Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove by complete induction a floor formula I cant prove by complete induction this formula:
$$(-1)^{\lfloor\frac{n+1}2\rfloor+1}=(-1)^{\lfloor\frac{n}2\rfloor+1+n}$$
I keep failing when I apply $(-1)^{\lfloor\frac{n+1+1}2\rfloor+1}=(-1)^{\lfloor\frac{n+1}2\rfloor+1+n+1}$. It does not give me the same values. For example if n=8 in the first one both parts will be odd but in the induction the first part is odd and the second is even. Probably I am ignoring some floor properties.
| Alt. hint (without induction): use Hermite's identity $\displaystyle\,\lfloor x \rfloor + \left\lfloor x + \frac{1}{2} \right\rfloor = \lfloor 2x \rfloor\,$ for $\displaystyle\,x=\frac{n}{2}\,$:
$$\left\lfloor\frac{n}{2}\right\rfloor + \left\lfloor\frac{n+1}2\right\rfloor = \left\lfloor 2 \cdot \frac{n}{2}\right\rfloor = \lfloor n \rfloor = n
\;\;\implies\;\; \left\lfloor\frac{n+1}2\right\rfloor = n - \left\lfloor\frac{n}2\right\rfloor
$$
Then, since $(-1)^k = (-1)^{-k}$ for all integer $k\,$:
$$
(-1)^{\left\lfloor\frac{n+1}2\right\rfloor+1}=(-1)^{n - \left\lfloor\frac{n}2\right\rfloor+1} = (-1)^{\left\lfloor\frac{n}2\right\rfloor+1+n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x=8+3\sqrt{7},$ then what is value of $\sqrt{x} -\frac{1}{\sqrt{x}}$?
If $x=8+3\sqrt{7},$ then what is value of $\sqrt{x} -\frac{1}{\sqrt{x}}$?
This question is somewhat different than I thought. I only know how to find the value when root is not given . Please help me.
| Note that $\left ( \sqrt{x}- \frac{1}{\sqrt{x}} \right )^2 = x + \frac{1}{x} - 2$. Further note that $8+3 \sqrt{7} > 1$, so $\sqrt{x}-\frac{1}{\sqrt{x}} > 0$.
$$x=8+3 \sqrt{7} \implies \frac{1}{x} = \frac{8-3\sqrt{7}}{8^2 - (3\sqrt{7})^2} = \frac{8-3\sqrt{7}}{64-63} = 8-3\sqrt{7}.$$
Thus, $x+\frac{1}{x} = 16$, which implies $x + \frac{1}{x} - 2 = 14$. Thus, $\sqrt{x}- \frac{1}{\sqrt{x}} = \sqrt{14}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3}$ for $ax^3 + bx^2 + cx + d$ Using Vieta's formulas, I can get $$\begin{align} \frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3} &= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\&= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\ &= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{\left (-\frac{d}{a} \right)^3}\end{align}$$
But then I don't know how to substitute the numerator.
| Let $y = 1/x$. So you ask for the value of $f(y) = y_1^3 + y_2^3 + y_3^3$ where the $y_i$ are solutions of $dy^3 + cy^2 + by + a = 0$. With this equation, write
$$- d f(y) = c (y_1^2 + y_2^2 + y_3^2) + b (y_1 + y_2 + y_3) + 3 a.$$
Note the identity $$y_1^2 + y_2^2 + y_3^2=(y_1+y_2+y_3)^2-2(y_1y_2+y_1y_3+y_2y_3).$$ So you have
$$- d f(y) = c (\color{red}{y_1+y_2+y_3})^2-2c (\color{green}{y_1y_2+y_1y_3+y_2y_3}) + b (\color{red}{y_1 + y_2 + y_3}) + 3 a.$$
Now Vieta's formulae give (note the order of the constants!)
$$y_1+y_2+y_3 = -\frac cd\quad \text{and}\quad y_1y_2+y_1y_3+y_2y_3 = \frac bd.$$
So you obtain
$$- d f(y) = c \left(\color{red}{-\frac cd}\right)^2-2c \left(\color{green}{\frac bd}\right) + b \left(\color{red}{-\frac cd}\right) + 3 a,$$
hence the final result
$$
f(y) = y_1^3 + y_2^3 + y_3^3 = - \frac{c^3}{d^3} + \frac{3bc}{d^2} - \frac{3a}d = \frac{-c^3+ 3 bcd - 3 ad^2}{d^3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2418954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$a= \frac{b^2-2b+16}{b^3} b= \frac{8a^3}{a^2-2a+16} a=? b=?$ it's part of my bigger homework, i got to state like that and i need some nice way to calculate a and b. I was trying to do this by setting b value from second equations to first equations, but it feels like so much work, can i do it smarter way?
If it's impossible i will send all my exercise, so you can see if i didn't noticed a better way to do whole thing.
$$ a= \frac{b^2-2b+16}{b^3}$$
$$ b= \frac{8a^3}{a^2-2a+16}$$
| HINT:
$8 a^3 b^3= 8 a^2(b^2-2b+16) = b^4(a^2-2a+16)$.
setting $a = x b$ you have
$8 x^3 b^6= 8 x^2 b^2(b^2-2b+16) = b^4(x^2 b^2-2 x b +16)$
Equating the second and third expression gives a quadratic equation in $x$ which you can solve for $x = x (b)$.
Equating the first and second expression gives
$ x = \frac{b^2-2b+16}{b^4}$ which you can now equate with your $x (b)$ obtained above, to obtain $b$. Here the story may become nasty....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving Inequality involving exponential function How can I solve the following inequality algebraicly for $x$?
$$\frac{\sqrt{5}}{\sqrt{t}}\cdot (\frac{x}{50})^\frac{5}{2}\cdot e^\sqrt{\frac{x}{50t}}<x\ \ \ \ \ \ \ (t>0)$$
I suppose it can be solved with using Lambert W function but I don't know how I can introduce Lambert W here.
| $$\frac{\sqrt{5}}{\sqrt{t}}\cdot (\frac{x}{50})^\frac{5}{2}\cdot e^\sqrt{\frac{x}{50t}}<x\ \ \ \left(t,x\in\mathbb{R}\right)$$
Reshape your inequality into an inequality of only one function. Bring for that all $x$ to the left-hand side of your inequality:
$$\frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{5}{2}}}{50^\frac{5}{2}}\cdot e^\sqrt{\frac{x}{50t}}-x<0$$
or
$$x<0:\ \frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{3}{2}}}{50^\frac{5}{2}}\cdot e^\sqrt{\frac{x}{50t}}>1$$
$$x>0:\ \frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{3}{2}}}{50^\frac{5}{2}}\cdot e^\sqrt{\frac{x}{50t}}<1$$
Consider the inequalities later and solve now one of the equations that are related to this inequalities:
$$\frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{5}{2}}}{50^\frac{5}{2}}\cdot e^\sqrt{\frac{x}{50t}}-x=0$$
or
$$\frac{\sqrt{5}}{\sqrt{t}}\cdot \frac{x^{\frac{3}{2}}}{50^\frac{5}{2}}\cdot e^\sqrt{\frac{x}{50t}}=1$$
On the left-hand side of this equation, you have a function in dependence of $x$. This function is an algebraic function in dependence of $x$ and $e^{\sqrt{x}}$. $x$ and $e^{\sqrt{x}}$ are algebraically independent for all $x\neq 0$. But $x=0$ is not a solution of your inequality. Therefore this function cannot be brought into a form of an algebraic function in dependence of only one transcendental argument. Therefore you cannot solve the equation by only applying elementary operations/functions.
Lambert W function is the inverse of the function $f\colon x\mapsto f(x)=xe^x.$ For applying only Lambert W and elementary functions, your equation should be in the form
$$f_1(f_2(x)e^{f_2(x)})=c,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
where $c$ constant and $f_1$ and $f_2$ are elementary functions with a suitable elementary partial inverse.
Collect all powers of $x$ and all powers of $e^{x}$:
$$\frac{\sqrt{5}}{\sqrt{t}\cdot 50^{\frac{5}{2}}}\cdot x^\frac{5}{2}\cdot e^{\frac{x^{\frac{1}{2}}}{\sqrt{50t}}}-x=0\ \ \ \ \ \ \ \ \ (2)$$
or
$$\frac{\sqrt{5}}{\sqrt{t}\cdot 50^{\frac{5}{2}}}\cdot x^\frac{3}{2}\cdot e^{\frac{x^{\frac{1}{2}}}{\sqrt{50t}}}=1.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$
Solve one of this equations for $x$ by applying Lambert W. Both are algebraically solvable by Lambert W.
Take e.g. equation (3):
$$a=\frac{\sqrt{5}}{\sqrt{t}\cdot 50^{\frac{5}{2}}}\in\mathbb{R},\ b=\sqrt{50t}\in\mathbb{R},\ x\in\mathbb{R}$$
$$ax^\frac{3}{2}e^{\frac{x^{\frac{1}{2}}}{b}}=1$$
$$z=x^{\frac{1}{2}}:$$
$$az^{3}e^{\frac{z}{b}}=1\ \ |\ ()^{\frac{1}{3}}$$
$$a^\frac{1}{3}ze^{\frac{z}{3b}}=1\ \ |\ :3a^{\frac{1}{3}}b$$
$$\frac{z}{3b}e^{\frac{z}{3b}}=\frac{1}{3a^{\frac{1}{3}}b}$$
$$\frac{z}{3b}=W\left(\frac{1}{3a^{\frac{1}{3}}b}\right)$$
$$z=3b\cdot W\left(\frac{1}{3a^{\frac{1}{3}}b}\right)$$
$$x=z^{2}:$$
$$x=9b^{2}\cdot W\left(\frac{1}{3a^{\frac{1}{3}}b}\right)^{2}$$
Solve the above inequalities then.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the determinant of this n by n matrix? Find the determinant of this n by n matrix.
$$
\begin{pmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{pmatrix}
$$
where, $$ k=n-1 $$.
I am new to matrices and determinants, but this is what I did:
I developed the determinant using the second column:
$$
(-1)^2*x_1
\begin{pmatrix}
x_1 & 0 & \cdots & 0 \\
x_2 & 1& \cdots & 0 \\
\vdots& \vdots& \ddots& \vdots \\
x_k & 0 & \cdots& 1 \\
\end{pmatrix} + (-1)^3 *1
\begin{pmatrix}
0 & x_2 & \cdots & x_k \\
x_2 & 1& \cdots & 0 \\
\vdots& \vdots& \ddots& \vdots \\
x_k & 0 & \cdots& 1 \\\end{pmatrix}
$$
the first determinant is triangular, so its equal to $ x_1 $ but this is where I got stuck. I don't know what to do with the second determinant. Any help is appriciated. Thanks
| If you denote:
$$D_k=\begin{vmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{vmatrix}$$
and expand it by the last column (or row), you get
\begin{align}D_k=\begin{vmatrix}
0 & x_1 & x_2 & \cdots& x_k \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_k & 0 & 0 & \cdots& 1 \\
\end{vmatrix}&=(-1)^kx_k\begin{vmatrix}
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_{k-1}&0 &0&\cdots&1\\
x_k & 0 & 0 & \cdots&0\\
\end{vmatrix}
+\begin{vmatrix}
0 & x_1 & x_2 & \cdots& x_{k-1} \\
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_{k-1} & 0 & 0 & \cdots& 1 \\
\end{vmatrix}\\[2ex]
&=(-1)^kx_k\begin{vmatrix}
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_{k-1}&0 &0&\cdots&1\\
x_k & 0 & 0 & \cdots&0\\
\end{vmatrix}
+ D_{k-1}
\end{align}
Now expand this determinant by the last row:
$$\begin{vmatrix}
x_1 & 1 & 0 & \cdots & 0 \\
x_2 & 0 & 1& \cdots & 0 \\
\vdots& \vdots& \vdots& \ddots & \vdots\\
x_{k-1}&0 &0&\cdots&1\\
x_k & 0 & 0 & \cdots&0\\
\end{vmatrix}=(-1)^{k-1}x_k\begin{vmatrix}
1&0&\cdots & 0 \\
0&1&\cdots & 0 \\
\vdots&\vdots& \ddots & \vdots\\
0&0&\cdots&1
\end{vmatrix}=(-1)^{k-1}x_k.
$$
Finally we obtain the recursion relation:
$$D_k=(-1)^{2k-1}x_k^2+
+ D_{k-1}=-x_k^2+
+ D_{k-1},$$
from which a trivial recurrence shows that
$$D_k=-\sum_{i=1}^k x_i^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Definite integral $\int_0^1\sqrt{x^2+1}\, dx$. Use trig substitution? I have an integral:
$$\int_0^1\sqrt{x^2+1}\, dx$$
but I have gotten stuck. Here's the work I have done already:
I'm not sure where to go from here. Using a trig identity doesn't seem like it would work. Integration by parts doesn't work (at least I think so)
Any ideas on what I should do next?
Thank you!
| You don't need a trigonometry substitution. Integrate by parts. Let $a\in\mathbb R$. I'm using $a^2$ instead of $a$ here because $x=a\tan t$, $x=a\cot t$ seem like possible substitutions. You've already tried $x=\tan \theta$ for your problem. At least one answer has shown how it can solve the problem.
$$I:=\int \sqrt{x^2+a^2}\, dx$$
$:=$ means "equal by definition".
$$\int u\, dv=uv-\int v\, du$$
$$u=\sqrt{x^2+a^2}$$
$$du=\frac{x}{\sqrt{x^2+a^2}}\, dx$$
$$dv=dx, v=x$$
$$\int \sqrt{x^2+a^2}\, dx=x\sqrt{x^2+a^2}-$$
$$-\int\frac{x^2}{\sqrt{x^2+a^2}}\, dx=$$
$$=x\sqrt{x^2+a^2}-\int \frac{(x^2+a^2)-a^2}{\sqrt{x^2+a^2}}\, dx=$$
$$=x\sqrt{x^2+a^2}-\int \sqrt{x^2+a^2}\, dx +$$
$$+a^2\int\frac{dx}{\sqrt{x^2+a^2}}=$$
$$=x\sqrt{x^2+a^2}-I+$$
$$+a^2\int \frac{d(x+\sqrt{x^2+a^2})}{x+\sqrt{x^2+a^2}}=$$
$$=x\sqrt{x^2+a^2}-I+$$
$$+a^2\ln|x+\sqrt{x^2+a^2}|$$
$$I=\frac{1}{2}(x\sqrt{x^2+a^2})+$$
$$+\frac{1}{2}(a^2\ln|x+\sqrt{x^2+a^2}|)+C$$
In the same way you could prove $$\int \sqrt{x^2-a^2}\, dx=\frac{1}{2}(x\sqrt{x^2-a^2})-$$
$$-\frac{1}{2}(a^2\ln|x+\sqrt{x^2-a^2}|)+C$$
We have
$$\int_0^1 \sqrt{x^2+a^2}\, dx=\frac{1}{2}(1\sqrt{1^2+a^2})+$$
$$+\frac{1}{2}(a^2\ln|1+\sqrt{1^2+a^2}|)-\frac{1}{2}(0\sqrt{0^2+a^2})-$$
$$-\frac{1}{2}(a^2\ln|0+\sqrt{0^2+a^2}|)$$
You used the substitution $x=\tan\theta$. You get $\int_0^1 \sec^3 \theta\, d\theta$. As one answer has shown, you can then use the substitution $u=\sin\theta$. Another possibility, which is always the case for any integral with all terms trigonometric terms $\sin x$, $\cos x$, etc., is the substitution $u=\tan\frac{\theta}{2}$, which turns the integral into a rational one using $\sin\theta = \frac{2u}{u^2+1}$, $\cos \theta = \frac{1-u^2}{1+u^2}$, $du=\frac{(1+u^2)}{2}\, d\theta$, etc. Then try using partial fractions. Read more about tangent half-angle substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Proving $\frac12\cdot\frac34\cdot\dots\cdot\frac{2n-1}{2n}\leq\frac1{\sqrt{3n+1}} ,\;\forall n\in\mathbb{N}$ using induction Base case. Let $n=1$, then $\frac12\leq\frac1{\sqrt{3+1}}$.
Induction step. Let's assume the inequality is true for some $k\in\mathbb{N}$.
We need to show that it's true for $k+1$, i.e. $\frac12\cdot\frac34\cdot\dots\cdot\frac{2k-1}{2k}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$.
From the assumption we get that $\frac12\cdot\frac34\cdot\dots\cdot\frac{2k-1}{2k}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}$.
So now I need to show that $\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$. How should I do this?
| squaring both sides of your inequality (the last one)
$$\frac{1}{3k+1}\cdot \frac{(2k+1)^2}{(2k+2)^2}\le \frac{1}{3k+4}$$
this is equivalent to
$$(3k+4)(2k+1)^2\le (3k+1)(2k+2)^2$$
simplifying this we obtain: $$k>0$$ and this is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\lim\limits_{x\to 4}\frac{2x\sqrt{x}+x-8\sqrt{x}-4}{x+\sqrt{x}-6}$ by just factoring Calculate $$\lim_{x\to 4}\dfrac{2x\sqrt{x}+x-8\sqrt{x}-4}{x+\sqrt{x}-6}$$ by just factoring
Factoring $2x\sqrt{x}+x-8\sqrt{x}-4$ gives us $(x-4)(2\sqrt{x}+1)$
$\color{red}{2x\sqrt{x}+x-8\sqrt{x}-4=x(2\sqrt{x}+1)-4(2\sqrt{x}+1)=(x-4)(2\sqrt{x}+1)}$
Factoring $x+\sqrt{x}-6$ gives us $(\sqrt{x}+3)(\sqrt{x}-2)$
Let $y=\sqrt{x}$, then $\color{blue}{y^2+y-6=(y+3)(y-2)=(\sqrt{x}+3)(\sqrt{x}-2)}$
So our limit becomes:
$$\lim_{x\to 4}\dfrac{(x-4)(2\sqrt{x}+1)}{(\sqrt{x}+3)(\sqrt{x}-2)}$$
but still this is indeterminate form, and so how would I further factor this, if posssible?
| Let $x=t^2$
If
$$x=4 $$
Then
$$t=2$$
We have to find
$$\lim_{t\to 2} \frac{2t^3+t^2-8t-4}{t^2+t-6} $$
$$\lim_{t\to 2} \frac{\color{blue}{(t-2)}(t+2)(2t+1)}{\color{blue}{(t-2)}(t+3)} $$
I think you can handle further.
$\color{red}{OR}$
In your equation factories as follows
$$(x-4)=(\sqrt x +2)\color{blue}{(\sqrt x-2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$5$ kids toss one die each. Find the number of ways that the sum of dice points is $22$. I have already done these following steps and I'm already lost. Can someone please help me?
$x_1 + x_2 + x_3 + x_4 + x_5 = 22$
when $1 \leq x_n \leq 6$ and $n = 1,2,3,4,5$.
Then,
\begin{align*}
f(x) & = (x+x^2+x^3+x^4+x^5+x^6)^5\\
& = \left(\frac{x(1-x^6)}{1-x}\right)^5\\
& = x^5 \cdot (1-x^6)^5 \cdot (1-x)^{-5}
\end{align*}
| As with the other answers, here is an expansion of the generating function.
$$
\begin{align}
x^5\left(\frac{1-x^6}{1-x}\right)^5
&=x^5\sum_{j=0}^5\binom{5}{j}\left(-x^6\right)^j\sum_{k=0}^\infty\binom{-5}{k}(-x)^k\tag{1}\\
&=\sum_{j=0}^5\sum_{k=0}^\infty(-1)^{j+k}\binom{5}{j}\binom{-5}{k}x^{k+6j+5}\tag{2}\\
&=\sum_{j=0}^5\sum_{k=0}^\infty(-1)^j\binom{5}{j}\binom{k+4}{k}x^{k+6j+5}\tag{3}\\
&=\sum_{j=0}^5\sum_{k=6j+5}^\infty(-1)^j\binom{5}{j}\binom{k-6j-1}{k-6j-5}x^k\tag{4}\\
&=\sum_{k=5}^\infty\color{#C00}{\sum_{j=0}^{\left\lfloor\frac{k-5}6\right\rfloor}
(-1)^j\binom{5}{j}\binom{k-6j-1}{4}}x^k\tag{5}
\end{align}
$$
Explanation:
$(1)$: Binomial Theorem
$(2)$: combine and rearrange
$(3)$: $\binom{-5}{k}=(-1)^k\binom{k+4}{k}$
$(4)$: $k\mapsto k-6j-5$
$(5)$: change order of summation and $\binom{n}{k}=\binom{n}{n-k}$
For $k=22$, we get from the red part of $(5)$:
$$
\begin{align}
\sum_{j=0}^2
(-1)^j\binom{5}{j}\binom{21-6j}{4}
&=\binom{5}{0}\binom{21}{4}-\binom{5}{1}\binom{15}{4}+\binom{5}{2}\binom{9}{4}\\[6pt]
&=420
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Algorithm to get the number of sorted combinations? Lets have a set of number S = {A,B,C,...,n} where A < B < C .... < n
how many ways can you sort their combinations?
Take this small example:
If the set is {1,2,8} (so n = 3) i can sort their combination in ascending order like this: 1,2,1+2,8,1+8,2+8,1+2+8 which is equal to: A,B,C,A+B,A+C,B+C,A+B+C
BUT if my set of numbers is 1,3,4 then my array will be equal to: A,B,A+B,C,A+C,B+C,A+B+C as you can see A+B will be greater or smaller sometimes.
So for N = 3 theres the follow possible combinations:
{A,B,C,A+B,A+C,B+C,A+B+C} and {A,B,A+B,C,A+C,B+C,A+B+C} theres only 2
Is there an algorithm that can take out for me this combinations(dont brute force)?
How many combinations are in N,
i want to know how many "combinations" can i get, so for N = 3 its 2 for N = 3 its ???.
Thanks.
| This is not a complete answer, but in order to get you started:
At least for small $N$ you can read off the cases from the Hasse diagram. E.g. for your example $N=3$ we have
$$\begin{matrix}
& & & & A+B+C \\
& & & \huge\diagup & \\
& & B+C \\
& & \huge| & & \\
& & A+C \\
& \huge\diagup & & \huge\diagdown \\
C & & & & A+B \\
& \huge\diagdown & & \huge\diagup & \\
& & B & & \\
& & \huge| & & \\
& & A & & \\
\end{matrix}$$
(where I've broken the symmetry because the hack I'm using to draw the diagram with MathJax has limitations around padding).
You can immediately see that the only incomparable elements are $A+B$ and $C$, and there are two ways to order two elements.
For $N=4$ it's more complicated:
$$\begin{matrix}
& & & & \mkern-36mu A+B+C+D \mkern-36mu & & & & \\
& & & & \huge| & & \\
& & & & B+C+D & & & & \\
& & & & \huge| & & \\
& & & & A+C+D & & & & \\
& & & \huge\diagup & & \huge\diagdown & & \\
& & C+D & & & & A+B+D \\
& & & \huge\diagdown & & \huge\diagup & & \huge\diagdown \\
& & & & B+D & & & & A+B+C \\
& & & \huge\diagup & & \huge\diagdown & & \huge\diagup \\
& & A+D & & & & B+C \\
& \huge\diagup & & \huge\diagdown & & \huge\diagup \\
D & & & & A+C \\
& \huge\diagdown & & \huge\diagup & & \huge\diagdown \\
& & C & & & & A+B \\
& & & \huge\diagdown & & \huge\diagup & \\
& & & & B & & \\
& & & & \huge| & & \\
& & & & A & & \\
\end{matrix}$$
The choices are not all independent. E.g. $A+B$ vs $C$ is the same choice as $C+D$ vs $A+B+D$, because there's an obvious symmetry: $x < y$ iff $A+B+C+D - x > A+B+C+D - y$. (You may wish to add $0$ to the diagrams to emphasise the symmetry).
More subtlely, the choices are no longer binary. $D$ is incomparable to the entire chain $A + C < B + C < A + B + C$, so there are four places we can insert it into the chain, and they make it collapse in different ways.
*
*If $D < A + C$ we're left with two decisions: $A + B$ vs $C < D$; and $A+D$ vs $B+C$.
*If $A + C < D < B + C$ we're left with two decisions: $C$ vs $A+B$ and $B+C$ vs $A + D$.
*If $B + C < D < A + B + C$ we're left with one decision: $C$ vs $A+B$.
*If $A + B + C < D$ we again have the one decision $C$ vs $A+B$.
All of the remaining decisions are independent, so there are 14 possible chains.
I did the above diagrams by hand. If you want to construct them algorithmically then you can do a transitive reduction from a set of simple but not minimal relations. Calling the values $A_1, A_2, \ldots A_N$:
*
*For each element $A_i + A_j + \ldots + A_m$, we remove in turn each variable $A_j$ and replace it with $A_k$ choosing the smallest $k > j$ such that $A_k$ is not already included. Then $A_i + A_j + \ldots + A_m < A_i + A_k + \ldots + A_m$.
*For each element $A_i + A_j + \ldots + A_m$ which doesn't include $A_1$, we add $A_1$. Then $A_i + A_j + \ldots + A_m < A_1 + A_i + A_j + \ldots + A_m$.
This gives a sparse graph, so the transitive reduction can be done in $2^N N$ time.
However, if you want to automate this rather than have a manual step, it may be simpler to construct the full transitive closure in $2^{3N}$ time (or slightly faster if you want to use a more complex matrix multiplication algorithm) and read off the incomparable pairs directly rather than reconstructing them from the forks in the transitive reduction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Distribute $N$ indistinguishable balls, where $N$ is even, into $n$ distinguishable boxes if each box can contain at most $N/2$ balls How many ways are there to distribute $N$ ($N$ is an even number) indistinguishable balls into $n$ distinguishable boxes if each box can contain at most $\frac{N}{2}$ balls.
My solution:
There are
$${N+n-1}\choose{n-1}$$
different ways to distribute $N$ indistinguishable balls into $n$ distinguishable boxes.
But we must subtract the number of those cases in which there are more than $\frac{N}{2}$ balls in one of the boxes.
If we put $\frac{N}{2}+1$ balls in the first box we get
$${\frac{N}{2}-1+n-2}\choose{n-2}$$
different ways to distribute $\frac{N}{2}-1$ balls into the boxes $2,\cdots , n$
Similary, for the case where we have $\frac{N}{2}+j$ balls in (exactly) one on the boxes, $j=1,..., \frac{N}{2}$ we get the formula
$${\frac{N}{2}-j+n-2}\choose{n-2}$$.
Then, the answer is
$${{N+n-1}\choose{n-1}}-n\sum_{j=1}^{\frac{N}{2}}{{\frac{N}{2}-j+n-2}\choose{n-2}}$$
| Since $N$ is even, let $N = 2k$. Then the question can be rephrased as follows: In how many ways can $2k$ indistinguishable balls be placed in $n$ boxes if at most $k$ balls can be placed in one box?
Clearly, $n \geq 2$, for otherwise we would not be able to distribute all the balls to the boxes given the restriction that at most $k$ balls may be placed in one box.
The number of ways of placing $2k$ balls in $n$ boxes is the number of solutions of the equation
$$x_1 + x_2 + x_3 + \ldots + x_n = 2k \tag{1}$$
in the nonnegative integers. Since a particular solution of equation 1 corresponds to the placement of $n - 1$ addition signs in a row of $2k$ ones, equation 1 has
$$\binom{2k + n - 1}{n - 1}$$
solutions since we must choose which of the $2k + n - 1$ positions (for $2k$ ones and $n - 1$ addition signs) will be filled with addition signs.
From these, we must exclude those cases in which one box receives more than $k$ balls. Notice that there can be at most one such box since $2(k + 1) = 2k + 2 > 2k = N$.
Suppose that box 1 contains more than $k$ balls. Then $x_1' = x_1 - (k + 1)$ is a nonnegative integer. Substituting $x_1' + k + 1$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + k + 1 + x_2 + x_3 + \ldots + x_n & = 2k\\
x_1' + x_2 + x_3 + \ldots + x_n & = k - 1 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{k - 1 + n - 1}{n - 1} = \binom{n + k - 2}{n - 1}$$
solutions. By symmetry, there are an equal number of solutions of equation 1 for each of the $n$ variables that could exceed $k$. Hence, we must exclude
$$n\binom{n + k - 2}{n - 1}$$
solutions.
Consequently, there are
$$\binom{2k + n - 1}{n - 1} - n\binom{n + k - 2}{n - 1}$$
ways to distribute $N = 2k$ indistinguishable balls to $n$ boxes if no box can contain more than $N/2 = k$ balls.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $\frac{b-a}{\sqrt{1-a^2}}<\sin^{-1}b-\sin^{-1}a<\frac{b-a}{\sqrt{1-b^2}}$ by using Mean Value Theorem? Given Question is Prove that
$$\frac{b-a}{\sqrt{1-a^2}}<\sin^{-1}b-\sin^{-1}a<\frac{b-a}{\sqrt{1-b^2}}$$
when $0<a<b<1$ using Mean Value Thoerem.
I considered $f(x)=\sin^{-1}x $ when $x\in(0,b)$ and used LMVT and established that $\frac{b}{\sqrt{a-c^2}}=\sin^{-1}b$ where $c \in (0,b)$.
But I don't know how to proceed after this. Please guide.
| Using $f (x)=arcsin (x) $ in the interval $[a,b] $ we have $\frac {\arcsin (b)-\arcsin (a)}{b-a}=\frac {1}{\sqrt {1-c^2} }$ now we have $a <c <b $ hence $-a^2>-c^2>-b^2$ adding 1 and recipeocating we have $\frac {1}{1-a^2}<\frac {1}{1-c^2}<\frac {1}{1-b^2} $ hence taking root completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2427411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Guessing the value of $n$ $A$ and $B$ play game, $A$ choose $n$ where $n \in \{1, 2,\ldots 1001\}=S$.
$B$ has to guess the value of $n$ by choosing a number of subsets of $S$, then $A$ will tell $B$ the number of subsets $B$ choose that contain $n$.
Do the same operation for $3$ times, let $k_1, k_2, k_3$ be the number of subsets that $B$ choose in the $1^{st}, 2^{nd}$ and $3^{rd}$ time respectively.
Find the minimum possible value of $ k_1 + k_2+ k_3$ such that $B$ always makes a correct guess.
My thought :
The $1^{st}$ time, $B$ choose $\{1\}, \{1,2\}, \{1, 2, 3\}, \ldots, \{1, 2, 3, \dots, 334\}$.
If $A$ says $334$, then $n=1$.
If $A$ says $1$, then $n=334$.
If $A$ says $0$, then $n\not\in \{1, 2, 3, \dots, 334\}$.
The $2^{nd}$ time, $B$ choose $\{335\}, \{335, 336\},\ldots, \{335, 336, 337 \dots, 671\}$.
The $3^{rd}$ time, $B$ choose $\{672\}, \{672, 673\},\ldots, \{672, 673, 674 \dots, 1001\}$.
| On the first round ask the six questions.
\begin{eqnarray*}
\{ i \mid i \equiv 1 \pmod 7 \} \\
\{ i \mid i \equiv 1 \pmod 7 \text{ or } i \equiv 2 \pmod 7 \} \\
\vdots \\
\{ i \mid i \equiv 1 \text{ or } 2 \text{ or } 3 \text{ or } 4 \text{ or } 5 \text{ or } 6\pmod 7 \} \\
\end{eqnarray*}
On the second round ask the $10$ questions
\begin{eqnarray*}
\{ i \mid i \equiv 1 \pmod {11} \} \\
\{ i \mid i \equiv 1 \text{ or } 2 \pmod {11} \} \\
\vdots \\
\{ i \mid i \equiv 1 \text{ or } 2 \text{ or } 3 \text{ or } 4 \cdots 10 \pmod {11} \} \\
\end{eqnarray*}
On the third round ask the $12$ questions
\begin{eqnarray*}
\{ i \mid i \equiv 1 \pmod {13} \} \\
\{ i \mid i \equiv 1 \text{ or } 2 \pmod {13} \} \\
\vdots \\
\{ i \mid i \equiv 1 \text{ or } 2 \text{ or } 3 \text{ or } 4 \cdots 12 \pmod {13} \} \\
\end{eqnarray*}
The value can be deduced using the Chinese remainder theorem. and the optimal value of $k_1+k_2+k_3$ is ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Eliminate $t$ from the equations: $x = \frac{1}{t} - t \, , \, y = \frac{1}{t^2} - 1$ $$x = \frac{1}{t} - t \tag{1}$$
$$y = \frac{1}{t^2} - 1 \tag{2}$$
$$\frac{1}{t^2} = y + 1$$
$$t^2 = \frac{1}{y + 1}$$
$$t = \pm\sqrt{\frac{1}{y + 1}}$$
$$x = \frac{1 - t^2}{t}$$
$$y = \frac{1 - t^2}{t^2}$$
$$\frac{x}{y} = \frac{1 - t^2}{t} \times \frac{t}{1 - t^2}$$
$$\frac{x}{y} = t \tag{3}$$
Substitute $(3)$ into $(1)$
$$x = \frac{y}{x} - \frac{x}{y}$$
$$x = \frac{y^2 - x^2}{xy}$$
$$x^2y = y^2 - x^2$$
$$x^2y + x^2 = y^2$$
$$x^2(y + 1) = y^2$$
$$x^2 = \frac{y^2}{y + 1}\tag{4}$$
$$x = \sqrt{\frac{y^2}{y + 1}}$$
From $(4):$
$$x^2y + x^2 = y^2$$
$$y^2 - x^2y = x^2$$
$$y(y-x^2) = x^2$$
I have no idea how to progress from that point. I've tried different methods in paper to make $y$ the subject of the formula, but none of them were fruitful.
How do I make $y$ the subject of the formula in $(4)$?
| from your second equation we get
$$t^2=\frac{1}{1+y}$$ for $$y> -1$$ and taking the square root we have
$$t_{1,2}=\pm\frac{1}{\sqrt{1+y}}$$ thus we get
$$x=\pm\sqrt{1+y}-\pm\frac{1}{\sqrt{1+y}}$$
from the first equation we get
$$0=t^2+tx-1$$ solving this we get
$$t_{1,2}-\frac{1}{2}x\pm\sqrt{\frac{x^2}{4}-1}$$
from here you will get
$$y=\frac{1}{\left(-\frac{1}{2}x+\sqrt{\frac{x^2}{4}+1}\right)^2}-1$$
or
$$y=\frac{1}{\left(-\frac{1}{2}x-\sqrt{\frac{x^2}{4}+1}\right)^2}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$ Question:
Prove $ \ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
My attempt:
Base case is trivial.
Suppose $ \ n \ge 2$ and $ \ 2^{n} + 3^{n} < 4^{n}$
Then,
$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n} $, by I.H.
I am stuck here. how do I show that this expression is $ < 4^{n+1}$?
| multiplying $$2^n+3^n<4^n$$ by $$4^n$$ we get
$$2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$ and we have
$$2^n\cdot 2<2^{3n}$$ since $$1<2^{2n-1}$$ for $n\geq 2$ and $$3^n\cdot 3<3^n\cdot 2^{2n}$$ since $$3<2^{2n}$$ therefore we have
$$2^{n+1}+3^{n+1}<2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find minimum value of sum of areas of squares
I tried using some trigonometry but couldn't arrive at any helpful inequality
| From trivial applications of similitude we get $y=\dfrac{15}{7}-\dfrac{3 x}{4}$
The sum of the area of the two squares is
$f(x)=x^2+\left(\dfrac{15}{7}-\dfrac{3 x}{4}\right)^2$
Derivative is
$f'(x)=2x+2\left(-\dfrac{3}{4}\right)\left(\dfrac{15}{7}-\dfrac{3 x}{4}\right)$
$f'(x)=0$ if $x=\dfrac{36}{35}$
This value is a minimum because $f'(x)<0$ for $x<\dfrac{36}{35}$ and $f'(x)>0$ for $x>\dfrac{36}{35}$
When $x=\dfrac{36}{35}$ we have $y=\dfrac{48}{35}$
The minimum value of the sum of the areas is $x^2+y^2=\left(\dfrac{36}{35}\right)^2+\left(\dfrac{48}{35}\right)^2=\dfrac{144}{49}$
$x+y=193$
Hope this helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
In the list of questions proposed in the "Meeting for Training for the Brazilian Olympiad", 2013. No answer provided. Could solve some problems in that list but got stuck in this one. My developments are going into very complicated expressions, and are most likely wrong. Hints or solutions are welcomed. Sorry if this is a duplicate.
| Let $I = (a+b)^7-a^7-b^7$.
$$a^7+b^7 = (a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)$$
and
$$(a+b)^6 = a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6$$
Then
\begin{eqnarray*}
I &=& (a+b)(7a^5b+14a^4b^2+21a^3b^3+14a^2b^4+7ab^5)\\
&=&7ab(a+b)(a^4+2a^3b+3a^2b^2+2ab^3+b^4)\\
& =& 7ab(a+b)(a^2+ab+b^2)^2
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
If $ax+by=7$, .
If
$$ax+by=7$$
$$ax^2+by^2=49$$
$$ax^3+by^3=133$$
$$ax^4+by^4=406$$
then find the value of
$$2014(x+y-xy) - 100(a+b)$$
My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$
| Hint $ $ Use $\ ax^{n+1}\!+by^{n+1} = (x\!+\!y)(ax^n\!+by^n) - xy(ax^{n-1}\!+by^{n-1})$ to solve for $\,xy,x\!+\!y$
Remark $ $ The recurrence is $\,(S-x)(S-y)f_n = (S^2 - (x+y)S+xy)f_n = 0\,$ where $\,Sf_n = f_{n+1}.\,$ It has solutions $\,f_n = x^n,\,y^n\,$ so also $\,f_n = ax^n + bx^n$ by linearity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Determining derivatives of trigonometric functions For what function $f$ and number $a$ is the limit
$$\lim_{x \to \pi/4} \frac {\tan x - 1}{4x-\pi}$$
the value of $f’(a)$?
All I’m asking is how I would begin solving this problem.
|
HINT, for the limit use:
$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}\tag1$$
So, we get:
$$\lim_{x\to\frac{\pi}{4}}\space\frac{\tan\left(x\right)-1}{4x-\pi}=\lim_{x\to\frac{\pi}{4}}\space\frac{\frac{\sin\left(x\right)}{\cos\left(x\right)}-1}{4x-\pi}=\lim_{x\to\frac{\pi}{4}}\space\frac{\cos\left(x\right)-\sin\left(x\right)}{\cos\left(x\right)\cdot\left(\pi-4x\right)}=$$
$$\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{1}{\cos\left(x\right)}\right\}\cdot\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{\cos\left(x\right)-\sin\left(x\right)}{\pi-4x}\right\}=$$
$$\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{1}{\cos\left(x\right)}\right\}\cdot\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{\frac{\text{d}}{\text{d}x}\left(\cos\left(x\right)-\sin\left(x\right)\right)}{\frac{\text{d}}{\text{d}x}\left(\pi-4x\right)}\right\}=$$
$$\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{1}{\cos\left(x\right)}\right\}\cdot\left\{\lim_{x\to\frac{\pi}{4}}\space\frac{\cos\left(x\right)+\sin\left(x\right)}{4}\right\}\tag2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2444772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is the limit $\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$ zero? I have this limit:$$\lim_{(x,y) \to \infty} \frac{x+2y}{x^2 - 2xy + 2y^2}$$
At first sight seems that limit equals 0. But WolframAlpha says that there is no limit. I tried to prove it. I considered cases $y = kx$, and so on. I never got to find subsequence, whose has limit $\neq 0$.
I think there is a problem in denominator. When $x\rightarrow \infty$ and $y \rightarrow \infty$ we got there $\infty - \infty + \infty$. It's unclear what to do with it and how to find necessary subsequence.
Maybe i'm on the wrong way to solve it. Please, give me a tip.
| It seems to me like the limit is zero. Write the function as
$$
\frac{(x-y)+y}{(x-y)^2+y^2}+\frac{2y}{(x-y)^2+y^2}
$$
The second fraction goes to zero since
$$
\left|\frac{2y}{(x-y)^2+y^2}\right|=\frac{|2y|}{(x-y)^2+y^2}\le \frac{2|y|}{y^2}=\frac2{|y|}\stackrel{y\to\infty}\to 0
$$
For the first fraction, use the Cauchy-Schwarz inequality:
$$
|(x-y)\cdot1+y\cdot1|\le \sqrt{(x-y)^2+y^2}\cdot \sqrt{1^2+1^2}
$$
Then
$$
\left|\frac{(x-y)+y}{(x-y)^2+y^2}\right|\le \sqrt{2}\cdot \frac{\sqrt{(x-y)^2+y^2}}{(x-y)^2+y^2}=\sqrt{2}\cdot\frac1{\sqrt{(x-y)^2+y^2}}\le \frac{\sqrt 2}{\sqrt{y^2}} \stackrel{y\to\infty}\to 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Given this matrix $A$, find $A^{144}$ I know this is a very common question to ask when one is making their way into Linear Algebra (i.e. given a matrix, find the result of that matrix to the nth-power).
I'm given this matrix: $$
A=
\left[ {\begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end{array} } \right]$$
I'm asked to compute, by hand, the matrix $A^{144}$
What I tried:
I calculated $A^2$, $A^3$, $A^4$ and $A^5$ and tried to find a pattern, so that I could first find the more general $A^n$ expression, and then make $n=144$. This is what I got:
$$
A^2=
\left[ {\begin{array}{cc}
-1 & 0 \\
0 & -1 \\
\end{array} } \right]$$
$$
A^3=
\left[ {\begin{array}{cc}
0 & 1 \\
-1 & 0 \\
\end{array} } \right]$$
$$
A^4=
\left[ {\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array} } \right]$$
$$
A^5=
\left[ {\begin{array}{cc}
0 & -1 \\
1 & 0 \\
\end{array} } \right]$$
So there seems to be kind of a "circular pattern", where $A^6=A^2$ and therefore, since $144/6=24$, I'm suspecting that: $$¿\,\boxed{A^{144}=A^2} \,?$$
However, I was hoping to find the more general $A^n$ matrix first and confirm the above result.
But, I can't seem to find a function such that:
$$f(n) =
\left\{
\begin{array}{ll}
0 & \mbox{if n is odd} \\
1 & \mbox{if n is even}
\end{array}
\right.
$$
This function would allow me to sort out the (1,1) and (2,2) elements of $A^n$ being 0 when n is odd (and the (1,2) and (2,1) being 0 when n is even).
I'm a bit confused at this point, your help is greatly appreciated.
| Note that $A^4 = I$, and $A^{144} = (A^4)^{36}$, and you're done. If you really want to go with that $A^6 = A^2$ instead, then we have that
$$A^{144} = (A^6)^{24} \\
= (A^2)^{24} = A^{48} \\
= (A^6)^8 = (A^2)^8 \\
= A^{16}= A^4\times A^{12}\\
= A^4 \times(A^6)^2 = A^4\times(A^2)^2\\
= A^8 = A^2\times A^6\\
= A^2\times A^2 = A^4$$
This seems a much more roundabout way to get the same result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
$f(x)=?$ if we have $f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$ $f(x)=?$
If we have $$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$$ to fractions are very similar. I don't have an idea to find $f(x)$. Can someone show me a clue ?
| Hint:
$$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}=\frac{1}{\frac{x^2-x+1}{x}}=\frac{1}{\frac{x^2+x+1-2x}{x}}=\frac{1}{\frac{x^2+x+1}{x}-2}$$
Note, that we have to assume $x\neq 0$ in this process. You will have to check if that is problematic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2450683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $\alpha,a,b$ are integers and $b\neq-1$, then prove that, if $\alpha$ satisfies the equation $x^2+ax+b+1=0$, $a^2+b^2$ must be composite.
Let $\alpha,a,b$ be integers such that $b\neq-1$. Assume that $\alpha$ satisfies the equation $x^2+ax+b+1=0$. Prove that the integer $a^2+b^2$ must be composite.
$\alpha=\frac{-a\pm\sqrt{a^2-4(b+1)}}{2}$is an integer. But manipulating this expression is leading me to nowhere. Please help.
| If $u=\alpha$ and $v$ are the roots
$$
\begin{align}
x^2+ax+b+1
&=(x-u)(x-v)\\
&=x^2-(u+v)x+uv
\end{align}
$$
Then $a=-(u+v)$ and $b=uv-1$. Since $u=\alpha\in\mathbb{Z}$, and $a\in\mathbb{Z}$, we know that $v\in\mathbb{Z}$. Therefore,
$$
\begin{align}
a^2+b^2
&=(u+v)^2+(uv-1)^2\\
&=u^2+v^2+u^2v^2+1\\
&=\left(u^2+1\right)\left(v^2+1\right)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How would I solve the following question about matrices? $A= \begin{pmatrix}-1&3\\ -2&6\end{pmatrix}$
Find two $2\times 2$ matrices $B$ and $C$ such that $AB=AC$ but $B\ne C$.
I have tried to do some row operations along with multiplication but I keep getting the wrong answer.
Any help?
| $B=\left(
\begin{array}{ll}
3 & 0 \\
2 & 1 \\
\end{array}
\right);\;C=\left(
\begin{array}{ll}
-6 & 9 \\
-1 & 4 \\
\end{array}
\right)$
$AB=AC=\left(
\begin{array}{ll}
3 & 3 \\
6 & 6 \\
\end{array}
\right)$
I did in this way
Called $B=\left(
\begin{array}{ll}
a & b \\
c & d \\
\end{array}
\right);\;C=\left(
\begin{array}{ll}
e & f \\
g & h \\
\end{array}
\right)$
I computed $AB$ and $AC$ and wrote an undeterminate system such that they were equal
I got
$g= -\dfrac{a}{3}+c+\dfrac{e}{3},\;h= -\dfrac{b}{3}+d+\dfrac{f}{3}$
And all the other unknowns arbitrary values.
Hope this can be useful
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Question regarding the proof of $\sum_{k=0}^{n} (-1)^k \binom{n}{k}^{2}$ Let $n$ be a positive integer. Prove that
$$\sum_{k=0}^{n} (-1)^k \binom{n}{k}^{2} = \begin{cases}
0 & \text{if $n$ is odd} \\
(-1)^m \binom{2m}{m}, & \text{if $n$ = 2m}
\end{cases}$$
So I wrote out a first few terms to get a feel of the problem
$\binom{n}{k} = \frac{n!}{k!(n-k)!}$
When n is odd (eg n =3): $$\binom{3}{0}^2 -\binom{3}{1}^2 +\binom{3}{2}^2 -\binom{3}{3}^2 = 0$$
When n is even (eg n = 4): $$\binom{4}{0}^2 - \binom{4}{1}^2 + \binom{4}{2}^2 - \binom{4}{3}^2 + \binom{4}{4}^2 = \binom{4}{2}^2 = \binom{2(2)}{2}^2 (-1)^2$$
So there is a hint that the textbook gave me, it is
For $n = 2m$, consider the coefficient of $x^n$ in $(1-x^2)^n = (1 +
x)^n(1-x)^n$
So I tried writing out what $(1+x)^n$ and $(1-x)^n$ are
$$(1+x)^n = \sum_{k=0}^{n} \binom {n}{k} x^k$$
$$(1-x)^n = \sum_{k=0}^{n} \binom {n}{k} (-1)^k x^k$$
And this is where I've been lost for the last few hours....
| If you have polynomials $f(x)=\sum_k a_kx^k$ and $g(x)=\sum_k b_kx^k$
then the $x^n$-coefficient of $f(x)g(x)$ is
$$c_n=a_0b_n+a_1b_{n-1}+\cdots+a_nb_0=\sum_{k=0}^n a_{k}b_{n-k}.$$
Here $a_k=\binom nk$ and $b_k=(-1)^k\binom nk$, and here
$$c_n=\sum_{k=0}^n (-1)^{n-k}\binom nk\binom{n}{n-k}.$$
Can you relate this to the sum you're studying?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2456925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove the inequality $a\frac{a+b}{a+c} + b\frac{b+c}{a+b} + c\frac{c+a}{b+c} \geq a+b+c$ for positive $a, b, c$ I faced problem proving this inequality for positive $a$, $b$, $c$:
$a\frac{a+b}{a+c} + b\frac{b+c}{a+b} + c\frac{c+a}{b+c} \geq a+b+c$
I tried to simplify it and I got that:
$bc^3 + a^3 c + a b^3 \geq a b^2 c + a^2 bc + a b c^2$
Then I tried to prove it using MMI, but it gave me nothing.
Please just give me a hint, maybe I am missing something important.
| An AM-GM proof
By AM-GM inequality $$ab^3+abc^2\geq2\sqrt{a^2b^4c^2}=2ab^2c\tag{1}$$ Similarly we get $$bc^3+bca^2\overset{\text{AM-GM}}\geq2abc^2\tag{2}$$ and
$$ca^3+cab^2\overset{\text{AM-GM}}\geq2a^2bc\tag{3}$$
Hence adding $(1),(2),(3)$ we get the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Magic squares in combinatorics Let $P_{3}(r)$ be the number of 3 x 3 magic squares that are symmetric to their main diagonal. Prove that $P_{3}(r) \leq (r+1)^3$.
$r$ in this problem seems to be the sum of each row and column
This is the first time I've dealt with magic squares, didn't even know of their existence before today. But from what I gathered off the internet they're square matrices containing non-negative integers where all row sums and column sums are equal to each other.
This question is from the chapter on permutations, strings over finite alphabets, and problems of choice. But I don't know how to prove this. Any ideas?
| Suppose that $a$ and $b$ have been selected:
$$\begin{pmatrix}
a & ? & ?\\
? & b & ? \\
? & ? & ?\\
\end{pmatrix}$$
We can immediately deduce the following value:
$$\begin{pmatrix}
a & ? & ?\\
? & b & ? \\
? & ? & (r-a-b)\\
\end{pmatrix}$$
Using symmetry we obtain these two values:
$$\begin{pmatrix}
a & ? & \frac{r-b}{2}\\
? & b & ? \\
\frac{r-b}{2} & ? & r-a-b\\
\end{pmatrix}$$
Using the sum of the first and last row we deduce:
$$\begin{pmatrix}
a & \frac{r-b-2a}{2} & \frac{r-b}{2}\\
? & b & ? \\
\frac{r-b}{2} & \frac{2a+b-r}{2} & r-a-b\\
\end{pmatrix}$$
Using the sum of the first and last column we deduce:
$$\begin{pmatrix}
a & \frac{r+b-2a}{2} & \frac{r-b}{2}\\
\frac{r+b-2a}{2} & b & \frac{3b+2a-r}{2} \\
\frac{r-b}{2} & \frac{2a+3b-r}{2} & r-a-b\\
\end{pmatrix}$$
It is not hard to see that the sum of the middle column is $3b$ so in fact $b=\frac{r}{3}$
Now notice $a=\frac{2a+3b-r}{2}$, so $P_3(r)=0$ for all $r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2458059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $x_{n} - 3x_{n-1} = -8$ with $n\geq 1$ and $x_0 = 2$ I tried two methods which gave different answers:
Method 1:
$$x_{n} - 3x_{n-1} = -8 \\ x_n = 3(3x_{n-2} - 8) - 8 \\ = 3^2 x_{n-2} -8 ( 1+3) \\ = 3^3 x_{n-3} - 8(1+3+3^2) \\ = 3^n x_{0} - 8(1+3+3^2 + \ldots + 3^{n-1}) \\ = 2\times 3^n - 8\left(\frac{3^n - 1}{3-1}\right)\\ = 2\times 3^n - 4(3^n - 1) \\ = -2\times 3^n +4.$$
Method 2:
Solving the homogenous equation using $x_n = r^n$,
$$r^n - 3r^{n-1} = 0 \\ \implies r = 3$$
So the homogenous solution is $h_n = a\times 3^n$ for some $a\in\mathbb{R}$.
With the initial condition, $h_n = 2\times 3^n$.
Also, by guessing the particular solution with $x_n = C$,
$$C - 3C = -8 \implies C = 4$$.
So, the final solution is
$$x_n = 2\times 3^n + 4.$$
I'm unsure why the two methods differ. It looks obvious that the first one gives the correct result.
| Alternatively, you can always double check the result with generating functions (like here and here for more examples):
$$f(t)=\sum\limits_{n=0}x_n\cdot t^{n}=2 + \sum\limits_{n=1}x_n\cdot t^{n}= 2 + \sum\limits_{n=1}(3x_{n-1}-8)\cdot t^{n}=\\
2 + 3\sum\limits_{n=1}x_{n-1}\cdot t^{n}-8\sum\limits_{n=1} t^{n}=2+3t\sum\limits_{n=1}x_{n-1}t^{n-1}-\frac{8}{1-t}+8$$
or
$$f(t)=10+3tf(t)-\frac{8}{1-t} \Rightarrow f(t)=\frac{10}{1-3t}-\frac{8}{(1-t)(1-3t)}=\\
\frac{10}{1-3t}+\frac{4}{1-t} -\frac{12}{1-3t}=\frac{4}{1-t}-\frac{2}{1-3t}$$
or
$$f(x)=4\sum\limits_{n=0}t^n-2\sum\limits_{n=0}(3t)^n=\sum\limits_{n=0}\left(\color{red}{4-2\cdot3^n}\right)t^n$$
Thus $$x_n=4-2\cdot3^n$$
Some of the shortcuts are explained here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Minimum of $\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} $. $\forall n \in \mathbf{N}$, prove that the function $f(x)=\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} $ attains its minimum at $x=\frac{1}{2}$.
Now it suffices to prove that
\begin{equation*}
\sum_{l=0}^{n} \binom{n}{l}^2 (x+y)^{2l} (x-y)^{2(n-l)} = \sum_{l=0}^{n} \binom{2l}{l} \binom{2(n-l)}{n-l} x^{2l}y^{2(n-l)}
\end{equation*}
| \begin{align*}
n &> 0 \text{.} \\
f'(1/2) &= \left. \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l)x^{2l-1}(1-x)^{2(n-l)} - 2(n-l)x^{2l}(1-x)^{2(n-l)-1} \right) \right|_{x = 1/2} \\
&= \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l)(1/2)^{2l-1}(1/2)^{2(n-l)} - 2(n-l)(1/2)^{2l}(1/2)^{2(n-l)-1} \right) \\
&= (1/2)^{2n-1}\sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} \left( (2l) - 2(n-l) \right) \\
&= 0 \text{.} \\
f(x) &= \sum_{l=0}^{n} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} \\
&= \frac{(1-x)^{2n}}{(n!)^2} + \frac{x^{2n}}{(n!)^2} + \sum_{l=1}^{n-1} \frac{1}{(l!)^2((n-l)!)^2} x^{2l} (1-x)^{2(n-l)} \text{.} \\
f''(x) &= \frac{2n(2n-1)\left(x^{2(n-1)} + (1-x)^{2(n-1)}\right)}{(n!)^2} + \\
{}&+ \sum_{l=1}^{n-1} \frac{1}{(l!)^2((n-l)!)^2} x^{2(l-1)}(1-x)^{2(n-l-1)} \left( 2n(2n-1)x^2 - 4l(2n-1)x + 2l(2l-1)\right) \\
&> 0 + \sum_{l=1}^{n-1} \frac{1}{(l!)^2((n-l)!)^2} x^{2(l-1)}(1-x)^{2(n-l-1)} \frac{2l(l-n)}{n} \\
&\geq 0 \text{.} \blacksquare
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2462830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluating limit using taylor expansions How can I evaluate this limit? Trying with Taylor doesn't seem to give me the right result, why ?
$$\lim_{x\to 0} \frac{\sqrt{1+4x} -1 -\sin(2x)}{\log(1+x^2)}$$
with Taylor I can approximate $\sin(2x)$ to $ 2x $
and $\log(1+x^2)$ to $ x^2 $. If I plug in these in the limits it does not give me the right limit, what am I doing wrong?
| binomial theorem:
$(a+b)^n) = a^n + na^{n-1}b + \frac {n(n-1)}{2} a^{n-1}b^2 \cdots$
works for fractional exponents
$(1+4x)^\frac 12 = 1 + \frac 12 (4x) - \frac 18 (4x)^2+ \cdots%$
$\sin x = x - \frac 16 x^2+\cdots\\
\sin 2x = (2x) - \frac16 (2x)^3+ \cdots$
numerator: $1 + 2x - 2x^2 - 1 - 2x + \frac 86 x^3 = -2x^2- \frac 16 x^3 + \cdots$
denominator:
$\ln (1 + x) = x - \frac 12 x^2 + \frac 13 x^3\\
\ln(1+x^2) = x^2 - \frac 12 x^4\cdots$
$-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Summation of $1\cdot 3\cdot 5\cdot 7 + 3\cdot 5\cdot 7\cdot 9 ...$
Find the sum of:
$1 \cdot 3\cdot 5\cdot 7 + 3\cdot 5\cdot 7\cdot 9+...$ till $n$ terms.
My attempt:
I got the $i^{th}$ term to be $(2i-1)(2i+1)(2i+3)(2i+5)$
Expansion gives: $16i^4 +64i^3+56i^2+-16i-15$
Required: $$\sum\limits_{i=1}^n (16i^4 +64i^3+56i^2+-16i-15) $$
Using summation identities, I got:
$\dfrac{16n(n+1)(2n+1)(3n^2+3n-1)}{30}+\dfrac{64n^2(n+1)^2}{4}+\dfrac{56(n)(n+1)(2n+1)}{6}- \dfrac{16n(n+1)}{2}- 15n$
However, answer given is simply $$\frac{1}{10}\{(2n-1)(2n+1)(2n+3)(2n+5)(2n+7)+1\cdot 3\cdot 5\cdot 7\}$$
| Let
$$
f(k) = (2k-1) \color{blue}{(2k+1)(2k+3)(2k+5)(2k+7)}
.
$$
Then
$$
f(k+1) = \color{blue}{(2k+1)(2k+3)(2k+5)(2k+7)}(2k+9)
,
$$
so
$$
f(k+1)-f(k) = 10 \, \color{blue}{(2k+1)(2k+3)(2k+5)(2k+7)}
,
$$
and your sum can be rewritten as the telescoping sum
$$
\frac{1}{10}\sum_{k=0}^{n-1} \Bigl( f(k+1)-f(k) \Bigr)
= \frac{1}{10} \Bigl( f(n)-f(0) \Bigr)
= \frac{f(n)- (-1)\cdot 1 \cdot 3 \cdot 5 \cdot 7}{10}
.
$$
(Which agrees with your answer, by the way, if you expand everything out and compare.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Arithmetic of complex numbers
If $z = \cos x + i\sin x$ , show that
$$\frac{2}{1+z} =1-i\tan\left(\frac{x}{2}\right),$$
I get up to:
$$\frac{1+\cos x -i\sin x}{1+\cos x}.$$
| Showing that the two numbers are equal can be done by showing that the magnitude and the argument for both numbers are the same.
Magnitude:
$$1+z = 1+\cos x + i\sin x$$
$$|1+z| = \sqrt{1 + 2 \cos x + \cos^2 x + \sin^2 x} = \sqrt{2+2\cos x}$$
$$| \frac{2}{1+z}| = \frac{\sqrt{2}}{\sqrt{1 + \cos x}} $$
$$|1-it| = \sqrt{1+t^2} = \sqrt{1 + \tan^2(x/2)} = \sec (x/2)$$
And the two are the same by trigonometric identity
Argument:
$$ \arg(1+z) = \tan^{-1} \frac{\sin x}{1+ \cos x} = \tan^{-1} \tan (x/2) = x/2$$
$$ \arg(\frac{2}{1+z}) = - x/2$$
$$ \arg(1-it) = \tan^{-1} (-\tan (x/2) = -x/2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving Recurrence Relation of $T(n)=4T(n-2)+2$ Question
Solve Recurrence Relation of $T(n)=4T(n-2)+2$
Base case-: $T(1)=1,T(2)=2$
My Approach/solution
$$T(n)=4T(n-2)+2$$
$$T(n-2)=4T(n-4)+2 \tag{1}$$
$$T(n-4)=4T(n-6)+2 \tag{2}$$
Using $(1)$ and $(2)$ in my equation
$$\begin{align*}
T(n)&=4\cdot (4T(n-4)+2)+2\\
&=4^{2}\cdot T(n-2\cdot 2)+2\cdot 4^{1}+2\cdot 4^{0}\\
&=4^{2}\cdot(4T(n-6)+2)+2\cdot 4^{1}+2\cdot 4^{0}\\
&=4^{3}\cdot T(n-2\cdot 3)+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}\\
\vdots \\
&=4^{k}\cdot T(n-2\cdot k)+2\cdot 4^{k-1}+...+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}
\end{align*}$$
Substituing $T(n-2\cdot k)$ by $2$, i.e $T(2)=2$
$$n-2\cdot k=2 \Rightarrow k=\frac{n-2}{2}$$
So our equation will look like
$$\begin{align*}
T(n)&=2\cdot 4^{k}+2\cdot 4^{k-1}+...+2\cdot 4^{2}+2\cdot 4^{1}+2\cdot 4^{0}\\
T(n)&=2\cdot \left(4^{0}+4^{1}+4^{2}+...+4^{k-1}+4^{k}\right)\\
T(n)&=2\cdot \left(4^{0}\cdot \frac{(4^{k+1}-1)}{4-1}\right)
\end{align*}$$
$k=\frac{n-2}{2}$
$$\begin{align*}
T(n)&=2\cdot \left(\frac{(4^{k+1}-1)}{4-1}\right)\\
T(n)&=2\cdot \frac{2^{n}-1}{3}
\end{align*}$$
Is it correct? Also if it is correct, can anyone hint me another approach as it is bit lengthy.
Thanks!
| i think the right solution is given by $$T(n)=\frac{1}{12} \left(9\ 2^n+(-1)^{n+1} 2^n-8\right)$$
at first solve the equation $$T(n)=4T(n-2)$$ with $$T(n)=q^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
minimize $x^4 - 6x^2 y^2 + y^4$ given $x^2 + y^2 \leq 1$ I have a constrained optimization problem. Can we maximize / minimize this function on the unit sphere?
$$ f(x,y,z) = x^4 - 6 x^2 y^2 + y^4 \quad\text{given that}\quad x^2 + y^2 + z^2 = 1$$
One idea could be to use the Cauchy-Schwartz inequality. Since I forget the proof:
$$ (x^2 + y^2)^2 \geq 0 \text{ so that }x^4 + y^4 \geq 2 x^2 y^2 \text{ and }f(x,y,z) \geq - 4 x^2 y^2 \geq - 4 $$
I could try other rearrangements as well. This one gives me an upper bound of $3$.
$$ x^4 - 6x^2 y^2 + y^4 \leq x^4 + 6x^2 y^2 + y^4
= (x^2 + y^2)^2 + 4x^2 y^2 \leq 3\, \big( x^2 + y^2 \big)^2 \leq
3\, \big( x^2 + y^2 + z^2 \big)^2 = 3$$
If I use some real analysis we know that the sphere as a subset of Euclidean space is compact, so that:
$$ -\infty < -4 \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) < 3 < +\infty$$
I'm trying to avoid Lagrange multipliers unless the're really natural here. Observer also that:
$$ \left[ x^2 + y^2 + z^2 = 1 \right] \to \left[ x^2 + y^2 \leq 1\right] $$
as the original problem was defined on the unit sphere but the $z$ is extraneous.
They might not be extraneous we could set spherical coordinates:
$$ (x,y,z ) = \big(\cos \theta \, \cos \varphi, \;\cos \theta \sin \varphi, \;\cos \varphi\big)$$
and we could put into our inequality:
\begin{eqnarray*} x^4 - 6x^2 y^2 + y^4 &=& \cos^4 \theta \cos^4 \varphi - 6 \cos^4 \theta \sin^2 \varphi \cos^2 \varphi+ \cos^4 \theta \sin^4 \varphi \\ \\
&=& \cos^4 \theta \,\big( \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi \big) \\ \\
&\leq & \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
\end{eqnarray*}
This looks promising as I have reduced a three-dimensional problem to a problem with only an angle $\varphi$, but I may have lost something with the final "$\leq$" sign.
Just a tiny bit more:
$$
\cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
= (\cos^2 \varphi - \sin^2 \varphi)^2 - 4 \cos^2 \varphi \sin^2 \varphi
= \cos^2 2\varphi - \sin^2 2\varphi
$$
and if we use the double-angle identity.
$$ 1 \geq \cos^2 2\varphi - \sin^2 2\varphi
= \cos^2 2\varphi - (1 -\cos^2 2\varphi)
= 2\, \cos^2 2\varphi - 1 \geq - 1$$
This is very similar to what I obtained before.
| For $x=y=\frac{1}{\sqrt2}$ we get a value $-1$.
We'll prove that it's a minimal value.
We need to prove that
$$x^4-6x^2y^2+y^4\geq-1.$$
Indeed,
$$x^4-6x^2y^2+y^4+1\geq x^4-6x^2y^2+y^4+(x^2+y^2)^2=2(x^2-y^2)^2\geq0$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2468432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
If $a+b+c=0$ prove that $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5} $$
I've tried squaring, cubing, etc. the $a+b+c=0$, but I've just dug myself in.
Is there a more elegant way to prove this, or a way to find the right "trick"?
| $$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = -2(ab + bc + ca)$$
$$a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc = 3abc$$
$$\begin{align*} a^5 + b^5 + c^5 & = (a + b + c)(a^4 + b^4 + c^4) - (ab + bc + ca)(a^3 + b^2 + c^3) + abc(a^2 + b^2 + c^2) = \\ & = -3abc(ab + bc + ca) - 2abc(ab + bc + ca) = -5abc(ab + bc + ca) \end{align*}$$
So both sides equal $-abc(ab + bc + ca)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How many bit strings of length $n$ contain exactly $k$ blocks of "$10$"? How many bit strings of length $n$ contain exactly $k$ blocks of "$10$"?
My attempt: Let $F(n, k)$ be the number of bit strings of length $n$ that contain exactly $k$ blocks of $10.$ Note that for $k \neq0, $ $F(0, k) = F(1, k)= 0.$
Consider a bit string of length $n$ where $ n \geq 2.$
Every bit string either begins with:
1) $10$,
or
2) $01$, $00$ or $11.$
In the former case, we get $F(n-2, k-1)$ since the remaining $n-2$ bits must contain $k-1$ blocks of $10$ in exactly $F(n-2, k-1)$ ways.
In the latter case, in either of the $3$ situations we get $F(n-2, k)$ since the remaining $n-2$ bits must contains $k$ blocks of $10$ in $F(n-2, k)$ ways.
Therefore for $n \geq 2, k \geq 1$ we have:
$F(n, k) = F(n-2, k-1)+ 3F(n-2, k)$.
For $k \geq 1$ let $A_k(x) = \displaystyle \sum_{n \geq2}F(n,k)x^n$ (and let $A_0(x)=1$) then multiplying the above recurrence by $x^n$ and summing over all $n \geq 2$ we get:
$A_k(x) = x^2 A_{k-1}(x) +3x^2A_{k}(x) $
$\Rightarrow A_k(x) = \dfrac{x^2}{1-3x^2} A_{k-1}(x)$
$\Rightarrow A_{k}(x) = \dfrac{x^{2k}}{(1-3x^2)^k} $
$\Rightarrow F(n, k) = [x^n] A_k(x)$
$\Rightarrow F(n, k) = [x^n] \dfrac{x^{2k}}{(1-3x^2)^k}$
$\Rightarrow F(n, k) = [x^{n-2k}] \displaystyle \sum_{r \geq 0} \binom{k+r-1}{r} 3^r x^{2r}$
which clearly gives a wrong answer for odd $n$. Could somebody point out where in the above calculation have I gone wrong? Thanks in advance.
| Thanks for the insight. I used a similar technique for a similar problem. I wrote this up for my assignment which is due in a week. Although I believe that I found the recurrence relation correctly, I am unable to go about extracting the coefficients from a quadratic raised to a negative k-th power. Could you please be able to help me how to proceed?
Let $f(n, k)$ represent the number of strings made of $\{a, b\}$ where $k$ is the number of occurrences of ''$ab$'' in a string of length $n$. Note that the string can begin with $ab, aa, bb$, or $ba$. If it begins with $ab$, then there are exactly $f(n-2, k-1)$ number of strings remaining. If it begins with $bb$, then there are exactly $f(n-2, k)$ strings remaining. If it begins with $aa$, or $ba$, then there are exactly $f(n-1, k)$ strings remaining. Thus we have the recurrence relation $f(n, k) = f(n-2, k-1) + f(n-2, k) + 2f(n-1, k)$. Trivially enough, we have that $f(0, k) = f(1, k) = 0$ for any non-zero $k$. Define $A_{k}(x) = \displaystyle \sum_{n=2}^{\infty} f(n, k)x^{n}$. Note that $A_{0}(x) = \displaystyle \sum_{n\geq 2} f(n, 0)x^{n} = \displaystyle \sum_{n \geq 2} (n+1)x^{n}.$
Thus $A_{k}(x) = \sum_{n=2}^{\infty} \Big[ f(n-2, k-1)x^{n} + f(n-2, k)x^{n} + 2f(n-1, k)x^{n}\Big].$
It follows that $A_{k}(x) = \displaystyle \sum_{n=2}^{\infty} x^{2} f(n, k-1)x^{n} + x^{2} f(n, k)x^{n} + 2x f(n, k)x^{n}$
Thus, \begin{align*}
A_{k}(x) = x^{2} A_{k-1}(x) + x^{2} A_{k}(x) + 2x A_{k}(x)\\
\implies A_{k}(x) = \displaystyle \frac{x^{2}}{1+2x-x^{2}}A_{k-1}(x)\\
\implies A_{k}(x) = \displaystyle \frac{x^{2}}{1+2x-x^{2}} \cdot \Big( \displaystyle \frac{x^{2}}{1+2x-x^{2}} A_{k-2}(x) \Big) \\
\therefore A_{k}(x) = \displaystyle \Bigg(\frac{x^{2}}{1+2x-x^{2}}\Bigg)^{k} \displaystyle (n+1) \sum_{n \geq 2} x^{n}\\
\therefore A_{k}(x) = (n+1) \displaystyle \frac{x^{2k+2}}{(1+2x-x^{2})^{k} (1-x)}\\
\end{align*}
But $f(n, k) = [x^{k}] A_{k}(x)$
\begin{align*}
\therefore f(n, k) = [x^{n-2-2k}] \displaystyle (n+1) \frac{1}{(1+2x-x^{2})^{k} (1-x)}\\
\therefore f(n, k) = [x^{n-2k-2}] (n+1) \displaystyle (1+2x-x^{2})^{-k} (1-x)^{-1}\\
\end{align*}
This is the part where I'm stuck at. Please help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2471895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $ab+bc+ca+abc=4$ then $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq3$, via AM-GM
Suppose, for positive reals $a$, $b$, $c$, that
$$ab+bc+ca+abc=4$$
Prove that
$$\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq3$$
I applied AM-GM on the first equality ie, $a$, $b$, $c$, and $abc$ to get $$ab+bc+ca \geq 3\qquad\text{and}\qquad abc \leq1$$
The exact equation is as follows
$$1=\frac{ab+bc+ca+abc}{4}\ge\sqrt[4]{(abc)^3}\implies 1\ge abc$$
However, I didn't manage to get any further than this after applying AM-GM to several other inequalities.
I'd like a solution for this that utilizes AM-GM only, as I'm very new to inequalities.
| Let $a=\frac{2x}{y+z}$ and $b=\frac{2y}{x+z}$, where $x$, $y$ and $z$ be positives.
Thus, the condition gives
$$\frac{4xy}{(x+z)(y+z)}+2c\left(\frac{x}{y+z}+\frac{y}{x+z}\right)+\frac{4xyc}{(x+z)(y+z)}=4$$ or
$$\frac{2c(x^2+y^2+xz+yz+2xy)}{(x+z)(y+z)}=4-\frac{4xy}{(x+z)(y+z)}$$ or
$$\frac{2c(x+y)(x+y+z)}{(x+z)(y+z)}=\frac{4z(x+y+z)}{(x+z)(y+z)}$$ or
$$c=\frac{2z}{x+y}$$ and we need to prove that
$$2\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq3, $$
which is AM-GM:
$$2\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq\sum_{cyc}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=$$
$$=\sum_{cyc}\left(\frac{x}{x+z}+\frac{z}{z+x}\right)=3. $$
Done!
There are proofs by trigonometry and $uvw$ but they are not easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Exercises EG.3 and EG.4 in Probability with martingales, by David Williams
Let $G$ be the free group with two generators $a$, $b$. Start at time 0 with unit element 1, the empty word. At each second multiple the current word on the right by one of the four elements $a$, $a^{-1}$, $b$, $b^{-1}$ choosing each with probability 1/4 independently of previous choices. For instance the choices
$a,a,b,a^{-1},a,b^{-1},a^{-1},a,b$ yield the reduced word of length 3, $aab$ at time 9.
EG.3 Prove that the probability that the reduced word $1$ ever occurs at a positive time is $1/3$.
EG.4 Now suppose elements $a$, $a^{-1}$, $b$, $b^{-1}$ are chosen with probability $\alpha$, $\alpha$, $\beta$, $\beta$ respectively, where $2\alpha+2\beta=1$. Prove that the conditional probability that the reduced word $1$ ever occurs at a positive time given that $a$ is chosen at time $1$ is the unique root in $(0,1)$ of the polynomial $3x^3 + (3-4\alpha^{-1})x^2+x+1$.
I know little about Markov Chains, and about limiting distributions but the chain is not of the form that a limiting distribution exists, so I am not able to make use of that facts. Chain I have in mind is of length of reduced word, states are $\{0,1,2,\dots,\}$. Now $p_{k\to k+1}=3/4,\ p_{k\to k-1}=1/4$ for $k\geq1$ and for $k=0,\ p_{0\to1}=1$.
| I have only been able to solve this through guidance of Did.
$\def\P{\mathbb{P}}$
$\def\Ex{\mathbb{E}}$
$\def\Z{\text{\ensuremath{\mathbb{Z}}}}$
$\def\I{\textrm{I}}$
$\def\as{\textrm{a.s.}}$
EG.3 Use the Markov Chain theory developed in Markov chains, by Norris,
to calculate the hitting time to reach sate $0$ from state of length
$1$. Specifically, we model the length of reduced word as Markov Chain with length as it states.
With $p:=\P\left(\left\{ \text{hitting }0\text{ given length is }1\right\} \right)$
we have
\begin{align*}
p= & \frac{1}{4}+\frac{3}{4}\P\left(\left\{ \text{hitting }0\text{ given length is }2\right\} \right)\\
= & \frac{1}{4}+\frac{3}{4}\P\left(\left\{ \text{hitting }0\text{ given length is }1,\text{hitting 1 given length is 2}\right\} \right)\\
\overset{\left(1\right)}{=} & \frac{1}{4}+\frac{3}{4}\P\left(\left\{ \text{hitting }0\text{ given length is }1\right\} \right)\P\left(\left\{ \text{hitting }1\text{given length is }2\right\} \right)\\
\overset{\left(2\right)}{=} & \frac{1}{4}+\frac{3}{4}p^{2}.
\end{align*}
where in $\left(1\right):$ Strong Markov Property; Markov chains, by Norris
is used and in $\left(2\right):$ we exploit symmetry of setup that
\begin{align*}
\P\left(\left\{ \text{hitting }0\text{ given length is 1}\right\} \right)=\P\left(\left\{ \text{hitting }1\text{ given length is }2\right\} \right).
\end{align*}
Solving $p=\frac{1}{4}+\frac{3}{4}p^{2}$ we get $p=\frac{1}{3},1$
where the latter is discarded by observing the Markov Chain (higher
probability to visit states down the chain).
EG.4
Similar to previous exercise let $p_{a}$ be the probability of interest;
i.e.
\begin{align*}
p_{a}=\P\left(\left\{ \text{hitting $0$ given state is }a\right\} \right).
\end{align*}
Likewise generalize the definitions above and the argument of Strong
Markov Property above to get
\begin{align*}
p_{aa}= & \P\left(\left\{ \text{hitting $0$ given state is }aa\right\} \right)=p_{a}^{2}\\
p_{ab}= & p_{a}p_{b}\\
p_{ab^{-1}}= & p_{a}p_{b}^{-1}=p_{a}p_{b}
\end{align*}
where the last equality is true by the symmetry of setup. Now note
that hitting time
\begin{align*}
p_{a}= & \alpha+\alpha p_{aa}+2\beta p_{ab}\\
= & \alpha+p_{a}^{2}\alpha+2\beta p_{a}p_{b}.
\end{align*}
Similarly, we find $p_{b}=\beta+\beta p_{b}^{2}+2\alpha p_{a}p_{b}$
but from equation above we have
\begin{align*}
p_{b}=\frac{p_{a}-\alpha\left(1+p_{a}^{2}\right)}{2\beta p_{a}}.
\end{align*}
Substiting this into $p_{b}=\beta+\beta p_{b}^{2}+2\alpha p_{a}p_{b}$
we get after simplification
\begin{equation}
p_{a}^{2}\left(2a^{2}-4b^{2}+1\right)+3a^{2}p_{a}^{4}-a^{2}-4ap_{a}^{3}=0.\label{eq:EG4}
\end{equation}
Now setting $p_{a}=1$ above we get
\begin{align*}
\left(2a^{2}-4b^{2}+1\right)+3a^{2}-a^{2}-4a=4\left(a^{2}-b^{2}-a\right)+1=4\left(\frac{1}{2}\left(a-b\right)-a\right)+1=0.
\end{align*}
Thus dividing original polynomial by $\alpha^{2}\left(p_{a}-1\right)$ (i.e. rejecting
this solution choice) we get $3p_{a}^{3}+\left(3-4\alpha^{-1}\right)p_{a}^{2}+p_{a}+1=0$.
This indeed has a unique solution in $\left(0,1\right)$ as can be
checked easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Write the parametric equation of the revolution surface generated by the line when it rotates around the axis $Oz$. Write the parametric equation of the revolution surface generated by the line whose equation is $2x - y + z = 1, x + y - 3z = 2$ when it rotates around the axis $Oz$.
If we let $z=t$ and we add the two equations together we get that $$x = 1 + \frac{2}{3}t \quad y = 1+\frac{7}{3}t$$
Then we multiply the vector by the rotation matix
$$\left(\begin{array}{ccc}
\cos\theta & -\sin\theta & 0 \\
\sin\theta & \cos\theta & 0 \\
0 & 0 & 1
\end{array}\right)
\left(\begin{array}{c}
1 + \frac{2}{3}t \\
1 + \frac{7}{3}t \\
t
\end{array}\right) =
\left(\begin{array}{c}
\cos\theta\left(1+\frac{2t}{3}\right)-\sin\theta\left(1+\frac{7t}{3}\right) \\
\cos\theta\left(1+\frac{7t}{3}\right)+\sin\theta\left(1+\frac{2t}{3}\right) \\ t
\end{array}\right)
$$
I am unsure how to carry on from here.
| The line has equation given by the system
$
\left\{
\begin{array}{l}
2x - y + z = 1\\
x + y - 3z = 2\\
\end{array}
\right.
$
The planes have normal vectors $(2,-1,1)$ and $(1,1,-3)$
The intersecting line is parallel to their cross product
$(2,-1,1)\times(1,1,-3)=(2,7,3)$
and passes through a point common to the two planes which can be found adding the two equations $3x-2z=3\to x=\frac{3+2z}{3}$ so for $z=0$ we have $x=1$ and plugging these values in one of the two planes get $y=2-x+3z=1$
So the line passes through $1,1,0$ and is parallel to $(2,7,3)$
Its parametric equation is $(x,y,z)=(1,1,0)+t(2,7,3)$
$
\left\{
\begin{array}{l}
x=1+2t\\
y=1+7t\\
z=3t\\
\end{array}
\right.
$
Any point of the line rotates and generates a circle whose parametric equation is
$
\left\{
\begin{array}{l}
x=\sqrt{(1+2t)^2+(1+7t)^2} \cos\theta\\
y=\sqrt{(1+2t)^2+(1+7t)^2} \sin\theta\\
\end{array}
\right.
$
at a $z$ value of $z=3t$
so the parametric equation of the surface of revolution is
$
\left\{
\begin{array}{l}
x=\sqrt{2 + 18 t + 53 t^2} \cos\theta\\
y=\sqrt{2 + 18 t + 53 t^2} \sin\theta\\
z=3t\\
\end{array}
\right.
$
for $t\in\mathbb{R};\;\theta\in[0,2\pi)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2473761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $n\in\mathbb N$ so that: $\sqrt{1+5^n+6^n+11^n}\in\mathbb N$ Find $n\in\mathbb N$ so that: $\sqrt{1+5^n+6^n+11^n}\in\mathbb N$
My attempt led me to have $\quad n=2k+1:\quad k\in\mathbb N$
The expression under square root is odd, so the square root's value is also odd.
I assumed $\sqrt{1+5^n+6^n+11^n}=2a+1:\quad a\in\mathbb N \Rightarrow$
$5^n+6^n+11^n=(2a+1)^2-1 \Rightarrow 5^n+6^n+11^n=4a(a+1)$
Then the expression on the left must be divisible by 8
$5^n\equiv 5\pmod 8$ if $n$ is odd, $5^n\equiv 1\pmod 8$ if $n$ is even.
$6^n\equiv 0\pmod 8$ if $n\ge3$.
$11^n\equiv 3\pmod 8$ if $n$ is odd, $11^n\equiv 1\pmod 8$ if $n$ is even.
Then $n$ must be odd and $\ge3$ for the expression to be divisible by 8.
I'm stuck here, would anyone give a hand please?
| It is never an integer. If $n\in\mathbb N$, then$$1+5^n+6^n+11^n=1+5^n+(5+1)^n+(10+1)^n\equiv3\pmod5,$$but every perfect square is congruent to $0$, $1$, or $4\pmod5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How could we express the space in an other way? I want to calculate a triple integral over the space $$D=\{(x,y,z)\mid |x|\leq 1, |y|\leq 1, z\geq 0, x^2+y^2+z^2\leq 1\}$$
$$$$
We have the following:
*
*$|x|\leq 1\Rightarrow -1\leq x\leq 1$
*$|y|\leq 1\Rightarrow -1\leq y\leq 1$
*$x^2+y^2+z^2\leq 1 \Rightarrow z^2\leq 1-x^2-y^2\Rightarrow -\sqrt{1-x^2-y^2}\leq z\leq \sqrt{1-x^2-y^2}$, since $z\geq 0$ we get $0 \leq z\leq \sqrt{1-x^2-y^2}$.
So that the square root is defined it must hold that $1-x^2-y^2 \geq 0 \Rightarrow x^2+y^2\leq 1 \Rightarrow x^2\leq 1-y^2 \Rightarrow -\sqrt{1-y^2}\leq x\leq \sqrt{1-y^2}$, right?
So, how could we express $D$ in an other way?
$$D=\{(x,y,z)\mid -1\leq x\leq 1, -1\leq y\leq 1, 0\leq z\leq \sqrt{1-x^2-y^2}\}$$ or $$D=\{(x,y,z)\mid -\sqrt{1-y^2}\leq x\leq \sqrt{1-y^2}, -1\leq y\leq 1, 0\leq z\leq \sqrt{1-x^2-y^2}\}$$ ?
| Note that $x^2+y^2+z^2\leq 1$ implies $|x|\leq 1, |y|\leq 1$:
$$|x|^2=x^2\leq x^2+y^2+z^2\leq 1 \implies |x|\leq 1.$$
Hence, $D$ is the upper-half unit ball centered at the origin
$$D=\{(x,y,z)\in\mathbb{R}^3\;:\; z\geq 0, x^2+y^2+z^2\leq 1\}.$$
By using the cartesian coordinates, we have (see your last line)
$$\iiint_D f(x,y,z)\, dxdydz =\int_{y=-1}^1\int_{x=-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\int_{z=0}^{\sqrt{1-x^2-y^2}} f(x,y,z)\, dxdydz.$$
By using the cylindrical coordinates, we get
$$\iiint_D f(x,y,z)\, dxdydz =\int_{z=0}^1\int_{\theta=0}^{2\pi}\int_{\rho=0}^{\sqrt{1-z^2}} f(\rho\cos\theta,\rho\sin\theta,z) \rho d\rho d\theta dz.$$
By using the spherical coordinates, we get
$$\iiint_D f(x,y,z)\, dxdydz \\=\int_{\rho=0}^1\int_{\theta=0}^{2\pi}\int_{\varphi=0}^{\pi/2} f(\rho\sin \varphi\cos\theta,\rho\sin \varphi\sin\theta,\rho\cos \varphi) \rho^2\sin(\varphi)d\rho d\varphi d\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is the polynomial $x^4+2x^3 y+4x^2 y^2+2 x y^3+y^4$ Schur positive? I tried to check that whether the symmetric polynomial $p=x^4+2x^3 y+4x^2 y^2+2 x y^3+y^4$ is Schur positive.
We have Schur polynomials $s_0 = 1$, $s_1 = x+y$, $s_2=x^2 + xy + y^2$, $s_3 = x^3 + x^2y + xy^2 + y^3$, $s_4 = x^4 + x^3y + x^2y^2 + xy^3 + y^4$.
Therefore
$$
p=s_4 + xys_2 + 2x^2y^2.
$$
Here the coefficients involve $x, y$. To express $p$ in terms of a linear combination of Schur polynomials, do we allow coefficients involve $x,y$? Is $p$ Schur positive? Thank you very much.
| The Jacobi-Trudi identity gives
$$
s_{(3,1)}=\left|\begin{array}\mbox{s}_3&s_4\\s_0&s_1\end{array}\right|=x^3y+x^2y^2+xy^3
$$
and
$$
s_{(2,2)}=\left|\begin{array}\mbox{s}_2&s_3\\s_1&s_2\end{array}\right|=x^2y^2
$$
So $p= s_4+s_{(3,1)}+2s_{(2,2)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2476956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
On the convergence of $\sum_{n=1}^{+\infty}\frac{1}{n}\,\cos\left(\frac{\pi n}{2}\right)$ Does the following series absolutely converge, conditionally converge or diverge?
$$\sum_{n=1}^{+\infty}\frac{1}{n}\,\cos\left(\frac{\pi n}{2}\right)$$
My answer:
$$ 0<\frac{1}{\sqrt{n}}<\left|\frac{\cos(\pi n/2)}{n}\right| $$
and $\sum_{n\geq 1}\frac{1}{\sqrt{n}}$ diverges by p-series test so by comparison test, the original series must also diverge.
| This series is conditionally convergent. Its terms are:
$$\frac{1}{n}\cos\left(\frac{n\pi}{2}\right) =
\left\{
\begin{array}{llll}
\frac{1}{n} & \mbox{if } n=4k , k>0\\
0 & \mbox{if } n=4k+1 , k\ge0\\
-\frac{1}{n} & \mbox{if } n=4k+2 , k\ge0\\
0& \mbox{if } n=4k+3, k\ge0\\
\end{array}
\right. $$
By removing zero expression, and just considering $n=4k$ and $n=4k+2$, we could replace and change indices. By write some sentences for this series:
$$\sum_{n=1}^{\infty}\frac{1}{n}\cos\left(\frac{n\pi}{2}\right)=0-\frac{1}{2}+0+\frac{1}{4}+0-\frac{1}{6}+0+\frac{1}{8}+0+...=\\-\frac{1}{2}\Big(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...\Big)=-\frac{\ln{2}}{2}\approx-0.3466$$
It's conditionally convergences because
$$\sum \left|\frac{\cos\left(\frac{n\pi}{2}\right)}{n}\right|=\frac{1}{2}\Big(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\Big)$$
doesn't converge.
References:
https://en.wikipedia.org/wiki/Alternating_series_test
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2478177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Compute the limit $\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n$ NOTE: L'Hopital's and Taylor series not allowed!
Taking the log and exponenting the entire thing I get $$\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n=\lim_{n\rightarrow\infty}e^{n\ln\left(\frac{1+\sqrt[n]{a}}{2}\right)}=\lim_{n\rightarrow\infty}e^{n\ln\left(\frac{1+\sqrt[n]{a}}{2}-1+1\right)\cdot\frac{\frac{1+\sqrt[n]{a}}{2}-1}{\frac{1+\sqrt[n]{a}}{2}-1}}.$$
Letting $k=\frac{1+\sqrt[n]{a}}{2}-1$ we see that the RHS can be simplified to
$$\lim_{n\rightarrow\infty}e^{nk\frac{\ln\left(k+1\right)}{k}}.$$
It follows that $k\rightarrow0$ as $n\rightarrow\infty,$ and $\ln(k+1)/k$ then tends to $1$ (standard limit). So we can write
$$\lim_{n\rightarrow\infty}(e^{k})^n=\lim_{n\rightarrow\infty}\left(\frac{e^{k}-1+1}{k}\cdot k\right)^n=\lim_{n\rightarrow\infty}\left( \frac{e^k-1}{k}\cdot k+1\right)^n=(1 \cdot0+1)^n=1.$$
The answer should be $\sqrt{a}.$ Why is my method wrong?
| $$\lim_{n\rightarrow\infty}\left(\frac{1+\sqrt[n]{a}}{2}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{\sqrt[n]{a}-1}{2}\right)^{\frac{2}{\sqrt[n]{a}-1}\cdot\frac{1}{2}\cdot\frac{\sqrt[n]a-1}{\frac{1}{n}}}=e^{\frac{1}{2}\ln{a}}=\sqrt{a}$$
I used the following.
We know that $\frac{\sqrt[n]{a}-1}{2}\rightarrow0$ for $n\rightarrow+\infty$.
Thus, if $\frac{\sqrt[n]{a}-1}{2}=y$ then $(1+y)^{\frac{1}{y}}\rightarrow e$.
Also, we know that $\lim\limits_{x\rightarrow0}\frac{e^x-1}{x}=1$.
Thus, $$\lim_{n\rightarrow+\infty}\frac{\sqrt[n]a-1}{\frac{1}{n}}=\lim_{x\rightarrow0}\frac{a^x-1}{x}=\ln{a}\lim_{x\rightarrow0}\frac{e^{x\ln{a}}-1}{x\ln{a}}=\ln{a}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 0
} |
Probability Problem: A box contains 100 balls A box contains 100 balls, numbered from 1 to 100. If two balls are selected at random and at the same time from the box, what is the probability that the numbers on the balls will be:
a) consecutives
b) 2 multiples of 6
c) odd and even
d) 2 divisors of 60
a) $P=\left(\dfrac{1}{100} \times \dfrac{1}{99}\right)100=\dfrac{1}{99}$
b) $P=\left(\dfrac{16}{100} \times \dfrac{15}{99}\right)= \dfrac{4}{165}$
c) $P=\left(\dfrac{50}{100} \times \dfrac{50}{99}\right)2=\dfrac{50}{99}$
d) $P=\left(\dfrac{10}{100} \times \dfrac{9}{99}\right)=\dfrac{1}{110}$
Is correct my answer?
|
A box contains $100$ balls, numbered from $1$ to $100$. If two balls are selected at random, what is the probability that the numbers on the balls will be consecutive?
There are $\binom{100}{2}$ ways to select two of the $100$ balls.
A pair of consecutive numbers is determined by the smaller of the numbers. There are $99$ pairs of consecutive numbers since the smaller number can be at most $99$. Hence, the desired probability is
$$\frac{99}{\dbinom{100}{2}}$$
What is the probability that two multiples of $6$ are selected?
There are $16$ multiples of six less than or equal to $100$. Hence, the number of favorable cases is $\binom{16}{2}$. Therefore, the desired probability is
$$\frac{\dbinom{16}{2}}{\dbinom{100}{2}}$$
Your answer is correct.
What is the probability that an even number and an odd number are selected?
We must select one of the $50$ even numbers and one of the $50$ odd numbers, so the desired probability is
$$\frac{\dbinom{50}{1}\dbinom{50}{1}}{\dbinom{100}{2}}$$
Your answer for this question is also correct.
What is the probability that $2$ divisors of $60$ are selected?
As @Mark pointed out in the comments, there are $12$ divisors of $60$. They are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$. Hence, the desired probability is
$$\frac{\dbinom{12}{2}}{\dbinom{100}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2483512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1}{xy}\ge4$ given that $x+y=1$ and conclude that $(1+\frac1{x^2})(1+\frac1{y^2})\ge2$
Let $x,y\in\mathbb R^+$ and $x+y=1$
1- Prove that $\frac{1}{xy}\ge4$
2- Conclude that $(1+\frac1{x^2})(1+\frac1{y^2})\ge25$
I have tried to start from $x+y=1$ or $x\ge0\land y\ge0$ and reach $\frac{1}{xy}\ge4$ but with no result.
Update:
I've proved the first part of the question (Thanks to Jack's comment).
Since the second question says "Conclude" that means I have to use the first proof. I tried to square the first proof and got it close to the second question, but
again no result. There is probably a trick that I don't know.
| given $X + Y = 1$, proof that $\frac{1}{XY} \ge 4$, where $X$ and $Y$ are real positive integers.
Since $X$ and $Y$ are positive integers then, $X \lt 1$ and $Y \lt 1$ for they to sum to $1$.
$X$ and $Y$ can be both equal
So that $X + Y = 1$
this then becomes $X$ or $Y$ $= \frac{1}{2} = 0.5$
But if $X$ and $Y$ are distinct,
then
$
\begin{align}
X \gt 0.5 & and & Y \lt 0.5 & or\\
X \lt 0.5 & and & Y \gt 0.5
\end{align}
$
Just depending of which is greater and lesser, for convenience let's assume $Y \gt X$.
So that a number $" a "$ must lie between both $X$ and $Y$
Such that....
$X = 0.5 - a$ and $Y = 0.5 + a$
where
$$
0 \le a \lt \frac{1}{2}
$$
because $X$ and $Y$ are positive integers. So that the product of $X$ and $Y$ is more pronounced as
$(0.5-a)×(0.5+a)$
which gives
$(0.5^2 - a^2)$
Therefore the inverse of there product
$\frac{1}{XY}$ becomes $\frac{1}{0.5^2 - a^2}$
Since $a \lt \frac{1}{2}$, $a$ runs from $\frac{1}{2}$ to approach $0$
$\frac{1}{\frac{1}{4} - a^2}$
its maximum value is when $a$ approach $0$. $\frac{1}{0.25 - a^2}$
$\frac{1}{XY} \ge \frac{1}{0.25}$ $\frac{1}{XY} \ge 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solve system $\cos^2x+\cos^2y+\cos^2z=1,$ $\cos x+\cos y+\cos z=1,$ $x+y+z=\pi$ I want to solve the system:
$$\cos^2(x)+\cos^2(y)+\cos^2(z)=1,$$
$$\cos(x)+\cos(y)+\cos(z)=1,$$
$$x+y+z=\pi.$$
I tried to prove that only one of cosines can be not a zero, but I just prove that one or three cosines can not be zero.
I get, that $$\cos(x)~\cos(y)+\cos(x)~\cos(z)+\cos(y)~\cos(z)=0$$
Also after substituting
$$\cos(x) = 1 - \cos(y)-\cos(z),$$
we get $$\cos^2(y) + \cos^2(z)=\cos(y)+\cos(z) - \cos(y)\cos(z)$$
| The hint.
Since $z=\pi-x-y$, we obtain
$$\cos{x}+\cos{y}-\cos(x+y)=1$$ or
$$2\cos\frac{x+y}{2}\cos\frac{x-y}{2}=2\cos^2\frac{x+y}{2}$$ or
$$\cos\frac{x+y}{2}\left(\cos\frac{x-y}{2}-\cos\frac{x+y}{2}\right)=0$$ or
$$\sin\frac{z}{2}\sin\frac{x}{2}\sin\frac{y}{2}=0.$$
I think the rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Algebraic proof of a simple inequality Prove $\frac{a+b+c}{2} ≥ \frac{bc}{b+c} + \frac{ca}{c+a} + \frac{ab}{a+b}$
I feel like this problem would have a much simpler solution than just expaning out and simplifying. Any pointers are appreciated
| By C-S
$$\sum_{cyc}\frac{ab}{a+b}=\sum_{cyc}\frac{1}{\frac{1}{a}+\frac{1}{b}}\leq\sum_{cyc}\frac{1}{\frac{(1+1)^2}{a+b}}=\sum_{cyc}\frac{a+b}{4}=\frac{a+b+c}{2}.$$
Another way:
It's $$\sum_{cyc}\frac{ab}{a+b}\leq\frac{a+b+c}{2}$$ or
$$\sum_{cyc}\left(\frac{a+b}{4}-\frac{ab}{a+b}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2}{a+b}\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof of an inequality by induction Prove using induction that
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{n^2} \le 2-\frac{1}{n}$$
for all positive whole numbers $n$.
I began by showing that it is true for $n=1$
I then assumed that it is true for $n=p$
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} = \sum_{k=1}^p \frac{1}{k^2} \le 2-\frac{1}{p}$$
I now want to show that it is true for $n=p+1$
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} + \frac{1}{(p+1)^2}= \sum_{k=1}^{p+1} \frac{1}{k^2} $$
If I add $\frac{1}{(p+1)^2}$ to $\sum_{k=1}^{p} \frac{1}{k^2}$, I will then get
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} + \frac{1}{(p+1)^2} \le 2-\frac{1}{p} + \frac{1}{(p+1)^2}$$
If this is true then
$$2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}$$
or
$$-\frac{1}{(p+1)}=-\frac{1}{p} + \frac{1}{(p+1)^2}$$
$$0= \frac{1}{(p+1)} -\frac{1}{p} + \frac{1}{(p+1)^2}$$
$$0= \frac{p(p+1)}{p(p+1)^2} -\frac{(p+1)^2}{p(p+1)^2} + \frac{p}{p(p+1)^2}$$
$$0= \frac{p(p+1)}{p(p+1)^2} -\frac{(p+1)^2}{p(p+1)^2} + \frac{p}{p(p+1)^2}$$
$$0= \frac{p(p+1)-(p+1)^2 + p}{p(p+1)^2}$$
$$0= \frac{(p^2 + p) - (p^2 + 2p + 1) + p}{p(p+1)^2}$$
$$0= \frac{-1}{p(p+1)^2}$$
This is invalid. I am not sure where I have made a mistake but I think it is
$$2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}$$
It then must be that
$$-\frac{1}{(p+1)} < -\frac{1}{p} + \frac{1}{(p+1)^2}$$
$$0 \le \frac{1}{(p+1)} - \frac{1}{p} + \frac{1}{(p+1)^2}$$
$$ 0< \frac{-1}{p(p+1)^2}$$
which is true for all positive whole numbers $p$.
I am pretty sure it is proved now but I would be happy if someone can confirm this.
| It can be much simpler.
You have equality for $n = 1$.
Adding a new $n+1$ term adds $\frac{1}{(n+1)^2}$ to LHS and $\frac{1}{n(n+1)}$ to the RHS. Guess which one is smaller for all $n>1$?!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can I quickly determine the eigenvalues of this matrix? I am working on observability and detectability in controls and I ran across this example that I didn't understand. The author deliberately sought the form of this matrix, because of its "block-form" in order to quickly find the eigenvalues
\begin{bmatrix}
l_{11} & -1 & -1 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 \\
0 & -1 & -1 & 0 & 0 \\
0 & -.1 & -2 & l_{42} & -.1 \\
0 & 1 & 2 & 0 & -.2
\end{bmatrix}
The author was then able to state the eigenvalues were $\{l_{11}, -1, l_{42}, -.2\}$
I was under the impression that I could only determine the eigenvalues via a matrix diagonal if the matrix was upper/lower triangular?
| Note that the characteristic polynomial is easy to evaluate by using the Laplace expansion of a determinant (along the blue column or row):
\begin{align}&\det\begin{bmatrix}
\color{Blue}{l_{11}-x} & -1 & -1 & 0 & 0 \\
\color{Blue}{0} & -1-x & 0 & 0 & 0 \\
\color{Blue}{0} & -1 & -1-x & 0 & 0 \\
\color{Blue}{0} & -.1 & -2 & l_{42}-x & -.1 \\
\color{Blue}{0} & 1 & 2 & 0 & -.2-x
\end{bmatrix}\\&=(l_{11}-x)\det\begin{bmatrix}
\color{Blue}{-1-x} & \color{Blue}{0} & \color{Blue}{0} & \color{Blue}{0} \\
-1 & -1-x & 0 & 0 \\
-.1 & -2 & l_{42}-x & -.1 \\
1 & 2 & 0 & -.2-x
\end{bmatrix}
\\&=(l_{11}-x)(-1-x)\det\begin{bmatrix}
\color{Blue}{-1-x} & \color{Blue}{0} & \color{Blue}{0} \\
-2 & l_{42}-x & -.1 \\
2 & 0 & -.2-x
\end{bmatrix}
\\&=(l_{11}-x)(-1-x)^2\det\begin{bmatrix}
\color{Blue}{l_{42}-x} & -.1 \\
\color{Blue}{0} & -.2-x
\end{bmatrix}\\&=(l_{11}-x)(-1-x)^2(l_{42}-x)(-.2-x).
\end{align}
Therefore the eigenvalues of the given matrix, which are the roots of the characteristic polynomial, are $$\{l_{11}, -1, l_{42}, -.2\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2488837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Ordinary generating function of $n\cdot 2^{n-1}$ demonstration How I could demonstrate that the ordinary generating function of $n\cdot 2^{n-1}$ is
$$\frac{x}{(1-2x)^2}?$$
| From $a_n=n2^{n-1}$ and the way generating functions are defined, we have
$$f(x)=\sum\limits_{n=0} a_n\cdot x^n=\sum\limits_{n=0}n2^{n-1}\cdot x^n=\sum\limits_{n=1}n2^{n-1}\cdot x^n=x\left(\sum\limits_{n=1}n2^{n-1}\cdot x^{n-1}\right)=\\
x\left(\sum\limits_{n=1}n(2x)^{n-1}\right)=\frac{x}{2}\left(\sum\limits_{n=1}2n(2x)^{n-1}\right)=\frac{x}{2}\left(\sum\limits_{n=1}(2x)^{n}\right)'=\frac{x}{2}\left(\frac{1}{1-2x}-1\right)'=\\
\frac{x}{2}\left(\frac{2x}{1-2x}\right)'=\frac{x}{2}\frac{2}{(1-2x)^2}=\frac{x}{(1-2x)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find roots of $x^4 -6x^3 + 12x^2 - 12 x + 4 = 0$ the original equation is:
$$(x^2 + 2)^2 -6x(x^2+2) + 8x^2=0.$$
cannot see how to go solving this. I tried following way to factorise:
$$(x^2+2)(x^2-6x+2) + 8x^2 = 0.$$
But this has no help to solve.
Thank you people, but I need the thinking process, not the answer.
| In the original version, let $y=x^2+2$. Then, we have $y^2-6xy+8x^2=0$, which gives us $(y-2x)(y-4x)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2494962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
On the summation $\sum \limits_{n=1}^{\infty} \arctan \left ( \frac{1}{n^3+n^2+n+1} \right )$ Here is a problem that I ran into. I seriously doubt if there is a closed form but you never know.
Evaluate the series
$$\mathcal{S} = \sum_{n=1}^\infty \arctan \left ( \frac 1 {n^3+n^2+n+1} \right) $$
I searched in vain to attack it using telescopic summation but I failed miserably. Then I remembered the following technique. Since ${\rm Im} \log (1+ix) = \arctan x$ we can express the sum as follows
\begin{align*}
\sum_{n=1}^\infty \arctan \left ( \frac{1}{n^3+n^2+n+1} \right ) &= \sum_{n=1}^\infty \arctan \left [ \frac{1}{(n+1)(n^2+1)} \right ] \\
&= \sum_{n=1}^\infty \operatorname{Im} \left [ \log \left ( 1 + \frac{i}{(n+1)(n^2+1)} \right ) \right ] \\
&= \operatorname{Im} \log \left [ \prod_{n=1}^\infty \left ( 1 + \frac{i}{(n+1)(n^2+1)} \right ) \right ]
\end{align*}
I tried to combine it with the famous Euler product
$$ \frac{\sin \pi z}{\pi z} = \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right) \tag{1} $$
but I see no connection. So, is there a possible way to evaluate it?
| Something tells me this might be difficult to do by hand. Plugging in your product formula into Mathematica and then Simplifying, this is what comes out:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Expand using binomial theorem Yesterday, my maths lecturer did this:
$[(\cos x+i\sin x)+(\cos x-i\sin x)]^4$
Expanded to:
$8\cos2x+2\cos4x+6$
I've tried using binomial theorem to get to this answer but I can never seem to get there. How can I do it?
When I do it I get to:
$2\cos4x+4(\cos3x+i\sin3x)(\cos x-i\sin x)+6(\cos^22x+\sin^2(2x))+4(\cos^2x-i^2\sin^2x)$
| If we combine like terms, we get $(2\cos x)^4$. The Double-angle formula for $\cos$ is: $$\cos 2x=2\cos^2x-1$$ We start with $(2\cos x)^4=16\cos^4x$ and get $4(2\cos^2x)^2=4(\cos2x+1)^2=4\cos^22x+8\cos 2x+4,$ and the first term gives $2(2\cos^22x)=2\cos4x+2$, for a grand total of $2\cos 4x+8\cos 2x+6$.
From where you left off, it is possible to use $\cos^2 a+\sin^2 a=1$ to reduce a good portion of your binomial results to a simple value of $10$, after which you can focus on the not-yet-distributed portion of your results.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
derivation of the cubic formula In my research I came across the following derivation of the cubic formula that was almost complete. However at the end it says to make the substitution $u=z^3$ which would determine the roots. I am not quite sure how to do this and then how to determine the roots and I am looking for some help.
A general form for the cubic equation is,
$$ax^3+bx^2+cx+d=0 \tag{1}$$
To find the roots of this equation we first try to get rid of the quadratic term $x^2$. The substitution $x=y-\dfrac{b}{3a}$ helps in achieving our goal. This results in, $$ay^3+\left(c-\dfrac{b^2}{3a}\right)y+\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)=0\tag{2}$$ which we transform into the following, $$y^3+\dfrac{1}{a}\left(c-\dfrac{b^2}{3a}\right)y+\dfrac{1}{a}\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)=0\tag{3}$$ Upon assuming $e=\dfrac{1}{a}\left(c-\dfrac{b^2}{3a}\right)$ and $f=\dfrac{1}{a}\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)$ we get the equation as, $$y^3+ey+f=0\tag{4}$$ We reduce this equation by the substitution $y=z+\dfrac{s}{z}$ and choosing $s=-\dfrac{e}{3}$ we obtain the simplified equation as, $$z^6+fz^3-\dfrac{e^3}{27}=0\tag{5}$$ What only remains is to make the substitution $u=z^3$.
| The cubic formula was derived from a series of substitutions. It's probably better to memorize the process of deriving the formula, rather than memorizing the actual formula.
Starting with the general cubic, make a substitution such that the squared term is removed. Let's denote the depressed cubic as$$x^3+qx+r=0$$Now, set $x=y+z$. Expanding and factoring gives$$y^3+z^3+(3yz+q)(y+z)+r=0$$Setting $3yz+q=0$ gives $$y=-\frac q{3z}$$If we substitute that in, we get a quadratic in $y^3$. Using the quadratic formula, we get the roots of $y^3$ and $z^3$ respectively. Therefore, we have the solution as$$x_1=\color{blue}{\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\tfrac 13}+\left\{-\frac r2-\sqrt{\frac {r^2}4-\frac {q^3}{27}}\right\}^{\tfrac 13}}$$
The other solutions can be found using the cube roots of unity.$$\omega=-\frac 12+\frac {i\sqrt{3}}2\qquad\omega^2=-\frac 12-\frac {i\sqrt3}2$$So$$\begin{align*} & x_2=\color{blue}{\omega\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\omega^2\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}}\\ & x_3=\color{blue}{\omega^2\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\omega\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2504590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the exact value of $\cos\frac{2\pi}{5}$ by solving equation There's this concept in the topic of complex numbers which I don't really understand much (and was devastated upon realising it'll appear in the topic often) - trigonometry!
I'm super lost to be honest.
We are asked to express $\cos3\theta$ and $\cos2\theta$ in terms of $\cos\theta$.
Then show that $\cos3\theta=\cos2\theta$ can be written as $4z^3-2z^2-3z+1=0$ where $z=\cos\theta$.
Lastly, by solving the equation above for $z$, we are to find out the value of $\cos\frac{2\pi}{5}$.
I attempted the question with just subbing in $z=\cos\theta$ into the equation and I got: $$4\cos^3\theta-3\cos\theta=\cos2\theta$$ and don't really know where to go next.
| $$\begin{align}
%
\cos(3\theta) - \cos(2\theta) &=
%
\mathcal R \left\{e^{3i\theta}\right\} - \left\{e^{2i\theta}\right\} \\ \\ &=
%
\mathcal R \left\{\left(e^{i\theta}\right)^3 - \left(e^{i\theta}\right)^2\right\} \\ \\ &=
%
\mathcal R \left\{\left(\cos(\theta) + \rm i \sin(\theta) \right)^3 - \left(\cos(\theta) + \rm i \sin(\theta) \right)^2\right\} \\ \\ &=
%
\mathcal R \left\{\begin{array} {ll}
& \left(\cos(\theta)^3 + 3~\rm i \cos(\theta)^2\sin(\theta) -
3~\rm \cos(\theta)\sin(\theta)^2 - \rm i \sin(\theta)^2 \right)
\\ - & \left(\cos(\theta)^2 + 2~\rm i \cos(\theta) \sin(\theta) - \sin(\theta)^2\right)
\end{array}\right\}
\\ \\ &=
%
\cos(\theta)^3 - 3\cos(\theta)\sin(\theta)^2 - \cos(\theta)^2 + \sin(\theta)^2 \\ \\ &=
%
z^3 - 3z(1 - z^2) - z^2 + 1 - z^2 \\ \\ &=
%
4z^3 - 2z^2 - 3z + 1
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Use the method of generating series to solve the recurrence: $b_n=2b_{n-1}-b_{n-2}$ with $b_0=0, b_1=5$ This is my attempt:
Let
\begin{equation}
B(x):=\sum^{\infty}_{n=0}b_nx^n.
\end{equation}
Multiplying our recurrence through by $x^n$ and then summing from $2$ to $\infty$
\begin{equation}
\sum^{\infty}_{n=2}b_nx^n=2\sum^{\infty}_{n=2}b_{n-1}x^n-\sum^{\infty}_{n=2}b_{n-2}x^n.
\end{equation}
Reindexing/shifting the appropriate sums
\begin{equation}
\sum^{\infty}_{n=0}b_nx^n-b_0-b_1x=2x\sum^{\infty}_{n=0}b_nx^n-2xb_0-x^2\sum^{\infty}_{n=0}b_nx^n.
\end{equation}
Substituting $b_0$ and $b_1$ values
\begin{equation}
\sum^{\infty}_{n=0}b_nx^n-5x=2x\sum^{\infty}_{n=0}b_nx^n-x^2\sum^{\infty}_{n=0}b_nx^n.
\end{equation}
Replacing the sums with $B(x)$
\begin{equation}
B(x)-5x=2xB(x)-x^2B(x),
\end{equation}
or equivalently
\begin{equation}
B(x)=\frac{5x}{1-2x+x^2}=\frac{5x}{(x-1)^2}.
\end{equation}
Decomposing into partial fractions
\begin{equation}
\frac{5x}{(x-1)^2}=\frac{p_1}{x-1}+\frac{p_2}{(x-1)^2},
\end{equation}
or equivalently
\begin{equation}
5x=p_1(x-1)+p_2.
\end{equation}
Letting $x=1$ gives us $p_2=5$ and letting $x=0$ gives
\begin{equation}
0=-p_1+5 \Rightarrow p_1=5.
\end{equation}
So we have
\begin{equation}
B(x)=\frac{5}{x-1}+\frac{5}{(x-1)^2}.
\end{equation}
Using the generalised binomial theorem to find $b_n$
\begin{equation}
5(x-1)^{-1}=5(1-x)^{-1}=-5\sum^{\infty}_{n=0}x^n.
\end{equation}
\begin{equation}
5(x-1)^{-2}=5(1-x)^{-2}=5\sum^{\infty}_{n=0}{-2\choose{n}}(-1)^nx^n=5\sum^{\infty}_{n=0}(n+1)x^n. \tag1
\end{equation}
We arrive at (1) due to
\begin{equation}
{-2\choose{n}}=\frac{\prod^{n-1}_{i=0}(-2-i)}{n!}=\frac{(-2)...(-n)(-1-n)}{n!}=\frac{(-1)^{n}n!(n+1)}{n!}=(-1)^{n}(n+1).
\end{equation}
Thus, the solution to the recurrence is
\begin{equation}
b_n=-5+5(n+1)=5n.
\end{equation}
However, this of course isn't right since $b_1$ should be $5$ and not $-15$. So where have I gone wrong here? I feel as though it was during that ${-2}\choose{n}$ step however I can't quite see what the mistake was exactly.
Just worth it to say what the "generalised binomial theorem" is
Let $q$ be any rational number and $1+a_1x$ be a formal power series. Then we have
\begin{equation}
(1+a_1x)^q=\sum^{\infty}_{n=0}{q \choose{n}}a_1^nx^n.
\end{equation}
EDIT: Thank you!
| Following your steps, we have that
\begin{align}b_n&=[x^n]\frac{5x}{(x-1)^2}=5[x^{n-1}](1-x)^{-2}=5[x^{n-1}]\sum_{k=0}^{\infty}\binom{-2}{k}(-x)^k\\
&=5\binom{-2}{n-1}(-1)^{n-1}=5(-1)^{n-1}\frac{(-2)(-3)\cdot (-2-(n-1)+1)}{(n-1)!}\\&=\frac{5n!}{(n-1)!}=5n.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Area of a square inside a square created by connecting point-opposite midpoint
Square $ABCD$ has area $1cm^2$ and sides of $1cm$ each.
$H, F, E, G$ are the midpoints of sides $AD, DC, CB, BA$ respectively.
What will the area of the square formed in the middle be?
I know that this problem can be solved by trigonometry by using Area of triangle ($\frac{1}{2}ab\sin{c}$) but,
is there another method or visual proof?
| I hope that the following makes sense. I would have been better if you had labelled points on the figure.
The side length of the square is $1$.
The area of one of the large triangles is $A_{\text{large}} = 1\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}$.
The hypotenuse of the large triangle is $\sqrt{1^2 + \frac{1}{4}} = \sqrt{5}/2$.
One of the smaller triangles is congruent to the the larger triangle and it has hypotenuse $1/2$. The side lengths of the smaller triangle is therefore
$$
a= (\frac{1}{2} /\frac{\sqrt{5}}{2}) \cdot 1 = \frac{1}{\sqrt{5}} \\
b = (\frac{1}{2} /\frac{\sqrt{5}}{2}) \cdot \frac{1}{2} = \frac{1}{2\sqrt{5}}.
$$
So the area of a smaller triangle is
$$
A_{\text{small}} = \frac{1}{20}.
$$
Now you can probably figure out the rest. I think maybe I get the area of the smaller square to be $\frac{4}{20}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 15,
"answer_id": 11
} |
Coefficient of $x^{n-2}$ in $(x-1)(x-2)(x-3)\dotsm(x-n)$ Question
Find the coefficient of $x^{n-2}$ in the expression $$(x-1)(x-2)(x-3)\dots(x-n)~~.$$
My approach
The coefficient of $x^n$ is $1$. The coefficient of $x^{n-1}$ is $- \frac{n(n+1)}{2}$
But I cannot proceed from here.
I would appreciate any help.
| Observe that the coefficient of $x^1$ in $$(x-1)(x-2)(x-3)$$ is $$2\cdot3+1\cdot2+1\cdot3$$
or the coefficient of $x^{4-2}$ in $$(x-1)(x-2)(x-3)(x-4)=(x-1)(x-4)(x-2)(x-3)=(x^2-(1+4)x+1\cdot4)(x^2-(2+3)x+2\cdot3)$$
is $$2\cdot3+(1+4)(2+3)+1\cdot4$$
So, the required sum
$$=\sum_{1\le r_1<r_2\le n}r_1r_2=\dfrac{(\sum_{r=1}^n r)^2-\sum_{r=1}^n r^2}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2527894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Derivative of $\sin^{-1}(x)$ I can find this using the fact that $\sin(\sin^{-1}(x)) = x$, for all $x\in[-1,1].$
Now, differentiate.
$$\frac{d}{d\sin^{-1}(x)}\sin(\sin^{-1}(x))\cdot \frac{d}{dx} \sin^{-1}(x)= \frac{d}{dx} x= 1$$
$$\cos(\sin^{-1}(x))\cdot \frac{d}{dx} \sin^{-1}(x) = 1$$
$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))}$$
$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-\sin^2(\sin^{-1}(x))}}$$
$$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$$
However, what if I wanted to differentiate this like $\ \sin^{-1}(\sin(x))$ without knowing the fact that $\ \frac{d}{dx}\sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$ ? Is there a solution for it? I keep getting stuck at a certain step when I try this...
| You could do it like this:
first, define $f$ by $f(x) = \sin^{-1}x,$
where $-1 \leq x \leq 1$ and
$-\frac\pi2 \leq \sin^{-1}x \leq \frac\pi2.$
(Recall that $\sin^{-1}$ is not a real function for $|x| > 1$
and that $\sin$ is not a one-to-one function, so we have to choose
an appropriate domain and range of $\sin^{-1}$; these are the
usual choices.)
Then for $-\frac\pi2 \leq \theta \leq \frac\pi2,$
$$
f(\sin \theta) = \theta. \tag1
$$
Take the derivative of each side of $(1)$ with respect to $\theta$:
$$
\frac{d}{d\theta}\left(f(\sin\theta)\right) = 1. \tag2
$$
Evaluate the left-hand side of $(2)$ using the chain rule:
$$
\frac{d}{d\theta}\left(f(\sin\theta)\right)
= f'(\sin\theta)\frac{d}{d\theta}\left(\sin\theta\right)
= f'(\sin\theta) \cos\theta. \tag3
$$
Combine $(2)$ and $(3)$:
$$
f'(\sin\theta) \cos\theta = 1. \tag4
$$
Divide by $\cos\theta$ on both sides of $(4)$:
$$
f'(\sin\theta) = \frac{1}{\cos\theta}. \tag5
$$
Let $x = \sin\theta.$ Then $1 - x^2 = \cos^2\theta,$
and since $-\frac\pi2 \leq \theta \leq \frac\pi2,$
it follows that $\cos\theta\geq 0,$ so we have
$\cos\theta = \sqrt{1 - x^2}.$
Making these substitutions in $(5)$,
$$
f'(x) = \frac{1}{\sqrt{1 - x^2}}.
$$
But $f'(x)=\frac{d}{dx}\sin^{-1}x,$ so
$$
\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Strange functional equation: $f(z)=\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left( \frac{1}{6} -4 \ f \left(\frac{-4\pi^2}{z}\right)\right)$ I would like to get some information about the following functional equation:
$$f(z)=\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ f \left (\frac{-4\pi^2}{z}\right )\right )$$
This functional relationship must hold only for $\mathfrak{R}(z)>0$. I would like to know wether it is possible to know what types of functions hold it.
To be clear, I do not know where to start from. I have barely worked with functional equations in my life.
Any help or bibliography will be welcomed.
Edit:
Answering the comments, we are supposed to know that $f(z)$ as $z \to 0^+$ is $ O\left ( \frac{1}{z^2} \right )$. Moreover, we know that $f(-2\pi)=f(2\pi)=\frac{1}{24} - \frac{1}{8\pi}$. Thank you for the help received.
Edit 2:
$f(z)$ is even: $f(z)=f(-z)$. From here, we could get the follwing functional equation for $\mathfrak{R}(z)<0$:
$$f(z)=\frac{1}{24} + \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ f \left (\frac{4\pi^2}{z}\right )\right )$$
So that we can remove the $-$ sign from the first equation to get for $\mathfrak{R}(z)>0$:
$$f(z)=\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ f \left (\frac{4\pi^2}{z}\right )\right )$$
| That formula puts no restrictions at all on your function on the half-plane $\Re(z) > 0$.
If you know $f(z_1)$ for some $z_1$ with $\Re(z_1) > 0$, then it tells you the value of $f(z_2)$ where $z_2 = \frac{-4\pi^2}{z_1}$. But $\Re(z_2) < 0$. And if you try to use the formula again, the new value is for $\frac{-4\pi^2}{z_2} = z_1$ (where it gives the same value for $f(z_1)$ that you started with).
So, you can choose any function $g$ on $\Re(z) > 0$, then define $$f(z) = \begin{cases} g(z) & \Re(z) > 0\\\frac{1}{24} - \frac{1}{2z} + \frac{\pi^2}{z^2} \left ( \frac{1}{6} -4 \ g \left(\frac{-4\pi^2}{z}\right )\right) & \Re(z) < 0\end{cases}$$
$f$ will be defined everywhere except the imaginary line and will satisfy your equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2534373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\displaystyle 1 <\cos A+\cos B+\cos C$ Prove that $\displaystyle 1 < \cos A+\cos B+\cos C \leq \frac{3}{2}$ where $A+B+C = \pi$
Attempt: $\cos A+\cos B+\cos C= 1+4\cos A\cdot \cos B\cdot \cos C$
Now $\displaystyle \cos A\cdot \cos B\cdot \cos C=\frac{1}{2}\cos A\left[\cos (B-C)-\cos A\right]\leq \frac{1}{2}\cos A(1-\cos A)\leq \frac{1}{2}\left[\frac{\cos A+1-\cos A}{2}\right]^2 = \frac{1}{8}$
so $\displaystyle \cos A+\cos B+\cos C\leq 1+\frac{1}{2} = \frac{3}{2}$
could some help me how to solve for minimum , thanks
| use that $$\cos(\alpha)+\cos(\beta)+\cos(\gamma)=1+\frac{r}{R}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find this maximum of the $\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$ Let $x\in \mathbb{R}$, find the function maximum of the value
$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
my attemp
$$x^2=5+4\sin{t},t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$
then
$$f=\dfrac{5\sqrt{3}}{4}+2\sin{\left(t+\frac{\pi}{6}\right)}\le 2+\dfrac{5}{4}\sqrt{3}$$
My Question:this function have other methods to find this maximum? such as AM-GM,Cauchy-Schwarz inequality and so on?
| We can simplify the problem by finding the minimun of
$$ g(t) = \frac{\sqrt{3}}{4}t + \frac{\sqrt{(9-t)(t-1)}}{4} $$
with $t = x^2, \ t \in [1,9]$
A straightforward approach is just to take the first derivative
$$ g'(t) = \frac{\sqrt{3}}{4} + \frac{5-t}{4\sqrt{-9+10t-t^2}} $$
Solving $g'(t)=0$ leads to
$$ \sqrt{16 - (t-5)^2} = \frac{t-5}{\sqrt{3}} \quad (t > 5) $$
which gives $t = 5 + 2\sqrt{3}$
Thus, the maximum is
$$g(5+2\sqrt{3}) = \frac{\sqrt{3}}{4}(5+2\sqrt{3}) + \frac{1}{2} = 2 + \frac{5\sqrt{3}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Show $\frac 1 3 C_0+\left(\frac 2 3 \right)\left(\frac 1 3 \right)^2 C_1+\left(\frac 2 3\right)^2 \left(\frac 1 3 \right)^3 C_2 +\dots =\frac 1 2$ From a puzzle involving Markov chains I have the following expression, and calculating $50$ terms of the sum I strongly believe $$\frac 1 3 C_0 + \left(\frac 2 3 \right) \left(\frac 1 3 \right)^2 C_1
+ \left(\frac 2 3 \right)^2 \left(\frac 1 3 \right)^3 C_2 + \dots = \frac 1 2$$
where $C_n$ is the nth Catalan number. I know each term of the sum
$$x_{n+1} = \left(\frac 2 9 \right) \frac{2(2n+1)}{n+2} x_n$$ so the sum converges, but I don't know how to prove it converges to $1/2$.
| Use the generating function for the Catalan numbers:
$$c(x) = \sum_{n=0}^\infty C_n x^n = \frac{2}{1+\sqrt{1-4x}}.$$
So, $$\begin{align}\frac 1 3 C_0 + \left(\frac 2 3 \right) \left(\frac 1 3 \right)^2 C_1
+ \left(\frac 2 3 \right)^2 \left(\frac 1 3 \right)^3 C_2 + \dots &= \frac13 \sum_{n=0}^\infty C_n \left(\frac29\right)^n \\
&= \frac{1}{3}\cdot \frac{2}{1+\sqrt{1-\frac89}} \\
&= \frac{1}{3} \frac{2}{\frac43} \\&= \frac12.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Equations with the floor function I am trying to solve the following problem:
For what real numbers $x$ is: $\lfloor2x-3\rfloor-3\lfloor x+2\rfloor=0$?
I have no idea how to do this, please help me
| Hint:
We know:
$x,y\in [z,z+1)$ then $\lfloor x\rfloor=\lfloor y\rfloor$
$\lfloor a+b\rfloor=\lfloor a\rfloor+b$ for integer $b$
With those 2:
$\lfloor2x-3\rfloor-3\lfloor x+2\rfloor=\lfloor2x\rfloor-3-3(\lfloor x\rfloor+2)=\lfloor2x\rfloor-3\lfloor x\rfloor-9=0\\\implies\lfloor2x\rfloor-3\lfloor x\rfloor=9$
Now divide it into 2 cases, when $\lfloor x\rfloor=\lfloor x+0.5\rfloor$ and when $\lfloor x\rfloor+1=\lfloor x+0.5\rfloor$
Can you continue from here?
Explanation about the cases:
I got to the equation $\lfloor2x\rfloor-3\lfloor x\rfloor=9$, let's look only on $\lfloor2x\rfloor$, if $\lfloor x+0.5\rfloor=\lfloor x\rfloor+1$ then $\lfloor2x\rfloor=2\lfloor x\rfloor+1$ elsewhere $\lfloor2x\rfloor=2\lfloor x\rfloor$
Solution to case one:
If $\lfloor x+0.5\rfloor=\lfloor x\rfloor+1$:
$$\lfloor2x\rfloor-3\lfloor x\rfloor=9\\2\lfloor x\rfloor+1-3\lfloor x\rfloor=9\\\lfloor x\rfloor=-8\\x=-8+c,c\in[0,1)\\\text{we know that $\lfloor x+0.5\rfloor=\lfloor x\rfloor+1$ so:}\\\lfloor -8+c+0.5\rfloor=\lfloor-8+c\rfloor+1=\lfloor-8+c+1\rfloor\\c\in[0.5,1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Generalization of Dirichlet integral Is it correct that
$$\lim_{L\to\infty} \int_0^L d x \int_0^L dy \int_0^L d z \frac{\sin(x+y)}{x+y}\frac{\sin(y+z)}{y+z}\frac{\sin(z+x)}{z+x} =\frac {\pi^3}{16}. $$
or does anybody has a reference for this? The value $\frac{\pi^3}{16}$ comes from numerics.
The identity
$$ \lim_{L\to\infty}\int_0^L \frac{\sin(x)}{x} = \frac \pi 2 $$
(which is sometimes also called Dirichlet integral) is well known and there are many ways to prove it. However, I need the slight generalization of this above.
| A lengthy method, but it works. First, we rewrite:
$$\sin(x+y) \sin(y+z) \sin(z+x)=\frac{1}{4} \left(\sin(2x)+\sin(2y)+\sin(2z)-\sin(2(x+y+z)) \right)$$
Now let's change the variables (keeping the same letters for simplicity):
$$x \to \frac{1}{2}x^2, \qquad y \to \frac{1}{2}y^2, \qquad z \to \frac{1}{2}z^2$$
$$I= 2 \int_0^\infty \int_0^\infty \int_0^\infty \frac{x y z dx dy dz}{(x^2+y^2)(y^2+z^2)(z^2+x^2)} \left(\sin(x^2)+\sin(y^2)+\sin(z^2)-\sin(x^2+y^2+z^2) \right)$$
Now we introduce spherical coordinates:
$$x= r \sin \theta \cos \phi, \qquad y = r \sin \theta \sin \phi, \qquad z= r \cos \theta$$
The limits are obvious:
$$I= 2 \int_0^\infty \int_0^{\pi/2} \int_0^{\pi/2} \frac{r^5 \sin^3 \theta \cos \theta \sin \phi \cos \phi dr d \theta d \phi}{r^6 \sin^2 \theta (\sin^2 \theta \sin^2 \phi+\cos^2 \theta)(\sin^2 \theta \cos^2 \phi+\cos^2 \theta)} \times \\ \times \left(\sin(r^2 \sin^2 \theta \cos^2 \phi)+\sin(r^2 \sin^2 \theta \sin^2 \phi)+\sin(r^2 \cos^2 \theta)-\sin(r^2) \right)$$
Simplifying and using:
$$\int_0^\infty \frac{\sin(a^2 r^2) dr}{r}=\frac{\pi}{4}$$
We obtain:
$$I= \pi \int_0^{\pi/2} \int_0^{\pi/2} \frac{\sin \theta \cos \theta \sin \phi \cos \phi d \theta d \phi}{(\sin^2 \theta \sin^2 \phi+\cos^2 \theta)(\sin^2 \theta \cos^2 \phi+\cos^2 \theta)}=\pi \int_0^{\pi/2} \int_0^{\pi/2} \frac{\sin 2 \theta \sin 2 \phi d \theta d \phi}{4 \cos^2 \theta+\sin^2 2 \phi \sin^4 \theta}$$
Changing the variables:
$$2 \theta \to \theta, \qquad 2 \phi \to \phi$$
$$I= \pi \int_0^{\pi} \int_0^{\pi} \frac{\sin \theta \sin \phi d \theta d \phi}{8 (1+\cos \theta)+\sin^2 \phi (1-\cos \theta)^2}$$
Substituting:
$$u= \cos \theta, \qquad v = \cos \phi$$
$$I= \pi \int_{-1}^1 \int_{-1}^1 \frac{d u d v}{8 (1+u)+(1-v^2) (1-u)^2}$$
Substituting:
$$1-u= t$$
$$I= \pi \int_0^2 \int_{-1}^1 \frac{d v d t}{(t-4)^2-t^2 v^2}=\pi \int_0^2 \int_{-1}^1 \frac{d v d t}{(t-4-tv)(t-4+tv)}$$
Taking the elementary integral w.r.t. $v$, we have:
$$I= \pi \int_0^2 \frac{dt}{t(t-4)} \ln \left(1-\frac{t}{2} \right)$$
Substituting:
$$t= 2s$$
$$I= \frac{\pi}{2} \int_0^1 \frac{ \ln (1-s) ds}{s(s-2)}=\frac{\pi}{4} \left( \int_0^1 \frac{ \ln (1-s) ds}{s-2} -\int_0^1 \frac{ \ln (1-s) ds}{s}\right)$$
$$I=\frac{\pi}{4} \left( \frac{\pi^2}{12}+\frac{\pi^2}{6}\right)=\frac{\pi^3}{16}$$
The last two logarithmic integrals are well known, and their values can be determined by various methods, for example the series expansion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_0^{\sqrt 3} \frac{1}{1+x^2} \sin^{-1} (\frac{2x}{1+x^2}) dx$ The question is to evaluate $$\int_0^{\sqrt 3} \frac{1}{1+x^2} \sin^{-1} (\frac{2x}{1+x^2}) dx$$
I used the substitution $x=\tan a$ and it led me to the answer $\pi^2 /9$.However the correct answer is $\frac{7 \pi^2}{72}$.I couldn't get why the substitution is leading to a wrong answer.Any ideas?
| $x=\tan y\implies0\le y\le\dfrac\pi3$
Using Principal values
$$\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=\sin^{-1}(\sin2y)=\begin{cases}2y&\mbox{if }0\le2y\le\dfrac\pi2 \\
\pi-2y & \mbox{if } \dfrac\pi2 \le2y\le\pi \end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The sum of all integers $n$ satisfying $\frac{1}{4} < \sin\frac{\pi}{n} < \frac{1}{3}$ Asked in China Junior Math Olympiad. (No calculators or tables allowed.)
Find the sum of all integers $n$ satisfying the following inequality:
$$\frac{1}{4} < \sin\frac{\pi}{n} < \frac{1}{3}$$
| $\dfrac{1}{4} < \sin\dfrac{\pi}{n}
\implies
\sin\dfrac{\pi}{n} > 0
$
$
-\pi < \dfrac{\pi}{n} < \pi
\implies
n > 0
$
$
n=1
$ does not work
$
\implies n \ge 2
$
$
-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}
\implies
\sin(x)
$ increasing
$
\sin\dfrac{\pi}{6} = \dfrac{1}{2} > \dfrac{1}{3}
\implies
n \ge 7
$
$\dfrac{1}{4} < \sin x < x
\implies
n < 4\pi
\implies
n \le 12
$
$
\dfrac{1}{3} > \sin x \ge x - \dfrac{x^3}{6}
\implies
n \ge 10
$
But we still need to prove that $n=10,11,12$ work.
It is enough to prove that $n=10$ and $n=12$ work. There are formulas for $\sin\dfrac{\pi}{n}$ in these two cases:
$$
\sin\dfrac{\pi}{10} = \dfrac{\sqrt5 - 1}{4},
\qquad
\sin\dfrac{\pi}{12} = \dfrac{\sqrt3 - 1}{2 \sqrt2}
$$
I don't know whether this whole approach is feasible in a test...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2545713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$ if $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$ The question:
Let $\alpha$ and $\beta$ be $2$ distinct real numbers which such that $\alpha^2+3 \alpha+1=\beta^2+3\beta+1=0$. Find the value of $\frac {\alpha}{\beta} + \frac {\beta}{\alpha}$.
This problem is seems to be related to Vieta's Theorem, but so far I have not used it. This is my working out:
\begin{align}
\frac {\alpha}{\beta} + \frac {\beta}{\alpha} & = \frac {\alpha^2}{\alpha \beta} + \frac {\beta^2}{\alpha \beta} \\
& = \frac {\alpha^2+\beta^2}{\alpha \beta}
\end{align}
Vieta's Theorem:
Let $p(x)=ax^2+bx+c$ be a quadratic polynomial with zeros $\alpha$,$~\beta$. Then
$$\frac {-b}{a}=\alpha + \beta \\
\frac{c}{a} = \alpha \cdot \beta $$
Well, it is clear that the question wants us to find $\alpha^2 + \beta^2$ and $\alpha \beta$.
\begin{align}
\alpha^2+\beta^2 & = \alpha^2+2\alpha \beta + \beta^2 -2\alpha\beta \\
& = (\alpha+\beta)^2-2(\alpha\beta)
\end{align}
We have:
\begin{align}
\alpha^2 + 3\alpha + 1 & = \beta^2 + 3\beta + 1 \\
0 & =\alpha^2 - \beta^2+3\alpha-3\beta \\
& = (\alpha+\beta)(\alpha-\beta)+3(\alpha-\beta) \\
& = (\alpha-\beta)(\alpha+\beta+3) \\
\therefore~ \alpha + \beta & = -3\tag{reject $\alpha=\beta$}
\end{align}
This is where I am stuck because I am unable to find $\alpha\beta$ by algebraic manipulation. I have thought about trying to find $\alpha$ and $\beta$ through the quadratic formula, but it seems quite tedious, so it is a last resort. Is there a method to finish this question off?
| You have already written down $\frac{c}{a} = \alpha \cdot \beta$, so there you have it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all prime numbers $p$ such that $16p+1$ is a perfect cube What I have attempted:
Suppose $16p+1=k^3$ where $k \in Z$ then $16p=k^3-1=(k-1)(k^2+k+1)$ so we can say that $k=17$ and thus $p=17^3+17+1=4931$ which is prime.
How would I find the remaining numbers?
| You had $$16p=k^3-1=(k-1)(k^2+k+1)$$
Because $k$ is odd, $k-1$ is even and $k^2+k+1$ is odd. If $k^2+k+1$ is odd, then $k-1$ must be a multiple of $16$. But for $k-1$ to be a multiple of $16$ other than $16$, $p$ would have to not be a prime. Therefore, $k-1 = 16$ and $k = 17$.
That means that $k^2+k+1$ must be our prime. So plug in $k=17$ to get $$p = 17^2 + 17 + 1 = 307$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Integral asked on MIT bee qualifier 2012 $\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}\,dx$ I have been trying to solve this one but I have no clue, I put it into wolfram and the result is absurd, considering this one is taken from MIT bee qualifier 2012, this is the integral:
$$\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \, dx$$
| You can rewrite the integrand as
$$ \frac{(x-1)(x+1)}{(x+1)^2\sqrt{x^2\left(x + \dfrac{1}{x} + 1\right)}} = \frac{x^2 - 1}{(x^3 + 2x^2 + x)\sqrt{x + \dfrac{1}{x} + 1}} \\
= \frac{1 - \dfrac{1}{x^2}}{\left( x + \dfrac{1}{x} + 2 \right)\sqrt{x + \dfrac{1}{x} + 1}} $$
Then make the substitution $u^2 = x + \dfrac{1}{x} + 1$ to get
$$ \int \frac{2}{u^2+1} du = 2\arctan u + C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Question about series involving zeta function I was observing the series $$\sum_{n=1}^{\infty} \frac{\zeta(2n+1)-1}{2n+1}$$
And Wolfram alpha says that it does not converge. But I'm convinced that this is wrong since $$\sum_{n=1}^{\infty} (\zeta(2n+1)-1) = \frac{1}{4}$$
I would imagine then, by comparison test, that the first series clearly converges. Am I wrong? If it does converge, do we have a value for it?
| Perhaps I can fill in the missing details for you to help show how
$$I = \int^\infty_0 \frac{\sinh x - x}{x} \cdot \frac{dx}{e^x (e^x - 1)} = 1 - \gamma - \frac{1}{2} \ln (2).$$
To do this we will follow Jack's suggestion and use Frullani's theorem together with the following integral representation for the Euler-Mascheroni constant of
$$\gamma = \int^\infty_0 \left (\frac{1}{e^x - 1} - \frac{1}{x e^x} \right ) \, dx.$$
We begin by observering that
$$\frac{1}{e^x (e^x - 1)} = \frac{1}{e^x - 1} - \frac{1}{e^x}.$$
So the integral can be rewritten as
\begin{align*}
I &= \int^\infty_0 \frac{\sinh x - x}{x} \left (\frac{1}{e^x - 1} - \frac{1}{e^x} \right ) \, dx\\
&= \int^\infty_0 \left [\frac{\sinh x - x}{x (e^x - 1)} - \frac{\sinh x - x}{x e^x} \right ] \, dx\\
&= \int^\infty_0 \left [\frac{\sinh x}{x(e^x - 1)} - \frac{1}{e^x - 1} - \frac{\sinh x}{x e^x} + \frac{1}{e^x} \right ] \, dx\\
&= \int^\infty_0 \left [\frac{\sinh x}{x(e^x - 1)} - \frac{1}{e^x - 1} - \frac{\sinh x + 1}{x e^x} + \frac{1}{xe^x} + e^{-x} \right ] \, dx\\
&= -\int^\infty_0 \left (\frac{1}{e^x - 1} - \frac{1}{x e^x} \right ) \, dx + \int^\infty_0 e^{-x} \, dx + \int^\infty_0 \left [\frac{\sinh x}{x(e^x - 1)} - \frac{\sinh x + 1}{x e^x} \right ] \, dx\\
&= -\gamma + 1 + \int^\infty_0 \frac{1}{x} \left [\frac{\sinh x}{e^x - 1} - \frac{\sinh x + 1}{e^x} \right ] \, dx\\
&= 1 - \gamma + I_\alpha.
\end{align*}
Now for the term appearing in the square brackets of the integral, it can be rewritten as
\begin{align*}
\frac{\sinh x}{e^x - 1} - \frac{\sinh x + 1}{e^x} &= \frac{e^x \sinh x - (e^x - 1)(\sinh x + 1)}{e^x (e^x - 1)}\\
&= \frac{\sinh x - e^x + 1}{e^x (e^x - 1)}\\
&= \frac{\frac{1}{2} (e^x - e^{-x}) - e^x + 1}{e^x (e^x - 1)}\\
&= -\frac{e^{-2x} - 2e^{-x} + 1}{2(e^x - 1)}\\
&= -\frac{(1 - e^{-x})^2}{2 e^x (1 - e^{-x})}\\
&= \frac{e^{-2x} - e^{-x}}{2}.
\end{align*}
Now as
$$I_\alpha = \int^\infty_0 \frac{e^{-2x} - e^{-x}}{x} \, dx,$$
is of the form of a Frullani integral, namely
$$\int^\infty_0 \frac{f(ax) - f(bx)}{x} \, dx = (f(0) - f(\infty)) \ln \left (\frac{b}{a} \right ),$$
where $f(x) = e^{-x}, a = 2, b = 1$, as $f(0) = 1$ and $f(\infty) = 0$ we have
$$I_\alpha = \ln \left (\frac{1}{2} \right ) = -\ln (2).$$
Thus
$$\int^\infty_0 \frac{\sinh x - x}{x} \cdot \frac{dx}{e^x (e^x - 1)} = 1 - \gamma - \frac{1}{2} \ln (2),$$
and is all thanks to Jack and his amazing insight!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$'s General Solution I was looking for the general solution of the differential equation $$(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$$
My work
We need to get the auxillary equation of the given differential equation above...
The auxillary equation would be:
$$r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$$
We need to get the roots of the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ using the rational zero theorem...
The rational zero theorem states that if $P(x)$ is a polynomial with integer coefficients and if is a zero of $P(x) ( P(\frac{p}{q} ) = 0 )$,
then $p$ is a factor of the constant term of $P(x)$ and $q$ is a factor of the leading coefficient of $P(x)$.
With that in mind....
The factors of $p$ are:
$$-1,12,1,-12$$
$$-2,6,2,-6$$
$$-3,4,3,-4$$
The factors of $q$ are:
$$1,1,-1,-1$$
Then the values of $\frac{p}{q}$ would be:
$$-1,1,2,-3,4,-4,12,-12,6,-6,-2,3$$
It was found out that only $r= -2$ and $r= 3$ could make the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ true, instead of $5$
Now....if I only got $2$ real and distinct roots out of possible $5$, then the solution of the differential equation $(D^5 + D^4 -7D^3 - 11D^2 - 8D - 12)y = 0$
would be:
$$ y = c_1e^{-2x} + c_2e^{3x} + (unknown \space term) + (unknown \space term) + (unknown \space term) $$
I'm stuck....How do you get the solution of the given differential equation above?
UPDATE
After some feedback from intrepid answerers....I was able to factor out the polynomial $r^5 + r^4 - 7r^3 - 11r^2 - 8r - 12 = 0$ into $(r+2)(r+2)(r-3)(r^2 +1)=0$, so I think the solution to the differential equation above would be:
$$ y = c_1e^{-2x} + c_2xe^{-2x} + c_3e^{3x} + e^{x}(c_4\cos (-x) + c_5\sin (-x)) $$
or
$$ y = c_1e^{-2x} + c_2xe^{-2x} + c_3e^{3x} + e^{x}(c_4\cos x - c_5\sin x) $$
Is my solution now correct?
| Hint: Since you know that two roots are $-2$ and $3$, you can exploit factoring by grouping or polynomial long division:
\begin{align}\begin{aligned}
&r^5+r^4-7r^3-11r^2-8r-12 \\&\quad= r^5+2r^4-r^4-2r^3-5r^3-10r^2-r^2-2r-6r-12\\
&\quad= (r+2)(r^4-r^3-5r^2-r-6)\\
&\quad= (r+2)(r^4-3r^3+2r^3-6r^2+r^2-3r+2r-6)\\
&\quad= (r+2)(r-3)(r^3+2r^2+r+2)
\end{aligned}\end{align}
That last cubic factors in a particularly nice way. (You can also try applying the rational roots theorem to the cubic $r^3+2r^2+r+2$ to try to pick out a root, and then use factoring by grouping again; since the last term is a +2, there's only 4 values to try, namely $\pm 1$ and $\pm 2$.)
EDIT:
It is true that this becomes
$$(r+2)^2(r-3)(r^2+1) = 0$$
which yields roots of $r = -2,-2,3,\pm i$
However, it is not true that your solution becomes
$$y = c_1 e^{-2x} + c_2 xe^{-2x} + c_3 e^{3x} + e^x(c_4 \cos(x)+c_5\sin(x)$$
The problem is with those last two terms. You can check that the function $$f(x) = e^x(c_4\cos(x) + c_5\sin(x))$$ is a solution to
$$((D-1)^2+1)f = 0$$
i.e., it corresponds to complex roots of $r=1\pm i$, not $r=\pm i$. The piece you're missing is Euler's equation:
$$e^{i\alpha} = \cos(\alpha) + i\sin(\alpha)$$
Using this you can conclude your solution is
$$y = c_1 e^{-2x} + c_2xe^{-2x} + c_3e^{3x} + c_4\cos(x) + c_5\sin(x)$$
Really, it all hinges on the solution to $(D^2+1)y = 0$, which one can derive in the following manner. We know that a candidate solution is (with $A$ and $B$ arbitrary real constants)
\begin{align}\begin{aligned}
y &= Ae^{ix} + Be^{-ix}\\
&= A(\cos(x)+i\sin(x)) + B(\cos(x)-i\sin(x))\\
&= (A+B)\cos(x) + (A-B)i\sin(x)
\end{aligned}\end{align}
and now we notice that if this is a solution, then so is the real part (i.e., $\cos(x)$) and the imaginary part (i.e., $\sin(x)$), hence arriving at the desired solution
$$y = c_1\cos(x) + c_2\sin(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
expected value, conflicting answers I am having a problem trying to understand why the two methods for finding the expected are resulting in different solutions. I am not sure if I am doing a mistake somewhere. Your help is appreciated
And I was also wondering how I could use the given ( a third method maybe?) $E(Y)$
Question in short
Given
$f_Y(y) = 2y$, $E(Y) = 2/3$, $W = (Y - 2/3)^2$
Find $E(W)$
My method 1
Let $g(y) = w \Rightarrow g(y)=(y-2/3)^2 = y^2-(4/3)y+4/9$
$E(g(y)) = \int^1_0g(y)f_Y(y)dy = 2\int^1_0(y^3-(4/3)y^2+(4/9)y)dy = 1/18$
My method 2
$F_Y(y)=\int^y_0f_Y(y) = \int^y_02y= y^2$
Notice that $F_W(w) = P((Y-2/3)^2 \leq W) = P(Y \leq \sqrt W + 2/3) = F_Y(\sqrt W + 2/3) = w + (4/3)\sqrt w + 4/9$
Then
$f_W(w) = \frac {d} {dw} F_W(w) =\frac {d} {dw} (w + (4/3) \sqrt w + 4/9) = 1+(2/3)w^{-1/2} $
Hence
$E(W) = \int^1_0wf_W(w)dw = \int^1_0(w + (2/3)w^{1/2}) = (1/2)w^2|^1_0 + (2/3)^2w^{3/2}|^1_0 = $ 17/18
As you can see, the first method is 1/18 while the second 17/18.
Also, I wanted to ask if there was a way I could use the given $E(Y)$. For example, using $E(aY+b)=aE(Y)+b$ where $a,b$ are some constants
Thank you.
| Method 1 is correct.
In Method 2, it is not true that $P((Y-2/3)^2 \leq w) = P(Y \leq \sqrt{w}+2/3)$ because $Y \leq \sqrt{w}+2/3$ does not imply $(Y-2/3)^2 \leq w$. You have to be careful when taking the square root. What is true is that
\begin{align*}
(Y-2/3)^2 \leq w &\Longleftrightarrow Y-2/3 \leq \sqrt{w} \text{ AND } (Y-2/3) \geq -\sqrt{w}\\
&\Longleftrightarrow -\sqrt{w}+2/3 \leq Y \leq \sqrt{w}+2/3.
\end{align*}
So you should have
\begin{align*}
P((Y-2/3)^2 \leq w) &= P(-\sqrt{w}+2/3 \leq Y \leq \sqrt{w}+2/3) \\
&= P(Y \leq \sqrt{w}+2/3)-P(Y < -\sqrt{w}+2/3)\\
&=F_Y(\sqrt{w}+2/3)-F_Y(-\sqrt{w}+2/3)
\end{align*}
Now you also have to be careful about the bounds of your variables. You did not say it, but $f_Y(y)=2y$ only for $0 \leq y \leq 1$, so actually
$$F_Y(y)=\begin{cases} 0 & y<0 \\ y^2 & 0 \leq y \leq 1 \\ 1 & y>1. \end{cases}$$
It is clear that for $w<0$, $P(W \leq w)=0$. If $0 \leq w\leq 1/9$, then $\sqrt{w}+2/3$ and $-\sqrt{w}+2/3$ are both between $0$ and $1$, so we can apply $F_Y(y)=y^2$. If $1/9 < W \leq 4/9$, then $\sqrt{w}+2/3$ is greater than $1$, so $F_Y(\sqrt{w}+2/3)=1$, but $-\sqrt{w}+2/3$ is still between $0$ and $1$ so we can apply $F_Y(y)=y^2$ to that. Finally, if $w >4/9$ then $\sqrt{w}+2/3>1$ and $-\sqrt{w}+2/3<0$. Bringing this all together, we get
\begin{align*}
F_W(w) &= \begin{cases} 0 & w<0 \\ (\sqrt{w}+2/3)^2-(-\sqrt{w}+2/3)^2 & 0\leq w\leq 1/9 \\ 1-(-\sqrt{w}+2/3)^2 & 1/9<w \leq 4/9 \\ 1 & w>4/9\end{cases}\\
&=\begin{cases} 0 & w<0 \\ \frac{8}{3}\sqrt{w} & 0\leq w\leq 1/9 \\ -w+\frac{4}{3}\sqrt{w}+5/9 & 1/9<w \leq 4/9 \\ 1 & w>4/9\end{cases}
\end{align*}
and then
$$f_W(w) = \begin{cases} 0 & w<0 \text{ or } w>4/9 \\ \frac{4}{3\sqrt{w}} & 0 \leq w \leq 1/9 \\ \frac{2}{3\sqrt{w}}-1 & 1/9<w \leq 4/9. \end{cases}$$
If you integrate this density (piecewise) you will again get $E[W]=1/18$.
There is no method that will give you $E[W]$ using $E[Y]$ alone, since the density of $Y$ matters. However, you could have saved yourself a bit of calculation in Method 1 by noting:
$$E[W]= \int_0^1 (y^2-(4/3)y+4/9)2y \ dy = \int_0^1 (y^2)(2y) \ dy - (4/3)E[Y]+4/9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Laurent Series for different domains I would like to determine the Laurent series for the function
$$f(z) = \frac{1}{z(z-2)^3}$$
for two different domains
$$\vert{z}\vert<2$$
and
$$\vert{z}\vert>2$$
But, I am unsure when I am computing the Laurent series when I am to take into consideration the different domains. I use standard tricks such as taking out the $\frac{1}{z}$ term and rearranging the remaining term into the sum formula of the geometric series to compute the general Laurent series.
Thanks for all help!
| For $|z|>2$ :
$$f(z) = \frac{1}{z(z-2)^3} = \frac{1}{z[z^3(1-\frac{2}{z})^3]}=\frac{1}{z^4(1-\frac{2}{z})^3}$$
A known geometric series is :
$$\frac{1}{1-w} = \sum_{n=0}^\infty w^n ,\quad |w| < 1$$
From that, we can derive :
$$\frac{1}{(1-w)^3} = \sum_{n=0}^\infty\frac{1}{2}(1+n)(2+n)w^n, \quad |w|<1$$
So, applying that for $w=\frac{2}{z}$, we get :
$$f(z) = \frac{1}{z^4} \sum_{n=0}^\infty\frac{1}{2}(1+n)(2+n)\bigg(\frac{2}{z}\bigg)^n=\sum_{n=0}^\infty2^{n-1}(1+n)(2+n)z^{-(n+4)}$$
for $|\frac{2}{z}| < 1 \Leftrightarrow |z| > 2 . $
For $|z|<2$, we :
$$f(z)=\frac{1}{z(z-2)^3}= - \frac{1}{z(2-z)^3}= - \frac{1}{z}\sum_{n=0}^\infty2^{-(n+4)}(1+n)(2+n)z^n$$
$$=$$
$$-\sum_{n=0}^\infty2^{-(n+4)}(1+n)(2+n)z^{n-1}$$
which hols for $|z|<2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{a_{n}} \right)^{a_{n}}=e$ if $\lim\limits_{n\rightarrow\infty} a_{n}=\infty$ What would be the nicest proof of the following theorem:
If $\lim\limits_{n \rightarrow \infty} a_{n} = \infty$, then $\lim\limits_{n \rightarrow \infty} \left(1 + \frac{1}{a_{n}} \right) ^ {a_{n} } = e$.
If $\lim\limits_{n \rightarrow \infty}b_{n} = 0$, then $\lim\limits_{n \rightarrow \infty} \left(1 + b_{n} \right) ^ {\frac {1} {b_{n}} } = e$.
I somehow failed to find a proof here on the website and in the literature.
| For any $a_n$ exists $x\in \mathbb N$ such that $x\le a_n\le x+1$ and
$$\left(1 + \frac{1}{x +1} \right)^{x} \leq \left(1 + \frac{1}{a_n} \right)^{a_n} \leq \left(1 + \frac{1}{x} \right)^{x+1}$$
and
$$\left(1 + \frac{1}{x +1} \right)^{x} =\frac{\left(1 + \frac{1}{x +1} \right)^{x+1}}{1 + \frac{1}{x +1} } \to \frac e 1=e$$
$$\left(1 + \frac{1}{x} \right)^{x+1}=\left(1 + \frac{1}{x} \right)^{x}\left(1 + \frac{1}{x} \right)\to e \cdot 1=e$$
then we can conclude by squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2554111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Calculate $\lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x}$ Beside using l'Hospital 10 times to get
$$\lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} = 3$$ and lots of headaches, what are some elegant ways to calculate the limit?
I've tried to write the functions as powers of $e$ or as power series, but I don't see anything which could lead me to the right result.
| $$\begin{array}{cl}
& \displaystyle \lim_{x\to 0} \frac{x(\cosh x - \cos x)}{\sinh x - \sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{xe^x + xe^{-x} - 2x\cos x}{e^x - e^{-x} - 2\sin x} \\
=& \displaystyle \lim_{x\to 0} \frac{x + x^2 + \frac12x^3 + o(x^4) + x - x^2 + \frac12x^3 + o(x^4) - 2x + x^3 + o(x^4)} {1 + x + \frac12x^2 + \frac16x^3 + o(x^4) - 1 + x - \frac12x^2 + \frac16x^3 + o(x^4) - 2x + \frac13x^3 + o(x^4)} \\
=& \displaystyle \lim_{x\to 0} \frac{2x^3 + o(x^4)} {\frac23x^3 + o(x^4)} \\
=& \displaystyle \lim_{x\to 0} \frac{3 + o(x)} {1 + o(x)} \\
=& 3
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2554448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to solve a system of linear equations modulo n? For example,
$4x - 10y \equiv 8\pmod {20}$
$7x + 2y \equiv 5\pmod {20}$
It resembles linear diophantine equations and the Chinese Remainder Theorem, but I don't know how to actually solve it..
| First you can multiply the system by any number that has an inverse, that is $\gcd(x,20)=1$.
So in particular you cannot multiply or divide by $2,4,5,10$ as you would not multiply or divide by $0$ in a normal non-modular system. Well you can do it, but you'll loose equivalence in the way.
For instance here we would like to have $4x-3x$. We notice that $7^{-1}=3\pmod {20}$ so let's multiply the second line by $9$.
$\begin{cases} 4x-10y\equiv 8\pmod{20}\\ 7x+2y\equiv 5\pmod{20}\end{cases}\iff\begin{cases} 4x-10y\equiv 8\pmod{20}\\ 3x+18y\equiv 5\pmod{20}\end{cases}$
Now we can subtract two lines like we do in normal systems
$\begin{cases} 4x-10y\equiv 8\pmod{20}\\ x-28y\equiv 3\pmod{20}\end{cases}\iff\begin{cases} 4x-10y\equiv 8\pmod{20}\\ x\equiv 8y+3\pmod{20}\end{cases}$
We now have $x$ and can report in the first equation
$\begin{cases} 32y+12-10y\equiv 8\pmod{20}\\ x\equiv 8y+3\pmod{20}\end{cases}\iff\begin{cases} 2y\equiv 16\pmod{20}\\ x\equiv 8y+3\pmod{20}\end{cases}$
$2$ is not invertible modulo $20$ but since all coefficients are even, we can divide everything by $2$ including the modulo.
$\begin{cases} y\equiv 8\pmod{10}\\ x\equiv 8y+3\pmod{20}\end{cases}$
Finally report in second equation to get $x$, by introducing a dummy variable $k$ for expressing $y$
$\begin{cases} y\equiv 8\pmod{10}\\ x\equiv 8(10k+8)+3\pmod{20}\equiv 64+3\equiv 7 \pmod{20}\end{cases}$
I have detailed a lot, since this is apparently your first time solving this.
Answering question in comment
$\begin{cases} 2x-6y\equiv 11\pmod{20}\\ 4x+8y\equiv 9\pmod{20}\end{cases}$
Rem: for parity reasons this has no solution, but let's do like we ignore that.
For this one $9$ and $11$ are invertible (and their own inverse), so let's multiply the lines by $9$ and $11$.
You get
$\begin{cases} 2x+14y\equiv 1\pmod{20}\\ 16x+12y\equiv 1\pmod{20}\end{cases}\iff \begin{cases} 2x+14y\equiv 1\pmod{20}\\ 14x-2y\equiv 0\pmod{20}\end{cases}$
Which reduce to $y\equiv 7x\pmod{10}$, and this system has no solution ($2x+14y$ is divisible by $20$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
The order of a modulo p.
If $p$is prime and $\operatorname{ord}_p(a)=4$, then $1+a+a^2+a^3≡ 0\bmod p$, where $\operatorname{ord}_p(a)$ is the order of $a$ modulo $p$.
I think it is true statement
$\operatorname{ord}_p(a)=4$, then $a^4≡1\bmod p$ so $a^4-1≡0\bmod p$
since $a^4-1=1+a+a^2+a^3$, then $a^4-1=1+a+a^2+a^3\equiv0\bmod p$
is that correct please?
| $a^4-1 \neq 1+a+a^2+a^3$, but it is true that $a^4-1=(a-1)(a^3+a^2+a+1)=0 \mod p $, and since there are no zero divisiors, either $a=1$, in which case its order cannot be $4$, or, as you conclude: $(a^3+a^2+a+1)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$ Given $2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$,
prove that $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$
Also we know that all the expressions are different from zero and defined. Including the expressions we received during the solution.
Tried to play with it, didn't seem to work for me.
| Applying sum to product rule on LHS and sine of the sum of 2 angles on RHS
$$2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$$
$$4\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha-\beta}{2}\right)=6\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha+\beta}{2}\right)$$
$$2\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)=3\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)$$
$$2\cos\left(\frac{\alpha}{2}\right)\cos \left(\frac{\beta}{2}\right)+2\sin \left(\frac{\alpha}{2}\right)\sin \left(\frac{\beta}{2}\right)=3\cos\left(\frac{\alpha}{2}\right)\cos \left(\frac{\beta}{2}\right)-3\sin \left(\frac{\alpha}{2}\right)\sin \left(\frac{\beta}{2}\right)$$
$$5\sin \left(\frac{\alpha}{2}\right)\sin \left(\frac{\beta}{2}\right)=\cos\left(\frac{\alpha}{2}\right)\cos \left(\frac{\beta}{2}\right)$$
$$\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)=\frac{1}{5} \quad \square$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Simplify expression $\frac{2\cos(x)+1}{4\cos(x/2+π/6)}$ How to simplify the following expression: $$\frac{2\cos(x)+1}{4\cos\left(\frac x2+\fracπ6\right)}$$
I got to: $ \dfrac{2\cos(x)+1}{4\cos\left(\dfrac x2\right)\cdot \dfrac{\sqrt3}2-\sin(x) \cdot \frac 12}$
| Let $\dfrac{x}{2}+\dfrac{\pi}{6}=y$. Then $x=2y-\dfrac{\pi}{3}$.
\begin{align*}
\frac{2\cos x+1}{4\cos\left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)}&=\frac{2\cos\left(2y-\dfrac{\pi}{3}\right)+1}{4\cos y}\\
&=\frac{2\cos2y\cos\dfrac{\pi}{3}+2\sin2y\sin\dfrac{\pi}{3}+1}{4\cos y}\\
&=\frac{2\cos^2y-1+2\sqrt{3}\sin y\cos y+1}{4\cos y}\\
&=\frac{\cos y+\sqrt{3}\sin y}{2}\\
&=\cos y\cos\frac{\pi}{3}+\sin y\sin\frac{\pi}{3}\\
&=\cos\left(y-\frac{\pi}{3}\right)\\
&=\cos\left(\frac{x}{2}-\frac{\pi}{6}\right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Evaluate the limit $\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$ Can someone please check my conclusion to the evaluation of the following limit?
$$\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$$
I got that the limit is equal to infinity. If limit is equal to infinity does this mean that limit does not exist?
| The limit is not infinity. If you need to do it the hard way, consider
$$
x(\sqrt[4]{16x^4+x^2+1}-2x)
=x\frac{\sqrt{16x^4+x^2+1}-4x^2}{\sqrt[4]{16x^4+x^2+1}+2x}
=\frac{x(x^2+1)}{(\sqrt[4]{16x^4+x^2+1}+2x)(\sqrt{16x^4+x^2+1}+4x^2)}
=\frac{x^3(1+\frac{1}{x^2})}
{x^3
\Bigl(\sqrt[4]{16+\frac{1}{x^2}+\frac{1}{x^4}}+2\Bigr)
\Bigl(\sqrt{16+\frac{1}{x^2}+\frac{1}{x^4}}+4\Bigr)
}
$$
However, it's much easier with a Taylor expansion. If you substitute $x=1/t$, with easy algebraic manipulations the limit becomes
$$
\lim_{t\to0^+}\frac{\sqrt[4]{16+t^2+t^4}-2}{t^2}=
2\lim_{t\to0^+}\frac{\sqrt[4]{1+t^2/16+t^4/16}-1}{t^2}
$$
and
$$
\sqrt[4]{1+t^2/16+t^4/16}=1+\frac{1}{4}\frac{t^2}{16}+o(t^2)
$$
About the case when the limit is infinity, it depends on the conventions you're following whether the limit “exists”. Check with your instructor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding $x$ values of $\cos 6x + 1 = \frac{3}{2} + \frac {1}{2} \cos 3x $
Solve $$\cos 6x + 1 = \frac{3}{2} + \frac {1}{2} \cos 3x $$ for $0^\circ<x<120^\circ$
I simplify it to
$$2 \cos 6x + 2 = 3 + \cos 3x $$
There is $\cos 6x $ and $\cos 3x$ how do I merge them together? To solve the equation from $0^\circ$ to $120^\circ$?
| Hint: Use the identity
$$\cos 2y=2\cos^{2}y-1$$
with $y=3x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2560925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do I evaluate $\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{ab}{(a+b)!}$
$$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{ab}{(a+b)!}$$
I'm not really comfortable with more than 1 sigma's and that's why this question is confusing me. I don't think it's possible to reduce the number of variables to 1 here.
The answer is $\frac{2}{3}e$
| Since all terms in the sum are non-negative, one can arbitrary change the order of summation without changing the value of the sum. In particular, we can group terms with same value of $k = a + b$ and sum over them first. Notice
$$\sum_{a+b=k, a, b \ge 1}ab
= \sum_{a=1}^{k-1} a(k-a)
= k \frac{(k-1)k}{2} - \frac{(k-1)k(2k-1)}{6}
= \frac{(k-1)k(k+1)}{6}
$$
We have
$$\begin{align}
\sum_{a=1}^\infty\sum_{b=1}^\infty \frac{ab}{(a+b)!}
&= \sum_{k=2}^\infty \frac{1}{k!}\sum_{a+b=k, a, b \ge 1}ab
= \frac16 \sum_{k=2}^\infty \frac{(k-1)k(k+1)}{k!}\\
&= \frac16 \sum_{k=2}^\infty \frac{k+1}{(k-2)!}
= \frac16 \sum_{k=0}^\infty \frac{k+3}{k!}
= \frac16 \left( \sum_{k=1}^\infty \frac{1}{(k-1)!} + \sum_{k=0}^\infty \frac{3}{k!}\right)\\
&= \frac16(e+3e) = \frac23 e
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Taylor series expansion of sin(x) I understand that Taylor series expansion for $\sin(x)$ is derived as follow:
$$
\sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-...
$$
Now, what exactly is the first, second, and third term?
Is the first term just $\sin(x) = x$?
Is the second term $\sin(x) = x-\frac{x^3}{3!}$?
| Terms are the members of a summation (whatever the formula) and they are enumerated in the order of their appearance (left to right, starting from first)
*
*$x$ is the first term,
*$-\dfrac{x^3}{3!}$ is the second term,
*$\dfrac{x^5}{5!}$ is the third term.
When speaking of a polynomial, a term is said of the $n^{th}$ degree depending on the power of the variable. So here there are no terms of even degree and
*
*$x$ is the first degree term,
*$-\dfrac{x^3}{3!}$ is the third degree term,
*$\dfrac{x^5}{5!}$ is the fifth degree term.
Now a Taylor expansion is written up to a remainder term, with as many terms as you like. The word order is used and equals the highest degree. So you can say
*
*$\sin(x)=x+r_1(x)$ is the first order expansion,
*$\sin(x)=x-\dfrac{x^3}{3!}+r_3(x)$ is the third order expansion,
*$\sin(x)=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+r_5(x)$ is the fifth order expansion.
Similarly, for the cosine you would have
First term $1$, second term $-\dfrac{x^2}2$, third term $\dfrac{x^4}{4!}$, and zero$^{th}$ order development/expansion $1$, second order $1-\dfrac{x^2}2$, fourth order $1-\dfrac{x^2}2+\dfrac{x^4}{4!}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find all natural numbers $n$ such that $2^n$ divides $3^n -1$
Find all natural numbers $n$ such that $2^n$ divides $3^n -1$
I think that the only solutions are $n = 0,1,2,4$, but I have no idea on how to prove it.
I tried to write $3^n-1$ as $1+3+3^2+...+3^{n-1}$ and manipulate the sum but found my self at the equally hard problem of finding the power of two dividing $3^k+1$
| Well $3^n - 1 = (3-1)(1 + 3 + 3^2 + ... + 3^{n-1}) = 2(1 + 3 + 3^2 + ... + 3^{n-1})$
So $2^n|3^n - 1$ if and only if $2^{n-1}|(1 + 3 + 3^2 + ... + 3^{n-1})$.
If $n$ is odd and greater than one $(1+3 + 3^2 + .... + 3^{n-1})$ is odd so we can assume $n$ is even.
Let $n = 2m$ then $2^{2m}|3^{2m} - 1=(3^m -1)(3^m+1)$. So $3^m \pm 1$ are both even and only one of them is is divisible by $4$.
So $2^{2m-1}|3^m \pm 1$ so $2^{2m-1} \le 3^m \pm 1$.
But $2^{2m-1} = \frac 12*4^{m} \le 3^m \pm 1$
So $(\frac 43)^m \le 2 \pm \frac 2{3^m} < 2\frac 23$
If $m \ge 3$ then $(\frac 43)^m \ge 2 \frac {10}{27} > 2 \frac 2{3^3}\ge 2 + \frac 2{3^m}$
So $m < 3$
So if $n > 1$ then $n= 2m; m\le 2$.
So solutions must be a subset of $\{0,1,2,4\}$.
And you have already determined that $\{0,1,2,4\}$ are all solutions.
====
There's probably a more elegant way.
My first thought was FTL that as $\gcd(3, 2^n) = 1$ and $\phi(2^n) = 2^{n-1}$ then $3^{2^{n-1}}\equiv 1 \mod 2^n$. So
If $3^m \equiv 1 \mod 2^n$ then $m$ is a multiple of a non-trivial factor of $2^{n-1}$ .i.e. even
but that didn't really get me closer.
Likewise $3^n = (2 + 1)^n = 2^n + \sum{n\choose k} 2^k$ and for $2^n|\sum{n\choose k} 2^k$ seemed like it should yeild something relevent but I wasn't able to put my finger on it exactly.
Similarly $3^n = (4 -1)^n$.
Its a enough to convince me the answers are related to powers of $2$ but not enough to actually prove it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Find the indefinite integral $\int\frac{dx}{(x^2+2x+5)^2}$ I need help with the indefinite integral
\begin{align}
& \int\frac{dx}{(x^2+2x+5)^2} \\[10pt]
= {} & \int\frac{dx}{((x+1)^2+4)^2} = \int\frac{du}{(u^2+4)^2} & & x+1=u,\quad du=dx \\[10pt]
= {} & \frac{1}{16} \int\frac{du}{(\frac{u^2}{4}+1)^2} \\[10pt]
= {} & \frac 1 8 \int \frac{ds}{(s^2+1)^2} = \text{?} & & \frac u 2 = s, \quad 2\,ds=du
\end{align}
or maybe there is an easier way ?
Any ideas ? thanks !
| If you write $s = \tan(t)$ then you obtain $\sec^2(t)\,\mathrm{d}t = \mathrm{d} s$ and
$$
\int \frac{1}{(1+s^2)^2}\,\mathrm{d}{s}=\int \cos^2(t)\,\mathrm{d}{t}
$$
you can then solve the second integral by using the double-angle formula for cosine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Binomial to the power of five, equality proof I want to find out when the equality $(x+y)^5=x^5+y^5$ for real numbers $y$ and $x$ holds.
Expanding this binomial yields $(x+y)^5=x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4$
Factoring the right side of the above equality gives $x^5+y^5+5xy(x^3+2x^2y+2xy^2+y^3)$
If $(x+y)^5=x^5+y^5$, then $5xy(x^3+2x^2y+2xy^2+y^3)=0$ which implies that
$xy=0$ or $x^3+2x^2+2xy^2+y^3=0$
if $xy\neq0$ then $x^3+2x^2y+2xy^2+y^3=0$ and thus $(x+y)^3=xy(x+y)$
This is where I'm stuck. How can the last equality help me to figure out when the original equality holds? I know that $xy\neq0$ and I can somehow see that showing that $x+y=0$ seems to be an appropriate step at this moment, but I'm now sure.
Thanks
| Since $$x^3+2x^2y+2xy^2+y^3=(x+y)(x^2-xy+y^2)+2xy(x+y)=$$
$$=(x+y)(x^2+xy+y^2)=(x+y)\left(\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\right),$$
we see that $x^3+2x^2y+2xy^2+y^3=0$ for $x+y=0$ only.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to proof: $\cos(10°)*\cos(30°)*\cos(50°)*\cos(70°)=\frac{3}{16}$ How to proof: $\cos(10°)*\cos(30°)*\cos(50°)*\cos(70°)=\frac{3}{16}$
I tried to: $\frac{8*3^{1/2}\sin(10°)*\cos(10°)*\sin(40°)*\sin(20°)}{16*\sin(10°)} $ and got nowhere
| $$\begin{align}\cos(10°)\cos(30°)\cos(50°)\cos(70°) &= \sin(80°)\cos(30°)\sin(40°)\sin(20°) &\\=& \cos(30°)\sin(80°)\cdot\dfrac{-1}{2}(\cos(60°) - \cos(20°))&\\=& \cos(30°)\cdot\dfrac{-1}{2}(\sin(80°)\cos(60°) - \cos(20°)\sin(80°))&\\=& \cos(30°)\cdot\dfrac{-1}{2}\left(\dfrac{1}{2}(\sin(140°)+\sin(20°)) - \\\dfrac12(\sin(100)+\sin(60°))\right)&\\=&\ \cos(30°)\cdot\dfrac{-1}{4}\left(\sin(40°)+\sin(20°) - \sin(80°)-\sin(60°)\right)&\\=&\ \cos(30°)\cdot\dfrac{-1}{4}(\sin(60°)\cos(20°) - \sin(20°)\cos(60°)\\&+\sin(20°) - \sin(60°)\cos(20°) -\sin(20°)\cos(60°)\\&-\sin(60°))&\\=&\ \cos(30°)\cdot\dfrac{-1}{4}\left( - 2\sin(20°)\cos(60°)+\sin(20°) -\sin(60°)\right)&\\=&\ \cos(30°)\cdot\dfrac{1}{4}\sin(60°) = \dfrac{3}{16}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving the Differential equation: $y'=\frac{2}{x}y+x^3$ We have the differential equation $$y'=\frac{2}{x}y+x^3$$ and we know $x \in (0, \infty)$.
My attempt with variation of constants
\begin{align}
\phi(x) &= \exp \left(\int \frac{2}{x} dx \right) \\
&= \exp(2\ln|x|) \\
&= x^2c
\end{align}
and
\begin{align}
\psi(x) &= (x^2c) \cdot \int \frac{x^3}{x^2} dx \\
&= (x^2c) \cdot \frac{x^2}{2}
\end{align}
but this solution is wrong. Where is the mistake?
| Multipying both side by $e^{-2\ln x}=1/x^2$ we get $$y'=\frac{2}{x}y+x^3\Longleftrightarrow (\frac{y}{x^2})'= x\Longleftrightarrow \frac{y}{x^2}=x^2/2+c\\\Longleftrightarrow y(x)= x^4/2+cx^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
How to prove that $\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$ without squaring both sides I have been asked to prove:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
Which I can easily do by converting the LHS to index form, then squaring it and simplifying it down to get 2, which is equal to the RHS squared, hence proved.
However I know you can't square a side during proof because it generates an extraneous solution. So: how do you go about this proof without squaring both sides? Or can my method be made valid if I do this:
$$\sqrt{2+\sqrt3}-\sqrt{2-\sqrt3}=\sqrt2$$
$$...=...$$
$$2=2$$
$$\lvert\sqrt2\rvert=\lvert\sqrt2\rvert$$
$$\sqrt2=\sqrt2\text{ hence proved.}$$
Cheers in advance :)
| Let $a=\sqrt{2+\sqrt3}\,$, $b = \sqrt{2-\sqrt3}\,$, then:
$$\require{cancel}
a^2+b^2 = 2+\bcancel{\sqrt{3}}+2 - \bcancel{\sqrt{3}} = 4 \\
ab = \sqrt{(2+\sqrt3)(2-\sqrt3)} = \sqrt{2^2 - (\sqrt{3})^2} = \sqrt{4-3} = \sqrt{1} = 1
$$
It follows that:
$$(a-b)^2 = a^2+b^2-2ab = 4 - 2 \cdot 1 = 2$$
Since $\sqrt{2+\sqrt3} \gt \sqrt{2-\sqrt3}\,$, $a-b \gt 0$ must be the positive root, so $a-b=\sqrt{2}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 12,
"answer_id": 7
} |
Show that the sequence $a_n = \frac{(n+1)^2 -n^2}{n}$ converges and give its limit. can somebody tell me if I did this somewhat correctly? First I estimated the limit of the sequence by calculating the first few results of the sequence and it looks like it is converging towards $2$. Using the Cauchy criterion we then have: $$\forall \epsilon > 0 \exists N\in\mathbb{N}_0 \forall n \geq N: \left|\frac{(n+1)^2 -n^2}{n} - 2\right| < \epsilon$$
and after doing a bit of math we have $$\left|\frac{(n+1)^2 -n^2}{n} - 2\right| < \epsilon \Leftrightarrow \left|\frac{1}{n}+2-2\right| < \epsilon \Leftrightarrow \frac{1}{n} < \epsilon \Leftrightarrow \frac{1}{\epsilon} < n $$ which means that we need a $N\in\mathbb{N}$ with $N > \frac{1}{\epsilon}$ and we have $\forall n\geq N$:$$\left|\frac{(n+1)^2 -n^2}{n} - 2\right| = \dots=\frac{1}{n}\leq\frac{1}{N}<\epsilon\,.$$
And hence our sequence $a_n \longrightarrow a_\infty$ with $a_\infty = 2$$._{\,\,\square}$
Did I do this right?
| To do this more cleanly but also rigorously, note that given $\epsilon > 0$, it is possible to find $N$ such that $N > \dfrac{1}{\epsilon}$. Then for all $n > N$, we have that $\dfrac{1}{n} < \epsilon$. From here, we can achieve the desired result:
$$
\epsilon > |{\dfrac{1}{n}}| = |\dfrac{2n + 1 - 2n}{n}| = |\dfrac{2n + 1}{n} - 2 | = |\dfrac{n^2 + 2n + 1 - n^2}{n} - 2 | = |\dfrac{(n + 1)^2 - n^2}{n} - 2|
$$
Thus, given any $\epsilon > 0$, if we choose any $n > N$, we have that
$$
|\dfrac{(n + 1)^2 - n^2}{n} - 2| < \epsilon
$$
Since such $N$ exists, this implies that the sequence $a_n$ converges to 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
| Let $x=(1+i\sqrt 7\;)/(2\sqrt 2\;).$ Since $x\bar x=x^2\bar x^2=1$ and $x+\bar x=1/\sqrt 2\;$ we have $$((1+i\sqrt 7\;)/2)^4+(1-i\sqrt 7\;)/2)^4=4(x^4+ \bar x^4)=$$ $$=4((x^2+\bar x^2)^2-2x^2\bar x^2)=4((x^2+\bar x^2)^2-2)=$$ $$=4( ((x+\bar x)^2-2x\bar x)^2-2)=$$ $$=4((1/\sqrt 2\;)^2-2)^2-2)=4((1/2-2)^2-2)=4(9/4-2)=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
} |
Expand $a^5 + b^5 + c^5$ in terms of Schur polynomials How to expand certain sums of powers in terms of Schur polyomials. I have been gaining proficiency with symmetric polynomials, today I would like to expand:
$$ a^5 + b^5 + c^5 = \sum_{\lambda_1 + \lambda_2 + \lambda_3 = 5} s_\lambda (a,b,c) \tag{$*$} $$
My first take is to write out all the partition of 5 into three parts, and I am going to write them both as sums and as "Frobenius partitions"
*
*$5 = 5 + 0 + 0 = (4|0)$
*$5 = 4 + 1 + 0 = (3|1)$
*$5 = 3 + 1 + 1 = (2|2)$
*$5 = 2 + 2 + 1 = (2,0|1,0)$ (not sure about this one)
So now I am going to write out all these determinants in terms of elementary symmetric functions:
$$ s_{5,0,0}(a,b,c) = \frac{1}{\Delta} \left|
\begin{array}{lll} a^7 & b^7 & c^7 \\
a & b & c \\
1 & 1 & 1 \end{array} \right| $$
$$ s_{4,1,0}(a,b,c) = \frac{1}{\Delta} \left|
\begin{array}{lll} a^6 & b^6 & c^6 \\
a^2 & b^2 & c^2 \\
1 & 1 & 1 \end{array} \right| $$
$$ s_{3,1,1}(a,b,c) = \frac{1}{\Delta} \left|
\begin{array}{lll} a^5 & b^5 & c^5 \\
a^2 & b^2 & c^2 \\
a & b & c \end{array} \right| $$
$$ s_{2,2,1}(a,b,c) = \frac{1}{\Delta} \left|
\begin{array}{lll} a^4 & b^4 & c^4 \\
a^3 & b^3 & c^3 \\
a & b & c \end{array} \right| $$
I am still working out what the Jacobi-Trudi identities say in these circumstances. In any case, there seem to be 4 Schur polynomials that I need (at this moment) and wish to do the expansion at $(*)$.
| The power sum symmetric function $p_n$ is the alternating sum of hooks of size $n$, that is:
$$p_n=\sum_{i=0}^{n-1}(-1)^{i}s_{(n-i,1^{i})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determine all p $\in \mathbb{R}$ such that the following integral converges The integral is:
$$\int_0^{+\infty} \frac{(\cos(x) - 1)x^2}{x^p + (x + 1)^6} dx$$
Now, what I did was this:
$$\frac{(\cos(x) - 1)x^2}{x^p + (x + 1)^6} \leq \frac{\cos(x)x^2}{x^p + (x + 1)^6} \leq \frac{\cos(x)x^2}{x^p + x^6} \leq \frac{x^2}{x^p + x^6} \leq \frac{x^2}{x^p}$$
And $\int_1^{+\infty} \frac{x^2}{x^p} dx$ converges when $p > 3$ because it is a $p$-integral. Is this approach right?
Edit: I only have to know what to do with $\int_0^1$
| Since
$$ \left|\int_0^{\infty}f(x)\ dx\right| \le \int_0^{\infty} |f(x)|\ dx $$
and $|\cos x-1| = 1-\cos x$, we consider
$$ \int_0^\infty \frac{(1-\cos x)x^2}{x^p + (x+1)^6}\ dx \le \int_0^\infty \frac{2x^2}{x^p + (1+x)^6}\ dx \le \int_0^1 2x^2 + \int_1^\infty \frac{2x^2}{x^p + x^6} $$
The first term clearly converges, so let's look at the other one
If $p > 6$
$$ \int_1^\infty \frac{2x^2}{x^p+x^6} \le \int_1^\infty \frac{x^2}{x^p}\ dx $$
which always converges
If $p \le 6$
$$ \int_1^\infty \frac{2x^2}{x^p+x^6} \le \int_1^\infty \frac{x^2}{x^6}\ dx $$
which always converges
So the integral converges for $\forall p$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding PDF of $Y = 1 - X^2$, piecewise monotonic Have been banging my head against this question from Casella and Berger for some time, not understanding where my gap in knowledge is here:
Given:
$f_X(x)=\frac38 (x+1)^2 \qquad Y = 1-X^2 $
Find the pdf of $Y$, seems simple enough...
Method: Given that $g(x)$ is not monotonic, we split it into two monotonic partitions, ignoring $A_0$:
$A_1 = (-1,0) \qquad g_1(x) = 1 - x^2 \qquad g^{-1}_1(y)=-\sqrt{1-y}$
$A_2= (0,1) \qquad\ \ \ g_2(x)=1-x^2 \qquad g^{-1}_2(y)=\sqrt{1-y}$
Using the theorem:
$f_Y(y) = \sum f_Xg^{-1}_i(y) \left\lvert {\frac{d}{dy}g^{-1}_i(y)} \right\rvert$
We have:
$f_Y(y) = \frac38(\sqrt{(1-y)}+1)^2
\left\lvert {\frac{1}{2\sqrt{1-y}}}\right\rvert
+
\frac38(-\sqrt{(1-y)}+1)^2
\left\lvert {-\frac{1}{2\sqrt{1-y}}}\right\rvert
$
Simplifying:
$f_Y(y) = \frac{3}{16\sqrt{1-y}}\left[ (\sqrt{1-y} + 1)^2 + (-\sqrt{1-y}+1)^2\right]$
$f_Y(y) = \frac{3}{16\sqrt{1-y}}\left[
1 - y +2\sqrt{1-y}+1 + 1-y - 2\sqrt{1-y}+1\right]$
$f_Y(y) = \frac{3}{16\sqrt{1-y}}\left[
4 - 2y\right]$
$f_Y(y) = \frac{3}{8\sqrt{1-y}}\left[
2 - y\right]$
Unfortunately it doesn't match the answer:
$f_Y(y)=\frac38(1-y)^{-\frac{1}{2}} + \frac38(1-y)^{\frac{1}{2}}$
I also tried doing this using the CDF of X, which is where the theorem comes from but giving it a shot anyway: $P(Y \le y) = P(-\sqrt{1-y} \le X \le \sqrt{1-y}) = F_X(\sqrt{1-y}) - F_X(-\sqrt{1-y})$
Then taking the derivative, but that gave me the same end result... Where am I going wrong?
EDIT: Forgot support of X, $-1 \lt x \lt 1$
| $$\frac{2-y}{\sqrt{1-y}} = \frac{1}{\sqrt{1-y}} + \frac{1-y}{\sqrt{1-y}} = \frac{1}{\sqrt{1-y}} + \sqrt{1-y}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An interesting integral with log: $\int_{0}^{\pi/2}\ln^2\left(\tan^2\left({x\over 2}\right)\right)\mathrm dx=\frac{\pi^3}2$
$$\int_{0}^{\pi/2}\ln^2\left(\tan^2\left({x\over 2}\right)\right)\mathrm dx=\frac{\pi^3}2\tag1$$
This is an interesting integral, I came aross it while I was solving a circuit board problem. My professor as me to find the current of the parallel circuit but but eventually I came up this is equation
Any hints how to tackle this problem.
I was thinking of frey-man trick but it prove too difficult.
| $\newcommand{\Log}{\operatorname{Log}}$First set $x=2\arctan(t)$ so that:
\begin{align}
I:=\int^{\pi/4}_0 \ln^2\left(\tan^2\left(\frac x 2\right)\right)\,dx=8\int^1_0 \frac{\ln^2(t)}{1+t^2}\, dt
\end{align}
Now we set $t=\frac{1}{u}$ so that:
\begin{align}
I=-8\int^1_\infty \frac{\ln^2(u)}{1+u^2}\,du = 8\int^\infty_1 \frac{\ln^2(u)}{1+u^2}\,du
\end{align}
Putting these things together yields:
\begin{align}
I = 4 \int^\infty_0 \frac{\ln^2(u)}{1+u^2}\,du
\end{align}
Now let me be stubborn and integrate a slightly different thing, namely:
\begin{align}
\int_{K_R} \frac{\Log^3(-z)}{1+z^2}\,dz
\end{align}
Where $\Log(\cdot)$ is the Principal Value of the logarithm and $K_R$ is the following contour consisting of $C_R$, $C_R^+$ and $C_R^-$ that can be seen in the picture below.
Let $R>1$, by The Residue Theorem we have:
\begin{align}
\int_{K_R} \frac{\Log^3(-z)}{1+z^2}\,dz = 2\pi i \left(\frac{\Log^3(-i)}{2i} +\frac{\Log^3(i)}{-2i}\right)=\frac{i\pi^4}{4}
\end{align}
That means:
\begin{align}
\int_{C_R} + \int_{C_R^-} +\int_{C_R^+} \frac{\Log^3(-z)}{1+z^2}\,dz = \frac{i\pi^4}{4}
\end{align}
When $R\to \infty$ we have the contribution from $C_R$ goes to zero ($\star$). On $C_R^+$ we have:
\begin{align}
\lim_{R\to\infty}\int_{C_R^+}\frac{\Log^3(-z)}{1+z^2}\,dz = \int^\infty_0 \frac{\ln^3(t)-3i\pi\ln^2(t)-3\pi^2\ln(t)+i\pi^3}{1+t^2}\,dt
\end{align}
On $C_R^-$ we have:
\begin{align}
\lim_{R\to\infty}\int_{C_R^-}\frac{\Log^3(-z)}{1+z^2}\,dz = -\int^\infty_0 \frac{\ln^3(t)+3i\pi\ln^2(t)-3\pi^2\ln(t)-i\pi^3}{1+t^2}\,dt
\end{align}
Now you understand why I was being stubborn because putting these things together yields something that looks like $I$, namely:
\begin{align}
\int^\infty_0 \frac{-6i\pi\ln^2(t)+2i\pi^3}{1+t^2}\,dt =\frac{i\pi^4}{4}
\end{align}
Hence:
\begin{align}
-6\int^\infty_0 \frac{\ln^2(t)}{1+t^2} \,dt =\frac{\pi^3}{4}-2\pi^2\int^\infty_0 \frac{1}{1+t^2}\,dt
\end{align}
On the RHS we have a popular integral which evaluates to $\pi/2$ (how?) so we get:
\begin{align}
-6\int^\infty_0 \frac{\ln^2(t)}{1+t^2}\,dt = -\frac{3\pi^3}{4}
\end{align}
That means:
\begin{align}
\int^\infty_0 \frac{\ln^2(t)}{1+t^2}\,dt = \frac{\pi^3}{8}
\end{align}
So finally:
\begin{align}
\int^{\pi/4}_0 \ln^2\left(\tan^2\left(\frac x 2\right)\right)\,dx = 4\int^\infty_0 \frac{\ln^2(t)}{1+t^2}\,dt = \frac{\pi^3}{2}
\end{align}
($\star$) We actually have some small part of a circle with radius $\epsilon$ in the origin, but that circle also have no contribution when $\epsilon\to 0$ using the ML-lemma.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ I am trying to find the derivative of $f\left(x\right)=\arctan\left(\sqrt{\frac{1+x}{1-x}}\right)$ by only using the formula $\arctan\left(u\left(x\right)\right)'=\frac{u'\left(x\right)}{u\left(x\right)^2+1}$. I don't honestly understand this formula, here are however my calculations. Is this the right way to find the derivative? And how do I proceed?
$$\LARGE
\frac{\frac{1}{2\sqrt{\frac{1+x}{1-x}}}\cdot \frac{\left(1-x\right)-\left(1+x\right)}{x^2-2x+1}}{\frac{x+1}{1-x}+1}
$$
I also appreciate if you can explain what that "formula" exactly is.
| Well done! Your answer is correct. To make it a bit more presentable, let us simplify this further to get: $$f’(x) = \frac{\frac{\sqrt{\frac{1-x}{1+x}}}{2}\times \frac{-2x}{x^2-2x+1}}{2}(1-x)$$ $$=-\frac{(1-x)^{\frac32}}{2\sqrt{1+x}(x^2-2x+1)}$$ $$=-\frac{1}{2\sqrt{1-x^2}}$$
EDIT:
Well, you have your derivative calculated wrong, it should be $$\frac{(1-x)-(1+x)(-1)}{x^2-2x+1} = \frac{2}{x^2-2x+1}$$
So, removing the minus sign, we have: $$f’(x) = \frac{1}{2\sqrt{1-x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Using the Left-Inverse to "Solve" an Impossible System of Equations I was working with the following system of equations:
$$\begin{split}
\begin{bmatrix}
4 & 0\\
0 & 5\\
0 & 0\\
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
\end{bmatrix}
& =
\begin{bmatrix}
1\\
1\\
1\\
\end{bmatrix}
\end{split}
$$
Clearly, this has no solution on account of the last rows in the coefficient and solution matrices.
However, multiplying by the left-inverse of the coefficient matrix seems to imply a solution:
$$\begin{split}
\begin{bmatrix}
\frac{1}{4} & 0 & b_{13}\\
0 & \frac{1}{5} & b_{23}\\
\end{bmatrix}
\begin{bmatrix}
4 & 0\\
0 & 5\\
0 & 0\\
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
\end{bmatrix}
& =
\begin{bmatrix}
\frac{1}{4} & 0 & b_{13}\\
0 & \frac{1}{5} & b_{23}\\
\end{bmatrix}
\begin{bmatrix}
1\\
1\\
1\\
\end{bmatrix} \\
\begin{bmatrix}
x_1\\
x_2\\
\end{bmatrix}
& =
\begin{bmatrix}
\frac{1}{4} + b_{13}\\
\frac{1}{5} + b_{23}\\
\end{bmatrix}
\end{split}
$$
Where $b_{13}$ and $b_{23}$ can be any number. As stated above, there is no solution to this system, so any solution obtained by the above method is wrong.
What I am not understanding is why, after applying the rules of matrix multiplication, does it seem possible that there is a solution? Did I make a mistake somewhere in the process, and/or is there some fundamental nuance of linear algebra that I'm missing here?
| Here is a simpler example of what you are doing. Consider the argument
\begin{align}
2 \times 3 &= 8 \\
\frac 12 \times 2 \times 3 &= \frac 12 \times 8 \\
3 &= 4
\end{align}
If you put $``="$ between two quantities that aren't equal, then you cannot believe any conclusions that may follow.
\begin{equation}
\begin{split}
\begin{bmatrix}
4 & 0\\
0 & 5\\
0 & 0\\
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
\end{bmatrix}
& =
\begin{bmatrix}
1\\
1\\
1\\
\end{bmatrix}
\end{split}
\end{equation}
Is equivalent to
\begin{equation}
\begin{split}
\begin{bmatrix}
4x_1\\
5x_2\\
0\\
\end{bmatrix}
& =
\begin{bmatrix}
1\\
1\\
1\\
\end{bmatrix}
\end{split}
\end{equation}
Which clearly has no solution.
Geometrically, the point $(1,1,1)$ is not on the plane spanned by the vectors $(4,0,0)$ and $(0,5,0)$.
Equations like this occur all of the time in statistics. Their, $``$solution$"$ would be to add an error term on the right end
$$\left[
\begin{array}{c}
4 & 0\\
0 & 5\\
0 & 0\\
\end{array}
\right]
\left[
\begin{array}{c}
x_1\\
x_2\\
\end{array}
\right]
=
\left[
\begin{array}{c}
1\\
1\\
1\\
\end{array}
\right]
+
\left[
\begin{array}{c}
\epsilon_1\\
\epsilon_2\\
\epsilon_3\\
\end{array}
\right]$$
and then find the values of $x_1$ and $x_2$ that will minimize the sum of the squares of the error: $\epsilon_1^2 + \epsilon_2^2 + \epsilon_3^2$
Geometrically, that would be the perpendicular projection of the point
$(1,1,1)$ onto the plane spanned by the vectors $(4,0,0)$ and $(0,5,0)$.
It turns out that, even if $Ax = b$ has no solution, $A^TAx = A^Tb$ does have a solution, namely $\hat x = \left(A^TA \right)^{-1}A^Tb$, and $\hat x$ minimizes the sum of the squares of the errors, $\|b- A\hat x \|^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 7,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.