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Maximum $\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$ where $y\in \left[0,1\right]$? How find maximum this integral $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$$ where $y\in \left[0,1\right]$?
| $$\begin{align}\frac{d}{dy}(f(y)) &= \frac{d}{dy} \Bigg(\int_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx\Bigg) \\&= \int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx + \sqrt{y^4+(y-y^2)^2} . 1\end{align}$$
Now $$\begin{align}&\int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx + \sqrt{y^4+(y-y^2)^2} = 0 \\&\Rightarrow \int_{0}^{y}\frac{(y -y^2)(1-2y)}{\sqrt{x^4+(y-y^2)^2}} dx = - \sqrt{y^4+(y-y^2)^2} \\&\Rightarrow \int_{0}^{y}\frac{1}{\sqrt{x^4+(y-y^2)^2}} dx = \frac{\sqrt{y^4+(y-y^2)^2}}{(y-y^2)(2y-1)} \end{align}$$
Check the sign of $\frac{d}{dy}f(y)$ to see if the function $f(y)$ is decreasing, increasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Sum of a series of a number raised to incrementing powers How would I estimate the sum of a series of numbers like this: $2^0+2^1+2^2+2^3+\cdots+2^n$. What math course deals with this sort of calculation? Thanks much!
| Interestingly, this sort of summation is not too difficult to prove either.
If we swap $2$ for a generic number $a$:
$$
\text{(1) } \sum\limits_{i = 1}^n {a^i} = a^0+a^1+a^2+a^3+...+a^n
$$
Multiplying $(1)$ by $a$ gives:
$$
\text{(2) } a\sum\limits_{i = 1}^n {a^i} = a^1+a^2+a^3+a^4+...+a^{n+1}
$$
Subtracting $(1)$ from $(2)$:
$$
\text{(3) } a\sum\limits_{i = 1}^n {a^i} - \sum\limits_{i = 1}^n {a^i} == (a-1)\sum\limits_{i = 1}^n {a^i}
$$
From $(1)$ and $(2)$ we can see this is the same as:
$$
\text{(4) } a^1+a^2+a^3+a^4+...+a^{n+1} - (a^0+a^1+a^2+a^3+...+a^n) = (a-1)\sum\limits_{i = 1}^n {a^i}
$$
Since all the middle bits cancel each other out, this simply leaves:
$$
\text{(5) } a^{n+1} - a^0 = (a-1)\sum\limits_{i = 1}^n {a^i}
$$
And since $a^0$ is always equal to $1$:
$$
\text{(6) } a^{n+1} - 1 = (a-1)\sum\limits_{i = 1}^n {a^i}
$$
Rearranging:
$$
\text{(7) } \frac{a^{n+1} - 1}{a-1} = \sum\limits_{i = 1}^n {a^i}
$$
And returning to the original post, where $a = 2$:
$$
\frac{2^{n+1} - 1}{2-1} = \sum\limits_{i = 1}^n {2^i} = 2^0+2^1+2^2+2^3+...+2^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
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} |
Coordinates in a rectangular triangle Numbers from $1$ to $\frac{n^2+n}{2}$ are formed into a triangle: numbers from $1$ to $n$ form first column, numbers from $n+1$ to $2n - 1$ form second and so on untill last column has just one number.
For example with $n = 5$ we get
$\begin{matrix}
1&&&&\\
2&6&&&\\
3&7&10&&\\
4&8&11&13&\\
5&9&12&14&15\\
\end{matrix}$
Is there a way to calculate coordinates of a number (it's row and column) given the number?
In other words: is there a formula $f_n: \mathbb R \rightarrow \mathbb R^2$ such that $f(x) = (a, b)$ where $a$ is column and $b$ is the row in which $x$ is (in a triangle as in example).
| If we add up entire columns we can form a trapezoid with the exact same area:
$$
\begin{align}
5&=1\cdot 5\\
9&=2\cdot\frac{5+4}2\\
12&=3\cdot\frac{5+3}2
\end{align}
$$
This can be continuously interpolated by $x=y\cdot\frac{n+(n+1-y)}{2}$ which then leads to
$$
y^2-(2n+1)y+2x=0
$$
which is a quadratic equation in $y$ with solutions
$$
y=\frac{2n+1\pm\sqrt{(2n+1)^2-8x}}{2}
$$
This $y$ should be the column number for the cases $x=5,9,12$ etc. so it has to be the solution with the minus sign since we want $y<\frac{2n+1}{2}$ for the column number to be within the range $1,2,...,n$. It follows that the column number will be given by
$$
a=\left\lceil\frac{2n+1-\sqrt{(2n+1)^2-8x}}{2}\right\rceil
$$
and then it should not be too hard to find an expression for the row number using the difference between the actual $x$ and the $\bar x$ that fills up the entire column $\bar x=a\cdot\frac{n+(n+1-a)}{2}$.
This Wolfram Alpha computation confirms that my formula for the column number works for your example with $n=5$ and thus $2n+1=11$ so that $a=\left\lceil\frac{1}{2}(11-\sqrt{11^2-8x})\right\rceil$.
The final function will be given by
$$
f(x)=\left(a(x),b(x)\right)
$$
where
$$
\begin{align}
a(x)&=\left\lceil\tfrac{1}{2}\left(2n+1-\sqrt{(2n+1)^2-8x}\right)\right\rceil\\
b(x)&=n+x-a(x)\cdot\frac{2n+1-a(x)}{2}
\end{align}
$$
Perhaps $b(x)$ can be reduced by plugging in the expression for $a(x)$ and the simplifying.
And just to demonstrate, here is a test of the $b(x)$ formula for your example of $n=5$ using Wolfram Alpha again. It works just perfectly ;).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this diophantine equation $(x^2-y)(y^2-x)=(x+y)^2$ have only three integer solution? HAPPY NEW YEAR To Everyone! (Now Beijing time 00:00 (2015))
Let $x,y$ are integer numbers,and such $xy\neq 0$,
Find this diophantine equation all solution
$$(x^2-y)(y^2-x)=(x+y)^2$$
I use Wolf found this equation only have two nonzero integer solution $(x,y)=(-1,1),(-1,-1)$,see wolf
$$\Longleftrightarrow x^3+y^3+x^2+y^2=xy(xy-1)$$
But How prove it?
and I found sometimes,and my problem almost similar with 2012 IMO shortlist:2012 IMO shortlist
| $$
\begin{array}{l}
\left( {x^2 - y} \right)\left( {y^2 - x} \right) = \left( {x + y} \right)^2 \\
\Leftrightarrow xy\left( {xy - 1} \right) = x^3 + y^3 + x^2 + y^2 \\
\Leftrightarrow xy\left( {xy - 1} \right) = x^2 \left( {1 + x} \right) + y^2 \left( {1 + y} \right) \cdots \left( * \right) \\
\left\{ \begin{array}{l}
x = y \\
\left( * \right) \Leftrightarrow x^2 - 1 = 2x\left( {1 + x} \right) \\
\left( * \right) \Leftrightarrow \left( {1 + x} \right)^2 = 0 \\
\left( * \right) \Leftrightarrow \left( {x;y} \right) \in \left\{ {\left( { - 1; - 1} \right)} \right\} \\
\end{array} \right.;\left\{ \begin{array}{l}
x = - y \\
\left( * \right) \Leftrightarrow x^2 + 1 = 2x^2 \\
\left( * \right) \Leftrightarrow \left( {x;y} \right) \in \left\{ {\left( {1; - 1} \right);\left( { - 1;1} \right)} \right\} \\
\end{array} \right. \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Does this polynomial exist? I'm looking for a polynomial $P(x)$ with the following properties:
*
*$P(0) = 0$.
*$P\left(\frac13\right) = 1$
*$P\left(\frac23\right) = 0$
*$P'\left(\frac13\right) = 0$
*$P'\left(\frac23\right) = 0$
From 1 and 3 we know that $P(x) = x\left(x - \frac23\right)Q(x)$. From 4 and 5 we know that $P'(x) = \alpha\left(x - \frac13\right)\left(x - \frac23\right)$.
$$
\begin{align}
P(x) & = \alpha\int\left(x - \frac13\right)\left(x - \frac23\right)\text{d}x \\
& = \alpha\left(\frac13x^3 - \frac12x^2 + \frac29x + \text{C}\right)
\end{align}
$$
Now 1 implies that the constant is $\text{C} = 0$, but 3 implies that the constant is $\text{C} = -\frac{2}{81}$. Am I doing something wrong or does this polynomial not exist?
If it doesn't exist, how close could I get to making a polynomial that satisfies these 5 conditions?
| This is the same solution as Ross's, but from a more linear algebra focused perspective.
You have five linearly independent conditions, so a polynomial with five parameters should work.
$P(x) = ax^4 + bx^3 + cx^2 + dx + e$
Now your five conditions can be written as such:
*
*$e = 0$
*$a(\frac{1}{3})^4 + b(\frac{1}{3})^3 + c(\frac{1}{3})^2 + d(\frac{1}{3}) + e = 1$
*$a(\frac{2}{3})^4 + b(\frac{2}{3})^3 + c(\frac{2}{3})^2 + d(\frac{2}{3}) + e = 0$
*$4a(\frac{1}{3})^3 + 3b(\frac{1}{3})^2 + 2c(\frac{1}{3}) + d = 0$
*$4a(\frac{2}{3})^3 + 3b(\frac{2}{3})^2 + 2c(\frac{2}{3}) + d = 0$
We can group these five equations together in matrix-vector form as
$\left( \begin{array}{ccccc}
0 & 0 & 0 & 0 & 1 \\
\frac{1}{81} & \frac{1}{27} & \frac{1}{9} & \frac{1}{3} & 1 \\
\frac{16}{81} & \frac{8}{27} & \frac{4}{9} & \frac{2}{3} & 1 \\
\frac{4}{27} & \frac{3}{9} & \frac{2}{3} & 1 & 0 \\
\frac{32}{27} & \frac{12}{9} & \frac{4}{3} & 1 & 0 \end{array} \right) \cdot \left( \begin{array}{c} a \\ b \\ c \\ d \\ e \end{array} \right) = \left( \begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{array} \right) $
Use a computerized linear algebra solver to find
$\left( \begin{array}{c} a \\ b \\ c \\ d \\ e \end{array} \right) = \left( \begin{array}{c} 81 \\ -108 \\ 36 \\ 0 \\ 0 \end{array} \right)$
So your polynomial is $P(x) = 81x^4 - 108x^3 + 36x^2$. This is the same one Ross Millikan found.
The advantage of this method is that it is more flexible. You could start with $P(x) = ax^{50} + bx^{42} + cx^9 + dx^2 + e$ if you wanted a find a degree 50 polynomial that would also work. You could even start with a function that wasn't polynomial at all. The disadvantage is that if you have more than 4 parameters or so then solving the system will be too cumbersome to do by hand, and you'll have to rely on a computer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How many sets of two factors of 360 are coprime to each other? My attempt:
$360=2^3\cdot3^2\cdot5^1$
Number of sets of two factor coprime sets for $2^3$ and $3^2$ only $=12+6=18$
With that if we add the effect of $ 5^1$, number of sets $=18+2\cdot 18-1=53$.
Is this ok?
The answer given is $56$.
| I'll assume you mean sets with exactly two elements, so the two factors $1$ and $1$ (represented by the singleton $\{1\}$) won't count.
Case 1: One of the factors is $1$.
$\gcd(1,n)=1$ for all $n$, so this is just the number of factors of $360$, minus $1$. The number of factors of $360=2^3\cdot3^2\cdot5$ is simply the product of one plus the exponents on the primes (think of the number of choices for the multiplicity of each prime factor), which is $4 \cdot 3 \cdot 2 = 24$.
This gives the number of sets in this case as $24-1=23$.
Case 2: Neither of the factors is $1$, but both are odd.
This means we choose a power of $3$ and a power of $5$. There are $2$ choices for a power of $3$ and $1$ for a power of $5$ (we don't count $3^0=5^0=1$ in this case). There are $2$ such sets.
Case 3: One of the factors is even, and the other is not $1$.
*
*If the even number has no odd prime factors, then there are $3$ choices for it: $2, 4,$ and $8$. Given this, there are $3 \cdot 2 -1$ choices for the other factor. This gives $15$ for this sub-case.
*If the even number is divisible by three, then there are $6$ choices for it. Given this, there is only $1$ choice for the other factor: $5$. There are $6$ possibilities in this sub-case.
*If the even number is divisible by five, then there are $3$ choices for it. Given this, there are $2$ choices for the other factor: $3$ or $9$. This gives $6$ possibilities for this sub-case.
There are a total of $15+6+6=27$ possibilites in this case.
Our total is $23+2+27=\boxed{52}$. This doesn't agree with your given answer.
| {
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"url": "https://math.stackexchange.com/questions/1087417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is meant by positive root of $x^3-x^3-1$? I am a bit confused. I think there must be a mistake.
In a text I read:
The entropy is $2\ln p$, where $$p=\frac{1}{3}\left(\sqrt[3]{\frac{29+9\sqrt{31/3}}{2}}+\sqrt[3]{\frac{29-9\sqrt{31/3}}{2}}+1\right)$$ is the positive root of $x^3-x^3-1$.
But $x^3-x^3-1=-1$... I think that must be a typo? From what is $p$ the positive root?
| There is no positive root, or any root at all, of $x^3-x^3-1=0$, since, as you noted, $x^3-x^3-1=-1$ for all $x$, and there is no $x$ that satisfies $-1=0$. This is definitely a typo.
The positive root of $$x^3-x^2-1=0$$
is exactly
$$\frac{1}{3}\left(\sqrt[3]{\frac{29+9\sqrt{31/3}}{2}}+\sqrt[3]{\frac{29-9\sqrt{31/3}}{2}}+1\right)$$
so that is almost certainly the intended polynomial.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}$ Good evening everyone,
how can I prove that
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = \frac{\pi}{\sqrt{3}}\;?$$
Well, I know that $\displaystyle\frac{1}{x^4+x^2+1} $ is an even function and the interval $(-\infty,+\infty)$ is symmetric about zero, so
$$\int_{-\infty}^{+\infty} \frac{1}{x^4+x^2+1}dx = 2\int_0^\infty \frac{1}{x^4+x^2+1}dx.$$
Then I use the partial fraction:
$$2\int_0^\infty \frac{1}{x^4+x^2+1}dx= 2\int_0^\infty \left( \frac{1-x}{2(x^2-x+1)} + \frac{x+1}{2(x^2+x+1)} \right)dx.$$
So that's all. What's next step?
| Another way to prove is to use series:
\begin{eqnarray}
\int_{-\infty}^{\infty} \frac{1}{x^4 + x^2 + 1} dx&=&2\int_{0}^{\infty} \frac{1}{x^4 + x^2 + 1} dx\\
&=&2\int_{0}^{1} \frac{1}{x^4 + x^2 + 1} dx+2\int_{1}^{\infty} \frac{1}{x^4 + x^2 + 1} dx\\
&=&2\int_{0}^{1} \frac{1}{x^4 + x^2 + 1} dx+2\int_{0}^{1} \frac{x^2}{x^4 + x^2 + 1} dx\\
&=&2\int_{0}^{1} \frac{1+x^2}{x^4 + x^2 + 1} dx\\
&=&2\int_{0}^{1} \frac{1-x^4}{1-x^6} dx\\
&=&2\int_{0}^{1}\sum_{n=0}^\infty x^{6n}(1-x^4)dx\\
&=&2\sum_{n=0}^\infty\left(\frac{1}{6n+1}-\frac{1}{6n+5}\right)\\
&=&8\sum_{n=0}^\infty\frac{1}{(6n+1)(6n+5)}\\
&=&\frac{1}{9}\sum_{n=-\infty}^\infty\frac{1}{(n+\frac{1}{2})^2+(\frac{i}{3})^2}\\
&=&\frac{1}{9}\frac{\pi\sinh(2\pi b)}{b\left(\cosh(2\pi b)-\cos(2\pi a)\right)}\bigg|_{a=-\frac12,b=\frac i3}\\
&=&\frac{\pi}{\sqrt3}.
\end{eqnarray}
Here we use a result from this post.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to solve $ \left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor $ I need some help to solve the next equation:
$$ \left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor $$
Where $ \left \lfloor \cdot \right \rfloor $ is the floor function.
What I've tried:
$$ x^2 - x - 2 - x < 1 $$
$$ x^2 - x - 2 \leq x < x^2 - x - 1 $$
When I try to solve this system, I don't get the right solution. Is it right what I'm doing? How can I do to solve this?
| $$\left \lfloor x^2 - x - 2 \right \rfloor = \left \lfloor x \right \rfloor \tag{1.}$$
Let's start with the observation that
$\left \lfloor y \right \rfloor = \left \lfloor x \right \rfloor
\iff 0 \le y-x < 1$ or $-1 \lt y-x \le 0$
The proof follows from the observation that, for any integer n,
$$ n \le x \le y < n+1 \iff n-x \le 0 \le y-x < 1 $$
Lets let $y = x^2-x-2$. One case where $(1.)$ must be true is when
\begin{align}
y &= x \\
x^2-x-2 &= x \\
x^2-2x-2 &= 0 \\
x &= 1 \pm \sqrt 3
\end{align}
In this case, $\left \lfloor y \right \rfloor=\left \lfloor x \right \rfloor = -1$
or $\left \lfloor y \right \rfloor =\left \lfloor x \right \rfloor = 2$.
Note that $y-x = x^2-2x-2$ is a decreasing function for $x \in (-\infty, -1]$ and an increasing function for $x \in [-1,-\infty)$. Note also that
$$y-x = (x-1)^2 - 3$$ and $$x-y = 3-(x-1)^2$$
We consider four cases.
*
*If $x \lt -1$, then $y-x \gt 1$.
*If $0\le x \le 1$, then $-2 \ge y-x \ge -3$.
*If $1\le x \le 2$, then $-3 \le y-x \le -2$.
*If $x \gt 3$, then $y-x \gt 1$.
In all four cases, $\left \lfloor x^2 - x - 2 \right \rfloor \ne \left \lfloor x \right \rfloor $
That leaves us with two possibilities.
If $\left \lfloor x \right \rfloor=\left \lfloor y \right \rfloor = -1$, then the possible values for $x$ are $x=-1 + h$ where $0 \le h < 1$.
Then we need to solve
\begin{array}{c}
&& \left \lfloor x^2 - x - 2 \right \rfloor &= &\left \lfloor x \right \rfloor \\
&& \left \lfloor h^2-3h \right \rfloor &=& -1 \\
&& \left \lfloor-1+(h^2-3h+1) \right \rfloor &=& -1 \\
0 &\le &h^2-3h+1 &<& 1 \\
-1 &\le &h^2-3h &<& 0 \\
\dfrac 54 &\le &h^2-3h+\dfrac 94 &<& \dfrac{9}{4} \\
\dfrac 54 &\le &\left(h-\dfrac 32\right)^2 &<& \dfrac{9}{4} \\
\dfrac{\sqrt 5}{2} &\le &\dfrac 32 - h &<& \dfrac 32 \\
\dfrac{\sqrt 5-3}{2} &\le& -h &<& 0 \\
0 &<& h &\le& \dfrac{3 - \sqrt 5}{2} \\
-1 &<& x &\le& -1 + \dfrac{3 - \sqrt 5}{2}\\
-1 &<& x &\le& \dfrac{1 - \sqrt 5}{2}\\
\end{array}
If $\left \lfloor x \right \rfloor=\left \lfloor y \right \rfloor = 2$, then the possible values for $x$ are $x=2 + h$ where $0 \le h < 1$.
\begin{array}{c}
&& \left \lfloor x^2 - x - 2 \right \rfloor &= &\left \lfloor x \right \rfloor \\
&& \left \lfloor h^2+3h \right \rfloor &=& 2 \\
&& \left \lfloor 2+(h^2+3h-2) \right \rfloor &=& 2 \\
0 &\le &h^2+3h-2 &<& 1 \\
2 &\le &h^2+3h &<& 3 \\
\dfrac{17}{4} &\le &h^2+3h+\dfrac 94 &<& \dfrac{21}{4} \\
\dfrac{17}{4} &\le &\left(h+\dfrac 32\right)^2 &<& \dfrac{21}{4} \\
\dfrac{\sqrt{17}}{2} &\le& h+\dfrac 32 &<& \dfrac{\sqrt{21}}{2} \\
\dfrac{\sqrt{17}-3}{2} &\le& h &<& \dfrac{\sqrt{21}-3}{2}\\
2 + \dfrac{\sqrt{17}-3}{2} &\le& x &<& 2+\dfrac{\sqrt{21}-3}{2} \\
\dfrac{1+\sqrt{17}}{2} &\le& x &<& \dfrac{1+\sqrt{21}}{2}
\end{array}
So the solution set is
$$x \in \left(-1, -\dfrac{\sqrt 5 - 1}{2} \right] \cup
\left[\dfrac{1+\sqrt{17}}{2}, \dfrac{1+\sqrt{21}}{2} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Coordinate of the excentre of a triangle I am just wondering that how the coordinate of the excentre comes out if we know the coordinates of vertices of the triangle.
| Consider the diagram:
$d=\overline{CE}=\overline{CF}$. Note that $c=\overline{AB}=(d-a)+(d-b)$. Therefore,
$$
d=\frac{a+b+c}2\tag{1}
$$
Furthermore, $d=\overline{CD}\cos(\theta/2)$ and $\overline{CH}=d\cos(\theta/2)$; therefore, $\overline{CH}=\overline{CD}\cos^2(\theta/2)$.
The Law of Cosines gives
$$
\cos(\theta)=\frac{a^2+b^2-c^2}{2ab}\tag{2}
$$
so that
$$
\cos^2(\theta/2)=\frac{\vphantom{b^2}1+\cos(\theta)}2=\frac{(a+b)^2-c^2}{4ab}\tag{3}
$$
$H$ is the mid-point of $\overline{EF}$; therefore,
$$
\begin{align}
H-C
&=\frac{d}2\left(\frac{B-C}a+\frac{A-C}b\right)\\[6pt]
&=\cos^2(\theta/2)(D-C)\tag{4}
\end{align}
$$
Putting together $(1)$, $(3)$, and $(4)$, we get
$$
\begin{align}
D
&=C+\frac{4ab}{(a+b)^2-c^2}\frac{a+b+c}4\left(\frac{B-C}a+\frac{A-C}b\right)\\
&=C+\frac{ab}{a+b-c}\left(\frac{B-C}a+\frac{A-C}b\right)\\
&=C+\frac{a(A-C)+b(B-C)}{a+b-c}\\[6pt]
&=\frac{aA+bB-cC}{a+b-c}\tag{5}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1091532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Calculate the limit of $(1+x2^x)/(1+x3^x)$ to the power $1/x^2$ when $x\to 0$ I have a problem with this: $\displaystyle \lim_{x \rightarrow 0}{\left(\frac{1+x2^x}{1+x3^x}\right)^\frac{1}{x^2}}$.
I have tried to modify it like this: $\displaystyle\lim_{x\rightarrow 0}{e^{\frac{1}{x^2}\ln{\frac{1+x2^x}{1+x3^x}}}}$ and then calculate the limit of the exponent: $\displaystyle \lim_{x\rightarrow 0}{\frac{1}{x^2}\ln{\frac{1+x2^x}{1+x3^x}}}$.
But I don't know what to do next. Any ideas?
| let $f(x)=\ln\left(\frac{1+x2^x}{1+x3^x}\right)$ and $g(x)=x^2$ then we have $0/0$ for $x$ tends to $0$ you must calculate $$\frac{f'(x)}{g'(x)}$$ and look if the limit exists.
you will get this here $$1/2\,{\frac {{2}^{x}\ln \left( 2 \right) {3}^{x}{x}^{2}-{2}^{x}{3}^{x
}\ln \left( 3 \right) {x}^{2}+x{2}^{x}\ln \left( 2 \right) -x{3}^{x}
\ln \left( 3 \right) +{2}^{x}-{3}^{x}}{ \left( 1+x{2}^{x} \right)
\left( 1+x{3}^{x} \right) x}}
$$
and now L'Hospital one more times
the next quotient is $${\frac {{2}^{x} \left( \ln \left( 2 \right) \right) ^{2}{3}^{x}{x}^{
2}+2\,{2}^{x}\ln \left( 2 \right) {3}^{x}x-{2}^{x}{3}^{x} \left( \ln
\left( 3 \right) \right) ^{2}{x}^{2}-2\,{2}^{x}{3}^{x}\ln \left( 3
\right) x+2\,{2}^{x}\ln \left( 2 \right) +x{2}^{x} \left( \ln
\left( 2 \right) \right) ^{2}-2\,{3}^{x}\ln \left( 3 \right) -x{3}^
{x} \left( \ln \left( 3 \right) \right) ^{2}}{2\,{2}^{x}\ln \left(
2 \right) {3}^{x}{x}^{3}+2\,{2}^{x}{3}^{x}\ln \left( 3 \right) {x}^{3
}+2\,{2}^{x}\ln \left( 2 \right) {x}^{2}+6\,{2}^{x}{3}^{x}{x}^{2}+2\,
{3}^{x}\ln \left( 3 \right) {x}^{2}+4\,x{2}^{x}+4\,x{3}^{x}+2}}
$$ and here you can set $x=0$ the result it $$\ln(2)-\ln(3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
For how many natural numbers $X(X+1)(X+2)(X+3)$ has exactly three different prime factors?
For how many natural numbers $X(X+1)(X+2)(X+3)$ has exactly three different prime factors?
My attempt:
I have used a hit and trial approach. I found out that only for x=2 and x=3 this is happening. But how can I be sure of it? Is there any general approach?
| you can use the following expission
$$\displaylines{
x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) = x\left( {x + 3} \right)\left( {\left( {x + 1} \right)\left( {x + 2} \right)} \right) \cr
= x\left( {x + 3} \right)\left( {x^2 + 3x + 2} \right) \cr}$$
$$\left( {x\left( {x + 3} \right) + 1} \right)^2 - 1 = x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1094874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplifying fractions with exponents I'm revising for an exam, which I have the solutions to. One of the questions asks me to prove that a sequence is a Cauchy sequence, sequence is written as:
$ a_n = \frac{2^{n+2}+1}{2^n},$
and then in the solutions, it has:
$\left | a_n - a_m \right | = \left | \frac{2^{n+2}+1}{2^n} - \frac{2^{m+2}+1}{2^m} \right | = \left | \frac {1}{2^n} - \frac {1}{2^m} \right |$
I am extremely rusty on my math skills, but could someone explain how the $2^{nd}$ and $3^{rd}$ absolute values are equal?
| The important equality is:
$$\frac{2^{n+2} + 1}{2^n} = \frac{2^{n+2}}{2^n} + \frac{1}{2^n} = 2^2 + \frac{1}{2^n}$$
The same process is performed on both fractions and the $2^2$ drops out. (One fraction contributes a $+4$ while the other fraction contributes a $-4$.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Trouble with definite integral calculating probabilities I cannot solve this:
$$\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \tan ^{-1}(a+\tan (x)) \, dx$$
it apeared when trying to find out the probability:
$$P\{\tan a - \tan b \leq 2x\},\ \ 0 < x < 1\sqrt{3}$$
Knowing that the joint distribution $f(a,b)$ is
$f(a,b) = \frac{2}{\pi^2}$ in the region $-\pi/2 < b < a < \pi/2$ (triangle)
"$a$" has marginal density:
$\frac{2x+\pi}{\pi^2}$ where $-\pi/2 < a < \pi/2$.
and "$b$" has density: $\frac{\pi-2x}{\pi^2}$ being $ -\pi/2 < b < a$.
| $$\tan{u} = a + \tan{x} \implies x=\arctan{(\tan{u}-a)} \implies dx = \frac{\sec^2{u}}{1+(\tan{u}-a)^2} du $$
Then the integral is equal to
$$\begin{align}\int_{-\pi/2}^{\pi/2} du \frac{u \sec^2{u}}{1+(\tan{u}-a)^2} &= \int_{-\pi/2}^{\pi/2} du \frac{u}{\cos^2{u} + (\sin{u}-a \cos{u})^2} \\ &= \frac12 \int_{-\pi}^{\pi} dv \frac{v}{2+a^2+a^2 \cos{v}-2 a \sin{v}} \end{align}$$
Let
$$J(\beta) = -\frac{i}{2} \int_{-\pi}^{\pi} dv \frac{e^{i \beta v}}{2+a^2+a^2 \cos{v}-2 a \sin{v}} $$
Then the integral we seek is $J'(0)$.
Consider the integral
$$-\oint_C dz \frac{z^{\beta}}{(a^2+i 2 a) z^2+(a^2+2) z+(a^2-i 2 a)} $$
where $C$ is the contour pictured below:
i.e., a unit circle with a keyhole about the negative real axis. The contour integral is then
$$J(\beta) - i 2\sin{\pi \beta} \int_0^1 dx \frac{x^{\beta}}{(a^2+i 2 a) x^2-(a^2+2) x+(a^2-i 2 a)}$$
This is also equal to $-i 2 \pi$ times the residue at the pole $z_0=-a/(a+i 2)$ inside $C$, which is $z_0^{\beta}/2$. Thus,
$$\begin{align}J'(0) &= i 2 \pi \int_0^1 \frac{dx}{(a^2+i 2 a) x^2-2 (a^2+2) x+(a^2-i 2 a)} - i \frac{\pi}{2} \log{z_0} \\ &= \frac{1}{2} \pi \arctan\left(\frac{a}{2}\right)-\frac{1}{4} i \pi \log \left(\frac{4}{a^2}+1\right) + \frac{1}{2} \pi \arctan\left(\frac{a}{2}\right)+\frac{1}{4} i \pi \log \left(\frac{4}{a^2}+1\right) \\ &= \pi \arctan{\left ( \frac{a}{2}\right )} \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factor the polynomial $z^5 + 32$ in real factors The question that I have trouble solving is the following:
Factor the polynomial $z^5 + 32$ in real factors. The answer should not use trigonometric functions. (Hint: you are allowed to use the fact that: $cos(\pi/5) = \frac{1+\sqrt{5}}{4}$ and $cos(3\pi/5) = \frac{1-\sqrt{5}}{4}$.
In the previous question you were supposed to find the complex roots to the equation $z^5 = -32$ in polar form, resulting in the roots:
$$z_1 = 2e^{i\frac{\pi}{5}}$$$$z_2 = 2e^{i\frac{3\pi}{5}}$$ $$z_3 = 2e^{i\pi} = -2$$ $$z_4 = 2e^{i\frac{7\pi}{5}}$$$$ z_5 = 2e^{i\frac{9\pi}{5}} $$
My attempt at a solution:
First we know that there were only one real root to the equation $z^5 = -32$, namely $-2$.
So if we write the polynomial as the equation: $z^5 + 32 = 0$ we know that $(z+2)$ must be a factor and we have four potential factors left to find. Since we know that the polynomial $z^5 +32$ only has real coefficients, the non real roots must come in pairs.
We should then find two roots that are composed of conjugates of the roots with imaginary components to produce the other two real roots... But I am stuck here.
| HINT:
$$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$$
$$a^4-a^3b+a^2b^2-ab^3+b^4=(a^2+b^2)^2-ab(a^2+b^2)-a^2b^2$$
If $ab\ne0,$
$$(a^2+b^2)^2-ab(a^2+b^2)-a^2b^2=a^2b^2\left[\left(\frac{a^2+b^2}{ab}\right)^2-\frac{a^2+b^2}{ab}-1\right]$$
So, we can express $a^2+b^2$ in terms of $ab$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1097856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Pythagorean triple problem I am doing research on perfect cuboids, and I'm looking for values $a,b,c$ such that the following is integer, and I'm not sure how to continue this. Any suggestions are appreciated!
$PED$ is a very large constant=$899231100768000$
$$
\begin{align}
&\exp\left(\sigma_1+\sigma_2+\frac{\ln(a^2+b^2+c^2)}{2}-\ln(PED)
\right)\in\mathbb Z\\
&\sigma_1=\ln a+\ln b+\ln c\\
&\sigma_2=\frac{\ln(a^2+b^2)}{2}+\frac{\ln(b^2+c^2)}{2}+\frac{\ln(a^2+c^2)}{2}
\end{align}
$$
Another way to write this is:
$$
\begin{align}
&\frac {abc*\sqrt{a^2+b^2}*\sqrt{a^2+c^2}*\sqrt{b^2+c^2}*\sqrt{a^2+b^2+c^2}}{899231100768000}\in\mathbb Z\\
\end{align}
$$
| assuming $a^2+b^2$, $a^2+c^2$, $b^2+c^2$, and $a^2+b^2+c^2$ are perfect squares, then it's impossible.
$\sqrt{(a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2)}$
must be an integer, therefor
$(a^2+b^2)(a^2+c^2)(b^2+c^2)(a^2+b^2+c^2)$
must be a square, which allows us to assume that all terms are squares, and that makes the first 3 terms Pythagorean triples.
one method of generating Pythagorean triples is
for $A^2+B^2=C^2$
$A=2mn$
$B=m^2-n^2$
$C=m^2+n^2$
$n$ and $m$ must be integers to insure that $A$, $B$, and $C$ are integers
Note: every Pythagorean triple must include $A$, $B$, and $C$
if $a=A$
and $b=B$
then, because $a^2+c^2$ must equal $A^2+B^2$, $c$ must be $B$
and because $b^2+c^2$ must equal $B^2+A^2$, $c$ must be $A$
therefor $c=A$ and $c=B$. Which means
$A=B$
$2mn=m^2-n^2$
divide both sides by $mn$, simplify,
$2=\frac mn-\frac nm$
and since
$\frac nm = \frac 1{\frac mn}$
you have
$2=\frac mn-\frac 1{\frac mn}$
then if you introduce a new variable $x$
$x=\frac mn$
substitute $\frac mn$
$2=x-\frac 1x$
then multiply both sides by $x$,
$2x=x^2-1$
$0=x^2-2x-1$
and use the quadratic equation, you get.
$x=\frac {2\pm\sqrt{8}}2$
which you can simplify further into
$1+\sqrt{2}$ and $1-\sqrt{2}$
which means
$\frac mn=1+\sqrt{2}$
or
$\frac mn=1-\sqrt{2}$
if we rearrange $A=2mn$, to get $\frac {A}{2n^2}=\frac mn$
then substitute $\frac mn$ and simplify, we see that
$A=2n^2(1+\sqrt{2})$
or
$A=2n^2(1-\sqrt{2})$
therefor, if $c$ is $A$ and $B$ (which it must be) then $c$ is irrational
you can do the same thing with $a$ and $b$, you will get the same contradiction.
ps. with the method of finding Pythagorean triples that I used, if you find one then you have found an infinite list of them by multiplying $a$, $b$, and $c$ by some integer $k$. If you do this for every integer pare $m$,$n$ you get almost all triples, the only ones missing are when $k=\frac {1}{2}$, but including that gets every triple that exists!
(including the ones that just flip $A$ and $B$ around) here is a video that can explain it better than I could.
ex: $m=2$, $n=1$
$3^2+4^2=5^2$, $k=1$
$6^2+8^2=10^2$, $k=2$
$9^2+12^2=15^2$, $k=3$
.
.
.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1099501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Closed form for ${\large\int}_0^\infty\frac{x-\sin x}{\left(e^x-1\right)x^2}\,dx$ I'm interested in a closed form for this simple looking integral:
$$I=\int_0^\infty\frac{x-\sin x}{\left(e^x-1\right)x^2}\,dx$$
Numerically,
$$I\approx0.235708612100161734103782517656481953570915076546754616988...$$
Note that if we try to split the integral into two parts, each with only one term in the numerator, then both parts will be divergent.
| This is another solution just for reference.
The laplace transform of $\displaystyle\frac{x-\sin{x}}{x^2}$ is given by
\begin{align}
\mathcal{L}_s\left(\frac{x-\sin{x}}{x^2}\right)
&=\int^s_\infty\int^t_\infty\frac{1}{u^2}-\frac{1}{1+u^2}\ du\ dt\tag1\\
&=\int^s_\infty-\frac{1}{t}-\arctan{t}+\frac{\pi}{2}\ dt\\
&=-\ln{s}+\frac{1}{2}\ln(1+s^2)-s\arctan{s}+\frac{\pi}{2}s-1\\
&=\frac{1}{2}\ln\left(1+\frac{1}{s^2}\right)+s\arctan\left(\frac{1}{s}\right)-1
\end{align}
Thus
\begin{align}
\int^\infty_0\frac{x-\sin{x}}{x^2(e^x-1)}dx
&=\sum^\infty_{n=1}\mathcal{L}_n\left(\frac{x-\sin{x}}{x^2}\right)\tag2\\
&=\frac{1}{2}\sum^\infty_{n=1}\ln\left(1+\frac{1}{n^2}\right)+\sum^\infty_{n=1}\left(n\arctan\left(\frac{1}{n}\right)-1\right)\\
&=\frac{1}{2}\ln\left.\frac{\pi z}{\pi}\prod^\infty_{n=1}\left(1+\frac{z^2}{n^2}\right)\right|_{z=1}+\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^k}{(2k+1)n^{2k}}\tag{3}\\
&=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\sum^\infty_{k=1}\frac{(-1)^k\zeta(2k)}{(2k+1)}\tag4\\
&=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\frac{1}{2i}\int^{i}_0\left(1-\pi z\cot(\pi z)\right)\ dz\tag5\\
&=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\frac{1}{2}-\frac{1}{8\pi}\int^{\exp(-2\pi)}_1\frac{\ln{u}(1+u)}{u(1-u)}du\tag6\\
&=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\frac{1}{2}-\frac{1}{8\pi}\left[2\mathrm{Li}_2(1-u)+\frac{\ln^2{u}}{2}\right]^{\exp(-2\pi)}_1\tag7\\
&=\color{#E2062C}{\frac{1}{2}-\frac{\pi}{4}+\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)-\frac{1}{4\pi}\mathrm{Li}_2\left(1-e^{-2\pi}\right)}
\end{align}
Explanation:
$(1)$: Differentiated under the integral twice.
$(2)$: Expanded $(e^{x}-1)^{-1}$.
$(3)$: Expanded $\arctan\left(n^{-1}\right)$.
$(4)$: Recognised the Weierstrass product for $\sinh$, summed in $n$.
$(5)$: Used the fact that $\displaystyle\pi z\cot(\pi z)=1-2\sum^\infty_{k=1}\zeta(2k)z^{2k}$.
$(6)$: Substitued $u=e^{2\pi iz}$.
$(7)$: $\displaystyle \frac{\ln{u}(1+u)}{u(1-u)}=\frac{2\ln{u}}{1-u}+\frac{\ln{u}}{u}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1100368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
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Taylor expansion about a point I need help with the following calculus problem:
Use completing the square and the geometric series to get the Taylor expansion about ${x=2}$ of ${\frac{1}{x^{2}+4x+3}}$
So far I have the following:
By completing the square. It can be shown that
${\frac{1}{x^{2}+4x+3} = \frac{1}{(x+2)^{2}-1}}$
We want to use the geometric series, so we recognize that
${\frac{1}{(x+2)^{2}-1} = -1\times\frac{1}{1-(x+2)^{2}}}$
Using what we know about the geometric series, it is seen that
${\frac{1}{1-(x+2)^2} = \sum\limits_{k=0}^\infty (x+2)^{2k}}$ and our Taylor series about ${x = -2}$ is ${-\sum\limits_{0}^{\infty}(x+2)^{2k}}$
I am stuck, however on figuring out how to use this expansion about ${x=-2}$ to find the expansion about ${x=2}$. How is this done, without calculating the derivative at ${x=2}$ of the original function and trying to find a pattern for the summation?
| $$\frac{1}{x^2+4x+3}=\frac{1}{(x+1)(x+3)}=\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}\right)=\frac{1}{2}\left(\frac{1}{(x-2)+3}-\frac{1}{(x-2)+5}\right)$$
Now just use:
$$\frac{1}{z+3}=\frac{\frac{1}{3}}{1+\frac{z}{3}}=\sum_{n\geq 0}\frac{(-1)^n}{3^{n+1}}z^n $$
to get:
$$\frac{1}{x^2+4x+3}=\frac{1}{2}\sum_{n\geq 0}\left(\frac{1}{3^{n+1}}-\frac{1}{5^{n+1}}\right)(-1)^n (x-2)^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1101836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove the following inequality without using differentiation Let $a,b,c$ be real numbers that satisfy $0\le a,b,c\le 1$. Show that
$$\frac a{b+c+1} + \frac b{a+c+1} + \frac c{a+b+1} + (1-a)(1-b)(1-c) \le 1.$$
I don't know where to start. Multiplying everything by the denominators creates extreme mess.
| WLOG: $a\le b\le c$, then
\begin{align*}&\dfrac{a}{b+c+1}+\dfrac{b}{c+a+1}+\dfrac{c}{a+b+1}+(1-a)(1-b)(1-c)\\
&\le\dfrac{a}{a+b+1}+\dfrac{b}{a+b+1}+\dfrac{c}{a+b+1}+(1-a)(1-b)(1-c)\\
&=\dfrac{a+b+c}{a+b+1}+\dfrac{(a+b+1)(1-a)(1-b)(1-c)}{a+b+1}\\
&\le\dfrac{a+b+c}{a+b+1}+\dfrac{(1+a)(1+b)(1-a)(1-b)(1-c)}{a+b+1}\\
&=\dfrac{a+b+c}{a+b+1}+\dfrac{(1-a^2)(1-b^2)(1-c)}{a+b+1}\\
&\le\dfrac{a+b+c}{a+b+1}+\dfrac{(1-c)}{a+b+1}\\
&=1
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Combination questions Q: A team of $11$ is to be chosen out of $15$ cricketers of whom $5$ are bowlers and $2$ others are wicket keepers. In how many ways can this be done so that the team contains at least $4$ bowlers and at least $1$ wicket keeper?
| $$\binom{5}{4}\binom{2}{1}\binom{8}{6}+\binom{5}{5}\binom{2}{1}\binom{8}{5}+\binom{5}{4}\binom{2}{2}\binom{8}{5}+\binom{5}{5}\binom{2}{2}\binom{8}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1108832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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My proof of: $|x - y| < \varepsilon \Leftrightarrow y - \varepsilon < x < y + \varepsilon$ Is it reasonable to prove the following (trivial) theorem?
If yes, is there a better way to do it?
Let $x, y \in \mathbb{R}$.
Let $\varepsilon \in \mathbb{R}$ with $\varepsilon > 0$.
$\textbf{Theorem.}$
We have
\begin{equation*}
\left| x - y \right| < \varepsilon
\quad \Longleftrightarrow \quad
y - \varepsilon < x < y + \varepsilon.
\end{equation*}
$\Large \textit{Proof.}$
$\large \textit{Subproof } \Rightarrow .$
$\textbf{Case: } 0 \le x - y . \quad$
Since both
$0 \le x - y$ and
$\left| x - y \right| < \varepsilon$, we have
$x - y < \varepsilon$.
That is, $x < y + \varepsilon$.
The rest of the case proves $y - \varepsilon < x$.
We have $0 \le x - y$.
That is, $y \le x$.
Thus, since $0 < \varepsilon$,
we have $0 + y < \varepsilon + x$.
That is, $y - \varepsilon < x$.
$\textbf{Case: } 0 > x - y . \quad$
Since both
$0 > x - y$ and
$\left| x - y \right| < \varepsilon$, we have
$y - x < \varepsilon$.
That is, $y - \varepsilon < x$.
The rest of the case proves $x < y + \varepsilon$.
We have $0 > x - y$.
That is, $x < y$.
Thus, since $0 < \varepsilon$,
we have $0 + x < \varepsilon + y$.
That is, $x < y + \varepsilon$.
$\large \textit{Subproof } \Leftarrow .$
$\textbf{Case: } 0 \le x - y . \quad$
We have $x < y + \varepsilon$.
That is, $x - y < \varepsilon$.
Also, since $0 \le x - y$, we have $x - y = |x - y|$.
Thus, $|x - y| < \varepsilon$.
$\textbf{Case: } 0 > x - y . \quad$
We have $y - \varepsilon < x$.
That is, $y - x < \varepsilon$.
Also, since $0 > x - y$, we have $y - x = |x - y|$.
Thus, $|x - y| < \varepsilon$.
QED
| This is correct. A neater way of putting it would be to note that
$$
\begin{align}
\mid x-y \mid < \epsilon &\iff x-y \lt \epsilon \text{ and } y-x \lt \epsilon\\
&\iff x-y \lt \epsilon \text{ and } x-y \gt -\epsilon\\
&\iff-\epsilon \lt x-y \lt \epsilon \\
&\iff y-\epsilon \lt x \lt y+\epsilon
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Integral $\int \frac{x+2}{x^3-x} dx$ I need to solve this integral but I get stuck, let me show what I did:
$$\int \frac{x+2}{x^3-x} dx$$
then:
$$\int \frac{x}{x^3-x} + \int \frac{2}{x^3-x}$$
$$\int \frac{x}{x(x^2-1)} + 2\int \frac{1}{x^3-x}$$
$$\int \frac{1}{x^2-1} + 2\int \frac{1}{x^3-x}$$
now I need to resolve one integral at the time so:
$$\int \frac{1}{x^2-1}$$ with x = t I have:
$$\int \frac{1}{t^2-1}$$
Now I have no idea about how to procede with this...any help?
| $$\begin{gathered}
\frac{{x + 2}}
{{{x^3} - x}} = \frac{{x + 2}}
{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{A}
{x} + \frac{B}
{{x - 1}} + \frac{C}
{{x + 1}} \hfill \\
= \frac{{A\left( {{x^2} - 1} \right) + Bx\left( {x + 1} \right) + Cx\left( {x - 1} \right)}}
{{x\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{\left( {A + B + C} \right){x^2} + \left( {B - C} \right)x - A}}
{{x\left( {x - 1} \right)\left( {x + 1} \right)}} \hfill \\
\Rightarrow \left\{ \begin{gathered}
A + B + C = 0 \hfill \\
B - C = 1 \hfill \\
- A = 2 \hfill \\
\end{gathered} \right. \Rightarrow \left\{ \begin{gathered}
A = - 2 \hfill \\
B = 3/2 \hfill \\
C = 1/2 \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$?
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$
This thing doesn't make sense how should I use first identity to find the second one.
| The response of @TonyK is perfect and exemplary. Since some seem not to follow it, let me reinterpret it. By using Euclidean Division on $X^2+X+1$ and $X^3$, you get the identity $X^3=(X^2+X+1)(X-1) + 1$. This identity is still valid when any number is substituted for $X$. In particular, when you substitute that $x$ for which $x^2+x+1=0$. So $x^3=1$, and $1/x^3=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1110433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Proof that a Limit equals a Function Let f(x) = $\frac{x}{|x|}$ if x $\neq$ 0, and define f(0)=0. Show that
f(x) = $\lim_{n \rightarrow \infty} \frac{2}{\pi} \tan^{-1}(nx)$
My work:
$\frac{x}{|x|}$ = $\lim_{n \rightarrow \infty} \frac{2}{\pi} \tan^{-1}(nx)$
$\frac{x}{|x|}$ = $\lim_{n \rightarrow \infty} \frac{2}{\pi} \frac{cos(nx)}{sin(nx)}$
$\lim_{n \rightarrow \infty} cos(n)$ = -1 and 1, likewise $\lim_{n \rightarrow \infty} sin(n)$ = -1 and 1.
So when $\frac{-1}{-1}$ and $\frac{1}{1}$ then the limit is equal to 1. Likewise when $\frac{-1}{1}$ and $\frac{1}{-1}$ then the limit is equal to -1.
f(x) always equals either 1 or -1 so they can be confirmed to be equivalent. However, I'm having problems with the $\frac{2}{\pi}$ and the effects of x in the limit.
Hints please! Thanks!
| If $x = 0$, $\frac{2}{\pi} \tan^{-1}(nx) = 0$ for all $n$, and thus $\lim_{n\to \infty} \frac{2}{\pi}\tan^{-1}(nx) = 0$ when $x = 0$. If $x > 0$, then $$\lim_{n\to \infty} \frac{2}{\pi} \tan^{-1}(nx) = \frac{2}{\pi}\lim_{u\to +\infty} \tan^{-1}(u) = \frac{2}{\pi}\cdot\frac{\pi}{2} = 1 = \frac{x}{|x|}.$$ If $x < 0$, then $$\lim_{n\to \infty} \frac{2}{\pi}\tan^{-1}(nx) = \frac{2}{\pi} \lim_{u\to -\infty} \tan^{-1}(u) = -1 = \frac{x}{|x|}.$$ Hence, for all $x\neq 0$, $\lim_{n\to \infty} \frac{2}{\pi}\tan^{-1}(nx) = \frac{x}{|x|}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit $\lim_{x\to\infty}(x(\log(1+\sqrt{1+x^2}-\log(x)))$ can someone give me a hint/solution for:
$$\lim_{x\to\infty}\left(x(\log(1+\sqrt{1+x^2}-\log(x))\right)$$
Shall I do a derivative ?
But there's no L'Hospital to use here..
Shall I change its form ?
$x(\log(1+(1+x^2)^{1/2})-\log(x))$ (minimal change)
Or what shall I do ?
Thanks
| for $x$ large
$\begin{align}
\ln(1+\sqrt{x^2 + 1}) - \ln x &= \ln \left( 1+ x(1+1/x^2)^{1/2} \right) - \ln x\\
&=\ln[ 1 + x(1 + \frac{1}{2x^2} + \cdots )] - \ln x\\
&= \ln[x(1 + \frac{1}{x} + \frac{1}{2x^2}+\cdots)] -\ln x\\
&= \ln(1 + \frac{1}{x} + \frac{1}{2x^2}+\cdots) \\
&= \frac{1}{x} +\cdots
\end{align}$
therefore $$\lim_{x \to \infty} x \{\ln(1+\sqrt{x^2 + 1}) - \ln x \} = 1 .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1112972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Limit $\lim_\limits{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)}$ Evaluate the given limit:
$$\lim_{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)} .$$
I've tried to evaluate it but I always get stuck... Obviously I need L'Hôpital's Rule here, but still get confused on the way. May someone show me what is the trick here?
Thanks.
| L'Hospital's rule:
$$\lim_{x\to0} \frac{\ln(x+\sqrt{1+x^2})-x}{\tan^3(x)}=\lim_{x\to0} \frac{\ln(x+\sqrt{1+x^2})-x}{x^3}\cdot\frac{\tan^3 x}{x^3}=$$
$$=\lim_{x\to0} \frac{(\ln(x+\sqrt{1+x^2})-x)'}{(x^3)'}=\lim_{x\to0}\frac{\frac{1}{\sqrt{1+x^2}}-1}{3x^2}=\lim_{x\to0}\frac{-x^2}{3x^2\sqrt{1+x^2}(1+\sqrt{1+x^2})}=-\frac{1}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1113338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
Determining $\gcd(94, 27)$ I want to determine $\gcd(94, 27)$. Using the Euclidean algorithm, I got
\begin{align}
94 &= 27 (3) + 13 \\
\implies 27 &= 13 (2) + 1 \\
\implies \;\;2 &= 2 (1)
\end{align}
Does this mean the GCD is $2$? Clearly $2$ doesn't divide $27$, so what am I doing wrong?
| As mentioned in the comments, the last non-zero remainder is the GCD, not the quotient corresponding to the expression with zero remainder. To highlight, that is the red/boxed number below:
\begin{align}
94 &= 27 (3) + 13 \\
27 &= 13 (2) + \color{red}{\boxed{1}} \\
13 &= 13 (1) + 0
\end{align}
For another example, consider $\gcd(45, 81)$:
\begin{align}
81 &= (1)45 + 36 \\
45 &= (1)36 + \color{red}{\boxed{9}} \\
36 &= (4)9 + 0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1114275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove relations between the roots of 3 quadratic equations
Let $x_1, x_2$ be the roots of the equation $x^2 + ax + bc = 0$, and $x_2, x_3$ the roots of
the equation $x^2 + bx + ac = 0$ with $ac \neq bc$. Show that $x_1, x_3$ are the roots of the
equation $x^2 + cx + ab=0$.
From Vieta's I have:
$\begin{cases} x_1+x_2=-a\\x_1x_2=bc\end{cases}$ $\begin{cases} x_2+x_3=-b\\x_2x_3=ac\end{cases}$
and I have to prove:
$\begin{cases} x_1+x_3=-a\\x_1x_3=ab\end{cases}$
| Subtract the first two equations to get $x_2$
$$\begin{align}x_2^2+ax_2+bc&=0\\x_2^2+bx_2+ac&=0\end{align}$$ to get $$(a-b)x_2-(a-b)c=0$$ from where $x_2=c$.
So, $$\begin{align}x_1&=b=-a-c\\x_3&=a=-b-c\end{align}$$ from where $$\begin{align}x_1+x_3&=a+b=c\\x_1x_3&=ab\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1116956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_0^a{f(x)}\, dx$ SMT 2013 Calculus #8:
The function $f(x)$ is defined for all $x\ge 0$ and is always nonnegative. It has the additional property that if any line is drawn from the origin with any positive slope $m$, it intersects the graph $y=f(x)$ at precisely one point, which is $\frac{1}{\sqrt{m}}$ units from the origin. Let a be the unique real number for which f takes on its maximum value at $x=a$ (you may assume that such an a exists). Find $\int_{0}^{a}f(x) \, dx$.
My work so far:
I tried interpreting the question in the following way:
"if any line is drawn from the origin with any slope $m$, it intersects the graph $y = f(x)$ at precisely one point, which is $\frac{1}{\sqrt{m}}$ units from the origin"
This can be parameterized as $$\begin{cases} y = mx \\ x^2 + y^2 = \frac{1}{m}\end{cases}$$
Then to find the maximum value $x = a$, you simply maximize the function $y = \sqrt{\frac{m}{1+m^2}}$ (which is the same as maximizing the function $\frac{m}{1+m^2}$).
$\left(\frac{m}{1+m^2}\right)' = 0 \implies m = 1 \implies$ the maximum point is $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$, so $ a=\frac{1}{\sqrt{2}}$.
After this point, I don't know how to compute the integral $$\int_0^\frac{1}{\sqrt{2}}{f(x)}\, dx.$$ (I tried using polar coordinates, but then I got confused.)
Edit 2: I have found an algebraic solution! It involved solving $x$ in terms $y$ (too tired to type out the entire solution). But if anyone can explain step-by-step a solution using polar coordinates, that would be greatly appreciated :)
| First Approach
If $y=mx$ and $x^2+y^2=\frac1m$, then $x^2+y^2=\frac xy$, which implies
$$
yx^2-x+y^3=0\tag{1}
$$
Taking the implicit derivative of $(1)$:
$$
y'=\frac{1-2xy}{x^2+3y^2}\tag{2}
$$
$y'=0$ means $1=2xy$; then applying $y=mx$, we get
$$
1=2xy=2mx^2\tag{3}
$$
applying $x^2+y^2=\frac1m$ gives
$$
1+2my^2=2mx^2+2my^2=2m\tag{4}
$$
using $y=mx$ yields
$$
m^2=\frac{y^2}{x^2}=2m-1\tag{5}
$$
therefore,
$$
m=1\tag{6}
$$
so we have $a=\frac1{\sqrt2}$.
Applying the quadratic formula to $(1)$:
$$
x=\frac{1-\sqrt{1-4y^4}}{2y}\tag{7}
$$
Integration by parts and substituting $z^2=1-4y^4$ gives
$$
\begin{align}
\int_0^{1/\sqrt2}y\,\mathrm{d}x
&=\frac12-\int_0^{1/\sqrt2}x\,\mathrm{d}y\\
&=\frac12-\int_0^{1/\sqrt2}\frac{1-\sqrt{1-4y^4}}{2y}\,\mathrm{d}y\\
&=\frac12-\frac18\int_0^{1/\sqrt2}\frac{1-\sqrt{1-4y^4}}{4y^4}\,\mathrm{d}4y^4\\
&=\frac12-\frac18\int_0^1\frac{1-z}{1-z^2}2z\,\mathrm{d}z\\
&=\frac12-\frac14\int_0^1\frac{z}{1+z}\,\mathrm{d}z\\
&=\frac12-\frac14\int_0^1\left(1-\frac1{1+z}\right)\,\mathrm{d}z\\
&=\frac14(1+\log(2))\tag{8}
\end{align}
$$
Polar Coordinate Approach
By virtue of the work done above, we know that the curve extends from $m=1$ to $m=\infty$, that is $\theta=\frac\pi4$ to $\theta=\frac\pi2$. There is also the triangle below that is not bounded by the curve.
Note that $r^2=x^2+y^2=\frac1m$ and $\tan(\theta)=\frac yx=m$. Therefore,
$$
r=\frac1{\sqrt{\tan(\theta)}}\tag{9}
$$
The region is given by $(9)$ for $\frac\pi4\le\theta\le\frac\pi2$, and for $0\le\theta\le\frac\pi4$ the region is a right triangle with legs of $\frac1{\sqrt2}$. Thus the area is
$$
\begin{align}
\color{#C000FF}{\frac12\cdot\frac1{\sqrt2}\cdot\frac1{\sqrt2}}+\color{#00A000}{\int_{\pi/4}^{\pi/2}\frac12r^2\,\mathrm{d}\theta}
&=\frac14+\frac12\int_{\pi/4}^{\pi/2}\frac{\mathrm{d}\theta}{\tan(\theta)}\\
&=\frac14+\frac12\left[\vphantom{\int}\log(\sin(\theta))\right]_{\pi/4}^{\pi/2}\\
&=\frac14(1+\log(2))\tag{10}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1117627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is there an obvious reason why $4^n+n^4$ cannot be prime for $n\ge 2$? I searched a prime of the form $4^n+n^4$ with $n\ge 2$ and did not find one with
$n\le 12\ 000$.
*
*If $n$ is even, then $4^n+n^4$ is even, so it cannot be prime.
*If $n$ is odd and not divisible by $5$ , then $4^n+n^4\equiv (-1)+1\equiv 0 \pmod 5$.
So, $n$ must have the form $10k+5$.
For $n=35$ and $n=55$, the number $4^n+n^4$ splits into two primes
with almost the same size.
So, is there an obvious reason (like algebraic factors) that there is no prime I am looking for ?
| An alternative to @labbhattacharjee’s exemplary response (for the case of odd $n$ only):
(1) $X^4 + 4=(X^2-2X+2)(X^2+2X+2)$; (2) $X^4+4a^4=(X^2-2aX+2a^2)(X^2+2aX+2a^2)$; (3) $X^4+4^{2k+1}=X^4+4\cdot2^{4k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Probability of Seven (Distinct) car accidents occurred on the same day Seven (Distinct) car accidents occurred in a week. What is probability that they all occurred on the same day?
My Solution:
All 7 accident occurs in 1 day in $\binom{7}{1}$ ways
All 7 accident occurs in 2 days in $\binom{7}{2}$ ways
All 7 accident occurs in 3 days in $\binom{7}{3}$ ways
All 7 accident occurs in 4 days in $\binom{7}{4}$ ways
All 7 accident occurs in 5 days in $\binom{7}{5}$ ways
All 7 accident occurs in 6 days in $\binom{7}{6}$ ways
All 7 accident occurs in 7 days in $\binom{7}{7}$ ways
Hence P(They all occurred on the Same day)$=\frac{\binom{7}{1}}{\binom{7}{1}+\binom{7}{2}+\binom{7}{3}+\binom{7}{4}+\binom{7}{5}+\binom{7}{6}+\binom{7}{7}}$
$=\frac{7}{2^7}$
Is my analysis correct or I've assumed something wrong
| The probability that $1$ accident occurs on a given day is $\frac17$
The probability that $7$ accidents occur on a given day is $(\frac17)^7$
The probability that $7$ accidents occur on the same day is $(\frac17)^7\cdot7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1121842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve this infinitely nested radical? $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}$
Apparently, the answer is 3.
| Hint: Observe $$(x+1)^2 - 1 = x(x+2),$$ or $$x+1 = \sqrt{1 + x(x+2)}.$$ Now recursively substitute: $$\begin{align*} x+1 &= \sqrt{1 + x\sqrt{1 + (x+1)(x+3)}} , \\ &= \sqrt{1 + x \sqrt{1 + (x+1)\sqrt{1 + (x+2)(x+4)}}}, \\ &= \sqrt{1 + x \sqrt{1 + (x+1)\sqrt{1 + (x+2)\sqrt{1 + (x+3)(x+5)}}}}, \ldots \end{align*}$$ Of course, you need to do something much more rigorous to complete the proof. I leave that to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1122097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $ Let $a,b,c\in\mathbb{R^+}$ such that $ a^2 + b^2 + c^2 = 3 $. Prove that
$$
(a+b+c)(a/b + b/c + c/a) \geq 9.
$$
My Attempt
I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.
| The homogeneous form of the inequality is
$$(a+b+c)^2(ca^2+ab^2+bc^2)^2\geq 27(a^2+b^2+c^2)\,a^2 b^2 c^2 $$
that is equivalent to:
$$ (a^2+b^2+c^2+2ab+2bc+2ac)(c^2 a^4+a^2 b^4 + b^2 c^4 + 2 a^3 b^2 c+2 b^3 c^2 a+2 c^3 a^2 b) \geq 27(a^4 b^2 c^2+a^2 b^4 c^2 + a^2 b^2 c^4) $$
that can be proved by combining Schur's inequality with Muirhead's inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Given that a,b,c are distinct positive real numbers, prove that (a + b +c)( 1/a + 1/b + 1/c)>9
Given that $a,b,c$ are distinct positive real numbers, prove that $(a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)>9$
This is how I tried doing it:
Let $p= a + b + c,$ and $q=\frac1{a}+ \frac1{b} + \frac1{c}$.
Using AM>GM for $p, q$, I get:
$$\frac{p+q}{2} > {(pq)}^{1/2}$$
$$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} < \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$
And for any $x\in \mathbb{R}, \space \space x+\frac1{x}≥2.$
Thus,
$$(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)<9, $$
which is the opposite of what had to be proven. What did I do wrong?
| Consider
$$(a-b)^2 \ge 0$$
$$a^2-2ab+b^2 \ge 0$$
$$a^2+b^2 \ge 2ab$$
$$\frac{a^2+b^2}{ab} \ge 2$$
$$\frac{a}{b}+\frac{b}{a}\ge 2$$
without a loss of generality, we can also say that
$$\frac{a}{c}+\frac{c}{a}\ge 2$$
and
$$\frac{b}{c}+\frac{c}{b}\ge 2$$
of course
$$3=1+1+1=\frac{a}{a}+\frac{b}{b}+\frac{c}{c}$$
So, putting it all together we have
$$\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}\ge 3+2+2+2$$
$$(a+b+c)\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)\ge 9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1124812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Where am I going wrong in my linear Diophantine solution? Let $-2x + -7y = 9$. We find integer solutions $x, y$.
These solutions exist iff $\gcd(x, y) \mid 9$. So,
$-7 = -2(4) + 1$
then
$-2 = 1(-2)$
so the gcd is 1, and $1\mid9$. OK.
In other words,
$(-7)(1) + (-2)(-4) = 1$
Multiply by 9 on each side to get
$(-7)(9) + (-2)(-36) = 9$
$\begin{align} a = -7\\
x_0 = 9\\
b = -2\\
y_0 = -36\\
\end{align}$
Then we end up with
$x' = 9 - 7t, y' = -36 + 2t$
But when $t = 2$, we get $x'(2) = 9 - 7(2) = -5$ and $y'(2) = -36 + 2(2) = -32$ but clearly that's not a solution....
Where am I going wrong?
| 1125588
$-2x-7y=9$
$2x+7y=-9$
$2x=-9-7y$
$x={-\frac{9+7y}2}=-5-4y+{\frac{1+y}2}$
New variable: $a={\frac{1+y}2}$
$y={2a-1}$
$x=-\frac{9+7(2a-1)}2=\frac{-9-14a+7}2=-1-7a$
Check it out: $-2(-1-7a)-7(2a-1)=2+14a-14a+7=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1125588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $f_{1}, f_{2}$ in $\mathbb{Z_{6}}[x]$ such that deg$(f_{1})$ = deg$(f_{2}) = 2$ and deg$(f_{1}+f_{2})=1$
*
*Find $f_{1}, f_{2}$ in $\mathbb{Z_{6}}[x]$ such that deg$(f_{1})$ = deg$(f_{2}) = 2$ and deg$(f_{1}+f_{2})=1$
*Find $g_{1}, g_{2}$ in $\mathbb{Z_{6}}[x]$ such that deg$(g_{1})$ = deg$(g_{2}) = 1$ and deg$(g_{1}.g_{2})=1$
| I have came up with this solution.
(1) Let $f_{1}(x) = \bar{3}x^2+\bar{2}x+\bar{1}$ and $f_{2}(x) = \bar{3}x^2+x+\bar{2}$, which are of degree $2$.
Then, $(f_{1}+f_{2}) = \bar{6}x^2 + \bar{3}x + \bar{3} = \bar{0}x^2 + \bar{3}x + 3$ is of degree $1$.
(2) Let $g_{1}(x) = 3+2x$ and $g_{2}(x) = 2+3x$, which are of degree $1$.
Then, $(g_{1}.g_{2}) = \bar{6} + \bar{13}x = \bar{0} + x$ is of degree $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Elementary number theory , when is $12n^2 + 1$ a square Prove that if
$$k = 2 + 2\sqrt{12n^2 + 1}$$
is an integer then it is a square.
Can anyone help me with this? All I know is that k is an integer if and only if ${12n^2 + 1}$ is a square. What do I do next?
| $m^2 = 12 n^2 + 1$ implies $m^2 - 12 n^2 = 1$ so
$$(m + n\sqrt{12} )= (7 + 2\sqrt{12} )^N$$
for some $N\ge 0$ (see also http://en.wikipedia.org/wiki/Pell%27s_equation, just mind you, their $n$ is our $12$).
We also have the conjugate equality:
$$(m - n\sqrt{12} )= (7 - 2\sqrt{12} )^N$$
and , therefore, for $\sqrt{12n^2 + 1} = m$ we get, as @Thomas Andrews indicated
$$\sqrt{12n^2 + 1} = m = \frac{1}{2}(\, (7 + 2\sqrt{12} )^N + (7 - 2\sqrt{12} )^N) $$
and so
$$2 + 2\sqrt{12n^2 + 1} = (7 + 2\sqrt{12} )^N + (7 - 2\sqrt{12} )^N) + 2$$
Note however that
$$7 \pm \sqrt{12} = (2 \pm \sqrt{3})^2 $$ and so we get
$$2 + 2\sqrt{12n^2 + 1} = (2 + \sqrt{3} )^{2N} + (2 - \sqrt{3} )^{2N}) + 2\cdot (2 + \sqrt{3} )^N \cdot (2 - \sqrt{3} )^N = \\ = (\,( 2 + \sqrt{3})^N + (2-\sqrt{3})^N)^2$$
and inside the bracket we have a natural number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1128563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluating Exponents
Answer choices:
22
15
10
358
None of the above
This is my solution:
| You are correct, the answer is 10.
\begin{align*} 2^2 + 9^ \frac{1}{2} + 2^1 + 225^0 &= 4 + 9^ \frac{1}{2} + 2^1 + 225^0 \\
&= 4 + {\sqrt{9}} + 2^1 + 225^0 \\
&= 4 + 3 + 2^1 + 225^0 \\
&= 4 + 3 + 2 + 225^0 \\
&= 4 + 3 + 2 + 1 \\
&= 10
\end{align*}
Did some extended working out for you.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
find the length of rectangle based on area of frame A rectangular picture is $3$ cm longer than its width, $(x+3)$. A frame $1$ cm wide is placed around the picture. The area of the frame alone is $42 \text{cm}^2$. Find the length of the picture.
I have tries:
$(x+6)(x+3) = 42 \\
x^2+3x+6x+12 = 42 \\
x^2+9x+12x-42 = 0 \\
\\
x^2 + 21 -42$
| Let $H$,$W$,$FH$,$FW$ be the height, width, frame height and frame width.
We're given the following equations:
$H = W+3$
$W + 2 = FW$
$H + 2 = FH$
$FW \cdot FH - W\cdot H = 42$
It follows:
$(W+2)\cdot (H+2) - W\cdot H = 42$
$\Leftrightarrow$ $((H-3)+2)\cdot (H+2) - ((H-3)\cdot H)= 42$
$\Leftrightarrow$ $H^2+2H-H-2-H^2+3H =42$
$\Leftrightarrow$ $4H = 44$
$\Leftrightarrow$ $H=11$
Alternatively, we can solve
$\left( \begin{array}{ccc}
1 & -1 & 0& 0 \\
0 & 1 & 0 & -1 \\
1 & 0 & -1 & 0 \\
2 & 2 & 0& 0 \end{array} \right)^{-1} \left( \begin{array}{c}
3 \\
-2 \\
-2 \\
46 \end{array} \right) = \left( \begin{array}{c}
H \\
W \\
FH \\
FW \end{array} \right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Inverse of diagonally dominant matrix with equal off-diagonal entries Is there an explicit expression for the inverse of strictly diagonally dominant matrix with identical off-diagonal elements?
For example:
$$ \begin{pmatrix} a & -b & -b \\
-b & c & -b \\
-b & -b & d \end{pmatrix} $$
where $|a|\gt 2|b|$, $|c|\gt 2|b|$, and $|d|\gt 2|b|$. I'm looking for an inverse for an arbitrary $n\times n$ matrix with the above property.
| The Sherman-Morrison formula gives the inverse of any rank one update of a matrix for which the inverse is known.
Here we can write your matrix as follows:
$$ \begin{pmatrix} a & -b & -b \\ -b & c & -b \\ -b & -b & d \end{pmatrix}
= \begin{pmatrix} a+b & 0 & 0 \\ 0 & c+b & 0 \\ 0 & 0 & d+b \end{pmatrix}
- b \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} $$
Since the first summand is an invertible diagonal matrix (regardless of the actual sign of $b$), we have that the Sherman-Morrison formula can be applied.
Let $A = \begin{pmatrix} a+b & 0 & 0 \\ 0 & c+b & 0 \\ 0 & 0 & d+b \end{pmatrix}$ and let $u = \begin{pmatrix} -b \\ -b \\ -b \end{pmatrix}$, $v^T = \begin{pmatrix} 1 & 1 & 1 \end{pmatrix}$. What you ask for is:
$$ (A + uv^T)^{-1} = A^{-1} - \left( \frac{1}{1+ v^T A^{-1} u} \right)
\left( A^{-1} uv^T A^{-1} \right) $$
Note that the first factor in the second term of the right hand side is just a scalar, obtained by taking the reciprocal of the scalar $1+ v^T A^{-1} u$. It multiplies the rank one matrix $A^{-1} uv^T A^{-1}$, showing that the inverse of the rank one update of $A$ is itself a rank one update of $A^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1132591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $x^3 - 9x^2 + 18x - 12$
If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $$x^3 - 9x^2 + 18x - 12.$$
This is not a homework problem. I'm not even a student. I'm going through an old textbook. I know this is a simple problem. Can't seem to crack it though.
| It makes sense to let $y = x - 3 = 3^{1 \over 3} + 3^{2 \over 3}$. You are to compute
$$(y + 3)^3 -9(y + 3)^2 + 18(y + 3) - 12$$
This works out to
$$y^3 - 9y - 12$$
Note that $y^3 = (3^{1 \over 3})^3(1 + 3^{1 \over 3})^3 = 3(1 + 3*3^{1 \over 3} + 3*3^{2 \over 3} + 3) = 3(4 + 3y) = 12 + 9y$. Thus $y^3 - 9y - 12 = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
An inequality with fractional parts $$
\frac{n^k-n}{2}
\leq
\left\{\sqrt[k]{1}\right\} + \left\{\sqrt[k]{2}\right\} + \dots + \left\{\sqrt[k]{n^k}\right\}
\leq
\frac{n^k-1}{2}
$$
how it can be proven?
| Checking at wolfram it actually seems to be the opposite that $\frac{n^k-1}{2}$ is a lower bound and I will prove it below.
We can separate the sum into intervals like this by using that $\{n\}=n-\lfloor n\rfloor$ and the only integers are the $k$'th powers so they are equal to $0$. $$\sum_{d=2}^{2^k-1}(\sqrt[k]{d}-1)+\sum_{d=2^k+1}^{3^k-1}(\sqrt[k]{d}-2)+\sum_{d=3^k+1}^{4^k-1}(\sqrt[k]{d}-3)+\cdots+\sum_{d=(n-1)^k+1}^{n^k-1}(\sqrt[k]{d}-n+1)$$
Now notice we can write as $$\sum_{d=2}^{2^k-1}\sqrt[k]d+\sum_{d=2^k+1}^{3^k-1}\sqrt[k]d+\cdots+\sum_{d=(n-1)^k+1}^{n^k-1}\sqrt[k]d=\sum_{d=1}^{n^k}\sqrt[k]d-\sum_{d=1}^nd$$Now notice that
$$\sum_{d=2}^{2^k-1}(-1)+\sum_{d=2^k+1}^{3^k-1}(-2)+\cdots+\sum_{d=(n-1)^k+1}^{n^k-1}(-n+1)\\=(-1)(2^k-1-2)+(-2)(3^k-1-(2^k+1))+\cdots+(n-1)(n^k-1-((n-1)^k+1)\\=\sum_{d=1}^{n-1}d((d+1)^k-d^k-2)$$
Now combining that information we get that the sum is equal to
$$=\sum_{d=1}^{n^k}\sqrt[k]d-\sum_{d=1}^nd-\sum_{d=1}^{n-1}d((d+1)^k-d^k-2)\\=\sum_{d=1}^{n^k}\sqrt[k]d-\frac{n(n+1)}{2}+2\sum_{d=1}^{n-1}d-\sum_{d=1}^{n-1}d((d+1)^k-d^k)\\=\sum_{d=1}^{n^k}\sqrt[k]d-\frac{n(n+1)}{2}+n(n-1)-\sum_{d=1}^{n-1}d((d+1)^k-d^k)$$
Now by telescoping like method we get that
$$\sum_{d=1}^{n-1}d((d+1)^k-d^k)=2^k-1+2\cdot 3^k-2\cdot 2^k+3\cdot 4^k-3\cdot 3^k+\cdots + (n-1)\cdot n^k-(n-1)\cdot (n-1)^k=(n-1)n^k-((n-1)^k+(n-2)^k+\cdots + 2^k+1)\\=n^{k+1}-\sum_{d=1}^nd^k$$ So we are left with $$\sum_{d=1}^{n^k}\sqrt[k]{d}+\frac{n(n-3)}{2}-\left(n^{k+1}-\sum_{d=1}^nd^k\right)\\=\sum_{d=1}^{n^k}\sqrt[k]d+\frac{(n-2)(n-1)}{2}-n^{k+1}+\sum_{d=1}^nd^k$$
Now lets say $n> 3$ and $k>3$ to avoid the edge cases. For the first inequality I used the formulas from Asymptotic behaviour of sums of consecutive powers $$\sum_{d=1}^nd^k> \frac{n^{k+1}}{k+1}+\frac{n^k}2$$
Also $$\sum_{d=1}^{n^k}\sqrt[k]{d}\geq\int_0^{n^k}\sqrt[k]xdx=\frac{kn^{k+1}}{k+1}$$
Now adding those estimates ignoring $(\frac{(n-1)(n-2)}{2})$ since it's positive we get
$$\left\{\sqrt[k]{1}\right\} + \left\{\sqrt[k]{2}\right\} + \dots + \left\{\sqrt[k]{n^k}\right\}=\sum_{d=1}^{n^k}\sqrt[k]d+\frac{(n-2)(n-1)}{2}-n^{k+1}+\sum_{d=1}^nd^k \\>\frac{kn^{k+1}}{k+1}+\frac{n^{k+1}}{k+1}+\frac{n^k}2-n^{k+1}=\frac{n^k}{2}> \frac{n^k-1}{2}$$
| {
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"url": "https://math.stackexchange.com/questions/1138006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Let $F$ be a field in which we have elements satisfying $a^2+b^2+c^2 = −1$. Show that there exist elements satisfying $d^2+e^2 = −1$. Let $F$ be a field in which we have elements $a, b$, and $c$ satisfying $a^2+b^2+c^2 = −1$. Show that there exist elements $d$ and $e$ of $F$, satisfying $d^2+e^2 = −1$.
Any hint?
This is an excercise from the book: The Linear Algebra a Beginning Graduate Student Should Know; Golan
| Hint $\ $ Let $\ d=1\ $ in $\,(a^2+b^2)(\overbrace{a^2+b^2+c^2+d^2}^{\large =\, 0})\, =\, (\overbrace{a^2+b^2}^{\large x})^2 + (\overbrace{ac-bd}^{\large y})^2 + (\overbrace{ad+bc}^{\large z})^2$
That yields $\ x^2+y^2+z^2 = 0\ $ which, divided by $\,x^2,\,$ yields the result.
Remark $\ $ The latter two summands arise from the Brahmagupta–Fibonacci identity
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sum of $\lfloor k^{1/3} \rfloor$ I am faced with the following sum:
$$\sum_{k=0}^m \lfloor k^{1/3} \rfloor$$
Where $m$ is a positive integer. I have determined a formula for the last couple of terms such that $\lfloor n^{1/3} \rfloor^3 = \lfloor m^{1/3} \rfloor^3$. For example if the sum is from 0 to 11 I can find the sum of the terms 8 through 11. I am stuck on what formula can be applied to find the sum of the terms before the final stretch however.
| Note: This answer provides a solution for the general case $\lfloor \sqrt[p]{k}\rfloor$ with $p\geq 1$.
The following is valid for $m\geq 1, p\geq 1$
\begin{align*}
\sum_{k=1}^{m}\lfloor \sqrt[p]{k}\rfloor
=m\lfloor \sqrt[p]{m}\rfloor^{p+1}
-\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j\lfloor \sqrt[p]{m}\rfloor^{p+1-j}+m\lfloor \sqrt[p]{m}\rfloor\tag{1}
\end{align*}
with $B_j$ the $j$-th Bernoulli Number.
The special cases $p=1,2,3$ give
\begin{align*}
\sum_{k=1}^m\lfloor k\rfloor&=\frac{1}{2}m^2+\frac{1}{2}m\\
\sum_{k=1}^m\lfloor \sqrt{k}\rfloor&=
m\lfloor\sqrt{m}\rfloor
-\frac{1}{3}\lfloor\sqrt{m}\rfloor^3-\frac{1}{2}\lfloor\sqrt{m}\rfloor^2+\frac{5}{6}\lfloor\sqrt{m}\rfloor\\
\sum_{k=1}^m\lfloor \sqrt[3]{k}\rfloor&=m\lfloor\sqrt[3]{m}\rfloor-\frac{1}{4}\lfloor\sqrt[3]{m}\rfloor^4
-\frac{1}{2}\lfloor\sqrt[3]{m}\rfloor^3-\frac{1}{4}\lfloor\sqrt[3]{m}\rfloor^2+\lfloor\sqrt[3]{m}\rfloor\tag{2}\\
\end{align*}
For convenient calculations we consider two aspects:
*
*We introduce a variable $a=\lfloor\sqrt[p]{m}\rfloor$. So, we have
$$a\leq \sqrt[p]{m} < a+1$$
*We use the Iverson Bracket notation, so we can replace the expression $\lfloor x\rfloor$ by
$$\lfloor x\rfloor=\sum_{j\geq 0}[1\leq j \leq x]$$
Special case: $m=a^p,a=\lfloor\sqrt[p]{m}\rfloor$
We start the calculation by conveniently assuming $m=a^p$. We obtain
\begin{align*}
\sum_{k=1}^{m}\lfloor\sqrt[p]{k}\rfloor&=\sum_{k=1}^m\sum_{j\geq 0}[1\leq j \leq \sqrt[p]{k}][0\leq k \leq a^p]\\
&=\sum_{j=1}^{a}\sum_{k=1}^{m}[j^p\leq k \leq a^p]\\
&=\sum_{j=1}^{a}(a^p-j^p+1)\\
&=a^{p+1}-\sum_{j=1}^{a}j^p+a\tag{3}\\
&=a^{p+1}-\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j\,a^{p+1-j}+a\tag{4}
\end{align*}
Comment:
*
*In (3) we use the known representation of the sum of $p$-th powers of natural numbers by the Bernoulli Numbers
$$\sum_{k=1}^{n}k^{p}=\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j\,n^{p+1-j}$$
General case: $m\geq a^p,a=\lfloor\sqrt[p]{m}\rfloor$
In the general case we let again $a=\lfloor\sqrt[p]{m}\rfloor$ and have additionally to consider the terms for
$a^p< k \leq m$.
They are all equal to $a$, so they sum up to $$(m-a^p)a.$$ Adding this to (4) we get the general formula with $a=\lfloor \sqrt[p]{m}\rfloor$
\begin{align*}
\color{blue}{\sum_{k=1}^{m}\lfloor\sqrt[p]{k}\rfloor}&\color{blue}{=ma-\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j\,a^{p+1-j}+a}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
Let's finally compute the special case $p=3$. We need the Bernoulli numbers $B_0=1, B_1=\frac{1}{2}, B_2=\frac{1}{6}$ and $B_3=0$.
\begin{align*}
\sum_{k=1}^{m}\lfloor\sqrt[3]{k}\rfloor&=na-\frac{1}{4}\sum_{j=0}^{3}\binom{4}{j}B_j\,a^{4-j}+a\\
&=na-\frac{1}{4}\left[\binom{4}{0}B_0a^4+\binom{4}{1}B_1a^3+\binom{4}{2}B_2a^2\right]+a\\
&=na-\frac{1}{4}a^4-\frac{1}{2}a^3-\frac{1}{4}a^2+a\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
The other cases ($p=1,2$) from the claim (2) follow similarly.
Note: This approach ($p=2$) can be found in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Equivalent of adding to a denominator? Given the inequality $\frac{n}{m} \ge \frac{1}{2}$, I want to add $1$ to both $n$ and $m$:
$$\frac{n+1}{m+1}.$$
What would be the equivalent operation on the RHS of the equation?
Adding $1$ to $n$ is equivalent to $+\frac{1}{m}$ on both sides, but what about adding $1$ to $m$?
| Hint 1
$$\frac{n}{m}=\frac{n+1}{m+1}+\left(\frac{n}{m}-\frac{n+1}{m+1}\right)$$
and
$$ a \leq b+c \iff a-b \leq c $$
Hint 2
$$\frac{n}{m}=\frac{n+1}{m+1}\cdot\left(\frac{n}{m}\cdot\frac{m+1}{n+1}\right)$$
and,
$$ a \leq b\cdot c \iff \frac{a}{b} \leq c \qquad \text{ if }\ b>0 \\ a \leq b\cdot c \iff \frac{a}{b} \geq c \qquad \text{ if }\ b<0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Pell Equations: $a^2+4=5b^2$ This is a challenge problem in the Pell Equations chapter of my number theory book, but I'm not seeing the connection to Pell Equations. The Pell Equation with the coefficient $5$ is $5b^2+1=a^2$, but it doesn't look like the one I have.
Thanks if you can help me.
| I know full well the problem was asked and answered 3 years ago. Yet I wish to contribute. Please critique and comment so I learn something.
$$\begin{align}
a^2+4&=5b^2 \\
a^2-5b^2&=-4 \quad \text{initial solution} \quad (a_1, b_1) =(1,1)\\
(a-b \sqrt5)(a+b \sqrt 5)&=-4\\
\text{since} \quad &(1-\sqrt 5)(1+\sqrt 5)=-4 \\
\text{then} \quad &(1-\sqrt 5)^2(1+\sqrt 5)^2=(-4)^2 \\
&(6-2 \sqrt 5)(6+2 \sqrt 5)=4^2 \quad \text{and}\\
(a-b \sqrt5)(6-2 \sqrt 5)(a+b \sqrt 5)(6+2 \sqrt 5)&=-4^3\\
(6a+10b)^2-5(2a+6b)^2&=-4^3 \\
\left(\frac{3a+5b}{2}\right)^2-5\left(\frac{a+3b}{2}\right)^2&=-4 \implies\\
&\begin{cases}
a_{k+1}=\frac{1}{2}(3a_k+5b_k) \\
b_{k+1}= \frac{1}{2}(a_k+3b_k)\end{cases}\\
&\implies \\
&(1,1) \to (4,2) \to (11, 5) \to (29, 13) \to \cdots
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve the following systems of equations ( Matrices) Solve the following systems of equations:
\begin{Bmatrix}
x_1 & -x_2 & -x_3 & +0x_4 & = 2 \\
-x_1 & +2x_2 & +0x_3 & +3x_4 &= 1 \\
x_1 & +0x_2 & +x_3 & +0x_4 & =5\\
\end{Bmatrix}
I try to make a zero triangle but always reach a step where I'm in a cycle trying to do the same thing over and over. I once got $x_3=1/3$
Working:
1-
\begin{Bmatrix}
1 & -1 & -1 & +0 & 2 \\
-1 & +2 & +0 & +3 & 1 \\
1 & +0 & +1 & +0 & 5\\
\end{Bmatrix}
2- Eq2 to Eq2 + Eq1
\begin{Bmatrix}
1 & -1 & -1 & +0 & 2 \\
0 & 1 & -1 & 3 & 3 \\
1 & +0 & +1 & +0 & 5\\
\end{Bmatrix}
3- Eq3 to Eq3 + Eq1
\begin{Bmatrix}
1 & -1 & -1 & +0 & 2 \\
0 & 1 & -1 & 3 & 3 \\
0 & 1 & 2 & 0 & 3\\
\end{Bmatrix}
4- Eq3 to Eq3 - Eq2
\begin{Bmatrix}
1 & -1 & -1 & +0 & 2 \\
0 & 1 & -1 & 3 & 3 \\
0 & 0 & 3 & 0 & 1\\
\end{Bmatrix}
Now I can say $3x_3=1$
Thus, $x_3=1/3$
After that i tried but couldn't complete...
Any suggestions would be appreciated
EDIT: The last step is incorrect.
It should be :
\begin{Bmatrix}
1 & -1 & -1 & +0 & 2 \\
0 & 1 & -1 & 3 & 3 \\
0 & 0 & 3 & -3 & 0\\
\end{Bmatrix}
So what I did next is:
From Third Row: $x_3+x_4=0$
so, $x_4=-x_3$
Then, From the Second row:
$x_2-x_3+3x_4=3$
$x_2-x_3+3(-x_3)=3$
$x_2-x_3=3$
$x_2=3+x_3$
And I stopped
| You should now do backwards elimination; first divide the third equation by $3$ and add the third equation to the second and the first, getting
$$
\begin{bmatrix}
1 & -1 & 0 & 0 & 7/3\\
0 & 1 & 0 & 3 & 10/3\\
0 & 0 & 1 & 0 & 1/3
\end{bmatrix}
$$
Now add the second equation to the first one:
$$
\begin{bmatrix}
1 & 0 & 0 & 3 & 17/3\\
0 & 1 & 0 & 3 & 10/3\\
0 & 0 & 1 & 0 & 1/3
\end{bmatrix}
$$
Now you know that you can give $x_4$ any value, say $x_4=h$, and determine consequently the other unknowns:
$$
\begin{cases}
x_1=\frac{17}{3}-3h\\
x_2=\frac{10}{3}-3h\\
x_3=\frac{1}{3}\\
x_4=h
\end{cases}
$$
| {
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"url": "https://math.stackexchange.com/questions/1144315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help me to solve this trigonometrical equation I want to solve the equation
$$\sin(3x+a)+\sin(3x-a)+\sin(a-x)-\sin(a+x)=2\cos a$$
I solved up to this part
$$\begin{align}
2 \sin 3x \cos a - 2 \cos a\sin x &= 2 \cos a \\
2 \cos a \left(\;\sin 3x -\sin x \;\right) &= 2 \cos a \\
\sin 3x - \sin x &= 1
\end{align}$$
| HINT:
$\sin3x=4\sin x-4\sin^3x$
Alternatively,
using Prosthaphaeresis Formula,
$$\sin3x-\sin x=2\sin x\cos2x=2\sin x(1-2\sin^2x)=2\sin x-4\sin^3x$$
Then you can use this
| {
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"answer_id": 0
} |
Proving $\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$ I dont have an idea to prove it because of exist $\sin(4n+2)^4$
$$\sum_{n=0}^{\infty }\frac{\sin^4(4n+2)}{(2n+1)^2}=\frac{5\pi ^2}{16}-\frac{3\pi }{4}$$
| One first calculates$$\sin\left(\theta\right)^{4}=\frac{1}{8}\cos\left(4\theta\right)-\frac{1}{2}\cos\left(2\theta\right)+\frac{3}{8}.$$
Then$$\sum_{n=0}^{+\infty}\frac{\sin\left(2\left(2n+1\right)\right)^{4}}{\left(2n+1\right)^{2}}=\frac{1}{8}\sum_{n=0}^{+\infty}\frac{\cos\left(8m\right)}{m^{2}}-\frac{1}{2}\sum_{n=0}^{+\infty}\frac{\cos\left(4m\right)}{m^{2}}+\frac{9}{32}\zeta\left(2\right)$$
where we changed the index $2n+1\mapsto m
.$
Now the answer can be found in this link :
Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Why is $\int$ $1\over(a^2+b^2)^{3/2}$ $da$ $ = $ $ a\over b^2\sqrt {(a^2+b^2)}$ $$\int\frac{da}{(a^{2}+b^{2})^{3/2}} =\frac{a}{b^{2}\sqrt{(a^{2}+b^{2})}}.$$
Found this in the solution to a problem in my physics textbook, and left clueless.
| Using the trig substitution $a = b\tan \theta$, we get $a^2 + b^2 = b^2\sec^2\theta$ (since $\tan^2\theta + 1 = \sec^2\theta$), so $(a^2 + b^2)^{3/2} = b^3\sec^3\theta$. Since $da = b\sec^2\theta\, d\theta$, we obtain
$$\int \frac{1}{(a^2 + b^2)^{3/2}}\, da = \int \frac{1}{b^3\sec^3\theta} b\sec^2\theta\, d\theta = \frac{1}{b^2}\int \frac{d\theta}{\sec \theta} = \frac{1}{b^2}\int \cos \theta\, d\theta = \frac{1}{b^2}\sin\theta + C = \frac{a}{b^2\sqrt{a^2 + b^2}} + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}<2-\frac{1}{n^2} $
| Find the fourier series of $f(x)=x^2$ on $(-1,1)$. It can give you the accurate solution which is $\dfrac {\pi ^2}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
For odd $m\ge3$, does it follow: $\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$ Unless I am making a mistake, I am calculating that:
$$\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$$
Here's my reasoning:
*
*$\dfrac{x^m + y^m}{x+y} = x^{m-1} - x^{m-2}y - xy^{m-2} + x^{m-3}y^2 + x^2y^{m-3} + \dots + x^{\frac{m-1}{2}}y^{\frac{m-1}{2}} + y^{m-1}$
*$\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-3} - x^{m-4}y - xy^{m-4} + x^{m-5}y^2 + x^2y^{m-5} + \dots + x^{\frac{m-3}{2}}y^{\frac{m-3}{2}} + y^{m-3}$
*$(xy)\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-2}y - x^{m-3}y^2 - x^2y^{m-3} + x^{m-4}y^3 + x^3y^{m-4} + \dots + x^{\frac{m-1}{2}}y^{\frac{m-1}{2}} + xy^{m-2}$
*So that, $\dfrac{x^m + y^m}{x+y} + (xy)\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$
*I am figuring that $\dfrac{x^{m-2} + y^{m-2}}{x+y}$ has exactly 2 terms less than $\dfrac{x^m + y^m}{x+y}$
Is this reasoning correct? For each value that I test, it seems correct.
Thanks,
-Larry
| $$\frac{x^m+y^m}{x+y}+(xy)\frac{x^{m-2}+y^{m-2}}{x+y}=\frac{x^m+y^m+(xy)(x^{m-2}+y^{m-2})}{x+y}= \\
=\frac{x^m+y^m+x^{m-1}y+xy^{m-1}}{x+y}=\frac{x^m+x^{m-1}y+xy^{m-1}+y^m}{x+y}\\
=\frac{x^{m-1}(x+y)+y^{m-1}(x+y)}{x+y}=x^{m-1}+y^{m-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\geq3\prod_{cyc}(a+b-c),$$ or
$$\sum_{cyc}(a^3-ab^2-ac^2+2abc)\geq\sum_{cyc}(-3a^3+3a^2b+3a^2c-2abc),$$ but it becomes very ugly.
| Since $(a,b,c)$ and $\left(\frac{1}{b+c-a},\frac{1}{a+c-b},\frac{1}{a+b-c}\right)$ are the same ordered, by Chebyshov and C-S we obtain:
$\sum\limits_{cyc}\frac{a}{b+c-a}\geq\frac{1}{3}(a+b+c)\sum\limits_{cyc}\frac{1}{b+c-a}=\frac{1}{3}\sum\limits_{cyc}(b+c-a)\sum\limits_{cyc}\frac{1}{b+c-a}\geq\frac{1}{3}\cdot9=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
} |
How to calculate derivative of $f(x) = \frac{1}{1-2\cos^2x}$? $$f(x) = \frac{1}{1-2\cos^2x}$$
The result of $f'(x)$ should be equals
$$f'(x) = \frac{-4\cos x\sin x}{(1-2\cos^2x)^2}$$
I'm trying to do it in this way but my result is wrong.
$$f'(x) = \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} =
\frac {1-2\cos^2x-(1-(2\cos^2x)')}{(1-2\cos^2x)^2} = $$
$$=\frac {-2\cos^2x + 2(2\cos x(\cos x)')}{(1-2\cos^2x)^2} =
\frac {-2\cos^2x+2(-2\sin x\cos x)}{(1-2\cos^2x)^2} = $$
$$\frac {-2\cos^2x-4\sin x\cos x}{(1-2\cos^2x)^2}$$
| The derivative of the function that is constantly $1$ should be zero, not $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Equation with complex numbers Solve the following equation in $ \mathbb{C} $:
$ |z - |z + 1|| = |z + |z - 1|| $
I started it but I don't know how to finish it. Here is what I did so far:
$ |z - |z + 1||^2 = (z - |z + 1|)(\bar{z} - \overline{|z + 1|}) = |z|^2 - z \cdot \overline{|z + 1|} - \bar{z} \cdot |z + 1| + |z + 1|^2 $
$ |z + |z - 1||^2 = (z + |z - 1|)(\bar{z} + \overline{|z - 1|}) = |z|^2 + z \cdot \overline{|z - 1|} + \bar{z} \cdot |z - 1| + |z - 1|^2 $
Now, $ |z - |z + 1|| = |z + |z - 1|| \Rightarrow |z - |z + 1||^2 = |z + |z - 1||^2 $
So we have: $ |z|^2 - z \cdot \overline{|z + 1|} - \bar{z} \cdot |z + 1| + |z + 1|^2 = |z|^2 + z \cdot \overline{|z - 1|} + \bar{z} \cdot |z - 1| + |z - 1|^2 $
$ z(\overline{|z + 1| + |z - 1|}) + \bar{z}(|z + 1| + |z - 1|) = (|z + 1| + |z - 1|)(|z + 1| - |z - 1|) $
We divide all by |z + 1| + |z - 1| and we get:
$ z + \frac{||z + 1| + |z - 1||^2}{(|z + 1| + |z - 1|)^2} + \bar{z} = |z + 1| - |z - 1|$
But $ |z + 1| + |z - 1| $ is a positive number so $ ||z + 1| + |z - 1|| = |z + 1| + |z - 1| $
So $ z + \bar{z} = |z + 1| - |z - 1| $.
And now I don't know what to do with this equation.
| $z+\overline{z}=|z+1|-|z-1|$
You might square twice
$$(z+\overline{z})^2=2z\overline{z}+2-2|z+1||z-1|\\
4(z+1)(\overline{z}+1)(z-1)(\overline{z}-1)=(2-z^2-\overline{z}^2)^2\\
4(z^2-1)(\overline{z}^2-1)=4-4z^2-4\overline{z}^2+z^4+2z^2\overline{z}^2+\overline{z}^4\\
0=(z^2-\overline{z}^2)^2$$
so $z^2$ is real, so $z=x$ or $z=iy$.
If $z=iy$, then $z+\overline{z}=0=|iy+1|-|iy-1|$, so the whole imaginary axis is included.
If $z=x$, then just the interval $[-1,1]$ is included.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
A Polygon is inscribed in a circle $\Gamma$ A regular polygon P is inscribed in a circle $\Gamma$. Let A, B, and C, be three consecutive vertices on the polygon P, and let M be a point on the arc AC of $\Gamma$ that does not contain B. Prove that
$MA\cdot MC=MB^2-AB^2$
I inscribed the polygon P in the unit circle and let B=1. Past this, I'm not really sure how to proceed. It might be helpful to have M at 1 as well but given that the polygon would have the n-th roots of unity in it B seemed like a good choice. What would be a good way to proceed from here?
| We can place the diagram in the complex plane so that $\Gamma$ is the unit circle, point $A$ goes to the complex number $a$, point $B$ goes to the complex number 1, and point $C$ goes to the complex number $\overline{a} = 1/a$. Let $m$ be the complex number corresponding to point $M$.
Then
\begin{align*}
MA \cdot MC &= |m - a| \cdot \left| m - \frac{1}{a} \right| \\
&= |a - m| \cdot \left| \frac{am - 1}{a} \right| \\
&= |a - m| \cdot |am - 1| \\
&= |(a - m)(am - 1)|.
\end{align*}
Also,
\begin{align*}
MB^2 - AB^2 &= |m - 1|^2 - |a - 1|^2 \\
&= (m - 1)(\overline{m} - 1) - (a - 1)(\overline{a} - 1) \\
&= (m - 1) \left( \frac{1}{m} - 1 \right) - (a - 1) \left( \frac{1}{a} - 1 \right) \\
&= 1 - m - \frac{1}{m} + 1 - 1 + a + \frac{1}{a} - 1 \\
&= \frac{a^2 m - am^2 - a + m}{am} \\
&= \frac{(a - m)(am - 1)}{am}.
\end{align*}
Since $M$ lies on the arc $AC$ that does not contain $B$, $MB^2 - AB^2$ is positive, so we can take the absolute value of this expression, to get
\begin{align*}
MB^2 - AB^2 &= \left| \frac{(a - m)(am - 1)}{am} \right| \\
&= \frac{|(a - m)(am - 1)|}{|am|} \\
&= |(a - m)(am - 1)|.
\end{align*}
Hence, $MA \cdot MC = MB^2 - AB^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Inequality involving exponential of square roots How can I show that:
$$ 2e^\sqrt{3} \leq 3e^\sqrt{2} $$
? (that's all I have)
Thank you so much!
| Note that
$$(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=3-2=1$$
and
$$(\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}>5+2\cdot 2=9$$
Thus $\sqrt{3}+\sqrt{2}>3$, which gives $\sqrt{3}-\sqrt{2}<\frac{1}{3}$.
Now we claim that $e^{\frac{1}{3}}<\frac{3}{2}$, this is because
$$e<3<\frac{27}{8}=\left(\frac{3}{2}\right)^3$$
It follows that
$$e^{\sqrt{3}-\sqrt{2}}<e^{\frac{1}{3}}<\frac{3}{2}$$
hence
$$2e^{\sqrt{3}}<3e^{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Integration of $\int \tan^2 x \ dx $ without the use of trig identities I am trying to integrate $\int \tan^2 x \ dx $ without the use of trig identities, although I know that using $ 1 + \tan^2 x \equiv \sec^2 x $ makes the question trivial.
Using the exponential form of tan(x) I arrived at
$$ -\int \dfrac{(e^{2ix}-1)^2}{(e^{2ix}+1)^2} \ dx $$
I have not studied complex integration, so I am wary of using substitutions of such as $ u = e^{2ix} + 1 $ because I do not know if they are valid.
| $$\int\tan^2xdx=\int\frac{\sin^2{x}}{\cos^2{x}}dx=\int\frac{1-\cos{2x}}{2\cos^2{x}}dx=*$$ here we used trigonometric formula $\sin^2{x}=1-\cos{2x},$ $$*= \int\frac{1-\cos{2x}}{2\cos^2{x}}dx=\frac{1}{2}\int\frac{1}{\cos{2x}}dx - \frac{1}{2} \int \frac{\cos{2x}}{\cos^2{x}}dx=*$$ now we will use this one $\cos{2x}=2\cos^2{x}-1,$ $$*= \frac{1}{2} \tan{x} - \frac{1}{2}×2\int\frac{\cos^2{x}}{cos^2{x}}dx + \frac{1}{2}\int\frac{1}{\cos^2{x}}dx= \frac{1}{2}\tan{x} -x + \frac{1}{2}\tan{x} +C=\tan{x} -x +C $$ Thats it. Just trigonometry.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
What am i doing wrong when solving this differential equation $$
f(x) = \frac{\frac{d^2}{dx^2}[e^y]}{\frac{d}{dx}[e^y]}
$$
Given that $f(x) = cx$
$$
\frac{c}{2}x^2 + k_1 = \ln(e^y y')
$$
$$
k_2\int e^{\frac{c}{2}x^2} dx = e^y
$$
$$
y = \ln(k_2\int e^{\frac{c}{2}x^2} dx)
$$
Therefore
$$
y'^2 - cxy' + y'' = 0
$$
$$
y' = \frac{c}{2}x \pm \sqrt{\frac{c^2}{4}- y''}
$$
$$
\frac{e^{\frac{c}{2}x^2}}{\int e^{\frac{c}{2}x^2} dx} = \frac{c}{2}x \pm \sqrt{\frac{c^2}{4}x^2- y''}
$$
Yet when $c=4$
$$
\frac{2\sqrt{\frac{2}{\pi}}e^{2x^2}}{\operatorname{erfi}(\sqrt{2}x)+c} = 2x \pm\sqrt{4x^2 + 8e^{2x^2}\frac{e^{2x^2}-\sqrt{2\pi}x\operatorname{erfi}(\sqrt{2}x)}{\pi \operatorname{erfi}(\sqrt{2}x)^2}}
$$
Is not correct, what am i doing wrong?
| Why does this not reduce to: y'' + (1-cx)y'=0?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$ How to calculate the following limit?
$$\lim\limits_{x \to \infty} \left(\frac{x^2+1}{x^2-1}\right)^{x^2}$$
| You can do it using L'Hopital's rule this way:
$$\begin{align}
\lim\limits_{x \to \infty} \Big(\frac{x^2+1}{x^2-1}\Big)^{x^2} = \lim\limits_{x \to \infty} \exp\ln\Big(\frac{x^2+1}{x^2-1}\Big)^{x^2} &= \lim\limits_{x \to \infty} \exp x^2\ln\Big(\frac{x^2+1}{x^2-1}\Big) \\ &= \exp \lim\limits_{x \to \infty} \frac{\ln(x^2+1)-\ln(x^2-1)}{x^{-2}} \\ &= \exp \lim\limits_{x \to \infty} \frac{2x(x^2+1)^{-1}-2x(x^2-1)^{-1}}{-2x^{-3}} \\ &= \exp \lim\limits_{x \to \infty} \frac{x^4(x^2+1)-x^4(x^2-1)}{(x^2+1)(x^2-1)} \\ &= \exp \lim\limits_{x \to \infty} \frac{2x^4}{(x^2+1)(x^2-1)} \\ &= e^2
\end{align}$$
It's permissible to swap the $\exp$ with the $\lim$ between lines one and two because the exponential function is continuous, and L'Hopital's rule is applied between lines two and three.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
CAL I - Quotient Rule to find Derivative I'm stuck on this problem $h(x) = \frac{\sqrt{x}}{x^3+1}$.
I get to here: $\frac{x^3+1-x^{1/2}(3x^2)}{2x^{1/2}(x^3+1)^2}$.
I started going this route: $\frac{x^3+1-3x^{5/2}}{2x^{1/2}(x^3+1)^2}$.
The book's solution is getting this in their step process: $\frac{x^3+1-6x^3}{2x^{1/2}(x^3+1)^2}$.
How are they getting the $-6x^3$ in the numerator?
| Using
$$
\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}
$$ we have
$$
\left(\frac{\sqrt{x}}{x^3+1}\right)'=\frac{\frac{1}{2\sqrt{x}}(x^3+1)-\sqrt{x}(3x^2)}{(x^3+1)^2}=\frac{1}{2\sqrt{x}}\frac{(x^3+1)-2x\:(3x^2)}{(x^3+1)^2}=\frac{1}{2\sqrt{x}}\frac{-5x^3+1}{(x^3+1)^2}
$$ as desired.
Hoping it's clear for you now.
| {
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"url": "https://math.stackexchange.com/questions/1169505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $ \cos x - \cos y = -2 \sin ( \frac{x-y}{2} ) \sin ( \frac{x+y}{2} ) $ Prove that $ \cos x - \cos y = -2 \sin \left( \frac{x-y}{2} \right) \sin \left( \frac{x+y}{2} \right) $ without knowing cos identity
We don't know that $ \cos0 = 1 $
We don't know that $ \cos^2 x + \sin^2 x = 1 $
I have managed to prove it using the above facts, but just realised that I can't use them. Now I have been going in circles for a while.
Any ideas how to prove this or even approach it?
Thanks !
| You may use the complex definition of the sin and cos functions derived from the Euler's formula.
$$\begin{array}{l}\cos x = \frac{{{e^{ix}} + {e^{ - ix}}}}{2},\sin x = \frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}};\\ - 2\sin (\frac{{x - y}}{2})\sin (\frac{{x + y}}{2})\\ = - 2\left( {\frac{{{e^{i(\frac{{x - y}}{2})}} - {e^{ - i(\frac{{x - y}}{2})}}}}{{2i}}} \right)\left( {\frac{{{e^{i(\frac{{x + y}}{2})}} - {e^{ - i(\frac{{x + y}}{2})}}}}{{2i}}} \right)\\ = \frac{1}{2}\left( {{e^{i(\frac{{x - y}}{2})}} - {e^{ - i(\frac{{x - y}}{2})}}} \right)\left( {{e^{i(\frac{{x + y}}{2})}} - {e^{ - i(\frac{{x + y}}{2})}}} \right)\\ = \frac{1}{2}\left( {{e^{ix}} - {e^{ - iy}} - {e^{iy}} + {e^{ - ix}}} \right)\\ = \frac{1}{2}\left( {{e^{ix}} + {e^{ - ix}}} \right) + \frac{1}{2}\left( { - {e^{ - iy}} - {e^{iy}}} \right)\\ = \frac{1}{2}\left( {{e^{ix}} + {e^{ - ix}}} \right) - \frac{1}{2}\left( {{e^{ - iy}} + {e^{iy}}} \right)\\ = \cos x - \cos y\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1170630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Number of distinct terms in the expansion of $\big(x+\frac{1}{x}+x^2+\frac{1}{x^2}\big)^{15}$
Number of distinct terms in the expansion of $\bigg(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\bigg)^{15}$ is equal to ?
We can write the above as,
$$ \bigg(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\bigg)^{15} = \dfrac{1}{x^{30}}(1+x+x^3+x^4)^{15} $$
Now my teacher says that we can expand the polynomial $(1+x+x^3+x^4)^{15} $ as,
$$ (1+x+x^3+x^4)^{15} = a_0 + a_1x + a_2x^2 + a_3x^3.....a_{60}x^{60} $$
Hence, as each term is divided by $x^{30}$, the number of distinct terms would be equal to 61.
But my question is how do we know that the expansion of the polynomial $(1+x+x^3+x^4)^{15} $ will contain all powers of $x$ from $x^0$ to $x^{60}$ ?
| Your expression is equal to
$$
\frac1{x^{30}}(x^3+1)^{15}(x+1)^{15}
$$
$(x^3+1)^{15}$ contains $16$ terms, each containing a power of $x^3$ with a positive coefficient. Since $(x+1)^{15}$ contains $16$ terms of consecutive powers of $x$ with positive coefficients, the product will fill in the gaps in $(x^3+1)^{15}$. Thus, in your expression, there are non-zero terms from $x^{30}$ to $x^{-30}$.
In fact, the coefficient of $x^n$ is the number of solutions to $4a+3b+c=n$ where $a+b+c\le15$ and $a,b,c\ge0$. Thus, there is always at least one solution for $0\le n\le60$ and no solutions for other values of $n$.
That means $61$ distinct terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Volume of Solid in Calculus Find the volume of the solid whose base is the region in the first quadrant bounded by $y=x^5, y=1$ and the $y$-axis and whose cross sections perpeddicular to the $x$ axis are semicircles
I am tutoring someone, and we keep getting the wrong answer, any help would be appreciated
| The diameter of the semicircular slice is $1 - x^5$, and its area then is $\pi(1-x^5)^2/8.$
The curve $y=x^5$ intersects the line $y=1$ at $x=1$ in the first quadrant, so your limits of integration are $[0,1]$.
Then, your infinitesimal slice has volume $(\pi(1-x^5)^2/8) dx$, and the volume then is
$$V = \frac{\pi}{8} \int_0^1 (1-x^5)^2 dx = \left.\frac{\pi}{8}(x - \frac{1}{3}x^6 + \frac{1}{11}x^{11})\right|_0^1 = \frac{25 \pi}{264}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of square root. We have
$$\int \sqrt{x} \text{ }dx=\int x^{1/2} \text{ }dx=\frac{2}{3}x^{3/2}+C.$$
But, let $u=\sqrt{x}$. Then $x=u^2$, and $dx=2udu$. Substituting, we have
$$\int \sqrt{x} \text{ }dx=\int 2u^2 \text{ }du=\frac{2}{3}x^3+C.$$
Which one is correct, and why?
| $$\int \sqrt{x} \text{ }dx=\int 2u^2 \text{ }du=\frac{2}{3}u^3+C.$$
Remember, after substituting, we need to calculate the integral with respect to $u$.
Once that's done, then "back"-substitute $\sqrt x = u$ to get $$\frac 23(x^{1/2})^3 + C = \frac 23 x^{3/2} + C$$
| {
"language": "en",
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"source": "stackexchange",
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How to show $n^5 + 29 n$ is divisible by $30$ Show that $n^5 + 29 n$ is divisible by $30$.
Attempt:
$n^4 ≡ 1 \pmod 5$ By Fermat Little Theorem
| You want to show that $n^5+29n$ is divisible by $2$, $3$ and $5$.
*
*$n^5+29n\equiv n^5+n\equiv n^2n^2n+n\equiv nnn+n\equiv nn+n\equiv n+n\equiv0\pmod{2}$
*$n^5+29n\equiv n^5-n\equiv n^3n^2-n\equiv nn^2-n\equiv n-n\equiv0\pmod{3}$
*$n^5+29n\equiv n^5-n\equiv n-n\equiv0\pmod{5}$
All use Fermat's little theorem: $n^p\equiv n\pmod{p}$ for $p$ prime.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For real numbers a and b, when is the equation |a + b| = |a – b| true? I put that the statement was true only when a = 0 and b = 0 but the correct answer was that it only held true for a = 0 OR b = 0. With 'and' I figured |0 + 0| = 0 and |0 - 0| = 0. Could someone explain why my answer was incorrect? Thank you.
| If we square both sides of the equation $|a + b| = |a - b|$, we obtain
\begin{align*}
|a + b|^2 & = |a - b|^2\\
a^2 + 2ab + b^2 & = a^2 -2ab + b^2\\
4ab & = 0\\
ab & = 0
\end{align*}
which holds if $a = 0$ or $b = 0$ since a product is equal to zero if and only if one of the factors is equal to zero.
The statement $a = 0$ or $b = 0$ means that at least one of the equations is true, that is, it means that $a = 0$ is true, $b = 0$ is true, or both $a = 0$ and $b = 0$ are true. The statement $a = 0$ and $b = 0$ means that both $a = 0$ and $b = 0$ are true.
As you showed, the statement $|a + b| = |a - b|$ is certainly true when $a = 0$ and $b = 0$. However, this is not the only solution. For instance, the equation is satisfied when $a = 1$ and $b = 0$ since
$$1 = |1| = |1 + 0| = |1 - 0| = |1| = 1$$
It is also satisfied when $a = 0$ and $b = 1$ since
$$1 = |1| = |0 + 1| = |0 - 1| = |-1| = 1$$
More generally, the statement $|a + b| = |a - b|$ is satisfied when $a = 0$ and $b$ is any real number or $b = 0$ and $a$ is any real number.
For the case $a = 0$ and $b \in \mathbb{R}$, observe that the equation $|a + b| = |a - b|$ holds since
$$|b| = |0 + b| = |0 - b| = |-b|$$
which, from a geometric point of view, is valid since a number and its additive inverse are equidistant from $0$ on the real number line and, from an algebraic point of view, is true since
$$|b| = \sqrt{b^2} = \sqrt{(-b)^2} = |-b|$$
For the case $b = 0$ and $a \in \mathbb{R}$, observe that the equation $|a + b| = |a - b|$ holds since
$$|a| = |a + 0| = |a - 0| = |a|$$
Thus, the statement $a = 0$ or $b = 0$ gives us the general solution, while the statement $a = 0$ and $b = 0$ gives us a particular solution.
| {
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Write down each of the terms in the expansion of $\sin(x^2)-(\sin(x))^2$. Write down each of the terms in the expansion of $\sin(x^2)-(\sin(x))^2$. Taylor's Theorem applies at the point $a=0$ and with $n=4$.
Got no idea how to proceed. My lecture notes have one example that I barely understand. I'd really appreciate a semi-detailed overview of what I should be doing. I can go to class and ask questions, but I need to have some idea of what I am looking at in order to ask intelligent questions.
| $$
\sin x = x - \frac{x^3}6 + \frac{x^5}{120} - \cdots
$$
So
$$
\sin (x^2) = x^2 - \frac{x^6}6+\frac{x^{10}}{120} - \cdots
$$
and
$$
(\sin x)^2 = x^2 - \frac{x^4}3 + \frac{2x^6}{45} -\cdots.
$$
To get the last series just look at
$$
\left(x - \frac{x^3}6 + \frac{x^5}{120} - \cdots\right)\left(x - \frac{x^3}6 + \frac{x^5}{120} - \cdots\right).
$$
Multiply the $x$ from one by the $x$ from the other to get $x^2$. Then multiply $x$ from the first by $-x^3/6$ from the second to get $-x^4/6$, and $x$ from the second by $-x^3/6$ from the first, and add those to get $-x^4/3$. Then $x$ from the first by $x^5/120$ from the second, plus $-x^3/6$ from the first by $-x^3/6$ from the second, plus $x^5/120$ from the first by $x$ from the second, adding up to $2x^6/45$. And so on.
So the first non-zero term in the expansion of $\sin(x^2)-(\sin x)^2$ is the fourth-degree term.
| {
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Proving $(|a|+|b|)^2(|c|+|d|)^2 \leq 4(a^2+b^2)(c^2+d^2)$. Given $a,b,c,d \in \Bbb R$, I want to prove that: $$(|a|+|b|)^2(|c|+|d|)^2 \leq 4(a^2+b^2)(c^2+d^2).$$
But the bound I'm getting is just too weak. I know that: $$|a| = \sqrt{a^2} \leq \sqrt{a^2+b^2} \implies |a|+|b| \leq 2\sqrt{a^2+b^2}\implies (|a|+|b|)^2\leq 4(a^2+b^2).$$ This gives me: $(|a|+|b|)^2(|c|+|d|)^2 \leq 16(a^2+b^2)(c^2+d^2)$. Not good. I tried reverse engineering it: $$\begin{align} (|a|+|b|)^2 &\leq 2(a^2+b^2) \\ |a|+|b| &\leq \sqrt{2}\sqrt{a^2+b^2} \\ |a|+|b| &\leq \sqrt{2a^2+2b^2} \\ |a|+|b|&\leq \sqrt{(a\sqrt{2})^2+(b\sqrt{2})^2},\end{align}$$ which suggests that I may try to prove that: $$|a| \leq \frac{\sqrt{2}}{2}\sqrt{a^2+b^2}.$$
I'm failing to see the step needed here. I just need a little push. Thanks.
| Here's another way of proving it. From the Cauchy-Schwarz Inequality we have:
$$(1 + 1)(a^2 + b^2) \ge \left(\sqrt{1\cdot a^2} + \sqrt{1\cdot b^2}\right)^2 = \left(\mid a \mid + \mid b \mid\right)^2$$
$$(1 + 1)(c^2 + d^2) \ge \left(\sqrt{1\cdot c^2} + \sqrt{1\cdot d^2}\right)^2 = \left(\mid c \mid + \mid d \mid\right)^2$$
Now multiply them to get:
$$4(a^2 + b^2)(c^2 + d^2) \ge (\mid a \mid + \mid b \mid)^2(\mid c \mid + \mid d \mid)^2$$
Hence the proof. We have equality when:
$$\frac {a^2}1 = \frac {b^2}1 \quad \text{ and } \quad \frac {c^2}1 = \frac {d^2}1$$
Or in other words when $a=b$ and $c=d$
| {
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$F_{2n} = F_{2n-2}+2F_{2n-4}+\dots+n$ rigorous proof Let $F_{n}$ be n-th fibonacci number($F_{0}$ = 0) and $g_{n} = F_{2n}$ if $n > 0$ $g_{0} = 1$.
I want to prove that $g_{n} = g_{n-1}+2g_{n-2}+\dots +ng_{0}$. It's obviously seen from direct evaluation but it's not so rigorous and my attempts to use induction end in failure..
| There are many ways to prove this. One way is to notice that
$$
F_{2n} = F_{2n-1} + F_{2n-2} = 2F_{2n-2} + F_{2n-3} = 2F_{2n-2} + F_{2n-3} + F_{2n-4} - F_{2n-4} = 3F_{2n-2} - F_{2n-4}.
$$
Therefore
$$ g_n = 3g_{n-1} - g_{n-2}. $$
Given this, it is easy to prove the claim by induction. Let's rephrase it slightly:
$$
g_n = \sum_{i=1}^{n-1} ig_{n-i} + n.
$$
In particular, when $n = 0$ we get $g_0 = 0$, and when $n = 1$ we get $g_1 = 1$, both obviously correct. Assuming the claim holds for $n-1,n-2$, we have
$$
\begin{align*}
g_n &= 3g_{n-1} - g_{n-2} \\ &=
3\sum_{i=1}^{n-2} ig_{n-1-i} + 3(n-1) - \sum_{i=1}^{n-3} ig_{n-2-i} - (n-2) \\ &=
\sum_{i=1}^{n-2} i(3g_{n-1-i} - g_{n-2-i}) + 2n-1 \\ &=
\sum_{i=1}^{n-2} ig_{n-i} + 2n-1 \\ &=
\sum_{i=1}^{n-1} ig_{n-i} - (n-1)g_1 + 2n-1 \\ &=
\sum_{i=1}^{n-1} ig_{n-i} + n.
\end{align*}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality: $\left|\sum_{i = n + 1}^{\infty} \frac{(-1)^i}{i + x}\right| \le \frac{1}{n + 1 + x}$ This (with $x > 0$ and $n \in \Bbb N$):
$$ \left|\sum_{i = n + 1}^{\infty} \frac{(-1)^i}{i + x}\right| \le \frac{1}{n + 1 + x}$$
Was used to prove that the series of general term the summand on the left hand side of the inequality is convergent uniformly on $]0, \infty[$. But I can't see why this should be true.
Using the $|\sum \cdot| \le \sum|\cdot|$ does nothing, because we get an upper bound which is greater than the RHS. It seems to be true after observing the first few terms of the sum. But how could it be proven? There's just some trick which I'm not seeing.
Thanks a lot.
| Hint 1: The terms $$ \frac{(-1)^i}{i + x} $$
are of alternating signs and their absolute value decreases.
Hint 2: Group the terms in the sum into pairs of two.
More detailed answer:
$$
\left| \sum_{i = n + 1}^{\infty} \frac{(-1)^i}{i + x}\right|
= | (-1)^{n+1} | \left| \Bigl( \frac{1}{n + 1 + x} - \frac{1}{n + 2 + x} \Bigr)
+ \Bigl( \frac{1}{n + 3 + x} - \frac{1}{n + 4 + x} \Bigr)
+ \dots \right|
$$
All terms in the parentheses are positive, so this is equal to
$$
\Bigl( \frac{1}{n + 1 + x} - \frac{1}{n + 2 + x} \Bigr)
+ \Bigl( \frac{1}{n + 3 + x} - \frac{1}{n + 4 + x} \Bigr)
+ \dots
$$
Now we group the sum differently:
$$
\frac{1}{n + 1 + x} + \Bigl( - \frac{1}{n + 2 + x} + \frac{1}{n + 3 + x} \Bigr)
+ \Bigl( - \frac{1}{n + 4 + x} + \frac{1}{n + 5 + x} \Bigr)
+ \dots
$$
All terms in the parentheses are now negative, and the result follows.
(A more rigorous proof would work with finite partial sums, but I hope this
demonstrates the idea.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ I know that $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ is correct. But having some hard time proofing it using trig relations. Some of the relations I used are
$$
\sin x \cos x= \frac{1}{2} \sin(2 x)\\
\sin^2 x + \cos^2 x= 1\\
\sin^2 x = \frac{1}{2} - \frac{1}{2} \cos(2 x)\\
\cos(2 x) = 2 \cos^2(x)-1
$$
And others. I tried starting with multiplying numerator and denominator with either $\cos x$ or $\sin x$ and try to simplify things, but I seem to be going in circles.
Any hints how to proceed?
| You have to prove that
$$
\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}.\tag{1}
$$
Note that $(1)$ may be expressed as follows by cross-multiplying:
$$
\underbrace{(\cos x)(1+\cos x+\sin x)}_{\text{LHS}}=\underbrace{(1+\sin x)(1+\cos x-\sin x)}_{\text{RHS}}.\tag{2}
$$
Now all you have to do is expand the LHS and RHS to see that they are equivlant:
We have
\begin{align}
\text{LHS}
&= (\cos x)(1+\cos x+\sin x)\tag{definition}\\[0.5em]
&= \cos x +\cos^2(x)+\cos x\sin x\tag{expand}
\end{align}
and
\begin{align}
\text{RHS} &= (1+\sin x)(1+\cos x-\sin x)\tag{definition}\\[0.5em]
&= 1+\cos x-\sin x+\sin x+\sin x\cos x-\sin^2(x)\tag{expand}\\[0.5em]
&= 1+\cos x+\sin x\cos x-\sin^2(x).\tag{simplify}
\end{align}
Now compare the LHS and RHS. We clearly have that
$$
\cos^2(x)=1-\sin^2(x)\Longleftrightarrow \cos^2(x)+\sin^2(x)=1,
$$
and we know this famous result to be true (or easily established otherwise). Hence, we have proved $(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Tricky 3d geometry problem We have a cube with edge length $L$, now rotate it around its major diagonal (a complete turn, that is to say, the angle is 360 degrees), which object are we gonna get?
Astoundingly the answer is D. And here is a demonstration:
Well now I'm required to calculate the volume of this monster. It's far beyond my capability. I don't know how to analytically describe the curve between the two cones (although I believe it is a hyperbola). And I'm still not sure why it should be a curve rather than a segment or something. Could you help me? Thanks in advance.
| Let's take the unit centered cube, with vertexes at $\pm 1$. To rotate it so that its main diagonal gets aligned with the $x$ axis (vertical axis in the figure) we can use two rotations along two axes, the first by 45 degrees, the second by $\tan^{-1}(\sqrt{1/2})=\sin^{-1}(\sqrt{1/3})$. We get then the rotation matrix:
$$ Q=
\begin{pmatrix}
\sqrt{\frac{1}{3}} & 0 & \sqrt{\frac{2}{3}} \\
0 & 1 & 0 \\
-\sqrt{\frac{2}{3}} & 0 & \sqrt{\frac{1}{3}} \\
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 \\
0 & \sqrt{\frac{1}{2}} & -\sqrt{\frac{1}{2}} \\
0 & \sqrt{\frac{1}{2}} & \sqrt{\frac{1}{2}} \\
\end{pmatrix}=
\begin{pmatrix}
\sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} & \sqrt{\frac{1}{3}} \\
0 & \sqrt{\frac{1}{2}} & -\sqrt{\frac{1}{2}} \\
-2\sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} & \sqrt{\frac{1}{6}} \\
\end{pmatrix}$$
Indeed, we can verify that the matrix is orthogonal and $Q \, (1, 1,1)'=(\sqrt{3} ,0, 0)'$
Lets consider first the upper part. That corresponds to the points spanned by the edges starting on the upper vertex, hence they correspond to the revolution of:
$$ \begin{pmatrix} x \\y \\ z \end{pmatrix} = Q \begin{pmatrix} 1 \\\alpha \\ 1 \end{pmatrix} =
\begin{pmatrix}
\frac{1}{\sqrt{3}}(\alpha+2) \\
\frac{1}{\sqrt{2}}(\alpha-1)\\
\frac{1}{\sqrt{6}}(\alpha-1)
\end{pmatrix}
$$
Then $\alpha=\sqrt{3} x-2$, in the range $\alpha \in[-1,1]$, or $x\in[1/\sqrt{3},\sqrt{3}]$.
The rotations along the $x$ axis will keep $r^2=y^2+z^2$ constant, and so
$$ r^2= \frac{2}{3}(\alpha-1)^2=\frac{2}{3}(3-\sqrt{3}x)^2$$
Or
$$r = \sqrt{6}\left(1-\frac{x}{\sqrt{3}}\right)$$
For the next part, we consider another edge, starting from a neighbour vertex, say from $(1,1,-1)'$:
$$ \begin{pmatrix} x \\y \\ z \end{pmatrix} = Q \begin{pmatrix} 1 \\\alpha \\ -1 \end{pmatrix} =
\begin{pmatrix}
\frac{1}{\sqrt{3}}\alpha \\
\frac{1}{\sqrt{2}}(\alpha+1)\\
\frac{1}{\sqrt{6}}(\alpha-3)
\end{pmatrix}
$$
with $\alpha=\sqrt{3} x$, in the range $\alpha \in[-1,1]$, or $x\in[-1/\sqrt{3},1/\sqrt{3}]$.
And here the radius is
$$ r^2= y^2+z^2=\frac{2}{3}\alpha^2+2=2x^2+2$$
Then the radius is
$$ r(x)=\begin{cases}
\sqrt{6}\left(1-\frac{x}{\sqrt{3}}\right) & \mbox{if } 1/\sqrt{3} \le x \le \sqrt{3}\\
\sqrt{2x^2+2} & \mbox{if } 0 \le x \le \sqrt{1/3}
\end{cases}$$
(This seems to agree with String's answer.)
So, yes the middle cross section is an hyperbola.
To compute the total volume you need to integrate : $V = 2 \int_0^\sqrt{3} \pi r(x)^2 dx $ and scale the result by multiplying it by $(L/2)^3$ (because our cube has edge length $2$, instead of $L$)
| {
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Find all possible values of $ a^3 + b^3$ if $a^2+b^2=ab=4$. Find all possible values of $a^3 + b^3$ if $a^2+b^2=ab=4$.
From $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(4-4)=(a+b)0$. Then we know $a^3+b^3=0$.
If $a=b=0$, it is conflict with $a^2+b^2=ab=4$.
If $a\neq0$ and $b\neq0$, then $a$ and $b$ should be one positive and one negative. This contradict with $ab$=4.
I don't know the solution.
Any help please.
| Suppose there were real solutions to $a^2 + b^2 = ab = 4$. Then $$0 \leq (a - b)^2 = a^2 - 2ab + b^2 = (a^2 + b^2) - 2ab = -4$$ which is a a contradiction. Thus there cannot be real $a, b$ satisfying the requirements.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Write coupled wave equations as a decoupled system Given the coupled wave equations:
$$
\frac{\partial }{\partial t} [u~~v] +
\begin{pmatrix}
a & c \\
c & a
\end{pmatrix} \frac{\partial }{\partial x} [u~~v] = [0~~0].
$$
I want to transform this problem into a decoupled system with variables [r,s]. Can someone give me an idea on where to start?
| The system
$$
\frac{\partial }{\partial t} \begin{pmatrix} u \\ v \end{pmatrix} +
\begin{pmatrix}
a & c \\
c & a
\end{pmatrix} \frac{\partial }{\partial x} \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.
$$
leads to the system
\begin{align}
\begin{pmatrix} u_{t} \\ v_{t} \end{pmatrix} = \begin{pmatrix} a u_{x} + c v_{x} \\ c u_{x} + a v_{x} \end{pmatrix}
\end{align}
Equating both sides leads to the two equations
\begin{align}
(1) \hspace{10mm} u_{t} &= a u_{x} + c v_{x} \\
(2) \hspace{10mm} v_{t} &= c u_{x} + a v_{x}.
\end{align}
Now from (1) it is seen that $u_{x} = (1/c)( v_{t} - a v_{x} )$. Differentiate this with respect to $t$ and also differentiate (1) with respect to $x$ to obtain
\begin{align}
\frac{1}{c} ( v_{xt} - a v_{xx} ) = a \cdot \frac{1}{c} ( v_{xt} - a v_{xx}) + c v_{xx}
\end{align}
which yields
\begin{align}
v_{tt} - 2 a v_{xt} = (c^{2} - a^{2} ) v_{xx}.
\end{align}
In a similar manor it can be shown that $u$ satisfies the same equation.
The decoupled set of equations, $u(x,t)$ and $v(x,t)$ both satisfy the equation
\begin{align}
\frac{\partial^{2} F}{\partial t^{2}} - 2 a \frac{\partial^{2} F}{\partial x \partial t} = (c^{2} - a^{2}) \frac{\partial^{2} F}{\partial x^{2}}
\end{align}
| {
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Find the derivative using the chain rule and the quotient rule
$$f(x) = \left(\frac{x}{x+1}\right)^4$$ Find $f'(x)$.
Here is my work:
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4\left(x+1\right)^3x^4}{\left(x+1\right)^8}$$
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4x^4\left(x+1\right)^3}{\left(x+1\right)^8}$$
I know the final simplified answer to be:
$${4x^3\over (x+1)^5}$$
How do I get to the final answer from my last step? Or have I done something wrong in my own work?
| it is easier to use logarithmic differentiation on this problem. here is how you do this. $$y = \left(\frac x{x+1}\right)^4 \to \ln y = 4 \ln x - 4 \ln x + 1\to \frac{dy}{y} =4\left(\frac{dx}{x} - \frac{dx}{x+1} \right) = \frac{4dx}{x(x+1)} $$ multiplying through by $y$ gives you $$\frac{dy}{dx} = \frac{4x^3}{(x+1)^5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$ I need to solve
$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$
I've chosen the substitution variables
$$u=\sqrt{x^2+2}$$
$$du=\frac{x}{\sqrt{x^2+2}}$$
However, I am completly stuck at
$$\int\sqrt{1+xu} dx$$
Which let me believe I've chosen wrong substitution variables.
I've then tried letting $u=x^2+2$ or simply $u=x$, but it does not help me at all solving it.
Would someone please give me an hint on this ?
Thanks.
| Consider the integral
\begin{align}
I = \int \sqrt{1 + x \sqrt{x^{2} + 2}} \, dx
\end{align}
Make the substitution $x = \sqrt{2} \, csch(t)$ to obtain the integral
\begin{align}
I = - \sqrt{2} \, \int \sqrt{1 + 2 \, csch(t) \, coth(t)} \cdot csch(t) \, coth(t) \, dt.
\end{align}
Now Wolfram can calculate the integral and provides the result
\begin{align}
- \frac{I}{\sqrt{2}} &= \frac{1}{2} \sinh(t) \sqrt{2 coth(t) csch(t)+1} \left[\frac{g(t)}{\sqrt{4 cosh(t)+cosh(2 t)-1}} - csch^2(t) \right]
\end{align}
where
\begin{align}
g(t) = \ln\left(\tanh^2\left(\frac{t}{2}\right)\right) - \ln\left(1 + \tanh^2\left(\frac{t}{2}\right) + \sqrt{-\tanh^4\left(\frac{t}{2}\right) + 2 \tanh^2\left( \frac{t}{2}\right) +1} \right)
\end{align}
By making a careful backward substitution the result becomes
\begin{align}
I &= \frac{1}{\sqrt{2}} \left[ x \sqrt{1+x\sqrt{x^{2}+2}} + \ln\left( 1 + \frac{(x + \sqrt{x^2 + 2})^{2}}{2} + \sqrt{\frac{(x + \sqrt{x^2 + 2})^{4}}{4} + (x + \sqrt{x^2 + 2})^{2} - 1} \, \right) \right]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Integration of fraction with square root I have a problem with integrating of fraction
$$
\int \frac{x}{x^2 + 7 + \sqrt{x^2 + 7}}
$$
I have tried to rewrite it as $\int \frac{x^3 + 7x - x \sqrt{x^2 + 7}}{x^4 + 13x^2 + 42} = \int \frac{x^3 + 7x - x \sqrt{x^2 + 7}}{(x^2 + 6)(x^2 + 7)}$ and then find some partial fractions, but it wasn't succesful.
| Consider the integral
\begin{align}
I = \int \frac{x}{x^{2} + 7 + \sqrt{x^2 + 7}} \, dx
\end{align}
Make the substitution $u = \sqrt{x^{2} + 7}$ for which $x = \sqrt{u^{2} - 7}$ and the integral becomes
\begin{align}
I &= \int \frac{\sqrt{u^{2}-7}}{u^{2} + u} \cdot \frac{u \, du}{\sqrt{u^{2}-7}} \\
&= \int \frac{du}{1+u} \\
&= \ln(1+u)
\end{align}
Making the reverse substitution yields
\begin{align}
\int \frac{x}{x^{2} + 7 + \sqrt{x^2 + 7}} \, dx = \ln(1 + \sqrt{x^{2} + 7}).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does every $9 \times 9$ Latin square contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?
Q: Does every $9 \times 9$ Latin square on the symbol set $\{1,2,\ldots,9\}$ contain a $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$?
This one has $1728$ such submatrices, which is as low as I've gotten:
$$
\begin{bmatrix}
6 & 7 & 8 & 9 & 1 & 4 & 2 & 3 & 5 \\
5 & 6 & 1 & 7 & 2 & 8 & 3 & 4 & 9 \\
9 & 1 & 6 & 2 & 4 & 3 & 7 & 5 & 8 \\
4 & 5 & 3 & 6 & 8 & 7 & 1 & 9 & 2 \\
1 & 2 & 4 & 8 & 3 & 5 & 9 & 6 & 7 \\
2 & 3 & 7 & 4 & 9 & 6 & 5 & 8 & 1 \\
8 & 9 & 2 & 3 & 5 & 1 & 6 & 7 & 4 \\
7 & 8 & 5 & 1 & 6 & 9 & 4 & 2 & 3 \\
3 & 4 & 9 & 5 & 7 & 2 & 8 & 1 & 6 \\
\end{bmatrix}$$
It doesn't seem likely that random Latin squares will help much; they average in the thousands of such submatrices. The one above is the best random Latin square I've found so far (although, I haven't busted a gut doing this; it seems like it won't work anyway).
The groups of order $9$ have lots ($C_9$ has $5832$ and $C_3 \times C_3$ has $19440$).
This question was motivated by answering this math.SE question which asks if any $9 \times 9$ Latin square can have its rows and columns permuted to give a sudoku square.
One way to find an explicit counterexample would be to find a $9 \times 9$ Latin square with no $3 \times 3$ submatrix containing each symbol in $\{1,2,\ldots,9\}$. But this attempt didn't work since I couldn't find one. Hence my question.
| How about this one?
$$
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
2 & 1 & 4 & 3 & 6 & 7 & 8 & 9 & 5 \\
3 & 4 & 2 & 1 & 8 & 9 & 5 & 6 & 7 \\
4 & 3 & 1 & 2 & 7 & 8 & 9 & 5 & 6 \\
5 & 6 & 9 & 8 & 2 & 3 & 4 & 7 & 1 \\
6 & 7 & 5 & 9 & 1 & 2 & 3 & 4 & 8 \\
7 & 8 & 6 & 5 & 9 & 1 & 2 & 3 & 4 \\
8 & 9 & 7 & 6 & 4 & 5 & 1 & 2 & 3 \\
9 & 5 & 8 & 7 & 3 & 4 & 6 & 1 & 2 \\
\end{bmatrix}$$
I discovered it by modifying a Latin Square solver to add the restriction that no 3x3 submatrix can contain 9 different values. It found this in about 1 hour.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the values of m and n(Trigononetry in series) $$\sin ^6(1)+\sin ^6(2)+... ...+\sin ^6(89)=\frac{m}{n}$$
Find $\frac{m}{n}$ in its simplest form , and hence find both values. (All angles are in degree)
I've no idea how to start to solve this questions. Need some guidance for it. Thanks in advance.
| $$\begin{align} 1 &= (\sin^2 t + \cos^2 t)^3 \\
&= \sin^6 t + 3\sin^4t \cos^2 t + 3 \sin^2t+ \cos^4t + \cos^6 t \\
&= \sin^6 t + \cos^6 t + 3\sin^2t \cos^2 t \\
&=\sin^6 t + \cos^6 t + \frac34\sin^2 (2t) \\
\end{align} $$
let $$ S= \sin^6 1^\circ + \sin^6 2\circ + \cdots + \sin^6 89^\circ. $$ then we have $$\begin{align}2S &= (\sin^6 1^\circ + \sin^6 89^\circ) + (\sin^6 2^\circ + \sin^6 87^\circ) + \cdots + (\sin^6 89^\circ + \sin^6 1^\circ)\\
&= (\sin^6 1^\circ + \cos^6 1^\circ) + (\sin^6 2^\circ + \cos^6 2^\circ) + \cdots + (\sin^6 89^\circ + \cos^6 89^\circ)\\
&= 89 - \frac34\left(\sin^2 2^\circ + \sin^2 4^\circ + \cdots+\sin^2178^\circ \right)\\
&=89 - \frac 34 - \frac32\left(\sin^2 2^\circ + \sin^2 4^\circ + \cdots+\sin^2 88^\circ \right)\\
&=89 - \frac 34 - \frac34\left((\sin^2 2^\circ + \sin^288^\circ) + (\sin^2 4^\circ + \sin^2 86) + \cdots+(\sin^2 88^\circ + \sin^2 2^\circ \right)\\
&=89 - \frac 34 - \frac34 44\\
\end{align}$$
so $$ S = \frac{221} 8$$
there is an easier way to this. thanks to claude. see the comments below. we can express $\sin^6 x $ as a $\cos$ series in the following way:
$$\sin^6 x = -\frac{15}{32}\cos 2x +\frac 3 {16} \cos 4x - \frac 1 {32} +\cos 6x + \frac 5{16} $$ the $\cos 2x,\cos 6x$ sum to zero and $\cos 4x$ terms sum to $-1$ therefore the
$$ S = -\frac{3}{16} + 89 \times \frac 5{16} = \frac{221}{8} .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Taylor series for the function $f(z) = \frac{1}{(z-5)(z-7)}$ on a disc centered at point $z_0=3$ I started by expressing the function as sum of two fractions using partial fraction decomposition to get $\frac{-1}{2(z-5)} + \frac{1}{2(z-7)}$
However I could only then end up writing that as the sum of two power series :
$$\frac{1}{10} \sum_{n=0}^\infty \frac{z^n}{3^n} - \frac{1}{14} \sum_{n=0}^\infty \frac{z^n}{7^n}$$ and I'm unsure how to write this in the form of a Taylor series.
| Hint: For example,
$$
\frac{1}{z-7} = \frac{1}{(z-3)-4} = -\frac{1}{4} \frac{1}{1-(z-3)/4} = -\frac{1}{4} \sum_{i=0}^\infty \left(\frac{z-3}{4}\right)^i.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is continuous and integrable function bounded? I have a function $f: \mathbb R \rightarrow \mathbb R$ continuous and integrable on $\mathbb R$. Is $f$ bounded?
| No! See this example here If $f\in L^1(\Bbb R,dx)$ then prove that for almost every $x\in\Bbb R$ $\lim\limits_{n\to \infty} f(nx) = 0.$
Where the function is a Polynomial on each interval $[n, n+\frac{1}{2^n}]$ and zero elsewhere.
Consider $f:\Bbb R\to \Bbb R$
\begin{align*}
f(x)= \begin{cases}
2^{n/2}P_n(x)& \text{if} ~~ n\le x\le n+\frac{1}{2^n} ~~\text{$n\in\mathbb{N}$}\\
0 &\text{ if}~~~j+\frac{1}{2^j}\le x\le j+1~~\text{ $j\in\mathbb{N}$}\\ f(-x)& \text{if }~~x <0.
\end{cases}
\end{align*}
where $P_n(x)= -2^{2n+2}(x-n)^2+2^{n+2}(x-n)$. One can check that $f$ is an even and continuous function. since $P_n(n)= P_n(n+\frac{1}{2^n})=0$. Next observe that $f\in L^1(\Bbb R)$ since
\begin{align*}
\int_\mathbb{R} f(x)\,dx= 2\sum_{n=0}^{\infty}2^{n/2}\int_{n}^{n+\frac{1}{2^n}} P_n(x) dx
\\= 2\sum_{n=0}^{\infty}-4\cdot 2^{n/2}
\left[ \frac{2^{2n}}{3}(x-n)^3-\frac{2^n}{2}(x-n)^2\right]_{n}^{n+\frac{1}{2^n}}\\
= \frac43\sum_{n=0}^{\infty} \frac{1}{2^{n/2}}<\infty.
\end{align*}
Moreover, for every $n\in\mathbb{N}$ one has
\begin{align*}
f(n+\frac{1}{2^{n+1}}) = 2^{n/2} P_n ( n+\frac{1}{2^{n+1}}) = 2^{n/2}\to \infty
\end{align*}
Therefore $f$ is unbounded.
Another example
\begin{align*}
f(x)= \begin{cases}
2^{\frac32(n+1)} x- n2^{\frac32(n+1)}& \text{if} ~~ n\le x\le n+\frac{1}{2^{n+1}} ~~\text{$n\in\mathbb{N}$}\\
-2^{\frac32(n+1)} x= n2^{\frac32(n+1)} +2^{\frac{n+3}{2}}& \text{if} ~~ n+\frac{1}{2^{n+1}}\le x\le n+\frac{1}{2^n} ~~\text{$n\in\mathbb{N}$}\\
0 &\text{ if}~~~j+\frac{1}{2^j}\le x\le j+1~~\text{$j\in\mathbb{N}$}\\ f(-x)& \text{if }~~x <0.
\end{cases}
\end{align*}
It is easy to check that $f\in L^1(\Bbb R)\cap C(\Bbb R)$
\begin{align*}
&\int_\mathbb{R} f(x)\,dx=2\sum_{n=0}^{\infty} 2^{-\frac{n+1}{2}}<\infty\\
&f(n)= f(n+\frac{1}{2^{n}})=0\qquad f(n+\frac{1}{2^{n+1}}) = 2^{\frac{n+1}{2}}\to \infty
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Continued Fraction Algorithm for 113/50 The numbers $a_k$ can be found for $\frac{113}{50}$ by using a continued fraction algorithm. Note that $\frac{113}{50}$ is rational, and as a result it will have to terminate. Can anyone help me determine the numbers $a_k$ for $\frac{113}{50}$?
What I have Done:
$\frac{113}{50} = 2.26$
$i=2$
$2.26 - 2 = .26$
$\frac{26}{100}$
$\frac{100}{26} = 3+ \frac{22}{26} $
3 is $a_1$
$\frac{26}{22} = 1 + \frac{4}{22} $
1 is $a_2$
$\frac{22}{4} = 5 + \frac{2}{4}$
$a_3 = 5$
$\frac{4}{2} = 2 $
$a_4 = 2 $
Is this correct, and what do I do next?
| Meanwhile, here is the calculation of the "convergents." It begins with the formal fractions $0/1$ and the fake (but necessary) $1/0.$ Then, separately for numerator and denominator, for each "digit" or "partial quotient" $a_k,$ you get then next number from the evident rule, now visible: for example, $0 + 2 \cdot 1 = 2,$ then $1 + 3 \cdot 2 = 7,$ then $2 + 1 \cdot 7 = 9,$ and so on. Similar for denominators. I have found this litle grid way of presenting the convergents very helpful. For on thing, you can clearly see how the "cross product" of two consecutive convergents is $\pm 1.$ As in: $1 \cdot 1 - 2 \cdot 0 = 1, \; \;$ $2 \cdot 3 - 7 \cdot 1 = -1, \; \;$ $7 \cdot 4 - 9 \cdot 3 = 1, \; \;$ $9 \cdot 23 - 52 \cdot 4 = -1, \; \;$ and so on.
$$
\begin{array}{cccccccccccc}
& & 2 & & 3 & & 1 & & 5 & & 2 & \\
\frac{0}{1} & \frac{1}{0} & & \frac{2}{1} & & \frac{7}{3} & & \frac{9}{4} & & \frac{52}{23} & & \frac{113}{50}
\end{array}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Critical points of a function $f(x) = x\sqrt{x-a}$ Find the critical points of a function $f(x) = x\sqrt{x-a}$.
A function $f(x)$ is said to have critical points at points $c$ such that $f^\prime(c)$ is $0$ or undefined.
For a function $f(x) = x\sqrt{x-a}$ :
$$\hspace{-0.5 in} f^\prime(x) = \frac{3x - 2a}{2\sqrt{x - a}}$$
It may be observed that the resulting derivative is a rational function $f^\prime(x) = \frac{g(x)}{h(x)}$ such that:
$$
\begin{align*}
f^\prime(c) &= 0 \hspace{0.825 in} \text{ when } g(c) = 0 \text{ and } h(c) \neq 0
\\ f^\prime(c) &\text{ is undefined} \hspace{0.2 in} \text{ when } h(c) = 0
\\
\\
\\ \end{align*}
$$
By this it is found that critical points will be at values $c$ where either $g(c) = 0$ or $h(c) = 0$.
$$
\begin{align*}
g(c) = 0 \implies 3c - 2a &= 0
\\ 3c &= 2a
\\ c &= \hspace{-.0325 in}\frac{2a}{3}
\\
\\ h(c) = 0 \implies 2\sqrt{c - a} &= 0
\\ c - a &= 0
\\ c &= a
\\
\\ \end{align*}
$$
Therefore, critical points will be at values $c = \{\frac{2a}{3}, a\}$.
However, my book contradicts my answer stating $c = \frac{2a}{3} \iff a < 0$.
$\hspace{.5 in}$
My answer is for a general $a$ of any real-value and includes the $c = a$ as a critical point. What's the issue here?
Edit:
In light of the redefined idea of a critical number being only those values $c$ which are within the domain of $f(x)$:
$$
\\ \begin{align*}
\\ x &\geq a \text{ for } a \geq 0
\\ x &\geq a \text{ for } a < 0
\\ \end{align*}
$$
Critical points must be within these domains when considering either case.
Assuming the domain of the function $f(x)$ was $(-\infty, \infty)$ there would be critical points at $x = \{ \frac{2a}{3}, a \} \text{ } \forall a$.
However, the function is only defined on the interval $[a, \infty)$ in either case.
$$
\\ \hspace{-.25 in} \frac{2a}{3} < a \text{, } \hspace{.5 in} a > 0
\\ \hspace{-.25 in} \frac{2a}{3} > a \text{, } \hspace{.5 in} a < 0
$$
This means that $x = \{ \frac{2a}{3}, a \}$ are critical points for $a < 0$, but $x = \{ a\}$, and $x \neq \{\frac{2a}{3}\}$ for $a \geq 0$.
This is not a selection, however. Insight?
| I normally don't consider a number a critical number unless it is part of the domain of the original function. So looking at $f$ and $f'$ domain we want $x>a$. You found $x=\frac{2a}{3}$ to be a critical number. So that means we want $\frac{2a}{3}>a$ What do you get when you solve that for $a$.
| {
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Show that $\pi =4-\sum_{n=1}^{\infty }\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}$ Show that $$\pi =4-\sum_{n=1}^{\infty }\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}$$
I found the formula of $\pi$ by using the numerical calculation but I dont have the proving. Any help would be appreciated.
| Note that $$4-\pi = \frac{4}{3}-\frac{4}{5}+\frac{4}{7}-\ldots$$ and by the Euler transform
$$
4-\pi = \frac{2}{3} + \frac{2}{3\cdot 5}+\frac{ 2\cdot 2}{3\cdot 5\cdot 7} +\frac{2\cdot 3 \cdot 2}{3\cdot 5\cdot 7\cdot 9}+\ldots=\sum_{n=1}^{\infty}\frac{2^{n+1}n!(n-1)!}{(2n+1)!}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1203368",
"timestamp": "2023-03-29T00:00:00",
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Limit of $\sqrt{25x^{2}+5x}-5x$ as $x\to\infty$ $\hspace{1cm} \displaystyle\lim_{x\to\infty} \left(\sqrt{25x^{2}+5x}-5x\right) $
The correct answer seems to be $\frac12$, whereas I get $0$.
Here's how I do this problem:
$$ \sqrt{25x^{2}+5x}-5x \cdot \frac{\sqrt{25x^{2}+5x}+5x}{\sqrt{25x^{2}+5x}+5x} = \frac{25x^2+5x - 25x^2}{\sqrt{25x^{2}+5x} +5x} = \frac {5x}{\sqrt{25x^{2}+5x}+5x} $$
$\sqrt{25x^{2}+5x}$ yields a bigger value than $5x$ as $x$ becomes a very big number. So the denominator is clearly bigger than numerator. So in this case, shouldn't the answer be $0$?
However, if I keep going and divide both numerator and denominator by $x$ I get:
$$ \frac{5}{ \frac{\sqrt{25x^2+5x}}{x} + 5 }$$
In the denominator, $\frac{\sqrt{25x^2+5x}}{x}$
yields a big number (because top is increasing faster than the bottom), in fact, it goes to infinity as $x$ goes to infinity.
In that case, it's just $5$ divided by something going to infinity, therefore, the answer should be $0$, but it's not, why?
| $$\sqrt{25x^{2}+5x}-5x \cdot \frac{\sqrt{25x^{2}+5x}+5x}{\sqrt{25x^{2}+5x}+5x} = \frac{25x^2+5x - 25x^2}{\sqrt{25x^{2}+5x} +5x} = \dfrac{5x}{\sqrt{25x^{2}+5x} +5x}$$
Your work is fine so far. Next factor out $x$ from denominator and cancel it with numerator
$$ \dfrac{5x}{\sqrt{25x^{2}+5x} +5x} = \dfrac{5x}{\sqrt{x^2(25+\frac{5}{x})} +5x} =\dfrac{5x}{x\left(\sqrt{25+\frac{5}{x}} +5\right)} = \dfrac{5}{\sqrt{25+\frac{5}{x}} +5} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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To calculate the expectation of a variable with a given pdf Let the pdf of a random variable X be given by $f(x)=ae^{-x^2-bx}, -\infty<x<\infty$. If $E(X)=-\frac{1}{2}$, then
(A)$a=\frac{1}{\sqrt{\pi}}e^{-1/4},b=1$
(B)$a=\frac{1}{\sqrt{\pi}}e^{-1/4},b=-1$
(C)$a=\sqrt{\pi}e^{-1/4},b=1$
(D)$a=\sqrt{\pi}e^{-1/4},b=-1$
My Steps:
$\begin{aligned}
E(X) & = \displaystyle\int\limits_{-\infty}^{\infty}xf(x)\;dx=-\frac{1}{2} \\
& = \displaystyle a \int\limits_{-\infty}^{\infty}xe^{-x^2-bx}\;dx \\
& = \displaystyle ae^{b/4} \int\limits_{-\infty}^{\infty}xe^{-(x+b/2)^2}\;dx \\
& = \displaystyle ae^{b/4} \int\limits_{-\infty}^{\infty}(x+b/2)e^{-(x+b/2)^2}\;dx - \displaystyle ae^{b/4}\frac{b}{2} \int\limits_{-\infty}^{\infty}e^{-(x+b/2)^2}\;dx \\
\end{aligned}$
I need hints to integrate the second part. Please advise.
| Recall that a normally distributed random variable $X$ with mean $\mu$ and standard deviation $\sigma$ has the probability density function $$f_X(x) = \frac{1}{\sqrt{2\pi} \sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad -\infty < x < \infty.$$ So if we set $\mu = -1/2$, then $\operatorname{E}[X] = \mu = -1/2$ as desired. Now write $$ae^{-x^2-bx} = ae^{b^2/4} e^{-x^2-bx-b^2/4} = ae^{b^2/4} e^{-(x+b/2)^2}.$$ So we now seek a choice of $\sigma > 0$ such that $$\frac{(x+1/2)^2}{2\sigma^2} = (x+b/2)^2,$$ and simultaneously, $$\frac{1}{\sqrt{2\pi} \sigma} = ae^{b^2/4}.$$ This suggests letting $b = 1$, $\sigma = 1/\sqrt{2}$, and solving for $a$ from these choices gives $$a = \pi^{-1/2} e^{-1/4},$$ which is answer choice A.
| {
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Does this series $\sum_{i=0}^n \frac{4}{3^n}$ diverge or converge? I a newbie to series, and I have not done too much yet. I have an exercise where I have basically to say if some series are convergent or divergent. If convergent, determine (and prove) the sum of the series.
This is the first series:
$$\sum_{i=0}^n \frac{4}{3^i}$$
I have heard about the ratio test, so I decided to try to apply it in this case to see if the series is or not convergent.
The ratio test is basically defined like this:
$$L = \lim{\left|\frac{a_{n+1}}{a_n}\right|}$$
Where $a_n$ is in this case $\frac{4}{3^n}$ and $a_{n+1}$ is consequently $\frac{4}{3^{n+1}}$. Thus, we have:
$$L = \lim{\left|\frac{\frac{4}{3^{n+1}}}{\frac{4}{3^n}}\right|} = \lim{\left|\frac{4 \cdot 3^n}{3^{n+1} \cdot 4}\right|} = \lim{\left|\frac{3^n}{3^{n+1}}\right|} = \lim{\left|\frac{3^n}{3^n\cdot 3}\right|} = \lim{\left|\frac{1}{3}\right|} = \frac{1}{3} < 1$$
From the ratio test, we know if $L< 1$, then the series converges.
Now, I need to find the sum. I have followed a video which explain how to find the sum using the ratio, which sincerely I have not understand well what it is. Can you explain a general rule to find the ratio, and what exactly is it?
Ok. So, following the same process done during the video, I have:
$$\sum_{i=0}^n \frac{4}{3^n} = 4 \sum_{i=0}^n \frac{1}{3^n} = 4 \sum_{i=0}^n \left(\frac{1}{3}\right)^n$$
Apparently, our ratio $r$ is $\frac{1}{3}$.
Now, to find the sum of the series, he uses a formula:
$$\frac{\text{first term}}{1 - r}$$
Where first term is the first term of the series. Can you explain where this formula comes from?
Applying that formula, we obtain:
$$\frac{\left(\frac{1}{3}\right)^0}{1 - \frac{1}{3}} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$$
I checked the sum using a online calculator, and apparently it is not $\frac{3}{2}$, so I decided to manipulate, and I thought I left a $4$ outside the series, so I multiply that $4$ by $\frac{3}{2}$, which produces $\frac{12}{2} = 6$, which is exactly the sum I found using 2 online calculators.
Now, my 3rd question is: why what I did is correct (or not correct)?
| Let $S_N=\sum_{i=0}^{N} r^i=1+r+r^2+\cdots r^N$. Note that
$$\begin{align}
rS_N&=\sum_{i=0}^{N} r^{i+1}\\
&=r+r^2+r^3+\cdots+r^N+r^{N+1}
\end{align}$$
So, if we form the difference $S_N-rS_N$ we find $S_N-rS_N=(1-r)S_N=1-r^{N+1}$ where upon solving for $S_N$ we find
$$\begin{align}
S_N&=\frac{1-r^{N+1}}{1-r}\\
\end{align}$$
If $r<1$, then $r^{N+1} \to 0$ as $N \to \infty$ and we have
$$\sum_{i=0}^{\infty} r^{i}=\frac{1}{1-r}$$.
If $r=\frac{1}{3}$, then $\sum_{i=0}^{\infty} (\frac13)^{i}=\frac{1}{1-\frac13}=\frac32$.
If the series is $4\sum_{i=0}^{\infty} (\frac13)^{i}$, then we have $4\sum_{i=0}^{\infty} (\frac13)^{i}=4\frac{1}{1-\frac13}=4\frac32 = 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1210267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the derivative of the function $F(x) = \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt$. $$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+t^4}}2x - \frac{1}{\sqrt{2+t^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+t^4}} - \frac{\sec^2{x}}{\sqrt{2+t^4}} \\
&= \frac{2x-\sec^2{x}}{\sqrt{2+t^4}} \\
\end{align}$$
Update:
Is this looking better??
$$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+(x^2)^4}}2x - \frac{1}{\sqrt{2+(\tan{x})^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+x^8}} - \frac{\sec^2{x}}{\sqrt{2+\tan^4{x}}} \\
\end{align}$$
Is this correct?
|
Hint to find derivative: Use FTC, http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements, $F'(x)=\frac{d}{dx}\int^x_a f(t)dt=f(x)$
Answer to original question: HINT to integrate: \begin{align}
\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt &= \frac{1}{\sqrt2} \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{1+\frac{t^4}{2}}}\,dt \\
&= \frac{1}{\sqrt2} \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{1+\frac{t^4}{2}}}\,dt \\
\end{align}
Then let $\tan^2 u=t^4/2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving $(n+1)!>2^{n+3}$ for all $n\geq 5$ by induction I am stuck writing the body a PMI I have been working on for quite some time.
Theorem: $∀n∈N ≥ X$, $(n+1)!>2^{n+3}$
I will first verify that the hypothesis is true for at least one value of $n∈N$.
Consider $n=3$: (not valid)
$$(3+1)!>2^{3+3} \implies 4!>2^{6} \implies 24>64$$
Consider $n=4$: (not valid)
$$(4+1)!>2^{4+3} \implies 5!>2^{7} \implies 120>128$$
Consider $n=5$: (valid)
$$(5+1)!>2^{5+3} \implies 6!>2^{8} \implies 720>256$$
Consider $n=6$: (valid)
$$(6+1)!>2^{6+3} \implies 7!>2^{9} \implies 5040>512$$
So clearly $X$ is $5$.
For the inductive assumption, we will assume the hypothesis holes from $n=5$ up to some arbitrary values $k$: $(k+1)!>2^{k+3}$
This is where I am lost. Originally I had written:
Now I will prove true for $k+1$ showing that: $(k+2)!>2^{k+4}$. Consider the $k+1$ term:
$$(k+2)(k+1)k! = (k+2)(k+1)2^k = (k+2)(k+1)2^k = (2)(2)(2^k) = (4)(2^k) = 2^{k+4}$$
by the inductive assumption since $k>5$ so $k+1>2$ and $k+2>2$ but I know this isn't correct.
I know the proof should look something like this, but I have no idea why:
$$(k+2)(k+1)!>(k+2)(2^{k+3})>2(2^{k+3})=2^{k+4}$$
| First, show that this is true for $n=5$:
$(5+1)!>2^{5+3}$
Second, assume that this is true for $n$:
$(n+1)!>2^{n+3}$
Third, prove that this is true for $n+1$:
$(n+2)!=\color{red}{(n+1)!}\cdot\color{blue}{(n+2)}>\color{red}{2^{n+3}}\cdot\color{blue}{(n+2)}>\color{red}{2^{n+3}}\cdot\color{blue}{2}=2^{n+4}$
Please note that the assumption is used only in the part marked red.
| {
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"url": "https://math.stackexchange.com/questions/1214914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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proving recursively defined sequence by induction I would like to prove the following recursively defined sequence from $n-1$ to $n$ by induction. Im not realy sure about it. Any help or alternative ways to understand and prove it are highly appreciated :
$0,1,4,12,35,98$
$a_0=0$, $a_1=1$, $a_n=a_{n-1}+5a_{n-2}+3$ for $n\geq2$
To prove $a_n\leq 3^n$
I thougt of it as: $a_{n-1}\leq 3^{n-1}$, $a_{n-2}\leq 3^{n-2}$
and thus:
$a_n\leq 3^{n-1} + 5\cdot 3^{n-2}+3$
$=3^{n-2} \cdot(3+5)+3$
$=3^{n-2} \cdot(8)+3$
$=3^{n-2} \cdot(9)+3$
$=3^{n-2} \cdot(3 \cdot 3)+3$
$=3^{n}+3$
$\leq 3^{n}$
| It will be easier to show $a_n \le 3^n-1$ (and hence $3^n$).
$$\begin{align}
a_n &= a_{n-1} + 5a_{n-2}+3 \\
&\le (3^{n-1}-1) + 5(3^{n-2}-1) + 3 \\
&= 3^{n-1} + 5\cdot 3^{n-2} -3\\
&= 3^{n-1} + 3\cdot 3^{n-2} + 2\cdot 3^{n-2} -3\\
&= 3^{n-1} + 3^{n-1} + 2\cdot (3^{n-2} -1)-1\\
&\le 3^{n-1} + 3^{n-1} + 2\cdot 3^{n-2} - 1\\
&\le \underbrace{3^{n-1} + 3^{n-1} + 3^{n-1}}_{3^n} - 1
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A sequence is defined by $f(x) = 4^{x-1}$, find the sum of the first $8$ terms. A sequence is defined by $f(x) = 4^{x-1}$, find the sum of the first $8$ terms.
$\dfrac{a(1-r^n)}{1-r}$
$\dfrac{1(1-4^7)}{1-4} = 5461$.
The answer in the book is $21845$. How is this so?
Thank you
| The sum of the first $n$ terms of a geometric sequence $\{a_n\}$ with initial term $a_1$ and common ratio $r$ is $$s_n = a_1\frac{1 - r^{n}}{1 - r}$$
For the sequence defined by $f(n) = 4^{n - 1}$, $a_1 = 4^{1 - 1} = 4^0 = 1$, and $$r = \frac{a_{n + 1}}{a_n} = \frac{4^n}{4^{n - 1}} = 4$$ You wish to find the sum of the first eight terms of the sequence, so $n = 8$. Thus, you should obtain
$$s_8 = 4^0 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6 + 4^7 = 1 \cdot \frac{1 - 4^8}{1 - 4} = 21845$$
Derivation: If a geometric sequence $\{a_n\}$ has initial term $a_1$ and common ratio $r$, the first few terms of the sequence are $\{a_1, a_1r, a_1r^2, a_1r^3, \ldots\}$. In general, the $n$th term of the sequence is $a_1r^{n - 1}$. The sum of the first $n$ terms of the sequence is
$$s_n = a_1 + a_1r + a_1r^2 + \cdots + a_1r^{n - 1}\tag{1}$$
Hence,
$$rs_n = a_1r + a_1r^2 + a_1r^3 + \cdots + a_1r^{n - 1} + a_1r^n\tag{2}$$
Subtracting equation 2 from equation 1 yields
$$(1 - r)s_n = a_1 - a_1r^n = a_1(1 - r^n)$$
If $r \neq 1$, then
$$s_n = a_1\frac{1 - r^n}{1 - r}$$
If $n = 1$, then the sequence is constant, with $a_n = a_1$ for each $n \in \mathbb{N}$, so the sum of the first $n$ terms of the sequence is $s_n = na_1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that equation $x^6+x^5-x^4-x^3+x^2+x-1=0$ has two real roots Prove that equation
$$x^6+x^5-x^4-x^3+x^2+x-1=0$$ has two real roots
and $$x^6-x^5+x^4+x^3-x^2-x+1=0$$
has two real roots
I think that:
$$x^{4k+2}+x^{4k+1}-x^{4k}-x^{4k-1}+x^{4k-2}+x^{4k-3}-..+x^2+x-1=0$$
and
$$x^{4k+2}-x^{4k+1}+x^{4k}+x^{4k-1}-x^{4k-2}-x^{4k-3}-..+x^2+x-1=0$$
has two real roots but i don't have solution
| Let $f(x)=x^6+x^5-x^4-x^3+x^2+x-1=0$. Then $f(0)=-1$. Note for $x>0$
\begin{eqnarray}
f'(x)&=&(6x^5+2x)+(5x^4+1)-4x^3-3x^2\\
&\ge& 4\sqrt3x^3+2\sqrt5x^2-4x^3-3x^2\\
&=&(4\sqrt3-4)x^3+(2\sqrt5-3)x^2\\
&>&0
\end{eqnarray}
hance $f(x)$ is strictly increasing. Noting that $\lim_{x\to\infty}f(x)=\infty$, we have a unique root in $(0,\infty)$. Similarly, there is a unique root in $(-\infty,0)$. Thus $f(x)$ has only two real roots in $(-\infty,\infty)$. Using the same argument for other function, you can get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there a natural number for which all the sums and differences of its factor pairs are prime? The 8 factor pairs of e.g. 462 are
$((1, 462), (2, 231), (3, 154), (6, 77), (7, 66), (11, 42), (14, 33), (21, 22))$.
Of the 16 non-negative integers which are the sums and differences of these pairs (such as $462+1=463$, $462-1=461$, $21+22=43$, and $22-21=1$), 15 of them are primes. (The only non-prime is $22-21=1$.)
Is there an integer for which all the $2n$ sums and differences of its $n$ factor pairs are primes?
Obviously any such integer needs be the even number between a twin prime pair, and needs to be $ = 2$ (mod 4), but that's all I figured out.
Amongst integers less than $10^7$, 462 seems to have the uniquely highest fraction of prime sums/differences of factor pairs, with $15/16$. If there is no integer with a fraction of 1 (the question above), can there be any with a higher fraction than 15/16?
| It seems the following.
There is a following recursive way to search such the integer $N$.
$N=1\cdot k_1$. Since all of numbers $k_1\pm 1$ are prime, one of these numbers is odd. So we have
$N=1\cdot 2\cdot k_2$. Since all of numbers $2k_2\pm 1$, $k_2\pm 2$ are prime, if $k_2$ is not divisible by $3$, then one of these numbers is divisible by $3$, so it equals $3$. This case is considered separately, and in the rest of the cases we have
$N=1\cdot 2\cdot 3\cdot k_3$. Since all of numbers $6k_3\pm 1$, $3k_3\pm 2$, $2k_3\pm 3$, $k_3\pm 6$ are prime, if $k_2$ is not divisible by $7$, then one of these numbers is divisible by $7$, so it equals $7$. This case is considered separately, and in the rest of the cases we have
$N=1\cdot 2\cdot 3\cdot 7\cdot k_4$. Since all of numbers $42k_4\pm 1$, $21k_4\pm 2$ are prime, if $k_4$ is not divisible by $5$, then one of these numbers is divisible by $5$, so it equals $5$. This case is considered separately, and in the rest of the cases we have
$N=1\cdot 2\cdot 3\cdot 7\cdot 5 \cdot k_5$, and so forth...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factorize $(x-2)^5+x-1$? This is a difficult problem.
How to factorize this?
$$(x-2)^5+x-1$$
we can't do any thing now and we should expand it first:
$$x^5-10x^4+40x^3-80x^2+81x-33$$
but I can't factorize it.
| Using the substitution $t=x-2$, it suffices to factor $t^5 + t+1$.
But $t^2+t+1$ divides $t^5+t+1$. Why?
$t^2+t+1 = (t-\zeta)(t-\zeta')$, where $\zeta,\zeta'$ are the primitive $3rd$ roots of unity, and thus $\zeta^5 + \zeta + 1 = \zeta^2 + \zeta + 1 = 0$, similarly for $\zeta'$.
Now, the quotient $(t^5+t+1)/(t^2+t+1)$ is easily determined via polynomial division.
Alternative solution. Using $t^3-1=(t^2+t+1)(t-1)$, we get:
$$t^5+t+1=(t^5-t^2)+(t^2+t+1)=(t^3-1)t^2 + (t^2+t+1)$$
$$=(t^2+t+1)(t-1)t^2 + (t^2+t+1) = (t^2+t+1)((t-1)t^2+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ax^2+bx+c=0$ has roots $x_1,x_2$. what are the roots of $cx^2+bx+a=0$. Given solution: Dividing the first equation by $x^2$ we get
$c(\frac{1}{x^2})+b(\frac{1}{x})+a=0$
so $(\frac{1}{x_1}),(\frac{1}{x_2})$ are the roots of $cx^2+bx+a=0$.{How?It is not obvious to me.}
The answers so far are proving retrospectively that the roots are indeed those given above.I would like to know how the relation of the roots is derived.
| Given $ax^2 + bx + c = 0$,
$$\begin{align} ax^2 + bx + c& = 0\\ a + b\left(\frac{1}{x}\right) + c\left(\frac{1}{x^2}\right) & = 0 \tag{Divide both sides by $x^2$}\\ cy^2 + by + a & = 0\tag{Let $y = \frac{1}{x}$}\end{align}$$
We'd discover that this polynomial has roots $y_1$ and $y_2$. We know that $y = \frac{1}{x}$. So, without loss of generality, assume $y_1 = \frac{1}{x_1}$. Then $y_2 = \frac{1}{x_2}$. These are thus the roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to differentiate $y=\sqrt{\frac{1+x}{1-x}}$? I'm trying to solve this problem but I think I'm missing something.
Here's what I've done so far:
$$g(x) = \frac{1+x}{1-x}$$
$$u = 1+x$$
$$u' = 1$$
$$v = 1-x$$
$$v' = -1$$
$$g'(x) = \frac{(1-x) -(-1)(1+x)}{(1-x)^2}$$
$$g'(x) = \frac{1-x+1+x}{(1-x)^2}$$
$$g'(x) = \frac{2}{(1-x)^2}$$
$$y' = \frac{1}{2}(\frac{1+x}{1-x})^{-\frac{1}{2}}(\frac{2}{(1-x)^2})
$$
| Another idea:
$$\frac{1+x}{1-x}=\frac2{1-x}-1\implies \left(\frac{1+x}{1-x}\right)'=\frac2{(1-x)^2}\implies$$
$$\left(\sqrt\frac{1+x}{1-x}\right)'=\sqrt\frac{1-x}{1+x}\cdot\frac1{(1-x)^2}$$
Added on request:
$$\sqrt\frac{1-x}{1+x}\cdot\frac1{(1-x)^2}=\frac1{\sqrt{1+x}}\cdot\frac1{\sqrt{1-x}\cdot(1-x)}=\frac1{\sqrt{1-x^2}(1-x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
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quadratic equation what am I doing wrong? solve
$$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} $$
$$ \sqrt{5x+19} = \sqrt{x+7} + 2\sqrt{x-5} \Rightarrow $$
$$ 5x+19 = (x+7) + 4\sqrt{x-5}\sqrt{x+7} + (x+5) \Rightarrow $$
$$ 3x + 17 = 4\sqrt{x-5}\sqrt{x+7} \Rightarrow $$
$$ 9x^2 + 102x + 289 = 16(x+7)(x-5) \Rightarrow $$
$$ 9x^2 + 102x + 289 = 16(x^2+ 2x - 35) \Rightarrow $$
$$ 7x^2 - 70x - 849 = 0 \Rightarrow $$
$$ b^2 - 4ac = (-70)^2 - 4 \cdot 7 \cdot (-849) = 28672=2^{12}\cdot7 $$
then I calculate the solution using the discriminant as
$$ 5 + 32\frac{\sqrt7}7 $$
and
$$ 5 - 32\frac{\sqrt7}7 $$
but when I plug in the values I find out that they are wrong,
does it have to do with the fact that I square the equation twice?
if so what is the best way to go about solving this equation?
| \begin{align}
\sqrt{5x+19}&=\sqrt{x+7}+2\sqrt{x-5}\\
5x+19&=x+7+4(x-5)+4\sqrt{(x+7)(x-5)}\\
32&=4\sqrt{(x+7)(x-5)}\\
8&=\sqrt{(x+7)(x-5)}\\
64&=x^2+2x-35\\
0&=x^2+2x-99
\end{align}
which gives $x=9$ and $x=-11$ are the solutions.
But both are giving something like this..
$x=9\implies \sqrt {5(9)+19}=\sqrt{9+7}+2\sqrt{9-5}$
$\implies\sqrt{64}=\sqrt{16}+2\sqrt{4}\implies 8=8$ correct know...!!!!
$x=-11\implies\sqrt {5(-11)+19}=\sqrt{-11+7}+2\sqrt{-11-5}$
$\implies\sqrt{-36}=\sqrt{-4}+2\sqrt{-16}\implies 6i=2i+4i$
But by the series of comments given by the well wishers.. i could understand $x=-11$ is not possible...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What will be the sum of the series of binomial co-efficients? What will be the sum of the following binomial co-efficent series
$$\binom{z+1}{z} + \binom{z+2}{z} + \binom{z+3}{z} + \dots + \binom{z+r}{z} = \sum\limits_{i=1}^r \binom{z+i}{z}$$
Thank you
| Hint Add $\binom{z+1}{z+1}$ to you sum and use $\binom{k}{z}+\binom{k}{z+1}=\binom{k+1}{z+1}$ repeatedly:
$$\binom{z+1}{z+1}+\binom{z+1}{z} + \binom{z+2}{z} + \binom{z+3}{z} + \dots + \binom{z+r}{z} \\
=\binom{z+2}{z+1}+\binom{z+2}{z} + \binom{z+3}{z} + \dots + \binom{z+r}{z} \\=...$$
Here I am using the standard notation $\binom{k}{z}$ for $_kC_z$.
P.S. You can also see this formula backwards: using
$$\binom{z+k}{z}=\binom{z+k+1}{z+1}-\binom{z+k}{z+1}$$
you sum becomes telescopic.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x + y + t \le 10$
For nonnegative integers, $x, y, t$ solve, $$x + y + t \le 10$$
This includes then: $x + y + t = 0$, ..., $x + y + t = 10$.
$x + y + t = 0$ has $1$ solution $= \binom{2}{2}$.
$x + y + t = 1$ has $3$ solutions, $= \binom{3}{2}$
$\cdots$
$x + y + t = 10$ has: $ $ solutions: $= \binom{12}{2}$
$$= \sum_{n=2}^{12} \binom{n}{2} = \binom{13}{3}$$
Total Solutions
| You can include a slack variable $s$, and then solve $$x+y+t+s=10$$
with each variable a nonnegative integer. You can solve this with stars and bars, with solution $${10+4-1\choose 4-1}={13\choose 3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
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