Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find all $Z\in C$ such that $|z|=2$ and $Im(z^6) = 8Im(z^3).$ Practice problem from a college textbook.
Find all $Z\in\mathbb C$ such that $|z|=2$ and $Im(z^6) = 8Im(z^3).$
My method is:
$2^6\sin6\phi = 8\times2^3\sin3\phi$
$\sin6\phi = \sin3\phi$
$\sin3\phi = 2\sin3\phi\cos3\phi$
$\cos3\phi = \frac{1}{2}$
$3\phi = \f... | The key equation here, as you write, is $$\sin 6\phi-\sin3\phi=0.$$
We can apply the formula for the difference of $\sin$ functions:
$$\sin 6\phi-\sin3\phi=2\sin(3\phi/2)\cos(9\phi/2)=0,$$
which immediately gives us all possible solutions:
$$\sin(3\phi/2)=0\Rightarrow \phi = \frac 23 \pi k,\quad k\in \Bbb Z, $$
$$\cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
show that $\frac{c^2}{a}+\frac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$ Let $a,c>0$ show that
$$\dfrac{c^2}{a}+\dfrac{a^2}{c}+16\sqrt{ac}\ge 9(a+c)$$
It seem AM-GM inequality.How?
| Certainly an Overkill
$$\frac{c^2}{a}+\frac{a^2}{c}+16\sqrt{ac} \geq 9(a+c)$$
$$\Longleftrightarrow \left(\frac{c^2}{a}-a\right)+\left(\frac{a^2}{c}-c\right)- 8(\sqrt{c}-\sqrt{a})^2 \geq 0$$
$$\Longleftrightarrow (\sqrt{c}-\sqrt{a})^2 \left[(\sqrt{c}+\sqrt{a})^2 (c+a) -8ac \right] \geq 0$$
$$\Longleftrightarrow (\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Diophantine equation: $x^2 + 4y^2- 2xy -2x -4y -8 = 0$ How many integer pairs $ (x, y) $ satisfy $$x^2 + 4y^2- 2xy -2x -4y -8 = 0 ?$$
As it seems at first glance, and hence, my obvious attempt was to make perfect squares but the $2xy$ term is causing much problem.
| As a comment has said, $$(x-y-1)^2 + 3(y-1)^2 = 12$$
We have $a^2\ge 0$ for all $a\in\mathbb R$, therefore $(x-y-1)^2\ge 0$ and $(y-1)^2\ge 0$.
Notice that if $(y-1)^2\ge 9$, then $$(x-y-1)^2 + 3(y-1)^2 \ge 0+3\cdot 9>12$$
You'll get $(y-1)^2\in\{0,1,4\}$, i.e. $y-1\in\{0,\pm 1,\pm 2\}$,
i.e. $y\in\{-1,0,1,2,3\}$. Chec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding roots of $3^x+4^x+5^x-6^x=0$.
How many real roots of $3^x+4^x+5^x-6^x=0$ exist?
Could anyone please tell both the graphical and a non graphical way?
For graphical way
I am not even able to find critical points.
Could just find one root by hit and trial that is $3$.
Answer is given to be one real root.
thanks ... | As indicated in Joey's comment, the function is monotonically decreasing from $\infty$ to $-1$, and so there is only
one solution, the $x=3$ you already found.
However, if the input data were different it might have been difficult to estimate a solution, around which to graph or start an iteration.
Therefore let me tak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Shank's Baby-Step Giant-Step for $3^x \equiv 2 \pmod{29}$ Problem: Solve $3^x \equiv 2 \pmod{29}$ using Shank's Baby-Step Giant-Step method.
I choose $k=6$ and calculated $3^i \pmod {29}$ for $i=1,2,...,6$.
$$3^1 \equiv 3 \pmod {29}$$
$$3^2 \equiv 9 \pmod {29}$$
$$3^3 \equiv 27 \pmod {29}$$
$$3^4 \equiv 23 \pmod {29}$$... | $$3^x \equiv 2 (\text{ mod } 29)$$
$\varphi(29) = 28$
$m = \lceil \sqrt{28} \rceil = 6$
$\begin{array}{c|ccccc}
j& 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
3^j & 1 & 3& 9 & 27 & 23 & \color{#f00}{11}
\end{array}$
$3^{-6}= 3^{28 -6 } = 3^{22} = 22$
$\begin{array}{c|ccccc}
i & 0 & 1 & 2\\
\hline
2\cdot 22^i & 2 & 15 & \color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Combinations with a condition of limited length If there was a mold of size $N$ and $N$ uniquely sized bricks $(N, N-1,\cdots, 1)$, in how many ways the bricks could be places into the mold? The order doesn't matter.
For examples, for $N = 5$ there are $9$ possible combinations:
$[5], [4+1], [4], [3+2], [3+1], [3], [2+... | To answer this question, start by considering how many combinations there are of non-negative integers $n$ and $m$ such that $n+m=N$:
If $N$ is even we have $\frac{N+2}{2}$ combinations. If $N$ is odd we have $\frac{N+1}{2}$ combinations.
To get all possible combinations sum together the result above for all numbers g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Two factoring problems that I can't figure out I have two problems I can't seem to factor.
*
*$8a^3+27b^3+2a+3b$
I first tried it by grouping. I grouped the first two terms together and used sum of two cubes to get
$$(2a+3b)(4a^2-6ab+9b^2)$$ and then I added the leftover $2a+3b$ to get:
$$(2a+3b)(4a^2-6ab+9b^2) + ... | .For the first question, you were very close to the answer. In fact, when you write $(2a+3b)(4a^2+9b^2-6ab) + 2a+3b$, all you need to notice is that the $2a+3b$ at the end can also be taken into the bracket, using the distributive law:
$$
(2a+3b)(4a^2+9b^2-6ab) + 2a+3b = (2a+3b)(4a^2+9b^2-6ab) + 1(2a+3b) \\ = (2a+3b)(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1918610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
For what value $k$ is $f(x) = \begin{cases} \frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\ k & x = 2 \end{cases}$ continuous at $x=2$? For what value $k$ is the following function continuous at $x=2$?
$$f(x) = \begin{cases}
\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\
k & x = 2
\end{cases}$$
All those square roo... | Hint :
$$\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2}=\frac1{\sqrt{2x+5}+\sqrt{x+7}}\xrightarrow[x\to2]{} ...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\thet... | hint: $a^4+b^4 = a^4+2a^2b^2+b^4-2a^2b^2=(a^2+b^2)^2 - 2a^2b^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
How do I prove that $\dfrac{(7+\sqrt{48})^n+(7-\sqrt{48})^n}{2}\equiv3\pmod4$ if and only if $n$ is odd? How do I prove that $\dfrac{(7+\sqrt{48})^n+(7-\sqrt{48})^n}{2}\equiv3\pmod4$ if and only if $n$ is odd?
I know that it will always be whole, and that it will never be $\equiv3$ if $n$ is even. But I want to know ho... | $(a \pm b)^n = a^n \pm n*a^{n-1}b + ...... + (-1)^i*{n \choose i}a^ib^{n-1}+ ....$.
So $(a -b)^n + (a+ b)^n = 2a^n + (n - n)a^{n-1}b + ...... (1 + (-1)^i){n \choose i}a^ib^{n-1}+....$.
$1 + (-1)^i = 0$ if $i$ is odd; $2$ if $i$ is even.
so $(a -b)^n + (a+ b)^n = 2(a^n + {n \choose 2}a^{n-2}b^2 + ...... {n \choose 2k}a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to find $\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$ without using L'Hospital's Rule? I have to find $$\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$$
We are not allowed to use L'Hospital's rule, any suggestions would be beneficial!
I have tried multiplying by the ... | Using the definition of the derivative of $\sin(x)$ at $x = \dfrac{\pi}{6}$, we get
\begin{align}
\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}
&=\lim_{x \to \frac{\pi}{6}}
\frac{\sin(x)-\sin(\frac{\pi}{6})}{x-\frac{\pi}{6}} \\
&= \sin'(\frac{\pi}{6}) \\
&= \cos(\frac{\pi}{6}) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Rational equation my question
İf $ x+\frac{1}{x^2}=3$
then find $( x^2 -\frac{1}{x})^2 $ .
I tried factoring, taking squares of both sides and some other things that did not work. what should i do?
| Contrary to $x+\frac{1}{x^2}=3$, the expresion $x-\frac{1}{x^2}$ cannot be constant because if $(x-\frac{1}{x^2})^2=a^2$ then one would have
$$x^3-3x^2+1=0\\x^3\mp ax^2-1=0$$ from which $$x=\pm\sqrt{\frac{2}{3\mp a}}$$ and this is not compatible with the three distinct possible (real) values of $x$.
An easy calculation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Pythagorean Identities Simplify.
$\tan^2\frac{\pi}{3}-\cot(\frac{\pi}{4})+2\csc(\frac{\pi}{6})$
I am trying to help my little sister with homework and am rusty on my Pythagorean Identities.
| $$
\tan^2\frac{\pi}{3}-\cot(\frac{\pi}{4})+2\csc(\frac{\pi}{6})=\tan^2\frac{\pi}{3}-\frac{1}{\tan\frac{\pi}{4}}+2\frac{1}{\sin\frac{\pi}{6}}=\\ =\sqrt{3}^2-1+2\cdot 2=3-1+4=6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Orthodiagonal quadrilateral Is it always possible to construct an orthodiagonal quadrilateral such that the diagonals and perimeter are fixed? More specifically, given 2 fixed diagonals, how is the perimeter of the quadrilateral bounded?
|
In the above configuration, let $OA=a,OB=b,OC=c,OD=d$ and assume $b>d$.
We may show that if both $b$ and $d$ are replaced by $\frac{b+d}{2}$ (that brings $B$ to $B'$ and $D$ to $D'$), the perimeter of $ABCD$ decreases. This is a consequence of the inequalities
$$ \sqrt{a^2+b^2}+\sqrt{a^2+d^2}\geq 2\sqrt{a^2+\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An inequality where we have to show that is works for all positive x, and describe equality and what Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality.
I was thinking about multiplying both sides by x and then squaring both sides, but I don't think I have the right... | Multiplying by $x$ and squaring both sides gives
$$
4x^3 + 4x^{\frac{3}{2}} + 1 \geq 9x^2 \Leftrightarrow
$$
$$
4x^3 + 4x^{\frac{3}{2}} + 1 - 9x^2 \geq 0.
$$
I do not see an easy way to factor this, but wolfram tells us that the factors are
$$
(\sqrt{x} - 1)^2(2\sqrt{x} + 1)(2x^{\frac{3}{2}} + 3x + 1) > 0.
$$
I'd s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $\tan{(\pi/7)} \tan{(2\pi/7)}\tan{(3\pi/7)}=\sqrt{7}$ I tried in this way.$\tan(a+b)=\frac{(\tan a + \tan b)}{1 - \tan a \tan b }$value of $\tan \frac{\pi}{7}$ is coming in decimal.what to do
| If $\theta = \frac{k\pi}{7}$ where $k = 1, 2, 3$, then $7\theta = k\pi$ and hence $4\theta = k\pi - 3\theta$. Thus $\tan(4\theta) = -\tan(3\theta)$. Expanding, and writing $t = \tan\theta$, we get
$$ \frac{4t-4t^3}{1-4t^2 + t^4} = -\frac{3t-t^3}{1-3t^2} $$
Simplifying we get
$$t^6 - 7t^4 + \cdots -7 = 0$$
The roots a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
| Suppose that $$ax^3+8x^2+bx+6=(x^2-2x-3)(ax-2).$$ And
$$(x^2-2x-3)(ax-2)=ax^3-(2a+2)x^2+(4-3a)x+6.$$
So clearly $a=-5, b=19$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 2
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Proving a determinant equality
Prove
$$\begin{vmatrix}
2bc-a^2 & c^2 & b^2 \\
c^2 & 2ac-b^2 & b^2 \\
b^2 & a^2 & 2ab-c^2\\
\end{vmatrix}
=(a^3+b^3+c^3-3abc)^2$$
My attempt:
I tried using the well-known result that
$$\begin{vmatrix}
bc-a^2 & ca-b^2 & ba-c^2 \\
ac-b^2&... | Hint: Observe
\begin{align}
\begin{bmatrix}
2bc-a^2 & c^2 & b^2\\
c^2 & 2ac-b^2 & a^2\\
b^2 & a^2 & 2ab-c^2
\end{bmatrix}
=
\begin{bmatrix}
a & b & c\\
b & c & a\\
c & a & b
\end{bmatrix}
\begin{bmatrix}
-a & -b & -c\\
c & a & b\\
b & c & a\\
\end{bmatrix}
\end{align}
Edit: Just take the determinant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Do the Taylor series of $\sin x$ and $\cos x$ depend on the identity $\sin^2 x + \cos^2 x =1$? I had this crazy idea trying to prove the Pythagorean trigonometric identity;$$\sin^2x+\cos^2x=1$$by squaring the infinite Taylor series of $\sin x$ and $\cos x$. But it came out quite beautiful, involving also a combinatoric... | Nice proof! To address your question,
Do Taylor's series of those functions depend on this identity? Because if they do, the proof will be circular.
You are right to be worried about circularity! However, which concepts or theorems depend on which others is a matter of certain flexibility. Often, we take one thing to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 3,
"answer_id": 0
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If $\cos A=\tan B$, $\cos B=\tan C$ … If $\cos A=\tan B$, $\cos B=\tan C$ and $\cos C=\tan A$, prove that $\sin A=\sin B=\sin C$.
My Attempt.
Let us consider $x$, $y$ and $z$ as:.
$$x = \tan^2A$$
$$y = \tan^2B$$
$$z = \tan^2C$$
$$\cos^2A = \tan^2B$$
$$\frac {1}{\sec^2A}= \tan^2B$$
$$\frac {1}{1 + \tan^2A} = \tan^2B$$
$... | If no two of $x,y,z$ are equal then all of them are unequal. Assume $x>y>z.$ From $(1) and (2)$ $y - z + y(x - z) = 0 $. This implies $ y < 0 $(since $y - z > 0, x - z > 0)$.....$(4)$. From $(2) and (3)$:$ z -x + z(y - x) = 0 $
This implies $z < 0$ (since $z - x < 0, y - x < 0)$ $......(5)$.
From $(3) and (1)$: $x - y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Combinatorial sum identity for a choose function $\sum\limits_{k=-m}^{n} \binom{m+k}{r} \binom{n-k}{s} =\binom{m+n+1}{r+s+1}$ I want to show that the following holds:
$$\sum_{k=-m}^{n} \binom{m+k}{r} \binom{n-k}{s} =\binom{m+n+1}{r+s+1}.$$
I have an idea of what is going on here. On the RHS we are selecting $r+s+1$ ele... | Suppose we seek to evaluate
$$\sum_{k=-m}^n {m+k\choose r} {n-k\choose s}
= \sum_{k=0}^{m+n} {k\choose r} {m+n-k\choose s}.$$
Here we may assume that $r$ and $s$ are non-negative. We introduce
$${m+n-k\choose s} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+n-k-s+1}} \frac{1}{(1-z)^{s+1}} \; dz.$$
This integral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Inequality $\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}\geq1+ \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ for positive $a$, $b$, $c$
If $A=\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}$ and $B = \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ and $a,b,c>0.$ Then prove that $A\geq 1+B$
$\bf{My\; Try::}... | More way.
We'll rewrite our inequality in the following form
$$3\sum\limits_{cyc}\frac{1}{a(1+a)}\geq\sum\limits_{cyc}\frac{1}{1+a}\sum\limits_{cyc}\frac{1}{a}$$
which is Rearrangement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit
I tried expanding out the denominator, and then using the comparison test with $\frac{1}{k^3}$ but I think... | Given that the rising and falling factorials are defined as
$$
\begin{array}{l}
n^{\,\overline {\,m\,} } = n\left( {n + 1} \right)\, \cdots \;\left( {n + m - 1} \right) \\
n^{\,\underline {\,m\,} } = n\left( {n - 1} \right)\, \cdots \;\left( {n - \left( {m + 1} \right)} \right) \\
n^{\,\underline {\, - \,m\,} }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1941502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc$ for $\frac1a+\frac1b+\frac1c=3$ and $a,b,c>0$
Let $a$, $b$, and $c$ be positive real numbers with $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that:
$$
ab(a+b) + bc(b+c) + ac(a+c) \geq \frac{2}{3}(a^{2}+b^{2}+c^{2})+ 4abc.
$$
... | We note that
\begin{align*}
ab(a+b)+bc(b+c)+ac(a+c) &= a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\\
&=\frac{a^2}{\frac{1}{b}}+\frac{b^2}{\frac{1}{a}} + \frac{b^2}{\frac{1}{c}}+
\frac{c^2}{\frac{1}{b}}+\frac{a^2}{\frac{1}{c}}+\frac{c^2}{\frac{1}{a}}\\
&\ge \frac{4(a+b+c)^2}{2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})} \qquad \mbox{(by th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
how to find analytic function?
If $v=\Large \frac{x-y}{x^2+y^2}$ then the analytic function $f(z)=u(x, y)+iv(x, y)$ is
a) $z+c\qquad $ b) $z^{-1}+c\qquad $ c) $\frac{1-i}{z}+c\qquad $ d) $\frac{1+i}{z}+c\qquad $
my try: i computed
$$\frac{\partial v}{\partial x}=\frac{y^2-x^2+2xy}{(x^2+y^2)^2}$$
$$\frac{\partial v}... | Let us suppose $u$ is of a similar form to $v$.
$$u = \frac{ax + by}{x^2 + y^2} + c$$
$$\frac{\partial u}{\partial x} = \frac{a(x^2 + y^2) - 2x(ax+by)}{(x^2 + y^2)^2} = \frac{ay^2 -ax^2 - 2bxy}{(x^2 + y^2)^2}$$
Equating coefficients with $\frac{\partial v}{\partial y}$ gives $a = 1,b = 1$. At this point it is worth che... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{b+c}}\geq\frac{9}{\sqrt{a+b+c+15}}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}$$
It seems nice enough.
I proved this inequality by Holder, but it quit... | $$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}\iff\sqrt{a+b+c+15}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\right)\ge9$$
Making $c=\frac{1 } {ab}$ the expression becomes
$$f(a,b)= \left(\sqrt{\frac{a^2b}{ab^2+1}}+\sqrt{\frac{ab^2}{a^2b+1}}+\sqrt{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Fourier series of $f(x)=\pi-x$ $$f(x)=\pi-x \qquad x \in [0,2 \pi[$$
$$a_0=\frac{1}{\pi} \ \int_0^{2 \pi}f(x) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi} (\pi-x) \ dx=0$$
$$a_n=\frac{1}{\pi} \ \int_0^{2 \pi} \cos(nx) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi}(\pi \ \cos(nx)-x \ \cos(nx)) \ dx=$$
$$\frac{1}{\pi} \ \Big( \ \Big[\frac... | Any function $F(x)$ defined on $[0,2\pi]$ expands as the Fourier series
$$ F(x) = \dfrac{A_0}2 + \sum^\infty_{n=1}A_n\cos(nx) + \sum^\infty_{n=1} B_n \sin(nx),$$
$$\text{where} \qquad A_0 = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)dx, \quad
A_n = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)\cos(nx)dx, \quad
B_n = \frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Algebraic inequalities with equality condition Let $a,b,c$ be positive real numbers such that $abc(a+b+c) =3$. Prove that $$(a+b)(b+c)(c+a)\ge 8$$ and determine when equality holds.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $9uv^2-w^3\geq8$, which is a linear inequality of $v^2$
and the condition does not depend on $v^2$.
Thus it remains to prove our inequality for an extremal value of $v^2$.
$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimum value of $2^{\sin^2x}+2^{\cos^2x}$ The question is what is the minimum value of
$$2^{\sin^2x}+2^{\cos^2x}$$
I think if I put $x=\frac\pi4$ then I get a minimum of $2\sqrt2$. But how do I prove this?
| To minimise
\begin{align}
y & = 2^{\cos^2 x} + 2^{\sin^2 x} \\
\frac{dy}{dx} & = \frac{d(2^{\cos^2 x})}{dx} + \frac{d(2^{\sin^2 x})}{dx}
\end{align}
we want $\frac{dy}{dx} = 0$. For the first summand we have
\begin{align}
f(x) & = 2^{\cos^2 x} \\
\ln f(x) & = \cos^2 x\ln 2 \\
\frac{d(\ln f(x))}{dx} & = \frac{d(\cos^2 x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
How to approach Inequalities that can be solved using Rearrangement Inequality? I am learning Inequalities and today I've encountered the famous Rearrangement Inequality. I'll state the theorem:
Consider any two collections of real numbers in increasing order, $$a_1\leq a_2\leq....a_n$$ and $$b_1\leq b_2\leq....b_n$$... | 1.61
$(a,b,c)$ and $(a^2,b^2,c^2)$ are the same ordered.
Thus, by Rearrangement $\sum\limits_{cyc}(a^2\cdot a)\geq\sum\limits_{cyc}a^2\cdot b$, which gives
$$a^3+b^3+c^3\geq a^2b+b^2c+c^2a$$
In another hand, $(ab,ac,bc)$ and $\left(\frac{1}{c^2},\frac{1}{b^2},\frac{1}{a^2}\right)$ are the same ordered.
Thus, by Rearr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \bino... | For any polynomial $f(x) = \sum\limits_{k=0}^{\deg f} a_k x^k$, we will use $[x^k] f(x)$ to denote $a_k$, the coefficient of $x^k$ in $f(x)$. Notice
$$\binom{n-k}{k} = [x^k](1+x)^{n-k} = [x^n] (x+x^2)^{n-k}$$
We have
$$\require{cancel}
\begin{align}\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\binom{n-k}{k}
&= \sum_{k=0}^n (-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
} |
Solving an ordinary differential equation nonlinear I have the following
$$ y' = \frac{ xy }{x^2 + y^2} $$
My approach would be to rewrite this in terms of $x$
$$ x' = \frac{x}{y} + \frac{y}{x} $$
And then let $u = \frac{y}{x} \implies u' = \frac{x + y x'}{x^2} = \frac{1}{x} + \frac{u}{x} x' = \frac{1}{x} + \frac{u}{x}... | $$y' = \frac{ xy }{x^2 + y^2}=\frac{\frac{y}{x}}{1+\frac{y^2}{x^2}}$$
let $y=ux$
$$y'=u+xu'$$
so
$$u+xu'=\frac{u}{1+u^2}$$
$$xu'=\frac{-u^3}{1+u^2}$$
$$-\frac{dx}{x}=\frac{u^2+1}{u^3}du$$
$$-\frac{dx}{x}=\frac{1}{u}du+\frac{1}{u^3}du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the sum $\sum_{n=1}^{\infty} \frac{4n}{n^4+2n^2+9}$ Find the sum
$$\sum_{n=1}^{\infty} \dfrac{4n}{n^4+2n^2+9}.$$
By calculator, we can predict that its sum is equal to $\dfrac{5}{6}$ so I think we should use inequalities to prove it. And I found that
$\dfrac{5}{6(n^4+n^2)} < \dfrac{4n}{n^4+2n^2+9}< \dfrac{5}{6(n^2... | HINT:
$$(n^2)^2+3^2+2n^2=(n^2+3)^2-(2n)^2=(n^2+2n+3)(n^2-2n+3)$$
$$(n^2+2n+3)-(n^2-2n+3)=?$$
Now if $f(m)=m^2-2m+3,$
$f(m+2)=(m+2)^2-2(m+2)+3=m^2+2m+3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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If $ X$ is the incentre of $\Delta ABY $ show that $∠CAD=90°$. In a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ intersect at $X$. Let the circumcircles of triangles $AXD$ and $BXC$ intersect again at $Y$. If $X$ is the incentre of triangle $ABY$ , show that $∠CAD = 90°.$
| Look at quadrilateral $XCYD$.
*
*$\angle\, XDY = \angle \, XAY = \alpha$ as inscribed in a circle.
*$\angle \, BAC = \angle \, BAX = \angle \, XAY = \alpha$ since $AC$ passes through the incenter $X$ of triangle $ABY$ and therefore $AC$ is the interior angle bisector of angle $\angle \, BAY$.
*$\angle \, XDC = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving indefinite integral $\int \frac{dx}{(x^4-1)^3}$ I'm trying to solve next integral, but I can't start. WolframAlpha gives me really terrible answer and can't give any step-by-step instructions, so I simply does not know how to start.
$$\int \frac{dx}{(x^4-1)^3}$$
Please give any hint or start point of solving th... | One step that can simplify the partial fraction decomposition is to first perform an integration by parts with the choice $$u = -(2x)^{-3}, \quad du = \frac{3}{8}x^{-4} \, dx, \quad dv = -\frac{(2x)^3}{(x^4-1)^{3}} \, dx, \quad v = (x^4-1)^{-2},$$ giving $$\int \frac{dx}{(x^4-1)^3} = -\frac{1}{8x^3(x^4-1)^2} - \frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How do I prove that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})$ I am unsuccessfully attempting a problem from Spivak's popular book 'Calculus' 3rd edition. The problem requires proof for the following equation:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}+\dotsb+xy^{n-2}+y^{n-1})$$
The solution to the problem, contai... | The simplest way is to prove first
$$1-t^n=(1-t)(1+t+\dots+t^{n-1})$$
by induction.
The formula is trivial for $n=1$. So suppose the formula is valid for some $n\ge 1$, and consider the exponent $n+1$.
\begin{align}
1-t^{n+1}&=(1-t^n)+(t^n-t^{n+1})\\
&=(1-t)(1+t+\dots+t^{n-1})+t^n(1-t)\\
&=(1-t)(1+t+\dots+t^{n-1+t^n}),... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Calculate limit involving $\sin$ function Calculate the following limit:
$$\lim_{x \rightarrow 0} \frac{x-\overbrace{\sin (\sin (...(\sin x)...))}^{150\ \text{times}\ \sin}}{x^3}$$
I tried applying L'Hospital's rule, but it got too messy.
Thank you in advance!
| Here is a more or less elementary calculation of the limit.
A quite commonly known limit is
$$
\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\sin x-\sin 0 }{x-0}=\sin'(0)=\cos(0)=1.
\tag{1}\label{1}
$$
Also, you may prove with L'Hopital that
$$
\lim_{x\to 0}\frac{x-\sin x}{x^3}=\lim_{x\to 0}\frac{1-\cos x}{3x^2}=\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Which of following is/are true? In expansion $(x^2+1+\frac{1}{x^2})^n$ , n $\in \mathbb{N}$
1.number of terms is $2n+1$
2.coefficient of constant term is $2^{n-1}$
3.coefficient of $x^{2n-2}$ is $n$
4.coefficient of $x^2$ is $n$
I tried by taking lcm and writing as $\frac{(1+x^2+x^4)^n}{x^{2n}}$. How do i proceed? tha... | Usign the Newton formula we have:
$\left(x^2+1+\frac{1}{x^2}\right)^n=(x^2+1)^n+n(x^2+1)^{n-1}\cdot\frac{1}{x^2}\cdots\left(\frac{1}{x^2}\right)^n$
$=x^{2n}+n(x^2)^{n-1}\cdot 1\cdots 1+n+n(x^2+1)^{n-1}\cdot\frac{1}{x^2}\cdots\left(\frac{1}{x^2}\right)^n$
$=x^{2n}+n(x)^{2n-2}\cdot 1\cdots 1+n+n(x^2+1)^{n-1}\cdot\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Fourier series for $\cosh(x)$ Find the odd Fourier series for the periodic function whose period is $2\pi$, and which is equal to $\cosh{x}$ in the range $0<x<\pi$. Hence show that
$\frac{\pi}{4}sech{(\frac{\pi}{2}})=\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)+(2n+1)^{-1}}$.
$a_n=\frac{1}{2\pi}\int^{\pi}_0(e^x+e^{-x})\co... | $$\int_{0}^{\pi}\cosh(x)\sin(nx)\,dx = \frac{n}{n^2+1}\left(1-(-1)^n \cosh\pi\right)\tag{1}$$
leads to
$$ \cosh(x) = \frac{2}{\pi}\sum_{n\geq 1}\frac{n\sin(nx)}{n^2+1}(1-(-1)^n\cosh \pi) \tag{2} $$
for any $x\in(0,\pi)$. By evaluating both sides of $(2)$ at $x=\frac{\pi}{2}$,
$$ \cosh\frac{\pi}{2}=\frac{2}{\pi}\sum_{m\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculation of product of cycles Let $n \in \mathbb{N}$, $n \geq 3$ and $n = 2m$ for some $m \in \mathbb{N}$. How do I efficently get the result of the product of cycles $$\begin{pmatrix} 2 & n\end{pmatrix}\begin{pmatrix} 3 & n-1\end{pmatrix}\cdots\begin{pmatrix} m & m+2\end{pmatrix}\begin{pmatrix} 1 & 2 & \dots & n\en... | We have $(k \; n-k+2)\; (n-k+2 \; n-k+3)\; (n-k+3 \;k-1) = (k \; k-1) $ (a middle transposition comes from a $(1 \; 2 \; \ldots \; n)$ cycle ) and it could be applied to compute this product:
$k$ goes to $k-1$ hence we have a cycle:
$ (n\; n-1 \; n-2\; \ldots \; 3 \; 2 \; 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1973134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find partial fraction $(2x^2-1)/((x^2-1)(2 x^2+3))$
Find partial fraction of $\cfrac{(2x^2-1)}{(x^2-1)(2 x^2+3)}$?
My attempt:
I did google and I tried to solved it as :
Let’s first get the general form of the partial fraction decomposition.
$\cfrac{(2x^2-1)}{(x^2-1)(2 x^2+3)}=\cfrac{A}{(x+1)}+\cfrac{B}{(x-1)}+\cfra... | you are mistaken in the second step, the expression must be simplified as
$$2x^2-1=A(x-1)(2x^2+3)+B(x+1)(2x^2+3)+(C+Dx)(x^2-1)$$
now, compare powers of $x^3, x^2, x$ & constant terms, you will get
$$2A+2B+D=0\tag 1$$
$$-2A+2B+C=2\tag 2$$
$$3A+3B-D=0\tag 3$$
$$-3A+3B-C=-1\tag 4$$
solve all four linear equations to ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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What is the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$ I've tried this but didn't get the answer :
Let $S=2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$
Using $n^3-(n-1)^3 = 3n^2-3n+1$,
\begin{align}
S &= 3(2014^2)-3(2014)+1+3(2012^2)-3(2012)+1+\ldots+3(2^2)-3(2)+1
\\&= 3\left ( 2014(... | \begin{align}&\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007
\\&=1007\left(\frac{12(1008)(2015)}{6}-\frac{2(1008)(3)}{2}+ 1\right)
\\&=1007\left(2(1008)(2015)-(1008)(3)+ 1\right)
\\&=1007\left(2(1007+1)(2015)-(1007+1)(3)+ 1\right)
\\&=1007\left(1007(2(2015)-3)+2(2015)-3+ 1\right)
\\&=1007\left(1007(2(20... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove that $\frac{\frac{1}{11}+\frac{1}{12}+\dots+\frac{1}{200}}{\frac{1}{10*11}+\frac{1}{11*12}+\dots+\frac{1}{19*20}}>19$ Prove that $$\dfrac{\dfrac{1}{11}+\dfrac{1}{12}+\dots+\dfrac{1}{200}}{\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+\dots+\dfrac{1}{19\cdot20}}>19$$
My attempt:The denominator is $\frac{1}{20}$ using ... | Your calculation for the denominator (telescopic sum and not series) was the hard part, since
\begin{align}\dfrac{\dfrac{1}{11}+\dfrac{1}{12}+\dots+\dfrac{1}{200}}{\dfrac{1}{20}}&>\dfrac{\dfrac{1}{200}+\dfrac{1}{200}+\dots+\dfrac{1}{200}}{\dfrac{1}{20}}=\dfrac{\dfrac{200-10}{200}}{\dfrac1{20}}=\frac{\dfrac{19}{20}}{\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the set of points on the complex plane for which $z^2 + z + 1$ is real and positive
Find the set of points on the complex plane for which $z^2 + z + 1$ is real and positive.
For being real I applied the condition: $z^2 + z + 1=\bar{z}^2 +
\bar{z} + 1$
For the expression to be positive I applied:
$z^2+z+1>0 \imp... | $$z^2+z+1=k,\ k>0$$
$$z^2+z+(1-k)=0$$
$$z=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot(1-k)}}{2\cdot1}$$
$$z=\frac{-1\pm\sqrt{4k-3}}2\tag1$$
When $0<k<\frac34$, $\sqrt{4k-3}$ is imaginary and $z=-\frac12\pm\frac{\sqrt{3-4k}}2i$; the range of the imaginary part is $(-\frac{\sqrt3}2,\frac{\sqrt3}2)$. When $k\ge\frac34$, $\sqrt{4k-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Showing equivalence $(A \Rightarrow (B\Rightarrow C))\Leftrightarrow (A\wedge B) \Rightarrow C)$ My original excercise is, to use the truthtable, but do i have to say anything here?
$\begin{array}{ccc|c@{}ccc@{}ccc@{}c@{}ccc@{}c@{}ccc@{}ccc@{}c}
A&B&C&(&A&\Rightarrow&(&B&\Rightarrow&C&)&)&\Leftrightarrow&(&(&A&\land&B&... | Yes, it is possible to do this without truth table.
$A \Rightarrow B$ means either $A$ is false (or) $B$ is true.
So, $(A \Rightarrow B)$ $\Leftrightarrow \overline{A} \lor B$
$A \Rightarrow (B \Rightarrow C) \Leftrightarrow (A \Rightarrow (\overline{B} \lor C)) \Leftrightarrow \overline{A} \lor \overline{B} \lor C$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$ $$P(x)=x^{4}-4x^{3}+12x^{2}-24x+24$$
I'd like to prove that $P(x)=|P(x)|$. I don't know where to begin. What would be the first step?
| .In this question, you have to attempt to complete two squares, namely:
$$
x^4-4x^3 + 12x^2-24x+24 = x^4 -4x^3+6x^2-4x+1 + 6x^2-20x+23
\\ = (x-1)^4 + 6x^2 -20x+23
$$
Now, as it turns out, we can complete the second square, and:
$$
6x^2-20x+23 = \frac{2}{3}(3x-5)^2 + \frac{19}{3}
$$
Hence, we can rewrite the whole e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 10,
"answer_id": 2
} |
What number must be subtracted from the denominator of $\frac{10}{23}$ to make the result $\frac13$? Is it $\frac{10}{23} - \frac{1}x = \frac13$ or something similar?
| Given that $\frac{10}{23} - \frac{1}{x} = \frac{1}{3}$ then multiply both sides by $3 \cdot 23 \cdot x$ to obtain $3\cdot 10 \cdot x - 3 \cdot 23 = 23 \cdot x$ which becomes $x = \frac{69}{7}$. This can be seen in the following:
\begin{align}
\frac{10}{23} - \frac{1}{x} &= \frac{10}{23} - \frac{7}{69} = \frac{30}{69} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Use a power series to approximate the definite integral to six decimal places. Use a power series to approximate the definite integral to six decimal places.
$\int_{0}^{0.3} \frac{x^2}{1+x^4}$
I'm not sure how to find the sum of this for solving when it has x's in the numerator, this is what I assumed however.
$\frac... | $$ I = \int_{0}^{\frac{3}{10}}\frac{x^2}{1+x^4}\,dx = \int_{0}^{\frac{3}{10}}\frac{x^2-x^6}{1-x^8}\,dx = \sum_{n\geq 0}\left(\frac{\left(\frac{3}{10}\right)^{8n+3}}{8n+3}-\frac{\left(\frac{3}{10}\right)^{8n+7}}{8n+7}\right) \tag{1}$$
and the last series is a series with positive terms. Since
$$ \sum_{n\geq 2}\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the minimum value of this Expression (Without Calculus!) Problem:
Suppose that $x,y$ and $z$ are positive real numbers verifying $xy+yz+zx=1$ and $k,l$ are two positive real constants. The minimum value of the expression:
$$kx^2+ly^2+z^2$$ is $2t_0$, where $t_0$ is the unique root of the equation $2t^3+(k+l+1)t-k... | Let $f = kx^2+ly^2+z^2,$ suppose $f_{\min} = 2t > 0.$ We need to prove
$$kx^2+ly^2+z^2 \geqslant 2t(xy+xz+yz),$$
equivalent to
$$kx^2+ly^2+z^2+t(x^2+y^2+z^2)\geqslant 2t(xy+xz+yz)+t(x^2+y^2+z^2),$$
or
$$(k+t)x^2+(l+t)y^2+(1+t)z^2 \geqslant t(x+y+z)^2.$$
Using the Cauchy-Schwarz inequality, we have
$$(k+t)x^2+(l+t)y^2+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What's the sum of all the positive integral divisors of $540$?
What's the sum of all the positive integral divisors of $540$?
My approach: I converted the number into the exponential form. And found out the integral divisors which came out to be $24$.
But couldn't find the sum...Any trick?
| Prime factorization of 540 is $2^2\cdot 3^3\cdot 5$. Also sum of divisors of 540 equals:
$$\begin{align}
\sigma\left(2^2\cdot 3^3\cdot 5\right) &=\frac {2^3-1}{2-1}\cdot\frac{3^4 - 1}{3-1}\cdot \frac{5^2 -1}{ 5-1}\\ &=7\cdot 40\cdot 6\\ &= 1680
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out this problem the following summation emerged:
$$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$
The is equivalent to
$$\begin{align}
\sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdo... | You may do the following to get rid of the powers of two and transform it into a standard sum:
\begin{equation} \binom{2m}{m}=\frac{(2m)!}{m!m!}=\frac{2^m m! 1\cdot3\cdot5\cdot \cdots (2m-1)}{m!m!}=2^{2m} \frac{(1/2)(3/2)\cdots (m-1/2)}{m!}=2^{2m}\binom{m-1/2}{m}~~~~~~ (1) \end{equation}
Then your sum becomes
$$\sum_{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1989966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Singapore math olympiad Trigonometry question: If $\sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ$, then $ab=$?
$$\text{If}\; \sqrt{9-8\sin 50^\circ} = a+b\csc 50^\circ\text{, then}\; ab=\text{?}$$
$\bf{My\; Try::}$ We can write above question as $$\sin 50^\circ\sqrt{9-8\sin 50^\circ} = a\sin 50^\circ+b$$
Now for Left si... | $$\sqrt{9-8\sin 50^\circ}$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin^350^\circ}$$
$$(\text{using }\sin^2x=1-\cos^x)$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin50^\circ(1-\cos^250^\circ)}$$
$$=\csc50^\circ\sqrt{9\sin^250^\circ-8\sin50^\circ+8\sin50^\circ\cos^250^\circ}$$
$$(\text{using }2\sin x\cos x=\sin2x)$$
$$=\cs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Surface area of sphere within a cylinder I have to
Compute the surface area of that portion of the sphere $x^2+y^2+z^2=a^2$ lying within the cylinder $\Bbb{T}:=\ \ x^2+y^2=by.$
My work:
I start with only the $\Bbb{S}:=\ \ z=\sqrt{a^2-x^2-y^2}$ part and will later multiply it by $2$.
$${\delta z\over \delta x}={-x\o... | Another approach in spherical coordinates: parametrize the surface with
\begin{cases}
x=a \sin\phi \cos\theta \\
y=a \sin\phi \sin\theta \\
z=a \cos\phi
\end{cases}
with $(\theta,\phi)\in [0,\pi]\times[0,\Phi]$, where $\Phi \in ]0,\pi]$ is the solution to
\begin{cases}
x^2+y^2+z^2=a^2 \\
x^2+y^2=by
\end{cases}
Substit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothi... | From what you did, we want to prove that
$$\frac{1}{1+\log_32}\lt \log_32,$$
i.e.
$$\log_32\gt \frac{\sqrt 5-1}{2},$$
i.e.
$$2\gt 3^{(\sqrt 5-1)/2},$$
i.e.
$$2^2\gt 3^{\sqrt 5-1},$$
i.e.
$$3\cdot 2^2\gt 3^{\sqrt 5}$$
It is sufficient to prove that
$$12\gt 3^{2.25},$$
i.e.
$$12^2\gt 3^4\sqrt 3,$$
i.e.
$$16\gt 9\sqrt 3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Mathematical Induction for Recurrence Relation I have solved the following recurrence relationship:
$T(1) = 1$
$T(n) = T(n-1) + n + 2$
so
$T(n) = \frac{1}{2}n^2+\frac{5}{2}n -2$
I am now trying to perform mathematical induction to prove this.
$Basis:$
$T(1)=1=3-2=\frac{1}{2} + \frac{5}{2} - 2 $
$Induction:$
$T(k+1) = T... | What can I do next?
Hint. Prove that, for $k=1,2,\cdots,$
$$
\frac{1}{2}k^2 + \frac{7}{2}k +1=\frac{1}{2}(k+1)^2+\frac{5}{2}(k+1)-2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum\limits_{cyc}\sqrt[3]{a^2+4bc}\geq\sqrt[3]{45(ab+ac+bc)}$
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$\sqrt[3]{a^2+4bc}+\sqrt[3]{b^2+4ac}+\sqrt[3]{c^2+4ab}\geq\sqrt[3]{45(ab+ac+bc)}$$
A big problem in this inequality there is around $(1,1,0)$.
I tried Holder:
$$\left(\sum\limits_{cy... | Assume that $ab+bc+ca > 0$. Rewrite the inequality as $\sqrt[3]{u} + \sqrt[3]{v} + \sqrt[3]{w} \ge 3$ where
$$u = \frac{27(a^2+4bc)}{45(ab+bc+ca)}, \ v = \frac{27(b^2+4ca)}{45(ab+bc+ca)}, \ w = \frac{27(c^2+4ab)}{45(ab+bc+ca)}.$$
We will use the fact that
$$\sqrt[3]{x} \ge \frac{3x(5+4x)}{5x^2+20x+2}, \ x \ge 0$$
whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
How to calculate eigenvalues? Given is differential equation: $2y''+5y'-3y=0$
Write this equations as 1. order system and calculate eigenvalues of the matrix of this differential equation.
My idea:
$z_{1}=y$
$z_{2}=y'$
$\frac{d}{dx}\begin{pmatrix} z_{1} \\ z_{2} \end{pmatrix} =\begin{pmatrix} z_{2} \\ \frac{3}{2}z_{1}... |
$$\frac{d}{dx}\begin{pmatrix} z_{1} \\ z_{2} \end{pmatrix} =\begin{pmatrix} z_{2} \\ \frac{3}{2}z_{1}-\frac{5}{2}z_{2} \end{pmatrix}$$
You're close, simply rewrite:
$$\begin{pmatrix} z_{2} \\ \frac{3}{2}z_{1}-\frac{5}{2}z_{2} \end{pmatrix}
= \begin{pmatrix} 0z_1+ 1z_{2} \\ \frac{3}{2}z_{1}-\frac{5}{2}z_{2} \en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the total sum of the cardinalities of all subsets of a set? I'm having a hard time finding the pattern. Let's say we have a set
$$S = \{1, 2, 3\}$$
The subsets are:
$$P = \{ \{\}, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\} \}$$
And the value I'm looking for, is the sum of the cardinalities o... | For the subsets of size $k$ one needs the number of ways how to draw $k$ distinguishable elements from $\lvert S \rvert$ available, which is given by the binomial coefficient. This is multiplied by the number of elements, thus $k$, and summed up over all possibilites:
$$
\sum_{A \in 2^S} \lvert A \rvert = \sum_{k=0}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 11,
"answer_id": 3
} |
Does there exist a basis for the set of $2\times 2$ matrices such that all basis elements are invertible? As the title says, I'm wondering whether there exists a basis for the set of $2\times 2$ matrices (with entries from the real numbers) such that all basis elements are invertible.
I have a gut feeling that it is fa... | Consider the counter-example basis:
$$\beta:=\left\{\begin{pmatrix} 1&0\\0&1\end{pmatrix},\begin{pmatrix} 0&1\\1&0\end{pmatrix},\begin{pmatrix} 1&0\\1&1\end{pmatrix},\begin{pmatrix} 0&1\\1&1\end{pmatrix}\right\}$$
Let $A=\begin{pmatrix} x_1&x_2\\x_3&x_4\end{pmatrix}$, let the following equation:
$$A=a\begin{pmatrix} 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
Prove: $\frac{1}{11\sqrt{2}} \leq \int_0^1 \frac{x^{10}}{\sqrt{1+x}}dx \leq \frac{1}{11}$ Prove: $\frac{1}{11\sqrt{2}} \leq \int_0^1 \frac{x^{10}}{\sqrt{1+x}}dx \leq \frac{1}{11}$
Hint: Use the (weighted) Mean Value Theorem for Integrals.
The MVT for Integrals:
Suppose that $u$ is continuous and $v$ is integrable and n... | Choosing $v(x) = \frac{1}{\sqrt{x+1}}$ doesn't work, I think:
I didn't use MVT directly. I used something from the proof of MVT:
$$m \int_0^1 \frac{1}{\sqrt{x+1}} dx \le \int_0^1 \frac{x^{10}}{\sqrt{x+1}} \le M \int_0^1 \frac{1}{\sqrt{x+1}} dx$$
The best $m$ and $M$ I got are $0$ and $1$ resp as $0 \le x^{10} \le 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the limit of $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}$ Find the limit of $q(0) = \lim\limits_{x\to 0}\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}$
For this I think I should use De l'Hopital's rule but it takes a lot time and I can't get to answer.
Can we use the De l'Hopital's rule twice?or thr... | Since we have a limit at $0$ of polynomials, only the term with the least degree matters. We have
$$
(x+1)(x+2)(x+3)(x+4)-24=24+(2\times3\times4+1\times3\times4+1\times2\times3+1\times2\times4)x+c_2x^2+c_3x^3+c_4x^4-24=50x+c_2x^2+c_3x^3+c_4x^4.
$$
Then
\begin{align}
\frac{(x+1)(x+2)(x+3)(x+4)-24}{x(x+5)}
&=\frac{50x+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Boolean Simplification using Algebra Having trouble showing the following relation:
$$A\cdot B + A'\cdot B' + B\cdot C = A\cdot B + A'\cdot B' + A'\cdot C
$$
using Boolean Algebra. Any help is appreciated.
| $$\begin{array}{rcl}(A+A')=1 &\mbox{ and } &(B+B')=1 \\
(A+A')\cdot(B+B')=1& \implies &(A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') =1\\
A\cdot B + A'\cdot B' +B\cdot C &=& (A\cdot B + A\cdot B' + A'\cdot B + A'\cdot B') \cdot (A\cdot B + A'\cdot B' +B\cdot C)\\
&=&(A\cdot B +0+A\cdot B\cdot C)+(0+0+0) \\
&&+ (0+0+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A square of a rational between two positive real numbers ?! Let $ a,b \in \mathbb{R}^+ ,(a<b) $. Prove that there is a rational number $ q $ such that $ a<q^2<b $, without using square root function.
Can anyone help me ?
| Write $q=r/s$ with positive integers $r$ and $s$. We need to show that
$$as^2\lt r^2\lt bs^2$$
for some choice of $r$ and $s$. If the interval $(as^2,bs^2)$ does not contain a square, then there is some integer $n$ such that $n^2\le as^2$ while $(n+1)^2\ge bs^2$. This implies $2n+1\ge(b-a)s^2$, which implies
$$as^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating an infinite series using partial fractions... I am having trouble evaluating an infinite series that uses partial fractions. The problem is as follows:
$$
\sum_{n = 1}^{\infty}{1 \over n\left(n + 1\right)\left(n + 2\right)}
$$
I realize that this is a telescoping series, but I am unable to find a general fo... | $$\frac{1}{n(n+1)(n+2)} = \frac{1}{(n+1)} \cdot \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2}\right) = \frac{1}{2} \left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right)$$
So
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} = \frac{1}{2} \left(\frac{1}{1(1+1)}-\frac{1}{(1+1)(1+2)}+\frac{1}{2(2+1)}-\frac{1}{(2+1)(2+2)}+\cdots\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Use euclidean algorithm to calculate the multiplicative inverse of $5$ in $\mathbb{Z}_{12}$ I really like to know the exact way how its done. Here is what I wrote:
$5$ must have a multiplicative inverse because $\text{ gcd }(12,5)=1$
So $5x \equiv 1 \text{ mod } 12 \Leftrightarrow x \equiv 5^{-1}(\text{mod } 12)$
$$12=... | I will show you how to do it through continued fractions, that is essentially the same. We have:
$$\frac{12}{5}=2+\frac{2}{5}= 2 +\frac{1}{2+\frac{1}{2}} = [2;2,2]\tag{1}$$
and if $\frac{p_n}{q_n},\frac{p_{n+1}}{q_{n+1}}$ are two consecutive convergents of the same continued fraction, the difference between them is $\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Card probability problem: Five cards will be dealt from a well-shuffled deck. Find the chance of getting an ace or a king among the 5 cards. Five cards will be dealt from a well-shuffled deck.
Find the chance of getting an ace or a king among the 5 cards.
I think question means getting at least a king or an ace, and he... | Using 1 -
$$1 - \frac {\binom{44}{5} } { \binom{52}{5} } = 0.5821374704$$
Using 1 + 2 + 3 + 4 + 5
$$\frac {\binom{8}{1} \binom{44}{4} + \binom{8}{2} \binom{44}{3} + \binom{8}{3} \binom{44}{2} + \binom{8}{4} \binom{44}{1} + \binom{8}{5}} { \binom{52}{5} } = 0.5821374704$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solving a set of linked recurrent relations I'm trying to find a method how to solve this set of linked recurrent relations.
$$
\left\{\begin{matrix}
a_{n} = 3*a_{n-1} + b_{n-1}\\
b_{n} = 2*a_{n-1} + 2*b_{n-1}\\
c_{n} = b_{n-1} + 3*c_{n-1}\\
d_{n} = a_{n-1} + 2*b_{n-1} + 3*c_{n-1}\\
e_{n} = 6*d_{n-1} + 3*e_{n-1} ... | It's not quite clear to me if this is equivalent to the method you suggested but in the case of nice, linear recurrences like you have, one can write:
$$ \vec{x}_n
= \begin{pmatrix}a_n\\b_n\\c_n\\d_n\\e_n\\f_n\\g_n\end{pmatrix}
= \begin{pmatrix}3 & 1 & 0 & 0 & 0 & 0 & 0
\\ 2 & 2 & 0 & 0 & 0 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2014376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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find $a,b,c$ such that $ \big| a + b\sqrt{2}+ c\sqrt{3} \big| <10^{-3} $ i know that linear combinations of $1$, $\sqrt{2}$, and $\sqrt{3}$ are dense in the real number line.
$$ \overline{ \mathbb{Z}[\sqrt{2},\sqrt{3}] } = \mathbb{R} $$
how quickly to these numbers converge to the entire number line. what are the... | Here are some lower bounds for the best you can do.
If $\alpha=a+b\sqrt{2}+c\sqrt{3}$ and $|\alpha|<1$ with $a,b,c$ integers, and $abc\neq 0$, then $$|\alpha|>\frac{1}{(2|b|\sqrt{2}+1)(2|c|\sqrt{3}+1)(2|a|+1)}\tag{1}$$
I get this lower bound since: $\alpha(\alpha-2b\sqrt{2})(\alpha-2c\sqrt{3})(\alpha-2a)$ must be an i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Quick solution of an equation in factorials Now I realize it is not that hard but it is a test question and should have a fast way to solve as indicated by the teacher. However, I fail to see it.
If both $m$ and $n$ are two-digit natural numbers and $m! = 156 \cdot n!$ find the value of $m-n$.
| $$
\binom{m}{n}=\frac{m!}{n!(m-n)!}=\frac{156}{(m-n)!}
$$
Thus $(m-n)!$ is a divisor of $156=2^2\cdot 3\cdot 13$. This forces $(m-n)!$ to be a divisor of $2^2\cdot3=12$ (the prime $13$ cannot appear for obvious reasons), leaving $m-n=1$, $m-n=2$ or $m-n=3$.
*
*If $m=n+1$, we have
$$
\binom{n+1}{n}=156=\binom{n+1}{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
If $\alpha = 2\arctan(2\sqrt{2}-1)$ and $\displaystyle \beta = 3\arcsin\left(\frac{1}{3}\right)+\arcsin\left(\frac{3}{5}\right)\;,$ Then prove that $\alpha>\beta$
$\bf{M... | On the one hand we have
$$\tan{\alpha\over2}=2\sqrt{2}-1>\sqrt{3}=\tan{\pi\over3}\ ,$$
hence ${\alpha\over2}>{\pi\over3}$, or $$\alpha>{2\pi\over3}\ .$$
On the other hand the convexity of $$t\mapsto\arcsin t\qquad(0\leq t\leq{\pi\over2})$$ implies
$$\arcsin{1\over3}<{2\over3}\arcsin{1\over2}={2\over3}\>{\pi\over6}={\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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factoring $(x^6 - y^6)$: what is going on here? I apologize if this question already exists, but it was quite difficult to word.
In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:
factor $x^6 - 64$
almost everyone other than me (including my teacher, a... | There are two errors in the factorization
$$x^6-64=(x^2-4)(x^4+4x-16)$$
The constant term in the quartic should be $+16$ instead of $-16$, and the linear term, $4x$, should be a quadratic term, $4x^2$. That is, the correct factorization is
$$x^6-64=(x^2-4)(x^4+4x^2+16)$$
This now agrees with the factorization $(x+2)(x... | {
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"source": "stackexchange",
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$f(z)=\frac{(iz+2)}{(4z+i)}$ maps the real axis in the $\mathbb{C}$-plane into a circle Find the center and radius of the circle. Also find the points on the complex plane which is mapped onto the center of the circle.
| The inverse $z'$ of a point $z$ with respect to the circle centred at $a$ with radius $r$ is given by $$z' = \frac{r^2}{\overline{z}-\overline{a}}+a$$
We have $$\frac{iz+2}{4z+i} = \frac{9}{4(4z+i)}+ \frac{i}{4} = \frac{9}{16 \left(z - \frac{-i}{4} \right)} + \frac{i}{4} =\frac{9}{16 \left(\overline{z} - \frac{-i}{4} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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$\lim_\limits{n\to \infty}\ (\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})$? what is the value given to this limit?
$$\lim_{n\to \infty}\ \bigg(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + \cdots + \frac{1}{n+1}\bigg)$$
Is it simply 0 because each term tends to 0 and you are just summing up ... | By using this,
we get $\lim_{n\to \infty}(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})=\ln{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
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Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$
Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$ subject to $x,y,z>0$ and $x+y+z=1$
This inequality is equivalent to;
$\left(\frac{x+1}{4}\cdot\frac1x\right)\left(\frac{y+1}{4}\cdot\frac1y\right)\left(\f... | By AM-GM we have:
$$
\frac{1}{3}=\frac{1}{3}(x+y+z)\geq\sqrt[3]{xyz}\implies\frac{1}{27}\geq xyz\implies\frac{1}{27xyz}\geq 1.
$$
Using AM-GM again, with 4 terms this time, we obtain
$$
1+\frac{1}{x}=1+\frac{1}{3x}+\frac{1}{3x}+\frac{1}{3x}\geq 4\sqrt[4]{\frac{1}{27x^3}}.
$$
Do that for the other 2 terms and multiply. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Determine the remainder when $f(x) = 3x^5 - 5x^2 + 4x + 1$ is divided by $(x-1)(x+2)$ This question arose while I was tutoring a student on the topic of the Remainder Theorem. Now, the Remainder Theorem tells us that when a polynomial $p(x)$ is divided by a linear factor $(x-a)$, the remainder is simply $p(a)$. However... | Hint: the remainder will be a polynomial of degree (at most) $1$ so:
$$f(x) = (x-1)(x+2)q(x) + ax + b$$
Substitute $x=1,-2$ in the above and you get two equations in $a,b$.
[ EDIT ] For a less conventional approach (justified in the answer here) note that $(x-1)(x+2)=0 \iff x^2=-x+2$. Repeatedly using the latter sub... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the $n$ derivative of $y= e^{2x}\sin^2 x$ We have
\begin{align*}
y&= e^{2x}\sin^2 x\\
&= e^{2x}\left(\frac{1-\cos 2x}{2}\right)\\
&= \frac{e^{2x}}{2} - \frac{e^{2x}\cos 2x}{2}
\end{align*}
Then
\begin{align*}
y^{(n)} &= \left(\frac{e^{2x}}{2}\right)^{(n)} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}\\
&= 2^{n-1}e... | We know that $$\sin^2(x) = \dfrac{1 - \cos(2x)}{2}= \dfrac{1}{2} -\dfrac{1}{2}\Re(\exp(2ix)), $$ where $\Re$ is the real part of a complex number.
It follows that
$$y =\exp(2x)\sin^2(x) = \dfrac{\exp(2x)}{2} - \dfrac{1}{2}\Re(\exp(zx)), $$ with $z = 2+2i.$
Using complex-valued derivatives:
$$\dfrac{dy}{dx} = \exp(2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 ... | Note that
$$f(x)=\frac{1}{x^2 + 2x + 2}=\frac{1/2}{1+(x+x^2/2)}\\=\frac{1}{2}\left(1-(x+x^2/2)+(x+x^2/2)^2-((x+x^2/2)^3+o(x^3)\right).$$
In general, for any positive integer $n$,
$$2f(x)=1+\sum_{k=1}^n x^k(1+x/2)^k+o(x^n)=
\sum_{k=1}^n (-1)^k\sum_{j=0}^{n-k} \frac{1}{2^j}\binom{k}{j}x^{j+k}+o(x^n)\\
=1+\sum_{k=1}^n \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof for rational roots Let $f(x) = a_0+a_1x+\cdots+a_nx^n$ be a polynomial of degree $n$ over $\Bbb Z$. |
Part A:
Given $\frac pq$ is a rational root of a polynomial$$f(x)=a_nx^n+x_{n-1}x^{n-1}+\ldots a_0\tag1$$
Where $a_n\in\mathbb{Z}$. We wish to show that $p|a_0$ and $q|a_n$. Since $\frac pq$ is a root$$0=a_n\left(\frac pq\right)^n+\ldots+a_0\tag2$$
Multiplying by $q^n$, we have$$a_np^n+\ldots+a_0q^n=0\tag3$$
Examinin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2026005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $G(t) = \sum_\limits{n=1}^{\infty} (\cos\frac{n\pi}{2}-1) \cos(\frac{nt\pi}{3}) $ Simplify $G(t) = \sum_\limits{n=1}^{\infty} (\cos\frac{n\pi}{2}-1) \cos(\frac{n\pi t}{3}) $
I am unsure how to simplify this in the best form, anyone have any ideas?
| Here is an alternate representation using the fact that
\begin{equation}
\cos\left(\dfrac{n\pi}{2}\right)-1=\begin{cases}
-1&\text{ for }n=4k-3,\,n=4k-1\\
-2&\text{ for }n=4k-2\\
\phantom{-}0&\text{ for }n=4k\\
\end{cases}
\end{equation}
\begin{eqnarray*}
G(t)&=&-\sum_{k=1}^{\infty}\left[\cos\left(\frac{4k-3}{3}\pi t\r... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does 1 not congruent imply Fermat n=4? A natural number is said to be congruent if it is the area of a right triangle with rational sides. I've been told that Fermat actually proved his last theorem for $n=4$ by proving that number 1 is not congruent, but I can't seem to find the connection!
It is probably very eas... | The key thing to realize is that $1$ being or not being a congruent number is not directly related to the Fermat equation $x^4 + y^4 = z^4$. It is related to $X^4 + Y^2 = Z^4$, where the second term on the left side is only a square, not a fourth power. The second equation is what Fermat used, and it is the one you'l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many positive integers less than 1,000,000 have the sum of their digits equal to 19? How many positive integers less than $1,000,000$ have the sum of their digits equal to $19$ ?
I tried to answer it by using stars and bars combinatorics method.
The question says that sum of $6$ digit numbers { considering the num... | The number is $<$ $1000000$ ,
$\Rightarrow$ it contains 6 digits.
Each of these digits can be one of $0,1,2,3....9$
$\Rightarrow$ problem reduces to no of integral solution to the following equation
$d_1+d_2+d_3+d_4+d_5+d_6 = 19$ where $0\leq d_i \leq 9$
Using generating function :
$$\begin{align*} & \ \ \ \left [ x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Finding the zeroes of the function. I would like to find the zeroes of the following function given that $3-i$ is a zero of $f$:
$f(x) = 2x^4-7x^3-13x^2+68x-30$
Please explain to me how to do this problem. Thanks!
| Let $x_1,x_2,x_3,x_4$ be the roots of:
$$f(x) = 2x^4-7x^3-13x^2+68x-30$$
Then by Vieta's formulas:
$$
\begin{cases}
x_1+x_2+x_3+x_4=\frac{7}{2} \\
x_1x_2x_3x_4 = -\frac{30}{2}
\end{cases}
$$
Now you know $x_1=3-i$, and its conjugate must be also a root $x_2=3+i$, so the above reduces to:
$$
\begin{cases}
x_3+x_4=\frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solution of this summation: $\sum_{i=1}^{10} {\frac{i}{i^4+i^2+1}}$ The summation in question:
$$\sum_{i=1}^{10} {\frac{i}{i^4+i^2+1}}$$
I have been able to factorize $i^4+i^2+1$ as $(i^2+i+1)(i^2-i+1)$ but I doubt this will help.
What is the solution?
| Writting $$\sum^{10}_{i=1}\frac{i}{i^4+i^2+1} = \frac{1}{2}\sum^{10}_{i=1}\bigg[\frac{(i^2+i+1)-(i^2-i+1)}{(i^2+i+1)(i^2-i+1)}\bigg]$$
$$ = \frac{1}{2}\sum^{10}_{i=1}\bigg[\frac{1}{i^2-i+1}-\frac{1}{i^2+i+1}\bigg]$$
Now Using Telescopic Sum (expanding summation), Then we get
$$ =\frac{1}{2}\bigg[\frac{1}{1^2-1+1}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$? How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$?
| The polynomial in invariant under any permutation of $a,b,c$.
One way to attack this sort of polynomial is express it in terms
of its elementary symmetric polynomials and try to locate/detect any patterns that are useful.
We have $3$ variables, so we have $3$ elementary symmetric polynomials.
Let us call them $A,B,C$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the eigenvalues of the following matrix. So I have to find the eigenvalues of this matrix: $\begin{bmatrix}-7/9&-4/9&8/9\\-4/9&-1/9&2/9\\2/9&-4/9&8/9\end{bmatrix}$.
What I did is start by writing it like this: $\begin{bmatrix}-7/9 - \lambda&-4/9&8/9\\-4/9&-1/9-\lambda&2/9\\2/9&-4/9&8/9-\lambda\end{bmatrix}... | Note: I recommend you factor out a $\dfrac{1}{9}$ to reduce the possibility of errors, so repeat this approach.
We want to find the determinant of
$$\begin{vmatrix}-7/9 - \lambda&-4/9&8/9\\-4/9&-1/9-\lambda&2/9\\2/9&-4/9&8/9-\lambda\end{vmatrix}=0$$
Expanding along the top row since there are no zeros to take advantag... | {
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"timestamp": "2023-03-29T00:00:00",
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Compute the characteristic of $R = Z[j]/(2-5j),$ where $j^3 = 1$ and $j^2 \ne 1.$ Compute the characteristic of $R = Z[j]/(2-5j),$ where $j^3 = 1$ and $j^2 \ne 1.$
Here i provide proof found in book. But i don't understand a lot from this proof.
$\overline x$ denotes an element in the quotient group involved.
Here we ... | $j^3 = 1$ impies $0=j^3-1=(j-1)(j^2+j+1)$.
$j^2 \ne 1$ implies $j \ne 1$ and so $j^2+j+1=0$.
$|\cdot|^2$ here is just the complex number absolute value, $|z|^2 = z \bar z$, which implies $|zw| = |z| \, |w|$.
Using that $\bar j = j^2 = -1-j$, it is easy to compute $|a + bj|^2 = a^2 + b^2 − ab$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show $\sum_{n=1}^\infty \left(\frac{n}{2n-1}\right)^n$ converges I'm trying to show that $\sum_{n=1}^\infty \left(\frac{n}{2n-1}\right)^n$ converges. Using the Limit Ratio Test for Series, we want to show that $\lim_{n\to \infty} \left\lvert \frac{a_{n+1}}{a_n}\right\lvert<1$. However, I'm having trouble finding said l... | If $a_n = \left(\frac{n}{2n-1}\right)^n$, then
\begin{align} \frac{a_{n+1}}{a_n} &= \left(\frac{n+1}{2(n+1)-1}\right)^{n+1}/\left(\frac{n}{2n-1}\right)^n \\
&= \left(\frac{n+1}{2n+1}\right)\frac{\left(\frac{n+1}{2n+1}\right)^n}{\left(\frac{n}{2n-1}\right)^n} \\
&= \left(\frac{n+1}{2n+1}\right)\left(\frac{(n+1)(2n-1)}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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52 cards, 5 picked, 3 of same suit 5 cards are drawn from a normal deck of cards (52). What is the probability that 3, and only 3, of the cards are of the same suit?
I'm wondering if my reasoning is sound:
C(13,1) * C(4,3) * C(12,2) * C(4,1)^2
-------------------------------------
C(52,2)
| 3 solutions.
Solution1: Exclude 1 suit from 52
Solution2: Consider cases for pair or not a pair for the 2 cards left
Solution3: Consider cases for cards of same suit or not of the same suit for the 2 cards left
Solution2:
4 ways for choose 3 cards from 13. (same suit)
for remaining 2 cards, we have 2 cases to consider:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the curve of intersection of the surfaces $x^2+y^2=z$ and $x+y+z=1$
Find the curve of intersection of the surfaces $x^2+y^2=z$ and $x+y+z=1$.
So if we substitute $z=1-x-y$ into the equation $x^2+y^2=z$ we get $x^2+y^2=1-x-y$ which can be written in the form $\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right... | After the coments, a possible parametrization is
$$r(t)=\left(\,-\frac12+\sqrt\frac32\,\cos t\,,\,\,-\frac12+\sqrt\frac32\,\sin t\,,\,\,2-\sqrt\frac32(\cos t+\sin t)\,\right)\;,\;\;0\le t\le 2\pi$$
Observe that we have above a general vector of the form $\;(x,y,z)\;$ , where for the third coordinate we have either $\;z... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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how can I prove this identity How can I prove this identity
$$\cos (3x/2)-\cos(x/2)=-2 \sin (x/2) \sin(x)$$
I try this but does not work with me $$\cos(3x/2) = \cos(x + x/2) = \cos (x) \cos (x/2) - \sin (x) \sin (x/2)$$
I do not know what to do after this step.
| We have $$\cos(A+B) -\cos(A+B) = (\cos A\cos B -\sin A\sin B)-(\cos A\cos B +\sin A\sin B) = -2\sin A\sin B. $$ Let $A+B = C$ and $A-B = D$. Our equation then becomes $\cos C -\cos D = -2\sin (\frac{C+D}{2})\sin (\frac{C-D}{2})$. With $C=\frac{3x}{2}$ and $D=\frac{x}{2}$, we have $$ \cos(\frac{3x}{2})-\cos(\frac{x}{2})... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:
Is $\ln(2)=\frac{1}{2}\ln(2)$??
The following argument seems suggesting that the answer is yes:
We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
which has a mathematically determined value $... | As others have pointed out, rearrangement is not allowed, so I will give you the most extreme case:
$$S=1-\frac12-\frac14-\frac16-\frac18+\frac13+\frac15+\frac17+\frac19+\frac1{11}+\frac1{13}+\dots$$
The general pattern is simple. Start at $1$. Then add up the even terms until the sum is less than $0$. Then add up t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
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Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$?
Why is $\log\left(\frac{2a+2b+3}{2a+2b+1}\right)>\frac{1}{a+b+1}$ ? with $a,b\ge 0$
If this is true (I don't know whether it is true, just inserted some values in wolfram alpha) then I can show that $\frac{1}{a+b+1}<\int_{a}^{a+1}\int_{b}^{b+1}\frac{1... | We have $$\frac{2a+2a+3}{2a+2b+1} = 1+ \frac{2}{2a+2b+1} = 1+\frac{1}{a+b+1/2}.$$
The Maclauren series for $\log(1-x)=-x-x^2/2-x^3/3-\cdots.$ So your left side
$$=\log\left(1-\frac{-1}{a+b+1/2}\right) = \frac{1}{a+b+1/2} - \frac{1}{2}\left(\frac{1}{a+b+1/2}\right)^2+\cdots . $$
The series is alternating so the above... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$
Find the limit of $a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}$ when $x\to\infty$, for given real numbers $a$, $b$, $c$.
I would like to see a solving method without l'Hopital or Taylor expansion.
| \begin{align*}
L &= \lim_{x\to \infty} (a\sqrt{x+1}+b\sqrt{4x+1}+c\sqrt{9x+1}) \\[5pt]
&= \lim_{x\to \infty} \sqrt{x}
\left(
a\sqrt{1+\frac{1}{x}}+2b\sqrt{1+\frac{1}{2x}}+3c\sqrt{1+\frac{1}{3x}}
\right) \\[5pt]
&= \lim_{x\to \infty} \sqrt{x}
\left[
a\left( 1+\frac{1}{2x} \right)+
2b\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054190",
"timestamp": "2023-03-29T00:00:00",
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Use Mathematical Induction to prove equation? Use mathematical induction to prove each of the following statements.
let $$g(n) = 1^3 + 2^3 + 3^3 + ... + n^3$$
Show that the function $$g(n)= \frac{n^2(n+1)^2}{4}$$ for all n in N
so the base case is just g(1) right? so the answer for the base case is 1, because 4/4 = 1
t... | Assuming you are asking what the inductive step is, you simply need to assume that $g(n) =\frac{n^2(n+1)^2}{4}$ and then use it to evaluate g(n+1) like so:
\begin{equation}
g(n+1) = g(n) + (n+1)^3
\end{equation}
\begin{equation}
g(n+1) = \frac{n^2(n+1)^2}{4} + (n+1)^3
\end{equation}
\begin{equation}
g(n+1) = \frac{4(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$.
We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$.
How can we continue?
| $x=0\iff y=0.$ And $x=y=0$ is a solution.
If $x\ne 0\ne y$ then $$2x^2+y^2=3x^2y\implies x^2(2-3y^2)=-y^2\ne 0.$$ This implies $x^2$ divides $y^2.$ Now $y^2/x^2$ is an integer for non-zero integers $x,y$ only if $x$ divides $y$. So we have $$y=zx$$ with integer $z.$
So $2x^2+z^2x^2=3x^3z$. Since $x\ne 0$ we have $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Distributing $8$ different articles among $7$ boys
Problem Statement:-
Find the number of ways in which $8$ different articles can be distributed among $7$ boys, if each boy is to receive at least one article.
Attempt at a solution:-
First, start by numbering the boys from $1$ to $7$, which can be done in $7!$ ways.... | The main problem here is that you have ordered the boys and also regarded the distribution of the presents as extra options.
Probably the simplest way to get to the answer is to select one boy to get two presents ($7$ options), he chooses those two ($\binom 82$) then each boy chooses in turn ($6!$ options) :
$$7\binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum\limits_{cyc}\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}\geq5$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\sqrt{\frac{8ab+8ac+9bc}{(2b+c)(b+2c)}}+\sqrt{\frac{8ab+8bc+9ac}{(2a+c)(a+2c)}}+\sqrt{\frac{8ac+8bc+9ab}{(2a+b)(a+2b)}}\geq5$$
I tried Holder, but without succ... | There is the following estimation of Nguyenhuyen_AG.
$$\sqrt{\frac{8a(b+c)+9bc}{(2b+c)(2c+b)}} \geqslant\frac{17a^2-b^2-c^2+20(ab+ac+bc)}{3(a^2+b^2+c^2+4(ab+bc+ca))}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove Fibonacci Identity using generating functions I have the following summation identity for the Fibonacci sequence.
$$\sum_{i=0}^{n}F_i=F_{n+2}-1$$
I have already proven the relation by induction, but I also need to prove it using generating functions, but I'm not entirely sure how to approach it.
I do know that th... | Since a generating function for the Fibonacci numbers $(F_n)_{n\geq 0}=(1,1,2,3,5,8,\ldots)$ is
\begin{align*}
\frac{1}{1-x-x^2}=1+x+2x^2+3x^3+5x^4+8x^5+\cdots
\end{align*}
and multiplication of a generating function $A(x)$ with $\frac{1}{1-x}$ results in summing up the coeffcients
\begin{align*}
\frac{1}{1-x}A(x)&=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$ The following is a problem from an older exam which the instructor didn't provide solutions to.
Evaluate $$\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$$
using only real-analytic techniques.
My work: this boils down to factoring $x^4 - x^3 + x^2 - x+ 1$ which I am... | Substituting $x\mapsto\frac1x$, we get
$$
\begin{align}
\int_1^\infty\frac{x^2\,\mathrm{d}x}{1+x^5}
&=\int_0^1\frac{x\,\mathrm{d}x}{1+x^5}
\end{align}
$$
Therefore,
$$
\begin{align}
\int_0^\infty\frac{x^2\,\mathrm{d}x}{1+x^5}
&=\int_0^1\frac{\left(x^2+x\right)\mathrm{d}x}{1+x^5}\\
&=\int_0^1\sum_{k=0}^\infty(-1)^k\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Combinatorial coefficients squared If $C_0,C_1,C_2,...,C_n$ are the combinatorial coefficients in the expansion of $(1+x)^n$ then prove that:
$$1C_0^2+3C_1^2+5C_3^2+...+(2n+1)C_n^2=\dfrac{(n+1)(2n)!}{n!n!}=(n+1)\binom {2n}n$$
I am able to compute the linear addition but not the squares of coefficients.
Thanks!
| Note that
$C_r=\binom nr$.
It is a well-known result that $\sum_{r=0}^n\binom nr^2=\binom {2n}n$ which can be proven easily using the Vandermonde identity.
For even $n$:
The number elements is odd, index from $0$ to $(\frac n2-1)$, $n$, and from $(\frac n2+1)$ to $n$
Since $\binom nr=\binom n{n-r}$, then $\sum_{r=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
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.