Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Easy way to decompose $\frac{1}{X^{3}\cdot (X-2)^{3}}$ into partial fractions? Is there any easy way or shortcut to decompose $$\frac{1}{X^{3}\cdot (X-2)^{3}}$$ into partial fractions ? because dealing with the usual way of replacing and giving values to $X$ is too clumsy in this case , so I was wondering if there is any quick way to find the decomposition of this rational function ?
| You can also do this as follows. If $R(x)$ is a rational function that tends to zero at infinity, then expanding $R(x)$ around each of its singularities in the complex plane, keeping only the singular parts of the expansion and adding up all the terms from all these expansions will yield the partial fraction expansion.
The reason why this works is simple, if you consider the difference between $R(x)$ and the sum of all the singular parts of the expansions, then the resulting function only has removable singularities, as you've subtracted precisely the singular behavior at all the singular points. Therefore what we have is a polynomial. Since $R(x)$ and all the terms subtracted from $R(x)$ tend to zero at infinity, this polynomial is identical to zero.
In this case there is a singularity at $x = 0$ and at $x = 2$. Expanding around $x = 0$ amounts to expanding the factor $\frac{1}{(x-2)^3}$ in positive powers of $x$, for our purpose we only need to keep terms up to $x^2$. This is easily done:
$$\frac{1}{(x-2)^3} = -\frac{1}{8}\left(1-\frac{x}{2}\right)^{-3}= -\frac{1}{8}\left[1+\frac{3}{2}x+\frac{3}{2}x^2+\cdots\right]$$
So, the part of the partial fraction expansion coming from the singularity at $x = 0$, is given by:
$$S_1(x) = -\frac{1}{8 x^3} - \frac{3}{16 x^2} - \frac{3}{16 x}$$
Expanding around the singularity at $x = 2$ yields the remaining part of the partial fraction expansion. Putting $x = t + 2$, we see that we need to expand the function
$$\widetilde{R}(t) = \frac{1}{t^3(t+2)^3}$$
around $t = 0$. But this looks similar to what we've done above to obtain $S_1(x)$. Note that $\widetilde{R}(t) = R(-t)$, therefore the singular terms from the expansion around $x = 2$ are given by:
$$\widetilde{S}_2(t) = S_1(-t) = \frac{1}{8 t^3} - \frac{3}{16 t^2} + \frac{3}{16 t}$$
The partial fraction expansion is thus given by:
$$-\frac{1}{8 x^3} - \frac{3}{16 x^2} - \frac{3}{16 x} + \frac{1}{8 (x-2)^3} - \frac{3}{16 (x-2)^2} + \frac{3}{16 (x-2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Taylor series expansion of $f(x) = \sin^3 \left(\ln(1+x) \right)$. How does one use Taylor series expansion to compute $f^{(3)}(0)$ in which $f(x) = \sin^3 \left(\ln(1+x) \right)$.
| Probably easier to use a linearized expression for $\sin^3(\theta)$ :
$$\sin^3(\theta)=\frac{3}{4}\sin(\theta)-\frac{1}{4}\sin(3\theta)$$
Now, for all $x>-1$ :
$$f(x)=\sin^3(\ln(1+x))=\frac{3}{4}\sin(\ln(1+x))-\frac{1}{4}\sin(3\ln(1+x))$$
and the Taylor expansion at $x=0$ is :
\begin{align}
f(x)&=&\frac{3}{4}\sin(x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3))-\frac{1}{4}\sin(3x-\frac{3x^2}{2}+x^3+o(x^3))\\
&=&\frac{3}{4}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^3}{6}+o(x^3)\right)-\frac{1}{4}\left(3x-\frac{3x^2}{2}+x^3-\frac{9x^3}{2}+o(x^3)\right)\\
&=&x^3+o(x^3)
\end{align}
Finally : $f^{(3)}(0)=3!\,\times\,1$, that is :$$\boxed{f^{(3)}(0)=6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a common direct complement for $K$ and $L$ in $M_n$ Given the matrices $$A=\begin{pmatrix} 1 & 0 \\ -1 & -1\end{pmatrix}, \ B=\begin{pmatrix} 1 & 2 \\ 0 & -1\end{pmatrix}$$ and subspaces of $M_2$:
$$K=\{X\in M_2: tr(XA)=0\}, L=\{X\in M_2: tr(XB)=0\}$$
Find bases for $K$ and $L$ and determine one common direct complement for $K$ and $L$ in $M_n$.
My attempt:
$$X\in K\Leftrightarrow tr(XA)=0,\ X=\begin{pmatrix} x_1 & x_2 \\ x_3 & x_4\end{pmatrix}\Leftrightarrow \begin{pmatrix} x_1 & x_2 \\ x_3 & x_4\end{pmatrix}\cdot\begin{pmatrix} 1 & 0 \\ -1 & -1\end{pmatrix}=\begin{pmatrix} x_1-x_2 & -x_2\\ x_3-x_4 & -x_4\end{pmatrix} $$
$$ x_1-x_2-x_4=0\Rightarrow x_1=x_2+x_4\Rightarrow X=\begin{pmatrix} x_2+x_4 & x_2 \\x_3 & x_4\end{pmatrix}\Rightarrow \\ X=\begin{pmatrix}x_2+x_4 & x_2 \\ x_3 & x_4\end{pmatrix}=x_2\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}+x_3\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}+x_4\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}$$
So one basis for $K$ is $\{\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix},\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}\}$
Similarly I found one basis for $L: \{\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}, \begin{pmatrix} 0 & 1\\0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0\\1 & 2\end{pmatrix}\}$
How do I now find a common direct complement?
In general, we get a direct complement as a span of those vectors that complement the base to the base of the whole vector space. So, would some direct complement of $K$ be $\{\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\0 & 1\end{pmatrix}$? And for $L$ $\{\begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix},\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix},\begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}\}$? And then I intersect those two to find a common one?
| First, a general observation. Let $V$ be an $n$-dimensional vector space and let $\varphi \colon V \rightarrow \mathbb{F}$ be a non-zero linear functional. By the rank-nullity theorem, we know that $\dim \ker(\varphi) = n - 1$. Let us choose any vector $w \in V$ such that $\varphi(w) \neq 0$. Then
$$ V = \ker(\varphi) \oplus \operatorname{span} \{ w \}. $$
To see this is true, not that if $v \in \ker(\varphi) \cap \operatorname{span} \{ w \}$ then $\varphi(v) = 0$ and $v = aw$ for some $a \in \mathbb{F}$. This implies that $\varphi(aw) = a\varphi(w) = 0$ but then $a = 0$ and $v = 0$. Since the dimensions add up, we have a direct sum decomposition.
Now if $\varphi_1, \varphi_2$ are two non-zero functionals and we want to find a direct sum complement for both $\ker(\varphi_1),\ker(\varphi_2)$, by the observation above we only need to choose $w \in V$ such that $\varphi_1(w) \neq 0$ and $\varphi_2(w) \neq 0$.
In your case, $V = M_2(\mathbb{F})$, $\varphi_1(X) = \operatorname{tr}(XA)$ and $\varphi_2(X) = \operatorname{tr}(XB)$. Hence, to find a direct sum complement for both $K = \ker(\varphi_1)$ and $L = \ker(\varphi_2)$, we need to find $X \in M_2(\mathbb{F})$ such that $\varphi_1(X) \neq 0$ and $\varphi_2(X) \neq 0$.
Let
$$ X = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. $$
Then for any $Y \in M_2(\mathbb{F})$, the matrix $XY$ will be a matrix whose first row is the first row of $Y$ and whose second row is zero. Hence,
$$ \operatorname{tr}(XA) = \operatorname{tr}(XB) = 1 $$
so $\operatorname{span}(X)$ complements both $K$ and $L$ to $M_2(\mathbb{F})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Identity of an inverse hyperbolic function Recently I stumbled upon the identity: $$\operatorname{csch}^{-1}(x) = \ln \left( \frac 1 x + \frac{\sqrt{1+x^2}}{|x|}\right)$$
So i decided to prove this identity by taking the integral of the derivative of inverse $\operatorname{csch}(x)$. Although i tried expressing the integral : $$-\int \frac{\mathrm dx}{ |x|\sqrt{1+x^2}}$$
into different forms, I still haven't found a way to write it in terms of $$\int \frac{f'(x)}{f(x)} \, \mathrm dx$$
**I know there is a solution which involves the application of the definition of hyperbolic functions (with exponentials), but I was wandering if it was possibile with integration
| The usual algebraic proof: Recall that $$\sinh x = \frac{e^{x} - e^{-x}}{2}.$$ Thus, $$\operatorname{csch} x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}.$$ It follows that $y = \operatorname{csch}^{-1} x$ is the solution to the equation $$x = \frac{2}{e^y - e^{-y}}.$$ To this end, we write $$e^y - e^{-y} = 2/x,$$ and multiplying through by $e^y$ gives the quadratic $$(e^y)^2 - (2/x) e^y - 1 = 0,$$ which has the solution $$e^y = \frac{(2/x) \pm \sqrt{4/x^2 + 4}}{2} = \frac{1}{x} \pm \sqrt{\frac{1}{x^2} + 1}.$$ Since $$\frac{1}{x^2} + 1 > \frac{1}{x^2},$$ if we require $y$ to be real-valued, we must choose the positive root of the quadratic; hence $$y = \operatorname{csch}^{-1} x = \log \left( \frac{1}{x} + \sqrt{\frac{1}{x^2} + 1} \right), \quad x \ne 0.$$
Consider $$I = \int \frac{dx}{x \sqrt{x^2+1}} = \int \frac{x \, dx}{x^2 \sqrt{x^2+1}}.$$ With the substitution $$u^2 = x^2 + 1, \quad 2u \, du = 2x \, dx,$$ we obtain $$\int \frac{u \, du}{u(u^2-1)} = \int \frac{du}{u^2-1} = \frac{1}{2}\left(\int \frac{du}{u-1} - \int \frac{du}{u+1}\right).$$ It follows that $$I = \frac{1}{2} \log \frac{u-1}{u+1} + C = \frac{1}{2} \log \frac{(u-1)^2}{u^2 - 1} + C = \log \frac{u-1}{x} + C = \log \frac{-1 + \sqrt{x^2+1}}{x} + C.$$ The details are left to you to complete, namely to demonstrate $C = 0$, and to refine the argument to take care of the absolute values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Intersection of 3 circles on equilateral triangle given the difference in their radii Imagine we are given three intersecting circles centered on the vertices of an equilateral triangle of side length $1$, with $(0,0)$ arbitrarily placed at the bottom left corner. The circles have radii $r$, $r+a$, and $r+b$ respectively (going clockwise again arbitrarily), where $r$ is unknown but $a$ and $b$ are given. For sake of ease we can assume $0<a<b$ and that $a$ and $b$ are "safe," aka a solution is possible. What is the intersection $(x,y)$ of the three circles (in terms of $a$ and $b$) inside the triangle?
I apologize for the crude nature of the diagram, but it was the best I could come up with using MS Paint. Additionally, I'm 99.9% there's a unique solution as long as we restrict our solution set to points inside the triangle, but this may not be the case. Thank you so much for any help.
Edit: if anybody has any bright ideas in this direction, the solution doesn't need to be exact; a close approximation with nicer algebra would be much appreciated.
| First, note that (x,y) is determined by the intersection of the circles with radii r and r+b, Given the "safe" values assumed, so we should be able to express both x and y in terms of r and b only.
Safe values values are those such that:
* 2r + a > 1
* 2r + b > 1
* r + a < 1
* r + b < 1
Since it was specified that a < b, this list can be shorted to:
* 2r + a > 1 (so that each circle intersects the other 2 circles)
* r + b < 1 (so even the largest circle has points in the triangle)
First, we find x in terms of r and b, from the equations for those circles.
Define the coordinates of the vertices as (0,0), $\left(\frac12,\frac{\sqrt3}2\right)$, (1,0) for the circles with radiui r, r+a, and r+b respectively.
These give the equations:
$$\eqalign{
x^2+y^2=r^2&\qquad\qquad(1)\cr
(x-1)^2+y^2=(r+b)^2&\qquad\qquad(2)\cr
(x-\tfrac12)^2+(y-\tfrac{\sqrt3}2)^2=(r+a)^2&\qquad\qquad(3)\cr}$$
We will use equations (1) and (2) to find x in terms of r and x.
Expanding (2) we get:
$$x^2 -2x + 1 + y^2 = r^2 + 2rb + b^2$$
Subtracting Equation (1) from this gives:
$$1 - 2x = 2rb +b^2$$
Solving for x we obtain:
$$ x = {1-b^2 \over 2} - rb$$
Substituting this for x gives:
$$
\begin{align}
y^2 & = r^2 - x^2 \\
& = r^2 - \left({1-b^2 \over 2} - rb \right)^2\\
& = {(1-b^2)^2 \over 4} - (1 - b^2)rb +r^2b^2 + r^2 \\
& = {(1-b^2)^2 \over 4} - (1 - b^2)rb +r^2(1 + b^2) \\
& = {(1-b^2)^2 \over 4} + r(b^3 +b) + r^2(1 +b^2) \\
& = {(1-b^2)^2 + 4r(b^3 +b) + 4r^2(1 +b^2) \over 4} \\
y & = {\sqrt{(1-b^2)^2 + 4r(b^3 +b) + 4r^2(1 +b^2) } \over 2} \\
\end{align}
$$
So $(x, y) = \left( {1-b^2 \over 2} - rb , {\sqrt{(1-b^2)^2 + 4r(b^3 +b) + 4r^2(1 +b^2) } \over 2}\right)$.
Done.
Remarks.
First, we know that since r and b are > 0, the last expression for $y^2$ is positive, so the square root gives a real number for y.
Second, we asserted at the beginning that the point (x,y) is determined only by r and b. We prove this by showing (x,y) in terms of only r and b, making no use of a. This also implies that a is determined by r and b.
By substituting the expressions for x and y in terms of r and b into Equation (3) then solving for a, an expression for a in terms of r and b can be found.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How do I continue solving this polynomial division? I had this equation as a bonus question on a quiz today. I got to a certain extend and was completely unsure how to continue.
*$$(10x^4-18x^3-94x^2-8x+2) \div (10x+2)$$
I figured the easiest first step would be to divide. (Actually, I assumed it was factorable, but I have no idea how to go about that)
$$\frac{(10x^4-18x^3-94x^2-8x+2)} {(10x+2)}$$
$$\frac{10x^4}{10x+2} + \frac{-18x^3}{10x+2} + \frac{-94x^2}{10x+2} + \frac{-8x}{10x+2} + \frac{2}{10x+2}$$
Simplifying some more (multiplying by $10x+2$)
$$10x^4-18x^3-94x^2-8x-2 = 0$$
But what do I do from here? Do I factor? Do do something else?
If I do factor How do I go about that? If not, what do I do instead?
| multiples of $10 x + 2$
$$
\begin{array}{rrrrrr}
x^3 \mapsto & 10 x^4 & + 2 x^3 &&& \\
-2x^2 \mapsto & & -20 x^3 & -4 x^2 && \\
-9x \mapsto & & & -90 x^2 & -18 x & \\
+1 \mapsto & & & & 10 x & + 2 & \\
&&&&& \\
& 10x^4 & - 18 x^3 & -94 x^2 & -8 x & + 2 & \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Value of the integral $\int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx$ Evaluate
$ \displaystyle \int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx$.
(I am relatively confident of my answer below but I would still like to make sure that it is correct.)
| \begin{align}
\int_1^\infty \frac{\lfloor x\rfloor}{x^3} \, dx &= \sum_{n=1}^\infty \int_n^{n+1} \frac n{x^3} \, dx \\
&= -\frac 12 \sum_{n=1}^\infty \frac n{x^2} \bigg\vert_n^{n+1} \\
&= \frac 12 \sum_{n=1}^\infty \left( \frac 1n - \frac n{(n+1)^2} \right) \\
&= \frac 12 \sum_{n=1}^\infty \frac{2n+1}{n(n+1)^2} \\
&= \sum_{n=1}^\infty \frac 1{(n+1)^2}+ \frac 12\sum_{n=1}^\infty \frac 1{n(n+1)^2} \\
&= \sum_{n=1}^\infty \frac 1{(n+1)^2}+ \frac 12\sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+1} \right) - \frac 12\sum_{n=1}^\infty \frac 1{(n+1)^2} \\
&= \frac 12\sum_{n=1}^\infty \frac 1{(n+1)^2} + \frac 12 \\
&= \frac 12\sum_{k=2}^\infty \frac 1{k^2}+\frac 12 \\
&= \frac 12\left(\frac{\pi^2}6-1 \right)+\frac 12 \\
&= \boxed{\frac{\pi^2}{12}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Triangle inequality in C Show that $\left|Im(2+ z^{c} -4z^2) \right| \leq 9.5$ When $ \left| z \right| \leq \frac {3}{2}$
$z^c$= compliment of z
$\left|Im(2+ z^{c} -4z^2) \right| \leq \left| 2+z^{c} - 4z^2 \right|$
I have tried to split it up directly i have tried to force complete the square i always get a weird value or a number bigger than 9.5.
EDIT: Why cant i write
$\left| Im(2+ z^{c} -4z^2) \right| = \left| Im(z^{c} -4z^2)\right| \leq \left| z^{c} - 4z^2 \right|= 2^{\frac {1}{2}}\left| \frac {z^{c}}{2} - 2z^2 \right| \leq 2^{\frac {1}{2}}(\left| \frac {z^{c}}{2}\right| +\left| 2z^2 \right|)$
and $2^{\frac {1}{2}}(\left| \frac {z^{c}}{2}\right| +\left| 2z^2 \right|)= 2^{\frac {1}{2}}(\frac {3}{4} + \frac {9}{2}) = \frac {21}{2*2^{\frac {1}{2}}} \leq 9.5$
| Credit: @MartinBladt is right that something is wrong with the question.
Let $z = \frac{\sqrt{5}}{2}+i$
$$|z|=\sqrt{\frac54+1}=\frac32$$
$$|\Im(2+z^c+-4z^2)|=|-\Im (z)-4(2) \Re(z)\Im(z)|=1+8\frac{\sqrt{5}}{2}=1+4\sqrt{5}>9.5$$
Edit:
The following statement is not valid:
$$\left| z^{c} - 4z^2 \right|= 2^{\frac {1}{2}}\left| \frac {z^{c}}{2} - 2z^2 \right| $$
In particular, we can let $z=1$, we can see that the left hand side is rational but the right hand side is irrational.
We do have the following equation:
$$\left| z^{c} - 4z^2 \right|= 2\left| \frac {z^{c}}{2} - 2z^2 \right| $$
| {
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"url": "https://math.stackexchange.com/questions/2105653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum of the digits of the four-digit number
It is known that $\overline{abcd}-\overline{dcba}=x$ and $a^3+b^3+c^3+d^3=x$. Find $a+b+c+d$.
From the given we get $999a+90b-90c-999d = a^3+b^3+c^3+d^3$. We could just test out the possibilities, but is there another way to find the sum of the digits?
| Partial Answer
As $x=999a+90b-90c-999d=9(111a+10b-10c-111d)$ then $x\equiv0\pmod{9}$. Looking at cubes modulo 9 we see:
$0^3\equiv0\pmod{9}$, $1^3\equiv1\pmod{9}$, $2^3\equiv8\pmod{9}$, $3^3\equiv0\pmod{9}$, $4^3\equiv1\pmod{9}$, $5^3\equiv8\pmod{9}$, $6^3\equiv0\pmod{9}$, $7^3\equiv1\pmod{9}$, $8^3\equiv8\pmod{9}$, $9^3\equiv0\pmod{9}$.
For the four cubes to be a multiple of $9$ we require:
*
*4 zeros $\big((a,b,c,d)\equiv(0,0,0,0)\pmod{3}\big)$, or
*2 zeros, a one and an eight $\big((a,b,c,d)\equiv(0,0,1,2)\pmod{3}\big)$, or
*2 ones, 2 eights $\big((a,b,c,d)\equiv(1,1,2,2)\pmod{3}\big)$
We have $999(a-d) = a^3+b^3+c^3+d^3+90(c-b)$. Next consider the maximum value this can take. This occurs when $(a,b,c,d)=(9,0,9,9)$. The right hand side in this case is equal $2997$ (with equality only if $a=d$). Thus $0\le a-d\le2$.
I'll keep adding to this...
Cheating
Via computer the answer is $a=4$, $b=6$, $c=8$, $d=3$. I'll keep thinking about how to get here without cheating. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factor $ a^4-3a^2-2ab+1-b^2$ In order to factor $a^4-3a^2-2ab+1-b^2$, I find that $a=1, b=-1$ makes the value of the expression 0. Thus, I assume $b=-a$.
I rewrite the expression on the assumption as:
$$a^4-3a^2-2ab+1-b^2$$
$$=a^4-3a^2+2a^2+1-a^2$$
$$=a^4-2a^2+1$$
$$=(a^2-1)^2$$
Then I insert $a+b$, which is another form of the assumption above, into one of the factors. Since $a+b=0$, this insertion should cause no change to whatever relationships in the original expression. I arbitrarily set its coefficient as $1$. A factor, therefore, would be $(a^2+a+b-1)$.
I then divide the original expression by the factor and find that it indeed can be divided without a remainder and I also find the other factor. Thus,
$$a^4-3a^2-2ab+1-b^2$$
$$=(a^2+a+b-1)(a^2-a-b-1)$$
Q.E.D.
The problem is that I don't know what I am doing. I tried to use the factor theorem for a bivariate (if this is the proper word for it) expression but I might be fabricating a fortuitous reasoning from the answer. Yes, I knew the answer before I tried, and no, I don't know how I can solve this in other ways. It would be great if someone can tell me if my reasoning can be made more clear.
| Here is probably the most simple method to factor it.
Note that $$a^2-b^2=(a-b)(a+b)$$So $$a^4-3a^2-2ab+1-b^2=a^4-2a^2+1-a^2-2ab-b^2=(a^2-1)^2-(a+b)^2$$
So we can factor it into $$(a^2+a+b-1)(a^2-a-b-1)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Write the quadratic function in the form $g(x)= a(x-h)^2 + k$ Write the quadratic function in the form $g(x)= a(x-h)^2 + k$
Then, give the vertex of its graph.
$g(x)= 2x^2-16x+35$
~
I tried, but still lost:
$(2x^2-16x)+35$
$2(x^2-4^2-16)+35$
$2(x^2-4^2-32+35$
| $2x^2-16x+35$
$=2(x^2-8x)+35$
$=2(x^2-8x+16)-2\cdot 16+35$
$=2(x-4)^2+3$
There you go. Completing the square.
The vertex is easy from this. It's just x=4 since squares are always nonnegative so to make the square part zero you let x=4. (to be more specific the vertex is (4, 3))
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $(f^{-1})''(x)$. If $f$ is strictly monotonous and differentiable for a given interval, we have
$$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$$
How much is $(f^{-1})''$?
| \begin{align*}
y &= f(x) \\
x &= f^{-1} (y) \\
f'(x) &= \frac{dy}{dx} \\
&= \frac{1}{\frac{dx}{dy}} \\
f''(x) &= \frac{d^{2}y}{dx^{2}} \\
&= \frac{d}{dy} \left( \frac{1}{\frac{dx}{dy}} \right) \times \frac{dy}{dx} \\
&= -\frac{\frac{d^{2}x}{dy^{2}}}{\left( \frac{dx}{dy} \right)^{3}} \\
\frac{d^{2}x}{dy^{2}} &=
-\frac{\frac{d^{2}y}{dx^{2}}}
{\left( \frac{dy}{dx} \right)^{3}} \\
[f^{-1}(y)]'' &= -\frac{f''(x)}{[f'(x)]^{3}} \\
&= -\frac{f''[f^{-1}(y)]}{\{f'[f^{-1}(y)]\}^{3}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim\limits_{n\to\infty} ( \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2})$ Find $\lim\limits_{n\to\infty} ( \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2})$
Please help me find this limit .
I cannot use cauchys limit theorem on this problem . So I am stuck and clueless
| Observe that
$$0 < \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2} < n\cdot\frac{1}{(n+1)^2}.$$
Since the limit of the right side is $0,$ the squeeze theorem shows our limit is $0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve $y'=-\frac{x+y-2}{x-y+4}$
$$y'=-\frac{x+y-2}{x-y+4}$$
I have used the substation:
$x=u-1$
$y=v+3$
And got $$\frac{dv}{du}=\frac{-1-\frac{v}{u}}{1-\frac{v}{u}}$$
$z=\frac{v}{u}$
And got:
$\frac{du}{u}\frac{1-z}{z^2-2z-1}$
$$u=e^{\frac{-z+2z+1}{4}+C}$$
What should I do next?
| $$\frac { dv }{ du } =\frac { -1-\frac { v }{ u } }{ 1-\frac { v }{ u } } \\ z=\frac { v }{ u } \\ v=uz\\ z+{ u }z^{ \prime }=\frac { -1-z }{ 1-z } =\frac { 1+z }{ z-1 } \\ u{ z }^{ \prime }=\frac { 1+z }{ z-1 } -z=\frac { 1+z-{ z }^{ 2 }+z }{ z-1 } =\frac { -{ z }^{ 2 }+2z+1 }{ z-1 } =\frac { { z }^{ 2 }-2z-1 }{ 1-z } \\ \int { \frac { 1-z }{ { z }^{ 2 }-2z-1 } } dz=\int { \frac { du }{ u } } \\ \int { \frac { d\left( { z }^{ 2 }-2z-1 \right) }{ { z }^{ 2 }-2z-1 } } =-2\int { \frac { du }{ u } } \\ \ln { \left| { z }^{ 2 }-2z-1 \right| } =-2\ln { \left| u \right| } \\ { z }^{ 2 }-2z-1=\frac { C }{ { u }^{ 2 } } \\ \frac { { v }^{ 2 } }{ { u }^{ 2 } } -2\frac { v }{ u } -1=\frac { C }{ { u }^{ 2 } } \\ { v }^{ 2 }-2vu-{ u }^{ 2 }=C\\ \\ $$
then plug substition $x$ and $y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find $x^2+y^2$ if $x^2-\frac{2}{x}=3y^2$ and $y^2-\frac{11}{y}=3x^2$
Find $x^2+y^2$ if $x^2-\frac{2}{x}=3y^2$ and $y^2-\frac{11}{y}=3x^2$.
My try:
$$\frac{y^2-\frac{11}{y}}{3}-\frac{2}{x}=3y^2$$
then?
| [ EDIT #3 ] The previously posted answer remains copied below, under Alternative Solution.
Write the equations as:
$$
\begin{cases}
\begin{align}
2 &= x\,\left(x^2-3y^2\right) \\
11 &= y\,\left(y^2-3x^2\right)
\end{align}
\end{cases}
$$
Square the two and add together:
$$
\begin{align}
125 = 2^2 + 11^2 &= x^2\,\left(x^2-3y^2\right)^2 + y^2\,\left(y^2-3x^2\right)^2 \\
&= x^6 - 6 x^4 y^2 + 9 x^2y^4 \;\;+\;\;y^6 - 6 y^4x^2 + 9 y^2x^4 \\
& = x^6 + 3 x^4y^2 + 3 x^2y^4+y^6 \\
&= \left(x^2+y^2\right)^3
\end{align}
$$
Therefore $\;x^2+y^2 = \sqrt[3]{125} = 5\,$.
[ Alternative Solution ]
As far as brute force goes, you had the right idea to try and eliminate one variable between the equations, but it works out better if first bringing one of the equations to contain only even powers.
From the the second equation:
$$x^2 = \frac{y^3-11}{3y} \tag{1}$$
Rearranging and squaring the first equation, then substituting $x^2$ from $(1)$:
$$
\begin{align}
\left(x^2-3y^2\right)^2 & = \frac{2^2}{x^2} \\[5px]
\frac{y^3-11}{3y} \left(\frac{y^3-11}{3y}-3y^2\right)^2 & = 4 \\[5px]
\left(y^3-11\right)\left(8y^3+11\right)^2 & = 108 y^3 \\[5px]
64 y^9 - 528 y^6 - 1923 y^3 - 1331 & = 0 \tag{2}
\end{align}
$$
Equation $(2)$ is a cubic in $t=y^3$ and has the "obvious" root $t=-1$ correspoding to $y=-1\,$. Dividing by $t+1$ leaves a quadratic in $t$ which gives the other two roots $t=\cfrac{37 \pm 30 \sqrt{3}}{8}\,$. Guessing that those might be of the form $(a \pm b\sqrt{3})^3$ and identifying coefficients turns out to work, which gives the corresponding $y = \cfrac{1 \pm 2\sqrt{3}}{2}$.
For each of the three roots $y$ we can calculate $x^2$ from $(1)$ then $x^2+y^2\,$:
$$
\begin{cases}
\begin{alignat}{3}
y &= -1 \quad\quad & x^2 &= 4 \quad\quad & x^2+y^2 &= 5 \\[5px]
y &= \frac{1 \pm 2\sqrt{3}}{2} \quad\quad & x^2 &= \frac{7 \mp 4 \sqrt{3}}{4} \quad\quad & x^2+y^2 &= 5
\end{alignat}
\end{cases}
$$
(Meta-comment: I didn't see an obvious way to derive $x^2+y^2$ directly, without actually solving the system, though the nice end result strongly suggests that there would be one.)
[ EDIT ] For a computer-assisted direct solution, note that the problem is equivalent to eliminating $x,y$ between the given equations and $z=x^2+y^2\,$, then solving the resulting equation in $z$.
Elimination can be done using polynomial resultants, and Wolfram Alpha gives the resultant as:
$$2^{24}\,\left(z^3-125\right)^6$$
Quite obviously, $z=5$ is the only real root so $x^2+y^2=5\,$.
[ EDIT #2 ] The following generalization explains the "magic constants" in the given problem:
$$x^2-\frac{a}{x}=3y^2\;, \quad y^2-\frac{b}{y}=3x^2 \quad \implies \quad x^2+y^2 = \sqrt[3]{a^2+b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Weird limit: $\lim_{n\to\infty}\left(\frac{f(1)+f(2)+...+f(n)}{n}\right)^n$ I couldn't find the right path for this limit to solve. $$\lim_{n\to\infty}\left(\dfrac{f(1)+f(2)+...+f(n)}{n}\right)^n$$
$$f:\mathbb{R}\to\mathbb{R}, f(x)=\sqrt[3]{x^3+3x^2+2x+1}-\sqrt[3]{x^3-x+1}$$
I know that we have the indeterminate $1^{\infty}$, but apart from that...I didn't get very far
| As N.S.John says in the comments, we can write $$\sqrt [3]{x^3+3x^2+2x+1} =\sqrt [3]{(x+1)^3-(x+1)+1}$$ and $$\sqrt [3]{x^3-x+1} =\sqrt [3]{(x-1+1)^3-(x-1+1)+1} $$ Thus $$f (1)=\sqrt [3]{2^3-2+1}-\sqrt [3]{1^3-1+1} $$ $$f (2)=\sqrt [3]{3^3-3+1}-\sqrt[3]{2^3-2+1}$$ $$\vdots $$ $$f (n)=\sqrt [3]{(n+1)^3-(n+1)+1}-\sqrt [3]{n^3-n+1} $$ $$\Rightarrow \sum_{i=1}^{n} f (i) = \sqrt [3]{(n+1)^3-(n+1)+1}-\sqrt [3]{1} $$ Can you take it from here? The answer is $\boxed {1}$ assuming $\lim_{n \to \infty}$. Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What number should be subtracted from What number should be subtracted from $4x^3+5x+3$
so that the resulting polynomial leaves remainder $-80$ when divided by $2x+5$?.
Let the required number to be subtracted be $K$.
let: $$P(x)=4x^3+5x+3-k$$
$$g(x)=2x+5=2(x+5/2)$$
Comparing $g(x)$ with $x-a $ , $a =\frac {-5}{2}$
Solving then gives $K=8$
but the answer in my book is $-152$. how?
| $$4x^3+5x+3=2x^2\underbrace{(2x+5)}-5x\underbrace{(2x+5)}+15\underbrace{(2x+5)}+3-75$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
convergence of $\sum\frac{\sin(\sqrt{n})}{n^a}$ For $a>0$, define $x_n=\dfrac{\sin(\sqrt{n})}{n^a}$.
Prove that $\sum x_n$ converges if and only if $a>\dfrac1 2$.
I proved the convergence when $a>\dfrac1 2$, but I can't derive the other direction. Could someone give me some help? Thanks a lot.
| A straightforward way to show that this series diverges if $a \leqslant 1/2$ is to show that the Cauchy criterion is not satisfied.
If $\sqrt{n} \in [c_1,c_2] = [\pi/6 + 2k\pi, 5\pi/6 + 2k\pi]$ then $\sin \sqrt{n} \geqslant 1/2$ and, with $a \leqslant 1/2$, we have
$$\begin{align} \left|\sum_{c_1 \leqslant \sqrt{n} \leqslant c_2}\frac{\sin \sqrt{n}}{n^a}\right| &= \left|\sum_{c_1^2 \leqslant n \leqslant c_2^2}\frac{\sin \sqrt{n}}{n^a}\right| \\ &\geqslant \frac{1}{2} \frac{c_2^2 - c_1^2}{c_2^{2a}} \\ &\geqslant \frac{1}{2}\frac{c_2^2 - c_1^2}{c_2} \\ &= \frac{1}{2}\frac{\frac{8 \pi^2k}{3} + \frac{2\pi^2}{3}}{\frac{5\pi}{6 }+ 2 k \pi} \end{align}$$
The expression on the right-hand side converges to $2 \pi/3$ as $k \to \infty$. For any $N \in \mathbb{N}$ there exists sufficiently large $k \in \mathbb{N}$ such that $c_2^2 > c_1^2 > N$
and
$$\left|\sum_{c_1^2 \leqslant n \leqslant c_2^2}\frac{\sin \sqrt{n}}{n^a}\right| > \frac{\pi}{3}.$$
Thus the series fails to satisfy the Cauchy criterion and is divergent if $a \leqslant 1/2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Any hints on how to show that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $ I am trying to prove that $$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $$
My attempt was to suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10}$, and show that it was absurd.
Here's what I did:
Suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10} $, then $2 - \sqrt{3} \geq \frac{2\sqrt{3}}{10}$ thus:
$$\frac{20 -10 \sqrt{3}}{10} \geq \frac{2 \sqrt{3}}{10} \implies 20 -\sqrt{300} \geq 2 \sqrt{3} \implies 10 - \sqrt{\frac{300}{4}}\geq \sqrt{3} \implies 10 - \sqrt{75} \geq \sqrt{3} \implies \frac{10}{\sqrt{3}}- \sqrt{25} \geq 1 \implies \frac{10}{\sqrt{3}}- 5 \geq 1$$
Now I elevate everything to the square:
$$\frac{100}{3} - \frac{100}{\sqrt{3}} + 25 \geq 1 \implies \frac{100}{3} - \frac{100 \sqrt{3}}{3} + \frac{75}{3} \geq 1 \implies \frac{100(1-\sqrt{3}) +75}{3} \geq \frac{3}{3}$$
All is left to show is that $$100(1-\sqrt{3}) +75 \geq 3$$
And here, I don't know where to go, and it doesn't seem as I simplified the problem.
Any hints of help will be appreciated
| multiplying by $10$ and $$2\sqrt{3}$$ we get $$20-10\sqrt{3}<2\sqrt{3}$$ and we get
$$20<12\sqrt{3}$$ dividing by $$4$$ and we have $$5<3\sqrt{3}$$ this is true since we have $$25<27$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Finding the number of $4$-digit numbers so no two adjacent digits are even Find the number of $4$-digit numbers with distinct digits chosen from the set $\{0,1,2,3,4,5\}$ in which no two adjacent digits are even.
| You can approach this thinking that you have two different subsets: the even numbers $\{0,2,4\}$ and the odd numbers $\{1,3,5\}$. In order to avoid adjacent even numbers you have the two following cases:
Alternating even and odd numbers, starting with an odd number:
$3 \cdot 3 \cdot 2 \cdot 2 = 36$ possible combinations
First you can place any of the digits from $\{1,3,5\}$, then any from $\{0,2,4\}$, in the third place any of the remaining digits of the odd numbers set (2 in total), and same case for the even numbers in the last digit.
Alternating even and odd numbers, starting with an even number:
$2 \cdot 3 \cdot 2 \cdot 2 = 24$ possible combinations
The trick here is that starting with $0$ doesn't lead to a 4 digit number (you can add the zeros you want to the left of a number and it remains the same number).
For even numbers at both extremes:
$2 \cdot 3 \cdot 2 \cdot 2 = 24$
For only one even number:
$2 \cdot 3 \cdot 2 \cdot 1 = 12$
$3 \cdot 3 \cdot 2 \cdot 1 = 18$
$3 \cdot 2 \cdot 3 \cdot 1 = 18$
$3 \cdot 2 \cdot 1 \cdot 3 = 18$
TOTAL:
Therefore, in total, there are $(24+36+18\cdot 3+12+24=150)$ possibilities
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the rank of the following matrix depending on $\lambda\in\Bbb R$ Find the rank of the following matrix depending on $\lambda\in\Bbb R$.
$$A=\begin{pmatrix}
1&2&3&4\\
2&\lambda&6&7\\
3&6&8&9\\
4&7&9&10
\end{pmatrix}$$
My attempt:
$$\begin{pmatrix}
1&2&3&4\\
2&\lambda&6&7\\
3&6&8&9\\
4&7&9&10
\end{pmatrix}\sim\begin{pmatrix}
1&2&3&4\\
0&\lambda-4&0&-1\\
0&0&-1&-3\\
0&-1&-3&-6
\end{pmatrix}\sim\begin{pmatrix}
1&0&-3&-8\\
0&\lambda-4&0&-1\\
0&0&-1&-3\\
0&-1&-3&-6\\
\end{pmatrix}$$
For $\lambda=4$ we have:
$$\begin{pmatrix}
1 &0&-3&-8\\
0&0&0&-1\\
0&0&-1&-3\\
0&-1&-3&-6
\end{pmatrix}\sim\begin{pmatrix}
1&0&0&1\\
0&0&0&-1\\
0&0&-1&-3\\
0&-1&0&3\end{pmatrix}\sim\begin{pmatrix}
1&0&0&0\\
0&0&0&-1\\
0&0&-1&0\\
0&-1&0&0\\
\end{pmatrix}$$
$\Rightarrow r(A)=4$
For $\lambda\neq 4$ we have:
$$\begin{pmatrix}
1&0&0&1\\
0&\lambda-4&0&-1\\
0&0&-1&-3\\
0&-1&0&3\\
\end{pmatrix}\sim\begin{pmatrix}
1&0&0&1\\
0&0&0&3\lambda-13\\
0&0&-1&-3\\
0&-1&0&3
\end{pmatrix}$$
For$\lambda=\frac{13}{3}\Rightarrow r(A)=3$ and for $\lambda\neq \frac{13}{3} \Rightarrow r(A)=4$
Is this correct? Thanks!
| I will point out that you can also check the result in Wolfram Alpha.
Here is my computation - rather similar to yours.
$$\begin{pmatrix}
1 & 2 & 3 & 4\\
2 &\lambda& 6 & 7\\
3 & 6 & 8 & 9\\
4 & 7 & 9 &10
\end{pmatrix}\overset{(1)}\sim
\begin{pmatrix}
1 & 2 & 3 & 4\\
2 &\lambda& 6 & 7\\
3 & 6 & 8 & 9\\
1 & 1 & 1 & 1
\end{pmatrix}\sim
\begin{pmatrix}
0 & 1 & 2 & 3\\
0 &\lambda-2& 4 & 5\\
0 & 3 & 5 & 6\\
1 & 1 & 1 & 1
\end{pmatrix}\overset{(2)}\sim
\begin{pmatrix}
0 & 1 & 2 & 3\\
0 &\lambda-2& 4 & 5\\
0 & 1 & 1 & 0\\
1 & 1 & 1 & 1
\end{pmatrix}\sim
\begin{pmatrix}
0 & 1 & 2 & 3\\
0 &\lambda-2& 4 & 5\\
0 & 1 & 1 & 0\\
1 & 1 & 1 & 1
\end{pmatrix}\sim
\begin{pmatrix}
0 & 0 & 1 & 3\\
0 & 0 & 6-\lambda & 5\\
0 & 1 & 1 & 0\\
1 & 1 & 1 & 1
\end{pmatrix}\sim
\begin{pmatrix}
0 & 0 & 1 & 3\\
0 & 0 & 0 & 3\lambda-13\\
0 & 1 & 1 & 0\\
1 & 1 & 1 & 1
\end{pmatrix}\sim
\begin{pmatrix}
1 & 1 & 1 & 1\\
0 & 1 & 1 & 0\\
0 & 0 & 1 & 3\\
0 & 0 & 0 & 3\lambda-13
\end{pmatrix}$$
$(1)$: Subtracted third row from the last one.
$(2)$: Subtracted twice the first row from the last one.
In both cases I did so because the resulting row seemed to be simple (only zeroes and ones), which made further row operations a bit easier.
From the final matrix we see that rank is three if $3\lambda-13=0$ and in all other cases rank in equal to four.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Solve the system ${x}^{2}+ \left( y-1 \right) ^{2}=4,z^{4}+y{z}^{2}+xz+1=0. $ Find all real solutions of the system of equation
\begin{cases}
{x}^{2}+ \left( y-1 \right) ^{2}=4,\\{z}^{4}+y{z}^{2}+xz+1=0.
\end{cases}
| This system of 2 equations has infinite solutions.
Let $x=2\cos\theta$ and $y=2\sin\theta+1$ which satisfy the first equation. Then
$$z^4+yz^2+xz+1=0$$
$$z^4+(2\sin\theta+1)z^2+(2\cos\theta)z+1=0$$
$$(z^2+\sin\theta)^2+(z+\cos\theta)^2=0$$
$z=-\cos\theta$ and $z^2=-\sin\theta$, this leads us to find solutions of $\theta$.
$z^2=-\sin\theta$ shows $\sin\theta<0$.
$\sin^2\theta-\sin\theta-1=0$ shows $\theta=\arcsin\dfrac{1-\sqrt{5}}{2}\sim-38.17^\circ$. Only two solution $(x,y,z)$ exist. one for $\theta=(180+38.17)^\circ$ and other for $\theta=(360-38.17)^\circ$ and then
$$(x,y,z)=(2\cos\theta,2\sin\theta+1,-\cos\theta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to integrate $\int\sqrt{x^2+1}dx$ $$\int\sqrt{x^2+1}dx$$
I've been attempting this problem for days and I've made very little progress. I understand that I should substitute $x=\tan(x)$ and $dx=\sec^2(x)dx$ but beyond that I am very much lost.
| You can also do:
$$\int\sqrt{x^2+1}\,dx = \int\frac{x^2+1}{\sqrt{x^2+1}}\,dx = \int x\cdot \frac{x}{\sqrt{x^2+1}}\,dx + \int\frac{1}{\sqrt{x^2+1}}\,dx = x\sqrt{x^2+1} - \int\sqrt{x^2+1}\,dx\,+ \ln(x+\sqrt{x^2+1}) \implies \int\sqrt{x^2+1}\,dx = \frac{1}{2}\big(x\sqrt{x^2+1}+\ln(x+\sqrt{x^2+1})\big) + C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve $y'+y=|x|, y(-1)=0$. Find y(1)=? This is my try.
$y(x)e^x=\int|x|e^xdx$
$=\int_x^0(-t)e^tdt+\int_0^xte^tdt$
$=2+2(x-1)e^x+c$
$y(x)=2e^{-x}+(x-1)2+ce^{-x}$.
with $y(-1)=0$ gives $c=1-\frac{3}{e}$
but later I find which is wrong. where is my mistake?
| The homogenous equation
$y' + y = 0\\
y = C_1 e^-t$
A particular solution
Suppose $y = Ax + B$
$y' = A\\
A + Ax + B = |x|\\
A = sgn x\\
B = -A$
$y = \begin{cases} C_1 e^{-t} + x - 1 & x\ge 0\\C_2 e^{-t} - x + 1 & x< 0 \end{cases}$
$y(-1) = 0$
$C_2 e + 2 = 0\\
C_2 = -2 e^{-1}$
$y(0) = $$-2e^{-1}+1 = C_1 -1\\
C_1 = -2e^{-1} + 2$
$y = \begin{cases} (-2e^{-1} + 2) e^{-t} + x - 1 & x\ge 0\\ -2e^{-t-1} - x + 1 & x< 0 \end{cases}$
$y(1) = (-2e^{-1} + 2) e^{-1} = -2e^{-2} + 2e^{-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the particular solution of $\sin(2x)dx + \cos(3y)dy = 0$ for $y(\pi/2) = \pi/3$ Find the particular solution of $\sin(2x)dx + \cos(3y)dy = 0$ for $y(\pi/2) = \pi/3$, plot the graph of the solution, and determine the interval in which the solution is defined.
Because the terms are separable, we integrate the expression $\sin(2x)dx = -\cos(3y)dy$ to yield $$-\frac{1}{2}\cos(2x) = -\frac{1}{3}\sin(3y) + C$$ We plug in our initial conditions and simplify as follows $$-\frac{1}{2}(-1) = -\frac{1}{3}(0) + C \rightarrow C = \frac{1}{2}$$ Thus, our implicit particular solution is of the form $-\cos(2x)/2 = -\sin(3y)/3 + 1/2$. Explicitly, this is $$y =\frac{\arcsin(\frac{3\cos(2x)+3}{2})}{3}$$ Utilizing the trigonometric identity $\cos^2(x) = 1/2 + \cos(2x)/2$, we simplify the above expression to $$y = \frac{\arcsin(3\cos^2(x))}{3}$$ However, this is not the correct answer. When we plug in our initial condition $x = \pi/2$, we get $0$ and not $\pi/3$. To accommodate for this shortcoming, we must look at the behavior of the function above. We note that $3\cos^2(2x) \leq 1$, because the domain of $\arcsin x$ is $0 \leq x \leq 1$.
That's all I got. This guy (http://www.math.drexel.edu/~tolya/example1.pdf) solves the problem, but beyond what I wrote above, I don't know where to go with it. Plotting the graph would be easy once solved, but I'm hooked up at the interval of definition as well.
Any help would be much appreciated.
| Your calculus is correct.
$$-\frac{1}{2}\cos(2x) = -\frac{1}{3}\sin(3y) + C$$
The condition $\quad y(\pi/2) = \pi/3\quad $ leads to $\quad C=\frac{1}{2}$
Note that, if the condition was $\quad y(\pi/2) = k\pi/3\quad $ the constant would be the same $\quad C=\frac{1}{2}\quad$ because $\quad \sin(3k\pi/3)=\sin(3\pi/3)=0$.
Thus, for a particular value of the constant $\quad C=\frac{1}{2}\quad$ the solution of the ODE is not only one particular function :
$$y =\frac{\arcsin(\frac{3\cos(2x)+3}{2})}{3} = \frac{\arcsin(3\cos^2(x))}{3}$$
but is a family of functions :
$$y =\frac{\arcsin(\frac{3\cos(2x)+3}{2})}{3} +\frac{k\pi}{3} = \frac{\arcsin(3\cos^2(x))}{3}+\frac{k\pi}{3}$$
where $k$ is any integer.
The particular solution according to the condition $\quad y(\pi/2) = \pi/3\quad $ has to be defined among this family of solutions, with a determined value of $k$, in the present case $k=1$.
$$y(x) =\frac{\arcsin(\frac{3\cos(2x)+3}{2})+\pi}{3}=\frac{\arcsin(3\cos^2(x))+\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve the following simultaneous trig equations? Equation 1
$X$ = $a_1\sin(\theta_1) + a_2\cos(\theta_2)$
Equation 2
$Y$ = $a_1\cos(\theta_1) + a_2\sin(\theta_2)$
where $X, Y, a_1,a_2$ are known.
WHAT I HAVE DONE SO FAR
Let
$sin(\theta_1) = u_1$
$cos(\theta_1) = v_1$
$sin(\theta_2) = u_2$
$cos(\theta_2) = v_2$
And hence according to trig identities,
$u_1^2 + v_1^2 = 1$ (3)
$u_2^2 + v_2^2 = 1$ (4)
substituting (3) and (4) in the original equations, I obtained,
$a_2^2v_2^2 + a_1^2v_1^2 - 2Xa_2v_2 = a_1^2 - X^2$ (Equation 5)
$a_1^2v_1^2 + a_2^2v_2^2 - 2Ya_1v_1 = a_2^2 - Y^2$ (Equation 6)
$a_1^2u_1^2 + a_2^2u_2^2 - 2Xa_1u_1 = a_2^2 - X^2$ (Equation 7)
$a_1^2u_1^2 + a_2^2u_2^2 - 2Ya_2u_2 = a_2^2 - Y^2$ (Equation 8)
How do I proceed from here?
These systems of 4 non-linear equation. Is developing an analytical solution possible? Or should try to find a numerical solution using some libraries?
Thanks,
Vino
| As I said in the comments, it looks like there is no nice closed-form solution for this. So I plugged your set of equations into Mathematica. The result is:
$$\begin{align}
\sin\theta_1=\frac{({a_1}^3-a_1{a_2}^2)X-\sqrt{-{a_1}^2 Y^2 \left({a_1}^4-2 {a_1}^2
\left({a_2}^2+X^2+Y^2\right)+\left(-{a_2}^2+X^2+Y^2\right)^2\right)}}{2 {a_1}^2 \left(X^2+Y^2\right)}+\frac X{2a_1}\\
\cos\theta_2=\frac{(a_1{a_2}^2-{a_1}^3)X+\sqrt{-{a_1}^2 Y^2 \left({a_1}^4-2 {a_1}^2
\left({a_2}^2+X^2+Y^2\right)+\left(-{a_2}^2+X^2+Y^2\right)^2\right)}}{2 {a_1} {a_2}
\left(X^2+Y^2\right)}+\frac X{2a_2}
\end{align}$$
To simplify it more, let
$$\Delta=2 {a_1}^2\left(X^2+Y^2+{a_2}^2\right)-\left(X^2+Y^2-{a_2}^2\right)^2-{a_1}^4$$
then
$$\begin{align}
\sin\theta_1&=X\left(\frac{a_1^2-a_2^2}{2a_1(X^2+Y^2)}+\frac{1}{2a_1}\right)-Y\frac{\sqrt{\Delta}}{2a_1(X^2+Y^2)}\\
\cos\theta_2&=X\left(\frac{a_2^2-a_1^2}{2a_2(X^2+Y^2)}+\frac{1}{2a_2}\right) +Y\frac{\sqrt{\Delta}}{2a_2(X^2+Y^2)}
\end{align}$$
I hope this helps.
| {
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"url": "https://math.stackexchange.com/questions/2128540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Polynomials with integer coefficients – find minimal $m$ Let $m$ be the minimal positive integer such that
$$(x+4)(x+5)(x+9)p(x) - (x-4)(x-5)(x-9)q(x) = m$$
where $p(x)$ and $q(x)$ are polynomials with integer coefficients. What is the value of $m$?
My work :
By Bezout's identity, we have, $\exists p(x), q(x) \in \mathbb{Z}[x]$ such that
$(x+4)(x+5)(x+9)p(x) - (x-4)(x-5)(x-9)q(x) = gcd((x+4)(x+5)(x+9), (x-4)(x-5)(x-9))$
Consider the order pair, $((p(x),-q(x))$, by Euclid's Algorithm,
$gcd((x+4)(x+5)(x+9), (x-4)(x-5)(x-9))$
$ = gcd(x^3-18x^2+101x-180, x^3+18x^2+101x+180)$
$ = gcd(x^3-18x^2+101x-180, 36x^2 + 360)$
Please suggest how to proceed.
| LEMMA: If $f(x),g(x)\in \mathbb Z[x]$ and a prime number $p$ divides every co-efficient of the product $f(x)g(x)$ then $p$ divides every co-efficient of $f(x)$ or of $g(x).$
From this we can show that if $n\in \mathbb N$ divides every co-efficient of $f(x)$ and $n$ is co-prime to some (any) co-efficient of $g(x)$ then $gcd (f(x),g(x))=gcd (f(x)/n,g(x)).$
From your last line therefore $$gcd (x^3-18x^2+101x-180, 36x^2+360)=gcd (x^3-18x^2+101x-180,x^2+10)=$$ $$=gcd (x^3-18x^2+101x-180-x(x^2+10),x^2+10)=gcd (-18x^2+91x-180,x^2+10)=$$ $$=gcd(-18x^2+91x-180+18(x^2+10),x^2+10)=$$ $$=gcd (91x,x^2+10)=gcd(x,x^2+10)=$$ $$=gcd(x,x^2+10-x(x))=gcd(x,10)=gcd(x,1)=1.$$ Note that the disappearance of the factors "$36$" and the "$10$" are due to the lemma.
PROOF OF LEMMA: Let $f(x)=\sum_i f_ix^i$ and $g(x)=\sum_j g_jx^j$ and $f(x)g(x)=\sum_kh_kx^k.$ Suppose by contradiction that the prime $p$ divides every $h_k$ but not every $f_i$ and not every $g_j.$ Let $i_1$ be the least $i$ such that $p\not | \;f_i$ and let $j_1$ be the least $j$ such that $p\not |\; g_j.$
We have $h_{(i_1+j_1)}=\sum_{i+j=i_1+j_1}f_ig_j.$ In this sum :(i) When $k<i_1$ we have $p|f_i,$ and (ii) When $k>i_1$ we have $j<j_1$ so $p|g_j.$ Therefore , modulo $p,$ we have $$0\equiv h_{(i_1+j_1)}\equiv f_{i_1}g_{j_1}\not \equiv 0,$$ a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution of $\lim_{x\to 0} \frac{(x \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $ using Mclaurin series The problem is
$$\lim_{x\to 0} \frac{(x \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $$
I know that
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}... $$
got this
$$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15}...$$
and suppose to use taylor expansion for $\ln {(1 + x)}$
Tried to expand $\sqrt{1 + \sin x}$ and $\ln{\sqrt{(1 + x^2)}}$ but it seems to be messy. Am I on the right direction?
| Note that $$x\sqrt{1+\sin x} = x\sqrt{1+x+O(x^3)} = x(1+\frac x2-\frac{x^2}8+O(x^3)) = x+\frac{x^2}{2}-\frac{x^3}8+O(x^4)$$
and
$$\ln\sqrt{1+x^2} = \ln(1+\frac{x^2}2+O(x^4)) = \frac{x^2}{2}+O(x^4)$$
so we have
$$\lim_{x\to 0}\frac{x\sqrt{1+\sin x}-\ln\sqrt{1+x^2}-x}{\tan^3x} = \lim_{x\to 0}\frac{-\frac{x^3}8+O(x^4)}{x^3+O(x^4)} = -\frac 18$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining if a subset of $\mathbb{C}$ is a subfield. I am trying to deduce whether or not
$$K=\{a+b\sqrt{2}|a,b\in\mathbb{Q}\text{ and }ab<\sqrt{2}\}$$
is a subfield of $\mathbb{C}$. I feel like I'm getting stuck. If I subtract two numbers $a+b\sqrt{2}$ and $c+d\sqrt{2}$, then we have to show that
$$(a-c)(b-d)=ab+cd-(ad+bc)<2\sqrt{2}-(ad+bc)<\sqrt{2}$$
Thus I have $ad+bc>\sqrt{2}$ and I'm not sure how to procede here. It I look at the multiplicative inverse of $a+b\sqrt{2}$,
$$(a+b\sqrt{2})^{-1}=\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}$$
and so the conditions must hold for
$$\frac{-ab}{(a^2-2b^2)^2}<\frac{-\sqrt{2}}{(a^2-2b^2)^2}<\sqrt{2}$$
$$\frac{1}{(a^2-2b^2)^2}>-1$$
Which is true for all $a,b$. I'm stuck....
| The key here is observing that, for $a,b,c,d\in\mathbb{Q}$,
$$
a+b\sqrt{2}=c+d\sqrt{2}
\quad\text{if and only if}\quad
a=c\text{ and }b=d
$$
which is easy, but should be noted before going on. One direction is obvious; for the reverse, if $b\ne d$, then
$$
\sqrt{2}=\frac{a-c}{d-b}
$$
which is a contradiction. Therefore $b=d$ and consequantly also $a=c$.
Now $2=2+0\sqrt{2}$ and $2\sqrt{2}=0+2\sqrt{2}$ both belong to $K$. However $2+2\sqrt{2}$ doesn't satisfy $2\cdot2<\sqrt{2}$ and there cannot be other rational numbers $a,b$ with $a+b\sqrt{2}=2+\sqrt{2}$. Thus $K$ is not closed under addition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Number of integral quotient of $n^2$ Is there a theorem which states the number of integral quotient of $n^2$ divided by $\{1,2,3, ... n^2\}$ is $2n-1$?
Example: If $n=4$,then $16 \div \{1,2,3, ... 16\} = 16,8,5,4,3,2,1$. There are $7$ integral quotients.
| The set $\{\,\lfloor \frac {n^2}d\rfloor\mid 1\le d\le n^2\,\}$ contains the numbers $1,2,\ldots,n$ from taking $d=n,\ldots, n^2$ because $d=n$ leads to $\lfloor\frac{n^2}{n}\rfloor=n$, and because $\lfloor \frac{n^2}{d+1}\rfloor$ is either $\lfloor \frac {n^2}d\rfloor$ or $\lfloor \frac {n^2}d\rfloor-1$ for $d\ge n$ (because $\frac{n^2}d-\frac{n^2}{d+1}=\frac{n^2}{d(d+1)}<1$). So that's $n$ values so far.
On the other hand, the values obtained from $1\le d\le n$ are pairwise distinct because $\frac{n^2}d-\frac{n^2}{d+1}=\frac{n^2}{d(d+1)}>1$ for $d<n$.
Thus we get a total of $2n-1$ distinct values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all matrices $X$ such that $ABXB^tA^t=I$ Find all matrices $X$ such that:
$$ABXB^tA^t=I$$ if
$A=\begin{pmatrix}
1 &-2 &2\\
3 &-5 &6\\
-1 &2 &-1
\end{pmatrix}$ and $B=\begin{pmatrix}
-3 &-2 &-2\\
2 & 1 &1\\
6 &3 &4
\end{pmatrix}$.
So I managed to get that $AB=\begin{pmatrix}
5 &2 &4\\
17 &7 &13\\
1&1&0
\end{pmatrix}$ and $B^tA^t=(AB)^t=\begin{pmatrix}
5 &17 &1\\
2 &7 &1\\
4 &13 &0
\end{pmatrix}$
So we have $\begin{pmatrix}
5 &2 &4\\
17 &7 &13\\
1&1 &0
\end{pmatrix}\cdot X\cdot\begin{pmatrix}
5 &17 &1\\
2&7&1\\
4&13&0
\end{pmatrix}=I$
Now how do I get $X$ from here?
| Matrices multiplication is not commutative. You are allowed to multiply an equation by a non-zero matrix, but you have to mind the side from which you are applying the multiplication.
Let us say that $P=AB$ and $Q=B^TA^T$.
Your equation takes the form of $P \cdot X \cdot Q = I$.
You can multiply each side by $P^{-1}$ from the left and $Q^{-1}$ from the right to get (note that $\det(P)\ne0\ne\det(Q)$, which makes $P$ and $Q$ invertible):
$A^{-1} \cdot P \cdot X \cdot Q \cdot Q^{-1} = P^{-1} \cdot I \cdot Q^{-1}$
This can be simplified based on $M^{-1} \cdot M = M \cdot M^{-1} = I$:
$I \cdot X \cdot I = P^{-1} \cdot I \cdot Q^{-1}$
This can further be simplified based on $M \cdot I = M$:
$X=P^{-1} \cdot Q^{-1}$
Mind that:
$M^{-1}=\frac{1}{\det(M)}\left(M^*\right)^T$
Where $M^*$ is the cofactor matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
What is the sum of $100$ terms of the sequence $a_{k} = \frac{1}{k} - \frac{1}{k+1}$? In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence?
The answer given is $\frac{100}{101}$, but I am not sure how.
So far I am have plugged in the values of $k's$ and have the following values $$\frac{1}{2}, \frac{1}{6}, \frac{1}{12}, \frac{1}{12}, \frac{1}{20}, \frac{1}{30},....$$
The numerator makes sense to me as it's just $1$ and $100 \times 1 = 100$, however, I am not sure about the denominator.
| $$a_{100} = \left( \frac11 - \frac12 \right)+\left( \frac12 - \frac13 \right)+
\left( \frac13 - \frac14 \right)+\cdots + \left( \frac1{100} - \frac1{101} \right)
\\= \frac11 + \left( -\frac12+\frac12\right)+\left( -\frac13+\frac13\right)
+\cdots +\left( -\frac1{100}+\frac1{100}\right) -\frac1{101} = 1 - \frac1{101}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Which quaternions are solutions of $x^2+1=0$? What would the final Quaternion Solutions look like for $x^2+1=0$?
I substituted in $x = a+bi+cj+dk$ and came up with a very long +/- square root.
| The solutions have the form
$$
bi + cj + dk
$$
where $b^2 + c^2 + d^2 = 1$. They are the "pure imaginary unit quaternions".
Why? Because $x^2 = -1$, so $|x|^2 = 1$ so $a^2 + b^2 + c^2 + d^2 = 1$. Since the real part of $x^2$ is $a^2 - b^2 - c^2 - d^2 = -1$, we get that $2a^2 = 0$, so $a = 0$.
Once you settle that, it's easy to check that all the other pure-imaginary unit quaternions do in fact square to $-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the maximum of positive integer $k$ so that for all positive real numbers $x$ we have: $x^6+x^5+x^4+x^2+x+4>kx^3$
Find the maximum of positive integer $k$ so that for all positive real numbers $x$ we have: $x^6+x^5+x^4+x^2+x+4>kx^3$
Since the power of the polynomial on LHS is greater than 3 I have no idea for it!
| Assume $k \ge 9$. Then we have $$9=1^6+1^5+1^4+1^2+1+4>k \times 1^3 \ge 9$$
Contradiction. So we have $k<9$.
Note that $$x^6+x^5+x^4+x^3+x^2+x+4=(x^6+4)+(x^5+x)+(x^4+x^2)>8x^3$$
As follows from $\text{AM-GM}$, and because the equality condition can not be satisfied, there is no equality symbol.
So since $8$ is possible, but $k$ is the maximum, we have $k \ge 8$.
So since $k$ is a positive integer that satisfies $$9>k \ge 8$$ we have $k=8$. The answer is $k=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\cos(\pi/7)$ is root of equation $8x^3-4x^2-4x+1=0$
Prove that $\cos\theta$ is root of equation $8x^3-4x^2-4x+1=0$, given $\theta=\frac{\pi}{7}$.
I put $\cos\theta$ in equation, but couldn't show the left-hand side to be zero.
| We have
$$ \cos{2\theta} = 2\cos^2{\theta}-1 \\
\cos{3\theta} = \cos{\theta}(2\cos^2{\theta}-1) - 2\sin^2{\theta}\cos{\theta} = 4\cos^3{\theta}-3\cos{\theta}, $$
so
$$ \cos^2{\theta} = \frac{1}{2}(\cos{2\theta}+1) \\
\cos^3{\theta} = \frac{1}{4}(\cos{3\theta}+3\cos{\theta}) $$
Putting these into the equation gives
$$ 8\cos^{3}{\theta}-4\cos^2{\theta}-4\cos{\theta}+1 \\
= 2\cos{3\theta}+6\cos{\theta} -2\cos{2\theta}-2-4\cos{\theta}+1 \\
= -1+2(\cos{3\theta}-\cos{2\theta}+\cos{\theta}) = \frac{\cos{(7x/2)}}{\cos{(x/2)}}, $$
the last part of which comes from the formula
$$ \sum_{k=-n}^n (-1)^k \cos{kx} = (-1)^n \frac{\cos{(n+1/2)x}}{\cos{(x/2)}}, $$
which can be proven by induction. It's then clear that this is zero if $x$ is a zero of $\cos{(7x/2)}$, but not $\cos{(x/2)}$, and the first one of these is $\theta=\pi/7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to solve this system of equations in Lagrange Multiplier problem Find the maximum and minimum values of ${x^{2} + y^{2} + z^{2}}$ subject to the conditions ${\frac{x^{2}}{4} + \frac{y^{2}}{5} + \frac{z^{2}}{25} = 1}$ and ${x + y - z = 0}$.
Using Lagrange multiplier method, I got following equations:
$$ {2x = \frac{\lambda_{1} x}{2} + \lambda_{2}}$$
$$ {2y = \frac{2 \lambda_{1} y}{5} + \lambda_{2}}$$
$$ {2z = \frac{2 \lambda_{1} z}{25} - \lambda_{2}}$$
$${\frac{x^{2}}{4} + \frac{y^{2}}{5} + \frac{z^{2}}{25} = 1}$$
$${x + y - z = 0}$$
I'm stuck after this. I've tried to solve this system of equations to get critical point many times. Any help will be greatly appreciated. Also is there any other way to approach this problem?
| I think Lagrange multipliers method is not necessary here.
We need to find a maximal and a minimal value of $x^2+y^2+(x+y)^2$, where $\frac{x^2}{4}+\frac{y^2}{5}+\frac{(x+y)^2}{25}=1.$
Let $2(x^2+xy+y^2)=k$.
Hence, the condition gives $29x^2+8xy+24y^2=100$ or
$$k(29x^2+8xy+24y^2)=200(x^2+xy+y^2),$$
which says that the equation
$$(29k-200)x^2+(8k-200)xy+(24k-200)y^2=0$$
has real solutions.
If $k=\frac{200}{29}$ we has solutions.
Let $k\neq\frac{200}{29}$.
Hence,
$$(4k-100)^2-(29k-200)(24k-200)\geq0,$$
which gives $$\frac{75}{17}\leq k\leq10$$
It's obvious that the equality occurs in both cases and we are done!
| {
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How could I know that $X^4+1$ is $(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1)$? I thought that $X^4+1$ was irreducible, but in fact, $$X^4+1=(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1).$$
In general, how can I have the intuition of such a factorisation if I don't know it ?
| We know that $$a^4+4=a^4+4a^2+4-4a^2=(a^2+2)^2-(2a)^2=(a^2+2a+2)(a^2-2a+2)$$
This is an well known identity, most easily identifiable from the difference between two squares. It is called the Sophie Germain Identity=
Putting in $x=\sqrt{2} a$, we have that $$x^2+1=(x^2+\sqrt{2} x+1)(x^2-\sqrt{2}x+1)$$
Though the first step is unnecessary, I added it as it is a generally useful formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 3
} |
show that $\lim_{x\rightarrow 0} \left(\frac{\sin(x)}{x}\right)^{1/x^2} = e^{-1/6}$ I see that this is indeterminate form, so I approach L'Hopital rule, but I can not find this limit. Please help me to find this limit.
| $$\lim _{ x\rightarrow 0 }{ { \left( \frac { \sin { x } }{ x } \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { \sin { x } }{ x } -1 \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { \sin { x-x } }{ x } \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =$$ hence $\lim _{ x\rightarrow 0 }{ \frac { \sin { x-1 } }{ x } } =0\quad $ we can use here a well known limit form $$\lim _{ x\rightarrow 0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e$$
then we can write it as $$=\lim _{ x\rightarrow 0 }{ { { \left[ { \left( 1+\frac { \sin { x } -x }{ x } \right) }^{ \frac { x }{ \sin { x } -x } } \right] }^{ \frac { \sin { x } -x }{ x } \cdot \frac { 1 }{ { x }^{ 2 } } } } } $$
the inner part of limit is clearly is $e$ so $$=\quad { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \sin { x } -x }{ x } \cdot \frac { 1 }{ { x }^{ 2 } } } }$$
now to find the exponent limit we use L'Hospital rule here three times $$\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \cos { x } -1 }{ 3{ x }^{ 2 } } } }=\\ \overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\sin { x } }{ 6{ x } } } }=\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\cos { x } }{ 6 } } }={ e }^{ -\frac { 1 }{ 6 } }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Prove: $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) > 10^{2009}$
Show that the number $ 3^{{4}^{5}} {+} 4^{{5}^{6}}$ can be expressed as the product of two integers greater than $ 10^{2009}$
By Sophie Germain:
$ 3^{{4}^{5}} {+} 4^{{5}^{6}}=\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)\cdot \left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)$
Now $\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)>\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)$
So need to prove $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)>10^{2009}$.
Please help with this, and I'm very enthusiast to know your method to solve above problem.
| $log_{10}(3^{4^5}+4^{5^6}) = 9407.1874$ then it has $9408$ digits. Since you express this number as a product of two numbers then each of those has $4704$ digits.
$log_{10}(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}) = 4703.5937$ as we can see both numbers have $4704$ digits then we conclude that :
$$3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6} > 10^{2009}$$ $$3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6} > 10^{2009}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to solve $2\cdot \left( \frac { 1 }{ 3 } \right) ^{ 0.3-x }-4=2$ How do I solve this equation:
$$2\cdot \left( \frac { 1 }{ 3 } \right) ^{ 0.3-x }-4=2$$
I've tried it myself but ended up with x = 0.3, eventhough the correct answer is 1.3
I followed these steps:
$$ 2*(1/3)^{(0.3-x)}-4=2$$
$$2*(1/3)^{(0.3-x)}=-2$$
$$(1/3)^{(0.3-x)}=-1$$
$$(1/3)^{(0.3-x)}=-(1/3)^0$$
$$0.3-x=0$$
$$x=0.3$$
| $$2\cdot \left( \frac { 1 }{ 3 } \right) ^{ 0.3-x }-4=2\\ 2\cdot \left( \frac { 1 }{ 3 } \right) ^{ 0.3-x }=6\\ \left( \frac { 1 }{ 3 } \right) ^{ 0.3-x }=3\\ { 3 }^{ x-0.3 }=3\\ x-0.3=1\\ x=1.3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Circle tangent to a parabola this is a calculus-geometry question that I found -
A circle of radius $1$ is tangent to the parabola $y = x^2 -2x + 1$ at two points. Find the coordinates of the center of the circle.
My attempt:
Let the center of the circle be $(a, b)$. Since there are two intersection points, we have:
$(x - a)^2 + (y - b)^2 - 1 = x^2 - 2x + 1$
$x^2 - 2ax + a^2 + y^2 - 2ay + b^2 = x^2 - 2x$
$-2ax + a^2 + y^2 - 2ay + b^2 = -2x$
I am unsure of how to proceed; I think I am approaching this problem wrong because I did not take the derivative. The correct answer is $(1, 5/4).$
| It is simpler to translate the parabola by $1$ towards the left, solving the problem for the parabola $y=x^2$ and then re-translating the final result by one unit towards the right. In this way, we can search a circle with its center on the $y $-axis, which typically has an equation of the form
$$x^2 +(y-k)^2=1$$
where $k $ is the $y $-coordinate of the lower point of the circle at its inferior intersection with the $y $-axis. The points of intersection between the parabola and the circle are the solutions of the system
$$\left\{\begin{array}{lll}
x^2 +(y-k)^2=1 \\
y=x^2
\end{array}\right.$$
The solutions for $y $ are
$$y = \frac {1}{2} \left(2 k \pm \sqrt{5 - 4 k} - 1 \right) $$
Because we are searching for points where the curves are tangent, the two solutions of $y $ have to coincide. This occurs for $k=\frac {5}{4} $. This leads to the circle
$$x^2 +(y-\frac {5}{4})^2=1$$
Re-translating towards the right we get the final equation of the circle
$$(x-1)^2 +(y-\frac {5}{4})^2=1$$
whose center is in $(1,\frac {5}{4}) $. Here is a graph of the two tangent curves.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $a^2+b^2+c^2=1$ so $\sum\limits_{cyc}\frac{1}{(1-ab)^2}\leq\frac{27}{4}$
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that:
$$\frac{1}{(1-ab)^2}+\frac{1}{(1-ac)^2}+\frac{1}{(1-bc)^2}\leq\frac{27}{4}$$
This inequality is stronger than $\sum\limits_{cyc}\frac{1}{1-ab}\leq\frac{9}{2}$ with the same condition,
which we can prove by AM-GM and C-S:
$$\sum\limits_{cyc}\frac{1}{1-ab}=3+\sum\limits_{cyc}\left(\frac{1}{1-ab}-1\right)=3+\sum\limits_{cyc}\frac{ab}{1-ab}\leq$$
$$\leq3+\sum\limits_{cyc}\frac{(a+b)^2}{2(2a^2+2b^2+2c^2-2ab)}\leq3+\sum\limits_{cyc}\frac{(a+b)^2}{2(a^2+b^2+2c^2)}\leq$$
$$\leq3+\frac{1}{2}\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)=\frac{9}{2},$$
but for the starting inequality this idea does not work.
By the way, I have a proof of the following inequality.
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=3$. Prove that:
$$\sum\limits_{cyc}\frac{1}{(4-ab)^2}\leq\frac{1}{3}$$
(we can prove it by SOS and uvw).
This inequality is weaker and it not comforting.
We can assume of course that all variables are non-negatives.
Thank you!
| The Buffalo Way works. (I think that Michael Rozenberg knew this.)
We only need to prove the case $a, b, c > 0$.
After homogenization and clearing the denominators, it suffices to prove that
$f(a, b,c)\ge 0$ where $f(a,b,c)$ is a homogeneous polynomial of degree $12$.
Due to symmetry, assume that $c\le b\le a$. Let $c = 1, \ b = 1+s, \ a = 1+s+t; \ s,t\ge 0$.
Then $f(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients. We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why does "$x^2 - 5x + 6 = 0$", which is the same as "$(x-3)(x-2) = 0$", represent a parabola? Consider the equation $x^2 - 5x + 6 = 0$. By factorising I get $(x-3)(x-2) = 0$. Which means it represents a pair of straight lines, namely $x-2 =0 $ and $x- 3 = 0$, but when I plot $x^2 - 5x + 6 = 0$, I get a parabola, not a pair of straight lines. Why?
Plotting: x^2 - 5x + 6 = 0
at Wolfram Alpha, I get the result:
|
I get $(x-3)(x-2) = 0$. which means it represent a pair of striaght
lines namely $x-2 =0 $ and $x- 3 = 0$
When we just write
$$
(x-3)(x-2) = 0
$$
and ask for $x$ we usually mean the set of solutions
$$
S = \{ x \mid (x-3)(x-2) = 0 \}
$$
Two vertical lines in two dimensions we can model as:
$$
L_1 = \{ (x,y) \mid (x-3) = 0 \} \\
L_2 = \{ (x,y) \mid (x-2) = 0 \}
$$
These are different sets. We sloppily write $x-3=0$ and $x-2=0$ but mean the above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 8,
"answer_id": 3
} |
If $a^2+b^2+c^2+d^2=4$ what is the range of $a^3+b^3+c^3+d^3$? I tried to used AM-GM but could not do.Setting any one as $2$ or $-2$ and the others $0$ we get the range as $[-8,8]$ but what is the formal way to do this?
| Without resorting to higher level of math such as the use of "compactness". Observe that $a^2 \le a^2+b^2+c^2+d^2 = 4\implies a \le 2\implies a^2(a-2) \le 0$, and similarly: $b^2(b-2) \le 0, c^2(c-2) \le 0, d^2(d-2) \le 0$. Thus $a^3+b^3+c^3+d^3 \le 2(a^2+b^2+c^2+d^2) = 2\cdot 4 = 8$. And the max is $8$ which occurs when exactly one of the variables is $2$ and the others are $0$. For the min, observe that $a \ge -2, b \ge -2, c \ge -2, d \ge -2$. Thus $a^2(a+2) \ge 0, b^2(b+2) \ge 0, c^2(c+2) \ge 0, d^2(d+2) \ge 0\implies a^3+b^3+c^3+d^3 \ge -2(a^2+b^2+c^2+d^2)=-8$, which is the min and this value is achieved when exactly one of the variables is $-2$, and the others are $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you find the derivative of $x\sqrt{x}$ with nonstandard analysis? Find the derivative of $x\sqrt{x}$. I'm doing this without power rules etc. , what I know beforehand are that
$$f'(x)=st\left(\frac{f(x+\Delta x)-f(x)}{\Delta x}\right)$$ and that
$$y'=st\left(\frac{\Delta y}{\Delta x}\right).$$
The correct answer is $y'=\frac{3}{2}\sqrt{x}$.
There are different approaches I have tried with without success. I have tried and tried, and do not know what to do to solve the problem. For example,
I do not know if it really is $\neq \frac{1}{(x+\Delta x)^{\frac{3}{2}}+x^{\frac{3}{2}}}$. But if I use that value for $\Delta y$, then I just get that $y'=\frac{1}{2x^{\frac{3}{2}}}$, and
$\frac{1}{2x^{\frac{3}{2}}} \neq \frac{3}{2} \sqrt{x}.$
PS the methods I am using are taught by Jerome H. Keisler in his book Elementary calculus: an infinitesimal approach.
Please help me,
Andreas
| Let's examine your proof.
Begin with
$$
\Delta y = (x + \Delta x)^{3/2} - x^{3/2}
$$
multiply numerator and denominator by a conjugate
$$
\big((x + \Delta x)^{3/2} - x^{3/2}\big) \cdot
\frac{(x + \Delta x)^{3/2} + x^{3/2}}{(x + \Delta x)^{3/2} + x^{3/2}}
$$
this is
$$
\frac{(x - \Delta x)^{3} - x^{3}}{(x + \Delta x)^{3/2} + x^{3/2}} =
\frac{x^3 + 3 x^2 (\Delta x) + 3 x^2 (\Delta x)^2 +(\Delta x)^3 - x^{3}}{(x + \Delta x)^{3/2} + x^{3/2}}
=
\frac{3 x^2 (\Delta x) + 3 x^2 (\Delta x)^2 +(\Delta x)^3}{(x + \Delta x)^{3/2} + x^{3/2}}
$$
I don't know how Keisler proceeds here, but I can say this is
$$
(\Delta x) \cdot \frac{3 x^2 + 3 x^2 (\Delta x) +(\Delta x)^2}{(x + \Delta x)^{3/2} + x^{3/2}}
$$
Then divide from the beginning
$$
\frac{\Delta y}{\Delta x} = \frac{3 x^2 + 3 x^2 (\Delta x) +(\Delta x)^2}{(x + \Delta x)^{3/2} + x^{3/2}}
$$
and compute that the right side is infinitely close to
$$
\frac{3 x^2 + 0 + 0}{(x+0)^{3/2}+x^{3/2}} = \frac{3 x^2}{2x^{3/2}} = \frac{3}{2}\;x^{1/2}
$$
Note the last step must be done differently when $x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the value of $\sum\binom{6n}{2k-1}(-3)^k$
Find the value of $\sum_{k=1}^{3n}\binom{6n}{2k-1}(-3)^k$.
My working:
\begin{align}
& \sum\binom{6n}{2k-1}(-3)^k\\
=& \sum\binom{6n}{2k-1}(i\sqrt3)^{2k}\\
=& i\sqrt3\sum\binom{6n}{2k-1}(i\sqrt3)^{2k-1}\\
=& \frac{i\sqrt3[(1+i\sqrt3)^{6n}-(1-i\sqrt3)^{6n}]}{2}\\
=& \frac{i\sqrt3[(1+i\sqrt3+1-i\sqrt3)^{3n}((1+i\sqrt3-1+i\sqrt3)^{3n}]}{2}\\
=& \frac{i\sqrt3\cdot2^{3n}2^{3n}(i\sqrt3)^{3n}}{2} \\
=& \frac{2^{6n}(i\sqrt3)^{3n+1}}{2}
\end{align}
But the correct answer is $0$ and I can't figure out how one gets that. Moreover, I want to know what's wrong with my solution.
| $\sum \binom{m}{2k-1}x^{2k-1}
+\sum \binom{m}{2k}x^{2k}
=\sum \binom{m}{k}x^k
=(1+x)^m
$.
$(1-x)^m
=\sum \binom{m}{k}(-1)^kx^k
$
so
$(1+x)^m+(1-x)^m
=\sum \binom{m}{k}x^k(1+(-1)^k)
=2\sum \binom{m}{2k}x^{2k}
$
and
$(1+x)^m-(1-x)^m
=\sum \binom{m}{k}x^k(1-(-1)^k)
=2\sum \binom{m}{2k-1}x^{2k-1}
$.
Therefore,
putting $x = i\sqrt{3}$
and $m = 6n$,
$\begin{array}\\
\sum \binom{6n}{2k-1}(-3)^{k}
&=\sum \binom{6n}{2k-1}(i\sqrt{3})^{2k}\\
&=i\sqrt{3}\sum \binom{6n}{2k-1}(i\sqrt{3})^{2k-1}\\
&=\frac{i\sqrt{3}}{2}((1-i\sqrt{3})^{6n}-(1+i\sqrt{3})^{6n})\\
&=0\\
\end{array}
$
since
$(1-i\sqrt{3})^3
=(1+i\sqrt{3})^3
=-8
$
and
$(1-i\sqrt{3})^6
=(1+i\sqrt{3})^6
=64
$.
Similarly,
$\begin{array}\\
\sum \binom{6n}{2k}(-3)^{k}
&=\sum \binom{6n}{2k}(i\sqrt{3})^{2k}\\
&=\frac{1}{2}((1-i\sqrt{3})^{6n}+(1+i\sqrt{3})^{6n})\\
&=\frac{1}{2}(64^n+64^n)\\
&=64^n\\
\end{array}
$
| {
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"url": "https://math.stackexchange.com/questions/2154481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Letter counting problem using generating functions Consider 4 different letters: a, b, c and d. Imagine you have access from 4 up to 12 letters, with at least 1 letter of each. For example, if I have 5 letters, one possible combination could be one $a$, one $b$, one $c$ and two $d$'s. Considering the case where it is irrelevant which letters are repeated, in how many ways can I choose the letters? By hand, I know the answer is 14, but the goal is to solve this problem using generating functions.
I know that the generating function $A(x)$ for the letter "a", for example, is
$$A(x) = x + x^2 + x^3 = x \left(\frac{1-x^3}{1-x}\right)$$
so, if the type of letters to be repeated were important, the total generating function would be
\begin{align}
A(x)B(x)C(x)D(x) & = \left(x \left(\frac{1-x^3}{1-x}\right)\right)^4 \\
& = x^4 + 4x^5 + 10x^6 + 16x^7 + 19x^8 + 16x^9 + 10x^{10} + 4 x^{11} + x^{12}
\end{align}
and the answer would be 81, the sum of all coefficients. The part where I'm struggling is, since the letters to be repeated are irrelevant, then the generating functions for the letters apart from $a$ should somehow depend on the numbers of $a$'s chosen in each case. I have to somehow subtract the extra irrelevant cases.
| We consider words of length $4$ up to $12$ built from a four character alphabet $\{a,b,c,d\}$ . The following holds
*
*Each word contains each of the four characters at least once
*The position of the characters don't matter: e.g. $$aabcd\equiv badca$$
*Multiplicity of characters matter, but not which character has a specific multiplicity: e.g. $$aabbcd\equiv abbcdd$$
With these rules, the problem boils down to look for integer partitions with exactly $4$ parts and size from $4$ up to $12$. If we consider e.g. the words of length $n=8$ there are $5$ different types
\begin{align*}
&5+1+1+1\qquad\longrightarrow &aaaaabcd\\
&4+2+1+1\qquad\longrightarrow &aaaabbcd\\
&3+3+1+1\qquad\longrightarrow &aaabbbcd\\
&3+2+2+1\qquad\longrightarrow &aaabbccd\\
&2+2+2+2\qquad\longrightarrow &aabbccdd\\
\end{align*}
The generating functions for the number of integer partitions with at most $k$ parts is given by
\begin{align*}
P^{(\leq k)}(z)=\prod_{m=1}^k\frac{1}{1-z^m}\tag{1}
\end{align*}
This can be found e.g. in Example I.7 in Analytic Combinatorics by P. Flajolet and R. Sedgewick.
Since we need a generating function which counts all integer partitions with exactly $4$ parts we obtain from (1)
\begin{align*}
P^{(= 4)}(z)&=P^{(\leq 4)}(z)-P^{(\leq 3)}(z)
=\prod_{m=1}^4\frac{1}{1-z^m}-\prod_{m=1}^3\frac{1}{1-z^m}\\
&=z^4\prod_{m=1}^4\frac{1}{1-z^m}\\
&=z^4 + z^5 + 2 z^6 + 3 z^7 + 5 z^8 + 6 z^9\\
&\qquad + 9 z^{10} + 11 z^{11} + \color{blue}{15} z^{12} + 18 z^{13} + \cdots
\end{align*}
Here the coefficient of $z^n, 4\leq n\leq 12$ gives (with some help of Wolfram Alpha) the number of wanted words.
$$ $$
Example: $n=15$
Let's look for example at the blue colored value which tells us, there are $\color{blue}{15}$ different partitions of $12$ with four parts.
\begin{align*}
9+1+1+1&&6+3+2+1&&5+3+2+2\\
8+2+1+1&&6+2+2+2&&4+4+3+1\\
7+3+1+1&&5+5+1+1&&4+4+2+2\\
7+2+2+1&&5+4+2+1&&4+3+3+2\\
6+4+1+1&&5+3+3+1&&3+3+3+3
\end{align*}
Note: With respect to some comments, the $q$-binomial coefficient comes into play when we are interested in the number of partitions with at most $k$ parts each at most $l$. See example I.16 in Analytic Combinatorics.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding an inverse matrix of an infinite matrix I tried to find the inverse matrix of
$$
\begin{pmatrix}
2 & -1 & 0 & \cdots& 0 \\
-1 & 2 & -1 & \cdots& 0\\
0 & -1 & 2 & \ddots& \vdots\\
\vdots & \vdots & \ddots & \ddots &-1\\
0 & 0 & \cdots & -1& 2\\
\end{pmatrix}
$$
unsuccesfuly.
*
*For $n=2$ the inverse matrix is $$\frac 1 3
\begin{pmatrix}
2 & 1 \\
1 & 2 \\
\end{pmatrix}
$$
*For $n=3$ the inverse matrix is
$$
\frac 1 4\begin{pmatrix}
3 & 2 & 1 \\
2 & 4 & 2 \\
1 & 2 & 3 \\
\end{pmatrix}
$$
*
*In general I figured out:
$$
\frac 1 {n+1} \begin{pmatrix}
n & n-1 & n-2 & \cdots& 1 \\
n-1 & ? & ? & ?& 2\\
n-2 & ? & ? & ? & \vdots\\
\vdots &? & ? & ? &n-1\\
1 & 2 & \cdots & n-1& n\\
\end{pmatrix}
$$
I just don't know how to write it generally even though I see those numbers in the middle of my matrices for $n = 2$ and $n = 3$. I tried it even for the larger $n$.
| Your examples suggest that the $(i,j)$-th entry in the upper triangular part (i.e. when $i\le j$) is $\frac{1}{n+1}i(n+1-j)$. It should be straightforward to verify this formula.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Calculate integral of fractional part I have to calculate $$I=\int _0^1\left\{nx\right\}^2dx , \:\:\:\:\: n \in \mathbb N, n \ge 1$$
Where {a} is $frac(a)$.
I know that $\left\{nx\right\}^2 = (nx - [nx])^2$ so
$$I\:=\int _0^1\:\left(n^2x^2-2nx\left[nx\right]+\left[nx\right]^2\right)dx=\frac{n^2}{3}-2n\int _0^1\:x\left[nx\right]dx+\int _0^1\:\left[nx\right]^2dx$$ What can I do next ? In my mind, because $x \in [0, 1] \implies [nx] = 0$ but the final answer is $\frac{1}{3}$.
| \begin{align}
\lfloor nx \rfloor = m
&\iff m \le nx < m+1 \\
&\iff \dfrac mn \le x < \dfrac{m+1}{n} \\
\end{align}
Let
\begin{align}
I_m &= \int_{x=\frac mn}^{\frac{m+1}{n}}\{nx\}^2 dx \\
&= \int_{x=\frac mn}^{\frac{m+1}{n}}(nx - m)^2 dx
& \left(\text{Let $y = x - \dfrac mn$.}\right) \\
&= n^2\int_{y=0}^{\frac 1n}y^2 dy \\
&= \dfrac{1}{3n} \\
\end{align}
Then
\begin{align}
I &= \int _0^1\left\{nx\right\}^2dx \\
&=\sum_{m=0}^{n-1} I_m \\
&=\dfrac 1n \sum_{m=0}^{n-1} \dfrac 13 \\
&= \dfrac 13
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Two right triangles of equal perimeter share a side. Given an acute angle of one triangle, find the acute angles of the other.
Two right triangles shown below have equal perimeters. The hypotenuse of the orange triangle is one leg of the green triangle stacked on top of it. If the smallest angle of the orange triangle is $20^\circ $, what are the angles of the green triangle?
Please help me. I didn't get any idea.
| I'll work through a slightly-more-general problem, replacing $20^\circ$ with $\alpha$. Writing $d$ for the hypotenuse of the lower triangle, and $\beta$ for the angle shown in the upper triangle ...
... the equal perimeters condition tells us:
$$d + d \sin\alpha + d \cos\alpha = d + d\tan\beta + d\sec\beta \tag{1}$$
That is,
$$\sin\alpha + \cos\alpha - \tan\beta = \sec\beta \tag{2}$$
Squaring both sides of (2), and noting the $\sec^2\theta = \tan^2\theta + 1$, we have
$$( \sin\alpha + \cos\alpha )^2 - 2 \tan\beta ( \sin\alpha + \cos\alpha ) + \tan^2\beta = \tan^2\beta + 1 \tag{3}$$
so that
$$\begin{align}
2\tan\beta (\sin\alpha + \cos\alpha) &= ( \sin\alpha + \cos\alpha )^2 - 1 \tag{4}\\
&= \sin^2\alpha + \cos^2\alpha + 2 \sin\alpha \cos\alpha - 1 \tag{5}\\
&= 1 + 2\sin\alpha\cos\alpha - 1 \tag{6}\\
&= 2\sin\alpha\cos\alpha \tag{7}
\end{align}$$
Therefore,
$$\tan\beta = \frac{\sin\alpha\cos\alpha}{\sin\alpha + \cos\alpha} \tag{$\star$}$$
In the particular case that $\alpha = 20^\circ$, we have
$$\tan\beta = 0.250753\ldots \quad\to\quad \beta = \operatorname{atan} 0.25073\ldots = 14.0769\ldots^\circ$$
I don't believe there's a "nice" form for this angle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find series representation of a function $ y=\frac{1}{1+x+x^{2}+x^{3}} $ Let $y = \frac{1}{1+x+x^{2}+x^{3}}$.
And I want to find the series representation of this function.
I've noticed that I can rewrite this like $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{1+x}*\frac{1}{1+x^{2}}$ or I can rewite this like a sum of two fractions $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{2(1+x)}+\frac{1}{2(1+x^{2})}$.
I know the series representations of these functions $(1+x)^{n}$ and $\frac{1}{1-x}$, and I guess I should use one of them. But the result looks weird. Can you help me, please, may be I can't see some easy way to solve it.
| There are two possibilities here: either we use the binomial theorem on $(1+x(1+x+x^2))^{-1}$, which looks awful, or we use a trick: multiplying and dividing by $1-x$, we have
$$ \frac{1-x}{(1-x)(1+x+x^2+x^3)} = \frac{1-x}{1-x^4}, $$
because the middle terms all cancel out. Now, we can expand the denominator,
$$ \frac{1-x}{1-x^4} = (1-x)\sum_{k=0}^{\infty} x^{4k}, $$
and it's pretty easy to multiply through and get the final answer from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Simplifying a quartic equation I have the following function to simplify and solve but there definitely is something wrong with my method as the initial conditions do not work with my final result so if anyone could pinpoint what I'm doing wrong, I would really appreciate it.
Solving: $\frac{(N-0.5)^4}{N^2(-N+1)^2}=Ae^t$ where A is a constant.
Essentially to solve for N my tutor recommended using the substitution $k=N-\frac{1}{2}$.
$\Rightarrow \frac{k^4}{(k+\frac{1}{2})^2(-k+\frac{1}{2})^2}=Ae^t$
$\Rightarrow \frac{k^2}{(k+(\frac{1}{2})(-k+\frac{1}{2})}=\sqrt{Ae^t}$
$\Rightarrow k^2=(k+\frac{1}{2})(-k+\frac{1}{2})\sqrt{Ae^t}$
$\Rightarrow k^2=(\frac{1}{4}-k^2)\sqrt{Ae^t}$
$\Rightarrow (1-\sqrt{Ae^t})k^2-\frac{1}{4}\sqrt{Ae^t}=0$
Then solving this like a quadratic gave me:
$k= \pm\frac{(\sqrt{Ae^t})^\frac{1}{4}}{2\sqrt(Ae^t)^\frac{1}{4}+1}$
Subbing back $N$ we get the following formula:
$N= \frac{1}{2}\pm\frac{(\sqrt{Ae^t})^\frac{1}{4}}{2\sqrt(Ae^t)^\frac{1}{4}+1}$
However, I was given the initial condition $N(0)=2$ which does not hold for either of my equations, so I have gone wrong somewhere but I'm not sure where personally.
Any insight would be much appreciated.
| Beginning with
\begin{equation}
\frac{k^4}{\left(k^2-\frac{1}{4}\right)^2}=Ae^t
\end{equation}
we get
\begin{equation}
\frac{k^2}{\left(k^2-\frac{1}{4}\right)}=\pm\sqrt{Ae^t}
\end{equation}
Solving for $k^2$ gives
\begin{equation}
k^2=\frac{\pm\sqrt{Ae^t}}{4\left(1\pm\sqrt{Ae^t}\right)}
\end{equation}
So
\begin{equation}
k=\pm\sqrt{\frac{\pm\sqrt{Ae^t}}{4\left(1\pm\sqrt{Ae^t}\right)}}
\end{equation}
giving
\begin{equation}
N=\frac{1}{2}+\sqrt{\frac{\pm\sqrt{Ae^t}}{4\left(1\pm\sqrt{Ae^t}\right)}}
\end{equation}
with the negative option being ruled out by the requirement that $N(0)=2$.
So when $t=0$ it must be the case that
\begin{eqnarray}
\frac{1}{2}+\sqrt{\frac{\pm\sqrt{A}}{4\left(1\pm\sqrt{A}\right)}}&=&2\\
\frac{\pm\sqrt{A}}{1\pm\sqrt{A}}&=&9\\
\frac{-\sqrt{A}}{1-\sqrt{A}}&=&9
\end{eqnarray}
Choosing the positive option would require $\sqrt{A}$ to be negative.
Thus $\sqrt{A}=\frac{9}{8}$ and $A=\frac{81}{64}$ which is verified when substituted into the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Number of triangles with integer sides Total number of right angle triangles whose inradius is $2013$ and sides are integer
Attempt: assuming that $a,b,c>0$ are the sides of a triangle.
so form a right angle triangle with sides $a,b,c$ and right angle at $C$
$\displaystyle r=(s-c)\tan \frac{C}{2}=(s-c)=\frac{a+b-c}{2}$ and $c^2=a^2+b^2$
$\displaystyle 2r = a+b-\sqrt{a^2+b^2}$ so $(\sqrt{a^2+b^2})^2 = (a+b-2r)$
$\displaystyle a^2+b^2 = (a+b)^2+4r^2-4r(a+b) = a^2+b^2+2ab-4r(a+b)$
$\displaystyle ab-2r(a+b)=0$
Could someone help me how to solve it, thanks.
| Remember the Pythagorian triple $(a,b,c)$ where
$$c=m^2+n^2, b=m^2-n^2,a=2mn. $$ We know that such triple defines a right triangle. The additional condition $$ab=2r(a+b)-2r^2 $$
implies that $$2mn(m^2-n^2)=2r(m^2-n^2+2mn)-2r^2 .$$
Now finding $m $ and $n$ answers your question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
given $I_{n}=\int^1_{0}x^n\sqrt{1-x^2}dx$, then finding value of $ \frac{I_{n}}{I_{n-2}}$ given $\displaystyle I_{n}=\int^{1}_{0}x^n\sqrt{1-x^2}dx$, then finding value of $\displaystyle \frac{I_{n}}{I_{n-2}}$
Attempt: put $x=\sin \theta$ and $dx = \cos \theta$
$\displaystyle I_{n} = \int^{\frac{\pi}{2}}_{0}\sin^{n}\theta \cdot \cos^2 \theta d \theta = \int^{\frac{\pi}{2}}_{0}\sin^{n}\theta (1-\sin^2 \theta)d \theta $
could some help me how to solve it, thanks
| $I_n=\dfrac12\beta(\dfrac{n+1}{2},\dfrac32)$ so
$$I_n=\dfrac12\beta(\dfrac{n+1}{2},\dfrac32)=\dfrac12\beta(\dfrac{n-1}{2},\dfrac32)\dfrac{n-1}{n+2}=\dfrac12\beta(\dfrac{n-3}{2},\dfrac32)\dfrac{n-3}{n}\dfrac{n-1}{n+2}=I_{n-2}\dfrac{n-3}{n}\dfrac{n-1}{n+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding a basis for a set of matrices in a vector space. I am trying to find a basis for the following vector space:
V = {2x2 matrices A | $\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$A = A$\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$}
So far, I have augmented $\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$ with the identity matrix to find that A = $\bigl( \begin{smallmatrix} -3 & 2 \\ 2 & -1 \end{smallmatrix} \bigr)$. Is the identity matrix the only basis for this question?
| Start with the definitions
$$
\mathbf{A} =
\left[
\begin{array}{cc}
1 & 2 \\
2 & 3 \\
\end{array}
\right],
\qquad
\mathbf{B} =
\left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right].
$$
Following the logic of @DonAntonio, compute the commutator:
$$
\left[ \mathbf{A}, \mathbf{B} \right] =
\mathbf{A} \mathbf{B} -
\mathbf{B} \mathbf{A} =
2\left[ \begin{array}{cc}
-b + c & -a - b + d \\
a+c-d & b -c
\end{array} \right]
= \mathbf{0}.
$$
From the diagonal terms we see that $b = c$. We now have matrix.
$$
\mathbf{B} =
\left[
\begin{array}{cc}
a & b \\
b & c \\
\end{array}
\right], \qquad
\left[ \mathbf{A}, \mathbf{B} \right] =
2\left[ \begin{array}{cc}
0 & -a - b + d \\
a+b-d & 0
\end{array} \right]
= \mathbf{0}.
$$
From this, $d = a+b$.
The set of matrices which commute with $\mathbf{A}$ are
$$
\mathbf{B} =
\left[
\begin{array}{cc}
a & b \\
b & a + b \\
\end{array}
\right], \quad \alpha, \beta \in \mathbb{C}.
$$
Notice the original matrix $\mathbf{A}$ has this form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Checking if two matrices are similar I have two matrices
$$ \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} $$
and $$ \begin{pmatrix}
2 & 2 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3
\end{pmatrix} $$
They are not diagonalizable. Share the same characteristic polynomial, the same trace, same determinant, eigenvalues, rank. What could I use more to say if they are similar or not?
| Although using Jordan Canonical forms are probably the fastest answer, it is possible to solve this question using brute force. We want to find out if there is an invertible matrix $P$ such that
$$P \cdot
\begin{pmatrix}
2 & 1 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{pmatrix} =
\begin{pmatrix}
2 & 2 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{pmatrix} \cdot P.$$
Denoting
$$P = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$$
this is equivalent with stating that
$$\begin{pmatrix}
2a & a + 2b & 3c\\
2d & d + 2e & 3f\\
2g & g + 2h & 3i
\end{pmatrix} =
\begin{pmatrix}
2a + 2d & 2b + 2e & 2c + 2f\\
2d & 2e & 2f\\
3g & 3h & 3i
\end{pmatrix}.$$
Comparing corresponding entries, we find that:
$$\begin{cases}
2a = 2a + 2d\\
a + 2b = 2b = 2e\\
3c = 2c + 2f\\
2d = 2d \\
d + 2e = 2e\\
3f = 2f\\
2g = 3g\\
g + 2h = 3h\\
3i = 3i
\end{cases}.$$
This is equivalent with
$$\begin{cases}
d = 0\\
a = 2e\\
c = 2f\\
0 = 0\\
e = e\\
f = 0\\
g = 0\\
h = 0\\
3i = 3i
\end{cases}.$$
And from equation 3 and 6, we also have that $c = 0$. Therefore, we find that
$$a = 2e, b = b, c = 0, d = 0, e = e, f = 0, g = 0, h = 0, i = i,$$
so we can take $b = 0$ and we find that
$$P = \begin{pmatrix}
2 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}$$
since we had a free choice for $b, i$ but we want $P$ to be invertible, so we can't take $i = 0$ (but any other value will do) and we take $b = 0$ (so $P$ is clearly invertible).
So this is a way to find out if both matrices are similar or not (if you would have found a contradiction in the system, then they are not (or if the resulting matrix $P$ is not invertible they are also not similar). However, as I already mentioned: if you know about Jordan canonical form, always use this approach, since it is clearly much shorter!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the equation of a circle tangent to a circle and x-axis, with center on a certain line. Find the equation of the circle, which is tangent to the circle $x^2 + y^2=4$ and $x-axis$, with center on the line $x=4$.
So far I have:$$(x-h)^2 + (y-k)^2=r^2$$
Since the center is on the line $x=4$, then the center is $(4;k)$.
$\therefore ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(x-4)^2 + (y-k)^2=r^2$
And because the circle touches the $x-axis$, $r=k$.
$\therefore~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(x-4)^2 + (y-k)^2=k^2$
Now, I am struggling to use $x^2 + y^2=4$ to find $k$.
Please help.
|
$$x^2+y^2=0$$
this circle have radius $=2$center$=(0,0)$
So, in $\triangle OO'A$ $$(OO')^2=(O'A)^2+(OA)^2\tag{Pythagoras theorem}$$
$$(OB+BO')^2=(O'A)^2+(OA)^2$$ $\therefore$ $$(2+k)^2=k^2+4^2$$ $$4+k^2+4k=k^2+16$$ $\therefore$ $$k=3$$
Hope it helps!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$ I am studying the book Equations and Inequalities by Herman et al, and am stuck on the following exercise:
Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$
The hint says to consider $(n+2)(n+1)S$.
Doing so, I get:
$(n+2)(n+1)S = \frac{1}{2}{n+2 \choose 0} + \frac{1}{3}{n+2 \choose 1} + \cdots + \frac{1}{n+2}{n+2 \choose n}$,
but I don't see how this is any easier to solve.
I wonder if somebody might help by expanding on the given hint, and maybe offering their suggestions as to how to use it properly.
Thanks.
| $$\dfrac1{k+2}\binom nk=\dfrac{k+1}{(n+1)(n+2)}\cdot\dfrac{(n+2)!}{(k+2)!\{n+2-(k+2)\}!}=\dfrac{k+1}{(n+1)(n+2)}\cdot\binom{n+2}{k+2}$$
Now $\displaystyle(k+1)\cdot\binom{n+2}{k+2}$
$\displaystyle=(k+2-1)\cdot\binom{n+2}{k+2}$
$\displaystyle=(n+2)\cdot\binom{(n+1)!}{(k+1)!\{n+1-(k+1)\}!}-\binom{n+2}{k+2}$
$\displaystyle=(n+2)\cdot\binom{n+1}{k+1}-\binom{n+2}{k+2}$
Now use $\displaystyle(1+1)^m=\sum_{r=0}^m\binom mr$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determine if 3 circles intersect at a common point Given three circles centered at points $A$, $B$, and $C$ with non-zero radii of lengths $R_A$, $R_B$ and $R_C$. Where the centers of the circles form a valid triangle, and where the distance between any two centers is less than or equal to the sum of their corresponding radii.
Question: Is it possible to determine if all three circles intersect at a common point using a calculation simpler than first determining the pair of intersection points between two pairs of circles then determining if any of the intersection points are equal?
| If we coordinatize, say, with
$$A = (0,0) \qquad B = (c, 0) \qquad c = (b\cos A,b\sin A)$$
and take suppose circles $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ (of respective radii $r_A$, $r_B$, $r_C$) meet at a point $P = (x,y)$, then we have three equations in two unknowns $x$ and $y$:
$$\begin{align}
x^2 + y^2 &= r_A^2 \\
x^2 + y^2 &= r_B^2 + 2 c x - c^2 \\
x^2 + y^2 &= r_C^2 + 2 b x \cos A x + 2 b y \sin A - b^2
\end{align}$$
We can eliminate $x$ and $y$ from these equations, leaving this relation:
$$\begin{align} a^2 b^2 c^2 + a^2 r_A^4 + b^2 r_B^4 + c^2 r_C^4 &= \left( a^2 r_A^2 + r_B^2 r_C^2 \right) \left(-a^2 + b^2 + c^2 \right) \\
&+ \left( b^2 r_B^2 + r_C^2 r_A^2 \right) \left(\phantom{-}a^2 - b^2 + c^2 \right)\\
&+ \left( c^2 r_C^2 + r_A^2 r_B^2 \right) \left(\phantom{-}a^2 + b^2 - c^2 \right)
\end{align} \tag{1}$$
The right-hand side seems to want to be re-written with cosines ...
$$\begin{align} a^2 b^2 c^2 + a^2 r_A^4 + b^2 r_B^4 + c^2 r_C^4 &= 2 b c \cos A \left( a^2 r_A^2 + r_B^2 r_C^2 \right) \\
&+ 2 c a \cos B \left( b^2 r_B^2 + r_C^2 r_A^2 \right)\\
&+ 2 a b \cos C \left( c^2 r_C^2 + r_A^2 r_B^2 \right)
\end{align} \tag{2}$$
... but this doesn't seem a great deal better. Perhaps if we use the Law of Sines to write
$$a = 2 r \sin A \qquad b = 2 r \sin B \qquad c = 2 r \sin C$$
where $r$ is the circumradius of $\triangle ABC$. With a little effort, we find this form for the relation:
$$\begin{align}
\frac{1}{64} \left(\;\begin{array}{c} r_A^2 \sin 2A + r_B^2 \sin 2B + r_C^2 \sin 2C \\
- 8 r^2 \sin A \sin B \sin C\end{array} \;\right)^2
= \frac{1}{16}&(\phantom{-}r_A \sin A + r_B \sin B + r_C \sin C ) \\
\cdot &(-r_A \sin A + r_B \sin B + r_C \sin C )\\
\cdot &(\phantom{-}r_A \sin A - r_B \sin B + r_C \sin C )\\
\cdot &(\phantom{-}r_A \sin A + r_B \sin B - r_C \sin C )
\end{align} \tag{3}
$$
The curious fractional coefficients are there to help us recognize the right-hand side as Heron's Formula for the square of the area of a triangle with side-lengths $r_A \sin A$, $r_B \sin B$, $r_C \sin C$. More precisely, when (and only when) the right-hand side of $(3)$ is non-negative, it gives the square of the area of the triangle with those side-lengths; when (and only when) the right-hand side is negative, those side-lengths fail to form a valid triangle. (Note: I consider a degenerate triangle of area $0$ to be valid.)
Since the left-hand side of $(3)$ is necessarily non-negative, we deduce that
$\bigcirc A$, $\bigcirc B$, $\bigcirc C$, with radii $r_A$, $r_B$, $r_C$, concur at a point only if $r_A \sin A$, $r_B \sin B$, $r_C \sin C$ are the edges of a valid triangle (ie, they satisfy the Triangle Inequality).
That gives you a way to weed-out bad candidates. To know for sure that the three circles concur, you'd need to check the full equality of $(3)$. (Is that "simpler" than the strategy you mentioned? I'm not sure.)
Note that $8 r^2 \sin A\sin B\sin C = 2 a b \sin C = 4 |\triangle ABC|$. If we call the "valid triangle" referenced above, say, $\triangle T$, then we can write $(3)$ as
$$r_A^2 \sin 2A + r_B^2 \sin 2B + r_C^2 \sin 2C \pm 8 |\triangle T|\;=\; 4 |\triangle ABC|
\tag{4}$$
To get at the behavior of the "$\pm$", consider $P$ at distance $p$ from the circumcenter of $\triangle ABC$ (and at distances $r_A$, $r_B$, $r_C$ from $A$, $B$, $C$, respectively). With the help of Mathematica, we get
$$\begin{align}
r_A^2 \sin 2A + r_B^2 \sin 2B + r_C^2 \sin 2C &= 4 (r^2 + p^2) \sin A \sin B \sin C \\
8|\triangle T| &= 4 |r^2-p^2|\sin A \sin B \sin C
\end{align}$$
We see, then, that "$\pm$" must be "$+$" when $r > p$ (that is, when $P$ is inside the circumcircle) and "$-$" when $r < p$ (when $P$ is outside the circumcircle); for $P$ on the circumcircle, $|\triangle T| = 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9$ Let $$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9,$$
$$\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32,$$
$$\frac{a^3}{b+c} + \frac{b^3}{c+a} + \frac{c^3}{a+b} = 122.$$
Find the value of $abc$.
Please check if my answer is correct or not.
$a+b+c + \frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b} = 32 +a+b+c$
$(a+b+c)\left(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\right) = 32 +a+b+c$
$a+b+c = 4$
$a^2+b^2+c^2 = 6$
$ab+bc+ca = 5$
$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} + 3= 12$
$(a+b+c)\left(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}\right) = 12$
$\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} = 3$
$\frac{a^2+b^2+c^2+3(ab+bc+ca)}{2abc+\sum{a^2b} + \sum{ab^2}} = 3$
$7 = (ab+bc+ca)(a+b+c)-abc$
$abc = 13$
| Your solution is correct.
You find $a^2+b^2+c^2=6$ using the same trick:
$$\frac{a^3}{b+c} + \frac{b^3}{c+a} + \frac{c^3}{a+b} = 122 \iff \\
(a+b+c)\left(\frac{a^2}{b+c} + \frac{b^2}{c+a} + \frac{c^2}{a+b}\right)-(a^2+b^2+c^2) = 122\Rightarrow \\
a^2+b^2+c^2=4\cdot 32-122=6.$$
Then:
$$ab+bc+ca=\frac12[(a+b+c)^2-(a^2+b^2+c^2)]=\frac12[4^2-6]=5.$$
For the following steps useful formulas:
$$\begin{align}(a+b+c)^3&=a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc \ \ (1)\\
(a+b+c)^3&=a^3+b^3+c^3+3(a+b)(b+c)(c+a) \ \ (2)\end{align}$$
From the first equation:
$$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 9\Rightarrow \\
9(a+b)(b+c)(c+a)=a^3+b^3+c^3+(a+b+c)(ab+bc+ca)$$
Considering this with $(2)$:
$$3(a+b+c)^3-3(a^3+b^3+c^3)=a^3+b^3+c^3+(a+b+c)(ab+bc+ca) \Rightarrow \\
a^3+b^3+c^3=3\cdot 4^3-4\cdot 5=43.$$
From $(1)$:
$$abc=\frac{43+3\cdot 4\cdot 5-4^3}{3}=13.$$
| {
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"url": "https://math.stackexchange.com/questions/2174348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find all primes $p$ for which the equation $x^6 \equiv -1$ mod $p$ has a solution. I'm not sure if i could apply Wilson's theorem here.
| Since $(x^3)^2 + 1^2 = 0$ mod $p$, $p$ is $1$ mod $4$ (Fermat's two squares)
$(x^2)^3 + 1^3 = (x^2 + 1)(x^4 + x^2 + 1) = 0 $ mod $p$
$\implies p | x^2 + 1$ as $p$ has to be of the form $a^2 + b^2$ (Fermat and skipping steps)
So if we take any $p$ such $p$ is a factor of $x^2 + 1$ then the equation $x^6 + 1 = 0$ mod $p$ holds.
| {
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"timestamp": "2023-03-29T00:00:00",
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then minimum number of number of roots of $p(x) = 0$ is let $p(x)=x^6+ax^5+bx^4+x^3+bx^2+ax+1.$ given that $x=1$ is a one rot of $p(x)=0$
and $-1$ is not a root. then minimum number of number of roots of $p(x) = 0$ is
Attempt: $x=0$ in not a root of $p(x)=0.$
So $\displaystyle \left(x^3+\frac{1}{x^3}\right)+a\left(x^2+\frac{1}{x^2}\right)+b\left(x+\frac{1}{x}\right)+1=0$
So $\displaystyle \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)+a\left(x+\frac{1}{x}\right)^2-2+b\left(x+\frac{1}{x}\right)+1=0$
So $\displaystyle t^3+at^2+(b-3)t+1=0,$ where $\displaystyle \left(x+\frac{1}{x}\right) = t$ and $|t|\geq 2$
Could some help me to solve it, thanks
| This answer assumes that you want to find the minimum number of the real roots of $p(x)$.
Since $x=1$ is a root of $p(x)$, we have
$$p(1)=0\iff 2a+2b+3=0\iff b=\frac{-2a-3}{2}$$
from which we can write
$$p(x)=(x-1)^2\left(x^4+(a+2)x^3+\left(a+\frac 32\right)x^2+(a+2)x+1\right)$$
So, we see that the number of the real roots of $p(x)$ is equal to or more than $2$.
By the way, for $a=-2$, we have
$$p(x)=(x-1)^2\left(x^4-\frac 12x^2+1\right)=(x-1)^2\left(\left(x^2-\frac 14\right)^2+\frac{15}{16}\right)$$
of which $x=-1$ is not a root.
The number of the real roots of $p(x)$ for $a=-2$ is $2$.
Therefore, the minimum number of the real roots of $p(x)$ is $\color{red}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $abc=1$ so $\sum\limits_{cyc}\frac{7-6a}{2+a^2}\geq1$
Let $a$, $b$ and $c$ be real numbers such that $abc=1$. Prove that:
$$\frac{7-6a}{2+a^2}+\frac{7-6b}{2+b^2}+\frac{7-6c}{2+c^2}\geq1$$
The equality occurs also for $a=b=2$ and $c=\frac{1}{4}$.
This inequality is a similar to the very many contest's inequalities, but nothing helps.
At least, I don't see how we can prove it.
An example of my trying.
We need to prove that
$$\sum_{cyc}\frac{7-6a}{2+a^2}\geq1$$ or
$$\sum_{cyc}\left(\frac{7-6a}{2+a^2}+1\right)\geq4$$ or
$$\sum_{cyc}\frac{(a-3)^2}{2+a^2}\geq4.$$
By C-S $$\sum_{cyc}\frac{(a-3)^2}{2+a^2}=\sum_{cyc}\frac{(a-3)^2(a+k)^2}{(2+a^2)(a+k)}\geq\frac{\left(\sum\limits_{cyc}(a-3)(a+k)\right)^2}{\sum\limits_{cyc}(2+a^2)(a+k)^2}$$
Now we'll find a value of $k$, for which the equality in the last inequality occurs for $a=b=2$ and $c=\frac{1}{2}$.
Since in all equality case we have
$$\frac{a-3}{(2+a^2)(a+k)}=\frac{b-3}{(2+b^2)(b+k)}=\frac{c-3}{(2+c^2)(a+k)},$$
we obtain:
$$\frac{2-3}{(2+2^2)(2+k)}=\frac{\frac{1}{4}-3}{(2+\left(\frac{1}{4}\right)^2)(\frac{1}{4}+k)},$$
which gives $k=-\frac{9}{4}$.
Thus, it remains to prove that
$$\left(\sum\limits_{cyc}(a-3)(4a-9)\right)^2\geq4\sum_{cyc}(2+a^2)(4a-9)^2,$$
which is wrong for $a=4$ and $b=c=\frac{1}{2}$.
Any hint?
Thank you!
| Clearly, that we only need to prove this inequality for case:$\ a , b, c >0$
Let $a=e^{t_1}, b= e^{t_2}, c= e^{t_3} $
$$f(t_1)+f(t_2)+f(t_3) \ge1 \ , \ t_1+t_2+t_3=0$$
$$f(t)=\dfrac{7-6e^t}{2+e^{2t}}$$
$f'(t)=\dfrac{2e^t(e^{t}-3)(3e^t+2)}{(e^{2t}+2)^2}$
Minimum of $f(t)$ is attained at $t=\ln(3)>0 \Rightarrow \min\left( f(t_1)+f(t_2)+f(t_3)\right)$, for $t_1+t_2+t_3=0$ is attained only for case : $t_1,t_2,t_3 \le t_*$
Since
$f''(t)=-\dfrac{2e^t(e^t-3+\sqrt{11})(e^t-3-\sqrt{11})(3e^t+2-\sqrt{22})(3e^t+2+\sqrt{22})}{9(e^{2t}+2)^3}$
we only need to consider the inequality in case : $t_1\le t_2=t_3 \le t_*$
$a={q^2} \ , \ b=c=\dfrac{1}{q}$
$\Leftrightarrow \dfrac{(3q^2+6q+5)(2q-1)^2(q-1)^2}{(2q^2+1)(q^4+2)}\ge 0$
Equality holdes for: $(a=b=c=1)\ $ and $\ (a=\dfrac{1}{4}, b=c=2)$
| {
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"url": "https://math.stackexchange.com/questions/2183109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving for $p$ and $k$. For the question,
$$F(x)= x^{2} + 6x + 20 + k(x^{2} -3x -12)$$
I have to find the value of k and the value of $p$ given that the minimum value of $F(x)$ is $p$ and $F(-2) = p$.
What I did is: I expanded the expression of $F(x)$ and reached a term for $p= 12 - 2p$. I tried to complete the square of the two quadratics of $F(x)$ separately then added the two terms while also correcting for coefficients. I doubt this operation however went for it anyway due to any alternatives seem wrong.
I reached the term $-\frac {39}4k + 11$ then promptly equated it to $12 - 2k$.
I reached the value of $-\frac{4}{31}$ for $K$.
My textbook has answers for $k = \frac{2}{7}$ and $p=\frac{80}{7}$.
| A quadratic $ax^2+bx+c$ with a positive leading term $a\gt0$ attains its minimum at $-\cfrac{b}{2a}\,$.
Rewriting $F(x) = (1+k)x^2+(6-3k)x+20-12k\,$ it follows that: $$-\cfrac{6-3k}{2(1+k)}=-2 \quad \implies \quad k = \cfrac{2}{7}$$
(Since the leading term $1+k = 1 + \cfrac{2}{7}\gt 0$ this is a valid solution.)
Substituting $k$ back gives $F(x)=\cfrac{9}{7} x^2+\cfrac{36}{7} x + \cfrac{116}{7}\,$, then $p=F(-2)=\cfrac{80}{7}\,$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Formula for number of ways to make $n$ units of currency if the coins we have are of value 1, 3, and 9 Suppose we live in a world where there are three forms of currency: each worth 1, 3, and 9 units. Can we find a formula for the number of ways to make $n$ units of currency, if we have an effectively unlimited supply of each? I am looking for a formula with finitely many binomial coefficients, the number of which does not grow with $n$.
| Let $f_1(n)$ be the # of ways to make $n$ units using only $1$; $f_3(n)$ be the # of ways to make $n$ units using $1$ and $3$; and $f_9(n)$ be the # of ways using $1$, $3$ and $9$.
If only $1$ and $3$ are allowed, $3$ can be used $0 \leq k \leq \lfloor \frac{n}{3} \rfloor$ times. For the remaining $n - 3k$ units, we fill it using only $1$. Therefore,
$$
f_3(n) = \lfloor \frac{n}{3} \rfloor + 1
$$
Suppose totally $k$ $9$'s are used, there are totally $f_3(n - 9k)$ ways to fill the remaining units. Thus
$$
f_9(n) = \sum_{k=0}^{\lfloor \frac{n}{9} \rfloor} f_3(n - 9k) = \sum_{k=0}^{\lfloor \frac{n}{9} \rfloor} \lfloor \frac{n}{3} \rfloor - 3k + 1 = (\lfloor \frac{n}{9} \rfloor + 1)(\lfloor \frac{n}{3} \rfloor + 1) - \frac{3\lfloor \frac{n}{9} \rfloor(\lfloor \frac{n}{9} \rfloor + 1)}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating a finite sum using a double sum I want to evaluate the sum $S_n = \sum_{k=0}^n k 3^k $.
Try:
Since $ \sum_{j=0}^{k-1} 1 = k $, then
$$ S_n = \sum_{k=0}^n \sum_{j=0}^{k-1} 3^k $$
Since we are summing over $(0 \leq k \leq n )$ and $(0 \leq j \leq k -1 )$, then we are summing over $(0 \leq j < k \leq n )$. Thus, we can write
$$ S_n = \sum_{j=0}^n \sum_{k=j}^n 3^k = \sum_{j=1}^n \frac{ 3^{n+1} - 3^j }{2} = \frac{n 3^{n+1}}{2} - \frac{1}{2} \sum_{j=0}^n 3^j = \boxed {\frac{n 3^{n+1} }{2} - \frac{1}{2} \cdot \left( \frac{ 3^{n+1} - 1 }{2} \right) } $$
Is this a correct argument?
|
$$ S_n = \sum_{j=0}^n \sum_{k=j}^n 3^k = \sum_{j=1}^n \frac{ 3^{n+1} - 3^j }{2} = \frac{n 3^{n+1}}{2} - \frac{1}{2} \sum_{j=0}^n 3^j = \boxed {\frac{n 3^{n+1} }{2} - \frac{1}{2} \cdot \left( \frac{ 3^{n+1} - 1 }{2} \right) } $$
Is this a correct argument?
You have an error.
Since $j\le k-1\iff k\ge j+1$, we have
$$\small S_n = \sum_{j=0}^n \sum_{k=\color{red}{j+1}}^n 3^k=\sum_{j=0}^{n}\frac{3^{n+1}-3^{j+1}}{2}=\frac{(n+1)\cdot 3^{n+1}}{2}-\frac 12\sum_{j=0}^{n}3^{j+1}=\frac{(n+1)\cdot 3^{n+1}}{2}-\frac 12\left(\frac{3^{n+2}-3}{2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The equations $x^3+5x^2+px+q=0$and $x^3+7x^2+px+r=0$ have 2 common roots, then find the third root of both equations
The equations $x^3+5x^2+px+q=0$and $x^3+7x^2+px+r=0$ have 2 common roots, then find the third root of both equations
From the first equation we can say, $\alpha\beta+\beta\gamma+\gamma\alpha=p/1=p$. Similarly from the second equation we know, $\alpha\beta+\beta\delta+\delta\alpha=p/1=p$
Hence,
$\alpha\beta+\beta\delta+\delta\alpha=\alpha\beta+\beta\gamma+\gamma\alpha$
$\delta(\beta+\alpha)=\gamma(\beta+\alpha)$
$\delta=\gamma$
Hence the third root of both equations should be equal, but $\alpha+\beta+\gamma=-5$ and $\alpha+\beta+\delta=-7$. Now, where did I go wrong?
| If $x^3 + 5x^2 + px + q = 0$ and $x^3 + 7x^2 + px + r = 0$ have two common roots then there must exists a common monic binomial factor, $F(x)$.
So $F(x)$ must be a divisor of
$$(x^3 + 7x^2 + px + r) - (x^3 + 5x^2 + px + q) = 2x^2 - (q - r) $$
So $F(x) = x^2 - \dfrac 12(q - r)$.
So, for some $s$
\begin{align}
x^3 + 7x^2 + px + r
&= F(x)(x-s) \\
&= \left[x^2 - \dfrac 12(q - r)\right](x-s)\\
&= x^3 - sx^2 - \dfrac 12(q - r)x + \dfrac 12(q - r)s\\
\hline
s &= -7 \\
p &= -\dfrac 12(q - r) \\
r &= -\dfrac 72(q - r) \\
\hline
s &= -7 \\
p &= \dfrac 15 q \\
r &= \dfrac 75 q
\end{align}
And, for some $t$
\begin{align}
x^3 + 5x^2 + px + q
&= \left[x^2 - \dfrac 12(q - r)\right](x-t)\\
&= x^3 - tx^2 - \dfrac 12(q - r)x + \dfrac 12(q - r)t\\
\hline
t &= -5 \\
p &= -\dfrac 12(q - r) \\
q &= -\dfrac 52(q - r) \\
\hline
t &= -5 \\
p &= \dfrac 15 q \\
r &= \dfrac 75 q
\end{align}
So the third roots are $-7$ and $-5$.
Reality check:
So, suppose $p = -\dfrac 15 q, r = \dfrac 75 q$. To get rid of the fractions, lets let q = 10k.
Then
*
*$p = 2k$
*$r = 14k$
*$q = 10k$
*$s = -7$
*$t = -5$
*$F(x) = x^2 + 2k$
*$x^3 + 7x^2 + px + r = x^3 + 7x^2 + 2kx + 14k$
*$x^3 + 5x^2 + px + q = x^3 + 5x^2 + 2kx + 10k$
*$F(x)(x-s) = (x^2 + 2k)(x+7) = x^3 + 7x^2 + 2kx + 14k$
*$F(x)(x-t) = (x^2 + 2k)(x+5) = x^3 + 5x^2 + 2kx + 10k$
| {
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"timestamp": "2023-03-29T00:00:00",
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$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$ let $$f(x)=\frac{1}{1+2\cos x}$$
prove that :
$$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$$
My Try :
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{4\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{6\pi}{7})}$$
$$L=\frac{1}{1+2\cos (\frac{6\pi}{7})}+\frac{1}{1+2\cos (\frac{4\pi}{7})}+\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
what now ?
| Let $2\cos\frac{2\pi}{7}=a$, $2\cos\frac{4\pi}{7}=b$ and $2\cos\frac{6\pi}{7}=c$.
Hence,
$$a+b+c=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{\sin\frac{\pi}{7}}=-1;$$
$$ab+ac+bc=4\left(\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}\right)=$$
$$=2\left(\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\right)=-2$$ and
$$abc=8\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=\frac{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{\sin\frac{2\pi}{7}}=\frac{\sin\frac{16\pi}{7}}{\sin\frac{2\pi}{7}}=1.$$
Id est, $$f\left(\frac{2\pi}{7}\right)+f\left(\frac{4\pi}{7}\right)+f\left(\frac{6\pi}{7}\right)=\sum_{cyc}\frac{1}{1+a}=\frac{\sum\limits_{cyc}(a+1)(b+1)}{\prod\limits_{cyc}(1+a)}=$$
$$=\frac{ab+ac+bc+2(a+b+c)+3}{1+a+b+c+ab+ac+bc+abc}=\frac{-2-2+3}{1-1-2+1}=1.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Regarding solutions of ordinary differential equations Got into a problem with the solution of this one:
$$
x\cos\left(\frac{y}{x}\right)(x\text{d}x+y\text{d}y)=y\sin\left(\frac{y}{x}\right)(x\text{d}y-y\text{d}x)
$$
I can identify differentials of $xy$ and $\frac{x}{y}$ and also of $\cos$ and $\sin$ functions but how should I manipulate the equation into a solvable form?
| We have:
$x \cos (\frac{y}{x})(x dx+ydy) = y \sin(\frac{y}{x})(xdy-ydx)$
$\Rightarrow x \cos (\frac{y}{x})(x + y\frac{dy}{dx}) = y \sin(\frac{y}{x})(x\frac{dy}{dx} - y)$
Let $u = \frac{y}{x}$
Then $\frac{du}{dx} = \frac{x\frac{dy}{dx} - y}{x^2} = \frac{\frac{dy}{dx} - u}{x}$
$\Rightarrow \frac{dy}{dx} = x\frac{du}{dx} + u$
Then we substitute this in:
$x\cos u [x+ux(x\frac{du}{dx} +u)] = ux\sin u(x^{2}\frac{du}{dx}+ux-ux)$
$\Rightarrow x^2\cos u [1+u(x\frac{du}{dx} +u)] = ux^3\sin u \frac{du}{dx}$
$\Rightarrow ux\sin u \frac{du}{dx} -\cos u - ux\cos u \frac{du}{dx}-u^2\cos u = 0$
$\Rightarrow ux(\sin u - \cos u)\frac{du}{dx}-(u^2+1)\cos u = 0$
$\Rightarrow ux(\sin u - \cos u)\frac{du}{dx}=(u^2+1)\cos u $
$\Rightarrow \int \frac{u(\sin u - \cos u)}{(u^2+1)\cos u}du = \int\frac{1}{x} dx$
Can you continue from there?
Edit: lol it appears I can't solve this integral after trying, perhaps someone could comment where I've gone wrong?
| {
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How to factor $r^6 -3r^4 +3r^2 - 1 = 0$ I tried two ways to factor $r^6 -3r^4 +3r^2 - 1 = 0$
When I factor $r^4$ out of $r^6 -3r^4$:
$r^4(r^2-3)+(3r^2-1) = 0$
When I factor $r^2$ out of $r^6 + 3r^2$:
$r^2(r^4+3)-(3r^4 + 1) = 0$
For both methods, I'm stuck at an equation that isn't factorable. But Wolfram Alpha says $r^6 -3r^4 +3r^2 - 1 = (r-1)^3(r+1)^3$. How do you get this?
| Let $x = r^{2}$
Then $x^{3} -3x^2+3x-1 = 0$
Notice that $x=1$ is a solution
Therefore we have $(x-1)(x^{2} +ax+1) = 0$ where $a$ is a constant we need to find
By comparing coefficients of $x$, we can see that $-a+1 = 3 \Rightarrow a = -2$
Then we have $(x-1)(x^2-2x+1)=0 \Rightarrow (x-1)(x-1)^{2}=0 \Rightarrow (x-1)^{3} = 0$
Then $(r^2-1)^3=0$
Alternatively, you could notice this is the binomial expansion (comes with experience) and thus factorise by inspection.
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"language": "en",
"url": "https://math.stackexchange.com/questions/2194225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Finding $\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$ Finding $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$
Attempt: Assume $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx =\int \frac{2x-x^{-2}}{\bigg[x^4-x^2+2x-\frac{2}{x}-\frac{1}{x}+\frac{1}{x^2}\bigg]}dx$
could some help me how to solve,thanks
| $$\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx=\frac{1}{3}\int\left(\frac{1}{x-1}+\frac{2x+1}{x^2+x-1}-\frac{3x^2+1}{x^3+x+1}\right)dx=$$
$$=\frac{1}{3}\left(\ln|x-1|+\ln|x^2+x-1|-\ln|x^3+x+1|\right)+C$$
| {
"language": "en",
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$3\mid a^3+b^3+c^3$ if only if $3\mid a+b+c $
Prove the following equivalence: $3\mid (a^3 + b^3 + c^3) $ if and only if $3\mid (a + b + c) $.
My try:
I know $a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc$, but I can't seem to proceed from here.
Thanks all!
| By Fermat's Little Theorem we know that $3$ divides $x^3-x$ for any $x \in \mathbb{Z}$
Which means $(a^3-a)+(b^3-b)+(c^3-c) = (a^3+b^3+c^3)-(a+b+c)$ is divisible by $3$.
The conclusion is easy to derive now.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $7p + 3^p -4$ is not a perfect square? How to prove that $7p + 3^p -4$ is not a perfect square?
I calculated: $\left(\frac{7p+3^p-4}{p}\right) = \left(\frac{-1}{p}\right)$. So if $p \equiv 3 \mod 4$, the result is $-1$. So in that case, $7p+3^p -4$ can't be a square. But what about the case $p \equiv 1 \mod 4$? Any hints? Thanks in advance.
| The squares modulo $4$ are
$$(4k)^2 = 4(4k^2) \equiv 0 \pmod{4},$$
$$(4k+1)^2 = 4(4k^2 + 2k) + 1 \equiv 1 \pmod{4},$$
$$(4k + 2)^2 = 4(4k^2 + 4k + 1) \equiv 0 \pmod{4},$$
and
$$(4k + 3)^2 = 4(4k^2 + 6k + 2) + 1 \equiv 1 \pmod{4},$$
but if $p = 4 \ell + 1$, then
$$7p + 3^p - 4\equiv (-1)(1) + (-1)^{4 \ell}(-1)^1 \pmod{4} \equiv -2 \pmod{4},$$ which is equivalent to neither $0$ nor $1$ modulo $4$, so $7p + 3^p - 4$ cannot be a square if $p = 4 \ell + 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can i find the sum? How can i calculate this serias?
$$\sum_{i=1}^n \frac{2^m}{2^i}\cdot i$$
I tried to do:
$$\ 2^m\cdot\left[\frac{1}{2}+\frac{2}{2^2}+...+\frac{n}{2^n}\right]$$
And i don't know how to continue..
Thank you.
| $$s(n)=\frac{1}{2}+\frac{2}{2^2}+...+\frac{n}{2^n}\quad (1)$$
multiply by $1/2$ and get
$$\frac{s(n)}{2}=\frac{1}{2^2}+\frac{2}{2^3}+...+\frac{n-1}{2^{n}}+\frac{n}{2^{n+1}}\quad (2)$$
Now make $(1)-(2)$ and get
$$\frac{s(n)}{2}=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n}}-\frac{n}{2^{n+1}}\\
s(n)=\left[\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-1}}\right]-\frac{n}{2^{n}}$$
Can you finish?
| {
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prove an inequality with conditions Let $a,b, c \in \mathbb{R}$ such that $0\le a\le1,0\le b\le1 , 0\le c\le1.$
If $$a+b\leq c+1,
\\ a+c \leq b+1,
\\b+c\leq a +1$$
can we prove that
$a^2+b^2+c^2\le 1+2abc$ ?
| Let
$$ a+b=(c+1) k_1,a+c=(b+1) k_2,b+c=(a+1) k_3 $$
and then
$$ a=-\frac{-k_2+k_3-k_1 \left(k_2 \left(k_3+2\right)+1\right)}{k_2+k_3+k_1 \left(1-k_2
k_3\right)+2},b=-\frac{k_2-k_3-k_1 \left(\left(k_2+2\right) k_3+1\right)}{k_2+k_3+k_1
\left(1-k_2 k_3\right)+2},c= -\frac{-k_3-k_2 \left(2 k_3+1\right)+k_1 \left(1-k_2 k_3\right)}{k_2+k_3+k_1 \left(1-k_2
k_3\right)+2}. $$
Here $0\le k_1,k_2,k_3\le 1$. An easy calculation shows
\begin{eqnarray}
&&a^2+b^2+c^2-1-2abc\\
&=&-\frac{4 (k_1+1)^2(k_2+1)^2(k_3+1)(2-k_1-k_2-k_3+k_1k_2 k_3)}{(k_2+k_3+k_1(1-k_2 k_3)+2)^3}.
\end{eqnarray}
Noting
\begin{eqnarray}
&&2-k_1-k_2-k_3+k_1k_2 k_3\\
&=&(1-k_1)+1-k_2-k_3+(k_1-1)k_2k_3+k_2k_3\\
&=&(1-k_1)(1-k_2k_3)+(1-k_2)(1-k_3)\ge0
\end{eqnarray}
one has
$$ a^2+b^2+c^2-1-2abc\le 0$$
or
$$ a^2+b^2+c^2\le1+2abc. $$
The job is done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Verify my proof of the irrationality of $\sqrt{6}$. Here's my proof for the irrationality of $\sqrt{6}$:
Proof:
Let $\sqrt{6}$ be a rational number. Then
$\frac{a}{b}$ = $\sqrt{6}$ , where $a$ and $b$ are coprime.
Therefore,
$a^2 = 6 b^2$.
Here $a^2$ is divisible by both $2$ and $3$. Which means, $a$ too is divisible by both $2$ and $3$ (fundamental theorem of arithmetic).
Thus, $6b^2=(3\cdot2\cdot k)^2$ ; where $k$ is an integer.
$6b^2=36k^2$ or
$b^2=6k^2$.
Here $b^2$ is divisible by both $2$ and $3$. Which means, $b$ too is divisible by both $2$ and $3$. Therefore, $a$ and $b$ have common factors, which is a contradiction. Therefore $\sqrt{6}$ is not rational.
Is the above proof correct? Is there an easier way to prove this?
| That works fine, but you don't need to do both $2$ and $3$. If:
$$a^2=6b^2$$
Then $a$ must be even, $a=2a_0$, and then: $4a_0^2=6b^2$ or $$2a_0^2=3b^2.$$
So $3b^2$ must be even, so by the fundamental theorem of arithmetic, $b$ must be even, and you are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the coefficient of $x^2y^3z^6$ in the expression $(x^2+y+z)^{10}$? How can find the coefficient of $x^2y^3z^6$ in the expression $(x^2+y+z)^{10}$?
I know this involves the binomial theorem, but I can't figure out how to solve it.
| We can apply the binomial theorem twice in order to find the coeffcient. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of an expression.
We obtain
\begin{align*}
[x^2y^3z^6](x^2+y+z)^{10}&=[x^2y^3z^6]\sum_{k=0}^{10}\binom{10}{k}x^{2k}(y+z)^{10-k}\tag{1}\\
&=[y^3z^6]\binom{10}{1}(y+z)^9\tag{2}\\
&=10[y^3z^6]\sum_{k=0}^9\binom{9}{k}y^kz^{9-k}\tag{3}\\
&=10\binom{9}{3}\tag{4}\\
&=840
\end{align*}
Comment:
*
*In (1) we apply the binomial theorem once to $(x^2+(y+z))^{10}$.
*In (2) we select the coefficient of $x^2$ which is given when the index $k=1$.
*In (3) apply the binomial theorem again.
*In (4) we select the coefficient of $y^3z^6$ corresponding to the index $k=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$x\sqrt{1-\frac{y(x+1)}{(x+y+1)^2}} = y\sqrt{1-\frac{z(y+1)}{(y+z+1)^2}} = z\sqrt{1-\frac{x(z+1)}{(z+x+1)^2}}$
If $x,y,z > 0$, and $$x\sqrt{1-\frac{y(x+1)}{(x+y+1)^2}} = y\sqrt{1-\frac{z(y+1)}{(y+z+1)^2}} = z\sqrt{1-\frac{x(z+1)}{(z+x+1)^2}}$$
Prove $x=y=z$.
I try it by Mathematica but can not solve, and Maple give a solution, but so complicated. I want to use inequality, but never find the solution.
| Let $$f(x,y)=x^2\Bigg(1-\frac{(x+1)y}{(x+y+1)^2}\Bigg)$$
Then the hypothesis says that $f(x,y)=f(y,z)=f(z,x)$. Note that for $x\neq y$ and $x,y$ positive,
$$
\frac{f(y,x)-f(x,y)}{y-x}=\frac{x^3+y^3+2(x^2+y^2)+xy(2(x+y)+3)+x+y}{(x+y+1)^2} > 0 \tag{1}
$$
Similarly, if we expand $\frac{f(x_2,y)-f(x_1,y)}{x_2-x_1}$, we find that it is equal to $\frac{P(x_1,x_2,y)}{(x_1+y+1)^2(x_2+y+1)^2}$ where $P$ is a polynomial in $x_1,x_2,y$ with positive coefficients. So for $x_1\neq x_2$ and $x_1,x_2,y$ positive, we have
$$
\frac{f(x_2,y)-f(x_1,y)}{x_2-x_1}> 0 \tag{2}
$$
We can now proceed with the proof. Suppose that $x<y$. Then $f(y,x)>f(x,y)$ by (1),but $f(x,y)=f(z,x)$ by hypothesis, so $f(y,x)>f(z,x)$. By (2) we deduce $y>z$. Then $f(y,z)>f(z,y)$ by (1),but $f(y,z)=f(x,y)$ by hypothesis, so $f(x,y)>f(z,y)$. By (2) we deduce $x>z$.
At this point we have $y>x>z$. Then $f(x,z)>f(z,x)$ by (1),but $f(z,x)=f(y,z)$ by hypothesis, so $f(x,z)>f(y,z)$ which contradicts (1).
This shows $x\geq y$. Similar arguments show that $y\geq z$ and $z\geq x$, and finally $x=y=z$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Conditional Poisson processes with Multiple conditions Let $\lbrace N(t)\rbrace_{t\geq 0}$ be a Poisson process with intensity $\lambda = 3$.
Compute $$P\left[N(6) = 2 \,|\, N(8) = 4, N(3) = 1\right].$$
I understand when there is only one condition i.e. $P\left[N(6) = 2 | N(8) = 4\right]$. Since that is $$P\left[N(6) = 2 , N(8) = 4\right]/ P\left[N(8) = 4\right] = P\left[N(6) = 2 , N(8)-N(6) = 2\right]/ P\left[N(8) = 4\right].$$ But I am unsure on how to go about doing this one
Any tips on where to begin? Thanks
| You can handle this using the definition of conditional probability, stationary increments and independent increments.
$$
\begin{align*}
P\left[N(6) = 2 \,|\, N(8) = 4, N(3) = 1\right] &= \frac{P[N(6) = 2,N(8) = 4, N(3) = 1]}{P[N(8) = 4, N(3) = 1]}\\
&=\frac{P[N(3) = 1, N(6) - N(3) = 1, N(8) - N(6) = 2]}{P[N(8) = 4, N(3) = 1]}\\
&=\frac{P[N(3) = 1]P[N(6) - N(3) = 1]P[N(8) - N(6) = 2]}{P[N(8) = 4, N(3) = 1]}\\
&=\frac{P[N(3) = 1]P[N(3)= 1]P[N(2)= 2]}{P[N(8) = 4, N(3) = 1]}\\
\end{align*}
$$
I'll let you finish it up from there.
| {
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How many positive integer solutions does the equation $a+b+c=100$ have if we require $aHow many positive integer solutions does the equation $a+b+c=100$ have if we require $a<b<c$?
I know how to solve the problem if it was just $a+b+c=100$ but the fact it has the restriction $a<b<c$ is throwing me off.
How would I solve this?
| First we count the number of triples of positive integers $(a, b, c)$ with $a + b + c = 100$. There are $\binom{99}{2}$ of them, which results from an elementary application of stars and bars. Just line up a hundred dots, place a bar in between the $a$th and $(a +1)$th dot, and then place another bar $b$ dots to the right of the first one. Then $c$ is the number of dots to the right of the second bar. There are 99 spaces in between a hundred dots and you're choosing two of them to place bars in.
This is an overcount since it doesn't account for the condition $a < b < c$. First we should subtract the number of triples in which $a, b, c$ are not all distinct. There are 49 triples such that $a = b$: to count them, choose $x$ with $ 0 < x<50$, set $a = b = x$ and $c = 100-2x$. This accounts for all possible cases since if $a = b \ge 50$ then $c$ is not positive. There are also 49 triples with $a = c$, or with $b = c$, respectively. There are no triples with $a = b = c$, so the total number of triples with $a , b, c$ not all distinct is $3*49 = 147$.
The number of triples of positive integers $(a, b, c)$ with $a, b, c$ distinct and $a + b + c = 100$ is therefore $\binom{99}{2} -147$. Given a triple $(a, b, c)$ with $a< b < c$ and $a + b + c = 100$, there are six distinct ways to permute $a, b, c$, all of which our current estimate accounts for. Therefore we should divide by 6 to throw out all the triples that are in the wrong order. The final answer is $\frac{1}{6}(\binom{99}{2} -147) = 784$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Area inside outer curve of limacon $r=1+2\cos\theta$ I need to find the area inside the outer curve of limacon $r=1+2\cos\theta$
Here is what I tried:
$$
0=1+2\cos\theta
$$
$$
\theta=\cos^{-1}(-\frac{1}{2})
$$
$$
\theta=\frac{2}{3}\pi , \theta=\frac{8}{3}\pi
$$
$$
A = 2\int^{\frac{8}{2}\pi}_{\frac{2}{3}\pi}r^2d\theta=\frac{r^3}{3}|^{\frac{8}{3}\pi}_{\frac{2}{3}\pi}=2\frac{(1+2\cos\theta)^3}{3}|^{\frac{8}{3}\pi}_{\frac{2}{3}\pi}=0
$$
so, 0 area is clearly wrong. Where was my mistake?
| You pretended that the $d\theta$ was a $dr$. Unfortunately you cannot do that.
$$2\int^{\frac{8}{2}\pi}_{\frac{2}{3}\pi}r^2d\theta\color{red}{\neq}\frac{r^3}{3}|^{\frac{8}{3}\pi}_{\frac{2}{3}\pi}$$
Now I'll let you try it. If you still need help, just ask!
| {
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curve on sphere from $(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ to $(-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$ Consider the set:
$$
A \equiv \left\{ (x,y,z) \in \mathbb{R}^{3} | x + y + z = 0\ \mathrm{and}\ x^{2} + y^{2} + z^{2} = 1 \right\}
$$
This is the intersection of the unit sphere with the above plane.
Is there a way to parametrize this set with a curve $\gamma : [0,1] \to A$ going from the point $(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ to $(-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$?
I simply can't figure it out since we have the extra constraint of being in the plane $x+y+z=0$
| When a plane intersects a sphere the curve of intersection will be a circle.
the origin is at the center of this cricle (because the plane goes through the origin, and the sphere is centered at the origin). Now you just need to parametarize the circle.
We have two points on this circle, but what we really need are orthogonal vectors. We can use one of the points, and find a vector in the plane that is orthogonal.
$(-\frac {\sqrt 2}{2},\frac {\sqrt 2}{2},0)$ is a vector orthogonal to $(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})$ that lies in the plane.
$(x,y,z) = (-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})\cos t + (-\frac {\sqrt 2}{2},\frac {\sqrt 2}{2},0)\sin t$
Will trace out the circle.
And for what bounds of $t$ will will that equation trace to the endpoints we seek?
$(-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}})\cdot(-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}) = 0.5$
$t = \frac{\pi}{3}$ or $\frac{-\pi}{3}$
$t\in[0,\frac {\pi}{3}]$
| {
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Finding value of $\lfloor b_{100} \rfloor $ in recursive relation
If $\displaystyle b_{n+1} = b_{n}+\frac{1}{b_{n}}$ and $b_{1} = 1,$ then find the value of $\lfloor b_{100}\rfloor$.
My attempt: $\displaystyle b_{n+1}b_{n} = b^2_{n}+1\Rightarrow b_{n}b_{n+1}-b^2_{n} = 1$
$$\frac{1}{b_{n}b_{n+1}-b^2_{n}} = 1\Rightarrow \frac{1}{b_{n}(b_{n+1}-b_{n})} = 1.$$
I am not able to go further, could some help me, thanks.
| Note that $b_n$ is strictly increasing and for $n\geq 2$, $b_n^2\ge 2n$, because $b_2=2$ and
$$b_{n+1}^2=\left(b_n+\frac{1}{b_n}\right)^2=b_n^2+2+\frac{1}{b_n^2}\geq 2n +2=2(n+1).$$
Moreover, for $n\geq 2$,
$$\begin{align}b_n^2&=b_2^2+\sum_{k=2}^{n-1}(b_{k+1}^2-b_k^2)=4+\sum_{k=2}^{n-1}\left(2+\frac{1}{b_k^2}\right)\le 2n+\frac{1}{2}\sum_{k=2}^{n-1}\frac1{k}\\&\leq 2n+\frac{1}{2}\int_1^{n-1}\frac{dx}{x}< 2n+\frac{\ln(n-1)}{2}.\end{align}$$
Hence
$$14^2=196<200\leq b_{100}^2< 200+\frac{\ln(99)}{2}< 203<225=15^2$$ which implies that $\lfloor b_{100}\rfloor={14}$.
| {
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Show that $\int_{0}^{\infty}{1\over x^{2}-x+1}{\mathrm dx\over \sqrt{x}}=\int_{0}^{\infty}{1\over (x^{2}-x+1)^2}{\mathrm dx\over \sqrt{x}}={\pi}$ Consider these integrals
$$\int_{0}^{\infty}{1\over x^{2}-x+1}\cdot{\mathrm dx\over \sqrt{x}}={\pi}\tag1$$
$$\int_{0}^{\infty}{1\over (x^{2}-x+1)^2}\cdot{\mathrm dx\over \sqrt{x}}={\pi}\tag2$$
An attempt:
Rewrite $(2)$ as
$$\int_{0}^{\infty}{16\over [(2x-1)^2+3]^2}\cdot{\mathrm dx\over \sqrt{x}}\tag3$$
$u=(2x-1)^{1/2}$, then $(3)$ becomes
$$\int_{i}^{\infty}{16u\over (u+3)^2}\mathrm du\tag4$$
$$\int_{i}^{\infty}\left({19\over 6(u+3)}-{3\over (u+3)^2}\right)\mathrm du\tag5$$
I don't think it makes sense to have the limit as an imaginary number.
Another sub: $u=2x-1$ then $(2)$ becomes
$$8\sqrt{2}\int_{-1}^{\infty}{1\over (u^2+3)^2}\cdot{\mathrm du\over \sqrt{u+1}}\tag6$$
Another sub: $u=x^2-x+1$ then $(2)$ becomes
$$\int_{1}^{\infty}{\mathrm du\over u^2\sqrt{u-3}}\cdot{\sqrt{2}\over \sqrt{1+\sqrt{u-3}}}\tag7$$
I am not getting anywhere!
How would one show that $(1)=(2)$ and verify its closed form?
| Alternative computation.
$\displaystyle J=\int_{0}^{\infty}{1\over x^{2}-x+1}\cdot{\mathrm dx\over \sqrt{x}}$
Perform the change of variable $y=\sqrt{x}$,
$\begin{align} J&=2\int_{0}^{\infty}{1\over x^{4}-x^2+1}dx\\
&=2\int_{0}^{\infty}{1+x^2\over x^{6}+1}dx\\
&=2\int_{0}^{\infty}{1\over 1+x^{6}}dx+2\int_{0}^{\infty}{x^2\over x^{6}+1}dx
\end{align}$
In the last two integrals perform the change of variable $y=x^6$,
$\begin{align} J&=\dfrac{1}{3}\int_{0}^{\infty} \dfrac{x^{-\tfrac{5}{6}}}{1+x}dx+\dfrac{1}{3}\int_{0}^{\infty} \dfrac{x^{-\tfrac{1}{2}}}{1+x}dx\\
&=\dfrac{1}{3}\text{B}\left(\tfrac{1}{6},\tfrac{5}{6}\right)+\dfrac{1}{3}\text{B}\left(\tfrac{1}{2},\tfrac{1}{2}\right)\\
&=\dfrac{1}{3}\Gamma\left(\tfrac{1}{2}\right)^2+\dfrac{1}{3}\Gamma\left(\tfrac{1}{6}\right)\Gamma\left(\tfrac{5}{6}\right)\tag 1\\
&=\dfrac{1}{3}\dfrac{\pi}{\sin\left(\tfrac{\pi}{2}\right)}+\dfrac{1}{3}\dfrac{\pi}{\sin\left(\tfrac{\pi}{6}\right)}\tag 2\\
&=\dfrac{1}{3}\pi+\dfrac{2}{3}\pi\\
&=\boxed{\pi}
\end{align}$
1)
https://en.wikipedia.org/wiki/Beta_function#Relationship_between_gamma_function_and_beta_function
2) https://en.wikipedia.org/wiki/Reflection_formula
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Singular value decomposition works only for certain orthonormal eigenvectors, not all? I'm trying to find the SVD of the following matrix:
$$A=
\begin{pmatrix}
1 & 1 \\
2 & -2 \\
2 & 2 \\
\end{pmatrix}
$$
I found the eigenvalues and vectors for $A'A$:
$$
\begin{array}{cc}
\lambda_1=10 & \lambda_2=8 \\
e_1'=(1,1) & e_2'=(-1,1). \\
\end{array}
$$
I find the eigenvalues and vectors for $AA'$:
$$
\begin{array}{ccc}
\lambda_1=10 & \lambda_2=8 & \lambda_3=0 \\
e_1=(1,0,2) & e_2=(0,1,0) & e_3=(-2,0,1), \\
\end{array}$$
and so my SVD should be:
$$\left(
\begin{array}{ccc}
\frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} \\
0 & 1 & 0 \\
\frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
\end{array}
\right).\left(
\begin{array}{cc}
\sqrt{10} & 0 \\
0 & \sqrt{8} \\
0 & 0 \\
\end{array}
\right).\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
However, this gives
$$\left(
\begin{array}{cc}
1 & 1 \\
-2 & 2 \\
2 & 2 \\
\end{array}
\right)$$ instead of $A$.
To get $A$ I need to decompose in the following way
$$\left(
\begin{array}{ccc}
\frac{1}{\sqrt{5}} & 0 & -\frac{2}{\sqrt{5}} \\
0 & 1 & 0 \\
\frac{2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\
\end{array}
\right).\left(
\begin{array}{cc}
\sqrt{10} & 0 \\
0 & \sqrt{8} \\
0 & 0 \\
\end{array}
\right).\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
\end{array}
\right)$$
This last decomposition is as if I had multiplied the first $e_2$ by $-1$.
These eigenvalues and vectors were derived from Mathematica, just to be sure I was using the correct elements.
Any help in explaining why my first decomposition doesn't work would be appreciated.
Edit: The book I'm using doesn't tell which of the orthonormal eigenvectors I have to use. For each eigenvalue, I have two orthonormal eigenvectors, $e_i$ and $-e_i$.
| The answer is yes. Since your $m\times n$ matrix $A$ is such that $m\geq n$, you first compute $\sigma_i$s and $v_i$s and than get the $u_i$s via relation
$$u_i = \frac{1}{\sigma_i}Av_i\text{.}$$
This follows from the proof of "SVD theorem": There exist unitary matrices $U$ and $V$ with the columns $u_i$ and $v_i$ (and matrix $\Sigma$ ...) for which $A = U\Sigma V^H$.
If you choose, for example $-u_i$s from the upper relation (which are still eigenvectors of $AA'$), you would get $U\Sigma V^H = -A$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int_0^\infty \frac{\sin^2(xu)du}{u^2}$ How do I evaluate$$\int_0^\infty \frac{\sin^2(xu)du}{u^2}$$
I tried integration by parts, but it kept getting messier.
| Here is a slight different line of reasoning: We begin by noting that
\begin{align*}
\frac{1}{a} &= \int_{0}^{\infty} e^{-ax} \, dx, \qquad (a > 0) \\
\frac{1}{a^2} &= \int_{0}^{\infty} x e^{-ax} \, dx, \qquad (a > 0) \\
\frac{a}{a^2+b^2} &= \int_{0}^{\infty} \cos(bx) e^{-ax} \, dx, \qquad (a > 0) \\
\end{align*}
which can be easily conformed by integration by parts. Then we can write
\begin{align*}
\int_{0}^{\infty} \frac{\sin^2(xu)}{u^2} \, du
&= \int_{0}^{\infty} \sin^2(xu)\left(\int_{0}^{\infty} s e^{-us} \, ds\right) \, du \\
&= \int_{0}^{\infty} s \left( \int_{0}^{\infty} \sin^2(xu) e^{-su} \, du\right) \, ds \\
&= \int_{0}^{\infty} s \left( \int_{0}^{\infty} \frac{1-\cos(2xu)}{2} e^{-su} \, du\right) \, ds \\
&= \int_{0}^{\infty} \frac{s}{2} \left( \frac{1}{s} - \frac{s}{4x^2+s^2}\right) \, ds \\
&= \int_{0}^{\infty} \frac{2x^2}{4x^2+s^2} \, ds.
\end{align*}
The final integral can be evaluated by the substitution $s = 2|x|\tan\theta$, yielding the same answer as others.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the value of $3+7+12+18+25+\ldots=$ Now, this may be a very easy problem but I came across this in an examination and I could not solve it.
Find the value of
$$3+7+12+18+25+\ldots=$$
Now here is my try
$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$
After that, I could not proceed.
| I'm adding another answer because people ask how to find $a_n$ without trial and error.
We note that:
$a_2 - a_1 = 4; a_3 - a_2 = 5; a_4 - a_3 = 6$,
which leads us to conclude that $a_n$ is given by the recurrence relation:
$a_{n+1} - a_n = n+3$
Let's start by solving the homogeneous equation:
$a_{n+1} - a_n = 0$
The associated polynomial is $P(r) =r - 1$. The root of this polynomial is $r = 1$. Therefore, a solution to the homogeneous equation is $a_n^{h} = 1^n = 1$.
Now, we want to find a particular solution, so let's try $a_n^{p} = An + B$.
Then:
$A(n+1) + B - An - B= n+3$. This does not work.
Let's try $a_n^{p} = An^2 + Bn + C$
Thus:
$A(n+1)^2 + B(n+1) + C - An^2 - Bn - C = n +3$
from which follows:
$2A = 1, A + B = 3, C \in \mathbb{R}$.
Hence:
$A = 1/2;
B = 5/2;
C \in \mathbb{R}$
Therefore, $a_n^p = 1/2n^2 + 5/2n + C$
and
$a_n = a_h + a_p = 1 + 1/2n^2 + 5/2n + C = 1/2n^2 + 5/2n + D$
Because $a_1 = 3$, it follows that $D = 0$.
We conclude that $a_n = 1/2n^2 + 5/2n \quad \triangle$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230870",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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"answer_id": 3
} |
Find the greatest common divisor of the polynomials $ \ f(x)=x^{3}+2 \ \ and \ \ g(x)=x- 1 \ $ over the field $ \ \ \mathbb{Z}_{3}$ Find the greatest common divisor of the polynomials $ \ f(x)=x^{3}+2 \ \ and \ \ g(x)=x- 1 \ $ over the field $ \ \ \mathbb{Z}_{3}$. $$ $$ We have $ \ f(x)=x^{3}+2 \ \ and \ \ g(x)=x-1=x+2 \ \ \ mod \ (3) \ $ . Now $ \ x^{3}+2=(x+2)(x^{2}-2x+4) \ =(x+2)(x^{2}+x+1)$. (since -2=1 and 4=1 in $ \ \mathbb{Z}_{3} $ ). So according to me the G.C.D =(x+2) . Am I right ? Any help please
| It's easier if you write $x^3+2=x^3-1=(x-1)(x^2+x+1)$, so it becomes apparent that $x-1$ is a divisor, hence the greatest common divisor.
Anyway, your computation is correct, using $x^3+2=x^3+8=(x+2)(x^2-2x+4)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$ If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$
My Attempt:
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2a.\dfrac {1}{a}$$
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2$$
How do I proceed further?
| You have $$a-\frac 1a=6$$ so squaring gives $$a^2-2+\frac{1}{a^2}=36$$ so...?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solutions of an equation Problem:
Find all real solutions for $x$ in
$$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $$
I know that you can move all the terms to one side, then divide by $2$ and factor, but what next?
| We seek all real solutions for $2(2^x-1)x^2+\left(2^{x^2}-2\right)x=2^{x+1}-2$.
Let's rewrite our equation into a very interesting form.
\begin{align}
2(2^x-1)x^2+\left(2^{x^2}-2\right)x&=2^{x+1}-2\\
2(2^x-1)x^2+\left(2^{x^2}-2\right)x-2^{x+1}+2&=0&&\text{Set equation = 0}\\
(2^x-1)x^2+\left(2^{x^2-1}-1\right)x-2^x+1&=0&&\text{Divide all terms by 2}\\
(2^x-1)x^2+\left(2^{x^2-1}-1\right)x-(2^x-1)&=0&&\text{Factor out $-1$}\\
(2^x-1)(x^2-1)+\left(2^{x^2-1}-1\right)x&=0&&\text{Factor out $2^x-1$}\\
(2^x-1)y+x(2^y-1)&=0\tag{1}\label{eq}&&\text{Replace $x^2-1$ with $y$}
\end{align}
Notice the symmetry of the left-hand side of Equation \ref{eq}. Also notice that if $x=0$, then $2^x-1=0$ as well.
The solutions of Equation \ref{eq} must be values such that $x=0$ or $y=x^2-1=0$. Therefore the solutions are $x=0, 1, -1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How to turn $12x^2-8x+1$ into $(2x-1)(6x-1)$ without quadratic equation? After almost seven years I recently started again learning math and have few holes in my algebra knowledge, so I apologize for the beginner question.
My question is:
Is there any simple trick to turn $12x^2-8x+1$ into $(2x-1)(6x-1)$ without using quadratic equation?
And if it's possible, what is step by step procedure?
| Write $x=y/12$, so the polynomial becomes
$$
12\frac{y^2}{12^2}-\frac{8y}{12}+1=
\frac{1}{12}(y^2-8y+12)=\frac{1}{12}(y-2)(y-6)
$$
The decomposition of $y^2-8y+12$ is obtained from the search for two numbers whose sum is $8$ and product is $12$.
Substituting back $y=12x$, we have $y-2=2(6x-1)$ and $y-6=6(6x-1)$, so the denominator $12$ cancels out.
Note that in this way the coefficient of $y$ is always $1$. In general, for $a\ne0$, we have, setting $y=x/a$,
$$
ax^2+bx+c=a\frac{y^2}{a^2}+\frac{by}{a}+c=
\frac{1}{a}(y^2+by+ac)
$$
so the trick “sum and product” can be attempted at.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 7
} |
Show this $|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\frac85$ Let $x\in R$ show that
$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{8}{5}$$
since
$$f(x)=f(x+\pi),$$it sufficient to show $x\in (0,\pi]$
| Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of $\sin$ on $[0, \pi]$ quickly reduces the problem to showing the inequality
$$2 \sin 1 > \frac{8}{5} .$$
From $\pi < \frac{22}{7}$ we deduce $\frac{3 \pi}{10} < \frac{66}{70} < 1$, and so
$$2 \sin 1 > 2 \sin \frac{3 \pi}{10} = 2 \cdot \frac{1}{4}(1 + \sqrt{5}) > \frac{8}{5}.$$ The last inequality (which itself is a reasonably tight bound on the Golden Ratio $\phi$) follows from rearranging and squaring.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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For $\Bbb K^{n \times n}$ matrices $A$ and $B$, $\exists p[p(A)=B] \implies \exists p[p(B)=A]$? For $\Bbb K^{n \times n}$ matrices $A$ and $B$, is it true that if there exists a matrix polynomial $p$ such that $p(A)=B$, then there exists another matrix polynomial $p$ (not necessarily the same) such that $p(B) = A$?
My attempt
Let $B = \displaystyle \sum_{k=0}^{n-1} a_k A^k$, where $a_k \in \Bbb K$.
From the Cayley-Hamilton theorem, there exists a polynomial $p$ of degree $n$ such that $p(A) = O$, so $A^m$ where $m \ge n$ can be expressed as linear combination of $\left\{ A^k \ | \ 0 \le k < n\right\}$.
So, we can express $B^m = \displaystyle \sum_{k=0}^{n-1} a_{mk} A^k$, where $a_{mk} \in \Bbb K$.
If $\{I, A, A^2, \cdots, A^{n-1}\}$ form a basis, then we can express a polynomial in $B$, $\displaystyle \sum_{k=0}^{n-1} b_k B^k$, as:
$$\begin{pmatrix}
a_{00} & a_{10} & a_{20} & \cdots & a_{(n-1)0} \\
a_{01} & a_{11} & a_{21} & \cdots & a_{(n-1)1} \\
a_{02} & a_{12} & a_{22} & \cdots & a_{(n-1)2} \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
a_{0(n-1)} & a_{1(n-1)} & a_{2(n-1)} & \cdots & a_{(n-1)(n-1)} \\
\end{pmatrix}
\begin{pmatrix}
b_0 \\ b_1 \\ b_2 \\ \vdots \\ b_{n-1}
\end{pmatrix}$$
Since we want the polynomial to equal $A$, we would need to solve:
$$\begin{pmatrix}
a_{00} & a_{10} & a_{20} & \cdots & a_{(n-1)0} \\
a_{01} & a_{11} & a_{21} & \cdots & a_{(n-1)1} \\
a_{02} & a_{12} & a_{22} & \cdots & a_{(n-1)2} \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
a_{0(n-1)} & a_{1(n-1)} & a_{2(n-1)} & \cdots & a_{(n-1)(n-1)} \\
\end{pmatrix}
\begin{pmatrix}
b_0 \\ b_1 \\ b_2 \\ \vdots \\ b_{n-1}
\end{pmatrix}
=
\begin{pmatrix}
0 \\ 1 \\ 0 \\ \vdots \\ 0
\end{pmatrix}$$
Until now I have transformed the problem into a problem in linear algebra, but is utterly useless. Firstly, $\{A^k \ | \ 0 \le k < n\}$ may not even be linearly independent.
| Take $A=\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$, and let $p(x)=x^2$. Then $p(A) = I$, but there's no polynomial $q$ with $q(I) = A$.
| {
"language": "en",
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Show that $\int_{0}^{1}\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right) dx=\int_{0}^{1}x\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right) dx=\pi$ Two integrals exhibit the same closed form
Motivated by this interesting question.
$$\int_{0}^{1}\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right)\mathrm dx=\pi\tag1$$
and
$$\int_{0}^{1}x\sqrt{x\over 1-x}\cdot\ln\left({x\over 1-x}\right)\mathrm dx=\pi\tag2$$
Making an attempt
$$u={x\over 1-x}\implies -(1-x)^2du=dx\tag a$$
$$u=\sqrt{x\over 1-x}\implies 2x(1-x)^2du=dx\tag b$$
I try on $(a)$ and $(b)$ it doesn't seem to be working. So I try another
$u=\sqrt{1-x}\implies -2\sqrt{1-x}du=dx$, apply to $(1)$ then we have
$$2\int_{0}^{1}\sqrt{1-u^2}\ln\left({1-u^2\over u^2}\right)\mathrm du=I_1-I_2\tag3$$
$$I_1-I_2=2\int_{0}^{1}\sqrt{1-u^2}\ln(1-u^2)\mathrm du-2\int_{0}^{1}\sqrt{1-u^2}\ln u^2\mathrm du\tag4$$
For $I_1$ make $v=\sqrt{1-u^2}\implies -{\sqrt{1-u^2}\over u}dv=du$, then
$$I_1=4\int_{0}^{1}{v^2\over \sqrt{1-v^2}}\cdot\ln v \mathrm dv\tag5$$
Recall $${1\over \sqrt{1+x}}=\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}(-x)^k\tag6$$
then $(5)$ becomes
$$I_1=4\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}\int_{0}^{1}v^4\ln v \mathrm dv\tag7$$
$$I_1=-\sum_{k=0}^{\infty}{(2k-1)!!\over (2k)!!}\tag8$$
Apply same substitution, $I_2$ becomes
$$I_2=2\int_{0}^{1}{v^2\over \sqrt{1-v^2}}\cdot\ln(1-v^2)\mathrm dv\tag9$$
$$I_2=2\sum_{0}^{\infty}{(2k-1)!!\over (2k)!!}\int_{0}^{1}v^4\ln(1-v^2)\mathrm dv\tag c$$
Surely this is going wrong because there is not sign of $\pi $ anywhere in $I_1$ nor $I_2$!
How do we show that $(1)=(2)$ and prove one of them?
| On the path of Latte,
$\displaystyle J=\int_0^1 \sqrt{\dfrac{x}{1-x}}\ln\left(\dfrac{x}{1-x}\right)dx$
Perform the change of variable $y=\sqrt{\dfrac{x}{1-x}}$ and then, use integration by parts,
$\begin{align} J&=4\int_0^{\infty} \dfrac{x^2\ln x}{(1+x^2)^2}dx\\
&=\left[-\dfrac{2x\ln x}{1+x^2}\right]_0^{\infty}+2\int_0^{\infty}\dfrac{\ln x}{1+x^2}dx+2\int_0^{\infty}\dfrac{1}{1+x^2}dx\\
&=2\int_0^{\infty}\dfrac{1}{1+x^2}dx\\
&=2\Big[\arctan x\Big]_0^{\infty}\\
&=2\times \dfrac{\pi}{2}\\
&=\boxed{\pi}
\end{align}$
$\begin{align}
\int_0^{\infty}\dfrac{\ln x}{1+x^2}dx=\int_0^{1}\dfrac{\ln x}{1+x^2}dx+\int_1^{\infty}\dfrac{\ln x}{1+x^2}dx
\end{align}$
In the latter integral perform the change of variable $y=\dfrac{1}{x}$,
$\begin{align}
\int_0^{\infty}\dfrac{\ln x}{1+x^2}dx&=\int_0^{1}\dfrac{\ln x}{1+x^2}dx-\int_0^{1}\dfrac{\ln x}{1+x^2}dx\\
&=0
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Cotangent expansion to compute infinite products At the end of class my professor remarked that the infinite product expansion of cotangent,
$$\text{cot}(z)=\frac{1}{z}+\prod \frac{2z}{z^2-n^2\pi^2}$$
can be used to compute infinite products like $\prod \frac{1}{n^2+1}$,$\prod \frac{1}{n^2+a^2}$. I'm assuming I'd substitute in some value for $z$ and the whole thing would pop up. But, I can't figure out how to get rid of the negative sign and its driving me crazy.
| Let $z = \pi i a$. Then
\begin{align*}
\cot (\pi i a) &= \frac{1}{\pi i a} + \prod \frac{2\pi i a}{i^2 \pi^2 a^2 - n^2 \pi^2}\\
- i \coth (\pi a) &= - \frac{i}{\pi a} - \frac{2 i a}{\pi} \prod \frac{1}{a^2 + n^2}\\
\frac{2 a}{\pi} \prod \frac{1}{a^2 + n^2} &= \coth (\pi a) - \frac{1}{\pi a} \\
\prod \frac{1}{a^2 + n^2} &= \frac{\pi}{2a} \coth (\pi a) - \frac{1}{2 a^2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2240313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $x$ and $y$ such that $15\sin(x+y)+7\sin x+7\sin y$ is maximized Question is: Find $x$ and $y$ such that
$$15\sin(x+y)+7\sin x+7\sin y$$
is maximized.
What I tried was
$$f(x,y)=15\sin(x+y)+7\sin x+7\sin y=15(\sin x\cos y+\cos x\sin y)+7(\sin x +\sin y)$$
$$\frac{\partial f}{\partial x}=15\cos(x+y)+7\cos x=0$$
$$\frac{\partial f}{\partial y}=15\cos(x+y)+7\cos y=0$$
$$\cos x=\cos y$$
$$y=-x$$
$$\cos x=-{15\over 7}\quad ??$$
| Hint:
Observe that
$$\frac{\partial f}{\partial x}=0=\frac{\partial f}{\partial y}\qquad\implies\qquad\cos x=\cos y\qquad\iff\qquad x=2n\pi\pm y\;\;\text{for some }n\in\mathbb{Z}$$
If $x=2n\pi+y$ we have
\begin{align*}
\frac{\partial f}{\partial x}=0&\quad\implies &15\cos(2n\pi+2y)+7\cos(2n\pi+y)&=0\\
&\quad\iff&15\cos (2y)+7\cos y&=0\\
&\quad\iff&15\left(2\cos^2 y-1\right)+7\cos y&=0\\
&\quad\iff&30\cos^2 y+7\cos y-15&=0\\
&\quad\iff&\cos y&\in\left\{\frac35,-\frac56\right\}\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let a general term $T_n$ be defined as $T_n =\left(\frac{1\cdot 2\cdot 3 \cdot 4 \cdots n}{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}\right)^2$ Let a general term $T_n$ be defined as
$$T_n =\left(\frac{1\cdot 2\cdot 3 \cdot 4 \cdots n}{1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n+1)}\right)^2$$
Then prove that
$\lim_{n\to\infty}(T_1 + T_2 +\cdots+T_n) \lt \frac{4}{27}.$
I tried finding pattern between terms ..
$T_1=\frac{1}{9} , \frac{T_2}{T_1}=(\frac{2}{5})^2, \frac{T_3}{T_2}=(\frac{3}{7})^2$
but could not think more of how to get a bound on the series.
Any help is appreciated.
| $$\dfrac{T_m}{T_{m-1}}=\left(\dfrac m{2m+1}\right)^2<\left(\dfrac m{2m}\right)^2=\dfrac14$$ for $m>0$
$$\sum_{r=1}^\infty T_r<\sum_{r=1}^\infty T_1\left(\dfrac14\right)^{r-1}=\dfrac{\dfrac19}{1-\dfrac14}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve a simple equation: $x+x\sqrt{(2x+2)}=3$
$x+x\sqrt{(2x+2)}=3$
I must solve this, but I always get to a point where I don't know what to do. The answer is 1.
Here is what I did:
$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1) \\
\frac{9-6x+x^{2}}{2x^{2}}&=x+1 \\
\frac{9-6x+x^{2}-2x^{2}}{2x^{2}}&=x \\
\frac{9-6x+x^{2}-2x^{2}-2x^{3}}{2x^{2}}&=0
\end{align}$$
Then I got:
$-2x^{3}-x^{2}-6x+9=0$
| As in the comment $x-1$ is a factor of $f(x) = -2x^3 -x^2-6x+9$. After an easy long division we get $f(x) = (x-1)(-2x^2-3x-9)$. From this we can use the quadratic equation to see the other two roots are not real.
We can do the long division in the following way:
We need to divide $-2x^3-x^2-6x+9$ by $x-1$ so we can ask what times $x-1$ gives the first term $-2x^3$? This is clearly $\boxed{-2x^2}$. But this also gives $+2x^2$ that we didn't want. So we need to take this away from what is left: we now have $-x^2-6x+9 - (+2x^2) = -3x^2-6x +9$. Again we look for a term that when multiplied with $x-1$ gives the highest order term ($-3x^2$). This is $\boxed{-3x}$ but this gives an extra $+3x$. Now we are left with $-6x+9 - (+3x) = -9x+9$. Here we see that this is $\boxed{-9}$ times $(x-1)$. Adding together the terms that we used along the way, we have $-2x^2-3x-9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos\left (\theta -\frac{\pi}{4}\right)$ If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos \left(\theta -\frac{\pi}{4}\right).$
I could not get any idea to solve. However I tried by using $\theta =0^\circ $. But could not get the answer.
| Rearrange into sines and cosines. Form a cosine addition identity.
\begin{align}
\tan(\pi\cos t)\tan(\pi\sin t) &= 1
\\
\sin(\pi\cos t)\sin(\pi\sin t) &= \cos(\pi\cos t)\cos(\pi\sin t)
\\
0 &= \cos(\pi\cos t)\cos(\pi\sin t) - \sin(\pi\cos t)\sin(\pi\sin t)
\\
0 &= \cos(\pi\cos t + \pi\sin t)
\\
0 &= \cos(\pi[\cos t + \sin t])
\end{align}
The argument of the cosine in the previous line must be an odd integer multiple of $\pi/2$, so we get
\begin{align}
\cos t + \sin t &= \frac{2n+1}{2}
\end{align}
where $n\in\mathbb{Z}$ (EDIT: More accurately, $n \in \{-1,0\}$ to be in the range of $\cos t + \sin t$). Use the cosine subtraction formula to achieve the desired result.
\begin{align}
\frac{1}{\sqrt2}\cos t + \frac{1}{\sqrt2}\sin t &= \frac{2n+1}{2\sqrt{2}}
\\
\cos (t)\cos(\pi/4) + \sin (t)\sin(\pi/4) &= \frac{2n+1}{2\sqrt{2}}
\\
\\
\cos(t -\pi/4) &= \frac{2n+1}{2\sqrt{2}}
\end{align}
EDIT: Following lab bhattacharjee's comment, the range of cosine restricts the possible solutions to
$$\cos(t -\pi/4) = \frac{\pm1}{2\sqrt{2}}.$$
It doesn't appear like the answer is unique to me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove inequality with integral Prove the following inequality
$$\ln \frac{\pi + 2}{2} \cdot \frac{2}{\pi} < \int \limits_0^{\pi/2} \frac{\sin\ x}{x^2 + x} < \ln \frac{\pi + 2}{2}$$
I can prove that $\frac{\sin\ x}{x^2 + x} < \frac{1}{x + 1} \ \forall x \in (0, +\infty) \Rightarrow \int \limits_0^{\pi/2} \frac{\sin\ x}{x^2 + x} < \int \limits_0^{\pi/2} \frac{1}{x + 1} = \ln \frac{\pi + 2}{2}$.
Unfortunately, I don't know what happens when $x = 0$.
I can prove $\leqslant$ on LHS using Mean Value Theorem, but I have no idea how to prove that the sign is strict.
| The given inequality just depends on the convexity inequality
$$ \forall x\in\left[0,\frac{\pi}{2}\right],\qquad \frac{2}{\pi}x\leq \sin(x)\leq x.$$
We may use another convexity inequality
$$ \forall x\in\left[0,\frac{\pi}{2}\right],\qquad x-\frac{1}{6}x^3\leq \sin(x)\leq x-\left(\frac{4}{\pi^2}-\frac{8}{\pi^3}\right)x^3$$
and derive a much better inequality:
$$\small \frac{1}{48} \left(4 \pi -\pi ^2+40 \log\left(\frac{2+\pi }{2}\right)\right)\leq I\\ I \leq \frac{1}{2\pi^3}\left(\pi(\pi-2)(4-\pi)+2 \left(8-4 \pi +\pi ^3\right) \log\left(\frac{2+\pi }{2}\right)\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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