Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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First-Order Differential Equation by Substitution Solve the differential equation by the appropriate substitution
$$x^2\frac{dy}{dx}+2xy=x^4y^2+1$$ .
| $x^2\dfrac{dy}{dx}+2xy=x^4y^2+1$
$x^2\dfrac{dy}{dx}=x^4y^2-2xy+1$
$\dfrac{dy}{dx}=x^2y^2-\dfrac{2y}{x}+\dfrac{1}{x^2}$
Note that this belongs to a Riccati equation and cannot find an obvious particular solution, so we should follow the method in http://en.wikipedia.org/wiki/Riccati_equation#Reduction_to_a_second_order_linear_equation.
Let $u=x^2y$ ,
Then $y=\dfrac{u}{x^2}$
$\dfrac{dy}{dx}=\dfrac{1}{x^2}\dfrac{du}{dx}-\dfrac{2u}{x^3}$
$\therefore\dfrac{1}{x^2}\dfrac{du}{dx}-\dfrac{2u}{x^3}=x^2\left(\dfrac{u}{x^2}\right)^2-\dfrac{2}{x}\dfrac{u}{x^2}+\dfrac{1}{x^2}$
$\dfrac{1}{x^2}\dfrac{du}{dx}-\dfrac{2u}{x^3}=\dfrac{u^2}{x^2}-\dfrac{2u}{x^3}+\dfrac{1}{x^2}$
$\dfrac{1}{x^2}\dfrac{du}{dx}=\dfrac{u^2}{x^2}+\dfrac{1}{x^2}$
$\dfrac{du}{dx}=u^2+1$
But luckily we get a separable ODE.
$\dfrac{du}{u^2+1}=dx$
$\int\dfrac{du}{u^2+1}=\int dx$
$\tan^{-1}u=x+C$
$u=\tan(x+C)$
$\therefore y=\dfrac{\tan(x+C)}{x^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of solutions for an equation with constraints on each variable in the equation I have to find the number of solutions for: $$x_1 + x_2 + x_3 + x_4 = 42$$
when given:
$$ (I) 12 <= x_1 <=13 $$
$$ (II) 3 <= x_2 <= 6 $$
$$ (III) 11 <= x_3 <= 18 $$
$$ (IV) 6 <= x_4 <= 10 $$
What I did so far, is define:
$$ y_1 = x_1 - 12 $$
$$ y_2 = x_2 - 3 $$
$$ y_3 = x_3 - 11 $$
$$ y_4 = x_4 - 6 $$
Thus we get:
$$ y_1 + y_2 + y_3 + y_4 = 10 $$
where
$$ 0 <= y_1 <= 1 $$
$$ 0 <= y_2 <= 3 $$
$$ 0 <= y_3 <= 7 $$
$$ 0 <= y_4 <= 4 $$
now I defined:
$\text{S = all solutions to}$ $y_1 + y_2 + y_3 + y_4 = 20$
$\text{where all}$ $y_i >=0$ $\text{without the upper limitation}$
and I know I have to subtract the rest of the cases, when $y_1 is >1$ and $y_2 is >3$ etc... but how do I count all those side cases that I should subtract?
| For the newer information on question, since $x_i$ are integral(which is same as saying that they are integers):
The answer will be coefficient of $x^{10}$ in the following expansion:
$$
\begin{align}
f(x) &= (x^0 + x^1)(x^0 + x^1 + x^2 + x^3)(x^0 + x^1 + x^2 + x^3 + x^4 + x^5 + x^6 + x^7)(x^0 + x^1 + x^2 + x^3 + x^4) \\
&= (1 + x)(1 + x + x^2 + x^3)(1 + x + x^2 + x^3 + x^4)(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/355622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is the coefficient of $x$ in $\frac{1}{x}=0$? I usually solve a quadratic equation:
$$ax^2+bx+c=0$$
Through a method I learned in school: For a monic quadratic, you make $x=y-\frac{b}{2}$.
The method is intended for a monic equation but in this case (non-monic equation), I divide all the equation by $a$ to transform it in a monic equation.
Now I'm trying to solve the cubic:
$$x^3+x^2-12x=0$$
For the problem in question, I was thinking if I couldn't do the same, I noticed that the coefficient of $x^2$ is $1$, then I thought about dividing everything by $x^3$ which would give me:
$$\frac{x^2}{x^3}-\frac{12x}{x^3}+1=0\tag{1}$$
$$-12x^{-2}+x^{-1}+1=0\tag{2}$$
$$-\frac{12}{x^2}+\frac{1}{x}+1=0\tag{3}$$
And now, trying to use the same procedure I demonstrated for quadratics, I guess I need to find the coefficient of $x$ in $\frac{1}{x}$ and at this part I'm stuck, If it was the coefficient of $\frac{x}{2}$ (for example), I would just need the coefficient of $x$ divided by $2$ which would be $\frac{1}{2}$, but for $\frac{1}{x}$ I am confused, I tried to use Wolfram|Alpha and it gave me this.
NOTE: I am aware of the existence of the quadratic and cubic formulas, I'm doing this just for curiosity.
| ADD With what you've done, you can let $u=1/x$ to move on, however, note that $x=0$ is a solution to the equation, so you can factor $x$ out and move on with a quadratic.
Note that the method you know about quadratic equations has an analog for cubics, however, what you do is anihilate the $x^2$ term. That is, suppose we're given $$x^3+ax^2+bx+c=0$$
Then by letting $x\mapsto y-\frac{a}{3}$ we get $$\eqalign{
& {\left( {y - \frac{a}{3}} \right)^3} + a{\left( {y - \frac{a}{3}} \right)^2} + b\left( {y - \frac{a}{3}} \right) + c = 0 \cr
& {y^3} - 3\frac{a}{3}{y^2} + 3\frac{{y{a^2}}}{9} - \frac{{{a^3}}}{{27}} + a{y^2} - \frac{{2{a^2}}}{3}y + \frac{{{a^3}}}{9} + by - \frac{{ba}}{3} + c = 0 \cr
& {y^3} + \left( {b - \frac{{{a^2}}}{3}} \right)y + \left( { - \frac{{ba}}{3} + c - \frac{{{a^3}}}{{27}} + \frac{{{a^3}}}{9}} \right) = 0 \cr
& y^3 + Ay + B = 0 \cr} $$ (I hope I made no mistakes up there)
The idea is that we can "depress" any cubic into the form $x^3+ax+b=0$ and there is a formula to solve this equation just as we can solve $x^2+ax+b=0$ using the famous Bhaskara formula. In fact, the first solutions to cubics were for cubics of the form
$$x^3=ax+c$$ $$x^3+ax=c$$
where we make sure the coefficients are kept positive by putting them in the appropriate "side" of $=$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/356497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving for three unknowns given three linear equations Have to admit, I forgot how to do simple 9th grade algebra problems. I need to know how to do this stuff for an exam tomorrow. I am doing recurrences and the only thing I need to do is find $a, b$, and $c$ from the following
$$\begin{align*}
a + b + c &= 3\\
4a + 2b+ c&= 6\\
9a + 3b + c& = 13
\end{align*}$$
Can someone refresh my memory on how to solve for $a, b$ and $c$
?
| You could represent your system as a single equation using matrices:
$$
\left[
\begin{matrix}
1 & 1 & 1\\
4 & 2 & 1\\
9 & 3 & 1
\end{matrix}
\right]
\left[
\begin{matrix}
a\\
b\\
c
\end{matrix}
\right]
=
\left[
\begin{matrix}
3\\
6\\
13
\end{matrix}
\right]
$$
To isolate the matrix of unknowns, multiply on the left by the inverse of the matrix of coefficients:
$$
\left[
\begin{matrix}
1 & 1 & 1\\
4 & 2 & 1\\
9 & 3 & 1
\end{matrix}
\right]^{-1}
\left[
\begin{matrix}
1 & 1 & 1\\
4 & 2 & 1\\
9 & 3 & 1
\end{matrix}
\right]
\left[
\begin{matrix}
a\\
b\\
c
\end{matrix}
\right]
=
\left[
\begin{matrix}
1 & 1 & 1\\
4 & 2 & 1\\
9 & 3 & 1
\end{matrix}
\right]^{-1}
\left[
\begin{matrix}
3\\
6\\
13
\end{matrix}
\right]
$$
which simplifies to
$$
\left[
\begin{matrix}
a\\
b\\
c
\end{matrix}
\right]
=
\left[
\begin{matrix}
2\\
-3\\
4
\end{matrix}
\right]
$$
That is, $a = 2$, $b = -3$, and $c = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/357964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2 } \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $
Evaluate
$$\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2
} \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $$
The question look's like there is a nice method to do it, but I can't figure out. Can someone provide some hint or answer?.
| Set R = 4 x3 + 5 x2 + 10 for brevity. Maybe based on Liouville's theorem, we can guess
$$\int \sqrt{R} \left( x^5 + x^4 + x^2 \right) = PR^{3/2} + c$$
where P is a polynomial of x and c is the constant of integration. Differentiate on both sides, we get
$$\sqrt{R} \left( x^5 + x^4 + x^2 \right) = P'R^{3/2} + PR' \sqrt{R}.$$
Rationalize the formula
$$\left( x^5 + x^4 + x^2 \right) = \left( 4x^3 + 5x^2 + 10 \right) P' + \left(
12x^2 + 10x \right) P.$$
P is cubic, so set P = a3x3 + a2x2 + a1x + a0 and solve the linear system. The system is overdeterined, but luckily there is still a solution
$$P = \frac{x^3}{30}.$$
As a result,
$$\int \sqrt{4x^3 + 5x^2 + 10} \left( x^5 + x^4 + x^2 \right) = \frac{x^3 \left( 4x^3 + 5x^2 + 10 \right)^{3/2}}{30} + c.$$
Therefore
$$\int_0^1 \sqrt{4x^3 + 5x^2 + 10} \left( x^5 + x^4 + x^2 \right) = \frac{19^{3/2}}{30}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Why can't you add terms with different exponents? When evaluating algebraic expressions,
1) you can add together like terms. $3x^5 + 6x^5 = 9x^5$, but you cannot add together different terms: $2x^4 + 3x^5$, because these have different exponents.
2) you can multiple different terms: $2x^4 \cdot 3x^5 = 6x^9$. When you multiple terms, the exponents are added together.
Why can't you add terms with different exponents?
Someone said it's because of the properties of algebra:
Commutative property: $a + b = b + a$ and $ab = ba$.
Associative property: $a + (b + c) = b + (c + a)$ and $a \cdot (b \cdot c) = b \cdot (a \cdot c)$.
Distributive property: $x(a+b) = xa + xb$.
So how do these properties suggest that you cannot add terms when exponents are different, but you can multiply terms with different exponents?
| If you pay for two bananas and then three bananas you have paid for five bananas ($2x^2 + 3x^2 \equiv 5x^2$). But if you pay for one banana and one peach, what have you paid for except one banana and one peach?! ($x^2+x^3 \equiv x^2+x^3$).
Even in primary school (elementary school in the US), children are taught the rule of BODMAS (or BIDMAS). We evaluate Brackets first, then Orders (or Indices). Then we do Division, Multiplication, Addition and finally Subtraction.
How can we hope to combine terms of the form $x^2$ and $x^3$? An expression of the form $(x \times x) + (x \times x \times x)$ has only one meaning, even if we take the brackets away, we still have to multiply first.
Think about what additions and multiplication mean:
\begin{array}{ccc}
2x^2 + 3x^2 &\equiv& 2(x \times x) + 3(x \times x) \\
&\equiv& [(x \times x) + (x \times x)] + [(x \times x) + (x \times x) + (x \times x)] \\
&\equiv& (x \times x) + (x \times x) + (x \times x) + (x \times x) + (x \times x) \\
&\equiv& 5(x \times x) \\
&\equiv& 5x^2
\end{array}
We can even verify this by experimentation: $2\times 4^2 + 3\times 4^2 = 32 + 48 = 80$ while $5 \times 4^2 = 80$. However, let's try to do this with terms of a different order, say $x^2$ and $x^3$:
\begin{array}{ccc}
x^2 + x^3 &\equiv& (x \times x) + (x \times x \times x) \\
&\equiv& \ldots\ldots?
\end{array}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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For vector x, y, when does |x+y| = |x|+|y|? In general $|x+y|\le|x|+|y|$. When does equality hold?
Spivak "Calculus on Manifolds" says
the answer is not "when x and y are linearly dependent."
However, that is the answer I get.
| One way to note the conditions for equality is to run through a proof of the triangle inequality
$$\|\mathbf{x}+\mathbf{y}\| \le \|\mathbf{x}\| + \|\mathbf{y}\|$$
This holds if and only if
$$\|\mathbf{x}+\mathbf{y}\|^2 \le \left(\|\mathbf{x}\| + \|\mathbf{y}\|\right)^2 = \|\mathbf{x}\|^2 + 2\|\mathbf{x}\|\|\mathbf{y}\|+\|\mathbf{y}\|^2$$
Rewriting in terms of dot products, we get
$$(\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y}) = \mathbf{x}\cdot\mathbf{x} + 2\mathbf{x}\cdot\mathbf{y} + \mathbf{y}\cdot\mathbf{y}\le \mathbf{x}\cdot\mathbf{x} + 2\|\mathbf{x}\|\|\mathbf{y}\|+\mathbf{y}\cdot\mathbf{y}$$
which reduces to
$$\mathbf{x}\cdot\mathbf{y} \le \|\mathbf{x}\|\|\mathbf{y}\|$$
Therefore the triangle inequality holds if and only if the above inequality holds. But the above inequality is immediately given by Cauchy-Schwarz since
$$\mathbf{x}\cdot\mathbf{y}\le|\mathbf{x}\cdot\mathbf{y}|\le\|\mathbf{x}\|\|\mathbf{y}\|$$
This proves the inequality. For the equality, we need both inequalities above to be equalities. The first equality holds if and only if $\mathbf{x}\cdot\mathbf{y} \ge 0$ and the second holds (via the equality case of Cauchy-Schwarz) if and only if $\mathbf{x} = c\mathbf{y}$ for some $c\in\mathbb{R}$. The first condition forces $c\ge 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all integer solutions of $35x^{31} + 33x^{25} + 19x^{21} \equiv 1 \pmod{ 55}$ Find set of all integers x for which the following holds: $35x^{31} + 33x^{25} + 19x^{21} \equiv 1 \pmod {55}$
Since $55 = 5\cdot 11$, simultaneous congruences:
$35x^{31} + 33x^{25} + 19x^{21} \equiv 1 \pmod 5$
$35x^{31} + 33x^{25} + 19x^{21} \equiv 1 \pmod {11}$
And then these can be simplified:
$3x^{25} - x^{21} \equiv 1 \pmod 5$
$2x^{31} - 3x^{21} \equiv 1 \pmod {11}$
How can I proceed from here?
| We will find Fermat's Theorem useful. Recall that it says that if $p$ is prime and $a$ is not divisible by $p$, then $a^{p-1}\equiv 1 \pmod{p}$.
We start as you did. Modulo $5$, we arrive at the congruence $3x^{25}-x^{21}\equiv 1\pmod{5}$. If $x$ is a solution, it cannot be divisible by $5$. Thus $x^4\equiv 1\pmod{5}$ by Fermat's Theorem. Thus $x^{24}=(x^4)^6\equiv 1 \pmod{5}$. Similarly, $x^{20}\equiv 1\pmod{5}$. So our congruence is equivalent to $3x-x\equiv 1\pmod{5}$. Solving $2x\equiv 1\pmod{5}$ is quick: we get $x\equiv 3\pmod{5}$.
For the second congruence, start from your $2x^{31}-3x^{21}\equiv 1\pmod{11}$. Note that $x$ cannot be divisible by $11$. Since $x^{10}\equiv 1\pmod{11}$ (Fermat's Theorem again), we get $x^{31}\equiv x\pmod{11}$ and $x^{20}\equiv 1\pmod{11}$. So our congruence is equivalent to $2x-3x\equiv 1\pmod{11}$, and the solution is $x\equiv -1\pmod{11}$.
So we want to solve $x\equiv 3\pmod{5}$, $x\equiv -1\pmod{11}$. For bigger numbers, we might use the machinery of the Chinese Remainder Theorem. But with these small numbers, it s easiest to look at the numbers that are congruent to $-1$ modulo $11$, and stop when we get a number congruent to $3$ modulo $5$. Start at the smallest positive candidate, $10$. Is it congruent to $3$ modulo $5$? No. Next try $21$ (no), $32$ (no), $43$. Bingo, got it!
There is a unique solution modulo $55$, so the general solution is $x\equiv 43\pmod{55}$, that is, all numbers of the form $55k+43$, where $k$ ranges over the integers, positive, negative, and $0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing the Semimajor and Semiminor axis of an Ellipse I have the equation of the ellipse which is $\frac {x^2}{4r^2}+\frac{y^2}{r^2}=1$ Putting the (4,2) point on the ellipse we get $r^2=8$ so we get the equation $\frac {x^2}{32}+\frac {y^2}8=1$ and the semi-major axis is $\sqrt {32}= 4\sqrt 2$, the semi-minor axis is $\sqrt 8=2\sqrt 2$.
My question is using the value of $r^2=8$ to $\frac {x^2}{4r^2}+\frac{y^2}{r^2}=1$ , how come the semi-major axis is $\sqrt {32}= 4\sqrt 2$ ? and semi-minor axis is $\sqrt 8=2\sqrt 2$?
I am puzzled on how to get that answer.
...
anyway this question is related to my previous question Finding the Width and Height of Ellipse given an a point and angle
| A circle centred at the origin, of radius $r$ has equation $x^2 + y^2 = r^2$ (alternatively, $\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1$).
An ellipse centred at the origin has equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a, b > 0$. The larger of the two numbers $a$ and $b$ is called the (length of) the semi-major axis, while the smaller of the two numbers is called the (length of) the semi-minor axis.
Note, the case where the (lengths of the) semi-major and semi-minor axes are the same is precisely when the ellipse is a circle.
| {
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"timestamp": "2023-03-29T00:00:00",
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Repeating Square Root Closed Form I've been thinking about repeating square roots:
$\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$. I wrote a program on my calculator to do it $n$ times and I found that, if $x = y^2 - y$ then $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ approaches $y$ as $n$ gets large. So, solving for $y$:
$$y^2 - y - x = 0$$
$$y = \frac{1 \pm \sqrt{1 + 4x}}{2} = \frac{\sqrt{4x + 1} + 1}{2}$$
$$\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} =\frac{\sqrt{4x + 1} + 1}{2} $$
This works, but where does the $y^2-y=x$ come from? I got here via plugging and guessing, but how can I prove that with an input of $x$ where $x = y^2 - y$, the answer will be $y$?
| Define
$$\alpha:=\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}=\sqrt{x+\alpha}\implies \alpha^2-\alpha-x=0\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/369852",
"timestamp": "2023-03-29T00:00:00",
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Simplifying $\sin(2\tan^{-1} x)$ I've been working on this for a while. The answer in the book is $\frac{2x}{x^2 + 1}$ Here's my workings:
$\sin(2\tan^{-1} x)$
Let $\alpha = \tan^{-1}x \Rightarrow \tan \alpha = x$
$\sin(2\alpha) = 2\sin\alpha\cos\alpha = 2\tan\alpha\cos^2\alpha = 2x\cos^2\alpha$
I'm not sure how to proceed to turn that $cos^2\alpha$ into $\frac{1}{x^2 + 1}$
| $\sin(2arctan(x))$
let $y=arctan(x)$
so $x=tan(y)$
$sin(2y)=2sin(y)cos(y)=2sin(arctan(x))\cdot cos(arctan(x))$
$sin(arctan(x)) = \frac{x}{\sqrt{1+x^2}}$ $cos(arctan(x))= \frac{1}{\sqrt{1+x^2}}$
$2\cdot\frac{x}{\sqrt{1+x^2}} \cdot \frac{1}{\sqrt{1+x^2}}=\frac{2x}{{1+x^2}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the values of $p$ such that $\left( \frac{7}{p} \right )= 1$ (Legendre Symbol)
Show that if $p$ is an odd prime coprime to $7$, then $\left( \frac{7}{p} \right) = 1$ if and only if $p \equiv \pm 1, \pm 3,$ or $\pm 9 \pmod{28}$. HINT: If $p$ is an odd prime, determine which values can $p$ take $\mod28$, and consider each of these values in turn. Note that if we know $p \mod 28$ then we know $p \mod 4$, and hence we know whether $\frac{p-1}{2}$ is odd or even.
Here, $\left( \frac{a}{b} \right)$ is the Legendre symbol.
The bit I don't understand in the hint is, what do they mean by consider the values that $p$ can take $\mod 28$. Do they mean the values that would make $p$ a quadratic residue $\mod 28$, i.e all the $x$ values satisfying $x^2 \equiv \mod 28$, because then isn't this just $1,4,9,16,25$?
What do they mean the to "consider each of these values in turn"?
| Here is the answer with minor details.
by Quadratic Reciprocity rule, $$\left( \frac{7}{p} \right)=(-1)^{\frac{7-1}{2} \cdot \frac{p-1}{2}} \left(\frac{p}{7} \right)=(-1)^{3 \cdot \frac{p-1}{2}} \left(\frac{p}{7} \right)=(-1)^{ \frac{p-1}{2}} \left(\frac{p}{7} \right).$$
Now we evaluate values of each factor on RHS:
\begin{align} (-1)^{\frac{p-1}{2}}&=\left\{ \begin{aligned}1~\text{if}~p \equiv 1 ~(\text{mod}~ 4) \\ -1~\text{if}~p \equiv 3~(\text{mod}~4) \end{aligned} \right. \\ \text{and by the previous answers, we have }~\left(\frac{p}{7}\right)&=\left\{ \begin{aligned}1~\text{if}~p \equiv 1,2,4 ~(\text{mod}~ 7) \\ -1~\text{if}~p \equiv 3,5,6~(\text{mod}~7) \end{aligned} \right. \end{align}
Now we can use Chinese Remainder Theorem, and combine the factors to obtain the final result:
\begin{align} \left(\frac{7}{p}\right)=\left\{ \begin{aligned}&1~\text{if}~p \equiv \pm 1, \pm 3, \pm 9 ~(\text{mod}~ 28) \\ -&1~\text{if}~p \equiv \pm 5, \pm 11, \pm 13~(\text{mod}~28) \end{aligned} \right. \end{align}
| {
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Geometry: Hyperbolas I was wondering how would I complete the square for this particular hyperbola?
$4x^2 - 5y^2 + 24y = 16$
I tried this technique but to no avail:
$$4x^2 - 5(y^2 + \frac{24}{5}y) = 16$$
$$ \Rightarrow 4x^2 - 5(y + \frac{12}{5})^2 = 16 + \left(\frac{12}{5}\right)^2$$
$$\Rightarrow 4x^2 - 5(y + \frac{12}{5})^2 = \frac{544}{25} .$$
Am I doing something wrong here? On my calculator it says that the equation should be a hyperbola
| First off, don't forget that when you factor out that "-5" to make the "completion of squares" easier, it also has to be factored out of +24y, so the equation becomes
$$4x^2 - 5(y^2 - \frac{24}{5}y ) = 16 $$
(as lab bhattacharjee already has).
You then complete the square by adding the term $\frac{12^2}{5^2}$ within the parentheses. However, since it is inside the parentheses, what you have just "added" to the left-hand side of the equation is really $-5 \cdot (\frac{12^2}{5^2})$, so the equation must be sustained by writing
$$4x^2 - 5(y^2 - \frac{24}{5}y + [\frac{12^2}{5^2}]) = 16 -5 \cdot (\frac{12^2}{5^2}) ,$$
now making the equation
$$4x^2 - 5(y -\frac{12}{5})^2 = 16 -(\frac{12^2}{5}) = \frac{16 \cdot 5 - 144}{5} = \frac{-64}{5} .$$
Putting the equation for this hyperbola in standard form gives us
$$\frac{-5 \cdot 4x^2}{64} - \frac{-5 \cdot 5(y -\frac{12}{5})^2}{64} = 1 \Rightarrow \frac{25(y -\frac{12}{5})^2}{64} - \frac{20x^2}{64} = 1.$$
[or, of course, $\frac{(y -\frac{12}{5})^2}{64/25} - \frac{x^2}{64/20} = 1$]
So this is a "vertical" hyperbola, with its focal axis along the y-axis, since it is the y-term that is positive. (Graphing the original and this standard-form equation confirms that they are identical.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Existence of real roots of a quartic polynomial Question
What is the minimum possible value of $a^{2}+b^{2}$ so that the polynomial $x^{4}+ax^{3}+bx^{2}+ax+1=0$ has at least 1 root?
Attempt
I divided by $x^{2}$ and got $$x^{2}+\frac{1}{x^{2}}+2+a\left(x+\frac{1}{x}\right)+b-2=0$$ by letting $$x+\frac{1}{x}=X$$
the equation becomes:
$$X^{2}+aX+(b-2)=0$$
$$\therefore X=\frac{-a\pm \sqrt{a^{2}-4b+8}}{2}$$ but I am not sure how to continue. If the polynomial has 1 root doesn't that is should be a double root? Are we counting multiplicity or not?
| It seems to me that requesting the smallest possible value for $a^2+b^2$ is almost a red herring.
Claim.
For $|a| \leq 4$ and $b > 2 |a| - 2$, the polynomial
$$
p(x) = x^4 + ax^3 + bx^2 + ax + 1
$$
has no real zeros.
Proof.
First set $b = 2|a| - 2 + \epsilon$, where $\epsilon > 0$. If we denote $\pm = \operatorname{sign} a$, so that $a = \pm |a|$, we have
$$
\begin{align}
p(x) &= x^4 + ax^3 + (2|a|-2+\epsilon)x^2 + ax + 1 \\
&= x^4 - 2x^2 + 1 + x(ax^2 + 2|a|x + a) + \epsilon x^2 \\
&= (x^2-1)^2 + ax(x \pm 1)^2 + \epsilon x^2 \\
&= (x \pm 1)^2\left[ (x \mp 1)^2 + ax \right] + \epsilon x^2 \\
&= (x \pm 1)^2\left[ x^2 \pm (|a|-2)x + 1 \right] + \epsilon x^2 \\
&\geq (x \pm 1)^2 \min\left\{x^2 - 2x + 1,x^2+2x+1\right\} + \epsilon x^2 \\
&= (x \pm 1)^2\min\left\{(x-1)^2,(x+1)^2\right\} + \epsilon x^2 \\
&> 0.
\end{align}
$$
Q.E.D.
The points on the boundary of the region
$$
A = \{(a,b) \,\colon |a| \leq 4 \,\,\,\text{and}\,\,\,b > 2 |a| - 2\}
$$
which have the smallest norm are
$$
(a,b) = \left(\pm \frac{4}{5},-\frac{2}{5}\right),
$$
as shown in the following image.
It only remains to demonstrate that one of these points yields a polynomial with a real zero. In fact they both do:
$$
x^4 + \frac{4}{5} x^3 - \frac{2}{5} x^2 + \frac{4}{5} x + 1
$$
has a zero at $x=-1$ and
$$
x^4 - \frac{4}{5} x^3 - \frac{2}{5} x^2 - \frac{4}{5} x + 1
$$
has a zero at $x=1$.
Thus
The smallest value of $a^2 + b^2$ for which we can find some $a,b\in \mathbb R$ such that $p(x)$ has a real zero is
$$
\left(\frac{4}{5}\right)^2 + \left(\frac{2}{5}\right)^2 = \frac{4}{5}.
$$
Below is a plot of the $(a,b)$-plane which shows the region (in blue) where the polynomial $p(x)$ has at least one real root. Note that for $|a| \leq 4$ the boundary of this region is precisely $b = 2|a| - 2$ (shown in black), but for $|a| > 4$ it curves inward.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Stuck on Epsilon proof.. Using the $\epsilon-M $ definition of the limit, calculate
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}.$$
Working so far:
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}=3$$
Given $\epsilon>0$, I want M s.t. $x>M \implies \left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$
$$\left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$$
$$\left|\frac{3x^2+7-3(x^2+x+8)}{x^2+x+8} \right|<\epsilon$$
$$\left|\frac{-3x-17}{x^2+x+8} \right|<\epsilon$$
And now I'm stuck.. Any help would be great, thanks.
| use the fact that $ \left | \frac 1 x - 0\right| < \delta $
$$\left| \frac{-3x - 17}{x^2 + x + 8}\right| < \left| \frac{-3x - 17}{x^2 }\right| \le 3\left |\frac 1 x \right | + 17\left |\frac 1 {x^2} \right | < 20 \delta = \epsilon $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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(4x^2+2kx-5)/(x+2) remainder is 3 find value of k? 2 methods - first is long division by $(x+2)$, 2nd is to use remainder theorem
let $f(x) = 4x^2+2kx-5$ and $g(x) = x+2$
to find the remainder of $\frac{f(x)}{g(x)}$ where $g(x) = (x+c)$ we need to evaluate $f(-c)$
$f(-2) = 4(-2)^2+2k(-2) -5$
Because the remainder is 3, we know that $f(-2)=3$
so $$16 - 4k -5 = 3$$
$$4k = 8$$
$$k=2$$
is that correct?
| Your method works well here. And indeed, $k = 2$. It's always a good idea to "check out" whether the equation, with $k = 2$, divided by $x + 2$, gives a remainder of $3$.
Substituting $\color{blue}{\bf k = 2}$ into $\dfrac{f(x)}{g(x)}$ gives us:
$$\dfrac{4x^2 + 2\cdot \color{blue}{\bf 2}(x) - 5}{x + 2} = \dfrac{4(x - 1)(x+2) + 3}{x+2},\text{ i.e.}\;\; f(x) = 4(x-1)(x+2) + 3$$
| {
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Calculating the sum of $f(x) = \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^n}$ In Calculus, how do I calculate this sum?
$$f(x) = \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^n}$$
This is what I did so far:
$$ f(x) = 2x \cdot \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^{n-1}} $$
Therefore:
$$ \frac{\int{f(x)}}{2x} = \sum_{n=0}^{\infty} {2^n \cdot x^n}$$
But I have no idea where to continue from here!
| Here is a more elementary proof, which doesn't use the differentiability/integrability property of power series:
Let
$$S_m= \sum_{n=0}^{m} {n \cdot 2^n \cdot x^n}$$
Then
$$2x S_m =\sum_{n=0}^{m} {n \cdot 2^{n+1} \cdot x^{n+1}}=\sum_{k=1}^{m+1} {(k-1) \cdot 2^{k} \cdot x^{k}}$$
$$=\left(\sum_{k=1}^{m+1} k \cdot 2^{k} \cdot x^{k}\right)-\left(\sum_{k=1}^{m+1} 2^{k} \cdot x^{k}\right)=\left(\sum_{k=0}^{m+1} k \cdot 2^{k} \cdot x^{k}\right)-\left(2x\cdot\frac{1-(2x)^{m+1}}{1-2x}\right)$$
$$2xS_m=\left(S_m+(m+1)2^{m+1}x^{m+1}\right)-\left(2x\cdot\frac{1-(2x)^{m+1}}{1-2x}\right)$$
Solving for $S_m$ yields:
$$S_m=\left(2x\cdot\frac{1-(2x)^{m+1}}{(1-2x)^2}\right)-\frac{(m+1)(2x)^{m+1}}{1-2x}$$
now, $S_m$ is convergent if and only if $(2x)^{m+1} \to 0$ if and only if $|2x|<1$.
In this case
$$\lim S_m= 2x\cdot\frac{1}{(1-2x)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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To split in to partial fractions, the expression $\frac{1}{x^2(x+a)^2}$ To split in to partial fractions, the expression $\frac{1}{x^2(x+a)^2}$
$\frac{1}{x^2(x+a)^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(x+a)}+\frac{D}{(x+a)^2}$
One method of finding the values of the constants A,B,C & D is as follows.
$Ax(x+a)^2+B(x+a)^2+C(x^2)(x+a)+Dx^2=1$
$Ax(x^2+a^2+2ax)+B(x^2+2ax+a^2)+C(x^3+ax^2)+Dx^2=1$
$A(x^3+a^2x+2ax^2)+B(x^2+2ax+a^2)+C(x^3+ax^2)+Dx^2=1$
$x^3(A+C)+x^2(2Aa+B+Ca+D)+x(Aa^2+2Ba)+Ba^2=1$
Equating coeffecients of $x^3,x^2,x^1,x^0$
$A+C=0$; $2Aa+B+Ca+D=0$; $Aa^2+2Ba=0$; $Ba^2=1$
$B=\frac{1}{a^2}$
$A=\frac{-2}{a^3}$
$C=\frac{2}{a^3}$
$D=\frac{1}{a^2}$
Is there a shorter method to find the values of A,B,C & D ?
| $$Ax(x+a)^2+B(x+a)^2+C(x^2)(x+a)+Dx^2=1$$
$x=0$ yields $b= \frac{1}{a^2}$.
$x=-a$ yields $d= \frac{1}{a^2}$.
Then you can derivate and set again $x=-a$ and $x=0$ to get $A,C$ or replace $B,D$ in that equation to get
$$Ax(x+a)^2+C(x^2)(x+a)+\frac{2x^2}{a^2}+\frac{2ax}{a^2}+1=1$$
After canceling the 1's, you can divide by $x(x+a)$ and get
$$A(x+a)+Cx+\frac{2}{a^2}=0$$
$x=0$ yields $A=-\frac{2}{a^3}$
and
$x=-a$ yields $C=\frac{2}{a^3}$
| {
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Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$
I can't seem to wrap my head around that.
| Actually, if you consider an integer division (so called Euclidian division) the proposition becomes:
$$
n^2 - \frac{n^2}{2} = \frac{n^2}{2} + \text{mod}{(n^2,2)}
$$
Indeed, by definition of the Euclidian division of the integer $n^2$ by the integer $2$, one has:
$$
n^2 = 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)}
$$
where $\text{mod}{(a,b)}$ denotes the remainder of the euclidian division of an integer $a$ by an integer $b$.
Now, by (cleverly enough) remarking that the term $2\times \frac{n^2}{2}$ equals to $\frac{n^2}{2} + \frac{n^2}{2}$ one can write:
$$
\begin{aligned}
n^2 &= 2\times\frac{n^2}{2} + \text{mod}{(n^2,2)} \\
n^2 &= \frac{n^2}{2} + \frac{n^2}{2} + \text{mod}{(n^2,2)} \\
\Leftrightarrow n^2 - \frac{n^2}{2}&= \frac{n^2}{2} + \text{mod}{(n^2,2)} \\
\end{aligned}
$$
that ends the proof ;)
| {
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"timestamp": "2023-03-29T00:00:00",
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How to express the following as a series? How you you express $$1+3+6+10+15+21+28+...$$ as a summation? I've tried different ways but they were wrong and always brought me back to square one. Thanks!
| $$S_n=1+3+6+10+15+\cdots+T_n$$
$$S_n=1+3+6+10+15+\cdots+T_{n-1}+T_n$$
On subtraction, $$0=1+(3-1)+(6-3)+(10-6)+(15-10)+\cdots+(T_n-T_{n-1})-T_n$$
$$\implies T_n=1+2+3+\cdots\text{up to }n\text{ terms}=\frac{n(n+1)}2$$
$$\text{Now, }(r+1)^3-r^3= 6\frac{r(r+1)}2+1$$ (Observe the Telescoping Sum in the Left Hand Side)
Putting $r=1,2,\cdots,n-1,n$ and adding them we get,
$$(n+1)^3-1=6S_n+n$$
$$\implies 6S_n=n^3+3n^2+3n+1-1-n=n^3+3n^2+2n=n(n+1)(n+2)$$
| {
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"url": "https://math.stackexchange.com/questions/381636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Applying the Mean Value Theorem to conclude a function has a zero
Consider the function $f$ given by $f(x)=(x-2)^4\cos(x^2-4x+4)$. Use the Mean Value Theorem to show that $f'$ has a zero on the interval on $[1,3]$.
I notice that to do this we must show $f'(c)=0$ where $c$ is real number in the interval $[1,3]$. Now by the Mean Value Theorem,
$$\frac{f(3)-f(1)}{3-1} =f'(c)\,.$$
Notice that $f'(c)$ is indeed $0$ on the left hand side.
| First note that $x^2-4x+4 = (x-2)^2.$
Then the left-hand side of your MVT equation gives
\begin{align}
\frac{f(3) - f(1)}{3-1} &= \frac{(3-2)^4 \cos (3-2)^2 - (1-2)^4 \cos (1-2)^2}{2} \\
&= \frac{(1)^4 \cos (1)^2 - (-1)^4 \cos (-1)^2}{2} \\
&= \frac{\cos(1) - \cos(1)}{2} \\
&= 0,
\end{align}
which equals $f'(c)$ for some $c \in (1,3).$
(Notice that my interval is open. I think this is what you intended to type.)
| {
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Triangle optimization problem Let $a,b,c$ be the sides of a triangle , then what is the maximum and minimum values (if exist) of the following quantities
(i) $\dfrac {a^2b^2c^2}{(a+2b)(a+2c)(b+2c)(b+2a)(c+2a)(c+2b)}$
(ii) $\dfrac {abc(a+b+c)^3}{(a+2b)(a+2c)(b+2c)(b+2a)(c+2a)(c+2b)}$
| (i)
\begin{align*}
&\dfrac {a^2b^2c^2}{(a+2b)(a+2c)(b+2c)(b+2a)(c+2a)(c+2b)}\\
=&\dfrac{1}{(1+2\frac{b}{a})(1+2\frac{a}{b})(1+2\frac{c}{a})(1+2\frac{a}{c})(1+2\frac{c}{b})(1+2\frac{b}{c})}
\end{align*}
For each two factors, for example, $(1+2\frac{b}{a})(1+2\frac{a}{b})=5+2(\frac{a}{b}+\frac{b}{a})\geqslant5+2\cdot2$
(ii)
\begin{align*}
&\dfrac {abc(a+b+c)^3}{(a+2b)(a+2c)(b+2c)(b+2a)(c+2a)(c+2b)}\\
=&\frac{(a+b+c)^3}{(1+2\frac{b}{a})(1+2\frac{c}{b})(1+2\frac{a}{c})(a+2c)(b+2a)(c+2b)}
\end{align*}
Note that $(1+2\frac{b}{a})(1+2\frac{c}{b})(1+2\frac{a}{c})=9+2(\frac{b}{a}+\frac{c}{b}+\frac{a}{c})+4(\frac{c}{a}+\frac{b}{c}+\frac{a}{b})\geqslant9+2\cdot3+4\cdot3$ and $(a+2c)(b+2a)(c+2b)$
| {
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finding the expansion of $\arcsin(z)^2$ Is there a fast and nice way to find the expansion of $\arcsin(z)^2$ without squaring expansion of $\arcsin(z)$ ?
For $|z|<1$ show that
$$(\sin^{-1}(z))^2 = z^2 + \frac{2}{3}\cdot \frac{z^4}{2} + \frac{2}{3}\cdot\frac{4}{5}\cdot \frac{z^6}{3}+ \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot \frac{z^8}{4} + \dots$$
It should have something like $c_{2n} = \frac{2^{2n}n!^2}{(2n+1)!n}$ as coefficient maybe we could use Residue theorem to evaluate it.
| A related problem. It seems we can go beyond this. Here is the Taylor series expansion of $ \arcsin(x)^3 $
$$\arcsin(x)^3 = \frac{3}{4\sqrt {\pi }}\sum _{m=1}^{\infty }\,{\frac { \left( {\pi }^{2}-2\,\psi'\left(
m+\frac{1}{2} \right) \right) \Gamma\left( m+\frac{1}{2} \right) {x}^{2\,m+1}}{
\left( 2\,m+1 \right) \Gamma \left( m+1 \right) }}.$$
Added: Here is the requested series
$$ \frac{\arcsin(x)}{\sqrt{1-x^2}}=\frac{\sqrt{\pi}}{2}\sum _{m=0}^{\infty}{\frac {\Gamma \left( m+1
\right)\, {x}^{2\,m+1}}{\Gamma \left( m+\frac{3}{2} \right) }}.$$
Added: Here are the power series of $\arcsin(x)^4$ and $\arcsin(x)^6$ respectively
$$ \frac{\sqrt {\pi }}{4}\sum _{m=0}^{\infty}\,{\frac { \left( {\pi }^{2}-6\,
\psi' \left( m+1 \right)\right) \Gamma \left( m+1\right){x}^{2\,
m+2}}{ \left( m+1 \right) \Gamma \left( m+\frac{3}{2} \right) }},$$
$$\frac{3\sqrt {\pi }}{32} \sum _{m=0}^{\infty }{}\,{\frac { \left( 60\,
\left( \psi' \left( m+1 \right) \right) ^{2}-20\,\psi' \left( m+1
\right) {\pi }^{2}+{\pi }^{4}+10\,\psi''' \left( m+1 \right) \right)
\Gamma \left( m+1 \right) {x}^{2\,m+2}}{ \left( m+1 \right) \Gamma\left( m+\frac{3}{2} \right) }}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $x-1$ divides $x^n-1$ In algebra & polynomials, how do we prove that
$$x-1 \mid x^n -1?$$
| Factorization method
Clearly, $x^{n} - 1 = (x - 1)(x^{n - 1} + x^{n - 2} + \dots + 1)$. Thus, $(x - 1) | (x^{n} - 1)$ since there exists the polynomial namely $(x^{n - 1} + x^{n - 2} + \dots + 1)$ that is multiplied by $(x - 1)$ to obtain $(x^{n} - 1)$.
Substitution method
Suppose that $x - 1 = 0 \rightarrow x = 1$. If we substitute that value for $(x^{n} - 1)$, then clearly $1^{n} - 1 = 0$. This shows that $(x^{n} - 1)$ has a factor $(x - 1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a+b+c=1$, find the minimum of $\frac{4+3abc}{ab+bc+ac}$ I came to ask this because I am really stuck at this problem. I have tried everything from arithmetic mean, geometric mean and harmonic mean. Also, I have tried playing with the variables and such, but it got me to nowhere.
If $a+b+c=1$; $a,b,c$ nonnegative, calculate the minimum of
$$\frac{4+3abc}{ab+bc+ac}$$
All I've got so far is:
$$\frac{3abc}{ab+bc+ac} \le \frac{1}{3}$$
But this is obviously on the wrong side of the inequality.
Also, I think that
$$\frac{1}{ab+bc+ac}\ge3$$
But I haven't been able to prove it.
Playing with the most possible and obvious values, one could think that the answer is 37/3, but the excercise is about proving it. Any help and little hints are greatly apprecieated.
| Proof
Denote $u = ab + bc + ca$, we have the following two results
(1) $0 \le 3 u \le 1$,
and
(2) $9abc \ge 4u - 1$.
Then,
$$
3abc+4
= \frac{9abc+12}{3}
\ge \frac{4u + 11}{3}
\ge \frac{4u + 11\times3u}{3} = \frac{37}{3} u.
$$
which is the desired result.
Lemma 1
To show (1), we expand
$$(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0$$
which gives
$$(a + b + c)^2 - 3 (ab+bc+ca) \ge 0$$
which means $1 \ge 3u$. The $u \ge 0$ part is obvious for $a$, $b$, and $c$ are nonnegative.
Lemma 2
For (2), we can use Schur's inequality
$$a^3+b^3+c^3+3abc
\ge
a^2(b+c)+b^2(c+a)+c^2(a+b).$$
Since $b + c = 1 - a$, etc, we get
$$2(a^3+b^3+c^3)+3abc \ge a^2+b^2+c^2.$$
Using
$$
a^2 + b^2 + c^2 = (a+b+c)^2 - 2 (ab+bc+ca),
$$
and
$$
a^3 + b^3 +c^3 - 3abc
= (a + b + c)^3 - 3 (a+b+c) (a b + b c + c a).
$$
we get
$$
2(1-3u)+9abc \ge 1-2u,
$$
which is Lemma 2.
Generalization
Generally, we can change the coefficients a bit,
$$
\frac{ 9 p \, a b c + q + p r + 3 q r } { a b + b c + c a + r }
\ge
p + 3q,
$$
and the problem is the special case of $p = \frac{1}{3}, q = 4, r = 0$.
This problem inspires another one, and the use of Schur's inequality is suggested by Dylan.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number theory: Prime powers and cubes Determine all triples $(p,a,b)$ of positive integers, where $p$ is prime and $a \leq b$ such that $$p^a+p^b$$ is a perfect cube.
I came across this question while looking at past maths Olympiad papers in my country. It has been frustrating me for a while. The only progress I've made is the following -
Let $p^a+p^b=k^3$.
Then $p^a(1+p^{b-a})=k^3$, but unless $p=2,a=b$, $\gcd(p^a,p^{b-a}+1)=1$.
Case 1: $b=a$, $p=2$.
Then $k^3=2^a+2^a=2^{a+1}$, so $a \equiv 2 \pmod3$. Hence $(2,a,a)$ is a solution $\forall \; \; a \equiv 2 \pmod3$.
Case 2: Case 1 is not true.
Then as $\gcd(p^a,p^{b-a}+1)=1$, $p^a$ and $1+p^{b-a}$ must both be cube and hence $3|a$.
Let $a=3m$,
I can't seem to get beyond this point although I did try ignoring the $p^a$ and focusing on making the other term a cube. I did some factorizing but it didn't help (as far as I could see).
Thanks in advance for any help.
| You took care of the case $a=b$, and obtained the infinite family of solutions
$2^{3k+2}+2{3k+2}$.
So assume that $a\lt b$. You showed that for $p^a(1+p^{b-a})$ a cube, $a$ should be divisible by $3$, and $1+p^c$ should be a cube, where $c=b-a$. We com[lete things from there.
If $b-a=1$, we want $1+p$ to be a cube, say $x^3$. The factorization $x^3-1=(x-1)(x^2+x+1)$ tells us that the only possibility is $x=2$, giving $p=7$.
This generates an infinite number of variants, namely
$$7^{3k}+7^{3k+1}.$$
So now we look at the case $1+p^c$ a cube, where $c\gt 1$. As Ross Millikan points out, there is no solution, by Mihailescu's Theorem. But let's see whether we can prove it without heavyweight machinery.
Suppose $p^c=x^3-1=(x-1)(x^2+x+1)$. Thus each of $x-1$ and $x^2+x+1$ is a power of $p$. Except in the trivial case $x=2$, $p$ must divide each of $x-1$ and $x^2+x+1$. Any common divisor of these must divide $(x^2+x+1)-(x-1)^2$, so it divides $3x$, and therefore $3$.
So we must have $p=3$. Moreover, since each of $x-1$ and $x^2+x+1$ is a proper power of $3$, the only remaining possibility is $x-1=3$. This doesn't work.
| {
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Show that $\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}=2$
Find $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$.
I found that, by calculator, it is actually $\bf{2}$.
Methods to denest something like $\sqrt{a+b\sqrt c}$ seems to be useless here, what should I do?
| $$S=\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$$
So, $$S^3=2+\frac {10} 9\sqrt 3+2-\frac {10} 9\sqrt 3+3\cdot \sqrt[3]{2+\frac {10} 9\sqrt 3}\cdot\sqrt[3]{2-\frac {10} 9\sqrt 3}\cdot S$$
$$\implies S^3=4+2S$$ as $(2+\frac {10} 9\sqrt 3)(2-\frac {10} 9\sqrt 3)=4-\frac{100\cdot3}{81}=\frac8{27}=(\frac23)^3$
$$\iff S^3-2S-4=0$$
$$\implies S^3-2^3-2(S-2)=0\implies (S-2)(S^2+S+2)=0$$
Observe that $2$ is the sole positive real root of the last eqaution
| {
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"source": "stackexchange",
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Weird Identities with Scalar Product & Transpose: $\vec{a}\cdot\vec{b} = \vec{b}^T \cdot {a}^T$, $\vec{a}^T \cdot \vec{b} = \vec{b}^T \cdot \vec{a} $? Let's say I have a row vector $\vec{a}$ and a column vector $\vec{b}$:
\begin{align}
\vec{a}= \begin{pmatrix}4 & 5 & 6\end{pmatrix} \qquad \vec{b} = \begin{pmatrix}1\\2\\3\end{pmatrix}
\end{align}
Now if I want to calculate a scalar product, it is easy for $\vec{a} \cdot \vec{b}= \begin{pmatrix}4 & 5 & 6\end{pmatrix}\begin{pmatrix}1\\2\\3\end{pmatrix} = \begin{pmatrix}4\cdot1+ 5\cdot 2 + 6\cdot 3\end{pmatrix} = 32.$
but it is not as easy for $\vec{b}\cdot \vec{a}$ where the scalar product is undefined:
\begin{align}
\vec{b}\cdot \vec{a} = \begin{pmatrix}1\\2\\3\end{pmatrix} \begin{pmatrix}4 & 5 & 6\end{pmatrix} \qquad{\scriptsize\text{Not defined!}}
\end{align}
So if I take a transpose of both of the vectors I get:
\begin{align}
\vec{b}{\hspace{0.4mm}}^\mathsf{T}\cdot \vec{a}{}^\mathsf{T} = \begin{pmatrix}1&2&3\end{pmatrix} \begin{pmatrix}4 \\ 5 \\ 6\end{pmatrix} = \begin{pmatrix}1\cdot4 + 2 \cdot 5 + 3 \cdot 6\end{pmatrix} = 32
\end{align}
So I get a relation that $\vec{a}\cdot\vec{b} = \vec{b}^T \cdot {a}^{T} = 32$. But why do I find a different equation on Wikipedia which says: $(\vec{a}\cdot\vec{b} ){}^\mathsf{T}= \vec{b}{}^\mathsf{T} \cdot \vec{a}{\hspace{0.4mm}}^\mathsf{T}$?
EDIT: If i do it like @Samuel says i define original vectors differently:
\begin{align}
\vec{a}= \begin{pmatrix} 4\\ 5 \\ 6 \end{pmatrix} \qquad \vec{b}= \begin{pmatrix} 1\\ 2 \\ 3 \end{pmatrix}
\end{align}
And now i try to calculate $\vec{a} \cdot \vec{b}$ which is not defined. I can calculate ${\vec{a}}^T \vec{b}$ though:
\begin{align}
{\vec{a}}^T \cdot \vec{b}= \begin{pmatrix} 4& 5 &6 \end{pmatrix} \begin{pmatrix} 1\\ 2 \\ 3 \end{pmatrix} = 4\cdot 1 + 5\cdot 2 + 6 \cdot 3 = 32
\end{align}
Now i try to calculate scalar product $\vec{b} \cdot \vec{a}$ and neither this one is defined. But i can calculate ${\vec{b}}^T\cdot \vec{a}$ though:
\begin{align}
{\vec{b}}^T \cdot \vec{a}= \begin{pmatrix} 1& 2 &3 \end{pmatrix} \begin{pmatrix} 4\\ 5 \\ 6 \end{pmatrix} = 1\cdot 4 + 2\cdot 5 + 3 \cdot 6 = 32
\end{align}
So now i can say ${\vec{a}}^T \cdot \vec{b} = {\vec{b}}^T \cdot \vec{a} $. So this now is a lot wierder and i cant find it on Wikipedia.
| A vector should always be a column vector. If you want to talk about a "row vector", you should write it as the transpose of some vector, or as a matrix with just one row.
I will add that, technically, the scalar product can not be written as a matrix product the way you are doing it. If $u,v$ are vectors, then the matrix product $u^T v$ is a $1\times 1$-matrix, whereas the scalar product $u\cdot v$ is a scalar. It is common to ignore this and consider them the same, but one should sometimes be careful when using this, for example when one wants to multiply the scalar product with a matrix.
Added: You cannot say that $a$ is a vector and then write $a=(4,5,6)$ as a row matrix. You can however say that $a$ is a $1\times 3$-matrix $(4,5,6)$, or that $a$ is a vector and $a^T=(4,5,6)$ (and in the latter case, $a$ is a column vector). Let us suppose that you defined the matrices $a=(4,5,6)$ and $b=\begin{pmatrix}1\\2\\3\end{pmatrix}$. Now, the matrix product $a\cdot b$ is equal to the $1\times 1$-matrix $[32]$, not the number 32, which is the scalar product of the vectors $\begin{pmatrix}4\\5\\6\end{pmatrix}$ and $\begin{pmatrix}1\\2\\3\end{pmatrix}$. Likewise, the matrix product $b^T\cdot a^T$ is equal to the $1\times 1$-matrix $[32]$. Then we can conclude that $a\cdot b=b^T\cdot a^T$ (note that you made a typo in your post and wrote $a^T\cdot b^T$ for the right-hand-side). This is consistent with Wikipedia, which states that $(a\cdot b)^T=b^T\cdot a^T$. However, your typo is also correct in this particular case, namely $(a\cdot b)^T=a^T\cdot b^T$, but this is only because both matrices are $1\times 1$-matrices, so they are equal to their own transposes.
Added after the edit: You get the identity $a^T\cdot b=b^T\cdot a$. This equation is not true for general matrices $a,b$, so you will not find it on Wikipedia. What is generally true, however, is the identity $(a^T\cdot b)^T=b^T\cdot a$. The reason your identity is true is because both matrices are $1\times 1$-matrices, and the transpose of a $1\times 1$-matrix does not change the matrix.
| {
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"timestamp": "2023-03-29T00:00:00",
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Laplace transform:$\int_0^\infty \frac{\sin^4 x}{x^3} \, dx $ I have a trouble with a integral:
Using this Laplace trasform equation:
$$\begin{align}
\int_0^\infty F(u)g(u) \, du & = \int_0^\infty f(u)G(u) \, du \\[6pt]
L[f(t)] & = F(s) \\[6pt]
L[g(t)] & = G(s)
\end{align}$$
Applying to compute this integral:
$$I = \int_0^\infty \frac{\sin^4 x}{x^3} \, dx $$
| Here is how we compute the LT of $\sin^4{t}$ directly. As mentioned in @Mhenni's solution, write as $\sin^4{t} = (e^{i t}-e^{-i t})^4/(2 i)^4$
$$F(u) = \frac{1}{16} \int_0^{\infty} dt \, (e^{i 4 t} - 4 e^{i 2 t} + 6 - 4 e^{-i 2 t} + e^{-i 4 t}) e^{-u t}$$
Evaluate separately and combine judiciously:
$$F(u) = \frac18 \left [ \frac{u}{u^2+16} - 4 \frac{u}{u^2+4} + \frac{3}{u}\right]$$
As also mentioned above, $g(u) = u^2/2$ (This is a direct look-up in the table, or it may be evaluated using residue theory.) Now, when we multiply this through, it looks like the integral will be divergent. However, we may cancel out the $u$ piece by using the fact that $u^2/(u^2+a^2) = (1/a^2) (1-1/(u^2+a^2))$. We then get
$$\begin{align}F(u) g(u) &= \frac{1}{16} \frac{u^3}{u^2+16} - \frac14 \frac{u^3}{u^2+4} + \frac{3}{16} u \\ &= \frac{u}{16} - \frac{u}{u^2+16} - \frac{u}{4} + \frac{u}{u^2+4}+\frac{3}{16}u\\ &= \frac{u}{u^2+4} - \frac{u}{u^2+16} \end{align}$$
To do the integral, note that we need to be careful because, taken individually, the integrals diverge - but the divergences cancel. Thus write
$$\begin{align}\int_0^{\infty} du \, F(u) g(u) &= \lim_{R \to \infty} \left (\int_0^R du\, \frac{u}{u^2+4} - \int_0^R du\, \frac{u}{u^2+16} \right ) \\ &= \frac12 \lim_{R \to \infty}\left [ \log{(R^2+4)} - \log{4} - \log{(R^2+16)} + \log{16}\right]\\ &= \frac12 [\log{2^4} - \log{2^2} ] \\ &= \log{2}\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve a cubic polynomial (given one root is four times a second root)? So, I've been stuck on a question for a long time now:
"Solve the equation $10x^3 + 23x^2 + 5x - 2 = 0$ given that one root is four times a second root."
How would you go about solving this?
Any help would be greatly appreciated.
| Solve $10x^3 + 23x^2 + 5x - 2 = (x-u)(x-4u)(x-v)$.
This can also be solved using synthetic division.
\begin{array}{r|rrrr}
& 10 & 23 & 5 & -2 \\
u & 0 & 10u & 23u+10u^2 & 5u+23u^2+10u^3 \\
\hline
& 10 & 23+10u & 5+ 23u+ 10u^2 & \color{red}{-2+5u+23u^2+10u^3 = 0} \\
4u & 0 & 40u & 92u+200u^2 \\
\hline
& 10 & 23+50u & \color{red}{5+115u+210u^2=0}
\end{array}
We find $5+115u+210u^2=0 \implies u \in \{-\frac 12, -\frac{1}{21} \}$
We compute that
$\left. -2+5u+23u^2+10u^3 \right|_{u=-\frac{1}{21}} \ne 0$
and
$\left. -2+5u+23u^2+10u^3 \right|_{u=-\frac 12} = 0$.
So $u = -\dfrac 12$ and $4u =-2$. Redoing the table above and continuing, we get
\begin{array}{r|rrrr}
& 10 & 23 & 5 & -2 \\
-\frac 12 & 0 & -5 & -9 & 2 \\
\hline
& 10 & 18 & -4 & 0 \\
-2 & 0 & -20 & 4 \\
\hline
& 10 & -2 & 0 \\
\frac 15 & 0 & 2 \\
\hline
& 10 & 0
\end{array}
So the roots are $\left\{ -\dfrac 12, -2, \dfrac 15 \right\}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to calculate $\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}}$ when $x>1$? Numerically, it looks that the limit is
$$\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}} = \frac{1}{\sqrt{2}}\log(1 - x) + cte $$,
but I have not been able to demonstrate it analytically. Does anyone have a idea on how to deal with this limit?
| I have found how to handle with this problem. The key point is to use the following integral representation of $\frac{1}{\sqrt{x-y}}$:
$\frac{1}{\sqrt{x-y}} = \frac{1}{\sqrt{\pi}}\int_{0}^{\infty}\frac{da}{\sqrt{\pi}} e^{-a(x-y)}$.
Using Mathematica, we can carry out the integral over $y$ and $a$ (in this order), obtaining a complex result in terms of generalized hypergeometric functions
$\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}} =
\frac{8 \sqrt{\frac{2}{\pi }} \left(2 x \Gamma \left(\frac{5}{4}\right) \Gamma \left(\frac{9}{4}\right) \, _3F_2\left(\frac{1}{4},\frac{3}{4},\frac{5}{4};\frac{1}{2},\frac{7}{4};\frac{1}{x^2}\right)+\Gamma \left(\frac{7}{4}\right)^2 \, _3F_2\left(\frac{3}{4},\frac{5}{4},\frac{7}{4};\frac{3}{2},\frac{9}{4};\frac{1}{x^2}\right)\right)}{15 x^{3/2}}.$
After that, the limit is easily obtained by Mathematica as
$\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}} = \frac{1}{\sqrt{2}}\log(1-x) + \frac{75 \log (2)-4 \left(5 \, _3F_2\left(\frac{1}{4},1,\frac{3}{2};\frac{5}{4},\frac{7}{4};1\right)+\, _3F_2\left(\frac{3}{4},1,\frac{3}{2};\frac{7}{4},\frac{9}{4};1\right)\right)}{15 \sqrt{2}}$,
which is the expected result.
Thanks by your help.
| {
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For every prime of the form $2^{4n}+1$, 7 is a primitive root. What I want to show is the following statement.
For every prime of the form $2^{4n}+1$, 7 is a primitive root.
What I get is that
$$7^{2^{k}}\equiv1\pmod{p}$$
$$7^{2^{k-1}}\equiv-1\equiv2^{4n}\pmod{p}$$
$$7^{2^{k-2}}\equiv(2^{n})^2\pmod{p}$$
Thus $(\frac{2^n}{p})=(\frac{7^{2^{k-2}}}{p})=1$.
I think that $7$ is important because $7$ is a primitive root but I don't know how to use $7$.
| Case $n=0$ is trivial and satisfies our assumption.
So consider all possible outcomes when the natural number $n = 3k + t$ for some $k \in \mathbb{Z}$ and $t \in \{\, 0,1,2 \,\}$:
\begin{align*}
& p = 2^{4\cdot 3k} + 1 = 2^{12k}+1 = (2^{4k} + 1)(2^{8k} - 2^{4k} + 1) \not\in \mathbb{P} \quad (\text{prime}); \\
& p = 2^{4\cdot (3k+1)} + 1 = 8^{4k} \cdot 16^1 + 1 \equiv 1^{4k} \cdot 2^1 + 1 = 3 \mod{7}; \\
& p = 2^{4\cdot (3k+2)} + 1 = 8^{4k} \cdot 16^2 + 1 \equiv 1^{4k} \cdot 2^2 + 1 = 5 \mod{7}.
\end{align*}
If $p \equiv 3 \mod{7}$ then we have $(p/7) = (3/7) = -(7/3) = -(1/3) = -1$.
If $p \equiv 5 \mod{7}$ then we have $(p/7) = (5/7) = (7/5) = (2/5) = -1$.
Since obviously $p \equiv 1 \mod{4}$, quadratic reciprocity law gives $(7/p) = (p/7)$ and
\begin{equation*}
-1 = (p/7) = (7/p) \equiv 7^{\frac{p-1}{2}} = 7^{2^{4n-1}}
\mod{p}.
\end{equation*}
This means that $\operatorname{ord}_p(7) \nmid 2^{4n-1}$ while at the same time $\operatorname{ord}_p(7) \mid \varphi(p) = 2^{4n}$.
Then the only possibility is that $\operatorname{ord}_p(7) = \varphi(p)$ and $7$ is a primitive root modulo $p$ by definition.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorize in R[x] I have the polynomial $x^8+1$, I know that there's no root for solve this in $\Bbb R[x]$ but i want to factorize this to the minimal expression. This is possible or this is irreducible?
| Fun fact:
$$x^4+1=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1).$$
This can be derived by setting $x^4+1$ equal to a product of two monic quadratics with unknown coefficients and then solving for said coefficients little by little. As a consequence,
$$x^8+1=(x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1).$$
We can go further. For example, set
$$x^4+\sqrt{2}x^2+1=(x^2+ax+b)(x^2-ax+1/b),$$
yielding
$$\begin{cases}-a^2+b+1/b & =\sqrt{2} \\ a/b-ab & = 0 \end{cases} $$
Rule out $a=0$ to obtain $b=\pm1$ from the second equation, then plug those candidates into the first equation yielding $a=\sqrt{\pm2-\sqrt{2}}=\sqrt{2-\sqrt{2}}$ with $b=1$. The second factor of $x^8+1$ that is written above is similarly reducible. I leave the details as an exercise. The full factorization is
$$\begin{array}{lll} x^8+1 & = & \left(x^2+\sqrt{2-\sqrt{2}}x+1\right) \times \left(x^2-\sqrt{2-\sqrt{2}}x+1\right) \\ & \times & \left(x^2+\sqrt{2+\sqrt{2}}x+1\right) \times\left(x^2-\sqrt{2+\sqrt{2}}x+1\right). \end{array}$$
over the real numbers $\bf R$. With the quadratic formula applied to the above you can get the roots to $x^8+1$ exactly (they are precisely the primitive $16$th roots of unity) in the form of nested radicals, and hence the full factorization in $\bf C$.
By the way, I should mention that the only nonlinear polynomials over $\bf R$ that are irreducible are quadratics with negative discriminant. This is because the nonreal roots of any real-coefficient polynomial can be paired off into conjugates, and then each conjugate pair of linear factors can be put together to obtain a real quadratic, thus every real-coefficient polynomial can be factored into real linear and quadratic factors.
Over $\bf Q$, as Belgi notes, a simple shift allows Eisenstein's criterion to apply.
| {
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What would be the value of $a$ and $b$ in following rational expression? If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,
How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?
| What was written was
$$\frac{5 + 2\sqrt{3}}{7 + \sqrt{3}} = a - \sqrt{3b},$$
but I'm going to boldly assume that's a typo and what was really intended was
$$\frac{5 + 2\sqrt{3}}{7 + \sqrt{3}} = a - \sqrt{3}b.$$
Multiply the numerator and denominator by the conjugate of $7+\sqrt{3}$, which is $7-\sqrt{3}$:
\begin{align}
& \frac{5 + 2\sqrt{3}}{7 + \sqrt{3}} = \frac{(5 + 2\sqrt{3})(7 - \sqrt{3})}{(7 + \sqrt{3})(7 - \sqrt{3})} = \frac{29 + 9\sqrt{3}}{7^2-3} = \frac{29 + 9\sqrt{3}}{46} \\[10pt]
& = \frac{29}{46} + \frac{9}{46}\sqrt{3} = \frac{29}{46} - \frac{-9}{46}\sqrt{3}.
\end{align}
So $a=\dfrac{29}{46}$ and $b=\dfrac{-9}{46}$.
| {
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Matrix Equation, Solving for Variables. I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question:
$
\begin{align}
\begin{bmatrix}
a-b & b+c\\
3d+c & 2a-4d
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 & 1\\
7 & 6
\end{bmatrix}
\end{align}
$
What I have done so far is:
$
\begin{align}
\begin{bmatrix}
a-b-8 & b+c-1\\
3d+c-7 & 2a-4d-6
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}
\end{align}
$
And the solving for variables,
$$
a-b-8 = 0
$$
$$
a-b = 8
$$
$$
a = \frac{8}{-b}
$$
$$
b+c-1=0
$$
$$
b+c=1
$$
$$
b=\frac{1}{c}
$$
$$
3d+c-7=0
$$
$$
3d+c=7
$$
$$
3d=\frac{7}{c}
$$
$$
d=\frac{7}{3c}
$$
$$
2a-4d-6=0
$$
$$
2a-4d=6
$$
$$
\frac{16}{-2b}-\frac{28}{12c}=6
$$
Am I going about this correctly? Or am I just doing this completely incorrect?
| Error 1: $a-b=8$ is not the same as $a = \frac{8}{-b}$ but $a = 8 + b$
Error 2: $b+c=1$ is not the same as $b = \frac{1}{c}$ but $b = 1-c$
Error 3: $3d+c=7$ ... same as errors 1 and 2
Error 4: $2a-4d=6$ Incorrect substitution because of previous errors. Fix the previous errors and then subsititue to fix this error
| {
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} |
worked conservative question. Can someone double check? Can you please check this question for me because my answer is different to my friends and they say it's wrong...
Given the equation: $G = (x^3 -3xy^2)\vec{i} +(y^3-3x^2y)\vec{j} +2\vec{k}$ is conservative, find the potential function for the fields that are conservative.
My answer
$f: \nabla f=F \\
fx=x^3-3xy^2 \\
fy=y^3-3x^2y \\
fz=z$
$\int(x^3-3xy^2)dx
=x^4/4-3x^2y^2/2 +h(y, z) \\
fy \text{ or } f'y= -3x^2 y +hy(y, z) \\
fy=y^3-3x^2y= -3x^2y+hy(y, z) \\
hy(y,z)=y^3$
$\int y^3 dy = y^4/4 +gz$
$fz=0$ because no $z$ term ^
$\int 0dz =c$
$F= x^4/4-3x^2y^2/2 +y^4/4+c$ <-potential function
my other friend solution was $F=x^4/4 -3x^2y^2/2 +z^2/2+y^4/4$.
| If we accept that the vector field $F = (x^3 -3xy^2)\vec{i} +(y^3-3x^2y)\vec{j} +2\vec{k}$ is conservative, so there is a function, say $f$ such that, $\nabla f=F$. So $$f_x=x^3-3xy^2,~~f_y=y^3-3x^2y,~~f_z=z$$ Assume that $f_x=x^3-3xy^2$, so $$f(x,y,z)=\frac{x^4}4-\frac{3}2x^2y^2+h(y,z)$$ Then $f_y=-3x^2y+h_y(y,z)$ but $f_y=y^3-3x^2y$, so we get $$-3x^2y+h_y(y,z)=y^3-3x^2y\Longrightarrow h_y(y,z)=y^3\to h(y,z)=\frac{y^4}4+g(z)$$ Therefore, we have up to now: $$f(x,y,z)=\frac{x^4}4-\frac{3}2x^2y^2+\frac{y^4}4+g(z)$$ Now take a differentiation with respect to $z$, we have: $$f_z=g'(z)$$ but $f_z=z$, so $g'(z)=z$ and then $g(z)=\frac{z^2}2$. Hence $$f(x,y,z)=\frac{x^4}4-\frac{3}2x^2y^2+\frac{y^4}4+\frac{z^2}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/399870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to solve this simultaneous equation of $3$ variables. I've stuck in this equation system. No clue how to start ?
$$\begin{eqnarray}
x+y+z &=&a+b+c\tag{1} \\
ax+by+cz &=&a^{2}+b^{2}+c^{2}\tag{2} \\
ax^{2}+by^{2}+cz^{2} &=&a^{3}+b^{3}+c^{3}\tag{3}
\end{eqnarray}$$
Find the value of $x,y,z$ is in the form of $a,b$ and $c$.
I want to know steps of solution.
|
I'm supposed to solve it using simultaneous eqn method like
elimination,substitution,cross-multiplication .NO knowledge of
matrices
By inspection we see that $(x,y,z)=(a,b,c)$ is a solution of the given system
$$
\begin{array}{l}
\text{Eq. 1}\qquad x+y+z=a+b+c \\
\text{Eq. 2}\qquad ax+by+cz=a^{2}+b^{2}+c^{2} \\
\text{Eq. 3}\qquad ax^{2}+by^{2}+cz^{2}=a^{3}+b^{3}+c^{3}.
\end{array}
\tag{0}$$
The other solution can be found as follows. Solve Eq. 1 for $z$. Multiply original Eq. 1 by $a$, subtract Eq. 2 and solve for $z$. This results in
$$
\begin{array}{l}
z=a+b+c-x-y,
\end{array}
\tag{1}$$
$$
\begin{array}{l}
z=\frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c} .
\end{array}
\tag{2}$$
Equate the right hand sides of $(1)$ and $(2)$
$$
\begin{array}{l}
\frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c}=a+b+c-x-y,
\end{array}
\tag{3}$$
and solve for $x$
$$
x=\frac{c-b}{a-c}y+\frac{-ac+b^{2}+a^{2}-bc}{a-c}.
\tag{4}$$
Substitute $x,z$ in $(0)$, Eq. 3, using $(4)$ and $(2)$
$$
a\left( \frac{c-b}{a-c}y+\frac{-ac+b^{2}+a^{2}-bc}{a-c}\right)
^{2}+by^{2}+c\left( \frac{b-a}{a-c}y+\frac{ab+ac-b^{2}-c^{2}}{a-c}\right)
^{2}=a^{3}+b^{3}+c^{3}.
\tag{5}$$
Solving for $y$ we get$^1$ the solution $y=b$ and the solution
$$y=\frac{B}{D},\tag{6}$$
where
$$
B=-2a^{3}c+2a^{3}b-a^{2}b^{2}-a^{2}bc+4a^{2}c^{2}-2acb^{2}-2ac^{3}+ab^{3}-abc^{2}+2bc^{3}+b^{3}c-c^{2}b^{2}$$
$$
D=a^{2}c+ac^{2}-6abc+a^{2}b+bc^{2}+ab^{2}+b^{2}c
$$
Finally substitute $y=b$ and $y=B/D$ in $(4)$ and $(2)$. We get the solutions $(x,z)=(a,c)$, and
$$(x,z)=\left(\frac{A}{D},\frac{C}{D}\right),\tag{7}$$
where
$$
A=a^{3}c+a^{3}b-2a^{2}bc-a^{2}b^{2}-a^{2}c^{2}+2ac^{3}-abc^{2}+2ab^{3}-acb^{2}-2bc^{3}-2b^{3}c+4c^{2}b^{2}
$$
$$
C=2a^{3}c-2a^{3}b+4a^{2}b^{2}-a^{2}c^{2}-a^{2}bc-acb^{2}-2ab^{3}+ac^{3}-2abc^{2}+bc^{3}+2b^{3}c-c^{2}b^{2}.
$$
So the two solutions of $(0)$ are: $$(x,y,z)=(a,b,c)\qquad\text{and }\qquad(x,y,z)=\left(\frac{A}{D},\frac{B}{D},\frac{C}{D}\right).$$
--
$^1$ Eq. $(5)$ is equivalent to
$$\begin{equation*}
\left( cb^{2}+c^{2}b+ac^{2}+ca^{2}+ab^{2}-6acb+ba^{2}\right) (y-b)\left( y-\frac{B}{D}\right) =0.
\end{equation*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding eccentricity of an ellipse from latus rectum The latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is the same as latus rectum of a parabola $y^2=4cx$ . Find eccentricity of the ellipse .
| HINT:
from this and this, the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $2a(1-e^2)$ and $b^2=a^2(1-e^2)$ where $a$ is Semi major Axis, $b$ is the Semi-minor Axis and $e$ is the Eccentricity
and the length of the latus rectum of the parabola $y^2=4ax$ is $4a$
Can you utilize the relations to find the value of $e$
Check separately for cases when $a\ge b$ and when $a<b$
EDIT: after a drastic change in the question $y^2=4ax$ to $y^2=4cx$
Now WLOG, we can choose $a\ge b$
As the focus of the parabola is $(c,0)$ and those of the ellipse are $(\pm ae,0)$
$c=ae$
Now, $2a(1-e^2)=4c=4ae\implies 1-e^2=2e\implies e^2+2e-1=0$
$\implies e=\frac{-2\pm{\sqrt{2^2-4(-1)1}}}{2\cdot1}=-1\pm\sqrt2$
As $0\le e<1, e=\sqrt2-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
addition with a variable (mod) Given $2+x \equiv 7 \pmod 3$.
$2 + 0 = 2$
$2 + 1 = 3$
$2 + 2 = 4$
.
.
.
$2 + 5 = 7$
so, the answer will be $x = 5, 8, 11, 14, 17,\dots$
Is this correct? Because somebody told me the answer should be $x = 2, 5, 8, 11, 14, 17,\dots$
| Hint: $\ x\equiv 7-2 \equiv \color{#c00}5 \pmod 3 = \color{#c00}5 + 3\,\Bbb Z = \{\,\ldots,-4, -1, 2, \color{#c00}5, 8, 11,14,\ldots\}$
Or: $\ x\equiv 5\pmod{3}\!\iff\! 3\mid x\!-\!5\!\iff\! 3n = x\!-\!5 \!\iff\! x = 5\! +\! 3n,\ $ for some $\, n\in \Bbb Z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the matrices of $R$ and $R\circ R$ with respect to the basis $(e_1,e_2,e_3,e_4)=(1,x,\frac{1}{2}x^2,\frac{1}{6}x^3)$ I am unsure how to calculate the basis matrices of the linear map defined below. I appreciate your help.
Let $V=\mathbb{Q}[x]_{\le3}$ be the set of polynomials over $\mathbb{Q}$ of degree at most $3$. Define $R:V \rightarrow V$ by $R(f)=xf''-5f$ where $f':=df/dx$. Prove that $R$ is linear (here the base field $K=\mathbb{Q}$). Calculate the matrices of $R$ and $R\circ R$ with respect to the basis $(e_1,e_2,e_3,e_4)=(1,x,\frac{1}{2}x^2,\frac{1}{6}x^3)$.
I proved that it's linear.
Now, $R(1) =-5$, $R(x) =-5x$, $R\left(\frac{1}{2}x^2\right) = x-\frac{5}{2}x^2$, $R\left(\frac{1}{6}x^3\right) = x^2-\frac{5}{6}x^2$.
Then the basis matrix of $R$ is as follows
$$\begin{pmatrix}
-5 & 0 & 0 & 0\\
0 & -5 & 1 & 0\\
0 & 0 & -5/2 & 1\\
0 & 0 & 0 & -5/6
\end{pmatrix}.$$
I want to find the matrices of $R \circ R$, so I thought of multiplying the matrix of R together
$$\begin{pmatrix}
-5 & 0 & 0 & 0\\
0 & -5 & 1 & 0\\
0 & 0 & -5/2 & 1\\
0 & 0 & 0 & -5/6
\end{pmatrix}\begin{pmatrix}
-5 & 0 & 0 & 0\\
0 & -5 & 1 & 0\\
0 & 0 & -5/2 & 1\\
0 & 0 & 0 & -5/6
\end{pmatrix} = \begin{pmatrix}
25 & 0 & 0 & 0\\
0 & 25 & -15/2 & 1\\
0 & 0 & 25/4 & -20/6\\
0 & 0 & 0 & 25/36
\end{pmatrix}.$$
I also tried of $R(R(1)) =25$, $R(R(x)) =25x$, $R\left(R\left(\frac{1}{2}x^2\right)\right) = -10x+\frac{25}{2}x^2$, $R\left(R\left(\frac{1}{6}x^3\right)\right) = -8x-5x^2-\frac{25}{6}x^3$ which results in the matrix
$$\begin{pmatrix}
25 & 0 & 0 & 0\\
0 & 25 & -10 & -8\\
0 & 0 & 25/2 & -5\\
0 & 0 & 0 & 25/6
\end{pmatrix}.$$
Which way is the correct one? Thank you for your time.
| The matrix of $R$ is wrong.
For example $R(e_3)=R\left(\frac{1}{2}x^2\right)=x-\frac52x^2=x-5\cdot\left(\frac12x^2\right)=e_2-5e_3$...
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
which parameters always make this rational equation evenly divisible? Hi guys I have the following equation:
$$x = \dfrac{a + b \times c - b}{c}$$
This is what I know about each variable:
$$a \ge 64$$
$$b \ge 0$$
$$8 \le c \le a$$
My questions is there a concise way for me to pick a, b, and c so that x will always be a positive integer >= 8.
For example if $a = 64$; $b= 0$; $c = 8$ (the smallest possible values for each)
$x = (64 + 0 \times 8 - 0)/8$
$x = 8$ which is a positive integer $\ge 8$
However, if let's say $a = 200$; $b = 2$; $c = 10$
$x = (200 + 2 \times 10 - 2)/10$
$x = \dfrac{218}{10}$ which is $21.8$ and is not a positive integer greater than or equal to $ 8$
How can I always know that $x$ will be a positive integer?? Thanks!
| As $$x = \frac{a+ b\cdot c -b}{c} = \frac{a-b}{c} + \frac{b\cdot c}{c} = \frac{a-b}{c} + b,$$ we can say that $x$ will be an integer whenever $a-b$ is divisible by $c$. That is, we need an integer $n$ such that $$a-b=cn \leftrightarrow a = cn + b.$$ Finally, as we want $x \geq 8$ this means we need $$\frac{a-b}{c} + b \geq 8.$$ By solving for $a$, we have $a \geq b -bc +8c$.
Combining our expressions involving $a$ results in $$cn+b \geq b-bc+8c$$ $$n+b\geq 8.$$ Now you may choose any $b \geq 0$ and any $n$ satisfying $n+b \geq 8$ and you will have $x \geq 8$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove that $a_{n}$ must be of the form $a^2+b^2$? let $a_{1}=1,a_{2}=2,a_{3}=5$,and
$$a_{n}=3a_{n-1}a_{n-2}-a_{n-3}$$
show that
$a_{n}=a^2+b^2,a,b\in N$
while $a_{1}=0^2+1^2,a_{2}=2=1^2+1^2,a_{3}=5=2^2+1^2,a_{4}=29=5^2+2^2,a_{5}=433=17^2+12^2$ and so on
| Hopefully you already know the very nice theorem from number theory that states that a natural number can be written as the sum of two squares iff every prime divisor of the number which is equal to $\,3\pmod 4\,$ appears to an even power (primes which are $\;1\pmod 4\;$ or $\;2\;$ make no problems).
You can now use some induction, proving/pointing out that:
-- If $\,p\,,\,q\,$ are two primes $\,p,q=1\pmod 4\implies pq=1\pmod 4\,$
-- The only primes dividing $\,a_n\,,\,\,n\ge 2\;$ , are $\,1\pmod 4\;$ or $\,2\,$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/408201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
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Solving $\int_0^{\infty} \sin(ax^2)\sin(b/x^2)\,\mathrm dx$ The integral is:
$$
\int_{0}^{\infty}\sin\left(ax^{2}\right)\sin\left(\frac b {x^{2}}\right)\,\mathrm{d}x
$$
I already know the answer but I cant prove it! ( You can get the answer using Maple ). I think it should be simplified to a couple of Gaussian integrals ( $\operatorname{erf}$ functions ) but it's not so obvious.
| $$\int_0^{\infty} \sin ax^2\sin bx^{-2}\,dx=\frac{1}{2}\int_0^{\infty} \cos \left(ax^2-\frac{b}{x^2}\right)\,dx-\frac{1}{2}\int_0^{\infty} \cos \left(ax^2+\frac{b}{x^2}\right)\,dx$$
In general:
$$\begin{aligned} \int_{0}^{\infty}f \left(p x-\frac{q}{x}\right)^2\,dx &= \sqrt{\frac{q}{p}}\int_{0}^{\infty }f \left(\sqrt{pq} x- \frac{\sqrt{pq}}{x}\right)^2\,dx \quad(\textstyle x\to \sqrt{\frac{q}{p}}x )\\ &=\sqrt{\frac{q}{p}}\left[ \int_{0}^{1} f \left(\sqrt{pq} x-\frac{\sqrt{pq}}{x} \right)^2\,dx+ \int_{1}^{\infty} f \left(\sqrt{pq}x-\frac{\sqrt{pq}}{x}\right)^2\,dx \right]\\&= \sqrt{\frac{q}{p}}\left[\int_{1}^{\infty} \left(1+\frac{1}{x^2}\right)f \left(\sqrt{pq} x-\frac{\sqrt{pq}}{x} \right)^2\,dx \right]\quad( x\to x^{-1})\\&=\frac{1}{p} \int_{0}^{\infty} f (t^2 )\,dt \quad \big(t=\sqrt{pq} x-\sqrt{pq}x^{-1}\big) \end{aligned}$$
Observe that:
$$\begin{aligned}\int_0^{\infty} \cos \left(ax^2+\frac{b}{x^2}\right)\,dx&=\cos 2\sqrt{ab}\int_{0}^{\infty} \cos \left(\sqrt{a}x-\frac{\sqrt{b}}{x}\right)^2\,dx\\&\;-\sin 2\sqrt{ab}\int_{0}^{\infty} \sin \left(\sqrt{a}x-\frac{\sqrt{b}}{x}\right)^2\,dx\end{aligned}$$
Now since $\displaystyle \int_0^{\infty}\cos x^2\,dx=\int_0^{\infty}\sin x^2\,dx=\sqrt{\frac{\pi}{8}}$ using our previous result we have:
$$\int_0^{\infty} \cos \left(ax^2+\frac{b}{x^2}\right)\,dx=\sqrt{\frac{\pi}{8a}}\left(\cos 2\sqrt{ab}-\sin 2\sqrt{ab}\right)$$
Similarly, writing:
$$\begin{aligned}\int_0^{\infty} \cos \left(ax^2-\frac{b}{x^2}\right)\,dx&=\cosh 2\sqrt{ab}\int_{0}^{\infty} \cos \left(\sqrt{a}x-i\frac{\sqrt{b}}{x}\right)^2\,dx\\&\;-\sinh 2\sqrt{ab}\int_{0}^{\infty} \sin \left(\sqrt{a}x-i\frac{\sqrt{b}}{x}\right)^2\,dx\end{aligned}$$
We get:
$$\int_0^{\infty} \cos \left(ax^2-\frac{b}{x^2}\right)\,dx=\sqrt{\frac{\pi}{8a}}\left(\cosh 2\sqrt{ab}-\sinh 2\sqrt{ab}\right)$$
Putting everything together:
$$\int_0^{\infty} \sin ax^2\sin bx^{-2}\,dx=\frac{1}{2}\sqrt{\frac{\pi}{8a}}\Bigg\{\big(\cosh 2\sqrt{ab}-\sinh 2\sqrt{ab}\big)-\big(\cos 2\sqrt{ab}-\sin 2\sqrt{ab}\big)\Bigg\}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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"answer_id": 3
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How do I find the exact value of $\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}$? I know that $\cos(6\phi)\equiv32c^6-48c^4+18c^2-1$ where $c=\cos\phi$.
I also know that when $\cos(6\phi)=0$, then $\phi=\frac{k\pi}{12}$ ($k = 1,3,5,7,9,11$).
How do I find the exact value of:
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$
| As $\frac{11\pi}{12}=\pi-\frac\pi{12}$ and $\frac{7\pi}{12}=\pi-\frac{5\pi}{12}$
and $\cos(\pi-x)=-\cos x,$
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$
$$=\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)\left( - \cos\left(\frac{5\pi}{12}\right)\right)\left( -\cos\left(\frac{\pi}{12}\right)\right)$$
$$=\left(\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)\right)^2$$
Now as $\frac{5\pi}{12}=\frac\pi2-\frac\pi{12}$ and $\sin2x=2\sin x\cos x,$
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)=\cos\left(\frac{\pi}{12}\right) \sin\left(\frac{\pi}{12}\right)=\frac{\sin \frac\pi6}2=\frac{\frac12}2=\frac14$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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3D Geometry Question In $3$-dimensional Geometry, if angle made of line segment $OP$ with $X,Y,Z$-axis are in
$1:2:3$, then what is the angle made by line segment with $Y$-axis?
My Solution:
Let $\alpha,\beta$ and $\gamma$ be the angle made by line with $X,Y,Z$-axis, respectively. Then
$$\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma = 1$$
Now given $\alpha = \theta$ and $\beta = 2\theta$ and $\gamma = 3\theta$,
So $\cos^2 \theta+\cos^2 2\theta+\cos^2 3\theta = 1$,
or
$$2\cos^2 \theta+2\cos^2 2\theta+2\cos^2 3\theta = 2,$$
or
$$1+\cos 2\theta+\cos^2 2\theta+1+\cos 6\theta = 2,$$
or
$$1+\cos 2\theta+2\cos^2 2\theta+1+4\cos^3 2\theta-3\cos 2 \theta = 2.$$
Let $\cos 2 \theta = y$, then
$$y+2y^2+4y^3-3y = 0,$$
$$4y^3+2y^2-2y = 0,$$
$$2y(2y^2-y-1) = 0.$$
Maybe I have missed something, but I feel that this method is very tedious.
So could anyone explain me a better method?
Thanks.
| say smallest angle is $\theta$
therefore $\theta+2\theta +3\theta= 180^\circ$.
Sum of the angles of a triangle equal $180^\circ$.
therefore $\theta=30^\circ$ (since $6\theta=180^\circ=> \theta=\dfrac{180}{6}=30^\circ$)
therefore angle made with y axis is $60^\circ$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding Maximum Under Constraint Suppose $a$,$b$,$c$ satisfy $a+b+c=1$ and $a$,$b$,$c\in [0,1]$
Find the maximum value of $(a-b)(b-c)(c-a)$
| Let us rename $a,b,c$ as $x,y,z$. From the constraint $x+y+z=1$ we may express $z$ as $z=1-x-y$ and study the function
$$
f(x,y) = (x-y)(y-z)(z-x) = (x-y)(2y+x-1)(1-2x-y)
$$
in the triangle
$$
T=\{(x,y)\in\mathbb R^2 : x,y\geq 0, ~x+y\leq1\}
$$
The partial derivatives of $f$ are
$$
\begin{cases}
f_x = -6x^2 - 6xy + 3y^2 + 6x -1\\
f_y = -3x^2 + 6xy - 6y^2 - 6y +1
\end{cases}
$$
The critical points of $f$ are the solutions of
$$
\begin{cases}
-6x^2 - 6xy + 3y^2 + 6x - 1 & = 0\\
-3x^2 + 6xy - 6y^2 - 6y + 1 & = 0
\end{cases}
~\Leftrightarrow~
\begin{cases}
-6x^2 - 6xy + 3y^2 + 6x - 1 & = 0\\
3x^2 + y^2 - 2x + 2y & = 0
\end{cases}
$$
from which you have $y=-1\pm\sqrt{1-3x^2+2x}$. Since $0\leq y\leq 1$, no solution $(x,y)$ can be inside $T$, therefore in $T$ $f$ does not have any critical point.
It follows that the maximum of $f$ is achieved on the boundary of $T$, i.e. either when $x=0$, $y=0$ or $y=1-x$:
1. $x=0$
$$
f(0,y) = y(y-1)(2y-1)
$$
when $y$ ranges in $[0,1]$ has maximum in $y=\frac{3-\sqrt3}6$, and
$$
f(0,\textstyle{\frac{3-\sqrt3}6}) ~=~ \frac{\sqrt3}{18}
$$
2. $x=0$
$$
f(x,0) = x(x-1)(1-2x)
$$
when $x$ ranges in $[0,1]$ has maximum in $x=\frac{3+\sqrt3}6$, and
$$
f(\textstyle{\frac{3+\sqrt3}6},0) ~=~ \frac{\sqrt3}{18}
$$
3. $y=1-x$
$$
f(x,1-x) = x(x-1)(2x-1)
$$
is analogous to case 1. w.r. to variable $y$.
Therefore we can conclude that the maximum is achieved when either $(a,b,c)$, $(b,c,a)$ or $(c,a,b)$ is $\big(\textstyle{\frac{3+\sqrt3}6},\textstyle{\frac{3-\sqrt3}6},0\big)$. In any case the maximum is
$$
(a-b)(b-c)(c-a) = \frac{\sqrt 3}{18}
$$
Edit (2): corrected 2 mistakes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Determining bounds for the sum $\sum\limits_{n=1}^\infty \frac{1}{2^n - 3^n }$ I have to give low and high bounds for the following:
$$
\sum_{n=1}^\infty \frac{1}{2^n - 3^n }
$$
How do I determine an upper bound? How can I show this sum exists?
edit: removed erroneous conclusion.
| $\begin{align}
\sum_{n=1}^\infty \frac{1}{3^n - 2^n }
&=\sum_{n=1}^\infty \frac{1}{3^n(1 - (2/3)^n) }\\
&=\sum_{n=1}^\infty \frac{1}{3^n} \sum_{k=0}^\infty(2/3)^{nk}) \\
&=\sum_{n=1}^\infty \frac{1}{3^n}+\sum_{n=1}^\infty \frac{1}{3^n} \sum_{k=1}^\infty(2/3)^{nk}) \\
&= \frac{1/3}{1-1/3}
+ \sum_{m=1}^{\infty} (2/3)^m \sum_{d|m} (1/3)^d\\
&= \frac{1}{2}
+ \sum_{m=1}^{\infty} (2/3)^m \sum_{d|m} (1/3)^d\\
\end{align}
$
Looking at the inner sum,
let
$S(m, r) = \sum_{d|m}r^d$,
where $0 < r < 1$.
$S(m, r) > r$,
since the term $d=1$ always occurs.
$S(m, r) < \sum_{d=1}^{m}r^d
= \frac{r}{1-r}
$,
since more terms are here than
in the original sum.
So
$S(m, 1/3) > 1/3$
and
$S(m, 1/3) < 1/2$.
Therefore
$\begin{align}
\sum_{n=1}^\infty \frac{1}{3^n - 2^n }
&= \frac{1}{2}
+ \sum_{m=1}^{\infty} (2/3)^m S(m, 1/3)\\
&> \frac{1}{2}
+ \sum_{m=1}^{\infty} (2/3)^m (1/3)\\
&> \frac{1}{2}
+ \frac1{3}\sum_{m=1}^{\infty} (2/3)^m\\
&= \frac{1}{2}
+ \frac1{3}\frac{2/3}{1-2/3}\\
&= \frac{1}{2}
+ \frac1{3}2\\
&= \frac{1}{2}
+ \frac{2}{3}\\
&= \frac{7}{6}\\
\end{align}
$
and
$\begin{align}
\sum_{n=1}^\infty \frac{1}{3^n - 2^n }
&< \frac{1}{2}
+ \frac1{2}2\\
&= \frac{3}{2}
\end{align}
$
More accurate bounds for
$S(m, r)$
would result in more accurate bounds
for the sum.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find the coefficient using binomial theorem. What is the coefficient of $x^{20}$ in the expression: $$(x+1)^{10}.(x^2 -1)^8$$
| $$(x+1)^{10}.(x^2 -1)^8=\sum_{i=0}^{10}\binom{10}{i}x^i\sum_{j=0}^{8}\binom{8}{j}x^{2j}(-1)^{8-j}$$
$$i+2j=20$$
$$(i,j)=(10,5),(8,6),(6,7),(4,8)$$
$$\binom{10}{10}\binom{8}{5}(-1)^{3}+\binom{10}{8}\binom{8}{6}(-1)^{2}+\binom{10}{6}\binom{8}{7}(-1)+\binom{10}{4}\binom{8}{8}(-1)^0=$$
$$=-\binom{8}{5}+\binom{10}{8}\binom{8}{6}-\binom{10}{6}\binom{8}{7}+\binom{10}{4}=-266$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 2
} |
Show that $\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$.
Given $x,y,z\geq0$ and $xy+yz+zx=1$. Show that $\displaystyle\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$.
I've tried many things but all failed. The only thing I know is that the equality holds when $x=y=1, z=0$.
Please help. Thank you.
| Without loss of generality, we assume that $x\ge y\ge z$.
Lemma 1:
$$\dfrac{1}{\sqrt{y^2+1}}+\dfrac{1}{\sqrt{z^2+1}}\ge 1+\dfrac{1}{\sqrt{(y+z)^2+1}}$$
Proof: this inequality is equivalent to
$$\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}+\dfrac{2}{\sqrt{(y^2+1)(z^2+1)}}\ge 1+\dfrac{1}{(y+z)^2+1}+\dfrac{2}{\sqrt{(y+z)^2+1}}.$$
Notice that
$$(y+z)^2+1-(y^2+1)(z^2+1)=yz(2-yz)\ge 0,$$
hence
$$\dfrac{2}{\sqrt{(y^2+1)(z^2+1)}}\ge\dfrac{2}{\sqrt{(y+z)^2+1}}.$$
Therefore, it suffices to prove that
$$\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}\ge1+\dfrac{1}{(y+z)^2+1}.$$
And this is equivalent to
$$\dfrac{yz[2-2yz-yz(y+z)^2]}{(y^2+1)(z^2+1)[(y+z)^2+1]}\ge 0.$$
Above is true because
$$2-2yz-yz(y+z)^2=2x(y+z)-yz(y+z)^2=(y+z)[2x-yz(y+z)]\ge (y+z)[2x-x^2(y+z)]=x(y+z)(2-xy-xz)\ge 0.$$
Lemma 2:
$$\dfrac{1}{x+y}+\dfrac{1}{\sqrt{x+z}}\ge\sqrt{y+z}+\sqrt{\dfrac{y+z}{(y+z)^2+1}}.$$
Proof:
\begin{align}\dfrac{1}{x+y}+\dfrac{1}{\sqrt{x+z}}&=\sqrt{y+z}\left(\dfrac{1}{\sqrt{(y+z)(x+y)}}+\dfrac{1}{\sqrt{(z+x)(z+y)}}\right)\\
&=\sqrt{y+z}\left(\dfrac{1}{\sqrt{1+y^2}}+\dfrac{1}{\sqrt{1+z^2}}\right)
\end{align}
and apply Lemma 1, then done!
Thus the original inequality is equivalent to
$$\dfrac{1}{\sqrt{y+z}}+\sqrt{y+z}+\sqrt{\dfrac{y+z}{(y+z)^2+1}}\ge 2+\dfrac{1}{\sqrt{2}}.$$
Letting $$t=\dfrac{1}{\sqrt{y+z}}+\sqrt{y+z}\ge 2,$$ the inequality becomes
$$t+\dfrac{1}{\sqrt{t^2-2}}\ge 2+\dfrac{1}{\sqrt{2}}.$$
This can be shown by the following:
\begin{align}
&t+\dfrac{1}{\sqrt{t^2-2}}-2-\dfrac{1}{\sqrt{2}}
\\
=&t+\dfrac{2\sqrt{2}}{2\sqrt{2}\cdot\sqrt{t^2-2}}-2-\dfrac{1}{\sqrt{2}}
\\
\ge& t+\dfrac{2\sqrt{2}}{2+t^2-2}-2-\dfrac{1}{\sqrt{2}}
\\
=&t+\dfrac{2\sqrt{2}}{t^2}-2-\dfrac{1}{\sqrt{2}}
\\
=&\dfrac{(t-2)(\sqrt{2}t^2-t-2)}{\sqrt{2}t^2}\ge 0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413683",
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"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 0
} |
How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
| This is the two-variable identity $(X+Y)^3 - (X^3 + Y^3) = 3XY(X+Y)$
presented as a formula for $X^3 + Y^3 + Z^3$ when $X+Y+Z=0$ (by setting $Z = -(X+Y)$),
and then using $(X,Y,Z)=(a-b,b-c,c-a)$ to parametrize solutions of $X+Y+Z=0$.
| {
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"source": "stackexchange",
"question_score": "16",
"answer_count": 12,
"answer_id": 0
} |
Prove $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$, given $x+y+z=3$ and $x,y,z\ge0$
Let $x+y+z=3,x,y,z\ge 0$,show that
$$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$$
Additional information
I have seen the following problem:
$x,y,z>0,x+y+z=3$, prove that
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge 6.$$
Without loss of generality we can let $x=\max{\{x,y,z\}}$
Proof: case 1
$x\ge y\ge z$
we can easily prove
$$\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge\sqrt{y^2+x+2}+\sqrt{z^2+z+2}$$
and
$$\sqrt{x^2+y+2}+\sqrt{y^2+x+2}\ge\sqrt{x^2+x+2}+\sqrt{y^2+y+2}$$
so we have
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^2+x+2}\ge \sqrt{x^2+x+2}+\sqrt{y^2+y+2}+\sqrt{z^2+z+2}.$$
Then use
$$\sqrt{x^2+x+2}\ge\dfrac{3}{4}x+\dfrac{5}{4}$$
$$\sqrt{y^2+y+2}\ge\dfrac{3}{4}y+\dfrac{5}{4}$$
$$\sqrt{z^2+z+2}\ge\dfrac{3}{4}z+\dfrac{5}{4}$$
to get the result. Whereas the case 2 when $x\ge z\ge y$ can be proved using the same methods.
Now,I have another idea: using Holder inequality
we have
$$\left(\sum\sqrt{x^2+yz+2}\right)^2\left(\sum\dfrac{x^2+2yz+9}{x^2+yz+2}\right)\ge 36^3$$
$$\Longleftrightarrow \sum\dfrac{x^2+2yz+9}{x^2+yz+2}\le 1296$$
and the following link has some discussion about this problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=538230
and
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=538752&p=3097872#p3097872
and Vasc gave the hint:
$$\sum\sqrt{8(a^2+bc+2)}\ge \sum\sqrt{(3a+b+c)^2+7}\ge 12\sqrt 2$$
How prove this hint?Thank you everyone.
and my other idea is as follows:
let $a=\min(a,b,c)$
we can prove
$$\sqrt{b^2+ca+2}+\sqrt{c^2+ab+2}\geq \sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\tag{1}$$
\begin{align*}
&\sqrt{a^2+bc+2}+\sqrt{(b+c)^2+2a(b+c)+8-(b-c)^2}\\
\geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8} \tag{2}
\end{align*}
Summing up
\begin{align*}
&\sum_{cyc}{\sqrt{a^2+bc+2}}\\
\geq &\sqrt{a^2+\frac{(b+c)^2}{4}+2}+\sqrt{(b+c)^2+2a(b+c)+8}\\
=&\sqrt{a^2+\frac{(3-a)^2}{4}+2}+\sqrt{(3-a)^2+2a(3-a)+8}
\end{align*}
By the way: someone said $(1)$ is wrong? why? can anyone give an example? And hopefully someone can use this method to prove this inequality? Thank you very much!
| Here is an explanation of Vasc's hint that you mentioned, which essentially solves the problem. Reproducing it here as:
$$\sum _{\text{cyc}} \sqrt{8\left(x^2+ \text{yz} + 2\right)}\geq \sum _{\text{cyc}} \sqrt{(3x+y+z)^2+7} = \sum _{\text{cyc}} \sqrt{(2x+3)^2+7}\geq 12\sqrt{2}$$
Now the last inequality follows immediately from Minkowski's inequality and $\sum x=3$, so I will focus only on the first inequality below.
The inequality $\displaystyle \sum _{\text{cyc}} \sqrt{8\left(x^2+ \text{yz} + 2\right)}\geq \sum _{\text{cyc}} \sqrt{(2x+3)^2+7}$ can be established through Schur-concavity of the vector function $f{(\color{blue}{u}}) = \sqrt{u_1}+\sqrt{u_2}+\sqrt{u_3}$. This function is symmetric and concave, hence Schur-concave. Alternately it is not difficult to establish the condition
$$\left(u_i- u_j\right)\left(\partial _{u_i}f-\partial _{u_j}f\right) = -\frac{\left(u_i- u_j\right){}^2}{\sqrt{u_i u_j}} \leq 0.$$
<< If you are not familiar with Schur-concave property, it should be possible to derive similar results using the lemma $\sqrt{x-\epsilon }+\sqrt{y+\epsilon }\geq \sqrt{x}+\sqrt{y}$. This is easily shown to hold if $x-y \ge \epsilon \ge 0$ by squaring. Using the sequence $$[x, y, z] \succ [x - u, y + u, z] \succ [x - u,
y + u - (v + u), z + (u + v)]$$ and applying the lemma twice, one can establish $$\sqrt{x-u}+\sqrt{y-v}+\sqrt{z+u+v}\geq \sqrt{x}+\sqrt{y}+\sqrt{z}$$ and the conditions needed.>>
As $f{(\color{blue}{u}})$ is Schur-concave, if we have $\color{blue}{a} \text{ and } \color{blue}{b}$ two vectors such that $\color{blue}{a}$ majorizes $\color{blue}{b}$, i.e. $\color{blue}{a} \succ \color{blue}{b} \implies f(\color{blue}{a})\leq f(\color{blue}{b})$. In this case we need to show that considered as vectors,
$$\left[(2x+3)^2+7,(2y+3)^2+7, (2z+3)^2+7\right] \succ \left[8\left(x^2+\text{yz}+2\right),8\left(y^2+\text{zx}+2\right), 8\left(z^2+\text{xy}+2\right)\right]$$
If $s = xy+yz+zx$, we have
$$\sum _{\text{cyc}} \sqrt{8\left(x^2+\text{yz}+2\right)}= \sum _{\text{cyc}} \sqrt{8x^2+8\text{yz} + (x+y+z)^2+7} = \sum _{\text{cyc}} \sqrt{(3x+y+z)^2+7+ 4(2\text{yz}- \text{xy}-\text{zx})} = \sum _{\text{cyc}} \sqrt{(2x+3)^2+7+ 4(3\text{yz}-s)}$$
Let $\displaystyle a = (2x+3)^2+7, b = (2y+3)^2 + 7 \text{ and } c = (2z+3)^2+7$. Also let $u = 4(s-3\text{yz}) \text{ and } v = 4(s-3\text{zx})$. Now $\displaystyle x \geq y \geq z \Longrightarrow \text{xy} \geq \text{zx} \geq \text{yz} \Longrightarrow u \geq 0$ and $u+v \geq 0$ and the deviations are in reverse order to the ordering of $a, b, c$- hence giving the conditions needed for majorization. It may be noted that in a crude sense, the deviations have mean $0$, and are ordered so as to bring the components "closer", hence increasing the concave function's value.
As $[a, b, c] \succ [a-u, b-v, c+(u+v)]$, we have $\displaystyle \sum _{\text{cyc}} \sqrt{8\left(x^2+ \text{yz} + 2\right)}\geq \sum _{\text{cyc}} \sqrt{(2x+3)^2+7}$.
Hope there is a simpler method to establish this. Another possible approach could be to directly check if $\displaystyle F(x, y, z) = \sum _{\text{cyc}} \sqrt{x^2+ \text{yz} + 2}$ is Schur-convex, in which case any allowable $[x, y, z]\succ [1, 1, 1]$ and the entire proof is done. $F$ is symmetric, but we also need to check if $x-y)\left(\partial _xF -\partial _yF\right)\geq 0$ for this.
| {
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"url": "https://math.stackexchange.com/questions/417573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 1
} |
Geometric multiplicity of repeated Eigenvalues I am still finding it difficult to determine the geometric multiplicity for repeated eigenvalues and the resultant eigenspace. For example, I am not quite sure what to do with the following matrix, where repeated Eigenvalues $\lambda_1 = \lambda_2 = 5$:
$$A=\begin{bmatrix}
5 & -4 & 0\\
1 & 0 & 2\\
0 & 2 & 5
\end{bmatrix}, [A-\lambda I] = \begin{bmatrix}
0 & -4 & 0\\
1 & -5 & 2\\
0 & 2 & 0
\end{bmatrix}$$
It is not obvious how to determine the Eigenvectors from this, as there are no free variables, and moving $e_2$ for example (for the first row) such that $-4e_2 = 0 \rightarrow e_2 = 0$ shows that all other values result to zero as well (which is not a valid eigenvector). How does one go about determining the geometric multiplicity and the Eigenspace with such a matrix?
| We are given:
$$A=\begin{bmatrix}
5 & -4 & 0\\ 1 & 0 & 2\\0 & 2 & 5
\end{bmatrix}$$
We form and solve: $|A-\lambda I|=\begin{bmatrix}0 & -4 & 0\\ 1 & -5 & 2\\0 & 2 & 0\end{bmatrix} = 0$
This yields a characteristic polynomial and eigenvalues as:
$$-(\lambda-5)^2 \lambda = 0 ~~~\rightarrow ~~~ \lambda_1 = 0, \lambda_{2,3} = 5$$
We have multiplicities of $1$ and $2$ for those eigenvalues.
To find the eigenvectors, we generally solve $[ A - \lambda_i I]v_i = 0$, but since we have a repeated eigenvalue, we may need to change that strategy and find a generalized eigenvalue.
So, for $\lambda_1 = 0$, we have:
$[A- 0I]v_1 = \begin{bmatrix}5 & -4 & 0\\ 1 & 0 & 2\\0 & 2 & 5\end{bmatrix}v_1 = 0$
Doing row-reduced-echelon-form (RREF), yields:
$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & \dfrac{5}{2} \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$
Thus, $b = -\dfrac{5}{2}c, a = -2c \rightarrow ~~\text{let}~~ c = 2 \rightarrow b = -5, a= -4, v_1 = (-4,-5,2)$.
Repeating this same process for the second eigenvalue, $\lambda_2 = 5$, we have as RREF:
$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}v_2 = 0$
So, $b = 0$, $a = -2c$, let $c = 1 ~~\rightarrow a = -2, v_2 = (-2,0,1)$
Unfortunately, we cannot get another linearly independent eigenvector, so need to get a generalized one, by doing $[A - \lambda_3 I]v_3 = v_2$ (this does not always work), so we have:
$\begin{bmatrix}0 & -4 & 0 \\ 1 & -5 & 2 \\0 & 2 & 0\end{bmatrix}v_3 = \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}$
After RREF, we arrive at:
$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}v_3 = \begin{bmatrix} \dfrac{5}{2} \\ \dfrac{1}{2} \\ 0 \end{bmatrix}$
So, we have: $a = \dfrac{5}{2} -2c, b = \dfrac{1}{2} \rightarrow ~~ \text{let} ~~ c = 0 \rightarrow a = \dfrac{5}{2}, b = \dfrac{1}{2}$, thus $v_3 = (\dfrac{5}{2},\dfrac{1}{2},0)$
You should get your hands around the above regarding your algebraic versus geometric multiplicities.
Putting all of this together, we have the eigenvalue/eigenvector pairs:
*
*$\lambda_1 = 0, v_1 = (-4, -5, 2)$
*$\lambda_2 = 5, v_2 = (-2, 0, 1)$
*$\lambda_3 = 5, v_3 = (\dfrac{5}{2},\dfrac{1}{2},0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417976",
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"question_score": "9",
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Every perfect cube is the difference of two perfect squares? How would you prove this without induction? I know that one easy way is using Al Kharchi's principle (namely that $1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2$), but are there other ways? Thanks!
| If $a^2-b^2=n^3=n^2\cdot n$
we can set $a+b= n^2$ and $a-b=n$
so that $a=\frac{n^2+n}2=\frac{n(n+1)}2$ which is an integer as $n(n+1)$ is even
Similarly, $b=\frac{n^2-n}2=\frac{n(n-1)}2$
More generally, if $a^2-b^2=n^{k+1}=n^k\cdot n$ for integer $k\ge2$
we can set $a+b= n^k$ and $a-b=n$ so that $a=\frac{n^k+n}2=\frac{n(n^{k-1}+1)}2$
Observe that $n$ and $n^{k-1}+1$ have opposite parities, making the product even
Similarly, $b$ can be handled
| {
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"source": "stackexchange",
"question_score": "5",
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How to integrate: $\int \frac{\mathrm dx}{\sin (x)-\sin(a)}$ How to integrate :
$$\int \frac{\mathrm dx}{\sin (x)-\sin(a)}$$
| Using Weierstrass substitution, $$\tan \frac x2=u$$
$$\implies \sin x=\frac{2u}{1+u^2}\text{ and } x=2\arctan u,dx=\frac{2du}{1+u^2}$$
$$I=\int\frac{dx}{\sin x-\sin \alpha} =\int\frac1{\frac{2u}{1+u^2}-\sin\alpha}\cdot\frac{2du}{1+u^2} =\int\frac{2du}{2u-(1+u^2)\sin\alpha}$$
Now,
$$2u-(1+u^2)\sin\alpha=-\sin\alpha (1+u^2-2u\csc\alpha)=-\sin\alpha \left((u-\csc\alpha)^2-(\cot\alpha)^2\right)$$
Using $\frac{dx}{x^2-a^2}=\frac1{2a}\ln \left|\frac{x-a}{x+a}\right|+C$
$$I=-\frac1{\sin\alpha}\frac1{\cot\alpha}\ln\left|\frac{u-\csc\alpha-\cot\alpha}{u-\csc\alpha+\cot\alpha}\right|+C$$
where $C$ is an arbitrary constant for indefinite integral
Using $\csc\alpha+\cot\alpha=\frac{1+\cos\alpha}{\sin\alpha}=\frac{2\cos^\frac\alpha2}{2\sin\frac\alpha2\cos\frac\alpha2}=\cot\frac\alpha2$ and similarly, $\csc\alpha-\cot\alpha=\tan\frac\alpha2$ (as $\sin2A=2\sin A\cos A,\cos2A=2\cos^2A-1$)
$$I=-\frac1{\cos\alpha}\ln\left|\frac{\tan\frac x2-\cot\frac \alpha2}{\tan\frac x2-\tan\frac\alpha2}\right|+C$$
Again, $$\ln\left|\frac{\tan\frac x2-\cot\frac \alpha2}{\tan\frac x2-\tan\frac\alpha2}\right|$$
$$=\ln\left|\frac{\cos\frac \alpha2\cos\frac x2\left(\sin\frac x2\sin \frac \alpha2-\cos\frac \alpha2\cos\frac x2\right)}{\sin\frac \alpha2\cos\frac x2\left(\sin\frac x2\cos\frac\alpha2-\sin\frac\alpha2\cos\frac x2\right)}\right|=\ln\left|-\cot\frac\alpha2\right|+\ln\left|\frac{\cos\frac{x+\alpha}2}{\sin\frac{x-\alpha}2}\right|$$
Clearly, $\ln\left|-\cot\frac\alpha2\right|$ is independent of $x,$ hence constant
| {
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"source": "stackexchange",
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Is there a deep reason why $(3, 4, 5)$ is pythagorean? The triple $(3, 4, 5)$ is a pythagorean triple - it satisfies $a^2 + b^2 = c^2$ and, equivalently, its components are the lengths of the sides of a right triangle in the Euclidean plane.
But of course, the first thing anybody notices is that the triple $(3, 4, 5)$ also happens to be an arithmetical succession of small numbers.
Is there a deep reason why choosing these three successive numbers just so happens to yield a pythagorean triple?
To anyone who feels the question is silly: consider $3^3+4^3+5^3$.
| $(3,4,5)$ is Phythagorean because $5$ is a prime of the form $4k+1$. Some known facts:
*
*Every prime $p$ of the from $4k+1$ can be rewritten as a sum of squares of two distinct positive integers:
$$\forall k \in \mathbb{Z}_{+}, p = 4k+1\text{ prime} \implies \exists \alpha, \beta \in \mathbb{Z}_{+} \text{ s.t. } \alpha \neq \beta \wedge p = \alpha^2 + \beta^2$$
*
*Every number $n$ that can be written as a sum of squares of two distinct positive integers
is part of a Pyhthagorean triplet because of an algebraic identity:
$$n = (\alpha^2+\beta^2) \implies n^2 = (\alpha^2+\beta^2)^2 = (\alpha^2-\beta^2)^2 + (2\alpha\beta)^2$$
*
*Every Phythagorean triplet $(a,b,c)$ has a parametrization of the form:
$$a^2 + b^2 = c^2 \implies \begin{cases}a = (\alpha^2-\beta^2)\mu\\b = 2\alpha\beta\mu\\c = (\alpha^2 + \beta^2)\mu\end{cases}\quad\quad\text{up to order of }a, b$$
*
*When $a, b$ is relative prime to each other, we can set $\mu$ above to 1.
Take $5 = 2^2+1^2$ as an example, we get:
$$\begin{cases}a = 2^2-1^2 = 3\\b = 2\cdot 2 \cdot 1 = 4\\c = 2^2 + 1^2 = 5\end{cases}
\quad\quad\text{is a Pythagorean triplet}$$
$c = 5$ is the smallest example of such Pythagorean triplet. Since there are only 4 numbers smaller than 5, it is just a coincidence that $(3,4,5)$ are successive integers.
| {
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} |
How to solve this system of non-linear equations My knowledge of algebra is still insufficient to solve this problem. Any help in solving the system of equations would be greatly appreciated.
$$
xy(x+y)=30\\
x^3+y^3=35
$$
| $(x+y)^3=x^3+y^3+3xy(x+y)$
$\Rightarrow (x+y)^3=35+90=125\Rightarrow (x+y)=5$
$\Rightarrow xy(x+y)=5xy=30\Rightarrow xy=6$
$(x-y)^2=(x+y)^2-4xy=25-24=1\Rightarrow (x-y)=\pm1$
case 1: $x-y=1$ and $x+y=5$ ,$\Rightarrow x=3,y=2$
case 2: $x-y=-1$ and $x+y=5$, $\Rightarrow x=2,y=3$
| {
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"source": "stackexchange",
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"answer_id": 0
} |
Generating function of partition with restriction
Let $c(m,n)$ denotes number of partitions $n$ into parts not greater than $m$, where order of elements does matter (so they are not classic partitions). Prove that:
$$\sum_{n\ge 0}c(m,n)x^n = \frac{1-x}{1-2x+x^{m+1}}$$
I don't know how to approach. Generating function of such partitions is (if I'm not crazy): $(x+x^2+x^3+...+x^m)(x+x^2+x^3+...+x^m)(x+x^2+x^3+...+x^m)...$ but I don't see how it can help here.
| Note 1: This incorporates the correction in the comments below.
Note 2: I have incorporated Marc van Leeuwen's suggestion about
simplifying the proof, though
I left in my original proof.
Working with your expression,
let
$f_m(x) =x+x^2+x^3+...+x^m
=x(1+x+x^2+x^3+...+x^{m-1})
=\frac{x(1-x^m)}{1-x}
$.
Then $f_m(x)$ is the GF (generating function)
of partitions into 1 term $\le$ m,
$f_m^2(x)$ is the GF
of partitions into 2 terms $\le$ m,
and, in general,
$f_m^k(x)$ is the GF
of partitions into $k$ terms $\le$ m.
Here is what I $should$ have done,
based on Marc van Leeuwen's comment:
Let $F_m(x, u, v)$
be the GF
of partitions into $u$ through $v$ terms (inclusive)
$\le$ m.
The interesting case is, of course,
$u=0$ and $v = \infty$.
Then
$\begin{align}
F_m(x, u, v)
&= \sum_{k=u}^{v} f_m^k(x)\\
&= f_m^u(x)\sum_{k=0}^{v-u} f_m^k(x)\\
&= f_m^u(x)\frac{1-f_m^{v-u+1}(x)}{1-f_m(x)}\\
\end{align}
$
If $u=0$,
$\begin{align}
F_m(x, 0, v)
&= \frac{1-f_m^{v+1}(x)}{1-f_m(x)}\\
&= \frac{1}{1-f_m(x)}(1-f_m^{v}(x))\\
&= \frac{}{1-\frac{x(1-x^m)}{1-x}}(1-f_m^{v}(x))\\
&= \frac{1-x}{1-x-(x(1-x^m))}(1-f_m^{v}(x))\\
&= \frac{1-x}{1-2x+x^{m+1}}(1-f_m^{v}(x))\\
\end{align}
$
If $u=0$ and $v=\infty$,
$F_m(x) = F_m(x, 0, \infty)
= \dfrac{1-x}{1-2x+x^{m+1}}
$
That is what I $should$ have done.
Instead, I did this:
So the total GF,
writing $f_m^0(x)=1$, is
$\begin{align}
F_m(x) &=
\sum_{k=0}^{\infty} f_m^k(x)\\
&= \sum_{k=0}^{\infty} \big(\dfrac{x(1-x^m)}{1-x}\big)^k\\
&= \sum_{k=0}^{\infty} \big(\dfrac{x}{1-x}\big)^k\sum_{j=0}^k \binom{k}{j}(-1)^j x^{jm}\\
&= \sum_{j=0}^{\infty}(-1)^jx^{jm}\sum_{k=j}^{\infty}\binom{k}{j} \big(\dfrac{ x}{1-x}\big)^k\\
\end{align}
$
Since
$\sum_{k=n}^{\infty}\binom{k}{n}x^{k} = \dfrac{x^n}{(1−x)^{n+1}}
$,
$\begin{align}
\sum_{k=j}^{\infty}\binom{k}{j} \big(\dfrac{ x}{1-x}\big)^k
&= \dfrac{ (x/(1-x))^j}{(1-(x/(1-x))^{j+1}}\\
&= \dfrac{ (x/(1-x))^j}{((1-2x)/(1-x))^{j+1}}\\
&= \dfrac{ x^j(1-x)}{(1-2x)^{j+1}}\\
&= \dfrac{ 1-x}{1-2x}\big(\dfrac{ x}{1-2x}\big)^{j}\\
\end{align}
$
Putting this in,
$\begin{align}
F_m(x) &=
\sum_{k=1}^{\infty} f_m^k(x)\\
&= \sum_{j=0}^{\infty}(-1)^jx^{jm}\dfrac{ 1-x}{1-2x}\big(\dfrac{ x}{1-2x}\big)^{j}\\
&= \dfrac{ 1-x}{1-2x}\sum_{j=0}^{\infty}(-1)^jx^{jm}\big(\dfrac{ x}{1-2x}\big)^{j}\\
&= \dfrac{ 1-x}{1-2x}\sum_{j=0}^{\infty}(-1)^j\big(\dfrac{ x^{m+1}}{1-2x}\big)^{j}\\
&= \dfrac{ 1-x}{1-2x}\frac1{1+\frac{ x^{m+1}}{1-2x}}\\
&= \dfrac{ 1-x}{1-2x+ x^{m+1}}\\
\end{align}
$
This is the same as your expression,
once I put in the correction suggested.
| {
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} |
Prove that $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b} \ge a+b+c$ If $a,b,c$ are positive , show that $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$$
Trial: Here I proceed in this way $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge \dfrac{2bc}{b+c}+\dfrac{2ca}{c+a}+\dfrac{2ab}{a+b}$$ then how I proceed. Please help.
| Potato has already answered this question elegantly. Here is another solution.
By Cauchy-Schwarz, we get
$$((b+c)+(c+a)+(a+b))\left(\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}\right)\ge \left(\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+\sqrt{a^2+b^2}\right)^2 $$
To prove the original inequality, it suffices to show that
$$\left(\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+\sqrt{a^2+b^2}\right)^2 \ge 2(a+b+c)^2$$
which boils down to showing
$$\sqrt{a^2+b^2}\sqrt{b^2+c^2} + \sqrt{c^2+a^2}\sqrt{a^2+b^2}+\sqrt{b^2+c^2}\sqrt{c^2+a^2} \ge 2(ab+bc+ca)$$
With one more application of Cauchy-Schwarz,
$$\begin{align}
(b^2+a^2)(b^2+c^2)\ge (b^2+ac)^2 \\
(c^2+b^2)(c^2+a^2)\ge (c^2+ba)^2 \\
(a^2+b^2)(a^2+c^2)\ge (a^2+bc)^2
\end{align}$$
Taking square roots of both sides, and adding them up, it suffices to prove
$$ a^2+b^2+c^2\ge ab+bc+ca$$
which is equivalent to
$$ (a-b)^2+(b-c)^2+(c-a)^2\ge 0$$
So we are done :) I agree that it is a bit overkill. But I like Cauchy-Schwarz inequality.
| {
"language": "en",
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"answer_count": 3,
"answer_id": 0
} |
failed application of magicry in Taylor expansion of $1/x^2$ near $x=2$ It's straightforward to find the Taylor expansion for $\frac{1}{x^2}$ near $x=2$ using the the Taylor series definition.
This is turns out to be $\frac{1}{4} - \frac{1}{4} (x-2) + \frac{3}{16}(x-2)^2 + \cdots$
I was trying to be cute by finding the expansion using the geometric series:
$$
\begin{align}
\frac{1}{x^2} &= \frac{1}{1 -(1 - x^2)} = \frac{1}{1-z} &\text{ where } z = 1-x^2\\
\end{align}
$$
This is where I run into trouble. My initial guess was to expand around $x=0$ and then shift over by 2.
$$
\begin{align}
&1 + z + z^2+ \cdots\\
&1 + (1-x^2) + (1-x^2) ^2 + \cdots \\
&1 + (\ 1-(x-2)^2\ ) + (\ 1-(x-2)^2\ )^2+\cdots
\end{align}
$$
I looked at the Wikipedia page on the geometric series and see that the formula is different when the common ratio is not 1.
$$a + ar + a r^2 + a r^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r}$$
This helped me to see that I had missed the $a$ term. I'm taking it on faith that the common ratio is less than one. The article points out that when $|r|<1$, the series will be $\frac{a}{1-r}$, which is the familiar geometric series.
Somehow, my substitution seems to be wrong. Any help will be appreciated.
| Make a calculation that starts exactly like the one by Don Antonio. We get
$$x=2+(x-2)=2\left(1+\frac{x-2}{2}\right).$$
Now the usual geometric series expansion gives
$$\frac{1}{x}=\frac{1}{2}-\frac{1}{4}(x-2)+\frac{1}{8}(x-2)^2-\frac{1}{16}(x-2)^3+\cdots.$$
Finally, differentiate with respect to $x$. On the left we get $-\frac{1}{x^2}$, so we will have to flip signs. On the right, differentiate term by term.
| {
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"url": "https://math.stackexchange.com/questions/426583",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Permutation and Combination - Algebraic Expansion I have come across the following question and solution
Question
In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively.
Solution given (http://www.askiitians.com/iit-jee-algebra/permutations-and-combinations/derangements-and-multinomial-theorem.aspx)
The number of ways in which they can denote Rs.10 is the same as the number of solutions of the equation $x_1 + x_2 + x_3 = 10$ subject to conditions $0 \leq x_1 \leq 6$, $0 \leq x_2 \leq 7$, $0 \leq x_3 \leq 8$.
Hence the number of solutions is
$$
\begin{align}
&=[x^{10}](1+x+x^2+...+x^6)(1+x+x^2+...+x^7)(1+x+x^2+...+x^8)\\
&=[x^{10}]\frac{(1-x^7)(1-x^8)(1-x^9)}{(1-x)^3}\\
&=[x^{10}](1-x^7-x^8-x^9)\left(1+\binom{3}{1}x + \binom{4}{2}x^2+ \binom{5}{3}x^3+...+\binom{12}{10}x^{10}\right)\\
&= \binom{12}{10} - \binom{5}{3} - \binom{4}{2} - \binom{3}{1}\\
&= 66 - 10 - 6 - 3 = 47.
\end{align}
$$
My Doubt is how coefficient of $x^{10}$ in $(1-x^7)(1-x^8)(1-x^9)/(1-x)^3$ is equal to coefficient of $x^{10}$ in $(1-x^7-x^8-x^9)(1+\binom{3}{1}x + \binom{4}{2}x^2+ \binom{5}{3}x^3+...+\binom{12}{10}x^{10})$ ?
Please help. Thanks in advance.
| There are two things at play here.
*
*Expansion of $(1-x)^{-3}$. Here, we use a (negative) binomial series:
$$
\begin{align}
(1-x)^{-3}&=\sum_{n=0}^{\infty}\binom{-3}{n}(-1)^nx^n\\
&=\sum_{n=0}^{\infty}\frac{(-3)(-3-1)(-3-2)\cdots(-3-(n-1))}{n!}(-1)^nx^n\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n3\cdot4\cdot5\cdots(3+n-1)}{n!}(-1)^nx^n\\
&=\sum_{n=0}^{\infty}\frac{(n+2)_{n}}{n!}x^n\\
&=\sum_{n=0}^{\infty}\binom{n+2}{n}x^n,
\end{align}
$$
where $(a)_b$ is the falling factorial. Notice, here, that all terms have a non-negative power of $x$.
*Expansion of $(1-x^7)(1-x^8)(1-x^9)$. When you multiply this out, you get
$$
1-x^7-x^8-x^9+x^{15}+x^{16}+x^{17}-x^{24}.
$$
Let us combine these two results. We can think of multiplication of two such sums as the sum of terms obtained by taking one element of the first and one element of the second, and multiplying them. Since neither sum involves any negative powers of $x$, we can never choose any term from either which involves $x^{11}$ or higher - such a term can never yield an $x^{10}$. So, we find
$$
\begin{align}
&[x^{10}](1-x^7-x^8-x^9+x^{15}+x^{16}+x^{17}-x^{24})\sum_{n=0}^{\infty}\binom{n+2}{n}x^{n}\\
=&[x^{10}](1-x^7-x^8-x^9)\sum_{n=0}^{10}\binom{n+2}{n}x^n,
\end{align}
$$
as claimed. From here, extracting the coefficient by $x^{10}$ should be no problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/427531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b} \ge 3$
If $a,b,c$ are positive numbers then show that
$$\dfrac{a^2+1}{b+c}+\dfrac{b^2+1}{a+c}+\dfrac{c^2+1}{a+b} \ge 3$$
I am stuck at the first stage. Please give me some hints so that I can solve the problem. Thanks in advance.
| By AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a^2+1}{b+c}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(a^2+1)}{\prod\limits_{cyc}(a+b)}}=3\sqrt[3]{\frac{\sqrt{\prod\limits_{cyc}\left((a^2+1)(1+b^2)\right)}}{\prod\limits_{cyc}(a+b)}}\geq$$
$$\geq3\sqrt[3]{\frac{\sqrt{\prod\limits_{cyc}\left(a+b\right)^2}}{\prod\limits_{cyc}(a+b)}}=3$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
A constrained extremum problem Find the maximum possible value of $$A = a^{333} + b^{333}+c^{333}$$ subject to the constraints $$a+b+c=0$$ and $$a^2+b^2+c^2=1,$$
where $a,b,c\in \mathbb{R}$
Thank you for helping me.
| let $A_{n}=a^n+b^n+c^n$ then we have
$$A_{n+3}=(a+b+c)A_{n+2}-(ab+bc+ac)A_{n+1}+abcA_{n}$$
and
$$ab+bc+ac=-\dfrac{1}{2}$$
so
$$A_{n+3}=\dfrac{1}{2}A_{n+1}+abcA_{n}$$
and use $$a+b+c=0,\Longrightarrow A_{3}=a^3+b^3+c^3=3abc$$
and $$A_{1}=0,A_{2}=1,A_{3}=3abc$$
so
$$A_{4}=\dfrac{1}{2}A_{2}+abcA_{1}=\dfrac{1}{2}$$
$$A_{5}=\dfrac{1}{2}A_{3}+abcA_{2}=\dfrac{3}{2}abc+abc=\dfrac{5}{2}abc$$
$$A_{6}=\dfrac{1}{2}A_{4}+abcA_{3}=\dfrac{1}{4}+3(abc)^2$$
$$A_{7}=\dfrac{1}{2}A_{5}+abcA_{4}=\dfrac{5}{4}abc+\dfrac{1}{2}abc=\dfrac{7}{4}abc$$
$$A_{8}=\cdots, A_{9}=\cdots,\cdots$$
$$A_{300}=f(abc)$$
and we have $|abc|\le\dfrac{1}{\sqrt{54}}$,see:$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$
| {
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"source": "stackexchange",
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Given the solution for a differential equation, find the corresponding differential equation The solution of a differential equation is:
$$\left \{ \begin{array}{lcl}
x_1(t) & = & e^{-2t}-e^{-5t} \\
x_2(t) & = & e^{-2t}+e^{-3t}+e^{-5t} \\
x_3(t) & = & e^{-3t} + e^{-5t}
\end{array}\right .$$
with initial conditions
$$\left \{ \begin{array}{lcl}
x_1(0) & = & 0 \\
x_2(0) & = & 3 \\
x_3(0) & = & 2
\end{array}\right .$$
What is the corresponding differential equation?
I had worked another linear algebra question using back substitution and I know that is what I am going to need to do here. In the previous problem I was substituting $A=QR$ and finding $\mathcal A$. Just not sure which approach to use in this situation.
$Ax=b$
$A=x^{-1}b$
Any help is greatly appreciated. :)
| Writing
$$ \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}=\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}\begin{pmatrix}e^{-2t} \\ e^{-3t} \\ e^{-5t}\end{pmatrix},$$
we have
$$\begin{array}{ll} \color{DarkOrange}{\frac{d}{dt}\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}} & = \begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}\begin{pmatrix}-2e^{-2t} \\ -3e^{-3t} \\ -5e^{-5t}\end{pmatrix} \\ & = \color{Purple}{\begin{pmatrix}-2 & 0 & 5 \\ -2 & -3 & -5 \\ 0 & -3 & -5\end{pmatrix}} \begin{pmatrix}e^{-2t} \\ e^{-3t} \\ e^{-5t}\end{pmatrix} \\ & = \color{Blue}{\begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}}\color{Green}{\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}}\begin{pmatrix}e^{-2t} \\ e^{-3t} \\ e^{-5t}\end{pmatrix} \\ & \color{DarkOrange}{= \begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}.} \end{array}$$
One may solve for the unknown matrix via
$$\begin{array}{ll} & \color{Purple}{\begin{pmatrix}-2 & 0 & 5 \\ -2 & -3 & -5 \\ 0 & -3 & -5\end{pmatrix}}=\color{Blue}{\begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}}\color{Green}{\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}} \\ \iff & \begin{pmatrix}-2 & 0 & 5 \\ -2 & -3 & -5 \\ 0 & -3 & -5\end{pmatrix}\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}^{-1}=\begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}. \end{array}$$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Evaluating an integral Gaussian-profile initial condition has the solution,
$$\phi (r,t)=\frac{R^{3}}{2}\frac{A}{\sqrt{\pi }}\int_{0}^{\infty
}ke^{-R^{2}k^{2}/4}\frac{\sin (kr)}{r}\cos (\omega t)\ dk,$$
where A is an arbitrary initial amplitude and the dispersion relation is $\omega= (k^2 + \omega^2_{mass})^\frac{1}{2}$, where the mass frequency $ \omega_{mass}= \sqrt{2}$.
Details : see the article equation 4.
Now how do we get the equation:
$$\phi (0,t)=\frac{A_{0}}{\left( 1+\frac{2t^{2}}{R^{4}}\right) ^{\frac{3}{4}}}\cos \left( \sqrt{2}t+\frac{3}{2}\tan ^{-1}\left[ \frac{\sqrt{2}t}{R^{2}}\right] \right) .$$
| They take the limit $r\to0$ to obtain
$$
\phi (0,t)=\frac{R^{3}}{2}\frac{A}{\sqrt{\pi }}\int_{0}^{\infty
}k^2e^{-R^{2}k^{2}/4}\cos (\sqrt{k^2+2} t)\ dk.
$$
Then, since the exponent $e^{-R^{2}k^{2}/4}$ goes very quickly to zero as $k$ goes to infinity, the authors say that the integral is dominated by small values of $k$, $k\sim 2R^{-1}$. For whose values the root in $\cos (\sqrt{k^2+2} t)$ is approximated as $\sqrt{k^2+2}=\sqrt2\sqrt{k^2/2+1}\approx \sqrt2(1+k^2/4)\ $. Now if to add an imaginary part,
$$
\cos(\sqrt2(t+tk^2/4))+i\sin (\sqrt2(t+tk^2/4))= e^{i\sqrt2(t+tk^2/4)},
$$
the integral turns into
$$
\frac{R^3 e^{i \sqrt{2} t} A}{2\sqrt{\pi }} \int_0^{\infty } k^2
e^{-\frac{1}{4} k^2 R^2} e^{\frac{1}{4} i \sqrt{2} k^2
t} \, dk=
\frac{R^3 e^{i \sqrt{2} t} A}{2\sqrt{\pi }} \int_0^{\infty } k^2
e^{-\frac{1}{4}( R^2-i \sqrt{2} t)k^2} \, dk=
$$
$$
\frac{AR^3 e^{i \sqrt{2} t}}{\left(R^2-i \sqrt{2}
t\right){}^{3/2}}.
$$
Taking the real part gives the answer:
$$
\phi (0,t)\approx\frac{A}{\left( 1+\frac{2t^{2}}{R^{4}}\right) ^{\frac{3}{4}}}\cos \left( \sqrt{2}t+\frac{3}{2}\tan ^{-1}\left[ \frac{\sqrt{2}t}{R^{2}}\right] \right).
$$
| {
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"source": "stackexchange",
"question_score": "3",
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Factor $x^6 +5x^3 +8$ I wanted to know, how can I factor $x^6 +5x^3 +8$, I have no idea. Is there any method to know if a polynomial is factored. Just some advice will do.
Help appreciated.
Thanks.
| Let $y = x^3$ to obtain $y^2 + 5y + 8 = 0$. This factors as $y = \frac{-5 \pm \sqrt{-7}}{2} = \frac{-5}{2} \pm \frac{\sqrt{7}}{2}i$. These have modulus $r = \sqrt{25/4 + 7/4} = 2\sqrt{2}$. Now solve $5/2 = 2\sqrt{2} \cos \theta$ to find the angle $\theta = \cos^{-1}(5/4\sqrt{2})$. Our two roots correspond to $re^{\pi-\theta}$ and $re^{\pi+\theta}$.
Now we have $x^3 = re^{\pi-\theta}$ and $x^3 = re^{\pi+\theta}$ to contend with. For the former, one root is $x_1 = \sqrt[3]{r}e^{(\pi-\theta)/3}$, so the other two are $x_2 = \sqrt[3]{r}\exp(\frac{\pi-\theta}{3} + 2\pi/3)$ and $x_3 = \sqrt[3]{r}\exp(\frac{\pi-\theta}{3} + 4\pi/3)$.
Similarly, the roots of the other equation are $x_4 = \sqrt[3]{r}e^{(\pi+\theta)/3}$, $x_5 = \sqrt[3]{r}\exp(\frac{\pi+\theta}{3} + 2\pi/3)$, and $x_6 = \sqrt[3]{r}\exp(\frac{\pi-\theta}{3} + 4\pi/3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/432223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow \infty} (\sqrt{n^2} - \sqrt[3]{n^3})\\
&= \lim_{n \rightarrow \infty} (n - n) = 0.
\end{align*}
| Although it seems like a pain, you can multiply by "the conjugate" here, too. But you want $a^6-b^6$ on too, so multiply by $a^5+a^4b+...+b^5$ on top and bottom. The too cancels out nicely. The bottom seems bad, but remember that every term on the bottom grows like $n^2$, and there are six terms, thus giving 1/6.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $f(-1) = 0$ and $f(2)=0$, and if $g(x)= 2x-1$, then find the value of $x$ for which $(f\circ g)(x) = 0$ I have found the following problem.
If $x = -1$ and $x=2$ then $f(x) = 0$. If $g(x)= 2x-1$, then find the value of $x$ for which $f\circ g(x) = 0$.
I have solved the above problem in the following way.
$f(x) = (x+1)(x-2)$
$f\circ g(x) = f(g(x)) = (2x)(2x-3)$
So $2x(2x-3) = 0 \implies x= 0$ or $x= 3/2$
So if $x = 0$ or $x=3/2$ then $f\circ g(x) = 0$
Is the above procedure right? I have solved the above problem with a bit of confusion. That's why I am asking this question here. Please, remove my confusion with a proper explanation.
| Yes: Assuming $f(x) = 0 $ if $x = -1,$ or $x = 2$, then one possible function for $f(x)$ will be
$f(x) = (x+1)(x-2)$.
And we are given $g(x) = 2x - 1$.
If this is the case, it follows that
$(f\circ g)(x) = f(g(x)) = (2x)(2x-3)$
And so: $2x(2x-3) = 0\implies x= 0$ or $x= 3/2$
But, since $f(x)$ isn't necessarily a polynomial, or may be a multiple of our guess at $f(x)$, so we would need to check that that at these two values, it holds that $g(x) = -1$, or $g(x) = 2$.
Can you see why knowing precisely what $f(x)$ isn't necessary? (In fact, we don't know for certain what $f(x)$ actually is.)
Without knowing $f(x)$, but only knowing $f(-1) = 0$ and $f(2) = 0$, we can find the values at which $g(x) = -1,$ and when $g(x) = 2$, because then we know for sure that $$f(g(x)) =f(-1) = 0\quad\text{and that }\quad f(g(x)) = f(2) = 0$$
$$g(x) = 2x - 1 = -1 \iff 2x = 0 \iff \bf x = 0$$
$$g(x) = 2x - 1 = 2\iff 2x = 3 \iff \bf x =\dfrac 32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/433724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Show that $\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =-\frac{\pi^2 \sqrt{2}}{16}$ I could prove it using the residues but I'm interested to have it in a different way (for example using Gamma/Beta or any other functions) to show that
$$
\int_{0}^{\infty}\frac{\ln\left(x\right)}{x^{4} + 1}\,{\rm d}x
=-\frac{\,\pi^{2}\,\sqrt{\,2\,}\,}{16}.
$$
Thanks in advance.
| Substitute $t=1/(1+x^4)$ then we get
$$\int_{0}^{\infty }\frac {\ln x}{x^4+1}\ dx =\frac{1}{16}\int_0^1 \ln\left(\frac{1-t}{t}\right)(1-t)^{-3/4}t^{-1/4}dt.$$
And $\mathrm{B}(x,y)=\Gamma(x)\Gamma(y)/\Gamma(x+y)$, we get
$$\frac{\partial}{\partial x}\mathrm{B}(x,y)=\mathrm{B}(x,y)[\psi(x)-\psi(x+y)]$$
where $\psi$ is digamma function. And by Euler integral of the first kind we get
$$\frac{\partial}{\partial x}\mathrm{B}(x,y)=\int_0^1 \ln t\cdot t^{x-1}(1-t)^{y-1}dt.$$
So
$$
\begin{array}{lcl}
&&\frac{1}{16}\int_0^1 \ln\left(\frac{1-t}{t}\right)(1-t)^{-3/4}t^{-1/4}dt \\ &=&\frac{1}{16}\int_0^1 \ln(1-t) (1-t)^{-3/4} t^{-1/4}dt-\frac{1}{16}\int_0^1 \ln(t)\cdot (1-t)^{-3/4} t^{-1/4}dt\\
&=& \frac{1}{16}\int_0^1 \ln (t)\cdot t^{-3/4} (1-t)^{-1/4}dt -\frac{1}{16}\int_0^1 \ln(t)\cdot (1-t)^{-3/4} t^{-1/4}dt\\
&=&\frac{1}{16}\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{1}{4}\right)-\psi(1)\right]-\frac{1}{16}\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{3}{4}\right)-\psi(1)\right] \\
&=&\frac{1}{16}\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right) \right]
\end{array}
$$
And $\mathrm{B}\left(\frac{1}{4},\frac{3}{4}\right)\left[\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right) \right]=-\pi^2\sqrt{2}$. (It can easily be derived from reflection formula of gamma and digamma function.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
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} |
Evaluating $\int \frac{\mathrm dz}{z^3 \sqrt{z^2 - 4}}$ $$\int \frac{dz}{z^3 \sqrt{z^2 - 4}}$$
$z = 4\sec\theta$
$dz = 4\sec\theta \tan d\theta$
$$\int \frac{\sec\theta \tan\theta}{4^3 \sec^3 \theta \tan \theta}$$
$$ \frac{1}{4^3} \int \frac{d \theta}{\sec^2 \theta}$$
I am again stuck, I have no idea how to proceed.
| Putting $z=2\sec\theta, dz=2\sec\theta\tan\theta d\theta$
$$\int \frac{dz}{z^3\sqrt{z^2-4}}=\int \frac{2\sec\theta\tan\theta d\theta}{(2\sec\theta)^32\tan\theta}=\frac18\int\cos^2\theta d \theta=\frac1{16}\int(1+\cos2\theta)d\theta$$
$$=\frac1{16}\left(\theta+\frac{\sin2\theta}2\right)+K$$
Now, $\cos\theta =\frac2z\implies \theta =\arccos \frac2z $
$\implies \sin\theta=\sqrt{1-\left(\frac2z\right)^2}=\frac{\sqrt{z^2-4}}z$
$\implies \sin2\theta=2\sin\theta\cos\theta=2\cdot \frac{\sqrt{z^2-4}}z\cdot \frac2z$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
three distinct positive integers $a, b, c$ such that the sum of any two is divisible by the third I need to determine three distinct positive integers $a, b, c$ such
that the sum of any two is divisible by the third.
I tried like with out loss of generality let $a<b<c$
As, $a\mid (b+c)$ so $b+c=ak_1$ for some $k_1\in\mathbb{N}$ similarly $$a+b=k_2 c$$ $$a+c=k_3b$$ so adding them I get $k_1+k_2+k_3=2$, could anyone help me to proceed?
| You should get $2(a+b+c)=k_1a+k_2b+k_3c $ - $(*)$
Now if $k_1=k_2=k_3=2$ we get $a=b=c$.
Suppose $k_1\le k_2\le k_3$, then we must have $k_3 \gt 2$ and $k_1 \lt 2$ else the two sides of $(*)$ cannot be equal.
Since $k_1$ is a non-zero positive integer less than $2$ is must be $1$.
So we have $a=b+c$, and $b+(b+c)=k_2c$, and $c+(b+c)=k_3b$ so that $$b=\frac {k_2-1}2c, \text{ } c=\frac {k_3-1}2b$$ whence $$(k_2-1)(k_3-1)=4$$ and we have $k_2=k_3=3$ with $b=c$, which is not allowed, or $k_2=2, \text { } k_3=5$.
In this case $c=2b$ and $a=b+c=3b$ gives the family of solutions suggeted in the comments.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
show that the function $z = 2x^2 + y^2 +2xy -2x +2y +2$ is greater than $-3$ Show that the function
$$z = 2x^2 + y^2 +2xy -2x +2y +2$$
is greater than $-3$
I tried to factorize but couldn't get more than $(x-1)^2 + (x+y)^2 +(y-1)^2 - (y)^2$.
Is there any another way to factorize or another method??
| HINT:
$$z = 2x^2 + y^2 +2xy -2x +2y +2=(x+y+1)^2+(x-2)^2+2-4-1$$
Derivation :
Let $$ 2x^2 + y^2 +2xy -2x +2y +2=(x+y+a)^2+(x+b)^2+2-a^2-b^2$$
$$\implies 2x^2 + y^2 +2xy -2x +2y +2=2x^2+y^2+2xy+2x(a+b)+2ay+2-a^2-b^2$$
Comparing the coefficients of $y, a=1$
Comparing the coefficients of $x, a+b=-1\implies b=-1-a=-2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Wolfram double solution to $\int{x \cdot \sin^2(x) dx}$ I calculated this integral :
$$\int{x \cdot \sin^2(x) dx}$$
By parts, knowing that $\int{\sin^2(x) dx} = \frac{1}{2} \cdot x - \frac{1}{4} \cdot \sin(2x) +c$. So I can consider $\sin^2(x)$ a derivative of $\frac{1}{2} \cdot x - \frac{1}{4} \cdot \sin(2x)$, and I get this result:
$$\frac{1}{4} \cdot x^2 - \frac{x}{4} \cdot \sin(2x) + \frac{1}{4} \cdot \sin^2(x) +c$$
I get the confirm on wolfram if I try to compute the derivative of $\frac{1}{4} \cdot x^2 - \frac{x}{4} \cdot \sin(2x) + \frac{1}{4} \cdot \sin^2(x) +c$, but here if I try to compute the integral of $\int{x \cdot \sin^2(x) dx}$ I get this result:
$$\frac{x^2}{4} -\frac{1}{4} \cdot x \cdot \sin(2x) -\frac{1}{8} \cdot \cos(2x) $$
But $\frac{1}{8} \cdot \cos(2x)$ isn't equal to $\frac{1}{4} \cdot \sin^2(x)$, which is the correct result?
| We have that $\sin^2 x = \frac{1-\cos (2x)}{2}$. Therefore, $\frac14 \sin^2 x = \frac{1-\cos (2x)}{8} = \frac18 - \frac18 \cos (2x)$. The extra $\frac18$ is taken care of by the constant of integration. In other words, if two functions differ by a constant, they have the same derivative. This is something that happens often when integrating trigonometric functions because of the various identities.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$
Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.
My (strange) proof:
$$
\begin{align*}
a^3+b^3+c^3 &\geq a^2b+b^2c+c^2a\\
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a^2 &\geq \sum\limits_{a,b,c} ab\\
a^2+b^2+c^2 &\geq ab+bc+ca\\
2a^2+2b^2+2c^2-2ab-2bc-2ca &\geq 0\\
\left( a-b \right)^2 + \left( b-c \right)^2 + \left( c-a \right)^2 &\geq 0
\end{align*}
$$
Which is obviously true.
However, this is not a valid proof, is it? Because I could just as well have divided by $a^2$ rather than $a$:
$$
\begin{align*}
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a &\geq \sum\limits_{a,b,c} b\\
a+b+c &\geq a+b+c
\end{align*}
$$
Which is true, but it would imply that equality always holds, which is obviously false. So why can't I just divide in a cycling sum?
Edit: Please don't help me with the original inequality, I'll figure it out.
| Without making any assumption, just simple $AM\ge GM$
$$a^3+a^3+b^3\ge3a^2b$$
$$b^3+b^3+c^3\ge3b^2c$$
$$c^3+c^3+a^3\ge3c^2a$$
$$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I write a sum of cosines as a product of sines? I am trying to prove that
$$\cos A+\cos B+\cos C=4\sin\frac A2\sin\frac B2\sin\frac C2$$ for ABC is a triangle.
I tried up to the stage of
$$-2\sin^2 C+2\cos\frac{180-C}2 \cos\frac{A+B}2$$
but how do I proceed from here?
| $$\cos A+\cos B+\cos C$$
$$=2\cos\frac{A+B}2\cos\frac{A-B}2+1-2\sin^2\frac C2$$
$$=2\sin\frac C2\cos\frac{A-B}2+1-2\sin^2\frac C2\text{ as } \cos\frac{A+B}2=\cos\left(\frac{\pi-C}2\right)=\sin \frac C2$$
$$=1+2\sin\frac C2\left(\cos\frac{A-B}2-\cos\frac{A+B}2\right)\text{ as }\sin \frac C2=\cos\frac{A+B}2$$
$$=1+2\sin\frac C22\sin\frac A2\sin\frac B2$$
$$=1+4\sin\frac A2\sin\frac B2\sin\frac C2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/439523",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2$ for odd $n$ In old popular science magazine for school students I've seen problem
Prove that $\quad $
$\dfrac{1}{\cos^2 20^\circ} +
\dfrac{1}{\cos^2 40^\circ} +
\dfrac{1}{\cos^2 60^\circ} +
\dfrac{1}{\cos^2 80^\circ} = 40. $
How to prove more general identity:
$$
\begin{array}{|c|}
\hline \\
\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2 \\
\hline
\end{array}
, \qquad \mbox{ where } \ n \ \mbox{ is odd.}$$
| From this answer we know that
$$
\sum\limits_{k=1}^{m}\tan^2\frac{\pi k}{2m+1}=m(2m+1)
$$
Similarly
$$
\sum\limits_{k=m+1}^{2m}\tan^2\frac{\pi k}{2m+1}=
\sum\limits_{l=1}^{m}\tan^2\frac{\pi (2m+1-l)}{2m+1}=
\sum\limits_{l=1}^{m}\tan^2\frac{\pi l}{2m+1}=m(2m+1)
$$
Since $\cos^{-2}\alpha=1+\tan^2\alpha$, then
$$
\begin{align}
\sum\limits_{k=0}^{2m}\frac{1}{\cos^2\frac{\pi k}{2m+1}}
&=\sum\limits_{k=0}^{2m} 1 +\sum\limits_{k=0}^{2m}\tan^2\frac{\pi k}{2m+1}\\
&=2m+1+\sum\limits_{k=1}^{m}\tan^2\frac{\pi k}{2m+1}+\sum\limits_{k=m+1}^{2m}\tan^2\frac{\pi k}{2m+1}\\
&=2m+1+m(2m+1)+m(2m+1)=(2m+1)^2
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/442715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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"answer_id": 3
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How to evaluate this double integral over a semicircle? $$\iint_{D}\frac{x^4+y^4}{1+e^{3x^2y-y^3}} dxdy$$
$D = \{(x,y):x^2+y^2 \leq 1,x>0\}$. From the shape of region $D$ it seems to me that it's better to convert it to polar coordinate,but with no luck.
| Hint 1:
$$x^4 + y^4 = r^4 - 2x^2y^2 = r^4-2r^4\cos^2(\theta)\sin^2(\theta)=\frac{1}{2}r^4\sin^2(2\theta)$$
Hint 2:
$$3x^2y - y^3 = y(3x^2 - y^2)= r^3\sin(\theta)\left(3\cos^2(\theta) - \sin^2(\theta)\right)$$
$$=2 \cos(2 \theta)+1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximize $(a-1)(b-1)(c-1)$ knowing that : $a+b+c=abc$. If : $a,b,c>0$, and : $a+b+c=abc$, then find the maximum of $(a-1)(b-1)(c-1)$.
I noted that : $a+b+c\geq 3\sqrt{3}$, I believe that the maximum is at : $a=b=c=\sqrt{3}$. (Can you give hints).
| Go for Lagrange Multipliers! Let $f(a,b,c)=(a-1)(b-1)(c-1)$, and $g(a,b,c)=abc-a-b-c$. Then you are trying to maximize $f(a,b,c)$ subject to $g(a,b,c)=0$.
Now, we have
$$
\nabla f(a,b,c)=\bigl\langle(b-1)(c-1),(a-1)(c-1),(a-1)(b-1)\bigr\rangle
$$
and
$$
\nabla g(a,b,c)=\langle bc-1, ac-1, ab-1\rangle.
$$
So, you want to find all tuples $(a,b,c,\lambda)$ such that
$$
\begin{align}
\tag{1} (b-1)(c-1)&=\lambda(bc-1)\\
\tag{2} (a-1)(c-1)&=\lambda(ac-1)\\
\tag{3} (a-1)(b-1)&=\lambda(ab-1)\\
\tag{4} abc-a-b-c&=0
\end{align}
$$
Go through and solve this system of equations. There are only 3 solutions!
| {
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Prove that $(n+1)(n+2)(n+3)$ is $O(n^3)$ Problem
*
*Prove that $(n+1)(n+2)(n+3)$ is $O(n^3)$
Attempt at Solution
*
*$f(n) = (n+1)(n+2)(n+3)$
*$g(n) = n^3$
*Show that there exists an $n_0$ and $C > 0$ such that $f(n) \le Cg(n)$ whenever $n > n_0$
*$f(n) = n^3+6n^2+11n+6 = n^3(1 + 6/n + 11/n^2 + 6/n^3)$
*$f(n) \le C*g(n)$ is
*$n^3(1 + 6/n + 11/n^2 + 6/n^3) \le C*n^3$ is
*$(1 + 6/n + 11/n^2 + 6/n^3) < C$
That's as far as I got. Should I plug in a value for n to find C? And then, would that value I plugged in for n be $n_0$?
Any help is appreciated.
Thank you in advance.
| Your multiplication isn't quite correct. You should get
$$(n + 1)(n + 2)(n + 3) = (n + 1)(n^2 + 5n + 6) = n^3 + 6n^2 + 11n + 6$$
Now proceeding as you did, we get
$$f(n) = n^3 (1 + \frac{6}{n} + \frac{11}{n} + \frac{6}{n^3})$$
Note that if $n \geq 1$, then $1/n \leq 1$; likewise, $1/n^2 \leq 1$ and $1/n^3 \leq 1$. Hence, try choosing
$$C = 1 + 6 + 11 + 6$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $x+3+\sqrt[3]{(x+1)(x^2-x+2)+\sqrt[3]{x^3+x+1}}+\sqrt[3]{x^3+x+1}=0$ solve the equalition
$$x+3+\sqrt[3]{(x+1)(x^2-x+2)+\sqrt[3]{x^3+x+1}}+\sqrt[3]{x^3+x+1}=0$$
I have seen some methods, for this problem.
my idea:
let $\sqrt[3]{x^3+x+1}=y,$ then
$\sqrt[3]{y^3+y+1}+y=-x-3$
and following have nice methods? thank you everyone.
| Hint: lhs of the equation is a strictly increasing function, so there is at most one root which is easy to guess.
| {
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Absolute convergence $\frac{(-1)^n}{n \ln n}$ $$\sum_2^\infty \frac{(-1)^n}{n \ln n}$$
So I claim that it is absolutely convergent because it only coverges when it is an absolute value. This is wrong but I don't see how.
| What is written below is really the Cauchy Condensation Test, without mention of Cauchy, and with calculations concretely carried out. We could work with partial sums, but we will do things somewhat more informally. Our series is
$$\left( \frac{1}{2\log 2}\right)+\left( \frac{1}{3\log 3}+\frac{1}{4\log 4}\right)+
\left( \frac{1}{5\log 5}+\frac{1}{6\log 6}+\frac{1}{7\log 7}+\frac{1}{8\log 8}\right)+\cdots$$
We have $1$ term of $\frac{1}{2\log 2}$, $2$ terms each $\ge \frac{1}{4\log 4}$, and $4$ terms each $\ge \frac{1}{8\log 8}$, and so on. So our sum is greater than
$$\frac{1}{2}\left(\frac{1}{\log 2}+\frac{1}{\log 4}+\frac{1}{\log 8}+\frac{1}{\log 16}+\cdots\right).$$
Note that $\log 4=2\log 2$, $\log 6=3\log 2$, $\log 16=4\log 2$, and so on. So our sum is greater than
$$\frac{1}{2\log 2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right).$$
But $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ (undramatically) blows up. It follows that our original series diverges.
| {
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How many special right triangles are there? We all learned in school about "special" right triangles. Special right triangles have integer side lengths. Examples include the $3$-$4$-$5$ right triangle, the $5$-$12$-$13$ right triangle, the $8$-$15$-$17$ right triangle, and their scalar multiples ($6$-$8$-$10$, $10$-$24$-$26$, $16$-$30$-$34$, etc).
How many are there? Is there a limit to the number of lowest-form (no scalar multiple) special right triangles? Are there any patterns that arise from the progression of integer side lengths?
| This goes back to Diophantus.
Let $a,b,c$ be positive integers such that $a^2+b^2=c^2$. Then $(a,b,c)$ is a Pythagorean triple. If $p$ is a prime common divisor of $a$ and $b$, then $p$ also divides $c$: if $a=pA$ and $b=pB$, then $c^2=p^2(A^2+B^2)$, so that $p$ divides $c^2$, hence $c$. Similarly, if $a$ and $c$ have a common prime divisor, this prime divides also $b$. Thus we can assume that $a$ and $b$ are coprime, by factoring out all prime common divisors. Such a triple is primitive.
Next we can show that $a$ and $b$ are of different parity: one is odd and the other is even. They can't be both even, because they are coprime. If they were both odd, we could write $a=2A+1$ and $b=2B+1$, so
$$
a^2+b^2=4(A^2+A+B^2+B)+2=c^2.
$$
This is impossible, because $c$ must be even, so $c=2C$ and we'd get
$$
2=4(A^2+A+B^2+B-C^2
$$
which is clearly impossible.
Assume, without loss of generality, that $a$ is odd and $b$ is even. Then we can write $b=2B$ and
$$
B^2=\frac{c+a}{2}\frac{c-a}{2}.
$$
Let $p$ be a prime dividing both $(c+a)/2$ and $(c-a)/2$. Then $p$ divides the sum, which is $c$, and the difference, which is $a$: absurd. Therefore $(c+a)/2$ and $(c-a)/2$ are coprime. Since their product is a square, both must be squares.
Thus
$$
\frac{c+a}{2}=u^2,\quad\frac{c-a}{2}=v^2
$$
from which we derive
$$
a=u^2-v^2,\quad b=2uv,\quad c=u^2+v^2.
$$
Moreover $u$ and $v$ are coprime; they can't be both odd, because otherwise $a$ would be even.
Conversely, any pair $(u,v)$ of coprime positive integers, one odd and the other even, with $u>v$, gives rise to a primitive Pythagorean triple.
For $u=2$, $v=1$ we get the triple $(3,4,5)$; for $u=3$, $v=2$ we get the next one $(5,12,13)$ and so on.
| {
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Distance between the product of marginal distributions and the joint distribution Given a joint distribution $P(A,B,C)$, we can compute various marginal distributions. Now suppose:
\begin{align}
P1(A,B,C) &= P(A) P(B) P(C) \\
P2(A,B,C) &= P(A,B) P(C) \\
P3(A,B,C) &= P(A,B,C)
\end{align}
Is it true that $d(P1,P3) \geq d(P2,P3)$ where d is the total variation distance?
In other words, is it provable that $P(A,B) P(C)$ is a better approximation of $P(A,B,C)$ than $P(A) P(B) P(C)$ in terms of the total variation distance? Intuitively I think it's true but could not find out a proof.
| I just find the following counter-example. Suppose $A,B,C$ are discrete variables. $A,B$ can each take two values while $C$ can take three values.
The joint distribution $P(A,B,C)$ is:
\begin{array}{cccc}
A & B & C & P(A,B,C) \\
1 & 1 & 1 & 0.1/3 \\
1 & 1 & 2 & 0.25/3 \\
1 & 1 & 3 & 0.25/3 \\
1 & 2 & 1 & 0.4/3 \\
1 & 2 & 2 & 0.25/3 \\
1 & 2 & 3 & 0.25/3 \\
2 & 1 & 1 & 0.4/3 \\
2 & 1 & 2 & 0.25/3 \\
2 & 1 & 3 & 0.25/3 \\
2 & 2 & 1 & 0.1/3 \\
2 & 2 & 2 & 0.25/3 \\
2 & 2 & 3 & 0.25/3 \\
\end{array}
So the marginal distribution $P(A,B)$ is:
\begin{array}{ccc}
A & B & P(A,B) \\
1 & 1 & 0.2 \\
1 & 2 & 0.3 \\
2 & 1 & 0.3 \\
2 & 2 & 0.2 \\
\end{array}
The marginal distributions $P(A), P(B)$ and $P(C)$ are uniform.
So we can compute that:
\begin{align}
d(P1,P3) &= 0.1 \\
d(P2,P3) &= 0.4/3
\end{align}
| {
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$a_n=\frac{1}{\pi}\left[\int_{-\pi}^{-\frac{\pi}{2}}(\sin 2x)(\cos nx)dx+\int_{0}^{\pi}(\sin 2x)(\cos nx)dx\right]$
$$a_n=\frac{1}{\pi}\left[\int_{-\pi}^{-\frac{\pi}{2}}(\sin 2x)(\cos nx)dx+\int_{0}^{\pi}(\sin 2x)(\cos nx)dx\right];\ldots (A)$$
The answer is
$$\frac{-2}{\pi}\frac{1+\cos(\frac{n\pi}{2})}{n^2-4}.$$
My attempt:
$\int(\sin 2x)(\cos nx)dx$
$=\frac{1}{2}\int[\sin(n+2)x-\sin(n-2)x]dx$
$=\frac{1}{2(n-2)}\cos(n-2)x-\frac{1}{2(n+2)}\cos(n+2)x$
S0,
$\int_{-\pi}^{-\frac{\pi}{2}}(\sin 2x)(\cos nx)dx$
$=\frac{1}{2(n-2)}\cos(n-2)(\frac{\pi}{2})-\frac{1}{2(n+2)}\cos(n+2)(\frac{\pi}{2})-\frac{1}{2(n-2)}\cos(n-2)(\pi)+\frac{1}{2(n+2)}\cos(n+2)(\pi)$
And
$\int_{0}^{\pi}(\sin 2x)(\cos nx)dx$
$=\frac{1}{2(n-2)}\cos(n-2)(\pi)-\frac{1}{2(n+2)}\cos(n+2)(\pi)-\frac{1}{2(n-2)}+\frac{1}{2(n+2)}$
form equation $(A)$,
$a_n=\frac{1}{\pi}[\frac{1}{2(n-2)}\cos(n-2)(\frac{\pi}{2})-\frac{1}{2(n+2)}\cos(n+2)(\frac{\pi}{2})-\frac{1}{2(n-2)}+\frac{1}{2(n+2)}]$
$\Rightarrow a_n=\frac{1}{\pi}[\frac{(n+2)\cos(n-2)(\frac{\pi}{2})-(n-2)\cos(n+2)(\frac{\pi}{2})-4}{2(n-2)(n+2)}]$
I couldn't come up with the result $\frac{-2}{\pi}\frac{1+\cos(\frac{n\pi}{2})}{n^2-4}$.
| $\cos{a-b}=\cos a \cos b + \sin a \sin b$
$\cos{\frac{(n-2)\pi}2}=-\cos{\frac{n\pi}2}$ since $\sin \pi=0$ and $\cos \pi=-1$
$(n-2)(n+2)=n^2-4$.
Your answer should simplify to the required answer.
| {
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A fair 6 sided dice is rolled 4 times. What is the probability that at least 3 of the numbers will be either 1 or 6? I'd really love a sanity check here as I walk through what I believe is the solution.
Total possible outcomes = $6^4 = 1296$
Possible combinations of 3 rolls being either 1 or 6 = $({}_4C_3)\cdot2 = (4)\cdot2 = 8$
Also take into account all 1's and all 6's = $1 + 1 = 2$
Answer = $\frac{8+2}{ 1296} = \frac{10}{1296} = \mathbf{\frac{5}{648}} $
Really appreciate the help! :)
| I would use a binomial probability:
$$ \begin{align*}
P(\text{at least 3 are 1 or 6}) &= P(\text{exactly 3 are 1 or 6}) + P(\text{exactly 4 are 1 or 6}) \\
&= {}_4C_3 \left(\dfrac{2}{6}\right)^3\left(\dfrac{4}{6}\right)^1 + {}_4C_4 \left(\dfrac{2}{6}\right)^4\left(\dfrac{4}{6}\right)^0 \\
&= 4 \left(\dfrac{1}{3}\right)^3\left(\dfrac{2}{3}\right) + \left(\dfrac{1}{3}\right)^4 \\
&= \dfrac{8+1}{81} \\
&= \dfrac{1}{9} \\
\end{align*} $$
| {
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What should be the method for solving and solution of $\frac{x^2+y^2}{x^2-y^2}$? If $\frac{x+y}{x-y}=3$
then $\frac{x^2+y^2}{x^2-y^2} = ?$
| $$\text{As }\frac{x+y}{x-y}=\frac31$$
Applying componendo and dividendo,
$$\frac{x+y+x-y}{x+y-(x-y)}=\frac{3+1}{3-1}\implies \frac xy=2$$
$$\implies \frac{x^2}{y^2}=\frac41$$
Again, applying componendo and dividendo, $$\frac{x^2+y^2}{x^2-y^2} = \frac{4+1}{4-1}=\frac53$$
| {
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Prove the inverse of the Hilbert matrix has integer entries
$1 \frac{1}{2} ... \frac{1}{n}$
$\frac{1}{2} \frac{1}{3} ... \frac{1}{n+1}$
$.$
$.$
$.$
$\frac{1}{n} \frac{1}{n+1} ... \frac{1}{2n-1}$
Does the inverse of this matrix have integer entries? Prove your
statement.
I thought of multiplying every line $i$ by $i!$, but it leads to a very complicated solution (if it leads to a solution at all).
Source: Linear Algebra, Kenneth Hoffman and Ray Kunze. Section 1.6, exercise 12.
| The answer to your first question is yes.
Do not dismiss the example referenced in the problem statement (the case where $n = 3$). Examples like the one I'll reproduce below help to build insight:
Given $\quad A=\begin{pmatrix}1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5}\end{pmatrix},\qquad$
$A^{-1}=\begin{pmatrix} 9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 & 180 \end{pmatrix}$.
I'd suggest you explore the matrix and its inverse for $n = 4$.
What we want to show is that
$B=\begin{pmatrix} 1 & \frac{1}{2} & \ldots & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & \ldots & \frac{1}{n+1} \\ \ldots & \ldots & & \ldots \\ \frac{1}{n} & \frac{1}{n+1} & \ldots & \frac{1}{2n-1}\end{pmatrix}$
is invertible and $B^{-1}$ has integer entries.
Hints to get you started: The matrix $B$ is known as a Hilbert matrix and the entries of its inverse can be represented as the product of binomial coefficients.
| {
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Summation and proof by induction question: $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$ I can't figure this out based on examples in textbooks, etc.
Show via induction that $\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$
So far, I have:
(a) base case
$P(1)= 1(1+1)(1+2) = \frac{1(1+1)(1+2)(1+3)}{4} = 6 = 6$
(b) inductive step $P(k)\rightarrow P(k+1)$
If $P(k)$ is true, then
$P(k+1) = (2)(3)+(2)(3)(4)+(3)(4)(5)+\cdots+(k+1)(k+2)(k+3)= \frac{(k+1)(k+2)(k+3)(k+4)}{4}$
But this doesn't seem to be going anywhere. Any ideas?
| We want to show that
$$\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$$
Base case, $n=0$. It is easy to see that the left-hand-side and right-hand-side are both equal to $0$, so the equality is true. You can also let the base case be $n=1$, and then both sides are equal to $6$.
Induction step. We are given that
$$\sum_{j=1}^{n}j(j+1)(j+2)=\frac{n(n+1)(n+2)(n+3)}{4}$$
and wish to prove
$$\sum_{j=1}^{n+1}j(j+1)(j+2)=\frac{(n+1)(n+2)(n+3)(n+4)}{4}$$
So we write
$$\sum_{j=1}^{n+1}j(j+1)(j+2) = \left(\sum_{j=1}^{n}j(j+1)(j+2)\right) + (n+1)(n+2)(n+3)$$
By assumption, this is equal to
$$= \frac{n(n+1)(n+2)(n+3)}{4} + (n+1)(n+2)(n+3) $$
$$= \frac{n(n+1)(n+2)(n+3)}{4} + \frac{4(n+1)(n+2)(n+3)}{4} $$
Factoring, this becomes
$$\frac{(n+4)(n+1)(n+2)(n+3)}{4} = \frac{(n+1)(n+2)(n+3)(n+4)}{4}$$
And the proof is complete.
| {
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Polynomial modulus Can anyone explain why the two solutions to $n^2+7n-2 = 0$ modulo $43$ are $n=13$ and $n=23$ and how they are found?
| The number $43$ is a prime, thus we can work like we always do, for example, with real numbers: all the same rules apply to the integers modulo $43$. Now, we seek to complete the square. All equalities are in $\Bbb Z/43\Bbb Z$, the integers modulo $43$:
$${n^2} + 7n - 2 = 0$$
$${n^2} + 2\frac 72n - 2 = 0$$
Note that $2$ has as an inverse the number $22$ since $2\times 22=44=1 \pmod {43}$. Thus $\dfrac 7 2=7\times 2^{-1}=7\times 22=25$. Then we get
$$\begin{align}n^2+2\cdot 25 n-2+25^2-25^2&=0\\
(n+25)^2&=25^2+2\\(n+25)^2&=25\\n+25&=\begin{cases}5\\-5\end{cases}\\n&=\begin{cases}-20=23\\-30=13\end{cases}\end{align}$$
| {
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Bernoulli numbers generating function Consider the following generating formula:
$$\frac{t}{e^t-1}=\sum_{n=1}^{\infty} B_n \frac{t^n}{n!}$$
There is some intuitive explanation about it?
I want to know because I need to proof to myself that the sum of the combination of the Bernoulli Numbers is $0$, like this:
$$\sum_{u=1}^\infty {{n+1}\choose u} B_u = 0$$
I've already understood the entire proof, but it assumes that $\frac{t}{e^t-1}=\sum_{n=1}^{\infty} B_n \frac{t^n}{n!}$ so I want to proof (or see how it was found) this last part.
Thanks!
| Let's assume that $g(x)$ is given and we try to find out $f(n)$
$$
f(n)=\sum_{i=1}^n g(i)
$$
$$
f(n+1)=\sum_{i=1}^{n+1}g(i)
$$
$$
f(n+1)-f(n)=g(n+1)
\tag 1$$
We know Taylor expansion
$$
f(x+h)=f(x)+hf'(x)+\frac{h^2 f''(x)}{2!}+\frac{h^3f'''(x)}{3!}+....
$$
Thus
$$
f(n+1)=f(n)+f'(n)+\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}+....
$$
If we put $f(n+1)$ taylor expansion in Equation $1$
$$f(n+1)-f(n)=g(n+1)$$
$$
f(n)+f'(n)+\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}+....-f(n)=g(n+1)
$$
$$
f'(n)+\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}+...=g(n+1)
\tag 2$$
$$
f(n)+\frac{f'(n)}{2!}+\frac{f''(n)}{3!}+\frac{f'''(n)}{4!}+...=\int g(n+1) dn
$$
We need $f(n)$ if so we need to cancel $f'(n)$ . So we need to
$$
-\frac{1}{2} ( f'(n)+\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}+...)=-\frac{1}{2}g(n+1)
$$
$$
f(n)+ (-\frac{1}{2.2} +\frac{1}{3!})f''(n)+(-\frac{1}{2.3!} +\frac{1}{4!})f'''(n)+...=\int g(n+1) dn-\frac{1}{2}g(n+1)
$$
$$
f''(n)+\frac{f'''(n)}{2!}+\frac{f^{4}(n)}{3!}+...=\frac{d(g(n+1))}{dn}
$$
If you continue in that way to cancel $f^{r}(n)$ terms step by step, you will get
$$
f(n)=\int g(n+1) dn-\frac{1}{2}g(n+1)+\frac{1}{12}\frac{d(g(n+1))}{dn}+a_4\frac{d^2(g(n+1))}{dn^2}+a_5\frac{d^3(g(n+1))}{dn^3}+...
$$
This is Euler-Maclaurin formula. (Please see also the Applications of the Bernoulli numbers). I just wanted to show Bernoulli numbers seen in one of the very important formulas in mathematics .
Where $$a_n= \frac{B_n}{n!}$$.
Because If you try to find out the coefficients of $\frac{t}{e^t-1}$ by polynomial division. You can get exactly same coefficients that seen in Euler-Maclaurin formula.
The Bernoulli numbers appear in Jacob Bernoulli's most original work "Ars Conjectandi" published in Basel in 1713 in a discussion of the exponential series.
You can also see that The Bernoulli numbers appears in the power series of $tan(x)$.
https://en.wikipedia.org/wiki/Taylor_series
(Check the List of Maclaurin series of some common functions)
Proof:
$$\frac{t}{e^t-1}=\frac{t}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1+\frac{(1-1)t-\frac{t^2}{2!}-\frac{t^3}{3!}-\frac{t^4}{4!}-...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$
$$\frac{t}{e^t-1}=1-\frac{t}{2}+\frac{+(\frac{1}{2}-\frac{1}{2!})t^2+(\frac{1}{2.2!}-\frac{1}{3!})t^3+(\frac{1}{2.3!}-\frac{1}{4!})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{t}{2}+\frac{(\frac{1}{2.2!}-\frac{1}{3!})t^3+(\frac{1}{2.3!}-\frac{t^4}{4!})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$
$$\frac{t}{e^t-1}=1-\frac{1}{2}t+\frac{\frac{1}{12}t^3+\frac{1}{24}t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{1}{2}t+\frac{1}{12}t^2+\frac{(\frac{1}{24}-\frac{1}{2.12})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$
| {
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Sequence and Series - If $a_n =\int^{\frac{\pi}{2}}_0 \frac{\sin^2nx}{\sin^2x}dx,$..... If $\displaystyle a_n =\int^{\frac{\pi}{2}}_0 \frac{\sin^2nx}{\sin^2x}dx, $ then find the value of
$$\begin{vmatrix}
a_1 & a_{51} & a_{101} \\ a_2 &a_{52} & a_{102}\\ a_3 & a_{53}&a_{103}\\
\end{vmatrix}.$$
My approach :
We know that $S_{n+1} - S_{n} = T_n$ where $S_{n+1} $ is sum of $n+1$ term and $S_n $ is sum of $n$ terms and $T_n $ is $n$th term.
Can we use this here somehow ..... as I used :
$\displaystyle \frac{\sin^2(n+1)x}{\sin^2x}- \frac{\sin^2nx}{\sin^2x} = \frac{\sin^2(n+1)x - \sin^2{nx}}{\sin^2x} $.... now what to do further... please suggest... thanks.
| Set $x=y/2$, then $$a_n=\frac{1}{2}\int_0^{\pi}\frac{\sin^2\frac{ny}{2}}{\sin^2\frac{y}{2}}dy.$$ If $y=-z$, you get that $$a_n=\frac{1}{2}\int_{-\pi}^0\frac{\sin^2\frac{nz}{2}}{\sin^2\frac{z}{2}}dz\Rightarrow a_n=\frac{1}{4}\int_{-\pi}^{\pi}\frac{\sin^2\frac{ny}{2}}{\sin^2\frac{y}{2}}dy=\frac{n}{4}\int_{-\pi}^{\pi}F_n(y)\,dy=\frac{2\pi n}{4},$$ where $F_n$ is the Fejer kernel, which has the property that its integral from $-\pi$ to $\pi$ is $2\pi$. So, your determinant is equal to $$\frac{(2\pi)^3}{4^3}\left|\begin{array}{c c c}1 & 51 & 101\\ 2 & 52 & 102\\ 3 & 53 & 103\end{array}\right|=\frac{(2\pi)^3}{4^3}\left|\begin{array}{c c c}1 & 50 & 101\\ 2 & 50 & 102\\ 3 & 50 & 103\end{array}\right|=\frac{(2\pi)^3}{4^3}\left|\begin{array}{c c c}1 & 50 & 100\\ 2 & 50 & 100\\ 3 & 50 & 100\end{array}\right|=0.$$
| {
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The Keys problem Another challenge:
A calculator has two special keys:
*
*A key transforms a number x in the number 2x.
*B key transforms a number x in the number 2x - 1.
Is it true that if you start with any positive integer, it is possible to press a special key sequence in such a way as to obtain finally the fifth power of an integer?
By simple inspection one realizes that the powers of 2 greater than 1 you can get a fifth power using the A key, but will be possible with other numbers?
| In this post, I shall interpret the problem as follows. For any positive integer $n$, if we start with $n$ and perform some sequence of the keystrokes $A,B$, we will reach a fifth power of a positive integer. It should be noted that Chris Culter resolved the interpretation with quantifiers reversed in another answer.
Hold $n$ fixed throughout.
Lemma 1. For any nonnegative integer $k$ and any integer $x$ in the interval $[2^kn-2^k+1,2^kn]$, there is a sequence of keystrokes which results in $x$ starting at $n$.
Proof. We induct on $k$. If $k=0$, then the interval is just $\{n\}$ consisting of only our starting value. Thus, assume $k>0$ and the inductive hypothesis for $k-1$. $$2^kn-2^k+1\le x\le 2^kn\implies 2^{k-1}n-2^{k-1}+1\le\left\lceil\frac x2\right\rceil\le 2^{k-1}n$$This means $\left\lceil\frac x2\right\rceil$ is the result of a sequence of keystrokes starting at $n$. If $x$ is even, $2\left\lceil\frac x2\right\rceil=2\frac x2=x$. If $x$ is odd, $2\left\lceil\frac x2\right\rceil-1=\frac{2(x+1)}2-1=x$.$\space\square$
Lemma 2. There exists an integer $k\ge 0$ such that for any positive integer $x$ satisfying $x^5\le 2^kn$, $$(x+1)^5-x^5\le 2^k-1.$$
Proof. First $(x+1)^5-x^5=1+5x+10x^2+10x^3+5x^4\le 31x^4\le 31\left(\sqrt[5]{2^kn}\right)^4$. Since $31x^4$ is an integer, we only need to find $k$ large enough that $\lfloor31(2^kn)^{4/5}\rfloor\le 2^k-1$ holds.
However, if $k>5\log_2(31n^{4/5})$, then $$31(2^kn)^{4/5}=2^{(4/5)k}(31n^{4/5})<2^{(4/5)k+(1/5)k}=2^k,$$ using $2^{(1/5)k}>31n^{4/5}$. It follows that $$(x+1)^5-x^5\le\lfloor31(2^kn)^{4/5}\rfloor\le 2^k-1.\space\square$$
Theorem. There is a positive integer $x$ such that $x^5$ can be reached by a sequence of keystrokes starting at $n$.
Proof. Let $k$ be as in Lemma 2. It suffices, by Lemma 1, to show that there is a fifth power in the interval $[2^kn-2^k+1,2^kn]$. There is a greatest integer $x$ such that $x^5<2^kn-2^k+1$. Then, $(x+1)^5-x^5\le2^k-1$, so $$(x+1)^5=(x+1)^5-x^5+x^5<2^k-1+2^kn-2^k+1=2^kn.$$ Hence, $(x+1)^5$ must be within the interval $[2^kn-2k+1,2^kn]$.$\space\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/456808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to force unitary Euclidean norm in a complex matrix by multiplication with a diagonal matrix I need to solve the following problem:
Suppose a non-sparse, non-singular complex matrix $\mathbf{P}$. If I want to force all rows in $\mathbf{P}$ to present unitary Euclidean norms by multiplying $\mathbf{P}$ with a diagonal matrix ($\mathbf{A}$) from the right side (it has to be on the right), such that:
\begin{align}
\mathbf{P} = \left[
\begin{array}{cc}
a & b\\
c & d
\end{array}
\right]
\end{align}
\begin{align}
\mathbf{A} = \left[
\begin{array}{cc}
x_1 & 0\\
0 & x_2
\end{array}
\right]
\end{align}
\begin{align}
\mathbf{P A} =
\left[
\begin{array}{cc}
a & b\\
c & d
\end{array}
\right]
\left[
\begin{array}{cc}
x_1 & 0\\
0 & x_2
\end{array}
\right] =
\left[
\begin{array}{cc}
ax_1 & bx_2\\
cx_1 & dx_2
\end{array}
\right]
\end{align}
then, the following system must be solved:
\begin{cases}
\left| a \right|^2 x_1^2 + \left| b \right|^2 x_2^2 = 1\\
\left| c \right|^2 x_1^2 + \left| d \right|^2 x_2^2 = 1\\
\end{cases}
In matrix notation
\begin{align}
\left[
\begin{array}{cc}
\left| a \right|^2 & \left| b \right|^2\\
\left| c \right|^2 & \left| d \right|^2
\end{array}
\right]
\left[
\begin{array}{c}
x_1^2\\
x_2^2
\end{array}
\right]
&=
\left[
\begin{array}{c}
1\\
1
\end{array}
\right]\\
\mathbf{T}
\left[
\begin{array}{c}
x_1^2\\
x_2^2
\end{array}
\right]
&=
\left[
\begin{array}{c}
1\\
1
\end{array}
\right]
\end{align}
where $\mathbf{T} = \left| \mathbf{P} \right|^2$, considering $\left| \cdot \right|$ to be element-wise.
The solution for $x_1^2$ and $x_2^2$ is easy to state:
\begin{align}
\left[
\begin{array}{c}
x_1^2\\
x_2^2
\end{array}
\right]
&=
\mathbf{T}^{-1}
\left[
\begin{array}{c}
1\\
1
\end{array}
\right]\\
&=
\left[
\begin{array}{c}
\sum \limits_{j=0}^1 t_{0j}\\
\sum \limits_{j=0}^1 t_{1j}
\end{array}
\right]
\end{align}
where $t_{ij}$ is the element from $\mathbf{T}^{-1}$ in the $i$-th line and $j$-th column.
However, if any of $x_1^2$ or $x_2^2$ happen to be negative, $x_1$ or $x_2$ will be complex and the unitary Euclidean norm fails.
Am I making any mistake?
| One way to look at this would be:
$$
|a|^2 x_1^2 + |b|^2 x_2^2 = 1
$$
is an ellipse. So is the other equation. Now, you have a valid solution when these ellipses intersect in a matching way. Since I am lazy, I used Wolfram Alpha to do this. In essence, the ellipses must intersect, and the resulting points' coordinates must provide a valid solution for the original system. That is, you may get a solution for an instance where $(x_1, x_2)$ work for $a = 2$, but not for $a = -2$, for example.
The cases where it would not work for any signs of $a, b, c$ and $d$ would be those where the ellipses don't intersect at all.
If someone can come up with a better intuition, I'd be happy to learn.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong
| Apart from the other answers, even at the last, $\sqrt{(5-\frac92)^2}=\pm(5-\frac92)$. with + it is wrong.
With $-(5-\frac92)$, that is $-5+\frac92$, adding the other $\frac92$ from the original equation, we do get $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 2
} |
Prove with Induction for $n\in \mathbb{N}$ and $n$ is even for $1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $ I want to prove by indection, for $n\in\mathbb N$ even:
$$1^2-3^2+5^2-7^2+\dots+(2n-3)^2-(2n-1)^2=-2n^2 $$
what I did first is to check the numbers, so if $n$ is even
lets take $n=2$ so $(2\cdot 2-3)^2-(2\cdot 2-1)^2=-2\cdot 4$
lets take $n=4$ so $(2\cdot 4-3)^2-(2\cdot 4-1)^2\neq-2\cdot 16$
I did something wrong?
Thanks!
| Let's put $n=2m$ we can then write the sum as:
$$(1^2-3^2)+(5^2-7^2) +\dots +((4m-3)^2-(4m-1)^2) =$$$$(1-3)\cdot(1+3)+(5-7)\cdot(5+7)+ \dots +((4m-3)-(4m-1))\cdot((4m-3)+(4m-1))=$$$$-2\cdot4-2\cdot12-\dots - 2\cdot(8m-4)=$$$$-8(1+3+\dots(2m-1))$$
The sum (in brackets) is now a linear one (well known to add to $m^2$ - proved by an easy induction, which I will leave to you). Since $-8m^2=-2n^2$ the desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Show $(3306 \cdot 3834)^2 - (11^2 \cdot 13 \cdot 17^2)^2 $ is divisible by $10875593$ Let $n = 10875593 $
Given $3306^2 - n = 11\cdot 17^3$ and $3834^2 - n = 11^3\cdot 13^2 \cdot 17$
Deduce that $(3306 \cdot 3834)^2 - (11^2 \cdot 13 \cdot 17^2)^2 $ is divisible by $n$. Then use a calculator and the Euclidean Algorithm to find a factor of $n$. What is the smallest $a$ such that $a^2 - n $ is a square?
Hi everyone, thanks in advance for the help. Here is what I have so far, first part seemed reasonably easy:
Given $3306^2 - n = 11\cdot 17^3$ and $3834^2 - n = 11^3\cdot 13^2 \cdot 17$
Then $(3306^2 - n)(3834^2 - n) = (11\cdot 17^3)(11^3\cdot 13^2 \cdot 17) $
so, $(3306 \cdot 3834)^2 - (11^2 \cdot 13 \cdot 17^2)^2 = 3306^2n + 3834^2n $
therefore, LHS is divisble by n and = $ 3306^2 + 3834^2 $
The next part, I used the Euclidean Algorithm, and as I understand it so far, it takes two arguments, so I used $n$ and $ 3306^2 + 3834^2 $, I obtained the gcd to be 1.
Which seems to me that $n$ could be prime? I understand that the Euclidean Algorithm does not imply this, but should I just pick random numbers to try with $n$ until I find otherwise? What sort of process should I use here?
For the final part of the question, I have tried to use Fermat's factorization method. By Fermat's factorization method, I know that $a >= \sqrt{n}$, so I have simply started from $ 3397^2 - n $ to try and find a perfect square.
I've tried enough iterations to feel like this is not the way to go about it, but we're in week 1 of the semester and I am a bit stuck.
Any hints or directions would be greatly appreciated!!
| Since $(3306 \cdot 3834)^2 - (11^2 \cdot 13 \cdot 17^2)^2= mn$, you need to calculate gcd of $(3306 \cdot 3834) - (11^2 \cdot 13 \cdot 17^2)$ and $n$ and gcd of $(3306 \cdot 3834) + (11^2 \cdot 13 \cdot 17^2)$ and $n$. One of these should give a factor of $n$. This is a standard technique of factoring $n$ where we try to express $n$ as $x^{2} - y^{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/458712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How to find the sum of this : $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....$ How to find the sum of the following :
$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+ \sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$
Please suggest as getting no clue on this... thanks..
| Hint:
Let us write it as follows:
$$\sum_{n=1}^{1999} \sqrt{1+ \frac{1}{n^{2}} + \frac{1}{(n+1)^{2}}}$$
We can rewrite the radical as:
$$\sqrt{\frac{n^{2}(n+1)^{2} + n^{2} + (n+1)^{2}}{n^{2}(n+1)^{2}}}$$
Which simplifies to:
$$\sqrt{\frac{(n^{2}+n+1)^{2}}{n^{2}(n+1)^{2}}} = \frac{n^{2}+n+1}{n^{2}+n} = 1 + \frac{1}{n^{2}+n}$$
So now our sum is:
$$\sum_{n=1}^{1999} 1 + \frac{1}{n^{2}+n}$$
Extracting the constant term gives
$$\sum_{n=1}^{1999} 1 + \frac{1}{n^{2}+n} = 1999 + \sum_{n=1}^{1999}\frac{1}{n(n+1)}$$
By partial fractions decomposition, we see:
$$\frac{1}{n(n+1)} = \frac{1}{n} -\frac{1}{n+1}$$
The series therefore telescopes, making the sum easy to compute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/462045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 1
} |
Solve : $\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$ Solve : $$\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$$
Can we take like this :
Let us take (k+1)th term = $$\frac{(k+1)^2(k+1)!}{2^{k+1}} ; \frac{(k+1)(k+1)!}{2^{k+1}}$$ and (k-1)th term is : $$\frac{(k-1)^2(k-1)!}{2^{k-1}}; \frac{(k-1)(k-1)!}{2^{k-1}}$$
What can we do further to this series... please suggest ......thanks.
| Possible way:
$$\sum^{2010}_{k=1} \frac{k^2k!}{2^k}+\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}=\sum^{2010}_{k=1} \frac{k!(k+1)k}{2^k}=\sum^{2010}_{k=1} \frac{k(k+1)!}{2^k}$$
$$=\sum^{2010}_{k=1} \frac{(k+2-2)(k+1)!}{2^k}=\sum^{2010}_{k=1} \frac{(k+2)!}{2^k}-\sum^{2010}_{k=1} \frac{(k+1)!}{2^{k-1}}$$
$$=\frac{2012!}{2^{2010}}-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/463794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given a rational number $x$ and $x^2 < 2$, is there a general way to find another rational number $y$ that such that $x^2Suppose I have a rational number $a$ and $a^2 < 2$. Can I find another rational number $B$ such that $a^2<B^2<2$?
Based on the answer to this question, I thought of doing the following:
$$ a^2 < 2 \implies a < \frac{2}{a}\\
\text{Let}\hspace{1cm} B=\frac{a+\frac{2}{a}}{2}=\frac{a^2+2}{2a}
$$
$B$ is greater than $a$ because:
$$ \begin{array} {aa} B>a & \implies \frac{a^2+2}{2a}>a \\
& \implies a^2 + 2 > 2a^2 \\
& \implies 2 > a^2 \\
& \implies a^2 < 2
\end{array}$$
If $B^2$ is less than $2$, then $B^2-2<0$, but:
$$\begin{array} {aa} B^2-2 < 0 & \implies \left( \frac{a^2+2}{2a} \right)^2 - 2 < 0 \\
& \implies \frac{a^4+4a^2+4}{4a} - \frac{8a^2}{4a^2} < 0 \\
& \implies \frac{(a^2-2)^2}{(2a)^2} < 0
\end{array}$$
Which is a contradiction since the left hand side of the inequality will be positive for all values of $a$.
But I think we must be able to find such a $B$ since based on my understanding of this answer, we can find a another rational number whose distance from $a$ is less than the distance between $a$ and $\sqrt{2}$
Therefore, I have 2 questions to ask:
*
*Why does this approach work in the case of $a^2>2$ but not when $a^2<2$?
*How should I approach these kind of questions since it seems that there are a few ways to construct a $B$ that satisfies a given set of restrictions? For example, see here (the proof is immediately before the section "13. The Completeness Axiom".
| Here $x<2$, so for all numbers $\delta\in(0,1)$ we have the estimate
$$
(x+\delta)^2=x^2+2x\delta+\delta^2=x^2+(2x+\delta)\delta<x^2+5\delta.
$$
So if $m=2-x^2$, then $0<m<2$, and the above estimate shows that we can select
$$
y=x+\frac m5=x+\frac{2-x^2}5.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
If $a\ge b\ge-c\ge0$, is $\sqrt[3]{a-b-c}\ge\sqrt[3]{a}-\sqrt[3]{b}-\sqrt[3]{c}$? Let $a\ge b\ge-c\ge0$. Is it true that $\sqrt[3]{a-b-c}\ge\sqrt[3]{a}-\sqrt[3]{b}-\sqrt[3]{c}$?
| Sure.
Since $a\geq 0$ and $-c \geq 0$, $a-c\geq 0$. Therefore
$$(a-b-c)-(\sqrt[3]{a}-\sqrt[3]{b}-\sqrt[3]{c})^3 = 3(\sqrt[3]{a}-\sqrt[3]{c})(\sqrt[3]{b}-\sqrt[3]{-c})(\sqrt[3]{a}-\sqrt[3]{b}) \geq 0$$
since all of the factors on the right are non-negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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can someone explain this limit i have,
$$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$$ the correct answer is $-\frac{3}{7}$ and in my case the result is $\frac{7}{3}$ i don't understand.
i tried this
$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$=$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}.\frac {\sqrt{x+49}+7}{\sqrt{x+49}+7}.\frac {3+\sqrt{x+9}}{3+\sqrt{x+9}}$=
$\lim_{x\to 0} \frac {x+49-49}{9-x+9}.\frac {\sqrt{x+49}+7}{3+\sqrt{x+9}}= \frac {x(\sqrt{x+49}-7)}{x(3-\sqrt{x+9})}=\frac{14}{3}=\frac{7}{3}$ i don't know what is the error!
| You made a couple mistakes. Observe that:
$$ \begin{align*}
\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}.\frac {\sqrt{x+49}+7}{\sqrt{x+49}+7}.\frac {3+\sqrt{x+9}}{3+\sqrt{x+9}}
&= \lim_{x\to 0} \frac {(x+49)-49}{9-(x+9)}.\frac {3+\sqrt{x+9}}{\sqrt{x+49}+7} \\
&= \lim_{x\to 0} \frac {x}{-x}.\frac {3+\sqrt{x+9}}{\sqrt{x+49}+7} \\
&= \lim_{x\to 0} -\frac {3+\sqrt{x+9}}{\sqrt{x+49}+7} \\
&= -\frac {3+\sqrt{0+9}}{\sqrt{0+49}+7} \\
&= -\frac {6}{14} \\
&= \frac{-3}{7}
\end{align*}$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/464866",
"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.