Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Total Derivative at a point Let $f(x,y)=x^{3}+y^{3}$ .
How do I find the total derivative at (0,0) by using the definition?
I am confused of what to take as the Error function.
| $$ (x+h)^3 +(y+k)^3 -x^3 -y^3 -3x^2 h -3y^2k = 3xh^2 +3yk^2 +h^3 +k^3 =\sqrt{h^2 +k^2} \left[(x+y)\sqrt{h^2 +k^2} +\frac{h^3 +k^3}{\sqrt{h^2 +k^2 }}\right] $$
setting
$r(h,k) =(x+h)^3 +(y+k)^3 -x^3 -y^3 -3x^2 h -3y^2k$
$\phi (h,k )=(x+y)\sqrt{h^2 +k^2} +\frac{h^3 +k^3}{\sqrt{h^2 +k^2 }}$
we have $$\frac{|f((x,y) +(h,k)) -f(x,y) -[3x^2 ,3y^2 ]\circ [h,k] |}{||(h,k)||} =\frac{|r(h,k)|}{|| (h,k)||} =\phi (h,k) \to 0 $$ as $(h,k)\to (0,0).$
Thus $f'(x,y) =[3x^2 , 3y^2] .$
| {
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"url": "https://math.stackexchange.com/questions/1786086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the set of values of x for which $\frac{x+1}{2x-3}<\frac{1}{x-3}$ Here's what I've done:
$\frac{x+1}{2x-3}<\frac{1}{x-3}$
$x+1<\frac{2x-3}{x-3}$
$(x+1)(x-3)<2x-3$
$x^2-2x-3<2x-3$
$x^2-4x<0$
$x(x-4)<0$
$0<x<4$
However this clearly fails because when $x$ is $2$, for example, the inequality fails.
Where have I gone wrong?
| Taking in account the comment of @SteamyRoot and resolve it again, you will have the answer $x\in I=]0,\frac{3}{2}[~\cup~]3,4[$.
Clearly the example you have taken ($x=2$) doesn't belong to I.
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction Proof of Taylor Series Formula I'm attempting to prove a formula for the taylor series of function from a differential equation.
The equation is
$$f(0)=1$$ $$f'(x) = 2xf(x)$$
I have found empirically that
$$f(x) = \sum_{k=0}^{\infty}\frac{x^{2k}}{k!}$$
I need to prove that this general formula works via induction.
Here is my attempt!
$$\mathrm{Show} \qquad1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+...+\frac{x^{2k}}{k!}=\sum_{n=0}^k \frac{f^{(2n)}(0)x^{2n}}{(2n)!}$$
Prove true for $k=0$
$$1=\frac{f^{(0)}(0)x^{0}}{(0)!}$$
$$1 = 1 \ \checkmark$$
Assume true for $k=c$
$$1+x^2+\frac{x^4}{2}+...+\frac{x^{2c}}{c!}=\sum_{n=0}^c \frac{f^{(2n)}(0)x^{2n}}{(2n)!}$$
Prove true for $k=c+1$
$$\begin{align} 1+x^2+\frac{x^4}{2}+...+\frac{x^{2c}}{c!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1} \frac{f^{(2n)}(0)x^{2n}}{(2n)!} \\
\sum_{n=0}^c \frac{f^{(2n)}(0)x^{2n}}{(2n)!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1} \frac{f^{(2n)}(0)x^{2n}}{(2n)!} \\ \frac{x^{2c+2}}{(c+1)!}&=\frac{f^{(2c+2)}(0)x^{2c+2}}{(2c+2)!}
\end{align}$$
From there I don't know how to proceed. Maybe I shouldn't have broken apart the sum on the right? Any help would be appreciated.
Note: I know the differential is easily separable and solvable, but the project involves comparing the solutions
| Note: This proof relies on a formula for the $n^{th}$ derivative of $f(x)$ that can be found HERE
The proof is rather extensive, and would roughly quadruple the length of this answer.
Proof of The Taylor Expansion of $f(x)$
$$\mbox{Prove true that } \ 1 + x^2 + \frac{x^4}{4} + \frac{x^6}{6} + \cdots + \frac{x^{2k}}{k!}=\sum_{n=0}^{k}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}$$
By Induction:
$\circ \text{ Prove true for }k=0$
$$\frac{f^{(0)}(0)x^{0}}{0!} \\ 1=1 \ \checkmark$$
$\circ \text{ Assume true for }k=c$
$$ 1 + x^2 + \frac{x^4}{4} + \frac{x^6}{6} + \cdots + \frac{x^{2c}}{c!}=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}$$
$\circ \text{ Prove true for }k=c+1$
\begin{equation}
\begin{split}
1 + x^2 + \frac{x^4}{4} + \frac{x^6}{6} + \cdots + \frac{x^{2c}}{c!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!} \\ \sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!} \mbox{By I.H}
\\ =\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(c+1)!} \cdot \frac{(2c+2)!}{(2c+2)!}
\\=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(2c+2)!} \cdot \frac{(2c+2)!}{(c+1)!}
\\=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(2c+2)!} \cdot \frac{(2c+2)!}{(\frac{2c+2}{2})!}
\\=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(2c+2)!} \cdot f^{(2c+2)}(0) \mbox{ From linked proof}
\\=\sum_{n=0}^{c+1}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!} \checkmark
\end{split}
\end{equation}
$$f(x)=\sum_{k=0}^{\infty} \frac{x^{2k}}{k!} \mathrm{Q.E.D}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Edwards Differential Calculus for Beginners, 1896. Chapter 4, Question 82 This question asks for an evaluation of an infinite series assuming only knowledge of basic differential calculus. I couldn't figure it out, but user 'Dr. MV' gave me a hint which was sufficient for me to find the answer. I hope this makes the question seem more relevant, I'm new around these parts and have yet to learn what makes for adequate structure in regards to asking questions.
Here is the question.
Prove that if $x <1$, then
$$\sum_{k=0}^\infty\frac{2^kx^{2^k-1}-2^{k+1}x^{2^{k+1}-1}}{1-x^{2^k}+x^{2^{k+1}}}=\frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac{4x^3-8x^7}{1-x^4+x^8} + ... =\frac{1+2x}{1+x+x^2}$$
| HINT:
$$\frac{2^kx^{2^k-1}-2^{k+1}x^{2^{k+1}-1}}{1-x^{2^k}+x^{2{k+1}}} =-\frac{d}{dx}\log\left(1-x^{2^k}+x^{2^{k+1}}\right)\tag 1$$
Then, write the argument of the logarithm in $(1)$ as
$$1-x^{2^k}+x^{2^{k+1}}=\frac{1+x^{2^{k+1}}+x^{2^{k+2}}}{1+x^{2^k}+x^{2^{k+1}}}$$
Then, telescope and differentiate.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $ \int \frac{1+x\cos(x)}{x(1-x^2(e^{2\sin(x)}))}dx $ $$ \int \frac{1+x\cos x}{x(1-x^2e^{2\sin x})}dx $$
Attempt:
I substituted $(1-xe^{2\sin(x)})$ by $u$ and tried from there by differentiating it. But I get stuck midway.
| Substitute $1-x^2e^{2\sin x} = u$
This gives $$-[2xe^{2\sin x} + 2x^2 \cos x e^{2\sin x}] dx = du \\
-2xe^{2 \sin x} (1+x\cos x) dx = du$$
Putting the value of $dx$ in your integral, we get
$$\int \frac{du}{-2x^2e^{2\sin x}(u)}$$
We know that $$\begin{align} 1-x^2 e^{2\sin x} &= u \\
x^2 e^{2\sin x} &= 1-u \\
-2x^2 e^{2\sin x} &= 2(u-1)\end{align}$$
This turns the integral to
$$\int \frac{du}{2u(u+1)}$$
Use partial fractions to go further.
| {
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"timestamp": "2023-03-29T00:00:00",
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Dividing an integer into a fixed number of integers What is the formula for dividing an integer into a fixed number of integers where the greatest distance between consecutive integers is 1.
Dividing 10 into 4 integers we can get:
[3, 3, 2, 2]
Dividing 7 into 4 integers we can get:
[2, 2, 2, 1]
Dividing 10 into 8 integers we can get:
[2, 2, 1, 1, 1, 1, 1, 1]
| Allowing fractions, you would split exactly as
$$\left[\frac{10}4,\frac{10}4,\frac{10}4,\frac{10}4\right].$$
Adding those fractions from left to right, you get
$$\left[\frac{10}4,\frac{20}4,\frac{30}4,\frac{40}4\right].$$
Then taking the integer parts,
$$[2,5,7,10].$$
And finally the pairwise differences
$$[2,3,2,3].$$
Other examples,
$$\left[\frac74, \frac74,\frac74,\frac74\right]\to\left[\frac74, \frac{14}4,\frac{21}4,\frac{28}4\right]\\
\to[1,3,5,7]\to[1,2,2,2].$$
$$\left[\frac{10}8,\frac{10}8,\frac{10}8,\frac{10}8,\frac{10}8,\frac{10}8,\frac{10}8,\frac{10}8\right]\to\left[\frac{10}8,\frac{20}8,\frac{30}8,\frac{40}8,\frac{50}8,\frac{60}8,\frac{70}8,\frac{80}8\right]\\
\to[1,2,3,5,6,7,8,10]\to[1,1,1,2,1,1,1,2].$$
This doesn't just give you the numbers, but a permutation that minimizes the cumulated error.
Algebraically, the numbers are
$$\left\lfloor\frac{n(i+1)}m\right\rfloor-\left\lfloor\frac{ni}m\right\rfloor.$$
They are in close relation with the Bresenham's line drawing algorithm that draws on oblique line on a square grid.
Note that by the definition of integer division,
$$m=qn+r=q(n-r)+(q+1)r$$ where $q=\lfloor n/m\rfloor$ is the integer quotient and $r=m\bmod n$ the remainder, such that $0\le r<n$. So among the $n$ numbers, $n-r$ of them equal $q$ and $r$ equal $q+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sum\limits_{cyc}\left(\frac{a^4}{a^3+b^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c^{\frac34}}{2^{\frac34}}$ When $a,b,c > 0$, prove
$$\left(\frac{a^4}{a^3+b^3}\right)^{\frac34}+\left(\frac{b^4}{b^3+c^3}\right)^ {\frac34}+\left(\frac{c^4}{c^3+a^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c ^{\frac34}}{2^{\frac34}}$$
I tried the substitution $x=a^4,\ldots$ but I have no idea how to deal with the left- hand side. I tried some C-S but it goes nowhere. I think Bernoulli's inequality may be the only way to prove this inequality.
| Define:
$$
x = \frac{a^{3/4}}{2^{3/4}} = \left(\frac{a}{2}\right)^{3/4} \quad ; \quad
y = \frac{b^{3/4}}{2^{3/4}} = \left(\frac{b}{2}\right)^{3/4} \quad ; \quad
z = \frac{c^{3/4}}{2^{3/4}} = \left(\frac{c}{2}\right)^{3/4}
$$
Then:
$$
a = 2\,x^{4/3} \quad ; \quad b = 2\,y^{4/3} \quad ; \quad c = 2\,z^{4/3}
$$
And:
$$
\left(a^4\right)^{3/4} = a^3 = 2^3 x^4 \quad ; \quad
\left(b^4\right)^{3/4} = b^3 = 2^3 y^4 \quad ; \quad
\left(c^4\right)^{3/4} = c^3 = 2^3 z^4
$$
Finally, when $x,y,z > 0$, prove:
$$
f(x,y,z) =
\frac{2^3 x^4}{\left(2^3 x^4 + 2^3 y^4\right)^{3/4}} +
\frac{2^3 y^4}{\left(2^3 y^4 + 2^3 z^4\right)^{3/4}} +
\frac{2^3 z^4}{\left(2^3 z^4 + 2^3 x^4\right)^{3/4}} \ge 1
$$
Where it can be assumed without loss of generality that: $\,x+y+z = 1$ .
The problem is thus reduced to a familiar one, quite similar to:
$x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$
Olympiad Inequality $\sum_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
And can be treated accordingly:
The minimum of our function inside the abovementioned triangle must shown to be greater or equal to one.
Another proof without words is attempted
by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as:
nivo := min + g/grens*(max-min); { grens = 20 ; g = 1..grens }
The whiteness of the isolines is proportional to the (positive) function values; they are almost black near the minimum and almost white near the maximum values.
Maximum and minimum values of the function are observed to be:
1.00001285611974E+0000 < f < 1.68177794816992E+0000
The little $\color{blue}{\mbox{blue}}$ spot in the middle is where $\,\left| f(x,y,z) - 1 \right| < 0.001$ . Due to symmetry,
an absolute minimum of the function is expected indeed at $(x,y,z) = (1/3,1/3,1/3)$.
Note.
Conditions similar to $\;x+y+z=1\;$ often occur in these inequalities, whether that is explicitly or implicitly.
An explicit example has been provided with another
HN_NH question :
Prove $\frac{xy}{5y^3+4}+\frac{yz}{5z^3+4}+\frac{zx}{5x^3+4} \leqslant \frac13$
The current inequality is an example of implicit occurrence. Let the function $f(x,y,z)$
be defined as above, then we have the equivalent inequality $\;f(x,y,z) \geqslant x+y+z$ .
It is clear that $f$ has the following property for all real $\lambda > 0$ :
$\;f(\lambda x,\lambda y,\lambda z) = \lambda f(x,y,z) \geqslant \lambda (x+y+z)$ .
Therefore $\lambda$ has no influence whatsoever on the inequality being true or false; we can always divide $x,y,z$ by a factor $\lambda$ such that $x+y+z=1$ .
Thus enabling a triangle mapping method once again.
| {
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"timestamp": "2023-03-29T00:00:00",
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$12^7+8^8$ divided by $13$ I Need to find what the remainder is when $12^7+8^8$ is divided by $13$
I have a solution, but don't know if it is right.
$12=-1\mod13$
$12^7=-1\mod13$
$8=8\mod13$
$8^2=6\mod13$
$8^4=10\mod13$
$8^8=9\mod13$
Then I did $-1\mod13+9\mod13=8\mod13$, so the remainder is $8$.
If someone could tell me if this is correct, it would be appreciated. Thanks
| By Fermat's Little Theorem we have $2^{12}=1\bmod13$. So $8^4=1\bmod13$ and hence $8^8=1\bmod13$. As you said, $12^7=(-1)^7=-1\bmod13$, so $12^7+8^8=0\bmod13$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding non-negative integer solutions to $3^x + 5^y = a^2$ $3^x+5^y=a^2$ ($x, y, a$ are non-negative integers)
Find all pairs $(x, y)$ which satisfy the equation.
I have found the trivial solution $x=1, y=0$, and I have tried with congruences, but it didn't really get me anywhere.
| By observing modulo 4 one could easily prove that $x$ is an odd number. If $x=1$, by observing modulo 3 it is easy to see that $y$ is an even number. If $x>1$, then $x\ge 3$. So, $27 \mid a^2-5^y$. Looking at modulo 27, we have $y$ is an even number (check every possibility for $0<y<18$ odd and $0<a<14$). Let $y=2t$, $3^x=a^2-5^{2t}=(a+5^t)(a-5^t)$, clearly one of $a+5^t$ or $a-5^t$ is not divisible by 3, so one of them is 1. It follows that $a=5^t+1$. So $2\cdot 5^t+1=3^x$, $2.5^t=3^x-1$. Clearly $x=1$ generate a solution and $x=0,2,3,4$ does not. Clearly $2\mid 3^1-1$ and $5\mid 3^4-1$. For $x>4$, from Zsigmondy's theorem it follows that there are other prime beside $2$ and $5$ that divides $3^x-1$. Hence, the only non-negative integer solution is $x=1$, $y=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $x+y^2+z^3 \geqslant x^2y+y^2z+z^2x$ for $xy+yz+zx=1$
Let $x,y,z \geqslant 0$ and $xy+yz+zx=1$. Prove that
$$x+y^2+z^3 \geqslant x^2y+y^2z+z^2x.$$
What I try:
$$x+y^2+z^3 \geqslant x^2y+y^2z+z^2x$$
$$\Leftrightarrow x(xy+yz+zx)+y^2+z^3- x^2y-y^2z-z^2x \geqslant 0$$
$$\Leftrightarrow \left(x^2z+\frac{z^3}{4}-z^2x \right)+ \left(y^2+\frac{3z^3}{4}+xyz-y^2z\right) \geqslant 0$$
I strongly believe that $\left(y^2+\frac{3z^3}{4}+xyz-y^2z\right) \geqslant 0$, but I have no proof.
| As you wrote, from the identity:
$$x(xy+yz+zx)+y^2+z^3-x^2y-y^2z-z^2x=z(\frac{z}{2}-x)^2+(y^2+\frac{3}{4}z^3-y^2z)+xyz$$
It suffices to check $y^2+\frac{3}{4}z^3\ge y^2z$ given $yz\le 1$. We take $3$ cases:
If $z\le 1$, then $y^2\ge y^2z$ already.
If $z\ge 1$ and $y\le \frac{1}{4}$, then $\frac{3}{4}z^3\ge \frac{3}{4}z\ge \frac{1}{16}z\ge y^2z$.
If $z\ge 1$ (so $y\le 1$) and $y\ge\frac{1}{4}$ (so $z\le 4$) then $y^2\ge \frac{1}{4}y^2z$ and $\frac{3}{4}z^3\ge \frac{3}{4}z\ge \frac{3}{4}y^2z$. Adding these gives the required.
This covers all cases, so the inequality is proved. "Equality" holds when $x=\frac{1}{\epsilon}, y=\epsilon, z=0$ and $\epsilon\to 0$.
| {
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Using L'hopital's rule to prove differentiability
Question: Define
$$
f(x) = \left\{\begin{aligned}
& \frac{2x\cos(x)}{x+\sin(x)} &&: x \ne 0 \\
&1 &&: x = 0
\end{aligned}
\right.$$
Show $f'(0)$ exist and find the value of it.
My Attempt:
After applying L'Hôpital's rule once I get $$\lim \limits_{h \to 0} \frac{2\cos(x)-2x\sin(x)}{2x+\sin(x)+x\cos(x)}$$
I cannot continue with L'hopital's rule as it is no longer in the $\frac{0}{0}$ form.
Where am I going wrong? Thank you.
| You can simply do it as follows
\begin{align}
f'(0) &= \lim_{x \to 0}\frac{f(x) - f(0)}{x}\notag\\
&= \lim_{x \to 0}\frac{2x\cos x - x - \sin x}{x^{2} + x\sin x}\notag\\
&= \lim_{x \to 0}\dfrac{2x\cos x - x - \sin x}{x^{2}\left(1 + \dfrac{\sin x}{x}\right)}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\dfrac{2x\cos x - 2x + x - \sin x}{x^{2}}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{x - \sin x}{x^{2}} - 2x\cdot\frac{1 - \cos x}{x^{2}}\notag\\
&= \frac{1}{2}\left(0 - 2\cdot 0\cdot\frac{1}{2}\right)\notag\\
&= 0\notag
\end{align}
The limit $$\lim_{x \to 0}\frac{x - \sin x}{x^{2}} = 0$$ can be evaluated easily by Taylor series, L'Hospital, or squeeze theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the following solvable equation? The following equation is solvable by radicals. How to get all the roots in terms of radicals?
$ x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1 = 0$
The transcendental solution is given by
$ \cos(k\pi/13)+i\sin(k\pi/13)$ for $ k = 1, 2,\dots.,12$. Only a solution by radicals is acepted as an answer.
| Let $\zeta = e^{2\pi i / 13}$.
$(\mathbb{Z}/13\mathbb{Z})^{\times}$ is cyclic of order $12$. You get something cubic over $\mathbb{Q}$ by adding $\zeta^k$ where $k$ runs through the subgroup of index $3$ or its cosets:
$$A := \zeta^1 + \zeta^5 + \zeta^8 + \zeta^{12}, \; B := \zeta^2 + \zeta^3 + \zeta^{10} + \zeta^{11}, \; C := \zeta^4 + \zeta^6 + \zeta^7 + \zeta^9$$ all have minimal polynomial $X^3 + X^2 - 4X + 1$, so they can be expressed by radicals by the cubic formula.
The next step is to consider sums of $\zeta^k$ where $k$ runs through the subgroup of index $6$ and cosets:
$$a = \zeta^1 + \zeta^{12}, \; b = \zeta^2 + \zeta^{11}, \; c = \zeta^3 + \zeta^{10},$$ $$d= \zeta^4 + \zeta^9, \; e = \zeta^5 + \zeta^8, \; f = \zeta^6 + \zeta^7.$$
These are determined by the equations $$a + e = A, \; ae = C, \; b+c = B, \; bc = A, \; d+f = C, \; df = B$$ which are quadratic equations over $\mathbb{Q}(A,B,C)$. For example, it follows that $$a = \frac{A \pm \sqrt{A^2 - 4C}}{2}.$$
Finally, once you know that $\zeta + \zeta^{12} = a$ and $\zeta \cdot \zeta^{12} = 1,$ you can solve $$\zeta = \frac{a \pm \sqrt{a^2 - 4}}{2} = \frac{a}{2} + i \sqrt{1 - a^2/4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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build absolute value equations know solution We have absolute value equations with unknown coefficients:
$$|x + a| = b$$
and we know the solutions:
$$x = 11 \text{ and } x = 5$$
We need to find $a$ and $b$. From
$$11 + a = b \\
5 + a = -b$$
we get $a = -8$ and $b = 3$.
But we can try another way:
$$11 + a = -b \\
5 + a = b$$
and get $a =8$ and $b = -3$, which is not correct, apparently we need take absolute value from $b$.
How can I formulate this rule and explain it to another person?
| Plug $x=11$ into the original equation and square both sides. Do the same with $x=5$. We obtain
$$(11+a)^2=b^2=(5+a)^2.\tag1$$
Expanding (1) and solving for $a$ we get
$$121+22a+a^2=25+10a+a^2\ \Longleftrightarrow\ a=-8$$
Plugging this value of $a$ back into the original equation, this gives
$$b=\vert 11+a\vert=\vert 5+a\vert=3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding a formula for a kth element in a sequence I've setup a recurrence relation as part of a numerical analysis problem, and found that $$x_{n+1} = \frac{x_n+1}{2}$$
The notes then say that for $x_0=0$, it is easy to show that $$x_k = 1 - 2^{-k}$$
Plugging in a few numbers shows this is clearly correct, and equating $x_{n+1}=x_n$ shows it converges to one - but how might one come up with $x_k$ formula in the first place (rather than just being able to verify it)?
| Here's one possible process for discovering the general formula: try applying the recurrence a few times:
$$
\begin{aligned}
x_1&=\frac{x_0+1}{2}=\frac{x_0}{2}+\frac{1}{2}\\
x_2&=\frac{x_1+1}{2}=\frac{x_1}{2}+\frac{1}{2}=\frac{x_0}{4}+\frac{1}{4}+\frac{1}{2}=\frac{x_0}{4}+\frac{3}{4}\\
x_3&=\frac{x_2+1}{2}=\frac{x_2}{2}+\frac{1}{2}=\frac{x_0}{8}+\frac{3}{8}+\frac{1}{2}=\frac{x_0}{8}+\frac{7}{8}\\
x_4&=\frac{x_3+1}{2}=\frac{x_3}{2}+\frac{1}{2}=\frac{x_0}{16}+\frac{7}{16}+\frac{1}{2}=\frac{x_0}{16}+\frac{15}{16}.
\end{aligned}
$$
The pattern is pretty clear—you're getting $x_n=\frac{x_0}{2^n}+\frac{2^n-1}{2^n}$—but you can also see why it's happening: every time you apply the recurrence, $x_0$ gets multiplied by $\frac{1}{2}$ one more time, while the constant becomes $\frac{1}{2}$ plus $\frac{1}{2}$ times the previous constant. So the constant in $x_2$ is $\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}$, the constant in $x_3$ is $\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}$, and so on. You may recognize the constant as a geometric series:
$$
\sum_{i=1}^n\left(\frac{1}{2}\right)^i=\frac{\frac{1}{2}-\left(\frac{1}{2}\right)^{n+1}}{1-\frac{1}{2}}=1-\left(\frac{1}{2}\right)^n,
$$
from which the formula we observed follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating the inverse trigonometric limit $\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}$
$$
\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}
$$
I was doing some questions on limits, I saw one in which there is $\arccos x$. I am stuck there, not able to proceed.
Can you give me some hint?
| That $\sqrt{1-x^2}$ shouts "Trigonometric functions!!!". So:
Put $x=\sin\theta$, which means that $\sqrt{1-x^2}=\cos\theta$. Then $$\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\arccos \left(2x\sqrt{1-x^2} \right)}{x-\frac{1}{\sqrt{2}}}\equiv \lim_{\theta \to \frac{\pi}{4}} \frac{\arccos \left(\sin 2\theta \right)}{\sin \theta-\frac{1}{\sqrt{2}}}\equiv \lim_{\theta \to \frac{\pi}{4}} \frac{\frac{\pi}{2}-2\theta}{\sin \theta-\frac{1}{\sqrt{2}}}$$
You can use l'Hôpital from here. Or...
Put $\phi=\theta-\frac{\pi}{4}$. The limit becomes $$\lim_{\phi \to 0} \frac{-2\phi}{\sin \phi \cos\frac{\pi}{4} + \cos \phi \sin\frac{\pi}{4}-\frac{1}{\sqrt{2}}} \equiv \lim_{\phi \to 0} \frac{-2\phi}{\frac{1}{\sqrt{2}}(\sin \phi + \cos \phi -1)}$$
…which you can l'Hôpital again, or note that for small $\phi$, $\sin\phi$ is practically $\phi$ and $\cos\phi$ is practically 1, so that the limit becomes $$\frac{-2}{\frac{1}{\sqrt{2}}}$$
…which I leave to you to work out. This is essentially reproducing what l'Hôpital does, but by doing it yourself you can actually see what you are doing.
| {
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A question about $ax = b$ I am studying inequality and come across the following statement. I don't understand it and want to believe the book must have made mistakes. I am going to copy what the book says here exactly.
A linear inequality with one variable is in the form: $ax>b$. When $a>0$, the solution is $x>b/a$, When $a<0$, the solution is $x<b/a$. When $a=0$, if $b<0$, there are infinite number of solutions; if $b>0$, there are no solutions.
| Your book is correct.
When $a>0$, multiply both sides of the inequality by $\frac{1}{a}$. Since this is a positive number, the direction of the inequality doesn't change:
\begin{align*}
ax&>b
\\\frac{1}{a}\cdot ax &> \frac{1}{a} \cdot b\\
x&>\frac{b}{a}
\end{align*}
When $a<0$, again multiply both sides by $\frac{1}{a}$. Now the direction of the inequality reverses since $\frac{1}{a}$ is negative.
\begin{align*}
ax&>b
\\\frac{1}{a}\cdot ax &< \frac{1}{a} \cdot b\\
x&<\frac{b}{a}
\end{align*}
When $a=0$, we cannot multiply by $\frac{1}{a}$. The inequality becomes $0\cdot x>b$, i.e., $0>b$. If $b>0$, there is a contradiction - $b$ cannot be both positive and negative. However, when $b<0$, every value of $x$ satisfies the equation $0 \cdot x >b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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proof that $\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$ diverges, by comparsion I need to prove that
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$$
diverges by comparsion.
The way I did was to use
$$\frac{n}{n^2+2n+1}>\frac{n}{n^2+2n^2+n^2} = \frac{n}{4n^2} = \frac{1}{4n}$$
which diverges. Can I do that? Because $$2n+1<2n^2+n^2\implies n^2+2n+1<n^2+2n^2+n^2\implies $$
$$\frac{1}{n^2+2n+1}>\frac{1}{n^2+2n^2+n^2}$$
for large $n$.
I also tried seeing
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1} = \sum_{n=1}^{\infty}\frac{n}{(n+1)^2}$$
but I couldn't find any comparsion for that. Am I right in the first demonstration? Can you find a comparsion for the second attempt I did?
| $$\sum_{n = 1}^{\infty} \frac{n}{(n+1)^2} \geq \sum_{n = 1}^{\infty} \frac{n}{(n+n)^2} = \sum_{n = 1}^{\infty} \frac{n}{4n^2} = \frac{1}{4}\sum_{n = 1}^{\infty} \frac{1}{n} = \infty. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How Do The Integration of The Given Expression Is $27\pi$ Basically, this is a double integration question from the book Calculus 2 by Howard Anton. Half of the question is resolved and now I'm at this stage where I can't prove the left hand side to the right hand side.
$$
\int_{-3}^3 (6\sqrt {9-x^2} - 2x \sqrt{9-x^2}) dx = 27\pi
$$
| Observe that $\sqrt{9 - x^2}$ is an even function and $x \sqrt{9 - x^2}$ is an odd function. Therefore, the given integral is equal to
\begin{equation*}
2 \times \int_0^3 6 \sqrt{9 - x^2} \,\mathrm dx = 12 \int_0^3 \sqrt{9 - x^2} \,\mathrm dx.
\end{equation*}
Now, substituting $x = 3 \sin t$, the integral becomes
\begin{align*}
12 \int_0^{\pi/2} (3 \cos t) (3 \cos t)\,\mathrm dt & = 12 \times 9 \int_0^{\pi/2} \cos^2 t\,\mathrm dt\\
& = 12 \times 9 \times \dfrac 1 2 \times \dfrac {\pi} 2\\
& = 27\pi.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$
$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove
$$\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$$
A natural though is that from the condition $x^2+y^2+z^2+xyz=4$, I tried a trig substitutions $x=2\cdot \cos A$, $y=2\cdot \cos B$ and $z=2\cdot \cos C$, where $A,B,C$ are angles of an acute triangle.
Our inequality becomes
$$\sqrt[2]{2(\cos A+ \cos B)}+\sqrt[3]{2(\cos B+\cos C)}+\sqrt[4]{2(\cos C+\cos A)} <4$$
I used formula $\cos(A)+\cos(B) < 2$ and estimate that $LHS <5$ but not $4$.
I do not know how to proceed.
| The condition $\;x^2+y^2+z^2+xyz=4\;$ can be rewritten as:
$$
z^2+xy\cdot z + (x^2+y^2-4) = 0 \quad \Longrightarrow \quad
z = - \frac{xy}{2} \pm \sqrt{\left(\frac{xy}{2}\right)^2-(x^2+y^2-4)}
$$
From $z \ge 0$ it follows that:
$$
z = - \frac{xy}{2} + \sqrt{\left(\frac{xy}{2}\right)^2-(x^2+y^2-4)} \quad \mbox{and} \quad x^2+y^2\le 4
$$
Therefore, after substitution of $\,z$ ,
we only have to investigate function behavior$f(x,y) = \sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} < 4$
in the area enclosed by x-axis, y-axis and a quarter of a circle, as depicted below.
Another proof without words is attempted
by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined
(in Delphi Pascal) as:
nivo := min + g/grens*(max-min); { grens = 40 ; g = 1..grens }
The blackness of the isolines is proportional to the (positive) function values; they are almost white
near the minimum and almost black near the maximum values.
Maximum and minimum values of the function are observed to be:
2.54387743763872E+0000 < f < 3.91477606446737E+0000
The $\color{blue}{\mbox{blue}}$ spot is where $\,\left| f(x,y) - 4 \right| < 0.9$ .
It is suggested by the rather large error $\,0.9\,$ that $\,4\,$ is not really the maximum.
Indeed, upon refinement of the grid we find for the maximum numerically (double precision)
a somewhat lower value:
3.91477205860402 < 4
| {
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"timestamp": "2023-03-29T00:00:00",
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Choosing Functions for the Squeeze Theorem
Evaluate $$\lim_{n\to \infty}\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$$
$$$$
I came across the the question on this site itself but had a few doubts on the given solution. As I do not yet have 50 reputation points, I cannot comment over there. Could somebody please help me?
$$$$
From what I understand of the Squeeze Theorem, the three functions are related as
$$g(x)\le f(x)\le h(x)$$
$$$$
Now in the selection of terms, the following inequality has been used:
$$\displaystyle \frac{1}{n+1} \leq \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{n} $$when $0 \leq k \leq 2n$
$$$$
This inequality lead to the one used as the three functions for the application of the Squeeze theorem:
$$\frac{2n+1}{n+1} \leq S(n) \leq \frac{2n+1}{n}$$
where $S(n)=\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$
$$$$
I don't understand how $$\displaystyle \frac{1}{n+1} \leq \frac{1}{\sqrt{n^2+k}} $$
$$$$
Isn't $$n^2+k\le n^2+2n<(n+1)^2 \Rightarrow n^2+k<(n+1)^2$$
$$\Rightarrow \sqrt{n^2+k}< (n+1)$$$$\Rightarrow \displaystyle \frac{1}{n+1} < \frac{1}{\sqrt{n^2+k}}$$
$$$$
Thus shouldn't the resulting set of inequalities be $$\displaystyle \frac{1}{n+1} < \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{n} $$
$$\Longrightarrow \frac{2n+1}{n+1} < S(n) \leq \frac{2n+1}{n}$$
$$$$
In this case there is a $<$ sign instead of the $\le$ sign. How then can the Squeeze Theorem be applied?
Many thanks in advance.
$$$$
EDIT: Also, since Limits preserve Inequalities, how can $$\lim_{n\to \infty} \frac{2n+1}{n+1} = \lim_{n\to \infty}\frac{2n+1}{n}$$ when $$\frac{2n+1}{n+1} < \frac{2n+1}{n}$$
| Limits don't necessarily preserve strict inequalities. For example, $1-\frac{1}{n}<1+\frac{1}{n}$, yet they have the same limit as $n$ goes to $\infty$.
| {
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If $a,b>0$ and $a+b=1\;,$ Then minumum value of $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ is If $a,b>0$ and $a+b=1\;,$ Then minumum value of $\displaystyle \left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2$ is
$\bf{My\; Try::}$ Let $a=\sin^2 \theta$ and $b=\cos^2 \theta\;,$ Then We have to minimize
$$\displaystyle f(\theta) = (\sin^2 \theta+\csc^2 \theta)^2+(\cos^2 \theta+\sec ^2 \theta) = \sin^4 \theta+\csc^4 \theta+\cos^4 \theta+\sec^4 \theta+4$$
So $$f(\theta) = 1-2\sin^2 \theta\cos^2 \theta+\frac{1}{\sin^2 \theta \cos^2 \theta}+4$$
So $$f(\theta) = 1-\frac{\sin^2 2\theta}{2}+\frac{4}{\sin^2 \theta}+4$$
Now Put $\sin^2 2\theta = t\;, t\in \left(0,1\right]$
So we get $$f(t)=5-\frac{t}{2}+\frac{4}{t}\;,$$ So we get $\displaystyle f'(t) = -\frac{1}{2}-\frac{4}{t^2}<0\;\forall t \in (0,1]$
So $$f(1)_{\min} = 5-\frac{1}{2}+4=9-\frac{1}{2}=\frac{17}{2}$$
But answer given as $$\frac{25}{2}$$
plz help me, Where I am wrong, Thanks
| Here is an alternative solution: take $f(x) = \left(x + \frac{1}{x}\right)^2$ then $f''(x) = 2 + \frac{6}{x^4} > 0$ so $f$ is a convex function and therefore satisfy
$$f(a) + f(b) \geq 2f\left(\frac{a+b}{2}\right) = 2f\left(\frac{1}{2}\right) = \frac{25}{2}$$
This argument can also be used to solve other similar types of problems. For example we have the following generalization: if $f:[0,1]\to\mathbb{R}$ is a convex function and $x_i\in[0,1]$ with $x_1+\ldots+x_n = 1$ then $\min\sum_{i=1}^nf(x_i) = nf\left(\frac{1}{n}\right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $\tan 3x=4$, find the value of $\tan^2 x+\tan ^2(120+x)+\tan^2 (60+x)$ Given $\tan 3x=4$, find the value of $S=\tan^2 x+\tan ^2(120+x)+\tan^2 (60+x)$
I expanded each of $\tan (120+x)$ and $\tan (60+x)$ getting as
$$S=\tan^2 x+\left(\frac{\tan 120+\tan x }{1-\tan 120 \tan x}\right)^2+\left(\frac{\tan 60+\tan x }{1-\tan 60 \tan x}\right)^2$$ so
$$S=\tan^2 x+\left(\frac{-\sqrt{3}+\tan x }{1+\sqrt{3} \tan x}\right)^2+\left(\frac{\sqrt{3}+\tan x }{1-\sqrt{3} \tan x}\right)^2$$
so
$$S=\tan^2 x+\frac{\left(4\tan x-\sqrt{3}(1+\tan^2x)\right)^2+\left(4\tan x+\sqrt{3}(1+\tan^2x)\right)^2}{(1-3\tan^2 x)^2}$$ so
$$S=\tan^2 x+\frac{\left(32 \tan^2 x+6(1+\tan^2 x)^2\right)}{(1-3\tan^2 x)^2}$$
but if i proceed further i dot think i will get an expression in terms of $\tan 3x$.
| Hint:
Observe the following :
$1)$ $3\tan(3x)=\tan(x)+\tan({x+60^{\circ}})+\tan({x+120^{\circ}})$
$2)$ $\tan{60^{\circ}=\tan[(x+60^{\circ})-60^{\circ}}]$ and similar results
$3)$ $\tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$
$4)$ $\tan(60^{\circ})=\sqrt{3}$ and $\tan(120^{\circ})=-\sqrt{3}$
From $(1)$, we get
$$\tan^2(x)+\tan^2({x+60^{\circ}})+\tan^2({x+120^{\circ}})=[\tan^2(3x)-2[\tan(x)\tan({x+60^{\circ}})+\tan({x+120^{\circ}})\tan({x+60^{\circ}})+\tan(x)\tan({x+120^{\circ}})]$$
From $(2),(3)$ and $(4)$ deduce that
$$\tan(x)\tan(x+60^{\circ})=\dfrac{\tan(x+60^{\circ})-\tan(x)-\sqrt{3}}{\sqrt{3}}$$ and similar results
You can surely take it from here.
| {
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Given $a_1,a_{100}, a_i=a_{i-1}a_{i+1}$, what's $a_1+a_2$? I've been given the following puzzle
Let $a_1, a_{100}$ be given real numbers. Let $a_i=a_{i-1}a_{i+1}$ for $2\leq i \leq 99$. Further suppose that the product of the first $50$ is $27$, and the product of all the $100$ numbers is also $27$.
Find $a_1+a_2$.
I tried the following, looking at the sequence for a moment we see:
$$
a_2=a_1\, a_3\\
a_3=a_2\,a_4\\
\vdots\\
a_{99}=a_{98}\,a_{100}
$$
So $$a_2=\frac 1 {a_{99}} \prod_i a_i =\frac {27}{a_{99}}$$
Looking at the other elements, we find that $$a_3=\frac {a_2} {a_1}, a_4=\frac {a_2}{a_1 a_2}, a_5=\frac {a_2}{a_1a_2a_3},\dots , a_n=\frac {a_2}{\prod_{i=1}^{n-2} a_i}$$
Then, $$27=\prod_i a_i=\prod_{1\leq i\leq 100} \frac {a_2}{\prod_{k=1}^{i-1}a_i}$$
The last product I believe is much more complicated than what I should have gotten...
Has anyone ran into this puzzle before?
| Such a sequence is periodic with period length 6, one cycle
$$ A, B, \frac{B}{A}, \frac{1}{A}, \frac{1}{B}, \frac{A}{B} $$
Note that the product of all six is $1.$ Therefore the product of the first 96 is one, and 27 is $\frac{B^2}{A}.$ The product of the first 48 is one, and 27 is also $AB.$ That is
$$ 27 A = B^2, $$
$$ 27 = AB.$$
From $$ 729 = 27AB = (27A)B = B^3$$ I get $729 = B^3 $ and
$$ B = 9. $$ Then
$$ A = 3. $$
The sum is $$ A + B =12. $$
The cycle of length six becomes
$$ 3, \; 9, \; 3, \; \frac{1}{3}, \; \frac{1}{9}, \; \frac{1}{3}. $$
Putting two together shows
$$ 3, \; 9, \; 3, \; \frac{1}{3}, \; \frac{1}{9}, \; \frac{1}{3}, \; 3, \; 9, \; 3, \; \frac{1}{3}, \; \frac{1}{9}, \; \frac{1}{3} $$
In the original notation, with cycles of six multiplying to one, we have
$$ a_{49} a_{50} = a_1 a_2 = 3 \cdot 9 = 27. $$
Also
$$ a_{97} a_{98} a_{99} a_{100} = a_1 a_2 a_3 a_4 = 3 \cdot 9 \cdot 3 \cdot \frac{1}{3} = 27. $$
| {
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Prove $12\int_{0}^{\infty}{x\over (e^{x^2}+1)(e^{x^2}+3)}dx=\ln(2)$ $$I=12\int_{0}^{\infty}{x\over (e^{x^2}+1)(e^{x^2}+3)}dx=\ln(2)$$
$u=e^{x^2}\rightarrow du=2xe^{x^2}dx$
$x\rightarrow \infty,\, u=\infty$
$x\rightarrow 0,\, u=1$
$$6\int_{1}^{\infty}{2x\over (u+1)(u+3)}\cdot{du\over 2xe^{x^2}}=6\int_{1}^{\infty}{1\over u(u+1)(u+3)}du$$
$6=A(u+1)(u+3)+Bu(u+3)+Cu(u+1)$
$u=0\rightarrow A=2$
$u=-1\rightarrow B=-3$
$u=-3\rightarrow C=1$
$$I=\int_{1}^{\infty}{2\over u}-{3\over u+1}+{1\over u+3}dx$$
$$I=\left.\ln{u^2(u+3)\over (u+1)^3}\right|_{1}^{\infty}$$
$$I=\lim_{u\to \infty}\ln{u^2(u+3)\over (u+1)^3}-\ln{4\over 8}$$
$$I=\ln(2)$$
Another quick method of approach at tackling I?
| Following Claude Leibovici's comment, assuming $\text{Re}(a),\text{Re}(b)>-1$,
$$\begin{eqnarray*}J(a,b)=\int_{0}^{+\infty}\frac{2x\,dx}{(e^{x^2}+a)(e^{x^2}+b)}&=&\int_{0}^{+\infty}\frac{du}{(e^{u}+a)(e^{u}+b)}\\&=&\frac{1}{b-a}\int_{0}^{+\infty}\left(\frac{1}{e^u+a}-\frac{1}{e^u+b}\right)\,du\\&=&\frac{1}{b-a}\int_{1}^{+\infty}\frac{dx}{x}\left(\frac{1}{x+a}-\frac{1}{x+b}\right)\\&=&\frac{1}{b-a}\int_{0}^{1}\left(\frac{1}{1+ax}-\frac{1}{1+bx}\right)\,dx\\&=&\color{red}{\frac{b\log(a+1)-a\log(b+1)}{ab(b-a)}}\end{eqnarray*}$$
and by taking the limit as $b\to a$,
$$ J(a,a) = \color{red}{\frac{(a+1)\log(a+1)-a}{a^2(a+1)}}.$$
| {
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How to integrate $\int \frac{dx}{(1+x^2)^2}$? I need to integrate this to finish an old STEP problem I'm doing, but I'm stuck here, at the very end:
$$\int_0^\infty \frac{dx}{(1+x^2)^2}$$
The result should be $\pi\over 4$ . I don't know how to approach this. *Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).
Also, Wolfram tells me:
$$\int \frac{dx}{(1+x^2)^2} = \frac{1}{2}\left(\frac{x}{x^2+1}+\tan^{-1}x\right)+c$$
but I don't see how one can derive this without knowing the result beforehand.
Please, help me!
Somehow, this question doesn't seem to've been posted here ever (at least I couldn't find it).
EDIT: If you're interested, the problem in question is: STEP II - problem 4 (year 2014).
| $$
\begin{aligned}
\int \frac{d x}{\left(1+x^2\right)^2} &=-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{1+x^2}\right) \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \int\left(-\frac{1}{x^2}\right) \frac{1}{1+x^2} d x \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \int\left(\frac{1}{1+x^2}-\frac{1}{x^2}\right) d x \\
&=-\frac{1}{2 x\left(1+x^2\right)}+\frac{1}{2} \tan ^{-1} x+\frac{1}{2 x}+C \\
&=\frac{1}{2}\left(\frac{x}{1+x^2}+\tan ^{-1} x\right)+C
\end{aligned}
$$
For definite integral, we have $$
\begin{aligned}
\int_0^{\infty} \frac{d x}{\left(1+x^2\right)^2} &=\frac{1}{2}\left[\frac{x}{1+x^2}+\tan ^{-1} x\right]_0^{\infty} =\frac{\pi}{4}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 7
} |
Find the element in $\mathbb{Z}/143\mathbb{Z}$ whose image is $(\overline{10},\overline{11})$ under the Chinese remainder theorem Find the element in $\mathbb{Z}/143\mathbb{Z}$ whose image is $(\overline{10},\overline{11})$ in $\mathbb{Z}/11\mathbb{Z} \times \mathbb{Z}/13\mathbb{Z}$ under the Chinese remainder theorem
So I figured we must find an element $\overline{x}$ in $\mathbb{Z}/143\mathbb{Z}$ that satisfies:
$x \equiv 10 \pmod {11} $
$x \equiv 11 \pmod {13} $
Applying the Chinese remainder theorem gives us $x=76$.
Can I then simply conclude that the element we're looking for in $\mathbb{Z}/143\mathbb{Z}$ is $\overline{76}$?
| By Bezout's identity we know that we can find $ x, y$ such that $ 11x + 13y = 1 $. Using the Euclidean algorithm, we have:
$$ 13 = 11\cdot 1 + 2 $$
$$ 11 = 2 \cdot 5 + 1 $$
$$ 2 = 2 \cdot 1 + 0 $$
and now we substitute back:
$$ 1 = 11 - 2 \cdot 5 = 11 - (13 - 11) \cdot 5 = 11 \cdot 6 - 13 \cdot 5 $$
This helps us because now we can explicitly produce the solution. Indeed, I claim that the solution is $ 11 \cdot 6 \cdot 11 - 13 \cdot 5 \cdot 10 = 76 $. We have that
$$ 11 \cdot 6 \cdot 11 - 13 \cdot 5 \cdot 10 \equiv 11 \cdot 6 \cdot 10 - 13 \cdot 5 \cdot 10 \equiv 10 \pmod{11} $$
and
$$ 11 \cdot 6 \cdot 11 - 13 \cdot 5 \cdot 10 \equiv 11 \cdot 6 \cdot 10 - 13 \cdot 5 \cdot 11 \equiv 11 \pmod{13} $$
Generally, if $ I $ and $ J $ are comaximal ideals of a ring $ R $, we can find $ i \in I $ and $ j \in J $ such that $ i + j = 1_R $. Then, the expression $ bi + aj $ lies in the cosets $ a + I $ and $ b + J $. In the case that $ R $ is an Euclidean domain, the Euclidean algorithm gives an easy way of computing $ i $ and $ j $ explicitly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$ Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$
According to Fermat's theorem:
$$1^7+7^7+13^7+19^7+23^7\equiv{1+7+13+19+23}\pmod{7}\equiv{63}\pmod{7}\equiv{0}\pmod{7}$$
Now we need to show: $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{9}$ , but how??
| Just as Fermat's theorem tells you that $x^6\equiv1\pmod{7}$, Euler's theorem (which generalizes Fermat's) tells you that $x^6\equiv1\pmod{9}$ whenever $\gcd(x,9)=1$. It follows that
$$1^7+7^7+13^7+19^7+23^7\equiv1+7+13+19+23\equiv0\pmod{9}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Finding the second derivative of$f(x)=x^2\sqrt{4-x}$
Find the second derivative of the function following:
$$f(x)=x^2\sqrt{4-x}$$
Here I go...
$$f(x)= x^2(4-x)^{1\over 2}$$
\begin{align*}
f'(x) &= 2x(4-x)^{1\over 2}+{1\over 2}x^2(4-x)^{-{1\over 2}}(-1)\\
&= 2x(4-x)^{1\over 2} - {1 \over 2}x^2(4-x)^{-{1\over 2}}\\
&={1\over 2}x(4-x)^{-{1\over 2}}[4(4-x)-x]\\
&= {1\over 2}x(4-x)^{-{1\over 2}}(16-5x)\\
f''(x)&= {-{1\over 4}}x(4-x)^{-{3\over 2}}(-1)({1\over 2})(16-5x)+(-5)[{1\over 2}x(4-x)^{-{1\over 2}}]\\
&= {-{1\over 8}}x(4-x)^{-{3\over 2}}(16-5x)-{5\over 2}(4-x)^{-{1\over 2}}\\
&= {-{1\over 8}}x(4-x)^{-{3\over 2}}[(16-5x)-20(4-x)]\\
&={-{1\over 8}}x(4-x)^{-{3\over 2}}(-64+15x^2)
\end{align*}
I think I messed up on the second derivative. Could anyone show me the steps for the second derivative? Thanks!
| Your first derivative is 100% correct. The second derivative indeed has an error, as you must do the product rule within a product rule if you don't simplify your first derivative as in Ángel Mario Gallegos answer.
$$f''(x)=(\frac{1}{2}x(4-x)^{\frac{-1}{2}})(-5)+(16-5x)(\frac{1}{2}x[\frac{-1}{2}(4-x)^{\frac{-3}{2}}(-1)]+(4-x)^{\frac{-1}{2}}\frac{1}{2})$$
$$=\frac{-5}{2}x(4-x)^{\frac{-1}{2}}+(16-5x)(\frac{1}{4}x(4-x)^{\frac{-3}{2}}+\frac{1}{2}(4-x)^{\frac{-1}{2}})$$
$$=\frac{-5}{2}x(4-x)^{\frac{-1}{2}}+4x(4-x)^{\frac{-3}{2}}+8(4-x)^{\frac{-1}{2}}-\frac{5}{4}x^2(4-x)^{\frac{-3}{2}}-\frac{5}{2}x(4-x)^{\frac{-1}{2}}$$
$$=\frac{\frac{-5}{2}x(4-x)+4x+8(4-x)-\frac{5}{4}x^2-\frac{5}{2}x(4-x)}{(4-x)^{\frac{3}{2}}}$$
$$=\frac{-10x+\frac{5}{2}x^2+4x+32-8x-\frac{5}{4}x^2-10x+\frac{5}{2}x^2}{(4-x)^{\frac{3}{2}}}$$
$$=\frac{\frac{15}{4}x^2-24x+32}{(4-x)^{\frac{3}{2}}}$$
$$=\frac{15x^2-96x+128}{4(4-x)^{\frac{3}{2}}}$$
P.S. I included every simplification step in case you decided to go this route, that's why it is so long
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sin \theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$, then $\sin 2\theta$ is a root of $x^2 -44x +36=0$ My own bonafide attempt.
$$ 0<\theta<\pi/2$$
and $$\sin\theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$$
then show that $\sin 2\theta$ is a root of the equation $$x^2 -44x +36=0$$
I tried to use the above given equation of all the trigonometric ratios, though I ended up with an expression of $\sin\theta\cos\theta$.
But this too is in the form of $\sin\theta$ and $\cos\theta$ which was $\sin\theta\cos\theta=\frac{1}{6-\sin\theta-\cos\theta}$.
When I put that in the quadratic equation, it would again transform into the form of $\sin\theta$ and $\cos\theta$, and hence at last I couldn't prove the thing.
Even hints would work, as I would like to solve the question myself.
| With an obvious short notation and setting $x=2cs$, we reduce to the common denominator
$$s+c+\frac sc+\frac cs+\frac 1s+\frac 1c=\frac{sx+cx+2+2c+2s}{x}=\frac{(x+2)(c+s)+2}x=7.$$
As we know that $x\ne0$ and with $(c+s)^2=x+1$, we rewrite
$$(x+2)^2(x+1)=(7x-2)^2,$$
$$x^3-44x^2+36x=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Equation of a tangent on a circle given the gradient and equation of the circle My maths teacher told me this problem was impossible without knowledge of implicit differentiation: is she right?
You are given the equation of the circle $\left(x+2\right)^2+\left(y-2\right)^2=16$ , what are the equations of the lines with a gradient of 2 which are tangents of this circle.
Here is a visualisation of the problem
The red circle is the one I described, and the green and blue lines are the equations I'm looking for.
I think the solution will have something to do with plugging in the value of 2 in some sort of equation for the gradient i.e. $2=\frac{x-x_0}{y-y_0}$ , and then the discriminant the equation you would use to find where they intersect can be solved to find values of $x$ , but I'm not sure how to find these values without using any implicit differentiation -can it be done?
EDIT:
Ok so now I'll try and run this through step by step until i find another problem:
So we can say that there's a line (the diameter) of gradient -1/2 which passes through the centre (-2,2):
$y=mx+c$
$y=-\frac{1}{2}x+c$
$2=1+c$
$y=-\frac{1}{2}x+1$
We can also say that a solution lies on the point $(a,b)$ such that
$\left(a+2\right)^2+\left(b-2\right)^2=16$ and $b=-\frac{1}{2}a+1$
So we can substitute b in the equation of the circle to arrive at
$\left(a+2\right)^2+\left(-\frac{1}{2}a-1\right)^2=16$
which simplifies down to
$\frac{5}{4}a^2+5a-11=0$
and solves to give $\frac{2}{5}\left(-5-4\sqrt{5}\right)$ and $\frac{2}{5}\left(4\sqrt{5}-5\right)$
so these are the possible values for $x$
Now if i substitute 1 of these back into the equation, I'll get two values of $y$ for the one value of $x$ because it's quadratic, but there's only 1 intersection point at each value of $x$ , what do i do at this point?
EDIT 2:
It looks like both the intersection points of the line $x=\frac{2}{5}\left(4\sqrt{5}-5\right)$ correspond to the $y$ values of the intersection points of the tangents, does this mean i only need to plug one of these $x$ values back into the equation?
Visualisation to show you what i mean (the blue line appears to intersect the circle at both the appropriate $y$ values)
| $$ (x+2)^2 + (y-2)^2 = 16 $$
$$ \Rightarrow x^2 + y^2+4x-4y+4+4 = 16 $$
$$ \Rightarrow x^2 + y^2+4x-4y-8= 0 $$
$$ \text{centre} =(-2,2), \text{ radius}=4 $$
for tangent the line must satisfies the condition, $r=d$
we have line $y=2x+c$ here gradient$=2$ need to find $c=$?
$$ \pm4 =\frac {2(-2)-2+c}{2^2+1} $$
$$ \pm 20 =-6 +c$$
$$ c= 26,-14$$
now the tangents are
$$y=2x+26, y=2x-14$$ are required tangents
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}}$ for $x,y,z \geq 0$ On the left we have arithmetic mean of pairwise quadratic means, which obeys:
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{ \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} } \geq \frac{x+y+z}{3}$$
On the right is quadratic mean of pairwise arithmetic means:
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{\frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}} } \geq \frac{x+y+z}{3}$$
By Wolfram Alpha it appears that:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}}$$
How to prove it? I've started, but it's not working out so far.
First, we can transform the RHS into a more familiar form:
$$\frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}}=\sqrt{\frac{x^2+y^2+z^2+xy+yz+zx}{6}}$$
It makes sense that this expression is very close to arithmetic mean, because we have the following inequality:
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \frac{x+y+z}{3} \geq \sqrt{\frac{xy+yz+zx}{3}}$$
Let's try to prove the highlighted inequality. It is equivalent to:
$$(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2 \geq 3(x^2+y^2+z^2+xy+yz+zx)$$
Expanding the LHS we get:
$$2\sum_{cyc}\sqrt{x^2+y^2}\sqrt{y^2+z^2} \geq 3(xy+yz+zx)+x^2+y^2+z^2$$
To prove this inequality let's bound the sum on the LHS:
$$ \sum_{cyc}\sqrt{x^2y^2+y^2z^2+z^2x^2+z^4} \geq 3\sqrt{x^2y^2+y^2z^2+z^2x^2} \geq \sqrt{3} (xy+yz+zx)$$
But this doesn't seem to prove the original inequality. So I'm not sure how to proceed.
| You're almost done! $\sqrt{2(x^2+y^2)} \ge x+y$, and so $2\sqrt{x^2+y^2}\sqrt{x^2+z^2} \ge x^2+xy+xz+yz$. Summing gives the inequality you want.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this $|a_{n}|\le 2$ Let sequence such $$|a_{n}-\dfrac{a_{n+1}}{2}|\le 1$$
if $|a_{n}|\le \dfrac{3^n}{2^n}$, show that
$$|a_{n}|\le 2$$
since
$$2a_{n}-2\le a_{n+1}\le 2a_{n}+2$$
so we
$$a_{n+1}-2\ge 2(a_{n}-2)\ge\cdots \ge 2^n(a_{1}-2)$$
| Suppose $a_N>2$. Put $a_N=2+k$ where $k>0$. Then we have $1+k\le\frac{1}{2} a_{N+1}$ and so $a_{N+1}\ge2+2k$. By a trivial induction $a_{N+m}\ge2+2^mk$. Now we claim that for sufficiently large $m$ we have $a_{N+m}>\frac{3^{N+m}}{2^{N+m}}$.
For $\left(\frac{4}{3}\right)^m>\frac{1}{k}\left(\frac{3}{2}\right)^N$ for sufficiently large $m$ and hence $2^mk>\left(\frac{3}{2}\right)^{N+m}$, which gives the required result since $a_{N+m}>2^mk$. But that gives us a contradiction, since we are given that $a_n\le\left(\frac{3}{2}\right)^n$ for all $n$. So we cannot have $a_N>2$.
Similarly, if $a_N<-2$. for then we can take $a_N=-2-k$ where $k>0$, and in a similar way we get $a_{N+m}<-2-2^mk$ and hence $a_{N+m}<-\left(\frac{3}{2}\right)^{N+m}$ for sufficiently large $m$, which again is a contradiction. So we cannot have $a_N<-2$.
So we have established that we must have $|a_N|\le2$ for all $N$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$ Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$
My Attempt:
Let the equation of the sphere be $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.As this sphere touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$.
So $1+4+4+2u+4v-4w+d=0\implies2u+4v-4w+d=-9.......(1)$
Also the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ passes through $(1,-1,0)$
So $1+1+2u-2v+d=0\implies2u-2v+d=-2......(2)$
I am stuck here.Please help me.
| Hints:
(1) The given sphere is $\;(x+1)^2+(y-3)^2+z^2=9\;$
(2) The line through the above sphere's center and the given tangency point is $\;(-1,3,0)+t(1,2,-2)\;,\;\;t\in\Bbb R\;$
(3) In a similar way as in plane geometry, two tangent spheres' centers are joined by a line passing through the tangency point.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $(x^2+y^2+z^2) \ \sin \frac{1}{\sqrt{x^2+y^2+z^2}}$ Differentiable in $(0,0,0)$? $$f(x,y,z)=\begin{cases} (x^2+y^2+z^2) \ \sin \frac{1}{\sqrt{x^2+y^2+z^2}} \qquad (x,y,z) \ne (0,0,0) \\ \\ 0 \qquad (x,y,z)=(0,0,0) \end{cases} $$
At first, I study the continuity in the origin.
I apply the concept of sequential continuity: $x_n \rightarrow x_0 \Rightarrow f(x_n) \rightarrow f(x_0)$
So, $x_n=\frac{1}{n} $
$$x_n \rightarrow 0 \Rightarrow f(x_n,x_n,x_n) \rightarrow f(0,0,0)=0$$
$$\lim_{n\rightarrow +\infty} \frac{3}{n^2} \ \sin \frac{1}{\frac{\sqrt{3}}{n}}=0 $$
How can I continue the study of differentiation?
Thanks!
| You can't study the continuity of $f$ at the origin by choosing one sequence $\mathbf{p_n} = (x_n,y_n,z_n)$ that approaches $\mathbf{0}$ in a very specific way ($x_n = y_n = z_n = \frac{1}{n}$) and checking that $f(\mathbf{p_n}) \rightarrow f(\mathbf{0})$. You need to check that this works for all sequences converging to $(0,0,0)$ or, equivalently, check the $\varepsilon-\delta$ definition. In your case, given $\varepsilon > 0$, if $|| \mathbf{p} - \mathbf{0} || = ||\mathbf{p}|| = ||(x,y,z)|| < \sqrt{\varepsilon}$ then
$$ |f(\mathbf{p}) - f(\mathbf{0})| \leq x^2 + y^2 + z^2 = ||(x,y,z)||^2 < \varepsilon $$
so indeed $f$ is continuous at $\mathbf{0}$. To check differentiability, check first whether the partial derivatives exist. For example,
$$ \frac{\partial f}{\partial x}(\mathbf{0}) = \lim_{h \to 0} \frac{f(h,0,0) - f(0,0,0)}{h} = \lim_{h \to 0} \frac{h^2 \sin \left( \frac{1}{h} \right)}{h} = \lim_{h \to 0} h \sin \left( \frac{1}{h} \right) = 0. $$
Similarly, the partial derivatives $\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ exist and vanish. This means that if $f$ is differentiable, we must have $df|_{\mathbf{0}} = \mathbf{0}$. Let us check that this is indeed the case (here, $\mathbf{p} = (x,y,z)$):
$$ \lim_{\mathbf{p} \to \mathbf{0}} \frac{f(\mathbf{p}) - f(\mathbf{0}) - 0 \cdot x - 0 \cdot y - 0 \cdot z}{||\mathbf{p}||} = \lim_{\mathbf{p} \to 0} \sqrt{x^2 + y^2 + z^2} \sin \left( \frac{1}{\sqrt{x^2 + y^2 + z^2}} \right) = 0 $$
and so $f$ is differentiable at $\mathbf{0} = 0$ with vanishing differential / gradient.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Calculating a limit with series Good evening to everyone. I have a limit that gave me a lot of trouble and I couldn't find a way to solve it. I tried solving it with series but I couldn't arrive at a result.
$$
\lim _{x\to 0+}\left(\frac{\left(e^{-\frac{1}{x^2}}\cos \left(\log _e\left(x\right)\right)+\cos \left(\arctan \left(x\right)\right)-e^{-\frac{x^2}{2}}\right)}{\log _e\left(1+x^2\right)-\sin \left(x^2\right)}\right)
$$
Thank you!
| tl;dr: This is going to be such a hoot.
*
*Let us start with the denominator:
$$\begin{align}
\ln\left(1+x^2\right)-\sin \left(x^2\right)
&=
x^2-\frac{x^4}{2} + o(x^4) - (x^2+o(x^4)) = -\frac{x^4}{2} + o(x^4)
\end{align}
$$
*Now, the meat — the numerator, piece by piece:
*
*No Taylor for the first, it does not apply (the arguments do not tend to $0$, for a start).
$$\begin{align}
\lvert e^{-\frac{1}{x^2}}\cos \left(\ln\left(x\right)\right)\rvert \leq
e^{-\frac{1}{x^2}} \xrightarrow[x\to 0]{} 0
\end{align}
$$
and this term is negligible in front of any power of $x$: it decays exponentially. So in particular it is $o(x^4)$, say. (Tell me if you need a proof of that. Edit: see $(\dagger)$ at the end)
*The second, we can do a series expansion:
$$\begin{align}
\cos \left(\arctan \left(x\right)\right)
&= \cos\left( x- \frac{x^3}{3} + o(x^4)\right)= 1 - \frac{x^2}{2}+\frac{3x^4}{8} + o(x^4)
\end{align}
$$
and the last term (series can be used as well):
$$\begin{align}
-e^{-\frac{x^2}{2}} &= -(1-\frac{x^2}{2}+\frac{x^4}{8}+o(x^4)) = -1+ \frac{x^2}{2}-\frac{x^4}{8}+o(x^4)
\end{align}
$$
so overall
$$\begin{align}
e^{-\frac{1}{x^2}}\cos \left(\ln\left(x\right)\right) +\cos \left(\arctan \left(x\right)\right)-e^{-\frac{x^2}{2}}
&= o(x^4) + 1 - \frac{x^2}{2}+\frac{3x^4}{8} + o(x^4) + -1+ \frac{x^2}{2}-\frac{x^4}{8}+o(x^4)\\
&= \frac{x^4}{4}+o(x^4)
\end{align}
$$
*Putting it all together:
$$
\frac{e^{-\frac{1}{x^2}}\cos \left(\log _e\left(x\right)\right)+\cos \left(\arctan \left(x\right)\right)-e^{-\frac{x^2}{2}}}{\log _e\left(1+x^2\right)-\sin \left(x^2\right)}
= \frac{\frac{x^4}{4}+o(x^4)}{-\frac{x^4}{2} + o(x^4)}
\xrightarrow[x\to0^+]{} -\frac{1}{2}.
$$
Edit:
$(\dagger)$ We used the fact that, for any fixed $\alpha\geq 0$,
$$
e^{-\frac{1}{x^2}} = o(x^\alpha)
$$
when $x\to 0$. Note that setting $u=\frac{1}{x^2}$, this is equivalent to proving that for every $\alpha \geq 0$
$
e^{-u} = o(u^{-\alpha})
$ when $u\to+\infty$, or, again, equivalently, that $$\lim_{u\to+\infty}u^\alpha e^{-u} = 0$$ for any fixed $\alpha \geq 0$. Do you see why this is true?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using AM-GM inequality to prove prove that $$x^4 + y^4 + z^4 \geq xyz(x+y+z)$$
This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.
| AM–GM is invoked in two steps as follows:
$\begin{array}{rcl}x^4+y^4+z^4 &=& \dfrac{x^4+y^4}2+\dfrac{y^4+z^4}2+\dfrac{z^4+x^4}2 \\\\ &\ge& x^2y^2+y^2z^2+z^2x^2 \\\\ &=& \dfrac{x^2y^2+y^2z^2}2+\dfrac{y^2z^2+z^2x^2}2+\dfrac{z^2x^2+x^2y^2}2 \\\\ &\ge& xy^2z+yz^2x+zx^2y \\\\ &=& xyz(x+y+z)\end{array}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $4$ divides $3^{2m+1} - 3$
Prove that $4$ divides $3^{2m+1} - 3$.
By plugging in numbers I can see this is true, but I can't figure out a way to prove this, I was thinking maybe proving first that it is divisible by $2$, and concluding its divisible by $4$.
| There are generally several ways to approach these types of divisibility problems. I am showing two of them.
By Mathematical Induction: Putting $m=0$ we have $3^{2m+1}-3=0$, which is obviously divisible by $4$. Now let $4\mid 3^{2m+1}-3$ for some $m\in N_0$. Let $3^{2m+1}-3=4k$, for some $k\in N_0$. Now $3^{2(m+1)+1}-3=9(3^{2m+1}-3)+24=9\cdot 4k+24=4(9k+6)$, which is also divisible by $4$. So we see that $4\mid 3^{2m+1}-3$ for all $m\in N_0$.
By Modular Arithmetic: We see that $3^2\equiv1\pmod{4}$. So for any $m\in N_0$, $3^{2m}\equiv1\pmod{4}$ and $3^{2m+1}\equiv3\pmod{4}$. That is, $4\mid 3^{2m+1}-3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\cos A \cdot \cos 2 A \cdot \cos 4 A \cdots \cos 2^{n-1} A = \frac{\sin 2^n A}{2^n \sin A}$ Just a bit of background on the question:
When proving:
$$\cos\frac{\pi}{15}\cdot \cos\frac{2\pi}{15} \cdot \cos\frac{3\pi}{15}\cdot \cos\frac{4\pi}{15} \cdot \cos\frac{5\pi}{15} \cdot \cos\frac{6\pi}{15}\cdot \cos\frac{7\pi}{15} = \frac{1}{128}$$
The following formula was used:
$$\cos A \cdot \cos 2A \cdot \cos 2^2A \cdot \cos 2^3A\cdot \cdots \cdot \cos2^{n-1}A = \frac{\sin 2^n A}{2^n \sin A}$$
I'm just interested to see how this is derived. The question assumes that this formula is memorised and known beforehand, but it looked interesting to me since I'd actually never come across it before.
If anyone's interested, I'll link full proof for the question here
Cheers in advance!
| By the sine duplication formula $\sin(2x)=2\sin(x)\cos(x)$ it follows that:
$$ \cos(2^n A) = \frac{\sin(2^{n+1} A)}{2\sin(2^n A)}\tag{1} $$
hence:
$$ \prod_{n=0}^{N-1}\cos(2^n A) = \frac{\sin(2^N A)}{2^N \sin(A)}\tag{2}$$
comes from a telescopic product.
| {
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"timestamp": "2023-03-29T00:00:00",
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System of diophantine equations $x^2+3y=u^2$, $y^2+3x=v^2$
Solve the following system of Diophantine equations(the unknowns are positive integers):
$$
\left\{
\begin{array}{c}
x^2+3y=u^2 \\
y^2+3x=v^2
\end{array}
\right.
$$
I worked as follows:
subtract the two equations to get: $4x^2-4y^2-12(x-y)=9y^2-9x^2\ \implies\ ... (x-y)(13x+13y-12)=0\implies x=y\ or\ 13x+13y-12=0$
The first equation has infinite answers and the second has none(since $gcd(13,13)$ does not divide $12$), am I right??
| We can without (much) loss of generality assume that $y\le x$. Note that $x^2+3y$ is a perfect square greater than $x^2$.
Thus we have $x^2+3y\ge (x+1)^2$. But $x^2+3y\lt (x+2)^2$. It follows that $3y=2x+1$.
Since $y^2+3x$ is a perfect square, so is $9y^2+27x$, that is, $4x^2+31x+1$. This has to be a square, so $4x^2+31x+1$ is equal to one of $(2x+1)^2$ or $(2x+2)^2$ and so on up to $(2x+7)^2$, since $(2x+8)^2$ is clearly too big.
The case $(2x+1)^2$ does not work, and neither does $(2x+2)^2$, nor $(2x+3)^2$. The case $(2x+4)^2$ gives $x=1$. The case $(2x+5)^2$ does not work. The case $(2x+6)^2$ gives $x=5$, which does not work because $y$ is not an integer, and the case $(2x+7)^2$ gives $x=16$.
We conclude that the solutions are $x=1,y=1$, $x=16, y=11$, (and $x=11,y=16$).
| {
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"timestamp": "2023-03-29T00:00:00",
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Re-write a quadratic equation in another form? $x^2 + \sqrt{2}x = \frac{1}{2}$
I need to find the real solutions for this equation and write it in this form:
$$\frac{-\sqrt{A} \pm B}{C}$$
So when I work the problem out with the quadratic equation I get: $x = 0.118121$
I had no idea how to even put that in the form described so I assumed they didn't want me to solve but just put it in that form so I put it as follows:
$A = x^2$
$B = \sqrt{2}x$
$C = \frac{1}{2}$
That was wrong also. I'm not really sure what they're asking me to do. I know the answer is: $x = 0.118121$ but is it even possible to put it in the form described? Any help would be appreciated.
| Form of any quadratic Equation is $ax^2+bx+c=0$
Here for you $ a=1 , b= \sqrt{2}$ and $c = -\frac{1}{2}$, $$x^2 + \sqrt{2}x + \left(-\frac{1}{2} \right)=0$$
Solution is $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-\sqrt{2}\pm \sqrt{(\sqrt{2})^2-4\cdot 1 \cdot \left(-\frac{1}{2}\right)}}{2 \cdot 1}$$
$$x=\frac{-\sqrt{2}\pm \sqrt{2+2}}{2 }$$
$$x=\frac{-\sqrt{2}\pm \sqrt{4}}{2 }$$
$$x=\frac{-\sqrt{2}\pm 2}{2 }$$
So $$A=2 ,B =2 ,C=2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Deriving the Airy functions from first principles I have just started reading about the Airy functions and am stuck on a particular step of their derivation. But first here is some background information to give this question some meaning, more information can be found from a previous question of mine:
The general solution to Bessel's differential equation: $$\fbox{$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y=0$}\tag{1}$$ is $$\fbox{$y=x^aZ_p\left(bx^c\right)$}\tag{2}$$ where $Z_p$ stands for $J_p$ or $N_p$ or any linear combination of them, and $a,b,c,p$ are constants. $J_p$ is called the Bessel function of the first kind of order $p$ and $N_p$ is any combination of $J_p$ and $J_{−p}$: $$N_p(x)=\frac{\cos(\pi p)J_p(x)-J_{-p}(x)}{\sin(\pi p)}\tag{3}$$ So equation $(2)$ can be written as $$y=x^{a}\left[AJ_{p}\left(bx^c\right)+BN_{p}\left(bx^c\right)\right]\tag{4}$$ or $$y=x^{a}\left[AJ_{p}\left(bx^c\right)+B\left(\frac{\cos(\pi p)J_p\left(bx^c\right)-J_{-p}\left(bx^c\right)}{\sin(\pi p)}\right)\right]\tag{5}$$
The Airy differential equation is $$y^{\prime\prime}-xy=0\tag{6}$$ and has solution $$y=x^{1/2}Z_{1/3}\left(\frac23 ix^{3/2}\right)\tag{7}$$
By using
$$I_p(x)=i^{-p}J_p(ix)\tag{8}$$
$$K_p(x)=\frac{\pi}{2}i^{p+1}\left[J_p(ix)+i\left(\frac{\cos(\pi p)J_p(ix)-J_{-p}(ix)}{\sin(\pi p)}\right)\right]\tag{9}$$
My objective is to show that $(7)$ can be written in terms of $I_{1/3}$ and $K_{1/3}$ to obtain
$$Ai(x)=\frac{1}{\pi}\sqrt{\frac{x}{3}}K_{1/3}\left(\frac23x^{3/2}\right)\tag{10}$$
$$Bi(x)=\sqrt{\frac{x}{3}}\left[I_{-1/3}\left(\frac23x^{3/2}\right)+I_{1/3}\left(\frac23x^{3/2}\right)\right]\tag{11}$$
Starting from $(7)$: $$\begin{align}y&=x^{1/2}Z_{1/3}\left(\frac23 ix^{3/2}\right)\\&=x^{1/2}\left[AJ_{1/3}\left(\frac23 ix^{3/2}\right)+BN_{1/3}\left(\frac23 ix^{3/2}\right)\right]\\&=x^{1/2}\left[AJ_{1/3}\left(\frac23ix^{3/2}\right)+B\left(\frac{\cos\left(\frac{\pi}{3}\right)J_{1/3}\left(\frac23ix^{3/2}\right)-J_{-1/3}\left(\frac23ix^{3/2}\right)}{\sin\left(\frac{\pi}{3}\right)}\right)\right]\\&=x^{1/2}\left[AJ_{1/3}\left(\frac23ix^{3/2}\right)+B\left(\frac{\frac12J_{1/3}\left(\frac23ix^{3/2}\right)-J_{-1/3}\left(\frac23ix^{3/2}\right)}{\left(\frac{\sqrt3}{2}\right)}\right)\right]\\&=x^{1/2}\left[AJ_{1/3}\left(\frac23ix^{3/2}\right)+\frac{BJ_{1/3}\left(\frac23ix^{3/2}\right)}{\sqrt3}-\frac{2BJ_{-1/3}\left(\frac23ix^{3/2}\right)}{\sqrt3}\right]\\&=x^{1/2}\left[Ai^{1/3}I_{1/3}\left(\frac23x^{3/2}\right)+\frac{i^{1/3}BI_{1/3}\left(\frac23x^{3/2}\right)}{\sqrt3}-\frac{2i^{-1/3}BI_{-1/3}\left(\frac23x^{3/2}\right)}{\sqrt3}\right]\tag{a}\end{align}$$
where in $(\mathrm{a})$ I used $(8)$ rearranged as $J_p(ix)=i^{p}I_p(x)$
I don't understand how to proceed with this calculation as I am unsure how to use $(9)$; I also have no idea what $i(x)$ means. Is $i$ really a function of $x$?
Is there anyone that could provide some hints or advice on how I can continue this calculation to obtain $(10)$ and $(11)$?
Below are some images showing some of the relevant formulae to this question:
| At first note, that $i=\sqrt{-1}$ is the imaginary unit in (9) and we see a plain multiplication with $i$.
We start with the representation (5) and use (7) to obtain
\begin{align*}
y&=x^{\frac{1}{2}}Z_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\\
&=x^{\frac{1}{2}}\left[AJ_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)
+B\left(\frac{1}{\sqrt{3}}J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)
-\frac{2}{\sqrt{3}}J_{-\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\right)\right]\tag{f}
\end{align*}
With $p=\frac{1}{3}$ we can write $K_p$ as
\begin{align*}
K_{\frac{1}{3}}\left(\frac{2}{3}x^{\frac{3}{2}}\right)=\frac{\pi}{2}i^{\frac{4}{3}}
\left[J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)+i\left(\frac{1}{\sqrt{3}}J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)
-\frac{2}{\sqrt{3}}J_{-\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\right)\right]
\end{align*}
Now we can write $Ai(x)$ using $K_{\frac{1}{3}}$ and see the structural connection with $(\mathrm{f})$.
\begin{align*}
Ai(x)&=\frac{1}{\pi}\sqrt{\frac{x}{3}}K_{\frac{1}{3}}\left(\frac{2}{3}x^{\frac{3}{2}}\right)\\
&=\frac{1}{2\sqrt{3}}i^{\frac{4}{3}}x^{\frac{1}{2}}
\left[J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)+i\left(\frac{1}{\sqrt{3}}J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)
-\frac{2}{\sqrt{3}}J_{-\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\right)\right]
\end{align*}
Comparison with $(\mathrm{f})$ results in
\begin{align*}
A=\frac{1}{2\sqrt{3}}i^{\frac{4}{3}}\qquad\qquad B=-\frac{1}{2\sqrt{3}}i^{\frac{1}{3}}
\end{align*}
We can see the connection of $B_i(x)$ with (5) by substituting $I_p(x)$ with $J_p(x)$. We obtain
\begin{align*}
Bi(x)&=\frac{1}{\sqrt{3}}x^{\frac{1}{2}}\left[I_{-\frac{1}{3}}\left(\frac{2}{3}x^{\frac{3}{2}}\right)
+I_\frac{1}{3}\left(\frac{2}{3}x^{\frac{3}{2}}\right)\right]\\
&=\frac{1}{\sqrt{3}}x^{\frac{1}{2}}\left[i^{\frac{1}{3}}J_{-\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)
+i^{-\frac{1}{3}}J_{\frac{1}{3}}\left(\frac{2}{3}ix^{\frac{3}{2}}\right)\right]
\tag{g}\end{align*}
Comparison of coefficients of $J_{-\frac{1}{3}}$ gives
\begin{align*}
B=-\frac{1}{2}i^{\frac{1}{3}}
\end{align*}
Comparison of coefficients of $J_{\frac{1}{3}}$ gives
\begin{align*}
A=\frac{1}{\sqrt3}\left(i^{-1/3}-\frac{i^{1/3}}{2}\right)
\end{align*}
Hint: Note that the Airy function is defined according to (10) in terms of $K_p$ which is defined according to (9). So, the right-hand side of the blue part is just the definition of the Airy function in terms of $J_p$ and $J_{-p}$. From this representation we can deduce $A$ and $B$ by a rather simple comparison with $(\mathrm{f})$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$ Prove that
$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$
$$\text{if and only if}$$
$$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after getting here i got stuck
$$\lim_{x\to\infty } \left({\sqrt{x+1}\left(a+b\sqrt{1+\frac{1}{x+1}}+c\sqrt{1+\frac{2}{x+1}}\right)}\right)$$
Got here by substituting $\sqrt{x+2}$ with $\sqrt{(x+1)(1+\dfrac{1}{x+1})}$
Edit: x tends to infinity, not to 0. I transcribed wrongly.
| Note that we have
$$\begin{align}
a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}&=\sqrt{x}\left(a\sqrt{1+\frac{1}{x}}+b\sqrt{1+\frac{2}{x}}+c\sqrt{1+\frac{3}{x}}\right)\\\\
&=a\sqrt{x}\left(1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)\right)\\\\
&+b\sqrt{x}\left(1+\frac{1}{x}+O\left(\frac{1}{x^2}\right)\right)\\\\
&+c\sqrt{x}\left(1+\frac{3}{2x}+O\left(\frac{1}{x^2}\right)\right)\\\\
&=(a+b+c)\sqrt{x}+\frac{(a+2b+3c)}{2\sqrt{x}}+O\left(\frac{1}{x^{3/2}}\right) \tag 1
\end{align}$$
from which we see that the limit is zero if and only if $a+b+c=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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problem proving: $(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$ I'm trying to prove this, and it is really frustrating, because it seems a really easy problem to prove, however, I'm having a little problem with exponents:
$$(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$$
Hypothesis
$F(x)=(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{x}}) = \frac{1-q^{{2}^{x+1}}}{1-q}$
The format can be a little problematic here, so, to clarify:
$(1+q^{{2}^{x}})$ = 1+(q^(2^x))
and
$1-q^{{2}^{x+1}}$ = 1 - (q^2^(x+1))
Proof:
$P1 | F(x) = (\frac{1-q^{{2}^{x+1}}}{1-q})(1+q^{{2}^{x+1}}) = \frac{1-q^{{2}^{x+2}}}{1-q} $
$P2| \frac{[(1)-(q^{{2}^{x+1}})][(1)+(q^{{2}^{x+1}})]}{(1-q)} = \frac{1-q^{{2}^{x+2}}}{1-q}$
Here I don't know if I should:
$P3 | \frac{(1-q^{{2}^{x+1}})^2}{1-q} = \frac{1-q^{{2}^{x+2}}}{1-q}| $ applying $ (a+b)(a-b) = a^2 - b^2 $
or
$P3 | \frac{1-q^{{2}^{x+1}+{2}^{x+1}}}{1-q} = \frac{1-q^{{2}^{x+2}}}{1-q} $
This arises because, suposedly, the property goes:
$(a^n)^m = a^{n*m}$
But, it seems, in this problem, the proponderance is different, kinda like this:
$a^{(n^{m})}$
Because, for example, when q=2:
$ F(0) = 1+2^{({2}^{0})} = 1 + 2^{1} = 3 $
which seems to be true, since when evaluating RHS:
$ F(0) = \frac{1-2^{{2}^{0+1}}}{1-2} = 3 $
| Induction on $n$.
For brevity let $\prod_{j=0}^n(1+q^{2^j})=P(q,n)$.
For the case $n=0$ we have $(1-q)P(q,0)=(1-q)(1+q)=1-q^2=1-q^{2^{0+1}}.$
If $(1-q)P(q,n)= 1-q^{2^{n+1}}$ then $$(1-q)P(q,n+1)=(1-q)P(q,n)(1+q^{2^{n+1}})=$$ $$=(1-q^{2^{n+1}})(1+q^{2^{n+1}})=1-(q^{2^{n+1}})^2=1-q^{2^{n+2}}.$$
For example $$(1-q)(1+q)(1+q^2)=(1-q^2)(1+q^2)=1-(q^2)^2=1-q^4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831623",
"timestamp": "2023-03-29T00:00:00",
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Solution of $(n+1)^{1/3}-n^{1/3}=\frac{1}{12}$ Solve the given equation for $n$
$(n+1)^{1/3}-n^{1/3}=\frac{1}{12}$
How to approach this particular question? Sorry cannot show any work because the only approach I can see is take cube on both sides but that is complicating the equation.
| Let $x=\sqrt[3]{n+1}$ and $y=\sqrt[3]{n}$. Note that $(x-y)^3=x^3-y^3-3xy(x-y)=1-3xy(x-y)$. So
$$\frac{1}{12^3}=1-\frac{xy}{4}.$$
Solve for $xy$. We get $xy=a$, where $a$ is a mildly messy number.
We also have $x-y=b$, where $b=\frac{1}{12}$.
So $(x+y)^2=b^2+4a$, and now we know that $x+y=\pm\sqrt{b^2+4a}$.
We know $x+y$ and $x-y$, so we know $y$. Finally, $n=y^3$.
Remark: The strategy used here is the one that Cardano used to find the roots of a reduced cubic.
| {
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Which number is greater, $11^{11}$ or $9^{12}$?
Which number is greater than $11^{11}$ or $9^{12}$?
My work so far:
$11^{11}=285311670611>9^{12}=282429536481$.
But to verify the validity of equality should be in the range of easily verifiable calculations.
| Just for the heck of it, here's another approach using the fact that
$$9^3=6\times11^2+3\qquad\text{ and }\qquad 6^4<11^3-3\times11.$$
It is by no means the slickest way;
\begin{eqnarray*}
9^{12}&=& (6\times11^2+3)^4=3^4\times(2\times11^2+1)^4\\
&=& 3^4\times(2^4\times11^8+4\times2^3\times11^6+6\times2^2\times11^4+4\times2\times11^2+1)\\
&=&6^4\times(11^8+2\times11^6+\tfrac{3}{2}\times11^4+\tfrac{1}{2}\times11^2+\tfrac{1}{16})\\
&<&(11^3-3\times11)\times(11^8+2\times11^6+\tfrac{3}{2}\times11^4+\tfrac{1}{2}\times11^2+\tfrac{1}{16})\\
&=&11^{11}+2\times11^9+\tfrac{3}{2}\times11^7+\tfrac{1}{2}\times11^5+\tfrac{1}{16}\times11^3\\
&\ &\quad\ \ \ -3\times11^9-\ 6\times11^7-\tfrac{9}{2}\times11^5-\ \tfrac{3}{2}\times11^3-\tfrac{3}{2}\times11
\end{eqnarray*}
The alignment of the last expression shows that it is smaller than $11^{11}$, as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
| Using the rule $a\sqrt b=\sqrt{a^2b}$ for $a,b>0$ one checks the signs of numerator and denominator to be both positive for the first fraction and both negative for the second fraction. This is therefore a comparison of two positive numbers (so no verdict yet), and by flipping the signs of numerator and denominator on the right we can ensure that all individual factors are positive: compare $\frac{3\sqrt3-4}{7-2\sqrt3}$ and $\frac{8-3\sqrt3}{2\sqrt3-1}$. Now multiply by the (positive!) product of the denominators to compare $(3\sqrt3-4)(2\sqrt3-1)$ and $(8-3\sqrt3)(7-2\sqrt3)$ which amounts to comparing $22-11\sqrt3$ and $74-37\sqrt3$ or finally $26\sqrt3$ and $52$. Since by happy coincidence $26\times2=52$ and $\sqrt3<2$ the number on the left is smaller.
I'm not sure whether many people could do this easily by mental arithmetic (I certainly did not). Each of the terms in the final comparison are obtained by adding $4$ products from the terms in the original expression. One product, namely $(2\sqrt3)(3\sqrt3)$, contributes twice with opposite signs and could therefore have been cancelled beforehand (leaving $7\times8-4\times1=52$ for that term), but that's all I can see for simplifications. Personally my main problem with mental computation is that I very often get the sign of one term wrong in my head, but maybe with a lot of training one can limit this kind of errors.
| {
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How to prove this inequality $x\sin^2{A}+y\sin^2{B}\ge xy\sin^2{C}$ In $\Delta ABC$,if $x,y>0$ and $x+y=1$.show that
$$x\sin^2{A}+y\sin^2{B}\ge xy\sin^2{C}$$
I have looked at the simpler methods,? Here is one solution
$$\dfrac{\sin^2{A}}{y}+\dfrac{\sin^2{B}}{x}\ge\dfrac{(\sin{A}+\sin{B})^2}{x+y}=\dfrac{\sin^2{(A+B)}}{1}=\sin^2{C}$$
where use this $\sin{A}+\sin{B}\ge \sin{(A+B)}$
other methods?
| It's $\frac{4S^2x}{b^2c^2}+\frac{4S^2y}{a^2c^2}\geq\frac{4S^2xy}{a^2b^2}$ or
$(x+y)(a^2x+b^2y)\geq xyc^2$ or $a^2x^2+b^2y^2+(a^2+b^2-c^2)xy\geq0$.
$\Delta=(a^2+b^2-c^2)^2-4a^2b^2=-16S^2<0$ and we are done!
| {
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Compute $\int \sqrt{1+4x^2} \, dx$ with Euler substitution In this post:
Computing $\int \sqrt{1+4x^2} \, dx$
someone mentioned Euler substitution to compute the following integral:
$$\int \sqrt{1+4x^2} \, dx$$
I tried to follow this advice and got very nice result, namely I substituted $\sqrt{1+4x^2}=t-2x$ which after raising to the square and reducing gives $x=\frac{t^2-1}{4t}$ and $t-2x=\frac{t^2+1}{2t}$, then derivative is equal to $\frac{dx}{dt}=\frac{t^2+1}{4t^2}$ and the whole integral:
$$\int \sqrt{1+4x^2} \, dx = \int \frac{(t^2+1)^2}{8t^3} \, dt$$
Could you please check my solution, because it seems a lot easier than all these trigonometric substitution (too easy which is suspicious...).
Thanks in advance.
| If you set $\sqrt{1+4x^2}=t-2x$, you have
$$
1+4x^2=t^2-4tx+4x^2
$$
so $4tx=t^2-1$ and therefore
$$
x=\frac{t^2-1}{4t}=\frac{t}{4}-\frac{1}{4t}
$$
Thus
$$
dx=\left(\frac{1}{4}+\frac{1}{4t^2}\right)\,dt=\frac{t^2+1}{4t^2}\,dt
$$
and
$$
\sqrt{1+4x^2}=t-\frac{t}{2}+\frac{1}{2t}=\frac{t^2+1}{2t}
$$
so the integral becomes
$$
\int\frac{(t^2+1)^2}{8t^3}\,dt=
\frac{1}{8}\int\left(t+\frac{2}{t}+\frac{1}{t^3}\right)\,dt
$$
that's elementary.
Yes, your computation is right.
The alternative way is setting $2x=\sinh(t/2)$, so
$$
\sqrt{1+4x^2}=\cosh\frac{t}{2},
\qquad
dx=\frac{1}{4}\cosh\frac{t}{2}\,dt
$$
and the integral is
$$
\frac{1}{4}\int\cosh^2\frac{t}{2}\,dt=
\frac{1}{8}\int(\cosh t-1)\,dt
$$
remembering that $2\cosh^2\frac{t}{2}+1=\cosh t$.
Another (tricky) way is to do $2x=t$, that reduces to computing, up to a scalar factor,
$$\DeclareMathOperator{\arsinh}{arsinh}
I=\int\sqrt{1+t^2}\,dt=\int\frac{1+t^2}{\sqrt{1+t^2}}\,dt=
\arsinh t+\int t\frac{t}{\sqrt{1+t^2}}\,dt=
\arsinh t+t\sqrt{1+t^2}-I
$$
(the last one by parts).
| {
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Solutions of $\sin^2\theta = \frac{x^2+y^2}{2xy} $ If $x$ and $y$ are real, then the equation
$$\sin^2\theta = \frac{x^2+y^2}{2xy}$$
has a solution:
*
*for all $x$ and $y$
*for no $x$ and $y$
*only when $x \neq y \neq 0$
*only when $x = y \neq 0$
| HINT:
$$(x/y)^2-2(x/y)\sin^2\theta+1=0$$
As $x/y$ is real,the discriminant is $$(2\sin^2\theta)^2-4\ge0\iff\sin^4\theta\ge1\iff\sin^2\theta\ge1\iff\cos^2\theta\le0$$
this is only possible if $\cos^2\theta=0\iff\sin^2\theta=?$
Can you take it from here?
| {
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How to calculate $∇(r^2/(2z(1+a/z^2)))$ in cylindrical coordinates How to calculate $$∇\bigg(\frac{(ρ^2)}{2z(1+\frac{a}{z^2})}\bigg)$$ where the function is in cylindrical coordinates $$ρ^2=x^2+z^2$$
$$∇\bigg(\frac{x^2+z^2}{2z(1+\frac{a}{z^2})}\bigg)$$
Is the answer in Cartesian coordinates a vector
$$[x/(z(1+a/z^2))),y/(z(1+a/z^2))),1/((a-z^2)/(a+z^2)^2)]$$
In cylindrical coordinates a vector ?
$$[ρ/(z(1+a/z^2))),0,1/((a-z^2)/(a+z^2)^2)]$$
| You need the gradient operator for cylindrical coordinates, e.g. see here.
$$
\nabla f =
\frac{\partial f}{\partial \rho} e_\rho +
\frac{1}{\rho}\frac{\partial f}{\partial \varphi} e_\varphi +
\frac{\partial f}{\partial z} e_z
$$
and apply it to
$$
f(\rho, \phi, z) = \frac{\rho^2}{2z \left( 1 + \frac{a}{z^2} \right)}
$$
so
\begin{align}
\nabla f =
\frac{\rho}{z \left( 1 + \frac{a}{z^2} \right)} e_\rho -
\frac{\rho^2}{\left( 2z \left( 1 + \frac{a}{z^2} \right) \right)^2}
\left( \left( 1 + \frac{a}{z^2} \right)
- z \left(2\frac{a}{z^3} \right) \right) e_z \\
=
\frac{\rho}{z \left( 1 + \frac{a}{z^2} \right)} e_\rho -
\rho^2 \frac{1 - \frac{a}{z^2}}{\left( 2z \left( 1 + \frac{a}{z^2} \right) \right)^2} e_z \\
\end{align}
In Cartesian coordinates it would be
$$
\nabla f =
(\partial_x, \partial_y, \partial_z) f(x,y,z) =
(\partial_x, \partial_y, \partial_z)
\frac{x^2 + y^2}{2z \left( 1 + \frac{a}{z^2} \right)}
$$
| {
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Find the Derivatives of $g(x) = \sqrt{3-2x^2}$ and $h(x) = \ln {(x^2 – x)}$ I am asked to find the derivatives of $g(x) =\sqrt{3-2x^2}$ and $h(x) = \ln{(x^2 – x)} $
For:
$g(x)h(x)$
and
$\dfrac{h(x)}{g(x)}$
and
$h^3 (x)$
First off I am not sure if my derivatives are correct. Here is what I have..
$g'(x) = \dfrac{2x} { \sqrt{3 - 2x^2}}$
$h'(x) =\dfrac {2x - 1} {x^2 - x}$
Thanks
Em
| Your $h'$ is correct, but your $g'$ should be $\dfrac 12 \cdot \dfrac{-4x}{\sqrt{3-2x^2}}=\dfrac {-2x}{\sqrt{3-2x^2}}$
So from the product rule we have $(h \cdot g)'=h' \cdot g + h \cdot g'=\dfrac {2x-1}{x^2-1} \cdot \sqrt{3-2x^2} + \ln (x^2-x) \cdot \dfrac {-2x}{\sqrt{3-2x^2}}$. You can't really do much to simplify it, except for making the whole second tern negative instead of having $-2x$ in the numerator.
$\dfrac hg$ is exactly the same, except a different formula.
For $h^3$, we use the chain rule: $(h^3)'=h' \cdot 3(h)^2=\dfrac {2x-1}{x^2-1} \cdot 3(\ln (x^2-x))^2$. Can't do much to simplify this either.
| {
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Solve $2^x+4^x=2$ This is the equation, but the result is different from wolframalpha:
$$2^x+4^x=2$$
$$2^x+2^{2x}=2^1$$
$$x+2x=1$$
$$x=\frac{1}{3}$$
WolframAlpha: $x=0$
Where is the error?
| $$2^x+2^{2x}=2$$
Now put $2^x=t$
$$t+t^{2}=2$$
$$t^{2}+t-2=0$$
$$(t-1)(t+2)=0$$
Thus $t=1$ or $t=-2$
$2^x=1$ or $2^x=-2$
Since $2^x>0 $ for all real $x$ , $2^x=1=2^0$
Therefore $x=0$
| {
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Inequality with square root $x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}}$ Good morning to everyone! The inequality is the following:$$ x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} $$. I don't know how to solve it. Here's what I tried: $$x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} \rightarrow 2x^2-11x+9+\sqrt{x^2-10x+9}\left(2x+1\right)\ge \:\sqrt{x^2-10x+9}$$ Therefore $ 2x+1 \ge 0 => x\ge \frac{1}{2} $ so $x$ belongs to $(\frac{1}{2},\infty) $ and $ x^2-10x+9 \ge 0 => x$ belongs to $ (-\infty,1] $ and $[9,\infty) $. Therefore the statements are contradicting each other.
| Hint:
The domain of the inequation is defined by
$$x^2-10x+9\ge 0\iff (x-5)^2\ge 16\iff x\ge 9\quad\text{or}\quad x\le 1.$$
$u\ge \sqrt u\iff u\ge 1$ or $u=0$. Here
$$u=0 \iff\sqrt{x^2-10x+9}=-x\iff -10x+9=0\quad\text{and}\quad x\le 0,$$
which has no solution. Hence the inequation comes down to
$$\sqrt{x^2-10x+9}\ge 1-x\iff x^2-10x+9\ge (1-x)^2\quad\textbf{or}\quad x\ge 1. $$
Can you take it from here?
| {
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Using the chain rule for cos and sin functions I am having issues with derivatives containing chain rules. I know there is multiple threads already but after reading a few, I still find myself confused. I also checked the actual answer following a step by step website without success.
Derivate: $$h(x)=\sin ( x^6 - cos^3 x^2)$$
Now I have $sin = f(x)$ and $( x^6 - cos^3 x^2) = g(x)$ so i have a $f(g(x))$ form
The general formula says $f(g(x))$ = $f'(g(x))*g'(x)$
First, let's derivate the first part using --> $sin f(x) = cos f(x) * f'(x)$
$$cos (x^6-cos^3x^2)*(6x^5 - ??)$$
I know that $cos^3 x$ is $(cos x)^3$ so that means $((cos x^2)^3)'= 3(cos x^2)^2*-sinx^2*2x$
So we have
$$cos (x^6-cos^3x^2)*(6x^5-(6x(cos x^2)^2*-sinx^2)))$$
Simplified
$$cos (x^6-cos^3x^2)(6x^5-6xcos^2 x^2*sinx^2)$$
Where i'm confused : The rule says that we need to do f'(g(x))*g'(x), this would mean that I would need to do:
$$(sin(x^6-cos^3x^2))'*(x^6-cos^3x^2)'$$
Which would make an even bigger answer
$$cos (x^6-cos^3x^2)*(6x^5-6xcos^2 x^2*sinx^2) * (6x^5-6xcos^2 x^2*sinx^2)$$
And then I don't understand.
Also, it seems that i have done a mistake because the actual answer is the following (I have a $-$ where it's suppose to be a $+$)
Real answer : $$[cos (x^6-cos^3x^2)](6x^5+6xcos^2 x^2sinx^2)$$
My answer : $$cos (x^6-cos^3x^2)(6x^5-6xcos^2 x^2*sinx^2)$$
| $f'(g(x))\cdot g'(x)$ means, in your case, $\sin'(x^6-\cos^3{x^2})\cdot (x^6-\cos^3(x^2))'$ and not what you have written. This would give the result to be $\cos(x^6-\cos^3(x^2))\cdot (6x^5-(3\cos^2(x^2)(-\sin (x^2))(2x))$ which equals the real answer given, i.e. $\cos(x^6-\cos^3(x^2))\cdot(6x^5+6x\cos^2(x^2)\sin(x^2))$
| {
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Proving that $\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+\cdots+\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}$ by induction I've proved the base case where $n=1$ and made the assumption that $n=k$ is true, but I'm stuck on the $n=k+1$ part. I just cannot seem to get the algebra to work in my favor.
Here is the original:
$\displaystyle\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+…+\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}\qquad \forall\, n \in \mathbb N$
I get it to the following form and just run in circles:
$\displaystyle\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+...+\frac{1}{(k+3)(k+4)} +\frac{1}{(k+4)(k+5)}=\frac{k+1}{4(k+1+4)}$
Simplifying
$\displaystyle\frac{k}{4(k+4)} +\frac{1}{(k+4)(k+5)}=\frac{k+1}{4(k+1+4)}$
Is this in the correct form?
| Let $S(n)$ be the statement: $\dfrac{1}{4(5)}+\dfrac{1}{5(6)}+\dfrac{1}{6(7)}+\cdots+\dfrac{1}{(n+3)(n+4)}=\dfrac{n}{4\hspace{1 mm}(n+4)}$; $n\in\mathbb{N}$
Basis step: $S(1)$:
LHS: $\dfrac{1}{\big((1)+3\big)\big((1)+4\big)}=\dfrac{1}{4\times{5}}$
$\hspace{47.5 mm}=\dfrac{1}{20}$
RHS: $\dfrac{n}{4\hspace{1 mm}(n+4)}=\dfrac{(1)}{4\hspace{1 mm}\big((1)+4\big)}$
$\hspace{29 mm}=\dfrac{1}{4\times{5}}$
$\hspace{29 mm}=\dfrac{1}{20}$
$\hspace{80 mm}$ LHS $=$ RHS (verified.)
Inductive step:
Assume $S(k)$ is true, i.e. assume that
$\dfrac{1}{4(5)}+\dfrac{1}{5(6)}+\dfrac{1}{6(7)}+\cdots+\dfrac{1}{(k+3)(k+4)}=\dfrac{k}{4\hspace{1 mm}(k+4)}$; $k\in\mathbb{N}$
$S(k+1)$: $\underline{\dfrac{1}{4(5)}+\dfrac{1}{5(6)}+\dfrac{1}{6(7)}+\cdots+\dfrac{1}{(k+3)(k+4)}}+\dfrac{1}{\big((k+1)+3\big)\big((k+1)+4\big)}$
$\hspace{13 mm}=\dfrac{k}{4\hspace{1 mm}(k+4)}+\dfrac{1}{(k+4)(k+5)}$
$\hspace{13 mm}=\dfrac{k\hspace{1 mm}(k+4)(k+5)+4\hspace{1 mm}(k+4)}{4\hspace{1 mm}(k+4)(k+4)(k+5)}$
$\hspace{13 mm}=\dfrac{(k+4)\big(k\hspace{1 mm}(k+5)+4\big)}{4\hspace{1 mm}(k+4)^{2}(k+5)}$
$\hspace{13 mm}=\dfrac{k^{2}+5k+4}{4\hspace{1 mm}(k+4)(k+5)}$
$\hspace{13 mm}=\dfrac{k^{2}+k+4k+4}{4\hspace{1 mm}(k+4)(k+5)}$
$\hspace{13 mm}=\dfrac{k\hspace{1 mm}(k+1)+4\hspace{1 mm}(k+1)}{4\hspace{1 mm}(k+4)(k+5)}$
$\hspace{13 mm}=\dfrac{(k+1)(k+4)}{4\hspace{1 mm}(k+4)(k+5)}$
$\hspace{13 mm}=\dfrac{k+1}{4\hspace{1 mm}(k+5)}$
So, $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $\dfrac{1}{4(5)}+\dfrac{1}{5(6)}+\dfrac{1}{6(7)}+\cdots+\dfrac{1}{(n+3)(n+4)}=\dfrac{n}{4\hspace{1 mm}(n+4)}$; $n\in\mathbb{N}$.
| {
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The value of $x^2+y^2+z^2+w^2$ Let$x,y,z,w$ satisfy
$$\frac{x^2}{2^2 - 1^2} +\frac{y^2}{2^2 - 3^2} +\frac{z^2}{2^2 - 5^2} +\frac{w^2}{2^2 - 7^2} =1$$
$$\frac{x^2}{4^2 - 1^2} +\frac{y^2}{4^2 - 3^2} +\frac{z^2}{4^2 - 5^2} +\frac{w^2}{4^2 - 7^2} =1$$
$$\frac{x^2}{6^2 - 1^2} +\frac{y^2}{6^2 - 3^2} +\frac{z^2}{6^2 - 5^2} +\frac{w^2}{6^2 - 7^2} =1$$
$$\frac{x^2}{8^2 - 1^2} +\frac{y^2}{8^2 - 3^2} +\frac{z^2}{8^2 - 5^2} +\frac{w^2}{8^2 - 7^2} =1$$
My work
$$\frac{x^2}{t - 1^2} +\frac{y^2}{t - 3^2} +\frac{z^2}{t - 5^2} +\frac{w^2}{t - 7^2} =1$$
where $t $ satisfy $4,16,36,64$
$$f(t)=0$$
$$f(t) = (t – 1)(t – 9)(t – 25)(t – 49)–x^2(t – 9)(t – 25)(t – 49) –y^2(t – 1)(t – 25)(t – 49) – z^2(t–1)(t–9)(t–49) – w^2(t–1)(t–9)(t–25)$$
then I compared the coefficient with different value of $t$ .
I want to know that is there any easier alternative methods for this .
| Set up a matrix equation
$$\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34}\\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{bmatrix} = \begin{bmatrix} 1\\1\\1\\1\end{bmatrix}$$
and then solve. You can essentially ignore the fact that your variables are squared, and just assume that means they must all be nonnegative.
| {
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Fibonacci summation Can anyone help me to prove the following relation.
$$\sum_{k=1}^{\infty} \frac{F_{2k}H^{(2)}_{k-1}}{k^2\binom{2k}{k}}=\frac{2\pi^4}{375\sqrt{5}}$$
I was studying recently about Fibonacci and Lucas numbers.
And I came through the above relationship. I tried applying golden ratio but nothing works.
Symbols have their usual meanings.
| It is well known that $$\sum_{k=1}^{\infty}\frac{x^{2k}}{k^{2}\dbinom{2k}{k}}H_{k-1}^{\left(2\right)}=\frac{2\arcsin^{4}\left(\frac{x}{2}\right)}{3},\,\left|x\right|\leq2$$ (see here or here for a proof), then from the Binet formula we get $$\sum_{k=1}^{\infty}\frac{F_{2n}}{k^{2}\dbinom{2k}{k}}H_{k-1}^{\left(2\right)}=\frac{1}{\sqrt{5}}\sum_{k=1}^{\infty}\frac{\left(1+\sqrt{5}\right)^{2n}}{2^{2n}k^{2}\dbinom{2k}{k}}H_{k-1}^{\left(2\right)}-\frac{1}{\sqrt{5}}\sum_{k=1}^{\infty}\frac{\left(1-\sqrt{5}\right)^{2n}}{2^{2n}k^{2}\dbinom{2k}{k}}H_{k-1}^{\left(2\right)}$$ $$=\frac{2\arcsin^{4}\left(\frac{1+\sqrt{5}}{4}\right)}{3\sqrt{5}}-\frac{2\arcsin^{4}\left(\frac{1-\sqrt{5}}{4}\right)}{3\sqrt{5}}=\color{red}{\frac{2\pi^{4}}{375\sqrt{5}}}.$$
| {
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Factoring $x^4-11x^2y^2+y^4$ I am brushing up on my precalculus and was wondering how to factor the expression
$$
x^4-11x^2y^2+y^4
$$
Thanks for any help!
| For emphasis, if you have a quadratic or quadratic form, $a x^2 + bx + c$ or $a x^2 + bxy + c y^2,$it factors over the integers if and only if the discriminant from the Quadratic Formula, $b^2 - 4 a c,$ is an integer squared.
Things are rather different for quartics with only EVEN exponents, either
$f^2 x^4 + g x^2 + h^2$ or $f^2 x^4 + g x^2 y^2 + h^2 y^4$ to keep it easy. No worry about the discriminant, but the only thing that can work (once no rational roots) is, where $fh$ could be positive or negative,
$$ (fx^2 + \gamma xy + h y^2 ) (fx^2 - \gamma xy + h y^2 ) = f^2 x^4 + g x^2 y^2 + h^2 y^4. $$
Here we just need to be able to solve for $\gamma$ in
$$ 2fh - \gamma^2 = g, $$
or
$$ 2fh - g = \gamma^2. $$
To repeat, we can negate $fh$ at need, useful if $g$ is negative. We see this in the original question and the answer by 6005, where we can choose either $ 2 + 11 = 13$ or $-2 + 11 = 9.$
Favorites
$$ (x^2 + 2 xy + 2 y^2)(x^2 - 2xy + 2 y^2) = x^4 + 4 y^4, $$
$$ (x^2 + xy + y^2)(x^2 - xy + y^2) = x^4 + x^2 y^2 + y^4. $$
The latter gives a real analytic alternative to the superellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Find the number of solutions to $ \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345$
Find the number of solutions of the equation
$$\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,$$
where $\lfloor\,\cdot\,\rfloor$ represents the floor function.
My work:
I use the fact that
$$\lfloor nx \rfloor =\sum_{k=0}^{n-1} \left\lfloor x +\frac kn \right\rfloor.$$
So the equation becomes
$$\lfloor x \rfloor +\sum_{k=0}^{1} \left\lfloor x +\frac k2 \right\rfloor +\sum_{k=0}^{3} \left\lfloor x +\frac k4 \right\rfloor +\sum_{k=0}^{7} \left\lfloor x +\frac k8 \right\rfloor \\
\qquad {}+\sum_{k=0}^{15} \left\lfloor x +\frac k{16} \right\rfloor +\sum_{k=0}^{31} \left\lfloor x+\frac k{32} \right\rfloor \\
= \lfloor x \rfloor + \left\lfloor x+\frac 12 \right\rfloor + \left\lfloor x+\frac 64 \right\rfloor + \left\lfloor x+\frac{28}{8} \right\rfloor \\
\qquad {}+ \left\lfloor x+\frac{120}{16} \right\rfloor + \left\lfloor x+\frac{496}{32} \right\rfloor$$
What should I do next?
| Let
$$
f(x) = \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor
+ \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor.
$$
For integers $n$ we have $f(n)=63n$, but for a small number to the left of $n$ all terms $\lfloor x \rfloor, \lfloor 2x \rfloor,\dots$ decrease by one. So at every integer $n$ we have that $f$ jumps from $63n-6$ to $63n$. By calculation we have $$f(196)=12348,$$ so at $n=196$ we find that $f$ jumps from $12342$ to $12348$. Since $f$ is increasing we find that $f(x)=12345$ has no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem
What is the remainder when $10^{35}$ is divided by $7$?
I used the binomial formula: $\dfrac {(7+3)^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot 3^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot + 35 \cdot 3^{34} \cdot 7}{7} + \dfrac {3^{35}}{7}$
Therefore, $10^{35}$ will have the same remainder as $3^{35}$ when divided by $7$.
I was stuck here, and I wanted to try to reverse engineer the answer. I know the remainder is $5$. Therefore I should be able to write $3^{35} -5 + 5$ as $7k+5$. However, I have no idea how to factor out a $7$ from $3^{35}-5$, or from $10^{35}-5.$ How could I find this $k$ non-explicitly?
| You can use the fact that the remainder of $ab$ is the remaineder of $a$ multiplied by the remaineder of $b$. So forgeting factors of $7$, we have $3^{35}=27*3^{32}=6*(9)^{16}=6*2^{16}=6*(16)^4=6*2^4=6*2=5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Polynomial ,divides and Induction Proof?
$\text{The polynomial } x-y \;\text{divides the polynomial}\; x^2-y^2 \text{ and } x^3-y^3 \text{because}\; x^2-y^2 = (x+y)(x-y) \text{ and } x^3-y^3=(x-y)(x^2+xy+y^2.) \; \text{for every natural number n } \quad x-y \;\text{ divides }\; x^n-y^n \text{ prove by induction.}$
What I am confused about in this example is when one makes $n=k$. Which variable is made equal to k.Where is $n$. Is it fair to assume that what the actual problem wants you to infer this. $x^n +y^n = x^n+y^n$
(i) Basis Step: $x^n-y^n = x^n-y^n$
$x-y = x-y$
$0=0$
(ii) Inductive step : $x^k-y^k = x^k-y^k$
WNTS: RHS: $n =k+1$
$x^{k+1} + y^{k+1}$
LHS: $x^{k+1} + y^{k+1} +x^k-y^k $
Is this a fair assumption to make I feel as if their is something fundamentally missing from my logic. Any suggestions would be good.
| Let say ${ x }^{ n }-{ y }^{ n }$ is dividing by $x-y$ then we have $${ x }^{ n }-{ y }^{ n }=\left( x-y \right) P\left( x \right) $$,where $P(x)$ is some polynomial.Now we should prove that ${ x }^{ n+1 }-{ y }^{ n+1 }$ is also dividing by $x-y$
$${ x }^{ n+1 }-{ y }^{ n+1 }={ x }^{ n+1 }-{ x }^{ n }y+{ x }^{ n }y-{ y }^{ n+1 }=\\{ x }^{ n }\left( x-y \right) +y\underbrace { \left( { x }^{ n }-{ y }^{ n } \right) }_{ \left( x-y \right) P\left( x \right) } ={ x }^{ n }\left( x-y \right) +y\left( x-y \right) P\left( x \right) =\\ =\left( x-y \right) \left( { x }^{ n }+yP\left( x \right) \right) $$
so it means
$${ x }^{ n+1 }-{ y }^{ n+1 }=\left( x-y \right) \left( { x }^{ n }+yP\left( x \right) \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$1+2+3+45+6+78+9=144$ what are other combinations Note that $$1+2+3+45+6+78+9 = 144$$ In how many other ways is it possible to make a total of $144$ using only $1, 2, 3, 4, 5, 6, 7, 8,$ and $9$ in that order and addition signs?
Sorry I am only in high school so dont over complicate the explanation. Thank you
| Inspired by the idea: "1+2+..9=45 therefore we need 99 more to get 144. The two digits 'ab'=10a+b so changing a+b to 'ab' adds 9a to the total. Therefore an extra eleven 9s are required. This means that the the possibilites are (3,8) (6,4,1) etc."
We need find all the combinations that add up to 11. But in these combinations, the difference of any two numbers must >=2. Otherwise, we can not make these two 2-digit numbers at the same time, for example, 5+6=11, but we can not make numbers 56 and 67 at the same time, as 6 has been used twice. Also we can not use number 9, as after 9, there is no more number to make 2-digit number 9x.
The following are all the possible combinations:
4+7=11
3+8=11
1+3+7=11
1+4+6=11
Therefore all the correspondent answers are:
1+2+3+45+6+78+9
1+2+34+5+6+7+89
12+34+5+6+78+9
12+3+45+67+8+9
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $a + b + c$, given that $(a+1)^{1/2} - a + (b+2)^{1/2} \cdot 2 - b + (c+3)^{1/2} \cdot 3 - c = \frac{19}{2}.$
Let $a,b,c$ be real number such that $$(a+1)^{1/2} - a + (b+2)^{1/2} \cdot 2 - b + (c+3)^{1/2} \cdot 3 - c = \frac{19}{2}.$$
Find $a + b + c$.
The answer is: $-\frac{5}{2}$. Please give me some clues or solution.
Thanks.
| The function $(a+1)^{1/2}-a$ has derivative $\frac{1}{2(a+1)^{1/2}}-1$, so reaches a max of $5/4$ at $a=-3/4$.
A similar calculation shows that $2(b+2)^{1/2}-b$ reaches a max of $3$ at $b=-1$, and the function $3(c+3)^{1/2}$ reaches a max of $21/4$ at $c=-3/4$.
Add up. The sum is $19/2$. What a coincidence! Thus $a+b+c=(-3/4)+(-1)+(-3/4)$.
Another way: In hindsight, it could have been done less messily. For example, let $b+2=y^2$. Then we are looking at $2y-(y^2-2)$, which is maximized at $y=1$. Similarly with the others.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \frac{\sqrt{64x^2-256}}{x}\,dx$ QUESTION
Evaluate $$\int \frac{\sqrt{64x^2-256}}{x}\,dx$$
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
MY ATTEMPT
*
*Typed
$\newcommand{\dd}{\; \mathrm{d}}\int \frac{\sqrt{64x^2-256}}x \dd x \to
\int \frac{\sqrt{64(x^2-4)}}x \dd x \to
\int \frac{8\sqrt{x^2-4}}x \dd x$
Use $x=a\sec\theta$, $\dd x=a\sec\theta \tan\theta \dd \theta$.
$a=2$ $\to$ $x=2\sec\theta$, $\dd x=2\sec\theta \tan\theta \dd \theta$.
$=\int \frac{8\sqrt{4\sec^2\theta-4}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to
\int \frac{8\sqrt{4(\sec^2\theta-1)}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta $
$=\int \frac{8\sqrt{4\tan^2\theta}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to
\int \frac{8(2\tan\theta)}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta$
$=\int 16\tan^2\theta \dd \theta \to
16\int\tan^2\theta \dd \theta \to
\underset{\text{trig. formula}}{\underbrace{16(\theta+\tan\theta)+C}}$
$\Rightarrow 16(\tan\theta-\theta)+C = 16\tan\theta-16\theta$
$x=2\sec\theta$, $\sec\theta= \frac x2$
$\boxed{16\tan\left(\frac{\sqrt{x^2-4}}2\right) -16\sec^{-1}\left(\frac x2\right)+C}$
*
*Handwritten
| $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,\mathrm{Li}}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With the
sub$\ds{\ldots\ t \equiv x - \root{x^{2} - 4}\ \imp\
x = {t^{2} + 4 \over 2t}}$:
\begin{align}
&\color{#f00}{\int{\root{64x^{2} - 256} \over x}\,\dd x} =
8\int{\root{x^{2} - 4} \over x}\,\dd x =
8\int\pars{{8 \over t^{2} + 4} - {16 \over t^{2}} + 3}\,\dd t
\\[3mm] = &\
32\arctan\pars{t \over 2} + {128 \over t} + 24t
\\[3mm] = &\
32\arctan\pars{x - \root{x^{2} - 4} \over 2} + {128 \over x - \root{x^{2} - 4}} + 24\pars{x - \root{x^{2} - 4}}
\\[3mm] = &\
32\arctan\pars{x - \root{x^{2} - 4} \over 2} +
32\pars{x + \root{x^{2} - 4}} + 24\pars{x - \root{x^{2} - 4}}
\\[3mm] = &\
\color{#f00}{32\arctan\pars{x - \root{x^{2} - 4} \over 2} +
56x + 8\root{x^{2} - 4}} + \pars{~\mbox{a constant}~}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show that $\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}n+\mathcal{O}\left(\frac{1}{n^{3/2}}\right)$
How can i prove that $$\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}=\frac{(-1)^n}{n} +\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)\tag{$*$}$$
using the following method :
note that :
$(1+x)^{\alpha}=1+\alpha x+\mathcal{O}(x^{2})\quad ( x\to 0) $
\begin{align}
\frac{(-1)^n}{n+(-1)^n\sqrt{n+1}}&=\frac{(-1)^n}{n} \left(1+\frac{(-1)^n\sqrt{n+1}}{n} \right)^{-1}\\
&\text{since } \sqrt{n+1}\sim \sqrt{n} \text{ then } \lim_{n\to+\infty}\frac{\sqrt{n+1}}{n}=0 \\
&\sim \frac{(-1)^n}{n}\left(1+\frac{(-1)^n\sqrt{n}}{n} \right)^{-1}\\
&= \frac{(-1)^{n}}n\left(1+\mathcal{O}\left( \frac{(-1)^n\sqrt{n}}{n}\right) \right)\\
&= \frac{(-1)^n}{n}+\mathcal{O}\left( \dfrac{(-1)^{2n}\sqrt{n}}{n^2}\right) \\
&=\frac{(-1)^n}{n}+\mathcal{O}\left(\dfrac{1}{n^{3/2}}\right)
\end{align}
*
*AM i right ?
| I would initially ignore the
$(-1)^n$
and do
$\begin{array}\\
\frac{1}{n+(-1)^n\sqrt{n+1}}-\frac{1}{n}
&=\frac{n-(n+(-1)^n\sqrt{n+1})}{n(n+(-1)^n\sqrt{n+1})}\\
&=\frac{-(-1)^n\sqrt{n+1}}{n(n+(-1)^n\sqrt{n+1})}\\
&=\frac1{n^{3/2}}\frac{(-1)^{n+1}\sqrt{1+1/n}}{1+(-1)^n\sqrt{1/n+1/n^2})}\\
\text{so}\\
\big|\frac{1}{n+(-1)^n\sqrt{n+1}}-\frac{1}{n}\big|
&=\frac1{n^{3/2}}\frac{\sqrt{1+1/n}}{1+(-1)^n\sqrt{1/n+1/n^2})}\\
&=O(\frac1{n^{3/2}})\\
\end{array}
$
since
$\sqrt{1+1/n} \to 1$
and
$\sqrt{1/n+1/n^2} \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Show that $(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\tfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\tfrac{1}{n^{3/2}} \right)$
I would like to show that :
$$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$
by starting from the left side and get the right side
My Proof:
\begin{align*}
(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)&=(-1)^{n}\left(\dfrac{1}{\sqrt{n+1}+\sqrt{n} }\right)\\
&=(-1)^{n}\left(\sqrt{n+1}+\sqrt{n} \right)^{-1}\\
&=(-1)^{n}\left(\sqrt{n}\left(1+\sqrt{1+\dfrac{1}{n}}\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{\sqrt{n}}\left(1+\left(1+\dfrac{1}{n}\right)^{\dfrac{1}{2}} \right)^{-1}\\
&=\mbox{Note that :}\left(1+x\right)^{\alpha}=1+\mathcal{O}\left( x\right) \\
&=\dfrac{(-1)^{n}}{\sqrt{n}}\left(1+\left(1+\mathcal{O}\left(\dfrac{1}{n} \right)\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{\sqrt{n}}\left(2+\mathcal{O}\left(\dfrac{1}{n}\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\dfrac{1}{n}\right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\mathcal{O}\left(\dfrac{1}{n}\right)\right) \right)\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}\left(1+\mathcal{O}\left(\dfrac{1}{n}\right) \right)\\
&=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)\\
\end{align*}
$$\fbox{$(-1)^{n}\left(\sqrt{n+1}-\sqrt{n} \right)=\dfrac{(-1)^{n}}{2\sqrt{n}}+\mathcal{O}\left(\dfrac{1}{n^{3/2}} \right)$}$$
*
*Am i right ?
| Simply using series expansion will do:
\begin{align*}
\sqrt{n+1}-\sqrt{n} &= \sqrt{n} \left( \sqrt{1+\frac{1}{n}}-1 \right) \\
&=\sqrt{n} \left( 1+\frac{1}{2n}-\frac{1}{8n^2}+\ldots-1 \right) \\
&= \frac{1}{2\sqrt{n}}-\frac{1}{8n^{3/2}}+\ldots
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find all even numbers that can be represented as a difference of squares in only two ways I am currently working on this proof. I am looking to find (with proof) all even numbers that can be represented as a difference of squares in only two ways.
My thoughts thus far.
I examined the first 40 natural numbers and found that 16, 24, 32, 40 have exactly two representations as a difference of squares. My claim is that all numbers $x$ for which $x=8k+16$ will have exactly two representations.
$8k+16=(k+a)^2 - (k+b)^2$ $\rightarrow$ $8k+16=(k+4)^2 - k^2$ which is the first representation.
My idea for finding the second representation is to express $8k+16$ as
$8k+16=(ak+b)^2 - (ck+d)^2$
Expanding and attempting to solve I get:
$$\left.\begin{matrix}
a^2 - c^2 = 0\\
2ab - 2cd= 8\\
b^2 - d^2= 16
\end{matrix}\right\}$$
This is where my problem arises, $b^2-c^2=(b-c)(b+c)=16=4^2$, I cannot seem to move on from here.
If anyone can offer a hint that would help with my understanding of the problem and the attempt at the proof, that would be extremely helpful. Thank you.
|
My claim is that all numbers $x$ for which $x=8k+16$ will have exactly two representations.
Counterexample:
\begin{align}
120 &= 31^2-29^2 \\
&= 17^2 - 13^2 \\
&= 13^2-7^2 \\
&= 11^2-1^2
\end{align}
Note that while I just picked a number divisible by $8$ with plenty of factors, the claim actually falls at $48$:
\begin{align}
48 &= 13^2-11^2 \tag{from (12,1)}\\
&= 8^2 - 4^2 \tag{from (6,2)}\\
&= 7^2-1^2 \tag{from (4,3)}\\
\end{align}
because $48/4=12$ has three pairs of factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 5
} |
Linear combination issue I have 4 vectors:
$u_1 = \begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix} $, $\; u_2 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{pmatrix} $, $\; u_3 = \begin{pmatrix}
1 \\
1 \\
0 \\
\end{pmatrix} $, $\; u_4 = \begin{pmatrix}
0 \\
1 \\
0 \\
\end{pmatrix} $
and I wanna express the following vector in terms of them:
$\; v = \begin{pmatrix}
2 \\
3 \\
4 \\
\end{pmatrix} $
I'm working this way:
First put vectors in a matrix and then put in rref:
$ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
1&1&1&1&3\\
1&0&0&0&4
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&0&0&1&1\\
0&-1&-1&0&2
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&-1&-1&0&2\\
0&0&0&1&1
\end{array}
\right] $ =>
$ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&1&1&0&-2\\
0&0&0&1&1
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&1&1&0&2\\
0&1&1&0&-2\\
0&0&0&1&1
\end{array}
\right] $ => $ \left[
\begin{array}{cccc|c}
1&0&0&0&4\\
0&1&1&0&-2\\
0&0&0&1&1
\end{array}
\right] $
but this result (4 -2 1) is meaningless to me (because $4u_1 -2u_2 + u_3 \ne v$)... does it make any sense? Can it represent some sort of coeficients of the linear combination?
If I swap $u_3$ for $u_4$ in the matrix, the (4 -2 1) is the same but it DOES make sense, because $4u_1 -2u_2 + u_4 = v$
How can I write v as combination of u's ? If I did the right way, how does this make sense?
| Note that $(4,-2,1)$ by itself cannot be the right answer since you need four coefficients to give a linear combination of your four vectors.
Since your reduced matrix has a non-leading (non-pivot) column, the system will have infinitely many solutions. You could take the third coefficient to be zero, giving the solution $(4,-2,0,1)$ which is correct since
$$4u_1-2u_2+0u_3+u_4=v\ .$$
Or you could take the second to be zero giving $(4,0,-2,1)$, which is also correct since
$$4u_1+0u_2-2u_3+u_4=v\ .$$
The general solution of your system is
$$(4,-2-\alpha,\alpha,1)$$
where $\alpha$ is a scalar, and this is also correct for any $\alpha$ since
$$4u_1+(-2-\alpha)u_2+\alpha u_3+u_4=v\ .$$
Hope this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Differentiate and simplify. $m(x) = \frac{x}{\sqrt{4x-3}}$ My work so far is:
\begin{align}
m'(x) &= \frac{(1)(\sqrt{4x-3})-(x)(1/2)(4x-3)^{-1/2}(4)}{(\sqrt{4x-3})^2} \\
&= \frac{\sqrt{4x-3} - 2x(4x-3)^{1/2}}{4x-3}
\end{align}
and now I'm stuck on how to simplify further
| In the right hand side of your second line, you wrote
$(4x-3)^{1/2}$
Note the exponent here is incorrect: it should still be $(4x-3)^{-1/2}$. Let's write it that way and try to proceed.
$$m'(x)=\frac{\sqrt{4x-3}-2x(4x-3)^{-1/2}}{4x-3}=\frac{\sqrt{4x-3}-\frac{2x}{\sqrt{4x-3}}}{4x-3}$$
Now, in the numerator, we can simplify via finding a common denominator:
$$\sqrt{4x-3}-\frac{2x}{\sqrt{4x-3}}=\frac{4x-3-2x}{\sqrt{4x-3}}=\frac{2x-3}{\sqrt{4x-3}}$$
Putting this back in the numerator, we get:
$$m'(x)=\frac{\frac{2x-3}{\sqrt{4x-3}}}{4x-3}=\frac{2x-3}{(4x-3)^{3/2}}$$
This is as simplified as it's likely expected to be.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
In AB + BC + AC = N, how can I find all possibilities for A, B and C in less than n³ computational time? The problem is the one on the title. Given a N, find all possibilies for A, B and C that make this true: $AB+BC+AC = N$when $A \ge B \ge C$.
This code in C do the job:
int ans = 0;
for(int a=1; a<=N; a++){
for(int b=0; b<=a and a*b <= N; b++){
if((N-a*b)%(a+b) == 0 and (N-a*b)/(a+b) <= b) ans++;
}
}
To those who don't know C, this code does the following check:
If $N-AB$ is divisible by $A+B$ and $(N-AB)/(A+B) \le B$, then this pair of A and B is a solution to $AB+BC+AC = N$.
| $$AB+AC+BC=N\tag{1}$$
is equivalent to:
$$(2A+B+C)^2-4A^2-(C-B)^2 = 4N\tag{2}$$
hence we may simply look for the integers $q\geq\sqrt{4N}$ that ensure that $q^2-4N$ is the sum of two squares, i.e. is not divided with odd multiplicity by a prime $p\equiv 3\pmod{4}$. For instance, let $N=10$. If we take $q=7$ we have $q^2-4N = 0^2+3^2$, leading to $2A+B+C=7, A=0, C-B=3$ and to $(0,2,5)$ as a trivial solution of $(1)$. If we take $q=8$, we have $q^2-4N=24$ with no associated solutions, since $3\parallel 24$. If we take $q=9$, we have $q^2-4N=4^2+5^2$, leading to $2A+B+C=9, A=2, C-B=5$ and the same trivial solution as before. We may also notice that $(1)$ is equivalent to
$$ (A+C)(B+C) = N+C^2 \tag{3} $$
and design a similar algorithm by factoring $N+C^2$ as the product of two integers in the range $[C,2C]$. That is not possible if $N\geq 3C^2$, hence we may stop searching at $C=\sqrt{\frac{N}{3}}$.
In the previous case, $N=10$, we only have trivial solutions since $11=10+\left\lfloor\frac{10}{3}\right\rfloor$ is a prime.
Moreover, the last algorithm clearly has complexity $\color{red}{O(N^{3/2})}$: we may simply test every integer in the range $[C,2C]$ for being a divisor of $N+C^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Problem 1, Ch. 6 in Piskunov's, Differential and Integral calculus
Find the curvature of the curve at indicated points
$b^2x^2+a^2y^2=a^2b^2$ at $(0,b)$ and $(a,0)$
My attempt
$\displaystyle{\kappa=\frac{|\frac{d^2{y}}{dx^2}|}{\left[1+\left(\frac{dy}{dx}\right)^2\right]^\frac{3}{2}}}$
Differentiating the implicit equation with respect to $x$,
$2b^2x+2a^2yy'=0\\
y'=-\frac{b^2}{a^2}{\cdot}\frac{x}{y}$
Differentiating again with respect to x,
$y''=-\frac{b^2}{a^2}{\cdot}\frac{y-xy'}{y^2}$
a) At $(0,b)$
$y'=0\\
y''=-\frac{b^2}{a^2}{\cdot}\frac{b-(0)(0)}{b^2}=-\frac{b}{a^2}$
$\displaystyle{\kappa=\frac{|-\frac{b}{a^2}|}{\left[1+\left(0\right)^2\right]^\frac{3}{2}}=\frac{b}{a^2}}$
b) At $(a,0)$
$y'=\infty\\
y''=-\frac{b^2}{a^2}{\cdot}\frac{\frac{y}{y'}-x}{\frac{y^2}{y'}}=\infty$
I am not sure, if my solution to part (b) is correct. The book says, the answer must be $\frac{a}{b^2}$.
| Use chain rule for double differentiation.
I think you should multiply by dy/dx in the expression for y''.
For example,let y=1/x
So y"=2/x^3
But the way you have done would give y"=-2y=-2/x
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Maximum Likelihood Estimate of a a discrete r.d - I spent more than 4 hours on this questions, help!! Suppose $X$ is a discrete r.d with the following p.d.f:
$$
\begin{array}{c|lc}
X & \text{0} & \text{1} & \text{2} & \text{3} \\
\hline
p(x) & 2\theta/2 & \theta/3 & 2(1-\theta)/3 & (1-\theta)/3
\end{array}
$$
where $0\leq\theta\leq1$ was taken from such a distribution $(3,0,2,1,3,2,1,0,2,1)$. What is the MLE of $\theta$?
Likelihood function:
$$(2\theta/2)^2.(\theta/3)^3.(2(1-\theta)/3)^3.((1-\theta)/3)^2$$
Log-Likelihood function:
$$2\log(2\theta/2)+3\log(\theta/3)+3\log(2(1-\theta)/3)+2\log((1-\theta)/3)$$
Set derivative to zero:
$$2/\theta+3/\theta-6/(2-2\theta)-2/(1-\theta)=0$$
$$\frac{5-5\theta}{\theta}-\frac{6+6\theta}{2-2\theta}=2$$
$$5-5\theta=2\theta+\frac{6\theta+6\theta^2}{2-2\theta}$$
$$2-2\theta(5-7\theta)=6\theta+6\theta^2$$
Finally I got something like this:
$$10+8\theta^2=30\theta$$
Don't know how to find $\theta$. But I guess I have made mistake because the answer is $\theta=0.5$.
I spent more than 4 hours on this, tried different ways, like simplify the likelihood function before $"\log"$ them and still cannot work out, anyone help please?
| Your $p(x)$ is wrong: its sum is $4/3$, not $1$. And who would write $2\theta/2$ instead of $\theta$? I suspect that $2\theta/2$ should be $2\theta/3$ and $2(1-\theta)/2$ should be $2(1-\theta)/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help solving this recurrence relation I wanted to resolve the determinant of the next (nxn) matrix via recurrence relations:
$$
\begin{vmatrix}
a & 1 & 0 & 0 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0\\
1 & a & 1 & 0 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0 \\
0 & 1 & a & 1 & 0 & 0 &.... 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & a & 1 & 0 &.... 0 & 0 & 0 & 0 & 0\\
.. & .. & .. & .. & .. & .. &..... & .. & .. & ..\\
0 & 0 & 0 & 0 & 0 & 0 & .... 0 & 1 & a & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & .... 0 & 0 & 1 & a & 1\\
0 & 0 & 0 & 0 & 0 & 0 & .... 0 & 0 & 0 & 1 & a\\
\notag
\end{vmatrix}
$$
After analyzing the matrix I found the recurrence relation:
$$
D_{n}-a*D_{n-1} +D_{n-2}=0
$$
So the polynomial that describes this recurrence is:
$$
P(\lambda) = \lambda^2 - a * \lambda + 1
$$
The roots will be:
$$
\lambda_1 = \frac{a}{2} + \frac{\sqrt{a^2-4}}{2}\\
\lambda_2 = \frac{a}{2} - \frac{\sqrt{a^2-4}}{2}
$$
To resolve the recurrence I need 2 constants (C1 & C2) that satisfy:
$$
D_n = C_1*(\frac{a}{2} + \frac{\sqrt{a^2-4}}{2})^n + C_2*(\frac{a}{2} - \frac{\sqrt{a^2-4}}{2})^n
$$
With the initial conditions
$$
D_1 = a\\
D_2 = a^2 - 1
$$
The problem is I don't know how to resolve the equation system generated by substituting the initial conditions on the function.
Any type of help is appreciated.
| You plug your initial conditions in, so you get
$$D_1=a = C_1*(\frac{a}{2} + \frac{\sqrt{a^2-4}}{2}) + C_2*(\frac{a}{2} - \frac{\sqrt{a^2-4}}{2})\\D_2=a^2-1 = C_1*(\frac{a}{2} + \frac{\sqrt{a^2-4}}{2})^2 + C_2*(\frac{a}{2} - \frac{\sqrt{a^2-4}}{2})^2$$
These are two equations in the two unknowns $C_1,C_2$. The usual substitution technique will work, though it will be messy. You can use the recurrence to evaluate $D_0=1=C_1+C_2$. Using that with the first gives $$a=(C_1+C_2)\frac a2+(C_1-C_2)\frac{\sqrt{a^2-4}}{2}\\C_1-C_2=\frac a{\sqrt{a^2-4}}\\C_1=\frac 12(1+\frac a{\sqrt{a^2-4}})\\C_2=\frac 12(1-\frac a{\sqrt{a^2-4}})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\prod\limits_{i=1}^{n}(x-4i+2)(x-4i+1)>\prod\limits_{i=1}^{n}(x-4i+3)(x-4i)$ for all $x\in\mathbb{R}$? I would like to prove that for $n\in\mathbb{N}$ we have $f_n(x):=\prod\limits_{r=1}^{n}(x-4r+2)(x-4r+1)>\prod\limits_{r=1}^{n}(x-4r+3)(x-4r)=:g_n(x)$ for all $x\in\mathbb{R}$ (actually it would suffice for $n$ even). My attempt was to form pairs on each side in such a way that we always obtain the same function plus a constant, i.e. for $n=2k$ even:
$$
f_n(x)=(x-2)(x-8k+1)\cdot(x-3)(x-4n+2)\cdot...\cdot(x-4k-2)(x-4k+1)=\\((x-4k-0.5)^2-(4k-1.5)^2)\cdot((x-4k-0.5)^2-(4k-2.5)^2)\cdot...\cdot((x-4k-0.5)^2-(1.5)^2)
$$
And similarly for $g_n(x)$. If we then substitute $x=0.5y+4k+0.5$ we see that the inequality is equivalent to:
$$
(y^2-3^2)(y^2-5^2)\cdot...\cdot(y^2-(2k-5)^2)(y^2-(2k-3)^2)>(y^2-1^2)(y^2-7^2)\cdot...\cdot(y^2-(2k-7)^2)(y^2-(2k-1)^2)
$$
for all $y\in\mathbb{R}$. I think now it would be enough to expand both sides and probably we would end up with a polynomial in $y^2$ where all coefficients are positive. However, computing all the coefficients will be quite tedious. How to prove it more elegantly?
Edit:
As several comments and answers use this same idea: simply reducing it to $(x-4r+2)(x-4r+1)=(x-4r)^2+3(x-4r)+2>(x-4r)^2+3(x-4r)=(x-4r+3)(x-4r)$ doesn't work as $a>c,b>d$ doesn't imply $ab>cd$ in general. But it was my first reflex too :)
| Let $a_r(x)=(x-(4r-2))(x-(4r-3)), b_r(x)=(x-(4r-1))(x-4r),
c_r(x)=\frac{a_r(x)}{b_r(x)}$,
and $A_n(x)=\prod_{r=1}^{n}a_r(x),B_n(x)=\prod_{r=1}^{n}b_r(x)$.
We wish to show that $A_n(x)>B_n(x)$ for any $n$ and $x$.
Note that when $n=1$, the result follows from $A_1-B_1=2$. Suppose now that
$n\geq 2$. For $k\in[|1,n|]$, let $I_k=[4k-2,4k-1], J_k=[4k-3,4k]$.
If $x$ is outside all the $J_k(1\leq k \leq n)$, then all the $a_r(x),b_r(x),c_r(x)$ are positive. From the $n=1$ case, we have $a_r>b_r$, whence $c_r>1$
and $\frac{A_n}{B_n}=\prod_{r=1}^n c_r >1$, so we are done.
If $x$ is in some $J_k\setminus I_k(1\leq k \leq n)$, then $A_n\geq 0 \geq B_n$
and $A_n$ and $B_n$ are not both zero, so we are done.
The only case left is therefore when $x$ is in some $I_k(1\leq k \leq n)$. And on that interval, $A_n$ and $B_n$ are both negative ; our job is then to show that
$\frac{A_n}{B_n}=\prod_{r=1}^n c_r$ is $\leq 1$ on $I_k$.
It is easy to see that the derivative of $c_r$ vanishes at $4r-\frac{3}{2}$ and
nowhere else. It follows that $c_r$ attains its maximum value on $I_k$ at a point
$m_k$ defined by
$$
m_k=\left\lbrace\begin{array}{lcl}
4k-2 & \text{when} & r \lt k, \\
4k-\frac{3}{2} & \text{when} & r=k,\\
4k-1 & \text{when} & r \gt k. \\
\end{array}\right.
$$
So
$$
\begin{array}{lcl}
\prod_{r=1}^n c_r
& = & \bigg(\prod_{r=1}^{k-1} c_r \bigg) c_k \bigg( \prod_{r=k+1}^{n} c_r \bigg) \\
&\leq & \bigg(\prod_{r=1}^{k-1} c_r(4k-2) \bigg) c_k(4k-\frac{3}{2}) \bigg( \prod_{r=k+1}^{n} c_r(4k-1) \bigg) \\
&\leq & \bigg(\prod_{t=1}^{k-1} c_{k-t}(4k-2) \bigg) c_k(4k-\frac{3}{2}) \bigg( \prod_{t=1}^{n} c_{k+t}(4k-1) \bigg) \\
& = & \bigg(\prod_{t=1}^{k-1} \frac{8t^2-2t}{8t^2-2t-1}\bigg) \frac{1}{9} \bigg( \prod_{t=1}^{n} \frac{8t^2-2t}{8t^2-2t-1} \bigg) \\
& \leq & \frac{1}{9} \bigg( \prod_{t=1}^{\infty} \frac{8t^2-2t}{8t^2-2t-1} \bigg)^2 \
( \ \text{because} \ \frac{8t^2-2t}{8t^2-2t-1} >1 ) \\
& \leq & \frac{1}{9} \bigg( \prod_{t=1}^{\infty} \frac{t^2+2t+1}{t^2+2t} \bigg)^2 \
( \ \text{because} \ 8t^2-2t-1\geq t^2+2t) \\
& = & \frac{1}{9} \bigg( \prod_{t=1}^{\infty} \frac{t+1}{t}\frac{t+1}{t+2} \bigg)^2 \\
& \leq & \frac{4}{9} \ ( \ \text{telescoping product} \ )
\\
\\
\end{array}
$$
This concludes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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I was trying to find out the intervals where $\sin ^{-1}x > \cos ^{-1}x$ I was trying to find out the intervals where $\sin ^{-1}x > \cos ^{-1}x$
The easiest way was to just look at the graph and I found out that the region is $x \in ({1\over \sqrt{2}} , 1]$
But I tried to prove the statement algebraically also but couldn't get it correctly.
For $$\sin ^{-1}x > \cos ^{-1}x$$
$$\Rightarrow \sin ^{-1}x -\cos ^{-1}x>0$$
$$\Rightarrow \sin ^{-1}x - \sin ^{-1} \sqrt{1-x^2}>0 \tag1$$
Using the identity $$\sin ^{-1}x -\sin ^{-1}y=\sin ^{-1} \left (x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)$$ Equation $(1)$ can be written as
$$\Rightarrow \sin ^{-1}(2x^2-1)>0$$
For this to be true $0<2x^2-1<1$ and also for the equation to be valid $-1<2x^2-1<1$
Hence we can take
$0<2x^2-1<1$ as the intersection of both the conditions
$$\Rightarrow 0\le 2x^2-1\le 1$$
$$\Rightarrow 1\le 2x^2\le 2$$
$$\Rightarrow 1/2\le x^2\le 1$$
The solution set of the inequality would be $x \in ({1\over \sqrt{2}} , 1] \cup [-1,{-1\over \sqrt{2}}),$
Can anybody tell me why I am getting the wrong answer.
| About your method you have to note that, unfortunately, $\arccos x=\arcsin\sqrt{1-x^2}$ is a false identity in general; it's true only for $0\le x\le 1$.
However, for $-1\le x<0$ we have $-\pi/2\le\arcsin x<0$ and $\pi/2<\arccos x\le\pi$, so we can exclude this interval, where the inequality certainly doesn't hold.
So we have
$$
\begin{cases}
\arcsin x>\arcsin\sqrt{1-x^2} \\[4px]
0\le x\le 1
\end{cases}
$$
that becomes, since the arcsine is increasing
$$
\begin{cases}
x>\sqrt{1-x^2} \\[4px]
0\le x\le 1
\end{cases}
$$
Due to the second limitation, we can square the first inequality
$$
\begin{cases}
x^2>1-x^2 \\[4px]
0\le x\le 1
\end{cases}
$$
and easily get $1/\sqrt{2} < x \le 1$.
On the other hand, if $\alpha=\arccos x$, we have $0\le\alpha\le\pi$, so $-\pi/2\le\pi/2-\alpha\le\pi/2$ and from $x=\cos\alpha$ we get $x=\sin(\pi/2-\alpha)$, so
$$
\frac{\pi}{2}-\alpha=\arcsin x
$$
and therefore
$$
\arccos x=\frac{\pi}{2}-\arcsin x
$$
which makes your inequality very easy to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Fourier transform property(uniformly converges) proof Suppose that f is a 2π-periodic function that satisfies the estimate
\begin{equation}
|f(x)-f(y)|\leqslant M|x-y|^\alpha
\end{equation}
for an 0< $\alpha$ <1
Show that $S_n(x)$ converges uniformly to f (x) for all real x.
\begin{equation}
S_n(x)=\int_0^{2\pi} f(y)\frac{sin((N+\frac{1}{2})(x-y))}{2\pi ~sin(\frac{1}{2}(x-y))} dy.
\end{equation}
My confuse is, $S_n(x)$ should $\longrightarrow$ 0 when n $\longrightarrow$ $\infty$ by using Riemann–Lebesgue lemma. So what mistake that I made?
| $$
f(x)-S_n(x) = \frac{1}{2\pi}\int_{x-\pi}^{x+\pi}\left[\frac{f(x)-f(y)}{\sin{\frac{1}{2}(x-y)}}\right]\sin((n+1/2)(x-y))dy
$$
Because $|f(x)-f(y)| \le M|x-y|^{\alpha}$, the expression enclosed by square brackets is absolutely integrable. So $S_n(x)\rightarrow f(x)$ for all $x$ by the Riemann-Lebesgue Lemma. The uniformity of the Lipshitz condition leads to uniform convergence.
To show how the above gives uniform convergence of $S_n$ to $f$, let $\epsilon > 0$ be given. Then, for small enough $\delta$, the following estimate holds independent of $x$:
$$
\left|\frac{1}{2\pi}\int_{x-\delta}^{x+\delta}\frac{f(x)-f(y)}{\sin(\frac{1}{2}(x-y))}\sin((n+1/2)(x-y))dy\right| \\
\le \frac{1}{2\pi}\int_{x-\delta}^{x+\delta}M\frac{|x-y|^{\alpha}}{|\sin\frac{1}{2}(x-y)|}dy \\
\le \frac{2M}{2\pi}\int_{x-\delta}^{x+\delta}|x-y|^{\alpha-1}dy \\
= \frac{4M}{2\pi}\int_{0}^{\delta}t^{\alpha-1}dt = \frac{4M}{2\pi\alpha}\delta
$$
So, by choosing $\delta$ small enough, the above is bounded by $\frac{\epsilon}{2}$ for all $x$. To bound the remaining piece by $\epsilon/2$ for large $n$, choose a function $g$ that is $2\pi$ periodic, continuously differentiable and satisfies
$$
\frac{1}{2\pi\sin(\delta/2)}\int_{\delta \le |y-x| \le \pi}|f(y)-g(y)|dy < \frac{\epsilon}{6} \tag{$\dagger$}
$$
Then
\begin{align}
&\left|\frac{1}{2\pi}\int_{\delta \le |x-y| \le \pi}\left[\frac{f(x)-f(y)}{\sin\frac{1}{2}(x-y)}\right]\sin((n+1/2)(x-y))dy\right| \\
& \le \left|f(x)\frac{1}{2\pi}\int_{\delta\le|u| \le \pi}\frac{\sin((n+1/2)u}{\sin\frac{1}{2}u}du\right| \\
& + \left|\frac{1}{2\pi}\int_{\delta \le |x-y|\le \pi}(g(y)-f(y))\frac{\sin((n+1/2)(x-y))}{\sin\frac{1}{2}(x-y)}dy\right| \\
& + \left|\frac{1}{2\pi}\int_{\delta \le |x-y| \le \pi}\frac{g(y)}{\sin\frac{1}{2}(x-y)}\sin((n+1/2)(x-y)dy\right|
\end{align}
The function $f$ is uniformly bounded because of the Lipschitz condition. Therefore, by the Riemann-Lebesgue lemma, there exists $N_1$ large enough that the first term on the right is bounded uniformly in $x$ by $\frac{\epsilon}{6}$ for $n \ge N_1$. The second term is automatically bounded by $\epsilon/6$ because of $(\dagger)$. The third term can be uniformly bounded for large $N$ by $\epsilon/6$ after a single integration by parts, which is permitted because $g$ is a smooth approximating function:
$$
\frac{1}{2\pi}\int_{\delta \le |x-y| \le \pi}\frac{g(y)}{\sin\frac{1}{2}(x-y)}\sin((n+1/2)(x-y)dy. \tag{$\dagger\dagger$}
$$
(Integrating the $\sin$ term gives $1/(n+1/2)$, leading to a uniform bound on the above for large enough $n$.) For example, the integral over $\delta < |x-y| \le \pi$ is over $y \in [x+\delta,x+\pi]$ and $y\in [x-\pi,x-\delta]$. The part of the integral over $[x+\delta,x+\pi]$ becomes
\begin{align}
&\int_{x+\delta}^{x+\pi}\frac{g(y)}{\sin\frac{1}{2}(x-y)}\sin((n+1/2)(x-y))dy\\
= &-\left.\frac{g(y)}{\sin\frac{1}{2}(x-y)}\frac{\cos((n+1/2)(x-y))}{(n+1/2))}\right|_{y=x+\delta}^{x+\pi} \\
+&\frac{1}{n+1/2}\int_{x+\delta}^{x+\pi}\frac{\partial}{\partial y}\left(\frac{g(y)}{\sin\frac{1}{2}(x-y)}\right)\cos((n+1/2)(x-y))dy
\end{align}
Now you can see that $(\dagger\dagger)$ tends to $0$ as $n\rightarrow\infty$. The evaluation terms tend to $0$ because of the presence of $1/(n+1/2)$; the integral term on the right tends to $0$ even without the $1/(n+1/2)$. And the convergence is uniform in $x$. Using a smooth $g$ to approximate $f$ in the integral sense simplifies the argument.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Why are the limits different? Consider $\lim_{ x \to -\infty} \sqrt{x^2-x+1}+x$
Rationalising, one will get, $\lim_{x \to -\infty} \frac{1-x}{\sqrt{x^2-x+1}-x}$, which after taking x common and cancelling out gives $-\infty$.
Now, replace $x$ by $-x$, so the limit becomes, $\lim_{x \to \infty} \frac{1+x}{\sqrt{x^2+x+1}+x}$, which evaluates to $1/2$. In the book, the answer given is $1/2$, I want to know, why did we use this change of variable to evaluate limit? and, why is the first method wrong?
| \begin{align*}
\lim_{x \to -\infty} \left(\sqrt{x^2 - x + 1} + x\right) & = \lim_{x \to -\infty} \left(\sqrt{x^2 - x + 1} + x\right) \cdot \frac{\sqrt{x^2 - x + 1} - x}{\sqrt{x^2 - x + 1} - x}\\
& = \lim_{x \to -\infty} \frac{x^2 - x + 1 - x^2}{\sqrt{x^2 - x + 1} - x}\\
& = \lim_{x \to -\infty} \frac{-x + 1}{\sqrt{x^2 - x + 1} - x}\\
& = \lim_{x \to -\infty} \frac{-x + 1}{|x|\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} - x}\\
& = \lim_{x \to -\infty} \frac{-x + 1}{-x\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} - x}\\
& = \lim_{x \to -\infty} \frac{1 - \frac{1}{x}}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + 1}\\
& = \frac{1}{2}
\end{align*}
where we have used the fact that $\sqrt{x^2} = |x| = -x$ if $x < 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution of Inequality $ \frac{1}{x-6}\le 3$
Solve the inequality: $\displaystyle \frac{1}{x-6}\le 3$
solution:
\begin{align*}\frac{1}{x-6}& \le 3 \\ x-6& \le \frac{1}{3}
\\x& \le 6+\frac{1}{3}\\
x&\le19/3\end{align*}
but, for values of $x\le 6$ also, inequality holds true as left hand side provides some negative value. but, i don't find this from the solution. what am i missing?
| Find the critical values: The values where the domain is undefined and when the function $f(x)= \frac{1}{x-6} -3$ is equal to $0$.
Here, the critical values are $6, \frac{19}{3} $
Use these values to divide the $x-axis$ into intervals: $(-\infty, 6), (6, \frac{19}{3}), (\frac{19}{3}, \infty) $
Test each intervals by plugging in the number within the interval and see if the value is lesser or equal than 3. If it is, then it's answer. If not, then you can eliminate that interval.
For example: $(-\infty, 6]$ pick $x=0$ then $ \frac{1}{0-6} = \frac{1}{-6} < 3$ so this is an answer.
$[6, \frac{19}{3}]$, pick $x= \frac{19}{3} $, then $\frac{1}{\frac{19}{3}-6} = 3$ So this is not an answer
Similarly, $[\frac{19}{3}, \infty) $, pick $x=7$, you can see that $\frac{1}{7-6} =1 < 3 $. Thus, this is an answer!
Therefore, the answer are the intervals: $(-\infty, 6)$ and $(\frac{19}{3}, \infty) $
Another way to do this is by graphing! Graph the function $f(x) = \frac{1}{x-6}$. This is the graph of the funcion $f(x) = \frac{1}{x}$ shifted right by 6 units. From the graph, you can see where the values are below the line $y=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Considering the complex number $z = m+i$ for which values of $m$ do we have $ \left|\overline{z}+\frac{2}{z}\right| \ge 1 $ Good evening to everyone. I have the following problem that I tried to solve but my mathematical instinct tells me that I didn't solve it right:
Considering the complex number $z = m+i$ for which values of $m$ do we have: $$ \left|\overline{z}+\frac{2}{z}\right| \ge 1 $$
Here's what I've tried: $$ \left|\overline{z}+\frac{2}{z}\right| \ge 1 \rightarrow \left|m-i+\frac{2}{m+i}\right|\ge 1 \rightarrow \left|\frac{m^2-1}{m+i}\right|\ge 1 \rightarrow -1\le \frac{m^2-1}{m+i}\le 1 \rightarrow\begin {cases} -1\le m^2-1\le 1 \\ -1\le m+i\le 1 \end{cases}\rightarrow \begin{cases} m \in [-2,2] \\ m \in [-1-i,1-i] \end{cases} $$
| you have mistakes ,it should be $$\left| \overline { z } +\frac { 2 }{ z } \right| \ge 1\rightarrow \left| m-i+\frac { 2 }{ m+i } \right| \ge 1\rightarrow \left| \frac { m^{ 2 }+3 }{ m+i } \right| \ge 1\rightarrow \frac { \left| { m }^{ 2 }+3 \right| }{ \sqrt { { m }^{ 2 }+1 } } \ge 1\\ { m }^{ 2 }+3\ge \sqrt { { m }^{ 2 }+1 } \Rightarrow { m }^{ 4 }+5{ m }^{ 2 }+8\ge 0 $$ which true for all real $m$
| {
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"url": "https://math.stackexchange.com/questions/1869818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does this $6\times 6$ matrix has a null determinant? about this matrix:
$$
\left(
\begin{array}{ c c c c c c}
1& 0& 0& \mathit{1}& 0& 0\\
0.5& 0.5& 0& 0.5& \mathit{0.5}& 0 \\
0.5& 0& 0.5& 0.5& 0& \mathit{0.5} \\
\mathit{0.25}& 0.5& 0.25& 0& 1& 0 \\
0 & \mathit{0.5}& 0.5& 0& 0.5& 0.5 \\
0& 0& \mathit{1}& 0& 0& 1\\
\end{array}
\right)
$$
I have calculated its determinant by two techniques and they all give 0.
I do not understand why, because:
*
*There is no entire row equal to zero
*Two rows or columns are not equal
*No row either column is a constant multiple of another row or column (or I cannot see)
*There is diagonal line (noted in italic) without any 0
Does someone has a idea why?
thanks.
| Looking at the first row, I notice that there are exactly two columns with an $1$ there, column 1 and column 4. Therefore my first thought was to take the difference of those:
$$\begin{pmatrix} 1\\0.5\\0.5\\0.25\\0\\0 \end{pmatrix}
- \begin{pmatrix} 1\\0.5\\0.5\\0\\0\\0 \end{pmatrix}
= \begin{pmatrix} 0\\0\\0\\0.25\\0\\0 \end{pmatrix}$$
Only a single entry remains. That's a good start! So let's see if we can get the same result from the other columns as well. Columns 2 and 5 indeed work:
$$\begin{pmatrix} 0\\0.5\\0\\0.5\\0.5\\0 \end{pmatrix}
- \begin{pmatrix} 0\\0.5\\0\\1\\0.5\\0 \end{pmatrix}
= \begin{pmatrix} 0\\0\\0\\-0.5\\0\\0 \end{pmatrix}$$
So now it's obvious that 2 times the first difference plus the second difference gives zero, therefore the determinant vanishes.
| {
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"url": "https://math.stackexchange.com/questions/1872345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$ I am working on the following exercise. Solve the equation $1-x+x^{2}-x^{3}+x^{4}=y^{4}$ in $\mathbb{Z}$.
I have a couple of ideas for going about this exercise.
$1)$ By moving $1$ to the other side of the equation we obtain:
$y^4-1=x^4-x^3+x^2-x \rightarrow (y^2-1)(y^2+1)=x(x-1)(x^2+1)$.
The LHS gives two consecutive integers I am able to see, but other than that I am stuck and not sure where to go from here.
$2)$ We notice that $1-x+x^2-x^3+x^4=\Phi_{10}(x)=y^4$.
We now have that $\Phi_{10}(x) \geq 0$ since $y=\pm \sqrt[4]{1-x+x^2-x^3+x^4}$.
In this case, with some computation, $x={0,1}$ are the only possibilities that will force $y \in \mathbb{Z}$. From there, I use induction to show that $x \geq 2$ and $x \leq -1$ do not yield a perfect fourth.
In either case, I keep running into some issues. If someone could please other a hint to help me continue through this exercise, that would be very helpful. Thank you.
Update:
Hello, my progress on the following problem is as follows:
Suppose that $x=y$, then we get $y^3-y^2+y-1=0 \rightarrow y=1, \pm i$. Thus we get two solutions $(x,y)={(1,1),(1,-1)}$, the $y=-1$ since $y^4$ was in the original equation.
Next, using @Batominovski hint we can write:
$(2x^2-x)^2 < (x^4-x^3+x^2-x+1) < 4(x^4-x^3+x^2-x+1) \leq (2x^2-x+2)^2 \forall x \in \mathbb{R}$ which is true for $x=0$.
Since the multiplication of a constant (in this case $k=4$) does not affect a solution from existing, we have that $x=0$ is a solution, consequently we find two more solutions are $(x,y)=(0,1),(0,-1)$.
| Hint: note that for $x$ big enough the following holds:
$$(x-1)^4 < x^4-x^3+x^2-x+1 < x^4.$$
Similarly, for $x$ small enough we have
$$(x-1)^4 > x^4-x^3+x^2-x+1 > x^4.$$
Therefore you are left with a finite number of possible values of $x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to determine the convergence behavior of the sequence $x_{n}=\sqrt{n^{2}+11n+21}-\sqrt{n^{2}+6}$? How to determine the convergence behavior of the sequence $x_{n}=\sqrt{n^{2}+11n+21}-\sqrt{n^{2}+6}$ and its limit?
My effort: I don't know where to begin.
| Hint: multiply and divide $x_n$ by $\sqrt{n^2+11n+21}+\sqrt{n^2+6}$
$\sqrt{n^2+11n+21}-\sqrt{n^2+6}=
(\sqrt{n^2+11n+21}-\sqrt{n^2+6})({\sqrt{n^2+11n+21}+\sqrt{n^2+6}\over {\sqrt{n^2+11n+21}+\sqrt{n^2+6}}}$
$={{11n+15}\over{\sqrt{n^2+11n+21}+\sqrt{n^2+6}}}$. So the limit is $11/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit
Prove $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit.
My question is, I've picked $\delta\lt1$, and I've found that $\delta \lt \min(1,\sqrt{\epsilon})$. Was picking $1$ problematic at all? and is my choice for $\delta$ correct?
Rest of Proof:
$$0\lt|x-1|\lt \delta \Rightarrow \left|\frac{(x-1)^2}{x+1}\right|\lt \epsilon$$
Picking $\delta \lt 1$:
$$|x-1|\lt 1 \Rightarrow -1\lt x-1 \lt 1$$
And we get from that $\frac13 \lt \frac{1}{x+1} \lt 1$ which leads to $\left|\frac{1}{x+1}\right| \lt 1$
Let's go back:
$$\left|\frac{(x-1)^2}{x+1}\right|\lt \left|1(x-1)^2\right|\lt \epsilon$$
Since $(x-1)^2\gt 0$ we can get rid of the absolute value
and we get
$$(x-1)^2\lt \epsilon \rightarrow x-1 \lt \sqrt{\epsilon}$$
Also: What is the difference between picking $\delta=1$ and $\delta \lt 1$
| Here is a more "gimmicky" kind of solution (Battani's is definitely more elegant and natural): We find a positive constant $C$ such that $\left|\frac{x-1}{x+1}\right|<C\Rightarrow |x-1|\left|\frac{x-1}{x+1}\right|<C|x-1|$, and we can make $C|x-1|<\epsilon$ by taking $|x-1|<\frac{\epsilon}{C}=\delta$. We restrict $x$ to lie in the interval $|x-1|<1$ and note the following:
\begin{align}
|x-1|<1&\implies 0<x<2\\[1em]
&\implies 1<x+1<3\\[1em]
&\implies 1>\frac{1}{x+1}>\frac{1}{3}\\[1em]
&\implies 3>\frac{x-1}{x+1}\\[1em]
&\implies C=3.
\end{align}
Thus, we should choose $\delta=\min\left\{1,\frac{\epsilon}{3}\right\}$. To see that this choice of $\delta$ works, consider the following: Given $\epsilon>0$, we let $\delta=\min\left\{1,\frac{\epsilon}{3}\right\}$. If $|x-1|<1$, then $\left|\frac{x-1}{x+1}\right|<3$. Also, $|x-1|<\frac{\epsilon}{3}$. Hence,
$$
\left|\frac{x^2+3}{x+1}-2\right|=|x-1|\left|\frac{x-1}{x+1}\right|<\frac{\epsilon}{3}\cdot3=\epsilon,
$$
as desired. $\blacksquare$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to evaluate the following integral: $\int_0^1\frac{\log(1+x)}{1+x^2}dx$ I would like to evaluate:
$$\displaystyle\int_0^1 \frac{\log(1+x)}{1+x^2}\, dx$$
I have tried to evaluate this using integration by parts but failed. How can I evaluate it?
| The previously presented solution is perhaps the simplest. Here is my favorite.
We define a function $$f(t) = \int^1_0 \frac{\ln (xt+1)}{x^2+1} \text{ d}x.$$ The goal is to evaluate $f(1)$. We notice that $f(0)=0$. Differentiating $f$ with respect to $t$ (and differentiating under the integral; allowable by Leibniz's rule) gives, $$f'(t) = \int^1_0 \frac{x}{(xt+1)(x^2+1)} \text{ d}t.$$ Using partial fractions, we see $$\frac{x}{(xt+1)(x^2+1)} = \frac{A}{xt+1} + \frac{Bx+C}{x^2+1} \,\,\, \implies \,\,\, x = A(x^2+1) + (Bx+C)(xt+1) $$ $$ \implies \,\,\, A+Bt = 0 \,\,\, , \,\,\, B+ Ct = 1 \,\,\, , \,\,\, A + C = 1.$$ This system is uniquely solved by $$A = \frac{-t}{t^2+1} \,\,\, , \,\,\, B = \frac{1}{t^2+1} \,\,\, , \,\,\, C = \frac{t}{t^2+1}.$$ Then \begin{align*} f'(t) &= \int^1_0 \left\{ \frac{A(t)}{xt+1} + \frac{xB(t)}{x^2+1} + \frac{C(t)}{x^2+1} \right \} \text{ d}x \\ &= \left. \left\{ \frac{A(t)}{t} \ln (xt+1) + \frac{B(t)}{2} \ln(x^2+1) + C(t) \tan^{-1}(x) \right\} \right|_{x=0}^{x=1} \\ &= -\frac{\ln(t+1)}{t^2+1} + \frac{\ln(2)}{2}\frac{1}{t^2+1} + \frac{\pi}{4} \frac{t}{t^2+1}.\end{align*} Integrating from $t=0$ to $t=1$ yields $$f(1) = -f(1) + \frac{\ln(2)}{2} \frac{\pi}{4} + \frac{\ln(2)}{2}\frac{\pi}{4}$$ $$\implies \,\, \boxed{f(1) = \frac{\pi \ln(2)}{8}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluation of given limit when $f(x)=\sum^{n}_{k=1} \frac{1}{\sin 2^kx}$ and $g(x)=f(x)+\frac{1}{\tan 2^nx}$
Question Statement:-
If $\displaystyle f(x)=\sum^{n}_{k=1} \frac{1}{\sin 2^kx}$ and $g(x)=f(x)+\dfrac{1}{\tan 2^nx}$, then find the value of
$$\lim_{x\to 0} \bigg( (\cos x)^{g(x)}+\bigg(\frac{1}{\cos x} \bigg)^{\frac{1}{\sin x}} \bigg)$$
I am not able to find value of $g(x)$. Could someone help me as how to calculate value of $g(x)?$
| The following holds:
$$g(x)=\sum^{n}_{k=1} \frac{1}{\sin 2^kx}+\dfrac{1}{\tan 2^nx},$$
$$g(x)=\sum^{n-1}_{k=1} \frac{1}{\sin 2^kx}+(\frac{1}{\sin 2^nx}+\dfrac{1}{\tan 2^nx}),$$
$$g(x)=\sum^{n-1}_{k=1} \frac{1}{\sin 2^kx}+\dfrac{1}{\tan 2^{n-1}x},$$
$$g(x)=\sum^{n-2}_{k=1} \frac{1}{\sin 2^kx}+(\dfrac{1}{\sin{2^{n-1}x}}+\dfrac{1}{\tan 2^{n-1}x}),$$
$$g(x)=\sum^{n-2}_{k=1} \frac{1}{\sin 2^kx}+\dfrac{1}{\tan 2^{n-2}x},$$
$$\vdots$$
$$g(x)=\frac{1}{\sin{2x}}+\frac{1}{\tan{2x}}=\tan{x}.$$
Therefore, we need to calculate $$\lim_{x\to 0} \bigg( (\cos x)^{\tan{x}}+\bigg(\frac{1}{\cos x} \bigg)^{\frac{1}{\sin x}} \bigg).$$
Since
$$(\cos x)^{\tan{x}}+\bigg(\frac{1}{\cos x} \bigg)^{\frac{1}{\sin x}}=e^{\tan{x}\ln{\cos{x}}}+e^{-\frac{1}{\sin{x}}\ln{\cos{x}}}$$
$$(\cos x)^{\tan{x}}+\bigg(\frac{1}{\cos x} \bigg)^{\frac{1}{\sin x}}=e^{\tan{x}\ln{(1+2\sin^2{\frac{x}{2}}})}+e^{-\frac{1}{\sin{x}}\ln{(1+2\sin^2{\frac{x}{2}}})}$$
and $$\ln{(1+2\sin^2{\frac{x}{2}}})\sim 2\sin^2{\frac{x}{2}},$$
we have
$$\lim_{x\to 0} \bigg( (\cos x)^{\tan{x}}+\bigg(\frac{1}{\cos x} \bigg)^{\frac{1}{\sin x}} \bigg)=\lim_{x\to 0} \bigg ( e^{\tan{x} *2\sin^2{\frac{x}{2}}} + e^{-\frac{1}{\sin{x}}*2\sin^2{\frac{x}{2}}} \bigg)$$
$$\lim_{x\to 0} \bigg( (\cos x)^{\tan{x}}+\bigg(\frac{1}{\cos x} \bigg)^{\frac{1}{\sin x}} \bigg)=\lim_{x\to 0} \bigg ( e^{\tan{x} *2\sin^2{\frac{x}{2}}} + e^{-\tan{\frac{x}{2}}} \bigg)=e^0+e^0=2$$
| {
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"url": "https://math.stackexchange.com/questions/1876867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find all rational solutions to $x^2+y^2=2$
Find all rational solutions to $x^2+y^2=2$
I rewrote the equation to $x= \sqrt{2-y^2}$ and thought that $x$ is rational if and only if $2-y^2$ is a square. So the only solution to the first problem is $x=1$ or $-1$ and $y=1$ or $-1$.
Is there another approach? Thank you.
Edit: Another question: Is there maybe a way to show this using elliptic curves?
| Given any rational $$ u^2 + v^2 = 1,$$ we find
$$ (u-v)^2 + (u+v)^2 = 2 $$
Meanwhile, given any integer Pythagorean triple $$ a^2 + b^2 = c^2, $$ we get rational
$$ \left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 = 1 $$
From the comments above, I guess I should add that all primitive solutions in integers to $x^2 + y^2 = 2 z^2,$ since this means $x \equiv y \pmod 2$ and $z$ odd, so we get
$$ \left( \frac{x-y}{2} \right)^2 + \left( \frac{x+y}{2} \right)^2 = z^2 $$
in integers and primitive. We can find all of these with the tandard parametrization for primitive Pythagorean triples.
| {
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"url": "https://math.stackexchange.com/questions/1877138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove inequality $\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$
Let $x,y,z>0$ and $x+y+z=1$. Prove that
$$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x\ge\frac{x^2+y^2+z^2+1}2$$
My work so far:
I use Titu's Lemma:
$$\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x=$$
$$=\frac{x^4}{xy}+\frac{y^4}{yz}+\frac{z^4}{zx}+\frac{x^2}{\frac zx}+\frac{z^2}{\frac yz}+\frac{y^2}{\frac xy}\ge$$
$$\ge\frac{\left(x^2+y^2+z^2+x+y+z\right)^2}{xy+yz+zx+ \frac zx+\frac yz+\frac xy}=\frac{\left(x^2+y^2+z^2+1\right)^2}{xy+yz+zx+ \frac zx+\frac yz+\frac xy}$$
I need help here.
| $$S=\frac{x^3}y+\frac{y^3}z+\frac{z^3}x+\frac{x^3}z+\frac{z^3}y+\frac{y^3}x=\frac{x^4+y^4}{xy}+\frac{y^4+z^4}{yz}+\frac{z^4+x^4}{zx},$$
$$S=\frac{(x^2+y^2)^2-2x^2y^2}{xy}+\frac{(y^2+z^2)^2-2y^2z^2}{yz}+\frac{(z^2+x^2)^2-2z^2x^2}{zx}.$$
Since $(a^2+b^2)^2=(a^2+b^2)(a^2+b^2)\geq 2ab(a^2+b^2)$ by AM-GM, we have
$$S\geq 4(x^2+y^2+z^2)-2(xy+yz+zx),$$
$$S\geq\frac{7}{2}(x^2+y^2+z^2)-2(xy+yz+zx)+\frac{x^2+y^2+z^2}{2}.$$
Since $$2(x^2+y^2+z^2)\geq 2(xy+yz+zx)$$ holds (because it is equivalent with $(x-y)^2+(y-z)^2+(z-x)^2$),
we have
$$S\geq\frac{3}{2}(x^2+y^2+z^2)+\frac{x^2+y^2+z^2}{2}.$$
Also, $$3(x^2+y^2+z^2)\geq (x+y+z)^2$$ by quadratic-arithmetic inequality.
Therefore,
$$S\geq\frac{1}{2}(x+y+z)^2+\frac{x^2+y^2+z^2}{2}$$
$$S\geq\frac{1}{2}+\frac{x^2+y^2+z^2}{2}=\frac{x^2+y^2+z^2+1}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Computing $\lim_{n \to \infty} \left(\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\cdots+\frac{n}{(n+n)^2}\right)$
Calculate $$\lim_{n \to \infty} \left(\dfrac{n}{(n+1)^2}+\dfrac{n}{(n+2)^2}+\cdots+\dfrac{n}{(n+n)^2}\right).$$
I tried turning this into a Riemann sum, but didn't see how since we get $\dfrac{1}{n} \cdot \dfrac{n^2}{(n+k)^2}$, which I don't see how relates.
| Just another way to do it, probably just for your curiosity.
$$\sum_{i=1}^n \frac 1{(n+i)^2}=\psi ^{(1)}(n+1)-\psi ^{(1)}(2 n+1)$$ where appears the first derivative of the digamma function.
Since, for large values of $x$ (see the Wikipedia page)
$$\psi^{(0)}(x) = \log(x) -\frac{1}{2 x}-\frac{1}{12
x^2}+O\left(\frac{1}{x^3}\right)$$
$$\psi^{(1)}(x) =\frac{1}{x}+\frac{1}{2 x^2}+\frac{1}{6
x^3}+O\left(\frac{1}{x^4}\right)$$ $$\psi ^{(1)}(n+1)-\psi ^{(1)}(2 n+1)=\frac{1}{2 n}-\frac{3}{8 n^2}+\frac{7}{48
n^3}+O\left(\frac{1}{n^4}\right)$$ $$\sum_{i=1}^n \frac n{(n+i)^2}=n (\psi ^{(1)}(n+1)-\psi ^{(1)}(2 n+1))=\frac{1}{2}-\frac{3}{8 n}+\frac{7}{48 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.
For illustartion purposes, if $n=10$, the exact value of the summation is $$\frac{502856614213805}{1083847519827072}\approx 0.4639551$$ while the above formula gives $$\frac{2227}{4800}\approx 0.4639583$$
| {
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"answer_id": 2
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The image of a circle under $ z \mapsto \frac{z- \frac{1}{2}}{\frac{1}{2}z-1} $ Let $C_1 = \{ z: |z| = \frac{1}{7} \}$ be a circle inside the unit circle $C_0 = \{ z: |z|=1\}$. The fractional linear transformation:
$$ \phi: z \mapsto \frac{z- \frac{1}{2}}{\frac{1}{2}z-1} $$
is a map from the unit cirle to itself, $\phi: C_0 \to C_0$. What is the image of $C_1$ ? Obviously it is a circle, but what is the radius and center of $\phi(C_1)$?
I hope this figure not too confusing. $\phi$ maps the light blue circle to the dark blue circle. Similarly it maps light green to dark green.
In regard to 3 points... I know there is formula for 3 points $(x_1, y_1), (x_2, y_2),(x_3, y_3)$ going through a circle using a determinant but I have not used it:
$$ \left|\begin{array}{cccc} x^2 + y^2 & x & y & 1 \\
x_1^2 + y_1^2 & x_1 & y_1 & 1 \\
x_2^2 + y_2^2 & x_2 & y_2 & 1 \\
x_3^2 + y_3^2 & x_3 & y_3 & 1 \\ \end{array} \right|=0$$
Which 3 points should I pick anyway? Even more alternatively, are there ways to solve this using power of a point from geometry?
I suspect @vvnitram's answer is wrong. Here is a plot of the image circle (computed numerically) and the circle he proposes. These are not quite the same.
| Since at this point I am solving my own question, let me note that:
$$ \phi: \frac{1}{7} \mapsto \frac{ \frac{1}{7}- \frac{1}{2}}{\frac{1}{2}\times\frac{1}{7}-1}, \hspace{0.25in}
-\frac{1}{7} \mapsto \frac{ -\frac{1}{7}- \frac{1}{2}}{-\frac{1}{2}\times\frac{1}{7}-1}, \hspace{0.25in}
\frac{i}{7} \mapsto \frac{ \frac{i}{7}- \frac{1}{2}}{\frac{1}{2}\times\frac{i}{7}-1}
$$
Those are fine and then I have to solve the determinant equation:
$$ \left|\begin{array}{cccc} x^2 + y^2 & x & y & 1 \\
x_1^2 + y_1^2 & x_1 & y_1 & 1 \\
x_2^2 + y_2^2 & x_2 & y_2 & 1 \\
x_3^2 + y_3^2 & x_3 & y_3 & 1 \\ \end{array} \right|=0 $$
Here's re-write in terms complex numbers $z = x+iy$ and $z = x - iy$. I did not to the change of basis, I just am guessing:
$$ \left|\begin{array}{rlll} |z|^2 & z & \overline{z} & 1 \\
|z_1|^2 & z_1 & \overline{z}_1 & 1 \\
|z_2|^2 & z_2 & \overline{z}_2 & 1 \\
|z_3|^2 & z_3 & \overline{z}_3 & 1 \\ \end{array} \right|=0 $$
In my case, $z_1 = \phi(\frac{1}{7})$ and $z_1 = \phi(-\frac{1}{7})$ and $z_1 = \phi(\frac{i}{7})$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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another inequality $\frac{1}{xy+z}+\frac{1}{yz+x}+\frac{1}{zx+y}\le\frac{1}{2}$
Let $x,y,z>0$ and such $xyz\ge 2+x+y+z$, show that
$$\dfrac{1}{xy+z}+\dfrac{1}{yz+x}+\dfrac{1}{zx+y}\le\dfrac{1}{2}\tag{1}$$
I have posted this problem. Unfortunately, it doesn't hold, but I guess $(1)$ may hold.
The theory basis of speculation Use the following classical results
$$xyz=2+x+y+z\Longrightarrow xy+yz+zx\ge 2(x+y+z)$$
| I agree with you. Your inequality is true!
Indeed, the condition gives $\sum\limits_{cyc}\frac{1}{x+1}\leq1$.
Let $x=a$, $y=b$ and $z=kc$, where $k>0$ and $\sum\limits_{cyc}\frac{1}{a+1}=1$. Hence,
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{kc+1}$$
which gives $k\geq1$.
Thus, $\sum\limits_{cyc}\frac{1}{xy+z}=\frac{1}{ab+kc}+\frac{1}{kac+b}+\frac{1}{kbc+a}\leq\frac{1}{ab+c}+\frac{1}{ac+b}+\frac{1}{bc+a}$.
Id est, it remains to prove that $\sum\limits_{cyc}\frac{1}{ab+c}\leq\frac{1}{2}$ for positives $a$, $b$ and $c$ such that $\sum\limits_{cyc}\frac{1}{a+1}=1$.
Let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y}$, where $x$, $y$ and $z$ are positive numbers.
Hence, $c=\frac{x+y}{z}$ and we need to prove that $\sum\limits_{cyc}\frac{1}{\frac{x+y}{z}\cdot\frac{x+z}{y}+\frac{y+z}{x}}\leq\frac{1}{2}$ or
$$\sum\limits_{cyc}\frac{xyz}{x(x+y)(x+z)+yz(y+z)}\leq\frac{1}{2}$$
or
$$\sum\limits_{cyc}\frac{xyz}{(x+y+z)(x^2+yz)}\leq\frac{1}{2}$$
or
$$\sum\limits_{cyc}\left(\frac{x}{2}-\frac{xyz}{x^2+yz}\right)\geq0$$
or
$$\sum\limits_{cyc}\frac{x(x^2-yz)}{x^2+yz}\geq0$$
or
$$\sum\limits_{cyc}\frac{x((x-y)(x+z)-(z-x)(x+y))}{x^2+yz}\geq0$$
or
$$\sum\limits_{cyc}(x-y)\left(\frac{x(x+z)}{x^2+yz}-\frac{y(y+z)}{y^2+xz}\right)\geq0$$
or
$$\sum\limits_{cyc}z(x-y)^2(x^2+y^2+xz+yz)(z^2+xy)\geq0$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$
Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$
I tried assuming it was equal to $\frac p q $.
I get $2^nq^2-q^2 = p^2 $
But I don't see where to go from there.
| To continue your idea, rather than to restart and do one of the other correct answers.
$2^nq^2 - q^2 = p^2$.
Assume $p = 1$
The $2^nq^2 - q^2 = 1$. Then $2^n - 1 = 1/q^2$ is an integer. So $q=1$. So $2^n - 1 = 1$. So $2^n = 2$ so $n = 1 \not > 1$.
Assume $p \ne 1$.
Let $k$ be a prime factor of $p$. Then either $k|q^2$ which isn't possible as we assumed (or should have assumed) $p/q$ is in lowest terms, or $k|2^n - 1$.
This is true for all prime factors so $p^2|2^n -1$.
So $q^2\frac{2^n - 1}{p^2} = 1$ and $\frac{2^n - 1}{p^2} \in \mathbb Z$ so $\frac{2^n - 1}{p^2} = 1/q^2 \in \mathbb Z$ so $q = \pm 1$ and
$2^n - 1 = p^2$.
$p$ is clearly odd so let $p = 2p' + 1$
$2^n = 4p'^2 + 4p' + 2$
$4 \not \mid 4p'^2 + 4p +2 = 2^n$ so $n < 2$. So $n \le 1$. So $n \not > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square
Prove that there exist infinitely many positive integers $n$ such that $\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square. Obviously, $1$ is the least integer having this property. Find the next two least integers with this property.
The given condition is equivalent to $(2n+2)(2n+1) = 12p^2$ where $p$ is a positive integer. Then since $\gcd(2n+2,2n+1) = 1$, we have that $2n+2 = 4k_1$ and $2n+1 = k_2$. We must have that $k_1$ is divisible by $3$ or that $k_2$ is divisible by $3$. If $k_1$ is divisible by $3$ and $k_2$ is not, then we must have that $k_1$ is divisible by $9$ and so $2n+2 = 36m$. Then we need $3mk_2$ to be a perfect square where $k_2+1 = 36m$. Thus if $3mk_2 = r^2$, we get $m = \dfrac{1}{72}\left(\sqrt{48r^2+1}+1\right)$.
| This can be solved by using the quadratic formula on $2n^2 + 3n + (1-6p^2)=0$ to convert to the Pell's equation $r^2 - 48p^2 = 1$. Standard techniques can be used to generate infinitely many solutions, though we need to be careful to keep only those with $r \equiv 3\pmod{4}$.
For example, we can arrive at infinitely many solutions by writing $r+p\sqrt{48} = (7+\sqrt{48})^{2k+1}$, beginning from the primitive solution $r=7, p=1$.
Here is a proof that results from that process:
If $(n,p)$ is a positive solution to $(n+1)(2n+1)=6p^2$, then $(97n+168p+72,56n+97p+42)$ is a larger solution, which we can see by computing directly:
$$((97n+168p+72) + 1)(2(97n+168p+72) + 1) - 6(56n+97p+42)^2$$
$$ = (n+1)(2n+1)-6p^2$$
Beginning from the solution $(n,p)=(1,1)$, this generates infinitely many solutions by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Solving the Integral $\int_0^{\infty} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) \, \mathrm{cos}(b \, x) \,\mathrm{d}x$ Is there any possibily to solve the following integral
$$\int_0^{\infty} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) \, \mathrm{cos}(b \, x) \,\mathrm{d}x$$
with $a>0$, $y>0$ and $-\pi/2<\mathrm{arg}(b)<0$.
I assume, the result is connected to Bessel and Struve functions. Thank you.
Edit:
Using integration by parts with
$$\int\mathrm{cos}(b \, x)\,\mathrm{d}x=\frac{\mathrm{sin}(b \, x)}{b}$$
$$\frac{\mathrm{d}}{\mathrm{d}x}( \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right))= - \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}}$$
and the limits
$$ \lim_{x\to0} \frac{\mathrm{sin}(b \, x)}{b} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) = 0$$
$$ \lim_{x\to\infty} \frac{\mathrm{sin}(b \, x)}{b} \, \mathrm{arcsinh} \left(\frac{a}{\sqrt{x^2+y^2}} \right) = 0 $$
Gives the integral
$$\int_0^{\infty} \, \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}} \, \frac{\mathrm{sin}(b \, x)}{b} \,\mathrm{d}x$$
if that makes anything simpler...
Edit2:
Mathematica tells me that the last integrand can be presented as the product of three G-functions. Inhere it is said that the integral of the product of three G-functions can be computed under certain restrictions. Sadly it is not mentioned which restrictions. Does anybody know anything about this?
It would be:
$$\frac{1}{x^2+y^2+a^2} = \frac{1}{\sqrt{\pi} \, \sqrt{y^2+a^2}} \,\mathrm{MeijerG}\left[\left\{\{\tfrac{1}{2} \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2+a^2}\right]$$
$$\frac{1}{x^2+y^2} = \frac{1}{y^2} \,\mathrm{MeijerG}\left[\left\{\{0 \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2}\right]$$
$$\mathrm{sin}(b\,x)= \sqrt{\pi} \, \mathrm{MeijerG}\left[\left\{\{ \},\{ \} \right\},\left\{\{\tfrac{1}{2} \},\{ 0\} \right\},\tfrac{x^2 \, b^2}{4}\right]$$
which finally results in
$\frac{a}{y^2 \, \sqrt{y^2+a^2}} \, \int_0^{\infty} \, x \, \mathrm{MeijerG}\left[\left\{\{\tfrac{1}{2} \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2+a^2}\right] \, \mathrm{MeijerG}\left[\left\{\{0 \},\{ \} \right\},\left\{\{0 \},\{ \} \right\},\tfrac{x^2}{y^2}\right] \, \mathrm{MeijerG}\left[\left\{\{ \},\{ \} \right\},\left\{\{\tfrac{1}{2} \},\{0 \} \right\},\tfrac{x^2 \, b^2}{4}\right] \, \mathrm{d}x$
The $\mathrm{MeijerG}$ are defined according to Mathematica syntax. Or (I hope I converted this correctly)
$$\frac{a}{y^2 \, \sqrt{y^2+a^2}} \, \int_0^{\infty} \, x \, G^{1,1}_{1,1}\left(\begin{array}{c|c}\begin{matrix}\frac{1}{2}\\ 0 \end{matrix}&\frac{x^2}{y^2+a^2}\end{array}\right) \, G^{1,1}_{1,1}\left(\begin{array}{c|c}\begin{matrix}0\\ 0 \end{matrix}&\frac{x^2}{y^2}\end{array}\right)\, G^{1,0}_{0,2}\left(\begin{array}{c|c}\begin{matrix}-\\\frac{1}{2},\, 0\end{matrix}&\frac{x^2 \,b^2}{4}\end{array}\right) \, \mathrm{d}x$$
| First, we simplify the parameters:
$$\int_0^{\infty} \, \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}} \, \frac{\mathrm{sin}(b \, x)}{b} \,\mathrm{d}x= \frac{a}{by} \int_0^\infty \frac{t \sin rt ~dt}{(1+t^2) \sqrt{c^2+t^2}}$$
$$c^2=1+\frac{a^2}{y^2}, \qquad r=b y$$
I assume $b$ is real for this calculation (which may not be the case for the OP).
So we want to find:
$$f(r,c)=\int_0^\infty \frac{t \sin rt ~dt}{(1+t^2) \sqrt{c^2+t^2}}$$
For $c=1$ (which means $a=0$), we have a closed form as a modified Bessel function:
$$f(r,1)=r K_0(r)$$
This gives the following idea. Write:
$$\sqrt{c^2+t^2}=\sqrt{1+\frac{a^2}{y^2}+t^2}=\sqrt{1+t^2} \sqrt{1+\frac{a^2}{y^2(1+t^2)}}$$
And expand the latter as a binomial series. Then we have:
$$f=\sum_{k=0}^\infty \frac{(-1)^k (2k)!}{k!^2} \left( \frac{a}{2y} \right)^{2k} \int_0^\infty \frac{t \sin rt ~dt}{(1+t^2)^{3/2+k} }$$
Amazingly enough all the integral have a closed form:
$$\int_0^\infty \frac{t \sin rt ~dt}{(1+t^2)^{3/2+k} }=\frac{r^{k+1}}{(2k+1)!!} K_k (r)$$
Here $!!$ means double factorial: $(2k+1)!!=1 \cdot 3 \cdot 5 \cdots (2k+1)$.
Finally we have for the original integral:
$$\int_0^{\infty} \, \frac{a \,x}{(x^2+y^2) \, \sqrt{x^2+y^2+a^2}} \, \frac{\mathrm{sin}(b \, x)}{b} \,\mathrm{d}x= a \sum_{k=0}^\infty \frac{(-1)^k (2k)!}{k!^2 (2k+1)!!} \left( \frac{a }{2}\sqrt{\frac{b }{y}} \right)^{2k} K_k (b y)$$
We could also attempt residues on the integral:
$$\int_{-\infty}^\infty \frac{t \sin rt ~dt}{(1+t^2) \sqrt{c^2+t^2}}$$
but it has a square root and so will involve the branch cuts, I'm not sure how to deal with it.
| {
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"url": "https://math.stackexchange.com/questions/1888943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the smallest value of $x$ such that the matrix $A$ is not invertible?
$$A = \begin{pmatrix}
-1 & -3 & x \\
-2 & -7 & -1 \\
x & -6 & 1
\end{pmatrix}$$
So far I tried taking the determinant by expanding down the last column but i still cant get the right answer.
x * det $ \begin{pmatrix}
-2 & -7 \\x & -6
\end{pmatrix} $ + 1 * det $ \begin{pmatrix}
-1 & -3 \\
x & -6
\end{pmatrix} $ +1*det $ \begin{pmatrix}
-1 & -3 \\
-2 & -7
\end{pmatrix} $
I end up with $7x^2+15x+7$$=$$0$
| You're on the right way continue by finding the roots
$x_1=-\frac{15}{14} - \frac{\sqrt{29}}{14}$
$x_2=-\frac{15}{14} + \frac{\sqrt{29}}{14}$
And taking the $\min\{x_1,x_2\} = -\frac{15}{14} - \frac{\sqrt{29}}{14}$
That's it that is your answer
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $\tan a = \frac{1}{7}$ and $\sin b = \frac{1}{\sqrt{10}}$, show $a+2b = \frac{\pi}{4}$.
Given
$$\tan a = \frac{1}{7} \qquad\text{and}\qquad \sin b = \frac{1}{\sqrt{10}}$$
$$a,b \in (0,\frac{\pi}{2})$$
Show that $$a+2b=\frac{\pi}{4}$$
Does exist any faster method of proving that, other than expanding $\sin{(a+2b)}$?
Thank You!
| $\sin b = \dfrac{1}{\sqrt{10}}$, $b\in(0,\pi/2)$ $\;\Rightarrow\;$ $b\in(0,\pi/6)$, i.e. $2b\in (0,\pi/3)$, since $1/\sqrt{10}<1/2$, and
$$\sin 2b = 2\sin b \cos b = 2 \dfrac{1}{\sqrt{10}} \dfrac{3}{\sqrt{10}} = \dfrac{3}{5}.$$
Then
$$
\tan 2b = \dfrac{\sin 2b}{\cos 2b} = \dfrac{3/5}{4/5} = \dfrac{3}{4}.
$$
Now use formula
$$
\tan(\alpha+\beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta},
$$
$$
\tan(a+2b) = \dfrac{\tan a + \tan 2b}{1-\tan a \tan 2b} = \dfrac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7}\cdot\frac{3}{4}} = \dfrac{4+21}{28-3}=\dfrac{25}{25}=1.
$$
Then , in general, $a+2b = \pi/4+\pi k$, $k \in \mathbb{Z}$.
But since $a$ and $2b$ are acute, then $a+2b=\pi/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to compute the sum $ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$ Could it be possible to find the solution for the following series?
$$ 1+a(1+b)+a^2(1+b+b^2)+a^3(1+b+b^2+b^3)+\cdots$$
Thanks in advance!
| A slight variant that requires no explicit cancelling of $1-b$ (so you needn't handle the $b=1$ case with a continuity argument) rearranges the double summation, viz. $$\sum_{n=0}^\infty\left(a^n\sum_{k=0}^n b^k\right)=\sum_{k=0}^\infty b^k\sum_{n=k}^\infty a^n=\sum_{k=0}^\infty b^ka^k\frac{1}{1-a}=\frac{1}{\left( 1-a\right)\left( 1-ab\right)}$$ provided $\left| a\right| < 1$ and $\left| ab\right| <1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Box contains 2 different coins, one is chosen randomly and tossed n times, head came all n times. Probability that n+1 toss is a head too? A box contains two coins: a regular coin and one fake two-headed coin (P(H)=1). One coin is choose at random and tossed $n$ times.
If the first n coin tosses result in heads, What is the probability that the $(n+1)^{th}$ coin toss will also result in heads?
My solution:
$$\text{Required Probability} = \frac{1}{2}\times \left(\frac{1}{2}\right)^n \times \frac{1}{2}+\frac{1}{2} \times 1^n \times 1$$
| First we find the probability that the coin is fair.
To do this we use Bayes theorem, we want $P(A|B)$, where $A$ is the probability the coin is fair and $B$ is the probability the first $n$ flips are heads.
By Bayes theorem:
$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$, the numerator is $\frac{1}{2^{n}}\times \frac{1}{2}$ and the denominator is $\frac{1}{2}+\frac{1}{2^n}\times \frac{1}{2}=\frac{2^n+1}{2^{n+1}}$. hence $P(A|B)=\frac{1}{2^{n}+1}$.
Therefore the final answer is:
$\color {green}{\frac{1}{2^n+1}\times \frac{1}{2}}+\color{\purple}{(1-\frac{1}{2^n+1})}=1-\frac{1}{2(2^n+1)}$
The green part is the probability that it is heads and that the coin is fair, and the purple part is the probability that it is heads and the coin is fake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893263",
"timestamp": "2023-03-29T00:00:00",
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$2^n > n^4$ proof by induction This is what I came up with so far:
Inductive step: assume $2^n > n^4$.
Need to prove $2^{n+1} > (n+1)^4$
$$
2^{n+1} = 2 \cdot 2^n > 2 \cdot n^4\\
(2 \cdot n^4)^{1/4} = (2)^{1/4} \cdot n > n+1 \implies 2n^4 > (n+1)^4 \implies 2^n > (n+1)^4
$$
Is there a better way to solve this problem?
| Here is a nice little non-induction proof that $n^4\lt2^n$ for $n\gt16$.
Note that
$$n^4=\left(\sqrt n\over2\right)^8\cdot2^8\cdot1^{n-16}$$
So if $n\gt16$, the arithmetic-geometric mean inequality tells us
$$\sqrt[n]{n^4}\lt{8\cdot\left(\sqrt n\over2\right)+8\cdot2+(n-16)\cdot1\over n}={4\over\sqrt n}+1\lt2$$
hence $n^4\lt2^n$.
(Remark: for $n=16$ both inequalities in the displayed expression become equalities. In particular, all $16$ terms in the AGM product $\left(\sqrt4\over2\right)^8\cdot2^8$ are equal to $2$. For $n\lt16$, the AGM connection falls apart, as it must, since the inequality goes the other way.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Matrix Inverse and Adjoint Method I have the following matrix:
$A = \begin{bmatrix}2 & 1 & 1\\ 1 & 2 & 1\\1 & 1 & 2 \end{bmatrix}$
If I use the Excel MINVERSE function it returns:
$A^{-1} = \begin {bmatrix}0.75 & -0.25 & -0.25\\ -0.25 & 0.75 & -0.25 \\ -0.25 & -0.25 & 0.75 \end{bmatrix}$
However if I calculate the minors and cofactors then $|A| = 10$. Using the Adjoint method yields:
$\frac{1}{10}\begin{bmatrix}5 & -3 & 3\\ -3 & 5 & -3\\ 3 & -3 & 5 \end{bmatrix}$ = $\begin{bmatrix}0.5 & -0.3 & 0.3\\ -0.3 & 0.5 & -0.3\\ 0.3 & -0.3 & 0.5 \end{bmatrix}$
Am I making a mistake in the calculation to derive the Adjoint or is this related to the fact that the matrix A is symmetrical?
| Determinant:
$det(A) = (2\cdot2\cdot 2 + 1\cdot1\cdot1 + 1\cdot1\cdot 1) - (1\cdot2\cdot1 + 1\cdot1\cdot2 + 2\cdot1\cdot1) = (8+1+1)-(2+2+2) = 10-6 = 4$
Cofactor Matrix:
$C_{11} = C_{22} = C_{33} = (-1)^{1+1}\,(2\cdot2 - 1\cdot1) = 3$
$C_{12} = C_{21} = C_{32} = C_{23} = (-1)^{1+2}\,(2\cdot1 - 1\cdot1) = -1$
$C_{13} = C_{31} = (-1)^{1+3}\,(1\cdot1 - 1\cdot2) = -1$
$C = \begin{bmatrix}3&-1&-1\\-1&3&-1\\-1&-1&3\end{bmatrix}$
$adj(A) = C^T = C$
Inverse Matrix:
$A^{-1} = \frac{1}{det(A)}\,adj(A) = \frac{1}{4}\begin{bmatrix}3&-1&-1\\-1&3&-1\\-1&-1&3\end{bmatrix} = \begin{bmatrix}0.75&-0.25&-0.25\\-0.25&0.75&-0.25\\-0.25&-0.25&0.75\end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $7$ Divides $\binom{2^n}{2}+1$ then $n =3k+2$ for some positive integer $k$ It is a straightforward question.
If $$ 7 \text{ }\Bigg | \binom{2^n}{2}+1$$ then $n=3k+2$ for some
positive integer $k$.
This is just curiosity no motivation just rummaging through some old question in a notebook. A simple counter example would work.
Note if $\mathfrak a(n) =\binom{2^n}{2}+1$ then $7$ divides $\mathfrak a(n)$ for $n=2,5,8,11,14,17,\ldots$ and from this I am guessing that $n=3k+2$
| $7|\binom{2^n}{2}+1\iff \binom{2^n}{2}\equiv -1 \bmod 7\iff 2^n(2^{n}-1)2^{-1}\equiv -1 \bmod 7\iff $
$2^n(2^n-1)\equiv -2\equiv 5\bmod 7$.
Now notice that $2^n\bmod 7$ repeats with period $3$, so $2^n(2^n-1)$ also repeats with period three.
In particular it goes: $2,5,0,2,5,0\dots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.