D^3)P^{-1}
$$
You have already found $P$ so once you have $D$ you just add.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does $\frac{\frac yx - \frac xy}{\frac1x - \frac1y} = x + y $ and $\frac{\frac xy - \frac yx}{\frac1x - \frac1y} = -(x + y)$? As the titles states: Why does $\frac{\frac yx - \frac xy}{\frac1x - \frac1y} = x + y $ and $\frac{\frac xy - \frac yx}{\frac1x - \frac1y} = -(x + y)$?
I am only able to find $\frac{y^2 - x^2}{y - x}$ and $\frac{x^2 - y^2}{y - x}$ respectively.
How I have done it so far:
$$\frac{\frac yx - \frac xy}{\frac 1x - \frac 1y} =
\frac{ \frac{y^2 - x^2}{xy} }{ \frac {y - x}{xy}} =
\frac{y^2 - x^2}{y - x}$$
$$\frac{\frac xy - \frac yx}{\frac 1x - \frac 1y} =
\frac{ \frac{x^2 - y^2}{xy}}{\frac {y - x}{xy}} =
\frac{x^2 - y^2}{y - x}$$
I reached $x + y$ and $-(x + y)$ respectively by using a sort of trial-and-error and looking at bigger/smaller than (each other) and negative/positive arbitrary values for $x $ and $y$ before I confirmed it on wolfram alpha. I can't wrap my head around how I would actually prove it though as I barely understand my own reasoning other than that it works. I got to $ x + y$ and $-(x + y)$ using common sense rather than math..
I hope I'm making sense and appreciate any and all help I can get!
| Notice that both numerators are differences of squares:
$$
\frac{y^2 - x^2}{y - x} = \frac{(y - x)(y + x)}{y - x} = y + x
$$
and likewise:
$$
\frac{x^2 - y^2}{y - x} = \frac{(x - y)(x + y)}{y - x} = \frac{-(y - x)(x + y)}{y - x} = -(x + y)
$$
| {
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Find the Sum of the Series: $1/(x+1) + 2/(x^2 + 1) + 4/(x^4 +1) +\cdots$ $n$ terms Find the sum of $n$ terms the following series:
$$\frac1{x+1} + \frac2{x^2 + 1} + \frac4{x^4 +1} +\cdots\qquad n\text{ terms}$$
$t_n$ seems to be $\dfrac{2^{n-1}}{x^{2^{n-1}} + 1}$
But after that I'm not sure as to how to proceed
| note: having just seen Andre's lovely solution this is very much a hammer-and-chisel job! however there was some devil in the detail, and i did learn something from the exercise!
define:
$$
D_m=\prod_{k=1}^m (1+x^{2^{m-1}}) = \sum_{k=0}^{2^m-1}x^k
$$
and
$$
N_m = \sum_{k=1}^{2^m-1} (2^m-k)x^{k-1} \tag{1}
$$
and now define the sum
$$
S_m=\frac{N_m}{D_m}
$$
then
$$
S_m+\frac{2^m}{x^{2^m}+1} = \frac{(x^{2^m}+1)N_m + 2^mD_m}{D_{m+1}}
$$
from (1) we have
$$
N_{m+1} = \sum_{k=1}^{2^{m+1}-1} (2^{m+1}-k)x^{k-1} \\
= \sum_{k=1}^{2^m-1} (2^m-k)x^{k-1} +\sum_{k=1}^{2^m-1} 2^m x^{k-1} + \sum_{k=2^m}^{2^{m+1}-1} (2^{m+1}-k)x^{k-1} \\
= N_m + 2^m(D_m -x^{2^m-1}) + \sum_{k=0}^{2^m-1} (2^m-k)x^{2^m+k-1} \\
= (1+x^{2^m}) N_m + 2^m D_m
$$
hence, with the obvious definition of $S_m$ we have the inductive step:
$$
S_m = \frac{N_m}{D_m} \Rightarrow S_{m+1} = \frac{N_{m+1}}{D_{m+1}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the integral $\int_{0}^{\infty }\frac{1}{(4x-3)(4x-1)}\,dx$. Which method that will be effective for solving this integral?
| The integral does not converge because of the singularities.
To compute the integral between $a$ and $\infty$ for $a>\frac 34$, you can use the following method:
\begin{align}
\int_{a}^{\infty }\frac{1}{(4x-3)(4x-1)} dx
&= \lim_{N\to\infty} \int_{a}^N \frac{1}{(4x-3)(4x-1)} dx
\end{align}
Now use the fact that $$
\frac{1}{(4x-3)(4x-1)} =
\frac 12 \left(
\frac 1{4x-3} - \frac 1{4x-1}
\right)
$$
so that
\begin{align}
\int_{a}^N \frac{dx}{(4x-3)(4x-1)}
&= \frac 12 \left(
\int_{a}^N \frac {dx}{4x-3} dx - \int_a^N \frac {dx}{4x-1} dx
\right)
\\&= \frac 18\left(
\log \frac{|4N-3|}{|4a-3|} - \log \frac{|4N-1|}{|4a-1|}
\right)
\\&= \frac 18\left(
\log \frac{4N-3}{4N-1} - \log \frac{4a-3}{4a-1}\right)
\to -\frac18 \log \frac{4a-3}{4a-1}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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$ab$ divides $3^a+1$ and $3^b+1$ Find all positive integers $a,b$ such that $ab$ divides $3^a+1$ and $3^b+1$.
It is clear that $3$ cannot divide either $a$ or $b$, because $3$ doesn't divide $3^a+1$ or $3^b+1$.
$(a,b)=(1,1),(2,1),(1,2)$ all work. If $b=1$, the condition reduces to $a$ divides $3^a+1$ and $4$, so $a=1,2,4$, but this has already been covered.
[Source: Korean competition problem]
| Let us start with a lemma.
Lemma For integer $m>2$, $gcd(m^{a}-1,m^{b}-1)=m^{gcd(a,b)}-1$
Proof Suppose wlg that $a>b$. Then
\begin{align*}
gcd(m^{a}-1,m^{b}-1)= \enspace&gcd(m^{a}-1,m^{a-b}(m^{b}-1))\\
=\enspace&gcd(m^{a}-1,m^{a-b}-1)\\
=\enspace&gcd(m^{b}-1,m^{a-b}-1)
\end{align*}
since $gcd(x,y)=gcd(x,x-y)$. By reiterating, $gcd(m^{a}-1,m^{b}-1)=gcd(m^{b}-1,m^{r}-1)$ with $r$ being the rest of the euclidiean division of $a$ by $b$. We can proceed so with Euclides' algortihm until we get $gcd(m^{a}-1,m^{b}-1)=gcd(m^{dk}-1,m^{d}-1)=m^{d}-1$ where $d=gcd(a,b)$.
Since $3^{a}+1$ divides $9^{a}-1$, here we get $ab | 9^{d}-1$ with $d=gcd(a,b)$. We shall prove that this implies $d=1$, hence solving the problem since then $ab | 8$. Suppose $d>1$ and let $p$ be the smallest prime factor of $d$. Then $9^{d}=1 \pmod p$. The order of $9$ must divide both $p-1$ and $d$ so that, since $p$ is the smallest prime factor of $d$, $9$ has order $1$ and $p=2$.
Now note that $3=-1 \pmod 4$ so that if $a$ is even $3^{a}+1=2 \pmod 4$.Consequently $a$ and $b$ can't be both even, otherwise $ab=0 \pmod 4$ and $3^{a}+1=2 \pmod 4$ which is impossible if $ab|3^{a}+1$. So $d$ cannot have $2$ as a prime factor. We get the desired contradiction, showing that $d=1$.
| {
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Finding $\int_{1}^{\infty} \frac{x 3^x}{(3^x-1)^2}$ How do we prove that $$\int_{1}^{\infty} \frac{x \cdot 3^x}{(3^x-1)^2} \mathrm{d}x=\dfrac{3\log{3}-2\log{2}}{2\log^2{3}}$$
I have tried some substitutions such as $3^x=t$, but it didn't work out. The denominator is already factored. I do not know what to do next. Please help me out. Thank you.
| Integrating by parts ($u=x$, $dv=\frac{3^x}{(3^x-1)^2}dx$), $$\int x\frac{3^x}{(3^x-1)^2}dx=x\int\frac{3^x}{(3^x-1)^2}dx-\int\left[\frac{dx}{dx}\int\frac{3^x}{(3^x-1)^2}dx\right]dx$$
Now for $\displaystyle\int\dfrac{3^x}{(3^x-1)^2}dx,$ set $3^x-1=u$
Finally $\displaystyle\int\dfrac{dx}{3^x-1},$ set $3^x-1=v$
| {
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How to solve $\left|\frac{x+4}{ax+2}\right| > \frac1x$ How to solve:
$$\left|\frac{x+4}{ax+2}\right| > \frac{1}{x}$$
What I have done:
I) $x < 0$:
Obviously this part of the inequation is $x\in(-\infty, 0), x\neq \frac{-2}{a}$
II) $x > 0$:
$$\left|\frac{x+4}{ax+2}\right| > \frac1x$$
$$\frac{|x+4|}{|ax+2|} > \frac1x$$
because $x > 0$ we can transform it to:
$x^2 + 4x > |ax + 2|$
$(x^2 + 4x)^2 > (ax + 2)^2$
$(x^2 + 4x)^2 - (ax + 2)^2 > 0$
$(x^2 + 4x - ax - 2) (x^2 +4x +ax + 2) > 0$
$(x^2 + (4-a)x - 2)(x^2 + (4+a)x + 2) > 0$
I am kind of stuck here. I also need to discuss the solution for various values of the parameter a. What is the easiest way out of this step? Something with the discriminants, perhaps?
| You have already identified and made use of the points where sides of the inequality are undefined. So all that is needed is to solve for equality and then make determinations about the subintervals between the points of equality and points where values are undefined.
Equality happens when $x=\left|\frac{ax+2}{x+4}\right|$. Note that $x$ and $x+4$ are certainly positive in any solution. So either
$$\begin{align}
x&=\frac{ax+2}{x+4}&{-x}&=\frac{ax+2}{x+4}\\
\end{align}$$
Visually, we are intersecting a diagonal line (either of slope $1$ or $-1$) with a simple rational function that has a singularity at $-4$ and a $y$-intercept at $\frac12$. The first of these equations clearly always has precisely one solution for $x$ with $x>0$ and $ax+2>0$ no matter what $a$ is (it's clear if you are considering the graphs). That solution must be the larger of the two solutions from the quadratic equation $x^2+(4-a)x-2=0$, namely $x=\frac{a-4+\sqrt{(a-4)^2+8}}{2}$.
The second of these equations has no positive solutions for $x$ unless $a$ is large enough negative, to make the initial slope of $\frac{ax+2}{x+4}$ steep enough negative to meet $y=-x$. When $a$ is large enough negative, these two curves will meet with multiplicity $2$, and then for larger negative $a$, there will be two points of intersection. The critical $a$ happens when $x^2+(4+a)x+2=0$ has a doubled root and $a$ is large enough negative: when $a=-\sqrt{8}-4$. For $a\leq-\sqrt{8}-4$, the two solutions are $x=\frac{-(a+4)\pm\sqrt{(a+4)^2-8}}{2}$. Using the $+$ sign from the $\pm$, this solution clearly has $ax+2<0$. But with the $-$ sign, for $a$ too negative, $ax+2$ might reach $0$.
$$\begin{align}
a\frac{-(a+4)-\sqrt{(a+4)^2-8}}{2}+2&=0\\
-(a^2+4a)-a\sqrt{(a+4)^2-8}+4&=0\\
-a\sqrt{(a+4)^2-8}+8&=(a+2)^2\\
\end{align}$$
A calculus investigation reveals that this equation in fact has no solutions with $a\leq-\sqrt{8}-4$.
So in summary, the list of critical values for $x$ for your relation is $\left\{-{\frac2a},0,\frac{a-4+\sqrt{(a-4)^2+8}}{2}\right\}$ And additionally, if $a\leq-\sqrt{8}-4$, $\left\{\frac{-(a+4)+\sqrt{(a+4)^2-8}}{2},\frac{-(a+4)-\sqrt{(a+4)^2-8}}{2}\right\}$
At these $3,4$, or $5$ $x$-values, the sign of $\left|\frac{x+4}{ax+2}\right| - \frac{1}{x}$ may change. Depending on what $a$ is, the critical values are ordered in different ways. But once you order them, you can test with a single value within each of the subintervals whether the difference is positive or negative.
| {
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"timestamp": "2023-03-29T00:00:00",
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Unable to find the sum of a series I am trying to find the sum of the following series:
$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}}$$
which I rewrote as
$$\sum_{n=1}^{\infty} \left(\frac{1}{9^n}+
\left(\frac{7}{9}\right)^n\right)$$
I am assuming that it is a geometric series and the initial value is
$$a_1=\frac{1}{9} + \frac{7}{9}$$
I also see that
$$a_2 = \frac{1}{9^2} + \frac{7^2}{9^2}$$
I know that in a geometric series the first term is $a$ and the second term is $ar$.
This allows me to see that
$$\left(\frac{1}{9}+\frac{7}{9}\right)r=\frac{1}{9^2}+\frac{7^2}{9^2}$$
which when solved for $r$ gives the value $\frac{25}{36}$.
Using the formula to find the sum of a geometric series $\frac{a}{1-r}$, I find that the sum is equal to $\frac{32}{11}$.
But this value is incorrect and the sum is actually $\frac{29}{8}$. How does one find that value?
| Hint: Try re-doing it as two separate series, $\sum(1/9)^n$ and $\sum(7/9)^n.$ The first has first term $1/9$ and common ratio $1/9$ while the second has first term $7/9$ with common ratio $7/9.$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Check: For each of the following permutations $\rho$ in $S(6)$ write $\rho$ as the product of transpositions For each of the following permutations $\rho$ in $S(6)$ write $\rho$ as the product of transpositions:
*
*$\begin{pmatrix} 1 & 3 & 5 & 4 & 6 & 2 \end{pmatrix}=\begin{pmatrix} 1 & 2 \end{pmatrix}\begin{pmatrix} 1 & 6 \end{pmatrix}\begin{pmatrix} 1 & 4 \end{pmatrix}\begin{pmatrix} 1 & 5 \end{pmatrix}\begin{pmatrix} 1 & 3 \end{pmatrix}$
*$\begin{pmatrix} 1 & 2 & 5 & 4 \end{pmatrix} \begin{pmatrix} 1 & 5 &
3 & 6 & 2 \end{pmatrix}=\begin{pmatrix} 1 & 2
\end{pmatrix}\begin{pmatrix} 2 & 5 \end{pmatrix}\begin{pmatrix} 5 &
4 \end{pmatrix}\begin{pmatrix} 1 & 5 \end{pmatrix}\begin{pmatrix} 5
& 3 \end{pmatrix}\begin{pmatrix} 3 & 6 \end{pmatrix}\begin{pmatrix}
6 & 2 \end{pmatrix}$
Are these correct?
| I'm assuming you're using the right-sided convention, in which case they are right.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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a two-variable cyclic power inequality $x^y+y^x>1$ intractable by standard calculus techniques If $x$ is in the open interval (0,1) and so is $y$, prove that $$x^y+y^x>1$$
A direct two-variable application of maxima and minima seems difficult.
| Let's suppose(if it's not true, change the place of $x,y$ in the following arguments) $$\dfrac{x^{y-1}}{y^{x-1}} \geq 1$$
Then we have $ (x^y + y^x)^\frac{1}{x} = (y^x (1+\dfrac{x^y}{y^x}))^\frac{1}{x} = y (1+\dfrac{x^y}{y^x})^\frac{1}{x} $
Then since $\frac{1}{x} > 1$, we have $$(x^y + y^x)^\frac{1}{x} = y (1+\dfrac{x^y}{y^x})^\frac{1}{x} \ge y(1+\dfrac{x^y}{y^xx}) = y(1+\dfrac{x^{y-1}}{y^x}) = y + \dfrac{x^{y-1}}{y^{x-1}} \geq y+1 > 1$$
since $\frac{1}{x} > 1$, we have $x^y + y^x > 1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Polynomial function question If $f(x)$ is equal to $\frac{1}{x^3 + 3x^2 + x}$, find the smallest value of $n$ for which $f(1) + f(2) + ... F(n) = \frac{503}{2014}$. I tried noting that first initial values of f sum to $\frac{1}{5}$, and $503$ divided by $2014$ is approx $\frac{1}{4}$, so we're aiming for values after $f(2)$ to be equal to $\frac{1}{12}$, we can test values and plug in values for $x$ until we find such a number $n$. This seems algorithmic and I'd try searching for a smarter solution. Is there any other solution? Thanks.
| Note that
$$0 < f(x) = \frac{1}{x^3 + 3x^2 + x} \leq \frac{1}{x} + \frac{1}{3x} + \frac{1}{x} = \frac{7}{3x},$$
since $x \in \mathbb{N}$.
Now,
$$f(1) = \frac{1}{5}$$
$$f(2) = \frac{1}{22}$$
$$f(3) = \frac{1}{57}$$
Consequently:
$$f(1) + f(2) = \frac{27}{110} = 0.2\overline{45}$$
$$f(1) + f(2) + f(3) = \frac{1649}{6270} > 0.262998405 > \frac{503}{2014} \approx 0.24975$$
Consequently, there is no such $n \in \mathbb{N}$ such that
$$f(1) + f(2) + \ldots + f(n) = \frac{503}{2014}$$
and
$$f(x) = \frac{1}{x^3 + 3x^2 + x}.$$
Perhaps there is a typo in the definition of the function $f$? =)
| {
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} |
A thinking problem of limit from my teacher. Please find the limit:$$\mathop {\lim }\limits_{n \to \infty } n\left[ {{{\left( {\frac{1}{\pi }\left( {\sin \left( {\frac{\pi }{{\sqrt {{n^2} + 1} }}} \right) + \sin \left( {\frac{\pi }{{\sqrt {{n^2} + 2} }}} \right) + \cdots+ \sin \left( {\frac{\pi }{{\sqrt {{n^2} + n} }}} \right)} \right)} \right)}^n} - \frac{1}{{\sqrt[4]{e}}}} \right].$$
| Since $n$ is supposed to be large, then the argument of the sine is small. So, start with $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ Now, replace $x$ by $\frac{\pi }{\sqrt{k+n^2}}$ and then $$\frac{1 }{\sqrt{k+n^2}}=\frac{1}{n}-\frac{k}{2 n^3}+\frac{3 k^2}{8
n^5}+O\left(\left(\frac{1}{n}\right)^6\right)$$ So, $$\sin \left(\frac{\pi }{\sqrt{k+n^2}}\right)\approx\frac{\pi }{n}-\frac{\frac{\pi k}{2}+\frac{\pi ^3}{6}}{n^3}$$ Now, sum over $k$ from $1$ to $n$ and the sum of sines is $$\pi-\frac{\pi }{4 n}-\frac{\pi \left(3+2 \pi ^2\right)}{12 n^2}\approx \pi-\frac{\pi }{4 n}$$ So, what is inside brackets write $$\left(1-\frac{1}{4 n}\right)^n-\frac{1}{\sqrt[4]{e}}$$
I am sure that you recognize the first term and that you can take from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find an equation of the tangent line to the curve
Find an equation of the tangent line to the curve $2·(x^2+y^2)^2=25·(x^2−y^2)$ at the point
$(3,1)$.
Any hints?
| The more general way to solve this type of problem is with implicit differentiation:
We have $2(x^2+y^2)^2=25(x^2-y^2)$. Differentiating both sides w.r.t. $x$, we get: $$4(x^2+y^2)(2x+2yy')=50x-50yy'$$ $$\implies (8(x^2+y^2)y+50y)y'=50x-4(x^2+y^2)\cdot 2x$$ $$\implies y'=\frac {50x-8x^3-8xy^2}{8x^2y+8y^3+50y}$$ $$\implies y'=-\frac{x(4x^2+4y^2-25)}{y(4x^2+4y^2+25}$$
Because we know that derivatives are just the slopes of tangent lines to graphs, we know that the slope of this implicit function at point $(3,1)$ is $y'|_{(3,1)}=-\frac{(3)(4(9)+4(1)-25)}{(1)(4(9)+4(1)+25)}=-\frac{9}{13}$.
We then just plug in that slope and point into the point-slope equation of a line: $$y-y_0=m(x-x_0)$$ and we have the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find minima of the multivariable function $\frac{4}{x^2+y^2+1}+2 xy$ $$f(x,y)=\frac{4}{x^2+y^2+1}+2 xy \\ \text{within the domain: }1/5\leq x^2+y^2\leq 4$$
I am able to find the maximum of the function at $x^2+y^2=4$ by substituting x,y for $\cos(t)$ and $\sin(t)$ and I am able to start working for a solution in the lower boundary:
$$ x=\frac{1}{\sqrt5}\cos(t), \ y=\frac{1}{\sqrt5}\sin(t) \\
\frac{4}{\frac{6}{5}}+\frac{2}{\sqrt5}(\cos(t)\sin(t))\\
\frac{df}{dt}=\frac{2}{\sqrt5}\cos(2t)=0\\
\tan(t)=1\\
x=\frac{1}{\sqrt5}\lvert\cos(\frac{\pi}{4})\rvert \ , \ \ y=\frac{1}{\sqrt5}\lvert\sin(\frac{\pi}{4})\rvert \\
x=\pm\frac{1}{\sqrt{10}},y=\pm\frac{1}{\sqrt{10}}$$
When i resubstitute this into the original function I get the wrong answer.
$$f\left(\frac{1}{\sqrt{10}},\frac{1}{\sqrt{10}}\right)=\frac{4}{\frac{12}{10}}+\frac{2}{10}=53/15
$$
| You know that $1/5\leq x^2+y^2\leq 4$, so:
$$\frac{4}{x^2+y^2+1} \geq \frac{4}{4+1}=\frac{4}{5}$$
In addiction by AM-GM:
$$2|xy| \leq x^2+y^2 \leq 4$$
$$|xy| \leq 2$$
And:
$$2xy \geq -4$$
So let's try to find such a $x,y$ that:
$$xy=-2,x^2+y^2=4$$
But for $x=-\sqrt{2}$ and $y=\sqrt{2}$ both equalities are satisfy, so minimum is:
$$\frac{4}{5}-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The limit of a difference between a cube root and square root I've been trying to evaluate the limit of $$\lim_{x \to \infty}[(x^3+x^2+1)^{1/3}-(x^2+x)^{1/2}]$$ I've tried using the property $$x^6-y^6 =(x-y)(x^5+x^4y+\cdots+xy^4+y^5)$$
where $x = (x^3+x^2+1)^{1/3}$ and $y=(x^2+x)^{1/2}$. But I can't seem to 'clean up' the denominator to get in a position to take the limit. Many thanks in advance.
| Using your property, you have
\begin{align}
(x^3+x^2+1)^{1/3}-(x^2+x)^{1/2}&=\frac{(x^3+x^2+1)^2-(x^2+x)^3}{(x^3+x^2+1)^{5/3}+\cdots+(x^2+x)^{5/2}}\\ \ \\
&=\frac{-x^5-2x^4+x^3+2x^2+1}{(x^3+x^2+1)^{5/3}+\cdots+(x^2+x)^{5/2}}\\ \ \\
&=\frac{-1-2/x+1/x^2+2/x^3+1/x^5}{(1+1/x+1/x^3)^{5/3}+\cdots+(1+1/x)^{5/2}}\\ \ \\
&\to-\frac16
\end{align}
(where in the last equality you divide numerator and denominator by $x^5$, and then use the fact that there are six terms in the denominator).
Also here is the way to do this limit that comes naturally to me. From the binomial series, $(1+a)^b\simeq 1+\frac ab$ for small $a$. Then, for small $x$,
\begin{align}
(x^3+x^2+1)^{1/3}-(x^2+x)^{1/2}&=x\,\left[ (1+1/x+1/x^3)^{1/3}-(1+1/x)^{1/2} \right]\\ \ \\
&\simeq x\,\left[1+\frac1{3x}+\frac1{3x^3}-1-\frac1{2x} \right]\ \\ \ \\
&=x\,\left[-\frac{1}{6x}+\frac1{3x^3} \right]=-\frac16+\frac1{3x^2}\\ \ \\
&\to-\frac16.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove complex equation Prove the following $$\frac{1}{z-1}*\frac{1}{z^n}= \dfrac{1}{z-1} - \sum_{k=1}^n\frac{1}{z^k}$$
for any integer n greater than 0. DO NOT USE ....
I believe that I can use mathematical induction. Any help would be greatly appreciated.
| You want to show that
$\frac{1}{z-1}*\frac{1}{z^n}= \dfrac{1}{z-1} - \sum_{k=1}^n\frac{1}{z^k}$
Multiply by
$z-1$ and it becomes
$\frac{1}{z^n}
= 1 - (z-1)\sum_{k=1}^n\frac{1}{z^k}
$
or
$1-\frac{1}{z^n}
= (z-1)\sum_{k=1}^n\frac{1}{z^k}
$.
But
$\begin{array}\\
(z-1)\sum_{k=1}^n\frac{1}{z^k}
&=z\sum_{k=1}^n\frac{1}{z^k}-\sum_{k=1}^n\frac{1}{z^k}\\
&=\sum_{k=0}^{n-1}\frac{1}{z^k}-\sum_{k=1}^n\frac{1}{z^k}\\
&=1-\frac1{z^n}\\
\end{array}
$
QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
area of figure in sector of intersecting circles I need to find an area of blue part of figure APBC. I draw line segments between B and C, between C and A, and got equilateral triangle. I'm stuck here. Please help. Thanks.
|AB| = a, P is midpoint of segment AB
| I believe we can use Descartes Circle Theorem. We can identify 4 circles:
*
*the circle $O_A$ with center at $A$
*circle with diameter $\overline{AP}$
*circle with diameter $\overline{PB}$
*unknown circle
Unfortunately, these 4 circles are not mutually tangent since the circle with diameter $\overline{AP}$ is not tangent to the circle $O_A$ with center at $A$.
The general Apollonius problem involves finding a circle tangent to 3 other circles (regardless of mutual tangency).
One, rather boring, solution is to solve 3 quadratic equations. Let $x,y,r$ be the center and radius of the unknown circle.
$$ (x-x_k)^2 + (y-y_k)^2 = (r \pm r_k)^2$$
where $k = 1,2,3$. Using the diagram, we put in the centers and radii.
$$ x^2 + y^2 = (r \pm 4)^2$$
$$ (x-1)^2 + y^2 = (r \pm 1)^2$$
$$ (x-3)^2 + y^2 = (r \pm 1)^2$$
In order to get the signs right, our circle is inside the first circle and outside the other two.
$$ x^2 + y^2 = (r - 4)^2$$
$$ (x-1)^2 + y^2 = (r + 1)^2$$
$$ (x-3)^2 + y^2 = (r + 1)^2$$
The fact that $x = 2$ should have been clear from the diagram. Writing the system again:
$$ 2^2 + y^2 = (r - 4)^2$$
$$ 1^2 + y^2 = (r + 1)^2$$
We can subtract the first and second equations, it is possible to solve the linear equation for $r$:
$$ 3 = 2^2 - 1^2 = (r-4)^2 - (r+1)^2 = (r^2-8r + 16) - (r^2 + 2r + 1) = -10r + 15$$
The answer is $r = \tfrac{6}{5}$. If we rescale so the radius with center $0$ is the unit circle, we get $\boxed{r=\tfrac{3}{10}}$
A more interesting solution would exploit the symmetry of the problem or use inversion somehow.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Non-linear system with all trajectories converging on the line $x=0$, rather than $(2,0)$? I have the following nonlinear system:
$$\begin{pmatrix}\dot{y}_1\\\dot{y}_2\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$
Which I set up to $F=\dot{y}$
Giving the jacobian of transformation being:
$$\dot{F}= \begin{pmatrix}2&0\\2y_1&0\end{pmatrix}=\begin{pmatrix}0&0\\2&0\end{pmatrix}\begin{pmatrix}F_1\\F_2\end{pmatrix}+\begin{pmatrix}2\\0\end{pmatrix}$$
Which gives me the eigenvalues $\lambda=0^2=0,0$ meaning a degenerate node since we have only the eigenvector:
$$F^{(1)}=\begin{pmatrix}0\\1\end{pmatrix}$$
So we have a straight line going up and down from $(2,0)$ I would have thought, but instead it is still coming from $x=0$ as shown below(from here):
| It would be better to attack the problem directly as Amzoti has said:
$$\begin{pmatrix}\dot{y}_1\\\dot{y}_2\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$
We want the critical points of this system, which we can find by taking:
$$\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$
$(0,\alpha)$ are the spots of the critical points, thus there are infinite critical points at $x=0$
Furthermore, you have made a mistake in your Jacobian.
Since we are taking $F=\dot{y}$, we need to be very clear that this means:
$$\begin{pmatrix}F_1\\F_2\end{pmatrix}=\begin{pmatrix}2y_1\\y_1^2\end{pmatrix}$$
And hence:
$$\dot{F}=\begin{pmatrix}\dot{F}_1\\\dot{F}_2\end{pmatrix}=\begin{pmatrix}2&0\\2y_1&0\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}F_1\\F_2\end{pmatrix}+\begin{pmatrix}2\\0\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simple Trigonometric Problem Here is a Trig problem, and I am missing some understanding of some basic algebraic rule:
$if \sin\theta= \frac{m^2 +2mn}{m^2+2mn+2n^2}$
then prove that $\tan\theta = \frac {m^2 +2mn}{2mn+2n^2},$ In the picture below can anyone explain me how did they achieve the step from 1 to 2 highlighted in red.
| $$\begin{align}
\sqrt{1-\frac{(m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}&=\sqrt{\frac{(m^2+2mn+2n^2)^2-(m^2+2mn)^2}{(m^2+2mn+2n^2)^2}}\\
&=\sqrt{\frac{(2n(m+n))^2}{(m^2+2mn+2n^2)^2}}\tag{1}\\
&=\frac{2n(m+n)}{m^2+2mn+2n^2}\\
\end{align}$$
$\text{Explanation } 1$ $$(a+b)^2-a^2 = 2ab+b^2=b(2a+b)$$
Where $a=m^2+2mn$ , $b=2n^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding lowest possible value please help me with this problem:
Find the lowest possible value of
$$
x+y^3
$$
where both x and y are positive and x*y=1.
I know how to solve this one using my method, but I was suggested to use geometric and arithmetic mean. I have no idea how to solve this using them.
| Since $xy = 1$, observe that $x + y^3 = y^{-1} + y^3 = y^3 + \frac{y^{-1}}{3} + \frac{y^{-1}}{3} + \frac{y^{-1}}{3} \geq 4\sqrt[4]{\frac{1}{3^3}}$.
The equality holds if and only if $y^3 = \frac{y^{-1}}{3}$, that means for $y = \frac{1}{\sqrt[4]{3}} > 0$, therefore your lowest possible value of $x + y^3$ is $4\sqrt[4]{\frac{1}{3^3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Problem involving cube roots of unity Given that $$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2\omega^2\;\;\;\;\;(1)$$ $$\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2\omega\;\;\;\;\;(2)$$
Find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=?\;\;\;\;\;\;(3)$$
I tried adding the given two equations, and simplified them. I'll show the working for one term here. $$\frac{1}{a+\omega}+\frac{1}{a+\omega^2}=\frac{2a-1}{a^2-a+1}=\frac{(2a-1)(a+1)}{(a^3+1)}=\frac{2a^2+a-1}{a^3+1}$$
$$\frac{1}{a+1}=\frac{a^2-a+1}{a^3+1}=\frac{1}{2}\left[\frac{(2a^2+a-1)-3(a-1)}{a^3+1}\right]=\frac12\left[\frac{2a^2+a-1}{a^3+1}\right]-\frac32\left[\frac{a-1}{a^3+1}\right]$$
Now, I cannot eliminate that extra term which I get at the end of the above expression. Is there hope beyond this? Or is there a better alternative?
| My original answer did not use the nice symmetry. Kelenner gave a better answer. Based on that, I have made yet another answer, using essentially the same method as Kelenner did, but with a different definition of the same function which is (for me) easier to understand. When I first wrote this answer, I did not realize I was using exactly the same function, otherwise I would not have written this down, but now that it is here I'll leave it, because it might be easier to read for some people.
Define the function $Q$ by
$$Q(x)=\left(\left(\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}\right) x-2\right)(a+x)(b+x)(c+x)$$
From the given equations:
$$Q(\omega)=0,\quad Q(\omega^2)=0.$$
The function $Q$ can be much simplified (with simple manipulations):
$$Q(x)=x^3-2abc$$
Because $\omega$ and $\omega^2$ are zeroes of $Q$, $Q$ must have a factor $(x^2+x+1)$. Following symbolic division, this can only be true if $(1-2abc)=0$, and then
$$Q(x)=(x^2+x+1)*(x-1)$$
So, $Q(1)=0$, and from the original definition of $Q$ it must follow that
$$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=2$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
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Evaluating $\displaystyle\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x} \frac{t^3\ln(1-t)}{t^4 + 4}\,dt$
Evaluate the following limit: $$\lim_{x\space\to\space0} \frac{1}{x^5}\int_0^{x}
\frac{t^3\ln(1-t)}{t^4 + 4}\,dt$$
Any advice on how to tackle this problem ?
| Substituting $t = xu$, we get
$$\int_0^x dt \frac{t^3 \log{(1-t)}}{t^4+4} = x^4 \int_0^1 du \frac{u^3 \log{(1-x u)}}{4+x^4 u^4} $$
so the limit is,
$$\lim_{x\to 0} \frac1{x}\int_0^1 du \frac{u^3 \log{(1-x u)}}{4+x^4 u^4} = \lim_{x\to 0} \frac1{x}\int_0^1 du \frac{-x u^4}{4} = -\frac1{20}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the limit $\lim_{x\to 0^-}| \left( 1+x^{3} \right)^{1/2}-1-x^{5} |/(\sin x-x)$ I am studying for my first calculus exam (well, it's half an exam), and of course we have to solve limits, without using L'Hospital rule, and using asymptotic analysis.
I can't solve this one $$\lim_{x\rightarrow 0^{-}}\frac{\left| \left( 1+x^{3} \right)^{\frac{1}{2}}-1-x^{5} \right|}{\sin x-x}$$
Can you guide me through the solution? Thanks a lot.
| Look at
$f(x)
=\frac{ ( 1+x^{3} )^{\frac{1}{2}}-1-x^{5} }{\sin x-x}
$.
By "asymptotic analysis",
as $x \to 0$,
$(1+x^3)^{\frac12}
\approx 1+\frac12 x^3 + \frac{\frac12 (-\frac12)}{2}(x^3)^2
= 1+\frac12 x^3 - \frac18 x^6
$
and
$\sin(x)
\approx x-\frac{x^3}{6}+\frac{x^5}{120}
$
so
$\sin(x)-x
\approx -\frac{x^3}{6}+\frac{x^5}{120}
$.
Therefore,
using the fact that
$ax^n+bx^m
\approx ax^n
$
as $x \to 0$
if $1 \le n < m$
and
$a\ne 0$,
$f(x)
\approx \frac{(1+\frac12 x^3- \frac18 x^6)-1-x^5}{-\frac{x^3}{6}+\frac{x^5}{120}} =\frac{\frac12 x^3 - \frac18 x^6-x^5}{-\frac{x^3}{6}+\frac{x^5}{120}}
\approx \frac{\frac12 x^3}{-\frac{x^3}{6}}
=-3
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\log_a b = \log_\sqrt a \sqrt b$ can be a basic logarithm law? Does the following equation is true for all $ a,b\in{\mathbb R}$?
$$\log_a b = \log_\sqrt a \sqrt b$$
I have tried to proove this, and I didnt find any contradiction. Is it true?
EDIT
Thanks guys for the answers. I've just found a different proof without using $\ln$, for the record, here it is :
$$\log_\sqrt{a} \sqrt{b} = \frac{ \log_a \sqrt{b}}{\log_a \sqrt{a}} = \frac{\frac{1}{2} \log_a b}{\frac{1}{2} \log_a a} = \frac{\log_a b}{1} = \log_a b$$
Second EDIT
Is it true to generelize this rule, like this?
$$\log_{a^n} {b^n} = \frac{ \log_a {b^n}}{\log_a {a^n}} = \frac{n \log_a b}{n \log_a a} = \frac{n\log_a b}{n} = \log_a b$$
| It's true for all positive $b$, $a \neq 1$ as
$$\log_\sqrt{a} \sqrt{b} = \frac{ \ln \sqrt{b}}{\ln \sqrt{a}} = \frac{\frac{1}{2} \ln b}{\frac{1}{2} \ln a} = \frac{\ln b}{\ln a} = \log_a b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Maximizing sin(a-b) given a trig relation Suppose $a$, $b$ are acute angle measures such that $\tan a = 5\tan b$. Find the maximum value of $\sin(a-b)$.
$\sin(a-b)=4\sin b \cos a$, but I don't know what to do from here.
| Since $a,b\in(0,\frac\pi2)$ and $\tan$ is positive and strictly increasing there, the assumption $\tan a=5\tan b$ implies $a>b$. Thus $a-b\in(0,\frac\pi2)$ also, so $\sin(a-b)>0$, and we might as well maximize $\sin^2(a-b)$ instead. And since
$$ \sin^2(a-b) = 1 - \frac1{1+\tan^2(a-b)} $$
we might as well maximize $\tan^2(a-b)$ instead.
$$ \tan^2(a-b)
= \left(\frac{\tan a-\tan b}{1+\tan a\tan b}\right)^2
= \left(\frac{4\tan b}{1+5\tan^2 b}\right)^2
= \left(\frac{2}{\frac12\bigl(\frac1{\tan b}+5\tan b\bigr)}\right)^2
\le \left(\frac{2}{\sqrt5}\right)^2 = \frac45 $$
by AM/GM, with equality iff $b=\arctan\frac1{\sqrt5}$ (and so $a=\arctan\sqrt 5$). Thus
$$\sin(a-b) \le \sqrt{1-\frac1{1+\frac45}} = \frac23 $$
with the same equality case.
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this diophantine equation $x^2+y^2+z^3=n$ always have integer solution
show that: For any postive ineteger $n$,then the equation
$$n=x^2+y^2+z^3$$
always have integer solution
My idea: such as $n=1$,then we have
$$1=0^2+0^2+1^3$$
$$2=0^2+1^2+1^3$$
$$3=1^2+1^2+1^3$$
$$4=2^2+0^2+0^3$$
$$5=1^2+2^2+0^3$$
$$6=1^2+2^2+1^3$$
$$7=2^2+2^2+(-1)^3$$
$$8=0^2+0^2+2^3$$
$$9=1^2+0^2+2^3$$
$$10=1^2+1^2+2^3$$
$\cdots\cdots\cdots$
But for general $n$, How prove it?
| My friend put this as an MAA Monthly problem, years ago. The comparison is that there are infinitely many numbers that have no expression as $x^2 + y^2 + z^9.$ This simple result defeated an existing conjecture; we sent it early to Robert C. Vaughan, so it made it into the second edition of his book The Hardy-Littlewood Method. It is likely that every number can be written as $x^2 + y^2 + z^5,$ but not certain.
See http://zakuski.utsa.edu/~jagy/Elkies_Kap.pdf
and related items at http://zakuski.utsa.edu/~jagy/inhom.html
| {
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"url": "https://math.stackexchange.com/questions/1039106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Simplifying $\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}$ I realize that is a basic math problem, but I am still having problems with it.
The expression $$\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}$$ equals one of the following:
*
*$2\sqrt{3}-1$
*$\dfrac{6}{11}$
*$4\sqrt{3}$
*$12$
How do I simplify this?
| Hint:
First Rationalize the denominator of first term by multiplying numerator and denominator with denominator's conjugate and then use
$$(a+b)(a-b)=a^2-b^2$$
$$\frac{\sqrt{3}}{2\sqrt{3}+1} + \frac{\sqrt{3}}{11}=\frac{\sqrt{3}}{2\sqrt{3}+1}\cdot\frac{2\sqrt{3}-1}{2\sqrt{3}-1} + \frac{\sqrt{3}}{11}=\frac{6-\sqrt3}{(2\sqrt3)^2-1}+ \frac{\sqrt{3}}{11}=\frac{6-\sqrt3}{11}+ \frac{\sqrt{3}}{11}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Quadratics with roots as integers; possible values of a Suppose $a$, $b$ are real numbers such that $a+b=12$ and both roots of the equation $x^2+ax+b=0$ are integers.
Determine all possible values of $a$.
I don't know how to go about doing this without long, messy casework. I tries $(x-s)(x-r)=x^2+ax+b$ and got $-r-s=a$ and $rs=b$, but was unable to find all solutionss based on only these and $a+b=12$. Could someone help me finish up? Thanks.
| Since $a + b = 12$
$$-(r+s) + rs = 12$$
Using Simon's favorite factoring trick:
$$(r-1)(s-1) - 1 = 12 \iff (r-1)(s-1) = 13$$
Only two possible solutions arise:
$r -1 = 1$ and $s -1 = 13$, therefore $r = 2$ and $s = 14$
$r -1 = -1$ and $s -1 = -13$, therefore $r = 0$ and $s = -12$
Therefore, $a = -16$ and $b = 28$
or
$a = 12$ and $b = 0$
| {
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"url": "https://math.stackexchange.com/questions/1039487",
"timestamp": "2023-03-29T00:00:00",
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How to integrate $\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$? How to integrate $$\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$$
I tried both $t=\sqrt{x^2-1}$ and $t=\sin x$ but didn't reach the right result.
| Apply the substitution $x=sec(u)$, then $dx=\frac{\tan \left(u\right)}{\cos \left(u\right)}du$.
$$\int \frac{1}{\sec ^2\left(u\right)\sqrt{\sec ^2\left(u\right)-1}}\frac{\tan \left(u\right)}{\cos \left(u\right)}du$$
$$\int \frac{1}{\sec ^2\left(u\right)\sqrt{\sec ^2\left(u\right)-1}}\frac{1}{\cos^2 \left(u\right)}du$$
Remember that $\frac {1}{\cos(u)}=\sec(u)$ :
$$\int \frac{\sin \left(u\right)}{\sqrt{\sec ^2\left(u\right)-1}}du$$
And $\sec ^2\left(x\right)=1+\tan ^2\left(x\right)$:
$$\int \frac{\sin \left(u\right)}{\sqrt{-1+1+\tan ^2\left(u\right)}}du$$
Then:
$$\int \frac{\sin \left(u\right)}{\sqrt{\tan ^2\left(u\right)}}du$$
$$if \int \frac{g\left(x\right)}{f\left(x\right)}=F\left(x\right)\mathrm{,\:then\:}\:\int \frac{g\left(x\right)}{\sqrt{f^2\left(x\right)}}=\frac{f\left(x\right)}{\sqrt{f^2\left(x\right)}}F\left(x\right)$$
$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\int \frac{\sin \left(u\right)}{\tan \left(u\right)}du$$
$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\int \cos \left(u\right)du$$
$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\sin \left(u\right)$$
Substitute back, $u=arcsec(u)$:
$$\frac{\tan \left(arcsec \left(x\right)\right)}{\sqrt{\left(\tan \left(arcsec \left(x\right)\right)\right)^2}}\sin \left(arcsec \left(x\right)\right)$$
Simplifies to:
$$\sqrt{1-\frac{1}{x^2}}+C$$
Now compute the boundaries and you will find $1-0=1$.
| {
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"url": "https://math.stackexchange.com/questions/1040128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 7
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Can all real polynomials be factored into quadratic and linear factors? So I understand how to do integration on rational functions with a linear and a quadratic denominator, and I understand how to do a partial fraction decomposition, but I was wondering what happens if the polynomial is higher degree and can't be factored. Later in the page, it says this:
However, it can be shown that any polynomial with real coefficients is a product of linear and/or irreducible quadratic factors with real coefficients.
And I was wondering, how do we know this and is this definitely true?
| For example, take $P(x) = x^n-1$. The factorization into complex linear factors is
$$(x^n-1) = \prod_{k=0}^{n-1}( x - (\cos \frac{2 k \pi}{n} + i \sin \frac {2 k \pi}{n}) )$$
From the pairs of non-real conjugate roots $\{ (\cos \frac{2 k \pi}{n} + i \sin \frac {2 k \pi}{n}) ,(\cos \frac{2 (n-k) \pi}{n} + i \sin \frac {2 (n-k) \pi}{n}) \}$ for $0< k < \frac{n}{2}$ we get the quadratic factors
$$( x - (\cos \frac{2 k \pi}{n} + i \sin \frac {2 k \pi}{n}) )(x-(\cos \frac{2 (n-k) \pi}{n} + i \sin \frac {2 (n-k) \pi}{n}))=\\=x^2 - 2\cos\frac{2 k \pi}{n}\, x + 1 $$
Thus we get the real factorization of $x^n-1$:
$$x^n-1 =(x-1)\prod_{k=1}^{\frac{n-1}{2}} (x^2 - 2\cos\frac{2 k \pi}{n}\, x + 1 )$$
if $n$ odd and
$$x^n-1 =(x-1)(x+1)\prod_{k=1}^{\frac{n}{2}-1} (x^2 - 2\cos\frac{2 k \pi}{n}\, x + 1 )$$
if $n$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
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} |
Recursive Sequence solving for $f(200)$ Let $f$ be defined recursively by: $f(0)=5$ and $f(n+1)=3f(n)-2$. Find $f(200)$
I'm really confused how to go about solving this. Can someone help? Thank you!
| Expanding on
OC-Sansoo's answer:
Since
$f(n+1)=3f(n)-2$,
subtracting one from each side,
$f(n+1)-1
=3f(n)-3
=3(f(n)-1)
$.
Letting
$g(n)
=f(n)-1
$,
$g(n+1)
=3g(n)
$.
By induction,
$g(n)
=3^n g(0)
$
or
$f(n)-1
=3^n(f(0)-1)
=4\cdot 3^n
$
or
$f(n)
=4\cdot3^n+1
$.
In general,
if
$f(n+1)
=af(n)+b
$,
we want to find $c$
such that,
if $g(n) = f(n)+c$,
then
$g(n+1) = ag(n)$
(so that
$g(n) = a^n g(0)$).
Since $f(n) = g(n)-c$,
$g(n+1)-c
=a(g(n)-c)+b
=ag(n)-ac+b
$
or
$g(n+1)
=ag(n)+c-ac+b
=ag(n)+c(1-a)+b
$.
If we choose
$c(1-a)+b=0$,
or
$c
=\frac{b}{a-1}
$,
then
$g(n) = a^n g(0)$
or
$f(n)+c = a^n(f(0)+c)$
or
$f(n)
= a^nf(0)+c(a^n-1)
= a^nf(0)+\frac{b(a^n-1)}{a-1}
$.
Note that does not work if $a=1$,
but then
$f(n)$ is linearly i ncreasing,
not growing by a power law.
| {
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Application of Mergesort We have $8$ players and we want to sort them in $24$ hours.
There is one stadium. Each game lasts one hour.
In how many hours can we sort them??
I thought that we could it as followed:
$$\boxed{P1} \ \boxed{P2} \ \boxed{P3} \ \boxed{P4} \ \boxed{P5} \ \boxed{P6} \ \boxed{P7} \ \boxed{P8} \\ \ \ \ \ \boxed{P1} \ \ \ \ \ \ \ \ \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P5} \ \ \ \ \ \ \ \ \ \ \ \boxed{P7} \\ \ \ \boxed{P1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P5} \\ \ \ \ \boxed{P1}$$
So, the best player is $P1$.
These games took place in $7$ hours.
We know that $P5$ is the best player among the players $P5$, $P6$, $P7$ and $P8$.
$$\boxed{P2} \ \boxed{P3} \ \boxed{P4} \ \boxed{P5}\\ \boxed{P2} \ \ \ \ \ \ \boxed{P5} \\ \ \ \boxed{P2}$$
So, the second best player is $P2$.
These games took place in $3$ hours.
$$\boxed{P3} \ \boxed{P4} \ \boxed{P5}\\ \ \ \ \ \ \boxed{P3}\ \ \ \ \boxed{P5} \\ \ \ \ \ \ \ \boxed{P3}$$
So, the third best player is $P3$.
These games took place in $2$ hours.
$$\boxed{P4} \ \boxed{P5}\\ \ \ \boxed{P4}$$
So, the $4^{th}$ best player is $P4$ and the $5^{th}$ best player is $P5$.
This game took place in $1$ hour.
$$\boxed{P6} \ \boxed{P7} \ \boxed{P8}\\ \ \ \ \ \ \boxed{P6}\ \ \ \ \boxed{P8} \\ \ \ \ \ \ \ \boxed{P6}$$
So, the $6^{th}$ best player is $P6$.
These games took place in $2$ hours.
$$\boxed{P7} \ \boxed{P8}\\ \ \ \ \ \boxed{P7}$$
So, the $7^{th}$ best player is $P7$ and $8^{th}$ best player is $P8$.
This game took place in $1$ hour.
Therefore, we sorted the players $P1 \geq P2 \geq P3 \geq P4 \geq P5 \geq P6 \geq P7 \geq P8$ in $7+3+2+1+2+1=16$ hours.
Is it correct??
Is it maybe an application of Mergesort??
EDIT:
In the case when we have more than one stadium is it as followed??
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P1} \ : \ \boxed{P2} \ \ \ \ \ \ \ \ \boxed{P3} \ : \ \boxed{P4} \ \ \ \ \ \ \ \ \boxed{P5} \ : \ \boxed{P6} \ \ \ \ \ \ \ \ \boxed{P7} \ : \ \boxed{P8} \Rightarrow 4 \text{ Stadiums } \\ \boxed{P1} \ > \ \boxed{P2} \ \ \ \ \ \ \ \ \boxed{P3} \ > \ \boxed{P4} \ \ \ \ \ \ \ \ \boxed{P5} \ > \ \boxed{P6} \ \ \ \ \ \ \ \ \boxed{P7} \ > \ \boxed{P8}$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P1} \ : \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P2} \ : \ \boxed{P4} \ \ \ \ \ \ \ \ \boxed{P5} \ : \ \boxed{P7} \ \ \ \ \ \ \ \ \boxed{P6} \ : \ \boxed{P8} \Rightarrow 4 \text{ Stadiums } \\ \boxed{P1} \ > \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P2} \ > \ \boxed{P4} \ \ \ \ \ \ \ \ \boxed{P5} \ > \ \boxed{P7} \ \ \ \ \ \ \ \ \boxed{P6} \ > \ \boxed{P8}$$
Now we know tht $P1 >P2>P4$ and $P1>P3>P4$, that means that we have to know whether $P2>P3$ or $P3>P2$.
Also, we know that $P5>P6>P8$ and $P5>P7>P8$, that means that we have to know whether $P6>P7$ or $P7>P6$.
So, we have the following:
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P2} \ : \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P4} \ : \ \boxed{P5} \ \ \ \ \ \ \ \ \boxed{P6} \ : \ \boxed{P7} \Rightarrow 3 \text{ Stadiums } \\ \boxed{P2} \ > \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P4} \ > \ \boxed{P5} \ \ \ \ \ \ \ \ \boxed{P6} \ > \ \boxed{P7}$$
So, we have sorted them $P1>P2>P3>P4>P5>P6>P7>P8$ in $3$ hours.
Is it correct??
| The idea of mergesort is to exploit the fact, that merging two pre-sorted lists is "cheap": if the lenght of list1 is $l_1$ and that of list2 is $l_2$ and $L=l_1+l_2$ then you need at most $L-1 $comparisions to make a sorted list. The minimum (in a simple algorithm) is just the length of the shorter list.
This idea can be "iterated": each of the lists may have been built from two shorter lists, say list1 = merge(list11,list12) and list2=merge(list21,list22) and the sum of the comparisions is again the sum of the four partial lists-lenghtes, which is again $L-1$. If you have at the beginning a list of 16 elements, we could generate it by at most 16 comparisions provided its two partial lists each of length 8 are already sorted. Thus if we have 4 lists each of them with length 4, presorted, then we need at most 2*16 comparisions. The same with one more iteration: if we have 8 lists each of 2 elements,presorted, we need 3*16 comparisions to sort the list completely. Now each list of 2 elements can be sorted by 1 comparision so we get, in the worst case 8+3*16=56 comparisions. (If the overall list length is not a perfect power of 2 the highest number of comparisions must be computed more seriously).
In this simple algorithm it can happen, that the same two elements are compared twice in the process. With real matches, one could then omit that match and look at the protocol instead. With 8 players we should have 4 initial matches to have 4 lists with two elements. Then two lists are merged, that needs two times at most 3 comparisions which is 6 comparisions, and the two remaining lists have 4 elements each which needs 7 comparisions at most and thus we can arrive at the result by 4+6+7=17 comparisions at most.
(Please check my computations of the worst and of the best cases! - I'm just typing from the hand without references)
The idea of the multiple stadiums adds then the possibility to let some matches happen at the same time (and should possibly only put your focus to the idea of generating partial lists (two lists = two stadiums) which shall later be merged)
If the length of the list is large, there are improvements possible: one can merge two long presorted lists using binary search for the initial elements-comparision, but that leads too far here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$ Prove by induction that
$$
\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}
$$
Let $n=r$, so that
$$
S_r=2-\frac{r+2}{2^r}
$$
Therefore
$$\begin{align}
S_{r+1}=S_r+\frac{r+1}{2^{r+1}}&=2-\frac{r+2}{2^r}+\frac{r+1}{2^{r+1}}\\
&=2-\frac{2^r(r+2)+r+1}{2^{r+1}}\\
&=2-\frac{2^{n-1}(n+1)+n}{2^n}
\end{align}$$
How does $2^{n-1}(n+1)\equiv2$? Or is my method wrong? I've probably made a stupid mistake!
| Your mistake seems to be in your second-to-last line.
You clearly wanted to get a common denominator of $2^{r+1}$, but $2^r\cdot2^r=2^{2r}$, not $2^{r+1}$. You should have multiplied numerator and denominator by $2$, not $2^r$. Also, be careful with your signs.
You should get $$S_{r+1}=2+\frac{-2(r+2)}{2^{r+1}}+\frac{r+1}{2^{r+1}}=2+\frac{-r-3}{2^{r+1}}=2-\frac{(r+1)+2}{2^{r+1}}$$
Also, remember that with induction you need to demonstrate that the base case ($n=1$) holds.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\int_{-1}^{1} 4x \sqrt{1-x^{2}}\, dx = -3\pi$ What is of interest is the assertion
$$\int_{-1}^{1} 4x \sqrt{1-x^{2}}\,dx = -3\pi.$$
Since
$$\pi = 2\int_{-1}^{1}\sqrt{1-x^{2}}\,dx,$$
i.e. the area of a unit circle,
it suffices to prove that
$$\int_{-1}^{1}x\sqrt{1-x^{2}}\,dx = -\frac{3}{2}\int_{-1}^{1}\sqrt{1-x^{2}}\,dx.$$
But I did not see the left-hand side can be equal to the right-hand side?
| $x=\sin \theta \implies dx= \cos \theta \ d\theta$
$\displaystyle\int x \sqrt{1-x^2}\ dx=\displaystyle\int \sin \theta \cos^2\theta\ d\theta$
$\cos^2 \theta=z \implies -2\sin \theta\cos \theta\ d\theta=dz$
$\therefore \displaystyle\int \sin \theta \cos^2\theta\ d\theta=-\dfrac{1}{2}\displaystyle\int \sqrt{z}\ dz$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differentiation with the quotient rule I have the question:
Given that $$y=\frac{e^x}{\sqrt{1+2x}}$$
Show that $$\frac{dy}{dx} = \frac{2xe^x}{\sqrt{(1+2x)^3}}$$
I've done the question but I got $2xe^x(\sqrt{(1+2x)^3})$. I feel like this is too similar to the sheet's answer, so am I wrong or is the sheet printed wrong?
Working:
Let $u = e^x$, $v = 1+2x$, $w = \sqrt{v}$.
Using the chain rule, $$\frac{dw}{dx} = \frac{dw}{dv} * \frac{dv}{dx}$$
So $$\frac{dw}{dx} = 2(\frac{1}{2}v^{-\frac{1}{2}}) = \frac{1}{\sqrt{v}} = \frac{1}{\sqrt{1+2x}}$$
And $\frac{du}{dx}$ is clearly $e^x$.
Therefore, using the quotient rule, $$\frac{dy}{dx} = \frac{e^x(\sqrt{1+2x})- e^x(\frac{1}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{e^x(\sqrt{1+2x}-\frac{1}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{e^x(\frac{2x}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{\frac{2xe^x}{\sqrt{1+2x}}}{(1+2x)^2}$$
Using the rule
$$\frac{x^m}{x^n} = x^{m-n}$$
We get
$$\frac{2xe^x}{(1+2x)^{-\frac{3}{2}}}$$
$$=2xe^x(\sqrt{1+2x})^3 $$
Have I done this correctly?
| The error is in
$$\frac1{\sqrt{2x+1}} \cdot \frac1{(2x+1)^2} = \frac1{(2x+1)^{2+\frac12}} = \frac1{(2x+1)^{\frac52}}$$
and
$$\frac1{\sqrt{2x+1}^2} \ne \frac1{(2x+1)^2}$$
so there is an error in the quotient rule.
(the exponent is positive in the denominator, not negative)
Tracing the errors through the equation we obtain
$$\frac1{\sqrt{2x+1}} \cdot \frac1{(\sqrt{2x+1})^2} = \frac1{\sqrt{2x+1}^3} = \frac1{(2x+1)^{\frac32}}$$
| {
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"url": "https://math.stackexchange.com/questions/1050478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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simple trigonometric functions how would you solve this
$$\cos^2 x - 2\sin x \cos x - \sin^2 x = 0$$
I tried to simplify it but I got $\cos 2x - \sin 2x$. I can't simplify that further.
| Based on what you have done,
\begin{align*}
\cos^2x - 2\sin x\cos x - \sin^2x & = 0\\
\cos^2x - \sin^2x - 2\sin x\cos x & = 0\\
\cos(2x) - \sin(2x) & = 0\\
\cos(2x) & = \sin(2x)\\
1 & = \tan(2x)
\end{align*}
Observe that if you want values of $x$ in the interval $[0, 2\pi)$ that satisfy the equation, then $0 \leq 2x < 4\pi$. Thus, we must find values of $2x$ in the interval $[0, 4\pi)$ such that $\tan(2x) = 1$. They are
\begin{align*}
2x & = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}\\
x & = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}
\end{align*}
Since $\tan(2x)$ has period $\dfrac{\pi}{2}$, the general solution is
$$x = \frac{\pi}{8} + n\frac{\pi}{2}, n \in \mathbb{Z}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Triple Integral to Find Volume Question: Use a triple integral to find the volume of the solid enclosed by the parabaloids $y=x^2+z^2$ and $y=8-x^2-z^2$.
My attempt: The best I can figure, this object looks kind of like a football oriented along the $y$-axis from $y=0$ to $y=8$ and is symmetric about the $y$-axis and the plane $y=4$.
It seems best to integrate first with respect to $y$, and $x^2+z^2 \le y \le 8-(x^2+z^2)$.
The widest part of the football is at $y=4$; substitute that into both of the equations above to find that the projection onto the $xz$-plane is $x^2+z^2=4$, or a circle of radius 2, so my bounds for $z$ are $-\sqrt{4-x^2} \le z \le \sqrt{4-x^2}$ and my bounds for $x$ are $-2 \le x \le 2$.
But I believe I can make this easier by integrating from $0\le z \le \sqrt{4-x^2}$ and multiplying by 2 and integrating from $0 \le x \le 2$ and multiplying by another 2. (I actually think I can integrate $y$ from $4 \le y \le 8-(x^2+z^2)$ and multiply by another 2, but that doesn't seem to simplify anything.)
Since I'm finding the volume, the function I integrate is one. I come up with this
$$4\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-(x^2+z^2)} 1\ dy\ dz\ dx.$$
(Is this right so far?)
If nothing's wrong yet, I still can't finish this integral
$$4\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-\left(x^2+z^2\right)} 1\ dy\ dz\ dx \\
4\int_0^2 \int_0^{\sqrt{4-x^2}} y \Big|_{x^2+z^2}^{8-\left(x^2+z^2\right)} dz\ dx \\
4\int_0^2 \int_0^{\sqrt{4-x^2}} \left(8-\left(x^2+z^2\right)\right)-\left(x^2+z^2\right) dz\ dx \\
4\int_0^2 \int_0^{\sqrt{4-x^2}} \left(8-2x^2-2z^2\right)\ dz\ dx \\
4\int_0^2 \left[\left(8-2x^2\right)z-\frac 23 z^3\right]_0^{\sqrt{4-x^2}} \ dx \\
4\int_0^2 \left[ \left(8-2x^2\right)\sqrt{4-x^2} - \frac 23 \sqrt{4-x^2}^3 \right]\ dx \\
-\frac{16}3\int_0^2 \left[ \left(x^2 -4\right)\sqrt{4-x^2} \right]\ dx \\
\vdots \\ ??$$
| The integral in your last line can be simplified to
$$
\frac{16}3\int_0^2 (4-x^2)^{3/2}\,dx
$$
To do this, first use the substitution $x=2\sin\theta$, obtaining
$$
\frac{16}3\int_0^{\pi/2}(4-4\sin^2\theta)^{3/2}\cos\theta\,d\theta=
8\cdot\frac{16}3\int_0^{\pi/2}\cos^4\theta\,d\theta
$$
To complete the integral, use the identity $\cos^2\phi = \frac{1+\cos(2\phi)}2$ to write
$$
\cos^4\theta = \left(\frac{1+\cos(2\theta)}{2}\right)^2=\frac12+\cos(2\theta)+\frac14\cos^22\theta=\frac12+\cos(2\theta)+\frac{1}{4}\left(\frac{1+\cos(4\theta)}{2}\right)
$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the sum of the series with unordered powers of $3$
Consider the following series: $$\sum_{n=1}^{\infty} a_n = 1/3+1+1/3^3+1/3^2+1/3^5+1/3^4+1/3^7+1/3^6 +\dots$$
Determine if it converges, and find the sum.
Here is what I got:
a) $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = (\frac{1}{3})^3 =1/27$
b) $\lim_{n\to\infty} \sqrt[n]{a_n} = 1/3$
Is it right?
| Since $a_n=\dfrac{1}{3^n}$ when n is odd and $a_n=\dfrac{1}{3^{n-2}}$ when n is even,
A)$\;\;$ $\dfrac{a_{n+1}}{a_n}=\begin{cases}\frac{1}{27},&\mbox{if n is odd}\\\; 3,&\mbox{if n is even}\end{cases}.$ Therefore $\displaystyle\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}$ does not exist.
B) $\;\;$$\displaystyle\left(a_n\right)^{\frac{1}{n}}=\begin{cases}\frac{1}{3},&\mbox{if n is odd}\\\left(\frac{1}{3}\right)^{1-\frac{2}{n}},&\mbox{if n is even}\end{cases}.$ $\;\;\;$ Therefore $\displaystyle\lim_{n\to\infty}\left(a_n\right)^{\frac{1}{n}}=\frac{1}{3}$.
As pointed out by Ross Millikan, this is just a rearrangement of the absolutely convergent series
$\;\;\;\displaystyle\sum_{n=0}^{\infty}\frac{1}{3^n}=\sum_{n=0}^{\infty}\left(\frac{1}{3}\right)^n$, so it converges and has the same sum.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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} |
Let $a, b, c$ be positive real numbers such that $a + 2b + 3c = 26$ and $a^2 + b^2 + c^2 = 52$. Find the largest possible value of $a$.
I used the Cauchy Schwarz inequality $(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$ as follows:
$a + 2b + 3c = 26$ is given; adding $a$ to both sides gives $2a + 2b + 3c = 26+a$. Then, we have $x = 2$, $y=2$, $z=3$. Putting this all into the inequality, we get:
$(ax+by+cz)^2 \leq (a^2+b^2+c^2)(x^2+y^2+z^2)$
$(2a+2b+3c)^2 \leq (a^2+b^2+c^2)(2^2+2^2+3^2)$
$(26+a)^2 \leq (52)(17)$
$(26+a)^2 \leq 884$
$26 + a \leq \sqrt{884}$
$a \leq 2\sqrt{221} - 26 \approx 3.73$
However, the listed answer gives $a \leq \frac{26}{7} \approx 3.71$. What am I doing wrong?
| Knowing the answer, it is not hard to construct the right CS inequality to solve your problem:
$$52\cdot (13^2+12^2+18^2) \ge (13a+12b+18c)^2 = (7a+6\cdot 26)^2 \implies a \le \frac{26}7$$
It is easy to check equality is attained when $a = \frac{26}7, b = \frac{24}7, c = \frac{36}7$.
If you don't know the answer to start with, your method can be modified a bit to give the result. We start with
$$52 \cdot \left(x^2+y^2+z^2 \right) \ge (x a + y b + z c)^2 \tag{1}$$
As the RHS needs to be of form $\left(\alpha a+\beta(a+2b+3c)\right)^2$ to isolate $a$, so we let $y = 2, z = 3$. Thus we have
$$52 \cdot \left(x^2+4+9 \right) \ge \left(x a + 2 b + 3 c \right)^2 =\left((x-1) a + 26 \right)^2$$
Equality conditions give
$$\frac{a}x=\frac{b}2=\frac{c}3 \implies b = \frac{2a}x, c = \frac{3a}x$$
and the initial conditions give
$$a=26\frac{x}{x+13}, \quad a^2 = 52\frac{x^2}{x^2+13} \implies x = \frac{13}6, a=\frac{26}7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
compute sums of x,y given a condition
Problem: given that $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=p$, then compute $x+y$
try: i tryed to solve by this way
$$\begin{align}
\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)&=p\\
1&=p\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)\\
p&=p^2\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)\\
\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)&=p^2\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)
\end{align}$$
however this only given
$$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=xy+x\sqrt{y^2+1}+y\sqrt{x^2+1}+\sqrt{(x^2+1)(y^2+1)}$$
$$\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)=xy-x\sqrt{y^2+1}-y\sqrt{x^2+1}+\sqrt{(x^2+1)(y^2+1)}$$
wich not help, i tryed a lot of think but i alway end like walking in circles, how i solve this problem
| It is not possible. We have that $f(x):\mathbb{R}\to\mathbb{R}^+$ given by:
$$f(x)=x+\sqrt{x^2+1}$$
is bijective. So, assume that $p=f(2)\cdot f(3)=6+5 \sqrt{2}+3 \sqrt{5}+2 \sqrt{10}$.
Obviously $x=2,y=3$ is a solution of $f(x)\cdot f(y)=p$, giving $x+y=5$.
But another solution exists, given by:
$$x=y=f^{-1}\left(\sqrt{6+5 \sqrt{2}+3 \sqrt{5}+2 \sqrt{10}}\right).$$
Since $f^{-1}(x)=\frac{x^2-1}{2x}$, the solution:
$$ x = y = \sqrt{\frac{5}{2} \left(1+\sqrt{2}\right)} $$
gives:
$$ x+y = \sqrt{10(1+\sqrt{2})} \neq 5. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Given a matrix, find a matrix that satisfies Let A be a matrix (3x4)
Prove that there does not exists a matrix X that satisfies
$$
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}X = \begin{pmatrix}
1 & 1 & 1 \\
0 & 2 & 0 \\
2 & 1 & 1 \\
\end{pmatrix}
$$
When I try to peform Gaussian elimination to get the reduced form of A, I always get a row of zeroes, e.g:
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}
$$ R_3 - R_1 \to R_3 $$
I get
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
What can I conclude from the fact that I got a zeroes row?
Does this help solving the problem?
| You can argue by contradiction, using the relation
$$\text{rank}(AB) \le \text{min}(\text{rang}(A),\text{rang}(B))$$
with $A= \begin{pmatrix}1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}$ and $B:=X$ a $4\times 3$ matrix with $\text{rank}(X)\le 3$.
Since then you would have
$$\text{rank}(AX) \le \text{min}(\text{rang}(A),\text{rank}(X)) \le \text{min}(2,3)=2,$$
while $\text{rank}\left(\begin{pmatrix}
1 & 1 & 1 \\
0 & 2 & 0 \\
2 & 1 & 1 \\
\end{pmatrix} \right)=3,$ since the determinant of the last matrix is different from $0$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Identities for hypergeometric functions ${}_2F_1$ with z=1/2 Is there a closed form (or approximation) for a hypergeometric function of form:
$_2F_1(1,b+c;c;\frac{1}{2}) \quad \text{where} \; b,c \in \mathbb{N}$ ?
I researched all identities in http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/04/ but nothing seems to work for me.
| Case $1$: $c\neq1$
Then $_2F_1\left(1,b+c;c;\dfrac{1}{2}\right)$
$=~_2F_1\left(b+c,1;c;\dfrac{1}{2}\right)$
$=\dfrac{1}{B(1,c-1)}\int_0^1\dfrac{(1-x)^{c-2}}{\left(1-\dfrac{x}{2}\right)^{b+c}}dx$
$=\dfrac{2^{b+c}(c-1)!}{0!(c-2)!}\int_0^1\dfrac{(1-x)^{c-2}}{(2-x)^{b+c}}dx$
$=2^{b+c}(c-1)\int_0^1\dfrac{(2-x-1)^{c-2}}{(2-x)^{b+c}}dx$
$=2^{b+c}(c-1)\int_0^1\sum\limits_{n=0}^{c-2}\dfrac{C_n^{c-2}(-1)^n(2-x)^{c-n-2}}{(2-x)^{b+c}}dx$
$=2^{b+c}(c-1)\int_0^1\sum\limits_{n=0}^{c-2}\dfrac{(-1)^n(c-2)!(2-x)^{-b-n-2}}{n!(c-n-2)!}dx$
$=2^{b+c}(c-1)\left[-\sum\limits_{n=0}^{c-2}\dfrac{(-1)^n(c-2)!(2-x)^{-b-n-1}}{n!(c-n-2)!(-b-n-1)}\right]_0^1$
$=\left[\sum\limits_{n=0}^{c-2}\dfrac{(-1)^n(c-1)!2^{b+c}}{n!(c-n-2)!(b+n+1)(2-x)^{b+n+1}}\right]_0^1$
$=\sum\limits_{n=0}^{c-2}\dfrac{(-1)^n(c-1)!2^{b+c}}{n!(c-n-2)!(b+n+1)}-\sum\limits_{n=0}^{c-2}\dfrac{(-1)^n(c-1)!2^{c-n-1}}{n!(c-n-2)!(b+n+1)}$
Case $2$: $c=1$
Then $_2F_1\left(1,b+1;1;\dfrac{1}{2}\right)$
$=~_2F_1\left(b+1,1;1;\dfrac{1}{2}\right)$
$=\left(1-\dfrac{1}{2}\right)^{-b-1}$
$=\dfrac{1}{2^{b+1}}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\sin\theta +\sin3\theta =0$
Solve the equation by first using a Sum-to-Product Formula.
$$\sin\theta +\sin3\theta =0$$
Steps I took:
$$\begin{align}0&=2\sin\frac { \theta +3\theta }{ 2 } \cos\frac { \theta -3\theta }{ 2 }\\
0&=2\sin\frac { 4\theta }{ 2 } \cos\frac { -2\theta }{ 2 } \\
0&=2\sin2\theta \cos(-\theta)\\
0&=2\sin2\theta \cos\theta\\
0&=2(2\sin\theta \cos\theta )\cos\theta\\
0&=4\sin\theta \cos^2\theta \\
0&=4\sin\theta \frac { 1+\cos2\theta }{ 2 } \\
0&=4\sin\theta \frac { 1+1-2\sin^2\theta }{ 2 } \\
0&=4\sin\theta \frac { 2-2\sin^{ 2 }\theta }{ 2 } \\
0&=\frac { 8\sin\theta -8\sin^{ 2 }\theta }{ 2 }\end{align}$$
| As far as I see, you didn't make any mistake in your calculation. You did, however, overcomplicate things.
For one, if $2\sin(\alpha)\cos(\beta) = 0$, you can get rid of the $2$ and just write $\sin(\alpha)\cos(\beta) = 0$.
The main mistake you made, however, was when you got to $2\sin(2\theta)\cos\theta$. From there on, you should ask yourself this:
If $a\cdot b = 0$, what can I say about $a$ and $b$?
An edit answering your question if you were completely wrong:
No, you were not wrong. The only place you were wrong is when you rewrote
$$4\sin(\theta)\frac{2-2\sin^2\theta}{2}$$
Into $$\frac{8\sin\theta - 8\sin^2\theta}{2}.$$
There is a small mistake here. Other than that, you are correct, but truly horribly inefficient.
For example, in row $6$, you wrote $4\sin\theta \cos^2\theta = 0$, then you took another $4$ rows before you got to $4\sin(\theta)\frac{2-2\sin^2\theta}{2} = 0$, when, in fact, you could just replace $\cos^2\theta$ with $1-\sin^2\theta$ and get the same result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the first two non vanishing maclaurin terms Find the first two nonvanishing terms in the Maclaurin series of $\sin(x + x^3)$.
Suggestion: use the Maclaurin series of $\sin(y)$ and write $y = x + x^3$
Using
this result, find
$\lim\limits_{x\to 0}\frac{\sin(x + x^3)−x}{x^3}$
$\sin(y)= y-\frac{y^3}{3!}+\frac{y^5}{5!}+\frac{y^7}{7!}$
$y= x+x^3$
$x+x^3-\frac{(x+x^3)^3}{3!}+\frac{(x+x^3)^5}{5!}-\frac{(x+x^3)^7}{7!}$
Now what do I do?
| jnh's answer is correct, other than a sign error in the middle of it which is corrected by the end.
But you can do this without taking any derivatives by hand. The trick is to partially expand the binomial. You have:
$$\sin(x+x^3)=\sum_{n=0}^\infty \frac{(x+x^3)^{2n+1} (-1)^n}{(2n+1)!}$$
Considering the binomial theorem, the first term contributes a term of order $1$ and a term of order $3$, while the second term contributes a term of order $3$ and higher order terms. More generally the $n$th term contributes a term of order $2n+1$ and higher order terms, so going up to $n=1$ retrieves the first two terms of the desired series.
The first two terms in the overall series will be order $1$ and $3$. The term of order $1$ can just be read off; it is just $x$. The term of order $3$ is a combination of the contribution from the $n=0$ term ($x^3$) and the contribution from the $n=1$ term, which by the binomial theorem is $\frac{-x^3}{6}$.
So we have $\sin(x+x^3)=x+\frac{5x^3}{6}+o \left ( x^3 \right )$. (The $o \left ( x^3 \right )$ is actually of order $5$, but this is not required to solve the problem.) Plugging this in we get
$$\frac{\sin(x+x^3)-x}{x^3}=\frac{\frac{5x^3}{6}+o \left ( x^3 \right )}{x^3} = \frac{5}{6} + o(1)$$
and so the desired limit is $\frac{5}{6}$. This approach of combining terms of the same order is common in perturbation theory problems in physics.
| {
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"url": "https://math.stackexchange.com/questions/1059711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Product of a Continuous and Discrete Distribution
Let $X \sim \mathcal{N}(0, 1)$ and $Y$ be a random variable independent of $X$ such that
\begin{align*}
P(Y=y) = \begin{cases}
\frac{1}{2} & y = -1\\
\frac{1}{2} & y = 1\\
0 & otherwise
\end{cases}
\end{align*}
If $ Z = XY$, show that $Z \sim \mathcal{N}(0,1)$.
Here is what I came up with:
\begin{align*}
P(Z<z) &= P(XY < z)\\
&=P(X < \frac{z}{Y})\\
&= P(X < \frac{z}{Y} \mid Y = -1)P(Y=-1) + P(X < \frac{z}{Y} \mid Y=1) P(Y=1)\\
&= \ldots\,?\\
&= P(X < z)
\end{align*}
Unfortunately, I'm not sure what to do, or whether I am even on the right track. I'd greatly appreciate any pointers!
| Your mistake: you can't divide by $Y$ without knowing its sign (if it's negative, the inequality would reverse).
\begin{align*}
P(XY<z)
&= P(XY<z \mid Y=1) P(Y=1) + P(XY<z \mid Y=-1) P(Y=-1)\\
&=\frac{1}{2} P(X<z) + \frac{1}{2} P(-X<z)\\
&= \frac{1}{2} P(X<z) + \frac{1}{2} P(X>-z)\\
&= P(X<z).
\end{align*}
The last step is due to symmetry: $P(X<z)=P(X>-z)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061955",
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"source": "stackexchange",
"question_score": "1",
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To prove $(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 \ge 9$ I used the following way but got wrong answer
$$A.M. \ge G.M.$$
$$ \frac{\sin \theta + \csc \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$
Squaring both sides,
\begin{equation*}
(\sin\theta + \csc\theta )^2 \ge 4 \tag{1}
\end{equation*}
Similarly
\begin{equation*}
(\cos\theta + \sec\theta )^2 \ge 4 \tag{2}
\end{equation*}
Adding equation (1) and (2)
\begin{equation*}
(\sin\theta + \csc\theta )^2+(\cos\theta + \sec\theta )^2 \ge 8
\end{equation*}
What is wrong?
| Without AM-GM, you could expand the expression which, after simplifications, write $$f(x)=(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 =\csc ^2(x)+\sec ^2(x)+5$$ The derivative write $$f'(x)=2 \tan (x) \sec ^2(x)-2 \cot (x) \csc ^2(x)=-8 \sin (4 x) \csc ^4(2 x)$$ and cancels for $x=\frac{\pi}{4}, \frac{3\pi}{4}, \cdots$ and, for these values, $f(x)=9$. The second derivative test shows that this is a minimum value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Problem integrating when attempting a solution with the Poincaré Lemma
d) This is part I am having troubles with.
I get that
$$
\hat{\mathbb{X}}_t = \left(\frac{\partial}{\partial t}\hat{\Phi}_t \right) \Phi_t^{-1}
= \left(\frac{\partial}{\partial t}\hat{\Phi}_t \right) (x/t,y/t,z)
= (x/t,y/t,0)
$$
so,
$$\begin{align}
i_{\hat{\mathbb{X}}_t}\beta &= i_{\hat{\mathbb{X}}_t} \left[\frac{3xz}{r^5} dy\wedge dz + \frac{3yz}{r^5} dz\wedge dx +\frac{2z^2-x^2-y^2}{r^5}dx \wedge dy \right] \\
&= \frac{3xz}{r^5}\left[\frac{y}{t}dz\right]+\frac{3yz}{r^5}\left[\frac{-x}{t}dz\right]+\frac{2z^2-x^2-y^2}{r^5}\left[\frac{x}{t}dy-\frac{y}{t}dx\right] \\
&= \frac{2z^2-x^2-y^2}{r^5}\left[\frac{x}{t}dy-\frac{y}{t}dx\right]
\end{align}$$
Then
and so
$$\begin{align}
\Phi_t^*\left(i_{\hat{\mathbb{X}}_t}\beta\right) &= \frac{2z^2-t^2x^2-t^2y^2}{(t^2x^2+t^2y^2+z^2)^{5/2}}\left[txdy-tydx \right] \\
&= \frac{2z^2-t^2x^2-t^2y^2}{(t^2x^2+t^2y^2+z^2)^{5/2}}txdy - \frac{2z^2-t^2x^2-t^2y^2}{(t^2x^2+t^2y^2+z^2)^{5/2}}tydx
\end{align}$$
However I cannot see how to integrate this. Help much appreciated.
UPDATE: Apparently the correct answer is $\alpha=(x dy -y dx)/r^3$. This would suggest I am on the right track.
The following integral may help $\displaystyle \int_0^w \frac{v^n}{(1+v)^s}dv=\frac{1}{1-s}\frac{w^n}{(1+w)^{s-1}}-\frac{n}{1-s}\int_0^w \frac{v^{n-1}}{(1+v)^{s-1}}dv$ $n>0, s\neq 1,w > -1$
I believe I may be missing something as I have had a number of problems with these type of questions. See my other question on the Poincaré Lemma application.
| Let $$I(x,y,z) = \int_0^1 \frac{2z^2 - t^2x^2-t^2y^2}{(t^2x^2+t^2y^2+z^2)^{5/2}}t\, dt.$$
Using the $u$-substitution $u = t^2x^2 + t^2y^2 + z^2$, we compute
\begin{align}I(x,y,z) &= \frac{1}{2(x^2+y^2)}\int_{z^2}^{r^2} \frac{2z^2 - (u - z^2)}{u^{5/2}}\, du\\
&= \frac{1}{2(x^2 + y^2)} \int_{z^2}^{r^2} \left(\frac{3z^2}{u^{5/2}} - \frac{1}{u^{3/2}}\right)\, du\\
&= \frac{1}{2(x^2 + y^2)} \left\{-\frac{2z^2}{u^{3/2}}\bigg|_{u = z^2}^{u = r^2} + \frac{2}{u^{1/2}}\bigg|_{u = z^2}^{u = r^2}\right\}\\
&= \frac{1}{2(x^2 + y^2)} \left\{\frac{2}{z} - \frac{2z^2}{r^3} + \frac{2}{r} - \frac{2}{z}\right\}\\
&= \frac1{2(x^2 + y^2)}\frac{2r^2 - 2z^2}{r^3}\\
&= \frac1{2(x^2 + y^2)}\frac{2(x^2 + y^2)}{r^3}\\
&= \frac{1}{r^3}.
\end{align}
Therefore, $$\alpha = I(x,y,z)(x\, dy - y\, dx) = \frac{x\, dy - y\, dx}{r^3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve $x^7-5x^4-x^3+4x+1=0$ for $x$
Solve for $x$
$$x^7-5x^4-x^3+4x+1=0$$
This equation has been bugging me since the past few days. I have found, using the Rational Root Theorem that $x=1$ is a root of this equation. However, after dividing, I cannot solve the six degree equation thus generated. I have also tried factorizing the equation, but it's not working.
| $x^4(x^3 + 1) + (x^3 + 1) = 0$
$\implies(x^3 + 1)(x^4 + 1) = 0$
$\implies(x + 1)(x^2 - x + 1)(x^4 + 2x^2 + 1 - 2x^2) = 0$
$\implies(x + 1)(x^2 - x + 1)((x^2 + 1)^2 - 2x^2) = 0$
$\implies(x + 1)(x^2 - x + 1)(x^2 + 1 + (\sqrt 2)(x))(x^2 + 1 - (\sqrt 2)(x)) = 0$
$ x^2 - x + 1 = 0$
$\implies x = \frac{1 \pm \sqrt{1 - 4\times1}}2 = \frac{1 \pm i \sqrt 3}2$
$ x^2 + \sqrt 2x + 1 = 0$
$\implies x = \frac{-\sqrt 2 \pm \sqrt{2 - 4 \times 1}}2 = \frac{-\sqrt 2 \pm i \sqrt 2}2$
$ x^2 - \sqrt 2x + 1 = 0$
$\implies 4x = \frac{\sqrt 2 \pm \sqrt{2 - 4 \times 1}}2 = \frac{\sqrt 2 \pm i sqrt 2}2$
The $7$ roots of $x$ are:
$x = -1, \frac{1 \pm i \sqrt 3}2, \frac{-\sqrt 2 \pm i \sqrt 2}2, \frac{\sqrt 2 \pm i \sqrt 2}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Finding the value of $y=b^2(3a^2+4ab+2b^2)$ if $a^2(2a^2+4ab+3b^2)=3$ and $a$ and $b$ are distinct zeros of $x^3-2x+c$
If $a$ and $b$ are distinct zeroes of the polynomial $x^3-2x+c$ and
$$a^2(2a^2+4ab+3b^2)=3$$
$$b^2(3a^2+4ab+2b^2)=y$$
Evaluate $y$
I tried for many hours but couldn't solve this question. This question is far tougher than it looks!
I made another equation using the fact that $a$ and $b$ are the distinct roots of the given polynomial. I got $a^2+ab+b^2=2$. I tried solving it as a quadratic in $a$ and then putting it in the other given equation, but the expression just got very ugly :(
Please Help!
Thanks!
| Since $a$ and $b$ are the solutions of $x^3-2x+c=0$ we have,
$a^3-2a+c=0$
&
$b^3-2b+c=0$
Subtracting the above two equations, we get,
$a^3-b^3-2(a-b)=0$
$\implies (a-b)(a^2+ab+b^2-2)=0$
$\implies a^2+ab+b^2=2$ (Since $a$ and $b$ are distinct)
Now,
$$a^2+ab+b^2=2$$
$$a^2(2a^2+4ab+3b^2)=3$$
$$b^2(3a^2+4ab+2b^2)=y$$
Squaring the first equation, we get,
$a^4+b^4+3a^2b^2+2a^3b+2ab^3=4$
$\implies 2a^4+2b^4+6a^2b^2+4a^3b+4ab^3=8$
Expanding and adding the other two equations, we get,
$y+3=2a^4+2b^4+6a^2b^2+4a^3b+4ab^3$
$\implies y+3=8$
$\implies \boxed{y=5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Trigonometric substitution and Integration of $\frac{1}{x^2\sqrt{x^2+1}} $ Regarding the integral
$$
\int \frac{dx}{x^2\sqrt{x^2 + 1}}
$$
I'm not sure what to do about the extra $x^2$ in the denominator. What can I do about it?
| $x=\cos \theta \implies dx=-\sin \theta\ d\theta$
$\therefore\displaystyle\int\dfrac{dx}{x^2\sqrt{x^2+1}}\\=-\displaystyle\int\dfrac{\sin \theta\ d\theta}{\cos^2 \theta\sqrt{\cos^2 \theta+1}}\\=-\displaystyle\int\dfrac{\sin \theta\ d\theta}{\cos^2 \theta\sqrt{2\cos^2 \left(\dfrac{\theta}{2}\right)}}\\=-\sqrt{2}\displaystyle\int\dfrac{\sin \left(\dfrac{\theta}{2}\right)\ d\theta}{2\cos^2\left(\dfrac{\theta}{2}\right)-1}\\$
$\cos\left(\dfrac{\theta}{2}\right)=z \implies-\dfrac{1}{2}\sin \left(\dfrac{\theta}{2}\right)d\theta=dz$
$\therefore-\sqrt{2}\displaystyle\int\dfrac{\sin \left(\dfrac{\theta}{2}\right)\ d\theta}{2\cos^2\left(\dfrac{\theta}{2}\right)-1}=\dfrac{1}{\sqrt{2}}\displaystyle\int\dfrac{dz}{2z^2-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Showing convergence of recursive sequence $A_{n+1}=\frac 1 {1+A_n}$
Given : $\forall n\in\Bbb N,\quad A_{n+1} = \frac 1 {1+A_n}$ and $A_1 = 0$
Show the sequence converges and find its limit.
Briefly what I did was to create two sub-sequences with an index difference of 2. Doing that I got two recursive formulas that depend on the initial A1, be it 0 for the odds or 1 for the evens.
After showing the sequences are bounded, I got stuck showing the sequence is monotonic.
Thanks,
| The recurrence is of the form
$$x_{n+1} = \phi(x_n)$$
where $\phi(x) = \frac{1}{1+x}$. The map $\phi$ has two fixed points $x_{1,2}$, roots of the equation $x^2 + x -1=0$, that is $\frac{-1\pm \sqrt{5}}{2}$. Now, we can write $\phi(x)$ in the form
$$\frac{\phi(x) -x_1}{\phi(x) - x_2} = k \cdot \frac{x-x_1}{x-x_2}$$
$k$ can be determined by givind a particular value for $x$, say $x=0$. We get
$$\frac{1-x_1}{1-x_2} = k \frac{x_1}{x_2}$$
and so
$$k = \frac{x_2(1-x_1)}{x_1(1-x_2)} = \frac{x_2 + 1}{x_1 + 1}=\frac{1-\sqrt{5}}{1+\sqrt{5}}$$
We get
$$\frac{\phi(x) -x_1}{\phi(x) - x_2} = \frac{1-\sqrt{5}}{1+\sqrt{5}} \cdot \frac{x-x_1}{x-x_2}$$
In other words, $\phi$ is conjugate to the map $t \mapsto \frac{1-\sqrt{5}}{1+ \sqrt{5}}\cdot t$. The monotony and convergence should be easy now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An exercise from Knuth's book - Proving a formula by induction I would like to find a formula for this sum:
$$
\frac{1^3}{1^4+4} - \frac{3^3}{3^4+4} + \frac{5^3}{5^4+4} - ... + \frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}
$$
The answer given (Knuth's book, The Art of Computer Programming Volume 1, Third Edition, Section 1.2.1, Exercise 11) is $$ \frac{(-1^n)(n+1)}{4(n+1)^2+1}$$
Please give me an approach or a solution to get this sum.
I tried to solve by pairing the consecutive terms. The last and the second last terms will correspond to $(2n+1)$ and $(2n-1)$ respectively. But this approach couldn't yield much.
The second part is to prove this sum by induction, i.e we've to prove this equation:
$$
\frac{1^3}{1^4+4} - \frac{3^3}{3^4+4} + \frac{5^3}{5^4+4} - ... + \frac{(-1)^n(2n+1)^3}{(2n+1)^4+4} = \frac{(-1^n)(n+1)}{4(n+1)^2+1}
$$
I approached it like this:
The sum is correct for $k=1$( and $k=0$). (k is a non-negative integer)
Assuming that the sum is correct for $k = 0,1,..,n$
we'll prove that its correct for $k=n+1$.
The sum for $k=n+1$ will be equal to: the sum for k=n + the last term, i.e. $$ \frac{(-1)^n(n+1)}{4(n+1)^2+1} + \frac{(-1)^{n+1}(2(n+1)+1)^3}{(2(n+1)+1)^4+4}$$
This should then be proved equal to $$\frac{(-1)^{n+1}((n+1)+1)}{4((n+1)+1)^2+1}$$
which again I'm not able to do. So, please help in proving this also.
P.S.: Someone please create a 'knuth' or 'taocp' tag for mathematical questions of this book.
| You are very close.
Take two consecutive sums and subtract them (actually you will be adding as the signs alternate):
$$\frac{n+1}{4(n+1)^2+1}+\frac{n+2}{4(n+2)^2+1}=\frac{8n^3+36n^2+54n+27}{16n^4+96n^3+216n^2+216n+85}.$$
The numerator is $(2n+3)^3$, and the denominator is $(2n+3)^4+4$.
To make this derivation more symmetric and manageable, you can use a half-integer index, $m=n+3/2$:
$$\frac{2m-1}{(2m-1)^2+1}+\frac{2m+1}{(2m+1)^2+1}=
\frac{2m-1}{(4m^2+2)-4m}+\frac{2m+1}{(4m^2+2)+4m}=
\frac{16m^3}{16m^4+4}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that the following polynomials are irreducible or not.
I want to show that:
1) $X^4+1$ is irreducible
The roots are the elements of
$$\left\{\frac{\pm 1+i}{\sqrt 2},\frac{\pm 1-i}{\sqrt 2}\right\}$$
therefore it's not a product of a polynomial of degree 1 and a polynomial of degree 3. Is there an easier way to prove that it's not of the form $(X^2+aX+1)(X^2+bX+1)$ than to solve the system $(X^2+aX+1)(X^2+bX+1)=X^4+1$ ?
2) $X^4-16X^2+4$ is irreducible
It's easy to prove that there is no integer roots, therefore it's not a product of a polynomial of degree 1 and a polynomial of degree 3. I suppose that
$$X^4-16X+4=(X^2+aX+b)(X^2+cX+d)$$
$$\iff X^4-16X^2+4= X^4+(a+c)X^3+(b+d+ac)X^2+(bc+da)X+bd$$
$$\iff\begin{cases}a+c=0,\quad(1)\\
b+d+ac=-16, \quad (2)\\
bc+da=0,\quad (3)\\
bd=4,\quad (4)\end{cases}$$
By $(4)$ we have that $(b,d)\in\{(1,4),(-1,-4), (2,2), (-2,-2)\}$. By $(1)$, $a=-c$. By $(3)$, $bc+da=bc-dc=c(b-d)=0$, therefore $c=0$ or $b=d$. If $c=0$, then by $(2)$, $b+d=-16$ and no elements of $\{(1,4),(-1,-4), (2,2), (-2,-2)\}$ satisfy this equation, then $c\neq 0$. Therefore $b=d$, and by $(4)$, $b^2$=4 therefore $b=d=\pm 2$. But by $(2)$, $$2b-c^2=-16\implies c^2=16+4=20\implies c=\pm\sqrt{20}\notin \mathbb Z,$$ therefore, the polynomial is irreducible on $\mathbb Q$.
It looks ok ? is there an easier proof ?
3) $X^4+X^2+\frac{1}{2}$
I have no idea. Actually, the real question is compute the degree of the extension $\mathbb Q(i,\alpha)/\mathbb Q$ where $\alpha$ is a root of $X^4+X^2+\frac{1}{2}$.
There is maybe an other way to solve the problem. If it can help, the root of this polynomial are the elements of
$$\left\{\frac{i}{\sqrt 2}\sqrt{1-i},-\frac{i}{\sqrt 2}\sqrt{1-i},\frac{i}{\sqrt 2}\sqrt{1+i},-\frac{i}{\sqrt 2}\sqrt{1+i}\right\}$$
| $1)$ : $$f(x)=x^4+1$$
$$f(x+1)=x^4+4x^3+6x^2+4x+2$$
and the criterion of Eisenstein shows that $f(x+1)$ is irreducible. Therefore
$f(x)$ is also irreducible.
| {
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"url": "https://math.stackexchange.com/questions/1074508",
"timestamp": "2023-03-29T00:00:00",
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Finding the points of intersection of a circle and a line In a test (of math in arabic language) we we're asked to find the points of intersection of a circle and a line. Their equation is given.
In the test I solved system of equations made of their equation and in the process I explain my line of thought using the words therefore, then, so, ... etc (in arabic) but in describing that process I acknowledge that the structure of my proof is done by equivalence, that can be easily seen from the context, we're solving an equation so we proceed by equivalence.
But my professor said that it was all wrong and that I should have used the statement is equivalent (in arabic) in each part of my proof and I got all of my exercises wrong, and there were some exercises in the test where in the process we had to solve an equation so he also said that I was wrong because of that reason so I only got 4 out of 20.
Is the professor wrong or I am?
My computer crashed so i couldn't edit so i made a new question. I've added the equations and system of equations with my solutions
Exercise 1
Circle has equation $x^2+y^2+(m+2)x-2my+m^2-36=0$, find center, radius
I found it to be center $(-\frac{m}{2}-1,m)$ radius r= $\sqrt{\frac{m^2}{4}+m+37}$
Exercise 2
Circle has equation $x^2+y^2+2 x - 2y - 4 = 0$ line has equation $x+y-1=0$ find points of intersection
i found $\left(\frac{-1+\sqrt{11}}{2},\frac{3-\sqrt{11}}{2}\right)$ and $\left(\frac{-1-\sqrt{11}}{2},\frac{3+\sqrt{11}}{2}\right)$
for exercise 3 and 4 it is the same and i checked my calculations again and again with the calculator there's no error but yet all of that was marked with 0 for not saying "is equivalent", i didn't use the symbol $\Rightarrow$ All i got was 4 from exercise 5 which had calculation of dot product and of area of triangle where we are given coordinates of two vectors in the plane
| The posted solutions are correct, below is a possible derivation.
Exercise 1
In this exercise you need to find the center and radius of this circle which depends on $m$.
$$x^2+y^2+(m+2)x-2my+m^2-36=0$$
Notice that $(y-m)^2=y^2+m^2-2my$ and rewrite as
$$x^2+(y-m)^2+(m+2)x-36=0$$
Add $0=1-1+m+\frac{m^2}4-m-\frac{m^2}4$
$$x^2+(y-m)^2+(m+2)x-37+1+m+\frac{m^2}4-m-\frac{m^2}4=0$$
Notice that $\left(x+\frac{m-2}2\right)^2=x^2+(m+2)x+\frac{m^2}4+m+1$ and rewrite as
$$\left(x+\frac{m+2}2\right)^2+(y-m)^2=37+m+\frac{m^2}4=0$$
We can now easily see that the center is
$$\left(-\frac{m+2}2,m\right)$$
And that the radius is
$$r=\sqrt{37+m+\frac{m^2}4}=\frac12\sqrt{m^2+4m+148}$$
Which is equivalent to your solution.
Exercise 2
Circle has equation line has equation find points of intersection
You need to find the intersections of the following:
$$
\begin{array}{l}
0=x^2+y^2+2 x - 2y - 4\\
0=x+y-1
\end{array}
$$
This simply means solving a system of equations:
From the second equation we get the solution for $x$ as
$$x=1-y$$
Now plug this into the first equation:
$$0=(1-y)^2+y^2+2 (1-y) - 2y - 4$$
Remove the paranthesises
$$0=2y^2-6y - 1$$
Solve for $y$ using the quadratic formula or complete the square
$$y_1=\frac12 (3+ \sqrt{11})$$
$$y_2=\frac12 (3- \sqrt{11})$$
Plug this into the formula for $x$ from earlier: $x=1-y$
$$x_1=1-\frac12 (3+ \sqrt{11})$$
$$x_2=1-\frac12 (3- \sqrt{11})$$
Simplify
$$\begin{array}l
x_1=-\frac12(1+\sqrt{11})\\
x_2=\frac12(\sqrt{11}-1)
\end{array}$$
This gives the two solutions:
$$\left(-\frac12(1+\sqrt{11}),\frac12 (3+ \sqrt{11})\right)$$
and
$$\left(\frac12(\sqrt{11}-1),\frac12 (3- \sqrt{11})\right)$$
Which is equivalent to the posted solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of a circumcenter triangle equals area of medial triangle Let $X$, $Y$, $Z$ be the midpoints of sides $BC$, $AC$, $AB$ respectively in triangle $ABC$. Let $O_{A}$, $O_{B}$, and $O_{C}$ be the circumcenters of triangles $AZX$, $BXY$, and $CYZ$ respectively. Prove that the area of triangle $O_{A} O_{B} O_{C}$ equals the area of triangle $XYZ$.
Basically, I've done a horrific coordinate bash, and the statement is definitely true. But I'd be very interested in seeing a purely Euclidean solution.
| As requested, here is a solution via coordinates. However, as mentioned above, I'd be interested in seeing a solution via pure geometry instead.
We assign coordinates $A(0,0)$, $B(b,0)$, $C(c,d)$ to the vertices of triangle $ABC$. This gives $X = ((b+c)/2, d/2)$, $Y = (c/2, d/2)$, $Z = (b/2, 0)$. $O_A$ has coordinates corresponding to the intersection of the perpendicular bisector of $AZ$ and $AX$. These bisectors have equations $x = b/4$ and $$y = -\frac{b+c}{d} \left(x-\frac{b+c}{2} \right)+ \frac{d}{4}$$ Solving the system gives $$O_{A} = \left(\frac{b}{4}, \frac{d}{4} - \frac{bc+c^2}{4d} \right)$$ Similarly, we compute $O_{B}$ and $O_{C}$: $$O_{B} = \left(\frac{2c+b}{4}, \frac{3bc+d^2-c^2-2b^2}{4d} \right)$$ $$O_{C} = \left(\frac{b^2 + 2(c^2 + d^2)}{4b}, -\frac{b^2 c+ (2c-3b)(c^2 + d^2)}{4bd} \right)$$ Plugging these in to the shoelace formula and simplifying with a CAS yields $$[O_{A} O_{B} O_{C}] = \frac{1}{4} \left(\frac{1}{2} bd\right) = \frac{1}{4} [ABC] = [XYZ]$$ as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$ Evaluating $$\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$$ using $ux=\sqrt{x^2-1}$
UPDATE 'official' solution
$$u^2x^2=x^2-1$$
$$x^2=\frac{-1}{u^2-1}$$
$$x^2+1=\frac{u^2-2}{u^2-1}$$
$$2xdx=\frac{-2u}{(u^2-1)^2}$$
$$\int{\frac{x}{x\sqrt{x^2-1}(x^2+1)}dx}$$
$$\int{\frac{1}{\frac{-1}{u^2-1}u\frac{u^2-2}{u^2-1}}\left(\frac{-u}{(u^2-1)^2}\right)du} $$
$$\int{\frac{1}{(u^2-2)}du} $$
$$\frac{\log \left(\sqrt{2}-x\right)-\log \left(x+\sqrt{2}\right)}{2 \sqrt{2}}$$
$$\frac{\log \left(\sqrt{2}-\sqrt{x^2-1}/x\right)-\log \left(\sqrt{x^2-1}/x+\sqrt{2}\right)}{2 \sqrt{2}}$$
However at mathematica I get $$\frac{\log \left(-3 x^2-2 \sqrt{2} \sqrt{x^2-1} x+1\right)-\log \left(-3 x^2+2 \sqrt{2} \sqrt{x^2-1} x+1\right)}{4 \sqrt{2}}$$
Where is the error?
| let me try a change of variable
$$x = {1 + t^2 \over 1 - t^2},\ dx = {-4t \ dt \over (1-t^2)^2},\ x^2+1 = {2(1+t^4) \over (1-t^2)^2}, \ x^2-1 = {4t^2 \over (1-t^2)^2}$$
$$\int {1 \over (x^2+1) \sqrt{x^2-1}} dx = \int {t^2 - 1 \over t^4 + 1} \ dt
= \int {1 \over t^2 + 1} dt - 2 \int {1 \over t^4 + 1} dt = \tan^{-1}t - 2\int{dt \over t^4+1}$$
i got to go now. i will see if can do the last integral later.
edit: Idris, thanks for the hint. i can now evaluate
$
\begin{eqnarray}
2\int {dt \over 1 + t^4} &=& 2\int {1/t^2 \over t^2 + 1/t^2}dt
= \int {1+1/t^2 \over t^2 + 1/t^2} \ dt -
\int {1-1/t^2 \over t^2 + 1/t^2} \ dt \\
&=&\int {d(t-1/t)\over (t-1/t)^2 + 2} -
\int {d(t+1/t)\over (t+1/t)^2 - 2} \\
&=&{1 \over \sqrt 2}\tan^{-1}\left({t - 1/t \over \sqrt 2} \right) -
{1 \over 2\sqrt 2}\ln\left({t + 1/t -\sqrt2 \over t + 1/t +\sqrt 2} \right)
\end{eqnarray}
$
so finally,
$$\int {1 \over (x^2+1) \sqrt{x^2-1}} dx = \tan^{-1}t -
{1 \over \sqrt 2}\tan^{-1}\left({t - 1/t \over \sqrt 2} \right) +
{1 \over 2\sqrt 2}\ln\left({t + 1/t -\sqrt2 \over t + 1/t +\sqrt 2} \right) + C$$
second edit: sorry, i made a mistake and i will fix it.
$\begin{eqnarray}
\int {1 \over (x^2+1) \sqrt{x^2-1}} dx &=& \int {t^2 - 1 \over t^4 + 1} \ dt
=\int{1 - 1/t^2 \over t^2 + 1/t^2}\ dt
= \int {d(t+1/t) \over (t+1/t)^2 - 2} \ dt \\
&=& {1 \over 2\sqrt 2}\ln\left({t + 1/t -\sqrt2 \over t + 1/t +\sqrt 2} \right) + C
\end{eqnarray}$
| {
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Maximising the Area of a Cyclic Quadrilateral In cyclic quadrilateral $ABCD$, $AB = AD$. If $AC = 6$ and $AB/BD = 3/5$, find the maximum possible value of $[ABCD]$.
(Source: SMT 2014)
If we let $AB=AD = 3x$ and $BD=5x$, from Ptolemy, we have $BC+CD=10$. From the triangle inequality, we can also get $BC+CD > BD \Rightarrow 10 > 5x \Rightarrow x < 2$. I'm stuck on where to go from here.
| I've been exposed to a solution to this problem.
Since $AB=AD$, $\angle ABD = \angle ADB$. Furthermore, from the proprerties of a cyclic quadrilateral, $\angle ACB = \angle ADB$ and $\angle ACD = \angle ABD$. Therefore, $\angle ACB = \angle ADB = \angle ABD = \angle ACD$. Most importantly, $\angle ACB = \angle ACD$. Note that the area of $ABCD$ is equal to the area of $ABC$ added to the area of $ADC$. We use the alternate area formula $\frac{1}2 ab \sin C$, as follows: (Note that we let $\angle ACB = \angle ADB = \angle ABD = \angle ACD = \alpha$)
\begin{align*}\\&[ABCD] \\&= [ABC]+[ADC] \\&=\frac{1}2(AC)(BC)\sin\alpha+\frac{1}2(AC)(DC)\sin\alpha \\&=\frac{1}2(AC)\sin\alpha(BC+CD)\end{align*}
We are given $BC+CD = 10$ and $AC = 6$, so that $[ABCD] =\frac{1}2 (10)(6)(\sin \alpha) = 30 \sin \alpha$. Since $\alpha = \angle ADB$, we can compute the $\cos$ of $\alpha$ easily: drop the perpendicular from A to BD, and let X be the foot of that perpendicular. Note that $AXD$ is a right triangle with $DX = \frac{5}{2} x$, so $\cos \alpha = \frac{\frac{5}2}3 = \frac{5}6$. Therefore, since $\sin^2 \alpha + \cos^2 \alpha = 1$ and $\sin \alpha$ is positive, $\sin \alpha = \sqrt{11}{6}$, and the answer is $\boxed{5\sqrt{11}}$. The area is the same no matter what.
Thanks to the active members of the Art of Problem Solving Forums for posting this solution. Original link was here. The problem was lovely.
| {
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"timestamp": "2023-03-29T00:00:00",
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primes of the form $4k+3$ and sums of squares It is well-known that if $p$ is a prime of the form $4k+3$ and $p|x^2+y^2$ then $p|x$ and $p|y$. I forget what is the name of this result, and where can I find a proof (please provide a link).
| EDIT: note that your binary quadratic form $x^2 + y^2$ has discriminant $\Delta = -4.$
Somewhere you need to find a proof of the fact that, for prime $q \equiv 3 \pmod 4,$ we always have
$$ (-1|q) = -1. $$
For instance,
Niven and Zuckerman (and Montgomery) page 132, Theorem 3.1 part (5), says
$$ (-1|p) = (-1)^{(p-1)/2}. $$ Same thing in Ireland and Rosen, Proposition 5.1.2, Corollary 3, top of page 52.
Given a binary quadratic form
$$ \color{blue}{f(x,y) = a x^2 + b xy+ c y^2} $$
with $a,b,c$ integers.
Given its discriminant
$$ \color{red}{ \Delta = b^2 - 4 a c}, $$
where we require that $\Delta,$ if non-negative, is not a square (so also $\Delta \neq 0,1$).
Proposition: given an odd prime $q$ such that $q$ does not divide $\Delta$ and, in fact, $$ (\Delta| q) = -1, $$
whenever $$ f(x,y) \equiv 0 \pmod q, $$ then BOTH
$$ x \equiv 0 \pmod q, \; \; \; \; \; y \equiv 0 \pmod q. $$
Proof: the integers taken $\pmod q$ form a field; every nonzero element has a multiplicative inverse. If either $a,c$ were divisible by $q,$ we would have $\Delta $ equivalent to $b^2$ mod $q,$ which would cause $(\Delta|q)$ to be $1$ rather than $-1.$ As a result, neither $a$ nor $c$ is divisible by $q.$
Next, $q$ is odd, so that $2$ and $4$ have multiplicative inverses mod $q,$ they are non zero in the field. Put togethaer, $$4a \neq 0 \pmod q$$
Now, complete the square:
$$ a x^2 + b xy+ c y^2 \equiv 0 \pmod q $$
if and only if
$$ 4a(a x^2 + b xy+ c y^2) \equiv 0 \pmod q, $$
$$ 4a^2 x^2 + 4abxy + 4ac y^2 \equiv 0 \pmod q, $$
$$ 4a^2 x^2 + 4abxy + (b^2 y^2 - b^2 y^2) + 4ac y^2 \equiv 0 \pmod q, $$
$$ 4a^2 x^2 + 4abxy + b^2 y^2 - (b^2 y^2 - 4ac y^2) \equiv 0 \pmod q, $$
$$ (4a^2 x^2 + 4abxy + b^2 y^2) - (b^2 - 4ac) y^2 \equiv 0 \pmod q, $$
$$ (2ax + by)^2 - (b^2 - 4ac) y^2 \equiv 0 \pmod q, $$
$$ (2ax + by)^2 - \Delta y^2 \equiv 0 \pmod q, $$
$$ (2ax + by)^2 \equiv \Delta y^2 \pmod q. $$
Now, ASSUME that $y \neq 0 \pmod q.$ Then $y$ has a multiplicative inverse which we are allowed to call $1/y,$ and we have
$$ \frac{(2ax + by)^2}{y^2} \equiv \Delta \pmod q, $$
$$ \left( \frac{2ax + by}{y} \right)^2 \equiv \Delta \pmod q. $$
However, the HYPOTHESIS that $ (\Delta| q) = -1 $ says that this is impossible, thus contradicting the assumption that $y \neq 0 \pmod q.$
So, in fact, $y \equiv 0 \pmod q.$ The original equation now reads
$$ a x^2 \equiv 0 \pmod q $$ with the knowledge that $a \neq 0 \pmod q,$ so we also get
$$ x \equiv 0 \pmod q, \; \; \; \; \; y \equiv 0 \pmod q. $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:
*
*$\sqrt{4x-1} = \sqrt{x+2}-3$
*$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
*$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
*$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x+2})^2-3\sqrt{x+2}-3\sqrt{x+2}+9$
*$\sqrt{(4x-1)^2} = \sqrt{(x+2)\times(x+2)}-6\sqrt{x+2}+9$
*$\underline{4x-1} = \sqrt{(x+2)^2}-6\sqrt{x+2}+9$
*$4x-1 = \underline{x+2}-6\sqrt{x+2}+9$
How did the underlined expressions in steps 6 and 7 not include $\pm$? I thought that $\sqrt{x^2} = \pm x$.
| After thinking a bit about some of what @Daniel Hast typed, I realized that the reason why, say, $\sqrt{(4x-1)^2} = 4x-1$ was that more generally, given that $\sqrt{x} = r$ such that $r^2 = x$, squaring $r$ in $\sqrt{x} = r$ produced $(\sqrt{x})^2 = r^2 = x$ and subsequently $(\sqrt{x})^2 = x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $I(a) = \int_0^{\pi/2} \frac{dx}{1-a\sin x}$ I have a problem with this integral. It seems that solution has to be simple, but I couldn't find out.
$$I(a) = \int_0^{\pi/2} \frac{dx}{1-a\sin x}$$
I tried using integration by parts and differentiating with regard to a, but neither helped.
| Observe that your integral is convergent in the usual sense for $|a|<1$. You may then write $t=\tan \dfrac{x}{2}$, giving $ \sin x=\dfrac{2t}{1+t^2}$.
Hence
$$
\begin{align}
I(a) &= \int_0^{\pi/2} \frac{dx}{1-a\sin x}\\\\
&= 2\int_0^{1} \frac{1}{1-a\dfrac{2t}{1+t^2}}\frac{dt}{1+t^2}\\\\
&= 2\int_0^{1} \frac{1}{(t-a)^2+(1-a^2)}dt\quad \left(t-a=\sqrt{1-a^2}\:u,\,dt=\sqrt{1-a^2}\:du\right)\\\\
&= \frac{2}{\sqrt{1-a^2}}\int_{-a/\sqrt{1-a^2}}^{(1-a)/\sqrt{1-a^2}} \frac{1}{u^2+1}du\quad \\\\
&= \frac{2}{\sqrt{1-a^2}}\arctan \left(\sqrt{\frac{1+a}{1-a}}\right)
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to evaluate residue of $\cot z/z^4$ at $z=0$? How to evaluate residue of $\cot z/z^4$ at $z=0$?
As we know :
$$f(x)=f(0)+f'(0)x+f''(0)x^2/2+...$$
but $\cot(0)\to\infty$ or is undefined? I know that:
$$\tan x=x+x^3/3+2x^5/15+...$$
|
$$\frac{\cos z}{z^4\sin z}$$
$$\newcommand{\f}[2]{\frac{#1}{#2}}
\newcommand{\r}[1]{\frac{1}{#1}}
\newcommand{\array}[2]{\begin{array}{#1}#2\end{array}}
\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\s}[1]{\left[#1\right]}
\newcommand{\d}[0]{\ldots}
\newcommand{\ul}[1]{\underline{#1}}
\array{r|lll}{
&&\r{z^5}&-\b{\r{3!}-\r{2!}}\r{z^3}&+\s{\b{\r{4!}-\r{5!}}+\r{3!}\b{\r{3!}-\r{2!}}}\r{z}&+\d\\\hline
z^5-\f{z^7}{3!}+\f{z^9}{5!}+\d&&1&-\f{z^2}{2!}&+\f{z^4}{4!}&+\d\\
&&1&-\f{z^2}{3!}&+\f{z^4}{5!}&+\d\\\hline
&&&\b{\r{3!}-\r{2!}}z^2&+\b{\r{4!}-\r{5!}}z^4&+\d\\
&&&\b{\r{3!}-\r{2!}}z^2&+\r{3!}\b{\r{3!}-\r{2!}}z^4&+\d\\\hline
&&&&\s{\b{\r{4!}-\r{5!}}+\r{3!}\b{\r{3!}-\r{2!}}}z^4&+\d\\
&&&&\s{\b{\r{4!}-\r{5!}}+\r{3!}\b{\r{3!}-\r{2!}}}z^4&+\d\\\hline
&&&&&+\d
}\\
\begin{align}
R&=\r{4!}\b{1-\r{5}}+\r{3!}\s{\r{2!}\b{\r{3}-1}}\\
&=\r{24}\b{\f45}+\r{12}\b{\f{-2}3}\\
&=\r{6}\b{\r{5}-\r{3}}\\
&=\f{-2}{6\times15}\\
&=-\r{45}
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
| Let
$$a=\lim_{x\to\infty}x\left(\sqrt{x^2+1}-x\right)$$
Now rationalize it:
$$a=\lim_{x\to\infty}\dfrac{x\left(x^2+1-x^2\right)}{\sqrt{x^2+1}+x}$$
Then simplify it:
$$a=\lim_{x\to\infty}\dfrac{x}{\sqrt{x^2+1}+x}$$
$$a=\lim_{x\to0}\dfrac{\dfrac1x}{\sqrt{\left(\dfrac1x\right)^2+1}+\dfrac1x}$$
$$a=\lim_{x\to0}\dfrac{1}{\sqrt{x^2+1}+1}$$
$$a=\dfrac{1}{\sqrt{0^2+1}+1}=\dfrac{1}{1+1}=\dfrac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $2(x+y)+xy=x^2+y^2$ where $x,y \in \mathbb{Z}$ Solve the equation: $$2(x+y)+xy=x^2+y^2$$
How should I go about solving this? Any guidance appreciated.
Thanks!
| $$8(x+y)=4(x^2-xy+y^2)\\8(x+y)=3(x-y)^2+(x+y)^2\\8z=3w^2+z^2\\16=3w^2+(z-4)^2$$
where $w=x-y$ and $z=x+y$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Quadratic congruences in which modulus is divisible by constant term There are two similar congruences:
$$
x^2-6x\equiv16\pmod{512}\\
y^2-y\equiv16\pmod{512}
$$
It is easy to see that the $\gcd$ of all three parts in both of them is $16$, $x$ and $x-6$ are even and one of $y$ and $y-1$ is odd. After also noticing $x\not\equiv x-6\pmod{4}$ we have
$$
x \equiv 0 \pmod{8}\\
or\\
x-6 \equiv 0 \pmod{8}
$$
$$
y \equiv 0 \pmod{16}\\
or\\
y-1 \equiv 0 \pmod{16}
$$
Making substitutions $x=8n,x=8n+6,y=16m,y=16m+1$ I obtained congruences with modulus $32$, but couldn't make it any further. I also discovered that the first congruence can be rewritten as $(x+2)(x-8) \equiv 0\pmod{512}$ while $-2$ and $8$ appear to be its solutions. However, it's still not clear to me whether there are more of them.
What am I missing? How can such congruences be solved?
| The first one is easy. You need $(x-8)(x+2)$ divisible by $512$. Both $x+8$ and $x-2$ are even, but only one of them can have higher powers of $2$ as factors. Thus, $x\equiv 8\pmod {256}$ or $x\equiv -2\pmod{256}$.
The second requires more care.
First assuming $y\equiv 0\pmod{16}$ you have $y=16y_0$ and $$16y_0^2=y_0+1\pmod{32}$$
Since $16y_0^2$ is even, $y_0+1$ is even, so $y_0$ is odd. When $y_0$ odd, $16y_0^2\equiv 16\pmod{32}$, so $y_0\equiv 15\pmod {32}=$ and $y\equiv 16\cdot 15=240\pmod{512}$.
Second, assume $y=16y_0+1$. Then $$16y_0^2 \equiv 1-y_0\pmod{32}$$ So $y_0$ is odd, $1-y_0\equiv 16\pmod {32}$, or $y_0\equiv 17\pmod{32}$. So $y\equiv 16\cdot 17+1=273\pmod{512}$
So the two solutions are $y\equiv 240,273\pmod{512}$.
Note that the sum of these two values is $1\pmod{512}$, as befits the two roots of any quadratic equation of the form $x^2-x+C=0$.
Indeed, we have $(y-240)(y-273)\equiv y^2-y-16\pmod{512}$, so these are the only solutions, since only one of $y-240$ and $y-273$ can be even.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of function at infinity: $ \lim_{n\to \infty} \frac{a^n-b^n}{a^n+b^n} $ I'm learning the basics of Calculus. I got stuck in a very (seemingly) simple problem of evaluating a limit. The problem given is:
$$
\lim_{n\to \infty} \frac{3^n-2^n}{3^n+2^n}
$$
The answer given is $1$ (I have no idea how, or why).
I think $\lim_{x \to a}\frac{x^m-a^m}{x^n-a^n}=\frac{m}{n}a^{m-n}$, isn't going to help me in any way because in this formula,
*
*$x \to a$, not $\infty$
*$x$ is not the exponent.
Sadly, there's no other formula in my book that seems close to the problem. So could anyone please show me how to solve such problems? Also, could you show me how to solve the general form of the problem, like:
$$
\lim_{n\to \infty} \frac{a^n-b^n}{a^n+b^n}
$$
Thanks!
| The other answers have already shown the standard trick. Here's how we might have proceeded if we hadn't noticed it applies.
One significant point we should understand is that as $n \to \infty$, $3^n$ dwarfs $2^n$ in size; the difference between $3^n$ and $3^n \pm 2^n$ should be negligible. So we estimate
$$ \frac{3^n - 2^n}{3^n + 2^n} \sim 1 $$
Of course, an estimate is not proof: so we look for the error term. We want to write
$$ \frac{3^n - 2^n}{3^n + 2^n} = 1 + x $$
But what is $x$? It's actually easy to find: just solve:
$$ x = \frac{3^n - 2^n}{3^n + 2^n} - 1 = -\frac{2 \cdot 2^n}{3^n + 2^n} $$
and so we want to find
$$ \lim_{n \to \infty} \left( 1 - \frac{2 \cdot 2^n}{3^n + 2^n} \right) $$
We might do this by recognizing that
$$ 0 < \frac{2^n}{3^n + 2^n} < \frac{2^n}{3^n} = \left( \frac{2}{3} \right)^n$$
so that we can use the squeeze theorem. But maybe we didn't notice that, nor did we notice we can use the standard trick. We could still repeat the idea above and estimate $x \sim -2 \cdot (2/3)^n$. Again finding the error term, we want
$$ \lim_{n \to \infty}\left( 1 - 2 \left(\frac{2}{3} \right)^n + \frac{2 \cdot 2^{2n}}{3^n (3^n + 2^n)} \right) $$
or rearranging,
$$ \lim_{n \to \infty}\left( 1 - 2 \left(\frac{2}{3} \right)^n + \left( \frac{2}{3} \right)^n \cdot \frac{2 \cdot 2^n}{3^n + 2^n} \right) $$
Now, the only 'trick' we need to compute the limit is to realize that
$$ 0 < \frac{2 \cdot 2^n}{3^n + 2^n} < 2 $$
because $2^n$ is always smaller than the denominator. Since $(2/3)^n \to 0$, this is enough to know that the third term vanishes.
The standard trick is, actually, not too dissimilar from the idea above. When we estimate $3^n \sim 3^n - 2^n$, the idea is rather than adding in an error term, we multiply in the error term: we want to write
$$ 3^n - 2^n = 3^n \cdot (1 + x) $$
and in this case, we get
$$ 3^n - 2^n = 3^n \cdot \left(1 - \left(\frac{2}{3}\right)^n\right) $$
Another way of thinking about this same idea is that we want to factor out the most significant part: in this case, $3^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of : $\lim_{x \to \infty}( \sqrt{4x^2+5x} - \sqrt{4x^2+x})$ $$\lim_{x \to \infty} \left(\sqrt{4x^2+5x} - \sqrt{4x^2+x}\ \right)$$
I have a lot of approaches, but it seems that I get stuck in all of those unfortunately. So for example I have tried to multiply both numerator and denominator by the conjugate $\left(\sqrt{4x^2+5x} + \sqrt{4x^2+x}\right)$, then I get $\displaystyle \frac{4x}{\sqrt{4x^2+5x} + \sqrt{4x^2+x}}$, but I can conclude nothing out of it.
| Set $h=\dfrac1x$
$$4x^2+ax=\frac{4+ah}{h^2}\implies\sqrt{4x^2+ax}=\frac{\sqrt{4+ah}}{\sqrt{h^2}}$$
Now as $h\to0^+,h>0\implies\sqrt{h^2}=|h|=h$
$$\implies\lim_{x \to \infty} (\sqrt{4x^2+5x} - \sqrt{4x^2+x})=\lim_{h\to0^+}\frac{\sqrt{4+5h}-\sqrt{4+h}}h$$
$$=\lim_{h\to0^+}\frac{4+5h-(4+h)}{h(\sqrt{4+5h}+\sqrt{4+h})}=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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find $\sum_{i=1}^\infty \frac{1}{n3^n}$ How to find $$\sum_{i=1}^\infty \frac{1}{n3^n}$$
Don't know how to start, any hints
A rigorous proof is also welcome
| Consider
\begin{align*}
\sum_{n=1}^\infty\frac{x^n}{n}.
\end{align*}
This series converges for all $|x|<1$. Differentiation wrt $x$ yields
\begin{align*}
\sum_{n=1}^\infty x^{n-1}=\sum_{n=0}^\infty x^n=\frac{1}{1-x},
\end{align*}
and integrating again gives
\begin{align*}
\sum_{n=1}^\infty\frac{x^n}{n}=\int\frac{1}{1-x}dx+c=-\log(1-x)+c.
\end{align*}
When we plug in $x=0$ we obtain $c=0$, so we eventually have
\begin{align*}
\sum_{n=1}^\infty\frac{x^n}{n}=-\log(1-x),\qquad |x|<1.
\end{align*}
Now, for $x=\frac{1}{3}$ we obtain
\begin{align*}
\sum_{n=1}^\infty\frac{1}{n3^n}=\log\left(\frac{3}{2}\right).
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^4+3x^3+6x+4=0$... easier way? So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$
Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but none of these work.
Because there were no rational linear factors, I had to assume that the quartic separated into two quadratic equations yielding either imaginary or irrational "pairs" of roots. My initial attempt was to "solve for the coefficients of these factors".
I assumed that $x^4+3x^3+6x+4=0$ factored into something that looked like this
$$\left(x^2+ax+b\right)\left(x^2+cx+d\right)=0$$
because the coefficient of the first term is one. Expanding this out I got
$$x^4+ax^3+cx^3+bx^2+acx^2+dx^2+adx+bcx+bd=0$$
$$x^4+\left(a+c\right)x^3+\left(b+ac+d\right)x^2+\left(ad+bc\right)x+bd=0$$
Equating the coefficients of both equations
$$a+c = 3$$
$$b+ac+d = 0$$
$$ad+bc = 6$$
$$bd = 4$$
I found these relationships between the various coefficients. Solving this system using the two middle equations:
$$\begin{cases} b+a\left(3-a\right)+\frac4b=0 \\ a\frac4b+b\left(3-a\right)=6 \end{cases}$$
From the first equation: $$a = \frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}$$
Substituting this into the second equation:
$$\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\cdot\frac4b+b\cdot\left(3-\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\right)=6$$
$$3\left(b-2\right)^2 = \left(b^2-4\right)\cdot\pm\sqrt{9+4b+\frac{16}b}$$
$$0 = \left(b-2\right)^2\cdot\left(\left(b+2\right)^2\left(9+4b+\frac{16}b\right)-9\left(b-2\right)^2\right)$$
So $b = 2$ because everything after $\left(b-2\right)^2$ did not really matter in this case. From there it was easy to get that $d = 2$, $a = -1$ and $c = 4$. This meant that
$$x^4+3x^3+6x+4=0 \to \left(x^2-x+2\right)\left(x^2+4x+2\right)=0$$
$$x = \frac12\pm\frac{\sqrt7}{2}i,\space x = -2\pm\sqrt2$$
These answers worked! I was pretty happy at the end that I had solved the equation which had taken a lot of work, but my question was if there was a better way to solve this?
| More generally:
Given a quartic equation with the form
$x^4+bx^3+cx^2+(br)x+r^2=0$
you guarantee factors with the form
$(x^2+px+r)(x^2+qx+r)=0$
For then the cubic and linear terms are both matched by rendering
$p+q=b$
and we have only to match the quadratic terms which oeadsto
$pq=c-2r^2$
From these equations $p$ and $q$ are roots of the quadratic equation
$u^2-bu+(c-2r^2)=0.$
Here $b=3,c=0,r=2$, from which
$u^2-3u-4=0.$
Thus $\{p,q\}=\{-1,4\}$ from which we obtain the factorization
$x^4+3x^3+6x+4=(x^2-x+2)(x^2+4x+2).$
The quartic equation is then solved by the roots of these quadratic factors.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "33",
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Find Minimum Value of: $P=x^2+y^2+2xy+2-\frac{1}{xy}$ Given: $x,y>0$ and $x^2y-x+xy^2-y-3xy=0$
Find Minimum Value of:
$P=x^2+y^2+2xy+2-\frac{1}{xy}$
I found $\min P =\frac{71}{4}$ at $x=y=2$ but I cant prove that
Could some one help me ?
| It is sure that, without using Lagrange multipliers, the problem is quite hard.
But let us try using brute force. The constraint $x^2y-x+xy^2-y-3xy=0$ gives a quadratic equation in $y$; so we can express $y$ as a function of $x$ $$y_{\pm}=\frac{-x^2+3 x+1\pm\sqrt{x^4-6 x^3+11 x^2+6 x+1}}{2 x}$$ Replace $y_+$ in $P$ (which is now an ugly function of $x$) and plot it (even better, compute $P'_x$ and plot it to find the root). You will effectively find $x=2$ to which corresponds $y=2$ and $P=\frac{71}{4}$.
Doing the same with $y_-$ shows a minimum for $x\approx 1.54426$ corresponding to $y \approx-0.399545$ which has to be discarded.
| {
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How prove this inequality $(y-z)^2>8(y+z)$
For three distinct postive integer $x,y,z$ such
$$(x+y)(x+z)=(y+z)^2$$
show that
$$(y-z)^2>8(y+z)$$
My some idea: since
$$x^2+(y+z)x+yz-(y+z)^2=0$$
so
$$\Delta_{x}=(y+z)^2-4yz+4(y+z)^2=5(y+z)^2-4yz=m^2$$then I can't
| WLOG $y<z$, so let $z=y+a$, where $a>0$. Then the discriminant is
\begin{align*}5(y+z)^2-4yz&=5(2y+a)^2-4y(y+a)=5(4y^2+4ya+a^2)-4y^2-4ya\\
&=4(4y^2+4ya+a^2)+a^2=4(2y+a)^2+a^2=(4y+2a)^2+a^2\end{align*}
So if it's a square, then it's either the next square
$$(4y+2a+1)^2=(4y+2a)^2+2(4y+2a)+1$$
and
$$a^2=4(2y+a)+1\implies 8y+5=a^2-4a+4=(a-2)^2,$$
but that's not possible since squares cannot be $5$ mod $8$ (the remainders can only be $0$, $1$, or $4$).
Or it must be at least
$$(4y+2a+2)^2=(4y+2a)^2+4(4y+2a)+4,$$
then
$$(z-y)^2=a^2>4(4y+2a)=8(2y+a)=8(y+z).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Deriving inverse matrix formula If matrix $A$ is given with dimensions $2 \times2 $ then,
A is invertible if, and only, if $ad - bc \neq 0$:
$$\begin{bmatrix}a & b\ \\c & d \ \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b\ \\-c & a \ \end{bmatrix}$$
How can this be derived?
I just need hints, I am not good at properties etc.. of matrices, please help!
| Notice that
$$\begin{align}\begin{bmatrix}a & b\ |\ 1 & 0 \\c & d \ |\ 0 & 1\end{bmatrix} &\Leftrightarrow\begin{bmatrix}a & b\ &|\ 1 & 0 \\0 & \frac{ad - bc}{a} \ &|\ -\frac{c}{a} & 1\end{bmatrix} \Leftrightarrow\begin{bmatrix}a & b\ &|\ 1 & 0 \\0 & 1 \ &|\ -\frac{c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix}\\&\Leftrightarrow \begin{bmatrix}a & 0\ &|\ \frac{ad}{ad-bc} & \frac{-ba}{ad-bc} \\0 & 1 \ &|\ \frac{-c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix} \Leftrightarrow \begin{bmatrix}1 & 0\ &|\ \frac{d}{ad-bc} & \frac{-b}{ad-bc} \\0 & 1 \ &|\ \frac{-c}{ad-bc} & \frac{a}{ad-bc}\end{bmatrix}\end{align}$$
Then
$$\begin{bmatrix}a & b\ \\c & d \ \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix}d & -b\ \\-c & a \ \end{bmatrix}$$
| {
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Efficient Way of finding possible sums in equation I am looking for an efficient way to find the following numbers numbers $a, b$
Let $x$ be 63 (as example)
$$\sum_{k = a}^b k = \sum_{l = 1}^b l - \sum_{j=1}^{a-1} j = \frac{b*(b+1)}{2}-\frac{a*(a-1)}{2} = x = 63$$
in this case for example
$$63 = 8+9+10+11+12+13 = \sum_{k = 8}^{13} k $$
| Given positive integer $x$, you want positive integers $a,b$ such that
$$ 2x = b(b+1) - a(a-1) $$
If $c = 2a-1$ and $d=2b+1$, this says
$$ (d-c)(d+c) = d^2 - c^2 = 8x $$
(with $c$ and $d$ both odd).
So look at factorizations of $8x$. If $8x = uv$ where $u$ and $v$ are positive integers, $u < v$,
we want $d-c = u$ and $d+c = v$, and then $d = (u+v)/2$, $c = (v-u)/2$.
Note that we can take $u \equiv 2 \mod 4$ and $v \equiv 0 \mod 4$ (or vice versa), and get $c$ and $d$ odd.
In your example, $x=63$, $8x = 2^3 \times 3^2 \times 7$, and the solutions are:
*
*$u = 2$, $v = 252$, $c = 125$, $d = 127$, $a = 63$, $b = 63$
*$u = 6$, $v = 84$, $c=39$, $d=45$, $a = 29$, $b=22$
*$u = 14$, $v = 36$, $c=11$, $d=25$, $a = 6$, $b=12$
*$u = 18$, $v = 28$, $c=5$, $d=23$, $a = 3$, $b=11$
*$u = 12$, $v = 42$, $c=15$, $d=27$, $a = 8$, $b=13$
*$u = 4$, $v = 126$, $c=61$, $d=65$, $a = 31$, $b=32$
| {
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Solve logarithmic equation for $x$ to find the inverse of $f(x)= \ln(x+\sqrt{x^2+1})$
Let $f(x)= \ln(x+\sqrt{x^2+1})$. Find $f^{-1}(x)$.
Here is what I got so far: $y= \ln(x+\sqrt{x^2+1})$, rewrite as $x= \ln(y+\sqrt{y^2+1})$,
then $$e^x= y+\sqrt{y^2+1}$$ $$e^x-y= \sqrt{y^2+1}$$ $$ y^2+ e^{2x}-2(e^x)y= 1$$
So if $e^x= a$, then $a^2-2ay-1= 0$
| $$y=\log(x+\sqrt{x^2+1})\implies x+\sqrt{x^2+1}=e^y\implies x^2+1=e^{2y}-2xe^y+x^2\implies$$
$$2e^yx=e^{2y}-1\implies x=\frac{e^{2y}-1}{2e^y}=\frac{e^y-e^{-y}}2=\sinh y (=\text{hyperbolic sine})$$
Also:
$$f'(x)=\left(1+\frac x{\sqrt{x^2+1}}\right)\frac1{x+\sqrt{x^2+1}}=\frac1{\sqrt{x^2+1}}$$
Thus, by the theorem of the derivative of the inverse function:
$$(f^{-1})'(x)=\left.\frac1{f'(x)}\right|_{x\leftrightarrow y}=\left.\sqrt{x^2+1}\right|_{x\leftrightarrow y} =\sqrt{\left(\frac{e^y-e^{-y}}2\right)^2+1}=\frac{e^y+e^{-y}}2=\cosh y$$
You can, of course, also differentiate directly the explicit formula for $\;f^{-1}\;$ .
| {
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integration using partial fraction with repeating denominator So i need to integrate this $1/[(x^3)(x-1)]$, that means it could be decomposed into:
$$A/x + B/x^2 + C/x^3 + D(x-1)$$
Furthermore the resulting equation would then be:
*
*$1 = A(x^2)(x-1) + Bx(x-1) + C(x-1) + Dx^3$
*if i let $x=0$ then $C=-1$
*if i let $x=1$ then $D=1$
But i don't know how to get the values for A and B can somebody help me?
| Here's an easy way to do it:
$$\frac{1}{x^3(x-1)}=\frac{Ax^2+Bx+C}{x^3}+\frac{D}{x-1}$$
I chose to start at $x^2$ for the first one since the degree of the denominator is $3$, and on the other $x^0$ because the denominator has degree $1$ (so go all the way down starting with $x^{\text{denominator degree}-1}$).
Combine the two fractions into a fraction with the original denominator:
$$\frac{(Ax^2+Bx+C)(x-1)+D(x^3)}{x^3(x-1)}$$
And recall that the numerator was equal to $1$:
$$(Ax^2+Bx+C)(x-1)+D(x^3)=1$$
Distribute:
$$Ax^3+Bx^2+Cx-Ax^2-Bx-C+Dx^3=1$$
Combine like terms:
$$(A+D)x^3+(B-A)x^2+(C-B)x-C=1$$
And guess what..?
$$(A+D)x^3+(B-A)x^2+(C-B)x-C=0x^3+0x^3+0x+1$$
Hence we have:
$$\begin{cases}A+D=0\\B-A=0\\C-B=0\\-C=1\end{cases}$$
Which gives you:
$$\begin{cases}A=-1\\B=-1\\C=-1\\D=1\end{cases}$$
And so you have:
$$\frac{-x^2-x-1}{x^3}+\frac1{x-1}=-\frac1{x}-\frac1{x^2}-\frac1{x^3}+\frac1{x-1}$$
Which is easy to integrate:
$$\begin{align}&\int\left(-\frac1{x}-\frac1{x^2}-\frac1{x^3}+\frac1{x-1}\right)dx\\=&-\ln x+\frac1{x}+\frac1{2x^2}+\ln(x-1)\end{align}$$
| {
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Permutation in discrete math Is the permutation
$$\begin{pmatrix} 1& 2 &3 &4 &5 &6&7 \\ 7 & 4 & 2 & 1 & 3 & 6 & 5 \end{pmatrix}$$
even or odd?
The product of disjoint cycles is $$\begin{pmatrix}1& 7&5&3&2&4\end{pmatrix}\begin{pmatrix}6\end{pmatrix}$$
and the transposition are $$\begin{pmatrix}1 & 7 \end{pmatrix}\begin{pmatrix}1 & 5 \end{pmatrix}\begin{pmatrix}1 & 3\end{pmatrix}\begin{pmatrix}1 & 2 \end{pmatrix}\begin{pmatrix}1 & 4 \end{pmatrix}\begin{pmatrix}6 \end{pmatrix}$$ is it correct?
and the answer is even??
I feel confused about this..
| Another way to check the parity of a permutation is to see how many pairs are out of order (this is the number of inversions). There are $11$:
$$
\{(7,4),(7,2),(7,1),(7,3),(7,6),(7,5),(4,2),(4,1),(4,3),(2,1),(6,5)\}
$$
Thus, this is an odd permutation.
| {
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If $2^x=9$ and $4^y=27$ then what will $\frac {x+2y}{2y-4x}$ be? If $2^x=9$ and $4^y=27$ then what will $\frac {x+2y}{2y-4x}$ be?
I have seen such problems many times but no specific method to solve them.
I hope I find the general rule to solve them here.
| since
$$2^{x+2y}=2^x\cdot 4^y=9\cdot 27=3^5$$
and note
$$2^{2y-4x}=4^y/(2^x)^4=27/9^4=3^{-5}$$
so
$$2^{x+2y}\cdot 2^{2y-4x}=2^{(x+2y)+(2y-4x)}=3^5\cdot 3^{-5}=1=2^{0}$$
so $$\Longrightarrow x+2y+2y-4x=0$$
$$\Longrightarrow x+2y=-(2y-4x)$$
so
$$\Longrightarrow \dfrac{x+2y}{2y-4x}=-1$$
| {
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Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}.$$
I know that $$\sum_{n=0}^\infty\frac{1}{n!}=e$$
so $$\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1$$
But, what can I do for $$\sum_{n=1}^\infty\frac{1}{(n+1)!}$$ ?
Am I allowed to start a sum for $n=-1$ ? How can I bring to a something similar to $$\sum_{n=0}^\infty\frac{1}{n!}$$?
| To evaluate the second series note:
$$ e=\sum_{n=0}^\infty\frac{1}{n!}=1+1+\frac{1}{2!}+\frac{1}{3!}+\dots$$
$$ \sum_{n=1}^\infty\frac{1}{(n+1)!}=\frac{1}{2!}+\frac{1}{3!}+\dots$$
Thus the second series is just $e-2$. If you combine the two series you get $1$.
| {
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If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ Problem :
If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$
Solution :
$f(x) f(\frac{1}{x})-f(x) =f(\frac{1}{x})$
$\Rightarrow f(x) =\frac{f(1/x)}{f(1/x)-1}$.....(i)
Also $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$
$\Rightarrow f(\frac{1}{x})=\frac{f(x)}{f(x)-1}$ ......(ii)
On multiplying (i) and (ii) , we get
$f(x) .f(\frac{1}{x})=\frac{f(1/x).f(x)}{(f(1/x)-1) ((f)(x)-1)}$
$\Rightarrow (f(\frac{1}{x}) -1)(f(x)-1)=1$
Please suggest how to proceed here since $f(x)-1 ; \& f(\frac{1}{x}-1)$ are reciprocal to each other
Thanks
| Let $f$ be an integer polynomial with $f(x)f(1/x) = f(x) + f(1/x)$. There are four solutions, $f = 0$, $f = 2$, and $f = 1 \pm x^n$.
Clearly $f = 0$ is a solution.
As in the question, we have:
$$ (f(x) - 1)(f(1/x) - 1) = 1 $$
Let $n$ be the degree of $f \neq 0$. Multiplying both sides of the above by $x^n$ gives:
$$ (f(x) - 1)(x^n f(1/x) - x^n) = x^n $$
Now both terms are now polynomials in $\mathbb Z [x]$ with degree $n$. Therefore, by unique prime factorization one of the factors on the left must be $x^n$ (up to units), and the other factor is some unit. This gives two possibilities, either:
$$ f(x) - 1 = \pm x^n $$
$$ f(x) = 1 \pm x^n $$
The other possibility:
$$ x^n f(1/x) - x^n = \pm x^n$$
$$ f(1/x) = 0, 2 \implies f(x) = 2 \text{ since $f \neq 0$}$$
Applying this to your question, we get two possibilities for $f(x)$. Either:
$$ 4^n + 1 = 65 $$
$$ 4^n - 1 = 65 $$
Since $n$ has to be a natural number, the only solution is $n=3$, giving the polynomial $f(x) = x^3 + 1 $.
| {
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How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?
| Knowing that $(x+2)$ is a factor, you know that the solution will be of the form
$$x^3+4x^2+x-6=(x+2)(ax^2+bx+c).$$
Execute the product
$$(x+2)(ax^2+bx+c)=ax^3+(2a+b)x^2+(2b+c)x+(2c).$$
Identify the coefficients
$$a=1,2a+b=4,2b+c=1,2c=-6.$$
Solve and get
$$a=1,b=2,c=-3,$$
so that
$$x^3+4x^2+x-6=(x+2)(x^2+2x-3).$$
| {
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Limit of $\left(\frac{x^2+x+1}{2x+1}\right)^{1/(x^2-1)}$ when $x\to1$ $$\displaystyle \lim_{x\to 1}\left(\frac{x^2+x+1}{2x+1}\right)^{\frac{1}{x^2-1}}$$
I know that the two-sided limit of $\frac{1}{x^2-1}$ does not exist. I don't know what to do with $\frac{x^2+x+1}{2x+1}$ to get something else than $1^\infty$
| Let $y=x-1$. Then we have $$\require{cancel}\lim_{y\to0}\left(\frac{y^2+3y+3}{y^2-(y+1)^2+4y+4}\right)^{\frac{1}{y(y+2)}}=\lim_{y\to0}\left(\frac{y^2+3y+3}{2y+3}\right)^{1/(2y)}=\\ \lim_{y\to0}\left(\frac{\cancel{y}(y+3+3/y)}{\cancel{y}(2+3/y)}\right)^{1/(2y)},$$ which, letting $z=1/y$, yields $$\lim_{z\to\infty}\left(\frac{3z+1/z+3}{3z+2}\right)^{z/2}=\lim_{z\to\infty}\left(1+\frac{1+1/z}{3z+2}\right)^{z/2}=\lim_{z\to\infty}\left(1+\frac{1}{3z+2}\right)^{z/2}.$$ Now we could do another substitution, but it is easy to see that $2$ next to $3z$ doesn't count when $z\to\infty$. Thus we can finally conclude the limit is $$\lim_{z\to\infty} \left(1+\frac{1}{3z}\right)^{z/2}=e^{1/6}.$$
| {
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Where does $r = 1 + 2\cos(\theta)$ have tangents? Where does:
$$r = 1 + 2\cos(\theta)$$
Have horizontal and vertical tangent lines?
$x = r\cos(\theta) = \cos(\theta) + 2\cos^2(\theta)$
$y = r\sin(\theta) = \sin(\theta) + 2\sin(\theta)\cos(\theta)$
$$dx/d\theta = -\sin(\theta) - 4\sin(\theta)\cos(\theta)$$
$$dy/d\theta = \cos(\theta) + 2\cos(2\theta)$$
By hand it is very hard to solve $dy/d\theta = 0, dx/d\theta = 0$
How can it be done?
| Note that $\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1$ so
$$\frac{dy}{d\theta} = 0 \Leftrightarrow \cos\theta + 4\cos^2 \theta - 2 = 0$$
Now compute the roots of $4x^2+x-2$ to get equations for $\cos \theta$ and apply $\arccos$ to find $\theta$.
| {
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Do I need to use partial fractions to find $\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2}-\frac{1}{(n+1)^2}$? I need to simplify $\frac{2n+1}{n^2(n+1^2)}$ as part of an exam question. The solution states $$\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2}-\frac{1}{(n+1)^2}$$
In the solution it does not state how this simplification was made, I figured this could be done in quite a long winded fashion, using partial fractions. But from how it's written in the solutions it seems like this should be an easy simplification.
Is there a simple trick to simplifying fractions like this?
| Note that $$\frac{2n+1}{n^2(n+1)^2} = \frac{(n^2 + 2n + 1)-n^2}{n^2(n+1)^2}$$
$$=\frac{(n+1)^2-n^2}{n^2(n+1)^2}$$
$$=\frac{(n+1)^2}{n^2(n+1)^2}-\frac{n^2}{n^2(n+1)^2}$$
$$=\frac1{n^2}-\frac1{(n+1)^2}$$
as desired. This approach is motivated by noticing that the original numerator is "almost" a recognizable perfect square, so one is tempted to add and subtract the missing quantity that completes the square.
| {
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Odd Primes Problem Proof Given the odd prime numbers,
Prove that if $x$ and $y$ are adjacent odd primes in this list, then $x + y$ has $3$ prime factors. The factors need not be distinct.
Here is an example I have provided:
$3 + 5 = 8 = 2 \cdot 2 \cdot 2$. Therefore, $8$ has $3$ repeated factors of $2$.
| Suppose $x<y$ are adjacent odd primes. Then $x+y = 2a$, $a\in\mathbb{Z}$.
$x<y \Rightarrow 2x<x+y<2y \Rightarrow 2x<2a<2z \Rightarrow x<a<y$
How $x$ and $y$ are adjacent primes, then $a$ is not prime. Suppose $a=b\cdot c$, with $b, c>1$.
So $x+y = 2\cdot b\cdot c$. Therefore $x+y$ has at least 3 prime factors
| {
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Evaluating the limit $\lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$ Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$
I have a question about the following solution:
We may write it in the form:
$$ \frac{1}{n} \left[ \frac{1}{1+(\frac{1}{n})^2} + \ldots + \frac{1}{1+(\frac{n}{n})^2} \right] $$
Somehow I need to figure out that the limit is actually the Riemann sum of $\frac{1}{1 + x^2}$ on $[0,1]$ for $\pi = 0 < \frac{1}{n} < \ldots < \frac{n}{n}$.
Can you explain to me to reach this conclusion?
| $$\frac1n\sum_{k=1}^n\frac1{1+\left(\frac kn\right)^2}\xrightarrow[n\to\infty]{}\int_0^1\frac{dx}{1+x^2}\left(=\frac\pi4\;.\;\;Why?\right)$$
Just note that the above sum is
$$\sum_{k=1}^n f\left(\frac kn\right)(x_{k+1}-x_k)\,,\,\,\text{with}\;\;f(x)=\frac1{1+x^2}\;$$
and the partition of the unit interval
$$\left\{0<\frac1n<\frac2n\ldots<\frac nn=1\right\}\;,\;\;\text{and}\;\;c_k:=\frac kn$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1105719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How prove this inequality $\frac{e^x}{x+1}>\frac{\cos{x}}{\sin{x}+\sqrt{2}}$ Today,when I use wolf found this following inequality
let $x>-1$, show that
$$\dfrac{e^x}{x+1}>\dfrac{\cos{x}}{\sin{x}+\sqrt{2}}$$
I found this
I want $$\Longleftrightarrow (\sin{x}+\sqrt{2})e^x-(x+1)\cos{x}>0$$
let
$$f(x)=(\sin{x}+\sqrt{2})e^x-(x+1)\cos{x}\Longrightarrow f'(x)=\sqrt{2}e^x+(x+e^x+1)\sin{x}+(e^x-1)\cos{x}$$
then I calculus $f''(x)$ and found ugly,so maybe this inequality have other methods?
| We know $\sqrt{2}+\sin x > 0$ since $|\sin x| \le 1$ and $\sqrt{2} > 1$.
$\displaystyle \frac{e^{x}}{x+1} \ge 1 \ge \frac{\cos x}{\sqrt{2}+\sin x}$, the later inequality is true since,
$1 \ge \sin (\frac{\pi}{4} - x) = \dfrac{\cos x - \sin x}{\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1106469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Implicit Differentiation: $(x/y)+(y/x) =1$ Hi can anyone please tell me where I goes wrong with this question:
Find $ \frac{dy}{dx} $ for the curves defines by this equation:
\begin{align}
\frac{x}{y} + \frac{y}{x} = 1
\end{align}
Here is what I did:
\begin{align}
&\frac{y-xy'}{y^2} + \frac{y'x - y}{x^2} =0 \\
&\frac{yx^2 - x^3 y'+y'y^2 x - y^3}{x^2 y^2} = 0 \\
&yx^2 + (-x^3 +y^2 x)y' -y^3 =0 \\
&\therefore y'= \frac{y^3 -yx^2}{-x^3 +y^2x}
\end{align}
The answer say it should be: $ y' = \frac{y}{x} $ but I had no clue how to proceed from there.
Please help, Thanks.
| we get by the quotient and the chaine rule
$$\frac{y-xy'}{y^2}+\frac{y'x-y}{x^2}=0$$ multiplying by $x^2y^2$ we obtain
$$x^2y-x^3y'+y'y^2x-y^3=0$$
solving for $y'$ we get
$$y'=\frac{x^2y-y^3}{x^3-xy^2}$$ if $x^3-y^2x\ne 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1107869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
find a function satisfying the recurrence find a function satisfying the recurrence
$$F (n) = 2F (\sqrt{n}) + 1$$
replace $n$ by $2^m$
Thus getting the answer as $$F(n)=\frac{1}{2}c \log(n) + \log(n) - 1$$
Is this correct
| Looks good:
\begin{align}
2 F(\sqrt{n}) + 1
&=
2\left(
\left(\frac{c}{2} + 1 \right) \log(\sqrt{n}) - 1
\right) + 1 \\
&=
2\left(
\left(\frac{c}{2} + 1 \right) \frac{1}{2} \log(n) - 1
\right) + 1 \\
&=
\left(\frac{c}{2} + 1 \right) \log(n) - 1 \\
&= F(n)
\end{align}
You could replace the constant factor $c/2 + 1$ by another constant $d$.
In case you asked about the derivation as well (the wording of your question is somewhat ambigious to me):
Setting $n = 2^m$ gives the equation
$$
F(2^m) = 2 F(2^{m/2}) + 1
$$
setting $F(2^m) = G(m)$ gives
$$
G(m) = 2 G(m/2) + 1 \iff
G(2k) = 2 G(k) + 1
$$
That looks pretty linear. Trying $G(k) = a k + b$ gives
$$
2 a k + b = 2 a k + 2 b + 1 \iff
b = - 1
$$
So $G(m) = a m - 1$ and $F(2^m) = a m - 1 \iff F(n) = a \log_2(n) - 1 = c \ln(n) - 1$ for some constant $c$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$ Understand the representation $$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$$=$$\frac{1}{2} \left(-\frac{2}{n+1}+\frac{1}{n+2}+\frac{1}{n}\right)$$
$$s_n=\frac{1}{2} \left(\left(\frac{1}{n+2}+\frac{1}{n}-\frac{2}{n+1}\right)+\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}\right)+\ldots +\left(1 \frac{1}{1}-\frac{2}{2}+\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{3}+\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{2}{7}+\frac{1}{8}\right)\right)$$
Im sorry for not being able to get the term in right order, but as you can see the first two terms in the parenthesis should be the last for a accurate representation.
$$\left(
\begin{array}{cc}
1 & 1-1+\frac{1}{3} \\
2 & \frac{1}{2}-\frac{2}{3}+\frac{1}{4} \\
3 & \frac{1}{3}-\frac{1}{2}+\frac{1}{5} \\
4 & \frac{1}{4}-\frac{2}{5}+\frac{1}{6} \\
5 & \frac{1}{5}-\frac{1}{3}+\frac{1}{7} \\
6 & \frac{1}{6}-\frac{2}{7}+\frac{1}{8} \\
\end{array}
\right)$$
I think that my pure understanding of the representation is what makes me confused. From the table I see that a pattern in the denominators emerge, when n=1, then (1-2+3), when n=2 then, 2-3+4. However given the part of sn where $$\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}$$ I fail to see how the first part of expression right above is not undefined when n=1.
I am convinced that it is my lack of knowledge of what the representation actually means. And I am also struggling to see what cancels to give$$\frac{1}{2} \left(\frac{1}{n+2}-\frac{1}{n+1}+\frac{1}{2}\right)$$
However given that this is true I am able to understand that the answer is (1/4). But as I said it is the representation I do not get. Could someone help me as this would be very useful for my next chapter in my textbook!
| $\dfrac{1}{n(n+1)(n+2)} = \dfrac{1}{2n(n+1)} - \dfrac{1}{2(n+1)(n+2)}$
therefore,
$$\sum_{n = 1}^{\infty} \dfrac{1}{n(n+1)(n+2)} = \dfrac{1}{2*1*2} = \dfrac{1}{4}$$
we used the telescoping series formula $\sum_{n=1}^\infty (f(n) - f(n+1)) = f(1) - \lim_{n \to \infty} f(n)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n>0\ \forall\ n\ge1$.
Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.
I can't show that $a_n > 0\ \forall n\ge1$. I tried using induction but it wouldn't work.
Attempt:
$$
\begin{align}
2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}&=(2\sqrt{n}-(\sqrt{n-1}+\sqrt{n+1}))\cdot {2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\\&={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\end{align}$$
(The calculations are true for sure. No check is desired.)
Denote
$$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\text{ and }b_n={1\over n^{2}}.$$
Then
$$\begin{align*}
\lim_{n\to \infty}{a_n\over b_n} & =\lim_{n\to \infty}n^{2}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=\lim_{n\to \infty}n^{2}\lim_{n\to \infty}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=0
\end{align*}$$
By the comparison test for series convergence, since $\lim_{n\to \infty}{a_n\over b_n}=0$, then if $b_n$ converges, which it does, so does $a_n$.
| Copied from my answer to this near-duplicate:
Hint
It's a telescoping sum. With $a_n := \sqrt n - \sqrt{n+1}$, $a_n - a_{n-1} = 2\sqrt n - \sqrt{n-1} - \sqrt{n+1}$ so
$$\sum_{i=1}^n 2\sqrt i - \sqrt{i-1} -\sqrt{i+1} = 1 + \sqrt n - \sqrt{n+1} \stackrel{n\to\infty}\longrightarrow 1$$
This even allows you to obtain the value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
recurrence relation with variable coefficients How to solve recurrence relation $y_n= y_{n-1}+ (n-1).y_{n-2}$ where $n$ is a variable ?
$y_n$ is a $n$th term, $y_{n-1}$ is $(n-1)$th term and $y_{n-2}$ is $(n-2)$th term.
| Hint
Let $y_{0}=a,y_{1}=b$,and let
$$f(x)=\sum_{n=0}^{\infty}y_{n}x^n$$
then we have
$$y_{n}x^n=xy_{n-1}x^{n-1}+x^2(n-2)y_{n-2}x^{n-2}+y_{n-2}x^n$$
$$\Longrightarrow \sum_{n=2}^{\infty}y_{n}x^n=x\sum_{n=2}^{\infty}y_{n-1}x^{n-1}+x^2\sum_{n=2}^{\infty}(n-2)y_{n-2}x^{n-2}+x^2\sum_{n=2}^{\infty}y_{n-2}x^{n-2}$$
$$\Longrightarrow f(x)-a-bx=x[f(x)-a]+x^3f'(x)+x^2f(x)$$
so
$$f'(x)+\dfrac{x^2+x-1}{x^3}f(x)=\dfrac{(a-b)x-a}{x^3}$$
so
$$f(x)=e^{-\int\frac{x^2+x-1}{x^3}dx}\left(\int\dfrac{(a-b)x-a}{x^3}e^{\int\frac{x^2+x-1}{x^3}dx}dx+C\right)$$
then I think you can do it,even it's ugly
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Inequalites of triangle side with $abc = 1$ Let $a,b,c$ be the sides of a triangle with $abc=1$.
Prove that $$ \frac{\sqrt{b+c−a}}{a} + \frac{\sqrt{c+a-b}}{b} + \frac{\sqrt{a+b−c}}{c} \ge a+b+c $$
| By Holder $\left(\sum\limits_{cyc}\frac{\sqrt{b+c-a}}{a}\right)^2\sum\limits_{cyc}a^2(b+c-a)^2\geq(a+b+c)^3$.
Hence, it remains to prove that $abc(a+b+c)\geq\sum\limits_{cyc}a^2(b+c-a)^2$, which is
$\sum\limits_{cyc}(a-b)^2(a+c-b)(b+c-a)\geq0$, which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Triple Integral of $1/\sqrt{2 + x^2 + y^2 + z^2}$ over unit sphere I'm studying triple integrals (physics major), and I'm having trouble solving this little beast:
$$
\iiint_V \frac{1}{\sqrt{2 + x^2 +y^2 +z^2}} \,dx\,dy\,dz$$
where V is $$x^2+y^2+z^2=1$$
Of course we use spherical coordinates:
$$I = \iiint_V \frac{r^2 \sin φ}{\sqrt{2+r^2}} \,dr\,dφ\,dθ$$
In order to solve the first integral over r I simplified the denominator using $$\sqrt{2+r^2} = \sqrt{2\left(1+\frac{r^2}{\sqrt{2}^2}\right)}$$
in order to substitute $\tan ω = \frac{r}{\sqrt{2}}$. However again even this integral leads to 2 pages of computations and I still haven't reached a correct result.
Is there any shortcut I'm not seeing?
PS: The limits of $r,θ,φ$ are the standard ones since we are on the unit sphere.
| Write
$$\frac{r^2}{\sqrt{a^2 + r^2}} = \sqrt{a^2 + r^2} - \frac{a^2}{\sqrt{a^2 + r^2}}$$
These two terms are now standard integrals, equal respectively to $\displaystyle \frac{1}{2} \left( r \sqrt{a^2 + r^2} + a^2 \ln(\sqrt{a^2 + r^2} + r) \right)$ and $-a^2 \cdot \ln(\sqrt{a^2 + r^2} + r)$
Hence
$$\int_0^1 \frac{r^2}{\sqrt{2 + r^2}} dr = \left[ \frac{1}{2} r \sqrt{r^2 + 2} - \ln(\sqrt{2 + r^2} + r) \right]_0^1 = \ ...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that {$r_n$} satisfies $r_n = 7r_{n-1} - 10r_{n-2}, n \geq 2$ Problem: For the sequence $r$ defined by
$$r_n = 3 \cdot 2^n - 4 \cdot 5^n, \ \ \ n \geq 0$$
Prove that {$r_n$} satisfies
$$r_n = 7r_{n-1} - 10r_{n-2}, \ \ \ n \geq 2$$
Can this problem be explained and broken down and show the process? I'd like to follow your steps on my own.
I've been studying this question but the opening multiplication is throwing me off.
Similar Problem
| To show that $r_n = 3\cdot 2^n - 4\cdot 5^n$ satisfy the equation then start by writing the equation as
$$r_n - 7r_{n-1} + 10r_{n-2} = 0$$
Now calculate $r_n, r_{n-1}$ and $r_{n-2}$ from the formula. This gives us
$$r_n = \color{red}{3\cdot 2^n - 4\cdot 5^n}$$
$$r_{n-1} = 3\cdot 2^{n-1} - 4\cdot 5^{n-1} = \color{blue}{\frac{3}{2}\cdot 2^{n} - \frac{4}{5}\cdot 5^{n}}$$
$$r_{n-2} = 3\cdot 2^{n-2} - 4\cdot 5^{n-2} = \color{green}{\frac{3}{2^2}\cdot 2^{n} - \frac{4}{5^2}\cdot 5^{n}}$$
where we have used the rule $x^{a+b} = x^a \cdot x^b$ and $x^{-a} = \frac{1}{x^a}$ to simplify (for example $2^{n-2} = 2^{n}\cdot 2^{-2} =2^n\cdot \frac{1}{4}$). Now we substitute this into the equation $r_n - 7r_{n-1} + 10r_{n-2} = 0$ to find
$$\left(\color{red}{3\cdot 2^n - 4\cdot 5^n}\right) - 7\cdot\left(\color{blue}{\frac{3}{2}\cdot 2^{n} - \frac{4}{5}\cdot 5^{n}}\right) + 10 \cdot \left(\color{green}{\frac{3}{2^2}\cdot 2^{n} - \frac{4}{5^2}\cdot 5^{n}}\right) = 0$$
and after rearranging we get
$$\left(\color{red}{3\cdot 2^n} - 7\cdot \color{blue}{\frac{3}{2}\cdot 2^n} + 10\cdot\color{green}{\frac{3}{2^2}\cdot 2^n }\right) + \left(-\color{red}{4\cdot 5^n} + 7\cdot \color{blue}{\frac{4}{5}\cdot 5^n} - 10\cdot\color{green}{\frac{4}{5^2}\cdot 5^n }\right) = 0$$
Now we take the $2^n$ and $5^n$ outside of the brackets to find
$$2^n\cdot\left(\color{red}{3} - 7\cdot \color{blue}{\frac{3}{2}} + 10\cdot\color{green}{\frac{3}{2^2}}\right) + 5^n\cdot \left(-\color{red}{4} + 7\cdot \color{blue}{\frac{4}{5}} - 10\cdot\color{green}{\frac{4}{5^2}}\right) = 0$$
By calculating he sums in the brackets we find that they are both $0$ so we are left with $0=0$ which shows that the formula for $r_n$ does indeed satify the equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$ How would I go about solving this following limit?
$$\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$$
My attempts:
Direct substitution yields the limit to be undefined, also ruling out the possibility of using L'Hospital's Rule.
I don't see any clever substitutions that can be made with this limit.
Would squeeze theorem help here? Maybe using the trig. identities:
$$-1 \le \cos x \le 1$$
and
$$-1 \le \cos x \le 1$$
EDIT
I attempted to break the limit down term by term.
So, for the first one:
$$y = \lim_{x\to 0} (1 + \tan x)^{1/x}$$
Taking the natural log of both sides:
$$\ln y = \lim_{x\to 0} \frac{\ln(1+\tan x)}{x}$$
Direct sub. yields $0/0$. Using L'Hospital's rule:
$$\ln y = \lim_{x\to 0} \frac{\frac{\sec^2{x}}{1+\tan x}}{1} = \frac{\sec^2{x}}{1+\tan x} = 1$$
Thus, $\ln y = 1$, so $y= e$
EDIT #2
Thanks to a random comment, it actually does help me:
$$\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$$
$$\lim_{x\to 0} \frac{e-e}{0} = \frac{0}{0}$$
Thus, we can use L'Hospitals here:
$$\lim_{x\to 0} \frac{(\tan x+1)^{1/x} \left(\frac{\sec^2 x}{x(\tan(x)+1)}-\frac{\ln(\tan(x)+1))}{x^2}\right)}{1} = (\tan x+1)^{1/x} \left(\frac{\sec^2 x}{x(\tan(x)+1)}-\frac{\ln(\tan(x)+1))}{x^2}\right)$$
I haven't made any further progress, sadly.
Any help would be appreciated.
| Let's proceed in the following manner $$\begin{aligned}L &= \lim_{x \to 0}\frac{(1 + \tan x)^{1/x} - e}{x}\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + \tan x)}{x}\right) - e}{x}\\
&= e\cdot\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + \tan x)}{x} - 1\right) - 1}{x}\\
&= e\cdot\lim_{x \to 0}\dfrac{\exp z - 1}{x}\text{ (putting }z = \dfrac{\log(1 + \tan x)}{x} - 1)\\
&= e\cdot\lim_{z \to 0}\dfrac{\exp z - 1}{z}\cdot\lim_{x \to 0}\frac{z}{x}\\
&= e\cdot 1\cdot\lim_{x \to 0}\frac{z}{x}\\
&= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - x}{x^{2}}\\
&= e\cdot\lim_{x \to 0}\left\{\frac{\log(1 + \tan x) - \tan x}{x^{2}} + \frac{\tan x - x}{x^{2}}\right\}\\
&= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{x^{2}} + e\cdot\lim_{x \to 0}\frac{\tan x - x}{x^{2}}\\
&= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{x^{2}} + e\cdot 0\\
&= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{\tan^{2}x}\cdot\frac{\tan^{2}x}{x^{2}}\\
&= e\cdot\lim_{x \to 0}\frac{\log(1 + \tan x) - \tan x}{\tan^{2}x}\cdot 1\\
&= e\cdot\lim_{t \to 0}\frac{\log(1 + t) - t}{t^{2}}\text{ (putting }t = \tan x)\\
&= e\cdot\lim_{t \to 0}\dfrac{\left(t - \dfrac{t^{2}}{2} + \cdots\right) - t}{t^{2}} = -\frac{e}{2}\end{aligned}$$ I have used the following limits $$\lim_{x \to 0}\frac{\log(1 + \tan x)}{x} = \lim_{x \to 0}\frac{\log(1 + \tan x)}{\tan x}\cdot \frac{\tan x}{x} = 1$$ so that $z \to 0$ $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}} = 0$$ which can be proved using inequalities $\sin x < x < \tan x$ for $x \in (0, \pi/2)$ (although the proof is slightly tricky but easily available in MSE). Finally in the last step Taylor series for $\log (1 + t)$ is used.
Update: Proof for $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}} = 0$$ Clearly we can consider only the case $x \to 0^{+}$ because the function under consideration is odd and the letting $x \to 0^{-}$ will only change the sign of the answer (which wont matter as the answer would come out as $0$). Now we can see that $$\begin{aligned}A &= \lim_{x \to 0^{+}}\frac{\tan x - x}{x^{2}}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x\cos x}{x^{2}\cos x}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x\cos x}{x^{2}\cdot 1}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} + \lim_{x \to 0^{+}}\frac{x - x\cos x}{x^{2}}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} + \lim_{x \to 0^{+}}x\cdot\frac{1 - \cos x}{x^{2}}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} + 0 \cdot\frac{1}{2}\\
&= \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}}\\\end{aligned}$$ Next we have the inequality $$\sin x < x < \tan x = \frac{\sin x}{\cos x}$$ for $0 < x < \pi/2$ and hence $$\cos x < \frac{\sin x}{x} < 1$$ or $$\frac{\cos x - 1}{x} < \frac{\sin x - x}{x^{2}} < 0$$ Now taking limits as $x \to 0^{+}$ and noting that $$\frac{\cos x - 1}{x} = -2\cdot\frac{\sin^{2}(x/2)}{(x/2)^{2}}\cdot\frac{(x/2)^{2}}{x} \to 0$$ we get $$A = \lim_{x \to 0^{+}}\frac{\sin x - x}{x^{2}} = 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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Triple Angle Condition Let $ABC$ be a triangle with integral side lengths such that $\angle A=3\angle B$. Find the minimum value of its perimeter.
Essentially we want sinb, sin3b, sin4b to have rational ratios (manipulate the sine law). But then we don't actually need sinb to be rational, so that's pretty mysterious. Afterwards just make them integers by multiplying by LCM of denominator.
| Let $x = \sin\angle B$. Using the identity $\sin 3\theta = -4\sin^3 \theta + 3\sin\theta$, $\sin\angle A = -4x^3+3x$.
Then $\sin\angle C = \sin (\pi - \angle B - \angle A) = \sin 4\angle B = 4\sin\angle B\cos\angle B(1-2\sin^2 \angle B) =4x\sqrt{1-x^2}(1-2x^2) $.
By the law of sines, $\frac{a}{b} = \frac{\sin\angle A}{\sin\angle B} = -4x^2+3$, etc.
The perimeter is therefore
$$ a+b+c = 4b\left[ 1-x^2+\sqrt{1-x^2}(1-2x^2) \right]$$
We want to minimize this subject to the quantity in brackets being rational.
Note also that $0 < \angle B < \pi/4$, so $0<x<1/\sqrt{2}$.
We need $x^2$ to be rational, so let $x^2 = p/q$. Since we also need $\sqrt{1-x^2}$ to be rational, both $q$ and $q-p$ must be squares. Let $q=a^2$, and $a^2-p = b^2$, with $2p<a^2$. Substituting this all back in, we get the following optimization problem:
$$ \min\; a^3-pa+ba^2-2bp $$
subject to $a$, $b$, $p$ being positive integers, $2p<a^2$, and $a^2-p=b^2$.
The objective is cubic in $a$, so we should try to find the smallest $a$. From the constraints, we see that $a\ge 3$, and the second constraint prevents $a=3$ from being a possibility. The next possibility is $a=4$, leading to $p=7$, and $b=3$. Plugging this in gives side lengths $b$, $a=\frac{5}{4}b$, and $c=\frac{3}{8}b$. Multiplying through by 8 gives 3, 8, 10 as side lengths, for a perimeter of 21.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1115813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$.
Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above.
$\mathbf{\color{red}{\text{Contest results:}}}$
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\
\text{3} & \text{mickep} & {} & {} & 2.81108 \\
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\
\end{array}$$
Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\
\text{6} & -4x\ln x & {} & {} & 3.21360 \\
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\
\end{array}$$
| As has already been explained at least twice, the best functions follow this pattern:
a continuous function $f$ with $f(0)=f(1)=0$
that approximates $y = h(x) = 1-\frac\pi8+\sqrt{x-x^2}$ for $0<x<1.$
I propose a family of functions for $n$ a positive integer,
$$f_n(x) = \sqrt{x-x^2} + \left(1-\frac\pi8\right)g_n(x),$$
where
$$g_n(x) = \left(1 + \frac{1}{2n}\right)\left(1-(1-2x)^{2n}\right).$$
Since
$$\int_0^1 1-(1-2x)^{2n}\; dx = \frac{2n}{2n+1},$$
we have $\int_0^1 g_n(x)\;dx = 1,$ and therefore $\int_0^1 f_n(x)\; dx = 1.$
The path integral is more difficult to compute than the area integral, but
$1-(1-2x)^{2n}$ takes on its maximum value, $1$, at $x=\frac12$.
So if we set $h_n(x) = h(x) + \frac{1}{2n}\left(1-\frac\pi8\right)$
we ensure that $f(x) \leq h_n(x)$ for $0 \leq x \leq 1.$
I claim the path length is less than the length of the bounding curve
consisting of the graph of $h_n(x)$ from $x=0$ to $x=1$
and the two segments joining $(0,0)$ to $(0,h_n(0))$ and $(1,h_n(1))$ to $(1,0)$.
The length of that bounding path is
$$2+\frac\pi4 + \frac1n\left(1-\frac\pi8\right) < 2+\frac\pi4 + \frac{0.607301}{n}.$$
Therefore if we pick, say $n = 1000000,$ the resulting path exceeds the
theoretical minimum by less than $6.074 \times 10^{-7},$
which is less than one part in $4.5 \times 10^6.$
To within the accuracy possible in any visual graph I could present here,
the graph of $f_n(x)$ for large $n$ is the same as
the graph of every other near-theoretical-minimum solution:
Alternatively, stealing an idea from robjohn, we have
$$\int_0^1 (x-x^2)^{1/n} = B\left(1+\frac1n, 1+\frac1n\right)
= \frac{\Gamma\left(1+\frac1n\right)^2}{\Gamma\left(2+\frac2n\right)},$$
so we can set
$$g_n(x) =
\frac{\Gamma\left(2+\frac2n\right)}{\Gamma\left(1+\frac1n\right)^2}(x-x^2)^{1/n}$$
and proceed as before.
This $n$th-root approach seems to converge faster than my $2n$th-power approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 12,
"answer_id": 5
} |
Find all primes $p$ such that $p^2-p+1$ is a perfect cube Find all primes $p$ such that $p^2-p+1$ is a perfect cube.
I found out that p is of the form $18n+1$ and $p=19$ is a solution but I am not getting anything further.
$p^2-p-(m^3-1)=0$
$1+4(m^3-1)=k^2$
$4m^3-3=k^2$
every square is $0,1,4 or 7(\mod9)$ and every cube is $0,1 or 8\mod9$
Using this fact we can conclude that $m^3\equiv{1\mod 9}$
Hence $k$ is either $1$ or $-1$ $\mod9$
But $k\equiv{-1\mod9}$ is not possible since $p$ is a prime. So putting $k\equiv{1\mod9}$ in the quadratic formula we get $p\equiv{1\mod9}$ and since $p$ is odd(as $p=2$ is not a solution), $p\equiv{1\mod18}$
| you can work in $Z[1/2+\sqrt{-3}/2]$.we have unique factorization in this ring,and units are $1,-1,1/2+\sqrt{-3}/2,1/2-\sqrt{-3}/2$,and 2 is prime,yo can easily check this facts by the norm
$N(a+b\sqrt{-3})=a^2+3b^2$
we have
$4m^3=k^2+3=(k-\sqrt{-3})(k+\sqrt{-3})$,we can see $gcd((k-\sqrt{-3}),(k+\sqrt{-3})|2\sqrt{-3}$
and because$(k,3)=1$ we have $gcd((k-\sqrt{-3}),(k+\sqrt{-3})=2 or 1 $and we have$4(a/2+\sqrt{-3}b/2)\not=k+\sqrt{-3}$ so we have this cases:
$$1:k+\sqrt{-3}=2(a/2+\sqrt{-3}b/2)^3,k-\sqrt{-3}=2(a/2-\sqrt{-3}b/2)^3$$
$$2:k+\sqrt{-3}=(1+\sqrt{-3})(a/2+\sqrt{-3}b/2)^3,k-\sqrt{-3}=(1-\sqrt{-3})(a/2-\sqrt{-3}b/2)^3$$
$$3:k+\sqrt{-3}=(1-\sqrt{-3})(a/2+\sqrt{-3}b/2)^3,k-\sqrt{-3}=(1+\sqrt{-3})(a/2-\sqrt{-3}b/2)^3$$
in case 1 by take imaginary part we have:$2(-3b^3/8+3a^2b/8)=1$ so case 1 is impossible.
in case 2 by take real and imaginary part we have :
$$-3b^3/8+3a^2b/8=1-k=2p$$
$$a^3/8-9ab^2/8=k+3$$
because $p\not=3$ case 2 is impossible
case 3:we have
$$-3b^3/8+3a^2b/8=1+k=2-2p$$
$$a^3/8-9ab^2/8=k-3=-2-2p$$
$$-3b^3/8+3a^2b/8+a^3/8-9ab^2/8=-4p$$
i cant go further in this case
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$x^2+3$ has two zeros over ${\Bbb F}_p$ provided that $x^2+x+1\in{\Bbb F}_p[x]$ has two? The following is an exercise in abstract algebra:
If $p=1\pmod{3}$, then $x^2+x+1\in\Bbb{F}_p[x]$ has two zeros. Prove in this case that $-3$ is a quadratic residue mod $p$.
Showing that $(x^2+x+1)\mid (x^p-x)$, one can get the first part done. But I don't see how the second part, namely, $x^2+3\in\Bbb{F}_p[x]$ has two zeros, can be deduced from the first one. Could one come up with some hints of how to get the first part? (I was thinking one might relate $x^2+3$ to $x^2+x+1$ to make an argument about the zeros of $x^2+3$. But I don't see how I can go on.)
| Complete the square. We have $x^2+x+1\equiv 0\pmod{p}$ iff $4x^2+4x+4\equiv 0\pmod{p}$ iff $(2x+1)^2\equiv -3\pmod{p}$.
Remark: Let $F$ be any field that has characteristic $\ne 2$. Then we can make precisely the same argument: We have $x^2+x+1=0$ if and only if $4x^2+4x+4=0$ if and only if $(2x+1)^2=-3$. So if $x^2+x+1=0$ has a solution, then $w^2=-3$ has a solution. And if $w^2=-3$, then by solving $2x+1=w$ we obtain a solution of $x^2+x+1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1123800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Parametrization of solutions of diophantine equation $x^2 + y^2 = z^2 + w^2$ I need integer solutions of $x^2 + y^2 = z^2 + w^2$ parametrized. Can it be done? Thanks.
| First let us solve equation
$$
xy-zt=0.
$$
Let $(x,y,z,t)$ be any solution with $x\neq 0$, and let $u=\text{gcd}(x,z)>0$. Then $x=uv$ and $z=uw$ with $v,w$ coprime and $v\neq 0$. Then $xy-zt=0$ implies that $uvy=uwt$, or $vy=wt$. Because $v$ and $w$ are coprime, this implies that $t$ is divisible by $v$, so we can write $t=vr$ for some integer $r$. Then $vy=w(vr)$, hence $y=wr$. In conclusion, any solution $(x,y,z,t)$ with $x\neq 0$ is representable in the form
$$
(x,y,z,t)=(uv,wr,uw,vr) \quad \text{for some integers} \,\, u,v,w,r.
$$
If $x=0$, then either (a) $z=0$ and $y,t$ are arbitrary, or (b) $t=0$ and $y,z$ are arbitrary. However, these cases are also covered by the above family: take $u=0$, $r=1$ and $v,w$ arbitrary in case (a) and $v=0$, $w=1$, and $u,r$ arbitrary in case (b).
Now we are ready to solve the equation
$$
x^2+y^2=z^2+t^2.
$$
If $x$ and $y$ are both even, then $x^2+y^2$ is divisible by $4$, hence so is $z^2+t^2$, which is possible only if $z$ and $t$ are both even. Similarly, if $x$ and $y$ are both odd, then so are $z$ and $t$, and if $x$ and $y$ are of opposite parities, then so are $z$ and $t$. Hence, by permuting $z$ and $t$ if necessary, we may assume that $x$ and $z$ are of the same parity, and so are $y$ and $t$. Then rewrite the equation as
$$
\frac{x-z}{2}\cdot \frac{x+z}{2} = \frac{t-y}{2}\cdot \frac{t+y}{2}.
$$
Hence
$$
\left(\frac{x-z}{2},\frac{x+z}{2},\frac{t-y}{2},\frac{t+y}{2}\right)=(uv,wr,uw,vr) \quad \text{for some integers} \,\, u,v,w,r.
$$
Thus, integer solutions to $x^2+y^2=z^2+t^2$ are
$$
(x,y,z,t)=\left(wr+uv,vr-uw,wr-uv,vr+uw\right), \quad u,v,w,r \in {\mathbb Z},
$$
as well as the solutions obtained from the above by swapping $z$ and $t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.