Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ I know that $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ is correct. But having some hard time proofing it using trig relations. Some of the relations I used are
$$
\sin x \cos x= \frac{1}{2} \sin(2 x)\\
\sin^2 x + \cos^2 x= 1\\
\sin^2 x = \frac{1}{2} - \frac{1}{2} \cos(2 x)\\
\cos(2 x) = 2 \cos^2(x)-1
$$
And others. I tried starting with multiplying numerator and denominator with either $\cos x$ or $\sin x$ and try to simplify things, but I seem to be going in circles.
Any hints how to proceed?
| $$\frac{1 + \cos x + \sin x}{1 + \cos x - \sin x}$$
$$= \frac{(1 + \cos x + \sin x)(1 + \cos x + \sin x)}{(1 + \cos x - \sin x)(1 + \cos x + \sin x)}$$
$$= \frac{1 + \cos^2 x + \sin^2 x + 2\cos x + 2\sin x + 2\sin x \cos x}{1 + 2\cos x + \cos^2 x - \sin^2 x}$$
$$= \frac{2(1 + \cos x + \sin x + \sin x \cos x)}{1 + 2\cos x + 2\cos^2 x - 1}$$
$$= \frac{2(1 + \cos x)(1 + \sin x)}{2\cos x(1 + \cos x)}$$
$$= \frac{1 + \sin x}{\cos x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find all possible values of $ a^3 + b^3$ if $a^2+b^2=ab=4$. Find all possible values of $a^3 + b^3$ if $a^2+b^2=ab=4$.
From $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(4-4)=(a+b)0$. Then we know $a^3+b^3=0$.
If $a=b=0$, it is conflict with $a^2+b^2=ab=4$.
If $a\neq0$ and $b\neq0$, then $a$ and $b$ should be one positive and one negative. This contradict with $ab$=4.
I don't know the solution.
Any help please.
| To give a different perspective, I can't find anywhere in the problem statement which states $a$ and $b$ must be real. Therefore, assume $a,b \in \mathbb{C}$. Then the equation $a^3+b^3=0$ along with the condition $ab=4$ implies, by multiplying both sides by either $a^3$ or $b^3$ that $a^6+64 = 0$ and $b^6+64=0$. Therefore,
$$ a = \pm\sqrt[6]{-64} = \pm2\sqrt[6]{-1},\\ b= \pm\sqrt[6]{-64} = \pm2\sqrt[6]{-1}.$$
As $a$ and $b$ are multiples of sixth roots of unity, there are six choices for each. Not all choices are valid since the condition $a^2+b^2=4$ must also be met. For example
$$a=b=\sqrt{3}+i$$
does not work since then $a^2+b^2=4+4\sqrt{3}i$. However, the choice
$$a=\sqrt{3}+i, b=\sqrt{3}-i$$
works fine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
If $\gcd(a,b)=1$ then $\gcd(a^2+b^2,a+2ab)=1$ or $5$ The question is already in the title.
Show that if $\gcd(a,b)=1$ then $\gcd(a^2+b^2,a+2ab)=1$ or $5$.
I saw yesterday this exercise in a book and I tried many things but I managed to show only that if some prime $p$ ($p$ must be $\equiv1\pmod{4}$)
divides $a^2+b^2$ and $a+2ab$ then $p\mid2b+1$ , $p\mid4a^2+1$ and $p\mid b-2a^2$ but all these seem pointless.
Am I missing something obvious here?
| suppose $gcd(a,b)=1$ and let $d=gcd(a^2+b^2,a+2b)$
the identity
$(a^2+b^2)+(a+2b)(2b-a)=5b^2$
shows that $d$ divides $5b^2$ and
since $5a^2=5(a^2+b^2)-5b^2$ , $d$ divides $5a^2$
since $a$ and $b$ are relatively prime, $d$ divides $5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find the derivative using the chain rule and the quotient rule
$$f(x) = \left(\frac{x}{x+1}\right)^4$$ Find $f'(x)$.
Here is my work:
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4\left(x+1\right)^3x^4}{\left(x+1\right)^8}$$
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4x^4\left(x+1\right)^3}{\left(x+1\right)^8}$$
I know the final simplified answer to be:
$${4x^3\over (x+1)^5}$$
How do I get to the final answer from my last step? Or have I done something wrong in my own work?
| You calculation of $f'(x)$ is correct, though it can be simplified. To do this, we factor out $4x^3(x+1)^3\require{cancel}$ in the numerator.
$$\begin{align}f'(x) &= \frac{4x^3(x+1)^4-4x^4(x+1)^3}{(x+1)^8}\\\\
&= \frac{\color{blue}{4x^3(x+1)^3}\Big((x+1) - x\Big)}{(x+1)^8}\\\\
&= \frac {(4x^3\cancel{(x+1)^3})(1)}{\cancel{(x+1)^8}^5}\\ \\
&= \frac{4x^3}{(x+1)^5}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Solve $\int\frac{x}{\sqrt{x^2-6x}}dx$ I need to solve the following integral
$$\int\frac{x}{\sqrt{x^2-6x}}dx$$
I started by completing the square,
$$x^2-6x=(x-3)^2-9$$
Then I defined the substitution variables..
$$(x-3)^2=9\sec^2\theta$$
$$(x-3)=3\sec\theta$$
$$dx=3\sec\theta\tan\theta$$
$$\theta=arcsec(\frac{x-3}{3})$$
Here are my solving steps
$$\int\frac{x}{\sqrt{(x-3)^2-9}}dx = 3\int\frac{(3\sec\theta+3)\sec\theta\tan\theta}{\sqrt{9(\sec^2\theta-1)}}d\theta$$
$$=\int\frac{(3\sec\theta+3)\sec\theta\tan\theta}{\tan\theta}d\theta$$
$$=\int(3\sec\theta+3)\sec\theta\tan\theta$$
$$=3\int\sec^2\theta\tan\theta d\theta + 3\int\sec\theta\tan\theta d\theta$$
$$u = \sec\theta, du=\sec\theta\tan\theta d\theta$$
$$=3\int udu + 3\sec\theta$$
$$=\frac{3\sec\theta}{2}+3\sec\theta+C$$
$$=\frac{3\sec(arcsec(\frac{x-3}{3}))}{2}+3\sec(arcsec(\frac{x-3}{3}))+C$$
$$=\frac{3(\frac{x-3}{3})}{2}+3(\frac{x-3}{3})$$
$$=\frac{x-3}{2}+x-3+C$$
However, the expected answer is
$$\int\frac{x}{\sqrt{x^2-6x}}dx=\sqrt{x^2-6x}+3\ln\bigg(\frac{x-3}{3}+\frac{\sqrt{x^2-6x}}{3}\bigg)$$
What did I misunderstood?
| $$
\begin{aligned}
\int \frac{x}{\sqrt{x^2-6 x}} d x & \int \frac{x}{x-3} d\left(\sqrt{x^2-6 x}\right) \\
= & \int\left(1+\frac{3}{x-3}\right)\left(\sqrt{x^2-6 x}\right) \\
= & \sqrt{x^2-6 x}+3 \int \frac{d\left(\sqrt{x^2-6 x}\right)}{\sqrt{\left(\sqrt{x^2-6 x}\right)^2+9}} \\
= & \sqrt{x^2-6 x}+3 \sinh^{-1} \left(\frac{\sqrt{x^2-6 x}}{3}\right)+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$ I need to solve
$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$
I've chosen the substitution variables
$$u=\sqrt{x^2+2}$$
$$du=\frac{x}{\sqrt{x^2+2}}$$
However, I am completly stuck at
$$\int\sqrt{1+xu} dx$$
Which let me believe I've chosen wrong substitution variables.
I've then tried letting $u=x^2+2$ or simply $u=x$, but it does not help me at all solving it.
Would someone please give me an hint on this ?
Thanks.
| Substitute $x=\frac{2-t}{2\sqrt t}$. Then, $dx=-\frac {2+t}{4t\sqrt t}\>dt$ and
\begin{align}
&\int\sqrt{1+x\sqrt{x^2+2}}\>dx \\
= & -\frac18 \int \frac{t+2}{t^2} \sqrt{ 4+4t-t^2}\>dt
= -\frac18 \int \frac{\sqrt{ 4+4t-t^2}}{t-2}d\left( \frac{(t-2)^2}t\right)\\
=& -\frac{t-2}{8t }\sqrt{ 4+4t-t^2}- \int \frac1{t\sqrt{ 4+4t-t^2}}dt\\
=& -\frac{t-2}{8t }\sqrt{ 4+4t-t^2}+\frac12 \tanh^{-1} \frac{t+2}{\sqrt{ 4+4t-t^2}}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Integration of fraction with square root I have a problem with integrating of fraction
$$
\int \frac{x}{x^2 + 7 + \sqrt{x^2 + 7}}
$$
I have tried to rewrite it as $\int \frac{x^3 + 7x - x \sqrt{x^2 + 7}}{x^4 + 13x^2 + 42} = \int \frac{x^3 + 7x - x \sqrt{x^2 + 7}}{(x^2 + 6)(x^2 + 7)}$ and then find some partial fractions, but it wasn't succesful.
| Since $x^2 + 7$ shows up twice it would make a good first substitution. If $u = x^2 + 7$ then
$$\int \frac{x}{x^2 + 7 + \sqrt{x^2 + 7}} \, dx = \frac 12 \int \frac{1}{u + \sqrt{u}} \, du.$$
Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given $n\equiv 1\pmod 8$, show that the number of subsets of a $n$-element set, whose size is $0\pmod 4$ is $2^{n-2}+2^ \frac{n-3}{2}$ Given $n\equiv 1\pmod 8$, show that the number of subsets of a $n$-element set, whose size is $0\pmod 4$ is $2^{n-2}+2^ \frac{n-3}{2}$
I don't get this question. If I have an n-element set, then it's size has to be $1\pmod 8$ and therefore cannot be $0\pmod 4$. Isn't this correct? How does this make sense? Any help would be appreciated!
| What you want is
$$\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots$$
for $n\equiv 1\pmod 8$.
Note that we have the followings :
$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\binom{n}{4}+\cdots+\binom{n}{n}=2^n$$
$$\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\binom{n}{4}-\cdots+(-1)^n\binom{n}{n}=0$$
$$\binom{n}{0}+\binom{n}{1}i-\binom{n}{2}-\binom{n}{3}i+\binom{n}{4}+\binom{n}{5}i-\cdots=(1+i)^n$$
$$\binom{n}{0}-\binom{n}{1}i-\binom{n}{2}+\binom{n}{3}i+\binom{n}{4}-\binom{n}{5}i-\cdots=(1-i)^n$$
Adding these gives us
$$4\left(\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots\right)=2^n+(1+i)^n+(1-i)^n$$$$=2^n+2^{n/2}\left(\cos (n\pi/4)+i\sin (n\pi/4)\right)+2^{n/2}\left(\cos (n\pi/4)-i\sin (n\pi/4)\right)$$$$=2^n+2\cdot 2^{n/2}\cos(n\pi/4).$$
Hence, we have
$$\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots=2^{n-2}+2^{(n-2)/2}\cos (n\pi/4)=2^{n-2}+2^{(n-3)/2}.$$
Here, note that $\cos (n\pi/4)=2^{-1/2}$ for $n\equiv 1\pmod 8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate $\int \frac{x^2 }{(x^2+1)^2} dx$? I'm trying to calculate $\int \frac{x^2 }{(x^2+1)^2} dx$ by using the formula:
$$ \int udv = uv -\int vdu $$
I supposed that $u=x$ s.t $du=dx$, and also that $dv=\frac{x}{(x^2+1)^2}dx$, but I couldn't calculate the last integral. what is the tick here?
the answer must be: $ \frac{1}{2}arctan\ x - \frac{x}{2(1+x^2)} $ + C
| $$\int\dfrac{x^2}{(x^2+1)^2}dx$$
$$=x\int\dfrac x{(x^2+1)^2}dx-\int\left[\frac{dx}{dx}\cdot\int\dfrac x{(x^2+1)^2}dx\right]dx$$
$$=x\cdot\frac{-1}{1+x^2}+\int\dfrac{dx}{1+x^2}=\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1198686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Continued Fraction Algorithm for 113/50 The numbers $a_k$ can be found for $\frac{113}{50}$ by using a continued fraction algorithm. Note that $\frac{113}{50}$ is rational, and as a result it will have to terminate. Can anyone help me determine the numbers $a_k$ for $\frac{113}{50}$?
What I have Done:
$\frac{113}{50} = 2.26$
$i=2$
$2.26 - 2 = .26$
$\frac{26}{100}$
$\frac{100}{26} = 3+ \frac{22}{26} $
3 is $a_1$
$\frac{26}{22} = 1 + \frac{4}{22} $
1 is $a_2$
$\frac{22}{4} = 5 + \frac{2}{4}$
$a_3 = 5$
$\frac{4}{2} = 2 $
$a_4 = 2 $
Is this correct, and what do I do next?
| $$\begin{gathered}
\frac{{113}}{{50}} = \frac{{100}}{{50}} + \frac{{13}}{{50}} = 2 + \frac{1}{{\frac{{39}}{{13}} + \frac{{11}}{{13}}}} = 2 + \frac{1}{{3 + \frac{{11}}{{13}}}} \hfill \\
= 2 + \frac{1}{{3 + \frac{{11}}{{11 + 2}}}} = 2 + \frac{1}{{3 + \frac{1}{{1 + \frac{2}{{10 + 1}}}}}} \hfill \\
= 2 + \frac{1}{{3 + \frac{1}{{1 + \frac{1}{{5 + \frac{1}{2}}}}}}} = [2;3,1,5,2] \hfill \\
\end{gathered}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the Inverse Laplace Transforms Find the inverse Laplace transform of:
$$\frac{3s+5}{s(s^2+9)}$$
Workings:
$\frac{3s+5}{s(s^2+9)}$
$= \frac{3s}{s(s^2+9} + \frac{5}{s(s^2+9)}$
$ = \frac{3}{s^2+9} + \frac{5}{s}\frac{1}{s^2+9}$
$ = \sin(3t) + \frac{5}{s}\frac{1}{s^2+9}$
Now I'm not to sure on what to do.
Any help will be appreciated.
| $\frac{3s+5}{s(s^2+9)} = -\frac{5}{9} \frac{s}{s^2+9} + \frac{2}{3}\frac{3}{s^2+9}+\frac{5}{9}\frac{1}{s}$ with ILT $-\frac{5}{9}\cos(3t)+\frac{2}{3}\sin(3t)+\frac{5}{9}step(t)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
(probability) Stuck on a proof in Grinstead and Snell This is probably (no pun intended) very simple, but it has fooled me. On page 294 of Grinstead and Snell's book on probability they derive the pdf of X+Y where both X and Y have the standard normal distribution. The step in the proof that stops me is from: $$\frac{1}{2\pi} \int_{-\infty}^{+\infty} exp(\frac{-(z-y)^2}{2})\;exp(\frac{-y^2}{2})\;dy $$ to
$$\frac{1}{2\pi}\;exp(\frac{-z^2}{4}) \int_{-\infty}^{+\infty} exp(-(y-z/2)^2)\;\;dy $$
(I have reproduced the second integral exactly as in the book).
There are lots of questions around this topic, and proofs for where the distribution is not standard (these befuddle me even more). It's obviously a very simple step but I am self-taught in maths and there are lots of very basic gaps in what I know!
| Suppose $X,Y \sim N(0,1)$ and $Z=X+Y$, and let the corresponding densities be $f_X,f_Y,f_Z$. Then
$$
\begin{align*}
f_Z(z) &= \int_{-\infty}^\infty f_X(z-y) f_Y(y) \, dy \\ &=
\int_{-\infty}^\infty \frac{1}{2\pi} \exp \left(-\frac{(z-y)^2}{2}\right) \exp \left(-\frac{y^2}{2}\right) \, dy \\ &=
\frac{1}{2\pi} \exp \left(-\frac{z^2}{4}\right) \int_{-\infty}^\infty \exp -\left(y - \frac{z}{2}\right)^2 \, dy \\ &=
\frac{1}{2\pi} \exp \left(-\frac{z^2}{4}\right) \int_{-\infty}^\infty \exp -y^2 \, dy \\ &= \frac{1}{2\sqrt{\pi}} \exp \left(-\frac{z^2}{4}\right).
\end{align*}
$$
The third step uses
$$
(z-y)^2 + y^2 = z^2 - 2yz + 2y^2 = \frac{z^2}{2} + \frac{z^2-4yz+4y^2}{2} = \frac{z^2}{2} + \frac{(2y-z)^2}{2}.
$$
After division by $2$, we get
$$
\frac{(z-y)^2}{2} + \frac{y^2}{2} = \frac{z^2}{4} + \frac{(2y-z)^2}{4} = \frac{z^2}{4} + \left(y-\frac{z}{2}\right)^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1204162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the area covered by any spherical rectangle? Is there any analytic or generalized formula to calculate area covered by any rectangle having length $l$ & width $b$ each as a great circle arc on a spherical surface with a radius $R$? i.e. How to find the area $A$ of rectangle in terms of length $l$, width $b$ and radius $R$
($A=f(l, b,R)$)?
Note: Spherical rectangle is a quadrilateral having equal opposite sides but non-parallel & all the interior angles are equal in magnitude & each one is greater than $90^\circ$.
| On a sphere of radius $R > 0$, a geodesic triangle with interior angles $\theta_{1}$, $\theta_{2}$, and $\theta_{3}$ has area $R^{2}(\theta_{1} + \theta_{2} + \theta_{3} - \pi)$. One way you might proceed, therefore, it to triangulate your geodesic polygon (whatever its actual shape) and sum the areas of its triangular pieces. For a quadrilateral with interior angles $\theta_{1}$, $\theta_{2}$, $\theta_{3}$, and $\theta_{4}$, the area is
$$
R^{2}(\theta_{1} + \theta_{2} + \theta_{3} + \theta_{4} - 2\pi).
$$
Added (in light of OP's clarifications): If $\theta \leq \pi$ is the interior angle at each vertex of the quadrilateral, then
$$
\theta = \pi - \cos^{-1}\left\lvert\tan\frac{\ell}{2R} \tan\frac{b}{2R}\right\rvert,
$$
so
$$
\text{Area}
= R^{2}(4\theta - 2\pi)
= R^{2} \left[2\pi - 4\cos^{-1}\left\lvert\tan\frac{\ell}{2R} \tan\frac{b}{2R}\right\rvert\right].
$$
(Particularly, the absolute value of the product of tangents inside the arccos does not exceed unity.)
To see this, it's convenient to work in Cartesian coordinates with the sphere centered at the origin. Denote the vertices of the quadrilateral by
$$
v_{1} = (A, B, C),\qquad
v_{2} = (A, -B, C),\qquad
v_{3} = (A, -B, -C),\qquad
v_{4} = (A, B, -C).
$$
Of course, $A^{2} + B^{2} + C^{2} = R^{2}$.
The plane through the origin, $v_{1}$, and $v_{2}$ has equation $Cx - Az = 0$, and so has unit normal vector
$$
n_{1} = \frac{(C, 0, -A)}{\sqrt{A^{2} + C^{2}}} = \frac{(C, 0, -A)}{\sqrt{R^{2} - B^{2}}}.
$$
The plane through the origin, $v_{1}$, and $v_{4}$ has equation $Bx - Ay = 0$, and so has unit normal vector
$$
n_{2} = \frac{(B, -A, 0)}{\sqrt{A^{2} + B^{2}}} = \frac{(B, -A, 0)}{\sqrt{R^{2} - C^{2}}}.
$$
The "large" angle between these planes is the "large" angle between the great circles they determine (because each normal vector $n_{i}$ is tangent to the sphere at $v_{1}$), i.e., the interior angle $\theta$ of the quadrilateral. Taking the ordinary dot product of the unit normals,
$$
\cos\theta = n_{1} \cdot n_{2}
= \frac{B}{\sqrt{R^{2} - B^{2}}}\, \frac{C}{\sqrt{R^{2} - C^{2}}}.
\tag{1}
$$
Let $2\psi_{1}$ denote the angle subtended at the center of the sphere by the side from $v_{1}$ to $v_{2}$, and let $\ell_{1} = 2R\psi_{1}$ denote the corresponding side length, so that $\psi_{1} = \ell_{1}/(2R)$. Thinking of $v_{1}$ and $v_{2}$ as vectors in space, the ordinary dot product gives
$$
\frac{v_{1} \cdot v_{2}}{R^{2}}
= \cos(2\psi_{1})
= 1 - 2\sin^{2} \psi_{1}.
$$
On the other hand, using the Cartesian components of these vectors, we have
$$
\frac{v_{1} \cdot v_{2}}{R^{2}}
= \frac{A^{2} - B^{2} + C^{2}}{A^{2} + B^{2} + C^{2}}
= 1 - 2\frac{B^{2}}{R^{2}}.
$$
Equating, it follows at once that $B/R = \pm\sin\psi_{1}$. (As a consistency check, the isoceles triangle with the origin, $v_{1}$, and $v_{2}$ as vertices has apex angle $2\psi_{1}$ and base $2B$.)
Similarly, letting $2\psi_{2}$ denote the angle subtended at the center of the sphere by the side from $v_{1}$ to $v_{4}$, and letting $\ell_{2} = 2R\psi_{2}$ denote the corresponding side length, we find $C/R = \pm\sin\psi_{2}$.
That is,
$$
B = \pm R\sin\frac{\ell_{1}}{2R},\qquad
C = \pm R\sin\frac{\ell_{2}}{2R},
\tag{2}
$$
and consequently
$$
\frac{B}{\sqrt{R^{2} - B^{2}}} = \pm\tan\frac{\ell_{1}}{2R},\qquad
\frac{C}{\sqrt{R^{2} - C^{2}}} = \pm\tan\frac{\ell_{1}}{2R}.
\tag{3}
$$
Substituting (3) into (1),
$$
\cos\theta
= \frac{B}{\sqrt{R^{2} - B^{2}}}\, \frac{C}{\sqrt{R^{2} - C^{2}}}
= \pm \tan\frac{\ell_{1}}{2R} \tan\frac{\ell_{2}}{2R}.
$$
To get the large angle, either take the arccos of the negative value, or $\pi$ minus the arccos of the positive value. (The angle/area formulas above do the latter.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 0
} |
How to prove through induction How can I prove by induction that
$$\binom{2n}n<4^n\;?$$
I have solved for the base case, $n=1$, and have formulated the induction hypothesis. I was thinking about Pascal's identity for the rest, but have not been able to come up with a way to use it.
| First, show that this is true for $n=1$:
$\binom{2}{1}<4^{1}$
Second, assume that this is true for $n$:
$\binom{2n}{n}<4^{n}$
Third, prove that this is true for $n+1$:
$\binom{2n+2}{n+1}=$
$\frac{(2n+2)!}{(n+1)!\cdot(n+1)!}=$
$\frac{(2n)!\cdot(2n+1)\cdot(2n+2)}{(n)!\cdot(n+1)\cdot(n)!\cdot(n+1)}=$
$\frac{(2n)!\cdot(2n+1)\cdot(2n+2)}{(n)!\cdot(n)!\cdot(n+1)\cdot(n+1)}=$
$\frac{(2n)!}{(n)!\cdot(n)!}\cdot\frac{(2n+1)\cdot(2n+2)}{(n+1)\cdot(n+1)}=$
$\color{red}{\binom{2n}{n}}\cdot\frac{(2n+1)\cdot(2n+2)}{(n+1)\cdot(n+1)}\color{red}<$
$\color{red}{4^{n}}\cdot\frac{(2n+\color{blue}{1})\cdot(2n+2)}{(n+1)\cdot(n+1)}\color{blue}<$
$4^{n}\cdot\frac{(2n+\color{blue}{2})\cdot(2n+2)}{(n+1)\cdot(n+1)}=$
$4^{n}\cdot\frac{2(n+1)\cdot2(n+1)}{(n+1)\cdot(n+1)}=$
$4^{n}\cdot2\cdot2=$
$4^{n}\cdot4=$
$4^{n+1}$
Please note that the assumption is used only in the part marked red.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
an operator question I know how the derivative operator $\Big(\frac{d}{dx}\Big)^n$ works. But then how does it work if I have $$\exp{\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)}f(x)$$ I thought to use $$\exp z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$ to have
\begin{align}
\exp{\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)}f(x)&=f(x)\\
&+\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)f(x)\\
&+\frac{1}{2!}\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)^2f(x)\\
&+\frac{1}{3!}\Big(a\frac{d}{dx}+b\frac{d^2}{dx^2}\Big)^3f(x)\\
&+\cdots
\end{align}
but then this looks strange ... I am not surer if I could do something like this.
I appreciate hints, references, ...
Many thanks in advance.
| For the operators $a \partial_{x} + b \partial_{x}^{2}$ it is seen that
\begin{align}
(a \partial_{x} + b \partial_{x}^{2})^{1} &= a \partial_{x} + b \partial_{x}^{2} \\
(a \partial_{x} + b \partial_{x}^{2})^{2} &= (a \partial_{x} + b \partial_{x}^{2})(a \partial_{x} + b \partial_{x}^{2}) \\
&= a^{2} \partial_{x}^{2} + 2 a b \partial_{x}^{3} + b^{2} \partial_{x}^{4} = \sum_{k=0}^{2} \binom{2}{k} \, a^{2-k} b^{k} \, \partial_{x}^{2+k} \\
(a \partial_{x} + b \partial_{x}^{2})^{3} &= a^{3} \partial_{x}^{3} + 3 a^{2} b \partial_{x}^{4} + 3 a b^{2} \partial_{x}^{5} + b^{3} \partial_{x}^{6} = \sum_{k=0}^{3} \binom{3}{k} \, a^{3-k} b^{k} \, \partial_{x}^{3+k} \\
\cdots &= \cdots \\
(a \partial_{x} + b \partial_{x}^{2})^{n} &= \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \, \partial_{x}^{n+k}
\end{align}
Now, for the exponential case,
\begin{align}
e^{a \partial_{x} + b \partial_{x}^{2}} \, f(x) &= \sum_{n=0}^{\infty} \frac{1}{n!} \,
\left( \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \, \partial_{x}^{n+k} \right) \, f(x)
\end{align}
This may also be obtained as follows:
\begin{align}
a \partial_{x} + b \partial_{x}^{2} = (a \partial_{x}) ( 1 + \frac{b}{a} \partial_{x})
\end{align}
for which
\begin{align}
e^{a \partial_{x} + b \partial_{x}^{2}} f(x) &= \sum_{n=0}^{\infty} \frac{1}{n!} [ (a \partial_{x}) (1 + \frac{b}{a} \partial_{x}) ] f(x) \\
&= \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \partial_{x}^{n+k} \, f(x)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $p$ for which all solutions of system/equation are real There is system of $5$ equations
$$
a+b+c+d+e = p; \\
a^2+b^2+c^2+d^2+e^2 = p; \\
a^3+b^3+c^3+d^3+e^3 = p; \\
a^4+b^4+c^4+d^4+e^4 = p; \\
a^5+b^5+c^5+d^5+e^5 = p, \\
\tag{1}
$$
where $p\in\mathbb{R}$.
One can prove (like here) that $a,b,c,d,e$ are roots of equation
$$
x^5-\binom{p}{1}x^4+\binom{p}{2}x^3-\binom{p}{3}x^2 + \binom{p}{4}x - \binom{p}{5} = 0.
\tag{2}
$$
I want to create related task for my son: to find polynomial $(2)$ for given system $(1)$. But it would be great to create task, where all $a,b,c,d,e$ are real and distinct. To have possibility to check manually all the sums (without complex numbers).
Question: Which values of $p$ provide $5$ real (pairwise) distinct $a,b,c,d,e$? And if there exists such one at all (for $5$ variables)?
| Using the mean value theorem repeatedly (by taking 3 derivatives), we get that $$60x^2-4 \dbinom{p}{1}x+6 \dbinom{p}{2}$$ has to have $2$ real roots. This only happens if $$\Delta_x > 0 \iff 0 < p < \frac{60}{59}.$$
Using Geogebra, it seems that none of the values in this range produce a function with $5$ real roots so we can be pretty confident in saying that no value of $p$ satisfies what you are looking for. I think we can use the fact that the quitic has a horizontal tangent at $x = 0$ for $p = 0,1$ to prove this observation but it seems messy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1207678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Predicting increasing income within a given time I have this game where player earns "gold" and "reputation". With higher reputation, the player nets more income (gold).
*
*Each reputation points yields 1% bonus in income (gold)
*For every 20 gold received, player receives 1 reputation points
*Player starts with receiving 10 gold every second. This increases as reputation point increases.
How much gold and reputation will the player receive after 1000 seconds?
So I'm asking for an equation/formula and how you came up with it. I tried simplifying the example so I can explain it easily. I think I need a quadratic equation of some sort, but i'm just speculating.
I'm not really that good with providing my own formula :(
EDIT:
Here's an excel simulation of what I described above: https://www.dropbox.com/s/vz5wizqg4x43599/offline%20progression%20simulation.xlsx?dl=0
The first row holds the initial value (Time = 0)
| Let $G(n)$ and $R(n)$ denote the amounts of gold and reputation after $n$ seconds, respectively. Then $G(0)=R(0)=0$ and $R(n)=\lfloor\tfrac{G(n)}{20}\rfloor$ and
$$G(n+1)=G(n)+10(1+0.1R(n))=G(n)+10(1+0.01\lfloor\tfrac{G(n)}{20}\rfloor).$$
If we ignore the floor function, this becomes $G(n+1)=1.005G(n)+10$, which can be written as
$$\begin{pmatrix}
G(n+1)\\1
\end{pmatrix}
=
\begin{pmatrix}
1.005&10\\
0&1
\end{pmatrix}
\begin{pmatrix}
G(n)\\1
\end{pmatrix}.$$
This gives us an explicit expression for $G(n+1)$ in terms of $n$;
$$\begin{pmatrix}
G(n+1)\\1
\end{pmatrix}
=
\begin{pmatrix}
1.005&10\\
0&1
\end{pmatrix}^{n+1}
\begin{pmatrix}
0\\1
\end{pmatrix}$$
Luckily, powers of this matrix are easy to compute because $\tbinom{0\ 10}{0\ \hphantom{1}0}^2=0$, so by the binomial theorem
\begin{eqnarray*}
\begin{pmatrix}
1.005&10\\
0&1
\end{pmatrix}^{n+1}
&=&\left(
\begin{pmatrix}
1.005&0\\
0&1
\end{pmatrix}
+\begin{pmatrix}
0&10\\
0&0
\end{pmatrix}\right)^{n+1}\\
&=&\begin{pmatrix}
1.005&0\\
0&1
\end{pmatrix}^{n+1}
+(n+1)\begin{pmatrix}
1.005&0\\
0&1
\end{pmatrix}^n
\begin{pmatrix}
0&10\\
0&0
\end{pmatrix}\\
&=&
\begin{pmatrix}
1.005^{n+1}&10(n+1)1.005^n\\
0&1
\end{pmatrix}
\end{eqnarray*}
This shows that
$$\begin{pmatrix}
G(n+1)\\1
\end{pmatrix}
=
\begin{pmatrix}
1.005^{n+1}&10(n+1)1.005^n\\
0&1
\end{pmatrix}
\begin{pmatrix}
0\\1
\end{pmatrix},$$
and hence that
$$G(n+1)=10(n+1)1.005^n.$$
In particular, after $n=1000$ seconds this gives us
$$G(1000)\approx1458464.$$
Keep in mind that we have ignored the floor function at the start; in this model you do not gain $1$ reputation for every $20$ gold in increments, but you simply gain $1/20$-th of your gold in reputation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does this series $\sum_{i=0}^n \frac{4}{3^n}$ diverge or converge? I a newbie to series, and I have not done too much yet. I have an exercise where I have basically to say if some series are convergent or divergent. If convergent, determine (and prove) the sum of the series.
This is the first series:
$$\sum_{i=0}^n \frac{4}{3^i}$$
I have heard about the ratio test, so I decided to try to apply it in this case to see if the series is or not convergent.
The ratio test is basically defined like this:
$$L = \lim{\left|\frac{a_{n+1}}{a_n}\right|}$$
Where $a_n$ is in this case $\frac{4}{3^n}$ and $a_{n+1}$ is consequently $\frac{4}{3^{n+1}}$. Thus, we have:
$$L = \lim{\left|\frac{\frac{4}{3^{n+1}}}{\frac{4}{3^n}}\right|} = \lim{\left|\frac{4 \cdot 3^n}{3^{n+1} \cdot 4}\right|} = \lim{\left|\frac{3^n}{3^{n+1}}\right|} = \lim{\left|\frac{3^n}{3^n\cdot 3}\right|} = \lim{\left|\frac{1}{3}\right|} = \frac{1}{3} < 1$$
From the ratio test, we know if $L< 1$, then the series converges.
Now, I need to find the sum. I have followed a video which explain how to find the sum using the ratio, which sincerely I have not understand well what it is. Can you explain a general rule to find the ratio, and what exactly is it?
Ok. So, following the same process done during the video, I have:
$$\sum_{i=0}^n \frac{4}{3^n} = 4 \sum_{i=0}^n \frac{1}{3^n} = 4 \sum_{i=0}^n \left(\frac{1}{3}\right)^n$$
Apparently, our ratio $r$ is $\frac{1}{3}$.
Now, to find the sum of the series, he uses a formula:
$$\frac{\text{first term}}{1 - r}$$
Where first term is the first term of the series. Can you explain where this formula comes from?
Applying that formula, we obtain:
$$\frac{\left(\frac{1}{3}\right)^0}{1 - \frac{1}{3}} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$$
I checked the sum using a online calculator, and apparently it is not $\frac{3}{2}$, so I decided to manipulate, and I thought I left a $4$ outside the series, so I multiply that $4$ by $\frac{3}{2}$, which produces $\frac{12}{2} = 6$, which is exactly the sum I found using 2 online calculators.
Now, my 3rd question is: why what I did is correct (or not correct)?
| Here is a simple proof for the geometric convergence:
Suppose we have the finite sum:
$$1+r+r^2+r^3+r^4+\dots+r^k$$ with $r\neq 1$
Since this is a finite sum, it will converge to some unknown quantity: $S$
So:
$$\begin{array}{lrl} S & =& 1+r+r^2+r^3+r^4+\dots+r^k\\
r\cdot S & =& ~~~~~~~ r+r^2+r^3+r^4+r^5+\dots+r^{k+1}\\
S-r\cdot S& = & 1 - r^{k+1}\\
S & = & \frac{1-r^{k+1}}{1-r}\end{array}$$
If $r=1$ then clearly $S = k+1$
Using this information for infinite series, it follows that $\sum\limits_{i=0}^\infty r^i = \lim\limits_{k\to\infty}\sum\limits_{i=0}^k r^i = \lim\limits_{k\to\infty} \frac{1-r^{k+1}}{1-r}$
In the case that $|r|>1$ this clearly diverges, and when $|r|<1$ then the $r^{k+1}$ term is insignificant, so this will converge. The case where $r=1$ followed the other form where $S = k+1$ and will clearly diverge, the case where $r=-1$ will also diverge since the sequence of partial sums will alternate between 1 and 0.
As for a general rule to find the ratio, in a geometric series, it will be of the form $\sum a \cdot r^n$. Try to algebraically manipulate the summand to be in that form. For example: $\sum 3\cdot \frac{5^n}{6^{n+1}} = \sum 3\cdot \frac{5^n}{6^n\cdot 6} = \sum \frac{3}{6}\cdot (\frac{5}{6})^n$, and your ratio will be $\frac{5}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find the derivative of the function $F(x) = \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt$. $$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+t^4}}2x - \frac{1}{\sqrt{2+t^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+t^4}} - \frac{\sec^2{x}}{\sqrt{2+t^4}} \\
&= \frac{2x-\sec^2{x}}{\sqrt{2+t^4}} \\
\end{align}$$
Update:
Is this looking better??
$$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+(x^2)^4}}2x - \frac{1}{\sqrt{2+(\tan{x})^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+x^8}} - \frac{\sec^2{x}}{\sqrt{2+\tan^4{x}}} \\
\end{align}$$
Is this correct?
| $\text{ Recall } \int_{a(x)}^{b(x)}f(t) dt=F(b(x))-F(a(x)) \text{ *note:} \text{( where } F'=f) \\ \frac{d}{dx}[F(b(x))-F(a(x))]=\frac{d}{dx}F(b(x))-\frac{d}{dx}F(a(x)) \text{ by difference rule } \\ \\ \text{ now you use chain rule and you are almost there } $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Proving $(n+1)!>2^{n+3}$ for all $n\geq 5$ by induction I am stuck writing the body a PMI I have been working on for quite some time.
Theorem: $∀n∈N ≥ X$, $(n+1)!>2^{n+3}$
I will first verify that the hypothesis is true for at least one value of $n∈N$.
Consider $n=3$: (not valid)
$$(3+1)!>2^{3+3} \implies 4!>2^{6} \implies 24>64$$
Consider $n=4$: (not valid)
$$(4+1)!>2^{4+3} \implies 5!>2^{7} \implies 120>128$$
Consider $n=5$: (valid)
$$(5+1)!>2^{5+3} \implies 6!>2^{8} \implies 720>256$$
Consider $n=6$: (valid)
$$(6+1)!>2^{6+3} \implies 7!>2^{9} \implies 5040>512$$
So clearly $X$ is $5$.
For the inductive assumption, we will assume the hypothesis holes from $n=5$ up to some arbitrary values $k$: $(k+1)!>2^{k+3}$
This is where I am lost. Originally I had written:
Now I will prove true for $k+1$ showing that: $(k+2)!>2^{k+4}$. Consider the $k+1$ term:
$$(k+2)(k+1)k! = (k+2)(k+1)2^k = (k+2)(k+1)2^k = (2)(2)(2^k) = (4)(2^k) = 2^{k+4}$$
by the inductive assumption since $k>5$ so $k+1>2$ and $k+2>2$ but I know this isn't correct.
I know the proof should look something like this, but I have no idea why:
$$(k+2)(k+1)!>(k+2)(2^{k+3})>2(2^{k+3})=2^{k+4}$$
| For $n\geq 5$, let $S(n)$ denote the statement
$$
S(n) : (n+1)! > 2^{n+3}.
$$
Base step $(n=5)$: $S(5)$ says $6!>2^8$, and this is true since $720>256$.
Inductive step: Fix some $k\geq 5$ and suppose that
$$
S(k) : (k+1)!>2^{k+3}
$$
is true. We must now prove that $S(k+1)$ follows where
$$
S(k+1) : (k+2)! > 2^{k+4}.
$$
Starting with the left-hand side of $S(k+1)$,
\begin{align}
(k+2)! &= (k+1)!\cdot(k+2)\tag{by definition}\\[0.5em]
&> 2^{k+3}(k+2)\tag{by $S(k)$}\\[0.5em]
&\geq 2^{k+3}\cdot 7\tag{since $k\geq 5$}\\[0.5em]
&> 2^{k+3}\cdot 2\\[0.5em]
&= 2^{k+4},
\end{align}
we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)\to S(k+1)$, completing the inductive step.
Thus, by mathematical induction, for all $n\geq 5$, the statement $S(n)$ is true. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1214914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Proof by induction: For all $n \geq 1$; $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots +(-1)^{n+1} \frac{1}{n} \leq 1$ Proof by induction: For all $n \geq 1$; $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots +(-1)^{n+1} \frac{1}{n} \leq 1$
This is what I have so far:
Base case: for $n = 1$
$(-1)^{1+1} \cdot \frac{1}{1} \leq 1$
$(-1)^2 \leq 1$
$1 \leq 1$
This holds true.
Inductive step: for $n = k$
Assume $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots + (-1)^{k+1} \frac{1}{k} \leq 1$ is true.
We want to show for $K=1$
$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots + (-1)^{k+1} \frac{1}{k} - (-1)^{k+2} \cdot \frac{1}{K+1} \leq 1$
$(-1)^{k+1} \frac{1}{k} - (-1)^{k+2} \cdot \frac{1}{K+1} \leq 1$
By doing some algebra I get this:
$-\dfrac{(-1)^k + (-1)^k k + (-1)^K k)}{k(1 + k)} \leq 1$
I am lost here, not sure if this says anything. Any help would be appreciated. Thanks in advance.
| Hint: rewrite the sum as
$$
1-\left(\frac12-\frac13\right)-\left(\frac14-\frac15\right)-\left(\frac16-\frac17\right)-\dots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the least positive integer with $24$ positive divisors.
Find the least positive integer with $24$ positive divisors.
My attempt:
$24=2^3.3$. We shall have to find out a positive integer (least) $n$ such that $N$ has $24$ positive divisors i.e we have to find $N$ where $\tau(N)=24.$.
We have, if $N=p_{1}^{\alpha_2}p_{2}^{\alpha_2}\dots p_{n}^{\alpha_n}$ then $\tau(N)=(\alpha_1+1)(\alpha_2+1)\dots (\alpha_n+1)$ where $p_i$, $i=1,2,\dots n$ are distinct primes and $\alpha_i$'s are all +ve.
Here $\tau(N)=(\alpha_1+1)(\alpha_2+1)\dots (\alpha_n+1)=24$. Now the problem is to find $\alpha_i$'s only.
The all possible factorizations are the following:
$24=1.24;~~ 24=2.12;~~ 24=3.8;~~ 24=4.6$. But how to get $\alpha_i$'s?
| We use the fact $p_1^{\alpha_1}p_2^{\alpha_2}\dots p_n^{\alpha_n}$ has $(\alpha_1+1)(\alpha_2+1)\dots(\alpha_n+1)$ divisors, and check minimum integer for each factorization, the possible factorizations for $24=2\cdot2\cdot2\cdot3$ are the following:
$2\cdot2\cdot2\cdot3$ minimum for this factorization is $2^2\cdot3\cdot5\cdot7=420$
$6\cdot2\cdot2$ minimum is $2^5\cdot3\cdot5=480$
$4\cdot2\cdot3$ minimum is $2^3\cdot3^2\cdot5=360$
$4\cdot 6$ minimum is $2^5\cdot3^3=864$
$12\cdot 2$ minimum is $2^{11}\cdot3>1000$
$8\cdot 3$ minimum is $2^{7}\cdot3^2>1000$
$24$ minimum is $2^{23}>1000000$
the minimum of the numbers in the right is $360$, which is our answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
limit of $\frac{xy-2y}{x^2+y^2-4x+4}$ as $(x,y)$ tends to $(2,0)$ Am I able to substitute $x$ by ($k$+$2$) with $k$ tending to $0$, then using polar coordinates to deduce its limit?!
\begin{equation*}
\lim_{(x,y)\to(2,
0)}{xy-2y\over x^2+y^2-4x+4}.
\end{equation*}
| Take polar coorinates: $$x=2+r \cos{\theta}$$ $$y=r \sin{\theta}$$
$r \rightarrow 0 \Rightarrow x \rightarrow 2$
Thus $$\lim_{r \rightarrow 0}f(r, \theta)=\lim_{r \rightarrow 0} \frac{(2+r\cos{\theta}-2)(r \sin{\theta})}{(2+r\cos{\theta})^2+r^2 \sin^2{\theta}-4(2+r\cos{\theta})+4}= \lim_{r \rightarrow 0} \frac{r^2 \cos{\theta} \sin{\theta}}{4+2r\cos{\theta}+r^2\cos^2{\theta}+r^2\sin{\theta}-8-4r\cos{\theta}+4}= \lim_{r \rightarrow 0} \frac{r^2\cos{\theta}\sin{\theta}}{r^2}=\cos{\theta}\sin{\theta}$$
This limit does not exist because for different values of $\theta$ we have different results.
Take for instance $$\theta=2 \pi$$ $$\theta= \frac{\pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1218520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integration By Parts (Logarithm) $$\int(2x+3)\ln (x)dx$$
My attempts,
$$=\int(2x\ln (x)+3\ln (x))dx$$
$$=2\int x\ln (x)dx+3\int \ln(x)dx$$
For $x\ln (x)$, integrate by parts,then I got
$$=x^2\ln (x)-\int (x) dx+3\int \ln(x)dx$$
$$=x^2\ln (x)-\frac{x^2}{2}+3\int \ln(x)dx$$
For $\ln(x)$, integrate by parts, then I got
$$=x^2\ln (x)-\frac{x^2}{2}+3x\ln (x)-3\int 1dx$$
$$=x^2\ln (x)-\frac{x^2}{2}+3x\ln (x)-3x+c$$
$$=\frac{1}{2}x(-x+2(x+3)\ln (x)-6)+c$$
But the given answer in book is $x^2\ln (x)-\frac{x^2}{2}+\frac{3}{x}+c$. What did I do wrong?
| Bookish answer is wrong:
$$\left(x^2\ln (x)-\frac{x^2}{2}+\frac{3}{x}+c\right)'=2x\log x-\frac3{x^2}$$
And yours:
$$\left(\frac{1}{2}x(-x+2(x+3)\ln (x)-6)+c\right)'=2x\log x+3\log x=(2x+3)\log x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $ 1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ for all $n\geq 2$ by induction Question:
Let $P(n)$ be the statement that $1+\dfrac{1}{4}+\dfrac{1}{9}+\cdots +\dfrac{1}{n^2} <2- \dfrac{1}{n}$. Prove by mathematical induction. Use $P(2)$ for base case.
Attempt at solution:
So I plugged in $P(2)$ for the base case, providing me with $\dfrac{1}{4} < \dfrac{3}{2}$ , which is true.
I assume $P(n)$ is true, so I need to prove $P(k) \implies P(k+1)$.
So $\dfrac{1}{(k+1)^2} < 2 - \dfrac{1}{k+1}$.
I don't know where to go from here, do I assume that by the Inductive hypothesis that it's true?
| Your base case is wrong. You should realize it's true since $\frac{5}{4}<\frac{3}{2}$ obtained from $$1+\frac{1}{2^2} = \frac{5}{4} < \frac{3}{2} = 2-\frac{1}{2}$$ For the induction step, suppose $P(n)$ is true for all $n \in \{1,2,\ldots,k\}$. Then $$\begin{align}1+\frac{1}{2^2}+\ldots+\frac{1}{k^2}+\frac{1}{(k+1)^2} = \left(1+\frac{1}{2^2}+\ldots+\frac{1}{k^2}\right)+\frac{1}{(k+1)^2} \\ < \left(2-\frac{1}{k} \right)+\frac{1}{(k+1)^2}\end{align}$$ Now if you can show that $$\left(2-\frac{1}{k} \right)+\frac{1}{(k+1)^2}<2-\frac{1}{k+1}$$ you are done. It is possible to get to that inequality simply starting with the fact that $\frac{1}{k+1}<\frac{1}{k}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Why is the sum of the digits in a multiple of 9 also a multiple of 9? The sum of the digits in $9 k$ (where $k$ is an integer) is a multiple of $9$: for example
$$9\cdot 1=9$$
$$9\cdot 7=63 \qquad \text{and } 6+3=9\cdot 1$$
$$9\cdot 11=99 \qquad \text{and } 9+9=9\cdot 2$$
But why?
| It is because $10$ gives $1$ as remainder when dividing by $9$. This can be expressed by the general notation of 'congruence'
$$10\equiv 1\pmod9$$
(that expresses that these two numbers give the same remainder when dividing by $9$).
Then, it easily follows from the property of these congruences that they behave like equality with respect to addition and multiplication, that $10^n\equiv 1^n=1\pmod9$.
Alternatively, of course $10^n$ will also give remainder $1$ modulo $9$, as the number $99\dots9$ is divisible by $9$.
Now, a number $n=\overline{abc\dots}$ is just $n=\left(((a\cdot10+b)\cdot10+c)\cdot 10+\dots\right)$, so modulo $9$ we have the following congruence:
$$n\equiv \left(((a\cdot 1+b)\cdot1+c)\cdot1+\dots\right)=a+b+c+\dots,$$
i.e., $n$ gives the same remainder modulo $9$ as the sum of its digits. In particular, it gives remainder $0$ (divisible by $9$) iff the sum of its digits is divisible by $9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
A bessel function integral $$\int_{-\infty}^{\infty} dy \frac{J_1 \left ( \pi\sqrt{x^2+y^2} \right )}{\sqrt{x^2+y^2}} = \frac{2 \sin{\pi x}}{\pi x} $$
How do I show this?
| This is going to be a little involved. Begin with the representation
$$\int_{-1}^1 du \frac{u}{\sqrt{1-u^2}} \sin{r u} = \pi J_1(r) $$
which may be obtained by differentiating the well-known relation
$$\int_{-1}^1 du \frac{\cos{r u}}{\sqrt{1-u^2}} = \pi J_0(r) $$
Then
$$\begin{align}\int_{-\infty}^{\infty} dy \frac{J_1 \left ( \sqrt{x^2+y^2} \right )}{\sqrt{x^2+y^2}} &= \frac1{\pi} \int_{-1}^1 du \frac{u}{\sqrt{1-u^2}} \int_{-\infty}^{\infty} dy \frac{\sin \left ( u\sqrt{x^2+y^2} \right )}{\sqrt{x^2+y^2}} \\ &= \frac{2}{\pi}\int_{-1}^1 du \frac{u}{\sqrt{1-u^2}} \int_1^{\infty} dv \frac{\sin{(u x v)}}{\sqrt{v^2-1}} \end{align}$$
One may show through a Laplace transform that the inner integral has a value of $\frac{\pi}{2} J_0(x u) \operatorname{sgn}{(x u)}$. The integral we seek is then
$$\begin{align}\int_{-\infty}^{\infty} dy \frac{J_1 \left ( \sqrt{x^2+y^2} \right )}{\sqrt{x^2+y^2}} &= 2 \int_0^1 du \frac{u}{\sqrt{1-u^2}} J_0(x u) \\ &= 2 \sum_{m=0}^{\infty} \frac{(-1)^m}{m!^2 2^{2 m}} x^{2 m} \int_0^1 du \frac{u^{2 m+1}}{\sqrt{1-u^2}}\\ &= 2 \sum_{m=0}^{\infty} \frac{(-1)^m}{m!^2 2^{2 m}} x^{2 m} \frac{2^{2 m}}{(2 m+1) \binom{2 m}{m}}\\ &= 2 \sum_{m=0}^{\infty} (-1)^m \frac{x^{2 m}}{(2 m+1)!} \\ &= \frac{2 \sin{x}}{x}\end{align} $$
ADDENDUM
It is straightforward to show that the Laplace transform of $J_0(t)$ is $(1+s^2)^{-1/2}$. (One may use the series representation of $J_0$.)
We may derive the integral above by considering the computation of the inverse Laplace transform of $(1+s^2)^{-1/2}$. Consider
$$\oint_C dz \frac{e^{z t}}{\sqrt{1+z^2}} $$
where $C$ is defined by the following contour:
where the Bromwich contour is deformed to avoid the branch points at $\pm i$. In the limit of the radius of the outer arcs goes to infinity and the radius of the smaller circles about the branch points goes to zero, we have that
$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{1+s^2}} + i \int_{\infty}^1 dy \frac{e^{i y t}}{\sqrt[+]{1-y^2}} + i \int_1^{\infty} dy \frac{e^{i y t}}{\sqrt[-]{1-y^2}} \\ - i \int_{\infty}^1 dy \frac{e^{-i y t}}{\sqrt[+]{1-y^2}} - i \int_1^{\infty} dy \frac{e^{-i y t}}{\sqrt[-]{1-y^2}} $$
where $\sqrt[+]{1-y^2} = i \sqrt{y^2-1}$ and $\sqrt[-]{1-y^2} = -i \sqrt{y^2-1}$ represent the different branches of the square root. We thus have, by Cauchy's theorem, that the contour integral is zero so that the ILT is
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{1+s^2}} = J_0(t) = \frac{2}{\pi} \int_1^{\infty} dy \frac{\sin{y t}}{\sqrt{y^2-1}} $$
for $t \gt 0$, as was to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1221767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
$ax^2+bx+c=0$ has roots $x_1,x_2$. what are the roots of $cx^2+bx+a=0$. Given solution: Dividing the first equation by $x^2$ we get
$c(\frac{1}{x^2})+b(\frac{1}{x})+a=0$
so $(\frac{1}{x_1}),(\frac{1}{x_2})$ are the roots of $cx^2+bx+a=0$.{How?It is not obvious to me.}
The answers so far are proving retrospectively that the roots are indeed those given above.I would like to know how the relation of the roots is derived.
| The definition says: $t$ is a root of a polynomial $P$ if and only if $P(t)=0$.
so let $P(x)=cx^2+bx+a$, we have :
$$P(\frac{1}{x_1})=c(\frac{1}{x_1^2})+b(\frac{1}{x_1})+a=\frac{c+bx_1+ax_1^2}{(x_1)^2}$$
and as we know $x_1$ is a root of $ax^2+bx+c=0$ hence $ax_1^2+bx_1+c=0$ and finally:
$P(\frac{1}{x_1})=0$
as a conclusion $\frac{1}{x_1}$ is a root of $cx^2+bx+a$.
You can do the same for $\frac{1}{x_2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $10n^8 - 9n^6 - n^2$ is divisible by $45$ Basically, I have to use Euler theorem to prove that
$10n^8 -9n^6 -n^2$ is divisible by $45$
So my approach so far is to say
$10n^8 - 9n^6 - n^2 = 0 \bmod 45$
Now $45$ can be factored into $5$ and $9$
So I figure I need to do something like
$10n^8 - 9n^6 - n^2 \bmod 5$
$10n^8 - 9n^6 - n^2 \bmod 9$
But I am not sure how to exactly prove it.
I know how to prove something like $n^{13} - n$ is a multiple of $2730$ using Fermat's theorem, but the proof works nicely because the exponent is a prime. My approach to $n^{13} - n$ would consist of proving that $n^{13} - n$ is divisible by $\{2, 3, 5, 7, 9, 13\}$.
What is throwing me off is the coefficients in $10n^8 - 9n^6 - n^2$ and the fact that the exponents are not primes.
| HINT
$$10n^8 - 9n^6 - n^2\equiv 0n^8-(-1)n^6 - n^2 \equiv n^6 - n^2 \pmod 5$$
By little fermat we have $n^5\equiv n \pmod 5$, therefore
$$n^6-n^2\equiv n\cdot n^5-n^2\equiv n\cdot n-n^2\equiv n^2-n^2\equiv 0\pmod{5}$$
See if you can work the other mod similarly
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\int_0^{\frac{\pi}{2}}\frac{x^2\cdot\log\sin x}{\sin^2 x}dx$ using a very cute way? Few days ago my friend gave me this integral and i cant get how to solve this. The integral is:$$\int_0^{\large \frac{\pi}{2}}\frac{x^2\cdot\log{{\sin{x}}}}{\sin^2{x}}dx$$
| Hint. You may observe that
$$
\frac{\log{{\sin{x}}}}{\sin^2{x}}=\left( -\frac{x^2}{2}-\log \sin x-\frac{1}{2} \log^2 \sin x\right)''
$$ thus integrating by parts twice, using $(x^2)''=2$, leads to
$$
\begin{align}
\int_0^{\pi/2}\frac{x^2\cdot\log{{\sin{x}}}}{\sin^2{x}}dx&=2\int_0^{\pi/2}\left( -\frac{x^2}{2}-\log \sin x-\frac{1}{2} \log^2 \sin x\right)dx\\\\
&=-\int_0^{\pi/2}x^2dx-2\int_0^{\pi/2}\log \sin x dx-\int_0^{\pi/2}\log^2 \sin x dx\\\\
&=-\frac{\pi ^3}{24}+\pi \ln 2-\frac{1}{24} \pi \left(\pi ^2+12 \ln^2 2\right)
\end{align}
$$ that is
$$
\int_0^{\pi/2}\frac{x^2\cdot\log{{\sin{x}}}}{\sin^2{x}}dx=-\frac{\pi ^3}{12}+\pi \ln 2-\frac{1}{2} \pi \ln^2 2,
$$
where we have used standard evaluations for the last integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
quadratic equation what am I doing wrong? solve
$$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} $$
$$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} \Rightarrow $$
$$ 5x+19 = (x+7) + 4\sqrt{x-5}\sqrt{x+7} + (x+5) \Rightarrow $$
$$ 3x + 17 = 4\sqrt{x-5}\sqrt{x+7} \Rightarrow $$
$$ 9x^2 + 102x + 289 = 16(x+7)(x-5) \Rightarrow $$
$$ 9x^2 + 102x + 289 = 16(x^2+ 2x - 35) \Rightarrow $$
$$ 7x^2 - 70x - 849 = 0 \Rightarrow $$
$$ b^2 - 4ac = (-70)^2 - 4 \cdot 7 \cdot (-849) = 28672=2^{12}\cdot7 $$
then I calculate the solution using the discriminant as
$$ 5 + 32\frac{\sqrt7}7 $$
and
$$ 5 - 32\frac{\sqrt7}7 $$
but when I plug in the values I find out that they are wrong,
does it have to do with the fact that I square the equation twice?
if so what is the best way to go about solving this equation?
| You forgot a factor $4$ and wrote $x+5$ instead of $x-5$ in the third line, which should be
$$
5x+19=x+7+4\sqrt{x+7}\,\sqrt{x-5}+4(x-5)
$$
giving
$$
4\sqrt{x+7}\,\sqrt{x-5}=32
$$
or
$$
\sqrt{x+7}\,\sqrt{x-5}=8
$$
that becomes, after squaring,
$$
x^2+2x-99=0
$$
The roots of this are $-11$ and $9$, but only the latter is a solution of the original equation, because the existence conditions on the radicals give
\begin{cases}
5x+19\ge0\\[3px]
x+7\ge0\\[3px]
x-5\ge0
\end{cases}
that is, $x\ge5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Limit of $x^2e^x $as $x$ approaches negative infinity without using L'hopital's rule I'm trying to find the limit of $e^x x^2$ as $x$ approaches negative infinity:
$$\lim_{x\to-\infty}e^xx^2$$
without using L'hopital's rule. Any suggestions?
| $$\lim_{x\rightarrow -\infty} x^2e^x\\y=-x\\so \\\lim_{y\rightarrow +\infty} (-y)^2e^{(-y)}=\\\lim_{y\rightarrow +\infty} \frac{y^2}{e^y}=?$$ as you know :$e^y=1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+...$ now see $$\lim_{y\rightarrow +\infty} \frac{y^2}{e^y}= \\lim_{y\rightarrow +\infty} \frac{y^2}{1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+...}$$ now $$0 \leq \frac{y^2}{1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+...} \leq \frac{y^2}{\frac{y^3}{3!}+\frac{y^4}{4!}+...}=\frac{y^2}{y^3(\frac{1}{3!}+\frac{y}{4!}+...)} \rightarrow 0 $$ so by squeeze theorem ,this lim goes to zero
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Polynomials mod prime $p$ The problem is $5m^2+m+4 \equiv 0\pmod 7$. I am supposed to first convert it to a quadratic whose first coefficient is $1$. But the polynomial cannot be factored, so I am unsure as to how to do this. Then, I am supposed to use the process of completing the square to convert it to a congruence of the form $y^2 \equiv a\pmod p$.
Can someone help me through this process?
| Multiply by the inverse of $5$, which is $3$:
$$ 3 \times ( 5 m^2+m+4) \equiv 0 \pmod{7}, $$
which becomes
$$ m^2+3m+5 \equiv 0 \pmod{7} $$
Completing the square means finding $a$ and $b$ so the following is true
$$ m^2+3m+5 \equiv (m+a)^2+b \pmod{7} $$
Expanding the bracket,
$$ 3m+5 \equiv 2am + (a^2+b) \pmod{7} $$
The inverse of $2$ is $4$, so
$$ a = 4 \times 3 \equiv 5 \pmod{7}, $$
and
$$ 5 \equiv 5^2 +b \equiv 4+b \pmod{7}, $$
Hence $b \equiv 1 \pmod{7}$, and the answer is
$$ (m+5)^2 \equiv -1 \equiv 6 \pmod{7} $$
Now you just have to find the square roots of $6$. At this point, as @hardmath pointed out, we have a problem, since $6$ is not a square modulo $7$ (easy to check by computing them all, or there's some theorem about $-1$ being a square only if the prime is $1 \pmod{4}$, IIRC). Therefore there are no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Maximum value of $ x^2 + y^2 $ given $4 x^4 + 9 y^4 = 64$ It is given that $4 x^4 + 9 y^4 = 64$.
Then what will be the maximum value of $x^2 + y^2$?
I have done it using the sides of a right-angled triangle be $2x , 3y $ and hypotenuse as 8 .
| One more stab at the problem. Instead of working with $x$, let's work with $u = x^2$ as it makes the algebra slightly easier.
As $4x^4 + 9y^4 = 64$, we have $y^2 = \frac{1}{3}\sqrt{64-4u^2}$, taking only the positive square roots as $y^2 \geq 0$. Hence want to maximize
$$f(u) = x^2 + y^2 = u + y^2 = u + \frac{1}{3}\sqrt{64-4u^2}$$
Setting the first derivative equal to zero,
$$0 = \frac{df}{du} = 1 - \frac{4}{3}u \frac{1}{\sqrt{64-4u^2}}$$
$$\Longrightarrow \ 64 - 4u^2 = \frac{16}{9}u^2 \hspace{30 mm}$$
$$\Longrightarrow \ \ u^2 = \frac{9}{52}\cdot 64 \ \ \text{ i.e., } u = \frac{12}{\sqrt{13}} \hspace{10 mm}$$
(We take only the positive square root in the last step as $u = x^2 \geq 0$.)
Using the earlier expression for $y^2$, we find for $u = x^2 = \frac{12}{\sqrt{13}}$ that
$$y^2 = \frac{1}{3}\sqrt{64 - 4\cdot \frac{144}{13}} = \frac{16}{3\sqrt{13}}$$
Hence
$$x^2 + y^2 = \frac{12}{\sqrt{13}} + \frac{16}{3\sqrt{13}} = \frac{52}{3\sqrt{13}} = \frac{4}{3}\sqrt{13}$$
This is a maximum as there are other points satisfying the constraint for which $x^2 + y^2 < \frac{4}{3}\sqrt{13}$. For example, for $(x,y) = (2,0)$, $$x^2 + y^2 = 2^2 + 0^2 = 4 < \frac{4}{3}\sqrt{13} \ \text{ as } 16 < \frac{16}{9}\cdot 13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Sufficient conditions to guarantee a series is Cesaro summable Is is true that every nonnegative, bounded series in $R$ is Cesaro summable?
Is there a list of sufficient conditions on series to guarantee that it is Cesaro summable?
| In the other case, if you mean the partial sums are non-negative and bounded, no, it is not true that every non-negative bounded series in R is Cesaro summable! This is from somewhere in Hardy's brilliant book 'Divergent Series', look for 'dilution'.
take $a_n = (-1)^n$ and $c_n = \begin{cases}a_k & n=2^k \\ 0 & \text{else} \end{cases} = \begin{cases} (-1)^k & n=2^k \\ 0 & \text{else} \end{cases} $.
We have $$ \sum c_n = 0 + 1 -1 + 0 + 1 + 0 + 0 + 0 -1 + 0 + …$$
By writing $\sigma_k = \sum^k_0 c_n$, we have
$$ \sigma_i = 0,1,0,0,1,1,1,1,0,0,...$$
That is, $$\sigma_1 = 1, \sigma_2 = 0, \sigma_4 = 1, \sigma_8 = 0, …$$ and in general for natural $k$, $\sigma_{2^k} = s_k := \begin{cases} 1 & k \text{ even} \\ 0 & k \text{ odd}\end{cases}$. Also, for each $j$ where $2^k \leq j < 2^{k+1}$, we have $\sigma_j = \sigma_{2^k} = s_k$ since the only non-zero term after the $2^k$th term is the $2^{k+1}$th term. That is,
$$ \sigma_n
= \begin{cases}1 & 2^{k}\leq n<2^{k+1}\text{ for some } k\text{ even} \\
0 & \text{else}\end{cases}
$$
Trying to obtain its $(C,1)$ sum causes problems as follows. Suppose $k=2\kappa+1$ is odd. Then by reasoning similar to the inclusion-exclusion principle from Combinatorics,
\begin{align*}
\frac{\sum_0^{2^k-1} \sigma_n}{2^k}
&=\frac{2^k - 2^{k-1} + \dots + 2 - 1}{2^k} \\
& = \frac{1}{3}\frac{2^{k+1}-1}{2^{k}} \xrightarrow[\kappa\to\infty]{} \frac{2}{3}
\end{align*}
but if $k=2\kappa$ is even, then $\sigma_{2^{k-1}},\sigma_{2^{k-1}+1},...\sigma_{2^k-1}$ are all zeros so we instead have
\begin{align}
\frac{\sum_0^{2^k-1} \sigma_n}{2^k}
&=\frac{2^{k-1} - 2^{k-2} + \dots + 2 - 1}{2^k} \\
& = \frac{1}{3}\frac{2^{k}-1}{2^{k}} \xrightarrow[\kappa\to\infty]{} \frac{1}{3}
\end{align}
So the average of the partial sums of $c_n$ will oscillate between $\frac{1}{3}$ and $\frac{2}{3}$, and the Cesaro sum cannot exist. Coincidentally this series is not Abel summable as well.
I see your edit now but I will leave this here in case it is helpful to someone else.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all possible two-way associations/relations between four numbers Given four numbers {1,2,3,4}, how to find all possible two-way associations/relations between them? I calculate them manually as in below (50 in total) but I would like to know whether a mathematical formula exists to find them? And how about generalizing the formula for n given numbers?
{1} --> {2}, {2} --> {1},
{1} --> {3}, {3} --> {1}
{1} --> {4}, {4} --> {1}
{2} --> {3}, {3} --> {2}
{2} --> {4}, {4} --> {2}
{3} --> {4}, {4} --> {3}
{1} --> {2,3}, {2,3} --> {1},
{1} --> {2,4}, {2,4} --> {1},
{1} --> {3,4}, {3,4} --> {1},
{2} --> {1,3}, {1,3} --> {2},
{2} --> {1,4}, {1,4} --> {2},
{2} --> {3,4}, {3,4} --> {2},
{3} --> {1,2}, {1,2} --> {3},
{3} --> {1,4}, {1,4} --> {3},
{3} --> {2,4}, {2,4} --> {3},
{4} --> {1,2}, {1,2} --> {4},
{4} --> {1,3}, {1,3} --> {4},
{4} --> {2,3}, {2,3} --> {4},
{1} --> {2,3,4}, {2,3,4} --> {1},
{2} --> {1,3,4}, {1,3,4} --> {2},
{3} --> {1,2,4}, {1,2,4} --> {3},
{4} --> {1,2,3}, {1,2,3} --> {4},
{1, 2} --> {3, 4}, {3, 4} --> {1, 2},
{1, 3} --> {2, 4}, {2, 4} --> {1, 3},
{1, 4} --> {2, 3}, {2, 3} --> {1, 4},
| Create the set $A$ by choosing at least $1$ element from $\{1, 2, 3, ..., n\}$ and leaving at least one for $B$. Suppose we chose $k$ elements ($1\leq k\leq n-1$). Now, from the remaining elements, we need to choose elements for $B$. There are $n-k\geq 1$ elements left and we must choose at least one of them. Say we choose $\ell\geq 1$ of them. There are $\binom{n}{k}\binom{n-k}{\ell}$ ways to do it as described. Now, we just need to vary $k$ and $\ell$ over the possible values they could take on (keeping in mind that the possibilities for $\ell$ depend on the choice of $k$). We end up with:
$$ \sum_{k=1}^{n-1} \binom{n}{k}\sum_{\ell=1}^{n-k}\binom{n-k}{\ell}.$$
To check that this works with $n=4$ as you gave, we have:
\begin{align*}
&\binom{4}{1}\left(\binom{3}{1}+\binom{3}{2}+\binom{3}{3}\right)+\binom{4}{2}\left(\binom{2}{1}+\binom{2}{2}\right) + \binom{4}{3}\binom{1}{1} \\ &= 4(3+3+1)+6(2+1)+4(1) \\
&=4(7)+6(3)+4 \\
&=28+18+4 \\
&= 50
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use the generating function to solve a recurrence relation We have the recurrence relation $\displaystyle a_n = a_{n-1} + 2(n-1)$ for $n \geq 2$, with $a_1 = 2$.
Now I have to show that $\displaystyle a_n = n^2 - n +2$, with $n \geq 1$ using the generating function.
The theory in my book is scanty, so with the help of the internet I have the following:
$\displaystyle \sum_{n = 2} ^\infty (a_n - a_{n-1}) x^n = \sum_{n = 2} ^\infty a_n x^n - \sum_{n = 2} ^\infty a_{n-1} x^n = \sum_{n = 2} ^\infty a_n x^n - x \sum_{n = 2} ^\infty a_{n-1} x^{n-1} = (a(x) - a_1 x) - x(a(x)) = (a(x) - 2 x) - x(a(x)) = a(x) (1-x) - 2x$
But how I have to work out $\displaystyle \sum_{n = 2} ^\infty 2(n-1) x^n$ ?
If I have this, an expression for $a(x)$ can be found. How should $a_n$ be found from $a(x)$?
| We have $a_n - a_{n-1} - 2n +2 = 0 \ (\star)$. Suppose the GF of $\langle a_n \rangle_{n\ge 1}$ is $f(x)$.
Then,
$$\begin{align*}
f(x) &= a_1 + &a_2 x& + a_3 x^2 + \cdots + a_n x^n + \cdots \\
-xf(x) &= &-a_1 x& - a_2 x^2 - \cdots - a_{n-1}x^n - \cdots\\
\frac{-2x}{(1-x)^2} &= &-2 x& - 4 x^2 \ \ - \cdots - 2n x^n + \cdots \\
\frac{2}{1-x} &= 2 + &2x& + 2x^2 \ \ + \cdots +2x^n + \cdots
\end{align*}$$
Adding these up:
$$f(x) -xf(x)-\frac{2x}{(1-x)^2} + \frac{2}{1-x} = (a_1 + 2) + \sum_{k=2}^\infty(a_k - a_{k-1} - 2k + 2)x^k.$$
But by $(\star)$, every term in the above infinite sum is $0$. Also our initial condition is $a_1 = 2$.
So,
$$f(x) (1-x) -\frac{2x}{(1-x)^2} + \frac{2}{1-x} = 4.$$
Solving for $f(x)$, we have
$$f(x) = \frac{-4x^2+4x-2}{(x-1)^3} = \frac{-4x^2}{(x-1)^3} + \frac{4x}{(x-1)^3} - \frac{2}{(x-1)^3}.$$
The power series representation of $f(x)$ is then
$$f(x) = \sum_{n=0}^\infty \left( n^2-n+2 \right) x^n.$$
But the generating function for $\langle a_n \rangle$ was $f(x)$. So we can conclude that
$$a_n = n^2 -n +2.$$
As you can see, using generating functions to solve recurrences is tedious, and requires a hefty amount of algebraic manipulation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Exploring $ \sum_{n=0}^\infty \frac{n^p}{n!} = B_pe$, particularly $p = 2$. I was exploring the fact that $$ \sum_{n=0}^\infty \frac{n^p}{n!} = B_pe,$$
where $B_n$ is the $n$th Bell number.
I found this result by exploring the series on wolframalpha and looking up the sequence of numbers generated. I have no experience with Bell numbers other than knowing that they represent the number of ways of partitioning a set.
What I tried
Looking at the base case we can verify
$$\sum_{n=0}^\infty \frac{n}{n!} = \sum_{n=1}^\infty \frac{1}{(n-1)!} = e. $$
But then I realize that $n^2$ is going to be difficult.
Specifically what I want
I am doing this for fun, so I just want to glean some sort of lesson out of this. Resources that let me learn for myself are just as good!
An accepted answer will have any one of the following:
*
*A proof that $\sum_{n = 0}^{\infty} \frac{n^2}{n!} = 2e$ (this interests me a lot)
*A description of why this series is related to Bell numbers and the number of partitions of a set
*A link to some resource that introduces Bell numbers and/or their connection to this series.
| Notice that
\begin{align}
S
& = \sum_{n = 0}^{\infty} \frac{n^{2}}{n!} \\
& = \frac{0^{2}}{0!} + \frac{1^{2}}{1!} + \frac{2^{2}}{2!} + \frac{3^{2}}{3!} +
\cdots \\
& = \frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \cdots.
\end{align}
Then
\begin{align}
S - e
& = \left(
\frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \cdots
\right) -
\left(
\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots
\right) \\
& = \frac{1}{1!} + \frac{2}{2!} + \frac{3}{3!} + \cdots \\
& = e. \qquad (\text{As shown in the OP.}) \\
\end{align}
Therefore, $ S = e + e = 2 e $.
In fact, using the same type of reasoning, it can be shown that $ \displaystyle \sum_{n = 0}^{\infty} \frac{n^{3}}{n!} = 5 e $.
Here is the connection with Bell numbers.
For each $ p \in \mathbb{N}_{0} $, let $ \displaystyle S_{p} \stackrel{\text{df}}{=} \sum_{n = 0}^{\infty} \frac{n^{p}}{n!} $. Notice that
\begin{align}
\forall p \in \mathbb{N}: \quad
S_{p}
& = \sum_{n = 0}^{\infty} \frac{n^{p}}{n!} \\
& = \sum_{n = 1}^{\infty} \frac{n^{p}}{n!} \qquad
\left( \text{As $ \dfrac{0^{p}}{0!} = 0 $.} \right) \\
& = \sum_{n = 1}^{\infty} \frac{n^{p - 1}}{(n - 1)!} \qquad
(\text{After canceling $ n $’s.}) \\
& = \sum_{n = 0}^{\infty} \frac{(n + 1)^{p - 1}}{n!}. \qquad
(\text{After re-indexing.})
\end{align}
Hence,
\begin{align}
\forall p \in \mathbb{N}_{\geq 2}: \quad
S_{p} - S_{p - 1}
& = \sum_{n = 0}^{\infty} \frac{(n + 1)^{p - 1}}{n!} -
\sum_{n = 0}^{\infty} \frac{n^{p - 1}}{n!} \\
& = \sum_{n = 0}^{\infty}
\frac{1}{n!} \left[ (n + 1)^{p - 1} - n^{p - 1} \right] \\
& = \sum_{n = 0}^{\infty}
\left[ \frac{1}{n!} \sum_{k = 0}^{p - 2} \binom{p - 1}{k} n^{k} \right]
\qquad (\text{By the Binomial Theorem.}) \\
& = \sum_{n = 0}^{\infty} \sum_{k = 0}^{p - 2}
\binom{p - 1}{k} \frac{n^{k}}{n!} \\
& = \sum_{k = 0}^{p - 2}
\left[ \binom{p - 1}{k} \sum_{n = 0}^{\infty} \frac{n^{k}}{n!} \right] \\
& = \sum_{k = 0}^{p - 2} \binom{p - 1}{k} S_{k}.
\end{align}
Therefore,
$$
\forall p \in \mathbb{N}_{\geq 2}: \quad
S_{p}
= \sum_{k = 0}^{p - 2} \binom{p - 1}{k} S_{k} + S_{p - 1}
= \sum_{k = 0}^{p - 1} \binom{p - 1}{k} S_{k}.
$$
As $ S_{0} = S_{1} = e $ by inspection, we see that the $ S_{p} $’s are the Bell numbers multiplied by $ e $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Differentiate $(x + 1)(x + 2)^2(x + 3)^3$ Obviously we can "brute force" this by multiplying the various terms and differentiating from there. But based upon the solution provided in the text where I found this problem, it looks like there's a "cleaner" method available.
Of note, this problem appears in the chapter introducing the chain rule.
Any ideas?
UPDATE: just to clarify, this problem appears in a very basic calculus text, so any/all "advanced" differentiation techniques would not be at the student's disposal. It's also worth pointing out the answer provided in the text:
$2(x + 2)(x + 3)^2(3x^2 + 11x + 9)$
| There is also the logarithmic differentiation approach: http://en.wikipedia.org/wiki/Logarithmic_differentiation
Note that $\ln((x+1)(x+2)^2(x+3)^3)=\ln(x+1)+2\ln(x+2)+3\ln(x+3)$.
Then $\frac{d}{dx}\ln(f(x))=\frac{f^{\prime}(x)}{f(x)}$ using the chain rule.
So take the derivative of the RHS (above) to get
$$\frac{f^\prime(x)}{f(x)} = \frac{1}{x+1}+\frac{2}{x+2}+\frac{3}{x+3}$$
Finish by multiplying both sides of the equation by the original function:
\begin{align*}f^\prime(x) &= f(x)\left(\frac{1}{x+1}+\frac{2}{x+2}+\frac{3}{x+3}\right) \\
&= (x+2)^2(x+3)^3 + 2(x+1)(x+2)(x+3)^3 + 3(x+1)(x+2)^2(x+3)^2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1244118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Contest Problem Let N be the set of positive integers whose decimal representation
is of the form 4ab4 (for example: 4174 or 4004). If a number x is
picked randomly from N, what is the probability that x is divisible
by 6, but not divisible by 12?
I got an answer of $\frac{17}{100}$. Could you someone verify that for me.
| $N$ has $100$ members.
Every member $x$ of $N$ is divisible by $2$; it is divisible by $3$ (and therefore by $6$) iff $a+b \equiv 1 \mod 3$. Of these, $x$ is divisible by $12$ iff $b$ is even. There are three cases $(b = 3, 5, 9$) with $b$ odd and three possible $a$'s ($a = 1,4, 7$ for $b = 3$ or $9$, $a = 2,5,8$ for $b=5$), and two ($b = 1, 7$) with $b$ odd and four possible $a$'s ($a = 0, 3, 6, 9$), for a total of $3 \times 3 + 2 \times 4 = 17$.
So the probability is indeed $17/100$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What's the solution to this binomial? what's the coefficient of $x^6$ in the expansion of $(1+X^2+X)^{-3}$?
I have factorized the term to $\left(\frac{1-x^{-3}}{1-x}\right)^3$
after this I'm having problem solving it
| According to Wolfram alpha, the Taylor series of your function is
$$1-3 x+3 x^2+2 x^3-9 x^4+9 x^5+3 x^6-18 x^7+18 x^8+4 x^9-30 x^{10}+\dots.$$
You can calculate this by solving linear equations. For example, here is how to find the first three coefficients. Let the beginning of the Taylor expansion be $A + Bx + Cx^2 + \dots$. Multiplying this by $(1+x+x^2)^3 = 1 + 3x + 6x^2 + \cdots$, we get
$$
A + (3A + B)x + (6A + 3B + C)x^2 + \dots.
$$
Since the result should be $1$, we get the system of equations
$$
A = 1 \\
3A + B = 0 \\
6A + 3B + C = 0
$$
This is a triangular system of equations, and so not difficult to solve. The first equation gives $A = 1$, the second $B = -3$, and the third $C = 3$. This matches what Wolfram alpha gave us above. In this way you can find all coefficients of the Taylor series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convert Function into Spherical Coordinates $$z=\sqrt{4-x^2-2y^2} $$
First thing I did was put the equation in standard form: $z^2+x^2+2y^2=4 $
Then I convert to spherical: $$\rho^2\cos^2(\phi)+\rho^2\sin^2(\phi)\cos^2(\theta)+2\rho^2\sin^2(\phi)\sin^2(\theta)=4$$
then I simplify:
$$\rho^2[\cos^2(\phi)+\sin^2(\phi)\cos^2(\theta)+2\sin^2(\phi)\sin^2(\theta)]=4$$
How do I simplify from here? As my answer has to be one among the following:
| There is a mistake in your last line, it should be
$$\rho^2[\cos^2\phi+\sin^2\phi\left[\cos^2\theta+2\sin^2\theta]\right]=4$$
so that
$$\rho^2[\cos^2\phi+\sin^2\phi\left(1+\sin^2\theta)\right]=4$$
$$\rho^2(\cos^2\phi+\sin^2\phi+\sin^2\phi\sin^2\theta)=4$$
$$\rho^2\left((\cos^2\phi+\sin^2\phi)+\sin^2\phi\sin^2\theta\right)=4$$
or
$$\rho^2(1+\sin^2\theta\sin^2\phi)=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many distinct possible forms for its Jordan canonical matrix are there? 4x4 non-diagonalizable matrix with two unique eigenvalues I know the sum of $A_m$ equals $4$ as $\dim(A) = 4$ and sum of $G_m$ can't equal $4$ as $A$ is non-diagonalizable.
After I write down all the cases, what should I do?
| After you write down the cases, count how many you have. I count 24. Following is my work where the first entry in an ordered pair refers to the first eigenvalue, and the second to the second distinct eigenvalue. Note that without a convention regarding Jordan canonical form, blocks can be permuted.
\begin{array}{cccrl}
\text{algebraic} & \text{geometric} & \text{Jordan}\\
\text{multiplicity} & \text{multiplicity} & \text{block sizes} & \text{permutations} \\ \hline
(1, 3) & (1, 1) & (1, 3) & 2! = 2 \\
(1, 3) & (1, 2) & (1, 1 \text{ and } 2) & 3! = 6 \\
(2, 2) & (1, 1) & (2, 2) & 2! = 2 \\
(2, 2) & (1, 2) & (2, 1 \text{ and } 1) & 3!/2! = 3 \\
(2, 2) & (2, 1) & (1 \text{ and } 1, 2) & 3!/2! = 3 \\
(3, 1) & (1, 1) & (3, 1) & 2! = 2 \\
(3, 1) & (2, 1) & (1 \text{ and } 2, 1) & 3! = 6 \\ \hline
& & & 24 & \text{total}
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1247574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determining the Laurent Series I need to determine the Laurent series of this function:
$$\frac{1}{(z-1)(z+5)}$$ Inside the annulus: $$\left\{z|1<|z-2|<6\right\}$$ Any help appreciated.
| Using partial fraction expansion, we can write
$$\frac{1}{(z-1)(z+5)}=\frac{1/6}{z-1}-\frac{1/6}{z+5}$$
We seek a series about the center of the annulus $z=2$. To that end,
$$\begin{align}
-\frac{1/6}{z+5}&=\frac16 \frac{1}{(z-2)+7}\\\\
&=-\frac{1}{42}\frac{1}{1+\frac{z-1}{7}}\\\\
&=-\frac{1}{42} \sum_{n=0}^{\infty} (-1)^n\left(\frac{z-2}{7}\right)^n
\end{align}$$
which converges within the entire annulus. We also can write
$$\begin{align}
\frac{1/6}{z-1}&=\frac16 \frac{1}{(z-2)+1}\\\\
&=\frac{1}{6}\frac{1/(z-2)}{1+(z-2)^{-1}}\\\\
&=\frac{1}{6} \sum_{n=1}^{\infty} (-1)^{n+1}(z-2)^{-n}
\end{align}$$
which converges for $|z-2|>1$, which completely covers the annulus. Thus, the appropriate Laurent series is
$$\frac{1}{(z-1)(z+5)}=\frac{1}{6} \sum_{n=1}^{\infty} (-1)^{n+1}(z-2)^{-n}-\frac{1}{42} \sum_{n=0}^{\infty} (-1)^n\left(\frac{z-2}{7}\right)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is $\sqrt{x^2} = (\sqrt x)^2$? Take $x=4$ for example:
$ \sqrt{(4)^2} = \sqrt{16} = \pm4 $
However:
$ (\sqrt{4})^2 = \sqrt{\pm2}$
Case 1: $ (-2)^2 = 4$
Case 2: $ (2)^2 = 4$
Solution : $+4$
How come the $ \sqrt{(4)^2} = \pm4$; but $ (\sqrt{4})^2 = 4 $ ?
What is missing?
| $\sqrt{x^2} = \sqrt{x \cdot x} = \sqrt{x} \cdot \sqrt{x} = \sqrt{x}^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
How to Find the Function of a Given Power Series? (Please see edit below; I originally asked how to find a power series expansion of a given function, but I now wanted to know how to do the reverse case.)
Can someone please explain how to find the power series expansion of the function in the title? Also, how would you do it in the reverse case? That is, given the power series expansion, how can you deduce the function $\frac{x}{1-x-x^3}$?
I've also rewritten the function as $$\frac{x}{1-x-x^3} = x \cdot \frac{1}{1-x(1-x^2)}$$ which is of the form of the Maclaurin series $\frac{1}{1-x} = 1+x+x^2+x^3+\ldots$
So the series expansion would then be $$x \cdot (1+(x-x^3)+(x-x^3)^2+(x-x^3)^3+\ldots)$$
However, expanding this to find the simplified power series expansion becomes complicated. According to WolframAlpha (link: http://www.wolframalpha.com/input/?i=power+series+of+x%2F%281-x-x%5E3%29) it eventually works out to $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots$$ which is a function defined recursively by $f_n = f_{n-1} + f_{n-3}$ for all $n\gt 3$ with the initial condition that $f_1=f_2=f_3=1$
Appreciate any and all help!
Thanks for reading,
A
Edit: I actually want to find the reverse of the original question. Given the recursively defined function (see above) $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots ,$$ how can I show that the function of this power series expansion is $\frac{x}{1-x-x^3}$?
| To answer both your old and your new question at the very same time, we can consider a surprising relationship between power series and recurrence relations through generating functions. As a simple example, consider representing $\frac{1}{1-x}$ as a power series. In particular, we want to discover an $f_n$ such that
$$\frac{1}{1-x}=f_0+f_1x+f_2x^2+f_3x^3+\ldots$$
How do we do it? It proves pretty easy; let's multiply both sides by $(1-x)$ to obtain:
$$1=(1-x)(f_0+f_1x+f_2x^2+f_3x^3+\ldots)$$
Now, if we distribute the $(1-x)$ over the infinite sum, we get:
$$\begin{align}1=f_0 &+ f_1x + f_2x^2 + f_3x^3+f_4x^4+\ldots\\& -f_0x-f_1x^2-f_2x^3-f_3x^4-\ldots \end{align}$$
and doing the subtractions in each column, we get to the equation:
$$1=f_0+(f_1-f_0)x+(f_2-f_1)x^2+(f_3-f_2)x^3+\ldots$$
What's clear here? Well, every coefficient of $x$ has to be $0$ - so we get that $f_1-f_0$ and $f_2-f_1$ and $f_3-f_2$ must all be zero. In other words, $f_{n+1}=f_n$. Then, the constant term, $f_0$, must be $1$. Hence $f$ is defined as:
$$f_0=1$$
$$f_{n+1}=f_n.$$
That's a very simple recurrence relation, solved as $f_n=1$ meaning
$$\frac{1}{1-x-x^2}=1+x+x^2+x^3+\ldots$$
Okay, that's pretty cool, but let's try something harder, like setting
$$\frac{1}{1-x-x^2}=f_0+f_1x+f_2x^2+f_3x^3+\ldots$$
Now, if we multiply through by $(1-x-x^2)$ and expand like before, we get
$$\begin{align}1=f_0&+f_1x+f_2x^2+f_3x^3+\ldots\\&-f_0x-f_1x^2-f_2x^3-\ldots\\ &\qquad\,-f_0x^2-f_1x^3-\ldots\end{align}$$
and gathering terms again we get:
$$1=f_0+(f_1-f_0)x+(f_2-f_1-f_0)x^2+(f_3-f_2-f_1)x^3+\ldots+(f_{n+2}-f_{n+1}-f_n)x^{n+2}+\ldots$$
where we see again that $f_0=1$ and that $f_1-f_0$ is zero, hence $f_1=1$. Then, for higher terms we get that the coefficient of $x^{n+2}$ must be zero, hence
$$f_{n+2}-f_{n+1}-f_n=0$$
$$f_{n+2}=f_{n+1}+f_n$$
Oh hey, that's the Fibonacci sequence, so we get the identity
$$\frac{1}{1-x-x^2}=1+1x+2x^2+3x^3+5x^4+8x^5+13x^6+\ldots$$
Your equation will work the same way: Multiply through by the denominator, gather like terms, and then equate the coefficients on the left with those on the right. You will discover exactly the recurrence you found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Evaluate $\text{k}$ from the given equation
If
$$ \int_{0}^{\infty} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \text{k} \times \int_{0}^{1} \dfrac{\ln (1-x)}{x} \mathrm{d}x =0$$
then find the value of $\text{k}$
My Approach :
Let
$\text{I}= \displaystyle \int_{0}^{\infty} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \displaystyle \text{k} \times \int_{0}^{1} \dfrac{\ln (1-x)}{x}\mathrm{d}x$
$=\displaystyle \int_{0}^{1} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \int_{1}^{\infty} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \displaystyle \text{k} \times \int_{0}^{1} \dfrac{\ln (1-x)}{x}\mathrm{d}x$
Now, for the second integral, let $x=\dfrac{1}{t}$,
$\implies \text{I}= \displaystyle \int_{0}^{1} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \displaystyle \int_{0}^{1} \left(\dfrac{\ln t}{1-t}\right)^{2} \mathrm{d}t + \displaystyle \text{k} \times \int_{0}^{1} \dfrac{\ln (1-x)}{x}\mathrm{d}x$
$= 2\displaystyle \int_{0}^{1} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \displaystyle \text{k} \times \int_{0}^{1} \dfrac{\ln (1-x)}{x}\mathrm{d}x$
$=2\displaystyle \int_{0}^{1} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \displaystyle \text{k} \times \int_{0}^{1} \dfrac{\ln x}{1-x}\mathrm{d}x$
However, I can't seem to think of a way to simplify it further and find the value of $\text{k}$.
Any help will be appreciated.
Thanks in advance.
| The crux to completing your work is to prove the following
$$\int_0^1 \frac{\log^2(1-x)}{x^2} \,\mathrm{d}x
= -2 \int_0^1 \frac{\log(1-x)}{x}\,\mathrm{d}x$$
You are close though, let me show you how to finish your work. I really liked this problem. Summarizing your work, you have proven that
$$
\int_0^\infty \left(\frac{\log x}{1-x}\right)^2 \,\mathrm{d}x
= 2\int_0^1 \left(\frac{\log x}{1-x}\right)^2 \,\mathrm{d}x
= 2\int_0^1 \frac{\log^2(1-x)}{x^2} \,\mathrm{d}x
$$
Where the substitution $u \mapsto 1-x$ was used in the last equality.
The idea to finish this problem is to use integration by parts with
$$
\begin{align*}
u & \ = \hspace{0.7cm} \log^2(1-x) \hspace{1.4cm} \mathrm{d}v \ = \ \frac{1}{u^2}\mathrm{d}x\\
\mathrm{d}u & \ = - 2\,\frac{\log^{\phantom{2}}(1-x)}{1-x}\,\mathrm{d}x \qquad v \ = \ - \frac{1}{x}
\end{align*}
$$
However using integration by parts we have to be extremely gentle. For reasons that shall become apparent later I replace the upper limit with $\varepsilon$. Then at the end we will let $\varepsilon \to 1$.
$$
\begin{align*}
\int_0^\varepsilon \frac{\log^2(1-x)}{x^2} \,\mathrm{d}x
& = \left[ - \frac{\log^2(1-x)}{x} \right]_0^\varepsilon - 2\int_0^\varepsilon \frac{\log(1-x)}{x(1-x)}\,\mathrm{d}x \\
& = -\frac{\log^2(1-\varepsilon)}{\varepsilon}
- 2 \int_0^\varepsilon \log (1-x)\left(\frac{1}{x}+\frac{1}{1-x}\right)\,\mathrm{d}x \\
& = -\frac{\log^2(1-\varepsilon)}{\varepsilon} - 2 \int_0^\varepsilon \frac{\log(1-x)}{x}\,\mathrm{d}x - 2\left[ -\frac{\log^2(1-x)}{2}\right]_0^\varepsilon \\
& = -2 \int_0^1 \frac{\log(1-x)}{x}\,\mathrm{d}x
\end{align*}
$$
Now the rest is elementary
$$
\int_0^\infty \left(\frac{\log x}{1-x}\right)^2 \,\mathrm{d}x
= 2\int_0^1 \frac{\log^2(1-x)}{x^2} \,\mathrm{d}x
= -4 \int_0^1 \frac{\log(1-x)}{x}\,\mathrm{d}x
$$
As wanted
To prove that
$$
\lim_{\varepsilon \to 1} \left(\log^2(1-\varepsilon) - \frac{\log^2(1-\varepsilon)}{\varepsilon}\right) = 0
$$
Can be proven for example by a clever taylor expansion or plain old l'hoptials rule
$$
\lim_{\varepsilon \to 1} \frac{\log^2(1-\varepsilon)}{\varepsilon/(\varepsilon-1)} \left[ \frac{0}{0}\right]
= \lim_{\varepsilon \to 1} -2(1-\varepsilon)\log(1-\varepsilon)
= -2 \lim_{x \to 0} x \log x
= 0
$$
Where the last limit is know (again l'hôpital or series expansion).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Could you explain the expansion of $(1+\frac{dx}{x})^{-2}$? Could you explain the expansion of $(1+\frac{dx}{x})^{-2}$?
Source: calculus made easy by S. Thompson.
I have looked up the formula for binomial theorem with negative exponents but it is confusing. The expansion stated in the text is:
$$\left[1-\frac{2\,dx}{x}+\frac{2(2+1)}{1\cdot2}\left(\frac{dx}{x}\right)^2 - \text{etc.}\right] $$
Please explain at a high school level.
| First, recall the expansion $\frac{1}{1-x}=1+x+x^2+\cdots$. If you don't believe this yet, let $z=1+x+x^2+\cdots$, then $(1-x)z=z-xz=(1+x+x^2+\cdots)-(x+x^2+x^3+\cdots)=1$, so $z=\frac{1}{1-x}$.
Next, we can derive an expansion of $\frac{1}{(1-x)^2}$ by squaring this sequence: $\frac{1}{(1-x)^2}=\left(1+x+x^2+\cdots\right)^2=1+2x+3x^2+\cdots+(i+1)x^i+\dots$ (count how many ways there are to get any particular power, for instance $x^2\cdot 1 + x\cdot x + 1\cdot x^2=3x^2$).
With this, let us put $-\frac{dx}{x}$ into the expansion:
$$1 + 2\left(-\frac{dx}{x}\right) + 3\left(-\frac{dx}{x}\right)^2 + \cdots $$
which is
$$1 - 2\frac{dx}{x} + 3\left(\frac{dx}{x}\right)^2 - 4\left(\frac{dx}{x}\right)^2 + \cdots$$
This doesn't explain why factorials show up in the derivation from the book, but it is an equivalent expansion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the solution to the differential equation Assume $x\gt 0$ and let
$$x(x+1)\frac{du}{dx} = u^2,$$
$$u(1) = 4.$$
I started off by doing some algebra to get:
$$\frac{1}{u^2}du = \frac{1}{x^2+x}dx.$$
I then took the partial fraction of the right side of the equation:
$$\frac{1}{u^2}du = \left(\frac{1}{x}-\frac{1}{x+1}\right).$$
I then took the integral of both sides:
$$-\frac{1}{u} = \log{x}-\log{(x+1)}+C.$$
From here I don't know what to do because we are solving for $u(x)$ and I'm not sure how to get that from $-\frac{1}{u}$.
| Solving for $\frac{du}{dx}$ we have
\begin{equation*}
\frac{du}{dx}=\frac{u^2}{x(x+1)}\\
\Rightarrow \frac{du}{dx}=\frac{u^2}{x^2+x}\\
\frac{\frac{du}{dx}}{u^2}=\frac{1}{x^2+x}.
\end{equation*}
Integrate both sides & evaluate the integrals:
\begin{equation*}
-\frac{1}{u}=\log(x)-\log(x+1)+C_1\\
\Rightarrow u=-\frac{1}{\log(x)-\log(x+1)+C_1}.
\end{equation*}
Now apply the initial condition:
\begin{equation*}
-\frac{1}{C_1-\log(2)}=4\Rightarrow C_1=\frac{1}{4}(-1+4\log(2)).
\end{equation*}
This gives the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{13}{24}$ by induction. Where am I going wrong? I have to prove that
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{2n}>\frac{13}{24}$$ for every positive integer $n$.
After I check the special cases $n=1,2$, I have to prove that the given inequality holds for the $n+1$ case by using the $n$ case. So, I have to prove:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}$$
what is equivavlent to
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}+\frac{1}{n}$$
By using the $n$ case, I know that
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+2}>\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2}$$
so I basically have to prove that
$$\frac{13}{24}+\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{13}{24}+\frac{1}{n}$$
what is equivavlent to
$$\frac{1}{2n+1}+\frac{1}{2n+2}\overset{?}{>}\frac{1}{n}$$
$$n(2n+2)+n(2n+1)\overset{?}{>}(2n+1)(2n+2)$$
$$4n^2+3n\overset{?}{>}4n^2+6n+2$$
$$-3n\overset{?}{>}2$$
$$n\overset{?}{<}-\frac{2}{3}$$
which can not be because $n$ is a positive integer. What am I doing wrong? Tnx!
| At first sight it seems to be false because 1/n tends to 0. But I trust you so your inequality is nice. Want I want to remark is you have at hand a pretty proof that the harmonic series is divergent (You have an infinity of "packages" each of them greater than 13/24)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Binomial sum identity Everyone knows that $\sum{n \choose 2k}=\sum{n \choose 2k+1}$
My question is follwing
What is the clear form of $\sum_{k=0} (-1)^k{n \choose 2k}$ and $\sum_{k=0} (-1)^k{n \choose 2k+1}$
For example, my exercise is
"calculate a+b where a=$\sum_{k=0}^{25} (-1)^k{50 \choose 2k}$ & b=$\sum_{k=0}^{24} (-1)^k{50 \choose 2k+1}$"
I found that it is relative to number 4.
If n has a form 4k+2 then a=0 and b=$(-1)^k2^{2k+1}$ or If n has a form 4k then b=0 and a=$(-1)^k2^{2k}$
But it's just my hypothesis.(I just wrote it when n=1,2,3,...,10 and found a rule)
| Start with
$$
(1 + i)^{n}
=
\sum_{k=0}^{n} \binom{n}{k} i^{k}
=
\sum_{h} \binom{n}{2 h} (-1)^{h}
+
i \sum_{h} \binom{n}{2 h + 1} (-1)^{h},
$$
and then note that
\begin{align}
(1 + i)^{n}
&=
\sqrt{2}^{n} \cdot \left( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right)^{n}
\\&=
\sqrt{2}^{n} \cdot \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right)^{n}
\\&=
\sqrt{2}^{n} \cdot \left( \cos\left(n \frac{\pi}{4}\right) + i \sin\left(n \frac{\pi}{4}\right) \right).
\end{align}
Note that this extends the argument for the case you mentioned
$$
0 = (1 + (-1))^{n}
=
\sum_{h} \binom{n}{2 h}
-
\sum_{h} \binom{n}{2 h + 1}.
$$
The download of generatingfunctionology is to be recommended.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding both max and min points by Lagrangian Using the method of the Lagrange multipliers, find the maximum and minimum values of the function
$$f(x,y,z) = x^2y^2z^2$$
where $(x,y,z)$ is on the sphere
$$x^2 +y^2 +z^2 = r^2$$
So $L(x,y,z;\lambda)=x^2y^2z^2 + \lambda (x^2 +y^2 +z^2-r^2)$.
After finding $L_{x,y,z,\lambda}=0$, I got $$\bigg(\pm \frac r {\sqrt 3},\pm \frac r {\sqrt 3},\pm \frac r {\sqrt 3}\bigg)$$ which seems like the maximum but what is the minimum?
| So the equations you find by setting the derivatives equal to zero are
$$ 2x( y^2 z^2 + \lambda) = 0 \\
2y (x^2 z^2 + \lambda) = 0 \\
2z (y^2 x^2 + \lambda) = 0 \\
x^2 + y^2 + z^2 - r^2 = 0
$$
Right, let's stop here and think. Suppose $x=0$. Then the first equation's obviously satisfied. The sphere equation can obviously be satisfied by various $y$ and $z$. The other two equations are
$$ 2y \lambda = 0 = 2z \lambda, $$
so can be satisfied by $\lambda=0$, whatever $y$ and $z$ are. Similarly starting from $y=0$ or $z=0$. If none of them are zero, you can cancel $x,y,z$ off the first 3 equations and carry on to find the other answers. Since $x^2y^2z^2$ is obviously at least zero, $x=0$ or $y=0$ or $z=0$ must give the minima (note these are not points, either, but circles $y^2+z^2=r^2$ and similar).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Compare $\sum_{k=1}^n \left\lfloor \frac{k}{\varphi}\right\rfloor$ ... Given two integer sequences
\begin{equation*}
\displaystyle A_n=\sum_{k=1}^n \left\lfloor \frac{k}{\varphi}\right\rfloor ,
\end{equation*}
\begin{equation*}
B_n=\left\lfloor\dfrac{n^2}{2\varphi}\right\rfloor-\left\lfloor \dfrac{n}{2\varphi^2}\right\rfloor
\end{equation*}
here: $\quad\varphi=\dfrac{1+\sqrt{5}}{2}\quad$ (golden ratio)
Prove that: $|A_n-B_n|\leq 1.$
I realized that the difference between $A_n$ and $B_n$ is very small but the failure in finding an exact formula for $A_n$
Could you help me?
| I can prove less; we have $$\frac {n^2 + n} {2 \varphi} - n < A_n \leqslant \frac {n^2 + n} {2 \varphi}$$ and $$\frac {n^2 - n/\varphi} {2 \varphi} - \frac {1} {2} < B_n \leqslant \frac {n^2 - n/\varphi} {2 \varphi}.$$ It follows that $$\frac {n (\varphi + 1 - 2 \varphi^2)} {2 \varphi^2} < A_n - B_n < \frac {(\varphi + 1) n} {2 \varphi^2} + \frac {1} {2}.$$ That is, we have $$- \frac {n} {2} < A_n - B_n < \frac {n + 1} {2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$
Find the equation whose roots are each six more than the roots of $x^2 + 8x - 1 = 0$
I must use Vieta's formulas in my solution since that is the lesson we are covering with our teacher.
My solution:
Let p and q be the roots of the quadratic.
$$\begin{align} p + q = & -8 \\ pq = & -1 \end{align}$$
If the roots are each six more than the roots of the quadratic, then we will have:
$$\begin{align} p + q + 12 = & -8 \\
p + q = & -20 \\
pq = & - 3 \end{align}$$
Also,
$$ \begin{align} (p+6)(q+6) = & pq + 6p + 6q + 36\\
=&pq + 6(p+q) + 36 \\
=&-3 + 6(-8) + 36 \\
=&-3 - 48 + 36 \\
&36 - 51 = -15 \end{align} $$
Hence, -15 = constant.
Thus, the quadratic equation is $x^2 + 20x - 15$
My worksheet gives an answer of $x^2 - 4x - 13$, how am I wrong?
Thanks!
| Let $r=p+6$ and $s=q+6$ be the roots of the quadratic you're trying to find. That quadratic's coefficients are $-(r+s)$ and $rs$. But
$$\begin{align}
r+s&=(p+6)+(q+6)\\
&=p+q+12\\
&=-8+12\\
&=4
\end{align}$$
and
$$\begin{align}
rs&=(p+6)(q+6)\\
&=pq+6(p+q)+36\\
&=-1+6(-8)+36\\
&=-1-48+36\\
&=-13
\end{align}$$
so the quadratic you're looking for is
$$x^2-(r+s)x+rs=x^2-4x-13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Limit without applying l'hopital's rule, $\lim_{x \rightarrow-\infty} \frac{|2x+5|}{2x+5}$. This is the question:
$$\lim_{x \rightarrow-\infty} \frac{|2x+5|}{2x+5}$$
I know the answer is $-1$, but can someone go through the steps and explaining it to me?
| You should better be aware of the definition of a modulus function.
$$|x| = \left\{\begin{matrix}
x & x > 0\\
0 & x = 0 \\
-x & x<0
\end{matrix}\right.$$
For this question,
$$|2x+5| = \left\{\begin{matrix}
2x+5 & 2x+5 > 0\\
0 & 2x+5 = 0 \\
-(2x+5) & 2x+5<0
\end{matrix}\right.$$
Here, since, $x \to -\infty$ , that is $2x + 5 <0 $ , so, $|2x+5| = -(2x+5)$ and you get the limit as $-1$.
To answer your question in specific that what will be its limit when $x$ tends to $-9/2$ or $-1/2$ . Well, here is the difference.
$x \to -1/2 \implies 2x \to -1 \implies 2x + 5 \to 4 $ ($2x+4 >0$ )
So, $\lim_{x \to -1/2} \cfrac{|2x+5|}{2x + 5} = 1 $
While for $x\to -9/2 \implies 2x \to -9 \implies 2x + 5\to -4 $ ($2x+5 <0$)
So, $\lim_{x \to -9/2} \cfrac{|2x+5|}{2x +5} = - 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What is the value of $\frac{a^3}{a^6+a^5+a^4+a^3+a^2+a+1}$ If $\frac{a}{a^{2}+1} = \frac{1}{3}$ Then find the value of $$\frac{a^3}{a^6+a^5+a^4+a^3+a^2+a+1}$$
Any hints to help me?
| Since $a+\frac 1a=3$, one has
$$a^2+\frac{1}{a^2}=\left(a+\frac 1a\right)^2-2=7$$
$$a^3+\frac{1}{a^3}=\left(a+\frac 1a\right)^3-3\left(a+\frac 1a\right)=18.$$
Now note that
$$\frac{a^3}{a^6+a^5+a^4+a^3+a^2+a+1}=\frac{1}{a^3+\frac{1}{a^3}+a^2+\frac{1}{a^2}+a+\frac 1a+1}=\frac{1}{18+7+3+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A possible dilogarithm identity? I'm curious to find out if the sum can be expressed in some known constants. What do you think about that? Is it feasible? Have you met it before?
$$2 \left(\text{Li}_2\left(2-\sqrt{2}\right)+\text{Li}_2\left(\frac{1}{2+\sqrt{2}}\right)\right)+\text{Li}_2\left(3-2 \sqrt{2}\right)$$
| Euler's reflection identity states:
$$\operatorname{Li}_{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(1-z\right)}=\zeta{(2)}-\ln{\left(z\right)}\ln{\left(1-z\right)}.$$
Landen's dilogarithm identity states:
$$\operatorname{Li}_{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(\frac{z}{z-1}\right)}=-\frac12\ln^2{\left(1-z\right)};~~~\small{z\notin(1,\infty)}.$$
Setting $z=2-\sqrt{2}$ in Euler's identity, we have
$$\operatorname{Li}_{2}{\left(2-\sqrt{2}\right)}=\zeta{(2)}-\ln{\left(2-\sqrt{2}\right)}\ln{\left(\sqrt{2}-1\right)}-\operatorname{Li}_{2}{\left(\sqrt{2}-1\right)}.$$
Setting $z=\frac{1}{2+\sqrt{2}}$ in Landen's identity, we have
$$\operatorname{Li}_{2}{\left(\frac{1}{2+\sqrt{2}}\right)}=-\frac12\ln^2{\left(\frac{1}{\sqrt{2}}\right)}-\operatorname{Li}_{2}{\left(1-\sqrt{2}\right)}.$$
Thus,
$$\begin{align}
I
&=2\left[\operatorname{Li}_{2}{\left(2-\sqrt{2}\right)}+\operatorname{Li}_{2}{\left(\frac{1}{2+\sqrt{2}}\right)}\right]+\operatorname{Li}_{2}{\left(3-2\sqrt{2}\right)}\\
&=2\left[\zeta{(2)}-\ln{\left(2-\sqrt{2}\right)}\ln{\left(\sqrt{2}-1\right)}-\operatorname{Li}_{2}{\left(\sqrt{2}-1\right)}\right]\\
&~~~~~+2\left[-\frac12\ln^2{\left(\frac{1}{\sqrt{2}}\right)}-\operatorname{Li}_{2}{\left(1-\sqrt{2}\right)}\right]+\operatorname{Li}_{2}{\left(3-2\sqrt{2}\right)}\\
&=2\zeta{(2)}-2\ln{\left(2-\sqrt{2}\right)}\ln{\left(\sqrt{2}-1\right)}-\frac14\ln^2{(2)}\\
&~~~~~-2\left[\operatorname{Li}_{2}{\left(\sqrt{2}-1\right)}+\operatorname{Li}_{2}{\left(-\left(\sqrt{2}-1\right)\right)}\right]+\operatorname{Li}_{2}{\left(3-2\sqrt{2}\right)}\\
&=2\zeta{(2)}-2\ln{\left((\sqrt{2}-1)\sqrt{2}\right)}\ln{\left(\sqrt{2}-1\right)}-\frac14\ln^2{(2)}\\
&~~~~~-\operatorname{Li}_{2}{\left(\left(\sqrt{2}-1\right)^2\right)}+\operatorname{Li}_{2}{\left(3-2\sqrt{2}\right)}\\
&=2\zeta{(2)}-2\ln^2{\left(\sqrt{2}-1\right)}-\ln{(2)}\ln{\left(\sqrt{2}-1\right)}-\frac14\ln^2{(2)}.\blacksquare\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Write 100 as the sum of two positive integers
Write $100$ as the sum of two positive integers, one of them being a multiple of $7$, while the other is a multiple of $11$.
Since $100$ is not a big number, I followed the straightforward reasoning of writing all multiples up to $100$ of either $11$ or $7$, and then finding the complement that is also a multiple of the other. So then
$100 = 44 + 56 = 4 \times 11 + 8 \times 7$.
But is it the smart way of doing it? Is it the way I was supposed to solve it? I'm thinking here about a situation with a really large number that turns my plug-in method sort of unwise.
| From Bezout's Lemma, note that since $\gcd(7,11) = 1$, which divides $100$, there exists $x,y \in \mathbb{Z}$ such that $7x+11y=100$.
A candidate solution is $(x,y) = (8,4)$.
The rest of the solution is given by $(x,y) = (8+11m,4-7m)$, where $m \in \mathbb{Z}$. Since we are looking for positive integers as solutions, we need $8+11m > 0$ and $4-7m>0$, which gives us $-\frac8{11}<m<\frac47$. This means the only value of $m$, when we restrict $x,y$ to positive integers is $m=0$, which gives us $(x,y) = (8,4)$ as the only solution in positive integers.
If you do not like to guess your candidate solution, a more algorithmic procedure is using Euclid' algorithm to obtain solution to $7a+11b=1$, which is as follows.
We have
\begin{align}
11 & = 7 \cdot (1) + 4 \implies 4 = 11 - 7 \cdot (1)\\
7 & = 4 \cdot (1) + 3 \implies 3 = 7 - 4 \cdot (1) \implies 3 = 7 - (11-7\cdot (1))\cdot (1) = 2\cdot 7 - 11\\
4 & = 3 \cdot (1) + 1 \implies 1 = 4 - 3 \cdot (1) \implies 1 = (11-7 \cdot(1)) - (2\cdot 7 - 11) \cdot 1 = 11 \cdot 2-7 \cdot 3
\end{align}
This means the solution to $7a+11b=1$ using Euclid' algorithm is $(-3,2)$. Hence, the candidate solution $7x+11y=100$ is $(-300,200)$. Now all possible solutions are given by $(x,y) = (-300+11n,200-7n)$. Since we need $x$ and $y$ to be positive, we need $-300+11n > 0$ and $200-7n > 0$, which gives us
$$\dfrac{300}{11} < n < \dfrac{200}7 \implies 27 \dfrac3{11} < n < 28 \dfrac47$$
The only integer in this range is $n=28$, which again gives $(x,y) = (8,4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 8,
"answer_id": 0
} |
What wolfram does to factor $x^6+x^2+2$? I am learning polynomials and I am trying to understand what wolfram did to obtain $$(x^2+1)(x^4-x^2+2)$$ from $$x^6+x^2+2$$
It does not show me the step-by-step option in this case and I got confused. I only see $$x^2(x^4+1)+2$$ for this case.
Thank you.
| Notice that $x^6+x^2+2=(x^6+1) + (x^2+1)$.
Since $x^6+1$ is a sum of cubes it factors like this: $$x^6+1=(x^2+1)(x^4-x^2+1).$$ Putting these together gives $$x^6+x^2+2=(x^2+1)(x^4-x^2+1)+(x^2+1)=(x^2+1)(x^4-x^2+2).$$
Not sure about Wolfram's actual algorithm though. For algorithms see Wikipedia. For an explanation of one algorithm that isn't too hard to follow, check out page 12 in
Milne's course notes on fields and Galois theory, it is remark 1.17.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluating a triple integral in spherical coordinates I need to evaluate the integral $\int \int \int \frac{x^2}{x^2+y^2}$ over the region $D$ where $D = {(x,y)} : 1\leq x^2+y^2+z^2 \leq 2, z^2>=x^2+y^2$ and $z\leq 0$
So I tried converting to spherical coordinates, therefore $1\leq r\leq\sqrt{2}$
and $0\leq \theta \leq 2\pi$ and $0 \leq \phi \leq \pi/4$
as from the equation of the cone, $\cos{^2}{\phi} = \sin{^2}{\phi}$.
The integrand becomes $r^2 \sin{\phi} \cos{^2}{\theta}drd\theta d\phi$. So we now evaluate the integral and using wolfram alpha the answer was $\frac{\pi}{6}(5\sqrt{2}-6)$ however this is not the answer that my professor gave us to verify our result.
I am not sure where I have made a mistake. I assumed by the symmetry of the problem that the integral is the same for $\pm z$, hence $\phi$ goes from $0$ to $\pi/4$.
| $$
I = \int\limits_D \frac{x^2}{x^2+y^2}\, dV
$$
$D$: $1 \leq x^2+y^2+z^2 \leq 2$ (spherical shell between $r=1$ and $r=\sqrt{2}$), $z^2>=x^2+y^2$ and $z\le 0$ (negative cone).
Going to spherical coordinates $dV = r \sin\theta\,dr\,d\theta\,d\phi$, $x = r \sin \theta \cos \phi$, $y = r \sin \theta \sin \phi$ gives
\begin{align}
I &= \int\limits_D \frac{r^2 \sin^2 \theta \cos^2 \phi}{r^2 \sin^2 \theta} \, dV \\
&= \int\limits_D \cos^2 \phi \, r \sin\theta\,dr\,d\theta\,d\phi \\
&=
\int\limits_0^{2\pi} \cos^2 \phi \, d\phi
\int\limits_1^\sqrt{2} r\, dr
\int\limits_{3\pi/4}^\pi \sin\theta\, d\theta \\
&=
\frac{1}{2}\left[\phi + \cos \phi \sin \phi \right]_0^{2\pi}
\frac{1}{2}\left[ r^2 \right]_1^\sqrt{2}
\left[ -\cos \theta\right]_{3\pi/4}^{\pi} \\
&= (\pi)(1/2)(1 -1/\sqrt{2}) \\
&= \frac{\pi}{2} \left(1 - \frac{1}{\sqrt{2}} \right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof using partial fractions I have to prove this formula:
$$\int \! \frac{1}{(x^2+\beta ^2)^{k+1}} \, \mathrm{d}x=\frac{1}{2k\beta ^2}\frac{x}{(x^2 +\beta^2)^k}+\frac{2k-1}{2k\beta^2}\int \! \frac{1}{(x^2+\beta ^2)^k} \, \mathrm{d}x
$$
I have to use partial fraction decomposition.
I started like this:
$$
\frac{a}{(x^2+\beta ^2)^k}+\frac{b}{x^2+\beta ^2}
$$
Now:
$$
1 = a(x^2+\beta^2) + b(x^2+\beta^2)^k
$$
Now I'm stuck. How do I find $a$ and $b$?
| Write
$$\begin{align}
\frac{1}{(x^2+\beta^2)^{k+1}}&=\frac{1}{\beta^2}\frac{(x^2+\beta^2)-x^2}{(x^2+\beta^2)^{k+1}}\\\\
&=\frac{1}{\beta^2}\left(\frac{1}{(x^2+\beta^2)^{k}}- \frac{x^2}{(x^2+\beta^2)^{k+1}}\right)
\end{align}$$
Now, integrate both sides and use integration by parts on the last term. Combine terms of like powers and you will have it.
NOTE: Integration by parts
$$-\int \frac{x^2}{(x^2+\beta^2)^{k+1}} dx = \frac{1}{2k}\frac{x}{(x^2+\beta^2)^k}-\frac{1}{2k}\int \frac{1}{(x^2+\beta^2)^{k+1}}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
To find jordan canonical form Which of the following matrices have Jordan canonical form of equal to the $3\times 3$ matrix
$$
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$$
a)$
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$
b)$
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}$
c)$
\begin{pmatrix}
0 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$
d)$
\begin{pmatrix}
0 & 1 & 1 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}$
Here characteristic equation of the matrix is $x^3$.Hence the 3 eigenvalues of the matrix are zero.
Do we want to find the eigenvalues of all the matrices in the options?Is there any other way?
| The given Jordan canonical form implies, minimal polynomial of corresponding matrix should be $x^2=0.$ Hence if matrix $A$ is having the property that $A \neq 0$ and $A^2=0 ,$ it will have desired Jordan canonical form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
transformation of a difference equation How can I translate the difference equation
$$x_{k+3}+4x_{k+2}+3x_{k+1}+x_k=2u_{k+2}$$
into a state-space representation of the following form (A and B are matrices)
$$x_{k+1}=Ax_k+Bu_k$$
| As always in such cases. Let
$$ X_k := \begin{pmatrix} x_k \\ x_{k+1}\\ x_{k+2} \end{pmatrix} $$
and
$$ U_k := \begin{pmatrix} u_k \\ u_{k+1} \\ u_{k+2}\end{pmatrix} $$
Then the above can be written as
\begin{align*}
X_{k+1} &= \begin{pmatrix} x_{k+1} \\ x_{k+2}\\ x_{k+3} \end{pmatrix} \\
&= \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -1 & -3 & -4\end{pmatrix}
\begin{pmatrix} x_k \\ x_{k+1}\\ x_{k+2} \end{pmatrix}
+ \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 2\end{pmatrix}
\begin{pmatrix} u_k \\ u_{k+1}\\ u_{k+2} \end{pmatrix}\\
&=: AX_k + BU_k
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$ Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$. I thought to write
$$\dfrac{2^2}{2^{\ln(x)}} + 2^2 \cdot 2^{\ln(x)} = 2^3 \implies \dfrac{2^2 + 2^2 \cdot \left(2^{\ln(x)}\right)^2}{2^{\ln(x)}} = 2^3$$
What should I do else?
| We have $2^{2-\ln(x)} = \dfrac4{2^{\ln(x)}}$ and $2^{2+\ln(x)} = 4\cdot2^{\ln(x)}$. Hence, setting $2^{\ln(x)} = a$, we obtain
$$\dfrac4a + 4a = 8 \implies a^2 + 1 =2a \implies a =1 \implies 2^{\ln(x)} = 1 \implies x = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
The limit of this function when x goes to minus infinity? I'm looking for the limit of $$\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} } \right)$$
I know it should be -1, but for some reason I always get to 1.
I'm not sure where the difference between $$- \infty$$ and $$\infty$$ is:
$$\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} {{\sqrt {1 + x + {x^2}} + \sqrt {1 - x + {x^2}} } \over {\sqrt {1 + x + {x^2}} + \sqrt {1 - x + {x^2}} }} = {{1 + x + {x^2} - \left( {1 - x + {x^2}} \right)} \over {\sqrt {1 + x + {x^2}} + \sqrt {1 - x + {x^2}} }} = {{2x} \over {\sqrt {1 + x + {x^2}} + \sqrt {1 - x + {x^2}} }}$$
$$ = {{2x} \over {\sqrt {1 + x + {x^2}} + \sqrt {1 - x + {x^2}} }}{{{1 \over x}} \over {{1 \over x}}} = {2 \over {\sqrt {{1 \over {{x^2}}} + {1 \over x} + 1} + \sqrt {{1 \over {{x^2}}} - {1 \over x} + 1} }} = {2 \over {\sqrt {0 + 0 + 1} + \sqrt {0 - 0 + 1} }} = 1$$
| Here is another and perhaps quicker way to proceed. Use
$$\sqrt{1+t}=1+\frac12 t +O(t^{-2})$$
Here we have
$$\sqrt{x^2\pm x+1}=\sqrt{x^2(1\pm x^{-1}+x^{-2})}=|x|\left(1\pm \frac1{2x}+O\left(\frac{1}{x^2}\right)\right)$$
whereupon subtracting the "upper" and "lower" signed expressions yields
$$|x|\left(\frac{1}{x}+O\left(\frac{1}{x^2}\right)\right)$$
$$\lim_{x \to -\infty}\left(|x|\left(\frac{1}{x}+O\left(\frac{1}{x^2}\right)\right)\right)=\lim_{x \to -\infty} \frac{|x|}{x}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Attempt:
I want to use the following theorem:
The largest exponent of $e$ of a prime $p$ such that $p^e$ is a divisor of $n!$ is given by $$e=[\frac{n}{p}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]+\cdots$$.
The number of times the prime divisor $5$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $5$ contained in $(5^{25}-1)!$, which is $e_1=[\frac{(5^{25}-1)}{5}]+[\frac{(5^{25}-1)}{5^2}]+[\frac{(5^{25}-1)}{5^3}]+[\frac{(5^{25}-1)}{5^4}]+\cdots$
The number of times the prime divisor $2$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $2$ contained in $(5^{25}-1)!$, which is $e_2=[\frac{(5^{25}-1)}{2}]+[\frac{(5^{25}-1)}{2^2}]+[\frac{(5^{25}-1)}{2^3}+[\frac{(5^{25}-1)}{2^4}]+\cdots$
Therefore, the number of zeros at the right end equals the greatest exponent of $10$ contained in $(5^{25}-1)!$$=\min{(e_1, e_2)}$.
I am unable to simplify $e_1, e_2$. Please help me.
| First in the définition of $e_1$ and $e_2$ there is no factoriel $!$ we have:
$$ e_1=\left\lfloor\frac{5^{25}-1}{5}\right\rfloor+\left\lfloor\frac{5^{25}-1}{5^2}\right\rfloor+\left\lfloor\frac{5^{25}-1}{5^3}\right\rfloor+\left\lfloor\frac{5^{25}-1}{5^4}\right\rfloor+\cdots+\left\lfloor\frac{{5^{25}-1}}{5^{24}}\right\rfloor
$$
And here you forgot to add the function : the greatest number less than $5^n-1$ each time because the numbers to sum must be all integers and finally we have to stop at $5^{24}$ because the rest are all null and we have:
$$e_1=5^{24}-1+5^{23}-1+\dots+5^0-1=\frac{5^{25}-1}{4}-25=\frac{5^{25}-101}{4}$$
and it's clear that the greatest power of $5$ dividing $(5^{25}-1)!$ is the smallest one. each time we have a $5$ n=in the factoriel we will have more then $3$ powers of $2$, you can prove this by comparing the formula of $e_1$ and $e_2$ for instance here is an example $1\cdot 2\cdot 3\cdot 4\cdot 5=5\cdot 3\cdot 2^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$f\left( x-1 \right) +f\left( x+1 \right) =\sqrt { 3 } f\left( x \right)$ Let f be defined from real to real
$f\left( x-1 \right) +f\left( x+1 \right) =\sqrt { 3 } f\left( x \right)$
Now how to find the period of this function f(x)?
Can someone provide me a purely algebraic method to solve this problem please?
Update:My method
An elementary algebraic approach to the problem :
$f(x-1)+f(x+1)=\sqrt { 3 } f(x)$
Replace $x$ with $x+1$ and $x-1$ respectively.
We get $f(x)+f(x+2)=\sqrt { 3 } f(x+1)$ and
$f(x-2)+f(x)=\sqrt { 3 } f(x-1)$
From these three equations we get $f(x-2)+f(x+2)=0$
Putting $x=x+2$ and adding with last equation we get $f(x-2)+f(x+4)=0$....(1)
Similarly $f(x-4)+f(x+2)=0$.....(2)
Put $x=x-6$ in (1)
We get $f(x-8)+f(x-2)=0$.....(3)
From (1) and (3) we get $f(x-8)=f(x+4)$
So the period of $f(x)$ is 12
| For each fixed $y \in [0,1)$, your equation is a second order linear recurrence relation with constant coefficients. This suggests that the first step should be to solve the characteristic equation $\lambda^2-\sqrt{3} \lambda + 1 = 0$. You find $\lambda_1,\lambda_2 = \frac{\sqrt{3} \pm i}{2}=e^{\pm i \pi/6}$. This means that, for $y \in [0,1)$ and $n \in \mathbb{Z}$:
$$f(y+n)=c_1 e^{i n \pi/6} + c_2 e^{-i n \pi/6}$$
where $c_1,c_2$ are specified by the choice of $f(y)$ and $f(y-1)$. This tells you that the period is no larger than $\frac{2 \pi}{\pi/6} = 12$. It can be smaller: for instance your equation has a trivial constant solution $f \equiv 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic manipulation is done between the first step and the second step, and I can't figure out what this manipulation could be.
The full breakdown comes from this solution
$$
\small\begin{align}
\frac1{x^2-5x+6}
&=\frac1{(x-2)(x-3)}
=\frac1{-3-(-2)}\left(\frac1{x-2}-\frac1{x-3}\right)
=\bbox[4px,border:4px solid #F00000]{-\frac1{x-2}+\frac1{x-3}}\\
&=\bbox[4px,border:4px solid #F00000]{\frac1{2-x}-\frac1{3-x}}
=\sum_{n=0}^\infty\frac1{2^{n+1}}x^n-\sum_{n=0}^\infty\frac1{3^{n+1}}x^n
=\bbox[4px,border:1px solid #000000]{\sum_{n=0}^\infty\left(\frac1{2^{n+1}}-\frac1{3^{n+1}}\right)x^n}
\end{align}
$$
Original image
| It's very easy. It comes by using factorization and simplification rules in general. In the case of your question, we have
$- \frac{1}{x-2} = \frac{(-1)}{(-1)(-x+2)}= \frac{(-1)}{(-1)}\times \frac{1}{(-x+2)}= \frac{1}{(-x+2)}$
and in the case of $\frac{1}{x-3}$, by multiplying both denominator and numerator with $(-1)$, we have that
$\frac{1}{x-3} = \frac{(-1)}{(-1)} \times \frac{1}{x-3} = \frac{(-1)}{(-x+3)}= -\frac{1}{(3 - x)}$. So we will have
$- \frac{1}{x-2} + \frac{1}{x-3} = \frac{1}{(2 -x)} - \frac{1}{(3 - x)} $.
And we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
How to compute $\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$? I have to compute the series $\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}}$.
$$\displaystyle\sum_{n=0}^{\infty}{\frac{3^n(n + \frac{1}{2})}{n!}} = \sum_{n=0}^{\infty}{\frac{3^n\frac{1}{2}}{n!}} + \sum_{n=0}^{\infty}{\frac{3^nn}{n!}} = \frac{e^3}{2} + \sum_{n=0}^{\infty}{\frac{3^nn}{n!}},$$ but how to compute the $\displaystyle\sum_{n=0}^{\infty}{\frac{3^nn}{n!}}$?
| As for integer $n>0, n!=n\cdot(n-1)!,$
$$\sum_{n=0}^{\infty}{\frac{3^nn}{n!}}=\sum_{n=1}^{\infty}{\frac{3^nn}{n!}}=3\sum_{n=1}^\infty\dfrac{3^{n-1}}{(n-1)!}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Find $\int \frac{1}{x^4+x^2+1} \,\, dx$ Find $$\int \frac{1}{x^4+x^2+1} \,\, dx$$
I tried to find like that:
$\int \frac{1}{x^4+x^2+1} = \int \frac{\frac{1}{2}x + \frac{1}{2}}{x^2+x+1} \,\, dx + \int \frac{-\frac{1}{2}x + \frac{1}{2}}{x^2-x+1} \,\, dx = \frac{1}{2} \Big(\int \frac{2x + 1}{x^2+x+1} - \int \frac{x}{x^2+x+1} \Big) - \frac{1}{2} \Big( \int \frac{2x - 1}{x^2-x+1} \,\, dx - \int \frac{x}{x^2-x+1} \,\, dx \Big)$
but then I don't know how to find integrals:
$\, \int \frac{x}{x^2-x+1} \,\, dx \,$ and $\, \int \frac{x}{x^2+x+1} \,\, dx \,$
Is there another way to integrate this function or way to end my calculations?
| simple methods,Note
$$1=\dfrac{1}{2}(x^2+1)+\dfrac{1}{2}(1-x^2)$$
so
$$\int\dfrac{1}{x^4+x^2+1}dx=\frac{1}{2}(I_{1}+I_{2})$$
$$I_{1}=\int\dfrac{x^2+1}{x^4+x^2+1}dx=\int\dfrac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1}dx=\int\dfrac{d(x-\frac{1}{x})}{(x-1/x)^2+3}$$
and simaler $I_{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Computing a limit similar to the exponential function I want to show the following limit:
$$
\lim_{n \to \infty}
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]
= \frac{1}{e^{2}}.
$$
I got the answer using WolframAlpha, and it seems to be correct numerically, but I am having trouble proving the result. My first instinct was to write the limit as
$$
\lim_{n \to \infty}
\frac
{
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
}
{1/n}.
$$
Then, I tried applying l'Hopital's rule, and I got
$$
\lim_{n \to \infty}
\frac
{
\left( 1 - \frac{1}{n} \right)^{2n}
\left( 2 \log\left( 1 - \frac{1}{n} \right) + \frac{2}{n-1} \right)
-
\left( 1 - \frac{2}{n} \right)^{n}
\left( \log\left( 1 - \frac{2}{n} \right) + \frac{2}{n-2} \right)
}
{-1/n^{2}}.
$$
This does not seem to have gotten me anywhere. My second attempt was to use the binomial theorem:
$$
\begin{align*}
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right]
&
=
n
\left[
\sum_{k=0}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k}}
- \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^{k} 2^{k}}{n^{k}}
\right]
\\ &
=
\sum_{k=2}^{n} \left[ \binom{2n}{k} - \binom{n}{k} 2^{k} \right] \frac{(-1)^{k}}{n^{k-1}}
+ \sum_{k=n+1}^{2n} \binom{2n}{k} \frac{(-1)^{k}}{n^{k-1}}.
\end{align*}
$$
At this point I got stuck again.
| $$
(1-\frac1{n})^{2n} =(1-\frac2{n}+\frac1{n^2})^n
$$
use the binomial expansion:
$$
(1-\frac2{n}+\frac1{n^2})^n =\sum_{k=0}^n(1-\frac2{n})^{n-k}\binom{n}{k}\frac1{n^{2k}}
$$
so
$$
n
\left[
\left( 1 - \frac{1}{n} \right)^{2n}
- \left( 1 - \frac{2}{n} \right)^{n}
\right] = \sum_{k=1}^n(1-\frac2{n})^{n-k}\binom{n}{k}\frac1{n^{2k-1}} \\
=(1-\frac2{n})^{n-1}+\sum_{k=2}^n(1-\frac2{n})^{n-k}\binom{n}{k}\frac1{n^{2k-1}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Area of region inequality What is the area of the region defined by the following set of inequalities?
$$\begin{array}{cc} (1) &-1 < xy < 1 \\ (2) &-1 < x^2-y^2 < 1 \end{array}$$
I think we use integration here
| The area that we want is:
Define $u = x y$ and $v = x^2-y^2$. Then, $ {du} = y {dx} + x {dy}$ and $ {dv} = 2x {dx} - 2y {dy}$.
Solve this system of equations for $ {dx}$ and $ {dy}$ to get:
$ {dx} = \frac{y}{x^2+y^2} {du} + \frac{x}{2x^2+2y^2} {dv}$ and $ {dy} = \frac{x}{x^2+y^2} {du} - \frac{y}{2x^2+2y^2} {dv}$.
Thus, $$ {dA} = {dx} \wedge {dy} = \left(\frac{y}{x^2+y^2} {du} + \frac{x}{2x^2+2y^2} {dv}\right) \wedge \left(\frac{x}{x^2+y^2} {du} - \frac{y}{2x^2+2y^2} {dv}\right) = \left(-\frac{x^2}{2 \left(x^2+y^2\right)^2}\right) {du} \wedge {dv} + \left(\frac{y^2}{2 \left(x^2+y^2\right)^2}\right) {dv} \wedge {du} = \left(-\frac{x^2}{2 \left(x^2+y^2\right)^2}-\frac{y^2}{2 \left(x^2+y^2\right)^2}\right) {du} \wedge {dv} = \left(-\frac{1}{2 \left(x^2+y^2\right)}\right) {du} \wedge {dv}$$.
Call $f = -\frac{1}{2 \left(x^2+y^2\right)}$, and recall that $u = xy$ and $v = x^2-y^2$.
Observe then that $f^2 = \frac{1}{4 (x^4+2x^2 y^2 + y^4)} = \frac{1}{4 (4 u^2 + v^2)} = \frac{1}{16u^2+4v^2}$. Thus, $f = \pm \frac{1}{\sqrt{16u^2+4v^2}}$, where each solution for $f$ corresponds to half of the problem region, as the two are symmetric over $y=x$.
So, our area is $$\displaystyle \iint {dA} = \iint {dx} \wedge {dy} = 2 \int _{-1}^1\int _{-1}^1\frac{1}{\sqrt{16 u^2+4 v^2}} {du} \, {dv} = \frac{1}{4} \int_{-2}^{2} \int_{-4}^{4} \frac{1}{\sqrt{a^2+b^2}} {da} \, {db} = \log \left(\frac{123+55 \sqrt{5}}{2}\right) \approx 4.812$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve for x in $2^{2x^2}+2^{x^2 + 2x + 2} =2^{5+4x}$ This is the question:
$$\large{2^{2x^2}+2^{x^2 + 2x + 2} =2^{5+4x}}$$
What I did was put $~\large{2^{x^{2}}=t}$
From this, I got, roots of the quadratic:
$$\large{-2^{x+1}\pm~\left( 2^{2x+1}\sqrt{3}\right)}$$
Now this is equal to $\boxed{\large{2^{x^2}}}$. How do i find the value of x from this?
| The given equation can be rewritten as $$2^{2x^2-4x+2}+2^{x^2-2x+4}=2^7$$Now take $2^{(x-1)^2}=z$ which gives you the equation $$z^2+8z=2^7\implies z=8,-16$$ Taking only the positive solution gives you, (assuming $x$ to be real)$$(x-1)^2=3\implies x=1\pm \sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Cannot understand an Integral $$\displaystyle \int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
I had to solve the integral and get it in this form.
My attempt:
$$\int _{ \pi /6 }^{ \pi /3 }{ \frac { dx }{ \sec x+\csc x } } $$
$$=\int _{\frac{\pi}{6}}^{ \frac{\pi}{3}} \dfrac{\sin x \cos x }{ \sin x+\cos x }dx $$
Substituting $t=\tan(\frac{x}{2})$,
$$\int_{\tan(\frac{\pi}{12})}^{\tan(\frac{\pi}{6})} \dfrac{2t}{1+t^2}\times\dfrac{1-t^2}{1+t^2}\times\dfrac{2}{1+t^2}dt$$
$$2\int_{2-\sqrt{3}}^{\frac{1}{\sqrt{3}}} \dfrac{2t(1-t^2)}{(1+t^2)^3}dt$$
Substituting $u=1+t^2$, $2t dt=du$, $1-t^2 = 2-u$
$$2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{(2-u)}{u^3}du$$
$$\displaystyle 4\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^3} \displaystyle -2\int_{8-4\sqrt{3}}^\frac{4}{3} \dfrac{du}{u^2}$$ Could somebody please tell me where I have gone wrong? Also could someone please tell me how to change the limits of the definite integral throughout?
| To evaluate the integral use the identity $$(\sin{x}+\cos{x})^2=1+2{\sin{x}}{\cos{x}}$$
And substitute $\sin(x)\cos(x) = u$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1288034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Proving $x\ln(\frac{x}{a})+y\ln(\frac{y}{b})\geq (x+y)\ln(\frac{x+y}{a+b})$
Let a,b,x,y be positive reals. Prove $x\ln(\frac{x}{a})+y\ln(\frac{y}{b})\geq (x+y)\ln(\frac{x+y}{a+b})$
I don't have any olympic background, so I may be missing some standard trick.
The inequality looks closely related to the concavity of $\ln$.
EDIT: the following is wrong as stated by Macavity.
Indeed using points $\frac{x}{a}$,$\frac{y}{b}$, and $\frac{x}{x+y}$,$\frac{y}{x+y}$ as weights, one has $$\frac{x}{x+y}\ln(\frac{x}{a})+\frac{y}{x+y}\ln(\frac{y}{b})\geq \ln(\frac{x^2}{x+y}\frac{1}{a}+\frac{y^2}{x+y}\frac{1}{b})$$
But I can't prove that $\displaystyle\frac{x^2}{x+y}\frac{1}{a}+\frac{y^2}{x+y}\frac{1}{b}\geq \frac{x+y}{a+b}$
| @math110 is right, but you have started with the wrong inequality. You can use concavity instead directly and get:
$$\frac{x}{x+y}\ln \left(\frac{a}x \right)+\frac{y}{x+y}\ln \left(\frac{b}y \right) \le \ln\left(\frac{a+b}{x+y} \right)$$
Now multiplying throughout by $-1$ (which flips the fractions in parentheses) gets you what you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$ Find the value of $$\sum^{n=k}_{n=1}\tan^{-1}\frac{4}{4n^2+3}$$
I tried multiplying numerator and denominator by $n^2$, but got nothing. How do I split the term inside $\tan^{-1}$?
| I tried again and got the answer. $\frac{4}{4n^2 + 3} = \frac{1}{n^2 + 3/4} = \frac{1}{1 + n^2 -1/4} = \frac{1}{1 + (n+1/2)(n-1/2)} = = \frac{(n+1/2) - (n-1/2)}{1 + (n+1/2)(n-1/2)}$
$$\sum^{n=k}_{n=1}\tan^{-1}\frac{4}{4n^2+3} = \sum^{n=k}_{n=1}\tan^{-1}\frac{(n+1/2) - (n-1/2)}{1 + (n+1/2)(n-1/2)} = \boxed{\tan^{-1}(k+1/2) - \tan^{-1}1/2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the parameterization of a curve for a line integral problem I have to calculate the work of a particle that travel along a curve, given the following vector field:
$F(x, y, z) = (2z-1, 0, 2y)$
and where the curve is the intersection between:
$s1: z = x^2 + y^2$ and $s2: 4x^2 + 4y^2 + 1 = 4x + 4y$
using the definition of line integrals.
What complicates me of this exercise is parameterize the curve, any help?
| There are many ways to parametrize a given curve, but I'll discuss one example here. The equation $4x^2 + 4y^2 + 1 = 4x + 4y$ can be rearranged:
\begin{align*}
4x^2 - 4x + 1 = -4y^2 + 4y -1 + 1
&\iff (2x-1)^2 = -(2y-1)^2 + 1 \\
&\iff (2x-1)^2 + (2y-1)^2 = 1.
\end{align*}
Knowing this, let us set $2x - 1 = \cos(t)$ and $2y-1 = \sin(t)$, where $t$ ranges from $0$ to $2\pi$. Using the equation $z = x^2 + y^2$, we can also find the parametrization of $z$:
$$
z = \left( \frac{\cos(t) + 1}{2} \right)^2 + \left( \frac{\sin(t) + 1}{2} \right)^2 = \frac{\sin(t)}{2} + \frac{\cos(t)}{2} + \frac{3}{4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge\frac32$ using derivatives
Let $a,b,c\in\mathbb{R}^+$ and $abc=1$. Prove that
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge\frac32$$
This isn't hard problem. I have already solved it in following way:
Let $x=\frac1a,y=\frac1b,z=\frac1c$, then $xyz=1$. Now, it is enought to prove that
$$L\equiv\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac32$$
Now using Cauchy-Schwarz inequality on numbers $a_1=\sqrt{y+z},a_2=\sqrt{z+x},a_3=\sqrt{x+y},b_1=\frac{x}{a_1},b_2=\frac{y}{a_2},b_3=\frac{z}{a_3}$ I got
$$(x+y+z)^2\le((x+y)+(y+z)+(z+x))\cdot L$$
From this
$$L\ge\frac{x+y+z}2\ge\frac32\sqrt[3]{xyz}=\frac32$$
Then I tried to prove it using derivatives. Let $x=a,y=b$ and
$$f(x,y)=\frac1{x^3\left({y+\frac1{xy}}\right)}+\frac1{y^3\left({x+\frac1{xy}}\right)}+\frac1{\left({\frac1{xy}}\right)^3(x+y)}$$
So, I need to find minimum value of this function. It will be true when
$$\frac{df}{dx}=0\land\frac{df}{dy}=0$$
After simplifying $\frac{df}{dx}=0$ I got
$$\frac{-y(3xy^2+2)}{x^3\left({xy^2+1}\right)^2}+\frac{1-x^2y}{y^2\left({x^2y+1}\right)^2}+\frac{x^2y^3(2x+3y)}{\left({x+y}\right)^2}=0$$
Is there any easy way to write $x$ in term of $y$ from this equation?
| why must you use derivatives? the proof is simple with Cauchy Schwarz:
we have
$$\frac{1}{a^3(b+c)}=\frac{\frac{1}{a^2}}{a(b+c)}=\frac{\frac{1}{a^2}}{\frac{b+c}{bc}}$$ thus we have
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\geq $$
$$\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{2}{a}+\frac{2}{b}+\frac{2}{c}}=\frac{\left(ab+bc+ca\right)^2}{2(ab+bc+ca)}\geq \frac{1}{2}3\sqrt[3]{(abc)^2}=\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Verifying $\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$ My math teacher gave us an equality involving trigonometric functions and told us to "verify" them. I tried making the two sides equal something simple such as "1 = 1" but kept getting stuck. I would highly appreciate if someone could show me (step by step) how to verify or solve this problem.
$$\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$$
| Use $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\sin(2x)=2\sin(x)\cos(x)$ to get $$\frac{\cos(2x)}{1+\sin(2x)}=\frac{\cos^2(x)-\sin^2(x)}{1+2\sin(x)\cos(x)}$$
Multiply top and bottom by $\frac{1}{\sin^2(x)}$ to get $$\frac{\cot^2(x)-1}{\frac{1}{sin^2(x)}+2\cot(x)}$$
Using $\sin^2(x)+\cos^2(x)=1$ we can get $$\frac{\cot^2(x)-1}{\frac{\sin^2(x)+\cos^2(x)}{sin^2(x)}+2\cot(x)}\ \ \ \ \ \ \ or\ \ \ \ \ \ \ \frac{\cot^2(x)-1}{1+\cot^2(x)+2\cot(x)}$$
By using $a^2-b^2=(a-b)(a+b)$ for the numerator and factoring the denominator, we get $$\frac{(\cot(x)-1)(\cot(x)+1)}{(\cot(x)+1)^2}$$ which equals to $$\frac{\cot(x)-1}{\cot(x)+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Derivative of $f(x) = \frac{\cos{(x^2 - 1)}}{2x}$ Find the derivative of the function $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}$$
This is my step-by-step solution: $$f'(x) = \frac{-\sin{(x^2 - 1)}2x - 2\cos{(x^2 -1)}}{4x^2} = \frac{2x\sin{(1 - x^2)} - 2\cos{(1 - x^2)}}{4x^2} = \frac{x\sin{(1 - x^2)} - \cos{(1 - x^2)}}{2x^2} = \frac{\sin{(1 - x^2)}}{2x} - \frac{\cos{(1-x^2)}}{2x^2}$$
and this is the output of WolphramAlpha: http://www.wolframalpha.com/input/?i=derivative+cos(x^2+-+1)%2F(2x)
Where is the mistakes?
| You need the chain rule:
$$
\frac d {dx} \cos(x^2-1) = -\sin (x^2-1)\cdot\frac d{dx}(x^2-1)=\cdots.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Solving the Diophantine equation $x^n-y^n=1001$ For all $n \in \mathbb{N}$, solve the Diophantine equation $x^n-y^n=1001$, where $x,y \in \mathbb{N}$.
The cases $n=1,2$ are trivial ones. But for $n>2$ I can't find any solutions. How could I prove that there are no integer solutions for $n>2$?
| We'll$\let\leq\leqslant\let\geq\geqslant$ use the following, which can easily be proved using for example Bézout's theorem:
If $p$ is prime, then $p\mid x^n-y^n$ implies $p\mid x^{\gcd(n,p-1)}-y^{\gcd(n,p-1)}$.
Note $1001=7\cdot11\cdot13$ and $x^n-y^n=(y+(x-y))^n-y^n\geq(x-y)^n$. If $x-y$ and $n$ are large enough, we might use the inequality $(x-y)^n>1001$.
*
*If $n=3$, then $11\mid x^3-y^3$ implies $11\mid x-y$, but $11^3>1001$.
*If $n=4$, then $77\mid x^4-y^4$ implies $77\mid x^2-y^2$. If $x-y>1$ we would have $x-y\geq7$, but $7^4>1001$. Hence $x-y=1$ and $77\mid x+y$. But then $x^4-y^4=(x-y)\cdot(x^3+\cdots)>x^3\geq34^3>1001$.
*If $n=5$, then $7\mid x^5-y^5$ implies $7\mid x-y$, but $7^5>1001$. (Also works with $13$ instead of $7$.)
*$n=6$ is covered by $n=3$.
*If $n=7$, then $13\mid x^7-y^7$ implies $13\mid x-y$, but $13^7>1001$. (Also works with $7$ or $11$ instead of $13$.)
*$n=8$ is covered by $n=4$.
*$n=9$ is covered by $n=3$.
*$n<10$, because $x^n-y^n\geq(y+1)^n-y^n\geq2^n-1>1001$ if $n\geq10$.
Note that some cases permit alternative approaches, for example $n=7$:
$7\mid x^7-y^7$ implies $7\mid x-y$. Using Newton's binomial theorem this is seen to imply $7^2\mid x^7-y^7$, but $7^2\nmid1001$.
Similarly for $n=11$ or $n=13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the number of possible values of $a$
Positive integers $a, b, c$, and $d$ satisfy $a > b > c > d, a + b + c + d = 2010$,
and $a^2 − b^2 + c^2 − d^2 = 2010$. Find the number of possible values of $a.$
Obviously, factoring,
$$(a-b)(a+b) + (c-d)(c+d) = 2010$$
$$a + b + c + d = 2010$$
Substituting you get:
$$(a-b)(a+b) + (c-d)(c+d) - (a + b) - (c + d) = 0$$
$$\implies (a - b - 1)(a + b) + (c - d- 1)(c+d) = 0$$
But I dont see anything else.
| Hint: examine the conditions you are given in the question - what can you say about the value of the expression in your last equation? Can you find a condition which makes it zero?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1295944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}}$ and so: $$\lim \limits _ {n \to \infty} n - n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}} = 0$$
Because both $\frac{a}{n}$ and $\frac{b}{n}$ tend to $0$.
What would give a correct answer:
Plotting the function $$f(x) = x - \sqrt{x+a} \sqrt{x+b}$$
Clearly indicates that it has an asymptote in $- \frac{a+b}{2}$. This result can be obtained multiplying the numerator and the denominator by $n + \sqrt{n+a} \sqrt{n+b}$:
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = $$
$$-\lim \limits _{n \to \infty} \frac {n(a+b)}{n + \sqrt{n+a} \sqrt{n+b}} - \lim \limits _{n \to \infty} \frac {ab}{n + \sqrt{n+a} \sqrt{n+b}}$$
The second limit is clearly $0$ and the first one gives the correct answer (dividing the numerator and denominator by $n$).
Why the first way I tried is wrong? I might have done something silly but I cannot find it.
| one way to see this is to use the binomial theorem. here is how it goes:
$$(n + a)^{1/2} = n^{1/2} + \frac12n^{-1/2}a + \cdots\\
\sqrt{n+a}\sqrt{n+b} = \left(\sqrt n + \frac a{2\sqrt n}+\cdots\right) \left(\sqrt n + \frac b{2\sqrt n}+\cdots\right) = n + \frac12(a+b)+\cdots $$
therefore $$\lim_{n \to \infty}\left(n - \sqrt{n+a}\sqrt{n+b}\right) =-\frac12(a+b). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 5
} |
How do I solve the trigonometric equation $1 - \sin^2x - \cos(2x) = \frac{1}{2}$? Solve for $x$ when $1-\sin^2x - \cos 2x = \dfrac{1}{2}$.
I can' t change it into a form I can work with. It is rather complicated.
| As given $$1-\sin^2 x-\cos 2x=\frac{1}{2} \implies \cos^2x-\left(2\cos^2 x-1\right)=\frac{1}{2}$$ $$ \cos^2 x=\frac{1}{2}=\cos^2 \frac{\pi}{4} \implies x=n\pi\pm \frac{\pi}{4}$$ Hence, the general solution of the equation is $x=n\pi\pm \frac{\pi}{4} $ where, $n$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1297353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
evaluate the sum $\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{1}{(n^2+n-1)(k^2+k-1)}$ I'm trying to evaluate this sum
$$\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{1}{(n^2+n-1)(k^2+k-1)}$$
I have no idea how to deal with it.
With one sum I can, with partial-fraction decomposition, express it as a function of Digamma function and I stuck there.
| As already pointed by the other answers, we just have to compute:
$$ S_1=\sum_{n\geq 1}\frac{1}{n^2+n-1},\qquad S_2=\sum_{n\geq 1}\frac{1}{(n^2+n-1)^2}.\tag{1}$$
and by exploiting the logarithmic derivative of the Weierstrass product for the cosine function we have, for any $\alpha\in\mathbb{R}^+\setminus\mathbb{N}$:
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^2-\alpha^2}=\frac{\pi\tan\frac{\pi\alpha}{2}}{4\alpha}\tag{2}$$
as well as:
$$ \sum_{n\geq 0}\frac{1}{\left((2n+1)^2-\alpha^2\right)^2}=\frac{\pi ^2 \sec^2\frac{\pi \alpha }{2}}{16\, \alpha ^2}-\frac{\pi\tan\frac{\pi \alpha }{2}}{8\, \alpha ^3}\tag{3}$$
hence it follows that:
$$ S_1 = 1+\frac{\pi}{\sqrt{5}}\,\tan\frac{\pi\sqrt{5}}{2}\tag{4} $$
and:
$$ S_2 = -1+\frac{\pi^2}{5}\,\sec^2\frac{\pi\sqrt{5}}{2}-\frac{2\pi}{5\sqrt{5}}\,\tan\frac{\pi\sqrt{5}}{2}\tag{5}$$
hence our double series equals:
$$\frac{\pi}{100}\,\sec^2\frac{\pi\sqrt{5}}{2}\left(15\pi-5\pi\cos(\pi\sqrt{5})+8\sqrt{5}\sin(\pi\sqrt{5})\right).\tag{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Remainder of $(1+x)^{2015}$ after division with $x^2+x+1$
Remainder of $(1+x)^{2015}$ after division with $x^2+x+1$
Is it correct if I consider the polynomial modulo $5$
$$(1+x)^{2015}=\sum\binom{2015}{n}x^n=1+2015x+2015\cdot1007x^2+\cdots+x^{2015}$$
RHS stays the same and then The remainder must be of the form $Ax+B$
$$x^{2015}+1\equiv Ax+B\pmod{1+x+x^2}$$
plug in $x=0\implies B=0$
plug in $x=1\implies A=-1\implies $ the remainder is $-x$
Is this a good way to solve the problem or were we lucky ?
| use the division algorithm to write $$(1 + x)^{2015} = q(x)(x^2 + x+ 1)+ax + b\tag 1$$ the roots of $x^2 + x+ 1 = 0$ are $\omega = e^{i2\pi/3}, \bar {\omega}.$ we need $$1 + \omega = e^{i\pi/3},1 + \bar \omega = e^{-i\pi/3} $$ subbing $e^{i\pi/3}, e^{-i\pi/3}$ in $(1)$ gives us $$e^{\pm i2015\pi/3} = ae^{\pm i\pi/3} + b\to e^{\mp i\pi/3} = ae^{\pm 2i\pi/3} + b$$
gives you $$ a = -1, b = 0, \text{ the remainder is } -x $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
General term of a sequence $(2-1)(2+1)(3-1)(3+1)...(n-1)(n+1)$ Can we use integrals, and are there some general methods for finding terms of a sequence?
| Let's reorder the terms and recognize the result:
$$
\begin{align}P&=(2-1)(2+1)(3-1)(3+1)\dots(n-1)(n+1)
\\&=(2-1)(3-1)\dots(n-1)\times(2+1)(3+1)\dots(n+1)
\\&=\frac{(n-1)!(n+1)!}{2}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, after that. I don't know how to do next in order to find the minimum value. Please help!
| You probably know about setting the derivative equal to $0$. The complexity of the equation we get may be discouraging. Your derivative is
$$\frac{x}{\sqrt{x^2+4}}-\frac{12-x}{\sqrt{x^2-24x+153}}.$$
I would rather write it as
$$\frac{x}{\sqrt{x^2+4}}-\frac{y}{\sqrt{y^2+9}}.$$
Nicer! Set this equal to $0$. So we get
$$\frac{x}{\sqrt{x^2+4}}=\frac{y}{\sqrt{y^2+9}}.$$
Square both sides and simplify a bit (cross-multiply). We get
$$x^2(y^2+9)=y^2(x^2+4).$$
There is some nice cancellation, and we get $9x^2=4y^2$. The rest should not be difficult.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
$\int_{\Gamma} \frac{dz}{z}$, $\Gamma = \gamma([0,2\pi])$, $\gamma(t) = a\cos(t) + i b \sin(t)$ I am solving an exercise:
Evaluate by two methods: $$\int_{\Gamma} \frac{dz}{z}; \ \text{where:} \ \Gamma = \gamma([0,2\pi]), \text{and:}$$
$$\gamma: [0,2\pi] \longrightarrow \mathbb C \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t \ \ \ \ \ \ \longrightarrow \gamma(t) = a\cos(t) + i b \sin(t)$$
Where $a$ and $b$ are fixed positive reals.
One method would be using Cauchy's integral formula:
$$\int_{\Gamma} \frac{dz}{z} = 2 \pi i $$
The second method would be by using the definition:
$$\int_{\Gamma} \frac{dz}{z} = \int_0^{2\pi} \frac{1}{\gamma(t)} \dot {\gamma(t)} dt$$
Which results in, after multiplying the numerator and denominator by the complex conjugate:
$$\int_{\Gamma} \frac{dz}{z} = \int_0^{2 \pi} \frac{(b^2 - a^2)\cos(t) \sin(t)}{a^2 \cos^2(t) + b^2 \sin^2(t)}dt + iab \int_0^{2\pi}\frac1{a^2 \cos^2(t) + b^2 \sin^2(t)}dt$$
Clearly, the first one should evaluate to $0$ because there is no real part in the above answer. So, this leaves us to just calculate the second one.
However, the question asks (in the second part) to deduce the value of:
$$\int_0^{2\pi}\frac1{a^2 \cos^2(t) + b^2 \sin^2(t)}dt$$
So, I shouldn't calculate that beforehand (supposing that it is possible to actually calculate it).
So, there must be some trick to be done to both of the above integrals in order to extract a result. Maybe a suitable change of variable or something.
What I tried:
*
*Reduction of boundaries using periodicity and parity, then a change of variable $\theta = \tan(t/2)$. It wasn't useful.
*Adding and subtracting random things. But no nice expressions emerged.
*Double angle formulas, random changes of variables, etc.
Any help?
Thank you.
| The first thing to do is use the double-angle formulae,
$$ \cos{2\theta} = 2\cos^2{\theta}-1 = 1-2\sin^2{\theta}, $$
which turns the denominator into
$$ \frac{1}{2} \left( (a^2+b^2) + (a^2-b^2)\cos{2\theta} \right) $$
Then we are dealing with something that has period $\pi$, so the transformation of the integral to the interval $[-\pi/2,\pi/2]$ is reasonable. Now setting $t=\tan{\theta}$, we have
$$ \cos{2\theta} = \frac{1-t^2}{1+t^2}, \quad d\theta = \frac{dt}{1+t^2}, $$
so the integral transforms to
$$ \int_0^{2\pi} \frac{d\theta}{a^2\cos^2{\theta}+b^2\sin^2{\theta}} = 2\int_{-\pi/2}^{\pi/2} \frac{d\theta}{a^2\cos^2{\theta}+b^2\sin^2{\theta}} \\
= 4\int_{-\pi/2}^{\pi/2} \frac{d\theta}{(a^2+b^2) + (a^2-b^2)\cos{2\theta}} \\
= 4\int_{-\infty}^{\infty} \frac{d\theta}{(a^2+b^2)(1+t^2)+(a^2-b^2)(1-t^2)} \\
= 2\int_{-\infty}^{\infty} \frac{d\theta}{a^2+b^2 t^2},
$$
which is easy enough to do using $\arctan$, and gives $\pi/(ab)$.
The real part integral is actually quite easy: you can either note that $\frac{d}{d\theta} \sin^2{\theta} = 2\sin{\theta}\cos{\theta} $, or just use the double-angle formulae:
$$ \int_0^{2\pi} \frac{(b^2-a^2)\sin{\theta}\cos{\theta}}{a^2\cos^2{\theta}+b^2\sin^2{\theta}} \, d\theta = \int_0^{2\pi} \frac{(b^2-a^2)\sin{2\theta}}{(a^2+b^2) + (a^2-b^2)\cos{2\theta}} \, d\theta = \left[ \frac{1}{2}\log{\left( (a^2+b^2) + (a^2-b^2)\cos{2\theta} \right)} \right]_0^{2\pi} $$
Now, $a,b$ are real and positive, so $a^2+b^2>\lvert a^2-b^2 \rvert$, so the logarithm is a legitimate continuous antiderivative, and the integral is therefore equal to zero.
Entirely possibly, the question is actually asking that you just get down to
$$ 2\pi i = \int_{\gamma} \frac{dz}{z} = iab\int_0^{2\pi} \frac{d\theta}{a^2\cos^2{\theta}+b^2\sin^2{\theta}}, $$
from which you then get the value of the latter integral by transitivity of equality (to get fancy with our deductive terminology).
Perhaps a sillier suggestion: substitute $e^{i\theta}=w$, so $d\theta = dw/(iw)$. Then you have
$$ \int_{0}^{2\pi} \frac{-a\sin{\theta}+ib\cos{\theta}}{a\cos{\theta}+b\sin{\theta}} d\theta = \int_{|w|=1} \frac{-ia(w-1/w)+b(w+1/w)}{a(w+1/w)-ib(w-1/w)}\frac{dw}{iw} \\
= \dotsb = \int_{|w|=1} \left( \frac{2(a+b)w}{a-b+(a+b)w^2}-\frac{1}{w} \right) dw$$
Right, now we have the fun of doing residues. The residue at $w=0$ is obviously $-1$. The problematic ones are $w=\pm \sqrt{(b-a)/(a+b)}$. First question: are they even inside $|w|=1$? Answer: $|b-a|<a+b$, so yes, both are. And it turns out the residues are actually reasonable to compute, because
$$ \frac{2w}{w^2-c^2} = \frac{1}{w-c}+\frac{1}{w+c}, $$
and hence both residues are $1$. Hence the sum of the residues is $1+1-1=1$, and the integral is
$$ \int_{0}^{2\pi} \frac{-a\sin{\theta}+ib\cos{\theta}}{a\cos{\theta}+b\sin{\theta}} d\theta = 2\pi i. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is this correct $(\cot x)(\sin x)=2(\cot x)^2$ $0≤x≤2\pi$ $x= \pi/2 , 3\pi/2$ $1.4371774 , 5.139467567$ Steps I took:
$$\cot x \sin x=2\cot^2 x$$
$$\cot x \sin x-2\cot^2 x=0$$
$$\cot x (\sin x-2\cot x)=0$$
$$\cot x \left(\frac{\sin^2 x}{\sin x} -2\frac{\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{\sin^2 x-2\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{1-\cos^2 x-2\cos x}{\sin x} \right)=0$$
$$\cot x \csc x (\cos x+2.414213562)(\cos -0.4142135624)=0$$
$$\cot x =0 \rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}$$
$\csc x$ cannot be $0$
$\cos x$ cannot be $-2.414213562$
$$\cos x = 0.4142135624$$
$$x=1.4371774 , 5.139467567$$
| Corrected solution:
Steps I took:
$$\cot x \sin x=2\cot^2 x$$
$$\cot x \sin x-2\cot^2 x=0$$
$$\cot x (\sin x-2\cot x)=0$$
$$\cot x \left(\frac{\sin^2 x}{\sin x} -2\frac{\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{\sin^2 x-2\cos x}{\sin x} \right)=0$$
$$\cot x \left(\frac{1-\cos^2 x-2\cos x}{\sin x} \right)=0$$
$$\cot x \csc x (\cos^2 x+2\cos x-1)=0$$
$$\cot x =0 \rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}$$
$$\csc x = 0 \rightarrow DNE$$
$$\cos^2 x +2\cos x -1=0 \rightarrow \cos x = \frac{-2\pm\sqrt{4-4*-1}}{2}=-1\pm\sqrt{2}$$
$$x=2\pi n \pm \cos^{-1} \left( -1\pm\sqrt{2}\right)$$
$$n \in \mathbb{Z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Roots of unity question. Question
Let $\omega=\cos\dfrac{4\pi}{7}+i\sin\dfrac{4\pi}{7}$. Show that
$\omega-1=2\sin\dfrac{2\pi}{7}\left(\cos\dfrac{11\pi}{14}+i\sin\dfrac{11\pi}{14}\right)$.
My attempt
Observe that $\omega$ is a seventh root of unity. Label the roots $1, \nu, \nu^2,\ldots,\nu^6$. Then $\omega=\nu^2$.
We have $1+ \nu+ \nu^2+\ldots+\nu^6=0$ and so $1+ \nu+ \omega+\ldots+\nu^6=0$.
Then $\omega-1=-2-(\nu+\nu^6)-(\nu^3+\nu^4)-\nu^5$.
But $\nu+\nu^6=2\cos\dfrac{2\pi}{7}$ and $\nu^3+\nu^4=2\cos\dfrac{6\pi}{7}$
I do not know how to continue.
| I would do it differently. Recall identities
$$\cos 2x = 1 - 2\sin^2 x \\[1ex]
\sin 2x = 2 \sin x \cos x$$
to obtain
$$\cos \frac{4 \pi}{7} = 1 - 2 \sin^2 \frac{2 \pi}{7} \\[1ex]
\sin \frac{4 \pi}{7} = 2 \sin \frac{2 \pi}{7} \cos \frac{2 \pi}{7}$$
and therefore
$$\omega - 1 = 2 \sin \frac{2 \pi}{7} \left( - \sin \frac{2 \pi}{7} + i \cos \frac{2 \pi}{7} \right).$$
But
$$
\begin{align*}
- \sin \frac{2 \pi}{7} + i \cos \frac{2 \pi}{7} & = i \left( \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7} \right) = \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) \left( \cos \frac{2 \pi}{7} + i \sin \frac{2 \pi}{7} \right) \\[2ex]
& = \cos \left( \frac{\pi}{2} + \frac{2 \pi}{7} \right) + i \sin \left( \frac{\pi}{2} + \frac{2 \pi}{7} \right) = \cos \frac{11 \pi}{14} + i \sin \frac{11 \pi}{14}.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
| Ever so slightly off-topic, but I can't resist reminding folks of the proof that $\sqrt[n]{2}$ is irrational for $n \ge 3$ using Fermat's Last Theorem:
Suppose that $\sqrt[n]{2} = a/b$ for some positive integers $a$ and $b$. Then we have $2 = a^n / b^n$, or $b^n + b^n = a^n$. But Andrew Wiles has shown that there are no nonzero integers $a, b$ satisfying the last equation. Thus $\sqrt[n]{2}$ must be irrational. [This proof is due to W. H. Schultz and appeared in the May, 2003 issue of the American Mathematical Monthly.]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 10
} |
Is this a valid partial fraction decomposition?
Write $\dfrac{4x+1}{x^2 - x - 2}$ using partial fractions.
$$ \frac{4x+1}{x^2 - x - 2} = \frac{4x+1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x-2} = \frac{A(x-2)+B(x+1)}{(x+1)(x-2)}$$
$$4x+1 = A(x-2)+B(x+1)$$
$$x=2 \Rightarrow 4 \cdot2 + 1 = A(0) + B(3) \Rightarrow B = 3$$
$$x = -1 \Rightarrow 4(-1) +1 = A(-3)+ B(0) \Rightarrow A = 1$$
Thus,
$$\frac{4x+1}{x^2-x-2} = \frac{1}{x+1} + \frac{3}{x-2}\textrm{.}$$
The substitution of $x$ ($x = 2, -1$) is a common method to find out the coefficient of the partial fractions. However, the equation on the third line is obtained by multiplying $(x+1)(x-2)$, which is assumed to be nonzero. Here we have a contradiction. Furthermore, the original function is not defined at $x=-1,2$.
How can we substitute these value for $x$? So is this method valid and rigorous? How to modify it so that it is rigorous?
| This is a classical source of confusion. But the the problem is avoided completely if one takes the limit of both sides as $x \to 2$, instead of substituting $x=2$.
Clearly the result is the same, but with this formulation there is no evaluation of functions at points where they are undefined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What is the spectrum of this matrix? $$A_n=\begin{bmatrix}
1 & 1 & 1 & \cdots & 1 & 1\\
1 & 2 & 2 & \cdots & 2 & 2\\
1 & 2 & 3 & \cdots & 3 & 3\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
1 & 2 & 3 & \cdots & n-1 & n-1\\
1 & 2 & 3 & \cdots & n-1 & n
\end{bmatrix}.$$
What are the eigenvalues and the corresponding eigenvectors of $A_n$?
| Since the inverse of $A_n$ is
$$
A^{-1}_n = \begin{pmatrix}
2 & -1 & 0 & \dots & 0 & 0\\
-1 & 2 & -1 & \dots & 0 & 0\\
0 & -1 & 2 & \dots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \dots & 2 & -1\\
0 & 0 & 0 & \dots & -1 & 1
\end{pmatrix}
$$
Solving $A_n^{-1} x = \lambda^{-1} x$ is the same that solving difference problem
$$\begin{aligned}
x_0 &= 0\\
-x_{k+1} + 2x_k -x_{k-1} &= \lambda^{-1} x_k, \quad k = 1, \dots, n-1\\
x_n - x_{n-1} &= \lambda^{-1} x_n
\end{aligned}
$$
Substituting $x_k = \sin \omega k$ gives following system of equations for $\omega, \lambda$
$$
4\sin^2\frac{\omega}{2} = \lambda^{-1}\\
(1-\lambda^{-1})\sin \omega n = \sin \omega (n-1)
$$
or
$$
1 - 4\sin^2\frac{\omega}{2} = \frac{\sin \omega(n-1)}{\sin \omega n}
$$
That equation has exactly $n$ different solutions for $\omega$ in $(0, \pi)$ since left is monotonically decreasing and right is increasing and has $n-1$ poles (see image for $n=5$). Values outside of $(0, \pi)$ produce the same values for $\lambda$ due to periodicity. So
$$
\lambda_j^{-1} = 4\sin^2\frac{\omega_j}{2}\\
(x_j)_k = \sin k \omega_j
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1314577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
First contest problem I downloaded a contest and worked the first problem which is:
There exists a digit Y such that, for any digit X, the seven-digit number 1 2 3 X 5 Y 7 is not a multiple of 11.
Compute Y.
My attempt:
$$\begin{align}
1+3+5+7&=2+X+Y \\
16&=2+X+Y \\
14&=X+Y \\
\implies\quad Y&=14-X \\
\end{align}$$
The values of Y for which the number can be divided by $11$ are: $5,6,7,8,9.$
How would I deduce that the only value of Y for which the number can't be divided by $11$ is only $4$?
| You say $Y =14-X$ if the number is divisible. Because X is a digit (in base ten), it ranges from $0$ to $9$; thus $Y$ ranges from $14-0=14$ to $14-9=5$. But note that $14$, $13$, $12$, $11$, and $10$ are not accepted values for a single digit. If the number has to be divisible by $11$, $Y$ must be $11$ less than the numbers above. Thus $Y$ can be 3, 2, 1, 0, -1 (this is still invalid), or from 5 to 9 as before.
To make the number not divisible, $Y$ must therefore be 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3 + xy^2$$
imaginary component
$$-2yx^2 i +yi + x^2 yi - y^3 i=0i$$
$$-2yx^2 +y + x^2 y - y^3 =0$$
...
can't solve this question
| If I started with your solution $$-2yx^2 +y + x^2 y - y^3 =0\implies y(x^2+y^2-1)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How to find $\lim_{x \to 0}\frac{\cos(ax)-\cos(bx) \cos(cx)}{\sin(bx) \sin(cx)}$ How to find $$\lim_\limits{x \to 0}\frac{\cos (ax)-\cos (bx) \cos(cx)}{\sin(bx) \sin(cx)}$$
I tried using L Hospital's rule but its not working!Help please!
| i will expand on the hint given by dr. mv. we will compute the numerator and denominators separately. we have
$$\begin{align}\cos ax - \cos bx \cos cx &= 1 - a^2x^2/2 + \cdots - (1 - b^2x^2/2+\cdots)(1-c^2x^2/2+\cdots)\\&= \frac12 (b^2 + c^2 - a^2) x^2 + \cdots\\
\sin bx \sin cx &= (bx + \cdots)(cx + \cdots) = bcx^2 + \cdots\end{align}$$
therefore, $$\lim_{x \to 0}\frac{\cos ax - \cos bx \cos cx}{\sin bx \sin cx} = \frac{b^2 + c^2 - a^2}{2bc}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Solving a linear congruence Use Euclids algorithm to find the multiplicative inverse of 11 modulo 59 and hence solve the linear congruence:
$11x \equiv 8 \mod59$
My working so far....
$ {11v + 51w = 1}$
Using Euclid's algorithm:
$ {59 = 5 \times 11 + 4}$
${11 = 2\times 4 + 3}$
${4 = 1 \times 3 + 1}$
$ {3 = 1 \times 3 + 0}$
$ {11v + 51w = 1}$
rearrange the equations to make the remainder the subject:
$ {4 = 59 - 5 \times 11}$
$ {3 = 11 -2 \times 4 }$
$ {1 = 4 - 1 \times 3} $
Using backwards substitution:
${1 = 4 -1 \times (11 -2 \times 4)}$
${1 = 3 \times 4 - 1 \times 11}$
$ {1 = 3 \times (59 - 5 \times 11) -1 \times 11}$
$ {1 = 3 \times 59 - 16 \times 11}$
$ {1 = 59(3) + 11(-16)}$
${59v + 11w = 1} $ where v = 3 and w = -16
I'm not sure where to go from here to solve the linear congruence.
| Good, so now you know $\displaystyle{59(3)+11(-16)=1}$. This tells you that
$$59(3)+11(-16)\equiv 1\pmod{\! 59}$$
$$\iff 11(-16)\equiv 1\pmod{\! 59}$$
Multiply both sides by $8$:
$$11(-16\cdot 8)\equiv 8\pmod{\! 59}$$
$$\iff 11(-128)\equiv 8\pmod{\! 59}$$
$$\iff 11(49)\equiv 8\pmod{\! 59}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent
First i subbed numbers in
$$\lim_{n \to \infty} \frac{(-1)^n}{1+\sqrt{n}} = \frac{-1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} - \frac{-1}{1+\sqrt{3}}$$
So it's divergent
1 LIMIT
$$\lim_{n \to \infty} \frac{1}{1+\sqrt{n}}=0 \quad \text{hence divergent} $$
2 $a_{n}$ and $a_{n+1}$
$$ \frac{1}{1+\sqrt{n}}>\frac{1}{1+\sqrt{n+1}} $$
$$ \frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}}> 0 $$
hence increasing
or
$$ \frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}} = \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n+1}} $$
$$ \therefore \sqrt{n+1}-\sqrt{n}> 0$$ hence decreasing
$ \sum_{n=1}^{\infty} \left\lvert \frac{-1^n}{1+\sqrt{n}} \right\rvert = \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} $
$$ \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} = \frac{1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} + \frac{1}{1+\sqrt{3}} $$
if this is convergent then the whole thing is absolutely convergent but i don't know how to prove this
UPDATE
I got something out for the second part
| Absolute convergence
If
$$ \sum \left|a_n\right| =\mbox{convergent}$$
Then
$$ \sum a_n =\mbox{absolutely convergent}$$
Note that an absolutely convergent series is also convergent.
Conditional convergence
If
$$ \sum \left|a_n\right| =\mbox{divergent}$$
And
$$ \sum a_n =\mbox{convergent}$$
Then
$$ \sum a_n =\mbox{conditionally convergent}$$
So now
$$\sum\limits_{n=1}^{\infty} \left|\frac{(-1)^n}{1+\sqrt{n}}\right|=\sum\limits_{n=1}^{\infty} \left|\frac{1}{1+\sqrt{n}}\right|$$
Since
$$ \sum\limits_{n=1}^{\infty}\left|\frac{1}{\sqrt{n}+\sqrt{n}}\right|\leq \sum\limits_{n=1}^{\infty}\left|\frac{1}{1+\sqrt{n}}\right|$$
And
$$ \sum\limits_{n=1}^{\infty}\left|\frac{1}{\sqrt{n}+\sqrt{n}}\right| =\frac12\sum\limits_{n=1}^{\infty}\left|\frac{1}{\sqrt{n}}\right|=\mbox{divergent}$$
Then by the comparison test
$$\sum\limits_{n=1}^{\infty}\left|\frac{1}{1+\sqrt{n}}\right|=\mbox{divergent}$$
Now for the following series
$$\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$$
Since
$$\lim\limits_{n\to\infty} \frac{1}{1+\sqrt{n}}=0$$
And
$$ \frac{1}{1+\sqrt{n}} \gt \frac{1}{1+\sqrt{n+1}} $$
Then by the alternating series test
$$\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}} =\mbox{convergent}$$
Putting everything together, we see that
$$\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}} =\mbox{conditionally convergent}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
System of equations with complex numbers-circles The system of equations
\begin{align*}
|z - 2 - 2i| &= \sqrt{23}, \\
|z - 8 - 5i| &= \sqrt{38}
\end{align*}
has two solutions $z_1$ and $z_2$ in complex numbers. Find $(z_1 + z_2)/2$.
So far I have gotten the two original equations to equations of circles,
$(a-2)^2 +(b-2)^2=23$ and $(a-8)^2+(b-5)^2=38$.
From here how do I find the solutions?
Thanks.
| The first thing to do is to make a drawing of the two circles. Label their centres $C_1$ and $C_2$, and the two intersections of the circles $I_1$ and $I_2$. Now connect these four points by straight lines, and label as $P$ the point half-way between $I_1$ and $I_2$, which is on the line connecting $C_1$ and $C_2$.
We recognize that we have four triangles, all with a $90$ degree angle. So we can apply Pythagoras. For convenience we label the line-piece $C_1$ to $P$ as $a$; $C_2$ to $P$ as $b$; $I_1$ (or $I_2$) to $P$ as $c$. Now:
$$(a+b)^2 = 45$$
$$a^2 + c^2 = 23$$
$$b^2 + c^2 = 38$$
We eliminate $c^2$ by subtracting the second equation from the third, yielding:
$$(b-a)(b+a) = 15$$
Dividing the first equation by this result gives
$$ \frac {b+a}{b-a} = 3$$
From which it follows that $b = 2a$. So $a$ is one-third of the distance between $C_1$ and $C_2$. Therefore the position of point $P$ is given by $2+2i$ + $(6+3i)/3$ = $4+3i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.