Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Factorize $x^4 + y^4 -x^2 y^2$ over $\mathbb C$ As in the title, I'd like to factorize $x^4 + y^4 - x^2y^2$ into irreducible factors over $\mathbb C$ (i.e. linear factors).
Attempts:
First I tried doing
$$x^4 + y^4 - x^2y^2 = (x^2 - y^2)^2 + x^2y^2 \\ = (x^2 - y^2) - (ixy)^2 \\ = (x^2 - y^2 - ixy)(x^2 - y^2 + ixy) $$
And I got stuck at this point.
Second (desperately), I expanded
$$(a_1 x + b_1y)(a_2 x + b_2 y)(a_3 x + b_3 y)(a_4 x + b_4 y)$$
and compared it to $x^4 + y^4 - x^2y^2$, but let's just say it didn't go smoothly.
I think there should be a nice way to do it. Anybody?
| $$x^4-x^2y^2+y^4=(x^2+y^2)^2-3x^2y^2=(x^2-\sqrt3xy+y^2)(x^2+\sqrt3xy+y^2)=$$
$$=\left(\left(x-\frac{\sqrt3}{2}y\right)^2+\frac{y^2}{4}\right)\left(\left(x+\frac{\sqrt3}{2}y\right)^2+\frac{y^2}{4}\right)=$$
$$=\left(x-\frac{\sqrt3}{2}y+\frac{1}{2}yi\right)\left(x-\frac{\sqrt3}{2}y-\frac{1}{2}yi\right)\left(x+\frac{\sqrt3}{2}y+\frac{1}{2}yi\right)\left(x+\frac{\sqrt3}{2}y-\frac{1}{2}yi\right).$$
| {
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"url": "https://math.stackexchange.com/questions/3258328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inverse of $\frac{x}{1-x^2}$ How did Munkres (page 104 – new international edition) invert the function
$$
F(x) = \frac{x}{1-x^2} \,, \quad x \in (-1,1)
$$
for
$$
G(y) = \frac{2y}{1 + \sqrt{1+4y^2}} \,, \quad y \in \mathbb{R} \,?
$$
Looks like some trig-substitution but cannot see it.
| The usual method to obtain the inverse is to let $F(x)=y$, interchange $x$ and $y$, and solve for $y$. Thus we solve
$$x=\frac{y}{1-y^2}$$
for $y$:
\begin{align*}
x&=\frac{y}{1-y^2}\\
\implies x-xy^2&=y\\
\implies xy^2+y-x&=0\\
\implies y^2+\tfrac1xy-1&=0\\
\implies(y+\tfrac1{2x})^2&=\frac{1+4x^2}{4x^2}\qquad\text{(completing the square)}\\[4pt]
\implies y+\frac{1}{2x} &=\frac{\sqrt{1+4x^2}}{2x}\\
\implies y&=\frac{-1+\sqrt{1+4x^2}}{2x},
\end{align*}
which is equivalent to $G(x)$ if you multiply the top and bottom by $-1-\sqrt{1+4x^2}$.
Think of this last extra step as similar to realising the denominator when you multiply by the complex conjugate. You can also get to it straight away by using the less well-known Muller's version of the quadratic formula:
$$ x= \frac{-2c}{b\mp\sqrt{b^2-4ac}}.$$
(This version has the advantage that if $a=0$, you still get valid roots).
| {
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"timestamp": "2023-03-29T00:00:00",
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value of $2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$
If $ x^3+bx^2+cx+1=0$ has only real root $\alpha $.
Where $(b<c)$. Then $\displaystyle 2\tan^{-1}(\csc \alpha)+\tan^{-1}(2\sin \alpha\sec^2\alpha)$ is
Plan
$$\tan^{-1}\bigg(\frac{2\csc \alpha}{1-\csc^2\alpha}\bigg)+\tan^{-1}\bigg(2\sin \alpha\sec^2\alpha\bigg)$$
$$\tan^{-1}\bigg(\frac{\frac{2\csc\alpha}{1-\csc^2\alpha}+2\sin\alpha\sec^2\alpha}{1-\frac{2\csc\alpha}{1-\csc^2\alpha}2\sin\alpha\sec^2\alpha}\bigg)$$
How do i solve it Help me please
| Let $f(x)=x^3+bx^2+cx+1$
$f(0)=1>0$
$f(-1)=(b-c)<0$
So, $\alpha$ lies between $f(0)$ and $f(-1)$ which is $-1,0$
$2\tan^{-1}(\csc\alpha)+\tan^{-1}(2\sin\alpha\sec^2\alpha)$
$2\tan^{-1}\alpha\left(\dfrac{1}{\sin\alpha}\right)+\tan^{-1}\left(\dfrac{2\sin\alpha}{\cos^2\alpha}\right)$
$2\tan^{-1}\alpha\left(\dfrac{1}{\sin\alpha}\right)+\tan^{-1}\left(\dfrac{2\sin\alpha}{1-\sin^2\alpha}\right)$
$2\left[\tan^{-1}\left(\dfrac{1}{\sin\alpha}\right)\right]+\tan^{-1}(\sin\alpha)$
$2\left(-\dfrac{\pi}{2}\right)=-\pi$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a,b,c>0$ and $ab+bc+ca=3$, prove that $\sum_{cyc} \frac{a}{\sqrt{a^3+5}} \leq \sqrt{6}/2$
If $a,b,c>0$ and $ab+bc+ca=3$, prove that $\displaystyle \sum_{cyc} \frac{a}{\sqrt{a^3+5}} \leq \sqrt{6}/2$.
My attempt was to use firstly AM-GM in the denominator, like $a^3+5 \geq 3a+3$ and the set $a=x^2-1$ but a lot of trouble occurred.
Then, I tried to use AM-HM, like $\sqrt{6 \cdot (a^3+5)} \geq ...$ but this didn't work either.
Any help?
I think it's completely obvious that if two problems have the same condition and are inequalities, are NOT the same.
| Another AM-GM helps:
$$\sum_{cyc}\frac{a}{\sqrt{a^3+5}}\leq\sum_{cyc}\frac{a}{\sqrt{\frac{3a^2-1}{2}+5}}=\sqrt{\frac{2}{3}}\sum_{cyc}\frac{a}{\sqrt{a^2+3}}=$$
$$=\sqrt{\frac{2}{3}}\sum_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq\frac{1}{2}\sqrt{\frac{2}{3}}\sum_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{\sqrt6}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260825",
"timestamp": "2023-03-29T00:00:00",
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Formation of commissions with generating functions Representatives of three research institutes should form a commission of 9 researchers. How many ways can this committee be formed such that no institute should have an absolute majority in the group?
My partial solution:
I will use the exponential generating function because the researchers of the three institutes are different. The maximum number of researchers at an institute is 8, so the generating function will be given by:
$$f(x) = \left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + \frac{x^7}{7!} + \frac{x^8}{8!}\right)^3$$
Is it correct here? How do I proceed?
| You should not use an exponential generating function here. You would use this if your were choosing an ordered lists of nine researchers. Here, we are just choosing how many representative appear from each institution, so ordinary generating functions are applicable.
The number of representatives can be anywhere between $0$ and $4$, because if there were $5$ or more, there would be an absolute majority. Therefore, the generating function for a single institution is $(1+x+\dots+x^4)=(1-x^5)/(1-x)$. Since there are three institutions, the generating function is the cube of that,
$$
(1-x^5)^3(1-x)^{-3}\tag{$*$}
$$
Finally, we want the coefficient of $x^9$ in the above expression. This is a product of two nice functions, $(1-x^5)^3$ and $(1-x)^{-3}$ so the coefficient of $x^9$ in the product is the convolution of these two sequences. Namely, since
$$
(1-x^5)^3= 1-3x^5+3x^{10}-x^{15},\quad \text{and}\\
(1-x)^{-3}=\sum_{k\ge 0}\binom{k+2}{2}x^k
$$
it follows that the coefficient of $x^9$ in $(*)$ is
$$
\binom{9+2}{2}-3\binom{4+2}2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\int^2_0 xe^{x^2}dx$ Well what I was thinking was to integrate the indefinite integral first.
$u=x^2$, $x=\sqrt u$
$du=2xdx = 2\sqrt {u} dx$
$dx= \frac{1}{2\sqrt{u}}du$
$\int xe^{x^2} dx = \int \sqrt{u}\frac{1}{2\sqrt{u}} du =\frac{1}{2}\int e^u du = \frac{1}{2}e^u =\frac{1}{2}e^{x^2} +C$
Now I can evaluate $\frac{1}{2}e^{x^2}\Big|_0^2= \frac{1}{2} e^{4} -\frac{1}{2} e^0 =\frac{1}{2}e^4-1$
so my answer should be $$\frac{1}{2}e^4-1$$
Is this correct? It's been a while since I've done stuff like this.
| You mean $\frac12 e^4-\frac12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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quadratic equation solving mistake I'm a student who started self learning quadratic equations for a youth university program. I'm busy at trying to solve such equation:
$$
(1 - 4x)^2 + 9x + 7 = 2(x+3)(1-x) + (x+4)^2
$$
this is my current progress:
\begin{align}
(1 - 4x)^2 + 9x + 7 &= 2(x+3)(1-x)+ (x+4)^2\\
(1 - 4x)(1 - 4x) + 9x + 7 &= (2x + 6)(1 - x) + (x + 4)(x + 4)\\
1 - 4x - 4x + 16x^2 + 9x + 7 &= 2x - 2x^2 + 6 - 6x + x^2 + 4x + 4x + 16\\
8 + 16x^2 + x &= 2x - x^2 + 6 - 6x + 8x + 16\\
8 + 16x^2 + x &= 4x - x^2 + 22\\
16x^2 + x &= 4x - x^2 + 14\\
16x^2 &= 3x - x^2 + 14\\
17x^2 &= 3x + 14
\end{align}
The solutions to this equation are $x = 1,~x=-14/17$.
So, where is my mistake? $x$ is negative, so I must be incorrect.
| Your working is all fine. Only the last step was remaining to be written down after factoring.
| {
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Circle and Locus _ ONLY PEN AND PAPER ALLOWED. Q) Let T be the line passing through the points P(–2, 7) and Q(2, –5). Let $F_{1}$ be the set of all pairs of circles $(S_{1}$, $S_{2}$) such that T is tangent to $S_{1}$ at P and tangent to $S_{2}$ at Q, and also such that $S_{1}$ and $S_{2}$ touch each other at a unique point, say, M. Let $E_{1}$ be the set representing the locus of M as the pair ($S_{1}$, $S_{2}$) varies in $F_{1}$. Let the set of all straight line segments joining a pair of distinct points of $E_{1}$ and passing through the point R(1, 1) be $F_{2}$. Let $E_{2}$ be the set of the mid-points of the line segments in the set $F_{2}$. Let $C$ be the circle $x^2+y^2+6(2y+7x)=53$. The number of times $C$ intersects $E_{1}$ and $E_{2}$ is (are):
| As the diameter of the first is $PQ$ $$E_1=\{(x,y)| (x+2)(x-2)+(y-7)(y+5)=0\}-\{P,Q\}$$
and the diameter of the second is $DR$, with $D$ the midpoint of $PQ$
$$E_2=\{(x,y)| x(x-1)+(y-1)^2=0\}-\{(4/5,7/5),(36/37,43/37)\}$$
where the disallowed points stem from the intersections with $l_{PR}$ and $l_{QR}$.
Now
$$\#((V((x+2)(x-2)+(y-7)(y+5))\cup V(x(x-1)+(y-1)^2))\cap C)=4$$
but three points are disallowed, leaving the other intersection $(400/409,349/409)$ so
$$\#((E_1\cup E_2)\cap C)=1$$
Edit
$E_2$ is part of a circle: let $y-1=m(x-1)$ be lines through $R$, then together with the equation for the circle $E_1$ is a part of, we get the relation $(m^2+1)y^2+(-2m^2+2m-2)y-38m^2-2m+1=0$ that gives us solutions $y_{1,2}$ with corresponding $x_{1,2}$. What we're looking for are points $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{1}{m^2+1},\frac{m^2-m+1}{m^2+1})$ which implicitizes to $x^2-x+y^2-2y+1=0.$
| {
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Prove $\lim\limits_{n\to \infty}\frac{1}{\sqrt n}\left|\sum\limits_{k=1}^n (-1)^k\sqrt k\right|= \frac{1}{2}$ I'm trying to show that $$\lim_{n\to \infty} x_n=\lim_{n\to \infty}\frac{1}{\sqrt n}\left|\sum_{k=1}^n (-1)^k\sqrt k\right|= \frac{1}{2}.$$
Assuming $\lim\limits_{n\to\infty} x_n=x$ exists, we have
$$x_{2n}=\frac{\sqrt{2n-1}(-x_{2n-1})+\sqrt {2n}}{\sqrt {2n}}$$
Letting $n\to \infty$,
$$\quad \quad x=-x+1$$
$$x=\frac{1}{2}$$
But I'm stuck on proving the existence of $\lim x_n$. Any idea?
Update: I just solved the problem using sandwich theorem + integral test. Still, I would like to see a continuation of my initial idea, i.e. proving
$\{x_{2n}\}$ is monotonically increasing (similarly, $\{x_{2n+1}\}$ is monotonically decreasing)
| $\displaystyle \lim_{n\to\infty}\frac{1}{\sqrt{n}}\left|\sum\limits_{k=1}^n (-1)^k\sqrt{k}\right|
= \lim_{n\to\infty}\frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n \frac{1}{\sqrt{2k-1}+\sqrt{2k}} $
$\displaystyle \frac{1}{2\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{ \sqrt{2k} } < \frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{\sqrt{2k-1} +\sqrt{2k} } $$\displaystyle < \frac{1}{2\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{\sqrt{2k-1} } < \frac{1}{2\sqrt{2n}}\left(1+\sum\limits_{k=1}^{n-1} \frac{1}{ \sqrt{2k} }\right)$
With Riemann we get:
$\displaystyle \frac{1}{\sqrt{2n}}\sum\limits_{k=1}^n\frac{1}{ \sqrt{2k} } = \frac{1}{2n}\sum\limits_{k=1}^n\frac{1}{\sqrt{k/n}} \to \frac{1}{2}\int\limits_0^1\frac{dx}{\sqrt{x}} = \sqrt{x}|_0^1 =1$
And therefore the confirmation of the claim.
About the monotonie.
$\displaystyle s_n := \sum\limits_{k=1}^n \frac{1}{\sqrt{2k-1}+\sqrt{2k}}$
Assumption about the monotonie: $\enspace \displaystyle \frac{s_n}{\sqrt{2n}} \enspace$ is strictly increasing.
$\displaystyle \frac{s_n}{\sqrt{2n}} < \frac{s_{n+1}}{\sqrt{2{n+2}}} \,$ leads to
$\displaystyle \left(\frac{1}{\sqrt{2n}} - \frac{1}{\sqrt{2n+2}}\right)s_n < \frac{1}{\sqrt{2n+2}}\frac{1}{ \sqrt{2n+1} + \sqrt{2n+2} }\,$ and therefore to
$\displaystyle s_n < a_n:=\frac{\sqrt{2n}}{2}\frac{ \sqrt{2n+2}+\sqrt{2n} }{\sqrt{2n+2} +\sqrt{2n+1} } $
For $n=1$ it’s o.k. . Assume it’s correct for $n$ . $(*)$
Then we have to show that it's also correct for $n \to n+1$ .
$\displaystyle s_{n+1} < \frac{\sqrt{2n}}{2}\frac{ \sqrt{2n+2}+\sqrt{2n} }{\sqrt{2n+2} +\sqrt{2n+1} } + \frac{1}{\sqrt{2n+1} +\sqrt{2n+2} } < a_{n+1}$
The first inequation follows from the assumption $(*)$ and the second inequation can be proved by some transformations, $2n$ replaced by $x$ and $x\geq 0$:
$(\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+4}+\sqrt{x+3}) <$
$<\sqrt{x+2}(\sqrt{x+4}+ \sqrt{x+2})( \sqrt{x+2}+\sqrt{x+1})$
Transformations:
$(\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+4}+\sqrt{x+2}) $$+ (\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+3}-\sqrt{x+2}) <$
$<\sqrt{x+2}(\sqrt{x+4}+ \sqrt{x+2})( \sqrt{x+2}+\sqrt{x+1})$
$(\sqrt{x}(\sqrt{x+2} +\sqrt{x}) + 2)(\sqrt{x+3}-\sqrt{x+2}) $$< (\sqrt{x+2}(\sqrt{x+2}+\sqrt{x+1})-(\sqrt{x}(\sqrt{x+2}+\sqrt{x})+2))(\sqrt{x+4}+\sqrt{x+2})$
$(\sqrt{x}\sqrt{x+2} +x+2)(\sqrt{x+3}-\sqrt{x+2}) $$< (\sqrt{x+2}\sqrt{x+4} +x+2)(\sqrt{x+1}-\sqrt{x})$
This is true because of:
$\sqrt{x}\sqrt{x+2} +x+2< \sqrt{x+2}\sqrt{x+4} +x+2$
$\sqrt{x+3}-\sqrt{x+2} < \sqrt{x+1}-\sqrt{x}$
Note: $\enspace\sqrt{x+1+a}-\sqrt{x+a}~$ is decreasing by growing $\,a>-x$
| {
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Stuck on this - In ABC right triangle AC= $2+\sqrt{3}$ and BC = $3+2\sqrt{3}$. circle touches point C and D, Find the Area of $AMD$ In ABC right triangle $AC= 2+\sqrt{3}$ and $BC = 3+2\sqrt{3}$. circle touches point C and D, Find the Area of $AMD$
Here's my strategy of solving this, I'm not sure if it's correct, if you find my explanation hard to understand you can just ignore and write the solution in your own way, thanks.
1. first area of main triangle, we know AC and CB so it'll be easy to calculate that,
2. to find the radius, we'll reflect triangle ABC on the left side of the circle, turning it into circle inscribed in isosceles triangle, and find it with the formula
3. to find the area of $AMD$ I'll subtract the area of sector $OMD$, triangle $OAD$ and triangle $CDB$ from triangle $AMD$,
4. $DBC$ is an isosceles triangle, so $CB=DB$, then to find the area, I split it into 2 right triangles(it becomes 90 30 60 triangle) and find its height. So we got the Area of $DBC$
5. Now similarly $OAD$ is isosceles, $OD=OC=radius$ of the circle which we "found"
also, split this in 2 to get right triangles and then calculate with Pythagorean theorem to find the height so we get Area of $OMD$ too, maybe we could find angles with trigonometry? I don't know that, and if we get the angle of $DOA$ we could find the sector $OMD$ as well and subtract it to the main triangle so we get the area of $AMD$.
| Clearly, $\angle CAB = 60$ because $BC = 3+2\sqrt{3} = \sqrt{3} \cdot (2 + \sqrt{3}) = \sqrt{3} \cdot AC$. Now, we need to find the lengths $AM$ and $AD$. Let $r$ be the radius of the circle. It is clear that $OC = r$, and $AO = \sqrt{3}r$ because of the 30-60-90 triangle $AOD$, as $OD$ is a tangent to $AB$. Therefore, $r (1 + \sqrt{3}) = AC = 2 + \sqrt{3} \implies r = \frac{\sqrt{3}+1}{2}$.
Now, we compute the area. Note that $AM = AO - MO = (\sqrt{3}-1)r$, and $AD = \frac{\sqrt{3}}{3}r$, so we have that $[AMD] = AM \cdot AD \cdot \frac{\sqrt{3}}{4} = (\sqrt{3}-1)(\frac{\sqrt{3}}{3})(\frac{\sqrt{3}}{4})r^2 = \frac{1+\sqrt{3}}{8}$.
| {
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Find the limit of $\lim \limits_{x \to 1} \frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}$ I need to find the limit of:$$\lim \limits_{x \to 1} \frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}$$ without L'Hospital.
I tried to rationalize the expression but I found nothing. Also I tried to use a substitution but it didn't work either.
Any hints?
| Let $5x-4=t^{12}$, then we have
:$$\lim \limits_{x \to 1} \frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}=\lim \limits_{t \to 1}\frac{t^4-1}{t^3-1}$$
and $$\lim \limits_{t \to 1}\frac{t^4-1}{t^3-1}=\lim \limits_{t \to 1}\frac{(t-1)(t+1)(t^2+1)}{(t-1)(t^2+t+1)}=\frac{4}{3}$$
| {
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Evaluate $\int\limits_0^{1}\frac{\sqrt{1+x^2}}{1+x}dx$ Evaluate:
$I=\int\limits_0^1 \frac{\sqrt{1+x^2}}{1+x}dx$
My try:
Let $x=\tan y$ then $dx=(1+\tan^{2} y)dy$
As for the integration limits: if $x=0$ then $y=0$ and if $x=1$ then $y=\frac{π}{4}$
So:
$I=\int\limits_0^{\frac{π}{4}}\frac{1+\tan^{2} y}{(1+\tan y)\cos y}\,dy$
$I=\int\limits_0^{\frac{π}{4}}\frac{1}{\cos^{3} y+\cos^{2} y\sin y}\,dy$
But I don't know how to continue.
| Substituting $x=\sinh t$, we get:
$$I=\int_0^{\sinh^{-1}(1)} \frac{\cosh^2 t}{1+\sinh t}dt$$
This integral can be dealt with using the substitution $t=\log u$:
$$I=\frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u+1/u)^2 du}{u(2+u-1/u)}=\frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u^2+1)^2 du}{u^2(2u+u^2-1)}= \\ = \frac{1}{2}\int_1^{1+\sqrt{2}} \frac{(u^4+2u^2+1) du}{u^2((u+1)^2-2)}=$$
$$=\frac{1}{2}\int_2^{2+\sqrt{2}} \frac{((v-1)^4+2(v-1)^2+1) dv}{(v-1)^2(v^2-2)}$$
This is a rational integral and we can use partial fractions to evaluate it:
$$\frac{1}{(v-1)^2(v-\sqrt{2})(v+\sqrt{2})}=\frac{A}{v-1}+\frac{B}{(v-1)^2}+\frac{C}{v-\sqrt{2}}+\frac{D}{v+\sqrt{2}}$$
It's a little ugly, but with elementary algebra I'm sure you can finish.
| {
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How to integrate $\int_{0}^{2\pi} \frac{1}{\sin^4x + \cos^4 x} \,dx$ So I followed the explanations made in this post and I got that:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \frac{\sqrt{2}}{2}\arctan\left(\frac{\sqrt{2}}{2}\tan\left(2x\right)\right) + C$$
But when I try to use the Leibniz-Newton formula and evaluate the integral from $0$ to $2\pi$ I get that it's $0$ because $\tan\left(2x\right)$ evaluates to $0$ at both $x=0$ and $x=2\pi$. Is there another way to solve this integral and get the correct answer ($2\pi\sqrt2$)?
| The given antiderivative, $$\frac{1}{\sqrt 2} \arctan \left( \frac{1}{\sqrt 2} \tan 2 x \right)$$ is not defined everywhere on the interval $[0, 2 \pi]$ of integration---it is undefined at $\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$, so the usual hypotheses of the Fundamental Theorem of Calculus are not satisfied for that interval. In retrospect this is not surprising, since the trigonometric substitution that leads to this formula is not defined at those values.
On the other hand, using the double angle formulas for $\sin$ and $\cos$ (or just their complex representations) shows that the integrand has period $\frac{\pi}{2}$; using this observation and the symmetry of the integrand gives
$$\int_0^{2 \pi} \frac{dx}{\sin^4 x + \cos^4 x} = 8 \int_0^{\frac{\pi}{4}} \frac{dx}{\sin^4 x + \cos^4 x} .$$ The given antiderivative is defined everywhere on $[0, \frac{\pi}{4})$, so we can write
$$\int_0^{2 \pi} \frac{dx}{\sin^4 x + \cos^4 x} = 8 \cdot \lim_{a \nearrow \frac{\pi}{4}} \int_0^a \frac{dx}{\sin^4 x + \cos^4 x}$$
apply the F.T.C., and compute the limit.
Of course, since $\sin$ and $\cos$ do not vanish simultaneously, the original integrand is defined everywhere, and hence it has a continuous antiderivative on all of $\Bbb R$. One can verify that the antiderivative $F$ satisfying $F(0) = 0$ is
$$
F(x) =
\left\{
\begin{array}{cl}
\cdots & \cdots \\
\frac{1}{\sqrt 2} \arctan \left( \frac{1}{\sqrt 2} \tan 2 x \right), & -\frac{\pi}{4} < x < \frac{\pi}{4} \\
\frac{\pi}{2 \sqrt 2}, & x = \frac{\pi}{4} \\
\frac{1}{\sqrt 2} \left(\arctan \left( \frac{1}{\sqrt 2} \tan 2 x \right) + \pi \right) , & \frac{\pi}{2} < x < \frac{3 \pi}{4} \\
\frac{3 \pi}{2 \sqrt 2}, & x = \frac{3 \pi}{4} \\
\frac{1}{\sqrt 2} \left(\arctan \left( \frac{1}{\sqrt 2} \tan 2 x \right) + 2 \pi \right) , & \frac{3 \pi}{2} < x < \frac{5 \pi}{2}\\
\cdots & \cdots
\end{array}
\right. .
$$
It is evidently less practical than appealing to periodicity and symmetry as above, but we can also evaluate the integral by finding this antiderivative---which does satisfy the hypotheses of the F.T.C. on the interval $[0, 2 \pi]$---and then applying the F.T.C.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Find $\,n\bmod 42\,$ given $\,7\mid 3^n+4n+1$ I have been working on a problem relating to remainders: "The number $3^n+4n+1$ is divisible by $7$. Find the remainder of n when it is divided by $42$." I have been unable to find a strategic way to solve this problem and so far, I have only found the remainders, $4$, $8$, $13$, $21$, $23$, and $24$. Can anyone confirm whether this is correct?
| Let's say that $n=42a+b$, with $a \in Z$ and $b \in \{0,1,2,3,...,41\}$, since $n\in Z$.
Substituing we get $ 3^n+4n+1=0 (mod 7) \Rightarrow 3^{42a+b}+4(42a+b)+1=0 (mod 7)$
$ \Rightarrow 3^b.(3^{42})^a+7.(24a)+4b+1=0 (mod 7) \Rightarrow 3^b.(3^{42})^a+4b+1=0 (mod 7)$
Note that $3^{42}=({3^2})^{21}=9^{21}=(7+2)^{21}=2^{21}=64=1 (mod 7)$.
So $3^b.(1)^a+4b+1=0 (mod 7) \Rightarrow 3^b+(7-3)b+1=0 (mod 7) \Rightarrow 3^b-3b+1=0 (mod 7)$
Now we only have to test all values of $b \in \{0,1,2,3,...,41\}$ which $ 3^b-3b+1$ is multiple of 7, remaining 4, 8, 13, 21, 23 and 24.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solve the following equation: $\sqrt {\sin x - \sqrt {\cos x + \sin x} } = \cos x$ Solve the following equation:
\begin{array}{l}{\sqrt{\sin x-\sqrt{\cos x+\sin x}}=\cos x} \\ \text{my try as follows:}\\{\sin x-\sqrt{\cos x+\sin x}=\cos ^{2} x} \\ {\sin x-\cos ^{2} x=\sqrt{\cos x+\sin x}} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x=\cos x+\sin x} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x-\cos x-\sin x=0} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x\left(1-\sin ^{2} x\right)-\cos x-\sin x=0} \\{\sin ^2}x + {\left( {1 - {{\sin }^2}x} \right)^2} - 2\sin x\left( {1 - {{\sin }^2}x} \right) - \cos x - \sin x = 0\\ {\sin ^{2} x+\sin ^{4} x-2 \sin ^{2} x+1-2 \sin x+2 \sin ^{3} x-\cos x-\sin x=0} \\ {\sin ^{4} x+2 \sin ^{3} x-\sin ^{2} x-3\sin x-\cos x+1=0}\end{array}
Now i think it gets more complicated , any help would be appreciated
| We see that $\sin{x}\geq0$ and $\cos{x}\geq0.$
Let $f(x)=\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}},$ where $x\in\left[0,\frac{\pi}{2}\right].$
Thus, $$f'(x)=\frac{\cos{x}-\frac{\cos{x}-\sin{x}}{2\sqrt{\sin{x}+\cos{x}}}}{2\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}}}>0$$ for $\cos{x}-\sin{x}<0.$
But for $\cos{x}-\sin{x}\geq0$ we obtain:
$$f'(x)=\frac{4\cos^2x(\sin{x}+\cos{x})-(\cos{x}-\sin{x})^2}{4\sqrt{\sin{x}+\cos{x}}\sqrt{\sin{x}-\sqrt{\sin{x}+\cos{x}}}(2\cos{x}\sqrt{\sin{x}+\cos{x}}+\cos{x}-\sin{x})}$$ and
$$4\cos^2x(\sin{x}+\cos{x})-(\cos{x}-\sin{x})^2=4\cos^2x(\sin{x}+\cos{x})+2\sin{x}\cos{x}-1=$$
$$=4\cos^2x(\sin{x}+\cos{x})+2\sin{x}\cos{x}-\sqrt{\sin^2x+\cos^2x}\geq$$
$$\geq2\cos^2x(\sin{x}+\cos{x})-\sin{x}-\cos{x}=(\sin{x}+\cos{x})^2(\cos{x}-\sin{x})\geq0,$$
which says that $f$ increases.
But $\cos$ decreases, which says that there is unique $\alpha\in\left[0,\frac{\pi}{2}\right]$, for which our equation has root.
$\frac{\pi}{2}$ is a root, which gives the answer:
$$\left\{\frac{\pi}{2}+2\pi k|k\in\mathbb Z\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
$\int_0^{\infty} \frac{\ln x}{x^6 + 1}dx$ using contour integration I am looking for the solution to this improper integral. This was my approach:
Step I. (We use the 'pie slice' $\frac{\pi}{3}$). We define our log with $\frac{-\pi}{2} < \arg z < \frac{3\pi}{2}$.
\begin{equation}
\int_C \frac{\log(z)}{z^6 + 1}dz = \int_{L_1}\frac{\log(z)}{z^6 + 1}dz + \int_{L_2}\frac{\log(z)}{z^6 + 1}dz + \int_{\gamma_\epsilon}\frac{\log(z)}{z^6 + 1}dz + \int_{\Gamma_R}\frac{\log(z)}{z^6 + 1}dz
\end{equation}
Step II. Finding residues.
The only singularity located within our contour is the singularity $e^\frac{\pi i}{6}$.
\begin{equation}
2\pi i \operatorname{Res}(f; e^\frac{\pi i}{6}) = 2 \pi i\lim_{z\rightarrow z_0}\frac{(z-z_0)\log(z)}{z^6 - 1} = \frac{\pi^2}{18}e^\frac{\pi i}{6}
\end{equation}
Step III. Show that the integrals over $\Gamma_R$ and $\gamma_\epsilon$ $\rightarrow 0$ as $R \rightarrow \infty$ and $\epsilon \rightarrow 0$. We omit this as it is not a necessary part for this question.
Step IV. Parametrise the integral over $L_2$.
\begin{equation}
\int_{L_2}\frac{\log z}{z^6 + 1}dz = -e^{\frac{\pi i}{3}}\int_{\epsilon}^{R} \frac{\ln(x) + i \frac{\pi}{3}}{x^6 + 1}dx
\end{equation}
Step V. When putting all of this together, we have
\begin{align}
\frac{\pi^2}{18}e^\frac{\pi i}{6} = (1-e^{\frac{\pi i}{3}})\int_{0}^{\infty} \frac{\ln(x)}{x^6 + 1}dx - i
e^\frac{\pi i}{3}\int_{0}^{\infty}\frac{\frac{\pi}{3}}{x^6 + 1}dx
\end{align}
Which is unfortunate because the answer is not real and therefore certainly not correct. Can you see what I did wrong?
| The formula you've obtained is correct. The formula contains imaginary units, but it's not yet the final answer. Except for assmuing that the formula is incorrect, you did nothing wrong.
You have (after multiplying by $e^{-\frac{i \pi}{6}}$):
$$\frac{\pi^2}{18} = -i \int_{0}^\infty \frac{\ln x}{x^6+1}dx + \frac{\pi(1-i\sqrt{3})}{6} \int_0^\infty \frac{dx}{x^6+1} $$
Knowing that the integrals have real values, you can take the real and the imaginary part of this equation
$$\frac{\pi^2}{18} = \frac{\pi}{6} \int_0^\infty \frac{dx}{x^6+1} $$
$$0 = - \int_{0}^\infty \frac{\ln x}{x^6+1}dx - \frac{\pi\sqrt{3}}{6} \int_0^\infty \frac{dx}{x^6+1} $$
Solving this set of equations gives you
$$ \int_0^\infty \frac{dx}{x^6+1} = \frac\pi 3$$
$$ \int_{0}^\infty \frac{\ln x}{x^6+1}dx = - \frac{\pi^2\sqrt{3}}{18}$$
If you want to obtain the values of these integrals without 'assuming' they are real, you can follow the method that Oscar Lanzi suggests in his answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Consider the polynomial expression $P(x) = (x+1)^5 + x$ Consider the polynomial expression $P(x) = (x+1)^5 + x$
*
*Find the quotient of $P(x)$ divided by $x^2 + x + 1$
*Prove that for each positive integer $n$, the integer number $P(n)$ is divisibile by at least two different prime numbers.
I've done the first task and came up with $P(x)= (x+1)^5 + x = (x^3+4x^2+5x+1)(x^2+x+1)$ which can be written as $A =BC +D$ where $D=0$. So, for the second task I think I have to prove that B or C or both are prime, but I don't know how to continue.
| $B$ and $C$ aren't necessarily prime, but they have distinct factors.
To show this, continue as follows:
$(x^3+4x^2+5x+1)=(x^2+x+1)(x+3)+x-2.$
$x^2+x+1=(x-2)(x+3)+7.$
Therefore, $\gcd(x^3+4x^2+5x+1,x^2+x+1)|7,$ so there are two possibilities.
Case ($1$): $\gcd(x^3+4x^2+5x+1,x^2+x+1)=1.$
In this case, since, for positive $x,$ $x^3+4x^2+5x+1>x^2+x+1>1$,
$x^3+4x^2+5x+1$ and $x^2+x+1$ have distinct prime factors.
Case ($2$): $\gcd(x^3+4x^2+5x+1,x^2+x+1)=7 .$
In this case $7|x-2.$ For $x=2, x^3+4x^2+5x+1=35$ has two distinct prime factors ($5$ and $7$).
For $x>2$, $x^3+4x^2+5x+1>x^2+x+1>7$, and $\gcd\left(\dfrac {x^3+4x^2+5x+1}7,\dfrac{x^2+x+1}7\right)=1,$
so $\dfrac {x^3+4x^2+5x+1}7$ and $\dfrac{x^2+x+1}7$ have distinct prime factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to choose your $u$ value properly to solve $\int\frac{1}{5+3\cos (x)} \cdot dx$. We are given the following facts:
*
*$\cos(x) = \frac{1-u^2}{1+u^2}$,
*$dx = \frac{2du}{1+u^2}$
Using 1. and 2. along with u substitution solve the following integral.
$$\int\frac{1}{5+3\cos (x)} \cdot dx$$
I re-wrote the integral in terms of $u$ and simplified it as much as possible.
$$\implies \int\frac{1}{5+3(\frac{1-u^2}{1+u^2})} \cdot \frac{2du}{1+u^2}= \int \frac{1}{5+\frac{3-3u^2}{1+u^2}} \cdot \frac{2du}{1+u^2} = \int \frac{1}{\frac{5+5u^2+3-3u^2}{1+u^2}} \cdot \frac{2du}{1+u^2}$$
$$=\int \frac{1}{\frac{8+2u^2}{1+u^2}} \cdot \frac{2du}{1+u^2}= \int \frac{1}{\frac{2(4+u^2)}{1+u^2}} \cdot \frac{2du}{1+u^2}= \int \frac{1+u^2}{2(4+u^2)} \cdot \frac{2du}{1+u^2}=\int \frac{1}{4+u^2} \cdot du$$
I know that $\int \frac{1}{1+x^2}= \text{arctan(x)}$ but in my case I have a $4$ and not a $1$. My guess is that I have to use $u$-subsitution again but I can't figure out what $v$ is going to be. I've seen a lot of videos from BlackPenRedPen about $u$-substitution and I've noticed that he's able to see what the end result should be and choose his $u$ based on that, however I have no such talent. Therefore I'm confused as to what my new $u$ which will actually be a $v$ since I've already used a $u$, will be.
| You need a substitution such that $$ 4+u^2 = 4(1+v^2)$$
that is $u=2v$. That gives you $$ \int \frac{du}{4+u^2} = \int\frac{2 dv}{4(1+v^2)} = \frac12 \arctan v + C = \frac12 \arctan\frac u 2 + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Smallest integer which leaves remainder 3, 2, 1 when divided by 17, 15, 13 Find the smallest positive integer so that the remainder when it is divided by $17,15,13$ is $3,2,1$ respectively.
This question can be solved using Chinese remainder theorem, but the theorem gives any integer, not the smallest.
For example, we have, letting $n_1, n_2, n_3 = 17, 15, 13$ and $a_1, a_2, a_3 = 3, 2, 1$ and $N_j = \frac{1}{n_j}\cdot\prod n_i$:
$$x = \sum a_i N_i^{\phi(n_i)}
\\x = 3\cdot(15\cdot 13)^{16} + 2\cdot(17\cdot13)^{8} + (17\cdot 15)^{12}$$
Essentially here $ N_1^{15} = (15\cdot 13)^{15}$ is root of $N_1 x \equiv 1(\mathrm{mod} \ 17)$
By choosing different roots, we get different answers, but how can we find the minimum $x$ that satisfies the condition?
Just so in this case the minimum is $1652$
| If $x\equiv3\mod 17$ and $x\equiv2\mod 15,$ then $17a+3\equiv 15b+2\mod 17\times15,$
so $17a+3\equiv2\mod 15,$ so $17a\equiv -1 \mod 15,$ so $2a\equiv -1\mod 15,$ so $a\equiv7 \mod 15,$
so $x=17(15c+7)+3=255c+122$.
If also $x\equiv1\mod 13$ then $255c+122\equiv13e+1$ so $255c+122\equiv1 \mod 13,$
so $255c\equiv-121\mod 13,$ so $8c\equiv9 \mod 13,$ so $c\equiv45\equiv6\mod 13,$ so $c=13d+6,$
so $x=255c+122=255(13d+6)+122=3315d+1652.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the minimum $K$ such that $\sum_{cyc} \frac{a}{\sqrt{a+b}} \leq K\sqrt{a+b+c}$ The question is to find the minimum $K$ such that $$\sum_{cyc} \frac{a}{\sqrt{a+b}} \leq K\sqrt{a+b+c}$$ holds true for all non-negative $a,b$ and $c$.
My attempt:
I used Cauchy-Schwarz to get
$$\left(\sum_{cyc} \frac{a}{\sqrt{a+b}}\right)^2=\left(\sum_{cyc} \sqrt{a}\sqrt{\frac{a}{{a+b}}}\right)^2\leq(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)$$
I now have to find the maximum value of $\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)$. Putting $\frac ba=x, \frac cb=y$ and $\frac ac=z$, this reduces to the problem of finding the maximum value of $$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$$ under the constraint $xyz=1$.
I didn't get any further than this, and I'm pretty sure that my approach is wrong. For the last expression, we find that the case where $x=y=z=1$ is not the maximum as the expression is greater than $\frac32$ for $x=\frac12,y=\frac12,z=4$ (It is equal to $\frac{23}{15}$).
I found this similar question which doesn't have an answer yet.
Any help would be appreciated!
(I found this question here)
In this similar question, the maximum doesn't appear to be attained in the case where $x=y=z$.
| Some thoughts
Let $x=b/a, y=c/b, z=a/c$. This transforms the inequality to
$$2≥f(x,y,z)≥1,f(x,y,z)=\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$$
Under the constraint $xyz=1$ (as you have shown).
Now
\begin{align}
f(x,y,z)&=\frac{(1+y)(1+z)+(1+z)(1+x)+(1+x)}{(1+y)(1+x)(1+y)(1+z)}\\
&=\frac{3+2(x+y+z)+(xy+yz+zx)}{1+(x+y+z)+(xy+yz+zx)+xyz}\\
&=\frac{3+2(x+y+z)+(xy+yz+zx)}{2+(x+y+z)+(xy+yz+zx)}\\
&=1+\left(\frac{1+(x+y+z)}{2+(x+y+z)+(xy+yz+zx)}\right)\\
&=2−\left(\frac{1+(xy+yz+zx)}{2+(x+y+z)+(xy+yz+zx)}\right).
\end{align}
The term in brackets on the last two lines are non-negative. Thus,
$$2≥f(x,y,z)≥1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximize $\int_{0}^{1} f(x)^5 dx$ over all $f\colon[0,1]\to[-1,1]$ with $\int_{0}^{1} f(x)^3 dx=\int_{0}^{1} f(x) dx= 0$.
Maximize $\int_{0}^{1} f(x)^5 dx$ over all $f\colon[0,1]\to[-1,1]$ with $\int_{0}^{1} f(x)^3 dx=\int_{0}^{1} f(x) dx= 0$.
I'm not even sure where to start with this problem. Any hints would be appreciated.
| The maximum value of $\int_0^1 f(x)^5 \mathrm{d}x$ is $\frac{1}{16}$.
We first prove that $\frac{1}{16}$ is an upper bound.
Note that if $t\le 1$ then $t^5\le \frac{5}{4} t^3 - \frac{5}{16} t + \frac{1}{16}$. Indeed,
$$t^5 - \left(\frac{5}{4} t^3 - \frac{5}{16} t + \frac{1}{16}\right) = (t-1)\left(t^2+\frac t2 - \frac 14 \right)^2 \le 0.$$
Putting $t=f(x)$ and integrating we obtain
$$\int_0^1 f(x)^5 \mathrm{d}x \le \int_0^1 \left(\frac 54 f(x)^3 - \frac{5}{16}f(x)+\frac{1}{16} \right) \mathrm{d}x = \frac{1}{16},$$
since by assumption $\int_0^1 f(x)\mathrm{d}x =\int_0^1 f(x)^3\mathrm{d}x=0$.
The following example shows that the bound $\frac{1}{16}$ is optimal:
$$f(x)=\begin{cases}
\frac 14 (-1-\sqrt 5) & \text{ for } 0\le x < \frac 25 \\
\frac 14(-1+\sqrt 5) & \text{ for } \frac 25 \le x < \frac 45 \\
1 & \text{ for } \frac 45 \le x \le 1 \\
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Sum of all the numbers in the grid.
A square containing numbers
$$
\begin{array}{|c|c|c|}
\hline 1 & 2 & 3 \\ \hline
1 & 2 & 2 \\ \hline
1 & 1 & 1 \\\hline
\end{array}
\qquad \qquad\qquad
\begin{array}{|c|c|c|c|}
\hline 1 & 2 & 3 & 4 \\ \hline
1 & 2 & 3 & 3 \\ \hline
1 & 2 & 2 & 2 \\ \hline
1 & 1 & 1 & 1 \\\hline
\end{array}
$$
Continue this pattern until the box is $10 \times 10$.
Then add all the numbers together.
So this was my attempt.
Set the square $1\times1$ as $a_1$ square $2 \times 2$ as $a_2$ and so on. $a_1$ to $a_2$'s d is $(1 \cdot 2)+2 a_2$ to $a_3$'s d is $(1 \cdot 2)+(2 \cdot 2)+3$ and so on. So then I can the individual ds from each square to the next square.
And because
$$
1+(1+2)+(1+2+3)\ldots+(1+2+3+\ldots+10)
= \sum_{k=1}^{1} 1 + \sum_{k=1}^{2} 1+ \ldots +\sum_{k=1}^{10} 1=\sum_{k=1}^{10} \frac{n(n+1)}{2}
$$
so
$$
(1\cdot2)+(1\cdot2)+(2\cdot2)\ldots+(1\cdot2)+(2\cdot2)\ldots+(10\cdot2)
= \sum_{k=1}^{1} 2k + \sum_{k=1}^{2} 2k+\ldots+\sum_{k=1}^{10} 2k
$$
which also means
$$
\sum_{k=1}^{10} 2 \cdot \frac{n(n+1)}{2}
=\sum_{k=1}^{10} {n^2+n}
$$
And these are my calculations
$$\frac{10 \cdot 11 \cdot 21}{6}+\frac{10 \cdot 11}{2}$$
$$385+55=440$$
$$1+2+3+4+\ldots+10=55$$
$$440+55=495$$
It all seems right to me but the answer says its 385 is there any thing I did wrong?
| To be honest I am not sure what you are doing. Let me add my solution. One way to write down a formula for the $n \times n$ version is $$\sum_{i = 1}^n i(2(n+1-i) - 1).$$
You will find $2n-1$ times the number $1$, $2(n-1)-1$ times the number $2$ etc. This means we want to add these up and then also add the values to the numbers by multiplying with $i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to obtain the function knowing its higher derivatives at $0$ does some one knows how to obtain $f(x)$ knowing that in x=0 they have the following value
$f^{n}(0)= \frac{1}{n-s}$ if $ n=1,3,5,\cdots$ and $f^{n}(0)=0$ otherwise
| Using the derivatives to fill in a Taylor series expansion...
$$
f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \\
\frac{x}{1! (1-s)} + \frac{x^3}{3! (3 - s)} + \frac{x^5}{5! (5 - s)} + \cdots \\
\sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!(2n-1-s)}
$$
This would be notoriously difficult to manually evaluate ot recognize as the expansion of a known function. Luckily, Wolfram Alpha gives us:
$$
\sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!(2n-1-s)}= \frac{_1F_2(\frac{1}{2}-\frac{s}{2};\frac{3}{2}, \frac{3}{2}-\frac{s}{2}; \frac{x^2}{4})x}{s-1}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to define function $f(n)$ for the number of integers in $\mathbb{O}_n$ that are evenly divisible by 11? Consider the set $\mathbb{O}_n$ which contains the odd integers less than or equal to $n$ that are not divisible by 3, 5 or 7 but includes 3, 5 and 7.
$\mathbb{O}_n$ = { 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, ... ,$n$ }
The integers 9, 15, 21, … , 3 + 6$i$ where $i$ is an integer, are not in the set since these numbers are divisible by 3.
Likewise, the integers 15, 25, 35, … 5 + 10$i$ and the integers 21, 35, 49, … 7 + 14$i$ where $i$ is an integer, are not in the set.
Let function $f(n)$ equal the number of integers in $\mathbb{O}_n$ that are divisible by 11, excluding 11.
Examples:
$f(127) = 1$ since there is 1 integer in $\mathbb{O}_{127}$ that is evenly divisible by 11 and is less than or equal to 127. That integer is 121.
$f(151) = 2$ since there are 2 integers in $\mathbb{O}_{151}$ that are evenly divisible by 11 and are less than or equal to 151. Those integers are 121 and 143.
$f(191) = 3$ since there are 3 integers in $\mathbb{O}_{191}$ that are evenly divisible by 11 and are less than or equal to 191. Those integers are 121, 143 and 187.
Define function $f(n)$ as a mathematical formula?
Also, prove that as $n \to\infty$, $\frac{f(n)}{|\mathbb{O}_{n}|} = \frac{1}{11}$.
Edit: Sorry for the mistake. The fraction should have been $\frac{f(n)}{|\mathbb{O}_{n}|}$, not $\frac{f(n)}{n}$.
| Using Inclusion/Exclusion, I believe this would be:
$$g(n) = \left\lfloor \dfrac{n}{11} \right\rfloor - \left( \left\lfloor \dfrac{n}{22} \right\rfloor + \left\lfloor \dfrac{n}{33} \right\rfloor + \left\lfloor \dfrac{n}{55} \right\rfloor + \left\lfloor \dfrac{n}{77} \right\rfloor\right) + \left(\left\lfloor \dfrac{n}{66} \right\rfloor + \left\lfloor \dfrac{n}{110} \right\rfloor + \left\lfloor \dfrac{n}{154} \right\rfloor + \left\lfloor \dfrac{n}{165} \right\rfloor + \left\lfloor \dfrac{n}{231} \right\rfloor + \left\lfloor \dfrac{n}{385} \right\rfloor\right) - \left(\left\lfloor \dfrac{n}{330} \right\rfloor + \left\lfloor \dfrac{n}{462} \right\rfloor + \left\lfloor \dfrac{n}{770} \right\rfloor + \left\lfloor \dfrac{n}{1155} \right\rfloor\right) + \left\lfloor \dfrac{n}{2310}\right\rfloor$$
However, this counts when 11 is in the set. So, we need to subtract 1 when $n\ge 11$:
$$f(n) = \begin{cases}g(n), & n<11 \\ g(n)-1, & n\ge 11\end{cases}$$
Testing this out with the numbers you have given:
$$f(127) = 11-(5+3+2+1)+(1+1+0+0+0+0)-(0+0+0+0)+0 - 1 = 1$$
$$f(151) = 13-(6+4+2+1)+(2+1+0+0+0+0)-(0+0+0+0)+0-1 = 2$$
$$f(191) = 17-(8+5+3+2) + (2+1+1+1+0+0) - (0+0+0+0) + 0 - 1 = 3$$
While I believe this is the correct formula for $f(n)$, it does not appear to give the limit you want.
$$\lim_{n \to \infty} \dfrac{f(n)}{n} = \dfrac{1}{11}-\left(\dfrac{1}{22}+\dfrac{1}{33}+\dfrac{1}{55}+\dfrac{1}{77}\right)+\left(\dfrac{1}{66}+\dfrac{1}{110}+\dfrac{1}{154}+\dfrac{1}{165}+\dfrac{1}{231}+\dfrac{1}{385}\right)-\left(\dfrac{1}{330}+\dfrac{1}{462}+\dfrac{1}{770}+\dfrac{1}{1155}\right)+\dfrac{1}{2310} = \dfrac{8}{385} \neq \dfrac{1}{11}$$
| {
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Inverse Function Verification
Are functions $f(x)=\frac{7x+4}{x+6}$ and $g(x)=\frac{6x-4}{7-x}$ inverses of each other?
I'm experiencing a very strange issue with this problem. When I find the inverse of $g$ I get $f(x)$. However, when I do the same thing for $f$ I do not get $g(x)$.
\begin{align*}
f(x)&=\frac{7x+4}{x+6} \equiv y=\frac{7x+4}{x+6} \\ x&=\frac{7y+4}{y+6} \\ x(y+6)&=7y+4 \\ xy-7y&=-6x+4 \\ y(x-7)&=-6x+4\\ y&=\frac{-6x+4}{x-7} \\ \therefore f^{-1}(x)&=\frac{-6x+4}{x-7}
\end{align*}
It's very close, only the signs deviate. Yet when I try this method on $g(x)$ I get
\begin{align*}
x&=\frac{6y-4}{7-y} \\ 7x-xy&=6y-4 \\ -y&=(\frac{-7x+4}{6+x}) \\ \therefore g^{-1}(x)&=\frac{7x+4}{x+6}
\end{align*}
Which says the functions are inverses of each other. What am I getting wrong with finding the inverse of function $f$?
| You got the correct result for $f^{-1}$. Just multiply top and bottom by $-1$:
$$f^{-1}(x)=\frac{-6x+4}{x-7}=\frac{6x-4}{7-x}=g(x)$$
| {
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Solve $y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$ & $[y+[y]]=2\cos x$ $[x]$ represents the greatest integer function
$y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$
&
$[y+[y]]=2 \cos x$
Find the number of solution
My approach is as follows
$\sin x \in (\pi,2\pi)$
$y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$
$y=-1$
$[-1+[-1]]=2 \cos x$
$\cos x=-1$ which is possible at $x=\pi$ hence NO SOLUTION
$\sin x \in {\frac{\pi}{2}}$
$y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$
$y=1$
$[1+[1]]=2 \cos x$
$\cos x=1$ which is possible at $x=0$ hence NO SOLUTION
Similary if $\sin x \in (0,\pi)-\frac{\pi}{2}$
$y=\frac{1}{3}[\sin x+[\sin x+[\sin x]]]$
$y=0$
$[0+[0]]=2 \cos x$
$\cos x=0$ which is possible at $x=\frac{\pi}{2}$ hence NO SOLUTION
My answer is zero but the official answer is 3 "three".
Please help me understand my mistake.
| I believe that your answer is correct. $[x+k]=[x]+k$ if $k$ is an integer, so the given pair of equations is just equivalent to $y=[\sin x]$ and $[y]=\cos x$.
This gives $\cos x=[[\sin x]]=[\sin x]$ which has no solution.
| {
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Finding $k$ such that $(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$
$(x^2 + kx + 1)$ is a factor of $(x^4 - 12 x^2 + 8 x + 3)$ . Find $k$
....couldnt figure out how to find $k$
I tried assuming $(x^2 + kx + 1)= (x - 1)^2 $ where $k = (-2) $ comsidering that $(x-1) $ is a factor of the above polynomial.....but it didnt help either.
| Let $x$ be a root of the polynomial $x^2+kx+1$, thus $x^2=-(kx+1)$. By substitute it into ${x}^{4}-12\,{x}^{2}+8\,x+3$ we get ${x}^{4}-12\,{x}^{2}+8\,x+3=\left( -{k}^{3}+14\,k+8 \right) x-{k}^{2}+16 \equiv 0$. Then solve system
\begin{cases}
-{k}^{3}+14\,k+8=0,\\
-{k}^{2}+16=0,
\end{cases}
we get that $k=4.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Volume generated by rotating around y-axis, curve $y=x^3$ and the lines $y=0$ and $x=2$
Find the volume of the solid generated by revolving about the y-axis the region bounded by the curve $y=x^3$ and the lines $y=0$ and $x=2$
I first found what $x=2$ would be in terms of y.
$$y= (2)^3
\\ = 8$$
And in terms of y, the original equation becomes:
$$y=x^3
\\ x= y^{1 \over 3}$$
So,
$$V= \int^8_0 (y^{1 \over 3})^2 \pi dy
\\ = \pi \int_0^8 y^{2\over 3}dy
\\ =\pi \bigg[ y^{5\over 3} ({3 \over 5}) \bigg]^8_0
\\ = (8^{5 \over 3})({3 \over 5})\pi
\\ = {96 \over 5} \pi$$
Therefore the answer is ${96 \over 5} \pi$ units $^3$
However, the answer is supposed to be ${64 \over 5} \pi$ units${^3}$
What went wrong?
| Washer method is quite easy. Express the washer area including its hole and extrude it along $y$.
$$
V=\pi\int_{0}^{8}\left(2^2-x^2\right)\,dy=\pi\int_{0}^{2}\left(2^2-x^2\right)\,3x^2 dx= $$
$$
V=3\pi\int_{0}^{2}\left(4-x^2\right)\, \cdot x^2 \, dx = ... = \frac{64 \pi}{5}$$
| {
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If $4 = \frac{3}{a_1} = \frac{3}{a_2}+a_1= \frac{3}{a_3} + a_2 = ... = \frac{3}{a_{n+1}} + a_n$ Prove that $a_n= \frac{3^{n+1} - 3}{3^{n+1}-1}$ If $$4 = \frac{3}{a_1} = \frac{3}{a_2}+a_1= \frac{3}{a_3} + a_2 = \space ... \space = \frac{3}{a_{n+1}} + a_n$$
Prove that $$a_n= \frac{3^{n+1} - 3}{3^{n+1}-1}$$
I tried to prove this via induction, but i't didn't work. Any hints with this problems?
| Proof by induction
Base case
$a_1 = \frac 43$
Suppose our proposition is true.
$\frac{3}{a_{n+1}} + a_n = 4\\
a_{n+1} = \frac{3}{4-a_n}\\
a_{n+1} = \frac{3}{4-\frac{3^{n+1} - 3}{3^{n+1} - 1}}$
By the inductive hypothesis
$a_{n+1} = \frac{3^{n+2} - 3}{3^{n+2}-1}$
| {
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Is there a closed-form solution for $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(3n+m)}$? I am seeking a closed-form solution for this double sum:
\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{3}n+m)}= ?.
\end{eqnarray*}
I will turn it into $3$ tough integrals in a moment. But first I will state some similar results:
\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)} &=& 2 \zeta(3) \\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{2}n+m)} &=& \frac{11}{8} \zeta(3) \\
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{4}n+m)} &=& \frac{67}{32} \zeta(3) -\frac{G \pi}{2}. \\
\end{eqnarray*}
where $G$ is the Catalan constant. The last result took some effort ...
Now I know most of you folks prefer integrals to sums, so lets turn this into an integral. Using
\begin{eqnarray*}
\frac{1}{n} &=& \int_0^1 x^{n-1} dx\\
\frac{1}{m} &=& \int_0^1 y^{m-1} dy\\
\frac{1}{3n+m} &=& \int_0^1 z^{3n+m-1} dz \\
\end{eqnarray*}
and summing the geometric series, we have the following triple integral
\begin{eqnarray*}
\int_0^1 \int_0^1 \int_0^1 \frac{z^3 dx dy dz}{(1-xz^3)(1-yz)}.
\end{eqnarray*}
Now doing the $x$ and $y$ integrations we have
\begin{eqnarray*}
I=\int_0^1 \frac{\ln(1-z) \ln(1-z^3)}{z} dz.
\end{eqnarray*}
Factorize the argument of the second logarithm ...
\begin{eqnarray*}
I= \underbrace{\int_0^1 \frac{\ln(1-z) \ln(1-z)}{z} dz}_{=2\zeta(3)} + \int_0^1 \frac{\ln(1-z) \ln(1+z+z^2)}{z} dz.
\end{eqnarray*}
So if you prefer my question is ... find a closed form for:
\begin{eqnarray*}
I_1 = - \int_0^1 \frac{\ln(1-z) \ln(1+z+z^2)}{z} dz.
\end{eqnarray*}
Integrating by parts gives:
\begin{eqnarray*}
I_1 = - \int_0^1 \frac{\ln(z) \ln(1+z+z^2)}{1-z} dz + \int_0^1 \frac{(1+2z)\ln(z) \ln(1-z)}{1+z+z^2} dz.
\end{eqnarray*}
and let us call these integrals $I_2$ and $I_3$ respectively.
All $3$ of these integrals are not easy for me to evaluate and any help with their resolution will be gratefully received.
| $$\boxed{I=\int_0^1 \frac{\ln(1-x) \ln(1-x^3)}{x}dx=\frac53\zeta(3) +\frac{2\pi^3}{27\sqrt 3} -\frac{\pi}{9\sqrt 3}\psi_1\left(\frac13\right)}$$
As mentioned in the question we have:
$$I=\int_0^1 \frac{\ln^2(1-x)}{x}dx+\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx=2\zeta(3)+J$$
We can make use of the following series:
$$ -\frac12 \ln(1-2x\cos t+x^2)=\sum_{n=1}^\infty \frac{\cos(nt)}{n} x^n,\quad |x|<1, t\in \mathbb R$$
$$\Rightarrow J=\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx=-2\sum_{n=1}^\infty \frac{\cos\left(\frac{2n \pi}{3}\right)}{n}\int_0^1 \ln(1-x) x^{n-1}dx $$
$$=2\sum_{n=1}^\infty \frac{\cos\left(\frac{2n \pi}{3}\right)}{n^2}H_n=2\Re \left(\sum_{n=1}^\infty \frac{z^n}{n^2}H_n\right),\quad z=e^{\frac{2\pi i}{3}}$$
Using the following generating function:
$$\sum_{n=1}^\infty \frac{x^n}{n^2}H_n=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\operatorname{Li}_2(1-x)\ln(1-x)+\frac{1}{2}\ln x \ln^2(1-x)+\zeta(3)$$
And by plugging in the values found in this post yields the announced result, as we obtain:
$$J=\int_0^1 \frac{\ln(1-x)\ln(1-x+x^2)}{x}dx=\frac{2\pi^3}{27\sqrt 3}-\frac13\zeta(3) -\frac{\pi}{9\sqrt 3}\psi_1\left(\frac13\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $x > 3$ and $y < 2$, then $x^2 − 2y > 5$
Suppose $x > 3$ and $y < 2$. Then $x^2 − 2y > 5$.
My attempt:
*
*$x>3 \implies x^2 > 9$
*$ y < 2 \implies 5 + 2y < 9$
It follows that $5+2y < 9 < x^2$. From this we can see that $5+2y < x^2 \implies x^2 -2y > 5$
Therefore, if $x>3$ and $y<2$, then $x^2 − 2y > 5$.
Is it accurate?
| Proof is fine:
Option:
1)$x >3$, then
$x^2= xx >3x >9$;
2) $y <2$, then
$x^2 - 2y >9-2y >9-2 \cdot 2=5$.
| {
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Comparing two close numbers How to compare these two numbers without using a calculator ?
$A=\left(\dfrac{11}{10}\right)^{\sqrt{5}}$ and $\;B=\left(\dfrac{12}{11}\right)^{\sqrt{6}}$.
Thanks for your help !
Here is what I tried for example : $$\left(\frac{A}{B}\right)^{\sqrt6-\sqrt5}=\frac{11}{10^{\sqrt{30}−5}12^{6−\sqrt{30}}}.$$
ln is concave, so $$10^{\sqrt{30}−5}12^{6−\sqrt{30}}\leq10(\sqrt{30}−5))+12(6−\sqrt{30})=22−2\sqrt{30}.$$ But $$22−2\sqrt{30}\approx11,05...$$
| Let $f(x)=\frac{\sqrt{x+1}\ln(1+x)}{x},$ where $x>0$.
Thus, $$f'(x)=\frac{\left(\frac{\ln(1+x)}{2\sqrt{1+x}}+\frac{1}{\sqrt{1+x}}\right)x-\sqrt{1+x}\ln(1+x)}{x^2}=\frac{2x-(x+2)\ln(1+x)}{2x^2\sqrt{1+x}}\leq0$$ because
$$\left(\ln(1+x)-\frac{2x}{x+2}\right)'=\frac{x^2}{(x+1)(x+2)^2}\geq0.$$
Id est, $f$ decreases and for all $n>0$ we obtain:
$$f\left(\frac{1}{n+1}\right)>f\left(\frac{1}{n}\right)$$ or
$$\frac{\sqrt{\frac{1}{n+1}+1}\ln\left(1+\frac{1}{n+1}\right)}{\frac{1}{n+1}}>\frac{\sqrt{\frac{1}{n}+1}\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}$$ or
$$\sqrt{(n+1)(n+2)}\cdot\ln\frac{n+2}{n+1}>\sqrt{n(n+1)}\cdot\ln\frac{n+1}{n}$$ or
$$\left(\frac{n+2}{n+1}\right)^{\sqrt{n+2}}>\left(\frac{n+1}{n}\right)^{\sqrt{n}}.$$
Now, take $n=10.$
| {
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How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
How can I prove that $\frac{n^2}{x_1+x_2+\dots+x_n} \le \frac{1}{x_1}+ \frac{1}{x_2} +\dots+ \frac{1}{x_n}$?
im trying to use AM-GM
$\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}}
\le \sum_{k=1}^n \frac{{\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}}{n}$
$ln\sqrt[n]{ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}} \le ln \frac{1}{n} \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$
$ \frac{1}{x_1} \frac{1}{x_2} \frac{1}{x_3} ..\frac{1}{x_n}\le \sum_{k=1}^n {\frac{1}{x_1} +\frac{1}{x_2}+ \frac{1}{x_3}+ ..\frac{1}{x_n}}$
im not sure is this right or not, however i dont know how to include the $n^2$ in the nominator?
is there also alternative proof using jensen inequality?
| It's wrong. Try $x_1\rightarrow0^-$
For positive variables it's true by C-S:
$$\sum_{k=1}^nx_k\sum_{k=1}\frac{1}{x_k}\geq\left(\sum_{k=1}^n\sqrt{x_k\cdot\frac{1}{x_k}}\right)^2=n^2.$$
Also, AM-GM works:
$$\sum_{k=1}^nx_k\sum_{k=1}\frac{1}{x_k}\geq n\sqrt[n]{\prod_{k=1}^nx_k}\cdot n\sqrt[n]{\prod_{k=1}^n\frac{1}{x_k}}=n^2.$$
Also, the Tangent Line method helps.
Since our inequality is homogeneous, we can assume that $\sum\limits_{k=1}^nx_k=n$ and we need to prove that
$$\sum_{k=1}^n\frac{1}{x_k}\geq n$$ or
$$\sum_{k=1}^n\left(\frac{1}{x_k}-1\right)\geq0$$ or
$$ \sum_{k=1}^n\left(\frac{1-x_k}{x_k}+x_k-1\right)\geq0$$ or
$$\sum_{k=1}^n\frac{(x_k-1)^2}{x_k}\geq0$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$.
I have tried:
$$\frac{10!}{(3!\times7!)} \times (-x + 2)^7 \times (x^2)^3 $$
But got an incorrect answer $-15360$.
| You can get $x^3$ only by multiplying $x^2$ with $-x$, or by multiplying three $(-x)$'s together where the rest of the terms you are multiplying with should be $2$'s.
We can multiply $x^2$ with $(-x)$ in $2\cdot \binom{10}{2}$ ways (We multiply by $2$ to take into account the order of $x^2$ and $(-x)$). And we can multiply three $(-x)$'s together in $\binom{10}{3}$ ways. Since both these yield a $(-1)$ coefficient, and taking into account that the rest of the terms are $2$'s, we get
$$-2^8\cdot 2\binom{10}{2}-2^7\cdot\binom{10}{3} = -38400$$
| {
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On the Clausen triple $8\rm{Cl}_2\left(\frac{\pi}2\right)+3\rm{Cl}_2\left(\frac{\pi}3\right)=12\,\rm{Cl}_2\left(\frac{\pi}6\right)$ While doing research on the Clausen function, I came across this nice identity,
$$8\operatorname{Cl}_2\left(\frac{\pi}2\right)+3\operatorname{Cl}_2\left(\frac{\pi}3\right)=12\operatorname{Cl}_2\left(\frac{\pi}6\right)$$
The two addends on the LHS are Catalan's constant and Gieseking's constants. It made me wonder if there were similar relations. Define,
$$\begin{aligned}
\text{Cl}_m\left(\frac{\pi}2\right) &= \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}2\right)}{k^m}=\sum_{n=0}^\infty\left(\frac1{(4n+1)^m}-\frac1{(4n+3)^m}\right)\\
\text{Cl}_m\left(\frac{\pi}3\right) &= \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}3\right)}{k^m}=\frac{2^{m-1}+1}{2^m}\sum_{n=0}^\infty\left(\frac{\sqrt3}{(3n+1)^m}-\frac{\sqrt3}{(3n+2)^m}\right)\\
\text{Cl}_m\left(\frac{\pi}6\right) &= \sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}6\right)}{k^m}
\end{aligned}$$
and,
$$\begin{aligned}
a &= 2^{m-1}(3^{m-1}+1)\\
b &= 3^{m-1}\\
c &= 2^m\,3^{m-1}
\end{aligned}$$
Q: How do we prove, with $a,b,c$ as defined above, that,
$$a\operatorname{Cl}_m\left(\frac{\pi}2\right)+b\operatorname{Cl}_m\left(\frac{\pi}3\right)=c\operatorname{Cl}_m\left(\frac{\pi}6\right)$$
or equivalently,
$$a\sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}2\right)}{k^m}+b\sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}3\right)}{k^m} = c\sum_{k=1}^\infty\frac{\sin\left(k\,\frac{\pi}6\right)}{k^m}$$
for all integer $m>1$?
| Let $~~m,n\in\mathbb{N}~~$ and $~~0<x<1~$ .
$\displaystyle \text{E}_m(x):=\text{Li}_m(e^{i2\pi x})~ , ~~ \text{Cl}_m(2\pi x)=\Im \text{E}_m(x)~ , ~~ \overline{\text{E}_m(x)}~$ is conjugated complex to $~\text{E}_m(x)$
We have:
$$ \text{E}_m(1-x) = \overline{\text{E}_m(x)}~~ , ~~~~ \sum\limits_{k=0}^{n-1}\text{E}_m \left(\frac{x+k}{n}\right) = n^{1-m}~\text{E}_m(x)$$
It follows
$\displaystyle (n,x):=\left(2,\frac{1}{6}\right) : \hspace{0.5cm} \text{E}_m \left(\frac{1}{12}\right) + \text{E}_m \left(\frac{7}{12}\right) = 2^{1-m}~\text{E}_m \left(\frac{1}{6}\right) \hspace{2.8cm}(\text{I})$
$\displaystyle (n,x):=\left(3,\frac{1}{4}\right) : \hspace{0.5cm} \text{E}_m \left(\frac{1}{12}\right) + \text{E}_m \left(\frac{5}{12} \right)+ \text{E}_m \left(\frac{3}{4}\right) = 3^{1-m}~\text{E}_m \left(\frac{1}{4}\right)$
$\hspace{4cm}$ and conjugated we get
$\displaystyle \hspace{4.2cm} \overline{\text{E}_m \left(\frac{1}{12}\right)} + \text{E}_m \left(\frac{7}{12}\right) + \text{E}_m \left(\frac{1}{4}\right) = 3^{1-m}~\overline{\text{E}_m \left(\frac{1}{4}\right)} \hspace{0.6cm}(\text{II})$
such that the difference $~\text{(I)}-\text{(II)}~$ of the equations yields the following relationship:
$$\text{E}_m \left(\frac{1}{12}\right) - \overline{\text{E}_m \left(\frac{1}{12}\right)} - \text{E}_m \left(\frac{1}{4}\right) = 2^{1-m}~\text{E}_m \left(\frac{1}{6}\right) - 3^{1-m}~\overline{\text{E}_m \left(\frac{1}{4}\right)}$$
We take the imaginary part of this equation, add $\displaystyle ~\Im\text{E}_m \left(\frac{1}{4}\right)~$, multiply with $~6^{m-1}~$ and receive the desired result:
$\displaystyle 2\Im\text{E}_m \left(\frac{1}{12}\right) - \Im\text{E}_m \left(\frac{1}{4}\right) = 2^{1-m} \Im\text{E}_m \left(\frac{1}{6}\right) + 3^{1-m} \Im\text{E}_m \left(\frac{1}{4}\right)$
$\displaystyle 2^{m-1} (3^{m-1}+1) \Im\text{E}_m \left(\frac{1}{4}\right) + 3^{m-1} \Im\text{E}_m \left(\frac{1}{6}\right) = 2^{m}3^{m-1} \Im\text{E}_m \left(\frac{1}{12}\right)$
$\displaystyle 2^{m-1} (3^{m-1}+1)\text{Cl}_m \left(\frac{\pi}{2}\right) + 3^{m-1}\text{Cl}_m \left(\frac{\pi}{3}\right) = 2^{m}3^{m-1}\text{Cl}_m \left(\frac{\pi}{6}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find $n$ if $\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$ Find $n$ if $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$$
In this video they show a shortcut and say $n=-1/2$ without any explanation.
Key observation here is that the geometric mean of $9$ and $4$ is $6$.
It seems numerator and denominator are partial sums of geometric series, but I don't know how to proceed. Any help?
| Start by dividing top and bottom by $4^n$:
$$\frac{\frac{9^{n+1}}{4^n} + \frac{4^{n+1}}{4^n}}{\frac{9^n}{4^n} + 1} = 6.$$
Substitute $u = \frac{9^n}{4^n}$ to get
$$\frac{9u + 4}{u + 1} = 6 \iff 9u + 4 = 6(u + 1) \iff u = \frac{2}{3}.$$
Therefore,
$$\left(\frac{9}{4}\right)^n = \frac{2}{3} \implies n = -\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
} |
Find $f(x) = x + \frac{2}{3}x^3 + \frac{2\cdot4}{3\cdot5}x^5 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^7+\cdots$ where $|x|<1$
Find $$f(x) = x + \frac{2}{3}x^3 + \frac{2\cdot4}{3\cdot5}x^5 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^7+\cdots +\infty\,,\quad|x|<1$$
My solution:
$$f'(x) = 1 + 2x^2 + \frac{2}{3}\cdot4x^4 + \frac{2\cdot4}{3\cdot5}\cdot6x^6 + \cdots + \infty $$
$$ = 1+x\left(\frac{d}{dx} xf(x)\right) $$ (By observation)
$$ = 1+x^2f'(x)+xf(x)$$
$\implies (1-x^2)\frac{dy}{dx} = 1 + xy$ (where $y=f(x), \frac{dy}{dx}=f'(x)$)
This reduces to a linear first order differential equation, which along with $f(0)=0$, gives $$f(x) = \frac{\sin^{-1}(x)}{\sqrt{1-x^2}}$$
Is there any other way to solve this question? Preferably along the lines of Taylor series, infinite GP, fubini theorem, etc.?
| The question asks for other ways to identity the function whose
power series expansion is
$$ f(x) := x + \frac{2}{3}x^3 + \frac{2\cdot4}{3\cdot5}x^5 +
\frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^7+
\ldots .$$
There are many potential methods but maybe finding numerical values of
the function will help to identify it. For example, $\, f(i) \approx
.62322...i \,$ is OEIS sequence A196525 which
is given as the value of $\,\ln(1+\sqrt{2})/\sqrt{2}.\,$ This is not much
of a clue. Let us try $f(1/2) \approx .60959... \,$ which is
OEIS sequence A073010 which is given as the
value of $\,\pi/\sqrt{27}.\,$ Recall that $\,\sin(\pi/6)=1/2,\,$ and
that $\,\cos(\pi/6)=\sqrt{3}/2.\,$ A possible result now is that
$\, f(1/2) = \frac{\pi/6}{\sqrt{1-1/4}} =
\frac{\sin^{-1}(1/2)}{\sqrt{1-(1/2)^2}}.\,$ Referring back to
$\, f(i) = \frac{\ln(1+\sqrt{2})}{\sqrt{2}}\,$ we can find that
$\, \ln(1+\sqrt{2}) \approx .88137... \,$ which is
OEIS sequence A091648 which is listed as
the value of $\, \sin^{-1}(i)/i.\,$ Thus,
$\, f(i) = \frac{\sin^{-1}(i)}{\sqrt{1-i^2}}.\,$
These two numerical evaluations suggest that
$\, f(x) = \frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\,$ and this can be
verified by several methods.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What $x$ solves $x^2 = R^2 - x^2 +x(R^2 - x^2)^\frac{1}{2}$? This is from "Calculus Made Easy", Exercise 9 problem 9c. The formula in the title is what I've gotten and I know from the answer in the back of the book that it reduces to $x = 0.8507R$, but I can't see how to reduce it further (other than grouping the two $x^2$ terms).
By the way, the formula results from differentiating $4\pi x(R^2-x^2)^\frac{1}{2}+2\pi x^2$ and setting to zero.
| Rearrange to get $$\frac{2x^2-R^2}{x} = (R^2-x^2)^{\frac{1}{2}}$$
Square both sides: $$\frac{4x^4-4x^2R^2+R^4}{x^2} = R^2-x^2$$
Multiply by $x^2$: $$4x^4-4x^2R^2+R^4 = x^2R^2-x^4$$
Rearrange some more: $$5x^4-5x^2R^2+R^4 = 0$$
This is a quadratic in $x^2$: $$x^2 = \frac{5R^2\pm \sqrt{25R^4-20R^4}}{10} = \frac{5R^2\pm R^2\sqrt{5}}{10} = (\frac{5 \pm \sqrt{5}}{10})R^2$$
Taking the square root finds $$x = \pm R\sqrt{\frac{5 \pm \sqrt{5}}{10}}$$
The non-extraneous solutions are $$\pm R\sqrt{\frac{5+\sqrt{5}}{10}} \approx \pm 0.8507R$$ for $x \ge 0$.
For $x \le 0$ the non-extraneous solutions are $$\pm R\sqrt{\frac{5-\sqrt{5}}{10}} \approx \pm 0.5257R$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the equation $a^n+b^n=c^n$ have positive integer solutions? We know that the equation $a^n+b^n=c^n$ has infinitely many positive integer solutions when $n=1,n=2$. Also, by Fermat's last theorem, we know that this equation has no positive integer solution for any integer value of $n$ greater than $2$.
My question is: Does the equation $a^n+b^n=c^n$ have positive integer solutions for non-integer $n>2$?
Say for example, $a^{2.5}+b^{2.5}=c^{2.5}$
| Well, [1] obviously not for every non-integer $n$ will have integer solutions, $a,b,c$.
But obviously [2] some (many) integer $a,b,c$ will have non-integer $n$ solutions.
[1]. It'd be rather bizarre and strange if none of the $n\in \mathbb Z; n> 2$ will have integer solutions to $a^n + b^n = c^n$ but that every $\x \not \in \mathbb Z; x > 2$ will have integer solutions to $a^x + b^x = c^x$.
And it's obviously not true as there are uncountably many $x$ but only countably many triplets $(a,b,c)$. [Each $x$ would have a unique set of triplets as $a^x + b^x \ne a^y + b^y$ if $x\ne y$.]
[2] If $a< c < b$ where $a^2 + b^2 > c^2$ there will be a value where $a^m + b^m < c^m$ and by intermediate value theorem the is an $x; 2 < m < x$ where $a^x + b^x = c^x$.
....
As for $a^{2.5} + b^{2.5} = c^{2.5}$, I highly suspect has no integer solutions.
==== old answer====
Suppose $a^k + b^k > c^k$ and $a^m + b^m < c^m$; $k, m \in \mathbb N$ and $k \ge 2$.
That would mean there is a $j\in \mathbb R; k < j < m$ so that $a^j + b^j = c^k$.
So for instance $4^2 + 5^2> 6^2$ but $4^3 + 5^3 = 189 <216 = 6^3$ so there must be a $j; 2 < j < 3$ where $4^j + 5^j = 6^j$.
As for any $x > 1$ we have $(a + b)^x > a^x + b^x$ we find can assure that for and $a < c < b$ we can have $a^x + b^x > c^x$ (although $x$ may be less than $2$) and there will be $y > x$ so that $a^y + b^y < c^y$ and $j; x < j < y$ where $a^j + b^j =c^j$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Laurent Expansion Coefficients of $\exp\left(\frac{z}{1-z}\right)$ Given the function $f(z)=\exp\left(\frac{z}{1-z}\right)$, I want to find the coefficients $a_0$, $a_{-1}$, and $a_{-2}$ of the Laurent expansion $f(z)=\sum_{n=-\infty}^{\infty}a_n(z+1)^n$ about $z=-1$, on the annulus $\{z\in\mathbb{C}:|z+1|>2\}$.
We know that the power series for $\exp(z)$ centered about $z=-1$ is $\sum_{n=0}^{\infty}\frac{(z+1)^n}{e\cdot n!}$, so
\begin{align*}
\exp\left(\frac{z}{1-z}\right) &= \sum_{n=0}^{\infty}\frac{1}{e\cdot n!}\left(\frac{z}{1-z}+1\right)^n\\
&= \sum_{n=0}^{\infty}\frac{1}{e\cdot n!}\left(\frac{1}{1-z}\right)^n\\
&= \sum_{n=-\infty}^0\frac{(-1)^n}{e\cdot (-n)!}(z-1)^n
\end{align*}
But this gives us the Laurent series of f(z) centered at 1 which does not seem to be helpful.
| Similar to José Carlos Santos'answer
Let $$z=x-1\implies \frac z {1-z}=\frac{1-x}{x-2}=-\frac 12 + \sum_{n=1}^\infty 2^{-(n+1)} x^n$$ which make
$$e^{\frac{1-x}{x-2}}=\frac{1}{\sqrt{e}}\left(1+\frac{x}{4 }+\frac{5 x^2}{32 }+\frac{37 x^3}{384
}+\frac{361 x^4}{6144 }\right)+O\left(x^5\right)$$
The numerator of the coefficients correspond to sequence $A025168$ in $OEIS$ and the denominator of the coefficients is just $4^n n!$.
Replave $x$ by $(z+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Interesting four-sum inequality $n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right) \ge...$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)$$ holds.
My work.In fact, I want to solve another problem ( Prove $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$ ) This another problem has already been solved, but I want to solve it by the method that the author of the problem intended. The fact is that in the original problem there was another inequality ( Prove the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ ). It seems to me that the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ is proved as the author of the problem wanted. I think that the author of the problem wanted us to prove inequality $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$ on the basis of inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$. To this end, it suffices to prove the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)$$ I checked this inequality by numerical methods on a computer. This inequality holds for all $n$. Interestingly, if $n \to \infty$ then this inequality (in the limit) is an equality.
Perhaps this will help to solve the problem: let $x_k=\frac{k+1}{k}$. Then $1<x_k \le 2$ and inequality takes the form $$ \left(\sum \limits_{k=1}^n (2k-1)x_k\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{1}{x_k}\right) \le n^2 \left(\sum \limits_{k=1}^n x_k\right) \left( \sum \limits_{k=1}^n \frac{1}{x_k}\right)$$
| Just a partial approach.
For the different sums, we have
$$\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}=n^2+2 n-H_n$$
$$\sum \limits_{k=1}^n (2k-1)\frac k{k+1}=3 H_{n+1}+(n-3) (n+1)$$
$$\sum \limits_{k=1}^n \frac{k+1}{k}=H_n+n$$
$$\sum \limits_{k=1}^n \frac k{k+1}=n+1-H_{n+1}$$
So, we need to prove that
$$f(n)=n^2 \left(H_n+n\right) \left(n+1-H_{n+1}\right)-\left(n (n+2)-H_n\right) \left(3
H_{n+1}+(n-3) (n+1)\right)\geq 0$$
At least, for large value of $n$, we have
$$f(n)=n^3-n^2 \left(\log \left({n}\right)+\gamma -2\right) \left(\log
\left({n}\right)+\gamma +3\right)+O(n)$$
For small values of $n$
$$f(n)=\frac{1}{12} \left(120-22 \pi ^2+\pi ^4\right) n^2+O\left(n^3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove the following inequality for any positive real number $x,y$ Prove the following inequality for any positive real number $x,y$
$xy^3 \leq \frac14x^4 + \frac34y^4$
I have tried subtracting $y^4$ from both sides to get:
$xy^3 - y^4 \leq \frac14x^4 - \frac14y^4$
$y^3(x - y) \leq \frac14(x - y)(x + y)(x^2 + y^2)$
$y^3 \leq \frac14(x + y)(x^2 + y^2)$
$0 \leq (x + y)(x^2 + y^2) - 4y^3$
$0 \leq x^3 + x^2y +xy^2 - 3y^3$
But I don't know where to go from here since I couldn't find a way to conveniently factor this. I have tried completing the square from this point, but I couldn't see any obvious way to prove the right side to be greater than zero.
I also tried completing the square from the beginning to get:
$0 \leq x^4 + 3y^4 - 4xy^3$
$0 \leq x^4 + 3y^2(y - \frac23x)^2 -\frac43x^2y^2$
But yet again I couldn't figure out where to go from here. Let me know what I'm doing wrong or what other methods I can try. Thanks!
| We need to prove that
$$x^4-4xy^3+3y^4\geq0$$ or
$$x^4-2x^3y+x^2y^2+2x^3y-4x^2y^2+2xy^3+3x^2y^2-6xy^3+3y^4\geq0$$ or
$$(x-y)^2(x^2+2xy+3y^2)\geq0,$$ which is true even for all reals $x$ and $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for $\forall m \ge 2$, $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid [(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}]$.
Let $(a, b, c)$ be a Pythagorean triple. Prove that for $\forall m \ge 2$, $$\large \dfrac{1}{2}(a^4 + b^4 + c^4) \mid \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}\right]$$
Let $P_{n} = a^{2^n} + b^{2^n} + c^{2^n}$, we have that $$4(abc)^{2^k}P_{m} + P^2_{m + 1} - P_{m + 2}$$
$$ = 4(abc)^{2^m} \cdot \left(a^{2^m} + b^{2^m} + c^{2^m}\right) + \left(a^{2^{m + 1}} + b^{2^{m + 1}} + c^{2^{m + 1}}\right)^2 - \left(a^{2^{m + 2}} + b^{2^{m + 2}} + c^{2^{m + 2}}\right)$$
$$ = 4 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m}\right] + 2 \cdot \left[(ab)^{2^{m + 1}} + (bc)^{2^{m + 1}} + (ca)^{2^{m + 1}} \right]$$
$$ = 2 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m} \right]^2$$
So now we need to prove that $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid 2 \cdot \left[(ab)^{2^m} + (bc)^{2^m} + (ca)^{2^m} \right]^2$
$\iff \dfrac{1}{2}(a^4 + b^4 + c^4) \mid 4(abc)^{2^k}P_{m} + P^2_{m + 1} - P_{m + 2}$. But I don't know about whether it is sufficient to prove that $\dfrac{1}{2}(a^4 + b^4 + c^4) \mid P_{n}$ for $\forall n \ge 2$.
And why does $(a, b, c)$ have to be a Pythagorean triple?
| As you did, this answer uses $P_m$ where
$$P_m=a^{2^m}+b^{2^m}+c^{2^m}$$
We may suppose that $c^2=a^2+b^2$.
Let $$Q_m=(ab)^{2^m}+(bc)^{2^m}+(ca)^{2^m},\qquad R=\frac{a^4+b^4+c^4}{2}$$
Now, let us prove by induction on $m$ that $R\mid P_m$ and $R\mid Q_m$.
We see that $R\mid P_2$ and $R\mid Q_2$ since
$$Q_2=(a^4+a^2b^2+b^4)^2\qquad\text{and}\qquad R=a^4+a^2b^2+b^4$$
Suppose that
$$R\mid P_m\qquad\text{and}\qquad R\mid Q_m$$
Then, since
$$Q_{m+1}=Q_m^2-2a^{2^m}b^{2^m}c^{2^m}P_m\qquad \text{and}\qquad P_{m+1}=P_m^2-2Q_m$$
we get
$$R\mid P_{m+1}\qquad\text{and}\qquad R\mid Q_{m+1}\qquad\blacksquare$$
| {
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"url": "https://math.stackexchange.com/questions/3315343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Indices - factorising Simplify: $$\frac {x^5y^2x^3 + x^4y^5 - y^5x^7y^4}{x^4y^3}$$
I know this is probably low level stuff but I need to be able to do this specific type of question and I have no way of checking my work. If anyone could offer a step by step working I'd be appreciative
| You have
\begin{align*}
\frac {x^5y^2x^3 + x^4y^5 - y^5x^7y^4}{x^4y^3}
&=\frac {x^8y^2 + x^4y^5 - x^7y^9}{x^4y^3} \\
&=\frac {x^4y^2\big(x^4 + y^3 - x^3y^7\big)}{x^4y^3} \\
&=\frac {x^4 + y^3 - x^3y^7}{y},\quad x\not=0.
\end{align*}
That's about as far as you can go.
| {
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"url": "https://math.stackexchange.com/questions/3316509",
"timestamp": "2023-03-29T00:00:00",
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Finding the surface area of a region which is generated by revolving a curve around a line The following problem is from the book, Calculus and Analytical Geometer by Thomas and Finney. It is early on in the book
so I would expect / hope any integral would be easy to solve.
Problem:
Find the area of the surface generated by revolving the following curve about the line $y = -1$. The curve is
$y = \frac{x^3}{3} + \frac{1}{4x}$ for $1 \leq x \leq 3$.
Answer:
Since we are revolving the curve about $y = -1$, I augment the function by adding $1$ to it and treating it as
revolving it around $y = 0$. The format of the integral for surface area revolved around the y-axis is:
$$ S = \int_a^b 2\pi x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \,\, dx $$
Now we need to find the bounds on $y$.
\begin{align*}
y(1) &= \frac{1^3}{3} + \frac{1}{4} = \frac{1}{3} + \frac{1}{4} \\
y &= \frac{7}{12} \\
y(3) &= \frac{3^3}{3} + \frac{1}{4(3)} = 9 + \frac{1}{12} \\
y(3) &= \frac{109}{12} \\
y &= \frac{x^3}{3} + \frac{x^{-1}}{4} \\
\frac{dy}{dx} &= x^2 - \frac{x^{-2}}{4} \\
\frac{dy}{dx} &=\frac{4x^2 - x^{-2}}{4} \\
\frac{dx}{dy} &= \frac{4}{4x^2 - x^{-2}} \\
S &= \int_{\frac{7}{12}}^{\frac{109}{12}} 2\pi x \sqrt{1 + \left( \frac{4}{4x^2 - x^{-2}} \right)^2} \,\, dx
\end{align*}
Now we need to integrate.
\begin{align*}
S &= \int_{\frac{7}{12}}^{\frac{109}{12}} 2\pi x \sqrt{1 + \frac{16}{\left(4x^2 - x^{-2}\right)^2 }} \,\, dx \\
S &= \int_{\frac{7}{12}}^{\frac{109}{12}} 2\pi x \sqrt{ \frac{\left(4x^2 - x^{-2}\right)^2 + 16 }{\left(4x^2 - x^{-2}\right)^2 }} \,\, dx \\
\end{align*}
This does not seem right to me.
Based upon the comments from the group, I updated my solution.
Since we are revolving the curve about $y = -1$, I augment the function by adding $1$ to it and treating it as
revolving it around $y = 0$. Let $S$ be the surface area we are trying to find.
\begin{align*}
y &= \frac{x^3}{3} + \frac{x^{-1}}{4} \\
y' &= x^2 - \frac{x^{-2}}{4} \\
S &= \int_1^3 2 \pi \left(y+1 \right) \sqrt{1 + \left( x^2 - \frac{x^{-2}}{4} \right) ^2 } \,\, dx \\
S &=
\int_1^3 2 \pi \left(\frac{x^3}{3} + \frac{x^{-1}}{4}+1 \right)
\sqrt{1 + \left( x^2 - \frac{x^{-2}}{4} \right) ^2 } \,\, dx \\
S &=
\int_1^3 2 \pi \left(\frac{x^3}{3} + \frac{x^{-1}}{4}+1 \right)
\sqrt{ \frac{16x^4 + 8 + x^{-4}}{16} } \,\, dx \\
\end{align*}
This does not seem right to me. How do I complete this integration?
| Consider a scalar function $f(x)$: $R$ $\rightarrow$ $R$. Then the surface area of the solid formed by revoling $f(x)$ about $y=0$ is S=$2\pi$$\int_a^b$$f(x)ds$, where $ds$ represents an infintesimal arclength element of the curve $f(x)$.
To calculate the surface area of the solid formed by revolving $f(x)$ about y=$-1$, you must add 1 to the integrand. So we have that S=$2\pi$$\int_a^b$$(f(x)+1)ds$= $2\pi$$\int_a^b$$g(x)$ds, where $g(x)$=$f(x)+1$=${x^3\over 3}$+${1\over 4x}$+$1$.
Note that $ds$=$\sqrt{1+({dg\over dx})^2}$$dx$=$\sqrt{{1}+(x^2-{{1\over4x^2}}})^2$$dx$.
Now substitute a=1, b=3, and the appropriate values for $g(x)$ and $ds$ into the integral S=$2\pi$$\int_1^3$$g(x)$ds. Calculating it will yield the desired surface area.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many real roots does $x^4 - 4x^3 + 4x^2 - 10$ have?
How many real roots does this polynomial have?
$$x^4 - 4x^3 + 4x^2 - 10$$
Because non-real roots come in pairs, it must have 4, 2 or 0 real roots.
Following Descartes' rules of signs, it either has one negative (real) number and one or three positive numbers.
How can I tell if it has 2 or 4 real roots?
Thank you very much in advance.
| As $x\to -\infty$ $f(x)\to +\infty>0$. And $f(-1)= -1$ so $f(x)$ "crosses" the $x$-axis at some $x < -1$. That's our one and only negative root.
Now $f(x) = x^2(x-2)^2 - 10$. If $x > 0$ then $x^2$ is strictly increasing and positive and $(x-2)^2$ is strictly increasing and non-negative so $f(x)$ is strictly increasing. As $x\to \infty$ we have $f(x)\to +\infty$ and $f(2)=-10 < 0$. For some $x > 2$, $f(x)$ "crosses" the axis, but as $f(x)$ is strictly increasing at that point it will only "cross" once and that will be our very last root.
So all that remains is to figure out if $f(x)$ has any roots between $0$ and $2$.
but for $0 \le 1$ we have $0 \le x^2 \le 1$ and $-3 \le x-2 \le -1$ and so $1\le (x-2)^2 \le 9$ and so $f(x) < 9-10 =-1 < 0$.
And if $1\le 2$ we have $1\le x^2 \le 4$ and $-1\le x-2 \le 0$ so $0 \le (x-2)^2 \le 1$ and so $f(x) < 4-10 =-6$.
So for $0\le x \le 2$ we have $f(x) < 0$ so there are no roots there.
So there is one negative root where $x<-1$ and one positive root where $x> 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate ${\int\sqrt{1 + \sin\frac{x}2}\,\mathrm{d}x}$ So I was doing a integral question and I stumbled upon this question.
$\displaystyle{\int\sqrt{1 + \sin\left(\frac x2\right)}\,dx}$
In order to solve it I did the following:
I took $u = \frac12x$
Then $\frac {du}{dx}$
Which gave me $2 du = dx$.
After that I substituted u in the equation to get $2\int \sqrt{1 + \sin(u)} du$.
After this I was stuck as I am new to integration of trigonometry so I checked my textbook which did the same just the same but the step after this was this one
$2\int{\sqrt{\sin^2 \frac12u + \cos^2\frac12u + 2\sin \frac 12u\cos \frac12u}\text{ du}}$
I am in a complete awe how the textbook got $2\int{\sqrt{\sin^2 \frac12u + \cos^2\frac12u + 2\sin \frac 12u\cos \frac12u}\text{ du}}$ from $2\int \sqrt{1 + \sin(u)} \text{ du} $
Can someone please explain me how the this is achieved? I am totally stuck
| Obviously, knowing the trigonometric identities as already covered is the far superior and simpler method. Here is a more 'generalised' approach. I hope it serves some value.
Here I will address the integral:
\begin{equation}
I = \int \sqrt{1 + \sin\left(\frac{x}{2}\right)}\:dx
\end{equation}
First let $u = \frac{x}{2}$:
\begin{equation}
I = \int \sqrt{1 + \sin(u)} \cdot 2\:du = 2\int \sqrt{1 + \sin(u)}\:du
\end{equation}
We now employ the Weierstrauss Substitution $t = \tan\left(\frac{u}{2}\right)$:
\begin{align}
I &= 2 \int \sqrt{1 + \frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2}\:dt = 4 \int \frac{t + 1 }{\left(t^2 + 1 \right)^{\frac{3}{2}}} \:dt = 4\left[ \int \frac{t}{\left(t^2 + 1\right)^{\frac{3}{2}}}\:dt+ \int \frac{1}{\left(t^2 + 1\right)^{\frac{3}{2}}}\:dt \right] \nonumber ]\\
&= 4\left[ -\frac{1}{\sqrt{t^2 + 1}} + J\right]
\end{align}
For the remaining integral $J$ let $t = \tan(s)$:
\begin{align}
J &= \int \frac{1}{\left(\tan^2(s) + 1\right)^{\frac{3}{2}}} \cdot \sec^2(s)\:ds = \int \cos(s) \:ds = sin(s) + C = \sin\left(\arctan(t)\right) + C
\end{align}
Where $C$ is the constant of integration.
Thus,
\begin{equation}
I = 4\left[ -\frac{1}{\sqrt{t^2 + 1}} + J\right] = 4\left[ -\frac{1}{\sqrt{t^2 + 1}} + \sin\left(\arctan(t)\right)\right] + C
\end{equation}
Now $t = \tan\left(\frac{u}{2} \right) = \tan\left(\frac{x}{4}\right)$
Thus,
\begin{align}
I &= 4\left[ -\frac{1}{\sqrt{t^2 + 1}} + \sin\left(\arctan(t)\right) \right] + C = 4\left[ -\frac{1}{\sqrt{\tan^2\left(\frac{x}{4}\right) + 1}} + \sin\left(\arctan\left(\tan\left(\frac{x}{4}\right)\right)\right) \right] + C \\
&= 4\left[ -\cos\left(\frac{x}{4} \right) + \sin\left(\frac{x}{4} \right) \right] = 4\cdot \sqrt{2}\sin\left(\frac{x}{4} = \frac{\pi}{4} \right) + C
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Prove $(x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $ Prove
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $$
if $x,y,z \ge 0$
I made this problem using Holders inequality, notice that
$$ (1+x)x^{3} + (1+y)y^{3} + (1+z)z^{3} \le \left[ (x^{3})^{2} + (y^{3})^{2} + (z^{3})^{2} \right]^{1/2} \left[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} \right]^{1/2} $$
then squaring both sides we get
$$ (x^{3} + x^{4} +y^{3} + y^{4} + z^{3} + z^{4})^{2} \le (x^{6} + y^{6} + z^{6})[ (1+x)^{2} + (1+y)^{2} + (1+z)^{2} ] $$
My question is, will this inequality be obvious to be solved without Holders inequality? if so then what is the alternative solution?
| This is just Cauchy inequality $$(a^2+b^2+c^2)(a'^2+b'^2+c'^2)\geq (aa'+bb'+cc')^2$$
where $a= 1+x$ and $a'=x^3$...
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $(1-\frac{1}{x-c})^{\epsilon x-c}\geq(1-\frac{1}{x})^{\epsilon x}$ How do I formally prove that $(1-\frac{1}{x-c})^{\epsilon x-c}\geq(1-\frac{1}{x})^{\epsilon x}$ which can be seen graphically. Here $\epsilon$ is a small positive constant $\ll1$, $x\gg0$ and $c>0$.
| Let us prove the inequality in a more precise manner.
The inequality is written as
$$(\epsilon x - c)\ln \big(1 - \frac{1}{x-c}\big) \ge \epsilon x \ln \big(1 - \frac{1}{x}\big)$$
or
$$\epsilon \le \frac{-c \ln \big(1 - \frac{1}{x-c}\big)}{x \ln \big(1 - \frac{1}{x}\big) - x \ln \big(1 - \frac{1}{x-c}\big) }.\tag{1}$$
To proceed, we need the following results. Their proof is given later.
Claim 1: Let $c > 0$ and $x - c > 1$. Then
$$\frac{-c \ln \big(1 - \frac{1}{x-c}\big)}{x \ln \big(1 - \frac{1}{x}\big) - x \ln \big(1 - \frac{1}{x-c}\big) }
\ge -(x-c-1)\ln \big(1 - \frac{1}{x-c}\big) .$$
Claim 2: $-(x-c-1)\ln \big(1 - \frac{1}{x-c}\big)$
is strictly increasing with $x-c$.
Let us proceed. From Claims 1 and 2, we can obtain some sufficient conditions for the inequality in (1) to be true: if $x-c \ge 2$ and $\epsilon \le \ln 2$, the inequality in (1) is true;
or if $x- c\ge \frac{11}{10}$ and $\epsilon \le \frac{\ln 11}{10} \approx 0.2397$, the inequality in (1) is true; etc.
$\phantom{2}$
Proof of Claim 1: It suffices to prove that
$$f(c) = \frac{c}{x-c-1} - x \ln \Big(1 - \frac{1}{x}\Big) + x \ln \Big(1 - \frac{1}{x-c}\Big)\ge 0.$$
Note that $\lim_{c\to 0} f(c) = 0$ and $f'(c) = \frac{c}{(x-c)(x-c-1)^2} > 0$. The desired result follows.
Proof of Claim 2: Let $g(u) = -(u-1)\ln \big(1 - \frac{1}{u}\big)$.
We have $g'(u) = -\frac{1}{u} - \ln \big(1 - \frac{1}{u}\big) > 0$ for $u > 1$
where we have used $\ln (1-v) < -v$ for $v\in (0, 1)$. We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the integral $\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}$ One may take $x= \cos t$ and get
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= -\frac{1}{4}\int \csc^4(t/2)~ dt=-\frac{1}{4} \int [\csc^2(t/2) +\csc^2(t/2) \cot^2(t/2)]~ dt.$$
$$\Rightarrow I=\frac{1}{2} \left [\cot (t/2)] +\frac{1}{3}\cot^3(t/2)\right]=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
The question is: How to obtain this integral by other methods not using the trigonometric substitution.
| Let us use the Euler transformation $$z=\frac{1+x}{1-x}\Rightarrow x =\frac{z-1}{z+1} \Rightarrow dx=\frac{2 dz}{(1+z)^2}$$ then
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}} dx = \frac{1}{4}\int (z^{-1/2}+z^{1/2}) dz=\frac{1}{2}z^{1/2}[1+\frac{z}{3}]=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327801",
"timestamp": "2023-03-29T00:00:00",
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Laurent series of $~\frac{1}{z+2}+\frac{1}{z^2}, ~~~~~~~0<|z+2|<2~$ Laurent series of $$~\frac{1}{z+2}+\frac{1}{z^2}, ~~~~~~~0<|z+2|<2~$$
I can't find a way to represent $z^2$ in terms of $z+2$. I've tried to do $(z+2)(z-2)+4$, but I'm stuck with the $+4$.
| Thanks for the answer !
I've tried the Taylor series, but because I was always simplifying the terms $\frac{2}{8}= \frac{1}{4}, \frac{4}{32}= \frac{1}{8}$ etc. to get it to look like the one in wolframalpha. I couldn't see the pattern to make the series looks like : $$\sum_{n=0}^\infty \frac{(n+1)}{2^{n+2}}(z+2)^n$$
Another solution is to do the Laurent series for $$\frac{1}{z} = \frac{1}{-2+(z+2)}$$ which is: $$-\sum_{n=0}^\infty \frac{1}{2^{n+1}}(z+2)^n$$ and take the derivative of both sides:
$(\frac{1}{z})^l = -\frac{1}{z^2}$ ; $(-\sum_{n=0}^\infty \frac{1}{2^{n+1}}(z+2)^n)^l = -\sum_{n=0}^\infty \frac{n+1}{2^{n+2}}(z+2)^n $ , multiplying by $-1$ and adding $\frac{1}{z+2}$, should look like:
$$\frac{1}{z+2}+\sum_{n=0}^\infty \frac{n+1}{2^{n+2}}(z+2)^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maclaurin expansion of $\arccos(1-2x^2)$ Maclaurin expansion of $\arccos(1-2x^2)$
This is what I tried.
$f'(x)=2(1-x^2)^{-1/2} \\
f''(x)=2(1-x^2)^{-3/2}+3 \cdot 2 x^2(1-x^2)^{-5/2} \\
f^{(3)}(x)=18x(1-x^2)^{-5/2}+2\cdot 3\cdot 5x^3(1-x^2)^{-7/2} \\
f^{(4)}(x)=18(1-x^2)^{-5/2}+180x^2(1-x^2)^{-7/2}+2\cdot 3\cdot 5\cdot 7x^4(1-x^2)^{-9/2}$
From this I get $f'(0)=2 $, $f''(0)=0 $, $f^{(3)}(0)=2 $, $f^{(4)}(0)=0 $
But I don't know how to find a general term. Can this be solved in easier steps?
| If you check your differentiation very carefully, you will find that $f(x)=\arccos(1-2x^2)$ is not continuously differentiable at $x=0$, so the MacLaurin expansion does not exist.
Details: if $x\ne0$ we have
$$\frac{d}{dx}\arccos(1-2x^2)=\frac{4x}{\sqrt{1-(1-2x^2)^2}}
=\frac{2x}{|x|\sqrt{1-x^2}}\ .$$
So
$$\lim_{x\to0^+}f'(x)=2\quad\hbox{but}\quad \lim_{x\to0^-}f'(x)=-2\ .$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can $S_n=n$ for the series $S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots$ How can $S_n=n$ for the series $$S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots$$
My try:
We can Re-write $S_n$ as
$$S_n=\frac{0+1}{2}+\frac{1+2}{2^2}+\frac{3+2^2}{2^3}+\frac{7+2^3}{2^4}+\cdots \frac{2^{n-1}-1+2^{n-1}}{2^n}$$
So we get
$$S_n=\frac{1}{2}+\frac{1}{2}+\cdots+\frac{1}{2}+\left(\frac{1}{2^2}+\frac{3}{2^3}+\frac{7}{2^4}+\cdots\right)$$
Hence
$$S_n=\frac{n}{2}+\frac{S_n}{2}$$
Hence
$$S_n=n$$
But the actual answer is entirely different. Whats wrong in my solution?
| Your solution is not quite correct because $S_n$ has terms from $\frac{1}{2}$ to $\frac{2^n - 1}{2^n}$, but with your adjusted value, the terms on the right are missing the last value of $S_n$. Thus, the appropriate expression is$$\begin{equation}\begin{aligned}
S_n & = \frac{n}{2} + \frac{S_n - \frac{2^{n} - 1}{2^{n}}}{2} \\
2S_n & = n + S_n - \left(1 - \frac{1}{2^{n}}\right) \\
2S_n & = n + S_n - 1 + \frac{1}{2^{n}} \\
S_n & = n - 1 + \frac{1}{2^{n}}
\end{aligned}\end{equation}\tag{1}\label{eq1}$$
As an alternate solution method, note you have
$$\begin{equation}\begin{aligned}
S_n & = \frac{2-1}{2} + \frac{4-1}{4} + \ldots + \frac{2^n-1}{2^n} \\
& = 1 - \frac{1}{2} + 1 - \frac{1}{4} + \ldots + 1 - \frac{1}{2^n} \\
& = (1 + 1 + \ldots + 1) - \left(\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^n}\right) \\
& = n - \frac{1}{2}\left(\frac{1 - \frac{1}{2^n}}{1 - \frac{1}{2}}\right) \\
& = n - 1 + \frac{1}{2^n}
\end{aligned}\end{equation}\tag{2}\label{eq2}$$
As you can see, this matches the corrected version of your solution attempt.
You can also check a few values to confirm, e.g., $S_1 = \frac{1}{2}$, $S_2 = 1 + \frac{1}{4} = \frac{5}{4} = \frac{1}{2} + \frac{3}{4}$, etc.
| {
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If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
If $a,b \in \mathbb{R}$ and $a+b\ge 0$, prove that $(a^2+b^2)^3\ge 32(a^3+b^3)(ab-a-b)$
This question in my opinion is difficult and have tried many things. I tried to expand both sides but that will not help since I will not be able to cancel anything out. I am also struggling with other methods like AM-GM since $a,b$ are not necessarily positive. Any help would be appreciated.
| We need to prove that
$$(a^2+b^2)^3+32(a^3+b^3)(a+b)\geq32(a^3+b^3)ab,$$ which is obviously true for $ab<0$.
Let $ab\geq0.$
Thus, by AM-GM three times we obtain:
$$(a^2+b^2)^3+32(a^3+b^3)(a+b)\geq2\sqrt{(a^2+b^2)^3\cdot32(a^3+b^3)(a+b)}=$$
$$=8\sqrt{2(a^2-ab+b^2+ab)^3(a^3+b^3)(a+b)}\geq8\sqrt{2\left(2\sqrt{(a^2-ab+b^2)\cdot ab}\right)^3(a^3+b^3)(a+b)}=$$
$$=32\sqrt{(a^3+b^3)^2\sqrt{(a^2-ab+b^2)a^3b^3}}\geq32\sqrt{(a^3+b^3)^2\sqrt{ab\cdot a^3b^3}}=32(a^3+b^3)ab.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to factorize $\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$? How to factorize and simplify the following?
$$\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$$
I've tried everything I know. Trying to factorize the numerator first then denominator, but I get no where. Usual identities like $(x+y)^2=x^2+2xy+y^2$ don't work either, and neither does long division. I'm pretty stuck.
The answer from wolfram is
$(2x-1)/((2x+1)(x-2))$.
But I can't get there.
| By the rational root test, $4x^3+4x^2-7x+2=0$ has roots $\frac{1}{2}$ and $-2$, so that
$$
4x^3+4x^2-7x+2=(2x - 1)^2(x + 2).
$$
In the same way we see that
$$
4x^4-17x^2+4=(2x + 1)(2x - 1)(x + 2)(x - 2).
$$
Now we can form the quotient and see the result.
| {
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Is there a simple pattern to memorize the sine of $0^\circ$, $15^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $75^\circ$, $90^\circ$? We know there is a nice pattern to memorize the sine of $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$ as follows.
\begin{align}
\sin 0^\circ &= \tfrac12\sqrt0\\
\sin 30^\circ &= \tfrac12\sqrt1\\
\sin 45^\circ &= \tfrac12\sqrt2\\
\sin 60^\circ &= \tfrac12\sqrt3\\
\sin 90^\circ &= \tfrac12\sqrt4
\end{align}
We also know that
\begin{align}
\sin 15^\circ &= \tfrac14(\sqrt6-\sqrt2)\\
\sin 75^\circ &= \tfrac14(\sqrt6+\sqrt2)
\end{align}
Question
If I want to combine these two groups, is there a simple nice pattern available for us to easily rote memorize them?
| This answer is inspired from Oleg567's answer. (I should say, I will explain why does this beautiful sequence appears.)
Firstly we know that: (It should be the really amazing and beautiful sequence)
$$\cos 0^\circ=\dfrac{\sqrt{4}}{2}\quad\cos 30^\circ=\dfrac{\sqrt{3}}{2}\quad\cos 45^\circ=\dfrac{\sqrt{2}}{2}\\\cos 60^\circ=\dfrac{\sqrt{1}}{2}\quad\cos 90^\circ=\dfrac{\sqrt{0}}{2}\quad\cos 120^\circ=-\dfrac{\sqrt{1}}{2}\\\cos 135^\circ=-\dfrac{\sqrt{2}}{2}\quad\cos 150^\circ=-\dfrac{\sqrt{3}}{2}\quad\cos 180^\circ=-\dfrac{\sqrt{4}}{2}\\$$
Then, we'll use the half-angle formula $\sin^2 \dfrac{\theta}{2} = \dfrac{1-\cos\theta}{2}$
$\because 0^\circ \le\theta\le 180^\circ \rightarrow 0^\circ \le\dfrac{\theta}{2}\le 90^\circ \\ \therefore \sin \dfrac{\theta}{2} = \sqrt{\dfrac{1-\cos\theta}{2}}\ge 0 \\ \text{The }\cos\theta\text{ we want are all in the form }\pm\dfrac{\sqrt{n}}{2}\text{ where }n=0,1,2,3,4 \\ \quad\sin \dfrac{\theta}{2}\\=\sqrt{\dfrac{1\mp\frac{\sqrt{n}}{2}}{2}}\\=\sqrt{\dfrac{2\mp\sqrt{n}}{4}}\\=\dfrac{1}{2}\sqrt{2\mp\sqrt{n}}$
Therefore, we get the the result below:
$$\sin 0^\circ=\sin \dfrac{1}{2}\left(0^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{4}}\quad\sin 15^\circ=\sin \dfrac{1}{2}\left(30^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{3}}\\\sin 22.5^\circ=\sin \dfrac{1}{2}\left(45^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{2}}\quad\sin 30^\circ=\sin \dfrac{1}{2}\left(60^\circ\right)=\dfrac{1}{2}\sqrt{2-\sqrt{1}}\\\sin 45^\circ=\sin \dfrac{1}{2}\left(90^\circ\right)=\dfrac{1}{2}\sqrt{2\mp\sqrt{0}}\quad\sin 60^\circ=\sin \dfrac{1}{2}\left(120^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{1}}\\\sin 67.5^\circ=\sin \dfrac{1}{2}\left(135^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{2}}\quad\sin 75^\circ=\sin \dfrac{1}{2}\left(150^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{3}}\\\sin 90^\circ=\sin \dfrac{1}{2}\left(180^\circ\right)=\dfrac{1}{2}\sqrt{2+\sqrt{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Remainders of powers of 10 when divided by prime While doing a number theory question, I observed that the repeating digits of decimal expansions of $\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}$ form a pattern. They are $142857,285714,428571,571428,714285,857142$. I wondered if there is a link between these and the fact that all of them represent the same permutation $(142857)$. Then, while trying to see how is that happening, I came upon the fact that $$\frac{10}{7}=10\times\frac{1}{7}$$
$$=1.428571...$$
and this is how we get the pattern for repeating values of $\frac{3}{7}$. Similarly, considering $\frac{100}{7},\frac{1000}{7},\frac{10000}{7},\frac{100000}{7}$ gives us the pattern for other four fractions. From $\frac{10^6}{7}$ onwards, the pattern starts repeating because of Fermat's little theorem, since,
$$10^{k+6}\equiv 10^{k}\pmod 7$$
The question finally reduces to:
Why do powers of 10 give distinct remainders upto $10^6$? Is this property satisfied by all primes other than $2$ and $5$?
Is this somehow related to the fact that $\mathbb Z_p$ is a field?
Please help.
| This works because $7$ is a factor of $10^{(7-1)}-1=10^6-1=999999$. We have that the period of the decimal expansion of $7$ will be a factor of $6$.
If we look at $999999=3^3\cdot 7\cdot 11 \cdot 13\cdot 37$ we find that
$3$ and $9$ are factors of $10^1-1=9$ and have period $1$
$11$ is a factor of $10^2-1=99$ and has period $2$ ($9$ we already know has period $1$)
$27$ and $37$ are factors of $10^3-1=999$ and have period $3$
$7$ and $13$ are factors of $10^6-1=999999$ not already considered, and have period $6$
First note that for any number there are only a finite number of possible remainders. If we exclude prime factors of $2$ and $5$ (these shift the decimal point rather than affecting the ultimate periodicity), a number $n$ will never divide $10^r$ without remainder. If you work through the calculation you will see that as soon as a remainder repeats, the calculation repeats and the decimal repeats.
Suppose you divide through and have a repeated remainder: $10^a=ns+r, 10^b=nt+r, b\gt a$, then subtracting these gives $10^b-10^a=n(t-s)=10^a\left(10^{b-a}-1\right)$ and $n$ is a factor of $10^{b-a}-1$. So the decimal repeats after $m$ digits precisely when $n$ is a factor of $10^m-1$ (it has no factors in common with $10^a$). If $m$ is the smallest positive integer for which this happens this is the shortest period over which the decimal (and the remainder) repeats.
Since we have completely factored $10^6-1$ we have identified all the possible numbers with period dividing $6$, including periods $1$, $2$ and $3$. The only other numbers which repeat with a period less than $6$ are found by factoring
$9999=9\times 11\times 101$ which gives $101$ and factors which contain it period $4$
$99999=9\times 41 \times 271$ with $41$ and $271$ and factors which contain them having period $5$
Other integers have longer periods and will therefore have no repetition of remainders in the first six.
There are plenty of interesting facts to explore, including precisely what does happen for $\frac 1n$ when $n$ has factors $2$ and/or $5$ and what happens in bases other than the arbitrary base $10$.
I am hoping that the examples here will prompt you to check that the periods are as advertised, and to explore a bit more what is going on.
| {
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"url": "https://math.stackexchange.com/questions/3340875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum $ 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + n \cdot n! $ Find the sum
$$ 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + ... + n \cdot n! $$
Attempt
$$ 2 \cdot 2! + 3 \cdot 3! = 2 \cdot 2! + 3^{2} \cdot 2! = 2! (2+3^{2})$$
$$ 4 \cdot 4! + 5 \cdot 5! = 4 \cdot 4! + 5^{2} \cdot 4! = 4! (4+5^{2})$$
$$ 6 \cdot 6! + 7 \cdot 7! = 6 \cdot 6! + 7^{2} \cdot 6! = 6! (6+7^{2})$$
then
$$ 2 \cdot 2! + 3 \cdot 3! + 4 \cdot 4! + 5 \cdot 5! + 6 \cdot 6! + 7 \cdot 7!$$
$$ = 2! (2+3^{2}) + 4! (4+5^{2}) + 6! (6+7^{2})$$
or perhaps we can calculate in total, for example
$$ 2 \cdot 2! + 3 \cdot 3! + 4 \cdot 4! + 5 \cdot 5!$$
$$ = 2!(2 + 3^{2}) + 4 \cdot 4! + 5 \cdot 5!= 2!(2 + 3^{2}) + 2! (4 \cdot 4 \cdot 3) + 5 \cdot 5! $$
$$ = 2! (2 + 3^{2} + 3 \cdot 4^{2}) + 2! (4 \cdot 3 \cdot 5^{2})$$
$$ = 2! (2 + 3^{2} + 3 \cdot 4^{2} + 3 \cdot 4 \cdot 5^{2})$$
| HINT:
$n\cdot n!=(n+1-1)n!=(n+1)!-n!$
Do you know about Telescoping Series?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $ If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $
Attempt:
Notice that $$ \tan(a+b+c) = \frac{\tan(a+b) + \tan(c)}{1 - \tan(a+b)\tan(c)} $$
$$ = \frac{\frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} + \tan(c)}{1 - \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \tan(c)} $$
$$ = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a) \tan(b) \tan(c)}{ (1-\tan(a)\tan(b)) - \tan(a) \tan(c) -\tan(b) \tan(c) } $$
Then the denomenator must be $0$, so it is proven?
Another way:
$$ \sin(a+b+c) = 1 \implies \sin(a+b) \cos(c) + \sin(c) \cos(a+b) = 1 $$
$$ \sin(a)\cos(b)\cos(c) + \sin(b)\cos(a)\cos(c) + \sin(c) \cos(a) \cos(b) - \sin(c) \sin(a) \sin(b) = 1$$
$$ \tan(a) \cos(a) \cos(b)\cos(c) + \tan(b) \cos(a) \cos(b) \cos(c) + \tan(c) \cos(a) \cos(b) \cos(c) - \sin(a) \sin(b) \sin(c) = 1$$
$$ \cos(a)\cos(b)\cos(c) (\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)) = 1 $$
| Another concise option: since $\tan c=\frac{1}{\tan(a+b)}=\frac{1-\tan a\tan b}{\tan a+\tan b}$, $\tan b\tan c+\tan c\tan a=1-\tan a\tan b$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Geometry - Rectangle ABCD with inside point E. Find the least possible value for sum of interger distances from E to 4 vertices.
The point E lies within the rectangle ABCD.
If the distances from the vertices to E are all distinct integers, what is the least possible value of AE + BE + CE + DE?
| Suppose we have such rectangle. Place it on a coordinate system so that the $E$ is on $(0,0)$ and $A = (-a, b); B= (-a,-c), C=(d,b); D=(d,-c)$ and we need for $\sqrt{a^2 + b^2}=m, \sqrt{a^2+c^2}=n, \sqrt{d^2 + b^2}=r, \sqrt{d^2+c^2}=s$ are all integers.
We must have $m^2 +s^2 = n^2 + r^2$ and that simply requires finding the smallest integer that is a sum of two distinct perfect squares in two possible ways that will have the smallest sum.
$m^2 + s^2 = n^2 + r^2 \implies m^2-n^2 = r^2 - s^2 \implies (m+n)(m-n)=(r+s)(r-s)$.
Now $m+n, m-n$ must be the same parity and and we could do $r+s=5, r-s=3$ and $m+n=15, m-n=1$ so $m=8;n=7;r=4;s=1$. This would be the smallest set of four distinct values for $m-n,r-s,r+s,m+n$. And so $s+r+n+m=20$.
And $8^2 + 1^2 = 7^2 + 4^2=65$ is a probably this smallest such integer.
(I'm not sure how to verify this other than trial and error. Oddly enough taking even values $m-n=2, r-s=4,r+s=6,m+n=12$ so $r=5;s=1;n=5;m=7$ gives us a smaller integer $7^2+1=5^2 + 5^2=50$ and as smaller sum of $18$ but the terms are not distinct.)
So we need $a^2 + b^2= 64; a^2+c^2=49; d^2+b^2=16; d^2 + c^2 =1$.
$d^2=1-c^2$ and so $a^2 + b^2 = 64; a^2+c^2=49; b^2-c^2=15$
$b^2 = 15+c^2$ and so $a^2+c^2=49$. So we need $0< c^2 <1$ the$1> d^2=1-c^2>0$, $b^2 = 15+c^2> 15$ and $a^2 = 49-c^2> 48$ are possible rectangles.
| {
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"timestamp": "2023-03-29T00:00:00",
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An application of the Banach Fixed Point Theorem to a system of equations I am solving a question whose first item is to demonstrate the Banach Fixed Point Theorem, and the second item is as follows:
Show that for any parameter $t \in \mathbb{R}$ the system
$$
\begin{cases}
x = \frac{1}{2}\sin(x+y) + t - 1 \\
y = \frac{1}{2}\cos(x-y) - t + \frac{1}{2}
\end{cases}
$$
has a unique solution that depends continuously on the parameter $t$.
Now, this seems clear to me that the question asks one to apply the just proved Banach Fixed Point Theorem to find the existence of fixed points for the map
$$
F(x, y) = (\frac{1}{2}\sin(x+y) + t - 1, \frac{1}{2}\cos(x-y) - t + \frac{1}{2})
$$
However, I am not able to show that such a map is a contraction. Any hints wil be the most appreciated.
Thanks in advance.
| By the mean value theorem, for any $a,b \in \mathbb{R}$, $|\sin(a)-\sin(b)| \le |a-b|$. Therefore, $$|(\frac{1}{2}\sin(x+y)+t-1)-(\frac{1}{2}\sin(x'+y')+t-1)| = \frac{1}{2}|\sin(x+y)-\sin(x'+y')|$$ $$\le \frac{1}{2}|(x+y)-(x'+y')| = \frac{1}{2}|(x-x')+(y-y')|.$$ Similarly, $$|(\frac{1}{2}\cos(x-y)-t+\frac{1}{2})-(\frac{1}{2}\cos(x'-y')-t+\frac{1}{2})| = \frac{1}{2}|\cos(x-y)-\cos(x'-y')|$$ $$\le \frac{1}{2}|(x-y)-(x'-y')| = \frac{1}{2}|(x-x')-(y-y')|.$$ Therefore, $$|F(x,y)-F(x',y')| \le \sqrt{\frac{1}{4}|(x-x')+(y-y')|^2+\frac{1}{4}|(x-x')-(y-y')|^2}$$ $$= \frac{1}{2}\sqrt{2|x-x'|^2+2|y-y'|^2} = \frac{\sqrt{2}}{2}|(x,y)-(x',y')|.$$ This shows that $F$ is a contraction with Lipschitz constant at most $\frac{\sqrt{2}}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve given differential equation $(x^3y^3+x^2y^2+xy+1)ydx+(x^3y^3-x^2y^2-xy+1)xdy=0$? The equation to solve is:
$$(x^3y^3+x^2y^2+xy+1)ydx+(x^3y^3-x^2y^2-xy+1)xdy=0$$
I tried putting $xy=t$ but that just gave me this:
$$\frac{t^3-t^2-t+1}{t^3+t^2+t+1}dt=\frac{dx}{x}$$
I suppose there must be some clever factoring involved somewhere but I can't see it so can someone guide me on how to advance or perhaps suggest an alternate method?
| The given equation is of the form $f_1(xy)ydx+f_2(xy)xdy=0.$
Here, $M=(x^3y^3+x^2y^2+xy+1)y$ & $N=(x^3y^3-x^2y^2-xy+1)x$
$$\therefore I.F.=\frac1{Mx-Ny}\\
=\frac1{2x^2y^2(xy+1)}$$
$\text{Multiplying the I.F. with the equation, we get}$
\begin{align}\\
&{\begin{aligned}\\
\frac{(x^3y^3+x^2y^2+xy+1)y}{2x^2y^2(xy+1)}dx &+\frac{(x^3y^3-x^2y^2-xy+1)x}{2x^2y^2(xy+1)}dy=0
\end{aligned}}\\
&{\begin{aligned}\\
\implies\frac{(x^2y^2+1)(xy+1)}{x^2y(xy+1)}dx &+\frac{(x^2y^2-1)(xy-1)}{xy^2(xy+1)}dy=0
\end{aligned}}\\
&{\begin{aligned}\\
\implies\frac{x^2y^2+1}{x^2y}dx &+\frac{(x^2y^2-1)(xy-1)}{xy^2(xy+1)}dy=0
\end{aligned}}\\
&{\begin{aligned}\\
\implies ydx+\frac{dx}{x^2y}+&\frac{(x^2y^2-1)(xy-1)}{xy^2(xy+1)}dy=0
\end{aligned}}\\
\end{align}
$\therefore\text{the solution is}$
\begin{align}
&{\begin{aligned}\\
\int_{(\text{y const})}Mdx+ &\int(\text{terms of N not containing x}) dy=c
\end{aligned}}\\
&{\implies \int ydx+\int\frac{dx}{x^2y}=c}\\
&{\implies xy-\frac{1}{xy}=c}\\
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculate the integral value using residues Hello I'm trying to solve this integral : $\\$
$$\int_{0}^{2\pi} \frac {\cos^2(x)}{4+3\cos(x)} dx.$$
I want to solve this integral using the theorem:
$$\int_{0}^{2\pi} R\bigl(\cos(\alpha),\sin(\alpha)\bigr) d\alpha =2\pi i \sum_{|z_{k}|<1} \operatorname*{Res}_{z=z_k}f(z)$$
where $\displaystyle\;f(z)=\frac {1}{iz}R\biggl(\frac{1}{2}\Bigl(z+\frac {1}{z}\Bigr),\frac {1}{2i}\Bigl(z-\frac {1}{z}\Bigr)\biggr).$
Let $R(x,y)=\frac {x^2}{4+3x}$, then i used the theorem to find $f(z)$.
So $f(z)=\frac{1}{i} \frac {(z+\frac {1}{z})^2}{16z+6z^{2}+6}$ then i find the zeros of ${16z+6z^{2}+6}$ which is $z_{1}=\frac{-8+\sqrt{28}}{6}$ and $z_{2}=\frac{-8-\sqrt{28}}{6}$.
Then i don't know but i should use only $z_{1}$ to answer the question?
$$I=\int_{0}^{2\pi} \frac {\cos^2(x)}{4+3\cos(x)} dx=2\pi i {Res}_{z=z_1}f(z)$$
I found ${Res}_{z=z_1}f(z)=\frac{64-8\sqrt{28}}{i(-8\sqrt{28}+28)}$
Using the theorem I mentioned, I found the singular points ($z_1=0$ essential point, $z_2=\frac{−8+sqrt(28)}{6}$ and $z_3=\frac{−8-sqrt(28)}{6}$ such as simple poles). But $z_3$ isn't in the disk $D(0,1)$. So the integral $=2\pi i(Res_{z=z1}f(z)+Res_{z=z2}f(z))$. I found the $Res_{z=z2} f(z)$ but i can't find $Res_{z=z1}f(z)$.
Can someone help me with that?
Thank you :)
| $$I=\int_{0}^{2\pi}\frac{\cos^2(x)}{4+3\cos x}\,dx = \int_{0}^{2\pi}\frac{\left(\frac{e^{ix}+e^{-ix}}{2}\right)^2}{4+3\left(\frac{e^{ix}+e^{-ix}}{2}\right)}\,dx=\int_{0}^{2\pi}\frac{e^{2ix}+e^{-2ix}+2}{16+6e^{ix}+6e^{-ix}}\,dx $$
equals, via $e^{ix}\to z$,
$$ -i\oint_{|z|=1}\frac{(z^2+1)^2}{z^2(6z^2+16z+6)}\,dz. $$
The only root of $6z^2+16z+6$ lying inside $|z|<1$ is at $\frac{\sqrt{7}-4}{3}$, so
$$ I = 2\pi\operatorname*{Res}_{z=0}\frac{(z^2+1)^2}{z^2(6z^2+16z+6)}+2\pi\operatorname*{Res}_{z=\frac{\sqrt{7}-4}{3}}\frac{(z^2+1)^2}{z^2(6z^2+16z+6)}$$
and this leads to
$$ I = 2\pi\left[-\frac{4}{9}+\frac{16}{9\sqrt{7}}\right]=\frac{2\pi}{9}\left(\frac{16}{\sqrt{7}}-4\right). $$
A real-analytic way deserves to be outlined, too. By symmetry
$$ I = \int_{-\pi}^{\pi}\frac{\cos^2(x)}{4-3\cos x}\,dx = \int_{0}^{\pi}\frac{1+\cos(2x)}{4-3\cos x}\,dx = 16\int_{0}^{\pi/2}\frac{\cos^2(x)}{16-9\cos^2(x)}\,dx$$
and the substitution $x=\arctan t$ turns the last integral into an elementary one (by partial fraction decomposition):
$$ I=16\int_{0}^{+\infty}\frac{dt}{(t^2+1)(16t^2+7)}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining $\lim_{h \to 0} \frac{f(x +h) - f(x)}{h}$ for the function $f(x) = 3x^2 + 2x$ Determine $\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}$ for the function $f(x)=3x^2+2x$.
Can anyone please confirm whether this is correct:
\begin{align*}
& = \frac{3x^2+6xh+3h^2+2x+2h-3x^2-2x}{h}\\
& = \frac{6xh+3h^2+2h}{h}\\
& = \frac{h(6x+3h+2)}{h}\\
& = 6x+3h+2
\end{align*}
Do I go a step further by saying:
\begin{align*}
\lim_{h \to 0} (6x+3h+2) & = 6x+(3 \cdot 0)+2\\
& = 6x+2
\end{align*}
I am confused about which answer is the final product and if I have gone a step too far.
Thanks for your help.
Apologies for the lack of information earlier.
| Given $f(x)=3x^2+2x$:
\begin{align}
\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}&=\lim\limits_{h\to0}\frac{3(x+h)^2+2(x+h)-2x^2-2x}{h}\\
&=\lim\limits_{h\to0}\frac{6xh+3h^2+2h}{h}\\
&=\lim\limits_{h\to0}\frac{h(6x+3h+2)}{h}\\
&=\lim\limits_{h\to0}\ (6x+3h+2)\\
&=6x+2
\end{align}
You have approached the problem correctly, but it is important to include all necessary notation in your solution (including the $\lim$ in each of your expressions, line after line, until you evaluate the limit by substitution).
| {
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Probabilities of selecting exactly $n$ white and $m$ black balls I stumbled upon a tricky problem that I don't know how to solve.
There $a$ white and $b$ black balls in an urn. We select one first ball at random (let's call its color $F$) and then return this ball back to the urn. We then add $c$ balls of the same color $F$.
From this urn we select $n + m$ balls without replacement ($n + m \leq a + b + c$). What is the probability that we end up with $n$ white and $m$ black balls?
And the second question is: how can we show that in the $k$-th step of selection the probability of the ball being white is $\dfrac{a}{a+b}$. It is quite clear intuitively, yet I would be interested to see a formal proof of this.
Any help would be greatly appreciated. I would also be interested in incomplete solutions.
| Part 1)
Let's say we have urn with $a$ white balls, $b$ black balls. After first draw, we return ball back to the urn and add $c$ balls of the same colour. We're asked about probability of event $E - $exactly $n$ white, $m$ black balls (We draw exactly $n+m$ balls ).
Let $X$ - be first draw, that is $X =$ white with probability $\frac{a}{a+b}$, and $X =$black with probability $\frac{b}{a+b}$, we'll condition on $X$.
$ \mathbb P(E) = \mathbb P(E|X=white)\mathbb P(X=white) + \mathbb P(E|X=black)\mathbb P(X=black) = \frac{{a+c \choose n} {b \choose m}}{{a+b+c \choose n+m}}\frac{a}{a+b} +\frac{{a \choose n} {b+c \choose m}}{{a+b+c \choose n+m}}\frac{b}{a+b}$.
Note that when both $a < n$ and $b < m$ this probability is simply $0$, due to that Newton Symbols.
Part 2)
I'll assume you count $k$ after that first draw when you add either black or white $c$ balls.
You want formal proof, so we'll use induction. Let $E_k$ - at $k'$th draw we get white ball. For $E_1$ we have:
$\mathbb P(E_1) = \mathbb P(E_1|X = white)\mathbb P(X=white) + \mathbb P(E_1 | X=black) \mathbb P(X = black) = \frac{a+c}{a+b+c}\frac{a}{a+b} + \frac{a}{a+b+c}\frac{b}{a+b} = \frac{a(a+b+c)}{(a+b)(a+b+c)} = \frac{a}{a+b}$
So assume it works for $E_k$, and let $Y$ be one draw after $X$ (that is the first one we count). Then we have:
$\mathbb P(E_{k+1}) = \mathbb P(E_{k+1} | X =white)\frac{a}{a+b} + \mathbb P(E_{k+1} | X = black)\frac{b}{a+b}$
Now, we'll condition once more, but on $Y$. We have:
$ \mathbb P(E_{k+1} | X = white ) = \mathbb P(E_{k+1} | (X,Y) = (white,white)) \mathbb P(Y=white | X = white) + \mathbb P (E_{k+1} | (X,Y) = (white,black)) \mathbb P(Y = black | X = white) = \frac{a+c-1}{a+b+c-1}\frac{a+c}{a+b+c} + \frac{a+c}{a+b+c-1}\frac{b}{a+b+c} = \frac{(a+c)(a+c-1+b)}{(a+b+c)(a+b+c-1)} = \frac{a+c}{a+b+c}$
Similarly:
$\mathbb P(E_{k+1} | X = black) = \frac{a-1}{a+b+c-1}\frac{a}{a+b+c} + \frac{a}{a+b+c-1}\frac{b+c}{a+b+c} = \frac{a(a+b+c-1)}{(a+b+c)(a+b+c-1)} =\frac{a}{a+b+c}$
Here I used our induction claim, but with different number of balls we started with ( for example $a+c-1$ white, $b$ black, or so. All that matters, after we perform $Y$ - draw, then we just have $k$ draws left and we can use it.)
Plugging it:
$\mathbb P(E_k) = \frac{(a+c)a + ab}{(a+b)(a+b+c)} = \frac{a(a+b+c)}{(a+b)(a+b+c)} = \frac{a}{a+b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349964",
"timestamp": "2023-03-29T00:00:00",
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What is the derivative of a function of the form $u(x)^{v(x)}$? So I have a given lets say $(x+1)^{2x}$ in addition to $\frac{\mathrm dy}{\mathrm dx}a^u=a^u\log(a)u'$. I still have to multiply this by the derivative of the inside function $x+1$ correct?
| This is what logarithmic differentiation is for. You start with writing the function as an equation
$$y = (x + 1)^{2x},$$
then take the natural log of both sides:
$$\ln y = \ln\left[(x + 1)^{2x}\right] = 2x \ln(x+1).$$
We then implicitly differentiate both sides with respect to $x$. By chain rule (remember, $y$ is a function of $x$), the left side comes to
$$\frac{1}{y} \cdot y'.$$
The right side can be differentiated as normal:
$$\frac{2x}{x + 1} + 2\ln(x + 1).$$
So,
\begin{align*}
&\frac{1}{y} \cdot y' = \frac{2x}{x + 1} + 2\ln(x + 1) \\
\implies \, &y' = y\left(\frac{2x}{x + 1} + 2\ln(x + 1)\right) \\
\implies \, &y' = (x + 1)^{2x}\left(\frac{2x}{x + 1} + 2\ln(x + 1)\right).
\end{align*}
| {
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Expression of $x^n+\frac1{x^n}$ by $x+\frac1{x}$ where $n$ is a positive odd number. There was a problem in a book:
Denote that $y=x+\dfrac{1}{x}$, express $x^7+\dfrac{1}{x^7}$ using $y$.
It's not a hard question, but I find a special sequence:
$x+\dfrac{1}{x}=y\\x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=y^3-3y \\ x^5+\dfrac{1}{x^5}=\left(x+\dfrac{1}{x}\right)^5-5\left(x^3+\dfrac{1}{x^3}\right)-10\left(x+\dfrac{1}{x}\right)=y^5-5\left(y^3-3y\right)-10y=y^5-5y^3+5y\\x^7+\dfrac{1}{x^7}=\left(x+\dfrac{1}{x}\right)^7-7\left(x^5+\dfrac{1}{x^5}\right)-21\left(x^3+\dfrac{1}{x^3}\right)-35\left(x+\dfrac{1}{x}\right)=y^7-7\left(y^5-5y^3+5y\right)-21\left(y^3-3y\right)-35y=y^7-7y^5+14y^3-7y$
I find that the coefficient has some relationship between the Pascal Triangle, such as $y^7-7y^5+14y-7y=y^7-7 \binom{2}{0}y^5+7\binom{2}{1}y^3-7\binom{2}{2}y$. That's strange but funny! However, I can't really prove this, or show that it is false. Hope there is someone who can answer me. Thank you!
| I don't see a clear pattern in the coefficients but there is a recurrence:
$$
y_{n+2} = y_2 y_n -y_{n-2} = (y^2-2)y_n -y_{n-2}
$$
where
$$
y_n = x^n+\dfrac{1}{x^n}
$$
| {
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"source": "stackexchange",
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Calculate the limit without l'Hopital rule I have the following limit:
$$
\lim_{x\to 7}\dfrac{x^2-4x-21}{x-4-\sqrt{x+2}}
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
| The numerator can be factored as $(x-7)(x+3)$. Now consider
$$
\lim_{x\to7}\frac{x-4-\sqrt{x+2}}{x-7}=\lim_{x\to7}\frac{x-7-(\sqrt{x+2}-3)}{x-7}=
1-\lim_{x\to7}\frac{x+2-9}{(x-7)(\sqrt{x+2}+3)}=1-\frac{1}{6}=\frac{5}{6}
$$
So your limit is
$$
\lim_{x\to7}\frac{x-7}{x-4-\sqrt{x+2}}(x+3)=\frac{6}{5}\cdot10=12
$$
| {
"language": "en",
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"source": "stackexchange",
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If $a$, $b$, $c$ are three positive integers such that $a^3+b^3=c^3$ then one of the integer is divisible by $7$ Let on contrary that none of the $a$, $b$, $c$ is divisible by $7$. Then either $a^3\equiv b^3\pmod{7}$ or $b^3\equiv c^3\pmod{7}$ or $c^3\equiv a^3\pmod{7}$.
Now how to go further?
| If none of a,b,c are divisible by 7,
then each of a$^3$, b$^3$ and c$^3$ = either 1 or -1 (mod 7).
A contradiction ensues.
| {
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About $e^{i z} = \cos z + i \sin z$ in Michael Spivak "Calculus 3rd Edition". I am reading "Calculus 3rd Edition" by Michael Spivak.
The author wrote as follows (p. 555):
Moreover, if we replace $z$ by $i z$ in the series for $e^z$, and make a rearrangement of the terms (justified by absolute convergence), something particularly interesting happens:
$$e^{i z} = 1 + i z + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots \\
=1 + iz - \frac{z^2}{2!} - \frac{i z^3}{3!} + \frac{z^4}{4!} + \frac{i z^5}{5!} + \cdots \\
= (1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots) + i (z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots),$$
so $$e^{i z} = \cos z + i \sin z.$$
But I think the author didn't use a rearrangement of the terms at all.
Am I right?
Let
$$c_n := 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots + (-1)^n \frac{z^{2 n}}{(2 n)!},$$
$$s_n := z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots + (-1)^n \frac{z^{2 n + 1}}{(2 n + 1)!},$$
$$e_n := 1 + i z + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \cdots + \frac{(iz)^n}{n!}.$$
Then, $$e_{2 n + 1} = c_n + i s_n,$$
$$\lim_{n \to \infty} e_{2 n + 1} = e^{i z},$$
$$\lim_{n \to \infty} c_n + i s_n = \lim_{n \to \infty} c_n + i \lim_{n \to \infty} s_n = \cos z + i \sin z,$$
so $$e^{i z} = \cos z + i \sin z.$$
| You are right that absolute convergence, while present as "security blanket", is not really used. The only re-arrangement that happens is that the exponential series is organized in groups of two subsequent terms, so that then the addition rule for convergent series can be applied:
$\sum_{n=0}^\infty a_n+\sum_{n=0}^\infty b_n=\sum_{n=0}^\infty(a_n+b_n)$, if two of the series converge, then so does the third.
This is also what you traced with the partial sums.
In this grouping procedure one reduces the sequence of partial sums of the exponential series to the sub-sequence of even indices. For this to be unproblematic requires that the terms of the series converge to zero, which for this power series is true for every $z$.
| {
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Derivation of 2$\pi$ in the period of a simple pendulum Most of us know that the period of a simple pendulum is given by
$$2 \pi \sqrt{\dfrac{\ell}{g}}.$$
But how did the $2\pi$ term get into that argument. From dimensional analysis, we can find the period but not the constant.
Is there a calculus argument as some limit is taken? Is it based on an energy equation? Surely there is a way to derive that $2 \pi$.
| The pendulum movement equation is given by
\begin{align*}
\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}} + \frac{g}{\ell}\sin(\theta) = 0
\end{align*}
For small values of $\theta$, we can make the approximation $\sin(\theta) \approx \theta$, from whence we obtain the equation
\begin{align*}
\ddot{\theta} + \frac{g}{\ell}\theta = 0
\end{align*}
whose associated characteristic equation is
\begin{align*}
x^{2} + \frac{g}{\ell} = 0 \Longrightarrow x = \pm i\sqrt{\frac{g}{\ell}}
\end{align*}
Thus the solutions are $\theta(t) = c_{1}\sin\left(\displaystyle t\sqrt{\frac{g}{\ell}}\right) + c_{2}\cos\left(\displaystyle t\sqrt{\frac{g}{\ell}}\right)$. Finally, for $T = 2\pi\sqrt{\displaystyle\frac{\ell}{g}}$, one has
\begin{align*}
\theta\left(t + 2\pi\sqrt{\frac{\ell}{g}}\right) = c_{1}\sin\left(t\sqrt{\frac{g}{\ell}} + 2\pi\right) + c_{2}\cos\left(t\sqrt{\frac{g}{\ell}} + 2\pi\right) = \theta(t)
\end{align*}
Therefore we conclude that $T$ is the period indeed.
| {
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If $a+b+c=2$ where $0 \leq {a,b,c} \leq 1$ find the range of $\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}$
If $a+b+c=2$ where $0 < {a,b,c} <1$ find the range of $$\frac{a}{1-a}\cdot\frac{b}{1-b}\cdot\frac{c}{1-c}.$$
I have tried using AM-GM but I was not able to solve it . I also assumed the expression equal to $k$ and tried to find the range of $k$ but I derived some absurd results. Please help me to solve it and guide me to the correct approach.
| Let $\frac{a}{1-a}=x$, $\frac{b}{1-b}=y$ and $\frac{c}{1-c}=z.$
Thus, $x$, $y$ and $z$ are positives, $a=\frac{x}{1+x},$ $b=\frac{y}{1+y},$ $c=\frac{z}{1+z}$ and the condition gives:
$$\sum_{cyc}\frac{x}{1+x}=2.$$
Id est, by AM-GM we obtain:
$$\prod_{cyc}\frac{x}{1+x}=\prod_{cyc}\left(2-\frac{y}{1+y}-\frac{z}{1+z}\right)=\prod_{cyc}\left(\frac{1}{1+y}+\frac{1}{1+z}\right)\geq$$
$$\geq\prod_{cyc}\frac{2}{\sqrt{(1+y)(1+z)}}=\frac{8}{\prod\limits_{cyc}(1+x)},$$
which gives $$xyz\geq8$$ or
$$\prod_{cyc}\frac{a}{1-a}\geq8.$$
The equality occurs for $a=b=c=\frac{2}{3},$ which says that $8$ is a minimal value.
Also, since $\prod\limits_{cyc}\frac{a}{1-a}$ is a continuous and for $a\rightarrow1^-$ this expression is closed to $+\infty,$ we got the answer:
$$[8,+\infty).$$
| {
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What is $\alpha^{4} + \beta^{4} + \gamma^{4}$? If
$$ \alpha + \beta + \gamma = 14 $$
$$ \alpha^{2} + \beta^{2} + \gamma^{2} = 84 $$
$$ \alpha^{3} + \beta^{3} + \gamma^{3} = 584 $$
What is $\alpha^{4} + \beta^{4} + \gamma^{4}$?
Attempt:
Notice that
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196$$
$$\alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196 \implies 84 + 2(\alpha \beta + \beta \gamma + \alpha \gamma) = 196 $$
$$ \alpha \beta + \beta \gamma + \alpha \gamma = 56 $$
Also
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{3} + \beta^{3} + \gamma^{3} + 3(\alpha^{2} \beta + \alpha^{2} \gamma + \beta^{2} \alpha + \beta^{2} \gamma + \gamma^{2} \alpha + \gamma^{2} \beta) + 3 \alpha \beta \gamma = 2744$$
$$ 584 = 2744 - 3(\alpha + \beta + \gamma)(\alpha \beta + \alpha \gamma + \beta \gamma) + 6 \alpha \beta \gamma$$
$$ \alpha \beta \gamma = 32$$
So that
$$ (\alpha + \beta + \gamma)^{4} = ( \alpha^{2} + \beta^{2} + \gamma^{2} + 2(\alpha \beta + \beta \gamma + \alpha \gamma) )^{2} $$
$$ 38416 = (\alpha^{2} + \beta^{2} + \gamma^{2})^{2} + 4 (\alpha \beta + \alpha \gamma + \beta \gamma)(\alpha^{2} + \beta^{2} + \gamma^{2}) + 4 (\alpha \beta + \alpha \gamma + \beta \gamma)^{2}$$
$$ 38416 = \left[ \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( (\alpha \beta)^{2} + (\beta \gamma)^{2} + (\alpha \gamma)^{2} \right) \right] + 4 (56)(84) + 4 (56^{2}) $$
$$ 7056 = \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( (\alpha \beta + \alpha \gamma + \beta \gamma)^{2} - 2\alpha \beta \gamma(\alpha + \beta + \gamma ) \right) $$
$$ 7056= \alpha^{4} + \beta^{4} + \gamma^{4} + 2 \left( 56^{2} - 2(32)(14) \right) $$
$$\alpha^{4} + \beta^{4} + \gamma^{4} = 2576$$
| Let $x=\alpha+\beta+\gamma$, $y=\alpha^2+\beta^2+\gamma^2$ and $z=\alpha^3+\beta^3+\gamma^3$ for any real numbers $\alpha,\beta,\gamma$. Then we have the identity:
$$\alpha^4+\beta^4+\gamma^4=\frac16\left(x^4-6x^2y+3y^2+8xz\right).\tag{1}$$
You can verify this by expanding everything. In our case, $x=14,y=84$ and $z=584$, so just substitute in the numbers to solve the problem. I get the answer of $4368$.
(Note: I came up with the expression $(1)$ with a computer running Mathematica. It might not be very viable in a competition setting.)
| {
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Contradiction when solving a linear system with Gauss-Jordan elimination Consider a linear system with an unknown constant $a$:
$$
\left\{
\begin{aligned}
x+(a-1)y+az&=1 \\
ax+ay+az&=1 \\
a^2x+y+z&=a
\end{aligned}
\right.
$$
This gives us an augmented matrix:
$$A= \left[
\begin{array}{ccc|c}
1&a-1&a&1\\
a&a&a&1\\
a^2&1&1&a
\end{array}
\right] $$
From the augmented matrix, it is clear to me that when $a = 1$, we will have one parameter, since there are three unknowns but only two equations.
Let $a = 1$:
$$ \left[
\begin{array}{ccc|c}
1&0&1&1\\
1&1&1&1\\
1&1&1&1
\end{array}
\right] \xrightarrow{R_3 - R_2}
\left[
\begin{array}{ccc|c}
1&0&1&1\\
1&1&1&1\\
0&0&0&0
\end{array}
\right] \xrightarrow{R_2 - R_1}
\left[
\begin{array}{ccc|c}
1&0&1&1\\
0&1&0&0\\
0&0&0&0
\end{array}
\right]
$$
So we get the solution $z = s, y = 0, x = 1 - s \implies \text{Infinite number of solutions}$.
However, if instead I don't consider $a = 1$ but only $a \ne 0$, then I can proceed with Gauss-Jordan elimination to eventually get the following RREF after these series of elementary row operations:
$$
\frac1aR_2\longrightarrow
R_1 \leftrightarrow R_2 \longrightarrow
R_2-R_1 \longrightarrow
R_1-\frac{1}{a^2}R_3 \longrightarrow
\frac{1}{a^2}R_3 \longrightarrow
R_3+R_1 \longrightarrow
a^2R_1 \longrightarrow
\frac{1}{a^2-1}R_1 \longrightarrow
R_2-aR_1 \longrightarrow
R_2+2R_1 \longrightarrow
R_3\leftrightarrow
R_1 \longrightarrow
R_2\leftrightarrow
R_1 \longrightarrow
R_3-R_2 \longrightarrow
R_2-R_1
$$
$$A_\text{RREF}=
\left[
\begin{array}{ccc|c}
1&0&0&\frac1a\\
0&1&0&\frac{-(a-1)}{a}\\
0&0&1&\frac{a-1}{a}
\end{array}
\right]
$$
I had assumed that there was an error in my Gauss-Jordan elimination because this RREF implies that when $a = 1$, we have one and only one unique solution of $x = 1, y = 0, z = 0$. However, I plugged this matrix into MatLab and got the same RREF.
How can it be that when $a = 1$, we get two different solutions from the two row-equivalent matrices?
| In the case $a=1$, the rows 2 and 3 of the $3\times 3$ matrix are the same and so the matrix has rank 2 and there are infinitely many solutions (the solution set is a line).
If $a\ne 1$, the $3\times 3$ matrix has rank 3 and there is unique solution.
| {
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How to prove the sequence of $p_n(x)$ are all integer polynomials on x? for $x \in \mathbb R ,|x|< \frac12$
define:
$A_n(x)$
solution of the relation:
$A_{n+1}(x) = \frac{x^2}{1-A_n(x)}$
$A_0(x) = 0$
(I found that $A_n(x) = 2x^2 \cdot \frac{(1+\sqrt{1-4x^2})^n - (1 - \sqrt{1-4x^2})^n}{(1+\sqrt{1-4x^2})^{n+1} - (1 - \sqrt{1-4x^2})^{n+1}}$)
prove that the sequence:
$p_0(x)=1$
$p_{n+1}(x) = p_n(x) \cdot (1-x^2 - A_n(x))$
consists of integer polynomials in $x$
| I will sketch a solution assuming the relation given by the OP for $A_n$ which looks correct. First some notations to simplify things: let $c=\sqrt{1-4x^2}$ and $Y_{1,2}=\frac{1 \pm c}{2}$. We will first prove that:
$Y_1^n-Y_2^n=cQ_n(x)$ where $Q_n(x), n \ge 0$ is an integral polynomial in $x$
Since $Y_{1,2}$ are the roots of the equation $Y^2-Y+x^2=0$ it follows by Newton's relations that any symmetric polynomial $\sigma(Y_1,Y_2)$ is an integral polynomial in $x$ and since $Y_1-Y_2=c$ the claim above follows by factoring $Y_1^n-Y_2^n$
But now $1-A_{n+1}(x) = \frac{1-A_n(x)-x^2}{1-A_n(x)}$ implies $(1-A_{n+1}(x))(1-A_{n}(x))=\frac{p_{n+1}(x)}{p_n(x)}$. Noting that also $1-A_{n}(x)=\frac{x^2}{A_{n+1}(x)}$, we get:
$\frac{p_{n+1}(x)}{p_n(x)}=\frac{x^2}{A_{n+2}} \frac{x^2}{A_{n+1}}$ so multiplying these relations gives us:
$p_{n+1}(x)=\Pi_{k=2}^{k=n+2}\frac{x^2}{A_k(x)}\Pi_{k=1}^{k=n+1}\frac{x^2}{A_k(x)}$
By the OP computation $\frac{x^2}{A_k(x)}=\frac{Y_1^{k+1}-Y_2^{k+1}}{Y_1^k-Y_2^k}$,
which immediately gives:
$p_{n+1}(x)=\frac{Y_1^{n+3}-Y_2^{n+3}}{Y_1^2-Y_2^2}\frac{Y_1^{n+2}-Y_2^{n+2}}{Y_1-Y_2}$
Now $Y_1-Y_2=c, Y_1^2-Y_2^2=c$ so:
$p_{n+1}(x)=Q_{n+3}(x)Q_{n+2}(x)$ integral so we are done!
| {
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A pattern on finding squares for which their sum is $\Big(\sum\limits_{i=0}^j x^i\Big)^2$ Apologies for the inactivity. I haven't been doing so well in life, lately, but I'm glad to be back at some maths! Here's a pattern I discovered. I'm not good at explaining with words, so hope you get it. It shouldn't be too difficult. Can it be proven?
For all $x\in\mathbb{R}$,
$$\begin{align}\bigg(\sum_{i=0}^1 x^i\bigg)^2&=x^2+2x+1\\ &=(x+1)^2\end{align}$$
$$\begin{align}\bigg(\sum_{i=0}^3 x^i\bigg)^2&=x^6+2x^5+3x^4+4x^3+3x^2+2x+1\\ &=(x+1)^2+2(x^2+x)^2+(x^3+x^2)^2\end{align}$$
$$\begin{align}\bigg(\sum_{i=0}^5 x^i\bigg)^2&= x^{10}
+2x^9+3x^8+4x^7+5x^6+6x^5+5x^4+4x^3+3x^2+2x+1\\ &=(x+1)^2+2(x^2+x)^2+3(x^3+x^2)^2+2(x^4+x^3)^2+(x^5+x^4)^2\end{align}$$
Clearly there is a pattern with the exponents in the square of each partial sum, as well as the coefficients. Can one prove this pattern continues ad infinitum?
For $\sum\limits^j$ I think the pattern also applies to even $j$ and not just odd $j$ as shown above, but I wrote this all on a napkin as I was eating at a restaurant, and now I've lost the napkin. (True story.)
Any help would be much appreciated. Thank you :)
| Induction.
We want to show that $$\left(\sum\limits_{i=0}^{2k+1}x^i\right)^2=(k+1)(x^{k+1}+x^k)^2+\sum_{j=1}^kj[(x^j+x^{j-1})^2+(x^{2k+2-j}+x^{2k+1-j})^2]$$ for all $k\ge1$. As you have shown the base case, suppose this is true for some $k\in\Bbb N$. Then consider $k+1$. \begin{align}\left(\sum\limits_{i=0}^{2k+3}x^i\right)^2&=\left(x^{2k+3}+x^{2k+2}+\sum\limits_{i=0}^{2k+1}x^i\right)^2\\&=(x^{2k+3}+x^{2k+2})^2+2(x^{2k+3}+x^{2k+2})\sum\limits_{i=0}^{2k+1}x^i+\left(\sum\limits_{i=0}^{2k+1}x^i\right)^2\\&=(x^{2k+3}+x^{2k+2})^2+x^{2k+2}+x^{4k+4}+2\sum_{i=2k+2}^{4k+3}x^i\\&\quad+(k+1)(x^{k+1}+x^k)^2+\sum_{j=1}^kj[(x^j+x^{j-1})^2+(x^{2k+2-j}+x^{2k+1-j})^2]\end{align} and we want to show that $$\left(\sum\limits_{i=0}^{2k+3}x^i\right)^2=(k+2)(x^{k+2}+x^{k+1})^2+\sum_{j=1}^{k+1}j[(x^j+x^{j-1})^2+(x^{2k+2-j}+x^{2k+1-j})^2].$$ Subtracting the two equations yields $$\sum_{i=k+2}^{2k+2}(x^i+x^{i-1})^2=x^{2k+2}+x^{4k+4}+2\sum_{i=2k+2}^{4k+3}x^i$$ which is true. $\square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all natural numbers $n$ such that $2n+1$ divides $n^{4}+n^{2}$.
Find all natural numbers $n$ such that $2n+1$ divides $n^{4}+n^{2}$.
My attempt:-
We need to find natural numbers, $n$, for which $2n+1\mid n^{4}+n^{2}$. We write, $$n^{4}+n^{2}=n^{2}(n^{2}+1)$$. It can be easily proved that $(n^{2},2n+1)=1$(so I think I need not type it here because that will increase the length of this post).This implies that, $$n^{2}+1\equiv 0\pmod {2n+1}\implies 4n^2+4\equiv 0\pmod{2n+1} \implies (2n+1)^{2}+3\equiv 4n\pmod{2n+1} \implies 4n\equiv 3\pmod{2n+1} \implies 2(2n+1)\equiv 5\pmod{2n+1}$$.
Hence, $2n+1\mid 5$ and this implies $n=2$, which is the only solution. Does this look good? I think this solution is correct but I'm not satisfied with it, because it took me quite a long time before I came up with this argument. Can this be solved without any such manipulations? Thank you.
| Divide $$ \frac{n^4 + n^2}{2n+1} = \frac{n^3}{2} - \frac{n^2}{4} + \frac{5n}{8} - \frac{5}{16} + \frac{5}{8 (2n+1)}$$ and note that the $\frac{5}{8 (2n+1)}$ term, if rewritten in lowest terms, will have an odd factor in the denominator unless $n = 2$, so the whole expression cannot be an integer. Then all that remains is checking that $n + 2$ is in fact a solution.
| {
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"source": "stackexchange",
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Find the sum to n terms $$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$
Now to solve this one has to calculate $\sum_{n=1}^{n}n^4$ which will be a very lengthy process, is there any shorter method to solve this question?
By the way I calculated $\sum_{n=1}^{n}n^4$ and it came as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(2\cdot n+1\right)\cdot\left(3\cdot n^2+3\cdot n-1\right)}{30}$, then I substituted this value into the original equation.
Then I got final answer as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(3\cdot n^2+6\cdot n+1\right)}{60}$
But it took me a very long time to calculate all of this, is there any shorter way to solve this problem?
| I change a little the notations from your original wording.
$T(n)=\dfrac{n(n+1)}{2}$
$S(n)=\sum\limits_{k=1}^nT(k)^2$
Cheating on the resulting formula for $f(n)=\dfrac{n(n+1)(n+2)(3n^2+6n+1)}{60}$
we notice that $f(0)=f(-1)=f(-2)=0$.
Is there a way to exploit these negative indices in order to find the $(3n^2+6n+1)$ part more easily?
In fact there is, $T(n)$ is perfectly defined for negative numbers so this part does not pose any issue.
But how to interpret $S(-n)$?
To be consistent with the summation identity $\sum\limits_a^{b-1}+\sum\limits_b^c=\sum\limits_a^c$ when $a<b<c\ $ one need to set $\ \sum\limits_M^m=-\sum\limits_{m+1}^{M-1}$ for the case $m<M$.
This results in $\displaystyle S(-n)=\sum\limits_{k=1}^{-n}T(k)^2=-\sum\limits_{k=-(n-1)}^0T(k)^2$
I let you do the calculation to verify that $\begin{array}{lcr}S(0)&=&0\\S(-1)&=&0\\ S(-2)&=&0\\ S(-3)&=&-1\\S(-4)&=&-10\\S(-5)&=&-46\end{array}$
Now by applying the identification method (I describe in my other post) to $$S(n)=a_0+a_1\,n+a_2\,n(n+1)+a_3\,n(n+1)(n+2)+\cdots+a_5\,n(n+1)(n+2)(n+3)(n+4)$$
You immediately get $a_0=a_1=a_2=0$
The others coefficients give you $$S(n)=n(n+1)(n+2)\bigg[\frac 16-\frac 14(n+3)+\frac 1{20}(n+3)(n+4)\bigg]=n(n+1)(n+2)\left(\dfrac{3n^2+6n+1}{60}\right)$$
| {
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"question_score": "2",
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For what values of $a$ does $b\in\mathbb{N}$? I've the following function:
$$b=\frac{1}{2}-\frac{1}{r-2}+\sqrt{\frac{1}{4}-\frac{r-3}{(r-2)^2}+\frac{a(1+a)}{r-2}+\frac{a(a^2-1)}{3}}$$
I know that:
*
*$\text{a}\ge1$ and $a\in\mathbb{N}$;
*$\text{r}\ge3$ and $r\in\mathbb{N}$.
If I set the value of $r$, what does $a$ has to be in order to let $b\ge1$ and $b\in\mathbb{N}$? Are there restrictions on $a$ in order to let $b\ge1$ and $b\in\mathbb{N}$?
| Let $s = r - 2$. Then $$b=\frac{1}{2}-\frac{1}{s}+\sqrt{\frac{1}{4}-\frac{s - 1}{s^2}+\frac{a(1+a)}{r-2}+\frac{a(a^2-1)}{3}}$$ $$b = \frac{1}{2}-\frac{1}{s}+\sqrt{\frac{1}{4}-\frac{1}{s}+\frac{1}{s^2}+\frac{a(1+a)}{s}+\frac{a(a^2-1)}{3}}$$ $$b = \frac{1}{2}-\frac{1}{s}+\sqrt{\left(\frac{1}{2}-\frac{1}{s}\right)^2+\frac{a(1+a)}{s}+\frac{a(a^2-1)}{3}}$$.
Let $t = \frac{1}{2}-\frac{1}{s}$ or $\frac{1}{s} = \frac{1}{2} - t$, such that $$b = t + \sqrt{t^2 - a(1 + a)t + \frac{a(1 + a)}{2} + \frac{a(1 + a)(a - 1)}{3}}$$
$$b = t + \sqrt{t^2 - a(1 + a)t + \frac{3a(1 + a) + 2(a - 1)a(1 + a)}{6}}$$
$$b = t + \sqrt{t^2 - 2t\frac{a(1 + a)}{2} + \frac{a(1 + a)(2a + 1)}{6}}$$
$$b = t + \sqrt{t^2 - 2t\sum_{i = 1}^{a} i + \sum_{i = 1}^{a} i^2}$$
As for all integers $a$, $$t < \sum_{i = 1}^{a} i$$
$$b = t + \sqrt{\left(-t + \sum_{i = 1}^{a} i\right)^2} = t - t + \sum_{i = 1}^{a} i = \sum_{i = 1}^{a} i = \frac{a(a + 1)}{2}$$
$b$ is only dependent upon $a$. If $a\in \mathbb{N}$ and $a \geq 1$, then $b\in \mathbb{N}$ and $b > 1$.
| {
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Solving equality of $3\times3$ determinant In Basic Mathematics, Lang, exercise of section 17.5...
Gives this answer...
My own calculations gave...
$(x_2-x_1)(x_3^2-x_1^2)-(x_3-x_1)(x_2^2-x_1^2)$
Is it correct?
| Personally, I would calculate in a slightly different way, in order to obtain directly a formula easier to memorise:
\begin{align}
\begin{vmatrix}
1&x_1&x_1^2 \\
1&x_2&x_2^2 \\
1&x_3&x_3^2 \\
\end{vmatrix}
&=
\begin{vmatrix}
0&x_1-x_2&x_1^2-x_2^2 \\
0&x_2-x_3&x_2^2-x_3^2 \\
1&x_3&x_3^2 \\
\end{vmatrix}
=(x_1-x_2)(x_2-x_3)
\begin{vmatrix}
0&1&x_1+x_2 \\
0&1&x_2+x_3 \\
1&x_3&x_3^2 \\
\end{vmatrix} \\
&=(x_1-x_2)(x_2-x_3)\bigl((x_2+x_3)-(x_1+x_2)\bigl) \\
&=\color{red}{(x_1-x_2)(x_2-x_3)(x_3-x_1)}.
\end{align}
| {
"language": "en",
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Area of a triangle ABC Triangle's $ABC$ angle $C=60°$, $AB = AC+2 = BC-1$.
Find the area of this triangle.
I've tried writing $AB$ as $x$ so $AC= x-2$ and $BC = x+1$.
Then i calculated the area with
$$\frac12\sin(60°)\cdot(x-2)\cdot(x+1)=\frac12\sin(60°)\cdot(x^2-x-2)$$
Now i have no idea what to do next.
| With the theorem of cosines we get
$$c^2=a^2+b^2-2ab\cos(\frac{\pi}{3})$$
substituting $$c=b+2,a=b+3$$ we get an equation for $b$:
$$(b+2)^2=(b+3)^2+b^2-2(b+3)b\cos(\frac{\pi}{3})$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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A basket has $10$ green, $10$ blue, and $10$ red balls. In how many ways can $10$ balls be selected from the basket if the number of red balls is odd? I tried to learn generating functions, but I am a way out of it.
And I know a way to solve problems like that by $S_0 - S_1 + S_2 +\cdots$
Please show me in this way because I don't know GF.
The question is:
A basket has $10$ green balls, $10$ blues and $10$ reds. In how many ways can $10$ balls be selected from the basket so that the number of red balls is odd?
Here is a picture of what I tried to do and obviously I'm getting negative number of solutions.
| Let $b$, $g$, and $r$ represent, respectively, the number of blue, green, and red balls that are selected from the basket. We know that
$$b + g + r = 10$$
However, we require that $r$ is odd, so there are five cases.
$r = 1$: Then
\begin{align*}
b + g + 1 & = 10\\
b + g & = 9
\end{align*}
which is an equation in the nonnegative integers. The number of solutions of the equation $b + g = 9$ in the nonnegative integers is
$$\binom{9 + 2 - 1}{2 - 1} = \binom{10}{1} = 10$$
More generally, the number of solutions of the equation
$$b + g = n$$
in the nonnegative integers is
$$\binom{n + 2 - 1}{2 - 1} = \binom{n + 1}{1} = n + 1$$
$r = 3$: Then
\begin{align*}
b + g + 3 & = 10\\
b + g & = 7
\end{align*}
which is an equation in the nonnegative integers with $7 + 1 = 8$ solutions.
$r = 5$: Then
\begin{align*}
b + g + 5 & = 10\\
b + g & = 5
\end{align*}
which is an equation in the nonnegative integers with $5 + 1 = 6$ solutions.
$r = 7$: Then
\begin{align*}
b + g + 7 & = 10\\
b + g & = 3
\end{align*}
which is an equation in the nonnegative integers with $3 + 1 = 4$ solutions.
$r = 9$: Then
\begin{align*}
b + g + 9 & = 10\\
b + g & = 1
\end{align*}
which is an equation in the nonnegative integers with $1 + 1 = 2$ solutions.
Total: Since the above cases are mutually exclusive and exhaustive, the number of ways of selecting $10$ balls from the basket if the number of red balls is odd is
$$10 + 8 + 6 + 4 + 2 = 30$$
Notice that fixing the value of $r$ reduces the number of variables by $1$, which is where you made your mistake in calculating $S_1$. Also, the problem asks you calculate what you are calling $S_1$. There is no need to apply the Inclusion-Exclusion Principle here.
| {
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"source": "stackexchange",
"question_score": "1",
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Find all integers such that $2n-1 | n^3 +1$ I try
$$2n-1| n^3 +1$$
$$\therefore 2n11 | 2n^3 + 2 -n^2(2n-1)$$
$$\therefore 2n-1| 4n-2$$
But but this is valid for all $n$. How to proceed? Thanks in advance
and I'm sorry if this is a duplicate, I don't see any similar questions
| Problem. Find all integers $n$ such that $2n - 1\, |\, n^3 + 1$.
Sol. Let $(a, b)$ denote the greatest common divisor (gcd) of $a$ and $b$. Then
$$\begin{array}{lll}
(2n - 1, n^3 + 1) &= (2n - 1, n^3 + 2n) \\
&= (2n - 1, n^2 + 2) \mbox{ since } (2n - 1, n) = 1 \\
&= (2n - 1, n^2 + 2 + 4n - 2) = (2n - 1, n(n + 4)) \\
&= (2n - 1, n + 4) \mbox{ since } (2n - 1, n) = 1 \\
&= (2n - 1 - 2(n + 4), n + 4) = (-9, n + 4).
\end{array}$$
Since $2n - 1\, |\, n^3 + 1$, $2n - 1\, |\, 9$. So, $n$ may be $-4, -1, 0, 1, 2, 5$.
| {
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Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$? Which is bigger, $3$ or $\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$
I am not sure how to do it. I thought of squaring both sides and moving the $2$ to the other side, then squaring again etc. But I don't see a pattern in the left side.
| Actually, there is a rather simple method to do this:
We can rewrite
$$a=\sqrt{2 + \sqrt{3 + \sqrt{4 + \sqrt{5 + \sqrt{... + \sqrt{100}}}}}}$$
into
$$b=\sqrt{2 + \sqrt{3 + \sqrt{100 + \sqrt{100 + \sqrt{100+\cdots}}}}}$$
And we can clearly see that
$$b>a$$
There is a classic method to evaluate continued root. Let
$$x=\sqrt{100 + \sqrt{100 + \sqrt{100+\cdots}}}$$
Then
$$x=\sqrt{100+x}$$
$$x^2=100+x$$
$$x=\frac{1+\sqrt{401}}{2}$$
Obviously,
$$20<\sqrt{401}<21$$
We see that
$$x<\frac{1+21}{2}=11$$
Then we have
$$b<\sqrt{2 + \sqrt{3+11}}$$
$$b<\sqrt{2 + \sqrt{14}}<\sqrt{6}<3$$
So
$$a<b<3$$
| {
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"source": "stackexchange",
"question_score": "7",
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Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$ Question :
Solve for natural number the equation :
$5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
My try :
Let : $X=5^{x}$ and $Y=2^{y}$ so above equation
equivalent :
$2X^{2}+(Y-4)X-6Y^{2}-Y+2=0$
We solve this equation for $X$
$\Delta =(7Y)^{2}$ mean : $X_{1}=\frac{3}{2}Y+1$ and $X_{2}=1-2Y$
From here how I can find $X$ and $Y$ , this is
all my effort ?
Thanks!
| Different substitutions, let
$$ u = 5^x \; , \; \; v = 2^{y-1} $$
Then
$$ \frac{1}{4} \left( (2u+v-2)^2 - 49 v^2 \right) = 0 \; , \; \; $$
$$ \left( 2u+8v-2 \right) \left( 2u-6v-2 \right) = 0 \; , \; \; $$
$$ \left( u+4v-1 \right) \left( u-3v-1 \right) = 0 \; , \; \; $$
Both $u,v > 0$ so we are left with
$$ 5^x = 3 \cdot 2^{y-1} + 1 $$
For example, $3 \cdot 8 + 1 = 25.$
Monday: in some cases there is an elementary proof that we have already found the largest solution. We already have $3 \cdot 8 + 1 = 25.$ If we were to have a larger solution, it would be of the form
$$ 24 \cdot 2^s + 1 = 25 \cdot 5^t $$
Subtract 25 from both sides,
$$ 24 \cdot 2^s - 24 = 25 \cdot 5^t - 25 \; , \; $$
$$ 24 \left( 2^s - 1 \right) = 25 \left( 5^t - 1 \right) $$
We will ASSUME $s,t \geq 1$ and get a contradiction.
Since $25 | (2^s - 1),$ we find $2^s \equiv 1 \pmod {25}$ and then $s$ is divisible by 20. This
$$ 2^{20} - 1 = 3 \cdot 5^2 \cdot 11 \cdot 31 \cdot 41. $$
The actual $2^s - 1$ is divisible by $2^{20} - 1,$ and therefore by the prime $41.$
Since $41 | (5^t - 1),$ we find $5^t \equiv 1 \pmod {41}$ and then $t$ is divisible by 20. This
$$ 5^{20} - 1 = 2^4 \cdot 3 \cdot 11 \cdot 13 \cdot 41 \cdot 71\cdot 521 \cdot 9161. $$
The actual $5^t - 1$ is divisible by $5^{20} - 1,$ and therefore by $2^4 = 16.$
We have arrived at $$ 16 | 24 (2^s - 1). $$ Divide throsgh by $8,$ we get
$$ 2 | 3 (2^s - 1) $$
This is a contradiction, as both $3$ and $2^s - 1$ are odd when $s \geq 1.$ The contradiction tells us the assumptions are wrong, and
$$ s = 0 $$
This completes the proof that $3 \cdot 8 + 1 = 25$ is the largest solution.
| {
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"source": "stackexchange",
"question_score": "3",
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Find all $n$ natural numbers such that $10\mid n^{10}+1$ Determine all natural numbers $n$ such that :
$10$ divisor of $n^{10}+1$
My attempt :
Let $n=r(\mod{10})$ so $n^{10}+1=(r^{10}+1)(\mod{10})$
This mean : $r^{10}+1=0(\mod{10})$
Now $r\in {0,1,2,3,4,5,6,7,8,9}$ after try I get $r=3,7$
So : $n=10k+3,10k+7$
Is my work correct ?
Please I need other simple method to computing
| You are correct but a few things to make you calculations much fewer.
1) if $n$ and $10$ are not relatively prime then any common divisor $a;a\ne 1$ will not divide $n^{10}+1$ so $10\not \mid n^{10}+1$.
So we need only test $1,3,7,9$
2) $(10-i)^{10}\equiv i^{10}\pmod{10}$ so we only check $1,3$.
3) Eulers Thereom says as $\phi(10) =4$ that for $n$ relatively prime to $10$ that $n^4\equiv 1 \pmod {10}$ so $n^{10}\equiv n^2 \pmod {10}$
So we just need to check whether $k^2 + 1\equiv 0 \pmod {10}$ for $k= 1,3$.
We get $k=3$.
and so our solutions are $n\equiv \pm 3 \pmod {10}$
or we could say $n\equiv 3,7 \pmod {10}$
or we could say $n = 10k +3$ or $n=10k +7$ for some integer $7$
or we could say "$n$ is any number with the last digit of $3$ or $7$".
| {
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Cauchy equation $ f ( x + y ) = f ( x ) + f ( y ) $ with an additional condition $ f \left( \frac 1 x \right) = \frac 1 { x ^ 2 } f ( x ) $
Let $ f : \mathbb R \to \mathbb R $ satisfy the following:
*
*$ f ( 1 ) = 1 $;
*$ f ( x + y ) = f ( x ) + f ( y ) $, $ \forall x , y \in \mathbb R $;
*$ f \left( \frac 1 x \right) = \frac 1 { x ^ 2 } f ( x ) , \forall x \in \mathbb R ^ * $.
Find $ f \left( \sqrt { 2019 } \right) $.
This question was a multiple choice one, so I simply observed that one such function is $ f ( x ) = x , \forall x \in \mathbb R $, so $ f \left( \sqrt { 2019 } \right) = \sqrt { 2019 } $.
This is definitely not a rigorous solution, it is based on pure observation. I wanted to know if there is some way to solve this thoroughly.
| Let $(1)$ denote the additive condition, and $(2)$ denote the condition that $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2} $. Now note that for all $x\notin \{-1,0\}$, $$\begin{align*}\frac{f\left(x^2+x\right)}{(x^2+x)^2}\stackrel{(2)}=f\left(\frac{1}{x^2+x}\right)&\stackrel{(1)}=f\left(\frac{1}{x}\right)-f\left(\frac{1}{x+1}\right) \\ &\stackrel{(2)}=\frac{f(x)}{x^2}-\frac{f(x+1)}{(x+1)^2} \\ &= \frac{(x+1)^2f(x)-x^2f(x+1)}{(x^2+x)^2} \\ &\stackrel{(1)}=\frac{(2x+1)f(x)-x^2}{(x^2+x)^2} \end{align*}$$
Equating the two, we find that $$\frac{f\left(x^2+x\right)}{(x^2+x)^2}=\frac{(2x+1)f(x)-x^2}{(x^2+x)^2} \implies f(x^2)+x^2=2xf(x)\tag{3}$$Now set $x+y$ in place of $x$ in the above to find $$f(x^2)+2f(xy)+f(y^2)+x^2+2xy+y^2=2(x+y)(f(x)+f(y))$$ and after simplifying using $(3)$, we find that $f(xy)-xy=y(f(x)-x)+x(f(y)-y)$, or $g(xy)=g(x)+g(y)$ where $g(x)=\frac{f(x)-x}{x},x\neq 0$. By $(2)$, $g(x)=g(\frac{1}{x})$, so $$g(xy)=g(x)+g(y)=g(x)+g(\frac{1}{y})=g\left(\frac{x}{y}\right)\tag{4}$$So for any positive reals $a$ and $b$, setting $x=\sqrt{ab}$ and $y=\sqrt{\frac{a}{b}}$ in $(4)$ gives $f(a)=f(b)$, hence $g$ is constant, and therefore $0$ over $\mathbb{R}^+$. Hence over $\mathbb{R}^+$, $f(x)\equiv x$. But by additivity, $f(-x)=-f(x)$, and $f(0)=0$.
Therefore, the only such function $f$ is indeed $f(x)=x$ for all real $x$, which clearly satisfies all conditions presented.
| {
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Find the explicit form of $ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1} $. Find the explicit form of
$$
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}.
$$
Let $S(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}$. It has radius of convergence $1$.
Let $S_1(x)=xS(x)$. Then $S_1'(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+2)}x^{n-1}$ for $|x|<1$.
Let $S_2(x)=x^3S_1'(x)$. Then $S_2'(x)=\sum_{n=1}^{\infty}(-x)^{n-1}=\frac{x^2}{1+x}$.
By integration, I obtained $S_1'(x)=\frac{1}{2x}-\frac{1}{x^2}+\frac{\ln (x+1)}{x^3}$. Then how to obtain $S(x)$? Or there is other method to do this problem?
| $$\log(1+x)=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}~~~(1)$$
Multipluy by $x$ on both sides and integrate w.r.t. $x$
$$\int x \log(1+x) dx= \sum_{n=1}^{\infty} (-1)^{n-1}\int \frac{x^{n+1}}{n} dx= \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n+2}}{n(n+2)} .$$
$$\implies S(x)=\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^{n-1}}{n(n+2)}= \frac{1}{x^3} \int x \log(1+x) ~dx.$$
Finally by carrying out integration by parts, we get the required sum
$$S(x)=\frac{1}{4x^3}[2x-x^2-2\log(1+x)+2x^2 \log(1+x)]$$
| {
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Express the other roots with the powers of $\alpha$. Let $f(x)=x^4+x+1 \in Z_2[x]$ and $\alpha$ one of its roots'.
Express the other roots using the powers of $\alpha$.
I believe I have to find the exact elements of its Galois group, and use them to express the other roots. The problem is, I could only find the identity and complex conjugation. I'm not even sure what its splitting field is, but i know that it has 2 complex conjugate root-pairs. Can this be done without exactly calculating the roots?
| Hint: Let $\alpha$ be a root of $x^4+x+1$ in $GF(16)$, the splitting field of the polynomial over $GF(2)$. Then $\alpha^4=\alpha+1$ in $GF(16)$ and so each nonzero element of $GF(16)$ can be expressed as a power of $\alpha$:
$$1, \alpha, \alpha^2, \alpha^3,\alpha^4=\alpha+1, \alpha^5=\alpha^2+\alpha, \alpha^6=\alpha^3+\alpha^2, \alpha^7 = \alpha^3+\alpha+1, \alpha^8=\alpha^2+1,\alpha^8=\alpha^3+\alpha,\alpha^{10} = \alpha^2+\alpha+1, \alpha^{11} = \alpha^3+\alpha^2+\alpha, \alpha^{12}=\alpha^3+\alpha^2+\alpha+1,\alpha^{13}=\alpha^3+\alpha^2+1,\alpha^{14}= \alpha^3+1,\alpha^{15}= 1.$$
Addendum:
$\alpha,\alpha^2,\alpha^4,\alpha^8$ are the roots of $x^4+x+1$.
*
*The zeros of an irreducible polynomial over $GF(2)$ are the powers of the Frobenius isomorphism: $x\mapsto x^2$. The zeros form a socalled conjugacy class.
$\alpha^7,\alpha^{14},\alpha^{13},\alpha^{11}$ are the roots of $x^4+x^3+1$, the conjugate polynomial of $x^4+x+1$.
*
*The zeros of the reciprocal polynomial contain $\alpha^{-1}=\alpha^{14}$.
$\alpha^3,\alpha^6,\alpha^{12},\alpha^9$ are the roots of $x^4+x^3+x^2+x+1$ which divides $x^5+1$.
*
*These zeros are 5th roots of unity.
$\alpha^5,\alpha^{10}$ are the roots of $x^2+x+1$ which divides $x^3+1$.
*
*These zeros are contained in $GF(4)$.
$1$ is the root of $x+1$.
| {
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Find the smallest number for which the number divides completely into 101
Find the smallest number $n,(n>4)$,
$A=\binom{3n-1}{11}+\binom{3n-1}{12}+\binom{3n}{13}+\binom{3n+1}{14}$
for which the number divides completely into $101$.
My solution:
$\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}\\\binom{3n-1}{11}+\binom{3n-1}{12}=\binom{3n}{12}\\\binom{3n}{12}+\binom{3n}{13}=\binom{3n+1}{13}\\\binom{3n+1}{13}+\binom{3n+1}{14}=\binom{3n+2}{14}\\\binom{3n+2}{14}=\frac{(3n+2)!}{14!(3n+2-14)!}=\frac{(3n+2)!}{14!(3n-12)!}$
and at the moment I don't know how to do the task efficiently
| Hint: You're almost there. Since $101$ is a prime number, it will divide $\ \frac{(3n+2)!}{14!(3n-12)!}\ $ evenly if and only if it divides one of the numbers $\ 3n+2,3n+1, 3n, \dots,3n-11\ $.
| {
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Compute the value of the $30$th derivative of the function $g(x) = \sinh{\left(x^4\right)}$ at the origin, i.e. $g^{(30)}(0)$. Compute the value of the $30$th derivative of the function $g(x) = \sinh{\left(x^4\right)}$ at the origin, i.e. $g^{(30)}(0)$.
So we have the (Macluarin) series
\begin{equation*}
\sinh{(x)} = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}.
\end{equation*}
at $x = 0$ so
\begin{equation*}
\sinh{(x^4)} = \sum_{n=0}^{\infty} \frac{(x^4)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{x^{8n+4}}{(2n+1)!}.
\end{equation*}
The Maclaurin series for $\sinh{(x^4)}$ is
\begin{equation*}
\sum_{n=0}^{\infty} \frac{(x^4)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{x^{8n+4}}{(2n+1)!} = \sum_{m=0}^{\infty} \frac{x^m}{\left(\frac{m}{4}\right)!}.
\end{equation*}
We can see that the coefficient for $x^m$ is $\frac{1}{\left(\frac{m}{4}\right)!}$. On the other hand, this is the Maclaurin series for $g(x)$, and so the coefficient on $x^m$ is equal to $\frac{g^{(m)}(0)}{m}$. Equating these two, we have
\begin{equation*}
\frac{g^{(m)}(0)}{m} = \frac{1}{\left(\frac{m}{4}\right)!} \Longleftrightarrow g^{(m)}(0) = \frac{m}{\left(\frac{m}{4}\right)!}.
\end{equation*}
Substituting $m = 30$ gives $g^{(30)}(0) = \frac{30}{7.5!}$. Am I on the right track here?
| Definitely on the right track. Last thing you need to know is that the coefficient of $x^n$ in the Maclaurin series is given by $a_n=\frac{f^{n}(0)}{n!}$
| {
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If each pair of equations $x^2=b_1x+c_1=0,x^2=b_2x+c_2 \text{ and } x^2+b_3x=c_3$ have a common root, prove following If each pair of equations $x^2=b_1x+c_1=0,x^2=b_2x+c_2 \text{ and } x^2+b_3x=c_3$ have a common root, prove that
$(b_1+b_2+b_3)^2=4(c_1+c_2+c_3+b_1b_2)$
My attempt is as follows:
For equations $x^2=b_1x+c_1,x^2=b_2x+c_2$ to have a common root:
$(c_2-c_1)^2=(b_1c_2-b_2c_1)(b_1-b_2)$
For equations $x^2=b_2x+c_2,x^2+b_3x=c_3$ to have a common root:
$(c_3-c_2)^2=(b_2c_3+b_3c_2)(b_3+b_2)$
For equations $x^2=b_1x+c_1,x^2+b_3x=c_3$ to have a common root:
$(c_3-c_1)^2=(b_1c_3+b_3c_1)(b_1+b_3)$
Adding all three equations:
$2({c_1}^2+{c_2}^2+{c_3}^2-c_1c_2-c_2c_3-c_3c_1)=({b_1}^2+b_2b_3)(c_2+c_3)+({b_2}^2+b_1b_3)(c_1+c_3)+({b_3}^2-b_1b_2)(c_1+c_2)$
But from here I was not able to proceed towards the proof. Please help me in this.
| I finally got this, thanks to @lab bhattacharjee
There are three quadratic equations
\begin{equation}
x^2-b_1x-c_1=0\tag{1}
\end{equation}
\begin{equation}
x^2-b_2x-c_2=0\tag{2}
\end{equation}
\begin{equation}
x^2+b_3x-c_3=0\tag{3}
\end{equation}
Suppose $(1)$ and $(2)$ have a common root as $p$, $(2)$ and $(3)$ have a common root as $q$, (1) and (3) have a common root as $r$.
\begin{equation}
p+r=b_1\tag{4}
\end{equation}
\begin{equation}
p+q=b_2\tag{5}
\end{equation}
\begin{equation}
-(q+r)=b_3\tag{6}
\end{equation}
$$b_1+b_2+b_3=2p$$
\begin{equation}
(b_1+b_2+b_3)^2=4p^2\tag{7}
\end{equation}
\begin{equation}
-pr=c_1\tag{8}
\end{equation}
\begin{equation}
-pq=c_2\tag{9}
\end{equation}
\begin{equation}
-qr=c_3\tag{10}
\end{equation}
Adding $(8),(9),(10)$
\begin{equation}
4(c1+c2+c3)=-4(pq+qr+pr)\tag{11}
\end{equation}
\begin{equation}
4b_1b_2=4(p+r)(p+q)\tag{12}
\end{equation}
Adding $(11),(12)$
$$4(c_1+c_2+c_3+b_1b_2)=-4(pq+qr+pr)+4(p^2+pq+pr+qr)$$
$$4(c_1+c_2+c_3+b_1b_2)=4p^2$$
Hence $4(c_1+c_2+c_3+b_1b_2)=(b_1+b_2+b_3)^2$
| {
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What is $\frac{d^2y}{dx^2}$ of $(y^5) + (x^5) = 8$ when you solve using implicit differentiation?
Find the Second derivative with respect to $x$ of :
$$ y^5 + x^5 = 8$$
When I solved it I got $\frac yx$ as the second derivative but I don’t think that I am right. Could someone explain it to me.
| Starting with
$y^5 + x^5 = 8, \tag 1$
we differentiate once to obtain
$5y^4 y' + 5x^4 = 0, \tag 2$
from which we may isolate $y'$:
$y' = -\dfrac{x^4}{y^4}; \tag 3$
differentiating this expression using the quotient rule yields
$y'' = -\dfrac{4x^3y^4 - 4x^4y^3y'}{y^8} = -\dfrac{4x^3y - 4x^4y'}{y^5}; \tag 4$
substitute (3) into (4):
$y'' = -\dfrac{4x^3y - 4x^4(-x^4/y^4)}{y^5} = -\dfrac{4x^3y^5 + 4x^8}{y^9}; \tag 5$
we can now in fact eliminate $y$ from the numerator using (1) in the form
$y^5 = 8 - x^5, \tag 6$
that is
$y'' = -\dfrac{4x^3y^5 + 4x^8}{y^9}$
$= -\dfrac{4x^3(8 - x^5) + 4x^8}{y^9} = -\dfrac{32x^3 - 4x^8 + 4x^8}{y^9} = -\dfrac{32x^3}{y^9}; \tag 7$
it is possible to use (6) once again to reduce the power of $y$ in the denominator, viz.
$y'' = -\dfrac{32x^3}{y^9} = -\dfrac{32x^3}{(8 - x^5)y^4}; \tag 8$
an expression which may have some utility, though I am somewhat skeptical as to its genuine advantage over, say, (7).
| {
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Showing continuity of $g(x, y)$ Let $g : \mathbb{R}^2 \to \mathbb{R}$. Let the domain be $y^2 \leq \frac{x}{3}$.
$g(x, y) = \begin{cases}
\frac{x^3}{y^2} \left (\sqrt{x} - \sqrt{y^2 + x} \right ) & y \neq 0\\
0 & y = 0
\end{cases} $
I want to show $g$ is continuous at $(0, 0)$.
We have $|g(x, y) - g(0,0)| = \frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right )$.
Fix some $\epsilon > 0$.
Now, I want to find a $\delta > 0$ such that $x^2 + y^2 < \delta$ $\implies $ $\frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right ) < \epsilon$
| $|\frac{x^3}{y^2} \left ( \sqrt{y^2 + x} - \sqrt{x} \right )|=|\frac{x^3}{\sqrt{x}+\sqrt{y^2+x}}| \leq |x|^{\frac{5}{2}}$
So the limit of the function at zero is zero.
| {
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Introduction of a phase term in the solution of the wave equation I would like to understand this step in the solution of the wave equation as presented here. The wave equation is formulated as
$$u_{tt}=c^2\Delta u$$
where $u$ is a function of time $t$ and space coordinates; $c^2$ is a positive constant, and $\Delta u$ the Laplacian.
Through separation of variables, the solution $u$ is hoped to be the product of a function of space $f$ and time $T,$ as in $u=Tf,$ in which case
$$T^{''}f = c^2T\Delta f,$$
implying then that the quotients below are constant:
$$\frac{T^{''}}{T}= c^2 \frac{\Delta f}{f} = -c^2 \omega^2,$$
leading to
$$T^{''} = -c^2 \omega^2 T$$
and
$$\Delta f = -c^2 \omega^2 f$$
From there,
$$T=A \sin(c\omega t) + B \cos(c\omega t)$$
and
$$\begin{align}
u &= \sum C f(x,y,z)\left(A \sin(c\omega t) + B \cos(c\omega t) \right)\\[3ex]
&=\sum C f(x,y,z) \sqrt{A^2+B^2} \sin(c\omega t + \beta) \\[3ex]
&= \sum \tilde C f(x,y,z) \sin(c\omega t + \beta)
\end{align}$$
with $\beta$ coming from $\sin\beta=\frac{B}{\sqrt{A^2 + B^2}}$ and $\cos\beta= \frac{A}{\sqrt{A^2+B^2}}.$
THE QUESTION: What trigonometric identity explains the last set of equations?
| Rewrite that expression in the following way:
$$A\sin(\omega t) + B\cos(\omega t) = \sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\sin(\omega t) + \frac{B}{\sqrt{A^2+B^2}}\cos(\omega t) \right)$$
Notice that the coefficients are not only numbers that have an absolute value less than $1$, they are also conveniently already ratios of sides of a right triangle. So define
$$\sin\beta = \frac{B}{\sqrt{A^2+B^2}} \hspace{20 pt} \cos\beta \frac{A}{\sqrt{A^2+B^2}}$$
Then we have
$$=\sqrt{A^2+B^2}\left(\sin(\omega t)\cos\beta + \cos(\omega t)\sin\beta\right) = \sqrt{A^2+B^2}\sin\left(\omega t + \beta\right)$$
by the angle addition identites. Notice that $\beta$ could have also been defined as
$$\tan\beta = \frac{A}{B}$$
so one heuristic way of interpreting this formula is that we've turned a sum of sine and cosine into their "polar" form and we found the "magnitude" and the "angle" relative to the $x$ axis assuming sine represented the $y$ component and cosine represented the $x$.
| {
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Prove inequalities $\frac {36}{25} \le A(a) < 2$ Given the expression,
$$ A(a) = \frac{\left( 1 +a +\frac 1a\right)^2 }{\left(\frac12+a+a^2\right)\left(\frac12+\frac 1a + \frac{1}{a^2}\right)} $$
with $a > 0$.
Prove the following inequalities:
$$\frac {36}{25} \le A(a) < 2$$
Note that the bounds are rather tight. I encountered this issue in determining a narrow range of an angle in a geometry problem. I was only able to examine certain limits and identified the correct answer. But, I did not manage to fully prove the above inequalities.
Since I am not all that versed in dealing with such type of problems and would appreciate if anyone could suggest solutions for the proof.
| In the form given, $A(a)$ is pretty hard to reason around. As a first step let's factor out a factor of $\frac{1}{a^2}$ and see if we can coerce the denominator to look more friendly.
$$
A(a) = \frac{\frac{1}{a^2} {(a^2 + a + 1)}^2}{(a^2 + a + 1 - \frac{1}{2})\frac{1}{a^2}(\frac{1}{2}a^2 + a + 1)} = \frac{{(a^2 + a + 1)}^2}{(a^2 + a + 1)(a^2 + a + 1 - \frac{1}{2}a^2) - \frac{1}{2}(\frac{1}{2}a^2 + a + 1)} \\
A(a) = \frac{{(a^2 + a + 1)}^2}{{(a^2 + a + 1)}^2 -\frac{1}{2}\left[a^2(a^2 + a + 1) + (\frac{1}{2}a^2 + a + 1)\right]} = \frac{{(a^2 + a + 1)}^2}{{(a^2 + a + 1)}^2 -\frac{1}{2}(a^4 + a^3 + \frac{3}{2}a^2 + a + 1)}
$$
Now to conclude anything from this let us examine ${(a^2 + a + 1)}^2$, expanded we get $a^4 + 2a^3 + 3a^2 + 2a + 1$, which compared to the negative term in the denominator is almost exactly twice as big. The $-\frac{1}{2}$ enables us to resolutely conclude that our denominator is greater than half the numerator for $a > 0$, so we have $A(a) < 2$.
Unfortunately for our other bound short of calculating the fairly convoluted derivative of $A(a)$ to conclusively prove $a = 1$ is a critical point, and in fact a minimum, I am not sure of an easier way way. However, proving $a =1$ is a minimum (it is) and plugging it in will give us $A(1) = \frac{36}{25}$, so we have $\frac{36}{25} \leq A(a) < 2$ for $a > 0$.
| {
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Can anyone prove this function is concave? Can anyone help me prove the following function $$f(x) = \frac{ax}{1-x^a}-\frac{x}{1-x}$$ is concave for any parameter $a \geq 1$ when $0<x<1$ ?
I have tried to prove its second derivative is negative, but it becomes another difficult problem.
| $f(x) = \frac{ax}{1-x^a} - \frac{x}{1-x} $ , $a\geq 1 , 0<x<1$
$$f'(x) =\frac{a}{1-x^a} + \frac{ax(ax^{a-1})}{(1-x^a)^2} - \frac{1}{(1-x)^2} $$
$$f'(x) = \frac{a}{1-x^a} + \frac{a^2 x^a}{(1-x^a)^2} - \frac{1}{(1-x)^2}$$
$$f''(x) = \frac{a^2x^{a-1}}{(1-x^a)^2} + \frac{a^3x^{a-1}}{(1-x^a)^2} + \frac{2a^3x^{2a-1}}{(1-x^a)^3}- \frac{2}{(1-x)^3}$$
$$f''(x) = \frac{a^2x^{a-1}(1+a)}{(1-x^a)^2}+ \frac{2a^3x^{2a-1}}{(1-x^a)^3}- \frac{2}{(1-x)^3}$$
$$f''(x) = \frac{a^2x^{a-1}(1+a)-a^2x^{2a-1}(1+a)+2a^3x^{2a-1}}{(1-x^a)^3} - \frac{2}{(1-x)^3}$$
$$f''(x) = \frac{a^2x^{a-1}(1+a)-a^2x^{2a-1}+a^3x^{2a-1}}{(1-x^a)^3} - \frac{2}{(1-x)^3}$$
$$f''(x) = \frac{a^2x^{a-1}(1+a-x^a+ax^a)}{(1-x^a)^3} -\frac{2}{(1-x)^3}$$
$$f''(x) = \frac{a^2x^{a-1}(1+a+x^a(a-1))}{(1-x^a)^3} -\frac{2}{(1-x)^3}$$
we need to show it is negative for the given conditions. For a fixed $x$ we can see that the first fraction goes to zero as $a$ increases because $0<x<1$. Can you continue?
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Maximum value of $a+b+c$ in an inequality Given that $a$, $b$ and $c$ are real positive numbers, find the maximum possible value of $a+b+c$, if
$$a^2+b^2+c^2+ab+ac+bc\le1.$$
From the AM-GM theorem, I have
$$a^2+b^2+c^2+ab+ac+bc\geq 6\sqrt[6]{a^4b^4c^4} = 6\sqrt[3]{a^2b^2c^2} \\
6\sqrt[3]{a^2b^2c^2} \le1 \\
a^2b^2c^2 \le \frac{1}{216} \\
abc \le \frac{\sqrt{6}}{36}$$
However, I don't know where to go from here.
| Hint: the inequality is equivalent to
$$
(a+b)^{2} + (b+c)^{2} + (c+a)^{2} \leq 2.
$$
Now put $x = b+c, y = c+a, z = a +b$ and then we need to find maximum of $a + b + c = \frac{1}{2}(x+y+z)$ under the condition $x^{2} + y^{2} + z^{2} \leq 2$ and you may use Cauchy-Schwartz inequality. Note that you need to check $x + y >z, y +z>x, z+x>y$ for $(x, y, z)$ that gives maximum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\angle ABC <$ $ 84^\circ$. Angles A, B and C meet the circle at P, Q and R respectively.
I made the picture below, but what would the solution be?
| $\begin{array}{} \text{Euler's relation} & IO=\sqrt{R(R-2r)} & IO=r=1 & R=1+\sqrt{2} \end{array}$
$\begin{array}{} sin(\frac{ π }{6})=\frac{1}{2} & AI=2 & AI^2=1+y_{A}^2 & A=(1,y_{A}=\sqrt{3}) \end{array}$
$\begin{array}{} {O}=λ ∩ ω & O=(h,k) \end{array}$
$\left\{ \begin{array}{} x^2+y^2=1 \\ (x-1)^2+(y-\sqrt{3})^2=(1+\sqrt{2})^2 \end{array} \right.$
$\begin{array}{} h=\frac{1-\sqrt{2}+\sqrt{3+6\sqrt{2}}}{4} & k=\frac{-3+3\sqrt{2}+\sqrt{3+\sqrt{2}}}{4\sqrt{3}} \end{array}$
$\begin{array}{} {C}=ψ∩t & \left\{ \begin{array}{} x=1 \\ (x-h)^2+(y-k)^2=(1+\sqrt{2})^2 \end{array} \right. \end{array}$
$\begin{array}{} y_{C}=k+\sqrt{2h-h^2+2\sqrt{2}+2} & y_{C}=\frac{3\sqrt{3}+3\sqrt{6}+\sqrt{9+18\sqrt{2}}}{6} \end{array}$
$\begin{array}{} AC=\sqrt{3}+y_{C} & AC=\frac{3\sqrt{3}+\sqrt{6}+\sqrt{1+2\sqrt{2}}}{2} \end{array}$
$\begin{array}{} ϕ= \angle(AOB)& θ=\frac{ϕ}{2} & AO=OC=1+\sqrt{2} \end{array}$
$\begin{array}{} AC^2=AO^2+OC^2-2·AO·OC·cos(ϕ) & cos(ϕ)=1-\left( \frac{AC}{OC} \right) ^2 \end{array}$
$cos(ϕ)=\frac{-7+4\sqrt{2}-5\sqrt{3+6\sqrt{2}}+3\sqrt{6+12\sqrt{2}}}{4}$
$\begin{array}{} θ=\angle(ABC) & θ=\frac{ϕ}{2} \end{array}$
$θ=\frac{1}{2}arccos\left( \frac{-7}{4}+\sqrt{2}-\frac{5}{4}\sqrt{3+6\sqrt{2}}+\frac{3}{4}\sqrt{6-12\sqrt{2}} \right)·\frac{180}{π}=83.905711$
$\begin{array}{} \text{checking} & θ<\frac{7π}{15} & ϕ=2θ<\frac{14π}{15} \end{array}$
$\begin{array}{} cos(ϕ)>cos(\frac{14π}{15}) & cos(ϕ)>cos(\frac{π}{15}) & cos(\frac{π}{15}>-cos(ϕ)) \end{array}$
$\frac{1}{8}\left( -1+\sqrt{5}+\sqrt{6(5+\sqrt{5}} \right) >\frac{-1}{4}\left( -7+4\sqrt{2}-5\sqrt{3+6\sqrt{2}}+3\sqrt{6-12\sqrt{2}} \right)$
how to prove it exactly?
$\begin{array}{} \text{True (numerical verification)} & \text{difference} & 6.896×10^{-4} \end{array}$
($λ$) Incircle of a $triangleABC$, (locus of point $O$) with center $I=(0,0)$ and radius $r=1$, $T=(1,0)$, ($t$) vertical line $x=1$ (locus of points $A$ and $C$), [$\frac{π}{3}$ angle construction] ($μ$) circle with center $I$ and radius=2, $A=(1,\sqrt{3})$ intersection $μ$ and $t$, $AI=2$, $D=(-2,0)$ intersection $μ$ and $x-axis$, $AD$, $\angle(IAT)=\frac{π}{6}$, $sin(\frac{π}{6})=\frac{1}{2}$, $\angle(DAI)=\frac{π}{6}$, ($ω$) circle with center $A$ and $R=1+\sqrt{2}$, (Euler's relation), $O=(h,k)$ intersection $λ$ and $ω$, ($ψ$) circle by $A$ with center in $O$ (locus points $A$, $B$ and $C$), $B$ intersection $AD$ and $ψ$, $C$ intersection $ψ$ and $t$, $BC$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Prove with the help of the $\varepsilon$-$n_0$ definition, that $\frac 1 4$ is the limit of the sequence $a_n=\frac{n^2-1}{4n^2+1}$
Prove with the help of the $\varepsilon$-$n_0$ definition, that $\frac 1 4$ is the limit of the number-sequence $a_n=\frac{n^2-1}{4n^2+1}$
Let $\varepsilon>0$. We need to show, that $\frac 1 4 - \varepsilon < \frac{n^2-1}{4n^2+1} < \frac 1 4 +\varepsilon$
"Upper bound": $\frac{n^2-1}{4n^2+1}\leq \frac{n^2+\frac 1 4}{4n^2+1}=\frac 1 4 < \frac 1 4 +\varepsilon$ if $\varepsilon > 0$.
"Lower bound": $\frac{n^2-1}{4n^2+1}\geq \frac{n^2-1}{4(n+1)^2}=\frac{(n-1)(n+1)}{4((n+1)(n+1))}=\frac{n-1}{4(n+1)}$ how to go on now? I can't seem to find a way to simplify this more. Can you help me a bit?
| We have $$\left|\frac{n^2-1}{4n^2+1}-\frac{1}{4}\right|=\left|\frac{4n^2-4-4n^2-1}{4(4n^2+1)}\right|=\frac{5}{4(4n^2+1)}=\frac{5}{4(4n^2+1)}<\frac{5}{16n^2}$$
I hope you can finish now!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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