Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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The sum of powers of $2$ between $2^0$ and $2^n$ Lately, I was wondering if there exists a closed expression for $2^0+2^1+\cdots+2^n$ for any $n$?
| Clue:
If you are searching closed formulas to a series like:
$a+ab+ab^2+ab^3+\ldots+ab^n = S_n$
You can use the follow trick:
$a+ab+ab^2+ab^3+\ldots+ab^n = S_n$
so,
$b(a+ab+ab^2+ab^3+\ldots+ab^n) = bS_n = ab+ab^2+ab^3+ab^4+\ldots+ab^n+ab^{n+1}=$
$a+ab+ab^2+ab^3+\ldots+ab^n -a +ab^{n+1}= bS_n \iff$
$S_n-a+ab^{n+1}=bS_n \iff S_n(b-1)=a(b^{n+1}-b) \iff$
$$S_n=\frac{a(b^{n+1}-b)}{b-1}$$
But, $b=1$ isn't covered by this formula, but if b=1, the series become
$S_n=a+a+a+...+a$ where $a$ appears $n$ times and $S_n=na$.
To your question,
$a=1=2^0$ and $b=2$,
so
$S_n=a+ab+ab^2+ab^3+\ldots+ab^n=2^0+2^1+\ldots+2^n=\frac{1(2^{n+1}-1)}{2-1}=2^{n+1}-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/86838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that if $a\equiv b \pmod{kn}$ then $a^k\equiv b^k \pmod{k^2n}$ What I have done is this:
$a\equiv b \pmod{2n}$,
$a=b+c\times2n$, for some $c$,
$a^2=b^2+2b\times c\times2n+c^2\times2^2n^2$,
$a^2-b^2=(b\times c+c^2n)\times4n$, then
$a^2\equiv b^2\pmod{2^2n}$.
I think that this is right: what I DON’T understand is how to generalize this to:
$a\equiv b\pmod{kn}\Rightarrow a^k\equiv b^k \pmod{k^2n}$.
Please give me a hint.
| Since
$a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\ldots b^{k-1})$
we need only prove that
$a^{k-1}+a^{k-2}b+\ldots + b^{k-1}$ is divisible by $k$. But since $a \equiv b \ (\text{mod} \ k)$ we see that
$ a^{k-1}+a^{k-2}b+\ldots + b^{k-1} \equiv a^{k-1}+a^{k-1}+\ldots +a^{k-1} \equiv ka^{k-1} \equiv 0 \ (\text{mod} \ k). $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Compositeness of number $k\cdot 2^n+1$? Every odd prime number can be expressed in the form $k \cdot 2^n+1$ ,where $k$ is an odd number .
For $n>2$ number $k \cdot 2^n+1$ is composite if :
$1.$ $k\equiv 1 \pmod {30} \land (n\equiv 2 \pmod 4 \lor n \equiv 1 \pmod 2 ) $
$2.$ $k\equiv 3 \pmod {30} \land n\equiv 3 \pmod 4$
$3.$ $k\equiv 5 \pmod {30} \land n\equiv 0 \pmod 2$
$4.$ $k\equiv 7 \pmod {30} \land n\equiv 1 \pmod 2$
$5.$ $k\equiv 9 \pmod {30} \land n\equiv 0 \pmod 4$
$6.$ $k\equiv 11 \pmod {30} \land n\equiv 0 \pmod 2$
$7.$ $k\equiv 13 \pmod {30} \land n\equiv 1 \pmod 2$
$8.$ $k\equiv 17 \pmod {30} \land (n\equiv 1 \pmod 4 \lor n\equiv 0 \pmod 2)$
$9.$ $k\equiv 19 \pmod {30} \land (n\equiv 0 \pmod 4 \lor n\equiv 1 \pmod 2)$
$10.$ $k\equiv 21 \pmod {30} \land n\equiv 2 \pmod 4$
$11.$ $k\equiv 23 \pmod {30} \land (n\equiv 3 \pmod 4 \lor n\equiv 0 \pmod 2)$
$12.$ $k\equiv 25 \pmod {30} \land n\equiv 1 \pmod 2$
$13.$ $k\equiv 27 \pmod {30} \land n\equiv 1 \pmod 4$
$14.$ $k\equiv 29 \pmod {30} \land n\equiv 0 \pmod 2$
Are there some other similar relations between coefficient $k$ and exponent $n$ that ensure compositeness of number $k \cdot 2^n+1$ ?
| There are zillions of such relations. For example: $2^6+1$ is a multiple of $13$, so $k2^n+1$ is composite if $n\equiv6\pmod{12}$ and $k\equiv1\pmod{13}$. You can make as many of these as you want.
EDIT: In general, take any numbers $q$ and $s$, and any prime $p$ dividing $q2^s+1$, and any $r$ such that $2^r\equiv1\pmod p$, then $k2^n+1$ is composite (since a multiple of $p$) when $k\equiv q\pmod p$ and $n\equiv s\pmod r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/89871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\sum \limits_{n=1}^{\infty}n(\frac{2}{3})^n$ Evalute Sum
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$
How can you compute the limit of
$\sum \limits_{n=1}^{\infty} n(2/3)^n$
Evidently it is equal to 6 by wolfram alpha but how could you compute such a sum analytically?
| $$
\begin{align*}
\sum_{n=1}^\infty n(2/3)^n &=
\sum_{m=1}^\infty \sum_{n=m}^\infty (2/3)^n \\ &=
\sum_{m=1}^\infty \frac{(2/3)^m}{1-2/3} \\ &=
\frac{2/3}{(1-2/3)^2} = 6.
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/90637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
} |
Sum of squares of roots of a polynomial $P(x)$ Well, I recently proved a formula (at least, I think) to the sum of the inverse of the roots $x_{1}, x_{2}, x_{3},\ldots, x_{n} \in \mathbb{C}$, and $\neq 0$. It starts:
Let a polynomial $P(x) = a_nx^n+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_1x + a_0$ of roots $x_{1}, x_{2}, x_{3},\ldots,x_{n} \in \mathbb{C^*}$ and $a_n \neq 0$. So, $$\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3}+\cdots+ \frac{1}{x_n} = -\frac{a_1}{a_0}, \qquad a_0 \neq 0.$$
Now, I am trying to prove another formula, the sum of the square of the roots, but I think it's getting pretty difficult to me.
Let $x_{1}^{2} + x_{2}^{2} + x_{3}^{2}+\cdots+x_{n}^{2}=u.$ So, if $x_1\cdot x_2\cdot x_3 \cdots x_n = (-1)^n\frac{a_0}{a_n}$, then
$$x_1 =(-1)^n\frac{a_0}{(x_2\cdot x_3 \cdots x_n)a_n},$$ and
$$x_1^2 =(-1)^nx_1\frac{a_0}{(x_2\cdot x_3 \cdots x_n)a_n}.$$
So,
$$\begin{align*}
x_{1}^{2} &+ x_{2}^{2} +\cdots+x_{n}^{2}=u\\
&=(-1)^nx_1\frac{a_0}{(x_2\cdot x_3 \cdots x_n)a_n} + (-1)^nx_2\frac{a_0}{(x_1\cdot x_3 \cdots x_n)a_n} \\
&\qquad +\cdots+(-1)^nx_n\frac{a_0}{(x_1\cdot x_2 \cdots x_{n-1})a_n}.
\end{align*}$$
It can be written as
$$(-1)^n\cdot \frac{a_0}{a_n}\left(\frac{x_1}{x_2\cdot x_3 \cdots x_n} + \frac{x_2}{x_1\cdot x_3 \cdots x_n} +\cdots+ \frac{x_n}{x_1\cdot x_2 \cdots x_{n-1}}\right) = u,$$
And I'm stuck here. That's my question. How can I write $u$ in function of the coefficients (or its impossibility)? Any help will be very appreciated. I'm young, and I do not have experience with proving things. That's all.
Thank you.
| Consider this:
Equate your functionto zero: $\large a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x^1+a_0x^0=0$
With Vieta's formula, we get the value
*
*$\large x_1+x_2+x_3+...+x_n=\frac{a_{n-1}}{a_n}$
and
*$\large\left(x_1x_2+x_1x_3+x_1x_4+...+x_1x_n\right)+\left(x_2x_3+x_2x_4+x_2x_5+...+x_2x_n\right)+...+\left(x_{n-1}x_{n-1}+x_{n-1}x_n\right)=\frac{a_{n-2}}{a_n}$
square both side of first formula and we get
$\large\left(x_1+x_2+x_3+x_4+...+x_n\right)^2=\left(\frac{a_{n-1}}{a_n}\right)^2$
expand and we get
$\large x_1^2+x_2^2+x_3^2+x_4^2+...+x_n^2+2\left(\left(x_1x_2+x_1x_3+x_1x_4+...+x_1x_n\right)+\left(x_2x_3+x_2x_4+x_2x_5+...+x_2x_n\right)+...+\left(x_{n-1}x_{n-1}+x_{n-1}x_n\right)\right)=\left(\frac{a_{n-1}}{a_n}\right)^2$
subtitute the second equation and we get
$\large x_1^2+x_2^2+x_3^2+x_4^2+...+x_n^2+2\left(\frac{a_{n-2}}{a_n}\right)=\left(\frac{a_{n-1}}{a_n}\right)^2$
so we get the answer
$\large x_1^2+x_2^2+x_3^2+x_4^2+...+x_n^2=\left(\frac{a_{n-1}}{a_n}\right)^2-2\left(\frac{a_{n-2}}{a_n}\right)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Are we unable to find the extrema of the function $f(x,y) = \frac{y}{x^2+y^2}$ using the second partial test?
$$\begin{align*}
&f(x,y) = \frac{y}{x^2+y^2}\\
&f_{xx} = \frac{∂}{∂x}\left(-\frac{2xy}{(x^2+y^2)^2}\right)=-\frac{2y(y^2-3x^2)}{(x^2+y^2)^3}\\
&f_{yy} = \frac{∂}{∂x}\left(\frac{x^2-y^2}{(x^2+y^2)^2}\right)=\frac{2y(y^2-3x^2)}{(x^2+y^2)^3}
\end{align*}$$
$a$ and $b$ must be $(0,0)$ or at least one of them has to be $0$ to be critical points of the function
if the point is $(0,0)$, $f_{xx}$ or $f_{yy}$ will be undefined and finding the point $(?,0)$ for which $f_x$ or $f_y$ would $= 0$ would be mighty hard.
Are we unable to find the extrema of this function using the second partial test?
| First, you’ve some errors in your third line of calculations. The correct partials are:
$$\begin{align*}
&f_x(x,y)=-\frac{2xy}{(x^2+y^2)^2}\\
&f_y(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}\\
&f_{xx}(x,y)=\frac{2y(3x^2-y^2)}{(x^2+y^2)^3}\\
&f_{xy}(x,y)=\frac{2x(3y^2-x^2)}{(x^2+y^2)^3}\\
&f_{yy}(x,y)=\frac{3y(y^2-2x^2)}{(x^2+y^2)^3}
\end{align*}$$
Clearly $f_x(x,y)=0$ only if $xy=0$, i.e., at least one of $x$ and $y$ is $0$. Similarly, $f_y(x,y)=0$ only if $0=x^2-y^2=(x-y)(x+y)$, i.e., only if $y=x$ or $y=-x$. The only way to satisfy both of these conditions is to have $x=y=0$, and the function and its partial derivatives aren’t defined at $(0,0)$. Thus, you’re quite right: there is no point at which the second derivative test applies.
Rewriting the function in polar coordinates as
$$f(r,\theta)=\frac{r\sin\theta}{r^2}=\frac{\sin\theta}r$$
may help to explain what’s going on. As we travel around the circle $C_r$ of radius $r$ centred at the origin, the function value is $0$ where $C_r$ crosses the $x$-axis, reaches a maximum of $\frac1r$ where $C_r$ crosses the positive $y$-axis, and reaches a minimum of $-\frac1r$ where $C_r$ crosses the negative $x$-axis. But that high point with value $\frac1r$ on $C_r$ can’t be a local maximum of the function, because the value of the function gets larger as you move down the $y$-axis towards the origin: $\frac1r$ increases as $r$ decreases. Similarly, the low point on $C_r$ can’t be a local minimum of $f$, because $-\frac1r$ gets smaller (more negative) as you move up the negative $y$-axis and $r$ decreases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/91654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Systems of equations finding right triangles I need help setting up the equation for the question, "Find all right triangles for which the perimeter is $24$ units and the area is $24$ square units."
I know that the area is $A = \frac12 b h$ and perimeter is $P = a + b + h$.
Using this, would the system look something like?
$a + b + c = 24$
$\frac12b c = 24$
(I'm using $c$ for consistency)
Then continue on using substitution?
| Let A=24. I am assuming c as hypotenuse and a, b the other two sides of the triangle $\implies c^2=a^2+b^2$.
We now have $1/2*a*b= A$ and $a+b+c=A$
$(a+b)^2=(A-c)^2$
$\implies a^2+b^2+2ab= A^2-2Ac+c^2$
$\implies c=\dfrac{A^2-2ab}{2A}=10$
There we have ab=48 and c=10. So the possible right triangle is 6,8,10.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Is $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$?
Is $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})$ ?
$$\mathbb{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbb{Q}\}$$
$$\mathbb{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbb{Q} \rbrace $$
So if an element is in $\mathbb Q(\sqrt{2},\sqrt{3})$, then it is in $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, because $\sqrt{6} = \sqrt{2}\sqrt{3}$.
How to conclude from there?
| $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is clear.
Now note that $$(\sqrt{2} + \sqrt{3})^{-1} = \frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = \sqrt{3} - \sqrt{2}$$ hence $\sqrt{3} - \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence
$\sqrt{2} + \sqrt{3} + \sqrt{3} - \sqrt{2} = 2 \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Note that by a similar argument you get $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3}) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/93463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "84",
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Proving by mathematical induction: $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>2(\sqrt{n+1}-1)$
Possible Duplicate:
Proof of an inequality: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$
Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction
I've tried to work on this equation for about 4-5 Hours.
I'm trying to show that the following equation is true with induction:
$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}>2(\sqrt{n+1}-1)$
after about 4-5 hours and numerous attempts I couldn't really get it to work out.
Any ideas ?
| There are ways other than induction (such as comparison with an integral) that have much clearer intuitive content. But let's stick with induction. We need to deal with the induction step. Suppose that we know for a given $k$ that
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{k}}>2(\sqrt{k+1}-1).\qquad\qquad(\ast)$$
We wish to show that
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{k}}+ \frac{1}{\sqrt{k+1}} > 2(\sqrt{k+2}-1).\qquad\qquad(\ast\ast)$$
By $(\ast)$, we have
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots +\frac{1}{\sqrt{k}}+ \frac{1}{\sqrt{k+1}} >2(\sqrt{k+1}-1)+ \frac{1}{\sqrt{k+1}}.$$
We will be finished if we can prove that
$$2(\sqrt{k+1}-1)+ \frac{1}{\sqrt{k+1}} >2(\sqrt{k+2}-1).$$
Equivalently, by a little algebra, we want to show that
$$2\sqrt{k+1}+ \frac{1}{\sqrt{k+1}} >2\sqrt{k+2}.$$
Equivalently, by a little more algebra, we want to show that
$$2(k+1)+1>2\sqrt{k+2}\sqrt{k+1}$$
(we multiplied throgh by the positive number $\sqrt{k+1}$).
So we want to show that
$$2k+3 >2\sqrt{k^2+3k+2}.$$
Since both sides are positive, this is equivalent to showing that
$$(2k+3)^2>4(k^2+3k+2).$$
Now we are finished, for the left side is $4k^2+12k+9$, while the right side is $4k^2+12k+8$, so the left side is greater than the right side.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/96444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Isolate yaw-pitch-roll from rotation I have a transformation that acts as such:
$$RX=Y$$
Where $R_{3\times3}$ is a yaw-pitch-roll rotation matrix, and $X, Y$ are 3D vectors.
Yaw ($\alpha$)-pitch ($\beta$)-roll ($\gamma$) rotations are the product of the 3 rotation matrices:
$$R_x(\gamma)R_y(\beta)R_z(\alpha)$$
Given $X$ and $Y$, how can I find $\alpha$, $\beta$ and $\gamma$?
| The first answer to any question on how to find Euler angles is: don't; use quaternions instead.
If external factors force you to use Euler angles, you could proceed like this: If $X$ and $Y$ are identical, the rotation can be a rotation around $X$ through any angle. Otherwise, the rotation can be a rotation through any axis $A$ in the plane $A\cdot X=A\cdot Y$. If $X$ and $Y$ are antipodal, the rotation must be through an angle $\pi$. Otherwise, that plane is spanned by the two vectors $X+Y$ and $X\times Y$, so you can parametrize the possible axes as
$$A(\xi)=\cos\xi\frac{X+Y}{|X+Y|}+\sin\xi\frac{X\times Y}{|X\times Y|}$$ with $\xi\in[0,\pi[$. The rotation angle must be the angle between the perpendicular projections of $X$ and $Y$ into the plane perpendicular to the axis, which are
$$
\begin{eqnarray}
X_\perp&=&
X - (A\cdot X)A\\
&=&X-\cos\xi\frac{X^2+X\cdot Y}{|X+Y|}A\\
&=&X-\frac12\cos\xi\frac{(X+Y)^2}{|X+Y|}A\\
&=&X-\frac12\cos\xi|X+Y|A
\end{eqnarray}$$
(where I used $X^2=Y^2$), and likewise for $Y_\perp$, so the angle between them is
$$
\begin{eqnarray}
\phi(\xi)
&=&\arccos\frac{X_\perp\cdot Y_\perp}{|X_\perp||Y_\perp|}\\
&=&\arccos\frac{X_\perp\cdot Y_\perp}{X_\perp\cdot X_\perp}\\
&=&\arccos\frac{X\cdot Y -\frac12\cos\xi|X+Y|A\cdot(X+Y)+\frac14\cos^2\xi |X+Y|^2}{X\cdot X -\frac12\cos\xi|X+Y|A\cdot(X+X)+\frac14\cos^2\xi |X+Y|^2}\\
&=&\arccos\frac{X\cdot Y -\frac14\cos^2\xi (X+Y)^2}{X\cdot X -\frac14\cos^2\xi (X+Y)^2}\\
&=&\arccos\frac{\cos^2\xi (X+Y)^2-4X\cdot Y}{\cos^2\xi (X+Y)^2-4X\cdot X}\\
&=&\arccos\frac{\sin^2\theta-\cos^2\theta\sin^2\xi}{\cos^2\theta\cos^2\xi-1}\;,\\
\end{eqnarray}
$$
where $\theta$ is the angle between $X$ or $Y$ and $X+Y$, that is, half the angle between $X$ and $Y$.
Using $A(\xi)$ and $\phi(\xi)$, you can construct the rotation matrices $R(\xi)$ and extract the Euler angles.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a sum of appropriate values of $\cos$ and $\sin$ to determine the value of a series The task is to find a sum of multiple values $\cos$ and $\sin$ to determine the value of
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!}$$
Since I had no clue how to approach this I consulted Wolfram|Alpha which returned this result:
$$\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \frac{\sin(1)}{2} + \cos(1) - \cos(0)$$
So I wrote down the partial sums of the given series and $\sin(1)$ and $\cos(1)$:
$$
\qquad\qquad\quad\sum_{n=1}^\infty (-1)^n \frac{n}{(2n+2)!} = \quad - \frac{1}{4!} + \frac{2}{6!} - \frac{3}{8!} \cdots
$$
$$
\;\;\sin(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n+1}}{(2n+1)!} = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} \cdots
$$
$$
\cos(1) = \sum_{n=0}^\infty (-1)^n \frac{1^{2n}}{(2n \qquad)!} = 1 - \frac{1}{2!} + \frac{1}{4!} - \frac{1}{6!} \cdots
$$
Looking at the numbers I can see that Wolfram|Alpha's result is correct: $\frac{1}{2}1 - \frac{1}{2!} = 0$ and $\frac{1}{2}\frac{-1}{3!} + \frac{1}{4!} = \frac{1}{4!}$, so the $\cos(1)$-series is shifted by $1$ since there is no $1$ at the beginning of the given series, so it needs to be subtracted from $\cos(1)$: $-cos(0)=-1$. But how do I get here without Wolfram|Alpha?
| No idea how Wolfram does it. But I guess you just start with power series and play around.
\begin{equation}
X=\sum_{n=1}^\infty(-1)^n\frac{n}{(2n+2)!}=-\frac{1}{4!}+\frac{2}{6!}-\frac{3}{8!}+\dots
\end{equation}
The annoying part is the numerators. Try doubling it so that the numerators "keep up" with the denominators.
\begin{equation}
2X=-\frac{2}{4!}+\frac{4}{6!}-\frac{6}{8!}+\dots
\end{equation}
Now the numerators are consistently off the denominators by 2.
\begin{equation*}
2X=(-\frac{4}{4!}+\frac{6}{6!}-\frac{8}{8!}+\dots)+2(\frac{1}{4!}+\frac{1}{6!}-\frac{1}{8!}+\dots)
\end{equation*}
That second part is pretty much $\cos(1)$. And the numerators of course cancel out in the first part.
\begin{equation}
2X=(-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\dots)+2(\cos(1)-1/2)
\end{equation}
Now the first part is pretty much $\sin(1)$.
\begin{equation}
2X=(\sin(1)-1)+2\cos(1)-1
\end{equation}
\begin{equation}
X=\frac{\sin(1)}{2}+\cos(1)-1
\end{equation}
Hope that didn't seem too random. My approach is to try to build the desired series out of ones I know using various manipulations like calculus, combining and rearranging.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/96571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Selecting multiple balls of same color from different urns Urn I contains 10 balls: 4 red and 6 blue. Urn II contains 20 balls: 16 red and 4 blue. A single ball is drawn from urn I, and two balls are drawn from urn II. What is the probability that all three balls are the same color?
| Numer of all possible ways of selecting 1 ball from urn 1 and 2 balls from urn 2 is ${\left( \begin{gathered}
10 \\
1 \\
\end{gathered} \right)\left( \begin{gathered}
20 \\
2 \\
\end{gathered} \right)}$.
Number of ways to select 1 red ball from urn 1 and 2 red balls from urn 2 is ${\left( \begin{gathered}
4 \\
1 \\
\end{gathered} \right)\left( \begin{gathered}
16 \\
2 \\
\end{gathered} \right)}$.
Number of ways to select 1 blue ball from urn 1 and 2 blue balls from urn 2 is ${\left( \begin{gathered}
6 \\
1 \\
\end{gathered} \right)\left( \begin{gathered}
4 \\
2 \\
\end{gathered} \right)}$.
Hence, number of ways of selecting 1 ball from urn 1 and 2 balls from urn 2 such that they are of the same color is ${\left( \begin{gathered}
4 \\
1 \\
\end{gathered} \right)\left( \begin{gathered}
16 \\
2 \\
\end{gathered} \right) + \left( \begin{gathered}
6 \\
1 \\
\end{gathered} \right)\left( \begin{gathered}
4 \\
2 \\
\end{gathered} \right)}$ since they are mutually exclusive.
Since probability of selecting balls of the same color is number of ways that can be done divided by number of all possible ways of selecting 1 ball from urn 1 and 2 balls from urn 2, we have that wanted probability is
$\frac{{\left( \begin{gathered}
4 \\
1 \\
\end{gathered} \right)\left( \begin{gathered}
16 \\
2 \\
\end{gathered} \right) + \left( \begin{gathered}
6 \\
1 \\
\end{gathered} \right)\left( \begin{gathered}
4 \\
2 \\
\end{gathered} \right)}}
{{\left( \begin{gathered}
10 \\
1 \\
\end{gathered} \right)\left( \begin{gathered}
20 \\
2 \\
\end{gathered} \right)}} = \frac{{129}}
{{475}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/98281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\sum\limits_{k=1}^{\infty}\frac{(18)[(k-1)!]^2}{(2k)!}$ Just as the topic ask how to evaluate $$\sum_{k=1}^{\infty}\frac{(18)[(k-1)!]^2}{(2k)!}.$$
| Notice that
$$
\frac{(k-1)!^2}{(2k)!} = \frac{\Gamma(k) \Gamma(k)}{ \Gamma(2k+1) } = \frac{1}{2k} B(k,k) = \frac{1}{2k} \int_0^1 x^{k-1} (1-x)^{k-1} \mathrm{d} x
$$
Thus
$$ \begin{eqnarray}
\sum_{k=1}^\infty \frac{(k-1)!^2}{(2k)!} &=& \int_0^1 \left( \sum_{k=1}^\infty \frac{1}{2k} x^{k-1} (1-x)^{k-1} \right) \mathrm{d} x = \int_0^1 \frac{-\ln(1-x+x^2)}{2x (1-x)} \mathrm{d} x \\
&=& -2 \int_{-1/2}^{1/2} \frac{\ln(3/4+u^2)}{1-4 u^2} \mathrm{d} u = -2 \int_0^1 \frac{\ln((3 +u^2)/4)}{1-u^2} \mathrm{d} u = \frac{\pi^2}{18}
\end{eqnarray}
$$
Added
In order to fill in on the last equality, define
$$
f(t) = -2 \int_0^1 \frac{\ln\left(1 - t(1-u^2) \right)}{1-u^2} \mathrm{d} u
$$
Clearly $f(0) = 0$, and we are interested in computing $f\left(\frac{1}{4} \right)$.
$$
f^\prime(t) = 2 \int_0^1 \frac{\mathrm{d} u}{1 + t(1-u^2)} \stackrel{{u = \sqrt{\frac{1-t}{t}} \tan(\phi)}}{=} \int_0^{\arcsin(\sqrt{t})} \frac{2 \mathrm{d} \phi}{\sqrt{t(1-t)}} = \frac{2 \arcsin(\sqrt{t})}{\sqrt{t(1-t)}} = \\ 2 \frac{\mathrm{d}}{\mathrm{d} t} \arcsin^2(\sqrt{t})
$$
Thus
$$
f\left(\frac{1}{4} \right) = \int_0^{\frac{1}{4}} 2 \frac{\mathrm{d}}{\mathrm{d} t} \arcsin^2(\sqrt{t}) = 2 \arcsin^2\left(\frac{1}{2}\right) = \frac{\pi^2}{18}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/98507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solve $\frac{4}{x+1}+\frac{2}{x-2} \leq 3$ Solve
$$\frac{4}{x+1}+\frac{2}{x-2} \leq 3$$
I must be making a very stupid mistake somewhere ... been stuck on this for 1hr+ or even more ...
| $$
\begin{eqnarray}
\frac{2(x-1)}{(x+1)(x-2)} &\leq& 1 \implies \\
\frac{2(x-1)}{(x+1)(x-2)} - 1 &\leq& 0 \\
\frac{(2x-2)-(x+1)(x-2)}{(x+1)(x-2)} &\leq& 0 \\
\frac{(2x-2)-(x^2-x-2)}{(x+1)(x-2)} &\leq& 0 \\
\frac{(2x-2)-(x^2-x-2)}{(x+1)(x-2)} &\leq& 0 \\
\frac{-(x^2-3x)}{(x+1)(x-2)} &\leq& 0 \\
\frac{-(x^2-3x)}{(x+1)(x-2)} &\leq& 0 \implies \\
\frac{x(x-3)}{(x+1)(x-2)} &\geq& 0 \\
\end{eqnarray}
$$
If we look at this on a number line,
first note that for $x$ large enough, the LHS is positive.
Next, note that the LHS changes sign each time $x$ crosses
a root of one of the terms in the top or bottom, i.e. when
$x \in \{-1,0,2,3\}$. Thus the LHS is nonnegative for
$x \in (-\infty,-1) \cup [0,2) \cup [3,\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/100092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Perfect Squares ending in 576 I want to find out perfect squares ending in 576, after the number 576.
Here is my derivation to arrive at such a number. Let the perfect square ending in $576$ be $1000k+576$. Every perfect square can be expressed as a the sum of a certain number of consecutive odd numbers. For eg: $2^2 = 1+3$, $3^2 = 1+3+5$, $4^2 = 1+3+5+7$, and so on..
Hence I can write my required perfect square ending in 576 as -
$$(1+3+5+7+\cdots+49) + \underbrace{(51+53+55+57+\cdots)}_{(n\text{ summands})}$$
Therefore,
$$(1+3+5+7+\cdots+49) + \underbrace{(51+53+55+57+\cdots)}_{
(n \text{ summands})} = 1000k +576.$$
Since $(1+3+5+7+ ....49) = 576$, the equation reduces to
$$\underbrace{(51+53+55+57+\cdots)}_{n\text{ summands})} = 1000k$$
Using formula for Arithmetic Progression starting with 51 and a common difference of 2,
$$\begin{align*}
\frac{n}{2}\left(2(51) + (n-1)2\right) &= 1000k\\
n(n+50) &= 1000k
\end{align*}$$
Put $n = 100$, $100\times 150 = 1000k$, hence $k = 15$.
Put $k = 15$ in the perfect square term $1000k+576$ we get the number $15576$.
But $15576$ is not a perfect square.
What is flawed in my derivation? Kindly help.
| Your question definitely deserves a more precise answer than this, see the answer by Peter Taylor.
But I'd like to point out that what you are looking for are squares that are $576 \pmod{1000}$, i.e. solutions to $x^2 = 24^2 \pmod{1000}$.
This reduces to the equations
$$ \begin{align*} x^2 &= 24^2 \pmod{125} \\ x^2 &= 24^2 \pmod{8} \end{align*}. $$
The first solution can be solved to $x \equiv \pm 24 \pmod{125}$; the second implies $x\equiv 0\pmod 4$. We may then use the Chinese Remainder Theorem to reassemble these and we find that $x \mod 1000 \in \{24,476,524,976\}$ (This comes from: $x \mod 500 = \pm 24$.) Thus the squares ending in $576$ are exacly the squares of the numbers ending in 024, 476, 524 and 976. For instance $1024^2 = 1048576$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
An expression of $1\cdot2 + 2\cdot3 + \cdots + n\cdot(n+1)$ I got a question in my homework, which is:
Find the following sum and prove your claim:
$$1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + n\cdot(n+1).$$
I want to prove this by mathematical induction, but I couldn't find an expression of the sum. If anyone has any idea, please share with me. Thank you.
|
Find the following sum and prove your claim:
$$1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + n\cdot(n+1).$$
$$1\cdot2 + 2\cdot3 + 3\cdot4 + \cdots + i(i+1) + \cdots + n(n+1) = \sum_{i=1}^n i(i+1)=\sum_{i=1}^n i^2+\sum_{i=1}^n i \tag1$$
It is easy to prove
$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$
But how you derive $\sum_{i=1}^n i^2$ here? Expand $(i-1)^3$:
$$(i-1)^3 = i^3 - 3i^2 + 3i -1 \ \Rightarrow \ i^3 -(i-1)^3 = + 3i^2 - 3i +1 \tag 2$$
Thus,
$$\sum_{i=1}^n \left(i^3 -(i-1)^3\right) = \sum_{i=1}^n 3i^2 - \sum_{i=1}^n 3i + \sum_{i=1}^n 1 \tag3$$
The RHS of the equation $(3)$ is telescopic series, sum of which is eqial to $n^3$. Thus,
$$3\sum_{i=1}^n i^2 = n^3 + 3\sum_{i=1}^n i - \sum_{i=1}^n 1 = n^3 +3 \frac{n(n+1)}{2} -n$$
$$\therefore \ \sum_{i=1}^n i^2 = \frac{1}{3} \left(n^3 +3 \frac{n(n+1)}{2} -n \right) = \frac{n}{6} \left(2n^2 +3 n + 1 \right) \\= \frac{n(2n+1)(n+1)}{6}$$
$$\therefore \ \sum_{i=1}^n i(i+1)=\sum_{i=1}^n i^2+\sum_{i=1}^n i = \frac{n(2n+1)(n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3} $$
To prove this is correct, check for $n=1$: $\sum_{i=1}^n i(i+1)= 1 \times 2 = 2$ and $\frac{n(n+1)(n+2)}{3} = \frac{1(1+1)(1+2)}{3} =2$, thus answer is true for $n=1$.
Therefore, soppose it is true for $n=k$. If it is true for $n=k+1$, the difference in summation should be $(k+1)(k+2)$:
$$\sum_{i=1}^{k} i(i+1)= \frac{k(k+1)(k+2)}{3} \tag4$$
$$\sum_{i=1}^{k+1} i(i+1)= \frac{(k+1)(k+2)(k+3)}{3} \tag5$$
The equations $(6)-(5)$:
$$\sum_{i=1}^{k+1} i(i+1) - \sum_{i=1}^{k} i(i+1) = \frac{(k+1)(k+2)(k+3)}{3} - \frac{k(k+1)(k+2)}{3}\\ = \frac{(k+1)(k+2)(k+3 -k)}{3} = (k+1)(k+2)$$
Thus solution is proved to be correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/106042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
How to find$\int x^2 \arctan(x) \;dx $ I was solving $\int x^2 \arctan(x) \;dx $
I set $u=x^2$, $dv= \arctan(x)$, so I could get $du=2x$, $v=x\arctan(x)-\frac12\ln(1+x^{2})$.
From $\int x^2 \arctan(x)\;dx = uv - \int v \; du$
I got
$$\begin{align*}&\int x^2 \arctan(x) \; dx =\\
&x^3\arctan(x) -\frac12\ln(1+x^{2})-\int 2x\left[x^{2}\arctan x -\frac12\ln(1+x^2)\right]\;dx \end{align*}$$
and simplified if; then I got
$$3\int x^2\arctan(x) \;dx = x^3 \arctan(x)-\frac12\ln(1+x^2)+\int x\ln(1+x^2) \;dx$$
after that I use $w=\ln(1+x^2) \; dv =dx$ to find $\int x\ln(1+ x^2)$
but I got
$$x^2 \arctan(x)-\int 2x\arctan(x) \; dx$$
If I got $$x^2 \arctan(x)-\int 2x^2 \arctan(x)\;dx$$ instead, it would be easy to solve the question....
How can I solve this question and if you find any my mistake could you post this wall ??
Thank you !
| Let $u=\arctan x$, $dv = x^2\,dx$. Then $du = \frac{dx}{1+x^2}$, $v = \frac{1}{3}x^3$, so
$$\begin{align*}
\int x^2\arctan x\,dx &= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int\frac{x^3}{1+x^2}\,dx\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int\left(\frac{x^3+x}{1+x^2}-\frac{x}{1+x^2}\,dx\right)\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{3}\int x\,dx + \frac{1}{3}\int\frac{x}{1+x^2}\,dx\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 +\frac{1}{3}\int\frac{\frac{1}{2}\,du}{u} &\quad&(u=1+x^2)\\
&=\frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\int\frac{du}{u}\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\ln|u| + C\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\ln|1+x^2|+C\\
&= \frac{1}{3}x^3\arctan x - \frac{1}{6}x^2 + \frac{1}{6}\ln(1+x^2)+C.
\end{align*}$$
If after integration by parts/substitution, the resulting integral is harder than the one you started with, then it's time to go back and try a different integration by parts/substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/106827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
What is the minimum possible value of $(a+b+c)$? $a, b$ and $c$ are real positive numbers satisfying $ \frac 13 \le ab+bc+ca \le 1 $ and $abc \ge \frac 1{27}$ then what is the minimum possible value of $(a+b+c)$?
Applying AM $\ge$ GM gives $(a+b+c) \ge 1$ and if we apply AM $\ge$ HM gives $(a+b+c) \ge\frac 13$ but apparently $1$ is the answer, so my question is why are we not taking the second one as minimum, (since $ \frac13 \lt 1) $?
| Answer is 1
just expand $(a+b+c)^2$
and use the AM-GM inequality for $a^2+b^2+c^2$
so $a^2+b^2+c^2$ has a minimum value of $2(abc)^{3/2}$ which is $1/3$
and $2(ab+bc+ca)$ has a minimum value of $2*1/3=2/3$
adding both you will get 1, which is minimum value of $(a+b+c)^2$ and hence $(a+b+c)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/107859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
For a polynomial $P$ for which $P(x+5) - P(x) = 2 ,\forall x$. What is the least possible value of $P(4) - P(2)$? There are infinite number of polynomials $P$ for which $P(x+5) - P(x) = 2,\forall x \in \mathbb{R}$.
How could we determine the least possible value of $P(4) - P(2)$?
| The condition $P(x+5)-P(x)=2$ for all $x$ can be rewritten as "$(x,y)$ is on the graph of $y=P(x)$ if and only if $(x+5,y+2)$ is." This implies that if $(x,y)$ is on the graph of $y=P(x)$, so are $(x+5k,y+2k)$ for all $k\in\mathbb{Z}$. All of those points lie on a line of the form $y=\frac{2}{5}x+b$ for some $b\in\mathbb{R}$, so that line intersects the graph of $y=P(x)$ infinitely many times, or equivalently, $P(x)=\frac{2}{5}x+b$ has infinitely many solutions.
The only way for that to happen with a polynomial $P$ is if $P(x)=\frac{2}{5}x+b$ for all $x\in\mathbb{R}$ (two distinct polynomials can only be equal at finitely many points—if two polynomials are equal at infinitely many points, they must be identically equal). Thus, $$P(4)-P(2)=\left(\frac{2}{5}\cdot 4+b\right)-\left(\frac{2}{5}\cdot 2+b\right)=\frac{4}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/108058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Constructing Isomorphism between finite field Consider $\mathbb{F}_3(\alpha)$ where $\alpha^3 - \alpha +1 = 0$ and $\mathbb{F}_3(\beta)$ where $\beta^3 - \beta^2 +1 =0$.
I know these two fields are isomorphic but I have difficulty buliding an isomorphism between them.
I know I have to determine where $\alpha$ is mapped to under the isomorphism map but I can't figure it out.
Any help is much appreciated.
| One way to do it is just to slug through it:
An isomorphism $\mathbb{F}_3(\alpha)\to\mathbb{F}_3(\beta)$ is completely determined by the image of $\alpha$; and, as you note in comments, the image of $\alpha$ must be of the form $a\beta^2 + b\beta + c$, with $a,b,c\in\mathbb{F}_3$. Moreover, the image of $\alpha$ must satisfy $\alpha^3 = \alpha - 1$; that is, you want to find $a,b,c$ such that
$$(a\beta^2 + b\beta + c)^3 = a\beta^2 + b\beta + c-1.$$
So you can just expand the left hand side, using $\beta^3 = \beta^2-1$, and figure out the coefficients. We have:
$$\begin{align*}
\beta^3 &= \beta^2 - 1\\
\beta^4 &= \beta^3-\beta\\
&= \beta^2 - \beta - 1\\
\beta^6 &= (\beta^3)^2
= \beta^4 - 2\beta^2 + 1\\
&= \beta^2 - \beta - 1 -2\beta^2 + 1\\
&= -\beta^2 - \beta.
\end{align*}$$
And so, since we are in characteristic $3$,
$$\begin{align*}
(a\beta^2 + b\beta + c)^3 &= a^3\beta^6 + b^3\beta^3 + c^3\\
&= a^3(-\beta^2-\beta) + b^3(\beta^2-1) + c^3\\
&= (b^3-a^3)\beta^2 + (-a^3)\beta + c^3-b^3\\
&= a\beta^2 + b\beta + c - 1.
\end{align*}$$
So we need to solve the equations
$$\begin{align*}
b^3-a^3 &= a\\
-a^3 &= b\\
c^3 -b^3&= c-1.
\end{align*}$$
Since $a,b,c\in\mathbb{F}_3$, where $x^3=x$ for all $x$, we get
$$\begin{align*}
b-a &= a\\
-a&=b\\
c-b&= c-1
\end{align*}$$
The first two equations both give $b=-a$; the last equation gives $b=1$. So $a=-1$, $b=1$, and $c$ is free (this gives the three roots of the polynomial). That is, $f(\alpha)$ can be any of $-\beta^2+\beta$, $-\beta^2+\beta+1$, $-\beta^2+\beta-1$. You can verify they all work.
| {
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"url": "https://math.stackexchange.com/questions/108717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Steps to solve this system of equations: $\sqrt{x}+y=7$, $\sqrt{y}+x=11$ I want to solve this system of equations, I have been out of Maths for a long!!
$$\sqrt{x}+y=7$$
$$\sqrt{y}+x=11$$
Just wondering easiest step to find values for $x$ and $y$ from the above equations?
| Let
$x=\tan^2\theta$
and $y=\sec^2\theta$
*
*$\tan\theta+\sec^2\theta=7$
*$\tan^2\theta+\sec\theta=11$
Take 1) $\tan\theta+1/\cos^2\theta=7$
Solve it and you will get
$$7\sin^2\theta+\sin\theta-6=0$\tag{A}$$
Now similarly take equation (2) solve it
you will get
$$12\cos^2\theta-\cos\theta-1=0\tag{B}$$
Take (A).
Let
$\sin\theta=t$, $7t^2+t-6=0$.
Solve it you will get
$t=-1$ or $t=6/7$
$\sin\theta=-1$ or $\sin\theta=6/7$.
Now
Take eq. (B)
Do similarly
As equation (A)
$12\cos^2\theta-\cos\theta-1=0$
Solve it as above
$\cos\theta=1/3$ or $\cos\theta=-1/4$
$\cos\theta$ value exists in 3 quadrant not -ve
So $-1/4$ not possible.
$\cos\theta=1/3$
$\sin\theta$ value exist in 3 quadrant not +ve so
$\sin\theta=-1$.
Now $\tan\theta=\sin\theta/\cos\theta=-1/(1/3)=-3$.
$x=\tan^2\theta=(-3)^2=9$
Put in equation 1 $y=4$
Therefore
$x=9$ and $y=4$ is answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/108962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How can I make the following 2 fractions integers? Let $m,n$ be integers. I want to find the possible values of $m,n$ such that $4(m+n)\over (2m+n)^2+3n^2$ and $4n\over (2m+n)^2+3n^2$ are both integers too. Would someone please help? Of course letting $(2m+n)^2+3n^2=4$ gives some good values, but is this all the $m,n$ I can get?
Added: I can see that the problem can be reduced to asking for $4k\over (2m+n)^2+3n^2$ to be an integer for both $k=m,n$
| Since $(2m+n)^2 + 3n^2 = 4m^2 + 4mn + n^2 + 3n^2 = 4m^2 + 4mn + 4n^2$, then
$$\frac{4(m+n)}{(2m+n)^2+3n^2} = \frac{4(m+n)}{4m^2+4mn+4n^2} = \frac{m+n}{m^2+mn+n^2}.$$
Similarly,
$$\frac{4n}{(2m+n)^2+3n^2} = \frac{4n}{4m^2+4mn+4n^2} = \frac{n}{m^2+mn+n^2}.$$
For both to be integers, you need $m^2+mn+n^2$ to divide both $m+n$ and $n$; hence, it must divide both $m$ and $n$.
If $mn\gt 0$, then this is clearly impossible, since then $m^2+mn+n^2$ is strictly larger than each of $m$ and $n$ (as they are all integers). So $m$ and $n$ cannot both be positive or both be negative.
If $m=0$ or $n=0$, then you have that $m^2$ divides $m$, or $n^2$ divides $n$, hence you must have $(m,n) = (0,\pm 1)$ or $(\pm 1,0)$.
So we are now reduced to the case where one of $m,n$ is negative and one is positive. Let us say, without loss of generality that $m\lt 0\lt n$. If $|m|\lt n$, then $mn\lt n^2$, so $m^2+mn+n^2 \gt m^2 \geq |m|$; so $m^2+mn+n^2$ cannot divide $m$. If $|m|\gt |n|$, then $-m\gt n$, so $-m^2\lt mn$; hence $m^2+mn\gt 0$, so $m^2+mn+n^2\gt n^2\geq n$; hence $m^2+mn+n^2$ cannot divide $n$.
Finally, if $|m|=n$, then $mn=-n^2$, so $m^2+mn+n^2 = m^2 = n^2$; this divides $n$ if and only if $n=1$. So another solution is $(m,n) = (\pm 1,\mp 1)$, in which case your first fraction is $0$, and your second fraction is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
unit circle, derive number for any degree, cosinus and sinus $\sin(90°)= \sin(\frac{1}{2}\pi)= 0$
$\cos(90°)= \cos(\frac{1}{2}\pi)= 1$
$\sin(60°)= \sin(\frac{1}{3}\pi)=\frac{\sqrt{3}}{2}$
$\cos(60°)= \cos(\frac{1}{3}\pi)=\frac{1}{2} $
$\sin(45°)= \sin(\frac{1}{4}\pi)=\frac{\sqrt{2}}{2}$
$\cos(45°)= \cos(\frac{1}{4}\pi)=\frac{\sqrt{2}}{2}$
$\sin(30°)= \sin(\frac{1}{6}\pi)=\frac{1}{2}$
$\cos(30°)= \cos(\frac{1}{6}\pi)=\frac{\sqrt{3}}{2}$
An heres my question (just for the purpose of curiosity):
What number (not with decimals, i want numbers like for those above, square roots and fractions allowed) would $x$ and $y$ be:
$\sin(1°)= \sin(\frac{1}{180}\pi)=\ x$
$\cos(1°)= \cos(\frac{1}{180}\pi)=\ y$
And how would I generally derive ANY degree, lets say $\sin(3°)$ or wathever.
| Using the usual formulas for $\sin(x+y),\cos(x+y)$, one obtains $\cos(3x) = 4\cos^3x-3\cos x$. Thus $\cos(10°)$ is a root of the polynomial $4x^3-3x-\sqrt{3}/2$. Cardano's formula now yields
$$\cos(10°) = \frac{1}{2}\big (\;\sqrt[3]{\frac{\sqrt{3} + i}{2}} + \sqrt[3]{\frac{\sqrt{3} - i}{2}}\;\big)$$
where the 3rd root is defined by the function $z \mapsto e^{z/3}$.
Applying the formula $\sin(x/2) = \sqrt{\frac{1-\cos x}{2}}$ then yields
$$\sin(5°) = \frac{1}{2} \sqrt{2 - \sqrt[3]{\frac{\sqrt{3} + i}{2}} - \sqrt[3]{\frac{\sqrt{3} - i}{2}}}\;\;.$$
Note: This also shows casus irreducibilis in action (cf. GEdgar's answer).
A closed formula for $\sin(1°)$ can be derived as follows: Express $\sin(5x)$ as polynomial in $\sin x$. Then $x:=6°$ yields a polynomial of degree 5 with linear factor $x-1/2$. Thus $\sin(6°)$ is a root of a polynomial of degree 4. That can be computed by Ferrari's formula. By using the $\sin(3x)$-formula from above and solving a degree 3 equation similar as above, one obtains an expression for $\sin(2°)$. Applying the $\sin(x/2)$-formula finally yields the searched formula for $\sin(1°)$.
But as the expression for $\sin(5°)$ already shows, the result will be a rather ugly formula. Therefore it's - in my opinion - waste of time to figure it out explicitely.
Added: The wikipedia article in Michael Hardy's comment above gives $\sin(3°), \cos(3°)$. In the same manner as $\sin(5°)$ before, one can compute $\cos(5°)$. Then one can use compute $\sin(2°) = \sin(5°-3°) = \sin(5°)\cos(3°)-\sin(3°)\cos(5°)$. Applying the $\sin(x/2)$-formula finally yields the searched formula for $\sin(1°)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$ I am attempting to prove the following problem:
Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$
I attempted to solve it by induction, but when proving for $n= x+1$ the algebra gets very messy very fast. I was wondering if this is the only way or if there is a quicker way to prove this. I guess I am a little unsure as to how to prove something is an integer.
I also noticed that letting $f(x) = \frac{x^5}5 + \frac{x^4}2 + \frac{x^3}3 - \frac x{30}$ and deriving $f(x)$ yields a fairly clean result, but I don't know if this helps me at all. Any help would be great.
| One quick way is to recognize that $$\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30}$$ is nothing but $1^4 + 2^4 +\cdots + n^4$.
Another way is to look at $6n^5 + 15n^4 + 10n^3 - n$ and prove that $30$ divides it i.e. prove that $2$, $3$ and $5$ divide $6n^5 + 15n^4 + 10n^3 - n$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 4
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Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
&= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
&= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2} - \cdots- x y^{n-1} - y^n \\
&= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n \\
&\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:
*
*Associate law for addition
*Existence of an additive identity
*Existence of additive inverses
*Commutative law for additions
*Associative law for multiplication
*Existence of a multiplicative identity
*Existence of multiplicative inverses
*Commutative law for multiplication
*Distibutive law
| You have everything right except the last line.
Maybe it is easier to do in this order:
$$(x−y)\left(x^{n−1}+x^{n−2}y+\cdots+xy^{n−2}+y^{n−1}\right)=\\
=x\cdot x^{n-1}-y\cdot x^{n-1} +x\cdot x^{n−2}y-
y\cdot x^{n−2}y+x\cdot x^{n−3}y^2-\cdots\\ \cdots -y\cdot x^2y^{n-3} +x\cdot xy^{n-2}-y \cdot y^{n-1}$$
The second term $y\cdot x^{n-1}$ is the same as the third term $x\cdot x^{n−2}y$ except the sign, similarly the 4th and the 5th terms are canceled... So the only terms left are: $x\cdot x^{n-1}$ and $y\cdot y^{n-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 9,
"answer_id": 3
} |
Limit of a Sequence involving $\frac {n^2} {\sqrt{n^{6}+k}}$ I am stuck on this problem:
Compute the limit of the sequence $(a_{n})_{n=1}^{\infty}$ defined by
$$a_{n}:=\frac {n^2} {\sqrt{n^{6}+1}}+\frac {n^2} {\sqrt{n^{6}+2}}+\cdot \cdot \cdot + \frac {n^2} {\sqrt{n^{6}+n}}=\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}$$
So I am trying to find:
$$\lim_{n \to \infty}\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}$$
In a situation like this should I be noting that the denominator is increasing in value faster than the numerator?
My first thought was to do some manipulation. I may have done something incorrectly. I began with the following.
$$\frac {n^2} {\sqrt{n^6+k}}=\frac {n^2} {\sqrt{n^6(1+k/n^6)}}=\frac {n^2} {\sqrt{n^6}\sqrt{1+k/n^6}}=\frac {1} {n} \cdot \frac {1} {\sqrt {1+k/n^6}}$$
So I now have
$$\lim_{n \to \infty} \frac {1} {n} \sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}$$
Now, I am unsure about the following. It looks to me as if $k/n^6$ goes to zero as $n \to \infty $. That would result in $\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}=n$. So I would be left with
$$\lim_{n \to \infty} \frac {n} {n}=1$$
| That looks great!
I'll show you another, slightly cleaner (in my opinion) method:
We seek $\displaystyle \lim_n \sum_{k = 1}^n \frac{n^2}{\sqrt{n^6 + k}}$
Each term in the sum shares bounds, namely $\dfrac{n^2}{\sqrt{n^6 + n}} \leq \dfrac{n^2}{\sqrt{n^6 + k}} < \dfrac{n^2}{\sqrt{n^6}}$
So then $\displaystyle 1 =\dfrac{n^3}{\sqrt{n^6 + n}} \leq \lim_n \sum_{k = 1}^n \frac{n^2}{\sqrt{n^6 + k}} \leq \lim \dfrac{n^3}{\sqrt{n^6}}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding a CDF given a PDF The PDF for $Y$ is
$$f_Y(y) = \begin{cases}
0 & |y|> 1 \\
1-|y| & |y|\leq 1
\end{cases}$$
How do I find the corresponding CDF $F_Y(y)$? I integrated the above piecewise function to get
$$F_Y(y)=\begin{cases}
1/2 -y/2-y^2/2 & [-1,0] \\
1/2-y/2+y^2/2 & [0,1]
\end{cases}
$$
by using the fact that $F_Y(y)=\int _{-\infty}^{y}{f_Y(y)}\,dy$, however my text claims the answer is
$$F_Y(y)=\begin{cases}
1/2 +y+y^2/2 & [-1,0] \\
1/2+y-y^2/2 & [0,1]
\end{cases}
$$
I am struggling with pdf and cdfs, so I asssume I did something wrong other than the simple integration. Who's correct? Me or the Text!?? $:)$
| We have that $F(y) = \displaystyle \int_{-\infty}^y f(x) dx$. In your case, we are given that $$f(x) = \begin{cases} 0 & x <-1\\ 1 + x & x \in[-1,0]\\ 1-x & x \in [0,1]\\ 0 & x > 1\end{cases}$$
*
*If $y < -1$, then we have $F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-\infty}^y 0 dx =0 $. We have the integrand $f(x) = 0$ since $x \leq y < -1$.
*If $y \in [-1,0]$, then we have $F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-\infty}^{-1} f(x) dx + \displaystyle \int_{-1}^{y} f(x) dx$. Since, $f(x) = 0$ for all $x < -1$, we get that $$F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-1}^{y} f(x) dx = \displaystyle \int_{-1}^{y} \left( 1+x \right) dx = \left( x + \frac{x^2}{2} \right)_{-1}^{y} $$
$$F(y) = \left(y + \frac{y^2}{2} \right( - \left( -1 + \frac12 \right) = \frac12 + y + \frac{y^2}{2}.$$
*If $y \in [0,1]$, then we have $F(y) = \displaystyle \int_{-\infty}^y f(x) dx = \displaystyle \int_{-\infty}^{-1} f(x) dx + \displaystyle \int_{-1}^{0} f(x) dx + \displaystyle \int_{0}^{y} f(x) dx$. Since, $f(x) = 0$ for all $x < -1$, we get that $$F(y) = \displaystyle \int_{-1}^{0} f(x) dx + \displaystyle \int_{0}^{y} f(x) dx = \displaystyle \int_{-1}^{0} \left( 1+x \right) dx + \displaystyle \int_{0}^{y} (1-x) dx$$
Hence, $$F(y) = \frac12 + \left( x - \frac{x^2}{2}\right)_0^{y} = \frac12 + y - \frac{y^2}{2}$$
*For $y > 1$, since $f(x) = 0$ for all $x>1$, we have that $F(y) = F(1)$ for all $y > 1$. Hence, $F(y) = F(1) = 1$.
Hence, $$F(y) = \begin{cases} 0 & y <-1\\ \frac12 + y + \frac{y^2}{2} & y \in[-1,0]\\ \frac12 + y - \frac{y^2}{2} & y \in[0,1]\\ 1 & y > 1\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/118742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Matrix-vector multiplication before evaluating the matrix inverse Suppose I need to evaluate the expression
$$\begin{bmatrix} 0 & 1 & 0 & 0 \end{bmatrix}
\begin{bmatrix} s+2 & -1 & 0 & 0\\
0 & s+3 & 0 & 0\\
-1 & -2 & s & -1\\
2 & -1 & 1 & 4 \end{bmatrix}^{-1}
\begin{bmatrix} 0\\
1\\
1\\
1 \end{bmatrix}
$$
If I compute the inverse of the $4 \times 4$ matrix first, then perform the matrix-vector multiplication, I'm left with the result
$$ \displaystyle \frac{1}{s+3}$$
If instead I perform the matrix-vector multiplication first, keeping in mind that I need to compute the inverse of whatever I'm left with, I get
$$(s+3)^{-1} = \displaystyle \frac{1}{s+3}$$
My question is this: is it safe to assume that I can always perform the matrix-vector multiplication first and then compute the inverse of the resulting scalar (assuming, of course, that the result is invertible) or did I just happen to get lucky in this case?
| The value of $$
\begin{bmatrix} s+2 & -1 & 0 & 0\\
0 & s+3 & 0 & 0\\
-1 & -2 & s & -1\\
2 & -1 & 1 & 4 \end{bmatrix}^{-1}$$
$$\text{is}$$
$$\begin{bmatrix}
\frac{1}{s + 2} & \frac{1}{s^2 + 5 s + 6} & 0 & 0 \\
0 & \frac{1}{s + 3} & 0 & 0\\ \frac{2}{4 s^2 + 9 s + 2} & \frac{9 s + 20}{4 s^3 + 21 s^2 + 29 s + 6} & \frac{4}{4 s + 1} & \frac{1}{4 s + 1}\\
-\frac{2 s + 1}{4 s^2 + 9 s + 2} & -\frac{ - s^2 + 2 s + 5}{4 s^3 + 21 s^2 + 29 s + 6} & -\frac{1}{4 s + 1} & \frac{s}{4 s + 1}
\end{bmatrix}$$
Multiplying with the row vector retains the 2nd row and eliminates all others.
$$\begin{bmatrix}
0 & \frac{1}{s + 3} & 0 & 0\\
\end{bmatrix}$$
Multiplying with the column vector picks up the last 3 columns and eliminates others
Answer = $\dfrac{1}{s+3}$
This is, however, merely a coincidence. This is not true in general.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of ways in which $38808$ can be expressed as a product of 2 coprime factors? Number of ways in which $38808$ can be expressed as a product of $2$ coprime factors ?
the answer given is $8$ ways,
what I did was,
$$38808 = 2^3 \times 3^2 \times 7^2 \times 11$$
so the number of ways of expressing $38808$ as product of $2$ co-prime factors should be
$$8 \cdot (9.49.11)$$
$$9 \cdot (8.49.11)$$
$$49 \cdot (9.8.11)$$
$$11 \cdot (9.49.8)$$
hence $4$, but the answer is $8$, am i missing some other co-prime factor pairs?
| No, 8 is correct. Look at the prime factorisation. Two factors would be coprime if the four exponent-expressions in the prime factorisation were partitioned between the two factors. There are 8 ways of doing this, as follows.
$$\{2^3 , 3^2\cdot 7^2\cdot 11\}$$
$$\{2^3\cdot 3^2, 7^2\cdot 11\}$$
$$\{2^3\cdot 7^2, 3^2\cdot 11\}$$
$$\{2^3\cdot 3^2\cdot 7^2, 11\}$$
$$\{2^3\cdot 11 , 3^2\cdot 7^2\}$$
$$\{2^3\cdot 3^2\cdot 11, 7^2\}$$
$$\{2^3\cdot 7^2\cdot 11, 3^2\}$$
$$\{2^3\cdot 3^2\cdot 7^2\cdot 11, 1\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does one prove that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}?$ I would like to show that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}$ holds for all natural numbers. I got stuck here:
$\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\cdot\frac{2n+1}{2n+2}\leq \frac{1}{\sqrt{3n+1}}\frac{2n+1}{2n+2}.$
I would appreciate your help.
| You want to show that
$$
\frac1{\sqrt{3n+1}}\;\frac{2n+1}{2n+2}\leq\;\frac1{\sqrt{3n+4}}.
$$
Since everything is positive, this inequality is the same as
$$
(3n+4)(2n+1)^2\leq (3n+1)(2n+2)^2.
$$
After expanding and cancelling the $n^3$ terms we get
$$
12n^2+19n\leq 24n^2+20n.
$$
This inequality holds trivially for any $n\in\mathbb{N}$, and now you can retrace the steps back to your desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119773",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Divide with remainder $\frac{x^2}{x^2 + x + 2}$ I am having a hard time long dividing:
$$\frac{x^2}{x^2 + x + 2}.$$
Could someone please show a step by step way to divide this, as I can only get it down to : $1 + \frac{x^2}{x + 2}$.
Thank you for your time!
| You can do long division of polynomials exactly like one does long division of integers on paper.
Set it:
___________________________
x^2+x+2 | x^2
Now: look at the leading term of the divisor: $x^2$; the leading term of the dividend: $x^2$; how many times does the leading term of the divisor go into the leading term of the dividend? $x^2$ goes into $x^2$ once. So we put that as our first "digit" of the quotient:
1
_______________________________
x^2+x+2 | x^2
Now, multiply $1$ by $x^2+x+2$, and subtract it from the dividend
1
_______________________________
x^2+x+2 | x^2
- (x^2 + x + 2)
---------------
- x - 2
Now, look at the leading term of what's left: $-x$; and the leading term of the divisor: $x^2$. How many times does $x^2$ go into $-x$? Zero times. So we're done with the division: the quotient is $1$, the remainder is $-x-2$. So we have that
$$\frac{x^2}{x^2+x+2} = 1 + \frac{-x-2}{x^2+x+2} = 1 - \frac{x+2}{x^2+x+2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given that $a+b\sqrt[3]{2} +c\sqrt[3]{4} =0$, where $a,b,c$ are integers. Show $a=b=c=0$ Given that $\displaystyle{a+b\sqrt[3]{2} +c\sqrt[3]{4} =0}$, where $a,b,c$ are integers. Show $a=b=c=0$
Do I use modular arithmetic?
| A decade late but here's a nice elementary proof:
Take $x = a$, $y = \sqrt[3]{2}b$ and $z = \sqrt[3]{4}c$
Note the identity
$$x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x + y + z)((x - y)^2 + (y - z)^2 + (x - z)^2)$$
If $x = y = z$, we have $a = b = c = 0$.
Otherwise $x + y + z = 0 \iff x^3 + y^3 + z^3 - 3xyz = 0$
Substitute back to obtain
$$\iff a^3 + 2b^3 + 4c^3 - 6abc = 0$$
Consider a solution $(a_0, b_0, c_0) \not = (0, 0, 0)$, take modulo $2$ and note $a_0 = 2a_1$ for some integer $a_1$.
Then,
$$8a_1^3 + 2b_0^3 + 4c_0^3 - 12a_1b_0c_0 = 0$$
Divide by 2 to obtain
$$b_0^3 + 2c_0^3 + 4a_1^3 - 6b_0c_0a_1 = 0$$
So $(b_0, c_0, a_1)$ is also a solution! We can keep applying this transformation 2 more times to get $(a_1, b_1, c_1)$ is also a solution, where similarly $(b_1, c_1) = (2b_0, 2c_0)$. However this means that we can divide factors of 2 out of our original solution an arbitrary amount of times, contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Evaluating $\int_{0}^{1} \sqrt{1+x^2} \text{ d}x$ I'm learning integral. Here is my homework:
$$\int_0^1 \sqrt{1+x^2}\;dx$$
I think this problem solve by change $x$ to other variable. Can you tell me how please. (just direction how to solve)
thanks :)
| If we choose to take a (purely) trigonometric route, we might start like this:
$$
\eqalign{
x&=\tan\theta\\
dx&=\sec^2\theta\,d\theta\\
I&=\int_0^1\sqrt{1+x^2}\,dx\\
&=\int_0^\frac{\pi}{4}\sec^3\theta\,d\theta\\
}
$$
However at this stage, we actually need the substitution
$t=\sec\theta+\tan\theta$, believe it or not:
$$
\eqalign{
t&=\sec\theta+\tan\theta\\
dt&=\left(\sec\theta\tan\theta+\sec^2\theta\right)\,d\theta
=t\sec\theta\,d\theta\\
\frac{dt}{t}&=\sec\theta\,d\theta
}
$$
And then we need some trigonometry inspirations:
$$
\eqalign{
t &
= \sec\theta+\tan\theta
= \frac{1+\sin\theta}{\cos\theta}
= \frac{\cos\theta}{1-\sin\theta}
\\
t-\frac1t &
= \frac{\cos\theta}{1-\sin\theta}
- \frac{\cos\theta}{1+\sin\theta}
= 2\tan\theta
\\
\tan\theta &
= \frac12 \left( t-\frac1t \right)
= \frac12 \left( t-t^{-1} \right)
\\
\tan^2\theta &
= \frac{t^2+t^{-2}}{4} - \frac12
\\
\sec^2\theta &
= \frac{t^2+t^{-2}}{4} + \frac12
}
$$
We can can then proceed as follows:
$$
\eqalign{
I &
= \int_0^\frac{\pi}{4}\sec^3\theta\,d\theta
= \int_1^{1+\sqrt{2}} \left( \frac{t^2+t^{-2}}{4}+\frac12 \right)\,\frac{dt}{t}
\\&
= \frac14 \int_1^{1+\sqrt{2}} \left(t+t^{-3}\right)\,dt
+ \frac12 \int_1^{1+\sqrt{2}} t^{-1}\,dt
\\&
= \frac18\left[t^2-t^{-2}\right]_1^{1+\sqrt2}
+ \frac12\left[\ln t \right]_1^{1+\sqrt2}
\\&
= \frac{\sqrt2+\ln{\left(1+\sqrt2\right)}}{2}
}
$$
where it is helpful to notice that
$$
\left(1+\sqrt2\right)^{-1} =
\frac{1}{1+\sqrt2} \cdot
\frac{1-\sqrt2}{1-\sqrt2} =
\frac{1-\sqrt2}{-1}
$$
so that
$$
\left(1+\sqrt2\right)^{-2} =
\left(1-\sqrt2\right)^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/120981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
Minimal polynomial of $\sqrt[3]{7-\sqrt{2}}$ I want to find the minimal polynomial (over $\mathbb{Q}$) of: $k:=\sqrt[3]{7-\sqrt{2}}$.
With simple 'tricks' I got that: $P=(x^3-7)^2+2$ is a polynomial such that $P(k)=0$.
But I don't know if, or how to prove that $P$ is the minimal polynomial. How can I prove this/find the minimal polynomial ?
| Let $\alpha = \sqrt[3]{7 - \sqrt{2}}$ and $P(x) = x^6 - 14x^3 + 47$. Since $P(\alpha) = 0$, in order to show $P$ is irreducible it suffices to show $[Q(\alpha):Q] = 6$. Since $7 - \alpha^3 = \sqrt{2}$, you have that $\sqrt{2} \in Q(\alpha)$ and thus
$$[Q(\alpha):Q] =[Q(\alpha):Q(\sqrt{2})][Q(\sqrt{2}):Q] = 2[Q(\alpha):Q(\sqrt{2})]$$
Thus our goal is to show that $[Q(\alpha):Q(\sqrt{2})] = 3$. Since $\alpha$ satisfies the third-degree polynomial equation $Q(x) = x^3 - 7 + \sqrt{2} = 0$ over $Q(\sqrt{2})$, our task is equivalent to showing $Q(x)$ is irreducible over $Q(\sqrt{2})$.
But if $Q(x)$ were not irreducible over $Q(\sqrt{2})$, it would have a linear factor. Thus it would have a root $a + b\sqrt{2}$, with $a$ and $b$ rational. But the roots of $Q(x) = 0$ are the numbers $\alpha$, $e^{2\pi i \over 3}\alpha$, and $e^{4\pi i \over 3}\alpha$. Of these, only $\alpha$ is real, so we have
$$\alpha = a + b\sqrt{2}$$
Cubing, we get
$$7 - \sqrt{2} = (a + b\sqrt{2})^3$$
Thus we must have
$$7 + \sqrt{2} = (a - b\sqrt{2})^3$$
Multiplying the above two equations together we get
$$47 = (a^2 - 2b^2)^3$$
Since $47$ is not the cube of a rational number, we have arrived at a contradiction and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$ Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$
if either
$(1) 0 \leq a,b \leq 1$
OR
$(2) ab \geq 3$
Since this question was under Trigonometry, I assumed the following.
Since $a,b$ are positive real numbers with $0 \leq a,b \leq 1$, I can assume that for some $x,y, a=\tan(x), b=\tan(y)$ and therefore it is to be shown that
$$\frac{1}{\sec x} + \frac{1}{\sec y} = \cos x+ \cos y \geq \frac{2\cos x \cos y}{\sqrt{cos(x-y)}}$$
(Originally posted without that $2$ on the right - Sorry!)
I do know that
$$\cos x + \cos y \geq 2 \sqrt{\cos x \cos y}$$
Now how to proceed? Just give me hints !
| The statement seems wrong: Put $a:=0.1$, $b:=1$. Then the left side is
$${1\over\sqrt{1.01}}+{1\over\sqrt{2}}\doteq1.702\ ,$$
and the right side is
$${2\over\sqrt{1.1}}\doteq 1.907\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Finding $x^2+y^2+z^2$ given that $x+y+z=0$, $x^3+y^3+z^3=3$ and $x^4+y^4+z^4=15$ I just ran into this:
$$\begin{align}
x^1+y^1+z^1&=0\\
x^3+y^3+z^3&=3\\
x^4+y^4+z^4&=15\\
x^2+y^2+z^2&=\text{?}
\end{align}$$
I know the answer, but can this be solved without guessing?
| Yes, this can be solved without guessing, using Newton's identities.
Since $x + y + z = 0$, they are the roots of $t^3 + at -b = 0$.
Newton's identities give us (in a straightforward mechanical manner) that
$$x^3 + y^3 + z^3 = 3b$$
$$x^4 + y^4 + z^4 = 2a^2$$
and
$$x^2 + y^2 + z^2 = -2a$$
This gives us $b=1$ and $2a^2 = 15$.
You can solve for $a$ and find the value of $x^2 + y^2 + z^2 = -2a$. Note that if you assume $x,y,z$ are real, then you need to pick $a \lt 0$ which gives us
$$x^2 + y^2 + z^2 = \sqrt{30}$$
Note that you did not even need to use the value of $x^3 + y^3 + z^3$.
You can also derive the above identities in a slightly tedious way, as in my answer here: Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/126069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Difficult conversion from polar equation to rectangular equation. How do we convert this into rectangular equation?
$r=5\theta$
| I get this easily, check that if i am wrong:
$r = a\theta$, so in terms of Cartesian coordinates…
$$x = a\theta\cos\theta,\qquad y =a\theta\sin\theta.$$
We can also solve for $x$ and $y$ through simple algebraic manipulation
We know: $r^2=x ^2+ y^2$. Let's solve for $x$ first:
$$\begin{align*}
r &= a\theta\\
r^2&=a^2\theta^2\\
x^2+y^2&=a^2\theta^2\\
x^2&=a^2\theta^2-y^2
\end{align*}$$
Square both sides and substitute $r^2=x^2+ y^2$, $y=r\sin\theta$ (hence $y^2=r^2\sin^2\theta$), $x^2=a^2\theta^2-r^2\sin^2\theta$ (hence $x^2=a^2\theta^2- a^2\theta^2\sin^2\theta$) we get
$$x^2=a^2\theta^2(1- \sin^2\theta).$$
$y$ follows easily
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/126130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding pairs of triangular numbers whose sum and difference is triangular The triangular numbers 15 and 21 have the property that both their sum and
difference are triangular. There are another 4 pairs less than 1000. To complete this problem,
I have done like this:
To satisfy the conditions, we have to find $u$, $v$, $w$, $x$ such that
$$\begin{align*}
\frac{u(u+1)}{2} + \frac{v(v+1)}{2} &= \frac{w(w+1)}{2}\\
\frac{v(v+1)}{2} - \frac{u(u+1)}{2} &= \frac{x(x+1)}{2}.
\end{align*}$$
Multiplying by 8 and completing the squares, we have;
$$\begin{align*}
\Bigl( (2u+1)^2 -1\Bigr) + \Bigl( (2v+1)^2-1\Bigr) &= (2w+1)^2-1\\
\Bigl( (2v+1)^2 - 1\Bigr) - \Bigl( (2u+1)^2 - 1\Bigr) &= (2x+1)^2 - 1.
\end{align*}$$
For convince representation, consider, $a = 2u +1$, $b = 2v + 1$, $c = 2w+1$ and $d = 2x +1$.
Then, we have; $a^2 + b^2 - 1 = c^2$ and $b^2 - a^2 + 1 = d^2$
by adding $2b^2 = c^2 + d^2$, hence $c^2 - b^2 = b^2 - d^2$
$$\frac{c-b}{b-d} = \frac{b+d}{b+c} = \frac{m}{n}$$
say, where $\gcd(m,n)=1$, $n\gt m$.
Hence
$$\begin{align*}
-nb +nc &= -md + mb\\
b(m+n) -nc &= md\\
nd + nb &= mb + mc\\
b(n-m) &= mc - nd
\end{align*}$$
I am so sorry...thereafter what to do and how to complete this problem, I don't know. If any one can solve, I am so grateful to them or discuss some other method to complete this problem.
| There are infinitely many solutions. I'll first give infinitely many, and then talk about describing the set of all solutions.
Let $(a,u)$ be a solution to Pell's equation
$$a^2-24u^2 =1$$
with $a$ and $u$ odd. All the integer solutions to this equation are of the form $a+u\sqrt{24} = (5+\sqrt{24})^n$ (using the methods in the Wikipedia link) and, when $n$ is odd, so are $a$ and $u$. The first few values of $a$ and $u$ are: $(5,1)$, $(485, 99)$, $(47525, 9701)$, $(4656965, 950599)$...
Then $(a,b,c,d) = (a, 5u, 7u, u)$ obeys
$$a^2-1 = 24u^2 = c^2-b^2 = d^2-c^2$$
and $(a,b,c,d)$ are all odd.
Let $p=c/b$ and $q=d/b$, so $p^2+q^2=2$. There is a standard method for describing rational points on a conic once you know one point. In this case, $(1,1)$ is on conic. Consider the line through $(1,1)$ with slope $k/\ell$. Its other intersection with the conic is at
$$\left( 1-\frac{2k(k+\ell)}{k^2+\ell^2}, 1-\frac{2k(k+\ell)}{k^2+\ell^2} \right) = \left( \frac{-k^2-2k\ell+\ell^2}{k^2+\ell^2}, \frac{k^2-2k\ell-\ell^2}{k^2+\ell^2} \right).$$
The line between two rational points always has rational slope, so every rational solution to $p^2+q^2=2$ is of the above form. Clearing denominators, we have
$$(b,c,d) = u \cdot (k^2+\ell^2, -k^2-2k\ell+\ell^2, k^2-2k\ell-\ell^2).$$
In order to get $(b,c,d)$ all odd, take $k$ and $\ell$ of different parities and $u$ odd. I found $(5,7,1)$ by taking $(k,\ell) = (1,2)$. (There are also some solutions where $u$ is a half integer and $k$ and $\ell$ are both odd. I didn't check this carefully, since my goal was to find infinitely many solutions, not to classify them all, but if you want to throughly understand the problem you should consider the possibility that the three terms inside parentheses have a common factor.)
In any case, my strategy was to think about fixing $(k, \ell)$ and looking for $u$'s such that the common difference
$$d^2-c^2 = c^2 - b^2 = u^2 \cdot 4 k \ell (k^2 - \ell^2)$$
would be of the form $a^2-1$. This is just Pell's equation
$$a^2 - Nu^2=1$$
with $N=4 k \ell (k^2 - \ell^2)$ and the auxilliary condition that $a$ and $u$ are odd.
Pell's equation will be solvable whenever $N>0$ and not a square. There are certainly plenty of examples when $N$ is not a square. In fact, it will never be a square unless it is $0$. If $N=m^2$ were square, then $c^2=b^2+(mu)^2$ and $d^2 = (mu)^2 + c^2$. Fermat showed that there are no solutions to $x^2+y^2=z^2$ and $y^2+z^2=t^2$ with $y \neq 0$, see here.
But will there be odd solutions to Pell's equation? Parity considerations show that, since $N$ is even, $a$ will always be odd. If the primitive solution $a_0 + u_0 \sqrt{N}$ to Pell's equation has $u_0$ odd, then so will the solutions of the form $(a_0+u_0 \sqrt{N})^{2m+1}$. But if $u_0$ is even, then every solution to Pell's equation will have $u$ even.
I don't know a principle which predicts which of these two cases will occur but, as the example above shows, $u_0$ is odd at least some of the time, and that's all you need for infinitely many solutions.
Just to show off, here's one more example. I'll take $(k, \ell) = (3,4)$. This leads to the discovery that
$$31^2 - 25^2 = 25^2-17^2$$
We have $N = 336$. The primitive solution to $a^2 - 336 u^2 =1$ is $55^2-3^2\cdot 336=1$. So there are infinitely many solutions of the form
$$(a,b,c,d) = (a, 25u, 31u, 17u)$$
the smallest of which is
$$(55, 75, 93, 51).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/127514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
"answer_id": 0
} |
How can I find power series of $f(x)$? $$f(x)=\dfrac{1}{1+\dfrac{x}{1+\dfrac{x^2}{1+\dfrac{x^3}{1+\dfrac{x^4}{\ddots}}}}}$$
How can a power series be found given the continued fraction $f(x)$?
I'm trying to find $f(x) =1+a_1x+\dfrac{a_2x^2}{2!}+\dfrac{a_3x^3}{3!}+\dfrac{a_4x^4}{4!}+\cdots$
I tried some ways to define $f(x)$ but I could not find the general patern. It goes to complex patern after $n=3$ in my approach. I think that I need another approach to problem.
$$\begin{align}
&f_1(x)=\frac{1}{1+x}=1-x+x^2-x^3+\ldots\\
&f_2(x)=\frac{1}{1+\frac{x}{1+x^2}}=\frac{1+x^2}{1+x+x^2}\\
&f_3(x)=\frac{1}{1+\frac{x}{1+\frac{x^2}{1+x^3}}}=\frac{1+x^2+x^3}{1+x+x^2+x^3+x^4}\\
&\lim_{n\to \infty} {f_n(x)}=f(x)
\end{align}$$
Could you please give me hand on how to find the patern of this function?
| You are looking for the reciprocal of the classic Rogers-Ramanujan Continued fraction. A listing of some of the power series coefficients can be found here: OEIS. You might be able to find more references there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/130194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\sum\limits_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$ I came across a question which required us to find $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^5-5n^3+4n}$. I simplified it to $\displaystyle\sum_{n=3}^{\infty}\frac{1}{(n-2)(n-1)n(n+1)(n+2)}$ which simplifies to $\displaystyle\sum_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$. I thought it might have something to do with partial fractions, but since I am relatively inexperienced with them I was unable to think of anything useful to do. I tried to check WolframAlpha and it gave $$\sum_{n=3}^{m}\frac{(n-3)!}{(n+2)!}=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}$$
From this it is clear that as $m\rightarrow \infty$ the sum converges to $\frac{1}{96}$, however I have no idea how to get there. Any help would be greatly appreciated!
| The Heaviside Method gives the partial fraction decomposition
$$
\begin{align}
&\frac{1}{(n-2)(n-1)n(n+1)(n+2)}\\[6pt]
&=\frac{1}{24(n-2)}-\frac{1}{6(n-1)}+\frac{1}{4n}-\frac{1}{6(n+1)}+\frac{1}{24(n+2)}\tag{1}
\end{align}
$$
Notice that
$$
\frac{1}{24}-\frac{1}{6}+\frac{1}{4}-\frac{1}{6}+\frac{1}{24}=0\tag{2}
$$
Therefore,
$$
\begin{align}
&\sum_{n=3}^m\left(\frac{1}{24(n-2)}-\frac{1}{6(n-1)}+\frac{1}{4n}-\frac{1}{6(n+1)}+\frac{1}{24(n+2)}\right)\\
&=\frac{1}{24}\sum_{n=1}^{m-2}\frac1n-\frac{1}{6}\sum_{n=2}^{m-1}\frac1n+\frac{1}{4}\sum_{n=3}^{m}\frac1n-\frac{1}{6}\sum_{n=4}^{m+1}\frac1n+\frac{1}{24}\sum_{n=5}^{m+2}\frac1n\tag{$\ast$}\\
&=\frac{1}{24}\left(\frac11+\frac12+\frac13+\frac14\right)\\
&-\frac{1}{6}\left(\frac12+\frac13+\frac14+\frac{1}{m-1}\right)\\
&+\frac{1}{4}\left(\frac13+\frac14+\frac{1}{m-1}+\frac{1}{m}\right)\\
&-\frac{1}{6}\left(\frac14+\frac{1}{m-1}+\frac{1}{m}+\frac{1}{m+1}\right)\\
&+\frac{1}{24}\left(\frac{1}{m-1}+\frac{1}{m}+\frac{1}{m+1}+\frac{1}{m+2}\right)\\
&=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}\tag{3}
\end{align}
$$
where $\displaystyle\sum_{n=5}^{m-2}\frac1n$ is cancelled out of each summation in $(\ast)$ due to $(2)$.
As $m\to\infty$ the $(3)$ tends to $\dfrac{1}{96}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/131283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve $\frac{4^{x+1}-9\cdot2^x+2}{4^x-5\cdot 2^x-24}\le 0$ Solve $$\frac{4^{x+1}-9\cdot2^x+2}{4^x-5\cdot 2^x-24}\le 0$$
So letting $\alpha = 2^x$,
$$\frac{4\alpha^2-9\alpha+2}{\alpha^2-5\alpha-24}\le 0$$
$$\frac{(4\alpha - 1)(\alpha-2)}{(\alpha+3)(\alpha-8)}\le 0$$
Multiply both sides by denominator squared:
$$(4\alpha - 1)(\alpha-2)(\alpha+3)(\alpha-8)\le 0$$
Is this step right?
If I do this, then I will get $\alpha < -3, \frac{1}{3} \le \alpha \le 2, \alpha > 8$?
Right answer is $−3 < α ≤ \text{ or } 2 ≤ α < 8
$
| $$\frac{f(\alpha)}{g(\alpha)}=\frac{(4\alpha - 1)(\alpha-2)}{(\alpha+3)(\alpha-8)} \leq 0$$
$1.$ $f(\alpha) \leq 0$ and $g(\alpha)>0$
Hence :
$\left(\alpha \in \left[\frac{-1}{4},2\right]\right) \cap (\alpha \in (-\infty,-3) \cup (8,+\infty)) \Rightarrow \alpha \in \emptyset$
$2.$ $f(\alpha) \geq 0$ and $g(\alpha) <0$
Hence :
$\left(\alpha \in \left(-\infty,\frac{-1}{4}\right] \cup [2,+\infty)\right)\cap (\alpha \in (-3,8)) \Rightarrow \alpha \in \left(-3, \frac{-1}{4}\right] \cup[2,8)$
So , final solution is :
$\alpha \in \left(-3, \frac{-1}{4}\right] \cup[2,8)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/131980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
| a) $\small n^5-n = n \cdot(n^4-1) $ and $\small (n^4-1) $ is divisible by 5 by little fermat.
b) $\small n^5-n = n \cdot(n^2-1)(n^2+1) $
$ \small \qquad \qquad =n \cdot (n-1)(n+1)(n^2+1) $
$\small \qquad \qquad = (n-1)n(n+1) \cdot (n^2+1) $
$\qquad$ and $\small (n-1)n(n+1) $ is divisble by 6 (because by 2 and by 3)
Combine a) AND b) to see that $\small n^5-n $ is divisble by 5 AND 6 so also by 30.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 21,
"answer_id": 18
} |
How to calculate the improper integral $\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx$ How to Prove: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$
| Put $x = \tan{\theta}$. Then you have
\begin{align*}
I &= \int_{0}^{\pi/2} \log(\tan\theta + \cot\theta) \ d\theta \\\ &= \int_{0}^{\pi/2} \log\frac{2}{\sin{2\theta}} \ d\theta \\\ &= - \int_{0}^{\pi/2} (\log{\sin\theta} +\log{\cos\theta}) d\theta \\\ &= -2 \int\limits_{0}^{\pi/2} \log\: {\sin\theta} \ d\theta = \pi \: \log{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding $\int e^{2x} \sin{4x} \, dx$ Finding $$\int e^{2x} \sin 4x \, dx$$
I think I should be doing integration by parts...
If I let $u=e^{2x} \Rightarrow du = 2e^{2x}$,
$dv = \sin{4x} \Rightarrow v = -\frac{1}{4} \cos{4x}$
$\int{ e^{2x} \sin{4x}} dx = e^{2x}(-\frac{1}{4}\cos 4x) - \color{red}{\int (-\frac{1}{4} \cos 4x) 2e^{2x} \, dx}$
Highlighted in red, I seem to integrating an exponential times trig expression again ... doesn't seem like te right way to go
So I let $u=\sin{4x} \Rightarrow du = 4\cos 4x \, dx$
$dv = e^{2x} dx \Rightarrow v = \frac{1}{2} e^{2x}$
$\sin{4x}(\frac{1}{2} e^{2x}) - \int (\frac{1}{2} e^{2x})(4\cos4x) \, dx$
Again, its an exponential times a trig function?
Am I using the wrong substitution?
| I would jump into complex values,
and noting that
$\sin 4x = \Im e^{4 i x}$
(I wrote \Im there, wanting to get the imaginary part, and got the $\Im$),
$$\int e^{2x} \sin 4x \, dx
= \Im \int e^{2x+4ix} \, dx
= \Im \int e^{x(2+4i)} \, dx
= \Im \frac{e^{x(2+4i)}}{2+4i}.
$$
Since $\frac{1}{2+4i} = \frac{2-4i}{20} = \frac{1-2i}{10}$
and $e^{x(2+4i)} = e^{2x}(\cos 4x + i \sin 4x)$,
I get, disregarding the $\frac{e^{2x}}{10}$ for a moment,
$\Im (1-2i)(\cos 4x + i \sin 4x)
= \Im ((\cos 4x + 2 \sin 4x) + i(-2\cos 4x + \sin 4x))
$
$
= -2\cos 4x + \sin 4x
$
so my final result is
$\frac{e^{2x}(-2\cos 4x + \sin 4x)}{10}$.
As a check the derivative of
$e^{2x}(-2\cos 4x + \sin 4x)$ (disregarding the 1/10 for now) is
$$
\begin{align}
& 2e^{2x}(-2\cos 4x + \sin 4x) + e^{2x}(8\sin 4x + 4\cos 4x) \\[4pt]
= {} & e^{2x}(-4\cos 4x + 2\sin 4x + 8\sin 4x + 4\cos 4x)
= e^{2x} (10 \sin 4x).
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/136595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Finding closed forms for $\sum n z^{n}$ and $\sum n^{2} z^{n}$
Using the identity $\frac{1}{1-z} = 1 + z + z^2 + \ldots$ for $|z| < 1$, find closed forms for the sums $\sum n z^n$ and $\sum n^2 z^n$.
My solution: Because $\displaystyle1 + z + z^{2} + \ldots = \frac{1}{1-z}$, $\displaystyle1 + 2z + 3z^2 + \ldots = \sum_{n=1}^\infty n z^{n-1} = \frac{1}{(1-z)^2}$ and $\displaystyle\sum_{n=1}^\infty n z^n = \frac{z}{(1-z)^2}.$ Similarly, $\displaystyle\sum_{n=1}^\infty n^2 z^{n-1}=\frac{1+z}{(1-z)^3}$ and $\displaystyle\sum_{n=1}^\infty n^2 z^n=\frac{z(1+z)}{(1-z)^3}$.
Could you please improve this exercise?
| I'd do the second one like this
$$
\begin{align}
\sum_{n=1}^\infty n^2 z^{n-1} & = z\sum_{n=2}^\infty n(n-1) z^{n-2} + \sum_{n=1}^\infty nz^{n-1} \\[10pt]
& = z\frac{d^2}{dz^2} \frac{1}{1-z} + \frac{d}{dz} \frac{1}{1-z}= \cdots\cdots
\end{align}
$$
(and then multiply by $z$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solving $x^3-y^3=xy+61$ in integers I'm trying to solve the following equation:
$$x^3-y^3=xy+61$$
I got:
$$(x-y)(x^2+xy+y^2)=xy+61$$
But I can't go any further. I am looking for a solution in integers.
I need some hints to proceed, please!
| If we restrict ourselves to positive solutions, we can do it by brute force. In that case clearly $x>y>0$.
If $x-y=1$, we have $x^2+y^2=61$, which has the unique solution $x=6,y=5$.
If $x-y=2$, we have $2(x^2+xy+y^2)=xy+61$, or $2(x^2+y^2)+xy=61$; clearly $x$ and $y$ must both be odd, with $x^2+y^2<30$. The only candidate is $x=3,y=1$, which clearly doesn't work.
If $x-y=3$, we have $3(x^2+y^2)+2xy=61$; here $x$ and $y$ must have opposite parity, with $x^2+y^2<20$, so $x=4,y=1$ is the only candidate, and it doesn't (quite) work.
If $x-y=4$, we have $4(x^2+y^2)+3xy=61$, so $x$ and $y$ must be odd, and $x^2+y^2<15$; there are no solutions in this case.
Suppose that $x-y\ge 4$. Then $(x-y)(x^2+xy+y^2)\ge 4(5^2+5\cdot1+1^2)>61$, so there are no solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Possible typo in problem involving quotients of powers Am I correct in stating that if $$\frac{x^n}{x^{n-1}}=\frac{x^n}{x^n\cdot x^{-1}}=\frac{1}{x^{-1}}=x$$ then
$$\begin{align*}
\left|\frac{(x^2-5x+2)^n}{2^{n+1}}\cdot\frac{2n}{(x^2-5x+2)^{n-1}}\right|&=\left|\frac{(x^2-5x+2)^n}{(x^2-5x+2)^{n-1}}\cdot\frac{2n}{2^{n+1}}\right|\\\\
&\neq\left|(x^2-5x+2)\right|\frac{1}{2}
\end{align*}$$
This would have been true if the second term was:
$$\frac{2^n}{2^{n+1}}$$
and not
$$\frac{2n}{2^{n+1}}$$
I'm trying to confirm that my textbook has a typo to make sure I am not screwing something up in the math.
| With the structure of the problem as you've given it, if the supposed answer is $\left|(x^2-5x+2)\right|\frac{1}{2}$, I'd agree that the $2n$ should have been $2^n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there a quicker way of doing this integral? I have an integral $\int_0^\infty {x^2\over 1+x^4} dx$. I gave it a go and it turned out quite messy, so I consulted Wolfram Alpha but the steps given there seem rather long winded too. Is there is a faster way of doing the integral?
| As SauravTomar pointed out
$$
\int\limits_{0}^\infty\frac{x^2}{x^4+1}dx=\int\limits_{0}^\infty\frac{1}{x^4+1}dx
$$
so
$$
\int\limits_{0}^\infty\frac{x^2}{x^4+1}dx=
\frac{1}{2}\int\limits_{0}^\infty\frac{x^2+1}{x^4+1}dx=
\frac{1}{2}\int\limits_{0}^\infty\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx
$$
$$
=\frac{1}{2}\int\limits_{0}^\infty\frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+2}dx=
\frac{1}{2}\int\limits_{-\infty}^\infty\frac{dt}{t^2+2}dx=
\frac{\pi}{2\sqrt{2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/146301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What's the sum of $\sum_{k=1}^{\infty} e^{-k(x-k)^{2}}$? Absolute convergence and uniform convergence are easy to determine for this power series. What could be a possible approach to find the sum of this series
$\sum_{k=1}^{\infty} e^{-k(x-k)^{2}}$
(the sum or an estimate of the sum)?
| The approach here is heuristic.
We find an asymptotic formula for the sum for large $x$, Eq. (1) below.
Let $f(x) = \sum_{k=1}^\infty f_k(x)$ where $f_k(x) = e^{-k(x-k)^2}$.
Each term $f_k(x)$ is an unnormalized Gaussian distribution with mean $k$ and standard deviation $\sigma_k = 1/\sqrt{2 k}$.
For $k \ge 18$ we find $6\sigma_k \le 1$, that is, the distance between the means of two adjacent Gaussians is six or more standard deviations.
Thus, for large $x$ the function is a sum of narrow, well separated Gaussian spikes whose width decreases as $1/\sqrt{2 k} \approx 1/\sqrt{2 x}$.
Notice that $\cos^2\pi x$ has almost the right behavior, but the width is wrong.
A reasonable ansatz is $g(x) = (\cos^2 \pi x)^{h(x)}$, where $h(x)$ modulates the width of the spikes.
In fact, if we expand
$$g(x) = (\cos^2 \pi x)^{x/\pi^2}$$
about $x = k$ we find
$$g(x) \sim e^{-k(x-k)^2}.$$
Let's study this expansion in a little detail.
Let $z=(x-k)/\sigma_k$.
Then,
$$\begin{eqnarray*}
g(z) &=& \exp\left(
-\frac{z^2}{2} - \frac{z^3}{2\sqrt{2}k^{3/2}} - \frac{\pi^2 z^4}{24k} + O\left(\frac{1}{k^{2}}\right)\right) \\
&=& \exp\left(
-\frac{z^2}{2} - O\left(\frac{1}{k}\right)\right),
\end{eqnarray*}$$
so in the limit we have a normal distribution with the appropriate mean and width.
Thus, for large $x$,
$$\begin{equation*}
\sum_{k=1}^\infty e^{-k(x-k)^2} \sim (\cos^2 \pi x)^{x/\pi^2}.\tag{1}
\end{equation*}$$
I tried to find such a solution from @Sasha's $\mathcal{F}(\omega)$ but had no luck.
It is likely something like this can be found by inverting $\mathcal{F}(\omega)$ in the appropriate limit.
Here's a plot of the sum and fit.
Figure 1. Plot of the sum (black) and fit (red).
Addendum: Series for $\log g(x)$
Let $x = k+z/\sqrt{2k}$ and expand about $k=\infty$,
$$\begin{eqnarray*}
\log g(k+z/\sqrt{2k}) &=& \frac{k+z/\sqrt{2k}}{\pi^2}
\log \cos^2\pi\left(k+\frac{z}{\sqrt{2k}}\right) \\
&=& \frac{k+z/\sqrt{2k}}{\pi^2}
\log \cos^2 \frac{\pi z}{\sqrt{2k}}
\hspace{10ex} (\textrm{sum formula for cosine, use }k\in\mathbb{Z}) \\
&=& \frac{k+z/\sqrt{2k}}{\pi^2}
\left[
-\left(\frac{\pi z}{\sqrt{2k}}\right)^2
- \frac{1}{6} \left(\frac{\pi z}{\sqrt{2k}}\right)^4
+ O\left(\frac{1}{k^3}\right)
\right] \\
&=&
-\frac{z^2}{2} - \frac{z^3}{2\sqrt{2}k^{3/2}} - \frac{\pi^2 z^4}{24k} + O\left(\frac{1}{k^{2}}\right).
\end{eqnarray*}$$
Notice that
$$\begin{eqnarray*}
\log \cos^2 \epsilon &=& \log\left[\left(1-\frac{\epsilon^2}{2}+\frac{\epsilon^4}{24} + O(\epsilon^6)\right)^2\right] \\
&=& \log\left(1-\epsilon^2 + \frac{\epsilon^4}{3} + O(\epsilon^6)\right) \\
&=& \left(-\epsilon^2+\frac{\epsilon^4}{3}\right)
- \frac{1}{2}\left(-\epsilon^2+\frac{\epsilon^4}{3}\right)^2 + O(\epsilon^6) \\
&=& -\epsilon^2 - \frac{\epsilon^4}{6} + O(\epsilon^6).
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/146375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Area and orthocenter I need help with the following proble.
Given a triangle ABC with orthocenter H and altitudes $CM= 2 \sqrt 2$ and $AN=3$. If H divides the altitude BP into segments with ratio 5:1, i.e.BH:HP=5:1, find the area of $\triangle ABC$.
| Let $CH=x$ and $HM=Y$. Let $AH=u$ and $HN=v$. Let $BH=5t$ and $HP=t$.
Triangles $HPC$ and $HMB$ are similar. It follows that $\frac{t}{x}=\frac{y}{5t}$ and therefore $xy=5t^2$. In a similar way we can show that $uv=5t^2$. So far we have the equations
$$x+y=2\sqrt{2},\qquad u+v=3,\qquad xy=uv.$$
We need another equation! One way of getting it goes as follows.
Let $K$ be the area of $\triangle ABC$. Note that $\triangle ABC$ can be decomposed into the $3$ triangles $\triangle AHC$, $\triangle BHA$, and $\triangle CHB$.
Look first at $\triangle AHC$. This has base $CA$ and height $t$, while our full triangle $ABC$ has base $CA$ and height $6t$. So the area of $\triangle AHC$ is $\frac{t}{6t}K$, that is, $\frac{K}{6}$.
Do a similar calculation with $\triangle BHA$. It has area $\frac{y}{x+y}K$, that is, $\frac{Ky}{2\sqrt{2}}$. Similarly, the area of $\triangle CHB$ is $\frac{Kv}{3}$.
The sum of the areas of these three triangles is $K$. That gives the equation
$$\frac{K}{6}+\frac{Ky}{2\sqrt{2}}+\frac{Kv}{3}=K,$$
which simplifies to
$$3y+2\sqrt{2}v=5\sqrt{2}.$$
The rest is mechanical. We have $3$ linear equations in our variables $x$, $y$, $u$, $v$. Use these to express $x$, $u$, and $v$ in terms of $y$. Then substitute in the equation $xy=uv$. We get a quadratic equation in $y$. Solve, then find $x$, $u$, $v$. Now we can determine $t$, and therefore everything.
Remark: There is a much simpler solution that I cannot recall. The annoying thing is that a number of years ago I gave a version of this problem in an assignment in a "problem solving" course!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/146858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Showing this sequence converges. Given a sequence $x_n=\left(\dfrac{2n^3+n}{n^3} \right)+ i\left(\dfrac{3n}{n+1}\right)$, how would I show it converges? How would I choose $N$?
I did the following.
Given $\epsilon >0,$ choose $N>[?]$. Then for $n>N$
\begin{align}
\left \lvert x_n-(2+3i) \right \rvert & = \left \lvert \frac{2n^3+n}{n^3}-2 \right \rvert+ \lvert \frac{3n}{n+1}-3 \rvert\\
& = \left \lvert \frac{1}{n^2} \right \rvert+ \left \lvert -\frac{3}{n+1} \right \rvert\\
& =\frac{1}{n^2}+\frac{3}{n+1} \\
& = \frac{n+1+3n^2}{n^2+n^3} <?<\epsilon
\end{align}
Hence $x_n \rightarrow 2+3i$.
| The sequence, as you have guessed right, converges to $2+3i$. Your argument is also almost fine. You need to clarify which norm you are using to prove convergence. Usually one uses the Euclidean norm, in which case, you get
\begin{align}
\left \lVert x_n - (2+3i) \right \rVert_2 & = \left \lVert \left( \frac{2n^3+n}{n^3} + \frac{3n}{n+1} i \right) - (2+3i) \right \rVert_2\\
& = \left \lVert \left( \frac{2n^3+n-2n^3}{n^3} \right) + i \left( \frac{3n-3(n+1)}{n+1} \right)\right \rVert_2\\
& = \left \lVert \left( \frac{1}{n^2} \right) - i \left( \frac{3}{n+1} \right)\right \rVert_2\\
& = \sqrt{\left( \frac1{n^2}\right)^2 + \left(\frac{3}{n+1} \right)^2}
\end{align}
Note that for $n \in \mathbb{Z}^+$, we have $\dfrac1{n^2} < \dfrac3{n+1}$. Hence, we get that $$\sqrt{\left( \frac1{n^2}\right)^2 + \left(\frac{3}{n+1} \right)^2} < \sqrt{\left(\frac{3}{n+1} \right)^2 + \left(\frac{3}{n+1} \right)^2} = \frac{3 \sqrt{2}}{n+1}$$
Now given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{3\sqrt{2}}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\dfrac{3 \sqrt{2}}{n+1} < \epsilon.$$
Hence, given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{3\sqrt{2}}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\lVert x_n - \left( 2+3i\right)\rVert_2 < \epsilon$$
Hence, $x_n \rightarrow 2+3i$.
EDIT
You could also use the $1$-norm as you have.
\begin{align}
\left \lVert x_n - (2+3i) \right \rVert_1 & = \left \lVert \left( \frac{2n^3+n}{n^3} + \frac{3n}{n+1} i \right) - (2+3i) \right \rVert_1\\
& = \left \lVert \left( \frac{2n^3+n-2n^3}{n^3} \right) + i \left( \frac{3n-3(n+1)}{n+1} \right)\right \rVert_1\\
& = \left \lVert \left( \frac{1}{n^2} \right) - i \left( \frac{3}{n+1} \right)\right \rVert_1\\
& = \left \lvert \left( \frac1{n^2} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert
\end{align}
Note that for $n \in \mathbb{Z}^+$, we have $\dfrac1{n^2} < \dfrac3{n+1}$. Hence, we get that $$\left \lvert \left( \frac1{n^2} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert < \left \lvert \left( \frac{3}{n+1} \right)\right \rvert + \left \lvert \left(\frac{3}{n+1} \right) \right \rvert = \frac{6}{n+1}$$
Now given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{6}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\dfrac{6}{n+1} < \epsilon.$$
Hence, given an $\epsilon > 0$, choose $N(\epsilon) = \dfrac{6}{\epsilon} - 1$. Now for all $n \geq N$, where $n \in \mathbb{Z}^+$, we have that $$\lVert x_n - \left( 2+3i\right)\rVert_1 < \epsilon$$
Hence, $x_n \rightarrow 2+3i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/147109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
General expression for determinant of a block-diagonal matrix Consider having a matrix whose structure is the following:
$$
A =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & a_{2,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{3,1} & a_{3,2} & a_{3,3} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & a_{4,4} & a_{4,5} & a_{4,6} & 0 & 0 & 0\\
0 & 0 & 0 & a_{5,4} & a_{5,5} & a_{5,6} & 0 & 0 & 0\\
0 & 0 & 0 & a_{6,4} & a_{6,5} & a_{6,6} & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & a_{7,7} & a_{7,8} & a_{7,9}\\
0 & 0 & 0 & 0 & 0 & 0 & a_{8,7} & a_{8,8} & a_{8,9}\\
0 & 0 & 0 & 0 & 0 & 0 & a_{9,7} & a_{9,8} & a_{9,9}\\
\end{pmatrix}
$$
Question.
What about its determinant $|A|$?.
Another question
I was wondering that maybe matrix $A$ can be expressed as a product of particular matrices to have such a structure... maybe using these matrices:
$$
A_1 =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}\\
\end{pmatrix}
$$
$$
A_2 =
\begin{pmatrix}
a_{4,4} & a_{4,5} & a_{4,6}\\
a_{5,4} & a_{5,5} & a_{5,6}\\
a_{6,4} & a_{6,5} & a_{6,6}\\
\end{pmatrix}
$$
$$
A_2 =
\begin{pmatrix}
a_{7,7} & a_{7,8} & a_{7,9}\\
a_{8,7} & a_{8,8} & a_{8,9}\\
a_{9,7} & a_{9,8} & a_{9,9}\\
\end{pmatrix}
$$
I can arrange $A$ as a compination of those: $A = f(A_1,A_2,A_3)$
Kronecker product
One possibility can be the Kronecker product:
$$
A=
\begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{pmatrix} \otimes A_1 +
\begin{pmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\\
\end{pmatrix} \otimes A_2 +
\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1\\
\end{pmatrix} \cdot A_3
$$
But what about the determinant??? There are sums in this case which is not good...
| As stated as a comment, the result is here. It really makes block diagonal matrices wonderful, hence finding canonical forms important.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/148532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
how to calculate this infinite integral of infinite product of cosine What is the value of this nontrivial itegral:
$$\int_0^{+\infty} \left( \prod_{n = 1}^{+\infty} \cos \frac{x}{n}\right) \, \mbox d x$$
I don't know if there is nice closed answer with known constants.
| Beginning of an answer. Use these:
$$\begin{align}
\cos x &= \prod_{k=0}^\infty \left(1-\frac{4x^2}{(2k+1)^2 \pi^2}\right)
\\
\frac{\sin x}{x} &= \prod_{k=1}^\infty \left(1-\frac{x^2}{k^2 \pi^2}\right)
\end{align}$$
so that
$$\begin{align}
f(x) &:= \prod_{n=1}^\infty \cos \frac{x}{n} = \prod_{n=1}^\infty \prod_{k=0}^\infty \left(1-\frac{4x^2}{(2k+1)^2n^2\pi^2}\right)
\\ &= \prod_{k=0}^\infty \prod_{n=1}^\infty \left(1-\frac{4x^2}{(2k+1)^2n^2\pi^2}\right) = \prod_{k=0}^\infty \frac{\sin\frac{2x}{2k+1}}{\frac{2x}{2k+1}} .
\end{align}$$
We have to check that the order can be reversed.
Now (at least for the first few $K$)$^*$ I get
$$
\int_0^\infty \prod_{k=0}^K \frac{\sin\frac{2x}{2k+1}}{\frac{2x}{2k+1}}\,dx
= \frac{\pi}{4}
$$
exactly. If we can find the right limit theorem, perhaps also
$$
\int_0^\infty f(x)\,dx = \int_0^\infty \prod_{k=0}^\infty \frac{\sin\frac{2x}{2k+1}}{\frac{2x}{2k+1}}\,dx
= \frac{\pi}{4}
$$
$^*$ added No, the answer $\pi/4$ is only true up to $K=6$, but fails for $7$ and up.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/151997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
The calculation of a series Calculate the series
\begin{equation}
\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}.
\end{equation}
| Hint:
$$\frac 1 {(4n+1)(4n+3)}=\frac 1 2\left(\frac{1}{4n+1}-\frac 1{4n+3}\right) $$
Then note that
$$\frac{1}{{4n + 3}} - \frac{1}{{4n + 1}} = \frac{1}{{2\left( {2n+1} \right) + 1}} - \frac{1}{{2\left( {2n} \right) + 1}}$$
from where
$$\sum\limits_{n = 1}^m {\left( {\frac{1}{{4n + 3}} - \frac{1}{{4n + 1}}} \right)} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + - \cdots + \frac{1}{{2(2m + 1) + 1}} - \frac{1}{{2(2m) + 1}} $$
then recall the series for $\tan ^{-1}x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/152636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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Calculating $\int \frac{dx}{1+3\sin x+\cos x}$ Can I get a detailed answer on integrating this? I am currently stumped.
$$
\int \frac{dx}{1+3\sin x+\cos x}
$$
Thanks.
| We can solve this example by taking trigonometric substitution.
Let $$\tan{\frac{x}{2}}=t$$
$$\therefore\frac{1}{2}\sec^{2}{\frac{x}{2}}dx=dt$$
$$\Rightarrow dx=\frac{2dt}{1+t^{2}}$$
Now,
$$\sin x=\frac{2\tan{\frac{x}{2}}}{1+\tan^{2}\frac{x}{2}}=\frac{2t}{1+t^{2}}$$
and
$$\cos x=\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}=\frac{1-t^{2}}{1+t^{2}}.$$
Therefore our integral becomes,
$$\int \frac{\frac{2dt}{1+t^{2}}}{1+\frac{6t}{1+t^{2}}+\frac{1-t^{2}}{1+t^{2}}}$$
Thus
$$\int \frac{2dt}{1+t^{2}+6t+1-t^{2}}$$
$$\Rightarrow \int\frac{2dt}{2+6t}=\frac{1}{3}\int\frac{dt}{\frac{1}{3}+t}=\frac{1}{3}\log |t+\frac{1}{3}|+C=\frac{1}{3}\log|\tan\frac{x}{2}+\frac{1}{3}|+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Showing inequality for harmonic series. I want to show that $$\log N<\sum_{n=1}^{N}\frac{1}{n}<1+\log N.$$ But I don't know how to show this.
| It is a good excercise to show that
$$\tag 1 \frac{x}{{x + 1}} \leqslant \log \left( {1 + x} \right) \leqslant x$$
One alternative is expanding $\log$ into powers of $x$ and $\dfrac{x}{x+1}$, which gives
$$\eqalign{
& \log \left( {1 + x} \right) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} +- \cdots \cr
& \log \left( {1 + x} \right) = \frac{x}{{x + 1}} + \frac{1}{2}{\left( {\frac{x}{{x + 1}}} \right)^2} + \frac{1}{3}{\left( {\frac{x}{{x + 1}}} \right)^3} + \frac{1}{4}{\left( {\frac{x}{{x + 1}}} \right)^4} + \cdots \cr} $$
From there we immediately get
$$\frac{x}{{x + 1}} \leqslant \log \left( {1 + x} \right) \leqslant x$$
Now let $x=\dfrac 1 N$
$$\eqalign{
& \frac{{\frac{1}{N}}}{{\frac{1}{N} + 1}}\leq\log \left( {1 + \frac{1}{N}} \right)\leq\frac{1}{N} \cr
& \frac{1}{{N + 1}}\leq\log \left( {\frac{{N + 1}}{N}} \right)\leq\frac{1}{N} \cr
& \frac{1}{{N + 1}}\leq\log \left( {N + 1} \right) - \log N\leq\frac{1}{N} \cr} $$
Now sum from $N=1$ to $N=M$
$$\sum\limits_{N = 1}^M {\frac{1}{{N + 1}}} \leq\log \left( {M + 1} \right)\leq\sum\limits_{N = 1}^M {\frac{1}{N}} $$
$$\sum\limits_{N = 1}^{M + 1} {\frac{1}{N}} - 1 \leqslant \log \left( {M + 1} \right) \leqslant \sum\limits_{N = 1}^M {\frac{1}{N}} $$
This gives
$$\sum\limits_{N = 1}^{M + 1} {\frac{1}{N}} \leqslant \log \left( {M + 1} \right) + 1$$
$$\log \left( {M + 1} \right) \leqslant \sum\limits_{N = 1}^M {\frac{1}{N}} \leqslant \sum\limits_{N = 1}^{M + 1} {\frac{1}{N}} $$
Another way to prove $(1)$ is to start from the alternative definition:
$$\log x = \mathop {\lim }\limits_{k \to +\infty} k({{x^{1/k}} - 1})$$
The note that, for $0<y<1$ and $y>1$ respectively
$$\eqalign{
& \sum\limits_{v = 0}^{k - 1} {{y^v}} \leqslant \sum\limits_{v = 0}^{k - 1} {1 = k} \cr
& \sum\limits_{v = 0}^{k - 1} {{y^v}} \geqslant \sum\limits_{v = 0}^{k - 1} {1 = k} \cr} $$
Thus for $y > 0$
$${y^k} - 1 = \left( {y - 1} \right)\sum\limits_{v = 0}^{k - 1} {{y^v}} \geqslant k\left( {y - 1} \right)$$
We now let $y^k =x$ and we get
$$x - 1 = \left( {y - 1} \right)\sum\limits_{v = 0}^{k - 1} {{y^v}} \geqslant k\left( {{x^{1/k}} - 1} \right)$$
From this is it trivial to get
$$\log x \geqslant 1 - \frac{1}{x}$$
since
$$\log x = - \log \frac{1}{x} \geqslant - \left( {\frac{1}{x} - 1} \right) = 1 - \frac{1}{x}$$
This last exposition is due to Edmund Landau.
| {
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Application of Cauchy's convergence criterion Use Cauchy's convergence criterion to prove convergence of $x_n$
$$x_n=1-\frac{1}{2}+\frac{1}{3}-\cdots+(-1)^{n+1}\frac{1}{n}$$
as far as I am concerned,supposing that $m>n$, it's apparently warranted that $|x_m-x_n|=\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k}$
but, furthermore, it was showed on the textbook that $\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k}<\dfrac{1}{n+1}$
I wonder how can we approach the last step
| My post will going to aim and clarify that inequality in your textbook by using some elementary approach. First of all, notice that each term in your series that is set on an odd position is positive. Let's fix n as an even number and fix an even number of terms of your series and then we get:
$$S=\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k} = \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3}-\frac{1}{n+4}+\frac{1}{n+5}-\frac{1}{n+6}+\cdots $$
$$= \frac{1}{(n+1)(n+2)}+\frac{1}{(n+3)(n+4)}+\frac{1}{(n+5)(n+6)}+ ...$$
Here we are going to use an inequality with a sum that telescopes (on the right side):
$$S< \frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)}+\frac{1}{(n+3)(n+4)}+ ...=$$
$$\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+2}-\frac{1}{n+3}+\frac{1}{n+3}-\frac{1}{n+4}+...<\frac{1}{n+1}.$$
Notice that if you have an odd number of terms, then when you write down the
inequality and
take out from both sides the term $\frac1{n+1}$ then on the left side you get an even number of terms and you can group them such that every pair is a negative sum. Moreover, if your sum starts with a negative number, then the sum is obviously negative because the terms of the sum decrease.
Therefore, the inequality holds.
The proof is complete.
| {
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"source": "stackexchange",
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Solving $\frac{dy}{dx} = xy^2$ This problem appears to be pretty simple to me but my book gets a different answer.
$$\frac{dy}{dx} = xy^2$$
For when y is not 0
$$\frac{dy}{y^2} = x \, dx$$
$$\int \frac{dy}{y^2} = \int x \, dx$$
$$\frac{-1}{y^1} = \frac{x^2}{2}$$
$$\frac{-2}{x^2} = y$$
Is there anything wrong with this solution? It is not what my book gets but it is similar to how they do it in the example.
| Continuing from what you did, we have
$$\int \frac{dy}{y^2} = \int x\,dx$$
$$-\frac{1}{y} + C_2= \frac{x^2}{2}+C_1$$
Where $C_1$ and $C_2$ are constants of integration. Letting $C_3=C_1-C_2$
$$-\frac{1}{y} + C_2= \frac{x^2}{2}+C_1=-\frac{1}{y}= \frac{x^2}{2}+C_3$$
Solving for $y$, we have:
$$-\frac{1}{y}= \frac{x^2}{2}+C_3= \frac{x^2+2C_3}{2} \implies -y=\frac{2}{x^2+2C_3} \\
\implies y=\frac{2}{2C_3-x^2}$$
Letting $2C_3=K$
$$y=\frac{2}{2C_3-x^2}=\frac{2}{K-x^2}$$
| {
"language": "en",
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a representation of $f''$ only using terms of $f$ and constants. Using Taylor formulare How can I prove the following result?
$
f''\left( x \right) \approx Af\left( x \right) + Bf\left( {x + h} \right) + Cf\left( {x + 2h} \right) + Df\left( {x + 3h} \right)\,\,h > 0
$
Obviously I have to consider the taylor expansion and "playing" with the terms, but that's my problem . Sorry for asking this stupid things
| Write
\begin{align}
f(x+h) & = f(x) + hf'(x) + \frac{h^2}2 f''(x) + \frac{h^3}{6}f'''(\zeta_1) \\\
f(x+2h) & = f(x) + 2hf'(x) + \frac{(2h)^2}2 f''(x) + \frac{(2h)^3}{6}f'''(\zeta_2) \\\
f(x+3h) & = f(x) + 3hf'(x) + \frac{(3h)^2}2 f''(x) + \frac{(3h)^3}{6}f'''(\zeta_3). \\\
\end{align}
Then after you work out the equations, you get
$$
\begin{align}
A f(x) + B f(x+h) & + C f(x+2h) + D f(x+3h) \\\
& = \\\
(A+B+C+D)f(x) + (B+2C + 3D)hf'(x) &+ \left( \frac B2 + 2C + \frac 92D \right)h^2 f''(x) + \text{error term}
\end{align}
$$
where the error term contains the third derivatives. So just solve for
$$
\begin{align}
A+B+C+D = 0 \\\
B+2C + 3D = 0 \\\
B +2C + 9D/2 = 1/h^2 \\\
\end{align}
$$
and you will find four functions of $h$ ($A(h)$, $B(h)$, $C(h)$ and $D(h)$) such that your approximation is as good as the upper bound you can put on the third derivative error term.
Hope that helps,
| {
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prove that $\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Assume: $a,b,c >0$ prove that :
$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
| I assume that by $x,y,z$ you mean $a,b,c$.
Without loss of generality, we can assume that $c$ is the smallest of the three i.e. $a, b \geq c >0$. Then
\begin{align}
\dfrac{b+c}{a^2} + \dfrac{c+a}{b^2} + \dfrac{a+b}{c^2} - \dfrac1a - \dfrac1b - \dfrac1c & = \dfrac{b+c-a}{a^2} + \dfrac{c+a-b}{b^2} + \dfrac{a+b-c}{c^2}\\
& = \dfrac{b^3 +b^2c - ab^2 + a^2c + a^3 - a^2b}{a^2b^2} + \dfrac{a+b-c}{c^2}\\
& = \dfrac{c(a^2+b^2)+a^3+b^3-a^2b - ab^2}{a^2b^2} + \dfrac{a+b-c}{c^2}\\
& = \dfrac{c(a^2+b^2)+(a+b)(a^2-ab+b^2)-ab(a+b)}{a^2b^2} + \dfrac{a+b-c}{c^2}\\
& = \dfrac{c(a^2+b^2)+(a+b)(a^2-2ab+b^2)}{a^2b^2} + \dfrac{a+b-c}{c^2}\\
& = \dfrac{c(a^2+b^2)}{a^2b^2}+\dfrac{(a+b)(a-b)^2}{a^2b^2} + \dfrac{a+b-c}{c^2}\\
\end{align}
Note that each term on the right side is non-negative. In fact, the first term (since $a,b,c >0)$ and last term (since $a,b \geq c > 0 \implies a+b > c$) are strictly positive. Hence, $$\dfrac{c(a^2+b^2)+(a+b)(a-b)^2}{a^2b^2} + \dfrac{a+b-c}{c^2} > 0$$ Hence, $$\dfrac{b+c}{a^2} + \dfrac{c+a}{b^2} + \dfrac{a+b}{c^2} - \dfrac1a - \dfrac1b - \dfrac1c > 0$$which gives us more than what we want.
| {
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How to prove $\frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$ How could we prove that
$$ \frac{4^{1/\log_4(3/4)}}{3^{1/\log_3(3/4)}} = \frac{1}{12}\ ?$$
I have reduced it the form
$$\frac{4^{\ln(4)/\ln(3/4)}}{3^{\ln(3)/\ln(3/4)}}$$
I am not sure what to do next to get snappy solution. Any ideas?
| This is another way, using the basics about logarithms.
You can use just the identity
\begin{equation}
b^{\rm{log}_{\frac{b}{a}}b}=b\cdot a^{\rm{log}_{\frac{b}{a}}b}
\end{equation}
Note that $\frac{1}{\rm{log}_x{\frac{3}{4}}} =$ log$_\frac{3}{4} \ x$, you can use that to get
\begin{equation}
\frac{4^{\frac{1}{\rm{log}_4(3/4)}}}{3^{\frac{1}{\rm{log}_3(3/4)}}} = \frac{4^{\rm{log}_{3/4}4}}{3^{\rm{log}_{3/4}3}}
\end{equation}
Using the identity we have that
\begin{equation}
3^{\rm{log}_{\frac{3}{4}}3}=3\cdot 4^{\rm{log}_{\frac{3}{4}}3}
\end{equation}
replacing we get
\begin{equation}
\frac{4^{\rm{log}_{3/4}4}}{3^{\rm{log}_{3/4}3}}=\frac{4^{\rm{log}_{3/4}4}}{3\cdot 4^{\rm{log}_{\frac{3}{4}}3}}=\frac{4^{\rm{log}_{3/4}4-\rm{log}_{3/4}3}}{3}=\frac{4^{\rm{log}_{3/4}(4/3)}}{3}=\frac{4^{-1}}{3}=\frac{1}{12}
\end{equation}
| {
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Computing the derivative from the definition Using the limit definition of the derivative which I know is:
$$f'(x)=\lim_{h\to0}\left(\frac{f(x+h)-f(x)}{h}\right)$$
I am trying to solve this problem
$$f(x)= \frac{x}{x+2} $$
How do I go about properly solving this, I seemed to get
$$\frac{x}{x+2}\ $$
as my answer again? What are the steps I should follow?
I am trying to find the derivative of $f(x)= \frac{x}{x+2}$ using the definition of the derivative.
| $f(x) = \frac{x+2 - 2}{x+2} = 1 - \frac{2}{x+2}$
$$\frac{f(x+h) - f(x)}{h} = \frac{\frac{2}{x+2} - \frac{2}{x+h+2}}{h} = \frac{2}{h} \cdot \frac{x+h+2 - (x+2)}{(x+2)(x+h+2)} = \frac{2}{(x+2)(x+h+2)} \xrightarrow{h\to 0} \frac{2}{(x+2)^2}$$
| {
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How to compute $\int\frac{x^2}{x^2+4}\,dx$ and $\int\frac{1}{\sqrt{4-x^2}}\,dx$? My calculus teacher gave us a list of integrals to solve through the substitution method. I've been trying everything for hours, and I just can't find out how to compute these:
$$∫\frac{x^2}{x^2+4}\,dx\qquad\text{and}\qquad \int\frac{1}{\sqrt{4-x^2}}\,dx.$$
Is it even possible to solve them through substitution?
| The first one can be solved by rewriting it first
$$\int\frac{x^2}{x^2+4}\,dx = \int\frac{x^2+4-4}{x^2+4}\,dx = \int\,dx - 4\int\frac{dx}{x^2+4}.$$
Then we can solve the second one of these integrals (first one is easy) by factoring out $4$ from the denominator and then using the substitution $u=\frac{x}{2}$:
$$\begin{align*}
\int\frac{dx}{x^2+4} & = \frac{1}{4}\int\frac{dx}{\left(\frac{x}{2}\right)^2+1}\\
&= \frac{1}{2}\int\frac{du}{u^2+1}.\end{align*}$$
The last integral has a direct antiderivative.
For the second integral, try $x=2u$. Then $dx = 2du$, so
$$\int\frac{1}{\sqrt{4-x^2}}\,dx = \int\frac{2du}{\sqrt{4-4u^2}} = \int\frac{du}{\sqrt{1-u^2}}.$$
Does the last one look familiar?
Alternatively, do the same trick: factor out $4$ in the denominator inside the square root:
$$\begin{align*}
\int\frac{1}{\sqrt{4-x^2}}\,dx &= \int\frac{1}{\sqrt{4(1 - (\frac{x}{2})^2}}\,dx\\
&= \int\frac{1}{2\sqrt{1 - (\frac{x}{2})^2}}\end{align*}$$
which may suggest the substitution $w=\frac{x}{2}$.
Added. In light of the comments...
My first impulse with the second integral would have been to use a trigonometric substitution. That's because integrals that involve $\sqrt{a^2-x^2}$ but which do not have an $x\,dx$ factor are the traditional proving grounds for the trigonometric substitution. Here, $a=2$, so we would try the substitution $x=2\sin\theta$, with $0\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, $dx = 2\cos\theta\,d\theta$. This wold turn the $\sqrt{4-x^2}$ factor into
$$\sqrt{4-x^2} = \sqrt{4-4\sin^2\theta} = \sqrt{4(1-\sin^2\theta)} = \sqrt{4\cos^2\theta} = 2|\cos\theta|=2\cos\theta$$
with the last equality because $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, so $\cos\theta\geq 0$. Then the integral would simplify to
$$\int\frac{dx}{\sqrt{4-x^2}} = \int\frac{\cos\theta\,d\theta}{2\cos\theta} = \frac{1}{2}\int\,d\theta,$$
which is trivial to do; finally, we would return to $x$ using $\theta=\frac{1}{2}\arcsin(x)$.
| {
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Show $|a|+|b|+|c|+|a+b+c| \geq |a+b|+|b+c|+|c+a|$ for complex $a$, $b$, $c$ How to prove for any complex numbers $a$, $b$, $c$, the inequality $$|a|+|b|+|c|+|a+b+c| \geq |a+b|+|b+c|+|c+a|$$ is correct?
| Both sides are non-negative, so it suffices to show that the square of the left-hand-side is at least the square of the right-hand-side. That is, we wish to show:
$$
|a|^2+|b|^2+|c|^2+|a+b+c|^2+2|ab|+2|bc|+2|ac+2(|a|+|b|+|c|)|a+b+c|
\geq\\
|a+b|^2+|b+c|^2+|a+c|^2+2(|a(a+b+c)+bc|+|b(a+b+c)+ac|+|c(a+b+c)+bc|)
$$
The square terms cancel:
$$
|a|^2+|b|^2+|c|^2+|a+b+c|^2 = 2|a|^2+2|b|^2+2|c|^2+2\operatorname{Re}(ab+bc+ac)=|a+b|^2+|b+c|^2+|a+c|^2
$$
and by the triangle inequality we have $|a(a+b+c)|+|bc|\geq |a(a+b+c)+bc|$ and cyclic permutations.
| {
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How to evaluate this definite integral: $\int_0^1 {(1-x^3)}/{(1-x^5)} \;\rm dx$ I need to compute:
$$\int_0^1 \frac{1-x^3}{1-x^5} dx$$
I tried integrating it by partial fractions but couldn't succeed. Is there any other way to integrate this?
| Use partial fraction:
$$
\frac{1-x^3}{1-x^5} = \frac{\sqrt{5}-1}{\sqrt{5}(2x^2+(1+\sqrt{5})x+2)}+\frac{\sqrt{5}
+1}{\sqrt{5}(2x^2+(1-\sqrt{5})x+2)}
$$
and then integrate both summands with
$$
\int 1/(ax^2+bx+c)dx= 2\frac{\tan^{-1}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)}{\left(\sqrt{4ac-b^2}\right)} + \text{constant},
$$
which is done by completing the square and scaling/translating the denominator to the form $A^2y^2+C^2$. You'll get $\int 1 /(A^2y^2+C^2)dy = \frac{\tan^{-1}(Ay/C)}{AC}+\text{constant}$.
Plugin your limits and you're done.
I leave the hard substitution work to you. Ship ahoi.
| {
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An infinite product for $\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^2$? Given the Dedekind eta function,
$$\eta(\tau) = q^{1/24} \prod_{n=1}^\infty (1-q^n)$$
where $q = \exp(2\pi i\tau)$. Consider the following "family",
$\begin{align}
\left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{24} &= \frac{u^8}{(-1+16u^8)^2},\;\;\; u = q^{1/8} \prod_{n=1}^\infty \frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})^2}\\[2.5mm]
\left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{12} &= \frac{c^3}{(1+c^3)(-1+8c^3)^2},\;\;c = q^{1/3} \prod_{n=1}^\infty \frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})^2}\\[2.5mm]
\left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^{6}\; &= \frac{r^5}{(r^5+u_5^5)(r^5-u_5^{-5})},\quad r\; =\; q^{1/5} \prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}\\[2.5mm]
\left(\frac{\eta(7\tau)}{\eta(\tau)}\right)^{4}\;&= \frac{h(h-1)}{1+5h-8h^2+h^3},\quad h = 1/q\, \prod_{n=1}^\infty \frac{(1-q^{7n-2})^2(1-q^{7n-5})^2(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{7n-1})^3(1-q^{7n-6})^3}\\[2.5mm]
\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} &=\frac{s}{(s-u_{13})(s+u_{13}^{-1})},\quad\; s =\; ???\\
\end{align}$
with fundamental units $u_n$ as $u_5 = \frac{1+\sqrt{5}}{2}$ and $u_{13} = \frac{3+\sqrt{13}}{2}$. The second-to-the-last appears in Chap 10 (10.2) of Duke's Continued Fractions and Modular Functions.
Question: What is the analogous infinite product, if any, for $\left(\frac{\eta\,(13\tau)}{\eta\,(\tau)}\right)^2$ similar to the ones above?
Postscript: This question has been modified before, as it was a bit unclear.
| Michael Somos just today found the identity,
$$
\left(\frac{\eta(13\tau)}{\eta(\tau)}\right)^{2} = \frac{s}{s^2-3s-1}$$
where,
$$s=\frac{1}{q}\; \prod_{n=1}^\infty \frac{ (1-q^{13n-2})(1-q^{13n-5})(1-q^{13n-6})(1-q^{13n-7})(1-q^{13n-8})(1-q^{13n-11}) }{(1-q^{13n-1})(1-q^{13n-3})(1-q^{13n-4})(1-q^{13n-9})(1-q^{13n-10})(1-q^{13n-12})} $$
thus completing the family for $N = 2,3,5,7,13$.
Does this address Matt E and Loeffler's comments? Is "s" a modular function?
| {
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proving inequality? Here is another inequality I am trying to prove:
Let $a,b,c$ be positive numbers. Prove that:
$$1) \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\geqslant (a+b+c)$$
$$2) \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}+\frac{1}{\sqrt{ab}}$$
In the book's hint, it uses the inequality: $$a^{2}+b^{2}+c^{2}\geq ab+bc+ca$$ (which is easy to prove), then it follows that : $$b^{2}c^{2}+a^{2}c^{2}+a^{2}b^{2}\geqslant abc(a+b+c)$$ which is equivalent to proving our claim. I need to know how the second inequality follows from the first one. Also, any suggestions for proving the second claim?
| $\frac {bc}{a}+\frac{ca}{b}+\frac{ab}{c}=\frac{1}{2}(\frac{ca}{b}+\frac{ab}{c})+\frac{1}{2}(\frac{ab}{c}+\frac{bc}{a})+\frac{1}{2}(\frac{bc}{a}+\frac{ca}{b}\geq\sqrt\frac{ca}{b}\sqrt\frac{ab}{c}+\sqrt\frac{ab}{c}\sqrt\frac{bc}{a}+\sqrt\frac{bc}{a}\sqrt\frac{ca}{b}=a+b+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/165761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$
This seems clear, but I don't know how to prove this..
I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ are divisible by $a,b$ respectively, but i think this is a wrong way..
| Here's a colored, edited version of user17762's answer.
The proof is by induction. Let us denote $\gcd(a,b) = d$. It's given $\gcd(a,b) = 1
\iff$
by Bezout's Lemma, there exist $x,y \in \mathbb{Z}$ such that $ax+by = d. (\dagger)$
Assume that the statement is true for all $n \leq k$. Scilicet $\gcd(a^n,b^n) = d^n$ for all $n \leq k$.
We now need to prove that $\gcd(a^{k+1},b^{k+1}) = d^{k+1}$.
Since $\gcd(a^k,b^k) = d^k$, there exists $x_k,y_k \in \mathbb{Z}$ such that $a^k x_k + b^k y_k = d^k$.
Multiply this with $(\dagger)$ to result in $\begin{align} (ax+by) \left( a^k x_k + b^k y_k \right)^2 & = d^{2k+1}. \\
\color{green}{a^{2k + 1}xx_k^2 + a^{k + 1}2b^kx_ky_kx} \color{blue}{+ ab^{2k}yk^2x} &= \\
+ \color{red}{a^{2k}bx_k^2y} + b^{k + 1}2a^kx_ky_ky + b^{2k+1}y_k^2y
\end{align}$
$LHS = \color{green}{a^{k+1} \left( a^k x_k^2 x \color{red}{+ a^{k-1} b x_k^2 y} + 2 b^k x_k y_k x \right)} + b^{k+1} \left( b^k y_k^2 y + \color{blue}{ab^{k-1} y_k^2 x} + 2 a^k x_k y_k y \right) $
Since $d|a$ and $d|b$, we have that $\color{brown}{a = d e}$ and $\color{brown}{b = df}$. $LHS = a^{k+1} \left( \color{brown}{d^k e^k} x_k^2 x + \color{brown}{d^k e^{k-1} f} x_k^2 y + 2\color{brown}{ d^k f^k} x_k y_k x \right) + b^{k+1} \left( \color{brown}{d^k f^k} y_k^2 y + \color{brown}{d^k ef^{k-1}} y_k^2 x + 2 \color{brown}{d^k e^k} x_k y_k y \right)$
Divide both sides by $d^k$ to result in
$$a^{k+1} \underbrace{( e^k x_k^2 x + e^{k-1} f x_k^2 y + 2 f^k x_k y_k x)}_{\huge{x_{k+1}}} + b^{k+1} \underbrace{\left( f^k y_k^2 y + ef^{k-1} y_k^2 x + 2 e^k x_k y_k y \right)}_{\huge{y_{k+1}}} = d^{k+1}.$$
Hence, we have found integers $x_{k+1}, \, y_{k+1}$ such that $a^{k+1} x_{k+1} + b^{k+1} y_{k+1} = d^{k+1}$
$ \iff \gcd(a^{k+1}, b^{k+1}) \vert d^{k+1}$.
It is also true that $d^{k+1} \vert a^{k+1}$ and $d^{k+1} \vert b^{k+1}$, since $d \vert a$ and $d \vert b$.
Hence, $d^{k+1} \vert \gcd(a^{k+1}, b^{k+1})$.
In all , we get that $\gcd(a^{k+1}, b^{k+1}) = \gcd(a,b)^{k+1}.$
Hence, by the principle of mathematical induction $\gcd(a^n,b^n) = \gcd(a,b)^n, \,\, \forall n \in \mathbb{N}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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} |
Find the value of a succession of additions $$1^2+2^2+3^2+...+10000$$
How do you find the exact value of that?
I'm studying induction, and I'm still not sure how to get that value.
| Let $$S=1+2+3+...+n$$ then, $$S=n+(n-1)+(n-2)+...+1$$ Adding corresponding terms of two equations gives , $$2S=(n+1)+(n+1)+(n+1)+...(n+1)=n(n+1)\implies S=n(n+1)/2$$ Now $$(n+1)^3-n^3=3n^2+3n+1$$ $$\implies 2^3-1^3=3(1)^2+3(1)+1$$ $$3^3-2^3=3(2)^2+3(2)+1$$ $$4^3-3^3=3(3)^2+3(3)+1$$ $$..$$$$..$$$$(n+1)^3-n^3=3(n)^2+3(n)+1$$ Adding corresponding terms gives, $$(n+1)^3-1^3=3(1^2+2^2+...+n^2)+3(1+2+...+n)+n$$ $$\implies n^3+3n^2+3n=3(ReqSum)+3n(n+1)/2+n$$ Solving for $ReqSum$ and factorization gives $ReqSum=n(n+1)(2n+1)/6$. For $n=100$, it is $100.101.201/6=338350$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/166870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluating $\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x$ I have to evaluate:
$$\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x. $$
I can't get the right answer! So please help me out!
| Let $I$ denote the integral and consider the substitution $u= \frac{\pi }{2} - x.$ Then $I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u}}{\sqrt{\cos u } + \sqrt{\sin u }} du$ and $2I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u} + \sqrt{\sin u }}{\sqrt{\cos u } + \sqrt{\sin u }} du = \frac{\pi }{2}.$ Hence $I = \frac{\pi }{4}.$
In general, $ \displaystyle\int_0^a f(x) dx = \displaystyle\int_0^a f(a-x) $ $dx$ whenever $f$ is integrable, and $\displaystyle\int_0^{\frac{\pi }{2}} \frac{\cos^a x}{\cos^a x + \sin^a x } dx = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sin^a x}{\cos^a x + \sin^a x } dx = \frac{\pi }{4}$ for $a>0$ (same trick.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/167409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
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} |
Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing
Show that $U_n:=\left(1+\dfrac{1}{n}\right)^n$, $n\in\Bbb N$, defines a monotonically increasing sequence.
I must show that $U_{n+1}-U_n\geq0$, i.e. $$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$
I am trying to go ahead of this step.
| We show that
$$
\left(1+\frac{1}{n-1}\right)^{n-1} \leq \left(1+\frac{1}{n}\right)^{n}
$$
for any $n \geq 2$.
For any $n \geq 2$ we have that
$$
n \int^n_{n-1} \frac{1}{x(x+1)}dx \leq \int^n_{n-1} \frac{1}{x}dx
$$
because $\frac{n}{x+1} \leq 1$ for all $x \in [n-1,n]$.
But this inequality is equivalent to
$$
n \int^n_{n-1} \frac{1}{x}-\frac{1}{x+1}dx \leq \int^n_{n-1} \frac{1}{x}dx \iff\\
\iff (n-1)\int^n_{n-1} \frac{1}{x}dx \leq n\int^n_{n-1} \frac{1}{x+1}dx \iff\\
\iff (n-1)\int^n_{n-1} \frac{1}{x}dx \leq n\int^{n+1}_{n} \frac{1}{x}dx \iff\\
\iff (n-1)\ln\left(\frac{n}{n-1}\right) \leq n \ln\left(\frac{n+1}{n}\right) \iff\\
\iff \left(1+\frac{1}{n-1}\right)^{n-1} \leq \left(1+\frac{1}{n}\right)^{n}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/167843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
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} |
Closed form solution of Fibonacci-like sequence Could someone please tell me the closed form solution of the equation below.
$$F(n) = 2F(n-1) + 2F(n-2)$$
$$F(1) = 1$$
$$F(2) = 3$$
Is there any way it can be easily deduced if the closed form solution of Fibonacci is known?
| Define $g(z) = \sum_{n \ge 0} F(n + 1) z^n$, write the recurrence as:
$$
F(n + 3) = 2 F(n + 2) + 2 F(n + 1) \qquad F(1) = 1, F(2) = 3
$$
Multiply the recurrence by $z$, sum over $n \ge 0$ and get:
$$
\frac{g(z) - F(1) - F(2)}{z^2} = 2 \frac{g(z) - F(1)}{z} + 2 g(z)
$$
Solve for $g(z)$:
$$
g(z) = \frac{1 + z}{1 - 2 z - 2 z^2}
= \frac{2 + \sqrt{3}}{2 \sqrt{3}} \cdot \frac{1}{1 - (1 + \sqrt{3}) z)}
+ \frac{2 - \sqrt{3}}{2 \sqrt{3}} \cdot \frac{1}{1 + (1 - \sqrt{3})z}
$$
Two geometric series:
$$
T(n+ 1) = \frac{2 + \sqrt{3}}{2 \sqrt{3}} \cdot (1 + \sqrt{3})^n
+ \frac{2 - \sqrt{3}}{2 \sqrt{3}} \cdot (1 - \sqrt{3})^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/167957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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} |
Solving a recurrence of a sequence The sequence $a_n$ is given by recurrence $$a_0=0,a_1=1,a_n=a_{n-2}+n+4, n>2$$ How to find the general term I am interested on different approach. Can generating functions help
| The recurrence can indeed be solved using generating functions. I prefer to write the recurrence as
$$a_n=a_{n-2}+n+4-4[n=0]-4[n=1]\;,$$
using Iverson brackets and assuming that $a_n=0$ for $n<0$. Then multiply through by $x^n$ and sum over $n\ge 0$ to get the generating function, $g(x)$:
$$\begin{align*}
g(x)&=\sum_{n\ge 0}a_nx^n\\
&=\sum_{n\ge 0}a_{n-2}x^n+\sum_{n\ge 0}nx^n+4\sum_{n\ge 0}x^n-4-4x\\
&=x^2g(x)+x\sum_{n\ge 0}nx^{n-1}+\frac4{1-x}-4-4x\\
&=x^2g(x)+x\left(\frac1{1-x}\right)'+\frac4{1-x}-4-4x\\
&=x^2g(x)+\frac{x}{(1-x)^2}+\frac4{1-x}-4-4x\\
&=x^2g(x)+\frac{x+4(1-x)-4(1-x)^2-4x(1-x)^2}{(1-x)^2}\\
&=x^2g(x)+\frac{4-3x-4(1+x)(1-x)^2}{(1-x)^2}\;,
\end{align*}$$
so
$$\begin{align*}
g(x)&=\frac{4-3x-4(1+x)(1-x)^2}{(1+x)(1-x)^3}\\
&=\frac{4-3x}{(1+x)(1-x)^3}-\frac4{1-x}\\
&=\frac18\left(\frac7{1+x}+\frac7{1-x}+\frac{14}{(1-x)^2}+\frac4{(1-x)^3}\right)-\frac4{1-x}\\
&=\frac18\left(\frac7{1+x}-\frac{25}{1-x}+\frac{14}{(1-x)^2}+\frac4{(1-x)^3}\right)\\
&=\frac18\left(7\sum_{n\ge 0}(-1)^nx^n-25\sum_{n\ge 0}x^n+14\sum_{n\ge 0}(n+1)x^n+4\sum_{n\ge 0}\binom{n+2}2x^n\right)\;,
\end{align*}$$
and
$$\begin{align*}
a_n&=\frac18\left(7(-1)^n-25+14(n+1)+4\binom{n+2}2\right)\\
&=\frac18\left(2n^2+20n-7+7(-1)^n\right)\\
&=\begin{cases}
\frac{n(n+10)}4,&\text{if }n\text{ is even}\\
\frac{n(n+10)-7}4,&\text{if }n\text{ is odd}\;.
\end{cases}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/170748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is the inradius of any triangle at most half its circumradius? Is there any geometrically simple reason why the inradius of a triangle should be at most half its circumradius? I end up wanting the fact for this answer.
I know of two proofs of this fact.
Proof 1:
The radius of the nine-point circle is half the circumradius. Feuerbach's theorem states that the incircle is internally tangent to the nine-point circle, and hence has a smaller radius.
Proof 2:
The Steiner inellipse is the inconic with the largest area. The Steiner circumellipse is the circumconic with the smallest area, and has 4 times the area of the Steiner inellipse. Hence the circumcircle has at least 4 times the area of the incircle.
These both feel kind of sledgehammerish to me; I'd be happier if there were some nice Euclidean-geometry proof (or a way to convince myself that no such thing is likely to exist, so the sledgehammer is necessary).
EDIT for ease of future searching: The internet tells me this is often known as "Euler's triangle inequality."
| Compute the area of a triangle (first method):
Consider the following diagram:
$\hspace{4.5cm}$
The area of the green triangle is
$$
A=\tfrac12ab\sin(\theta)\tag{1}
$$
By the Inscribed Angle Theorem, the angle that $c$ subtends at the origin, $o$, is $2\theta$. Therefore, we get that
$$
c=2R\sin(\theta)\tag{2}
$$
Combining $(1)$ and $(2)$ yields
$$
4AR=abc\tag{3}
$$
Compute the area of a triangle (second method):
Consider the following diagram:
$\hspace{4.5cm}$
Note that the areas of the the red ($\color{#C00000}{\triangle iyz}$), green ($\color{#00A000}{\triangle izx}$), and blue ($\color{#0000FF}{\triangle ixy}$) triangles are $\frac12r$ times $a$, $b$, and $c$, respectively. Therefore, we get
$$
2A=r(a+b+c)\tag{4}
$$
Compute $d$:
Translate the circumcenter, $o$, of $\triangle xyz$ to the origin. Then
$$
|x|=|y|=|z|=R\tag{5}
$$
Furthermore, using $a=|y-z|$, $b=|z-x|$, and $c=|x-y|$, we get
$$
\begin{align}
2y\cdot z&=2R^2-a^2\tag{6a}\\
2z\cdot x&=2R^2-b^2\tag{6b}\\
2x\cdot y&=2R^2-c^2\tag{6c}
\end{align}
$$
Explanation:
$\text{(6a)}$: $a^2=(y-z)\cdot(y-z)=|y|^2+|z|^2-2y\cdot z$, then apply $(5)$
$\text{(6b)}$: $b^2=(z-x)\cdot(z-x)=|z|^2+|x|^2-2z\cdot x$, then apply $(5)$
$\text{(6c)}$: $c^2=(x-y)\cdot(x-y)=|x|^2+|y|^2-2x\cdot y$, then apply $(5)$
As mentioned above, the areas of the the red ($\color{#C00000}{\triangle iyz}$), green ($\color{#00A000}{\triangle izx}$), and blue ($\color{#0000FF}{\triangle ixy}$) triangles are proportional to $a$, $b$, and $c$, respectively. Thus, the barycentric coordinates of the incenter, $i$, are the mean of the vertices weighted by the lengths of the opposite sides:
$$
i=\frac{ax+by+cz}{a+b+c}\tag{7}
$$
and therefore, using $(3)$-$(7)$ yields that $d$, the distance between the incenter and circumcenter, satisfies
$$
\begin{align}
d^2
&=\frac{a^2R^2+b^2R^2+c^2R^2+2abx\cdot y+2bcy\cdot z+2caz\cdot x}{(a+b+c)^2}\tag{8a}\\
&=\frac{a^2R^2+b^2R^2+c^2R^2+ab(2R^2-c^2)+bc(2R^2-a^2)+ca(2R^2-b^2)}{(a+b+c)^2}\tag{8b}\\
&=\frac{(a+b+c)^2R^2-(a+b+c)abc}{(a+b+c)^2}\tag{8c}\\[3pt]
&=R^2-\frac{abc}{a+b+c}\tag{8d}\\[6pt]
&=R^2-2Rr\tag{8e}\\[12pt]
&=R(R-2r)\tag{8f}
\end{align}
$$
Explanation:
$\text{(8a)}$: take the dot product of $(7)$ with itself and apply $(5)$
$\text{(8b)}$: apply $(6)$ to the dot products
$\text{(8c)}$: collect terms
$\text{(8d)}$: simplify
$\text{(8e)}$: apply $(3)$ and $(4)$
$\text{(8f)}$: factor
Since $d^2\ge0$, and $R>0$, we immediately get that
$$
r\le\tfrac12R\tag9
$$
| {
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"url": "https://math.stackexchange.com/questions/170853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Solving $\cos^2 \theta + \cos \theta = 2$ Solve the following for $\theta$:
$\cos^2 \theta + \cos \theta = 2$ [Hint: There is only one solution.]
I started this out by changing $\cos^2\theta$ to $\dfrac{1+\cos(2\theta)}{2}+\cos\theta=2$
$1+\cos(2\theta)$ turns into $1+\cos^2\theta-\sin^2\theta$ which all becomes; $\dfrac{1+\cos^2\theta-\sin^2\theta}{2}+\cos\theta=2$
Not to sure what to do after this. I was going to try a power reducing rule for $\sin^2\theta$ but that would make $\dfrac{1+\cos^2\theta- \left(\frac{1-\cos(2\theta)}2 \right)}2+\cos\theta=2$. Please do help.
| Replacing $\cos^2\theta$ with and expression involving $\cos2\theta$ is not necessarily a good idea; then you have to deal with cosines of two different angles.
A better approach is to realize that what we have is a quadratic equation: let us define $y$ to be $y=\cos\theta$. Then we can rewrite the equation as
$$y^2 + y = 2$$
or $y^2 + y - 2 = 0$.
We know how to solve quadratic equations: the solutions are
$$\begin{align*}
y_1 &= \frac{-1+\sqrt{1+8}}{2} = \frac{-1+3}{2} = 1\\
y_2 &= \frac{-1-\sqrt{1+8}}{2} = \frac{-1-3}{2} = -2.
\end{align*}$$
However, now we remember that $y$ is actually $\cos\theta$, so now we want to find the solutions to $\cos\theta = 1$ and of $\cos\theta=-2$.
Since $-1\leq\cos\theta\leq 1$, the latter equation has no solutions.
So the answer is that the solutions are exactly the $\theta$ for which $\cos(\theta)=1$.
(Which we could have figured out cleverly by making the observation made by David Mitra in comments, but I wanted to give you an idea of how to approach this kind of equation if the answer is not so obvious.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Write each expression in the form $ca^pb^q$ Write each expression in the form $ca^pb^q$
c) $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$
\begin{align*}
&= \frac{a\left(\frac{2}{b}\right)}{1}*\frac{\left(\frac{a}{3}\right)}{3}=\dfrac{a^2\left(\frac{2}{b}\right)}{3}=\frac{a^2}{1}*\frac{2}{b}*\frac{1}{3}=\frac{2a^2}{3b}*\frac{b}{1}=\frac{2a^2b}{3}=\frac{2}{3}a^2b^1
\end{align*}
e) $\dfrac{a^{-1}}{(b^{-1})\sqrt{a}}$
\begin{align*}
&= \frac{1}{(b^{-1})a\sqrt{a}}=\frac{1b}{1a^1a^{\frac{1}{2}}}=\frac{1b^1}{1a^{\frac{2}{3}}}=1a^{\frac{-2}{3}}b^1
\end{align*}
These are my steps. Any corrections help.
| I'll do the first in two ways, $\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}$, and we'll see what we think. First, I'm going to separate all the 'numbers'.
*
*$\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}}= \dfrac{2a\frac{1}{b}}{3\frac{1}{a}} = \frac{2}{3}\dfrac{a\frac{1}{b}}{\frac{1}{a}}$
Now I'll do the $a$ terms.
$\frac{2}{3}\dfrac{a \frac{1}{b}}{\frac{1}{a}} = \frac{2}{3} a \frac{1}{b} \cdot \frac{a}{1} = \frac{2}{3}a^2 \frac{1}{b}$.
Finally, we know that $\frac{1}{b} = b^{-1}$. So we have $\frac{2}{3} a^2 b^{-1}$.
*Let's do it a different way. We can get rid of the $\frac{3}{a}$ on the bottom, as dividing by a fraction is the same as multiplying by its reciprocal.
$\dfrac{a\left(\frac{2}{b}\right)}{\frac{3}{a}} = a\left(\frac{2}{b}\right) \cdot \frac{a}{3} = a \frac{2a}{3b} = 2a^2 \frac{1}{3b} = \frac{2}{3} a^2 b^{-1}$.
And if I had any doubt, I could check my answer by plugging in some numbers and making sure that both sides give me the same number.
| {
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How to solve this equation involving $()^x$? I have the equation:
$\left (\sqrt{3+2\sqrt{2}} \right )^x- \left (\sqrt{3-2\sqrt{2}} \right )^x=\frac{3}{2}$
I wrote the left side of the equation as square roots.
$(1+\sqrt{2})^x-(1-\sqrt{2})^x=\frac{3}{2}$
How do I found out the final solution? Thank you very much!
P.S. The answers I can choose from are:
a) $x=1$
b) $x=2$
c) $x=\frac{2\lg2}{\lg(3+2\sqrt2)}$
d) $x=\frac{2\lg2}{\lg(3-2\sqrt2)}$
e) no solution
f) $x=2\lg2$
| First of all, $\sqrt{3-2\sqrt{2}}$ is $\sqrt{2} - 1$.
Now, Let $(\sqrt{2}+1)^x$ be y.
$y - \frac{1}{y} = \frac{3}{2}$.
So $2y^2 - 3y - 2 = 0$.
So y is 2 or $-\frac{1}{2}$. As y = $(\sqrt{2}+1)^x$, y must be positive and thus 2.
$(\sqrt{2}+1)^x = 2$
So $x = \frac{log(2)}{log(\sqrt{2}+1)}$, which is same as option c) here as you can see by simplifying the expression in c). I guess they are also taking lg instead of log, but as you can see it makes no difference whatsoever.
| {
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Proving $\frac{x}{\sqrt{y^2+yz+z^2}}+\frac{y}{\sqrt{z^2+zx+x^2}}+\frac{z}{\sqrt{x^2+xy+y^2}} \ge \sqrt{3}$
Let $x,y,z >0$. Prove
that:$$\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}}
\ge \sqrt{3}$$
My solution:
By Hölder,
$$\left(\sum\frac{x}{\sqrt{4y^2+yz+4z^2}}\right)^2(\sum x(4y^2+yz+4z^2)) \ge (x+y+z)^3$$
Let us denote $\sum_{sym} x^2y = X$, so we have to prove
$$(x+y+z)^3 \ge 4X+3xyz \iff \sum x^3 +3xyz \ge X$$
which is Schur.
How to prove it in a different way ?
| We need to prove that
$$\sum_{cyc}x\sqrt{(x^2+xy+y^2)(x^2+xz+z^2)}\geq\sqrt{3\prod_{cyc}(x^2+xy+y^2)}.$$
Now, by C-S
$$(x^2+xy+y^2)(x^2+xz+z^2)=\left(\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\right)\left(\left(x+\frac{z}{2}\right)^2+\frac{3}{4}z^2\right)\geq$$
$$\geq\left(\left(x+\frac{y}{2}\right)\left(x+\frac{z}{2}\right)+\frac{3}{4}yz\right)^2=\left(x^2+\frac{x(y+z)}{2}+yz\right)^2.$$
Thus, it remains to prove that
$$\left(\sum_{cyc}(2x^3+x^2y+x^2z+2xyz)\right)^2\geq12\prod_{cyc}(x^2+xy+y^2).$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Since $\prod\limits_{cyc}(x^2+xy+y^2)$ is a linear expression of $w^3$, we see that
$$\left(\sum_{cyc}(2x^3+x^2y+x^2z+2xyz)\right)^2-12\prod_{cyc}(x^2+xy+y^2)=81w^6+A(u,v^2)w^3+B(u,v^2),$$
which says that the inequality
$$\left(\sum_{cyc}(2x^3+x^2y+x^2z+2xyz)\right)^2-12\prod_{cyc}(x^2+xy+y^2)\geq\left(\sum_{cyc}(x^3-x^2y-x^2z+xyz)\right)^2$$
is a linear inequality of $w^3$, which says that it remains to prove the last inequality for an extremal value of $w^3$, which happens in the following cases.
*
*$w^3\rightarrow0^+$.
Let $z\rightarrow0^+$ and $y=1$.
We obtain $$(x-1)^2(3x^4+14x^3+12x^2+12x+4)\geq0,$$
which is obvious;
*$y=z=1$, which gives
$$(x-1)^2(x^3+6x^2+9x+8)x\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the divisors of the number $p^3q^6$ My text says that $p^3q^6$ has 28 divisors. Could anyone please explain to me how they got 28 here?
Edit:
$p$ and $q$ are distinct prime numbers Sorry for the late addition..
| Let's take $p=3$ and $q=2$ as an example. Then the number is $3^32^6 = 1728$, and the 28 divisors of 1728 are:
$$\begin{matrix}
1&2&4&8&16&32&64 \\
3&6&12&24&48&96&192 \\
9 & 18 & 36 & 72 & 144 & 288 & 576 \\
27 & 54 & 108 & 216 & 432 & 864 & 1728
\end{matrix}$$
These values are, respectively:
$$\begin{matrix}
2^03^0 & 2^13^0 & 2^23^0 & 2^33^0 & 2^43^0 & 2^53^0 & 2^63^0 \\
2^03^1 & 2^13^1 & 2^23^1 & 2^33^1 & 2^43^1 & 2^53^1 & 2^63^1 & \\
2^03^2 & 2^13^2 & 2^23^2 & 2^33^2 & 2^43^2 & 2^53^2 & 2^63^2 & \\
2^03^3 & 2^13^3 & 2^23^3 & 2^33^3 & 2^43^3 & 2^53^3 & 2^63^3
\end{matrix}$$
| {
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"url": "https://math.stackexchange.com/questions/174055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Inequality for cosines Is the following inequality in a triangle known?
$$4(\cos A + \cos B + \cos C) \le 3 + \cos \left(\frac{B-C}{2}\right) + \cos \left(\frac{C-A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$$
It looks correct to me but I would appreciate if someone confirm it.
| $\cos A+\cos B+\cos C=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$
$=2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}\quad$ as $\quad A+B+C=\pi$ $\implies\cos\frac{A+B}{2}=\sin\frac{C}{2}$
$\leq 2\sin\frac{C}{2}+1-2\sin^2\frac{C}{2}\quad$ as $\quad\cos\frac{A-B}{2}\leq 1$
$=-\frac{1}{2}(2\sin\frac{C}{2}-1)^2+1+\frac{1}{2}$
The maximum value will come if $2\sin\frac{C}{2}=1\quad$ and if $\quad\cos\frac{A-B}{2}=1$
or if $\quad C=\frac{\pi}{3}\quad$ and $\quad A=B$. Then $A+B=\frac{2\pi}{3}\implies A=B=\frac{\pi}{3}=C$
The maximum value of $\cos A+\cos B+\cos C\quad$ thus $\quad\frac{3}{2}$
(i)Now, $4(\cos A+\cos B+\cos C)=2(\cos A+\cos B+\cos C)+2(\cos A+\cos B+\cos C)$
$\leq 2(\cos A+\cos B+\cos C)+2.\frac{3}{2}$
$=3+(\cos A+\cos B)+(\cos B+\cos C)+(\cos C+\cos A)$
Now, $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}=2\sin\frac{C}{2}\cos\frac{A-B}{2}$
So, the problem reduces to $\sum 2\sin\frac{C}{2}\cos\frac{A-B}{2}≤ \sum \cos\frac{A-B}{2}$
(ii) Now, $4(\cos A+\cos B+\cos C)=(\cos A+\cos B+\cos C)+3(\cos A+\cos B+\cos C)$
$\leq (\cos A+\cos B+\cos C)+3.\frac{3}{2}$
$=3+\frac{1}{2}\sum (\cos A+\cos B + 1)$
$=3+\frac{1}{2}\sum (2\sin\frac{C}{2}\cos\frac{A-B}{2} + 1)$
So, the problem reduces to
$\sum( \sin\frac{C}{2}\cos\frac{A-B}{2} + \frac{1}{2}) ≤ \sum \cos\frac{A-B}{2} $
I'm trying to prove this.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Confusing double angle identity How would I solve the following double angle identity.
$$\cos^4x=\frac{3}{8}+\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$$
So far my work is
$$\frac{3}{8}+\frac{2\cos^x-1}{2}+\frac{1}{8}(2\cos^2x-1)$$
But how would I proceed.
| \begin{align*}
\cos^4(x) &= \left(\frac{e^{ix}+e^{-ix}}{2}\right)^4\\
&= \frac{e^{4xi} + 4e^{2xi} + 6 + 4e^{-2xi} + e^{-4xi}}{16}\\
&= \frac{3}{8} + \frac{1}{2} \frac{e^{2xi} + e^{-2xi}}{2} + \frac{1}{8} \frac{e^{4xi}+e^{-4xi}}{2}\\
&= \frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$ I have to prove this most difficult trigonometric identity.
$$\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}.$$
I know
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
My problem is with the extra $Y$ in this problem. What can I do about I think I know a solution which is to do $\tan(A+B)$ then $\tan(B+Y)$ but I am not sure how to apply it.
| Replacing $B$ by $B+Y$ in your expression for $\tan(A+B)$ gives
$$\dfrac{\tan A + \tan(B+Y)}{1 - \tan A \tan(B+Y)}$$
Expanding gives
$$\dfrac{\tan A + \frac{\tan B + \tan Y}{1 - \tan B \tan Y}}{1 - \tan A \frac{\tan B + \tan Y}{1 - \tan B \tan Y}}$$
Multiplying through by $1 - \tan B \tan Y$ gives
$$\frac{\tan A(1 - \tan B \tan Y) + \tan B + \tan Y}{1 - \tan B \tan Y - \tan A(\tan B + \tan Y)}$$
And simplifying gives you what you want.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Polynomial coefficients in exponential-series: how can I convert this into a composite of $\exp(x)$? Assume we have the exponential-series $ \small \exp(x) = 1+ {x \over 1!} + {x^2 \over 2! } + \cdots = \sum\limits_{k=0 }^\infty {x^k \over k!} $ modified with a polynomial in the coefficients
say $ \qquad \displaystyle f_1(x)= \sum_{k=0}^\infty { k^2 + k \over 2} {x^k \over k!} $
or $ \qquad \displaystyle f_2(x)= \sum_{k=0}^\infty ( 15 k^3 + 15 k^2 - 10k - 8) {x^k \over k!} $
or in general
$ \qquad \displaystyle f_3(x)= \sum_{k=0}^\infty ( d k^3 + c k^2 + b k + a) {x^k \over k!} $
-: is there a good formula/algorithm/scheme how this has to be expressed as composition of the $\exp(x)$-function? (I know this can be solved using the derivatives and cancelling of k's in the polynomial with the factorials in the denominator - I'm asking for a handy/memorizable translation-formula )
Because I can factor my examples under study: is there possibly a special handy scheme, if the polynomials are given in a form like this
$ \qquad \displaystyle f_2(x)= \sum_{k=0}^\infty ( k-1)(k-2) {x^k \over k!} $ ?
| \begin{eqnarray}
\sum_{k=0}^\infty k\frac{x^k}{k!}&=&\sum_{k=1}^\infty \frac{x^k}{(k-1)!}=x\sum_{k=0}^\infty\frac{x^k}{k!}=xe^x;\cr
\sum_{k=0}^\infty k^2\frac{x^k}{k!}&=&x\sum_{k=1}^\infty k\frac{x^{k-1}}{(k-1)!}=x\sum_{k=0}^\infty(k+1)\frac{x^k}{k!}=x(x+1)e^x\cr
\sum_{k=0}^\infty k^3\frac{x^k}{k!}&=&\sum_{k=1}^\infty k^2\frac{x^k}{(k-1)!}=x\sum_{k=0}^\infty(k+1)^2\frac{x^k}{k!}\cr
&=&x\sum_{k=0}^\infty(k^2+2k+1)\frac{x^k}{k!}=x[x(x+1)+2x+1]e^x\cr
&=&x(x^2+3x+1)e^x.
\end{eqnarray}
Hence
\begin{eqnarray}
\sum_{k=0}^\infty (dk^3+ck^2+bk+a)\frac{x^k}{k!}&=&d\sum_{k=0}^\infty k^3\frac{x^k}{k!}+c\sum_{k=0}^\infty k^2\frac{x^k}{k!}+b\sum_{k=0}^\infty k\frac{x^k}{k!}+a\sum_{k=0}^\infty \frac{x^k}{k!}\cr
&=&dx(x^2+3x+1)e^x+c(x^2+x)e^x+bxe^x+ae^x\cr
&=&[a+bx+c(x^2+x)+d(x^3+3x^2+x)]e^x\cr
&=&[a+(b+c+d)x+(c+3d)x^2+dx^3]e^x
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2}$ How would I prove the following trig identity?
$$\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2} $$
My work thus far has been:
$$\dfrac{2\cos\dfrac{A+B}{2} \cos\dfrac{A-B}{2}}{-2\sin\dfrac{A+B}{2} \sin\dfrac{A-B}{2}} =-\cot\dfrac{A+B}{2} \cot\dfrac{A-B}{2} \ .$$
| First, note that $\dfrac{\cos(A + B)}{\cos A - \cos B} \neq -\cot \left( \dfrac{A-B}{2} \right) \cot \left(\dfrac{A+B}{2} \right)$
In particular, if we use something like $A = \pi/6, B = 2\pi/6$, then the left is $0$ as $\cos(\pi/2) = 0$ and the right side is a product of two nonzero things.
I suspect instead that you would like to prove:
$$\dfrac{\cos A + \cos B}{\cos A - \cos B} = -\cot \left( \dfrac{A-B}{2} \right) \cot \left(\dfrac{A+B}{2} \right)$$
HINTS
And this looks to me like an exercise in the sum-to-product and product-to-sum trigonometric identities (wiki reference). In fact, if you just apply these identities to the top and the bottom, you'll get the result.
| {
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Use the given equalities to derive trigonometric functions.
(A) $\sin(-x)=-\sin x$
(B) $\cos(-x)=\cos x$
(C) $\cos(x+y)=\cos x\cos y-\sin x\sin y$
(D) $\sin(x+y)=\sin x\cos y+\cos x\sin y$
Use these equalities to derive the following important trigonometric functions:
f) $\left|\cos\dfrac{x}{2}\right|=\sqrt{\dfrac{1+\cos x}{2}}$
g) $\left|\sin\dfrac{x}{2}\right|=\sqrt{\dfrac{1-\cos x}{2}}$
This is for (f): Since this is a half-angle identity I replace $x$ with $\frac{\pi}{2}$. And I'll use (C). $\cos(\frac{\pi}{2}+\frac{\pi}{2})=\cos\frac{\pi}{2}\cos\frac{\pi}{2}-\sin\frac{\pi}{2}\sin\frac{\pi}{2}\Rightarrow \cos2\frac{\pi}{2}=\cos^2\frac{\pi}{2}-\sin^2\frac{\pi}{2}$
Using power reduction identity of: $\cos^2\theta=\dfrac{1+\cos2\theta}{2}$ yields $\cos2\frac{\pi}{2}=\dfrac{1+\cos2\frac{\pi}{2}}{2}$.
I do not believe this is correct because $\cos^2\theta\ne \cos2\theta$. Please help, but no answers.
| Note that the identity $$\cos^2\theta=\frac{1+\cos 2\theta}2$$ is equivalent to (f), so you shouldn’t be using it: your argument will necessarily be circular. Using (C) is fine, however: just set apply it to $\cos\left(\frac{x}2+\frac{x}2\right)$. You’ll also need the Pythagorean identity $\sin^2x+\cos^2x=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| Note that $$a+b+c≥3, \tag{1}$$ by AM-GM inequality
Now, by Titu's lemma $$a^2+b^2+c^2 ≥\frac{(a+b+c)^2}3 ≥ \frac{(a+b+c)^2}{a+b+c}$$ by using (1)
then $$a^2+b^2+c^2 ≥a+b+c$$ Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$
So, LS=
$$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$
$$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\right)-\left(\dfrac{1}{\tan\theta}\cdot \dfrac{\sin\theta}{1}\right)$$
$$\dfrac{\tan\theta}{\sin\theta}-\dfrac{\sin\theta}{\tan\theta}$$
Now, considering the fact that I must have a common denominator to subtract, would this be correct:
$$\dfrac{\tan^2\theta}{\tan\theta\sin\theta}-\dfrac{\sin^2\theta}{\tan\theta\sin\theta}\Rightarrow \dfrac{\tan^2\theta-\sin^2\theta}{\tan\theta\sin\theta}$$
I feel like I'm close to the answer because the denominator is the RS of the OP. Please help. Do not give me the answer.
| Brian's answer is short, sweet, and correct. But you might wonder if you can finish it from where you left off. You can.
*
*$\dfrac{\tan^2 \theta - \sin ^2 \theta}{\sin \theta \tan \theta}$
*Expand everything into $\sin$ and $\cos$ and simplify into a fraction. Along the way, we get $\dfrac{(\sin^2 \theta - \sin^2 \theta \cos^2 \theta)\cos \theta}{\cos^2 \theta\sin^2 \theta} = \dfrac{\sin^2 \theta \cos \theta(1 - \cos^2 \theta)}{\sin^2 \theta \cos^2 \theta} = \dfrac{(1 - \cos^2 \theta)}{\cos \theta}$
*Remember that $\sin^2 \theta + \cos^2 \theta = 1$, and use this to end with the same result of Brian's answer: $\dfrac{\sin^2 \theta}{\cos \theta}$.
*Finish it from there.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let, both $a$ and $b$ belong to the set {1,2,3,4}. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots. Let, both $a$ and $b$ belong to the set {1,2,3,4}. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots.
for real roots, $a \gt 0$, $b^2-4{a}{c} \ge 0$
Here we have $c=1$, and $a \ge 0$
Now we need to have $b^2-4a \ge 0$
i.e. $(-2\sqrt{a} \ge b) \cup (b \ge 2\sqrt{a})$
| There are of course 16 equations of the form $ax^2+bx+1=0$ with $a,b \in X:= \{1,2,3,4\}$. Which of them have real roots?
As you note, we have to determine for which $a,b\in X$ we have $b^2-4a\ge 0$.
So if $a=1$, we have $b=2,3,4$.
If $a=2$, we have $b=3,4$, if $a=3,4$ then we must have $b=4$. Summing up there are 7 equations.
| {
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Equilateral triangle inscribed in a triangle Consider triangle ABC. Point D is on AC, E is on BC and F is on AB. Given that triangle DEF is equilateral and that segments CD, BE and AF are equal in length, prove that triangle ABC must also be equilateral. I've played with this one for long enough using circles, sine law, cosine law, etc., and have almost reached the point of capitulation. Any hints would be appreciated. I'm not actually sure if the statement is true, but I haven't been able to find a counterexample, either, so I'm stuck both ways.
| Let $a$, $b$, $c$ be the lengths of the sides opposite $A$, $B$, $C$, and let points $D$, $E$, $F$ lie on the sides opposite $A$, $B$, $C$. (The latter is slightly different than in the original question). Let $r = |DE| = |EF| = |FD|$ be the length of the sides of the equilateral triangle; let $s = |AF| = |BD| = |CE|$ be the lengths of the congruent sub-segments of the sides.
The Law of Cosines for each $r$-side in $\triangle AEF$, $\triangle BFD$, $\triangle CDE$ gives
$$\begin{align}
r^2 = s^2 + (b-s)^2 - 2 s (b-s) \cos A \\
r^2 = s^2 + (c-s)^2 - 2 s (c-s) \cos B \\
r^2 = s^2 + (a-s)^2 - 2 s (a-s) \cos C
\end{align}$$
Setting the right-hand sides equal and subtracting $2s^2$ yields:
$$b^2-2bs-2s(b-s)\cos A = c^2-2cs-2s(c-s)\cos B = a^2 - 2 a s - 2 s(a-s) \cos C$$
Using the Law of Cosines (in $\triangle ABC$) to re-write each of $\cos A$, $\cos B$, $\cos C$, and then simplifying a bit, gives
$$\begin{align}
&\phantom{=}\;a\left( s^2\left(-a^2+b^2+c^2\right)+ s b\left(a^2-\left(b+c\right)^2\right)+b^3 c\right) \\
&= b\left( s^2\left(\phantom{-}a^2-b^2+c^2\right)+ s c\left(b^2-\left(c+a\right)^2\right)+c^3 a\right) \\
&= c\left( s^2\left(\phantom{-}a^2+b^2-c^2\right)+ s a\left(c^2-\left(a+b\right)^2\right)+a^3 b\right)
\end{align}$$
We can break the triple equality into a system of two quadratic equations in the parameter $s$. Eliminating $s$ gives a polynomial equation, $p(a,b,c)=0$; the factors of $p$ provide these equations:
$$\begin{align}
abc (a+b+c) &= 0 &(1) \\
a^2 b\left(a-b\right)+b^2c\left(b-c\right)+c^2 a\left(c-a\right) &= 0 & (2) \\
a b\left(a-b\right)\left(a\left(b-c\right)\left(a^2-b^2\right)-b^2c^2\right) \\
+b c\left(b-c\right)\left(b\left(c-a\right)\left(b^2-c^2\right)-c^2a^2\right) \\
+c a\left(c-a\right)\left(c\left(a-b\right)\left(c^2-a^2\right)-a^2b^2\right) &= 0 & (3)
\end{align}$$
Solutions in each case correspond to ostensible solutions to the problem.
We can ignore (1), which has only trivial solutions. I believe (2)'s only non-trivial solutions (with the Triangle Inequality in play) require $a=b=c$, but I need to double-check this.
As for (3), clearly $a=b=c$ works; and, indeed, if (say) $a=b$, then $a=b=c$. So any non-equilateral solutions would need to be strictly scalene; it turns out that (3) admits scalene solutions.
Below is Mathematica's plot of (3), with $c=1$; the horizontal axis is $a$, and the vertical is $b$. The lines are $a+b=c$ and $b+c=a$ and $c+a=b$, which bound the region of feasibility according to the Triangle Inequality. The point $(1,1)$ seems to have been missed by the implicit plotter, but lots of other points appear in blue. (I believe there are even more. Zooms show an island near the $a+b=c$ border, for instance.)
Here's the figure for $a=8$, $b=8.58667$, $c=1$:
Equilateral $\triangle DEF$ extends beyond the confines of $\triangle ABC$. This may be why @Brian was having difficulty ruling-out non-equilateral $\triangle ABC$s.
I suspect equation (3) can be expressed in a way that's more illuminating; it's not clear to me, for instance, whether every solution to (3) corresponds to an "external" triangle. There's probably a whole other approach to this problem that makes everything obvious, but this is the best I have at the moment.
| {
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Triple "Pythagorean identity" It's not hard to find multiple trigonometric functions of period $2\pi$ that added to self shifted by some constant offset result in a constant.
In classic pythagorean identity, you have
$$F(x)+F\left(x+\frac{\pi}{2} \right) = 1 $$
where $F(x)=\sin^2 x$
Or you can use symmetry of the sine wave and create
$F(x)+F(x+\pi ) = 0$
where $F(x) = \sin x$
Now what I'm looking for is a transformation of the sine function that while retaining the $2\pi$ period, gives you identity if repeated three times, with $\frac 2 3 \pi$ shift in each appearance:
$$F(x) + F\left(x + \frac 2 3 \pi \right) + F\left(x+\frac 4 3 \pi\right) = \mathrm{const}$$
Can you find such a function?
Reason and purpose:
I've been trying to develop a better RGB$<=>$HSV color space conversion - all the common ones use sawtooth style variant functions with variant equations, and I think using trigonometric functions could result in more smooth color passages, never mind much simpler algorithm.
| Let $F(x)=\sin(ax+b)$ where a,b are indeterminate constants.
$F(x) + F\left(x + \frac 2 3 \pi \right) + F\left(x+\frac 4 3 \pi\right) $
$=\sin(ax+b)+\sin(a(x+\frac{2\pi}{3})+b)+\sin(a(x+\frac{4\pi}{3})+b)$
$=\sin(ax+b)+\sin(a(x+\frac{4\pi}{3})+b)+\sin(a(x+\frac{2\pi}{3})+b)$
$=(1+2\cos\frac{2\pi a}{3})\sin(a(x+\frac{2\pi}{3})+b)$
This expression can not be constant for all $x$ unless $1+2\cos\frac{2\pi a}{3}=0$
Or, $\cos\frac{2\pi a}{3}=-\frac{1}{2}=\cos\frac{2\pi}{3}$
$\implies \frac{2\pi a}{3}=2m\pi±\frac{2\pi}{3}$ where $m$ is an integer,
$\implies a=3m±1$
$F(x)=\sin(ax+b)$ where $a$ is an integer with $(a,3)=1$.
As $b$ is an indeterminate constant, we can have some other indeterminate constant $c=b-\frac{\pi}{2}$
$F(x)=\sin(ax+\frac{\pi}{2}+c)=\cos(ax+c)$
| {
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"url": "https://math.stackexchange.com/questions/189832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A matrix inequality If $A=(a_{ij})$ is a real positive definite symmetric matrix of order $n$. How to show $(n-1)\prod_{i=1}^na_{ii}+\det A\ge \sum_{i=1}^na_{ii}\det A(i)$? $A(i)$ means the submatrix of $A$ by deleting the $i$th row and $i$th column.
| if $n=3$. consider the two matrix below:
$$A=\begin{pmatrix}1 & \frac{1}{4} & -\frac{1}{4}\cr \frac{1}{4} & 1 & -\frac{1}{4}\cr -\frac{1}{4} & -\frac{1}{4} & 1\end{pmatrix}$$
$$B=\begin{pmatrix}1 & -\frac{1}{4} & -\frac{1}{4}\cr -\frac{1}{4} & 1 & -\frac{1}{4}\cr -\frac{1}{4} & -\frac{1}{4} & 1\end{pmatrix}$$
they are both positive definite matrix. but for matrix $A$, we have:$$2\prod_{i=1}^3a_{ii}+\det A> \sum_{i=1}^3a_{ii}\det A(i)$$ hold.
but for matrix $B$, the inequality don't hold. and matrix $B$ satisfy:$$2\prod_{i=1}^3b_{ii}+\det B< \sum_{i=1}^3b_{ii}\det B(i)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Proving the inequality $a^2+b^2+c^2+ab+bc+ca\ge6$ Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:
$a^2+b^2+c^2+ab+bc+ca\ge6$
| By squaring $a+b+c=3$ we get
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=9.$$
From the AM-GM inequality
(or from the fact that $(x-y)^2=x^2+y^2-2xy\ge 0$, i.e. $2xy\le x^2+y^2$)
we have
$$ab+ac+bc \le \frac{a^2+b^2}2+\frac{a^2+c^2}2+\frac{b^2+c^2}2=a^2+b^2+c^2,$$
i.e. $\frac12(a^2+b^2+c^2) \ge \frac12(ab+ac+bc)$, which is equivalent to
$\frac12(a^2+b^2+c^2) - \frac12(ab+ac+bc) \ge0$.
By adding the above equality and inequality together you get
$$\frac32(a^2+b^2+c^2+ab+ac+bc)\ge9,$$
which is equivalent to
$$a^2+b^2+c^2+ab+ac+bc\ge6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/193140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ I am trying to prove $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$
This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.
$$I_n := \int\cos^n x \ dx = \int\cos^{n-1} x \cos x \ dx \tag{1}$$
First question: why rewrite the original instead of immediately integrating by parts of $\int \cos^n x \ dx$?
Integrate by parts with
$$u = \cos^{n-1} x, dv = \cos x \ dx \implies du = (n-1)\cos^{n-2} x \cdot -\sin x, v = \sin x$$
which leads to
$$I_n = \sin x \ \cos^{n-1} x +\int\sin^2 x (n-1) \ \cos^{n-2} x \ dx \tag{2}$$
Since $(n-1)$ is a constant, we can throw it out front of the integral:
$$I_n = \sin x \ \cos^{n-1} x +(n-1)\int\sin^2 x \ \cos^{n-2} x \ dx\tag{3}$$
I can transform the integral a bit because $\sin^2 x + cos^2 x = 1 \implies \sin^2 x = 1-\cos^2 x$
$$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$
According to Wikipedia as noted here, this simplifies to:
$$I_n = \sin x \ \cos^{n-1} x + (n-1) \int \cos^{n-2} x \ dx - (n-1)\int(\cos^n x) \ dx \tag{5}$$
Question 2: How did they simplify the integral of $\int(1-\cos^2 x) \ dx$ to $\int(\cos^n x) \ dx$?
Assuming knowledge of equation 5, I see how to rewrite it as
$$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x - (n-1) I_{n} \tag{6}$$
and solve for $I_n$. I had tried exploiting the fact that
$$\cos^2 x = \frac{1}{2} \cos(2x) + \frac{1}{2} $$
and trying to deal with $\int 1 \ dx - \int \frac{1}{2} \cos (2x) + \frac{1}{2} \ dx$
which left me with $\frac{x}{2} - \frac{1}{4} \sin(2x)$ after integrating those pieces. Putting it all together I have:
$$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x \left(-(n-1) (\frac{x}{2} - \frac{1}{4} \sin 2x) \right) \tag{7}$$
but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of
$$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$
| How about verifying using differentiation? We are asked to show
$$\int_0^x \cos^n t dt = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n} \int^x_0 \cos^{n-2} t dt + C$$
for some constant $C$. This is true iff the derivatives of each side are equal.
Differentiating the right side, we get using the product rule
$$RHS = \frac{n-1}{n} \cos^{n-2} x (-\sin x) \sin x + \frac{1}{n} \cos^{n-1} x \cos x + \frac{n-1}{n} \cos^{n-2} x,$$
then combining,
$$RHS = \frac{n-1}{n} (1-\sin^2 x) \cos^{n-2} x + \frac{1}{n} \cos^n x$$
and finally using $\cos^2 x + \sin^2 x = 1$,
$$RHS = \frac{n-1}{n}\cos^n x+\frac{1}{n} \cos^nx = \cos^n x,$$
which is the what we want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/193435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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} |
Inequality.$\frac{ab}{\sqrt{ab+2c^2}}+\frac{bc}{\sqrt{bc+2a^2}}+\frac{ca}{\sqrt{ca+2b^2}} \geq \sqrt{ab+bc+ca}$ Let $a,b,c > 0$. Prove that (using Hölder's inequality):
$$\frac{ab}{\sqrt{ab+2c^2}}+\frac{bc}{\sqrt{bc+2a^2}}+\frac{ca}{\sqrt{ca+2b^2}} \geq \sqrt{ab+bc+ca}.$$
Thanks :)
I tried to apply Hölder's inequality how I apply in this exercise but I didn't obtained anything.
| Using Hölder's inequality, we get
$$
\left(\sum_{cyc} \frac {ab}{\sqrt{ab + 2c^2}}\right) \left(\sum_{cyc} \frac {ab}{\sqrt{ab + 2c^2}}\right) \left(\sum_{cyc} ab (ab + 2c^2)\right) \geq (ab + bc + ca)^3
$$
The above inequality can be rewritten as
$$
\left(\sum_{cyc} \frac {ab}{\sqrt{ab + 2c^2}}\right)^2 (ab + bc +ca)^2 \geq (ab + bc + ca)^3
$$
And so
$$
\sum_{cyc} \frac {ab}{\sqrt{ab + 2c^2}} \geq \sqrt{ab + bc + ca}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/193820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$ I try to find $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$.
Left side:
$$\sin(4 \alpha + \frac{\pi}{6}) =$$
$$= \sin4\alpha \times \cos \frac{\pi}{6} + \cos 4\alpha \times \sin\frac{\pi}{6} =$$
$$= \sin4\alpha \times \frac{\sqrt{3}}{2} + \cos 4\alpha \times \frac{1}{2} = $$
$$= 2\sin2\alpha \times \cos2\alpha \times \frac{\sqrt{3}}{2} + (\cos^2 2\alpha - \sin^2 2\alpha) \times \frac{1}{2} = $$
$$= 2\sin\alpha \times \cos \alpha \times \cos2\alpha \times \sqrt{3} + \frac{\cos^2 2\alpha - \sin^2 2\alpha}{2} = $$
$$= 2\sin\alpha \times \cos \alpha \times (\cos^2\alpha - \sin^2\alpha) \times \sqrt{3} + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha) - \sin^2 2\alpha}{2} = $$
$$= 2\sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha) - (2\sin\alpha \times \cos\alpha)(2\sin\alpha \times \cos\alpha)}{2} = $$
$$= 2\sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha)}{2} - 2\sin^2\alpha \times \cos^2\alpha = $$
$$= 2 \times \sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{\cos^4 \alpha - (\sin^2\alpha \times \cos^2\alpha) - (\sin^2\alpha \times \cos^2\alpha) + \sin^4 \alpha}{2} - 2\sin^2\alpha \times \cos^2\alpha = $$
$$= 2 \sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2\alpha - \sin^2\alpha)^2}{2} - 2\sin^2\alpha \times \cos^2\alpha = $$
$$= 2\sqrt{3}(\sin\alpha \times \cos^3 \alpha - \sin^3\alpha \times \cos\alpha) + \frac{(1 - 2\sin^2\alpha)^2}{2} - 2\sin^2\alpha \times \cos^2\alpha = $$
$$= 2\sqrt{3}(\sin\alpha \times \cos^3 \alpha - \sin^3\alpha \times \cos\alpha) + \frac{1}{2}-2\sin^2\alpha+2\sin^4\alpha - 2\sin^2\alpha \times \cos^2\alpha = $$
$$= 2\sin\alpha(\sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha) = $$
Right side:
$$\sin(2\alpha + \frac{\pi}{5}) = \sin2x \times \cos\frac{\pi}{5} + \cos2\alpha \times \sin\frac{\pi}{5} = $$
$$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + (\cos^2\alpha - \sin^2\alpha)\sin\frac{\pi}{5} = $$
$$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + (1 - 2\sin^2\alpha)\sin\frac{\pi}{5} = $$
$$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + \sin\frac{\pi}{5} - 2\sin^2\alpha \times \sin\frac{\pi}{5} = $$
$$= 2\sin\alpha(\cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5})$$
I can't get any further on either. Bringing them together I get:
$$2\sin\alpha(\sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha) = 2\sin\alpha(\cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5})$$
$$\iff \sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha = \cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5}$$
Where do I go from here? Om am I already dead wrong?
| Hint:
if $\sin A=\sin B$,then we have $A=k\pi +(-1)^kB$,it is easy to prove.
| {
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"url": "https://math.stackexchange.com/questions/195229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\sum\limits_{n=1}^{9999} \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})} $ How we can find
$$\sum_{n=1}^{9999} \frac{1}{(\sqrt{n+1}+\sqrt{n})(\sqrt[4]{n}+\sqrt[4]{n+1})} $$
| $\frac {1}{(\sqrt {n+1}+\sqrt {n})(\sqrt[4] {n+1}+\sqrt[4] {n})}=\frac {(\sqrt[4] {n+1}-\sqrt[4] {n})}{(\sqrt {n+1}+\sqrt {n})(\sqrt[4] {n+1}+\sqrt[4] {n})(\sqrt[4] {n+1}-\sqrt [4]{n})}=\frac {(\sqrt[4] {n+1}-\sqrt [4]{n})}{n+1-n}=\sqrt[4] {n+1}-\sqrt[4] {n}$ so the sum is transformed to $\sum _{n=1}^{9999} \sqrt[4] {n+1}-\sqrt[4] {n}=\sqrt[4] {10000}-\sqrt[4] {1}=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/196156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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$\int \frac{\sin^3x}{\sin^3x + \cos^3x)}$? Is it possible to evaluate the following integral:$$\int \frac{\sin^3x}{(\sin^3x + \cos^3x)} \, dx$$
| Put $$I = \displaystyle\int{\dfrac{\sin^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x , \quad J = \displaystyle\int{\dfrac{\cos^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x.$$
We have $$I + J = \displaystyle\int{\mathrm{d}x} = x+ C.$$
and
\begin{equation*}
I - J = \displaystyle\int{\dfrac{\sin^3 x - \cos^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x = \displaystyle\int{\dfrac{(\sin x - \cos x)(1 + \sin x \cdot \cos x)}{(\sin x + \cos x)(1 - \sin x \cdot \cos x)}\mathrm{d}x}
\end{equation*}
Put $t = \sin x + \cos x$, then $\mathrm{d}t = -(\sin x - \cos x) \mathrm{d}x$ and $ \sin x \cdot \cos x = \dfrac{ t^2-1}{2}.$
We get
\begin{equation*}
I - J = \displaystyle\int{\dfrac{t^2 + 1}{t(t^2-3)}\mathrm{d}t} = \dfrac{2}{3}\ln{(t^2-3)}-\dfrac{1}{3}\ln t + C'.
\end{equation*}
and then
\begin{equation*}
I - J = \dfrac{2}{3}\ln{((\sin x + \cos x)^2-3)}-\dfrac{1}{3}\ln (\sin x + \cos x) + C'.
\end{equation*}
From $I + J $ and $I - J$, we can calculate $I$.
| {
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"url": "https://math.stackexchange.com/questions/198083",
"timestamp": "2023-03-29T00:00:00",
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Solving an equation with three quadratic radicals in the set of real numbers How do we solve the following equation in the set of real numbers?
$$(26-x)\cdot\sqrt{5x-1} -(13x+14)\cdot\sqrt{5-2x} + 12\sqrt{(5x-1)\cdot(5-2x) }= 18x+32.$$
I tried putting $a=\sqrt{5x−1}$ and $b=\sqrt{5−2x}$ and then $2a^2+5b^2=23$.
| This is another solution.
Put $a=\sqrt{5x-1}$ and $b=\sqrt{5-2x}$.
We have
\begin{equation*}
26-x = \dfrac{47}{23}(5x - 1 ) + \dfrac{129}{23}(5-2x) = \dfrac{47}{23}a^2 + \dfrac{129}{23}b^2,
\end{equation*}
\begin{equation*}
13x+14= \dfrac{93}{23}(5x - 1 ) + \dfrac{83}{23}(5-2x) = \dfrac{93}{23}a^2 + \dfrac{83}{23}b^2,
\end{equation*}
\begin{equation*}
18x+32= \dfrac{154}{23}(5x - 1 ) + \dfrac{178}{23}(5-2x) = \dfrac{154}{23}a^2 + \dfrac{178}{23}b^2.
\end{equation*}
The given equation become
\begin{equation*}
\begin{cases}
47a^3+129ab^2-93ba^2-83b^3-154a^2-178b^2+276ab = 0,\\
2a^2+5b^2=23.
\end{cases}
\end{equation*}
This system of equations is solved here
How do we solve the system of equations?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the all integer solutions How to Find the all integer solutions for:
$$x+y+z=3$$
$$x^3+y^3+z^3=3$$
| We have $x^3+y^3=3-z^3$ and $x+y=3-z$. Since $x+y$ divides $x^3+y^3$, we conclude that $z-3$ divides $z^3-3$, and therefore $z-3$ divides $24$. Similar considerations apply to $x$ and $y$.
So we are down to a finite and indeed fairly short list of candidates. We can use further little tricks to winnow the list.
Remark: Let's throw in some number theory. It is a sometimes useful fact that $a^3$ is always congruent to $0$, $1$, or $-1$ modulo $9$. Thus if $x^3+y^3+z^3=3$, we must have $x^3$, $y^3$, and $z^3$ all congruent to $1$ modulo $9$. It follows that all of $x$, $y$, and $z$ are congruent to $1$ modulo $3$, and hence so are $x-3$, $y-3$, and $z-3$. The only divisors of $24$ that satisfy this condition are $1$, $4$, $-2$, and $-8$. So our only candidates for $x$, $y$, and $z$ are $4$, $7$, $1$, and $-5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/202848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the solution of this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$
Possible Duplicate:
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$.
Help me solve this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$
| HINT: Note that $2-\sqrt{3} = \dfrac1{2+\sqrt{3}}$.
Denote $\left(\sqrt{2-\sqrt{3}} \right)^x$ as $t$ and proceed to solve the quadratic in $t$ and hence solve for $x$.
| {
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"url": "https://math.stackexchange.com/questions/202997",
"timestamp": "2023-03-29T00:00:00",
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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.