Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Recursive sequence depending on the parameter
For the given parameter $\mathbb R\ni t\geq 1$, the sequence is
defined recursively: $$a_1=t,\;\;a_{n+1}a_n=3a_n-2$$ $(a)$ Let $t=4$.
Prove the sequence $(a_n)$ converges and find its limit.
$(b)$ Which parameters $t\geq 1$ is the sequence $(a_n)$ increasing
for?
My attempt:
Bolzano-Weierstrass:A sequence converges if it is monotonous and
bounded
$$a_{n+1}a_n=3a_n-2\implies a_{n+1}=3-\frac{2}{a_n}$$
$(a)$
First few terms: $a_1=4,a_2=\frac{5}{2},a_3=\frac{11}{5}$
Assumption: the sequence is decreasing
Proof by induction:
the basis (n=1) is trivial: $\frac{5}{2}<4$
Assumption: $a_n<a_{n-1},\;\forall n\in\mathbb N$
Step: $$a_n<a_{n-1}\implies\frac{1}{a_n}\geq\frac{1}{a_{n-1}}\Bigg/\cdot(-2)$$
$$\iff-\frac{2}{a_n}\leq-\frac{2}{a_{n-1}}\iff \underbrace{3-\frac{2}{a_n}}_{a_{n+1}}\leq\underbrace{3-\frac{2}{a_{n-1}}}_{a_n}$$
The limit: $$L=3-\frac{2}{L}\implies L^2-3L+2=0$$
I take into account only $2$ because the parabola is convex and $$a_n\to L^-.$$
Then I have to prove: $a_n\geq 2\;\forall n\in\mathbb N$ after the formal computing: $a_{n+1}\geq 3-\frac{2}{2}=2$
$\underset{\implies}{\text{Bolzano-Weierstrass theorem}}(a_n)\to 2$
$(b)$ Since the sequence doesn't have to be convergent, only increasing:
$$a_2=3-\frac{2}{t}\geq t\implies t\in[1,2]$$
Then, it should follow inductively,analogously to $(a)$, this time it is increasing.
Is this correct?
| If you have already proven $a_n > 2, \forall n$ , then
Let $b_n=a_n-2$, $b_n>0$. $a_{n+1} = 3 - \frac{2}{a_n} \Rightarrow b_{n+1} + 2=3-\frac{2}{b_n+2} \Rightarrow b_{n+1} = \frac{b_n}{2+b_n} < \frac{b_n}{2}$.
Therefore as $n\to \infty, b_n \to 0, a_n \to 2.\blacksquare$
The next is pure hindsight based on dan_fulea's solution but I believe it can be useful when the two fixed points are distinct.
$a_{n+1} - 1 = 2-\frac{2}{a_n} = \frac{2(a_n - 1)}{a_n}$
$a_{n+1} - 2 = 1-\frac{2}{a_n} = \frac{a_n - 2}{a_n}$
Therefore $\frac{a_{n+1}-1}{a_{n+1}-2} = 2\cdot \frac{a_n-1}{a_n-2}$
$\frac{a_n-1}{a_n-2}$ is a geometric sequence with initial value $\frac{3}{2}$,
so $\frac{a_n-1}{a_n-2} = 2^{n-1} \frac{3}{2} = 1+\frac{1}{a_n-2} \Rightarrow a_n = 2+ \frac{1}{2^{n-1}\frac{3}{2}-1} = \frac{6\cdot 2^{n-1}-2}{3\cdot 2^{n-1}-2}.\blacksquare$
In general if there are two distinct fixed points $r$ and $s$ then the ratio $\frac{a_n-r}{a_n-s}$ is a geometric sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$ If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$
My attempt is as follows:-
$$a_0+a_1(x-2)+a_2(x-2)^2+\cdots a_{2n}(x-2)^{2n}=b_0+b_1(x-3)+b_2(x-3)^2+\cdots b_{2n}(x-3)^{2n}$$
Let's see the term on R.H.S with coefficient as $b_n$
$$T_{n+1}=b_n(x-3)^n$$
Comparing the coefficients of $x^0$
$$a_0-2a_1+(-2)^2a_2+(-2)^3a_3+\cdots a_{2n}(-2)^{2n}=b_0-3b_1+(-3)^2b_2+(-3)^3b_3+\cdots+b_n(-3)^n+\cdots b_{2n}(-3)^{2n}$$
Comparing the coefficients of $x^1$
$$a_1+2a_2(-2)^1+3a_3(-2)^2\cdots 2n\cdot a_{2n}(-2)^{2n-1}=b_1+2a_2(-3)^1+3a_3(-3)^2\cdots nb_n(-3)^{n-1} \cdots (2n)a_{2n}(-3)^{2n-1}$$
Now I am not getting any way to find $b_n$ , its so mixed up in these expressions, not able to get any breakthrough. Any inputs?
| We have
\begin{eqnarray*}
\sum_{r=0}^{2n} (x-2)^r = \sum_{r=0}^{2n} b_r (x-3)^r.
\end{eqnarray*}
Expand the $(x-2)$
\begin{eqnarray*}
(x-2)^r=(x-3+1)^r= \sum_{s=0}^{r} \binom{r}{s} (x-3)^s.
\end{eqnarray*}
Invert the order of the plums
\begin{eqnarray*}
\sum_{r=0}^{2n} \sum_{s=0}^{r} \cdots = \sum_{s=0}^{2n} \sum_{r=s}^{2n} \cdots.
\end{eqnarray*}
So we need to perform the following sum, which is the hockey stick identity
\begin{eqnarray*}
b_s= \sum_{r=s}^{2n} \binom{r}{s} =\binom{2n+1}{s+1}.
\end{eqnarray*}
In particular for $b_n$ we have
\begin{eqnarray*}
b_n =\binom{2n+1}{n+1}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3
If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.
I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.
*
*If $x$ or $y$ is divisible by $3$, but not both, then:
$\begin{align}x \equiv \pm 1 \pmod{3} \land y \equiv 0 \pmod{3}
&\iff x^2 \equiv 1 \pmod{3} \land y^2 \equiv 0 \pmod{3}
\\
&\iff x^2 - y^2 \equiv 1 \pmod{3}\end{align}$
So if $x$ or $y$ is divisible by $3$ but not both then no values for $x$ and $y$ will ever make $x^2 - y^2$ divisible by $3$.
*If $x$ and $y$ are divisible by $3$:
Since both $x$ and $y$ are divisible by $3$ they can be expressed $x = 3m$ and $y = 3n$ for some integers $m$, $n$.
$(3m)^2 - (3n)^2 = 1995 \iff 9m^2 - 9n^2 = 1995 \iff 3(m^2-n^2) = 665$
And it's easy to see that there are no whole number solutions for m and n, and thus there are no whole number solutions for x and y such that $x^2 - y^2 = 1995$ where x or y (or both) is divisible by 3.
Is this a reasonable proof? Are there more elegant ways to do it?
| Another method is $x^2-y^2=1995$ where $x,y \in I$ can be written as
$$(x+y)(x-y)=3×5×133$$
Hence $x+y=1995,665,399,133$ and $x-y=1,3,5,15$
Hence $(x,y)=(\pm 998,\pm 997),(\pm 334,\pm 331),(\pm 202,\pm 197),(\pm 74,\pm 59)$ where none is multiple of $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Expectation and variance of the squared distance between $X$ and $Y$ Given that $X$ and $Y$ are two independent univariate random variables sampled uniformly from the unit interval [0,1]. I am trying to find the expected value and the variance of the random variable $Z = \mid X - Y \mid ^2 $ (the squared distance between $X$ and $Y$.
So far, I'm assuming that $X$ and $Y$ are uniformly distributed and this is what I have:
\begin{align}
Z &= \mid X - Y\mid ^2 \\
Z &= \mid X^2 - 2XY + Y^2 \mid \\
E[Z] &= \mid E[X^2] - 2E[XY] + E[Y^2]\mid \\
&= \mid E[X^2] - 2E[X]E[Y] + E[Y^2]\mid \\
&= \frac{1}{3}\frac{1}{b-a}(b^3-a^3)-2(\frac{a+b}{2})(\frac{a+b}{2})+\frac{1}{3}\frac{1}{b-a}(b^3-a^3)
\end{align}
But I am not sure if this is the right approach or the right direction and what is E[Z] = $\mid X - Y\mid ^2$. If so, can I do something similar to the variance?
| Variace is $E[(X-Y)^4] -[E[(X-Y)^2]]^2$
=$ E(X^4) + E(Y^4) + 6E(X^2)E(Y^2) -4E(X^3)E(Y) - 4E(X)E(Y^3) -\frac{1}{36}$
= $\frac{2}{5} + \frac{6}{9} - 1 - \frac{1}{36}$
=$\frac{7}{180}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to get expectation of a table $X$ and $Y$ are random variables with joint distribution table given as
$$\begin{array}{l|ccc}
& x=1 & x=2 & x=3 \\ \hline
y=1 & 3/12 & 1/12 & 3/12 \\
y=2 & 1/12 & 3/12 & 1/12
\end{array}$$
Calculate $\mathrm{Cov}(X, Y)$ and explain why $X$ and $Y$ are independent.
Attempt:
I know
$$\mathrm{Cov}(X, Y) = E(XY) - E(X)E(Y)$$
But I'm not sure how to calculate the expectations
$$E(X) = 1 \times \frac3{12} + 2 \times \frac1{12} + 3 \times \frac3{12}$$
I don't think thats right because of the y value. Could someone help
| Your first intiution is correct, however the expected value of the random variable $X$ is given by the expected value of the marginal probability density function of X. It means that in order to get $E(X)$, you have to sum up on X, to get $E(Y)$ you have to do the same but on $Y$.
$$
\begin{array}{l|ccc}
& x=1 & x=2 & x=3 & P(Y=y) \\ \hline
y=1 & 3/12 & 1/12 & 3/12 & 7/12\\
y=2 & 1/12 & 3/12 & 1/12 & 5/12\\
P(X=x) & 4/12 & 4/12 & 4/12 & 1
\end{array}
$$
In that case you can calculate E(X) and E(Y) as follows:
$$
E(X) = 1 \times \frac{4}{12} + 2 \times \frac{4}{12} + 3\times \frac{4}{12} = 2
$$
$$
E(Y) = 1 \times \frac{7}{12} + 2 \times \frac{5}{12} = \frac{17}{12}
$$
You can get $E(XY)$ by simply multiplying each probability value at the table by the corresponding $x$ and $y$ values:
$$
E(XY) = 1 \times 1 \times \frac {3}{12} + 1 \times 2 \times \frac{1}{12} + 1 \times 3 \times \frac{3}{12} + 2 \times 1 \times \frac{1}{12} + 2 \times 2 \times \frac{3}{12} + 2 \times 3 \times \frac{1}{12} = \frac{17}{6}
$$
Hence the covariance is as follows:
$
Cov(X,Y) = E(XY) - E(X)E(Y) = \frac{17}{6} - 2 \times \frac{17}{12} = 0
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int_{0}^{\pi} \frac{1}{1+3^{\cos x}} dx.$ I am not able to solve the following integration, $$\int_{0}^{\pi} \frac{1}{1+3^{\cos x}} dx.$$ I have tried in different ways but most good one I think,
\begin{align*} \int_{0}^{\pi} \frac{1}{1+3^{\cos x}} dx &= \int_{0}^{\pi} \frac{\sin x}{\sin x (1+3^{\cos x})} \\ &= \int_{-1}^{1} \frac{dz}{\sqrt{1-z^2} (1+3^{z})}.\end{align*} Now how could I proceed. Please help me.
| Just for the fun of it.
Dr Zafar Ahmed DSc provided the good solution.
For the fun of it, I did consider the more general problem of
$$\int_{a}^{\pi-a} \frac{dx}{1+k^{\cos x}} $$
Expanding the integrand as a Taylor series built around $x=\frac \pi 2$ we have
$$\frac{1}{1+k^{\cos x}}=\frac{1}{2}+\frac{\log (k)}{4} \left(x-\frac{\pi }{2}\right) -\frac{\log (k) \left(\log
^2(k)+2\right)}{48}
\left(x-\frac{\pi }{2}\right)^3 +\frac{\log(k)\left(\log
^4(k)+5 \log ^2(k)+1\right)}{480} \left(x-\frac{\pi }{2}\right)^5 +O\left(\left(x-\frac{\pi }{2}\right)^7\right)$$ which, as expected, shows only odd powers of $\left(x-\frac{\pi }{2}\right)$.
As a result
$$\int_{a}^{\pi-a} \frac{dx}{1+k^{\cos x}}= \frac{\pi }{2}-a\qquad \forall k >0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find angle in triangle $ABC$ with cevian line $AD$, such that $AB=CD$.
As you can see in the picture, there is a triangle $ABC$ with $∠C=30°$ and $∠B=40°$. Now we assuming that $AB=CD$, try to find the exact value of $∠CAD$.
My attempt: Denote $∠CAD$ by $x$, we know that
$$\frac{\sin C}{AD}=\frac{\sin{x}}{CD},\quad\frac{\sin B}{AD}=\frac{\sin{(x+C)}}{AB}$$
Then we have ($∠C=30°=\frac{\pi}{6},∠B=40°=\frac{2\pi}{9}$)
$$\frac{\sin{\frac{\pi}{6}}}{\sin{x}}=\frac{\sin{\frac{2\pi}{9}}}{\sin{(x+\frac{\pi}{6})}}=\frac{AD}{AB}$$
Everything looks ok so far, but I have trouble solving the equation. What's more, Wolfram tells me that the answer is $x=\frac{5\pi}{18}$.
This exercise is in my sister's assignment, so I think this exercise should have a high-school (or high-school olympic) level answer.
More: The exercise appears in geometry part, so a pure geometric method will be better.
| An algebraic proof.
Consider the triangle $ACD$ and label $\varphi=\angle\, CAD$. Both $AC$ and $AD$ can be related to $AB$, i.e., $CD$, by the law of sines: $\frac{CD}{\sin \angle CAD}=\frac{AD}{\sin \angle ACD}=2AD$, thus $AD=\frac{AB}{2\sin \varphi}$. Also, $\frac{AC}{\sin \angle ABC}=\frac{AB}{\sin \angle ACB}=2AB$, thus $AC=2\sin 40^\circ AB$. By the law of cosines,
$$CD^{2}=AC^{2}+AD^{2}-2AC\cdot AD\cos\varphi
\\
\Rightarrow
\left(4\sin^{2}40^\circ+\frac{\csc^{2}\varphi}{4}-1-{2\cot\varphi\sin40^\circ}\right)AB^{2}=0
$$
Since $AB>0$, this equation has a solution iff $\varphi$ is a root of the parenthetical term. Let $\cos\varphi=x$, $\sin40^\circ=\lambda$. Thus, equating the parenthetical term to zero gives the quadratic in $x^2$,
$$16\left(16\lambda^{4}-4\lambda^{2}+1\right)x^{4}+8\left(-64\lambda^{4}+20\lambda^{2}-3\right)x^{2}+\left(256\lambda^{4}-96\lambda^{2}+9\right)=0$$
which can be solved for
$$\begin{align}
x&=\pm\frac{1}{4}\sqrt{\frac{256\lambda ^{4}-80\lambda ^{2}+12\pm\ 8\sqrt{3}\lambda }{\left(16\lambda ^{4}-4\lambda ^{2}+1\right)}}
\\
x&\in\left\{\pm0.6428,\,\pm0.9740\right\}
\\
\varphi&\in\left\{13.12^\circ,\,50.00^\circ,\,130.00^\circ,\,166.91^\circ\right\}
\end{align}
$$
It suffices to see that $\varphi=50.00^\circ$ is the correct solution, since the angle is acute and $\varphi=13.12^\circ$ would make $\angle\,DAB=96.88^\circ$, which is absurd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the real value(s) solution of the equation $3^x = 3(x+6)$ I tried using Lambert W function but I got stuck while trying to set the question in the form $xe^x =y$.
| $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
Consider
\begin{align}
3^x &= 3(x+6)
\tag{1}\label{1}
\end{align}
as a special case of
\begin{align}
a^x&=b\,x+c
,\quad a=3,b=3,c=18
\tag{2}\label{2}
\end{align}
and transform it as follows:
\begin{align}
\exp(\ln(a)\,x)&=b\,x+c
,\\
\exp(\tfrac 1b\,\ln(a)\,(b\,x))&=b\,x+c;
,\\
\exp(\tfrac 1b\,\ln(a)\,(b\,x+c-c))&=b\,x+c;
,\\
\exp(\tfrac 1b\,\ln(a)\,(b\,x+c))
\,\exp(-\tfrac cb\,\ln(a))&=b\,x+c;
,\\
\exp(-\tfrac cb\,\ln(a))&=(b\,x+c)\,\exp(-\tfrac 1b\,\ln(a)\,(b\,x+c))
,\\
a^{-c/b}&=(b\,x+c)\,\exp(-\tfrac 1b\,\ln(a)\,(b\,x+c))
,\\
-\tfrac 1b\,\ln(a)\,a^{-c/b}
&=
-\tfrac 1b\,\ln(a)\,(b\,x+c)\,\exp(-\tfrac 1b\,\ln(a)\,(b\,x+c))
.
\end{align}
At this point we have the right-hand side in a proper form $y\e^y$
and can apply the Lambert $\W$ function:
\begin{align}
\W\Big(-\tfrac 1b\,\ln(a)\,a^{-c/b}\Big)
&=
\W\Big(-\tfrac 1b\,\ln(a)\,(b\,x+c)\,\exp(-\tfrac 1b\,\ln(a)\,(b\,x+c))\Big)
,\\
-\tfrac 1b\,\ln(a)\,(b\,x+c)
&=
\W\Big(-\tfrac 1b\,\ln(a)\,a^{-c/b}\Big)
,\\
x&=-\frac 1{\ln(a)}\,\W(-\tfrac 1b\,\ln(a)\,a^{-c/b})-\frac cb
.
\end{align}
Substitution of the values from \eqref{2} gives
\begin{align}
x&=-\frac 1{\ln 3}\,\W(-\tfrac 1{2187}\ln 3)-6
.
\end{align}
The argument of $\W$ is
\begin{align}
-\tfrac 1{2187}\ln 3
&\approx -0.0005
\end{align}
fits inside the interval $(-\tfrac1\e,0)$, where the Lambert $\W$ function has two real
branches, $\Wp$ and $\Wm$, so we must have two real solutions to \eqref{1}:
\begin{align}
x_0&=-\frac 1{\ln 3}\,\Wp(-\tfrac 1{2187}\ln 3)-6
\approx -0.5025901139
,\\
x_1&=-\frac 1{\ln 3}\,\Wm(-\tfrac 1{2187}\ln 3)-6
=3.
\end{align}
In the second case the value of $x_1=3$ is exact integer, since the argument of $\W$
can also be represented in a form $y\,\e^y$:
\begin{align}
-\tfrac 1{2187}\ln 3
&=
-\tfrac 1{3^7}\ln 3
=
-\tfrac 9{3^9}\ln 3
=
\tfrac 1{3^9}\,\ln(\tfrac 1{3^9})
\\
&=
\ln(\tfrac 1{3^9})\,\exp\Big(\ln(\tfrac 1{3^9})\Big)
=
-9\,\ln3\,\exp(-9\,\ln3)
.
\end{align}
The numeric value
\begin{align}
-9\,\ln3&\approx -9.88751
\end{align}
is less than $-1$, and belongs to the range of $\Wm$,
so in this case we have
\begin{align}
x_1&
=-\frac 1{\ln 3}\,\Wm(-9\,\ln3\,\exp(-9\,\ln3))-6
=-\frac 1{\ln 3}\,(-9\,\ln3)-6
=3.
\end{align}
$\endgroup$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove $a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$ for the recursive sequence $a_{n+1}=\frac{3a_n-1}{3-a_n}$
Prove the statement:
$$a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$$
for the given recursive sequence:
$$a_{n+1}=\frac{3a_n-1}{3-a_n}. $$
My attempt:
Proof by induction:
(1) Base: $\tau(1)$.
$$
\begin{align}
a_2 &=\frac{(2+1)a_1-2+1}{2+1-(2-1)a_1} \\
&=\frac{3a_1-1}{3-a_1}.
\end{align}
$$
(2) Assumption: Let
$$ a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1} $$
hold for some $n\in\mathbb N$.
(3) Step $\tau(n+1)$:
$$
\begin{align}
a_n &= \frac{3\cdot\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}-1}{3-\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}} \\
&= \frac{3\cdot2^{n-1}a_1+3a_1-3\cdot2^{n-1}+3-2^{n-1}-1+2^{n-1}a_1-a_1}{3\cdot 2^{n-1}+3-3\cdot 2^{n-1}a_1+3a_1-2^{n-1}a_1-a_1+2^{n-1}-1} \\
&= \frac{(4\cdot 2^{n-1}+2)a_1-4\cdot 2^{n-1}+2}{4\cdot 2^{n-1}+2-(4\cdot 2^{n-1}+2)a_1} \\
&= \frac{(2^n+1)a_1-2^n+1}{2^n+1-(2^n+1)a_1}
\end{align}
$$
Is this correct and is there a more efficient method than induction?
| Let
$$a_n=\frac{x_n}{y_n}$$
Then
$$\frac{x_{n+1}}{y_{n+1}}=\frac{3x_n-y_n}{3y_n-x_n}$$
Assume
\begin{align}
&x_{n+1}=3x_n-y_n&
&y_{n+1}=3y_n-x_n
\end{align}
Then $y_n=3x_n-x_{n+1}$ from which $3x_{n+1}-x_{n+2}=3(3x_n-x_{n+1})-x_n$ which gives
$$x_{n+2}-6x_{n+1}+8x_n=0$$
This linear recurrence has general solution
$$x_n=2^{n-1}(4x_0-x_1)+2^{2n-1}(x_1-2x_0)$$
By symmetry, we have
$$y_n=2^{n-1}(4y_0-y_1)+2^{2n-1}(y_1-2y_0)$$
hence
\begin{align}
a_n
&=\frac{2^{n-1}(4x_0-x_1)+2^{2n-1}(x_1-2x_0)}{2^{n-1}(4y_0-y_1)+2^{2n-1}(y_1-2y_0)}\\
&=\frac{(4x_0-x_1)+2^{n}(x_1-2x_0)}{(4y_0-y_1)+2^{n}(y_1-2y_0)}\\
&=\frac{(2^n-1)x_1-4x_0(1-2^{n-1})}{(2^n-1)y_1-4y_0(1-2^{n-1})}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
limit of $\lim_{x\to 7}(\frac{x}{7})^{(\frac{x^2-18x+80}{x-7})}$ $$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{x^2-18x+80}{x-7}\right)}$$
It is $1^{\infty}$
$$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}$$
I tried to take
$$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}}=\lim_{x\to 7}e^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)}$$
Now it is $e^{(0\cdot \infty)}$ which we can not conclude about the limit
| $\frac{(x-8)(x-10)}{x-7}\ \frac{\ln(1+\frac{x}{7}-1)}{\frac{x}{7}-1}\times (\frac{x}{7}-1)$
as $\frac{x}{7}\to 1\Rightarrow\displaystyle\lim_{x\rightarrow7} \frac{\ln(1+\frac{x}{7}-1)}{\frac{x}{7}-1}=1$
So we are left with $\frac{(x-8)(x-10)}{7(x-7)}\times(x-7)$ can you do from here.
If you have to find limit of $1^{\infty}$ which is $f(x)^{g(x)}$ , you can always remember it would $e^{g(x)\times(f(x)-1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to compute an upper bound of $1 + \frac{1}{2^a} + \frac{1}{3^a} + \cdots$? Suppose $a>2$.
How to compute an upper bound of $1 + \frac{1}{2^a} + \frac{1}{3^a} + \cdots$ ?
Is $\frac{1}{a-2}$ an upper bound?
| When a = 2 we can say
$1 + \frac 14 (1 + \frac {4}{9}) + \frac {1}{16}( 1 + \frac {16}{25} + \frac {16}{36} + \frac {16}{49}) + \cdots < 1 + \frac 14 (2) + \frac 1{16} (4) + \cdots = 2$
when $a> 1,$
$\sum_\limits{n=0}^{\infty} \frac {1}{n^a} < \sum_\limits{n=0}^{\infty} (\frac {1}{2^{a-1}})^n = \frac {2^{a-1}}{2^{a-1}-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Problem with computing $\int\frac{dx}{2x^2+5} $ by trigonometric substitution I so close to can solve this problem but I don't find the correct response:
$$\int\frac{dx}{2x^2+5}
$$
Always get the answer:
$$ \frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}}
$$
But the correct answer have more one square root multiplying the square root of $5:$
(this is the correct answer):
$$
\frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}\mathbf{\sqrt{2}}}
$$
I'm using the follow propertie to try solve the problem:
$$
\int{\frac{dx}{u^2+a^2}} = \frac{1}{a}\arctan{\frac{u}{a}}
$$
--My steps for the solution:
Before to apply the properties get square root from $~2x^2~$ and $5$, staying that way:
$$\int\frac{dx}{(x\sqrt{2})^2+(\sqrt{5})^2}
$$
So when I apply the previous properties get my wrong answer
| Setting $x\mapsto\ \sqrt{\frac{5}{2}}u$ yields:
$$\int_{ }^{ }\frac{dx}{2x^{2}+5}=\sqrt{\frac{1}{10}}\int_{ }^{ }\frac{1}{u^{2}+1}du=\sqrt{\frac{1}{10}}\arctan\left(u\right)+C=\sqrt{\frac{1}{10}}\arctan\left(\sqrt{\frac{2}{5}}x\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$ I need to solve the following problem:
$$\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$$
I tried with trigonometric substitution, but I couldn't solve it. The square root is messing me up. I searched it up and nothing appears to work well so I hope someone can help. Thanks.
| Integrate as follows\begin{align}
& \int{\frac{x}{(2x^2-3) \sqrt{3x^2-2x+1}}}\ dx\\
=& \ \frac12\int \left(\frac{1}{2 x+\sqrt{6}}+\frac{1}{2 x-\sqrt{6}}\right)\frac1{\sqrt{3
x^2-2 x+1} }\ dx\>\>\>\>\>\>\>t_{\pm}=\frac1{2x\pm \sqrt6}\\
=& -\frac14\int \frac {1}{\sqrt{3-2(2+3\sqrt6)t_+ +2(11+2\sqrt6)t_+^2}}\ dt_+ \\
&\>-\frac14\int \frac {1}{\sqrt{3-2(2-3\sqrt6)t_- +2(11-2\sqrt6)t_-^2}}\ dt_-\\\
=& \ - \frac{\sinh^{-1}\frac{2(11+2\sqrt6)t_+-(2+3\sqrt6)}{2\sqrt2}}{4\sqrt{2(11+2\sqrt6)}}- \frac{\sinh^{-1}\frac{2(11-2\sqrt6)t_+-(2-3\sqrt6)}{2\sqrt2}}{4\sqrt{2(11-2\sqrt6)}}+C\\
= & \ \frac{\sinh^{-1}\frac{(\sqrt2+3\sqrt3)x -(\sqrt2+\sqrt3)}{2x+\sqrt6}}{4\sqrt{2(11+2\sqrt6)}}+\frac{\sinh^{-1}\frac{(\sqrt2-3\sqrt3)x-(\sqrt2-\sqrt3)}{2x-\sqrt6}}{4\sqrt{2(11-2\sqrt6)}}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Finding supremum & infimum of a given set
If exist, find, $\sup S,\inf S$\begin{equation}
S:=\left\{\left\lfloor\sqrt[|n|]{\frac{7}{3}}\right\rfloor\cdot
\frac{n^{2}-2 n-4}{n^{2}-n-6}: n \in
\mathbb{Z}\setminus\{-2,0,3\}\right\} \end{equation}
My attempt:
\begin{equation}f(n):=\left\lfloor\sqrt[|n|]{\frac{7}{3}}\right\rfloor\cdot \frac{n^{2}-2 n-4}{n^{2}-n-6}\end{equation}
$n=1, f(1)=\left\lfloor\frac{7}{3}\right\rfloor\cdot\frac{5}{6}=\frac{5}{3}$
$n=-1,f(-1)=\left\lfloor\frac{7}{3}\right\rfloor\cdot\frac{1}{4}=\frac{1}{2}$
$n=2,f(2)=\left\lfloor\sqrt{\frac{7}{3}}\right\rfloor\cdot\frac{1}{2}=1$
\begin{equation}\lim_{n\to -2}f(n)=\pm\infty=\lim_{{n\to 3}}f(n)\end{equation}
\begin{equation}\lim_{n\to\infty}f(n)=\left\lfloor\left(\frac{7}{3}\right)^{\left|\frac{1}{n}\right|}\right\rfloor\cdot\frac{1-\frac{2}{n}-\frac{4}{n^2}}{1-\frac{1}{n}-\frac{6}{n^2}}=1\end{equation}
$$\implies\inf S=\frac{1}{2},\;\sup S=\frac{5}{3}$$
Is this correct?
| The points $n=0$, $n=-2$, and $n=3$ are excluded by definition. When $|n|\geq4$ the integer part of the square root is $1$ and the quotient is between $2/3 $ and $3/2$.
Besides the values you calculated, you have $f(-3)=11/6$. So the supremum is actually a maximum, and it is equals $11/6$. The infimum is a minimum and it equals $1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
2018 MathCounts: Let $D(k)$ be the number of diagonals for a polygon with $k$ sides. If $D(m) + D(n) = 125$, what is the value of $m+n$? The question was asked in the 2018 Raytheon MATHCOUNTS National Competition. It appears in the video around the 35:34 mark:
https://www.youtube.com/watch?v=dSnOLW_W6og&t=2134
Let $D(k)$ be the number of diagonals for a polygon with $k$ sides. If
$D(m) + D(n) = 125$, what is the value of $m+n$?
The formula for the number of diagonals is well-known and easily derived as $k(k-3)/2$.
I then computed several values, and found a suitable pair of numbers to solve the problem.
But is there any easier way to solve this equation:
$$\frac{n(n-3)}{2} + \frac{m(m-3)}{2} = 125$$
$$n,m \in {3, 4, ...}$$
The students were supposed to solve this in minutes without a calculator.
My guess is there is not a simpler way--in MathCounts the students may have just memorized many values.
| Let $m=x-n$. Then,
$$\frac{n(n-3)}{2}+\frac{m(m-3)}{2}=125$$
is equivalent to
$$x^2-(2n+3)x+2n^2- 250 = 0$$
which implies
$$x=\frac{2n+3\pm\sqrt{1018-(2n-3)^2}}{2}$$
So, there has to be an integer $k$ such that $$1018=(2n-3)^2+k^2$$
Here, both $2n-3$ and $k$ are odd.
Since the rightmost digit of $(\text{odd})^2$ is either $1,9$ or $5$, both the rightmost digit of $(2n-3)^2$ and that of $k^2$ are $9$.
So, we only need to consider
$$3^2=9, 7^2=49, 13^2=169, 17^2=289, 23^2=529, 27^2=729$$
So, we see that
$$27^2+17^2=1018$$
is the only possible sum.
So, $2n-3=27$ implies $n=15$, and $2n-3=17$ implies $n=10$.
Therefore, $$\color{red}{m+n=x=25}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given positives $x, y , z$ such that $x + y + z = xyz$. Calculate the minimum value of $\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$.
Given positives $x, y , z > 1$ such that $x + y + z = xyz$. Calculate the minimum value of $$\large \frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}$$
We have that $x + y + z = xyz \implies \dfrac{1}{yz} + \dfrac{1}{zx} + \dfrac{1}{xy} = 1$ and $$\left(\frac{x - 1}{y^2} + \frac{y - 1}{z^2} + \frac{z - 1}{x^2}\right) \cdot \left(\frac{1}{x - 1} + \frac{1}{y - 1} + \frac{1}{z - 1}\right)$$
$$ \ge \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)^2 \ge 3 \cdot \left(\frac{1}{yz} + \frac{1}{zx} + \frac{1}{xy}\right) = 3$$
Now we need to find the maximum value of $\dfrac{1}{x - 1} + \dfrac{1}{y - 1} + \dfrac{1}{z - 1}$.
Let $\dfrac{1}{x} = a, \dfrac{1}{y} = b, \dfrac{1}{z} = c$ which implies that $a, b, c \in (0, 1)$.
It could be observed that $ab + bc + ca = 1$ and $$\dfrac{1}{x - 1} + \dfrac{1}{y - 1} + \dfrac{1}{z - 1} = \frac{a}{1 - a} + \frac{b}{1 - b} + \frac{c}{1 - c}$$
$$ = 3 - \left(\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}\right) \le 3 - \frac{9}{3 - (a + b + c)} = \frac{-3(a + b + c)}{3 - (a + b + c)}$$
Then I'm stuck.
| First of all, as you observed, the condition can be written as
$$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1$$
Let's call the expression $S$. Write $S$ as:
$$\sum_{cyc}(x-1)\left(\frac{1}{y^2}+\frac{1}{x^2}\right) - \frac{1}{x}-\frac{1}{y}-\frac{1}{z}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
Since $x,y,z > 1$, we can apply AM-GM as:
$$\frac{1}{y^2}+\frac{1}{x^2} \geq \frac{2}{xy}$$
to get that:
$$S \geq (x-1)\cdot \frac{2}{xy}+(y-1)\cdot \frac{2}{yz}+(z-1)\cdot \frac{2}{zx} - \frac{1}{x}-\frac{1}{y}-\frac{1}{z}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
$$=2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right) - \frac{1}{x}-\frac{1}{y}-\frac{1}{z}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=$$
$$=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-2+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
Now, just use the trivial inequalities $a^2+b^2+c^2 \geq ab+bc+ca$ and $(a+b+c)^2\geq 3(ab+bc+ca)$ to get
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \sqrt{3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)} = \sqrt{3}$$
and
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1$$
This gives the final result $S\geq \sqrt{3}-1$, attained when $x=y=z=\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Three-variable generating series for strings I need to find the generating series of strings from $C = \{a, b, c\}$ without including the strings $aac$ and $baa$. Also, all $b$ substrings must have an odd length (so $bbb, cbc, aaab$ are all valid but $bbcb$ is not) and the $a$ substrings must have a minimum length of 3 (so $aaab, aaaac$ are both valid but $acaca$ isn't).
I need to find a three-variable generating series for it, $f(x, y, z) = \sum_{\delta \in C}x^{l(a)}y^{l(b)}z^{l(c)}$ where $l(a)$ is the number of $a$'s in the string, etc.
I'm not sure how to do this; I especially don't know how to incorporate the fact that there are forbidden strings. At first I was thinking I'd use a tactic similar to how one would find a generating series for compositions so that $a$ comes from the set $\{3, 4, 5, ...\}$ with generating series $\frac{x^2}{1-x}$, $b$ from $\{1, 3, 5, ...\}$ with generating series $\frac{x}{1-x^2}$ and $c$ from $\{1, 2, 3, ...\}$ so generating series $\frac{x}{1-x}$ except this doesn't really take into account the forbidden substrings/the fact that based on how the strings are combined, the number of a given letter varies, ie $bcb$ is allowed since the length of the b blocks are odd but their total count is 2 which is even.
If anyone could help me out with how to solve/approach this problem I'd really appreciate it!
| Here's a derivation of the one-variable generating function $f(z,z,z)$. Maybe it will help you.
Let $a_n$, $b_n$, and $c_n$ be the number of strings that start with $a$, $b$, and $c$, respectively. Then $a_0=b_0=c_0=1$, $a_1=a_2=0$, and, by conditioning on the length $k$ of the current run, we see that
\begin{align}
a_n &= \sum_{k=3}^n b_{n-k} &&\text{for $n \ge 3$}\\
b_n &= \sum_{\substack{k=1\\\text{$k$ odd}}}^n c_{n-k} &&\text{for $n \ge 1$}\\
c_n &= \sum_{k=1}^n (a_{n-k}+b_{n-k}-[k=n]) &&\text{for $n \ge 1$}
\end{align}
Let $A(z)=\sum_{n=0}^\infty a_n z^n$, $B(z)=\sum_{n=0}^\infty b_n z^n$, and $C(z)=\sum_{n=0}^\infty c_n z^n$. Then the recurrence relations imply
\begin{align}
A(z)-1 &=\frac{z^3}{1-z} B(z) \\
B(z)-1 &=\frac{z}{1-z^2} C(z) \\
C(z)-1 &=\frac{z}{1-z} (A(z)+B(z)-1)
\end{align}
Solving for $A(z)$, $B(z)$, and $C(z)$ yields
\begin{align}
A(z) &= \frac{1-2z-z^2+4z^3-z^4-3z^5+z^6}{1-2z-z^2+3z^3-z^4-z^5}\\
B(z) &= \frac{1-z-2z^2+3z^3-z^4}{1-2z-z^2+3z^3-z^4-z^5}\\
C(z) &= \frac{1-z-z^2+z^3+z^4-z^6}{1-2z-z^2+3z^3-z^4-z^5}
\end{align}
So the desired generating function (subtracting twice the $1z^0=1$ term for the empty string that is otherwise counted three times) is
$$f(z,z,z)=A(z)+B(z)+C(z)-2=\frac{1-2z^2+2z^3+z^4-z^5}{1-2z-z^2+3z^3-z^4-z^5}.$$
Update: Here's the general solution. Let $f_a(x,y,z)$, $f_b(x,y,z)$, and $f_c(x,y,z)$ be the generating functions for the sequences that start with $a$, $b$, and $c$, respectively. Then
\begin{align}
f_a &= \frac{x^3}{1-x}(f_b+1)\\
f_b &= \frac{y}{1-y^2}(f_c+1)\\
f_c &= \frac{z}{1-z}(f_a+f_b+1)
\end{align}
Solving these equations yields
\begin{align}
f_a &= \frac{-x^3 (y^2 - y - 1) (z - 1)}{x^3 y z + x y^2 z - x y^2 - x y z - x z + x - y^2 z + y^2 + y z + z - 1}\\
f_b &= \frac{-x^3 y z + x y - y}{x^3 y z + x y^2 z - x y^2 - x y z - x z + x - y^2 z + y^2 + y z + z - 1}\\
f_c &= \frac{(x^3 - x + 1) (y^2 - y - 1) z}{x^3 y z + x y^2 z - x y^2 - x y z - x z + x - y^2 z + y^2 + y z + z - 1}
\end{align}
Now (including 1 for the empty string), we have $$f = f_a + f_b + f_c + 1=\frac{(x^3 - x + 1) (y^2 - y - 1)}{x^3 y z + x y^2 z - x y^2 - x y z - x z + x - y^2 z + y^2 + y z + z - 1}.$$
As a sanity check, substituting $z$ in place of $x$ and $y$ does yield the one-variable generating function obtained earlier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3542005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Distinct Spanning Trees Say that you are asked to produce a labelled graph $G$ that has $E$ edges and exactly $k$ distinct spanning trees. In general, how much do we know about whether such a graph exists? For example, a labelled graph with $6$ edges and exactly $7$ distinct spanning trees seems to be impossible to create. Beyond brute force/exhaustion, what tools do we have at our disposal to answer questions of this nature?
| Take for example the graph $G=(V,E)$ where $V = \{a,b,c,d\}$ and $E = \{ab, bc, bd, cd\}$. Then the adjacency matrix is
$$
A = \left(
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 \\
\end{array}
\right)
$$
and the degree matrix
$$
D = \left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 3 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2 \\
\end{array}
\right).
$$
Hence the Laplacian matrix is
$$
D - A = \left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 3 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 2 \\
\end{array}
\right) - \left(
\begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 \\
0 & 1 & 1 & 0 \\
\end{array}
\right) = \left(
\begin{array}{cccc}
1 & -1 & 0 & 0 \\
-1 & 3 & -1 & -1 \\
0 & -1 & 2 & -1 \\
0 & -1 & -1 & 2 \\
\end{array}
\right).
$$
This matrix has eigenvalues $\lambda=4,3,1,0$ and hence by Kirchhoff's theorem it has $\frac14(4\cdot3\cdot 1) = 3$ spanning trees. We can verify this by nothing that a spanning tree of $G$ has as edges $ab$ and exactly two of $bc, bd$, and $cd$, for a total of $\binom 32=3$ spanning trees.
More generally, note that for any tree, $|V|= |E|+1$. Cayley's formula tells us that there are $(|E|+1)^{|E|-1}$ trees on graphs with $|E|$ edges. So the first thing to check would be whether $k>(|E|+1)^{|E|-1}$, in which case no such graph exists. If this inequality does not hold, however, then there is more work to be done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Which of these answers is the correct indefinite integral? (Using trig-substitution or $u$-substitution give different answers) Answers obtained from two online integral calculators:
$$\begin{align}\int\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}\,\mathrm dx &= -\sqrt{\dfrac{x + 1}{1 - x}} + \sqrt{\dfrac{x + 1}{1 - x}}x - 2\arcsin\left(\dfrac1{\sqrt2}\sqrt{1 - x}\right) + C \\
\int\dfrac{\sqrt{1 + x}}{\sqrt{1 - x}}\,\mathrm dx &= 2\sin^{-1}\left(\dfrac{\sqrt{x + 1}}{\sqrt2}\right) - \sqrt{1 - x^2} + \text{ constant}
\end{align}$$
Answers from online calculator shown above, and my answers shown in the link:
Update:
I realized that the substitution for $\theta$ was supposed to be $\arcsin$ not $\arccos$, so the answer would have been the same as the right hand side.
But I also noticed that using the initial substitution to plug $x$ back in the final answer will not always give the correct answer because in a similar question:
$$\int \frac{\sqrt{x^2-1}}x dx$$
has a trig-substitution of $x = \sec\theta$, and the answer in terms of $\theta$ would be: $\tan \theta - \theta + C$. Then the final answer in terms of $x$ should be : $\sqrt{x^2-1} - \operatorname{arcsec}(x) + C$.
But online integral calculators give the answer: $\sqrt{x^2-1} - \arctan(\sqrt{x^2-1}) + C$, which doesn't match the original substitution of:
$$x = \sec\theta \to \theta = \operatorname{arcsec}(x)$$
Anyone know why the calculator gives that answer which doesn't match the original trig-substitution of $x = \sec \theta \to \theta = \operatorname{arcsec}(x)$?
| Now, both answers are correct. They merely look different. They differ by a constant.
Note 1...
$$
-\sqrt{\frac{x+1}{1-x}}+\sqrt{\frac{x+1}{1-x}}\;x
=
-\sqrt{\frac{x+1}{1-x}}\;(1-x)
=
-\frac{\sqrt{x+1}\;(1-x)}{\sqrt{1-x}}
\\=
-\sqrt{x+1}\sqrt{1-x}
=-\sqrt{(1+x)(1-x)}
=-\sqrt{1-x^2}
$$
Note 2...
$$
2\,\arcsin \left( \frac{\sqrt {1+x}}{\sqrt {2}} \right) =\pi-2\,\arcsin
\left( \frac{\sqrt {1-x}}{\sqrt {2}} \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
On the alternating quadratic Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$ My question is:
Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$.
$$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$$
What I have managed to do so far is to convert $S$ to two rather difficult integrals as follows.
Starting with the result
$$\frac{H_{2n}}{2n} = -\int_0^1 x^{2n - 1} \ln (1 - x) \, dx \tag1$$
Multiplying (1) by $(-1)^n H_n/n$ then summing the result from $n = 1$ to $\infty$ gives
$$S = -2 \int_0^1 \frac{\ln (1 - x)}{x} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n}. \tag2$$
From the following generating function for the harmonic numbers
$$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \frac{1}{2} \ln^2 (1 - x) + \operatorname{Li}_2 (x),$$
replacing $x$ with $-x^2$ leads to
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n} = \frac{1}{2} \ln^2 (1 + x^2) + \operatorname{Li}_2 (-x^2).$$
Substituting this result into (2) yields
$$S = -2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx,$$
or, after integrating the first of the integrals by parts twice
$$S = -\frac{5}{2} \zeta (4) + 4 \zeta (3) \ln 2 - 8 \int_0^1 \frac{x \operatorname{Li}_3 (x)}{1 + x^2} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx. \tag3$$
I have a slim hope the first of these integrals can be found (I cannot find it). As for the second of the integrals, it is proving to be a little difficult.
Can someone find each of the integrals appearing in (3)? Or perhaps an alternative approach to the sum will deliver the closed-form I seek, I am fine either way.
Update
Thanks to Ali Shather, the first of the integrals can be found. Here
\begin{align}
\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \ dx &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{2n-1}\ln(1-x)\ dx\\
&= -\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n^3}\\
&=-4\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}\\
&=-4 \operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3}.
\end{align}
And using the result I calculated here, namely
$$\operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3} = \frac{5}{8} \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{195}{256} \zeta (4) + \frac{5}{192} \ln^4 2 - \frac{5}{32} \zeta (2) \ln^2 2 + \frac{35}{64} \zeta (3) \ln 2,$$
gives
\begin{align}
\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx &= -\frac{5}{2} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{195}{64} \zeta (4) - \frac{5}{48} \ln^4 2\\
& \qquad + \frac{5}{8} \zeta (2) \ln^2 2 - \frac{35}{16} \zeta (3) \ln 2.
\end{align}
| A Third (Magical) Solution by Cornel Ioan Valean
This time we assume we have in hand that $\displaystyle \int_0^1 \frac{\arctan^2(x)}{x}\log\left(\frac{x}{(1-x)^2}\right)=G^2$, which is calculated extremely easily at this link. Furthermore, we also assume we have at our disposal the value of the powerful harmonic series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{2n}}{n^3}$ which is derived in a simple manner in this answer. Now, as seen in this post, the integral is reducible to the previously mentioned harmonic series and $\color{red}{\text{the main harmonic series}}$.
Using the simple facts stated above, we conclude that
$$\color{red}{\sum _{n=1}^{\infty} (-1)^{n-1}\frac{ H_n H_{2 n}}{n^2}}$$
$$\color{red}{=2 G^2-2\log(2)\pi G-\frac{1}{8}\log^4(2)-\frac{21}{8}\log(2)\zeta(3)+\frac{1}{4}\log^2(2)\pi ^2+\frac{773}{5760}\pi ^4}$$
$$\color{red}{-4 \pi \Im\biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}-3 \operatorname{Li}_4\left(\frac{1}{2}\right)},$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Prove equilateral triangle if vertices $a,b,c$ satisfy $|a|=|b|=|c|$ and $\sum_{cyc}\frac1{8-a/b-b/a-a/c-c/a}=0.3$
If $a,b,c\in \mathbb{C}$ are different complex numbers in pairs and $|a|=|b|=|c|=1$ and $P(a),Q(b),R(c)$ are the vertices of a triangle and $\displaystyle \sum_{\bf{cyc}}\frac{1}{8-\frac{a}{b}-\frac{b}{a}-\frac{a}{c}-\frac{c}{a}}=0.3.$ Then show that $PQ=QR=RP.$
What I tried
Let $a=e^{i\alpha}=\cos \alpha+i\sin \alpha$
and $b=e^{i\beta}=\cos \beta+i\sin \beta$
and $c=e^{i\gamma}=\cos \gamma+i\sin \gamma$
Now $\displaystyle \frac{a}{b}+\frac{b}{a}=e^{i(\alpha-\beta)}+e^{-i(\alpha-\beta)}=2\cos (\alpha-\beta)$
and $\displaystyle \frac{a}{c}+\frac{c}{a}=e^{-i(\gamma-\alpha)}+e^{i(\gamma-\alpha)}=2\cos(\gamma-\alpha)$
$\displaystyle \Longrightarrow \sum_{\bf{cyc}}\frac{1}{8-2\cos(\alpha-\beta)-2\cos(\gamma-\alpha)}=0.3$
$\displaystyle \Longrightarrow \sum_{\bf{cyc}}\frac{1}{4-\cos(\alpha-\beta)-\cos(\gamma-\alpha)}=0.6$
Let $\alpha-\beta=A,\beta-\gamma=B,\gamma-\alpha=C$
$\displaystyle \Longrightarrow \sum_{\bf{cyc}}\frac{1}{4-\cos A-\cos C}=0.6$
How do I solve it after that
Please help.
| Apply the Jensen's inequality
$$f(x_1) + f(x_2) + f(x_3) \ge 3f\left(\frac{x_1+x_2+x_3}3\right)$$
for the convex function $f(x)=\frac1x$ to get
$$0.3=\sum_{{cyc}}\frac{1}{8-\frac{a}{b}-\frac{b}{a}-\frac{a}{c}-\frac{c}{a}}
\ge \frac{1}{24-2\left(\frac{a}{b}+\frac{b}{a}+\frac bc+\frac cb+\frac{c}{a}+\frac{a}{c}\right)}
$$
or,
$$\frac{a}{b}+\frac{b}{a}+\frac bc+\frac cb+\frac{c}{a}+\frac{a}{c}\le -3\tag 1$$
Given that $|a|=|b|=|c|$, assume, without loss of generality $\alpha, \ \beta, \ \gamma >0$,
$$\frac ba =e^{i\alpha},\>\>\>\>\>\frac cb =e^{i\beta},\>\>\>\>\>\frac ac=e^{i\gamma}
\>\>\>\>\>\alpha + \beta + \gamma = 2\pi$$
Then, the inequality (1) becomes,
$$\cos\alpha + \cos\beta + \cos \gamma \le -\frac32\tag 2$$
Note that,
$$\cos\alpha + \cos\beta + \cos \gamma
= 2\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2 + 2\cos^2\frac\gamma2-1$$
$$= 2\cos^2\frac\gamma2 - 2\cos\frac{\gamma}2\cos\frac{\alpha-\beta}2 -1
\ge 2\cos^2\frac\gamma2 - 2\cos\frac{\gamma}2 -1 $$
$$= 2\left(\cos\frac\gamma2 - \frac12\right)^2 -\frac32 \ge-\frac32\tag 3$$
From (2) and (3), we have
$$\cos\alpha + \cos\beta + \cos \gamma=-\frac32$$
and, according to (3), the equality occurs at $\cos\frac\gamma2 - \frac12=0$ and $\cos\frac{\alpha-\beta}2 =1$, which leads to $\alpha=\beta =\gamma = 120^\circ$. Thus, the neighboring vertexes of equal module all have the same argument angle between them, hence, forming an equilateral triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding all angles that satisfy $8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
$\text { Hence, solve the equation } 8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
The previous part was to prove that $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \quad \text { by replacing } 3 \theta \text { by }(2 \theta+\theta)$.
So, I used this to simplify the equation to
$2 \cos 3 \theta +1 = 0$
$\implies \cos 3 \theta =\frac{-1}{2}$
Since $\cos^{-1} \frac{-1}{2} = \frac{2\pi}{3} + 2 \pi n$,$\implies \theta =\frac{2 \pi}{9},\frac{8 \pi}{9},\frac{-8 \pi}{9}$ or $\frac{-2 \pi}{9}$. However, the graph seems to be showing another root which is $\frac{4 \pi}{9}$. Why did I miss this root? How should I find more angles that satisfy the equation in a given range. In general, how does one find all angles that satisfy an equation even after adding $2 \pi n$
| Note that, if $\cos x >0$, $x$ is in the 1st or 4th quadrant; if $\cos x< 0$, $x$ is in the 2nd or 3rd quadrant.
So, given $3\theta \in [-3\pi, 3\pi]$ and $\cos 3\theta = -\frac12< 0$, the angle $3\theta$ has two sets of values, with one set in the 2nd quadrant, i.e.
$$3\theta =\frac{2\pi}3 + 2\pi k = -\frac{4\pi}3 , \>\frac{2\pi}3, \> \frac{8\pi}3$$
and the other set in the 3rd quadrant, i.e.
$$3\theta = -\frac{2\pi}3 + 2\pi k =-\frac{8\pi}3,\> -\frac{2\pi}3, \>\frac{4\pi}3$$
Thus, all eligible angles are $ \pm\frac{2\pi}9, \> \pm\frac{4\pi}9, \> \pm\frac{8\pi}9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving $\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}=\frac{4}{n}\sum _{k=1}^n \frac{1}{2 k-1}$ How can one prove
$$
\sum_{k = 1}^{n}\frac{\left(-1\right)^{k - 1}\, 16^{k}\,
\left(k - 1\right)!\, k!\, \left(k + n - 1\right)!}
{\left[\left(2k\right)!\right]^{\, 2}\,\left(n - k\right)!} =
\frac{4}{n}\sum_{k = 1}^{n}\frac{1}{2k - 1}
$$
I was given this without proof, but only a hint (to evaluate $\int_{0}^{\pi/2}\frac{2}{n}\,\frac{1 - \cos\left(2nx\right)}{\sin\left(x\right)} \, dx$ in $2$ ways) instead. By induction the integral is easily seen equivalent to RHS, but I wonder how on earth it's related to LHS. Any help will be appreciated.
| Remarking that
\begin{equation}
\sum_{k=1}^n\sin\left( \left( 2k-1 \right)x \right)=\frac{\sin^2nx}{\sin x}
\end{equation}
the proposed integral
\begin{align}
I_n&= \frac{2}{n}\int_0^{\frac{\pi }{2}}\frac{1-\cos (2 n x)}{\sin (x)} \, dx\\
&= \frac{4}{n}\int_0^{\frac{\pi }{2}}\frac{\sin^2nx}{\sin x},dx\\
&=\frac{4}{n}\sum_{k=1}^n\int_0^{\frac{\pi }{2}}\sin\left( \left( 2k-1 \right)x \right)\,dx\\
&=\frac{4}{n}\sum_{k=1}^n\frac{1}{2k-1}\\
&=\text{rhs}
\end{align}
which shows that the integral is equal to the rhs of the identity.
This decomposition suggests the use of the Chebyshev polynomials to evaluate the lhs,
\begin{equation}
\text{lhs}=\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}
\end{equation}
Indeed, the Chebyshev polynomials of the first kind reads
\begin{equation}
T_n(z)=n\sum_{k=0}^n(-2)^k\frac{(k+n-1)!}{(n-k)!(2k)!}(1-z)^k
\end{equation}
and thus
\begin{equation}
\sum_{k=1}^n(-1)^{k-1}\frac{(n+k-1)!}{(n-k)!(2k)!}\left( 2(1-z) \right)^k=\frac{1}{n}\left( 1-T_n(z) \right)
\end{equation}
and with $Z=2(1-z)$,
\begin{equation}
\sum_{k=1}^n(-1)^{k-1}\frac{(n+k-1)!}{(n-k)!(2k)!}Z ^k=\frac{1}{n}\left[ 1-T_n(1-\frac{Z}{2}) \right]
\end{equation}
This summation is very similar to the proposed one. To introduce the missing factor $\frac{(k-1)!k!}{(2k)!}=\mathrm{B}(k,k+1)$ (here, $\mathrm{B}(k,k+1)$ is the Beta function), we use the integral representation:
\begin{equation}
\int_{0}^{\pi/2}{\sin^{2a-1}}\theta{\cos^{2b-1}}\theta\mathrm{d}\theta=\tfrac{1}{2}\mathrm{B}\left(a,b\right)
\end{equation}
with $a=k,b=k+1$, to express
\begin{align}
\mathrm{B}(k,k+1)&=2\int_{0}^{\pi/2}{\sin^{2k-1}}\theta{\cos^{2k+1}}\theta\,d\theta\\
&=2^{1-2k}\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta}\sin^{2k}2\theta\,d\theta
\end{align}
Thus
\begin{align}
\text{lhs}&=\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k+n-1)!}{(2 k)! (n-k)!}\mathrm{B}(k,k+1)\\
&=2\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta}\,d\theta\sum _{k=1}^n \frac{(-1)^{k-1} (k+n-1)!}{(2 k)! (n-k)!}16^k2^{-2k}\sin^{2k}2\theta\\
&=\frac{2}{n}\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta}
\left[ 1-T_n(1-2\sin^22\theta) \right]\,d\theta\\
&=\frac{2}{n}\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta}
\left[ 1-T_n(\cos4\theta) \right]\,d\theta
\end{align}
But $T_n(\cos4\theta)=\cos 4n\theta$ and $1-\cos 4n\theta=2\sin^22n\theta$. We obtained then
\begin{equation} \text{lhs}=\frac{4}{n}\int_{0}^{\pi/2}\frac{\cos\theta}{\sin\theta}\sin^22n\theta\,d\theta\ \end{equation}
By changing $\theta=u/2$ in the above integral and using simple trigonometric manipulations,
\begin{align}
\text{lhs}&=\frac{2}{n}\int_{0}^{\pi}\frac{\cos\frac{u}{2}}{\sin\frac{u}{2}}\sin^2nu\,du\\
&=\frac{4}{n}\int_{0}^{\pi}\frac{\cos^2\frac{u}{2}}{\sin u}\sin^2nu\,du\\
&=\frac{2}{n}\int_{0}^{\pi}\frac{\sin^2nu}{\sin u}\left( 1+ \cos u\right)\,du\\
&=\frac{2}{n}\int_{0}^{\pi}\frac{\sin^2nu}{\sin u}\,du+\frac{2}{n}\int_{0}^{\pi}\frac{\sin^2nu}{\sin u} \cos u\,du
\end{align}
By symmetry, the second integral vanishes and using symmetry for the first one,
\begin{align}
\text{lhs}&=\frac{4}{n}\int_{0}^{\pi/2}\frac{\sin^2nu}{\sin u}\,du\\
&=I_n
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Solving $\frac{5}{8} \cot36^\circ = \cos^3x$ without substituting the trig values for $36^\circ$
Find the value of $x$ such that
$$\frac{5}{8} \cot36^\circ = \cos^3x$$
The answer is $x=18^\circ$.
It's really messy to plug in the standard values of $\cos36^\circ$, $\sin36^\circ$ and miraculously guess a suitable value of $x$ and prove that our guess is correct.
Is there any nice way to find the value of $x$?
| Yet an other way to proceed, using algebraic only manipulations after a quick reshape...
We want to show
$$
\frac 58\cdot \frac{\cos 36^\circ}{\sin 36^\circ}
\overset{!}=\sin^3 72^\circ \ .
$$
Let us denote by $c,s$ respectively the values for $\cos 36^\circ$ and $\sin 36^\circ$. Then we want:
$$
\frac 58\cdot\frac cs
\overset{!}=
8s^3c^3\ ,\text{ or equivalently }
5 - 64s^4c^2=0\ .
$$
We start with $(s+ic)^5=\cos (5\cdot 36^\circ)+i\sin(5\cdot 36^\circ)=-1$, and consider only the imaginary part, thus getting
$$
\begin{aligned}
0
&= 5sc^4-10s^3c^2+s^5\ ,\text{ and since $s\ne 0$ we get}\\
0 &= 5c^4-10s^2c^2+s^4\\
&=5(1-2s^2+s^4)-10s^2(1-s^2)+s^4\\
&=5-20s^2+16s^4\ .
\qquad\text{From here:}
\\[3mm]
5-64s^4c^2
&=5 - 64s^4(1-s^2)
\\&=
5-64s^4+64s^6\\
&=\underbrace{(5-20s^2+16s^4)}_{=0}(1+4s^2)=0\ .
\end{aligned}
$$
$\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What does this series converge to? $\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$ What does the following expression converge to?
$$\sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)$$
(It looks like this problem)
$\displaystyle\sum_{k=0}^n a_{2k+1}+a_{2k+2}-a_{k+1}$ ;
$\displaystyle\sum_{k=0}^n a_{k+1}$ does not converge.
| $$\mathop {\lim }\limits_{n \to + \infty } \sum_{k=0}^n \left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+1}\right)=$$
$$=\mathop {\lim }\limits_{n \to + \infty } \sum_{k=0}^n\left(\frac{1}{4k+1}-\frac{1}{4k+2}\right)-\mathop {\lim }\limits_{n \to + \infty } \sum_{k=0}^n\left(\frac{1}{4k+2}-\frac{1}{4k+3}\right)=$$
$$=\frac{\pi}{8}+\frac{\ln2}{4}-\left(\frac{\pi}{8}-\frac{\ln2}{4}\right)=\frac{\ln2}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Can a Sum of distinct squares ever equal power of two? Does there exist $2^t,\ t\in\mathbb{Z}_+$ which can be express as Sum of two or more distinct square number.
Or
Can it be shown that
$$\begin{split}2^t &\ne \sum a_i^2 = a_1^2+ a_2^2+\cdots+a_n^2\end{split}$$
Where $n\ge 2$ and $\{a_i,t\}\in\mathbb{Z}_+$ and $a_i \ne a_j$ for $1\le i,j \le n$
Example: $2^6=64=7^2+3^2+2^2+1^2+1^2$ here $1^2$ repeat two times so this is not allowed.
My incomplete attempt for arithmetic squares
Edit: check related new post, Can a sum arithmetic square ever equal to power of two?
Let $n,u,d\in\mathbb{Z}_+$
$$\begin{split}\sum_{q=0}^u (n+qd)^2 &=n^2+(n+d)^2+(n+2d)^2+\cdots+(n+ud)^2\\ &=n^2(u+1)+\frac{(u+1)u}{2}(2nd+d^2)+\frac{(u+1)u(u-1)}{3}d^2 \end{split}$$
Let
$$\begin{split}2^t &=\sum_{q=0}^u (n+qd)^2 \\ \implies 3\cdot 2^{t+1}&=6n^2(u+1)+3(u+1)u(2nd-d^2)+(u+1)u(u-1)2d^2 \\ &= (u+1)(6n^2+3u(2nd+d^2)+u(u-1)2d^2)\\ &(in\ case,\ u+1= 3) \\
\implies 2^t&= 3n^2+3(2nd+d^2)+2d^2\\ &= n^2+(n+1)^2+(n+2d)^2 \end{split}$$
Now we need to simplify for case, $6n^2+3u(2nd+d^2)+u(u-1)2d^2=3\cdot2^x$ and $u+1=2^y$ where $x+y=t+1$ but I'm stuck here. Thank you.
Related post:
Can a sum of consecutive $n$th powers ever equal a power of two?
| Counterexample: For $t=8, n = 5$ and $(a_1,a_2,a_3,a_4,a_5)=(1,5,7,9,10)$ we have:
$$2^8=1^2+5^2+7^2+9^2+10^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
For $n>1,$ prove that, for $\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$ For $n>1$ prove that for $$\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$$
I know that $\dfrac{1}{x}>\dfrac{1}{(x+1)}$ and I'm trying to break the summation into smaller sums to work with, but I'm just not making that final bridge to the $\dfrac{5}{4n}$.
| Since
$$
\frac{1}
{{k^2 }} < \int\limits_{k - 1}^k {\frac{1}
{{x^2 }}} dx\,\,\,\,\,\,\forall k \geqslant 2
$$
we have that
$$
\sum\limits_{k = n}^{3n - 1} {\frac{1}
{{k^2 }} < } \int\limits_{n - 1}^{3n - 1} {\frac{1}
{{x^2 }}} dx = \left[ { - \frac{1}
{x}} \right]_{n - 1}^{3n - 1} = \frac{1}
{{n - 1}} - \frac{1}
{{3n - 1}}
$$
We have that
$$
\frac{1}
{{n - 1}} - \frac{1}
{{3n - 1}} < \frac{5}
{{4n}}
$$
is true if and only if
$$
\frac{{2n}}
{{\left( {n - 1} \right)3n - 1}} < \frac{5}
{{4n}}
$$
which is true if and only if
$$7n^2-20n+5>0$$. This is true for every $n \geq 3$. On the other side, a direct computation shows that the inequality is true for $n=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq\frac{3n-2}{2n-1}$. Let $$x_n=\frac{1}{1^2}+\frac{1}{2^2}+\ldots+\frac{1}{n^2},\;\forall n\geq 1.$$ Show that $$\frac{1}{x_1^2}+\frac{1}{4\cdot x_2^2}+\ldots+\frac{1}{n^2\cdot x_n^2}\leq \frac{3n-2}{2n-1}.$$
I tried to prove it by induction but I got a contradiction.
| It is trivial if you use $$\sum_{n=1}^\infty\dfrac{1}{n^2} = \dfrac{\pi^2}{6}$$
Note that $x_1=1$ and $x_i\ge\dfrac{5}{4}\forall i\ge 2$. Hence, for $n\ge 4$,
$$LHS = \sum_{i=1}^n\dfrac{1}{i^2x_i^2}$$
$$\le1+\dfrac{16}{25}\sum_{i=2}^n\dfrac{1}{i^2}$$
$$<1+\dfrac{16}{25}\left(\dfrac{\pi^2}{6}-1\right)$$
$$<1.42<RHS$$
(using the fact that RHS is an increasing sequence which takes value $\dfrac{10}{7}$ at $n=4$). Verify for $n=1,2,3$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
May I know is there quick way to factorize multi-variables equations? I want to factorize such equation:
$$b^2c + bc^2 +a^2c+ac^2+a^2b+ab^2 +2abc$$ into product of linear factors.
May I know is there any quick way/trick to do so?I am very confused with such equation.
Hello users, my main point here is what to think when we looking at factoring such such multi-variable questions. When I was doing this, I can't even start writing, since I really don't know what to do. Is there any formulae to remember? Or is just experience. I knew it might be very easy for some of you, but it is very difficult for me to factorize such equation.
| There is a nice article on this, Gary Brookfield (2016) Factoring Forms, The American Mathematical Monthly,
volume 123 number 4, pages 347-362. The main theorem was proved by Aronhold in 1849.
The general theorem, easy to state, is that a homogeneous cubic factors completely over the complexes if and only if the determinant of its Hessian matrix is a constant multiple of the original form. Here it is convenient to take exactly half the Hessian,
$$
\frac{H}{2} =
\left(
\begin{array}{ccc}
b+c & a+b+c & a+b+c \\
a+b+c & a+c & a+b+c \\
a+b+c & a+b+c & a+b \\
\end{array}
\right)
$$
and the determinant of this is precisely the original polynomial. So, it factors completely, as the others have already shown. In this case, everything can be done with real numbers
In this case, the determinant calculation gives us extra information, as it says (take the original plynomial as $f$)
$$ f = (b+c)(a+c)(a+b) + 2 (a+b+c)^3 - 2(a+b+c)(a+b+c)^2 = (b+c)(a+c)(a+b) $$
I learned about this while attempting a different factoring problem
$$ (x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right) $$
The roots of $\eta^3 - 3 \eta - 1 = 0$ are
$$ A = 2 \cos \left( \frac{7 \pi}{9} \right) \approx -1.532 \; \; \; , B = 2 \cos \left( \frac{5 \pi}{9} \right) \approx -0.347 \; \; \; , C = 2 \cos \left( \frac{ \pi}{9} \right) \approx 1.879 \; \; \; . $$
We get identity
$$ \color{red}{(x+y+z)^3 - 9 \left( x^2 y + y^2 z + z^2 x \right)} = \color{magenta}{(Ax+By+Cz)(Bx+Cy+Az)(Cx+Ay+Bz) } $$
That one was at How to show that if $x, y, z$ are rational numbers satisfying $(x + y + z)^3 = 9(x^2y + y^2z +z^2x)$, then $x = y = z$
Here is one I made up yesterday for a different problem,
$$ f(x,y) = x^3-3x^2 y - 3 x y^2 + y^3 - 6x^2z -6 y^2z + 16 z^3 $$
In factoring, all coefficients are real, mostly rational but we do need to throw in a $\sqrt 3$ as part of some of the coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve $\cos(z) =3/4+i/4$ I need to solve the complex trinometric equation
$$\cos(z) =\frac{3}{4}+\frac{i}{4} $$
What I've done so far is:
Using the cos formula I got $e^{iz} +e^{-iz} =\frac{3}{2}+\frac{i}{2}$
Making $t=e^{iz} $ we have $t+\frac{1}{t}=\frac{3}{2}+\frac{i}{2}$
Multiplying by $t^2$ we get
$$t^2-\frac{3+i}{2}t+1=0$$
Solving that we get
$$t=\frac{(\frac{3}{2}+\frac{i}{2}) \pm \sqrt{(\frac{3}{2}+\frac{i}{2}) ^2-4}} {2} = \frac{3+i \pm \sqrt{-8+6i} } {4} $$
Converting 3+i to polar we get $\sqrt{10} e^{0.3218i}$
Converting $\sqrt{-8+6i}$ to polar we get $\sqrt{10} e^{-0.3218i}$
So $t=\frac{\sqrt{10} e^{0.3218i} \pm \sqrt{10} e^{-0.3218i}} {4} $
Which means $e^{iz} =\frac{\sqrt{10} e^{0.3218i} \pm \sqrt{10} e^{-0.3218i}} {4} = \frac{\sqrt{10}} {2} (\frac{e^{0.3218i} \pm e^{-0.3218i} }{2})$
And I dont know where to go from there
| I guess the following helps you more:
$$\cos z=\cos(x+iy)=\cos x\cosh y-i\sin x \sinh y$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve $y''+y=p(x)$ where $p$ is a polynomial function The equation is $$y'' + y=p(x)$$ Let $p(x)=\sum\limits_{k=0}^n a_k x^k$
The homogeneous solution to the equation is $y_h = a\cos(x)+b\sin(x)$. I can look for a particular solution of the form $y_p=\sum\limits_{k=0}^n b_k x^k$.
We get $$\sum\limits_{k=0}^{n-2} ((k+2)(k+1)b_{k+2}+b_k)x^k+b_{n-1}x^{n-1}+b_nx^n=\sum\limits_{k=0}^n a_kx^k$$ Hence, if $0\le k \le n-2$, $(k+2)(k+1)b_{k+2}+b_k=a_k$ and $b_{n-1}=a_{n-1}$, $b_n=a_n$
I'm trying to show that $y_p=\sum\limits_{n=0}^{\infty}(-1)^n p^{(2n)}(x)$ from here, how do I proceed?
| As you stated, you have $b_n=a_n$ and $b_{n-1}=a_{n-1}$. You also have that, for $0 \le k \le n - 2$ that
$$(k+2)(k+1)b_{k+2}+b_k=a_k \implies b_k = a_k - (k+2)(k+1)b_{k+2} \tag{1}\label{eq1A}$$
Thus, you can start from $k = n - 2$ and work down to $k = 0$ to get each value of $b_k$ in turn. For example with the next $3$ values of $b_k$, you have
$$\begin{equation}\begin{aligned}
b_{n-2} & = a_{n-2} - (n)(n-1)b_{n} \\
& = a_{n-2} - (n)(n-1)a_{n}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
$$\begin{equation}\begin{aligned}
b_{n-3} & = a_{n-3} - (n-1)(n-2)b_{n-1} \\
& = a_{n-3} - (n-1)(n-2)a_{n - 1}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
$$\begin{equation}\begin{aligned}
b_{n-4} & = a_{n-4} - (n-2)(n-3)b_{n-2} \\
& = a_{n-3} - (n-2)(n-2)(a_{n-2} - (n)(n-1)a_{n})
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
With these coefficients $b_k$ determined, as Mick's question comment suggests, you can then calculate the coefficients of each $x^k$ of what you're trying to show, i.e., $y_p=\sum\limits_{n=0}^{\infty}(-1)^n p^{(2n)}(x)$, to see whether or not they match the determined $b_k$ coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
On various angular velocities of a straight line $y$ = $\frac{a}{100}\times x$, with $a$ varying from $-1000$ to $+1000$. Let a straight line be defined by $y$ = $\frac{a}{100}\times x$.
Let $a$ vary from $-1000$ to $+1000$
Desmos : https://www.desmos.com/calculator/xs2sowsaal
When $a$ is big ( in absolute value) important changes in $a$ produce moderate angular velocities.
But when $a$ is small ( say, between $-100$ and $+100$ ) the angular velocity is much bigger.
That seems astonishing ( at first sight, at least to me).
Is this observaton correct? And how to analyze this apparent fact?
| Consider the circunference $\gamma:\:x^2+y^2=1$ and let $A(x_A,y_A)$ the intersection of $\gamma$ and $t$ where $t: \: y=\frac{a}{10^2}x$.
Now $(x_A,y_A)$ are the solution to: $$\left\{\begin{matrix}
x^2+y^2=1
\\ y=\frac{a}{10^2}x
\end{matrix}\right.$$
So, I have: $$\left\{\begin{matrix}
x=\sqrt{\frac{10^4}{10^4+a^2}}=10^2\sqrt{\frac{1}{10^4+a^2}}
\\ y=\sqrt{1-\frac{10^4}{10^4+a^2}}=a\sqrt{\frac{1}{10^4+a^2}}
\end{matrix}\right.$$
Now, let $\theta$ the angle formed by $t$ with the $x$-axis; we have:$$\theta=tan^{-1}\left(\frac{y_A}{x_A}\right)=tan^{-1}\left(\frac{a\sqrt{\frac{1}{10^4+a^2}}}{10^2\sqrt{\frac{1}{10^4+a^2}}}\right)=tan^{-1}\left(\frac{a}{10^2}\right)$$
In this graph I skecth the trend of $\theta$ with $m=\frac{a}{\sqrt{10}}$ (because with $a=\frac{a}{10^2}$ it's very difficult to see the trend but the conclusion it's essentially the same) and we have:
From her, it's very easy to see that if $a$ is small $\theta$ grows very fast: Whenever $a$ is bigger the $tan^{-1}\left(\frac{a}{10^2}\right)$ grows not so fast.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$ Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$
I don't really know how to approach this. I was thinking doing something like squaring or cubing $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$ would help, but it doesn't really work out...Any help???
| There's a remarkable identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$
Applying this to $x=\sqrt[3]a,y=\sqrt[3]b,z=\sqrt[3]c$ gives that
$$a+b+c-3\sqrt[3]{abc}=0\implies a+b+c=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Egyptian fraction representation of $1$ where all denominators of the fractions are odd. Question: Is there an Egyptian fraction representation for $1$ where all the fractions have odd denominators?
I tried to generate one below:
$$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{23}+\frac{1}{721}+\frac{109}{106711605}.$$
The last term can be further decomposed to: $$\frac{1}{979007}+\frac{158}{1.04471\cdot 10^{14}}.$$
or, it is impossible for any collection of $\frac{1}{n}$ where $n$ is odd to produce $1$?
| $$\frac 13+\frac 15+\frac 17+\frac 19+\frac{1}{15}+\frac{1}{19}+\frac{1}{21}+\frac{1}{25}+\frac{1}{173}+\frac{1}{1294257}+\frac{1}{2233466948475}$$ is a solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
prove that $\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)} \ge \frac{27}{(a+b+c)^2}$
Prove that $$\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)} \ge \frac{27}{(a+b+c)^2},$$ where $a,b,c$ are positive reals.
After applying AM-GM I got
$$
\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)}
\ge \frac{36}{(a+b+c)^2+a^2+b^2+c^2},
$$
which is similar to original expression but not quite enough. Any hints??
| By Holder $$\sum_{cyc}\frac{2}{b(a+b)}=\frac{1}{(a+b+c)^2}\sum_{cyc}b\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{b(a+b)}\geq\frac{27}{(a+b+c)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving a sequence is a Cauchy sequence using definition I am intending to show that the sequence $<f_n>$ where $f_n$ = $ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}........+\frac{(-1)^{n-1}}{n}$ is a Cauchy sequence using the definition of a Cauchy sequence.
My attempt: Let $\epsilon>0$ and $n>m$
$|f_n-f_m|=|\frac{(-1)^m}{m+1}+\frac{(-1)^{m+1}}{m+2}........\frac{(-1)^{n-1}}{n}|$
I am stuck here for how do I find a particular m for each $\epsilon>0$ such that the conditions required for being a Cauchy sequence is satisfied.
| If $(n-m)$ is odd, we may assume $m$ is even without loss of generality, since $\left\lvert k \right\rvert = \left\lvert -k \right\rvert$. Then \begin{align*}\left\lvert f_n - f_m\right\rvert &= \left\lvert \frac{1}{m+1} - \frac{1}{m+2} + \frac{1}{m+3} - \frac{1}{m+4} + \dotsb + \frac{1}{n-2} - \frac{1}{n-1} + \frac{1}{n}\right\rvert \\ &= \frac{1}{m+1} - \frac{1}{m+2} + \dotsb + \frac{1}{n-2} - \frac{1}{n-1} + \frac{1}{n}\end{align*} since we can see that each pair of terms, of the form $\left(\frac{1}{k} - \frac{1}{k+1}\right)$, is positive, and the final term, $\frac{1}{n}$, is also positive, so the whole expression inside the absolute value signs is positive. Now, we can rewrite this as \begin{align*}\left\lvert f_n - f_m \right\rvert = \frac{1}{m+1} + \left(\frac{1}{m+3}-\frac{1}{m+2}\right) + \dotsb + \left(\frac{1}{n}-\frac{1}{n-1}\right)\end{align*} where each bracket is clearly negative (or zero), from which it follows that \begin{equation*}\left\lvert f_n - f_m \right\rvert \leq \frac{1}{m+1} \leq \frac{1}{m}\end{equation*}
In the case where $(n-m)$ is even, again assume $m$ is even without loss of generality. We can now re-use our working above, as we have \begin{align*}\left\lvert f_n - f_m \right\rvert &= \left\lvert f_{n} - f_{m+1} + f_{m+1} - f_{m}\right\rvert \\ &\leq \left\lvert f_{n} - f_{m+1}\right\rvert + \left\lvert f_{m+1}-f_{m}\right\rvert \qquad \text{(by the triangle inequality)} \\ &= \left\lvert f_{n} - f_{m+1}\right\rvert + \frac{1}{m+1} \\ &\leq \frac{1}{m+1} + \frac{1}{m+1} \qquad \text{(by the above working, as $(n-(m+1))$ is odd)} \\ &\leq \frac{2}{m}\end{align*}
Since $\frac{1}{m} \leq \frac{2}{m}$, it follows that you can make $\left\lvert f_n - f_m \right\rvert$ less than any desired $\epsilon > 0$ by taking $n > m > \frac{2}{\epsilon}$, so you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Double integral over square shaped region $$\iint_R(y-2x^2)dxdy$$ where $R$ is the region inside the square $|x|+|y|=1$.
So the area is:
\begin{align}
4×\left[\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}(y-2x^2)dydx\right]
&=4×\left[\int_0^1\int_{y=0}^{y=1-x}ydydx-2\int_0^1\int_{y=0}^{y=1-x}x^2dydx\right]\\
&=4×\left[\frac{1}{2}\int_0^1(1-x)^2dx-2\int_0^1x^2(1-x)dx\right]\\
&=4×\left[\int_0^1(1-x)^2dx-2\int_0^1x^2dx+2\int_0^1x^3dx\right]\\
&=4×\left[-\frac{1}{6}[(1-x)^3]_0^1-\frac{2}{3}[x^3]_0^1+\frac{2}{4}[x^4]_0^1\right]\\
&=4×\left[\frac{1}{6}-\frac{2}{3}+\frac{1}{2}\right]\\
&=0
\end{align}
I can't find my fault please check this..
| First, let's break it into two integrals:
$$
\underbrace{\iint_R y - 2x^2 \ \ dxdy}_{I} = \underbrace{\iint_R y \ \ dxdy}_{I_1} -2 \underbrace{\iint x^2\ \ dxdy}_{I_2}
$$
*
*As $y$ is an odd function, then the integral over the upper part is the negative of lower part:
\begin{align*}
I_{1} & = \iint_{R} y \ \ dxdy \\
& = \iint_{lower} y \ dx dy + \iint_{upper} y \ \ dx dy \\
& = \iint_{lower} y \ dx \ dy + \left(-\iint_{lower} y \ \ dx dy\right) \\
& = 0
\end{align*}
*As $x^2$ is an even function, then the integral over the left part is the same as the right part:
\begin{align*}
I_{2} & = \iint_{R} x^2 \ \ dxdy \\
& = \iint_{left} x^2 \ dx dy + \iint_{right} x^2 \ \ dx dy \\
& = \iint_{right} x^2 \ dx \ dy + \iint_{right} x^2 \ \ dx dy \\
& = 2\iint_{right} x^2 \ dx \ dy \\
& = 2 \int_{0}^{1} \left( \int_{-1+x}^{1-x} x^2 \ dy \right) dx \\
& = 2 \int_{0}^{1} x^2 \cdot \left([1-x] - [-1+x]\right) \ dx \\
& = 4 \int_{0}^{1} x^2 (1-x) \ dx \\
& = 4 \int_{0}^{1} x^2 -x^3 \ dx \\
& = 4 \left[\dfrac{x^3}{3} - \dfrac{x^4}{4}\right]_{0}^{1} \\
& = 4 \left[\dfrac{1}{3} - \dfrac{1}{4}\right] = \dfrac{1}{3}
\end{align*}
Then
$$
\boxed{I = I_1 - 2I_2 = 0 - 2\cdot \dfrac{1}{3} = \dfrac{-2}{3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Complex number proof involving angles I need to show that (3 + i)^3 = 18 + 26i and use this to show that the angle AOC = 3AOB, where O, A, B, C are points in the plane given by O = (0, 0), A = (1, 0), B = (3, 1) and C = (18, 26).
This is what I have done:
Expand (3 + i)^3
(3 + i)^3 = (3 + i)(3 + i)(3 + i)
= 9 + 3i + 3i + i^2 * (3 + i)
= 9 + 6i - 1 * (3 + i)
= (8 + 6i) * (3 + i)
= 24 + 8i + 18i + 6i^2
= 24 + 26i - 6
= 18 + 26i
Find the magnitude of OB = 3 + i and OC = 18 + 26i to determine ->OB and ->OC
|OB| = sqrt(3^2 + 1^2) = sqrt(10)
|OC| = sqrt(18^2 + 26^2) = 10 * sqrt(10)
Therefore,
->OB = 3 + i = sqrt(10) * cis(AOB)
->OC = 18 + 26i = 10 * sqrt(10) * cis(AOC)
Cube ->OB and show to show AOC = 3AOB
->OB^3 = (3 + i)^3 = (sqrt(10))^3 * cis^3(AOB)
= 18 + 26i = 10 * sqrt(10) * cis^3(AOB)
We found that (3 + i)^3 = 18 + 26i and by cubing ->OB we find the magnitude is the same as ->OC. Since this is the case, this shows that the angle AOC = 3AOB.
Would this be the correct way to solve this question?
| It is actually easier to use polar co-ordinates on this one.
Let's say that $B$ had the polar co-ordinates $(r_1, \theta_1)$ and $C$ had $(r_2, \theta_2)$. Then by the multiplication all you have to show is this:
$r_2=r_1^3 \implies r_2^2=r_1^6 \ldots (1)$
$\theta_2=3\theta_1 \ldots (2)$
Calculating $r_2$ would be the distance $OC$, which can be easily done with the Pythagorean theorem and the coordinates of $C$
$r_2^2=18^2+26^2=10^3=(3^2+1^2)^3=(r_1^2)^3=r_1^6$
For the next one we'll have to use this:
$\tan 3x=\dfrac{3\tan x -\tan^3 x}{1-3\tan^2 x}$
So, what's the tangent of $\theta_2$
$\tan \theta_2=\dfrac{26}{18}$
And by the way $\tan \theta_1=\dfrac{1}{3}$
$=\dfrac{26}{27 \cdot \dfrac{2}{3}}=\dfrac{\dfrac{26}{27}}{\dfrac{2}{3}}$
$=\dfrac{1-\dfrac{1}{27}}{1-\dfrac{1}{3}}=\dfrac{3 \cdot \dfrac{1}{3}-(\dfrac{1}{3})^3}{1-3 \cdot (\dfrac{1}{9})^2}$
$=\dfrac{3\tan \theta_1-\tan^3 \theta_1}{1-3\tan^2 \theta_1}=\tan 3\theta_1$
Therefore $3\theta_1=\theta_2$
To be honest I would've never done the identity part practically. That was just for fun. The point was that $\dfrac{\tan^{-1} \dfrac{26}{18}}{\tan^{-1} \dfrac{1}{3}}=3$
Anyway, peace!
Edit: If you show that $(1)$ is true then the one for angles falls in perfectly going by how polar co-ordinates in the complex plane are used in multiplication. But a little algebra doesn't hurt.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3595535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show that the largest root of $f$ is greater than $5n$ where $n(\ge 3)\in \mathbb N$. Given $f=x^3+9x^2+24x-40n^3+40xn^2+94n^2-12x^2n-62nx-74n+20$ has real roots, show that the largest root of $f$ is greater than $5n$ where $n(\ge 3)\in \mathbb N$.
I tried to do it by directly finding the roots in wolfram alpha for any $n\ge 3$, however I am getting the roots as complex which is against the hypothesis of the problem.
Also the roots obtained in Wolfram Alpha are very nasty or bad which is making my life my difficult.
Kindly help me out on how to find all the real roots of $f$ and show that they are $>5n$
| Your polynomial equation is
$$\begin{equation}\begin{aligned}
f(x)&=x^3+9x^2+24x-40n^3+40xn^2\\
& \; \; \; +94n^2-12x^2n-62nx-74n+20
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note you have
$$\begin{equation}\begin{aligned}
f(5n) & = (5n)^3 + 9(5n)^2 + 24(5n) - 40n^3 \\
& \; \; \; \; \; + 40(5n)n^2 + 94n^2 - 12(5n)^2n - 62n(5n) - 74n + 20 \\
& = 125n^3 + 225n^2 + 120n - 40n^3 + 200n^3 \\
& \; \; \; \; \; + 94n^2 - 300n^3 - 310n^2 - 74n + 20 \\
& = -15n^3 + 9n^2 + 46n + 20
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
At $n = 3$, you thus get
$$\begin{equation}\begin{aligned}
f(15) & = -15(3)^3 + 9(3)^2 + 46(3) + 20 \\
& = -166
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
However, since the coefficient of the highest power of $f(x)$ is $1$ in $x^3$, this means that $\lim_{x \to \infty}f(x) = \infty$. Since $f(x)$ is continuous, there must be a root larger than $5n$ for when $n = 3$.
To confirm this is also true for all $n \gt 3$, one way is take the derivative of \eqref{eq2A}, as shown below
$$\frac{df(5n)}{dn} = -45n^2 + 18n + 46 \tag{4}\label{eq4A}$$
Using the quadratic formula to get the roots gives
$$\begin{equation}\begin{aligned}
n & = \frac{-18 \pm \sqrt{18^2 - 4(-45)(46)}}{2(-45)} \\
& = \frac{3 \mp \sqrt{3^2 + 3(46)}}{15} \\
& = \frac{3 \mp \sqrt{3(3 + 46)}}{15} \\
& = \frac{3 \mp 7\sqrt{3}}{15} \\
& \approx -0.61, 1.01
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
The quadratic polynomial in \eqref{eq4A} being a concave-down parabola means its values are only positive with $n$ in the approximate range of $(-0.61,1.01)$, with it being negative everywhere else. Thus, for $n \ge 3$, the derivative is negative, so the value of \eqref{eq2A} would keep decreasing, confirming there's always a real root $\gt 5n$ for \eqref{eq1A}.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Number of real roots of $3x^4+6x^3+x^2+6x+3$ How many real roots does the following quartic polynomial have?
$$3x^4+6x^3+x^2+6x+3$$
After dividing both sides by $x^2$, we get
$$3x^2+6x+1+\dfrac6x+\dfrac3{x^2}=0$$
Or,$$3\left(x^2+\dfrac1{x^2}\right)+6\left(x+\dfrac1x\right)+1=0$$
Taking $x+\dfrac1x$ as $t$
$$3t^2-2+6t+1=0$$
Or,$$3t^2+6t-1=0$$
On solving I got the roots
$$\dfrac{-3+2\sqrt6}3$$ and $$\dfrac{-3-2\sqrt6}3$$
Then I plugged in the values and found only 2 roots are real use discriminant
| $$x = \frac{-3 - 2 \sqrt{6} \pm \sqrt{12 \sqrt{6} - 3}}{6}$$
There are $2$ more complex roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\lim_{(x;y)\rightarrow (0;0)} \frac{xy(x-y)}{x^4+y^4} = 0$ using the squeeze theorem My attempt:
$$0 \le \frac{|xy(x-y)|}{|x^4+y^4|} = \frac{|x^2y-xy^2|}{|x^4+y^4|} \le \frac{|x^2y|+|xy^2|}{x^4+y^4} \le \frac{|x|^2|y|+|x||y|^2}{x^4+y^4} = \frac{|x|^2|y|+|x||y|^2}{(x^2+y^2)^2-(\sqrt{2}xy)^2} = ?$$
What do I do next?
| The main reason you're having trouble is the limit doesn't exist. For example, if you approach $(0,0)$ along the line $x = 2y$, you have
$$\begin{equation}\begin{aligned}
\frac{xy(x-y)}{x^4+y^4} & = \frac{(2y)y(2y-y)}{(2y)^4+y^4} \\
& = \frac{2y^3}{17y^4} \\
& = \frac{2}{17y}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This goes to $\infty$ as $y \to 0^{+}$ and $-\infty$ as $y \to 0^{-}$.
Note a fairly easy way to see this is that the powers of $x$ and $y$ are $4$ in the denominator, but just $3$ in the numerator.
However, for example, if the actual fraction to check in the limit is
$$\frac{xy(x-y)}{x^2+y^2} \tag{2}\label{eq2A}$$
instead, then the limit does go to $0$. You can see this using the squeeze theorem by using the AM-GM inequality, or by just noting $(x - y)^2 = x^2 - 2xy + y^2 \ge 0 \implies x^2 + y^2 \ge 2xy$, plus $(x + y)^2 = x^2 + 2xy + y^2 \ge 0 \implies x^2 + y^2 \ge -2xy$, so when combined it results in $x^2 + y^2 \ge \left|2xy\right| \implies \frac{1}{x^2 + y^2} \le \left|\frac{1}{2xy}\right|$. This gives
$$0 \le \left|\frac{xy(x-y)}{x^2+y^2}\right| \le \left|\frac{xy(x-y)}{2xy}\right| \le \left|\frac{x-y}{2}\right| \tag{3}\label{eq3A}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I get from $x^4+2x^3y+3x^2y^2+2xy^3+y^4$ to $(x^2+xy+y^2)^2$? I was doing an example $$(x+y)^4+x^4+y^4$$ and I need to factor it. I've tried and couldn't really do much, so I checked if there was anything to help, and I came across a post asking about the same thing. But my question is how do I know that $$x^4+2x^3y+3x^2y^2+2xy^3+y^4=(x^2+xy+y^2)^2$$ without knowing the answer.
I know about $$(a+b+c)^2$$ but then how would i chose my $a, b$ and $c$?
There are many $xy$ combinations here, with different powers, so would I chose the lowest as my $b$, and $x^2$ as my $a$ and $y^2$ as my $c$, or perhaps is there another formula that can help me.
| In cases like $\,x^4+2x^3y+3x^2y^2+2xy^3+y^4\,$ where the difference between consecutive coefficients is "simpler" than the original sequence, it is worth trying to multiply by $\,x-y\,$:
$$
(x-y)(x^4+2x^3y+3x^2y^2+2xy^3+y^4) \,=\, x^5 + x^4 y + x^3 y^2 - x^2 y^3 - x y^4 - y^5
$$
The same pattern of "simpler" differences applies here, so multiplying by $\,x-y\,$ one more time:
$$
\begin{align}
(x-y)^2(x^4+2x^3y+3x^2y^2+2xy^3+y^4) \,&=\, (x-y)(x^5 + x^4 y + x^3 y^2 - x^2 y^3 - x y^4 - y^5)
\\ &=\, x^6 - 2 x^3 y^3 + y^6
\\ &=\, (x^3-y^3)^2
\\ &=\, (x-y)^2(x^2+xy+y^2)^2
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Finding the Laurent series without contour integration I would like to expand the following series around $x=0$. I know the answer already (one way is by simply plugging it into Mathematica) but I would like to understand how to get to the result.
$$\frac{1}{\sqrt{1+x} - 1} = \frac{2}{x} + \frac{1}{2} - \frac{x}{8} + \frac{x^2}{16} + ...$$
I'm not too familiar with Laurent series but from what I understand, the solution is rarely found using the contour integration of Laurent's theorem - rather it usually involves some algebraic manipulation till one reaches a geometric sum.
However, the square root is throwing me off and most introductory examples only involve polynomials of $x$.
| How about using the expansion (for $|x| < 1$)
$$
\sqrt{1+x} = 1 + x/2 - x^2/8 + x^3/16 - (5 x^4)/128 + (7 x^5)/256 + O(x^6)
$$
Hence
$$
\frac{x}{\sqrt{1+x} - 1} = \frac{1}{1/2 - x/8 + x^2/16 - (5 x^3)/128 + (7 x^4)/256 + O(x^5)}
$$
which can be expanded
$$
\frac{x}{\sqrt{1+x} - 1} =2 + x/2 - x^2/8 + x^3/16 - (5 x^4)/128 - (7 x^5)/128 + O(x^6)
$$
Now divide by $x$ and you're done
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that when $p$ varies, the midpoint of $PR$ lies on the curve $y^2=2x-4$. The line passing through the point $P$ ($p^2,2p$) on the curve $y^2=4x$ and the point $Q$(2,0) intersects the curve once again at point $R$, find the coordinates of point $R$ in terms of $p$.
I am able to solve this part of the question and found that the coordinates of $R$ in terms of $p$ is $R\Big(\displaystyle\frac{4}{p^2},-\frac{4}{p}\Big)$.
Working for the first part of the question: Let coordinates of R be ($x,y$).
$m_{PQ}=m_{QR}$
$\displaystyle\frac{0-2p}{2-p^2}=\frac{y-0}{x-2}$
$-2px+4p=2y-p^2y \tag{1}$
Substitute $\displaystyle\frac{y^2}{4}$ into $(1)$
$-2p\Big(\displaystyle\frac{y^2}{4}\Big)+4p=y(2-p^2)$
$py^2+(4-2p^2)y-8p=0$
$y=\displaystyle\frac{-(4-2p^2)\pm\sqrt{(4-2p^2)^2-4p(-8p)}}{2p}$
$y=\displaystyle\frac{-4+2p^2\pm(2p^2+4)}{2p}$
$y=2p$ or $\displaystyle\frac{-4}{p}$
When $y=\displaystyle\frac{-4}{p},x=\frac{4}{p^2}$
Therefore, coordinates of R = $\Big(\displaystyle\frac{4}{p^2},\frac{-4}{p}\Big)$
However, I am not sure how to prove the second part of the question, that is, to show that when $p$ varies, the midpoint of $PR$ lies on the curve $y^2=2x-4$.
| The midpoint of $PR$ in terms of $p$ is:
$$\left(\frac{p^2+4/p^2}{2}, \frac{2p-4/p}{2} \right)$$
and you can surely show that $\left(\frac{2p-4/p}{2} \right)^2 = 2 \left( \frac{p^2+4/p^2}{2} \right)-4$, right?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $a,b,c$ be side lengths of a triangle, $a+b+c=1$. Prove that $P=a^3+b^3+c^3+3abc<\frac{1}{4}$.
Let $a,b,c$ be side lengths of a triangle such that $a+b+c=1$. Prove that $$a^3+b^3+c^3+3abc<\frac{1}{4}\,.$$
I solved this question. However, I'd like to know if there's a neater solution that doesn't involve the substitutions that I used.
My solution:
$$P=a^3+b^3+c^3-3abc+6abc$$ $$=(\sum a) (\sum a^2 -\sum ab)+6abc$$ $$=\sum a^2 -\sum ab+6abc$$ $$= (\sum a)^2-3\sum ab+6abc$$ $$=1-3\sum ab+6abc.$$
We have that $abc=4pRr, \sum ab=p^2+r^2+4Rr$ and $p=\frac{1}{2}$ where $p,R,r$ are the semiperimeter, circumradius, and incenter of the triangle, respectively. Plugging these into $P$:
$$P=1-3p^2-3r^2-12Rr+24pRr$$
$$=\frac{1}{4}-3r^2<\frac{1}{4}.$$
| Because it is a somewhat rare occurrence that the inequality could be derived easily/naively from the equality case, I'm posting this solution which is essentially the same as Michael's written backwards.
Observe that when $ a = b = 0.25, c = 0.5$, we get equality.
This suggests that we need to look at the triangle inequality $ a + b - c > 0 $, and put it into play.
And of course, cyclic versions of it.
Naively, we look at $ \prod ( a + b - c ) > 0$.
Expanding this out, we get
$$ a^3 + b^3 + c^3 + 2abc < a^2b + b^2 c + c^2 a + a^2 c + b^2 a + c^2 b $$
This looks very promising, given that we've bounded the high power terms.
Of course, the next step is to use $ a + b + c = 1$. Naturally, we cube this to get
$$ a^3 + b^3 + c^3 + 3a^2b + 3a^2 c + 3b^2 c + 3b^2 a + 3c^2 a + 3c^2 b + 6 abc = 1.$$
This only uses terms that appear on both sides, so that's again very promising. Put them together to get
$$ a^3 + b^3 + c^3 + 2abc < a^2b + b^2 c + c^2 a + a^2 c + b^2 a + c^2 b = \frac{ 1 - (a^3 + b^3 + c^3 + 6abc)}{3} $$
Hence, we obtain
$$ 4a^3 + 4b^3 + 4c^3 + 12 abc < 1$$
We might have considered $ \sum \prod(a+b+c)(a+b-c)(a-b+c) > 0$, which expands to the weaker inequality
$$ a^3 + b^3 + c^3 - 6abc < a^2b+ b^2 c + c^2a + a^2c + b^2a + c^2 b. $$
Thinking about why it's so much weaker, this is because only 1 of the triangle inequalities can become an equality for fixed values, so this inequality can never become an equality (in the limiting case).
So, we need all 3 triangle inequalities to be in play at the same time, which is why $ \prod ( a + b - c)$ is so natural to consider.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Choosing numbers without replacement What is the probability of choosing 4 numbers from a bag of numbers from 1-10 without replacement, so that the smallest number choosen is 4?
I was thinking $\left(\frac{7}{10}\cdot\frac{6}{9}\cdot\frac{5}{8}\cdot\frac{4}{7}\right)$.
Is it correct or am I wrong?
Thank you in advance!
| After $4$ is fixed, the remaining three numbers have to be from $5$ to $10$ inclusive. So the number of draws whose smallest number is $4$ is the number of draws from $5$ to $10$, or $\displaystyle\binom{6}{3}=20$. Then the desired probability follows as $\frac{20}{\displaystyle\binom{10}{4}}=\frac{20}{210}=\frac2{21}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculus: $\lim_{x \to 1} \frac{x^3 - 1}{x-1}$ Is the limit $0$ or $3$?
$x^3 -1$ can be $(x-1)(x^2 +x +1)$, with the transformation, the limit will be $3$?
Why cannot we just say $x$ is to be $1$, $x^3 -1$ is going to be $0$ so the limit is $0$?
| Use $$\frac{x^3-1}{x-1}=x^2+x+1$$ for $x\neq1$.
Also, by the definition of the limit if $x$ is closed to $1$, so $x\neq1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3623628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Verifying Stoke's Theorem using path integration along a parametrized ellipse
Verify the Stoke's Theorem by showing that the line integral
$$ \int\limits_{C} -y^3 dx + x^3 dy - zdz = \frac{3\pi}{2} \tag{1},$$
through direct computation, where $C$ is the intersection of the cylinder $x^2 + y^2 =1$ and the plane $x+y+z =1$. (Hint : Find a parametrization of $C$ then compute the line integral.)
My Attempt
The ellipse that results from the intersection of the cylinder and the plane is parameterized as
\begin{align}
C(t) &= (r\cos\theta, r\sin\theta, 1-r\cos\theta - r\sin\theta) \\
\implies C'(t) &= (-r\sin\theta, r\cos\theta, r\sin\theta - r\cos\theta).
\end{align}
Then using the definition of the line integral we get
\begin{align}
\int\limits_{C} -y^3 dx + x^3 dy - zdz &= \int_{0}^{2\pi} \int_{0}^{1} F(C(t)) \cdot C'(t) \ drd\theta \\
&= \int_{0}^{2\pi} \int_{0}^{1}(-r^{3}\sin^{3}\theta , r^{3}\cos^{3}\theta, -1+4\cos\theta + r\sin\theta) \\
&\quad \cdot (-r\sin\theta, r\cos\theta , r\sin\theta - r\cos\theta) \ drd\theta \\
&= \int_{0}^{2\pi} \int_{0}^{1} r^{4}(1-2\sin^{2}\theta \cos^{2}\theta)-r\sin\theta+r\cos\theta - r^2 \cos(2\theta) \ drd\theta
\end{align}
Plugging the above integral expression in Wolfram alpha gives $\dfrac{3\pi}{10}$ , but the correct answer is $\dfrac{3\pi}{2}$. I don't understand where I have gone wrong ,does anyone have an idea of what's going on ?
EDIT
THe correct parametrization is
\begin{align}
C(\theta) &= (\cos\theta, \sin\theta, 1-\cos\theta - \sin\theta) \\
\implies C'(\theta) &= (-\sin\theta, \cos\theta, \sin\theta - \cos\theta).
\end{align}
Then using the definition of line integral we get
\begin{align}
\int\limits_{C} -y^3 dx + x^3 dy - zdz &= \int_{0}^{2\pi} F(C(\theta)) \cdot C'(\theta) \ d\theta \\
&= \int_{0}^{2\pi}(\sin^{3}\theta , \cos^{3}\theta, -1+\cos\theta + \sin\theta) \\
&\quad \cdot (-\sin\theta, \cos\theta , \sin\theta - \cos\theta) \ d\theta \\
&= \int_{0}^{2\pi} (1-2\sin^{2}\theta \cos^{2}\theta)-\sin\theta+\cos\theta - \cos(2\theta) \ d\theta = \frac{3\pi}{2}.
\end{align}
| Assuming that the $ z$ in $ \vec F$ is a typo, found in an exercise of integral transformation, I consider $ \vec F=-y^3\hat i + x^3\hat j - z^3\hat k$
Stoke's theorem tells us that $ \int_C \vec F.d\vec{r}=\iint_S (\vec\nabla\times\vec F) d\vec S$ where $ C$ is the curve bounded by the region $ S$.
So $ \vec\nabla\times\vec F=
\begin{vmatrix}
\hat i & \hat j & \hat k \\
\dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\
-y^3 & x^3 & -z \\
\end{vmatrix}
=3(x^2+y^2)\hat k$
We take S to be the region in
the plane $ h(x, y, z) = x + y + z = 1$ with boundary $ C$.
A unit normal to $ S$ is
given by $ \hat n= \dfrac{\vec\nabla.f}{\vert \vec\nabla.f \vert}=\dfrac{\hat i+\hat j+\hat k}{\sqrt 3}$
Therefore \begin{align} \iint_S (\vec\nabla\times\vec F) d\vec S &=\iint_S3(x^2+y^2)\hat k.\dfrac{(\hat i+\hat j+\hat k)}{\sqrt 3}\;\dfrac{dxdy}{\vert\hat n.\hat k\vert}\\
&=\iint_S3(x^2+y^2)\;dxdy\\
&= 3\int_{\theta=0}^{2\pi}\int_{r=0}^1r^2\cdot r\cdot dr dθ\;[\text{putting} \quad x=r\cos\theta, y=r\sin\theta]\\
&=6\pi\cdot\dfrac{r^4}{4}\Big |_0^1\\
&=\dfrac{3\pi}{2}
\end{align}
Parameterizing the curve $ C$ we can
write $ x = \cos t, y = \sin t$ and $ z = 1 − x − y = 1 − \cos t − \sin t, 0 \le t \le 2\pi$ and write
\begin{align}
\int_C −y^3dx + x^3dy − z^3dz &= \int_0^{2\pi} \left(−y^3\dfrac{dx}{dt} + x^3\dfrac{dy}{dt} − z^3\dfrac{dz}{dt}\right) dt\\
&= \int_0^{2\pi}\sin^4 t+\cos^4t+(1 − \cos t − \sin t)^3(\sin t − \cos t) dt\\
\end{align}
But this is too much work to calculate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem, but $\frac{x}{(x^2+a^2)}$ and $\frac{x^2}{(x^2+a^2)}$ aren't in $L^1(\mathbb{R})$.
Any hints hold be appreciated.
| $$\begin{align}
&\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx\\
=&\int^{\infty}_{0}\frac{2dx}{x^2+\frac{a^2b^2}{x^2}+(a^2+b^2)}\\
=& \int^{\infty}_{0}\frac{d(x+\frac{ab }x)}{(x+\frac{ab}{x})^2+(a-b)^2}
+ \int^{\infty}_{0}\frac{d(x-\frac{ab }x)}{(x-\frac{ab}{x})^2+(a+b)^2}\\
=& \int^{\infty}_{\infty}\frac{dt}{t^2+(a-b)^2}
+ \int^{\infty}_{-\infty}\frac{dt}{t^2+(a+b)^2}\\
=& \>0+\frac\pi{a+b}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is: If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$
My attempt is as follows:-
$$\dfrac{1}{2}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right)$$
Let $y=\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}$
$$A.M\ge H.M$$
$$\dfrac{\dfrac{s-a}{a}+\dfrac{s-b}{b}+\dfrac{s-c}{c}}{3}\ge \dfrac{3}{y}$$
$$\dfrac{\dfrac{s\cdot (ab+bc+ca)}{abc}-3}{3}\ge\dfrac{3}{y}$$
$$y\ge\dfrac{9}{\dfrac{(a+b+c)(ab+bc+ca)}{2abc}-3}\tag{1}$$
Let $z=\dfrac{(a+b+c)(ab+bc+ca)}{abc}$
Equation $(1)$ will give us $y_{min}$, so for that we need to find maximum value of $z$
But unfortunately I was able to find minimum value of $z$ in the following way
$$\dfrac{a+b+c}{3}\ge \dfrac{3abc}{ab+bc+ca}$$
$$\dfrac{(a+b+c)(ab+bc+ca)}{abc}\ge 9$$
But nevertheless, I tried plugging this minimum value of $z$ into equation $(1)$ and I got $y\ge 6$ and as the original expression was $\dfrac{y}{2}$, so $\dfrac{y}{2}\ge 3$ and surprisingly this answer is correct. What am I missing here?
| If $x,y,z>0$, then the sides of any triangle are : Then $$a=y+z, b=x+z,c=x+y \implies b+c-a=2x, c+a-b=2y, a+b-c=2x$$
Then the LHS of the required inequality becomes:
$$\frac{y+z}{2x}+\frac{x+z}{2y}+\frac{x+y}{2z}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}\right)\ge \frac{1}{2}(2+2+2)\ge 3$$
Here we have used $AM+GM: P+Q\ge 2\sqrt{PQ}$. The equality holds when the triangle is equilateral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to find/prove the maximum of a symmetric homogeneous function involving sides of a triangle? Let a, b and c be the side lengths of a triangle. Consider the following function
$$ f(a,b,c)=\frac{\sqrt{b\,c\,(a+c-b)(a+b-c)}}{a\,(a+b+c)} + \frac{\sqrt{a\,c\,(b+c-a)(a+b-c)}}{b\,(a+b+c)} + \frac{\sqrt{a\,b\,(b+c-a)(a+c-b)}}{c\,(a+b+c)} $$.
It is a symmetric and homogeneous function with degree 0. With the help of the computer algebra system Maple, I learn that it has a maximum 1 when a=b=c, but I don't know how to prove that the maximum of f(a,b,c) is 1. Maple only provides the answer when I use the command "Maximize" within the package "Optimization". Could anyone provide an analytic(readable) proof for the conclusion? Thanks
One more observation: Since it is a homogeneous function with degree 0, f(a, b, c)=f(a.1, a.(b/a), a.(c/a))=f(1, b/a, c/a). So, WLOG, we can assume a=1. Consequently the function is reduced to f(1, b, c) with only two variables.
| By AM-GM, we have $\sqrt{(a+c-b)(a+b-c)} \le \frac{a+c-b + a+b-c}{2} = a$
(or, $(a+c-b)(a+b-c) = a^2 - (b-c)^2 \le a^2$).
So, the first term is less than or equal to $\frac{\sqrt{bc}}{a+b+c}$.
Similarly, the second term is less than or equal to $\frac{\sqrt{ac}}{a+b+c}$,
and the third term is less than or equal to $\frac{\sqrt{ab}}{a+b+c}$.
So, $f(a,b,c) \le \frac{\sqrt{bc} + \sqrt{ac} + \sqrt{ab}}{a+b+c} \le 1$
(by letting $a = x^2, b = y^2, c = z^2$, it is just $x^2 + y^2 + z^2 \ge xy + yz + zx$
which is obvious).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to determine the convergence of the series $\sum_{n=1}^\infty\frac{e^nn!}{n^n}$? $$\sum_{n=1}^\infty\frac{e^nn!}{n^n}$$
we cant use the ratio test nor the root test, because the limit will be 1.
the main question was to determine whether or not the series
$$\sum_{n=1}^\infty\frac{a^nn!}{n^n}$$
converge, for every $a\in\mathbb R$. I found that for every $-e < a < e$ the series absoulutly converges. For every $a>e$ or $a<-e$ the series does not converge (also not conditional convergence) so now I have to determine if for $a=e$ or $a=-e$ the series will converge. (hopefully it will converge for $a=e$ and therefor also for $a=-e$. other wise i'll have to find out also for $a=-e$) thanks.
by the way, im not familier with tools like Rabbe's test or Sterling's formula, so solutions using other ideas will be great. thanks
| Without Stirling's formula
When the ratio test is inconclusive, we can try Gauss's test. (An extension of the ratio test.)
Let $u_n > 0$ and suppose
$$
\frac{u_n}{u_{n+1}} = 1 + \frac{h}{n} + O(n^{-r}), \qquad h \in \mathbb R,r>1
$$
If $h>1$, then $\sum u_n$ converges; if $h \le 1$ then $\sum u_n$ diverges.
In our case,
$$
u_n := \frac{e^n n!}{n^n},
\\
\frac{u_n}{u_{n+1}}= \frac{1}{e}\left(\frac{n+1}{n}\right)^n
= 1-\frac{1}{2n} +O(n^{-2})
$$
So we apply Gauss's test with $h=-1/2, r=2$ and conclude the series diverges.
Explanation for the approimation.
$$
\lim_{n\to \infty} \frac{1}{e}\left(\frac{n+1}{n}\right)^n
=\frac{1}{e}\;\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \frac{e}{e} = 1
$$
so the first term is $1$. Next I claim
$$
\left(\frac{1}{e}\left(\frac{n+1}{n}\right)^n - 1\right)n \to -\frac{1}{2}
$$
so the next term is $-\frac{1}{2n}$. For that computation:
$$
\left(\frac{1}{e}\left(\frac{n+1}{n}\right)^n - 1\right)n =
\frac{(1+\frac{1}{n})^n-e}{e/n}
$$
This is indeterminate of the form $0/0$. L'Hopital suggests this limit is the same as the limit of
$$
\frac{(1+\frac{1}{n})^n\left[\log(1+\frac{1}{n})-\frac{1}{n(1+1/n)}\right]}{-\frac{e}{n^2}}
= -\frac{(1+\frac{1}{n})^n}{e}\;\cdot\;
\left[n^2\log\left(1+\frac{1}{n}\right)-\frac{n^2}{n+1}\right];
$$
Now $-\frac{(1+\frac{1}{n})^n}{e} \to -1$, and to compute the other factor:
\begin{align*}
\log\left(1+\frac{1}{n}\right) &= \frac{1}{n} - \frac{1}{2n^2} + O(n^{-3})
\\
n^2\log\left(1+\frac{1}{n}\right) &= n - \frac{1}{2} + O(n^{-1})
\\
\frac{n^2}{n+1} &= n - 1 + O(n^{-1})
\\
n^2\log\left(1+\frac{1}{n}\right)- \frac{n^2}{n+1}
&= \frac{1}{2} + O(n^{-1})
\\
n^2\log\left(1+\frac{1}{n}\right)- \frac{n^2}{n+1}
&\to \frac{1}{2}
\end{align*}
so that
$$
-\frac{(1+\frac{1}{n})^n}{e}\;\cdot\;
\left[n^2\log\left(1+\frac{1}{n}\right)-\frac{n^2}{n+1}\right]
\to -\frac{1}{2}
$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3628189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to find the solution to a multi-inifite summation? I'm familiar with techniques to solve a single infinite sum such as:
$$
\sum_{i=1}^{\infty} (1 + i) \cdot \frac{1}{2^i}
$$
Which ends up being equal to $3$, but I'm having trouble figuring out how to tackle a multi summation such as this:
$$
\sum_{i=0}^\infty \sum_{j=0}^\infty (3 + i +j) \cdot \left( \frac{1}{3} \right)^{(1+i)} \cdot \left( \frac{2}{3} \right)^{(1+j)}
$$
For the first sum I was able to find a pattern in the partial sums. But with this double summation, I don't think that will work as easily. Does anyone know of a technique to solve this second summation analytically? Thanks
| A key to calculate this double sum lies in the fact that for absolutely convergent series $\sum_{i=0}^{\infty}a_i$ and $\sum_{j=0}^{\infty}b_j$ we have
$$\left(\sum_{i=0}^{\infty}a_i\right)\left(\sum_{j=0}^{\infty}b_j\right)=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_ib_j = \sum_{j=0}^{\infty}\sum_{i=0}^{\infty}a_ib_j$$
Now, you can just split up the given series and apply above fact
$$\sum_{i=0}^\infty \sum_{j=0}^\infty (3 + i +j) \left( \frac{1}{3} \right)^{(1+i)} \left( \frac{2}{3} \right)^{(1+j)}$$ $$= \underbrace{3\sum_{i=0}^\infty \sum_{j=0}^\infty \left( \frac{1}{3} \right)^{(1+i)} \left( \frac{2}{3} \right)^{(1+j)}}_{=S_1} + \underbrace{\sum_{i=0}^\infty \sum_{j=0}^\infty i \left( \frac{1}{3} \right)^{(1+i)} \left( \frac{2}{3} \right)^{(1+j)}}_{=S_2} + \underbrace{\sum_{i=0}^\infty \sum_{j=0}^\infty j\left( \frac{1}{3} \right)^{(1+i)} \left( \frac{2}{3} \right)^{(1+j)}}_{=S_3}$$
$$S_1 = 3\cdot \frac 13 \cdot \frac 23 \left(\sum_{i=0}^{\infty}\left( \frac{1}{3} \right)^{i}\right)\left(\sum_{j=0}^{\infty}\left( \frac{2}{3} \right)^{j}\right) = \frac 23\cdot \frac 32 \cdot 3=3$$
$$S_2 = \frac 23 \left(\sum_{i=0}^{\infty}i\left( \frac{1}{3} \right)^{(1+i)}\right)\left(\sum_{j=0}^{\infty}\left( \frac{2}{3} \right)^{j}\right) \stackrel{\sum_{i=0}^{\infty}i x^{i+1} =\frac{x^2}{(1-x)^2}}{=} \frac 23\cdot \frac 14 \cdot 3 = \frac 12$$
$$S_3 = \frac 13 \left(\sum_{i=0}^{\infty}\left( \frac{1}{3} \right)^{i}\right)\left(\sum_{j=0}^{\infty}j\left( \frac{2}{3} \right)^{(1+j)}\right) \stackrel{\sum_{j=0}^{\infty}j x^{j+1} =\frac{x^2}{(1-x)^2}}{=} \frac 13\cdot \frac 32 \cdot 4 = 2$$
So, you get
$$\sum_{i=0}^\infty \sum_{j=0}^\infty (3 + i +j) \left( \frac{1}{3} \right)^{(1+i)} \left( \frac{2}{3} \right)^{(1+j)} = S_1 +S_2 + S_3 = \boxed{\frac{11}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
When does $\sqrt{x+\sqrt{x+1+\sqrt{x+2+...}}}=0$? Consider the function $f$ defined as the limit of the functions
$$f_0(x)=\sqrt{x}$$
$$f_1(x)=\sqrt{x+\sqrt{x+1}}$$
$$f_2(x)=\sqrt{x+\sqrt{x+1+\sqrt{x+2}}}$$
$$...$$
so that $f(x)$ is defined iff $f_n(x)$ is defined for some $n$. The unique root $x_0$ of the function $f$ satisfies $f(x_0)=0$, and it can be alternatively expressed as the limit of the roots of $f_0, f_1, f_2, ...$. See the graphs below:
Can anyone find an expression equal to this limit? I realize that the chances of something nice and closed-form are slim - can we find a series, integral, or even nested radical representation of the real root of $f(x)$?
| This problem is equivalent to proving that there exists some real number $r$ such that
\begin{equation}
\lim_{n \to \infty} f_n(r) = 0
\end{equation}
where $f_n = g_n^0(x)$ and $g^k_n$ is defined by the recurrence
\begin{equation}
g_n^k(x) =
\begin{cases} \sqrt {x+k} & \text{if }n=1 \\
\sqrt{x + k + g^{k+1}_{n-1}(x)} & \text{o.w.} \\
\end{cases}.
\end{equation}
Therefore, by definition, we have
\begin{equation}
f_n(x) =
\begin{cases} \sqrt {x} & \text{if }n=1 \\
\sqrt{x + g^{1}_{n-1}(x)} & \text{o.w.} \\
\end{cases},
\end{equation}
which after proving the following:
\begin{equation}
g_n^1(x) = g_n^0(x+1) ,
\end{equation}
by a simple induction, gives us that
\begin{equation}
f_n(x) =
\begin{cases} \sqrt {x} & \text{if }n=1 \\
\sqrt{x + f_{n-1}(x+1)} & \text{o.w.} \\
\end{cases}.
\end{equation}
Let us suppose that $r_n - f_{n-1}(r_n+1) = \epsilon_n$ and that $\lim_{n \to \infty} \epsilon_n = 0 $ then we have that
\begin{equation}
\lim_{n \to \infty} f_{n}(r_n) = 0.
\end{equation}
One possible route to find this is to consider the growth of the sequence
\begin{equation}
\delta_n(x) = f_n(x+1) - f_n(x).
\end{equation}
Therefore given a good approximation to $r_n$, an educated guess for $r_{n+1} $ should be some
\begin{equation}
r_{n+1} \in [r_{n} - 1, r_{n}],
\end{equation}
this is because
\begin{equation}
f_{n+1}^2(r_{n} - \epsilon) = r_{n} - \epsilon + f_n(r_{n} - \epsilon) +\delta_n(r_{n} - \epsilon)
\end{equation}
and
\begin{equation}
\delta_n(r_{n} - \epsilon) = f_n(r_{n} +1- \epsilon) - f_n(r_{n} - \epsilon) = f_n(r_{n} +(1-\epsilon)) - f_n(r_{n} - \epsilon)
\end{equation}
gives us that we should be looking for some
\begin{equation}
-\delta_n(r_{n} - \epsilon) \approx r_{n} - \epsilon + f_n(r_{n} - \epsilon)
\end{equation}
which is equivalent to saying
\begin{equation}
-\delta_n(r_{n} - \epsilon) - f_n(r_{n} - \epsilon) \approx r_{n} - \epsilon
\end{equation}
or
\begin{equation}
-f_n(r_{n} + (1- \epsilon)) \approx r_{n} - \epsilon,
\end{equation}
or
\begin{equation}
f_n(r_{n} + (1- \epsilon)) \approx \epsilon - r_n .
\end{equation}
which is a "sort-of" fixed point equation. Now we are getting somewhere!
We now have a method to recursively approximate solutions from previous solutions; i.e.
\begin{equation}
f_n(r_{n} + (1- \epsilon)) = \epsilon -r_{n} \implies f_{n+1}(r_{n} - \epsilon) = 0 .
\end{equation}
Let us compute $r_n$ for small values to verify the solution.
*
*$f_1(x) = \sqrt x $ is the base case and it has the obvious solution $r_1 = 0$.
*Since $f_1(0) = 0$ and $\sqrt{1- \frac{\sqrt5 -1}{2}} = \frac{\sqrt5 -1}{2} $ we have an exact solution in the form of $r_1 = 0 $, $\epsilon = \frac{\sqrt5 -1}{2} $, and $r_2 = r_1 - \epsilon $; i.e. $f_2 (r_1 - \epsilon) = f_2\left(\frac{ 1-\sqrt5 }{2}\right) = 0$.
*Let $\epsilon = \frac{1}{2} (3 - \sqrt{5})$ then one can easily check that both $f_2(r_2 +1 - \epsilon) = \epsilon - r_2 $ and $f_3(r_2 - \epsilon) =f_3(-1) = 0 . $
*Let $\epsilon = 0.153761 $ then one can easily check that both $f_2(r_3 + 1 - \epsilon) \approx \epsilon - r_3 $ and $f_4(r_3 - \epsilon) = f_3(-1.153761) \approx 0 . $
In conclusion, we have the following recursion relation for the roots
\begin{equation}
r_{n+1} =
\begin{cases} 0 & \text{if }n=1 \\
r_n - \epsilon & \text{if } f_n(r_n +1 - \epsilon) = r_n - \epsilon \\
\end{cases};
\end{equation}
however, it is possible that a recursive analysis of the following two functions
*
*$f'_n(x)$
*$\delta_n(x)$
could give a better recursive bound $b_n < b_{n-1}<1 $ on the intervals
\begin{equation}
r_{n+1} \in [r_n-b, r_n]
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 0
} |
If $a+b+c=0$, then $a^3+b^3+c^3$ is ... $0$? $1$? $a^3b^3c^3$? $3abc$? Many mistakes in this post. ( See comments below). I let it as it is, as an example of what shouldn't be done.
If $a+b+c=0$, then $a^3+b^3+c^3 = \ldots $
A. $\;0\quad$ B. $\;1\quad$ C. $\;a^3b^3c^3\quad$ D. $\;3abc$
Source: 4/12/2020, Competitive Exams Reasoning Sample Paper 3- Translation in Hindi, Kannada, Malayalam, Marathi, Punjabi, Sindhi, Sindhi, Tamil, Telgu - Examrace. Downloaded from examrace.com
I can only see the pitfall consisting in inferring that all 3 numbers must be equal to 0.
What I can conclude from the premise is that one of the 3 numbers is the additive inverse of the sum of the 2 others.
Admitting it is number $c$, we get
$$a+b+c = 0= (a+b) + \left( - (a+b) \right) \tag{1}$$
In that case
$$c^3 = [- (a+b)]^3 = - (a+b) (a+b)(a+b) = - ( a^3 +2a^2b+2ab^2+b^3) \tag{2}$$
So
$$\begin{align}
a^3+b^3+c^3 &= a^3+b^3 - ( a^3 +2a^2b+2ab^2+b^3) \tag{3} \\
&= a^3+b^3 - a^3 - 2a^2b- 2ab^2- b^3 \tag{4}\\
&= 2a^2b - 2ab^2 \tag{5} \\
&=2 ( a^2b - b^2a) \tag{6} \\
&= 2 ( a) (ab-b^2) \tag{7} \\
&= 2 ( a) (b) (a-b) \tag{8} \\
&= 2 ( a) (b) (- c) \quad\text{[ Since $c = -(a+b) = b - a = - (a-b) $]} \tag{9} \\
&= - 2 ( a) (b) (c) \tag{10}
\end{align}$$
However, this isn't one of the possible answers.
What did I miss? Was I wrong in supposing that I could take any number $a$, $b$, or $c$ to play the role of additive inverse of the sum of the two others?
| Let $a=1$, $b=1$, $c=-2$. So $a+b+c=0$ and $a^3+b^3+c^3=-6$.
Therefore, choices (1), (2), and (3) (whose values are $0$, $1$, and $-8$, respectively) are wrong. But choice (4) has $3abc=-6$. Therefore choice (4) is the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3636449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Inequality $a^ab^bc^c \geq (a+b-c)^a(b+c-a)^b(c+a-b)^c$ I found a problem where now I essentially have to show the following inequality
$$a^ab^bc^c \geq (a+b-c)^a(b+c-a)^b(c+a-b)^c$$ where $a,b,c$ are the sides of a triangle.
I have tried a lot of approaches but am not making headway.
Any hint in the right direction would be highly appreciated.
: Could it be the above inequality is false and what was meant to be proved was what Calvin Lin proved below....If so, can anybody help provide a counterexample?
| The inequality to be proven is equivalent to
$$\left(\frac{a+b-c}{a}\right)^{\frac{a}{a+b+c}}\left(\frac{b+c-a}{b}\right)^{\frac{b}{a+b+c}}\left(\frac{c+a-b}{c}\right)^{\frac{c}{a+b+c}}\leq 1\,.$$
By the Weighted AM-GM Inequality,
$$\begin{align}&\left(\frac{a+b-c}{a}\right)^{\frac{a}{a+b+c}}\left(\frac{b+c-a}{b}\right)^{\frac{b}{a+b+c}}\left(\frac{c+a-b}{c}\right)^{\frac{c}{a+b+c}}\\&\phantom{abcde}\leq \frac{a}{a+b+c}\left(\frac{a+b-c}{a}\right)+\frac{b}{a+b+c}\left(\frac{b+c-a}{b}\right)+\frac{c}{a+b+c}\left(\frac{c+a-b}{c}\right)\,.\end{align}$$
Can you guess what the right-hand side of the inequality above is equal to?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral with negative radicand $ \int \frac{dx}{\sqrt{-a^2-x^2}}$ I saw this integral:
$$
\int \frac{dx}{\sqrt{-a^2-x^2}}.
$$
I tried to solve it by taking $-1$ as common factor which leaded me to:
$\frac{1}{i\sqrt{x^2+a^2}}$ which leads to:
$-i\ln(\sqrt{x^2+a^2}+x)$
However I placed it on Wolfram Alpha and it says the result is:
$$
\arctan\left(\frac{x}{\sqrt{-a^2-x^2}}\right)
$$
So my question would be..is it possible that both expressions are equivalent??? Sorry if this a dumb question....
Thanks in advance.
| Note $\arctan (i z) = i \text{arctanh(}z) =i \ln \sqrt{\frac{1+z}{1-z}}$.
Thus
$$
\arctan\left(\frac{x}{\sqrt{-a^2-x^2}}\right)
=\arctan\left(-i \frac{x}{\sqrt{a^2+x^2}}\right)\\
= -i \ln \sqrt{\frac{\sqrt{a^2+x^2}+x}{\sqrt{a^2+x^2}-x}}
= -i \ln {\frac{\sqrt{a^2+x^2}+x}{a^2}}\\
\hspace{-4cm}=-i\ln(\sqrt{x^2+a^2}+x)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A series of multiplication leads to $\frac{1}{2} = 2$ I'm presented with the equation $\frac{a+b}{a} = \frac{b}{a+b}$
Performing cross multiplication yields $a^2+2ab+b^2 = ab$
Subtracting $ab$ from both sides, we get $a^2+ab+b^2 = 0$
Multiplying both sides by (a-b) and simplifying:
$(a-b)(a^2+ab+b^2) = 0 * (a-b)$
$a^3 - b^3 = 0$
$a^3 = b^3$
$a = b$
Substituting into the original equation, we finally arrive at $2 = \frac{1}{2}$ ... something has obviously gone terribly wrong. Where did I mess up?
Also it is worth noting that the original problem explicitly states that $a ≠ b$
| You multiplied with $0=a-b$. That is wrong. You are just recovering this solution. Just to give you a similar example, say you have $$x-1=0$$
If I multiply both sides by $x+1$, I get $$x^2-1=0$$ or $$x=\pm 1$$Obviously $x=-1$ is not a solution. You introduced it when you multiplied with $x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Generating function expansion I am working on a generating function for a sequence problem and I am stuck with expanding.
I have
$$\frac{1}{4(1-x)^2} - \frac{1}{4(x+1)^2} $$
From my notes, the first term will be $\sum_{n=0}^{\infty}\frac{1}{4}(n+1)x^n.$
I am stuck on the second term where the denominator is $(x+1)^2$ Am I able to make it look like $(1-(-x))^2$ and do something similar to the first term and get
$$\sum_{n=0}^{\infty}\frac{1}{4}(-1)^n(n+1)x^n
$$
| You can observe
$\frac{1}{4(1-x)^2} - \frac{1}{4(x+1)^2}=\frac{x}{(1-x^2)^2}=\frac{x}{1-x^2}\sum_{n=0}^\infty x^{2n}$
$=\sum_{n=0}^\infty \frac{x^{2n+1}}{1-x^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $x^4y^{\prime\prime} = (y-xy^\prime)^3, y(1) = y^\prime(1) = 1$ $\lambda^4x^4\lambda^{n-2}y^{\prime\prime} = (\lambda^ny-\lambda x\lambda^{n-1}y^\prime)^3 \Rightarrow \lambda^{n+2}x^4y^{\prime\prime} = (\lambda^n(y-xy^\prime))^3$.
$\lambda^{n+2} = \lambda^{3n} \Rightarrow n+2 = 3n \Rightarrow n = 1$.
Let $x = e^t, y = ue^{nt} = ue^t$.
Now, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{u^\prime e^t + ue^t}{e^t} = u^\prime + u$. And, $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}}{\frac{dx}{dt}}(\frac{dy}{dx}) = \frac{\frac{d}{dt}(u^\prime + u)}{e^t} = e^{-t}(\frac{d^2u}{dt^2} + \frac{du}{dt})$.
Thus,
$e^{4t}(e^{-t}(\frac{du^2}{dt^2} + \frac{du}{dt})) = (ue^{t}-e^t(u+\frac{du}{dt}))^3 \Rightarrow e^{3t}(\frac{d^2u}{t^2} + \frac{du}{dt}) = e^{3t}(u-(u+\frac{du}{dt}))^3 \Rightarrow (\frac{d^2u}{dt^2} + \frac{du}{dt}) = (\frac{du}{dt})^3$.
Let $p = \frac{du}{dt}, p^\prime = \frac{d^2u}{dt^2} = \frac{dp}{du}\frac{du}{dt} = p\frac{dp}{du}$.
Thus,
$(p + p\frac{dp}{du}) = p^3 \Rightarrow 1 + \frac{dp}{du} = p^2 \Rightarrow \frac{dp}{du} = p^2-1 \Rightarrow u + c_1 = \int\frac{dp}{p^2-1} = \int\frac{dp}{(p-1)(p+1)} = \int(\frac{1}{2(p-1)} - \frac{1}{2(p+1)})dp = \frac{1}{2}\ln(p-1)-\frac{1}{2}\ln(p+1) = \ln(\frac{\sqrt{p-1}}{\sqrt{p+1}}) \Rightarrow c_1e^u = \frac{\sqrt{p-1}}{\sqrt{p+1}} \Rightarrow c_1e^{2u} = \frac{p-1}{p+1} \Rightarrow p-1 = c_1e^{2u}p + c_1e^{2u} \Rightarrow p = \frac{c_1e^{2u}+1}{1-c_1e^{2u}} = \frac{du}{dt}$.
But then I think that I have to use the initial conditions to get $c_1$, but I don know how to do that.
| Since $x=e^t$, $x=1$ exactly when $t=0$. So, since $y=ue^{t}$, the boundary condition $y(1)=1$ translates into $u(0)e^0=1$ which means $u(0)=1$.
Then use the fact that $y'=u'+u$. The boundary condition $y'(1)=1$ means $u'(0)+u(0)=1$. Since we already have that $u(0)=1$, this means $u'(0)=0$.
You have already partially solved for $u'$, because you have $u'=\frac{c_1e^{2u}+1}{1-c_1e^{2u}}$. This gives you that $u'(0)=\frac{c_1e^{2u(0)}+1}{1-c_1e^{2u(0)}}=\frac{c_1e^2+1}{1-c_1e^2}$. Setting this last expression equal to $0$, you get $c_1=-\frac{1}{e^2}$.
(Keep in mind that terms like $y'$ and $u'$ implicitly denote derivatives with respect to $x$ and $t$, respectively).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Stuck when transforming and solving this Given abc=1 ( all positive real numbers). Prove that:
$$\frac ab + \frac bc + \frac ca +3( \frac ba +\frac cb +\frac ac) \ge 2(a +b +c+\frac 1a+ \frac 1b +\frac1c)$$
My attempt:
$$\frac ab + \frac bc + \frac ca +3( \frac ba +\frac cb +\frac ac) \ge 2(a +b +c+\frac 1a+ \frac 1b +\frac1c)$$
Remove denominator
$$a^2c+ab^2+bc^2+3(b^2c+c^2a+a^2b) + a +b + c\ge 3(a+b+c) + 2(ab+bc+ca)$$
Turn this to
$$(a^2+1)c +(b^2+1)a + (c^2+1)b +3(b^2c+c^2a+a^2b)\ge 3(a+b+c) + 2(ab+bc+ca)$$ (1)
We have: $$(a^2+1)c \ge 2ac$$ (2)
From (1) and (2) i observe what is left to prove is $$3(b^2c+c^2a+a^2b)\ge 3(a+b+c)$$
Meaning to prove $$b^2c+c^2a+a^2b\ge a+b+c$$ given abc = 1. Which is where i stuck
| The equality $a^2b+b^2c+c^2a\geq a+b+c$ we can prove by AM-GM:
$$\sum_{cyc}a^2b=\frac{1}{3}\sum_{cyc}(2a^2b+c^2a)\geq\frac{1}{3}\sum_{cyc}3\sqrt[3]{(a^2b)^2c^2a}=$$
$$=\sum_{cyc}\sqrt[3]{a^5b^2c^2}=\sum_{cyc}a.$$
Another way.
Let $a=\frac{x}{y}$ and $b=\frac {y}{z},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{z}{x}$ and we need to prove that:
$$\sum_{cyc}\left(\frac{\frac{x}{y}}{\frac{y}{z}}+3\frac{\frac{y}{z}}{\frac{x}{y}}-\frac{2x}{y}-\frac{2y}{x}\right)\geq0$$ or
$$\sum_{cyc}(x^3y^3+3x^4yz-2x^3y^2z-2x^3z^2y)\geq0,$$ which is true by Muirhead or by SOS:
$$\sum_{cyc}(2x^3y^3-x^3y^2z-x^3z^2y+6x^4yz-3x^3y^2z-3x^3z^2y)\geq0$$ or
$$\sum_{cyc}(z^3(x^3-x^2y-xy^2+y^3)+3xyz(x^3-x^2y-xy^2+y^3))\geq0$$ or
$$\sum_{cyc}(x-y)^2(x+y)z(z^2+3xy)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\log_{a} c + \log_{b} c = \log_{a+b} c$ if and only if $1 + \log_{b} a = \log_{a+b} a$ If a, b and c are positive numbers, than equality $$\log_{a} c + \log_{b} c = \log_{a+b} c$$ is true if and only if $$1 + \log_{b} a = \log_{a+b} a$$
Prove it!
I have looked at the solution but it is not clear for me.
We will prove that if $$\log_{a} c + \log_{b} c = \log_{a+b} c$$
than it is
$$1 + \log_{b} a = \log_{a+b} a$$
$$\log_{a} c + \log_{b} c = \frac{\log_{a+b} c}{\log_{a+b} a} +\frac{\log_{a+b} c}{\log_{a+b} b} =\log_{a+b} c $$
So this is only thing that is not clear for me how is this equal.
$$\frac{\log_{a+b} c}{\log_{a+b} a} +\frac{\log_{a+b} c}{\log_{a+b} b} =\log_{a+b} c $$
| WE have:
$$1+ log_b (a)= log_{a+b} (a) $$
Let $a=c$ then:
$$ log _a(c)= log _c (c)=1$$
therefore:
$$log _a (c) + log _b (c)= log _{a+b} (c)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
I substituted $3x^2 = 1-x$ in $6x^3-x^2-3x+2010$ and it simplified to
$$\frac{-8x}{3}+2010$$
I know this is a simple problem but I can't solve it. I think there's some method I'm not trying. Please help me with this.
| by using the long division
$$\frac{6x^3-x^2-3x+2010}{1}=\frac{6x^3-x^2-3x+2010}{3x^2+x}=(3x^2+x)(2x-1)-\frac{2x-2010}{3x^2+x}$$
$$=(1)(2x-1)-\frac{2x-2010}{1}=2009$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3654124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Calculating matrix determinant This is a very easy determinant to calculate, but I get two different results when I calculate it in two different ways.
\begin{equation}
A = \begin{bmatrix} 1 & 2 & 3 \\
0 & 1 & 2 \\
2 & 2 & 4
\end{bmatrix}
\end{equation}
When I used Laplace expansion right away I got:
\begin{equation}
\det(A) = 1 \cdot \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} - 0 \cdot \begin{bmatrix} 2 & 3 \\ 2 & 4 \end{bmatrix} + 2 \cdot \begin{bmatrix} 2 & 3 \\ 1 & 2\end{bmatrix} = 1 \cdot (4 - 4) + 2 \cdot (4 - 3) = 2
\end{equation}
But when I rearrange the rows in the matrix and then try to calculate the determinant:
\begin{equation}
A = \begin{bmatrix} 1 & 2 & 3 \\
0 & 1 & 2 \\
2 & 2 & 4 \\
\end{bmatrix} \overset{r_1 \leftarrow 2 \cdot r_1 - r_3}{\longrightarrow} \begin{bmatrix} 0 & 2 & 2 \\
0 & 1 & 2 \\
2 & 2 & 4 \\
\end{bmatrix}\\
\det(A) = 0 \cdot \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} - 0 \cdot \begin{bmatrix} 2 & 2 \\ 2 & 4 \end{bmatrix} + 2 \cdot \begin{bmatrix} 2 & 2 \\ 1 & 2\end{bmatrix} = 2 \cdot (4 - 2) = 4
\end{equation}
I have probably made a simple mistake, but I can't figure out where and I really want to get the basics down, before I move to harder examples.
| The operation $r_1 \leftarrow 2r_1 - r_3$ does not preserve the determinant. Indeed, if you check the rules, the determinant is preserved if you subtract a multiple of a row from the actual other row (and not a multiple of it).
Suppose you did $r_1 \leftarrow r_1 - r_3$, then did Laplace expansion, things would work. Suppose you did $r_1 \leftarrow r_1 - 2r_3$ followed by Laplace expansion, then things would work. However, because you did changed $r_1$ to $2r_1$ during the row change, the determinant got multiplied by exactly $2$ , which came out to be the decisive factor in the end, since $\frac 42 = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3656372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove and give geometric meaning of $|\sqrt{a^2+b^2} - \sqrt{a^2+c^2} | < |b-c| $
Carefully, prove that
$$ | \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | \leq |b-c| $$
and give a geometric interpretation.
pf
We observe that $| \sqrt{a^2+b^2} - \sqrt{a^2+c^2} | = \dfrac{ |b^2 - c^2 | }{\sqrt{a^2+b^2} + \sqrt{b^2+c^2} } \leq \dfrac{ |b^2-c^2| }{\sqrt{b^2} + \sqrt{c^2} } = \dfrac{ |b+c | }{|b| + |c| } \cdot |b-c| $
Since by the triangle ineqality $|b+c| \leq |b| + |c| \implies \dfrac{ |b+c| }{|b| + |c| } \leq 1$, then the required result is proved!
Now, as for the hardest part, I dont see a geometric interpretation of this inequality. Can someone enlighten me?
| As pointed out in the comments, the points $(0,0), (a,b), (a,c)$ form a triangle. Using the triangle inequality,
$$\sqrt{a^2 + b^2} + |b-c| \gt \sqrt{a^2 + c^2}$$
and
$$\sqrt{a^2 +c^2} + |b-c| \gt \sqrt{a^2 +b^2}$$
$$\implies \left|\sqrt{a^2+b^2}-\sqrt{a^2+c^2} \right| \lt |b-c|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Homework Problem, Power Series Limit I am looking to find the solution for:
$$\lim_{x\rightarrow 0} {5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$$
A hint was provided:
transform: ${5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$ to $5y{5^{y-1} - 1 \over y-1}, y = {1\over cos^s(x)} \rightarrow 1$
The transformation is straigt forward:
$tan^2(x) + 1 = {1\over cos^2(x)} = y \text{ and } \\1-cos^2(x) = ycos^2(x) - cos^2(x) = cos^2(x)(y-1)$ combined we have:
$$5y{(5^{y-1} - 1) \over y-1}$$
as $y \rightarrow 1$ $5y{(5^{y-1} - 1) \over y-1}$ is not defined. Since both, nominator and denominator are $0$ I tried L'Hopital but ended at:
$5 5^{y-1} + 5y(y-1)5^{y-2}-5$ with $\lim_{y \rightarrow 1} = 5 + 0 - 5 = 0$ and $(y-1)' = 1$
Here I have to stop with no solution. I have also tried to use the quotient rule to differentiate the expression which did not get me anywhere.
| Hint:
$$5\cdot\dfrac{5^{\tan^2x}-1}{\tan^2x}\cdot\dfrac{\tan^2x}{1-\cos^2x}=5\cdot\dfrac{5^{\tan^2x}-1}{\tan^2x}\cdot\dfrac1{\cos^2x}$$
Now as $a=e^{\ln a}$ for $a>0$
$$\lim_{h\to0}\dfrac{a^h-1}h=\cdots=\ln a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Primes dividing an integer polynomial $x^8 + 8x^6y^2 -2x^4y^4 + 8x^2y^6 + y^8$ are congruent to 1 mod $4$ I need to prove that if $p\neq 2$ divides the integer polynomial
$$x^8 + 8x^6y^2 -2x^4y^4 + 8x^2y^6 + y^8\in\mathbb{Z}[x,y],$$ $gcd(x,y)=1,$ then $p\equiv 1\pmod{4}.$ Any ideas?
Edit (the OPs reaction to the hint that this is a sum of two squares):
If $p|a^2+b^2=(a+bi)(a−bi)$, then wlog $p|a+bi$ which means that $a+bi≡0\pmod p$, so $i=\sqrt{−1}$ is defined modulo $p$. That means that $-1$ is a quadratic residue mod $p$, so $p\equiv1 \mod p$.
| A proof can be based on the following
Fact. If an odd prime $p$ is a factor of the sum $x^2+y^2$ such that $x$ and $y$ are not multiples of $p$, then $p\equiv1\pmod4$.
Proof. Without loss of generality $p\nmid y$. Therefore we can divide the congruence
$$
x^2\equiv -y^2\pmod p
$$
by $y^2$, and conclude that there exists an integer $z\equiv xy^{-1}$ such that $z^2\equiv-1\pmod p$. This is well known
to be possible only when $p\equiv1\pmod4$.
The octic polynomial in the question can be written as a sum of two squares:
$$
x^8 + 8x^6y^2 -2x^4y^4 + 8x^2y^6 + y^8=(x^2+y^2)^4+4x^2y^2(x^2-y^2)^2.
$$
If neither $a=(x^2+y^2)^2$ nor $b=2xy(x^2-y^2)$ is divisible by $p$, then the other cannot be either. Hence the Fact bites, and allows us to conclude that $p\equiv1\pmod4$.
But if $p\mid a$, then $p\mid x^2+y^2$, and we can repeat the argument. Observe that the assumption $\gcd(x,y)=1$ rules out the possibility that both $x$ and $y$ would be divisible by $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3660015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Square root of prime is irrational. Is this a valid proof? I know similar questions exist, but I want to know, if this is a valid proof.
A prime number has 1 and it self as divisors. So the subset of a prime number is $ D(p) = \{1, p\}$
Now I want to prove $\sqrt{p} = \frac{a}{b}$ , to get the opposite.
Also $gcd(a,b)=1$.
Let's square it.
$p = \frac{a^2}{b^2}$
And divide by $b^2$
$\frac{p}{b^2} = \frac{a^2}{b^4}$
In this case ${b^2}$ is also a divisor of $p$, which doesn't fit the definition of prime numbers. So the square root of a prime number must be irrational.
| We have that $\sqrt{p}=\dfrac{a}{b}$ with $a$ and $b$ being relatively prime integers. Thus, $p=\dfrac{a^2}{b^2}$ and so $a^2=p\cdot b^2$.
$a^2$ must be divisible by $p$ because $p, a^2,$ and $b^2$ are all integers and so $a$ must be divisible by $p$ as well. If we define $a=pc$ where $c$ is also an integer, then $p=\dfrac{p^2\cdot c^2}{b^2} \rightarrow 1 = \dfrac{p\cdot c^2}{b^2} \rightarrow b^2 = p \cdot c^2$.
$b^2$ must also be divisible by $p$ because $p, b^2,$ and $c^2$ are all integers and so $b$ must be divisible by $p$ as well. However, $a$ and $b$ were defined to be relatively prime and yet they have a common factor of $p$, which means that they are not relatively prime.
Thus, the claim that $\sqrt{p}$ can be written as the division of two relatively prime integers $a$ and $b$ is contradicted as $a$ and $b$ cannot be relatively prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integral $\lim _{n\to \infty}\int _0^1\sqrt{\frac{1}{x}+n^2x^{2n}}\,dx$ Evaluate $\displaystyle \lim \limits _{n\to \infty}\int \limits _0^1\sqrt{\frac{1}{x}+n^2x^{2n}}\,dx$.
The integral is non asymptotic. The convergence is non uniform at both $x=0$ and $x=1$. I'm not sure how to proceed.
| We have
\begin{align}
I_n &= \int_0^1 \sqrt{\frac{1}{x} + n^2 x^{2n}} \mathrm{d} x \\
&\le \int_0^1 \left(\sqrt{\frac{1}{x}} + nx^n\right) \mathrm{d} x \\
&\le \int_0^1 \left(\sqrt{\frac{1}{x}} + (n+1)x^n\right) \mathrm{d} x \\
&= \left(2\sqrt{x} + x^{n+1}\right)\Big\vert_0^1 \\
&= 3.
\end{align}
Also, we have
\begin{align}
I_n &= \int_0^{1-\frac{1}{\sqrt{n}}} \sqrt{\frac{1}{x} + n^2 x^{2n}}\ \mathrm{d} x
+ \int_{1-\frac{1}{\sqrt{n}}}^1 \sqrt{\frac{1}{x} + n^2 x^{2n}}\ \mathrm{d} x\\
&\ge \int_0^{1-\frac{1}{\sqrt{n}}} \sqrt{\frac{1}{x}}\ \mathrm{d} x
+ \int_{1-\frac{1}{\sqrt{n}}}^1 nx^n \mathrm{d} x \\
&= 2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right].
\end{align}
Combining the above results, we have
$$2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right] \le I_n \le 3.$$
Note that
$$\lim_{n\to \infty} 2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right] = 3.$$
By the squeeze theorem, we have
$\lim_{n\to \infty} I_n = 3$. We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Find $\sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$
For $\displaystyle\omega =
\exp\left({2\pi \over 15}\,\mathrm{i}\right),\quad$ find
$\displaystyle\ \sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$.
I tried to write $x^{15}-1$=$(x-1) (x-\omega).....(x-\omega^{14})$
And took log and differentiate thrice but it's very lenghty.
| Strigthforward approach:
First we combine $\frac{1}{(\omega^{k}-1)^3}$ with $\frac{1}{(\omega^{15-k}-1)^3}$ to get rid of $i$ in the denominator.
$$\sum\limits_{k=1}^{14}\frac{1}{(\omega^k-1)^3}=
\sum\limits_{k=1}^{7}\frac{(\omega^{15-k}-1)^3+(\omega^{k}-1)^3}
{\left((\omega^k-1)(\omega^{15-k}-1)\right)^3}=$$
$$\sum\limits_{k=1}^{7}\frac{
\left((\omega^{15-k}-1)+(\omega^{k}-1)\right)
\left(
(\omega^{15-k}-1)^2-(\omega^{15-k}-1)(\omega^{k}-1)+(\omega^{k}-1)^2
\right)
}
{\left(2-\omega^k-\omega^{15-k}\right)^3}=$$
$$-\sum\limits_{k=1}^{7}\frac{
(\omega^{15-k}-1)^2-(\omega^{15-k}-1)(\omega^{k}-1)+(\omega^{k}-1)^2
}
{\left(2-\omega^k-\omega^{15-k}\right)^2}=$$
$$-\sum\limits_{k=1}^{7}\frac{
\omega^{30-2k}-2\omega^{15-k}+1-2+\omega^k+\omega^{15-k}+\omega^{2k}-2\omega^{k}+1
}
{\left(2-\omega^k-\omega^{15-k}\right)^2}=$$
$\displaystyle-\sum\limits_{k=1}^{7}\frac{
\omega^{30-2k}+\omega^{2k}-\omega^{15-k}-\omega^{k}
}
{\left(2-\omega^k-\omega^{15-k}\right)^2}=$
$\displaystyle-\sum\limits_{k=1}^{7}\frac{\omega^{2k} + \omega^{k} + 1}{\omega^{2k} - 2 \omega^{k} + 1}
$
$=\displaystyle-7-3\sum\limits_{k=1}^{7}\frac{1}{\omega^{k} - 2 + \omega^{-k}}=$
$$-7-\frac32\sum\limits_{k=1}^{7}\frac{1}{\cos\frac{2k\pi}{15}-1}=
-7+\frac34\sum\limits_{k=1}^{7}\frac{1}{\sin^2\frac{k\pi}{15}}=$$
with a large overkill
$\displaystyle -7+\frac34\cdot\frac23\cdot 7 \cdot 8=21$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
How do I show that $\sum_{\text{cyc}}\frac{3a^2-2ab+3b^2}{(a+b)^2} < \frac{9}{4}(\frac{a}{c} + \frac{c}{a}) - \frac{3}{2}$? Let the real numbers $a,b,c \in \mathbb(0, \infty\ )$, $a\leq b\leq c$. Prove that $$\frac{3a^2-2ab+3b^2}{(a+b)^2}+\frac{3b^2-2bc+3c^2}{(b+c)^2}+ \frac{3c^2-2ac+3a^2}{(c+a)^2} \leq\frac{9}{4}(\frac{a}{c}+\frac{c}{a}) -\frac{3}{2}.$$
I tried to prove that $$\frac{3a^2-2ab+3b^2}{(a+b)^2}\leq\frac{3}{4}\left(\frac{a}{b}+\frac{b}{a}\right)-\frac{1}{2},$$ but it didn't work.
| By your work $$\sum_{cyc}\frac{3a^2-2ab+3b^2}{(a+b)^2}\leq\sum_{cyc}\left(\frac{3}{4}\left(\frac{a}{b}+\frac{b}{a}\right)-\frac{1}{2}\right)\leq\frac{9}{4}\left(\frac{a}{c}+\frac{c}{a}\right)-\frac{3}{2}$$
because $$\frac{a}{b}+\frac{b}{a}\leq\frac{a}{c}+\frac{c}{a}$$ (it's just $bc-a^2\geq0$) and
$$\frac{b}{c}+\frac{c}{b}\leq\frac{a}{c}+\frac{c}{a}$$ (it's just $c^2-ab\geq0$).
The first inequality we can prove by the following way:
$$\frac{3}{4}\left(\frac{a}{b}+\frac{b}{a}\right)-\frac{1}{2}-\frac{3a^2-2ab+3b^2}{(a+b)^2}=$$
$$=\frac{3a^2-2ab+3b^2}{4ab}-\frac{3a^2-2ab+3b^2}{(a+b)^2}=\frac{(a-b)^2(3a^2-2ab+3b^2)}{4ab(a+b)^2}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate logarithmic integral $\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$ I need help to evaluate the integral with the residue theorem:
$$\int_{-\infty}^{\infty}\frac{\ln{(x^2+a^2)}}{x^2+b^2}\,dx$$
where a,b>0 real numbers.
I think I could consider the contour integral where C is the half circle in the first two quadrant.
But I'm not sure how to continue.
Could someone help me?
| Let $ a,b $ be reals such that $ 0<b<a $, we have the following :
\begin{aligned}\int_{0}^{+\infty}{\frac{\ln{\left(1+a^{2}x^{2}\right)}}{1+b^{2}x^{2}}\,\mathrm{d}x}&=\int_{0}^{+\infty}{\int_{0}^{1}{\frac{a^{2}x^{2}}{\left(1+b^{2}x^{2}\right)\left(1+a^{2}x^{2}y\right)}\mathrm{d}y}\,\mathrm{d}x}\\&=\int_{0}^{+\infty}{\int_{0}^{1}{\left(\frac{1}{\left(y-\frac{b^{2}}{a^{2}}\right)\left(1+b^{2}x^{2}\right)}-\frac{1}{\left(y-\frac{b^{2}}{a^{2}}\right)\left(1+a^{2}x^{2}y\right)}\right)\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{0}^{+\infty}{\int_{0}^{1}{\frac{\mathrm{d}y\,\mathrm{d}x}{\left(y-\frac{b^{2}}{a^{2}}\right)\left(1+b^{2}x^{2}\right)}}}-\int_{0}^{+\infty}{\int_{0}^{1}{\frac{\mathrm{d}y\,\mathrm{d}x}{\left(y-\frac{b^{2}}{a^{2}}\right)\left(1+a^{2}x^{2}y\right)}}}\\ &=\left(\int_{0}^{+\infty}{\frac{\mathrm{d}x}{1+b^{2}x^{2}}}\right)\left(\int_{0}^{1}{\frac{\mathrm{d}y}{y-\frac{b^{2}}{a^{2}}}}\right)-\int_{0}^{1}{\frac{1}{y-\frac{b^{2}}{a^{2}}}\int_{0}^{+\infty}{\frac{\mathrm{d}x}{1+a^{2}x^{2}y}}\,\mathrm{d}y}\\ &=\frac{\pi}{2b}\ln{\left(\frac{a^{2}}{b^{2}}-1\right)}-\frac{\pi}{a}\int_{0}^{1}{\frac{\mathrm{d}y}{2\sqrt{y}\left(y-\frac{b^{2}}{a^{2}}\right)}}\\ &=\frac{\pi}{2b}\ln{\left(\frac{a^{2}}{b^{2}}-1\right)}-\frac{\pi}{a}\int_{0}^{1}{\frac{\mathrm{d}y}{x^{2}-\frac{b^{2}}{a^{2}}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \textrm{We substituted : }y=x^{2}\\ &=\frac{\pi}{2b}\ln{\left(\frac{a^{2}}{b^{2}}-1\right)}-\frac{\pi}{2b}\left(\int_{0}^{1}{\frac{\mathrm{d}y}{x-\frac{b}{a}}}-\int_{0}^{1}{\frac{\mathrm{d}y}{x+\frac{b}{a}}}\right)\\ &=\frac{\pi}{2b}\ln{\left(\frac{a^{2}}{b^{2}}-1\right)}-\frac{\pi}{2b}\ln{\left(\frac{\frac{a}{b}-1}{\frac{a}{b}+1}\right)}\\ \int_{0}^{+\infty}{\frac{\ln{\left(1+a^{2}x^{2}\right)}}{1+b^{2}x^{2}}\,\mathrm{d}x}&=\frac{\pi}{b}\ln{\left(1+\frac{a}{b}\right)}\end{aligned}
Thus : \begin{aligned}\int_{0}^{+\infty}{\frac{\ln{\left(a^{2}+x^{2}\right)}}{b^{2}+x^{2}}\,\mathrm{d}x}&=\frac{\ln{\left(a^{2}\right)}}{b}\int_{0}^{+\infty}{\frac{\frac{1}{b}\,\mathrm{d}x}{1+\left(\frac{x}{b}\right)^{2}}}+\frac{1}{b^{2}}\int_{0}^{+\infty}{\frac{\ln{\left(1+\frac{x^{2}}{a^{2}}\right)}}{1+\frac{x^{2}}{b^{2}}}\,\mathrm{d}x}\\ &=\frac{\pi\ln{a}}{b}+\frac{\pi}{b}\ln{\left(1+\frac{b}{a}\right)}\\ \int_{0}^{+\infty}{\frac{\ln{\left(a^{2}+x^{2}\right)}}{b^{2}+x^{2}}\,\mathrm{d}x}&=\frac{\pi\ln{\left(a+b\right)}}{b}\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3671716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $2z+p^2+qy+2y^2=0$ using Charpit's method Show by solving $2z+p^2+qy+2y^2=0$ using Charpit's method that $y^2[(x-a)^2+y^2+2z]=b$.
My efforts :
The given equation is
\begin{equation}
F=2z+p^2+qy+2y^2=0.
\end{equation}
Charpit's equation:
Charpit's auxiliary equations are
$ \dfrac{dp}{F_x+pF_z}=\dfrac{dq}{F_y+qF_z}=\dfrac{dz}{-pF_p-qF_q}=\dfrac{dx}{-F_p}=\dfrac{dy}{-F_q}$. This implies
$ \dfrac{dp}{2p}=\dfrac{dq}{3q+4y}=\dfrac{dz}{-2p^2-qy}=\dfrac{dx}{-2p}=\dfrac{dy}{-y}$
Taking 1 st and 4 th fraction we get,
$
\Longrightarrow \frac{d p}{2 p}=\frac{d x}{-2 p}
$
So that $, p=-x+a$.
Now taking 2 nd and 5 th fraction ,we get
\begin{align*}
& \frac{d q}{3 q+4 y}=\frac{d y}{-y} \\
\implies & \frac{d q}{d y}=-\frac{3 q+4 y}{y} \\
\implies & \frac{d q}{d y}+\frac{3}{y} q=-4
\end{align*}
Which is a linear equation in first order.
$
\therefore I . F=e^{\int \frac{3}{y} d y}=e^{3 \log y}=e^{\log y^{3}}=y^{3}
$.
Now, $ q y^{3}=\int(-4) y^{3} d y+b=-y^{4}+b$, where $b$ is a constant. This implies
$ q=-y+b/y^3 $.
Now putting the value of $p$ and $q$ in.
$\begin{array}{l}
d z=p d x+q d y \text { we get } d z=-x(a-x)dx+(b/y^3-y) dy \\
\Longrightarrow z=-(a-x)^2/2-b/(2y^2)-y^2/2+d
\end{array}
$,
where $d$ is a integration constant.
This solution is different from the solution what was given. How can I prove the result?
| $$2z+p^2+qy+2y^2=0 \tag 1$$
You found :
$$z=-\frac12(a-x)^2-\frac{b}{2y^2}-\frac12 y^2+d \tag 2$$
In order to check your result, put your result (2) into equation (1) :
$p=a-x$
$q=\frac{b}{y^3}$
$$2\left(-\frac12(a-x)^2-\frac{b}{2y^2}-\frac12 y^2+d\right)+(a-x)^2+\frac{b}{y^3}y+2y^2=2d$$
This is not $=0$. So your result is false in general (except if $d=0$ ). Thus
$$z=-\frac12(a-x)^2-\frac{b}{2y^2}-\frac12 y^2 \tag 3$$
$2z=-(a-x)^2-\frac{b}{y^2}-y^2$
$2z+(a-x)^2+y^2=\frac{-b}{y^2}$
$y^2[(x-a)^2+y^2+2z]=-b$
$b$ is an arbitrary constant. You can change it into any other constant. Thus you can change $-b$ into $b$ :
$$y^2[(x-a)^2+y^2+2z]=b \tag 4$$
Your solution (2) with $d=0$ is the expected solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3678660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factorizing symmetric expressions Factorize $$\sum_{\text{cyc}}a^4(b-c)$$ I used cyclic/symmetric method to find that $(a-b), (b-c), (c-a)$ are factors of the above equation but I am not able to completely factorize this. I used long division method but I found it quite lengthy so is there any shorter method than to apply long division method after find the three factors above. Cheers!
| $$\sum_{cyc}a^4(b-c)=a^4(b-c)+bc(b^3-c^3)-a(b^4-c^4)=$$
$$=(b-c)(a^4+b^3c+b^2c^2+bc^3-a(b^3+b^2c+bc^2+c^3))=$$
$$=(b-c)(a(a^3-b^3)+b^2c(b-a)+bc^2(b-a)+c^3(b-a))=$$
$$=(b-c)(a-b)(a^3+a^2b+ab^2-bc^2-b^2c-c^3)=$$
$$=(b-c)(a-b)((a-c)(a^2+ac+c^2)+b(a-c)(a+c)+(a-c)b^2)=$$
$$=(a-b)(a-c)(b-c)(a^2+b^2+c^2+ab+ac+bc).$$
Another way:
Easy to see that for $a=b$ our expression is equal to $0$, which gives a factor $a-b$.
The same words we can say about $a=c$ and about $b=c$.
Thus, we got a factor $$(a-b)(a-c)(b-c)$$ and the last factor should be something quadratic, cyclic and homogeneous, which says that it should be something symmetric.
Now, from the the coefficient before $a^4b$ we obtain something like the following:
$$(a-b)(a-c)(b-c)(a^2+b^2+c^2+k(ab+ac+bc))$$ and from here we can get the answer.
I prefer the first way for the Schur's polynomials of a big degree.
For example, factoring $$\sum_{cyc}(a^5b-a^5c)=(a-b)(a-c)(b-c)\sum_{cyc}\left(a^3+a^2b+a^2c+\frac{1}{3}abc\right)$$ much more better to get by the first way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving for an unknown $c$ in relation to a ratio of gamma functions I have been working with ratios of gamma functions and I am surprised how difficult it is to make even elementary conclusions. I am hoping it is just the learning curve.
Consider the following problem. Find a positive real $c$ for an integer $x > 1$ such that:
$$\frac{\Gamma(2x+3-c)}{\Gamma(2x+1)} = x^2$$
Here's my reasoning for why I am confident that a solution exists for each integer $x$.
$\dfrac{\Gamma(2x+3)}{\Gamma(2x+1)} = \dfrac{(2x+2)!}{(2x)!}=(2x+2)(2x+1) = 4x^2 + 6x + 2 > x^2$
I am completely at a loss how to tackle what appears to me to be such a simple use of the Gamma function.
I would greatly appreciate if someone can either solve this problem for some $x > 1$ or help me to understand the methods that could be applied to this problem to provide an estimate for $c$ with, ideally, an upper and lower bound.
For the estimate, I am looking for something more interesting than the trivial:
$$0 < c < 1$$
| One may use Newton's method to solve $\ln \Gamma(2x+3-c) = 2\ln x + \ln \Gamma(2x+1)$ to find $c$ (see [2]):
Choose the initial $c_0$,
$$c_{k+1} = c_k - \frac{\ln \Gamma(2x+3-c_k) - 2\ln x - \ln \Gamma(2x+1)}{-\Psi(2x+3-c_k)}, \ k=0, 1, 2, \cdots$$
where $\Psi(x) = (\ln \Gamma(x))'$ is the digamma function.
(Remark: $c \mapsto \ln \Gamma(2x+3-c)$ is a convex function.)
For bounds, one may use good bounds for the gamma function. For example, in the following, we give some simple bounds.
First, we deal with the case when $x > 2$. Clearly, $0 < c < 1$.
The equation is written as
$$\frac{\Gamma(2x+3-c)}{\Gamma(2x+3)} = \frac{x^2}{(2x+1)(2x+2)}$$
or
$$\frac{\Gamma(2x+3)}{\Gamma(2x+3-c)} = \frac{(2x+1)(2x+2)}{x^2}. \tag{1}$$
Recall Gautschi's inequality [1]: for $y > 0$ and $s \in (0, 1)$,
$$y^{1-s} < \frac{\Gamma(y+1)}{\Gamma(y+s)} < (y+1)^{1-s}.$$
By letting $y = 2x + 2$ and $s = 1-c$ in Gautschi's inequality, we have
$$(2x+2)^c < \frac{\Gamma(2x+3)}{\Gamma(2x+3-c)} < (2x+3)^c.\tag{2}$$
From (1) and (2), we have
$$(2x+2)^c < \frac{(2x+1)(2x+2)}{x^2} < (2x+3)^c$$
which gives
$$\frac{\ln (2x+1) + \ln (2x+2) - 2\ln x}{\ln (2x+3)} < c < \frac{\ln (2x+1) + \ln (2x+2) - 2\ln x}{\ln (2x+2)}.$$
Second, we deal with the case when $x = 2$. Clearly, $1 < c < 2$.
The equation is written as
$$\frac{\Gamma(2x+3-c)}{\Gamma(2x+2)} = \frac{x^2}{2x+1}$$
or
$$\frac{\Gamma(2x+2)}{\Gamma(2x+3-c)} = \frac{2x+1}{x^2}. \tag{3}$$
Using Gautschi's inequality, we get
$$1 + \frac{\ln 5 - 2\ln 2}{\ln 6} < c < 2 - \frac{2\ln 2}{\ln 5}.$$
Reference
[1] https://en.wikipedia.org/wiki/Gautschi%27s_inequality
[2] Folitse Komla Amenyou, "Properties and Computation of the Inverse of the Gamma function".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
On the nested square roots $\sqrt{1^2+\sqrt{2^2+\sqrt{3^2 ...+\sqrt{(n-1)^2+\sqrt{n^2}}}}}$ I saw this on quora:
Use induction to show that
$\sqrt{1^2+\sqrt{2^2+\sqrt{3^2
...+\sqrt{(n-1)^2+\sqrt{n^2}}}}}
\le 2
$.
My question is simpler.
If
$x_n
=\sqrt{1^2+\sqrt{2^2+\sqrt{3^2
...+\sqrt{(n-1)^2+\sqrt{n^2}}}}}
$,
how can $x_{n+1}$
be expressed in terms of
$x_n$?
No ellipsis
("...")
or iterations
are allowed.
This is all I've come up with,
but it doesn't seem to be
of much use.
Let
$\begin{array}\\
f_{n}(x)
&=\sqrt{1^2+\sqrt{2^2+\sqrt{3^2
...+\sqrt{n^2+x}}}}\\
f_{n+1}(x)
&=\sqrt{1^2+\sqrt{2^2+\sqrt{3^2
...+\sqrt{n^2+\sqrt{(n+1)^2+x}}}}}\\
&=f_n(\sqrt{(n+1)^2+x})\\
\end{array}
$
so if
$y
=\sqrt{(n+1)^2+x}
$,
$f_n(y)
=f_{n+1}(x)
$.
$y^2
=(n+1)^2+x
$,
so
$x
=y^2-(n+1)^2
$
or
$f_{n+1}(y^2-(n+1)^2)
=f_n(y)
$.
| [Not intended as a complete answer. It doesn't relate $x_n$ and $x_{n+1}$ as requested by OP, in part because that's not the "right" thing to look at.]
It is often harder to deal with inserting the variable just at the very extreme end of the nested roots. This makes intuitive sense in part because the function is so un-sensitive to changing just that value. E.g. I expect that $f_n (2^{2^n}) - f_n (0) < 1 $ (but am not fully certain).
It is generally more useful to define the recursive nature by "shifting" the terms, which can be done in the following way:
Let $g_n(x) = \sqrt{ (x+1)^2 + \sqrt{ (x+2)^2 + \sqrt{ \ldots + \sqrt{ (x+n)^2 }}}}$
Then, $ g_{n} (x) = \sqrt{ (x+1)^2 + g_{n-1}(x+1)}$.
We are asked to show that $g_n(0) \leq 2$.
Why is this possibly the "right" thing to look at?
One naive/obvious/intuitive/brute-force approach is to "square both sides, subtract terms, and repeat till we get everything". In the language of this notation, we can write it as:
WTS $g_n(0) \leq 2 $
$\Leftarrow g_{n-1} (1) \leq 2^2 - 1^2 = 3 $
$\Leftarrow g_{n-2} (2) \leq 3^2 - 2^2 = 5 $
$\Leftarrow g_{n-3} (3) \leq 5^2 - 3^2 = 16 $
$\Leftarrow g_{n-4} (4) \leq 16^2 - 4^2 = 240 $
$\Leftarrow \vdots $
$ \Leftarrow g_1 (n-1) \leq $ ??
Now, $g_1 (n-1) = n$, and it is clear that the RHS is huge (almost squaring each time), so the last inequality should almost always be true, especially in a hand-waving context.
Rigorizing this simply requires finding a good bound on the RHS. By staring and conjecturing really hard, one might come up with:
$g_1 (n-1) < n+1 $
$\Rightarrow g_2 (n-2) < \sqrt{ (n-1)^2 + n+1} < n$ for $n-2 \geq 0$
$\Rightarrow g_3 (n-3) < \sqrt{ (n-2)^2 + n} < n-1$ for $n-3 \geq 0$
$\Rightarrow \vdots $
$\Rightarrow g_{n-1} (1) < \sqrt{ 2^2 + 4} < 3$
$\Rightarrow g_n(0) < \sqrt{1^2 + 3 } = 2 $.
Note: It's somewhat surprising that we have a linear bound (though that makes the math nice). There are other possibilities that make use of the exponential growth.
To proceed by induction, one would have to guess that the strengthened hypothesis is
$g_n(m) < m+2$ for $ m \geq 0$.
This can be easily proven by inducting on $n$, and is essentially that reversed chain
The base case is $ g_1 (m) = m+1 < m+2 $.
The induction step is $ g_{n+1} (m) = \sqrt{ (m+1)^2 + g_n(m+1) } < \sqrt{(m+1)^2 + m+3 } < m+2. $
Hence $g_n(0) < 2 $.
There could be other ways to prove the problem by induction. For example, another strengthened induction approach would be to show that $ x_n < 2 - h(n)$, in which case we want to show that $x_{n+1} - x_n < h(n) - h(n+1) $.
This might be why studying how $x_{n+1}$ is related to $x_n$ could be helpful. However, as explained earlier, it's very hard to get a handle on this difference, because of how nested everything is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the limit of $ \frac{\tan^2x-x^2}{x^2\tan^2x} $ How to find the limit of $ \frac{\tan^2x - x^2}{x^2\tan^2x}$ as $x$ approaches $0$.
I don't know how to approach and starting differentiating the numerator and denominator many times which after the 4th time gave me the result $\frac{2}{3}$. I want to know if there is any good simplification which can lead me to a simple and quick solution. Please give some ideas.
Thank you.
| Note that as $x \to 0$, we have $\tan(x) = x + \dfrac{x^3}{3} + O(x^5)$ (this is really a 2 line calculation). So,
\begin{align}
\tan^2(x) &= x^2 + \dfrac{2x^4}{3} + O(x^6)
\end{align}
Hence,
\begin{align}
\dfrac{\tan^2x - x^2}{x^2 \tan^2x} &= \dfrac{\frac{2x^4}{3} + O(x^6)}{x^4 + O(x^6)} \\
&= \dfrac{\frac{2}{3} + O(x^2)}{1 + O(x^2)},
\end{align}
where I divided top and bottom by $x^4$. Thus, if we take the limit $x \to 0$, the answer is $\dfrac{2}{3}$.
By the way, observe that it is no coincidence you had to differentiate $4$ times, because as you can see in the above derivation, the Taylor polynomials of the numerator and denominator begin at the $x^4$ term, so it is precisely the ratio of these coefficients which you're calculating if you use something like L'Hopital's rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What is the value of $\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}$? Let $x,y,z$ are real numbers such that $x,y,z \geq -1$, $xyz\not= 0$ and
\begin{matrix} \sqrt{x+1} + \sqrt{y+2} +\sqrt{z+3}= \sqrt{y+1} +\sqrt{z+2} + \sqrt{x+3} \\ \sqrt{y+1} + \sqrt{z+2} +\sqrt{x+3} = \sqrt{z+1}+\sqrt{x+2} + \sqrt{y+3} \end{matrix}
What is the value of $\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}$ ?
Thanks for helping solving this problem.
EDIT: I have done
$(\sqrt{x+1} + \sqrt{y+2} +\sqrt{z+3})^2$=$ (\sqrt{y+1} +\sqrt{z+2} + \sqrt{x+3})^2$ and $( \sqrt{y+1} + \sqrt{z+2} +\sqrt{x+3})^2 = (\sqrt{z+1}+\sqrt{x+2} + \sqrt{y+3})^2$ but haven't recived any helpful things.
| The only answer that will satisfy the two equations involving square roots is that $$x=y=z$$. $$x$$,$$y$$,and$$z$$ can be any real numbers between -1 and 0 and above 0. If $$x=y=z$$, $$x-y=0$$, $$y-z=0$$, and $$z-x=0$$. So the result will be $$0/z+0/x+0/y$$ and since $$x,y, and z$$ do not equal zero. The resulting equation will equal zero $$0/z+0/x+0/y=0$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $|R_{n,0}| \leq \frac{|x|^{n+1}}{(n+1)(x+1)}$
The problem is: Prove that if $-1 < x \leq 0$, the the Taylor's remainder $R_{n,0}$ corresponding to $f(x) = \log(x+1)$ satisfies $$\displaystyle |R_{n,0}| \leq \frac{|x|^{n+1}}{(n+1)(x+1)} $$
First, I know that Taylor's remainder formula is $$\displaystyle R_{n,a} (x) = \frac{f^{(n+1)}(s)}{(n+1)!}(x-a)^{n+1}$$
Knowing that $$ f^{(n+1)}(x) = \frac{(-1)^n n!}{(x+1)^{n+1}}$$
So, for $\log(x+1)$, $$ R_{n,0}(x) = \frac{(-1)^n x^{n+1}}{(n+1)(s+1)^{n+1}} $$
Then, $$ |R_{n,0}|= |\frac{(-1)^n x^{n+1}}{(n+1)(s+1)^{n+1}}| \leq \frac{|(-1)^n| |x^{n+1}|}{|(n+1)||(s+1)^{n+1}|} = \frac{|x^{n+1}|}{|(n+1)||(s+1)^{n+1}|}$$
I got stuck here, because I don't see where to go from this. Even if I rename $s$ as $x$, I see that $\displaystyle \frac{|x|^{n+1}}{(n+1)(x+1)}$ does not limit $\displaystyle \frac{|x|^{x+1}}{(n+1)(x+1)^{n+1}}$.
Any suggestion/help?
| Here is an easy way
to get the series
for
$\ln(1-t)$.
$\sum_{k=0}^{n-1} t^k
=\dfrac{1-t^n}{1-t}
$
so
$\dfrac1{1-t}
=\sum_{k=0}^{n-1}t^k+\dfrac{t^n}{1+t}
$.
Integrating from
$0$ to $x$,
$\begin{array}\\
\int_0^x\dfrac1{1-t}dt
&=\sum_{k=0}^{n-1}\int_0^x t^kdx+\int_0^x \dfrac{t^n}{1-t}dt\\
\text{or}\\
-\ln(1-x)
&=\sum_{k=0}^{n-1}\dfrac{x^{k+1}}{k+1}+\int_0^x \dfrac{t^n}{1-t}dt\\
\end{array}
$
so
$\begin{array}\\
-\ln(1-x)
-\sum_{k=0}^{n-1}\dfrac{x^{k+1}}{k+1}
&=\int_0^x \dfrac{t^n}{1-t}dt\\
&=E_n(x)\\
\end{array}
$
$\begin{array}\\
E_n(x)
&=\int_0^x \dfrac{t^n}{1-t}dt\\
&\gt \int_0^x t^ndt\\
&=\dfrac{x^{n+1}}{n+1}\\
E_n(x)
&=\int_0^x \dfrac{t^n}{1-t}dt\\
&\lt \int_0^x \dfrac{t^n}{1-x}dt\\
&=\dfrac{x^{n+1}}{(n+1)(1-x)}\\
\end{array}
$
Put $n=1$.
Then
$E_1(x)
=-\ln(1-x)-x
$
and
$\dfrac{x^2}{2}
\lt E_1(x)
\lt \dfrac{x^2}{2(1-x)}
$.
If
$0 < x \le \frac12$,
then
$E_1(x)
\lt x^2
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find sum of the series $ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$ I have to find the sum of the given series $$ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$$
My attempt
$$ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$$
or,
$$ S=1 + \left( \frac{1}{3}-\frac{1}{2}+\frac{1}{5} \right)+\left(\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}\right)+\left(\frac{1}{11}-\frac{1}{6}+\frac{1}{13}\right)+\left( \frac{1}{15}-\frac{1}{8}+\frac{1}{17}\right)+ \cdots$$
or,
$$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{4n-1}-\frac{1}{2n}+\frac{1}{4n+1}\right)$$
or,
$$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{4n-1}-\frac{1}{4n}-\frac{1}{4n}+\frac{1}{4n+1}\right)$$
or,
$$S= 1+ \sum_{n=1}^\infty \left( \frac{1}{(4n-1)(4n)}-\frac{1}{(4n)(4n+1)} \right)$$
Since rearrrangement is allowed now, we have
$$S= 1+\sum_{n=1}^\infty \frac{1}{(4n-1)(4n)}-\sum_{n=1}^\infty\frac{1}{(4n)(4n+1)}$$
or,
$$ S = 1+ I_1 - I_2$$
But even after many trials, I was unable to find the value of either $I_1$ or $I_2$, any help regarding this will be much appreciated.
| Let $s_n$ be the sum of the first $n$ terms of the series. It is clear that if $s_{3n}$ converges, then the series converges to the same sum. $$\begin{align}
s_{3n}&=\sum_{k=1}^{2n}\frac1{2k-1}-\sum_{k=1}^n\frac1{2k}\\
&=\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1{2k}-\sum_{k=1}^{n}\frac1{2k}\\
&=\sum_{k=1}^{4n}\frac1k-\frac12\sum_{k=1}^{2n}\frac1{k}-\frac12\sum_{k=1}^{n}\frac1{k}
\end{align}$$
Can you continue from here? Use the standard approximation for $\sum\frac1k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the integral $\int_\Gamma(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz$ I need to calculate the integral
$$\int_\Gamma(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz$$
being $\Gamma=S_1\cap S_2$, given:
*
*$S_1=\{(x,y,z)\in\mathbb{R}:2x+2y+z=3\}$
*$S_2=\{(x,y,z)\in\mathbb{R}:z=9-x^2-y^2\}$
In my problem i'm asked to solve it using both direct integration and the Stokes Theorem. I've started trying the Stokes part, calculating the rotational of $F$:
$$F=((y^2-z^2),(z^2-x^2),(x^2-y^2))\Longrightarrow \text{rot}(F)=-2(y+z,x+z,x+y)$$
Now, I know that the normal vector $N$ is $(2,2,1)$ (because $S_1\subset\Gamma$ and that's $S_1$'s normal vector at any point).
So now I using Stokes I have that (after simplifying)
$$\int_\Gamma(y^2-z^2)dx+(z^2-x^2)dy+(x^2-y^2)dz=\frac{-2}{3}\iint_S(3x+3y+4z)d\sigma$$
I'm stucked here. I don't get what $S$ am I supposed to use in the double integral (I guess I must do some variable change).
For the direct integration part, i don't know where to start. I will thank any help.
Edit: I need two different solutions, one using direct integration, and another using Stokes Theorem.
| Direct approach:
Let's study the intersection by equating the $z$:
$$3-2x-2y=9-x^2-y^2$$
$$x^2+y^2-2x-2y=9-3=6$$
$$(x-1)^2+(y-1)^2=(2\sqrt2)^2$$
Hence I would use the parametrization $x=1+2\sqrt2\cos \theta, y = 1+2\sqrt2\sin \theta, z=-1-4\sqrt2\sin \theta - 4\sqrt2 \cos \theta.$
$$\frac{dx}{d\theta} = -2\sqrt2 \sin \theta, \frac{dy}{d\theta}= 2\sqrt2\cos \theta, \frac{dz}{d\theta}= -4 \sqrt2\cos \theta + 4\sqrt2\sin \theta.$$
\begin{align}
&\int_{\Gamma} (y^2-z^2) \, dx + (z^2-x^2) \, dy + (x^2-y^2) \, dz \\
&= \int_0^{2\pi}y^2\left( \frac{dx}{d\theta} - \frac{dz}{d\theta} \right) +z^2\left( \frac{dy}{d\theta} - \frac{dx}{d\theta} \right) + x^2 \left( \frac{dz}{d\theta} - \frac{dy}{d\theta} \right)\, d\theta \\
&= \int_0^{2\pi} \left( 1+4\sqrt2\sin \theta + 8\sin^2 \theta\right)\left( -6\sqrt2\sin \theta + 4\sqrt2\cos \theta\right) \\
&+\left( 1+32 \sin^2 \theta + 32 \cos^2 \theta + 8\sqrt2 \sin \theta + 8\sqrt2 \cos \theta + 32 \sin \theta \cos \theta\right)\left( 2\sqrt2\cos \theta + 2\sqrt2\sin \theta\right) \\
&+\left( 1+4\sqrt2\cos \theta + 8\cos^2 \theta\right)\left( -6\sqrt2\cos \theta + 4\sqrt2\sin\theta\right) \, d\theta \\
&= \int_0^{2\pi} -16 \sin^2 \theta -16 \cos^2 \theta \, d\theta \\
&=\int_0^{2\pi} -16 \, d\theta \\
&= -32 \pi
\end{align}
Remark: There is no need to work out the details in the middle as we know that for $\{\sin^3 \theta, \cos^3 \theta, \sin^2 \theta \cos \theta, \sin \theta \cos^2 \theta, \sin \theta \cos \theta, \sin \theta, \cos \theta\}$, they vanish when we integrate form $0$ to $2 \pi$. We just have to keep track of $\sin^2 \theta$ and $\cos^2 \theta$.
Alternatively, we can use the Stoke's theorem,
$$curl(F)=-2(y+z, x+z, x+y)$$
The normal vector is $(2,2,1)$.
The dot product is $-2(3x+3y+4z)=-2(3x+3y+4(3-2x-2y))=-2(12-5x-5y)$.
Let's integrate using the parametrization $x=1+r\cos \theta, y = 1+r \sin \theta, 0 \le \theta \le 2\pi, 0 \le r \le 2\sqrt2$:
\begin{align}
&\int_{\Gamma} (y^2-z^2) \, dx + (z^2-x^2) \, dy + (x^2-y^2) \, dz \\
&= -2\int_0^{2\sqrt2} \int_0^{2\pi} (12-5(1+r\cos \theta)-5(1+r\sin \theta))r\, d\theta \,dr \\
&= -2 \int_0^{2\sqrt2} \int_0^{2\pi} (2r-5r^2\cos \theta-5r^2\sin \theta)\, d\theta \,dr \\
&= -2 \int_0^{2\sqrt2} \int_0^{2\pi} (2r)\, d\theta \,dr \\
&= -4\pi \int_0^{2\sqrt2} 2r \, dr \\
&= -32 \pi
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}$ How do I show convergence/divergence of the series
$$\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}?$$
I begin by writing $\left(\cos\frac{1}{n}\right)^{n^3} = e^{n^3\ln\left(\cos\frac{1}{n}\right)}$ and continue by Taylor expanding around $0$; first cosine, then ln. But I get nowhere. I would appreciate any help.
| The ratio test works fine using your way
$$a_n=\cos ^{n^3}\left(\frac{1}{n}\right)\implies \log(a_n)=n^3\log\left(\cos \left(\frac{1}{n}\right) \right)$$
$$\cos \left(\frac{1}{n}\right)=1-\frac{1}{2 n^2}+\frac{1}{24 n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log\left(\cos \left(\frac{1}{n}\right) \right)=-\frac{1}{2 n^2}-\frac{1}{12 n^4}+O\left(\frac{1}{n^6}\right)$$
$$\log(a_n)=n^3\left(-\frac{1}{2 n^2}-\frac{1}{12 n^4}+O\left(\frac{1}{n^6}\right) \right)=-\frac{n}{2}-\frac{1}{12 n}+O\left(\frac{1}{n^3}\right)$$ Now, apply twice and continue with Taylor series
$$\log(a_{n+1})-\log(a_n)=-\frac{1}{2}+\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1}}{a_{n}}=e^{\log(a_{n+1})-\log(a_n) }=\frac{1}{\sqrt{e}}\left(1+\frac{1}{12 n^2}\right)+O\left(\frac{1}{n^3}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find $n$ and $m$, if $n,m$ are natural numbers, such that $m^6 + 279 = 2^n$. Suppose $n, m \in \mathbb{N}$. If $m^{6}+279=2^{n}$, find $n, m$.
What are some ways to approach this question? Instinct told me to take logarithm but doesn't work to well.
| Suppose that $n$ is odd. Then
$$2=2*1^k\equiv 2*4^k=2^{2k+1}=2^n=m^6+279\equiv m^6=m^2*m^2*m^2(\text{mod }3)$$
If $m$ is divisible by $3$, then this is $0$ modulo $3$. If not, then
$$\equiv 1*1*1=1\ (\text{mod }3)$$
Since this is a contradiction either way, we conclude $n$ is even. Then we can simplify the equation to
$$m^6+279=4^k$$
for some $k\in\mathbb{N}$. Rearranging, we get
$$279=4^k-m^6=(2^k)^2-(m^3)^2=(2^k-m^3)(2^k+m^3)$$
Now, the prime factorization of $279$ is
$$279=3*3*31$$
This means that these prime numbers ($3$ twice and $31$ once) and only these prime numbers can divide
$$(2^k-m^3)(2^k+m^3)$$
Now, we have that
$$31>9=3*3$$
$$2^k+m^3>2^k-m^3$$
Thus, we know
$$31|(2^k+m^3)$$
However, since the other divisors are either $3$ or $9$, we may conclude
$$2^k+m^3\in\{31,63,279\}$$
Since $k>9$ and $m>7$ imply
$$2^k+m^3>279$$
we only need to check a finite number of combinations for possible solutions. In fact, we find one candidate:
$$2^2+3^3=31$$
where $k=2$ and $m=3$. If we plug these into the other factor, we get
$$2^k-m^3=2^2-3^3=4-27=-23$$
Since this is not divisible by $3$, we may conclude that the original equation has no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
calculate $\sum_{n=0}^\infty \frac{3^n}{n!(n+3)}$ using power series let $f(x)=\frac{e^x-1-x-\frac{x^2}{2}}{x}$, because $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$, $f$ can be expressed as $$f(x) = \frac{\sum_{n=0}^\infty \frac{x^n}{n!}-1-x-\frac{x^2}{2}}{x}=\frac{\sum_{n=3}^\infty \frac{x^n}{n!}}{x}=\sum_{n=0}^\infty \frac{x^{n+2}}{(n+3)!}$$
the power series converge in $(-\infty, \infty)$ because $\lim_{n\to\infty} \sqrt[n]{\frac{1}{(n+3)!}}=0$ and let $f_n(x) = \frac{x^{n+2}}{(n+3)!} \Longrightarrow f'_n(x) = \frac{x^{n+1}}{(n+1)!(n+3)}$, $\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)!(n+3)}= \sum_{n=0}^\infty f'_n(x)$ also converge in $(-\infty, \infty)$ (for the same reason), hence $$f'(x) = \sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)!(n+3)}$$ by repeating this process once more I get $$f''(x) = \sum_{n=0}^\infty \frac{x^n}{n!(n+3)}$$ and if $x=3$ I get $$\sum_{n=0}^\infty \frac{3^n}{n!(n+3)} = f''(3)$$ which is what was looking for. my problem is that $f$ isn't defined for $x=0$ yet the series does converge for it as $\sum_{n=0}^\infty \frac{0^n}{n!(n+3)}=0$, so was the function $f$ I used wrong? or could it be that I can't differentiate $f$ the way I did?
| Try using the following instead so you avoid that mess:
$$x^2e^x=\sum_{n=0}^{\infty} \frac{x^{n+2}}{n!}$$
$$\int x^2e^x \; dx=x^2e^x-2xe^x+2e^x+C=\sum_{n=0}^{\infty} \frac{x^{n+3}}{n!(n+3)}$$
At $x=3$:
$$9e^3-6e^3+2e^3-2=5e^3-2=\sum_{n=0}^{\infty} \frac{3^{n+3}}{n!(n+3)}$$
Notice that the series on the right is what you're looking for but multiplied by $3^3$, so divide both sides by $27$.
I understand that this approach is slightly different than yours, but I believe this approach is faster and easier to understand so I thought you might appreciate it. I'm sure you can refer to other answers posted here regarding your confusion with $f(0)$.
$$\boxed{\frac{5e^3-2}{27}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse [CONT..]
Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ such that all lines touch a standard ellipse. Find $e$ for the ellipse
Let the point $(h,k)$ lie on the given circle
The chord of the contact drawn to the given ellipse is
$$\frac{hx}{4}+\frac{ky}{9}-1=0$$
This line is coincident with the the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$$y=mx\pm \sqrt{a^2m^2+b^2}$$
Then comparing the two equations
$$m=\frac{-9h}{4k}$$
And $$\frac{81}{k^2}=a^2m^2+b^2$$
$$\frac{81}{k^2}=\frac{81a^2h^2}{16k^2}+b^2$$
$$(81)(16)=81a^2h^2+16k^2b^2$$
How do I proceed from here? Simply substituting $h^2=100-k^2$ doesn’t give any details for $a$ and $b$
| According to $\frac{hx}{4}+\frac{ky}{9}=1$, the horizontal and vertical lines corresponding to circular points $(h,k) = (0,10),\> (10,0)$ are $ y= \frac9{10}$, $ x = \frac4{10}$. which also corresponds to the elliptical axes $a= \frac4{10}$ and $b= \frac9{10}$. Thus, the equation of the standard ellipse is
$$\frac{x^2}{(\frac4{10})^2}+\frac{y^2}{(\frac9{10})^2}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\mathrm{tr}(AX^{-1}A'X)$ Given $A=\begin{bmatrix}
a_1 & & \\
& a_2 & \\
&&a_2
\end{bmatrix}$ and $B\in\mathbb{R}^{3\times2}$, where $a_1,a_2>1$, and $X=AX(I+BB'X)^{-1}A$.
1) Find $\mathrm{tr}(AX^{-1}AX)$ in terms of $a_i$.
2) Show that $\frac{x_1\cdot\begin{vmatrix}
x_2 & x_{23}\\
x_{23}&x_3
\end{vmatrix}}{|X|}$ is independent of $B$, where $|X|=\mathrm{det}X$.
My attempt: $X$ is the solution to discrete-time algebraic Riccati equation.
Let $X=\begin{bmatrix}
x_1 & x_{12} & x_{13}\\
x_{12} & x_2 & x_{23}\\
x_{13}&x_{23}&x_3
\end{bmatrix}$ and $z=x_1\cdot\begin{vmatrix}
x_2 & x_{23}\\
x_{23}&x_3
\end{vmatrix}$, then I found that
\begin{align}
\mathrm{tr}(AX^{-1}AX)
=(a_1-a_2)^2\frac{z}{|X|}+(2a_1a_2+a_2^2).
\end{align}
I did simulation by fixing $A$ and found out that $\frac{z}{|X|}$ doesn't depend on $B$, thus $\mathrm{tr}(AX^{-1}AX)$ also doesn't depend on $B$ (as long as $B$ is full rank).
Observation 1: $A=\begin{bmatrix}
7 & & \\
& 2 & \\
&&2
\end{bmatrix}$ and $A=\begin{bmatrix}
2 & & \\
& 7 & \\
&&7
\end{bmatrix}$ has equal $\frac{z}{|X|}$.
Observation 2: The larger the gap between $a_1$ and $a_2$, the smaller $\frac{z}{|X|}$ and vice-versa.
| Assume that $\det X \ne 0$.
From $X = AX(I + BB^\mathsf{T}X)^{-1}A$,
we have $I + BB^\mathsf{T}X = AX^{-1}AX$
and $BB^\mathsf{T} = AX^{-1}A - X^{-1}$.
Let $Y = X^{-1}$. We have $BB^\mathsf{T} = AYA - Y$.
Let
$$Y=\begin{bmatrix}
y_1 & y_{12} & y_{13}\\
y_{12} & y_2 & y_{23}\\
y_{13}&y_{23}&y_3
\end{bmatrix}, \quad
B = \begin{bmatrix}
b_1 & b_2 \\
b_3 & b_4 \\
b_5 & b_6
\end{bmatrix}.
$$
From $BB^\mathsf{T} = AYA - Y$, we solve $Y$ uniquely
\begin{align}
y_1 &= \frac{b_1^2 + b_2^2}{a_1^2 - 1}, \quad y_{12} = \frac{b_1b_3 + b_2b_4}{a_1a_2 - 1}, \quad
y_{13} = \frac{b_1b_5 + b_2 b_6}{a_1a_2 - 1}, \\
y_2 &= \frac{b_3^2 + b_4^2}{a_2^2 - 1}, \quad y_{23} = \frac{b_3b_5 + b_4b_6}{a_2^2-1}, \quad
y_3 = \frac{b_5^2+b_6^2}{a_2^2-1}.
\end{align}
By using the well-known relation $M^{-1} = \frac{1}{\det M} \mathrm{adj}(M)$, from $Y = X^{-1}$ and $X = Y^{-1}$, we have
\begin{align}
y_1 &= \frac{\begin{vmatrix}
x_2 & x_{23}\\
x_{23}&x_3
\end{vmatrix}}{\det X}, \\
x_1 &= \frac{\begin{vmatrix}
y_2 & y_{23}\\
y_{23}&y_3
\end{vmatrix}}{\det Y}.
\end{align}
Thus, we have
$$\frac{z}{\det X} = y_1x_1 = y_1 \frac{\begin{vmatrix}
y_2 & y_{23}\\
y_{23}&y_3
\end{vmatrix}}{\det Y} = \frac{(a_1a_2-1)^2}{(a_1-a_2)^2}.$$
Then, we have
$$\mathrm{Tr}(AX^{-1}AX) = (a_1-a_2)^2\frac{z}{\det X}+(2a_1a_2+a_2^2) = a_1^2a_2^2 + a_2^2 + 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that $\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$ Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that,
$$\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$$
Well from their sum we do get that $\frac{1}{4} \geq \sqrt[4]{abcd}$
$$\Rightarrow \frac{1}{4^8} \geq a^2b^2c^2d^2$$
and applying AM-GM on LHS of the given inequality we get ,
$$\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq 4 \cdot \sqrt[4]{\frac{a^3b^3c^3d^3}{(a+b)(b+c)(c+d)(d+a)}}$$
and $(a+b)(b+c)(c+d)(d+a) \geq 16 \cdot abcd$ or
$$\frac{a^3b^3c^3d^3}{(a+b)(b+c)(c+d)(d+a)} \leq 16 \cdot a^2b^2c^2d^2 \leq 4^2 \cdot \frac{1}{4^8}= \frac{1}{4^6}$$
$$\Rightarrow \frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq 4 \cdot\frac{1}{8}=\frac{1}{2} > \frac{1}{8} \blacksquare.$$
Is this proof correct ? Did i miss any details? My doubt really stems from the fact that i didnt get $\frac{1}{8}$ directly but $\frac{1}{2}$, which makes my resultant inequality strict instead of being $\geq$ and it makes me wonder whether my proof is right. Thanks.
EDIT: Well guys i haven't read Titu's Lemma or Holder's inequality just yet though both of them do seem very powerful. I guess i'll just come to this question later when m done with those topics. Thanks for your help. Also I was just wondering whether it is possible to do it purely using AM-GM or maybe WAM-WGM ? Thanks again.
| Credits to Calvin Lin for the hint to apply Titu's Lemma, but I don't find the remaining part of the proof obvious.
Applying Titu's lemma yields:
\begin{align*}
\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} &= \frac{a^4}{ab+ac}+\frac{b^4}{bc+bd}+\frac{c^4}{cd+ca}+\frac{d^4}{da+db} \\
&\geq \frac{(a^2 + b^2 + c^2 + d^2)^2}{ab + ac + bc + bd + cd + ca + da + db}
\end{align*}
Applying Cauchy-Schwarz Inequality to the denominator yields:
\begin{align*}
ab + ac + bc + bd + cd + ca + da + db &\leq \sqrt{\left(2a^2 + 2b^2 + 2c^2 + 2d^2 \right)^2} \\
&= 2(a^2 + b^2 + c^2 + d^2)
\end{align*}
Therefore:
\begin{align*}
\frac{(a^2 + b^2 + c^2 + d^2)^2}{ab + ac + bc + bd + cd + ca + da + db} \geq \frac{1}{2}(a^2 + b^2 + c^2 +d^2)
\end{align*}
Applying Cauchy-Schwarz inequality again with $(a^2,b^2,c^2,d^2) \cdot (1,1,1,1)$ yields::
\begin{align*}
4(a^2 + b^2 + c^2 + d^2) \geq (a + b + c + d)^2 = 1
\end{align*}
Therefore:
\begin{align*}
\frac{1}{2}(a^2 + b^2 + c^2 +d^2) \geq \frac{1}{2}\left(\frac{1}{4}\right) = \frac{1}{8}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compositeness test using $S_k=2S_{k-1}-3S_{k-2}+2S_{k-3}$ recurrence relation Can you prove or disprove the following claim:
Let $S_k=2S_{k-1}-3S_{k-2}+2S_{k-3}$ with $S_0=0$ , $S_1=1$ , $S_2=1$ . Let $n$ be an odd natural number greater than $2$. Let $\left(\frac{D}{n}\right)$ be the Jacobi symbol where $D$ represents the discriminant of the characteristic polynomial $x^3-2x^2+3x-2$, and let $\delta(n)=n-\left(\frac{D}{n}\right)$ , then:
$$\text{If } n \text{ is a prime then } S_{\delta(n)} \equiv 0 \pmod{n}$$
You can run this test here. Note that $D=-28$ .
I have verified this claim for all $n$ up to $200000$ . I was searching for counterexample using the following PARI/GP code:
rec(m,P,Q,R)={s0=0;s1=1;s2=1;l=3;while(l<=m,s=P*s2+Q*s1+R*s0;s0=s1;s1=s2;s2=s;l++);return(s);}
RPT(n1,n2)={forprime(n=n1,n2,d=n-kronecker(-28,n);if(Mod(rec(d,2,-3,2),n)!=0,print(n);break))}
P.S.
Wolfram Alpha gives the following closed form for $S_k$ : $$S_k=\frac{i2^{-k} \cdot \left(\left(1-i\sqrt{7}\right)^k-\left(1+i\sqrt{7}\right)^k\right)}{\sqrt{7}}$$
| The roots of $x^3-2x^2+3x-2=0$ are ${1, \frac{1+\sqrt{-7}}{2}, \frac{1-\sqrt{-7}}{2}}$
Let $\alpha = \frac{1+\sqrt{-7}}{2},\quad \beta = \frac{1-\sqrt{-7}}{2}$
Thus, the formula can be written as $S_n = A\cdot1^n + B\cdot\alpha^n + C\cdot\beta^n$ with $A,B,C.$
We can determine (A,B,C) using $S_0,S_1,S_2$ below
$A = \frac{-(\beta-1+\alpha)}{(-1+\beta)(-1+\alpha)}=0$
$B = \frac{\beta}{-\beta+\beta\alpha+\alpha-\alpha^2}=\frac{1}{\sqrt{-7}}$
$C = \frac{-\alpha}{(-\alpha+\beta)(-1+\beta)}=\frac{-1}{\sqrt{-7}}$
Hence we get $S_n = \frac{1}{\sqrt{-7}}((\frac{1+\sqrt{-7}}{2})^n-(\frac{1-\sqrt{-7}}{2})^n)$
Let $p$ be an odd prime.
\begin{align}
S_{p+1} &= \frac{1}{\sqrt{-7}} \left(\middle(\frac{1+\sqrt{-7}}{2}\middle)^{p+1}-\middle(\frac{1-\sqrt{-7}}{2}\middle)^{p+1} \right)\\
&= \frac{1}{2^{p+1}}\frac{1}{\sqrt{-7}} \left(\middle(1+\sqrt{-7}\middle)^{p}\middle(1+\sqrt{-7}\middle)-\middle(1-\sqrt{-7}\middle)^{p}\middle(1-\sqrt{-7}\middle)\right)\\
\end{align}
Since coefficients $\binom{p}{k}$ with $1\leqq k \leqq p-1$ are divisible by $p$, then we get,
$2^{p}S_{p+1}\equiv(-7)^{\frac{p-1}{2}}+1 \pmod{p}$
Using $(-7)^{\frac{p-1}{2}}\equiv ({\frac{-7}{p}}) \pmod{p},$ then
$2^{p}S_{p+1}\equiv({\frac{-7}{p}})+1 \pmod{p}$
Hence if $({\frac{-7}{p}})=-1$ then $S_{p+1} \equiv 0 \pmod{p}$
$$ $$
$$ $$
$$ $$
\begin{align}
S_{p-1} &= \frac{1}{\sqrt{-7}} \left(\middle(\frac{1+\sqrt{-7}}{2}\middle)^{p-1}-\middle(\frac{1-\sqrt{-7}}{2}\middle)^{p-1} \right)\\
&= \frac{1}{2^{p-1}}\frac{1}{\sqrt{-7}} \left(\middle(1+\sqrt{-7}\middle)^{p}\middle(1+\sqrt{-7}\middle)^{-1}-\middle(1-\sqrt{-7}\middle)^{p}\middle(1-\sqrt{-7}\middle)^{-1}\right)\\
\end{align}
Since $(1+\sqrt{-7})^{-1}=(1-\sqrt{-7}/8$, $(1-\sqrt{-7})^{-1}=(1+\sqrt{-7}/8$, we get
$2^{p+1}S_{p-1}\equiv(-7)^{\frac{p-1}{2}}-1 \pmod{p}$
Using $(-7)^{\frac{p-1}{2}}\equiv ({\frac{-7}{p}}) \pmod{p},$ then $2^{p+1}S_{p-1}\equiv({\frac{-7}{p}})-1 \pmod{p}$
Hence if $({\frac{-7}{p}})=1$ then $S_{p-1} \equiv 0 \pmod{p}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can I solve $f(x) = \int \frac{\cos{x}(1+4\cos{2x})}{\sin{x}(1+4\cos^2{x})}dx$? $$f(x) = \int \frac{\cos{x}(1+4\cos{2x})}{\sin{x}(1+4\cos^2{x})}dx$$
I have been up on this problem for an hour, but without any clues.
Can someone please help me solving this?
| It is quite straight forward after rewriting the integrand:
$$\frac{\cos x (1+4\cos 2x )}{\sin x (1+4\cos^2 x)}= \frac{\cos x}{\sin x}\left(1 - \frac{4\sin^2 x}{1+4\cos^2 x}\right)$$
$$= \frac{\cos x}{\sin x} - \frac{2\sin 2x}{3+2\cos 2x}$$
Note, that $(3+2\cos 2x)' = -4\sin 2x$ and integrate:
$$\int \frac{\cos{x}(1+4\cos{2x})}{\sin{x}(1+4\cos^2{x})}dx =\log |\sin x| + \frac 12 \log(3+2\cos 2x) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Sum of Squares for $a^2+b^2+c^2+d^2+abcd+1\ge ab+bc+cd+da + ac+bd$ $\color{red}{\textrm{Update}}$
To clarify: I hope to see a (simple) SOS solution similar to the one of the following 3-variable problem, no other assumptions e.g. $a\ge b\ge c\ge d$.
Let us see an example. In the link below, by letting $y = \mathrm{mid}(x, y, z)$,
two SOS solutions are given. They are not what I want. It is more difficult to get a SOS solution without the assumption $y = \mathrm{mid}(x, y, z)$ for that problem.
Prove $2\left(x^2+y^2+z^2+1)(x^3y+y^3z+z^3x+xyz\right) \le \left(x^2+y^2+z^2+3xyz\right)^2.$
Another example: Mongolian TST 2008, day 2 problem 3, Ji Chen gave a SOS solution with the assumption $x = \min(x, y, z)$.
Actually, I gave a SOS solution without any assumption.
https://artofproblemsolving.com/community/c6h205316p11219067
$\phantom{2}$
Problem 1. Let $a, b, c, d \ge 0$. Prove that $a^2+b^2+c^2+d^2+abcd+1\ge ab+bc+cd+da + ac+bd$.
There are quite a few solutions (including my solution). Here, I am particularly interested in $\color{blue}{\textrm{(simple) Sum of Squares (SOS) solutions}}$.
Denote $f(a, b, c, d) = \mathrm{LHS} - \mathrm{RHS}$.
Then, $f(x^2, y^2, z^2, w^2) \ge 0$ for all real numbers $x, y, z, w$.
However, $f(x^2, y^2, z^2, w^2)$ may not be expressed as SOS (polynomial).
I found that $(x^2+y^2+z^2+w^2)^2f(x^2, y^2, z^2, w^2)$ may be expressed as SOS,
since numerically $(x^2+y^2+z^2+w^2)^2f(x^2, y^2, z^2, w^2) \approx u^\mathsf{T}Q u$ where $Q$ is $185\times 185$ matrix
and $u$ is a vector containing monomials in $x, y, z, w$. But $Q$ is quite large, I have not yet proceeded.
Any comments and solutions are welcome.
Relevant information
For 3-variable problem below, there are quite a few SOS solutions.
Problem 2. Let $a, b, c\ge 0$. Prove that $a^2+b^2+c^2+2abc+1 \ge 2(ab+bc+ca)$.
My SOS solution is
\begin{align}
&a^2+b^2+c^2+2abc+1 - 2(ab+bc+ca) \\
=\ & \frac{1}{2(a+b)^2}\Big[(a^2-ac-b^2+bc-a+b)^2 +(a^2-2ab-ac+b^2-bc+a+b)^2\\
&\qquad\qquad\quad + 4ab(a-b)^2 + 4ab(c-1)^2 + 4abc(a+b-2)^2\Big].
\end{align}
FYI, my non-SOS solution for Problem 1:
WLOG, assume that $a\ge b\ge c\ge d$.
If $cd \ge 1$, then $ab \ge 1$ and $(ab-1)(cd-1) \ge 0$, i.e., $ab+cd \le abcd+1$.
Also, $bc + da + ac + bd \le \frac{b^2+c^2}{2} + \frac{d^2+a^2}{2}
+ \frac{a^2+c^2}{2} + \frac{b^2+d^2}{2} = a^2+b^2+c^2+d^2$. Add them up to get the desired result.
If $cd < 1$, let $u = a - \frac{c+d}{cd+1}, \ v = b - \frac{c+d}{cd+1}$. We have
\begin{align}
\mathrm{LHS} - \mathrm{RHS} &= u^2 + v^2 - (1-cd)uv +\frac{c^3d+cd^3-c^2d^2-2cd+1}{cd+1}\\
&\ge 2|uv| - (1-cd) |uv| + \frac{c^3d+cd^3-c^2d^2-2cd+1}{cd+1}\\
&\ge \frac{c^3d+cd^3-c^2d^2-2cd+1}{cd+1}\\
&\ge \frac{2c^2d^2-c^2d^2-2cd+1}{cd+1}\\
&= \frac{(cd-1)^2}{cd+1}\\
&\ge 0.
\end{align}
(Q.E.D.)
| A very nice SOS solution by KaiRain@AoPS:
We have
\begin{align*}
&a^2+b^2+c^2+d^2+abcd+ 1 - ( ab+bc+cd+da + ac+bd)\\
=\,& \frac{\left(2a^2-2ab+2b^2-ac-ad-bc-bd+2 \right)^2+3 \left(a-b \right)^2 \left(c-d \right)^2 }{4 \left(a^2-ab+b^2+1 \right)}\\
&\quad + \frac{abcd \left(a-b \right)^2+cd \left(ab-1 \right)^2 +\left(c-d \right)^2}{a^2-ab+b^2+1 }.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can any cyclic polynomial in $a, b, c$ be expressed in terms of $a^2b+b^2c+c^2a$, $a+b+c$, $ab+bc+ca$ and $abc$? Problem. Let $f(a,b,c)$ be a cyclic polynomial in $a, b, c$.
Can $f$ always be expressed as $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)$ for some polynomial $g(p, q, r, Q)$?
Motivation: If we want to prove $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)\ge 0$, and $g(p, q, r, Q)$ is non-increasing with $Q$,
by using the known inequality $a^2b + b^2c + c^2a \le \frac{4}{27}(a+b+c)^3 - abc$(see How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$),
sometimes, we may prove $g(a+b+c, ab+bc+ca, abc, \frac{4}{27}(a+b+c)^3 - abc)\ge 0$.
As a result, we deal with a symmetric inequality rather than the original cyclic inequality.
Is it known? Is it easily to prove?
Let us see some examples. Denote $p = a+b+c, q = ab+bc+ca, r = abc$ and $Q = a^2b+b^2c+c^2a$.
1) $ab^2 + bc^2 + ca^2 = (a+b+c)(a^2+b^2+c^2) - a^3-b^3-c^3 - (a^2b+b^2c+c^2a)$
2) From the known identity $a^3b+b^3c+c^3a +(ab+bc+ca)^2 = (a+b+c)(a^2b+b^2c+c^2a+abc)$, we have
$a^3b+b^3c+c^3a = p(Q + r) - q^2$.
3) $a^3b^2+b^3c^2+c^3a^2 = Qq - r(p^2-2q) - rq$.
4) $(a^2+b)(b^2+c)(c^2+a) = \cdots$
Any comments and solutions are welcome.
| Let $R$ be the ring generated by the symmetric polynomials and $f=a^2b+b^2c+c^2a$.
Certainly $f$ also contains $g=ab^2+bc^2+ca^2$ since $f+g$ is symmetric.
Let
$$u_{m,n}=a^mb^n+b^mc^n+c^ma^n.$$
To prove that $R$ contains all cyclic polynomials,
it suffices to prove that $u_{m,n}\in R$ for all $m$ and $n$. Indeed it
suffices to prove this when $m>n\ge1$.
Then,
$$(a+b+c)u_{m,n}=u_{m+1,n}+u_{m,n+1}+abc u_{m-1,n-1}\tag1$$
(when $n\ge1$)
and
$$(ab+bc+ca)u_{m,n}=u_{m+1,n+1}+abcu_{m-1,n}+abc u_{m,n-1}\tag2$$
(when $n\ge1$).
If we proceed by induction on $m+n$, (2) shows that if all $u_{m,n}\in R$
for $r\le k$ then so do all $u_{m,n}$ with $m+n=k+1$ save perhaps $u_{k,1}$.
But applying (1) for $(m,n)=(k-1,1)$ gives
$$u_{k,1}=(a+b+c)u_{k-1,1}-u_{k-1,2}-abcu_{k-2,0}.$$
As we have $u_{k-2,2}\in R$, then $u_{k,1}\in R$ too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Show that $\int_{0}^{\pi/2} (\frac{1}{\sin^3{\theta}} - \frac{1}{\sin^2{\theta}})^{1/4} \cos{\theta} d\theta = \frac{(\Gamma(1/4))^2}{2\sqrt{\pi}}$ Well, I have shown that $B(n, n+1) = \frac{(\Gamma(n))^2}{2\sqrt{2n}}$
From there I could deduce that $B(1/4, 5/4) = \frac{(\Gamma(1/4))^2}{2\sqrt{\pi}}$, then $n=1/4$.
I also know that $B(x, y) = \frac{(\Gamma(x))(\Gamma(y))} {\Gamma(x+y)} = 2 \int_{0}^{\pi/2} \sin^{2x-1}{\theta} \cos^{2y-1}{\theta} d\theta$.
So I suppose I should reduce the given integral to a form similar to the one above. Is there any trigonometric property that can help me that or am I seeing it wrong?
| Since you already received a good answer, another possible solution using the substitution proposed by @gt6989b in comments.
$$\int \left(\frac1{u^3} - \frac1{u^2}\right)^{1/4}\, du=2 \sqrt[4]{(1-u) u}+2 \sqrt[4]{u} \,\,\, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};u\right)$$
$$\int_0^1 \left(\frac1{u^3} - \frac1{u^2}\right)^{1/4}\, du=2\,_2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};1\right)=\frac{2 \sqrt{2 \pi }\, \Gamma \left(\frac{5}{4}\right)}{\Gamma
\left(\frac{3}{4}\right)}=\frac{\left(\Gamma(1/4)\right)^2}{2\sqrt{\pi}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3714871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Determine all pairs of positive integers $(a,b)$ such that $a^2+b^2+ab$ is a perfect square. Consideration via mod $4$ shows that $a,b=0$ mod $4$ or one of between $a,b$ is $=1$ mod $4$ while the other is $=0 $ mod $4$.
Considering $(a+b)^2,a^2+an+b^2,(a+b-1)^2$ as we can deduce that $a,b>2$.
| Primitive solutions to $a^2+ab+b^2=c^2.$
Let $x=a/c, y=b/c.$
$x^2+xy+y^2=1\tag{1}$
Substitute $x = t , y=1+kt$ to equation $(1)$, then we get $$t = -\frac{(1+2k)}{(1+k+k^2)}$$
Thus, we get a parametric solution.
$(x,y,z) = (-\frac{(1+2k)}{(1+k+k^2)}, -\frac{(k-1)(k+1)}{(1+k+k^2)})$
Hence $(a,b,c)=(1+2k, k^2-1, 1+k+k^2)$
$k$ is arbitrary.
Further, take $k=\frac{\alpha}{\beta}$ then we get
$(a,b,c)=(\beta^2+2\alpha\beta,\alpha^2-\beta^2
,{\beta}^{2}+\alpha\,\beta+{\alpha}^{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$\frac{dy}{dx} - {8} -{2}x^2+{4}y^2+y^2x^2 = 0.$ how should I procced from here having the equation $$\frac{dy}{dx} - {8} -{2}x^2+{4}y^2+y^2x^2 = 0.$$
I am getting to the following $$\frac{1}{2^{\frac{3}{2}}}\ln \left(y+\sqrt{2}\right)-\frac{1}{2^{\frac{3}{2}}}\ln \left(y-\sqrt{2}\right)=\frac{x^3}{3}+4x+c$$ from here I can do $\frac{1}{2^{\frac{3}{2}}}(ln\frac{y+\sqrt(2)}{y-\sqrt(2)})=\frac{x^3}{3}+4x+c$
how should I get none implicit function of $y$?
| $$\frac{y+\sqrt{2}}{y-\sqrt{2}}=\exp\left[2^{3/2}\left(\frac{x^3}{3}+4x+c\right)\right]$$
to make a few steps easier I will call the RHS $f(x)$:
$$y+\sqrt{2}=(y-\sqrt{2})f(x)$$
$$y(1-f(x))=-\sqrt{2}(1+f(x))$$
and so:
$$y=-\sqrt{2}\frac{1+f(x)}{1-f(x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Question about area of circle word problem Picture of problem
Please refer to the picture of the problem
I am trying to figure out: What percent of the total area is worth ten points?
So this is my attempt: the area of the 10 point region is pi * (4)^2 = 50.24
the total radius is 4+ 3+3+3 = 13
so the total area is 13^2 * pi = 530.66
So the percent is 50.23/530.66 = 10%
However, the answer is 30% which I do not get.
| there's three areas worth $10$ points.
The bullseye has area $\pi r_1^2 = \pi 4^2 = 16\pi$.
The area of the entire dart board has area $\pi r_2^2 = \pi (4+3*3)^2= 13^2\pi =169\pi$
The area of the three inner rings is $\pi r_3^2 = \pi (4+2*3)=10^2\pi =100\pi$.
So the area the $4$th ring only is the total area of the dart board minus the three inner rings, which is $169\pi -100\pi = 69\pi$.
The areas worth $10$ points are $\frac 14$ of that ring so area of those areas worth $10$ are are each $\frac {69}4 \pi$.
So the total area worth $10$ point if we add them up is $16\pi + \frac {69}4\pi + \frac {69}4\pi = 50.5 \pi$
So the percentage of the dart board is $\frac {50.5\pi}{169\pi} =\frac {101}{338}\approx 30\%$
....
Now if I wanted to be a smart-aff I could note the radius of the dart board is $13$ inches. The bullseye has radius of $4$ inches or $\frac 4{13}$ of the radius.
So the bullseye has $(\frac 4{13})^2$ of the area.
And the last ring has $\frac 3{13}$ of the radius. So it has $(\frac 3{13})^2$ of the area. And the parts worth ten points are half of that so,
The areas worth ten points are $(\frac 4{13})^2 + \frac{(\frac 3{13})^2}2 \approx 30\%$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Base conversion problems Let $b$ be an integer greater than 2, and let $N_b = 1_b + 2_b + \cdots + 100_b$ (the sum contains all valid base $b$ numbers up to $100_b$). Compute the number of values of $b$ for which the sum of the squares of the base $b$ digits of $N_b$ is at most 512.
Since we are adding up to $100$ in base $b$, I set up the equation
$$N=\frac{b^2(b^2+1)}{2},$$
and then tried to find the values that satisfy this.
For even values, of $b$, i got 16 values, but I don't know about odd values.
Now I am stuck.
| Note that from your formula, we can deduce that $N_b = \frac{10100_b}{ 2}$. Let us carry out the division in base $b$, distinguishing two cases:
*
*If $b$ is even, then we can note that since
$$N_b = \frac{10100_b}{ 2} = \frac{10000_b}{2} + \frac{100_b}{2} = \frac{b^4}{2} + \frac{b^2}{2} = \frac{b}{2}(b^3) + \frac{b}{2} (b)$$
the resulting quotient is $\left(\frac b2 0\frac {b}{2} 0\right)_b$. (Here the digits are the digits of $b/2$, followed by a zero, then followed by the digits of $b / 2$, then followed by a final zero. So for example, if $b = 6$, then the result is $3030_6$.) So the sum of squares of the digits is less than $512$ if and only if
$$\left( \frac{b}{2}\right)^2 + \left( \frac{b}{2}\right)^2 = \frac{b^2}{2} \leq 512 \iff b \leq 32$$
Indeed, there are $15$ such even values of $b$ that satisfy this inequality (remember we have counted $b = 2$ in a separate case).
*If $b$ is odd, then we can note that since
$$N_b = \frac{10100_b}{ 2_b} = \frac{b^4}{2} + \frac{b^2}{2} = \frac{b - 1}{2}(b^3) + \frac{b + 1}{2} (b^2)$$
the resulting quotient is $\left(\frac{b-1}{2} \frac{b + 1}{2} 0 0\right)_b$. Note that both $(b - 1) / 2$ and $(b + 1) / 2$ are integers as $b$ is odd. The sum of the squares of digits is less than $512$ if and only if
$$\frac14 \left[(b - 1)^2 + (b + 1)^2\right] \leq 512 \iff b^2 + 1 \leq 1024 \iff b \leq 31$$
And so you can count $15$ possible values of $b$ here.
In total, there are $\boxed{31}$ possible values of $b$, corresponding to $2 \leq b \leq 32$. $\square$
Follow-Up: What if we want the sum of squares of digits to be less than $512_b$?
Answer: As pointed out by mwt, for this question to make sense, we require $b \geq 6$. To answer this is not difficult; instead of having $512$ on the right-hand side of the above inequalities, we note that since
$$512_b = 5b^2 + b + 2$$
we instead want
$$\begin{cases}
\frac{b^2}{2} \leq 5b^2 + b + 2 \iff 9b^2 + 2b + 4 \geq 0 & b \text{ is even} \\
\frac{b^2 + 1}{2} \leq 5b^2 + b + 2 \iff 9b^2 + 2b + 3 \geq 0 & b \text{ is odd}
\end{cases}$$
You can solve these cases to determine the allowed values of $b$. Curiously, now every value of $b$ is satisfying!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Probability question on Properties of Expectation Problem
There are $x$ red and $y$ blue candies in a jar. A child randomly eats one candy everyday. If a child eats a red candy, the child will put a blue candy into the jar.
If a child eats a blue candy, the child doesn't put another candy into the jar.
Let $X$ denote the number of blue candies in the jar, after the child ate the last red candy and puts another blue candy into the jar. Find $\mathbb{E}[X]$.
Why I'm Stuck
This question has just completely struck my brain.
I've approached the quick-sort algorithm and have defined $x+y$ indicator variables.
I'm stuck in the process of writing the formula... Anybody can help?
| The expected value of $B$, the number of remaining blue candies just after all red candies are replaced, if we start with $x$ red candies and $y$ blue candies appears to be
$$\mathbb{E}_{x,y}[B] = \frac{y}{x+1} + H_x \approx \frac{y}{x + 1} + \log(x)$$
where $H_x$, the $x$th harmonic number, is equal to $\sum_{i = 1}^x \frac{1}{i}$.
To prove its correctness, note that the expectation $\mathbb{E}_{x, y} [B]$ satisfies the recurrence
$$\mathbb{E}_{x, y}[B] = \frac{x}{x + y} \mathbb{E}_{x - 1, y + 1} [B] + \frac{y}{x + y} \mathbb{E}_{x, y - 1} [B] $$
And indeed,
\begin{align}
\frac{x}{x + y} \mathbb{E}_{x - 1, y + 1} [B] + \frac{y}{x + y} \mathbb{E}_{x, y - 1} [B] &= \frac{x}{x+y} \left[\frac{y + 1}{x} + H_{x-1} \right] + \frac{y}{x+y}\left[\frac{y - 1}{x+1} + H_x\right] \\ \\
&= \frac{(y + 1)(x+1) + y(y-1)}{(x + y)(x + 1)} + H_{x-1} + \frac{y}{x(x + y)} \\ \\
&= \frac{x^2y + xy^2 + x^2 + x + xy + y}{x(x+y)(x+1)} + H_{x - 1} \\ \\
&= \frac{(x + y)(x + 1) + x^2y + y^2x}{x(x+y)(x+1)} + H_{x - 1} \\ \\
&= \frac{(x + y)xy}{x(x+y)(x+1)} + H_{x-1} + \frac{1}{x} \\ \\
&= \frac{y}{x + 1} + H_x = \mathbb{E}_{x, y}[B]
\end{align}
I'm not completely sure how to solve the recurrence without guessing though!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I solve this definite integral: $\int_{0}^{a}\frac{x^4dx}{\sqrt{a^2-x^2}}$
Evaluate $$\int_{0}^{a}\dfrac{x^4dx}{\sqrt{a^2-x^2}}$$
I tried taking $t$ as
$$t = \sqrt{a^2-x^2}$$
Thus my final integral became
$$\int_{0}^{a}(a^2-t^2)^{3/2}dt$$
but I couldn't go any further in solving this integral.
I also tried by taking $t$ as
$$t = a\sin^{-1}{x}$$
But I don't know how to solve the resulting integrand.
Also, can the king's rule be applied here? If yes then how?
| There are already 4 solutions. instead, I am going to evaluate the general integral $$I(k)=\int_{0}^{a} \frac{x^{k}}{\sqrt{a^{2}-x^{2}}} d x, \quad \textrm{ where }k=0,1,2, \cdots$$
by establishing a reduction formula on $k$.
$$
\displaystyle \begin{aligned}
I(k) &=\int_{0}^{a} \frac{x^{k}}{\sqrt{a^{2}-x^{2}}} d x \\
&=-\int_{0}^{a} x^{k-1} d \sqrt{a^{2}-x^{2}} \\
&=-\left[x^{k+1} \sqrt{a^{2}-x^{2}}\right]_{0}^{a}+(k-1) \int_{0}^{a} x^{k-2} \sqrt{a^{2}-x^{2}} \\
&=(k-1) \int_{0}^{a} \frac{x^{k-2}\left(a^{2}-x^{2}\right)}{\sqrt{a^{2}-x^{2}}} d x \\
&=(k-1) a^{2} I(k-2)-(k-1) I(k) \\
\displaystyle I(k) &=\dfrac{(k-1) a^{2}}{k} I(k-2)\\ &= \vdots \\& = \displaystyle \left\{\begin{array}{l}
\dfrac{k-1}{k} a^{2} \cdot \dfrac{k-3}{k-2} a^{2} \cdots \dfrac{1}{2}a^2 I(0) \quad \text { if }k \textrm{ is even .} \\
\dfrac{k-1}{k} a^{2} \cdot \dfrac{k-3}{k-2} a^{2} \cdots \dfrac{2}{3} a^2I(1) \quad \text { if }k \textrm{ is odd. }
\end{array}\right.
\\ \\ \displaystyle \displaystyle &= \left\{\begin{array}{ll}
\dfrac{a^k(k-1) ! ! \pi}{2(k ! !)} & \text { if } k \text { is even. } \\
\dfrac{a^k(k-1) ! !}{2(k ! !)} & \text { if } k \text { is odd. }
\end{array}\right.\end{aligned}
$$
Now go back to the original integral
$$I(k)=\int_{0}^{a} \frac{x^{4}}{\sqrt{a^{2}-x^{2}}} dx =I(4)=\frac{3 \times 1 a^{4} \pi}{2(4 \times 2)}=\frac{3 a^{4} \pi}{16} \quad \blacksquare $$
Footnotes:
*
*$\displaystyle I(0)=\int_{0}^{a} \frac{d x}{\sqrt{a^{2}-x^{2}}}=\left[\sin ^{-1} \frac{x}{a}\right]_{0}^{a}=\frac{\pi}{2}$
*$ \displaystyle I(1) =\int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} d x =-\frac{1}{2}\left[\sqrt{a^{2}-x^{2}}\right]_{0}^{a} =\frac{a}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Inequality 6 deg For $a,b,c\ge 0$ Prove that $$4(a^2+b^2+c^2)^3\ge 3(a^3+b^3+c^3+3abc)^2$$
My attempt: $$LHS-RHS=12(a-b)^2(b-c)^2(c-a)^2+2(ab+bc+ca)\sum_{sym} a^2(a-b)(a-c)$$
$$+\left(\sum_{sym} a(a-b)(a-c)\right)^2+14\left(\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\right)+\sum_{sym} c^2(a-b)^2[2(a+b)^2+(a-b)^2]$$
Since $\sum_{sym} x^3+3xyz-\sum_{sym} xy(x+y)\ge 0$, set $(x,y,z)\rightarrow (ab,bc,ca)$ we obtain $$\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\ge 0$$
Thus $LHS-RHS\ge 0$
I think this's a complicated solution and hard to find it by hand :">
Please give me a simplier solution. Thank you very much.
| We have$:$ \begin{align*} \text{LHS}-\text{RHS}\\&=\frac{1}{3} \sum\limits_{cyc} \left( 2\,{a}^{4}+10\,{a}^{2}{b}^{2}+2\,{a}^{2}bc+5\,{a}^{2}{c}^
{2}+2\,a{b}^{3}+4\,ab{c}^{2}+{b}^{4}+4\,{b}^{3}c+24\,{c}^{4} \right)
\left( a-b \right) ^{2}\end{align*}
Another solution.
Let $p=a+b+c,\,q=ab+bc+ca,\, r=abc.$ We have$:$
\begin{align*} \text{LHS}-\text{RHS}&=\frac{4}{3}\, \left( 5\,{p}^{2}-9\,q \right) \left( {q}^{2} -3\,pr\right) \\&\,\,\,+\frac{1}{3}\, \left( {p}^{2}-3\,q \right) \left( 3\,{p}^{4}-9\,{p}^{2}q+4\,{q} ^{2} \right)\\&\,\,\, -16\,{p}^{3}r+4\,{p}^{2}{q}^{2}+72\,pqr-16\,{q}^{3}-108\, {r}^{2}\\&\geqq 0\end{align*}
Note that$:$ $$-16\,{p}^{3}r+4\,{p}^{2}{q}^{2}+72\,pqr-16\,{q}^{3}-108\, {r}^{2}=4(a-b)^2(b-c)^2(c-a)^2 \geqq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\int_{}{} \frac{x^2}{\sqrt{x^2-x+1}} dx$ A physics problem says that the following integral is said to evaluate to: $$\int_{}{} \frac{x^2}{\sqrt{x^2-x+1}} dx = \frac{2x+3}{8}\sqrt{x^2-x+1} + \frac{1}{16}\ln(2x+1+2\sqrt{x^2-x+1})$$
I can not seem to find a way that evaluates to this. I know that a integral of this form can be broken down to:
$$\int_{}{} \frac{P(x)}{\sqrt{ax^2+bx+c}}dx = Q(x)\sqrt{ax^2+bx+c} + t\int{}{}\frac{1}{\sqrt{ax^2+bx+c}}dx$$
Where $Q(x)$ is a polynomial of degree one less than $P(x)$. Can anyone see a solution from this angle?
| Equate numerator as $x^2=A(x^2-x+1)+B(2x-1)+C$
which gives $A=1, B=\frac12$ & $C=-\frac12$. Now proceed as follows
$$\int_{}{} \frac{x^2}{\sqrt{x^2-x+1}} dx$$
$$=\int \frac{(x^2-x+1)+\frac12(2x-1)-\frac12}{\sqrt{x^2-x+1}} dx$$
$$=\int\left( \sqrt{x^2-x+1} +\frac12\frac{(2x-1)}{\sqrt{x^2-x+1}}-\frac12\frac{1}{\sqrt{x^2-x+1}} \right)dx$$
$$=\int \sqrt{\left(x-\frac12\right)^2+\frac34}\ dx +\frac12\int\frac{d(x^2-x+1)}{\sqrt{x^2-x+1}} -\frac12\int \frac{dx}{\sqrt{\left(x-\frac12\right)^2+\frac34}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3725807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.