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Deriving Taylor series without applying Taylor's theorem. First, a neat little 'proof' of the Taylor series of $e^x$.
Start by proving with L'Hospital's rule or similar that
$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$
and then binomial expand into
$$e^x=\lim_{n\to\infty}1+x+\frac{n-1}n\frac{x^2}2+\dots+\frac{(n-1)(n-2)(n-3)\dots(n-k+1))}{n^{k-1}}\frac{x^k}{k!}+\dots$$
Evaluating the limit, we are left with
$$e^x=1+x+\frac{x^2}2+\dots+\frac{x^k}{k!}+\dots$$
which is our well known Taylor series of $e^x$.
As dxiv mentions, we can exploit the geometric series:
$$\frac1{1-x}=1+x+x^2+\dots$$
$$\ln(1+x)=x-\frac12x^2+\frac13x^3-\dots$$
$$\arctan(x)=x-\frac13x^3+\frac15x^5-\dots$$
which are found by integrating the geometric series and variants of it.
I was wondering if other known Taylor series can be created without applying Taylor's theorem. Specifically, can we derive the expansions of $\sin$ or $\cos$?
| Alan Turing, at a young age, derived the series expansion of $\arctan$ without using (and, purportedly without knowing) calculus whatsoever.
Using the identity
$$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x},$$
he obtained
$$\tan(2 \arctan x) = \frac{2x}{1-x^2},$$
and
$$2 \arctan x = \arctan\left( \frac{2x}{1-x^2}\right).$$
Using the geometric series for $|x| < 1$,
$$\tag{1}2 \arctan x =\arctan [2x(1 + x^2 + x^4 + x^6 + \ldots)]$$
Assuming $\arctan x = a_0 + a_1x + a_2x^2 + \ldots$ and matching coefficients in the expansions of each side of (1), he obtained
$$\arctan x = a_1\left(x - \frac{1}{3} x^3 + \frac{1}{5}x^5 - \ldots \right).$$
Some basic trigonometry reveals
$$a_1 = \lim_{x \to 0} \frac{\arctan x}{x} = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Finding Polynomial Equations for $1^4 + 2^4 + 3^4 + \ldots n^4$ Find a polynomial expression for
$$ 1^4 + 2^4 + 3^4 + ... + n^4 $$
I know you have to use the big theorem but I can't figure out how you would start to compute the differences. Suggestions?
| Even though this might be too much for simple task you want (you can assume the polynomial of fifth degree and calculate coefficients) I think it is worth mentioning that there is a general formula for finding polynomial of sum of $p$-th powers $1^p+2^p+\dots+n^p$, and it is called Faulhaber's formula.
In short
$$1^p+2^p+\dots+n^p = \frac{1}{p+1} \sum_{j=0}^{p}(-1)^j\binom{p+1}{j}B_j n^{p+1-j}$$
where $B_j$ are called Bernoulli numbers, which you can compute for example as
$$
B_n = \sum_{k=0}^{n}\frac{1}{k+1}\sum_{r=0}^{k}(-1)^r \binom{k}{r} r^n
$$
giving first couple values $1,-1/2$,$1/6$,$0$,$-1/30$,$\dots$.
So in your case this becomes
\begin{align}
1^4+2^4+\dots+n^4 &= \frac{1}{5} \sum_{j=0}^{4}(-1)^j\binom{5}{j}B_j n^{5-j}\\
&= \frac{1}{5} ((-1)^0\binom{5}{0}B_j n^{5}+(-1)^1\binom{5}{1}B_1 n^{5-1}\\
&+(-1)^2\binom{5}{2}B_2 n^{5-2}+(-1)^3\binom{5}{3}B_3 n^{5-3}+(-1)^4\binom{5}{4}B_4 n^{5-4})\\
&= \frac{1}{5} (B_0 n^{5}-5 B_1 n^{4}+10 B_2 n^{3}-10 B_3 n^{2}+5 B_4 n)\\
&= \frac{1}{5} (n^{5}+\frac{5}{2}n^{4}+\frac{10}{6} n^{3}-\frac{5}{30} n)\\
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2006979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If we are given $3$ positive integers $a,b,c$ such that $a>b>c$ , and $91b>92c>90a$ . What is the minimal value of $a+b+c$? If we are given $3$ positive integers $a,b,c$ such that $a>b>c$ , and $91b>92c>90a$ . What is the minimal value of $a+b+c$?
I am getting the bounds of the fractions $\frac{a}{b},\frac{b}{c},\frac{c}{a}$..
But I dont know what to do next
| This follows up on my former comment and proves $413=139+138+136$ is the minimal sum.
$92 c \gt 90 a \implies 2c > 90(a-c) \ge 90 \cdot 2 = 180$ therefore $c \gt 90 \iff c \ge 91$.
$91 b \gt 92 c \implies 91(b-c) \gt c \ge 91$ therefore $b-c \gt \frac{91}{91} = 1 \iff b-c \ge 2 \iff b \ge c+ 2$.
Since $b \ge c+2$ and $a \gt b$ it follows that $a-c \ge 3$ and the first inequality $2c \gt 90(a-c) \ge 270$ gives the stronger bound $c \gt 135 \iff c \ge 136$.
Using the lowest values allowed by $c \ge 136$ and $b \ge c+2$ gives $c=136$, $b=c+2=138$, and $a=b+1=139$. The triplet satisfies $91 \cdot 138 = 12558 \gt 92 \cdot 136 = 12512 \gt 90 \cdot 139 = 12510$, so the minimal sum is $139+138+136 = 413$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving ${\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=\log_{\frac{1}{2-\sqrt{3}}}(x^2-4x-3)}$ This was given to me by my Math Teacher almost a year ago and I've not been able to make much progress on it. I am hoping to see it resolved by our community members. $$\large{\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=\log_{\frac{1}{2-\sqrt{3}}}(x^2-4x-3)}.$$
| Hint
If $x^2-4x-2<0$ or $x^2-4x-3<0$ then the corresponding $x$ cannot be a solution. But forget about that for a second and explain the meaning of the equation below:
$$\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=\log_{\frac{1}{2-\sqrt{3}}}(x^2-4x-3).$$
If for an $x$ the equation holds then there must exist some $A$ for which
$$\log_{\frac{2}{\sqrt{2-\sqrt{3}}}}(x^2-4x-2)=A$$
and
$$\log_{\frac{1}{2-\sqrt{3}}}(x^2-4x-3)=A.$$
By the definition of the logarithm one can say that for such an $A$:
$$\left(\frac{2}{\sqrt{2-\sqrt{3}}}\right)^A=x^2-4x-2$$
and
$$\left(\frac{1}{2-\sqrt{3}}\right)^A=x^2-4x-3.$$
Or
$$x^2-4x-\left(2+\left(\frac{2}{\sqrt{2-\sqrt{3}}}\right)^A\right)=0$$
and
$$x^2-4x-\left(3+\left(\frac{1}{2-\sqrt{3}}\right)^A\right)=0.$$
These are two quadratic equations which can be solved for $x$. The solutions will depend on the parameter $A$. Since the corresponding solutions will have to be equal one will have equations for the possible $A$'s.
Given a suitable $A$ an $x$ can be calculated. (Finally test the negativity of the argument of the logarithm.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all real solutions of this equation $x^2=2y-1$,$x^4+y^4=2$. Find all real solutions of this equation $x^2=2y-1$,$x^4+y^4=2$.
My attempt:I put the value of $x^2$ in the second equation.I get:
$(2y-1)^2+y^4=2 \Rightarrow [(2y-1)^2-1^2]+(y^4-1^4)=0 \Rightarrow 4y(y-1)+(y-1)(y+1)(y^2+1)=0 \Rightarrow (y-1)(y^3+y^2+5y+1)=0$
Now one solution is $y=1,x=1or-1$ but what about others is there another real solution?
| There are no other real solutions.
From $2y-1=x^2\ge 0,$ $y$ has to satisfy $y\ge \frac 12$.
However, if $y\gt 0$, then $y^3+y^2+5y+1\gt 0$.
| {
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Rewrite rational function $f(x)$ as a series if the quadratic expression in the denominator has no roots A function of the type
$$f(x)=\frac{ex+f}{ax^2+bx+c}$$
with $b^2-4 a c \geq 0$ can be written as a series using partial fraction decomposition and geometric series.
But if one has the same function with $b^2-4 a c < 0$? Partial fraction decomposition does not help and geometric series neither, so what is the strategy to rewrite $f(x)$ as a series in that case?
Example
$$f(x)=\frac{x+1}{x^2+x+3}$$
| More than likely, too simplistic !
You wrote the function $$F(x)=\frac{ex+f}{ax^2+bx+c}$$ Since you look for an expansion around $x=0$, let us rewrite it as $$F(x)=\frac{f+ex}{c+bx+ax^2}$$ and now use the long division to get $$F(x)=\frac{f}{c}+\frac{ c e-b f}{c^2}x+\frac{ b^2f-a c f-b c
e}{c^3}x^2+O\left(x^3\right)$$ I did not write the next terms because the expression becomes too long.
Using your example, we should get $$\frac{x+1}{x^2+x+3}=\frac{1+x}{3+x+x^2}=\frac{1}{3}+\frac{2 x}{9}-\frac{5 x^2}{27}-\frac{x^3}{81}+\frac{16
x^4}{243}+O\left(x^5\right)$$ I am sure that you notice the powers of $3$ as denominators of the coefficients.
Let us try an easy value $x=\frac 1{10}$. The exact value is $\frac{110}{311}\approx 0.3536977492$ while the expansion gives $\frac{429743}{1215000}\approx 0.3536979424$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Doolittle transformation is non-unique for singular matrices Decomposing the singular matrix $$A = \begin{bmatrix}
1 & 2 \\
1 & 2
\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 2 \\ 0 & 0\end{bmatrix}=LU$$
by Doolittle decomposition seems to be unique for this case. But how to proove that this is not necessarily unique?
| The row of zeroes in your $U $ allows you to play with the second column of $L $. You have
$$
\begin{bmatrix}
1 & 2 \\
1 & 2
\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 1 & x\end{bmatrix}\begin{bmatrix}1 & 2 \\ 0 & 0\end{bmatrix}
$$ for any choice of $x $.
If you need $x=1$, there is no other choice and the decomposition is unique for that matrix. Such is the case for any singular $2\times2$ matrix: if
$$
A=\begin{bmatrix}r&s\\ tr&ts\end{bmatrix}
=
\begin{bmatrix}1&0\\
x&1\end{bmatrix}
\begin{bmatrix}a&b\\
0&c\end{bmatrix},
$$ it follows immediately that $a=r $, $b=s $, $x=t $, $c=0$.
For $3\times3$, here is an example where you are free to choose $z $:
$$
A=\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}
=
\begin{bmatrix}1&0&0\\
1&1&0\\
1&z&1\end{bmatrix}
\begin{bmatrix}1&1&1\\
0&0&0\\ 0&0&0\end{bmatrix}.
$$
On the other hand, the decomposition is always unique when $A$ is non-singular and we require one of the two triangular matrices to have all ones in the diagonal, like
$$
A=\begin{bmatrix}1&2&3\\4&6&9\\5&8&11\end{bmatrix}
=
\begin{bmatrix}1&0&0\\
4&1&0\\
5&1&1\end{bmatrix}
\begin{bmatrix}1&2&3\\
0&-2&-3\\ 0&0&-1\end{bmatrix}
$$
is unique.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the integral: $\int_{0}^{\large\frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}$ What is
$$\int_{0}^{\large\frac{\pi}{4}}\frac{\cos(x)\:dx}{a\cos(x)+b \sin(x)}?$$
$a,b \in \mathbb{R}$ appropriate fixed numbers.
| $$\mathcal{I}\left(\text{a},\text{b}\right)=\int_0^{\frac{\pi}{4}}\frac{\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}=$$
$$\frac{\text{a}}{\text{a}^2+\text{b}^2}\int_0^{\frac{\pi}{4}}1\space\text{d}\text{x}-\frac{\text{b}}{\text{a}^2+\text{b}^2}\int_0^{\frac{\pi}{4}}\frac{\text{a}\sin\left(\text{x}\right)-\text{b}\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}$$
Now, use:
$$\int_0^{\frac{\pi}{4}}1\space\text{d}\text{x}=\frac{\pi}{4}$$
So, we get:
$$\mathcal{I}\left(\text{a},\text{b}\right)=\frac{\pi}{4}\cdot\frac{\text{a}}{\text{a}^2+\text{b}^2}-\frac{\text{b}}{\text{a}^2+\text{b}^2}\int_0^{\frac{\pi}{4}}\frac{\text{a}\sin\left(\text{x}\right)-\text{b}\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}$$
Now, substitute $\text{u}=\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)$ and $\text{d}\text{u}=\left(\text{b}\cos\left(\text{x}\right)-\text{a}\sin\left(\text{x}\right)\right)\space\text{d}\text{x}$:
$$\int_0^{\frac{\pi}{4}}\frac{\text{a}\sin\left(\text{x}\right)-\text{b}\cos\left(\text{x}\right)}{\text{a}\cos\left(\text{x}\right)+\text{b}\sin\left(\text{x}\right)}\space\text{d}\text{x}=-\int_\text{a}^{\frac{\text{a}+\text{b}}{\sqrt{2}}}\frac{1}{\text{u}}\space\text{d}\text{u}=-\left(\ln\left|\frac{\text{a}+\text{b}}{\sqrt{2}}\right|-\ln\left|\text{a}\right|\right)$$
So, we get:
$$\mathcal{I}\left(\text{a},\text{b}\right)=\frac{\pi}{4}\cdot\frac{\text{a}}{\text{a}^2+\text{b}^2}+\frac{\text{b}}{\text{a}^2+\text{b}^2}\left(\ln\left|\frac{\text{a}+\text{b}}{\sqrt{2}}\right|-\ln\left|\text{a}\right|\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of Maclaurin series Find the sum of the infinite series
\begin{equation}
\sum_{n=2}^\infty\frac{7n(n-1)}{3^{n-2}}
\end{equation}
I think it probably has something to do with a known Maclaurin series, but cannot for the life of me see which one.. Any hints would be appreciated!
Edit: Using your hints, I was able to solve the problem. Solving it here in case someone is wondering about the same thing:
Manipulate \begin{equation}\sum_{n=2}^\infty x^n=\frac{1}{1-x}-x-1 \end{equation}
by differentiating both sides:
\begin{equation}\sum_{n=2}^\infty nx^{n-1}=\frac{1}{(1-x)^2}-1 \end{equation}
differentiate again:
\begin{equation}\sum_{n=2}^\infty (n-1)nx^{n-2}=\frac{2}{(1-x)^3} \end{equation}
plugging in $\frac{1}{3}$ for x:
\begin{equation}\sum_{n=2}^\infty n(n-1)(\frac{1}{3})^{n-2}=\frac{2}{(1-\frac{1}{3})^3} \end{equation}
which is equivalent to
\begin{equation}\sum_{n=2}^\infty \frac{n(n-1)}{3^{n-2}} \end{equation}
finally, multiplying by 7 gives us
\begin{equation}\begin{split}\sum_{n=2}^\infty \frac{7n(n-1)}{3^{n-2}}&=7\frac{2}{(1-\frac{1}{3})^3}\\
&=\underline{\underline{\frac{189}{4}}} \end{split} \end{equation}
| Hint. We have that for $x\not=1$, and $N\geq 2$,
$$\frac{d^2}{dx^2}\left(\frac{1-x^{N+1}}{1-x}\right)=\frac{d^2}{dx^2}\left(\sum_{n=0}^N x^n\right)=\sum_{n=2}^N n(n-1)x^{n-2}.$$
P.S. for the downvoters. I considered a finite sum because it is not so straightforward to say that we can interchange the differentiation and the infinite sum: for $|x|<1$,
$$\frac{d^2}{dx^2}\left(\frac{1}{1-x}\right)=\frac{d^2}{dx^2}\left(\sum_{n=0}^{\infty} x^n\right)=
\sum_{n=0}^{\infty} \frac{d^2(x^n)}{dx^2}
=\sum_{n=2}^{\infty} n(n-1)x^{n-2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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An differential equation$(x^{2}＋y^{2}＋3)\frac{dy}{dx}＝2x(2y-\frac{x^{2}}{y})$ How to solve this ODE $$(x^{2}＋y^{2}＋3)\frac{dy}{dx}＝2x(2y-\frac{x^{2}}{y})$$
I tried to find its integral factor, but failed.
Many thanks for your help.
| $(y^2+x^2+3)(y^2)'=4x(2y^2-x^2)$
$(y^2+x^2+3)(y^2+x^2+3)'=4x(2y^2-x^2)+2x(y^2+x^2+3)$
$(y^2+x^2+3)(y^2+x^2+3)'=10x(y^2+x^2+3)-4x(3x^2+6)$
With $\enspace z:=y^2+x^2+3\enspace$ we get $\enspace zz'=10xz-12x^3-24x$ .
I don't know how to solve this, but Wolfram solves it in a closed form with
http://www.wolframalpha.com/input/?i=z(dz%2Fdx)%3D10xz-12x%5E3-24x .
| {
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In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
$\bf{My\; Attempt:}$
Using Sin formula:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
So $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin 60^0}\Rightarrow \sin A = \frac{a}{b}\cdot \frac{\sqrt{3}}{2}$ and $\displaystyle \frac{c}{\sin C} = \frac{b}{\sin 60^0}\Rightarrow \sin C = \frac{c}{b}\cdot \frac{\sqrt{3}}{2}$
So $$\sin A\sin C = \frac{3}{4}\cdot \frac{ac}{b^2}$$
Now using Cosine formula:
$$\cos B = \cos 60^0 = \frac{1}{2}= \frac{a^2+c^2-b^2}{2ac}\Rightarrow b^2=a^2+c^2-ac$$
So $$\sin A\sin C = \frac{3}{4}\bigg[\frac{ac}{a^2+c^2-ac}\bigg] = \frac{3}{4}\bigg[\frac{1}{\frac{a}{c}+\frac{c}{a}-1}\bigg]\leq \frac{3}{4}$$
Using $\bf{A.M\geq G.M},$ We get $$\frac{a}{c}+\frac{c}{a}\geq 2\Rightarrow \frac{a}{c}+\frac{c}{a}-1\geq 1$$
$\bf{ADDED::}$ Using Jensen Inequality:: For $f(x) = \ln(\sin x)\;,$
$$\ln(\sin A)+\ln(\sin C)\leq 2\cdot \ln \sin \left(\frac{A+C}{2}\right) = \ln \cos 30^0 = 2\cdot \ln \frac{\sqrt{3}}{2} = \ln \frac{3}{4}$$
But I do not understand how to calculate the lower bound for $\sin A\sin C$.
Thanks in advance!
| Clearly $\sin(A)\sin(C)\geq 0$ since $A$ and $C$ are between $0^{\circ}$ and $180^{\circ}$. Let $A$ approach $0^{\circ}$. Then $\sin(A)$ approaches $0$ as well, while $\sin(C)$ is bounded above by $1$. This shows that
$$\sin(A)\sin(C)\rightarrow 0$$
if $A\rightarrow 0^{\circ}$ so that $\sin(A)\sin(C)$ can be as small as needed. The lower bound is $0$.
| {
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For $f(x)=4x^3+3x^2-x-1$, The Range of values $\frac{f(x_1)-f(x_2)}{x_1-x_2}$ can take is- My Attempt :-
$f'(x)=12x^2+6x-1$ where $f'(x) \ge \frac{-7}{4}$. So (I think) from LMVT we can directly say that
$$\frac{f(b)-f(a)}{b-a} \ge \frac{-7}{4}$$ But the answer Given is $$\frac{f(b)-f(a)}{b-a} > \frac{-7}{4}$$
So my Question is Is This Application of LMVT correct ,if yes then why $\frac{-7}{4}$ is excluded from the given range ?
| Intuitively, as long as $f'(x)$ attains its minimum at one single point, there won't exist two distinct points where the secant matches that slope.
To formalize, assume $f'(x)$ has a global minimum $m$, then it follows from MVT that:
$$\frac{f(b)-f(a)}{b-a} \ge m \quad \text{for} \;\;\forall a \ne b$$
If $a,b$ points exist where the equality holds, then by MVT for the intervals $(a,\frac{a+b}{2})$ and $(\frac{a+b}{2},b)$:
$$m = \frac{f(b)-f(a)}{b-a} = \frac{1}{2}\left(\frac{f(b)-f(\frac{a+b}{2})}{\frac{b-a}{2}} +
\frac{f(\frac{a+b}{2})-f(a)}{\frac{b-a}{2}}\right) \ge \frac{1}{2}(m+m) = m
$$
It follows that the middle inequality must be an equality as well, so:
$$m = \frac{f(b)-f(a)}{b-a} = \frac{f(b)-f(\frac{a+b}{2})}{\frac{b-a}{2}} = \frac{f(\frac{a+b}{2}) - f(a)}{\frac{b-a}{2}}
$$
From which:
$$f\left(\frac{a+b}{2}\right) = \frac{f(a) + f(b)}{2}$$
It follows that $f(x)$ is both midpoint convex and concave, thus linear by continuity (see for example Function that is both midpoint convex and concave). But the given $f(x)$ is a $3^{rd}$ degree polynomial which is obviously not linear, so such points $a,b$ do not exist. Therefore the strict inequality holds.
| {
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Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$. Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$.
By distinct it means that $(1, 0, 0)$ is a solution, but $(0, \pm 1, 0)$ counts as the same solution.
I can only seem to find two; namely $(1, 0, 0)$ and $( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3})$. Is there a method to finding a third or is it still just trial and error?
| Here's a start that shows that
any other solutions
would have to have
distinct $a, b, $ and $c$.
In
$a^2+b^2+c^2 =1$
and
$a+b+c= \pm 1$,
if $a=b$,
these become
$2a^2+c^2 = 1,
2a+c = \pm 1$.
Then
$c = -2a\pm 1$,
so
$1
= 2a^2+(-2a\pm 1)^2
=2a^2+4a^2\pm 4a+1
=6a^2\pm 4a+1
$
so
$0 = 6a^2\pm 4a
=2a(3a\pm 2)
$.
Therefore
$a=0$ or
$a = \pm \frac23$.
If $a=b=0$,
then
$c = \pm 1$;
if $a=b=\pm \frac23$,
then
$c = -2a\pm 1
=\mp \frac43 \pm 1
=\pm \frac13
$
and these are the solutions
that you already have.
Therefore
any other solutions
would have to have
distinct $a, b, $ and $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find a general form of a sequence and its sum I have a problem to find a general form of the sequence
\begin{align}
- \frac{{n\left( {n - 1} \right)}}{{2\left( {2n - 1} \right)}},\frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{2 \cdot 4 \cdot \left( {2n - 1} \right)\left( {2n - 3} \right)}}, - \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)}}{{2 \cdot 4 \cdot 6 \cdot \left( {2n - 1} \right)\left( {2n - 3} \right)\left( {2n - 5} \right)}}, \cdots :=a_n(k),\qquad n\ge 2
\end{align}
and then to find the sum $\sum |a_n(k)|^2$, $1\le k\le n$.
I have tried as follows:
$a_n(k)=\frac{(-1)^kP(n,2k)}{(2k)!!A_n(k)}$, where $P(n,2k)=\frac{n!}{(n-2k)!}$ and $\,\,A_n(k):=(2n-1)(2n-3)(2n-5)\cdots (2n-2k+1),\qquad k\ge1$
Is this true. If yes, can we write $A_n(k)$ in a closed form?! After all of that what is the sum of $|a_n(k)|^2$, $1\le k\le n$.
| Here is a more compact representation as sum formula, most of it was already stated in the comment section.
Since
\begin{align*}
a_n(k)=\frac{(-1)^kn(n-1)\cdots (n-2k+1)}{2\cdot4\cdots (2k)\cdot(2n-1)(2n-3)\cdots(2n-2k+1)}\qquad\qquad 1\leq k\leq n
\end{align*}
We obtain
\begin{align*}
a_n(k)&=(-1)^k\frac{n!}{(n-2k)!}\cdot\frac{1}{(2k)!!}\cdot\frac{(2n-2k-1)!!}{(2n-1)!!}\tag{1}\\
&=(-1)^k\frac{n!}{(n-2k)!}\cdot\frac{1}{(2k)!!}\cdot\frac{(2n-2k)!}{(2n-2k)!!}\cdot\frac{(2n)!!}{(2n)!}\tag{2}\\
&=(-1)^k\frac{n!}{(n-2k)!}\cdot\frac{1}{2^kk!}\cdot\frac{(2n-2k)!}{2^{n-k}(n-k)!}\cdot\frac{2^nn!}{(2n)!}\tag{3}\\
&=(-1)^k\frac{n!n!}{(2n)!}\cdot\frac{1}{k!(n-k)!}\cdot\frac{(2n-2k)!}{(n-2k)!}\\
&=\frac{(-1)^k\binom{n}{k}\binom{2n-2k}{n}}{\binom{2n}{n}}
\end{align*}
Comment:
*
*In (1) we use double factorials $(2n)!!=(2n)(2n-2)\cdots4\cdot 2$
*In (2) we use
$(2n)!=(2n)!!(2n-1)!!$
*In (3) we use
$(2n)!!=2^nn!$
I don't think that the series
\begin{align*}
\sum_{k=1}^n\left|a_n(k)\right|^2=\binom{2n}{n}^{-2}\sum_{k=1}^{n}\binom{n}{k}^2\binom{2n-2k}{n}^2\qquad\qquad n\geq 1
\end{align*}
has a nice closed formula. The first few terms are
\begin{align*}
0,4,144,3636,82000,1764400,37164736,\ldots
\end{align*}
but they are not known to OEIS.
I've also tried some standard techniques and checked for instance section 2.9 in Riordan Array Proofs of Identities in Gould’s Book which contains binomial identities of the type we need, but without success.
Wolfram Alpha provides following representation via hypergeometric series
\begin{align*}
\sum_{k=1}^n\left|a_n(k)\right|^2
&={}_{4}F_{3}\left(\frac{1}{2}-\frac{n}{2},\frac{1}{2}-\frac{n}{2},-\frac{n}{2},-\frac{n}{2};1,\frac{1}{2}-n,\frac{1}{2}-n;1\right)-1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
line integral hard function to differentiate $$\int_\gamma \frac{(x^2+y^2-2)\,dx+(4y-x^2-y^2-2) \, dy}{x^2+y^2-2x-2y+2}$$
where $\gamma = 2\sin\left(\frac{\pi x}{2}\right)$ from $(2,0)$ to $(0,0)$.
I think it should be a shortcut to this problem that I cannot see , if that is not the case I will keep trying to simplify it .
Thnaks in advance
| Complete the squares in the denominator $(x-1)^2+(y-1)^2$ and change the variables $x-1\mapsto x$ and $y-1\mapsto y$. You get the vector field
\begin{align}
&\left[\frac{x^2+2x+y^2+2y}{x^2+y^2},\frac{4y-x^2-2x-y^2-2y}{x^2+y^2}\right]=\\
&=\left[1+\frac{2x}{x^2+y^2},-1+\frac{2y}{x^2+y^2}\right]-2\left[\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right].
\end{align}
The first term is conservative (with an easy potential), the second term has zero curl (easy to check).
P.S. Actually the second term has a potential too if you restrict you domain e.g. to the plain without the negative $y$-axis, but the potential is less obvious (easier to get in the polar coordinates).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1-\cos^2x$ on both sides and then comparing the results.
Are there other solutions, simpler approaches?
I found some other questions about the same expression, but they simplify this to another form: Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here?, Alternative proof of $\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$ and Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$. Perhaps also Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ can be considered similar. The expression $\cos^6x+\sin^6x$ also appears in this integral Find $ \int \frac {\tan 2x} {\sqrt {\cos^6x +\sin^6x}} dx $ but again it is transformed to a different from then required here.
Note: The main reason for posting this is that this question was deleted, but I still think that the answers there might be useful for people learning trigonometry. Hopefully this new question would not be closed for lack of context. And the answers from the deleted question could be moved here.
| Any symmetric polynomial in $X$ and $Y$ can be expressed as a polynomial in $S=X+Y$ and $P=XY$. If $X=\cos^2x$ and $Y=\sin^2x$, then $S=\cos^2x+\sin^2x=1$, so a symmetric polynomial expression in $\cos^2x$ and $\sin^2x$ can be written as a polynomial in $P=\cos^2x\sin^2x$.
If the symmetric polynomial is also homogeneous, all terms in the new expression are homogeneous as well.
In this case, $P$ appears at most with degree $1$, because $P$ counts for degree $2$ in $X$ and $Y$.
Hence $X^3+Y^3=a+b(X+Y)^3+c(X+Y)XY$; evaluating at $X=Y=0$ entails $a=0$. With $X=1$ and $Y=0$ we get $b=1$; with $X=1$ and $Y=1$ we get
$$
1+1=8+2c
$$
so $c=-3$; therefore
$$
\cos^6x+\sin^6x=1-3\cos^2x\sin^2x
$$
As another example, $X^4+Y^4=a(X+Y)^4+bXY(X+Y)^2+c(XY)^2$. Evaluating at $X=1$ and $Y=0$ gives $1=a$; evaluating at $X=1$ and $Y=-1$ gives $2=c$; evaluating at $X=Y=1$ gives $2=16a+4b+c$, so $b=-4$.
Therefore
$$
\cos^8x+\sin^8x=1-4\cos^2x\sin^2x+2\cos^4x\sin^4x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 11,
"answer_id": 8
} |
Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$
To be proven: $((k)(k+1)/2)^2 + (k+1)^3 = ((k+1)(k+2)/2)^2$
My problem is that I do not know how I can put the $ + (k+1)^3$ inside $((k)(k+1)/2)^2$.
Simplifying the left and right part of the statement does not help:
Simplifying the left side: $((k)(k+1)/2)^2 + (k+1)^3 = ((k^2+k)/2)^2 + (k+1)^3 $
Simplifying the right side: $((k+1)(k+2)/2)^2 = ((k^2+3k+2)/2)^2$
So i am left with: $((k^2+k)/2)^2 + (k+1)^3 = ((k^2+3k+2)/2)^2$
That is the same as: $1/4 (k^2+k)^2 + (k+1)^3 = 1/4((k^2+3k+2))^2$
Going further with the left side:$1/4 (k^2+k)^2 + (k+1)^3 = (1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1$
Going further with the right side: $1/4((k^2+3k+2))^2 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am stuck with: $(1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am kind of left with garbage. Am I missing something? What do I do wrong? Where can I find a good resources to learn how to solve this issue?
| \begin{align}
\underbrace{1^3+2^3+\ldots+n^3}_{\left[\frac{n(n+1)}{2} \right]^2}+(n+1)^3
=&
\left[\frac{n(n+1)}{2} \right]^2+(n+1)^3
\\
=&
\frac{n^2\color{red}{(n+1)^2}}{4}
+
(n+1)\color{red}{(n+1)^2}
\\
=&
\left\lgroup
\frac{n^2}{4}
+
(n+1)
\right\rgroup
\color{red}{(n+1)^2}
\\
=& \left\lgroup \frac{n^2 +4(n+1)}{4} \right\rgroup \color{red}{(n+1)^2}\\
=&\left\lgroup
\frac{(n+2)^2}{4}
\right\rgroup \color{red}{(n+1)^2}
\\
=&\left\lgroup
\frac{(n+2)}{2}
\right\rgroup^2 \color{red}{(n+1)^2}
\\
=&\left\lgroup
\frac{\big(\color{red}{(n+1)}+1\big)}{2}
\right\rgroup^2 \color{red}{(n+1)^2}
\\
=&\left[
\frac{\color{red}{(n+1)}\big(\color{red}{(n+1)}+1\big)}{2}
\right]^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
heterogeneous recurrence with f(n) as constant How to solve this $s_{n+1}=4s_{n-1}-3s_n+5$ where f(n)=5 conditions $s_0=-3$ $s_1=3$
I calculated the general solution $s_n=c_1*(-4)^n+c_2*1^n$ of this recurrence. The roots are $q_1=-4$ and $q_2=1$ but I have problem with particular solution with method of prediction .
I have problem with this 5 as constant
solution(homo+heterogeneous) is $s(n) = n - (-4)^n - 2$
I learn in Polish so I can't find too much words to describe this
| $s_{n+1}=4s_{n-1}-3s_n+5$
$q^2-4+3q=0$
$\Delta=3^2-4*1*(-4)=25$
$\sqrt\Delta=5$
$q_1=\frac{-3-5}2=-4$
$q_2=\frac{-3+5}2=1$
homo general $s_n=c_1*1^n+c_2*(-4)^n$
$k=1$ where $k$ is multipiclity of root
hetero particular $s_n=Q(n)*q^n*n^k$
$s_n=A*1^n*n^1=An$; $1^n=1$ because $1^0=1$, $1^1=1$ and so on
$A(n+1)=4(A(n-1))-3An+5$
$An+A=4An-4A-3An+5$
$An+A=An-4A+5$
$A=1$
hetero particular $s_n=n$
hetero general $s_n=c_1*1^n+c_2*(-4)^n+n$
$\begin{cases} -3=s_0=c_1*1^0+c_2*(-4)^0=c_1+c_2 => c_1=-3-c_2=-3+\frac{6}{5}=\frac{-9}{5}
3=s_1=c_1*1^1+c_2*(-4)^1=-3-c_2-4*c_2 => -5c_2=6 => c_2=\frac{-6}{5}
\end{cases}$
hetero general $s_n=\frac{-9}{5}*1^n-\frac{6}{5}*(-4)^n+n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inequality with $x,y,z\geq 0$, $x+y+z=1.$ With $x,y,z\geq 0$, $x+y+z=1$.Prove that $$\sqrt{x+y^2}+\sqrt{y+z^2}+\sqrt{z+x^2}\geq 2 \tag{i}$$
The hint is using a lemma:
If $a,b,c,d\geq 0 $satisfying $a+b=c+d$ and$|a-b|\leqslant|c-d|$ then we have $\sqrt{a}+\sqrt{b}\geq \sqrt{c}+\sqrt{d}$
How to prove this lemma? And is there a different way to prove the inequality (i)?
| There is the following Vo Quoc Ba Can's solution.
We need to prove that
$$\sum\limits_{cyc}\left(\sqrt{x+y^2}-y\right)\geq1$$ or
$$\sum\limits_{cyc}\frac{x}{\sqrt{x+y^2}+y}\geq1.$$
Now, by AM-GM
$$\sum\limits_{cyc}\frac{x}{\sqrt{x+y^2}+y}=\sum\limits_{cyc}\frac{x(x+y)}{(x+y)\sqrt{x+y^2}+y(x+y)}\geq\sum\limits_{cyc}\frac{x(x+y)}{\frac{1}{2}((x+y)^2+x+y^2)+y(x+y)}.$$
Thus, it remains to prove that
$$\sum\limits_{cyc}\frac{x(x+y)}{2x^2+4y^2+5xy+xz}\geq\frac{1}{2},$$
which is $$\sum\limits_{cyc}(4x^4y^2+16x^4yz+3x^3y^2z-19x^3z^2y-4x^2y^2z^2)\geq0,$$
which is obvious.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Advice on proof in linear algebra. I just wrote my first proof in linear algebra so I'd love some advice on the things that go well and what could be improved upon. It's a proof by induction.
Theorem:
Let $A_n$ be a $n\times n$ matrix of the form:
$\begin{pmatrix}
2 & 1 & 0 & 0 && & \cdots & 0\\
1 & 2 & 1 & 0 && & \cdots & 0\\
0 & 1 & 2 & 1 && & \cdots & 0\\
0 & 0 & 1 & 2 && & \cdots & 0\\
\vdots& & & & \ddots & && \vdots\\
0&\cdots&&&&2&1&0\\
0&\cdots&&&&1&2&1\\
0&\cdots&&&&0&1&2\\
\end{pmatrix}$
Then $det(A_n)=3(n-1)$.
Proof:
We'll give a proof by mathematical induction.
Let $n=2$:
$\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}=4-1=3$.
Let $n=3$:
$\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}-\begin{vmatrix}
1&0\\
1&2\\
\end{vmatrix}=8-2=6$
Let $n=4$:
$\begin{vmatrix}
2&1&0&0\\
1&2&1&0\\
0&1&2&1\\
0&0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}-\begin{vmatrix}
1&0&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}-\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}=12-3=9$
Notice how due to the repetitive nature of our matrix, we'll candevise a recursive formula for our determinant: $|A_n|=2|A_{n-1}|-|A_{n-2}|$.
If we assume: $|A_n|=3(n-1)$, $|A_{n-1}|=3(n-2)$, $|A_{n-2}|=3(n-3)$, then:
$|A_n|=6(n-2)-3(n-3)=3(n-1)$
Since we checked $n=2, 3$; we conclude that $|A_n|=3(n-1)$ for all $n \in \mathbb{N} | n>1$
| You could remark some of the easy steps you did, making the proof more understandable yet (even though they are very simple). For example, you could write $3=3(2-1)$ (same with the other 2) to show that the formula holds for the smaller values of $n$.
Moreover, and more important, you should explain how you used Laplace's formula to derive the recurrent relationship (which involves another induction proof). It is not enough to show that the relationship is true for some values of $n$, but, of course, it will help you to generalize the expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Help with $\int \cos^6{(x)} \,dx$ Problem:
\begin{eqnarray*}
\int \cos^6{(x)} dx \\
\end{eqnarray*}
Answer:
\begin{eqnarray*}
\int \cos^4{(x)} \,\, dx &=& \int { \cos^2{(x)}(\cos^2{(x)}) } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=& \int { \frac{(1+\cos(2x))^2}{4} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \frac{\cos^2(2x)^2 + 2\cos(2x)+1}{4} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \frac{(\frac{1+\cos(4x)}{2} + 2\cos(2x)+1}{4} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \frac{1+\cos(4x) + 4\cos(2x)+2}{8} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \frac{\cos(4x) + 4\cos(2x)+3}{8} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=& \frac{\sin(4x)+ 8 \sin(2x)+12x}{32} \\
\text{Let }I_6 &=& \int \cos^6{(x)} \,\, dx \\
\end{eqnarray*}
To perform this integration, I use integration by parts with
$u = \cos^5(x)$ and $dv = \cos(x) dx$.
\begin{eqnarray*}
I_6 &=& \sin(x)\cos^5(x) - \int \sin(x) 5\cos^4(x)(-\sin(x)) \,\, dx \\
I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x)(\sin(x))^2 \,\, dx \\
I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x)(1 - \cos(x))^2 \,\, dx \\
I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x) \,\, dx - 5I_6 \\
6I_6 &=& \sin(x)\cos^5(x) + \int 5\cos^4(x) \,\, dx \\
6I_6 &=& \sin(x)\cos^5(x) + \frac{5\sin(4x)+ 40 \sin(2x)+60x}{32} + C_1 \\
6I_6 &=& \frac{32\sin(x)\cos^5(x) + 5\sin(4x)+ 40 \sin(2x)+60x}{32} + C_1 \\
I_6 &=& \frac{32\sin(x)\cos^5(x) + 5\sin(4x)+ 40 \sin(2x)+60x}{192} + C \\
\end{eqnarray*}
I believe that the above result is wrong. Using an online integral
calculator, I get:
\begin{eqnarray*}
I_6 &=& \frac{\sin(6x) + 9\sin(4x) + 45 \sin(2x) + 60x}{192} + C \\
\end{eqnarray*}
I am hoping that somebody can tell me where I went wrong.
Bob
| In your solution, when substituting the already known expression for $I_4$ (in the third line from the bottom), you forgot to multiply it by $5$. That's the only error there. Put it back in there, and you'll have a correct answer. Your answer would still look different from the output of that online integrator, but the two are in fact equivalent via trigonometric identities.
On a side note, this integral can also be found without integration by parts, but by using the same approach that worked for $I_4$ if you write $\cos^6(x)=(\cos^2(x))^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of angles of a triangle In a triangle $ABC$, if $\sin A+\sin B+\sin C\leq1$,then prove that $$\min(A+B,B+C,C+A)<\pi/6$$
where $A,B,C$ are angles of the triangle in radians.
if we assume $A>B>C$,then $\sum \sin A\leq 3 \sin A$,and $ A\geq \frac{A+B+C}{3}=\pi/3$.also $\sum \sin A\geq 3\sin C$ and $ C\leq \frac{A+B+C}{3}=\pi/3$.But I could not proceed with this. Please help me in this regard.Thanks.
| Since you assumed $A\geq B\geq C$, it must be that $\dfrac{A}{2}+C\leq\dfrac{\pi}{2}.$ Hence, $\sin\tfrac{A}{2}<\sin(\tfrac{A}{2}+C) = \cos(\tfrac{B-C}{2})$.
Finally, $$1\geq \sin A+\sin B+\sin C = \sin A+2\sin\tfrac{B+C}{2}\cos\tfrac{B-C}{2} = 2\cos\tfrac{A}{2}\big(\sin\tfrac{A}{2}+\cos\tfrac{B-C}{2}\big)>4\cos\tfrac{A}{2}\sin\tfrac{A}{2} = 2\sin A\Rightarrow$$
$\sin A< \tfrac{1}{2}$, which means that $A\gt\frac{5\pi}{6}$, because $A$ is the largest angle, by assumption. Then, $B+C<\tfrac{\pi}{6}$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Integrate $\int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}\,dx$ I don't know how to deal with this integral
$$I=\displaystyle\int_0^{\pi/2}\frac{\cos^2x}{a\cos^2x + b\sin^2x}\,dx$$
I reached the step
$$I
=\displaystyle\ \int_0^{\pi/2}\frac{1}{a + b\tan^2x}dx$$
Now what should I do? Please help.
| Now substitute
$\displaystyle\ u=\tan x \implies x= tan^{-1} u \implies dx=\frac{1}{1+u^2} du$
$\displaystyle\ =\int_0^{\infty}\frac{1}{(1+u^2)(a + bu^2)}du$
Partial Fractions
$\displaystyle\ =\int_0^{\infty}\frac{1}{(1+u^2)(a + bu^2)}du= \frac{1}{a-b} \left(\int_{0}^{\infty} \frac{du}{1+u^{2}} - b \int_{0}^{\infty} \frac{du}{a+bu^{2}} \right)$
Evaluating
$\displaystyle\ =\frac{1}{a-b} \left(\int_{0}^{\infty} \frac{du}{1+u^{2}} - b \int_{0}^{\infty} \frac{du}{a+bu^{2}} \right)$
$\displaystyle\ =\frac{1}{a-b} \left([\tan x]_0^{\infty} - b \left(\frac{\tan^{-1}\left[\frac{\sqrt{b}x}{\sqrt{a}}\right)}{\sqrt{ab}} \right]_0^{\infty}\right)$
Put the Values
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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4th root question/guidance Find all 4th roots of $-8 + 8i\sqrt 3$
$a=-8$
$b=8\sqrt 3$
$r= \sqrt{a^2+b^2}= \sqrt {(-8^2)+(8\sqrt{3})^2)}=\sqrt{64+192}=\sqrt {256} =16$
$\frac ar= cos\theta=\frac{-1}{2}$ $\space $ $\frac br= sin\theta$=$\frac {\sqrt3}{2}$
This gives me a different $\theta$ one being 120 degrees and the other 60 degrees
I can get the roots I just don't see were I am making the mistake with $\theta$, after that point I can finish it off myself.
| Polar form:
$z_1= 2(\cos 30 + i \sin 30)$
$z_2= 2(\cos 120 + i \sin 120)$
$z_3= 2(\cos 210 + i \sin 210)$
$z_4= 2(\cos 300 + i \sin 300)$
Rectangular Form:
$z_1= 2(\frac {\sqrt{3}}{2} + i \frac 12) = \sqrt {3} + i$
$z_2= 2(-\frac 12 + i\frac {\sqrt{3}}{2} ) =-1 +i \sqrt {3}$
$z_3= 2(-\frac {\sqrt{3}}{2} - i \frac 12)=-\sqrt {3} -i$
$z_4= 2(\frac 12 - i\frac {\sqrt{3}}{2}) = 1 -i\sqrt {3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum of $\displaystyle\sum_{n=3}^\infty \frac{2^n-1}{3^n}$ My work:$$\sum_{n=3}^\infty \frac{2^n-1}{3^n}$$
$$a_1 = \frac{2^3-1}{3^3}$$
$$a_1 = \frac{7}{27}, r=\frac{2}{3}$$
$$S_N=\frac{\frac{7}{27}}{1-\frac{2}{3}} = \frac{7}{9}$$
The correct answer is $\frac{5}{6}$
| Both geometric series are convergent, so
$$S=\sum_{n=3}^{+\infty}(\frac{2}{3})^n-\sum_{n=3}^{+\infty}\frac{1}{3^n}$$
$$\frac{2^3}{3^3}\frac{1}{1-\frac{2}{3}}-\frac{1}{3^3}\frac{1}{1-\frac{1}{3}}$$
$$=\frac{8}{9}-\frac{1}{18}=\frac{5}{6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given a series defined by recursion. Prove that there are integers $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$ I have this problem from an old exam that I can't solve.
Let $\{a_i\}_{i \geq 0}$ be the series define by recursion as:
$a_0 = 2$
$a_1 = 3$
$a_{n+1} = \frac{a_n^2 +5}{a_{n-1}}$ , $\forall n \in \Bbb N$
Prove that there are integers $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$
I've been trying with induction but I guess that there must be a direct, and more algebraic, way of proving it.
Any thoughts?? Thanks!
| Define a series $a_{n+1} = \frac{a_n^2 +k}{a_{n-1}}$ with some $a_0, a_1$ for $n\ge2$. Assume there are $S$ and $T$ such that $a_{n+1} = S a_n + T a_{n-1}$.
$$a_{n+1} = \frac{a_n^2+k}{a_{n-1}} \\
S a_n + T a_{n-1} = \frac{a_n^2+k}{a_{n-1}} \\
S a_n a_{n-1} + T a_{n-1}^2 = a_n^2+k \tag{1}
$$
now
$$a_{n+1} = \frac{a_n^2+k}{a_{n-1}} \\
a_{n+1}a_{n-1} = a_n^2+k \\
a_{n+1}a_{n-1} - k = a_n^2
\tag{2}
$$
combine the two:
$$
S a_n a_{n-1} + T(a_n a_{n-2} - k) = a_n^2+k \\
a_n(Sa_{n-1} + Ta_{n-2}) - Tk = a_n^2+k \\
a_n^2 - Tk = a_n^2+k \\
T = -1
$$
now find $S$:
$$
a_{2} = \frac{a_1^2+k}{a_{0}} \\
S a_1 - a_{0} = \frac{a_1^2+k}{a_{0}} \\
S = \frac{a_0^2+a_1^2+k}{a_{0}a_{1}} \\
$$
and in this case: $$S = \frac{2^2+3^2+5}{2\times3} = \frac{18}{6} = 3$$
So $S$ is an integer if it exists.
Prove by induction that $$S a_n - a_{n-1} = \frac{a_n^2+k}{a_{n-1}}$$ (note that the base case $n=2$ is covered by how we set $S$). Assume the equality is correct for all $n+1\ge k\ge3$.
$$
a_{n+2} = \frac{a_{n+1}^2+k}{a_{n}} \\
a_{n+2} = \frac{(S a_n-a_{n-1})^2+k}{a_{n}} = \frac{S^2 a_n^2-2S a_n a_{n-1}+a_{n-1}^2+k}{a_{n}} = S^2 a_n-2Sa_{n-1}+\frac{a_{n-1}^2+k}{a_{n}} \\
a_{n+2} = S(S a_n-a_{n-1})-Sa_{n-1}+\frac{a_{n-2}}{a_{n}}\frac{a_{n-1}^2+k}{a_{n-2}} = S a_{n+1} -Sa_n + \frac{a_{n-2}}{a_{n}} a_n \\
a_{n+2} = S a_{n+1} -Sa_n + a_{n-2} = S a_{n+1} -Sa_n + (S a_{n-1} - a_n) = S a_{n+1} - a_n
$$
As required.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Elementary Algebra Problem (in 8th grade) The exercise is to prove that $$ \forall x \in [0,3] $$ : $$ f(x)=\sqrt{18 + 3x -x^{2}} + \sqrt{9-x^{2}} + \sqrt{9-6x+x^{2}} + \sqrt{9x-3x^{2}} \le 12 $$
I notice that when $$ x=0 \implies f(x) = 3\sqrt{2} + 3 + 3 \le 12 $$
and when $$ x=3 \implies f(x) = 3\sqrt{2} \le 12 $$, but how further?
| $$f(x)=\sqrt{(6-x)(3+x)}+\sqrt{(3-x)(3+x)}+\sqrt{(3-x)^2}+\sqrt{3x(3-x)}$$
$\forall x\in [0,3]$, the factors $6-x$, $3+x$, $3-x$ and $3x$ in the radicands are non-negative.
By $GM\le AM$,
$$f(x)\le \frac{(6-x)+(3+x)}{2}+\frac{(3-x)+(3+x)}{2}+3-x+\frac{3x+(3-x)}{2}=12$$
Note that $12$ is not the global maximum, but a resonable upper bound.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating the convergence radius of a power series I've tried to calculate the convergence radius of the following power series:
$$\sum_{n=1}^{\infty}\frac{3^n+4^n}{5^n+6^n}x^n$$
The Cauchy–Hadamard theorem doesn't help in this situation (I think).
So what I did is I tried to apply the d'Alembert ratio test to it and got the following limit:
$$\lim_{n\to\infty}\frac{\frac{3^n+4^n}{5^n+6^n}}{\frac{3^{n+1}+4^{n+1}}{5^{n+1}+6^{n+1}}}=\lim_{n\to\infty}\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})}$$
but I haven't mannaged to solve in any way. I tried to calculate the limit of the function $$\lim_{x\to\infty}\frac{(3^x+4^x)(5^{x+1}+6^{x+1})}{(5^x+6^x)(3^{x+1}+4^{x+1})}$$ but of course that Lhospital's rule doesn't help (because it's in the power of n) so I was wondering:
*
*Is there a different way to find the convergence radius by using something other than the ration test?
*Might there be a identity regarding $$a^n+b^n=?$$ or $$\frac{a^n+b^n}{a^{n+1}+b^{n+1}}=?$$
| $$\begin{align}
\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})} &=\frac{6\cdot24^n+6\cdot18^n+5\cdot20^n+5\cdot15^n}{4\cdot24^n+4\cdot20^n+3\cdot18^n+3\cdot15^n}\\
&=\frac{6+6(3/4)^n+5(5/6)^n+5(5/8)^n}{4+4(5/6)^n+3(3/4)^n+3(5/8)^n}\to\frac32
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove using mean value theorem. $|\arctan(\frac{a}{4})^{4}-\arctan(\frac{b}{4})^{4}|\leq \pi^{3}\cdot|a-b|$
I started with $f(x)=\arctan(\frac{x}{4})^{4}$.
The function is continuous on $[a,b]$ and differentiable in (a,b)
So there exists a $c\in (a,b)$ such that $f'(c)=\frac{\arctan(\frac{b}{4})^{4}-\arctan(\frac{a}{4})^{4}}{b-a}$. From this I have
$$|\frac{-1}{1+c^{2}}|= \frac{1}{1+c^{2}}=\frac{|\arctan(\frac{b}{4})^{4}-\arctan(\frac{a}{4})^{4}|}{|b-a|}$$
and $a<c<b$. How to get the inequality with $\frac{1}{1+c^{2}}$from it?
| Let $f(x) = x^4$, and suppose $|x| \le M$, then $|f'(x)| \le 4 M^3$ for $|x| \le M$.
Using the mean value theorem
we have $|f(x)-f(y)| \le 4 M^3 |x-y|$.
Let $M = {\pi \over 2}$ and note that $|\arctan x| \le M$ and
$|\arctan' x| \le 1$, hence
$|(\arctan x)^4 - (\arctan y)^4| \le {1 \over 2} \pi^3 |\arctan x -\arctan y | \le {1 \over 2} \pi^3 |x-y|$.
Now let $x = {a \over 4}, y = {b \over 4}$ to get.
$|(\arctan {a\over 4})^4 - (\arctan {b \over 4})^4| \le {1 \over 2} \pi^3 |{a \over 4} -{b \over 4}| \le { \pi^3 \over 8} |x-y| $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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General solution to $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$.
My Approach:
Considering the equation as
$$
a\cos x+b\sin x=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\Big)=\sqrt{a^2+b^2}\big(\sin y.\cos x+\cos y.\sin x\big)=\sqrt{a^2+b^2}.\sin(y+x)=2
$$
$\frac{a}{\sqrt{a^2+b^2}}=\sin y$ and $\frac{b}{\sqrt{a^2+b^2}}=\cos y$.
$$
{\sqrt{a^2+b^2}}=\sqrt{8}=2\sqrt{2}\\\tan y=a/b=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}-\frac{1}{2}.\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}+\frac{1}{2}.\frac{1}{\sqrt{2}}}=\frac{\sin(\pi/3-\pi/4)}{\sin(\pi/3+\pi/4)}=\frac{\sin(\pi/3-\pi/4)}{\cos(\pi/3-\pi/4)}=\tan(\pi/3-\pi/4)\implies y=\pi/3-\pi/4=\pi/12
$$
Substituting for $y$,
$$
2\sqrt{2}.\sin(\frac{\pi}{12}+x)=2\implies \sin(\frac{\pi}{12}+x)=\frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\\\implies \frac{\pi}{12}+x=n\pi+(-1)^n\frac{\pi}{4}\implies x=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{12}
$$
What's going wrong with the approach ?
| Our hint is: $a\cos \theta +b\sin \theta =c$.
Given: $(\sqrt{3}-1)\cos \theta +(\sqrt{3}+1)\sin \theta =2$.
Let $(\sqrt{3}-1) = r\cos \alpha$ and $(\sqrt{3}+1) =r\sin \alpha$.
Then $r\cos \alpha \cos \theta + r\sin \alpha \sin \theta =2 \Rightarrow r\cos(\theta-\alpha) =2 \Rightarrow \cos(\theta-\alpha) =\frac{2}{r}$.
Now, $r =\sqrt{(\sqrt{3}-1)^2 +(\sqrt{3}+1)^2} = \sqrt{8} =2\sqrt{2}$.
Thus, $\cos(\theta-\alpha) =\frac{1}{\sqrt{2}} = \cos \frac{\pi}{4}$.
Also, $\tan \alpha =\frac{\sqrt{3}+1}{\sqrt{3}-1} = \tan(\frac{\pi}{2}-\frac{\pi}{3} +\frac{\pi}{4}) \Rightarrow \alpha =\frac{5\pi}{12}$.
Thus:$(\theta-\alpha) =2n\pi \pm \frac{\pi}{4}$. Giving, $\theta = 2n\pi \pm \frac{\pi}{4} +\frac{5\pi}{12}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On solvable octic trinomials like $x^8-5x-5=0$ Solvable quintic trinomials
$$x^5+ax+b=0$$
have been completely parameterized. Finding $6$th-deg versions is relatively easy to do such as,
$$x^6+3x+3=0$$
which factors over $\sqrt{-3}$. No $7$th-deg are known, but surprisingly there are octic ones, such as the simple,
$$x^8-5x-5=0$$
which factors over $\sqrt{5}$. And the not-so-simple ones,
$$x^8-11(4x+3)=0\\x^8+16(4x+7)=0\\x^8 + 5\cdot23^2(12 x+43) =0$$
which factors over a quartic extension (and needs the cube root of unity).
Q: Any other octic examples, if possible parametric?
$\color{green}{Update:}$
Klajok in his answer below has found a family for the class of octic trinomials that factor over a quadratic extension. However, another class needs a quartic extension. For example,
$$x^8-44x-33=0\tag1$$
which factors into four quadratics,
$$x^2 + v x - (2v^3 - 7v^2 + 5v + 33)/13=0$$
and where $v$ is any root of $v^4 + 22v + 22=0$. More generally, eliminating $v$ between
$$x^2 + v x + (pv^3 +qv^2 + rv + s)=0$$
$$v^4+av^2+bv+c=0$$
easily done by the resultant function of Mathematica will result in an irreducible but solvable octic and judicious choice of rational coefficients will yield a trinomial. However, it is not known if this second class of trinomials like $(1)$ has a parametric family as well.
| A result of Harris [1] is that every monic palindromic polynomial of degree-8 can be factored into two monic palindromic polynomials of degree-4.
$$
\begin{align}
f(x) & = x^8 + ax^7 + bx^6 + cx^5 + dx^4 + cx^3 + bx^2 + ax + 1 \\
& = (x^4 + px^3 + qx^2 + px + 1)(x^4 + rx^3 + sx^2 + rx + 1) \\
& = x^8+x^7 (p+r)+x^6 (pr+q+s)+x^5 (p(s+1)+qr+r)+x^4 (2pr+qs+2)+x^3 (p(s+1)+qr+r)+ x^2 (pr+q+s)+x (p+r)+1
\end{align}
$$
Equating the coefficients:
$$
\begin{align}
a &= p+r \\
b &= pr + q + s \\
c &= p(s+1) + qr + r \\
d &= 2pr + qs + 2
\end{align}
$$
For the subset of degree-8 monic palindromic polynomials of the form
$$
\begin{align}
f(x) & = x^8 + 0x^7 + 0x^6 + 0x^5 + dx^4 + 0x^3 + 0x^2 + 0x + 1 \\
& = x^8 + dx^4 + 1 \\
\text{we have }
0 &= p+r \\
0 &= pr + q + s \\
0 &= p(s+1) + qr + r \\
d &= 2pr + qs + 2
\end{align}
$$
We have the parametric solutions:
$$
r = 0 ∧ s = -q ∧ p = 0 ∧ d = 2 - q^2, \\
r = \sqrt{2} \sqrt{q} ∧ s = q ∧ p = -\sqrt{2} \sqrt{q} ∧ d = q^2 - 4 q + 2, \\
r = -\sqrt{2} \sqrt{q} ∧ s = q ∧ p = \sqrt{2} \sqrt{q} ∧ d = q^2 - 4 q + 2. \\
$$
Note that all of the above are parametrized on $d$.
On similar lines, we can factor the monic quartic palindromic polynomials into two monic quadratic palindromic polynomials each and then use the quadratic formula to get the roots.
References
[1]: J. R. Harris, "96.31 Palindromic Polynomials," The Mathematical Gazette, vol. 96, no. 536, p. 266–69, 2012. https://doi.org/10.1017/S0025557200004526
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "14",
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Curious limits with tanh and sin These two limits can be easily solved by using De l'Hopital Rule multiple times (I think), but I suspect that there could be an easier way... Is there?
\begin{gather}
\lim_{x\to 0} \frac{\tanh^2 x - \sin^2 x}{x^4} \\
\lim_{x\to 0} \frac{\sinh^2 x - \tan^2 x}{x^4}
\end{gather}
Thanks for your attention!
| From the standard Taylor series expansions, as $x \to 0$,
$$
\begin{align}
\sin x&=x-\frac{x^3}{6}+\frac{x^5}{120}+O(x^6)
\\\tanh x&=x-\frac{x^3}{3}+\frac{2 x^5}{15}+O(x^6)
\end{align}
$$ ones gets
$$
\begin{align}
\left(\sin x\right)^2&=x^2-\frac{x^4}{3}+O(x^6)
\\\left(\tanh x\right)^2&=x^2-\frac{2 x^4}{3}+O(x^6)
\end{align}
$$ giving, as $x \to 0$,
$$
\frac{\left(\tanh x\right)^2-\left(\sin x\right)^2}{x^4}=\frac{-\frac{x^4}{3}+O(x^6)}{x^4}=-\frac13+O(x^2).
$$
| {
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Integration with Half space Gaussian I have a problem to solve and I have something that i don't know how to do.
The half-space Gaussian integral is given :
$$\int_{0}^\infty \exp(-ax^2)dx = \frac{1}2 \sqrt{\frac{\pi}{a}}$$
I have to calculate
$$\int_{0}^\infty \exp \left(-ax^2 - \frac{b}{x^2} \right)dx$$
a and b are real and positive
Is someone have an idea ? :)
Thanks for your answers
Mathieu
|
Note that in THIS ANSWER, I presented a solution to a more general version of the integral of interest herein.
Let $I(a,b)$ be the integral given by
for $a>0$ and $b>0$.
$$I(a,b)=\int_0^\infty e^{-\left(ax^2+\frac{b}{x^2}\right)}\,dx \tag 1$$
Enforcing the substitution $x\to \sqrt[4]{b/a}x$ into $(1)$ reveals
$$\begin{align}
I(a,b)&=\sqrt[4]{\frac{b}{a}}\int_0^\infty e^{-\sqrt{ab}\left(x^2+\frac{1}{x^2}\right)}\,dx \tag 2
\end{align}$$
Next, noting that $x^2+\frac{1}{x^2}=\left(x-\frac1x\right)^2+2$, we can write $(2)$ as
$$\begin{align}
I(a,b)&=\sqrt[4]{\frac{b}{a}}e^{-2\sqrt{ab}}\int_0^\infty e^{-\sqrt{ab}\left(x-\frac{1}{x}\right)^2}\,dx\tag 3
\end{align}$$
Enforcing the substitution $x\to 1/x$ in $(3)$ yields
$$I(a,b)=\sqrt[4]{\frac{b}{a}} e^{-2\sqrt{ab}}\int_0^\infty e^{-\sqrt{ab}\left(x-\frac{1}{x}\right)^2}\,\frac{1}{x^2}\,dx\tag4$$
Adding $(3)$ and $(4)$, we obtain
$$\begin{align}
I(a,b)&=\frac12\sqrt[4]{\frac{b}{a}} e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-\sqrt{ab}\left(x-\frac{1}{x}\right)^2}\,d\left(x-\frac{1}{x}\right)\\\\
&=\frac12\sqrt[4]{\frac{b}{a}} e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-\sqrt{ab}\,x^2}\,dx\\\\
&=\frac1{2\sqrt a} e^{-2\sqrt{ab}}\int_{-\infty}^\infty e^{-x^2}\,dx\\\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{\sqrt{\pi}}{2\sqrt a}e^{-2\sqrt{ab}}}
\end{align}$$
| {
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If $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$, find the value of $a+b$
$\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$ where $a,b$ are natural numbers. Find the value of $a+b$.
I am not able to proceed with solving this question as I have no idea as to how I can calculate $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}$. A small hint would do.
| √(31+√(31+√(31....))) = s
s = √(31+s)
s² = s+31
s = (5√5 + 1)/2 (by the quadratic formula)
√(1+√(1+√(1....))) = k
k = √(1+k)
k² = k+1
k = (√5 + 1)/2 (by the quadratic formula)
s/k = (5√5 + 1)/(√5 + 1)
= (5√5 + 1)(√5 - 1)/4 (multiplying the numerator and denominator both by conjugate √5 - 1)
= (24-4√5)/4
= 6-√5
Since 6 - √5 = a - √b,
a = 6, b = 5
a + b = 11
Thus, the answer is 11.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059779",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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show $\frac{y}{x^2+y^2} $ is harmonic except at $y=0,x=0$ Let $f(z)=u(x,y)+iv(x,y) $
where $$ f(z)=u(x,y)=\frac{y}{x^2+y^2}$$
show $u(x,y)$ is harmonic except at $z=0$
Attempt
$$ u=\frac{y}{x^2+y^2}=y(x^2+y^2)^{-1} $$
Partial derivatives with x
$$\begin{aligned}
u_x&= y *(x^2+y^2)^{-2}*-1*2x
\\ &= -y*2x(x^2+y^2)^{-2}=-2xy(x^2+y^2)^{-2}
\\ u_{xx} &= -2xy*(x^2+y^2)^{-3}*-2*2x+-2y*(x^2+y^2)^{-2}
\\&= \frac{-2xy}{(x^2+y^2)^3}*-4x +\frac{-2y}{(x^2+y^2)^2}
\\&=\frac{8x^2y}{(x^2+y^2)^3}+\frac{-2y}{(x^2+y^2)^2}
\\
\end{aligned} $$
Partial Derivatives with y
$$\begin{aligned}
u_y&=1(x^2+y^2)^{-1}+y*(x^2+y^2)^{-2}*-1*2y
\\ &=(x^2+y^2)^{-1}-2y^2(x^2+y^2)^{-2}
\\ u_{yy}&=-1(x^2y^2)^{-2}*2y -2*2y(x^2+y^2)^{-2}-2y^2*-2(x^2+y^2)^{-3}*2y
\\ &=-2y(x^2+y^2)^{-2}-4y(x^2+y^2)^{-2}+8y^3(x^2+y^2)^{-3}
\\ &=\frac{-6y}{(x^2+y^2)^2} + 8y^3(x^2+y^2)^{-3}
\end{aligned} $$
From here need to show that $u_{xx}+u_{yy}=0$ and technically say why the other partials are continous right?? This was a test question whith 3 lines of paper by the way
| While I believe that the "right" answers are those already given, I would like to add yet another one, based on polar coordinates. Introduce
$$
\begin{cases}
x=r\cos \phi\\
y=r\sin \phi
\end{cases}
$$
The given function $f(x, y)=\frac{y}{x^2+y^2}$ is harmonic if and only if
$$
\left(\partial_r^2 +r^{-1}\partial_r +r^{-2}\partial_\phi^2\right)\left(\sin (\phi) r^{-1}\right)=0, $$
which is, of course, true.
This computation is slightly faster than the Cartesian one because $f$ is separable in polar coordinates. ("Separable" here means that a function is expressed as a product of functions of a single variable).
| {
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Showing $\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le1\Rightarrow x\le0$
$\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le1\Rightarrow x\le0$
How is the proof. If I separate the denominator with triangle inequality,
$\lvert \frac{x^2-2x+3}{x^2-4x+3}\rvert\le \frac{\lvert x^2-2x+3\rvert}{\lvert x^2-2x+3 \rvert-\lvert 2x\rvert}\rvert \ge1$
Which means that the triangle inequality should stay a strict inequality, so $x^2-2x+3$ and $2x$ have opposite signs. And the discriminant of the former is negative, therefore the roots are complex, in particular the graph doesn't touch the x-axis. One can insert any value to determine its sign, for example at $0$ gives $3$, so $x$ has to be negative.
Is there another possibility, without much text to prove it ?
| What you want is
$$\left|\frac{x^2-2x+3}{x^2-4x+3}\right|\le1\iff-1\le\frac{x^2-2x+3}{x^2-4x+3}\le1$$
Beginning with the left inequality:
$$-1\le\frac{x^2-2x+3}{x^2-4x+3}\iff\frac{x^2-2x+3}{x^2-4x+3}+1\ge0\iff\frac{2x^2-6x+6}{x^2-4x+3}\ge0\iff$$
$$\frac{x^2-3x+3}{(x-1)(x-3)}\ge 0\iff (x-1)(x-3)>0\;\text{ (why?)}\implies x<1\;\text{or}\;x>3$$
Now you do the other inequality and take the intersection of both solution sets.
| {
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If $a(n)=n^2+1$ then $\gcd(a_n,2^{d(a_n)})=1\text{ or }2$? Let $n\in\mathbf{N}$. I write $a_n=n^2+1$ and let $d(a_n)$ count the number of divisors of $a_n$. Set $$\Phi_n=\gcd\left(a_n,2^{d\left(a_n\right)}\right)$$ I would like to show and I believe it to be true that
$$\Phi_n =
\begin{cases}
1, & \text{if $n$ is even} \\[2ex]
2, & \text{if $n$ is odd}
\end{cases}$$
My gut instinct is two beak it down by parity and then use Euclid's lemma. But I am not sure how to use Euclid's lemma.
To see a working example consider $n=15$. Then $a_n=226$, $d(a_n)=4$ and $$\text{ }\Phi_n=\gcd(226,2^{4})=\gcd(226,16)=2$$
| Note that $2^{d(a_n)}$ can only be divisible by $1$ and powers of $2$.
If $n$ is even then $n^2+1$ is odd and in that case $\gcd=1$.
If $n$ is odd, then $n^2+1 \equiv 2 \pmod{4}$. Thus $n$ is not divisible by $4$, hence the $\gcd=2$.
Added explanation:
Using the division algorithm, we can write any integer $n=4k+r$, where $r \in \{0,1,2,3\}$. So when $n$ is odd, then it can only be of the form $4k+1$ or $4k+3$. Now consider the case when $n=4k+1$, then
$$n^2+1=(4k+1)^2+1=16k^2+8k+2=4(\text{some integer})+2.$$
Likewise when $n=4k+3$,we have
$$n^2+1=(4k+3)^2+1=16k^2+24k+8+2=4(\text{some integer})+2.$$
This means $n^2+1$ will always leave a remainder of $2$, when divided by $4$.
| {
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How to prove $\sqrt[n+1]{n+1}-\sqrt[n]{n}\sim-\frac{\ln{n}}{n^2}$
Show that
$\sqrt[n+1]{n+1}-\sqrt[n]{n}\sim-\frac{\ln{n}}{n^2}$,
when $n\to+\infty$
I'm learning Taylor's Formula. The given solution is:
$\sqrt[n+1]{n+1}-\sqrt[n]{n}=e^{\frac{\ln(n+1)}{n+1}}-e^{\frac{\ln{n}}{n}}$
and use Taylor's Formula:
$e^{\frac{\ln(n+1)}{n+1}}=1+\frac{\ln(n+1)}{n+1}+\frac{1}{2}(\frac{\ln(n+1)}{n+1})^2+\frac{1}{6}(\frac{\ln(n+1)}{n+1})^3+o(\frac{(\ln n)^3}{n^3})$ and
$e^{\frac{\ln(n)}{n}}=1+\frac{\ln(n)}{n}+\frac{1}{2}(\frac{\ln(n)}{n})^2+\frac{1}{6}(\frac{\ln(n)}{n})^3+o(\frac{(\ln n)^3}{n^3})$,
then $\sqrt[n+1]{n+1}-\sqrt[n]{n}=-\frac{\ln{n}}{n^2}+o(\frac{\ln{n}}{n^2})\sim\frac{\ln n}{n^2}$
I don't understand the last step: why is it $=-\frac{\ln{n}}{n^2}+o(\frac{\ln{n}}{n^2})$ ? And any other solutions?
| You just need to handle the subtraction of series for $\sqrt[n+1]{n+1}$ and $\sqrt[n] {n} $ term by term in a proper manner upto 3 terms. The first term $1$ cancels out in both series. The second terms upon subtraction lead to
\begin{align}
A &= \frac{\log(n+1)}{n+1}-\frac{\log n} {n} \notag\\
&= \frac{\log(n+1)}{n+1}-\frac{\log n} {n+1} +\log n\left(\frac{1}{n+1}-\frac{1}{n}\right) \notag\\
&= \frac{1}{n+1}\log\left(1+\frac{1}{n}\right)-\frac{\log n}{n(n+1)}\notag\\
&= - \frac{\log n} {n^{2}}+\left(\frac{1}{n}-\frac{1}{n^{2}}+o\left(\frac{1}{n^{2}}\right)\right)\left(\frac{1}{n}-\frac{1}{2n^{2}}+o\left(\frac{1}{n^{2}}\right)\right)+\frac{\log n} {n} \left(\frac{1}{n}-\frac{1}{n+1}\right)\notag\\
&= - \frac{\log n} {n^{2}}+\frac{1}{n^{2}}+\frac{\log n} {n^{2}(n+1)}+o\left(\frac{1}{n^{2}}\right)\notag\\
&= - \frac{\log n} {n^{2}}+o\left(\frac{\log n} {n^{2}}\right)\notag
\end{align}
The third terms on subtraction lead to
\begin{align}
B &= \frac{A} {2}\left(\frac{\log(n+1)}{n+1}+\frac{\log n} {n} \right) \notag\\
&= o\left(\frac{\log n} {n^{2}}\right)\notag
\end{align}
and therefore we have $$\sqrt[n+1]{n+1}-\sqrt[n]{n}=-\frac{\log n} {n^{2}}+o\left(\frac{\log n} {n^{2}}\right)$$
| {
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"url": "https://math.stackexchange.com/questions/2065510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The sum of the fourth powers of the first $n$ positive integers I am studying mathematical induction and most of the times I have to prove something. Like, for example:
$1 + 4 + 9 + ...+ n^2 = \frac{n(n+1)(2n+1)}{6}$
This time I found a question that ask me to find a formula for
$1 + 16 + 81 + .... + n^4$
How can I do this with induction? And is there really a formula for this sum?
| As $S_0=0$ and $S_n-S_{n-1}=n^4$, $S_n$ must be a polynomial of the fifth degree with no independent term, let
$$S_n=an^5+bn^4+cn^3+dn^2+en.$$
Then
$$S_n-S_{n-1}=\\
a(n^5-n^5+5n^4-10n^3+10n^2-5n+1)+
\\b(n^4-n^4+4n^3-6n^2+4n-1)+\\
c(n^3-n^3+3n^2-3n+1)+\\
d(n^2-n^2+2n-1)+\\
e(n-n+1)=\\
a(5n^4-10n^3+10n^2-5n+1)+
\\b(4n^3-6n^2+4n-1)+\\
c(3n^2-3n+1)+\\
d(2n-1)+\\
e.
$$
By identification with $n^4$,
$$\begin{cases}5a=1\\-10a+4b=0\\10a-6b+3c=0\\-5a+4b-3c+2d=0\\a-b+c-d+e=0.\end{cases}$$
This is a triangular system, which readily gives
$$a=\frac15,b=\frac12,c=\frac13,d=0,e=-\frac1{30}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$
Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ for $n>1$
I can use $$(n+1)^n>(2n)!!=n!2^n$$ but in the my case, the exponent is always decreased by $1$, for the moment I don't care about it, I apply the same for $n+2$
$(n+2)^{n+1}>(2n+2)!!=(n+1)!2^{n+1}$
gathering everything together,
$(n+1)^{n-1}(n+2)^n=\frac{(n+1)^n(n+2)^{n+1}}{(n+1)(n+2)}>\frac{(n+1)(n!)^22^{2n+1}}{(n+1)(n+2)}$
$\iff(n+1)^{n-1}(n+2)^n>(n!)^2\times\frac{2^{2n+1}}{(n+2)}$
but $\frac{2^{2n+1}}{(n+2)}>3^n$ is not true for $n=2$
can you suggest another approach ?
| EDIT: This answer is wrong, because I mixed up my left and right-hand sides right at the end. I think it is salvageable, but it'll be quite a bit of work.
I'll do it without induction.
Rearrange: we want $\left(\frac{(n+1)(n+2)}{3}\right)^{n-1} \frac{n+2}{3} > (n!)^2$
We'll show that this actually holds if we remove the $\frac{n+2}{3}$ term (which is always $\geq 1$ anyway).
The right-hand side of the modified desired inequality is is $$2^2 \times 3^2 \times \dots \times n^2$$ with $n-1$ terms.
The left-hand side is $$\frac{(n+1)(n+2)}{3} \times \dots \times \frac{(n+1)(n+2)}{3}$$ with $n-1$ terms again.
But $\frac{(n+1)(n+2)}{3}$ is bigger than or equal to the $i^2$ term whenever $i$ is less than or equal to $\sqrt{\frac{(n+1)(n+2)}{3}}$, and that's $\leq n$ whenever $n > 2$. So if $n>2$, for every $i \leq n$ we have each right-hand term less than its corresponding left-hand term.
So we only need to check $n=1$ and $n=2$, and they're very easy.
| {
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A natural number is a perfect square as well as a perfect cube. Show that it is $0$ or $1$ $ ($mod $7$ $)$. A natural number is a perfect square as well as a perfect cube. Show that it is $0$ or $1$ $($mod $7$$)$.
I tried the following.
There are integers numbers $x,y$ such that $n=x^{2}=y^{3}.$ By using Euclidean division, then $x$ and $y$ can be written as $7k,7k+1,7k+2,7k+3,7k+4,7k+5$ or $7k+6$. I am trying some contradictory stuff.
I don't know from where to go from here. Please suggest some hints.
| You know that all the squares are equals to $0,1,2$ or $4$ mod $7$ because:
$$0^2=0\pmod 7$$
$$1^2=1\pmod 7$$
$$2^2=4\pmod 7$$
$$3^2=2\pmod 7$$
$$4^2=2\pmod 7$$
$$5^2=4\pmod 7$$
$$6^2=1\pmod 7.$$
And all the cubes are equals to $0,1$ or $6$ mod $7$ because:
$$0^3=0\pmod 7$$
$$1^3=1\pmod 7$$
$$2^3=1\pmod 7$$
$$3^3=6\pmod 7$$
$$4^3=1\pmod 7$$
$$5^3=6\pmod 7$$
$$6^3=6\pmod 7.$$
So if your number is a square and a cube at the same time, then it is necessarily equal to $0$ or $1$ mod $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069374",
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"source": "stackexchange",
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Determinant of a non-square block matrix $M_{n\times k}$ is defined as a matrix whose all elements are '-1'.
The following block matrix is as such:
$A =\begin{bmatrix}m\cdot I_{n-1} & M_{n-1\times m}\\M_{m\times n-1} & n\cdot I_{m}\end{bmatrix}$
prove the following: $det A = n^{m-1}\cdot m^{n-1}$
| Let more generally let $C(\gamma) = \gamma 1_{n-1}1_{m}^T $ denote the ${n-1}\times m$ matrix with each element equal to $\gamma$ (here $1_k$ denotes the k-dimensional column vector of all ones) and let $A(\gamma) = \begin{pmatrix} mI_{n-1} & C(\gamma) \\ C(\gamma)^T & nI_m\end{pmatrix}$. Your problem is to compute the determinant of $A(-1)$.
Using the following identity
$$ \begin{pmatrix} I_{n-1} & 0 \\ - \dfrac{C(\gamma)^T}{m} & I_m \end{pmatrix} \begin{pmatrix} mI_{n-1} & C(\gamma) \\ C(\gamma)^T & nI_m\end{pmatrix} = \begin{pmatrix} mI_{n-1} & C(\gamma)\\ 0 & nI_m - \dfrac{C(\gamma)^TC(\gamma)}{m} \end{pmatrix}.$$
we have on taking determinants $$
\begin{align}
\operatorname{det}(A(\gamma)) &=\operatorname{det}(mI_{n-1})\operatorname{det}(nI_m - \dfrac{C(\gamma)^TC(\gamma)}{m})\\ &= m^{n-1} \times n^{m} \det(I_m - \dfrac{C(\gamma)^TC(\gamma)}{mn})\\ &= m^{n-1}n^{m} \det(I_{m} - \gamma^2 \dfrac{(n-1)}{nm} 1_{m}1_{m}^{T}) \tag{*}
\\
&= m^{n-1} n^m \det(I_m - \delta 1_m 1_m^T)
\end{align}
$$
where $\delta = \gamma^2\dfrac{n-1}{nm}.$
In (*) we have used $C(\gamma) = \gamma 1_{n-1}1_{m}^T$.
Consider the following identities:
$$
\begin{pmatrix} I_m & 0 \\ -1_m^T & 1 \end{pmatrix} \begin{pmatrix} I_m & \delta 1_m \\ 1_m^T & 1 \end{pmatrix} = \begin{pmatrix} I_m & \delta 1_m \\ 0 & 1 - m \delta \end{pmatrix},$$
$$\begin{pmatrix} I_m & -\delta 1_m \\ 0 & 1 \end{pmatrix} \begin{pmatrix} I_m & \delta 1_m \\ 1_m^T & 1 \end{pmatrix} = \begin{pmatrix} I_m - \delta 1_m 1_m^T & 0 \\ 1_m^T & 1 \end{pmatrix}. $$
On taking determinants we have $\det\begin{pmatrix} I_m - \delta 1_m 1_m^T \end{pmatrix} = 1 - m \delta.$
This gives $\det(A(\gamma))= m^{n-1}n^m(1 - m \delta) = m^{n-1}n^m (1 - \gamma^2\dfrac{n-1}{n})$ so $ \det(A(-1))=m^{n-1}n^{m-1}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $\int_{-\infty}^x\frac{1}{\sqrt{2\pi}} e^{-(\frac{x^2}{2}+2x+2)}\,dx$ I want to solve this integral from $-\infty$ to $x$ .
$$f_X(x)=\frac{1}{\sqrt{2\pi}} e^{-(\frac{x^2}{2}+2x+2)}, -\infty<x<\infty$$
I have searched as much as I could and I found a solution in wikipedia
$$\int_{-\infty}^{\infty} x e^{-a(x-b)^2} dx=b \sqrt{\frac{\pi}{a}}$$
In that solution there is a $x$ before $e$ but in my problem there is not any $x$ before $e$ and it is the only deference also the answer in that solution is a number but I need a function which contains $x$ .
I really need this answer as soon as possible
| Complete the square in the exponential function, then let $y=\frac{x+2}{\sqrt{2}}$
\begin{align}
\frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{z} \mathrm{e}^{-(\frac{1}{2}x^{2}+2x+2)} dx &=
\frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{z} \mathrm{e}^{-\frac{1}{2}(x+2)^{2}} dx \\
&= \frac{1}{\sqrt{\pi}} \int\limits_{-\infty}^{(z+2)/\sqrt{2}} \mathrm{e}^{-y^{2}} dy \\
&= \frac{1}{\sqrt{\pi}} \frac{\sqrt{\pi}}{2} \mathrm{erf}(y) \Big|_{-\infty}^{(z+2)/\sqrt{2}} \\
&= \frac{1}{2} \mathrm{erf}\left(\frac{z+2}{\sqrt{2}} \right) + \frac{1}{2}
\end{align}
Where
$$\int \mathrm{e}^{-y^{2}} dy = \frac{\sqrt{\pi}}{2} \mathrm{erf}(y) + C$$
is the error function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072617",
"timestamp": "2023-03-29T00:00:00",
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How many ways can 8 teachers be distributed among $4 $ schools? There are several ways that the teachers can be divided amongst $4$ schools, namely here are the possible choices I came up with:
$1) 1 1 1 5$
$2) 1 1 2 4$
$3) 1 1 3 3$
$4) 1 2 2 3$
$5) 2 2 2 2$
now given the fact that say $2213$ is the same as $1 2 2 3$ it was omitted. With out repeats I believe these 5 are the only possibilities.
1)
${8 \choose 5} \times {3 \choose 1} \times {2 \choose 1} \times {1 \choose 1}$:
$\frac{8!}{5!3!} \times \frac{3!}{1!2!} \times \frac{2!}{1!1!} \times 1$
Which comes out to
$56 \times 3 \times 2 \times 1 = 336$
2)
${8 \choose 4} \times {4 \choose 2} \times {2 \choose 1} \times {1 \choose 1}$:
$\frac{8!}{4!4!} \times \frac{4!}{2!2!} \times \frac{2!}{1!1!} \times 1$
Which comes out to
$70 \times 6 \times 2 \times 1= 840$
3)
${8 \choose 3} \times {5 \choose 3} \times {2 \choose 1} \times {1 \choose 1}$
$\frac{8!}{3!5!} \times \frac{5!}{3!2!} \times \frac{2!}{1!1!} \times 1$
Which comes out to
$56 \times 10 \times \times 2 = 1,120$
4)
${8 \choose 3} \times {5 \choose 2} \times {3 \choose 2} \times {1 \choose 1}$
$\frac{8!}{3!5!} \times \frac{5!}{2!3!} \times \frac{3!}{2!1!} \times \frac{1!}{1!0!}$
Which comes out to:
If 8 new teachers are to be divided amongst 4 new schools how many divisions are possible?
$56 \times 10 \times 3 \times 1= 1,680$
5)
${8 \choose 2} \times {6 \choose 2} \times {4 \choose 2} \times {2 \choose 2}$
$\frac{8!}{2!6!} \times \frac{6!}{2!4!} \times \frac{4!}{2!2!} \times \frac{2!}{2!0!}$
Which comes out to:
$28 \times 15 \times 6 \times 1 = 2,520$
What am I missing?
| Assuming distinct teachers, distinct schools, and having identified the $5$ patterns, a foolproof mechanical way is to sum up the product of two multinomial coefficients for each case, one for the pattern, the other for the frequencies of singletons, doubles, triples, etc, viz.
$\binom{8}{1,1,1,5}\binom{4
}{3,1} + \binom{8}{1,1,2,4}\binom{4}{2,1,1} + \binom{8}{1,1,3,3}\binom{4}{2,2} + \binom{8}{1,2,2,3}\binom{4}{1,2,1} + \binom{8}{2,2,2,2}\binom44 = 40824$
And if preferring permutations to multinomial coefficients, the equivalent expression
$\frac{8!}{1!1!1!5!}\cdot\frac{4!}{3!1!} + \frac{8!}{1!1!2!4!}\cdot\frac{4!}{2!1!1!} + \frac{8!}{1!1!3!3!}\cdot\frac{4!}{2!2!}+\frac{8!}{1!2!2!3!}\cdot\frac{4!}{1!2!1!}+ \frac{8!}{2!2!2!2!}\cdot\frac{4!}{4!} = 40824 $
| {
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Given $\lim_{n\to\infty} a_n = a$ what is the limit $\lim \limits_{n \to \infty}\frac{a_n}{3^1}+\frac{a_{n-1}}{3^2}+\ldots+\frac{a_1}{3^n}$?
Given $\lim \limits_{n \to \infty}a_n = a$ then I need to find the limit $\lim \limits_{n \to \infty} \frac{a_n}{3} + \frac{a_{n-1}}{3^2} + \frac{a_{n-2}}{3^3} + \dotso + \frac{a_1}{3^n}$.
It seems this problem can be tackled by Stolz–Cesàro theorem. Unfortunately, I don't know how to pick $x_n$ and $y_n$.
| Fix $\epsilon > 0$. Since $\displaystyle\lim_{n \to \infty}a_n = a$, there exists an $N \in \mathbb{N}$ such that $|a_n-a| < \epsilon$ for all $n \ge N$.
Let $S_n := \dfrac{a_n}{3}+\dfrac{a_{n-1}}{3^2}+\cdots+\dfrac{a_2}{3^{n-1}}+\dfrac{a_1}{3^n}$. Then, $S_{n+1} = \dfrac{1}{3}S_n+\dfrac{1}{3}a_{n+1}$ for all $n \in \mathbb{N}$.
We can rewrite this as $S_{n+1} - \dfrac{a}{2} = \dfrac{1}{3}(S_n-\dfrac{a}{2})+\dfrac{1}{3}(a_{n+1}-a)$. Then, for $n \ge N$ we have:
\begin{align}
\left|S_{n+1} - \dfrac{a}{2}\right| & = \left|\dfrac{1}{3}(S_n-\dfrac{a}{2})+\dfrac{1}{3}(a_{n+1}-a)\right|
\\
&\le \dfrac{1}{3}\left|S_n-\dfrac{a}{2}\right|+\dfrac{1}{3}|a_{n+1}-a|
\\
&\le \dfrac{1}{3}\left|S_n-\dfrac{a}{2}\right|+\dfrac{1}{3}\epsilon.
\end{align}
Now, use induction to show that $\left|S_n-\dfrac{a}{2}\right| \le \left(\left|S_N-\dfrac{a}{2}\right|-\dfrac{1}{2}\epsilon\right) \cdot 3^{-(n-N)}+\dfrac{1}{2}\epsilon$ for all $n \ge N$.
Then, pick $N' > N$ such that $\left|\left|S_N-\dfrac{a}{2}\right|-\dfrac{1}{2}\epsilon\right| \cdot 3^{-(n-N)} \le \dfrac{1}{2}\epsilon$ for all $n \ge N'$.
With this choice of $N'$, we have $\left|S_n-\dfrac{a}{2}\right| \le \epsilon$ for all $n \ge N'$.
This can be done for any $\epsilon > 0$. Thus, $\displaystyle\lim_{n \to \infty}S_n = \dfrac{a}{2}$.
| {
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"url": "https://math.stackexchange.com/questions/2075424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\sqrt{1000} < x < 1000$? I have been given that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac {996}{995} \times \frac{998}{997} \times \frac {1000}{999}$$
How can I prove that $\sqrt{1000} < x < 1000$?
| \begin{align}
x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{1000}{999} \times \frac {1000}{999}\right) \\
&\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{1000}{999} \times \frac {1001}{1000}\right) \\
&= 1001
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the inequality and show solution sets on the real line $$-\frac{x+5}{2} \le \frac {12+3x}{4}$$
I always have issues with problems like this. I chose to ignore the negative sign at the beginning and got the answer. Is that a good method? When solving this inequality what is the best method for no mistakes?
My answer/ method:
$$-(2)(12+3x) \le (4)(x+5)$$
$$-24-6x \le 4x+20$$
$$-24-20 \le 4x+6x$$
$$-44\le10x$$
$$-\frac{44}{10}\le x$$
$$-\frac{22}{5} \le x$$
$$\left(\infty,-\frac{22}{5}\right]$$
| $$-\frac{x+5}{2} \le \frac {12+3x}{4}$$
Multiply $4$ both sides:
$$-2(x+5) \le 12+3x$$
$$-2x-10 \le 12 +3x$$
$$-22 \le 5x$$
$$x \geq \frac{-22}{5}$$
Your mistakes:
$-2(12+3x)=-24-6x$ rather than $-24+6x$. If this is just a typo, the next line is fine.
After this mistake, surprisingly in the next line, you corrected yourself.
$$x \ge -\frac{22}{5} \iff x \in [\frac{-22}{5},\infty)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A question about neighboring fractions. I have purchased I.M. Gelfand's Algebra for my soon-to-be high school student son, but I am embarrassed to admit that I am unable to answer seemingly simple questions myself.
For example, this one:
Problem 42. Fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are called neighbor fractions if their difference $\dfrac{ad - bc}{bd}$ has numerator $\pm1$, that is, $ad - bc = \pm 1$.
Prove that
(a.) in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);
(b.) if $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are neighbor fractions then $\dfrac{a + b}{c + d}$ is between them and is a niehgbor fraction for both $\dfrac{a}{b}$ and $\dfrac{c}{d}$; moreover, ...
Here is the snapshot from the book online (click on Look Inside on the Amazon page):
So, (a) is simple, but I have no idea how to prove (b). It just does not seem right to me. Embarrassing. Any help is appreciated.
| We have to prove that for positive integers $a,b,c,d$, we either have
$$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$
or
$$\frac{c}{d}<\frac{a+c}{b+d}<\frac{a}{b}$$
First of all, $\frac{a}{b}=\frac{a+c}{b+d}$ is equivalent to $ab+ad=ab+bc$, hence $ad=bc$, which contradicts the assumption $ad-bc=\pm 1$. We can
disprove $\frac{a+c}{b+d}=\frac{c}{d}$ in the same manner.
In the case of $\frac{a}{b}<\frac{a+c}{b+d}$, we get $ad<bc$, hence $ad-bc=-1$. The condition $\frac{a+c}{b+d}<\frac{c}{d}$ is equivalent to $ad+cd<bc+cd$, hende $ad<bc$, which implies $ad-bc=-1$ again. So, $\frac{a}{b}<\frac{a+c}{b+d}$ is equivalent to $\frac{a+c}{b+d}<\frac{c}{d}$.
So, if we have $\frac{a}{b}>\frac{a+c}{b+d}$, we must have $\frac{a+c}{b+d}>\frac{c}{d}$ (which can also be proven directly analogue to the calculation above)
To show that the middle fraction is a neighbor-fraction to both fractions, just use the definition of neighbor-fractions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx $ In the following integral:
$$I = \int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx $$
I thought of making partial fractions , then solve it .
But I am not able to make partial fractions.
| $\displaystyle \int\frac{5x^4+4x^5}{(x^5+x+1)^2}dx = \int\frac{5x^4+4x^5}{x^{10}(1+x^{-4}+x^{-5})^2}dx = \int\frac{5x^{-6}+4x^{-5}}{(1+x^{-4}+x^{-5})^2}dx$
put denominator is $=t$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\cos{\frac{\pi}{n}}\notin\mathbb Q$ for any postive integer $n\ge 4$ Let $n\ge 4$ postive integer,show that
$$\cos{\dfrac{\pi}{n}}\notin\mathbb Q$$
Now I have solve for a case:
Assmue that
$$\cos{\dfrac{\pi}{n}}=\dfrac{q}{p},(p,q)=1,p,q\in N^{+}$$ use Chebyshev polynomials?
$$T_{n}(\cos{x})=\cos{(nx)}$$
so we have
$$T_{n}\left(\dfrac{q}{p}\right)=-1$$
then we have
$$2^{n-1}\left(\dfrac{q}{p}\right)^n+a_{n-1}\left(\dfrac{q}{p}\right)^{n-1}+\cdots+a_{0}+1=0$$
so we have
$$q|\le a_{0}+1$$
since
$|a_{0}|=0,1,-1$,
(1):$a_{0}\neq -1$,so we have
$$\dfrac{q}{p}\le\dfrac{2}{3}$$,but
$$\cos{\dfrac{\pi}{n}}\ge\cos{\dfrac{\pi}{4}}=\dfrac{\sqrt{2}}{2}>\dfrac{2}{3}$$
contradiction。
(2):But for $a_{0}=-1$, I can't prove it
| An elementary proof. From deMoivre's Theorem, if $k\in \Bbb N$ then $\cos kx=\sum_{(0\le 2j\le k)}\binom {k}{2j}(-1)^j(\cos x)^{k-2j}(1-\cos^2 x)^j.$ So if $\cos (\pi/n)\in \Bbb Q$ and $m|n$ then $\cos (\pi/m)\in \Bbb Q.$ So it suffices to show $\cos (\pi/m)\not \in \Bbb Q$ when (i)$\,m=4,$ or (ii) $\,m=6,$ or (iii) $\,m=9,$ or (iv) $\,m \ge 5$ is prime.
For $m=9,$ let $y=2\cos (\pi/9).$ Then $4(y/2)^3-3(y/2)=\cos (\pi/3)=1/2,$ so $y^3-3y-1=0.$ If $y=A/B$ with $A,B\in \Bbb N$ and $\gcd (A,B)=1$ then $A^3-3AB^2-B^3=0 ,$ so we have $$1\equiv 1+A^3-3AB^2-B^3\equiv 1+A+AB+B\equiv (1+A)(1+B) \pmod 2$$ implying $1+A$ and $1+B$ are odd, implying $A$ and $B$ are even, contrary to $\gcd(A,B)=1.$
For prime $m\ge 5,$ if $\cos \pi/m=A/B$ with $A,B\in \Bbb N$ and $\gcd(A,B)=1$ then by deMoivre's Theorem we have $$(\bullet)\quad -B^m=\sum_{j=0}^{(m-1)/2}\binom {m}{2j}A^{m-2j}(B^2-A^2)^j(-1)^j.$$ The RHS of $(\bullet)$ is divisible by $A$ so $A|(-B^m).$ But $\gcd(A,B)=1$ so we must have $A=1.$
$\quad$ Now $B^m\equiv B \mod m,$ as $m$ is prime. And if $m$ is an odd prime and $0< 2j<m$ then $m$ divides $\binom {m}{2j}.$ So from $(\bullet)$ we have $$-B\equiv -B^m\equiv A^m\equiv 1 \pmod m.$$ So $B\ge m-1.$
But now $1/2=\cos (\pi/3)<\cos (\pi/m)=A/B=1/B\le 1/(m-1)\le 1/4,$ a paradox.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Interesting log series Calculate the sum of series:
$$\sum_{k=1}^{\infty} k\left(\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-\log\left(\frac{k}{\sqrt{k^2+1}}\right)\right)$$
| We write
$$ k\left(\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-\log\left(\frac{k}{\sqrt{k^2+1}}\right)\right) = (k+1) \log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-k \log\left(\frac{k}{\sqrt{k^2+1}}\right) +\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right).$$
The sum of the first two terms is telescoping and we obtain that your sum $S$ is given by
$$S= \overbrace{\lim_{k\to\infty} k \log\left(\frac{k}{\sqrt{k^2+1}}\right)}^{=0} - \overbrace{\lim_{k\to 1} k \log\left(\frac{k}{\sqrt{k^2+1}}\right)}^{=-\frac12 \ln2} - \overbrace{\sum_{k=1}^\infty \log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)}^{=S'}.$$
Now for $S'$, we use the rules of the logarithm, to obtain
$$S' = \lim_{n\to\infty} \sum_{k=1}^n \left(\log(k+1) -\frac12 \log\left((k+1)^2+1\right) \right).$$
The latter sum is evaluated as $S'=\frac12 \log\left(\frac{2 \pi}{\sinh{\pi}}\right), see below.$
Thus, we obtain the final result
$$S= \frac{1}{2} \log\left(\frac{\sinh \pi}{\pi}\right).$$
Edit:
we are interested in the value of $$
S' = \lim_{n\to\infty} \sum_{k=1}^n \left(\ln(k+1) -\frac12 \log\left((k+1)^2+1\right) \right) =
\lim_{n\to\infty} \left(\log \Gamma(n+2) - \frac12\sum_{k=1}^n\left[ \log\left(k+1+i\right) + \log\left(k+1-i\right) \right] \right)
= \lim_{n\to\infty} \left( \log\Gamma(n+2) - \frac12 \log\left[\Gamma(2+i +n)/\Gamma(2+i) \right]- \frac12 \log\left[\Gamma(2-i +n)/\Gamma(2-i) \right] \right) = \frac12 \left(\lim_{n\to\infty}
\log\left( \frac{\Gamma^2(n+2)}{\Gamma(2+i+n) \Gamma(2-i +n)} \right) + \log[\Gamma(2+i)\Gamma(2-i) ]\right). $$
Now the ratio of the $\Gamma$ function in the first term approaches 1 and thus this term vanishes. Whereas in the second term, we use Euler's reflection formula to obtain the result quoted above.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple inequality ( or false) Given $a$, $b$ and $c > 0$ such that $abc=1$, prove that
$$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b} \leq 8abc$$
I don't know if its true or not but i need help in both cases.
Thanks
| If $a$ and $b$ are "small" and very very close together (but $a$ slightly larger), then $c$ is "large" and $c/(a-b)$ can be made arbitrarily large.
$a/(b-c)$ is negative but a it can be a "small" negative (as $c$ is "large" compared to $a$ and $b$. $b/(c-a)$ is positive (but small). So that sum can be arbitrarily large.
Example:
Fix $M$ to be a "large" number.
Let $a = 1/M$ and $b = 1/2M$ are two small numbers close together so.....
Then $c = 1/ab = 2M^2$
Then $\frac a{b-c} = \frac 1{M(\frac 1{2M} - 2M^2)}=\frac 1{\frac 12 - 2M^3} \approx 0$
And $\frac b{c-a} = \frac{1}{2M(2M^2 - \frac 1M}=\frac{1}{4M^3- 1} \approx 0$
And $\frac c{a-b} = \frac{2M^2}{\frac {1}{2M}} =4M^3$
Then $\frac a{b-c} + \frac b{c-a} + \frac c{a-b} \approx 4M^3 > 8 = 8abc$.
$M$ doesn't even have to be that large. Say $M=2$ so $a= 1/2$ and $b=1/4$ and $c = 8$.
Then $\frac a{b-c} = -\frac 1{2(7\frac 34)}= -\frac {2}{31}$
$\frac b{c-a} = \frac 1{4(7 \frac 12)} = \frac 1{30}$
$\frac c{a-b} = \frac 8{1/4} = 32$ and
$-\frac {2}{31}+ \frac 1{30} + 32 = 32 - \frac {2*30 - 31}{31*30}=32 - \frac{29}{930} = 31 \frac{901}{930} > 8*\frac 12*\frac 14*8 = 8 $
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometry and Quadratics If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
My Attempt:
Using the sum and product formulae we have,
$q=\tan A\tan B, $ $-p=\tan A+\tan B$
And,
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \Rightarrow \tan (A+B)=\frac{-p}{1-q}$
| You can proceed easily after finding $\tan (A+B)$.
So we have $$p =\tan (A+B)(q-1) $$ So substituting into the LHS we get, $$\sin^2 (A+B)+ (q-1)\tan (A+B)\sin (A+B)\cos (A+B) +q \cos^2 (A+B) = \sin ^2 (A+B) +(q-1)\sin ^2(A+B) +q\cos ^2 (A+B) =q\sin ^2 (A+B) +q\cos ^2 (A+B) =q $$ Hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the numbers $n$ such that $2n+5$ is composite. Let $n$ be a positive integer greater than zero. I write
$$a_n =
\begin{cases}
1 , &\text{ if } n=0 \\
1 , &\text{ if } n=1 \\
n(n-1), & \text{ if $2n-1$ is prime} \\
3-n, & \text{ otherwise}
\end{cases}$$
The sequence goes like this $$1,1,2,6,12,-2,30,42,-5,72,90,-8,132,-10,-11,\ldots$$ I would like to prove the following two claims.
claim 1 : If $a_n>0$ and
${a_n \above 1.5 pt 3} \notin \mathbb{Q}$ then $\sqrt{4a_n+1}$ is prime.
The table below illustrates what I am seeing:
\begin{array}{| l | l | l | l }
\hline
n & a_n & {a_n \above 1.5 pt 3} & \sqrt{4a_n+1}\\ \hline
0 & 1 & .333333.. & 2.2360679.. \\
1 & 1 & .333333.. & 3 \\
2 & 2 & .666666.. & 3 \\
3 & 6 & 2 & 5 \\
4 & 12 & 4 & 7 \\
6 & 30 & 10 & 11 \\
7 & 42 & 14 & 13 \\
9 & 72 & 24 & 17 & \\
10 & 90 & 30 & 19 \\
12 & 132 & 44 & 23 \\
15 & 210 & 70 & 29 \\
16 & 240 & 80 & 31 \\
19 & 342 & 114 & 37 \\
21 & 420 & 140 & 41 \\
22 & 462 & 154 & 43 \\
\hline
\end{array}
claim 2: If $a_n<0$ then $2a_n+5$ is composite
| Ignore $n = 0, 1$ since they're kind of irrelevant. Then $a_n \leq 0$ for all $n$ unless $2n - 1$ is prime, by the definition. In that case $a_n = n(n - 1)$. So,
$$4a_n + 1 = 4(n^2 - n) + 1 = (2n-1)^2$$
So, then $\sqrt{4a_n + 1} = 2n - 1$, which is prime.
| {
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Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$
Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.
This is my proof.
Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The left-hand side is $a\cos x+b\sin x=\frac{ab}{c}+\frac{ab}{c}=2\frac{ab}{c}$. The right-hand side is $\sqrt{a^2+b^2}\cos(x-\alpha)=c\left(\cos{x}\cos{\alpha}+\sin{x}\sin{\alpha}\right)=c\left(\frac{ab}{c^2}+\frac{ab}{c^2}\right)=2\frac{ab}{c}$.
Is my proof valid? Is there a more general way to prove it?
For the second part of the question, I think it should be $\alpha=\arccos\frac{a}{\sqrt{a^2+b^2}}=\arcsin\frac{b}{\sqrt{a^2+b^2}}$. Is this correct?
| We can prove this by using properties of complex numbers:
$$ \begin{align}a\cos(x) + b\sin(x) &= a \cos(x) + b \cos\left(x-\frac{\pi}{2}\right)\\
&=a \operatorname{Re}\left\{e^{ix}\right\} + b \operatorname{Re}\left\{e^{i(x-\pi/2)}\right\}\\
&= \operatorname{Re}\left\{ae^{ix} + be^{i(x-\pi/2)}\right\}\\
&= \operatorname{Re}\left\{e^{ix}(a-bi) \right\} \\
&= \operatorname{Re}\left\{e^{ix}\sqrt{a^2+b^2}e^{i\cdot\operatorname{atan2}(-b,a)} \right\}\\
&= \operatorname{Re}\left\{\sqrt{a^2+b^2}e^{i(x+\operatorname{atan2}(-b,a))}\right\}\\
&= \sqrt{a^2+b^2} \cos(x+\operatorname{atan2}(-b,a))
\end{align}
$$
where $\operatorname{atan2}(y_0,x_0)$ is the angle of the complex number $x_0+iy_0$.
This formula works for all real numbers $a$ and $b$ except when $a=b=0$.
Generalization
Following the same procedure, we can find a general formula for the sum of two sinusoids with same frequency but (possibly) different phases:
$$ \begin{align}a\cos(x+\alpha) + b \cos(x+\beta) &= \operatorname{Re}\left\{ae^{i(x+\alpha)} + be^{i(x+\beta)}\right\}\\
&= \operatorname{Re}\left\{e^{ix}\left(ae^{i\alpha}+be^{i\beta}\right) \right\}\\
&= \operatorname{Re}\left\{e^{ix}ce^{i\gamma} \right\}\\
&= \operatorname{Re}\left\{ce^{i(x+\gamma)} \right\}\\
&= c \cos(x+\gamma)
\end{align}
$$
Where $ce^{i\gamma} = ae^{i\alpha}+be^{i\beta}$.
We have
$$\begin{align}ce^{i\gamma}&=a (\cos(\alpha)+i\sin(\alpha)) +b (\cos(\beta)+i\sin(\beta))\\
&= (a\cos(\alpha)+b\cos(\beta)) + (a\sin(\alpha)+b\sin(\beta))i.\\
\end{align}
$$
So
$$c=\sqrt{(a\cos(\alpha)+b\cos(\beta))^2+(a\sin(\alpha)+b\sin(\beta))^2}$$
and
$$\gamma=\operatorname{atan2}\big(a\sin(\alpha)+b\sin(\beta),a\cos(\alpha)+b\cos(\beta)\big).$$
Sidenotes:
*
*This result is equivalent to
phasor addition.
*We can get similar formula involving $\sin$ by taking the imaginary
part.
| {
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Find all positive integer solutions to the equation $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ Find all positive integers to the equation $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
Multiply both sides with $(abc)^2$ to get $(bc)^2 + (ac)^2 = (ab)^2$.
I then tried some pythagorean triples and nothing worked so I assumed that there arent any solutions. Thank you for your responses.
| Rewrite the equation as
$$
\frac{c^2}{a^2}+\frac{c^2}{b^2}=1
$$
and set $c/a=x/z$, $c/b=y/z$, where $x$ and $y$ are coprime.
Then $(x,y,z)$ is a primitive Pythagorean triple, so, without loss of generality,
$$
x=u^2-v^2,\quad y=2uv,\quad z=u^2+v^2
$$
where $\gcd(u,v)=1$, one among $u$ and $v$ being even.
Thus
$$
\frac{c}{a}=\frac{u^2-v^2}{u^2+v^2},\quad
\frac{c}{b}=\frac{2uv}{u^2+v^2}
$$
Therefore, for some positive integers $h$ and $k$,
$$
c=h(u^2-v^2)=2kuv,\quad
a=h(u^2+v^2),\quad
b=k(u^2+v^2)
$$
Note however that $u^2-v^2$ and $2uv$ are coprime, so $h=2muv$ and $k=m(u^2-v^2)$, so
$$
a=2muv(u^2+v^2),\quad
b=m(u^4-v^4),\quad
c=2muv(u^2-v^2)
$$
It's easy to see that any choice of $u$ and $v$ as before and any $m$ produces a solution of the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Maximum value of $|x-y|$ Given $x,y\in\mathbb R$ such that
$$5x^2+5y^2-6xy\ =\ 8$$
find the maximum value of $|x-y|$.
My attempt
$5x^2 - 6yx + (5y^2-8)\ =\ 0$
$x\ =\ \dfrac{6y\pm\sqrt{(6y)^2-4(5)(5y^2-8)}}{10} = \dfrac{6y\pm\sqrt{160-64y^2}}{10} = \dfrac{3y\pm2\sqrt{10-4y^2}}{10}$
$5y^2 - 6xy + (5x^2-8)\ =\ 0$
$y\ =\ \dfrac{3x\pm2\sqrt{10-4x^2}}{10}$
$|x-y|\ =\ \dfrac{\left|3(y-x)\pm2\sqrt{10-4y^2}\mp2\sqrt{10-4x^2}\right|}{10}$
What do I do from here?
| You tried to solve for both $x$ in terms of $y$ and $y$ in terms of $x$.
Try expressing everything as either a function of $x$ or of $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, find the period of $f(x)$. If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, then the period of $f(x)$ is?
I tried replacing $x$ with $x-1$ and stuff, but didn't lead me to anything...
| Note that $\sqrt{2} = \frac{2}{\sqrt{2}}$. Hence,
$$f(x + 1) + f(x - 1) = \frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}f(x)$$
$$\begin{align}f(x + 1) &= \frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}\left(f(x) - \sqrt{2}f(x - 1)\right)\\
&=\frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}\left(-f(x - 2)\right)\\
&= \frac{1}{\sqrt{2}}\left(f(x) - f(x - 2)\right)\end{align}$$
$$\sqrt{2}f(x + 1) = f(x) - f(x - 2)$$
$$\sqrt{2}f(x) = f(x - 1) - f(x - 3)$$
Hence,
$$f(x - 1) - f(x - 3) = f(x + 1) + f(x - 1)$$
$$f(x + 1) = -f(x - 3)$$
$$f(x) = -f(x - 4)$$
Therefore,
$$f(x) = -f(x - 4) = -(-f(x - 8)) = f(x - 8)$$
thus the period is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Prove that the sequence $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)$ is not increasing Let $x_1\ge1$ and $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)$. Prove that the sequence $\{x_n\}$ is not increasing, that is, $x_{n+2}\le x_{n+1}$.
My attempt was as follows: we have to prove that $1\le \frac{x_{n+1}}{x_{n+2}}$. Now:
$$\frac{x_{n+1}}{x_{n+2}}=\frac{x_n+\frac{1}{x_n}}{x_{n+1}+\frac{1}{x_{n+1}}}=\frac{x_n+\frac{1}{x_n}}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)+\frac{1}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)}}$$
let $a=x_n+\frac{1}{x_n}$ for the sake of convenience, then:
$$ \frac{x_n+\frac{1}{x_n}}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)+\frac{1}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)}}=\frac{a}{\frac{1}{2}a+\frac{2}{a}}$$
so for the claim that $1\le \frac{x_{n+1}}{x_{n+2}}$to hold, we must have:
$$ 1\le\frac{a}{\frac{1}{2}a+\frac{2}{a}}\rightarrow \frac{1}{2}a+\frac{2}{a}\le a\rightarrow \frac{1}{2}+\frac{2}{a^2}\le 1\rightarrow a^2\ge 4\rightarrow a\ge 2\rightarrow x_n+\frac{1}{x_n}\ge 2$$
which follows by AM-GM.
My concern is, is this reasoning correct? I skipped some cases, assuming that we are dealing with positive numbers only - is my reasong rigorous? Or is there a serious flaw?
Either way, maybe there is a cleaner and perhaps more clever way to prove the claim?
| The reasoning seems correct until the end when you write
$$x_n+\frac{1}{x_n}\geq 2$$ Now here you should prove that $$x_n+\frac{1}{x_n}\geq 2$$
Which is trivial by AM-GM.
Here's a similar approach note that by AM-GM
$$x_{n+1}=\frac{1}{2}(x_n+\frac{1}{x_n})\geq 1$$
hence we have that $x_n\geq 1$.
Now $$\frac{x_n}{x_{n+1}}=\frac{2x_n}{x_n+\frac{1}{x_n}}$$
But we have that $x_n\geq\frac{1}{x_n}$ because $x_n\geq 1$ that means $$\frac{x_n}{x_{n+1}}=\frac{2x_n}{x_n+\frac{1}{x_n}}\leq 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Finding the coefficient of $x^m$ in $(x+x^2+x^3+.....+x^i)^n$ EDIT:
My doubt may be silly but if the expansion of $(1-x)^{-n}$ is infinite, how come wolfram displayed 11 terms?
Wolfram result
Can anybody help to solve the whole equation?
I'm trying to solve this question.
I have an equation $(x+x^2+x^3+x^4+x^5+x^6)^2$ [ Taken 6-sided die x 2].
I want to find the coefficient of $x^5$.
I have reduced the equation to $x^2(1-x^6)^2(1-x)^{-2}$.
For $(1-x)^{-2}$,
$$(1-x)^{-2}=\sum_{k\ge0}(-1)^k\binom{-2}{k}x^k=\sum_{k\ge0}\binom{2+k-1}{k}x^k $$
I got -> $1+2x+3x^2$
For $(1-x^6)^2$
$$(1-x^6)^2=\sum_{k\ge0}(-1)^k\binom{2}{k}x^{6k}$$
I got -> $1-2x^6+x^{12}$
So my equation will be$$x^2(1+2x+3x^2)(1-2x^6+x^{12})$$
Am I correct till here? Because when I multiply all the terms I'm getting
$$x^2+2x^3+3x^4-2x^8-4x^9-6x^{10}+x^{14}+2x^{15}+3x^{16}$$
Where am I going wrong?
| This GitHub repository solved my problem.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Number of ordered triples with given condition We are given that there are $6250k$ ordered triples $(m,n,p)$ where $m,n,p$ are integers such that $$1≤m≤100,\;1≤n≤50,\;1≤p≤25\quad \& \quad 3\,|\,2^m+2^n+2^p$$ Find $k$.
With the given range of values of $m, n,p$, I guess the possible numbers of ordered triples without any conditions would be $25\times (50-1)\times (100-2)$, i.e., $25\times 49\times 98$. But after this I am unable to detect the relationship between m, n and p for the given divisibility condition.
Please help.
| Given that $1≤m≤100,\;1≤n≤50,\;1≤p≤25$ and $ 3\,|\,2^m+2^n+2^p$ where $m,n,p$ are integers.
Now if $2^m+2^n+2^p$ is divisible by $3$, then each of $2^m$, $2^n$, $2^p$ is divided by $3$.
Remainder of ${2^m}$ when it is divisible by $3$ is $(-1)^m$.
So remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}$. Now if ${(-1)^m+(-1)^n+(-1)^p}$ is divisible by $3$, the we are done.
Case I: If $m$ is odd and $n$ and $p$ are even, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={-1+1+1}={1}$ which is not divisible by $3$.
Case II: If $m$ is even and $n$ and $p$ are odd, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={1-1-1}={-1}$ which is not divisible by $3$.
Case III: If $m$, $n$ and $p$ are even, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={1+1+1}={3}$ which is divisible by $3$.
Case IV: If $m$, $n$ and $p$ are odd, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={-1-1-1}={-3}$ which is divisible by $3$.
So we are interested in the last two cases.
Now $1≤m≤100,\;1≤n≤50,\;1≤p≤25$
In the given range, the number of possible choice that all $m, n, p$ are even is $50, 25, 12$ respectively and the number of possible choice that all $m, n, p$ are odd is $50, 25, 13$ respectively.
So the total number of ordered triples $(m,n,p)$ where $m,n,p$ are integers that $ 3\,|\,2^m+2^n+2^p$ is $(50 \times 25 \times 12)+(50 \times 25 \times 13)=50 \times 25 \times 25= 31250 = 5\times 6250$.
So the possible value of $k=5$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of first $n$ terms of the series: $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$
I have the series $$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$$
I know the following formulas: $$1+2+3+\cdots +n=\frac {n (n+1)}{2}\tag1$$ $$1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}\tag2$$ $$1^3+2^3+3^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2\tag3$$ But none of $(1)(2)$ and $(3)$ worked.
Please help___.
| Try to observe that $$\frac{1}{n\times (n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ $\therefore$ The given series can be written as $$1-\frac12+\frac12-\frac13+\frac13+\cdots -\frac{1}{n}+\frac1n-\frac{1}{n+1}$$ Each term will cancel out other term except $1$ and $\frac{1}{n+1}$ .
$\therefore$ $$=1-\frac1{n+1}$$ $$=\frac{n}{n+1}$$ Hope it helps!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Maximum of $f(x) =(1+\cos x)\cdot \sin(\frac{x}{2})$ on $x \in (0, \pi)$ I've attempted to solve this question by using $$f(x) = 2\sin\frac{x}{2}\cos^2\frac{x}{2} \leq \frac{(2\sin\frac{x}{2}+\cos^2\frac{x}{2})^2}{2}$$ but it results in the wrong answer every time. Is there another way to solve this question and is there anything I can do to change my method to make it work.
| By AM-GM $$\left(1+\cos{x}\right)\sin\frac{x}{2}=2\cos^2\frac{x}{2}\sin\frac{x}{2}=2\left(\sin\frac{x}{2}-\sin^3\frac{x}{2}\right)=$$
$$=-2\left(-\sin\frac{x}{2}+\sin^3\frac{x}{2}+\frac{1}{3\sqrt3}+\frac{1}{3\sqrt3}\right)+\frac{4}{3\sqrt3}\leq$$
$$\leq-2\left(-\sin\frac{x}{2}+3\sqrt[3]{\sin^3\frac{x}{2}\cdot\frac{1}{3\sqrt3}\cdot\frac{1}{3\sqrt3}}\right)+\frac{4}{3\sqrt3}=\frac{4}{3\sqrt3}.$$
THe equality occurs for $\sin\frac{x}{2}=\frac{1}{\sqrt3}$, which says that the answer is $\frac{4}{3\sqrt3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Complex Roots with improper fraction I'm having trouble with the following:
$(-16i)^{5/4}$
My calculations for the Principal root is:
$32(\cos (3\pi/2) * 5/4) + i \sin (3\pi/2)* 5/4))$
$=32(Cis (15\pi/8))$
This answer does not agree with the online calculators. It gives a positive real value and the online calculators show a principal angle in Quadrant 3.
Confused on what happened here.
| Let $z=-16i$ and $n=\dfrac54$. For solving you have to compute $r=|z|$ and argument $\theta$ where $\tan\theta=\dfrac{y}{x}$.
then
$r=|z|=|-16i|=16$ and argument $\theta=\dfrac{3\pi}{2}$. Then write
$$z_k=r^n(\cos n\theta+i\sin n\theta)$$
But argument adds with $2k\pi$ so we have
$$z_k=r^n\Big(\cos n(\theta+2k\pi)+i\sin n(\theta+2k\pi)\Big)$$
so
$$z_k=16^\dfrac54\Big(\cos\dfrac54(\dfrac{3\pi}{2}+2k\pi)+i\sin\dfrac54(\dfrac{3\pi}{2}+2k\pi)\Big)$$
and for $k=0,1,2,3$ write
$$z_k=32\Big(\cos(\dfrac{15\pi}{8}+\frac{5k\pi}{2})+i\sin(\dfrac{15\pi}{8}+\frac{5k\pi}{2})\Big)$$
finally for $k=0,1,2,3$ we conclude that
\begin{eqnarray}
k=0 &\Rightarrow& z_0=32\Big(\cos\dfrac{15\pi}{8}+i\sin\dfrac{15\pi}{8}\Big)=19.28+25.5i\\
k=1 &\Rightarrow& z_1=32\Big(\cos\dfrac{35\pi}{8}+i\sin\dfrac{35\pi}{8}\Big)=12.24+29.5i\\
k=2 &\Rightarrow& z_2=32\Big(\cos\dfrac{55\pi}{8}+i\sin\dfrac{55\pi}{8}\Big)=-29.5+12.24i\\
k=3 &\Rightarrow& z_3=32\Big(\cos\dfrac{75\pi}{8}+i\sin\dfrac{75\pi}{8}\Big)=-12.24-29.5i
\end{eqnarray}
This was step by step solving.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A Ramanujan infinite series $$ 1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1 \cdot 3}{2 \cdot 4}\right)^3-13\left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right)^3+\cdots $$
I went on evaluating the above series and encountered that solving $\displaystyle \sum_{n\ge 0}\left(\binom{2n}{n}\right)^3x^n$ would suffice.
But how do we make a generating function for the third power of a central binomial coefficient using the fact $\displaystyle \sum_{n\ge 0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$
| Modifying @Claude Leibovici's answer a little.
$\frac{1}{\left(1-z\right)^{a}}=\,_{1}F_{0}\left(a;;z\right)={\displaystyle \sum_{n=0}^{\infty}\frac{a_{n}}{n!}z^{n}}$
and
$\frac{1}{\sqrt{1-4\cdot x}}={\displaystyle \sum_{n=0}^{\infty}\left(\begin{array}{c}
2n\\
n
\end{array}\right)x^{n}}$
Let $z=4x,x=\frac{z}{4}$
$\frac{1}{\left(1-4x\right)^{a}}=\,_{1}F_{0}\left(a;;4x\right)={\displaystyle \sum_{n=0}^{\infty}\frac{a_{n}}{n!}\,\left(4x\right)^{n}}$
$4^{n}\frac{\left(\frac{1}{2}\right)_{n}}{n!}=\left(\begin{array}{c}
2n\\
n
\end{array}\right)$
And we have the generating function for $\left(\begin{array}{c}
2n\\
n
\end{array}\right)^{k}\left(4^{k}x\right)^{n}$
i.e.
$\,_{k}F_{k-1}\left({\displaystyle \prod_{i=1}^{k}\left(\frac{1}{2}\right)_n};{\displaystyle \prod_{i=1}^{k-1}\left(1\right)_n};4^{k}z\right)={\displaystyle \sum_{n=0}^{\infty}}\left(\begin{array}{c}
2n\\
n
\end{array}\right)^{k}\left(4^{k}x\right)^{n}$
Where we “borrowed” the $n!=\left(1\right)_{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$ I was looking back in my junk, then I found this:
$$x + z + y = 5$$
$$x^2 + z^2 + y^2 = 21$$
$$x^3 + z^3 + y^3 = 80$$
What is the value of $xyz$?
A) $5$
B) $4$
C) $1$
D) $-4$
E) $-5$
It's pretty easy, any chances of solving this question? I already have the
answer for this, but I didn't fully understand.
Thanks for the attention.
| Consider the polynomial
$$p(t) = (1-x t)(1-y t)(1-z t)$$
Let's consider the series expansion of $\log\left[p(t)\right]$:
$$\log\left[p(t)\right] =-\sum_{k=1}^{\infty}\frac{S_k}{k} t^k$$
where
$$S_k = x^k + y^k + z^k$$
Since we're given the $S_k$ for $k$ up to $3$ we can write down the series expansion of $\log\left[p(t)\right]$ up to third order in $t$, but that's sufficient to calculate $p(t)$, as it's a third degree polynomial. The coefficient of $t^3$ equals $-xyz$, so we only need to focus on that term. We have:
$$\log\left[p(t)\right] = -\left(5 t +\frac{21}{2} t^2 + \frac{80}{3} t^3+\cdots\right)$$
Exponentiating yields:
$$p(t) = \exp(-5t)\exp\left(-\frac{21}{2}t^2\right)\exp\left(-\frac{80}{3}t^3\right)\times\exp\left[\mathcal{O}(t^4)\right]$$
The $t^3$ term can come in its entirety from the first factor, or we can pick the linear term in $t$ from there and then multiply that by the $t^2$ term from the second factor or we can take the $t^3$ term from the last factor. Adding up the 3 possibilities yields:
$$xyz = \frac{5^3}{3!} -5\times\frac{21}{2} + \frac{80}{3} = -5$$
It is also easy to show that $x^4 + y^4 + z^4 = 333$ by using that the coefficient of the $t^4$ term is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Proving that $\cos(\arcsin(x))=\sqrt{1-x^2}$ I am asked to prove that $\cos(\arcsin(x)) = \sqrt{1-x^2}$
I have used the trig identity to show that $\cos^2(x) = 1 - x^2$
Therefore why isn't the answer denoted with the plus-or-minus sign?
as in $\pm \sqrt{1-x^2}$.
Thank you!
| Let $\arcsin x = \theta$. Then, by definition of the arcsine function, $\sin\theta = x$, where $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, and
$\cos(\arcsin x) = \cos\theta$. Using the Pythagorean Identity $\sin^2\theta + \cos^2\theta = 1$, we obtain
\begin{align*}
\sin^2\theta + \cos^2\theta & = 1\\
\cos^2\theta & = 1 - \sin^2\theta\\
\cos^2\theta & = 1 - x^2\\
|\cos\theta| & = \sqrt{1 - x^2}
\end{align*}
Since $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, $\cos\theta \geq 0$. Thus, $|\cos\theta| = \cos\theta$, whence
\begin{align*}
\cos\theta & = \sqrt{1 - x^2}\\
\cos(\arcsin x) & = \sqrt{1 - x^2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Find all complex numbers $z$ satisfying the equation $z^{4} = -1+\sqrt{3}i$ Since any complex number can be of polar form. We set that $z = r\cos \theta +ir\sin \theta$. Now by de Moivre's Theorem, we easily see that $$z^{4} = r^{4}\cos4 \theta + ir^{4}\sin4 \theta$$
Since $$z^{4} = -1+\sqrt{3}i$$
We equip accordingly and see that $$r^{4}\cos4 \theta = -1$$
$$r^{4}\sin4 \theta = \sqrt{3}$$
Solving the above 2 equations we have $$\tan 4\theta = -\sqrt{3} \Rightarrow 4\theta = -\dfrac{\pi}{3} \Rightarrow \theta = -\dfrac{\pi}{12}$$
Hence we have $$\text{arg}z = -\dfrac{\pi}{12} + k\pi$$
Furthermore, $$\text{Arg}z = -\dfrac{\pi}{12} \text{ or } \dfrac{11\pi}{12}$$
However, i cannot find out what $r$ is, as when i substitute in to solve, $r$ become a complex number?? Is my answer correct, how can i make my steps better? Thanks!
| Observe that $-1+\sqrt{3}i=2e^{\frac{2\pi i}{3}}$. Then the roots of the equation are
$$
z=\sqrt[4]{2}e^{\frac{\pi i}{6}+\frac{k\pi}{2}},
$$
where $k=0,1,2,3$. Thus $r=\sqrt[4]{2}$. Substitute $k$, then we get $z=\sqrt[4]{2}e^{\frac{\pi i}{6}}$, $z=\sqrt[4]{2}e^{\frac{2\pi i}{3}}$, $z=\sqrt[4]{2}e^{\frac{7\pi i}{6}}$, and $z=\sqrt[4]{2}e^{\frac{5\pi i}{3}}$. There is no root whose argument is $\frac{11\pi}{12}$. What happened? Because your attempt has a flaw: Since $r^4 \cos 4\theta=-1$ and $r^4\sin 4\theta=\sqrt{3}$, we get $\tan 4\theta=-\sqrt{3}$, but it doesn't imply that $4\theta = -\frac{\pi}{3}$. If it is true, then $r^4 \cos4\theta$ cannot be negative!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Finding the limit of the area of a Koch Snowflake this is my first question for this site and I made this account specifically for help with the following topic.
I am doing a research presentation on the Koch Snowflake, specifically, the area.
So far, I have been attempting to generalize a formula for finding the area of the snowflake at n iterations, and I am now trying to find the limit as n tends toward infinity.
So, basically, what is the limit for the following?:
$$\lim_{n\to\infty} \sum_{r=2}^{n} \frac{3 \cdot 4^{r-2}}{9^{r-1}} \cdot \frac{s^2 \sqrt{3}}{4} $$
| The following forms a GP with $a=\frac{1}{9}$ and $r=\frac{4}{9}$.
$$\begin{align}
&\lim_{n\to\infty} \sum_{r=2}^{n} \frac{3 \cdot 4^{r-2}}{9^{r-1}} \cdot \frac{s^2 \sqrt{3}}{4}\\
=\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\lim_{n\to\infty} \sum_{r=2}^{n} \frac{ 4^{r-2}}{9^{r-1}} \\
=\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\cdot\frac{\frac{1}{9}}{1-\frac{4}{9}}\\
=\ &\frac{ 3\sqrt{3}\cdot s^2}{4}\cdot\frac{1}{5}\\
=\ &\frac{ 3\sqrt{3}\cdot s^2}{20}
\end{align}\\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $y'$ given $y\,\sin\,x^3=x\,\sin\,y^3$? The problem is
$$y\,\sin\,x^3=x\,\sin\,y^3$$
Find the $y'$
The answer is
Can some explain how to do this, please help.
| I have found that it helps students to understand implicit differentiation if first they think of both $x$ and $y$ as functions of some third variable such as $t$ and take the derivative of both sides with respect to $t$, being careful to use the product rule, chain rule, etc when needed. Then as a final step, multiply both sides by $dt$. Since the object usually is to solve for $y^\prime=\frac{dy}{dx}$ you can then divide both sides by $dx$ and solve.
\begin{eqnarray}
\frac{d}{dt}\left[y\sin(x^3)\right]&=&\frac{d}{dt}\left[x\sin(y^3)\right]\\
\sin(x^3)\frac{dy}{dt}+y\frac{d\sin(x^3)}{dt}&=&\sin(y^3)\frac{dx}{dt}+x\frac{d\sin(y^3)}{dt}\\
\sin(x^3)\frac{dy}{dt}+3x^2y\cos(x^3)\frac{dx}{dt}&=&\sin(y^3)\frac{dx}{dt}+3xy^2\cos(y^3)\frac{dy}{dt}\\
\sin(x^3)dy+3x^2y\cos(x^3)dx&=&\sin(y^3)dx+3xy^2\cos(y^3)dy\\
\sin(x^3)\frac{dy}{dx}+3x^2y\cos(x^3)&=&\sin(y^3)+3xy^2\cos(y^3)\frac{dy}{dx}\\
\left[\sin(x^3)-3xy^2\cos(y^3)\right]\frac{dy}{dx}&=&\sin(y^3)-3x^2y\cos(x^3)\\
\frac{dy}{dx}&=&\frac{\sin(y^3)-3x^2y\cos(x^3)}{\sin(x^3)-3xy^2\cos(y^3)}
\end{eqnarray}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ \sin\theta + \cos\theta = \frac 1 2$, what does $\tan\theta + \cot\theta$ equal? A SAT II question asks:
If $ \sin\theta + \cos\theta = \dfrac 1 2$, what does $\tan\theta + \cot\theta$ equal?
Which identity would I need to solve this?
| Hint
$$\sin\theta+\cos\theta=\frac{1}{2} \implies \left( \sin\theta+\cos\theta \right)^2 = \frac{1}{4} \iff \color{blue}{\cos\theta\sin\theta} = \cdots$$
and
$$\tan\theta+\cot\theta = \frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta} = \frac{\cos^2\theta+\sin^2\theta}{\cos\theta\sin\theta}= \frac{1}{\color{blue}{\cos\theta\sin\theta}} = \cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability with balls and a box - complementary event of "exactly" There are $12$ balls in a box: $b$ blue, $y$ yellow and $3$ red balls. Three balls are randomly chosen. If the probability of choosing one blue, one yellow and one red is $3/11$, find the number of yellow balls in a box.
Attempt:
If $t$ is the total number of balls in a box, then:
$t=3+b+y$
$A$: "We choose exactly one blue, yellow and red ball."
$$P(A)=\frac{b}{t}\cdot\frac{y}{t-1}\cdot\frac{3}{t-2}=\frac{3}{11}$$
Substituting $t$ gives
$$P(A)=\frac{b}{3+b+y}\cdot\frac{y}{2+b+y}\cdot\frac{3}{1+b+y}=\frac{3}{11}$$
Now we have one equation with two unknowns, so we need to define another event.
Because we already know $P(A)$, that new event should be the complementary event of $P(A)$.
How to define that event and how to evaluate it?
| Remember that you are given $t=12$.
Also,
$$\frac{b}{12}*\frac{y}{11}*\frac{3}{10}=\frac{3}{11}$$ is not correct, as the order in which you draw the balls doesn't matter. There are 6 total ways to draw three distinctly colored balls, so your equation should be:
$$6*\frac{b}{12}*\frac{y}{11}*\frac{3}{10}=\frac{3}{11}$$
This leads to:
$$by=20$$
Now, since we also know that $b+y=9$, we can conclude there are either 4 or 5 yellow balls.
| {
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"timestamp": "2023-03-29T00:00:00",
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Combinatorics: throwing a dice three times to get an even number. Suppose you throw a six face dice three times, how many times will be the sum of the faces even?
I approached it this way:
You either get all three times even face, or twice odd and once even.
As there are only 3 faces that are even, you have $3^3$ possibilities.
Then, for the second situation, you have $3$ choices for then face, twice $3$ choices for the odd one. Thus giving again $3^3$ possibilities.
Overall, there will be $3^3 + 3^3$ possibilities, yet my textbook shows $3^3 + 3^4$ possibilities. What's wrong with my reasoning?
| P(all even)$=\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=\dfrac{1}{8}$
P(one even and two odd)$=\dbinom{3}{2}\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)=\dfrac{3}{8}$
Required probability $=\dfrac{1}{8}+\dfrac{3}{8}=\dfrac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Explain why $(a−b)^2 = a^2 −b^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(a−b)^2 = a^2 −b^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(a−b)^2$ to $(a−b)^2 = (a-b)(a-b) = a^2 -2ab +b^2$. To Prove that $(a−b)^2 = a^2 −b^2 $ if b = 0 I substituted b with zero both in the expanded expression and the original simplified and I got $(a−b)^2 = (a-0)^2 = (a-0)(a-0) = a^2 - a(0)-a(0)+0^2 = a^2$ and the same with $a^2 -2ab +b^2$ which resulted in $a^2 - 2a(0) + 0^2 = 2a$ or if I do not substite the $b^2$ I end up with $a^2 + b^2$. That's what I got when I try to prove the expression true for $b=0$.
As for the part where $b=a$, $(a−b)^2 = (a-b)(a-b) = a^2-2ab+b^2$, if a and b are equal, let $a=b=x$ and I substite $a^2-2ab+b^2 = x^2-2(x)(x) + x^2 = x^2-2x^2+x^2 = 1-2+1=0$ I do not see where any of this can be reduced to $a^2-b^2$ unless that equals zero......I do see where it holds but I do not see how would a solution writting out look.After typing this it seems a lot clearer but I just can't see how to phrase a "solution".
P.S: This is my first time asking a question here so whatever I did wrong I am sorry in advance and appreciate the feedback.
| It might just be easier to use that $a^2-b^2=(a-b)(a+b)$.
So if $a-b=0$ then $(a-b)^2=(a-b)(a+b)$, and if $a-b\neq 0$ then $(a-b)^2=(a-b)(a+b)$ if and only if $a-b=a+b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$ How to prove, using the definition of limit of a sequence, that:
$$\lim_{n\to\infty}\frac{6n^4+n^3+3}{2n^4-n+1}=3$$
Subtracting 3 and taking the absolute value of the function I have:
$$<\frac{n^3+3n}{2n^4-n}$$
But it's hard to get forward...
| $$\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3\right|<\epsilon\Leftrightarrow \left|\frac{n^3+3n}{2n^4-n+1}\right|<\epsilon.$$ Now, $4n^3\ge n^3+3n$ and $2n^4-n+1\ge n^4$ for all $n$ positive integer. So $$\left|\frac{n^3+3n}{2n^4-n+1}\right|\le \frac{4n^3}{n^4}=\frac{4}{n},$$ and choosing $n_0=\lfloor 4/\epsilon \rfloor+1$ we have $\dfrac{4}{n}<\epsilon$ if $n\ge n_0,$ as a consequence $$\left|\frac{6n^4+n^3+3}{2n^4-n+1}-3\right|<\epsilon\text{ if }n\ge \left\lfloor \frac{4}{\epsilon} \right\rfloor+1.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculating the $100th$ term of a series I came across a series
$1,5,13,25,41,61.......$
I have to calculate the $100^{th}$ term of this series. How do I do it?
Is there any formula to calculate the $n^{th}$ term of a series?
| The difference of successive terms forms an Arithmetic progression: $4,8,\cdots $.
We can write the terms of our sequence as: $$0\times 4+1,1\times 4+1, 3\times 4+1, 6\times 4+1, 10\times 4+1, \cdots $$ $$=\frac {0\times 1}{2}\times 4+1, \frac {1\times 2}{2}\times 4+1, \frac {2\times 3}{2}\times 4+1, \frac {3\times 4}{2}\times 4+1, \cdots $$
We can see a pattern emerging. Thus the $n$th term of the sequence is $$\frac {n (n-1)}{2}\times 4+1 =4\binom {n}{2} +1$$ The hundredth term is thus $\boxed {4\binom {100}{2}+1=19801}$. Hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find coefficient of generating function f(x). Find coefficient of generating function.
$ f(x) = \frac{2x}{1-x^{2}} +x$
MY WAY OF SOLVING SIMILAR PROBLEM:
1) $ g(x) = \frac{2x}{1-x^{2}}$
2) partial fraction $g(x) = \frac{A}{1-x} + \frac{B}{1+x} $
3) $ g(x) = \sum\limits_{n=0}^\infty Ax^{n} + \sum\limits_{n=0}^\infty B (-1)^nx^{n} = \sum\limits_{n=0}^\infty (A+(-1)^nB)x^{n} $ -solution
But what can I do with $f(x)$? I can't use my method because:
$f(x) = \frac{2x+x(1-x^2)}{1-x^2} $
$\frac{-x^3 +3x}{1-x^2} = \frac{A}{1-x} + \frac{B}{1+x}$
$ -x^3+3x = A(1+x) + B(1-x) $
$-x^3 = 0 \cdot x^3 $
$ -1 =0 $
| There is no reason to add up the right hand terms, since the term $x$ is simple and convenient. The other term can be expanded using the geometric series expansion
\begin{align*}
\frac{1}{1-y}=\sum_{n=0}^\infty y^n\qquad\qquad |y|<1
\end{align*}
with $y=x^2$.
We obtain
\begin{align*}
f(x)&=\frac{2x}{1-x^2}+x\\
&=2x\sum_{n=0}^\infty x^{2n}+x\\
&=3x+2\sum_{n=1}^\infty x^{2n+1}
\end{align*}
We conclude the coefficient $[x^n]$ of $f(x)$ is
\begin{align*}
[x^n]f(x)=
\begin{cases}
3&n=1\\
2&n>1, odd\\
0&n\geq 0, even
\end{cases}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find two numbers whose $AM+...$ Find two numbers whose $AM + GM =25$ and $AM:GM=5:3$.
My Attempt;
Given,
$\frac {AM}{GM}=\frac {5}{3} = k (let) $
$\AM=5k,
GM=3k$.
Also,
$AM+GM=25$
$5k+3k=25$
$8k=25$
$k=\frac {25}{8}$.
Am I going right? Or, is there any other simple alternative.?
| we have $$AM=\frac{5}{3}GM$$ from here we get with the first equation:
$$\frac{5}{3}GM+GM=25$$ thus we have $$\frac{8}{3}GM=25$$ and $$GM=\frac{75}{8}$$
from here we get $$ab=\left(\frac{75}{8}\right)^2$$ and $$a+b=\frac{375}{12}$$ you can solve one equation e.g. for $a$ and plug these equation in the other one
ok with $$b=\frac{375}{12}-a$$ we get
$$a\left(\frac{375}{12}-a\right)=\left(\frac{75}{8}\right)^2$$
this is equivalent to $$0=a^2-\frac{375}{12}a+\left(\frac{75}{8}\right)^2$$
solveing this we get $$a=\frac{225}{8}$$ or $$a=\frac{25}{8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\tan {\frac{A}{2}} + \tan {\frac{B}{2}} +\tan{\frac{C}{2}} \geq 4 - \sqrt {3} $
In a triangle $ABC$ with one angle exceeding $\frac {2}{3} \pi$, prove that
$\tan {\frac{A}{2}} + \tan {\frac{B}{2}} + \tan{\frac{C}{2}} \geq 4 - \sqrt {3} $
I tried expanding that half angle, applying AM-GM on various sets, using Sine rule and Napier's Analogy, but without success.
Can anyone provide a hint ?
Also, how does the left hand side of the inequality behave when the condition of one angle exceeding $\frac {2} {3 }\pi$ is removed?
Thanks in advance :) .
| Let $\gamma\geq\frac{2\pi}{3}$ and $\tan\frac{\gamma}{4}=x$.
Hence, $\frac{1}{\sqrt3}\leq x<\frac{1}{\sqrt2}$ and since $\tan$ is a convex function on $\left[0,\frac{\pi}{2}\right)$, by Jensen we obtain:
$$\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}\geq2\tan\frac{\alpha+\beta}{4}=2\tan\left(\frac{\pi}{4}-\frac{\gamma}{4}\right)=\frac{2(1-x)}{1+x}.$$
Thus, it remains to prove that
$$\frac{2(1-x)}{1+x}+\frac{2x}{1-x^2}\geq4-\sqrt3$$ or
$$(\sqrt3x-1)\left(x+\frac{1}{4+3\sqrt3}\right)\geq0,$$
which is obvious.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0 = 1$ with error smaller than $10^{-4}$ I should find $\sqrt{1.1}$ using Taylor series of the function $\sqrt{x+1}$ in $x^{}_0=1$ with error smaller than $10^{-4}$.
The first derivatives are
$$f'(x)=\frac{1}{2\sqrt{x+1}}$$
$$f''(x)=\frac{-1}{4\sqrt{x+1}^ 3}$$
$$f'''(x)=\frac{3}{8\sqrt{x+1}^5}$$
Applying $x^{}_0$ we have:
$$f(1)=\sqrt{2}$$
$$f'(1)=\frac{1}{2\sqrt{2}}$$
$$f''(1)=\frac{-1}{4\sqrt{2}^ 3}$$
$$f'''(1)=\frac{3}{8\sqrt{2}^5}$$
And we can build the Taylor polynomial
$$T(x)=\sqrt2 + \frac{1}{2\sqrt{2}}(x+1)+\frac{-1}{2!·4\sqrt{2}^3}(x+1)^2+\frac{3}{3!·8\sqrt{2}^5}(x+1)^3+R(\xi)$$
Is everything right until here?
What I don't understand is how can I check that $R(\xi) > 10^{-4}$
| The Taylor Theorem tells us that the estimation error after $n$ terms is given by $f'(c)\frac{(x-a)^{n+1}}{(n+1)!}$, for som $c \in (x-a)$. Now you should be able to find an upper bound on the derivative in that interval, which should give you an upper bound on the error.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all prime solutions of equation $5x^2-7x+1=y^2.$ Find all prime solutions of the equation $5x^2-7x+1=y^2.$
It is easy to see that
$y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$
In the same way from $y^2+2x=1 \mod 5$ we have that $y^2=1 \mod 5$ and $x=0 \mod 5$ or $y^2=-1 \mod 5$ and $x=4 \mod 5.$
How put together the two cases?
Computer find two prime solutions $(3,5)$ and $(11,23).$
| Try working mod $3$ and mod $8$. Assuming $x, y>3$, we have $x,y = \pm 1$ mod $3$. Since $x, y$ are odd we have $x^2, y^2=1$ mod $8$, so
$$x^2, y^2 = 1 \text{ mod } 24.$$
Substituting in the equation gives $$x = 24k+11 $$ for some integer $k$.
Rearranging the original equation we get
$$x(5x-7)=(y-1)(y+1), \tag{1}$$
therefore $x |y-1$ or $x|y+1$, since $x$ is a prime number.
Solving for $x$ gives
$$ x = \frac{17}{10} + \frac{1}{10}\sqrt{20y^2+29}>\frac{1}{3}(y+1).$$
Note that $x$ is odd and $y \pm 1$ is even, so $x \ne y\pm1$. This forces $x = \frac{1}{2} (y \pm1)$, or
$$y = 2x \pm 1 = 48k + 22 \pm 1 \Rightarrow y = 48k+23.$$
$48k+21$ is rejected being divisible by 3.
Plugging these in $(1)$ gives the solution $k=0$ or
$$x = 11, \space y = 23.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to factorize this cubic equation? In one of the mathematics book, the author factorized following term
$$x^3 - 6x + 4 = 0$$
to
$$( x - 2) ( x^2 + 2x -2 ) = 0.$$
How did he do it?
|
Note: I understand that there is already an accepted answer for this question, so this answer may be useless, but regardless, I'm still posting this to spread knowledge!
A simple way to factorize depressed cubic polynomials of the form$$x^3+Ax+B=0\tag1$$
Is to first move all the constants to the RHS, so $(1)$ becomes$$x^3+Ax=-B\tag2$$
Now, find two factors of $B$ such that one fact minus the square of the other factor is $A$. We'll call them $a,b$ so$$\begin{align*} & a-b^2=A\tag3\\ & ab=-B\tag4\end{align*}$$
Multiply $(2)$ by $x$, add $b^2x^2$ to both sides and complete the square. Solving should give you a value of $x$ and allow you to factor $(1)$ by Synthetic Division.
Examples:
*
*Solve $x^3-6x+4=0$ (your question)
Moving $4$ to the RHS and observing its factors, we have $-2,2$ as $a,b$ since$$-2-2^2=A\\-2\cdot2=-4$$Therefore, we have the following:$$x^4-6x^2=-2\cdot2x$$$$x^4-6x^2+4x^2=4x^2-4x$$$$x^4-2x^2=4x^2-4x$$$$x^4-2x^2+1=4x^2-4x+1\implies(x^2-1)^2=(2x-1)^2$$$$x^2=2x\implies x=2$$
Note that we do have to consider the negative case when square rooting, but they lead to the same pair of answers. So it's pointless.
*Solving $x^3+16x=455$
A factor of $455$ works, namely when $a=65,b=7$.$$65-7^2=16$$$$65\cdot7=455$$
Therefore,$$x^4+16x^2=65\cdot7x$$$$x^4+65x^2=49x^2+455x$$$$\left(x^2+\dfrac {65}{2}\right)^2=\left(7x+\dfrac {65}{2}\right)^2$$$$x=7$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$ Prove that:$$\frac {\sec (16A) - 1}{\sec (8A) - 1}=\frac {\tan (16A)}{\tan (4A)}$$.
My Attempt,
$$L.H.S= \frac {\sec (16A)-1}{\sec (8A)-1}$$
$$=\frac {\frac {1}{\cos (16A)} -1}{\frac {1}{\cos (8A)} -1}$$
$$=\frac {(1-\cos (16A)).(\cos (8A)}{(\cos (16A))(1-\cos (8A))}$$.
What should I do next?
| $\frac{\sec 16A -1}{\sec 8A -1}$
= $\frac{\frac{1}{\cos 16A}-1}{\frac{1}{\cos 8A}-1}$
= $\frac{\frac{1 - \cos 16A}{\cos 16A}}{\frac{1 - \cos 8A}{\cos 8A}}$
= $\frac{2 \sin^2 8A}{\cos 16A} × \frac{\cos 8A}{2 \sin^2 4A}$
= $\frac{2 \sin 8A \cos 8A}{\cos 16A} × \frac{\sin 8A}{2 \sin^2 4A}$
= $\frac{\sin 16A}{\cos 16A} × \frac{ 2 \sin 4A \cos 4A}{2 \sin^2 4A}$
= $\tan 16A × \frac{\cos 4A}{\sin 4A}$
= $\tan 16A × \cot 4A$
= $\frac{\tan 16A}{\tan 4A}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this limit: $\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac1x}{e^2 +x}\right)^\frac1x$ $$\lim\limits_{x \to 0}\left(\frac{(1+2x)^\frac{1}{x}}{e^2 +x}\right)^\frac{1}{x}=~?$$
Can not solve this limit, already tried with logarithm but this is where i run out of ideas. Thanks.
| Using L'Hospital rule twice we get $$\lim _{ x\to 0 } \left( \frac { (1+2x)^{ \frac { 1 }{ x } } }{ e^{ 2 }+x } \right) ^{ \frac { 1 }{ x } }=~ { e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \ln { \left( \frac { (1+2x)^{ \frac { 1 }{ x } } }{ e^{ 2 }+x } \right) } } }={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ x } \left[ \frac { 1 }{ x } \ln { \left( 1+2x \right) -\ln { \left( { e }^{ 2 }+x \right) } } \right] } }=\\ ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { \ln { \left( 1+2x \right) -x\ln { \left( { e }^{ 2 }+x \right) } } }{ { x }^{ 2 } } } }\overset { L'Hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { \frac { 2 }{ 1+2x } -\ln { \left( { { e }^{ 2 }+x } \right) -\frac { x }{ { e }^{ 2 }+x } } }{ 2{ x } } } }\overset { L'hospital }{ = } { e }^{ \lim _{ x\rightarrow 0 }{ \frac { -\frac { 4 }{ { \left( 1+2x \right) }^{ 2 } } -\frac { 1 }{ { e }^{ 2 }+x } -\frac { { e }^{ 2 } }{ { \left( { e }^{ 2 }+x \right) }^{ 2 } } }{ 2 } } }={ e }^{ -\frac { 4{ e }^{ 2 }+2 }{ 2{ e }^{ 2 } } }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Quadratic equation, find $1/x_1^3+1/x_2^3$ In an exam there is given the general equation for quadratic: $ax^2+bx+c=0$.
It is asking: what does $\dfrac{1}{{x_1}^3}+\dfrac{1}{{x_2}^3}$ equal?
| If $x_{1}$ and $x_{2}$ are the roots then
$x_{1} + x_{2} = -\frac{b}{a}$ and $x_{1} \cdot x_{2}=\frac{c}{a}$, now $\frac{1}{x_{1}}+\frac{1}{x_{2}} = -\frac{b}{c}$ and $$\frac{1}{x_{1}^{3}}+\frac{1}{x_{2}^{3}} = \left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)^3- 3\cdot\frac{1}{x_{1}.x_{2}}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right) = \frac{3abc-b^{3}}{c^{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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How to use finite differences to determine an equation of a polynomial given consecutive integer $x$ and corresponding $y$ coordinates of the graph? This chart is given:
for $x=-3$, $y=-9$
for $x=-2$, $y=3$
for $x=-1$, $y=3$
for $x=0$, $y=-3$
for $x=1$, $y=-9$
for $x=2$, $y=-9$
for $x=3$, $y=3$
I found the finite differences to be 6 and degree of the polynomial with the given points to be 3 ($n=3$). But how do I find the polynomial function rule out of this information? Given no factors? Please help with a short solution. Thanks a lot in advance. I also did $y=kx^3$, $-9=k(-3)^3$, $k=1/3$. But of course the rule isn't $y=1/3x^3$.
| Treat your numbers as a sequence with $g(0)$ being the first term, corresponding to $f(-3)$,
$$f(x-3):=g(x) : \color{green}{-9},3,3,-3,-9,-9,3$$
$$\Delta g=g(x+1)-g(x) : \color{green}{12},0,-6,-6,0,12$$
$$\Delta^2 g : \color{green}{-12},-6,0,6,12$$
$$\Delta^3 g : \color{green}{6},6,6$$
Now assume all else is $0$. To get an umbral Taylor series representation,
$$f(x-3)=\color{green}{-9}+\color{green}{12}{x \choose 1}\color{green}{-12}{x \choose 2}+\color{green}{6}{x \choose 3}$$
So that with $x \mapsto x+3$,
$$f(x)=-9+12{{x+3} \choose 1}-12{{x+3} \choose 2}+6{{x+3} \choose 3}$$
$$=x^3-7x-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Fourier Series/ fourier transform What is the Fourier series of the following piece-wise function?
$$
f(x) = \begin{cases}
0 & -1 \leq x < -0.5 \\
\cos (3 \pi x) & -0.5 < x < 0.5 \\
0 & 0.5 \leq x < 1
\end{cases}
$$
| Given f(x) = \begin{cases}
0 & -1 \leq x < -0.5 \\
\cos (3 \pi x) & -0.5 < x < 0.5 \\
0 & 0.5 \leq x < 1
\end{cases}
Its nth Fourier polynomial is $S_n(x)=\sum_{v=-n}^{n}\alpha_ve^{ivx}$, where $\alpha_v=\int_{-\pi}^{\pi}f(x)e^{ivx}dx=\int_{-1/2}^{1/2} \cos(3\pi x)e^{ivx}dx$. Notice that $\dfrac{e^{3\pi ix}+e^{-3\pi ix}}{2}=cos(3 \pi x)$, so $$\alpha_v=\int_{-1/2}^{1/2}\dfrac{e^{(v+3\pi) ix}+e^{(v-3\pi) ix}}{2}dx=\dfrac{e^{(v+3\pi)ix}}{2i(v+3\pi)}+\dfrac{e^{(v-3\pi)ix}}{2i(v-3\pi)}\mid_{-1/2}^{1/2}=\dfrac{e^{(v+3\pi)i/2}}{2i(v+3\pi)}+\dfrac{e^{(v-3\pi)i/2}}{2i(v-3\pi)}-\dfrac{e^{-(v+3\pi)i/2}}{2i(v+3\pi)}-\dfrac{e^{-(v-3\pi)i/2}}{2i(v-3\pi)}=\dfrac{\sin(\dfrac{v+3\pi}{2})}{2(v+3\pi)}+\dfrac{\sin(\dfrac{v-3\pi}{2})}{2(v-3\pi)}=\dfrac{3\pi\cos(\dfrac v2)}{v^2-9\pi^2}$$
Notice that if we regard $\alpha_v$ as a function of $v$, it is an even function. That is, $\alpha_v=\alpha_{-v} \;\forall |v|\le n$. Furthermore, $\alpha_0=-\dfrac1{3\pi}$. Hence $S_n(x)=-\dfrac1{3\pi}+3\pi\sum_{v=1}^{n}\dfrac{e^{ivx/2}+e^{3ivx/2}}{v^2-9\pi^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate a limit involving definite integral Evaluate the following limit:
$$\lim_{n \to \infty} \left[n - n^2 \int_{0}^{\pi/4}(\cos x - \sin x)^n dx\right]$$
I've tried to rewrite the expression as follows:
$$\lim_{n \to \infty} \left[n - n^2 \sqrt{2}^n \int_{0}^{\pi/4}\sin^n \left( \frac{\pi}{4} - x \right) dx\right]$$
However, this doesn't seem to help too much. Thank you!
| Writing
$$ (\cos x - \sin x)^n = \color{blue}{\frac{\cos x - \sin x}{\cos x + \sin x}} \cdot \color{red}{ (\cos x + \sin x)(\cos x - \sin x)^{n-1}} $$
and applying integrating by parts, we have
$$ \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)^n \, dx
= \frac{1}{n} - \frac{2}{n} \int_{0}^{\frac{\pi}{4}} \frac{(\cos x - \sin x)^n}{(\sin x + \cos x)^2} \, dx. $$
Plugging this back,
$$ n - n^2 \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)^n \, dx = 2n \int_{0}^{\frac{\pi}{4}} \frac{(\cos x - \sin x)^n}{(\sin x + \cos x)^2} \, dx. $$
Now from the observation
$$ n \int_{0}^{\frac{\pi}{4}} (\sin x + \cos x)(\cos x - \sin x)^n \, dx
= \frac{n}{n+1}, $$
we can apply the usual approximation-to-the-identity argument to obtain
\begin{align*}
\lim_{n\to\infty} \left[ n - n^2 \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x)^n \, dx \right]
&= \lim_{n\to\infty} 2n \int_{0}^{\frac{\pi}{4}} \frac{(\cos x - \sin x)^n}{(\sin x + \cos x)^2} \, dx \\
&= \lim_{x \to 0^+} \frac{2}{(\sin x + \cos x)^3} \\
&= \color{red}{\boxed{2}}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Finding a geometric interpretation I recently solved a question of complex numbers which was this:
$A\left( \frac{2}{\sqrt{3}} e^\frac{i\pi}{2}\right)$, $B\left( \frac{2}{\sqrt{3}} e^\frac{-i\pi}{6}\right)$, $C\left( \frac{2}{\sqrt{3}} e^\frac{-5i\pi}{6}\right)$ are the vertices of an equilateral triangle. If $P$ be a point on the incircle of the triangle, prove that $AP^2 + BP^2 + CP^2 = 5$.
My approach:
The point $P$ is given by $z = \frac{1}{\sqrt{3}}e^{i\theta}$, since the radius of the incircle of an equilateral triangle is half its circumradius. Then,
$$
AP² = |A - z|^2 \\
= (A - z)(A^* - z^*) \\
= AA^* - Az^* - A^*z + zz^* \\
= |A|² + |z|² - Az^* - A^*z \\
= \frac43 + \frac13 - Az^* - A^*z \\
= \frac53 - Az^* - A^*z.
$$
Now,
$$
AP^2 + BP^2 + CP^2 \\
= 3\times\frac53 - z^*(A+B+C) - z(A^*+B^*+C^*) \\
= 5
$$
since $A+B+C = A^*+B^*+C^* = 0$ because the position vectors $\vec{A}$, $\vec{B}$ and $\vec{C}$ are coplanar and are mutually separated by $120^\circ$.
This, I guess, is a pretty neat solution. But what I'm looking for is a more intuitive solution, rather a geometric interpretation. Does anyone know such an approach?
| The sides of your triangle are $2$. E.g. the distance of $A$ and $B$ is the length of $$\frac{2}{\sqrt{3}}(e^{-\frac{i\pi}{6}}-e^{\frac{i\pi}{2}})=\frac{2}{\sqrt{3}}(\frac{\sqrt{3}}{2}-\frac{3}{2}i)=1-\sqrt{3}i.$$
We can calculate $PA^2+PB^2+PC^2$ using elementary methods, if $P$ is a point on the incircle of an equilateral triangle $ABC$ of side $a$.
In this case the radius of the incircle is $r=\frac{a\cdot\sqrt{3}}{6}$. Let $O$ be the incenter. Then $OA=OB=OC=\frac{a\sqrt{3}}{3}$. Let $\angle POC$ be $\varphi$. Then using the cosine theorem for the triangle $POC$ $$PC^2=r^2+OC^2-2r\cdot OC\cdot\cos\varphi = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos\varphi.$$
Since $\angle POA=120^{\circ}-\varphi$ and $\angle POB=120^{\circ}+\varphi$, similarly we get
$$PA^2 = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos(120^{\circ}-\varphi)$$
and
$$PB^2 = \frac{a^2}{12}+\frac{a^2}{3}-\frac{a^2}{3}\cos(120^{\circ}+\varphi).$$
Summarizing and using the addition theorem
$$PA^2+PB^2+PC^2=a^2(\frac{1}{4}+1-\frac{1}{3}\cos\varphi-\frac{1}{3}(-\frac{1}{2}\cos\varphi+\frac{\sqrt{3}}{2}\sin\varphi)-\frac{1}{3}(-\frac{1}{2}\cos\varphi-\frac{\sqrt{3}}{2}\sin\varphi))=\frac{5}{4}a^2.$$
Since in our case $a=2$, this gives $PA^2+PB^2+PC^2=5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How many distinct groups of 5 possible from 10 seniors and 6 freshman? A school is forming a group of 5. There are 10 freshman and 6 seniors, and the group must have at least 2 freshman and at least 1 senior. How many distinct groups are possible?
My approach (that I think is definitely flawed) -
This would be a combination problem because order does not matter. There has to be at least 2 freshmen and 1 senior so there are only 2 open spots left with a pool of 8 freshmen and 5 seniors, or 13 people. Would it just be C(13,2) or 13!/(2*11!)?
| As you say the order doesn't matter. Then you have to think of all allowed groups, that is 2 freshmen 3 seniors, 3 freshmen 2 seniors, and 4 freshmen 1 senior.
2 freshmen 3 seniors:
$\dfrac{10 \cdot 9}{2!} \cdot \dfrac{6 \cdot 5 \cdot 4}{3!} = \dfrac{10!}{2!(10-2)!}\cdot \dfrac{6!}{3!(6-3)!} = 900 $ possible combinations
3 freshmen 2 seniors:
$\dfrac{10 \cdot 9 \cdot 8}{3!} \cdot \dfrac{6 \cdot 5}{2!} = \dfrac{10!}{3!(10-3)!}\cdot \dfrac{6!}{2!(6-2)!} = 1,800 $ possible combinations
4 freshmen 1 senior:
$\dfrac{10 \cdot 9 \cdot 8 \cdot 7}{4!} \cdot 6 = \dfrac{10!}{(10-4)!4!}\cdot \dfrac{6!}{1!(6-1)!} = 1,260 $ possible combinations
TOTAL:
Therefore, in total, there are $3,960$ possibilities
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}$$
I think I should be using recurrence relations here but I'm not entirely sure how that method works. I tried this:
Multiplying the first row by $(-1)$ and adding it to all rows:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\end{vmatrix}=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
0 &0 &0 &\ldots &0 &-6 &6\\
0 &0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &0 &-6 &\ldots &0 &0 &6\\\
0 &-6 &0 &\ldots &0 &0 &6\\
-6 &0 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
Applying Laplace's method to the first column
$$D=5\begin{vmatrix}
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}+6\begin{vmatrix}
5 &5 &\ldots &5 &5 &-1\\
0 &0 &\ldots &0 &-6 &6\\
0 &0 &\ldots &-6 &0 &6\\
\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
0 &-6 &\ldots &0 &0 &6\\
-6 &0 &\ldots &0 &0 &6
\end{vmatrix}$$
I can see that this one is $D$ but of order $99$...Is this leading anywhere? How would you solve this?
| you start with a $2×2$ matrix and see the eigenvalues they are $4,6$.
Then see for $3×3$ matrix the eigenvalues are $9,6,-6$
for $4×4$ they are $14,6,-6,6.$.
Hence whenever the order is even the eigenvalues 6 exceeds the eigenvalue $-6$ by 1 in multiplicity.
and hence for even $n$ the snswer is
$det= (5(n-1)-1).6^{n/2}.(-6)^{\frac{n}{2}-1}$
In your case the answer is $494.(6)^{50}.(-6)^{49}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Prove: If $x+y+z=xyz$ then $\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$ If $x+y+z=xyz$, prove that:
$$\frac {x}{1-x^2} +\frac {y}{1-y^2} + \frac {z}{1-z^2}=\frac {4xyz}{(1-x^2)(1-y^2)(1-z^2)}$$.
My Attempt:
$$L.H.S=\frac {x}{1-x^2}+\frac {y}{1-y^2}+\frac {z}{1-z^2}$$
$$=\frac {x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)}{(1-x^2)(1-y^2)(1-z^2)}$$
$$=\frac {x+y+z-xz^2-xy^2+xy^2z^2-yz^2-yx^2+x^2yz^2-zy^2-zx^2+zx^2y^2}{(1-x^2)(1-y^2)(1-z^2)}$$.
I could not move on from here.Please help.
Thanks
| Continuing from where you left, expressing terms of the numerator as:
$$-xz^2-x^2z+x^2yz^2 =-xz (z+x-xyz) =-xz (-y) $$
$$-xy^2-x^2y+x^2y^2z =-xy (x+y-xyz) =-xy (-z) $$
$$-yz^2-y^2z+xy^2z^2 =-yz (y+z-xyz)=-yz (-x) $$
$$x+y+z=xyz $$
Now add everything and the result follows. Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the greatest common divisor of $2^{2004}-1$ and $2^{2002}-1$. Find the greatest common divisor of $2^{2004}-1$ and $2^{2002}-1$.
Using Euclidean algorithm:
$$2^{2004}-1=4(2^{2002}-1)+3$$
$$2^{2002}-1=x\cdot 3+y$$
The solution manual says that $2^{2002}-1$has the remainder $0$ when divided by $3$, that is $y=0$ so GCD is $3$. But how do I find that remainder?
| We have,
$2^{2002} - 1$
$= (2^4)^{500}.2^2 - 1$
Now $2^4 \equiv 1 (\mod 3)$
From above,
$= (1)^{500} .2^2 - 1$
$= 1.2^2 - 1$
$= 4 - 1 = 3$
Divisible by 3.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Let $a,b,c$ be positive real numbers such that $abc =1$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$a^2+b^2+c^2\geq a+b+c$$.
Also, state the condition for equality.
My Attempt,
$a,b,c$ are real and positive numbers, then
$$(a-1)^2+(b-1)^2+(c-1)^2\ge 0$$
$$a^2-2a+1+ b^2-2b+1+c^2-2c+1\ge 0$$
$$a^2+b^2+ c^2-2(a+b+c)+3\ge 0$$.
I have made a start in this way, but I am not sure if this works. Please help me, with any simple and beautiful method.
| By AM-GM $$6(a^2+b^2+c^2)=\sum_{cyc}(4a^2+b^2+c^2)\geq6\sum_{cyc}\sqrt[6]{a^8b^2c^2}=6(a+b+c)$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Two inequalities involving the rearrangement inequality Well, there are two more inequalities I'm struggling to prove using the Rearrangement Inequality (for $a,b,c>0$):
$$
a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2\ge 6a^2b^2c^2
$$
and
$$a^2b+ab^2+b^2c+bc^2+ac^2+a^2c\ge 6abc
$$
They seems somewhat similar, so I hope there'a an exploitable link between them. They fall easily under Muirhead, yet I cannot figure out how to prove them using the Rearrangement Inequality.
Any hints greatly appreciated.
| by AM-GM we get $$\frac{a^4b^2+a^4c^2+b^4a^2+b^4c^2+c^4a^2+c^4b^2}{6}\geq \sqrt[6]{a^{12}b^{12}c^{12}}=a^2b^2c^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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How to prove that $\sum_{n \, \text{odd}} \frac{n^2}{(4-n^2)^2} = \pi^2/16$? The series:
$$\sum_{n \, \text{odd}}^{\infty} \frac{n^2}{(4-n^2)^2} = \pi^2/16$$
showed up in my quantum mechanics homework. The problem was solved using a method that avoids evaluating the series and then by equivalence the value of the series was calculated.
How do I prove this directly?
| First, the partial fraction of the summand can be written
$$\begin{align}
\frac{n^2}{(4-n^2)^2}&=\frac14\left(\frac{1}{n-2}+\frac{1}{n+2}\right)^2\\\\
&=\frac14 \left(\frac{1}{(n-2)^2}+\frac{1}{(n+2)^2}+\frac{1/2}{n-2}-\frac{1/2}{n+2}\right)
\end{align}$$
Second, we note that
$$\begin{align}
\sum_{n\,\,\text{odd}}\frac{1}{(n\pm 2)^2}&=\sum_{n=-\infty}^\infty \frac{1}{(2n-1)^2}\\\\
&=2\sum_{n=1}^\infty \frac{1}{(2n-1)^2}\\\\
&=2\left(\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty \frac{1}{(2n)^2}\right)\\\\
&=\frac32 \sum_{n=1}^\infty \frac{1}{n^2}\\\\
&=\frac{\pi^2}{4}
\end{align}$$
Third, it is easy to show that
$$\sum_{n=-\infty}^\infty \left(\frac{1}{2n-3}-\frac{1}{2n+1}\right)=0$$
Putting it all together we have
$$\sum_{n,\,\,\text{odd}}\frac{n^2}{(4-n^2)^2}=\frac{\pi^2}{8}$$
If we sum over the positive odd only, then the answer is $(1/2)\pi^2/8=\pi^2/16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If:$(\sqrt{x + 9})^{\frac{1}{3}} - (\sqrt{x-9})^{\frac{1}{3}} = 3$, Find $x^2$
If:$$\sqrt[3]{(x + 9)} - \sqrt[3]{(x-9)} = 3$$
Find $x^2$
I can't seem to solve this question. Any hints or solutions is welcomed.
| If $(9+x)^\frac 13 + (9-x)^\frac 13 = 3$,
then there is a $y$ such that $(9+x)^\frac 13 = (\frac 32+y)$ and $(9-x)^\frac 13 = (\frac 32-y)$.
Taking cubes of both equations you get $9 \pm x = (\frac {27} 8+\frac 92y^2) \pm (\frac {27} 4 +y^2)y$,
and so $9 = \frac {27} 8+\frac 92y^2$ and $x = (\frac {27} 4 +y^2)y$.
This gives you $y^2 = \frac 54$, then $x = \frac {32} 4 y = 8y$, and so $x^2 = 64y^2 = 16 \times 5 = 80$.
So $(9 + \sqrt {80})^\frac 13 + (9 - \sqrt {80})^\frac 13 = (\frac {3 + \sqrt 5}2) + (\frac {3 - \sqrt 5}2) = 3$ .
As a bonus, you also know the values of the individual cube roots.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A closed form for a triple integral with sines and cosines $$\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin(x)\sin(y)\sin(z)}{xyz(x+y+z)}(\sin(x)\cos(y)\cos(z) + \sin(y)\cos(z)\cos(x) + \sin(z)\cos(x)\cos(y))\,dx\,dy\,dz$$
I saw this integral $I$ posted on a page on Facebook . The author claims that there is a closed form for it.
My Attempt
This can be rewritten as
$$3\small\int^\infty_0 \int^\infty_0 \int^\infty_0 \frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z)}{xyz(x+y+z)}\,dx\,dy\,dz$$
Now consider
$$F(a) = 3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz(x+y+z)}\,dx\,dy\,dz$$
Taking the derivative
$$F'(a) = -3\int^\infty_0 \int^\infty_0 \int^\infty_0\frac{\sin^2(x)\sin(y)\cos(y)\sin(z)\cos(z) e^{-a(x+y+z)}}{xyz}\,dx\,dy\,dz$$
By symmetry we have
$$F'(a) = -3\left(\int^\infty_0 \frac{\sin^2(x)e^{-ax}}{x}\,dx \right)\left( \int^\infty_0 \frac{\sin(x)\cos(x)e^{-ax}}{x}\,dx\right)^2$$
Using W|A I got
$$F'(a) = -\frac{3}{16} \log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)$$
By integeration we have
$$F(0) = \frac{3}{16} \int^\infty_0\log\left(\frac{4}{a^2}+1 \right)\arctan^2\left(\frac{2}{a}\right)\,da$$
Let $x = 2/a$
$$\tag{1}I = \frac{3}{8} \int^\infty_0\frac{\log\left(x^2+1 \right)\arctan^2\left(x\right)}{x^2}\,dx$$
Question
I seem not be able to verify (1) is correct nor find a closed form for it, any ideas ?
| Another approach to break down the last integral might be to consider the integral of $\displaystyle \frac{\log^3 (1-iz)}{z^2}$ along a positively oriented semi-circular contour $\gamma_R = [-R,R]\cup Re^{i[0,\pi]}$ in the upper half-plane. (We choose the branch of logarithm $\log (1-iz)$ in the lower half-plane along $[-i,-i\infty)$).
The integral along the arc is $\displaystyle \mathcal{O}\left(\frac{\log^3 R}{R}\right)$, which vanishes as $R \to +\infty$ we have,
\begin{align*}0 = \lim\limits_{R \to \infty} \int\limits_{\gamma_R} \frac{\log^3 (1-iz)}{z^2}\,dz = \int_{-\infty}^{\infty} \frac{\log^3\left((1+x^2)^{1/2} - i\arctan x\right)}{x^2}\,dx\end{align*}
Comparing the real parts on both sides,
\begin{align*} \int_{-\infty}^{\infty} \frac{\log(1+x^2)(\arctan x)^2}{x^2}\,dx &= \frac{1}{12}\int_{-\infty}^{\infty} \frac{\log^3(1+x^2)}{x^2}\,dx \\&= \frac{1}{6}\int_{0}^{\infty} \frac{\log^3(1+x^2)}{x^2}\,dx\\&\underset{\text{(IBP)}}{=} \int_0^{\infty} \frac{\log^2 (1+x^2)}{1+x^2}\,dx \\&= \int_0^{\pi/2} \log^2 (\cos \theta)\,d\theta \\&= \lim\limits_{b \to \frac{1}{2}}\frac{1}{2}\frac{\partial^2}{\partial b^2} B\left(\frac{1}{2},b\right)\\&= \frac{\pi}{2}\left(4\log^2 2 + \frac{\pi^2}{3}\right)\end{align*}
Hence, $$\displaystyle \int_{0}^{\infty} \frac{\log(1+x^2)(\arctan x)^2}{x^2}\,dx = \frac{\pi^3}{12} + \pi\log^2 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128300",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "18",
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Proving non-existence of rational points in a simple equation Working on Chapter 6.20 of Hammack Book of Proof
Show that $x^{2} + y^{2} - 3 = 0$ has no rational points.
First prove: IF $3\not\vert m$ THEN $m^{2}\equiv1(mod 3)$
If $3\not\vert m$ then either of two cases are true:
CASE $m\equiv 1 (mod3)$
$m=3k+1$ where $k \in Z$
$m^{2}= 9K^{2}+6k+1 = 3(3K^{2}+2K)+1$
Thus $m^{2}\equiv1(mod3)$
CASE $m\equiv 2 (mod3)$
$m=3k+2$ where $k \in Z$
$m^{2}= 9K^{2}+12k+4 = 3(3K^{2}+4K+1)+1 $
Thus $m^{2}\equiv1(mod3)$
So in both cases $m^{2}\equiv1(mod3)$
Therefore IF $3\not\vert m$ THEN $m^{2}\equiv1(mod 3)$
Part 2: Show general case no rational points for $a^{2}+b^{2}=3c^{2}$
$a,b,c$ are relatively prime to each other (no common factors aside 1)
By inspection, $3\vert (a^{2}+b^{2})$ thus $(a^{2}+b^{2}= 3K+0$ where $k \in Z$
`
This means $(a^{2}+b^{3})$ must have no remainder when divided by three. If $3\not\vert a$ or $3\not\vert a$ then $(a^{2}+b^{2})$ would have a remainder of $(1+0, 0+1, or 1+1)$ and violate the statement. Therefore $3\vert a$ AND $3\vert b$.
Because $3\vert a$ AND $3\vert b$ we can redefine $a=3m$ and $b=3n$ where $m,n \in Z$.
Then we rewrite $(a^{2}+b^{2})=3c^{2}$ as:
$(3m)^{2}+(3n)^2=3c^{2}$
$9m^{2}+9n^{2}=3c^{2}$
$3(3m^{2}+3n^{2})=3c^{2}$
$(3m^{2}+3n^{2})=c^{2}$
$3(m^{2}+n^{2})=c^{2}$
which means: $3\vert c^{2}$ which means $3\vert c^{2}$
This gives us a contradiction -- a, b, c all are divided by three, but we stated they were relatively prime and should only have 1 as a common factor!
Thus conclude there are no rational point solution for $a^{2}+b^{2}=3c^{2}$
final section
$a^{2} + b^{2} - 3 =0$
$a^{2} + b^{2} =3$
replace the rationals $a,b$ with rationals $p,q,m,n \in Z$
$\left(\frac{p}{q}\right)+\left(\frac{m}{n}\right)=3$
$(pn)^{2}+(mq)^{2}=3(qn)^{2}$
rename $a=pn$, $b=mq$, and $c=qn$ and we get $a^{2}+b^{2}=3c^{2}$, which we know there is no rational point solution for.
QED?
Further I'm confused by Hammack's solution/hint, as he says I should be inspecting $mod4$ results, while I believe I solved this using $mod3$
| Just do the hint.
If $n=2k $ then $n^2=4k^2\equiv 0\mod 4$
If $n=2k+1$ then $n^2=4k^2+4k+1\equiv 1 \mod 4$
So $3c^2 \equiv 0|3\mod 4$
And $a^2+b^2=0,1,2\mod 4$
So if $a^2 +b^2 =3c^2$ then all $a,b,c$ are even. But if we let $a=\gcd (a,b,c)a';b=\gcd (a,b,c)b';c=\gcd (a,b,c)c'$. $a',b',c'$ can't all be even (unless they are all $0$). But $a'^2+b'^2=3c'^2$ so they must be all even. A contradiction.
So $a^2+b^2=3c^2$ has no integer solutions (except $(0,0,0) $).
Let $r=n/m;s=p/q \in \mathbb Q$ and let $(r, s) $ be a solution to $r^2+s^2-3=0$. Then $(nq)^2+(pm)^2=3 (mq)^2$ . But that is impossible.
So we aren't solving with mod 4. We are using a specific property that the sum of square integer which is true for any problem involving sums of squares.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Lower bound for $\frac{(x+y+z)^3}{xy+yz+zx}$ Let $x,y,z\geq 0$ and $x^2+y^2+z^2\geq 3$. What is the minimum value of $$D(x,y,z)=\frac{(x+y+z)^3}{xy+yz+zx}?$$
When $x=y=z=1$, $D(x,y,z)=9$. We have $(x+y+z)^2\leq 3(x^2+y^2+z^2)$ and $xy+yz+zx\leq x^2+y^2+z^2$, but these do not help directly with bounding $D(x,y,z)$.
| Let $x^2+y^2+z^2=k(xy+xz+yz)$.
Hence, by AM-GM we obtain:
$$\frac{(x+y+z)^3}{xy+xz+yz}\geq\frac{(x+y+z)^3}{xy+xz+yz}\sqrt{\frac{3}{x^2+y^2+z^2}}=$$
$$=\sqrt{\frac{3(x+y+z)^6}{(xy+xz+yz)^2(x^2+y^2+z^2)}}=\sqrt{\frac{3(k+2)^3}{k}}\geq\sqrt{\frac{3(3\sqrt[3]k)^3}{k}}=9.$$
The equality occurs for $x=y=z=1$, which says that the answer is $9$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Subsets and Splits