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If $x=2+i$, $gcd(a,b,c)=1$, and $ax^4+bx^3+cx^2+bx+a=0$, then what is $|c|$?
Suppose
$$a(2+i)^4 + b(2+i)^3 + c(2+i)^2 + b(2+i) + a = 0,$$
where $a,b,c$ are integers whose greatest common divisor is $1$.
Determine $|c|$.
So I first simplified the exponents and combined like terms.
I received $$a(-6+24i)+b(4+12i)+c(3+4i)=0.$$
I don't really know how to progress. What is the answer $|c|$?
| A polynomial of this symmetric form can be reduced in degree by the following trick. Divide by $x^2$ and get:
$\begin{align}
a (x^2 + x^{-2}) + b (x + x^{-1}) + c
&= a (x + 1/x)^2 + b (x + 1/x) + c - 2 a
\end{align}$
The substitution $y = x + 1/x$ finishes this off.
This works because the powers of $x + 1/x$ turn out to be combinations of $x^k + x^{-k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proof by induction that $5|11^n-6$ for all positive integers $n$
Prove by induction that $5|11^n-6$ for all positive integers $n$
Let $p(n) = 11^n-6.$ We have $p(1) = 5$, thus it holds for $p(1)$. Assume it holds for $p(k)$. We will prove that it's true for $p(k+1)$. We have $p(k+1) = 11^{k+1}-6$. But $6 = 11^k-f(k).$ Thus
$$f(k+1) = 11\cdot 11^k-11^k+f(k) = 10\cdot 11^k+f(k). $$
$5|(10\cdot 11^k+f(k))$ by hypothesis thus the statement is true for $p(k+1)$. Hence it's true for $p(n)$.
Is the above correct?
| Let $x=11^n-6$ then $$11^{n+1}-6=11(x+6)-6=11x+60$$ and this later is divisible by $5$ if $x$ is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Has this equation appeared before? I want to know if the following equation has appeared in mathematical literature before, or if it has any important significance.
$$\sqrt{\frac{a+b+x}{c}}+\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{c+a+x}{b}}=\sqrt{\frac{a+b+c}{x}},$$
where $a,b,c$ are any three fixed positive real and $x$ is the unknown variable.
| This provides the explicit polynomial in $x$ (for those curious), though I'm not aware if the equation has appeared in the mathematical literature. We get rid of the square roots by multiplying out the $8$ sign changes,
$$\prod^8 \left(\sqrt{\frac{a+b+x}{c}}\pm\sqrt{\frac{b+c+x}{a}}\pm\sqrt{\frac{a+c+x}{b}}\pm\sqrt{\frac{a+b+c}{x}}\right)=0$$
then collecting powers of $x$. It turns out the $8$th-deg equation factors into a linear (cubed), a quadratic, and a cubic. For simplicity, let,
$$\begin{aligned}
p &= a+b+c\\
q &= ab+ac+bc\\
r &= abc
\end{aligned}$$
Then,
$$(p+x)^3=0\tag1$$
$$r^2 - 2 q r x + (q^2 - 4 p r) x^2 = 0\tag2$$
$$p r^2 + r (-2 p q + 9 r) x + (p q^2 - 4 p^2 r + 6 q r) x^2 + (q^2 - 4 p r) x^3 = 0\tag3$$
Example:
Let $a,b,c = 1,2,4$, then
$$(7+x)^3=0\\
-16 + 56 x + 7 x^2 = 0\\
-112 + 248 x - 119 x^2 + 7 x^3 = 0$$
The roots of the quadratic solve,
$$\sqrt{\frac{a+b+x}{c}}\pm\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{a+c+x}{b}}-\sqrt{\frac{a+b+c}{x}}=0$$
while a root of the cubic solves,
$$\sqrt{\frac{a+b+x}{c}}-\sqrt{\frac{b+c+x}{a}}-\sqrt{\frac{a+c+x}{b}}+\sqrt{\frac{a+b+c}{x}}=0$$
and two others, while the linear root takes care of the remaining three sign changes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$ Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$
My try
My book gives as a hint to move everything to the left hand side of the inequality and then factor and see what I get in the long factorization process and to lookout for squares.
So that's what I have tried:
\begin{array}
((a^7+b^7)(a^2+b^2) &\ge (a^5+b^5)(a^4+b^4) \\\\
(a^7+b^7)(a^2+b^2)-(a^5+b^5)(a^4+b^4) &\ge 0 \\\\
(a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)(a^2+b^2)-(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)(a^4+b^4) &\ge 0 \\\\
(a+b)\left[(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)((a+b)^2-2ab)-(a^4+b^4)(a^4-a^3b+a^2b^2-ab^3+b^4)\right] &\ge 0
\end{array}
Now it's not clear what I have to do next.I am stuck.
Note: My book doesn't teach any advanced technique for solving inequality as AM-GM ,Cauchy inequality etc..
| We may suppose that $a\geq b>0$. Hence dividing by $b^9$ and putting $x=a/b$, we have to show that for $x\geq 1$, we have $(x^7+1)(x^2+1)\geq (x^5+1)(x^4+1)$ or
$$\frac{x^7+1}{x^5+1}\geq \frac{x^4+1}{x^2+1}$$
For $x\geq 1$ fixed, put $\displaystyle f(u)=\frac{x^2u+1}{u+1}$. It is easy to see that $f$ is increasing on $[0,+\infty[$; as $x^5\geq x^2$, we get $f(x^5)\geq f(x^2)$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$
It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?
| I think I have got close to proving this but I'm not sure if this is a valid proof - but here goes...
Using AM-GM we can show that:
$$a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\tag{1}$$
$$ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{2}$$
We can also show that:
$$a^2+b^2+c^2\ge ab+bc+ca\tag{3}$$
We can therefore infer that:
$$a^2+b^2+c^2\ge ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\tag{4}$$
Using these facts we can say that:
$$a^2+b^2+c^2=3\sqrt[3]{a^2b^2c^2}+\delta_1\text{ where }\delta_1\ge0\tag{5}$$
$$ab+bc+ca=3\sqrt[3]{a^2b^2c^2}+\delta_2\text{ where }\delta_2\ge0\tag{6}$$
We can then use (4) to further infer that:
$$\delta_1\ge\delta_2\ge0$$
Finaly we can state that:
$$\begin{align}
a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}&=6\sqrt[3]{a^2b^2c^2}+\delta_1\\
&\ge6\sqrt[3]{a^2b^2c^2}+\delta_2\\
&\ge2\left(3\sqrt[3]{a^2b^2c^2}+\frac{\delta_2}{2}\right)\\
&\ge2\left(ab+bc+ca-\frac{\delta_2}{2}\right)\\
\end{align}$$
I am hoping this aproach triggers a thought in someones brain who can then come up with the final proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$
Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then I would get $\geq$ instead of $>$.
| Converting your expression so that all that is cubed is positive:
$$(z-x)^3 - (z-y)^3 - (y-x)^3$$
$$(A+B)^3-A^3-B^3$$
with $A,B$ positive. Since cubing is concave up with $0^3=0$, this expression must be positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Umbilics on the ellipsoid Show that, if p, q and r are distinct positive numbers, there are
exactly four umbilics on the ellipsoid $$\frac{x^2}{p^2}+\frac{y^2}{q^2}+\frac{z^2}{r^2}=1$$
What happens if $p$, $q$ and $r$ are not distinct?
$$$$
To show that I considered the parametrization $\sigma (u,v)=(p\cos u\sin v, q\sin u\sin v, r\cos v)$.
Then the principal curvatures are the roots of
$\begin{vmatrix}
L-\kappa E & M-\kappa F\\
M-\kappa F & N-\kappa G
\end{vmatrix}=0$.
I found the following:
$$\kappa^2 \sin^2 v \\ \left [((p^4+q^4)\sin^2 u\cos^2 u+p^2q^2(\cos^4 u+\sin^4 u))\cos^2 v+r^2\sin^2{v}(p^2\sin^2 u+q^2\cos^2 u)-(q^2-p^2)\sin^2u \cos^2u \cos^2v\right ]-\kappa \frac{pqr\sin^2 v}{\sqrt{\sin^2 v(q^2r^2\cos^2 u+p^2r^2\sin^2 u)+p^2q^2\cos^2 v}} \left [(p^2\cos^2 u+q^2\sin^2 u)(\cos^2 v+1)+r^2\sin^2 v)\right ]+\frac{p^2q^2r^2\sin^2 v}{\sin^2 v(q^2r^2\cos^2 u+p^2r^2\sin^2 u)+p^2q^2\cos^2 v}=0 $$
Can this be correct? Can we simplify it?
$$$$
$$$$
| Consider a parametrization $f$ :
$$ x=a\cos\ t \cos\ s,\ y=b\cos\ t\sin\ s,\ z=c\sin\ t $$
Define $$ A:=\cos\ t,\ B=\sin\ t,\ C=\cos\ s,\ D=\sin\ s $$
Then $$
f_t=(-aBC,-b BD, c A ) ,$$ $$ f_s=(-aAD, b AC, 0 )
$$
$$ f_{tt} =(-aAC,-b AD,-cB) $$
$$ f_{st} = (aB D, -bBC,0) $$
$$ f_{ss} = (-aAC,-bA D,0 ) $$
Hence we have first fundamental forms : $$ E=a^2 B^2 C^2 +
b^2 B^2 D^2 + c^2 A^2
$$
$$ F=(a^2-b^2) ABCD $$
$$ G=A^2 ( a^2 D^2 + b^2C^2) $$
so that $$ EG-F^2 = A^2\{ a^2b^2 B^2D^2 + a^2c^2 A^2D^2 + b^2 c^2
A^2 C^2 \}
$$
Then unit normal is
$$ N = \frac{f_t\times f_s}{|f_t\times f_s|} =
(-bcA^2 C, ac A^2 D, -ab AB ) l $$ where $$ l= \frac{1}{ |A| \sqrt{
(bc AC)^2 + (ac AD)^2 + (abB)^2 } } $$ And second
fundamental forms are
$$ e= (N,f_{tt})= abc\ A\{ A^2 C^2 - A^2 D^2 +B^2 \} l $$
$$ f=(N, f_{st})= -2abc\ A^2BCD l $$
$$ g= (N, f_{ss})= abc \ A^3(C^2-D^2) l $$
Then
$$ \frac{ eg-f^2 }{(abcA^2l)^2} = A^2 ( C^2-D^2)^2 +
4 B^2C^2D^2 + B^2 (C^2-D^2) :=m $$
And $$ \frac{1}{2} \frac{gE -fF+ eG }{abc\ A^3l} = \frac{1}{2} \{ (
a^2 D^2 + b^2C^2) \{ A^2 C^2 - A^2 D^2 +B^2 \} $$ $$+2(a^2-b^2)
(BCD)^2 +( a^2 B^2 C^2 + b^2 B^2 D^2 + c^2 A^2)
(C^2-D^2) \} $$
$$ =\frac{1}{2}\bigg[ a^2 \{ C^2 D^2 - A^2D^4+B^2C^4 + B^2D^2 \} $$ $$+ b^2 \{ -C^2D^2 +
A^2C^4 + B^2C^2-B^2 D^4 \} + c^2A^2 (C^2-D^2) \bigg] :=n$$
By $H^2=K$, we have $$ \bigg( \frac{n \ abc\ A^3l}{EG-F^2 } \bigg)^2
= \frac{m\ (abc\ A^2l)^2}{EG-F^2} \Rightarrow n^2A^2 =m(EG-F^2) $$
$$\frac{1}{A^2} m(EG-F^2) -n^2$$
$$= ( a^2b^2 B^2D^2 + a^2c^2 A^2D^2 + b^2 c^2 A^2 C^2 )( A^2
(C^2-D^2)^2 + 4 B^2C^2D^2 + B^2(C^2-D^2) )$$ $$ - \frac{1}{4} ( a^2
( C^2 D^2 -A^2D^4+B^2C^4 + B^2D^2 ) $$ $$+ b^2 ( -C^2D^2 +A^2C^4 +
B^2C^2-B^2 D^4 )+c^2A^2 (C^2-D^2) )^2
$$
$a=b,\ C=0,\ D=1$ Case : We will consider $z\geq
0$ Then $(0,0,c)$ is umbilic. Assume that $0< A\leq 1$ Then
$$\frac{1}{A^2} m(EG-F^2) -n^2 = -\frac{1}{4} \{
a^2(-3A^2+2) + A^2c^2 \}^2$$
Hence $$ c= \frac{a\sqrt{3A^2-2}}{A} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$
Let $a,b,$ and $c$ be positive real numbers with $ab+bc+ca = 1$. Prove that $$\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{3}{2}$$
Attempt
The $ab+bc+ca = 1$ condition reminds of the rearrangement inequality. Thus, I would say that $a^2+b^2+c^2 \geq ab+bc+ca = 1$ then rewrite the given inequality as $4(a+b+c)^2 = 4(a^2+b^2+c^2) + 8(ab+bc+ca) = 4(a^2+b^2+c^2) \leq 9(a^2+1)(b^2+1)(c^2+1)$
I don't know what to do next.
| By AM-GM $\sum\limits_{cyc}\frac{a}{\sqrt{a^2+1}}=\sum\limits_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq\frac{1}{2}\sum\limits_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Tangent line Exercise. $f (x) = 5e^{−(x−2)^2}$ . Find the coordinates of points where the hill is the most steep. I have this exercise, look easy but I don't know where to start, I think that I need a extra function or value to continue with the calculus.
Imagine that you are riding over a hill having its profile given by
the function $f(x) = 5e^{−(x−2)^2}$ Find the coordinates of points where
the hill is the most steep.
Can you help me?. The answer is $x = 2 \pm\frac{1}{\sqrt{2}}$
| Notice, the slope of the tangent to the curve: $f(x)=5e^{-(x-2)^2}$ is given as
$$f'(x)=\frac{d}{dx}(f(x))=\frac{d}{dx}(5e^{-(x-2)^2})$$
$$=5e^{-(x-2)^2}(-2(x-2))$$$$=-10(x-2)e^{-(x-2)^2}$$
$$f''(x)=-10(x-2)\frac{d}{dx}(e^{-(x-2)^2})-10e^{-(x-2)^2}\frac{d}{dx}(x-2)$$
$$=20(x-2)^2e^{-(x-2)^2}-10e^{-(x-2)^2}$$
$$f'''(x)=-40(x-2)^3e^{-(x-2)^2}+60(x-2)e^{-(x-2)^2}$$
Now, the hill will be most steep at the points where the slope $f'(x)$ is maximum i.e. $f''(x)=0$ hence, $$20(x-2)^2e^{-(x-2)^2}-10e^{-(x-2)^2}=0$$
$$10e^{-(x-2)^2}(2(x-2)^2-1)=0$$
but $e^{-(x-2)^2}\ne 0$, hence
$$2(x-2)^2-1=0$$
$$(x-2)^2=\frac 12$$
$$x-2=\pm \frac{1}{\sqrt 2}$$
$$x=2\pm \frac{1}{\sqrt 2}$$
Now, substituting these values of $x$ in $f'''(x)$, one should get
$$f'''\left(2+\frac{1}{\sqrt 2}\right)>0, \ \ f'''\left(2-\frac{1}{\sqrt 2}\right)<0$$
hence, the slope $f'(x)$ is maximum at the point where $x=2-\frac{1}{\sqrt 2}$,hence corresponding y-coordinate of the point $$f\left(2-\frac{1}{\sqrt 2}\right)=5e^{-\left(2-2+\frac{1}{\sqrt 2}\right)^2}=\frac{5}{\sqrt e}$$
hence, the coordinates of the point where the hill is most steep i.e. slope is maximum is $\color{red}{\left(2-\frac{1}{\sqrt 2}, \frac{5}{\sqrt e}\right)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Least squares problem: am I solving it correctly? So I have this question:
In $\mathbb R^3$ with inner product calculate all the least square solutions, and choose the one with shorter length, of the system:
$ x + y + z = 1 $
$ x + z = 0 $
$ y = 0 $
My attempt: So I applied the formula $ A^T A x = A^T b $ with A as being the matrix with row 1 (1,1,1) row 2 (1,0,1) and row 3 (0,1,0); x being the column $(x_1,x_2,x_3)$ and b being the column (1,0,0).
So I did it and I reached to the solution $(x_1, \frac {1}{3}, \frac {1}{3} + x_1)$
And I expanded this solution in two vectors $(0, \frac {1}{3}, \frac {1}{3}) $ and $(1,0,1)$.
So these are the least square solutions and the one with shorter length is the first one.
My doubt is if I'm doing this correctly or if I made any mistake because I used an online calculator that only give one least square solution. Can someone help me to verify my attempt? Thanks!
| The reduced row echelon form is
$$
\begin{align}
\mathbf{A} &\mapsto \mathbf{E}_{\mathbf{A}} \\
%
\left[
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 0 \\
\end{array}
\right]
%
&\mapsto
%
\left[
\begin{array}{ccc}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
%
\end{align}
$$
There is a rank defect of $1$. Row 1 = row 2 + row 3.
The linear system is
$$
\begin{align}
\mathbf{A} x &= b \\
%
\left[
\begin{array}{ccc}
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 0 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
x \\
y \\
z
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
1 \\
0 \\
0
\end{array}
\right]
%
\end{align}
$$
The least squares solution is resolved to $\color{blue}{range}$ and $\color{red}{null}$ space components
$$
\begin{align}
x_{LS} &=
\color{blue}{\mathbf{A}^{\dagger}b} +
\color{red}{\left( \mathbf{I}_{3} - \mathbf{A}^{\dagger} \mathbf{A} \right) \xi} \\
%
&=
%
\color{blue}{\frac{1}{6}
\left[
\begin{array}{crr}
1 & 2 & -1 \\
2 & -2 & 4 \\
1 & 2 & -1 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
1 \\
0 \\
0
\end{array}
\right]}
%
+
%
\color{red}{
\frac{1}{2}
\left[
\begin{array}{rcr}
1 & 0 & -1 \\
0 & 0 & 0 \\
-1 & 0 & 1 \\
\end{array}
\right] \xi}, \quad \xi \in \mathbf{C}^{n} \\[3pt]
%
&=
\color{blue}{
\frac{1}{6}
\left[
\begin{array}{r}
1 \\
2 \\
1
\end{array}
\right]}
+
\color{red}{
\alpha
\left[
\begin{array}{r}
-1 \\
0 \\
1
\end{array}
\right]}, \quad \alpha \in \mathbb{C}
%
\end{align}
$$
where $\xi$ is an arbitrary vector in the domain $\mathbb{C}^{3}$.
The residual error vector is
$$
\color{red}{r} = \mathbf{A} \color{blue}{x_{LS}} - b = \color{red}{\frac{1}{3}
\left[
\begin{array}{r}
-1 \\
1 \\
0
\end{array}
\right]},
$$
and the total error is $r^{2} = \frac{2}{9}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\int (\tan x)^{1/ 6} \,\text{d}x$? How do I compute the following integral
$$
I=\int (\tan x)^{1/ 6} \,\text{d}x
$$
| Given you said you "won't even try" to solve this integral, try Mathematica:
$$\frac{-2 \left(\sqrt{3}-1\right) \tan ^{-1}\left(\frac{-2 \sqrt{2} \sqrt[6]{\tan
(x)}+\sqrt{3}-1}{1+\sqrt{3}}\right)+4 \tan ^{-1}\left(1-\sqrt{2} \sqrt[6]{\tan
(x)}\right)-4 \tan ^{-1}\left(\sqrt{2} \sqrt[6]{\tan (x)}+1\right)+2
\left(\sqrt{3}-1\right) \tan ^{-1}\left(\frac{2 \sqrt{2} \sqrt[6]{\tan
(x)}+\sqrt{3}-1}{1+\sqrt{3}}\right)-2 \left(1+\sqrt{3}\right) \tan ^{-1}\left(-2
\sqrt{2+\sqrt{3}} \sqrt[6]{\tan (x)}+\sqrt{3}+2\right)+2 \left(1+\sqrt{3}\right) \tan
^{-1}\left(\left(\sqrt{2}+\sqrt{6}\right) \sqrt[6]{\tan (x)}+\sqrt{3}+2\right)-2 \log
\left(\sqrt[3]{\tan (x)}-\sqrt{2} \sqrt[6]{\tan (x)}+1\right)+2 \log
\left(\sqrt[3]{\tan (x)}+\sqrt{2} \sqrt[6]{\tan (x)}+1\right)-\left(1+\sqrt{3}\right)
\log \left(2 \sqrt[3]{\tan (x)}+\sqrt{2} \left(\sqrt{3}-1\right) \sqrt[6]{\tan
(x)}+2\right)+\left(\sqrt{3}-1\right) \log \left(2 \sqrt[3]{\tan (x)}-2
\sqrt{2+\sqrt{3}} \sqrt[6]{\tan (x)}+2\right)+\left(1+\sqrt{3}\right) \log \left(2
\sqrt[3]{\tan (x)}+\left(\sqrt{2}-\sqrt{6}\right) \sqrt[6]{\tan
(x)}+2\right)-\left(\sqrt{3}-1\right) \log \left(2 \sqrt[3]{\tan
(x)}+\left(\sqrt{2}+\sqrt{6}\right) \sqrt[6]{\tan (x)}+2\right)}{4 \sqrt{2}}$$
...just in case you need the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric contour integral I cannot figure out what I'm doing wrong:
$$\int_0^{2\pi} \frac{1}{a+b\sin\theta} d\theta\quad a>b>0$$
$$\int_{|z|=1} \frac{1}{a+\frac{b}{2i}(z-z^{-1})} \frac{dz}{iz}$$
$$\int_{|z|=1} \frac{2i}{2ia+b(z-z^{-1})} \frac{dz}{iz}$$
$$\int_{|z|=1} \frac{2}{2iza+bz^2-b} dz$$
$$2iz_0a+bz_0^2-b=0$$
$$z_0=\frac{-2ia\pm\sqrt{-4^2+4b^2}}{2b}$$
$$z_0=\frac{a\pm\sqrt{a^2 - b^2}}{bi}$$
where only $z_0=\frac{a-\sqrt{a^2 - b^2}}{bi}$ is within $C$.
So $Res(z_0) = \frac{-2b}{-2\sqrt{a^2-b^2}}$
So the integral is $\frac{2\pi bi}{\sqrt{a^2-b^2}}$
But this is wrong. The $bi$ should not be there. However I cannot see what I'm missing here? I also checked $Res(0)$ for order 1 and it was equal to $0$.
| The integral $I=\int_0^{2\pi}\frac{1}{a+b\sin \theta}\,d\theta$ is simply $2\pi i$ times $2/b$ times the residue of $z^2+i 2(a/b) z-1$ at $z=i\left(-\frac ab +\sqrt{(a/b)^2-1}\right)$. Thus, we have
$$I=(2\pi i)\frac2b \frac{1}{2i\sqrt{(a/b)^2-1}}=\frac{2\pi}{\sqrt{a^2-b^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$ Let $a,b,c,x,y,z$ be positive real numbers such that $x+y+z=1$. Prove that
$$ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$$.
my try:
$2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{\frac{2(a+b+c)}{3}}$
But this is not the right choice because
$ax+by+cz\le{\frac{a+b+c}{3}}$ is not always true.
| By Cauchy-Schwarz,
$(LHS)\le \sqrt{(a^2+b^2+c^2)(x^2+y^2+z^2)}+\sqrt{2(ab+bc+ca)2(xy+yz+zx)}=f(a,b,c,x,y,z)$
However, $$f(a,b,c,x,y,z)^2\le (a^2+b^2+c^2+2ab+2bc+2ca)(x^2+y^2+z^2+2xy+2yz+2zx)=(RHS)^2(\because Cauchy)$$
Therefore, $(LHS)^2\le (RHS)^2$. Our proof is done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding Laurent's series of a function I am trying express the function $$f(z)=\frac{z^3+2}{(z-1)(z-2)}$$ like a Laurent's series in each ring centering in $0$, but I do not now how could I express it, in first I said that $$f(z)=(z^3+2)\left[\frac{1}{z-2}-\frac{1}{z-1} \right] $$
Ok, now, I see two posibilities: $A\equiv open\ ring=A(C;r,R)$ where $$C\equiv center\\ r<R$$
so, I see:
$a)$ $A(0;1;2)$
$b)$ $D(0,1)\cup\{\mathbb{C}-\overline{D}(0,2)$
It is correct? Ok, then I suposse that $a)$ I think that I have to try express like series of potence with $\frac{1}{z}$, and in the second case, $b)$, in $\overline{D}(0,1)$ like potences of $z$ and in $\mathbb{C}-\overline{D}(0,2)$ like potences of $\frac{1}{z}$ but I can not, I need someone help, please.
All I know is that $\sum_{0}^{\infty} z^n=\frac{1}{1-z}$ when $|z|<1$
I do not know if it is so, I need someone clarify my doubts
| You have to find possible Laurent series expansion for each case !
Case I : $0<|z|<1$
Case II : $1<|z|<2$
Case III : $2<|z|$
Case I : $0<|z|<1$
$$f(z)=(z^3+2)\left[\frac{1}{z-2}-\frac{1}{z-1} \right] $$
$$\Rightarrow f(z)=(z^3+2)\left[\frac{1}{1-z} -\frac{1}{2}\cdot \frac{1}{1-\frac{z}{2}}\right] $$
For all $|z|<1$ , $$\displaystyle \frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$
And $\displaystyle |z|<1 \Rightarrow \frac{|z|}{2}<1$.
So $$\frac{1}{1-\frac{z}{2}}=\sum_{n=0}^{\infty}\frac{z^n}{2^n}$$
Therefore for each $|z|<1$ , $$f(z)=(z^3+2)\left[ \sum_{n=0}^{\infty}z^n-\frac{1}{2}\cdot\sum_{n=0}^{\infty}\frac{z^n}{2^n} \right]$$
$$f(z)=\left[ \sum_{n=0}^{\infty}z^{n+3}-\frac{1}{2}\cdot\sum_{n=0}^{\infty}\frac{z^{n+3}}{2^n} +2\sum_{n=0}^{\infty}z^{n}-\cdot\sum_{n=0}^{\infty}\frac{z^{n}}{2^n}\right]$$
Case II : $1<|z|<2$
Now consider $1<|z|<2$ ,
Then $$f(z)=(z^3+2)\left[\frac{1}{1-z} -\frac{1}{2}\cdot \frac{1}{1-\frac{z}{2}}\right] $$
$$\Rightarrow f(z)=(z^3+2)\left[-\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}} -\frac{1}{2}\cdot \frac{1}{1-\frac{z}{2}}\right] $$
Observe that $1<|z| \Rightarrow \frac{1}{|z|}<1$
So $$\frac{1}{1-\frac{1}{z}}=\sum_{n=0}^{\infty}\frac{1}{z^n}$$
And clearly for $|z|<2$ , $$\frac{1}{1-\frac{z}{2}}=\sum_{n=0}^{\infty}\frac{z^n}{2^n}$$
Thus for each $1<|z|<2$
$$f(z)=(z^3+2)\left[-\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}} -\frac{1}{2}\cdot \frac{1}{1-\frac{z}{2}}\right] $$
$$f(z)=(z^3+2)\left[-\frac{1}{z}\cdot\sum_{n=0}^{\infty}\frac{1}{z^n} -\frac{1}{2}\cdot \sum_{n=0}^{\infty}\frac{z^n}{2^n}\right] $$
Case III : $2<|z|$
Now consider $2<|z|$.
Clearly $2<|z|\Rightarrow \frac{2}{|z|}<1 \Rightarrow \frac{1}{|z|}<1$
Now change $f(z)$ .
$$f(z)=(z^3+2)\left[\frac{1}{z-2}-\frac{1}{z-1} \right] $$
$$f(z)=(z^3+2)\left[\frac{1}{z}\cdot \frac{1}{1-\frac{2}{z}}-\frac{1}{z}\cdot \frac{1}{1-\frac{1}{z}} \right] $$
Clearly for each $2<|z|$ ,
$$\frac{1}{1-\frac{2}{z}}=\sum_{n=0}^{\infty}\frac{2^n}{z^n}$$
So
$$f(z)=(z^3+2)\left[\frac{1}{z}\cdot \sum_{n=0}^{\infty}\frac{2^n}{z^n}-\frac{1}{z}\cdot \sum_{n=0}^{\infty}\frac{1}{z^n} \right] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to
$(A)\frac{1}{2}\hspace{1cm}(B)8\hspace{1cm}(C)2\hspace{1cm}(D)3$
$(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$
$(x,y)$ satisfies $(x-2)^2+(y+1)^2=2^2$
The expression $x^2+y^2-10x-6y+34$ can be written as $(x-5)^2+(y-3)^2$
But i do not know how to further solve it.Please help.
| Since you have simplified the given equation as $(x-2)^2+(y+1)^2=2^2$, then taking the parameter $\theta$, put $x=2+2\cos\theta$ and $y=-1+2\sin\theta$.
As $\theta$ varies, you can clearly see that the point $(x,y)=(2+2\cos\theta,-1+2\sin\theta)$ is on the circle $(x-2)^2+(y+1)^2=2^2.$
Now use this parametrization in $x^2+y^2-10x-6y+34$.
So we start as,
$(2+2\cos\theta)^2+(-1+2\sin\theta)^2-10(2+2\cos\theta)-6(-1+2\sin\theta)+34$
$=29-(12\cos\theta+16\sin\theta)$
Now use the result $-\sqrt {a^2+b^2} \le a\cos\theta+b\sin\theta \le \sqrt{a^2+b^2}$.
Hence $-\sqrt{12^2+16^2} \le 12\cos\theta+16\sin\theta \le \sqrt{12^2+16^2}$
$\Rightarrow -20 \le 12\cos\theta+16\sin\theta \le 20$
$\Rightarrow 29-20 \le x^2+y^2-10-6y+34 \le 29+20$
$\Rightarrow 9 \le x^2+y^2-10x-6y+34 \le 49.$
Therefore all the given options are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4)$. The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4).$ Find 3 more solutions in positige integers. [hint. look for solutions of the form $(a,b,c) = (xz, yz, z^2)$
Attempt:
So I tried to use the hint in relation to the triple that they gave that worked. So I observed that the triple that worked $(2,2,4)$ could also be written as $$2(1,1,2)$$ So i attempted to generalize it to $$n(1,1,2)$$ I thought this may work because I am using the "origin" of where the triple $(2,2,4)$ came from. This did not work unfortunately. So I am asking what else could I comsider to devise some system to find these triples? Because I highly doubt they're asking me to just shoot in the dark and pick random numbers
| Here is an
infinite number of solutions:
For any positive integer $c$,
$x = (1+c^{3})$,
$y
=c(1+c^{3})$,
and
$z = (1+c^{3})^2
$.
Check:
$x^3+y^3
=(1+c^3)^3+(c(1+c^3))^3
=(1+c^3)^3(1+c^3)
=(1+c^3)^4
$
and
$z^2
=((1+c^{3})^2)^2
=(1+c^{3})^4
$.
This is gotten from the following,
which is a re-creation
of some algebra of mine
from many years ago:
To find solutions to
$x^n+y^n = z^m$
where
$(n, m) = 1$.
Since $(n, m) = 1$,
there are $a$ and $b$
such that
$am-bn = 1$
or
$am=bn + 1$.
If
$x = u^b$,
$y = v^b$,
$z = w^a$,
then
$u^{bn}+v^{bn}
=w^{am}
=w^{bn+1}
$.
Let
$v = cu$.
Then
$u^{bn}+v^{bn}
=u^{bn}+(cu)^{bn}
=u^{bn}+c^{bn}u^{bn}
=u^{bn}(1+c^{bn})
$.
If
$w = 1+c^{bn}$,
we want
$w^{bn+1}
=w\,w^{bn}
=wu^{bn}
$
or
$u = w$.
Therefore,
a solution is
$x = (1+c^{bn})^b$,
$y = cu
=c(1+c^{bn})^b$,
and
$z = (1+c^{bn})^a
$.
If
$n=3$ and $m=2$,
we want
$2a-3b = 1$.
This is satisfied by
$a=2, b=1$.
We get
$x = (1+c^{3})$,
$y
=c(1+c^{3})$,
and
$z = (1+c^{3})^2
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \ \ \ d.)\ 7 $
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta \\
=\sin^{2} \theta +\dfrac{1}{\sin^{2} \theta }+\cos^{2} \theta+\dfrac{1}{\cos^{2} \theta }+\tan^{2} \theta+\dfrac{1}{\tan^{2} \theta } \\
\color{blue}{\text{By using the AM-GM inequlity}} \\
\color{blue}{x+\dfrac{1}{x} \geq 2} \\
=2+2+2=6 $
Which is not in options.
But I am not sure if I can use that $ AM-GM$ inequality in this case.
I look for a short and simple way .
I have studied maths upnto $12$th grade .
| Let $\sin^2 \theta =u$ and $\cos^2 \theta =v.$ Then $u+v=1.$ The expression exists only when $0<u<1$ and $0<v<1,$ when it is $$(u+v)+\left(\frac {1}{u}+\frac {1}{v}\right)+\left(\frac {u}{v}+\frac {v}{u}\right)=$$ $$=1+\frac {u+v}{uv}+\frac {u^2+v^2}{uv}=$$ $$=1+\frac {1}{uv}+\frac {u^2+v^2}{uv}=$$ $$=1+\frac {1}{uv}+\frac {(u+v)^2-2uv}{uv}=$$ $$=1+\frac {1}{uv}+\frac {1-2uv}{uv}=$$ $$=-1+\frac {2}{uv}=$$ $$=-1+\frac {2}{u(1-u)}.$$ The maximum value of $u(1-u)$ for $0<u<1$ is $\frac {1}{4}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Prove that $\left | \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a} \right | < \frac{1}{8}.$
Let $a,b,$ and $c$ be the lengths of the sides of a triangle. Prove that $$\left | \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} \right | < \dfrac{1}{8}.$$
The best idea I had was to expand the fractions to get something nicer. So we get $\dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} = \dfrac{a^2 b-a^2 c-a b^2+a c^2+b^2 c-b c^2}{(a+b) (a+c) (b+c)}.$ Then I would try to relate this to side lengths of a triangle to prove the desired inequality but that is the step where I am unsure.
| I'll prove a stronger inequality.
Let $a,b,$ and $c$ be the lengths-sides of a triangle. Prove that $$\left | \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} \right | < \dfrac{1}{22}$$
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence,
$$\sum_{cyc}\frac{a-b}{a+b}=\frac{\sum\limits_{cyc}(a-b)(a+c)(b+c)}{\prod\limits_{cyc}(a+b)}=\frac{\sum\limits_{cyc}(a-b)(c^2+ab+ac+bc)}{\prod\limits_{cyc}(a+b)}=$$
$$=\frac{\sum\limits_{cyc}(a-b)c^2}{\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{\prod\limits_{cyc}(a+b)}=\frac{(y-x)(z-x)(z-y)}{\prod\limits_{cyc}(2x+y+z)}.$$
Thus, we need to prove that
$$(2x+y+z)(2y+x+z)(2z+x+y)\geq22\left|(x-y)(x-z)(y-z)\right|$$
Since the last inequality is symmetric, we can assume
$x\leq y\leq z$ and $y=x+u$, $z=x+v$, $v=ku$,
where $x>1$ (if $k=1$ so our inequality is obviously true).
Id est, we need to prove that
$$(4x+u+v)(4x+2u+v)(4x+u+2v)\geq22uv(v-u)$$
and since $$(4x+u+v)(4x+2u+v)(4x+u+2v)>(u+u)(2u+v)(u+2v),$$
it remains to prove that
$$(k+1)(2k+1)(k+2)\geq22(k-1)k$$ or
$$2k^3-15k^2+29k+2\geq0,$$
which is true by AM-GM:
$$2k^3+29k+2>2k^3+29k\geq2\sqrt{2\cdot29}k^2>15k^2.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
If $\tan A+\tan B+\tan C=6$ and $\tan A\tan B=2 $ in $\triangle ABC$, then find the type of triangle.
In $\triangle ABC$, $\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2
$
Then the triangle is
$a.)\text{Right-angled isosceles} \\
b.) \text{Acute-angled isosceles}\\
\color{green}{c.)\text{Obtuse-angled}} \\
d.)\text{equilateral} $
$\ \ \ $
$\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2 \\
A+B=180-C\\
\dfrac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C\\
\tan A+\tan B=\tan C\\
\tan C=3 \\
\tan A\tan B=2\ \text{and} \ \tan A+\tan B=3 \\
\implies \tan A=2, \tan B=1\ \ \text{or}\ \ \tan A=1, \tan B=2
$
Now I am stucked, I look for a short and simple way.
I have studied maths upto $12$th grade .
Note:- Calculator is not allowed.
| For any triangle, we have
$$\tan(A) + \tan(B) + \tan(C) = \tan(A) \tan(B) \tan(C)$$
This means we have
$$\tan(A) \tan(B) \tan(C) = 6$$
Hence, we have $\tan(C) = 3$. Hence, we have
$$\tan(A) + \tan(B) = 3 \text{ and }\tan(A) \tan(B) = 2$$
This gives us that $\tan(A) = 2$ and $\tan(B) = 1$. Hence, the triangle is an acute angled triangle with one angle being $\dfrac{\pi}4$ and others being $\arctan(2)$ and $\arctan(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I simplify and evaluate the limit of $(\sqrt x - 1)/(\sqrt[3] x - 1)$ as $x\to 1$? Consider this limit:
$$ \lim_{x \to 1} \frac{\sqrt x - 1}{ \sqrt[3] x - 1}
$$
The answer is given to be 2 in the textbook.
Our math professor skipped this question telling us it is not in our syllabus, but how can it be solved?
| Another way : change variable $x=1+y$; so $$A=\frac{\sqrt x - 1}{ \sqrt[3] x - 1}=\frac{\sqrt{1+y} - 1}{ \sqrt[3] {1+y} - 1}$$ Now, using the fact that, close to $y=0$ (using the generalized binomial theorem as lulu commented) $$(1+y)^a=1+a y+\frac{1}{2}a \left(a-1\right) y^2+O\left(y^3\right)$$ which makes $$A=\frac{1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)-1}{1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)-1}\approx \frac{\frac{y}{2}-\frac{y^2}{8}}{\frac{y}{3}-\frac{y^2}{9}}=\frac{\frac{1}{2}-\frac{y}{8}}{\frac{1}{3}-\frac{y}{9}}$$ Now make $y\to0$ to get the result.
You can even get more if you know long division. Omitting the high order terms, the last expression is $\sim\frac{3}{2}+\frac{y}{8}$ which reveals not only the limit but also how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
} |
induction to prove the equation $3 + 9 + 15 + ... + (6n - 3) = 3n^2$ I have a series that I need to prove with induction. So far I have 2 approaches, though I'm not sure either are correct.
$$3 + 9 + 15 + ... + (6n - 3) = 3n^2$$
1st attempt:
\begin{align*}
& = (6n - 3) + 3n^2\\
& = 3n^2 + 6n - 3\\
& = (3n^2 + 5n - 4) + (n + 1)
\end{align*}
That seems way wrong lol ^^^
2nd attempt:
\begin{align*}
f(n) & = 3 + 9 + 15 + ... + (6n - 3)\\
f(n + 1) & = 6(n + 1) - 3\\
f(n + 1) & = 6(n - 3) + 6(n + 1) - 3\\
& = ?
\end{align*}
I don't know I feel like I'm headed in the wrong direction
I guess another attempt I have would be:
\begin{align*}
f(n) & = 3n^2\\
f(n+1) & = 3(n + 1)^2\\
& = 3(n^2 + 2n + 1)\\
& = 3n^2 + 6n + 3\\
& = f(n) + (6n + 3)
\end{align*}
| An other way
$$\sum_{k=1}^n(6k-3)=6\sum_{k=1}^n k-3n=6\cdot\frac{n(n+1)}{2}-3n=3n^2.$$
The only proof you need to do (by induction if you need to use induction) is that $$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Series expansion of $\frac{1}{\sqrt{x^3-1}}$ near $x \to 1^{+}$ How can I arrive at a series expansion for $$\frac{1}{\sqrt{x^3-1}}$$ at $x \to 1^{+}$? Experimentation with WolframAlpha shows that all expansions of things like $$\frac{1}{\sqrt{x^y - 1}}$$ have $$\frac{1}{\sqrt{y}\sqrt{x-1}}$$ as the first term, which I don’t know how to obtain.
| I like to expand around zero.
So, in
$\frac{1}{\sqrt{x^3-1}}
$,
let $x = 1+y$.
Then
$\begin{array}\\
x^3-1
&=(1+y)^3-1\\
&=1+3y+3y^2+y^3-1\\
&=3y+3y^2+y^3\\
&=y(3+3y+y^2)\\
\end{array}
$
so
$\begin{array}\\
\frac{1}{\sqrt{x^3-1}}
&=\frac{1}{\sqrt{y(3+3y+y^2)}}\\
&=\frac1{\sqrt{y}}\frac{1}{\sqrt{3+3y+y^2}}\\
&=\frac1{\sqrt{3y}}\frac{1}{\sqrt{1+y+y^2/3}}\\
&=\frac1{\sqrt{3y}}(1+y+y^2/3)^{-1/2}\\
\end{array}
$
Now apply the generalized binomial theorem
in the form
$(1+a)^b
=\sum_{n=0}^{whatever}\binom{b}{n}a^n
$
with
$a=y+y^2/3$
and
$b = -\frac12$.
Remember that
$\binom{b}{n}
=\frac{\prod_{k=0}^{n-1} (b-k)}{n!}
$,
so that
$\binom{-\frac12}{0}
=1
$,
$\binom{-\frac12}{1}
=-\frac12
$,
and
$\binom{-\frac12}{1}
=\frac12(-\frac12)(-\frac12-1)
=\frac12(-\frac12)(-\frac32)
=\frac38
$.
This will give you the first few terms
in terms of $y$.
To get them in terms of $x$,
replace $y$ by
$x-1$ and expand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2} \, \mathrm{d}\theta$ I have some trouble in how to evaluate this integral:
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right)
\,\mathrm{d}\theta
$$
I think it maybe has another form
$$
\int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right)
\,\mathrm{d}\theta
=
\sum_{n=1}^{\infty}{1 \over n^{2}}
\left[\psi\left(n + {1 \over 2}\right) - \psi\left(1 \over 2\right)\right]
$$
| Here's another way to solve the integral.
$$\int_{0}^{\pi}\theta \ln\tan\frac{\theta}{2}\mathrm{d}\theta = 4\int_{0}^{\frac{\pi}{2}}x\ln\tan{x}\mathrm{d}x = 4\left(\int_{0}^{\frac{\pi}{2}}x\ln\sin{x}\mathrm{d}x-\int_{0}^{\frac{\pi}{2}}x\ln\cos{x}\mathrm{d}x\right)$$
We solve the later integral first. But before that, recall, $\cos{x} = \frac{e^{ix}+e^{-ix}}{2}$
\begin{align*}
\int_{0}^{\frac{\pi}{2}}x\ln\cos{x}\mathrm{d}x &= \operatorname{Re}\left[\int_{0}^{\frac{\pi}{2}}x\ln\left(\frac{e^{ix}+e^{-ix}}{2}\right)\mathrm{d}x\right]\\
&= \operatorname{Re}\left[\int_{0}^{\frac{\pi}{2}}x\ln\left(\frac{1+e^{2ix}}{2e^{ix}}\right)\mathrm{d}x\right] \\
&= \operatorname{Re}\left[\int_{0}^{\frac{\pi}{2}}x\ln(1+e^{2ix})-x\ln(2e^{ix})\mathrm{d}x\right] \\
\end{align*}
Here, we use the maclaurin expansion of $\ln(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k$
\begin{align*}
\int_{0}^{\frac{\pi}{2}}x\ln\cos{x}\mathrm{d}x &= \operatorname{Re}\left[\int_{0}^{\frac{\pi}{2}}x\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}(e^{i2kx})\mathrm{d}x -\int_0^{\frac{\pi}{2}} x\ln(2) \mathrm{d}x - \int_0^{\frac{\pi}{2}} ix^2\mathrm{d}x\right] \\
\end{align*}
Interchanging the integral and sum, we get
\begin{align*}
\int_{0}^{\frac{\pi}{2}}x\ln\cos{x}\mathrm{d}x &= \operatorname{Re}\left[\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\left(\int_{0}^{\frac{\pi}{2}} x(e^{i2k})^x \mathrm{d}x \right) - \frac{\pi^2}{4}\ln 2 - i\frac{\pi^3}{24}\right] \\
\end{align*}
Now observe that $$\int xa^x \mathrm{d}x = \frac{xa^x}{\ln{a}} - \frac{a^x}{(\ln{a})^2} + C$$ (by Integration by Parts)
\begin{align*}
\int_{0}^{\frac{\pi}{2}}x\ln\cos{x}\mathrm{d}x &= \operatorname{Re}\left[\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\left(\frac{x(e^{2ki})^x}{i2k}+\frac{(e^{i2k})^x}{4k^2}\right|_0^\frac{\pi}{2} - \frac{\pi^2}{4}\ln 2 - i\frac{\pi^3}{24}\right] \\
&= \operatorname{Re}\left[\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\left(\frac{-i\pi}{4k}e^{ik\pi} + \frac{1}{4k^2} (e^{ik\pi}-1)\right) - \frac{\pi^2}{4}\ln 2 - i\frac{\pi^3}{24}\right] \\
&= \operatorname{Re}\left[-i\frac{\pi}{4}\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}e^{ik\pi} + \sum_{k=1}^\infty \frac{(-1)^{k+1}}{4k^3} (e^{ik\pi}-1) - \frac{\pi^2}{4}\ln 2 - i\frac{\pi^3}{24}\right] \\
\end{align*}
Since,
$$ e^{ik\pi} =
\begin{cases}
1 & k \equiv 0 \pmod{2} \\
-1 & k \equiv 1 \pmod{2} \\
\end{cases}
$$
And taking only the real part, we finally arrive at
$$\int_{0}^{\frac{\pi}{2}}x\ln\cos{x}\mathrm{d}x = \sum_{k=1\\ \text{k odd}}^\infty \frac{(-1)^k}{2k^3} - \frac{\pi^2}{4}\ln 2 $$
Evaluate the sum as,
\begin{align*}
\sum_{k=1\\ \text{k odd}}^\infty \frac{(-1)^k}{2k^3} &= \frac{-1}{2} \sum_{k=1}^\infty \frac{1}{(2k-1)^3} \\
&= \frac{-1}{2} \left(\zeta(3) - \sum_{k=1}^\infty \frac{1}{(2k)^3} \right) \\
&= \frac{-7}{16}\zeta(3)
\end{align*}
Therefore,
$$\int_0^\frac{\pi}{2} x\ln\cos{x} \mathrm{d}x = \frac{-7}{16}\zeta(3) - \frac{\pi^2}{8}\ln2$$
In a similar way, you can show that
$$\int_0^\frac{\pi}{2} x\ln\sin{x} \mathrm{d}x = \frac{7}{16}\zeta(3) - \frac{\pi^2}{8}\ln2$$
And hence you get –
$$\boxed{\int_{0}^{\pi}\theta \ln\tan\frac{\theta}{2}\mathrm{d}\theta = \frac{7}{2}\zeta(3)}$$
P.S. I am not sure as to how should I justify for the interchange of integral and sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Choosing a substitution to evaluate $\int \frac{x+3}{\sqrt{x+2}}dx$ Is there any other value you can assign to the substitution variable to solve this integral?
$$\int \frac{x+3}{\sqrt{x+2}}dx$$
Substituting $u = x + 2$:
$$du = dx; u +1 = x+3 ,$$
and we get this new integral that we can then split into two different ones:
$$\int \frac{u + 1}{\sqrt{u}}du = \int \frac{u}{\sqrt{u}}du + \int \frac{1}{\sqrt{u}}du .$$
We can substitute again $s = \sqrt u$ and get two immediate integrals:
$$s = \sqrt{u}; \quad ds = \frac{1}{2\sqrt{u}}du; \quad 2s^2 =u .$$ Substituting back $u$ to $s$ and $x$ to $u$ we get this result,
$$s^2 + \ln{\left | \sqrt{u} \right |} = u + \ln{\left | \sqrt{u} \right |} = x+2+\ln{\left | \sqrt{x+2} \right |},$$ which doesn't look quite to be right. What am I doing wrong? I'm pretty unsure about the second substitution, $2s^2 = u$. Is it correct?
| An other way is to write
$$\int\frac{x+3}{\sqrt{x+2}}dx=\int\frac{x+2+1}{\sqrt{x+2}}dx$$
$$=\int\frac{x+2}{\sqrt{x+2}}dx+\int\frac{1}{\sqrt{x+2}}dx=\int\sqrt{x+2}dx+\int\frac{1}{\sqrt{x+2}}dx$$
$$=I_1+I_2.$$
In $I_1$ we put the change of variable $u=x+2,\ du=dx$ and in $I_2$ we put $w=\sqrt{x+2},\ dw=\frac{1}{2\sqrt{x+2}}dx$, after calculations we obtain
$$\int\frac{x+3}{\sqrt{x+2}}dx=\frac{2}{3}\left(x+2\right)^{1/2}(x+5)+C,$$ as mentioned in the above result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove that $a(x+y+z) = x(a+b+c)$ If $(a^2+b^2 +c^2)(x^2+y^2 +z^2) = (ax+by+cz)^2$
Then prove that $a(x+y+z) = x(a+b+c)$
I did expansion on both sides and got:
$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2(abxy+bcyz+cazx) $
but can't see any way to prove $a(x+y+z) = x(a+b+c)$. How should I proceed?
| HINT: To do it without linear algebra, expand both sides and subtract like terms to leave
$$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2abxy+2acxz+2bcyz\;.$$
Notice that you can rearrange this as
$$(a^2y^2-2abxy+b^2x^2)+(a^2z^2-2acxz+c^2x^2)+(b^2z^2-2bcyz+c^2y^2)=0\;,$$
or
$$(ay-bx)^2+(az-cx)^2+(bz-cy)^2=0\;.$$
*
*What can you conclude about $ay-bx$, $az-cx$, and $bz-cy$?
*What can you conclude about $a(x+y+z)$ and $x(a+b+c)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find all pairs of nonzero integers $(a,b)$ such that $(a^2+b)(a+b^2)=(a-b)^3$
Find all pairs of nonzero integers $(a,b)$ such that
$(a^2+b)(a+b^2)=(a-b)^3$
My effort
Rearranging the equation I have
\begin{array}
\space (a^2+b)(a+b^2)-(a-b)^3 &=0 \\
a^2(b^2+3b)+a(-3b^2+b)+2b^3 &=0 \\
\end{array}
Letting $a=x$, we have the polynomial $Q(x)$ such that $$Q(x)=x^2(b^2+3b)+x(-3b^2+b)+2b^3 $$
So I must have that the roots $a_1,a_2$ must be such that
\begin{array}
\space a_1+a_2=-\cfrac{-3b+1}{b+3} \\
a_1\cdot a_2=\cfrac{2b^2}{b+3} \\
\end{array}
I am stuck now,should I go by brute force and verify some values or
there's still something I can do here to simplify the problem ?
| The general solution to $\alpha x^2+\beta x+\gamma=0$ is $x=\frac{-\beta\pm \sqrt{\beta^2-4\alpha\gamma}}{2\alpha}$.
Here, after factoring out the common $b$, you get $\alpha = b+3, \beta=-(3b-1),\gamma=2b^2$. In particular, you need $(3b-1)^2-4(b+3)(2b^2)=-8b^3-15b^2-6b+1$ to be a perfect square to even get a rational $a$.
But then a miracle occurs, because $-8b^3-15b^2-6b+1=(b+1)^2(1-8b)$. So this is a square exactly when $1-8b$ is a square.
Let $1-8b = (2c+1)^2$. Then $b=-\frac{c(c+1)}{2}$, and the formula:
$$ \begin{align}a&=\frac{3b-1 \pm (b+1)\sqrt{1-8b}}{2(b+3)}\\
&=\frac{3 \pm \sqrt{1-8b}}{2} +\frac{-10\pm 2\sqrt{1-8b}}{2(b+3)}\\
&=\frac{3\pm (2c+1)}{2} + \frac{-10 \pm 2(2c+1)}{-c^2-c+6}\\
&=\frac{3\pm (2c+1)}{2}+\frac{-10\pm(4c+2)}{(2-c)(c+3)}
\end{align}$$
Note that $3\pm(2c+1)$ is even so you need the second have to be an integer.
But this just means you need $4c+12=4(c+3)$ divisible by $(c+3)(c-2)$, or you need $4c-8=4(c-2)$ divisible by $(c+3)(c-2)$. So you need either $c-2$ or $c+3$ a factor of $4$. We can restrict to $c$ non-negative, so we get $c=0,1,3,4,6$.
The case $c=0$ gives $b=0$, disallowed.
The case $c=1$ gives $a=b=-1$.
The case $c=3$ gives $a=9,b=-6$.
The case $c=4$ gives $(a,b)=(8,-10)$.
The case $c=6$ gives $(a,b)=(9,-21)$.
You should also check specifically the case where $b+3=0$, since then the equation is not quadratic. There is no solution in this case.
--
Wondering if we write the polynomial in terms of $b$:
$$2b^2 + (a^2-3a)b + (3a^2+a)=0$$
The discriminant of this is $a^4-6a^3-15a^2-8a$. While we have an easy factor of $a$ there, that just returns us to a cubic $a^3-6a^2-15a-8=(a+1)^2(a-8)$. But it isn't obvious to me how to see that factorization if you stumbled in this direction first.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
About Factorization I have some issues understanding factorization.
If I have the expression $x^{2}-x-7$ then (I was told like this) I can put this expression equal to zero and then find the solutions with the quadratic formula, so it gives me $x_{0,1}= 1 \pm 2\sqrt{2}$ then $$x^{2}-x-7 = (x-1-2\sqrt{2})(x-1+2\sqrt{2}).$$
That is correct I have checked it.
Now for the expression $3x^{2}-x-2$ if I do the same I have $x_{0} = 1$ and $x_1=\frac{-2}{3}$ so I would have $$3x^{2}-x-2 = (x-1)(x+\frac{2}{3})$$
but this is not correct since $(x-1)(x+\frac{2}{3}) = \frac{1}{3}(3x^{2}-x-2)$,
the correct factorization is $3x^{2}-x-2 = (3x+2)(x-1)$.
So I guess finding the roots of a quadratic expression is not sufficient for factorizing.
| Look at what happens when we complete the square:
$$\begin{array}{lll}
ax^2+bx+c &=& \frac{1}{4a}(4a^2x^2 + 4abx + 4ac)\\
&=& \frac{1}{4a}((2ax)^2 + 2b(2ax) + 4ac)\\
&=& \frac{1}{4a}((2ax)^2 + 2b(2ax) +b^2-b^2+ 4ac)\\
&=& \frac{1}{4a}((2ax+b)^2-(b^2 - 4ac))\\
&=& \frac{1}{4a}((2ax+b)^2-\bigg(\sqrt{b^2 - 4ac}\bigg)^2)\\
&=& \frac{1}{4a}(2ax+b+\sqrt{b^2-4ac})(2ax+b-\sqrt{b^2-4ac})\\
&=& \frac{4a^2}{4a}\bigg(\frac{2ax+b+\sqrt{b^2-4ac}}{2a}\bigg)\bigg(\frac{2ax+b-\sqrt{b^2-4ac}}{2a}\bigg)\\
&=& a\bigg(x-\frac{-b-\sqrt{b^2-4ac}}{2a}\bigg)\bigg(x-\frac{-b+\sqrt{b^2-4ac}}{2a}\bigg)\\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Trigonometric Equation Simplification $$3\sin x + 4\cos x = 2$$
To solve an equation like the one above, we were taught to use the double angle identity formula to get two equations in the form of $R\cos\alpha = y$ where $R$ is a coefficient and $\alpha$ is the second angle being added to $x$ when using the double angle identity.
Why can't we use the identity $\sin(x) = \cos(x-90)$ to get $3\cos(x-90) + 4\cos x = 2$? Is this equation difficult to simplify further?
Additionally, why was the relationship between $sinx$ and $cosx$ in the pythagorean theorem (modified for the unit circle) not put to use? I did the following:
$$\sin^2x = 1 - \cos^2x$$
$$\therefore \sin x = ±\sqrt{1-\cos^2x}$$
If:
$$\sin x = y$$
Then,
$$y = ±\sqrt{1-\cos^2x}$$
Meaning that,
$$\cos x = ±\sqrt{1-y^2}$$
Inputting this into the original equation,
$$3y + 4\sqrt{1-y^2} = 2$$
We see, $$3y-2=-4\sqrt{1-y^2}$$
So, $$(3y-2)^2=16-16y^2$$
Therefore, $$9y^2-12y+4=16-16y^2$$
Rearranging gives, $$25y^2-12y-12=0$$
And so the solutions are, $$y_0,y_1=\frac{12}{50}\pm\frac{1}{50}\sqrt{144+1200}$$
And simplifying yields, $$y_0,y_1=\frac{6\pm4\sqrt{21}}{25}$$
Checking these solutions will give us the unique solution: $$y_0=\frac{6-4\sqrt{21}}{25}$$
$$q.e.d.$$
Can the above method be generalised? Has it been generalised?
| By complex numbers:
By the complex definition of the trigonometric functions, setting $z=e^{ix}$,
$$3\frac{z+z^{-1}}{2i}+4\frac{z+z^{-1}}2=2.$$
Multiplying by $z$ and rearranging,
$$\left(2+i\frac32\right)z^2-2z+\left(2-i\frac32\right)=0.$$
Then solving the quadratic equation
$$z=\frac{(8+6i)\pm\sqrt{21}(3-4i)}{25}.$$
The real and imaginary parts give $\cos x$ and $\sin x$ and $x$ is the imaginary part of the logarithm of $z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Modeling with Markov Chains and one-step analysis
I have set up the following model:
Let $X_n$ be the number of heads in the $n$-th toss and $P(X_0=0)=1$. I can calculate the transition matrix $P$. Define
$$
T=\min\{n\geq 0\mid X_n=5\}.
$$
Then $P(X=1)=P(X_{T-1}=4)$. Noting that $X_n$ is a Markov chain and defining
$$
u_i=P(X_{T-1}=4\mid X_0=i)
$$
we have
$$
u_i=\sum_{j=0}^5P(X_{T-1}=4\mid X_1=j)\cdot P_{ij}.
$$
I ended up with $u_0=0$ which seems nonsense.
*
*[Edited:]What is wrong with my model?
*Would anybody come up with a better one?
[Added:] Here is possibly where I get the calculation wrong:
$$
P(X_{T-1}=4\mid X_1=i)=u_i
$$
and $u_5=0$.
[Last Update:] What is wrong is
$$
P(X_{T-1}=4\mid X_1=4)=u_4.
$$
One should get $P(X_{T-1}=4\mid X_1=4)=1$ instead.
| It might simplify slightly to say we just require the chain to reach state $4$ at any point, starting at $0$, and we can forget about the variable $T$. So re-define $u_i$ as
$$u_i = P(X_n = 4 \text{ for some $n$}\mid X_0=i).$$
Our transition probability matrix:
$$
\begin{matrix}
\qquad 0 \qquad & 1 \qquad & 2 \qquad & 3 \quad & 4 \qquad & 5 \\
\end{matrix} \\
P = \begin{bmatrix}
1/32 & 5/32 & 10/32 & 10/32 & 5/32 & 1/32 \\
0 & 1/16 & 4/16 & 6/16 & 4/16 & 1/16 \\
0 & 0 & 1/8 & 3/8 & 3/8 & 1/8 \\
0 & 0 & 0 & 1/4 & 1/2 & 1/4 \\
0 & 0 & 0 & 0 & 1/2 & 1/2 \\
0 & 0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}
$$
The values for $u_i$ can be calculated in turn, using first-step analysis, until we get the required value, $u_0$:
\begin{align}
u_3 &= \frac{1}{4}u_3 + \frac{1}{2} \\
\therefore\quad u_3 &= \frac{2}{3} \\
& \\
u_2 &= \frac{1}{8}u_2 + \frac{3}{8}u_3 + \frac{3}{8} \\
\therefore\quad u_2 &= \frac{5}{7} \\
& \\
u_1 &= \frac{1}{16}u_1 + \frac{1}{4}u_2 + \frac{3}{8}u_3 + \frac{1}{4} \\
\therefore\quad u_1 &= \frac{76}{105} \\
& \\
u_0 &= \frac{1}{32}u_0 + \frac{5}{32}u_1 + \frac{5}{16}u_2 + \frac{5}{16}u_3 + \frac{5}{32} \\
\therefore\quad u_0 &= \frac{157}{217}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Improper Integral $\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx$ $$I=\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx\stackrel?=\frac{5}{24}\pi^3-\frac{\pi}2\log^2 2-2\pi\chi_2\left(\frac1{\sqrt 2}\right)$$
This result seems to me digitally correct?
Can we prove that the equality is exact?
| We have:
$$ \arcsin^2(z^2)=\sum_{n\geq 0}\frac{2^{2n+1} n!^2}{(2n+2)!}z^{4n+4} \tag{1}$$
hence:
$$ I = \frac{\pi}{4}\sum_{n\geq 0}\frac{n!^2 (4n+3)!}{2^{2n+1}(2n+2)!^2 (2n+1)!}=\frac{3\pi}{16}\cdot\phantom{}_5 F_4\left(1,1,1,\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{3}{2},2,2;1\right).\tag{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Proving $\tan A=\frac{1-\cos B}{\sin B} \;\implies\; \tan 2A=\tan B$
If $\tan A=\dfrac{1-\cos B}{\sin B}$, prove that $\tan 2A=\tan B$.
My effort:
Here
$$\tan A=\frac{1-\cos B}{\sin B}$$
Now
$$\begin{align}\text{L.H.S.} &=\tan 2A \\[4pt]
&=\frac{2\tan A}{1-\tan ^2A} \\[6pt]
&=\frac{(2-2\cos B)\over\sin B}{1-\frac{(1-\cos B)^2}{\sin^2 B}}
\end{align}$$
On simplification from here, I could not get the required R.H.S.
| You can proceed in this way:
$$\tan A=\frac{1-\cos B}{\sin B}=\frac{2\sin^2 \frac{B}{2}}{2\sin\frac{B}{2}\cos\frac{B}{2}}=\frac{\sin\frac{B}{2}}{\cos\frac{B}{2}}=\tan \frac{B}{2}$$
And hence comparing, we can write that $A=n\pi + \frac{B}{2}$ where $n$ is any integer.
So we can say that $2A=2n\pi + B \Rightarrow \tan 2A = \tan(2n\pi + B) = \tan B$
Hence proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to derive the equation of tangent to an arbitrarily point on a ellipse?
Show that the equation of a tangent in a point $P\left(x_0, y_0\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, could be written as: $$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$$
I've tried implicit differentiation $\to \frac{2x\frac{d}{dx}}{a^2}+\frac{2y\frac{d}{dy}}{b^2} = 0$, but not sure where to go from here. Substituting $P$ doesn't seem to help me much, and solving for $y$ from original equation seems to cause me more trouble than help. Please don't give me the solution, rather just give a slight hint or two:)
Thanks in advance!
Edit:
Solving for $\frac{dy}{dx}$ and putting it into point slope gives me:
$\frac{d}{dx} \left[\frac{x^2}{a^2} + \frac{y^2}{b^2}\right] = \frac{d}{dx}\left[1\right] \to \frac{2x}{a^2} + \frac{y}{b^2}\frac{dy}{dx}= 0\to \frac{dy}{dx}=-\frac{xb^2}{ya^2} \to \frac{dy}{dx}(P)=-\frac{x_0b^2}{y_0a^2}$
Then we get:
$y-y_0=\frac{dy}{dx}\left(x-x_0\right)\to y = -\frac{x_0b^2}{y_0a^2}(x-x_0)+y_0\to y = \frac{x_0^2b^2}{y_0a^2}-\frac{xx_0b^2}{y_0a^2} + y_0\to /:b^2,*y_0\to \frac{yy_0}{b^2}+\frac{xx_0}{a^2}=\frac{y_0^2}{b^2}$
Which looks close, but not exactly the expression i wanted. Where is the error?
| Let the parametric equation of the tangent be
$$x=x_0+t\cos(\theta),y=y_0+t\sin(\theta),$$ where $\theta$ is unknown.
Plug in the equation of the ellipse to get
$$\frac{(x_0+t\cos(t))^2}{a^2}+\frac{(y_0+t\sin(t))^2}{b^2}=1\\
=\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}+2\left(\frac{x_0\cos(\theta)}{a^2}+\frac{y_0\sin(\theta)}{b^2}\right)t+\left(\frac{\cos^2(\theta)}{a^2}+\frac{\sin^2(\theta)}{b^2}\right)t^2.$$
After simplification, this equation is of the form $\alpha t^2+\beta t=0$. As the line is tangent, the root $t=0$ must be double. This occurs if $\beta=0$,
$$\frac{x_0\cos(\theta)}{a^2}+\frac{y_0\sin(\theta)}{b^2}=0.$$
Then for any $t$
$$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=\frac{(x_0+t\cos(\theta))x_0}{a^2}+\frac{(y_0+t\sin(\theta))y_0}{b^2}=\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the smallest positive value taken by $a^3+b^3+c^3-3abc$
Find the smallest positive value taken by $a^3+b^3+c^3-3abc$ for positive integers $a,b,c$. Find all integers $a,b,c$ which give the smallest value.
Since it is generally hard to find the minimum of a multivariate polynomial, I tried factoring it at first. The $abc$ reminds me of AM-GM, but I wasn't sure how to relate the $a^3+b^3+c^3$ to it. Also the condition that we are working with integers may make it easier to find the minimum.
| $$\quad F=a^3+b^3+c^3-3 a b c=(a+b+c)(a^2+b^2+c^2- a b -b c- c a)=$$ $$= \frac {1}{2}(a+b+c)(\;[a-b]^2+[b-c]^2+[c-a]^2\;).$$
If $a=b=c$ then $F=0.$
If $a,b,c$ are not all equal then $a+b+c\geq 1+1+2=4,$ and also at least two of $|a-b|, |b-c|, |c-a|$ are non-zero, giving $(a-b)^2+(b-c)^2+(c-a)^2\geq 1^2+1^2+0^2=2 .$
Therefore $F>0\implies F\geq \frac {1}{2}(2)(4)=4.$ Which is attained when $a=b=1$ and $c=2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I find the difference between the gradients of two lines represented by an equation I want to find the difference between the gradients (or slopes?) of two lines. The equation of the lines is $$x^2(\tan^2 \theta+\cos^2 \theta)-2xy\tan\theta+y^2 \sin^2 \theta=0$$
I have assumed the gradients are $m_1$ and $m_2$.
I compared the given equation with the general equation $ax^2+2hxy+by^2$ and found out the respective values and plugged it in the formula I know ie
$m_1-m_2=2\sqrt{h^2-ab}=2\sqrt{\tan^2 \theta-(\tan^2 \theta+\cos^2 \theta) \sin^2 \theta}$
Proceeding from here is difficult and the answer given in my book is $2$. How do I do this or is that an error? Thanks in advance!
| I'm going to guess wildly at what you actually mean because you do not write what you actually mean. I expect that if you clarify your question and it turns out I have guessed wrongly, others will vote this answer into oblivion, as they should...
Say we have two lines in point-slope form: $$\begin{align}
y &= m_1 x + b_1 , \\
y &= m_2 x + b_2 .
\end{align}$$ We may combine these equations as $$
(y-m_1 x - b_1)(y - m_2 x - b_2) = 0
$$ and expand, giving $$
y^2 -(m_1 + m_2) x y + m_1 m_2 x^2 -(b_1 + b_2) y + (b_1 m_2 + b_2 m_1) x + b_1 b_2 = 0 .
$$ Writing your equation in the same form, we have $$
y^2 -2\frac{\tan \theta}{\sin^2 \theta} x y + \frac{\tan^2 \theta + \cos^2 \theta}{\sin^2 \theta} x^2 - 0 y + 0 x + 0 = 0 .
$$ Comparing coefficients of the constant and linear terms, we discover $b_1 + b_2 = 0$ and $b_1 b_2 = 0$, so $b_1 = b_2 = 0$. Comparing coefficients of the degree $2$ terms, we find the system $$\begin{align}
m_1 + m_2 = 2\frac{\tan \theta}{\sin^2 \theta} \text{ , and} \\
m_1 m_2 = \frac{\tan^2 \theta + \cos^2 \theta}{\sin^2 \theta} .\\
\end{align}$$
It's a useful fact that $(m_1 + m_2)^2 - 4 m_1 m_2 = (m_1 - m_2)^2$, which one can easily verify by expanding. Therefore, $$\begin{align}
(m_1 - m_2)^2 &= 4 \frac{\tan^2 \theta}{\sin^4 \theta} - 4 \frac{(\tan^2 \theta + \cos^2 \theta)}{\sin^2 \theta} \\
&= 4 \frac{\tan^2 \theta - \sin^2 \theta (\tan^2 \theta + \cos^2 \theta)}{\sin^4 \theta} \\
&= 4 \frac{\tan^2 \theta (1 - \sin^2 \theta) - \sin^2 \theta \cos^2 \theta}{\sin^4 \theta} \\
&= 4 \frac{(\tan^2 \theta - \sin^2 \theta) \cos^2 \theta}{\sin^4 \theta} \\
&= 4 \frac{(\sec^2 \theta - 1) \cos^2 \theta}{\sin^2 \theta} \\
&= 4 \frac{\tan^2 \theta \cos^2 \theta}{\sin^2 \theta} \\
&= 4 ,
\end{align}$$ as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $n_{1}+n_{2}+n_{3}+n_{4}+n_{5} = 20.$ Then number of such distinct arrangements of $(n_{1},n_{2},n_{3},n_{4},n_{5})$
Let $n_{1}<n_{2}<n_{3}<n_{4}<n_{5}$ be the positive integers such that $n_{1}+n_{2}+n_{3}+n_{4}+n_{5} = 20$
Then number of such distinct arrangements of $(n_{1},n_{2},n_{3},n_{4},n_{5})$ is
$\bf{My\; Try::}$ Let $n_{1} = x+1\;,n_{2} = y+2,n_{3}=z+3,n_{4}=t+4\;,n_{5}=u+5$
where $x,y,z,t,u\geq 0$ and $x\leq y \leq z\leq t \leq u$
So our equation convert into $x+y+z+t+u=5$
Now Put $x=0\;,$ We get $y+z+t+u=5\;,$ Then put $y=0\;,$ We get $z+t+u=5$
But using that way our task is very Complicated.
So plz explain me a better way to solve that equation.
Thanks
| Here is an elementary argument.
The minimum possible $n_1$ is $1$.
Since the sum is $20$, dividing $20$ by $5$ we get $4$, which means $n_1$ cannot be $4$, since the numbers are strictly increasing.
If $n_1=3$, increasing each following number by $1$, we get $3+4+5+6+7=25$.
So the maximum possible $n_1$ is $2$. With $2+3+4+5+6=20$ we get the only possible arrangement when $n_1=2$.
Now come back to $n_1=1$. We consider what is the maximum possible $n_2$. Using the same argument as above, we see that $4$ does not work. $n_2=3$ only gives one arrangement $1+3+4+5+7=20$.
So now $n_1=1, n_2=2$. We consider the maximum possible $n_3$. We find that it is $4$.
Now two cases: $n_1=1, n_2=2, n_3=4$ and $n_1=1, n_2=2, n_3=3$. In each case you can discuss as above what is the maximum possible $n_4$. With the first case, we get $1+2+4+6+7$ and $1+2+4+5+8$. With the second case we get other $3$ arrangements.
| {
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Proving for all integer $n \ge 2$, $\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$ Prove the following statement by mathematical induction:
For all integer $n \ge 2$, $$\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$$
My attempt: Let the given statement be $p(n)$ .
1.\begin{align*} \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} & =\frac{\sqrt 2 +1}{\sqrt 2} \\
2 &< \sqrt 2 +1 \\
\sqrt 2 &< \frac{\sqrt 2 +1}{\sqrt 2}=\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} \end{align*}
Hence, $p(2)$ is true.
2.For an arbitrary integer $k \ge 2$, suppose $p(k)$ is true.
That is, $$\sqrt k < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}$$
Then we must show that $p(k+1)$ is true.
We're going to show that $$\sqrt {k+1} < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}+\frac{1}{\sqrt {k+1}}$$
I'm stuck on this step. I can't develop it further. How can I complete this proof?
| Hint: Note that $$\sqrt{k+1}-\sqrt{k}=\frac{1}{\sqrt{k+1}+\sqrt{k}}<\frac{1}{\sqrt{k+1}}$$
| {
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Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are, of course, brute force techniques for showing this, but I'm hoping for something elegant.
| Here's the symmetric polynomial approach.
$a^n+b^n+c^n$ is a symmetric homogeneous polynomial of degree $n$. So it can be expresses as a linear combination of polynomials $s_1^is_2^js_3^k$ where $i+2j+3k=n$, and $s_1=a+b+c,$ $s_2=ab+bc+ac,$ $s_3=abc$ are the elementary symmetric polynomials.
Now, $s_1=a+b+c=0$ implies that $a^n+b^n+c^n$ can be written as a linear combination of $s_2^js_3^k$ with $2j+3k=n$.
It the case where $n=2,3,4,5,7$, there is only one pair $j,k$ such that $2j+3k=n$, so $a^n+b^n+c^n$ becomes a monomial in $s_2,s_3$.
When $n=2$, this means $a^2+b^2+c^2=k_2s_2$ for some constant $k_2$. Setting $(a,b,c)=(2,-1,-1)$, we see $k_2=-2$.
When $n=3$, this means $a^3+b^3+c^3=k_3s_3$ for some constant $k_3$. Setting $a=2,b=c=-1$, we get $k_3=3$.
Similarly $a^5+b^5+c^5=k_5s_2s_3$, and again we use $(a,b,c)=(2,-1,-1)$ to get that $k_5=-5$.
Likewise, $a^7+b^7+c^7=7s_2^2s_3$.
Similar formula under the same condition:
$$\frac{a^2+b^2+c^2}{2}\frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}\\
\frac{a^2+b^2+c^2}{2}\frac{a^5+b^5+c^5}{5} = \frac{a^7+b^7+c^7}{7}
$$
More generally, if $a,b,c\in\mathbb Z$ with $a+b+c=0$ and $n$ is relatively prime to $6$ then
$a^n+b^n+c^n$ is divisible by $s_2s_3=(ab+ac+bc)abc$.
There's actually an expression for $a^n+b^n+c^n$ with explicit coefficients:
$$a^n+b^n+c^n = \sum_{i+2j+3k=n} (-1)^j\frac{n}{i+j+k}\binom{i+j+k}{i,j,k} s_1^is_2^js_3^k\tag{1}$$
I confess I found that formula when looking at Fermat's Last many many years ago. It occurred to me at the time that Fermat can be expresses, for odd $n$, as
Given odd positive integer $n>1$. Then for integers $a,b,c$, $a^n+b^n+c^n=0$ if and only if $a+b+c=0$ and $abc=0$.
which seemed to imply it is a question about symmetric polynomials.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x,y,z$ are positive real number number, Then minimum value of $\frac{x^4+y^4+z^2}{xyz}$
If $x,y,z$ are positive real number number, Then minimum value of $\displaystyle \frac{x^4+y^4+z^2}{xyz}$
$\bf{My\; Try::}$ Given $x,y,z>0.$ So Using $\bf{A.M\geq G.M\;,}$ We get
$$\displaystyle x^4+y^4\geq 2x^2y^2$$ and then Using $$2x^2y^2+z^2\geq 2\sqrt{2}|xyz|\geq 2\sqrt{2}xyz$$
So we get $$x^4+y^4+z^4\geq 2x^2y^2+z^2 \geq 2\sqrt{2}xyz$$
and equality hold when $x^4=y^4$ and $2x^2y^2=z^2$
So we get $x=y$ and $z=\sqrt{2}x^2$, Now Minimum occur at $\left(x,x,\sqrt{2}z^2\right)$
So at that point we get $\displaystyle \left[\frac{x^4+y^4+z^4}{xyz}\right]_{\bf{\min}} = \sqrt{2}+2x^2$
and Answer given is $2\sqrt{2}$, I did not understand how this can happen
olz explain me, Thanks
My Question is
| We have, with AM-GM, that \begin{align}
x^4+y^4+z^2&=4\cdot\frac{x^4+y^4+\tfrac12z^2+\tfrac12z^2}{4}\\
&\geq 4\sqrt[4]{x^4y^4z^4\cdot\tfrac14}\\
&=xyz\cdot 2\sqrt{2}
\end{align}
So that $$\frac{x^4+y^4+z^2}{xyz}\geq 2\sqrt{2}$$
With equality iff $x^4=y^4=\frac12z^2$, so for example $x=y=1$ and $z=\sqrt{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that a series converges I am given the series $\Sigma_{n=1}^{\infty} \frac{\sqrt{a_n}}{n}$, where $a_n \geq 0 \forall n$ , and $\Sigma_1^{\infty} a_n $ converges.
I was advised to expand $|\sqrt{a_n} - \frac{1}{n}|^2 $ . Doing this gives me expressions like $-\frac{2 \sqrt{a}}{n}+a+\frac{1}{n^2}$ , but I don't see how this is useful.
| Writing $(x-y)^2 = x^2 - 2xy + y^2$, we see that
$$ xy \leq \frac{1}{2} \left( x^2 + y^2 \right). $$
This implies that
$$ \frac{\sqrt{a_n}}{n} \leq \frac{1}{2} \left( a_n + \frac{1}{n^2} \right). $$
Since both $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converge, the series $\sum_{n=1}^{\infty} \frac{1}{2} \left( a_n + \frac{1}{n^2} \right)$ also converges and so by the comparison test, the series $\sum_{n=1}^{\infty} \frac{\sqrt{a_n}}{n}$ also converges.
| {
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"timestamp": "2023-03-29T00:00:00",
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Make $2^8 + 2^{11} + 2^n$ a perfect square Can someone help me with this exercise? I tried to do it, but it was very hard to solve it.
Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.
It is the same thing like $4=2^2$.
| $2^8 + 2^{11} + 2^n = 2^8(1 + 8 + 2^{n-8})=2^8(9 + 2^{n-8})$
Therefore, $9 + 2^{n-8}$ has to be a perfect square.
Clearly, $9 + 16 = 25$ is a perfect square.
So, $2^{n-8} = 2^4$ giving, $$n = 12$$
| {
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Given two sets, is there a formula to know the quantity of injective function realationships? For example.
\begin{align*}
&A\quad =\quad \{ 1,2\} \\
&B\quad =\quad \{ a,b\} \\
&\texttt{Then 2 injective functions from } A\rightarrow B \\
&\left\{ (1,a),(2,b) \right\} \quad \{ (1,b),(2,a)\} \\ \\
&A\quad =\quad \{ 1,2\} \\
&B\quad =\quad \{ a,b,c\} \\
&\texttt{Then 6 injective functions from } A\rightarrow B \\
&\left\{ (1,a),(2,b) \right\} \quad \{ (1,a),(2,c)\} \quad \\
&\{ (1,b),(2,a)\} \quad \{ (1,b),(2,c)\} \\
&\{ (1,c),(2,a)\} \quad \{ (1,c),(2,b)\} \\ \\
&A\quad =\quad \{ 1,2,3\} \\
&B\quad =\quad \{ a,b,c\} \\
&\texttt{Then 12 injective functions from } A\rightarrow B \\
&\left\{ (1,a) (2,b) (3,a) \right\}
\left\{ (1,a) (2,b) (3,c) \right\}\\
&\left\{ (1,a) (2,c) (3,a) \right\}
\left\{ (1,a) (2,c) (3,b) \right\}\\
&\left\{ (1,b) (2,a) (3,b) \right\}
\left\{ (1,b) (2,a) (3,c) \right\}\\
&\left\{ (1,b) (2,c) (3,a) \right\}
\left\{ (1,b) (2,c) (3,b) \right\}\\
&\left\{ (1,c) (2,a) (3,b) \right\}
\left\{ (1,c) (2,a) (3,c) \right\}\\
&\left\{ (1,c) (2,b) (3,a) \right\}
\left\{ (1,c) (2,b) (3,c) \right\}\\
\end{align*}
| Let $A$ and $B$ be nonempty, finite sets such that $A$ has at most as many elements as $B$. Let $m$ be the number of elements in $A$ and let $n$ be the number of elements in $B$. WLOG assume that $A = \{ 1, 2, \ldots, m \}$ and $B = \{1, 2, \ldots, n \}$. Then there are precisely $\binom{n}{m} \cdot m!$ many injections $A \to B$.
To see this, let $[B]^m$ be the set of all subsets of $B$ with $m$ elements. And let $S_m$ be the permutation group on $m$ elements. Let $\mathcal F$ the the set of all injections $A \to B$. Then
$$\pi \colon \mathcal F \to [B]^m \times S_m, f \mapsto (\operatorname{im}(f), f^*)$$
is a bijection, where $f^* \colon m \to m$ is the unique permutation such that for all $x,y \in \{1, \ldots, m\}$ we have $x < y$ iff $f \circ f^*(x) < f \circ f^*(y)$.
Since $[B]^m$ has $\binom{n}{m}$ many elements and $S_m$ has $m!$ many elements, the result follows.
| {
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How does this factoring work? $$ (z^2 - 2i ) = (z -1 -i)(z + 1 +i) $$
I see if you multiply out the right-hand side, you obtain the left-hand side, but how does one know to factor like that or this?
$$ (z^2 − 3iz − 3 + i) = (z − 1 − i)(z + 1 − 2i) $$
| The first one
is using
$z^2-a^2
=(z-a)(z+a)
$
where
$a = \sqrt{2i}
=1+i
$
since
$(1+i)^2
=1+2i-1 = 2i
$.
The second one
just uses the quadratic formula,
which works for
complex as well as real coefficients,
to solve
$z^2 − 3iz − 3 + i
= 0
$.
If the roots are
$u$ and $v$,
then
$z^2 − 3iz − 3 + i
= (z-u)(z-v)
$.
(This was added later)
Using the quadratic formula,
the roots are
$\begin{array}\\
\dfrac{3i\pm\sqrt{(-3i)^2-4(-3+i)}}{2}
&=\dfrac{3i\pm\sqrt{-9+12-4i}}{2}\\
&=\dfrac{3i\pm\sqrt{3-4i}}{2}\\
&=\dfrac{3i\pm(2-i)}{2}
\qquad\text{since }\sqrt{3-4i} = 2-i\\
&=\dfrac{3i+(2-i), 3i-(2-i)}{2}\\
&=\dfrac{2+2i, -2+4i}{2}\\
&=1+i, -1+2i\\
\end{array}
$
| {
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If $b$ is an odd composite number and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, what happens when $q = 2^{r + 1} - 1$? (Note: An improved version of this question has been cross-posted to MO.)
Let $\sigma(X)$ be the sum of the divisors of $X$. For example, $\sigma(2) = 1 + 2 = 3$, and $\sigma(4) = 1 + 2 + 4 = 7$.
My question is:
If $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, what happens when $q = 2^{r + 1} - 1$ (with $r \geq 1$)?
Without the restriction on $q = 2^{r + 1} - 1$ being prime, I only know that $M = {2^r}{b^2}$ must be an even almost perfect number. (That is, it must satisfy $\sigma(M) = 2M - 1$.)
I guess my question can be rephrased as follows:
(1) If $b$ is an odd composite, how often is $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ a prime number?
(2) If $b$ is an odd composite, and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, how often is $q$ a Mersenne prime?
| The only thing that I can say is that if $b=3^k$ for $k>1$ then $q=2$.
Up to $10^8$ there are no other values of $b$ that make $q$ prime.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taking implicit derivative of $(x^2 + y^2)^3 = 5x^2 \cdot y^2$ I am a bit confused about taking implicit derivative of $(x^2 + y^2)^3 = 5x^2 \cdot y^2$.
$$\frac{d(x^2 + y^2)^3}{dx} = \frac{d(5x^2 y^2)}{dx} $$
Edit: Incorrect step
$$= 3(x^2 + y^2) \left(2x + 2y\frac{dy}{dx}\right) = 10xy^2 + 5x^22y\frac{dy}{dx}$$
$$= 3(x^2 + y^2) \left(2x + 2y\frac{dy}{dx}\right) = 5 \left(2xy^2 + x^22y\frac{dy}{dx}\right)$$
...
So I don't get how we got $2y\frac{dy}{dx}$ from $y^2$. Why it is not $2\frac{dy}{dx}$???
| An expression $f(x,y)=g(x,y)$ is implicitly differentiated by
$$ \frac{\partial f}{\partial y} y' + \frac{\partial f}{\partial x} x' =\frac{\partial g}{\partial y} y' + \frac{\partial g}{\partial x} x'$$
but with $x'=1$ in your case.
What I get is
$$ \left(6 y (x^2+y^2)^2 \right) y' + \left( 6 x (x^2 + y^2)^2 \right) 1 = \left( 10 x^2 y \right) y' + \left(10 x y^2 \right) 1$$
which needs some simplifications.
The specific question as to why $\frac{{\rm d} }{{\rm d} x}y^2 = 2 y y'$ use the rule above to get
$$ \frac{\partial y^2}{\partial y} y' + \frac{\partial y^2}{\partial x} 1 = (2 y) y' + (0) 1 = 2 y y' $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate $\operatorname{taxicab}(3,8,2)$ (sum of 8 cubes in two different ways) Could someone explain me how I can calculate $\operatorname{taxicab}(3,8,2)$?
$\operatorname{taxicab}(3,8,2)$ is the smallest natural number that can be written in $2$ different ways as a sum of $8$ powers $3$.
For instance:
\begin{align}\operatorname{taxicab}(4,3,2) &= 2673\\
&= 7^4 + 4^4 + 2^4 (or 2401+256+16)
&= 6^4 + 6^4 + 3^4 (or 1296+1296+81)
\end{align}
How can I calculate $\operatorname{taxicab}(3,8,2)$?
This notion is a generalization of the notion of Taxicab number, which is mentioned in the famous story about Hardy and Ramanujan. See also Proof that $1729$ is the smallest taxicab number
| $$\eqalign{132=1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 (or 1+1+1+1+1+1+1+125) =1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 (or 1+8+8+8+8+8+27+64)}$$
This can be found by a recursive computation.
Let $N(x,m)$ be the number of different ways to write $x$ as the (unordered) sum of $m$ positive cubes. Then $N(x,m) = M(x,m,\lfloor x^{1/3} \rfloor)$ where $M(x,m,n)$ is the number of different ways to write $x = a_1^3 + \ldots + a_m^3$ with $1 \le a_1 \le \ldots \le a_m \le n$.
$$\eqalign{M(x,0,n) &= \cases{1 & if $x=0$\cr 0 & otherwise}\cr
M(x,m,n) &= \sum_{y=1}^{\min(n, \lfloor x^{1/3} \rfloor)} M(x - y^3, m-1, y)}$$
| {
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Compute $\sum_{k=1}^{n+1} \binom{n}{k-2} \frac{2^k}{3^{2k-1}}$ Compute $$\sum_{k=1}^{n+1} \binom{n}{k-2} \frac{2^k}{3^{2k-1}}$$
$\sum_{k=1}^{n+1} \binom{n}{k-2} \frac{2^k}{3^{2k-1}}$
$k - 2 = t , k = t + 2$
$$\sum_{t=0}^{n} \binom{n}{t} \frac{2^{t+2}}{3^{2t+1}} - \binom {n}{n} \frac{2^n}{3^{2n}}$$
Let's simplify this:
$$\sum_{t=0}^{n} \binom{n}{t} \frac{2^{t+2}}{3^{2t+1}} = \frac{4}{3}\sum_{t=0}^{n} \binom{n}{t} \frac{2^t}{3^{2t}}$$
Now let's calculate this: $\sum_{t=0}^{n} \binom{n}{t} \frac{2^t}{3^{2t}}$
$a = 1 , b = 2/9$
$$\sum_{t=0}^{n} \binom{n}{t} \frac{2^t}{3^{2t}} = (1+\frac{2}{9})^n$$
My final answer: $\frac{4}{3}(\frac{11}{9})^n - \frac{2^n}{3^{2n
}}$
Opinions?
| Let $k-2=u$
$$=3\sum_{u=-1}^{n-1}\binom nu\left(\dfrac2{3^2}\right)^{u+2}=3\left(\dfrac2{3^2}\right)^2\sum_{u=0}^{n-1}\binom nu\left(\dfrac2{3^2}\right)^u$$
Now,
$$\sum_{u=0}^{n-1}\binom nu\left(\dfrac2{3^2}\right)^u=\left(1+\dfrac2{3^2}\right)^n-\binom nn\left(\dfrac2{3^2}\right)^n=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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recursive function: non-recursive form possible? Can the following recursive function be converted to a non-recursive form?
$$f(x,c,\ell)=\frac{c-c^\ell}{1-c}+(c-2)\sum \limits_{k=1}^{\ell-1}f(x,c,k)$$
$$f(x,c,1)=c$$
$$c= \text{constant}$$
$$\ell=\text{length}$$
If so, where do I start?
| I use $n$ instead of $\ell$, hope you do not mind. Let's prove by (strong) induction that, for $n \geq 2$, it holds $f(c,n) = c^n - c(c-1)^{n-2}$. For $n=2$ it is clear, since $f(c,2) = c+(c-2)c = c^2-c$. If the assumption holds for every $2 \leq j \leq n$, then
\begin{align*}
f(c,n+1) &= \frac{c^{n+1}-c}{c-1} + (c-2)\sum_{j=1}^n f(c,j) = \frac{c^{n+1}-c}{c-1} + (c-2)\left [ \sum_{j=2}^n \left (c^j - c(c-1)^{j-2}\right ) + c \right ] \\ &= \frac{c^{n+1}-c}{c-1} + (c-2)\sum_{j=2}^n c^j -c(c-2)\sum_{j=2}^n (c-1)^{j-2} + (c-2)c \\ &= \frac{c^{n+1}-c}{c-1} + (c-2)\sum_{j=0}^n c^j -(c-2)(1+c) - c(c-2)\sum_{j=0}^{n-2} (c-1)^j + (c-2)c \\ &= \frac{c^{n+1}-c}{c-1} + (c-2)\frac{c^{n+1}-1}{c-1} - c(c-2)\frac{(c-1)^{n-1}-1}{c-2} -c+2 \\ &= \frac{c^{n+1}-c+c^{n+2}-2c^{n+1}-c+2+2c-2}{c-1} - c(c-1)^{n-1} = c^{n+1} - c(c-1)^{n-1}.
\end{align*}
| {
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Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+Ct+D=At+At^3+B+Bt^2 +Ct+D$$
for this I have that $A=0, B=2, C=0, D=-2$
so now I have
$I=\int \frac{2t^2}{(1+t^2)^2}dt= \int\frac{2}{1+t^2}dt - \int\frac{2}{(1+t^2)^2}dt$
Now,
$$ \int\frac{2}{1+t^2}dt = 2\arctan t$$
and
$$\int\frac{2}{(1+t^2)^2}dt$$
using partial integration we have:
$$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{1+t^2}$$
and $$dt=dv \Rightarrow t=v$$
so now we have:
$$\int\frac{2}{(1+t^2)^2}dt =\frac{t}{(1+t^2)^2} + 4\int\frac{t^2}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4\int\frac{t^2 + 1 -1}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4t -4\arctan t$$
so, the final solution should be:
$$I=2\arctan t - \frac{t}{(1+t^2)^2} - 4t +4\arctan t$$
since the original variable was $x$ we have
$$I= 6\arctan \sqrt{x} - \frac{\sqrt{x}}{(1+x)^2} - 4\sqrt{x} $$
But, the problem is that the solution to this in my workbook is different, it says that solution to this integral is $$I=\arctan \sqrt{x} - \frac{\sqrt{x}}{x+1}$$
I checked my work and I couldn't find any mistakes, so i am wondering which solution is correct?
| Alternative Approach:
Let $x=\tan^2{\theta}$,$dx=2\tan{\theta} \sec^2{\theta} d\theta$
\begin{align}
I&=\int{\frac{\tan{\theta}\cdot 2\tan{\theta}\sec^2{\theta} d\theta}{ \sec^4{\theta} }}\\&=2\int{\sin^2{\theta}}d\theta\\&=\int{1-\cos{(2\theta)}}d\theta\\&=\theta-\frac12 \sin{(2\theta)}+C\\&=\arctan{\sqrt x}-\frac{\sqrt x}{1+x}+C
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral inequality $\int_0^1 f(x)\,dx \ge 4 \int_0^{1\over2} f(x)\,dx$? If $f: [0, 1] \to \mathbb{R}$ is a convex and integrable function with $f(0) = 0$, does it necessarily follow that$$\int_0^1 f(x)\,dx \ge 4 \int_0^{1\over2} f(x)\,dx?$$
| For $0\le x\le \frac{1}{2}$, by convexity one has $$f\left(\frac{1}{2}\right)\le \frac{1}{2}\left[f\left(\frac{1}{2} + x\right) + f\left(\frac{1}{2} -x\right)\right].$$ Also, $$\int_0^1f(x)dx=\int_0^{\frac{1}{2}}f(x)dx +\int_{\frac{1}{2}}^1f(x)dx=\int_0^{\frac{1}{2}}f(x)dx +\int_0^{\frac{1}{2}}f\left(x+\frac{1}{2}\right)dx.$$ From the first inequality and the above, one has $$\int_0^1f(x)dx\ge\int_0^{\frac{1}{2}}f\left(x\right)dx + 2f\left(\frac{1}{2}\right)\int_0^{\frac{1}{2}}dx - \int_0^{\frac{1}{2}}f\left(\frac{1}{2}-x\right)dx.$$ The last and the first integral in the above inequality cancel. Finally note that by convexity and the condition $f(0)=0$ $$\int_0^{\frac{1}{2}}f(x)dx\le \frac{1}{2}\frac{1}{2}f(\frac{1}{2}).$$ The required inequality follows from above.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove the trigonometric identity $\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi - \tan\varphi)^2$ $$\frac{1-\sin\varphi}{1+\sin\varphi}$$
I have no idea how to start this, please help. This problem is essentially supposed to help me solve the proof of
$$\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi - \tan\varphi)^2$$
| If we solve the Pythagorean identity
$$\sin^2\varphi + \cos^2\varphi = 1$$ for $\cos^2\varphi$
we obtain
$$\cos^2\varphi = 1 - \sin^2\varphi$$
Since
$$\sec\varphi = \frac{1}{\cos\varphi}$$
and
$$\tan\varphi = \frac{\sin\varphi}{\cos\varphi}$$
we need to obtain $\cos^2\varphi$ in the denominator. We can do this by multiplying the denominator by $1 - \sin\varphi$ to obtain $1 - \sin^2\varphi = \cos^2\varphi$. However, what we do to the denominator, we must also do to the numerator in order to preserve equality.
\begin{align*}
\frac{1 - \sin\varphi}{1 + \sin\varphi} & = \frac{1 - \sin\varphi}{1 + \sin\varphi} \cdot \frac{1 - \sin\varphi}{1 - \sin\varphi}\\
& = \frac{1 - 2\sin\varphi + \sin^2\varphi}{1 - \sin^2\varphi}\\
& = \frac{1 - 2\sin\varphi + \sin^2\varphi}{\cos^2\varphi}\\
& = \frac{1}{\cos^2\varphi} - \frac{2\sin\varphi}{\cos^2\varphi} + \frac{\sin^2\varphi}{\cos^2\varphi}\\
& = \sec^2\varphi - 2\frac{\sin\varphi}{\cos\varphi} \cdot \frac{1}{\cos\varphi} + \tan^2\varphi\\
& = \sec^2\varphi - 2\tan\varphi\sec\varphi + \tan^2\varphi\\
& = (\sec\varphi - \tan\varphi)^2
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Trying to find volume by rotating a region..getting a negative value I'm working on a review assignment for volumes of solids obtained by rotating a region bounded by given curves around a certain line (using washers or shells) and seem to be getting a negative answer for this one problem...
Find volume of solid obtained by rotating region formed by x-y=3 and
$y^2 = x-1$ about the line $y=-2$
I tried to solve using washers with large radius $y^2+3$ and small radius y+5.
I end up trying to solve the following integral, and get $\frac{-207\pi}{5}$ , which I don't believe is the correct answer since it is negative..
$$\pi\int\limits_{-1}^{2} (y^2+3)^2-(y+5)^2 \, dy$$
Points of intersection are at $(2,-1)$ and $(5,2)$.
Any advice on what is going on here? Thanks in advance!
| First of all, let us have a look at the region. The blue line is $y=-2$, and the region is clearly visible.
So we will have $x$ ranging over $1$ and the $x$ of the rightmost intersection of the curves. Let us find the intersections: $x-y=3$ or $y=x-3$, so subbing into $y^2=x-1$ we get $(x-3)^2=x-1$, or $x^2-6x+9=x-1$, or $x^2-7x+10=0$, which yields $x=\frac{7\pm\sqrt{49-40}}{2}=\frac{7\pm3}{2}$, which means $2$ and $5$. So $x\in[1,5]$. What about $y$? $y$ will range from $\max\{-\sqrt{x-1},x-3\}$ to $\sqrt{x-1}$. As can be seen from the picture, the maximum in question is $-\sqrt{x-1}$ from $1$ to the $2$, and $x-3$ from there on.
Since we are rotating about $y=-2$, we should consider $y+2$ instead of $y$. So for $x\in[1,2]$, $y+2\in[2-\sqrt{x-1},2+\sqrt{x+1}]$, while for $x\in[2,5]$, $y+2\in[x-1,2+\sqrt{x-1}]$.
To compute the volume of the solid obtained by rotating, we have to integrate the areas of the circular coronas that compose it. Breaking this into the two regions where the left extreme of $y$ has an analytical expression not involving a $\max$, we should get that our volume $V$ is:
$$V=\int\limits_1^2\pi((2+\sqrt{x-1})^2-(2-\sqrt{x-1})^2)\mathrm{d}x+\int\limits_2^5\pi((2+\sqrt{x-1})^2-(x-1)^2)\mathrm{d}x.$$
I guess you can take it from here.
For the curious, Wolfram (1 and 2) evaluates the two bits as:
$$V=\frac{16}{3}\pi+\frac{103}{3}\pi=\frac{16+103}{3}\pi=\frac{119}{3}\pi.$$
| {
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solve the following system of equations in real $x$, $y$ solve for real $x,y$
$$2^{x^2+y}+2^{x+y^2}=8 \tag{1}$$
$$\sqrt{x}+\sqrt{y}=2 \tag{2}$$
Trivially $x=y=1$ Now Equation $(1)$ can be written as
$$2^{x^2+(\sqrt{y})^2}+2^{x+(\sqrt{y})^4}=8$$ so we get
$$2^{x^2+(2-\sqrt{x})^2}+2^{x+(2-\sqrt{x})^4}=8$$ so
$$2^{x^2+4+x-4\sqrt{x}}+2^{x^2+25x+16-32\sqrt{x}-8x\sqrt{x}}=8$$
But i have no clue to proceed..
| We will show that the solution $(x,y)=(1,1)$ is unique.
$$2^{x^2+y}+2^{x+y^2}=8 \tag{1}\\ln(2^{x^2+y}+2^{x+y^2})=ln8⇒$$
We assume that $x^2+y>y^2+x$, (similarly you can assume that $x^2+y>y^2+x$ and reach the same result) and it follows that:
$$ln(2^{x^2+y}+2^{x+y^2})=\Big(ln2^{x^2+y}\Big(1+\frac{2^{y^2+x}}{2^{x^2+y}}\Big)\Big)=(x^2+y)ln2+ln\Big(1+\frac{2^{y^2+x}}{2^{x^2+y}}\Big)$$
And since $x^2+y>y^2+x$, we have:
$${\frac{2^{y^2+x}}{2^{x^2+y}}}<1$$
So, $$ln\Big(1+\frac{2^{y^2+x}}{2^{x^2+y}}\Big)<ln2⇒\\(x^2+y)ln2+ln\Big(1+\frac{2^{y^2+x}}{2^{x^2+y}}\Big)=ln8<(x^2+y)ln2+ln2⇒\\(x^2+y)+1>\frac{ln8}{ln2}=3$$
And we get,
$$x^2+y>2$$
and our second equation which we did not use till now,
$$\sqrt{x}+\sqrt{y}=2$$
You can show that these two lead to the solutions $(0,4)$ and $(4,0)$, which do not satisfy our initial condition.
Therefore it must hold that:$$x^2+y=y^2+x\\\sqrt{x}+\sqrt{y}=2$$
Solving this last system will involve a cubic polynomial with one real and two complex solutions. The real solution leads to $(x,y)=(1,1)$
| {
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Sum of two squares and the rearrangement Let $a,b,c,d$ be four nonzero integers such that $ab=cd$. If $a^2+b^2\ne c^2+d^2$, then what is the minimal value of $|(a^2+b^2)-(c^2+d^2)|$? Surely it must be bigger than or equal to $1$, but I guess it is strictly bigger than $1$. How can I prove(or disprove) this conjecture?
| You can write $|a^2+b^2-c^2-d^2|=|a^2+2ab+b^2-c^2-2cd-d^2|=|(a+b)^2-(c+d)^2|$ and note that the difference of squares is never $1$ unless they are $0,1$. To have them $0,1$ we would need $c=-d, a=1-b$ but then we would need $-c^2=b-b^2$ or $b(b-1)=c^2$, which requires $c=0$
Added: WOLOG we can ask $a+b \gt c+d$ and get rid of the absolute value signs. We can also require that all the variables be positive. Then $a^2+b^2-c^2-d^2=(a+b+c+d)(a+b-c-d)$ and we note the two factors have the same parity so the product cannot be $2$ or $6$. If the product is $3$, we must have $c+d=1$, which is not possible. If the product is $4$, both factors are $2$ and $c+d=0$. If the product is $5$, we have $c+d=2, c=d=1, ab \neq cd$. If the product is $7$, $c+d=3, c=2, d=1, ab \neq cd$. If the product is $8$, the factors must be $4,2$, $c+d=1$, again not possible. So $9$ is the smallest product and you have found an example.
| {
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The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$ ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$ $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and this is where i got stuck at. I tried to substitute $u =x^7$
but then the integral become $\int \frac{u^2}{28\sqrt{u+1}}du$
the final answer that I found using wolffram calculator is $\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)$
| Set $u=x^7$ and $du=7x^6dx$
$$=\frac 1 7 \int u \sqrt{u+1}du$$
Set $\nu=u+1$ and $d\nu=du$
$$\begin{align}
\frac 1 7\int(\nu-1)\sqrt{\nu}d\nu&=\frac 1 7\int\nu ^{3/2}d\nu -\frac 1 7\int\sqrt{\nu}d\nu\\\\
&=\frac{2}{35}(x^7+1)^{5/2}-\frac{2}{21}(x^7+1)^{3/2}+\mathcal C\\\\&=\color{red}{\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)+\mathcal C}\end{align}$$
| {
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If $p>3$ a prime number then $\binom {p-1}{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} 4^{p-1} \pmod {p^3}$ Here is one of Morley's theorem in number theory.
My idea is to begin in $\mathbb{Z}/p\mathbb{Z}$ :
$\binom {p-1}{\frac{p-1}{2}} = \frac{(p-1)!}{(\frac{p-1}{2})!(\frac{p-1}{2})!}=\frac{(p-1)!}{(-1)^{\frac{p-1}{2}}(p-1)!}=\frac{1}{(-1)^{\frac{p-1}{2}}}\equiv (-1)^{-\frac{p-1}{2}} \pmod p.$
Now, I want to build a solution for $p^2$ then for $p^3$ with maybe Hensel's lemma but I don't know if it's the right way or if it can work.
Thanks in advance !
| Just a partial answer, for now. By Wolstenholme's theorem we have $\binom{2p-1}{p-1}\equiv 1\pmod{p^3}$, and
$$\begin{eqnarray*} \binom{2p-1}{p-1} &=& \frac{\color{red}{(2p-1)\cdot (2p-3)\cdot\ldots\cdot(p+2)}}{\color{blue}{(p-2)\cdot(p-4)\cdot\ldots\cdot 1}}\cdot\binom{p-1}{\frac{p-1}{2}}\\ &=&(-1)^{\frac{p-1}{2}}\binom{p-1}{\frac{p-1}{2}}\prod_{k=1}^{\frac{p-1}{2}}\left(1-\frac{2p}{2k-1}\right)\tag{1}\end{eqnarray*} $$
but $q(z)=\prod_{k=1}^{\frac{p-1}{2}}\left(1-\frac{2z}{2k-1}\right)$ is a polynomial that fulfills $q(0)=1$ with roots at $z\in\left\{\frac{1}{2},\frac{3}{2},\ldots,\frac{p-2}{2}\right\}$. By Viète's theorem, the coefficient of $z$ in $p(z)$ just depends on:
$$ \frac{2}{1}+\frac{2}{3}+\ldots+\frac{2}{p-2} = 2H_{p-1}-H_{\frac{p-1}{2}}\tag{2}$$
while the coefficient of $z^2$ just depends on:
$$ \frac{1}{2}\left[\left(2H_{p-1}-H_{\frac{p-1}{2}}\right)^2-\left(4H_{p-1}^{(2)}-H_{\frac{p-1}{2}}^{(2)}\right)\right] \tag{3}$$
so Morley's theorem ultimately boils down to understanding the behaviour of $H_{\frac{p-1}{2}}\pmod{p^2}$ and $H_{\frac{p-1}{2}}^{(2)}\pmod{p}$, just like Wolstenholme's theorem is equivalent to $H_{p-1}\equiv 0\pmod{p^2}$ and $H_{p-1}^{(2)}\equiv 0\pmod{p}$.
| {
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"url": "https://math.stackexchange.com/questions/1679895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove $\forall x,y\in\mathbb{R} : x^3+x^2y=y^2+xy \Leftrightarrow y=x^2\lor y=-x?$ Let $x,y\in\mathbb{R}$
Assume $x^3+x^2y=y^2+xy$
Then $x^2(x+y)=y(x+y)$
Then either $(x+y)=0$ or $(x+y)\ne0$
Assume $x+y=0$
Then $y=-x$
Assume $(x+y)\ne0$
Then $y=x^2$
Then $x^3+x^2y=y^2+xy \Rightarrow y=x^2\lor y=-x$
Now assume $y=x^2\lor y=-x$
Assume $y=x^2$
Then $(x+y)y=(x+y)x^2$ #By definition
Then $xy+y^2=x^3+yx^2$
Then $x^3+x^2y=y^2+xy$
Assume $y=-x$
Then $y+x=0$
Also $x^2\times 0=y\times 0$ #By definition
Then $x^2(x+y)=y(x+y)$
Then $x^3+x^2y=y^2+xy$
Then $x^3+x^2y=y^2+xy \Leftarrow y=x^2\lor y=-x.$
Then $\forall x,y\in\mathbb{R} : x^3+x^2y=y^2+xy \Leftrightarrow y=x^2\lor y=-x.$
I am not sure if the proof is well structured. Could someone check?
| This is not feedback on your proof, but instead an alternative proof, which seems a lot shorter and simpler.$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\equiv}{\Leftrightarrow}
\newcommand{\then}{\Rightarrow}
\newcommand{\when}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$
Let's just calculate, for all real $\;x,y\;$:
$$\calc
x^3+x^2y=y^2+xy
\op\equiv\hint{LHS: factor out common $\;x^2\;$; RHS: factor out common $\;y\;$}
x^2(x+y) = y(y+x)
\op\equiv\hint{simplify: divide by $\;x+y\;$, avoiding division by zero}
x^2 = y \;\lor\; x+y=0
\op\equiv\hint{arithmetic -- to achieve the stated goal}
y = x^2 \;\lor\; y = -x
\endcalc$$
This completes the proof.
Note how this proves both directions at the same time, instead of having to do both $\;\then\;$ and $\;\when\;$.
It looks like you wanted to go more into detail for the middle step. Is this perhaps because you were given a list of real number axioms that you must use?
| {
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"url": "https://math.stackexchange.com/questions/1683089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Nested radical $\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ I am studying the $f(x) = \sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ for $x \in (0,\infty)$ and I am trying to get closed form formula for this, or at least some useful series/expansion. Any ideas how to get there?
So far I've got only trivial values, which are
$$f(1)=\sqrt{1+\sqrt{1+\cdots\sqrt{1}}}=\frac{\sqrt{5}+1}{2}$$
$$f(4) = 3$$
Second one follows from
$$2^n+1 = \sqrt{4^n+(2^{n+1}+1)} = \sqrt{4^n+\sqrt{4^{n+1}+(2^{n+2}+1)}} = \sqrt{4^n+\sqrt{4^{n+1}+\cdots}}$$
I have managed to compute several derivatives in $x_0=1$ by using chain rule recursively on $f_n(x) = \sqrt{x^n + f_{n+1}(x)}$, namely:
\begin{align*}
f^{(1)}(1) &= \frac{\sqrt{5}+1}{5}\\
f^{(2)}(1) &= -\frac{2\sqrt{5}}{25}\\
f^{(3)}(1) &= \frac{6\sqrt{5}-150}{625}\\
f^{(4)}(1) &= \frac{1464\sqrt{5}+5376}{3125}\\
\end{align*}
These gave me Taylor expansion around $x_0=1$
\begin{align*}
T_4(x) &= \frac{\sqrt{5}+1}{2} + \frac{\sqrt{5}+1}{5} (x-1) - \frac{\sqrt{5}}{25} (x-1)^2 + \frac{6\sqrt{5}-150}{3750} (x-1)^3 \\
&\ \ \ \ \ + \frac{61\sqrt{5}+224}{3125} (x-1)^4
\end{align*}
However this approach seems to be useful only very closely to the $x=1$. I am looking for something more general in terms of any $x$, but with my limited arsenal I could not get much further than this. Any ideas?
This was inspiring but kind of stopped where I did
http://integralsandseries.prophpbb.com/topic168.html
Edit:
Thanks for the answers, i will need to go through them, looks like the main idea is to divide by $\sqrt{2x}$, so then I am getting
$$\frac{\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}}{\sqrt{2x}} = \sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{16x}+\sqrt{\frac{1}{256x^4}+...}}}}$$
Then to make expansion from this. This is where I am not yet following how to get from this to final expansion.
| For small $x$ and $n>4$,
$$\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\ldots+\sqrt{x^{n}}}}}=
x^{n/2^{n}}+\frac{1}{2}x^{1-n/2^{n}}+\frac{1}{8}x^{2-n/2^{n}}+\ldots$$
Thus $$\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\ldots}}} \approx
1+\frac{x}{2}+\frac{x^{2}}{8}$$
For large $x$,
$$\sqrt{x+\sqrt{x^{2}+\sqrt{x^{3}+\ldots}}}=
\sqrt{2x}+\frac{1}{4\sqrt{2}}-\frac{5}{64\sqrt{2x}}+\frac{85}{512x\sqrt{2}}-\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
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} |
Finding Maximum Area of a Rectangle in an Ellipse
Question: A rectangle and an ellipse are both centred at $(0,0)$.
The vertices of the rectangle are concurrent with the ellipse as shown
Prove that the maximum possible area of the rectangle occurs when the x coordinate of
point $P$ is $x = \frac{a}{\sqrt{2}} $
What I have done
Let equation of ellipse be
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Solving for y
$$ y = \sqrt{ b^2 - \frac{b^2x^2}{a^2}} $$
Let area of a rectangle be $4xy$
$$ A = 4xy $$
$$ A = 4x(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) $$
$$ A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) + 4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}} \times \frac{-2b^2x}{a^2} \right) $$
$$ A'(x) = 4\sqrt{ b^2 - \frac{b^2x^2}{a^2}} + \frac{-8x^2b^2}{\sqrt{ b^2 - \frac{b^2x^2}{a^2}}a^2} = 0 $$
$$ 4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 8x^2b^2 = 0 , \sqrt{ b^2 - \frac{b^2x^2}{a^2}a^2} \neq 0 $$
$$ 4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 8x^2b^2 = 0 $$
$$ 4a^2b^2 - 4b^2x^2 - 8x^2b^2 = 0 $$
$$ 4a^2b^2 - 12x^2b^2 = 0 $$
$$ 12x^2b^2 = 4a^2b^2 $$
$$ x^2 = \frac{a^2}{3} $$
$$ x = \frac{a}{\sqrt{3}} , x>0 $$
Where did I go wrong?
edit:The duplicate question is the same but both posts have different approaches on how to solve it so I don't think it should be marked as a duplicate..
| Let equation of ellipse be
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Solving for y
$$ y = \sqrt{ b^2 - \frac{b^2x^2}{a^2}} $$
Let area of a rectangle be $4xy$
$$ A = 4xy $$
$$ A = 4x(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) $$
$$ A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) + 4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}} \times \frac{-b^2x}{a^2} \right) $$
$$ A'(x) = 4\sqrt{ b^2 - \frac{b^2x^2}{a^2}} + \frac{-4x^2b^2}{\sqrt{ b^2 - \frac{b^2x^2}{a^2}}a^2} = 0 $$
$$ 4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 4x^2b^2 = 0 , \sqrt{ b^2 - \frac{b^2x^2}{a^2}a^2} \neq 0 $$
$$ 4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 4x^2b^2 = 0 $$
$$ 4a^2b^2 - 4b^2x^2 - 4x^2b^2 = 0 $$
$$ 4a^2b^2 - 8x^2b^2 = 0 $$
$$ 8x^2b^2 = 4a^2b^2 $$
$$ x^2 = \frac{a^2}{2} $$
$$ x = \frac{a}{\sqrt{2}} , x>0 $$
The mistake is in third step while differentiating.
differentiating $\sqrt x$ will give you $\frac{1}{2\sqrt x}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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How to evaluate the integral $\int_0^{2\pi}\mathrm{d}\theta e^{ia\cos(\theta-\theta_1)}\cos^2(\theta-\theta_2)$ I have an integral:
$$
\int_{0}^{2\pi}
\mathrm{e}^{\mathrm{i}a\cos\left(\theta - \theta_{1}\,\right)}
\,\,\,\cos^{2}\left(\theta - \theta_{2}\right)\,
\mathrm{d}\theta,
$$
where $a, \theta_1$ and $\theta_2$ are reals. Any idea on how to evaluate this integral.
| Let $\theta \mapsto \theta + \theta_1$, so that the integral becomes
$$\frac12 \int_0^{2 \pi} d\theta \, e^{i a \cos{\theta}} \, \left [ 1+\cos{(2 (\theta+\theta_1-\theta_2))} \right ]$$
The first term produces
$$\frac12 \int_0^{2 \pi} d\theta \, e^{i a \cos{\theta}} $$
Let $z=e^{i \theta}$; then the integral is
$$-\frac{i}{2} \oint_{|z|=1} \frac{dz}{z} e^{i (a/2) \left ( z+z^{-1} \right )} $$
To deal with the essential singularity in the exponential, we form the Laurent series of the integrand:
$$-\frac{i}{2} \sum_{n=0}^{\infty} \frac{i^n}{n!} \oint_{|z|=1} \frac{dz}{z} \left ( \frac{a}{2} \right )^{n} \left ( z+\frac1z \right )^n $$
The only nonzero terms will be the constant term of the expansion of the binomial terms, and those only appear in the even terms. Thus, the integral is
$$-\frac{i}{2} i 2 \pi \sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n)!} \binom{2 n}{n} \left ( \frac{a}{2} \right )^{2 n} = \pi \sum_{n=0}^{\infty} \frac{(-1)^n}{(n!)^2} \left ( \frac{a}{2} \right )^{2 n} = \pi \, J_0(a)$$
The next term may be split up as $\cos{(2 (\theta+\theta_1-\theta_2))} = \cos{2 \theta} \cos{2 (\theta_1-\theta_2)} - \sin{2 \theta} \sin{2 (\theta_1-\theta_2)} $.
$$\begin{align} \frac12 \int_0^{2 \pi} d\theta \, e^{i a \cos{\theta}} \cos{2 \theta} &= -\frac{i}{4} \oint_{|z|=1} \frac{dz}{z} e^{i (a/2) \left ( z+z^{-1} \right )} \left (z^2+\frac1{z^2} \right )\\ &= -\frac{i}{4} \sum_{n=0}^{\infty} \frac{i^n}{n!} \oint_{|z|=1} \frac{dz}{z} \left ( \frac{a}{2} \right )^{n} \left ( z+\frac1z \right )^{n} \left (z^2+\frac1{z^2} \right ) \\ &= -\frac{i}{4} i 2 \pi \sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n)!} \left [ \binom{2 n}{n-1} + \binom{2 n}{ n+1}\right ]\left ( \frac{a}{2} \right )^{2 n} \\ &= \pi \sum_{n=0}^{\infty} \frac{(-1)^n}{(n-1)! (n+1)!} \left ( \frac{a}{2} \right )^{2 n} \\ &= -\pi \left ( \frac{a}{2} \right )^{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{n! (n+2)!} \left ( \frac{a}{2} \right )^{2 n} \\ &= -\pi J_2(a)\end{align}$$
Note that the integral involving $\sin{2 \theta}$ has, upon expansion of the exponential, terms such as $\binom{2 n}{n-1} - \binom{2 n}{ n+1}$, which are all zero. Thus,
$$\int_0^{2 \pi} d\theta \, e^{i a \cos{(\theta-\theta_1)}} \, \cos^2{(\theta-\theta_2)} = \pi J_0(a) - \pi J_2(a) \cos{2 (\theta_1-\theta_2)} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$ where $x$, $y$, $z$ are angles of a triangle Can any one give me a hint to find value of $x$
where:
$$\frac{\sin x}{4}=\frac{\sin y}{3}=\frac{\sin z}{2}$$
and $x$, $y$, $z$ are angles of a triangle.
I tried to use sine law but got nothing.
| Unless $x, y, z$ represent the angles of a triangle, there'll be infinitely many solutions.
If three sides of a triangle are given, it's better to use cosine law:
\begin{align*}
\cos x &= \frac{3^{2}+2^{2}-4^{2}}{2(3)(2)} \\
\cos y &= \frac{4^{2}+2^{2}-3^{2}}{2(4)(2)} \\
\cos z &= \frac{3^{2}+4^{2}-2^{2}}{2(3)(4)} \\
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ Solve this equation :
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
Such that $a+b+c=\pi$
I don't have any idea. I can't try anything.
| Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^2B+\cos^2C-1=\cos^2B-\sin^2C=\cos(B+C)\cos(B-C)$$
$$x^2+2x\cos B\cos C+\cos^2B+\cos^2C-1=0$$
$$\iff x^2+x\{\cos(B+C)+\cos(B-C)\}+\cos(B+C)\cos(B-C)=0$$
Now use $y^2+(a+b)y+ab=(y+a)(y+b)$ and $\cos(B+C)=\cos(\pi-A)=-\cos A$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find x for $\sqrt{(5x-1)}+\sqrt{(x-1)}=2$ Solve:
$$\sqrt{(5x-1)}+\sqrt{(x-1)}=2$$
When $x=1$, we get the following equation to equal to $2$
I've been trying to solve this problem but when I square both sides and simplify I end up with:
$$x^2+6x+2=0$$ and of course $x=1$ cannot be a solution. So im not sure what im doing wrong. Any help on this problem?
| Yes, squaring both sides gives:
$$ (5x - 1) + 2\sqrt{5x-1}\sqrt{x-1} + (x-1) = 4 $$
Then:
$$ (5x - 1) + 2\sqrt{(5x-1)(x-1)} + (x-1) = 4 $$
Which simplifies to:
$$ (5x - 1) + 2\sqrt{(5x^2-6x+1)} + (x-1) = 4 $$
And thus:
$$ \sqrt{(5x^2-6x+1)} = 3 - 3x $$
Now squaring both sides again:
$$ 5x^2-6x+1 = 9- 18x +9x^2 $$
Which gives the quadratic equation:
$$4x^2 -12x +8 = 0$$
And thus:
$$4(x^2 -3x +2) = 0$$
The potential solutions are thus the roots of $x^2 -3x +2$, i.e. $x=1$ or $x=2$. Plugging these values into the original equation we see that only $x=1$ is a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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how to show that $\tan^2 y=-\csc 2x$ If $\csc y=\sin x -\cos x$ how to show that $\tan^2 y=-\csc 2x$
Can anyone explain to me? What identity I should use?
| Question: Show $tan^2(y) = -csc(2x)$, given that $csc(y) = sin(x) - cos(x)$.
Identities/Knowledge:
*
*$csc(\theta) = \frac {1}{sin(\theta)}$
*$sin(2\theta) = 2sin(\theta)cos(\theta)$
*$cos^2(\theta) + sin^2(\theta) = 1$
*$tan(\theta) = \frac {1}{csc(\theta)cos(\theta)}$
Working Out:
Lets' begin with what the right hand side is equal too:
$$R.H.S = -csc(2x) = -\frac {1}{sin(2x)}$$
That was simple, now we move onto the left hand side (the fun part, I'll annotate as I go on):
$$L.H.S = \frac {sin^2(y)}{cos^2(y)} = (\frac {1}{csc(y)cos(y)})^2 $$
$$= \frac {1}{csc^2(y)cos^2(y)}$$
But $csc(y) = sin(x) - cos(x)$, so (equation 1);
$$ \frac {1}{(sin(x) - cos(x))^2 cos^2(y)}$$
Note that $cos^2(\theta) = 1 - sin^2(\theta)$
$$cos^2(y) = ? = 1 - (\frac {1}{sin(y)})^{-2} = 1 - (\frac {1}{\frac {1}{sin(y)}})^2$$
But $\frac {1}{sin(y)} = csc(y)$
$$ cos^2(y) = 1 - (\frac {1}{sin(x) - cos(x)})^2$$
Sub $cos^2(y)$ into equation 1:
$$\frac {1}{(sin(x) - cos(x))^2 (1 - \frac {1}{(sin(x) - cos(x))^2})}$$
$$ = \frac {1}{(sin(x) - cos(x))^2 (\frac {(sin(x) - cos(x))^2 -1}{(sin(x) - cos(x))^2})}$$
Let $a = sin(x) - cos(x)$
$$\frac {1}{\frac {a^2(a^2-1)}{a^2}}$$
The $a^2$ cancels out, so you're left with:
$$\frac {1}{a^2 - 1}$$
But $a = sin(x) - cos(x)$
$$\frac {1}{(sin(x) - cos(x))^2 - 1}$$
Note that $(sin(x) - cos(x))^2 = cos^2(x) + sin^2(x) - 2sin(x)cos(x)$, but because of the identities listed above we can simplify it to; $1 - sin(2x)$.
$$\therefore \frac {1}{1 - sin(2x) - 1} = - \frac {1}{sin(2x)} = R.H.S$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Radius of convergence - Complex plane A example in my textbook explain that $x^2-2x+2=0$ has the solutions $x = 1± i$. The distance between $x=0$ and $x = 1± i$ is $\sqrt{2}$ in the complex plane. So the radius of convergence of the Taylor series for $(x^2-2x+2)^{-1}$ is $\sqrt{2}$ around $x=0$.
I looked for why this result is true, but I didn't find it. Is anyone could explain to me why this is true or just give me the theorem related?
| Hint: Compute the partial fraction decomposition of $\frac{1}{x^2-2x+2}$ and consider the radius of convergence of the power series for $\frac{1}{1-x}$.
More precisely, we can write
$$
\frac{1}{x^2-2x+2}=\frac{A}{x-(1+i)}+\frac{B}{x-(1-i)}.
$$
Using standard techniques from calculus, one can find the constants $A$ and $B$.
Now, the power series expansion of
$$
\frac{A}{x-(1+i)}
$$
can be computed as follows:
$$
\frac{A}{x-(1+i)}=\frac{-A}{1+i}\cdot\frac{1}{1-\frac{x}{1+i}}.
$$
Using the substitution $y=\frac{x}{1+i}$, we have that
$$
\frac{A}{x-(1+i)}=\frac{-A}{1+i}\cdot\frac{1}{1-y}
$$
which is a common function from calculus whose power series centered at $y=0$ converges for $|y|<1$. By substitution, $|x|<|1+i|=\sqrt{2}$. Approach the other fraction similarly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $\tan(x) = 2\sqrt2 $ , $ x\in[ \pi , \frac{3\pi}{2}] $ , What is the exact value of $\sin(3x)$?
Given $\tan(x) = 2\sqrt2 $ , $ x\in[ \pi , \frac{3\pi}{2}] $
What is the exact value of $\sin(3x)$?
What I have done:
Given $\tan(x) = 2\sqrt2 $ , I drew a right angled triangle and found the hypotenuse to be $3$ so $\sin(x) = \frac{2\sqrt2}{3}$
Recalling that $$\sin(3x) = 3\sin(x) - 4\sin^3(x)$$
Subbing in $\sin(x) = \frac{2\sqrt2}{3}$ and $\sin^3(x) = \frac{16\sqrt2}{27}$
$$\sin(3x) = 3(\frac{2\sqrt2}{3}) - 4(\frac{16\sqrt2}{27})$$
$$ \sin(3x) = {2\sqrt2} - \frac{64\sqrt2}{27} $$
$$ \sin(3x) = \frac{-10\sqrt2}{27} $$
Is this correct? If so would there be any other way I could have gone about solving this?
| Your strategy is correct, but you failed to take into account that $x \in \left[\pi, \frac{3\pi}{2}\right] \implies \sin x < 0$.
As for an alternative method:
We use the identity
$$\sin(3x) = 3\cos^2x\sin x - \sin^3x$$
Using the identity $\sec^2x = \tan^2x + 1$ yields
$$\sec^2x = (2\sqrt{2})^2 + 1 = 8 + 1 = 9 \implies \cos^2x = \frac{1}{\sec^2x} = \frac{1}{9}$$
Using the Pythagorean identity $\sin^2x + \cos^2x = 1$ yields
$$\sin^2x = 1 - \cos^2x$$
Since $x \in \left[\pi, \frac{3\pi}{2}\right]$, $\sin x < 0$. Hence,
$$\sin x = -\sqrt{1 - \cos^2x} = -\sqrt{1 - \frac{1}{9}} = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3}$$
Substitution into the identity $\sin(3x) = 3\cos^2x\sin x - \sin^3x$ yields
$$\sin(3x) = 3\left(\frac{1}{9}\right)\left(-\frac{2\sqrt{2}}{3}\right) - \left(-\frac{2\sqrt{2}}{3}\right)^3 = -\frac{6\sqrt{2}}{27} + \frac{16\sqrt{2}}{27} = \frac{10\sqrt{2}}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
indefinite integration $ \int \frac { x^2 dx} {x^4 + x^2 -2}$ problem : $ \int \frac { x^2 dx} {x^4 + x^2 -2}$
solution : divide numerator and denominator by $x^2$
$ \int \frac { dx} {x^2 + 1 -\frac{1}{x^2}}$
Now whats the next step $?$
Am I doing right $?$
| Hint: $x^4+x^2-2$ can be factored as $(x-1)(x+1)(x^2+2)$. then find $A,B,C$, and $D$ which are constants, such that
$$
\frac{x^2}{x^4+x^2-2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+2}
$$
and integrate termwise.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the expected value of the largest of the three dice rolls?
You toss a fair die three times. What is the expected value of the largest of the three outcomes?
My approach is the following:
calculate the probability of outcome when $\max=6$, which is
$$P(\text{at least one $6$ of the three rolls}) = 1 - P(\text{no }6) = 1 - (5/6)^3$$
and then calculate the probability of outcome when $\max=5$, which is
$$P(\text{at least one $5$ of the three rolls & $5$ is max}) = 1 - P(\text{no $5$ & $5$ is max}) = 1 - (4/6)^3.$$
I wonder if this approach is right.
| Let X denote the largest value, then:
*
*$P(X=1)=\frac{ 1}{216}$
*$P(X=2)=\frac{ 7}{216}$
*$P(X=3)=\frac{19}{216}$
*$P(X=4)=\frac{37}{216}$
*$P(X=5)=\frac{61}{216}$
*$P(X=6)=\frac{91}{216}$
Hence the expected value is:
$$1\cdot\frac{1}{216}+2\cdot\frac{7}{216}+3\cdot\frac{19}{216}+4\cdot\frac{37}{216}+5\cdot\frac{61}{216}+6\cdot\frac{91}{216}=\frac{119}{24}$$
Details:
Let $C_n$ denote the number of combinations with largest value $n$.
Observe that $C_n=n^3-\sum\limits_{k=1}^{n-1}C_{n-1}$, therefore:
*
*$C_1=1^3=1$
*$C_2=2^3-1=7$
*$C_3=3^3-1-7=19$
*$C_4=4^3-1-7-19=37$
*$C_5=5^3-1-7-19-37=61$
*$C_6=6^3-1-7-19-37-61=91$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 0
} |
The right way to cancel out the terms in the following telescoping series So how do I cancel and simplify the terms in the following telescopic series.
Been at it for hours, cant seem to figure it out.
$\sum\limits_{k = 1}^n \frac{1}{2(k+1)} -\frac{1}{k+2}+\frac{1}{2(k+3)} $
Any help would be deeply appreciated.
P.S: I need to show that its equal to the following,
$\frac{1}{12} - \frac{1}{2(n+2)} + \frac{1}{2(n+3)}$
But I cant seem to figure out the right cancellation method.
| We have:
$$
a_n=\sum\limits_{k = 1}^n \left(\frac{1}{2(k+1)} -\frac{1}{k+2}+\frac{1}{2(k+3)}\right)= a_{n-1}+ \left(\frac{1}{2(n+1)} -\frac{1}{n+2}+\frac{1}{2(n+3)}\right)=a_{n-1}+\frac{1}{(n+1)(n+2)(n+3)}
$$
Does this solve your problem?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$ Problem : Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$
My Approach : I would assume that we can prove by induction.
Base case $n=2$.
$$=2{2 \choose 2}+ {2 \choose 1}= (2\cdot 1)+2 =4$$
$$n^2 =2^2 = 4.$$
Assume for $n\ge 2$, $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$ for $n \le k$.
Let $n=k+1$
$$2{k+1 \choose 2}+{k+1 \choose 1} = (k+1)^2$$
And, that's as far as I got.
I get stuck with the ${k+1 \choose 2}$ part.
| We can also give this a combinatorial interpretation:
*
*$\binom{n}{2}$ is the number of unordered pairs of distinct elements of $\underline{n}$
*So $2\binom{n}{2}$ is the number of ordered pairs of distinct elements of $\underline{n}$
*So $2\binom{n}{2}+\binom{n}{1}$ is the number of ordered pairs of $\underline{n}$
*So $2\binom{n}{2}+\binom{n}{1}=|\underline{n}^2|$
*So $2\binom{n}{2}+\binom{n}{1}=n^2$
With enough care, this can be made into a rigorous proof (these are called "combinatorial proofs").
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Number of common roots of $x^3 + 2 x^2 +2x +1 = 0$ and $x^{200} + x^{130} + 1 = 0 $ The equations $x^3 + 2 x^2 +2x +1 = 0$ and $x^{200} + x^{130} + 1 = 0 $ have
*
*exactly one common root;
*no common root;
*exactly three common roots;
*exactly two common roots.
I factored the first equation. I think the roots are $-1$, $\omega$ and $\omega^2$.
| $$x^3+2x^2+2x+1=(x+1)(x^2+x+1)$$
The roots are $-1,\omega,$ and $\omega^2$, where $\omega,\omega^2$ are non real cube roots of unity.
Substituting in the other equation,
$$(-1)^{200}+(-1)^{130}+1=1\ne0$$
$$\omega^{200}+\omega^{130}+1=\omega^2+\omega+1=0$$
$$(\omega^2)^{200}+(\omega^2)^{130}+1=\omega+\omega^2+1=0$$
Thus, two roots are common.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1700560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove the sequence converge and find its limit Consider the following sequence:
$a_{n+2} = \frac{a_{n+1}+a_{n}}{2}$, $a_{1}$ and $a_{2}$ are given.
Write $a_{n}$ as a function of $a_{1}$ and $a_{2}$ and show that its limit is $\frac{a1 + 2a_{2}}{3}$
I think I am loosing myself on algebra here. I can't even do the first part. Any help is welcome, thanks a lot!
| $A = \pmatrix{1/2&1/2\\1&0}$
$\pmatrix{a_{n}\\a_{n-1}}=A\pmatrix{a_{n-1}\\a_{n-2}}$
$\pmatrix{a_{n+2}\\a_{n+1}}=A^n\pmatrix{a_2\\a_1}$
$A = PDP^{-1}; A^n =PD^nP^{-1}$
$A = \pmatrix{1&1\\1&-2}\pmatrix{1&0\\0&-1/2}\pmatrix{2/3&1/3\\1/3&-1/3}$
$A^n = \pmatrix{1&1\\1&-2}\pmatrix{1&0\\0&-2^{-n}}\pmatrix{2/3&1/3\\1/3&-1/3}$
Limit as $n\to \infty
= \pmatrix{1&1\\1&-2}\pmatrix{1&0\\0&0}\pmatrix{2/3&1/3\\1/3&-1/3}$
$\pmatrix{2/3&1/3\\2/3&1/3}$
$\lim\limits_{n\to\infty} a_n = 2/3a_2 + 1/3 a_1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of the Powers of $2$ Suppose I have a sequence consisting of the first, say, $8$ consecutive powers of $2$ also including $1$: $1,2,4,8,16,32,64,128$. Why is it that for example, $1 + 2 + 4 = 7$ is $1$ less than the next term in the series, $8$? Even if one was to try, for instance, $524,288 + 1,048,576$ ($2^{19}$ and $2^{20}$), the next term ($2^{21}$) will be $2,097,152$, which is one greater than the previous sum. I found this whilst solving the "if you are given a penny each day and it doubles" problem. Can someone please explain this to me? Thank you.
| If you think about the list of the positive integers written in base 2, it goes like this:
$$\begin{align*}
1&=1\cdot 2^0 &=1_2\\
2&=1\cdot 2^1 + 0\cdot 2^0 &=\textbf{10}_2\\
3&=1\cdot 2^1+1\cdot 2^0 &=11_2\\
4&=1\cdot 2^2+0\cdot 2^1+0\cdot 2^0 &=\textbf{100}_2\\
5&=1\cdot 2^2+0\cdot 2^1+1\cdot 2^0 &=101_2\\
6&=1\cdot 2^2+1\cdot 2^1+0\cdot 2^0 &=110_2\\
7&=1\cdot 2^2+1\cdot 2^1+1\cdot 2^0 &=111_2\\
8&=1\cdot 2^3+0\cdot 2^2+0\cdot 2^1+0\cdot 2^0 &=\textbf{1000}_2\\
9&=1\cdot 2^3+0\cdot 2^2+0\cdot 2^1+1\cdot 2^0 &=1001_2\\
\vdots&=\vdots &\vdots\\
2^{n}-1&=1\cdot2^{n-1}+1\cdot 2^{n-2}+\cdots + 1\cdot 2^0 &=111\ldots 1_2\\
2^n&=1\cdot 2^n + 0\cdot 2^{n-1} + \cdots + 0\cdot 2^0 &=\textbf{100}\ldots\textbf{00}_2
\end{align*}$$
The number of digits in the expression in base 2 changes exactly at the powers of 2, and right before changing the number (in base 2) looks like a sequence of 1's, which corresponds to the sum $2^0+\ldots+2^{n-1}$ for some $n$.
Note that a similar idea will work for any base $b$.
| {
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"url": "https://math.stackexchange.com/questions/1704740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
What's wrong with my infinite series expansion for $\log(x)$? Here, log is natural log.
Looking at $f(x)=\frac{1}{x}$, I tried to put $f(x)$ in the form $\frac{a}{1-r}$ that an infinite geometric series $\sum_{n=0}^\infty (a \cdot r^n)$ converges to when $\mid r \mid < 1$.
That gave me $f(x) = \frac{1}{1-(1-x)}$, so that $\sum_{n=0}^\infty (1-x)^n$ should converge to $\frac{1}{x}$ for $0<x<2$.
Now, taking the anti-derivative of $\frac{1}{x}$ gives me $\log(x) + C$, and anti-differentiating $\sum_{n=0}^\infty (1-x)^n$ should give me a power series expansion for $\log(x)$, no?
But I end up with $$\int 1 + (1-x) + (1-x)^2 + ...dx$$
$$= x + x - \frac{1}{2}x^2 + x - x^2 + \frac{1}{3}x^3-...+C$$
And this doesn't look right.
| Your problem is from expanding all the terms in parentheses; while it's not wrong, it's obfuscating the point, which is to end up with an expression involving powers of $x - 1$.
So don't expand. Get
$$C + x - \frac{(1 - x)^2}{2} - \frac{(1 - x)^3}{3} - ...$$
I'll leave it to you to check that $C = -1$, so the series is
$$- \left((1 - x) + \frac{(1 - x)^2}{2} + \frac{(1 - x)^3}{3} + \dots\right)$$
After rearranging, this is
$$(x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \dots$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve the following trigonometric inequalities I'm solving the following inequality:
$$\cos(2x+\frac \pi 3) \ge - \frac 1 2 \text{ with } -\frac \pi 2<x<\frac \pi 2$$
I found the following range of solutions:
$$\frac 4 3 \pi \leq2x+\frac \pi 3 \leq \frac 2 3 \pi$$
That I simplified with
$$\frac \pi 3 \leq 2x\leq\pi \rightarrow \frac \pi 6 \leq 2x<\frac \pi 2$$
Though, my book suggests the following solutions:
$$-\frac \pi 2< x \leq \frac \pi 6$$
Any hints on why my solution is wrong?
| Let $\theta = 2x + \dfrac{\pi}{3}$.
Observe that the requirement that $x$ satisfies the inequalities
$$-\frac{\pi}{2} < x < \frac{\pi}{2}$$
implies that
$$\theta = 2x + \frac{\pi}{3} > 2\left(-\frac{\pi}{2}\right) + \frac{\pi}{3} = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3} \tag{1}$$
and that
$$\theta = 2x + \frac{\pi}{3} < 2\left(\frac{\pi}{2}\right) + \frac{\pi}{3} = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \tag{2}$$
Combining inequalities 1 and 2 yields
$$-\frac{2\pi}{3} < \theta < \frac{4\pi}{3}$$
Hence, the inequality
$$\cos\left(2x + \frac{\pi}{3}\right) \geq -\frac{1}{2}, -\frac{\pi}{2} < x < \frac{\pi}{2} \iff \cos\theta \geq -\frac{1}{2}, -\frac{2\pi}{3} < \theta < \frac{4\pi}{3}$$
In the interval $(-2\pi/3, 4\pi/3)$,
$$\cos\theta \geq -\frac{1}{2} \implies -\frac{2\pi}{3} < \theta \leq \frac{2\pi}{3}$$
Substituting $2x + \dfrac{\pi}{3}$ for $\theta$ yields
\begin{align*}
-\frac{2\pi}{3} & < 2x + \frac{\pi}{3} \leq \frac{2\pi}{3}\\
-\pi & < 2x \leq \frac{\pi}{3}\\
-\frac{\pi}{2} & < x \leq \frac{\pi}{6}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $x,y,z>0$ and $x+y+z=1$, then find the least value of ${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$ Let $x,y,z>0$ and $x+y+z=1$, then find the least value of
$${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$
I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am not good at inequalities. Please help me.
I wrote $x$ as $1-(y+z)$ and I took $x+y$ as $a$ and the others as $b$ and $c$. And I am trying.
| You can also use Cauchy-Schwarz:
$$\left(x(2-x)+y(2-y)+z(2-z)\right)\left(\frac{x}{2-x}+\frac{y}{2-y}+\frac{z}{2-z}\right)\ge (x+y+z)^2=1.$$
Furthermore,
$$\begin{aligned}x(2-x)+y(2-y)+z(2-z) &= 2(x+y+z)-(x^2+y^2+z^2)\\
&\le 2-\frac{(x+y+z)^2}{3}=\frac53.
\end{aligned}$$
Thus the minimum is $3/5$ attained at $x=y=z=1/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1708395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Finding the error in an evaluation of the limit $\lim_{x\to0} \frac{e^x-x-1}{x^2} $ \begin{align}
\lim_{x\to0} \frac{e^x-x-1}{x^2}
&= \lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} \\
&= \lim_{x\to0} \frac{e^x-1}{x}\lim_{x\to0} \frac{1}{x} - \lim_{x\to0} \frac{1}{x} \\ &= \lim_{x\to0} \frac{1}{x} - \lim_{x\to0} \frac{1}{x} \\
&= 0
\end{align}
| The error starts here:
$$\lim_{x\to0} \frac{e^x-x-1}{x^2} $$ $$\color{red}{\neq\lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x}
=\dots}$$
See Basic Limit Laws
Hint: Applaying L'Hopital twice gives $\frac 1 2 \lim\limits_{x\to 0} e^x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Inequality with complex numbers involving modules Let $z_1,z_2,z_3 \in \Bbb C$ so as $|z_1|=|z_2|=|z_3|=1$ and let $a=|z_1-z_2|, b=|z_2-z_3|, c=|z_3-z_1|$. Show, using algebraic methods, that:
$$ \frac {1} {(a+b-c)^2}+ \frac {1} {(b+c-a)^2}+ \frac {1} {(a+c-b)^2} \ge 1.$$
| This solution is a bit geometric. I'll use the fact that $a,b,c$ are the sides of a triangle inscribed in a circle of radius $1$. It's clear that $$a+b+c\le 3\sqrt3.$$ Now call $a+b-c=x$, $a+c-b=y$, $b+c-a=z$. We have that $x,y,z>0$ and $x+y+z=a+b+c$. By Cauchy-Schwarz and AM-HM
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{3}\ge\frac{3^3}{(x+y+z)^2}\ge\frac{3^3}{(3\sqrt3)^2}=1$$
and this is the desired inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\csc a +\cot a = \cot\frac{a}{2}$
Prove: $$\csc a + \cot a = \cot\frac{a}{2}$$
All I have right now, from trig identities, is
$$\frac{1}{\sin a} + \frac{1}{\tan a} = \frac{1}{\tan(a/2)}$$
Where do I go from there?
| We start with the following identities: $\quad\sin(2a) = 2\sin a \cos a\quad$ $\quad\cos(2a) = 1-2\sin^2 a\quad$
We solve these to get the half-angle identities: $\quad \sin(a) = 2\sin \frac a2 \cos \frac a2\quad$$\quad\sin^2 \frac a2 = \frac 12 (1-\cos a)$
We now tackle the problem
$$\frac{1}{\sin a} + \frac{\cos a}{\sin a} = \frac{\cos \frac a2}{\sin \frac a2}$$
multiplying out both sides we get that
$$2\sin^2 \frac a2(\cos a + 1) =2\sin \frac a2\cos \frac{a}{2} \sin a$$
Using the identities above we get that
$$(1-\cos a)(1+\cos a) = \sin^2 a$$
$$\implies 1 - \cos^2 a = \sin^2 a$$
$$\implies 1 = \sin^2 a + \cos^2 a$$
We now have a trivial trigonometric identity, so the equivalence is proved
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$ I am trying to prove that
$$I=\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$$
where $\beta(s)$ is the Dirichlet Beta function and $G$ is the Catalan's constant. I managed to derive the following series involving polygamma functions but it doesn't seem to be of much help.
$$
\begin{align*}
I &=\frac{1}{64}\sum_{n=0}^\infty \frac{\psi_2 \left(\frac{n}{2}+1 \right) -\psi_2\left(\frac{n+1}{2} \right)}{2n+1} \\
&= \frac{1}{8}\sum_{n=1}^\infty \frac{\psi_2(n)}{2n-1}-\frac{1}{32}\sum_{n=1}^\infty\frac{\psi_2\left(\frac{n}{2}\right)}{2n-1}
\end{align*}
$$
Numerical calculations show that $I \approx 0.235593$.
| I was able to solve this problem on my own.
Using integration by parts,
$$\begin{align*}
&\; \int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx \\
&= -2\int_0^1 \frac{\log(x)\tan^{-1}(x)\tanh^{-1}(x)}{x}dx-\int_0^1 \frac{\log^2(x)\tan^{-1}(x)}{1-x^2}dx \tag{1}
\end{align*}$$
I posted the solution to both these integrals on another forum. Here are the links:
*
*http://integralsandseries.prophpbb.com/topic711.html#p3975
*http://integralsandseries.prophpbb.com/topic245.html#p1680
$$\begin{align*}\int_0^1\frac{\log(x)\tan^{-1}(x)\tanh^{-1}(x)}{x}dx &= \frac{\pi^2}{16}G-\frac{7\pi\zeta(3)}{32} \tag{2}\\
\int_0^1\frac{\log^2(x)\tan^{-1}(x)}{1-x^2}dx &= -\beta(4)-\frac{\pi^2}{24}G+\frac{7\pi}{16}\zeta(3)\tag{3}
\end{align*}$$
$G$ denotes the Catalan's constant and $\beta(4)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}$. Substituting these two results in equation (1) gives:
$$\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G \tag{4}$$
Proof sketch of integrals (2) and (3) : (Please see the above links for a more detailed answer)
The idea behind evaluating (2) and (3) is breaking them down into Euler Sums. Using the taylor series expansion $\tan^{-1}(x)=\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{2n+1}$ and integrating term-wise, we obtain the following relations:
\begin{align*}
\int_0^1\frac{\log(x)\tan^{-1}(x)\tanh^{-1}(x)}{x}dx &= -\log(2)\frac{\pi^3}{32}-\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}\left( \gamma+\psi_0(n+1)\right) \\ &\;+\frac{1}{4}\sum_{n=0}^\infty \frac{(-1)^n \psi_1(n+1)}{(2n+1)^2} \tag{5}\\
\int_0^1 \frac{\log^2(x)\tan^{-1}(x)}{1-x^2}dx &=-\frac{1}{8}\sum_{n=0}^\infty\frac{(-1)^n\psi_2(n+1)}{2n+1}\tag{6}
\end{align*}
These Euler Sums can be evaluated using the techniques shown in the paper "Euler Sums and Contour Integral Representations" by Philippe Flajolet and Bruno Salvy. Here is it's link.
\begin{align*}
\sum_{n=0}^\infty\frac{(-1)^n\psi_2(n+1)}{2n+1} &= 8\beta(4)+\frac{\pi^2}{3}G-\frac{7\pi}{2}\zeta(3) \\
\sum_{n=0}^\infty\frac{(-1)^n\psi_1(n+1)}{(2n+1)^2} &= 6\beta(4)+\frac{\pi^2}{4}G-\frac{7\pi}{4}\zeta(3) \\
\sum_{n=0}^\infty \frac{(-1)^n\left( \gamma+\psi_0(n+1)\right)}{(2n+1)^3} &= 3\beta(4)-\frac{7\pi}{16}\zeta(3)-\frac{\pi^3}{16}\log(2)
\end{align*}
Substituting these into equations (5) and (6) gives us the integrals (2) and (3).
A related integral
Using similar techniques, we can show that
$$\displaystyle \int_0^1 \frac{\log^2(x)\tan^{-1}(x)}{x\left(1-x^2 \right)}dx=\beta(4)+\frac{7\pi \zeta(3)}{64}+\frac{\pi^3 \log(2)}{32}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
} |
When is matrix $A$ diagonalizable? I got the following matrix:
$$ A =
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & 0 \\
1 & 2 & 1 \\
\end{pmatrix}
$$
I need to answer when this matrix is diagonalizable.
Its characteristic polynomial is $ t(t-a)(t-1) $. So its 3 eigenvalues are 0, 1 and a. Both the algebraic and geometry multiplicities of those values are 1 (for all of them).
Let's look at the matrices for those eigenvalues:
$$ A - 0I =
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & 0 \\
1 & 2 & 1 \\
\end{pmatrix}
$$
$$ A - I =
\begin{pmatrix}
a -1 & 0 & 0 \\
b & -1 & 0 \\
1 & 2 & 0 \\
\end{pmatrix}
$$
$$ A - aI =
\begin{pmatrix}
0 & 0 & 0 \\
b & -a & 0 \\
1 & 2 & 1 - a \\
\end{pmatrix}
$$
$\rho (A - 0I) = 2 $
$\rho (A - 1I) = 2 $
$\rho (A - aI) = 2 $
It seems that for every $a$ and $b$ this matrix would be diagonalizable.
But it's not. Where am I wrong?
| Your matrix has at least $2$ eigenvalues, namely $0$ and $1$, and maybe a third, namely$~a$ it it is different from those two others. In the latter case we have $3$ simple roots of the characteristic polynomial, and $A$ is automatically diagonalisable. So the remaining interesting case is $a\in\{0,1\}$.
In that case $A$ is diagonalisable if and only if the polynomial $(X-0)(X-1)=X^2-X$ annihilates $A$, in other words if $A^2-A=0$ (the kernel of $A^2-A$ is the sum of the eigenspaces of $\lambda=0$ and $\lambda=1$). Now
$$
A^2-A =
\begin{pmatrix}
a^2-a & 0 & 0 \\
b(a-1) & 0 & 0 \\
a+2b & 0 & 0 \\
\end{pmatrix}
$$
so in the remaining cases $A$ is diagonalisable only if either $a=0$ and $b=0$, or $a=1$ and $b=-\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Three questions about the form $X^2 \pm 3Y^2 = Z^3$ and a related lemma In Ribenboim’s Fermat’s Last Theorem for Amateurs, he gives the following lemma [Lemma 4.7, pp. 30–31].
Lemma. Let $E$ be the set of all triples $(u, v, s)$ such that $s$ is odd, $\gcd(u,v) = 1$ and $s^3 = u^2 + 3v^2$. Let $F$ be the set of all pairs $(t,w)$ where $\gcd(t, w) = 1$ and $t \not\equiv w\!\pmod{2}$. The mapping $\Phi : F \rightarrow E$ given by $\Phi (t, w) = (u, v, s)$ with
\begin{cases}
\, u = t(t^2−9w^2), \\
\, v = 3w(t^2-w^2),\\
\, s = t^2 + 3w^2,
\end{cases}
is onto $E$.
Q1: Am I understanding “onto” correctly by interpreting this to mean that this is a complete integer parameterization of the form $X^2 + 3Y^2 = Z^3$?
Q2: Is there a similar solution for the form $X^2-3Y^2 = Z^3$? It appears that
\begin{cases}
\, x = t(t^2+9w^2), \\
\, y = 3w(t^2+w^2),\\
\, z = t^2 - 3w^2,
\end{cases}
satisfies, but I want to be sure it’s complete.
Q3: Is this generalizable to the form $X^2 \pm kY^2 = Z^3$ for any [possibly non-square or squarefree] integer $k$?
| To describe the solutions of the equation. $$x^2+qy^2=z^3$$
I think best would be to describe a solution using $3$ parameters.
$$x=p^6+q(b^2+8bs-5s^2)p^4+q^2(s^2-b^2)(b^2-8bs-5s^2)p^2+q^3(s^2-b^3)^3$$
$$y=2p(q^2(2s+b)(s^2-b^2)^2+2qb(b^2-3s^2)p^2-(2s-b)p^4)$$
$$z=p^4+2q(s^2+b^2)p^2+q^2(s^2-b^2)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the general solution to differential equation $x(x+1)^2(y'-\sqrt x)=(3x^2+4x+1)y$ Equation can be transformed to linear differential equation:
$$LHS=x^3y'+2x^2y'+xy'-x^{7/2}-2x^{5/2}-x^{3/2}$$
$$RHS=3x^2y+4xy+y$$
$$\Rightarrow y'(x^3+2x^2+x)+y(-3x^2-4x-1)=x^{7/2}+2x^{5/2}+x^{3/2}$$
After dividing by $(x^3+2x^2+x),x\neq 0\land x\neq -1$
$$\Rightarrow y'+\frac{-3x^2-4x-1}{x^3+2x^2+x}y=\frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}$$
The general solution is
$$y=e^{-\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx}\left(c+\int \frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}{e^{\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx}}\mathrm dx\right)$$
Integral $\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx$ can be found using partial fractions, $$\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx=\ln(|x|(x+1)^2)+c$$
How to evaluate integral
$$\int \frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}{e^{\int \frac{-3x^2-4x-1}{x^3+2x^2+x}\mathrm dx}}\mathrm dx=\int \frac{x^{7/2}+2x^{5/2}+x^{3/2}}{x^3+2x^2+x}e^{\ln(|x|(x+1)^2)}\mathrm dx?$$
Is there an easier method than transforming to linear equation?
| What seems to be interesting is to define $$y=z\, x\,(1+x)^2$$ which makes the differential equation to be $$x (x+1)^2 \,z'-\sqrt{x}=0$$ which is separable and "quite" simple to integrate. $$z=\int \frac{\sqrt{x}}{x (x+1)^2}\,dx$$ Make $x=t^2$ to get $$z=2\int \frac{dt}{\left(1+t^2\right)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integer solutions for $n$ for $|{\sqrt{n} - \sqrt{2011}}| < 1$ $$|{\sqrt{n} - \sqrt{2011}}| < 1$$
What is the number of positive integer $n$ values, which satisfy the above inequality.
My effort:
$
({\sqrt{n} - \sqrt{2011}})^2 < 1 \\n + 2011 -2\sqrt{2011n} < 1\\ n+2010<2\sqrt{2011n}\\ n^2+2 \times 2010 \times n +2010^2<4 \times 2011n \\n^2 -4024n +2010^2 < 0 $
But it seems this won't lead me for desired answer.
| Brute force says that $(\sqrt{2011}-1)^2\leq n\leq (\sqrt{2011}+1)^2$. So you can use a calculator to find the lower and upper bound integer values: $1923\leq n\leq
2101$, do there are $2101-1923+1$ integer values $n$.
Algebraically, we see that $(\sqrt{2011}-1)^2 = 2012-2\sqrt{2011}$ and $(\sqrt{2011}+1)^2=2012+2\sqrt{2011}$. So a non-calculator value is $1+2\lfloor2\sqrt{2011}\rfloor$.
These both give the value $179$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do I find the solution to the equation $z^2=-81i$? This question is from the Powers of Complex Numbers, Precalculus section of KhanAcademy
Find the solution to the following equation whose argument is between $90°$ and $180°$
$$z^2=-81i$$
What I understand thus far:
I am going to set $r$ and $\theta$ to be the modulus and argument of $z$, respectively.
Therefore, $z^{ 2 }=r^{ 2 }[cos(2\cdot \theta )+isin(2\cdot \theta )]$
Now, I can understand how the modulus is $81$, but I do not understand how it was determined that the argument is $270°$ plus any multiple of $360°$. I am quite confused at this point and a hint in the right direction would be the best thing to help me figure out the solution to this problem and ones like it that I will encounter in the future.
| Write $-81i$ in trigonometric form:
$$
-81i=81\cdot(-i)=81\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right)
$$
so by De Moivre its square roots are
$$
9\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)
$$
and
$$
9\left(\cos\left(\frac{3\pi}{4}+\pi\right)+
i\sin\left(\frac{3\pi}{4}+\pi\right)\right)
=
9\left(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}\right)
$$
Since $\pi/2<3\pi/4<\pi$, the first one is what you're looking for. Thus the answer is
$$
9\left(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)
$$
If you use degrees, $\frac{3\pi}{2}=270^\circ$ and $\frac{3\pi}{4}=135^\circ$.
In general, if $z=r(\cos\alpha+i\sin\alpha)$, the $n$-th roots of $z$ are
$$
\sqrt[n]{r}\left(
\cos\left(\frac{\alpha}{n}+\frac{2k\pi}{n}\right)
+i\sin\left(\frac{\alpha}{n}+\frac{2k\pi}{n}\right)
\right),
\qquad k=0,1,\dots,n-1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1716905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove $\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^{n+1}} = \frac{(1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)} \pi \ \ \ \forall n \in \mathbb{N}$ My attempt starts with a contour integral in the half disk, I let the radius -> infinity and so the contour integral
\begin{equation}
\int_{\gamma} \frac{dz}{(1+z^2)^{n+1}} = 2 \pi i \ res_{z_0 = i} f
\end{equation}
reduces to
\begin{equation}
\int^{\infty}_{-\infty} \frac{dx}{(1+x^2)^{n+1}} = 2 \pi i \ res_{z_0 = i} f
\end{equation}
Since the poles are of order n+1, the residual is
\begin{equation}
res_{z_0 = i} f = \lim_{z\to i} \frac{1}{n!} \bigg ( \frac{d}{dz} \bigg)^n (z+i)^{-[n+1]} = \frac{(-1)^n (2n)! (2i)^{-[2n+1]}}{(n!)^2} = \frac{(2n)!}{(n!)^2 2i 2^{2n}}
\end{equation}
and so I have
\begin{equation}
\frac{(2n)! \pi}{(n!)^2 2^{2n}}
\end{equation}
now the part that I'm struggling with, showing that
\begin{equation}
\frac{(2n)!}{(n!)^2 2^{2n}} = \frac{(1)(3)(5)...(2n-1)}{(2)(4)(6)...(2n)}
\end{equation}
| $$\begin{equation}
\frac{(2n)!}{(n!)^2 2^{2n}} = \frac{(1)(2)(3)...(2n)}{1 \cdot 2 \cdot ... \cdot n \cdot 2 \cdot 2 ...\cdot 2} \times \frac{1}{1 \cdot 2 \cdot ... \cdot n \cdot 2 \cdot 2 ...\cdot 2}
\end{equation}$$
Now, rewrite
$$1 \cdot 2 \cdot ... \cdot n \cdot 2 \cdot 2 ...\cdot 2=2 \cdot 4 \cdot 6 \cdot ... \cdot (2n)$$
in the first denominator, and cancel.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that the group is abelian Let $M$ be a field and $G$ the multiplicative group of matrices of the form $\begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}$ with $x,y,z\in M$.
I have shown that all the elements of the center $Z(G)$ are the matrices of the form $\begin{pmatrix}
1 & 0 & \tilde{y} \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$.
How could I show that $G/Z(G)$ is abelian?
We have that $$G/Z(G)=\{gZ(G)\mid g\in G\}$$
Do we have to take $A=g_1Z(G)$ and $B=g_2Z(G)$ and show that $AB=BA$ ?
Or do we have to take the same $g$ just an other element of the center?
| Pick an arbitrary
$$g = \begin{pmatrix}
1 & x & y \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix} \in G$$
In the quotient $G/Z(G)$, $g$ is a representative of the coset
$$gZ(G) = \left\{ \begin{pmatrix}
1 & x & y+ \tilde{y} \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}: \tilde{y} \in M \right\}$$
so you can choose another (more clever) representative for $gZ(G)$, namely
$$g' = \begin{pmatrix}
1 & x & 0 \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix}$$
(i.e. $gZ(G)=g'Z(G)$). This will make computations easier.
Now, pick two arbitrary matrices
$$a=\begin{pmatrix}
1 & x & 0 \\
0 & 1 & z \\
0 & 0 & 1
\end{pmatrix} , b=\begin{pmatrix}
1 & x' & 0 \\
0 & 1 & z' \\
0 & 0 & 1
\end{pmatrix}$$
and compute
$$ab= \begin{pmatrix}
1 & x+x' & xz' \\
0 & 1 & z+z' \\
0 & 0 & 1
\end{pmatrix} , ba= \begin{pmatrix}
1 & x+x' & x'z \\
0 & 1 & z+z' \\
0 & 0 & 1
\end{pmatrix}$$
it is easily verified that $abZ(G) = baZ(G)$. This is enough to conclude that $G/Z(G)$ is abelian.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$
In $\triangle ABC$, if
$$\cos A \cos B \cos C=\frac{1}{3}$$
then can we find value of
$$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$
Please give some hint. I am not sure if $\tan A \tan B+\tan B \tan C+\tan C \tan A$ will be constant under given condition.
| Rephrasing Mathematics's answer:
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ and $\cos(A+B)=\cdots=-\cos C$
$\cos^2A+\cos^2B+\cos^2C=1+\cos^2A-\sin^2B+\cos^2C=1-2\cos A\cos B\cos C$
If $\cos A\cos B\cos C=S$
$\iff\cos^2A+\cos^2B+\cos^2C=1-2S$
let $y=\tan A\tan B\iff y-1=-\dfrac{\cos(A+B)}{\cos A\cos B}=\dfrac{\cos^2C}S$
$\iff\cos^2C=S(y-1)$
$\sum S(y-1)=\sum\cos^2C=1-2S\iff y=3+\dfrac{1-2S}S$
Here $S=\dfrac13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find the exponential generating function for the number of ways to distribute $r$ distinct objects into five different boxes Find the exponential generating function for the number of ways to distribute $r$ distinct objects into five different boxes when $b_1<b_2\le 4$, where $b_1,b_2$ are the numbers of objects in boxes $1$ and $2$, respectively.
I understand that boxes $3, 4$, and $5$ will all just have the fundamental exponential generating function and that their combined generating function will be $e^{3x}$. What I don't understand is the first two boxes with the equality parameter given.
Can someone explain how to do this?
| Here is a very painstaking approach that may help you to see exactly what’s going on.
The possible values of $b_1$ are $0,1,2$, and $3$, so far starters we try
$$1+x+\frac{x^2}2+\frac{x^3}6$$
to account for $b_1$. Similarly, the possible values of $b_2$ are $1,2,3$, and $4$, so we try
$$y+\frac{y^2}2+\frac{y^3}6+\frac{y^4}{24}$$
to account for $b_2$. I’m using different indeterminates for now, because at this point I still need to keep the $b_1$ and $b_2$ contributions separate.
The product of these polynomials is
$$\begin{align*}
&y+\frac{y^2}2+\frac{y^3}6+\frac{y^4}{24}+\\
&xy+\frac{xy^2}2+\frac{xy^3}6+\frac{xy^4}{24}+\\
&\frac{x^2y}2+\frac{x^2y^2}4+\frac{x^2y^3}{12}+\frac{x^2y^4}{48}+\\
&\frac{x^3y}6+\frac{x^3y^2}{12}+\frac{x^3y^3}{36}+\frac{x^3y^4}{144}\;;
\end{align*}$$
however, we don’t want the terms in $x^ky\ell$ with $k\ge\ell$, since they correspond to having $b_1\ge b_2$. After we throw them away, we have
$$y+\frac{y^2}2+\frac{y^3}6+\frac{y^4}{24}+\frac{xy^2}2+\frac{xy^3}6+\frac{xy^4}{24}+\frac{x^2y^3}{12}+\frac{x^2y^4}{48}+\frac{x^3y^4}{144}\;.$$
Now replace $y$ by $x$, collect terms, and adjust the denominators to match the exponents to get
$$\frac{x}{1!}+\frac{x^2}{2!}+\frac{4x^3}{3!}+\frac{5x^4}{4!}+\frac{15x^5}{5!}+\frac{15x^6}{6!}+\frac{35x^7}{7!}\;,$$
which is the egf for boxes $1$ and $2$ combined. Multiply this by $e^{3x}$, and you’re done.
(And now that I’ve written this, I see that Markus has given you the abbreviated version of it.)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A conjecture about traces of projections Let $M_n$ denote the space of all $n\times n$ complex matrices. Define $\tau:M_n\rightarrow \mathbb{C}$ by $$\tau(X)=\frac{1}{n}\sum_{i=1}^n x_{ii},$$ where of course $X=[x_{ij}]\in M_n$. Recall that a matrix $P\in M_n$ is called an "orthogonal projection" if $P=P^*=P^2$. Let $A, B, C$ be orthogonal projections in $M_n$ and define two quantities - $$x=\frac{1}{3}\tau(A+B+C)$$ and $$y=\frac{1}{3}\tau(AB+BC+CA).$$
It is easy to see that for any orthogonal projections $A,B,C\in M_n$ the value of $x$ lies in $[0,1]$. I want to investigate the case when $x\in [\frac{1}{3},\frac{1}{2}]$. My goal is to compute the infimum of $y$ under this constrain on $x$.
My observation is that the minimum value of $y$ is $\frac{3x-1}{4}$ when $x\in [\frac{1}{3},\frac{1}{2}]$ and for all $n\in \mathbb{N}$.
However I am unable to prove this and I wondered if people here may be able to see why this might hold, or provide a counterexample to this conjecture.
| I prove only $y\ge \frac x2(3x-1)$. We know that $A+B+C\ge 0$, so let $\lambda_1,\dots,\lambda_n$ be its eigenvalues. Then
$$
\frac 1n Tr(A+B+C)=\frac 1n \sum \lambda_i=\bar \lambda= 3x
$$
and
$$
Tr((A+B+C)^2)=\sum \lambda_i^2\ge n \bar \lambda^2=9nx^2.
$$
But
$$
(A+B+C)^2=A^2+B^2+C^2+AB+AC+BA+BC+CA+CB
$$
and so
$$
Tr((A+B+C)^2)=Tr(A+B+C)+2Tr(AB+BC+CA)=3nx+6ny.
$$
Hence $3nx+6ny\ge 9nx^2$ which yields $y\ge \frac x2(3x-1)$.
${\bf{Edit:}}$
The bound given by the OP is attained for every $n$ and every admissible $x$:
Set
$$
A_0=\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix},\quad B_0=\begin{pmatrix} 1/4&-\sqrt{3}/4\\ -\sqrt{3}/4&3/4 \end{pmatrix}\quad\text{and}\quad C_0=\begin{pmatrix} 1/4&\sqrt{3}/4\\ \sqrt{3}/4&3/4 \end{pmatrix}.
$$
Then $Tr(A_0B_0+B_0C_0+C_0A_0)=\frac 34$.
Let now $n$ be given, and $x=\frac {n+k}{3n}$ be admissible, i.e., $0\le k\le n/2$. Define
$A$ to have $k$ times the block matrix $A_0$ on the diagonal, and complete the diagonal with $n-2k$ times 1:
$$
A=\begin{pmatrix} A_0&&&&& \\&\ddots&&& 0& \\ && A_0&&&\\ &&&1 && \\ &0&&& \ddots &\\ &&&&& 1\end{pmatrix}.
$$
Define
$B$ to have $k$ times the block matrix $B_0$ on the diagonal, and complete the diagonal with $n-2k$ times 0:
$$
B=\begin{pmatrix} B_0&&&&& \\&\ddots&&& 0& \\ && B_0&&&\\ &&&0 && \\ &0&&& \ddots &\\ &&&&& 0\end{pmatrix}.
$$
Define
$C$ to have $k$ times the block matrix $C_0$ on the diagonal, and complete the diagonal with $n-2k$ times 0:
$$
C=\begin{pmatrix} C_0&&&&& \\&\ddots&&& 0& \\ && C_0&&&\\ &&&0 && \\ &0&&& \ddots &\\ &&&&& 0\end{pmatrix}.
$$
Then $x=\frac{1}{3n}Tr(A+B+C)=\frac{n+k}{3n}$ (which implies $\frac kn=3x-1$) and
$$
y=\frac{1}{3n}Tr(AB+BC+CA)=\frac{1}{3n}Tr(k(A_0B_0+B_0C_0+C_0A_0))=\frac{1}{3n}\frac 34k=
\frac 14 \frac kn=\frac{3x-1}{4},
$$
hence the bound is attained.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorization of polynomials with degree higher than 2 I need help to factorize $x^4-x^2+16$. I have tried to take $x^4$ as $(x^2)^2$ and factorize it in the typical way of factorizing a quadratic expression but that did not help. Can someone help me to factor this and also introduce me to the procedure that i need to follow to factorize expressions with degree higher than two?
| Can be done by making perfect squares
$$
Let\ ax^2 +bx + c=0
\\Try\ to\ make\ the\ equation\ look\ in\ the\ form
\\ax^2 +2\sqrt {ca}x +c-(2\sqrt{ca}-b)x=0
\\You\ will\ see\ that\ ax^2 +2\sqrt {ca}x +c\ makes\ a \ perfect\ square
\\\therefore \quad ax^2 +2\sqrt {ca}x +c=(\sqrt ax + \sqrt c )^2
\\Thus\ the\ equation\ will\ convert\ to\ (\sqrt ax + \sqrt c )^2-(2\sqrt{ca}-b)x=0
\\Consider \ the\ term\ (2\sqrt{ca}-b)x
\\We\ can\ write\ it\ as\ {[\sqrt{(2\sqrt{ca}-b)x}]}^2
\\Thus\ the\ equation\ is\ converted\ to\ form\ p^2-q^2=0
\\where\ p=\sqrt ax + \sqrt c \quad and\quad q=\sqrt{(2\sqrt{ca}-b)x}
\\Equation\implies [\sqrt ax + \sqrt c]^2 - {[\sqrt{(2\sqrt{ca}-b)x}]}^2=0
$$
Now you can apply $p^2-q^2=(p+q)(p-q)$ to get the required factorization.
Regarding your question Consider
$$ a= 1,b=-1 ,c=16 \ and\ usual \ substitution\ x=x^2\ (both\ x\ are\ different)
\\ \therefore p=x^2+4 \quad q=3\sqrt {x^2}=3x
$$
After this its easy mathematics,you have to factorize 2 more easy equations thus giving you 4 roots.I hope , I did well to explain you :)
| {
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"url": "https://math.stackexchange.com/questions/1726026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Last Digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
Given $x$ and $p$. Find the last digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
I need a general formula. I can find that the sum is equal to
$\dfrac{x^{p+1}-1}{x-1}$
But how to find the last digit.
P.S: $x\leq 999999$ and $p \leq 10^{15}$
| Let $s = \sum_{k=0}^{p}x^k$. Calculate $0 \leq a_1, a_2 < 5$, where
$$
a_1 \equiv s \pmod 2 \\
a_2 \equiv s \pmod 5
$$
$a_1$ is easy. If $x \equiv_2 0$, then $a_1 = 1$, else $a_1 \equiv_2 (1+p)$, cause it's $1 + 1^1 + 1^2 + 1^3 + \dots + 1^p$.
$a_2$ isn't harder, $x \equiv_5 0, 1$ are same. For $2, 3, 4$ you can easily observe regularity. Eg.
$$
x \equiv_5 2 \Longrightarrow\\ s \equiv_5
1 + 2^1 + 2^2 + \dots + 2^{p} \equiv_5 \\
1 + 2 + 4 + 8 + \dots + 2^{p} \equiv_5 \\
1 + 2 + 4 + 3 + 1 + 2 \dots + 2^{p} \equiv_5\\
(1 + 2 + 4 + 3) + 1 + 2 \dots + 2^{p} \equiv_5\\
2 \cdot 5 + 1 + 2 \dots + 2^{p}
$$
If you know $a_1, a_2$ you can use Chinese remainder theorem, to calculate number $r \equiv_{10} s$ and it's last digit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Cascading Summation $\sum_{i=1}^n\sum_{j=i}^n\sum_{k=j}^n \frac {i(j+2)(k+4)}{15} $ Evaluate
$$\sum_{i=1}^n\sum_{j=i}^n\sum_{k=j}^n \frac {i(j+2)(k+4)}{15} $$
Background
Many basic summation questions on MSE relate to a single index - it might be interesting to devise a question where the summand is a product of the three indices but can be solved easily.
| Our sum depends on three sums:
$$ S_1 =\!\!\!\!\sum_{1\leq i \leq j \leq k\leq n}\!\!\!ijk, \qquad S_2 = \!\!\!\!\sum_{1\leq i \leq j \leq k\leq n}\!\!\!ij, \qquad S_3 =\!\!\!\! \sum_{1\leq i \leq j \leq k\leq n}\!\!\!ik $$
that can be evaluated by using standard symmetry tricks. For instance:
$$ \left(\sum_{i=1}^{n} i\right)\cdot\left(\sum_{j=1}^{n} j\right)+\sum_{j=1}^{n}j^2 = 2\cdot\!\!\!\!\sum_{1\leq i\leq j\leq n}\!\!ij $$
holds, and we have a similar identity in three variables. As an alternative, $S_1$ is:
$$ S_1 = \!\!\!\!\!\!\sum_{\substack{a,b,c,d\geq 0\\a+b+c+d=n-1}}\!\!\!\!(a+1)\cdot(a+b+1)\cdot(a+b+c+1) $$
so it can be computed by expanding the general term as a sum of monomials, then going along the following lines:
$$\begin{eqnarray*} \sum_{\substack{a+b+c+d=n-1}}\!\!\!\!\!\!\!\!a^2 b^1 c^1 d^0 &=& [x^{n-1}]\left(\sum_{n\geq 0}n^2 x^n\right)\cdot\left(\sum_{n\geq 0}n^1 x^n\right)\cdot\left(\sum_{n\geq 0}n^1 x^n\right)\cdot \left(\sum_{n\geq 0}n^0 x^n\right)\\[0.2cm]&=&[x^{n-1}]\left(\frac{x(1+x)}{(1-x)^3}\cdot\frac{x}{(1-x)^2}\cdot\frac{x}{(1-x)^2}\cdot\frac{1}{1-x}\right)\\[0.2cm]&=&[x^{n-4}]\frac{1+x}{(1-x)^8}=[x^{n-4}]\frac{1}{(1-x)^8}+[x^{n-5}]\frac{1}{(1-x)^8}\\[0.2cm]&=&\color{red}{\binom{n+3}{7}+\binom{n+2}{7}}.\end{eqnarray*}$$
One way or another, the final outcome is:
$$ \sum_{i=1}^{n}\sum_{j=i}^{n}\sum_{k=j}^{n}\frac{i(j+2)(k+4)}{15}=\color{red}{\binom{n+5}{6}}.$$
Since it is obvious that the final outcome is a polynomial in $n$ having degree $3+1+1+1=6$, also without computing it, the answer can be derived also through Lagrange interpolation: it is enough to compute the given triple sum for $n\in\{0,1,2,3,4,5,6\}$.
Addendum: the explicit way.
$$\begin{eqnarray*}\sum_{1\leq i\leq j\leq k\leq n}\!\!\!i(j+2)(k+4)&=&\!\!\!\!\!\sum_{\substack{a,b,c,d\geq 0\\a+b+c+d=n-1}}\!\!\!\!\!(a+1)(a+b+3)(a+b+c+5),\end{eqnarray*} $$
$$\begin{eqnarray*}\sum_{\substack{a,b,c,d\geq 0\\a+b+c+d=n-1}}\!\!\!\!\! a^3 b^0 c^0 d^0 &=& [x^{n-1}]\frac{1}{(1-x)^3}\sum_{n\geq 0}n^3 x^n=[x^{n-1}]\frac{x(1+4x+x^2)}{(1-x)^7}\\[0.1cm]&=&[x^{n-2}]\frac{1}{(1-x)^7}+4[x^{n-3}]\frac{1}{(1-x)^7}+[x^{n-4}]\frac{1}{(1-x)^7}\\[0.25cm]&=&\binom{n+4}{6}+4\binom{n+3}{6}+\binom{n+2}{6},\\[0.3cm]
\sum_{\substack{a,b,c,d\geq 0\\a+b+c+d=n-1}}\!\!\!\!\! a^2 b^1 c^0 d^0 &=& [x^{n-1}]\frac{x}{(1-x)^4}\sum_{n\geq 0}n^2 x^n=[x^{n-1}]\frac{x^2(1+x)}{(1-x)^7}\\[0.1cm]&=&[x^{n-3}]\frac{1}{(1-x)^7}+[x^{n-4}]\frac{1}{(1-x)^7}\\[0.25cm]&=&\binom{n+3}{6}+\binom{n+2}{6},\\[0.3cm]
\sum_{\substack{a,b,c,d\geq 0\\a+b+c+d=n-1}}\!\!\!\!\! a^1 b^1 c^1 d^0 &=& [x^{n-1}]\frac{1}{(1-x)}\left(\sum_{n\geq 0}n x^n\right)^3=[x^{n-1}]\frac{x^3}{(1-x)^7}\\[0.1cm]&=&[x^{n-4}]\frac{1}{(1-x)^7}\\[0.25cm]&=&\binom{n+2}{6},\end{eqnarray*}$$
the remaining part is easy.
| {
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"url": "https://math.stackexchange.com/questions/1729231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Determinant of $N \times\ N$ matrix So the question asks:
For $n \geq 2$, compute the determinant of the following matrix:
$$
B =
\begin{bmatrix}
-X & 1 & 0 & \cdots & 0 & 0 \\
0 & -X & 1 & \ddots & \vdots & \vdots \\
\vdots & \ddots & \ddots & \ddots & 0 & \vdots \\
\vdots & & \ddots & \ddots & 1 & 0 \\
0 & \cdots & \cdots & 0 & -X & 1 \\
a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X)
\end{bmatrix}
$$
Looking at the $2 \times 2$ and $3 \times 3$ forms of this matrix:
$\det \begin{bmatrix} -X & 0 \\ 0 & (a_1-X) \end{bmatrix} = -X(a_1-X) - 0 = X^2 - a_1X $
by expansion along the first row:
$\det \begin{bmatrix} -X & 1 & 0 \\ 0 & -X & 0 \\ 0 & 0 & (a_2-X) \end{bmatrix} = (-X) \times\det \begin{bmatrix} -X & 0 \\ 0 & a_2-X \end{bmatrix} - 1 \det\begin{bmatrix} 0 & 0 \\ 0 & a_2-X \end{bmatrix}$
$= (-X)[(-X)(a_2-X) -0] - 0 = X^3 - a_2X^2 $
So it looks like:
$\det \begin{bmatrix}
-X & 1 & 0 & \cdots & 0 & 0 \\
0 & -X & 1 & \ddots & \vdots & \vdots \\
\vdots & \ddots & \ddots & \ddots & 0 & \vdots \\
\vdots & & \ddots & \ddots & 1 & 0 \\
0 & \cdots & \cdots & 0 & -X & 1 \\
a_0 & a_1 & \cdots & \cdots & a_{n-2} & (a_{n-1} - X)
\end{bmatrix}
= X^{n} - a_{n-1}X^{n-1} - a_{n-2}X^{n-2} ... - a_1X$
Does this look right? Is "prove by induction" valid to use here?
| Let $v= \begin{pmatrix} 1 \\ x \\ x^{2} \\ \cdot \\ \cdot \\ \cdot \\ x^{n-2} \\x^{n-1} \end{pmatrix}$.
Then $Bv = (x-X)v \iff a_{0} + a_{1}x + ... + a_{n-1}x^{n-1} -x^{n} = p(x) =0$. Thus, all the roots $\alpha$ of the monic polynomial $p(x)$ of degree $n$ noted here give us our eigenvectors $v_{\alpha}$, which are linearly independent since they are columns of a Vandermonde matrix. The associated eigenvalues are $\lambda = \alpha - X$.
| {
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"url": "https://math.stackexchange.com/questions/1732914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove $\sum_{i=1}^na_i> n- \frac{9}{4}$ Given that $a_1=\frac{1}{2}$ and $a_{n+1}=\sqrt[]{\frac{1}{3}a_n^{2}+\frac{2}{3}a_n}$ ,prove that
$$\sum_{i=1}^na_i> n- \frac{9}{4}$$Thanks.
| Let $f(x)=\sqrt{\frac{1}{3}(x^2+2x)}$, for $x\ge0$. Clearly $f$ is increasing and it is staightforward to check that
$$0\le x\le 1\Longrightarrow 0\le f(x)\le x\le 1$$
This alows us to prove that the sequence $(a_n)$ is increasing and bounded by $1$, so it must converge to the positive solution of the equation $f(x)=x$ which is $1$. Moreover, $\frac12\le a_n\le1$ for every $n$.
Now,
$$f'(x)=\frac{1}{\sqrt{3}}\frac{x+1}{\sqrt{x^2+2x}}=\frac{1}{\sqrt{3}}\sqrt{1+\frac{1}{x^2+2x}}$$
So, $f'$ is decreasing on $[\frac12,1]$. So,
$$0\le f'(x)\le f'(1/2)=\sqrt{3/5}\quad\text{for $x\in[0.5,1]$.}$$
The mean value theorem allows us to write
$$0\le 1-f(x)\le (1-x) \sqrt{3/5}\quad\text{for $x\in[0.5,1]$}$$
So, if $\delta_n=1-a_n$, then we have, for every $n$,
$$\delta_{n+1}\le \sqrt{3/5}\delta_n$$
Consequently,
$$\delta_n\le\left(\frac35\right)^{n/2}\delta_0$$
Thus
$$\sum_{k=1}^n\delta_k<\frac12\sum_{k=1}^\infty\left(\frac35\right)^{k/2}=
\frac{\sqrt3}{2(\sqrt5-\sqrt3)}\approx 1.72<\frac{9}{4}$$
This is equivalent to the desired inequality.
| {
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"url": "https://math.stackexchange.com/questions/1736072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of solutions of $x_1+2\cdot x_2+2\cdot x_3 = n$ I have to find number of solutions of $x_1+2\cdot x_2+2\cdot x_3 = n$. I guess it would be $[x^n](1+x+x^2 \dots)(1 + x^2 + x^4 \dots)^2$, but how to compute it? I know only that $\frac{1}{1-x} = 1+x+x^2 \dots$.
| Generating Function
The generating function is
$$
\begin{align}
\frac1{(1-x)\left(1-x^2\right)^2}
&=\frac1{(1-x)^3(1+x)^2}\\
&=\sum_{j=0}^\infty\binom{-3}{j}(-x)^j\sum_{k=0}^\infty\binom{-2}{k}x^k\\
&=\sum_{j=0}^\infty\binom{j+2}{2}x^j\sum_{k=0}^\infty\binom{k+1}{1}(-x)^k\\
&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{k+1}{1}\binom{n-k+2}{2}x^n\tag{1}
\end{align}
$$
Thus, the number of solutions is
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=0}^n(-1)^k\binom{k+1}{1}\binom{n-k+2}{2}=\frac{2n^2+10n+11+(-1)^n(2n+5)}{16}}\tag{2}
$$
Alternating Sums of Powers
To compute the sum in $(2)$ we have used the following
$$
\begin{align}
\sum_{k=0}^n(-1)^k1&=\frac{(-1)^n+1}2\tag{3}\\
\sum_{k=0}^n(-1)^kk&=\frac{(-1)^n(2n+1)-1}4\tag{4}\\
\sum_{k=0}^n(-1)^kk^2&=\frac{(-1)^n\left(n^2+n\right)}2\tag{5}\\
\sum_{k=0}^n(-1)^kk^3&=\frac{(-1)^n\left(4n^3+6n^2-1\right)+1}8\tag{6}
\end{align}
$$
Recursion for the Alternating Sums of Powers
The formulas in $(3)$-$(6)$ were derived by induction and $(9)$.
Combining the following two sums by reindexing the first gives
$$
\begin{align}
\sum_{k=0}^n(-1)^{k+1}(k+1)^m+\sum_{k=0}^n(-1)^kk^m
&=\sum_{k=1}^{n+1}(-1)^kk^m+\sum_{k=0}^n(-1)^kk^m\\
&=(-1)^{n+1}(n+1)^m+2\sum_{k=0}^n(-1)^kk^m\tag{7}\\
\end{align}
$$
Combining the same two sums without reindexing gives
$$
\sum_{k=0}^n(-1)^{k+1}(k+1)^m+\sum_{k=0}^n(-1)^kk^m
=\sum_{k=0}^n(-1)^{k+1}\left((k+1)^m-k^m\right)\tag{8}
$$
Equating the right sides of $(7)$ and $(8)$ gives
$$
\sum_{k=0}^n(-1)^kk^m
=\frac12\left[(-1)^n(n+1)^m-\sum_{k=0}^n(-1)^k\left((k+1)^m-k^m\right)\right]\tag{9}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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What can we do to solve the following equation with $6$ variables with some information provided?
Q) There are unique integers $a_2, a_3, a_4, a_5, a_6, a_7$ such that $$\frac{a_2}{2!}+\frac{a_3}{3!}+\frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!}=\frac 57$$,where $0\le a_i < i$. Then the value of $a_2+a_3+a_4+a_5+a_6+a_7$(1)$8$(2)$9$(3)$10$(4)$11$
Answer: $(2)$
My workout
all these are unique integers hence all are having different values.
now multiplying the big one equation by $7!$
$$7\left\{6!\frac{a_2}{2!}+6!\frac{a_3}{3!}+6!\frac{a_4}{4!}+6!\frac{a_5}{5!}+6!\frac{a_6}{6!}\right\}+a_7= 6!.5$$
which is like the remainder theorem which implies, $6!.5=3600$, when divided by 7 leaves remainder $a_7$ so dividing 3600 by 7 gives 2 hence $a_7=2$ . Now from inequality equation we can say that
$a_2\in\{0,1\}$
$a_3\in\{0,1,2\}$
$a_4\in\{0,1,2,3\}$
$a_5\in\{0,1,2,3,4\}$
$a_6\in\{0,1,2,3,4,5\}$
$a_7\in\{0,1,2,3,4,5,6\}$
so from all the data above i can conclude that $a_2=0, a_3=1, a_4=3, a_5=4, a_6=5, a_7=2$ hence summing them up $0+1+3+4+5+2=15$ which is none of the above option
| It is easy to see that $(a_2,\ldots,a_7)=(1,1,1,0,4,2)$ is a solution. Indeed, we have
$$
\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{4}{720}+\frac{2}{5040}=\frac{5}{7}.
$$
In which case we have $a_2+\cdots +a_7=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $x,y$ so that $\dfrac{x+y+2}{xy-1}$ is an integer. I am trying to find the integers $x,y$ so that
$\dfrac{x+y+2}{xy-1}$ is an integer.
What I have done:
I suppose there exists $t$ such that $$t=\dfrac{x+y+2}{xy-1}$$ where $xy\neq 1$ then consider the following scenarios:
$$x=y$$ $$x>y>0$$ $$x>0>y$$ ,etc.
This approach helps me find the solutions but it is very long. Any simpler method?
| For this to happen we need $|x+y+2|\geq |xy-1|$ or $x+y+2=0$.
If $x+y+2=0$ we get solutions $(a,-a-2),(-a-2,a)$.
otherwise we have $|x+y|+2\geq |xy|-1\rightarrow |x|+|y|+3\geq |x||y|$.
Notice that if $x$ or $y$ is $0$ it holds trivially.
We now find all possible values $n,m> 0$ so that $n+m+3\geq nm\iff \frac{n+3}{n-1}\geq m$
If $n=1$ any $m$ works.
If $n=2$ we get $m\leq 5$
If $n=3$ we get $m \leq 3$
If $n=4$ we get $m\leq 2$
If $n=5$ we get $m\leq 2$.
If $n\geq 6$ we get $m=1$.
So we first analyse the case when $|x|=1$.
If $x=1$ we get $y-1$ divides $y+3$, so $y-1$ divides $(y+3)-(y-1)=4$, hence $y-1$ can be $-4,-2,-1,1,2$ or $4$. It is easy to see all of them work. So we get $(1,-3),(1,-1),(1,0),(1,2),(1,3),(1,5)$.
If $x=-1$ we get $-(y-1)$ divides $y-1$. So $y+1$ divides $y-1$ and hence $y+1$ divides $y+1-(y-1)=2$. So $y+1$ can be $-2,-1,1,2$. It is easy to see all of them work. So we get $(-1,-3),(-1,-2),(-1,0),(-1,1)$.
Since $x$ and $y$ play symmetric roles in the equation, this gives all solutions with $|x|=1$ or $|y|=1$.
Now we analyse when $|x|=2$.
If $x=2$ we get $2y-1$ divides $y+4$, we need only try with $y=-5,-4,-3,-2,0,2,3,4,5$. We see only $(2,-4),(2,0),(2,2),(2,5)$ work.
If $x=-2$ we get $-2y-1$ divides $y$, so $2y+1$ divides $y$, and so $2y+1$ divides $2y+1-2(y)=1$. So we only try with $2y+1=1$ and $2y+1=-1$. So $(-1,-1)$ and $(-1,1)$ are solutions.
So we only need to check when $|x|,|y|>2$.
If $|x|=3$ we need only check if $|y|=3$. There are four cases, and out of these only $(3,3)$ works.
This finishes all the cases.
(I'll try to make this a bit cleaner when I get home).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Conditional Probability (dice) A die is rolled 7 times.
What is the probability that all outcomes are odd, given that the first outcome was greater than 3?
My approach: If the first outcome if > 3, then the dice rolled is either 4,5 or 6. So, we have a probability of $\frac{1}{3}$ to get a odd number.
So for the remaining 6 rolls, we must get all odd numbers which has probability of $(\frac{1}{2})^6 = \frac{1}{64}$. Now, I'm stuck with this approach.
EDIT: P(all outcomes odd | first outcome > 3) = $\frac{1}{3} \cdot \frac{1}{64}$
The other approach:
A = event that all outcomes are odd numbers,
B = event that first outcome > 3
$$P(A) = (\frac{1}{2})^7$$
$$P(B) = \frac{1}{2}$$
$P(A | B) = \frac{\text{P(A $\cap$ B)}}{P(B)}$
P(A $\cap$ B) = $(\frac{1}{2})^7 \cdot \frac{1}{2}$
__
Would like any feedback . Thank you!
| P(all outcomes are odd AND first outcome >3)=# (ways to get first outcome as 5 & remaining outcomes all odd)/n(S)=(1*3^6)/3*6^6=3^5/6^6. Regarding cardinality of the sample space, is my following logic correct? We want first outcome >3 which are 4,5,6 so there 3 possibilities for the first roll. For the remaining 6 rolls, we can have any of 1,2,3,4,5,6 and hence there are 6 ways each in 2nd,3rd,...,7th roll and hence 3*6*6*..*6=3*6^7 ways. Now, P(first outcome >3)=(3*6^6)/6^7. Thus, required probability=P(all outcomes are odd AND first outcome >3)/P(All outcomes >3)=(3^5/6^6)÷ (3/6)=3^4/6^5.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Lagrange Multiplier in 3D Find the minimum and maximum values of the function $f(x,y,z) = x+2y+3z$ where $(x,y,z)$ is on the sphere $x^2+y^2+z^2=1$ using Lagrange multiplier.
So I put them into the Lagrange form and got
$L(x,y,z,\lambda) = x+2y+3z+ \lambda(x^2+y^2+z^2-1)$
And you then get simultaneous equations
$L_x = 1+ 2\lambda x$,
$L_y = 2+ 2\lambda y$,
$L_z = 3+ 2\lambda z$,
$L_\lambda = x^2 +y^2 + z^2-1$.
However, when I solved these simultaneously, I got some very strange solutions, so I don't know if I've done something wrong up to this point on my solving was wrong.
Thanks
| $x+2y+3z\leq\sqrt {x^2+y^2+z^2}\sqrt {1+4+9} = \sqrt {14}$, by Cauchy-Schwarz, so $\max$ occurs at $x = \dfrac{y}{2} = \dfrac{z}{3} = \dfrac{1}{\sqrt{14}}$ and $\min$ occurs at $x = \dfrac{y}{2} = \dfrac{z}{3} = -\dfrac{1}{\sqrt{14}}$.
You should get a same answer if you solve the Lagrange equations correctly.
| {
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"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the Min of P(x,y) Find the Minimum of the following function :
$$P(x,y) = \frac{(x-y)}{(x^4+y^4+6)}.$$
This is a math problem I found in an internet math competition but it is really complex to me !!!
| It is not difficult to see that the minimum is located at
$x<0$ and $y>0$, so by the symmetry, it suffices to consider the problem
$$\max P(x,y)\quad\mbox{s.t.}\quad x,y\geq 0.$$
Now we use the polar coordinate as below.
Let $x=r\cos\theta$ and $y=r\sin\theta$, where $r\geq 0$ and
$0\leq\theta\leq\frac{\pi}{2}$. Then
\begin{align*}
P(x,y)\equiv\bar{P}(r,\theta)&=\frac{r(\cos\theta+\sin\theta)}{r^4(\cos^4\theta+\sin^4\theta)+6}\\
&=\frac{\sqrt{2}r\cos(\theta-\frac{\pi}{4})}{r^4[(\cos^2\theta+\sin^2\theta)^2-2\sin^2\theta\cos^2\theta]+6}\\
&=\frac{\sqrt{2}r\cos(\theta-\frac{\pi}{4})}{r^4[1-\frac{1}{2}\sin^2(2\theta)]+6}\\
&=\frac{\sqrt{2}r\cos(\theta-\frac{\pi}{4})}{\frac{r^4}{4}[3+\cos(4\theta)]+6}.
\end{align*}
We see that when fixing $r$, $\bar{P}$ reaches to maximum possibly when $\theta=\frac{\pi}{4}$. That is,
$$\bar{P}(r,\theta)\leq\frac{\sqrt{2}r}{\frac{r^4}{2}+6}=\frac{2\sqrt{2}r}{r^4+12}\equiv Q(r).$$
Finally, we calculate
$$Q'(r)=\frac{2\sqrt{2}(r^4+12)-8\sqrt{2}r^4}{(r^4+12)^2}=0
\quad\Longrightarrow\quad r=\sqrt{2}.$$
By checking $Q''(\sqrt{2})<0$, we conclude that
$P(1,1)=\bar{P}(\sqrt{2},\frac{\pi}{4})$ is the maximum. Hence $P(-1,1)=-\frac{1}{4}$ is the minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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