Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Simplify $(\sqrt{x}) + x + 2 = (\sqrt{y}) + y + 2$ $\sqrt{x} + x + 2 = \sqrt{y} + y + 2$
I've simplified as follows:
$\sqrt{x} + x = \sqrt{y} + y$, square both sides
$x + x^2 = y + y^2$
It seems obvious that $x = y$ but I can’t get to that solution by algebraic means.
| both $\sqrt{x}$ and $\sqrt{y}$ are roots of a quadratic equation:
$$
t^2+t + c=0
$$
hence either $\sqrt{x}=\sqrt{y}$ or $\sqrt{x}+\sqrt{y}=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$. I tried it this way:
$$ 40\cosθ+9\sinθ=41 $$
$$ 9\sinθ=41-40\cos\theta $$
Squaring both the sides:
$$81\sin^2\theta=1681+1600\cos^2\theta-2\cdot 40\cdot 41 \cos\theta$$
$$81-81 \cos^2\theta= 1681+1600\cos^2\theta-3280 \cos\theta$$
$$1681\cos^2\theta-3280\cos\theta+1600=0$$
Solving the quadratic equation gives $\cos\theta=\dfrac{40}{41}$. It is not easy to solve the quadratic equation without calculator so there must be some other method, if yes then please explain.
P.S: I've found the other method so I am self-answering the question.
| I noted that $9,40$ and $41$ are pythagorean triple, i.e $9^2+40^2=41^2$. This fact can be used to solve the question easily. I'll show the solution for the general case:
$a\cos\theta+b\sin\theta=c,$ given that $a^2+b^2=c^2$
$$a\cos\theta+b\sin\theta=c$$
$$b\sin\theta=c-a\cos\theta$$
Squaring both the sides:
$$b^2\sin^2\theta=c^2+a^2\cos^2\theta-2ca\cos\theta$$
Substituting $c^2$ with $a^2+b^2$ gives:
$$b^2\sin^2\theta=a^2+b^2+a^2\cos^2\theta-2ca\cos\theta$$
$$a^2+b^2\cos^2\theta+a^2\cos^2\theta-2ca\cos\theta=0$$
$$a^2+(a^2+b^2)\cos^2\theta-2ca\cos\theta=0$$
$$a^2+(c\cos\theta)^2-2a(c\cos\theta)=0$$
$$(a-c\cos\theta)^2=0$$
$$\cos\theta=\dfrac{a}{c}$$
Substituting $c=41$ and $a=40$ will give the solution of the particular equation discussed in the question.
An interesting thing is that $\sin\theta=\sqrt{1-\dfrac{a^2}{c^2}}=\dfrac{b}{c}, $ , i.e the equation actually belongs to a right angled triangle having $a$ as its base, $b$ as its prependicular and $c$ as its hypotenuse.
There is another method in which we divide both sides of the equation with $\sqrt{a^2+b^2}$ and then use the identity:
$\sin (A+B)=\sin A \cos B+\cos A \cos B$
This method will give the same results.
Converse
Consider a right angled triangle having base $a$, perpendicular $b$ and hypotenuse $c$. We have:
$$\cos\theta=\dfrac{a}{c}\ \ \text{and}\ \ \ \sin\theta=\dfrac{b}{c}$$
$$\implies a=c\cos\theta \ \ \ \text{and}\ \ \ b=c\sin\theta$$
By the pythagoras theorem:
$$a^2+b^2=c^2 $$
$$\implies\ c^2\cos^2\theta+c^2\sin^2\theta=c^2$$
$$\implies c\cos\theta \cdot c\cos\theta + c\sin\theta \cdot c\sin\theta=c^2$$
$$\implies a\cos\theta+b\sin\theta=c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Why is $2^{n/2}(5/3)^n(\cos(n\pi/4)+\sin(n\pi/4))$ an alternate form of the complex number $(5/3)^n(1+i)^n$? How do I get to that point?
I am aware of the formula
$$z = r (\cos \alpha + i \sin \alpha)$$
and that
$$z^n = r^n(\cos n\alpha+ i\sin n\alpha)$$
But I don't know how to get to what is in the title, particularly why is the angle $n\pi /4$?
| Let
$$z = 1+i = \sqrt{2}\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right) = 2^{1/2}\left(\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\right)$$
Now use de Moivre:
$$z^n = 2^{n/2}\left(\cos\left(\frac{n\pi}{4}\right)+i\sin\left(\frac{n\pi}{4}\right)\right)$$
And finally:
$$\left(\frac{5}{3}\right)^n (1+i)^n = \left(\frac{5}{3}\right)^n z^n = 2^{n/2} \left(\frac{5}{3}\right)^n \left(\cos\left(\frac{n\pi}{4}\right)+i\sin\left(\frac{n\pi}{4}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let a,b $\in$ $\mathbb{R}.$ Show that $a^4+b^4+8\ge 8ab.$
Let $a,b \in \mathbb{R}.$ Show that $a^4+b^4+8\ge 8ab.$
The question is from the inequalities section of An Excursion in mathematics by Bhaskaraycharya Pratisthanan. My heuristics include using the AM-GM inequality. I am unable to design the problem to proceed further.
Please provide me some hints.
| You're on right path just a small trick using AM-GM
Write $$a^4+b^4+8=a^4+b^4+4+4$$
And then apply AM-GM
$$\frac{a^4+b^4+4+4}{4}\ge \sqrt[4]{16a^4b^4}$$
$$\frac{a^4+b^4+4+4}{4}\ge 2ab$$
$${a^4+b^4+8}\ge 8ab$$
Or You can just SOS it
$$(a^2-b^2)^2\ge0\implies a^4+b^4-2a^2b^2\ge0$$
$$2(ab-2)^2\ge0\implies2a^2b^2+8-8ab\ge0$$
Add above two inequalities to get
$$ a^4+b^4+8\ge8ab$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove $\forall n\in\mathbb{N}, \exists m\in\mathbb{N}; n=\pm1^2\pm2^2\pm\cdots\pm m^2.$ And we choose the positive and negative signs in a way that the equation becomes true.
I think it can be proved with mathematical induction. So here's how I begin:
For $n=1$, $1=+1^2$ which is true. It's also true for 2 and 3, e.g. $2=-1^2-2^2-3^2+4^2$.
Let it be true for $n$ and now I need to show it holds for $n+1$. I've got some ideas, but are a bit fuzzy. Do you have any ideas to prove the statement with or without using mathematical induction?
|
Lemma A. We may represent $0,1,2,3$.
$$\begin{eqnarray*}0 &=& 1^2 + 2^2 - 3^2 + 4^2 - 5^2 - 6^2 + 7^2\\ 1 &=& 1^2\\2&=&-1^2-2^2-3^2+4^2\\3&=&-1^2+2^2.\end{eqnarray*}$$
$\phantom{}$
Lemma B. If we are able to represent some $n$, we may represent $n+4$ as well.
Since for any $m$ we have $(m+1)^2-(m+2)^2-(m+3)^2+(m+4)^2=4$, we may simply take the representation of $n$ and, assuming it ends with $\pm m^2$, and append to it $+(m+1)^2-(m+2)^2-(m+3)^2+(m+4)^2$, getting a representation of $n+4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sqrt [3]{2}-\sqrt [3]{4}$ is algebraic How do I show, step by step, that $\sqrt [3]{2}-\sqrt [3]{4}$ is a root of $x^3+6x+2$?
Start with $x=\sqrt [3]{2}-\sqrt [3]{4}$ do not use the cubic, the cubic is given for convenience.
( This is example 4.1.3 from Introductory ANT by Alaca/Williams )
| Let $a = 2^{1/3}$ and $b = 4^{1/3} = (2^2)^{1/3} = (2^{1/3})^2 = a^2$. Then $$(a-b)^3 = a^3 - b^3 - 3ab(a-b) = a^3 - b^3 - 3a^3(a-b),$$ where upon substituting values we obtain $$(a-b)^3 = 2-4 - 3(2)(a-b) = -2 - 6(a-b).$$ Thus if $x = a-b = 2^{1/3} - 4^{1/3}$, then $$x^3 + 6x + 2 = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)} $ Let $x,y,z$ be real numbers in the range of $(0,1)$.
Prove that
$$\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)}.$$
| \begin{eqnarray}
&&[\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)}][xyz+(1-x)(1-y)(1-z)]\\
&=&[\frac{yz}{1-y}+\frac{(1-x)(1-z)}{x}]+[\frac{xz}{1-z}+\frac{(1-x)(1-y)}{y}]+[\frac{xy}{1-x}+\frac{(1-y)(1-z)}{z}]\\
&\ge&[\frac{z}{1-y}+\frac{1-z}{x}-1]+[\frac{x}{1-z}+\frac{1-x}{y}-1]+[\frac{y}{1-x}+\frac{1-y}{z}-1]\\
&\geq& 6\sqrt[6]{\frac{z}{1-y}\cdot\frac{1-z}{x}\cdot\frac{x}{1-z}\cdot\frac{1-x}{y}\cdot\frac{y}{1-x}\cdot\frac{1-y}{z}}-3=3
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
convergence series geometric test Prove if this converges:
$$\sum_{n=1}^\infty \frac{2^n+3}{3^n-1}$$
pf: using geometric
$$0 < \frac{2^n+3}{3^n-1} \leq \frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}} = \cdots $$ and so on
I know how to do the rest but my question is that where in the world did my teacher get
$$\frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}}$$
| You teacher increased the numerator (from $2^n+3$ to $2^n+ 2\cdot 2^n$, since $3\leq 2\cdot 2^n$), and decreased the denominator (from $3^n-1$ to $3^n-3^n/2$, since $1<3^n/2$), thus obtains a fraction (i.e. $\frac{2^n+2\cdot 2^n}{3^n-3^n/2}$) which is bigger than the original (i.e. $\frac{2^n+3}{3^n-1}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove or disprove: $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$ Is it true that $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$ for any positive integer $n$?
We cannot prove the statement using induction as it is, because the left-hand side is increasing while the right-hand side stays constant. So we need to modify the right-hand side to something like $3-\dfrac{1}{n}$. But it is hard to quantify how much the left-hand side increases when we go from $n$ to $n+1$.
| We want to show that
$$\sum_{k=1}^{\infty} \dfrac{\ln(k+1)}{2^{k}} < \ln(3)$$
We have $\ln(1+k) \leq k$ for all $k \in \mathbb{N}$. Hence, we have
$$\sum_{k=1}^{\infty} \dfrac{\ln(k+1)}{2^k} = \sum_{k=1}^{10} \dfrac{\ln(k+1)}{2^k} + \sum_{k=11}^{\infty} \dfrac{\ln(k+1)}{2^k} \leq \sum_{k=1}^{10} \dfrac{\ln(k+1)}{2^k} + \sum_{k=11}^{\infty} \dfrac{k}{2^k}$$
$$\sum_{k=1}^{10} \dfrac{\ln(k+1)}{2^k} \approx 1.013 < 1.02$$
$$\sum_{k=11}^{\infty} \dfrac{k}{2^k} = \dfrac3{256} = 0.01171875$$
Hence, the sum is less than $1.04$, which is less than $\ln(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How do you find the imaginary roots of a fourth degree polynomial that cannot be simplified? I started out with $f(x)=16x^6-1$, and I got it down to $64x^4+16x^2+4$ by synthetically dividing by roots $0.5$ and $-0.5$ How should I continue in order to find the other roots?
| Note that $f\left(\pm \frac{1}{2}\right) = 16\left(\pm \frac{1}{2}\right)^6 - 1 = -\frac{3}{4}$, so $\pm \frac{1}{2}$ are not roots of $f$.
One way to proceed with the original problem is to write $v^6 := 16 x^6$, that is, $v = \sqrt[3]{4} x$, so that the original polynomial is $v^6 - 1$. This clearly has roots $v = \pm 1$, which (substituting) gives corresponds to
$$x = \pm 4^{-1/3}.$$ Using synthetic division gives
$$v^6 - 1 = (v - 1)(v + 1)(v^4 + v^2 + 1).$$
One can find the roots of the latter by writing $w = v^2$ and applying the quadratic formula to the resulting expression $w^2 + w + 1$; this only has complex roots, $w = -\frac{1}{2} \pm \frac{\sqrt{3}}{2} i,$ and we can then solve readily for $v$ and then $x$.
We can also attempt to factor $v^4 + v^2 + 1$. By the Rational Root Theorem it has no roots in $\mathbb{Q}$, so if it has a nontrivial factorization, it must be into a product of quadratics. One can set this up manually---writing down a product of general quadratics, distributing, and equating coefficients---and find the factorization
$$(v^2 + v + 1)(v^2 - v + 1).$$
(On the other hand, if one knows about the cyclotomic polynomials $\Phi_n$, one can write down the factorization of any polynomial $v^m - 1$ as a product of such; in our case,
$$v^6 - 1 = \Phi_1(v) \Phi_2(v) \Phi_3(v) \Phi_6(v) = (v - 1) (v + 1) (v^2 + v + 1) (v^2 - v + 1).)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
The inequality $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+3/4 >0$ holds for all $x\in\mathbb R$
Show
$\forall \ x \in \mathbb{R}:\quad x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+\dfrac{3}{4}>0$
My attemps:
Case $x=-1$
That is true for this case
Then for $x \neq - 1$:
$$\dfrac 3 4 - x + x^2 - x^3 + x^4 - x^5 + x^6 = \dfrac{1 + x^7}{1 + x} - \dfrac 1 4$$
let $g(x)=\dfrac{1 + x^{7}}{1 + x}-\dfrac{1}{4} \quad \forall x\in \mathbb{R} \backslash \{-1\}$
then
$
\begin{align*}
g'(x)&=\dfrac {(1+x^7)'(1+x)-(1+x)'(1+x^7)}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}\\
g'(x)&=\dfrac {(7 x^6)(1+x)-(1+x^7)}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}\\
g'(x)&=\dfrac {7 x^6+7 x^7-1-x^7)}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}\\
g'(x)&=\dfrac{7 x^6+6 x^7-1}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}
\end{align*}
$
To determine the sign of the numerator $(7 x^6+6 x^7-1)$
once time let :$ h(x)=7 x^6+6 x^7-1$
then
$$h'(x)=42x^5+42x^6=42(1+x)x^5$$
$$h'(x)=0 \Longleftrightarrow x=-1 \text{or} x=0$$
thus $h$ admits a minimum on the point $x=0$
and a maximum on the point $x=-1$
or $h(-1)=0$ and $h(0)=-1$
$\lim_{x\to -\infty}h(x)=-\infty$ and $\lim_{x\to +\infty}h(x)=+\infty$
as $h(-1)=0$ and $h(0)=-1$ by Intermediate value theorem there is $u \in(-1, 0)$ such that h(u) = 0.
i'm stuck here
*
*am i on my way ?
*is there any other ways to solve it
| Another way would be to prove $(1 + x^7) - {1 \over 4}(1 + x) > 0$ for $x > -1$, and $(1 + x^7) - {1 \over 4}(1 + x) < 0$ for $x < -1$, using calculus on $f(x) = (1 + x^7) - {1 \over 4}(1 + x) = x^7 - {1 \over 4}x + {3 \over 4}$. Since $\lim_{x \rightarrow \infty} f(x) = \infty$ and $\lim_{x \rightarrow -\infty} f(x) = -\infty$, we need show the minimum of $f(x)$ for $x > -1$ is positive and the maximum of $f(x)$ for $x < -1$ is negative.
One has $f'(x) = 7x^6 - {1 \over 4}$, which is equal to zero at $x = \pm (28)^{-{1 \over 6}}$, both of which are greater than $-1$. Hence $f$ is increasing for $x > -1$, so since $f(-1) = 0$, this gives the $x < -1$ part.
As for $x > -1$, note that $f''(x) = 28x^5$, which is positive at $x = (28)^{-{1 \over 6}}$ and negative at $-(28)^{-{1 \over 6}}$. So $f$ has a local minimum at $x = (28)^{-{1 \over 6}}$ and a local maximum at $x = -(28)^{-{1 \over 6}}$. So we just have to plug in the values $x = -1$ (the left endpoint) and $x = (28)^{-{1 \over 6}}$ (the minimum). We saw $f(-1) = 0$, and plugging in $x = (28)^{-{1 \over 6}}$ gives $(28)^{-{7 \over 6}} - {1 \over 4} (28)^{-{1 \over 6}} + {3 \over 4} = .62..$ which is positive. So $f(x)$ is positive for $x > -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods can have solution of $x^4-3x^3+2x^2-3x+1=0$ using only high school methods???
i only know quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ i tried many algebraic manipulations and i get $(x^2+1)^2=3(x^3+x)$, so can we have solution using only high school methods? i guess i have to do an algebraic substitution to reduce $x^3+x$ to polynomial of degree 2?? i also know $$(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1r_2$$ where $r_1,r_2$ are roots of polynomial of degree 2
| Note that the polynomial can be written as
$$x^4+2x^2+1 - 3x(x^2+1) = (x^2+1)^2 - 3x(x^2+1) = (x^2+1)(x^2-3x+1)$$
I trust you can finish the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
$8x +9y = 5$ where $x,y \in \mathbb{Z}$
Solve the following Diophantine equation algebaically: $$8x+9y=5$$
Give 3 possible solutions for the equation
I have the following:
The Diophantine equation has solutions $x,y \iff 8x=5\mod{9}$ has a solution $x \equiv\mod{9}$
Since $\gcd(8,9)=1$, by Bezout's Lemma, for $r,t \in \mathbb{Z}, \gcd(8,9)=1=r(8)+t(9)$ and $x\equiv r(5)\mod{9}$ is a solution for the linear congruence above.
By Euclid's algorithm for determining $\gcd(8,9)$ we have
\begin{align}9 &= 1(8) +1 \\ 8 &=9(1)+0\end{align}
so $1=(-1)8 + 1(9)$ and $r=-1 \implies x \equiv(-1)5\mod{9}$.
Now \begin{align}[-5]_9 &= \{-5 + 9k \ | k\in\mathbb{Z} \} \\ &= \{ ..., -5,4,13,... \} \\ &=[4]_9\end{align}
$\therefore x \equiv 4 \mod{9}$, that is $x=4+9k$ for all $k \in \mathbb{Z}$ upon which it can be seen that $y= -3 -8k$.
Is this correct?
| $8x+9y=5\iff8~(x+y)+y=5=8\cdot0+5\iff x+y=0$ and $y=5\iff x=-5$. Then all numbers of the form $x=-5-9k$ and $y=5+8k$ are solutions to the above equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$ How does one evaluate the following integral?
$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$
This is a homework problem and I have been evaluating this integral for hours yet no success so far. I have tried to rationalize the integrand by multiplying it with
$$\frac{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}$$
but the integrand is getting worse. I have tried to use trigonometric substitutions like $x=\tan\theta$ and $x=\sqrt{2}\sin\theta$, but I cannot rid off the square root form. I have also tried to use hyperbolic trigonometric substitutions but the thing does not get any easier neither also substitution $y=x^2$ nor $y=\sqrt{(2-x^2)(1+x^2)}$. Using integration by parts is almost impossible for this one. I have also tried to use the tricks from this thread, but still did not get anything. No clue is given. My professor said, we must use clever substitutions but I cannot find them. Any idea or hint? Any help would be appreciated. Thanks in advance.
Edit :
The answer I got from my Prof is $\dfrac{3-2\sqrt{2}}{6}$.
| It's easy. Differentiate
$$\frac{1}{4 (-1 + 2 x^2)} - \frac{\sqrt{2 + x^2 - x^4}}{6 (-1 + 2 x^2)} -
{1\over 4} \log[3 + 2 \sqrt{2 + x^2 - x^4}]$$
To get your integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 6
} |
Prove that $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2>ab+bc+ca$ How can I prove $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2>ab+bc+ca$?
We have
$(a+b-c)^2+=a^2+b^2+b^2+2ab-2bc-2ca$,
$(b+c-a)^2+=b^2+c^2+a^2+2bc-2ca-2ab$
$(c+a-b)^2+=c^2+a^2+b^2+2ca-2ab-2bc$
But I don't know how to show the required result?
Please give hint. Thank you
| this inequality is equivalent to $a^2+b^2+c^2\geq ab+bc+ca$ which is
$(a-b)^2+(b-c)^2+(c-a)^2\geq 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
dividing polynomials using long division I'm not following logic of using long division on polynomials.
If you are using regular long division, we would do the following:
+----------
2 | 86
How many times does 2 go into 8? 4. Multiply 2 by 4 and drop down the next tenth power. So we end up with 6.
4
+----------
2 | 86
8
-----------
6
How many times does 2 go into 6. 3. So we add that to the next tenth place, and drop it down and continue doing this until we get 0 or a remainder.
43
+----------
2 | 86
8
-----------
6
6
-----------
0
Now I am asked to do the same thing but with polynomials.
+----------
x + 3 | x^2 + 10x + 21
In order to solve this, we are told to determine how many times does x go into x^2. The answer is x because x*x=x^2. Then we are told to multiply that x by the whole divisor x + 3, so we get a result that looks like this:
x
+---------------
x + 3 | x^2 + 10x + 21
x^2 + 3x
----------------
7x + 21
And then again we ONLY check how many times x goes into 7x, which is 7 times, so our result looks like this:
x + 7
+---------------
x + 3 | x^2 + 10x + 21
x^2 + 3x
----------------
7x + 21
7x + 21
----------------
0
In the polynomial division, we are only checking to see how many times x goes into x^2. Shouldn't we be checking how many times x + 3 goes into x^2 + 10x? After all, if this was arithmetic, we don't break up the divisor into pieces, like how we are doing with the polynomials. If this was arithmetic, if we couldn't divide it, then we add a decimal point and a zero until we could. How come in polynomials, we are allowed to break up the divisor in pieces like this?
| Let's take more easily comparable examples such as $\dfrac{156}{13}$ and $\dfrac{x^2+5x+6}{x+3}$
So for the numerical example, you first subtract $10\times 13$ and then $2 \times 13$ to get $156 = (10+2)\times 13$ so $\dfrac{156}{13}=12$, or laying it out
10+2
+----------
13 | 156
130
-----------
26
26
-----------
0
while for the polynomial
x + 2
+-------------
x+3 | x^2 + 5x + 6
x^2 + 3x
--------------
2x + 6
2x + 6
--------------
0
and the correspondence between $156 = (10+2)\times 13$ and $x^2+5x+6=(x+2)(x+3)$ is clear (though numerically the usual practice is write $12$ rather than $10+2$)
Now let's consider the slightly different $\dfrac{x^2+x-6}{x+3}$, to answer your question about why we only look at the leading terms: $x$ divides $x^2$ to give $x$, but $x+3$ would divide $x^2+x$ to give less, so when we do the subtraction $(x^2+x-6)-(x^2+3x)$ we get the apparently negative remainder $-2x-6$. No matter: we can still match leading terms and $x$ divides $-2x$ to give $-2$. This time you get
x - 2
+-------------
x+3 | x^2 + x - 6
x^2 + 3x
--------------
-2x - 6
2x + 6
--------------
0
and indeed $x^2 + x - 6 = (x-2)(x+3)$. In polynomial long division, each step removes the leading term of the remainder from the previous step, and that only requires comparison of the leading terms of that remainder with the leading term of the divisor
So long as there is a final remainder of $0$, this polynomial division approach will work, since you are subtracting and recording multiples of the divisor at each step (in these examples $x+3$). If there is a final non-zero remainder, that can be part of the result too: compare $\frac{159}{13}=12 + \frac{3}{13}$ with $\frac{x^2+5x+9}{x+3}=x+2+\frac{3}{x+3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\sum_{n = 1}^{\infty} \frac{2}{2^{n}}$
Evaluate $$\sum_{n = 1}^{\infty} \frac{2}{2^{n}}$$
This is a geometric series and since $a = \dfrac{1}{2}$ Then the infinite sum is jsut $S = \dfrac{1}{1-\frac{1}{2}} = 2$ Then I multiply by $2$ to get $4$ right? But the actual answer should just be $2$. Am I missing something?
| $$\sum\limits_{n = 1}^\infty {\frac{2}{{{2^n}}}} = 2\sum\limits_{n = 1}^\infty {{{\left( {\frac{1}
{2}} \right)}^n}} = 2.\frac{{1/2}}
{{1 - 1/2}} = 2.1 = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Pair of die and probability A pair of dice is loaded. The probability that a 2 appears on the first
die is 3/13 and the probability that a 4 appears on the second die is
3/13. Other outcomes for each die appear with probability 2/13. What
is the probability of 6 appearing as the sum of the numbers when the
two dice are rolled?
| Add up the following:
*
*Probability of $1$ on 1st dice and $5$ on 2nd dice: $\frac{2}{13}\cdot\frac{2}{13}$
*Probability of $2$ on 1st dice and $4$ on 2nd dice: $\frac{3}{13}\cdot\frac{3}{13}$
*Probability of $3$ on 1st dice and $3$ on 2nd dice: $\frac{2}{13}\cdot\frac{2}{13}$
*Probability of $4$ on 1st dice and $2$ on 2nd dice: $\frac{2}{13}\cdot\frac{2}{13}$
*Probability of $5$ on 1st dice and $1$ on 2nd dice: $\frac{2}{13}\cdot\frac{2}{13}$
And you get $\frac{25}{169}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$ \mathop {\lim }\limits_{n \to + \infty } \frac{{v_{n + 1} }}{{v_n }} = 2$ help me please
true or fulse
(1)$$\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1?
$$
\begin{array}{l}
u_n = \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}}\quad ;u_n > 0 \\
\Leftrightarrow \ln \left( {u_n } \right) = \ln \left( {n\left( {n + 1} \right) \cdots \left( {n + n} \right)} \right)^{\frac{1}{n}} \\
\Leftrightarrow \ln \left( {u_n } \right) = \frac{1}{n}\ln \left( {n\left( {n + 1} \right) \cdots \left( {n + n} \right)} \right) \\
\Leftrightarrow \ln \left( {u_n } \right) = \frac{{\ln \left( n \right)}}{n} + \frac{{\ln \left( {n + 1} \right)}}{n} + \cdots + \frac{{\ln \left( {n + n} \right)}}{n} \\
\Leftrightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {\ln \left( {u_n } \right)} \right) = \mathop {\lim }\limits_{n \to + \infty } \left[ {\frac{{\ln \left( n \right)}}{n} + \frac{{\ln \left( {n + 1} \right)}}{n} + \cdots + \frac{{\ln \left( {n + n} \right)}}{n}} \right] = 0 \\
\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1 \\
\end{array}
(2)
$$
\begin{array}{l}
t_n = \frac{{1 \times 3 \times \cdots (2n - 1)}}{{n^n }} \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \frac{{t_{n + 1} }}{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \frac{{1 \times 3 \times \cdots (2n - 1)(2n + 1)}}{{\left( {n + 1} \right)^{n + 1} }} \times \frac{{n^n }}{{1 \times 3 \times \cdots (2n - 1)}} \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \frac{{t_{n + 1} }}{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \frac{{(2n + 1)}}{{\left( {n + 1} \right)^{n + 1} }} \times \frac{{n^n }}{1} \\
= \mathop {\lim }\limits_{n \to + \infty } \left( {2\left( {\frac{n}{{n + 1}}} \right)^{n + 1} + \frac{{n^n }}{{\left( {n + 1} \right)^{n + 1} }}} \right) = 2e^{ - 1} ;\quad \left( {\mathop {\lim }\limits_{n \to + \infty } \left( {1 + \frac{x}{n}} \right)^n = e^x } \right) \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{\frac{{1 \times 3 \times \cdots (2n - 1)}}{{n^n }}}} = 2e^{ - 1} \\
\end{array}
$$
(3) $$\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^2 }}\sqrt[n]{{\frac{{3n!}}{{n!}}}}$$
$$\left( 4 \right)\mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = !?
$$
| $$
\begin{array}{l}
n! = \sqrt {n\pi e} \left( {\frac{n}{e}} \right)^n \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\sqrt {\left( {2n} \right)\pi e} \left( {\frac{{2n}}{e}} \right)^{2n} }}{{\sqrt {\left( {3n} \right)\pi e} \left( {\frac{{3n}}{e}} \right)^{3n} }}}}} \right) \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{4}{{27}}e\sqrt[n]{{\sqrt {\frac{2}{3}} }}} \right) = \frac{4}{{27}}e\quad \\
;\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt[n]{{\sqrt {\frac{2}{3}} }}} \right) = \mathop {\lim }\limits_{n \to + \infty } \left( {\frac{2}{3}} \right)^{\frac{1}{{2n}}} = 1 \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding the generating function of a series with a binomial coefficient and a exponential coefficient So I am given this series
$$2^8, 2^7 \binom{8}{1}, 2^6 \binom{8}{2}, 2^5 \binom{8}{3}, 2^4 \binom{8}{4}, 2^3 \binom{8}{5}, 2^2 \binom{8}{6}, 2^1 \binom{8}{7}, \binom{8}{8}, 0, 0, 0, 0, ...$$
which I converted to the summation
$$\sum_{n=0}^\infty \frac{(2^8 \cdot \binom{8}{n} \cdot x^n)}{2^n} $$
The problem is to find the closed form generating function for this series
I know that the closed form generating function for
$$\sum_{n=0}^\infty \binom{8}{n}x^n = (1+x)^8$$ and
$$\sum_{n=0}^\infty \frac{(2^8 \cdot x^n)}{2^n} = \frac{2^8}{1-x/2}$$
I just can't find anything in my book or online on how to deal with the combination of the two
| $$\sum_{n=0}^\infty \frac{2^8\binom 8n x^n}{2^n}=\sum_{n=0}^8\binom 8n 2^{8-n}x^n
=(2+x)^8\qquad \blacksquare$$
OR
$$(a+x)^N=a^N\left(1+\frac xa\right)^N\\
=a^N\sum_{n=0}^N \binom Nn \left(\frac xa\right)^n\\
=\sum_{n=0}^\infty a^N\binom Nn \left(\frac xa\right)^n$$
Put $a=2, N=8$:
$$(2+x)^8=\sum_{n=0}^\infty 2^8\binom 8n \left(\frac x2\right)^n\qquad \blacksquare$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does $x,y,z>0$ and $x+y+z=1$ imply $\left(1+\frac 1x\right)\left(1+\frac 1y \right)\left(1+\frac 1z \right)\ge 64$? If $x,y,z$ are positive real numbers such that $x+y+z=1$ then is it true that
$\left(1+\dfrac 1x\right)\left(1+\dfrac 1y \right)\left(1+\dfrac 1z \right)\ge 64$ ?
| By inequality between harmonic and geometric mean we have:
$$
\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\ge\left(\frac{3}{\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}}\right)^3
$$
Now if we prove that
$$
\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\le\frac{3}{4}
$$
we are done. But it is the same to prove that
$$
\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{3}{\frac{x+1+y+1+z+1}{3}}=\frac{9}{4}.
$$
The last inequality is the inequality between harmonic and arithmetic mean.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Finding inverse of a function $h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$ I have a function:
$$h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$$
With just pen and paper, how can I determine if there exists an inverse function? Am I supposed to sketch it on paper to see if it can have an invers? Or is there another/simplier way to do it?
How would I go about solving it? This is what I did:
$$h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}} = y$$
$$y(1 + \sqrt{x}) = 1-\sqrt{x}$$
$$y + y\sqrt{x} = 1-\sqrt{x}$$
$$\text{Cancel out roots:}$$
$$y^2 + y^2x = 1+x$$
$$y^2 + y^2x - x = 1$$
$$y^2 + x(y^2 - 1) = 1$$
$$x(y^2 - 1) = 1 - y^2$$
$$x = \frac{1 - y^2}{y^2 - 1}$$
$$\text{Swap x and y and we get the inverse function:}$$
$$f^{-1}(x) = \frac{1 - x^2}{x^2 - 1}$$
$$\text{But the correct answer is supposed to be:}$$
$$f^{-1}(x) = \frac{(1 - x)^2}{(1 + x)^2}, -1< x \le 1$$
What am I doing wrong?
| Nice work!
But this is how you should have done it!
$$\begin{align}1-y&=y\sqrt x+\sqrt x\\
1-y&=(y+1) \sqrt x\\
\frac{1-y}{y+1}&=\sqrt x\\
\sqrt x&=\frac{1-y}{y+1}\\
x&=\frac{(1-y)^2}{(y+1)^2}\\
\end{align} $$
Hence $f^{-1}(x)$
$$f^{-1}(x) = \frac{(1 - x)^2}{(x+ 1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Given $f(x)= \frac{1}{4}(x+4)^2-2$ Find vertex, $ y$ intercept etc. Given $f(x)= \frac{1}{4}(x+4)^2-2$
Find: vertex, $y$-intercept, $x$-intercepts (if any), axis of symmetry
What I have so far:
Vertex: $(-4,-2)$
$y$-intercept: $(0,2)$
$x$-intercept: $2$
Axis of symmetry $x=-4$
If you could please tell me if these are right or not that would be great. If they are not right please correct them and tell me why they are wrong.
| Because the function is in parabola vertex-form the vertex is at $\left( -4,-2\right)$
The axis of symmetry is about the $x-$value of the vertex, because the function is a vertical parabola.
$$ \therefore x_{axis}=-4 $$
for the $y-$intercept you need $f\left( 0\right)$
$$ f\left( 0\right) = \frac{1}{4}\left( 0+4\right) ^2-2 $$
$$ f\left( 0\right) =\frac{4^2}{4}-2 $$
$$ f\left( 0\right) = 4-2 $$
$$ f\left( 0\right) = 2 $$
For $x-$intercepts the line $y=0$ intersects the function $y=\frac{1}{4}\left( x+4\right) ^2-2$
$$\therefore \quad 0=\frac{1}{4}\left( x+4\right) ^2-2$$
$$ 2= \frac{1}{4}\left( x+4\right) ^2$$
$$ 8= \left( x+4\right) ^2$$
$$ x+4=\pm\sqrt8 $$
$$ x=-4\pm\sqrt8 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Number of real solutions of the equation $1+8^x+27^x = 2^x+12^x+9^x$
Find the number of real solutions $x\in\mathbb{R}$ of the equation
$$
1+8^x+27^x = 2^x+12^x+9^x
$$
My Attempt:
Let $2^x=a>0$ and $3^x=b>0$ where $x\in \mathbb{R}$. This allows us to change the equation to
$$
1+a^3+b^3 = a+a^2b+b^2
$$
This can be rewritten as
$$
(a+b)^3-3ab(a+b)+1 = a+ab(a+b)
$$
How can I solve the problem from this point?
| Hint: by rearrangement inequality
$$a^3+b^3+c^3 = a^2b+b^2c+c^2a$$
happens iff $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Why is $(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000\cdots$? Apparently,
$$(1-\cot 37^\circ)(1-\cot 8^\circ)=2.00000000000000000\cdots$$
Since it is a $2.0000000000\cdots$ instead of $2$, it isn't exactly $2$.
Why is that?
| $$\begin{align}
1-\cot{\left(x\right)}
&=1-\frac{\cos{\left(x\right)}}{\sin{\left(x\right)}}\\
&=\frac{\sin{\left(x\right)}-\cos{\left(x\right)}}{\sin{\left(x\right)}}\\
&=-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-x\right)}}{\sin{\left(x\right)}}.\\
\end{align}$$
Therefore,
$$\begin{align}
1-\cot{\left(\frac{\pi}{4}-x\right)}
&=-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-\left(\frac{\pi}{4}-x\right)\right)}}{\sin{\left(\frac{\pi}{4}-x\right)}}\\
&=-\sqrt{2}\frac{\sin{\left(x\right)}}{\sin{\left(\frac{\pi}{4}-x\right)}}.\\
\end{align}$$
Hence,
$$\begin{align}
\left(1-\cot{\left(x\right)}\right)\cdot\left(1-\cot{\left(\frac{\pi}{4}-x\right)}\right)
&=\left[-\sqrt{2}\frac{\sin{\left(\frac{\pi}{4}-x\right)}}{\sin{\left(x\right)}}\right]\cdot\left[-\sqrt{2}\frac{\sin{\left(x\right)}}{\sin{\left(\frac{\pi}{4}-x\right)}}\right]\\
&=2.\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Find all solutions in positive integers of the diophantine equation $w^2+x^2+y^2=z^2$ It's an exercises of the text book Elementary Number Theory and It's Applications 6th Edition by Kenneth H.Rosen. I wanted to solve it using the method in solving the diophantine equation $x^2+y^2=z^2$. But some difficult gaps showed up. And I found that the rational solutions of $w^2+x^2+y^2=1$ is
$$w = \frac{2s}{1+s^2+t^2}$$
$$x = \frac{2t}{1+s^2+t^2}$$
$$y = \frac{1-s^2 - t^2}{1+s^2+t^2}$$
where $s$ and $t$ are both rational numbers. From this I found out that
$$(2mnq^2)^2 + (2pqn^2)^2 + (n^2q^2 - m^2q^2-n^2p^2)^2 = (n^2q^2 + m^2q^2+n^2p^2)^2.$$
But it helps little because I cannot prove that the solutions have to be in such form.
Can any one help?
| Here I have a solution just using Pythagorean triples. we know that any solution of $$x^2+y^2=z^2$$ can be written as $$x=2ab,\,\,\,\,\, y=a^2-b^2,\,\,\,\,\ z=a^2+b^2.$$
Therefore for your equation, we can choose
$$w=4abc,\,\,\,\,\, x=2(a^2-b^2)c,\,\,\,\,\ y=(a^2+b^2)^2-c^2,\,\,\,\,\,z=(a^2+b^2)^2+c^2.$$
Isn't it interesting? :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is $s_3$ and $s_4$ for $x$ $\sum_{i=0}^n i^k = s_k(n)$,
$s_k$ polynomial from degree $k+1$
I have already shown for $s_2(x) = \frac{x(x+1)(2x+1)}6$
How from the sum and $s_2(x)$ can be shown for $s_3(x)$ and $s_4(x)$ respectively.
| if $f(x)$ be the generator function of ${(a_n)}_{n \in \mathbb N}$ then $\frac{f(x)}{1-x}$ is generator function of $\sum_{i=1}^{k}a_i $.
$\frac{1}{1-x}= 1+x+ x^2+x^3 +x^4 +... $ $\rightarrow $ $\frac{x}{(1-x)^2}=x+ 2x^2+ 3x^3+4x^4+... $ $\rightarrow $ $A+\frac{x(1-x)^2-2x(1-x)}{(1-x)^4}=1+2^2x^2+3^2x^3+... $
then coefficient $x^n$ in $\frac{A}{1-x}$ is $s_2(n)$.
so $s_3(n)$ is coefficient $x^n$ in $B=\frac{xA^{\prime}}{1-x}$
and $s_4(n)$ is coefficient $x^n$ in $\frac{xB^{\prime}}{1-x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to determine without calculator which is bigger, $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ or $\left(\frac{1}{3}\right)^{\frac{1}{2}}$ How can you determine which one of these numbers is bigger (without calculating):
$\left(\frac{1}{2}\right)^{\frac{1}{3}}$ , $\left(\frac{1}{3}\right)^{\frac{1}{2}}$
| Since $3^3=27 > 4=2^2$, we have
$$
\left(\frac{1}{2}\right)^2
>
\left(\frac{1}{3}\right)^3
$$
Take square roots and get
$$
\left(\frac{1}{2}\right)
>
\left(\frac{1}{3}\right)^{3/2}
$$
Take cube roots and get
$$
\left(\frac{1}{2}\right)^{1/3}
>
\left(\frac{1}{3}\right)^{1/2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 9,
"answer_id": 5
} |
Show that $\sum\limits_{i=0}^{n/2} {n-i\choose i}2^i = \frac13(2^{n+1}+(-1)^n)$ While doing a combinatorial problem, with $n$ being even, I came up with the expression
$$\sum_{i=0}^{n/2} {n-i\choose i}2^i$$
for which I used wolfram to get a closed form expression of $\dfrac{1}{3}\left(2^{n+1}+(-1)^n\right)$.
Is there an easy way to obtain this closed-form?
Also, if there are any good references for binomial coefficient identities like these I'd appreciate it. I searched some but did not find any similar to this.
| Permit me to contribute a proof complex variables, for variety's sake,
which is an instructive exercise.
Suppose we seek to verify that
$$\sum_{q=0}^{\lfloor n/2 \rfloor}
{n-q\choose q} 2^q
= \frac{1}{3} \left(2^{n+1} + (-1)^n\right)$$
with $n$ a positive integer.
Introduce the integral representation
$${n-q\choose q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n-q}}{z^{q+1}} \; dz$$
Note that this zero for $\lfloor n/2 \rfloor \lt q\le n$ so we may
extend the sum to $n,$ getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\sum_{q=0}^n
\frac{2^q}{(1+z)^q z^{q}} \; dz.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\frac{2^{n+1}/(1+z)^{n+1}/z^{n+1}-1}
{2/(1+z)/z-1} \;dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\frac{2^{n+1}/(1+z)^{n}/z^{n}-z(1+z)}
{2-z(1+z)} \;dz.$$
This has two pieces, the first is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{2^{n+1}}{2-z(1+z)} \;dz
\\ = \frac{2^{n+1}}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(\frac{1}{3}\frac{1}{1-z} + \frac{1}{6}\frac{1}{1+z/2}\right)
\; dz.$$
Extracting coefficients we find
$$2^{n+1}
\left(\frac{1}{3} + \frac{1}{6}\frac{(-1)^n}{2^n}\right)
= \frac{1}{3}
\left(2^{n+1} + (-1)^n\right).$$
The second piece is
$$-\frac{1}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{n+1}
\frac{1}{2-z(1+z)} \;dz$$
which gives a contribution of zero.
Collecting everything we obtain
$$\frac{1}{3}
\left(2^{n+1} + (-1)^n\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Continuous polar equation of Sqrt spiral Please help find polar coordinate equation $ r =f(\theta) $ or $ \theta =g(r) $
where $r$ and $\theta$ are continuous functions of $n$ as given:
SqrtSpiral
EDIT1:
n is a discrete (discontinuous) variable ( n = 1,2,3..) with
$$ r_n = \sqrt n ;$$
$$ r_{n+1}^2 -r_n^2 = 1 ; \\ r_n = \cot\Delta\theta =\cot (\theta_{n+1} - \theta_n );$$
All known spirals in the plane are either
$ r= f(\theta) $ or parametrically $ r(t),\theta(t), $
into which form Sqrt spiral is not castable.
I tried to adjust Pythogorean relation
$$ m^2+n^2 , m^2-n^2 , 2 m n $$
with no luck, so I posted here for help.
| $r_{n+1}^2 -r_n^2 = 1 ; \\ r_n = \cot\Delta\theta =\cot (\theta_{n+1} - \theta_n ) ;$
I'll write $t$ for $\theta$ since I'm lazy.
From your equations,
$\cot^2 (t_{n+2} - t_{n+1} )
=\cot^2 (t_{n+1} - t_n ) +1
$,
so
$t_{n+2}$
is a function of
$t_{n+1}$
and
$t_{n}
$.
Since
$\cot(a-b)
=\dfrac{1+\tan a \tan b}{\tan a -tan b}
$,
letting
$u_n
=\tan(t_n)
$,
$\dfrac{(1+u_{n+2} u_{n+1})^2}{(u_{n+2}-u_{n+1})^2}
=\dfrac{(1+u_{n+1} u_{n})^2}{(u_{n+1}-u_{n})^2}+1
$.
This is the relation among
the $u_i$.
Letting
$g
=\dfrac{(1+u_{n+1} u_{n})^2}{(u_{n+1}-u_{n})^2}+1
$,
$(1+u_{n+2} u_{n+1})^2
=g(u_{n+2}-u_{n+1})^2
$
or,
letting
$p = u_{n+2}$
and
$q = u_{n+1}$,
$(1+pq)^2 = g(p-q)^2$
or
$1+2pq+p^2q^2
=g(p^2-2pq+q^2)
$
or
$p^2(q^2-g)
+p(2-2gq)
+gq^2-1
=0
$,
or
$p^2(q^2-g)
+2p(1-gq)
+gq^2-1
=0
$.
The discriminant of this
divided by $4$
is
$\begin{array}\\
((1-gq)^2-(q^2-g)(gq^2-1)
&=(1-2gq+g^2q^2)-(gq^4-q^2(g^2+1)+g)\\
&=(1-2gq+g^2q^2)-(gq^4-q^2(g^2+1)+g)\\
&=1-2gq+g^2q^2-gq^4+q^2g^2+q^2-g\\
&=1-2gq+2g^2q^2-gq^4+q^2-g\\
\end{array}
$
I don't know where to go from here,
so I'll stop.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$
Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.
Is it correct? Is there another way to solve?
| We already have that
$$\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$$
If $x$ is an acute angle strictly between $0$ and $\pi/2$, then $\sin x \cos x > 0$. Hence $\tan x + \cot x \geq 2$ if and only if $$1 \geq 2\sin x \cos x \ \ \ \ \ \text{ or alternatively } \ \ \ \ \ \ \sin(2x) \leq 1$$
and this last result is true because $\sin\theta \leq 1$ for all angles $\theta$.
In other words, you almost had it. In your second line, you just had to realize that $$2\sin x \cos x$$ could be written as $\sin 2x$ and then bounded above by $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
Find closed form formula for $c(n,n-4)$. Find closed form formula for $c(n,n-4)$.
Where $c(n,k)$ are signless stirling numbers of first kind.
I need help.This is my last question all others problems of my exercise I have solved this is last one please help me.
Thanks.
| Here is a solution by generating functions for verification purpose
until something simpler appears.
Observe that there cannot be a cycle of length at least six because
that leaves $n-6$ items which can form at most $n-6$ cycles for a
total of $n-5$ cycles. Similarly for a cycle of length seven and so
on.
The species of permutations with cycles of length at most five is
is
$$\mathfrak{P}
(\mathcal{A}_1 \mathfrak{C}_{=1}(\mathcal{Z})
+ \mathcal{A}_2 \mathfrak{C}_{=2}(\mathcal{Z})
+ \cdots
+ \mathcal{A}_5 \mathfrak{C}_{=5}(\mathcal{Z})).$$
This gives the generating function
$$G(z) = \exp\left(\sum_{q=1}^5 a_q \frac{z^q}{q}\right)
= \prod_{q=1}^5 \exp\left( a_q \frac{z^q}{q} \right).$$
Now there are several cases.
First case.
A five-cycle and $n-5$ fixed points.
This gives the generating function
$$\frac{1}{(n-5)!} \left(\frac{z}{1}\right)^{n-5}
\exp\left(a_2 \frac{z^2}{2}\right)
\exp\left(a_3 \frac{z^3}{3}\right)
\exp\left(a_4 \frac{z^4}{4}\right)
\frac{1}{1} \left(\frac{z^5}{5}\right)^1.$$
We can drop the contributions in $a_2, a_3, a_4$ as no such cycles
appear because there aren't any items left over, for an answer of
$$\frac{1}{(n-5)!} \left(\frac{z}{1}\right)^{n-5}
\left(\frac{z^5}{5}\right)^1.$$
Second case.
A four-cycle, a two-cycle and $n-6$ fixed points, for a
contribution of
$$\frac{1}{(n-6)!} \left(\frac{z}{1}\right)^{n-6}
\left(\frac{z^2}{2}\right)^1
\left(\frac{z^4}{4}\right)^1.$$
Third case. Two three-cycles and $n-6$ fixed points, for a
contribution of
$$\frac{1}{(n-6)!}\left(\frac{z}{1}\right)^{n-6}
\frac{1}{2} \left(\frac{z^3}{3}\right)^2.$$
Fourth case. A three-cycle, two two-cycles and $n-7$ fixed
points for a contribution of
$$\frac{1}{(n-7)!} \left(\frac{z}{1}\right)^{n-7}
\frac{1}{2} \left(\frac{z^2}{2}\right)^2
\left(\frac{z^3}{3}\right)^1.$$
Fifth case. Four two-cycles and $n-8$ fixed
points for a contribution of
$$\frac{1}{(n-8)!}\left(\frac{z}{1}\right)^{n-8}
\frac{1}{24} \left(\frac{z^2}{2}\right)^4.$$
Adding the contributions from these generating functions we obtain
$$\frac{z^n}{(n-5)!} \times
\left(\frac{1}{5}
+ \frac{n-5}{8}
+ \frac{n-5}{18}
+ \frac{(n-5)(n-6)}{24}
+ \frac{(n-5)(n-6)(n-7)}{384} \right)
\\ = \frac{z^n}{(n-5)!} \times
\left(\frac{1}{384} n^3 - \frac{1}{192} n^2
+ \frac{1}{1152} n + \frac{1}{2880}\right).$$
Performing coefficient extraction on this we obtain the answer
$$n! [z^n]
\frac{z^n}{(n-5)!} \times
\left(\frac{1}{384} n^3 - \frac{1}{192} n^2
+ \frac{1}{1152} n + \frac{1}{2880}\right)
\\ = \frac{n!}{(n-5)!}
\left(\frac{1}{384} n^3 - \frac{1}{192} n^2
+ \frac{1}{1152} n + \frac{1}{2880}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Lagrange Multipliers: Find $\min$ of $f(x,y)=3(x+1) +2(y-1)$ subject to the constraint $x^2+y^2=4$
Find the minumum value of the function $f(x,y)=3(x+1) +2(y-1)$, subject to the constraint that $x^2+y^2=4$.
The problem states to use Lagrange Multipliers. In doing so I obtained the point $(\frac 6{\sqrt{13}},\frac 4{\sqrt{13}})$. But when checking if it is a maximum I found that the result was inconclusive. What did I do wrong?
| Write: $f(x,y) = 3x+2y + 1$. For a quick answer, using Cauchy-Schwarz inequality:
$|3x+2y| \leq \sqrt{3^2+2^2}\sqrt{x^2+y^2} = 2\sqrt{13}$. Thus:
$-2\sqrt{13}\leq 3x+2y \leq 2\sqrt{13} \to 1-2\sqrt{13} \leq f(x,y) \leq 1+2\sqrt{13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Intersection multiplicity of the curves I want to find the intersection multiplicity of the curves $f(x,y)=x^5+x^4+y^2$ and $g(x,y)=x^6-x^5+y^2$ at the point $P=(0,0)$.
That`s what I have tried:
$f$ and $g$ have a common tangent, the $y=0$.
So $I(P, f\cap g) > m_P(f) \cdot m_P(g)=4$
$$f(x, 0)=x^5+x^4 \Rightarrow s=\deg f(x, 0)=5$$
$$g(x, 0)=x^6-x^5 \Rightarrow r =\deg g(x, 0)=6$$
$$s \leq r$$
So we consider $h(x, y)=g(x, y)-x f(x, y)$
$$h(x, y)=x^6-x^5+y^2-x(x^5+x^4+y^2)=x^6-x^5+y^2-x^6-x^5-xy^2 \\ \Rightarrow h(x, y)=-2x^5+y^2-xy^2$$
$\deg h(x, 0)=5<r$
So $I(P, f\cap g)=I(P,f\cap h)$
$$f(x,0)=x^5+x^4\Rightarrow \deg f(x,0)=5=s$$
$$h(x,0)=-2x^5\Rightarrow \deg h(x,0)=5=p$$
They have a common tangent, $x=0$, so they don`t intersect traverrsally.
We consider the polynomial $$h_1(x,y)=h(x,y)+2f(x,y)=3y^2-xy^2+2x^4$$
$deg h_1(x,0)=4<s,p$
So, $I(P, f\cap h)=I(P,f\cap h_1)$
$f(x,0)=x^5+x^4 \Rightarrow \deg f(x,0)=5=s$
$h_1(x,0)=2x^4\Rightarrow \deg h_1(x,0)=4=t$
They have a common tangent $x=0$,so they don`t intersect traversally.
We consider the polynomial $h_2(x,y)=2f(x,y)-xh_1(x,y)=2x^4+2y^2-3xy^2+x^2y^2$
$\deg h_2(x,0)=4<s$
So $I(P, f\cap h_1)=I(P, h_1\cap h_2)$
$h_1(x,0)=2x^4\Rightarrow \deg h_1(x,0)=4=s$
$h_2(x,0)=2x^4\Rightarrow \deg h_2(x,0)=4=m$
They have a common tangent $x=0$, so they don`t intersect traversally.
We consider the polynomial $h_3(x,y)=h_1(x,y)-h_2(x,y)=y^2(1+2x-x^2)$
$\deg h_3(x,0)=0<s,m$
So $I(P,h_1\cap h_2)=I(P,h_2\cap h_3)$
$h_2(x,0)=2x^4\Rightarrow \deg h_2(x,0)=4=m$
$h_3(x,0)=0\Rightarrow \deg h_3(x,9)=9=n$
So $I(P,h_2\cap h_3)=I(P,h_2\cap y^2)+I(P,h_2\cap (1+2x-x^2))$
$I(P,h_2\cap y^2)=8$
$I(h_2\cap (1+2x-x^2))=0$
Therefore, $I(P, f \cap g)=8$.
Is it right? Do we find that $f$ and $h$ have a common tangent from $f(x,0)$ and $h(x,0)$ ?
| $ f(x,y) =0 $ (red curve) has no real common tangent with $ g(x,y) =0 $ (blue curve) at $ P(0,0) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1052007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Determine whether or not the following sequence converges. If so, find the limit. Im not sure if I am evaluating the convergence of the following sequence correctly, and I am unsure of how to determine the limit. All help is greatly appreciated.
$a_n = {1 \over (\sqrt{n^2-1}-\sqrt{n^2 +n})}$
My attempt at the convergence:
$ {1 \over (\sqrt{n^2-1}-\sqrt{n^2 +n})} < {1 \over n} $
$ \lim_{x\to \infty} {1 \over n} = 0$
Therefore, ${1 \over (\sqrt{n^2-1}-\sqrt{n^2 +n})}$ converges to as $ {n \to \infty} $
| We have: $\sqrt{n^2-1} - \sqrt{n^2+n} = (\sqrt{n^2-1} - n) - (-n +\sqrt{n^2+n}) = -\dfrac{1}{\sqrt{n^2-1} + n} - \dfrac{1}{\sqrt{1+\dfrac{1}{n}} + 1} \to -\dfrac{1}{2}$. Thus $a_n \to -2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $\sin45°+\sin15°=\sin75°$ Steps I took:
1) Finding the value of the left hand side
$$\sin45=\sin\frac { 90 }{ 2 } =\sqrt { \frac { 1-\cos90 }{ 2 } } =\sqrt { \frac { 1 }{ 2 } } =\frac { \sqrt { 2 } }{ 2 } $$
$$\sin15=\sin\frac { 30 }{ 2 } =\sqrt { \frac { 1-\cos30 }{ 2 } } =\sqrt { \frac { 1-\frac { \sqrt { 3 } }{ 2 } }{ 2 } } =\frac { \sqrt { 2-\sqrt { 3 } } }{ 2 } $$
So, $\sin45+\sin15=\frac { \sqrt { 2 } +\sqrt { 2-\sqrt { 3 } } }{ 2 } $
2) Rewriting $\sin75$
$$\sin75=\sin(45+30)=\sin45\cos30+\sin30\cos45$$
$$=\frac { \sqrt { 2 } }{ 2 } \cdot \frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } \cdot \frac { \sqrt { 2 } }{ 2 } $$
$$=\frac { \sqrt { 6 } +\sqrt { 2 } }{ 4 } $$
So now I have these two expressions that are obviously equal (when I compute them) but how do I show them as being equal to each other. I assume the question is implying for the resulting expressions to look equal.
| Both of them are positive, so let's square both of them:
first one will give:
\begin{gather}
\frac{1}{4} \left( 4 - \sqrt{3} + 2 \sqrt{4-2\sqrt{3}}\right) = \frac{1}{4} \left( 4 - \sqrt{3} + 2 \sqrt{1-2\sqrt{3}+3}\right) =\frac{1}{4} \left( 4 - \sqrt{3} + 2 \sqrt{(1-\sqrt{3})^2}\right) = \frac{1}{4} \left( 4 - \sqrt{3} + 2 (\sqrt{3}-1) \right) = \frac{1}{4} \left( 2 + \sqrt{3} \right) = 0.5 + \frac{\sqrt{3}}{4}
\end{gather}
the second one:
$$\frac{1}{16} \left( 6+2+2\sqrt{12}\right) = 0.5+\frac{\sqrt{3}}{4} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
compute an integral $\int {\frac{dx}{(x-1)\sqrt{x^2-3x+2}}}$ How we can compute this integral:
$$\int {\frac{dx}{(x-1)\sqrt{x^2-3x+2}}}$$
i know that the solution is:
$$I=\frac{2 (x-2)}{\sqrt{x^2-3x+2}}$$
| Substitute $\displaystyle t=\frac{\sqrt{x^2-3x+2}}{x-1}$ to get $x=\dfrac{t^2-2}{t^2-1}$ and $dx=\dfrac{2t}{(1-t^2)^2}dt$; then
$\displaystyle\int {\frac{dx}{(x-1)\sqrt{x^2-3x+2}}}=\int\frac{1}{\frac{1}{(1-t^2)^2}\cdot t}\cdot\frac{2t}{(1-t^2)^2} dt=2\int dt=2t+C$
$\displaystyle=\frac{2\sqrt{x^2-3x+2}}{x-1}+C=\frac{2(x-2)}{\sqrt{x^2-3x+2}}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Maximum area of a isosceles triangle in a circle with a radius r As said in the title, I'm looking for the maximum area of a isosceles triangle in a circle with a radius $r$.
I've split the isosceles triangle in two, and I solve for the area $A=\frac{bh}{2}$*. I have made my base $x$, and solve for the height by using the Pythagorean theorem of the smaller triangle (seen in picture).
$h=r+\sqrt{r^2-x^2}$
So my the formula, I think, for both triangles should be $A=x(r+\sqrt{r^2-x^2})$
But after I solved for the derivative, when put "$= 0$", and checked on my calculator, I got the maximum to be about $4.3301r$, which differs a lot from my book's answer of $\frac{3\sqrt{3}}{4}r^2$*. Is my formula for the area right? Am I going about this the wrong way, or is it just my derivative that is wrong? Thanks in advance
*
Edited from original post
$A=rx+x\sqrt{r^2-x^2}$
$A'=r+x(\frac{1}{2})(r^2-x^2)^{-\frac{1}{2}}(-2x)+\sqrt{r^2-x^2}$
$r+\sqrt{r^2-x^2}=\frac{x^2}{\sqrt{r^2-x^2}}$
$r\sqrt{r^2-x^2}+(r^2-x^2)=x^2$
$r^2+r\sqrt{r^2-x^2}=2x^2$
$r^2(r^2-x^2)=(2x^2-r^2)^2$
$r^4-r^2x^2=4x^2-4x^2r^2+r^4$
$4x^4=3x^2r^2$
$x=\frac{\sqrt{3}}{2}r$
| Let $\theta$ be one-half of the vertex angle (less than a right angle) of the isosceles triangle.
Exercise: Show that the area of the inscribed triangle is
$A(\theta) = \dfrac{h b}{2} = \dfrac{(r + r \cos\theta)}{2} (2 r \sin\theta) = r^2 \,(1+\cos\theta) \, \sin\theta $
Differentiating $A$ and setting it to $0$, you will be left with the problem of solving
$\cos\theta = -\cos2 \theta.$
Exercise: Show that the only $\theta$ in $\left(0, \dfrac{\pi}{2}\right)$ that works is $\dfrac{\pi}{3}$.
Plugging these into $A$ we get the answer: $\dfrac{3\sqrt 3 r^2}{4}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Solve for $x$ in the following trigonometric equation $$\sqrt 2\cos^2 x-\cos x=0$$
Solve for $x$ algebraically, where $x$ is greater than or equal to zero, and less than $2\pi$. Answer must be an exact solution.
To be honest, I don't know where to start with this one. I know I need to isolate $\cos x$, but I have little idea as to what I need to do to get there. Is subtracting $\cos x$ from both sides the best way to go about this?
Here is one thing I tried.
Am I completely on the wrong track here?
EDIT:
$\sqrt{2}\cos x - 1 = 0$
$\cos x = \dfrac{1}{\sqrt{2}}$
$\dfrac{1}{\sqrt{2}} = 45^\circ$
$360 - 45 = 315^\circ = \dfrac{7\pi}{4}$
$\cos x = 0$
$x = 0$ at $90^\circ$, or $\dfrac{\pi}{2}$ and $270^\circ$ or $\dfrac{3\pi}{2}$
So:
$x = \dfrac{\pi}{2}, \dfrac{\pi}{4}, \dfrac{3\pi}{2}, \dfrac{7\pi}{4}$
| We have: $\sqrt{2}\cos^{2}(x)-\cos(x)=0$
Let's factor out a $\cos(x)$ to get:
$\Rightarrow \cos(x)\big(\sqrt{2}\cos(x)-1\big)=0$
As we have a product that equals to zero, either one of the multiples must equal to zero:
$\Rightarrow \cos(x)=0 \Rightarrow x=\arccos(0)=\dfrac{\pi}{2}$, $\hspace{1 mm}\dfrac{3\pi}{2}$
$\Rightarrow \sqrt{2}\cos(x)-1=0 \Rightarrow x=\arccos\bigg(\dfrac{\sqrt{2}}{2}\bigg)=\dfrac{\pi}{4}$, $\hspace{1 mm}\dfrac{7\pi}{4}$
Therefore, the solutions to the equation are $x=\dfrac{\pi}{2}$, $x=\dfrac{3\pi}{2}$, $x=\dfrac{\pi}{4}$ and $x=\dfrac{7\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1056545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I prove the pattern $\sqrt{1 + 155555…5} = 2 \sqrt{3888…89}?$ How can I prove this
$$\sqrt{1+155}=2\sqrt{39}$$
$$\sqrt{1+1555}=2\sqrt{389}$$
$$\sqrt{1+15555}=2\sqrt{3889}$$
$$\sqrt{1+155555}=2\sqrt{38889}$$
| The expressions on the right are
$$
\sqrt{4\left( 3\cdot 10^{k-1} + \frac{8}{9}\cdot 10^{k-1} - \frac{80}{9} + 9\right) }$$
for $k \geq 2$. I'm going to compare that to the expressions on the left, which have $k$ 5's.
$$
\sqrt{4\left( 3\cdot 10^{k-1} + \frac{8}{9}\cdot 10^{k-1} - \frac{80}{9} + 9\right) }
= \sqrt{4\left( \frac{35}{9} \cdot 10^{k-1} + \frac{1}{9}\right)} = \sqrt{\frac{140}{9}\cdot 10^{k-1} + \frac{4}{9} } = \sqrt{\frac{14}{9} \cdot 10^k + 1 - \frac{5}{9} }= \sqrt{1 + 10^k + \frac{5}{9}\cdot 10^k - \frac{5}{9} }
$$
and the latter is the expression on the left.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1057951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$ Proving the roots of
$$x^4-2(a+b)x^2+(a-b)^2=0$$
are......
$$x=\sqrt{a}+\sqrt{b}$$
$$x=\sqrt{a}-\sqrt{b}$$
$$x=-\sqrt{a}+\sqrt{b}$$
$$x=-\sqrt{a}-\sqrt{b}$$
When $a$ and $b$ are real numbers(negative or positive)
I proved this by substituting $x=\sqrt{a}+\sqrt{b}$ in the main equation , but I think this way not effective because It needs to be repeated four times to complete the proving.
Is there another effective proving.
| One way to verify this with minimal computations is to expand the product $P(x) = \prod (x\pm\sqrt{a}\pm\sqrt{b})$ (where the product runs over all combinations of signs).
Taking as a variable $t_1 = x+\sqrt{a}$ and $t_2 = x-\sqrt{a}$, this becomes
$$P(x) = (t_1 + \sqrt{b})(t_1-\sqrt{b})(t_2 + \sqrt{b})(t_2-\sqrt{b}) = (t_1^2 - b)(t_2^2 - b).$$
Now $t_1t_2 = x^2-a$ and $t_1^2 + t_2^2 = 2(x^2 + a)$. So
$$P(x) = (x^2-a)^2 - 2b(x^2+a) + b^2 = x^4 - 2(a+b)x^2 +a^2 - 2ab + b^2. $$
This being said, the from math110's answer is nicer. In fact if you read it like this, it is even more obvious.
$$ \begin{aligned} x = \sqrt{a} + \sqrt{b} & \overset{\mathrm{square}}\implies x^2 = a^2 + b^2 + 2\sqrt{ab} \\ &\overset{\mathrm{rewrite}}\implies x^2 - a^2 - b^2 = 2\sqrt{ab} \\ &\overset{\mathrm{square}}\implies (x^2-a^2-b^2)^2 = 4ab. \end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Prove if $\left\{x_n\right\}$ converges to $2$, then $\left\{\frac{1}{x_n}\right\}$ converges to $\frac{1}{2}$ We know that for all $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $\lvert x_n - 2 \rvert < \epsilon$ for all $n \geq N$, and we want to show that for all $\varepsilon' > 0$, there exists an $N' \in \mathbb{N}$ such that $\left| \frac{1}{x_n} - \frac{1}{2} \right| < \epsilon'$ for all $n \geq N'$.
Let $\varepsilon = \varepsilon' - \frac{3}{2}$. Then by the triangle inequality, we have $$\left| \frac{1}{x_n} - \frac{1}{2} \right| \leq \left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right|.$$
Because we proved earlier that $x_n > 1$ for all $n \geq N$, we know that $$\left| \frac{1}{x_n}\right| + \left| -\frac{1}{2}\right| < \left| x_n\right| + \left| -\frac{1}{2}\right|.$$
By the triangle inequality again, we know that $$\left| x_n - \frac{1}{2}\right| \leq \left| x_n\right| + \left| -\frac{1}{2}\right|.$$
Note that $\left| x_n - \frac{1}{2}\right| = \left| (x_n - 2) + (2 - \frac{1}{2})\right|$, so we get $$\left| (x_n - 2) + (2 - \frac{1}{2})\right| \leq \left| x_n - 2\right| + \left| 2 - \frac{1}{2}\right|.$$
Then we have $$\left| x_n - 2\right| + \left|2 - \frac{1}{2}\right| < \varepsilon + \frac{3}{2} = \left(\varepsilon' - \frac{3}{2}\right) + \frac{3}{2} = \varepsilon',$$ so $\frac{1}{x_n} \rightarrow \frac{1}{2}$.
I just realized that I don't know how to make sure that $\varepsilon' - \frac{3}{2}$ is a positive number. How can I do this?
| In fact there's no reason why $\varepsilon' - \frac{3}{2}$ should be positive.
Instead notice that for all natural numbers $n$ such that $x_n\neq 0$, it holds that $$\left|\dfrac{1}{x_n}-\dfrac1 2\right|=\left|\dfrac{x_n-2}{2x_n}\right|.$$
Relating the RHS to the convergence of $(x_n)_{n\in \mathbb N}$ and to $(x_n)_{n\in \mathbb N}$ being greater (eventually) than $1$ should get you there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Computing the trigonometric sum $ \sum_{j=1}^{n} \cos(j) $ I have a task to compute such a sum:
$$ \sum_{j=1}^{n} \cos(j) $$
Of course I know that the answer is $$ \frac{1}{2} (\cos(n)+\cot(\frac{1}{2}) \sin(n)-1) = \frac{\cos(n)}{2}+\frac{1}{2} \cot(\frac{1}{2}) \sin(n)-\frac{1}{2} $$
but I don't have any idea how to start proving it.
| Using the formula for a geometric series:
$$\sum_{k=1}^Ne^{ik}=\frac{e^{i(N+1)}-1}{e^{i}-1}-1=\frac{e^{i(N+1)}-1}{e^{i}-1}\frac{e^{-i}-1}{e^{-i}-1}-1$$
Taking real parts :
\begin{align}
\sum_{k=1}^N\cos(k)&=\Re\left[\frac{\left(e^{i(N+1)}-1\right)\left(e^{-i}-1\right)}{2-2\cos(1)}\right]-1\\
&=\Re\left[\frac{e^{iN}-e^{i(N+1)}-e^{-i}+1}{2-2\cos(1)}\right]-1\\
&=\frac{\cos(N)-\cos(N+1)-\cos(1)+1}{2-2\cos(1)}-1\\
&=\frac{\cos(N)-\cos(N+1)}{2-2\cos(1)}-\frac{1}{2}\\
&=\frac{\cos(N)-\cos(N)\cos(1)+\sin(N)\sin(1)}{2-2\cos(1)}-\frac{1}{2}\\
&=\frac{\sin(N)\sin(1)}{2-2\cos(1)}-\frac{1}{2}+\frac{1}{2}\cos(N)\\
&=\frac{2\sin(N)\sin\left(\frac{1}{2}\right)\cos\left(\frac{1}{2}\right)}{4\sin^2\left(\frac{1}{2}\right)}-\frac{1}{2}+\frac{1}{2}\cos(N)\\
&=\frac{1}{2}\cot\left(\frac{1}{2}\right)\sin(N)-\frac{1}{2}+\frac{1}{2}\cos(N)\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$ If $2x^2+ 3y^2+4z^2 =1$ find the maximum of $4x+3y+2z$.
This is a question from a regional math olympiad and thus there must exist solutions without application of calculus.
I have no idea how to begin.
| WLOG, we can set $\sqrt2x=\cos A,\sqrt3y=\sin A\cos B,2z=\sin A\sin B$
$\implies4x+3y+2z=2\sqrt2\cos A+\sqrt3\sin A\cos B+\sin A\sin B$
$=2\sqrt2\cos A+\sin A(\sqrt3\cos B+\sin B)$
$=2\sqrt2\cos A+2\sin A\cos\left(B-\dfrac\pi6\right)$
If $\sin A\ge0,$
$4x+3y+2z\le2\sqrt2\cos A+2\sin A$ as $\cos\left(B-\dfrac\pi6\right)\le1$
The equality occurs if $\cos\left(B-\dfrac\pi6\right)=1$
$\iff B-\dfrac\pi6=2m\pi\iff B=2m\pi+\dfrac\pi6$ where $m$ is any integer
Again, $2\sqrt2\cos A+2\sin A=2(\sqrt2\cos A+\sin A)=2\sqrt3\sin\left(A+\arcsin\dfrac1{\sqrt3}\right)\le2\sqrt3$
The equality occurs if $\sin\left(A+\arcsin\dfrac1{\sqrt3}\right)=1$
$\iff A+\arcsin\dfrac1{\sqrt3}=2n\pi+\dfrac\pi2$
$\iff A=2n\pi+\dfrac\pi2-\arcsin\dfrac1{\sqrt3}=2n\pi+\arccos\dfrac1{\sqrt3}$
where $n$ is any integer
Similarly, if $\sin A<0,$
$4x+3y+2z\le2\sqrt2\cos A+2\sin A$ as $\cos\left(B-\dfrac\pi6\right)\ge-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1070626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
How to solve these equations for x and y.. equations are
$(x-y)(x+2y)(2x+y) = 20$
and
$x^2+xy+y^2 = 7$
i want the METHOD not the solutions
| Do the operations to result to $$2x^3 + 3x^2y- 3xy^2 - 2y^3 - 20 = 0 \iff \\ 2(x^3 - y^3) +3xy(x-y) - 20 = 0 \iff \\ 2(x-y)(x^2 + xy + y^2) + 3xy(x-y) - 20 = 0 \iff \\ 14(x-y)+3xy(x-y) - 20 = 0$$
That simplifies the solution so keep substituting (2) into (1) to simplify the system
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
How to compute $\sum_{n\ =\ 1}^{\infty}\arctan\left(\,\frac{3n^{2}}{ 2n^{4} - 1}\,\right)$ I find this problem on facebook group.
$$\mbox{Is it possible to find exact value of}\quad
\sum_{n\ =\ 1}^{\infty}\arctan\left(\,\frac{3n^{2}}{ 2n^{4} - 1}\,\right)\ {\large ?}.
$$
I think this is not telescope sum. And Wolfram Alpha can not find it.
Thank in advances.
| @Jack DAurizio answer is nice,and I have two solution for this
\begin{align*}\arctan{\dfrac{1}{2n^2}}&=\arctan{\dfrac{2}{4n^2-1+1}}=\arctan{\dfrac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}}\\
&=\arctan{(2n+1)}-\arctan{(2n-1)}
\end{align*}
solution 2:
\begin{align*}\arctan{\dfrac{1}{2n^2}}&=\arctan{\dfrac{n^2-(n^2-1)}{(n^2-n)+(n^2+n)}=\arctan{\dfrac{\dfrac{n}{n-1}-\dfrac{n+1}{n}}{1+\dfrac{n}{n-1}\cdot\dfrac{n+1}{n}}}}\\
&=\arctan{\dfrac{n}{n-1}}-\arctan{\dfrac{n+1}{n}}
\end{align*}
and for $$\sum_{n=1}^{\infty}\arctan{\dfrac{1}{n^2}}=\arctan{\dfrac{\tan{\frac{\pi}{\sqrt{2}}}-\tanh{\frac{\pi}{\sqrt{2}}}}{\tan{\frac{\pi}{\sqrt{2}}}+\tanh{\frac{\pi}{\sqrt{2}}}}}$$
see this (AMM E3375) post: prove this $\sum_{n=1}^{\infty}\arctan{\left(\dfrac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)\cdots\right)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Probability of winning a rigged coin-flipping game Betsy and Katie are playing a game with an unfair coin. The coin is rigged to come up heads with probability $\frac35$ and tails with probability $\frac25$.
Betsy goes first. The two take turns. The first player to flip a tail wins. What is Betsy's probability of winning?
What confuses me is that the game can continue for an unlimited amount of times (at least until Betsy flips a tail). How would I solve this?
| Let $p$ be the probability that Betsy wins the game. We check some cases. If Betsy flips tails on her first move, then she wins and the game ends; this happens with probability $2/5$. Otherwise, Katie gets a chance to play; this happens with probability $3/5$. If Katie gets a chance to play, she wins with probability $2/5$, or resets the game with probability $3/5$. Therefore, $$p = \frac{2}{5} \cdot 1 + \frac{3}{5} \left(\frac{2}{5} \cdot 0 + \frac{3}{5} p \right) \implies p = \frac{5}{8}$$
Alternatively, the probability that Betsy wins equals the probability she wins on her first move, plus the probability she wins on her second move, plus ...
The probability Betsy wins on her $k$th move is $$\left(\frac{3}{5} \right)^{2(k-1)} \cdot \frac{2}{5}$$ since every flip before her winning move must have been a heads. So the answer is $$\frac{2}{5} \cdot \sum_{k=1}^{\infty} \left(\frac{3}{5} \right)^{2(k-1)} = \frac{5}{8}$$
Note that both of these solutions are essentially the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1074949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
If we know $x+y+z=1$, $x^2+y^2+z^2=2$, and $x^3+y^3+z^3=3$, how to find $x^4+y^4+z^4$? Let $x$, $y$, and $z$ be such that
$$\begin{align*}
x+y+z&=1\\
x^2+y^2+z^2&=2 \\
x^3+y^3+z^3&=3
\end{align*}$$
Then $x^4+y^4+z^4=?$
| An additional solution is using Groebner bases. For this problem, consider the ring $R=\mathbb{Q}[x,y,z]$ and the ideal
$$
I=\langle x+y+z-1,x^2+y^2+z^2-2,x^3+y^3+z^3-3\rangle.
$$
Using a computer algebra system we can find a Groebner basis (with the grRevLex order) as
$$
\{x+y+z-1,2y^2+2yz+2z^2-2y-2z-1,6z^3-6z^2-3z-1\}
$$
Then, computing the remainder of $x^4+y^4+z^4$ under division by this basis gives
$$
\frac{25}{6},
$$
as others have found through more elementary means.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How find this integral $I=\int_{-1}^{1}\frac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}$ Show this integral
$$I=\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}$$
where $0<a,b<1$
my idea:
Let
\begin{align*}&(-2ax+a^2+1)(-2bx+b^2+1)=4abx^2-2(a+b+a^2b+b^2a)x+(a^2+1)(b^2+1)\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a^2+1)(b^2+1)-\dfrac{4ab(a+b+a^2b+b^2a)^2}{4ab}\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a^2+1)(b^2+1)-(a+b+a^2b+b^2a)^2\\
&=4ab\left(x-\dfrac{a+b+a^2b+b^2a}{2\sqrt{ab}}\right)^2+(a+b)^2+(ab-1)^2+(a+b)^2(1+ab)^2
\end{align*}
so I think this idea is not good, maybe this have good methods,because this result is nice
| \begin{eqnarray}
& &\int_{-1}^{1}\dfrac{1}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}dx\\
&=&-\frac{1}{\sqrt{ab}}\tanh^{-1}\sqrt{\frac{{b(a^2+1)-2abx}}{{a(b^2+1)-2abx}}}\bigg|_{-1}^1\\
&= &\frac{1}{\sqrt{ab}}\tanh^{-1}\dfrac{2\sqrt{ab}}{1+ab}
=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Writing $1+3x^2+8x^4+21x^6+\cdots$ as a power series representation How would I write the power series $$1+3x^2+8x^4+21x^6+\cdots$$ as a power series representation (something neat similar to $\frac{1}{1-x}$)?
This reminds me of the power series $1+x^2+x^4+x^6+\cdots$ where the power series representation for that is $\frac{1}{1-x^2}$, but how would I add the Fibonacci numbers as coefficients into that?
Hints only!!
| Let $f(x) = 1+2x+3x^2+5x^3+8x^4+13x^5+21x^6+\cdots$.
You probably already know a closed form for $f(x)$.
Then, $f(-x) = 1-2x+3x^2-5x^3+8x^4-13x^5+21x^6-\cdots$.
Do you see how to get the series you want from $f(x)$ and $f(-x)$?
To get a closed form for $f(x)$ try combining the following equations in a way that leaves a finite number of terms on the right side:
$f(x) \ \ \ \ = 1+2x+3x^2+5x^3+8x^4+13x^5+21x^6+\cdots$
$xf(x) \ \ = \ \ \ \ \ \ \ 1x+2x^2+3x^3+5x^4+ \ \ 8x^5+13x^6+\cdots$
$x^2f(x) = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1x^2+2x^3+3x^4+ \ \ 5x^5+ \ \ 8x^6+\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1076089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Help identifying the singularities of $\csc(\cos z) = \frac{1}{\sin(\cos z)}$ I am really stuck with this one:
$\frac{1}{\sin(\cos z)}$ has a singularity when $\cos z = k\pi $ since $\sin(k\pi) = 0$ but how do I solve for the value of $z$, how can i evaluate $\cos z = k\pi $?
| Solve $\cos z = w$, let $y=e^{iz}$. Then $\cos z = \frac 12\left(y+y^{-1}\right)$. So $y+\frac{1}{y} = 2w$, or $y^2-2wy+1=0$ or $$y=\frac{2w\pm\sqrt{4w^2-4}}{2}= w \pm \sqrt{w^2-1}$$
So $iz = \log\left(k\pi\pm \sqrt{k^2\pi^2-1}\right)$.
So:
$$z = i\log\left(k\pi\pm \sqrt{k^2\pi^2-1}\right)$$
Since $(k\pi+ \sqrt{k^2\pi^2-1})(k\pi- \sqrt{k^2\pi^2-1})$=1, you actually have:
$$z=\pm i\log\left(k\pi+ \sqrt{k^2\pi^2-1}\right)$$
Now, $\log$ is a multivalued function, so for each $k$ you get infinitely many poles.
When $k=0$, the roots are $z=\frac{\pi}{2}+m\pi$ for $m\in\mathbb Z$. Those are the "real" roots.
When $k>0$, $k\pi+\sqrt{k^2\pi^2-1}$ is positive real, so $z$ is necessarily an imaginary number plus some multiply of $2\pi$.
When $k<0$, we can take any pole for $-k>0$ and add $\pi$.
So for $k>0$, the polls corresponding to $k$ and $-k$ are:
$$\pm i\ln\left(k\pi+ \sqrt{k^2\pi^2-1}\right) + m\pi$$ for some $m\in\mathbb Z$, and $\ln$ is the standard real natural log.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding the basis and dimension of a vector space Find the basis and dimension of vector space $ L_{1}$ spanned by
vectors $ a_{1} ,a_{2},a_{3} $, the basis and dimension of vector space $ L_{2}$ spanned by
vectors $ b_{1} ,b_{2},b_{3} $
and also $ L_{1} + L_{2} , L_{1} \cap
L_{2}$
$ a_{1} = (-10,10,-6,-2) ,a_{2} = (-6,4,-4,-4) ,a_{3} = (-2,-12,-4,-20) $
$ b_{1} = (6,3,-6,4) ,b_{2} = (-4,0,0,-3) ,b_{3} = (-13,-7,9,-12) $
I've tried this:
\begin{pmatrix}-10 & 10 & -6 & -2\\
-6 & 4 & -4 & -4\\
-2 & -12 & -4 & -20
\end{pmatrix}
\begin{pmatrix}-2 & -12 & -4 & -20\\
0 & 70 & 14 & 98\\
0 & 0 & 0 & 0
\end{pmatrix}
So basis for $ L_{1} $ is $ (-2,-12,-4,-20), (0,70,14,98) $ ?
| For the first, form a matrix $A$ whose columns are the vectors $a_i$ and row reduce it:
$$\begin{bmatrix}
-10 & -6 & -2 \\
10 & 4 & -12 \\
-6 & -4 & -4 \\
-2 & -4 & -20
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & 0 & -4 \\
0 & 1 & 7 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
$$
There are leading ones in the first and second column of the row reduced matrix, so that means the first and second columns of $A$ (i.e., the vectors $a_1$ and $a_2$) are linearly independent and span the space $\text{span}\{a_1,a_2,a_3\}$. Thus $L_1$ is two dimensional and a basis for $L_1$ is $\{a_1,a_2\}$.
Adapt this technique to answer your other questions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving an algebraic inequality For any $a$, $b$, and $c$ prove
$$3a^2+3b^2-2b+2a+1>0$$
I tried the following
$$(a+1)^2+(b-1)^2+2(a^2+b^2)-1>0\\
(a+1)^2-1+(b-1)^2+2(a^2+b^2)>0\\
(a+1-1)(a+1+1)+(b-1)^2+2(a^2+b^2)>0\\
(a^2+2a)+(b-1)^2+2(a^2+b^2)>0\\
$$
$a^2$ and $(b-1)^2$ and $2(a^2+b^2)$ are always positive and greater than zero, when $a>0$ our inequality is proved. However when a is negative I can not say $2a$ is always less than or equal to $a^2$. I think this problem should be solved by algebraic identities alone and not by discussing the different signs of $a$? I appreciate your help
| $$3a^2+3b^2−2b+2a+1 = (3 a^2 + 2a +\frac{1}{3}) + (3 b^2 - 2b + \frac{1}{3}) + \frac{1}{3} = $$
$$ = (\sqrt{3}a + \frac{1}{\sqrt{3}})^2 + (\sqrt{3}b - \frac{1}{\sqrt{3}})^2 + \frac{1}{3} > 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
prove conjecture; the limit of iterating is $\sqrt{z^2 - 2}$ $$\lim_{n \to \infty} f_n(x)=x-\frac{1}{nx}\;\;\; g(x)=f_n^{on}(x)$$
The conjecture is for values of $|x|>\sqrt{2}$: $g(x) = \sqrt{z^2 - 2}$
This question comes from another matstack question/answer.
If $n=2^m$, then convergence is much quicker by starting with $f_n(x)=x-\frac{1}{nx}$ and then iterating the Taylor/Laurent series for $f(x) \mapsto f(f(x))$ m times. I start by generating the formal Taylor series after moving the fixed point from infinity to zero, $1/f(\frac{1}{x})\;$. Then we start iterating with $f_n$
$$f_n(x) = \frac{x}{1 - x^2/n} = x + \frac{x^3}{n} + \frac{x^5}{n^2} + \frac{x^7}{n^3} + \frac{x^9}{n^4} + \frac{x^{11}}{n^5} ...$$
Using this speedup with $m=\log_2(n)$, one can iterate the Taylor series for $f(x) \mapsto f(f(x))$ m times, rather then iterating $f^{on}$, but the two are of course identical. Its just that otherwise convergence is pretty slow, with n iterations to get accuracy to 1/n.
The formal Taylor series coefficients of $1/g(\frac{1}{x}) = \sqrt{z^2/(1-2z^2)}$ are:
$x + x^3 +
\frac{3 x^5 }{2 } +
\frac{5 x^7 }{2 } +
\frac{35 x^9 }{8 } +
\frac{63 x^{11} }{8 } +
\frac{231 x^{13} }{16 } +
\frac{429 x^{15} }{16 } +
\frac{6435 x^{17} }{128 } +
\frac{12155 x^{19} }{128 } +
\frac{46189 x^{21} }{256 } ...$
Empirically, for the $2^n$th iteration starting with $f_n$ in the limit above, the Taylor coefficients are accurate to approximately $O 2^{-n}$, so there is pretty good numerical computation evidence for the conjecture, but I have no idea how to prove it.
EDIT
I (https://math.stackexchange.com/users/39261/mick) will place a bounty for the following problem :
Basicly the inverse :
Suppose we are given $g(x)=\sqrt{x^2-2}$ and we are asked to find $f_n$ such that :
$$\lim_{n \to \infty} \;\;\; g(x)=f_n^{on}(x)$$
How do we solve such problems ??
EDIT
EDIT 2
As Sheldon's comment says $f_n(x)=\sqrt{x^2-\frac{2}{n}}$ is also a solution but I want to find the $f_n$ from the OP : $x - \frac{1}{nx}$ or another $f_n$ that is real-meromorphic on the entire complex plane.
EDIT 2
| Pick an $x > \sqrt{2}$ and $n > \dfrac{1}{x^2-2}$ and define a sequence by $x_0 = x$, $x_{k+1} = f_n(x_{k})$ for $k \ge 0$.
Rearrange $x_{k+1} = f_n(x_k) = x_k - \dfrac{1}{nx_k}$, to get $x_k(x_k-x_{k+1}) = \dfrac{1}{n}$.
Clearly, $x_0 \ge \sqrt{x^2}$. Now, suppose that $x_{m-1} \ge \sqrt{x^2-\dfrac{2(m-1)}{n}}$ for some $1 \le m \le n$.
Then, $x_{m-1} \ge \sqrt{x^2-2} > \dfrac{1}{\sqrt{n}}$, and so, $x_m = x_{m-1} - \dfrac{1}{nx_{m-1}} > 0$. Furthermore, we have:
$x^2-x_m^2 = \displaystyle\sum_{k = 0}^{m-1}(x_k^2-x_{k+1}^2) = \sum_{k = 0}^{m-1}\left[(2x_k-(x_k-x_{k+1}))(x_k-x_{k+1})\right]$ $= \displaystyle\sum_{k = 0}^{m-1}\left[2x_k(x_k-x_{k+1}) - (x_k-x_{k+1})^2\right] = \sum_{k = 0}^{m-1}\left[\dfrac{2}{n} - \dfrac{1}{n^2x_k^2}\right] = \dfrac{2m}{n} - \dfrac{1}{n^2}\sum_{k = 0}^{m-1}\dfrac{1}{x_k^2} \le \dfrac{2m}{n}$.
Hence, $x_m^2 \ge x^2 - \dfrac{2m}{n}$. Then, since $x_m > 0$, we have $x_m \ge \sqrt{x^2-\dfrac{2m}{n}}$.
So by induction, we have $x_n \ge \sqrt{x^2-2}$. (Note: all of that was necessary to ensure that $x_n > 0$ , otherwise, $x^2-x_n^2 \le 2$ doesn't imply that $x_n \ge \sqrt{x^2-2}$.)
Then, since $x^2-x_n^2 = 2 - \dfrac{1}{n^2}\displaystyle\sum_{k = 0}^{n-1}\dfrac{1}{x_k^2} \ge 2 - \dfrac{1}{nx_n^2} \ge 2 - \dfrac{1}{n(x^2-2)}$, we have $x_n^2 \le x^2-2 + \dfrac{1}{n(x^2-2)}$, i.e. $x_n \le \sqrt{x^2-2 + \dfrac{1}{n(x^2-2)}}$.
Therefore, $\sqrt{x^2-2} \le f_n^n(x) \le \sqrt{x^2-2 + \dfrac{1}{n(x^2-2)}}$ for all $n > \dfrac{1}{x^2-2}$.
Taking the limit as $n \to \infty$ and squeezing yields $g(x) = \displaystyle\lim_{n \to \infty}f_n^n(x) = \sqrt{x^2-2}$, as desired.
Note that the proof that $g(x) = \sqrt{x^2-2}$ for $x < -\sqrt{2}$ is similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
What are the intermediate fields of $\mathbb Q(\sqrt[3]2,e^{\frac{2i\pi}{3}})$ (Galois group) The elements of Galois group are
\begin{align*}
\sigma _1:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _2:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{2i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _3:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{4i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _4:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _5:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{2i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _6:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{4i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}}.
\end{align*}
We remark that
\begin{align*}
\sigma _2^3=1\\
\sigma _3^3=1\\
\sigma _4^2=1\\
\sigma _5^2=1\\
\sigma_6^2=1
\end{align*}
But $$\sigma _2\sigma _4(\sqrt[3]2)=\sigma _2(\sqrt[3]2)=\sqrt[3]2e^{\frac{2i\pi}{3}}$$
and
$$\sigma _4\sigma _2(\sqrt[3]2)=\sigma _4(\sqrt[3]2e^{\frac{2i\pi}{3}})=\sqrt[3]2e^{\frac{4i\pi}{3}},$$
therefore $\{\sigma _i\}_{i=1}^6$ is not a commutatif group and thus
$$\text{Gal}(E/\mathbb Q)=\{\sigma _i\}_{i=1}^6\cong \mathfrak S_3.$$
My problem is that it should have a $\sigma _i$ such that $\sigma _i^6=1$ and $\sigma _i^n\neq 1$ for $i\in\{1,2,3,4,5\}$ because one of my intermediate field must be $\mathbb Q$. And if not, is there a mistake in my applications $\sigma _i$?
| The subfields are the fields fixed by a subgroup of the Galois group. Not every subgroup is generated by a single element, whence it is no problem that there is no element of order $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Simplifying sum of 2 square roots I might be a bit slow on math, but I haven't seen this explicitly described anywhere, I'm reading a paper and in the paper it has the equation:
$$\Delta d=\sqrt{y^2+(x+D/2)^2}-\sqrt{y^2+(x-D/2)^2}$$
After which it says
After some simplifications, this equation can be rewritten in a more compact form as
$$\frac{x^2}{a}-\frac{y^2}{b}=\frac{x^2}{\Delta d^2/4}-\frac{y^2}{D^2/4-\Delta d^2/4}=1$$
How does it go from the first equation to the second one?
(the paper equation 4 and 5)
| $$\Delta d=\sqrt{y^2+(x+D/2)^2}-\sqrt{y^2+(x-D/2)^2}$$$$\therefore\sqrt{y^2+(x+D/2)^2}=\Delta d+\sqrt{y^2+(x-D/2)^2}$$$$\therefore y^2+(x+D/2)^2=\left(\Delta d+\sqrt{y^2+(x-D/2)^2}\right)^2$$$$\therefore y^2+x^2+Dx+D^2/4=\Delta d^2+y^2+(x-D/2)^2+2\Delta d\sqrt{y^2+(x-D/2)^2}$$$$=\Delta d^2+y^2+x^2-Dx+D^2/4+2\Delta d\sqrt{y^2+(x-D/2)^2}$$$$\therefore2\Delta d\sqrt{y^2+(x-D/2)^2}=2Dx-\Delta d^2$$$$\therefore4\Delta d^2(y^2+(x-D/2)^2)=4D^2x^2-4Dx\Delta d^2+\Delta d^4$$$$\therefore4\Delta d^2(y^2+x^2-Dx+D^2/4)=4D^2x^2-4Dx\Delta d^2+\Delta d^4$$$$\therefore4\Delta d^2y^2+4\Delta d^2x^2-4\Delta d^2Dx+\Delta d^2D^2=4D^2x^2-4Dx\Delta d^2+\Delta d^4$$$$\therefore4\Delta d^2y^2+4(\Delta d^2-D^2)x^2=\Delta d^2(\Delta d^2-D^2)$$$$\therefore\frac{4y^2}{\Delta d^2-D^2}+\frac{4x^2}{\Delta d^2}=1$$Hopefully you can see how to get to their expression from here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
solving for sin of the sum of two angles of a triangle In triangle $ABC$,
$3\sin B+4\cos C=6$ and $4\sin C+3\cos B=1$.
Show that $\sin(B+C)=0.5$.
Can we assume $\angle A = 180 - ( B + C)$ and use sum formula.
| Squaring both equations yields:
$9\sin^2 B + 24 \sin B\cos C + 16 \cos^2 C = 36$
$9\cos^2 B + 24 \cos B\sin C + 16 \sin^2 C = 1$
Now what happens when you add the two together?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Summation identity involving the floor function (Kömal November B. 4666)
Prove that $\sum_{k=1}^n (2k-1) [\frac{n}{k}]=\sum_{k=1}^n [\frac{n}{k}]^2$ for every positive integer $n$, where $[n]$ is the largest integer greater than or equal to $n$.
| The trick here is to introduce a second summation and then reverse the order of summation. Introducing the second summation is easy:
$$\sum_{k=1}^n(2k-1)\left\lfloor\frac{n}k\right\rfloor=\sum_{k=1}^n(2k-1)\sum_{j=1}^{\lfloor n/k\rfloor}1=\sum_{k=1}^n\sum_{j=1}^{\lfloor n/k\rfloor}(2k-1)\;.\tag{1}$$
Reversing the order of summation is a little trickier than usual. Since $\left\lfloor\frac{n}1\right\rfloor=n$, it’s clear that we want the outer sum to be $\sum_{j=1}^n$; the trick is to get the inner sum to catch exactly the values of $k$ for which $\left\lfloor\frac{n}k\right\rfloor\ge j$. Since we’re talking about positive integers, $\left\lfloor\frac{n}k\right\rfloor\ge j$ iff $n\ge kj$ iff $\left\lfloor\frac{n}j\right\rfloor\ge k$, so
$$\sum_{k=1}^n\sum_{j=1}^{\lfloor n/k\rfloor}(2k-1)=\sum_{j=1}^n\sum_{k=1}^{\lfloor n/j\rfloor}(2k-1)=\sum_{j=1}^n\left\lfloor\frac{n}j\right\rfloor^2\;,\tag{2}$$
since the sum of the first $m$ odd positive integers is $m^2$.
Added: To get a more intuitive idea of where the trick comes from, consider the case $n=8$, say:
$$\begin{array}{rccc}
k:&1&2&3&4&5&6&7&8\\
2k-1:&1&3&5&7&9&11&13&15\\
\lfloor 8/k\rfloor:&8&4&2&2&1&1&1&1
\end{array}$$
Now think of the bottom line as giving a frequency count for the entry in the line $2k-1$ line, and replace it with that many copies of the corresponding $2k-1$ entry:
$$\begin{array}{rccc}
k:&1&2&3&4&5&6&7&8\\
2k-1:&1&3&5&7&9&11&13&15\\
\lfloor 8/k\rfloor:&8&4&2&2&1&1&1&1\\ \hline
&1&3&5&7&9&11&13&15\\
&1&3&5&7\\
&1&3\\
&1&3\\
&1\\
&1\\
&1\\
&1\\
\end{array}$$
The sum $(1)$ corresponds to summing the columns below the line and then adding those sums. Reversing the order of summation as in $(2)$ corresponds to summing the rows below the line and then adding those sums.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to solve $y=(xy'+2y)^2$? What kind of differential equation is this thing and how to solve it?
$$y=(xy'+2y)^2$$
$$y=x^2y'^2+4xyy'+4y^2$$
| Lets try a substitution $y = x^{\alpha}u$
this results in
$$
x^{\alpha}u = x^2\left(\alpha x^{\alpha-1}u + x^{\alpha}u'\right)^2 + 4x\left(\alpha x^{\alpha-1}u + x^{\alpha}u'\right)x^{\alpha}u + 4x^{2\alpha}u^2
$$
dividing by $x^{\alpha}$ and collecting like terms we find
$$
u = \left(\alpha^2+4\alpha + 4\right)u^2 + (2\alpha + 4)x^{\alpha+1}uu' + x^{\alpha+2}u'^2
$$
if we set $\alpha = -2$
then we have
$$
\alpha^2+4\alpha + 4 = 4 -8 + 4 = 0\\
2\alpha + 4 = -4 + 4 = 0
$$
thus we have
$$
u = u'^2
$$
this results in
$$
u' = \pm\sqrt{u} \implies 2\sqrt{u} = \pm x + C
$$
integrating leads to
$$
2u^{1/2} = \pm x + C
$$
replacing the subs, we obtain
$$
y = x^{-2}\left(\pm x + C\right)^2 = \left(\frac{C}{x}\pm 1\right)^2
$$
lets try the solution (always do this)
$$
xy' + 2y = -2\left(\frac{C}{x}+1\right)\frac{C}{x^2}x + 2\left(\frac{C}{x}+ 1\right)^2\\
(xy'+2y)^2 = \left(\frac{C}{x}+1\right)^2\left[-\frac{2C}{x} + 2\left(\frac{C}{x}+ 1\right)\right]^2 = \left(\frac{C}{x}+1\right)^2 = y
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Synthetic division by $ax^2+bx+c$. I know that synthetic division can be used in order to find quotient $q(x)$ and remainder $r(x)$ of a polynomial $p(x)$ when it is divided by some linear polynomial like $x-c$. Now, does exist some procedure (another than long division) in order to find $q(x)$ and $r(x)$ when the divisor is $ax^2+bx+c$? Suppose $b^2-4ac<0$.
| Assume your divisor is $x-d$ rather than $x-c$ since $c$ is already used in the formation of the original polynomial. You can write the dividend $y$ as:
$y = a(x-d)^2 + b(x-d) + 2adx - ad^2 + c + bd = a(x-d)^2 + b(x-d) + 2ad(x-d) + 2ad^2 -ad^2 + c + bd = a(x-d)^2 + (b+2ad)(x-d) + ad^2 + bd + c = (x-d)(a(x-d) + b+2ad) + ad^2+bd+c = (x-d)(ax + ad + b) + ad^2+bd+c \Rightarrow q(x) = ax+ad+b, r(x) = ad^2+bd+c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Sum $\sum_{n=1}^\infty \frac{n^2}{(n+2)!}$ Problem is to find sum
$$\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots$$
What I knew doesn't apply on this problem
Some series are telescoping, some types are solvable using binomial , both look useless here
Binomial gives
$$n (n-1) (n-2)x^3=1$$
$$n (n-1)( n-2)(n-3)x^4=4$$
What approach to use here?
| Replace the range so that it starts at $n = 0$ and substitute $n^2 = (n+2)(n+1) - 3(n+2) + 4$ to obtain
$$
\begin{align*}
\sum_{n=0}^\infty \frac{n^2}{(n+2)!} &=
\sum_{n=0}^\infty \frac{(n+2)(n+1)}{(n+2)!} -
\sum_{n=0}^\infty \frac{3(n+2)}{(n+2)!} +
\sum_{n=0}^\infty \frac{4}{(n+2)!} \\ &=
\sum_{n=0}^\infty \frac{1}{n!} -
3\sum_{n=0}^\infty \frac{1}{(n+1)!} +
4\sum_{n=0}^\infty \frac{1}{(n+2)!} \\ &=
\sum_{n=0}^\infty \frac{1}{n!} -
3\sum_{n=1}^\infty \frac{1}{n!} +
4\sum_{n=2}^\infty \frac{1}{n!} \\ &=
(1-3+4) \sum_{n=0}^\infty \frac{1}{n!} -3\left(-\frac{1}{0!}\right) +4\left(-\frac{1}{0!}-\frac{1}{1!}\right) \\ &=
2e+3-8 \\ &= 2e-5.
\end{align*}
$$
More generally, this shows that for every polynomial $P(n)$ over the rationals and $k \geq 0$, we have
$$
\sum_{n=0}^\infty \frac{P(n)}{(n+k)!} = \alpha e + \beta
$$
for some rationals $\alpha,\beta$, and these can be calculated by presenting $P(n)$ as a linear combination of the polynomials
$$ 1, n+k, (n+k)(n+k-1), (n+k)(n+k-1)(n+k-2), \ldots. $$
In your case, we needed to represent $n^2$ as a linear combination of $1,n+2,(n+2)(n+1)$. The general form of such a combination is
$$ n^2 = A+B(n+2)+C(n+2)(n+1) = Cn^2 + (3C+B)n + (2C+2B+A). $$
Equating coefficients, we get the linear system $C = 1$, $3C+B = 0$, $2C+2B+A = 0$, which we can solve by back-substitution to obtain $C=1$, $B=-3$, $A=4$. Back-substitution works in general, and we obtain a simple and efficient algorithm for evaluating the general form of this series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1094852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove inequality $\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2 $ I cannot prove the following inequality.
Let $a,b,c$ be positive numbers.Prove that:
$\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}\geq(a-b)^2+(b-c)^2+(c-a)^2. $
I tried to use Cauchy inequality. I think the functions $f(a,b,c)=\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}$ and $g(a,b,c)=(a-b)^2+(b-c)^2+(c-a)^2$ are symmetric with respect to $a,b,c.$
| $$\begin{align}&\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b} - (a-b)^2-(b-c)^2-(c-a)^2 \\ =& \sum\limits_{cyc} \left(\frac{a^3}{b+c} - 2a^2 +a(b+c)\right) \\ =& \sum\limits_{cyc} \frac{a(a-b-c)^2}{b+c} \ge 0 \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1095720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Matrices, determinants, and applications to identities involving Fibonacci numbers Preamble
It is well known that since:
$$
\begin{pmatrix}
F_{n+1} \\
F_n \\
\end{pmatrix} =
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}
\begin{pmatrix}
F_n & F_{n-1} \\
\end{pmatrix}
$$
it is valid that:
$$
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}^n =
\begin{pmatrix}
F_{n+1} & F_n \\
F_n & F_{n-1} \\
\end{pmatrix}
$$
By calculating determinants, it immediately follows that:
$$F_{n-1}F_{n+1} - {F_n}^2
=\det\left[\begin{matrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{matrix}\right]
=\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]^n
=\left(\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]\right)^n
=(-1)^n$$
or
$$F_{n-1}F_{n+1} - {F_n}^2=(-1)^n$$
This is known as Cassini's identity.
I find both Cassini's identity and this proof exceptionally attractive.
Now, there are other identities that resemble Cassini's identity:
$$F_{n-2}F_{n-1}F_{n+3} - {F_n}^3=(-1)^nF_{n-3}$$
$$F_{n+2}F_{n+1}F_{n-3} - {F_n}^3=(-1)^nF_{n+3}$$
$${F_{n-3}}{F_{n+1}}^2-{F_{n-2}}^2{F_{n+3}}=4 (-1)^n{F_{n}}$$
$${F_{n-1}}^2{F_{n+1}}^2-{F_{n-2}}^2{F_{n+2}}^2=4 (-1)^n{F_{n}}^2$$
$$F_{n-2}F_{n-1}F_{n+1}F_{n+2} - {F_n}^4=-1$$
Question
Is there a proof of five identities above that relies on matrices, the proof that would be attractive and similar to the mentioned proof
of Cassini's identity?
Also, is there any other Fibonacci identity that follows from a suitable correspondant matrix equation?
Trivia
Not directly related to any regular (meaning having the form of an equality) Fibonacci identity, I found also some fairly surprising matrix identities involving Fibonacci numbers, and among them the strangest is:
$$\pmatrix{
3&6&-3&-1\\
1&0&0&0\\
0&1&0&0\\
0&0&1&0
}^n\pmatrix{
2197&512&125&27\\
512&125&27&8\\
125&27&8&1\\
27&8&1&1
}=\pmatrix{
F_{n+7}^3&F_{n+6}^3&F_{n+5}^3&F_{n+4}^3\\
F_{n+6}^3&F_{n+5}^3&F_{n+4}^3&F_{n+3}^3\\
F_{n+5}^3&F_{n+4}^3&F_{n+3}^3&F_{n+2}^3\\
F_{n+4}^3&F_{n+3}^3&F_{n+2}^3&F_{n+1}^3\\
}
$$
Here, one more unusual fact: $8$, $27$, and $125$ are also cubes of a Fibonacci number.
| Following statement and proof are interesting:
Prove that
$$F_{2n-1} = {F_{n}}^2 + {F_{n-1}}^2$$
Let us start with
$$
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}^n =
\begin{pmatrix}
F_{n+1} & F_n \\
F_n & F_{n-1} \\
\end{pmatrix}
$$
If we split $n$ into $p$ and $q$, this can be rewriten as:
$$
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}^p
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}^q
=
\begin{pmatrix}
1 & 1 \\
1 & 0 \\
\end{pmatrix}^{p+q}
$$
which is actually the same as:
$$
\begin{pmatrix}
F_{p+1} & F_{p} \\
F_{p} & F_{p-1} \\
\end{pmatrix}
\begin{pmatrix}
F_{q+1} & F_{q} \\
F_{q} & F_{q-1} \\
\end{pmatrix}
=
\begin{pmatrix}
F_{p+q+1} & F_{p+q} \\
F_{p+q} & F_{p+q-1} \\
\end{pmatrix}
$$
and, by calculating the right bottom element of the right hand side matrix, follows that:
$$F_{p+q-1} = F_p F_q + F_{p-1} F_{q-1}$$
Since we can assume this is valid for any natural $p$ and $q$, it will be valid for $p=q=n$ too:
$$F_{2n-1} = {F_{n}}^2 + {F_{n-1}}^2$$
and this is what we wanted to prove.
Moreover, from matrix equation, this is also valid:
$$F_{p+q} = F_{p+1}F_q + F_{p} F_{q-1}$$
and this leads to identity:
$$F_{2n} = F_{n+1}F_n + F_{n} F_{n-1}$$
There is also a possibility to split $n$ into $p$, $q$, and $r$, and obtain different identities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the value of $ \sum _{r=0} ^{2n} r ( ^{2n}C _r) ( \frac 1{r+2} ) $
Find the value of $$ \sum _{r=0} ^{2n} r ( ^{2n}C _r ) ( \frac 1{r+2} )$$
In order to solve this I am trying to make the term(s) of the series independent of $r$. However I'm unable to solve this further: $$ \sum _{r=0} ^{2n} 2n (^{2n-1}C_{r-1})( \frac 1{r+2} ) $$
Any help would be appreciated. :)
| HINT:
$$r\binom{2n}r\cdot\frac1{r+2}=\frac{r(r+1)}{(2n+2)(2n+1)}\cdot\frac{(2n+2)!}{[(2n+2)-(r+2)]!\cdot(r+2)!}$$
$$=\frac{r(r+1)}{(2n+2)(2n+1)}\cdot\binom{2n+2}{r+2}$$
Now let $r(r+1)=(r+2)(r+1)+A(r+2)+B$
$r^2+r=r^2+r(3+A)+2+2A+B$
$\implies A+3=0\iff A=-3$
and $B+2A+2=0\iff B=-2A-2=4$
$$\implies(2n+2)(2n+1)r\cdot \binom{2n}r\cdot\frac1{r+2}=[(r+2)(r+1)+(-3)(r+2)+4]\binom{2n+2}{r+2}$$
$$=(r+2)(r+1)\binom{2n+2}{r+2}-3(r+2)\binom{2n+2}{r+2}+4\binom{2n+2}{r+2}$$
$$=(2n+2)(2n+1)\binom{2n}r-3(2n+1)\binom{2n+1}{r+1}+4\binom{2n+2}{r+2}$$
Finally, $\sum_{r=0}^{2n}\binom{2n}r=(1+1)^{2n}$
and $\sum_{r=0}^{2n}\binom{2n+1}{r+1}=(1+1)^{2n+1}-\binom{2n+1}0$
and $\sum_{r=0}^{2n}\binom{2n+2}{r+2}=(1+1)^{2n+2}-\binom{2n+2}0-\binom{2n+2}1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solve the ODe with five variables How to solve $$\frac{dx}{2p}=\frac{dy}{2q}=\frac{du}{2(p^2+q^2}=\frac{dp}{2up}=\frac{dq}{2uq}=dt$$
as functions $$x=x(t), y=y(t), u=u(t), p=p(t), q=q(t)$$
My method is use the last three equalities to deduce $$\frac{d^2u}{dt^2}+u\frac{du}{dt}=0$$
But this nonlinearity troubles me...
| Based on your last line:
Consider $F = u^2$
$$ \frac{df}{dt} = 2 u \frac{du}{dt} $$
Thus suppose we re-arrange
$$ \frac{d^2u}{dt^2} + u \frac{du}{dt} = 0 $$
Into
$$ \frac{d^2u}{dt^2} = - u \frac{du}{dt} $$
Then we can re-write it as
$$ \frac{d}{dt} \left[ \frac{du}{dt} \right] = \frac{d}{dt}\left[ - \frac{1}{2}u^2\right] $$
Thus we have
$$ \frac{du}{dt} = C_1-\frac{1}{2}u^2 $$
From here we will use the standard algorithm for solving linear ODEs
$$ \frac{1}{C_1 - \frac{1}{2}u^2} du = 1 dt$$
Since the choice of C is arbitrary this can be re-written as
$$2 \frac{1}{C_1 - u^2} du = 1 dt$$
We can integrate both sides:
$$2 \int \frac{1}{C_1 - u^2} du = t + C_2$$
The left hand side can be attacked with Partial Fractions: We decompose the integrand into:
$$ \frac{1}{C_1 - u^2} = \frac{1}{(\sqrt{C_1} - u)(\sqrt{C_1} + u)} = \frac{A}{\sqrt{C_1} - u} + \frac{B}{\sqrt{C_1} + u}$$
We find the constants A,B
$$ B(\sqrt{C_1} - u) + A(\sqrt{C_1} + u) = 0 $$
Which (based on grouping like terms of $C_1$ and $u$ tells us
$$ \begin{pmatrix} Au - Bu = 0 \\ \sqrt{C_1}(A + B) = 1 \end{pmatrix} \rightarrow \begin{pmatrix} A = B \\ (A + B) = \frac{1}{\sqrt{C_1}} \end{pmatrix} $$
Giving us
$$ A = B = \frac{1}{2 \sqrt{C_1}} $$
So our integral is now
$$2 \int \left[ \frac{\frac{1}{2 \sqrt{C_1}}}{\sqrt{C_1} -u} + \frac{\frac{1}{2 \sqrt{C_1}}}{\sqrt{C_1} + u} \right] du $$
We factor the top parts giving us
$$ \frac{1}{\sqrt{C_1}} \int \left[ \frac{1}{\sqrt{C_1} -u} + \frac{1}{\sqrt{C_1} + u} \right] du $$
Which integrates to
$$ \frac{1}{\sqrt{C_1}} \left( \ln(\sqrt{C_1} + u) - \ln(\sqrt{C_1} - u) \right) $$
So we have
$$2 \int \frac{1}{C_1 - u^2} du = t + C_2$$
Gives rise to
$$ \frac{1}{\sqrt{C_1}} \left( \ln(\sqrt{C_1} + u) - \ln(\sqrt{C_1} - u) \right) = t + C_2$$
Which resolves to
$$ \ln \left( \frac{\sqrt{C_1} + u}{\sqrt{C_1} - u} \right) = C_1 t + C_2 $$
We exponentiate both sides to find
$$ \left( \frac{\sqrt{C_1} + u}{\sqrt{C_1} - u} \right) = C_2 e^{C_1t} $$
And now solve for $u$ as
$$ \sqrt{C_1} + u = (C_2 e^{C_1t})(\sqrt{C_1} - u) \rightarrow $$
$$\left(1 - C_2 e^{C_1t} \right)u= (C_2 e^{C_1t}-1)\sqrt{C_1} $$
$$ u = \frac{(C_2 e^{C_1t}-1)\sqrt{C_1}}{\left(1 - C_2 e^{C_1t} \right)} $$
Alternate Trig Approach
$$ \int \frac{2}{C_1 - u^2} du = \frac{2}{\sqrt{C_1}} arctanh^{-1} \left( \frac{u}{\sqrt{C_1}} \right) $$
So our solution then becomes
$$\frac{2}{\sqrt{C_1}} arctanh^{-1} \left( \frac{u}{\sqrt{C_1}} \right) = t + C_2 $$
$$ arctanh^{-1} \left( \frac{u}{\sqrt{C_1}} \right) = C_1t + C_2$$
$$ u = \sqrt{C_1}tanh\left( C_1t + C_2\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the derivative $f(x)=\sqrt{x^2 -9}$, I need to find the slope at a=5, using the definition for the function $f(x)=\sqrt{x^2 -9}$,
$$f'(x) = \lim_{\Delta x \to 0} {f(x+\Delta x)\over \Delta x}$$
The answer book says the slope is ${1\over 4}$
Here's what I did,
$$f'(x) = \lim_{\Delta x \to 0} {(\sqrt{(x+\Delta x)^2 -9} - \sqrt {x^2 -9} )(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)})\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(1)\\=\lim_{\Delta x \to 0} {(x+\Delta x)^2 -9 -x^2 +9\over \Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(2)\\=\lim_{\Delta x \to 0}{x^2 +2x \Delta x+ \Delta x^2 -x^2\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(3)\\=\lim_{\Delta x \to 0}{2x\Delta x +\Delta x^2\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(4)\\=\lim_{\Delta x \to 0} {2x+\Delta x\over(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(5)\\={2x\over \sqrt{x^2-9+x^2-9}}
(6)\\={2x\over2 \sqrt{x^2 -9}}
(7)\\={x\over \sqrt {x^2 -9}} (8)$$
Now I substitute 5, and I don't get 1/4!!
What have I done wrong??
Thanks
| (5)to (6) step !!!you have error $$\sqrt{(x+\Delta x)^2-9}+\sqrt{(x)^2-9} \neq \sqrt{(x)^2-9+(x)^2-9} $$in fact $$\sqrt{a+b} \neq \sqrt{a} +\sqrt{b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the integral $\int_0^\infty \frac{x (\ln(x))^2}{x^4 + x^2 + 1}\text{ d}x$ What is the value of $\displaystyle\int_0^\infty \frac{x (\ln(x))^2}{x^4 + x^2 + 1}\text{ d}x$?
This is a question I came up with myself. It is not homework.
I constructed this example to make the following technique work:
Integrate $\frac{z (\log(z))^3}{z^4 + z^2 + 1}$ along a "key-hole" contour. The argument can be made rigorous by splitting the contour into two parts, and using two different branch cuts for each part. Warning: This method is time-consuming and not for the faint-hearted
| Let's actually do the integral using the keyhole contour. It may be time-consuming but it is not as bad as it looks.
We can begin by simplifying the integral using the substitution $u=x^2$:
$$I = \frac18 \int_0^{\infty} du \frac{\log^2{u}}{u^2+u+1} $$
Consider
$$\oint_C dz \frac{\log^3{z}}{z^2+z+1}$$
where $C$ is the keyhole contour, as pictured below.
The integral over the circular arcs vanish, and the contour integral is equal to
$$i \left ( -6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} + 8 \pi^3 \int_0^{\infty} dx \frac{1}{x^2+x+1}\right ) + 12 \pi^2 \int_0^{\infty} dx \frac{\log{x}}{x^2+x+1}$$
We can easily show that
$$\int_0^{\infty} dx \frac{\log{x}}{x^2+x+1} = 0$$
by splitting up the integration interval into $[0,1]$ and $[1,\infty)$ and subbing $x=1/u$ in the latter subinterval.
Now, we can evaluate the other integral any way we want, but let's stay consistent within our chosen methodology, and evaluate the integral using the residue theorem, all the same. Let the poles of the denominator be $z_{\pm}$; here
$$z_+ = e^{i 2 \pi/3} \quad z_-=e^{i 4 \pi/3} $$
Then
$$\int_0^{\infty} dx \frac{1}{x^2+x+1} = - \left (\frac{\log{z_+}}{2 z_++1} +\frac{\log{z_-}}{2 z_-+1}\right ) = -\frac{i 2 \pi/3}{i \sqrt{3}} + \frac{i 4 \pi/3}{i \sqrt{3}} = \frac{2 \pi}{3 \sqrt{3}}$$
The contour integral is of course equal to $i 2 \pi$ times the sum of the residues at $z=z_{\pm}$. Thus we have
$$-3 \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} + 4 \pi^2 \frac{2 \pi}{3 \sqrt{3}} = \left (\frac{\log^3{z_+}}{2 z_++1} +\frac{\log^3{z_-}}{2 z_-+1}\right ) = \frac{56 \pi^3}{27 \sqrt{3}}$$
Thus, from above, we have
$$\int_0^{\infty} dx \frac{x \log^2{x}}{x^4+x^2+1} = \frac{2 \pi^3}{81 \sqrt{3}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Surface integral of $x^4+y^4+z^4$ over the sphere $x^2+y^2+z^2=a^2$ After doing regular methodology have reached upto integral shown in figure , but when i eliminate z from it it becomes very complicated to solve .Is there any other way to solve this .Thanks
| It seems the following.
Use spherical coordinates $$x=a\sin\varphi\cos\theta,$$ $$y=a\sin\varphi\sin\theta,$$ $$z=a\cos\varphi$$ ($0\le\varphi\le\pi$, $0\le\theta\le 2\pi$). Then $$dS=a^2\sin\varphi d\varphi d\theta$$ and
$$x^4+y^4+z^4=$$
$$a^4(\sin^4\varphi\cos^4\theta+\sin^4\varphi\sin^4\theta+\cos^4\varphi)=$$
$$a^4(\sin^4\varphi( (\cos^2\theta+ \sin^2\theta)^2-2\cos^2\theta\sin^2\theta)+\cos^4\varphi)=$$
$$a^4(\sin^4\varphi\left( 1-\frac{\sin^2 2\theta}2\right)+\cos^4\varphi)=$$
$$a^4\left(1-\frac{\sin^4\varphi \sin^2 2\theta}2\right).$$
So the initial integral is equal to
$$a^6\int_0^{2\pi} \int_0^{\pi}\left(1-\frac{\sin^4\varphi \sin^2 2\theta}2\right) \sin\varphi d\varphi d\theta=$$
$$a^6\int_0^{2\pi} \int_0^{\pi}\sin\varphi d\varphi d\theta-a^6\int_0^{2\pi} \int_0^{\pi}\frac{\sin^4\varphi \sin^2 2\theta}2 \sin\varphi d\varphi d\theta=$$
$$2\pi a^6 \int_0^{\pi}\sin\varphi d\varphi-\frac{a^6}2\int_0^{2\pi} \sin^2 2\theta d\theta \int_0^{\pi}\sin^5\varphi d\varphi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1098863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify $\cos 1^\circ + \cos 3^\circ + \cdots+ \cos 43^\circ$? I am currently working on a problem and reduced part of the equations down to
$$\cos(1^\circ)+\cos(3^\circ)+\cdots+\cos(39^\circ)+\cos(41^\circ)+\cos(43^\circ)$$
How can I calculate this easily using the product-to-sum formula for $\cos(x)+\cos(y)$?
| let $S = \cos(1^\circ)+\cos(3^\circ)+.....+\cos(39^\circ)+\cos(41^\circ)+\cos(43^\circ).$ then
\begin{align} 2S\sin 1^\circ &= 2\cos 1^\circ \sin 1^\circ + 2 \cos 3^\circ \sin 1^\circ+\cdots + 2 \cos 43^\circ \sin 1^\circ \\
& = (\sin 2^\circ - \sin 0^\circ)+(\sin 4^\circ - \sin 2^\circ ) + \cdots
+(\sin 44^\circ- \sin 42^\circ))\\
& = \sin44^\circ
\end{align}
therefore $$ S = \dfrac{\sin 44^\circ}{ 2\sin 1^\circ} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)^3}$ using complex analysis Evaluate:
$$S = \sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)^3} \space \text{using complex analysis}$$
This my question: we need to consider a $f(z)$ such that,
$$\frac{1}{2\pi i} \cdot\oint_{C_N} f(z) dz = \text{something (maybe residue)} + \sum_{n=1}^{N} \frac{1}{(n+1)(n+2)^3}$$
How do we do this? I believe we consider, a square with vertices:
We also need to prove that as $N \to \infty$ that $\displaystyle \oint_{C_N} f(z) dz \to 0$
Can someone give me an idea?
| Here is an unorthodox answer that needs some more work to make it rigorous.
Consider $$f(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}.$$
We can certainly integrate around a circular contour as ML gives $2\pi R\times \log R/R^4.$
Now we have
$$\mathrm{Res}_{z=0} f(z) = \frac{1}{8},$$
$$\mathrm{Res}_{z=-1} f(z) = -\gamma,$$
$$\mathrm{Res}_{z=-2} f(z) = \frac{\pi^2}{6} +\zeta(3) - 3 +\gamma,$$
and finally
$$\mathrm{Res}_{z=n} f(z) = \frac{1}{(n+1)(n+2)^3},$$
This means that the sum is
$$\frac{23}{8} - \frac{\pi^2}{6} -\zeta(3).$$
Addendum.
To see how to calculate the residue of $f(z)$ at $z=-2$ use the Cauchy
Integral Formula for the series coefficients which is
$$\frac{1}{2\pi i}
\int_{|z+2|=\epsilon}
\frac{\psi(-z)}{(z+2)^{n+1}} \; dz.$$
Put $z+2 = -w$ to get
$$- \frac{(-1)^{n+1}}{2\pi i}
\int_{|w|=\epsilon}
\frac{\psi(w+2)}{w^{n+1}} \; dw$$
which is
$$\frac{(-1)^n}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}}
\left(\frac{1}{w+1} + \psi(w+1)\right) \; dw$$
The first component can be evaluated to give the series
$$1+(z+2)+(z+2)^2+(z+2)^3+\cdots$$
(remember that the CIF integral gives the coefficient on $(z+2)^n.$)
The second component is
$$\frac{(-1)^n}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}}
\psi(w+1) \; dw$$
The rational zeta series for the digamma function is
$$\psi(z+1) = -\gamma - \sum_{k\ge 1} \zeta(k+1) (-z)^k$$
for $|z|<1.$
It follows that the second component gives the series
$$-\gamma - \sum_{k\ge 1} \zeta(k+1) (z+2)^k.$$
This finally yields
$$\psi(-z)
= 1-\gamma + \sum_{k\ge 1} (1-\zeta(k+1)) (z+2)^k.$$
Now since
$$\frac{1}{z+1} = \frac{1}{-1+z+2}
= -\frac{1}{1-(z+2)}
= -1 - (z+2) - (z+2)^2 - (z+2)^3 -\cdots$$
The desired residue is
$$-(1-\zeta(3)) - (1-\zeta(2)) - (1-\gamma)
= -3 + \gamma + \frac{\pi^2}{6} + \zeta(3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
ODE Separable Equation Let $y = Φ(x)$ be a solution to $y' = y(5-y)(8-y)$ subject to $y(0) = 7$. Determine $\lim_{x \to ∞} Φ(x)$.
Workings:
I'm thinking I have to solve the differential equation.
$y' = y(5-y)(8-y) dy$
$y' = y(y-5)(y-8) dy$
$∫\frac{1}{(y-5)(y-8)y} dy = ∫1 dx$
$∫\frac{1}{(y-5)(y-8)y} dy$
$\frac{1}{(y-5)(y-8)y} = \frac{A}{(y-5)} + \frac{B}{(y-8)} + \frac{C}{y}$
Greatest Common Divisor gives:
$1 = Ay(y-8) + By(y-5) + C(y-5)(y-8)$
$y = 0:$
$1 = A(0)(0-8) + B(0)(0-5) + C(0-5)(0-8)$
$1 = 40C$
$C = \frac{1}{40}$
$y= 5:$
$1 = A(5)(5-8) + B(5)(5-5) + C(5-5)(5-8)$
$1 = -15A$
$A = \frac{-1}{15}$
$y = 8:$
$1 = C(8)(8-8) + B(8)(8-5) + C(8-5)(8-8)$
$1 = 24B$
$B = \frac{1}{24}$
$\frac{A}{(y-5)} + \frac{B}{y} + \frac{C}/{(y-8)}$
$= \frac{(-1/15)}{(y-5)} + \frac{(1/24)}{y} + \frac{(1/24)}{(y-8)}$
$= \frac{-1}{(15(y-5))} + \frac{1}{40y} + \frac{1}{(24(y-8))}$
$∫\frac{-1}{(15(y-5))} + \frac{1}{40y} + \frac{1}{(24(y-8))} dy$
$= \frac{-1}{15} \ln|y-5| + \frac{1}{40} \ln|y| + \frac{1}{24} \ln|y-8|$
$∫1 dx$
$= x + c$
$\frac{-1}{15} \ln|y-5| + \frac{1}{40} \ln|y| + \frac{1}{24} \ln|y-8| = x + c$
(I'm not too sure if I should solve for y here or not. There is another question later in the assignment where it states it can be expressed implicitly)
Letting $y(0) = 7$:
$\frac{-1}{15} \ln |(7)-5| + \frac{1}{40} \ln|(7)| + \frac{1}{24} \ln|(7)-8| = (0) + c$
$\frac{-1}{15} \ln|2| + \frac{1}{40} \ln|7| + \frac{1}{24} \ln|-1| = c$
$\frac{-1}{15} \ln|y-5| + \frac{1}{40} \ln|y| + \frac{1}{24} \ln|y-8| = x - 1/15 \ln|2| + 1/40 \ln|7| + 1/24 \ln |-1|$
I'm not sure if I did this correctly. Or what to do next. Any help will be appreciated.
| to find the limit there is no need to solve the equation. look at the steady state solutions $y = 0, y = 5 y = 8.$
this is called the phase portrait:
(a) on a number line mark these points.
(b) on each region bounded by these points, determine the sign of $y^\prime$ by evaluating $y^\prime$ at a test point.
here is what it looks like for your problem:
sign of $y^\prime = y(5-y)(8-y):$ (-) (+) (-) (+) from left to right.
$-\infty$--------<<--------0-------->>------5------<<---------8------>>---------$\infty$
if the initial $y$ value is $7,$ the solution will approach $5$ as $t \to \infty$ and $y \to 8$ as $t \to -\infty$
by uniqueness theorem for differential equation, the solutions may never reach, let alone cross these special solutions.
p.s. appreciate if anyone can augment this post with a line diagram.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1103415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Approximate $\coth(x)$ around $x = 0$ I'm trying to approximate $\coth(x)$ around $x = 0$, up to say, third order in $x$. Now obviously a simple taylor expansion doesn't work, as it diverges around $x = 0$. I'm not quite sure how to proceed from here. I had a look at
Series expansion of $\coth x$ using the Fourier transform
but this is a bit technical for my purposes, and I also don't need the entire series. Is there an easy way of getting some terms, or do I really have to solve for the entire series? And is there a simpler way to do so than with the Fourier transform?
| The series representation must reflect that divergence at $x=0$. Thus,
$$\begin{align}\coth{x} &= \frac{1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}{\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}+O(x^7)} \\ &= \frac1{x} \frac{1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)}{1+\frac{x^2}{3!}+\frac{x^4}{5!}+O(x^6)}\\&= \frac1{x} \left (1+\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6) \right ) \left [1 - \left (\frac{x^2}{3!}+\frac{x^4}{5!}+O(x^6) \right )+\left (\frac{x^2}{3!}+O(x^4) \right )^2+O(x^6) \right ] \\ &= \frac1{x} \left [1+\left (\frac1{2!}-\frac1{3!} \right ) x^2 +\left (\frac1{(3!)^2} + \frac1{4!}-\frac1{5!} - \frac1{2! 3!}\right ) x^4+O(x^6) \right ]\\ &= \frac1{x}+\frac{x}{3} - \frac{x^3}{45}+O(x^5) \end{align}$$
Note that the divergence at $x=0$ is reflected in the $1/x$ term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Show that if $a$ is an integer, then 3 divides $a^3 - a $ Show that if $a$ is an integer, then 3 divides $a^3 - a $
we can write, where $k$ is an integer;
$a^3 - a = 3k $
$a(a^2 - 1) = 3k $
Now if $a = k$ then
$a^2 -1 = 3$ and $a= \pm2 $ so $ a^3 - a = 24 = 3 \times 8$
If $ a $ is not equal to $k$;
then
$a(a^2 - 1) = a(a+1)(a-1) = 3k$
since $a(a+1)(a-1)$ is the product of 3 consecutive integers, the expression is divisible by 3.
Is this ok?, just a check. I'm not really all that good at number theory.
| In fact, divisible by $6$. The product of three consecutive integers is divisible by $3!$. Might as well take them naturals. Note that the quotient is
$$\frac{(a+1)a(a-1)}{1\cdot 2\cdot 3} = \binom{a+1}{3}$$
Similarly you can show that the product of $n$ consecutive integers is divisible by $n!$.
Another important generalization is the fact discovered by Fermat
$p$ divides $a^p-a$ for any prime $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Problem with the rational root theorem Consider this polynomial: $f(x)=(2x+5)(x-3)(x+8/3)=0$. Then $f(x)=2x^3+...+(-40)$
Here is a list of all factors of $40$ and $2$:
$40$: $±1$, $±2$, $±4$, $±5$, $±8$, $±10$, $±20$
$2$: $±2$, $±1$
Now, $3$ is clearly a root, but there is no combination between factors of $40$ divided by factors of $2$ that gives $3$ as a result. What am I doing wrong?
| The problem is that for the rational root theorem to work, we need all the coefficients to be integral.
Expanding, we have
$$f(x)=(2x+5)(x-3)(x+8/3)=2x^3+\frac{13}{3}x^2-\frac{53}{3}x-40$$
Now notice that the solutions to $f(x)=0$ are the same as the solutions to $3f(x)=0$, so we can safely multiply through by $3$ to obtain the equation
$$6x^3+13x^2-53x-120=0$$
Now because all of the coefficients are integral, we can apply the rational root theorem and find that $3$ may be a root (since $3\mid120$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1109724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help me find the following limit : $\lim_{{n}\to{\infty}} (\frac{2^x+3^x+\cdots+n^x}{n-1})^\frac{1}{x} = ?$ I have no idea where to start.$$\begin{align}\lim_{{n}\to{\infty}} \left(\dfrac{2^x+3^x+\cdots+n^x}{n-1}\right)^{1/x} = ?, n >1\\\end{align}$$
| Hint: Assume that $x > 0$, $\ln f(x) = \dfrac{1}{x}\cdot \ln\left(\dfrac{2^x+3^x+\cdots n^x}{n-1}\right)=\dfrac{1}{x}\cdot \left(\ln\left(\dfrac{2^x+3^x+\cdots n^x}{n^x}\right)+x\ln\left(\dfrac{n}{n-1}\right)\right)$.
Now we find the limit as $n \to \infty$ of the quotient: $q_n(x)=\dfrac{2^x+3^x+\cdots n^x}{n^x}$ by using Stolz-Cesaro's theorem:
$\displaystyle \lim_{n\to\infty} q_n(x) = \displaystyle \lim_{n\to \infty} \dfrac{(n+1)^x}{(n+1)^x-n^x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Integral of $\cos^4(2t)\,dt$ with bounds from $0$ to $\pi$ $$\int_0^\pi\cos^4(2t)\,dt=?$$ I have attempted this problem two different ways and got two different answers that are nowhere near the correct answer. Could you please show me detailed steps on how to work this problem out. The final answer is $\frac{3\pi}{8}$.
| Expanding on Alex Silva's answer:
Since
$\cos^2(x) = \frac{1+\cos(2x)}{2}$,
$\begin{array}\\
\cos^4(x)
&= \frac{(1+\cos(2x))^2}{4}\\
&= \frac{1+2\cos(2x)+\frac{1+\cos(4x)}{2}}{4}\\
&= \frac{2+4\cos(2x)+1+\cos(4x)}{8}\\
&= \frac{3+4\cos(2x)+\cos(4x)}{8}\\
\end{array}
$
so
$\cos^4(2x)
=\frac{3+4\cos(4x)+\cos(8x)}{8}
$.
The integrals of the
$\cos(4x)$
and
$\cos(8x)$
terms are zero,
so the result is
$\frac{3\pi}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1111240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Proving that if $xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}}$, then $x+y+z\geq \sqrt{3}$ If $x, y, z$ are positive real numbers such that
$$xy + yz + zx \geq \frac{1}{\sqrt{x^2+y^2+z^2}},$$
then prove that $x+y+z\geq \sqrt{3}$.
| $(x+y+z)^2 \geq 3 \iff x^2+y^2+z^2 + 2(xy+yz+zx) \geq 3 \iff x^2+y^2+z^2+ \dfrac{2}{\sqrt{x^2+y^2+z^2}}\geq 3 \iff t^2 + \dfrac{2}{t} \geq 3$. By AM-GM inequality: $t^2+\dfrac{2}{t} =t^2+ \dfrac{1}{t}+\dfrac{1}{t}\geq 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Conditional Distributions Choose a random integer $X$ from the interval
$[0, 4]$. Then choose a random integer $Y$ from the interval
$[0, x]$, where $x$ is the observed value of $X$. Make assumptions
about the marginal pmf $f_X(x)$ and the conditional
pmf $h(y | x)$ and compute $P(X + Y > 4)$.
I tried to make a sample space of the ordered pairs and came up with:
$S=$ { $(0,0),(1,0),(1,1),(2,0),(2,1),(2,2),(3,0),(3,1),(3,2),(3,3),(4,0),(4,1),(4,2),(4,3),(4,4)$ }
(a total of 15 ordered pairs)
Next, I defined the pmf $f_X(x)$ as:
$f_X(x) =$
$\frac{1}{15} , x=0;\frac{2}{15} , x=1;\frac{3}{15} , x=2;\frac{4}{15} , x=3;\frac{5}{15} , x=4;$
(sorry for the messy piecewise display)
My idea ended up being wrong. Can someone help me with this problem?
| If $X=3$, then we get sum $\gt 4$ with probability $\frac{2}{4}$, and if $X=4$ we get sum $\gt 4$ with probability $\frac{4}{5}$, so the required probability is $\frac{1}{5}\cdot \frac{2}{4}+\frac{1}{5}\cdot \frac{4}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Symmetric and homogeneous three variable inequality with radicals. While trying to solve a problem, I got the following inequality which appears correct, but I cannot prove. For positive $x, y, z$,
$$\sum_{cyc} \frac{x}{y^2+z^2} \ge \sum_{cyc} \sqrt{\frac{y^2+z^2}{2(x^2+y^2)(z^2+x^2)}}$$
Equality is obviously when $x=y=z$. Tried AM-GM (takes it in opposite directions), Cauchy Schwarz (not tight enough) etc. with no luck. Adding constraints doesn't seem to simplify either. Any hints welcome.
| By C-S $\left(\sum\limits_{cyc}\sqrt{\frac{y^2+z^2}{2(x^2+y^2)(x^2+z^2)}}\right)^2\leq\sum_(y^2+z^2)\sum\limits_{cyc}\frac{1}{2(x^2+y^2)(x^2+z^2)}=\frac{2(x^2+y^2+z^2)^2}{\prod\limits_{cyc}(x^2+y^2)}$.
Thus, it remains to prove that
$\left(\sum\limits_{cyc}(x^5+x^3y^2+x^3z^2+x^2y^2z)\right)^2\geq2(x^2+y^2+z^2)^2(x^2+y^2)(x^2+z^2)(y^2+z^2)$.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where
$f(w^3)=(81u^5-135u^3v^2+54uv^4+9u^2w^3-5v^2w^3)^2-$
$-2(3u^2-2v^2)^2(81u^2v^4-54v^6-54u^3w^3+36uv^2w^3-w^6)$.
We see that $f'(w^3)=2(81u^5-135u^3v^2+54uv^4+9u^2w^3-5v^2w^3)(9u^2-5v^2)+$
$+2(3u^2-2v^2)^2(54u^3-36uv^2+2w^3)\geq0$.
Id est, $f$ is an increasing function, which says that $f$ gets a minimal value for extremal value of $w^3$, which happens in two following cases only.
*
*$y=1$ and $z\rightarrow0^+$, which gives $(x^2+1)^2(x^4+2x^3+x^2+2x+1)(x-1)^2$;
*$y=z=1$, which gives $(x^2+1)^2(x^3+2x^2+x+8)x(x-1)^2\geq0$. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Polynom equality modulo p I found these two equations:
(a) $$X^4 + 1 \equiv (X + 1)^4 \mod \ 2$$
(b) $$X^4 + 1 \equiv (X^2 - X - 1)(X^2 + X - 1) \mod \ 3$$
I would like to understand the concept of modulo for Polynoms.
How were they made? And how to verify these equality ?
Thanks in Advanced!
| For $(a),$
$y^{m+1}-y=y(y^m-1)$ is divisible by $y(y-1)$ which being the product of two consecutive integers is even
$\implies y^{m+1}\equiv y\pmod2$ for $m\ge1$
$\implies y^4\equiv y\pmod2$ set $y=X, X+1$
For $(b),$
$(x^2-1-x)(x^2-1-x)=(x^2-1)^2-x^2=x^4-3x^2+1\equiv x^4+1\pmod3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$ Let a and b be natural numbers so that $2a-1,2b-1$ and $a+b$ are prime numbers. Prove that $a+b$ can't divide $a^a+b^b$ nor $a^b+b^a$.
I get that $gcd(a,b)=1$.
I haven't got anything special for now but if I do I will update the question.
| Since $2a-1, 2b-1$ are primes, then $a,b\neq 1$. Since $a+b$ is a prime, then we can suppose $a$ is even and $b$ is odd.
Suppose $p=a+b|a^a+b^b$. We have: $$a^a+b^b\equiv a^a-a^b=a^b(a^{a-b}-1) \pmod {a+b}$$
Since $\gcd(a^b,a+b)=1$, then $p|a^{a-b}-1$. Let $h=ord_p(a)$, then $h|p-1,h|a-b$, then $h|p-1-(a-b)=2b-1$, thus $h=1$ or $h=2b-1$.
If $h=1$, then $a+b|a-1$, but this is impossible since $a+b>a-1$.
Then $h=2b-1$, which means $p|a^{2b-1}-1$. Hence $p|a^{2a-1}(a^{2b-1}-1)=a^{2a+2b-2}-2^{2a-1}$, then $p|2^{2a-1}-1$. Thus $2b-1|2a-1$, or $2a-1=2b-1$ (since they are both primes), or $a=b$, which is impossible.
Proving $a+b \not|a^b+b^a$ is similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate $\lim_{n\to\infty}\int_0^n\frac{x^2+a^2}{x^4+b^2x^2+b^4}dx$ Evaluate this limit:
$$\lim_{n\to\infty}\int_0^n\dfrac{x^2+a^2}{x^4+b^2x^2+b^4}dx$$
I tried to simplify this fraction. I noticed that $x^4+b^2x^2+b^4$ can be written as
$$\dfrac{x^6-b^6}{x^2-b^2}$$
Then limit will be
$$\lim_{n\to\infty}\int_0^n\dfrac{\left(x^2+a^2\right)\left(x^2-b^2\right)}{x^6-b^6}dx$$
But what next? Is it possible to more simplify this fraction? Can we find indefinite integral of this fraction first?
| Maple does the indefinite integral in terms of arctangents and logarithms. So, yes, you can do it by partial fractions, where you first factor the denominator ... to factor it, complete the square, so it becomes a difference of squares
Then take limit to get the answer. Maple says:
$$
\frac{(a^2+b^2)\pi}{b^3 2\sqrt{3}}
$$
assuming $b > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question about series convergence $\sum_{n=1}^\infty \frac{1}{n}$ and $\sum_{n=1}^\infty \frac{1}{n^2}$ So I have been playing around with convergent series recently and I still have a hard time understanding why $\sum_{n=1}^\infty \frac{1}{n}$ diverges and $\sum_{n=1}^\infty \frac{1}{n^2}$ converges.
I can write out the first couple of terms of $\sum_{n=1}^\infty \frac{1}{n}$ and compare them to a second series:
$\color{red}{1}+\color{green}{\frac{1}{2}}+\color{blue}{\frac{1}{3}+\frac{1}{4}}+\color{maroon}{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}+...$
$\color{red}{1}+\color{green}{\frac{1}{2}}+\color{blue}{\frac{1}{4}+\frac{1}{4}}+\color{maroon}{\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}}+...$
The colored terms in the second series always add up to $\frac{1}{2}$ and the corresponding colored terms in the first series always add up to more than $\frac{1}{2}$. Since the second series diverges the first one must also diverge. This still makes perfect sense. However if I write out the first couple of therms of
$\sum_{n=1}^\infty \frac{1}{n^2}$
$\frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}+\frac{1}{49}+\frac{1}{64}...$
then how does this converge?
Why is it not possible to group them similar to the first series? It seems to me that if I have infinitely many numbers I will be able to group $n$ of them to get to some number $k$ that I can add up forever.
Thanks in advance.
| Think about Zeno's paradox in reverse.
$$ \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots $$
You want to walk a mile.
First you walk half of it. Then take a bit of rest. Then walk half of the remaining, then take a bit of rest. And so on.
Will you ever go beyond the mile?
You have an infinite number of numbers (of the form $\frac{1}{2^n}$), but you can never go beyond the mile. Basically, the numbers are decreasing too fast, that no matter how many you take, you still are bounded.
In this case, you can prove that (using $\frac{1}{n^2} \lt \frac{1}{(n-1)(n+1)}$)
$$ 1 + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} \lt 2 -\frac{1}{n}$$
So it is in fact bounded, and you cannot go beyond $2$, no matter how many terms you take.
(And any bounded monotonic sequence converges).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why do we put absolute brackets for ln? When writing out the final answer in $\ln$ form, why is it necessary to put absolute brackets? How does it affect the answer?
I have this answer of $-3\ln|\frac{3+\sqrt{9-x^2}}{x}|$, but why does it suddenly become
$3\ln|\frac{\sqrt{9-x^2}-3}{x}|$?
| In this case it's exactly the same:
\begin{align}
\left|\frac{3+\sqrt{9-x^2}}{x}\right|^{\!-1}
&=
\left|\frac{x}{3+\sqrt{9-x^2}}\right|\\[2ex]
&=
\left|\frac{x}{3+\sqrt{9-x^2}}\frac{3-\sqrt{9-x^2}}{3-\sqrt{9-x^2}}\right|\\[2ex]
&=\left|\frac{x(3-\sqrt{9-x^2})}{9-9+x^2}\right|\\[2ex]
&=\left|\frac{3-\sqrt{9-x^2}}{x}\right|\\[2ex]
&=\left|\frac{\sqrt{9-x^2}-3}{x}\right|
\end{align}
Thus
$$
-3\log\left|\frac{3+\sqrt{9-x^2}}{x}\right|=
3\log\left|\frac{3+\sqrt{9-x^2}}{x}\right|^{\!-1}=
3\log\left|\frac{\sqrt{9-x^2}-3}{x}\right|
$$
Why putting the absolute value when writing
$$
\int\frac{1}{x}\,dx=\log|x|+c
$$
and not leaving just $x$? Because this works independently whether the interval where the integral is done is a subset of $(-\infty,0)$ or of $(0,\infty)$. However, one should always recall that such a notation has a meaning only if the integrand is considered defined on an interval.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
What is the general term of $a_{n+1}=\frac{2a_n-1}{5a_n-1} \ , \ \ a_1=1$? I've struggled to solve this exercise
$$a_{n+1}=\frac{2a_n-1}{5a_n-1}\ , \ \ a_1=1$$
$$b_{n+1}=(5a_n-1)b_n \ , \ \ b_1=1$$
Find $b_{\ 40}$ .
$$$$
I thought 'taking inverse' will be useful, but not yet... :-(
$\color{red}{01.}$ How can I find $b_{\ 40}$ ?
$\color{red}{02.}$ How can I find the general term of $a_n$, $b_n$ ?
$$$$
Thank you for your attention to this matter.
$$$$
| here is a partial answer:
the eigenvalues of $A = \pmatrix{2 & -1\\5&-1}$ are $\lambda_{1,2} = \dfrac{1}{2} \pm i\dfrac{\sqrt{11}}{2}=\sqrt 3(\cos t \pm i\sin t)$ and the corresponding eigenvectors are
$u_{1,2}=\pmatrix{2\\3\mp i\sqrt{11}}$
let $a, b, c$ be complex numbers such that
$z = au_1 + bu_2 = a \pmatrix{2\\3 -i\sqrt{11}} + b \pmatrix{2\\3 + i\sqrt{11}} = 2c\pmatrix{1\\1}$
we need $a + b = c, 3a + 3b +i\sqrt{11}(b-a)=2c$ that means
$i\sqrt{11}(a-b) = c, a + b = c$ we can choose
$c = 2\sqrt{11}, a = \sqrt{11} - i, b = \sqrt{11} + i$
$\begin{align}
A^k z &= A^k(au_1 +bu_2) = a\lambda_1^ku_1+b \lambda_2^k u_2 \\
& =3^{k/3} \{ a(\cos kt + i \sin kt)u_1 + b(\cos kt - i \sin kt)u_2 \}\\
& =3^{k/3} \pmatrix{2(a+b)\cos kt + 2(a-b)i \sin kt)\\
(a+b)(3\cos kt + \sqrt{11} \sin kt+i(a-b)(3\sin kt -\sqrt{11}\cos kt)} \\
& = 3^{k/3} \pmatrix{4(\sqrt{11}\cos kt + \sin kt)\\
2\sqrt{11}(3\cos kt + \sqrt{11} \sin kt)+2(3\sin kt -\sqrt{11}\cos kt) } \\
&=3^{k/3} \pmatrix{4(\sqrt{11}\cos kt + \sin kt)\\
4\sqrt{11}\cos kt + 28 \sin kt)} \\
\end{align}$
so $$a_k = \dfrac{\sqrt{11}\cos kt + \sin kt}{\sqrt{11}\cos kt + 7 \sin kt} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1118343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Integrating: $ \;\int \frac{1}{x^2+3x+2} dx $ How can I solve the following integral:
$$ \int \frac{1}{x^2+3x+2} dx $$
Should I proceed by changing the variable (substitution)? or should I use integration by parts? Or another method altogether?
Thank you!
| This integral is "screaming": Use partial fraction decomposition!, especially after we note that the denominator factors nicely $$x^2 + 3x + 2 = (x + 2)(x+1)$$
So our integral will take the form:
$$\int \frac 1{x^2 + 3x + 2} \,dx = \int \left(\frac A{x+2} + \frac B{x+1}\right)\,dx$$
Now, we solve for $A, B$ knowing that $$A(x+1) + B(x+2) = 0\cdot x + 1$$
If $x= -2$, then we have $A(-1) + 0\cdot B = 1\implies A = -1$.
If $x = -1,$ then we have $0\cdot A+ B = 1 \implies B = 1$.
So, the integral becomes $$\int \frac 1{x^2 + 3x + 2} \,dx = \int \left(\frac {-1}{x+2} + \frac 1{x+1}\right)\,dx =\int \left(\frac 1{x+1} - \frac 1{x+2}\right)\,dx$$
I trust you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the sum $x + \frac{x^3}{3} + \frac{x^5}{5} + ... $
Evaluate the sum $$x + \frac{x^3}{3} + \frac{x^5}{5} + ... $$
I was able to notice that: $$ \sum_{n=0}^\infty \frac{x^{2n-1}}{2n-1} = \sum_{n=0}^\infty \int x^{2n-2}dx = \lim_{N\to\infty} \sum_{n=0}^N \int x^{2n-2} dx $$
Where should I take it from here? (assuming I'm not the right way)
EDIT
Following the anwer:
$$\int \sum_{n=0}^\infty x^{2n-1} dx = \int \frac{\sum_{n=0}^\infty (x^2)^n}{x} dx = \int \frac{\frac{1}{1-x^2}}{x} dx= \int \frac{1}{x(1-x^2)}dx = \int \frac{1-x^2+x^2}{x(1-x^2)} = \int \frac{1}{x} dx + \int \frac{x}{1-x^2}dx = \ln(x) + \int \frac{1}{2}\frac{1}{1-t}dt + C = \ln(x) - \frac{\ln(x^2)}{2} + C$$
*
*Is that right?
*How to evaluate $C$?
| You got the index wrong. The actual sum is
$$ \sum_{n = 0}^{\infty} \frac{x^{2n+1}}{2n+1} = \int \sum_{n = 0}^{\infty} x^{2n} \,dx = \int \frac{1}{1-x^2}\,dx $$
This sum converges for $|x| < 1$ and the integral evaluates to
$$ f(x) = \frac{1}{2}\, \ln \left| \frac{1+x}{1-x} \right| + C $$
The constant $C$ can be found by looking at the series expansion (what's $f(0)$?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1120721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to show that $\sqrt[3]{-1+\sqrt{-7}}+\sqrt[3]{-1-\sqrt{-7}}$ is a real number at a time before the invention of complex numbers I have read this PDF from ocw.mit.edu about complex numbers. There is one interesting question: Imagine yourself at the time, when complex numbers had to be invented yet. How to show that $$\sqrt[3]{-1+\sqrt{-7}}+\sqrt[3]{-1-\sqrt{-7}}$$ is a real number.
| Let $$r = (-1 + \sqrt{-7})^{1/3} + (-1 - \sqrt{-7})^{1/3},$$ where by $\sqrt{-7}$ we mean some number $x$ whose square is $-7$; i.e., $x^2 = -7$. Recall the identity $$(a^{1/3} + b^{1/3})^3 = a + b + 3(ab)^{1/3}(a^{1/3} + b^{1/3}).$$ Consequently, $$r^3 = -2 + 3((-1)^2 + 7)^{1/3}r = 6r - 2.$$ Now consider the trigonometric identity $$\begin{align*} \cos 3\theta &= \cos \theta \cos 2\theta - \sin \theta \sin 2\theta \\ &= \cos^3 \theta - \cos \theta \sin^2 \theta - 2 \sin^2 \theta \cos \theta \\ &= \cos^3 \theta - 3 \cos\theta \sin^2 \theta \\ &= \cos^3 \theta - 3 \cos\theta (1 - \cos^2 \theta) \\ &= 4 \cos^3 \theta - 3 \cos \theta. \end{align*}$$ This suggests the choice $r = 2 \sqrt{2} \cos \theta$ gives $$\begin{align*} r^3 - 6r + 2 &= 16 \sqrt{2} \cos^3 \theta - 12 \sqrt{2} \cos \theta + 2 \\ &= 4 \sqrt{2} (4 \cos^3 \theta - 3 \cos \theta) + 2 \\ &= 4 \sqrt{2} \cos 3\theta + 2 = 0. \end{align*}$$ Consequently, $$\theta = \frac{1}{3}\cos^{-1}\left( - \frac{1}{2 \sqrt{2}} \right),$$ hence $$r \in 2 \sqrt{2} \cos \left( \frac{1}{3} \cos^{-1} \left( - \frac{1}{2\sqrt{2}} \right) + \frac{2\pi k}{3}\right), \quad k = 0, 1, 2,$$ all of which are real values. Of course, this result presupposes that the rules that allow us to work with real numbers also work for numbers like $\sqrt{-7}$. To be completely rigorous, in my opinion, requires an axiomatic treatment of the complex numbers. The above really only demonstrates that $r$ is real if the rules we use for arithmetic are extended to such values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Product of Divisors of some $n$ proof The function $d(n)$ gives the number of positive divisors of $n$, including n itself.
So for example, $d(25) = 3$, because $25$ has three divisors: $1$, $5$, and $25$.
So how do I prove that the product of all of the positive divisors of $n$ (including $n$ itself) is $n^{\frac{d(n)}{2}}$.
For example, the divisors of $12$ are $1$, $2$, $3$, $4$, $6$, and $12$. $d(12)$ is $6$, and
$1 · 2 · 3 · 4 · 6 · 12 = 1728 = 12^3 = 12^{\frac{6}{2}} = 12^{\frac{d(n)}{2}}$
| The solution by Pedro Tamaroff is far more compact, and better. We will give the same proof in a more long-winded way.
First we mention a fact about the private lives of the divisors of $n$. It is easy to see that $d$ is a divisor of $n$ if and only if $\frac{n}{d}$ is a divisor of $n$. It turns out that $d$ and $\frac{n}{d}$ are a couple, or to put it in the language of today, they are partners. In your example, $1$ and $12$ are partners, as are $2$ and $6$, as are $3$ and $4$. Note that the product of any $2$ partnered numbers is $n$.
Then if $n$ is not a perfect square, the divisors of $n$ are divided into couples. If $n$ is a perfect square, then $\frac{n}{\sqrt{n}}=\sqrt{n}$, so in that case all the divisors of $n$ are coupled except for $\sqrt{n}$.
Let us look first at the case $n$ not a perfect square. Then there are $\frac{d(n)}{2}$ couples. The product of the elements in any couple is $n$, so the product of all the divisors of $n$ is $n^{d(n)/2}$.
Now suppose that $n$ is a perfect square. Then there are $\frac{d(n)-1}{2}$ couples plus a solitary individual $n^{1/2}$.
The product of the elements in any couple is $n$, so the product of all the coupled elements is $n^{(d(n)-1)/2}$. Multiply by $n^{1/2}$. We get $n^{d(n)/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
If $a,b,c\in\mathbb{R^+}$ such that $ abc = 1 $ and $ ab + bc + ca = 5 $. Prove that $ 17/4 \leq (a+b+c)\leq 1+ \sqrt{32}. $ If $a,b,c\in\mathbb{R^+}$ such that $ abc = 1 $ and $ ab + bc + ca = 5 $. Prove that
$$
\frac{17}{4} \leq (a+b+c)\leq 1+ \sqrt{32}.
$$
My attempt
Tried using Vieta but it didn't work. Also I used some standard inequalities but got $ \geq\sqrt{15} $.
| WLOG, using the symmetry, we may assume $a \le b \le c$, so $a \le 1, \; c \ge 1$.
For the left inequality, note that
$$a+b+c-\frac{17}4 = \frac{5-1/c}{c}+c-\frac{17}4 = \frac{(c-2)^2(4c-1)}{4c^2} \ge 0$$
and as equality is achieved when $a=\frac14, b=c=2$ (or any permutation), this in fact gives the minimum.
Similarly, for the right inequality, note
$$a+b+c-(1+4\sqrt2) = a+\frac{5-1/a}{a}-(1+4\sqrt2) \\ = -\frac{(a+1-\sqrt2)^2(3+2\sqrt2-a)}{a^2} \le 0$$
and as equality is possible when $a=b=\sqrt2-1, c= 3+2\sqrt2$, this gives the maximum.
P.S. For a bonus, the above expressions also indicate that $\min(a, b, c) \ge \frac14$ and $\max(a, b, c) \le 3+\sqrt2$ (why?).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Integration of the following What is the definite integral of
$$
\int_0^1 \left(\frac{g(x)}{f(x)}\right)'\cdot\frac{1}{g(x)}\,dx,
$$
where the conditions are as follows:
$f(0) = 2 $
$f(1) = 3 $
$f'(x) $ is continuous
For all $x$ in the range $\{0,1\}, f(x)^2-g(x)^2=1$.
Thank you very much.
| By integration by parts,
\begin{align}\int_0^1 \left(\frac{g}{f}\right)' \frac{1}{g}\, dx &= \frac{g}{f}\cdot \frac{1}{g}\bigg|_{t = 0}^{t = 1} - \int_0^1 \frac{g}{f}\frac{d}{dx}\left(\frac{1}{g}\right)\, dx\\
&= \frac{1}{f(1)} - \frac{1}{f(0)} + \int_0^1 \frac{g'}{fg}\, dx\\
&= -\frac{1}{6} + \int_0^1 \frac{g'}{fg}\, dx.
\end{align}
Differentiating both sides of the equation $f^2 - g^2 = 1$, we get $2ff' - 2gg' = 0$, or $ff' = gg'$. So $g'/f = f'/g$, in which case $$\int_0^1 \frac{g'}{fg}\, dx = \int_0^1 \int_0^1 \frac{f'}{g^2}\, dx = \int_0^1 \frac{f'}{f^2 - 1}\, dx = \int_{2}^{3} \frac{du}{u^2 - 1} = \frac{1}{2}\log\left|\frac{u - 1}{u + 1}\right|\bigg|_{u = 2}^{u = 3} = \frac{1}{2}\log\frac{3}{2}.$$
Hence $$\int_0^1 \left(\frac{g}{f}\right)'\frac{1}{g}\, dx = -\frac{1}{6} + \frac{1}{2}\log\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1126643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Relation between two sets of generators of SO(3) I am working with the spin 1 representation of SU(2), which is just SO(3). The ordinary generators used in quantum mechanics are:
$J_x = \left(
\begin{array}{ccc}
0 & \frac{1}{\sqrt{2}} & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
0 & \frac{1}{\sqrt{2}} & 0 \\
\end{array}
\right)$; $J_y = \left(
\begin{array}{ccc}
0 & -\frac{i}{\sqrt{2}} & 0 \\
\frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\
0 & \frac{i}{\sqrt{2}} & 0 \\
\end{array}
\right)$; $J_z = \left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & -1 \\
\end{array}
\right)$.
But I can also work with just the ordinary 3D rotation matrices that rotate vectors specified by (x,y,z) components. From taking the infinitesimal limit of these matrices, I find that the generators are:
$J_x' = \left(
\begin{array}{ccc}
0 & 0 & 0 \\
0 & 0 & -i \\
0 & i & 0 \\
\end{array}
\right); J_y' = \left(
\begin{array}{ccc}
0 & 0 & i \\
0 & 0 & 0 \\
-i & 0 & 0 \\
\end{array}
\right); J_z' = \left(
\begin{array}{ccc}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right)$.
After exponentiating, I get rotation operators $D(\psi,\theta,\phi)$ and $D'(\psi,\theta,\phi)$, where the $D$ operators act on vectors written in the j=1, m =-1,0,1 basis and the $D'$ operators act on vectors written in a different basis.
What is the relation between these two operators (or between the generators)? Are they related by a unitary transformation? If so, how does one go about finding that unitary transformation? (That is what I first suspected, but I was unable to find any unitary transformation that works.) Or are they not related in a direct way?
EDIT:
Corrected sign error in generators.
| As per the suggestion of Phoenix87, we can find the unitary transformation by noticing that the transformation that diagonalizes $J_z'$ will also transform $J_z'$ to into $J_z$ (since $J_z$ is diagonal). So we find the eigenvectors of $J_z'$ and then use them as the columns of our unitary transformation. We are free, however, to choose the phase of the eigenvectors, and so we make the choice that gives $U^\dagger J_x' U = J_x$ and $U^\dagger J_y' U = J_y$. When all is said and done, we obtain:
$U = \left(
\begin{array}{ccc}
-\frac{i}{\sqrt{2}} & 0 & \frac{i}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
0 & i & 0 \\
\end{array}
\right)$.
Which gives $U^\dagger J_i' U = J_i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1129400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inverse Laplace Transform of $\frac{s-3}{s[(s-3)^2+9]}$ I am having problems solving this inverse Laplace transform:
ℒ$^{-1}\Large [\frac{s-3}{s[(s-3)^2+9]}]$
I did partial fraction decomposition, but ended up with complex expressions in some denominators, and I have no idea what do do from here or if this is even how to "solve" it.
Could anyone give me some helpful hints or references? I would appreciate it a lot!
Edit:
My problem is, specifically, finding the inverse Laplace of $\frac{1}{s^2-6s+18}$.
| Note $\dfrac{s-3}{s\left[(s-3)^2+9\right]}$ can be expressed as
\begin{align*}
\frac{s-3}{s\left[(s-3)^2+9\right]}&=\frac{s-3}{s(s^2-6s+18)}\\
&=\frac{A}{s}+\frac{Bs+C}{s^2-6s+18}
\end{align*}
For some real numbers $A$, $B$ and $C$, that means
\begin{align*}
s-3&=A(s^2-6s+18)+(Bs+C)s\\
&=(A+B)s^2+(-6A+C)s+18A\\
\end{align*}
$$\iff \left\{\begin{matrix}A+B&=0\\-6A+C&=1\\18A&=-3\end{matrix}\right.$$
Then $A=-\frac{1}{6}$, $B=\frac{1}{6}$ and $C=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simple algebra involving trigonometry, but confusing How do I get from
$$\frac{\sqrt3}{2} + \frac12 \tan x = 2 \tan x \cdot \frac{\sqrt3}{2}$$
to
$$\frac{\sqrt3}{2 \sqrt3 - 1} = \tan x$$
and then to
$$11 \tan x = 6 + \sqrt3$$
| I will try to make it a bit more elaborate since you are going to need to be able to manipulate equations like this all the time.
Multiply both sides by 2:
$\implies \sqrt{3}+\tan(x)=2\tan(x)\sqrt{3}$
Subtract $\tan(x)$ from both sides in order isolate $\tan(x)$
$\implies \sqrt{3}=2\tan(x)\sqrt{3}-\tan(x)$
You have two terms with $\tan(x)$ on the RHS so you can factor out $\tan(x)$
$\implies \sqrt{3}=\tan(x)(2\sqrt{3}-1)$
Divide both sides by $(2\sqrt{3}-1)$
$\implies \tan(x)=\frac{\sqrt{3}}{2\sqrt{3}-1}$
Now you can multiply the fraction on the RHS by $\frac{2\sqrt{3}+1}{2\sqrt{3}+1}$ in order to get rid of the sqare root in the denominator. Remember you are just multiplying by 1 so it doesn't change the equation
$\implies \tan(x)=\frac{\sqrt{3}}{2\sqrt{3}-1}\cdot \frac{2\sqrt{3}+1}{2\sqrt{3}+1}$
Just do a bit of multiplication and you will get
$\tan(x)=\frac{6+\sqrt{3}}{11} \iff 11\tan(x)=6+\sqrt{3} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Taylor Series $(x+2)/(2-3x)$ at $x=2$ How can I find Taylor series for $$\frac{(x+2)}{(2-3x)}$$ at $x=2$?
| Hints:
$$\frac{x+2}{2-3x}=(x+2)\;\frac13\;\frac1{2-x-\frac43}=-\frac{x+2}3\;\frac34\;\frac1{1+\frac{3(x-2)}4}=$$
$$=-\frac{x+2}4\left(1-\frac{3(x-2)}4+\frac{9(x-2)^2}{16}-\ldots\right)$$
Well, now find the general expression for the above and for what values of $\;x\;$ it is valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the solutions in positive integers $m^2 + 615 =2^n$ Find all positive integer solutions $m$ and $n$ of the equation
$$ m^2 + 615 =2^n. $$
| Note that $3 \mid 615$ and $3$ is never a factor of $2^n$, so $3 \nmid m$.
Therefore $m^2 \equiv 1 \bmod 3$ and thus also $2^n\equiv 1 \bmod 3$.
Odd powers of $2$ are $\equiv 2 \bmod 3$ and even powers of $2$ are $\equiv 1 \bmod 3$, so we know that $n$ is even. So say $n=2k$.
Now we have $m^2 +615 =2^{2k}$ so $615= 2^{2k}-m^2 = (2^k)^2-m^2 = (2^k+m)(2^k-m)$
The factor pairs of $615$ are $\{(1,615),(3,205),(5,123),(15,41)\}$ so the possible values of $2^k$ from those are $\{308,104,64,28\}$. Only $64$ is a power of $2$ among these, so this gives $k=6$ and
$\hspace{2in}\boxed{ n=12 \\ m=123-64=59}$
Check: $59^2+615 = 3481+615= 4096=2^{12} \quad\bigstar$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Under what conditions to have $ (b+1)^n \bmod 2b = (b+1)$ Under What conditions, for any integer $n \geq 1$, we will have:
$$ (b+1)^n \bmod 2b = (b+1)$$
I tested it with a couple of numbers, and it is true for $3, 5, 7$, etc.
Please prove your theory.
It is based on a comment of this question
| For $b$ odd, $b=2k+1$ and $$b^2=b(2k+1)=2b\cdot k+b \equiv b \bmod (2b)$$
Now $$(b+1)^2 = b^2+2b+1 \equiv b^2+1 \equiv (b+1) \bmod (2b)$$ as required.
Then given $(b+1)^k \equiv (b+1)\bmod(2k)$,
clearly $(b+1)^{k+1} = (b+1)^k(b+1) \equiv (b+1)(b+1) \equiv (b+1) \bmod (2b)$,
and the result holds for all $n$.
For $b$ even, $b=2k$, $b^2 \equiv 0 \bmod (2b)$ and $(b+1)^2 \equiv 1 \bmod (2b)$,
so the original requirement does not hold.
Note that with even $b$ and odd $n$, the original condition will hold, since all even powers of $(b+1)$ will reduce to $1 \bmod (2b)$ So we could say:
$$ (b+1)^n \equiv (b+1) \bmod (2b) \qquad \text{whenever }b\text{ or }n\text{ odd}$$
We can also make some observations on $(b-1)^n \bmod (2b)$. For odd $b$, powers of $(b-1)$ alternate between $(b+1)$ and $(b-1) \bmod (2b)$, and for even $b$, powers of $(b-1)$ alternate between $1$ and $(b-1) \bmod (2b)$. So it turns out that
$$(b-1)^n \equiv (b-1) \bmod (2b) \qquad \text{for }n\text{ odd}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}$ The limit is:
$$
\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}
$$
When I use calculator and substitute $h$ with $0.000001$ and $-0.000001$, the result is:
$$
\frac{1}{5x^{\frac45}}
$$
My question is:
*
*How to do it without calculator.
*Show me the steps on how it's being done.
| Remember that
$$
a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4).
$$
Then,
$$
a-b = \frac{a^5-b^5}{a^4 + a^3b + a^2b^2 + ab^3 + b^4}
$$
Taking $a = (x+h)^{1/5}$ and $b=x^{1/5}$ we get that
\begin{align}
\lim_{h \to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h} &= \lim_{h \to 0}\frac{(x+h) -x}{h(a^4 + a^3b + a^2b^2 + ab^3 + b^4)} \\
&= \lim_{h \to 0}\frac{1}{(a^4 + a^3b + a^2b^2 + ab^3 + b^4)} \\
&= \frac{1}{5x^{4/5}},
\end{align}
since $\lim_{h\to 0}a = \lim_{h\to 0}b = x^{1/5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1137385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Prove $\lim_{x\to \infty} \frac{4x^2 - 7}{2x^3 - 5} = 0$ using $\epsilon$-$N$ limit definition I am having difficulties manipulating the problem so that I can find a $N$ value to choose.
Suppose $x > N$, then
$$\left|\frac{4x^2 - 7}{2x^3 - 5}\right| \leq \frac{4x^2}{|2x^3 - 5|} + \frac{7}{|2x^3 - 5|} $$
This is basically as far as I gone which I know is correct. I am not sure what would be the next step. Any help/hints would be appreciated!
Edit:
I found an interesting way to approach it, not sure if it is right.
Assume $x > 2$, then
$$\left|\frac{4x^2 - 7}{2x^3 - 5}\right| < \frac{4x^2}{2x^3} = \frac{2}{x} < \frac{2}{N} < \epsilon$$
That means I can choose $N = \max\left\lbrace 2, \frac{2}{\epsilon}\right\rbrace$
Is that correct?
| Dividing out an $x^2$, you have to bound
$$\left | \frac{4-\frac{7}{x^2}}{2x-\frac{5}{x^2}} \right |.$$
Suppose $x>\sqrt{5}$. Then $4-\frac{7}{x^2}>0$, so $\left | 4 - \frac{7}{x^2} \right | = 4 - \frac{7}{x^2}<4.$ Additionally $\left | 2x-\frac{5}{x^2} \right | = 2x - \frac{5}{x^2} > 2x-1.$ So the whole thing is less than
$$\frac{4}{2x-1}.$$
Additionally we will have $x>1$ here, so $2x-1>x$. Hence the whole thing is less than
$$\frac{4}{x}.$$
Can you take it from here?
The hard part was finding the $\sqrt{5}$. The idea is ultimately that (after factoring the $x^2$) the numerator is very nearly $4$ and the denominator is very nearly $2x$, provided $x$ is large. We just have to choose $x$ large enough to take advantage of that, and be sure to replace terms with things that make the quantity smaller. In the numerator this was easy, because that term makes the entire expression smaller. In the denominator it was harder, because the term we wanted to remove was actually making the entire expression bigger. So we had to tweak the "good" term to balance out removing the "bad" term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is $4 \times 6$ defined as $4 + 4 + 4 + 4 + 4 + 4$ or $6 + 6 + 6 + 6$? There are long debates among Indonesian netizens about this http://www.globalindonesianvoices.com/15785/is-4x6-the-same-as-6x4-this-primary-school-math-made-controversy-in-social-media/
| Are you aware, the multiplication operation is commutative over $\mathbb{N}$.
As such, $6 \times 4 = 4 \times 6$.
And we have also $4+4+4+4+4+4=24=6+6+6+6= 4 \times 6 = 6 \times 4$
EDIT: I forgot the following:
$6+6+6+6=(4+2)+(4+2)+(4+2)+(4+2)=(4+4+4+4)+(2+2+2+2)=(4+4+4+4)+(4+4)=4+4+4+4+4+4$
Thanks to other properties of the additive operation, such as commutativity and associativity...
EDIT2: Changed to $\mathbb{N}$ from $\mathbb{R}$ per suggestion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.