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Manipulating the left-hand side into a right-hand side of a Hypergeometric sequence expansion
How do you show that$$F\left(\frac 13,\frac 23;\frac 32;\frac {27}4x^2(1-x^2)^2\right)=\frac {2\sin\left[\tfrac 13\arcsin\left(\tfrac {3\sqrt3}2x(x^2-1)\right)\right]}{x\sqrt3(x^2-1)}$$
I'm not sure where to begin. I thought about using the expansion of $\sin^{-1}$ since its expansion is equivalent to$$\arcsin z=z\, F\left(\frac 12,\frac 12;\frac 32;z^2\right)$$But am not sure how to incorporate that. I also thought about using some identity to make a transformation, but I'm not sure which specific one to use.
| With some magic one can show the following.
Let
\begin{align}
F(t) = {}_{2}F_{1}\left(\frac{1}{3}, \frac{2}{3}; \frac{3}{2}; t^{2}\right)
\end{align}
and use
\begin{align}
\left(\frac{1}{3} \right)_{n} \, \left(\frac{2}{3}\right)_{n} &= \frac{1}{(1)_{n}} \, \left(\frac{1}{3} \right)_{n} \, \left(\frac{2}{3}\right)_{n} \, \left(\frac{3}{3}\right)_{n} = \frac{(1)_{3n}}{3^{3 n} \, (1)_{n}} \\
\left(\frac{3}{2}\right)_{n} &= \frac{1}{(1)_{n}} \, \left(\frac{2}{2}\right)_{n} \, \left(\frac{3}{2}\right)_{n} = \frac{(2)_{2 n}}{2^{2 n} \, (1)_{n}}
\end{align}
to obtain
\begin{align}
F(t) &= \sum_{n=0}^{\infty} \frac{\left(\frac{1}{3} \right)_{n} \, \left(\frac{2}{3}\right)_{n}}{\left(\frac{3}{2}\right)_{n} \, n!} \, t^{2 n} \\
&= \sum_{n=0}^{\infty} \frac{(1)_{3n}}{n! \, (2)_{2n}} \, \left(\frac{4 \, t^{2}}{27}\right)^{n} \\
&= \sum_{n=0}^{\infty} \frac{(3n)!}{n! \, (2n+1)!} \, \left(\frac{4 \, t^{2}}{27}\right)^{n}.
\end{align}
The series expansion, about $x=0$, of $\sin\left(\frac{1}{3} \, \sin^{-1}(x)\right)$ is given by
$$ \sin\left(\frac{1}{3} \, \sin^{-1}(x)\right) = 3 \, x \, \sum_{n=0}^{\infty} \frac{(3n)!}{n! \, (2n+1)!} \, \left(\frac{4 \, x^{2}}{27}\right)^{n}.$$
Now, by comparison it is determined that
$${}_{2}F_{1}\left(\frac{1}{3}, \frac{2}{3}; \frac{3}{2}; t^{2}\right) = \frac{1}{3 \, t} \, \sin\left(\frac{1}{3} \, \sin^{-1}(t)\right).$$
By setting $2 \, t = 3 \, \sqrt{3} \, x \, (1-x^{2})$ leads to the desired expression. The region of convergence is $0 \leq x \leq \sqrt{3}/3$.
With much magic, or confusion, one may show that
$${}_{2}F_{1}\left(\frac{1}{3}, \frac{2}{3}; \frac{3}{2}; t^{2}\right) = \frac{1}{3 \, t} \, \sin\left(\frac{1}{3} \, \sin^{-1}(t)\right) = \frac{1}{1-x^{2}},$$
where $2 \, t = 3 \, \sqrt{3} \, x \, (1-x^{2})$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of $\frac{1}{1\times3\times5}+\frac{1}{3\times5\times7}+\cdots + \frac{1}{13\times15\times17}$ I solved like this, but it took me lots of time and I have to use the calculator. But in exam I can't use the calculator and I need to be fast.
(1) Is there any alternative to solve this easily?
(2) How to solve this if given for $n$ terms?
My work:
\begin{align}
\frac{1}{1\times3\times5}&+\frac{1}{3\times5\times7}+\cdots +\frac{1}{13\times15\times17}\\
&=\frac{1}{1\times3\times5}+\frac{1}{3\times5\times7}+\frac{1}{5\times7\times9}+\frac{1}{7\times9\times11}+\frac{1}{9\times11\times13}\\
&\ \ \ \ \ +\frac{1}{11\times13\times15}+\frac{1}{13\times15\times17}\\
&=\frac{1}{15}[1+1/7+1/21]+\frac{1}{11\times9}[1/7+1/13]+\frac{1}{15\times17}[1/11+1/17]\\
&=7/85
\end{align}
| Hint:
$$\dfrac4{(2n-1)(2n+1)(2n+3)}$$
$$=\dfrac{2n+3-(2n-1)}{(2n-1)(2n+1)(2n+3)}=\dfrac1{(2n-1)(2n+1)}-\dfrac1{(2n+1)(2n+3)}=f(n)-f(n+1)$$
where $$f(m)=\dfrac1{(2m-1)(2m+1)}$$
See Telescoping series
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\tan\left( \frac {a+b}{2}\right) = \frac{\sin a+\sin b} {\cos a + \cos b } $
Prove $$\tan\left( \frac {a+b}{2}\right) = \frac{\sin a+\sin b} {\cos a + \cos b } $$
Can someone help? I separated $\tan$ and did double-angle, but I just went into a circle and couldn't get the trig functions without the halves.
| Use the formulas
$$\sin a+\sin b=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right),$$
$$\cos a+\cos b=2\cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right).$$
Therefore, we have
$$\frac{\sin a+\sin b}{\cos a+\cos b}=\frac{\sin\displaystyle\left(\frac{a+b}{2}\right)}{\cos\displaystyle\left(\frac{a+b}{2}\right)}=\tan\left(\frac{a+b}{2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2500784",
"timestamp": "2023-03-29T00:00:00",
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Prove this inequality $\frac{n}{3^n-2^n}<\left(\frac{2}{5}\right)^{n-1}$ Let $n\ge 3$. How to prove show that
$$\dfrac{n}{3^n-2^n}<\left(\dfrac{2}{5}\right)^{n-1}\tag1$$
or
$$3^n\cdot 2^{n-1}-2^{2n-1}-5^{n-1}\cdot n>0,n\ge 3\tag2$$
It seem right:see wolfram
when I do this problem found it:How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}<\frac{5}{3}$
if this inequality (1) has prove it,then
$$\sum_{k=1}^{n}\dfrac{k}{3^k-2^k}<1+\dfrac{2}{5}+\sum_{k=3}^{+\infty}\dfrac{2^{k-1}}{5^{k-1}}=\dfrac{5}{3}$$
Question: How to prove inequality (1) or (2)
| Some thoughts-not a full answer!
Let's try induction. For $n=3$ we readily verify that it holds: $$\frac{3}{3^3-2^3}\lt\Big(\frac25\Big)^2\Rightarrow\frac3{19}\lt\frac4{25}\Rightarrow75\lt76$$
We assume that the statement is true for $n$. We wish to show that $$\dfrac{n+1}{3^{n+1}-2^{n+1}}<\left(\dfrac{2}{5}\right)^{n}$$
But we know that $$\dfrac{n}{3^n-2^n}\lt\left(\dfrac{2}{5}\right)^{n-1}$$
We multiply this with $\Large\frac25$ to obtain $\Big(\dfrac25\Big)\dfrac{n}{3^n-2^n}\lt\left(\dfrac{2}{5}\right)^{n}$.
Thus it suffices to show that $$\Big(\dfrac25\Big)\dfrac{n}{3^n-2^n}\gt\dfrac{n+1}{3^{n+1}-2^{n+1}}\Rightarrow\\ \\ $$ $$ \Big(\dfrac25\Big)\dfrac{3^{n+1}-2^{n+1}}{3^{n}-2^{n}}\gt\frac{n+1}{n}\Rightarrow\\ $$ $$\large\Big(\dfrac25\Big)\dfrac{\frac{3^{n+1}}{3^{n+1}}-(\frac23)^{n+1}}{\frac{3^{n}}{3^{n+1}}-\frac{2^n}{3^{n+1}}}\gt1+\frac1n\Rightarrow\\ $$ $$\large\Big(\dfrac25\Big)\dfrac{1-(\frac23)^{n+1}}{\frac{1}{3}-\frac13(\frac{2}{3})^n}\gt1+\frac1n\Rightarrow\\$$
$$ \Big(\dfrac65\Big)\dfrac{{1-(\frac23)^{n+1}}}{1-(\frac23)^n}\gt1+\frac1n$$
Now if we show that the function $$f(x)=\Big(\dfrac65\Big)\dfrac{{1-(\frac23)^{x+1}}}{1-(\frac23)^x}-1-\frac1x, x\in[1,+\infty)$$ is strictly increasing we are done since $f(1)=0$ and we will have $f(x)\gt f(1)=0$.
This appears to be the case-here is a graph illustrating it
Unfortunately I can't seem to go far when trying to prove that $f$ is strictly increasing.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a + b = 20$, then what is the maximum value of $ab^2$? Knowing that
$$a + b = 20$$
and $a,b \in \mathbb{Z}_{+}$
What is the maximum value that $ab^2$ can take?
If $b$ is even then so is $a$. Hence, let $b = 2k$ and $a = 20 - 2k$, where $k \in \{1, 2, 3, \cdots, 9\}$
$$ab^2 = (20 - 2k)(2k)^2 = 80k^2 - 8k^3$$
The value of $k$ which maximizes $ab^2$ is therefore $k = 7$ which gives $ab^2 = 1176$.
However this only shows that this is the maximum for even values of $a$ and $b$. It turns out that the maximum of $ab^2$ is when both are odd.
How do I solve this question since my approach is clearly not very elegant?
My solution:
$$a = 20 - b \Longrightarrow ab^2 = (20 - b)b^2$$
Differentiate to find the maximum value of $b$
$$40b - 3b^2 = b(40-3b) = 0$$
$$\Longrightarrow b = \left\lfloor \frac{40}{3} \right\rfloor$$ $$\Longrightarrow b = \left\lceil \frac{40}{3} \right\rceil$$
Therefore $a, b = 7, 13$ or $a, b = 6, 14$. Inspecting these $2$ cases yields the former.
So the maximum value $ab^2$ can take is $1183$.
| Comment: Let $p = ab^2.$ For integers, it is neither imaginative nor difficult to try all possibilities:
a b p
1 19 361
2 18 648
3 17 867
4 16 1024
5 15 1125
6 14 1176
7 13 1183 # <-----
8 12 1152
9 11 1089
10 10 1000
11 9 891
12 8 768
13 7 637
14 6 504
15 5 375
16 4 256
17 3 153
18 2 72
19 1 19
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this ( and general )logarithmic Equation $$ 4^{x+1} - 6^x - 2*9^{x+1} = 0$$
I recently stumbled across this logarithmic Equation and really I have no clue how to solve this. Also , please provide a tactical way to approach such questions
| Hint:
$$2^{x+1}\times 2^{x+1} -2^x\times 3^x-2\cdot3^{x+1}\times 3^{x+1}=0$$
$$2\cdot 2^x\cdot 2^x\cdot 2-2^x\cdot 3^x-2\cdot 3^x\cdot3\cdot3^x\cdot 3=0$$
$$a:=2^x,\qquad b:=3^x$$
$$4\cdot a^2-a\cdot b-18\cdot b^2=0$$
$$(a+2b)(4a-9b)=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Questions on the tent map, $T_c (x)$ Suppose we have the tent map, defined by $$T_c(x) = \begin{cases} cx & 0\leq x \leq 1/2\\ c-cx& 1/2 \leq x \leq 1. \end{cases}$$
What are the prime period 3 orbits for this map? I have seen them for the tent map when $c=2$, but I cannot find them in terms of $c$ for the life of me.
| The line segment of $T_c^3$ have equations and fixed points
\begin{array}{c|rl|c}
0&c(c(c(x)))&=c^3x&\\
1&c(1-c(c(x)))&=c-c^3x&\dfrac{c}{1+c^3}\\
2&c(c(1-c(x)))&=c^2-c^3x&\dfrac{c^2}{1+c^3}\\
3&c(1-c(1-c(x)))&=c-c^2+c^3x&\dfrac{c}{1+c+c^2}\\
4&c(c(c(1-x)))&=c^3-x^3&\dfrac{c^3}{1+c^3}\\
5&c(1-c(c(1-x)))&=c-c^3+c^3x&\dfrac{c+c^2}{1+c+c^2}\\
6&c(c(1-c(1-x)))&=c^2-c^3+c^3&\dfrac{c^2}{1+c+c^2}\\
7&c(1-c(1-c(1-c)))&=c-c^2+c^3-c^3x&\dfrac{c}{1+c}
\end{array}
where one can recognize the two 3 cycles
$$
\dfrac{c}{1+c^3}\to\dfrac{c^2}{1+c^3}\to\dfrac{c^3}{1+c^3}
$$
and
$$
\dfrac{c}{1+c+c^2}\to\dfrac{c^2}{1+c+c^2}\to\dfrac{c+c^2}{1+c+c^2}
$$
These exist for
$$
\dfrac{c}{1+c+c^2}\le \frac12\le\dfrac{c^2}{1+c+c^2}\\\iff\\
2c\le1+c+c^2\le 2c^2\\\iff\\
-c^2+c-1\le 0 \le c^2-c-1
$$
which is true for $c\ge\frac{1+\sqrt5}2$, see also Prove that the critical value for the appearance of period 3-orbits in the tent map is $\frac{1+\sqrt5}{2}$.
| {
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Partial fractions decomposition of ${\frac{2x}{(x+2)^2}}$ Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g. $$2x = A\cdot (x+2) + B$$
Substitute $x=-2$
$2\cdot(-2)$ = $A\cdot (2-2) +B$
$-4 = B$
In other questions there is always another factor to multiply by at this stage.
| There is a nice method for rational functions that only have one repeated factor at the denominator: let the fraction be $f(x)=\frac{N(x)}{D(x)}$ where $D(x)=d(x)^k$.
The general form of the partial fraction decomposition is
$$\displaystyle f(x)=\frac{a_1(x)}{d(x)}+\frac{a_2(x)}{d(x)^2}+\dots+\frac{a_k(x)}{d(x)^k}$$
Then, if you perform the division of $N(x)$ by $d(x)$, you can write $f(x)$ as
$$\displaystyle f(x)=q(x)d(x)+r_1(x)$$
where $\deg(r_1)<\deg(d)$. We then got
$$\displaystyle f(x)=\frac{q(x)d(x)+r_1(x)}{d(x)}=\frac{r_1(x)}{d(x)}+\frac{q(x)}{d(x)^{k-1}}$$
so $r_1$ is actually $a_1$. By repeating the process with $q$, we can get all the other functions $a_2$,...,$a_k$.
Example with $\frac{2x}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$: $2x=2(x+2)-4$, so $A=-4$ and $q(x)=2$. As $q(x)=0(x+2)+2$, we obviously have $B=2$. We obtain
$$f(x)=\frac{2(x+2)-4}{(x+2)^2}=\frac{2}{(x+2)}+\frac{-4}{(x+2)^2}$$
When $d(x)$ has degree $1$, we can use Synthetic Division to perform the division, so it is very fast.
| {
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$\mathbb Z_n[x]$, $x+(n-1)$ is a factor of $x^m+(n-1)$ for all $m$ and $n$
Prove that: in $\mathbb Z_3[x]$, $x+2$ is a factor of $x^m+2$. In $\mathbb Z_n[x]$, $x+(n-1)$ is a factor of $x^m+(n-1)$ for all $m$ and $n$.
You can use long division and get that $x^m+2=(x+2)(x^{m-1}+...+x+1)$. Does this prove that $x+2$ is then a factor or do I need to prove something more?
Similarly, using long division on $x^m+(n-1)=(n-1)(x^{m-1}+...+x+1)$ (I believe).
| The natural surjective homomorphism
$\phi_n: \Bbb Z \to \Bbb Z_n \tag1$
induces a homomorphism, also surjective, and also denoted here by $\phi_n$,
$\phi_n: \Bbb Z[x] \to \Bbb Z_n[x]. \tag 2$
Now in $\Bbb Z[x]$, we have
$x^m - 1 = (x - 1)(x^{m -1} + x^{m - 2} + \ldots + 1); \tag 3$
thus
$\phi_n(x^m - 1) = \phi_n((x - 1)(x^{m -1} + x^{m - 2} + \ldots + 1))$
$= \phi_n(x - 1) \phi_n(x^{m -1} + x^{m - 2} + \ldots + 1). \tag 4$
The image of $x^m - 1$ under $\phi_n$ is
$\phi_n(x^m - 1) = x^m - \phi_n(1) = x^m + (n -1) \in \Bbb Z_n[x], \tag 5$
and the image of $x - 1 \in \Bbb Z[x]$ is
$\phi_n(x - 1) = x - \phi_n(1) = x + (n - 1) \in \Bbb Z_n[x]; \tag 6$
by virtue of (5) and (6), (4) becomes
$x^m + (n - 1) = (x + (n - 1))\phi_n(x^{m -1} + x^{m - 2} + \ldots + 1) \in \Bbb Z_n[x], \tag 7$
which shows that
$x + (n - 1) \mid x^m + (n - 1) \tag 8$
in $\Bbb Z_n[x]$.
Note that we don't need to explicitly present $\phi_n(x^{m -1} + x^{m - 2} + \ldots + 1)$ to complete this result, but it is nevertheless worth noting that
$\phi_n(x^{m -1} + x^{m - 2} + \ldots + 1) = x^{m - 1} + x^{m - 2} + \ldots + 1 \in \Bbb Z_n[x]. \tag 9$
The preceding argument, by virtue of (8), demonstrates that
$x + 2 \mid x^m + 2 \tag{10}$
in $\Bbb Z_3[x]$. Long division is not really necessary once one has (3), which is easy to verify by direct multiplication:
$(x - 1)(x^{m -1} + x^{m - 2} + \ldots + 1)$
$= x(x^{m -1} + x^{m - 2} + \ldots + x + 1) - 1(x^{m -1} + x^{m - 2} + \ldots + x + 1)$
$= x^m + x^{m -1} + \ldots + x - x^{m - 1} - \ldots - x -1 = x^m -1 \tag {11}$
(11) avoids the use of long division.
In the above we have made use of the fact that
$n - 1 \equiv -1 \; \mod n. \tag {12}$
| {
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Find the minimum distance from the point Find the minimum distance from the point $ \ (5,0) \ $ to the curve $ \ y=\sqrt x+2 \ $.
Answer:
Let $ \ (x,y) \ $ be the closest point on $ \ y=\sqrt x+2 \ $ from $ (5,0) $ .
Then the distance is given by
$ d(x,y)=\sqrt{(x-5)^2+y^2} \ $
We will minimize the function $ \ g(x,y)=(x-5)^2+y^2 \ $
replacing $ y \ \ by \ \ \sqrt x+2 \ $ , we get
$ g(x)=(x-5)^2+(\sqrt x+2)^2 \ $
The extreme points \ are
$ f'(x)=0 \\ 2(x-5)+\frac{\sqrt x+2}{\sqrt x} =0 \\ 2x \sqrt x-9 \sqrt x+2=0 $
This becomes complicated . I am unable to calculate the closest point.
Help me out
| Now, let $\sqrt{x}=t$.
Thus,$$2x\sqrt{x}-9\sqrt{x}+2=2t^3-9t+2=2t^3-4t^2+4t^2-8t-t+2=$$
$$=(t-2)(2t^2+4t-1).$$
Thus, $x_{min}=4$ and $x_{max}=\left(\sqrt{1.5}-1\right)^2.$
Also, we need to check, what happens for $x=0$.
| {
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Proof that pre-defined $a_n$ satisfies recursion formula. (problems with understanding notation) problem
We have recursion formula for sequence which is:
$$ a_{n+2}=2a_{n+1}+3a_{n} $$
Now I want to prove that:
$$ a_n=A\cdot 3^n+B\cdot(-1)^n $$
satisfies the recursion formula.
my approach
I would try to just insert $a_n$ to right side of the recursion formula.
$$ a_{n+2}=2\cdot(A\cdot 3^n+B\cdot(-1)^n)+3\cdot(A\cdot 3^n+B\cdot(-1)^n) $$
but I doubt this works in this way since we have $2a_{n+1}$ there is $+1$ in lower index and I don't know how to deal with lower index with arithmetic in it. I would need to first solve what is $a_{n+1}$? since I do know what $a_{n}$ but $a_{n+1}$ is unknown?
Now if someone could provide some insight on this that would be highly appreciated.
| When we say that $a_n = A\cdot 3^n + B\cdot (-1)^n$ then it means that $a_\color{red}{1} = A\cdot 3^\color{red}{1} + B\cdot (-1)^\color{red}{1}$, $a_\color{red}{2} = A\cdot 3^\color{red}{2} + B\cdot (-1)^\color{red}{2}$, and so on.
Likewise,
$a_\color{red}{n+1} = A\cdot 3^\color{red}{n+1} + B\cdot (-1)^\color{red}{n+1}$ and $a_\color{red}{n+2} = A\cdot 3^\color{red}{n+2} + B\cdot (-1)^\color{red}{n+2}$. Now you have to prove that
$$a_{n+2} = 2 a_{n+1} + 3a_n.$$
Observe that
$$2 a_{n+1} + 3a_n = 2(A\cdot 3^{n+1} + B\cdot (-1)^{n+1}) + 3(A\cdot 3^{n} + B\cdot (-1)^{n}),$$
which can be rearranged as
$$2 a_{n+1} + 3a_n = A\cdot 3^{n+1}(2+1) + B\cdot (-1)^{n+2}(-2+3).$$
That is,
$$2 a_{n+1} + 3a_n = A\cdot 3^{n+2} + B\cdot (-1)^{n+2} = a_{n+2}.$$
| {
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Prove that $2^{58}+1$ can be written as a product of 3 positive integers, each greater than 1. I first attempted to factor it using SFFT, but I got 3 terms left over that wasn't a product. Am I headed in the right direction, or is the correct answer using a completely different method?
| There is this paper on Cyclotomic and Aurifeuillian polynomials with a section dedicated to the problem of finding factors of large numbers. So
$$2^{58}+1=4^{29}+1=(4+1)\cdot(4^{28}-4^{27}+4^{26}-4^{25}+4^{24}-...+4^{2}-4^{1}+1)$$
$5$ is one factor. Now applying Aurifeuillian factorisation
$$2^{58}+1=2^2\cdot (2^{14})^4+1=\left(2\cdot (2^{14})^2\right)^2+1=\left(2\cdot 2^{28}\right)^2+2\cdot \left(2\cdot 2^{28}\right) +1-2\cdot 2\cdot 2^{28}=\\
\left(2\cdot 2^{28}+1\right)^2-\left(2^{15}\right)^2=\left(2\cdot 2^{28}+1-2^{15}\right)\cdot \left(2\cdot 2^{28}+1+2^{15}\right)=\\
\left(2^{29}+1-2^{15}\right)\cdot \left(2^{29}+1+2^{15}\right)$$
So we found two more factors $2^{29}+1-2^{15}$ and $2^{29}+1+2^{15}$ both clearly larger than $5$. From Euclid lemma, $5$ (which is a prime) will divide one of them, thus, there will be (at least) 3 factors.
| {
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"timestamp": "2023-03-29T00:00:00",
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General solution of $(\sqrt3 - 1)\sin\theta + (\sqrt3 + 1)\cos\theta =2 $ Find general solution of the equation $(\sqrt3 - 1)\sin\theta + (\sqrt3 + 1)\cos\theta =2 $.
My approach:
Squared on both sides, formed a quadratic equation in $\cos\theta$ and finally got two solutions for theta,
$$\theta = 2n\pi \pm \frac{\pi}{6}$$
$$\theta = 2n\pi \pm \frac{\pi}{3}$$
But the answer given in my book is $$2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$$
Pretty strange
can anyone help me?
| \begin{align}
(\sqrt3 - 1)\sin\theta + (\sqrt3 + 1)\cos\theta &=2\\
\left(\frac{\sqrt3}{2} -\frac{1}{2}\right)\sin\theta+\left(\frac{\sqrt3}{2} +\frac{1}{2}\right)\cos\theta&=1\\
\left(\cos30 -\sin(30)\right)\sin\theta+\left(\cos30 +\sin(30)\right)\cos\theta&=1\\
\sin\theta\cos30 -\sin\theta\sin(30)+\cos\theta\cos30 +\cos\theta\sin(30)&=1\\
\sin(\theta+30) +\cos(\theta+30) &=1\\
\sin(\theta+\pi/6) +\cos(\theta+\pi/6) &=1\\
\text{Clearly the solution set includes $\pi/3 + 2\pi k$ and $-\pi/6+2\pi k$}\\
\text{The trick now to simplify is to take the average value}\\ (\pi/3 + -\pi/6)/2&= \pi/12\\
\text{The two solutions show up every $2\pi k$}\\
\text{The distance for either solution from $\pi/12$ is $\pi/4$ }\\
\theta &= \pi/12 + 2\pi k \pm \pi/4
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the number of real solutions of the equation below
Find the number of real solutions of the equation
$$\sec(\theta) + \csc(\theta) = \sqrt{15}$$
lying between $0$ and $\pi$.
My Approach :
Converted the equation in the form of $\sin\theta$ and $\cos\theta$ and eventually got $\sin\theta + \cos\theta = -\sqrt\frac{3}{5}$. Don't know how to proceed from here. Am I going wrong?
The answer given in my book is 4.
Can anyone help me?
| I got something other.
Let $\sin{\theta}+\cos{\theta}=t$.
Thus, $$\sin \theta \cos \theta=\frac{t^2-1}{2}$$ and we obtain
$$t=\frac{\sqrt{15}(t^2-1)}{2}$$ or
$$\sqrt{15}t^2-2t-\sqrt{15}=0,$$
which gives $t=\sqrt{\frac{5}{3}}$ or $t=-\frac{4}{\sqrt{15}}$ and since $\left|\sqrt{\frac{5}{3}}\right|<\sqrt2$ and $\left|-\frac{4}{\sqrt{15}}\right|<\sqrt2$,
we obtain $2$ solutions.
Indeed, $$\sin\theta+\cos\theta=\sqrt{\frac{5}{3}}$$ gives
$$\sin\theta\cdot\frac{1}{\sqrt2}+\cos\theta\cdot\frac{1}{\sqrt2}=\sqrt{\frac{5}{6}}$$ or
$$\sin\left(\theta+45^{\circ}\right)=\sqrt{\frac{5}{6}}$$ and since $0^{\circ}<\theta<180^{\circ},$ we obtain
$$\theta_1=\arcsin\sqrt{\frac{5}{6}}-45^{\circ}$$ and $$\theta_2=-\arcsin\sqrt{\frac{5}{6}}+135^{\circ}.$$
The equation $$\sin\theta+\cos\theta=-\sqrt{\frac{16}{15}}$$ has no solutions for $0^{\circ}<\theta<180^{\circ}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometric identity of finite terms Prove that:
$$\dfrac{1}{\cos x+\cos {3x}} + \dfrac{1}{\cos x+ \cos {5x}}+\dots+\dfrac{1}{\cos x+ \cos {(2n+1)x}} \\= \frac{1}{2}\csc x \,[ \tan{(n+1)x}-\tan{x}]$$
I tried to prove this using the regular formulas. But failed. Please help me.
| In the case $n=0$ there is no summand on the left-hand side and the right-hand side is $0$, so the base case of the induction holds.
Suppose it holds for $n-1$; then the left-hand side with $n$ can be written, by the induction hypothesis,
$$
\frac{1}{2\sin x}\bigl(\tan nx-\tan x\bigr)+\frac{1}{\cos x+\cos(2n+1)x}
$$
and you want to prove this equals
$$
\frac{1}{2\sin x}\bigl(\tan(n+1)x-\tan x\bigr)
$$
which is equivalent to
$$
\frac{\tan nx}{2\sin x}+\frac{1}{\cos x+\cos(2n+1)x}=
\frac{\tan(n+1)x}{2\sin x}
$$
By the sum-to-product formulas, this becomes
$$
\frac{\sin nx}{2\sin x\cos nx}+\frac{1}{2\cos nx\cos(n+1)x}=
\frac{\sin(n+1)x}{2\sin x\cos(n+1)x}
$$
Reduce to the same denominator and conclude the equality holds.
The relation $$\sin nx\cos(n+1)x+\sin x=\sin(n+1)x\cos nx$$ is true, because it is equivalent to $$\sin x=\sin(n+1)x\cos nx-\cos(n+1)x\sin nx$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why does this strategy sieve out composites without factors $2,3$? Take
$$
3 + 2 = 5 \\
3 + 4 = 7 \\
5 + 6 = 11 \\
5 + 8 = 13 \\
7 + 10 = 17 \\
7 + 12 = 19 \\
9 + 14 = 23 \\
9 + 16 = 25 \\
\cdots
$$
I'm having trouble writing down what the pattern is to start with.
I think it's $2n + 1 + 2n + 2$ and $2n + 1 + 4n$ interwoven so the list essentially enumerates the two sets $4\Bbb{N} + 3$ and $6\Bbb{N} + 1$. So how do I go from there to say that the list of right hand side is the ordered list of all numbers not divisible by $2,3$?
Secondly, how do we generalize this summation procedure to filter out more composites?
| Working modulo $6$, we know that numbers divisible by $2$ or $3$ will have remainders $0, 2, 3$ or $4$. Therefore, we have to show that the numbers in this list have remainders $1$ or $5$.
We have:
$(3)+(2) \equiv 5 \pmod 6$
$(3)+(2+2) \equiv 1 \pmod 6$
$(3+2)+(2+2+2) \equiv 5 \pmod 6$
$(3+2)+(2+2+2+2) \equiv 1 \pmod 6$
and so on. Notice that from line 1 to line 2, we add $2$ to $5$, which is $1$ modulo $6$. Then, from line 2 to line 3, we add $4$ to $1$, which is $5$ again, and the cycle repeats.
Therefore, the only remainders in the sequence are $1$ and $5$, which are also the remainders which are not divisible by $2$ or $3$ modulo $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that$\frac{\cos\left(x\right)}{1+\sin\left(x\right)} = \frac{1-\sin\left(x\right)}{\cos\left(x\right)}$ How can I prove that:$$\frac{\cos\left(x\right)}{1+\sin\left(x\right)} = \frac{1-\sin\left(x\right)}{\cos\left(x\right)}$$
My attempt:
$$\cos^2(x/2) =\frac{\cos x +1}{2}$$
$$\sin x = 2 \sin(x/2)\cos(x/2)$$
$$\frac{\cos\left(x\right)}{1+\sin\left(x\right)} =\frac{2\cos^2(x/2) -1}{1+2 \sin(x/2)\cos(x/2)}= \frac{(\sqrt2\cos(x/2) -1)\sqrt2\cos(x/2) +1)}{1+2 \sin(x/2)\cos(x/2)}$$
Can anyone help me to simply this last expression or provide me with tip?
| Multiplying the $\mathrm{LHS}$ by the conjugate of the denominator:
$$\frac{\cos\left(x\right)}{1+\sin\left(x\right)}\cdot\frac{1-\sin\left(x\right)}{1-\sin\left(x\right)}$$
$$\frac{\cos\left(x\right)\left(1-\sin\left(x\right)\right)}{1-\sin^{2}\left(x\right)}$$
Using the the Pythagorean identity:
$$\frac{\cos\left(x\right)\left(1-\sin\left(x\right)\right)}{\cos^{2}\left(x\right)}$$
Canceling $\cos(x)$:
$$\frac{1-\sin\left(x\right)}{\cos\left(x\right)}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify nth Roots $6\sqrt[3]{9000} + 7\sqrt[3]{576}$ I'm having some concerns about how I am going about simplifying this radical expression. I wanted to know if this would be an accurate method of solving.
Simplify:
$$6\sqrt[3]{9000} + 7\sqrt[3]{576}$$
Radical expression product rule & greatest common factor:
$$6\sqrt[3]{125}\sqrt[3]{72} + 7\sqrt[3]{8}\sqrt[3]{72}$$
Principle of nth root:
$$6 \cdot 5\sqrt[3]{72} + 7 \cdot 2\sqrt[3]{72}$$
Simplify:
$$30\sqrt[3]{72} + 14\sqrt[3]{72}$$
Addition of radical rule:
$$30 + 14\sqrt[3]{72}$$
Simplify:
$$44\sqrt[3]{72}$$
Is $44\sqrt[3]{72}$ completely simplified?
I'm positive that 72 has no whole number $a$ that satisfies $72 = a^{3}.$
| $$ \sqrt[3]{9000} = (9 \times 10^3)^{\frac{1}{3}} = 10 \times 9^{\frac{1}{3}}$$
$$ \sqrt[3]{576} = \sqrt[3]{(2^3 \times 3)^2} = 4 \times 9^{\frac{1}{3}}$$
$$ 6\sqrt[3]{9000} = 60 \times 9^{\frac{1}{3}}$$
$$ 7\sqrt[3]{576} = 28 \times 9^{\frac{1}{3}} $$
$$ 7\sqrt[3]{576} + 6\sqrt[3]{9000} = (60 \times 9^{\frac{1}{3}}) + (28 \times 9^{\frac{1}{3}}) = 88 \times 9^{\frac{1}{3}} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $d\mid2n^2$, could $n^2+d$ be a square of a natural number? Here $d, n \in \Bbb N$. If $d\mid2n^2$, could $n^2+d$ be a square of a natural number? Here $d, n \in \Bbb N$.
What I tried so far: $2n^2 = kd, k \in N$ $$d = 2\frac{n^2}{k}$$
$$n^2 + d = $$
$$n^2 +2\frac{n^2}{k}= n^2\left(\frac{k+2}{k}\right) = \frac{kd}{2}\left(\frac{k+2}{k}\right) = \frac{d(k+2)}{2}$$
I don't see where this could lead me.
| I am assuming, for this answer, that $0\not\in\mathbb{N}$.
The Reasoning in the Question
The question asks whether there are $m,n,d\in\mathbb{N}$ so that $d\mid2n^2$, and $n^2+d=m^2$. This is equivalent to asking if there are $m,n\in\mathbb{N}$ so that $m\gt n$ and $m^2-n^2\mid2n^2$.
We can assume that $(m,n)=1$. If not, then divide $m$ and $n$ by $(m,n)$.
Suppose that
$$
m^2-n^2\mid2n^2\tag1
$$
This means there is a $k$ so that
$$
k\left(m^2-n^2\right)=2n^2\tag2
$$
which means that
$$
(k+2)n^2=km^2\tag3
$$
Finishing with Bezout
Since $(m,n)=1$, this answer says we also have $\left(m^2,n^2\right)=1$. Bezout gives us $a,b$ so that $am^2+bn^2=1$. Therefore,
$$
\begin{align}
k
&=akm^2+bkn^2\tag{4a}\\
&=a(k+2)n^2+bkn^2\tag{4b}\\
&=(ak+2a+bk)n^2\tag{4c}
\end{align}
$$
Explanation:
$\text{(4a):}$ multiply $am^2+bn^2=1$ by $k$
$\text{(4b):}$ apply $(3)$
$\text{(4c):}$ collect multiples of $n^2$
$(4)$ says that $n^2\mid k$. Then, dividing $(2)$ by $n^2$ gives
$$
\frac{k}{n^2}\left(m^2-n^2\right)=2\tag5
$$
That is, $m^2-n^2\mid2$. This means that either $m^2-n^2=2$ or $m^2-n^2=1$.
If $2=m^2-n^2=(m+n)(m-n)$, we must have $m+n=2$ and $m-n=1$ which means $m=\frac32$ and $n=\frac12$.
If $1=m^2-n^2=(m+n)(m-n)$, we must have $m+n=1$ and $m-n=1$ which means $m=1$ and $n=0$.
Since neither $\frac12$ nor $0$ are in $\mathbb{N}$, we cannot have $m,n\in\mathbb{N}$ and $m^2-n^2\mid2n^2$.
A Comment on the Other Answer
While the claim made in G Tony Jacobs' answer, that if $\frac ab=\left(\frac pq\right)^2$ and $(a,b)=1$ then $a$ and $b$ are squares of integers, is true, I think it requires some justification.
Suppose that $(a,b)=1$ and $(p,q)=1$ and
$$
\frac ab=\left(\frac pq\right)^2\tag6
$$
Bezout gives us $x,y$ such that
$$
ax+by=1\tag7
$$
We have
$$
\begin{align}
p^2
&=ap^2x+bp^2y\tag{8a}\\
&=ap^2x+aq^2y\tag{8b}\\
&=a\left(p^2x+q^2y\right)\tag{8c}
\end{align}
$$
Explanation:
$\text{(8a):}$ multiply $(7)$ by $p^2$
$\text{(8b):}$ apply $(6)$
$\text{(8c):}$ factor out $a$
Similarly, we have
$$
\begin{align}
q^2
&=aq^2x+bq^2y\tag{9a}\\
&=bp^2x+bq^2y\tag{9b}\\
&=b\left(p^2x+q^2y\right)\tag{9c}
\end{align}
$$
As mentioned above $(p,q)=1\implies\left(p^2,q^2\right)=1$. Thus, since $(8)$ and $(9)$ say that $p^2x+q^2y$ is a common factor of $p^2$ and $q^2$, we must have
$$
p^2x+q^2y=1\tag{10}
$$
Then $(8)$, $(9)$, and $(10)$ say that $a=p^2$ and $b=q^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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rewrite $\sum_{n=0}^\infty \frac{(-1)^nx^{n-1}}{2^{n+1}}$ as $-\sum_{n=0}^\infty (x+1)^{2n}$ Rewrite $\sum_{n=0}^\infty \frac{(-1)^n x^{n-1}}{2^{n+1}}$ as $-\sum_{n=0}^\infty (x+1)^{2n}$.
These two series are equal because they can both be derived from $\frac{1}{x^2+2x}$. To get the first one, you split it into $\frac{1}{x}\cdot \frac{1}{x+2}$ and then you use the power series. To get the second series, you complete the square with $\frac{1}{x^2+2x+1-1}$.
The point is those two are obviously equal but I can't figure out how to manipulate them while leaving them in series form to make them both equal each other (without using $\frac{1}{x^2+2x}$ as an in between equals). This was a problem that came up while my teacher was lecturing and he couldn't figure it out.
| $\sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{2^{n+1}}
=\sum_{n=0}^\infty (-1)^n(x/2)^{n+1}
$
converges for
$-2 \lt x \lt 2$
and
$-\sum_{n=0}^\infty (x+1)^{2n}
$
converges for
$-1 < x+1 < 1$
or
$-2 < x < 0$.
They are representations of
$\frac{1}{x^2+2x}
$
in different regions.
You can get another representation by
$\frac{1}{x^2+2x}
=\frac1{x^2}\frac{1}{1+2/x}
=\frac1{x^2}\sum_{n=0}^{\infty} (-1)^n (2/x)^n
$
which converges for
$|x| > 2$.
Look up analytic continuation
for fun and profit.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum and Minimum of complex number $z$ where $|z^4+z^3-z-1| = |z^4-z^3+z-1|$.
If $z\in C$ satisfies $|z^4+z^3-z-1| = |z^4-z^3+z-1|.$ Then minimum value of $|z-1-2i|$ and maximum value of $|z^2-1-2i|$ is . . .
$\bf{Attempt}$ from $|z^4+z^3-z-1| = |z^4-z^3+z-1|$
$$|(z+1)(z^3-1)| = |(z-1)(z^3+1)|$$
$$\bigg|\frac{z-1}{z+1}\bigg|=\bigg|\frac{z^3-1}{z^3+1}\bigg|=k$$
Could some help me how to solve it, thanks
| Hint: Your factorizations are incomplete.
$$ z^3 - 1 = (z-1)(z^2+z+1) $$
$$ z^3 + 1 = (z+1)(z^2-z+1) $$
This leaves $|z^2 -z +1| = |z^2 + z + 1|$ (having already extracted the factors $z-1$ and $z+1$).
| {
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Simplifying absolute value expression with square roots I'm attempting to simplify the expression $|(|\sqrt2+ \sqrt3|-|\sqrt5-\sqrt7|)|$
Previously I've shown that if $0<a<b$ then $\sqrt{a}<\sqrt{b}$
Thus we have $\sqrt5-\sqrt7<0$ and therefore $|\sqrt5-\sqrt7|=\sqrt7-\sqrt5$
This means $|\sqrt2+\sqrt3|-|\sqrt5-\sqrt7|=\sqrt2+\sqrt3+\sqrt5-\sqrt7$
Now I've clearly simplified the expression, but I want to find out if it is possible to simplify further. As stated about $\sqrt5-\sqrt7<0$, and $\sqrt2+\sqrt3>0$. Intuitively I somehow see that this whole expression has to be greater than zero, but how do I prove it?
Thanks
| You seem to be asking two questions: (1) can this be simplified further, and (2) is the expression positive or negative?
*
*No. There isn't really anything that you can do to make this expression simpler. Each radical expression has a different radicand, so there are no like terms to combine, and the radicals themselves are pretty simple.
*One possible approach, among many, is as follows:
\begin{align}
\sqrt{2} + \sqrt{3} + \sqrt{5} - \sqrt{7} \ge 0
&\iff \sqrt{2} + \sqrt{3} + \sqrt{5} \ge \sqrt{7} \\
&\iff \left( \sqrt{2} + \sqrt{3} + \sqrt{5} \right)^2 \ge 7. && (\text{$x \mapsto x^2$ is increasing on $[0,\infty)$})
\end{align}
Multiplying out the expression on the left, we get
\begin{align}
\left(\sqrt{2} + \sqrt{3} + \sqrt{5}\right)^2
&= 2 + \sqrt{6} + \sqrt{10} + \sqrt{6} + 3 + \sqrt{15} + \sqrt{10} + \sqrt{15} + 5 \\
&= 10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}.
\end{align}
Therefore we have
$$ \left( \sqrt{2} + \sqrt{3} + \sqrt{5} \right)^2 = 10 + \underbrace{2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}}_{\ge 0} \ge 10 > 7, $$
from which it follows that
$$ \sqrt{2} + \sqrt{3} + \sqrt{5} - \sqrt{7} > 0. $$
| {
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Evaluating $\int_1^\infty \frac{1}{x(x^2+1)}\ dx$ $$I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx$$
I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.
$$\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i}\right)$$
So
$$I = \frac{1}{2}\int_1^\infty\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i} \ dx \\ = \frac{1}{2}\left[\log |x| - \log(|x+i|) - \log(|x-i|)\right]_1^\infty
$$
which evaluates to be $\infty$.
| By enforcing the substitution $x=\frac{1}{z}$ the result is straightforward to find:
$$ \int_{1}^{+\infty}\frac{dx}{x(x^2+1)}=\int_{0}^{1}\frac{z\,dz}{z^2+1}=\left[\frac{1}{2}\log(1+z^2)\right]_{0}^{1}=\color{red}{\frac{\log 2}{2}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Minima and maxima of a trig function let $f(x) = \sqrt {4\sin^4x - \sin^2x\cos^2x + 4\cos^4x}$
what is the product of the minimum and maximum value of the function?
I tried turning them all to $\sin x$
$-\sin^2x\cos^2x = \sin^4x-\sin^2x ; 4\cos^4x = (1-\sin^2x)^2 = 4sin^4x - 4sin^2x +4 $
which gives us :
$9\sin^4x - 2\sin^2x +4$
then i dont know how to proceed
EDIT:
The equation above is wrong, i got if you try solving for the ones i have given :
$9\sin^4x - 5\sin^2x +4$
then I tried making $sin^2x = a$ (Thanks @Greninja for the info)
where in the the min is when $a = 5/18$ and the value is 47/36,
then the max is when $a=1$ and the value is 8
Min x Max = $12\sqrt{94}$ which looks far fetched
I dont know where I went wrong with my solution Please help.
And yes, I have been told that the answer is $\sqrt{7}$
| Use Indentity $$\sin^4 x+\cos^4 x = (\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x = 1-2\sin^2 x\cos^2 x$$
So let $$y = \sqrt{4(\sin^4 x+\cos^4 x)-\sin^2 x\cos^2 x} = \sqrt{4-9\sin^2 x\cos^2 x} = \sqrt{4-\frac{9\sin^2(2x)}{4}}$$
Now use $0 \leq \sin^2(2x)\leq 1$
| {
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Difference between Fractional Exponents and Fractions? how would you explain the difference between exponential multiplication and fractional multiplication? $${x^{1/3}}{^{}{}} * {x^{1/3}}{^{}{}} *{x^{1/3}}{^{}{}} * {x^{1/3}}{^{}{}} = {x^{4/3}}$$ Why is this the same as $$4 * {^{1/3}}$$ On the other hand $$1/3 * 1/3 * 1/3 * 1/3 = 1/81$$ So in this case, this is not the same as $$4 * {{1/3}}$$
| Depending on the level of student, the explanation might go as follows:
First, note that if $m$ and $n$ are integers, then we might regard exponentiation as repeated multiplication
$$ x^m \cdot x^n
= (\underbrace{x\cdot x\cdot \dotsb \cdot x}_{\text{$m$ times}})\cdot (\underbrace{x\cdot x\cdot \dotsb \cdot x}_{\text{$n$ times}})
= \underbrace{x\cdot x\cdot \dotsb \cdot x}_{\text{$m+n$ times}}
= x^{m+n}.
$$
This is a little bit of a lie (or not—again, it depends on the mathematical maturity of the audience). This can be worked out explicitly with small $m$ and $n$, and/or with fixed values of $x$ (say, work out $2^3\cdot 2^5$ by hand).
Then, when fractional exponents are introduced, it is reasonable to extend the operation above: if $p$ and $q$ are rational, then we should have
$$ x^p\cdot x^q = x^{p+q}. $$
Again, with small $x$, $p$, and $q$, this could be justified by hand. Consider
$$ 3^{\frac{1}{2}} \cdot 3^{\frac{5}{2}}
= \sqrt{3}^1 \cdot \sqrt{3}^5
= \sqrt{3}^{5+1}
= 3^{\frac{5+1}{2}}
= 3^{\frac{5}{2} + \frac{1}{2}}.
$$
This isn't really rigorous, but it should make the idea clear to students how have not seen it before. The moral of the story is that when terms with like bases are multiplied, the exponents are added.
This is in contrast to something like
$$ \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}
= \left( \frac{1}{3} \right)^4 = \frac{1}{3^4} = \frac{1}{81}.
$$
Here we are again using the idea that exponentiation can be viewed as repeated multiplication. Compare this to something like
$$ 3+3+3+3 = 4\cdot 3,$$
which is not the same as
$$ 3+3+3+3 = 4+3. $$
Writing the second identity would be the same kind of error as writing
$$ \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}
= 4 \cdot \frac{1}{3}. $$
| {
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Evaluate $\sum_{p=1}^{32}(3p+2)\left[\sum_{q=1}^{10}\left(\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}\right)\right]^p$ Evaluate$$\sum_{p=1}^{32}(3p+2)\left[\sum_{q=1}^{10}\left(\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}\right)\right]^p$$
I wanted to convert this problem in the form $e^{i\theta}$ but not able to proceed
| Noting $z=\cos\frac{2\pi}{11}+i\cdot \sin\frac{2\pi}{11}$ then
$$\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}=(-i)\left(\cos\frac{2q\pi}{11}+i\sin\frac{2q\pi}{11}\right)=(-i)z^q$$
and
$$\sum\limits_{q=1}^{10}\left(\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}\right)=(-i)\sum\limits_{q=1}^{10}z^q=(-i)\left(-1+\sum\limits_{q=0}^{10}z^q\right)=\\
(-i)\left(-1 + \frac{z^{11}-1}{z-1}\right)=(-i)\left(-1+\frac{\cos\frac{2\cdot 11\pi}{11}+i\sin\frac{2\cdot 11\pi}{11}-1}{z-1}\right)=\\
(-i)\left(-1+\frac{1-1}{z-1}\right)=i$$
Then
$$\sum_{p=1}^{32}(3p+2)\left[\sum_{q=1}^{10}\left(\sin\frac{2q\pi}{11}-i\cos\frac{2q\pi}{11}\right)\right]^p=
\sum_{p=1}^{32}(3p+2)i^p=\\
3i\sum_{p=1}^{32}pi^{p-1}+2i\sum_{p=1}^{32}i^{p-1}=3i\left.\left(\sum_{p=1}^{32}px^{p-1}\right)\right|_{x=i}+2i\sum_{p=0}^{31}i^{p}=\\
3i\left.\left(\sum_{p=0}^{32}x^{p}\right)'\right|_{x=i}+2i\frac{i^{32}-1}{i-1}=
3i\left.\left(\frac{x^{33}-1}{x-1}\right)'\right|_{x=i}=
3i\left.\left(\frac{1 - 33 x^{32} + 32 x^{33}}{(x-1)^2}\right)\right|_{x=i}=
3i\left(-16-16i\right)=48(1-i)$$
| {
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AM>HM Problem $\frac{1}{n+1}+...+\frac{1}{3n+1}>1$ I am having difficulty solving one of the problems from "Problems in Mathematical Analysis I" - W. J. Kaczor;M. T. Nowak .
It's a problem 1.2.5 b), and it goes like this:
1.2.5. For $n \in \mathbb{N}$, verify the following claims:
$$\tag{b} \qquad \dfrac{1}{n + 1} + \dfrac{1}{n + 2} + \dfrac{1}{n + 3} + \ldots + \dfrac{1}{3n + 1} \, > \, 1$$
In solutions it says: "Use the arithmetic-harmonic mean inequality!"
I tried to apply it on whole inequality but got:
\begin{align}
& \dfrac{\frac{1}{n+1}+\ldots+\frac{1}{3n+1}}{2n}\, >\, \dfrac{2n}{n+1+\ldots+3n+1} \\
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{8n^2}{2n(n+1+3n+1)} \\
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{2n}{2n+1}
\end{align}
| You claim that $(n+1)+\cdots+(3n+1) = 2n((n+1)+(3n+1))$, which is not correct.
First of all, there are $2n+1$ terms, not $2n$ (as user8734617 says in a comment).
Also, you need to multiply with the average of the first and the last term.
So you actually get $$\frac{(2n+1)^2} { (2n+1) \left( \frac{(n+1)+(3n+1)}{2} \right)} = \frac{(2n+1)^2}{(2n+1)^2} = 1$$
For an alternative solution, try induction.
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Basic Integral Calculus Evaluate:
$$\int\frac{–4x^3(x^4–2)^{1/3}}{(x^2+2)(x^2–2)}dx$$
I had so many attempt in solving this equation but I can't find the correct answer .
I am asking anybody's help. Please help me.
Thank you!
| Expanding from @szw1710's hint
Let $x^4-2 = t^3$ then $4x^3\ dx = 3t^2\ dt$ and so
$$\begin{align}
\int\frac{(x^4-2)^{1/3}}{x^4 - 4}\ 4x^3dx &= \int \frac{3t^3}{t^3-2}\ dt \\
&= \int \left(3 + \frac{6}{t^3-2}\right)\ dt \\ &= 3t + \int \frac{6}{t^3-2} \ dt
\end{align}$$
Then let $t = \sqrt[3]{2}u$ we get
$$ \int \frac{6}{t^3 -2}\ dt = \int\frac{3\sqrt[3]{2}}{u^3-1}\ du$$
Now use partial fractions to decompose
$$ \frac{1}{u^3-1} = \frac{A}{u-1} + \frac{B(2u+1)}{u^2+u+1} + \frac{C}{\left(u+\frac{1}{2}\right)^2+\frac{3}{4}} $$
which integrates to
$$ \int \frac{1}{u^3-1}\ du = A\ln \big|u-1\big| + B\ln \big|u^2 + u + 1\big| + \frac{2}{\sqrt{3}}C\arctan \left( \frac{2u+1}{\sqrt{3}} \right) $$
The result is, how you say, not pretty... But it's doable.
| {
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Prob. 4, Chap. 7, in Baby Rudin: On what intervals does the series converge uniformly? Here is Prob. 4, Chap. 7, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Consider
$$ f(x) = \sum_{n=1}^\infty \frac{1}{1+n^2 x }. $$
For what values of $x$ does the series converge absolutely? On what intervals does it converge uniformly? On what intervals does it fail to converge uniformly? Is $f$ continuous wherever the series converges? Is $f$ bounded?
My Attempt:
Fix a real number $\delta > 0$.
If $x \in [ \delta, +\infty)$, then we see that
$$ 0 < \frac{1}{1+n^2 x} < \frac{1}{n^2 x } \leq \frac{1}{n^2 \delta } = \frac{1}{\delta} \frac{1}{n^2}, $$
and the series $\sum \frac{1}{n^2} $ converges. So, by Theorem 7.10 in Rudin, our series converges uniformly on $[ \delta , +\infty)$.
And, if $$x \in (- \infty, - \delta ]\setminus \left\{ \ -1, -\frac{1}{2^2}, -\frac{1}{3^2}, - \frac{1}{4^2}, \ldots \ \right\},$$
then $x < 0$, and, for all large enough $n$, we also have $1 + n^2 x < 0$, and for those $n$, we have
$$ \left\lvert \frac{1}{1+n^2 x} \right\rvert = \frac{1}{ \left\lvert 1+n^2 x \right\rvert} = \frac{1}{ -1 - n^2 x} < \frac{ 1}{ \frac{n^2 x}{2} - n^2 x} = -\frac{2}{x} \frac{1}{n^2} = \frac{2}{\delta} \frac{1}{n^2}. $$
Here we have used the fact that, as $x < 0$, so as $n$ gets larger and larger, we eventually have $1 + n^2 x < -1$, so $n^2 x < -2$, which implies that $\frac{n^2 x}{2} < -1$ and hence $\frac{n^2 x}{2} - n^2 x < -1 - n^2 x$; also from $1 + n^2 x < -1$, we obtain $1 < -1 - n^2 x$ and hence $0 < 1 < -1 - n^2 x$. And, we have also used the fact that, as $x \leq - \delta$, so $-x \geq \delta > 0$, and hence $$ 0 < -\frac{1}{x} \leq \frac{1}{\delta}.$$
As the series $\sum \frac{1}{n^2}$ converges, so it follows from Theorem 7.10 in Baby Rudin that our series converges uniformly on the following subset of $\mathbb{R}$:
$$ (-\infty, -\delta ] \setminus \left\{ \ -1, -\frac{1}{2^2}, - \frac{1}{3^2}, - \frac{1}{4^2}, \ldots \ \right\}.$$
Is what I have done so far correct? If not, then where have I erred?
What if $x \in (- \delta, \delta)$?
And, what about the (uniform) convergence of this series for complex values of $x$?
| Solution I
for values of $x>0$ we have
$$\sum\left|\frac{1}{1 + n^2x}\right|= \left|\frac{1}{x}\right|\sum\left|\frac{1}{\frac1x + n^2}\right| \leq \left|\frac{1}{x}\right|\sum\left|\frac{1}{ n^2}\right|$$
By the comparison test this shows that $f(x)$ converges absolutely. If $x = 0$ then we have
$$\sum\left|\frac{1}{1 + n^2x}\right| = \sum\left|1\right|= \infty$$
This series clearly doesn’t converge, absolutely or otherwise. If $x < 0$ things get more complicated: If $x = −1/n^2$
for any $n \in \mathbb{N}$ then the nth term of the series is undefined and therefore $f(x)$ is undefined. If $x \neq −1/n^2$ for any $n \in \mathbb{N}$ then we can use the fact that
For what intervals does the function converge uniformly?
If $E$ is any interval of the form $[a, b]$ with $a > 0$ then we have
$$\sup\left|f_n(x)\right| = f_n(a) = \frac{1}{1 + n^2a}$$
And therefore we have
$$\sum\sup\left|f_n(x)\right| = \sum\frac{1}{1 + n^2a} \leq \frac1a\sum\frac1{n^2}$$
The complete Answer you can find here (Page. 124)
Here is a second solution to the problem, which I think is more accurate than the previous
Solution II See link
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Unrationalize Denominator What is the method to unrationalize or reverse a rationalized fraction?
For example: How do you simplify $\frac{1}{2\sqrt\frac{1}{2}}$ = $\frac{1}{\sqrt{2}}$
| $$\frac{1}{2\sqrt{\frac{1}{2}}}=\frac{1}{\sqrt{4}\sqrt{\frac{1}{2}}}=\frac{1}{\sqrt{\frac{4}{2}}}=\frac{1}{\sqrt{2}}$$
Now you can multiply by "1":
$$\frac{1}{\sqrt{2}}\frac{1}{1}=\frac{1}{\sqrt{2}} \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{(\sqrt{2})^2}=\frac{\sqrt{2}}{2}$$
| {
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Linear Algebra Problem row echelon
Is there anyone could solve rhis question?
I try to minimize the matrix to row echelon form to find inconsistent and determinant of it but i cannot reduce it some how
try 1 =
$$\left[\begin{array}{cccc|c} 2&-1 & -2a & 1 & b \\ -2 & a & -3 & 0 & 4\\ 2 & -1 & 2a+1 & a+1 & 0 \\ -2 &-1 & 1-2a & -2 & -2b-2 \end{array} \right]$$
$$\to r_1 +r_2$$
$$ \to -r_1+r_3$$
$$ \to r_1+r_4$$
$$\left[\begin{array}{cccc|c} 2 & -1 & -2a & 1 & b \\ 0 & a-1 & -2a-3 & 1 & b+4\\ 0 & 0& 4a+1 & a & -b \\ 0 & 0 & 1-4a & -1 & -b-2 \end{array} \right]$$
$$\to r_3+r_4$$
$$\left[\begin{array}{cccc|c} 2 & -1 & -2a & 1 & b\\ 0 & a-1 & -2a-3 & 1 & b+4\\ 0 & 0 & 4a+1 & a & -b\\ 0 & 0 & 2 & a-1 & -2b-2 \end{array}\right]$$
I am stuck?
| Guide:
Great, now I will consider the following cases:
Case $1$: $a=1$. In this case, the matrix on the left is completely determined and it is certainly singular.
$$\left[\begin{array}{cccc|c} 2 & -1 & -2 & 1 & b\\ 0 & 0 & -5 & 1 & b+4\\ 0 & 0 & 5 & 1 & -b\\ 0 & 0 & 2 & 0 & -2b-2 \end{array}\right]$$
Check if there are rows where the left hand side is zero but the right hand side is not. If such case exists, then there is no solution, otherwise, you have infinitely many solution.
Case $2$:
Now, let's switch the third and fourth row:
$$\left[\begin{array}{cccc|c} 2 & -1 & -2a & 1 & b\\ 0 & a-1 & -2a-3 & 1 & b+4 \\ 0 & 0 & 2 & a-1 & -2b-2 \\ 0 & 0 & 4a+1 & a & -b\end{array}\right]$$
Perform $-\frac{(4a+1)}2r_3+r_4$
$$\left[\begin{array}{cccc|c} 2 & -1 & -2a & 1 & b\\ 0 & a-1 & -2a-3 & 1 & b+4 \\ 0 & 0 & 2 & a-1 & -2b-2 \\ 0 & 0 & 0 & -\frac{(4a+1)(a-1)}2+a & (4a+1)(b+1)-b\end{array}\right]$$
Find value of $a$ such that $(4,4)$ entry is zero, then depending on the value of $b$, you either have no solution of infinitely many solution. Otherwise, the solution is unique.
Edit: I realized there was a typo earlier, but fortunately it cancel out. when you first write down your matrix, it should be
$$\left[\begin{array}{cccc|c} 2&-1 & -2a & 1 & b \\ -2 & a & -3 & 0 & 4\\ 2 & -1 & 2a+1 & a+1 & 0 \\ -2 & \color{red}1 & 1-2a & -2 & -2b-2 \end{array} \right]$$
| {
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Evaluating $\int_0^{\pi /2} \frac{ \log (1+\cos a \cos x)}{\cos x} dx$ The question is to evaluate $$\int_0^{\pi /2} \frac{ \log (1+\cos a \cos x)}{\cos x} dx$$
I tried using leibnitz rule
$$F'(a)=\int_0^{\pi /2} \frac{ -\sin a}{(1+\cos a \cos x)}dx$$ Now I used the substitution $\tan(x/2)=t$ to get $$-2 \sin a \int_0^1 \frac{ dt}{1+t^2 +\cos a (1-t^2)} $$ which can be rewritten as
$$-2\frac{\sin a} {1- \cos a}\int_0^{1} \frac{ dt}{t^2 +\frac{1+ \cos a}{1-\cos a}} $$ which evaluates to $-a$.i am not sure where I went wrong.Any ideas?
| Well, as you did:
$$\mathscr{I}:=\int\frac{1}{1+\text{n}\cdot\cos\left(x\right)}\space\text{d}x\tag1$$
Substitute $\text{u}:=\tan\left(\frac{x}{2}\right)$:
$$\mathscr{I}=\frac{2}{1+\text{n}}\cdot\int\frac{1}{1+\frac{1-\text{n}}{1+\text{n}}\cdot\text{u}^2}\space\text{d}\text{u}\tag2$$
Substitute $\text{s}:=\text{u}\cdot\sqrt{\frac{1-\text{n}}{1+\text{n}}}$:
$$\mathscr{I}=\frac{2}{\sqrt{1-\text{n}}\cdot\sqrt{1+\text{n}}}\cdot\int\frac{1}{1+\text{s}^2}\space\text{d}\text{s}=\frac{2\cdot\arctan\left(\text{s}\right)}{\sqrt{1-\text{n}}\cdot\sqrt{1+\text{n}}}+\text{C}=$$
$$\frac{2\cdot\arctan\left(\tan\left(\frac{x}{2}\right)\cdot\sqrt{\frac{1-\text{n}}{1+\text{n}}}\right)}{\sqrt{1-\text{n}}\cdot\sqrt{1+\text{n}}}+\text{C}\tag3$$
So, for the definite integral:
$$\int_0^\frac{\pi}{2}\frac{1}{1+\text{n}\cdot\cos\left(x\right)}\space\text{d}x=\frac{2\cdot\arctan\left(\sqrt{\frac{2}{\text{n}^2}-1}\right)}{\sqrt{1-\text{n}^2}}\tag4$$
| {
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Modular arithmetic - How to find failure case theoretically I am unable to find the wrong intersection point of two modular classes theoretically, for the below question:
If $n$ is an odd integer, and $n^2 + 2n$ and $n$ have the same residue $\pmod 8$. What can be the remainder on division by $8$?
Residues of $n \mod 8$, for odd $n$, are: $\{1,3,5,7\}$. And of $2n \mod 8$ the residues are: $(2,6,2,6)$, and of $n^2 \mod 8$ are found by simply finding $r^2 \mod 8$:$(1,1,1,1)$, or $\{1\}$. $n^2 + 2n$ have the possible residues found by adding up the corresponding values of residues for $2n \pmod 8$ and $ n^2 \pmod 8$: $(3,7,3,7)$, or the set: $\{3,7\}$. So, $n$ values can be $8k+3, 8k+7, \forall k \in \mathbb {Z}$. So, the remainders $\mod 8$ can be : $3, 7$.
But, there is a flaw in this approach: you cannot figure out that $8k+3$ case will not work before-hand. One need see this case manually, with $n \pmod 8$ yielding residue $3$ and $n^2 + 2n \pmod 8$ yielding residue (using residue rather than actual value of $8k+3$):
$3^2 + 6 => 15 \pmod 8 => 7$.
This is also substantiated by a value of $n=19$ that falls in the class $8k+ 3$, with the value of $n^2 + 2n = 399 \equiv 7 \pmod 8$. However, $19 \equiv 3\pmod 8$.
Addendum Based on the answers given, have modified the answer to :
Need to equate the solutions given (theoretically, or algebraically) for the two modular classes:
$n^2 + 2n \equiv n\pmod 8 => n + 2 \equiv 1\pmod 8 => n \equiv -1 \pmod 8$.
Addendum -2 My approach for modification is to theoretically factor in the residue(remainder) from the start, as follows for the OP case, with $k$ representing not $n$ but the residue.
$n^2 \mod 8$ is given for odd $n$ as: $(2k+1)^2 \mod 8 => (4k^2 + 4k + 1) \mod 8 $. On adding $2n \mod 8$ to it, get: $((4k^2 + 4k + 1) + (4k+2)) \mod 8 => (4k^2 + 8k + 3) \mod 8$. This should be equal to $n \pmod 8$.
$4k^2 + 8k + 3 \equiv 2k+1 \pmod 8 => 2k^2 + 3k + 1 \equiv 0 \pmod 8 => 2k(k+1) + 1.(k+1) \equiv 0 \pmod 8 => (2k+1)(k+1) \equiv 0 \pmod 8$.
Only answer for integer value of $k$ (residue) possible is: $-1$ in modulo $8$ system
| All odd numbers square to $1$ mod $8$ so your equation actually reads $$1 + 2n \equiv n \, \text{mod} \, 8,$$ with solution $n \equiv 7$.
| {
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Solve the equations $(n+1)^n=2n^k+3n+1$ Solve the equations in positive integers
$$(n+1)^n=2n^k+3n+1$$
My attempts:
$$(n+1)^n=2n^k+3n+1$$
$$(n+1)^n-1=n(2n^{k-1}+3)$$
$$(n+1)^{n-1}+(n+1)^{n-2}+(n+1)^{n-3}+...+n+1=2n^{k-1}+3$$
$$(n+1)^{n-1}+(n+1)^{n-2}+(n+1)^{n-3}+...+n=2(n^{k-1}+1)$$
| [Note:I'm assuming for large $n$ that $k \le n$.]
If $n=1$ we get $2=6$ which is false. If $n=2$ we have $9=2\cdot 2^k+7$, which is true for $k=0$. Otherwise there are at least three terms in the binomial expression for $(n+1)^n\ge n^n+n^n+\frac {n-1}2n^{n-1}$
So we need $$\frac {n-1}2n^{n-1}\le 3n+1$$ or $$\frac n4\cdot n^{n-1}\lt 4n$$ (using crude estimates on both sides $\frac n4\lt \frac {n-1}2$ and $3n+1\lt 4n$) or $$n^{n-1}\lt 16$$
The left-hand side is increasing with $n$ and $4^3=64\gt 16$ so that $n\le 3$
If $n=3$ we have $64=2\cdot 3^k+10$, which gives $k=3$ .
[For $k\le n$, suppose for contradiction $k\ge n+1$ then divide through by $n^n$ to get $(1+\frac 1n)^n=2n{k-n}+\frac 1{n^n}$ The left-hand side is $\lt e$ and for $n\ge 2$ the right-hand side is $\gt 4$]
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The convergence of the series $\frac{n}{2(n+1)(n+2)}$ using difference method Checking the convergence of the series:
$$\sum\frac{n}{2(n+1)(n+2)}$$
I proved that the series is divergent using limit comparison test with the simple harmonic series:
$$lim_{n\to\infty}\frac{n^2}{2(n+1)(n+2)}=1/2$$
The series diverges since the harmonic series diverges.
My problem is that when I try the difference method , I find that the series converges
$$\frac{n}{2(n+1)(n+2)}=\frac{1}{2}(\frac{-1}{n+1}+\frac{2}{n+2})$$
The partial sum
$$S_n=\frac{1}{2}[(\frac{-1}{2}+\frac{2}{3})+(\frac{-1}{3}+\frac{1}{2})+(\frac{-1}{4}+\frac{2}{5})+(\frac{-1}{5}+\frac{1}{3})+(\frac{-1}{6}+\frac{2}{7})+(\frac{-1}{7}+\frac{1}{4})+...+(\frac{-1}{n+1}+\frac{2}{n+2})]$$
$$S_n=0.5 [ \frac{2}{3}+\frac{2}{5}+\frac{-1}{n+1}]$$
I got the sum of the series by taking the limit of the partial sum when n goes to infinity:
$$S=lim_{n\to\infty}S_n=\frac{8}{15}$$
I would like to know what is wrong with the difference method ( telescoping series) .
| The problem is that the terms do not telescope. While the partial fraction decomposition
$$
\frac{n}{2(n+1)(n+2)}=\frac12\left(\frac2{n+2}-\frac1{n+1}\right)
$$
is valid,
$$
\begin{align}
\sum_{n=0}^\infty\frac{n}{2(n+1)(n+2)}
&=\sum_{n=0}^\infty\frac12\left(\frac2{n+2}-\frac1{n+1}\right)\\
&=\frac12\left(\sum_{n=0}^\infty\frac1{n+2}+\color{#C00}{\sum_{n=0}^\infty\left(\frac1{n+2}-\frac1{n+1}\right)}\right)\\
&=\frac12\left(\sum_{n=0}^\infty\frac1{n+2}\color{#C00}{-1}\right)
\end{align}
$$
That is, it is divergent since it is the sum of a divergent, harmonic series and a convergent, telescoping series (in red).
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For $\text {HCF}(x,y,z)=1$ and $x^2+y^2=2z^2$, prove the following Let $x,y,z$ be positive integers such that $\text {HCF}(x,y,z)=1$ and $x^2+y^2=2z^2$. Prove the following
$1.$ $3$ divides $x+y$ or $3$ divides $x-y$
$2.$ $5$ divides $z(x^2-y^2)$
From $x^2+y^2=2z^2$, it is clear that $x$ and $y$ will be both odd or both even.
If both are even then $z$ will be odd so that $\text {HCF}(x,y,z)=1$ holds true but I think $x^2+y^2=2z^2$ could give more information which I am not able to observe and hence not able to prove the given statements. Could someone give some hint to proceed.
| b) By Fermat little theorem we have if $5\nmid a$ then $a^4 \equiv_5 1$ thus $a^2 \equiv_5 \pm1$.
If $5|z$ we are done, if $5\nmid 5$ then $2z^2\equiv_5 \pm 2$ so $x^2 \equiv_5 y^2 \equiv_5 1$ or $x^2 \equiv_5 y^2 \equiv_5 -1$ thus $5\mid x^2-y^2$. And we are done.
a) The same procedure can be done with mod $3$ of $3\nmid z$. If $3\mid z$ then $3\mid x^2+y^2$ then $3\mid x$ and $3\mid y$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2553985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying an expression: Stuck with it I have to prove that the expression
$$\frac{\omega C - \frac{1}{\omega L}}{\omega C - \frac{1}{\omega L} + \omega L - \frac{1}{\omega C}}$$
is equal to
$$\frac{1}{3-( (\frac{\omega_r}{\omega})^2 + (\frac{\omega}{\omega_r})^2)}$$
where $\omega_r= \frac{1}{\sqrt{LC}}$.
My attempt:
What I started to do was to get rid of the denominators in the fraction and put everything together.
$$\frac{\omega^2C^2L-C}{\omega^2C^2L-C+\omega^2CL^2-L}$$
Then I divided the denominator by the numerator
$$\frac{1}{1+\frac{\omega^2CL^2-L}{\omega^2C^2L-C}}$$
And I'm kind of stuck now. Can someone give an hint on how should I proceed next?
Or is there any easier way to start the proof? I'm just looking for a hint, thanks.
| The expressions are NOT equal when $\omega = 2$ and $L=1$ and $C=1$:
$$\frac{\omega C - \frac{1}{\omega L}}{\omega C - \frac{1}{\omega L} + \omega L - \frac{1}{\omega C}} \;\; = \;\; \frac{2\cdot1\; - \; \frac{1}{2\cdot 1}}{2\cdot 1 \; - \; \frac{1}{2\cdot1} \; + \; 2\cdot1 \; - \; \frac{1}{2\cdot 1}} \;\; = \;\; \frac{\frac{3}{2}}{\;4 - 1\;} \;\; = \;\; \frac{1}{2} $$
and
$$\frac{1}{3 \; - \; \left[ \left(\frac{\omega_r}{\omega}\right)^2 + \left(\frac{\omega}{\omega_r}\right)^2\right]} \; = \; \frac{1}{3 \; - \; \left[ \left(\frac{1}{2}\right)^2 + \left(\frac{2}{1}\right)^2\right]} \; = \; \frac{1}{\;3 \; - \; \frac{1}{4} \; - \; 4\;} \; = \; \frac{1}{\;-\frac{5}{4}\;} \; = \; -\frac{4}{5}, $$
where I've used the fact that $\;\omega_r= \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1\cdot 1\;}} = 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the equation of hyperbola whose foci are $F_1 = (3, 4)$ and $F_2 = (-1,-2)$ and $a=1$?
Find the equation of hyperbola whose foci are $F_1 = (3, 4)$ and $F_2 = (-1,-2)$ and $a=1$?
I need some help with this exercise. I know that this hyperbola is not centered at the origin, but I don't know its orientation and consequently the form of its equation. Does the distance between the foci still $2c$ in this case?
| Apply the definition, directly
$PF_1-PF_2=2a$
$$\sqrt{(x-3)^2+(y-4)^2}-\sqrt{(x+1)^2+(y+2)^2}=2$$
$$\sqrt{x^2-6 x+y^2-8 y+25}=\sqrt{x^2+2 x+y^2+4 y+5}+2$$
and square both sides
$$x^2-6 x+y^2-8 y+25=x^2+2 x+y^2+4 y+9+4 \sqrt{(x+1)^2+(y+2)^2}$$
Rearrange and divide both sides by $4$
$$4 - 2 x - 3 y=\sqrt{x^2+2 x+y^2+4 y+5}$$
and square again
$$\color{red}{3 x^2+8 y^2+12 x y-18 x-28 y+11=0}$$
is the equation of the wanted hyperbola.
Hope this can be useful
$$...$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor Remainder, $\arcsin(x)$ Given that the $0$ degree Taylor polynomial centered at $x=\frac{1}{2}$ for $f(x)=\arcsin(x)$ is given by $T_{0,\frac{1}{2}}(x)=\frac{\pi}{6}$ how would I use this to show that $1.2(x-\frac{1}{2})\leq R_{0,\frac{1}{2}}(x) \leq(x-\frac{1}{2})$ on $[0.4,0.5]$
I know that by Taylor's theorem $R_{0,\frac{1}{2}}(x)=\frac{1}{\sqrt{1-c^2}}(x-\frac{1}{2}$) and since $\frac{1}{\sqrt{1-x^2}}$ is an increasing function on $[0.4,0.5]$ that $R_{0,\frac{1}{2}}(x) \leq \frac{1}{\sqrt{1-0.5^2}}(x-\frac{1}{2})$
| Since as you notice
$$
\frac{1}{\sqrt{1-0.4^2}}\le\frac{1}{\sqrt{1-c^2}}\le\frac{1}{\sqrt{1-0.5^2}},
$$
all you need to prove is that
$$
\color{red}{1\le\frac{1}{\sqrt{1-0.4^2}}}\le\frac{1}{\sqrt{1-c^2}}\le\color{blue}{\frac{1}{\sqrt{1-0.5^2}}\le 1.2}
$$
and then multiply by $x-0.5$, which is negative for $0.4\le x\le 0.5$, so the inequalities become reversed.
| {
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"url": "https://math.stackexchange.com/questions/2559112",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Calculate $\lim_{n \to +\infty} \int_{0}^{+\infty} \frac{n \sin(\frac{x}{n})}{1 + x^2} dx$ The question is as follows:
Calculate $\lim_{n \to +\infty} \int_{0}^{+\infty} \frac{n \sin(\frac{x}{n})}{1 + x^2} dx$.
$\textbf{Some ideas:}$
We can use the fact that $\sin(\frac{x}{n}) \simeq \frac{x}{n} $.But then we find that
$\lim_{n \to +\infty} \int_{0}^{+\infty} \frac{n \sin(\frac{x}{n})}{1 + x^2} dx \simeq \lim_{n \to +\infty} \int_{0}^{+\infty} \frac{n \times \frac{x}{n}}{1 + x^2} = \lim_{n \to +\infty} \int_{0}^{n} \frac{ x }{1 + x^2} dx $
$ \hspace{9.1cm} = \lim_{n \to +\infty} \frac{1}{2}\int_{0}^{n} \frac{ 2x }{1 + x^2} dx $
$ \hspace{9.1cm} \text{take } x^2=y$
$ \hspace{9.1cm} = \lim_{n \to +\infty} \frac{1}{2}\int \frac{ dy }{1 + y} $
$ \hspace{9.1cm} = \lim_{n \to +\infty} \frac{\ln(y)}{2} $
$\hspace{9.1cm} = \lim_{n \to +\infty} \frac{\ln(x^2)}{2} \mid_{0}^{n}$
$\hspace{9.1cm} = \lim_{n \to +\infty} \frac{\ln(n^2)}{2} = +\infty$
But someone said me that the final result should be $\frac{\pi}{2}$?
Can you please let me know where is my mistake?
Thanks!
| This is too long to be in the comment section. The purpose of this post is to compute
\begin{align}
\int^\infty_0 \frac{n\sin^2\frac{x}{n}}{1+x^2}\ dx
\end{align}
exactly (not the limit). This is a harder way to show that the limit of my proposed revision of the problem is indeed $\pi/2$.
First, note that the integrand is integrable for all values of $n$. Next, observe
\begin{align}
\int^\infty_0 \frac{n\sin^2 \frac{x}{n}}{1+x^2}\ dx =&\ \frac{n}{2}\int^\infty_0 \frac{1}{1+x^2}\ dx -\frac{n}{2}\int^\infty_0\frac{\cos \frac{2x}{n}}{1+x^2}\ dx \\
=&\ \frac{n\pi}{4} - \frac{n}{4} \int^\infty_{-\infty} \frac{\exp(i\frac{2x}{n})}{1+x^2}\ dx.
\end{align}
Now, using contour integration, we can show that
\begin{align}
\int^\infty_{-\infty} \frac{\exp(i\frac{2x}{n})}{1+x^2}\ dx = \pi e^{-2/n}.
\end{align}
Hence it follows
\begin{align}
\int^\infty_0 \frac{n\sin^2\frac{x}{n}}{1+x^2}\ dx = \frac{n\pi}{4}\left( 1- e^{-2/n}\right) = \frac{\pi}{2}ne^{-1/n}\sinh n^{-1}.
\end{align}
Finally, let us make the observation that for $n$ large we have that
\begin{align}
1-e^{-2/n} = \frac{2}{n}+\mathcal{O}\left(\frac{1}{n^2}\right)
\end{align}
which means
\begin{align}
\int^\infty_0 \frac{n\sin^2\frac{x}{n}}{1+x^2}\ dx = \frac{\pi}{2}+\mathcal{O}\left(\frac{1}{n}\right).
\end{align}
Thus, as $n\rightarrow \infty$, we see that the integral approaches $\pi/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{i=1}^{n} \frac{n+i}{i+1} \le 1 + n(n-1) \ \forall n \in \Bbb{N}$ (without calculus) I'm trying to prove the following proposition (I'm not supposed to use calculus):
$$\sum_{i=1}^{n} \frac{n+i}{i+1} \le 1 + n(n-1) \ \forall n \in \Bbb{N}$$
(I'm assuming that $0 \notin \Bbb{N}$)
This is what I've tried so far:
$\sum_{i=1}^{n} \frac{n+i}{i+1} = \sum_{i=1}^{n}[ \frac{n}{i+1} + \frac{i}{i+1} ] = [\sum_{i=1}^{n} \frac{n}{i+1}] + \sum_{i=1}^{n} \frac{i}{i+1} = n[\sum_{i=1}^{n} \frac{1}{i+1}] + \sum_{i=1}^{n} \frac{i}{i+1}$
Let $j= i +1$ . So now we have:
$\sum_{i=1}^{n} \frac{n+i}{i+1} = n[\sum_{j=2}^{n+1} \frac{1}{j}] + \sum_{j=2}^{n+1} \frac{j-1}{j}$
$= n[\sum_{j=2}^{n+1} \frac{1}{j}] + \sum_{j=2}^{n+1} \frac{j}{j} - \sum_{j=2}^{n+1} \frac{1}{j}$
$= n[\sum_{j=2}^{n+1} \frac{1}{j}]- \sum_{j=2}^{n+1} \frac{1}{j} + \sum_{j=2}^{n+1} 1 $
$= (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + [\sum_{j=1}^{n+1} 1 ] -1 $
$= (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n+1 -1 $
$= (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n$
Therefore proving the following inequality is the same as proving the original:
$$ (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n \le 1 + n(n-1)$$
If $n=1$ then $ (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n = 1$ and $1 + n(n-1) = 1$
Now I assume that $n>1$:
$ (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n \le 1 + n(n-1) \iff (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] + n \le 1 + n^2 - n$
$\iff (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] \le 1 + n^2 - 2n$
$\iff (n-1)[\sum_{j=2}^{n+1} \frac{1}{j}] \le (n-1)^2$
$\iff \sum_{j=2}^{n+1} \frac{1}{j} \le \frac{(n-1)^2}{n-1}$
$\iff \sum_{j=2}^{n+1} \frac{1}{j} \le n-1$
$\iff 1 + \sum_{j=2}^{n+1} \frac{1}{j} \le n$
$\iff \sum_{j=1}^{n+1} \frac{1}{j} \le n$
And here is where I'm stuck. I know that an upper bound for the harmonic sum can be found using an integral test but I'm not supposed to use calculus. Is there a discrete way of proving that $\sum_{j=1}^{n+1} \frac{1}{j} \le n$ for all natural $n \ge 2$ ? Or maybe another way of proving the proposition without ending up with the harmonic sum?
| You can use $$\frac{1}{j} \le \frac 12 ~ \forall j \ge 2$$
Therefore $$\sum_{j=2}^{n+1}\frac 1j \le n \times \left(\frac{1}{2}\right) \implies \color{blue}{1} +\sum_{j=2}^{n+1}\frac 1j \le \frac{n+2}{2} \le n \quad \forall \;n \ge 2$$
For, $n=1$ it is easy to verify, that it's incorrect.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the function $\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}$ Continuous at $(0,0)$ The function is defined piece-wise with $f(0,0) = 0$, so I took the limit $(x,y) \rightarrow (0,0)$
I proceeded like this:
$$\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2} \leq \frac{|y|(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2} \leq \frac{|y|(x^4+4x^2y^2+y^4)}{(x^2+y^2)^2}
$$
$$= \frac{|y|(x^2+y^2)^2+|y|2x^2y^2}{(x^2+y^2)^2} = |y| + \frac{|y|2(xy)^2}{(x^2+y^2)^2} \leq |y| + \frac{|y|2(\frac{x^2+y^2}{2})^2}{(x^2+y^2)^2} = |y| + \frac{|y|}{2} \rightarrow0$$
Hence by the sandwich/squeeze theorem, the original function also goes to $0$
Is this answer ok?
| Yes your proof looks good, very good.
Your problem is a special case of this result: Suppose $p(x,y),q(x,y)$ are homogeneous polynomials with $\deg p > \deg q,$ and $q(x,y)\ne 0$ except at $(0,0).$ Then
$$\lim_{(x,y)\to (0,0)}\frac{p(x,y)}{q(x,y)}=0.$$
A popular way of proving this is to go to polar coordinates.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given $0Z$
Given $x\in \mathbb R$, $0<x<1$, $\{m,n\}\subset \mathbb N$,$0<n<m$,
$$Z=x^m-\frac{1}{x^m}~~\text{and}~~Y=x^n-\frac{1}{x^n}.$$
Show that: (a) $Z$ and $Y$ are negative; (b) $Y>Z.$
Edit: the original exponents for $Y$ and $Z$ were switched in the original post. Now they are correct.
Problem from a book on inequalities at contest level. I got stuck on part (b), not able to find an argument for the general case.
My attempt:
Part (a): developing $Z$ it follows that
$$Z=\frac{x^{2m}-1}{x^m}<0$$
as $0<x<1$, $x^{2m}-1<0$ and $x^m>0$. Similar argument holds for $Y$.
Part (b): only able to show for restricted cases, as $m=2$ and $n=1$, for instance:
$$Y=x-\frac{1}{x}~~\text{and}~~Z=x^2-\frac{1}{x^2}=(x-\frac{1}{x})(x+\frac{1}{x})=Y(x+\frac{1}{x})$$
as $Y<0$ and $x+\frac{1}{x}>1$, it follows that $Y>Z$, as required. I've got other restricted cases but I was not able generalize for all $0<n<m$. My intuition is suggesting an argument constructed on the identity
$$x^n-\frac{1}{x^n}=(x-\frac{1}{x})(x^{n-1}+x^{n-2}\frac{1}{x}+\ldots+\frac{1}{x^{n-1}})$$
but I have not find it so far.
Hints and solutions are appreciated. Sorry if this is a duplicate.
| Since $0<n<m$ then we have $x^n>x^m$ because $0<x<1$.
Now suppose that we have $Y \geq Z$.
$Y \geq Z \implies x^m-\dfrac{1}{x^m} \geq x^n-\dfrac {1}{x^n} \implies \dfrac {x^{2m}-1}{x^m} \geq \dfrac {x^{2n}-1}{x^n} \implies \dfrac {1-x^{2m}}{x^m} \leq \dfrac{1-x^{2n}}{x^n} \implies x^n(1-x^{2m}) \leq x^m(1-x^{2n}) \implies x^n-x^{2m+n} \leq x^m - x^{2n+m} \implies x^n-x^m \leq x^{2m+n}-x^{2n+m}=x^{m+n}(x^m-x^n)$,
which is a contradiction, since $x^n-x^m$ is positive and $x^{m+n}(x^m-x^n)$ is negative, so we have $Y<Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a way to solve for the missing angle?
I was working on this problem. I tried to draw ${AC}$, ${BD}$ as isosceles triangle and divide into cases to find the missing angle $??$, but I got stuck. Can someone help me please or give me a clue?
|
$$AB=AD=BC=x,AC=t\\t^2=x^2+x^2-2x^2\cos166\\\to t^2=2x^2(1-\cos166)=2x^2(2\sin^2(83))=4x^2\cos^2(7)\\t=2x\cos 7$$so
$$\frac{t}{\sin(120-c)}=\frac{x}{sin(c-7)}\\\frac{2x\cos 7}{\sin(120-c)}=\frac{x}{sin(c-7)}\\2\sin(c-7)\cos 7=\cos(30-c)$$using product to sum formula
$$\sin(c-14)+\sin c=\cos(30-c)\\
\sin 30 \cos c+ \cos 30 \sin c+\sin c=\cos 30 \cos c + \sin30 \sin c\\\sin c(1-\frac 12 +\cos 14)=\cos c(\cos 30+\sin14)\\\tan c=\frac{\cos 30+\sin 14}{\frac12+\cos 14}=\frac{\sin 60+\sin 14}{\cos 60+\cos 14}=\frac{2\sin37 \cos 23}{2\cos37 \cos 23}=\tan(37)$$finally
$$\tan(c)=\tan(37) \to c=37$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do you find the values that $x$ can take by squaring? $$|x+4| \cdot |x-4| = x+4$$
How do you find the values that $x$ can take by squaring?
How I've tried
$$|x+4|^2 \cdot |x-4|^2 = (x+4)^2$$
and
$$|x^2+16| \cdot |x^2+16| = x^2+16$$
I think I went wrong.
| You can split the problem into 3 cases: $x\lt-4$, $-4\le x\le4$, and $4\lt x$.
For $x \lt -4$:
*
*There are no solutions for this case because $|x+4|*|x-4|$ is always positive, and $x+4$ will always be negative for $x<-4$.
For $-4\le x\le 4$:
*
*$|x-4| = 4-x$, and $|x+4| = x+4$, so solve $(4-x)*(x+4)=x+4$.
*$-x^2+16 = x+4$
*$0=x^2+x-12$
*$x = 3, -4$
For $4\lt x$:
*
*$|x-4| = x-4$, and $|x+4| = x+4$, so solve $(x-4)*(x+4)=x+4$.
*$x^2-16 = x+4$
*$0=x^2-x-20$
*$x = 5, -4$
All solutions: $x = -4, 3, 5$
| {
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Evaluate $\int_{1}^{4} \int_{-1}^{2z} \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy \; dx\; dz$ $\int_{1}^{4} \int_{-1}^{2z} \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy \; dx\; dz$
Consider $ \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy = $ $\int_{0}^{\sqrt{3}x} \frac{x}{x^2 +y^2} - \frac{y}{x^2+y^2} \;dy$
$\int_{0}^{\sqrt{3}x} \frac{x}{x^2 +y^2} \; dy = \arctan(\sqrt{3}) = \pi/3$
$x^2 + y^2 = t \Rightarrow y \; dy = dt/2$
$ \int_{0}^{\sqrt{3}x} \frac{y}{x^2 +y^2} \;dy = \int_{x^2}^{3+x^2} \frac{dt}{t} = \frac{\ln(3+x^2)}{2} - \ln(x)$
$ \int_{0}^{\sqrt{3}x} \frac{x-y}{x^2 +y^2} \;dy = \pi/3 + \ln(x) - \frac{\ln(3+x^2)}{2}$
Now it will be very lengthy and difficult to integrate this with respect to x and z.
So I am stuck here. Is there any easy method to solve such questions ?
| $$\int\frac{x-y}{x^2+y^2}\,dy=\arctan\left(\frac{y}{x}\right)-\frac{1}{2}\log(x^2+y^2).$$
Therefore, your integral becomes
$$\begin{align}&\int_1^4\int_{-1}^{2z}\left(\arctan(\sqrt{3})-\frac{1}{2}\log(4x^2)-\arctan0+\frac{1}{2}\log(x^2)\right)\,dx\,dz\\
=&\int_1^4\int_{-1}^{2z}\left(\arctan(\sqrt{3})-\frac{1}{2}\log(4)-\frac{1}{2}\log(x^2)-\arctan0+\frac{1}{2}\log(x^2)\right)\,dx\,dz\\
=&\int_1^4\int_{-1}^{2z}\left(\frac{\pi}{3}-\log(2)\right)\,dx\,dz\\
=&\left(\frac{\pi}{3}-\log(2)\right)\int_1^4(2z+1)dz\\
=&18\left(\frac{\pi}{3}-\log(2)\right)\\
=&6\pi-18\log 2
\end{align}$$
Your mistake was in calculating
$\int\frac{y}{x^2+y^2}\,dy.$ Note that
$$\int\frac{y}{x^2+y^2}\,dy=\frac{1}{2}\log(x^2+y^2).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof $\sum^a_{n=1}\prod^b_{x=1}(n+x)=\frac{\prod^b_{y=0}a+y}{b+1}$ I have found this formula to a sum of multiples:
$$\sum^a_{n=1}\prod^{b-1}_{x=0}(n+x)=\frac{\prod^{b}_{x=0}(a+x)}{b+1}$$For example, when $a=5$ and $b=3$, the sum would be $1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+5\cdot6\cdot7$, and the formula would give $\frac{5\cdot6\cdot7\cdot8}{4}=420$. It seems to work for many values of $a$ and $b$, but how do I prove this formula?
| As Tiwa Aina suggests, the formula
$$\sum^a_{n=1}\prod^b_{x=1}(n+x)=\frac{\prod^b_{y=0}a+y}{b+1}$$
is wrong.
This answer proves that
$$\sum^a_{n=1}\prod^{\color{red}{b-1}}_{x=\color{red}{0}}(n+x)=\frac{\prod^b_{y=0}(a+y)}{b+1}$$
Considering binomial coefficients should help.
$$\begin{align}\sum_{n=1}^{a}\prod_{x=0}^{b-1}(n+x)&=\sum_{n=1}^{a}\frac{(n+b-1)!}{(n-1)!}\\\\&=b!\sum_{n=1}^{a}\frac{(n+b-1)!}{b!(n-1)!}\\\\&=b!\sum_{n=1}^{a}\binom{n+b-1}{b}\\\\&\color{red}{=}b!\binom{b+a}{b+1}\\\\&=\frac{b!(b+a)!}{(b+1)!(a-1)!}\\\\&=\frac{1}{b+1}\prod_{y=0}^{b}(a+y)\end{align}$$
The proof for the equality in red can be seen here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
Sum of fifth power of the roots of equation $x^3-x^2+1=0$ The equation $x^3-x^2+1=0$ has three roots $\alpha$, $\beta$ and $\gamma$. Find the value of $\alpha^5 + \beta^5 + \gamma^5$
I tried it this way:
$x^3=x^2-1$
$\alpha + \beta + \gamma = 1$
$\alpha \cdot \beta \cdot \gamma = -1$
$\alpha \cdot \beta + \beta \cdot \gamma + \alpha \cdot \gamma = 0$
So, $\alpha^3=\alpha^2-1$
$\alpha^5=\alpha^4-\alpha^2$
And similarly for $\beta$ and $\gamma$
Now I did add them but I am unable to find something useful in it.
| $$\alpha^5+\beta^5+\gamma^5=(\alpha+\beta+\gamma)^5-5(\alpha+\beta+\gamma)^3(\alpha\beta+\alpha\gamma+\beta\gamma)+5(\alpha\beta+\alpha\gamma+\beta\gamma)^2+$$
$$+5(\alpha+\beta+\gamma)^2\alpha\beta\gamma-5(\alpha\beta+\alpha\gamma+\beta\gamma)\alpha\beta\gamma=1+5\cdot(-1)=-4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How to find $\sum \left(1 + 1/2 + \dots + 1/(n + k + 1)\right)\frac{1}{n(n + 1)...(n + k + 1)}$? I've got two series that look alike,
\begin{align}
S_1 = \sum_{n = 1}^\infty \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right).\frac{1}{n(n + 1)}\tag{1}\\
S_2 =\sum_{n = 1}^\infty \left(1 + \frac{1}{2} + \dots + \frac{1}{n+ 1}\right).\frac{1}{n(n + 1)}.\tag{2}\\
\end{align}
For the first one I can find the sum, it equals to $\pi^2/6$ by writing $$S_1 = \sum_{n = 1}^\infty \left(\frac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n} -\frac{1 + \frac{1}{2} + \dots + \frac{1}{n+ 1}}{n + 1}+ \frac{1}{(n + 1)^2}\right)$$
and the sum telesope.
As for the second one, by substracting the two series we got
$$S_2 - S_1 = \sum_{n = 1}^\infty \frac{1}{n(n + 1)^2} = 2 - \frac{\pi^2}{6},$$ hence $S_2 = 2$.
I got curious if there exist any general formula for the series $$ \sum_{n = 1}^\infty \left(1 + \frac{1}{2} + \dots + \frac{1}{n + k}\right)\frac{1}{n(n + 1)...(n + k+1)}$$
and
$$ \sum_{n = 1}^\infty \left(1 + \frac{1}{2} + \dots + \frac{1}{n + k+ 1}\right)\frac{1}{n(n + 1)...(n + k+1)}$$ with $k \geq 0$.
I think it would be similar to the case $k = 0$ above. But I just can't find my way.
| $$\begin{eqnarray*}\sum_{n\geq 1}\frac{H_{n+k+1}}{n(n+1)\cdots(n+k+1)}&=&\frac{1}{k!}\sum_{n\geq 1}\left(\frac{\binom{k}{0}}{n}-\frac{\binom{k}{1}}{n+1}+\ldots\pm\frac{\binom{k}{k}}{n+k}\right)\frac{H_{n+k+1}}{n+k+1}\\&=&\frac{1}{k!}\sum_{n\geq 1}\frac{H_{n+k+1}}{n+k+1}\int_{0}^{1}\sum_{h=0}^{k}\binom{k}{h}(-1)^h x^{n-1+h}\,dx\\&=&\frac{1}{k!}\int_{0}^{1}(1-x)^k\sum_{n\geq 1}\frac{H_{n+k+1}}{n+k+1}x^{n-1}\,dx\end{eqnarray*}$$
can be computed from Euler's Beta function,
$$ \sum_{m\geq 1}\frac{H_m}{m}\, x^m =\frac{1}{2}\log^2(1-x)+\text{Li}_2(x),$$
$$ \int_{0}^{1}(1-x)^k \log^2(1-x)\,dx = \frac{2}{(k+1)^3}, $$
$$ \int_{0}^{1}(1-x)^k\text{Li}_2(x)\,dx \stackrel{\text{IBP}}{=}\frac{\zeta(2)}{k+1}-\frac{H_{k+1}}{(k+1)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2570478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$. I'm looking for help with (b) and (c) specifically. I'm posting (a) for completeness.
(a) Show convergence for $a_n=\sqrt{n+1}-\sqrt{n}$ towards $0$ and test $\sqrt{n}a_n$ for convergence.
(b) Show $b_n=\sqrt[k]{n+1}-\sqrt[k]{n}$ converges towards $0$ for all $k \geq 2$.
(c) For which $\alpha\in\mathbb{Q}_+$ does $n^\alpha b_n$ converge?
I'm pretty sure I solved (a). I have proven the convergence of $a_n$ by using the fact that $$\sqrt{n}<\sqrt{n+1}\leq\sqrt{n}+\frac{1}{2\sqrt{n}}$$ which holds true since $$(\sqrt{n}+\frac{1}{2\sqrt{n}})^2=n+1+\frac{1}{4n}\geq n+1\,.$$
This gives us $$0<\sqrt{n+1}-\sqrt{n}\leq\frac{1}{2\sqrt{n}}$$
and after applying the squeeze theorem with noting that $\frac{1}{2\sqrt{n}}\longrightarrow0$ we can tell that also $a_n\longrightarrow0$.
Now $x_n=\sqrt{n}a_n=\sqrt{n}(\sqrt{n+1}-\sqrt{n})$.
We have \begin{align*}\sqrt{n}(\sqrt{n+1}-\sqrt{n})&=\sqrt{n}\sqrt{n+1}-\sqrt{n}\sqrt{n}\\&=\sqrt{n(n+1)}-n\\&=\sqrt{n^2+n}-n\\&=\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n}\\&=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}\\&=\frac{n}{\sqrt{n^2+n}+n}\\&=\frac{n}{n\sqrt{1+\frac{1}{n}}+n}\\&=\frac{1}{\sqrt{1+\frac{1}{n}}+1}\end{align*}
and hence since the harmonic sequence $\frac{1}{n}$ converges towards 0 we have $$\text{lim}_{n\rightarrow\infty} \frac{1}{\sqrt{1+\frac{1}{n}}+1} = \frac{1}{1+1} = \frac{1}{2}\,._{\,\,\square}$$
| Hint: It might be useful applying mean-value theorem for continuous function $f(x)=\sqrt[k]{x}$ on $[1,\infty)$ then there exists $\xi\in[n,n+1]$ such that
$$\sqrt[k]{n+1}-\sqrt[k]{n}=\dfrac{1}{k\sqrt[k]{\xi}}\leq\dfrac{1}{k\sqrt[k]{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prime numbers yield from Pythagoras triples Pythagoras theorem $$a^2+b^2=c^2$$
we got $$P_{prime}(a,b)={a^4+b^4+(a+b)^4\over a^2+b^2+(a+b)^2}$$
Where $(a,b,c)$ are Pythagoras theorem triples, this function $P_{prime}(a,b)$ always produce a prime number for the values of a and b.
Examples: $P_{prime}(3,4)=37$, $P_{prime}(5,12)=229$, $P_{prime}(68,285)=105229$ and so on...
I have checked a lot of values, it seem to be prime so far.
My question is: Does the function $P_{prime}(a,b)$ always produce prime numbers?
| This is a more abstract take on my computational answer.
The fraction cancels out to $a^2+ab+b^2$.
Now suppose that we have a prime $p=12n+1$. Modulo $p$ the multiplicative group is cyclic of order $12n$. We work modulo $p$, so we can find $w$ and $t$ with $w^3=1, w\neq 1$ and $t^2=-1$.
Now choose $y$ and let $x=(w+w^2t)y$
If we now set $a=x^2-y^2=(w^2+2t-w-1)y^2=2w^2(1+wt)y^2$ and $b=2xy=2w(1+wt)y^2$ whence $a=wb$ and we have $$a^2+ab+b^2=b^2(1+w+w^2)=0$$So we have solved the congruence modulo $p$.
We can choose $\pm t$ and $w$ or $w^2$ for this construction, but once selected they give an infinite family of solutions modulo $p$. These grow indefinitely, and since divisible by $p$ the larger ones are not prime.
As an example, for $p=13$ we have $5^2\equiv-1$ and $3^3\equiv 1$.
Choose $y=1$ to give $x=3+45\equiv-4$ and $a=15, b=-8$ which gives a value of $169$.
We could also choose $y=1, x=9$ (we can choose equivalents modulo $13$) to give $a=80, b=18$
Choosing $y=2$ gives $x=2\times -4\equiv5$ and this gives $a=21, b=20$ giving $1261=13 \times 97$ and again we can choose any representatives of the residue classes for $x$ and $y$ giving an infinite family of solutions.
Note that this can be written in terms of a primitive twelfth root of unity modulo $p$ as $x=(u-u^2)y$ with $a=2u^3(u-1)y^2, b=2u(1-u)y^2$. Since there are four such roots $(u, u^5, u^7, u^{11})$, this encompasses the four choices which could be made above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\left(\frac {-1+\sqrt {-3}}{2}\right)^n + \left(\frac {-1-\sqrt {-3}}{2}\right)^n$ follows this pattern
Prove that:
$$\left(\dfrac {-1+\sqrt {-3}}{2}\right)^n + \left(\dfrac {-1-\sqrt {-3}}{2}\right)^n=\begin{cases}
2, & \textrm { if } n \textrm { is a multiple of 3},\\
-1, & \textrm { if } n \textrm { is any other integer}
\end{cases}$$
My Attempt:
$$\dfrac {-1+\sqrt {-3}}{2}=\dfrac {-1+i\sqrt {3}}{2}$$ which is a complex cube root of unity.
Let $\omega = \dfrac {-1+i\sqrt {3}}{2}$.
Similarly, $\omega^2=\dfrac {-1-i\sqrt {3}}{2}$
| Using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i*\sin(\varphi)$?,
$$-\dfrac12+\dfrac{\sqrt3i}2=\cos120^\circ+i\sin120^\circ=e^{i120^\circ}$$
$$\left(-\dfrac12+\dfrac{\sqrt3i}2\right)^n=\cos(120^\circ n)+i\sin(120^\circ n)$$
Similarly,
$$-\dfrac12-\dfrac{\sqrt3i}2=\cos(-120^\circ)+i(-\sin120^\circ)=e^{-i120^\circ}$$
$$\left(-\dfrac12-\dfrac{\sqrt3i}2\right)^n=\cos(-120^\circ n)+i\sin(-120^\circ n)=\cos(120^\circ n)-i\sin(120^\circ n)$$
$$\left(-\dfrac12+\dfrac{\sqrt3i}2\right)^n+\left(-\dfrac12-\dfrac{\sqrt3i}2\right)^n=2\cos(120^\circ n)$$
Now $n$ can be written as one of $3m,3m+1,3m+2$ where $m$ is any integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\left(\frac{-2}{p}\right)$
Prove that for $p\ge 3$, a prime: $$\left(\frac{-2}{p}\right) = \begin{cases}
1 & p\equiv1,3\pmod{8}\\
-1 & p\equiv-1,-3\pmod{8}
\end{cases}$$
I already now that:
$$\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p^{2}-1}{8}}=(-1)^{\frac{(p-1)(p^{2}-1)}{8}}=(-1)^{\frac{p-1}{2}\cdot\frac{p+1}{2}\cdot\frac{p-1}{4}}$$
And I'm trying to show when the exponent is even (for the $1$ case).
We have:
*
*$\frac{p-1}{4} = 2k \implies p- 1 = 8k \implies p = 8k + 1$. So we can infer that if $p\equiv 1\pmod{8}$ then $\left( \frac{-2}{p} \right)=1$
*$\frac{p-1}{2}$ - doesn't yield anything new to us (I think)
*$\frac{p+1}{2} = 2t \implies p+1 = 4t \implies p = 4t-1$
but I don't really see how to infer the other option $(p \equiv 3 \pmod{8})$
| There's an error in your computation:
$$\binom{-2}p=(-1)^{\tfrac{p-1}{2}}(-1)^{\tfrac{p^{2}-1}{8}}=(-1)^{\tfrac{(p-1)\color{red}{+}(p^{2}-1)}{8}}=(-1)^{\tfrac{p-1}{2}\bigl(1+\tfrac{p+1}{4}\bigr)}=(-1)^{\tfrac{p-1}{2}\tfrac{p+5}{4}}$$
This is equal to $1$, i.e. $-2$ is a square mod. $p$, if and only if $\;\dfrac12\dfrac{p-1}2\,\dfrac{p+5}2\,$ is even. Observe that
$$\dfrac{p-1}2\;\text{is}\:\begin{cases}\text{even}\\\text{odd}\end{cases}\iff \dfrac{p+5}2\;\text{is}\:\begin{cases}\text{odd}\\\text{even}\end{cases}$$
so we have the following cases:
*
*if $\dfrac{p-1}2$ is even, we have $\;\dbinom{-2}p=1\iff\dfrac{p-1}2\equiv 0\mod 4\iff p\equiv 1\mod 8$,
*if $\dfrac{p-1}2$ is odd, $\;\dbinom{-2}p=1\iff\dfrac{p+5}2\equiv 0\mod 4\iff p\equiv -5\equiv 3\mod 8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to prove the value of an trigonometric term to zero? Given that , in $\triangle ABC$ , $AC \neq BC$ . We have to prove that , $\dfrac{BC\cos C-AC\cos B}{BC\cos B-AC\cos A}+\cos C=0$
My trying:
As $AC \neq BC \Rightarrow b \neq a \Rightarrow a=c \Rightarrow A=C$ . So for the following term , $$ \\ $$
\begin{align}
\frac{BC\cos C-AC\cos B}{BC\cos B-AC\cos A}+\cos C
&= \frac{a\cos C-b\cos B}{a\cos B-b\cos A}+\cos C = \frac{a\cos A-b\cos B}{a\cos B-b\cos A}+\cos A \\[1ex]
&= \frac{a\cos A-b\cos B+a\cos A\cos B=b{\cos }^2A}{a\cos B-b\cos A}
\end{align}
Then I got stuck and have no clue how to go ahead . Can anyone please help me to solve this problem ? It will be of great help .
| for the first numerator i have got $$1/2\,{\frac {{a}^{2}+{b}^{2}-{c}^{2}}{b}}-1/2\,{\frac {b \left( {a}^{2
}-{b}^{2}+{c}^{2} \right) }{ac}}
$$
for the denominator i have got
$$1/2\,{\frac {{a}^{2}-{b}^{2}+{c}^{2}}{c}}-1/2\,{\frac {-{a}^{2}+{b}^{2
}+{c}^{2}}{c}}
$$
and the sum is given by
$${\left( 1/2\,{\frac {{a}^{2}+{b}^{2}-{c}^{2}}{b}}-1/2\,{\frac {b
\left( {a}^{2}-{b}^{2}+{c}^{2} \right) }{ac}} \right) \left( 1/2\,{
\frac {{a}^{2}-{b}^{2}+{c}^{2}}{c}}-1/2\,{\frac {-{a}^{2}+{b}^{2}+{c}^
{2}}{c}} \right) ^{-1}}+1/2\,{\frac {{a}^{2}+{b}^{2}-{c}^{2}}{ab}}
$$
have you got this?
the simplified numerator is given by $$1/2\,{\frac {{a}^{3}c-{a}^{2}{b}^{2}+a{b}^{2}c-a{c}^{3}+{b}^{4}-{b}^{2
}{c}^{2}}{bac}}
$$
and the simplified denominator is given by $${\frac { \left( a-b \right) \left( a+b \right) }{c}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $4^{3x+1} + 2^{3x+1}+ 1$ is divisible by 7 I want to show that $4^{3x+1} + 2^{3x+1} + 1$ is divisible by 7, I am trying to show this with modular arithmetic.
If I break up each part of the equation, I can see that
$4^{3x+1} = 4$ x $2^{6x}$
which implies that $4^{3x+1}mod(7) = 4$
I can't quite find a nice factorization of $2^{3x+1}$
Any help specifically on how to treat $2^{3x+1}$ would be appreciated.
| Hint:
$$2^{3x+1} = 8^x\times 2$$
$$8 \equiv 1\pmod{7} \implies 8^x \equiv 1 \pmod{7}\implies 8^x \times 2\equiv 2\pmod{7}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Fallacy in showing that $\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$ Given a cyclic quadrilateral $ABCD$, $AB=a, BC=b, CD=c, \angle ABC=120^\circ, \angle ABD=30^\circ$, then, show that $$\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$$
I tried to do it using some trig bashing.
What I did was basically assigned the angle $\angle BCA = \theta$ and used trigonometry.
It's not hard to see that the radius of the circle is $\frac{c}{2}$.
I've squared the equation on both sides and obtained: $$c+2a+2b=2\sqrt{(c+a)(c+b)}$$.
And using $\text {Sine Rule}$, the equation can be reduced further.
I used
*
*$\frac{a}{\sin\theta}=c$
*$\frac{b}{\sin(60^\circ - \theta)}=c$
And reduced the equation to: $$1+2\sin\theta+2\sin(60^\circ - \theta)=2\sqrt{(1+\sin\theta)(1+\sin(60^\circ - \theta))}$$.
Then I continued to reduce it using the addition-subtraction formulae of trigonometry.
And at the end of the day what I get is $$\boxed{\cos(30^\circ-\theta)=\frac{1}{2}}$$ after all those addition-subtraction of trigonometric equations.
And that's not true I believe.
May I get rectified?
| By law of cosines $$AC=\sqrt{a^2+b^2-2ab\cos120^{\circ}}=\sqrt{a^2+b^2+ab}.$$
In another hand, since $\measuredangle CBD=90^{\circ},$ we see that $CD$ is a diameter of the circle.
Thus, $$\cos30^{\circ}=\frac{\sqrt{a^2+b^2+ab}}{c}$$ or
$$3c^2=4(a^2+b^2+ab),$$ which is $\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$ exactly!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Limit $\lim_{x\to \frac{\pi}{3}} \frac{\sin(x-\frac{\pi}{3})}{1-2 \cos{x}}$ I need to evaluate following limit:
$$\lim_{x\to \frac{\pi}{3}} \frac{\sin(x-\frac{\pi}{3})}{1-2 \cos{x}}$$
Tried multiplying with the argument inside the sinus function but finished with this limit:
$$\lim_{x\to \frac{\pi}{3}} \frac{x - \frac{\pi}{3}}{1-2\cos{x}}$$
I know this limit is $\frac{1}{\sqrt{3}}$, but I have no steps which would show how I obtained this value
PS: l'Hospital is forbidden
| $$ \frac {\sin(x-\frac{\pi}{3})} {1-2 \cos{x}} = \frac {\sin(x-\frac{\pi}{3})} {x-\frac{\pi}{3}} \frac {x-\frac{\pi}{3}} {1-2 \cos{x}}=\frac {\sin y} {y} \frac {{y}} {1-2 \cos{\left(y+\frac{\pi}{3}\right)}}=\frac {\sin y} {y} \frac {{y}} {1-\cos y+\sqrt{3}\sin y}\to \frac{1}{\sqrt{3}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find: $ \lim_{x\to \infty}\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
Find: $\displaystyle \lim_{x\to \infty} f(x)=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
The answer provided in the book is 0 (also checked in Wolfram Alpha), but I can't find a good argument (without L'Hopital), to prove that. I end up in a $\infty \times 0$ situation which makes me uncomfortable.
Attempt: $$f(x)=(x-1)\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{x}(\sqrt{1+1/x}-1)}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{1+1/x}-1}{\sqrt{x+1}}$$
$$f(x)=x(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{x}\sqrt{1+1/x}}=\sqrt{x}(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{1+1/x}}.$$
Therefore the original limit is equivalent to
$$\lim_{x\to \infty} \sqrt{x}(1-1/x)\left(1-\frac{1}{\sqrt{1+1/x}}\right)$$
that appears to me a situation like $\infty\times 0$. How can I proceed to conclude that this limit is indeed $0$.
Hints and solutions are appreciated. Sorry if this is a duplicate.
| $f(x):= \dfrac {x-1}{√x} -\dfrac{x-1}{\sqrt{x+1}}=$
$√x -\dfrac{1}{√x} -\sqrt{x+1} + \dfrac{2}{\sqrt{x+1}}$.
Note:
$√x-\sqrt{x+1}= \dfrac{-1}{√x +\sqrt{x+1}}$ .
Combining:
$|f(x)| \le |\dfrac{-1}{√x+\sqrt{x+1}}| +|\dfrac{1}{√x}|$
$+ |\dfrac{2}{\sqrt{x+1}}|.$
$|f(x)| \lt \dfrac{1}{2√x} +\dfrac{1}{√x} + \dfrac{2}{2√x}.$
$0\le |f(x)|\lt \dfrac{5}{2√x}.$
$0\le \lim_{x \rightarrow \infty} |f(x)| \le$
$\lim_{x \rightarrow \infty} (5/2)\dfrac{1}{√x} = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 7
} |
Find: $\lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
Find: $\displaystyle \lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
After developing the expression, by multiplying the fraction by the conjugates and rearranging, I found:
$$f(x)=x\times \frac{1+\sqrt{1+1/x^6}}{1+\sqrt{1+1/x^4}}$$
I can solve it for the limit when $x\to \infty$, with answer $\infty$ which is correct (by the book and Wolfram Alpha). But when the limit is $x\to -\infty$ the answer is zero, but from this expression, I get $-\infty$ as the answer.
I'm certainly missing something.
Hints and answers appreciated. Sorry if this is a duplicate.
| Note that if $x\lt 0$, we have
$$\sqrt{x^6\left(1+\frac{1}{x^6}\right)}=|x^3|\sqrt{1+\frac{1}{x^6}}=\color{red}{-x^3}\sqrt{1+\frac{1}{x^6}}$$
$$\begin{align}&\lim_{x\to -\infty}\frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}\\\\&=\lim_{x\to -\infty}\frac{(x^2-\sqrt{x^4+1})(x^2+\sqrt{x^4+1})(x^3+\sqrt{x^6+1})}{(x^3-\sqrt{x^6+1})(x^2+\sqrt{x^4+1})(x^3+\sqrt{x^6+1})}\\\\&=\lim_{x\to -\infty}\frac{x^3+\sqrt{x^6+1}}{x^2+\sqrt{x^4+1}}\\\\&=\lim_{x\to -\infty}\frac{x^3+\sqrt{x^6(1+\frac {1}{x^6})}}{x^2+\sqrt{x^4(1+\frac{1}{x^4})}}\\\\&=\lim_{x\to -\infty}\frac{x^3+|x^3|\sqrt{1+\frac{1}{x^6}}}{x^2+|x^2|\sqrt{1+\frac{1}{x^4}}}\\\\&=\lim_{x\to -\infty}\frac{x^3+(\color{red}{-x^3})\sqrt{1+\frac{1}{x^6}}}{x^2+x^2\sqrt{1+\frac{1}{x^4}}}\\\\&=\lim_{x\to -\infty}x\cdot\frac{\left(1-\sqrt{1+\frac{1}{x^6}}\right)\left(1+\sqrt{1+\frac{1}{x^6}}\right)}{\left(1+\sqrt{1+\frac{1}{x^4}}\right)\left(1+\sqrt{1+\frac{1}{x^6}}\right)}\\\\&=\lim_{x\to -\infty}\frac{-1}{x^5\left(1+\sqrt{1+\frac{1}{x^4}}\right)\left(1+\sqrt{1+\frac{1}{x^6}}\right)}=0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
Find determinant Find determinant of A.
$${det}A =\left( \begin{array}{cc}
1 & 1 & 1 & 1 & 1\\
a & b & c & d & x\\
a^2 & b^2 & c^2 & d^2 & x^2\\
a^3 & b^3 & c^3 & d^3 & x^3\\
a^4 & b^4 & c^4 & d^4 & x^4\end{array} \right)$$
I was thinking that I should reduce ones on top, but then I would be stuck with this:
$${det}A =\left( \begin{array}{cc}
b-a & c-a & d-a & x-a\\
b^2-a^2 & c^2-a^2 & d^2-a^2 & x^2-a^2\\
b^3-a^3 & c^3-a^3 & d^3-a^3 & x^3-a^3\\
b^4-a^4 & c^4-a^4 & d^4-a^4 & x^4-a^4\end{array} \right)$$
Any suggestions?
| One of the tricks to solving this is using the fact that $\text{det }A^T = \text{det }A$.
Use this once to get
$\text{det }A = \begin{vmatrix}1&1&1&1&1 \\ a&b&c&d&x \\ a^2&b^2&c^2&d^2&x^2 \\ a^3&b^3&c^3&d^3&x^3 \\ a^4&b^4&c^4&d^4&x^4 \end{vmatrix} = \begin{vmatrix}1&a&a^2&a^3&a^4 \\ 1&b&b^2&b^3&b^4 \\ 1&c&c^2&c^3&c^4 \\ 1&d&d^2&d^3&d^4 \\ 1&x&x^2&x^3&x^4 \end{vmatrix}$
Reduce this to get $\text{det }A = \begin{vmatrix}1&a&a^2&a^3&a^4 \\ 0&b-a&b^2-a^2&b^3-a^3&b^4-a^4 \\ 0&c-a&c^2-a^2&c^3-a^3&c^4-a^4 \\ 0&d-a&d^2-a^2&d^3-a^3&d^4-a^4 \\ 0&x-a&x^2-a^2&x^3-a^3&x^4-a^4 \end{vmatrix}$
Now use the initial property with transposes again to get $\text{det }A = \begin{vmatrix}1&0&0&0&0 \\ a&b-a&c-a&d-a&x-a \\ a^2&b^2-a^2&c^2-a^2&d^2-a^2&x^2-a^2 \\ a^3&b^3-a^3&c^3-a^3&d^3-a^3&x^3-a^3 \\ a^4&b^4-a^4&c^4-a^4&d^4-a^4&x^4-a^4 \end{vmatrix}$
Row reduce yet again to get $\text{det }A = \begin{vmatrix}1&0&0&0&0 \\ 0&b-a&c-a&d-a&x-a \\ 0&b(b-a)&c(c-a)&d(d-a)&x(x-a) \\ 0&b^2(b-a)&c^2(c-a)&d^2(d-a)&x^2(x-a) \\ 0&b^3(b-a)&c^3(c-a)&d^3(d-a)&x^3(x-a) \end{vmatrix}$
And expand to get $\text{det }A = \begin{vmatrix} b-a&c-a&d-a&x-a \\ b(b-a)&c(c-a)&d(d-a)&x(x-a) \\ b^2(b-a)&c^2(c-a)&d^2(d-a)&x^2(x-a) \\ b^3(b-a)&c^3(c-a)&d^3(d-a)&x^3(x-a) \end{vmatrix}$
Now, since we know that $\text{det }A^T = \text{det A}$ and that the row operation of multiplying by a scalar multiplies the determinant by that scalar, we can use "column operations" to reduce this to
$$
\text{det} A = (b-a)(c-a)(d-a)(x-a) \begin{vmatrix} 1&1&1&1 \\ b&c&d&x \\ b^2&c^2&d^2&x^2 \\ b^3&c^3&d^3&x^3 \end{vmatrix}
$$
This matrix looks a lot like the beginning matrix, and you're going to solve it in basically the exact same way so that your answer comes out to
$$
\text{det} A = (b-a)(c-a)(d-a)(x-a)(c-b)(d-b)(x-b)(d-c)(x-c)(x-d)
$$.
On another note, this generalizes to what others were talking about in the comments, the Vandermonde matrix (or the transpose of the Vandermonde matrix in your case, but they have the same determinant). If we define the Vandermonde determinant $\Delta$ to be
$$
\Delta = \begin{vmatrix} 1&1&\cdots&1 \\ x_1&x_2&\cdots&x_n \\ x_1^2&x_2^2&\cdots&x_n^2 \\ \vdots&\vdots&\ &\vdots \\x_1^{n-1}&x_2^{n-1}&\cdots&x_n^{n-1} \end{vmatrix}
$$
Then $$\Delta = \prod_{1 \leq i < j \leq n} {x_j-x_i}$$
You could use this result to quickly do your problem, but you should at least see how one would go about proving this result.
| {
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"source": "stackexchange",
"question_score": "1",
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Find $\lim_{x \to 0} \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}$
Find $\lim_{x \to 0} \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)}$
By taylor polynomials we get: $e^{x^4}=1+x^4+\frac{x^8}{2}+\mathcal{O}(x^{12})$
$\sin(x^2)=x^2-\frac{x^6}{6}+\mathcal{O}(x^{10})$
$\cos(x^3)=1-\frac{x^6}{2}+\mathcal{O}(x^{12})$
so putting these together:
$$ \frac{x^2e^{x^4}-\sin(x^2)}{1-\cos(x^3)} = \frac{x^2+x^6+\frac{x^{10}}{2}-x^2+\frac{x^6}{6}-\mathcal{O}(x^{10})}{\frac{x^6}{2}-\mathcal{O}(x^{12})}=\frac{\frac{7}{6}x^6+\frac{1}{2}x^{10}-\mathcal{O}(x^{10})+\mathcal{O}({x^{12}})}{\frac{1}{2}x^6-\mathcal{O}(x^{12})}$$
Now I am not too familiar with the Big-Oh notation for limits so I am stuck here.
How does arithmetic work with them, can I simplify the oh's in the numerator and can I take $x$'s out?
| Assuming the limr $x\to0$
$$\dfrac{\lim_{x\to0}\dfrac{e^{x^4}-1}{x^4}+\lim_{x\to0}\dfrac{x^2-\sin(x^2)}{x^6}}{\left(\lim_{x\to0}\dfrac{\sin x^3}{x^3}\right)^2}\cdot\lim_{x\to0}(1+\cos x^3)$$
$$=\dfrac{\left(1+\dfrac16\right)(1+\cos0)}{1^2}$$
using Are all limits solvable without L'Hôpital Rule or Series Expansion
| {
"language": "en",
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"source": "stackexchange",
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How to solve (in)equality of three variables with trigonometric solutions I'm working through a set of inequality problems and I'm stuck on the following question:
Find all sets of solutions for which $$(a^2+b^2+c^2)^2=3(a^3b+b^3c+c^3a)$$ holds. Note that $a,b,c\in\mathbb{R}$.
Firstly, one can easily see that when $a=b=c$ then the equality holds: $$\text{LHS}=(a^2+a^2+a^2)^2=(3a^2)^2=9a^4$$ and $$\text{RHS}=3(a^4+a^4+a^4)=3(3a^4)=9a^4.$$
A hint is given to use the substitutions $a=x+2ty$, $b=y+2tz$ and $c=x+2tz$ for real $t$. The LHS is rather nice in that it simplifies to $$(4t(xy+xz+yz)+(1+4t^2)(x^2+y^2+z^2))^2$$ but I can't find a similar simplification for the RHS. I guess this is a type of uvw question but I don't know where to start.
There are actually four sets of solutions: $$a=b=c,$$ $$\frac{a}{\sin^2\frac{4\pi}7}=\frac{b}{\sin^2\frac{2\pi}7}=\frac{c}{\sin^2\frac{\pi}7},$$ $$\frac{b}{\sin^2\frac{4\pi}7}=\frac{c}{\sin^2\frac{2\pi}7}=\frac{a}{\sin^2\frac{\pi}7},$$ and $$\frac{c}{\sin^2\frac{4\pi}7}=\frac{a}{\sin^2\frac{2\pi}7}=\frac{b}{\sin^2\frac{\pi}7}$$ I have no idea how the trigonometric expressions are obtained. How could the equality be solved?
| The hint.
Let $b=a+u$,$c=a+v$ and $u=xv$.
Hence, $$(a^2+b^2+c^2)^2-3(a^3b+b^3c+c^3a)=\sum_{cyc}(a^4+2a^2b^2-3a^3b)=$$
$$=(u^2-uv+v^2)a^2+(u^3-5u^2v+4uv^2+v^3)a+u^4-3u^3v+2u^2v^2+v^4\geq0$$
because
$$(u^3-5u^2v+4uv^2+v^3)^2-4(u^2-uv+v^2)(u^4-3u^3v+2u^2v^2+v^4)=$$
$$=-3(u^3-u^2v-2u^2+v^3)^2=-3v^6(x^3-x^2-2x+1)^2\leq0.$$
The equality occurs for $x^3-x^2-2x+1=0$,
which gives $$x\in\left\{2\cos\frac{\pi}{7},-2\cos\frac{2\pi}{7},2\cos\frac{3\pi}{7}\right\}.$$
For example, $x=2\cos\frac{\pi}{7}$ gives the following case.
$$\frac{b}{\sin^2\frac{4\pi}7}=\frac{c}{\sin^2\frac{2\pi}7}=\frac{a}{\sin^2\frac{\pi}7}.$$
My solution by $uvw$ see here:
https://artofproblemsolving.com/community/c6h6026p5329091
There is also the following nice solution.
https://gbas2010.wordpress.com/2010/01/08/problem-19vasile-cirtoaje/
| {
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If in ,$r=1$,$R=3$ and $s=5$ then find value of $a^2+b^2+c^2$ If in $\triangle ABC$,$r=1$,$R=3$ and $s=5$, then find value of $a^2+b^2+c^2$
(where $r$=inradius,$R$=circumradius and $s$=semi-perimeter)
The answer given is $24$
| $a+b+c=10$.
Let $S$ be an area of the triangle.
Thus, $$\frac{2S}{a+b+c}=1,$$ which gives $$S=5$$ or
$$\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}=5$$ or
$$(a+b-c)(a+c-b)(b+c-a)=40.$$
Now, by AM-GM
$$40=(a+b-c)(a+c-b)(b+c-a)\leq\left(\frac{a+b-c+a+c-b+b+c-a}{3}\right)^3=\frac{1000}{27},$$
which is impossible.
Thus, this triangle does not exist.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $a^4+b^4+c^4$ The problem:
The sum of three numbers is $6$, the sum of their squares is $8$, and the sum of their cubes is $5$. What is the sum of their fourth powers?
Based on the above information, we have:
\begin{align}
a + b + c &= 6 \\
a^2 + b^2 + c^2 & = 8 \\
a^3 + b^3 + c^3 & = 5 \\
\end{align}
I had a feeling that this vaguely had to do with Viete's theorem, which states for a cubic polynomial $f(x) = x^3 - px^2 + qx - r$ which has roots $\alpha , \beta , \gamma$,
\begin{align}
p & = \alpha+\beta+\gamma \\
q & = \alpha \beta+\alpha\gamma+\beta\gamma \\
r & = \alpha\beta\gamma
\end{align}
Notice that we already have $p$, because $a+b+c=6=\alpha + \beta + \gamma = p$. Then to find $q$:
\begin{align}
2q& = 2\alpha\beta+2\alpha\gamma+2\beta\gamma \\
& = [(a+b+c)^2-(a^2+b^2+c^2)] \\
& = (a^2+ab+ac+b^2+ab+bc+c^2+ac+bc)-a^2-b^2-c^2 \\
& = 2ab+2ac+2bc
\end{align}
\begin{align}
q & = ab+ab+bc \\
& = \frac{1}{2}[(a+b+c)^2-(a^2+b^2+c^2)] \\
& = \frac{1}{2}[6^2-8] \\
& = 14
\end{align}
So now we have $f(x) = x^3-6x^2+14x-r$. And it follows that $f(\alpha)=f(\beta)=f(\gamma)=0$.
\begin{align}
f(\alpha) & = \alpha^3-6\alpha^2+14\alpha-r = 0\\
f(\beta) & = \beta^3-6\beta^2+14\beta-r = 0\\
f(\gamma) & = \gamma^3-6\gamma^2+14\gamma-r = 0\\
\end{align}
\begin{align}
0 & = f(\alpha) + f(\beta) + f(\gamma) \\
& = (\alpha^3+\beta^3+\gamma^3)-6(\alpha^2+\beta^2+\gamma^2)+14(\alpha+\beta+\gamma)-3r\\
& = 5-6(8)+14(6)-3r\\
3r& = 41\\
r & = \frac{41}{3} \\
\end{align}
Now we have that $f(x)=x^3-6x^2+14x-\frac{41}{3}$. Here's where I am stuck. Of course, the above working out was the culmination of hours of trying things out, and eventually we have this equation. If you are reading this now, I'd appreciate if you gave any hints as to how I should continue the problem.
I've had the idea of multiplying $f(x)$ by $x$ to get fourth powers, but I haven't tried that yet. Perhaps that might yield some results?
| Euler's method works as follows. We put
$$
\begin{split}
a_1 &= a\\
a_2 &= b\\
a_3 &= c
\end{split}
$$
and
$$S_r = \sum_{j=1}^3 a_j^r$$
Consider the function:
$$f(u) = \sum_{j=1}^3\log\left(1-\frac{a_j}{u}\right)\tag{1}$$
Expanding this around infinity yields the series:
$$f(u) = -\sum_{k=1}^{\infty}\frac{S_k}{k u^k}\tag{2}$$
From (1) it is clear that $\exp\left[f(u)\right]$ is a third degree polynomial in $u^{-1}$. So, if we exponentiate the series (2), then terms of order 4 and higher in $u^{-1}$ will vanish. We can therefore compute $S_4$ by equating the fourth order term equal to zero. It's then convenient to write the exponential in factored form:
$$\exp\left[f(u)\right] = \exp\left(-\frac{S_1}{u}\right) \exp\left(-\frac{S_2}{2 u^2}\right) \exp\left(-\frac{S_3}{3u^3}\right) \exp\left(-\frac{S_4}{4u^4}\right) +\mathcal{O}(u^{-5})\tag{3}$$
Setting the coefficient of $u^{-4}$ equal to zero yields::
$$\frac{S_1^4}{24} - \frac{S_1^2 S_2}{4} + \frac{S_1 S_3}{3} +\frac{S_2^2}{8} - \frac{S_4}{4} = 0 $$
So, we have:
$$S_4 = \frac{S_1^4}{6} - S_1^2 S_2 + \frac{4}{3} S_1 S_3 +\frac{S_2^2}{2}$$
which equals zero for the case at hand. To compute the sum over higher powers, we can derive a recursion relation using that the third degree polynomial $u^3\exp\left[f(u)\right]$ has zeroes at $u = a_j$. This is therefore also the characteristic polynomial for the recursion relation for $S_n$.
Expanding (3) to third order yields:
$$u^3\exp\left[f(u)\right] = u^3 - S_1 u^2 + \left(\frac{S_1^2-S_2}{2}\right)u - \frac{S_1^3}{6} + \frac{S_1 S_2}{2}-\frac{S_3}{3}$$
This implies that:
$$S_{n+3} = S_1 S_{n+2} + \left(\frac{S_2 - S_1^2}{2}\right) S_{n+1} + \left(\frac{S_1^3}{6} - \frac{S_1 S_2}{2}+\frac{S_3}{3}\right) S_n$$
| {
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"question_score": "4",
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Solve $\frac{1}{x-1}-\frac{4}{x-2}+\frac{4}{x-3}-\frac{1}{x-4}<\frac{1}{30} \forall x\in R$ I plugged this into WolframAlpha. It gave the factorized form as:
$$\frac{(x-7)(x-6)(x+1)(x+2)}{(x-4)(x-3)(x-2)(x-1)}>0$$
Once we've reached this point, I know how to find the solution for $x$. But, currently, I am having trouble converting the given question expression into the factorized form above. I started off strong with:
$$(\frac{1}{x-1}-\frac{1}{x-4})+(\frac{4}{x-3}-\frac{4}{x-2})<\frac{1}{30}$$
which simplifies to:
$$\frac{-3}{(x-1)(x-4)}+\frac{12}{(x-3)(x-2)}<\frac{1}{30}$$
which simplifies to:
$$\frac{3x^2-15x+10}{(x-1)(x-2)(x-3)(x-4)}<\frac{1}{90}$$
and now this is a dead end! As you can see, I got to the denomiator, but the numerator is not what I expected it to be. I could transfer $1/30$ to the LHS but that would bring the deadly fourth power into the numerator. If this was a competitive exam question, I would not be sitting in the hall factorizing fourth powers...
Is there a simpler method to factorize the LHS? Or is there an altogether different approach to solve this question?
| Looking at those denominators, there seems to be some symmetry around $x=5/2$. Replacing $u=x-5/2$ we get
$$\frac{1}{u+\frac{3}{2}}-\frac{4}{u+\frac{1}{2}}+\frac{4}{u-\frac{1}{2}}-\frac{1}{u-\frac{3}{2}}$$
Now, $\frac{1}{u+B/2}-\frac{1}{u-B/2}=-\frac{4B}{4u^2-B^2}$ so the above reduces to
$$ -\frac{12}{4u^2-9}+\frac{16}{4u^2-1}$$
Letting know $s=4u^2 $ we have
$$-\frac{12}{s-9}+\frac{16}{s-1} < \frac{1}{30}$$ which produces a quadratic inequation, which you should be able to solve - with the additional restriction $s\ge 0$. From that, you recover $x = 5/2 + u = (5\pm \sqrt{s})/2 $
| {
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"url": "https://math.stackexchange.com/questions/2591614",
"timestamp": "2023-03-29T00:00:00",
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Factoring the polynomial $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$
can you help me factoring the following polynomial?
$6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$
I don't know where to start from.
| Regroup first,
\begin{align}
(6a^4+5a^2b^2-b^4)+(a^3b+ab^3) &= (6a^2-b^2)(a^2+b^2)+ab(a^2+b^2) \\
&= (6a^2+ab-b^2)(a^2+b^2) \\
&= (3a-b)(2a+b)(a^2+b^2)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $T$ and $T^{-1}$ from Matrix to solve differential equation We have the differential equation $y'= \begin{pmatrix}
0 & 1 & 0 \\
-4 & 4 & 0 \\
-2 & 1 & 2
\end{pmatrix} \cdot y$
I found out that the eigenvalues are 2 and the jordan normal form is $\begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}$
For my next step I need to find $T$ such that $\begin{pmatrix}
0 & 1 & 0 \\
-4 & 4 & 0 \\
-2 & 1 & 2
\end{pmatrix}$ = $T \cdot \begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix} \cdot T^{-1}$
How can I find this $T$ (Change of basis)? I got the problem that the eigenvector is$ \begin{pmatrix}
x_1 \\
2x_1 \\
x_3
\end{pmatrix}$
| The characteristic polynomial is
\begin{align}
\mbox{det}\begin{pmatrix}
\lambda & -1 & 0 \\
4 & \lambda-4 & 0 \\
2 & -1 & \lambda-2
\end{pmatrix} &= \lambda(\lambda-4)(\lambda-2)+4(\lambda-2) \\
&= (\lambda-2)(\lambda^2-4\lambda+4) =(\lambda-2)^3
\end{align}
This affords a lot of simplification. The solution of the vector ODE is
$$
y = e^{tA}y_0,\;\;\; \mbox{where } y_0 =y(0).
$$
The solution may be expressed as a truncated power series:
$$
y = e^{t(A-2I)}e^{2t}=e^{2t}\left(I+t(A-2I)+\frac{t^2}{2}(A-2I)^2\right)y_0.
$$
| {
"language": "en",
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Integrate $\int \frac{\sin(3x)}{1-\cos^2(\frac{3x}{2})}dx$ Integrate $$\int \frac{\sin(3x)}{1-\cos^2(\frac{3x}{2})}dx$$
My attempt,
Let $u=1-\cos^2(\frac{3x}{2})$. Then
$$du=-2\cdot\cos\left(\frac{3x}{2}\right)\cdot -\sin\left(\frac{3x}{2}\right)\cdot \frac{3}{2}=3\sin\left(\frac{3x}{2}\right)\cos\left(\frac{3x}{2}\right)\,dx$$
How should I continue? Thanks in advance.
| $$\int \frac{\sin3x}{1-\cos^2(\frac{3x}{2})}\,dx=\int\frac{2\sin\frac{3x}2\cos\frac{3x}2}{\sin^2(\frac{3x}{2})}\,dx=2\int \cot\frac{3x}2\,dx=\frac43\ln\bigg|\sin\frac{3x}2\bigg|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Show that $\sum_{n = 1}^\infty\frac{2^{n-1}-1}{2^{n-1}n^2} = (\ln 2)^2$. Show that $\sum_{n = 1}^\infty\frac{2^{n-1}-1}{2^{n-1}n^2} = (\ln 2)^2$.
I've been trying to find a solution this for a while now, but I can't make any progress. The problem comes from a math competition, so there should be an elementary solution to this. Can someone please help me?
| As the series converges absolutely we can write
$$S = \sum_{n = 1}^\infty \left [\frac{1}{n^2} - \frac{2}{2^n n^2} \right ] = \sum_{n = 1}^\infty \frac{1}{n^2} - 2 \sum_{n = 1}^\infty \frac{(1/2)^n}{n^2}.$$
The first sum to the right corresponds to the Riemann zeta function $\zeta (s)$, namely
$$\zeta (s) = \sum_{n = 1}^\infty \frac{1}{n^s},$$
when $s = 2$ while the second corresponds to the dilogarithm function $\text{Li}_2 (x)$, namely
$$\text{Li}_2 (x) = \sum_{n = 1}^\infty \frac{x^n}{n^2},$$
when $x = 1/2$. Thus
$$S = \zeta(2) - 2\text{Li}_2 \left (\frac{1}{2} \right ).$$
Now using the (reasonably?) well-known result for the value of dilogarithm function when its argument is equal to one half, namely
$$\text{Li}_2 \left (\frac{1}{2} \right ) = \frac{1}{2} \zeta (2) - \frac{1}{2} \ln^2 (2),$$
we immediately see the series sums to $\ln^2 (2)$.
| {
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Finding the area between the two ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ I'm trying to find the area inside the intersection of the region bounded by the ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ using vector calculus methods.
I understand how to calculate area of regions by taking the line integral around the boundary and using Green's Theorem. However what I'm stuck with is finding a paramterization of the boundary! Can anyone point me in the right direction?
| The figure will be symmetric across the line $x = y$ (along with $x = 0$ and $y = 0$)
We can analyze $\frac 18$ the figure, and then use this symmetry to our advantage.
Suppose $a<b$
$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ will be on the "inside" from $0$ to the point of intersection.
Parameterize the area inside the curve:
$x = ar\cos\theta\\
y = br\sin\theta\\
dx\ dy = ab r\ dr\ d\theta$
The last line is the Jacobean for this coordinate system.
And the point of intersection is where $x = y$
$a\cos\theta = b \sin \theta\\
\tan \theta = \frac {a}{b}$
$8\int_0^{\arctan \frac {a}{b}}\int_0^{1} abr\ dr\ d\theta$
$4 ab \arctan\frac {a}{b}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Whats the result of double integral $\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$ Whats the result of double integral
$$\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$$
I was trying to get this $$\int_{0}^{1}\sqrt{1+x^2}\arctan {\frac{1}{\sqrt{1+x^2}}}dx=\int_{0}^{1}(1+x^2)\frac{1}{\sqrt{1+x^2}}\arctan {\frac{1}{\sqrt{1+x^2}}}dx$$
$$=\int_{0}^{1}(1+x^2)\int_{0}^{1}\frac{1}{1+x^2+y^2}dxdy=\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy$$
So far I don't know what to do next.any helps are to be grateful.
| $\begin{align} J&=\int_{0}^{1}\int_{0}^{1}\frac{1+x^2}{1+x^2+y^2}dxdy\\
&=\int_0^1 \left[\sqrt{1+x^2}\arctan\left(\frac{y}{\sqrt{1+x^2}}\right)\right]_{y=0}^{y=1}\,dx\\
&=\int_0^1 \sqrt{1+x^2}\arctan\left(\frac{1}{\sqrt{1+x^2}}\right)\,dx
\end{align}$
Perform the change of variable $\displaystyle x=\tan \theta$,
$\begin{align} J&=\int_{0}^{\frac{\pi}{4}}\frac{1+\tan^2\theta}{\cos\theta}\arctan\left(\cos \theta\right)d\theta\\
&=\int_{0}^{\frac{\pi}{4}}\frac{\arctan\left(\cos \theta\right)}{\cos^3\theta}\,d\theta\\
&=\left[\frac{\tan\theta\arctan\left(\cos \theta\right)}{\cos\theta}\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\tan \theta\left(\frac{\sin \theta\arctan\left(\cos \theta\right)}{\cos^2 \theta}-\frac{\sin x}{\cos x(1+\cos^2 x)}\right)\,d\theta\\
&=\sqrt{2}\arctan\left(\frac{1}{2}\right)-\int_{0}^{\frac{\pi}{4}}\frac{(1-\cos^2x)\arctan\left(\cos \theta\right)}{\cos^3 \theta}\,d\theta+\int_{0}^{\frac{\pi}{4}}\frac{\tan^2\theta}{\cos^2\theta(2+\tan^2 \theta)}\,d\theta\\
&=\sqrt{2}\arctan\left(\frac{1}{2}\right)-J+\int_{0}^{\frac{\pi}{4}}\frac{\arctan\left(\cos \theta\right)}{\cos \theta}\,d\theta+\int_0^1 \frac{x^2}{2+x^2}\,dx\\
&=\sqrt{2}\arctan\left(\frac{1}{2}\right)-J+\int_{0}^{\frac{\pi}{4}}\frac{\arctan\left(\cos \theta\right)}{\cos \theta}\,d\theta+\left[x-\sqrt{2}\arctan\left(\frac{x}{\sqrt{2}}\right)\right]_0^1\\
&=1-J+\int_{0}^{\frac{\pi}{4}}\frac{\arctan\left(\cos \theta\right)}{\cos \theta}\,d\theta\\
\end{align}$
Perform the change of variable $x=\cos \theta$,
$\begin{align}J&=\frac{1}{2}+\frac{1}{2}\int_{\frac{1}{\sqrt{2}}}^1\frac{\arctan x}{x\sqrt{1-x^2}}\,dx\\
&=\frac{1}{2}+\frac{1}{2}\int_{0}^1\frac{\arctan x}{x\sqrt{1-x^2}}\,dx-\frac{1}{2}\int_0^{\frac{1}{\sqrt{2}}}\frac{\arctan x}{x\sqrt{1-x^2}}\,dx\end{align}$
But, it is well known that,
$\displaystyle \int_{0}^1\frac{\arctan x}{x\sqrt{1-x^2}}\,dx=\frac{\pi}{2}\ln(1+\sqrt{2})$
(see Definite integral $\int_0^1 \frac{\arctan x}{x\,\sqrt{1-x^2}}\,\text{d}x$ )
and, thanks to Vladimir Reshetnikov for,
$\displaystyle \int\frac{\arctan x}{x\,\sqrt{1-x^2}}\,dx=\frac i2\left[\operatorname{Li}_2\left(\frac{-1-\sqrt2}{1+\sqrt{1-x^2}}\,ix\right)-\operatorname{Li}_2\left(\frac{1+\sqrt2}{1+\sqrt{1-x^2}}\,ix\right)\\
-\operatorname{Li}_2\left(\frac{1-\sqrt2}{1+\sqrt{1-x^2}}\,ix\right)+\operatorname{Li}_2\left(\frac{-1+\sqrt2}{1+\sqrt{1-x^2}}\,ix\right)\right]\color{gray}{+C}$
(see Definite integral $\int_0^1 \frac{\arctan x}{x\,\sqrt{1-x^2}}\,\text{d}x$)
Therefore,
$\begin{align}J&=\frac{1}{2}+\frac{\pi}{4}\ln(1+\sqrt{2})-\frac{i}{4}\left[\operatorname{Li}_2(-i)-\operatorname{Li}_2(i)-\operatorname{Li}_2\Big(-(3-2\sqrt{2})i\Big)+\operatorname{Li}_2\Big(\left(3-2\sqrt{2}\right)i\Big)\right]\\
&=\frac{1}{2}+\frac{\pi}{4}\ln(1+\sqrt{2})+\frac{1}{2}\Im\left(\operatorname{Li}_2(-i)\right)+\frac{1}{2}\Im\left(\operatorname{Li}_2\Big((3-\sqrt{2})i\right)\Big)\\
&=\boxed{\frac{1}{2}+\frac{\pi}{4}\ln(1+\sqrt{2})-\frac{1}{2}\text{G}+\frac{1}{2}\Im\left(\operatorname{Li}_2\Big((3-\sqrt{2})i\right)\Big)}
\end{align}$
($\text{G}$ is the Catalan constant)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
If 2 matrices are such that $(A+B)^k=A^k+B^k$ for $k=2,3$, show that $(A+B)^m=A^m+B^m $ for all $m \in \mathbb{N}$ Let $A,B \in M_n(C) $. The matrix $A-B$ is invertible and $(A+B)^k=A^k+B^k $, $k \in {2,3} $. Prove that $(A+B)^m=A^m+B^m $ for every $m \in N $.
PS. I obtained $AB+BA=0$ and $A^2B+B^2A=0$, but I need your help, please :(
| Hint For $k = 4$, we have
$$(A + B)^4 = (A + B)^3 (A + B) = (A^3 + B^3)(A + B) = A^4 + \color{red}{A^3 B + B^3 A} + B^4 .$$
On the other hand, using both of the identities you deduced gives
$$B^3 A = B (B^2 A) = B (-A^2 B) = BABA = -B^2 A^2 = A^2 B A = -A^3 B ,$$ so the red terms vanish.
Can you generalize this manipulation to arbitrary $k$?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the intersection curve between a plane $x+y=1$ and a sphere $x^{2}+y^{2}+z^{2}=1$. Given two equations:
$\left\{\begin{matrix}
x^{2}+y^{2}+z^{2}=1\\
x+y=1
\end{matrix}\right.$
Find the set of points in $3$-space represented by this pair.
My work so far:
The equation $x^{2}+y^{2}+z^{2}=1$ represents a sphere with radius $1$ centered at the origin. The equation $x+y=1$ represents a plane in the $xy$-plane that has the points (1, 0, 0) and (0, 1, 0).
I am not sure how to find the equation of the curve (the circle) that is formed from the intersection of these two equations. Do I use substitution?
Let $x = 1 - y$.
$(1-y)^{2}+y^{2}+z^{2}=1$
$1-2y+y^{2}+y^{2}+z^{2}=1$
$2\left (y-\frac{1}{2} \right )^{2}+z^{2}=\frac{1}{2}$
I am close to the answer in the textbook, which is a circle with radius $\frac{1}{\sqrt{2}}$ centered at (1/2, 1/2, 0).
I know that the circle is in the $xy$-plane, so $z = 0$. However, how did they get the $x = \frac{1}{2}$ coordinate when I don't see $x$ in the equation...?
| Points of intersection will satisfy both equations, thus $x^2+y^2+z^2=x+y=1$. Simplifying $x^2+y^2+z^2=x+y$ we get $x^2-x+y^2-y+z^2=0$ or $(x-\frac{1}{2})^2+(y-\frac{1}{2})^2+z^2=\frac{1}{2}$. Don't be alarmed that it looks like another sphere, there is no general equation for a circle in 3d, we can only write it on a particular plane, which is $x+y=1$ in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On commuting matrices Consider the complex matrix $$A=\begin{pmatrix}i+1&2\\2&1\end{pmatrix}$$ and the linear map $$f:M(2,\mathbb{C})\to M(2,\mathbb{C}),\qquad X\mapsto XA-AX.$$
I want to find a basis of $\ker f$.
I already know the canonical basis $\{E_{11},E_{12},E_{21},E_{22}\}$ and computed $$f(E_{11})=\begin{pmatrix}0&2\\-2&0\end{pmatrix},f(E_{12})=\begin{pmatrix}2&0\\0&-2\end{pmatrix},f(E_{21})=\begin{pmatrix}-2&0\\0&2\end{pmatrix},f(E_{22})=\begin{pmatrix}0&-2\\2&0\end{pmatrix}$$
Does this help to find the basis?
| Yes, it helps. The matrix of $f$ with respect to this basis is
$$
\begin{bmatrix}
0 & 2 & -2 & 0 \\
2 & 0 & 0 & -2 \\
-2 & 0 & 0 & 2 \\
0 & -2 & 2 & 0
\end{bmatrix}
\xrightarrow{\text{Gaussian elimination}}
\begin{bmatrix}
1 & 0 & 0 & -1 \\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
$$
The RREF tells us that a basis of the null space of the matrix is
$$
\left\{
\begin{bmatrix}0\\1\\1\\0\end{bmatrix},
\begin{bmatrix}1\\0\\0\\1\end{bmatrix}
\right\}
$$
so that a basis of the kernel of $f$ is
$$
\{E_{12}+E_{21},E_{11}+E_{22}\}=
\left\{
\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
\right\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which points on the curve $5x^2+4xy+2y^2-6=0$ are closest to the origin.
Which points on the curve $5x^2+4xy+2y^2-6=0$ are closest to the origin.
I have solved countless of problems like this but this one is just giving me such a hard time. I'm supposed to solve this with Lagrange's method. So I want to minimize $f(x,y)=x^2+y^2$ due to the constraint $g(x,y)=5x^2+4xy+2y^2-6=0$.
Ok easy: Find $x,y$ so the following equations are satisfied:
$2x+\lambda(10x+4y)=0$
$2y+\lambda(4y+4x)=0$
$5x^2+4xy+2y^2-6=0$
Right? But however i do, i get very complicated equations with root terms to solve, getting me nowhere. I would love to see how you would solve this. Thanks.
| Let $x^2+y^2=k$.
Thus, we need to find all points $(x,y)$ such that $$5x^2+4xy+2y^2=6$$ for which $k$ gets a minimal value, which says the equation
$$x^2+y^2=k\cdot\frac{5x^2+4xy+2y^2}{6}$$ has solutions or
$$(5k-6)x^2+4kxy+(2k-6)y^2=0,$$
which for $k\neq\frac{6}{5}$ gives
$$4k^2-(5k-6)(2k-6)\geq0$$ or
$$k^2-7k+6\leq0,$$ which gives
$$1\leq k\leq6.$$
Since $1<\frac{6}{5}$, we see that $1$ is a minimal value of $k$ because the equality occurs for
$$-x^2+4xy-4y^2=0,$$ which gives $x=2y$ and we got two following points:
$\left(\frac{2}{\sqrt5},\frac{1}{\sqrt5}\right)$ and $\left(-\frac{2}{\sqrt5},-\frac{1}{\sqrt5}\right)$
| {
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"url": "https://math.stackexchange.com/questions/2602265",
"timestamp": "2023-03-29T00:00:00",
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Show that the function $f(x) = x^5 - 10x^3 + 50x - 21$ is an increasing function for all x values. I have worked out that $dy/dx = 5x^4 - 30x^2 + 50$ and this must be greater than $0$ if it is an increasing function.
I simplified this to $x^4 - 6x^2 + 10 > 0$. I know that I must prove this inequality true but I cannot work out how.
I am able to input various values of x and show them to be larger than 0 but am unsure how to do it algebraically.
I know for a quadratic, I can put it into a 'complete the square' form and then you can prove it to be larger than 0 as squaring something = positive and then adding a positive will remain positive. But am unsure how to do it where power of x > 2.
| We can proof it also by definition.
Indeed, let $a>b$.
Thus, by AM-GM we obtain:
$$f(a)-f(b)=a^5-10a^3+50a-21-(b^5-10b^3+50b-21)=$$
$$=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4+50-10(a^2+ab+b^2))\geq$$
$$=(a-b)\left(2\sqrt{50(a^4+a^3b+a^2b^2+ab^3+b^4)}-10(a^2+ab+b^2)\right)=$$
$$=\frac{10(a-b)(2(a^4+a^3b+a^2b^2+ab^3+b^4)-(a^4+2a^3b+3a^2b^2+2ab^3+b^4))}{\sqrt{2(a^4+a^3b+a^2b^2+ab^3+b^4)}+a^2+ab+b^2}=$$
$$=\frac{10(a-b)(a^4-a^2b^2+b^4)}{\sqrt{2(a^4+a^3b+a^2b^2+ab^3+b^4)}+a^2+ab+b^2}>0$$ and we are done!
| {
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Where is the flaw in my proof? Possible proof of the conjecture that there always exists a prime between $n^2$ and $(n+1)^2$?
Assume the statement is invalid. Therefore, for some $n \in \mathbb{N}$, there does not exist a prime that ranges from $n^2$ and $(n+1)^2$ exclusive. Let $\pi (x)$ denote the amount of prime numbers less than or equal to $x$, then we have that if $\pi ((n + 1)^2) - \pi (n^2) = k$, there exists $k$ primes that range between $n^2$ and $(n + 1)^2$ exclusive. Therefore, by assuming the conjecture is false, we consider $k = 0$, which consequently gives us the following: $$\begin{align} \pi ((n + 1)^2) - \pi (n^2) &= 0 \\ \Leftrightarrow \pi ((n+1)^2) &= \pi (n^2). \end{align}$$ It is well known that $\pi (x) \approx \dfrac{x}{\ln x}$ so by order of substitution, $$\frac{n^2}{\ln n^2} \approx \frac{(n + 1)^2}{\ln (n + 1)^2}.$$ For some $m$, one can see that $\ln x^2 = m \Rightarrow e^m = x^2 \Rightarrow \sqrt {e^m} = e^{m/2} = x \Rightarrow \ln x = m/2$. Therefore, we conclude that $\ln x^2 = 2\ln x$. $$\begin{align} \therefore \frac{n^2}{2\ln n} &\approx \frac{(n + 1)^2}{2\ln (n + 1)} \\ \\ \Leftrightarrow \bigg(\frac n2\bigg)\bigg(\frac{n}{\ln n}\bigg) &\approx \bigg(\frac{n + 1}{2}\bigg)\bigg(\frac{n + 1}{\ln (n + 1)}\bigg) \\ \\ \Leftrightarrow \bigg(\frac n2\bigg)\pi (n) &\approx \bigg(\frac{n + 1}{2}\bigg)\pi (n + 1) \\ \\ \Leftrightarrow \frac{n}{n + 1} &\approx \frac{\pi (n + 1)}{\pi (n)}. \end{align}$$
Now, by taking the reciprocal of the equation, it follows then that, $$\frac{n + 1}{n } = 1 + \frac 1n \approx \frac{\pi (n)}{\pi (n + 1)}.$$ However, although true, $$\square \ \dfrac{n + 1}{n} > \dfrac{\pi (n)}{\pi (n + 1)}.$$
By definition of the function $\pi$, we have that $\pi (n) = \pi (n + 1)$ if and only if $n$ and $n + 1$ are not prime and $\pi (n) < \pi (n + 1)$ if and only if $n + 1$ is prime, therefore $\pi (n) \leqslant \pi (n + 1)$. We thus arrive at the following conclusion: $$\begin{align} \frac{\pi (n)}{\pi (n + 1)} &\leqslant 1 \\ \\ \Leftrightarrow \frac{n + 1}{n} &> \frac{\pi (n)}{\pi (n + 1)} \\ \\ \Leftrightarrow \pi ((n + 1)^2) &> \pi (n^2) \\ \\ \Leftrightarrow \pi ((n + 1)^2) - \pi (n^2) &\geqslant 1 \\ \\ \therefore k &= 1. \end{align}$$ Thus, the conjecture is true.$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\bigcirc$
This looks correct to me, but this is a famous conjecture because it is apparently very difficult to prove and if true, it will tell us more about how the prime numbers are distributed across the positive integers. Therefore, I must have some kind of flaw in my proof, because it was too easy (only one page long). Surely this cannot be the case.
Thank you in advance.
| You write $1+\frac{1}{n} \approx \frac{\pi(n)}{\pi(n+1)}$, which is equivalent to
$$
\lim_{n\to \infty}\frac{1+\frac{1}{n}}{\frac{\pi(n)}{\pi(n+1)}}=1,
$$
which is true because both numerator and denominator have limit $1$.
| {
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optimization with strict inequality of variables Maximize $f(x_1,x_2, x_3) = x_{2}+x_{3} - (x_{2}^2+x_{3}^2)$ given $\sum_{i=1}^{3}x_{i} = 1$ and $x_{i}>0$ for $i=1,2,3$. I f I assume that $x_{i}\geq0$ for $i=1,2,3$ then the solution is $x_2 = x_3 = 1/2, x_1 = 0$. How to get the solution when we have strict inequlity $x_{i}>0$ for $i=1,2,3$ ?
| By C-S and AM-GM $$x_2+x_3-(x_2^2+x_3^2)=x_2+x_3-\frac{1}{2}(1^2+1^2)(x_2^2+x_3^2)\leq x_2+x_3-\frac{(x_2+x_3)^2}{2}=$$
$$=\frac{1}{2}\cdot(x_2+x_3)(2-x_2-x_3)\leq\frac{1}{2}\left(\frac{x_2+x_3+2-x_2-x_3}{2}\right)^2=\frac{1}{2}.$$
The equality occurs for $x_2=x_3$ and when $x_2+x_3=2-x_2-x_3,$
which gives $x_2=x_3=\frac{1}{2}$ and $x_1=0.$
If $x_1>0$ then the maximum does not exist, but $\sup{f}=\frac{1}{2}.$
| {
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prove that $\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$
prove that $$\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$$
My attempt:
C is semicircle in upper half complex plane
Simple poles = $e^{i\frac{\pi}{8}}, e^{i\frac{3\pi}{8}},e^{i\frac{5\pi}{8}},e^{i\frac{7\pi}{8}}$ lie in upper semi-circle C and real axis
Given integral value $= 2\pi i \cdot (\text{sum of residues}) = 2 \pi i \left(\frac{-1}{8}\right) \left[e^{i\frac{5\pi}{8}}+e^{i\frac{15\pi}{8}}+e^{i\frac{25\pi}{8}}+e^{i\frac{35\pi}{8}}\right] = 0.27059 \pi$
This is numerically equal to $\frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$. But without using calculator, how to get this expression.
| An alternative approach using complex analysis.
It is not nice to have to calculate many residues. Just reduce it by noting the following:
\begin{align}
I=\int^{\infty}_{-\infty}\frac{x^4}{x^8+1}\,dx=\Re\left(2\int^\infty_{0}\frac{1}{x^4-i}\,dx\right)
\end{align}
Now calculate the RHS by considering:
\begin{align}
\oint_C\frac{1}{z^4-i}\,dz
\end{align}
Where $C$ is a quarter circle with radius $R>1$ in the first quadrant. There is only one pole inclosed instead of four! Call that number $\alpha$. So you only have to find 1 residue (one fourth of the original number!). By the residue theorem we have:
\begin{align}
\oint_C\frac{1}{z^4-i}\,dz=2\pi i \text{Res}_{z=\alpha}\frac{1}{z^4-i}
\end{align}
On the other hand we have:
\begin{align}
\oint_C\frac{1}{z^4-i}\,dz=-i\int^R_0\frac{1}{x^4-i}\,dx+\int^R_0 \frac{1}{x^4-i}\,dx+\int_{B_R}\frac{1}{z^4-i}\,dz
\end{align}
where $B_R$ is the circular part. After letting $R\to\infty$ you can verify that:
\begin{align}
(1-i)\int^{\infty}_0 \frac{1}{x^4-i}\,dx=2\pi i \text{Res}_{z=\alpha}\frac{1}{z^4-i}
\end{align}
Hence:
\begin{align}
I=\Re\left[4\pi \frac{i}{1-i}\text{Res}_{z=\alpha}\frac{1}{z^4-i} \right]
\end{align}
The calculation are left for you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Logic or meaning of $5 \equiv \frac{4}{8}\pmod {12}$ In modular division, what is the meaning that should be ascribed to the notation exemplified below (also given on p. 5 of this)?
$$\begin{align}
\implies & 5\cdot8 \equiv 4\pmod {12} \tag{i} \\[2ex]
\implies & 5 \equiv \frac{4}{8}\pmod {12} \tag{ii} \\[2ex]
\implies & 8 \equiv \frac{4}{5}\pmod {12} \tag{iii}
\end{align}$$
I think in terms of values reached by different residue classes, but I am unable to get any clue. As a very simple example, values taken by $4 \pmod{12}$ residue class are: $4, 16, 28, 40$values taken by $5 \pmod{12}$ residue class are: $5, 17, 29, 41$values taken by $8 \pmod{12}$ residue class are: $8, 20, 32, 44$
This lends no meaning to eqns. $\text{(ii), (iii)}$ above.
| *
*$5 \cdot 5 =25 \equiv 1 \pmod{12}$ so the multiplicative inverse of $5 \bmod 12\,$ is $5^{-1}=5$.
Therefore $5 \cdot 8 \equiv 4 \implies 5^{-1}\cdot5\cdot8 \equiv 5^{-1} \cdot 4 \implies 8 \equiv 5^{-1}\cdot 4 \pmod{12}\,$. The latter may sometimes be written as $8 \equiv \frac{4}{5} \pmod{12}\,$ but that's arguably an abuse of notation, unless such notation was very explicitly and narrowly defined before being used.
*$8$ has no multiplicative inverse $\bmod 12$, so $5 \equiv \frac{4}{8} \pmod{12}$ makes no sense whatsoever.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2607253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute a division with integer and fractional part I have a problem that I don't know how to solve:
Compute $[\frac{\sqrt{7}}{frac(\sqrt{7})}]$
Here's what I've tried: $[\sqrt{7}]=2 \rightarrow frac(\sqrt{7}) = \sqrt{7}-2 \rightarrow [\frac{\sqrt{7}}{frac(\sqrt{7})}] =[\frac{\sqrt{7}}{\sqrt{7}-2}] = [\frac{\sqrt{7}(\sqrt{7}+2)}{7-4}] = [\frac{7+2\sqrt{7}}{3}]$
But from here I don't know what to do anymore.
| $$x\le \frac{7+2\sqrt{7}}{3}<x+1$$
$$3x\le 7+2\sqrt{7}<3x+3$$
$$3x-7\le 2\sqrt{7}<3x-4$$
$$\frac{3x-7}{2}\le \sqrt{7}<\frac{3x-4}{2}$$
$$\frac{9x^2-42x+49}{4}\le 7<\frac{9x^2-24x+16}{4}$$
$$9x^2-42x+49\le 28<9x^2-24x+16$$
where $x\in N$. There are integer solution is $x=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2607814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
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} |
Find the greatest and least values of $(\sin^{-1}x)^2+(\cos^{-1}x)^2$ Find the upper and lower limit of
$$
(\sin^{-1}x)^2+(\cos^{-1}x)^2
$$
My Attempt:
$$
\frac{-\pi}{2}\leq\sin^{-1}x\leq \frac{\pi}{2}\quad\&\quad0\leq\cos^{-1}x\leq\pi\\(\sin^{-1}x)^2\leq\frac{\pi^2}{4}\quad\&\quad(\cos^{-1}x)^2\leq\pi^2\\
0\leq(\sin^{-1}x)^2+(\cos^{-1}x)^2\leq\frac{\pi^2}{4}+\pi^2=\frac{5\pi^2}{4}
$$
Here, I can see the upper limit is $\frac{5\pi^2}{4}$ which is fine. But, $0$ is one lower limit not the lower limit.
Why am I not getting the lower limit in my approach ?
How do I approach similar problems involving max and min, when you don't get the lower or upper limits ?
| You can do it with calculus: if $f(x)=(\arcsin x)^2+(\arccos x)^2$, then
$$
f'(x)=\frac{2\arcsin x}{\sqrt{1-x^2}}-\frac{2\arccos x}{\sqrt{1-x^2}}
$$
(for $-1<x<1$). The derivative vanishes for $\arcsin x=\arccos x$, that is, $x=1/\sqrt{2}$. Since
$$
f(1/\sqrt{2})=\frac{\pi^2}{8}
\qquad
f(-1)=\frac{5\pi^2}{4}
\qquad
f(1)=\frac{\pi^2}{4}
$$
you can easily draw the conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Find largest possible value of $x^2+y^2$ given that $x^2+y^2=2x-2y+2$
Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$.
My attempt:
$x^2+y^2=2x-2y+2$
$(x^2-2x)+(y^2+1)=2$
$(x-1)^2+(y+1)^2=4$
I have no idea how to continue here. Any help?
| Try Lagrange Multipliers. Set
\begin{align}
f(x,y) &= x^2 + y^2 \\
g(x,y) &= x^2 + y^2 -2x + 2y = 2 \\
\nabla f(x,y) &= \lambda\nabla g(x,y)
\end{align}
Then
\begin{align}
\nabla f(x,y) &= (2x,2y) \\
\nabla g(x,y) &= (2x-2,2y+2)
\end{align}
That is
\begin{align}
2x &= \lambda(2x-2) \tag1\label1 \\
2y &= \lambda(2y+2) \tag2\label2 \\
x^2+y^2-2x+2y &= 2 \tag3\label3
\end{align}
Add \eqref{1} and \eqref{2}: $(\lambda-1)(x+y)=0$
*
*$\lambda = 1$: we get $0 = -2$ in \eqref{1}, so this is rejected.
*$x + y = 0$: $g(x,y) = 2x^2-4x = 2 \iff x^2-2x-1=0$. Solve it to get $(x,y) = (1 \pm \sqrt2, -1 \mp \sqrt2)$.
\begin{align}
f(1 - \sqrt2, -1 + \sqrt2) &= 2(\sqrt2-1)^2 \\
f(1 + \sqrt2, -1 - \sqrt2) &= 2(\sqrt2+1)^2
\end{align}
Hence the maximum of $f$ is attained at $(x,y) = (1 + \sqrt2, -1 - \sqrt2)$ and the maximum of $f$ is $2(\sqrt2+1)^2 = 6 + 4\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Specific steps in applying the Chinese Remainder Theorem to solve modular problem splitting modulus I am trying to get an idea of how the Chinese Remainder Theorem (CRT) can be used to finish up this problem, in which the problem
$$7^{30}\equiv x\pmod{ 100}$$
is attempted by splitting the modulus into relatively prime factors $25$ and $4,$ arriving at
$$\begin{align}
7^{30}&\equiv1\pmod4\\
7^{30}&\equiv-1\pmod{25}
\end{align}$$
I understand that the CRT may be called upon because $m=\prod m_i,$ and we have the same $7^{30}$ value on the LHS, but I don't know how to carry it out.
The question was touched upon in this post as the second entry:
How do I efficiently compute $a^b \pmod c$ when $b$ is less than $c.$ For instance $5^{69}\,\bmod 101.$
However, I don't see this particular point clearly worked out, perhaps because it is a multi-pronged question.
Following this presentation online, this seems to be the verbatim application of the CRT without any added concepts or shortcuts:
From @gimusi's answer (upvoted):
$$\begin{cases}
x \equiv 7^{30} \pmod4\\
x\equiv 7^{30} \pmod{25}
\end{cases}$$
rearranged into
\begin{cases}
x \equiv 1 \pmod4\\
x\equiv -1 \pmod{25}
\end{cases}
Given the general form of the equations above as $x\equiv a_i \pmod {m_i},$ the CRT states $x\equiv a_1 b_1 \frac{M}{m_1}+a_2 b_2 \frac{M}{m_2}\pmod M$ with $M=\prod m_i,$ and with
$$b_i =\left(\frac{M}{m_i}\right)^{-1}\pmod {m_i}.$$
The inverse of $\frac{M}{m_i}$ is such that $\frac{M}{m_i}\left(\frac{M}{m_i}\right)^{-1}\pmod {m_i}\equiv 1.$
Calculating the components:
$$\begin{align}
a_1&=1\\
a_2&=-1\\
M&=4\times 25 =100\\
\frac{M}{m_1} &= \frac{100}{4}=25\\
\frac{M}{m_2} &= \frac{100}{25}=4\\
b_1 &= \left(\frac{M}{m_1}\right)^{-1} \pmod 4 = (25)^{-1}\pmod 4 =1\\
b_2 &= \left(\frac{M}{m_2}\right)^{-1} \pmod {25}= (4)^{-1} \pmod{25}=19
\end{align}$$
Hence,
$$x=1\cdot 25 \cdot 1 + (-1)\cdot 4 \cdot 19 = -51 \pmod{100}\equiv 49.$$
| Welcome to Math SX! You have to use Euler's theorem as $\varphi(4)=2$, $\;\varphi(25)=20$ we have
$$ 7^{30}\equiv7^{30\bmod2}=1\mod 4,\qquad 7^{30}\equiv7^{30\bmod20}=7^{10}\mod 25$$
To find the latter power, you can use the modular fast exponentiation algorithm, but here, it will be simpler: modulo $25$,
$$7^2\equiv -1\enspace\text{so}\enspace 7^4=1,\enspace\text{hence } \;7^{30}\equiv 7^{30\bmod 4}=7^2\equiv -1.$$
Finally, since $\;25-6\cdot 4=1$ (Bézout's identity),
$$7\equiv \begin{cases}\phantom{-}1\mod4\\-1\mod 25\end{cases}\iff 7\equiv 1\cdot 25-(-1)\cdot 6\cdot 4=49\mod 100.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Solving $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$? (1983 AIME problem 3)
What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?
I know it is a messy/bad idea, but I first started off by squaring both sides and moving everything to one side to get $$x^4 + 36x^3 + 384x^2 + 1080x + 900 - 4x^2 - 72x - 180 = x^4 + 36x^3 + 380x^2 + 1008x + 720 .$$
And by (generalisation) of Vieta's formula, the product of the real roots should be $\frac{720}{1} = 720$, but that is wrong, and I don't understand why.
| Write $t=x^2+18x +30$, then we get $$t^2=4t+60$$ so $(t-10)(t+6)=0$ and thus
*
*case $x^2+18x+36=0$ so $x_1=x_2=-6$ which is impossible (left side is negative)
*case $x^2+18x+20=0$ so the product of real roots is $20.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Let the sequence $x_n = (1-\frac{1}{2})(1-\frac{1}{4})\dots(1-\frac{1}{2^n})$ Let the sequence $x_n$ be given by
$$x_n = (1-\frac{1}{2})(1-\frac{1}{4})\dots(1-\frac{1}{2^n})$$
Prove that the sequence $x_n$ converges and that the limit is not $0$.
Attempt:
The convergence part is easy, it is evident that $x_{n+1} < x_n$ and $0<x_n<\frac{1}{2}$. A strictly decreasing sequencing sequence bounded below is convergent hence $x_n$ is convergent. But I am not sure how to prove $x_n $ is not convergent to $0$.
| Hint For $x,y \geq 0$ you have
$$(1-x)(1-y) \geq 1-x-y$$
Then
$$(1-\frac{1}{4})(1-\frac{1}{8}) \geq 1-\frac{1}{4}-\frac{1}{8} \\
(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16}) \geq (1-\frac{1}{4}-\frac{1}{8})((1-\frac{1}{16}) \geq 1-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}\\
............. \mbox{ ( i.e. by induction)}\\\
(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16}).. (1-\frac{1}{2^n}) \geq 1-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}- ...- \frac{1}{2^n}\\
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the convergence of $\sum\limits_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$ Hints preferred:
I need to prove convergence and determine the limit of following series: $$ \sum_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$$
I need to use partial fraction decomposition to simplify the term, which I attempted with following:
$$
\frac 1 {(2k+1)(2k+5)} = \frac a {(2k+1)} + \frac b {(2k+5)} \\~\\
\frac 1 {(2k+1)(2k+5)} = \frac {a(2k+5)} {(2k+1)(2k+5)} + \frac {b(2k+1)} {(2k+1)(2k+5)} \\~\\$$
Then we have:
$$
k: 2a + 2b = 0 \\
k^0: 5a + b = 1 \\~\\
a = \frac 1 4 ~ \land b = -\frac 1 4
$$
So we have:
$$
\frac 1 {(2k+1)(2k+5)} = \frac { \frac 1 4 } {(2k+1)} - \frac { \frac 1 4} {(2k+5)} = \\ \frac { 1 } {4(2k+1)} - \frac { 1 } {4(2k+5)}
$$
Which can be defined as:
$$
\frac 1 4 (\frac { 1 } {2k+1} - \frac { 1 } {2k+5} )
$$
I'm not sure how to turn the result into a telescopic series (if the partial fraction decomposition is right in the first place).
| To prove the convergence of the series we can use the limit comparison test. Indeed
$$
\frac{1}{(2k+1)(2k+5)}\sim\frac{1}{4k^2}.
$$
Since $\sum_{1}^\infty k^{-2}$ converges it follows that
$$
\sum_{k=2}^\infty\frac1{(2k+1)(2k+5)}<\infty.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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How to find the Taylor series of order $2$ at $x=1$ of the function $f(x)=\frac{x-1}{\ln (x)}.$ How to find the Taylor series of order $2$ at $x=1$ of the function $$f(x)=\frac{x-1}{\ln (x)}.$$
I don't want to use differentiation because I want to be prepared to find the Taylor series for higher orders.
My Attempt: Write $$f(x)=\frac{1-x}{1-(1+\ln(x))}.$$ Now I consider the auxiliary function $$\frac{1}{1-u(x)}=1+(u(x)-1)+(u(x)-1)^2+o((u(x)-1)^2)$$ and since $1+\ln(x)\to 1$ as $x\to 1$ we have that $$f(x)=(1-x)(1+(x-x^2/2)+(x-x^2/2)^2+o(x^2))$$ $$=1-x^2/2+o(x^2).$$ However this answer is not correct. Where am I going wrong?
| $$\begin{align}\ln(1+x)&=x-x^2/2+x^3/3-\cdots\\ \ln(x)&=(x-1)-(x-1)^2/2+(x-1)^3/3\\ \frac1{\ln x}&=\frac{1}{x}\left(\frac1{1-(x-1)/2+(x-1)^2/3-\cdots}\right)\\&=\frac1{x-1}\cdot \left(1+\frac{x-1}2-\frac{(x-1)^2}3+\frac{(x-1)^2}4+O((x-1)^3)\right)\end{align}$$
So $$\frac{x-1}{\ln(x)}=1+\frac{x-1}2-\frac{(x-1)^2}{12}+O((x-1)^3)$$
Here we used the fact that $$\frac1{1-t}=1+t+t^2+\cdots$$
which is valid for $|t|<1$, which is true close to $x=1$ in the example we use ($x-1$ is small in this region).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many nondecreasing functions $f: \{1, 2, 3, \ldots, 15\} \to \{1, 2, 3, 4, 5\}$ are there? Consider the equation $x_1 + x_2 + x_3 + x_4 + x_5 = 15$. How many nonnegative integer solutions does it have? How about positive integer solutions? How many nondecreasing functions $f : \{1, 2, 3, \ldots, 15\} \to \{1, 2, 3, 4, 5\}$ are there?
Nonnegative and positive integer solutions I just used combinatorics (bars and stars or pirates and gold) which is not too bad. But for the last question of nondecreasing function i suppose i can rewrite it as $x_1 < x_2 < ... < x_5$. Where do i go from there?
| A non-decreasing function from $\{1, 2, 3, \ldots, 15\} \to \{1, 2, 3, 4, 5\}$ is completely determined by how many numbers in the domain are assigned to each element in the codomain. For instance, $(15, 0, 0, 0, 0)$ corresponds to the constant function $f(k) = 1$ for $1 \leq k \leq 15$, while $(3, 4, 5, 2, 1)$ corresponds to the function
\begin{align*}
f(1) & = f(2) = f(3) = 1\\
f(4) & = f(5) = f(6) = f(7) = 2\\
f(8) & = f(9) = f(10) = f(11) = f(12) = 3\\
f(13) & = f(14) = 4\\
f(15) & = 5
\end{align*}
Let $x_k$ be the number of elements in the domain that are assigned to $k$ in the codomain. Then the number of decreasing functions is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 15 \tag{1}$$
in the nonnegative integers. A particular solution of equation corresponds to the placement of four addition signs in a row of $15$ ones. For instance,
$$1 1 1 1 + 1 1 1 1 1 1 + + 1 1 1 + 1 1$$
corresponds to the solution $x_1 = 4$, $x_2 = 6$, $x_3 = 0$, $x_4 = 3$, $x_5 = 2$. There are
$$\binom{15 + 4}{4} = \binom{19}{4}$$
such solutions since we must choose which $4$ of the $19$ positions required for $15$ ones and $4$ addition signs will be filled with addition signs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{1}{6} - 1) (x^\frac{1}{6} + 1) } $$
$$\lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{2}{6} - 1) } $$
| By the L'Hospital's rule we obtain: $$\lim_{x\rightarrow1}\frac{x^{\frac{1}{5}}-1}{x^{\frac{1}{6}}-1}=\lim_{x\rightarrow1}\frac{\frac{1}{5}x^{-\frac{4}{5}}}{\frac{1}{6}x^{-\frac{5}{6}}}=\frac{\frac{1}{5}}{\frac{1}{6}}=\frac{6}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I solve this fractional indices equation $\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}$? Solve the equation $\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}$
I thought that the best way of approaching this would be to rewrite everything using $3$ as the base of the exponents, hence creating an equivalence which would allow me to equate numerators to numerators, and denominators to denominators. Doing this yields:
$$\frac{3^{5x+2}}{3^{2(1-x)}}=\frac{3^{3(4+3x)}}{3^6}$$
Equating the exponents of each numerator:
$$ 5x+2=3(4+3x) $$
$$ 5x+2=12+9x $$
$$ -4x=10 $$
$$ x = \frac{10}{-4}=-\frac{5}{2} $$
Doing this for the denominator yields a different value of $x$:
$$ 2(1-x)=6 $$
$$ 2-2x=6 $$
$$ -2x=4 $$
$$ x = -2 $$
Why is that I'm obtaining two different values of $x$?
Further to this, the solution in the book states the answer as $x=-3$, what am I doing wrong?
| \begin{align}
\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}
\implies &
\frac{3^{5x+2}}{(3^2)^{(1-x)}}=\frac{(3^3)^{(4+3x)}}{3^6}
\\
\implies &
\frac{3^{5x+2}}{3^{2\cdot (1-x)}}=\frac{3^{3\cdot(4+3x)}}{3^6}
\\
\implies &
3^{5x+2-2\cdot (1-x)}=3^{3\cdot(4+3x)-6}
\\
\implies &
5x+2-2\cdot (1-x)=3\cdot(4+3x)-6
\\
\implies &
5x+2-2+2x=12+9x-6
\\
&
7x=6+9x
\\
&
-2x=6
\\
&
x=-3
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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The sum of two positive numbers is 1. The sum of their cubes is a maximum. What are the numbers? I set this up and end up finding the minimum (the two numbers would both be $1/2$). To find a maximum value, I could reflect the functions and use $y^3-x^3$ but I still end up finding $1/2$ as the two numbers. It does make sense that two two values to produce a max would be 0 and 1 but I can't figure out how to set up the problem from the start. What I have is...
$x+y=1 \\
x^3 + y^3 = max$
Subsitution...
$x^3 - (1-x)^3 = max \\
x^3 - 1 + 3x - 3x^2 + x^3 = max \\
2x^3 - 3x^2 + 3x - 1 = max \\
6x^2 - 6x + 3 = 0 \\
x = 1/2 \text{ which makes }y = 1/2 \\
$
That is where my issue is. How do I set it up to find the sum of the cubes to be a max?
| If $x\ge 0$ and $y\ge 0$ and $x+y=1$, then $y=1-x$ and $0 \le x \le 1$
So $x^3+y^3=x^3 + (1-x)^3 = 3x^2-3x+1$. This is a parabola with axis of symmetry, and minimum, at $x = \frac 12$. It follow that the max values will be at $x=0$ and $x=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with $x^2+y^2+z^2=3$ If x,y,z are positive real numbers and $x^2+y^2+z^2=3$ prove that
$$\sum_\text{cyc}\frac{1}{x^3+y^2+z}\le\frac{1}{xyz}.$$
I have tried to prove it this way:
$$\sum_\text{cyc}\frac{1}{x^3+y^2+z}\le\frac{1}{xyz}$$
$$\Leftrightarrow \sum_\text{cyc}\frac{xyz}{x^3+y^2+z}\le1$$
$$\Leftrightarrow \sum_\text{cyc}\frac{1}{\frac{x^2}{yz}+\frac{y}{xz}+\frac{1}{xy}}\le1$$
But we know that $$\frac{x^2}{yz}+\frac{y}{xz}+\frac{1}{xy}\ge3\sqrt[3]{\frac{x^2}{yz}\cdot\frac{y}{xz}\cdot\frac{1}{xy}}=\frac{3}{\sqrt[3]{z^2y}}$$
$$\Rightarrow \sum_\text{cyc}\frac{1}{\frac{x^2}{yz}+\frac{y}{xz}+\frac{1}{xy}}\le\sum_{cyc}\frac{\sqrt[3]{z^2y}}{3}$$
We now have to prove that $\sum_\limits\text{cyc}\frac{\sqrt[3]{z^2y}}{3}\le1$.
$$\frac{\sqrt[3]{z^2y}}{3}+\frac{\sqrt[3]{x^2z}}{3}+\frac{\sqrt[3]{y^2x}}{3}=\frac{\sqrt[3]{1\cdot z^2\cdot y}}{3}+\frac{\sqrt[3]{1\cdot x^2\cdot z}}{3}+\frac{\sqrt[3]{1\cdot y^2\cdot x}}{3}$$
$$\le\frac{\frac{1+ z^2+ y}{3}+\frac{1+ x^2+ z}{3}+\frac{1+ y^2+ x}{3}}{3}=\frac{3+3+x+y+z}{9}$$
And the last part: $\;\dfrac{3+3+x+y+z}{9}\le1 \Leftrightarrow x+y+z\le3$ or
$$\frac{x+y+z}{3}\le \sqrt{\frac{x^2+y^2+z^2}{3}}$$ which is true.
I hope that it's correct. Any suggestions are welcome.
| Alternative solution, by AM-GM
$$\frac{x^3+y^2+z}{3}\ge\sqrt[3]{x^3y^2z}\implies \frac{1}{x^3+y^2+z}\le\frac{1}{3\sqrt[3]{x^3y^2z}}=\frac{\sqrt[3]{z^2y}}{3xyz}$$
thus
$$\sum_\text{cyc}\frac{1}{x^3+y^2+z}\le \sum_\text{cyc} \frac{1}{3\sqrt[3]{x^3y^2z}}\le\sum_\text{cyc}\frac{\sqrt[3]{z^2y}}{3xyz} \le 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Roots to an equation, $\frac{3-2x}{x-1}-\sqrt[4]{x^3}$
Find the number of roots to $\frac{3-2x}{x-1}-\sqrt[4]{x^3}=0$.
My method:
$\frac{3-2x}{x-1}-\sqrt[4]{x}$, we multiply $\sqrt[4]{x}$ and $x-1$ to both sides, getting $3\sqrt[4]{x}-2x\sqrt[4]{x}-x(x-1)=0$. We let $\sqrt[4]{x}=k$. Then we have $3k-2xk-x^2+x=0$. Finding the discriminant ($b^2-4ac$), getting $4k^2+8k+1$, evidently greater than $0$, which means there are 2 roots.
Plotting this on Desmos gives 1 root only. How am I wrong?
| Your analysis is wrong becase $k$ depends on $x$: $k=\sqrt[4]x$. Instead, write $x=k^4$:
$$3k-2k^5-k^8+k^4=0$$
$$k^8+2k^5-k^4-3k=0$$
Descartes's rule of signs tells us there is exactly one positive root for this last polynomial. $k$ cannot be negative as it is the fourth root of $x$, so the lone root of the polynomial in $k$ translates to only one root for the original equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $\int \frac{ \sqrt{(a^2 - x^2)}}{x^2}$? I tried the following trig substitution:
$x = a\sin \theta$
$$ \int \frac{ \sqrt{(a^2 - x^2)}}{x^2} = \int \frac{\sqrt{a^2 - a^2 \sin^2 \theta}}{a^2 \sin^2 \theta} = \int \frac{a \cos \theta}{a^2 \sin^2}$$
Setting $u = \sin x$ yields:
$$\frac{1}{a} * \int u^{-2} = - \frac{1}{a \sin \theta}$$
However my textbook states that the answer should really be:
$$ -\frac{\sqrt{a^2 - x^2}+x\tan^{-1}(\frac{x}{\sqrt{a^2-x^2}})}{x}$$
Where did I mess up?
| Say no to trig substitution for easy integrals.
$$\int \dfrac{\sqrt{a^2 - x^2}}{x^2} dx = \dfrac{-1}{x} \sqrt{ a^2-x^2} - \int\dfrac{1}{x}\dfrac{x}{\sqrt{a^2 - x^2}}\ dx \\= \dfrac{-1}{x} \sqrt{x^2 - a^2} - \sin^{-1} \left(\dfrac xa\right) +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Checking the differentiability of the two variables function Let:
$f(x,y)=\frac{x^3+y^3}{x^2+y^2}$
It is required to test the differentiability of the given two variables function at $(0,0)$. The value of function at $(0,0)$ is $0$.
My approach
I conclude that function is not differentiable since:
$f_{x}(0,0)=f_{y}(0,0)=1$ and
$f(x,y)=x+y+x(\frac{-y^2}{x^2+y^2})+y(\frac{-x^2}{x^2+y^2})$. Here,
$A(x,y)=\frac{-y^2}{x^2+y^2}$ and $B(x,y)=\frac{-x^2}{x^2+y^2}$. These two functions of $x$ and $y$ are not tending towards zero as $(x,y)\rightarrow(0,0)$. Hence, by definition the function is not differentiable.
I always make mistake in the differentiability test of two variables. I am also not sure whether my conclusion is correct or wrong. So, please help me identify any mistakes.
Thanks.
| $\require{cancel}$
$$1.\,\,\,\,\,\partial_xf(0,0)=\lim_\limits{h\to0}\frac{f(0+h,0)-f(0,0)}{h}=\lim_\limits{h\to0}\frac{h^3/h^2}{h}=1.$$
$$2.\,\,\,\,\,\partial_yf(0,0)=\lim_\limits{h\to0}\frac{f(0,0+h)-f(0,0)}{h}=\lim_\limits{h\to0}\frac{h^3/h^2}{h}=1.$$
$$3.\,\,\,\,\,\lim_\limits{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-\partial_xf(0,0)x-\partial_yf(0,0)y}{\sqrt{x^2+y^2}}=\lim_\limits{(x,y)\to(0,0)}\frac{\frac{x^3+y^3}{x^2+y^2}-x-y}{\sqrt{x^2+y^2}}=\lim_\limits{(x,y)\to(0,0)}\frac{\cancel{x^3}+\cancel{y^3}-\cancel{x^3}-x^2 y-x y^2-\cancel{y^3}}{\sqrt{(x^2+y^2)^3}}=\lim_\limits{(x,y)\to(0,0)}\frac{-xy\,(x+y)}{\sqrt{(x^2+y^2)^3}}=\lim_\limits{r\to0}\frac{-\cancel{r^3}\cos(\theta)\sin(\theta)\,(\cos(\theta)+\sin(\theta))}{\cancel{r^3}\sqrt{\cancel{(\cos^2(\theta)+\sin^2(\theta))^3}_1}}=\cos(\theta)\sin(\theta)\,(\cos(\theta)+\sin(\theta))=\text{Indeterminate.}$$
Since the last limit is not $0$ then the function is not differentiable at $(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Exponentiation of a symmetric $2 \times 2$ matrix Why does the following hold?
$$\left[\begin{matrix}a & b\\b & a\end{matrix}\right]^k=\frac{1}{2}\left[\begin{matrix}\left(a - b\right)^{k} + \left(a + b\right)^{k} & - \left(a - b\right)^{k} + \left(a + b\right)^{k}\\- \left(a - b\right)^{k} + \left(a + b\right)^{k} & \left(a - b\right)^{k} + \left(a + b\right)^{k}\end{matrix}\right]$$
| Guide:
Given a symmetric $2 \times 2$ matrix $A$, one way to commpute $A^k$ is as follows:
$$\begin{bmatrix} a & b \\ b & a \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = (a+b) \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
$$\begin{bmatrix} a & b \\ b & a \end{bmatrix}\begin{bmatrix} 1 \\ -1 \end{bmatrix} = (a-b) \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$
Diagonalize the matrix in the form of $A=UDU^T$ where $U$ is orthogonal and $D$ is diagonal, then $A^k=UD^kU^T$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Evaluating $\int x \cdot \sqrt{x-1} \cdot \mathrm{d}x$ using substitution I know that my result is wrong. What did I do wrong?
Let $u = \sqrt{x-1}$. Then we find first:
$$ u^2 = x - 1 \Rightarrow x = u^2 + 1$$
$$ \Rightarrow \mathrm{d}x = 2u \cdot \mathrm{d}u $$
So:
$$\int x \cdot \sqrt{x-1} \cdot \mathrm{d}x = \int (u^2 + 1) \cdot u \cdot 2u \cdot \mathrm{d}u = 2 \cdot \int (u^4 + u^2) \mathrm{d}u$$
$$ = 2 \cdot (\frac{1}{5} u^5 + \frac{1}{2} u^2) + c = \frac{2}{5} u^5 + u^2 + c$$
$$ = \frac{2}{5} (\sqrt{x-1})^5 + x + c$$
Differentiating this does not give me $ x \cdot \sqrt{x-1}$, so it must be wrong. Which step went wrong and why?
| You made just a small error: $\int u^2 du=\frac{u^3}{3}+C$. Therefore following your approach,
$$\begin{align}\int x \cdot \sqrt{x-1} dx&=2\int (u^4 + u^2) \cdot \mathrm{d}u=\frac{2u^5}{5}+\frac{2u^3}{3}+C\\&=\frac{2(x-1)^{5/2}}{5}+\frac{2(x-1)^{3/2}}{3}+C\\
&=\frac{2(3x^2-x-2)\sqrt{x-1}}{15}+C.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of terms involving binomial coefficients and floor functions I would like to know how to prove the following identity.
Let m be a positive integer. Then
$$\sum_{i=0}^{m+1} (-1)^i \Bigl\lfloor \frac{i}{2} \Bigr\rfloor {2m+1 \choose m+i}=2^{2m-2}$$
Here $\Bigl\lfloor \frac{i}{2} \Bigr\rfloor$ denotes the floor function.
A computational proof is welcome. On the other hand, is it possible to give a combinatorial proof? For example, we interpret the left-hand side as coefficients of a variable in a polynomial.
| We seek to evaluate
$$\sum_{q=0}^{m+1} (-1)^q
\Big{\lfloor} \frac{q}{2} \Big{\rfloor} {2m+1\choose m+q}
= \sum_{q=0}^{m+1} (-1)^q
\Big{\lfloor} \frac{q}{2} \Big{\rfloor}
{2m+1\choose m+1-q}.$$
Observe that
$$\Big{\lfloor} \frac{q}{2} \Big{\rfloor}
= [z^q] \left(\sum_{p\ge 0} p z^{2p}
+ \sum_{p\ge 0} p z^{2p+1}\right)
= [z^q] (1+z) \frac{z^2}{(1-z^2)^2}
\\ = [z^q] \frac{z^2}{(1-z)^2 (1+z)}.$$
We thus have for the sum
$$\sum_{q=0}^{m+1} (-1)^q [z^q] \frac{z^2}{(1-z)^2 (1+z)}
[w^{m+1-q}] (1+w)^{2m+1}
\\ = [w^{m+1}] (1+w)^{2m+1}
\sum_{q=0}^{m+1} w^q (-1)^q [z^q] \frac{z^2}{(1-z)^2 (1+z)}.$$
Now when $q\gt m+1$ the sum term does not contribute to the coefficient
extractor $[w^{m+1}]$ and hence we may extend $q$ to infinity,
getting
$$[w^{m+1}] (1+w)^{2m+1}
\sum_{q\ge 0} w^q (-1)^q [z^q] \frac{z^2}{(1-z)^2 (1+z)}.$$
Here we may apply the substitution rule or alternatively, an
annihilated coefficient extractor (ACE) to obtain
$$[w^{m+1}] (1+w)^{2m+1}
\frac{w^2}{(1+w)^2 (1-w)}
= [w^{m-1}]\frac{1}{1-w} (1+w)^{2m-1}.$$
This is
$$\sum_{q=0}^{m-1} {2m-1\choose q}
= \frac{1}{2} 2^{2m-1} = 2^{2m-2}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the following by two different methods, one combinatorial and one algebraic Reading through my textbook I came across the following problem, and I am looking for some help solving it. I am asked to prove the following by two different methods, one combinatorial and one algebraic. If I could get help with either or both it would be great, thanks!
Prove that this identity is true,
$$\binom{n}{k} -\binom{n-3}{k} =\binom{n-1}{k-1} + \binom{n-2}{k-1} + \binom{n-3}{k-1}$$
| The following algebraic approach is some sort of overkill and for curiosity only. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[x^k](1+x)^n
\end{align*}
We start with the left-hand side and obtain
\begin{align*}
\color{blue}{\binom{n}{k}-\binom{n-3}{k}}&=[x^k]\left((1+x)^n-(1+x)^{n-3}\right)\\
&=[x^k]\left((1+x)^3-1\right)(1+x)^{n-3}\\
&\color{blue}{=[x^k](x^3+3x^2+3x)(1+x)^{n-3}}\tag{1}
\end{align*}
The right-hand side admits the representation
\begin{align*}
\color{blue}{\binom{n-1}{k-1} }&\color{blue}{+ \binom{n-2}{k-1} + \binom{n-3}{k-1}}\\
&=[x^{k-1}]\left((1+x)^{n-1}+(1+x)^{n-2}+(1+x)^{n-3}\right)\\
&=[x^{k-1}]\left((1+x)^2+(1+x)+1\right)(1+x)^{n-3}\\
&=[x^{k-1}]\left(x^2+3x+3\right)(1+x)^{n-3}\\
&\,\,\color{blue}{=[x^k]\left(x^3+3x^2+3x^2\right)(1+x)^{n-3}}\tag{2}\\
\end{align*}
Since (1) and (2) coincide the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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