Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find $X, Y \in \mathbb{Z}$ such that $2^a X + (2^b - 1) Y = 1$ (coprimality) I've been wrecking my brain trying to solve this exercise. Is this answer wrong?
$$X= (2^{a})^{b-1}, Y= (-1) (2^b +1) \ [(2^b -1)(2^b +1)]^{a-1}$$
| Supposing $a,b>0$.For sure there exists a number $m\in \Bbb{Z}$ such that:
$$mb\geq a\Rightarrow m\geq \frac{a}{b} \Rightarrow m\geq \left \lceil{\frac{a}{b}}\right \rceil=M $$
Notice that if:
$$Y=-\sum_{i=0}^{M} 2^{bi}=\frac{-2^{b(M+1)}+1}{2^b-1}$$
Then the equation becomes:
$$2^a X-2^{b(M+1)}+1=1$$
Clearly this is satisfied by:
$$X=2^{bM+b-a}$$
So our particolar solutions are:
$$X=2^{b\left(\left \lceil{\frac{a}{b}}\right \rceil+1\right)-a} \ , \ Y=\frac{1-2^{b\left(\left \lceil{\frac{a}{b}}\right \rceil+1\right)}}{2^b-1}$$
So the general solutions are:
$$X=2^{b\left(\left \lceil{\frac{a}{b}}\right \rceil+1\right)-a}+(2^b-1)k \ , \ Y=\frac{1-2^{b\left(\left \lceil{\frac{a}{b}}\right \rceil+1\right)}}{2^b-1}-2^ak$$
:)
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $2x^2-5x+2=$ $\frac{5-\sqrt{9+8x}}{4}$ Solve $2x^2-5x+2$= $\frac{5-\sqrt{9+8x}}{4}$
I simply do square both sides solve it and I get two value of x one is 2 and other is $\frac{3-√5}{2}$ but this approach it take more time so is there any approach for solving this equation.
| I don't know another way then going through the algebra, I got the same solutions.
$$\begin{align} 2x^2-5x+2 &= \frac{5-\sqrt{9+8x}}{4} \\
8x^2-20x+3 &= -\sqrt{9+8x} \\
(8x^2-20x+3)^2 &= (-\sqrt{9+8x})^2 \\
64x^4-320x^3+448x^2-120x+9 &= 9 +8x\\
64x^4-320x^3+440x^2-120x &= 0 \\
64x(x-2)(x^2-3x+1) &=0 \\
\end{align}$$
We can see the solutions to that are $0$, $x=2$, $x = \frac{3 + \sqrt{5}}{2}$, $x = \frac{3 - \sqrt{5}}{2}$. Then plugging those into the original equation we get $x=2$ and $x = \frac{3 - \sqrt{5}}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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When recreating the quadratic formula by completing the square of $ax^2+bx+c=0$ I cannot shorten the right hand side I am attempting to derive the quadratic formula by completing the square on the generic generic expression:
$$ax^2+bc+c=0$$
I'm struggling with the right hand side of the equation which, for the step I'm on I know should be $\frac{b^2-4ac}{4a^2}$. However, I arrive at $\frac{b^2a-4a^2c}{4a^3}$
Here's my working:
(Approach copied largely from textbook)
Start with:
$ax^2+bx+c=0$
Move constant term to the right:
$ax^2+bx=-c$
Divide by $a$ to ensure leading coefficient is 1:
$x^2+\frac{b}{a}x=-\frac{c}{a}$
Calculate the amount needed to complete the square and add to both sides:
$(\frac{1}{2}*\frac{b}{a})^2$ = $(\frac{b}{2a})^2$ = $\frac{b^2}{4a^2}$
Now add this to both sides:
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}+-\frac{c}{a}$
Write the left side as a perfect square:
$(x^2+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}$
Simplify the right hand side by finding a common denominator:
This is where I'm tripping up
$\frac{b^2}{4a^2}-\frac{c}{a}$
The common denominator will be the product of the denominators so $4a^3$
This doesn't "feel" right and I suspect I should be looking for a "least common denominator" but I don't know what that would be given the existence of the radical.
Rewriting using the common denominator $4a^3$ I multiply the numerator and denominator of left side of the minus sign by just $a$. I then multiple the numerator and denominator on the right side of the minus sign by $4a^2$:
$\frac{b^2a}{4a^3}-\frac{4a^2c}{4a^3}$ = $\frac{b^2a-4a^2c}{4a^3}$
How can I arrive at $\frac{b^2-4ac}{4a^2}$?
I know that I'm not done yet after figuring out the above, but it's this in between step I'm tripping up on.
| In your last step $\frac{b^2a - 4a^2c}{4a^3}$ cancel $a$ from numerator and denominator, to get $ \frac{b^2 - 4ac}{4a^2}$
or take the common denominator as $4a^2$ in the beginning itself.
| {
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"timestamp": "2023-03-29T00:00:00",
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Elementary Proof Relating to Pythagorean Triples Please could I get some feedback on the correctness and overall quality of the following proof, which is meant to show that all Pythagorean triples contain unique integers.
To Prove: all Pythagorean triples contain unique integers.
First we note that in a right-angled triangle, if $c$ is the hypotenuse and $a,b$ are the remaining sides, $c > a$ and $c > b$ by triangle inequality. Now it remains to prove that
$\forall (a,b,c) \in \mathbb{N}$, where $a^2+b^2=c^2$, $a \ne b$
We use contradiction.
Assume that $a = b$. Then we have $a^2 + a^2 = c^2$
Thus, $c = \sqrt{2a^2} = a\sqrt{2}$
Since an integer multiple of an irrational number remains irrational, $c \notin \mathbb{N}$, which is a contradiction of our original assumption.
Therefore $a \ne b$
Since $a \ne b$ and $c > a$ and $c > b$,
$a,b,c$ are distinct integers and theorem is proven.
$\square$
| We are given$\quad A=m^2-n^2\quad B=2mn\quad C=m^2+n^2.$ and we wish to prove that that $A,B,C$ are unique integers.
If $A=B$ then$\quad(m^2-n^2)^2+(m^2-n^2)^2=C^2\Rightarrow C=(m^2-n^2)\sqrt{2}$ so it is not Pythagorean triple where all sides are integers.
If $B=A$ then$\quad (2mn)^2+(2mn)^2=8m^2n^2=C\Rightarrow C=2mn\sqrt{2}$ so it is also not Pythagorean triple where all sides are integers.
If $A=C$ then$\quad m^2-n^2=m^2+n^2\Rightarrow 0=2n^2$ and any zero value of $n$ will produce a trivial triple where $B=0$ such as $(1,0,1)$.
If $B=C$ then $\quad 2mn=m^2+n^2\Rightarrow m^2-2mn+n^2=(m-n)^2=0$. This can only be true if $m=n$ and that will always produce a trivial triple where $A=0$ such as $(0,1,1)$
None of these conditions produce a Pythagorean triple. Therefore, $A,B,C$ are always distinct integers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Having the identity: $\int_0^1 \frac{x^{2n}}{1+x^2} dx= \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1}$, how do I square the fraction inside the sum? I have the integral:
\begin{align}
\int_0^1 \frac{x^{2n}}{1+x^2} dx &= \int_0^1 x^{2n} \sum_{k=0}^{\infty}(-1)^kx^{2k} dx\\
&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1}
\end{align}
For $n=0$ we have $\pi/4$. So, for some positive integer $n$, the integral gives something in terms of $\pi/4$.
I wish the fraction on right hand to have a squared denominator inside the sum, like this:
$$ \sum_{k=0}^{\infty} \frac{(-1)^k}{(2n+2k+1)^2} $$
Because for a positive integer $n$ it gives something in terms of the Catalan constant $(1-1/3^2+1/5^2-\cdots)$.
one way to do this is the following (due to Beukers):
\begin{align}
\int_0^1 \frac{x^{2n+y}}{1+x^2} dx &= \int_0^1x^{2n+y} \sum_{k=0}^{\infty}(-1)^k x^{2k} dx\\
&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2n+2k+1+y}
\end{align}
Differentiate inside the integral with respect to $y$:
\begin{align}
& \int_0^1 \frac{\partial}{\partial y} \frac{x^{2n+y}}{1+x^2} dx
= \sum_{k=0}^{\infty} \frac{\partial}{\partial y} \frac{(-1)^k}{n+k+1+y} \\
& \int_0^1 \frac{x^{2n+y}}{1+x^2} \ln x \: dx = \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(n+k+1+y)^2}
\end{align}
for $y=0$
\begin{align}
\int_0^1 \frac{x^{2n}}{1+x^2} \ln x \: dx = \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{(n+k+1)^2}
\end{align}
The problem is the $\ln(x)$ because it makes the integral way too larger than before. If we chose some polynomial $(P_{2n}(x))$ with integer coefficients and even powers:
$$\int_0^1 P_{2n}(x)\ln (x)/(1+x^2)dx \gg \int_0^1 P_{2n}(x)/(1+x^2)dx$$
for some integer $n$. By the sign $\gg$ I mean that the integral is very "sensitivity" for values of $n$, even if the difference between the integrals are $0.01$ this already makes a big difference.
Does anyone have another idea?
| Really what you want to do is define the function
$$f_n(x)=\int_0^x\frac{t^{2n}}{1+t^2}dt=\int_0^xt^{2n}\sum_{k\geq0}(-1)^kt^{2k}dt\\
=\sum_{k\geq0}\frac{(-1)^k}{2n+2k+1}x^{2n+2k+1}$$
Then notice that the sum you seek is given by
$$g(n)=\int_0^1 f_n(x)\frac{dx}{x}=\sum_{k\geq0}\frac{(-1)^k}{(2n+2k+1)^2}=\frac14\sum_{k\geq0}\frac{(-1)^k}{(\frac{2n+1}2+k)^2}$$
To compute this, we recall the definition
$$\psi_m(x)=\left(\frac{d}{dx}\right)^{m+1}\log\Gamma(x)=(-1)^{m+1}m!\sum_{k\geq0}\frac1{(x+k)^{m+1}}$$
So we have that
$$\begin{align}
\sum_{k\geq0}\frac{(-1)^k}{(x+k)^m}&=\sum_{k\geq0}\frac{(-1)^{2k}}{(x+2k)^m}+\sum_{k\geq0}\frac{(-1)^{2k+1}}{(x+2k+1)^m}\\
&=2^{-m}\sum_{k\geq0}\frac1{(\frac{x}2+k)^m}-2^{-m}\sum_{k\geq0}\frac1{(\frac{x+1}2+k)^m}\\
&=\frac{(-1)^m}{2^m(m-1)!}\left[\psi_{m-1}\left(\frac{x}2\right)-\psi_{m-1}\left(\frac{x+1}2\right)\right]\tag{1}
\end{align}$$
So just plug in $m=2$ and $x=\frac{2n+1}{2}$ into $(1)$ to find $g(n)$:
$$\begin{align}
g(n)&=\frac14\sum_{k\geq0}\frac{(-1)^k}{(\frac{2n+1}2+k)^2}\\
&=\frac14\cdot\frac{(-1)^2}{2^2(1)!}\left[\psi_{1}\left(\frac{2n+1}4\right)-\psi_{1}\left(\frac{2n+3}4\right)\right]\\
&=\frac1{16}\left[\psi_{1}\left(\frac{2n+1}4\right)-\psi_{1}\left(\frac{2n+3}4\right)\right]
\end{align}$$
Alternatively, you could note that
$$f_n(x)=\int_0^x \frac{t^{2n}}{1+t^2}dt=x^{2n+1}\int_0^1 \frac{t^{2n}}{1+(xt)^2}dt$$
So you'd have the double integral
$$g(n)=\int_0^1\int_0^1\frac{(uv)^{2n}}{1+(uv)^2}dudv$$
Although evaluating such a double integral would require the use of the $\psi_1$ function, and would eventually lead us back to the result we just obtained.
| {
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"url": "https://math.stackexchange.com/questions/3216113",
"timestamp": "2023-03-29T00:00:00",
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Find tangents to a circle parallel to the straight line Find tangents to a circle $x^2+y^2=5$ parallel to the straight line $2x-y+1=0$
My solution:
$$x^2+y^2=5$$
$$S=(0,0)$$
$$r=\sqrt{5}$$
$$y=2x-1$$
$$a=2$$
Searching for b
$$y=2x+b$$
Using following formula:
$$d=\frac{Ax_0+By_0+C}{\sqrt{A^2+B^2}}$$
$$-2x+y-b$$
$$\sqrt{5}=\frac{-2\cdot0+1\cdot0+b}{\sqrt{(-2)^2+1^2}}$$
$$\sqrt{5}=\frac{b}{\sqrt{5}}$$
$b= 5$ or $b= -5$
Tangents:
$$y=2x+5$$
$$y=2x-5$$
==============================
Seems like it should be right since we got $\sqrt{5}$ at the end, but could someone look into it to make sure?
| *
*The correct distance formula between a point $P_0(x_0,y_0)$ and the straight line $Ax+Bx+C=0$ is $$d=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}.\tag{1}$$
*From $(1)$ and since the center of the given cirle ( $ x^2+y^2=5 $ ) is $(x_0,y_0)=(0,0)$, we obtain
$$d=\frac{|C|}{\sqrt{A^2+B^2}}.\tag{2}$$
*The straight line $2x−y+1=0$ (thick blue, in the figure below) is perpendicular to the straight line $(1/2)x+y=0$ (red), because two straight lines with equations
$$Ax+By+C=0\qquad\text{ and } \qquad A'x+B'y+C'=0\tag{3}$$ are perpendicular if and only if
$$AA'+BB'=0.\tag{4}$$
This line $(1/2)x+y=0$ defines the two tangent points to the circle, $(2,-1)$ and $(-2,1)$, which are the two solutions of the system of simultaneous equations, one representing the given circle $x^2+y^2=5$ (thick black) and the other, the straight line $(1/2)x+y=0$:
$$
\left\{
\begin{aligned}
x^2+y^2 &=5 \\
\frac{1}{2}x+y &=0
\end{aligned}
\right. \tag{5}
$$
*The family of the parallel lines to the given line $2x-y+1=0$ is defined by equation
$$2x−y+K=0,\tag{6} $$
where $K$ is a parameter.
*The two tangent lines, whose equations have been found by you ( $y=2x\pm 5$ ) correspond to $K=\pm 5$.
$$\mathbf{Figure.}\text{ Circle } x^2+y^2=5, \text{given line (thick blue) } 2x−y+1=0, \text{and the two tangents at the points } (2,−1) \text{ and }(−2,1) \text{ (thin blue).} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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general solution to a Non homogeneous differential equation Determine the general solution to the following differential equation:
$$y'' + 9y = t^2e^{3t}$$
The homogeneous solution is
$$y(t)=C_1\cos(3t) +C_2\sin(3t)$$
Solving the non homogeneous solution I let
$$y(t)=(At^2+Bt+C)e^{3t}$$
I found $A=1/36$, $B=-5/216$, $C=7/864$
Does anybody know if this is correct?
| $y" + 9y = t^2e^{3t}$
$(D^2 + 9)y = t^2e^{3t}$
Auxiliary equation: $m^2 + 9 = 0$ or $m = \pm3i$
$CF = C_1\cos(3t) + C_2\sin(3t)$
$PI = \frac{1}{D^2 + 9}t^2e^{3t}$
Isolating $e^{3t}$ and replacing $D$ by $D+3 $ (coeff. of t)
$PI = e^{3t}\frac{1}{(D+3)^2 + 9}t^2$
$PI = e^{3t}\frac{1}{D^2+6D+18}t^2$
$PI = e^{3t}\frac{1}{18}\frac{1}{1 + \frac{D^2+6D}{18}}t^2$
$PI = \frac{e^{3t}}{18} \bigg[1-\frac{D^2+6D}{18}+ (\frac{D^2+6D}{18})^2 \cdots\bigg]t^2$
$\bigg[D(t^2) = 2t \ , \ D^2(t^2) = 2 \ , \ D^3(t^2) = D^4(t^2) = \cdots = 0\bigg]$
$PI = \frac{e^{3t}}{18} \bigg[1-\frac{D^2+6D}{18}+ \frac{D^4+12D^3 + 36D^2}{18^2} \cdots\bigg]t^2$
$PI = \frac{e^{3t}}{18} \bigg[t^2 - \frac{2+6(2t)}{18} + \frac{0 + 0 + 36(2)}{18^2}\bigg]$
$PI = \frac{e^{3t}}{18}[t^2 - \frac{2t}{3} + \frac{1}{9}]$
$$PI = e^{3t}\bigg[\frac{t^2}{18} - \frac{t}{27} + \frac{1}{162}\bigg]$$
Thus, $$A = \frac{1}{18} \ , \ B = -\frac{1}{27} \ , \ C = \frac{1}{162}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for all $n \in \mathbb{N}$, either 3 or 13 divides $3^n + 13n^2 + 38$ Let $a\in \{3,13\}.$ I'm having trouble with this proof. I know that
$$3^{n+1} + 13(n+1)^2 + 38 = (3^n + 13n^2 + 38) + (2\cdot 3^n + 26n + 13)$$
But I can't prove that $a \mid 2\cdot3^n + 26n + 13$. I know that 13 doesn't divide this because $13 \nmid 2\cdot3^n$. How can I prove that $3 \mid 26n + 13$?
| Define $f(n) = 3^n + 13n^2 + 38$.
Using algebra it can be shown that for all $n \ge 0$ we have the following identities,
$\tag 1 f(n+1) = 3 f(n) -26 n^2 + 26 n -63$
$\tag 2 f(n+2) = 9 f(n) -104 n^2 + 52 n -252$
$\tag 3 f(n+3) = 27 f(n) -338 n^2 + 78 n -871 =$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \; \; \, \quad \quad 27 f(n) -13 (26 n^2 - 6 n + 67)$
By $\text{(3)}$, whenever $13 \mid f(n)$ it must also be true that $13 \mid f(n+3)$. Snce $f(0)) = 39$,
$\tag 4 13 \text{ divides every number in } \{f(0), f(3), f(6), \dots\}$
If we take $n = 3k$ in $\text{(1)}$ with $k \ge 0$, we can conclude that
[note that $3$ divides $3$, ${(3k)}^2$, $3k$ and $63$]
$\tag 5 3 \text{ divides every number in } \{f(1), f(4), f(7), \dots\}$
If we take $n = 3k$ in $\text{(2)}$ with $k \ge 0$, we can conclude that
[note that $3$ divides $9$, ${(3k)}^2$, $3k$ and $252$]
$\tag 6 3 \text{ divides every number in } \{f(2), f(5), f(8), \dots\}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How many Taylor series terms are needed to accurately approximate $\sqrt{a+x}-\sqrt{a}$? Naive evaluation of $\sqrt{a + x} - \sqrt{a}$ when $|a| >> |x|$ suffers from catastrophic cancellation and loss of significance.
WolframAlpha gives the Taylor series for $\sqrt{a+x}-\sqrt{a}$ as:
$$\frac{x}{2 \sqrt{a}} - \frac{x^2}{8 a^{3/2}} + \frac{x^3}{16 a^{5/2}} - \frac{5 x^4}{128 a^{7/2}} + \frac{7 x^5}{256 a^{9/2}} + O(x^6)$$
which (I think) equals:
$$\sqrt{a} \left( \frac{1}{2} \left(\frac{x}{a}\right) - \frac{1}{8} \left(\frac{x}{a}\right)^2 + \frac{1}{16} \left(\frac{x}{a}\right)^3 - \frac{5}{128} \left(\frac{x}{a}\right)^4 + \frac{7}{256} \left(\frac{x}{a}\right)^5 + O\left(\left(\frac{x}{a}\right)^6\right) \right)$$
How quickly do the coeffients decrease?
How many terms are needed to reach $53$ bits of accuracy (IEEE double precision) in the result given that $10^{-300} < \left|\frac{x}{a}\right| < 1$ is known?
Alternatively, what are the threshold values of $\left|\frac{x}{a}\right|$ where the number of terms changes?
What about rounding errors, assuming each value is stored in double precision?
| It was pointed out by Robert Israel that the series does badly when $|x| \approx |a|$, but in that case the loss of significance of the naive evaluation is small.
It was also suggested by Winther (and a since-deleted answer) to rewrite as $$\frac{x}{\sqrt{a+x}+\sqrt{a}}$$
The series for the denominator is similar to the series in the question, only with a leading constant term. This means that when $\left|\frac{x}{a}\right|$ is small enough, the addition of terms eventually becomes insignificant in double arithmetic.
If $\left|\frac{x}{a}\right| < 2^{-52}$, $1$ term is sufficient. Otherwise
If $\left|\frac{x}{a}\right| < 2^{-25}$, $2$ terms are sufficient. Otherwise
If $\left|\frac{x}{a}\right| < 2^{-16}$, $3$ terms are sufficient. Otherwise
If $\left|\frac{x}{a}\right| < 2^{-11}$, $4$ terms are sufficient. Otherwise
If $\left|\frac{x}{a}\right| < 2^{-9}$, $5$ terms are sufficient. Otherwise
If $\left|\frac{x}{a}\right| > 2^{-9}$, the loss of precision in the addition $a + x$ is relatively small.
But in fact, perhaps $$\frac{x}{\sqrt{a + x} + \sqrt{a}}$$ evaluated in double precision is good enough for all $|x| << |a|$ and series are unnecessary?
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Explicit formula for a sequence of alternating differences Imagine this sequence: $1, 11, 12, 22, 23, 33, 34, 44$, ...
Obviously the sequence is alternating in adding 10 and adding 1. You can find the formula for the differences easily with $5*(-1)^x+6$, but I can't find a way to find the exact explicit formula for the sequence, assuming there is one.
| If you try to find a formula for the terms in positions $2,4,6,\ldots$ which are $11,22,33,\ldots$ you get $11 \frac{x}2 = \frac{11}{2}x$
If you try to find a formula for the terms in positions $1,3,5,\ldots$ which are $1,12,23,\ldots$ you get $11 \frac{x-1}2 +1 = \frac{11}{2}x-\frac{9}{2}$
These expressions are of the form $\frac{11}{2}x-\frac{9}{4} \pm \frac{9}{4}$ and you can see that the $x$th term is $$\frac{11}{2}x-\frac{9}{4} + \frac{9}{4}(-1)^x$$
| {
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"url": "https://math.stackexchange.com/questions/3224967",
"timestamp": "2023-03-29T00:00:00",
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Confused on an algebraic step of inductive proof
The part I've boxed is what is throwing me - why is it 6(k+1) instead of 6(k+1)^2?
| Let's add some steps in:
\begin{align*}
1^2 + 2^2 + \cdots + k^2 + (k + 1)^2 &= \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \\
&= \frac{(k+1)}{6} \cdot\frac{k(2k+1)}{1} + (k+1)(k+1) \\
&= \frac{(k+1)}{6}k(2k+1) + \frac{6(k+1)}{6}(k+1) \\
&= \frac{(k+1)}{6}[k(2k+1) + 6(k+1)].
\end{align*}
It's not squared, because a factor of $(k + 1)$ needed to be factorised out, leaving only one of the factors of $(k + 1)$ behind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $(x-1)^2$ is a factor of $x^n -nx +n-1$
Show that $(x-1)^2$ is a factor of $x^n -nx +n-1$
By factor theorem we know that $(x-a)$ is a factor of $f(x)$ if $f(a)=0$.
In this case, $f(x)=x^n -nx +n-1 \implies f(1)=0$
Hence we conclude that $(x-1)$ is a factor. From hereon, how can I say that $(x-1)^2$ is a factor?
Can we approach the problem without calculus approach? This problem was taken from a book of pre-calculus algebra.
| By division theorem we have:
$$x^n -nx +n-1 = k(x)(x-1)^2+ax+b $$
So for $x=1$ we have $0 = 0+a+b$ so $a=-b$, so $$x^n -nx +n-1 = k(x)(x-1)^2+a(x-1) $$
$$(x-1)(x^{n-1}+...+x^2+x+1) -n(x-1)= k(x)(x-1)^2+a(x-1) $$
so, after dividing by $x-1$ we get $$(x^{n-1}+...+x^2+x+1) -n= k(x)(x-1)+a $$
Now puting $x=1$ again we have $$\underbrace{1+1+...+1}_n -n = 0+a\implies a=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
} |
An indefinite Integral Problem with algebric numerator and trigonometric denominator $$\int \frac{x^2+(n(n-1))}{(x\sin x +n\cos x)^2 } dx$$
I know this is an homework problem, but I really couldn't think of any way to solve it. Like DI Method (No go) , What kind of substitution as denominator is trigonometric whereas Numerator is algebric. Thought of n(n-1) can come by double differentiating but.. like how would we have it here ... etc confusing and weird thoughts. Please help me out
| $I = \int\frac{x^2 + n(n-1)}{(x\sin x + n\cos x )^2}dx$
Now we'll try to convert it into the form of $\frac{a}{y} + \frac{b}{y^2}$, where $a,b$ are functions of $x$ and $y$ is the denominator.
$$x^2+n(n-1) =(x\sin x + (n-1)\cos x)(x\sin x+n\cos x)-((1-n)\sin x + x\cos x)(n\sin x - x\cos x)$$
$$$$
$I = \int \big[\frac{(x\sin x + (n-1)\cos x)(x\sin x+n\cos x)}{(x\sin x + n\cos x )^2} - \frac{((1-n)\sin x + x\cos x)(n\sin x - x\cos x)}{(x\sin x + n\cos x )^2}\big]dx$
Now,
$I = \int\big[\frac{(x\sin x + (n-1)\cos x)}{(x\sin x + n\cos x )} - \frac{((1-n)\sin x + x\cos x)(n\sin x - x\cos x)}{(x\sin x + n\cos x )^2}\big]dx$
Let $I_1 = \int\frac{(x\sin x + (n-1)\cos x)}{(x\sin x + n\cos x )}dx$ , $I_2 = \frac{((1-n)\sin x + x\cos x)(n\sin x - x\cos x)}{(x\sin x + n\cos x )^2}dx$
In $I_2$,
let $u = n\sin x - x\cos x$, $dv = \frac{(1-n)\sin x + x\cos x}{(x\sin x + n\cos x )^2}dx$
$du = (n\cos x - \cos x + x\sin x)dx$ ,
[In $v$ , $t = x\sin x + n\cos x$, $dt = (x\cos x + \sin x - n\sin x )dx = (x\cos x + (1-n)\sin x)dx$]
$v = \int\frac{(1-n)\sin x + x\cos x}{(x\sin x + n\cos x )^2}dx = \int\frac{dt}{t^2} = -\frac{1}{t} = - \frac{1}{x\sin x + n\cos x}$
So,
$I_2 = uv - \int vdu = -(n\sin x - x\cos x)\frac{1}{x\sin x + n\cos x} + \int (n\cos x - \cos x + x\sin x).\frac{1}{x\sin x + n\cos x}dx + c$
$I_2 = - \frac{n\sin x - x\cos x}{x\sin x + n\cos x} + \int \frac{x\sin x + (n-1)\cos x}{x\sin x + n\cos x}dx +c$
$I_2 = - \frac{n\sin x - x\cos x}{x\sin x + n\cos x} + I_1 +c $
$$I = I_1 - I_2 = \frac{n\sin x - x\cos x}{x\sin x + n\cos x} + k$$
($k = -c$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
find limit of a multivariable function I have to show continuity at $(0,0)$ of $f(x,y)=\frac{\sin(x^2) + \sin(y^2)}{\sqrt{x^2 +y^2}}$ for $(x,y)\ne(0,0)$ and $f(0,0)=0$.
I tried to find the limit using polar coordinates
$ \frac{\sin(r^2\cos^2 α) + \sin(r^2\sin^2 α)}{r} $
but it's still $\frac00$
| \begin{align*}T&:=\frac{\sin \ x^2+\sin\ y^2}{\sqrt{x^2+y^2}}
\\&=\frac{\sin \ x^2 }{x^2}\frac{x^2}{
\sqrt{x^2+y^2}} +
\frac{\sin\ y^2}{y^2}\frac{y^2}{
\sqrt{x^2+y^2}}
\\&\leq \frac{\sin \ x^2 }{x^2}\frac{x^2 +y^2}{
\sqrt{x^2+y^2}} +
\frac{\sin\ y^2}{y^2}\frac{x^2+y^2}{
\sqrt{x^2+y^2}} \rightarrow 0 \end{align*}
When $x,\ y\rightarrow 0$, then $T\geq 0$ so that $T$ has a limit
$0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve recurrence with characteristic polynomial $a_n=7\cdot a_{n-1} -7\cdot a_{n-2}+175\cdot a_{n-3}+450\cdot a_{n-4}+(5+13\cdot n)\cdot9^n $ The equation
$$a_n=7\cdot a_{n-1} -7\cdot a_{n-2}+175\cdot a_{n-3}+450\cdot a_{n-4}+(5+13\cdot n)\cdot9^n \enspace,$$
where
$a_0=148, a_1=144, a_2=-55, a_3=-61$.
I assume that a solution will look like $a^s_n+a^h_n$ , where $a^s_n$. I solve like
$$x^4-7x^3+7x^2-175x-450=0 \enspace,$$
then get $x_1=-2,\ \ x_2=9,\ \ x_3=-5i,\ \ x_4=5i\ \ $
and don't know what next.
But the real problem is with the second part $(5+13\cdot n)\cdot9^n$. I completely don't know what do with that.
| Considering the homogeneous
$$
a_n-7\cdot a_{n-1}+7\cdot a_{n-2}-175\cdot a_{n-3}-450\cdot a_{n-4}=0
$$
after substituting $a_n = C_0 \beta^n$ we obtain
$$
C_0(\beta -9) (\beta +2) \left(\beta ^2+25\right) \beta ^{n-4}=0
$$
so we have the set of exponent solutions (for $n > 4$ assuming $\beta\ne 0$)
$$
\{-2,9,\pm i 5\}
$$
then
$$
a_n^h = C_1(-2)^n+C_29^n+C_3 (i 5)^n+C_4 (-i 5)^n
$$
for the particular we adopt
$$
a_n^p = (c_1+c_2n+c_3n^2)9^n
$$
and after substitution we obtain
$$
a_n^h = 9^n \left(c_1+\frac{16996635 n}{1359556}+\frac{9477 n^2}{2332}\right)
$$
and finally
$$
a_n = C_1(-2)^n+C_29^n+C_3 (i 5)^n+C_4 (-i 5)^n + 9^n \left(c_1+\frac{16996635 n}{1359556}+\frac{9477 n^2}{2332}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Angle formed by orthocenter, incenter and circumcenter of a triangle $>135^\circ$? If $H$ is the orthocenter, $I$ the incenter and $O$ the circumcenter of a triangle , the I need to show that the angle $HIO>135^\circ$
With the assumptions of $OI^2=R^2-2Rr$, $OH^2=9R^2-(a^2+b^2+c^2)$, $HI^2=2r^2-4R^2\cos A\cos B \cos C$ and $R^2\cdot8(1+\cos A\cos B\cos C)= a^2+b^2+c^2$
Applying the $cosine$ rule I got to something like $$\cos (HIO)=\frac{2r^2-2Rr+R^2\cdot 4\cos A\cos B\cos C}{2(2r^2-4R^2\cos A\cos B\cos C)(R^2-2Rr)}$$
I need to show that the LHS is between $\left[-\frac{1}{\sqrt{2}},-1\right]$
How t0 proceed?
Is there any simpler way to prove the question (without applying cosine rule) ?
| This formula calculates the $θ$ angle formed by Orthocenter(H),Incenter(I) and circumcenter(O) of ABC triangle (sides:a,b,c).
$θ(a,b,c)=acot\left [\left [ \frac{(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}{(-a+b+c)(a-b+c)(a+b-c)}-2abc+(-a+b+c)(a-b+c)(a+b-c) \right ] \sqrt{\frac{(-a+b+c)(a-b+c)(a+b-c)}{4(a+b+c)^3(a-b)^2(a-c)^2(b-c)^2}} \right ] $
$θ(3,10,11)= π -acot(\frac{523}{1344}\sqrt{6})=133.627°$
defining $θ$ as a function in the interval of existence of the ABC triangle to use the Calculus tools
$\begin{array}{} |b-c|<a<b+c & a≠b & a≠c \end{array} $
$Range(θ)=(\frac{ π }{2}, π )$
$θ=\frac{ π }{2}$ in the right triangle, $θ(\sqrt{b^2+c^2-2bc·cos(\frac{ π }{2})},b,c)$ whith $c→∞$ or $c→0$. $θ=π$ for $a=|b-c|$, $a=b$, $a=c$ and $a=b+c$.
maximum and minimum
$\begin{array}{} \frac{d}{da}θ(a,b,c)=0 & solve&for&a & \end{array}$
$\begin{array}{} r1=2.43938 & r2=10.54162 & r3=12.39471 \end{array}$
$θ(r3,b,c)=132.292°$
$\begin{array}{} θ(a,10,11)=135°=\frac{3}{4}π & a1=1.82913 & a2=3.27918 \end{array}$
$\begin{array}{} \frac{R}{r}<4 & ? \end{array}$
$k=\frac{R}{r}=\frac{2abc}{(-a+b+c)(a-b+c)(a+b-c)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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} |
Square valued integer polynomial For what integer values of $n$ is the expression $n^{6}+n^{4}+1$ a square?
This is a square when $n=2$; when $n$ is odd, this expression is $3 \ mod(8)$ and so cannot be a square.
| The only value that works is $n=2$.
Indeed, if there is another value that works, then there exists integers $n > 2$ and $k>0$ such that
$$n^6 + n^4 +1 = (n^3+k)^2$$
which implies that there exists exists integers $n > 2$ and $k>0$ such that
$$n^4+1 = 2n^3+k^2$$
However, note that on the one hand, for $k=n/2$ that $2n^3+k^2 = n^4+n^2/4$ which is strictly greater than $n^4+1$ for $n > 2$ [and equal to $n^4+1$ for $n=2$]. On the other hand, or $k=n/2-1$, note that
$2n^3+k^2 < n^4-n^3+n^2/4 < n^4+1$ for $n > 2$.
Can you see how this implies no integral $k$ will solve the equation $n^4+1 = 2n^3+k^2$, and then in turn $n^4+1 = 2n^3+k^2$ for integral $n >2$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x + y + z = 2$, then show $\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$, added a second question(Problem 2). Problem number 1:
The problem is that $x, y, z$ are proper fractions, and each one of them is greater than zero.
Given $x + y + z = 2$, prove
$$\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$$
I have tried to solve this using AM $\geq$ GM inequality.
Attempt :
$$\frac{\dfrac{1-x}{x} + \dfrac{1-y}{y} + \dfrac{1-z}{z}}{3}\geq\left(\frac{(1-x)(1-y)(1-z)}{xyz}\right)^{1/3}$$
What should I do to calculate the value of $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$?
Doubt:
*
*Let $a, b, c, d$ all be positive numbers.
*Let $a > c$ and $b > d$.
*Can we say that $\dfrac {a}{b} > \dfrac{c}{d}$?
Problem number 2:
$x,y,z$ are unequal positive quantities, prove that
$(1+x^3)(1+y^3)(1+z^3) > (1+xyz)^3$
My attempt:
$ \frac{x^3+y^3+z^3}{3} > xyz$
$ => (1+x^3)+(1+y^3)+(1+z^3) > 3(1+xyz)$
now by cubing the both sides we get,
$ => \frac{(1+x^3)+(1+y^3)+(1+z^3)}{27} > (1+xyz)^3$ ---(eqn. 1)
$ \frac{(1+x^3)+(1+y^3)+(1+z^3)}{3} > {\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)}^{\frac{1}{3}}$
$=> \frac{(1+x^3)+(1+y^3)+(1+z^3)}{27} > {\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)}$ ---(eqn. 2)
So, now if we do : $\frac{(eqn. 1)}{(eqn. 2)}$
then the result will be :
${\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)} > (1+xyz)^3$
and this is the correct result for all $x,y,z$ greater than zero.
But if we do $\frac{(eqn. 2)}{(eqn. 1)}$ then the result will be :
${\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)} < (1+xyz)^3$
So, my question is, how to prove it correctly without doing the division operation, cause when we divide the two equations then we get two separate results.
| We have $x+y+z=2$ so $1-x+1-y+1-z=3-(x+y+z)=1$ using Am-GM inequality twice once for $$x+y+z\geq 3(xyz)^{\frac{1}{3}}$$ and second for $$1-x+1-y+1-z\geq 3((1-x)(1-y)(1-z))^{\frac{1}{3}}$$,cubing both inequalities and dividing the two should yield the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum_{0\leq k\leq r} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$ where $m,n,r\in \mathbb{N} $ $\sum_{0\leq k\leq r} \binom{n+k}{k} \binom{m+n+k}{n+k} = \binom{m+n}{n} \binom{m+n+r+1}{m+n+1}$ where $m,n,r\in \mathbb{N} $.
Exam problem which stayed unproven for me. I tried induction/binomial properties together and separately.I also tried to find a power serie which could use these binomials. Could someone provide at least a hint?
| We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n }{k}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^r}&\color{blue}{\binom{n+k}{k}\binom{m+n+k}{n+k}}\\
&=\sum_{k=0}^r\frac{(n+k)!}{n!k!}\cdot\frac{(m+n+k)!}{m!(n+k)!}\\
&=\binom{m+n}{n}\sum_{k=0}^r\binom{m+n+k}{k}\\
&=\binom{m+n}{n}\sum_{k=0}^r[z^k](1+z)^{m+n+k}\tag{2}\\
&=\binom{m+n}{n}[z^0](1+z)^{m+n}\sum_{k=0}^r\left(\frac{1+z}{z}\right)^k\tag{3}\\
&=\binom{m+n}{n}[z^0](1+z)^{m+n}\frac{\left(\frac{1+z}{z}\right)^{r+1}-1}{\frac{1+z}{z}-1}\tag{4}\\
&=\binom{m+n}{n}[z^r](1+z)^{m+n}\left((1+z)^{r+1}-z^{r+1}\right)\tag{5}\\
&\,\,\color{blue}{=\binom{m+n}{n}\binom{m+n+r+1}{r}}\tag{6}
\end{align*}
and the claim follows.
Comment:
*
*In (2) we use the coefficient of operator according to (1).
*In (3) we do some rearrangements and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (4) we use the finite geometric series expansion.
*In (5) we do some simplifications and apply the rule from (3) again.
*In (6) we select the coefficient of $z^r$ from $(1+z)^{m+n+r+1}$ noting that the other term $-(1+z)^{m+n}z^{r+1}$ does not contribute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove or disprove: If $a+b \leq \frac{1}{2}$, then $\frac{(1-a)(1-b)}{ab} \geq 1$ for positive $a,b$
Let $a,b$ be two positive numbers. Prove or disprove the statement:
If $a+b \leq \frac{1}{2}$, then $\dfrac{1-a}{a} \dfrac{1-b}{b} \geq 1$.
True. Assume $a+b \leq \frac{1}{2}$. Then
$$\dfrac{1-a}{a} \dfrac{1-b}{b}=\dfrac{1-b-a+ab}{ab}=\dfrac{1}{ab}-\dfrac{a+b}{ab}+1=-\dfrac{a+b}{ab}+\dfrac{1}{ab}+1\geq \dfrac{-1}{ab}+\dfrac{1}{ab}+1=1. $$
Can you check my answer?
| $$\frac{1- a}{a}\times \frac{1- b}{b}> \frac{(\,a+ b\,)- a}{a}\times \frac{(\,a+ b\,)- b}{b}= 1$$
$$\because\,1> a+ b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the probability that the first 1 comes in even trials of a six faced fair dice? A six faced fair dice is thrown until 1 comes,
then what is the probability that 1 comes in even no. of trials?
I tried my best and my efforts include:
I find the
probability of getting a 1 on 2nd throw $p(2) = (5/6) (1/6)$
probability of getting a 1 on 4th throw $p(4) = (5/6)^3 (1/6)$
probability of getting a 1 on 6th throw $p(6) = (5/6)^5 (1/6)$
then I combine them like $p = p(2)+p(4)+p(6)+ \ldots$
now, which formula should I apply to solve this problem?
| Observe that if we define $\Pr(2k)$ to be the probability that the first 1 appears on the $2k$th throw, then
$$\Pr(2k) = \left(\frac{5}{6}\right)^{2k - 1}\left(\frac{1}{6}\right)$$
Hence, the probability that the first one appears on an even-numbered throw is
$$\sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 1}\left(\frac{1}{6}\right)$$
By extracting a factor of $\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)$, we can obtain a convergent geometric series.
\begin{align*}
\sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 1}\left(\frac{1}{6}\right) & = \frac{1}{6} \sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 1}\\
& = \left(\frac{1}{6}\right)\left(\frac{5}{6}\right)\sum_{k = 1}^{\infty} \left(\frac{5}{6}\right)^{2k - 2}\\
& = \frac{5}{36} \sum_{k = 1}^{\infty} \left[\left(\frac{5}{6}\right)^2\right]^{k -1}\\
& = \frac{5}{36} \sum_{k = 1}^{\infty} \left(\frac{25}{36}\right)^{k - 1}\\
& = \frac{5}{36} \cdot \frac{1}{1 - \frac{25}{36}} & \text{geometric series with $r = \frac{25}{36}$}\\
& = \frac{5}{36} \cdot \frac{1}{\frac{11}{36}}\\
& = \frac{5}{36} \cdot \frac{36}{11}\\
& = \frac{5}{11}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/3246798",
"timestamp": "2023-03-29T00:00:00",
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A limit with sum $S_n=1+\frac{n-1}{n+2}+\frac{n-1}{n+2}\cdot \frac{n-2}{n+3}+\cdots +\frac{n-1}{n+2}\cdot \frac{n-2}{n+3}\cdots \frac{1}{2n}$ $$S_n=1+\frac{n-1}{n+2}+\frac{n-1}{n+2}\cdot \frac{n-2}{n+3}+\cdots +\frac{n-1}{n+2}\cdot \frac{n-2}{n+3}\cdots \frac{1}{2n}$$ Then $S_n/\sqrt{n}$ tends to $\frac{\pi}{2}$.
How to show then? the general term in $S_n$ is $$\frac{n-1}{n+2}\frac{n-2}{n+3}\cdots \frac{n-k}{n+k+1}=\frac{C_{n+1}^{k+2}}{C_{n+k+1}^{k+2}}$$ where $C_n^k=\frac{n!}{k!(n-k)!}$. Then how to do?
| The logarithm of a term is roughly
$$\int_{n-k-\frac12}^{n-\frac12}\log t dt - \int_{n+\frac32}^{n+k+\frac32}\log t dt$$
Find the leading order behaviour of that, then approximate the sum $S_n by an integral.
| {
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"url": "https://math.stackexchange.com/questions/3247505",
"timestamp": "2023-03-29T00:00:00",
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Prove for all positive a,b,c that $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$ Prove for all positive a,b,c $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$$
My Try
I tried taking common denominator of the expression,
$\frac{a^2b+ab^2+b^2c+c^2b+ac^2+a^2c}{abc}$
How to proceed? Is there a way to write them as perfect squares to get the least value? a Hint is much appreciated. Thanks!
| Notice that $$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge 3 \left( \frac{a}{b} \frac{b}{c} \frac{c}{a}\right)^{1/3}=3~~~~ \mbox{by AM-GM}$$
and $$\frac{b}{a}+ \frac{c}{b}+ \frac{a}{c} \ge 3 \left( \frac{b}{a} \frac{c}{b} \frac{a}{c}\right)^{1/3}=3~~~~ \mbox{by AM-GM}.$$ Adding these two we get the required inequality.
In each case and overall equality occurs when $a=b=c$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $x^2-8x+17>0$ for all real values of x Could someone please help me with this question. I thought I had to use the discriminant, which I found to be -4 but now I'm not sure if i'm just meant to factorise, as the discriminant would show there are no real solutions.
Thank you :)
| If the discriminant is negative then there are no values where the expression is equal to zero and as values can't "jump" from positive to negative without "passing through" zero then all values must be either all positive or all negative.
And as for $x=0$ this takes a positive value then all values are positive.
In general, if the determinate is $D = b^2 - 4ac$ and $a\ne 0$ then
$ax^2 + bx + c = $
$a(x + 2\frac {b}{2a}x + \frac {b^2}{4a^2}) + c - \frac {b^2}{4a} =$
$a(x + \frac b{2a})^2 + \frac {4ac - b^2}{4a} = $
$a(x+\frac b{2a})^2 - \frac D{4a}$.
Now $(x + \frac b{2a})^2 \ge 0$ because it is a square.
So if $a > 0$ then $ax^2 + bx + c = a(x+\frac b{2a})^2 - \frac D{4a} \ge -\frac D{4a}$.
If $D < 0$ then this is always a positive value.
If $a < 0$ then $ax^2 + bx + c = a(x+\frac b{2a})^2 -\frac D{4a} \le -\frac D{4a}$. If $D < 0$ then this is always a negative value. (Note: $D$ and $a$ are both negative so the "cancel each other out").
(Note: if $D = b^2 - 4ac < 0$ then $ac > \frac {b^2}4 \ge 0$. So $a$ and $c$ must either both be positive of both be negative.)
...
But it's easier to just complete the square:
$x^2 -8x + 17 = $
$(x^2 - 8x) + 17 = $
$(x^2 - 8x + 16) +17 - 16 = $
$(x-4)^2 + 1$.
And $(x-4)^2 \ge 0$ so $(x-4)^2 + 1 \ge 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Double Integration Problem $\int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$
Compute
$$I = \int_{0}^{1} \int_0^1 \frac{1}{1+y(x^2-x)}dydx$$
Here are my steps:
$$\begin{split}
I &=\int_{0}^{1} \left(\int_0^1 \frac{dy}{1+y(x^2-x)}\right)dx\\
&=\int_{0}^{1} \left[\frac{\ln(1+y(x^2-x))}{x^2-x}\right]_0^1dx\\
&=\int_{0}^{1} \left[\frac{\ln(1+(1)(x^2-x))}{x^2-x}
-\frac{\ln(1+(0)(x^2-x))}{x^2-x}\right]dx\\
&=\int_{0}^{1} \frac{\ln(1+x^2-x)}{x^2-x}dx
\end{split}
$$
And here I can't find any substitution to solve this integral.
Can anyone help me?
By the way, I also used Simpson's 3/8 method to find the approximation and got $1.063$.
But I want to find it using Calculus.
| Switching the order of integration, we get $$4 \int_{y=0}^1 \frac{1}{\sqrt{y(4-y)}} \tan^{-1} \sqrt{\frac{y}{4-y}}\, dy.$$ Then the substitution $$u = \tan^{-1} \sqrt{y/(4-y)}, \quad du = \frac{1}{1+\frac{y}{4-y}} \cdot \frac{2}{(4-y) \sqrt{y(4-y)}} \, dy = \frac{1}{2 \sqrt{y (4-y)}} \, dy,$$ gives $$4 \left[\left(\tan^{-1} \sqrt{\frac{y}{4-y}} \right)^2\right]_{y=0}^1 = \frac{\pi^2}{9}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplifying $\sum_{cyc}\tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)$. I get $0$, but the answer is $\pi$. So the question is
$$ \tan^{-1}\left(\sqrt{\frac{x(x+y+z)}{yz}}\right)+\tan^{-1}\left(\sqrt{\frac{y(x+y+z)}{xz}}\right)+\tan^{-1}\left(\sqrt{\frac{z(x+y+z)}{yx}}\right) =\ ? $$
So my take on the question is to rewrite it as
$$ \tan^{-1}\left(x\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(y\sqrt{\frac{(x+y+z)}{xyz}}\right)+\tan^{-1}\left(z\sqrt{\frac{(x+y+z)}{yzx}}\right) $$
Then say $$\frac{x+y+z}{yzx}= a^2.$$
We get
$$ \tan^{-1}\left( \frac{a((x+y+z)-a^2xyz)}{1-a^2(xy+yz+zx)}\right)$$
And since $ (x+y+z) = a^2xyz $ , this is just equal to $\tan^{-1}(0)= 0 $ but the answer given is $\pi.$
| Let $x$, $y$ and $z$ be positive numbers. We consider
a triangle $ABC$ with side lengths $a=BC=y+z$, $b=CA=x+z$ and $c=AB=x+y$. The semi-perimeter $s=x+y+z$ inradius $r$. Now, by Heron’s formula we have
$$\eqalign{\cot(A/2)&=\frac{s-a}{r}=\frac{s(s-a)}{{\rm Area}(ABC)}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\cr
&=\sqrt{\frac{x(x+y+z)}{yz}}}$$
So,
$$\eqalign{\tan^{-1}\sqrt{\frac{x(x+y+z)}{yz}}&=\frac{\pi-A}{2}\cr
\tan^{-1}\sqrt{\frac{y(x+y+z)}{zx}}&=\frac{\pi-B}{2}\cr
\tan^{-1}\sqrt{\frac{z(x+y+z)}{xy}}&=\frac{\pi-C}{2}}$$
Adding we get $\pi$ as a sum.
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation Involving Product Of Two Identical Polynomials. Recently I stuck, to a problem. However I rarely think that there is some proper formula for this problem, but here I am in search of algorithm's or theorem that relate to this problem or can solve this problem.
We have three integers positive integers $a, b, n$ and we need to compute this summation:
$$\sum\limits_{k = 1}^n{(k^a - {(k - 1)}^a)(k^b - {(k - 1)}^b)}$$
It looks simple, if we try to solve it using computer, but real problem lies in it's constraints:
$$n < 10^{12}$$
$$a < 10^4$$
$$b < 10^4$$
Clearly, we cannot iterate from $k = 1$ to $k = n$, thus we need to think of an algorithm that solve it in constant time
or in complexity in terms of $a$ and $b$
One can easily find relation like this $\sum\limits_{k = 1}^n{(k^a - {(k - 1)}^a)} = n^a$ but the problem has product of two such terms. So, Please give me some suggestion to solve this problem.
| This is not a full answer because the bounds for $a$ and $b$ given by OP imply that there are situations where $a>>n$ and $b>>n.$ (For example, $n=10, a=1000, b=2000.$) The following asymptotic formula works well if $n>>a$ and $n>>b.$ Given the square of bounds for the parameters for $a$ and $b$ relative to $n$, the formula actually covers about all but 1 part in 10^16 of the domain.
$$ S:=\sum_{k=1}^n \big(k^a -(k-1)^a)\big(k^b -(k-1)^b) \sim
\frac{n^{a+b-1} a\,b}{a+b-1} \Big( 1 - \frac{(a-1)(b-1)}{12 n^2} \frac{a+b-1}{a+b-3)} \Big) $$
The proof is easily enough constructed by using a few terms of the Faulhaber formula,
$$ (*) \quad \sum_{k=1}^n k^m = \frac{n^{m+1}}{m+1} + \frac{n^m}{2} + m \, \frac{n^{m-1}}{12} + \binom{m+1}{3}\frac{B_3}{m+1}n^{m-2} + ... $$
and the binomial theorem. Note that the Bernoulli number $B_3$ is 0. To go beyond the two terms given in the asymptotic formula, you'll need more terms in the binomial expansion and the Faulhaber formula. By a simple series arrangement,
$$ S= - n^{a+b} + \sum_{k=1}^n 2 k^a - k^a(k-1)^b - k^b(k-1)^a $$ Define a symbol $$(a,b)_k := \binom{a}{k} + \binom{b}{k}.$$ Then to a 4th order binomial expansion,
$$S=-n^{a+b} + (a+b)\sum_{k=1}^n k^{a+b-1} - (a,b)_2\sum_{k=1}^n k^{a+b-2}
+ (a,b)_3\sum_{k=1}^n k^{a+b-3} - (a,b)_4\sum_{k=1}^n k^{a+b-4} $$
Use (*):
$$ S= - n^{a+b} + (a+b)\Big(\frac{n^{a+b}}{a+b} + \frac{n^{a+b-1}}{2} + \frac{a+b-1}{12}n^{a+b-2} + 0 \cdot n^{a+b-3} + ... \Big) $$
$$- (a,2)_n \Big( \frac{n^{a+b-1}}{a+b-1} + \frac{n^{a+b-2}}{2} + \frac{a+b-2}{12}n^{a+b-3} + ... \Big)$$
$$+(a,3)_n \Big( \frac{n^{a+b-2}}{a+b-2} + \frac{n^{a+b-3}}{2} + ...\Big) -
(a,4)_n \Big( \frac{n^{a+b-3}}{a+b-3} + ...\Big)$$
The ellipses indicate terms that are not shown, because the powers of $n$ become too small to appear in the final answer. The first 2 terms cancel. Organize the rest as a descending series in powers of $n:$
$$ S = n^{a+b-1}\Big( \frac{a+b}{2} - \frac{(a,b)_2}{a+b-1} \Big) + n^{a+b-2}\Big(
\frac{(a+b)(a+b-1)}{12} - \frac{(a,b)_2}{2} + \frac{(a,b)_3}{a+b-2} \Big)$$
$$+n^{a+b-3}\Big(-\frac{(a+b-2)}{12}(a,b)_2 + \frac{(a,b)_3}{2} - \frac{(a,b)_4}{a+b-3} \Big) + ...$$ Simplification of the coefficients gives the answer of the final form.
It can be seen that once $a$ or $b$ become comparable to $n,$ the second term will be as large as the first and so the asymptotic solution begins to fail. Also, we must have a+b>3.
As an example, for n=5000, a=20, b=60: The first term of the asymptotic expansion gives about 5 digits precision, and both terms give about 10 digits precision.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $(y^2 + xy)(x^2 - x + 1) = 3x - 1$ over the integers.
Solve $$(y^2 + xy)(x^2 - x + 1) = 3x - 1$$ over the integers.
There are many solutions to this problem, and perhaps I chose the worst one possible. I hope that someone could come up with a better answer.
This problem is adapted from a recent competition (which is different than all of the "recent competitions" that I have mentioned before.
| We have that $$(y^2 + xy)(x^2 - x + 1) = 3x - 1 \implies x^2 - x + 1 \mid 3x - 1$$
$$\implies x^2 - x + 1 \le |3x - 1| \iff x \in [-2, 0] \cup [2 - \sqrt 2, 2 + \sqrt 2]$$
However, $x$ is an integer $\implies x \in \{0, \pm 1, \pm 2, 3\}$
We can set up a table for different values of $x$ and $y^2 + xy$.
$$\begin{matrix} x& -2& -1& 0& 1& 2& 3\\ y^2 + xy= \dfrac{3x - 1}{x^2 - x + 1}& -1& -\dfrac{4}{3}& -1& 2& \dfrac{5}{3}& \dfrac{8}{7} \end{matrix}$$
$\implies \left[ \begin{align} x = -2 &\text{ and } y^2 - 2y = -1\\ x = 0 &\text{ and } y^2 = -1\\ x = 1 &\text{ and } y^2 + y = 2 \end{align} \right.$$\implies \left[ \begin{align} x = -2 &\text{ and } (y - 1)^2 = 0\\ x = 1 &\text{ and } (y + 2)(y - 1) = 0\end{align} \right.$
$\implies (x, y) \in \{(-2, 1), (1, -2), (1, 1)\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the maximum value of $\sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}$ where $x + 2y + 3z = 2$
$x$, $y$ and $z$ are positives such that $x + 2y + 3z = 2$. Calculate the maximum value of $$ \sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}$$
This problem is brought to you by a recent competition. There should be different answers that are more creative than the one I have provided. Oh well...
| Let $x=\frac{2}{3}a$, $y=\frac{1}{3}b$ and $z=\frac{2}{9}c$.
Thus, $a+b+c=3$ and by AM-GM we obtain:
$$\sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}=\sum_{cyc}\sqrt{\frac{bc}{bc+3a}}=$$
$$=\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{a}{b+a}\right)=\frac{3}{2}.$$
The equality occurs for $a=b=c=1$, which says that we got a maximal value.
| {
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"timestamp": "2023-03-29T00:00:00",
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Fibonacci power series and fraction decomposition In a recent problem I have established that if $|x|<\frac{1}{2}$ then
$$f(x)=\sum_{n=1}^{\infty}a_{n}x^{n-1}=\frac{-1}{x^2+x-1}$$
where $a_{1}=a_{2}=1$ and $a_{n+1}=a_{n}+a_{n-1}$
(To work out the sum to the fraction, simply note that if $f(x)$ represents the sum then $f(x)-xf(x)-x^2f(x)=1)$)
I am now asked to find another series representation using the fraction $\frac{-1}{x^2+x-1}$
By fraction decomposition we have that
$$\frac{-1}{x^2+x-1}=\frac{\frac{1}{\sqrt{5}}}{x-\frac{-1-\sqrt{5}}{2}}+\frac{\frac{-1}{\sqrt{5}}}{x-\frac{-1+\sqrt{5}}{2}}$$
I now thought that i could transform the expressions into the form $$\frac{1}{1-(\frac{x}{a})}=\sum_{n=0}^{\infty}\left(\frac{x}{a}\right)^n$$
which I went about as follows (starting from the decomposite version):
\begin{align*}
\frac{\frac{1}{\sqrt{5}}}{x-\frac{-1-\sqrt{5}}{2}}+\frac{\frac{-1}{\sqrt{5}}}{x-\frac{-1+\sqrt{5}}{2}}
=&\frac{1}{\sqrt{5}}\left(\frac{1}{x-\frac{-1-\sqrt{5}}{2}}-\frac{1}{x-\frac{-1+\sqrt{5}}{2}}\right) \\
=&\frac{1}{\sqrt{5}} \left(\frac{1}{\frac{2x}{-1-\sqrt{5}}-1}-\frac{1}{\frac{2x}{-1+\sqrt{5}}-1}\right) \\
=&\frac{1}{\sqrt{5}} \left(\frac{-1}{1-\frac{2x}{-1-\sqrt{5}}}+\frac{1}{1-\frac{2x}{-1+\sqrt{5}}}\right)\\
=& \frac{1}{\sqrt{5}}\left[-\sum_{n=0}^{\infty}\left(\frac{2}{-1-\sqrt{5}}\right)^n x^n+\sum_{n=0}^{\infty}\left(\frac{2}{-1+\sqrt{5}}\right) x^n\right] \\
=& \frac{1}{\sqrt{5}}\sum_{n=0}^{\infty}\left[\left(\frac{2}{-1+\sqrt{5}}\right)^n-\left(\frac{2}{-1-\sqrt{5}}\right)^n\right]x^n
\end{align*}
However wolfram says that the last sum evaluates to $\frac{-x}{x^2+x-1}$ which is one x to much. I would appreciate any help on where I went wrong!
| On the second line of your equalities, you multiplied denominators but not numerator by $$\frac2{-1\pm\sqrt5}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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least value of $\lfloor \frac{a+b}{c}\rfloor+\lfloor \frac{c+b}{a}\rfloor+\lfloor \frac{a+c}{b}\rfloor$
If $a,b ,c>0$ . Then least value of $$\bigg\lfloor \frac{a+b}{c}\bigg\rfloor+\bigg\lfloor \frac{c+b}{a}\bigg\rfloor+\bigg\lfloor \frac{a+c}{b}\bigg\rfloor$$
Where $\lfloor x\rfloor$ is floor function of $x$
Plan
Using $$x-1< \lfloor x\rfloor\leq x$$
$$\frac{a+b}{c}-1< \bigg\lfloor \frac{a+b}{c}\bigg\rfloor \leq \frac{a+b}{c}$$
$$\frac{b+c}{a}-1< \bigg\lfloor \frac{b+c}{a}\bigg\rfloor \leq\frac{b+c}{a}$$
$$\frac{c+a}{b}-1< \bigg\lfloor\frac{c+a}{b}\bigg\rfloor \leq \frac{c+a}{b}$$
How do i solve it Help me please
| $$= 3 \bigg\lfloor \frac{a+b}{c} \bigg\rfloor$$
And given all variables are positive, the minimum occurs when $a+b$ is as close to $0$ as possible and $c$ is as large as possible.
| {
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Find all possible integers $n$ such that $\sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.
Find all possible integers $n$ such that $m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.
Guess what? This problem is adapted from a recent competition. There have been a solution below for you to check out. I am aware of the fact that there are other solutions that are more practical and suitable in test setting.
| We have that $n+2 = a^2, n + a = b^2$, where $a$ and $b$ are non-negative integers.
If $a > 2$, then $$a^2 = n+2 < n+a = a^2 + a - 2 < a^2 + 2a + 1.$$
This means that $n+a$ cannot be a perfect square as it's bounded between 2 consecutive perfect squares, which is a contradiction. Hence $ 0 \leq a \leq 2$.
If $ a= 0$, then $n = -2$ but $n+a = -2$ is not a perfect square. No solution.
If $a = 1$, then $n = - 1$ and $n+a = 0$ is a pefect square. This gives $m=1$.
If $a=2$, then $n=2$ and $n+a = 4$ is a perfect square. This gives $m=4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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quadratic equation solving mistake I'm a student who started self learning quadratic equations for a youth university program. I'm busy at trying to solve such equation:
$$
(1 - 4x)^2 + 9x + 7 = 2(x+3)(1-x) + (x+4)^2
$$
this is my current progress:
\begin{align}
(1 - 4x)^2 + 9x + 7 &= 2(x+3)(1-x)+ (x+4)^2\\
(1 - 4x)(1 - 4x) + 9x + 7 &= (2x + 6)(1 - x) + (x + 4)(x + 4)\\
1 - 4x - 4x + 16x^2 + 9x + 7 &= 2x - 2x^2 + 6 - 6x + x^2 + 4x + 4x + 16\\
8 + 16x^2 + x &= 2x - x^2 + 6 - 6x + 8x + 16\\
8 + 16x^2 + x &= 4x - x^2 + 22\\
16x^2 + x &= 4x - x^2 + 14\\
16x^2 &= 3x - x^2 + 14\\
17x^2 &= 3x + 14
\end{align}
The solutions to this equation are $x = 1,~x=-14/17$.
So, where is my mistake? $x$ is negative, so I must be incorrect.
| $$
(1 - 4)^2 + 9 + 7 = 2(1+3)(1-1) + (1+4)^2
$$
seems true ($25$ in both members), and with a little more effort
$$
\left(1 - 4\frac{\overline{14}}{17}\right)^2 + 9\frac{\overline{14}}{17} + 7 = 2\left(\frac{\overline{14}}{17}+3\right)\left(1-\frac{\overline{14}}{17}\right) + \left(\frac{\overline{14}}{17}+4\right)^2
$$
is
$$(17+56)^2-9\cdot17\cdot14+7\cdot17^2=2(-14+51)(17+14)+(-14+68)^2$$which is $5210$ on both sides.
As a quadratic equation has at most two roots, your work is right.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding general solution of $3{\times}3$ system I am given the following:
$$
x'=
\begin{bmatrix}
2 &0 &0 \\
-7&9 &7 \\
0&0 &2
\end{bmatrix}
x
$$
Solving $\det(A-\lambda I)$, I get $\lambda = 2,2,9$. Solving $\det(A-2\lambda)$, I get
\begin{bmatrix}
0&0 &0 \\
-7&7 &7 \\
0&0 &0
\end{bmatrix}
So we have geom multi. = alg. multi, so our matrix is complete.
Take $v_1 =
\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}
$
Similarly, solve $\det(A-9\lambda)$ to get
$
\begin{bmatrix}
-7&0 &0 \\
-7&0 &7 \\
0&0 &-7
\end{bmatrix}
$
So take $v_2 =
\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
$
So the general solution should be $x(t) = C_1e^{2t}\begin{bmatrix}
1\\
1\\
0
\end{bmatrix}
+C_2e^{2t} \begin{bmatrix}
1\\
1\\
0
\end{bmatrix}
+ C_3e^{9t}\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}$
However, according to the back of the book solution, this is incorrect. What am I missing here? Thank you.
| You are right that $v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $2$; i.e $ v_1 \in \ker(A-2I)$. But like you mentioned, the dimension of this kernel is $2$, so you need to find another linearly independent eigenvector. It is easy to verify that $\xi = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} $ is such a vector. In other words, $\{v_1, \xi\}$ form a basis for $\ker(A-2I)$. Hence, your general solution will be
\begin{equation}
x(t) = C_1 e^{2t} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} +
C_2 e^{2t} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} +
C_3 e^{9t} \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
\end{equation}
| {
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How to I find a formula for $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)}$ using the given method? The following is problem $6$ $(iii)$ from chapter $2$ of Spivak's Calculus:
The formula for $1^{2} + \cdots + n^{2}$ may be derived as follows. We begin with the formula $$ (k+1)^{3} - k^{3} = 3k^{2}+3k +1$$.
Writing this formula for $k=1, \cdots , n$ and adding
\begin{align*}
2^{3}-1^{3}&=3\cdot1^{2} + 3\cdot1 + 1 \\
3^{3}-2^{3}&=3\cdot2^{2} + 3\cdot2 + 1 \\
&\vdots\\
(n+1)^{3}-n^{3}&=3\cdot n^{2} + 3\cdot n +1 \\
\end{align*}
we obtain
$$(n+1)^{3}-1 = 3[1^{2} + \cdots + n^{2}] + 3[1 + \cdots + n] + n.$$
Thus we can find $$\sum_{k=1}^n k^{2}$$ if we already know $$\sum_{k=1}^n k .$$ Use this method to find
$$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)}$$.
So originally, I attempted $$\sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n \frac{1}{k+1}.$$ This means that we have
$$(1 + \frac{1}{2} + \cdots + \frac{1}{n}) - (\frac{1}{2} + \frac{1}{3} + \cdots \frac{1}{n}+\frac{1}{n+1})$$
which gives us $1-\dfrac{1}{n+1}$. Although, I'm not sure if the method I've used suffices as the method required for the problem; that is, it's supposed to look similar to the example using $(k+1)^{3}$ above.
| It is a perfectly fine answer. The above method which is being illustrated is known as Telescopic method of summation. You can read more about it here.
https://en.wikipedia.org/wiki/Telescoping_series
https://brilliant.org/wiki/telescoping-series/
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the following equation: $\sqrt {\sin x - \sqrt {\cos x + \sin x} } = \cos x$ Solve the following equation:
\begin{array}{l}{\sqrt{\sin x-\sqrt{\cos x+\sin x}}=\cos x} \\ \text{my try as follows:}\\{\sin x-\sqrt{\cos x+\sin x}=\cos ^{2} x} \\ {\sin x-\cos ^{2} x=\sqrt{\cos x+\sin x}} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x=\cos x+\sin x} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x \cos ^{2} x-\cos x-\sin x=0} \\ {\sin ^{2} x+\cos ^{4} x-2 \sin x\left(1-\sin ^{2} x\right)-\cos x-\sin x=0} \\{\sin ^2}x + {\left( {1 - {{\sin }^2}x} \right)^2} - 2\sin x\left( {1 - {{\sin }^2}x} \right) - \cos x - \sin x = 0\\ {\sin ^{2} x+\sin ^{4} x-2 \sin ^{2} x+1-2 \sin x+2 \sin ^{3} x-\cos x-\sin x=0} \\ {\sin ^{4} x+2 \sin ^{3} x-\sin ^{2} x-3\sin x-\cos x+1=0}\end{array}
Now i think it gets more complicated , any help would be appreciated
| In any solution, $\cos(x)$ is nonnegative, since it equals some square root.
In any solution, $\sin(x)$ is nonnegative, since the equation includes the square root of $\sin(x)$ minus some nonnegative quantity.
So $\cos(x)+\sin(x)\geq\sin(x)$, and therefore $$\sqrt{\cos(x)+\sin(x)}\geq\sqrt{\sin(x)}\geq\sin(x)$$
where the last inequality is because $\sin(x)\leq1$. And note that equality only holds if $\sin(x)=1$ or $\sin(x)=0$.
So $\sin(x)-\sqrt{\cos(x)+\sin(x)}\leq\sin(x)-\sin(x)=0$, and therefore $$\sin(x)-\sqrt{\cos(x)+\sin(x)}\leq0$$
In the equation, we take a square root of this expression. So it is not defined unless we have equality, which we noted is only possible if $\sin(x)=1$ or $\sin(x)=0$.
No solution can have $\sin(x)=0$, or else $\sin(x)-\sqrt{\cos(x)+\sin(x)}<0$, and we cannot take its square root in the equation.
Any $x$ with $\sin(x)=1$ implies $\cos(x)=0$, and such $x$ do indeed make a solution, by inspection.
So the given equation is equivalent to $\sin(x)=1$, and the solution set is $\frac{\pi}{2}+2k\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Why can I simplify radicals? (eg, $\sqrt{153} = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{17}$) I know it might sound like a ridiculously easy question to answer, but I just can't put two and two together for some reason.
Say for example you have:
$$\sqrt{153}$$
You can break it down to
$$\sqrt{3}\cdot \sqrt{3}\cdot \sqrt{17}$$
(and of course you can further simplify it, but we'll just leave it at that for now).
Why can I do that?
| Ignoring some (ahem) complexities, we can say, for non-negative(!) $x$,
$\sqrt{x}\;$ is the non-negative(!) number that multiplies by itself to give $\;x$.
With that in mind, consider what happens when we multiply $\sqrt{x}\cdot\sqrt{y}$ by itself:
$$\begin{align}
\left(\;\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\;\right)\cdot\left(\;\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\;\right) &=
\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\cdot\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}} \\
&= \color{red}{\sqrt{x}}\cdot\color{red}{\sqrt{x}}\cdot\color{blue}{\sqrt{y}}\cdot\color{blue}{\sqrt{y}} \\
&= \left(\;\color{red}{\sqrt{x}}\cdot\color{red}{\sqrt{x}}\;\right)\cdot\left(\;\color{blue}{\sqrt{y}}\cdot\color{blue} {\sqrt{y}}\;\right) \\
&= \color{red}{x}\cdot\color{blue}{y}
\end{align}$$
Since $\sqrt{x}$ and $\sqrt{y}$ are non-negative, so is their product. Therefore, we have shown that
$\sqrt{x}\cdot\sqrt{y}\;$ is the non-negative(!) number that multiplies by itself to give $\;x\cdot y$.
More simply,
$$\sqrt{x\cdot y} = \sqrt{x}\cdot\sqrt{y}$$
The idea extends to show that
$$\sqrt{x\cdot y\cdot z} = \sqrt{x}\cdot\sqrt{y}\cdot\sqrt{z}$$
and so forth. (By a completely-analogous argument, this kind of thing works for cube-roots, and fourth-roots, and $n$-th roots in general.) This is precisely what allows you to simplify radicals.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Wolfram Alpha can't find an explicit form (or even an explicit form of an approximation of) this innocuous series? I suppose my question could be restated as "why?" to the above statement.
I'll rephrase my question if this breaks any rules asking it this way, but the expression is simply:
$$\sum_{m=2}^\infty \frac{1}{m^{2n}(m^2-1)} = ??$$
with the partial sum being:
$$\sum_{m=2}^a \frac{1}{m^{2n}(m^2-1)} = ??$$
$m, n \in \mathbb N$
Are the above two somehow more insidious than they look?
| $$
\begin{align}
\sum_{m=2}^\infty\frac1{m^{2n}(m^2-1)}
&=\sum_{m=2}^\infty\sum_{k=n+1}^\infty m^{2k}\\
&=\sum_{k=n+1}^\infty\sum_{m=2}^\infty m^{2k}\\
&=\sum_{k=n+1}^\infty(\zeta(2k)-1)\\
&=\bbox[5px,border:2px solid #C0A000]{\frac34-\sum_{k=1}^n(\zeta(2k)-1)}\tag1
\end{align}
$$
since
$$
\begin{align}
\sum_{k=1}^\infty(\zeta(2k)-1)
&=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac1{n^{2k}}\\
&=\sum_{n=2}^\infty\sum_{k=1}^\infty\frac1{n^{2k}}\\
&=\sum_{n=2}^\infty\frac1{n^2-1}\\
&=\frac12\sum_{n=2}^\infty\left(\frac1{n-1}-\frac1{n+1}\right)\\[3pt]
&=\frac34\tag2
\end{align}
$$
In $(1)$, $\zeta(2k)$ is a rational multiple of $\pi^{2k}$, as shown in this answer.
Here are the first several sums:
$$
\begin{array}{c|l}
n&\frac34-\sum\limits_{k=1}^n(\zeta(2k)-1)\\\hline
1&\frac74-\frac{\pi^2}6\\
2&\frac{11}4-\frac{\pi^2}6-\frac{\pi^4}{90}\\
3&\frac{15}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}\\
4&\frac{19}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}-\frac{\pi^8}{9450}\\
5&\frac{23}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}-\frac{\pi^8}{9450}-\frac{\pi^{10}}{93555}\\
6&\frac{27}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}-\frac{\pi^8}{9450}-\frac{\pi^{10}}{93555}-\frac{691\pi^{12}}{638512875}
\end{array}\tag3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the value of series $$\sum^\infty_{k=0 }\frac{2^k}{5^{2^k}+1}$$
The hint is expanding it to double series.
So, $$\frac{2^k}{5^{2^k}+1}=\frac{2^k}{5^{2^k}}\frac{1}{1+5^{-2^k}}=\frac{2^k}{5^{2^k}}\sum_{n=0}^\infty(-1)^n5^{-n2^k}=\sum_{n=0}^\infty(-1)^n\frac{2^k}{5^{2^k(n+1)}}$$
Thus $$\sum^\infty_{k=0 }\frac{2^k}{5^{2^k}+1}=\sum_{k=0}^\infty\sum_{n=0}^\infty(-1)^n\frac{2^k}{5^{2^k(n+1)}}$$
Since $$\sum^\infty_{k=0 }\frac{2^k}{5^{2^k}+1}\lt \sum_{k=0}^\infty\frac{2^k}{5^k}=\frac{5}{3}$$
It is always increasing and has an upper bond. So it converges.
Therefore the order of the double series is exchangeable. But what to do next?
| Telescopic Solution
Notice that
$$
\frac{2^k}{5^{2^k}-1}-\frac{2^k}{5^{2^k}+1}=\frac{2^{k+1}}{5^{2^{k+1}}-1}
$$
Therefore, we get a telescoping series:
$$
\begin{align}
\sum_{k=0}^\infty\frac{2^k}{5^{2^k}+1}
&=\sum_{k=0}^\infty\left(\frac{2^k}{5^{2^k}-1}-\frac{2^{k+1}}{5^{2^{k+1}}-1}\right)\\[6pt]
&=\frac14
\end{align}
$$
Analysis of the Solution Provided
Let $\Xi=\sum_{k=0}^\infty\frac{2^k}{5^{2^k}+1}$. Note that
$$
\frac{2^k}{5^{2^k}+1}=\frac{2^k}{5^{2^k}}\frac1{1+5^{-2^k}}=\sum_{n=0}^\infty(-1)^n\frac{2^k}{5^{2^k(n+1)}}.\tag1
$$
$(1)$ uses the sum of a geometric series
Therefore, we have
$$
\Xi=\sum_{k=0}^\infty\sum_{n=0}^\infty(-1)^n\frac{2^k}{5^{2^k(n+1)}}
=\sum_{N=1}^\infty\frac1{5^N}\sum_{N=(n+1)2^k}(-1)^n2^k.\tag2
$$
In $(2)$, we have collected all the terms with a given power of $5$ in the denominator. Now comes the part that seems to be what is causing the difficulty:
Given any $N\in\mathbb{N}$, let $N=a2^m$ where $a$ and $m$ are non-negative integers and $a$ is odd. Then, for fixed $N$, we have
$$
\sum_{N=(n+1)2^k}(-1)^n2^k=\sum_{k=0}^m(-1)^{a2^{m-k}-1}2^k
=2^m-\sum_{k=0}^{m-1}2^k=1.\tag3
$$
In other words, for any positive $N\in\mathbb{N}$, we can uniquely factor $N=a2^m$ where $a$ is odd. In $(2)$, we want to use all $n$ and $k$ so that $N=a2^m=(n+1)2^k$. For $0\le k\le m$, we have $n+1=a2^{m-k}$. Since $a$ is odd, $n$ is even for $k=m$ and odd for all $k\lt m$. Therefore,
$$
\sum_{N=(n+1)2^k}(-1)^n2^k=2^m-2^{m-1}-2^{m-2}-\cdots-1=1\tag4
$$
Plugging $(4)$ into $(2)$ yields
Therefore,
$$
\Xi=\sum_{N=1}^\infty\frac1{5^N}=\frac14.\tag5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Division of 11 people into 3 groups with at least 2 people in each
How many ways can 11 people be divided into three teams where each
team must have at least two members?
We are supposed to use multinomial coefficients and number of integer solutions. I have tried this
The number of ways to divide 5 people into three groups
and this
Arranging $10$ people in $2$ lanes. Each lane has to have at least $2$ people. and more specifically this Number of ways to divide n people into k groups with at least 2 people in each group, but the last one in particular I could not understand at all (since I do not know what the brackets { } mean).
I have tried several things so far, but I am not convinced any of them are correct. Here is my best attempt:
If each group has at least two people, I initially choose 6 from 11:
$$\binom{11}{6}=42$$
Then, these 6 people need to be put in the groups of 2 people:
$$\frac{6!}{2!2!2!}=90$$
But, since order does not matter, we must divide by 3!:
$$\frac{6!}{2!2!2!3!} = 15$$
So for the choice of the two people in each group we have 90*15 = 1350 possibilities.
Now we need to consider the 5 remaining people. Let $x_1,x_2,x_3$ be the number of people in each group. Then we have a total of
$$\binom{5+3-1}{3-1} = \binom{7}{2}=21$$
(non negative) integer solutions for $x_1+x_2+x_3=5$.
However, the possible cases are:
$$(0,1,4),(0,2,3),(0,0,5),(1,1,3),(1,2,2),$$
where the first two triplets appear a total of 3! each (order does not matter) and the last three apear $\frac{3!}{2!} =3$ times (due to permutation of terms with same amount of people).
Case $(0,1,4)$:
$\binom{5}{1}\binom{4}{4}=5,$ giving a total of $3!5 = 30$ possibilities.
Case $(0,2,3)$:
$\binom{5}{2}\binom{3}{3}=10,$ giving a total of $3!10 = 60$ possibilities.
Case $(0,0,5)$:
$\binom{5}{5}=5,$ giving a total of $3\times 5 = 15$ possibilities.
Case $(1,1,3)$:
$\binom{5}{1}\binom{4}{1}\binom{3}{3}=2,0$ giving a total of $3\times 20 = 60$ possibilities.
Case $(1,2,2)$:
$\binom{5}{1}\binom{4}{2}\binom{2}{2}=6,$ giving a total of $3\times 6 = 18$ possibilities.
Then we will have a total of
$1350(30+60+15+60+18) = 247050$
possibilities.
Can anyone help with the logical reasoning here to see if this is correct? In case it is wrong, where am I going wrong?
| Looking at the number of persons in the groups we discern the following possibilities:
*
*for $(2,2,7)$ there are $\frac{11!}{2!2!7!}\cdot\frac12$ possibilities.
*for $(2,3,6)$ there are $\frac{11!}{2!3!6!}$ possibilities.
*for $(2,4,5)$ there are $\frac{11!}{2!4!5!}$ possibilities.
*for $(3,3,5)$ there are $\frac{11!}{3!3!5!}\cdot\frac12$possibilities.
*for $(3,4,4)$ there are $\frac{11!}{3!4!4!}\cdot\frac12$ possibilities.
Note the factor $\frac12$ in the cases where exactly two groups have the same size.
This factor repairs double counting.
In all other cases the groups are distinguishable by the number of members.
The summation of these numbers is the answer on your question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $3x(2+\sqrt{9x^2 + 3}) + (4x-2)(\sqrt{x^2 - x +1}+1) = 0$ for $x\in \mathbb{R}$
Solve the equation in $ \mathbb{R}$: $$3x(2+\sqrt{9x^2 + 3}) + (4x-2)(\sqrt{x^2 - x +1}+1) = 0$$
I've been tried to solve this question for 3 hours, but can't find out any answers.
Just like I running in the maze, if I represent $\sqrt{9x^2 + 3} = A$, $ \sqrt{x^{2}-x+1}= B$
Finally we've got $2A^{2} -15 = 18B^2 + 9\sqrt{4B^2 +1}$ , which is not help me to find relation between $A$ and $B$ anymore.
I just wonder if we have a nice solution approaches to the problem.
I appreciate any suggestion and help. Thank you.
| It is a nicely set up problem:
$$(3x)(2+\sqrt{9x^2 + 3}) + (4x-2)(\sqrt{x^2 - x +1}+1) = 0 \iff \\
(6x)(1+\sqrt{\frac94x^2 + \frac34}) = (2-4x)(1+\sqrt{x^2 - x +1}) \\
\begin{cases}6x=2-4x\\ \frac94x^2 + \frac34=x^2-x+1\end{cases} \Rightarrow \begin{cases}x=\frac15\\ 5x^2+4x-1=0 \Rightarrow x=\frac15;-1\end{cases} \Rightarrow x=\frac15.$$
Addendum: The left hand side function is:
$$\begin{cases}f(x)<0,x\le 0\\ f(x)>0,x\ge \frac12\end{cases}\\
f'(x)=3(2+\sqrt{9x^2+3})+3x\cdot \frac{9x}{\sqrt{9x^2+3}}+\\
4(\sqrt{x^2-x+1}+1)+(2x-1)\cdot \frac{2x-1}{\sqrt{x^2-x+1}}>0,0<x<\frac12$$
Hence, the single root is $x=\frac15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Is it possible to improve on the bounds for $\varphi(N)/N$, if $N = q^k n^2$ is an odd perfect number with special prime $q$? Let $N = q^k n^2$ be an odd perfect number given in Eulerian form. That is, $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
From a comment underneath this earlier question, we have the equation (and corresponding inequalities)
$$1 < \frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1} \leq \frac{5}{4}$$
since $q$ is prime with $q \equiv 1 \pmod 4$ implies that $q \geq 5$.
This implies that
$$\frac{4}{5} \leq \dfrac{\dfrac{\varphi(N)}{N}}{\dfrac{\varphi(n)}{n}} = \frac{q-1}{q} < 1.$$
But from the following source:
Advanced Problem H-661, On Odd Perfect Numbers, Proposed by J. L´opez Gonz´alez, Madrid, Spain and F. Luca, Mexico (Vol. 45, No. 4, November 2007), Fibonacci Quarterly, we have the bounds
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2}.$$
However, we also have
$$\frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}.$$
Notice that
$$\frac{4}{5} \leq \frac{\varphi(q^k)}{q^k} = \frac{q^k \bigg(1 - \frac{1}{q}\bigg)}{q^k} = \frac{q - 1}{q} < 1.$$
Therefore, we have the bounds
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{\varphi(n)}{n},$$
and
$$\frac{4}{5}\cdot\frac{\varphi(n)}{n} \leq \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{1}{2},$$
which implies that
$$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8}.$$
WolframAlpha gives the rational approximation
$$\frac{120}{217\zeta(3)} \approx 0.4600409433626.$$
Here is my question:
Is it possible to improve on the bounds for $\varphi(N)/N$, if $N = q^k n^2$ is an odd perfect number with special prime $q$?
MOTIVATION FOR THE INQUIRY
It can be shown that the equation
$$\frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1}$$
together with the bounds
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2}$$
and
$$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8}$$
imply
$$0.92 \approx \dfrac{\dfrac{120}{217\zeta(3)}}{\dfrac{1}{2}} < \frac{q}{q-1} < \dfrac{\dfrac{5}{8}}{\dfrac{120}{217\zeta(3)}} \approx 1.358574729,$$
from which we obtain trivial bounds.
Nonetheless, it can be shown that the equation
$$\frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1}$$
together with the upper bound $\varphi(N)/N < 1/2$ implies that
$$q < \frac{x}{x-1}$$
where
$$x = \frac{2\varphi(n)}{n}.$$
Thus, if we can improve the upper bound for $\varphi(N)/N$ to something smaller than $1/2$ (say $1/2 - \varepsilon$ for some tiny $\varepsilon > 0$), then we can improve the coefficient of $\frac{\varphi(n)}{n}$ in $x$ to some number bigger than $2$. Likewise, if we can get a better lower bound for $\varphi(N)/N$, then we will be able to get an improved lower bound for $\varphi(n)/n$. Together, they would translate (hopefully!) to a numerical upper bound for the special/Euler prime $q$!
| Eureka!!!
Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$.
From the equation and lower bound for $\varphi(N)/N$
$$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}$$
and the equation
$$\frac{\varphi(q^k)}{q^k} = \frac{q - 1}{q},$$
we get the lower bound
$$2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1} < \frac{2\varphi(n)}{n} = x.$$
This implies that we have the upper bound
$$q < \frac{x}{x-1} < \frac{2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}}{\bigg(2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}\bigg) - 1}$$
which can be solved using WolframAlpha, yielding the upper bound
$$q < \frac{217\zeta(3)}{217\zeta(3) - 240} \approx 12.5128,$$
from which it follows that $q=5$, since $q$ is a prime satisfying $q \equiv 1 \pmod 4$. QED
| {
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"url": "https://math.stackexchange.com/questions/3284645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What is the problem here (all integers are irrational proof...I think so)? Let us assume $a$ is an integer which is rational which implies $a=p/q$ (where $p$ and $q$ are integers and $q$ not equal to $0$). If $p$ and $q$ are not coprime, let us simplify the fraction so this it is (I don't know how to talk like mathematicians).
Which implies, $$a=b/c$$ (where $b$ and $c$ are coprime integers). Squaring on both sides,
\begin{align}
a^2&=b^2/c^2\\
a^2c^2&=b^2
\end{align}
So $a^2$ is a factor of $b^2$, and also of $b$, due to the uniqueness of the fundamental theorem of arithmetic.
So,
\begin{align}
b &=a^{2}d \tag{where $d$ is an integer}\\
b^2 &= a^{4}d^{2}
\end{align}
But $b^2=a^2c^2$ So,
\begin{align}
a^2c^2 &= a^4d^2\\
c^2 &= a^2d^2
\end{align}
So, $a^2$ is a factor of $c^2$ and $c$ due to the fundamental theorem of arithmetic. So $b$ and $c$ have $a^2$ as a common factor. But this contradicts the fact that $b$ and $c$ are coprime. This is because we have taken $a$ as a rational integer, so $a$ cannot be a rational integer.
What's wrong here (genuinely asking)?
| The problem in the proof is that $a^2|b^2\nRightarrow a^2|b$. For instance, take $a=2$ and $b=6$. Clearly, $4|36$ but $4\nmid 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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The integral $\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}}dx =\frac{\pi}{2}\ln (1+\sqrt{2})$ At Mathematica the numerical value of the integral
$$\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}} dx$$
equals 1.3844.., which is nothing but $\frac{\pi}{2}\ln (1+\sqrt{2})=z$. Also, one of its transformed forms is evaluated to be $z$ by Mathematica. The question is: How to do it by hand?
| Consider the following integral:
$$I(a)=\int_0^\infty \frac{\operatorname{arccot}(a\sqrt{1+x^2})}{\sqrt{1+x^2}}dx\Rightarrow I'(a)=-\int_0^\infty \frac{1}{1+a^2+a^2x^2}dx$$
$$=-\frac{1}{a\sqrt{1+a^2}}\arctan\left(\frac{ax}{\sqrt{1+a^2}}\right)\bigg|_0^\infty=-\frac{\pi}{2}\frac{1}{a\sqrt{1+a^2}} $$
Since $I(\infty)=0$ and we are looking for $I(1)$ we have:
$$I(1)=-(I(\infty)-I(1))=\frac{\pi}{2}\int_1^\infty \frac{1}{a\sqrt{1+a^2}}da\overset{a=\frac{1}{x}}=\frac{\pi}{2}\int_0^1 \frac{1}{\sqrt{1+x^2}}dx$$
$$\Rightarrow \boxed{\int_0^\infty \frac{\operatorname{arccot}(\sqrt{1+x^2})}{\sqrt{1+x^2}}dx=\frac{\pi}{2}\ln(1+\sqrt 2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve $x^{3} = 6+ 3xy - 3 ( \sqrt{2}+2 )^{{1}/{3}} , y^{3} = 9 + 3xy(\sqrt{2}+2)^{{1}/{3}} - 3(\sqrt{2}+2)^{{2}/{3}}$ Solve the system of equations for $x,y \in \mathbb{R}$
$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}} $
$ y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}$
I just rearranged between those equations and get $ \frac{y^{3}-9}{x^{3} -6} = (\sqrt{2}+2)^{\frac{1}{3}}$ then I don't know how to deal with it.
Please give me a hint or relevant theorem to solve the equation.
Thank you, and I appreciate any help.
Furthermore I get an idea
how about we subtract two equation and get
$y^{3}-x^{3} = 3 + 3xy((\sqrt{2}+2)^{\frac{1}{3}} -1) - (3(\sqrt{2}+2)^{\frac{2}{3}} - 3(\sqrt{2}+2)^{\frac{1}{3}})$
$(y-x)(x^2+xy+y^2)= 3[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)]$
$y-x = 3$ and $x^2 +xy+y^2 =[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $
or,
$y-x = [1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $ and $x^2 +xy+y^2 = 3$
Am I on the right track?
| Solving your first equation for $y$ we get
$$y=\frac{x^3+3 \sqrt[3]{2+\sqrt{2}}-6}{3 x}$$
Plugging this in the second equation we get
$$\frac{x^6}{27}-\frac{2}{3}
\left(1+\sqrt[3]{2+\sqrt{2}}\right)
x^3+\frac{-6+\sqrt{2}+12 \sqrt[3]{2+\sqrt{2}}-6
\left(2+\sqrt{2}\right)^{2/3}}{x^3}-2
\left(2+\sqrt{2}\right)^{2/3}+2
\sqrt[3]{2+\sqrt{2}}-5=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Splitting a population, probability of 2 people landing in same subpopulation.
$10$ people has been split into $3$ groups $A,B,C$ of $5,3,2$ people respectively. What is the probability that $2$ predetermined people $x,y$ land in the same group?
My attempt:
There is $\binom{10}{5, 3, 2} = \frac{10!}{5!3!2!}$ ways to split the population into mentioned above subpopulations.
Let's count all the partitions in which $x,y\in A$.
There's $\binom{10-2}{3}$ ways to choose the rest of people in $A$, $\binom{5}{3}$ ways to choose people in $B$ from the remaining 5, and $\binom{2}{2}$ ways to choose people in $C$. Repeating the above for the cases $x,y\in B$ and $x,y \in C$ we get:
$$P(A)=\frac{ \binom{8}{3}\binom{5}{3} +\binom{8}{1}\binom{7}{5}+\binom{8}{0}\binom{8}{5}}{\binom{10}{5,3,2}},$$
which is wrong.
The given answer is
$$P(A)=\frac{ \binom{8}{3}\binom{2}{2} +\binom{8}{1}\binom{2}{2}+\binom{2}{2}}{\binom{10}{5,3,2}},$$
and I'm having a hard time engineering it back. Could anyone point where I made mistake?
| I agree with your answer. In particular, your term for $x,y \in A$ exceeds the whole book answer and your logic is sound.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3290263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $x+y+z \ge xy+yz+zx$ Given $x,y,z \ge 0$ and $x+y+z=4-xyz$ Then Prove that
$$x+y+z \ge xy+yz+zx$$
My try:
Letting $x=1-a$, $y=1-b$ and $z=1-c$ we get
$$(1-a)+(1-b)+(1-c)+(1-a)(1-b)(1-c)=4$$
$$-(a+b+c)-(a+b+c)+ab+bc+ca-abc=0$$
$$ab+bc+ca-abc=2(a+b+c)$$
Where $a, b,c \le 1$
is there a clue here?
| Suppose $x+y+z < xy+yz+zx$, then from Schur's inequality we have
$$\begin{align}
\frac{9xyz}{x+y+z} &\geqslant 4(xy+yz+zx)- (x+y+z)^2 \\
&> (x+y+z)\left( 4- (x+y+z)\right) \\
&= (x+y+z) \cdot xyz \\
\end{align}$$
This gives $x+y+z< 3$, further we have from $4= x+y+z+xyz \geqslant 4\sqrt{xyz} \implies xyz \leqslant 1$, so $4 = x+y+z+xyz < 3+1$, a contradiction.
Equality is when $x=y=z=1$ or when $(x, y, z)=(2, 2, 0)$ or a permutation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Finding constant $a$ so that differential equation becomes correct Question:
Find the value of $a$ so that the function
$$y = \sqrt{x} \ln{x}$$
is a solution to the differential equation
$$y' - \frac{a}{x} \cdot y = \frac{1}{\sqrt{x}}$$
Attempted solution:
My basic approach would be to take the derivative of y, put y and y' into the differential equation and solve for a.
The derivative of y is gotten with the product rule and then the chain rule for the square root:
$$y' = \sqrt{x} \cdot (\ln{x})' + (\sqrt{x})' \cdot \ln{x} = \frac{\sqrt{x}}{x} + \frac{\ln{x}}{2 \cdot \sqrt{x}}$$
Solving for a:
$$y' - \frac{a}{x} \cdot y = \frac{1}{\sqrt{x}} \Rightarrow a = \frac{y' - \frac{1}{\sqrt{x}}}{y} = \frac{xy' - \sqrt{x}}{y}$$
Putting y and y' into the expression for a gives:
$$a = \frac{x(\frac{\sqrt{x}}{x} + \frac{\ln {x}}{2\sqrt{x}})}{\sqrt{x}\ln {x}} = \frac{1+\frac{\ln{x}}{2}}{\ln{x}} = \frac{1}{\ln{x}} + \frac{1}{2}$$
The expected answer is just $\frac{1}{2}$, so for some reason I have gotten an extra $\frac{1}{\ln{x}}$ somwhere.
| It is $$\frac{\ln(x)}{2\sqrt{x}}+\frac{\sqrt{x}}{x}-\frac{a}{x}\sqrt{x}\ln(x)=\frac{1}{\sqrt{x}}$$ multiplying by $\sqrt{x}$ we get
$$\frac{\ln(x)}{2}-a\ln(x)=0$$ so $$a=\frac{1}{2}$$ for $\ln(x)\ne 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I prove that starting from $0$, and repeatedly taking $+10,-10,\times10$ or $\div10$, requires at least $9$ operations to get to $2019$? Sometimes, I teach my sister math homework. Today, when she was doing her math problem solving summer homework, I teached her how to do the problems except the final problem that I don't know how to do.
A machine is showing the number $0$. It has $4$ buttons: $\boxed{+10}, \boxed{-10}, \boxed{×10},$ and $\boxed{÷10}$. If you press a button, the number that the machine is showing will be changed by the operation on the button. At least how many times do you need to press the buttons in order to make $2019$?
I think that the answer is 9, and the sequence is $\boxed{+10}, \boxed{+10}, \boxed{×10},\boxed{×10},\boxed{+10}, \boxed{+10}, \boxed{×10},\boxed{-10},\boxed{÷10}$.
But is it correct and how can I explain to my sister that it is the least possible value?
| You can say that every button $\boxed{+10}$ or $\boxed{-10}$ has a different worth depending on how many buttons $\boxed{\times 10}$ or $\boxed{÷10}$ are pressed at any point after it. Each $\boxed{\times 10}$ increases the values of any previous $\boxed{+10}$ or $\boxed{-10}$ by a factor of $10$, and each $\boxed{÷10}$ decreases the values of any previous $\boxed{+10}$ or $\boxed{-10}$ by a factor of $10$.
For example, if after you press $\boxed{+10}$ and then sometimes after you press $\boxed{\times 10}$ three times, and $\boxed{÷10}$ one time, the original $\boxed{+10}$ will at the end be worth $10\times 10\times 10 \times 10 ÷10 = 1000$.
Therefore, in the end, each pressing of $\boxed{+10}$ may be worth $10$, $100$, $1000$... or $1$, $0.1$, $0.01$... depending on how many $\boxed{\times 10}$ and $\boxed{÷10}$ are pressed after them. Similarily, $\boxed{-10}$ may be worth $-10$, $-100$, $-1000$... or $-1$, $-0.1$, $-0.01$...
The required number needs to be constructed from these final worths of $\boxed{+10}$ and $\boxed{-10}$. The easiest way to write $2019$ in this way is $$ 2019 = 1000 + 1000 +10 + 10 -1$$
That means you need:
*
*Two instances of pressing $\boxed{+10}$ that have their value increased 2 times
*Two instances of pressing $\boxed{+10}$ that have their value neither increased or decreased
*One instances of pressing $\boxed{-10}$ that have their value decreased 1 time
Therefore in total you'll need to press $\boxed{+10}$ four times, and $\boxed{-10}$ one time. Somewhere between them you'll need to press $\boxed{\times 10}$ and $\boxed{÷10}$ appropriate number of times, so that two of the $\boxed{+10}$ had their value increased twice, two $\boxed{+10}$ had their value unchanged, and $\boxed{-10}$ had their value decreased once.
There are actually three ways to do this in a minimal number of operations. One is the one you've found:
$$\boxed{+10}\boxed{+10}\boxed{\times 10}\boxed{\times 10}\boxed{+10}\boxed{+10}\boxed{\times 10}\boxed{-10}\boxed{÷10} $$
The other two are
$$\boxed{+10}\boxed{+10}\boxed{\times 10}\boxed{\times 10}\boxed{+10}\boxed{\times 10}\boxed{-10}\boxed{÷10}\boxed{+10} $$
$$\boxed{+10}\boxed{+10}\boxed{\times 10}\boxed{\times 10}\boxed{\times 10}\boxed{-10}\boxed{÷10}\boxed{+10}\boxed{+10} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Show that if $a\sin x + b\cos x + ce^x$ is the zero function for constants $a, b, c$ then $a = b = c = 0$
Suppose that $a\sin x + b\cos x + ce^x$ is the zero function. Prove that $a=b=c=0$.
Does the zero function simply mean that $a\sin x + b\cos x + ce^x = 0$? I am going under the assumption that it is.
Geometrically speaking, I think it is pretty clear that $a=b=c=0$ must be true if $a\sin x + b\cos x + ce^x = 0$ as there are no x-values that make the equation true otherwise.
Is there a purely algebraic way of saying this that does not utilize showing a graph?
| Given
$a\sin x + b\cos x + ce^x = 0, \tag 1$
we multiply through by $e^{-x}$, which yields
$ae^{-x}\sin x + be^{-x}\cos x + c = 0; \tag 2$
letting $x \to \infty$, we find that
$c = 0; \tag 3$
we are left with
$a\sin x + b\cos x = 0; \tag 4$
now set
$x = \dfrac{\pi}{2}, \tag 5$
and obtain
$a = a\sin \dfrac{\pi}{2} + b\cos \dfrac{\pi}{2} = 0; \tag 6$
thus
$a = 0; \tag 7$
in a similar manner, choosing
$x = 0 \tag 8$
we find
$b = a \sin 0 + b \cos 0 = 0 \tag 9$
as well.
We have in fact established that $\sin x$, $\cos x$, and $e^x$ are linearly independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find all extrema of a complicated trigonometric function Problem
Find all local extrema for
$$f(x) = \frac{\sin{3x}}{1+\frac{1}{2}\cos{3x}}$$
Attempted solution
My basic approach is to take the derivative, set the derivative equal to zero and solve for x.
Taking the derivative with the quotient rule and a few cases of the chain rule for the trigonometric functions with a final application of the Pythagorean identity:
$$f'(x) = \frac{(1+\frac{1}{2}\cos{3x})(3\cos{3x})+1.5\sin{3x}\sin {3x}}{(1+\frac{1}{2}\cos{3x})^2} = \frac{3\cos 3x+1.5\cos^2 3x + 1.5\sin^2 3x}{(1+\frac{1}{2}\cos 3x)^2} = \frac{3 \cos 3x + 1}{(1+\frac{1}{2}\cos 3x)^2}$$
Putting it equal to zero and solving for x:
$$3\cos 3x + 1 = 0 \Rightarrow x = \frac{\arccos{\Big(-\frac{1}{3}\Big)}}{3} = \frac{\pi}{6} + \frac{2\pi n}{3}$$
...however the expected answer is $\pm\frac{2\pi}{9} + \frac{2\pi n}{3}$
So I must have gone wrong somewhere.
| It suffices to cancel the numerator of the derivative,
$$\cos(3x)(2+\cos(3x))+\sin(3x)\sin(3x)=2\cos(3x)+1=0$$
and
$$3x=2k\pi\pm\frac{2\pi}3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Prove ${_4 F_3} \left(\frac12, \frac12, 1, 1; \frac34, \frac54, \frac32; \frac14 \right)= \frac14 \left(\frac{\pi^2}{4}+\log^2 (2+\sqrt{3} ) \right)$ I was experimenting with series and numerically found this gem:
$$S=\sum_{n=0}^\infty \frac{2^{2n+1}}{(2n+1)^2 \binom{4n+2}{2n+1}}= \frac14 \left(\frac{\pi^2}{4}+\log^2 (2+\sqrt{3} ) \right)$$
Or, rewriting in hypergeometric form:
$${_4 F_3} \left(\frac12, \frac12, 1, 1; \frac34, \frac54, \frac32; \frac14 \right)= \frac14 \left(\frac{\pi^2}{4}+\log^2 (2+\sqrt{3} ) \right)$$
How can we prove this result?
I know that:
$$\int_0^\infty \frac{t^{2n}}{(1+t)^{4n+2}}dt=\frac{2}{(2n+1) \binom{4n+2}{2n+1}}$$
This gives us:
$$S=\int_0^\infty \frac{dt}{(1+t)^2} \sum_{n=0}^\infty \frac{(2t)^{2n}}{(2n+1) (1+t)^{4n}}$$
$$S=\frac12 \int_0^\infty \frac{dt}{t} \tanh^{-1} \left(\frac{2t}{(1+t)^2} \right)$$
Not sure how to find the closed form from here.
| $$I=\int_0^\infty \operatorname{actanh}\left(\frac{2x}{(1+x)^2}\right)\frac{dx}{x}=\frac12 \int_0^\infty \frac{\ln(1+4x+x^2)-\ln(1+x^2)}{x}dx$$
Now we will consider the following integral:
$$I(a)=\frac12 \int_0^\infty \frac{\ln(1+ax+x^2)-\ln(1+x^2)}{x}dx\Rightarrow I'(a)=\frac12 \int_0^\infty \frac{dx}{1+ax+x^2}$$
$$=\frac{1}{\sqrt{4-a^2}}\arctan\left(\frac{a+2x}{\sqrt{4-a^2}}\right)\bigg|_0^\infty =\frac{1}{\sqrt{4-a^2}}\arctan\left(\frac{\sqrt{4-a^2}}{a}\right)$$
We have $I(0)=0$ and we are looking to find $I(4)$, then:
$$I=\int_0^4 \frac{1}{\sqrt{4-a^2}}\arctan\left(\frac{\sqrt{4-a^2}}{a}\right)da=-\frac12 \arctan^2\left(\frac{\sqrt{4-a^2}}{a}\right)\bigg|_0^4$$
$$=\lim_{a\to 4}\frac12 \operatorname{arctanh}^2\left(\frac{\sqrt{a^2-4}}{a}\right)+\lim_{a\to 0}\frac12 \arctan^2\left(\frac{\sqrt{4-a^2}}{a}\right)$$
$$\Rightarrow \boxed{\int_0^\infty \operatorname{actanh}\left(\frac{2x}{(1+x)^2}\right)\frac{dx}{x}=\frac12 \ln^2(2+\sqrt 3)+\frac{\pi^2}{8}}$$
Another way to differentiate under the integral sign:
$$I= \int_0^\infty \operatorname{arctanh} \left(\frac{2t}{(1+t)^2} \right)\frac{dt}{t}\overset{t=\tan \frac{x}{2}}=\int_0^\pi \operatorname{arctanh} \left(\frac{\sin x}{\sin x+1}\right)\frac{dx}{\sin x}$$
$$\operatorname{arctanh} x=\frac12 \ln\left(\frac{1+x}{1-x}\right)\Rightarrow I=\frac12 \int_0^\pi \frac{\ln(1+2\sin x)}{\sin x}dx$$
$$I(a)=\frac12 \int_{0}^\pi \frac{\ln(1+\sin a\sin x)}{\sin x}dx\Rightarrow I'(a)=\frac12 \int_0^{\pi}\frac{\cos a}{1+\sin a\sin x}dx$$
$$\overset{\tan \frac{x}{2}=t}=\int_0^\infty \frac{\cos a}{(t+\sin a)^2+\cos^2 a}dt=\arctan\left(\frac{t+\sin a}{\cos a}\right)\bigg|_0^\infty =\frac{\pi}{2}-a$$
$$I(0)=0\Rightarrow I(a)=\int_0^a \left(\frac{\pi}{2}-x\right)dx=\frac{a}{2}(\pi-a)$$
$$\Rightarrow \boxed{I=\frac{\arcsin 2}{2}(\pi-\arcsin 2)=\frac{\pi^2}{8}+\frac12 \ln^2(2+\sqrt 3)}$$
| {
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"url": "https://math.stackexchange.com/questions/3297691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
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Solve $\int_{-\infty}^{\infty} \text{sech}^2(x-a)\text{sech}^2(x+a)\ \mathrm{d}x$ I am wondering if anyone knows a trick on how to solve this integral $$f(a)=\int_{-\infty}^{\infty} \text{sech}^2(x-a)\text{sech}^2(x+a)dx.$$ The answer should be a function of $a$.
Basically I am trying to reproduce some results from a paper. I've tried Matlab/Wolfram/Mathematica to no avail.
| Use the exponential definition, $\operatorname{sech}(x)=\frac{2}{e^{x}+e^{-x}}$, to rearrange the integrand.
$$\begin{aligned}\operatorname{sech}^2(x-a)\operatorname{sech}^2(x+a)&=\left(\frac{2}{e^{x-a}+e^{-(x-a)}}\right)^2\left(\frac{2}{e^{x+a}+e^{-(x+a)}}\right)^2
\\
&=\frac{16}{\left(e^{2x}+e^{-2x}+c\right)^2}
\end{aligned}$$
where $c=e^{2a}+e^{-2a}=\frac{2}{\operatorname{sech}(2a)}$. The integrand is an even function, so $f(a)=32\int_{0}^{+\infty}\frac{1}{\left(e^{2x}+e^{-2x}+c\right)^2}\ \mathrm{d}x$. Substitute $t=e^{2x}$, $\mathrm{d}x=\frac1{2t}\mathrm{d}t$.
$$\begin{aligned}f(a)&=32\int_1^{+\infty}\frac{1}{(t+1/t+c)^2}\cdot\frac{1}{2t}\ \mathrm{d}t
\\
&=16\int_1^{+\infty}\frac{t}{(t^2+ct+1)^2}\ \mathrm{d}t
\end{aligned}$$
It then suffices to use the standard integrals of $\frac{x}{(Ax^2+Bx+C)^n}$ and $\frac{1}{Ax^2+Bx+C}$. To avoid confusion (since unfortunately we've chosen the same letters), I'll only refer to the constants from the integral table in capitals.
$$\begin{aligned}f(a)&=16\left[-\frac{ct+2}{(4-c^2)(t^2+ct+1)}-\frac{c}{4-c^2}\int\frac{1}{(t^2+ct+1)}\ \mathrm{d}t\right]_1^{+\infty}
\\
&=16\left[-\frac{ct+2}{(4-c^2)(t^2+ct+1)}-\frac{c}{4-c^2}\left(\frac{1}{\sqrt{c^2-4}}\ln\left|\frac{2t+c-\sqrt{c^2-4}}{2t+c+\sqrt{c^2-4}}\right|\right)\right]_1^{+\infty}
\end{aligned}
$$
Note that $4-c^2<0$ except when $a=0$. Hence, we used the integral in the table for the case $4AC-B^2<0$.
As $t\to+\infty$, the $t^2$ in the denominator of the first term dominates, so the term tends to $0$. Likewise, the argument of the logarithm tends to $1$, so the second term tends to $0$. Hence, we just substitute $t=1$ and remember to take the negative sign since it's the integral's lower bound.
$$
\begin{aligned}
f(a)&=-16\left[-\frac{c+2}{(4-c^2)(1+c+1)}-\frac{c}{4-c^2}\left(\frac{1}{\sqrt{c^2-4}}\ln\left|\frac{2+c-\sqrt{c^2-4}}{2+c+\sqrt{c^2-4}}\right|\right)\right]
\\
&=
\frac{16}{(4-c^2)}\left(1+\frac{c}{\sqrt{c^2-4}}\ln\left(\frac{c-\sqrt{c^2-4}}{2}\right)\right)
\\
\end{aligned}$$
Note that $c>0$, which allowed us to simplify the absolute value in the logarithm.
Now, to find $f(a)$ in terms of $a$, we substitute back $c=e^{2a}+e^{-2a}$ and find
$$
\begin{aligned}
f(a)&=\frac{16}{-(e^{4a}-2+e^{-4a})}\left(1+\frac{e^{2a}+e^{-2a}}{\sqrt{(e^{4a}-2+e^{-4a})}}\ln\left(\frac{e^{2a}+e^{-2a}-\sqrt{(e^{4a}-2+e^{-4a})}}{2}\right)\right)
\\
&=\frac{-16}{(e^{2a}-e^{-2a})^2}\left(1+\frac{e^{2a}+e^{-2a}}{e^{2a}-e^{-2a}}\ln\left(\frac{e^{2a}+e^{-2a}-e^{2a}+e^{-2a}}{2}\right)\right)
\\
&=\frac{16}{(e^{2a}-e^{-2a})^2}\left(2a\frac{e^{2a}+e^{-2a}}{e^{2a}-e^{-2a}}-1\right)
\\
&=4\left(2a\frac{e^{2a}+e^{-2a}}{e^{2a}-e^{-2a}}-1\right)\left(\frac{2}{e^{2a}-e^{-2a}}\right)^2
\\
&=4\left(2a\coth(2a)-1\right)\operatorname{csch}^2(2a)
\end{aligned}$$
Where, in the second line, we have used $(e^{4a}-2+e^{-4a})=(e^{2a}-e^{-2a})^2$. Hence, the solution to the integral agrees with the one given by Wolfram Alpha.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining whether a base is a Generator in $Z^*_{11}$ as well as differentiating Primitive Root and/or Primitive Element? Given $Z^*_{11}$, specifically looking at the bases $2$ and $3$:
\begin{array} {|r|r|}\hline 2^1 \pmod {11}=2 & 3^1 \pmod {11}=3 \\ \hline 2^2 \pmod {11}=4 & 3^2 \pmod {11}=9 \\ \hline 2^3 \pmod {11}=8 & 3^3 \pmod {11}=5 \\ \hline 2^4 \pmod {11}=5 & 3^4 \pmod {11}=4 \\ \hline 2^5 \pmod {11}={10} & 3^5 \pmod {11}=1 \\ \hline 2^6 \pmod {11}=9 & 3^6 \pmod {11}=3 \\ \hline 2^7 \pmod {11}=7 & 3^7 \pmod {11}=9 \\ \hline 2^8 \pmod {11}=3 & 3^8 \pmod {11}=5 \\ \hline 2^9 \pmod {11}=6 & 3^9 \pmod {11}=4 \\ \hline 2^{10} \pmod {11}=1 & 3^{10} \pmod {11}=1 \\ \hline \end{array}
$ord_{11}(2)=10$ which is equivalent to $φ(11)=10$, so $k$ is the smallest integer
in $2^k \equiv 1 \pmod {11}$.
$ord_{11}(3)=5$, where $5$ is the smallest $k$
in $3^k \equiv 1 \pmod {11}$.
I'm fairly certain that $2$ is a generator, however I don't think $3$ is a generator because it doesn't generate every power in $Z^*_{11}$? But I've seen some posts that conflict with this. If $2$ is a generator, is it also a primitive root? I've also seen the term primitive element and wonder if it is synonomous as well?
| $x$ is a primitive root (synonymously, a primitive element) mod $n$ if and only if
$x$ is a generator of the multiplicative group of integers modulo $n$.
Modulo $11$, $2$ is a primitive root and $3$ is not.
Modulo $7$, on the other hand, $3$ is a primitive root and $2$ is not.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all $x$ such that $\sin x = \frac{4}{5}$ and $\cos x = \frac{3}{5}$.
Let
$$
\left\{
\begin{array}{c}
\sin x = \frac{4}{5} \\
\cos x = \frac{3}{5}
\end{array}
\right.
$$
Find all of the possible values for $x$.
My try: By dividing the equations we obtain $\tan x = \frac{4}{3}$ and then $$x = \arctan\frac{4}{3} + k\pi$$
But WolframAlpha gives
$$x = 2k\pi + 2\arctan\frac{1}{2}$$
Using $\arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$, we get $$2\arctan\frac{1}{2} = \arctan\frac{4}{3}$$
but the answers are different still.
Why does this happen? And what is the correct answer?
| Your original (set of) equations implies $\tan x=\frac{4}{3}$ but not the other way around. When you solve $\tan x=\frac{4}{3}$ you get the solutions of your original equation and the solutions $\sin x=\frac{-4}{5}$, $\cos x=\frac{-3}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Does this function hit every odd number for some integers $n$ and $m$? I have recently been looking into a problem and created a function that looked interesting and I wondered if it would hit all odd numbers on a graph.
For integer $n$, define
$$g_n(x) = \frac13\cdot \begin{cases}
(3n-1) \cdot 2^{2x-1} - 1, & n \text{ even} \\[4pt]
(3n-2) \cdot 2^{2x\phantom{-1}} - 1, & n \text{ odd}
\end{cases} \tag{$\star$}$$
Is it true that, for any positive odd integer $k$, there are integers $n$ and $m$ such that $g_n(m)=k$?
PS: I think is easier to see as a graph: desmos graph
Note: The form of $(\star)$ is dramatically different than what appeared in the original version of this question. See answer by @Blue for the derivation, which matches coincident work by @automaticallyGenerated.
| It can be seen that $f_n = 3n+\frac{cos(\pi n)-3}{2}$. $\frac{cos(\pi n)-3}{2}$ will be equal to $-2$ if $n$ is odd and $-1$ if $n$ is even.
Thus, $f_n = 3n-(
n\pmod 2)-1$.
From that, we get
$$g_n(x) = \frac{(3n-(
n\pmod 2)-1)*2^{2x+1-(((3n-(
n\pmod 2)-1) \space \text{mod} \space 3))}-1}{3}$$
Simplifying this finds $$g_n(x) = \frac{(3n-(
n\pmod 2)-1)*2^{2x+1-(((-(
n\pmod 2)-1) \space \text{mod} \space 3))}-1}{3}$$
Simplifying this further finds $$g_n(x) = \frac{(3n-(
n\pmod 2)-1)*2^{2x+n\pmod 2 -1}-1}{3}$$
Let's say that $n = 2k$, with $k$ an integer.
Then,
$$g_n(x) = \frac{(6k-1)*2^{2x-1}-1}{3}$$
If we fix $x = 1$, we get all values $y$, such that $y = 3 \pmod 4$
Similarly, if we fix $x = 2$, we get all values $y$, such that $y = 13 \pmod {16}$.
If we fix $x = 3$, all values such that $y = 53 \pmod {64}$. In general, if we fix $x$ as a positive integer, we get all values such that $y = \frac{5*4^{x}-2}{6} \pmod {4^x}$
So far, we have proved that the odd integers of the form $y = $: $$3 \pmod {4}$$ $$13 \pmod {16}$$ $$53 \pmod{64}$$ etc exist.
Covering the other case, where $n = 2k+1$, we get $$g_n(x) = \frac{(6k+1)*2^{2x}-1}{3}$$ If we similarly fix $x$ as a positive integer, we get values $y$ such that $y = \frac{4^x-1}{3} \pmod {2^{2x+1}}$
This yields $y$ such that $y = $:
$$1 \pmod {8}$$
$$5 \pmod {32}$$
$$21 \pmod {128}$$
etc exist.
Both of these sets cover all positive odd integers. The reason for this is that $3 \pmod 4$ covers all odd values except for $y = 1 \pmod 4$. $1 \pmod 8$ then covers all values that are not already covered except for $y = 5 \pmod 8$. Then $13 \pmod {16}$ covers everything except for $5 \pmod {16}$. This process can be extended ad infinitum until all odd positive integers are "covered".
Therefore, this function does cover all positive odd integers when $n, m$ are integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Alternative methods for solving a system of one linear one non linear simultaneous equations Take the equations $$x+y=5$$ $$x^2 + y^2 =13$$
The most basic method to solve this system is to first express the linear equation in terms of one of the variables and then sub that into the non-linear equation.
But I am curious if there are other methods to solve such a system ?
| The quadratic equation can be used.
Given:
x + y = 5, then y = 5 -4
Given
x^2 + y^2 = 13
then x^2 + (4-x)^2 = 13
and x^2 + x^2 - 10x + 25 -13 = 0
2x^2 + (-10x) + 12 = 0
Then the co-factors are a = 2, b = -10, c = 12
y = [-b (+-) sqrt(b^2 - 4ac)]/[2a] <-- Quadratic Formula
y = [-(-10) (+-) sqrt((-10)^2 - 4(2*12))]/(2*2)
y = [10 (+-) sqrt(100-96)]/4
y = [10 + 2]/4 and y = [10-2]/4
y = 12/4 and y = 8/4
y = 3 and y = 2
given x + y = 5
When y = 3, x + 3 = 5, x = 5-3, x = 2
when y = 2, x+2 = 5, x = 5-2, x = 3
Answers: x = 3, y = 2 and x = 2, y = 3
Try your answers in all of the original equations and against any given or implied restrictions to make sure they work. They do! (Always check for 'extraneous' answers.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 10,
"answer_id": 2
} |
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$.
I have tried:
$$\frac{10!}{(3!\times7!)} \times (-x + 2)^7 \times (x^2)^3 $$
But got an incorrect answer $-15360$.
| For a problem of this size, I would just ploddingly take a few derivatives:
$$\begin{align}
f(x)=(x^2-x+2)^{10}
&\implies f'(x)=10(x^2-x+2)^9(2x-1)\\
&\implies f''(x)=90(x^2-x+2)^8(2x-1)^2+20(x^2-x+2)^9\\
&\implies f'''(x)=720(x^2-x+2)^7(2x-1)^3+360(x^2-x+2)^8(2x-1)+180(x^2-x+2)^8(2x-1)\\
&\implies f'''(0)=-720\cdot2^7-360\cdot2^8-180\cdot2^8=-(720+1080)2^7=-1800\cdot128\\
\end{align}$$
The coefficient of $x^3$ is $f'''(0)/6=-300\cdot128=-38400$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How to show the matrix has Rank $\le 5$
I want to show that the following matrix has Rank $\le 5$.
The matrix is
\begin{bmatrix}
2&1&1&1&0&1&1&1\\
1&2&1&1&1&0&1&1\\
1&1&2&1&1&1&0&1\\
1&1&1&2&1&1&1&0\\
0&1&1&1&2&1&1&1\\
1&0&1&1&1&2&1&1\\
1&1&0&1&1&1&2&1\\
1&1&1&0&1&1&1&2
\end{bmatrix}
I found that there is a submatrix in the matrix which has rank $ =4$ given by
$[2,1,1,1],[1,2,1,1],[1,1,2,1],[1,1,1,2]$.
I need to show the given matrix has at least 3 zero rows in order to show that Rank $\le 5$..
But I dont know how to show it.
Can someone help.
| We will show that the rank of the given matrix is exactly $5$.
The matrix can be written as a block matrix such that (see LU-decomposition),
$$M:=\begin{bmatrix} A & B \\B &A \end{bmatrix}=\begin{bmatrix} I & 0 \\BA^{-1}&I \end{bmatrix}\cdot \begin{bmatrix} A&0\\0&A-BA^{-1}B \end{bmatrix}\cdot\begin{bmatrix} I&A^{-1}B \\ 0 & I\end{bmatrix}.$$
with
$$A = \begin{bmatrix} 2 & 1 & 1 & 1 \\ 1 & 2 & 1 &1 \\ 1 &1&2&1\\1&1&1&2\end{bmatrix}, \quad B = \begin{bmatrix} 0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0 \end{bmatrix}.$$
Now
$$A^{-1}=\frac{1}{5}\begin{bmatrix} 4 & -1 & -1 & -1 \\ -1 & 4 & -1 &-1 \\ -1 &-1&4&-1\\-1&-1&-1&4\end{bmatrix}\quad
\text{and}\quad
A-BA^{-1}B=\frac{4}{5}\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 &1 \\ 1 &1&1&1\\1&1&1&1\end{bmatrix}.$$
Therefore we may conclude that
$$\text{rank}(M)=\text{rank}\left(\begin{bmatrix} A&0\\0&A-BA^{-1}B \end{bmatrix}\right)=\text{rank}(A)+\text{rank}(A-BA^{-1}B)
=4+1=5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find $n$ if $\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$ Find $n$ if $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$$
In this video they show a shortcut and say $n=-1/2$ without any explanation.
Key observation here is that the geometric mean of $9$ and $4$ is $6$.
It seems numerator and denominator are partial sums of geometric series, but I don't know how to proceed. Any help?
| Another idea: $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n}=\frac{5\cdot9^n+4\left(9^n+4^n\right)}{9^n+4^n}=5\frac{9^n}{9^n+4^n}+4\implies$$
$$\frac{9^{n+1}+4^{n+1}}{9^n+4^n}=6\iff5\frac{9^n}{9^n+4^n}=2\implies\frac1{\frac{9^n+4^n}{9^n}}=\frac25\implies\frac1{1+\left(\frac49\right)^n}=\frac25\implies1+\left(\frac49\right)^n=\frac52\implies$$
$$\left(\frac49\right)^n=\frac32=\left(\frac23\right)^{-1}\iff\left(\frac23\right)^{2n}=\left(\frac23\right)^{-1}\implies 2n=-1\implies n=-\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$. Let $n \in \mathbb{N}$. Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$.
I began this problem giving some values for $n$ and I found that $\gcd(n^2+3, (n+1)^2+3)=1$ for most of $n$ I tried, but if $n=6$, then $\gcd=13$.
Then I tried to prove that only for $n=6$, $\gcd \neq 1$ but I couldn't.
Can someone help me with this problem?
| We assume the Lemma that if $\gcd(a,b)=1$ and $c$ is an integer, then $\gcd(a,bc)=\gcd(a,c).$
Also, that in general: $\gcd(a,b)=\gcd(a,b-ak)$ for any integer $k.$
Then $$\begin{align}
\gcd(n^2+3,(n+1)^2+3)&=\gcd(n^2+3,(n+1)^2+3-(n^2+3))\\
&=\gcd(n^2+3,2n+1)\\
&=\gcd(n^2+3-3(2n+1),2n+1)\\
&=\gcd(n(n-6),2n+1)\\
&=\gcd(n-6,2n+1)\quad \quad\text{ since }\gcd(n,2n+1)=\gcd(n,2n+1-2n)=1\\
&=\gcd(n-6,2n+1-2(n-6))\\
&=\gcd(n-6,13)
\end{align}$$
So the GCD is $13$ when $n-6$ is divisible by $13$ and $1$ otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Evaluating $\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx$ My first instinct was to evaluate the indefinite form of the integral, which I did by substituting $x=\tan t$, therefore yielding
\begin{align}
\int \frac{x\ln x}{(1+x^2)^2} \,dx
&= \int \frac{\tan t \sec^2 t \ln\tan t}{(1+\tan^2 t)^2} \,dt
&& \text{by substitution} \\
&= \int \tan t \cos^2 t \ln \tan t \,dt \\
&= \int \sin t \cos t \ln \tan t \,dt \\
&= -\frac{1}{2} \cos^2 t \ln \tan t + \frac{1}{2} \int \cot t \, dt
&& \text{by parts} \\
&= -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t + k
\end{align}
I run into a wall when I introduce the limits of the integral, since I get
\begin{align}
\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx
&= \bigg[ -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t \bigg] ^\frac{\pi}{2} _0
\end{align}
I'm not too sure how to evaluate the limit of the final equation as $t \rightarrow 0$. I feel like the solution is something very trivial, but I can't quite put my finger on what I'm forgetting.
| Set $\dfrac1x=y$
$$\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx=I=\int_\infty^0\dfrac{y^4\ln(y^{-1})}{y(1+y^2)^2}\left(\dfrac{dy}{-y^2}\right)=-\int_0^\infty\dfrac{y\ln y}{(1+y^2)^2}\ dy =-I$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
Find the asymptote of the function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ We have a function $f(x) = \sqrt{\frac{x^3}{x - 3}} - x$ and when $x$ goes towards $-\infty$, we have an asymptote
$y = -2x - 3/2$.
How we get this asymptote?
| Let
$$
f(x)
=\sqrt{\frac{x^3}{x-3}}-x
=|x|\sqrt{\frac{x}{x-3}}-x
$$
For $x<0$ we have $|x|=-x$, thus
$$
f(x)
=-x\sqrt{\frac{x}{x-3}}-x
=\Bigl(-\sqrt{\frac{x}{x-3}}-1\Bigr)x
=\Bigl(-\sqrt{\frac{1}{1-3/x}}-1\Bigr)x.
$$
For $x<0$;
$$\lim_{x\to-\infty}\frac{f(x)}{x}
=\lim_{x\to-\infty}\Bigl(-\sqrt{\frac{1}{1-3/x}}-1\Bigr)
=-1-1=-2.
$$
Now, study
\begin{align*}
f(x)-(-2x)
&
=f(x)+2x
=-x\sqrt{\frac{x}{x-3}}-x+2x
=-x\sqrt{\frac{x}{x-3}}+x
=\Bigl(-\sqrt{\frac{x}{x-3}}+1\Bigr)x.
\end{align*}
Maclaurin expansion gives
\begin{align*}
\sqrt{\frac{x}{x-3}}
&=\sqrt{\frac{x-3+3}{x-3}}
=\sqrt{1+\frac{3}{x-3}}
=1+\frac{1}{2}\cdot\frac{3}{x-3}+\Bigl(\frac{3}{x-3}\Bigr)^2L(1/x)
\end{align*}
and we have
\begin{align*}
\Bigl(-\sqrt{\frac{x}{x-3}}+1\Bigr)x
&=\Bigl(-\Bigl(1+\frac{1}{2}\cdot\frac{3}{x-3}+\Bigl(\frac{3}{x-3}\Bigr)^2L_1(1/x)\Bigr)+1\Bigr)x
\\&=\Bigl(-1-\frac{1}{2}\cdot\frac{3}{x-3}-\frac{1}{(x-3)^2}L_2(1/x)+1\Bigr)x
\\&=\Bigl(-\frac{3}{2}\cdot\frac{1}{x-3}-\frac{1}{(x-3)^2}L_2(1/x)\Bigr)x
\\&=-\frac{3}{2}\cdot\frac{x}{x-3}-\frac{x}{(x-3)^2}L_2(1/x)
\\&=-\frac{3}{2}\cdot\frac{1}{1-3/x}-\frac{x}{(x-3)^2}L_2(1/x)
\\&\to-\frac{3}{2}
\end{align*}
as $x\to-\infty$, where $L_{1,2}(1/x)$ is limited for large $x$.
Hence
$$
\lim_{x\to-\infty}\bigl(f(x)+2x\bigr)=-\frac{3}{2}
$$
showing that
$$
f(x)\,\,\mathrm{"}=\mathrm{"}\,-2x-\frac{3}{2}
$$
for large negative $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Interesting four-sum inequality $n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right) \ge...$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)$$ holds.
My work.In fact, I want to solve another problem ( Prove $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$ ) This another problem has already been solved, but I want to solve it by the method that the author of the problem intended. The fact is that in the original problem there was another inequality ( Prove the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ ). It seems to me that the inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$ is proved as the author of the problem wanted. I think that the author of the problem wanted us to prove inequality $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$ on the basis of inequality $\sum \limits_{k=1}^n \frac{k+1}{k} \cdot \sum \limits_{k=1}^n \frac{k}{k+1} \le \frac{9}{8}n^2$. To this end, it suffices to prove the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le n^2 \left(\sum \limits_{k=1}^n \frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n \frac{k}{k+1}\right)$$ I checked this inequality by numerical methods on a computer. This inequality holds for all $n$. Interestingly, if $n \to \infty$ then this inequality (in the limit) is an equality.
Perhaps this will help to solve the problem: let $x_k=\frac{k+1}{k}$. Then $1<x_k \le 2$ and inequality takes the form $$ \left(\sum \limits_{k=1}^n (2k-1)x_k\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{1}{x_k}\right) \le n^2 \left(\sum \limits_{k=1}^n x_k\right) \left( \sum \limits_{k=1}^n \frac{1}{x_k}\right)$$
| Here's a probabilistic interpretation and solution to this problem, although the method itself may be negligibly faster than the existing answers. Let $X$ be a discrete random uniform random variable on $\{1, \cdots, n\}$. Then the problem can be rewritten as
\begin{align*}
\mathbb{E}\left[(2X-1)\frac{X+1}{X}\right]\mathbb{E}\left[(2X-1)\frac{X}{X+1}\right] = \mathbb{E}[2X-1]\mathbb{E}\left[\frac{X+1}{X}\right]\mathbb{E}[2X-1]\mathbb{E}\left[\frac{X}{X+1}\right]
\end{align*}
Rewritting the LHS in terms of $\mathbb{E}[f(X)g(X)] = \text{Cov}(f(X), g(X)) + \mathbb{E}[f(X)]\mathbb{E}[g(X)]$ and dividing by the RHS,
\begin{align*}
\left(1 + \frac{\text{Cov}\left(2X-1, \frac{X+1}{X}\right)}{n\mathbb{E}\left[\frac{X+1}{X}\right]}\right)\left(1 + \frac{\text{Cov}\left(2X-1, \frac{X}{X+1}\right)}{n\mathbb{E}\left[\frac{X}{X+1}\right]}\right) \le 1
\end{align*}
Note that
\begin{align*}
\text{Cov}\left(2X-1, \frac{X+1}{X}\right) = 2\text{Cov}\left(X, \frac{X+1}{X}\right) = n+3 - (n+1)\mathbb{E}\left[\frac{X+1}{X}\right] \\
\text{Cov}\left(2X-1, \frac{X}{X+1}\right) = 2\text{Cov}\left(X+1, \frac{X}{X+1}\right) = n+1 - (n+3)\mathbb{E}\left[\frac{X}{X+1}\right]
\end{align*}
So we have to prove
\begin{align*}
\left(-\frac{1}{n} + \left(1 + \frac{3}{n}\right)\frac{1}{\mathbb{E}[\frac{X+1}{X}]}\right)\left(-\frac{3}{n} + \left(1 + \frac{1}{n}\right)\frac{1}{\mathbb{E}[\frac{X}{X+1}]}\right) \le 1
\end{align*}
Let $a_n = \mathbb{E}[\frac{X+1}{X}]$ and $b_n = \mathbb{E}[\frac{X}{X+1}]$. Reorganizing, the statement we need to prove is
\begin{align*}
f(n) = (a_n b_n - 1)n^2 + (3b_n + a_n - 4)n + (a_n + 9b_n - 3a_nb_n - 3) \ge 0
\end{align*}
It is straightforward to prove the bounds
\begin{align*}
1 &< a_n = 1 + \frac{H_n}{n} \\
1 + \frac{1}{n+1} - \frac{H_n}{n} &= b_n < 1 \\
a_n b_n &= 1 - \text{Cov}\left(\frac{X+1}{X}, \frac{X}{X+1}\right) \\
&= 1 + \text{Cov}\left(\frac{1}{X}, \frac{1}{X+1}\right) \\
&= 1 + \mathbb{E}\left[\frac{1}{X(X+1)}\right] - \mathbb{E}\left[\frac{1}{X}\right]\mathbb{E}\left[\frac{1}{X+1}\right] \\
&= 1 + \frac{1}{n}\left(1 - \frac{1}{n+1}\right) - \mathbb{E}\left[\frac{1}{X}\right]\mathbb{E}\left[\frac{1}{X+1}\right] \\
&\begin{cases} < 1 + \frac{1}{n} \\
> 1 + \frac{1}{n} - \frac{1}{n(n+1)} - \frac{H^2_n}{n^2}
\end{cases}
\end{align*}
Therefore,
\begin{align*}
f(n) &> \left(\frac{1}{n} - \frac{1}{n(n+1)} - \frac{H^2_n}{n^2}\right)n^2 + \left(\frac{3}{n+1} - 2\frac{H_n}{n}\right)n + \left(4 + \frac{9}{n+1} - \frac{8H_n}{n} - \frac{3}{2n}\right) \\
&= n-\frac{n}{n+1} - H_n^2 + 3\frac{n}{n+1} - 2H_n + \left(4 + \frac{9}{n+1} - \frac{8H_n}{n} - \frac{3}{2n}\right) \\
&= n + \underbrace{2\frac{n}{n+1}}_{\ge 1.9 \text{ for } n \ge 20} - H_n^2 - 2H_n + \underbrace{\left(4 + \frac{9}{n+1} - \frac{8H_n}{n} - \frac{3}{2n}\right)}_{\ge 2.9 \text{ for } n \ge 20} \\
&> n + 4.8-2H_n - H_n^2
\end{align*}
and the last expression is $> 0$ and increasing for $n \ge 20$. We can manually check the inequality is satisfied for $1 \le n \le 19$. With some better bounding, we can decrease the number of cases we need to check.
| {
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"url": "https://math.stackexchange.com/questions/3312678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Rules of inequality and modulus inequality. I am very confused whenever I encounter an inequality .
Like suppose
$ \frac{| \sqrt {1 -4x^{2}}|}{|x|} \le 1 $
Now I am confused whether to square or not .
I am very well aware of the rule at we do not square if the sign of $x$ is not known . But what if the range of $x$ is given as
$ \frac {-1}{2} \le x \le \frac 12 $ .
Now here I am confused whether to square or not as $ x $ can take either negative value.
If someone could give me a link to a set of rules of inequalities under such conditions , it would be appreciated .
|
Note: All of what follows is based on the axiom that if $a < b$ (regardless of the signs of either) and $c > 0$ then $ac < bc$ and likewise if $d < 0$ then $ad > bd$.
I'm assuming that is known and familiar.
You can do whatever you want. You can take them out to dinner for all I care. You just have to know what you are doing.
If you have $x < 3$ and $x$ is negative, say, then you can not say $x^2 < 9$ because: If we multiple both sides of the equation $x < 3$ by $x$ we flip the inequality to get $x^2 > 3x$ and if we multiply both sides of $x < 3$ by $3$ we get $3x < 9$. So we get $x^2 > 3x$ and $9> 3x$ but we have no way of comparing $x^2$ to $9$.
That's not wrong. It's just not what we'd expect if we were stupid and just monkeying around with rules we didn't understand and "squaring both sides" without thinking.
Anyway.... If we do know the signs say $0 < x < 3$ then we can square both sides because $x< 3$ and $x > 0$ means $x^2 < 3x$ and $x < 3$ and $3>0$ so $3x < 9$. So $0 < x^2 < 9$.
We can actually do if $-a < x < b$ where $-a < 0$ and $b < 0$ we can get.
If $x < 0$ then $-a < x \implies -ax > x^2; (-a)^2 = a^2 > -ax$ so $x^2 < a^2$. But if $x > 0$ we have $-ax < x^2$ and $(-a)^2 =a^2 > -ax$ and we have no way of compare $a^2$ to $x^2$.
And if $x < 0$ we get $x < b$ means $x^2 > bx$ and $bx < b^2$ but we have no way to compare $x^2$ to $b^2$. But if $x > 0$ we have $x^2 < b^2$.
So we have either $x^2 < a^2$ or $x^2 < b^2$ so we do have $x^2 < \max(a^2,b^2)$.
And note. If $a = b$ and you have $-a < x < a$ we have $0 \le x^2 < a^2$.
.....
But to your problem.....
Well, yours is simple. $|K| \ge 0$ so you have
$0 \le \frac {|1-4x^2|}{|x|} \le 1$
So we know the signs and we can square both sides.
$0 \le \frac {1-4x^2}{x^2} \le 1$ so
$0 \le 1-4x^2 \le x^2$ so
$4x^2 \le 1 \le 5x^2$ so
$0 \le 4x^2 \le 1 \le 5x^2$ so
$0 \le 2|x| \le 1 \le \sqrt 5|x|$. So $|x| \le \frac 12$ and $\frac 1{\sqrt 5} \le |x|$ so $\frac 1{\sqrt {5} }\le |x| \le \frac 12$.
We don't know the sign of $x$ so
Either $-\frac 12 \le x \le -\frac 1{\sqrt 5}$ or $\frac 1{\sqrt {5} }\le x \le \frac 12$
| {
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"url": "https://math.stackexchange.com/questions/3313361",
"timestamp": "2023-03-29T00:00:00",
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Indices - factorising Simplify: $$\frac {x^5y^2x^3 + x^4y^5 - y^5x^7y^4}{x^4y^3}$$
I know this is probably low level stuff but I need to be able to do this specific type of question and I have no way of checking my work. If anyone could offer a step by step working I'd be appreciative
| An alternate way is to break the problem down into simple workable parts:
$$\frac{x^5y^2x^3+x^4y^5−y^5x^7y^4}{x^4y^3}$$
$$=\frac{x^5y^2x^3}{x^4y^3}+\frac{x^4y^5}{x^4y^3}-\frac{y^5x^7y^4}{x^4y^3}$$
$$=\frac{x^8y^2}{x^4y^3}+\frac{x^4y^5}{x^4y^3}-\frac{x^7y^9}{x^4y^3}$$
$$=\frac{x^4}{y}+y^2-x^3y^6,\quad x\not=0.$$
This is equivalent to the answers given above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I evaluate $\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$ without using L'Hopital or series?
How do I evaluate $\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$ without using L'Hopital or series?
I've tried expanding the variable such as $x = 2y$ or $x = 3y$, but seemed to still get stuck.
| First, read this answer. This answer shows that, for $x$ close to $0$, we have the following:
$$\sin x \leq x \leq \tan x$$
Thus, since $\sin x \leq x$ and $\sin x$ has the same sign as $x$ (i.e. either both are positive, both are negative, or both are $0$), we know that $\frac{1}{\sin x} \geq \frac{1}{x}$. Thus, $\frac{1}{\sin x}-\frac{1}{x}\geq 0$.
Also, since $x \leq \tan x$ and $\tan x$ has the same sign as $x$, we know that $\frac{1}{x} \geq \frac{1}{\tan x}$. This implies that $\frac{1}{\sin x}-\frac{1}{x} \leq \frac{1}{\sin x}-\frac{1}{\tan x}$,
Now, we have:
$$0 \leq \frac{1}{\sin x}-\frac{1}{x} \leq \frac{1}{\sin x}-\frac{1}{\tan x}$$
Simplify the trig expressions:
$$0 \leq \frac{x-\sin x}{x\sin x} \leq \frac{1-\cos x}{\sin x}$$
Now, $\lim_{x\to 0} 0=0$ for obvious reasons. Thus, let's consider $\lim_{x\to 0} \frac{1-\cos x}{\sin x}$. Multiply numerator and denominator by $1+\cos x$:
$$\lim_{x\to 0}\frac{1-\cos^2 x}{\sin x(1+\cos x)}=\lim_{x\to 0}\frac{\sin^2 x}{\sin x(1+\cos x)}=\lim_{x\to 0}\frac{\sin x}{1+\cos x}=\frac{\sin 0}{1+\cos 0}=0$$
Thus, $\lim_{x\to 0} 0=\lim_{x\to 0} \frac{1-\cos x}{\sin x}=0$, so by Squeeze Theorem, $\lim_{x\to 0}\frac{x-\sin x}{x\sin x}=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Getting different answers for an integral: $\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$ vs $\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$ Problem:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
Using two different methods I am getting two different answers and have trouble finding why.
Method 1:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
$$\int\frac{1}{2}-\frac{3}{2}\left(\frac{1}{x+2}\right)dx$$
$$\int\frac{1}{2}dx-\frac{3}{2}\int\frac{1}{x+2}dx$$
$$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{x+2}dx$$
$$x+2=u$$
$$dx=du$$
$$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$
$$\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$$
Method 2:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
$$\int\frac{1}{2}-3\left(\frac{1}{2x+4}\right)dx$$
$$\int\frac{1}{2}dx-3\int\frac{1}{2x+4}dx$$
$$\frac{1}{2}x-3\int\frac{1}{2x+4}dx$$
$$2x+4=u$$
$$dx=\frac{du}{2}$$
$$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$
$$\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$$
| $$\begin{align}\frac12 x + \frac 32 \ln |2x + 4| + C_1 &= \frac12x + \frac32\ln(2\cdot|x+2|) + C_1\\&=\frac12x + \frac32(\ln 2 + \ln|x+2|)+C_1\\&=\frac12x + \frac32 \ln|x+2| + (\frac 32 \ln 2 + C_1)\\&=\frac12x + \frac32\ln|x+2| + C_2\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321992",
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"source": "stackexchange",
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Rate of change with $f(x)=4x^2-7$ on $[1,b]$ As part of a textbook exercise I am to find the rate of change of $f(x)=4x^2-7$ on inputs $[1,b]$.
The solution provided is $4(b+1)$ and I am unable to arrive at this solution.
Tried:
$f(x_2)=4b^2-7$
$f(x_1)=4(1^2)-7=4-7=-3$
If the rate of change is $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ then: $\frac{(4b^2-7)-3}{b-1}$ = $\frac{4b^2-10}{b-1}$
This is as far as I got. I tried to see if I could factor out the numerator but this didn't really help me:
$(4b^2-10)=2(2b^2-5)$
If I substitute this for my numerator I still cannot arrive at the provided solution.
I then tried isolating b in the numerator:
$4b^2-10=0$
$4b^2=10$
$b^2=10/4$
$b=\frac{\sqrt{10}}{\sqrt{4}}=\frac{\sqrt{10}}{2}$
This still doesn't help me arrive at the solution.
How can I arrive at $4(b+1)$?
| A small sign-mistake! See the highlighted parts in red and blue:
Tried:
$f(x_2)=4b^2-7$
$\color{blue}{f(x_1)}=4(1^2)-7=4-7=\color{blue}{-3}$
If the rate of change is $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ then: $\frac{(4b^2-7)\color{red}{-3}}{b-1}$ = $\frac{4b^2-10}{b-1}$
Which should be:
$$\frac{f(x_2)\color{red}{-}\color{blue}{f(x_1)}}{x_2-x_1} = \frac{(4b^2-7)\color{red}{-}\left(\color{blue}{-3}\right)}{b-1} = \frac{4b^2-4}{b-1}$$
Then proceed with $4b^2-4=4\left(b^2-1\right)=4\left(b-1\right)\left(b+1\right)$ and simplify.
| {
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How can we not use Muirhead's Inequality for proving the following inequality? There was a question in the problem set in my math team training homework:
Show that $∀a, b, c ∈ \mathbb{R}_{≥0}$ s.t. $a + b + c = 1, 7(ab + bc + ca) ≤ 2 + 9abc.$
I used Muirhead's inequality to do the question (you can try out yourself):
By Muirhead's inequality, $$\begin{align}7(ab+bc+ca)&=7(a+b+c)(ab+bc+ca)\\&=21abc+6\big(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\big)+\sum_{sym}a^2b\\&\le21abc+6\big(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\big)+\sum_{sym}a^3\\&=2(a+b+c)^3+9abc\\&=2+9abc\end{align}$$As $(3,0,0)$ majorizes $(2,1,0)$.
Is above proof correct? Also, can we find a proof not using Muirhead's inequality?
Any help is appreciated!
| Yes, your proof is correct. I have a similar proof where, for the last step, we need just the AM-GM inequality.
So it suffice to show that
$$2(a+b+c)^3 + 9abc \geq 7(a+b+c)(ab+bc+ca)$$
that is
$$2(a^3+b^3+c^3) \geq (a^2b +b^2c+c^2a)+ (ab^2 +bc^2+ca^2).$$
Now the inequality
$$a^3 +b^3 +c^3 \ge a^2b +b^2c+c^2a$$
follows from AM-GM inequality
$$x + y + z \geq 3 \sqrt[3]{xyz}.$$
By letting $x = y = a^3$ and $z = b^3$ we get
$$a^3 + a^3 + b^3 \geq 3 \sqrt[3]{a^3a^3b^3} = 3a^2b.$$ Similarly, we find
\begin{align*}
b^3 + b^3 + c^3 &\geq 3 \sqrt[3]{b^3b^3c^3} = 3b^2c \\
c^3 + c^3 + a^3 &\geq 3 \sqrt[3]{c^3c^3a^3} = 3c^2a
\end{align*}
Add all together and we are done. By symmetry also the other one holds
$$a^3+b^3+c^3 \geq ab^2 +bc^2+ca^2.$$
| {
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Solving the integral $\int\frac{1}{\sqrt{x^2+1}}\,dx$ I need to solve
$$\int\frac{1}{\sqrt{x^2+1}}$$
but I'm nowhere near the solution. I tried substituting
$u = x^2 + 1$
such that $dx = \frac{1}{2x}$
yielding
$$\int u^{-1/2} du = \frac{1}{2x}\frac{u^{1/2}}{1/2}=\frac{\sqrt{u}}{x} =\frac{\sqrt{x^2+1}}{x}$$
but this is nothing near the solution of
$$\int\frac{1}{\sqrt{x^2+1}} = ln|\sqrt{x^2+1}+x|+C$$
| Hint: Substituting $$x=\tan(t)$$ then we get $$x^2+1=\tan^2(t)+1=\frac{\sin^2(t)+\cos^2(t)}{\cos^2(t)}=\frac{1}{\cos^2(t)}$$
and $$dx=(\tan^2(t)+1)dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the integral $\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}$ One may take $x= \cos t$ and get
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= -\frac{1}{4}\int \csc^4(t/2)~ dt=-\frac{1}{4} \int [\csc^2(t/2) +\csc^2(t/2) \cot^2(t/2)]~ dt.$$
$$\Rightarrow I=\frac{1}{2} \left [\cot (t/2)] +\frac{1}{3}\cot^3(t/2)\right]=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
The question is: How to obtain this integral by other methods not using the trigonometric substitution.
| Let us first find evaluate the indefinite integral $ \displaystyle \int \dfrac1{ (1-x)\sqrt{1-x^2} } \, dx $.
$$\begin{array} {r c l} \displaystyle \int \dfrac1{ (1-x)\sqrt{1-x^2} } \, dx &=& \displaystyle \int \dfrac{1+x}{ (1-x^2)\sqrt{1-x^2} } \, dx \\ &=& \displaystyle \int \dfrac{1}{ (1-x^2)\sqrt{1-x^2} } \, dx + \displaystyle \int \dfrac{x}{ (1-x^2)\sqrt{1-x^2} } \, dx \end{array} $$
Using the substitution $y = \sqrt{1-x^2}$ for both these integrals, we can find that they evaluate to $ \frac x{\sqrt{1-x^2}} $ and $ \frac1{\sqrt{1-x^2}} $, respectively. Thus, $$ \displaystyle \int \dfrac1{ (1-x)\sqrt{1-x^2} } \, dx = \frac{x+1}{\sqrt{1-x^2}} + C .$$
Now we integrate by parts,
$$ \begin{array} {r c l} \displaystyle I := \int \dfrac1{ (1-x)^2\sqrt{1-x^2} } \, dx &=& \displaystyle \int \underbrace{\dfrac1{1-x}}_{=u} \cdot \underbrace{\dfrac1{ (1-x)\sqrt{1-x^2} } \, dx}_{=dv} \\ &=& \displaystyle \int u \, dv = uv - \int v \, du \\
&=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} - \displaystyle \int \dfrac{x+1}{\sqrt{1-x^2}} \cdot \dfrac1{(1-x)^2} \, dx \\
&=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} - \displaystyle \int \dfrac{x}{(1-x)^2 \sqrt{1-x^2}}\, dx - \underbrace{\displaystyle \int \dfrac{1}{\sqrt{1-x^2}} \cdot \dfrac1{(1-x)^2} \, dx}_{=I} \\
2I &=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} + \displaystyle \int \dfrac{1-x - 1}{(1-x)^2 \sqrt{1-x^2}}\, dx \\
&=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} + \displaystyle \int \dfrac{1}{(1-x)\sqrt{1-x^2}}\, dx - I \\
3I &=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} + \dfrac{x+1}{\sqrt{1-x^2}} + C \\
I &=& \dfrac{(x+1)(x-2)}{3(x-1)\sqrt{1-x^2}} + C \\
\end{array} $$
| {
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Evaluate $\int_{0}^{2\pi} \frac{cos \theta}{2 + cos \theta} d\theta$ using the residue theorem My attempt to a solution for $I = \int_{0}^{2\pi} \frac{cos \theta}{2 + cos \theta} d\theta$ is as follows.
On the unit circle we have $z=e^{i\theta} \implies dz = izd\theta \iff d\theta = \frac{dz}{iz}$, and furthermore $cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2} = \frac{z+1/z}{2}.$
By this variable change and the residue theorem, the integral becomes, $$I = \int_{|z|=1}^{} \frac{(z+1/z)\frac{1}{2}}{(2+\frac{z+1/z}{2})} \frac{dz}{iz} = \frac{1}{i} \int_{|z|=1}^{} \frac{z^2+1}{z(z^2+4z+1)}dz = \frac{1}{i} \int_{|z|=1}^{} \frac{z^2+1}{z(z+2-\sqrt{3})(z+2+\sqrt{3})}dz = 2\pi i \ \sum_{j=1}^{2}\text{Res}\left[\frac{z^2+1}{z(z+2-\sqrt{3})(z+2+\sqrt{3})}, z_j\right],$$
where, on the unit disc, the integrand $f(z) = \frac{z^2+1}{z(z+2-\sqrt{3})(z+2+\sqrt{3})}$, has a simple pole at $z_1=0$, and one at $z_2=-2+\sqrt{3}$.
Evaluating the residues we obtain $\text{Res}\left[f(z),0\right] = \lim_{z \to 0} \left[\frac{z^2+1}{(z+2-\sqrt{3})(z+2+\sqrt{3})}\right] = \frac{1}{(2-\sqrt{3})(2+\sqrt{3})} = 1$, and $\text{Res}\left[f(z),-2+\sqrt{3}\right] = \lim_{z \to -2+\sqrt{3}} \left[\frac{z^2+1}{(z+2-\sqrt{3})(z+2+\sqrt{3})}\right] = \frac{(-2+\sqrt{3})^2 +1}{(-2+\sqrt{3})(-2+\sqrt{3} + 2 + \sqrt{3}))} = \frac{8-4\sqrt{3}}{6-4\sqrt{3}} = \frac{4-2\sqrt{3}}{3-2\sqrt{3}}.$
Finally we obtain, $$I = 2\pi i \left(1 + \frac{4-2\sqrt{3}}{3-2\sqrt{3}}\right) = \frac{2\pi i (7-4\sqrt{3})}{3-\sqrt{3}}.$$
However the solution in the text claims the answer to be $I = \pi(1-\frac{2}{\sqrt{3}}).$ And I do not doubt the validity of the texts solution since I agree with how they derived it. However they solved it using a different way which I would never apply my self.
All help is appreciated. Thanks!
| Your calculation for the residue at $z=-2+\sqrt3$ is wrong. In fact
$$ \text{Res}f(z)\bigg|_{z=-2 + \sqrt3}=\frac{z^2+1}{z(z+2+\sqrt3)}\bigg|_{z=-2 + \sqrt3}=-\frac{2}{\sqrt3}.$$
| {
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Show that $\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1$
Show that $$\left\vert\frac{\pi}{4} - \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9}\right)\right\vert < 0.1 .$$
I know that $\arctan 1 = \frac{\pi}{4}$ and that the sequence being subtracted is a partial sum of its Taylor series. I believe you use the alternating series test to explain, but all I get from it is that the series will converge on $[-1,1]$.
| Consider the function
$$\frac{x^{10}}{x^2+1} \equiv 1-x^2+x^4-x^6+x^8-\frac{1}{x^2+1}$$
The integral:
$$J = \int_0^1 \frac{x^{10}}{x^2+1} \text{d}x = \int_0^1 1-x^2+x^4-x^6+x^8-\frac{1}{x^2+1} \text{d}x$$
is clearly positive.
Since $1\leqslant x^2+1 \leqslant 2$, we have:
$$\frac{1}{2} \int_0^1 x^{10} \text{d}x < J < \int_0^1 x^{10} \text{d}x$$
$$\frac{1}{22} < J < \frac{1}{11}$$
Then since $|-J|=J$ and $\frac{1}{11} < \frac{1}{10}$, the desired inequality follows.
| {
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How to draw graph of $|x+y-1| + |2x + y + 1|=1$?
How to draw graph of $|x+y-1| + |2x + y + 1|=1$ ?
My attempt
I can operate modulus in equations like $|x-1| + |2x + 1|=1$ but I don't know how to proceed in this question since it contains $2$ variable, $x$ and $y$.
| If you know already a bit of linear algebra you may proceed as follows:
You surely can draw $|u|+|v|=1$.
Now, consider the mapping
$$\begin{pmatrix}u \\ v\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}+\begin{pmatrix}-1 \\ 1\end{pmatrix}$$
The inverse mapping (just solve for $\begin{pmatrix}x \\ y\end{pmatrix}$) is
$$\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}-1 & 1 \\ 2 & -1\end{pmatrix}\begin{pmatrix}u\\ v\end{pmatrix}+\begin{pmatrix}-2 \\ 3\end{pmatrix}$$
Since we are dealing with a linear mapping plus a shift (a so-called affine mapping), you only need to find the images of the corners of $|u|+|v|=1$:
*
*$\begin{pmatrix}1 \\ 0\end{pmatrix},\begin{pmatrix}-1 \\ 0\end{pmatrix},\begin{pmatrix}0 \\ 1\end{pmatrix},\begin{pmatrix}0 \\ -1\end{pmatrix}$
Plugging these into the inverse mapping you get as graph the sides of the parallelogram spanned by
$$\boxed{\begin{pmatrix}-3 \\ 5\end{pmatrix},\begin{pmatrix}-1 \\ 1\end{pmatrix},\begin{pmatrix}-1 \\ 2\end{pmatrix},\begin{pmatrix}-3 \\ 4\end{pmatrix}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$n$th derivative of $e^{ax}\sin(bx+c)$
*
*How can we substitute $r \cos \alpha$ and $r \sin \alpha$ for $a$ and $b$?
*How, on successive differentiation, is there another $r$ multiplied?
| First, note that the title in the excerpt is incorrect. It should be "$n$th derivative of $e^{ax}\sin(bx+c)$".
(1) Here we introducing new quantities $r$ and $\alpha$, so we may define them however we want. Essentially, this amounts to writing the pair $(a, b)$ in polar coordinates, as $(r, \alpha)_{\textrm{polar}}$.
More explicitly, any such $r$ satisfies $$a^2 + b^2 = (r \cos \alpha)^2 + (r \sin \alpha)^2 = r^2 (\cos^2 \alpha + \sin^2 \alpha) = r^2 ,$$
so, $$r = \pm \sqrt{a^2 + b^2}.$$
Now the length of $$\left(\frac{a}{r}, \frac{b}{r}\right)$$ is
$$\sqrt{\left(\frac{a}{r}\right)^2 + \left(\frac{b}{r}\right)^2} = \sqrt{\frac{a^2 + b^2}{r^2}} = \sqrt{\frac{r^2}{r^2}} = 1 ,$$
and so it lies on the unit circle. In particular, since $\theta \mapsto (\cos \theta, \sin \theta)$ parameterizes all of unit circle, there is some $\alpha$ such that $$\frac{a}{r} = \cos \alpha, \qquad \frac{b}{r} = \sin \alpha,$$
and rearranging gives the equations in the text, namely,
$$a = r \cos \alpha, \qquad b = r \sin \alpha .$$
Since $(r \cos (\alpha + \pi), r \sin(\alpha + \pi)) = (-r \cos \alpha, -r \sin \alpha)$, by possibly adding $\pi$ to $\alpha$ we may as well assume that $r \geq 0$, that is that $r = \sqrt{a^2 + b^2}$.
(2) The computation for $y_1$ gives that
$$y_1 = r e^{\alpha x} \sin (b x + c + \alpha) .$$
If we want to compute the next derivative, we have
$$y_2 = (r e^{\alpha x} \sin (b x + c + \alpha))' = r (e^{\alpha x} \sin (b x + c + \alpha))$$
But if we replace $c$ with $c + \alpha$ in our rule, it tells us that $(e^{\alpha x} \sin (b x + c + \alpha) = r e^{\alpha x} \sin (b x + c + 2 \alpha)$, and substituting back in the previous display equation gives
$$y_2 = r (r e^{\alpha x} \sin (b x + (c + \alpha) + \alpha)) = r^2 e^{\alpha x} \sin (b x + c + 2 \alpha)$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Find value for $k$ such that $(x^2-k)$ is a factor for $f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$ Find value for $k$ such that $(x^2-k)$ is a factor for, $$f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$$
My Try
Since $x^2-k=0$ when we substitute $x=\pm k$ to $f(x)$ it should be equal to $0.$
But this gives a polynomial of k where it has $\sqrt{k}$ terms as well. Is my approach correct or is there a simpler way? Please any hint would be highly appreciated.
| Factor out $x^2-k$:
$$f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6=\\
[2x^4-2x^2k+2x^2k-2k^2+\color{red}{2k^2}]+\\
[(3k-4)x^3-(3k-4)xk+\color{red}{(3k-4)xk}]+\\
[(2k^2-5k-5)x^2-(2k^2-5k-5)k+\color{red}{(2k^2-5k-5)k}]+\\
\color{red}{(2k^3-2k^2-3k-6)x+6}\Rightarrow \\
\begin{cases}2k^3-2k^2-3k-6+(3k-4)k=0\\
2k^2+(2k^2-5k-5)k+6=0 \end{cases} \Rightarrow \\
\begin{cases}2k^3+k^2-7k-6=0\\
2k^3-3k^2-5k+6=0\end{cases} \Rightarrow \\
4k^2-2k-12=0 \Rightarrow \\
2k^2-k-6=0 \Rightarrow \\
k_{1,2}=\frac{1\pm 7}{4}=-\frac32,2.$$
Verification: WA answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
How to equalize correctly? If i have this number:
$2 \sqrt{2-\sqrt{3}}$ and i want to find some $x,y$ nonzero real numbers such that $2\sqrt{2-\sqrt{3}} = \sqrt{x} + \sqrt{y}$
And for that, i do this:
$(2 \sqrt{2-\sqrt{3}})^2 = x + 2\sqrt{xy} + y$
$4(2-\sqrt{3})=(x+y)+2\sqrt{xy}$
$(8)+(-4\sqrt{3})=(x+y)+(2\sqrt{xy})$
Then:
$i) 8 = x+y$,
$ii)-4\sqrt{3} = 2\sqrt{xy} => -2\sqrt{3}=\sqrt{xy}$
$ii) = (-2\sqrt{3})^2= (\sqrt{xy})^2 => 4\cdot 3=xy , x = 12/y$
And solving the equation $y^2-8y+12=0$ gives $y_{1,2} = \{6,2\}$
But $2\sqrt{2-\sqrt{3}} \neq \sqrt{6} + \sqrt{2}$
I know that the correct value must be $\sqrt{6} - \sqrt{2}$ but my result is different. What is wrong with my development?
| I can write $2\sqrt{2-\sqrt{3}} = \sqrt{y}+\sqrt{x}$ as: $2\sqrt{(\frac{\sqrt3}{\sqrt2}-\frac{\sqrt2}{2})^2} = \sqrt{y}+\sqrt{x}=2(\frac{\sqrt3}{\sqrt2}-\frac{\sqrt2}{2})$ or $2\sqrt{(\frac{1}{\sqrt2}-\frac{\sqrt6}{2})^2} = \sqrt{y}+\sqrt{x}=2(\frac{1}{\sqrt2}-\frac{\sqrt6}{2})$. This can be easily obtained by the system:
$$\left\{\begin{matrix}
a^2+b^2=2
\\2ab=-\sqrt{3}
\end{matrix}\right.$$
So: $\sqrt{x}=-\frac{\sqrt2}{2}$, but this is impossibile because the square root can't be negative: same arguments for $\sqrt{y}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that complex numbers form a regular pentagon Let there be five nonzero complex numbers having the same absolute value and such that zero is equal to their sum, which is equal to the sum of their squares. Prove that the affixes of these numbers in the complex plane form a regular pentagon.
I am familiar with the topic of roots of unity. We have to prove that a,b,c,d,e are roots of $x^5 = 1$. I do not know how to prove starting from $a+b+c+d+e = 0$ and $a^2+b^2+c^2+d^2+e^2=0$
| Let $|a| = R$. Then $\frac{R^2}{a} = \overline a$, and thus
$$ 0 = \overline a + \overline b + \overline c + \overline d + \overline e = \frac{R^2}{a} + \frac{R^2}{b} + \frac{R^2}{c} + \frac{R^2}{d} + \frac{R^2}{e} = \frac{R^2}{abcde}\left(bcde + acde + abde + abce + abcd\right).$$
Thus, $\sum_{sym}abcd = 0$. Similarly, we obtain
$$ 0 = \overline a^2 + \overline b^2 + \overline c^2 + \overline d^2 + \overline e^2 = \frac{R^4}{a^2} + \frac{R^4}{b^2} + \frac{R^4}{c^2} + \frac{R^4}{d^2} + \frac{R^4}{e^2} = R^4\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} + \frac{1}{e^2}\right).$$
Subtracting the given identites, we infer that
$$0 = \frac{1}{2}\left(\left(a+b+c+d+e\right)^2-(a^2+b^2+c^2+d^2+e^2)\right) = \sum_{sym} ab,$$
and by an anologous trick,
$$ 0 = \frac{1}{2}\left(\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{e}\right)^2-\left(\frac{1}{a} + \frac{1}{b^2} + \frac{1}{c^2} + \frac{1}{d^2} + \frac{1}{e^2}\right) \right) = \sum_{sym}\frac{1}{ab} = \frac{1}{abcde}\sum_{sym}cde.$$
Altogether, we now know that $0 = \sum_{sym}a = \sum_{sym}ab = \sum_{sym}abc = \sum_{sym}abcd$. Finally, $abcde = R^5e^{5i \theta }$ for some $\theta \in [0,2\pi)$. Therefore, by Vieta's theorem, $a,b,c,d,e$ are the solutions of
$$ x^5 - R^5e^{5i \theta} = 0,$$
and these are $\{Re^{i\theta + 2\pi i\frac{k}{5}}, k = 1,\dots,5\}$, the vertices of a regular pentagon.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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For which $a$, $b$, $c$ does this linear system have exactly one solution? $x+ay+a^2z=0$, $x+by+b^2z=0$, $x+cy+c^2z=0$
For which $a$, $b$, $c$ does this linear system have exactly one solution?
$$x + ay + a^2z = 0$$
$$x + by + b^2z = 0$$
$$x + cy + c^2z = 0$$
I started this problem by recognizing that if the RREF of a linear system's augmented matrix has a leading 1 in every column except the last, then the system must have exactly one solution. The system's augmented matrix reduces to
$$
\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{matrix}
$$
which indicates that no matter the values of $a, b, c$, $x = 0, y = 0,$ and $z = 0$. But one can see that if $a = 0$ and $b = 0$, then $x = 0$ and the third equation of the system becomes the equation of a line, indicating an infinite number of solutions, meaning that no two of $a,b,c$ van be equal to $0$.
So my best guess is that the system has exactly one solution when $a,b,c \in \mathbb{R}$ and $a, b \neq 0$. Is this correct? Close at all? How would I go about proving this?
| The other solution uses determinants which is going to be the standard route to solving this problem. You mentioned RREF which I don't think is impossible, but you should be careful with putting it in that form:
$$\begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix}$$
subtracting row 1 from row 2 and row 1 from row 3:
$$\begin{pmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{pmatrix}$$
Now we immediately see that if $b=a$ or $a=c$ then RREF will have at least one row which is all 0's. What does that mean about the number of solutions?
If not then we can divide by $b-a$ and $c-a$:
$$\begin{pmatrix}1&a&a^2\\0&1&b+a\\0&1&c+a\end{pmatrix}$$
Subtracting $a$ times row 2 from row 1 and row 2 from row 3:
$$\begin{pmatrix}1&0&-ba\\0&1&b+a\\0&0&c-b\end{pmatrix}$$
We again notice that if $b=c$ then there is a row of all 0s. If not then we can divide row 3 by $c-b$ to get:
$$\begin{pmatrix}1&0&-ba\\0&1&b+a\\0&0&1\end{pmatrix}$$
now adding $ba$ times row 3 to row 1 and subtracting $b+a$ times row 3 from row 2:
$$\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$$
What does this mean?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\tan 195^{\circ}$ without using the calculator How to evaluate $\tan 195^{\circ}$ without using the calculator, and how to give the answer in the form $a+b \sqrt{3}$, where $a$ and $b$ are integers?
| Using the identity $\tan \left(180^\circ+\theta\right)=\tan\theta$ and $\tan \left(\alpha-\beta\right)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$,
$\quad\tan195^\circ \\=\tan 15^\circ\\=\tan \left(60^\circ-45^\circ\right)\\=\dfrac{\tan60^\circ-\tan45^\circ}{1+\tan60^\circ\tan45^\circ}\\=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\\=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\=\dfrac{4-2\sqrt{3}}{2}\\=\boxed{2-\sqrt{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a way to calculate the enclosed volume of two unit balls centered at $(0,0,0)$ and $(1,1,1)$ using a triple integral?
Let $\Omega \subset \mathbb{R}^3$ be the subset $$\Omega = B((0,0,0),1) \cap B((1,1,1),1)$$ where $B(x,r)$ stands for the ball centered in $x$ with radius $r$.
I was looking for an easy way to calculate the volume with a triple integral.
Also because the exercise asks, after this, to calculate $$\int_{\Omega} (x+2y+z) dV$$ and $$\int_{\partial \Omega} (x+2y+z) d\Sigma$$
Since this was an exercise from a test that does not last long, there must be some shortcut that I'm missing here.
Does someone have some ideas? Thanks!
| Due to the perfect symmetry of the two-ball configuration around the axis (1,1,1), these three integrals can be carried out relatively straightforwardly.
Such symmetry is obvious for the simple volume integral,
$$I_1=\int_{\Omega} dV$$
For the other two integrals, we observe that
$$\int_{\Omega} xdV = \int_{\Omega} ydV = \int_{\Omega} zdV$$
and rewrite them in symmetric forms explicitly
$$I_2=\int_{\Omega} (x+2y+z) dV=\frac{4}{3}\int_{\Omega} (x+y+z) dV$$
$$I_3=\int_{\partial \Omega} (x+2y+z) d\Sigma=\frac{4}{3}\int_{\partial\Omega} (x+y+z) d\Sigma$$
Observe further that $x+y+z = a$ is a plane intersecting all three axes at $a$ and $a$ is proportional to the altitude $h$ from the origin to the plane. In fact, $a = \sqrt{3}h$. Therefore, $x+y+z$ in the integrands can be conveniently replaced by $\sqrt{3}h$, i.e.
$$I_2=\frac{4}{3}\int_{\Omega} \sqrt{3}h dV$$
$$I_3=\frac{4}{3}\int_{\partial\Omega} \sqrt{3}h d\Sigma$$
Now rotate the $z$-axis to be align with the direction (1,1,1) and reparametrize the two balls as,
$$x^2+y^2+z^2=1$$
$$x^2+y^2+(z-\sqrt{3})^2=1$$
All three integrals can then be performed readily along the $z$-axis, albeit in two regions corresponding to the two spherical caps. In the rotated ordinates, the altitude is simply $h=z$, and the three integrals become,
$$I_1 = \int_{\sqrt{3}-1}^{\sqrt{3}/2} \pi[1-(z-\sqrt{3})^2]dz + \int_{\sqrt{3}/2}^1 \pi(1-z^2)dz=\pi\left(\frac{4}{3}-\frac{3\sqrt{3}}{4} \right).$$
$$I_2 = \frac{4}{3}\int_{\sqrt{3}-1}^{\sqrt{3}/2} \pi\sqrt{3}z[1-(z-\sqrt{3})^2]dz + \frac{4}{3}\int_{\sqrt{3}/2}^1 \pi\sqrt{3}z(1-z^2)dz=\pi\left(\frac{8}{3}-\frac{3\sqrt{3}}{2} \right).$$
$$I_3 = \frac{4}{3} \int_{\sqrt{3}-1}^{\sqrt{3}/2} 2\pi\sqrt{3}z\sqrt{1-(z-\sqrt{3})^2}dz + \frac{4}{3}\int_{\sqrt{3}/2}^1 2\pi\sqrt{3}z\sqrt{1-z^2}dz=\pi\left(\frac{2\pi}{3}-\sqrt{3} \right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Density of $ Y = X + \frac{1}{X}$ when $X\sim U(a,b)$ Let $ X $ be a continuous random variable uniformly distributed in $ \left[a, b\right] $, that is $X \sim U(a, b)$. Suppose $ 0 < a < b $.
We wish to find the density function of $$ Y = X + \frac{1}{X} $$ Therefore, we want $ f_Y(y) $.
It is obvious that for $$ y\in (-2, 2) => f_Y(y) = P(Y = y) = P\left(X + \frac{1}{X} = y\right) = 0 $$ because $X$ takes real values, and therefore it is known that the function $ f(x) = x + \frac{1}{x} $ ranges in $(-\infty, -2]\cup[2, \infty)$.
I tried using the definition for $y\in(-\infty, -2]\cup[2, \infty)$: $$ P(Y = y) = P\left(X + \frac{1}{X} = y\right) = P(X^2 - yX + 1 = 0) = \cdots = \\P\left(\left(X - \frac{y + \sqrt{y^2 - 4}}{2} \right)\left(X - \frac{y - \sqrt{y^2 - 4}}{2} \right) = 0\right) = \\
P\left(X = \frac{y + \sqrt{y^2 - 4}}{2} \right) + P\left(X = \frac{y - \sqrt{y^2 - 4}}{2} \right) $$ which seems awfully dull. I am not even sure if it works.
I tried considering the random variable $Z = \frac{1}{X}$. It can easily be derived that $ f_Z(z) = \frac{z^{-2}}{b -a} $ for $b^{-1} < z < a^{-1} $. Therefore we can say $ Y = X + Z $. From here we can proceed:
$$ P(Y = y) = P(X + Z = y) = \int_{0}^{y} P(X = y - k, Z = k)dk $$
But firstly the random variables $X$ and $Z$ are dependant, and secondly, it seems wrong. Any ideas?
| A possible path
Let $F_X(x)$ and $f_X(x)$ be the c.d.f and p.d.f. of $X\sim\mathcal U(a,b)$. Let also
$$g(x) = \frac1x + x,$$
$$\alpha(y) = \frac{y-\sqrt{y^2-4}}{2},$$
$$\beta(y) = \frac{y+\sqrt{y^2-4}}{2}.$$
Case 1 ($a\geq 1$)
$g(x)$ in monotonically increasing in $[a,b]$. Thus
\begin{eqnarray}
P(Y<y) &=& \begin{cases}
P(X<\beta(y)) & (g(a) \leq y \leq g(b))\\
0 & \mbox{(otherwise)}
\end{cases}\\
&=&\begin{cases}
F_X(\beta(y)) & (g(a) \leq y \leq g(b))\\
0 & \mbox{(otherwise)}
\end{cases}
\end{eqnarray}
Consequently, by differentiation you derive the p.d.f. of $Y=\frac1X+X$ as
\begin{eqnarray}
f_Y(y) &=& \begin{cases}
f_X(\beta(y)) \cdot \frac{d\beta}{d y} & (g(a) \leq y \leq g(b))\\
0 &\mbox{(otherwise)}
\end{cases}\\
&=&\begin{cases}
\frac1{b-a}\left(
\frac12+\frac{y}{2\sqrt{y^2-4}}\right) & (g(a) \leq y \leq g(b))\\
0 &\mbox{(otherwise)}
\end{cases}
\end{eqnarray}
Case 2 ($b\leq 1$)
You proceed in a very similar manner, by noting that now $g(x)$ in monotonically decreasing in $[a,b]$. Thus
\begin{eqnarray}
P(Y<y) = \begin{cases}
P(X<\alpha(y)) & (g(b) \leq y \leq g(a))\\
0 & \mbox{(otherwise)}
\end{cases}
\end{eqnarray}
Case 3 ($0<a < 1 <b$ and $ab>1$)
In this case you have $g(b) > g(a)>2$, so
$$
P(Y<y) =
\begin{cases}
P(\alpha(y) <x< \beta(y)) & (2<y< g(a))\\
P(x<\beta(y)) & (g(a)<y<g(b))\\
0 & (\mbox{otherwise})
\end{cases}
$$
Case 4 ($0<a < 1 <b$ and $0<ab<1$)
In this case you have $2<g(b) < g(a)$, so
$$
P(Y<y) =
\begin{cases}
P(\alpha(y) <x< \beta(y)) & (2<y< g(b))\\
P(x<\alpha(y)) & (g(b)<y<g(a))\\
0 & (\mbox{otherwise})
\end{cases}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Knowing that $m$ and $n$ are positive integers satisfying $mn \mid m^2 + n^2 + m$, prove that $m$ is a square number.
Knowing that $m$ and $n$ are positive integers satisfying $$\large mn \mid m^2 + n^2 + m$$, prove that $m$ is a square number.
We have that $mn \mid m^2 + n^2 + m \implies mn \mid (m^2 + n^2 + m)(n + 1)$
$\implies mn \mid m^2n + n^3 + mn + m^2 + n^2 + m$ $\implies \left\{ \begin{align} &mn \mid n^3\\ &mn \mid n^3 + m^2 + n^2 + n \end{align} \right.$
$\implies \left\{ \begin{align} m &\mid n^2\\ mn &\mid m(m + 1) + n^2(n + 1) \end{align} \right.$ $\implies \left\{ \begin{align} mm' &= n^2\\ mn &\mid m(m + 1) + mm'(n + 1) \end{align} \right. (m \in \mathbb Z^+)$
$\implies \left\{ \begin{align} mm' &= n^2\\ m'n &\mid m'(m + 1) + m'^2(n + 1) \end{align} \right.$ $\implies \left\{ \begin{align} mm' &= n^2\\ m'n &\mid n^2 + m' + m'^2 \end{align} \right.$
That also means that if the $m$ is proven to be a square number, $m'$ is also a perfect square. Moreover, $mm' = n^2 \implies (m, m') = 1$ needs to be proven. But I don't know how to.
| Note $\ kmn = m^2+m+n^2\,\Rightarrow\, x^2+(1\!-\!kn)\,x + n^2\,$ has roots $\,x = m,m'\in \Bbb Z\,$ satisfying $ \underbrace{mm' = \color{#0a0}{n^2}}_{\textstyle\small p\mid m\,\Rightarrow\, p\mid\color{#c00} n}\!$ and $\: \underbrace{m+m' = kn-1}_{\textstyle\small p\mid m,m', \color{#c00}n\,\Rightarrow\,p\mid 1 }\,$ thus $\,\color{}{m,m'}\,$ are coprime, hence both are also $\color{#0a0}{\text{squares}}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $ If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $
Attempt:
Notice that $$ \tan(a+b+c) = \frac{\tan(a+b) + \tan(c)}{1 - \tan(a+b)\tan(c)} $$
$$ = \frac{\frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} + \tan(c)}{1 - \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \tan(c)} $$
$$ = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a) \tan(b) \tan(c)}{ (1-\tan(a)\tan(b)) - \tan(a) \tan(c) -\tan(b) \tan(c) } $$
Then the denomenator must be $0$, so it is proven?
Another way:
$$ \sin(a+b+c) = 1 \implies \sin(a+b) \cos(c) + \sin(c) \cos(a+b) = 1 $$
$$ \sin(a)\cos(b)\cos(c) + \sin(b)\cos(a)\cos(c) + \sin(c) \cos(a) \cos(b) - \sin(c) \sin(a) \sin(b) = 1$$
$$ \tan(a) \cos(a) \cos(b)\cos(c) + \tan(b) \cos(a) \cos(b) \cos(c) + \tan(c) \cos(a) \cos(b) \cos(c) - \sin(a) \sin(b) \sin(c) = 1$$
$$ \cos(a)\cos(b)\cos(c) (\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)) = 1 $$
| Since $\tan(a+b+c)$ is undefined, anything where you put it in an equation is suspect. But if you start with $\cot(a+b+c)$ then you can get
$1-\tan(a)\tan(b) - \tan(a) \tan(c) -\tan(b) \tan(c)$
in the numerator and show that it must be zero.
I don't see exactly how you intended to finish the proof that starts with $\sin(a+b+c).$ I would start with $\cos(a+b+c)=0$ instead.
Expand this using the rule for cosine of a sum twice, then divide by
$\cos a \cos b \cos c$ (which must all be non-zero, otherwise the tangents in the formula are not all defined--this really should be an explicit condition of the proposition).
| {
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Volume of a solid created by an extended tetrahedron
Every edge of a tetrahedron with length $p$ is extended through the vertices by $p$.
Now all 12 points create a new solid $J$ of which I seek the volume dependent on the volume of the tetrahedron in the centre.
With some help the solution becomes clear:
The whole Volume of all pyramids is:
$$V_P=4\cdot\frac{1}{6}\cdot\left(\frac{a}{\sqrt{2}}\right)^3 +4\cdot\frac{1}{6}\cdot\left(\sqrt{2}a\right)^3 =\frac{3}{\sqrt{2}}a^3 \tag{1}$$
The side of the large cube is
$$s=\frac{3}{\sqrt{2}}a\tag{2}$$
and the Volume is respectively
$$V_C=\left(\frac{3}{\sqrt{2}}a\right)^3=\frac{27\sqrt{2}}{4}\cdot a^3 \tag{3}$$
The last step is subtracting the Volume of the pyramids.
$$V_J=V_C-V_P= \frac{27\sqrt{2}a^3}{4}-\frac{3}{\sqrt{2}}a^3=\frac{21\sqrt{2}a^3}{4} \tag{4}$$
Thx for the help.
| Here is an answer - complementing your 3d object to a cube with side - $3p/2^{0.5}$
Subtract 4 right angle pyramids with diagonal $p$ and 4 right angle pyramids with diagonal $2p$. The pyramids sides are, $p/2^{0.5}$ and $2^{0.5}p$.
The volume of the cube is given by:$(3p/2^{0.5})^3$
The volume of the pyramids - $4X(1/6)X(p/2^{0.5})^3+4X(1/6)X(2p/2^{0.5})^3$
I will leave you to subtract the volumes.
| {
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Matrix diagonalization. Is $A = PDP^{-1} = P^{-1}DP$? Diagonalization of a square matrix $A$ consists in finding matrices $P$ and $D$ such that $A=PD P^{-1}$ where $P$ is a matrix composed of the eigenvectors of $A$, $D$ is the diagonal matrix constructed from the corresponding eigenvalues, and $P^{-1}$ is the matrix inverse of $P$
I wonder if $PDP^{-1} = P^{-1}DP$ ?
Can I say $A=P^{-1}DP$ too?
| No,
$$\begin{pmatrix}1&2\\1&0\end{pmatrix}
\begin{pmatrix}2&0\\0&1\end{pmatrix}
\begin{pmatrix}0&1\\\frac12&-\frac12\end{pmatrix}
=\begin{pmatrix}1&1\\0&2\end{pmatrix}$$
while
$$\begin{pmatrix}0&1\\\frac12&-\frac12\end{pmatrix}
\begin{pmatrix}2&0\\0&1\end{pmatrix}
\begin{pmatrix}1&2\\1&0\end{pmatrix}=\begin{pmatrix}1&0\\\frac12&2\end{pmatrix}.$$
| {
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Find $p,q$ such that $p+q-pq+1=0$ where $p,q$ are primes with $p
Find $p,q$ such that $p+q-pq+1=0$ where $p,q$ are primes with $p<q$.
My try:
If we take $p=2,q=3$ the equality is satisfied.
But how can I show that this is the only solution?
| $p + q-pq +1=0$ means $pq - q=p+1$ so $q=\frac {p+1}{p-1}$.
And $pq-p = q+1$ so $p =\frac {q+1}{p-1}$.
To solve $\frac {x+1}{x-1}\in \mathbb Z$, we can get $\gcd(x+1,x-1)= \gcd(x+1, (x+1)-(x-1)) = \gcd (x,2) = 1,2$ so $x-1=1,2$ (assuming $x-1 >0$) and $x+1=3,4$ and $x = 2,3$.
Alternatively we could have assigned $m = x-1$ and realized that $m|m+2$ so $m|2$ so $m = 1,2$
So $p$ and $q$ are equal to $2$ and $3$ and as $p > q$ we have $p=3; q=2$.
| {
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Having trouble solving $\int x(2x+5)^8dx$. I am not sure how to solve $\int x(2x+5)^8dx$.
I have tried some different things, but nothing seems to give me the cancelling effect I need to solve the integral.
My Process:
Let $u=2x+5$
Then $du = 2dx$ and $dx = \frac{1}{2}du$
so $\int x u^8 \frac{1}{2} \;du = \frac{1}{2}\int x u^8 \; du$
This is where I get stuck.
The answer should be $\frac{1}{40}(2x+5)^{10}-\frac{5}{36}(2x+5)^{9} + C$ according to my textbook.
How can I solve this problem?
| Let $\displaystyle I=\int (2x+5)(2x+5)^8\,dx=\int (2x+5)^9\,dx=\frac{1}{20}(2x+5)^{10}$
And let $J=\displaystyle\int(2x+5)^8
\,dx=\frac{1}{18}(2x+5)^9$
I exclude constants until the end.
Then, $\displaystyle \int x(2x+5)^8\,dx=\frac{1}{2}(I-5J)+C=\frac{1}{40}(2x+5)^{10}-\frac{5}{36}(2x+5)^9+C$.
| {
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Cubic diophantine equation with a prime $x^3 + y^3 + z^3 - 3xyz = p$. Question: Find all triple positive integers $(x, y, z)$ so that
$$x^3 + y^3 + z^3 - 3xyz = p,$$
where $p$ is a prime number greater than $3$.
I have tried the following: The equation factors as
$$(x + y + z) (x^2 + y^2 + z^2-xy-yz-zx) = p.$$
Since $x + y + z> 1$, we must have $x + y + z = p$ and
$$x^2 + y^2 + z^2-xy-yz - zx = 1.$$
The last equation is equivalent to
$$(x-y)^2 + (y-z)^2 + (z-x)^2 = 2.$$
Without loss of generality you can assume that $x ≥ y ≥ z$, we have $xy ≥ 1$ and $xz ≥ 2$, implying
$$(xy)^2 + (yz)^2 + (zx)^2 ≥ 6> 2.$$
Who can help me and correct me, thank you.
| Result: If $p>3$ is a prime number and $x$, $y$ and $z$ are positive integers such that
$$x^3 + y^3 + z^3 - 3xyz = p,$$
then if $p\equiv1\pmod{3}$ we have, after permuting $x$, $y$ and $z$, that
$$(x,y,z)=\left(\tfrac{p-1}{3},\tfrac{p-1}{3},\tfrac{p+2}{3}\right),$$
and if $p\equiv2\pmod{3}$ we have, after permuting $x$, $y$ and $z$, that
$$(x,y,z)=\left(\tfrac{p+1}{3},\tfrac{p+1}{3},\tfrac{p-2}{3}\right).$$
Proof: As you already note, the equation can be expressed as
$$(x + y + z) (x^2 + y^2 + z^2-xy-yz-zx) = p,$$
which immediately shows that, because $x$, $y$ and $z$ must be positive,
$$x+y+z=p\qquad\text{ and }\qquad x^2 + y^2 + z^2-xy-yz-zx=1.\tag{1}$$
The latter can be rewritten as
$$(x-y)^2+(x-z)^2+(y-z)^2=2,$$
which shows that two of the three numbers are the same, and the third differs from them by only $1$. That is to say, without loss of generality we have
$$x=y=z\pm1.$$
Plugging this back into the first equation found at $(1)$ shows that
$$p=x+y+z=3x\pm1,$$
and so we find that $x=\tfrac{p\mp1}{3}$. As $x$ must be an integer we see that only one of the two choices of the $\pm$-sign is possible, depending on whether $p\equiv1\pmod{3}$ or $p\equiv2\pmod{3}$.
Conversely, a routine check shows that if $p\equiv\pm1\pmod{3}$ then the triplet of positive integers
$$(x,y,z)=\left(\tfrac{p\mp1}{3},\tfrac{p\mp1}{3},\tfrac{p\pm2}{3}\right),$$ and its three distinct permutations satisfy the equation
$$x^3 + y^3 + z^3 - 3xyz = p.$$
| {
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Are there integer solutions such that the sum of the roots of x and y equal the root of a prime p? How could one prove whether or not integer pairs (x, y) exist such that $\sqrt{x} + \sqrt{y} = \sqrt{p}$, where p is prime? `
| Using elementary methods.
We'd have
$$x + 2\sqrt{xy} + y = p$$
and hence
\begin{align}
x + y - p = -2\sqrt{xy}
&\implies
x^2+y^2+p^2+2xy-2xp-2yp = 4xy
\\&\iff
(x^2-2xy+y^2) + p^2= 2p(x+y)
\\&\iff
(x-y)^2 +p^2 = 2p(x+y)\tag{1}
\\&\iff
(x-y)^2 +2p(x-y) +p^2 = 2p(x+y) + 2p(x-y)\tag{2}
\\&\iff
(x-y+p)^2 = 4px.
\end{align}
This implies $x-y+p = \pm 2\sqrt{px}$, so $x = pa^2$ for some integer $a$.
At $(2)$, we added $2p(x-y)$ to both sides of equation $(1)$.
If we instead had subtracted $2p(x-y)$, we'd get
\begin{align}
x + y - p = -2\sqrt{xy}
&\implies
(x-y)^2 -2p(x-y) +p^2 = 2p(x+y) - 2p(x-y)\tag{3}
\\&\iff
(x-y-p)^2 = 4py.
\end{align}
Like before, we obtain that $x-y-p = \pm 2\sqrt{py}$, and hence $y = pb^2$ for some integer $b$.
The initial equation can then be rewritten:
$$|a|\sqrt p + |b|\sqrt p = \sqrt p \iff |a| + |b| = 1.$$
This can happen only when $|a| = 1$ and $b=0$ or vice versa, that is, either $x = p$ and $y=0$ or $x=0$ and $y=p$.
| {
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Expression of $x^n+\frac1{x^n}$ by $x+\frac1{x}$ where $n$ is a positive odd number. There was a problem in a book:
Denote that $y=x+\dfrac{1}{x}$, express $x^7+\dfrac{1}{x^7}$ using $y$.
It's not a hard question, but I find a special sequence:
$x+\dfrac{1}{x}=y\\x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=y^3-3y \\ x^5+\dfrac{1}{x^5}=\left(x+\dfrac{1}{x}\right)^5-5\left(x^3+\dfrac{1}{x^3}\right)-10\left(x+\dfrac{1}{x}\right)=y^5-5\left(y^3-3y\right)-10y=y^5-5y^3+5y\\x^7+\dfrac{1}{x^7}=\left(x+\dfrac{1}{x}\right)^7-7\left(x^5+\dfrac{1}{x^5}\right)-21\left(x^3+\dfrac{1}{x^3}\right)-35\left(x+\dfrac{1}{x}\right)=y^7-7\left(y^5-5y^3+5y\right)-21\left(y^3-3y\right)-35y=y^7-7y^5+14y^3-7y$
I find that the coefficient has some relationship between the Pascal Triangle, such as $y^7-7y^5+14y-7y=y^7-7 \binom{2}{0}y^5+7\binom{2}{1}y^3-7\binom{2}{2}y$. That's strange but funny! However, I can't really prove this, or show that it is false. Hope there is someone who can answer me. Thank you!
| Let $s_1:=x+\dfrac1x$. Then
$$x=\frac{s_1\pm\sqrt{s_1^2-4}}2$$ and
$$s_n=\left(\frac{s_1+\sqrt{s_1^2-4}}2\right)^n+\left(\frac{s_1-\sqrt{s_1^2-4}}2\right)^n.$$
Then by the Binomial development, after cancellation of the odd terms
$$s_n=\frac1{2^{n-1}}\sum_{2k=0}^n\binom n{2k} s_1^{n-2k}\left(s_1^2-4\right)^{k}.$$
Then, developing again,
$$s_n=\frac1{2^{n-1}}\sum_{2k=0}^n\sum_{j=0}^k\binom n{2k}\binom kj(-4)^{k-j} s_1^{n-2k+2j}$$
and you can regroup the terms by equal powers of $s_1$.
| {
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Formulas for Sequences Removing Multiples of 2, 3, and 5 First off, I am a programmer so please excuse if some of the terms I use are not the correct mathematical terms. I was working on devising a function to improve one of my prime number generation algorithms. With this in mind, I first set out to find the formulas for a sequence removing multiples of 2 and 3:
\begin{array}{c|c}
x&y\\ \hline
0&5\\ \hline
1&7\\ \hline
2&11\\ \hline
3&13\\ \hline
4&17\\ \hline
5&19\\ \hline
6&23\\ \hline
7&25\\ \hline
8&29\\ \hline
\vdots&\vdots
\end{array}
The equations that I came up for for this sequence are as follows:
$$y = 3x + 5 - x \bmod 2$$
$$x = \left\lfloor\frac{y - 5 + [y \bmod 3 \neq 0]}{3}\right\rfloor$$
After this, I tried to do the same for a sequence removing multiples of 2, 3, and 5:
\begin{array}{c|c}
x&y\\ \hline
0&7\\ \hline
1&11\\ \hline
2&13\\ \hline
3&17\\ \hline
4&19\\ \hline
5&23\\ \hline
6&29\\ \hline
7&31\\ \hline
8&37\\ \hline
9&41\\ \hline
10&43\\ \hline
11&47\\ \hline
12&49\\ \hline
13&53\\ \hline
\vdots&\vdots
\end{array}
While I think I found an equation to get $y$ from a value of $x$, I cannot find a way to get the value of $x$ from a given value $y$.
$$y = 4x + 7 - 2\left\lfloor\frac{1}{8}x\right\rfloor - 2\left[\{2, 3, 6\} \ \text{contains}\ (x \bmod 8)\right] - 4\left[\{4, 5, 7\} \ \text{contains}\ (x \bmod 8)\right]$$
$$x =\ ?$$
I am wondering if an equation that produces the corresponding value of $x$ for a given value of $y$ for the aforementioned sequence exists, and if indeed it does, what the equation is.
| If this is for a programming task, you don't want a fancy formula. That will only slow you down. Instead, for the first table, just cycle through the values $6n+1,6n+5$ for $n=0,1,2,\ldots$ And for the second table, cycle through the values $30n+1,30n+7,30n+11,$ etc. (there are eight of them).
Updated to add:
Perhaps the following is more in the spirit of what the OP is looking for. This is for the case $2,3,5$. Declare a short array:
offset[8] = { 1,7,11,13,17,19,23,29 }
Then the nth number that is not divisible by $2, 3,$ or $5$ is simply
30 * (n / 8) + offset[n % 8]
where n/8 is understood to be rounded down to an integer.
| {
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Find the determinant whose result is $(x-n)^{n+1}$
Find the determinant
$$
\left|\begin{array}{cccccc}{x} & {1} & {} & {} & {} & {} \\ {-n} & {x-2} & {2} & {} & {} & {} \\ {} & {-(n-1)} & {x-4} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {-2} & {x-2 n+2} & {n} \\ {} & {} & {} & {} & {-1} & {x-2 n}\end{array}\right|
$$
I know the answer is $(x-n)^{n+1}$, and I tried to find it with Gaussian elimination but failed. (That is to reduce the matrix to a matrix with all $(x-n)$ on the diagonal line.)
How to find the determinant? Any method will be appreciated.
| Here's a possible method:
From up to down, add the row above to each row:
$$
D_n=\left|\begin{array}{cccccc}{x} & {1} & {} & {} & {} & {} \\ {x-n} & {x-1} & {2} & {} & {} & {} \\ {x-n} & {x-n} & {x-2} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {x-n} & {x-(n-1)} & {n} \\ {x-n} & {x-n} & {} & {} & {x-n} & {x- n}\end{array}\right|
$$
From left to right, subtract each column from the right column, thus
$$
D_n=\left|\begin{array}{cccccc}{x-1} & {1} & {} & {} & {} & {} \\ {-(n-1)} & {x-1-2} & {2} & {} & {} & {} \\ {} & {-(n-2)} & {x-1-4} & {\ddots} & {} & {} \\ {} & {} & {\ddots} & {\ddots} & {n-1} & {} \\ {} & {} & {} & {-1} & {x-1-2(n-1)} & { } \\ {} & {} & {} & {} & { } & {x- n}\end{array}\right|
=(x-n)D_{n-1}$$
As $D_0=x-n$, we deduce that $D_n=(x-n)^{n+1}$.
| {
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If $a$, $b$, $c$ are three positive integers such that $a^3+b^3=c^3$ then one of the integer is divisible by $7$ Let on contrary that none of the $a$, $b$, $c$ is divisible by $7$. Then either $a^3\equiv b^3\pmod{7}$ or $b^3\equiv c^3\pmod{7}$ or $c^3\equiv a^3\pmod{7}$.
Now how to go further?
| There's 36 combinations of $a$ and $b$ in the integers mod 7, where they're not already $0$. Just try checking them.
| {
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A box contains 13 four-sided dice, 36 six-sided dice, and 8 eight-sided dice. A box contains 13 four-sided dice, 36 six-sided dice, and 8 eight-sided dice. One die is chosen at random and then rolled. The probability that the die rolled was a six-sided die, given that the outcome was 5 is / where and are relatively prime positive integers. What is +?
On the solutions: ℙ(6∩5)=/ 6⋅57 and ℙ(5)=(36/6)+(8/8)/57.
ℙ(6|5)=4⋅36/ 4⋅36+3⋅8= 6/7.
Could someone explain how the answer is 13?
I know total dice is 57, but how is the total dice 7/57 and not 44/57. There are 36 six-sided dice and 8 eight-sided dice, and the outcome is 5. So 36 + 8 = 44 or am I forgetting something? Thank you for any help.
| There are 57 dice. If one is chosen at random (with uniform probability) then the probability of choosing a six-sided die and rolling a five is:
$\frac{36}{57} \frac{1}{6} = \frac{6}{57}$.
The probability of rolling a 5 in general is:
$\frac{13}{57} \frac{0}{4} + \frac{36}{57} \frac{1}{6} + \frac{8}{57} \frac{1}{8} = \frac{6}{57} + \frac{1}{57} = \frac{7}{57}$
So the conditional probability is:
$\frac{6}{57}$ divided by $\frac{7}{57}$, which is $\frac{6}{7}$.
6 and 7 are coprime, and $6+7=13$
| {
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What is the probability of being dealt a hand where all $5$ cards are different denominations? In a game of poker you are dealt $5$ cards at random from a standard deck of $52$. A standard deck of $52$ contains $13$ different denominations of cards (Ace, 2, 3, ..., 10, jack, queen, king), each in four different suits
(a) What is the probability of being dealt a hand where all $5$ cards are different denominations?
(b) What is the probability of being dealt a hand that includes at least one Ace?
My attempts:
(a) Let D = the event where a hand is dealt such that all $5$ cards are different denominations.
There are ($\binom{13}{5}$ ways to pick the denominations $\times \binom{4}{1}$ ways to pick the suit)/ ($\binom{52}{5}$ total combinations)
(b) Let A = the event where a hand is dealt such that it includes at least one ace.
$P(a) =$ ($\binom{4}{1}$ ways to pick the ace $\times \binom{13}{4}$ ways to pick the other cards $\times \binom{4}{1}$ ways to pick the suit)/ ($\binom{52}{5}$ total combinations)
If someone could walk me through a problem similar to this it'd be really appreciated this stuff is sort of confusing. (poor cs major)
|
What is the probability of being dealt a hand where all five cards are of different denominations?
There are $\binom{52}{5}$ ways to select a five-card hand.
There are $\binom{13}{5}$ ways to select five different denominations. For each such denomination, there are $\binom{4}{1}$ ways to choose the suit of the card of that denomination. Therefore, there are $\binom{13}{5}\binom{4}{1}^5$ favorable cases.
Hence,
$$\Pr(\text{five different denominations}) = \frac{\dbinom{13}{5}\dbinom{4}{1}^5}{\dbinom{52}{5}}$$
As Lord Shark the Unknown indicated in the comments, you failed to pick a suit for each denomination.
What is the probability of being dealt a hand with at least one ace?
Method 1: Since there are four aces in the deck, such a hand either contains one ace and four other cards, two aces and three other cards, three aces and two other cards, or four aces and one other card. Since there are $4$ aces and $48$ other cards in the deck,
$$\Pr(\text{at least one ace}) = \frac{\dbinom{4}{1}\dbinom{48}{4} + \dbinom{4}{2}\dbinom{48}{3} + \dbinom{4}{3}\dbinom{48}{2} + \dbinom{4}{4}\dbinom{48}{1}}{\dbinom{52}{5}}$$
Method 2: We subtract the probability of selecting no aces from $1$. If no aces are selected, we must select five of the other $48$ cards. Hence,
$$\Pr(\text{no aces}) = \frac{\dbinom{48}{5}}{\dbinom{52}{5}}$$
Thus,
$$\Pr(\text{at least one ace}) = 1 - \frac{\dbinom{48}{5}}{\dbinom{52}{5}}$$
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If $n$ numbers are generated, what is the probability that the product of all those numbers is a multiple of 10? A computer generates random numbers from the set $\{1,2,3,4,5,6,7,8,9\}$ (each has equal probability). If $n$ numbers are generated (with replacement), what is the probability that the product of all those numbers is a multiple of 10?
Attempt:
Let us see easier case, when $n=4$. The total number of possible random numbers can be counted as the number of non-negative solutions to
$$ x_{1} + x_{2} + ... + x_{8} + x_{9} = 4$$
This is because, for example: $\{1,2,3,9\}$ can be seen as $x_{1}=x_{2}=x_{3}=x_{9}=1$ and all others are 0, then again $\{5,5,2,4\}$ as $x_{5}=2, x_{2}=x_{4}=1$ and the others 0. The number of solutions is $\binom{8+4}{4}$.
Now we have to count the total number of possibilities when the product of the generated numbers is multiple of 10. If it is multiple 10, then it must contain a 5 and at least one from $\{2,4,6,8\}$. We can count by the following:
*
*2 generated numbers are $5$ and $2$, with no restriction for the other 2. (note that this can contain $4,6,8$). This will be the same as counting nonnegative solutions: $x_{1} + ... + x_{9} = 2$. Count is $\binom{8+2}{8}$
*2 generated numbers are $5$ and $4$, but this time there is restriction for the other 2: it cannot contain 2 because this was counted earlier. This is the number of nonegative solutions for: $x_{1} + x_{3} + ... + x_{9} = 2$. Count is $\binom{7+2}{7}$
*2 generated numbers are $5$ and $6$, but restriction for the other 2: cannot contain a 2 or a 4. Similarly we consider: $x_{1} + x_{3} + x_{5} + ... +x_{9} = 2$. Count is $\binom{6+2}{6}$
*2 generated numbers are $5$ and $8$, restriction for other 2: cannot contain a 2, 4, or 6. Similarly consider: $x_{1} + x_{3} + x_{5} + x_{7} + x_{8} + x_{9} = 2$. Count is $\binom{5+2}{5}$
So when $n=4$, the probability of interest is $\frac{\binom{8+2}{8} + \binom{7+2}{7} + \binom{6+2}{6} + \binom{5+2}{5}}{\binom{8+4}{8}}$
I want to double check my approach, and also looking for better answers.
| To begin, I am making the assumption (that most people would do) that each of the digits is selected in sequence uniformly and independently at random. This makes each of the $9^n$ possible sequences of choices equally likely to occur. This is in direct contrast to only considering the number of each digit selected which results in outcomes that we count which are not equally likely to occur. For instance, the outcome where we have one each of the digits $1,2,3,4,5$ is $5!$ times more likely to occur than the outcome where we have five $1$'s.
To continue, let us look at the opposite outcome: The product of the numbers is not a multiple of $10$
This occurs if and only if at least one of the two events occurs: No $5$'s were included, No even numbers were included.
Let $A$ be the event that no fives were included and $B$ be the event that no even numbers were included.
We have then $Pr(\text{product is multiple of 10})=1-Pr(A\cup B) = 1-Pr(A)-Pr(B)+Pr(A\cap B)$
From here, you should be able to continue.
$1-\left(\frac{8}{9}\right)^n-\left(\frac{5}{9}\right)^n + \left(\frac{4}{9}\right)^n$
| {
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Factorizing $x^5+1$ as a product of linear and quadratic polynomials. I am encountering some trouble with this question:
Factorize $$x^5+1$$ as a product of real linear and quadratic polynomials.
I know that if we subtract 1 from $x^5+1$, we get that $x^5 = -1$, but I am unsure where to go from here.
Can anyone help with this? Thanks.
| First you seem to know that $(x+1)|(x^5+1)$. Carry out this division with polynomial long division to get
$x^5+1=(x+1)(x^4-x^3+x^2-x+1)$
The remaining fourth degree polynomial is symmetric or palindromic, its coefficients $1,-1,1,-1,1$ read the same forward and backwards. When this happens there will be quadratic factors that are palindromic too, so you must have
$x^4-x^3+x^2-x+1=(x^2+ax+1)(x^2+bx+1)$
Expand:
$x^4-x^3+x^2-x+1=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1$
Matching like powers of $x$ gives $b=-1-a$ and $ab=-1$. Eliminating $b$ between these equations gives a quadratic equation for $a$ which you can solve.
I leave you to prove: (1) Both roots for $a$ give the same factors for $x^4-x^3+x^2-x+1$, only the order is different. (2) The quadratic factors you find have negative discriminants and thus cannot be broken down to real linear factors.
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Given $0 < x \le 1 \le y$. Calculate the minimum value of $\frac{x}{y + 1} + \frac{y}{x + 1} + \frac{x^2 + 3y^2 + 2}{6xy(xy + 1)}$.
Given $0 < x \le 1 \le y$. Calculate the minimum value of $$\large \dfrac{x}{y + 1} + \dfrac{y}{x + 1} + \dfrac{x^2 + 3y^2 + 2}{6xy \cdot (xy + 1)}$$
We have that $$\frac{x^2 + 3y^2 + 2}{6xy \cdot (xy + 1)} \ge \left[ \begin{align} \frac{2 \cdot (x + 3y + 2)^2}{12xy \cdot (6xy + 6)}\\ \frac{2 \cdot (2xy + 4y)}{12xy \cdot (xy + 1)} \end{align} \right. = \frac{8 \cdot y(\sqrt{x} + 2)^2}{9 \cdot (3xy + 1)^2}$$
But that's all I have.
| For $x=y=1$ we'll get a value $\frac{3}{2}$.
We'll prove that it's a minimal value.
Indeed, let $y=1+a$.
Thus, $a\geq0$ and we need to prove that:
$$6x^2a^4+(3+9x+21x^2-9x^3)a^3+3(4+9x+3x^2-10x^3+2x^4)a^2+$$
$$+(1-x)(17+37x+14x^2-12x^3)a+2(x+1)^2(1-x)(5-3x)\geq0,$$ which is obvious.
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