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Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$ How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$
What's the approach to it?
Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education ).I just know simple elementary results of definite and indefinite integration. Substitutions and all those works good. :)
| Consider
$$
I(c)=\int_{0}^{\frac{\pi}{2}} \ln \left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x+c\right) d x, \textrm{ where } c\geq 0.
$$
Using my post, we have
$$
I(c)=\pi \ln \left(\frac{\sqrt{a^{2}+c}+\sqrt{b^{2}+c}}{2}\right)
$$
Differentiating $I(c)$ w.r.t. $c$ yields
$$
I^{\prime}(c)=\frac{\pi}{2 \sqrt{\left(a^{2}+c\right)\left(b^{2}+c\right)}}
$$
Once more gives
$$
I^{\prime \prime}(c)=-\frac{\pi}{4\left[\left(a^{2}+c\right)\left(b^{2}+c\right)\right]^{\frac{3}{2}}\left(2 c+a^{2}+b^{2}\right)}
$$
Hence
$$
\boxed{\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x+c\right)^{2}}= \frac{\pi\left(2 c+a^{2}+b^{2}\right)}{4\left[\left(a^{2}+c\right)\left(b^{2}+c\right)\right]^{\frac{1}{2}}}}
$$
Putting $c=0$ yields the result
$$\boxed{\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x+c\right)^{2}}= \frac{\pi\left(a^{2}+b^{2}\right)}{4 a^{3} b^{3}} }$$
Furthermore, in general,
$$
\boxed{\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x+c\right)^{n}}=\frac{(-1)^{n} \pi}{2(n-1) !} \frac{d^{n-1}}{d c^{n-1}}\left[\left(a^{2}+c\right)\left(b^{2}+c\right) \right] ^{-\frac{1}{2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Prove ${\frac {1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{ \frac {1+{c}^{3}}{1+c{a}^{2}}}\ge 3 $
Let $a,b,c \ge0$, prove the on equality: $${\frac
{1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{ \frac
{1+{c}^{3}}{1+c{a}^{2}}}\ge 3 $$
I tried: $$LHS = \sum\frac 1{1+ab^2}+\sum \frac {a^4}{a+a^2b^2} \ge\frac 9{3+\sum ab^2} + \frac {(a^2+b^2+c^2)^2}{\sum a+ \sum (ab)^2}\ge...$$
but it seem like useless.
Look like better inequality is $ \left( 1+{a}^{3} \right) \left( 1+{b}^{3} \right) \left( 1+{c}^{3}
\right) \ge \left( 1+a{b}^{2} \right) \left( 1+b{c}^{2} \right)
\left( 1+c{a}^{2} \right)$ (but still can't prove it)
| Without loss of generality, let $a\geqslant b\geqslant c\geqslant0$. The target inequality is equivalent to: $$\left(\frac{1+{a}^{3}}{1+a{b}^{2}}-1\right)+\left(\frac{1+{b}^{3}}{1+b{c}^{2}}-1\right)+\left(\frac{1+{c}^{3}}{1+c{a}^{2}}-1\right)\geqslant0$$
That is
\begin{align*}
a\frac{a^2-b^2}{1+ab^2}+b\frac{b^2-c^2}{1+bc^2}+c\frac{c^2-a^2}{1+ca^2}\geqslant0\Leftrightarrow\\
a\frac{a^2-b^2}{1+ab^2}+b\frac{b^2-c^2}{1+bc^2}+c\left(\frac{c^2-b^2}{1+ca^2}+\frac{b^2-a^2}{1+ca^2}\right)\geqslant0\Leftrightarrow\\
(a^2-b^2)\left(\frac{a}{1+ab^2}-\frac{c}{1+ca^2}\right)+(b^2-c^2)\left(\frac{b}{1+bc^2}-\frac{c}{1+ca^2}\right)\geqslant0
\end{align*}
Think of the numerators of $\frac{a}{1+ab^2}-\frac{c}{1+ca^2}$ and $\frac{b}{1+bc^2}-\frac{c}{1+ca^2}$ are $$(a-c)+ac(a^2-b^2)\geqslant0~\mbox{and}~(b-c)+bc(a^2-c^2)\geqslant0$$Hence we complete the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show that $n \ge \sqrt{n+1}+\sqrt{n}$ (how) Can I show that:
$n \ge \sqrt{n+1}+\sqrt{n}$ ?
It should be true for all $n \ge 5$.
Tried it via induction:
*
*$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.
*$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$
Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)
Is this the right way?
| What I would try first if I wanted it to be true - no tricks:
$n \ge \sqrt{n + 1} + \sqrt{n} \iff \frac{n}{\sqrt{n}} \ge \frac{\sqrt{n+1}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{n}} \iff \sqrt{n} \ge \sqrt{1 + \frac{1}{n}} + 1 \Leftarrow n \ge 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 0
} |
Show that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$ cannot be a square when $x\neq3$ and $x\neq2$ respectively. As the title says, I need help showing that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$ cannot be a square when $x\neq3$ and $x\neq2$ respectively, where $x$ is a natural number.
In order to do this, I have to use the fact that if $x$ is any natural number >1, any natural number $n\geq x$ can be represented uniquely in the form
$n=c_0+c_1x+c_2x^2+...+c_mx^m$,
where $0\leq c_i \leq x-1$ for $i=0,1,2,...,m-1$ and
$0< c_m \leq x-1$.
| If $x > 3$ and $n = x^4+x^3+x^2+x+1$ is a square then $4n$ is also a square. However
$$ \begin{eqnarray}4(x^4+x^3+x^2+x+1) &=& (2x^2 + x)^2 + 3x^2 + 4x+4 \\ &=& (2x^2+x+1)^2 - (x+1)(x-3) \end{eqnarray}$$
which shows that
$$(2x^2 + x)^2 < 4n < (2x^2+x+1)^2.$$
Since $4n$ is between two consecutive squares it cannot be a square itself. In the same way
$$ \begin{eqnarray}x^4+2x^3+2x^2+2x+5 &=& (x^2 + x)^2 + x^2 + 2x+5 \\ &=& (x^2+x+1)^2 - (x+2)(x-2) \end{eqnarray}$$
which shows that $x^4+2x^3+2x^2+2x+5$ is not a square when $x>2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/403473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to show that $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$ $\sqrt{1+\sqrt{2+\sqrt{3+\cdots\sqrt{2006}}}}<2$.
I struggled on it, but I didn't find any pattern to solve it.
| Idea: You can unwrap like this:
$$\sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}}<2$$
if
$$\sqrt{2+\sqrt{3+\cdots+\sqrt{2006}}}<2^2-1$$
if
$$\sqrt{3+\cdots+\sqrt{2006}}<(2^2-1)^2-2$$
and so on, so we want to show
$$2006 < (((2^2-1)^2-2)^2-\cdots)^2-2005$$
might as well prove it by induction for all $n$ rather than just 2006, so we need to show that
$$n+1 < (((2^2-1)^2-2)^2-\cdots)^2-n$$
implies
$$n+2 < ((((2^2-1)^2-2)^2-\cdots)^2-n)^2-(n+1)$$
but thats just
$$2n+3 < ((((2^2-1)^2-2)^2-\cdots)^2-n)^2$$
which holds since
$$2n+3 < (n+1)^2$$
for all $n>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
| $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$
since priority of multiplication $\times$ or $\cdot$ is greater than addition $+$ so expression will be:
$1+1+1+1+1+1+1+1+1+1+1+0+1\implies 12$
This is the order of operation:
$1$ B:- Brackets first
$2$ O:- Orders (i.e. Powers and Square Roots, etc.)
$3$ DM:- Division and Multiplication (left-to-right)
$4$ AS:- Addition and Subtraction (left-to-right)
I think this will helpful :
http://www.magicalmaths.org/bodmas-starter-5-5-5-5-5-5-5-5-x-0-i-bet-more-people-will-answer-it-wrongp-give-a-try/
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Trigonometric identities using $\sin x$ and $\cos x$ definition as infinite series Can someone show the way to proof that $$\cos(x+y) = \cos x\cdot\cos y - \sin x\cdot\sin y$$ and $$\cos^2x+\sin^2 x = 1$$ using the definition of $\sin x$ and $\cos x$ with infinite series.
thanks...
| In both cases, you'll want to use the Cauchy product, and the binomial theorem will be useful (for at least the first one), too. I leave the second one to you.
For the first one, $$\begin{align}\sin x\cdot\sin y &= \left(\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\right)\cdot\left(\sum_{n=0}^\infty\frac{(-1)^ny^{2n+1}}{(2n+1)!}\right)\\ &= \sum_{n=0}^\infty\left(\sum_{k=0}^n \frac{(-1)^kx^{2k+1}}{(2k+1)!}\cdot\frac{(-1)^{n-k}y^{2(n-k)+1}}{(2(n-k)+1)!}\right)\\ &= \sum_{n=0}^\infty\left(\sum_{k=0}^n \frac{(-1)^nx^{2k+1}y^{2(n+1)-(2k+1)}}{(2k+1)!(2(n+1)-(2k+1))!}\right)\\ &= \sum_{n=0}^\infty\frac{(-1)^n}{(2(n+1))!}\left(\sum_{k=0}^n\frac{(2(n+1))!x^{2k+1}y^{2(n+1)-(2k+1)}}{(2k+1)!(2(n+1)-(2k+1))!}\right)\\ &= \sum_{n=0}^\infty\frac{(-1)^n}{(2(n+1))!}\left(\sum_{k=0}^n\binom{2(n+1)}{2k+1}x^{2k+1}y^{2(n+1)-(2k+1)}\right)\\ &= -\sum_{n=0}^\infty\frac{(-1)^{n+1}}{(2(n+1))!}\left(\sum_{k=0}^n\binom{2(n+1)}{2k+1}x^{2k+1}y^{2(n+1)-(2k+1)}\right)\\ &= -\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left(\sum_{k=0}^{n-1}\binom{2n}{2k+1}x^{2k+1}y^{2n-(2k+1)}\right),\end{align}$$ whence $$-\sin x\cdot\sin y=\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left(\sum_{k=0}^{n-1}\binom{2n}{2k+1}x^{2k+1}y^{2n-(2k+1)}\right).$$
You can do some of the same sorts of manipulations to see that $$\cos x\cdot\cos y = 1+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left(\sum_{k=0}^n\binom{2n}{2k}x^{2k}y^{2n-2k}\right).$$ On the other hand, the binomial theorem shows us that for $n\ge 1$ we have $$(x+y)^{2n} = \sum_{j=0}^{2n}\binom{2n}jx^jy^{2n-j},$$ and splitting the right-hand side into two sums--one for even $j$ and one for odd $j$--gives us $$(x+y)^{2n} = \left(\sum_{k=0}^{n}\binom{2n}{2k}x^{2k}y^{2n-2k}\right)+\left(\sum_{k=0}^{n-1}\binom{2n}{2k+1}x^{2k+1}y^{2n-(2k+1)}\right).$$ Can you put the pieces together and fill in the omitted steps/justifications?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Prove inequality: $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)} + xyz$
Let $x,y,z\in \mathbb R^+$ prove that:
$$\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \ge xyz + \sqrt[3]{xyz(x^2+yz)(y^2+xz)(z^2+xy)}$$
The inequality $\sqrt {(x^2y+ y^2z+z^2x)(y^2x+z^2y+x^2z)} \overset{C-S}{\ge} 3xyz$ will not help because $3xyz \le RHS$. How to prove above inequality ? It is too hard for me.
| The inequality is homogeneous, so we may assume that $xyz=1$. Then if we expand left and right hand side, we see that we have to show that
$$\sqrt{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 3} \ge 1 + \sqrt[3]{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 2}.$$
Let
$$u = \sqrt[3]{x^3 + y^3 + z^3 + \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3} + 2}.$$
If we square both sides of the expanded inequality, we get the equivalent inequality
$$u^3 \ge 2u + u^2,$$
or,
$$u^2 - u - 2 \ge 0.$$
Therefore it is enough to show that $u \ge 2$. But this is certainly true since $x^3 + \frac{1}{x^3}, y^3 + \frac{1}{y^3}, z^3 + \frac{1}{z^3} \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/406779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Equilibrium distribution of a Markov Chain Can anyone please tell me or help me with this question shown below?
A drunken chess grandmaster dials a long string of digits on a standard telephone keypad (laid out as shown below). It takes more than alcohol to make a grandmaster forget the rules of chess, so each digit he dials is a knight's move away from the previous one (e.g. 4 can be followed by 3, 9, or 0). The choices are made at random, independently of previous ones, with all available knight's moves being equally likely to be chosen for each digit.
Find the equilibrium distribution of this Markov chain. (Hint: it helps to use the symmetry, e.g. 1 and 3 must have the same equilibrium probability.)
1 2 3
4 5 6
7 8 9
0
I seem not to understand the question at all. Unlike finding other equilibrium distribution of a Markov Chain question where it is shown as a transition matrix with state space given, could anyone please help on trying to "read" and "identify" this question?
Your helps would be much appreciated.
| Hint: the transition matrix looks like this (1 is the first entry and 0 is the last in each row/column). I have assumed that if the grand master presses $5$ first, we will randomly pick the next digit uniformly.
$$\left[\begin{array}{cccccccccc}
0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0\\
0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 & 0\\
0 & 0 & \frac{1}{3} & 0 & 0 & 0 & 0 & 0 & \frac{1}{3} & \frac{1}{3}\\
\frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10}\\
\frac{1}{3} & 0 & 0 & 0 & 0 & 0 & \frac{1}{3} & 0 & 0 & \frac{1}{3}\\
0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & 0\\
\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & 0
\end{array}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/407154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to solve integrals of type $ \int\frac{1}{(a+b\sin x)^4}dx$ and $\int\frac{1}{(a+b\cos x)^4}dx$ $$\displaystyle \int\frac{1}{(a+b\sin x)^4}dx,~~~~\text{and}~~~~\displaystyle \int\frac{1}{(a+b\cos x)^4}dx,$$
although i have tried using Trg. substution. but nothing get
| Another method is the following:
$$
\frac{1}{(a+b\cos x)^4} = -\frac{1}{6}\frac{\partial^3}{\partial a^3}\frac{1}{a+b\cos x} \\
\begin{align}
\int\frac{\text{d}x}{(a+b\cos x)^4} & = -\frac16\int\frac{\partial^3}{\partial a^3}\frac{\text{d}x}{a+b\cos x} \\
& = -\frac16\frac{\partial^3}{\partial a^3}\int\frac{\text{d}x}{a+b\cos x}
\end{align}
$$
The resulting integral can be solved using "trying"'s answer to receive:
$$
\int\frac{\text{d}x}{(a+b\cos x)^4} = \frac 13\frac{\partial^3}{\partial a^3}\frac 1{\sqrt{b^2-a^2}}\text{artanh}\left[\frac{(a-b)}{\sqrt{b^2-a^2}}\tan\left(\frac x2\right)\right] + C
$$
Following a similar process for the other integral yields
$$
\int\frac{\text{d}x}{(a+b\sin x)^4} = -\frac 13\frac{\partial^3}{\partial a^3}\frac1{\sqrt{a^2-b^2}}\arctan\left[\frac{a\tan(x/2)+b}{\sqrt{a^2-b^2}}\right] + C
$$
Also:
If you're interested in how exactly you use the hint "trying" gave, the following gives an outline:
$$
\int\frac{\text{d}x}{a+b\cos x} = \int\frac{\text{d}x}{a+b\left(\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}\right)} \\
\text{sub } v = \tan(x/2)\text{ and simplify} \\
\int\frac{\text{d}x}{a+b\left(\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}\right)} = 2\int\frac{\text{d}v}{(a-b)v^2+(a+b)}
$$
The resulting integral can be solved using $\arctan(x) = \int\frac{\text{d}x}{x^2+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Exponential equations involving natural numbers at power "x" Find x : $$4^x+15^x=9^x+10^x(2^x-3^x)(2^x-3^x-5^x)$$
| Oops - I made a mistake:
$x=0$ is NOT a root
since
$3^0-2^0 = 0$, not $1$.
Please retract my points.
I'll write it as
$$4^x+15^x=9^x+10^x(3^x-2^x)(5^x+3^x-2^x)$$
We see that $x=0$ is a root,
since
both sides are $2$
(because all the $n^x$ are 1).
(No it's not, but I talked fast, didn't I?)
I'll submit this and then look for more.
If $x=1$,
the two sides are
19 and 69, so the right side is larger.
If $x \ge 1$,
the right side is at least
$9^x+10^x 5^x
= 9^x+50^x$
and this is larger than the left side.
For the fun of it,
I'll try $x=-1$.
The left side is
$\frac1{4}+\frac1{15}
= \frac{19}{60}
$
and the right side is
$\frac1{9}+\frac1{10}(\frac1{3}-\frac1{2})(\frac1{5}+\frac1{3}-\frac1{2})
= \frac1{9}+\frac1{10}(-\frac1{6})(\frac{6+10-15}{30})
=\frac1{9}-\frac1{10}\frac1{6}\frac{1}{30}
=\frac1{9}-\frac1{1800}
$.
These are not equal, so -1 is not a root.
I'll stop here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/411394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there a contradiction is this exercise? The following exercise was a resolution to this problem
Let $\displaystyle\frac{2x+5}{(x-3)(x-7)}=\frac{A}{(x-7)}+\frac{B}{(x-3)}\space \forall \space x \in \mathbb{R}$. Find the values for $A$ and $B$
The propose resolution was:
In order to isolate $A$ on the right side, multiply all the equation by $x-7$
$\displaystyle\frac{(2x+5)(x-7)}{(x-3)(x-7)}=\frac{A(x-7)}{(x-7)}+\frac{B(x-7)}{(x-3)}$
Now is my doubt. The resolution suggests that $x-7$ cancel out.
$\displaystyle\frac{(2x+5)}{(x-3)}=A+\frac{B(x-7)}{(x-3)}$
But, $x-7$ can be equal zero. In this situation, is allowed to perform this operation? One can say "for every $x\neq7$", but on the next step the resolution says "for $x=7$ we have".
$\displaystyle\frac{(14+5)}{(7-3)}=A+\frac{B(0)}{(7-3)} \Leftrightarrow A=\frac{19}{4}$
I think there is a contradiction is this resolution.
| I would do it in another way. the equalation is $$\frac {2x+5}{(x-3)(x-7)}=\frac A {x-3}+\frac B {x-7}$$. what we can do is to take the common domniator on the right side getting $$\frac {2x+5}{(x-3)(x-7)}=\frac {A(x-7)+B(x-3)} {(x-7)(x-3)}$$. since the numerator in both sides is condcuted from numbers and x, we can conclude that: $$\begin{cases}
A+B=2 & \\
-7A-3B=5 & \
\end{cases}
$$. solving this system is relatively simple and by substituting $A=2-B$ in the second equaltion you get $$B=4.75,A=-2.75$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/411658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Maximize area of a triangle with fixed perimeter If perimeter of a triangle is $2d$, what is the length of sides so the triangle has maximal area?
I found some solution using circle and angles, but I think I have to use derivatives.
I need help.
| Let $a$, $b$ and $c$ be the sides of a triangle. The perimeter, $p=a+b+c$, is fixed and we want to find the values of $a$, $b$ and $c$ that give the triangle maximum area. Heron's formula says that the triangle's area is
$$A=\sqrt{s(s-a)(s-b)(a-c)}$$
where s is the semiperimeter $\frac{a+b+c}{2}=\frac{p}{2}.$
Because p is fixed, we can write $c=p-a-b$. Substituting this into the equation above and squaring we find that
\begin{eqnarray}
16A^2=p(p-2a)(p-2b)(2a+2b-p). \quad\quad(1)
\end{eqnarray}
In the first part, we fix $a$ and see what we can do with $b$ to get a maximum. To this end, differentiating with respect to $b$ gives
\begin{align*}
\nonumber 32A\frac{dA}{db}=p(p-2a)\left[(p-2b)(2)+(2a+2b-p)(-2)\right]
\nonumber =4p(p-2a)(p-2b-a).
\end{align*}
If we set this equal to $0$ to find the critical points we find there are two possibilities. The first is that $p=2a$ which leads to $a=\frac{p}{2}$ and $b=c=\frac{p}{4},$ which do not make a proper triangle.
The more interesting possibility is that $p-2b-a=0$, or that $b=\frac{p-a}{2}.$ The significance of this value for $b$ becomes apparent when we see that $c=p-a-b=p-a-\frac{p-a}{2}=\frac{p-a}{2}=b.$ Thus we have established that the triangle is at least iscosceles.
In the second part, we use the value of $b$ just obtained, and see what we can do with $a$. Substituting for $b$ in (1) we find that
$$16A^2=p(p-2a)a^2$$
which we differentiate with respect to $a$ to get
$$32A\frac{dA}{da}=p\left[(p-2a)(2a)+a^2(-2)\right] = 2ap(p-3a).$$
Setting this equal to 0 again gives us $a=\frac{p}{3}.$
Substituting back, we find that $b = \frac{p-a}{2}=\frac{p}{3}$ and finally $c=p-a-b=\frac{p}{3}$ as well, proving that maximum area is achieved when the triangle is equilateral.
| {
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"timestamp": "2023-03-29T00:00:00",
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Estimating the nested radical with $1,2,3,4,\dots$ under square root signs I require some assistance in proving the following inequality:
$$\sqrt{1 + \sqrt{2 + \sqrt{5}}} < \sqrt{ 1 + \sqrt{2 + \sqrt{3 + \sqrt{4 + \cdots}}}} < \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{5}}}}$$
I'm not quite sure how to create a rigorous enough argument here, help would be appreciated.
| I was just curious if this was a way it could be resolved:
$0 < \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}} < \sqrt{ 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}$
This is true because each term in the middle radical is less than $1$.
$\implies 1 < \sqrt{1 + \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}}} < \sqrt{ 1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}$
$\implies 1 < \sqrt{1 + \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}}} < \dfrac{1 + \sqrt{5}}{2}$
$\implies 2 < 2\sqrt{1 + \sqrt{\frac{5}{2^4}+\sqrt{\frac{6}{2^8}+\sqrt{\frac{7}{2^{16}}+\cdots}}}} < 1 + \sqrt{5}$
$\implies 2 < \sqrt{4 + \sqrt{5+\sqrt{6+\sqrt{7+\cdots}}}} < 1 + \sqrt{5}$
$\implies 5 < 3 + \sqrt{4 + \sqrt{5+\sqrt{6+\sqrt{7+\cdots}}}} < 4 + \sqrt{5}$
$\implies \sqrt{1 + \sqrt{2 + \sqrt{5}}} < \sqrt{ 1 + \sqrt{2 + \sqrt{3 + \sqrt{4 + \cdots}}}} < \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{5}}}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor Series Remainder Use Taylor's Theorem to estimate the error in approximating $\sinh 2x$ by $2x + 4/3x^3$ on the interval $[-0.5,0.5]$.
For this question, I use the Taylor's remainder formular, $$
R_n(x)= \frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!}$$
and I get $R_4(0.5) = 0.012\,85$.
Is this correct?
| Generally, you want to use:
$$\displaystyle |R_{n}(x)| \le \frac{M_{n+1}}{(n+1)!}|x-a|^{n+1}$$
where, $\displaystyle R_{n}(x) = f(x) -T_n(x)$ is the remainder term and $T_n(x)$ is the Taylor polynomial of degree $n$ for $f(x)$, centered at $x = a$.
For this problem, we have an odd function, $f(x) = \sinh 2x$ on the interval $\displaystyle \left[-\frac{1}{2},\frac{1}{2}\right]$, so we can take:
$$\displaystyle T_4(x) = P_0(x) + P_1(x) + P_2(x) + P_3(x) + P_4(x) = f(0) + f'(0)(x-0) + \frac{f''(0)}{2!}(x-0)^2 + \frac{f'''(0)}{3!}(x-0)^3 + \frac{f^{(4)}(0)}{4!}(x-0)^3= 0 + 2x + 0 + \frac{8 x^3}{6} + 0 = 2x + \frac{4}{3} x^3$$
We also have: $f^{(5)}(x) = 32 \cosh x$, so
Max $\displaystyle |f^{(5)}(x)|$ on $\displaystyle -\frac{1}{2} \le x \le \frac{1}{2}$ occurs on the endpoints, that is $\displaystyle x = \pm \frac{1}{2}$, so:
$$\displaystyle M_5 = \max_{-\frac{1}{2} \le x \le \frac{1}{2}} \left|f^{(5)}(x)\right| = 32 \cosh(1) \approx 49.37858$$
So, the upper error bound is given by:
$$\displaystyle |f(x) -T_4(x)| = |R_4(x)| \le \frac{M_5}{5!}|x-0|^5 = \frac{49.37858}{5!}\left|\frac{1}{2}\right|^5 = 0.0128590052083333$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Geometric Inequality Related To Median, Altitude For a triangle $ABC$, let $m_{a}$, $h_{a}$ be $A$-median, $A$-altitude.
Define $m_{b}$,$h_{b}$ and $m_{c}$,$h_{c}$ likewise.
Prove that $\dfrac{h_{a}}{m_{b}}+\dfrac{h_{b}}{m_{c}}+\dfrac{h_{c}}{m_{a}}\leq 3$
I have no solution.
| Let $a^2=x$, $b^2=y$ and $c^2=z$.
Thus, by C-S $$\sum_{cyc}\frac{h_a}{m_b}\leq\sqrt{\sum_{cyc}h_a^2\sum_{cyc}\frac{1}{m_b^2}}=\sqrt{\sum_{cyc}\frac{4S^2}{a^2}\sum_{cyc}\frac{4}{2a^2+2c^2-b^2}}=$$
$$=\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)\sum\limits_{cyc}\frac{1}{a^2}\sum_{cyc}\frac{1}{2a^2+2c^2-b^2}}=$$
$$=\sqrt{\frac{\sum\limits_{cyc}(2xy-x^2)\sum\limits_{cyc}xy\sum\limits_{cyc}(2x+2y-z)(2x+2z-y)}{xyz\prod\limits_{cyc}(2x+2y-z)}}=3\sqrt{\frac{\sum\limits_{cyc}(2xy-x^2)\left(\sum\limits_{cyc}xy\right)^2}{xyz\prod\limits_{cyc}(2x+2y-z)}}$$ and it's enough to prove that
$$xyz\prod\limits_{cyc}(2x+2y-z)\geq\sum\limits_{cyc}(2xy-x^2)\left(\sum\limits_{cyc}xy\right)^2$$ or
$$\sum_{sym}(x^4y^2-x^4yz-x^3y^3+x^3y^2z-x^2y^2z^2)\geq0$$ or
$$(x-y)^2(x-z)^2(y-z)^2\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving there are no integer solutions for $3x^2=9+y^3$
Prove there are no $x,y\in\mathbb{Z}$ such that $3x^2=9+y^3$.
Initial proof
Let us assume there are $x,y\in\mathbb{Z}$ that satisfy the equation, which can be rewritten as $$3(x^2-3)=y^3.$$ So, $$3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3.$$ As $3 \mid -3$, it follows that $$3 \mid x^2 \Rightarrow 3^2 \mid x^2.$$ Say $x^2=3^2 \cdot a^2$ for some $a \in \mathbb{Z}$. Then $$\begin{align*}x^2-3 &= 3^2 \cdot a^2 - 3 \\&= 3 \cdot (3 \cdot a^2 - 1)\end{align*}$$ As $3^2 \mid x^2-3$, it follows that $3 \mid 3 \cdot a^2 - 1$. It is obvious that $3 \mid 3 \cdot a^2$, so it follows that $3 \mid -1$, which is false. Therefore, the assumption that there are $x,y\in\mathbb{Z}$ was wrong.
Alternative proof
$$
\begin{align*}
3x^2&=9+y^3\\
3(x^2-3)&=y^3
\end{align*}
$$
So, $3 \mid y \Rightarrow 3^3 \mid y^3 \Rightarrow 3^2 \mid x^2 - 3 \Rightarrow 3^2 \mid x^2 \Rightarrow 3^2 \not\mid x^2 - 3$ and we are done.
Both proofs are essentially the same, except for the fact that the second proof is much shorter. I think the first proof is much more clear, but it takes a bit longer to write down.
*
*Would the second proof suffice in a math competition, or is it too short?
*This problem was part of the Dutch finals of 1978. I figured it out in under a minute, which is quite unusual for me, so now I wonder: is my proof correct?
| Suppose there is a solution to $3x^2 = 9 + y^3$.
Then $3$ divides the RHS hence $3|y^3$, so $3|y$. Let $y = 3a$ for some integer $a$.
Then $3x^2 = 9 + 27a^3$. Hence $9 | 3x^2$ and so $3|x^2$, implying $3|x$. Let $x = 3b$ for some integer $b$.
Then $27b^2 = 9 + 27a^3$, and after cancelling $9$'s we get $3b^2 = 1 + 3a^2$ or $3(b^2 - a^2) = 1$, which clearly has no solutions since the LHS is divisible by $3$.
| {
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Examples of how to calculate $e^A$ I'm trying to learn the process to discover $e^A$
For example, if $A$ is diagonalizable is easy:
$$A =\begin{pmatrix}
5 & -6 \\
3 & -4 \\
\end{pmatrix}$$
Then we have the canonical form $$J_A =\begin{pmatrix}
2 & 0 \\
0 & -1 \\
\end{pmatrix}$$
because the auto-values are $2$ and $-1$.
Am I right? so I continue
$$e^A =P\begin{pmatrix}
e^2 & 0 \\
0 & e^{-1} \\
\end{pmatrix}P^{-1}$$
Where $$P=\begin{pmatrix}
2 & 1 \\
1 & 1 \\
\end{pmatrix}$$
Because the auto-vectors are (2,1) and (1,1).
If the auto-values the things become more complicated:
For example:
$$B =\begin{pmatrix}
0 & 1 \\
-2 & -2 \\
\end{pmatrix}$$
The auto-values are $-1+i$ and $-1-i$, then the canonical form is:
$$J_B =\begin{pmatrix}
-1 & -1 \\
1 & -1 \\
\end{pmatrix}$$
I don't know how to discover $e^B$ in the complex case. How do I have to proceed in this case?
I would like to know also if there are some pdfs or books with examples or problems with solutions about this subject.
I really need help
Thanks a lot.
EDIT
I found another example of a matrix whose some auto-values are complexes:
$$
C=\begin{pmatrix}
1 & 0 & -2 \\
-5 & 6 & 11 \\
5 & -5 & -10 \\
\end{pmatrix}
$$
Its canonical form
$$
J_C=\begin{pmatrix}
1 & 0 & 0 \\
0 & -2 & 1 \\
0 & -1 & -2 \\
\end{pmatrix}
$$
Why $J_A$ is the matrix $A$ in the canonical form? the author doesn't use complexes numbers, why? How do I find $e^A$ in this case? in the same way?
EDIT 2
The book I'm using says that the complex Jordan block related to the auto-value $a+bi$ is
$$
\begin{pmatrix}
a & b \\
-b & a \\
\end{pmatrix}
$$
The book also says that it's using the real Jordan canonical form in contrast with the complex Jordan canonical form (see answer below).
| You might want to have a look at Nineteen Dubious Ways to Compute the Exponential of a Matrix, Twenty-Five Years Later
Other references can be found in The matrix exponential: Any good books?
We have:
$$A = \begin{bmatrix}0 & 1 \\-2 & -2\end{bmatrix}$$
We want to find the characteristic polynomial and eigenvalues by solving
$$|A -\lambda I| = 0 \rightarrow \lambda^2+2 \lambda+2 = (\lambda+(1-i)) (\lambda+(1+i))= 0 \rightarrow \lambda_1 = -1+i, ~~\lambda_2 = -1-i $$
If we try and find eigenvectors, we setup and solve: $[A - \lambda I]v_i = 0$.
So, we have:
$$[A - \lambda_1 I]v_1 = 0 = \begin{bmatrix}1-i & 1\\-2 &-1-i\end{bmatrix}v_1 = 0$$
This leads to a RREF of:
$$\begin{bmatrix}1 & \dfrac{1}{2}(1 +1)\\0 & 0\end{bmatrix}v_1 = 0 \rightarrow v_1 = \left(1, -\dfrac{1}{2}(1 + i)\right)$$
Doing the same process for the second eigenvalue yields: $v_2 = \left(1, -\dfrac{1}{2} + \dfrac{i}{2}\right)$
From all of this information, we can write the matrix exponential using the Jordan Normal Form.
We have the diagonal form:
$$A = P \cdot J\cdot P^{-1} = \begin{bmatrix}-1/2+i/2 & -1/2-i/2 \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}-1-i & 0 \\ 0 & -1+i\end{bmatrix} \cdot \begin{bmatrix}-i & 1/2-i/2 \\ i & 1/2+i/2 \end{bmatrix}$$
So, now we can take advantage of the diagonal form as:
$\displaystyle e^A = P \cdot e^J \cdot P^{-1} = \begin{bmatrix}-1/2+i/2 & -1/2-i/2 \\ 1 & 1\end{bmatrix} \cdot e^{\begin{bmatrix}-1-i & 0 \\ 0 & -1+i\end{bmatrix}} \cdot \begin{bmatrix}-i & 1/2-i/2 \\ i & 1/2+i/2 \end{bmatrix} = \begin{bmatrix}-1/2+i/2 & -1/2-i/2 \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}e^{-1-i} & 0 \\ 0 & e^{-1+i}\end{bmatrix} \cdot \begin{bmatrix}-i & 1/2-i/2 \\ i & 1/2+i/2 \end{bmatrix} = \begin{bmatrix} (1/2+i/2) e^{-1-i}+(1/2-i/2) e^{-1+i} & (1/2) i e^{-1-i}-(1/2) i e^{-1+i} \\-i e^{-1-i}+i e^{-1+i} & (1/2-i/2) e^{-1-i}+(1/2+i/2) e^{-1+i} \end{bmatrix}$
Other methods are shown in that paper I referenced above.
If we compare this for the real case, the process is the same and we end up with:
$\displaystyle e^A = P \cdot e^J \cdot P^{-1} = \begin{bmatrix}1 & 2\\1 & 1\end{bmatrix} \cdot e^{\begin{bmatrix}-1 & 0\\0 & 2\end{bmatrix}} \cdot \begin{bmatrix}-1 & 2\\1 & -1\end{bmatrix}= \begin{bmatrix}1 & 2\\1 & 1\end{bmatrix} \cdot {\begin{bmatrix}e^{-1} & 0\\0 & e^2\end{bmatrix}} \cdot \begin{bmatrix}-1 & 2\\1 & -1\end{bmatrix} = \begin{bmatrix}\dfrac{-1+2 e^3}{e} & -\dfrac{2 (-1+e^3)}{e}\\ \dfrac{-1+e^3}{e} & \dfrac{2-e^3}{e}\end{bmatrix}$
It is also worth noting that things can also get very ugly, for example, see:
generalized eigenvector for 3x3 matrix with 1 eigenvalue, 2 eigenvectors
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x+{1\over x} = r $ then what is $x^3+{1\over x^3}$? If $$x+{1\over x} = r $$ then what is $$x^3+{1\over x^3}$$
Options:
$(a) 3,$
$(b) 3r,$
$(c)r,$
$(d) 0$
| Given, $x+1/x=r$
$x^3+\frac1{x^3}=\left(x+\frac1x\right)^3-3\cdot x\cdot\frac1x\left(x+\frac1x\right)$
which gives us, $x^3+\frac1{x^3}$ = $ r^3-3r$
| {
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"timestamp": "2023-03-29T00:00:00",
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$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
| Form an equation with $a,b,c$ as its roots.
We have $ab+bc+ca=-\frac{1}{2}$ from the given relation.
So the equation with $a,b,c$ as its roots is,
$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0$
$\Rightarrow x^3-\frac{1}{2}x-abc=0$
If all of $a,b,c$ are of same sign then we have $a=b=c=0$ in which case the inequality is trivially true.So we may consider without loss of generality that $c>0,b<0$ then we have
$a=-(c+b)\Rightarrow a^2=c^2+b^2+2bc\le c^2+b^2=1-a^2\Rightarrow 2a^2\le 1\Rightarrow \frac{1}{2}-a^2\ge 0$
Putting $a$ in the above equation we have $a^3-\frac{1}{2}a+abc=0$
So we have ,
$abc=a(a^2-\frac{1}{2})$
$\Rightarrow a^2b^2c^2=a^2(a^2-\frac{1}{2})^2=a^2(\frac{1}{2}-a^2)^2$
By applying A.M. G.M. on $2a^2(\frac{1}{2}-a^2)^2$(Note that $a^2\ge 0,\frac{1}{2}-a^2\ge 0$ so we can apply A.M.-G.M.) we have,
$2a^2(\frac{1}{2}-a^2)^2\le \left(\frac{2a^2+\frac{1}{2}-a^2+\frac{1}{2}-a^2}{3}\right)^{3}=\frac{1}{27}\Rightarrow a^2b^2c^2=a^2(\frac{1}{2}-a^2)^2\le \frac{1}{27}$
Equality will hold iff $2a^2=\frac{1}{2}-a^2\Rightarrow 3a^2=\frac{1}{2}\Rightarrow a=\frac{1}{\sqrt{6}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. Can anyone give me a hint to evaluate this integral?
$$\int_0^\infty \frac{dx}{1+x^4}$$
I know it will involve the gamma function, but how?
| let $I=\int \dfrac{dx}{1+x^4}$
I=$\dfrac {1}{2}\int \dfrac {x^2+1-(x^2-1)}{1+x^4} $
splitting the fraction,
I=$\dfrac {1}{2} (\int \dfrac {x^2+1}{1+x^4} - \int \dfrac {x^2-1}{1+x^4} )$
let $I_1$ =$ \int \frac {x^2+1}{1+x^4}$
Dividing the numerator and denominator by $x^2$,
$I_1$= $ \int \dfrac {1+\dfrac {1}{x^2} }{x^2 +\dfrac {1}{x^2} } $
put $x-\dfrac{1}{x}=t $ for $I_1$ (observe that the derivative of $x-\dfrac{1}{x}=t $ is $1+\dfrac{1}{x^2}$)
Let $I_2 = \int \dfrac {x^2-1}{1+x^4}$
Dividing the numerator and denominator by $x^2$,
$I_2 = \int \dfrac {1-\dfrac{1}{x^2}}{x^2+\dfrac {1}{x^2}}$
put $x+\dfrac{1}{x}=t $ for $I_2$ (observe that the derivative of $x+\dfrac{1}{x}=t $ is $1-\dfrac{1}{x^2}$)
So, $ I=\dfrac {1}{2} (I_1-I_2) $
Finally, substitute the limits.
Hope this helps.. Cheers!
| {
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Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$ We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$
I have tried and it gets confusing.
| $$\frac{\sin\theta -\cos\theta +1}{\sin\theta +\cos\theta -1}= \frac{1}{\sec\theta - \tan\theta}$$
By taking $$\mbox{L.H.S ( Left hand side )} = \frac{\sin\theta -\cos\theta +1}{\sin\theta +\cos\theta -1} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}-2\sin^2\frac{\theta}{2}}$$
[By applying $1-\cos\theta = 2\sin^2\frac{\theta}{2}$]
Which after simplification gives : $$\frac{\sin\frac{\theta}{2}+ \cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}$$
Now taking R.H.S we get :
$$
\begin{align}
\frac{1}{\sec\theta - \tan\theta} &= \frac{\cos\theta}{1-\sin\theta}\\
&= \frac{\cos^2\theta - \sin^2\theta}{\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \\
&= \frac{(\cos^2\frac{\theta}{2}-\sin^\frac{\theta}{2})}{(\cos\frac{\theta}{2}-\sin\frac{\theta}{2})^2} \\ &= \frac{\sin\frac{\theta}{2}+ \cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}\\ &= \mbox{L.H.S ( Left hand side )}\end{align},$$
where the second equality comes from applying $\cos\theta = \cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}$ and $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$.
| {
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Norm of a linear transformation Let $T:\mathbb R^2\to \mathbb R^2$ be given by the matrix $\begin{pmatrix}a&b\\ c& d\end{pmatrix}$. Let $u:=a^2+b^2+c^2+d^2+2(ad-bc)$ and $v:=a^2+b^2+c^2+d^2-2(ad-bc)$.
I need to show that $\mid\mid T\mid\mid=\frac 12 (\sqrt u+\sqrt v)$.
I tried using the definition $\mid\mid T\mid\mid =\sup\{\mid \mid Tx\mid\mid\ : x\in \mathbb R^2, \mid\mid x\mid\mid=1 \}$, and trying to express each $x$ as $(\cos \theta,\sin \theta)$ for $\theta\in [0,2\pi)$. This just led to a big mess, though, eventually reaching that
$$\mid\mid T(\cos\theta,\sin \theta)\mid\mid^2 < a^2+b^2+c^2+d^2+2(ab+cd)$$
This is kind of similar to $u$ and $v$ but not quite good enough, so I'm not sure now if this approach with $\theta$ and the trig functions is the right one.
I also don't know how to get started for the other inequality.
| You have to find largest eigenvalue of $T^tT$ and show that it is equal to $\frac{u+v+2\sqrt{uv}}{4}.$ You get into the following equations $$(x-a^2-c^2)(x-b^2-d^2)-(ab+cd)^2=0\\\implies x^2-(a^2+b^2+c^2+d^2)x+(ad-bc)^2=0\\\implies x_\max=\frac{a^2+b^2+c^2+d^2+\sqrt{(a^2+b^2+c^2+d^2)^2-4(ad-bc)^2}}{2}\\=\frac{u+v+2\sqrt{uv}}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Matrix to power $2012$ How to calculate $A^{2012}$?
$A = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$
How can one calculate this? It must be tricky or something, cause there was only 1 point for solving this.
| $$A = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$
$$A^2=\left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]\cdot \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$
$$A^2 = \left[\begin{array}{ccc}{9-2-4}&{-3+2}&{-6+2+2}\\{6-4}&{-2+2}&{-4+2}\\{6-2}&{-2+1}&{-4+2+1}\end{array}\right]$$
$$A^2 = \left[\begin{array}{ccc}3&-1&-2\\2&0&-2\\2&-1&-1\end{array}\right]$$
$$A^2=A$$
$$A=A^2=A^3=\cdots =A^n,n\in N$$
so $$A^{2012}=A$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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A problem of AM>GM
If $n$ is a positive integer with $n> 1$, prove that
$$2^{n(n+1)}>(n+1)^{(n+1)}\cdot\left(\frac{n}{1}\right)^n\cdot\left(\frac{n-1}{2}\right)^{(n-1)}\cdots\left(\frac{2}{n-1}\right)^2\cdot\frac{1}{n}$$
For solving it I have considered the numbers $\displaystyle{n+1,\,n,\,\frac{n-1}{2},\,\ldots,\frac{2}{n-1},\,\frac{1}{n}}$ with associated weights $n+1,n,n-1,\ldots,2,1$ and applied AM>GM
$$\left[{(n+1)^2+n^2+\frac{(n+1)^2}{2}+\ldots+\frac{2^2}{(n-1)}+\frac{1}{n}}\over{\frac{(n+1)\cdot(n+2)}{2}}\right]^{\tfrac{(n+1)\cdot(n+2)}{2}}$$
but cannot solve it please help
| Hint:
$$\left(\frac{n}{1}\right)^n\cdot\left(\frac{n-1}{2}\right)^{(n-1)}\cdots\left(\frac{2}{n-1}\right)^2\cdot\frac{1}{n} = \binom{n}{0} \binom{n}{1} \cdots \binom{n}{n}$$
where $\binom{n}{k}$ is the usual binomial coefficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/431476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$ If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
| From the content of this thread follows,
$x^2 + 3x +1 = 0 \Leftrightarrow x^2 +2x+1=-x \Leftrightarrow \boxed{(x+1)^2 =-x} (*)$
$\frac{a^2}{(b+1)^2}+\frac{b^2}{(a+1)^2} \Leftrightarrow
\left(\frac{b^2}{b^2}\right)\frac{a^2}{(b+1)^2}+\left(\frac{a^2}{a^2}\right)\frac{b^2}{(a+1)^2} \\
\hspace{3.05cm}\Leftrightarrow \underset{\scriptsize -1 \, by \, (*)}{\boxed{\frac{b}{(b+1)^2}}}\frac{a^2
b}{b^2} + \underset{\scriptsize -1}{\boxed{\frac{a}{(a+1)^2}}}
\frac{b^2 a}{a^2} \\
\hspace{3.05cm}\Leftrightarrow -\frac{a^2}{b}-\frac{b^2}{a}\\
\hspace{3.05cm}\Leftrightarrow \frac{- a^3 - b^3}{ab} \\
\hspace{3.05cm}\Leftrightarrow \boxed{- a^3 - b^3}$
$-a^3-b^3 = -(a+b)(a^2+b^2-ab+2ab-2ab) \\
\hspace{1.85cm} = 3((a+b)^2-3ab)\\
\hspace{1.85cm} = 3(9-3) \\
\hspace{1.85cm} = \boxed{18}
$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove that $(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}>2$ for $x > 0$
Let $x>0$. Show that
$$(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}>2.$$
Do you have any nice method?
My idea $F(x)=(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}$
then we hvae $F'(x)=\cdots$ But it's ugly.
can you have nice methods? Thank you
by this I have see this same problem
let $0<x<1$ we have $$x+\dfrac{1}{x^x}<2$$
this problem have nice methods:
becasue we have
$$\dfrac{1}{x^x}=\left(\dfrac{1}{x}\right)^x\cdot 1^{1-x}<x\cdot\dfrac{1}{x}+(1-x)\cdot 1=2-x$$
| First to prove $f(x)=x^{\frac{1}{x}}$ have a max $f(e)$.
Edit:2nd version
when $x<e-1, f(x+1)>f(x)$
when $x>e,f(x)>f(x+1)$
for $f(x+1)>f(x)$, it is trivial .
when $e-1\le x\le e$, :
$f(x) $ is mono increasing, so $f(x)^{-1}_{min}=f(e)$, $f(x+1)$ is mono decreasing, so $ f(x+1)_{min}=f(e+1)$, thus,$f(x+1)+f(x)^{-1} > (e+1)^{\frac{1}{e+1}}$$+\dfrac{1}{e^{\frac{1}{e}}}=2.11 >2$
for $x>e$, let $g(x)=(x+1)^{\frac{1}{x+1}}+x^{-\frac{1}{x}}$, I will prove g(x) is mono decreasing,so $g(x)_{min}=g(+\infty)=2$
$g'(x)=x^{-2 - \frac{1}{x}} (-1 + Ln(x)) - (1 + x)^{-2 + \frac{1}{1 + x}} (-1 + Ln(1 + x))$, now to prove:
$x^{-2 - \frac{1}{x}}<(1 + x)^{-2 + \frac{1}{1 + x}} \iff \dfrac{x^2x^{\frac{1}{x}}(x+1)^{\frac{1}{x+1}}}{(x+1)^2}>1 \iff \dfrac{x^2(x+1)^{\frac{1}{x+1}}(x+1)^{\frac{1}{x+1}}}{(x+1)^2}>1 \iff \left(\dfrac{x*(x+1)^{\frac{1}{x+1}}}{x+1} \right)^2>1 \iff\dfrac{x*(x+1)^{\frac{1}{x+1}}}{x+1}>1 \iff \dfrac{1}{x+1}Ln(x+1)>Ln(x+1)-Ln(x) \iff (x+1)Ln(x)>xLn(x+1) \iff x^{\frac{1}{x}} > (x+1)^{\frac{1}{x+1}} $
so $g'(x)<0$ when $x>e$. Done
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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How to integrate these integrals $$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \cos x}$$
$$\int^{\frac {\pi}2}_0 \frac {dx}{1+ \sin x}$$
It seems that substitutions make things worse:
$$\int \frac {dx}{1+ \cos x} ; t = 1 + \cos x; dt = -\sin x dx ; \sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (t-1)^2} $$
$$ \Rightarrow
\int \frac {-\sqrt{1 - (t-1)^2}}{t} = \int \frac {-\sqrt{t^2 + 2t}}{t} = \int \frac {-\sqrt t \cdot \sqrt t \cdot \sqrt{t +
2}}{\sqrt t \cdot t} $$
$$= \int \frac {- \sqrt{t + 2}}{\sqrt t } = \int - \sqrt{1 + \frac 2t} = ? $$
What next? I don’t know. Also, I’ve tried another “substitution”, namely $1 + \cos x = 2 \cos^2 \frac x2) $
$$ \int \frac {dx}{1+ \cos x} = \int \frac {dx}{2 \cos^2 \frac x2} = \int \frac 12 \cdot \sec^2 \frac x2 = ? $$
And failed again. Help me, please.
| Use this simple trigonometry manipulation:
$$
\frac{1}{1+\cos x}=\frac{1}{1+\cos x}\cdot\frac{1-\cos x}{1-\cos x}=\frac{1-\cos x}{\sin^2 x}.
$$
Therefore
$$
\begin{align}
\int\frac{1}{1+\cos x}\;dx&=\int\frac{1-\cos x}{\sin^2 x}\;dx\\
&=\int\frac{1}{\sin^2 x}\;dx-\int\frac{\cos x}{\sin^2 x}\;dx\\
&=-\int\;d(\cot x)-\int\frac{1}{\sin^2 x}\;d(\sin x)\\
&=-\cot x+\frac{1}{\sin x}+C\\
&=\frac{1-\cos x}{\sin x}+C\\
&=\frac{\sin^2\frac{1}{2}x+\cos^2\frac{1}{2}x-\cos^2\frac{1}{2}x+\sin^2\frac{1}{2}x}{2\sin\frac{1}{2}x\cos\frac{1}{2}x}+C\\
&=\tan\frac{1}{2}x+C.
\end{align}
$$
Similarly with
$$
\frac{1}{1+\sin x}=\frac{1}{1+\sin x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin x}{\cos^2 x}.
$$
$$
\text{# }\mathbb{Q.E.D.}\text{ #}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Expressing polynomial roots expression in terms of coefficients This is my first question on MSE. Apologies in advance for any textual or LaTeX errors.
I'm stuck with this problem:
Given $x^3 - bx^2 + cx - d = 0$ has roots $\alpha$, $\beta$, $\gamma$, find an expression in terms of $b$, $c$ and $d$ for:
(i) $\alpha^2 + \beta^2 + \gamma^2$
(ii) $\alpha^3 + \beta^3 + \gamma^3$
(iii) $(1 + \alpha^3)(1 + \beta^3)(1 + \gamma^3)$
I had no trouble with (i) or (ii), but got stuck on (iii) as follows: Expanding,
$$\begin{align*}
(1 + \alpha^3)(1 + \beta^3)(1 + \gamma^3) & = (1 + \alpha^3 + \beta^3 + \alpha^3\beta^3)(1 + \gamma^3)\\
& = 1 + (\alpha^3 + \beta^3 + \gamma^3) + (\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3) + \alpha^3\beta^3\gamma^3
\end{align*}$$
The first, second and fourth RHS terms are no problem, leaving us with:
$$\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3 = \left(\frac{1}{\gamma^3} + \frac{1}{\alpha^3} + \frac{1}{\beta^3} \right)\alpha^3\beta^3\gamma^3$$
So now we are left with the term in brackets. My next thought was to transform the original polynomial to one with roots $\frac{1}{\alpha}$, $\frac{1}{\beta}$ and $\frac{1}{\gamma}$ and then use the answer to (ii) above. Will this work? Or is there a better approach?
| Let $y=1+\alpha^3$
Again, $\alpha^3=b\alpha^2-c\alpha+d$
$$\implies y-1-d=b\alpha^2-c\alpha$$
Cubing we get $$ (y-1-d)^3=b^3\alpha^6-c^3\alpha^3-3bc\alpha^2\cdot \alpha(b\alpha^2-c\alpha)$$
$$(y-1-d)^3=b^3(y-1)^2-c^3(y-1)-3bc(y-1)(y-1-d)$$ as $\alpha^3=y-1,b\alpha^2-c\alpha=y-1-d$
Arrange as $y^3+By^2+Cy+D=0$ whose roots are $1 + \alpha^3,1 + \beta^3,1 + \gamma^3$
Using Vieta's formulas,
$$(1 + \alpha^3)(1 + \beta^3)(1 + \gamma^3)=-D$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/434362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Confused about harmonic series and Euler product So Euler argued that
$$1 + \frac{1}{2} + \frac{1}{3} + \frac {1}{4} + \cdots = \frac {2 \cdot 3 \cdot 5 \cdot 7 \cdots} {1 \cdot 2 \cdot 4 \cdot 6 \cdots} $$
which you can rearrange to
$$ \left( \frac {1 \cdot 2 \cdot 4 \cdot 6 \cdots} {2 \cdot 3 \cdot 5 \cdot 7 \cdots} \right) \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac {1}{4} + \cdots \right) = 1$$
which in turn you might write as
$$\prod_{n=1}^\infty \frac{p_n-1}{p_n} \times \sum_{n=1}^\infty \frac{1}{n} = 1.$$
I'm confused about how this works rigorously. How do the primes 'align themselves' with the naturals? For example, is the correct statement something like
$$\prod_{n=1}^z \frac{p_n-1}{p_n} \times \sum_{n=1}^z \frac{1}{n} \to 1 \text{ as } z \to \infty$$
where you're taking the first $z$ primes and the first $z$ naturals, or is it something like
$$\prod_{p \le z}^z \frac{p-1}{p} \times \sum_{n=1}^z \frac{1}{n} \to 1 \text{ as } z \to \infty$$
where you're taking all the primes less than or equal to $z$ and the first $z$ naturals, or is it something else entirely? I tried to get a clue with numerical programming but didn't get very far.
| $$\prod_1^r{p_n-1\over p_n}=\prod_1^r\left(1-{1\over p_n}\right)^{-1}=\prod_1^r\left(1+{1\over p_n}+{1\over p_n^2}+\cdots\right)=\sum{1\over m}$$ where the sum is over all $m$ divisible by no primes other than $p_1,\dots,p_n$. Formally, the limit as $r\to\infty$ gives the first displayed equation in the question. Rigorously, it can be shown that $$\prod_1^{\infty}\left(1-{1\over p_n^s}\right)^{-1}=\sum_1^{\infty}{1\over n^s}$$ for $s\gt1$ (or even for real part of $s$ exceeding $1$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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where this series converges Given the series $$\sum_{j=0}^{\infty}\frac{1}{6j^2-5j+1}$$
I am completely stuck and do not understand the answer from my book which is $\pi^2/36-1$. I need explanation and different approach how this result is gained. Thanks
| Rewrite your series as :
\begin{align}
\tag{1}\sum_{j=0}^{\infty}\frac{1}{6j^2-5j+1}&=\sum_{j=0}^{\infty}\frac{1}{(3j-1)(2j-1)}\\
&=\sum_{j=0}^{\infty}\frac2{2j-1}-\frac3{3j-1}\\
\tag{2}&=-2+3+\sum_{j=1}^{\infty}\frac1{j-\frac 12}-\frac1{j-\frac 13}\\
&=1-\psi\left(1-\frac 12\right)+\psi\left(1-\frac 13\right)\\
\tag{3}&=1+2\ln(2)-\frac{3\ln(3)}2+\frac{\pi}{2\sqrt{3}}\\
&\approx 1.645275610234835007\\
\end{align}
using the special values of the digamma function (or the Gauss digamma sum) and the resolution method exposed in the excellent Abramowitz and Stegun $(6.8)$.
Here a more 'elementary' derivation is possible if we observe that for any integer $n>1$ :
\begin{align}
\sum_{j=1}^{\infty}\frac1{j-\frac 1n}-\frac1{j}&=\sum_{j=1}^{\infty}\int_0^1x^{j-\frac 1n-1}-x^{j-1}\;dx\\
&=\int_0^1 \sum_{k=0}^{\infty}\left(x^{k-1/n}-x^{k}\right)\;dx\\
&=\int_0^1 \frac{x^{-1/n}-1}{1-x}\;dx\\
\text{setting}\ x:=y^n\ \text{ we get :}\\
&=n\int_0^1 \frac{y^{-1}-1}{1-y^n}y^{n-1}\;dy\\
&=n\int_0^1 \frac{y^{n-2}-y^{n-1}}{1-y^n}\;dy\\
\text{that may be solved using}&\text{ partial fractions.}
\end{align}
From $(2)$ we need :
\begin{align}
\sum_{j=1}^{\infty}\frac1{j-1/2}-\frac1{j-1/3}&=2\int_0^1 \frac{1-y}{1-y^2}\;dy-3\int_0^1 \frac{y-y^2}{1-y^3}\;dy\\
&=2\ln(2)-3\int_0^1 \frac{y}{1+y+y^2}\;dy\\
&=2\ln(2)-\frac 32\left(\int_0^1 \frac{1+2y}{1+y+y^2}\;dy-\int_0^1 \frac 1{(3/4)+(y+1/2)^2}\;dy\right)\\
&=2\ln(2)-\frac 32\left|\ln(1+y+y^2)-\frac 2{\sqrt{3}}\arctan\left(\frac{1+2y}{\sqrt{3}}\right)\right|_0^1\\
&=2\ln(2)-\frac {3\ln(3)}2+\sqrt{3}\left(\arctan\left(\sqrt{3}\right)-\arctan\left(\frac1{\sqrt{3}}\right)\right)\\
\end{align}
Adding $1$ from $(2)$ we get again the result $(3)$ :
$$\boxed{\displaystyle \sum_{j=0}^{\infty}\frac{1}{6j^2-5j+1}=1+2\ln(2)-\frac{3\ln(3)}2+\frac{\pi}{2\sqrt{3}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/435299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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How find this value $\frac{a^2+b^2-c^2}{2ab}+\frac{a^2+c^2-b^2}{2ac}+\frac{b^2+c^2-a^2}{2bc}$ let $a,b,c$ such that
$$\left(\dfrac{a^2+b^2-c^2}{2ab}\right)^2+\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2+\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2=3,$$
find the value
$$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}$$
is true?
Yes, I tink this problem can prove
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$
so $$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}=1or -3$$
How many nice methods prove $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$ ?
and I see this easy problem
http://zhidao.baidu.com/question/260913315.html
| I think the answer must be $1$ if we impose a triangle inequality. The individual terms are all cosines of angles of a triangle. Thus, the sum of those angles must be $\pi$. But the sum of the squares of their cosines is $3$; therefore each cosine must be $\pm 1$. But the sum of the angles is, again, $\pi$, so that two of the angles must be zero and the third $\pi$. Thus, the sum of the cosines, which is sought, must be $1+1-1=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/435376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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If $\sin a+\sin b=2$, then show that $\sin(a+b)=0$ If $\sin a+\sin b=2$, then show that $\sin(a+b)=0$.
I have tried to solve this problem in the following way :
\begin{align}&\sin a + \sin b=2 \\
\Rightarrow &2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)=2\\
\Rightarrow &\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)=1
\end{align}
What will be the next ?
| HINT:
As $-1\le \sin x\le 1$ for real $x$
$\implies -2\le \sin a+\sin b\le 2$
The equality occurs if $\sin a=\sin b=1$
$a=2n\pi+\frac\pi2, b=2m\pi+\frac\pi2,$ for some integer $m,n$
So, $a+b=2\pi(m+n)+\pi=\pi(2m+2n+1)$ and we know $\sin r\pi=0$ for integer $r$
Alternatively, if $\sin x=1,\cos x=\pm\sqrt{1-1^2}=0\implies $ here $\cos a=\cos b=0$
and we know $\sin(a+b)=\sin a\cos b+\cos a \sin b$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given that $a^2(a+k)=b^2(b+k)=c^2(c+k)$, find the value of $1/a+1/b+1/c$
Given $$a^2(a+k)=b^2(b+k)=c^2(c+k)$$
find the value of $1/a+1/b+1/c$.
I tried to derive a relation from the equality but it did not help my cause.
| HINT:
Let $$a^2(a+k)=b^2(b+k)=c^2(c+k)=d$$
So $a,b,c$ are the roots of $x^3+x^2k-d=0$
Using Vieta's Formula, $a+b+c=-k, ab+bc+ca=0,abc=d$
$$\text{Now,}\frac1a+\frac1b+\frac1c=\frac{ab+bc+ca}{abc}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factor Equations Please check my answer in factoring this equations:
Question 1. Factor $(x+1)^4+(x+3)^4-272$.
Solution: $$\begin{eqnarray}&=&(x+1)^4+(x+3)^4-272\\&=&(x+1)^4+(x+3)^4-272+16-16\\
&=&(x+1)^4+(x+3)^4-256-16\\
&=&\left[(x+1)^4-16\right]+\left[(x+3)^4-256\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)^2-16\right]\\
&=&\left[(x+1)^2+4\right]\left[(x+1)^2-4\right]+\left[(x+3)^2+16\right]\left[(x+3)-4\right]\left[(x+3)+4\right]\end{eqnarray}.$$
Question 2. Factor $x^4+(x+y)^4+y^4$
Solution: $$\begin{eqnarray}&=&(x^4+y^4)+(x+y)^4\\
&=&(x^4+y^4)+(x+y)^4+2x^2y^2-2x^2y^2\\
&=&(x^4+2x^2y^2+y^4)+(x+y)^4-2x^2y^2\\
&=&(x^2+y^2)^2+(x+y)^4-2x^2y^2
\end{eqnarray}$$
I am stuck in question number 2, I dont know what is next after that line.
| Let $w = x + 2$
\begin{align}
(x+1)^4+(x+3)^4-272
&= (w-1)^4+(w+1)^4-272\\
&= 2w^4 + 12w^2 - 270\\
&= 2(w^4 + 6w - 135)\\
&= 2(w^4 +6w^2 + 9 - 144)\\
&= 2(w^2 + 3)^2 - 12^2)\\
&= 2(w^2 + 15)(w^2 - 9)\\
&= 2(w^2 + 15)(w - 3)(w + 3)\\
&=2(x^2 + 4x + 19)(x - 1)(x + 5)\\
\end{align}
\begin{align}
(x^4+y^4)+(x+y)^4
&= 2x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 2y^4\\
&= 2(x^4 + 2x^3y + 3x^2y^2 + 2xy^3 + y^4)\\
&= 2((x^4 + x^3y + x^2y^2) +
(x^3y + x^2y^2 + xy^3) +
(x^2y^2 + xy^3 + y^4))\\
&= 2(x^2(x^2 + xy + y^2) +
xy(x^2 + xy + y^2) +
y^2(x^2 + xy + y^2))\\
&= 2(x^2 + xy + y^2)^2
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating function of the number of integer partitions of $n$ into all distinct parts Let $p_d (n)$ denote the number of integer partitions of $n$ into all distinct parts. I am given the following equation, but I can't figure out why it holds:
$$\sum_{n \ge 0} p_d(n)x^n = \prod_{i \ge 1}(1+x^i).$$
P.S. This example is from Miklos Bona's A Walk Through Combinatorics (2nd edition).
| Consider first the finite product
$$\prod_{i = 1}^n (1+x^i).$$
If you expand it, you get
$$\prod_{i = 1}^n (1+x^i) = \sum_{\varepsilon \in \{0,\,1\}^n} x^{\sum_{i=1}^n \varepsilon_i \cdot i}.$$
The term $x^k$ occurs as many times in that sum, as there are ways to write $k$ as a sum of distinct positive integers $\leqslant n$.
For $k \leqslant n$, the coefficient doesn't change anymore when you multiply further factors $(1+x^i),\, i > n$.
Perhaps expanding the product in full for $n = 4$ helps a little:
$$\begin{align}
(1+&x)(1+x^2)(1+x^3)(1+x^4)\\
&= 1\cdot 1\cdot 1 \cdot 1 + 1\cdot 1\cdot 1\cdot x^4 + 1\cdot 1 \cdot x^3 \cdot 1 + 1\cdot 1 \cdot x^3 \cdot x^4\\
&\;+ 1\cdot x^2 \cdot 1\cdot 1 + 1\cdot x^2\cdot 1\cdot x^4 + 1\cdot x^2 \cdot x^3 \cdot 1 + 1\cdot x^2 \cdot x^3 \cdot x^4\\
&\;+ x^1\cdot 1\cdot 1 \cdot 1 + x^1\cdot 1\cdot 1\cdot x^4 + x^1\cdot 1 \cdot x^3 \cdot 1 + x^1\cdot 1 \cdot x^3 \cdot x^4\\
&\;+ x^1\cdot x^2 \cdot 1\cdot 1 + x^1\cdot x^2\cdot 1\cdot x^4 + x^1\cdot x^2 \cdot x^3 \cdot 1 + x^1\cdot x^2 \cdot x^3 \cdot x^4\\
&= 1 + x^4 + x^3 + x^7\\
&\; + x^2 + x^6 + x^5 + x^9\\
&\;+ x^1 + x^5 + x^4 + x^8\\
&\;+ x^3 + x^7 + x^6 + x^{10}\\
&= 1\cdot x^0 + 1\cdot x^1 + 1\cdot x^2 + 2\cdot x^3 + 2\cdot x^4 + 2\cdot x^5 + 2\cdot x^6 + 2\cdot x^7 + 1\cdot x^8 + 1\cdot x^9 + 1\cdot x^{10}.
\end{align}$$
From each factor, we can choose either the $1 (= x^0)$ or the $x^i$ term. That gives $2^n$ products in total. Then we group these products by the resulting exponent of $x$.
Each of the $2^n$ products corresponds to one partition of an exponent into a sum of distinct positive integers, and each partition of $k$ into a sum of distinct positive integers $\leqslant n$ corresponds to one choice of the $x^0$ or the $x^i$ term of each factor.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Continuity and Finding Values Find value of a,b,c such that F is continuous on the real number system:
$$
f(x) = \left\{
\begin{array}{lr}
-1 & : x\le-1\\
ax^2+bx+c & : |x|<1,x\ not\ equal\ to\ 0\\
0 &:x=0\\
1 &:x\ge1
\end{array}
\right.
$$
We went over the following solution in class:
Show $f(x)$ is continuous at -1:
For all $\epsilon > 0, \exists\ \delta$ such that for all $x, |x+1|$ < $\delta$ and $|f(x)-f(-1)|=$
if $x\le-1: f(x)=f(-1)=-1$ and $|f(x)-f(-1)|=0<\epsilon$
if $x>-1: |f(x)-f(-1)|=|ax^2+bx+c-(-1)|=|a[x^2-(-1)^2]+b[x-(-1)]+c-(-1)+a(-1)^2+b(-1)|<\epsilon.$
so $a-b+c+1=0$
Show $f(x)$ is continuous at 0:
For all $\epsilon > 0, \exists\ \delta\ $ such that for all $x, |x-0|<\delta\ $ and $|f(x)-f(0)|=|f(x)|$=$|ax^2+bx+c| = x$
$c=0$$a+b+c-1=0$
$a-b=-1,a+b=1$, thus $a = 0$ and $b = 1$I am having difficulty processing through this solution. I am first confused with why for showing the continuity at -1, it was broken down into two cases ($x\le-1$ and $x>-1$).
Then, how did we simplify into $a-b+c+1=0$ in that second case?Finally, in showing continuity at $0$, how did we know that $c = 0$?
| The reason they break it down into two cases at $x=-1$ is that the function is fundamentally different on $\{x<-1\}$ than it is on $\{x>-1\}$.
The way they simplify into $a-b+c+1=0$ is by simply plugging $x=-1$ into $ax^2 + bx + c - (-1)$.
The way we knew that $c=0$ is that as $x$ gets small, $ax^2 + bx + c$ will get arbitrarily close to $c$. If $f$ is continuous, it also must get arbitrarily close to $f(0) = 0$. Therefore, $c=0$.
| {
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Exponentials of a matrix I just was working with matrix exponentials for solving problems in control theory. Suppose $A $ is a square matrix. How can we interpret $A_1 = e^ {\textstyle-A\log(t) }$, where $\log$ is natural logarithm? Is there a formula for extending the scalar case of $e^ {\textstyle-a\log(t) }$ which gives $\dfrac{1}{t^a}$? How do we evaluate $A_1$?
Here is how I proceeded along the lines of scalar case:
$$A_1 = \exp(-A\log t)) = \exp(-\log {t^A}) \overset{\textstyle?}{=}\;( t^A)^{-1}$$
Is the final equality correct?
| With the matrix exponential, more care is necessary. One approach is to multiply each item in the matrix $A$ by $\ln t$ and then find the diagonal matrix and take advantage of its structure.
Try working these two examples and see if you get them.
Example 1: Matrix is in diagonal form
$A = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$.
For a matrix of this type, we can take advantage of the fact that it is a diagonal matrix, and right away write out:
$\displaystyle e^{-A \ln t} = e^{\begin{bmatrix}-1 & 0\\0 & -1 \end{bmatrix}*\ln t} = \begin{bmatrix} 1/t & 0 \\ 0 & 1/t \end{bmatrix}$
Example 2: Diagonalizable Matrix
$A = \begin{bmatrix}-1 & -3\\-1 & -3\end{bmatrix}$, so
$\displaystyle e^{-A \ln t} = e^{\begin{bmatrix}1 & 3\\1 & 3 \end{bmatrix}*\ln t} = \begin{bmatrix}-3 & 1 \\ 1 & 1 \end{bmatrix} \cdot e^{\begin{bmatrix}0 & 0 \\ 0 & 4 \ln t \end{bmatrix}}\cdot \begin{bmatrix} -1/4 & 1/4 \\ 1/4 & 3/4 \end{bmatrix} = \begin{bmatrix}-3 & 1 \\ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 \\ 0 & t^4 \end{bmatrix}\cdot \begin{bmatrix} -1/4 & 1/4 \\ 1/4 & 3/4 \end{bmatrix} = \begin{bmatrix} \dfrac{t^4}{4}+ \dfrac{3}{4} & \dfrac{3 t^4}{4}-\dfrac{3}{4} \\ \dfrac{t^4}{4}-\dfrac{1}{4} & \dfrac{3 t^4}{4} + \dfrac{1}{4}\end{bmatrix}$
Example 3: Non-diagonalizable Matrix
For this matrix, we will write it using the Jordan Normal Form as $A = P \cdot J \cdot P^{-1}$ and take advantage of the Jordan block.
$A = \begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}$, so
$\displaystyle e^{-A \ln t} = e^{\begin{bmatrix}-1 & -2\\0 &-1 \end{bmatrix}*\ln t} = \begin{bmatrix}1 & 0 \\ 0 & -\dfrac{1}{2} \end{bmatrix} \cdot e^{\begin{bmatrix}-\ln t & \ln t \\ 0 & - \ln t \end{bmatrix}}\cdot \begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & -\dfrac{1}{2} \end{bmatrix} \cdot \begin{bmatrix}\dfrac{1}{t} & \dfrac{\ln t}{t} \\ 0 & \dfrac{1}{t} \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & -2 \end{bmatrix} = \begin{bmatrix} \dfrac{1}{t} & -\dfrac{2 \ln t}{t} \\ 0 & \dfrac{1}{t} \end{bmatrix} $
Example 4
$A = \begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 2 \\ 2 & 1 & 0\end{bmatrix}$
Give it a go. The matrix is diagonalizable and you should get:
$$e^{-A \ln t} = \frac{1}{15 t^3}\begin{bmatrix}4 t^5+5 t^2+6 & 2 t^5-5 t^2+3 & 6 (1-t^5) \\ ~~2 (2 t^5-5 t^2+3) & 2 t^5+10 t^2+3 & 6 (1-t^5) \\ ~~6 (1-t^5) & 3 (1-t^5) & 3 (3 t^5+2)\end{bmatrix}$$
| {
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"source": "stackexchange",
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Calculus Implicit Differentiation I'm learning implicit differentiation and I've hit a snag with the following equation.
$$
f(x, y) = x + xy + y = 2
$$
$$
Dx(x) + Dx(xy) + Dx(y) = Dx(2)
$$
$$
1 + xy' + y + y' = 0
$$
$$
xy' + y' = -1 - y
$$
$$
y'(x + 1) = 1 + y
$$
$$
y' = \dfrac{(1 + y)}{(x + 1)}
$$
$$
y'' = \dfrac{(x + 1)y' - (1 + y)}{(x + 1)^2}
$$
$$
y'' = \dfrac{(x + 1)\dfrac{(1 + y)}{(x + 1)} - (1 + y)}{(x + 1)^2}
$$
Ok now what according to this y'' = 0 which is wrong.
| Your mistake seems to originate when moving from this:
$$
xy' + y' = -1 - y
$$
...to this, where you "lost the sign":
$$
y'(x + 1) = 1 + y$$
We need $$y'(x + 1) = -(1 + y)$$
Let's back up: $$\begin{align} 1 + xy' + y + y' & = 0 \\ \\ xy' + y' & = -1 - y\\ \\ & = -(1 + y)\end{align}$$
Then we factor out $y'$ on the left-hand side, giving us:
$$y'(x + 1) = -(1 + y)$$
Fixing for that, then, we get:
$$\begin{align}
y' &= \dfrac{-(1 + y)}{(x + 1)} \\ \\
y'' & = \frac{(x + 1)(-y') - [-(1 + y)]}{(x + 1)^2} \\ \\
& = \frac{-(x + 1)\dfrac{-(1 + y)}{(x + 1)} + (1 + y)}{(x + 1)^2} \\ \\
& = \frac{(1 + y) + (1+y)}{(x+1)^2}\\ \\
& = \frac{2(y+1)}{(x+1)^2}
\end{align}$$
| {
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Big $\Omega$ question! Prove $(n-1)(n-2)(n-3)$ is $\Omega(n^3)$ Problem
Prove $(n-1)(n-2)(n-3)$ is $\Omega(n^3)$.
Attempt @ Solution
*
*$f(n) = n^3(1-6/n+11/n^2-6/n^3)$
*$g(n) = n^3$
*Show that there exists a $C > 0$ and $n_0$ such that $f(n) \ge Cg(n)$ for all $n > n_0$.
*I tried plugging in different numbers for $n$ that would make $f(n) > n^3$. I found that setting $n = 7$ makes sure that $f(n)$ is greater than $g(n)$. So, is that my answer? Evaluating the expression with $n=7$ to solve for $C$, and setting $n_0$ as $7$? Is that a sufficient proof? Also, Does my constant have to be a Natural number, or can it simply be a Rational number?
| You will not be able to show that $f(n)\gt g(n)$, because it is in fact smaller.
What I would suggest is that if $n\ge 6$, then $n-3\ge \frac{n}{2}$, as are $n-2$ and $n-1$. Thus for $n\ge 6$, we have $f(n)\ge \frac{1}{8}g(n)$. So we can take $C=\frac{1}{8}$.
And $C$ certainly does not have to be an integer. In our particular problem, we cannot even find a positive integer $C$ with the desired property.
Remark: Dividing by $n^3$ like you did was a good idea, expanding was not. When we divide by $n^3$ we get
$$\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3}{n}\right).$$
Now you can take your favourite $n\ge 4$. Let's pick $6$. Then if $n\ge 6$, the above expression is $\ge \frac{5}{6}\cdot \frac{4}{6}\cdot \frac{3}{6}$. We can pick this for our $C$.
| {
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Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The inductive step can be proved as follows.
$2^k < \binom{2k}{k} \implies 2^{k+1} < 2\binom{2k}{k} = \frac{2(2k)!}{k!k!} = \frac{2(2k)!(k + 1)}{k!k!(k + 1)} = \frac{2(2k)!(k+2)}{(k+1)!k!}<\frac{(2k)!(2k+2)(2k+1)}{(k+1)!k!(k+1)} = \binom{2(k+1)}{k+1}$
Second part: $2^{2n} > \binom{2n}{n}$. Again, the base is trivial. We can assume that for some $k$ our statement is satisfied and prove that inductive step as follows:
$2^{2k} > \binom{2k}{k} \implies 2^{2k + 2} > 2^2\binom{2k}{k} = \frac{2\cdot2(2k)!}{k!k!} = \frac{2\cdot2(2k)!(k+1)(k+1)}{k!k!(k+1)(k+1)} = \frac{(2k)!(2k+2)(2k+2)}{(k+1)!(k+1)!} > \frac{(2k)!(2k+1)(2k+2)}{(k+1)!(k+1)!} = \binom{2(k+1)}{k+1}$
Is there a non-inductive derivation for the inequality?
| Let $A$ be the set of all binary words containing $n$ bits, $B$ be the set of all binary words containing $2n$ bits, exactly $n$ of which are $1's$, $C$ be the set of all binary words containing $2n$ bits.
Then as $B$ is a proper subset of $C$ we have $\binom{2n}{n} <2^{2n}$.
We define a function $E : B \to A$ by $E(x)=$erase the last $n$ digits of $x$. This function is onto but not one-to-one, thus $B$ has more elements than $A$.
This proves $\binom{2n}{n} >2^{n}$.
| {
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Interpreting the ; in a series This question is linked to this question.
So, suppose I set $n=5$. Given the following formula:
$$\frac{1}{n}, \dots , \frac{n-1}{n} $$
Am I suppose to get:
$$
\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \hspace{8.2cm}(1)
$$
Or
$$
\frac{1}{5}, \frac{1}{4}, \frac{1}{3}, \frac{2}{5}, \frac{1}{2},
\frac{3}{5}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5} \hspace{5cm} (2)
$$
?
In other words, what purpose to the ";" in:
$$ \frac{1}{2}; \frac{1}{3}, \frac{2}{3}; \frac{1}{4}, \frac{3}{4}; \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}; \dots ; \frac{1}{n}, \dots , \frac{n-1}{n}.$$
serve?
Also, the formula does not mention anything about skipping numbers that are not in lowest common terms. Is skipping this assumed given the definition of $f$? For example, in (2), there is no $\frac{2}{4}$ because it is equal to $\frac{1}{2}$ which is already listed earlier.
Thank you in advance for any help provided.
| The commas separate values for a single given value of $n$. The semi-colons separate for different values of $n$:
$$\underbrace{\frac12}_{n=1};\underbrace{\frac{1}{3}, \frac{2}{3}}_{n=3};\underbrace{\frac14,\frac{2}{4},\frac{3}{4}}_{n=4};\ldots$$
The intention was that this form a single set of unique rational numbers; the $\frac{2}{4}$ can be omitted. The semi-colons distinguish by values of $n$ only for ease of visualization.
| {
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If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Problem: If $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $. Find $f(x)$
Solution: $\int f(x) \sin{x} \cos{x}\,\mathrm dx = \frac {1}{2(b^2 - a^2)} \log f(x) +c $
Differenting both sides,we get
$ f(x) \sin{x} \cos{x} = \frac {f'(x)}{2(b^2 - a^2)f(x)} $
Am I doing right ?
| Yes and to complete: we have
$$\int f'(x)(f(x))^{-2}\;dx=C\int\sin(2x)\;dx$$
where $C=b^2-a^2$ so
$$-\frac{1}{f(x)}=-\frac{C}{2}\cos(2x)+C'$$
and you can take $f(x)$ from it.
| {
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If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$.
If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$. How do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
I know how to solve a problem like
"If $\cos\alpha = \frac{\sqrt{3}}{2}$ , find $\sin2\alpha$"
by using the 'double angle' formula: $\sin2\alpha = 2\sin\alpha\cos\alpha$ like this:
Start by computing $\sin\alpha$
$$\sin^2\alpha = 1 -\cos^2\alpha = 1-(\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$$
so
$$\sin\alpha = \pm\frac{1}{2}$$
then it's just a simple matter of plugging $\sin\alpha = \pm\frac{1}{2}$ and $\cos\alpha=\frac{\sqrt{3}}{2}$ into $$\sin2\alpha = 2\sin\alpha\cos\alpha$$ to get $$\sin2\alpha = \pm\frac{\sqrt{3}}{2}$$
Where I can not make progress with the question
"If $\sin\alpha + \cos\alpha = 0.2$, find the numerical value of $\sin2\alpha$".
Is how do I find a value for $\sin\alpha$ or $\cos\alpha$ so I can use a double angle formula?
What I have tried:
If $\sin\alpha+\cos\alpha = 0.2$
then $\sin\alpha=0.2-\cos\alpha$
and $\cos\alpha=0.2-\sin\alpha$. Should I start by by computing $\sin\alpha$ using
$\sin^2\alpha = 1 -\cos^2\alpha = 1-(0.2-\cos\alpha)^2$?
| If $\sin\alpha + \cos\alpha = 0.2$, then squaring both sides and simplifying will produce $2 \sin \alpha \cos \alpha = -0.96$.
It follows that $\sin^2 \alpha - 2 \sin(\alpha) \cos(\alpha) + \cos^2 \alpha = 1.96$, from which we conclude $\sin\alpha - \cos\alpha = \pm 1.4$
From
$\left\{
\begin{array}{l}
\sin\alpha + \cos\alpha = 0.2\\
\sin\alpha - \cos\alpha = 1.4
\end{array}
\right \}$
we find $(\sin \alpha, \cos \alpha) = (0.8, -0.6)$
From
$\left\{
\begin{array}{l}
\sin\alpha + \cos\alpha = 0.2\\
\sin\alpha - \cos\alpha = -1.4
\end{array}
\right \}$
we find $(\sin \alpha, \cos \alpha) = (-0.6, 0.8)$
Note that neither of these solutions is extraneous.
| {
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Need help calculating this determinant using induction This is the determinant of a matrix of ($n \times n$) that needs to be calculated:
\begin{pmatrix}
3 &2 &0 &0 &\cdots &0 &0 &0 &0\\
1 &3 &2 &0 &\cdots &0 &0 &0 &0\\
0 &1 &3 &2 &\cdots &0 &0 &0 &0\\
0 &0 &1 &3 &\cdots &0 &0 &0 &0\\
\vdots &\vdots &\vdots&\ddots &\ddots &\ddots&\vdots &\vdots&\vdots\\
0 &0 &0 &0 &\cdots &3 &2 &0 &0\\
0 &0 &0 &0 &\cdots &1 &3 &2 &0\\
0 &0 &0 &0 &\cdots &0 &1 &3 &2\\
0 &0 &0 &0 &\cdots &0 &0 &1 &3\\
\end{pmatrix}
The matrix follows the pattern as showed.
I have to calculate it using induction (we haven't learnt recursion so far).
Thanks
| You have a tridiagonal matrix.
It's determinant can be written as a recurrence relation:
$$\det A \stackrel{\textrm{def}}{=} f_n = a_nf_{n-1} + c_{n-1}b_{n-1}f_{n-2}.$$
Define $f_0 = 1, f_{-1} = 0$.
However, your $a_n, b_n, c_n$ values are constant, so
$$ \det A = 3f_{n-1}+ 2 f_{n-2}.$$
Thus, you have
$$\begin{align*}
n = 1: & f_1 = 3f_0 + 2f_{-1} = 3 \\
n = 2: & f_2 = 3f_1 + 2f_0 = 3\cdot 3 + 2 = 11 \\
n = 3: & f_3 = 3f_2 + 2f_1 = 33 + 6 = 39 \\
\vdots &
\end{align*}
$$
| {
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not both $2^n-1,2^n+1$ can be prime. I am trying to prove that not both integers $2^n-1,2^n+1$ can be prime for $n \not=2$. But I am not sure if my proof is correct or not:
Suppose both $2^n-1,2^n+1$ are prime, then $(2^n-1)(2^n+1)=4^n-1$ has precisely two prime factors. Now $4^n-1=(4-1)(4^{n-1}+4^{n-2}+ \cdots +1)=3A$. So one of $2^n-1, 2^n+1$ must be $3,$ which implies $n=1$ or $n=2$ (rejected by assumption). Putting $n=1$, we have $2^n-1=1,$ which is not a prime. Hence the result follows.
I also wanna know if there is alternative proof, thank you so much.
| If $n$ is even $=2m,2^n-1=2^{2m}-1=4^m-1$ is divisible by $4-1=3$ and $4^m-1>3$ if $m\ge1\iff n\ge2$
If $n$ is odd $=2m+1,2^n+1=2^{2m+1}+1$ is divisible by $2+1=3$ and $2^{2m+1}+1>3$ if $m\ge1\iff n\ge3$
alternatively,
$$(2^n-1)(2^n+1)=4^n-1$$ is divisible by $4-1=3$
So, at least one of $2^n-1,2^n+1$ is divisible by $3$
Now, $2^n+1>2^n-1>3$ for $n>2$
$\implies $ for $n>2,$ one of $2^n-1,2^n+1$ must be composite
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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If $\sin\theta+\sin\phi=a$ and $\cos\theta+ \cos\phi=b$, then find $\tan \frac{\theta-\phi}2$. I'm trying to solve this problem:
If $\sin\theta+\sin\phi=a$ and $\cos\theta+ \cos\phi=b$, then find $\tan \dfrac{\theta-\phi}2$.
So seeing $\dfrac{\theta-\phi}2$ in the argument of the tangent function, I first thought of converting the left-hand sides of the givens to products which gave me:
$$2\sin\frac{\theta+\phi}2\cos\frac{\theta-\phi}2=a\quad,\quad2\cos\frac{\theta+\phi}2\cos\frac{\theta-\phi}2=b$$
But then, on dividing the two equations (assuming $b\ne0$), I just get the value of $\tan\dfrac{\theta+\phi}2$.
I don't know how else to proceed.
Any help would be appreciated!
| Method $1:$
Squaring & adding what you have derived $$4\cos^2\frac{\theta-\phi}2=a^2+b^2$$
$$\implies \sec^2\frac{\theta-\phi}2=\frac4{a^2+b^2}$$
$$\implies \tan^2\frac{\theta-\phi}2=\frac4{a^2+b^2}-1=\frac{4-a^2-b^2}{a^2+b^2}$$
Method $2:$
$$\text{As }\quad2\cos\frac{\theta+\phi}2\cos\frac{\theta-\phi}2=b,$$
$$\implies \sec\frac{\theta-\phi}2=\frac2{b\sec\frac{\theta+\phi}2}$$
$$\implies \sec^2\frac{\theta-\phi}2=\frac4{b^2\sec^2\frac{\theta+\phi}2} =\frac4{b^2\left(1+\tan^2\frac{\theta-\phi}2\right)}=\frac4{b^2\left(1+\frac{a^2}{b^2}\right)}$$ as $\tan\frac{\theta+\phi}2=\frac ab$
$$\implies \sec^2\frac{\theta-\phi}2=\frac4{b^2+a^2}$$
$$\text{Now, }\tan^2\frac{\theta-\phi}2=\sec^2\frac{\theta-\phi}2-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the number of real solutions of the following? $ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $ What is the number of real solutions of the following?
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
My solution:
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
$$ \implies \sqrt{(\sqrt{x-1}-2)^2} + \sqrt{(\sqrt{x-1}-3)^2} = 1 $$
$$ \implies (\sqrt{x-1}-2) + (\sqrt{x-1}-3) = 1 $$
$$ \implies \sqrt{x-1} = 3$$
So, $ x = 10$ is the only solution.
But the answer key (and Wolfram alpha too) says there are infinite number of solutions to this equation. Where I am going wrong?
| Separate the square roots and square both sides. If you isolate the term $\sqrt{x+3-4\sqrt{x-1}}$, after simplifying you will obtain:
$$\sqrt{x+8-6\sqrt{x-1}}=3-\sqrt{x-1}$$
Squaring both sides again gives:
$$x+8-6\sqrt{x-1}=9-6\sqrt{x-1}+x-1$$
Which holds for all valid $x$. It follows that the original equation is true for any $x$ in the domain of the left hand side. This is $5\le x\le 10$.
| {
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"timestamp": "2023-03-29T00:00:00",
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show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
| Let $$f(y) = \int_{0}^{\infty} \frac{\sin^3{yx}}{x^3} \mathrm{d}x$$
Then,
$$f'(y) = 3\int_{0}^{\infty} \frac{\sin^2{yx}\cos{yx}}{x^2} \mathrm{d}x = \frac{3}{4}\int_{0}^{\infty} \frac{\cos{yx} - \cos{3yx}}{x^2} \mathrm{d}x$$
$$f''(y) = \frac{3}{4}\int_{0}^{\infty} \frac{-\sin{yx} + 3\sin{3yx}}{x} \mathrm{d}x$$
Therefore,
$$f''(y) = \frac{9}{4} \int_{0}^{\infty} \frac{\sin{3yx}}{x} \mathrm{d}x - \frac{3}{4} \int_{0}^{\infty} \frac{\sin{yx}}{x} \mathrm{d}x$$
Now, it is quite easy to prove that $$\int_{0}^{\infty} \frac{\sin{ax}}{x} \mathrm{d}x = \frac{\pi}{2}\mathop{\mathrm{signum}}{a}$$
Therefore,
$$f''(y) = \frac{9\pi}{8} \mathop{\mathrm{signum}}{y} - \frac{3\pi}{8} \mathop{\mathrm{signum}}{y} = \frac{3\pi}{4}\mathop{\mathrm{signum}}{y}$$
Then,
$$f'(y) = \frac{3\pi}{4} |y| + C$$
Note that, $f'(0) = 0$, therefore, $C = 0$.
$$f(y) = \frac{3\pi}{8} y^2 \mathop{\mathrm{signum}}{y} + D$$
Again, $f(0) = 0$, therefore, $D = 0$.
Hence, $$f(1) = \int_{0}^{\infty} \frac{\sin^3{x}}{x^3} = \frac{3\pi}{8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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matrix representation of polynomial Here is a polynomial $p(x,y) = (ax + by)^2$, it can be written like this $$p(x,y) = \left(\left[ \begin{array}{cc}
a & b \\
\end{array} \right] \left[ \begin{array}{c}
x\\
y\\
\end{array} \right]\right)^2$$
and I know that it can also be written as something like $v^TMv$, here $v = [x,y]^T$, and $$M = \left[ \begin{array}{cc}
a^2 & ab \\
ab & b^2\\
\end{array} \right]$$. But how could I find out $M$, any technique?
Furthermore, here $p(x,y)$ has degree $2$, and it can be represented with the multiplication of matrix and vector, what if the degree is $3,4...$?
| $$\left[ \begin{array}{cc}
a & b \\
\end{array} \right] \left[ \begin{array}{c}
x\\
y
\end{array} \right] = \left[ \begin{array}{cc}
x & y \\
\end{array} \right] \left[ \begin{array}{c}
a\\
b
\end{array} \right]$$
Then $M= \left[ \begin{array}{c}
a & b
\end{array}\right] \left[ \begin{array}{c}
a\\
b
\end{array}\right]$
Letting $v=\left[ \begin{array}{c}
x\\
y
\end{array} \right]$ and $A=\left[ \begin{array}{cc}
a & b \\
\end{array} \right]$ we use the first line to note that $Av=v^TA^T$., and hence that $$(Av)^2 = (Av)^T(Av) = v^TA^TAv = v^T(A^TA)v$$
So $M=A^TA$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/454007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If a,b,c are sides of a triangle, prove: $ \sqrt{a+b-c} + \sqrt{b+c-a} + \sqrt{c+a-b} \le \sqrt{a} + \sqrt{b} + \sqrt{c} $ I did substitute $a=x+y, b=x+z, c=y+z$ and I arrived at $\sqrt{2x} + \sqrt{2y} + \sqrt{2z} \le \sqrt{x+y} + \sqrt{x+z} + \sqrt{y+z}$.
However, after this, I tried various methods like AM-GM and Cauchy-Schwarz inequality for hours and I still can't prove it. Can someone help please? Thanks.
| Let $a\geq b\geq c$ and $f(x)=\sqrt{x}$.
Hence, $a+b-c\geq a+c-b\geq b+c-a$, $(a+b-c,a+c-b,b+c-a)\succ(a,b,c)$ and since $f$ is a concave function, the starting inequality it's just Karamata.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove $$\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$$ by induction.
I have seen it many times and proved it before but can't remember what it was I did. I see that for the first two terms $n = 1, n=2$ I get:
for $n = 1$, $\frac{1}{1^2} = 1 < 2$
for $n = 2$, $\frac{1}{1^2} + \frac{1}{2^2} = \frac{5}{4} < 2$
Now I am stumped, I know I want to show this works for the $n+1$ term and am thinking, let the series $\sum_{n=1}^\infty \frac{1}{n^2} = A(n)$ Then look to show the series holds for $A(n+1)$ But $A(n+1) = A(n) + \frac{1}{(n+1)^2}$ But now what? If I tried $A(n+1) - A(n) = \frac{1}{(n+1)^2}$ , but would have to show that this is less than $2 - A(n)$. I am stuck.
Thanks for your thoughts,
Brian
| $$\sum_{n=1}^\infty{1\over n^2}<1+\sum_{n=2}^\infty{1\over n^2-{1\over4}}= 1+\sum_{n=2}^\infty\left({1\over n-{1\over2}}-{1\over n+{1\over2}}\right)=1+{2\over3}={5\over3}\ .$$
By the way: Since the left hand side is $={\pi^2\over6}$ you obtain from this the estimate $\pi^2<10$, which is not bad. In fact $\pi^2\doteq9.87$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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"answer_id": 5
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Find $a,b\in\mathbb{Z}^{+}$ such that $\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$ find positive intergers $a,b$ such that
$\large (\sqrt[3]{a}+\sqrt[3]{b}-1)^2=49+20\sqrt[3]{6}$
Here i tried plugging
$x^3=a,y^3=b$
$(x+y-1)^2=x^2+y^2+1+2(xy-x-y)=49+20\sqrt[3]{6} $
the right hand part is a square hence can be written as $(p+q)^2$
| This one is interesting.
Note that not both $a$ and $b$ are perfect cubes. Why? because otherwise the lhs would be a perfect cube!
Now, let's tackle the case where one of them is a perfect cube, say $a$. In that case, $(\sqrt[3]{a} - 1)$ is an integer. By elementary property of surds, square of this integer must match $49$, integer part of rhs, giving us
$$(\sqrt[3]{a} - 1)^2 = 49$$
And thus, $a = 8^3 = 512$.
The remaining part then reduces to $(7 + \sqrt[3]{b})^2 = 49+20\sqrt[3]{6}$, i.e. $b^{2/3} + 14b^{1/3} = 20(6)^{1/3}$
Edit: This quadratic has no integer solution. I was wrong in assuming $6$ as a solution.
This means neither $a$ nor $b$ is a perfect cube
Edit 2: I first thought integer solutions must somewhere involve integer cubes. Although neither $a$ nor $b$ are perfect cubes, it turns out that $ab$ is a perfect cube and that is a key to the solution. See Calvin's solution to prove that $ab$ is a perfect cube and find $a$, $b$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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seeming ugly limit i want to compute the limit
$$\lim_{x \rightarrow 0} \frac{e^x-1-x-\frac{x^2}{2}-\frac{x^3}{6}-\frac{x^4}{24}-\frac{x^5}{120}-\frac{x^6}{720}}{x^7}$$
Instead of doing some messy calculation, I think if there is some ingenious way to compute this limit, but i don't know how to do. thank you so much.
| Apply the L'Hôpital rule $7$ times (until the derivatives kill the polynomial $1+x+\ldots+\frac{x^6}{6!}$) and we find
$$\lim_{x \rightarrow 0} \frac{e^x-1-x-\frac{x^2}{2}-\frac{x^3}{6}-\frac{x^4}{24}-\frac{x^5}{120}-\frac{x^6}{720}}{x^7}=\lim_{x \rightarrow 0}\frac{e^x}{7!}=\frac{1}{7!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How find the of minimum $ab+bc+\frac{\sqrt{2}}2ac$ for $a^2+b^2=4,b^2+c^2=8$ let $a,b,c$ are real numbers and such $a^2+b^2=4,b^2+c^2=8$, find the minimum of
$$ab+bc+\dfrac{\sqrt{2}}{2}ac$$
if this problem ask find the maximum
we can use $AM-GM$
$$ab+bc+\dfrac{\sqrt{2}}{2}ac=\dfrac{1}{Ax}Aa\cdot xb+\dfrac{1}{Bz}Bb\cdot zc+\dfrac{\sqrt{2}}{2Cv}va\cdot Cc\le \dfrac{1}{Ax}(A^2a^2+x^2b^2)+\dfrac{1}{Bz}(B^2b^2+z^2c^2)+\dfrac{\sqrt{2}}{2Cv}(v^2a^2+C^2c^2)$$
But for mimimum I can't solve it.Thank you
for mimimum I have some idea
let $$a=2\cos{x},b=2\sin{x},b=2\sqrt{2}\cos{y},c=2\sqrt{2}\sin{y}$$
and $\sin{x}=\sqrt{2}\cos{y}$
then $$ab+bc+\dfrac{\sqrt{2}}{2}ac=4\sin{x}\cos{x}+4\sqrt{2}\sin{x}\cos{y}+4\sin{y}\cos{x}$$
my nice methods
$$ab+bc+\dfrac{\sqrt{2}}{2}ac=\dfrac{\sqrt{2}}{4}(a+\sqrt{2}b+c)^2-3\sqrt{2}$$
| Looks like a job for Lagrange multipliers...
let $$F(a,b,c,\lambda,\mu) = ab + bc + \frac{\sqrt{2}}{2} ac + \lambda (a^2 + b^2 - 4) + \mu(b^2 + c^2 - 8)$$
and solve $\dfrac{\partial F}{\partial a} = \dfrac{\partial F}{\partial b} = \dfrac{\partial F}{\partial c} = \dfrac{\partial F}{\partial \lambda} = \dfrac{\partial F}{\partial \mu} = 0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Computing invariant factors from Smith normal form The goal is to find the Jordan Canonical Form of the matrix
$$A=\begin{bmatrix}2&1&1&2\\0&2&0&1\\0&0&2&-1\\0&0&0&1\end{bmatrix}$$
Since the matrix is already upper-triangular, it's obvious that the eigenvalues are 2 and 1, where 1 has geometric and algebraic multiplicity 1, so that I could easily find the JCF by computing $\operatorname{rank}{(A-2I)^{i}}$ for each $i$. However, I thought I would instead try to do it by computing the invariant factors by finding the Smith normal form of the characteristic matrix $xI-A$. The problem is that using elementary row and column operations, I only seem able to obtain the matrix
$$\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&(x-2)^2&0\\0&0&0&(x-1)(x-2)\end{bmatrix}$$
My questions are:
*
*Even though this matrix is not in Smith normal form, is it valid to conclude that the elementary divisors are the powers of the irreducible factors that appear in each diagonal entry, i.e. $(x-1)$, $(x-2)^2$, and $(x-2)$? It happens to be true for this $A$, but would it always be true?
*How do I coax the above matrix into Smith normal form?
| I'll answer your second question first.
Let
$$P(x) = \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&(x-2)^2&0\\0&0&0&(x-1)(x-2)\end{bmatrix}.$$
We need to fix the bottom right $2 \times 2$ principal submatrix. I have explained how to do that here, so I'll just present the results here:
\begin{align*}
\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&(x-2)^2&0\\0&0&0&(x-1)(x-2)\end{bmatrix} &\mapsto \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&(x-2)^2&(x-1)(x-2)\\0&0&0&(x-1)(x-2)\end{bmatrix} \\
&\mapsto \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&(x-2)^2&(x-2)^2+(x-2)\\0&0&0&(x-1)(x-2)\end{bmatrix} \\
&\mapsto \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&(x-2)^2&x-2\\0&0&0&(x-1)(x-2)\end{bmatrix} \\
&\mapsto \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&x-2&(x-2)^2\\0&0&(x-1)(x-2)&0\end{bmatrix} \\
&\mapsto \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&x-2&0\\0&0&(x-1)(x-2)&-(x-1)(x-2)^2\end{bmatrix} \\
&\mapsto \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&x-2&0\\0&0&0&-(x-1)(x-2)^2\end{bmatrix} \\
&\mapsto \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&x-2&0\\0&0&0&(x-1)(x-2)^2\end{bmatrix}
\end{align*}
Irreducible factors are $x-2$ and $x-1$, while the elementary divisors are $x-2$, $(x-2)^2$ and $x-1$.
I think, but I'm not sure, that what you suggest in your first question can always be done to obtain the correct result. One would have to prove it to actually use it. Personally, I find it easier to just properly reduce the matrix polynomial or to fix it to become the Smith normal form, like I did above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Quadratic sum of Gauss integers Let $A$ be a set, define $nA=\{x\mid x=a_1+a_2+\cdots a_k,a_i\in A,1\leq k\leq n\}$.
Denote $G=\{z\mid z=(a+bI)^2,a,b\in \mathbb Z,I=\sqrt{-1}\},K=\{z\mid z=a+2bI,a,b\in \mathbb Z,I=\sqrt{-1}\}$
What's the smallest integer $n$ such that $K=nG$ (if there exist)?
Lagrange's four-square theorem states that any natural number can be represented as the sum of four integer squares: $a=x_1^2+x_2^2+x_3^2+x_4^2,$ hence $a+2bI=x_1^2+x_2^2+x_3^2+x_4^2+b\cdot(1+I)^2$.
$a+bI~(2\not\mid b)$ cannot be represented as the sum of several integer squares, because $2\mid \Im(x+yI)^2=2xy$.
| (This is not a full answer, but too long for a comment)
Claim: $n\leq 8$.
Given $a+2bi\in\mathbb Z[i]$ for some $a,b\in\mathbb Z$. By Langrange's theorem, we have:
$$|a|=x_1^2+x_2^2+x_3^2+x_4^2$$
$$|b|=y_1^2+y_2^2+y_3^2+y_4^2$$
for some $x_i,y_i\in\mathbb Z$. Now, define
$$c_a=\begin{cases}1&a\geq 0\\i&a<0\end{cases}\qquad c_b=\begin{cases}1+i&b\geq 0\\1-i&b<0\end{cases}$$
Then
$$\begin{align}
&\quad (c_ax_1)^2+(c_ax_2)^2+(c_ax_3)^2+(c_ax_4)^2+(c_by_1)^2+(c_by_2)^2+(c_by_3)^2+(c_by_4)^2 \\
&=c_a^2\left(x_1^2+x_2^2+x_3^2+x_4^2\right) + c_b^2\cdot\left(y_1^2+y_2^2+y_3^2+y_4^2\right) \\
&=c_a^2\cdot |a| + c_b^2\cdot |b| \\
&=a+2bi
\end{align}$$
(Check, that the last equation holds for all four cases of $(c_a,c_b)$.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\,\frac{1}{\sqrt{1}+ \sqrt{3}} +\frac{1}{\sqrt{5}+ \sqrt{7} }+\ldots+\frac{1}{\sqrt{9997}+\sqrt{9999}}\gt 24$
Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}}
+\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
My work:
Rationalizing the denominator gives
$$\frac{\sqrt{3}-1}{2} +\frac{\sqrt{7}-\sqrt{5}}{2}+......+\frac{\sqrt{9999}-\sqrt{9997}}{2} .$$
Now by taking two as common and separating the positive and negative terms gives
$$\frac{1}{2} [ \{\sqrt{3} +\sqrt{7}+\dots +\sqrt{9999}\} - \{1+\sqrt{5} +\dots+\sqrt{9997}\}].$$
Can we do like this please suggest. Thanks.
| Note that $\frac{\sqrt{3}-1}{2}>\frac{\sqrt{5}-\sqrt{3}}{2}$, etc. because $\sqrt{\phantom{x}}$ is concave down. So twice your left-hand side is greater than a telescoping sum.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorize $(x+1)(x+2)(x+3)(x+6)- 3x^2$ I'm preparing for an exam and was solving a few sample questions when I got this question -
Factorize : $$(x+1)(x+2)(x+3)(x+6)- 3x^2$$
I don't really know where to start, but I expanded everything to get :
$$x^4 + 12x^3 + 44x^2 + 72x + 36$$
I used rational roots test and Descartes rule of signs to get guesses for the roots. I tried them all and it appears that this polynomial has no rational roots.So, what should I do to factorize this polynomial ?
(I used wolfram alpha and got the factorization : $(x^2 + 4x + 6) (x^2 + 8x + 6)$ But can someone explain how to get there ?)
| $(x+1)(x+2)(x+3)(x+6)-3x^2$
$\rightarrow(x+1)(x+6)(x+2)(x+3)-3x^2$
$\rightarrow(x^2+x+6x+6)(x^2+2x+3x+6)-3x^2$
$\rightarrow(x^2+7x+6)(x^2+5x+6)-3x^2$
$\rightarrow(x^2+6+7x)(x^2+6+5x)-3x^2$
Put :- $x^2+6=y$ eq($1$) . So :-
$(y+7x)(y+5x)-3x^2$
$\rightarrow y^2+7xy+5xy+35x^2-3x^2$
$\rightarrow y^2+12xy+32x^2$
$\rightarrow y^2+4xy+8xy+32x^2$
$\rightarrow y(y+4x)+8x(y+4x)$
$\rightarrow (y+4x)(y+8x)$
Put value of $y$ from eq(1), we get :-
$\rightarrow (x^2+6+4x)(x^2+6+8x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 4,
"answer_id": 3
} |
Given $\sec \theta + \tan \theta = 5$ , Find $\csc \theta + \cot \theta $. The question is to find the value of $ \csc \theta + \cot \theta $ if $\sec \theta + \tan \theta = 5$ .
Here is what I did :
$\sec \theta + \tan \theta = 5$
$\sec \theta = 5 - \tan \theta $
Squaring both sides ,
$$\sec^2 \theta = 25 + \tan^2 \theta -10\tan \theta$$
Substituting $1+\tan^2 \theta$ for $\sec^2 \theta$ ,
$$1+\tan^2 \theta = 25 + \tan^2 \theta -10\tan \theta$$
Thus , $$\tan \theta=24/10$$
So , $\cot \theta = 10/24 $ and $\csc \theta=26/24$
Thus $ \csc \theta + \cot \theta =3/2$ .
But I checked the answer sheet and the answer is not 3/2 but $(3+\sqrt5 )/2$ .
Where have I went wrong ? Please help.
| Here is a simpler solution to this problem:
$$\left(\sec(\theta)+\tan(\theta) \right)\left(\sec(\theta)-\tan(\theta) \right)=\sec^2(\theta)-\tan^2(\theta)=1$$
Since $\sec(\theta)+\tan(\theta)=5$ you get $\sec(\theta)-\tan(\theta)=\frac{1}{5}$.
Adding and subtracting these two relations you get
$$2\sec(\theta)=5+\frac15=\frac{26}{5} \,;\, 2\tan(\theta)=5-\frac15=\frac{24}{5}$$
Thus $\tan(\theta)=\frac{24}{10}$ and
$$\sin(\theta)=\frac{\tan(\theta)}{\sec(\theta)}=\frac{24}{26} \,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Evalute $\lim_{x\to-\infty} \frac{\sqrt{x^2+4x^4}}{8+x^2}$ Having a hard time with this. So far I have:
$$ \frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \frac{x\sqrt{1+4x^2}}{8+x^2}$$
| This is a match-in-heaven for high-school level Non-Standard Analysis (see Keisler textbook).
Since $x$ goes to infinity, we can replace it by the "larger than any real number" $H = \frac {1}{\epsilon} $ where $\epsilon$ is the infinitesimal (smallest than any real number), and dispense with the $\lim$ altogether.
So we can write
$$\lim_{x\to -\infty}\frac{\sqrt{x^2(1+4x^2)}}{8+x^2} = \lim_{x\to -\infty} \frac{x\sqrt{1+4x^2}}{8+x^2} = \frac {H\sqrt{1+4H^2}}{8+H^2}$$
Note that we continue raising $H$ to powers as though it was a number. To by-pass the square-root, consider the square root of the expression squared:
$$\sqrt {\left(\frac {H\sqrt{1+4H^2}}{8+H^2}\right)^2} = \sqrt {\frac {H^2(1+4H^2)}{64+16H^2 + H^4}} = \sqrt {\frac {H^2+4H^4}{64+16H^2 + H^4}}$$
Take out $H^4$ as common factor,
$$= \sqrt {\frac {H^4(\frac {1}{H^2}+4)}{H^4(\frac {64}{H^4}+\frac {16}{H^2} + 1)}}$$
The term $H^4$ simplifies, and the terms $\frac {1}{H^2}$, $\frac {64}{H^4}$, $\frac {16}{H^2}$ all equal zero. So we ar left with
$$= \sqrt {\frac 41} = 2$$
Who said that you cannot treat infinity as a number? Abraham Robinson formalized it once and for ... the time being.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Creating sequences from natural numbers How much different arithmetic sequences can you make from the numbers 1 to 51 ?
Note:
The sequence length has to be 3 numbers.
The difference between each 2 numbers is positive.
| There are several ways to count in an organized manner.. We describe three of them.
(Our favourite way is the first, and relatives.)
First way: We need to choose two numbers $a$ and $b$ to serve as the ends of our sequence. The numbers $a$ and $b$ determine a three-term increasing arithmetic sequence precisely if $\frac{a+b}{2}$ is an integer. So we want $a$ and $b$ to have the same parity.
There are $25$ even numbers in our interval. We can choose $2$ of them in $\binom{25}{2}$ ways. There are $26$ odd numbers in the interval. We can choose $2$ of them in $\binom{26}{2}$ ways.
Thus the total is $\binom{25}{2}+\binom{26}{2}$.
Second way: There is one three-term increasing arithmetic sequence with "middle" number $2$, there are $2$ with middle number $3$, there are $3$ with middle number $4$, and so on up to $24$ with middle number $25$.
There are $25$ with middle number $26$.
And by symmetry there are just as many with middle number $\gt 26$ as there are with middle number $\lt 26$.
Thus our count is $2(1/2)(24)(25)+25$, that is, $25^2$.
Third way: There is $1$ (three-term increasing) arithmetic sequence with common difference $25$, there are $3$ with common difference $24$, there are $5$ with common difference $23$, and so on up to $49$ with common difference $1$.
The sum $1+3+5+\cdots+49$ of the odd numbers up to $49$, is, by a standard formula, equal to $25^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to write $\textrm{cosec}^2\left(\frac{4\pi}7\right)$ as $\textrm{cosec}^2\left(\frac{3\pi}7\right)$. Then converted in $\sin$... But in vain.. Is there any other approach?
| Consider the polynomial $$P(z):=z^6+z^5+z^4+z^3+z^2+z+1\,.$$ Let $t:=z+\dfrac{1}{z}$. Therefore, $P(z)=z^3\,Q(t)$, where
$$Q(t):=t^3+t^2-2t-1\,.$$
Let $\omega:=\exp(\text{i}\theta)$, where $\theta:=\dfrac{2\pi}{7}$. Then, $\omega$, $\omega^2$, $\omega^3$, $\omega^4$, $\omega^5$, and $\omega^6$ are all the roots of $P(z)$. That is, $t_1:=2\cos(\theta)$, $t_2:=2\cos(2\theta)$, and $t_3:=2\cos(3\theta)$ are all the roots of $Q(t)$. Note that, for an arbitrary $\phi$,
$$\text{csc}^2(\phi)=\frac{1}{1-\cos^2(\phi)}=\frac{4}{4-\big(2\cos(\phi)\big)^2}\,.$$
That is,
$$S:=\text{csc}^2(\theta)+\text{csc}^2(2\theta)+\text{csc}^2(4\theta)=\text{csc}^2(\theta)+\text{csc}^2(2\theta)+\text{csc}^2(3\theta)=\sum_{j=1}^3\,\frac{4}{4-t_j^2}\,.$$
Since $Q(t_j)=0$ for each $j$, we get that
$$t_j^3+t_j^2-2t_j-1=0\text{ or }(t_j^2-4)(t_j+1)+2t_j+3=0\,.$$
That is,
$$t_j+1=\frac{2t_j+3}{4-t_j^2}=\frac{2(t_j-2)+7}{4-t_j^2}=\frac{7}{4-t_j^2}-\frac{2}{t_j+2}\,.$$
Furthermore,
$$0=t_j^3+t_j^2-2t_j-1=(t_j+2)(t_j^2-t_j)-1\text{ or }\frac{1}{t_j+2}=t_j^2-t_j\,.$$
Hence,
$$\frac{4}{4-t_j^2}=\frac{4}{7}\,\left(2(t_j^2-t_j)+t_j+1\right)=\frac{4}{7}\,\left(2t_j^2-t_j+1\right)\,.$$
Finally,
$$S=\sum_{j=1}^3\,\frac{4}{4-t_j^2}=\frac{4}{7}\,\sum_{j=1}^3\,\left(2t_j^2-t_j+1\right)\,.$$
Since $\sum\limits_{j=1}^3\,t_j=-1$ and $\sum\limits_{j=1}^3\,t_j^2=(-1)^2-2(-2)=5$ by Vieta's Formulas, we end up with
$$S=\frac{4}{7}\,\big(2\cdot 5-(-1)+3\big)=8\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/470614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The asymptotic expansion for the weighted sum of divisors $\sum_{n\leq x} \frac{d(n)}{n}$ I am trying to solve a problem about the divisor function. Let us call $d(n)$ the classical divisor function, i.e. $d(n)=\sum_{d|n}$ is the number of divisors of the integer $n$. It is well known that the sum of $d(n)$ over all positive integers from $n=1$ to $x$, when $x$ tends to infinity, is asymptotic to $$x \ln(x) + (2 \gamma-1) x + O(\sqrt{x})$$
I would like to calculate a similar asymptotic expression for the sum of $d(n)/n$, again calculated from $n=1$ to $x$ and for $x$ that tends to infinity. I have made some calculations and obtained the formula
$$1/2 (\ln(x))^2 + 2 \gamma \ln (x) + O(1)$$
where gamma is the Euler-Mascheroni constant. I am interested in the constant term of the expression, which seems to be around $0.48$. I suspect that it could correspond to $\gamma^2 - 2\gamma_1$, where $\gamma_1$ is the first Stieltjes constant ($-0.072...$). Could someone confirm this to me?
As an additional question, I would be very interested in obtaining similar asymptotic formulas, with explicitly given constant terms, for the same sum of $d(n)/n$ calculated over all odd integers from $1$ to $x$, and for that calculated over all even integers from $1$ to $x$.
Many thanks.
| I will do the case of the sum being restricted to even integers for the sake of completing the original problem as stated by the OP.
Start with $\Lambda(s)$ for $\sum_{n \; \text{odd}} d(n)/n^s$ which is
$$\Lambda(s) = \left(1-\frac{1}{2^s}\right)^2\zeta(s)^2.$$
It follows that for $\sum_{n \; \text{even}} d(n)/n^s$ we have
$$\Lambda(s) = \left(\frac{2}{2^s} + \frac{3}{2^{2s}} + \frac{4}{2^{3s}}
+ \cdots + \frac{k+1}{2^{ks}} + \cdots \right)
\left(1-\frac{1}{2^s}\right)^2\zeta(s)^2.$$
With the closed form of the sum we obtain
$$\Lambda(s) = \frac{1/2^s(2-1/2^s)}{(1-1/2^s)^2}
\left(1-\frac{1}{2^s}\right)^2\zeta(s)^2
= \frac{1}{2^s}\left(2-\frac{1}{2^s}\right)\zeta(s)^2.$$
It follows that the Dirichlet series for $\sum_{n \; \text{even}} d(n)/n/n^s$ is
$$\Lambda(s) = \frac{1}{2^{s+1}}\left(2-\frac{1}{2^{s+1}}\right)\zeta(s+1)^2.$$
Now using the same technique as in the previous two cases we have
$$\operatorname{Res}
\left( \frac{1}{2^{s+1}}\left(2-\frac{1}{2^{s+1}}\right)\zeta(s+1)^2
\frac{n^s}{s(s+1)}; s=0\right)= \frac{3}{8}\log^2 n \\
+ \left(\frac{3}{2}\gamma - \frac{1}{2}\log 2 - \frac{3}{4}\right) \log n
+ \frac{3}{4} (\gamma-1)^2 + \frac{1}{2}\log 2 -\gamma \log 2 - \frac{3}{2}\gamma_1.$$
Finally compute $q(n)-q(n-1)$ as in the previous two posts to get
$$\frac{3}{8}\log^2 n
+ \left(\frac{3}{2}\gamma - \frac{1}{2}\log 2\right) \log n
+ \frac{3}{4} \gamma^2 -\gamma \log 2 - \frac{3}{2}\gamma_1.$$
The remainder term is $\mathcal{O}(1/\sqrt{n})$ as before. What is very nice about this computation is that it is precisely the difference of the previous two results as predicted by Eric in his comment.
The Mellin integral converges because
$$ \frac{1}{\sqrt{2}} \left(2 - \frac{1}{\sqrt{2}}\right) \le
\left|\frac{1}{2^{s+1}}\left(2-\frac{1}{2^{s+1}}\right)\right| \le
\frac{1}{\sqrt{2}} \left(2 + \frac{1}{\sqrt{2}}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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$f$ holomorphic from unit disc to itself. $f\left(\frac{1}{2}\right) = f\left(-\frac{1}{2}\right) = 0$. Show that $|f(0)| \le 1/3$. I'm studying for a qual exam. I cannot solve the following problem:
Let $f$ be holomorphic from the unit disc to itself. $f\left(\frac{1}{2}\right) = f\left(-\frac{1}{2}\right) = 0$. Show that $|f(0)| \le \frac{1}{3}$.
With an application of Schwarz lemma, I showed that $|f(0)| \le \frac{1}{2}$. I can't prove the tighter bound. Any ideas?
| Consider the function
$$g(z) = f(z) \frac{1 - r^2 z^2}{r^2 - z^2}$$
where $r = \frac{1}{2}$, which is holomorphic on $\mathbb{D}$ since $f$ has roots at $\pm r$. When $|z| = 1$, $|f(z)| \leq 1$ and
$$|g(z)| \leq \left|\frac{1 - r^2 z^2}{r^2 - z^2} \right|$$
We have
$$16 |1 - r^2 z^2|^2 = |4 - z^2|^2 = (4 - z^2) (4 - \overline{z}^2) = 16 - 4z^2 - 4\overline{z}^2 + 1$$
and
$$16|r^2 - z^2|^2 = |1 - 4z^2|^2 = (1 - 4z^2)(1 - 4\overline{z}^2 = 1 - 4z^2 - 4\overline{z}^2 + 16$$
We conclude that $|1 - r^2 z^2| = |r^2 - z^2|$, so $|g(z)| \leq 1$ whenever $|z| = 1$.
So by the Maximum Modulus Principle, $|g(0)| \leq 1$; in particular,
$$|f(0)| \frac{1}{1/4} \leq 1 \implies |f(0)| \leq \frac{1}{4}$$
which is a stronger result.
The approach was suggested by an answer here, as indicated in Clayton's comment.
| {
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"url": "https://math.stackexchange.com/questions/471608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How prove this $\frac{1}{4}<\sum_{n=1}^{\infty}g\left(\frac{1}{n}\right)<1$ let $x>0$, and such
$$(1+x^2)f'(x)+(1+x)f(x)=1,g'(x)=f(x),f(0)=g(0)=0$$
show that
$$\dfrac{1}{4}<\sum_{n=1}^{\infty}g\left(\dfrac{1}{n}\right)<1$$
my idea: we can find
$$f(x)=e^{-\int\dfrac{1+x}{1+x^2}dx}\left(\int\dfrac{1}{1+x^2}e^{\int\dfrac{1+x}{1+x^2}dx}dx+C\right)$$
since
$$\int\dfrac{1+x}{1+x^2}dx=\arctan{x}+\dfrac{1}{2}\ln{(1+x^2)}$$
then following is very ugly,can someone have good methods? Thank you
| Notice that
$$(1+x^2)f'(x)+(1+x)f(x)=1\tag1$$
implies
$$ \left(e^{\arctan x}\sqrt{1+x^2}f(x)\right)'=\frac{e^{\arctan x}}{\sqrt{1+x^2}},\tag2$$
and $f(0)=0$. Thus, for $x>0$, it holds that
\begin{align*}
g'(x)&=f(x)\\&=\frac{1}{e^{\arctan x}\sqrt{1+x^2}}\int_0^x \frac{e^{\arctan t}}{\sqrt{1+t^2}}{\rm d}t\\
&\le \frac{1}{e^{\arctan x}}\int_0^x e^{\arctan x}{\rm d}t\\
&= x.\tag3
\end{align*}
Therefore
$$g(x)=\int_0^x g'(t){\rm d}t\le \int_0^x t{\rm d}t= \frac{x^2}{2}.\tag4$$
On the other hand, for $x\geq 0$, it also holds that
\begin{align*}
f'(x)=\frac{1-(1+x)f(x)}{1+x^2}\ge \frac{1-(1+x)x}{1+x^2}\ge 1-2x,\tag5
\end{align*}
which implies
$$ f(x)=\int_0^x f'(t){\rm d}t\ge \int_0^x (1-2t) {\rm d}t=x-x^2.\tag6$$
Thus
$$ g(x)=\int_0^x f(t){\rm d}t\ge \int_0^x (t-t^2){\rm d}t=\frac{x^2}{2}-\frac{x^3}{3}.\tag7$$
Consequently
$$ \frac{1}{4}\leq\frac{\pi^2}{36}=\sum_{n=1}^{\infty}\frac{1}{6n^2}\leq\sum_{n=1}^{\infty}\left(\frac{1}{2n^2}-\frac{1}{3n^3}\right) \leq\sum_{n=1}^{\infty}g\left(\frac{1}{n}\right)\leq \sum_{n=1}^{\infty}\frac{1}{2n^2}=\frac{\pi^2}{12}\le 1,\tag8$$
which is desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/471894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Taylor Series Expansion of $f(x) = 2/(1-x)$ centered at $x=3$. Give the interval of convergence. Find the Taylor Expansion for the function f(x) = 2/(1-x) centered at x = 3.
Give the interval of convergence for this series.
So if I remember correctly, we first take the first four or so derivatives.
$f(x) = 2/(1-x)$
$f'(x) = 2/(1-x)^2$
$f''(x) = 4/(1-x)^3$
$f'''(x) = 12/(1-x)^4$
$f''''(x) = 48/(1-x)^5$
Now at x = 3 for those derivatives are:
$f(3) = -1$
$f'(3) = .5$
$f''(3) = -.5$
$f'''(3) = (3/4)$
$f''''(3) = -(3/2)$
Is this the Taylor Expansion?
$f(x) = -1 + (1/2)(x-3) - (1/2)( (x-3)^{2} / 2! ) + (3/4)( (x-3)^{3} / 3! ) ...$
I need to find a general solution for this, but I can't really nail down a pattern. To do the interval of convergence portion I need to figure out that pattern so this is where I am stuck.
| The idea is to take $y=x-3$ so $x=y+3$ and then
$$f(x)=\frac{2}{1-x}=\frac{-2}{2+y}=\frac{-1}{1+(y/2)}=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{y}{2}\right)^n=\sum_{n=0}^\infty(-1)^{n+1}\left(\frac{x-3}{2}\right)^n$$
The ratio test gives $R=2$ so the interval of convergence centred at $3$ is $(1,5)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Use the identity $(r+1)^3-r^3\equiv3r^2+3r+1$ to find $\sum_{r=1}^nr(r+1)$ Use the identity $(r+1)^3-r^3\equiv3r^2+3r+1$ to find $$\sum_{r=1}^nr(r+1)$$
I can obtain $$\sum_{r=1}^n3r^2+3r+1=(n+1)^3-1$$
and I think the next step is
$$3\sum_{r=1}^nr(r^2+1)+\frac13=\left((n+1)^3-1\right)$$
But how do I deal with the constant 3 at the end of the LHS and also tidy up the answer to get $\frac13n(n+1)(n+2)$?
| $$3r^2+3r=3r(r+1)$$ and $$n(n+1)(n+2)=\dfrac{1}{4}n(n+1)(n+2)[(n+3)-(n-1)]\\=\dfrac{1}{4}n(n+1)(n+2)(n+3)-\dfrac{1}{4}(n-1)n(n+1)(n+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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I need to calculate $x^{50}$ $x=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}$, I need to calculate $x^{50}$
Could anyone tell me how to proceed?
Thank you.
| This is a very elementary approach based on finding the general form. If we do $x^2$, we find $$x^2=\begin{pmatrix}1&0&0\\1&1&0\\1&0&1\end{pmatrix},~~x^4=\begin{pmatrix}1&0&0\\2&1&0\\2&0&1\end{pmatrix}$$ so I guess that we have $$x^{2k}=\begin{pmatrix}1&0&0\\k&1&0\\k&0&1\end{pmatrix}$$ An inductive approach adimits this general form is valid for integers $k>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
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Can I get a hint on solving this recurrence relation? I am having trouble solving for a closed form of the following recurrence relation.
$$\begin{align*}
a_n &= \frac{n}{4} -\frac{1}{2}\sum_{k=1}^{n-1}a_k\\
a_1 &= \frac{1}{4}
\end{align*}$$
The first few values are $a_1=\frac{1}{4},a_2=\frac{3}{8},a_3=\frac{7}{16},a_4=\frac{15}{32}...$ So it seems the pattern is $a_n = \frac{2^{n}-1}{2^{n+1}}$, but I have been unable to show this algebraically. Here is what I tried:
$$\begin{align*}
2a_n &= \frac{n}{2} - \sum_{k=1}^{n-1}a_k\\
a_n + \sum_{k=1}^n a_k &= \frac{n}{2}\\
a_{n-1} + \sum_{k=1}^{n-1} a_k &= \frac{n-1}{2} \\
2a_n - a_{n-1} & = \frac{1}{2} \\
a_n = \dfrac{2a_{n-1} + 1}{4}
\end{align*}$$
I am so close, I can taste the closed form. Can someone nudge me in the right direction without giving too much away?
| By inspection we determine a particular solution to $2a_n-a_{n-1}=1/2$ is given by $a_n=1/2$ trivially -- try an ansatz of the form $a_n=k$ and thus we get $k=2k-k=1/2$.
Considering the homogeneous case, $2a_n-a_{n-1}=0$, let $a_n=\lambda^n$ hence:$$2\lambda^n-\lambda^{n-1}=0\\2\lambda-1=0\\\lambda=\frac12$$... and so it follows that $a_n=(1/2)^n=1/2^n$ is a solution to our general equation and further so is any scalar multiple (since our equation is linear) i.e. $a_n=C/2^n$. Adding our particular equation to the mix we get a solution of the form $a_n=C/2^n+1/2$. Impose your initial conditions to determine $C$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot[{(x+\sin{\frac{1}{x}})}^{\frac{1}{3}} -x^{\frac{1}{3}}]}$ I need to find the limit of
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
| $$Limit=\lim_{x\rightarrow\infty}x^2[(1+\frac{\sin \frac{1}{x}}{x})^\frac{1}{3}-1]=\lim_{x\rightarrow\infty}x^2\frac{\sin \frac{1}{x}}{3x}=\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480463",
"timestamp": "2023-03-29T00:00:00",
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Show that $\int \limits_{0}^{\infty}\frac{x}{\sinh ax}dx=\left(\frac{\pi}{2a}\right)^2$ Show that
$$\int \limits_{0}^{\infty}\frac{x}{\sinh ax}dx=\left(\frac{\pi}{2a}\right)^2$$
using 2 ways: the first using contour integration and the second using real analysis.
| $$a>0:$$
$$\int_0^{\infty} \frac{x\,dx}{\sinh ax}=\frac{1}{a^2}\int_0^{\infty}\frac{x\,dx}{\sinh x}=\frac{2}{a^2}\int_0^{\infty} \left(\frac{x}{e^{x}}\right)\frac{dx}{1-e^{-2x}}=\frac{2}{a^2}\int_0^{\infty}x\sum_{k=0}^{\infty}e^{-(2k+1)x}\,dx$$
Now, since
$$\int_0^{\infty} xe^{-kx}\,dx=\frac{1}{k^2}$$
We have:
$$\int_0^{\infty} \frac{x\,dx}{\sinh ax}=\frac{2}{a^2}\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}=\frac{\pi^2}{4a^2}$$
The latter sum follows from:
$$\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}=\sum_{k=1}^{\infty} \frac{1}{k^2}-\sum_{k=1}^{\infty} \frac{1}{(2k)^2}=\frac{3}{4}\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{8}$$
The case $a<0$ is dealt with by adding a negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Find the roots of the given equation : $2^{x+2}.3^{\frac{3x}{x-1}} =9$ - Logarithm problem Find the roots of the given equation :
$2^{x+2}.3^{\frac{3x}{x-1}} =9$
My working :
Taking log on both sides we get : $$\log (2^{x+2}.3^{\frac{3x}{x-1}}) =\log 3^2 \Rightarrow (x+2)(\log2) + \frac{3x}{x-1}\log 3 = 2\log 3$$
Now how to proceed further in this problem... please suggest thanks....
| we could argue that 9 is divisible by 3 and not 2
so wlog
$\large2^{x+2}3^{\frac{3x}{x-1}}=9$
$2^{x+2}$ should be one,so $x=-2$
also
$ \large 3^{\frac{3x}{x-1}}=3^2$
${\frac{3x}{x-1}}=2\implies x=-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Nonlinear ordinary differential equation (Elsgolts) Please, help me to solve the following non-linear ODEs:
\begin{align}
\tag 1 y &= (y')^4 -(y')^3 -2 \\
\tag 2 y' &= \dfrac{y}{x+ y^3}
\end{align}
Thanks.
| $$ y = y'^4 - y'^3 - 2 $$
$$ y' = p \Rightarrow y = p^4 - p^3 - 2 $$
$$ \Rightarrow y' = p = p^2(4p-3)\frac{dp}{dx} $$
$$ \Rightarrow p(4p-3) \ dp = dx \Rightarrow \int 4p^2 - 3p \ dp = \int dx $$
$$ \Rightarrow \frac{4p^3}{3} - \frac{3p^2}{2} + c_1 = x $$
$$ \Rightarrow \left\{\begin{matrix}
y = p^4 - p^3 - 2 \\
\frac{4p^3}{3} - \frac{3p^2}{2} + c_1 = x \\
\end{matrix}\right. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/480733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculate $ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $ How evaluate the following limit?
$$ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $$
I cannot apply L'Hopital because $ f(x) = 3 -\sqrt{5 -x} \neq 0 $ at $x = 5$
| I think the problem is $$\lim_{x\to4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$$
HINT:
$$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\frac{3^2-(5+x)}{1^2-(5-x)}\cdot\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}=-\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}$$ if $x\ne4$
Now here $x\to4\implies x\ne4$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\forall x \in \Bbb R, 0 \lt \frac{1}{ x^2+6x+10} \le 1$ I am having trouble understanding the meaning of this pictorially.
Do I just have to multiply across the inequality by $x^2+6x+10$ since $x^2+6x+10 \gt 0$ for all real $x$, giving:
$0 \lt1 \le x^2+6x+10$, giving that $0 \lt 1 $ and $x^2+6x+10 \ge 1$? $(x+3)^2 \ge 0$
Am I missing the point in anyway?
| Let $\displaystyle y=\frac1{x^2+6x+10}\ \ \ \ (1)$
On rearrangement, $x^2y+6yx+10y-1=0\ \ \ \ (2)$ which is a Quadratic Equation is $x$
As $x$ is real, the discriminant must be $\ge0$
$\implies (6y)^2-4y(10y-1)\ge0\iff y^2-y\le0\iff y(y-1)\le 0$
$\implies 0\le y\le1$
But if $y=0,(2)\implies -1=0$ for finite $x$ which is impossible
If $y=1, x^2+6x+9=0\implies (x+3)^2=0$
$\implies 0<y\le1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/481381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Congruence equation - Find all possible $k$ Let $x,\ y,\ z > 1 \in \mathbb{Z_+}$ and $\gcd(x,y,z)=1$.
Find all possible $k$ such :
$$ \sum_{\mathrm{cyc}}{x}\equiv\sum_{\mathrm{cyc}}{x^2y}\equiv\sum_{\mathrm{cyc}}{xy^2}\equiv 3xyz\equiv 0\pmod{k} $$
PS: It's a part of bigger task (I reduce it by myself).
| First, if $p$ is a prime not equal to 3 that divides $k$, then we have $xyz \equiv 0 \pmod{p}$, so WLOG $x \equiv 0 \pmod{p}$. Then, since $ x + y + z \equiv 0 \pmod{p}$, hence $y = - z \pmod{p}$. Substituting this into the other equation, we get that $ y^3 \equiv 0 \pmod{p}$, so $y \equiv 0 \pmod{p}$. But this gives $p \mid \gcd(x,y,z)$.
Hence, $ k = 3^n$ for some $n$. Consider $n \geq 2$. From $3xyz \equiv 0 \pmod{3^n}$, we know that one of the terms must be a multiple of 3. If 2 terms are a multiple of 3, then from $x+y+z \equiv 0 \pmod{3^n}$, hence all 3 terms are, contradicting $\gcd(x,y,z)=1$. WLOG, let $x$ be the term that is a multiple of $3^{n-1}$. If $ x \equiv 0 \pmod{3^n}$, then we likewise have $ y \equiv -z \pmod{3^n} \Rightarrow y \equiv 0 \pmod{3^n}$, so there are no solutions.
WLOG, we may assume that $ x \equiv 3^{n-1} \pmod{3}$ (otherwise, multiply each term by 2). From $x + y + z \equiv 0 \pmod{3^n}$, we can set $ y \equiv 3^{n-1} +k, z \equiv 3^{n-1} -k \pmod{3^n}$. From the other equality, we get that $k^3 \equiv 0 \pmod{3^n}$, which implies that $k$ is a multiple of 3. This contradicts $\gcd(x,y,z)=1$.
Hence, we must have $k=3$ as the only possibility. A quick check shows that this works, e.g. $(x,y,z)= (4,7,10).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why does this series converge? My question is:
Why does the series $$ \sum_{j,k=1}^\infty \frac{1}{j^4+k^4} $$ converge?
I tested the convergence with Mathematica and Octave, but I can't find an analytical proof. In fact, numerical computations suggest that the value of the series is $<1$.
One obvious thing to do would be to use the generalized harmonic series to see that \begin{align} \sum_{j,k=1}^\infty \frac{1}{j^4+k^4} &= \sum_{k=1}^\infty \frac{1}{2k^4} + \sum_{j,k=1; j\neq k} \frac{1}{j^4+k^4} \\ &= \frac{\pi^4}{180} + \sum_{j=1}^\infty \sum_{k=1}^{j-1} \frac{1}{j^4+k^4} + \sum_{j=1}^\infty \sum_{k=j+1}^{\infty} \frac{1}{j^4+k^4}\\ &\leq \frac{\pi^4}{180} + \sum_{j=1}^\infty \sum_{k=1}^{j-1} \frac{1}{(j-k)^4} + \sum_{j=1}^\infty \sum_{k=j+1}^{\infty} \frac{1}{(j-k)^4} \end{align} but unfortunately the last two (double-)series do not converge.
The problem arises when one tries to estimate the Hilbert-Schmidt norm of the Laplacian in $H^2(\mathbb{T}_\pi^2)$.
| Grouping the terms by $i = (j + k)$, we get
$$
\sum_{i = 2}^\infty\:\: \sum_{j, k \gt 0; j + k = i} \frac{1}{j^4 + k^4}
$$
we can estimate that the inner sum, $\sum_{j, k \gt 0; j + k = i} \frac{1}{j^4 + k^4} \leq\frac{i - 1}{(i/2)^4}$, since we have $j^4 + k^4 \geq \left(\frac{j + k}{2}\right)^4 = (i/2)^4$. So we need to check whether the sum $$\sum_{i = 2}^\infty \frac{i-1}{(i/2)^4}$$
converges. But we see that
$$
\frac{i-1}{(i/2)^4} = \frac{2(i - 1)}{i(i/2)^3} = \frac{2(1 - 1/i)}{(i/2)^3} = \frac{2 - 1/i}{i^3/8} \leq \frac{16}{i^3}
$$
for $i \geq 2$. This is easily seen to converge, and so the original series must converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/485524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Evaluating an indefinite integral I need help evaluating the following indefinite integral explicitly
$$\int \frac{1}{1+t^{2^{-n}}} dt$$
I would appreciate any help
| $\frac{1}{1+t^{\frac{1}{2^n}}} = \frac{1}{t^{\frac{1}{2^n}}}\frac{1}{1+\frac{1}{t^{\frac{1}{2^n}}}} = \frac{1}{t^{\frac{1}{2^n}}} \sum_{k=0}^\infty (-1)^k ( \frac{1}{t^{\frac{1}{2^n}}})^k = \sum_{k=0}^\infty (-1)^k \frac{1}{t^{{\frac{1}{2^n}}(k+1)}}$, and this convergence is uniform for $t \ge 2$.
Hence
\begin{eqnarray}
\int_2^x \frac{1}{1+t^{\frac{1}{2^n}}} dt &=& \sum_{k=0}^\infty (-1)^k \int_2^x \frac{1}{t^{{\frac{1}{2^n}}(k+1)}} dt \\
&=& \sum_{k=0,\, $k \ne 2^n-1}^\infty (-1)^k \frac{1}{1-{\frac{1}{2^n}}(k+1)} ( \frac{1}{x^{{\frac{1}{2^n}}(k+1)-1}} - \frac{1}{2^{{\frac{1}{2^n}}(k+1)-1}} ) + (-1)^{2^n-1} \ln \frac{x}{2}\\
&=& \sum_{k=0,\, $k \ne 2^n-1}^\infty (-1)^k \frac{1}{1-{\frac{1}{2^n}}(k+1)} ( \frac{x}{x^{{\frac{1}{2^n}}(k+1)}} - \frac{2}{2^{{\frac{1}{2^n}}(k+1)}} ) + (-1)^{2^n-1} \ln \frac{x}{2}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding a non-recursive formula for a recursively defined sequence So I have a recursive definition for a sequence, which goes as follows:
$$s_0 = 1$$
$$s_1 = 2$$
$$s_n = 2s_{n-1} - s_{n-2} + 1$$
and I have to prove the following proposition: The $n$th term of the sequence defined above is $s_n = \frac{n(n+1)}{2} + 1$.
To prove this, I started it off by induction. The base case is for $n = 0$ and its true since the non-recursive and recursive results match. I assumed the following hypothesis to be true for some $n = k$: that the $n$th term of the sequence is $s_k = \frac{k(k+1)}{2} + 1$. Then, in the induction step, we need to show that $s_{k + 1} = \frac{(k+1)(k+2)}{2} + 1$.
Using the recursive definition of the sequence, we have that $s_{k+1} = 2s_k - s_{k-1} + 1$, so we can use the hypothesis to replace $s_k$ and $s_{k-1}$ by their non-recursive formulas:
$$ s_{k+1} = 2 (\frac{k(k + 1)}{2} + 1) - (\frac{(k-1)k}{2} + 1) + 1$$
After simplifying i get
$$ s_{k+1} = \frac{k(k+3)}{2} + 2 $$
which is clearly wrong. Can someone point out what I'm doing wrong and where I can go with this?
EDIT: The answer I have given is correct, except that we need to simplify further to get the form we want:
$$ \frac{k(k+3)}{2} + 2 = \frac{k^2 + 3k + 4}{2}$$
after expansion and common denominators, and then this is clearly equal to
$$ \frac{(k+1)(k+2)}{2} + 1 = \frac{k^2 + 3k + 2}{2} + 1 = \frac{k^2 + 3k + 4}{2} $$
| At the end of this answer is a brief description of inverting finite difference operators. In the case here
$$
\begin{align}
\Delta^2s_k
&=s_n-2s_{n-1}+s_{n-2}\\
&=1
\end{align}
$$
Is a second order finite difference operator. Inverting the operator by summing twice, we get that the solution is of the form
$$
s_n=\frac12n^2+c_1n+c_0
$$
We can compute the constants by plugging in known values:
$$
\frac12\cdot0^2+c_i\cdot0+c_0=s_0=1\\
\frac12\cdot1^2+c_i\cdot1+c_0=s_1=2
$$
Thus, $c_0=1$ and $c_1=\frac12$. Therefore,
$$
\begin{align}
s_n
&=\frac12n^2+\frac12n+1\\
&=\frac{n(n+1)}{2}+1
\end{align}
$$
The only problem in your solution is that you left out the $+1$
$$
\begin{align}
s_{k+1}
&=2\left(\frac{k(k+1)}{2}+1\right)-\left(\frac{(k-1)k}{2}+1\right)\color{#C00000}{+1}\\
&=(k^2+k+2)-\frac12(k^2-k+2)+1\\
&=\frac{k^2+3k+4}{2}\\
&=\frac{(k+1)(k+2)}{2}+1
\end{align}
$$
However, since the answer you got was $\frac{k(k+3)}{2}+2$, perhaps the omission of the $+1$ was simply a typo. In that case, you got the right answer since
$$
\frac{k(k+3)}{2}+2=\frac{(k+1)(k+2)}{2}+1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int\frac{1}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}dx$
Solve the indefinite integral
$$
I=\int\frac{1}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}\;dx
$$
My Attempt:
$$
\begin{align}
I&=\int\frac{1}{\sin x+\cos x+\frac{1}{\sin x \cos x}+\frac{\sin x +\cos x}{\sin x\cos x}}\;dx\\
\\
&=\int\frac{\sin x\cos x}{\left(\sin x+\cos x\right)\left(\sin x\cos x \right)+1+\left(\sin x+\cos x\right)}\;dx
\end{align}
$$
How can I complete the solution from here?
| You want
$\begin{align}
\int\frac{dx}{\sin x+\cos x+\tan x+\cot x+\csc x+\sec x}
&= \int\frac{dx}{\sin x+\cos x+\frac{\sin x}{\cos x}
+\frac{\cos x}{\sin x}+\frac1{\sin x}+\frac1{\cos x}}\\
&= \int\frac{\sin x \cos x\ dx}{\sin^2 x \cos x+\cos^2 x \sin x+\sin^2 x
+\cos^2 x+\cos x+\sin x}\\
&= \int\frac{\sin x \cos x\ dx}{\sin x \cos x(\sin x+ \cos x)+1+\cos x+\sin x}\\
&= \int\frac{\sin x \cos x\ dx}{(\sin x \cos x+1)(\sin x+ \cos x)+1}\\
\end{align}
$.
Applying substitutions
$\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}, dx = \frac{2 dt}{1+t^2}$,
this becomes
$\begin{align}
\int\frac{\sin x \cos x\ dx}{(\sin x \cos x+1)(\sin x+ \cos x)+1}
&=\int\frac{\frac{(2t)(1-t^2)(2dt)}{(1+t^2)^3}}
{(\frac{(2t)(1-t^2)}{(1+t^2)^2}+1)(\frac{2t}{1+t^2}+ \frac{1-t^2}{1+t^2})+1}\\
&=\int\frac{(2t)(1-t^2)(2dt)}
{((2t)(1-t^2)+(1+t^2)^2)(1+2t-t^2)+(1+t^2)^3}\\
&=\int\frac{4t(1-t^2)dt}
{((2t-2t^3)+1+2t^2+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\
&=\int\frac{4t(1-t^2)dt}
{(1+2t+2t^2-2t^3+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\
&=\int\frac{4t(1-t^2)dt}
{2 (t+1) (2 t^4-3 t^3+3 t^2+t+1)} \quad \text{ (according to Wolfram Alpha)}\\
&=\int\frac{2t(1-t)dt}
{2 t^4-3 t^3+3 t^2+t+1}\\
\end{align}
$.
According to Wolfram Alpha,
that quartic can be factored
as the product of two quadratics,
but the coefficients look irrational.
I'll leave it at this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 2
} |
Integrating $\int \cos^3(x)\cos(2x) \, dx$ How would Integrate the following?
$$\int \cos^3(x)\cos(2x) \, dx.$$
I did
$$\int \cos^3(x)(1-2\sin^2(x)) \, dx = 2\int \cos^3(x)-\cos^3x\sin^2x \, dx$$
But I find myself stuck....
| \begin{align}
\int \cos^3(x) \cos(2x) dx &= \int \cos^3(x)(1-2\sin^2(x)) dx \\
&= \int \cos^2(x) \cdot \cos(x) (1-2\sin^2(x)) dx \\
&= \int (1-\sin^2x)\cdot \cos(x) \cdot(1-2\sin^2(x)) dx \\
&= \int ( \cos(x)- 3 \sin^2x \cdot \cos(x) + 2 \sin^4x \cdot \cos(x) ) dx.
\end{align}
Now using inverse chain rule which is
$$\int f^n(x) \cdot f'(x) dx= f^{n+1}(x) +C$$
then
$$ \int \cos^3(x) \cos(2x) dx = \sin(x)- \sin^3x + \frac{2}{5} \, \sin^5x + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Radius of convergence of product
Let $\sum_{i=0}^\infty a_nz^n$ and $\sum_{i=0}^\infty b_nz^n$ be power series, and define the product $\sum_{i=0}^\infty c_nz^n$ by $c_n=a_0b_n+a_1b_{n-1}+\ldots+a_nb_0$. Find an example where the first two series has radius of convergence $R$, while the third (the product) has radius of convergence larger than $R$.
The radius of convergence of $\sum_{i=0}^\infty a_nz^n$ is given by $1/R=\limsup{|a_n|^{1/n}}$. I tried some sequences like $a_0=a_1=\ldots=b_0=b_1=\ldots=1$. Then the two sequences have radius $1$. But $c_i=i+1$, and $\lim_{i\rightarrow\infty}(i+1)^{1/i}=1$. So the radius is the same as the original two sequences, which doesn't work.
| A simpler example: let
$$
f(z) = \frac{1+z}{1-z} = \frac{1}{1-z} + \frac{z}{1-z}.
$$
Note that the first term is just the formula for the geometric sum with first term 1, $$\frac{1}{1-z} = 1 + z + z^2 + z^3 + \cdots, \qquad |z| < 1,$$ and the second term is the formula for a geometric sum with first term equal to the common ratio $z$:
$$
\frac{z}{1-z} = \frac{1}{1-z} - 1 = z + z^2 + z^3 + \cdots, \qquad |z| < 1.
$$
Then the power series for $f(z)$ is given by
$$
f(z) = \frac{1+z}{1-z} = 1 + 2z + 2z^2 + 2z^3 + \cdots = 1 + 2\sum_{n=1}^\infty z^n, \qquad |z| < 1,
$$
and has radius of convergence $R_f = 1$.
If we form a new power series $g(z)$ by making the substitution $z \mapsto -z$, we have
$$
g(z) = \frac{1-z}{1+z} = 1 - 2z + 2z^2 - 2z^3 + \cdots = 1 + 2\sum_{n=1}^\infty (-z)^n, \qquad |z| < 1,
$$
also with radius of convergence $R_g = 1$. However, the product series is
$$
f(z)g(z) = \left( \frac{1+z}{1-z} \right) \left( \frac{1-z}{1+z} \right) = 1 = 1 + 0z + 0z^2 + 0z^3 + \cdots, \qquad \forall z\in\mathbb{C}
$$
and has radius of convergence $R_{fg} = \infty$, which is strictly larger than $R_f = R_g = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found the pattern:
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$$
But I have yet to figure out how to prove it algebraically.
Suggestions?
| The matrix
$$N=\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$$
is nilpotent with index 2 of nilpotency: $N^2=0$ so by the binomial formula we have
$$\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}^n=(I_2+N)^n=\sum_{k=0}^n {n\choose k}N^k={n\choose 0}I_2+{n\choose 1}N=I_2+nN=\begin{pmatrix} 1& n\\ 0 & 1 \end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
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} |
Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$
Problem:Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ where $a,b,c$ are distinct real numbers
Solution:$(x-a)^3+(x-b)^3+(x-c)^3=0$
$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0$
By Descartes rule of sign,number of positive real roots $=3$
But are they distinct $?$
Answer :- number of distinct real roots $ =1$
| Set $m=(a+b+c)/3$, $A=a-m$, $B=b-m$, $C=c-m$ and $x=y+m$. Then your equation becomes
$$
(y-A)^3 + (y-B)^3 + (y-C)^3 = 0
$$
and, since $A+B+C=0$, your expansion applies to give
$$
y^3+(A^2+B^2+C^2)-\frac{A^3+B^3+C^3}{3}=0
$$
which is a suppressed cubic, whose discriminant is
$$
\frac{1}{4}\biggl(-\frac{A^3+B^3+C^3}{3}\biggr)^2+\frac{1}{27}(A^2+B^2+C^2)^3>0
$$
so the equation has only one real root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to prove: $a+b+c\le a^2+b^2+c^2$, if $abc=1$? Let $a,b,c \in \mathbb{R}$, and $abc=1$. What is the simple(st) way to prove inequality
$$
a+b+c \le a^2+b^2+c^2.
$$
(Of course, it can be generalized to $n$ variables).
| A).
If values $~a,b,c~$ are positive, then
denote $m = \dfrac{a+b+c}{3},~$ ($m\ge \sqrt[3]{abc}=1$, AM-GM);
$\left\{\begin{array}{l}
a=m+\alpha,\\
b=m+\beta, ~~~~~~(\alpha+\beta+\gamma=0).\\
c=m+\gamma.
\end{array}\right.~~~
$
Then
$$
a^2+b^2+c^2 ~=~ 3m^2 + \alpha^2+\beta^2+\gamma^2 ~\ge~ 3m ~=~ a+b+c.
$$
B). If some (two) of values $~a,b,c~$ are negative, then
$$
a^2+b^2+c^2 \ge ~|a|+|b|+|c| ~>~ a+b+c.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
A question by Ramanujan about a relational expression of a triangle I found the following question in a book without any proof:
Question : Suppose that each length of three edges of a triangle $ABC$ are $BC=a, CA=b, AB=c$ respectively. If
$$\frac1a=\frac1b+\frac1c, \frac2a=\frac{1}{c-b}-\frac{1}{c+b},$$
then prove
$$\sqrt[3]{\cos{2A}}+\sqrt[3]{\cos{2B}}+\sqrt[3]{\cos{2C}}=\sqrt[3]{\frac{5-3\color{blue}{\sqrt[3]{7}}}{2}}\ .$$
This book says that this is the question by Ramanujan.
I've tried to prove this, but I'm facing difficulty. Can anyone help? If you have any helpful information, please let me know it.
| $$\begin{align}\text{Partial answer:}\qquad\qquad\quad\frac1a=\ \frac1b+\frac1c\quad\iff\quad&a=\frac1{\frac1b+\frac1c}=\frac{bc}{b+c}\qquad\qquad(1)\\\\\\\\\\\frac2a=\frac1{c-b}-\frac1{c+b}\quad\iff\quad\frac2a=\frac{2b}{c^2-b^2}\quad\iff\quad&a=\frac{c^2-b^2}b\ \qquad\qquad\qquad\quad\ (2)\end{align}$$
$$\begin{align}&\overset{(1)}{\underset{(2)}\iff}\quad\frac{bc}{b+c}=\frac{c^2-b^2}b\quad&\iff&\quad b^2c=(c-b)(b+c)^2\ \qquad\ |:(b^2c)\\\\\\\\&\iff\quad1=\frac{c-b}c\left(\frac{b+c}b\right)^2\quad&\iff&\quad1=\left(1-\frac1t\right)(1+t)^2\qquad|\cdot t\end{align}$$
$$\iff\quad t=(t-1)(t+1)^2\quad\iff\quad t^3+t^2-2t-1=0\quad\iff\quad t=\frac cb=\ldots$$
We are dealing with a so-called irreducible case , with a single positive real root. Furthermore, from the Generalized Pythagorean Theorem, also known as the Law of Cosines, we deduce the following three relationships, and their implications:
$$a^2+b^2-2ab\cos C=c^2\iff\cos C=\frac{a^2+b^2-c^2}{2ab}\underset{(2)}=\frac{a^2-ab}{2ab}=\frac12\left(\frac ab-1\right)\underset{(2)}=\frac{t^2-1}2$$
$$a^2+c^2-2ac\cos B=b^2\iff\cos B=\frac{a^2+c^2-b^2}{2ac}\underset{(2)}=\frac{a^2+ab}{2ac}=\frac{a+b}{2c}\ \underset{(2)}=\ \frac c{2b}=\frac t2$$
$$b^2+c^2-2bc\cos A=a^2\iff\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac12\left(\frac bc+\frac cb-\frac{a^2}{bc}\right)\underset{(1)}=\underset{(1)}=\frac12\left(\frac bc+\frac cb-\frac{a}{b+c}\right)\underset{(2)}=\frac12\left(\frac bc+\frac cb-\frac{c-b}b\right)=\frac12\left(\frac bc+1\right)=\frac12\left(\frac1t+1\right)$$
$$\text{Then we use:}\qquad\cos2x=\cos^2x-\sin^2x=\cos^2x-(1-\cos^2x)=2\cos^2x-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve equation? How to solve this equation in the set of real numbers?
$$(x^{2}+3y^{2}-7)^{2} + \sqrt{3-xy-y^2}=0$$
I tried to solve $x^{2}+3y^{2}-7=0$ and $\sqrt{3-xy-y^2}$=0 for x. But it did not help.
| $(x^2 + 3y^2 - 7)^2=0$ and $3-xy-y^2=0$
$x^2 + 3y^2 - 7=0$ and $3-xy-y^2=0$
$x=\pm (7-3y^2)^2$... So you have two cases:
You put $x=(7-3y^2)$ in the equation $3-xy-y^2=0$ and you solve
You put $x=-(7-3y^2)$ in the equation $3-xy-y^2=0$ and you solve . . .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Approaching modular arithmetic problems I'm a little stumbled on two questions.
How do I approach a problem like $x*41 \equiv 1 \pmod{99}$.
And given $2$ modulo, $7x+9y \equiv 0 \pmod{31}$ and $2x−5y \equiv 2 \pmod{31}$ (solve for $x$ only)?
When I solve for $x$ for the latter, I got a fraction as the answer and I'm not sure if I can have a fraction as an answer? I'm not sure how to approach the first problem either.
| Finding the solution to $$x \times 41 \equiv 1 \pmod {99}$$ is equivalent to asking for the multiplicative inverse of $41$ modulo $99$. Since $\gcd(99,41)=1$, we know $41$ actually has an inverse, and it can be found using the Extended Euclidean Algorithm:
\begin{align*}
99-2 \times 41 &= 17 \\
41-2 \times 17 &= 7 \\
17-2 \times 7 &= 3 \\
7-2\times 3 &= 1 &=\gcd(99,41). \\
\end{align*}
Going back, we see that
\begin{align*}
1 &= 7-2\times 3 \\
&= 7-2\times (17-2 \times 7) \\
&= 5 \times 7-2\times 17 \\
&= 5 \times (41-2 \times 17)-2\times 17 \\
&= -12 \times 17+5 \times 41 \\
&= -12 \times (99-2 \times 41)+5 \times 41 \\
&= 29 \times 41-12 \times 99 \\
\end{align*}
Hence $29 \times 41 \equiv 1 \pmod {99}$ and thus $x=29$.
In the second case, we have $$7x+9y \equiv 0 \pmod {31}$$ and $$2x-5y\equiv 2 \pmod {31}.$$ Here we want to take $7x+9y=0 \pmod {31}$ and rearrange it to get $x \equiv ?? \pmod {31}$, then substitute it into the other equation and solve for $y$. This requires finding the multiplicative inverse of $7$ modulo $31$ (which we can do as above). It turns out $7 \times 9 \equiv 1 \pmod {31}$. Hence
\begin{align*}
& 7x+9y=0 \pmod {31} \\
\iff & 7x \equiv -9y \pmod {31} \\
\iff & x \equiv -9y \times 9 \pmod {31} \\
\iff & x \equiv 12y \pmod {31}.
\end{align*}
We then substitute this into the equation $2x-5y\equiv 2 \pmod {31}$, which implies $$2 \times 12y-5y \equiv 2 \pmod {31}$$ or equivalently $$19y \equiv 2 \pmod {31}.$$ Yet again, we find a multiplicative inverse, this time of $19$ modulo $31$, which turns out to be $18$. So
\begin{align*}
& 19y \equiv 2 \pmod {31} \\
\iff & y \equiv 2 \times 18 \pmod {31} \\
\iff & y \equiv 5 \pmod {31}.
\end{align*}
Hence $$x \equiv 12y \equiv 29 \pmod {31}.$$ Thus we have the solution $(x,y)=(29,5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the smallest value of $x^2+y^2$ when $x+y=6$? If $ x+y=6 $ then what is the smallest possible value for $x^2+y^2$?
Please show me the working to show where I am going wrong!
Cheers
| We have
$$2(x^2+y^2)=(x+y)^2 +(x-y)^2=36+(x-y)^2.$$
But $36+(x-y)^2$ is smallest when $x=y$. Thus the minimum value of $2(x^2+y^2)$ is $36$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 4
} |
How to integrate a function that has no exact integration and cannot be expanded by a Taylor series Today I posted a question about the integral
$$\int\frac{(\cos{c x})^2}{a+bx}dx$$
for which a Taylor series can be built and the integral solved for the desired approximation.
Another term in the stiffness matrix has the integral:
$$\int\frac{x^2(\cos{c x})^2}{a+bx}dx$$
whose Taylor series cannot be used since it does not converge:
$$x^4 \left(\frac{b^2}{a^3}-\frac{c^2}{a}\right)-\frac{b x^3}{a^2}+x^6 \left(\frac{b^4}{a^5}-\frac{b^2 c^2}{a^3}+\frac{c^4}{3 a}\right)+x^5 \left(\frac{b c^2}{a^2}-\frac{b^3}{a^4}\right)+\frac{x^2}{a}+O[x]^7$$
In this case, which method could be used to calculate the integral?
| Maple finds a closed form for your antiderivative:
$$ 1/4\,{\frac {x\sin \left( 2\,cx \right) }{bc}}+1/8\,{\frac {\cos
\left( 2\,cx \right) }{{c}^{2}b}}-1/4\,{\frac {a\sin \left( 2\,cx
\right) }{c{b}^{2}}}+1/2\,{a}^{2}{\it Si} \left( 2\,cx+2\,{\frac {ac}
{b}} \right) \sin \left( 2\,{\frac {ac}{b}} \right) {b}^{-3}\\+1/2\,{a}^
{2}{\it Ci} \left( 2\,cx+2\,{\frac {ac}{b}} \right) \cos \left( 2\,{
\frac {ac}{b}} \right) {b}^{-3}+1/4\,{\frac {{x}^{2}}{b}}-1/2\,{\frac
{ax}{{b}^{2}}}+1/2\,{\frac {{a}^{2}\ln \left( bcx+ac \right) }{{b}^{3
}}}
$$
where Si and Ci are the Sine-integral and Cosine-integral functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/504792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do you factor this? $x^3 + x - 2$ How do you factor $x^3 + x - 2$?
Hint: Write it as $(x^3-x^2+x^2-x+2x-2)$ to get $(x-1)(x^2+x+2)$
Note the factored form here. Thanks!
| By inspection, we see that $1$ is a root of $x^3 + x - 2$, so it is divisible by $x - 1$; alternatively, the rational roots theorem would suggest this too.
Now $x^2 + x + 2 = x^2 + x + \frac{1}{4} + \frac{7}{4} = (x + \frac{1}{2})^2 + \frac{7}{4}$ has no real roots, and is irreducible. If you're factoring over $\Bbb{C}$, then it's got roots at $\pm \sqrt{\frac 7 4}i - \frac{1}{2}$; denoting these as $\alpha_+$ and $\alpha_-$, the original polynomial then splits as
$$(x - 1)(x - \alpha_+)(x - \alpha_-)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/507467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital $$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$
I tried using $\lim\limits_{x\to0} \frac{\sin x}x=1$.
But it doesn't work :/
| $$
L=\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=\lim_{x\to0}\frac{x-\sin x}{x\cos x-\sin x}\cos x = \lim_{x\to0}\frac{2x-\sin2x}{2x\cos2x-\sin2x}\cos2x\\
= \lim_{x\to0}\frac{x-\cos x\sin x}{x(1-2\sin^2x)-\cos x\sin x}\cos2x=\lim_{x\to0}\frac{x-\cos x\sin x}{x-\cos x\sin x-2x\sin^2x}\cos2x
$$
Which, noting that $\lim_{x\to0}\cos2x=1$, we can then write as
$$
\lim_{x\to0}\frac{1}{1-\frac{2x\sin^2x}{x-\cos x\sin x}} = \frac{1}{1-2\lim_{x\to0}\frac{x\sin^2x}{x-\cos x\sin x}}=\frac{1}{1-2M}
$$
Now, we turn our attention to that new limit...
$$
\frac1M=\lim_{x\to0}\frac{x-\cos x\sin x}{x\sin^2x}=\lim_{x\to0}\frac{1-\cos x\frac{\sin x}x}{1-\cos^2x}=1+\lim_{x\to0}\frac{1-\frac{\sin x}{x\cos x}}{1-\cos^2x}\cos^2 x\\
=1+\lim_{x\to0}\frac{x-\tan x}{x\sin^2x}
$$
But we also have
$$
\frac1M = \lim_{x\to 0} \frac{2x-\sin2x}{2x\sin^2x}=2\lim_{x\to 0} \frac{2x-\sin2x}{2x(1-\cos2x)}=2\lim_{x\to0}\frac{x-\sin x}{x(1-\cos x)}\\
=2\lim_{x\to0}\frac{x-\sin x}{x(1-\cos^2 x)}(1+\cos x)=4\lim_{x\to0}\frac{x-\sin x}{x(1-\cos^2 x)}=4\lim_{x\to0}\frac{x-\sin x}{x\sin^2x}\\
=4\lim_{x\to0}\frac{x-\sin x}{x-\tan x}\cdot\frac{x-\tan x}{x\sin^2x}
$$
So, we have
$$
\frac1M = 4L\left(\frac1M-1\right)
$$
or $1=4L(1-M)$... but $L=\frac{1}{1-2M}$ (or $1=L(1-2M)$).
Therefore, we have that
$$
1-2=4L-4LM-2L+4LM = 2L = -1
$$
Therefore, $L=-\frac12$. No use of $\lim_{x\to0}\frac{\sin x}x$ required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/508733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
How to solve inequalities with infinite terms Consider the following inequality:
$x + 2 < 1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} ... $ with $x>0$.
Is there a general way to solve such an inequality with infinite terms?
The best I can do is some conjectures:
For $x = 2$ the right hand side equals 2, so I know that $x < 2$. Logically $ 0 < x \leq 1$, so what remains open is the case $1 < x < 2 $.
But I was thinking if there exists for example an algebraic way of solving this stuff, instead of what I'm doing.
| If $|\frac1x|<1 \iff |x|>1$
$$1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} ... =\frac1{1-\frac1x}=\frac x{x-1}$$
So, we need $\displaystyle x+2<\frac x{x-1}$
Multiplying either sides by $(x-1)^2,$
$\displaystyle\iff (x+2)(x-1)^2< x(x-1)$
If $x>1,$ we have $(x+2)(x-1)<x\iff x^2-2<0\iff -\sqrt2<x<\sqrt2$
$\implies 1<x<\sqrt2$
If $x<1,$ we have $(x+2)(x-1)>x\iff x^2-2>0\iff x<-\sqrt2$ or $x>\sqrt2$
$\implies -\sqrt2<x<1$ as $x<1$
But given that $x>0\implies 0<x<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/508803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How prove this $a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$ let $a,b,c>0$, show that
$$a+b+c+3\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$$
I know this
$$a+b+c\ge 3\sqrt[3]{abc}$$
so
$$\Longleftrightarrow 6\sqrt[3]{abc}\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})$$
But this maybe not true?
| WLOG suppose that $a\geq b\geq c$.
First see that using weighted AG-inequality, we have:
$$
c+3\sqrt[3]{abc}\geq 4\sqrt c \sqrt[4]{ab} (*)
$$
Then we have:
$$
a+b+c+3\sqrt[3]{abc}- 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=\\
(\sqrt{a}+\sqrt{b}-\sqrt{c})^2-4\sqrt{ab}+3\sqrt[3]{abc}\\
\geq (2\sqrt[4]{ab}-\sqrt{c})^2-4\sqrt{ab}+3\sqrt[3]{abc}\\
= c- 4\sqrt c \sqrt[4]{ab}+3\sqrt[3]{abc}\geq 0\\
$$
The last one is positive due to $(*)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Let P be a moving point such that if $PA$ and $PB$ are two tangents drawn from $P$ to the circle $x^2+y^2=1 ( $ A ,B being the points of contact) ,... Problem :
Let $P$ be a moving point such that if $PA$ and $PB$ are two tangents drawn from $P$ to the circle $x^2+y^2=1$ ( $A$, $B$ being the points of contact) , then $\angle AOB = 60^{\circ}$, where $O$ is origin.Then find the locus of $P$.
My approach :
We know that tangents drawn from same point to the same circle are congruent. Also, line joining the centre of this circle to the point of tangency is perpendicular i.e. angle formed by tangent and radius in $90^{\circ}$
OA and OB = 1 unit.
But I am unable to proceed further , I know that locus is a circle but unable to find the radius of this circle which is OP. Please guide how to find the radius of this circle. Thanks...
| First note that the quadrilaterial $PAOB$ is a kite, this follows from the fact that the tangent from one point to a circle are from a same length, and the $AO=OB=r$.
From the fact that $\angle AOB = 60^{\circ}$ and using $AO=OB$ we get that the $\triangle AOB$ is equilaterial. This leads to conclusion $AB=r=1$, which is the smaller diagonal from the kite. Now we need to find the other one.
Let $D$ be the point when both diagonals intersect, and from the propertiy of the kite we know that they are normal to each other. So we can split the diagonal $PO$ into $PO = OD + PD$. We know that the $OD$ is the height of the equilaterial triangle so we have:
$$OD = \frac{r\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$
We know that the angle between the tangent line to a circle and the radius at the touching point is $90^{\circ}$, so $\angle PAO = 90^{\circ}$. We can denote $\angle PAO$ as sum of two angles: $\angle PAO = \angle OAB + \angle BAP = 60^{\circ} + \angle BAP$, because $\angle OAB$ is an angle in an equilaterial triangle. So from this we have:
$$\angle BAP = 30^{\circ}$$
Also we know that $AD = \frac r2 = \frac 12$. Now in the right triangle $PAD$ we have:
$$\tan \angle BAP = \frac{PD}{AD}$$
$$\tan \angle 30^{\circ} = \frac{PD}{\frac 12}$$
$$ \frac {1}{\sqrt{3}} = \frac{PD}{\frac 12}$$
$$ PD = \frac{1}{2\sqrt{3}}$$
Now we have: $PO = OD + PD$, i.e.:
$$PO = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}}$$
$$PO = \frac{3}{2\sqrt{3}} + \frac{1}{2\sqrt{3}}$$
$$PO = \frac{4}{2\sqrt{3}}$$
$$PO = \frac{2}{\sqrt{3}} \approx 1.154$$
So the distance from $O$ to $P$ is constant, that means that $P$ can be any point that's $\frac{2}{\sqrt{3}}$ units away from $(0,0)$, i.e. $P$ lies on a circle with this equation:
$$x^2 + y^2 =\left(\frac{2}{\sqrt{3}}\right)^2$$
$$x^2 + y^2 = \frac 43$$
So the locus of the point $P$ is:
$$x^2 + y^2 = \frac 43$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
integral of $\int\frac{2\sin(2x)-\cos(x)}{6-\cos^2x -4\sin x}\mathrm{d}x$ so i got this problem $$\int\dfrac{2\sin 2x-\cos x}{6-\cos^2x -4\sin x}\mathrm{d}x$$
now this is what i tried $=\int\dfrac{4\sin(x)\cos(x)-\cos(x)}{6-(1-\sin^2x) -4\sin x}\mathrm{d}x=\int\dfrac{\cos(x)(4\sin(x)-1)}{5+\sin^2x-4\sin x}\mathrm{d}x$
Substituting $\sin x=t$
implying $\int\dfrac{(4t-1)}{5+t^2-4t}\mathrm{d}t$=$\int\dfrac{(4t)}{5+t^2-4t}\mathrm{d}t+\int\dfrac{(-1)}{5+t^2-4t}\mathrm{d}t$
we may now solve second part by substuting for $\arctan( x)$ but what to do with $\int\dfrac{(4t)}{5+t^2-4t}\mathrm{d}t$ ??
| $\large Hint:$ $u \equiv \sin\left(x\right)$.
\begin{align}
&
\int
{2\sin\left(2x\right) - \cos\left(x\right)
\over
6 - \cos^{2}\left(x\right) - 4\sin{x}}\,{\rm d}x
=
\int
{4\sin\left(x\right) - 1
\over
\sin^{2}\left(x\right) - 4\sin{x} + 5}\,\cos\left(x\right)\,{\rm d}x
\\[3mm]&=
\int{4u - 1 \over u^{2} - 4u + 5}\,{\rm d}u
=
\int{4\left(u - 2\right) + 7 \over \left(u - 2\right)^{2} + 1}\,{\rm d}u
=
2\ln\left(\left[u - 2\right]^{2} + 1\right)
+
7\arctan\left(u - 2\right)
\\[3mm]&=
2\ln\left(\sin^{2}\left(x\right) - 4\sin{x} + 5\right)
+
7\arctan\left(\sin\left(x\right) - 2\right)
\end{align}
$+$ a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Derivation of the quadratic equation So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
| $ax^2+bx+c=0 (a\neq0)$
$x^2+\frac{b}{a}x+\frac{c}{a}=0$
$x^2+\frac{b}{a}x=-\frac{c}{a}$
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}$
$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$
$(x+\frac{b}{2a})=\pm\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}$
$\therefore x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Definite integrals: Evaluate the integral
Evaluate:
$$ \int ^{1/2}_{1/4} \frac{dx}{ \sqrt{x-x^2}}dx$$
can u help me with this?
What is meant by the dx in the numerator?
EDIT:
ANSWER AS GIVEN IN THE BOOK
$$ \int ^{1/2}_{1/4} \frac{dx}{ \sqrt{x-x^2}}dx$$
$$=\int ^{1/2}_{1/4} \frac{1}{\sqrt {-(x^2-x+\frac{1}{4}- \frac{1}{4})}}dx$$
$$=\int ^{1/2}_{1/4} \frac{1}{\sqrt{({\frac{1}{2}})^2-(x-\frac{1}{2})^2}}$$
$$=\int ^{1/2}_{1/4} \frac{dx}{\sqrt{({\frac{1}{2}})^2-(x-\frac{1}{2})^2}}$$
$$= [ \arcsin\frac{x-1/2} {1/2} ]^{1/2}_{1/4}= \arcsin0-\arcsin\frac{-1}{2}$$
$$=\pi/6$$
Please offer your assistance :)
| $$I=\int_{\frac14}^{\frac12}\frac{dx}{\sqrt{\left(\frac12\right)^2-\left(x-\frac12\right)^2}} =\int_{\frac14}^{\frac12}\frac{2dx}{\sqrt{1^2-(2x-1)^2}}$$
Put $2x-1=\sin y\implies 2dx=\cos y dy,$
When $x=\frac12,\sin y=0\implies y=0$ and $x=\frac14,\sin y=-\frac12\implies y=-\frac\pi6$
$$\implies I=\int_{-\frac\pi6}^0\frac{\cos ydy}{\cos y}=0-\left(-\frac\pi6\right)=\frac\pi6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/516181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\frac{a+b+c}{x+y+z}$ given $a^2+b^2+c^2$, $x^2+y^2+z^2$ and $ax+by+cz$. We are given $a^2+b^2+c^2=m$, $x^2+y^2+z^2=n$ and $ax+by+cz=p$ where $m,n$ and $p$ are known constants. Also, $a,b,c,x,y,z$ are non-negative numbers.
The question asks to find the value of $\dfrac{a+b+c}{x+y+z}$.
I have thought a lot about this problem, but I can't solve it.
I can write $(a+b+c)^2$ as $a^2+b^2+c^2+2(ab+bc+ca)=m+2(ab+bc+ca)$, but then I don't know how to determine $ab+bc+ca$. I'm also not certain how to use the equation $ax+by+cz=p$ to help me in finding the given expression's value.
Any help would be much appreciated.
| I don't think this can be solved. Think of $A=(a, b, c)$ and $X=(x, y, z)$ as vectors. Then we know $A\cdot A$, $X\cdot X$ and $A\cdot X$. So all we know is the lengths of $A$ and $X$, and the angle between them. Now let $C=(1,1,1)$. What we want is
$$\frac{A\cdot C}{X\cdot C}$$
But if we hold $A$ and $C$ fixed we can still vary $X$ whilst keeping $X\cdot X$ and $A\cdot X$ the same: Just rotate $X$ around the axis given by $A$. This changes $X\cdot C$ without changing $A\cdot C$ and so it has to change the value of the expression we want to calculate. So this expression is not determined purely by the quantities given.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/518225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
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