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Help with $\int \frac{\sqrt{x^2 + 1}}{x}\:dx$ I have been starting at the following integral for the past 2 hours and can not see where I have gone wrong. Can anyone please assist in isolating where I've made a mistake.
My sanity thanks you in advance,
\begin{equation*}
I = \int \frac{\sqrt{x^2 + 1}}{x}\:dx
\end{equation*}
Let $x = \tan(s)$:
\begin{equation*}
\frac{dx}{ds} = \sec^2(s) \rightarrow dx = \sec^2(s)\:ds
\end{equation*}
Thus,
\begin{align*}
I &= \int \frac{\sqrt{x^2 + 1}}{x}\:dx = \int \frac{\sqrt{\tan^2(s) + 1}}{\tan(s)}\cdot \sec^2(s)\:ds = \int \frac{\sec(s)}{\tan(s)}\sec^2(s)\:ds = \int \frac{\sec^3(s)}{\tan(s)}\:ds \\
&= \int \frac{\frac{1}{\cos^3(s)}}{\frac{\sin(s)}{\cos(s)}}\:ds = \int \frac{1}{\cos^3(s)} \cdot \frac{\cos(s)}{\sin(s)}\:ds = \int \frac{1}{\cos^2(s)\sin(s)}\:ds = \int \frac{1}{\left(1 - \sin^2(s)\right)\sin(s)}\:ds \\
&= \int \frac{1}{\left(1 + \sin(s)\right)\left(1 - \sin(s)\right)\sin(s)}\:ds
\end{align*}
Applying a Partial Fraction Decomposition we see:
\begin{equation*}
\frac{1}{\left(1 + \sin(s)\right)\left(1 - \sin(s)\right)\sin(s)} = \frac{1}{\sin(s)} - \frac{1}{2}\cdot \frac{1}{1 + \sin(s)} - \frac{1}{2}\cdot \frac{1}{1 - \sin(s)}
\end{equation*}
Thus,
\begin{align*}
I &= \int \frac{1}{\left(1 + \sin(s)\right)\left(1 + \sin(s)\right)\sin(s)}\:ds = \int \left[\frac{1}{\sin(s)} - \frac{1}{2}\cdot \frac{1}{1 + \sin(s)} - \frac{1}{2}\cdot \frac{1}{1 - \sin(s)} \right]\:ds \\
&= \int \frac{1}{\sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 + \sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 - \sin(s)}\:ds
\end{align*}
We now employ the Weierstrass Substitution $t = \tan\left(\frac{s}{2} \right)$:
\begin{equation*}
ds = \frac{2}{1 + t^2}\:dt, \quad \sin(s) = \frac{2t}{1 + t^2}
\end{equation*}
Thus,
\begin{align*}
I&= \int \frac{1}{\sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 + \sin(s)}\:ds - \frac{1}{2} \int \frac{1}{1 - \sin(s)}\:ds \\
&= \int \frac{1}{\frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2}\:dt - \frac{1}{2} \int \frac{1}{1 + \frac{2t}{1 + t^2}}\cdot \frac{2}{1 + t^2}\:dt - \frac{1}{2} \int \frac{1}{1 - \frac{2t}{1 + t^2}}\cdot \frac{2}{1 + t^2}\:dt \\
&= \int \frac{1}{t}\:dt - \int \frac{1}{1 + t^2 + 2t}\:dt - \int \frac{1}{1 + t^2 - 2t}\:dt \\
&= \int \frac{1}{t}\:dt - \int \frac{1}{\left(t + 1\right)^2}\:dt - \int \frac{1}{\left(t - 1\right)^2}\:dt = \ln\left|t\right| - -\frac{1}{t + 1} - - \frac{1}{t - 1} + C \\
&= \ln\left|t\right| +\frac{1}{t + 1} + \frac{1}{t - 1} + C = \ln\left|t\right| +\frac{2t}{t^2 - 1} + C
\end{align*}
Where $C$ is the constant of integration. Here:
\begin{equation*}
t = \tan\left(\frac{s}{2} \right) = \tan\left(\frac{\arctan(x)}{2} \right) = \frac{\sqrt{x^2 + 1} - 1}{x}
\end{equation*}
Thus,
\begin{align*}
I&= \ln\left|t\right| +\frac{2t}{t^2 - 1} + C = \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| +\frac{2\left(\frac{\sqrt{x^2 + 1} - 1}{x}\right)}{\left(\frac{\sqrt{x^2 + 1} - 1}{x} \right)^2 - 1} + C \\
&= \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| + 2 \cdot \frac{\sqrt{x^2 + 1} - 1}{x} \cdot \frac{1}{\left(\frac{x^2 + 1 - 2\sqrt{x^2 + 1} + 1}{x^2} \right) - 1} + C \\
&= \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| + 2 \cdot \frac{\sqrt{x^2 + 1} - 1}{x} \cdot \frac{x^2}{\left(x^2 + 1 - 2\sqrt{x^2 + 1} + 1\right) - x^2} + C \\
&= \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| + 2 \cdot \frac{\sqrt{x^2 + 1} - 1}{x} \cdot \frac{x^2}{2\left(1 - \sqrt{x^2 + 1} \right)} + C = \ln\left|\frac{\sqrt{x^2 + 1} - 1}{x}\right| - x + C \\
\end{align*}
| The first problem is where you took $1-\sin^2 s=(1+\sin s)(1+\sin s).$ I believe this is just a slip and you know the right factorization.
The second one is that the decomposition into partial fractions is incorrect. We have that $$\frac1z-\frac12\left(\frac{1}{1+z}+\frac{1}{1-z}\right)\ne\frac{1}{z(1-z^2)}.$$
And to evaluate the $$\int\frac{\sec s}{\tan s}\sec^2s\mathrm ds,$$ you don't need to go to all this trouble. A little rewriting of the integrand helps, as follows: $$\frac{\sec s}{\tan s}\sec^2s=\frac{\sec s}{\tan s}(1+\tan^2s)=\frac{\sec s}{\tan s}+\sec s\tan s=\csc s+\sec s\tan s.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int \cos^2x\sin^4x\mathrm{d}x$
Evaluate integral $$\int \cos^2x\sin^4x\mathrm{d}x.$$
Attempt. Setting $\tan x=t$, gives:
$$\int \cos^2x\sin^4x\mathrm{d}x =\int \frac{1}{1+t^2} \,\left(\frac{t^2}{1+t^2}\right)^2 \frac{\mathrm{d}t}{1+t^2}=\int \frac{t^4}{(1+t^2)^4} \mathrm{d}t,$$
which does not seem to be elementary.
Thank in advance for the help.
| For trigonometric monomials, the standard method is linearisation when both exponents are even, substitution when an exponent is odd ($u=\sin x$ or $u=\cos x$, depending on which exponent is odd).
You can linearise either with trigonometry formulæ or via the complex exponential. I'll show both:
*
*with trigonometry: use repeatedly the linearisation formulæ
$$\cos^2x=\tfrac12(1+\cos 2x),\qquad\sin^2x=\tfrac12(1-\cos 2x).$$
So
\begin{align}
\cos^2x\sin^4x&=\tfrac12(1+\cos 2x)\tfrac14(1-\cos 2x)^2=\tfrac 18(1+\cos 2x)(1-2\cos 2x+\cos^22x)\\
&=\tfrac 18(1+\cos 2x)\bigl(\tfrac32-2\cos 2x+\tfrac12\cos4x)\\
&=\tfrac1{16}(3-4\cos 2x+\cos 4x+3\cos 2x-4\cos^2 2x+\underbrace{\cos2x\cos 4x}_{\frac12(\cos 6x+\cos 2x)}) \\
&=\tfrac1{32}(2-\cos 2x-2\cos 4x+\cos 6x).
\end{align}
*with the complex exponential: to have lighter computations, set $u=\mathrm e^{ix}$. We'll use the definitipn of sine and cosine in terms of $u$:
$$\sin x=\tfrac1{2i}(u-\bar u),\qquad \cos x =\tfrac 12(u+\bar u),$$
so that, taking into accoount that $u\bar u=1$,
\begin{align}
\cos^2x\sin^4x &=\tfrac14(u+\bar u)^2\cdot\tfrac1{16}(u-\bar u)^4=\tfrac1{64}(u^2+\bar u^2+2)(u^4-4^2+6-4\bar u^2+\bar u^4) \\
&=
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Non-calculus ways to get higher-order terms in $(1 + k/n)^n$? There's a well-known limit $$\lim_{n \to \infty} \left( 1 + \frac{k}{n} \right)^n = e^k $$ that can be proved either by computing limits of individual terms in the binomial expansion, or by defining $$f(x) = (1 + kx)^{1/x}$$ and using calculus to compute $$\lim_{x \to 0} \ln f(x) = \lim_{x \to 0} \frac{\ln (1+kx)}{x} = \lim_{x \to 0} \frac{k}{1+kx} = k.$$
I was working on a problem for which the rate of convergence of this limit is relevant: computing the expected size of the image of a random function from $\{1, \ldots, n\}$ to itself. The chance that any particular number is included in the image is $1 - \left(\frac{n-1}{n}\right)^n$, so the expected size, by linearity of expectation, is $n - n\left( \frac{n-1}{n} \right)^n = \left( 1 - \frac{1}{e} \right) n + O(1).$ WolframAlpha showed that the error term was a constant $\frac{2}{e} + O(n^{-1})$, and I proved this myself by developing $f$ as a power series:
\begin{align*}
\ln f(x) &= \frac{\ln (1+kx)}{x} \\
\frac{d}{dx} \ln f(x) &= -\frac{\ln (1+kx)}{x^2}+ \frac{k}{x (1+kx)} \\
f'(x) &= f(x) \frac{d}{dx} \ln f(x) \\
&= (1 + kx)^{1/x} \left[-\frac{\ln (1+kx)}{x^2} + \frac{k}{x (1+kx)}\right] \\
\lim_{x \to 0} f(x) &= e^k \lim_{x \to 0} \left[-\frac{\ln (1+kx)}{x^2} + \frac{k}{x (1+kx)}\right] \\
&= e^k \lim_{x \to 0} \frac{-(1+kx) \ln (1+kx) + kx}{x^2 (1+kx)} \\
&= e^k \lim_{x \to 0} \frac{-k \ln (1+kx) }{3kx^2 + 2x} \\
&= e^k \lim_{x \to 0} \frac{-k^2}{(6kx+2) (1+kx)} \\
&= \frac{-e^k k^2}{2}
\end{align*}
and therefore $$\left(1 + \frac{k}{n}\right)^n = e^k - \frac{e^k k^2}{2} \frac{1}{n} + O\left( \frac{1}{n^2} \right).$$
(This procedure can be continued, cumbersomely, to get higher-order terms, each of which is a $e^k$ times a polynomial in $k$ with rational coefficients that don't seem to match any OEIS sequence.) I'd like to know, though: Is there any way of getting the $-e^k k^2/2n$ term purely through series manipulation, without resorting to calculus? I tried finding a way for a bit, but the computations quickly get unpleasant.
| Hint:
$$\left(1+\frac kn\right)^n=\\
1+k+\left(1-\frac1n\right)\frac{k^2}{2}
+\left(1-\frac1n\right)\left(1-\frac2n\right)\frac{k^3}{3!}
+\left(1-\frac1n\right)\left(1-\frac2n\right)\left(1-\frac3n\right)\frac{k^4}{4!}+\cdots$$
The terms in $\dfrac1n$ are
$$-\frac{k^2}2-3\frac{k^3}{3!}-6\frac{k^4}{4!}-\cdots \frac{m(m-1)}2\frac{k^m}{m!} -\cdots\\
=-\frac12k^2-\frac12k^3-\frac12\frac{k^4}{2!}-\cdots \frac{1}2\frac{k^m}{(m-2)!} -\cdots\\
=-\frac{k^2}2\left(1+k+\frac{k^2}2+\frac{k^3}{3!}+\cdots\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving an inequality with three variables Find the largest integer $\lambda$ such that :
$$\frac{\lambda (xyz)}{x+y+z} \le(x+y)^2 + (x+y+4z)^2.$$
Here $x,y$ and $z$ are positive real numbers
The answer is only achieved using calculus. Can anyone use AM-GM and get $\lambda$ ?
| I think it is $$\frac{x+y+z}{xyz}\left((x+y)^2+(x+y+4z)^2\right)\geq 100$$ and the equal sign holds if $$x=y=1,z=\frac{1}{2}$$
This is true since we have $$\frac{x^3+3x^2y+3xy^2+y^3+5x^2z+5y^2z+12xz^2+12yz^2+8z^3}{9}\geq \sqrt[9]{259200x^9y^9z^9}$$
Now it must
$$9\times\sqrt[9]{259200}\times 2\geq 40$$
I forgot the factor $9$!!!
$$\sqrt[9]{259200}\times 9\times 2>40$$!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algebraic Expression making perfect square For what value of $a$ and $b$ would $$x^4+2x^3+ax^2+bx+9$$
Would give perfect square?
I have found $a=7$ and $b=6$.
How many more are there?
Can we find all of it ?
| The given polynomial is a perfect square if and only if
$$x^4+2x^3+ax^2+bx+9=(x^2+Ax+B)^2$$
for some $A,B$. That is, after expanding and comparing the coefficients,
$$2x^3+ax^2+bx+9=2Ax^3+(2B+A^2)x^2+2ABx+B^2
\Leftrightarrow
\begin{cases}
A=1\\
2B+A^2=a\\
2AB=b\\
B^2=9
\end{cases}$$
Now it should be easy to find ALL the values of $a$ and $b$ that give you a perfect square.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convert $31213111332_{4}$ to hexadecimal I did
$$31213111332_{4} = \\
2*4^0+3*4^1+3*4^2+1*4^3+1*4^4+1*4^5+3*4^6+1*4^7+2*4^8+1*4^9+3*4^{10} = \\
2*16^0+3*4^1+3*16^1+1*4*16^1+1*16^2+1*4*16^2+3*16^3+1*4*16^3+2*16^4+1*4*16^4+3*16^5 = \\
2*16^0 + 3*4^1+7*16^1+5*16^2+7*16^3+6*16^4+3*16^5 = \\
36757E_{16}$$
Which is correct but I have a question. What if $2*16^0+3*4^1$ was greater than 15 (F) ?
| It's easier if you convert groups of two base-4 digits into one base-16 digit, from right to left:
$$
\begin{array}{rcl}
31213111332_{4}
&=&
03_{4} \ 12_{4} \ 13_{4} \ 11_{4} \ 13_{4} \ 32_{4}
\\&=&
3_{10} \ 6_{10} \ 7_{10} \ 5_{10} \ 7_{10} \ 14_{10}
\\&=&
3_{16} \ 6_{16} \ 7_{16} \ 5_{16} \ 7_{16} \ E_{16}
\\&=&
36757E_{16}
\end{array}
$$
Two base-4 digits correspond to $4$ bits, the size of a base-16 digit.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given three non-negative numbers $a,b,c$. Prove that $\frac{a+b+c}{k}\geqq\sum\limits_{cyc}\frac{a-b}{b+ k}$ for $k= constant$ so that $k> 0$ .
Given three non-negative numbers $a, b, c$. Prove that
$$\frac{a+ b+ c}{k}\geqq \frac{a- b}{b+ k}+ \frac{b- c}{c+ k}+ \frac{c- a}{a+ k}$$
for $k= constant$ so that $k> 0$ .
For $k= 2$, we can use
$$\frac{a+ b}{2}\geqq \frac{a- b}{b+ 2}$$
but for $k= constant$ ? And I think it would be really really easy if expands. Thanks for help that a lot
| By adding $3$ to both sides of the given inequality, it is equivalent to
$$\frac{(a+k)+(b+k)+(c+k)}{k}\geq \frac{a+k}{b+k}+\frac{b+k}{c+k}+\frac{c+k}{a+k}.\ \ \ \ \ (1)$$
This is also equivalent to
$$\frac{A+B+C}{k}\ge \frac{A}{B}+\frac{B}{C}+\frac{C}{A}\ \ \ \ \ (2)$$
for all $A,B,C\geq k$ and $k>0$ (by setting $A=a+k$, $B=b+k$, and $C=c+k$). But this is again equivalent to
$$x+y+z\geq \frac{x}y+\frac{y}z+\frac{z}x\ \ \ \ \ (3)$$
for all $x,y,z\ge 1$ (by setting $x=A/k$, $y=B/k$, and $z=C/k$).
However this is trivial as
$$x-x/y=x(1-1/y)\ge 0$$
and similarly $y-y/z\ge 0$ and $z-z/x\ge 0$. Therefore (3) is true (with equality case $x=y=z=1$). Thus (2) is true with equality case $A=B=C=k$. Hence (1) and the required inequality is also true with $a=b=c=0$ being the equality case.
In fact, one can improve this solution:
$$\frac{a}{k}-\frac{a-b}{b+k}=(a+k)\left(\frac1k-\frac{1}{b+k}\right)\ge 0,$$
we have two other similar inequalities.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify this expansion : $\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$ Find a simple closed form of :
$\omega=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}+(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
My try :
Let :
$A=(28+(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
And
$B=(28-(\frac{5290}{3})^{\frac{3}{2}})^{\frac{1}{3}}$
Now :
$A^{3}+B^{3}=56$
But how I can now find $A$ and $B$ ?
| I would write
$$(28+x)^{1/3}+(28-x)^{1/3}=s$$ and now raise this to the power three.
| {
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"timestamp": "2023-03-29T00:00:00",
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Greatest prime divisor of $ n^2+1$ Prove that there exist infinitely many positive integers $n$ such that the greatest prime divisor of $n^2+1$ is less than $n \cdot \pi^{-2019}.$
What did I do: $n = 2a^2$, then $n^2+1 = 4a^4+1 = ((a-1)^2+a^2)((a+1)^2+a^2)$.
Funny that it's trivial for $P (n ^ 2 + 1) <n$, because all n satisfy. But $\pi^{-2019}.$ goes down a lot .... how do I prove it?
| Let us prove that for every $\varepsilon > 0$ there exist infinitely many integers $n > 0$ such that $P(n^2 + 1) < \varepsilon n$.
Obviously, it is enough to prove that there exists one such $n$. As you noticed, if $n = 2m^2$, for some integer $m > 0$, then $n^2 + 1 = (2m^2 - 2m + 1)(2m^2 + 2m + 1)$. Hence, it is enough to prove that we can find $m$ such that $P(2m^2 - 2m + 1) < 2\varepsilon m^2$ and $P(2m^2 + 2m + 1) < 2\varepsilon m^2$.
It is well known and not difficult to prove that for every polynomial $f \in \mathbb{Z}[x]$ the set of prime numbers $p$ such that $p$ divides $f(k)$ for some integer $k > 0$ is infinite. Therefore, we can find an integer $\ell > 0$ such that $p_1 := P(2\ell^2 - 2\ell + 1) > 2/\varepsilon$ and $p_2 := P(2\ell^2 + 2\ell + 1) > 2/\varepsilon$.
Now letting $m := \ell + t p_1 p_2$, for some integer $t > 0$, we get that $p_1$ divides $2m^2 - 2m + 1$ and $p_2$ divides $2m^2 + 2m + 1$.
Hence, $P(2m^2 - 2m + 1) < \max(p_1, (2m^2 - 2m + 1) / p_1)$ and $P(2m^2 + 2m + 1) < \max(p_2, (2m^2 + 2m + 1) / p_2)$.
Taking $t$ sufficiently large, we get that $P(2m^2 + 2m + 1) < (2m^2 + 2m + 1) / p_2 < \varepsilon(2m^2 + 2m + 1) / 2 < \varepsilon 4m^2 / 2 < 2\varepsilon m^2$ and, similarly, $P(2m^2 - 2m + 1) < 2\varepsilon m^2$, as required.
| {
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Number Of Solutions (Find number of solutions) Let $x_i \in Z$ , such that $|x_1| + |x_2| + \dots + |x_{10}| = 100$. Find number of solutions
I think the answer is in the form of an alegebric summation instead of a number, right?
If yes then count the number of solutions in terms of K when K terms of out those ten are zero( you may get an idea from MathWiz's answer) and then add all the cases as k varies from 0 to 9.
What would the next cases look like? Or how to find the solutions?
The amount of positive integer solutions would be $\frac{110!}{9! X 100 !} $, at home solution I could or could not put a minus sign in the number (because it's in module)
For a 10-double I would have $2 ^ {10}$. Am I right?
| Case 1: All $x_{i}\neq 0$: There are $\binom{100-10+9}{9}=\binom{99}{9}$ ways to distribute the 100, since $x_{i}$ can be negative, you have $2^{10}$ ways to distribute the negative sign. So this gives you a total of $2^{10}\binom{99}{9}$ ways.
Case 2: Only one out of $x_{i}$ is $0$: There are $\binom{10}{1}$ ways to make any one of $x_{i}$ equal to $0$ and $\binom{99}{8}$ ways to distribute the 100 amongst the remaining $x_{i}$ and $2^{9}$ ways to make any of them negative.So this gives you a total of $2^{9}\binom{10}{1}\binom{99}{8}$ ways.
Case 3: Two out of $x_{i}$ is $0$: There are $\binom{10}{2}$ ways to make any two of $x_{i}$ equal to $0$ and $\binom{99}{7}$ ways to distribute the 100 amongst the remaining $x_{i}$ and $2^{8}$ ways to make any of them negative.So this gives you a total of $2^{8}\binom{10}{2}\binom{99}{7}$ ways.
................
Adding them up, you get the number of solutions:
$$2^{10}\binom{99}{9}+2^{9}\binom{10}{1}\binom{99}{8}+2^{8}\binom{10}{2}\binom{99}{7}+2^{7}\binom{10}{3}\binom{99}{6}+2^{6}\binom{10}{4}\binom{99}{5}+2^{5}\binom{10}{5}\binom{99}{4}+2^{4}\binom{10}{6}\binom{99}{3}+2^{3}\binom{10}{7}\binom{99}{2}+2^{2}\binom{10}{8}\binom{99}{1}+2^{1}\binom{10}{9}$$.
Note: I can't edit ! :mad: : All signs in the expression above are positive !
In General,
$$\left | x_{1} \right |+\left | x_{2} \right |+...+\left | x_{n} \right |= m$$Stars and bars give that there are $\binom{n}{k}\binom{m-1}{k-1}$ solutions where $k$ numbers are positive and the rest are zero. Each of these give rise to $2^{k}$ integer solutions. Hence, the total number of solutions is
$$\sum_{k=1}^{n}2^{k}\binom{n}{k}\binom{m-1}{k-1}$$
| {
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Cyclic inequality $\sum_{\text{cyc}}a^ab^bc^c\le1$ how can I prove that for all positive reals $a,b,c$ with $a+b+c=1$ we have
$$a^ab^bc^c+a^bb^cc^a+a^cb^ac^b\le 1$$
It tried using weighted am.gm. on each term and got $$a^2+b^2+c^2+2ab+2bc+2ac\le 1$$ but now I don‘t know how to proceed.
| By weighted AM-GM we can wtite
$$(a^a~b^c~c^c)^{(a+b+c)^{-1}} \le \frac{a.a+b.b+c.c}{a+b+c}~~~(1)$$
and
$$(a^b~b^c~c^a)^{(a+b+c)^{-1}} \le \frac{a.b+b.c+c.a}{a+b+c}~~~(2)$$
and
$$(a^c~b^a~c^b)^{(a+b+c)^{-1}} \le \frac{a.c+b.a+c.b}{a+b+c}~~~(3)$$
Adding all these equations and using $a+b+c=1$. we get
$$a^ab^bc^c+a^bb^cc^a+a^cb^ac^b \le (a^2+b^2+c^2+2(ab+bc+ca))=(a+b+c)^2=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f(2x-3) = 4x-2$, then what is $f(x)$? I have this statement:
If $f(2x-3) = 4x-2,$ the function $f(x)$ is ...?
My attempt was:
Move the function $3$ units to the left
$f(2x) = 4(x+3) -2 = 4x+10$
Divide $x$ by $2$
$f(x) = 2x+10$
Verifiy $f(x) = 2x+10 \to f(2x) = 4x+10 \to f(2x-3) = 4(x-3)+10 = \underbrace{4x-2}_{f(2x-3)}$
But according to the guide the correct answer is $2x+4$ and i don't know why. Thans in advance.
| If $x\in\mathbb R$, then\begin{align}f(x)&=f\left(2\frac x2\right)\\&=f\left(2\left(\frac x2+\frac32\right)-3\right)\\&=4\left(\frac x2+\frac32\right)-2\\&=2x+4.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3387446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomials that induce the zero function mod $n$
*
*Which polynomials induce the zero function mod $n$?
In particular:
*
*What is the polynomial of least degree that induces the zero function mod $n$?
*What is the monic polynomial of least degree that induces the zero function mod $n$?
These are not vacuous questions because of the following general result:
If $r$ is the maximum exponent in the prime factorization of $n$, then $x \mapsto x^{r+\lambda (n)}-x^r$ is the zero function mod $n$. [Wikipedia]
Here, $\lambda$ is the Carmichael function.
*
*When is $x^{r+\lambda (n)}-x^r$ the monic polynomial of least degree that induces the zero function mod $n$?
Fermat's theorem implies that $x^n-x$ is the answer for $n$ prime: all polynomials that induce the zero function mod $n$ are a multiple of $x^n-x$.
How can this be generalized to composite $n$?
Here are some other examples:
$$
\begin{array}{rll}
n & L_n: \text{least degree} & M_n: \text{least degree monic}
\\ 2 & x^2+x
\\ 3 & x^3-x
\\ 4 & 2(x^2+x) & x^4-x^2
\\ 5 & x^5-x
\\ 6 & 3(x^2+x) & x^3-x
\\ 7 & x^7-x
\\ 8 & 4(x^2+x) & x^4+2x^3+3x^2+2x = x(x+1)(x^2+x+2)
\\ 9 & 3(x^3-x) & x^8-x \quad (???)
\\ 10 & 5(x^2+x) & x^5-x
\\ 11 & x^{11}-x
\\ 12 & 6(x^2+x) & x^4+5x^2+6x = x(x+1)(x^2-x+6)
\\ 13 & x^{13}-x
\\ 14 & 7(x^2+x) & ???
\\ 15 & 5(x^3-x) & x^5-x
\\ 21 & 7(x^3-x) & ???
\\ 24 & 12(x^2+x) & x^4+2x^3+11x^2+10x = x(x+1)(x^2+x+10)
\end{array}
$$
It seems clear that $L_{2m}=m(x^2+x)$, because $x^2+x$ is always even.
More generally,
*
*Is $L_{pq} = qL_p$ and $M_{pq}=L_q$ for $p<q$ primes?
*If $n=pm$ and $p$ is smallest prime divisor of $n$, then is $L_{pm}=mL_p$?
Corrections and additions welcome.
Please collect partial results as answers.
| Not complete answers, but too large for comment. I have the results, when $n=p_1^{\alpha_1}\cdots p_{k}^{\alpha_k}$:
Non-monic case: The minimal degree is exactly $\min p_i.$
Monic case: The minimal degree is the maximum of the degrees for each $p_i^{\alpha_i}.$ The degree of $p_i^{\alpha_i}$ must be divisible by $p_i$ and is at most $\alpha_ip_i.$ This is not the absolute minimum, for example when $p=2$ and $\alpha=3,$ the minimal degree for $8$ is $4<6=p\alpha.$
My conjecture is that the minimal monic degree $d$ for $p^k$ is the smallest $d$ such that $\nu_p(d!)\geq k.$
If that is true, then if $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ then the minimal monic degree is the minimal $d$ such that each $p_i^{\alpha_i}$ divides $d!.$
Non-monic case:
If $d(n)$ is the smallest degree for $n$ in the non-monic case, we can get, for any $m.n,$ $d(mn)=\min(d(m),d(n)).$
This is because if $p_m(x)$ is a minimal polynomial for $m$ then we take $np_m(x).$
This means that if $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots$ then $d(n)=\min_i d\left(p_i\right).$ But when $p$ is prime, $d(p)=p,$ so we get $d(n)=\min_i p_i.$
Monic case:
If $D(n)$ is the smallest degree of a monic, then when $\gcd(m,n)=1$ you have that $D(mn)=\max (D(m),D(n)).$
This is because if $p_m,p_n$ are the corresponding monic polynomials with $D(m)\geq D(n)$ you can apply Chinese remainder theorem coefficient by coefficient to find a polynomial $P_{mn}$ so that:
$$\begin{align}P_{mn}(x)&\equiv p_m(x)\pmod{m}\\
P_{mn}(x)&\equiv x^{D(m)-D(n)}p_n(x)\pmod{n}\end{align}$$
Since both polynomials are monic and of degree equal to the $D(m),$ we get $P_{mn}$ monic and $P_{mn}(x)$ satisfies your conditions.
We also have that $D\left(p^{\alpha}\right)\leq p\alpha$ since $(x^p-x)^{\alpha}$ is monic of degree $p\alpha$ and satisfies our conditions.
As noted in my comments above, if $p(x)$ is always zero modulo $n$ then so is $p(x+1)-p(x).$
In the monic case, this means if $p(x)$ is minimal then $d=\deg p(x)$ must have a common factor with $n,$ since otherwise $q(x)=p(x+1)-p(x)$ is of smaller degree with leading coefficient $d$ so we solve $du-nv=1$ and take $r(x)=uq(x)-nvx^{d-1}$ which is monic of smaller degree and satisfies our condition.
To finish the question, one needs to compute a value of $D(p^{\alpha}),$ which we know is divisible by $p$ and $\leq p\alpha.$
Conjecture
My guess is that $D\left(p^{\alpha}\right)$ is the smallest $d$ such that $\nu_p(d!)\geq \alpha.$ In particular, if $\alpha\leq p$ then $d=\alpha p.$
If $\alpha=p+1$ then $\nu_p\left((p^2)!\right)=p+1.$ Do $D(p^p)=D(p^{p+1})=p^2.$
This is definitely an upper bound, because the falling factorial $(x)_d$ is monic and of degree $d$ and is always divisible by $p^{\nu_p(d)}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $a = b$ (Leningrad 1990) Let a and b be natural numbers that $b ^ 2 + ba + 1$ divides $a ^ 2 + ab +1$. Prove that $a = b$
I thought of the limitation property $b ^ 2 + ba + 1 ≤ a ^ 2 + ab + 1$ and hence $b ≤ a$, but I don't know how to process. How can I prove it differently?
| If $b^2+ba+1$ divides $a^2+ab+1$ then $b^2+ba+1 \le a^2+ab+1$ and so $b \le a$.
Suppose $b<a$. Then $b^2+ba+1$ divides the difference $(a^2+ab+1) - (b^2+ba+1) = a^2-b^2$.
But $b^2+ba+1 = b(a+b)+1 = 1 \mod a+b$ and $a^2-b^2=(a-b)(a+b) = 0 \mod a+b$.
So either $a+b=1$, in which case $b=0, a=1$; or $b \not \lt a$ in which case $a=b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3389610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does $x \cdot \sin \left( \frac{1}{x} \right) = 1$ has solutions? A look to the plot show to you that the function $f(x) = x \cdot \sin \left( \frac{1}{x} \right) - 1 $ has no zero near the origin. Wolframalpha software says that $ x \cdot \sin \left( \frac{1}{x} \right) = 1$ if $x \approx 5.16\cdot 10^{15}$ but I suspect this huge number is a wrong solution due to numerical problems. Is it true that for some $x$, far away from the origin, the equation $$ x \cdot \sin \left( \frac{1}{x} \right) = 1 $$ is satisfied? Can, far away, $f(x)= x \cdot \sin \left( \frac{1}{x} \right)$ exceeds 1?
| Notice that:
$$x \cdot \sin \left( \frac{1}{x} \right) = \frac{ \sin \left( \frac{1}{x} \right)}{\left( \frac{1}{x} \right)}.$$
Moreover, observe that:
$$\lim_{x \to +\infty} \frac{ \sin \left( \frac{1}{x} \right)}{\left( \frac{1}{x} \right)} = 1.$$
This means that your function "is equal to $1$" for $x \to +\infty$. Numerically, $x \to +\infty$ stands for a huge number. The solution provided by Wolfram ($x \approx 5.16\cdot 10^{15}$) is indeed a huge number.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For $a \in \mathbb R$, find polynomials $P$ such that $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$ Find all polynomials $P(x)$ with real coefficients such that:- $(x^2-ax+18)P(x)-(x^2+3x)P(x-3)=0$
This is $a$ creating some big problems for me. I don't know what to do.
I am not able to figure out anything because of that $a$. The best I can figure out is that I will find the roots of $P(x)$ because $a$ is not computable as there is no way of finding out the zeros of $P(x-3)$. If I would have been able to find the roots of $(x^2-ax+18)$ then I would have been able to figure out what to do.
If there would have been no $a$ I would have found of the roots of $P(x)$ like for example $\alpha, \beta$ then I would have written out $P(x)$ in the form of $(x-\alpha)(x-\beta)Q(x)$ for $Q(x)$ being any polynomial. Then I would have tried to calculate the answer.
Any help would be appreciated
| At first we have $(x^2-ax+18)P(x)=(x^2+3x)P(x-3)=x(x+3)P(x-3)$. Now since $x \nmid (x^2-ax+18)$, we have $2$ cases:
$(x^2-ax+18)$ divisible by $(x+3)$:
$$(x+3) | (x^2-ax+18) \Rightarrow (x^2-ax+18)=(x+3)(x+6) \Rightarrow$$
$$\Rightarrow (x+6)P(x) = xP(x-3) \Rightarrow$$
$$\Rightarrow \left\{\begin{array}{c} x | P(x) \\ (x+6) | P(x-3) \end{array}\right\} \Rightarrow \left\{\begin{array}{c} (x-3) | P(x-3) \\ (x+9) | P(x) \end{array}\right\} \Rightarrow$$
$$\Rightarrow (x+9)Q(x) = (x-3)Q(x-3) \Rightarrow \dots$$
That means $P(x)$ must have infinitely many factor that's not the case!
$(x^2-ax+18)$ not divisible by $(x+3)$:
$$(x+3) \nmid (x^2-ax+18) \Rightarrow \left\{\begin{array}{c} x(x+3) | P(x) \\ (x^2-ax+18) | P(x-3) \end{array}\right\} \Rightarrow$$
$$\Rightarrow \left\{\begin{array}{c} (x-3)x | P(x-3) \\ (x^2+(6-a)x+(27-3a)) | P(x) \end{array}\right\} \Rightarrow x^2 | P(x) \Rightarrow \dots$$
That means $P(x)$ must have infinitely many $x$ that's not the case!
So there is no $P(x)$ except $P(x)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove geometric sum with induction I'm not certain how to complete the proof:
Question:
Prove by induction that $1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^n = (\frac{3}{4})[1 − (\frac{−1}{3})^{n+1}]$, for every non negative integer $n$.
Solution
Base Step:
Verify that:
$LHS = 1 = (\frac{3}{4})[1 − (\frac{−1}{3})^{0+1}] = RHS$.
$RHS = (\frac{3}{4})[1 − (\frac{−1}{3})^1]\\
= (\frac{3}{4})[1 − (\frac{−1}{3})]\\
= (\frac{3}{4})(1 + \frac{1}{3})\\
= (\frac{3}{4})(\frac{4}{3}) = 1 = LHS.$
Inductive Step:
Assume that:
$1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^k = (\frac{3}{4})[1 − (\frac{−1}{3})^{k+1}]$, for some integer k.
We try to deduce that:
$1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} = (\frac{3}{4})[1 − (\frac{−1}{3})^{k+2} ]$.
$LHS
= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} \\
= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k} + (\frac{−1}{3})^{k+1}\\
= \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + (\frac{-1}{3})^{k+1}\\
... $
Lost from this point onwards.
| After a small amount of work, my final solution is:
\begin{align}
LHS &= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k+1} \\
&= 1 − \frac{1}{3} + \frac{1}{9} − · · · + (\frac{−1}{3})^{k} + (\frac{−1}{3})^{k+1}\\
&= \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + (\frac{-1}{3})^{k+1}\\
&= \frac{3}{4}[1-(\frac{-1}{3})^{k+1}] + \frac{3}{4}[\frac{4}{3}(\frac{-1}{3})^{k+1}]\\
&= \frac{3}{4}[1-(\frac{-1}{3})^{k+1} + \frac{4}{3}(\frac{-1}{3})^{k+1}]\\
&= \frac{3}{4}[1+(\frac{4}{3}-1)(\frac{-1}{3})^{k+1}]\\
&= \frac{3}{4}[1 - (\frac{-1}{3})^{k+2}]\\
&= RHS
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergent series??? Determine for each of the following series whether or not it is convergent. You must justify your answer.
1. $\sum_{n=1}^{\infty} \left(\left(n^3+1\right)^{\frac{1}{3}}-n\right)$.
Could I use $a^3-b^3 = (a-b)(a^2+ab+b^2))$? If so how?
Using the hint given we get that
\begin{equation*}
\begin{split}
\left(n^3+1\right)^{\frac{1}{3}}-n &= \frac{\left[\left(n^3+1\right)^{\frac{1}{3}}\right]^3-n^3}{\left[\left(n^3+1\right)^{\frac{1}{3}}\right]^2+\left(n^3+1\right)^{\frac{1}{3}}\cdot n+n^2} \\
&= \frac{n^3+1-n^3}{\left(n^3+1\right)^{\frac{2}{3}}+n\left(n^3+1\right)^{\frac{1}{3}}+n^2} \\
&= \frac{1}{\left(n^3+1\right)^{\frac{2}{3}}+n\left(n^3+1\right)^{\frac{1}{3}}+n^2} \\
&\leq \frac{1}{n^2}.
\end{split}
\end{equation*}
But we know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges ($p$-series with $p = 2 > 1$). Thus, by the comparison test $\sum_{n=1}^{\infty} \left(\left(n^3+1\right)^{\frac{1}{3}}-n\right)$ converges as well.
2. $\sum_{n=2}^{\infty} \frac{1}{n\ln{(n)}^3}$.
The function $f:[2,\infty) \mapsto \mathbb{R}$ defined by $f(x) := \frac{1}{x\ln{(x)}^3}$ is continuous, positive and monotonely decreasing by examining the derivative $f'(x) = -\frac{\ln{(x)}+3}{x^2(\ln{(x)})^4} < 0$ for all $x\geq 2$.
Since $f(n) = \frac{1}{n\ln{(n)}^3}$, to determine whether or not the series converges it suffices to see what happens to $\int_{2}^{\infty} f(x) \; dx$. Doing so, letting $u = \ln{(x)} \Rightarrow du = \frac{dx}{x}$ gives
\begin{equation*}
\begin{split}
\int_{2}^{t} \frac{1}{x\ln{(x)}^3} \; dx &= \lim_{t\to\infty} \int_{\ln{(2)}}^{t} \frac{1}{u^3} \; du \\
&= \lim_{t\to\infty} \left[-\frac{1}{2u^2}\right]_{\ln{(2)}}^{t} \\
&= \frac{1}{2\ln{(2)}^2}
\end{split}
\end{equation*}
which is finite. This integral converges so by the integral test $\sum_{n=2}^{\infty} \frac{1}{n\ln{(n)}^3}$ converges as well.
3. $\sum_{n=1}^{\infty} \frac{\sin{(\pi+n)}}{\sqrt{n}+2^n}$.
First we note that
\begin{equation*}
\frac{\sin{(\pi+n)}}{\sqrt{n}+2^n} = -\frac{\sin{(n)}}{\sqrt{n}+2^n}.
\end{equation*}
This series is absolutely convergent if and only if
\begin{equation*}
\sum_{n=1}^{\infty} \left|-\frac{\sin{(n)}}{\sqrt{n}+2^n}\right| = \sum_{n=1}^{\infty} \frac{\sin{(n)}}{\sqrt{n}+2^n}
\end{equation*}
converges. We'll prove that this series does converge by using the comparsion test.
Observe that for any $k\geq 1$,
\begin{equation*}
\frac{\sin{(n)}}{\sqrt{n}+2^n}\leq \frac{1}{\sqrt{n}+2^n}\leq \frac{1}{2^n}.
\end{equation*}
But we know that the series $\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n$ converges as it's just a geometric series with ratio $\frac{1}{2} < 1$. So the comparison test tells us that $\sum_{n=1}^{\infty} \frac{\sin{(\pi+n)}}{\sqrt{n}+2^n}$ converges, and thus the original series is absolutely convergent.
| We have that for $\sum_{n=1}^{\infty} \left(\left(n^3+1\right)^{\frac{1}{3}}-n\right)$ we can use that
$$\left(\left(n^3+1\right)^{\frac{1}{3}}-n\right)=n\left(\left(1+1/n^3\right)^{\frac{1}{3}}-1\right)$$
then binomial series, then limit comparison test with $\sum \frac1{n^2}$.
For $\sum_{n=2}^{\infty} \frac{1}{n\ln{(n)}^3}$ we can use Cauchy condesation test.
For $\sum_{n=1}^{\infty} \frac{\sin{(\pi+n)}}{\sqrt{n}+2^n}$use absolute convergence and comparison test with $\sum \frac1{2^n}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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What's the correct definition of "conjugate", and do we identify them?
What does a "conjugate" in math mean?
I know the definition says if we have $x+y$, then $x-y$ is its conjugate.
So, $5+ \sqrt{5}$ and $5-\sqrt{5}$ are conjugate of one another.
Are $8$ and $2$ conjugate of one another as they can be $5+3$ and $5-3$?
Now, $5$ and $\sqrt{2}$ are not conjugate. How? We can write them like $\frac{5 + \sqrt{2}+(5-\sqrt{2})}{2}$ and $\frac{5 + \sqrt{2}-(5-\sqrt{2})}{2}$.
Now, they are.
So, what's the concept?
| Conjugates are usually of two different domains. For example, in your case of $5 \pm \sqrt{2}$, the first number is rational, the second is not. The reason why this is useful is when the product and sum of conjugates is easier to deal with than the conjugates themselves. Again, in your example, the sum of the two, and their product, are rational (and actually are integers).
Another such example are complex conjugates. E.g. $a \pm bi$ where $a,b$ are reals and $i = \sqrt{-1}$. Then, sum of the conjugates is real, their difference is imaginary (still easier to deal with than general complex) and their product is real:
$$
(a+bi)(a-bi) = a^2 + b^2.
$$
Of course you can write any two reals as conjugates, but generally that does not buy you much advantage unless you are trying to get rid of some property which either conjugates have but their sum or product will not.
Another example. Consider simplifying
$$
\frac{1}{a + b\sqrt{c}}
= \frac{1}{a + b\sqrt{c}} \times \frac{a - b\sqrt{c}}{a - b\sqrt{c}}
= \frac{a - b\sqrt{c}}{a^2-b^2c}
= \frac{a}{a^2-b^2c} - \frac{b}{a^2-b^2c} \sqrt{c},
$$
which for real $a,b,c>0$ ends up with a square root in the numerator instead of a square root in the denominator, typically a much simpler expression to manage. To illustrate on specific numbers,
$$
\frac{1}{3 + 2\sqrt{2}} = 3 - 2\sqrt{2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A marble is dropped straight down from point $A$
A marble is dropped straight down from point $A$. The time required
for the ball to travel the distance $h_1$ between $A$ and $B$ is equal
to the required time to travel the distance $h_2$ between $B$ and $C$.
Find $\dfrac{h_1}{h_2}$ by using $g \approx 10$ $m/s^2$.
I am not sure how to start the solution. All the distance is: $h=\dfrac{gt^2}{2}$, where $g\approx$ $10$ $m/s^2$ and $h=\dfrac{10(t+t)^2}{2}=\dfrac{10(2t)^2}{2}=\dfrac{10\cdot 4t^2}{2}=\dfrac{40t^2}{2}=20t^2$. We can calculate $h_1$ :
$h_1=\dfrac{a_1t^2}{2}$. Here ($h_2$) we have initial speed, that according to me is equal to $a_1t$. Therefore, $h_2=\dfrac{a_1t^2+a_2t^2}{2}$ and$h_1+h_2=\dfrac{a_1t^2}{2}+\dfrac{a_1t^2+a_2t^2}{2}=\dfrac{2a_1t^2+a_2t^2}{2}=20t^2$. It seems like I can't finish the problem in this way.
| From first principles. The equation for velocity is
$ v = u + ft $
The equation for distance is
$s = ut + \frac{1}{2} f t^2$
So from $A$ to $B$
$v1 = gt $
$h1 = \frac{1}{2} g t^2$
And from $B$ to $C$
$h2 = gt^2 + \frac{1}{2} g t^2$
So
$ \frac {h 1}{h 2} = \frac{\frac{1}{2} g t^2}{gt^2 + \frac{1}{2} g t^2}$
Cancelling out
$ \frac {h 1}{h 2} = \frac{\frac{1}{2}}{1 + \frac{1}{2}} = \frac{1}{3}$
| {
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Intersection of curve $ y = x^4 – 6x^3 + 12x^2 + cx + 1$ The number of integers in the range of 'c' such that there exists a line which intersects the curve $ y = x^4 – 6x^3 + 12x^2 + cx + 1$ at four distinct points.
My approach we need to intersect with line $y=mx+C$
Substituting we get $x^4 – 6x^3 + 12x^2 + cx + 1-mx-C=0$
Now this is an equation of polynomial of degree 4
$x^4 – 6x^3 + 12x^2 + (c-m)x + 1-C=0$
All four roots needs to be real.
I don't have any idea how to check whether all the roots are real or not.
| Consider equation
$$
x^4 – 6x^3 + 12x^2 - 9x + 2 = 0.\tag{1}
$$
It is equivalent to
$$
(x - 2) (x - 1) (x^2 - 3 x + 1) = 0,\tag{2}
$$
and has $4$ real solutions:
$$
x_{1} = 1,\\
x_{2} = 2, \\
x_{3} = \dfrac{3-\sqrt{5}}{2}, \\
x_{4} = \dfrac{3-\sqrt{5}}{2}.\tag{3}
$$
Then for any $c\in\mathbb{R}$ we can rewrite $(1)$ in the form
$$
0 = x^4 – 6x^3 + 12x^2 -9x + 2 \\
= (x^4 – 6x^3 + 12x^2 +cx + 1) - ((9+c)x-1),
$$
which means that equation
$$
x^4 – 6x^3 + 12x^2 +cx + 1 = (9+c)x-1
$$
has those four real solutions: see $(3)$.
Other words, given curve and line $$y=(9+c)x-1$$ intersect in four distinct points with abscissas from list $(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Factoring $2^{2x} +2^{-2x}-2$ I have been stumped by this question asking to solve for $x$ in an equation with negative exponents. The combination of positive and negative exponents is what is confusing me so much.
So, how would one factor this?
$$2^{2x} +2^{-2x}-2$$
I've tried substituting $2^{2x}$ into another variable like $a$ but haven't had any success. Thank you.
| Suppose $a = 2^x$, $b=2^{-x}$. First we should note that $ab=1$.
Then the equation becomes $$2^{2x}+2^{-2x}-2 = (2^{x})^2 + (2^{-x})^2 - 2 = a^2 + b^2 -2ab = (a-b)^2 = (2^x - 2^{-x})^2$$
Alternatively, we could just write
$$2^{2x}+2^{-2x}-2 = 2^{2x} - 2 + 2^{-2x} = (2^{x})^2 - 2(2^{x}\cdot 2^{-x}) + (2^{-x})^2 = (2^x - 2^{-x})^2$$
Some people may find it helpful to write $2^{2x} - 2 + 2^{-2x}$ as $2^{2x} - 2\cdot 1 + 2^{-2x} = 2^{2x} - 2\cdot 2^{0x} + 2^{-2x}$. This may make it easier to see that you can factor out $2^{-2x}$, as follows:
$$2^{2x} - 2\cdot 2^{0x} + 2^{-2x} = 2^{-2x}(2^{4x} - 2\cdot 2^{2x} + 1) = 2^{-2x}((2^{2x})^2 - 2\cdot 2^{2x} + 1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3399683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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} |
Show: $\int_0^1\int_0^1\frac{dx\,dt}{(xt)^2+xt+1}=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{2\pi}{3}\right)$
How do you prove that $$\int_0^1\int_0^1\frac{dx\,dt}{(xt)^2+xt+1}=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{2\pi}{3}\right)?$$
My proof:
Suppose we have a sequence $(q_n)_{n\ge0}$ defined by the recurrence $$q_{n+2}=aq_{n+1}+bq_n,\qquad q_0,q_1,a,b \text{ given}.$$
We define the generating function
$$q(t)=\sum_{n\ge0}q_nt^n,$$
and see that
$$q(t)-q_1t-q_0=at(q(t)-q_0)+bt^2q(t),$$
so that $$q(t)=\frac{(q_1-aq_0)t+q_0}{1-at-bt^2}.$$
Consider the case
$$\begin{align}
q_0&=0\\
q_1&=1\\
a&=-1\\
b&=-1,
\end{align}$$
i.e. $q_{n+2}+q_{n+1}+q_n=0$ for all $n\in\Bbb Z_{\ge0}$, with $q_0=0,q_1=1$.
This has the generating function
$$q(t)=\frac{t}{t^2+t+1}.$$
Note that if $q_n=0$ and $q_{n+1}=1$ for any $n\in\Bbb Z_{\ge0}$, then $q_{n+2}=-1$. This implies that $q_n=q_{n+3}$ for all $n$, that is, the sequence is periodic with a period of $3$. The first few values are $0,1,-1,0,1,-1,0,1,-1,..$. Because of the alternating behavior of the sequence, we can represent it as a sine function:
$$q_n=\alpha \sin(n\beta).$$
Since $1=q_1=\alpha\sin\beta\ne 0$ and $-1=q_2=\alpha\sin2\beta\ne0$, both $\alpha$ and $\beta$ are nonzero, and additionally $\beta/\pi\not\in\Bbb Z$. Since $q_n=0$ for $3|n$, $3\beta/\pi\in\Bbb Z$. Finally, if we ensure that $\alpha$ and $\beta$ are both positive, we get that $\beta=2\pi/3$ and $\alpha=2/\sqrt3$. Thus
$$\frac{1}{t^2+t+1}=\frac{2}{\sqrt3}\sum_{n\ge1}\sin\left(\tfrac{2\pi}{3}n\right)t^{n-1}.$$
If we replace $t$ by $xt$ and integrate over $(x,t)\in[0,1]\times[0,1]$, we get
$$\begin{align}
\int_0^1\int_0^1\frac{dxdt}{(xt)^2+xt+1}&=\frac{2}{\sqrt3}\sum_{n\ge1}\sin\left(\tfrac{2\pi}{3}n\right)\int_0^1x^{n-1}dx\int_0^1 t^{n-1}dt\\
&=\frac{2}{\sqrt3}\sum_{n\ge1}\frac{\sin\left(\tfrac{2\pi}{3}n\right)}{n^2}\\
&=\frac{2}{\sqrt3}\mathrm{Cl}_2\left(\tfrac{2\pi}{3}\right).
\end{align}$$
$\square$
From @LeBlanc's comment,
$$\int_0^1 \int_0^1\frac{dx\,dt}{(xt)^2-2xt\cos\theta+1}=\frac1{\sin\theta}\mathrm{Cl}_2(\theta).$$
| \begin{align}
I&=\int_0^1\int_0^1\frac{dx\ dt}{(xt)^2+xt+1}\overset{xt=y}{=}\int_0^1\int_0^x\frac{dx\ dy}{x(y^2+y+1)}\\
&=\int_0^1\frac{1}{y^2+y+1}\left(\int_y^1\frac{dx}{x}\right)dy=-\int_0^1\frac{\ln y}{y^2+y+1}\ dy\\
&=\int_0^1\frac{y-1}{1-y^3}\ln y\ dy=\int_0^1\frac{y\ln y}{1-y^3}\ dy-\int_0^1\frac{\ln y}{1-y^3}\ dy\\
&=\frac19\int_0^1\frac{x^{-1/3}\ln x}{1-x}\ dx-\frac19\int_0^1\frac{x^{-2/3}\ln x}{1-x}\ dx\\
&=-\frac19\psi_1(2/3)+\frac19\psi_1(1/3)
\end{align}
using $\psi(1-x)-\psi(x)=\pi\cot(\pi x)$, differentiate both sides with respect to $x$ then set $x=2/3$
we get $\psi_1(2/3)=\frac43\pi^2-\psi_1(1/3)$.
Thus
$$\boxed{I=\frac29\psi_1(1/3)-\frac4{27}\pi^2}$$
Note: $\psi_n(z)=-\int_0^1\frac{x^{z-1}\ln^n x}{1-x}\ dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Markov Chain with dice rolling board game Simple board game.
S---1---2---3---4---5---H
You start at S. Each turn, your roll a standard six-sided die, and move forward that number of spaces. Your goal is to reach H. You can only get to H on an exact roll; if you roll a number that would take you past H, you do not move. For example, suppose you are sitting at 4. If you roll a 1, you move to 5. If you roll a 2, you move to H. If you roll anything else, you do not move. Once you reach H, you remain there forever.
What is the transition matrix?
| The transition matrix is straightfoward; enumerate the states as $S,1,2,3,4,5,H$, then we have
$$
P= \begin{pmatrix}
0 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6\\
0 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6\\
0 & 0 & 1/3 & 1/6 & 1/6 & 1/6 & 1/6\\
0 & 0 & 0 & 1/2 & 1/6 & 1/6 & 1/6\\
0 & 0 & 0 & 0 & 2/3 & 1/6 & 1/6\\
0 & 0 & 0 & 0 & 0 & 5/6 & 1/6\\
0 & 0 & 0 & 0 & 0 &0 & 1
\end{pmatrix}.
$$
Note that the states $S$, $1$, $2$, $3$, $4$, and $5$ are transient and $H$ is absorbing, so $P$ is in the canonical form
$$
P = \begin{pmatrix}Q&R\\0&I\end{pmatrix}
$$
where $Q$ is the substochastic matrix corresponding to transitions between transient states, $R$ is the substochastic matrix corresponding to transitions from a transient state to the absorbing state, and $I$ the identity matrix. The probability of transitioning from a transient state $i$ to a transient state $j$ in exactly $k$ steps is simply $Q^k$; summing this for all $k$ yields the fundamental matrix $N=\sum_{k=0}^\infty Q^k$. Now, since $Q$ has norm strictly less than one, it can be shown that $\sum_{k=0}^\infty Q^k = (I-Q)^{-1}$ (recall the formula for a geometric series), and hence
$$
N = \left(
\begin{array}{cccccc}
1 & \frac{1}{5} & \frac{3}{10} & \frac{1}{2} & 1 & 3 \\
0 & \frac{6}{5} & \frac{3}{10} & \frac{1}{2} & 1 & 3 \\
0 & 0 & \frac{3}{2} & \frac{1}{2} & 1 & 3 \\
0 & 0 & 0 & 2 & 1 & 3 \\
0 & 0 & 0 & 0 & 3 & 3 \\
0 & 0 & 0 & 0 & 0 & 6 \\
\end{array}
\right).
$$
Here the $(i,j)$ entry of $N$ is the expected number of visits to state $j$, given that the chain started in state $i$. Moreover, the expected number of steps before being absorbed is given by
$$
N\cdot\mathbf 1 = \left(
\begin{array}{cccccc}
1 & \frac{1}{5} & \frac{3}{10} & \frac{1}{2} & 1 & 3 \\
0 & \frac{6}{5} & \frac{3}{10} & \frac{1}{2} & 1 & 3 \\
0 & 0 & \frac{3}{2} & \frac{1}{2} & 1 & 3 \\
0 & 0 & 0 & 2 & 1 & 3 \\
0 & 0 & 0 & 0 & 3 & 3 \\
0 & 0 & 0 & 0 & 0 & 6 \\
\end{array}
\right)\begin{pmatrix}1\\1\\1\\1\\1\\1\\1\end{pmatrix} = \begin{pmatrix}6\\6\\6\\6\\6\\6\end{pmatrix},
$$
which is the same $6$ regardless of the starting state. (An interesting result, in my opinion.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many ways can you seat $12$ people at two round tables? In how many ways can you seat $12$ people at two round tables with $ 6$ places each? Think of possible ways of defining when two seatings are different, and find the answer for each.
Attempt:
Two considerations:
*
*Do we count equivalent rotations?
*Does the table matter?
Case $1$: Rotations & table matters
There are $12!$ ways of seating the group
Case $2$: Rotations matter, table doesn't
There are two ways to switch the tables. So there are $\frac{12!}{2}$ ways to seat the group.
Case $3$: Rotations don't matter, table does
Consider seating relative to special guest. There are $11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 $ ways to seat people around the first table. Seat the next special guest. There are $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ ways to seat the rest. The total is $11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 + 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.$
Case $4$: Rotations and tables don't matter
$\dfrac{(11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 + 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{2}$
| We can choose the group that will be at each table: $\frac{C_ {12}^{6}}{2} $ ways
After that, consider a circular permutation of both tables: $5!5!$
Then there will be $ \frac{C_{12}^{6}}{2}5!5! = 6,652,800 $ ways
In case the tables are different, simply multiply the answer by 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Argument of a complex fraction, why different answers? I want to take the argument of the following complex fraction. Using the second method I get a different answer, why is that?
$$
G(\omega)=
\frac{1}{(1+2\omega i)^2}
\tag 1
$$
Method 1:
\begin{align}
\arg\frac{1}{(1+2\omega i)^2}
&=\arg1-\arg\Big((1+2\omega i)^2\Big) \tag 2\\
&=\arg1-\arg\Big((1+2\omega i)(1+2\omega i)\Big) \tag 3\\
&=\arg1-\arg(1+2\omega i)-\arg(1+2\omega i) \tag 4\\
&=\arctan\frac{0}{1}-\arctan\Big(\frac{2\omega}{1}\Big)-\arctan\Big(\frac{2\omega}{1}\Big) \tag 5\\
&=-2\arctan(2\omega) \tag 6
\\
\end{align}
Method 2:
Expand the denominator:
$$
(1+2\omega i)^2=1-4\omega^2+4\omega i
$$
So I have
\begin{align}
\arg \frac{1}{(1+2\omega i)^2}
&=\arg\frac{1}{1-4\omega^2+4\omega i} \tag 7\\
&=\arg1-\arg(1-4\omega^2+4\omega i)\tag 8\\
&=-\arctan\bigg (\frac{4\omega}{1-4\omega^2}\bigg) \tag 9
\end{align}
So $(6)$ is not equal to $(9)$, What is wrong with method 2?
| There is nothing wrong with either method (besides not being cautious about range restrictions). The tangent double angle identity:
$$\tan(2A) = \frac{2\tan A}{1-\tan^2 A}$$
should answer your questions if you let $A = \arctan(2\omega)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$a(x)= x^4 - b^2x^3 - bx^2 -2x$ is divisible by $x-2$. Find $b$. $a(x)= x^4 - b^2x^3 - bx^2 -2x$ is divisible by $x-2$ for certain value $b$. Calculate all possible values $b$ can assume.
According to the factor theorem $x=2$ is a root when $a(x)=0$
$a(x)= x^4 - b^2x^3 - bx^2 -2x=0$
$a(2)= 2^4 - b^22^3 - b2^2 -2*2=0$
$8b^2 + 4b- 12=0$
$b^2 + 0.5b- 1.5=0$
After solving the quadratic equation, I get $b_1=-1,5$ and $b_2=1$
Is this right? Would you have solved it differently?
| As for how I might have solved it differently, I would've literally just done long division and tried to figure out which $b$ might work out - or at least that's my gut instinct. But just the thought of that now seems like a chore at best, and your method proves superior and is totally valid.
I guess if I really had to nitpick, I wouldn't have divided by $8$ in solving the quadratic, since the quadratic formula allows for your quadratic to have a leading coefficient which is not $1$ - I would've just jumped right into using it. But that's a personal thing; if you prefer to do the division, more power to you!
I would be interested in seeing other methods of solving the problem though.
Anyhow, to see the validity of these answers, simply plug each in for $b$ and check if $a(2) = 0$. If so, then $x-2$ divides the expression for $a(x)$ generated by that substitution of $b$, verifying the calculation.
$$\begin{align}
b=-1.5= -\frac 32 &\implies a(x) = x^4 - \frac 9 4 x^3 + \frac 3 2 x^2 - 2x \\
&\implies a(2) = 16 - \frac 9 4 \cdot 8 + \frac 3 2 \cdot 4 - 2 \cdot 2 \\
&\implies a(2) = 0 \\
b = 1 &\implies a(x) = x^4 - x^3 - x^2 - 2x \\
&\implies a(2) = 16 - 8 - 4 - 4 \\
&\implies a(2) = 0
\end{align}$$
Both solutions are thus valid. That $a(2)=0$ gives you a quadratic in $b$ for your original expression also ensures you have at most two possible values for $b$, ensuring that these two are the only solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\lim_{(x,y)\to (0,0)}\frac{x^4y^3}{x^2+y^4} = 0$. Show that
\begin{equation*}
\lim_{(x,y)\to (0,0)}\frac{x^4y^3}{x^2+y^4} = 0.
\end{equation*}
Let's show this formally using an $\epsilon-\delta$ proof. For $(x,y)\neq (0,0)$, let $\epsilon > 0$. Then if $(x,y)\in \mathbb{R}^2$ and $|(x,y)| < \frac{\epsilon}{3}$, then
\begin{equation*}
|y|^3\leq x^2+y^4 < \epsilon^3,
\end{equation*}
so $|y| < \sqrt[3]{\epsilon}$. Thus,
\begin{equation*}
\begin{split}
\left|\frac{x^4y^3}{x^2+y^4}-0\right| &= \left|\frac{x^4y^3}{x^2+y^4}\right| \\
&= \frac{x^4|y|^3}{x^2+y^4} \\
&\leq \frac{x^4y^3}{x^4} \\
&= |y|^3 \\
&< \epsilon
\end{split}
\end{equation*}
and we are done.
Where have I gone wrong? Thanks.
| By AM-GM we get $$x^2+y^4\geq 2|x|y^2$$ so we get
$$\frac{x^4y^3}{x^2+y^4}\le \frac{x^4|y|^3}{2|x|y^2}=\frac{|x|^3|y|}{2}$$ this tends to zero if $x,y$ tends to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos\sqrt{x}}}$ without using the l'Hospital rule The task is to evaluate
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}}.$$
I tried to use some trigonometric identities such as
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}= \lim_{x \to \ 0} \frac{1-
\sqrt{\cos\left(x\right)}}{1-
\cos\left(\sqrt{x}\right)}\cdot\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}}$$$$= \lim_{x \to \ 0} \frac{1-
\cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=
\lim_{x \to \ 0} \left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}
\right)^{2}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=\frac{1}{2}\lim_{x \to \
0}\left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}\right)^{2}$$
and this is where I have a problem.
| $$\lim_{x\to0}\dfrac{1-\sqrt{\cos x}}{1-\cos\sqrt x}$$
$$=\lim_{x\to0}\dfrac{1-\cos x}{1-\cos^2\sqrt x}\cdot\lim_{x\to0}\dfrac{1+\cos\sqrt x}{1+\sqrt{\cos x}}$$
$$=\left(\lim_{x\to0}\dfrac{\sin x}{\sin\sqrt x}\right)^2\cdot\lim_{x\to0}\dfrac{1+\cos\sqrt x}{(1+\sqrt{\cos x})(1+\cos x)}$$
Now $\lim_{x\to0}\dfrac{\sin x}{\sin\sqrt x}=\dfrac{\lim_{x\to0}\dfrac{\sin x}x}{\dfrac{\lim_{x\to0}\sin(\sqrt x)}{\sqrt x}}\cdot\lim_{x\to0}\dfrac x{\sqrt x}=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Find $\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$ without L'Hopital or Taylor series.
$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$$
My try is as follows:
$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=$$$$ \lim_{x\to
∞}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$$$=\lim_{x\to ∞}x\lim_{x\to
∞}\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$
which is $∞×0$ , but clearly this zero is not exactly zero. I was thinking about generalized binomial
theorem, but seems it will make the limit difficult, so how this kind of limits can be solved without using Taylor series or L'Hopital's rule?
| We first note that for any positive integer $n$ and any real $a$,
$$\lim_{x\to \infty}x\left(\sqrt[n]{1+\frac{a}{x}}-1\right)=
\lim_{s\to 1}a\frac{s-1}{s^n-1}=\lim_{s\to 1}\frac{a}{s^{n-1}+s^{n-2}+\dots +s +1}=\frac{a}{n}$$
where $s=\sqrt[n]{1+a/x}$ and therefore $a/x=s^n-1$, and $x=a/(s^n-1)$.
Hence, from your work, we split the limit in two:
$$\begin{align}\lim_{x\to +\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})
&=\lim_{x\to +\infty}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)
\\&=\lim_{x\to +\infty}x\left(\sqrt[3]{1+\frac{3}{x}}-1\right)-\lim_{x\to \infty}x\left(\sqrt[2]{1 +\frac{-2}{x}}-1\right)\\&=\frac{3}{3}-\frac{-2}{2}=1+1=2.
\end{align}$$
P.S. Note that on the other side,
$$\lim_{x\to -\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=-\infty$$
| {
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"url": "https://math.stackexchange.com/questions/3408325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
what is $\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}- 2\sqrt{x^{2}+x}+x\right)$? $$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}+x\right)$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-1-2\sqrt{x^{2}+x}+1+x\right)$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\left(\sqrt{x^{2}+x}+1\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\lim\limits_{x\to\infty}\left(x\sqrt{1+\frac{2}{x}}-1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\lim\limits_{x\to\infty}\left(x\sqrt{1+\frac{1}{x}}+1\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\left(1\right)+\lim\limits_{x\to\infty}\left(-2x\right)\left(\frac{1}{2}\right)+\lim\limits_{x\to\infty}x^{2}= ∞$$
another way I tried:$$\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}\right)+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}x\left(\sqrt{x^{2}+2x}-2\sqrt{x^{2}+x}\right)\cdot\frac{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{x\left(-2x^{2}-2x\right)}{\sqrt{x^{2}+2x}+2\sqrt{x^{2}+x}}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{x\left(-2x^{2}-2x\right)}{x\left(\sqrt{1+\frac{2}{x}}+2\sqrt{1+\frac{1}{x}}\right)}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{\left(-2x^{2}-2x\right)}{\left(\sqrt{1+\frac{2}{x}}+2\sqrt{1+\frac{1}{x}}\right)}+\lim\limits_{x\to\infty}x^{2}$$$$=\lim\limits_{x\to\infty}\frac{-2x^{3}-2x}{3}+\lim\limits_{x\to\infty}x^{2}=\lim\limits_{x\to\infty}\frac{x^{3}\left(-2+\frac{3}{x}-\frac{2}{x^{2}}\right)}{3}=-∞$$
and none of them are the answer, why the solutions are not right and can someone use an elementary way to solve the limit?
| Use the binomial theorem for fractional exponents to get $\sqrt{x^2+2x} = x\sqrt{1+\frac{2}{x}}=x\left(1+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{2x^3}-\dots\right)=x+1-\frac{1}{2x}+\frac{1}{2x^2}+\dots).$
Similarly, $\sqrt{x^2+x}=x(1+\frac{1}{x})^{1/2}=x(1+\frac{1}{2x}-\frac{1}{8x^2}+\frac{1}{16x^3}\dots)=x+\frac{1}{2}-\frac{1}{8x}+\frac{1}{16x^2}-\dots.$
Now, using these approximations, the limit can be rewritten as $\lim\limits_{x\to\infty}x((x+1-\frac{1}{2x}+\frac{1}{2x^2}-\dots)-2(x+\frac{1}{2}-\frac{1}{8x}+\frac{1}{16x^2}+\dots)+x),$ which yields a limit of $-\frac{1}{4}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Urn has 5 black and 5 white balls. X is RV that is number of balls you draw till first white. Calculate the PMF, E[X] and Var(X) An urn contains 5 black and 5 white balls. You draw without replacement till you draw your first white. X is a RV representing the total balls you have drawn including the first white.
a) What is the PMF?
$5\choose1$/$10\choose 1$ for $1\le X<2$
$5\choose1$$5\choose1$/$10\choose2$ for $2 \le X <3$
$5\choose 2$$5\choose 1$/$10\choose3$ for $3 \le X < 4$
$5\choose 3$$5\choose 1$/$10\choose4$ for $4 \le X < 5$
$5\choose 4$$5\choose 1$/$10\choose5$ for $5 \le X < 6$
$5\choose 5$$5\choose 1$/$10\choose6$ for $6 \le X < 7$
b)What is E[X]?
$5\choose1$/$10\choose 1$+2[$5\choose1$$5\choose1$/$10\choose2$]+3[$5\choose 2$$5\choose 1$/$10\choose3$]+4[$5\choose 3$$5\choose 1$/$10\choose4$]+5[$5\choose 4$$5\choose 1$/$10\choose5$]+6[$5\choose 5$$5\choose 1$/$10\choose6$]
c)What is Var(X)?
$E[X]^2-E[X^2]$
$E[X]^2-$$5\choose1$/$10\choose 1$+2[$5\choose1$$5\choose1$/$10\choose2$]+3[$5\choose 2$$5\choose 1$/$10\choose3$]+4[$5\choose 3$$5\choose 1$/$10\choose4$]+5[$5\choose 4$$5\choose 1$/$10\choose5$]+6[$5\choose 5$$5\choose 1$/$10\choose6$]
Is this all correct? I feel unconfident about the PMF, but that's what everything else relies on.
| You have not accounted for the order of selection. If the first white ball is drawn on the $k$th draw, then black balls must be drawn during the first $k - 1$ draws. The probability that this occurs is found by multiplying the probability that black balls are drawn in each of the first $k - 1$ draws by the probability that the $k$th ball drawn is white given that the first $k - 1$ balls were black. The probability that the first $k - 1$ balls drawn are black is
$$\frac{\dbinom{5}{k - 1}}{\dbinom{10}{k - 1}}$$
since each of the first $k - 1$ balls must be drawn from among the five black balls when drawing $k - 1$ balls from the $5 + 5 = 10$ balls in the urn. The probability that the $k$th ball drawn is white given that the first $k - 1$ balls were black is
$$\frac{5}{10 - (k - 1)} = \frac{5}{11 - k}$$
since $5$ of the remaining $10 - (k - 1) = 11 - k$ balls are white. Therefore,
the probability that the first white ball is drawn on the $k$th draw, where $1 \leq k \leq 6$, is
$$\Pr(X = k) = \frac{\dbinom{5}{k - 1}}{\dbinom{10}{k - 1}} \cdot \frac{5}{11 - k}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the $1000$th digit after the decimal point of $\sqrt{n},$ where $n=\underbrace{11\dots1}_{1998 \text{ 1's}}$
Find the $1000$th digit after the decimal point of $\sqrt{n}$, where $n=\underbrace{11\dots1}_{1998 \text{ 1's}}$.
Obviously, $\underbrace{11\dots1}_{1998 \text{ 1's}}=\dfrac{1}{9}\left(9\cdot10^{1997}+9\cdot 10^{1996}+\dots+9\right),$ so we want to find $\left(\dfrac{10^{1998}-1}{9}\right).$ If only there was some way to convert this expansion into some closed form. I'm not sure if calculus would be useful. The problem asks for a single digit, so if we consider repeating digits, everything will be a lot easier. There seems to be a pattern in the decimal expansions of numbers consisting of only $1.$ For instance,
$$\sqrt{1}=1,$$
$$\sqrt{11}=3.3166247...$$
$$\sqrt{111}=10.5356537...$$
$$\sqrt{1111}=33.3316666...$$
$$\sqrt{11111}=105.408728...$$
$$\sqrt{111111}=333.333166...$$
$$\sqrt{1111111}=1054.09250...$$
Every term of the form $\displaystyle\sum_{n=0}^{2k+1}10^n$ has $k+1$ $3$'s at the beginning and $k+1$ 3's right after the decimal expansion, followed by one $1,$ and $2(k+1)$ $6$'s. Proving this would prove that the $1000$th digit is $1.$ This is the same as showing that $\sqrt{\left(\dfrac{10^{2m}-1}{9}\right)}=\dfrac{10^{m}-1}{3}+\dfrac{1}{3}-\dfrac{1}{6}\cdot 10^{-m}+\epsilon_m$ where $|\epsilon_m|<10^{-2m}.$
Edit: the previous question I asked was inspired by the current one, but the previous question seemed to have a rather unpleasant answer, so I changed it.
| It seems you actually are on the right track.
Solving "smaller" problems of the "same kind" often pays off,
and this is one of the times it does.
You already found that
$$ n = \frac{10^{1998}-1}{9} = \frac{10^{1998}}{9} - \frac19.$$
The binomial expansion for $(a+b)^{1/2}$
with $a = \frac{10^{2m}-1}{9}$ and $b = -\frac19$ gives us
\begin{multline}
\left(\frac{10^{2m}}{9} - \frac19\right)^{\!1/2} =
\left(\frac{10^{2m}}{9}\right)^{\!1/2}
+ \frac12 \left(\frac{10^{2m}}{9}\right)^{\!-1/2} \left(-\frac19\right)\\
- \frac18 \left(\frac{10^{2m}}{9}\right)^{\!-3/2} \left(-\frac19\right)^{\!2}
+ \frac1{16} \left(\frac{10^{2m}}{9}\right)^{\!-5/2}
\left(-\frac19\right)^{\!3}
+ \cdots.\end{multline}
Now try the following comparisons:
\begin{align}
\left(\frac{10^{2m}}{9}\right)^{\!1/2} && \text{vs.} &&&
\dfrac{10^{m}-1}{3}+\dfrac13, \\
\end{align}
\begin{align}
\frac12 \left(\frac{10^{2m}}{9}\right)^{\!-1/2} \left(-\frac19\right)
&& \text{vs.} &&& - \dfrac16 10^{-m} \\
\end{align}
\begin{align}
- \frac18 \left(\frac{10^{2m}}{9}\right)^{\!-3/2} \left(-\frac19\right)^{\!2}
+ \frac1{16} \left(\frac{10^{2m}}{9}\right)^{\!-5/2}
\left(-\frac19\right)^{\!3} + \cdots
&& \text{vs.} &&& 10^{-2m}
\end{align}
You should be able to confirm the formula that you worked out from the pattern of digits in $\sqrt{11},$ $\sqrt{1111},$ and $\sqrt{111111}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int \sqrt{x^{2}+5}\,dx$ I tried:
But the answer is: $$\int \sqrt{x^{2}+5}\,dx=\frac{x\sqrt{x^{2}+5}}{2}+\frac{5}{2}\ln(x+\sqrt{x^{2}+5})+c$$
Where is my mistake? I checked my answer several times But still cant find out why my answer is wrong
| Your final answer is also correct. We have
\begin{align*}
\dfrac{5}{2}\ln\left|\dfrac{x+\sqrt{x^2+5}}{\sqrt{5}}\right|&=\dfrac{5}{2}\left(\ln|x+\sqrt{x^2+5}|-\ln\sqrt{5}\right)\\&=
\dfrac{5}{2}\ln(x+\sqrt{x^2+5})-\dfrac{5}{2}\ln\sqrt{5}\\&=
\dfrac{5}{2}\ln(x+\sqrt{x^2+5})-5\ln5.
\end{align*}
The absolute-value sign is not necessary since
$$\forall x\in\mathbb{R}:\;\; x+\sqrt{x^2+5}>0$$
and the constant number $-5\ln 5$ can be absorbed in the contant of integration $C$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to express $f(x)=\frac{1}{x-1} + \frac{1}{x-3}$ as a power series? I am working through my Calculus 2 course problems, and I have reached..
Express the function $f(x)=\frac{2x-4}{x^2-4x+3}$ as the sum of a power series by first using partial fractions.
I can do the partial fractions..
$$\frac{2x-4}{x^2-4x+3}=\frac{2(x-2)}{(x-1)(x-3)} =\frac{A}{x-1}+\frac{B}{x-3}$$
$$2(x-2)=A(x-3)+B(x-1)$$
$$\to -2=-2A \to1=A$$
$$\to2=2B\to 1 = B$$
$$\frac{1}{x-1} + \frac{1}{x-3}$$
I am not sure how to approach creating a power-series with two fractions like this.
| As written in the comments for your question, for $\vert r\vert<1$ we know that:
$$ \sum\limits_{n=0}^\infty r^n= \frac{1}{1-r} $$
You can rewrite for all $a\neq 0$:
$$ \frac{1}{x-a}=-\frac{1}{a} \cdot \frac{1}{1-\frac{x}{a}}= -\frac{1}{a} \sum\limits_{n=0}^\infty \Big( \frac{x}{a} \Big)^n= -\sum\limits_{n=0}^\infty \frac{x^n}{a^{n+1}} $$
Which is true for all $\vert x\vert<\vert a\vert$. And you can apply that to your fractions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $f$ by $f(x)=17+\int_0^x\frac{f(t)}{(t+2)(t+3)}\,dt$ Solve $f(1)$ by $f(x)=17+\int_0^x\frac{f(t)}{(t+2)(t+3)}\,dt$ when $x>0$.
My attempt was to differentiate both side and get $f'(x)=\frac{f(x)}{(x+2)(x+3)}$. However, I can't continue going on. Need help.
| Use the first part of the Fundamental Theorem of Calculus. From the first part
$$ f(x) = 17 + \int_0^x \frac{f(t)}{(t + 2)(t + 3)}dt $$
differentiating both sides give
$$ f’(x) = \frac{f(x)}{(x + 2)(x + 3)} $$
divide both sides by $f(x)$
$$ \frac{f’(x)}{f(x)} = \frac{1}{(x + 2)(x + 3)}$$
Partial Fraction Decomposition gives the RHS
$$ \frac{A}{x + 2} + \frac{B}{x + 3} $$
So when $x = -2$, $A = 1$ and when $x = -3$, $B = -1$ and therefore, RHS would be
$$ \frac{f’(x)}{f(x)} = \frac{1}{x + 2} - \frac{1}{x + 3} $$
Now, Integrate both sides with respect to x. Your equation now becomes
$$ \int\frac{f’(x)}{f(x)}dx = \int \frac{1}{x + 2} - \frac{1}{x + 3} dx $$
$$ = ln |f(x)| = ln |x + 2| - ln |x + 3| + C $$
Clean things up
$$ ln|f(x)| = ln\left|\frac{x + 2}{x + 3}\right| + C $$
exponentiating both sides
$$ e^{f(x)} = e^{ln\left|\frac{x + 2}{x + 3}\right| + C} $$
$$ f(x) = \frac{x + 2}{x + 3} C $$
We can say $e^C$ is also equal to $C$ because it’s the same as the constant of integration. Now let us try finding $f(0)$ by relating $f(x)$ into our current equation. In our original equation;
$$f(x) = 17 + \int_0^x \frac{f(t)}{(t + 2)(t + 3)}dt$$
$$\frac{(x + 2)}{(x + 3)}C = \int_0^x\frac{f(t)}{(t + 2)(t + 3)}dt$$
So when $x = 0$ we are left with $17$ alone in the right side of our equation because any integral that contains both limits equal to one another evaluates straightly to zero.
$$f(0) = 17$$
$$\frac{2}{3}C = 17$$
$$C = 17\left(\frac{3}{2}\right)$$
$$C = \frac{51}{2}$$
So, when x = 1
$$ f(1) = \frac{3}{4} C$$
$$ f(1) = \left(\frac{3}{4}\right) \left(\frac{51}{2}\right)$$
$$ f(1) = \frac{153}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417638",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Rationalize nested radical expression $\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$ I have a college task to rationalize this fraction.
$$\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}$$
I do not know how to simplify this fraction.
Please, explain how to remove the radical from the denominator. Thanks, for your help.
| As shown below, the expression can be rationalized and simplified to,
$$\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}} =6\sqrt2 + 2\sqrt{10}+4\sqrt{5+\sqrt5}$$
First, apply the denesting formula $\sqrt{a-\sqrt c}=\sqrt{\frac{a+\sqrt{a^2-c}}{2}}-\sqrt{\frac{a-\sqrt{a^2-c}}{2}}$ to the denominator,
$$\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}
=\frac{\sqrt{3+\sqrt5}}{2} -\frac{\sqrt{5-\sqrt5}}{2} $$
The expression then becomes,
$$A=\frac{8}{\sqrt{2-\sqrt{\frac{5+\sqrt{5}}{2}}}}
= \frac{16}{\sqrt{3+\sqrt5}-\sqrt{5-\sqrt5}}
$$
Next, apply the conjugate $\sqrt{3+\sqrt5}+\sqrt{5-\sqrt5}$ to the denominator,
$$A= \frac{8(\sqrt{3+\sqrt5}+\sqrt{5-\sqrt5})}{\sqrt5 -1}
$$
Apply the conjugate $\sqrt5 +1$ to the denominator again to obtain,
$$A= 2\left(\sqrt{3+\sqrt5}+\sqrt{5-\sqrt5}\right)(\sqrt5 + 1)
$$
Recognize $\sqrt{3+\sqrt5} = \frac{\sqrt5+1}{\sqrt2}$ to simplify,
$$A= \sqrt2 (\sqrt5+1)^2+2\sqrt{(5-\sqrt5)(\sqrt5 + 1)^2}$$
$$=\sqrt2 (6+2\sqrt5)+2\sqrt{(5-\sqrt5)(6+2\sqrt5 )}$$
$$=6\sqrt2 + 2\sqrt{10}+4\sqrt{5+\sqrt5}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is $| \lceil \frac{a}{2} \rceil - \lceil \frac{b}{2} \rceil |\geq \lfloor |\frac{a - b}{2}| \rfloor $? Let $a$ and $b$ be integers. Is it true that
$$
\left | \left \lceil \frac{a}{2} \right \rceil - \left \lceil \frac{b}{2} \right \rceil \right |\geq \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor
$$
Where $\lceil \cdot \rceil$ is the ceiling function, $\lfloor \cdot \rfloor$ the floor function and $|\cdot|$ is the absolute function.
The inequality seems to be true when I check it programatically but I would like to get a proof (or disproof) for this inequality.
| Assume without loss of generality that $a\ge b$. Then the inequality is
$$
\left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil
\ge \left\lfloor \frac {a-b}2 \right\rfloor
$$
If either $a$ or $b$ is an even integer, then we can pull the whole number $\frac a2$ or $\frac b2$ out of the floor function, and the inequality reduces to $\left\lceil \frac a2 \right\rceil \ge \left\lfloor \frac {a}2 \right\rfloor$ or $-\left\lceil \frac b2 \right\rceil \ge \left\lfloor -\frac {b}2 \right\rfloor$ (where the first is trivial and the second is actually an equality).
Assume therefore that neither of $a$ and $b$ is an even integer. Let $2m<a<2(m+1)$ and $2n<b<2(n+1)$, for some $m,n\in \mathbb Z$. Then
$$
\left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil
= (m+1)-(n+1)
= m-n
$$
On the other hand
$$
m-n-1<\frac a2 - \frac b2 < m-n+1
$$
which means that
$$
\left\lfloor \frac {a-b}2 \right\rfloor
\le m-n
= \left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil
$$
so we are done.
EDIT: I didn't notice you assumed $a$ and $b$ to be integers. Well, my answer works for all real numbers.
| {
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Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$ Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$
Finding the domain:
$$x^2+x+1>=0 \text { it is always true }$$
$$-1<=\sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}>=-1 \text { and } \sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}<=1$$
$$x^2+x+1<=1$$
$$x(x+1)<=0$$
$$x\in[-1,0]$$
Finding the range
$$f(x)=\sin^{-1}\left(\sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left([-1,0]+\frac{1}{2}\right)^2+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[\frac{-1}{2},\frac{1}{2}\right]^2+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[0,\frac{1}{4}\right]+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[\frac{3}{4},1\right]}\right)$$
$$f(x)=\sin^{-1}\left({\left[\frac{\sqrt{3}}{2},1\right]}\right)$$
$$f(x)\in \left[2m\pi+\dfrac{\pi}{3},2m\pi+\dfrac{2\pi}{3}\right] \text { where m is integer }$$
but actual answer is $\left[\dfrac{\pi}{3},\dfrac{\pi}{2}\right]$
|
$$f(x)\in \left[2m\pi+\dfrac{\pi}{3},2m\pi+\dfrac{2\pi}{3}\right] \text { where m is integer }$$
This isn't true because the range of the inverse sine function is from $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Let $$\theta=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$$
then
$$\sin{\theta}=\frac{\sqrt{3}}{2}$$
with the restriction that $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. The possible triangles with this ratio of $y$ to hypotenuse (i.e. $\sin$) are shown below
where only $\pi/ 3$ falls within the restricted range for the inverse sine function. Similarly, letting
$$\theta=\sin^{-1}\left(1\right)$$
means
$$\sin{\theta}=1$$
where only $ \pi/ 2$ falls within the restricted range for the inverse sine function.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof of a Zeta function identity How do I show that
$$
\sum_{n\geq 1} \frac{\zeta(2n)}{n(2n+1)}=\ln\frac{2\pi}{e}.
$$
I found this equation in my homework. I tried to integrate Zeta function's generating function twice, but the result has Li function in it. Is there any simple method to prove it?
| Alternative solution:
For convenience, set $0\ln 0 = 0$. We have
\begin{align}
\sum_{n\ge 1} \frac{\zeta(2n)}{n(2n+1)} &= \sum_{n\ge 1} \frac{1}{n(2n+1)}\sum_{k\ge 1}\frac{1}{k^{2n}}\tag{1}\\
&= \sum_{k\ge 1} \sum_{n\ge 1} \frac{1}{n(2n+1)} \frac{1}{k^{2n}}\tag{2}\\
&= \sum_{k\ge 1}\Big((k-1)\ln (k-1) - (k+1)\ln (k+1) + 2\ln k + 2 \Big)\tag{3}\\
&= \lim_{n\to \infty} \left[ \sum_{k=1}^n \Big((k-1)\ln (k-1) - (k+1)\ln (k+1) + 2\ln k + 2 \Big)\right]\tag{4}\\
&= \lim_{n\to \infty}\Big( - n\ln n - (n+1)\ln (n+1) + 2\ln(n!) + 2n\Big)\tag{5} \\
&= \ln \frac{2\pi}{\mathrm{e}}.\tag{6}
\end{align}
Here, in $(2) \Rightarrow (3)$, we have used Fact 1 given later;
in $(5) \Rightarrow (6)$, we have used Stirling's formula
$$\ln n! = \ln \sqrt{2\pi n} + n\ln \frac{n}{\mathrm{e}} + \mathrm{O}\big(\frac{1}{n}\big).$$
$\phantom{2}$
Fact 1: Let $y > 1$. Then
$$\sum_{n\ge 1}\frac{1}{n(2n+1)}\frac{1}{y^{2n}}
= (y-1)\ln (y-1) - (y+1)\ln (y+1) + 2\ln y + 2.$$
Since the series in LHS converges at $y=1$ and the limit of RHS as $y$ approaches $1$ exists, from Abel's theorem,
the above formula works also for $y=1$.
Proof of Fact 1: First, we have
$$\sum_{n\ge 1} \frac{1}{n(2n+1)}\frac{1}{y^{2n}} = \sum_{n\ge 1} \frac{1}{n} \frac{1}{y^{2n}} - \sum_{n\ge 1} \frac{2}{2n+1} \frac{1}{y^{2n}}.$$
Second, we have
$$\sum_{n\ge 1} \frac{1}{n} \frac{1}{y^{2n}} = -\ln \Big(1 - \frac{1}{y^2}\Big).$$
Third, let
$$g(x) = \sum_{n\ge 1} \frac{2}{2n+1} \frac{1}{y^{2n}}x^{2n+1}, \quad x\in [0, 1].$$
We have
$$g'(x) = \sum_{n\ge 1} 2\frac{1}{y^{2n}}x^{2n} = \frac{2x^2}{y^2-x^2} = -2 + \frac{y}{y-x} + \frac{y}{y+x}.$$
Since $g(0)=0$, we have
$$g(x) = -2x - y\ln (y-x) + y\ln (y+x)$$
which enables us to obtain (by letting $x=1$)
$$\sum_{n\ge 1} \frac{2}{2n+1} \frac{1}{y^{2n}} = -2 - y\ln (y-1) + y\ln (y+1).$$
The desired result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it true that $\lceil (3/2)^n \rceil - (3/2)^n > (3/4)^n$ for all $n>1$? It is easy to check that
$$
\lceil (3/2)^n \rceil - (3/2)^n \ge 1/2^n
$$
(to see this, just look at the binary representation of $(3/2)^n$).
Prove or disprove the stronger statement that
$$
\lceil (3/2)^n \rceil - (3/2)^n > (3/4)^n
$$
for all integer $n>1$. Here $\lceil x \rceil$ is the ceiling function; e.g.
$\lceil 3/2 \rceil = 2$.
Note: Using PARI/GP, the statement has been verified up to $n=20000$. (For this verification, default precision may need to be increased and/or the original inequality should be restated in terms of the fractional part $\{(3/2)^n\}$, as Gottfried Helms explains below.)
| From $3^n=2^n\cdot k+ r, 0\leq r<2^n-1$
$$\left \lceil \frac{3^n}{2^n} \right \rceil - \frac{3^n}{2^n} > \frac{3^n}{2^{2n}} \iff
2^n\left \lceil \frac{3^n}{2^n} \right \rceil - 3^n > \frac{3^n}{2^n} \iff \\
2^n\left(\left \lfloor \frac{3^n}{2^n}\right \rfloor+1\right)-3^n>\frac{3^n}{2^n} \iff
2^n(k+1)-3^n> \frac{3^n}{2^n} \iff ...$$
but $3^n+2^n=2^n\cdot(k+1)+r$, thus
$$...\iff 2^n-r > \frac{3^n}{2^n} \tag{1}$$
It is worth noting that $r=3^n-2^n\left \lfloor \frac{3^n}{2^n}\right \rfloor$ and it is known that
$$r>2^n-\left \lfloor \frac{3^n}{2^n}\right \rfloor-2$$
admits only finitely many solutions for $n$ (although, none known). I.e. from some large enough $n_0$ onwards
$$r\leq 2^n-\left \lfloor \frac{3^n}{2^n}\right \rfloor-2 \Rightarrow
2^n-r\geq 2^n - 2^n+\left \lfloor \frac{3^n}{2^n}\right \rfloor+2=\\
\left \lfloor \frac{3^n}{2^n}\right \rfloor+2 > \frac{3^n}{2^n}$$
and, via the equivalences from $(1)$, we have that the original inequality is valid, from some $n_0$ onwards.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\;\sum_{i=1}^n \frac{p_i}{p_{i+1}+p_{i+2}} \ge \frac{n}{2}$ using Jensen's inequality I want to prove the inequality in a general case ($n\ge3$): $\;\sum\limits_{i=1}^n \dfrac{p_i}{p_{i+1}+p_{i+2}} \ge \dfrac{n}{2}$ for any positive numbers $p_i,\,i=\overline{1,n}$, where $p_{n+1} = p_1$ and $p_{n+2} = p_2$. It seems that there is no counter-example...
In particular cases when $n=3$, $n=4$ and $n=5$ there is no problem with use of Jensen's inequality for the function $f(x) = \dfrac{1}{x}$:
$$\sum\limits_{i=1}^n \dfrac{m_i}{x_i} \ge \dfrac{\left(\sum_{i=1}^n m_i\right)^2}{\sum_{i=1}^n m_i x_i}$$
One can choose $m_i=p_i$ and $x_i=p_{i+1}+p_{i+2}$. Then, for example, for $n=3$ we have:
$$\dfrac{p_1}{p_2+p_3} + \dfrac{p_2}{p_3+p_1} + \dfrac{p_3}{p_1+p_2} \ge \dfrac{(p_1 + p_2 + p_3)^2}{p_1(p_2+p_3) + p_2(p_3+p_1) + p_3(p_1+p_2)}.$$
RHS is equal to
$$\dfrac{(p_1 + p_2 + p_3)^2}{p_1^2 + p_2^2 + p_3^2 + 2(p_1p_2 + p_2p_3 + p_3p_1) - (p_1^2 + p_2^2 + p_3^2)} = \dfrac{(p_1 + p_2 + p_3)^2}{(p_1 + p_2 + p_3)^2 - (p_1^2 + p_2^2 + p_3^2)}.$$
According to Cauchy–Bunyakovsky inequality for $a_1=a_2=a_3=1;\; b_i=p_i$:
$$(1^2+1^2+1^2)\cdot (p_1^2 + p_2^2 + p_3^2) \ge (p_1 + p_2 + p_3)^2,$$
finally we have the following:
$$\dfrac{p_1}{p_2+p_3} + \dfrac{p_2}{p_3+p_1} + \dfrac{p_3}{p_1+p_2} \ge \dfrac{(p_1 + p_2 + p_3)^2}{(p_1 + p_2 + p_3)^2 - \frac{1}{3} (p_1 + p_2 + p_3)^2} = \dfrac{3}{2}.$$
But there is a problem with a plenty of $p_i p_j$ in cases starting with $n=6$. Can someone help to prove the inequality for $n=6$ or greater?
| For $n=6$ we need to prove that
$$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+f}+\frac{e}{f+a}+\frac{f}{a+b}\geq3,$$ where $a$, $b$, $c$, $d$, $e$ and $f$ are positives.
Indeed, by C-S we obtain:
$$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c+d+e+f)^2}{\sum\limits_{cyc}(ab+ac)}=$$
$$=\frac{(a+d+b+e+c+f)^2}{(a+d)(b+e)+(a+d)(c+f)+(b+e)(c+f)}\geq3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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inequality under condition $x+y+z=3$ $x$, $y$ and $z$ being three positive real numbers such that $x+y+z=3$
It is asked to prove that
$$ \dfrac{\sqrt x}{y + z}+ \dfrac{\sqrt y}{x + z} + \dfrac{\sqrt z}{y + x} \ge \dfrac 3 2$$
I tried relating it to Nesbitt but achieved no progress
Thanks for any ideas.
| Let $x=a^2,$ $y=b^2$ and $z=c^2$, where $a$, $b$ and $c$ are positives.
Thus, $a^2+b^2+c^2=3$ and
$$\sum_{cyc}\frac{\sqrt{x}}{y+z}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{3-a^2}-\frac{1}{2}\right)=\frac{1}{2}\sum_{cyc}\frac{(a-1)(a+3)}{3-a^2}=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{(a-1)(a+3)}{3-a^2}-(a^2-1)\right)=\frac{1}{2}\sum_{cyc}\frac{(a-1)^2(a+2)a}{3-a^2}\geq0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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if $s=\sqrt{x^2+6x+9} +\sqrt{x^2+24x+144}$ and $-10 < x < -8$ there follows that...
if $s=\sqrt{x^2+6x+9} + \sqrt{x^2+24x+144}$ and $-10 < x < -8$ there follows that $s=9$
After noticing that the first radical is $x+3$ and the second one is $x+12$ all I get is $s=2x+15$, which btw. is suggested as an alternative answer in the exam.
So, how to go about this?
| It is not true that $\sqrt{x^2 + 6x + 9} = \sqrt{(x + 3)^2} = x + 3$ since $x + 3 < 0$ if $x < -3$. Remember that $\sqrt{x}$ is the principal (nonnegative) square root of $x$. Actually,
$$\sqrt{x^2 + 6x + 9} = |x + 3|$$
Similarly,
$$\sqrt{x^2 + 24x + 144} = |x + 12|$$
Since $x + 3 \geq 0$ if $x \geq -3$ and $x + 3 < 0$ if $x < -3$,
$$
|x + 3| = \begin{cases}
x + 3 & \text{if $x \geq -3$}\\
-x - 3 & \text{if $x < -3$}
\end{cases}
$$
Since $x + 12 \geq 0$ if $x \geq -12$ and $x + 12 < 0$ if $x < -12$
$$
|x + 12| = \begin{cases}
x + 12 & \text{if $x \geq -12$}\\
-x - 12 & \text{if $x < -12$}
\end{cases}
$$
Thus, if $-10 < x < -8$,
\begin{align*}
s & = \sqrt{x^2 + 6x + 9} + \sqrt{x^2 + 12x + 24}\\
& = |x + 3| + |x + 12|\\
& = -x - 3 + x + 12\\
& = 9
\end{align*}
| {
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Fibonacci sum: $\sum\limits_{k\ge0}\frac{F_{2k+1}}{2k+1}\left(\frac{2+2\sqrt{2}}{1+\sqrt{\frac{17+8\sqrt{2}}{5}}}\right)^{2k+1}(-\frac{1}{5})^k$
Prove $$\frac{3\pi}{8}=\sum\limits_{k\ge0}\left(-\frac{1}{5}\right)^k\frac{F_{2k+1}}{2k+1}q^{2k+1}$$
where $q=\frac{2+2\sqrt{2}}{1+\sqrt{\frac{17+8\sqrt{2}}{5}}}$, and $F_n$ are the Fibonacci numbers. This result is from Wolfram.
My ideas as to a proof are limited. I know that this is some sort of series for the inverse tangent because $3\pi/8=\arctan(1+\sqrt{2})$, but I have never seen any series for $\arctan$ involving the Fibonacci numbers. Essentially, it comes to the explicit evaluation of the function $$f(x)=\sum_{k\ge0}\frac{F_{2k+1}}{2k+1}x^{k}.$$
Indeed, the sum in question is the value $qf(-q^2/5)$.
Potentially connected is the closed form
$$\sum_{k\ge0}F_kx^k=\frac{x}{1-x-x^2},$$
perhaps we could get some sort of $\arctan$-related integral out of this.
Could I have some help evaluating $f$? Thanks.
| Let $$f(x) = \sum_{k\geq 0}F_{k+1}x^k$$ so that $$(1-x-x^2)f(x)=F_1+(-F_1+F_2)x$$ or $$f(x) =\frac{1}{1-x-x^2}$$ Integrating the above we get $$F(x) =\int_{0}^{x}\frac{dt}{1-t-t^2}=\sum_{k\geq 0}\frac{F_{k+1}}{k+1}x^{k+1}$$ The integrand on left side can be written $$\frac{1}{(1-at)(1-bt)}=\frac{A}{1-at}+\frac{B}{1-bt}$$ where $a, b$ are roots of $z^2-z-1=0$ and $$A=\frac{a} {a-b}, B=-\frac{b} {a-b} $$ Letting $$a=\frac{1+\sqrt{5}}{2},b=\frac{1-\sqrt {5}}{2}$$ we get $a-b=\sqrt{5}$ and hence we have $$F(x) =\frac{1}{\sqrt {5}}\log\frac{1-bx}{1-ax}$$ and $$G(x) =F(x) - F(-x) =2\sum_{k\geq 0}\frac{F_{2k+1}}{2k+1}x^{2k+1}$$ Replacing $x$ by $ix$ we get $$\sum_{k\geq 0}(-1)^{k}\frac{F_{2k+1}}{2k+1}x^{2k+1}=-i\frac{G(ix)} {2}$$ The sum in question is $$-i\sqrt {5}\frac{G(iq/\sqrt{5})}{2}$$ Now $$G(x) =\frac{1}{\sqrt{5}}\log\frac{1+\sqrt {5}x+x^2}{1-\sqrt{5}x+x^2}$$ and hence $$G(iq/\sqrt{5})=\frac{1}{\sqrt{5}}\log\frac{5+5iq-q^2}{5-5iq-q^2}=\frac{2i}{\sqrt {5}}\arctan\frac{5q}{5-q^2}$$ The desired sum is thus equal to $$\arctan\frac{5q}{5-q^2}$$ The fact that $5q/(5-q^2)=1+\sqrt{2}$ needs some effort and I have not been able to do so efficiently.
As pointed out in comments the series for $G$ diverges if $|x|>1/\phi$ or equivalently $|q| >\sqrt{5}/\phi$. The value of $q$ used in question is greater than $\sqrt{5}/\phi$ and hence the sum in question diverges.
| {
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If $a^{7!} +b^{8!} +c^{9!} +d^{10!} =x$ where a,b,c and d are natural numbers that are not multiples of 10, the..... If $a^{7!} +b^{8!} +c^{9!} +d^{10!} =x$ where a,b,c and d are natural numbers that are not multiples of 10, then how many distinct values of unit's digit of x are possible ?
How to proceed in such question, I don't have any idea on this. .. please guide thanks a lot ...
| Notice $4$ divides all of $7!, 8!, 9!,10!$.
Euler's theorem says that if $\gcd(k,10) = 1$ then $k^4 \equiv 1 \pmod {10}$ so the last digit of $k^{v!}$ is $1$ if $k$ is an odd number that doesn't end with $5$.
If you don't know Euler's theorem or modular arithmetic notice that if $k = 10w + v$ where $v=\pm 1, \pm 3$ then $k^4 = 10^4w^4 + 4*10^3w^3v + 6*10^2w^2v^2 + 4*10w*v^3 + v^4$ so the last digit if $k^4$ is the same as the last digit of $v^4$ and $v^4 =1$ or $v^4 = 81$. So the last digit is one. So $k^{4m}$ will have the last digit of $1$ if $k$ ends with $1, 3, 10-3=7, $ or $10-1=9$.
The other things to worry about are if $k$ is even or $k$ ends in $5$.
If $k$ is ends with $5$ then $k^m$ ends with $5$ (that's obvious isn't it?)
And if $k$ is even... well by the chinese remainder theorem $k^w\equiv 0 \pmod 2$ and $k^4\equiv 1 \pmod 5$ but Euler Th so $k^4 \equiv 6\pmod {10}$.
If you don't know Euler's th or CRT... well,... if $k =2j$ is even then $k^4 = 2^4 j^4 = 16*j^4$. If $j$ is odd and not a multiple of $5$ then $j^4$ ends with $1$ and $k^4$ ends with $6$. If $j$ is even just repeat: $j = 2l$ and $k^4 = 16*16*l^4$ and that ends with $6$ if $l$ is odd and if $l$ is even repeat as often as necessary.
So you have the last digits are $1, 5$ or $6$.
So we can have $4$ ones and end with $4$.
We can have $3$ ones and a five and end with $8$
We can have $3$ ones and a six and end with $9$.
We can have $2$ ones and $2$ fives and end with $2$ and .... so on.
Edit: an improved way of enumerating the possibile sums $\bmod 10$
Once you render the units digit of each term as $\in\{1,5,6\}$, you can consider the sum separately $\bmod 5$ and $\bmod 2$.
In $\bmod 5$ the residue matches the number of $1$s plus the number of $6$s in the sum. This can be any of $0,1,2,3$ or $4$.
In $\bmod 2$ you can have either residue $0$ or $1$ by swapping a $1$ for a $6$ or vice versa, provided that at least one of these is included. But that requires an overall residue of $1,2,3$, or $4\bmod 5$ from the above. The combination with residue $0\bmod 5$, $5+5+5+5$, does not allow this swapping and thus forces a residue of $0\bmod 2$.
So the can all combinations of resudues $\bmod 2$ and $\bmod 5$ except $(1\bmod 2, 0\bmod 5)$ allowing to all units digits except $5$.
| {
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$2\arctan(\phi^{-n})=\arctan\frac{p}{q}$ or $\arctan\frac{p\sqrt{5}}{q}$, where $\phi$ is the Golden Ratio. Is there a pattern in the $\frac{p}{q}$s? It is very interesting to know that
$$\arctan\frac{1}{\phi} + \arctan\frac{1}{\phi^3}= \arctan 1 = \frac{\pi}{4}$$
where Golden ratio $\phi = \frac12(\sqrt5 +1)$ is in association with circle constant $\pi$.
More interesting phenomenon is evaluation of inverse tan functions of inverse of $\phi$ in its consecutive powers as follows
$$\begin{align}
2\arctan\frac{1}{\phi} &= \arctan 2 &
2\arctan\frac{1}{\phi^2} &= \arctan\frac{2\sqrt{5}}{5}\\
2\arctan\frac{1}{\phi^3} &= \arctan\frac{1}{2} &
2\arctan\frac{1}{\phi^4} &= \arctan\frac{2\sqrt{5}}{15}\\
2\arctan\frac{1}{\phi^5} &= \arctan\frac{2}{11} &
2\arctan\frac{1}{\phi^6} &= \arctan\frac{\sqrt5}{20} \\
2\arctan\frac{1}{\phi^7} &= \arctan\frac{2}{29} &
2\arctan\frac{1}{\phi^8} &= \arctan\frac{2\sqrt5}{105} \\
2\arctan\frac{1}{\phi^9} &= \arctan\frac{1}{38} &
2\arctan\frac{1}{\phi^{10}} &= \arctan\frac{2\sqrt5}{275} \\
2\arctan\frac{1}{\phi^{11}} &= \arctan\frac{2}{199}
\end{align}$$
Here are the observations
*
*Odd powers of inverse $\phi$ in double arctan functions lead to arctan of well defined fractions
*Even powers of inverse $\phi$ in double arctan functions lead to arctan of fractions involving $\sqrt5$.
My curiosity is to know, is there any pattern in these interesting series?
I will be grateful to understand more, if anyone has come across such evaluations.
| Hint:
$$ \arctan(x)\pm\arctan(y) = \arctan(z) $$
where $z$ is:
$$ z = \frac{x\pm y}{1\mp xy} $$
We will take a look at this:
$$ 2\arctan\phi^{-n} = \arctan \phi^{-n}+\arctan\phi^{-n} = \arctan(\frac{\phi^{-n}+\phi^{-n}}{1-\phi^{-n}\phi^{-n}}) = \arctan(\frac{2\phi^{-n}}{1-\phi^{-2n}}) = $$
$$ = \arctan(\frac{2}{\phi^{n}-\phi^{-n}}) = \arctan(\frac{2}{e^{\ln\phi^{n}}-e^{\ln\phi^{-n}}}) =\arctan(\frac{1}{\sinh(n\ln\phi)})$$
| {
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$(p-1)(p+1)/24 \in \mathbb N$ for all primes $p \geq 5$ I want to show
\begin{align}
\frac{(p-1)(p+1)}{24} \in \mathbb N \quad \text{for all primes} \quad p \geq 5 \tag{1}.
\end{align}
I can show $(1)$, if the following statement is true.
Let $a,b,c,d \in \mathbb N$ and $a \geq b \cdot c \cdot d$.
\begin{align}
\text{If} \quad \frac{a}{b},\frac{a}{c},\frac{a}{d} \in \mathbb N, \quad \text{then} \quad \frac{a}{b \cdot c \cdot d} \in \mathbb N \tag{2}.
\end{align}
Given $(2)$ we show that $(1)$ is true for $a = (p-1)(p+1)$, $b = 2$, $c = 3$ and $d = 4$. Since $p$ is a prime $(p-1)$ and $(p + 1)$ are even, implying $(p-1)/2 \in \mathbb N$, $(p+1)/2 \in \mathbb N$ and thus $(p-1)(p+1)/2 \in \mathbb N$ and $(p-1)(p+1)/4 \in \mathbb N$. One of the three numbers $(p-1)$, $p$ and $(p+1)$ must be divisible by 3. Since $p$ is a prime either $(p-1)$ or $(p+1)$ is divisible by 3, implying $(p-1)(p+1)/3 \in \mathbb N$.
Question Is $(2)$ true?
| There are two cases : Either $p = 4k+1$ or $4k+3$ ,
Case 1 :
$$ \begin {align}p &=4k+1 \\
p^2-1 &= 16k^2 +1+8k-1 = \color{red}{8k\,(2k+1)} \end{align}$$
which is clearly divisible by $8$.
Now $K $can be of two form $k = 3m \,, 3m+1$.If $k = 3m+2$ , then $p$ is not prime.
If $k = 3m$ , it is clear that $p^2-1$ is divisible by $24$.
If $k = 3m+1$ , then $(2k+1)$ is divisible by $3$
Case 2 :
$$\begin{align} p &=4k+3 \\
p^2-1 &= 16k^2 +9+24k-1 = 8(2k^2+3k+1) = \color{blue}{8\,(k+1)\,(2k+1)} \end{align}$$
which is clearly divisible by $8$.
Now $K $can be of two form $k =3m+1 ,3m+2 $.
If $k = 3m+1$ , it is clear that $2k+1$ is divisible by $3$.
If $k = 3m+2$ , then $(k+1)$ is divisible by $3$
Which completes all the cases and proves your result.
| {
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Issue with the following limit $\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$ Calculate the following limit:
$\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$
When I calculate it I get to different answers.
First way (Edit: this is where I did the mistake): $$\bigg(2 * \sqrt{1 + \frac{1}{n}}\bigg)^n = \bigg({4 + \frac{4}{n} \bigg)^\frac{n}{2}} = \bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4} \cdot \frac{4}{n}\cdot\frac{n}{2}}}$$
When we do $\lim_{n \to \infty}\bigg(\bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4}\cdot \frac{4}{n}\cdot\frac{n}{2}}}\bigg)$ we get $e^2$
Now the second way:
$$\bigg(2 \cdot \sqrt{1 + \frac{1}{n}}\bigg)^n = 2^n\cdot (1 + \frac{1}{n})^{n \cdot \frac{1}{2}}$$
When we do limit out of this we get $2^\infty \cdot \sqrt{e}$ which is of course $\infty$.
Could someone point out the mistake I made?
Edit:
I just realised where my mistake lies! I mistakenly thought that $(4 + \frac{4}{n})^\frac{n}{4} = e$ which is false, actually $(1 + \frac{4}{n})^\frac{n}{4} = e$. The second way of calculating this limit is the correct one!
| $$
\lim\limits_{x\to a} f(x)g(x) \ne \lim\limits_{x\to a} f(x).\lim\limits_{x\to a} g(x)
$$
if the individual limits don't exist
Therefore, to say
$$
\lim\limits_{n\to \infty} \bigg(2 * \sqrt{1 + \frac{1}{n}}\bigg)^n = \lim\limits_{n\to \infty} 2^n * \lim\limits_{n\to \infty}(1 + \frac{1}{n})^{\frac{n}{2}}
$$
is incorrect since the first limit on the RHS doesn't exist.
The second way is therefore inapplicable. @user has given the excellent observation which gives the answer
$$
\bigg(2 * \sqrt{1 + \frac{1}{n}}\bigg)^n\ge 2^n \to \infty
$$
| {
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Is it possible to swap only two elements on a 3x3 grid by swapping rows and columns? Take a 3x3 grid with a different number in each spot, like one cell of a Sudoku puzzle:
\begin{array}{|c|c|c|}
\hline
1&2&3\\
\hline
4&5&6\\
\hline
7&8&9\\
\hline
\end{array}
Suppose you can rearrange the number by swapping two rows or two columns.
$$\left.\begin{array}{|c|c|c|}\hline 1&2&3\\ \hline 4&5&6\\ \hline7 & 8 & 9\\ \hline
\end{array} \longrightarrow \begin{array}{|c|c|c|}\hline 1&2&3\\ \hline7&8&9\\ \hline4&5&6\\ \hline
\end{array} \longrightarrow \begin{array}{|c|c|c|}\hline2&1&3\\ \hline8&7&9\\ \hline5&4&6\\ \hline
\end{array}
\right.$$
Is it possible to manipulate the grid such that an arbitrary pair of numbers is swapped (with the rest returned to their original positions)?
e.g.
$$\left.\begin{array}{|c|c|c|}\hline 1&2&6\\ \hline 4&5&3\\ \hline7&8&9\\ \hline
\end{array} \qquad\text{or}\qquad \begin{array}{|c|c|c|}\hline 8&2&3\\ \hline4&5&6\\ \hline7&1&9\\ \hline
\end{array} \qquad\text{or}\qquad \begin{array}{|c|c|c|}\hline1&2&3\\ \hline4&7&6\\ \hline5&8&9\\ \hline
\end{array}
\right.$$
| Four numbers that are in the "corners" of a rectangle (like 1379 or 1278) will remain in the corners of a rectangle no matter how many or which swaps happen. So you cannot swap just two elements while keeping the rest fixed.
| {
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Suppose $a,b$ are real numbers and $0 < a < b$. Prove $ba^n + ab^n < a^{n+1} + b^{n+1}$
$a,b$ are real numbers and $0 < a < b$
Prove that for all $n ≥ 1$, $ba^n + ab^n < a^{n+1} + b^{n+1}$
My attempt:
By contradiction. Suppose
$$\tag{* }ba^n + ab^n ≥ a^{n+1} + b^{n+1}$$
Rearranging $(*)$ gives
$$\begin{align}
0 & ≥ a^{n+1} + b^{n+1} - ba^n - ab^n \\
& ≥ a^{n+1} - ba^n + b^{n+1} - ab^n \\
& ≥ a^n(a - b) - b^{n}(a - b) \\
& ≥ (a - b)(a^n - b^n)
\end{align}$$
But since $0 < a < b$, we have $(a-b) < 0$ and $(a^n - b^n) < 0$, and thus $(a-b)(a^n - b^n) > 0$.
Hence a contradiction. $\Box$
*
*Is it correct?
*Is there a better way? (Preferably direct proof)
| According to the Rearrangement inequality:
$$0<a<b \Rightarrow 0<a^n<b^n\\
ba^n + ab^n < b\cdot b^n +a\cdot a^n=a^{n+1} + b^{n+1}.$$
| {
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$(z + 1/z)^2 + (z^2 + 1/z^2)^2 + (z^3 + 1/z^3)^2 + (z^4 + 1/z^4)^4 + (z^5 + 1/z^5)^2 + (z^6 + 1/z^6)^2$ if $z^2 + z + 1 = 0$
Solve $(z + \frac{1}{z})^2 + (z^2 + \frac{1}{z^2})^2 + (z^3 + \frac{1}{z^3})^2 + (z^4 + \frac{1}{z^4})^2 + (z^5 +\frac{1}{z^5})^2 + (z^6 + \frac{1}{z^6})^2$
if $z^2 + z + 1 = 0$
I tried this problem and come up with this solution:
$z^2 + z + 1 = 0$
$z^2 = -z -1$
$(z^2)^2 = (-z -1)^2$
$z^4 + (z^2 + 1) = z^2+2z+1+ (z^2 + 1)$
$z^4 + z^2 + 1 = 2(z^2 + z + 1) = 0$
and this help me to figure out that the answer is 12. However the step
$(z^2)^2 = (-z -1)^2$ isn't quite valid because it turns a quadratic equation to a quartic equation, which will lead to extreneous roots.
Therefore, I want to ask for a better solution to this. Thanks in advance.
| $z+\dfrac1z=-1$
$z^2+\dfrac1{z^2}=\left(z+\dfrac1z\right)^2-2=?$
$z^3-1=(z-1)(z^2+z+1)=?$
| {
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Friend squares numbers
Two perfect squares are friends if one is obtained from the other
adding the digit $1$ at the left.
For instance, $1225 = 35^2$ and $225 = 15^2$ are friends.
Prove that there are infinite pairs of odd perfect squares that are
friends.
Solution: Suppose that $n^2$ and $m^2$ are friends and both odd, with the former being greater than the latter. Then $n$ and $m$ are both odd, and $n^2 - m^2$ is a power of ten.
Then, because of the difference-of-squares identity $n^2 - m^2 = (n-m)(n+m)$, this means that $n+m = 2^a 5^b$ for some $a, b$ and $n-m = 2^c 5^d$ for some $c, d$, where $a + c = b + d$.
Adding the two equations, we obtain $2n = 2^a 5^b + 2^c 5^d$. The right hand side has to be even, but not a multiple of 4. This means that $a$ and $c$ have to be both positive or zero. If they are both positive, they cannot be both at least two (since the right hand side would be a multiple of 4), but they cannot be both one (since $2^a 5^b + 2^c 5^d = 2(5^b + 5^d)$ would be a multiple of 4), so one of them has to be one and the other has to be at least two. If $a$ and $c$ are both zero, then $2^a 5^b + 2^c 5^d = 5^b + 5^d \equiv 1^b + 1^d \equiv 2\pmod{4}$, so all such cases work.
Similarly, subtracting the two equations, we obtain $2m = 2^a 5^b - 2^c 5^d$. The right hand side has to be even, but not a multiple of 4. This means that $a$ and $c$ have to be both positive or both zero. If they are both positive, they cannot be both at least two, and they cannot be both one, so one of them has to be one and the other has to be at least two. However, if $a$ and $c$ are both zero, then $2^a 5^b - 2^c 5^d = 5^b - 5^d \equiv 1^b - 1^d \equiv 0\pmod{4}$, so no such cases work.
In conclusion, we have demonstrated that $n^2$ and $m^2$ are friends and both odd iff $n+m = 2^a 5^b$ for some $a, b$ and $n-m = 2^c 5^d$ for some $c, d$, where $a + c = b + d$ and where $a = 1, c \geq 2$ or vice versa. This clearly covers infinitely many cases, so we are done.
Why is $ n ^ 2 - m ^ 2 $ a power of ten?
| You may be interested in a relatively simple method of generating an infinite sequence of such numbers.
First we shall prove that if $A$ and $B$ are powers of $5$ and $C$ a power of $2$ such that
$\,\,\,\,\,\,2A^2BC$ is a power of $10$ and $1\le\frac{C}{B}\le \frac{5}{2},$
then a solution is given by $n=A(5B+C)$ and $m=A(5B-C)$.
$n^2-m^2=20A^2BC$ is a power of $10$ and so we require $$20A^2BC>A^2(5B-C)^2\ge 2A^2BC.$$ This gives two inequalities which are both satisfied when $\frac{C}{B}$ is in the above range $$C^2-12BC+25B^2\ge 0 \text { and } C^2-30BC+25B^2<0. $$
$A=5,B=1,C=2$ is an example of numbers satisfying the condition given above. From any such solution we can generate a larger solution as follows.
If $\frac{C}{B}< \frac{25}{16}$ then $(A',B',C')=(5A,5B,8C)$
If $\frac{C}{B}\ge \frac{25}{16}$ then $(A',B',C')=(5A,25B,16C)$
The first few solutions generated from $(5,1,2)$ are as follows.
$n^2=1225$ and $m^2=225$
$n^2=15405625$ and $m^2=5405625$
$n^2=12127515625$ and $m^2=2127515625.$
| {
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Evaluating $\lim_{x\to0}{\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}}}$ without L'Hopital's rule I'm currently struggling with this task:
$$\lim_{x\to0}{\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}}}$$
By now I have only come up with the idea to use the formula of sum of cubes, so, the numerator would become $\cos{4x}-\cos{5x}$ and the denominator would be $3(1-\cos{3x})$:
$$\frac{\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}}}{1-\cos{3x}} = \frac{(\sqrt[3]{\cos{4x}}-\sqrt[3]{\cos{5x}})*(\sqrt[3]{cos^2{4x}} + \sqrt[3]{\cos{4x}}*\sqrt[3]{\cos{4x}} + \sqrt[3]{\cos^2{5x}})}{(1-\cos{3x})*(\sqrt[3]{cos^2{4x}} + \sqrt[3]{\cos{4x}}*\sqrt[3]{\cos{4x}} + \sqrt[3]{\cos^2{5x}})} = \frac{\cos{4x}-\cos{5x}}{3(1-\cos{3x})}$$
In this expression it looks like $\cos{4x}-\cos{5x}$ and $(1-\cos{3x})$ increase with the same speed (if I can say it this way) and the limit is likely to be $\frac{1}{3}$.
Still I have no idea how to prove that without using L'Hopital's rule which is prohibited by the task.
| You can use the Taylor Expansion of $\cos(x)$.
The numerator is $\dfrac{(5x)^2-(4x)^2}{2!} + O\left({x^4}\right)$
The denominator is $\dfrac{(3x)^2}{3\cdot2!} + O\left({x^4}\right)$
Dividing by $x^2$ and using $3^2+4^2=5^2$ gives the limit as $1/3$.
| {
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Find the given limit of $S$?
Find the given limit of $S$ ?
$$S=\lim _{n \rightarrow \infty} \left (\frac{\sin \frac{\pi}{ n+1}} { 1} +\frac{\sin \frac{2\pi}{ n+1}} { 2} + + ....+ \frac{\sin \frac{n\pi}{ n+1}} { n} \right )$$
My attempt :
i construct modified the given series
$\frac{\pi}{ n+1}\left (\frac{\sin \frac{\pi}{ n+1}} { \frac{1\pi}{ n+1}} +\frac{\sin \frac{2\pi}{ n+1}} { \frac{2\pi}{ n+1}} + ... + ....+ \frac{\sin \frac{n\pi}{ n+1}} { \frac{n\pi}{ n+1}} \right )$
After that im not able to proceed further
Any hints/solution will be appreciated
thanks u
| This converges to:
$$Si(\pi) = \int_{0}^{\pi} \frac{sin(x)}{x} dx \approx 1.851937051982468 $$
by numerical analysis.
HTH
| {
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$\lim_{x\to 1} (x^3+2x^2-2)=1$ using the definition to prove that $\lim_{x\to 1} (x^3+2x^2-2)=1$
let $\varepsilon>0$, and $x\in\mathbb{R}$
we must find $\delta>0$ such that if $|x-1|<\delta $ then $|x^3+2x^2-2-1|<\varepsilon$
i do :
$$
|x^3-1+2x^2-2|\leq |x^3-1|+2|x^2-1|=|x^3-1|+2 |(x-1)| |(x+1)|=
$$
$$
|x-1|(|x^2+x+1|+2|x+1|)\leq \delta (|x^2+x+1|+2|x+1|)
$$
how to continue?
| You can go the other way
$|x - 1|<\delta$
$-\delta < x-1 <\delta$
$1-\delta < x < 1+\delta$
If we assume $0 < \delta < 1$ then
$(1-\delta)^3 + 2(1-\delta)^2 - 2 < x^3 + 2x^2-2 < (1+\delta)^3 + 2(1+\delta)^2 - 2$
$-\delta^3 +4\delta^2 -5\delta + 1 < x^3 + 2x^2-2<\delta^3 + 4\delta^2 +5\delta + 1$
Now if we assume $0 < \delta<1$ then $0< \delta^3 <\delta^2 < \delta < 1$ so
$-\delta^3 + 4\delta^2 - 5\delta + 1 > -\delta^3 + 4\delta^3 -5\delta + 1=$
$-3\delta^3 - 5\delta + 1 > -3\delta - 5\delta + 1 = -8\delta + 1$.
And $\delta^3 + 4\delta^2 +5\delta + 1 < \delta + 4 \delta + 5 \delta + 1=10\delta + 1$.
So $-10\delta + 1<-8\delta+1 < x^3 + 2x^2-2< 10\delta + 1$
So $-10\delta < x^3 + 2x^2 -2 - 1 < 10 \delta$
$|(x^3 + 2x^2 - 2)-1|< 10\delta$.
So if we set $\delta = \min(\frac \epsilon {10}, 1)$ then
$|x-1| < \delta \implies |(x^3 + 2x^2 - 2)-1|< 10\delta \le \epsilon$
| {
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"timestamp": "2023-03-29T00:00:00",
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Bulgarian Mathematical Olympiad 1987 The sequence $x_1,x_2,\dots$ is defined by the equalities $x_1=x_2=1$ and $$x_{n+2}=14x_{n+1}-x_n-4, n\geq 1$$
Prove that each number of the given sequence is a perfect square.
I used the standard way to solve recurrence relations and arrived at
$$x_n=\frac{1}{6}\bigg( (2+\sqrt{3})^{n-\frac{3}{2}} + (2-\sqrt{3})^{n-\frac{3}{2}} \bigg)^2$$
which I think should be quite close. Any idea how to proceed?
| Long, but hopefully useful version ... Solving recurrence using characteristic polynomials (e.g. like here)
$$\color{red}{x_n}=\frac{1}{6}\left((26-15\sqrt{3})\cdot(7+4\sqrt{3})^n + (26+15\sqrt{3})\cdot(7-4\sqrt{3})^n + 2\right)=\\
\frac{1}{6}\left((26-15\sqrt{3})(2+\sqrt{3})^{2n} + (26+15\sqrt{3})(2-\sqrt{3})^{2n} + 2\right)=\\
\frac{1}{6}\left((2-\sqrt{3})(2+\sqrt{3})^{2(n-1)} + (2+\sqrt{3})(2-\sqrt{3})^{2(n-1)} + 2\right)=\\
\frac{1}{6}\left(\sqrt{2-\sqrt{3}}\cdot(2+\sqrt{3})^{n-1} + \sqrt{2+\sqrt{3}}\cdot(2-\sqrt{3})^{n-1}\right)^2=\\
\frac{1}{6}\left(\frac{\sqrt{3}-1}{\sqrt{2}}\cdot(2+\sqrt{3})^{n-1} + \frac{\sqrt{3}+1}{\sqrt{2}}\cdot(2-\sqrt{3})^{n-1}\right)^2=\\
\frac{1}{6}\left(\frac{3-\sqrt{3}}{\sqrt{6}}\cdot(2+\sqrt{3})^{n-1} + \frac{3+\sqrt{3}}{\sqrt{6}}\cdot(2-\sqrt{3})^{n-1}\right)^2=\\
\left(\frac{3-\sqrt{3}}{6}\cdot(2+\sqrt{3})^{n-1} + \frac{3+\sqrt{3}}{6}\cdot(2-\sqrt{3})^{n-1}\right)^2=\color{red}{(z_{n-1})^2}$$
where
$$z_{n+2}=4z_{n+1}-z_n, \space z_1=1, z_2=3 \tag{1}$$
Note that
$$\left(3-\sqrt{3}\right)\left(2+\sqrt{3}\right)=3+\sqrt{3}$$
and the recurrence in $(1)$ isn't really guessing, since both $2+\sqrt{3}$ and $2-\sqrt{3}$ are roots of $x^2-4x+1=0$, which is the characteristic polynomial of $(1)$.
| {
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"source": "stackexchange",
"question_score": "5",
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Using De Moivre's Theorem to prove things So I need to use ($\cos\theta$ + i$\sin\theta$)^4 to prove;
1) $\cos4\theta$ = 8$\cos^4\theta$ - 8$\cos^2\theta$ + 1
2) $\sin4\theta$ = 4$\sin\theta$$\cos\theta$ ($\cos^2\theta$ - $\sin^2\theta$)
The farthest I can get is equating;
$\cos4\theta$ = $\cos^4\theta$ - 6$\cos^2\theta$$\sin^2\theta$ + $\sin^4\theta$
and
$\sin4\theta$ = 4$\cos^3\theta$$\sin\theta$ - 4$\cos\theta$$\sin^3\theta$
How do it prove 1) and 2) from this? (assuming I didn't mess up my expansions and the equating)?
| You're almost there.
For $\cos4\theta$, use that $\sin^2\theta=1-\cos^2\theta$ and simplify.
For $\sin4\theta$, factor out $4\sin\theta \cos\theta$.
| {
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Solve inequality $\cos(x)+2\tan(x)\le2+\sin(x)$ Solve inequality
$\cos(x)+2\tan(x)\le2+\sin(x)$
My proof:
$\cos(x)+2\tan(x)\le2+\sin(x)\\\cos^2(x)+2\sin(x)\le2\cos(x)+\sin(x)\cos(x)\\\cos(x)\left(\cos(x)-2 \right )+\sin(x)\left(2-\cos(x) \right )\le0\\\left(\cos(x)-\sin(x) \right )\left(\cos(x)-2 \right )\le0\\\sqrt{2}\left(\sin\left(x-\frac{\pi}{4} \right ) \right )\left(\cos(x)-2 \right )\le0\\-\sqrt{2}\cos\left(x+\frac{\pi}{4} \right )\left(\cos(x)-2 \right )\le0$
I stopped at this moment and I have no idea what to do now
| $\cos x + 2 \tan x < 2 + \sin x$
$\cos x + 2 \frac {\sin x}{\cos x} - \sin x - 2 < 0$
$\frac {\cos^2 x + 2\sin x - \sin x\cos x - 2\cos x}{\cos x} < 0\\
\frac {\cos x (\cos x - 2) + \sin x(2-\cos x)}{\cos x} < 0\\
\frac {(\cos x-\sin x) (\cos x - 2)}{\cos x} < 0$
$\cos x - 2 < 0$
$\frac {(\cos x-\sin x)}{\cos x} > 0$
$\cos x > \sin x$ and $\cos x > 0$
or $\cos x < \sin x$ and $\cos x < 0$
Over the interval $[0,2\pi)$
$[0 , \frac {\pi}{4}) \cup (\frac {\pi}{2},\frac {5\pi}{4})\cup (\frac {3\pi}{2}, 2\pi)$
| {
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Find maximum lattice-point triangle with orthocenter at origin I was wondering if someone could help me understand this problem:
Let $S$ be the set of points $(x,y)$ in the Cartesian plane for which $\left(\tfrac{x}{x^2+y^2},\tfrac{y}{x^2+y^2}\right)$ is a lattice point. Find the largest possible area of a triangle with vertices in $S$ and orthocenter at the origin.
So I assume that means find all $(x,y)\in\mathbb{R}:\left(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}\right)\in\mathbb{Z}\backslash\{0,0\}$ correct? If that's the case, then it's easy to identify these points as solutions to the equations
$$
\begin{align}
\frac{x}{x^2+y^2}=n\\
\frac{y}{x^2+y^2}=m
\end{align}
$$
with $(n,m)\in\mathbb{Z}$. This turns out to be the intersections of four (infinitely) concentric circles shown in the plot below with the dot in $S$ and so the matter then becomes finding the largest triangle with such vertices and orthocenter at the origin.
Is this the correct way of interpreting this problem? I get $3/10$ as the answer if so but that's only by brute-force checking and I at present do not have a rigorous proof.
Thanks guys.
:
| $$\frac{x}{x^2+y^2}=n \qquad \frac{y}{x^2+y^2}=m$$
$$x^2 = n^2(x^2+y^2)^2 \qquad y^2 = m^2(x^2+y^2)^2$$
$$(x^2+y^2)=(m^2+n^2)(x^2+y^2)^2$$
$$(m^2+n^2)(x^2+y^2) = 1$$
$$x^2+y^2 = \dfrac{1}{m^2+n^2}$$
So $(x,y)$ can be any point on the circle with center at the origin and radius of $\dfrac{1}{\sqrt{m^2+n^2}}$
With $m,n \in \mathbb Z$, the largest possible radius is $\dfrac{1}{2}$.
From this answer the largest triangle will be any equilateral triangle inscribed on the circle $x^2+y^2=\dfrac 12$. Which, as you commented, has an area of $\dfrac 38 \sqrt 3$
| {
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Sum of the series $\sum_{n=0}^{\infty} \lfloor nr \rfloor x^n$ where $r$ is rational? Can we find the exact sum of the series $\sum_{n=0}^{\infty} \lfloor nr \rfloor x^n$ where $r$ is rational?
There is a special case given here but I don't know how to prove it and can we get the sum for the general case?
| First, this is not a full answer but enough significant progress that I felt it was worth typing up.
Note that the sum only converges if $|x|<1$. As such, for the remainder of this post we shall assume $-1<x<1$.
Let $r=a/b$ be rational. If $r=a/b$ is an integer, then the sum is easily computed to be
$$\sum_{n=0}^\infty\left\lfloor n \frac{a}{b}\right\rfloor x^n=\sum_{n=0}^\infty n rx^n=\frac{r x}{(x-1)^2}$$
Thus, we may as well assume $r=a/b$ is not an integer from here on out. Now, we can write $n$ as $n=bq+s$ where $q\in\{0,1,2,3,...\}$ and $0\leq s <b$. Then
$$\left\lfloor n \frac{a}{b}\right\rfloor=\left\lfloor (bq+s) \frac{a}{b}\right\rfloor=qa+\left\lfloor s \frac{a}{b}\right\rfloor$$
Then the sum can be rewritten as
$$\sum_{n=0}^\infty \left\lfloor n \frac{a}{b}\right\rfloor x^n=\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left\lfloor (bq+s) \frac{a}{b}\right\rfloor x^{bq+s}=\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left(qa+\left\lfloor s \frac{a}{b}\right\rfloor\right) x^{bq+s}$$
We can split this sum into two parts:
$$\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left(qa+\left\lfloor s \frac{a}{b}\right\rfloor\right) x^{bq+s}=\sum_{q=0}^\infty\sum_{s=0}^{b-1}qa x^{bq+s}+\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^{bq+s}$$
The first infinite sum is easily found to be
$$\sum_{q=0}^\infty\sum_{s=0}^{b-1}qa x^{bq+s}=\frac{a x^b}{(x-1) \left(x^b-1\right)}$$
Since the second infinite sum converges absolutely, we may as well sum the $q$ sum before the $s$ sum. That is
$$\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^{bq+s}=\sum_{s=0}^{b-1}\sum_{q=0}^\infty\left\lfloor s \frac{a}{b}\right\rfloor x^{bq+s}$$
$$=\sum_{s=0}^{b-1}\left(\left\lfloor s \frac{a}{b}\right\rfloor x^s\sum_{q=0}^\infty x^{bq}\right)=\sum_{s=0}^{b-1}\left(\left\lfloor s \frac{a}{b}\right\rfloor x^s\frac{1}{1-x^b}\right)$$
$$=\frac{1}{1-x^b}\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^s$$
The question then becomes: what is the sum
$$\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^s?$$
Now, we can slightly simplify this if we assume $a/b$ is in its most reduced form. That is, $\gcd(a,b)=1$ (if $a=0$, then the original sum is clearly $0$ so we may ignore this case). We may then write $a=mb+t$ for $m\in\mathbb{Z}$, $\gcd(t,b)=1$, and $0< t<b-1$. Note that if $t=0$, then $a/b$ is an integer. Otherwise, the sum is
$$\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^s=\sum_{s=0}^{b-1}\left\lfloor s \frac{mb+t}{b}\right\rfloor x^s=\sum_{s=0}^{b-1}\left\lfloor s \frac{mb+t}{b}\right\rfloor x^s$$
$$=\sum_{s=0}^{b-1}smx^s+\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s=\frac{m \left(b x^{b+1}-x^{b+1}-b x^b+x\right)}{(x-1)^2}+\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s$$
We have now further simplified the question to: what is the sum
$$\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s$$
where $\gcd(t,b)=1$ and $0<t<b$? Unfortunately, this seems like a difficult problem. Here are the first few such sums, for $b=1,2,\cdots ,6$:
$$\left(
\begin{array}{c}
\{0\} \\
\{0\} \\
\left\{0,x^2\right\} \\
\left\{0,2 x^3+x^2\right\} \\
\left\{0,x^4+x^3,2 x^4+x^3+x^2,3 x^4+2 x^3+x^2\right\} \\
\left\{0,4 x^5+3 x^4+2 x^3+x^2\right\} \\
\end{array}
\right)$$
Overall, the sum is
$$\frac{a x^b}{(x-1) \left(x^b-1\right)}+\frac{1}{1-x^b}\left( \frac{m \left(b x^{b+1}-x^{b+1}-b x^b+x\right)}{(x-1)^2}+\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s \right)$$
where $m=\lfloor a/b\rfloor$, $\gcd(t,b)=1$, and $0<t<b$.
Just to show that this answer is correct, we can rederive the special case given in the link above. In this case, $a=1$, implying the second term is $0$ as
$$m=\lfloor a/b\rfloor=\lfloor 1/b\rfloor=0$$
and
$$ \sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s =\sum_{s=0}^{b-1}\left\lfloor s \frac{1}{b}\right\rfloor x^s =\sum_{s=0}^{b-1}\left\lfloor \frac{s}{b}\right\rfloor x^s =0$$
as $0\leq s<b$. Thus, the sum is simply
$$\frac{x^b}{(x-1) \left(x^b-1\right)}.$$
In the example, they use $k=1/x$, which gives us
$$\frac{ (1/k)^b}{((1/k)-1) \left((1/k)^b-1\right)}=\frac{ (1/k)^bk^{b+1}}{((1/k)-1) \left((1/k)^b-1\right)k^{b+1}}=\frac{k}{(k-1)(k^b-1)}$$
which is what they got.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
First 10 terms of a sequence My problem:
Give the first 10 terms of $f(5,b)$ in the recursive sequence defined by
\begin{equation}
f(a,b) = \begin{cases}
-2 & \text{if } b = 0\\
3 & \text{if } b = 1\\
1 & \text{if } b = 2\\
f(f(b+2, b-3), b-1)-a & \text{if } b \geq 3\\
\end{cases}
\end{equation}
I think the answer may be -4, -9, -14, -19, -24, -29, -34, -39, -44, -49. Can anyone confirm this?
| We have $f(a,3)=f(f(5,0),2)-a=-a+1$
$f(a,4)=f(f(6,1),3)-a=f(3,3)-a=-a-2$
$f(a,5)=f(f(7,2),4)-a=f(1,4)-a=-a-3$
$f(a,6)=f(f(8,3),5)-a=f(-7,5)-a=-a+4$
$f(a,7)=f(f(9,4),6)-a=f(-11,6)-a=-a+15$
$f(a,8)=f(f(10,5),7)-a=f(-13,7)-a=-a+28$
$f(a,9)=f(f(11,6),8)-a=f(-7,8)-a=-a+35$
$f(a,10)=f(f(12,7),9)-a=f(3,9)-a=-a+32$
$f(a,11)=f(f(13,8),10)-a=f(15,10)-a=-a+17$
$f(a,12)=f(f(14,9),11)-a=f(21,11)-a=-a-4$
So substituting $a=5$ we get:
$f(5,0)=-2; f(5,1)=3; f(5,2)=1; f(5,3)=-4$
$f(5,4)=-7,f(5,5)=-8,f(5,6)=-1,f(5,7)=10$
$f(5,8)=23,f(5,9)=30,f(5,10)=27,f(5,11)=12,f(5,12)=-9$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 1,
"answer_id": 0
} |
Triangle $ABC$ where $a+c=12$, $b+c=13$, $\angle A = 60^{\circ}$, what is $a^{2} + b^{2} + c^{2}$? Triangle $ABC$ where $a+c=12$, $b+c=13$, $\angle A = 60^{\circ}$, what is $a^{2} + b^{2} + c^{2}$?
By the cosine rule
$$ a^{2} = b^{2} + c^{2} - bc$$
then since $(b+c)^{2}=169=b^{2}+c^{2}+2bc$, the above becomes
$$ a^{2} = 169 - 3bc$$
We know $b = a+1$. Also
$$ (a+c)(b+c) = 156 $$
$$ ab + ac + bc + c^{2} = 156$$
Also, $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab+ac+bc)$, so
$$ ab +ac + bc = \frac{(25-c)^{2}- (a^{2} + b^{2}+ c^{2})}{2}$$
$$ = \frac{(25-c)^{2}- (169 + b^{2}+ c^{2} - 3bc)}{2} $$
$$ = \frac{(25-c)^{2}- (169 + (13-c)^{2}+ c^{2} - 3(13-c)c)}{2} $$
How to solve this?
*Without Heron's formula?
| Hint
$$b=a+1,c=12-a$$
$$2bc\cos A=b^2+c^2-a^2$$
Replace the values of $b,c,A=60^\circ$ to find $a$
| {
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} |
$\int\frac1{(x^2+1)^2}\ dx$ by partial fraction decomposition Is there any possible way to calculate the integral of $\frac{1}{(x^2+1)^2}$ by partial fraction decomposition? I do not know the formulas for the trigonometric method.
Thank you!
| $$\Re\frac1{(x+i)^2}=\Re\frac{(x-i)^2}{(x+i)^2(x-i)^2}=\frac{x^2-1}{(x^2+1)^2}=\frac1{x^2+1}-\frac2{(x^2+1)^2}$$
and by integration,
$$\Re\frac{-1}{x+i}=\arctan x-2\int\frac{dx}{(x^2+1)^2}.$$
Hence
$$2\int\frac{dx}{(x^2+1)^2}=\arctan x+\frac{x}{x^2+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3451875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\lim_{n\to\infty}{\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+\cdots+\frac{a^n}{n}\right)}$ where $a>1$
$$\underset{n\rightarrow\infty}\lim{\frac{n}{a^{n+1}}\left(a+\frac{a^2}{2}+\frac{a^3}{3}+\cdots+\frac{a^n}{n}\right)}=?, \;\;a>1$$
In Shaum's Mathematical handbook of formulas and tables I've seen:
$$\;\;\;\;\;\;\;\;\;\;\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\;,x\in\langle-1,1]\;\;\;\;\;\;\;$$
$$\frac{1}{2}\ln{\Bigg(\frac{1+x}{1-x}\Bigg)}=1+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\cdots\;\;\;,x\in\langle-1,1\rangle$$
The term in parentheses reminded me of the harmonic series. I thought of using the Taylor series. Is that a good idea?
It says $a>0$ so I probably can't use these two formulas.
On the other hand:
$$e^x=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\;\;\;\;\;\;,$$
but there are no factorials in the denominators.
Source in Croatian: 2.kolokvij, matematička analiza
| $$
\begin{align}
\lim_{n\to\infty}\frac{n}{a^{n+1}}\left(a+\frac{a^2}2+\cdots+\frac{a^n}n\right)
&=\lim_{n\to\infty}\left(\frac1a+\frac{n}{n-1}\frac1{a^2}+\frac{n}{n-2}\frac1{a^3}+\cdots\right)\tag1\\
&=\frac1a+\frac1{a^2}+\frac1{a^3}+\cdots\tag2\\
&=\frac1{a-1}\tag3
\end{align}
$$
The series on the right side of $(1)$ is dominated by
$$
\frac1a+\frac2{a^2}+\frac3{a^3}+\cdots=\frac{a}{(a-1)^2}\tag4
$$
which validates $(2)$.
| {
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"url": "https://math.stackexchange.com/questions/3454372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Prove that the minimum values of $x^2+y^2+z^2$ is $27$ with given condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$. Question: Prove that the minimum values of $x^2+y^2+z^2$ is $27$, where $x,y,z$ are positive real variables satisfying the condition $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$.
From AM$\ge$ GM, we have $\left( \dfrac{x^2+y^2+z^2}{3}\right)^3\ge (xyz)^2=(xy+yz+zx)^2$. Is it possible to show thae result from this relation?
| To maximize $x^2+y^2+z^2$ subject to the constraint $\dfrac1x+\dfrac1y+\dfrac1z=1$, you could use the Lagrange multiplier method and look at $x^2+y^2+z^2-\lambda\left(\dfrac1x+\dfrac1y+\dfrac1z-1\right)$. Take derivatives and see that $2x+\dfrac\lambda{x^2}=2y+\dfrac\lambda{y^2}=2z+\dfrac\lambda{z^2}=\dfrac1x+\dfrac1y+\dfrac1z-1=0,$ so $2x^3+\lambda=2y^3+\lambda=2z^3+\lambda,$ so $x=y=z$ and $x=3$ and $x^2+y^2+z^2=27$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
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A bound for $\sqrt\frac{b+c-a}a+\sqrt\frac{c+a-b}b+\sqrt\frac{a+b-c}c$ in a triangle Assume that $ABC$ is a triangle with $a\geq b\geq c$, where the angle $A$ has a fixed value. We denote by $\Sigma$ the sum
$$\sqrt\frac{b+c-a}a+\sqrt\frac{c+a-b}b+\sqrt\frac{a+b-c}c.$$
Then the only possible values of $A$ are $\pi/3\leq A<\pi$ and we have:
(i) The smallest possible value $\Sigma$ is
$$\frac{4\sin\frac A2+\sqrt{2\left( 1-\sin\frac A2\right)}}{\sqrt{2\sin\frac A2}}.$$
(ii) If $\pi/3\leq A<\pi/2$ then the largest possible value of $\Sigma$ is
$$\frac{4\cos A+\sqrt{2\left( 1-\cos A\right)}}{\sqrt{2\cos A}}.$$
(iii) If $\pi/2\leq A<\pi$ then there is no finite upper bound for $\Sigma$.
My question is how to prove (i), (ii), and (iii).
I firstly tried to square the $LHS$, but nothing. I also tried the Radulescu-Maftei theorem, but it didn't help. Hence, I am looking forward to seeing your ideas.
| I will prove (i) for now. Use the Ravi substitution:
$$a = y+z,\,\, b = x+z,\,\, c = x+y\implies x\leq y\leq z.$$
And then use the following substitution:
$$\dfrac{2x}{y+z} = X^2,\,\, \dfrac{2y}{x+z} = Y^2,\,\, \dfrac{2z}{x+y} = Z^2\implies X\leq Y\leq Z.$$
After these two, the LHS is simply $X+Y+Z.$ For the RHS, we can easily see:
$$\sin^2\frac A2=\dfrac{1-\cos A}{2} = \dfrac{a^2-b^2-c^2+2bc}{4bc} = \dfrac{(a-b+c)(a+b-c)}{4bc} = \dfrac{yz}{(x+y)(x+z)}=\dfrac{Y^2Z^2}{4}.$$
and so our inequality will be equivalent to:
$$X+Y+Z\geq 2\sqrt{YZ} + \sqrt{\dfrac{2-YZ}{YZ}}.$$
Now I am going to switch to lower-case $X\to x,Y\to y,Z\to z$ for ease of typing. One can check that our $x,y,z$ satisfies:
$$x^2y^2+y^2z^2+z^2x^2 + x^2y^2z^2 = 4$$
and that $yz\geq 1$ because $yz = 2\sin\frac A2\geq 1$ is fixed.
Finally, we have to prove the following inequality that is equivalent to our original:
$$\sqrt{\dfrac{(2-yz)(2+yz)}{y^2+z^2+y^2z^2}}+y+z\geq 2\sqrt{yz}+\sqrt{\dfrac{2-yz}{yz}}.$$
Subtracting the RHS from the LHS and taking out $(\sqrt{y} - \sqrt{z})^2,$ we are left with:
$$(\sqrt{y} - \sqrt{z})^2\left(1 - \sqrt{\dfrac{2-yz}{yz}}\cdot\dfrac{y+z+2\sqrt{yz}}{\sqrt{yz(2+yz)} + y^2+z^2+y^2z^2}\right)\geq 0,$$
because the two fractions are clearly less than or equal to $1$ due to the fact that $yz\geq 1.$
(iii) is kind of obvious. Take $b = a - \epsilon$ and $c = 2\epsilon.$ Then, the last term on the expression:
$$\dfrac{a+b-c}{c} =\dfrac{2a-3\epsilon}{2\epsilon}\to\infty$$
as $\epsilon\to 0.$
(ii) becomes much more complicated by the same, completing the square approach. Lagrange multiplier works but still more tedious than the nice one in the other answer.
| {
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"url": "https://math.stackexchange.com/questions/3456307",
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"source": "stackexchange",
"question_score": "10",
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How to solve a system of equations? How can I solve this system of equations?
$$ y^2 = x^3 - 3x^2 + 2x$$
$$ x^2 = y^3 - 3y^2 + 2y$$
| $$y^2 = x^3 - 3x^2 + 2x$$
$$x^2 = y^3 - 3y^2 + 2y$$
Subtracring $(1.)$ from $(2.)$ ,
$$\begin{align} x^2 - y^2 & = y^3 - x^3 + 3x^2 - 3y^2 + 2y-2x
\\ \color{#f04}{(x+y)(x-y)}& \color{#f04}{= (y-x)(y^2+x^2+xy) +3(x+y)(x-y) + 2(y-x)} \\
\end{align}$$
If $(x-y)=0 \implies x=y$ , Which gives $\color{#39f}{x^2 = x^3 - 3x^2 + 2x}$
, Which yields $\color{#4b9}{x =0 , 2\pm \sqrt 2}$
If $x \ne y$ , Then :
$$x+y = -(x^2 + y^2 +xy) + 3(x+y) - 2$$
$$(x^2+y^2+xy) = 2(x+y-1)$$
$$x^2 + (y-2)x + (y^2-2y+2) = 0$$
Using Quadratic formula , we have obtain the discriminant :
$$D = (y-2)^2 - 4(y^2-2y+2)$$
$$D = -3y^2 + 4y - 4 \implies \boxed{D = -3 \left(y - \frac23 \right)^2 - \frac83}$$
We can observe that $\color{#da0}{D\lt 0}$ for all values of $y$. Hence the equation has no solutions for real $x,y$.
Hence the only solutions are $(x , y) = (0,0) \,\, , (2\pm\sqrt2,2\pm\sqrt2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I show equivalence between $\frac{(x^2 + 1)^{5/2}}{5} - \frac{(x^2 + 1)^{3/2}}{3}$ and $\frac{(x^2 + 1)^{3/2}(3x^2 - 2)}{15}$? I wanted to find $\int x^3 \sqrt{x^2 + 1} dx$ and came to this answer:
$$\frac{(x^2 + 1)^{5/2}}{5} - \frac{(x^2 + 1)^{3/2}}{3}$$
Using software to verify, I find Maple does not show equivalence between this and the actual answer, yet an online integral calculator does show equivalence between this and the actual answer with a footnote that further simplification does yield the expected answer.
For reference, the actual answer is:
$$\frac{(x^2 + 1)^{3/2}(3x^2 - 2)}{15}$$
How do I show algebraically that these two are equivalent? I could plug in $x$ for an arbitrary amount of numbers and compare but I want to be more declarative and show equality algebraically.
I've tried to combine this into one fraction:
$$\begin{align}
& \frac{(x^2 + 1)^{5/2}}{5} - \frac{(x^2 + 1)^{3/2}}{3} \\
=& \frac{3\times(x^2 + 1)^{5/2}}{3\times5} - \frac{5\times(x^2 + 1)^{3/2}}{5\times3} \\
=& \frac{3(x^2 + 1)^{5/2}}{15} - \frac{5(x^2 + 1)^{3/2}}{15} \\
=& \frac{3(x^2+1)^{5/2} - 5(x^2+1)^{3/2}}{15}
\end{align}$$
But I am not sure where or how else to go from here.
| Factor out $(x^2+1)^{3/2}$:
$$\frac{(x^2 + 1)^{5/2}}{5} - \frac{(x^2 + 1)^{3/2}}{3}$$
$$=(x^2+1)^{3/2}\left(\dfrac{x^2+1}5-\dfrac13\right)$$
$$=(x^2+1)^{3/2}\;\dfrac{[3(x^2+1)-5]}{15}$$
$$=(x^2+1)^{3/2}\;\dfrac{(3x^2-2)}{15}$$
Or from where you left off:
$$\frac{3(x^2+1)^{5/2} - 5(x^2+1)^{3/2}}{15}$$
$$=(x^2+1)^{3/2}\;\dfrac{[3(x^2+1)-5]}{15}$$
$$=(x^2+1)^{3/2}\;\dfrac{(3x^2-2)}{15}$$
| {
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Taylor expansion of $\exp(\sin x)$ around $c=0$
Consider the Taylor's expansion around $c=0$
and find the first 4 terms for the function $\exp(\sin x)$.
I have done this but I'm not sure if is correct.
$$e^{\sin x}=\left(1+x+\frac{x^2}{2!}+\dots+\frac{x^{n-1}}{(n-1)!}\right)^{x-\frac{x^3}{3!}+\frac{x^5}{5!}+\dots+(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}}$$
| You want terms up to $x^3$. The exponential is $e^t=1+t+\frac{t^2}2+\cdots$ and the sine $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$. Hence you only need to expand from two terms of the sine, as the others already exceed the degree $3$.
Now, only expanding the necessary terms
$$\begin{align}1&\to1\\
t&\to x-\frac{x^3}6\\
\frac{t^2}2&\to\frac{x^2}2-\cdots\\
\frac{t^3}6&\to\frac{x^3}6-\cdots
\end{align}$$
and in total
$$1+x+\frac{x^2}2+0\,x^3.$$
With a little more courage, up to degree $5$:
$$\begin{align}1&\to1\\
t&\to x-\frac{x^3}6+\frac{x^5}{120}\\
\frac{t^2}2&\to\frac{x^2}2-\frac{x^4}6\cdots\\
\frac{t^3}{6}&\to\frac{x^3}{6}-\frac{x^5}{12}\cdots\\
\frac{t^4}{24}&\to\frac{x^4}{24}\cdots\\
\frac{t^5}{120}&\to\frac{x^5}{120}\cdots
\end{align}$$
gives
$$1+x+\frac{x^2}2-\frac{x^4}8-\frac{x^5}{15}.$$
| {
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"url": "https://math.stackexchange.com/questions/3461284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Eigenvalues of a special stochastic matrix I'm trying to find an explicit formula of all the eigenvalues for the following $n$ by $n$ stochastic matrix (sum of each row/column is one):
\begin{bmatrix}0&\frac{1}{n-1}&\frac{1}{n-1}&\cdots&\frac{1}{n-1}\\
\frac{1}{n-1}&\frac{n-2}{n-1}&0&\cdots&0
\\ \frac{1}{n-1}&0&\frac{n-2}{n-1}&0&\cdots \\ \cdots&\cdots&\cdots&\cdots&\cdots
\\ \frac{1}{n-1}&0&0&\cdots&\frac{n-2}{n-1} \\
\\
\end{bmatrix}
Since this is a stochastic matrix, it is clear that $1$ is an eigenvalue. After doing some numerical experiments, I believe there are only three distict eigenvalues of this special matrix: $1$, $\frac{-1}{n-1}$, and $\frac{n-2}{n-1}$(with multiplicity $n-2$). I want to show this conclusion formally, but I cannot really decompose the matrix into the sum of identity matrix and zero diagonal matrix, since the first element in the matrix is $0$.
Update: I just realize $e_2-e_i$ will always be an eigenvector with eigenvalue $\frac{n-2}{n-2}$. There are $n-2$ pairs of them so we are done.
| Let
$$\mathrm M := \begin{bmatrix} 0 & \frac{1}{n-1} 1_{n-1}^\top\\ \frac{1}{n-1} 1_{n-1} & \frac{n-2}{n-1} \mathrm I_{n-1}\end{bmatrix}$$
whose characteristic polynomial is
$$\begin{aligned} \det \left( s \mathrm I_n - \mathrm M \right) &= \det \begin{bmatrix} s & -\frac{1}{n-1} 1_{n-1}^\top\\ -\frac{1}{n-1} 1_{n-1} & \left( s - \frac{n-2}{n-1} \right) \mathrm I_{n-1}\end{bmatrix}\\ & = \left( s - \frac{n-2}{n-1} \right)^{n-1} \cdot \left( s - \frac{1}{n-1} \left( s - \frac{n-2}{n-1} \right)^{-1} \right)\\ &= \left( s - \frac{n-2}{n-1} \right)^{n-2} \cdot \left( s \left( s - \frac{n-2}{n-1} \right) - \frac{1}{n-1} \right)\\ &= \left( s - \frac{n-2}{n-1} \right)^{n-2} \cdot \left( s^2 - \left(\frac{n-2}{n-1} \right) s - \frac{1}{n-1} \right)\\ &= \left( s - \frac{n-2}{n-1} \right)^{n-2} \cdot (s - 1) \cdot\left( s + \frac{1}{n-1} \right)\end{aligned}$$
where the Schur complement was used.
| {
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Show that the system of trigonometric functions is an orthogonal system The space consisting of all continuous functions on the interval $[a, b]$ equipped with the inner product $\langle f, g \rangle := \int^{b}_{a} f(t)g(t)dt$ is given. Show that the system of trigonometric functions
$$1, \cos \Big (\frac{2\pi nt}{b - a} \Big), \sin \Big (\frac{2\pi nt}{b - a} \Big) \quad \quad n \in \mathbb{N} \setminus \{0 \}$$
is an orthogonal system.
We have to show that the above elements are mutually perpendicular, that is that $\langle x, y \rangle = 0$ for each pair of elements in the above system.
When I try solving $\langle x, y \rangle = 0$, I get stuck on the following:
\begin{equation}
\begin{split}
\langle 1, \cos \Big (\frac{2\pi nt}{b - a} \Big) \rangle & = \int^{b}_{a} \cos \Big (\frac{2\pi nt}{b - a} \Big) dt \\
& = \Big[ \sin \Big (\frac{2\pi nt}{b - a} \Big) \Big ]^{b}_{a} \frac{b - a}{2 \pi n} \\
& = \Big ( \sin \Big (\frac{2\pi nb}{b - a}\Big ) - \sin \Big (\frac{2\pi na}{b - a} \Big )\Big ) \frac{b - a}{2 \pi n} \\
& = \Big ( \sin \Big (\frac{2\pi nb}{b - a}\Big ) + \sin \Big (\frac{-2\pi na}{b - a} \Big )\Big ) \frac{b - a}{2 \pi n}.
\end{split}
\end{equation}
Similarly,
\begin{equation}
\begin{split}
\langle 1, \sin \Big (\frac{2\pi nt}{b - a} \Big) \rangle & = \int^{b}_{a} \sin \Big (\frac{2\pi nt}{b - a} \Big) dt \\
& = \Big[ -\cos \Big (\frac{2\pi nt}{b - a} \Big) \Big ]^{b}_{a} \frac{b - a}{2 \pi n} \\
& = \Big ( -\cos \Big (\frac{2\pi nb}{b - a}\Big ) + \cos \Big (\frac{2\pi na}{b - a} \Big )\Big ) \frac{b - a}{2 \pi n}.
\end{split}
\end{equation}
And for $u = \sin \Big (\frac{2\pi nt}{b - a} \Big)$, $du = \cos \Big (\frac{2\pi nt}{b - a} \Big)\frac{2 \pi n}{b-a}dt$, $\cos \Big (\frac{2\pi nt}{b - a} \Big) dt = \frac{b - 1}{2 \pi n}du$ we have
\begin{equation}
\begin{split}
\langle \cos \Big (\frac{2\pi nt}{b - a} \Big), \sin \Big (\frac{2\pi nt}{b - a} \Big) \rangle & = \int^{b}_{a} \cos \Big (\frac{2\pi nt}{b - a} \Big)\sin \Big (\frac{2\pi nt}{b - a} \Big) dt \\
& = \int^{u(a)}_{u(b)}\frac{b - a}{2 \pi n}u du \\
& = \frac{b - a}{2 \pi n}\frac{1}{2} \Big [ u^{2} \Big ]^{u(b)}_{u(a)} \\
& = \frac{b - a}{4 \pi n} \Big [ \sin^{2}\Big (\frac{2\pi nt}{b - a} \Big) \Big ]^{b}_{a} \\
& = \frac{b - a}{4 \pi n} \Big ( \sin^{2}\Big (\frac{2\pi nb}{b - a}\Big ) - \sin^{2}\Big (\frac{2\pi na}{b - a} \Big )\Big ) \\
& = \frac{b - a}{4 \pi n} \Big ( \sin^{2}\Big (\frac{2\pi nb}{b - a}\Big ) + \sin^{2}\Big (-\frac{2\pi na}{b - a} \Big )\Big ).
\end{split}
\end{equation}
It seems to me that the inner products above are only equal to zero when $a = b$ which would give division by zero.
However, when I take $b = \pi$ and $a = -\pi$, the orthogonality follows. I think that I am missing something trivial about trigonometric functions (i.e. a property).
| Note that for all $a\not=b$, we have that
$$
\sin \Big (\frac{2\pi nb}{b - a}\Big )
=\sin \Big (\frac{2\pi n[(b-a)+a]}{b - a}\Big )
=\sin \Big (2\pi n+\frac{2\pi n a}{b - a}\Big )
=\sin \Big (\frac{2\pi n a}{b - a}\Big ).
$$
Similarly for the cosine function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
solve $(D^2-2D+1)y=x\sin x$
Find a particular solution to the equation $$(D^2-2D+1)y=x\sin x$$
\begin{align}
\text{P.I.}&=\frac{1}{(D^2-2D+1)}x\sin x\\
&=\frac{1}{(D-1)^2}x\sin x\\
&=\text{I.P.}\left[\frac{1}{(D-1)^2}xe^{ix}\right]\\
&=\text{I.P.}\left[e^{ix}\frac{1}{(D+i-1)^2}x\right]\\
&=\text{I.P.}\left[e^{ix}\frac{1}{2}(1-\frac{iD^2}{2}+D+iD+\cdots)x\right]\\
&=\text{I.P.}\left[(\cos x+i\sin x)(\frac{1}{2}x+\frac{1}{2}+\frac{1}{2}i)\right]\\
&=\text{I.P.}\left[\frac{1}{2}x\cos x+\frac{1}{2}\cos x+\frac{1}{2}i\cos x+\frac{1}{2}xi\sin x+\frac{1}{2}i\sin x-\frac{1}{2}\sin x\right]\\
&=\frac{1}{2}\cos x+\frac{1}{2}x\sin x+\frac{1}{2}\sin x\\
\text{I.P.}&=\text{Imaginary Part}
\end{align}
But the solution provided by Mathematica is:
Is my solution is wrong$?$ Or I am getting correct answer in different form$!$
| Mathematica's answer is correct. Maybe you made a sign mistake somewhere. Here I used Operator Method too. A bit different.
$$\begin {align}
y_p&=\frac 1 {(D-1)^2} x\sin x \\
y_p&=(x-\frac 2 {(D-1)})\frac 1 {(D-1)^2} \sin x \\
y_p&=(x-\frac 2 {(D-1)})\frac 1 {-2D} \sin x \\
y_p&=\frac 1 2 (x-\frac 2 {(D-1)})\cos x \\
y_p&=\frac 1 2 x\cos x -\frac 1 {(D-1)}\cos x \\
y_p&=\frac 1 2 x \cos x-\frac {D+1}{(D^2-1)}\cos x \\
&=\frac 1 2 x\cos x +\frac 1 2 {(D+1)}\cos x\\
&=\frac 1 2 x \cos x+\frac 1 2 (-\sin x +\cos x) \\
y_p&=\frac 1 2 (x \cos x-\sin x +\cos x)
\end{align}
$$
Edit:
For your solution it's $2i$ that you should have at the denominator not 2...
$$P(D)=(D+i-1)^2=D^2+2D(i-1)-2i=-2i(1+iD^2/2-D(1+i))$$
So you have to take all the terms that have a i factor...You are close to the solution
$$P(D)^{-1}=\frac i 2(1+D(1+i)+.......)$$
$$P(D)^{-1}x=\frac i 2(x+1+i)$$
$$y_p=\mathcal {Im} \{e^{ix}P(D)^{-1}x\}$$
$$y_p=\frac 1 2\mathcal {Im} \{(\cos x +i \sin x)(ix+i-1)\}$$
$$y_p=\frac 1 2 \{(x\cos x+\cos x - \sin x) \}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3469625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
is there a real solution to this differential equation? This is the equation:
$\frac {d^4y}{dx^4}=-y$
I know the solution to this equation:
$\frac {d^2y}{dx^2}=-y$
is this:
$y(x)=a\cdot \cos(x)+b\cdot \sin(x)$
and it involves $e^{ix}$, but I wondered if there was any solution to a higher-order version of this.
| $$y''''+y=0$$
Characteristic equation is
$$(r^4+1)=0$$
$$(r^2-i)(r^2+i)=0$$
$$(r^2-i)=(r-(\frac {\sqrt 2}{2}+i\frac {\sqrt 2}{2}))(r+(\frac {\sqrt 2}{2}+i\frac {\sqrt 2}{2}))$$
$$(r^2+i)=(r-i(\frac {\sqrt 2}{2}+i\frac {\sqrt 2}{2}))(r+i(\frac {\sqrt 2}{2}+i\frac {\sqrt 2}{2}))$$
So the solution is
$$y_1(x)=c_1e^{\frac {\sqrt 2}{2}x}\cos(\frac {\sqrt 2}{2}x)+c_2e^{\frac {\sqrt 2}{2}x}\sin(\frac {\sqrt 2}{2}x)$$
And
$$y_2(x)=c_3e^{-\frac {\sqrt 2}{2}x}\cos(\frac {\sqrt 2}{2}x)+c_4e^{-\frac {\sqrt 2}{2}x}\sin(\frac {\sqrt 2}{2}x)$$
$$y(x)=y_1(x)+y_2(x)$$
You can also use exponetials:
$$r^4=-1 \implies (\rho e^{i\beta})^4=e^{i\pi}$$
We have that $\rho=1$ and:
$$4\beta=\pi +2k\pi ,k=0,1,2.....$$
$$\beta =\frac {\pi(1+2k)}4$$
We find four values for $\beta$ :
$$\beta=\frac {\pi}4,\frac {3\pi}4,\frac {5\pi}4,\frac {7\pi}4$$
The solution for example for $\beta=\frac {\pi}4$
$$r=e^{i\pi/4}=\cos \frac {\pi}4+ i \sin \frac {\pi}4=\frac {\sqrt 2}{2}+i\frac {\sqrt 2}{2}$$
$$y_1(x)=c_1e^{\frac {\sqrt 2}{2}x}\left (\cos(\frac {\sqrt 2}{2}x)+i\sin(\frac {\sqrt 2}{2}x) \right)$$
Do this for all values of $\beta$ to find the complete answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What is the radial function of a Minkowski sum of a circle and an ellipse?
Let $C$ denote the unit circle in the two-dimensional plane, centered at the origin (the blue circle in the pic). Let $E$ an ellipse whose equation is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. The set $C+E$, known as the Minkowski sum of $C$ and $E$, is defined by
$$C+E=\{\vec{\alpha}+\vec{\beta}: \vec{\alpha}\in C,\ \ \vec{\beta}\in E\}$$
In the picture above, the orange oval-shaped enclosing figure is the boundary of the sum $C+E$,
(where for the sake of this example, $E$ is the ellipse with $a=2,b=1$). The green ellipse in the picture is
$Q+E$, where for the sake of example, $Q=(1/\sqrt{2},1/\sqrt{2})$. The point $P$ represents a typical boundary point of the sum $C+E$, which is the orange oval, which is the geometric location of all points obtained by rotating around the green ellipse with its center varying along the blue circle.
The point $O$ is the origin $(0,0)$ and the point $M$ is the orthogonal projection of $P$ onto the $x$ axis.
Problem: Express the length of $\vec{OP}$ in terms of the angle POM.
I can prove that if $$F(\varphi)=\arctan\left(\tan\varphi \frac{\sqrt{a^2\cos^2\varphi+b^2\sin^2\varphi}+b^2}{\sqrt{a^2\cos^2\varphi+b^2\sin^2\varphi}+a^2}\right),\quad (0\leq\varphi\leq \pi/2)$$
Then if the angle POM is $\theta$, then the angle QOM is $F^{-1}(\theta)$ (the inverse function), and then it is not too difficult to express the $\vec{QP}$ in terms of $F^{-1}(\theta)$ and consequently the length of $\vec{OP}$ in terms of $\theta$ (and $a,b$ of course). However, the result seems to be too complicated; I suspect there might be a simpler geometric argument that escapes me.
| HINT
Since Minkowsky sum is commutative, then the alternative order of the calculations can be proposed.
The point of the circle farthest from the ellipse is located on the normal to the ellipse.
Denote
$$\angle xOQ = \varphi,\quad \angle xOP = \theta.$$
Parametric equation of the ellipse is
$$x=a\cos\varphi,\quad y=b\sin\varphi,$$
the guide vector of the tangent line is $\{-a\sin\varphi,b\cos\varphi\},$ and the angle coefficient of the line $QP$ is
$$k=\dfrac ab\tan\varphi.$$
Then
\begin{align}
&\overline{OQ}=\{a\cos\varphi,b\sin\varphi\}, \\[8pt]
&\overline{QP}=\dfrac1{\sqrt{k^2+1}}\{1,k\}
=\dfrac{\{b\cos\varphi,a\sin\varphi\}}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}},\\[8pt]
&\overline{OP} = \overline{OQ}+\overline{QP}\\
&=\left\{\left(a+\dfrac{b}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)\cos\varphi,
\left(b+\dfrac{a}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)\sin\varphi\right\}\\
&=\sqrt{\left(a+\dfrac{b}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)^2\cos^2\varphi
+ \left(b+\dfrac{a}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)^2\sin^2\varphi}\\
&\times\{\cos\theta,\sin\theta\},\\[8pt]
&\left(a+\dfrac{b}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)^2\cos^2\varphi
=\cos^2\theta\\
&\times\left(\left(a+\dfrac{b}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)^2\cos^2\varphi
+ \left(b+\dfrac{a}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}\right)^2\sin^2\varphi\right),\\[8pt]
&\left(a\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}+b\Large\mathstrut\right)
\cos\varphi\sin\theta
=\left(b\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}+a\right)\sin\varphi\cos\theta,\\[8pt]
&(a\cos\varphi\tan\theta-b\sin\varphi)\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi} =a\sin\varphi - b\cos\varphi\tan\theta,\\[8pt]
\end{align}
$$\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}
=\dfrac{a\sin\varphi - b\cos\varphi\tan\theta}{a\cos\varphi\tan\theta-b\sin\varphi},\tag1$$
$$(a\tan\theta-b\tan\varphi)^2(a^2\tan^2\varphi+b^2)
= (a\tan\varphi - b\tan\theta)^2(1+\tan^2\varphi),\tag2$$
and fourth-order algebraic equation has known exact solution
$$\tan\varphi = f(\tan\theta).$$
Also from $(1)$ should
$$a+\dfrac{b}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}
= a+b\,\dfrac{a\cos\varphi\tan\theta-b\sin\varphi}
{a\sin\varphi - b\cos\varphi\tan\theta}
= (a^2-b^2)\dfrac{\sin\varphi}{a\sin\varphi - b\cos\varphi\tan\theta},$$
$$b+\dfrac{a}{\sqrt{a^2\sin^2\varphi+b^2\cos^2\varphi}}
= b+a\,\dfrac{a\cos\varphi\tan\theta-b\sin\varphi}
{a\sin\varphi - b\cos\varphi\tan\theta}
= (a^2-b^2)\dfrac{\cos\varphi\tan\theta}{a\sin\varphi - b\cos\varphi\tan\theta},$$
$$OP = (a^2-b^2)\dfrac{\sin\varphi\cos\varphi\sqrt{1+\tan^2\theta}}{a\sin\varphi - b\cos\varphi\tan\theta},$$
$$OP = (a^2-b^2)\dfrac{\sin\varphi}{a\tan\varphi\cos\theta - b\sin\theta}.\tag3$$
Formulas $(2),(3)$ define required result.
$\color{brown}{\mathbf{About\ equation\ (2).}}$
Substitutions
$$t=a\tan\varphi-b\tan\theta,\quad \tan\theta = p\tag4$$
present equation $(2)$ in the form of
$$(a^2p - b(t+bp))^2((t+bp)^2+b^2) = t^2(a^2+(t+bp)^2),$$
$$\Bigl(((a^2-b^2)p - bt)^2-t^2\Bigr)((t+bp)^2+b^2) = (a^2-b^2)t^2,$$
$$(b^2-1)t^2((t+bp)^2+b^2) = (a^2-b^2)\Bigl(t^2+p(2bt-a^2+b^2)((t+bp)^2+b^2)\Bigr),$$
$$(b^2-1)t^4+2bp(2b^2-a^2-1)t^3
+\Bigl(b^2(b^2-1)(p^2+1)+p(a^2-b^2)^2-a^2+b^2\Bigr)t^2$$
$$+2bp(a^2-b^2)(p(a^2-b^2)-b^2(p^2+1))t +pb^2(a^2-b^2)^2)=0,\tag5$$
If formula $(5)$ is correct for the given parameters' values, then it should be reduced to the form of
$$g(t)= t^4+2pt^3\pm q^2t^2+2rs+\pm s^2=0.\tag6$$
And the presentation
$$g(t) = (t^2+pt)^2 - (rt+s)^2,$$
taking in account
$$q^2 = q^2(\cos^2\beta+\sin^2\beta)= q^2(\cosh^2\gamma-\sinh^2\gamma),$$
allows to obtain unknown $\gamma$ via suitable cubic equation and to deal with the quadratic ones.
Proposed approach looks hard, but I do not see more effective way.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove $2^x≥x^2$ I could not understand this proof. How are we able to divide both sides of the inequality with different things and claim that the inequality stays the same? Plus, is this an induction proof?
If $x \in \mathbb{Z}$ and $x \ge 4$ then $2^x \ge x^2$ holds.
When $x = 4$ we have $2^4 = 16 \ge 16 = 4^2$.
We $x > 4$ we have $\frac{2^{x + 1}}{2^x} = 2$ and
$$
\frac{(x + 1)^2}{x^2}
= \left(1 + \frac{1}{x}\right)^2
\le \left(\frac{5x}{4}\right)^2
$$
since $2 \le 1.5625$
| Yes , it is sort of an induction proof . We first show that the inequality holds for $x = 4$ . Now we find the ratios of the subsequent terms of $x^2$ and $2^x$ :
$$\dfrac{2^{x+1}}{2^x} = 2$$
And $$\dfrac{(x+1)^2}{x^2} = \left(1 + \dfrac{1}{x}\right)^2$$
Note that for $x\ge 4$ , the expression$\color{#d05}{\left(1 + \dfrac{1}{x}\right)^2 \le 2}$ . Since $2^x$ increases faster than $x^2$ , we can induct that $\color{#20f}{2^{x+1}\ge(x+1)^2}$for all $x\ge 4$ and the claim follows .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! = n(n + 1)!$ Show that
$$2 · 1! + 5 · 2! + 10 · 3! + . . . + (n
^2 + 1)n! = n(n + 1)!$$
I got stuck after the n+1 part.
$$((n+1)^2 + 1)(n+1)! = (n+1)(n+2)!.$$
I'm not sure how to proceed from this part.
| proof by Induction $$P(1):=2\cdot1!=1\cdot(1+1)!$$
Now we assume it is true for $P(n)$ then we proof it is true for $P(n+1)$, $$P(n+1):=2 · 1! + 5 · 2! + 10 · 3! + . . . + (n
^2 + 1)n! + ((n+1)^2+1)(n+1)! = (n+1)((n + 1)+1)!$$
Use the hypothesis of induction $P(n)$ and LHS of above expression reduces to $$n(n+1)!+((n+1)^2+1)(n+1)! = (n+1)!(n+n^2+2n+2) \\ =(n+1)!(n^2+3n+2)=(n+1)!(n+2)(n+1) \\ =(n+1)(n+2)! $$
Hence we get the RHS of the $P(n+1)$. So we have proved $P(n+1)$ is true whenever $P(n)$ is true.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Jordan normal form of $A=\left(\begin{smallmatrix} 0 & 4 & 2 \\ -3 & 8 & 3 \\ 4 & -8 & -2 \\ \end{smallmatrix}\right)$
Find a matrix in Jordan normal form that is similar to
$$A=\begin{pmatrix}
0 & 4 & 2 \\
-3 & 8 & 3 \\
4 & -8 & -2 \\
\end{pmatrix}$$
The characteristic equation of $A$ is $(\lambda-2)^3=0$, hence, $\lambda=2$ is a eigenvalue of algebraic multiplicity three. So eigenvector corresponds to $\lambda=2$,
$$(A-2I)=\begin{pmatrix}
-2 & 4 & 2 \\
-3 & 6 & 3 \\
4 & -8 & -4 \\
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\left\{\begin{pmatrix}1\\0\\1\end{pmatrix},\begin{pmatrix}2\\1\\0\end{pmatrix} \right\}$$For one more eigenvector I tried to get one general eigenvector. $$(A-2I)=\begin{pmatrix}
-2 & 4 & 2 \\
-3 & 6 & 3 \\
4 & -8 & -4 \\
\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\0\\1\end{pmatrix}\text{ or }\begin{pmatrix}2\\1\\0\end{pmatrix}$$ But both of them are inconsistent system and I stuck. Then how to get complete eigenspace to find $P$ such that $A=PJP^{-1}$
| The way you are trying to find the third vector is not going to work, as you have demanded the result of $(A-2I)w$ to be one of the two eigenvectors you found.
I prefer confirming the minimal polynomial, here $(\lambda - 2)^2.$ And $(A - 2 I)^2 = 0.$ So, take any column vector that is not an eigenvalue, for example $w = (0,0,1)^T.$ The eigenvector that is now forced is $v =(A-2I)w,$ so that $v = (2,3,-4).$ CHECK that it really is an eigenvector. Finally, choose an eigenvector $u$ that is not a multiple of $v.$ I like the difference of your two listed eigenvectors, so I chose $u = (1,1,-1)^T$ since the determinant comes out nicely.
We are going to construct $J = Q^{-1} A Q$ having chosen
$$
Q =
\left(
\begin{array}{ccc}
1&2&0 \\
1&3&0 \\
-1&-4 &1 \\
\end{array}
\right)
$$
The determinant of $Q$ is one, the inverse is
$$
Q^{-1} =
\left(
\begin{array}{ccc}
3&-2&0 \\
-1&1&0 \\
-1& 2 &1 \\
\end{array}
\right)
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\left(1+\sqrt2\right)^{2011}=a+b\sqrt{2}$, for integers $a$ and $b$, then what is $\left(1-\sqrt2\right)^{2010}$ expressed using $a$ and $b$?
Being $ a $ and $ b $ integers such that $\left(1+\sqrt{2}\right)^{2011} =a+b\sqrt{2}, \left(1-\sqrt{2}\right)^{2010}$ equals:
a) $a+2b+(a-b)\sqrt{2}$
b) $a-2b+(a-b)\sqrt{2}$
c) $a+2b+(b-a)\sqrt{2}$
d) $2b-a+(b-a)\sqrt{2}$
e) $a+2b-(a+b)\sqrt{2}$
Solution:
Not going with any of the alternatives
| Note that if $(1+\sqrt{2})^n=a_n+b_n \sqrt{2}$ then $(1-\sqrt{2})^n=a_n-b_n \sqrt{2}$.
We have
\begin{eqnarray*}
(1+\sqrt{2})^{2011}=a+b \sqrt{2} \\
(1-\sqrt{2})^{2010}=x+y \sqrt{2}. \\
\end{eqnarray*}
Multiply the second equation by $(1-\sqrt{2})$ & do a bit of algebra ... this gives
\begin{eqnarray*}
x-2y=a\\
x-y=b .\\
\end{eqnarray*}
Should be easy from here ?
d)
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Definite integral of $1/(5+4\cos x)$ over $2$ periods Question:
$$\int_0^{4\pi}\frac{dx}{5+4\cos x} $$
My approach:
First I calculated the antiderivative as follows:
Using: $\cos\theta= \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$ we have:
$\int\frac{dx}{5+4\cos x}=\int\frac{dx}{5+4\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}=\int\frac{1+\tan^2\frac{x}{2}}{5+5\tan^2\frac{x}{2}+4-4\tan^2\frac{x}{2}}dx=\int\frac{\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx$
Using substitution we have:
$u=\tan\frac{x}{2}$
$du=\frac{1}{2}\frac{1}{\cos^2\frac{x}{2}}dx$
$2\int\frac{\frac{1}{2}\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx=2\int\frac{du}{3^2+u^2}=\frac{2}{3}\arctan\frac{u}{3}+\mathscr{C}=\frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}+ \mathscr{C}$
Now we can calculate the definite integral as follows:
$\int_0^{4\pi}\frac{dx}{5+4\cos x} = \frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}\bigl|_0^{4\pi}=\frac{2}{3}(\arctan\frac{\tan\frac{4\pi}{2}}{3}-\arctan\frac{\tan\frac{0}{2}}{3})=0$
The result I get is $0$ but the correct one is $\frac{4\pi}{3}$. Can someone explain me why?
Here it shows that the correct answer is $\frac{4\pi}{3}$.
| Just observe that $I=\int\limits_0^{4\pi}\frac{dx}{5+4\cos x} = 4\int\limits_0^{\pi}\frac{dx}{5+4\cos x} $.
Then you can use tangent half-angle substitution to get
$I=\frac{8}{3}\int_\limits0^{\infty}\frac{(1/3)dx}{1+{(u/3)}^2}=\frac{8}{3}\cdot\tan^{-1}(u/3)|_0^\infty =\frac{4\pi}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to find an ellipse equation with non-symmetrical foci values? I have tried a lot but i do not seem to find an equation.
Question is the following:
Find an equation of the specified curve, that is, of the locus of points the sum of whose distances from the points $(2,3)$ and $(4,1)$ is $8$.
I first tried to translate and rotate the axis to see the ellipse like its center lies on the origin $(0,0)$. It seemed to be tough. I know I can use the general equation with center $(h,k)$ but then there will not be any rotation though the major axis is rotated.
| I don't know. For problems like this, I just use the distance formula and grind through the algebra. (Then double-check my solution against Wolfram Alpha.)
$$\sqrt{(x-2)^2+(y-3)^2}=8-\sqrt{(x-4)^2+(y-1)^2}\\
(x-2)^2+(y-3)^2=64+(x-4)^2+(y-1)^2-16\sqrt{{(x-4)^2+(y-1)^2}}\\
(x^2-4x+4)+(y^2-6y+9)=64+(x^2-8x+16)+(y^2-2y+1)-16\sqrt{{(x^2-8x+16)+(y^2-2y+1)}}\\
4x-4y-68=-16\sqrt{x^2+y^2-8x-2y+17}\\
x-y-17=-4\sqrt{x^2+y^2-8x-2y+17}\\
(x-y-17)^2=16(x^2+y^2-8x-2y+17)\\
x^2+y^2+289-2xy-34x+34y=16x^2-128x+16y^2-32y+272\\
15x^2+2xy+15y^2-94x-66y-17=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Show that $a^2+b^2+c^2$ is a square when $\frac{1}{a}+\frac{1}{b} = \frac{1}{c}$ and $a,b,c\in\mathbb{Q}$
Knowing that $$\dfrac1a+ \dfrac1b=\dfrac1c$$ Prove that $a^2+b^2+c^2$ is a square, where $a,b,c\not=0$ are rational numbers.
It can probably be solved by a quick factoring trick, but I really can’t figure it out.
| Below we show how to simply derive the value of $\,\color{#90f}x,$ using only $\rm\color{#0a0}{\textit{difference of squares}}\,$ factorization
$$\begin{align}
a^2+b^2+\color{#c00}c^2\, &=\, \color{#90f}{x^2},\ \ \ {\rm let}\ \ \ X = (a+b) x\\
\smash[t]{\overset{\!\!\!\large \times\ (a+b)^2}\iff}\ (a+b)^2(a^2+b^2) + (\color{#c00}{ab})^2 &=\, X^2\ \ \ \ {\rm by}\ \ \ \left[\,(a+b)\color{#c00}c = \color{#c00}{ab}\,\right]^2\\[.3em]
\iff\qquad\quad\ \ (a+b)^2(a^2+b^2) &=\, \color{#0a0}{X^2 - (\color{c00}{ab})^2}\ \\[.3em]
\smash[b]{\iff\ \, (\underbrace{a^2+b^2+2ab}_{})(\underbrace{a^2+b^2}_{})} &= {(X-ab)(X+ab)}\\
X^{\phantom{|^{|}}}\!\!\!+ab\ \ \ \ \ \ \ \ \ X-ab,\ \!\! &\ \ \ X = a^2+b^2+ab
\end{align}\qquad\qquad\quad\ $$
Thus $\ \color{#90f}x = \dfrac{X}{a+b} = \dfrac{a^2+b^2+ab}{a+b}\ = \dfrac{(a+b)^2-ab}{a+b} = a+b-c\ $ is a solution.
Remark $ $ With that insight we can easily derive the key identity at the heart of it, which is simply the symmetric polynomial Newton identity $\ p_2 = \color{#0a0}{p_1^2} - 2\, \color{#c00}{e_2},\ $ with $\,C\to -C,\,$ i.e.
$$A^2+B^2+C^2 = \color{#0a0}{(A+B-C)^2} + 2(\color{#c00}{(A+B)C-AB})\qquad$$
hence we infer that $\ A^2+B^2+C^2\ $ is a $\,\rm\color{#0a0}{square},\,$ when $\,\color{#c00}{(A+B)C = AB}$.
Proof $ $ expanding $\,\color{#0a0}{((A+B)-C)^2}$ yields RHS $= (A+B)^2\!-2AB+C^2 = $ LHS
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Proof by Induction $2\cdot 7^n + 3\cdot 5^n - 5$ is multiple of $24$ Proof by induction that $2\cdot 7^n + 3\cdot 5^n - 5$ is a multiple of $24$.
I tried solving but got stuck.
Show that it is true for n=1
$$2\cdot 7^1 + 3\cdot 5^1 - 5 = 14 + 15 - 5 = 24$$
Assume it true for $n = k$
$$2\cdot 7^k + 3\cdot 5^k - 5 = 24g$$
Show it is true for $n= k + 1$
$$2\cdot 7^{k+1} + 3\cdot 5^{k+1} - 5$$
is a multiple of 24
$$(2\cdot 7)(2\cdot 7^k) + (3\cdot 5^k)(3\cdot 5) - 5$$
$$2\cdot7(24g + 5)3\cdot 5 - 5$$
I'm stuck and don't know how to proceed
| In essence, the induction step relies on the identity $$(ax^n + by^n)(x+y) = (ax^{n+1} + by^{n+1}) + xy(ax^{n-1} + by^{n-1}).$$ You can verify this by simple multiplication. So if we define $$f_n(a,b,x,y) = ax^n + by^n,$$ then we have the recursion relation $$f_{n+1}(a,b,x,y) = (x+y)f_n(a,b,x,y) - xy f_{n-1}(a,b,x,y).$$ In your case, it is natural to choose $a = 2$, $b = 3$, $x = 7$, $y = 5$ to obtain $$f_{n+1} = 12 f_n - 35 f_{n-1}$$ (where I have omitted the arguments of the function). So if we let $g_n = f_n - 5$, then $$g_{n+1} + 5 = 12(g_n + 5) - 35(g_{n-1} + 5)$$ or $$g_{n+1} = 12g_n - 35 g_{n-1} - 120.$$ Since $120 = 5(24)$, it follows that $24 | g_{n+1}$ whenever $24 | g_n$ and $24 | g_{n-1}$; since $g_0 = 0$ and $g_1 = 24$, the proof is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Evaluate $\int \cos 2\theta \ln\left(\frac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta$ $$\int \cos 2\theta \ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)d\theta$$
My attempt is as follows:-
$$\ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)=t\tag{1}$$
$$\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\cdot\dfrac{\left(\cos\theta-\sin\theta\right)^2-(-\sin\theta-\cos\theta)(\cos\theta+\sin\theta)}{(\cos\theta-\sin\theta)^2}=\dfrac{dt}{d\theta}$$
$$\dfrac{2}{\cos2\theta}=\dfrac{dt}{d\theta}$$
Let's calculate $\cos2\theta$ from equation $1$
$$\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}=e^t$$
$$\dfrac{1+\tan\theta}{1-\tan\theta}=e^t$$
Applying componendo and dividendo
$$\dfrac{2}{2\tan\theta}=\dfrac{e^t+1}{e^t-1}$$
$$\dfrac{e^t-1}{e^t+1}=\tan\theta$$
$$\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$$
$$\cos2\theta=\dfrac{(e^t+1)^2-(e^t-1)^2}{(e^t+1)^2+(e^t-1)^2}$$
$$\cos2\theta=\dfrac{4e^t}{2(e^{2t}+1)}$$
$$\cos2\theta=\dfrac{2e^t}{e^{2t}+1}\tag{2}$$
So integral will be
$$\dfrac{1}{2}\cdot\int \left(\dfrac{2e^t}{e^{2t}+1}\right)^2dt$$
$$\dfrac{1}{2}\cdot\int \dfrac{4e^{2t}}{(1+e^{2t})^2}$$
$$e^{2t}+1=y$$
$$2e^{2t}=\dfrac{dy}{dt}$$
$$2e^{2t}dt=dy
$$\int \dfrac{dy}{y^2}$$
$$-\dfrac{1}{y}+C$$
$$-\dfrac{1}{1+e^{2t}}+C$$
$$-\dfrac{1}{1+e^{\ln\left(\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\right)^2}}+C$$
$$-\dfrac{1}{1+\dfrac{1+\sin2\theta}{1-\sin2\theta}}+C$$
$$-\dfrac{1-\sin2\theta}{2}+C$$
$$\dfrac{\sin2\theta}{2}+C'$$
And this should be actually wrong because if we differentiate the result, it will give $\cos2\theta$, but integrand is $\cos 2\theta \ln\left(\dfrac{\cos \theta+\sin\theta}{\cos\theta-\sin\theta}\right)$
What am I missing here, checked multiple times, but not able to get the mistake. Any directions?
| If you multiply the fraction inside ln top and bottom by $\cos(2\theta)+\sin(2\theta)$ and use the trig identities
$\cos^2(t)-\sin^2(t)=\cos(2t)$, $\sin(2t)=2\sin(t) \cos(t)$, the integral becomes
$$\int\cos(2\theta) \ln(\sec(2\theta)+\tan(2\theta))d\theta$$
Now use the substitution $u=\sec(2\theta)+\tan(2\theta)$ that gives you $du=2u \sec(2\theta)d\theta$ and express $\cos(2\theta)$ as a function of $u$:
$$\cos(2\theta)=\frac{2u}{u^2+1}$$
When you substitute this into the integral, you can integrate by parts using the fact that $$\frac{d}{du}\left(\frac{2}{u^2+1}\right)=-\frac{4u}{(u^2+1)^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
What's wrong in my calculation of $\int_0^{3 \pi/4} \frac{\cos x}{1 + \cos x}dx$? What's wrong in my calculation of $$\int_0^{3 \pi/4} \dfrac{\cos x}{1 + \cos x} dx\,?$$
I have to find:
$$\displaystyle\int_0^{3 \pi/4} \dfrac{\cos x}{1 + \cos x} dx$$
and I can't seem to get the right answer. This is what I did:
I decided to use the Weierstrass substitution with:
$$t = \tan \dfrac{x}{2}$$
$$\cos x = \dfrac{1 - t^2}{1 + t^2}$$
$$\sin x = \dfrac{2t}{1 + t^2}$$
$$dx = \dfrac{2}{1+ t^2}$$
I am pretty new to this type of substitution.
Anyway, the boundaries become:
$$t_1= \tan 0 = 0$$
$$t_2 = \tan \dfrac{3 \pi }{8}$$
We have
$$\displaystyle\int_0^{3 \pi/4} \dfrac{\cos x}{1 + \cos x} dx = $$
$$=\displaystyle\int_0^{\tan 3 \pi/8} \dfrac{\dfrac{1 - t^2}{1 + t^2}}{1 + \dfrac{1 - t^2}{1 + t^2}} \cdot \dfrac{2}{1 + t^2} $$
$$=\displaystyle\int_0^{\tan 3 \pi/8} \dfrac{1-t^2}{1+t^2} dt$$
$$=\displaystyle\int_0^{\tan 3 \pi/8} \dfrac{1}{1+t^2}dt - \displaystyle\int_0^{\tan 3 \pi/8} \dfrac{t^2}{1+t^2} dt $$
$$=\displaystyle\int_0^{\tan 3 \pi/8} \dfrac{1}{1+t^2} dt - \displaystyle\int_0^{\tan 3 \pi/8} \dfrac{t^2 + 1 - 1}{1+t^2} dt$$
$$= \displaystyle\int_0^{\tan 3 \pi/8} \dfrac{1}{1+t^2} dt - \displaystyle\int_0^{\tan 3 \pi/8} 1 dt + \displaystyle\int_0^{\tan 3 \pi/8} \dfrac{1}{1+t^2} dt $$
$$= 2 \bigg [\arctan(t) \bigg ]_0^{\tan 3\pi/8} - \bigg [ t \bigg ]_0^{\tan 3\pi/8}$$
$$= 2\arctan \bigg ( \tan \dfrac{3 \pi}{8} \bigg ) - \tan \dfrac{3\pi}{8}$$
$$= 2 \dfrac{3 \pi}{8} - \tan \dfrac{3 \pi}{8}$$
$$=\dfrac{3 \pi}{4} - \tan \dfrac{3 \pi}{8}$$
So that's the answer that I got. However, my textbook claims that the correct answer is in fact $\dfrac{\pi}{4} + \tan \dfrac{3 \pi}{8} - 2$. So, what did I do wrong?
| Here is the step-by-step solution for the indefinite integral:
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int \frac{1}{\cot \frac{x}{2}\cdot\cot\frac{x}{3}\cdot\cot\frac{x}{6}}\text{d}x$ $$\int \dfrac{1}{\cot \dfrac{x}{2}\cdot\cot\dfrac{x}{3}\cdot\cot\dfrac{x}{6}}dx$$
My multiple attempts are as follows:-
Attempt $1$:
$$\int \tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\tan\dfrac{x}{6}dx$$
$$\int \tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\tan\left(\dfrac{x}{2}-\dfrac{x}{3}\right)dx$$
$$\int \tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\dfrac{\tan\dfrac{x}{2}-\dfrac{x}{3}}{1+\tan\dfrac{x}{2}\cdot\tan\dfrac{x}{3}}dx$$
$$\int \dfrac{\tan^2\dfrac{x}{2}\cdot\tan\dfrac{x}{3}\cdot\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{3}}{\cos\dfrac{x}{6}}-\dfrac{\tan\dfrac{x}{2}\cdot\tan^2\dfrac{x}{3}\cdot\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{3}}{\cos\dfrac{x}{6}}dx$$
$$\int \dfrac{\sin^2\dfrac{x}{2}\cdot\sin\dfrac{x}{3}}{\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{6}}-\dfrac{\sin\dfrac{x}{2}\cdot\sin^2\dfrac{x}{3}}{\cos\dfrac{x}{3}\cos\dfrac{x}{6}}dx$$
$$2\left(\int \dfrac{\sin^2\dfrac{x}{2}\cdot\sin\dfrac{x}{6}}{\cos\dfrac{x}{2}}-\dfrac{2\sin\dfrac{x}{2}\cdot\sin^2\dfrac{x}{6}\cos\dfrac{x}{6}}{\cos\dfrac{x}{3}}\right)dx$$
This doesn't seem to be going anywhere.
Attempt $2$:
$$\int \dfrac{\sin\dfrac{x}{2}\cdot\sin\dfrac{x}{3}\cdot\sin\dfrac{x}{6}}{\cos\dfrac{x}{2}\cdot\cos\dfrac{x}{3}\cdot\cos\dfrac{x}{6}}dx$$
$$\int \dfrac{\cos\dfrac{x}{6}\cdot\sin\dfrac{x}{6}-\cos\dfrac{5x}{6}\cdot\sin\dfrac{x}{6}}{\cos^2\dfrac{x}{6}+\cos\dfrac{x}{6}\cdot\cos\dfrac{5x}{6}}dx$$
$$\int \dfrac{\sin\dfrac{x}{3}-\sin x+\sin\dfrac{2x}{3}}{1+\cos\dfrac{x}{3}+\cos x+\cos \dfrac{2x}{3}}dx$$
As we know $\cos \alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+(n-1)\beta)=\dfrac{\sin\dfrac{n\beta}{2}}{\sin\dfrac{\beta}{2}}\cos\left(\dfrac{\alpha+\alpha+(n-1)\beta}{2}\right)$
$$\int \dfrac{\sin\dfrac{x}{3}-\sin x+\sin\dfrac{2x}{3}}{\cos(0)+\cos\left(0+\dfrac{x}{3}\right)+\cos\left(0+\dfrac{2x}{3}\right)+\cos\left(0+\dfrac{3x}{3}\right)}dx$$
$$\int \dfrac{\sin\dfrac{x}{3}-\sin x+\sin\dfrac{2x}{3}}{\dfrac{\sin\dfrac{4x}{6}}{\sin\dfrac{x}{6}}\cdot\cos\dfrac{x}{2}}dx$$
$$\int \dfrac{\sin\dfrac{x}{3}+\sin x+\sin\dfrac{2x}{3}-2\sin x}{\dfrac{\sin\dfrac{4x}{6}}{\sin\dfrac{x}{6}}\cdot\cos\dfrac{x}{2}}dx$$
$$\int \dfrac{\dfrac{\sin\left(\dfrac{3x}{6}\right)}{\sin\dfrac{x}{6}}\cdot\sin\dfrac{2x}{3}-2\sin x}{\dfrac{\sin\dfrac{4x}{6}}{\sin\dfrac{x}{6}}\cdot\cos\dfrac{x}{2}}dx$$
$$\int \tan\dfrac{x}{2}dx-\int \dfrac{4\sin\dfrac{x}{2}\sin\dfrac{x}{6}}{\sin\dfrac{2x}{3}}dx$$
$$\int \tan\dfrac{x}{2}dx-2\int \dfrac{\cos\dfrac{x}{3}-\cos\dfrac{2x}{3}}{\sin\dfrac{2x}{3}}dx$$
$$\int \tan\dfrac{x}{2}dx-\int \mathrm{cosec}\dfrac{x}{3}dx+2\int \cot\dfrac{2x}{3}dx$$
$$2\ln\left|\sec \dfrac{x}{2}\right|-3\ln\left|\mathrm{cosec}\dfrac{x}{3}-\cot\dfrac{x}{3}\right|+3\ln\left|\sin\dfrac{2x}{3}\right|+C$$
But this got too long,any short and better approach.
| I think your first substitution was on the right track, but instead note that since
$$\tan\left(\frac{x}{6}\right) = \frac{\tan\left(\frac{x}{2}\right) - \tan\left(\frac{x}{3}\right)}{1+\tan\left(\frac{x}{2}\right)\tan\left(\frac{x}{3}\right)} \implies \tan\left(\frac{x}{2}\right)\tan\left(\frac{x}{3}\right) = \frac{\tan\left(\frac{x}{2}\right) - \tan\left(\frac{x}{3}\right)}{\tan\left(\frac{x}{6}\right)} - 1$$
so the integral becomes
$$\int \tan\left(\frac{x}{2}\right) - \tan\left(\frac{x}{3}\right) - \tan\left(\frac{x}{6}\right) \: dx$$
$$ = 2\log\left(\sec\left(\frac{x}{2}\right)\right) - 3\log\left(\sec\left(\frac{x}{3}\right)\right) - 6 \log\left(\sec\left(\frac{x}{6}\right)\right) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Can we say that if $x\to 0^+$ then $\left \lfloor{\frac{1}{x}}\right \rfloor \sim \frac{1}{x}$?
Can we say that if $x\to 0^+$ then $\left \lfloor{\frac{1}{x}}\right \rfloor \sim \frac{1}{x}$?
Since $x\to 0^+\implies \frac{1}{x}\to +\infty$ therefore $\left \lfloor{\frac{1}{x}}\right \rfloor \to +\infty$.
So can we conclude that $\left \lfloor{\frac{1}{x}}\right \rfloor \sim \frac{1}{x}$ as $x\to 0^+$?
Thanks for your help.
| The fact that both go to infinity is not enough to conclude that they are asymptotically equivalent. For example, $\frac{1}{x}$ and $\frac{1}{x^2}$ a lso both go to $\infty$ as $x\rightarrow0^+$, but their ratio goes to $0$. However,
$$\lim_{x\rightarrow0^+}\frac{\frac{1}{x}}{\left\lfloor\frac{1}{x}\right\rfloor}=\lim_{x\rightarrow0^+}\frac{\left\lfloor\frac{1}{x}\right\rfloor+\left\{\frac{1}{x}\right\}}{\left\lfloor\frac{1}{x}\right\rfloor}=\lim_{x\rightarrow0^+}\left(1+\frac{\left\{\frac{1}{x}\right\}}{\left\lfloor\frac{1}{x}\right\rfloor}\right)=1.$$
The last equality follows from the squeeze theorem, since
$$0\le\frac{\left\{\frac{1}{x}\right\}}{\left\lfloor\frac{1}{x}\right\rfloor}\le\frac{1}{\left\lfloor\frac{1}{x}\right\rfloor}\stackrel{x\rightarrow0^+}{\rightarrow}0.$$
Therefore, $\frac{1}{x}\sim\left\lfloor\frac{1}{x}\right\rfloor$ as $x\rightarrow0^+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Expanding brackets to natural powers When expanding a bracket with a power to get terms in the form $a^n+b^n$, there is a pattern that appears similar to binomial theorem(if not a special case of). For example,
$$a^2+b^2=(a+b)^2-2ab$$
$$a^3+b^3=(a+b)^3-3(ab)(a+b)$$
$$a^4+b^4=(a+b)^4-4(ab)(a+b)^2+2(ab)^2$$
2, 3, 4 are binomial coefficients from pascals triangle, which makes sense as there are two terms in the bracket, so n choose 2, but I can't really get my head around the $ab$, $a+b$ terms. Is there a name for this where I can find out more about this, or a general method for any finding $a^n + b^n$.
| We derive a formula for $a^n+b^n$ with the help of the generating function
$Q(z)=\sum_{n=0}^\infty \left(a^n+b^n\right)z^n$.
We obtain
\begin{align*}
Q(z)&=\sum_{n=0}^\infty\left(a^n+b^n\right)z^n=\sum_{n=0}^\infty (az)^n+\sum_{n=0}^\infty(bz)^n\\
&=\frac{1}{1-az}+\frac{1}{1-bz}\tag{2}\\
&=\frac{2-(a+b)z}{1-(a+b)z+abz^2}\\
&=\left(2-(a+b)z\right)\sum_{k=0}^\infty\left((a+b)-abz\right)^kz^k\tag{3}
\end{align*}
In (2) and (3) we apply a geometric series expansion. We see in (3) a representation in terms of $a+b$ and $ab$ and use this relationship to derive a formula for $a^n+b^n$. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series.
We obtain from (3) for $n\ge 0$:
\begin{align*}
\color{blue}{a^n+b^n}&=[z^n]Q(z)\\
&=[z^n]\sum_{k=0}^{\infty}\left((a+b)-abz\right)^kz^k\left(2-(a+b)z\right)\\
&=\sum_{k=0}^n[z^{n-k}]\sum_{j=0}^k\binom{k}{j}(-abz)^j(a+b)^{k-j}\left(2-(a+b)z\right)\tag{4}\\
&=\sum_{k=0}^n[z^k]\sum_{j=0}^{n-k}\binom{n-k}{j}(-abz)^j(a+b)^{n-k-j}\left(2-(a+b)z\right)\tag{5}\\
&=2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k}{k}(-ab)^k(a+b)^{n-2k}\\
&\qquad-\sum_{k=1}^{\lfloor n/2\rfloor}\binom{n-k}{k-1}(-ab)^{k-1}(a+b)^{n-2k+2}\tag{6}\\
&=2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k}{k}(-ab)^k(a+b)^{n-2k}\\
&\qquad-\sum_{k=0}^{\lfloor n/2\rfloor-1}\binom{n-k-1}{k}(-ab)^{k}(a+b)^{n-2k}\tag{7}\\
&\,\,\color{blue}{=\sum_{k=0}^{\lfloor n/2\rfloor}\left(2\binom{n-k}{k}-\binom{n-k-1}{k}\right)(-ab)^k(a+b)^{n-2k}}\tag{8}
\end{align*}
Comment:
*
*In (4) we expand the binomial and apply the rule $[z^p]z^qA(z)=[z^{p-q}]A(z)$. We also restrict the upper limit of the sum with $n$ since indices with $k>n$ do not contribute.
*In (5) we change the order of summation $k\to n-k$.
*In (6) we select the coefficient of $[z^k]$. We also restrict the upper limit of the sum with $\lfloor n/2\rfloor$ since greater indices do not contribute.
*In (7) we shift the index by one to start with $k=0$ in the second sum, too.
*In (8) we collect the sums by noting that we use $\binom{p}{q}=0$ if $p,q$ are non-negative integer with $p<q$.
The sequence $\left(q_{n,k}\right)_{n,k}$ of coefficients in
\begin{align*}
&Q(z)=\sum_{n=0}^\infty \left(a^n+b^n\right)z^n=\sum_{n=0}^\infty \sum_{k=0}^{\lfloor n/2\rfloor} q_{n,k} (ab)^k (a+b)^{n-2k}z^n\\
\\
&\left(q_{n,k}\right)_{n\geq 0, 0\leq k\leq \lfloor n/2\rfloor}=(1;1;1,-2;1, -3;1, -4, 2;1, -5, 5;\\
&\qquad \qquad \qquad \qquad \qquad1,-6, 9, -2;\color{blue}{1, -7, 14, -7};\ldots)
\end{align*}
is archived in OEIS as A132460. We get for instance for $n=7$ the entries $\color{blue}{1,-7,14}$ and $\color{blue}{-7}$, so that
\begin{align*}
a^7+b^7=(a+b)^7-7ab(a+b)^5+14(ab)^2(a+b)^3-7(ab)^3(a+b)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a limit in order to find an asymptote I need to draw the graph of the function $y=\sqrt[3]{(x^2)(x+9)}$. Suppose $y=k\cdot x + b$ is the equation for the asymptote.
I managed to find $$k=\lim_{x\to\infty}\frac{y(x)}{x}=\lim_{x\to\infty}\sqrt[3]{\frac{x+9}{x}}=1$$
But then I got stuck finding $$b=\lim_{x\to\infty}y(x)-k\cdot x$$
Any ideas on how to deal with it?
| So the task is to find the following limit:
$$b=\lim_{x\to\infty}\left(y(x)-kx\right)=\lim_{x\to\infty}\left(\sqrt[3]{x^3+9x^2}-x\right).$$
To rationalize a cube root, we can take advantage of the formula
$$A^3-B^3=(A-B)(A^2+AB+B^2).$$
In this example we'll apply it to $A=\sqrt[3]{x^3+9x^2}$ and $B=x$. So let's multiply and divide out expression by $(A^2+AB+B^2)$:
$$\lim_{x\to\infty}\left(\sqrt[3]{x^3+9x^2}-x\right)=\lim_{x\to\infty}\frac{\left(\sqrt[3]{x^3+9x^2}-x\right)\left(\sqrt[3]{(x^3+9x^2)^2}+x\sqrt[3]{x^3+9x^2}+x^2\right)}{\sqrt[3]{(x^3+9x^2)^2}+x\sqrt[3]{x^3+9x^2}+x^2}=\lim_{x\to\infty}\frac{(x^3+9x^2)-x^3}{\sqrt[3]{(x^3+9x^2)^2}+x\sqrt[3]{x^3+9x^2}+x^2}=\lim_{x\to\infty}\frac{9x^2}{\sqrt[3]{(x^3+9x^2)^2}+x\sqrt[3]{x^3+9x^2}+x^2}=\frac{9}{3}=3,$$
where the last step follows by simultaneously dividing the numerator and denominator by $x^2$ (or, equivalently, by comparing the highest degree terms and their coefficients, so to speak).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof that the sum of the reciprocals of primes diverges Reading the proof-sketch (as set out here) that the sum of the reciprocals of the prime numbers is divergent, I get stuck. The proof starts with Euler's product formula $\sum _{n=1}^{\infty } \frac{1}{n}=\prod_p \frac{1}{1-p^{-1}}$. Then it proceeds as follows:
"Euler considered the above product formula and proceeded to make a sequence of audacious leaps of logic. First, he took the natural logarithm of each side, then he used the Taylor series expansion for $\ln x$ as well as the sum of a converging series:
$$\ln \left(\sum _{n=1}^{\infty } \frac{1}{n}\right)=\ln \left(\prod_p \frac{1}{1-p^{-1}} \right)$$
$$=\sum_p \frac{1}{p} +\frac{1}{2p^2}+\frac{1}{3p^3}+...$$
$$=\sum_p \frac{1}{p}+\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+...$$
$$=A+\frac{1}{2}B+\frac{1}{3}C+...$$
$$=A+K$$
for a fixed constant $K<1$."
I don't follow the last step. Clearly, $B=\sum_p \frac{1}{p^2}$, $C=\sum_p \frac{1}{p^3}$, $D=\sum_p \frac{1}{p^4}$ and so on, and each of these sums is convergent. But why is $\frac{1}{2}B+\frac{1}{3}C+\frac{1}{4}D+...<1$ given that $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...$ is divergent?
I get that the value of $\sum_p \frac{1}{p^k}$ becomes rapidly smaller with increasing $k$, but how does one prove the convergence of $\frac{1}{2}B+\frac{1}{3}C+\frac{1}{4}D+...$ to a value less than $1$?
| Could be following from this (for $k\ge 2$): $$\sum_p\frac{1}{p^k}\lt\sum_{n=2}^{\infty}\frac{1}{n^k}\lt\int_1^{\infty}\frac{dx}{x^k}=\frac{1}{k-1}$$ so $\frac{1}{2}B+\frac{1}{3}C+\frac{1}{4}D\cdots$ becomes bounded by a simple telescopic series $\sum_{k=2}^\infty\frac{1}{(k-1)k}$ with the sum $1$?
| {
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"source": "stackexchange",
"question_score": "3",
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Sequence of functions inequality I need some help to prove that $f_n(x)\leq 1$ , where $f_n:[0,1]\to\mathbb{R}, f_n(x)=\frac{1+nx^2}{(1+x^2)^n}, n\in\mathbb{N}$.
I can write $(1+x^2)^n=\binom{0}{n}x^n+\binom{1}{n}x^{n-1}y+\binom{2}{n}x^{n-2}y^2+...+\binom{n-2}{n}x^2y^{n-2}+\binom{n-1}{n}xy^{n-1}+\binom{n}{n}y^n=\\x^n+nx^{n-1}y+\frac{n(n-1)}{2!}x^{n-2}y^2+...+\frac{n(n-1)}{2!}x^2y^{n-2}+nxy^{n-1}+y^n$
$\\$
I unsuccessfully tried to find $1+nx^2$ in $(1+x^2)^n$ to show that $1+nx^2\leq(1+x^2)^n$.
| $1+nx^2\le(1+x^2)^n$ is true by Bernoulli's inequality, since $x^2\ge0$. Thus $f_n(x)\le1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve this equations system: $x^{5} + y^{5} + z^{5} = 1, x + y + z = 1$
Solve this equations system: \begin{cases} x^{5} + y^{5} + z^{5} =
1 \\ x + y + z = 1 \end{cases}
Sketch of solutions:
If x,y,z is solution of this equations system then $$(x+y+z)^{5} - (x^{5} + y^{5}+z^{5}) = 0$$ when $x+y= 0$, then equation system is true for any z.
So we have
$$(x+y+z)^5 - (x^{5}+y^{5}+z^{5})= A(x+y)(y+z)(x+y)(x^{2}+xy+xz+y^{2}+yz+z^{2})=$$
for $x=1, y=1, z=0$ we have
$$2^{5}-1-1 = 6A$$
$$ A = 5 $$
So we have
$5(x+y)(y+z)(x+z)(x^{2}+xy+xz+y^{2}+yz+z^{2})=0$
So the solutions:
1.
$$x+y= 0, z = 1 $$
2.
$$y+z= 0, x = 1 $$
3.
$$x+z= 0, y = 1 $$
4.
$$x^{2}+xy+xz+y^{2}+yz+z^{2} =$$
$$ \frac{1}{2}(x^{2}+2xy+y^{2}+x^{2}+2xy+z^{2}+y^{2}+2yz+z^{2}$$
$$ \frac{1}{2}((x+y)^{2}+(y+z)^{2}+(x+z)^{2})==0$$
this imply that (0,0,0) is solution. It's wrong because we have $0=1$
Do you see mistake?
Solutions from Mathematica:
$\left\{\{x\to 1,z\to -y\},\{y\to 1,z\to -x\},\{y\to -x,z\to 1\},\left\{y\to \frac{1}{2} \left(-\sqrt{-3 x^2+2 x-3}-x+1\right),z\to \frac{1}{2} \left(\sqrt{-3 x^2+2 x-3}-x+1\right)\right\},\left\{y\to
\frac{1}{2} \left(\sqrt{-3 x^2+2 x-3}-x+1\right),z\to \frac{1}{2} \left(-\sqrt{-3 x^2+2 x-3}-x+1\right)\right\}\right\}$
| If we have $x^5+y^5=A$ and $x+y=B$ we have
$$ A = x^5+(B-x)^5 $$
such that
$$ x = \frac{B}{2}\pm\frac{\sqrt{5} \sqrt{\pm 2 \sqrt{5} B \sqrt{B \left(4 A+B^5\right)}-5 B^4}}{10 B},\quad y = \frac{B}{2}\mp\frac{\sqrt{5} \sqrt{\pm 2 \sqrt{5} B \sqrt{B \left(4 A+B^5\right)}-5 B^4}}{10 B}$$
because $(B/2+z)^5+(B/2-z)^5$ is a biquadratic polynomial. Now you may just set $A=1-z^5, B=1-z$ and notice that the intersection (in $\mathbb{R}^3$) of $x+y+z=1$ and $x^5+y^5+z^5=1$ is just given by three lines:
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Recognise factors for fractional exponents Hi is there a general rule to factorising these type of exponent fractions? The answer makes sense but I wouldn't have thought of it.
$\ x^{ \frac{2x}{1-2x} } - 2x^{ \frac{1}{1-2x} }$
Answer:
$\ x^{ \frac{2x}{1-2x} } (1- 2x$)
| This has more to do with the manipulation of rational expressions.
We have $$\frac1{1-2x}=\frac{1\color{blue}{-2x+2x}}{1-2x}=1+\frac{2x}{1-2x}.$$
As such, \begin{align}
x^{\frac{2x}{1-2x}}-2x^{\frac1{1-2x}}&=x^{\frac{2x}{1-2x}}-2x^{1+\frac{2x}{1-2x}}\\
&=x^{\frac{2x}{1-2x}}-2x\cdot x^{\frac{2x}{1-2x}}\\
&=x^{\frac{2x}{1-2x}}(1-2x).
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that $\cos 2x + \cos 2y + \cos 2z = -4(\cos x \cos y \cos z) -1$ when $x+y+z=\pi$ I found on List of trigonometric identities on wikipedia that $\cos 2x + \cos 2y + \cos 2z = -4(\cos x \cos y \cos z) -1$ if $x+y+z=\pi = 180°$. I couldn't find a proof.
| $$
\text{LHS}{=\cos 2x+\cos 2y+\cos (2x+2y)
\\=\cos 2x+\cos 2y+\cos 2x\cos 2y-\sin 2x\sin 2y
}
$$
$$
\text{RHS}{=4\cos x\cos y\cos(x+y)-1
\\=4\cos^2 x\cos^2 y-4\cos x\cos y\sin x\sin y-1
\\=(\cos 2x+1)(\cos 2y+1)-\sin 2x\sin 2y-1
\\=\cos 2x+\cos 2y+\cos 2x\cos 2y-\sin 2x\sin 2y
}
$$
which completes the proof $\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution verification: $\lim_{x\to \infty}\Bigg(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\Bigg)^{x\ln x}$
Find: $$\displaystyle\lim_{x\to
\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}$$
My attempt:
$\displaystyle\lim_{x\to\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln(x^2+3x+4)-\ln(x^2+2x+3)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\\\displaystyle\lim_{x\to\infty}\left(1+\frac{\ln\left(\frac{x^2+3x+4}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln\left(1+\frac{x+1}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}$
What I used:
$\displaystyle\lim_{x\to\infty}\ln\left(1+\frac{x+1}{x^2+2x+3}\right)=0\;\;\&\;\;\lim_{x\to\infty}\ln(x^2+2x+3)=+\infty$
In the end, I got an indeterminate form: $\displaystyle\lim_{x\to\infty}1^{x\ln x}=1^{\infty}$
Have I made a mistake anywhere? It seems suspicious.
added: replacement $\frac{x+1}{x^2+2x+3}$ by $\frac{1}{x}$ wasn't appealing either.
Would:$$\lim_{x\to\infty}\Big(\Big(1+\frac{1}{x}\Big)^x\Big)^{\ln x}=x=\infty$$
be wrong?
//a few days after users had provided hints and answered the question,we discussed this with our assistant and he suggested a table formula that can also be applied (essentialy the last step in methods provided in the answers I recieved).:$$\lim_{x\to c}f(x)=1\;\&\;\lim_{x\to c}g(x)=\pm\infty$$then$$\lim_{x\to c}f(x)^{g(x)}=e^{\lim_{x\to c}(f(x)-1)g(x)}//$$
| By your work and since $e^x$ and $\ln$ are continuous function, we obtain: $$\lim_{x\rightarrow+\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln{x}}=\lim_{x\rightarrow+\infty}\left(1+\frac{\ln\frac{x^2+3x+4}{x^2+2x+3}}{\ln(x^2+2x+3)}\right)^{\frac{\ln(x^2+2x+3)}{\ln\frac{x^2+3x+4}{x^2+2x+3}}\cdot\frac{x\ln{x}\ln\frac{x^2+3x+4}{x^2+2x+3}}{\ln(x^2+2x+3)}}=$$
$$=e^{\lim\limits_{x\rightarrow+\infty}\frac{\ln{x}\ln\left(1+\frac{x+1}{x^2+2x+3}\right)^x}{2\ln{x}+\ln\left(1+\frac{2}{x}+\frac{3}{x^2}\right)}}=e^{\frac{1}{2}\ln\lim\limits_{x\rightarrow+\infty}\left(1+\frac{x+1}{x^2+2x+3}\right)^{\frac{x^2+2x+3}{x+1}\cdot\frac{x(x+1)}{x^2+2x+3}}}=e^{\frac{1}{2}\ln{e}}=\sqrt{e}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\log (\log n + 1) \leq \log \log n + \frac{1}{\log (n + 1)}$ How does one show that $\log (\log n + 1) \leq \log \log n + \dfrac{1}{\log (n + 1)}$ for $n \geq 3$?
For context, I saw this inequality in https://arxiv.org/pdf/math/0008177v2 (formula 3.8).
| Let's show that
For $x\in\left[0,\frac{1}{2}\right]$ we have
$$\ln(1+x) \leq x-\frac{x^2}{3} \tag{1}$$
Indeed, $f(x)=\ln(1+x)-x + \frac{x^2}{3}$ has first derivative $f'(x)=\frac{2x}{3}+\frac{1}{1+x}-1=\frac{x(2x-1)}{3(x+1)}\leq 0$ when $x\in \left[0,\frac{1}{2}\right]$.
So $f(x)$ is descending on $x\in \left[0,\frac{1}{2}\right]$ or for $0\leq x\leq \frac{1}{2} \Rightarrow f(0)\geq f(x)$ which is $0\geq \ln(1+x)-x + \frac{x^2}{3}$ and the result follows.
This can also be reformulated as, for $x \geq 2$
$$\ln\left(1+\frac{1}{x}\right) \leq \frac{1}{x}-\frac{1}{3x^2} \iff
\color{blue}{x\ln\left(1+\frac{1}{x}\right) \leq 1-\frac{1}{3x}} \tag{2}$$
I will also use the fact that
$$\left(1+\frac{1}{x}\right)^x<e \tag{3}$$
Now let's change the original inequality as
$$\log (\log n + 1) \leq \log \log n + \frac{1}{\log (n + 1)}\iff \\
\log\left(1+\frac{1}{\log{n}}\right)\leq \frac{1}{\log(n+1)} \iff \\
\log(n+1)\cdot\log\left(1+\frac{1}{\log{n}}\right)\leq 1 \iff \\
\left(\log{n}+\log\left(1+\frac{1}{n}\right)\right)\cdot\log\left(1+\frac{1}{\log{n}}\right)\leq 1 \iff $$
$$\left(\log{n}+\frac{\log\left(1+\frac{1}{n}\right)^n}{n}\right)\cdot\log\left(1+\frac{1}{\log{n}}\right)\leq 1 \tag{4}$$
And finally
$$\left(\log{n}+\frac{\log\left(1+\frac{1}{n}\right)^n}{n}\right)\cdot\log\left(1+\frac{1}{\log{n}}\right) \overset{(3)}{<} \\
\left(\log{n}+\frac{1}{n}\right)\cdot\log\left(1+\frac{1}{\log{n}}\right) = \\
\left(1+\frac{1}{n\log{n}}\right)\cdot \color{red}{\log{n} \cdot\log\left(1+\frac{1}{\log{n}}\right)} \overset{(2)}{<} \\
\left(1+\frac{1}{n\log{n}}\right)\cdot \left(1-\frac{1}{3\log{n}}\right)=\\
1-\frac{1}{3n\log^2(n)}-\frac{1}{3\log{n}}+\frac{1}{n\log{n}} < 1$$
since
$$-\frac{1}{3\log{n}}+\frac{1}{n\log{n}} <0$$
and for $\log{n}>2 \iff n >e^2 > 7$. So $(4)$ is true for $n>7$. Cases $n\in\{3,4,5,6,7\}$ can be validated manually or with a computer program.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find supremum and infimum for $C={\frac{x^2+1}{x^2+2}: x \in \mathbb{R}}.$ Find supremum and infimum for
$C={\frac{x^2+1}{x^2+2}: x \in \mathbb{R}}.$
We can easy see that $\frac{x^2+1}{x^2+2}$ is bounded from below with $m=\frac{1}{2}.$ This is also infimum. To show this I proceed like this.
Let us assume that $\frac{1}{2}$ is infinum (is not the largest bound for $\frac{x^2+1}{x^2+2}$), i.e there exists $m>\frac{1}{2}$ such that $\frac{x^2+1}{x^2+2}\geq m$. Let us take $x=0$. We have $\frac{0^2+1}{0^2+2}=\frac{1}{2}\geq m>\frac{1}{2}.$vSince we get $\frac{1}{2}>\frac{1}{2}$ which is not true, so we conclude that $\frac{1}{2}$ is infinum.
For supremum, we can see that $\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}$ is bounded from above with $M=1$ and I think this is supremum (the smallest bound). I do not know how to prove it.vI would be grateful for any help.
| Attempt:
$y:=x^2$; $y\ge 0$;
$f(y):=\dfrac{y+1}{y+2}= 1-\dfrac{1}{y+2}$.
1) $\dfrac{1}{y+2} \le 1/2$.
Equality for $y=0$.
$\min f(y)= \inf f(y) =1/2$;
2) $\dfrac{1}{y+2} \gt 0$;
Hence $1$ is an upper bound for $f(y)$.
Assume there is a smaller upper bound $b >0$, i.e.
$f(y)= 1- \dfrac{1}{y+2} \le b <1$, $y \ge 0$.
$0<1-b\le \dfrac{1}{y+2}$;
$y+2 \le \dfrac{1}{1-b}$, $y \ge 0$;
a contradiction.
Hence $\sup f(y)=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a second order Taylor polynomial for a function $z=\arctan\left( \frac{1-x+y}{1+x-y} \right)$ First we make substitution, so that $\arctan\left(\frac{1-u}{1+u} \right)$, where $u=x-y$. Then $u_x=1; u_y=-1$ Then $z_u$ from there is expressed as $$z_u=\frac{1}{1+\left(\frac{1-u}{1+u} \right)^2}=-\frac{1}{1+u^2}$$
Then $z_x=z_u \cdot u_x=z_u$ and $z_y=z_u \cdot u_y=-z_u$. Then it is noted, that because $z$ can be written at neighbourhood of $\left[ \begin{matrix} 0 \\ 0 \end{matrix}\right], \text{ then } z=\frac{\pi}{4}-x+y$. Why is it so?
$z_{xx}=z_{uu} \cdot u_x, z_{yy}=z_{uu} \cdot u_y, z_{xy}=z_{uu} \cdot u_y$
$$z_{uu}=\frac{2}{(1+u^2)^2}$$
| If you have
$$z=\arctan\left(\frac{1-u}{1+u} \right)$$ let $t=\frac{1-u}{1+u}$
$$\frac{dz}{du}=\frac{dz}{dt} \frac{dt}{du}=\frac1{1+t^2 } {\frac{-2}{(u+1)^2} }$$ and
$$\frac1{1+t^2 }=\frac{(u+1)^2}{2 \left(u^2+1\right)}$$
$$\frac{dz}{du}=\frac{(u+1)^2}{2 \left(u^2+1\right)}{\frac{-2}{(u+1)^2} }=-\frac{1 } {u^2+1 }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral with partial fractions I'm trying to calculate the following integral:
$$I=\int \frac{\arctan(x)}{x^4}dx$$
My steps so far are:
Per partes:
$$\frac{-\arctan(x)}{3x^3}+\int{\frac{1}{1+x^2} \frac{1}{3x^3}}dx
=\frac{-\arctan(x)}{3x^3}+\frac{1}{3}\int{\frac{1}{1+x^2} \frac{1}{x^3}}dx$$
and now I want to do partial fractions. However, with this integral, I fail to do partial fractions. Could you help me?
Thanks
| Or more simply:
$$A=\frac{1}{1+x^2}*\frac{1}{x^3}$$
$$A=-\frac 1 x \left (\frac{1}{1+x^2}-\frac{x}{x^3} \right )$$
$$A=-\frac{1}{x(1+x^2)}+\frac{1}{x^3}$$
$$B=\frac{1}{x(1+x^2)}=\frac{1}{x}-\frac x {(1+x^2)}$$
We have that:
$$I=\int \left \{\frac x {(1+x^2)}-\frac{1}{x}+\frac{1}{x^3} \right \}dx$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $a^b \pmod {c}$, $a^b \pmod {d}$, $a^b \pmod {e}$, how can $a^b mod (c*d*e)$ be deduced? This question covered large exponents on the $b$ side. What about the $m$ side?
Given:
$$a^b \pmod m$$
where $m$ is a large compound number.
For example:
Given
$$5^{2003} \pmod {7} \equiv 3$$
$$5^{2003} \pmod {11} \equiv 4$$
$$5^{2003} \pmod {13} \equiv 8$$
how can one quickly deduce:
$$5^{2003} mod (7*11*13)$$
| Let the number be $x$.
Then we get from the Chinese Remainder Theorem:
$$5^{2003}\pmod{7\cdot 11\cdot 13}\equiv x\iff\begin{cases}x\pmod 7\equiv 3\\x\pmod{11}\equiv 4\\ x\pmod{13}\equiv 8\end{cases}$$
The following is one method to apply the Chinese Remainder Theorem.
From the 3rd equation:
$$x=13k+8\tag 4$$
Combine with the 1st equation:
$$13k+8\equiv -k+1\equiv 3\pmod 7\implies k=7l-2$$
Substitute in (4):
$$x=13(7l-2)+8=13\cdot7l-18\tag 5$$
Combine with 2nd equation:
$$13\cdot7l-18\equiv 3l+4\equiv 4\pmod{11} \implies l=11m$$
Substitute in (5):
$$x=13\cdot7\cdot 11m -18 \equiv -18\pmod{7\cdot 11\cdot 13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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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.