Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
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_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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:
$$ y_1 = x_1... | For the newer information on question, since $x_i$ are integral(which is same as saying that they are integers):
The answer will be coefficient of $x^{10}$ in the following expansion:
$$
\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) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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 transfor... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/356497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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
$$\begin{align*}
a + b + c &= 3\\
4a +... | You could represent your system as a single equation using matrices:
$$
\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]
$$
To isolate the matrix of unknowns, multiply on the left by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/357964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2 } \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $
Evaluate
$$\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2
} \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $$
The question look's like there is a nice method to do ... | Set R = 4 x3 + 5 x2 + 10 for brevity. Maybe based on Liouville's theorem, we can guess
$$\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{... | {
"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 m... | 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 ru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/361710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
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}\|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/365063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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} + 19... | 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}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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... | 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-majo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/369050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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 ... | 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",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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 pro... | $\sin(2arctan(x))$
let $y=arctan(x)$
so $x=tan(y)$
$sin(2y)=2sin(y)cos(y)=2sin(arctan(x))\cdot cos(arctan(x))$
$sin(arctan(x)) = \frac{x}{\sqrt{1+x^2}}$ $cos(arctan(x))= \frac{1}{\sqrt{1+x^2}}$
$2\cdot\frac{x}{\sqrt{1+x^2}} \cdot \frac{1}{\sqrt{1+x^2}}=\frac{2x}{{1+x^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
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 consid... | 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:
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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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:
$$4x^2 - 5(y^2 + \frac{24}{5}y) = 16$$
$$ \Rightarrow 4x^2 - 5(y + \frac{12}{5})^2 = 16 + \left(\frac{12}{5}\right)^2$$
$$\Rightarrow 4x^2 - 5(y + \frac{12... | First off, don't forget that when you factor out that "-5" to make the "completion of squares" easier, it also has to be factored out of +24y, so the equation becomes
$$4x^2 - 5(y^2 - \frac{24}{5}y ) = 16 $$
(as lab bhattacharjee already has).
You then complete the square by adding the term $\frac{12^2}{5^2}$ within th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Existence of real roots of a quartic polynomial Question
What is the minimum possible value of $a^{2}+b^{2}$ so that the polynomial $x^{4}+ax^{3}+bx^{2}+ax+1=0$ has at least 1 root?
Attempt
I divided by $x^{2}$ and got $$x^{2}+\frac{1}{x^{2}}+2+a\left(x+\frac{1}{x}\right)+b-2=0$$ by letting $$x+\frac{1}{x}=X$$
the equa... | It seems to me that requesting the smallest possible value for $a^2+b^2$ is almost a red herring.
Claim.
For $|a| \leq 4$ and $b > 2 |a| - 2$, the polynomial
$$
p(x) = x^4 + ax^3 + bx^2 + ax + 1
$$
has no real zeros.
Proof.
First set $b = 2|a| - 2 + \epsilon$, where $\epsilon > 0$. If we denote $\pm = \operator... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/374505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Stuck on Epsilon proof.. Using the $\epsilon-M $ definition of the limit, calculate
$$\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}... | 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",
"url": "https://math.stackexchange.com/questions/375372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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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,... | 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:
$$\dfrac{4x^2 + 2\cdot \color{blue}{\bf 2}(x) - 5}{x + 2} = \dfrac{4(x - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/375469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Calculating the sum of $f(x) = \sum_{n=0}^{\infty} {n \cdot 2^n \cdot x^n}$ In Calculus, how do I calculate this sum?
$$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}^{... | 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
$$2x S_m =\sum_{n=0}^{m} {n \cdot 2^{n+1} \cdot x^{n+1}}=\sum_{k=1}^{m+1} {(k-1) \cdot 2^{k} \cdot x^{k}}$$
$$=\left(\sum_{k=1}^{m+1} k \cdot 2^{k} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
To split in to partial fractions, the expression $\frac{1}{x^2(x+a)^2}$ To split in to partial fractions, the expression $\frac{1}{x^2(x+a)^2}$
$\frac{1}{x^2(x+a)^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(x+a)}+\frac{D}{(x+a)^2}$
One method of finding the values of the constants A,B,C & D is as follows.
$Ax(x+a)^2+B(x+a)^... | $$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$$
After canceling the 1's, you can divide b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$
I can't seem to wrap my head around that.
| Actually, if you consider an integer division (so called Euclidian division) the proposition becomes:
$$
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)}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 15,
"answer_id": 14
} |
How to express the following as a series? How you you express $$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)
Put... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/381636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
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 b... | 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 so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/381926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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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$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/383007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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... | A related problem. It seems we can go beyond this. Here is the Taylor series expansion of $ \arcsin(x)^3 $
$$\arcsin(x)^3 = \frac{3}{4\sqrt {\pi }}\sum _{m=1}^{\infty }\,{\frac { \left( {\pi }^{2}-2\,\psi'\left(
m+\frac{1}{2} \right) \right) \Gamma\left( m+\frac{1}{2} \right) {x}^{2\,m+1}}{
\left( 2\,m+1 \right) \Ga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/383068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Prove that $x-1$ divides $x^n-1$ In algebra & polynomials, how do we prove that
$$x-1 \mid x^n -1?$$
| Factorization method
Clearly, $x^{n} - 1 = (x - 1)(x^{n - 1} + x^{n - 2} + \dots + 1)$. Thus, $(x - 1) | (x^{n} - 1)$ since there exists the polynomial namely $(x^{n - 1} + x^{n - 2} + \dots + 1)$ that is multiplied by $(x - 1)$ to obtain $(x^{n} - 1)$.
Substitution method
Suppose that $x - 1 = 0 \rightarrow x = 1$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/383379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 1
} |
If $a+b+c=1$, find the minimum of $\frac{4+3abc}{ab+bc+ac}$ I came to ask this because I am really stuck at this problem. I have tried everything from arithmetic mean, geometric mean and harmonic mean. Also, I have tried playing with the variables and such, but it got me to nowhere.
If $a+b+c=1$; $a,b,c$ nonnegative, c... | Proof
Denote $u = ab + bc + ca$, we have the following two results
(1) $0 \le 3 u \le 1$,
and
(2) $9abc \ge 4u - 1$.
Then,
$$
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 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
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 ma... | 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, s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
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}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/386488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 5,
"answer_id": 3
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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... | 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 mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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:
$$I... | Here is how we compute the LT of $\sin^4{t}$ directly. As mentioned in @Mhenni's solution, write as $\sin^4{t} = (e^{i t}-e^{-i t})^4/(2 i)^4$
$$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 \... | {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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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 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
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 ana... | 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), obtai... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
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^... | 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^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/395652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
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,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396087",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
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}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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}
$
Wha... | 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
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/398802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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 conse... | 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 $f_y=-3x^2y+h_y(y,z)$ bu... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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... |
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} \\... | {
"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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/401889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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",
"answer_count": 2,
"answer_id": 0
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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 mo... | 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",
"url": "https://math.stackexchange.com/questions/406621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"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$$
My questions is there a concise way for me to pick a, b, and c so that x will always be a... | As $$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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/407327",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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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 s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
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 integral... | $$\int_0^{\infty} \sin ax^2\sin bx^{-2}\,dx=\frac{1}{2}\int_0^{\infty} \cos \left(ax^2-\frac{b}{x^2}\right)\,dx-\frac{1}{2}\int_0^{\infty} \cos \left(ax^2+\frac{b}{x^2}\right)\,dx$$
In general:
$$\begin{aligned} \int_{0}^{\infty}f \left(p x-\frac{q}{x}\right)^2\,dx &= \sqrt{\frac{q}{p}}\int_{0}^{\infty }f \left(\sqrt{... | {
"language": "en",
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"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
How do I find the exact value of $\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}$? I know that $\cos(6\phi)\equiv32c^6-48c^4+18c^2-1$ where $c=\cos\phi$.
I also know that when $\cos(6\phi)=0$, then $\phi=\frac{k\pi}{12}$ ($k = 1,3,5,7,9,11$).
How do I find the exact value of:
$$\cos\left(\... | As $\frac{11\pi}{12}=\pi-\frac\pi{12}$ and $\frac{7\pi}{12}=\pi-\frac{5\pi}{12}$
and $\cos(\pi-x)=-\cos x,$
$$\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right) \cos\left(\frac{7\pi}{12}\right) \cos\left(\frac{11\pi}{12}\right)$$
$$=\cos\left(\frac{\pi}{12}\right) \cos\left(\frac{5\pi}{12}\right)\left( - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/409396",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
3D Geometry Question In $3$-dimensional Geometry, if angle made of line segment $OP$ with $X,Y,Z$-axis are in
$1:2:3$, then what is the angle made by line segment with $Y$-axis?
My Solution:
Let $\alpha,\beta$ and $\gamma$ be the angle made by line with $X,Y,Z$-axis, respectively. Then
$$\cos^2 \alpha+\cos^2 \beta+\co... | 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 $60^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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
$$
\begin{cases}
f_x = -6x^2 - 6xy + 3y^... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Determining bounds for the sum $\sum\limits_{n=1}^\infty \frac{1}{2^n - 3^n }$ I have to give low and high bounds for the following:
$$
\sum_{n=1}^\infty \frac{1}{2^n - 3^n }
$$
How do I determine an upper bound? How can I show this sum exists?
edit: removed erroneous conclusion.
| $\begin{align}
\sum_{n=1}^\infty \frac{1}{3^n - 2^n }
&=\sum_{n=1}^\infty \frac{1}{3^n(1 - (2/3)^n) }\\
&=\sum_{n=1}^\infty \frac{1}{3^n} \sum_{k=0}^\infty(2/3)^{nk}) \\
&=\sum_{n=1}^\infty \frac{1}{3^n}+\sum_{n=1}^\infty \frac{1}{3^n} \sum_{k=1}^\infty(2/3)^{nk}) \\
&= \frac{1/3}{1-1/3}
+ \sum_{m=1}^{\infty} (2/3)^m \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find the coefficient using binomial theorem. What is the coefficient of $x^{20}$ in the expression: $$(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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"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... | 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
$$\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}+\dfrac{2}{\sqrt{(y^2+1)(z^2+1)}}\ge 1+\dfrac{1}{(y+z)^2+1}+\dfrac{2}{\sqrt{(y+z)^2+1}}.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
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How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
| This is the two-variable identity $(X+Y)^3 - (X^3 + Y^3) = 3XY(X+Y)$
presented as a formula for $X^3 + Y^3 + Z^3$ when $X+Y+Z=0$ (by setting $Z = -(X+Y)$),
and then using $(X,Y,Z)=(a-b,b-c,c-a)$ to parametrize solutions of $X+Y+Z=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Prove $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$, given $x+y+z=3$ and $x,y,z\ge0$
Let $x+y+z=3,x,y,z\ge 0$,show that
$$\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^... | 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 immedi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 1
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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 $\lambda_1 = \lambda_2 = 5$:
$$A=\begin{bmatrix}... | We are given:
$$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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
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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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/418500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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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,
$$2u-(1+u^2)\sin\alpha=-\sin\alpha (1+u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/420318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is there a deep reason why $(3, 4, 5)$ is pythagorean? The triple $(3, 4, 5)$ is a pythagorean triple - it satisfies $a^2 + b^2 = c^2$ and, equivalently, its components are the lengths of the sides of a right triangle in the Euclidean plane.
But of course, the first thing anybody notices is that the triple $(3, 4, 5)$ ... | $(3,4,5)$ is Phythagorean because $5$ is a prime of the form $4k+1$. Some known facts:
*
*Every prime $p$ of the from $4k+1$ can be rewritten as a sum of squares of two distinct positive integers:
$$\forall k \in \mathbb{Z}_{+}, p = 4k+1\text{ prime} \implies \exists \alpha, \beta \in \mathbb{Z}_{+} \text{ s.t. } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/421521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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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 $x+y=5$, $\Rightarrow x=2,y=3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 functi... | 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 t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/423008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge \dfrac{2bc}{b+c}+\d... | Potato has already answered this question elegantly. Here is another solution.
By Cauchy-Schwarz, we get
$$((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 $$
To prove the original inequality, it suffices to show t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/424150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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failed application of magicry in Taylor expansion of $1/x^2$ near $x=2$ It's straightforward to find the Taylor expansion for $\frac{1}{x^2}$ near $x=2$ using the the Taylor series definition.
This is turns out to be $\frac{1}{4} - \frac{1}{4} (x-2) + \frac{3}{16}(x-2)^2 + \cdots$
I was trying to be cute by finding the... | Make a calculation that starts exactly like the one by Don Antonio. We get
$$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.$$
Finally, differentiate with respect to $x$. On the left we ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Permutation and Combination - Algebraic Expansion I have come across the following question and solution
Question
In how many different ways three persons A, B, C having 6, 7 and 8 one rupee coins respectively can donate Rs.10 collectively.
Solution given (http://www.askiitians.com/iit-jee-algebra/permutations-and-com... | There are two things at play here.
*
*Expansion of $(1-x)^{-3}$. Here, we use a (negative) binomial series:
$$
\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)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/427531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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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:
$$\sum_{cyc}\frac{a^2+1}{b+c}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(a^2+1)}{\prod\limits_{cyc}(a+b)}}=3\sqrt[3]{\frac{\sqrt{\prod\limits_{cyc}\left((a^2+1)(1+b^2)\right)}}{\prod\limits_{cyc}(a+b)}}\geq$$
$$\geq3\sqrt[3]{\frac{\sqrt{\prod\limits_{cyc}\left(a+b\right)^2}}{\prod\limits_{cyc}(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/428315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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A constrained extremum problem Find the maximum possible value of $$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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/428804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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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
$$\l... | Writing
$$ \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 & -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/430236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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}... | 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 $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"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-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 ... | 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",
"answer_count": 8,
"answer_id": 5
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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) ... | 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... | {
"language": "en",
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"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
$$
\int_{0}^{\infty}\frac{\ln\left(x\right)}{x^{4} + 1}\,{\rm d}x
=-\frac{\,\pi^{2}\,... | Substitute $t=1/(1+x^4)$ then we get
$$\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)]$$
where $\psi$ is di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
} |
Evaluating $\int \frac{\mathrm dz}{z^3 \sqrt{z^2 - 4}}$ $$\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$$
Now, $\cos\theta =\frac2z\imp... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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three distinct positive integers $a, b, c$ such that the sum of any two is divisible by the third I need to determine three distinct positive integers $a, b, c$ such
that the sum of any two is divisible by the third.
I tried like with out loss of generality let $a<b<c$
As, $a\mid (b+c)$ so $b+c=ak_1$ for some $k_1\in\m... | You should get $2(a+b+c)=k_1a+k_2b+k_3c $ - $(*)$
Now if $k_1=k_2=k_3=2$ we get $a=b=c$.
Suppose $k_1\le k_2\le k_3$, then we must have $k_3 \gt 2$ and $k_1 \lt 2$ else the two sides of $(*)$ cannot be equal.
Since $k_1$ is a non-zero positive integer less than $2$ is must be $1$.
So we have $a=b+c$, and $b+(b+c)=k_2c$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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show that the function $z = 2x^2 + y^2 +2xy -2x +2y +2$ is greater than $-3$ Show that the function
$$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$$
Comparing the coefficients of $y, a=1$
Comparing the coefficients of $x, a+b=-1\implies b=-1-a=-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Wolfram double solution to $\int{x \cdot \sin^2(x) dx}$ I calculated this integral :
$$\int{x \cdot \sin^2(x) dx}$$
By parts, knowing that $\int{\sin^2(x) dx} = \frac{1}{2} \cdot x - \frac{1}{4} \cdot \sin(2x) +c$. So I can consider $\sin^2(x)$ a derivative of $\frac{1}{2} \cdot x - \frac{1}{4} \cdot \sin(2x)$, and I... | We have that $\sin^2 x = \frac{1-\cos (2x)}{2}$. Therefore, $\frac14 \sin^2 x = \frac{1-\cos (2x)}{8} = \frac18 - \frac18 \cos (2x)$. The extra $\frac18$ is taken care of by the constant of integration. In other words, if two functions differ by a constant, they have the same derivative. This is something that happens ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/437681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$
Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.
My (strange) proof:
$$
\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... | 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",
"answer_count": 4,
"answer_id": 1
} |
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?
| $$\cos A+\cos B+\cos C$$
$$=2\cos\frac{A+B}2\cos\frac{A-B}2+1-2\sin^2\frac C2$$
$$=2\sin\frac C2\cos\frac{A-B}2+1-2\sin^2\frac C2\text{ as } \cos\frac{A+B}2=\cos\left(\frac{\pi-C}2\right)=\sin \frac C2$$
$$=1+2\sin\frac C2\left(\cos\frac{A-B}2-\cos\frac{A+B}2\right)\text{ as }\sin \frac C2=\cos\frac{A+B}2$$
$$=1+2\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/439523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2$ for odd $n$ In old popular science magazine for school students I've seen problem
Prove that $\quad $
$\dfrac{1}{\cos^2 20^\circ} +
\dfrac{1}{\cos^2 40^\circ} +
\dfrac{1}{\cos^2 60^\circ} +
\dfrac{1}{\cos^2 80^\circ} = 40. $
How to prove mo... | From this answer we know that
$$
\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{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 3
} |
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",
"url": "https://math.stackexchange.com/questions/444624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/445293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
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(... | 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 $1/n^3 \leq 1$. Henc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/447696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
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? t... | Hint: lhs of the equation is a strictly increasing function, so there is at most one root which is easy to guess.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/447922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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)+
\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
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... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
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... | 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 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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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}.$$
... | $\cos{a-b}=\cos a \cos b + \sin a \sin b$
$\cos{\frac{(n-2)\pi}2}=-\cos{\frac{n\pi}2}$ since $\sin \pi=0$ and $\cos \pi=-1$
$(n-2)(n+2)=n^2-4$.
Your answer should simplify to the required answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/452536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A fair 6 sided dice is rolled 4 times. What is the probability that at least 3 of the numbers will be either 1 or 6? I'd really love a sanity check here as I walk through what I believe is the solution.
Total possible outcomes = $6^4 = 1296$
Possible combinations of 3 rolls being either 1 or 6 = $({}_4C_3)\cdot2 = (4)\... | I would use a binomial probability:
$$ \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)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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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, $$\frac{x^2+y^2}{x^2-y^2} = \frac{4+1}{4-1}=\frac53$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
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 $... | 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} & \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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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 ... | 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... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
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:
$$\sum_{u... | Let's assume that $g(x)$ is given and we try to find out $f(n)$
$$
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!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
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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 t... | 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 $$a_n=\frac{1}{2}\int_{-\pi}^0\frac{\sin^2\frac{nz}{2}}{\sin^2\frac{z}{2}}dz\Rightarrow a_n=\frac{1}{4}\int_{-\pi}^{\pi}\frac{\sin^2\frac{ny}{2}}{\sin^2\frac{y}{2}}dy=\frac{n}{4}\int_{-\pi}^{\pi}F_n(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
The Keys problem Another challenge:
A calculator has two special keys:
*
*A key transforms a number x in the number 2x.
*B key transforms a number x in the number 2x - 1.
Is it true that if you start with any positive integer, it is possible to press a special key sequence in such a way as to obtain finally the f... | In this post, I shall interpret the problem as follows. For any positive integer $n$, if we start with $n$ and perform some sequence of the keystrokes $A,B$, we will reach a fifth power of a positive integer. It should be noted that Chris Culter resolved the interpretation with quantifiers reversed in another answer.
H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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How to force unitary Euclidean norm in a complex matrix by multiplication with a diagonal matrix I need to solve the following problem:
Suppose a non-sparse, non-singular complex matrix $\mathbf{P}$. If I want to force all rows in $\mathbf{P}$ to present unitary Euclidean norms by multiplying $\mathbf{P}$ with a diagon... | One way to look at this would be:
$$
|a|^2 x_1^2 + |b|^2 x_2^2 = 1
$$
is an ellipse. So is the other equation. Now, you have a valid solution when these ellipses intersect in a matching way. Since I am lazy, I used Wolfram Alpha to do this. In essence, the ellipses must intersect, and the resulting points' coordinates ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/457348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$2+2 = 5$? error in proof $$\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(\fra... | 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",
"answer_count": 9,
"answer_id": 2
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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\cd... | 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 $m^2$ - ... | {
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"url": "https://math.stackexchange.com/questions/458179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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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 Algorith... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/458712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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+\fr... | 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:
$$\sqrt{\frac{(n^{2}+n+1)^{2}}{n^{2}(n+1)^{2}}} = \frac{n^{2}+n+1}{n^{2}+n} = 1 + \frac{1}{n^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/462045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
Solve : $\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$ Solve : $$\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... | Possible way:
$$\sum^{2010}_{k=1} \frac{k^2k!}{2^k}+\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}=\sum^{2010}_{k=1} \frac{k!(k+1)k}{2^k}=\sum^{2010}_{k=1} \frac{k(k+1)!}{2^k}$$
$$=\sum^{2010}_{k=1} \frac{(k+2-2)(k+1)!}{2^k}=\sum^{2010}_{k=1} \frac{(k+2)!}{2^k}-\sum^{2010}_{k=1} \frac{(k+1)!}{2^{k-1}}$$
$$=\frac{2012!}{2^{201... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Given a rational number $x$ and $x^2 < 2$, is there a general way to find another rational number $y$ that such that $x^2Suppose I have a rational number $a$ and $a^2 < 2$. Can I find another rational number $B$ such that $a^2<B^2<2$?
Based on the answer to this question, I thought of doing the following:
$$ a^2 < 2 \... | 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",
"answer_count": 1,
"answer_id": 0
} |
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 {... | 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}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/464866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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