Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find the derivative of $(2x+5)^3(3x-1)^4$ How to find a derivative of the following function?
$$\ f(x)=(2x+5)^{3} (3x-1)^{4}$$
So I used:
$$(fg)'= f'g + fg'$$ and $$(f(g(x)))'= f'(g(x)) + g'(x)$$
Then I got:
$$ f(x)= 6(2x+5)^{2} + 12(3x-1)^{3}$$
and I don't know how to get the solution from this, which is: $$ 6(... | I assume that you are supposed to have $$f(x)=(2x+5)^3(3x-1)^4,$$ instead of what you've written. By product rule, we have $$f'(x)=(2x+5)^3\bigl[(3x-1)^4\bigr]'+\bigl[(2x+5)^3\bigr]'(3x-1)^4.$$ Applying chain rule (which you've incorrectly stated, by the way) gives us $$f'(x)=(2x+5)^3\cdot 4(3x-1)\cdot(3x-1)'+(3x-1)^4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/580881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integrate: $ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $ without using complex analysis methods Can this integral be solved without using any complex analysis methods: $$ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $$
Thanks.
| $x \to 1/x$. We then have
$$I = \int_0^{\infty}\dfrac{\log(x)}{(1+x^2)^2}dx = \int_0^1\dfrac{\log(x)}{(1+x^2)^2}dx + \int_1^{\infty}\dfrac{\log(x)}{(1+x^2)^2}dx$$
$$\int_1^{\infty}\dfrac{\log(x)}{(1+x^2)^2}dx = \int_1^0 \dfrac{x^2\log(x)}{(1+x^2)^2}dx$$
Hence,
$$I = \int_0^1 \dfrac{1-x^2}{(1+x^2)^2}\log(x)dx = \int_0^1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/581155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 4
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Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.
Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$a^2 + b^2 + c^2 \ge a + b + c$.
I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?
| Using C-S and then AM-GM gives something shorter:
$$3(a^2+b^2+c^2) \ge (a+b+c)^2 \Rightarrow a^2+b^2+c^2 \ge (a+b+c)\frac{(a+b+c)}{3} $$
$$\ge (a+b+c)\sqrt[3]{abc}=a+b+c$$
And we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/581992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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Proof that $\sum_{k=2}^{\infty} \frac{H_k}{k(k-1)} $ where $H_n$ is the sequence of harmonic numbers converges? How to prove that $$\displaystyle \sum_{k=2}^{\infty} \dfrac{H_k}{k(k-1)} $$ where $H_n$ is the sequence of harmonic numbers converges and that $\dfrac{H_n}{n(n-1)}\to 0 \ $
I have already proven by inductio... | This just begs to be telescoped:
$$\sum_{k=2}^\infty\sum_{n=1}^k \frac{1}{k(k-1)n} = \sum_{k=2}^\infty \frac{1}{k(k-1)} + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty\frac{1}{k(k-1)}$$
$$= \sum_{k=2}^\infty (\frac{1}{k-1} -\frac{1}{k}) + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty(\frac{1}{k-1} -\frac{1}{k})$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/583536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Proving $a^ab^b + a^bb^a \le 1$, given $a + b = 1$ Given $a + b = 1$, Prove that $a^ab^b + a^bb^a \le 1$;
$a$ and $b$ are positive real numbers.
| Equation $(2)$ is the same as Don Anselmo's proof, but the rest is different.
Bernoulli's Inequality says that for $0\le x\le1$
$$
\left(1+\frac{a-b}b\right)^x\le1+x\frac{a-b}b\tag{1}
$$
multiply by $b$
$$
a^xb^{1-x}\le xa+(1-x)b\tag{2}
$$
substitute $x\mapsto1-x$ in $(2)$
$$
a^{1-x}b^x\le (1-x)a+xb\tag{3}
$$
add $(2)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/586465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 1
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A question on greatest common divisor I had this question in the Maths Olympiad today. I couldn't solve the part of the greatest common divisor. Please help me understand how to solve it. The question was this:
Let $P(x)=x^3+ax^2+b$ and $Q(x)=x^3+bx+a$, where $a$ and $b$ are non-zero real numbers. If the roots of $P(x)... | It was not clear to me how the given four relations of the roots gave that $a=b=1$, so here is a different approach.
We have that
$$
Q(x)=x^3+bx+a=0\tag{1}
$$
and
$$
P\left(\frac1x\right)=\frac1{x^3}+a\frac1{x^2}+b=0\tag{2}
$$
have the same roots. Multiplying $(2)$ by $\dfrac{x^3}{b}$ yields
$$
R(x)=x^3+\frac abx+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/588176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Maximum of |sin x| + |sin y| + |sin z| If $x$, $y$ and $z$ are real numbers with the property $x+y+z= \pi$, then the maximum of $\sin x+\sin y+\sin z$ is $3\sqrt{3}/2$.
Now, if $x+y+z=0$ then is the maximum of $|\sin x| + |\sin y| + |\sin z|$ again $3\sqrt{3}/2$?
| $$f(x,y):=|\sin x|+|\sin y|+|\sin(-x-y)|=|\sin x|+|\sin y|+|\sin(x+y)|$$
In order to maximize $f$, let's consider $$\nabla f=\left\langle\frac{\sin x}{|\sin x|}\cos x+\frac{\sin (x+y)}{|\sin(x+y)|}\cos(x+y),\frac{\sin y}{|\sin y|}\cos y+\frac{\sin (x+y)}{|\sin(x+y)|}\cos(x+y)\right\rangle$$
If we require $\cos x-\cos(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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If $U = (1, 2)$, $V = (3, -4)$, is the answer to $2U + \frac{1}{2}V$ the vector $(3.5, 2)$? Check my answer. If $U = (1, 2)$, $V = (3, -4)$, is the answer to $2U + \frac{1}{2}V$ the vector $(3.5, 2)$?
I did the following:
\begin{align*}
U : ( ( 2 * 1 ), ( 2 * 2 ) ) &= ( 2 , 4 )\\
V : ( ( 0.5 * 3 ), ( 0.5 *... | Your final answer is correct, but some of your intermediate steps could be considered wrong (or at least unclear).
First of all, you write
$$U : (2 * 1, 2 * 2) = (2, 4).$$
It's not clear at all what you mean by this. What you have done is calculated $2U$. A much better way of writing this is
$$2U = (2 * 1, 2 * 2) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/591864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question about Jordan form - Linear algebra Quick question,
We are given that the characteristic polynomial of a matrix $A$ is $P_A(x)=(x-1)^4$
We are asked to find all the possible jordan forms of $A$.
Obviously the minimal polynomial of $A$ can be $m_A(x)=(x-1)^k$, $ 1 \leq k \leq 4$
I understand that the size of the... | The possibilities are
$$\begin{pmatrix}1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\end{pmatrix}\;\;,\;\;\begin{pmatrix}1&1&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\end{pmatrix}$$
$$\begin{pmatrix}1&1&0&0\\
0&1&0&0\\
0&0&1&1\\
0&0&0&1\end{pmatrix}\;\;,\;\;\begin{pmatrix}1&1&0&0\\
0&1&1&0\\
0&0&1&0\\
0&0&0&1\end{pmatrix}$$
$$\begin{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Fast way to prove that $(a^2+b^2+c^2-ab-ac-bc)^2=(a-b)^2\times (a-c)^2 + (b-c)^2\times(b-a)^2 + (c-b)^2\times(c-a)^2$ What is the most simplest way to prove that $$(a^2+b^2+c^2-ab-ac-bc)^2=(a-b)^2\times (a-c)^2 + (b-c)^2\times(b-a)^2 + (c-b)^2\times(c-a)^2$$
Please!! Thanxx
| The simplest way is probably just to multiply out both sides separately and see that they are equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/593121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
coloring a tetrahedron (abstract algebra) Consider a regular tetrahedron. Each of its faces can either be painted blue or red. Up to rotation how many ways can the tetrahedron be painted?
I am thinking that applying Burnside's counting formula will work, but it has been so long since I have used Burnside's counting for... | For the general case, we can give an elementary combinatorial solution.
General Case: We have $n$ different colors. The vertices of a regular tetrahedron will be painted using these colors. The colorings obtained from one another as a result of rotating the tetrahedron are considered identical. The number of different ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/593655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Indented Path Integration The goal is to show that
$$\int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx = \frac{\pi^2}{6}$$
and that
$$\int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx = \frac{\pi}{\sqrt{3}}.$$
So, we start with the function
$$f(z) = \frac{z^{1/3}\log(z)}{z^2 + 1}.$$
Let $c_r$ be the upper semicircle with radius r ... | In this case, I would use a keyhole contour about the positive real axis. Consider the contour integral
$$\oint_C dz \frac{z^{1/3}}{1+z^2} \log{z}$$
where $C$ is the keyhole contour. As it is clear that the integral about the circular arcs, large and small, vanish as their respective radii go to infinity and zero, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/595015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Deriving the formula of a summation Derive the formula for
$$
\sum_{k=1}^n k^2
$$
The solution's that I was given has $k^3 + (k-1)^3$ as the first step but doesn't say how it got to that. Any help?
| Note $(k+1)^3-k^3=3k^2+3k+1$ so $$\begin{align} n^3=\sum_{k=0}^{n-1} (k+1)^3-k^3&=\sum_{k=0}^{n-1} 3k^2+3k+1\\ &=3\sum_{k=0}^{n-1}k^2+3\sum_{k=0}^{n-1}k+\sum_{k=0}^{n-1}1\\
&=3\sum_{k=0}^{n-1}k^2+3\frac{(n-1)n}{2}+n\\
\frac{n^3}{6}-\frac{n(n-1)}{2}-\frac n3&=\sum_{k=0}^{n-1}k^2\\
\frac{n(n-1)(2n-1)}{6}&=\sum_{k=0}^{n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/600199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Spring Calculation - find mass
A spring with an $-kg$ mass and a damping constant $9$ can be held stretched $2.5 \text{ meters}$ beyond its natural length by a force of $7.5 \text{ Newtons}$. If the spring is stretched $5 \text{ meters}$ beyond its natural length and then released with zero velocity, find the mass tha... | 1) find the $k$
2)$\zeta=c/(2\sqrt{k/m})=1 $ for critical damping. So solve for $m$ as you alreaday know $c$ and $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/603745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove the following trigonometric identity. $$\frac{\sin x - \cos x +1}{\sin x + \cos x -1}=\frac{\sin x +1}{\cos x}$$
I tried substituting $\sin^2x+\cos^2x = 1$ but I cannot solve it.
| I will start like Timotej, but finish differently.
$$\cos x(\sin x-\cos x+1)=\cos x(1+\sin x)-\cos^2x=\cos x(1+\sin x)-(1-\sin^2x)$$
$$=(1+\sin x)\{\cos x-(1-\sin x)\}$$
$$\implies \cos x(\sin x-\cos x+1)=(1+\sin x)(\sin x+\cos x-1)$$
Now change the sides of $\displaystyle \cos x, \sin x+\cos x-1$
Let me derive some ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/604169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Inverse Trig & Trig Sub Can someone explain to me how to solve this using inverse trig and trig sub?
$$\int\frac{x^3}{\sqrt{1+x^2}}\, dx$$
Thank you.
| You can use hyperbolic substitution, i.e. let $x=\sinh t$ then
\begin{align*}
\int\frac{x^3}{\sqrt{1+x^2}}\, dx&=\int \sinh^3 t\, dt\\
&=\int (\cosh^2t -1)\sinh t\, dt\\
&=\frac{\cosh^3 t}{3}-\cosh t+C\\
&=\frac{(\sqrt{1+x^2})^3}{3}-\sqrt{1+x^2}+C
\end{align*}
Also you can use triangle substitution : $x=\tan \theta$
\b... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Simple examples of $3 \times 3$ rotation matrices I'd like to have some numerically simple examples of $3 \times 3$ rotation matrices that are easy to handle in hand calculations (using only your brain and a pencil). Matrices that contain too many zeros and ones are boring, and ones with square roots are undesirable. A... | Here are some:
$$\left[ \begin {array}{ccc} 1/3&2/3&2/3\\ 2/3&-2/3&1/3\\ 2/3&1/3&-2/3\end {array} \right]
$$
$$\left[ \begin {array}{ccc} 2/7&3/7&6/7\\ 3/7&-6/7&2/7\\ 6/7&2/7&-3/7\end {array} \right]
$$
$$
\left[ \begin {array}{ccc} \frac{2}{11}&{\frac {6}{11}}&{\frac {9}{11}}\\ -{\frac {6}{11}}&-{\frac {7}{11}}&{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
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Inequality: if $\cos^2a+\cos^2b+\cos^2c=1$, prove $\tan a+\tan b+\tan c\geq 2(\cot a+\cot b+\cot c)$ Let $a,b,c$ be in $\left(0;\dfrac{\pi}2\right)$ such that $\cos^2a+\cos^2b+\cos^2c=1$. I am trying to prove the following inequality: $\tan a+\tan b+\tan c\geq 2\left(\cot a+\cot b+\cot c\right)$, but I do not know how.... | Since $a,b,c\in\left(0,\dfrac{\pi}{2}\right)$, and \[\cos^2a+\cos^2b+\cos^2c=1.\]We have\[a+b>\dfrac{\pi}{2},b+c>\dfrac{\pi}{2},c+a>\dfrac{\pi}{2}.\]Without loss of generality, we can assume that $a\geqslant b\geqslant c$, then\[1-3\cos^2a\geqslant1-3\cos^2b\geqslant 1-3\cos^2c,\]and\[\sin2a\leqslant \sin2b\leqslant \s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Improper Integral $\int\limits_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx$ How can I find a closed form for the following integral
$$\int_0^{\frac12}(2x - 1)^6\log^2(2\sin\pi x)\,dx?$$
| I tried the partial integration, with
$\frac{d \log^2(2\sin(\pi x))}{dx} = \frac{1}{3}\log^3(2\sin(\pi x))\frac{1}{\frac{1}{2\sin(\pi x)}2cos(\pi x)\pi} = \frac{1}{3}\log^3(2\sin(\pi x))\frac{sin(\pi x)}{\pi cos(\pi x)} = \frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}$
$\int_0^{\frac{1}{2}}(2x-1)^2\log^2(2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 5
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Limit with missing variables
Find the values for $a$ and $b$ such that
$$\lim_{x \to 0} \frac{\sqrt{a + bx} - \sqrt{3}}{x} = 3$$
Basically what I did so far was I started by multiplying by the conjugate. and obtained $$\frac{a+bx-3}{x(\sqrt{a+bx}+\sqrt 3)}$$ I don't know what to do after this.
| $$ \lim_{x \to 0}\frac{\sqrt{a+bx}-\sqrt{3}}{x}=3 \Longrightarrow \lim_{x \to 0}x\cdot\frac{\sqrt{a+bx}-\sqrt{3}}{x}=0 \cdot3=0$$
$$ i.e. \quad\lim_{x \to 0} (\sqrt{a+bx}-\sqrt{3})=0 \quad \therefore \sqrt{a}=\sqrt{3} ,\quad a=3$$
then
$$ \lim_{x \to 0}\frac{\sqrt{3+bx}-\sqrt{3}}{x}=\lim_{x \to 0}\frac{(3+bx)-3}{x(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/613546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the ratio of areas produced by perpendiculars from the $3$ sides of an equilateral triangle.
A point O is inside an equilateral triangle $PQR$ and the perpendiculars $OL,OM,\text{and } ON$ are drawn to the sides $PQ,QR,\text{and } RP$ respectively. The ratios of lengths of the perpendiculars $OL:OM:ON \text{ i... | HINT:
Let $OL, OM, ON$ be $x, 2x, 3x$ respectively . Now you will observe
$$area(\triangle PQR) = area(\triangle ONR) +area(\triangle ORM)+area(\triangle OMQ)+area(\triangle OQL)+area(\triangle OPL)+area(\triangle OPN)$$
since these are right angled and assuming $k$ be the side of the triangle you will get the resulta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Help finding the $\lim\limits_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$ I need help finding the $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$$
I did the following:
$$\begin{align*}
\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \... | $$F=\lim_{x \to \infty} \frac{\sqrt[3]x - \sqrt[5]x}{\sqrt[3]x + \sqrt[5]x}=\lim_{x \to \infty} \frac{x^{\frac13} - x^{\frac15}}{x^{\frac13} + x^{\frac15}}$$
As lcm$(3,5)=15$ I will set $\displaystyle x^{\frac1{15}}=y\implies x^{\frac13}=y^5$ and $\displaystyle x^{\frac15}=y^3$
$$\implies F=\lim_{y\to\infty}\frac{y^5-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Recurrence relation $f(n)=5f(n/2)-6f(n/4) + n$ I have been trying to solve this recurrence relation for a week, but I haven't come up with a solution.
$$f(n)=5f\left(\frac n2\right)-6f\left(\frac n4\right) + n$$
Solve this recurrence relation for $f(1)=2$ and $f(2)=1$
At first seen it looks like a divide and conquer eq... | Suppose we start by solving the following recurrence:
$$T(n) = 5 T(\lfloor n/2 \rfloor) - 6 T(\lfloor n/4 \rfloor) + n$$
where $T(0) = 0.$
Now let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
be the binary representation of $n.$
We unroll the recursion to obtain an exact formula for all $n:$
$$T(n) = \sum_{j=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Finding the orthogonal family of curves to a given family of curves. I am missing some. Given the family of curves: $F(x,y,x_0)=0$ where $F(x,y,x_0) = (x-x_0)^2 + y^2 - R^2\ ,\ x_0 \in \mathbb{R}$ find the orthogonal family.
This is my attempt:
I first get the differential equation satisfied by the curves $F(x,y,x_0)... | Remember that $\displaystyle \int\frac{dy}y = \log|y|$. It seems to be that easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that at least two of these inequalities are true: $|a-b|\le2$, $|b-c|\le2$, $|c-a|\le2$. It's given that: $$\begin{cases}a,b,c>0\\a+b+c\le4\\ab+bc+ac\ge4\end{cases}$$
Prove without using calculus that it's true that at least two of these are correct inequalities:$$\begin{cases}|a-b|\le2\\|b-c|\le2\\|c-a|\le2\end{... | Suppose two of these are false. Then
$$8<(a-b)^2+(a-c)^2+(b-c)^2=2(a^2+b^2+c^2)-2(ab+ac+bc)\le 16-8$$
i.e. $8<8$, a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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If $a,b\in\mathbb R\setminus\{0\}$ and $a+b=4$, prove that $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2\ge12.5$. If $a,b\in\mathbb R\setminus\{0\}$ and $a+b=4$, prove that $$\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge12.5$$
I could expand everything: $$a^2+2+\frac{1}{a^2}+b^2+2+\frac{1}{b^2}\ge12.5$$
Subtract ... | You can do this with the quadratic-arithmetic mean: (this is possible, because $a^2\geq 0$.)
$$
\sqrt{\frac{a^2+b^2}2}\geq\frac{a+b}2\\
a^2+b^2\geq \frac{(a+b)^2}2=8
$$
Now, you only have to proof $\frac 1{a^2}+\frac 1{b^2}\geq \frac 12$.
Just as before, we know that
$$
\frac 1{a^2}+\frac 1{b^2}\geq \frac{(\frac 1a+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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} |
Limits of square root $$\lim_{x\to\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x + \sqrt x} }}-\sqrt x\right) $$
(original screenshot)
Compute the limit
Can you please help me out with this limit problem
| $$
\begin{align}
\lim_{x\to\infty}\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}-\sqrt{x}
&=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}}{\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}+\sqrt{x}}\\
&=\lim_{x\to\infty}\frac{\sqrt{1+\frac1x\sqrt{x+\sqrt{x+\dots}}}}{\sqrt{1+\frac1x\sqrt{x+\sqrt{x+\dots}}}+1}\\
&=\frac12
\end{align}
$$
To s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Finding the maximum value of a function on an ellipse Let $x$ and $y$ be real numbers such that $x^2 + 9 y^2-4 x+6 y+4=0$. Find the maximum value of $\displaystyle \frac{4x-9y}{2}$.
My solution: the given function represents an ellipse. Rewriting it, we get $\displaystyle (x-2)^2 + 9(y+\frac{1}{3})^2=1$. To find the ma... | Here is yet another way to approach the problem, which fits somewhere between the methods described by Fly By Night and mathlove .
The expression we seek to maximize can be treated as a function of two variables $ \ z \ = \ f(x,y) \ = \ 2x \ - \ \frac{9}{2}y \ $ , which will produce as its level curves a family of p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How can I calculate this series? How can I calculate the series
$\displaystyle{%
\sum_{n=0}^{\infty }{\left(-1\right)^{n}
\over \left(2n + 1\right)\left(2n + 4\right)}\,\left(1 \over 3\right)^{n + 2}\
{\large ?}}$
| For $x\in (0,1)$ we have
\begin{eqnarray}
f(x)&=&\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(2n+4)}x^{2n}=\frac13\sum_{n=0}^\infty\left(\frac{(-1)^n}{2n+1}x^{2n}-\frac{(-1)^n}{2n+4}x^{2n}\right)\\
&=&\frac{1}{3x}\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}-\frac{1}{3x^4}\sum_{n=0}^\infty\frac{(-1)^n}{2n+4}x^{2n+4}\\
&=&\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is the smallest natural number n? What is the smallest natural number n for which there is a natural k, such that, the lasts 2012 digit in the representation decimal of $n^k$ are equal to 1?
I don't even know how to start with it ...
| $n^k\equiv-1\bmod4$ so $k$ and $n$ are odd.
If n ends in 2012 zeroes we have $n^k\equiv\frac{-1}{9}\bmod10^{2012}$ such a $k$ exists if and only if there is a $t$ such that $n^t\equiv-9\bmod10^{2012}$.
$n^t\equiv1\bmod5,n^t\equiv7\bmod16$ .Since $\lambda(5),\lambda(16)=4$ we have $t=1$ or $3$ ($t$ is odd). Check this ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Sum of Binomial Coefficients I am trying to find an exact formula for the following:
$\sum_{k=0}^n \binom{k+n}{k}$
Maple 16 says: $\binom{2n+1}{n+1}$
I can conclude with a induction but I would like to know if I can use a direct calculus ?
Thank you in advance.
| We want to choose $n+1$ numbers from the numbers $1,2,\dots,2n+1$. Clearly there are $\binom{2n+1}{n+1}$ ways to do this.
We count the number of choices in another way. First note that $\binom{k+n}{k}=\binom{n+k}{n}$.
Maybe the smallest number chosen is $n+1$. Then we must choose $n$ numbers from the remaining $n+0$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
evaluation of $\lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $ (1) $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$
(2)$\displaystyle \lim_{x\rightarrow \infty}\l... | $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} =\lim_{x\rightarrow \infty}\frac{lnx^n}{\lfloor x\rfloor}-1=\lim_{x\rightarrow \infty}\frac{nlnx}{\lfloor x\rfloor}-1$
Now $\displaystyle\frac{nlnx}{x} \le\frac{nlnx}{\lfloor x \rfloor}\le \frac{nlnx}{x-1}$. Hence $\displaysty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find coefficient of $x^8$ in $\frac{1}{(x+3)(x-2)^2}$ Using which way can we find coefficient of $x^8$ in $\frac{1}{(x+3)(x-2)^2}$? I have used binomial theorem but failed to find an answer for it.
| use $$(1-x)^{-1}=1+x+x^2 +x^3...$$
and $$(1+x)^{-1}=1-x+x^2-x^3....$$
your question is of the form
$$3^{-1}.(-2)^{-2}(\frac{x}{3}+1)^{-1}(1-\frac{x}{2})^{-2}$$
generally
$$(1-x)^{-k}=1+kx+\frac{k(k+1)}{2!}x^2+ \frac{k(k+1)(k+2)}{3!}x^3+.... $$
now to find coefficient of $x^8$ . group all possible terms so that the resu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$ Calculation of remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$
$\bf{My\; Try}::$ Using Division Algorithm:: $p(x) = q(x)\cdot g(x)+r(x)$
Now Let $r(x) = ax^2+bx+c$
So $(x+1)^n=q(x)\cdot (x-1)^3+ax^2+bx+c.................... | Your method is correct. I have not checked your calculations but you can try some small values of $n$. At minimum check $n=0,1,2$ where you know that $q(x)=0$.
Is there a simpler method? I prefer to write
$$
(x+1)^n = q(x) (x-1)^3 + a \, (x-1)^2 + b (x-1) + c
$$
and the proceed as you did. You can always expand out the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the improper integral $\int^\infty_0\frac{1}{x^3+1}\,dx$ $$\int^\infty_0\frac{1}{x^3+1}\,dx$$
The answer is $\frac{2\pi}{3\sqrt{3}}$.
How can I evaluate this integral?
| Using Partial Fraction Decomposition,
$$\frac1{1+x^3}=\frac A{1+x}+\frac{Bx+C}{1-x+x^2}$$
Multiply either sides by $(1+x)(1-x+x^2)$ and compare the coefficients of the different powers of $x$ to find $A,B,C$
As $\displaystyle1-x+x^2=\frac{4x^2-4x+4}4=\frac{(2x-1)^2+3}4$
Using Trigonometric substitution, set $\displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/625821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving an irrational equation Solve for $x$ in:
$$\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=\sqrt{5}$$
I used the property of proportions ($a=\sqrt{3+x}$, $b=\sqrt{3-x})$:
$$\frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{2a}{2b}=\frac{a}{b}$$
I'm not sure if that's correct.
Or maybe the notations $a^3=3+x$, $b^3=3-x$... | Yes you have applied Componendo and dividendo to get
$$\frac{\sqrt{3+x}}{\sqrt{3-x}}=\frac{\sqrt5+1}{\sqrt5-1}$$
Squaring we get $$\frac{3+x}{3-x}=\frac{6+2\sqrt5}{6-2\sqrt5}$$
Again apply Componendo and dividendo to get $$\frac3x=\frac6{2\sqrt5}\implies x=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $. My attempt:
We prove that $$\displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $$
It is sufficient to show that for an arbitrary real number $\epsilon\gt0$, there is a $K$
... | $\displaystyle\lim_{n\to \infty} \frac{n(23 + \frac{2}{n})}{n(4+\frac{1}{n})} = \cdots$
Use the fact that $\frac{\alpha}{n}$ tends to $0$ when $n$ tends to infinity, and theorems of limits of sequences.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/631462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove or disprove the implication: Prove or disprove the implication:
$a^2\cdot \tan(B-C)+ b^2\cdot \tan(C-A)+ c^2\cdot \tan(A-B)=0 \implies$
$ ABC$ is an isosceles triangle.
I tried to break down the left hand side in factors, but all efforts were in vain.
Does anyone have a suggestion?
Thank you very much!
| The claim holds if we replace the function $\tan(B-C)$ (and cyclic permutations) with $\tan^*(B-C)=\frac{\sin(B-C)}{\cos(B+C)}$:
Since:
$$\tan^*(B-C)=\frac{\sin B \cos C - \sin C \cos B}{\cos B \cos C - \sin B \sin C}$$
and $2ab\cos(C) = a^2+b^2-c^2, 2R\sin A=a, 2ab\sin C=4\Delta$, we have:
$$2R\tan^*(B-C)= \frac{2abc(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
How find this integral $\int_{0}^{\pi}\frac{2t+2\cos{x}}{t^2+2t\cos{x}+1}dx$? Find this follow integral
$$F(t)=\int_{0}^{\pi}\dfrac{2t+2\cos{x}}{t^2+2t\cos{x}+1}dx$$
where $t\in R$
my try:
$$F(t)=\int_{0}^{\frac{\pi}{2}}\dfrac{2t+2\cos{x}}{t^2+2t\cos{x}+1}dx+\int_{\frac{\pi}{2}}^{\pi}\dfrac{2t+2\cos{x}}{t^2+2t\cos{x}... | Any integral of a function that is rational in $\sin(x)$ and $\cos(x)$ can be transfored to an integral of a rational function using Weierstrass substitution. You might try that and see what happens.
$$\begin{align}
\int_0^\pi\frac{2t+2\cos x}{t^2+2t\cos(x)+1}\,dx
&=\int_0^\infty\frac{2t+2\frac{1-u^2}{1+u^2}}{t^2+2t\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/634380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\prod_{n=2}^\infty {n^3-1\over n^3+1}$
The value of the infinite product $$P = \frac 79 \times \frac{26}{28} \times \frac{63}{65} \times \cdots \times \frac{n^3-1}{n^3+1} \times \cdots$$ is
(A) $1$
(B) $2/3$
(C) $7/3$
(D) none of the above
I wrote first 6 terms and tried to cancel out but did not get any ... | Hint:
$$
\begin{align}
\prod_{n=2}^\infty\frac{n^3-1}{n^3+1}
&=\lim_{m\to\infty}\prod_{n=2}^m\frac{n-1}{n+1}\frac{n^2+n+1}{n^2-n+1}\\
\end{align}
$$
The limit of partial products overcomes the difficulty of multiplying two divergent products that an old answer had.
Telescoping Products
Look for terms that cancel:
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/634973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Why can't I find any solution of this three equations? $$2x+6y-z=10$$
$$3x+2y+2z=1$$
$$5x+4y+3z=4$$
I do not find any solution of this three equations.I know that if two equations or planes are parallel then we do't find any solution.But I can't understand what is the problem here.Are they parallel.and if they are para... | Note that
$$\ det\begin{pmatrix}
2 & 6 & -1 \\
3 & 2 & 2 \\
5 & 4 & 3
\end{pmatrix}= 0$$ while $$rank \begin{pmatrix}
2 & 6 & -1 & 10\\
3 & 2 & 2 & 1\\
5 & 4 & 3 & 4
\end{pmatrix}=3$$
because $$\begin{vmatrix}
6 & -1 & 10 \\
2 & 2 & 1\\
4 & 3 & 4
\end{vmatrix} \neq 0$$ so the system has no ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Why does $2^{-n} = 5^n \times 10^{-n}$? If we look at the decimal equivilents of $2^{-n}$, we see they resemble $5^n$ with a decimal point in front of them:
$\begin{align}
2^{-1} &= 0.5 \\
2^{-2} &= 0.25 \\
2^{-3} &= 0.125 \\
2^{-4} &= 0.0625 \\
2^{-5} &= 0.03125 \\
...
\end{align}$
It looks like it's as simple as sayi... | $\dfrac{1}{2} = \dfrac{5}{10}$ (or $\dfrac{2}{4}$, $\dfrac{3}{6}$, or $\dfrac{4}{8}$ for that matter)
Then just raise to the power of $n$ on both sides and use the fact that $a^{-n} = \dfrac{1}{a^n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the volume between two surfaces
Find the volume between $z=x^2$ and $z=4-x^2-y^2$
I made the plot and it looks like this:
It seems that the projection over the $xy$-plane is an ellipse, because if $z=x^2$ and $z=4-x^2-y^2$ then $2x^2+y^2=4$ which means that $\displaystyle\frac{x^2}{(\sqrt{2})^2}+\displaystyle\f... | The mistake in your solution is that the volume of the set $E$ isn't
$$\iiint\limits_{E} z \, dV, \text{ but } \iiint\limits_{E} \, dV.$$
So we have
$$V(E) = \iint \hspace{-5pt} \int_{x^2}^{4-x^2-y^2} \, dz \, dA = \iint 4-2x^2 -y^2 \, dA.$$
This can be tackled in a couple of ways. One is using cartesian coordinates, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/650264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
how to find $ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$ How can I find this?
$ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$
| $ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right) =\lim\limits_{x \to \infty} \left(\sqrt{x^2 +1}-x +\sqrt{4x^2 + 1}-2x - (\sqrt{9x^2 + 1}-3x)\right) = $
$\lim\limits_{x \to \infty}\frac{1}{\sqrt{x^2 +1}+x}+\lim\limits_{x \to \infty}\frac{1}{\sqrt{4x^2 + 1}+2x} -\lim\limits_{x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/651148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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how many ways to go from place a to place b through 9 squares
Please see the image. How many ways are there from M to N without passing through the sqaure more than once...
I counted upto 6 ways...is it the right answer??
| The answer is $E$. Here are the paths:
*
*$(0,0) \rightarrow (1,1) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rightarrow (1,1) \rightarrow (2,0) \rightarrow (3,1) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rightarrow (1,1) \rightarrow (0,2) \rightarrow (1,3) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Sum from $k=1$ to $n$ of $k^3$ $$\sum_{k=1}^n k^3 = \left(\frac{1}{2}n(n+1) \right)^2$$
I want to prove this using induction. I start with $(\frac{n}{2}(n+1))^2 + (n+1)^3$ and rewrite $(n+1)^3$ as $(n+1)(n+1)^2$, then factor out an $(n+1)^2$ from the expression:
$(n+1)^2((\frac{n}{2})^2 + (n+1))$
I'm confused where to ... | Observe that
$$ \left( \left( \tfrac{n}{2} \right)^2 + n + 1 \right) = \left( \frac{n^2 + 4n + 4}{4} \right) = \left( \frac{n+2}{2} \right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/658956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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When is the sum of squares of two consecutive integers also an integer? For example:
*
*two consecutive integers $3$ and $4$, with $\sqrt{3^2+4^2}=5$
*and for $20$ and $21$ answer the square root of $20^2+21^2$ is $29$
*for $0$ and $1$ the square root is $1$.
Any other examples?
| Consider the sequence $(a_n)$ given by
$$ a_0=0, \qquad a_1=3, \qquad a_{n}=6a_{n-1}-a_{n-2}+2\text{ for }n\ge 2$$
and the sequence $(c_n)$ given by
$$ c_0=1, \qquad c_1=5, \qquad c_{n}=6c_{n-1}-c_{n-2}\text{ for }n\ge 2.$$
Then $a_n^2+(a_n+1)^2=c_n^2$ is immediately verified for $n=0$ and $n=1$.
Then by induction t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/661331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove trig identity: $\csc \tan - \cos = \frac{\sin^2}{\cos}$ I keep hitting seeming dead-ends.
\begin{align*}
\csc\ x \tan\ x - \cos\ x &= \left(\frac{1}{\sin\ x}\right)\left(\frac{\sin\ x}{\cos\ x}\right) - \cos\ x \\
&= \frac{\sin\ x}{(\sin\ x)(\cos\ x)} - \cos\ x \\
&= \frac{\sin\ x}{(\sin\ x)(\cos\ x)} - \frac{(\... | Following your first line, just write
\begin{align*}
\csc t \tan t - \cos t &= \frac{1}{\cos t} - \cos t \\
&= \frac{1 - \cos^2 t}{\cos t}
\end{align*}
from a common denominator. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/661495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Uniform convergence of recursively defined function sequence For all $t>0$, define $f_0(t) = \dfrac{1}{e^t-1}$ and $f_{n+1}(t)=\dfrac{3}{4t} + \dfrac{t^3}4 (f_n(t))^4.$
Does this sequence converge uniformly to $\dfrac 1t$ on the positive real numbers?
Remarks: It is not hard to see that the sequence converges monoto... | Let us write
$$d_n(t) = \frac{1}{t} - f_n(t).$$
Then we find
$$\begin{align}
f_{n+1}(t) &= \frac{3}{4t} + \frac{t^3}{4}f_n(t)^4\\
&= \frac{3}{4t} + \frac{t^3}{4}\left(\frac{1}{t} - d_n(t)\right)^4\\
&= \frac{3}{4t} + \frac{t^3}{4}\left(\frac{1}{t^4} - \frac{4}{t^3}d_n(t) + \frac{6}{t^2}d_n(t)^2 - \frac{4}{t}d_n(t)^3 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/665260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the LCM of 3 numbers given HCF of each 2.
Answer is
I was totally confused when I saw the question. I never encountered a question like this.
Can anyone tell me the way to solve this.
I tried every method I could find but hard luck.
| Hint: Proceed one prime at a time. Let's deal with $2$.
The highest power of $2$ that divides $a$ and $b$ is $2^3$.
But the highest power of $2$ that divides $b$ is at least $2^4$, since $\gcd(b,c)$ is divisible by $2^4$.
So the highest power of $2$ that divides $a$ is $2^3$.
Since $2^4$ divides both $b$ and $c$, it fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/665823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solve the following recurrence exactly. $$t(n)=\begin{cases}n&\text{if }n=0,1,2,\text{ or }3\\t(n-1)+t(n-3)+t(n-4)&\text{otherwise.}\end{cases} $$
Express your answer as simply using the theta notation.
I don't know where to go with this.
$$t(n) - t(n-1) - t(n-3) + t(n-4) = 0$$
Is the characteristic polynomial $x^3 - x... | OK, time to whip out generating functions. Define $T(z) = \sum_{n \ge 0} t(n) z^n$. Take the recurrence written as:
$$
t(n + 4) = t(n + 3) + t(n + 1) + t(n) \qquad t(0) = 0, t(1) = 1, t(2) = 2, t(3) = 3
$$
multiply by $z^n$, add over $n \ge 0$ and express in terms of $T(z)$ do get:
$$
\frac{T(z) - t(0) - t(1) z - t(2) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/668852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\frac{(a_1 a_2\cdots a_n)^2-1}{8}\equiv\sum_{i=1}^n\frac{a^2_i -1}{8}\pmod 8$ Let $a_1,a_2,\cdots,a_n$ be odd numbers, show that
$$\frac{(a_{1}a_{2}\cdots a_{n})^2-1}{8}\equiv\sum_{i=1}^{n}\dfrac{a^2_{i}-1}{8} \pmod 8$$
Special cases
$n=1$: It is obvious that
$$\frac{a^2_{1}-1}{8}\equiv\dfrac{a^2_{1}-1}{8... | We prove the congruence by induction on the number of odd factors.
For $n = 1$, the congruence is even an equality, so the base case is settled.
Now assume that $n > 1$ and, as the induction hypothesis, that the congruence
$$\frac{\left(\prod_{k=1}^{n-1}a_k\right)^2 - 1}{8} \equiv \left(\sum_{k=1}^{n-1} \frac{a_k^2-1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/669653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ It is known that
\begin{align}
\arcsin x + \arcsin y =\begin{cases}
\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 \le 1 &\text{or} &(x^2+y^2 > 1 &\text{and} &xy< 0);\\
\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\t... | Let $a=\sin^{-1}x,$ $b=\sin^{-1}y\implies\sin a=x,$ $\sin b=y$ and $a,b\in[-\pi/2,\pi/2]\implies a+b\in[-\pi,\pi]$
$$
\sin(a+b)=\sin a\cos b+\cos a\sin b=x\sqrt{1-y^2}+y\sqrt{1-x^2}\\=\sin\bigg[\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)\bigg]\\
\implies a+b=\color{red}{\sin^{-1}x+\sin^{-1}y=n\pi+(-1)^n\sin^{-1}\Big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/672575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 0
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How can I find all integers $x≠3$ such that $x−3|x^3−3$ How can I find all integers $x≠3$ such that $x−3|x^3−3$?
I tried expand $x^3−3$ as a sum but I couldn't find a way after that.
| $$x - 3 | x^3 - 3$$
$$\frac{x^3 - 3}{x - 3} = \frac{x^3 - 3 + 27 - 27}{x - 3}$$
$$ = \frac{x^3 - 27 + 24}{x - 3}$$
$$ = \frac{x^3 - 3^3 + 24}{x - 3}$$
$$ = \frac{(x - 3) \times (x^2 + 3x + 9) + 24}{x - 3}$$
$$ = x^2 + 3x + 9 + \frac{24}{x - 3}$$
The result is integer when $x - 3$ is a factor of $24$ and $x \ne 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/672854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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proving that $S_{2n}=3n^2+n,S_{2n+1}=3n^2+5n+1$ Given $a_n=\frac{1}{2}(-1)^n(2-n)+\frac{3n}{2}$
Need to show that $S_{2n}=3n^2+n,S_{2n+1}=3n^2+5n+1$.
I tired to separate $a_n$ to odd and even but something went wrong.
Thanks.
| We have
$$A_{2n-1}=1+3+\dots+2n-1=n^2,$$
$$A_{2n}=2+4+\dots+2n=n^2+n.$$
Also,
\begin{eqnarray}
a_n&=&\frac{(-1)^n}{2}(2-n)+\frac{3n}{2},\\
a_n&=&\frac{(-1)^n}{2}\cdot 2-\frac{(-1)^n}{2}\cdot n+\frac{3n}{2},\\
a_n&=&\frac{(-1)^n}{2}\cdot 2+\frac{3-(-1)^n}{2}\cdot n,\\
a_n&=&(-1)^n+\frac{3-(-1)^n}{2}\cdot n,
\end{eqnarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/672929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $\lim\limits_{n \to\infty} z_n = A$ implies $\lim\limits_{n \to\infty} \frac{1}{n} (z_1 + z_2 + \ldots + z_n) = A$ In what follows let all values be in $\mathbb{C}$. I'm trying to show that if
$$\lim z_n = A,$$
that then
$$
\lim_{n \to \infty} \frac{1}{n} (z_1 + z_2 + \ldots + z_n) = A.
$$
For ease of no... | I think the problem with your strategy appears in your first step, when you are trying to approximate $A - \frac{1}{n}z_1 - \cdots - \frac{1}{n}z_n$. The problem is that you really aren't ever considering the difference between $A$ and $z_n$ when you are coming up with your bounds; remember, the only information you ha... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Incongruent solutions to $7x \equiv 3$ (mod $15$) I'm supposed to find all the incongruent solutions to the congruency $7x \equiv 3$ (mod $15$)
\begin{align*}
7x &\equiv 3 \mod{15} \\
7x - 3 &= 15k \hspace{1in} (k \in \mathbb{Z}) \\
7x &= 15k+3\\
x &= \dfrac{15k+3}{7}\\
\end{align*}
Since $x$ must be an integer, we mus... | $$x = 9+15m, m\in \mathbb{Z}.$$
is the same as
$$x \equiv 9 \pmod{15} \,.$$
Equations of the type $ax \equiv b \pmod{n}$ sometimes have no solution, sometimes have one solution $\pmod{n}$ and sometimes they have two or more (incongruent solutions) $\pmod{n}$.
In this case, there is only one, the one you found. A muc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/675867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How to solve this simple trignometric problem? So this is the question that was given in a textbook and i attempted to win from the book which was saying i was wrong?
If $$\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{5}{4}$$ then what is te value of$$\frac{\tan^2\theta + 1}{\tan^2\theta - 1}$$
so ... | Applying Componendo and dividendo on
$$\dfrac{\sin\theta +\cos\theta}{\sin \theta -\cos\theta}=\dfrac54$$
to get $$\frac{\sin\theta}{\cos\theta}=\frac{5+4}{5-4}$$
$$\implies \frac{\tan^2\theta}1=\frac{9^2}1$$
Again apply Componendo and dividendo
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area Between Two Curves I have been trying to solve a homework problem that reads: "The region between the graphs of $f(x) = x^{2}+2$ and $g(x) = -5x+2$ has what area?" I have been trying to solve this problem and have come up with the answer of $185/6$ square units, but it has come to my attention that that is not cor... | The two graphs intersect at the points where their $x$-coordinates are equal and their $y$-coordinates are equal. Their $y$-coordinates are $x^2+2$ and $-5x+2$. These are equal when
$$
x^2+2=-5x+2.
$$
Hence $x^2=-5x$. If $x\ne 0$ then we can divide both sides by $x$ and get $x=-5$. If $x=0$, then that is also a sol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving Recurrence Relation by Generating Function Method Im trying to solve
an-7a(n-1)+10a(n-2)
Im at the point where ∈aX^n-7∈a(n-1)X^n+10∈a(n-2)x^n=0
(terms of n are subscript)
After this step it is given as replace the infinite sum by an expression from table of eq. expressions
and it becomes(Summation from 2 to i... | I'm guessing your recurrence is: $a_n -7a_{n-1} +10a_{n-2}=0 \longleftrightarrow a_n = 7a_{n-1}-10a_{n-2} \longleftrightarrow a_{n+2} = 7a_{n+1} -10a_n $. Let's say $a_0=0$, $a_1=1$ for sake of example.
Set $F(x) = a_0 + a_1x + a_2x^2+\ldots = \sum_{n=0}^{\infty} a_nx^n $
So $\frac{F(x) - a_0}{x} = a_1 + a_2x + a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this sum$\sum_{k=0}^{\infty}\frac{(-1)^{\frac{k(k+1)}{2}}}{(2k+1)^2}=\frac{\sqrt{2}\pi^2}{16}$ How show that
$$\sum_{k=0}^{\infty}\dfrac{(-1)^{\frac{k(k+1)}{2}}}{(2k+1)^2}=\dfrac{\sqrt{2}\pi^2}{16}$$
My idea:I know how to prove the following sum
$$\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^2}=\int_{0}^{1}\dfr... | The pattern of the signs of the sum are
$$a_0-a_1-a_2+a_3+a_4-a_5-a_6+a_7+a_8-\cdots$$
so that each term of the form $8 k+3$ and $8 k+5$ is negative, and $8 k+1$ and $8 k+7$ is positive. Thus we may write the sum as
$$\sum_{k=0}^{\infty} \left [\frac1{(8 k+1)^2} - \frac1{(8 k+3)^2} - \frac1{(8 k+5)^2}+\frac1{(8 k+7)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/683418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Conic Sections Question - Hyperbolas & Circles
*
*So, if you have a hyperbola with foci at $(4,0)$ & $(-2,0)$, and the slopes of the asymptotes are $+4$ and $-4$, what would the equation for this hyperbola be? I know that the center would be $(1,0)$, and that would mean that $c$ is $3$ (and $c^2$ is $9$), but I don't... | 1. You are correct that the center is at $(1, 0)$, and so the focal length is indeed $c = 3$.
Recall that you can draw a rectangle, centered at the center of the hyperbola such that the asymptotes pass through the corners and the left and right sides of the rectangle are tangent to the vertices of the hyperbola. The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/684320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Trying to get a bound on the tail of the series for $\zeta(2)$ $$\frac{\pi^2}{6} = \zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2}$$ I hope we agree.
Now how do I get a grip on the tail end $\sum_{k \geq N} \frac{1}{k^2}$ which is the tail end which goes to zero?
I want to show that $\sqrt{x}\cdot \mathrm{tailend}$ is bound... | We can actually compute the complete asymptotic expansion of the remainder term. Introduce
$$S(x) = \sum_{k\ge 1} \left(\frac{1}{k^2}-\frac{1}{(x+k)^2}\right)$$
so that our answer is given by $$\frac{\pi^2}{6}-S(N).$$
Re-write $S(x)$ as follows:
$$S(x) = \sum_{k\ge 1} \frac{1}{k^2} \left(1-\frac{1}{(x/k+1)^2}\right).$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/685435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
Find permutation that solves $\;\tau \circ X = \sigma$ I need to find a permutation $X$ that solves $\;\tau \circ X = \sigma,\;$ given
$$\tau = \begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 4 & 5 & 2 & 1
\end{bmatrix} = (1,3,5)(2,4)$$
$$\sigma = \begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 2 & 1 & 5 & 4
\end{bmatrix} = (1,3)... | Assuming you compose left to right:
$$\tau = \begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 4 & 5 & 2 & 1
\end{bmatrix} = (1,3,5)(2,4)$$
times
$$X = \begin{bmatrix}
3 & 4 & 5 & 2 & 1\\
3 & 2 & 1 & 5 & 4
\end{bmatrix} = (1,4,2,5)(3)$$
equals
$$\sigma = \begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 2 & 1 & 5 & 4
\end{bmatrix} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/689056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Show that $(x^2-yz)^3+(y^2-zx)^3+(z^2-xy)^3-3(x^2-yz)(y^2-zx)(z^2-xy)$ is a perfect square and find its square root. Show that $(x^2-yz)^3+(y^2-zx)^3+(z^2-xy)^3-3(x^2-yz)(y^2-zx)(z^2-xy)$ is a perfect square and find its square root.
My work:
Let, $x^2-yz=a,y^2-zx=b,z^2-xy=c$. So, we can have,
$a^3+b^3+c^3-3abc=(a+b+... | HINT:
$$(x^2-yz)^3-(x^2-yz)(y^2-zx)(z^2-xy)$$
$$=(x^2-yz)[(x^2-yz)^2-(y^2-zx)(z^2-xy)]$$
$$=(x^2-yz)x(x^3+y^3+z^3-3xyz)$$
$$=(x^3-xyz)\underbrace{(x^3+y^3+z^3-3xyz)}$$
As the terms under the brace is symmetric wrt $x,y,z$
we shall reach at the similar expressions from
$$(y^2-zx)^3-(x^2-yz)(y^2-zx)(z^2-xy)$$ and $$(z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/691078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Long division, explain the result! I have this: $$ \frac{x^2}{x^2+1} $$ Wolfram Alpha suggests that I should do long division to get this: $$ 1- \frac{1}{x^2+1} $$ But I don't understand how it can be that, please explain.
| This is the calculation taking place:
$$\frac{x^2}{x^2 + 1} = \overbrace{\frac{x^2 + 1 -1}{x^2 + 1}}^{Adding\,+1-1} = \underbrace{\frac{x^2 + 1}{x^2 + 1}}_{=1} - \frac{1}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/694163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Volume with triple integrals Calculate integral $$\iiint_V \frac{e^{-x^2-y^2-z^2}}{\sqrt{x^2+y^2+z^2}} dV$$ Where $V\subset\mathbb{R}^3$ is the exterior of a origocentered sphere with radius of 2
\begin{align*}
V=&\iiint_V \frac{e^{-x^2-y^2-z^2}}{\sqrt{x^2+y^2+z^2}} dV \\
=& \int_0^{2\pi}\int_0^{\pi}\int_0^2 \frac{e^{... | Converting to polar coordinates yields
$$
\begin{align}
\int_0^2\frac{e^{-r^2}}{r}4\pi r^2\,\mathrm{d}r
&=2\pi\int_0^4 e^{-t}\,\mathrm{d}t\\
\end{align}
$$
using the substitution $t=r^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/694394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Eigenvalues of vectors with irrational entries I have been trying to find eigenvalues and eigenvectors of this matrix: $\begin{bmatrix}3 & -2\\1 & -1\end{bmatrix}$. So far I have got $\lambda_1=1+\sqrt2$ and $\lambda_2=1-\sqrt2$. I am stuck at finding eigenvectors at this point. Regular row-reduction method gives me ha... | You've gotten the eigenvalues correctly. Now, given a matrix $A$ with an eigenvalue $\lambda$, an eigenvector for $\lambda$ is just a non-zero element of the null space of $A-\lambda I$.
So, let's find the null space of $$\begin{bmatrix}3 \strut& -2\\1 & -1\end{bmatrix}-\begin{bmatrix}1+\sqrt{2} & 0\\0 & 1+\sqrt{2}\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/695340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Determinant algebra If $A$ and $B$ are $4 \times 4$ matrices with $\det(A) = −2$, $\det(B) = 3$, what is $\det(A+B)$?
At first I approached the problem that $\det(A+B) = \det(A) + \det(B)$ but this general rule would not hold true, so I do not know how to approach the problem from here.
| Let
$$
A
=
\begin{pmatrix}
-2 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\quad
B=
\begin{pmatrix}
3 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\quad
C=
\begin{pmatrix}
-3 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/697259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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solve a trigonometric equation $\sqrt{3} \sin(x)-\cos(x)=\sqrt{2}$ $$\sqrt{3}\sin{x} - \cos{x} = \sqrt{2} $$
I think to do :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{2}}$$
but i dont get anything.
Or to divied by $\sqrt{3}$ :
$$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{3}}$$
| $$\sqrt{3} \sin x - \cos x = \sqrt{2}$$
Dividing both sides by 2, we get
$$\frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x = \frac{\sqrt{2}}{2}$$
By substituting $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ & $\sin 30^{\circ} = \frac{1}{2}$, we get
$$\sin x \cos30^{\circ} - \cos x \sin 30^{\circ} = \frac{\sqrt{2}}{2}$$
Using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/698964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
$\forall x\in\mathbb R$, $|x|\neq 1$ it is known that $f\left(\frac{x-3}{x+1}\right)+f\left(\frac{3+x}{1-x}\right)=x$. Find $f(x)$.
$\forall x\in\mathbb R$, $|x|\neq 1$ $$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{3+x}{1-x}\right)=x$$Find $f(x)$.
Now what I'm actually looking for is an explanation of a solution to th... | It's a bit unclear, but if you look for what $t$'s each of the equalities hold, it starts to make sense.
Note that if $x\ne-1$ and $t_0\ne1$, then $t_0=\frac{x-3}{x+1}\Longleftrightarrow(x+1)t_0=x-3\Longleftrightarrow t_0+3=x(1-t_0)\Longleftrightarrow x=\frac{3+t_0}{1-t_0}$
So the first equality holds for all $t_0\in\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/699323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
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For all square matrices $A$ and $B$ of the same size, it is true that $(A+B)^2 = A^2 + 2AB + B^2$? The below statement is a true/false exercise.
Statement:
For all square matrices A and B of the same size, it is true that
$(A + B)2 = A^2 + 2AB + B^2$.
My thought process: Since it is not a proof, I figure I can show by ... | $$\begin{align*}
(A+B)^2 &= (A+B)(A+B) \\
&= AA+AB+BA+BB \\
&= A^2 + AB+BA+B^2.\end{align*}$$
Is it always true that $AB+BA = 2AB$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/699505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
} |
Equation with an infinite number of solutions I have the following equation: $x^3+y^3=6xy$. I have two questions: 1. Does it have an infinite number of rational solutions?
2. Which are the solutions over the integers?($ x=3 $ and $ y=3 $ is one)
Thank you!
| If $x$ and $y$ are rational, then so is $y/x=\alpha$. Then $x^3+y^3=6xy$ becomes
$$
(\alpha^3+1)x^3-6\alpha x^2=0\tag{1}
$$
and then $x=0$ or $x=\dfrac{6\alpha}{\alpha^3+1}$. Thus, for any rational $\alpha$, we have the rational solutions
$$
\left(\frac{6\alpha}{\alpha^3+1},\frac{6\alpha^2}{\alpha^3+1}\right)\tag{2}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/702936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Find the limit of $\lim_{x\to 0}\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}$ Can someone help me solve this limit?
$$\lim_{x\to0}\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}$$
with $a>0$ and $b>0$.
| Without using L'Hopital
$$\lim_{x\to0}\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}\cdot\frac{\sqrt{x^2+a^2}+a}{\sqrt{x^2+b^2}+b}\cdot\frac{\sqrt{x^2+b^2}+b}{\sqrt{x^2+a^2}+a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/703116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proof: $ \lfloor \sqrt{ \lfloor x\rfloor } \rfloor = \lfloor\sqrt{x}\rfloor $. I need some help with the following proof:
$ \lfloor \sqrt{ \lfloor x\rfloor } \rfloor = \lfloor\sqrt{x}\rfloor $.
I got:
(1) $[ \sqrt{x} ] \le \sqrt{x} < [\sqrt{x}] + 1 $ (by definition?).
(2) $[ \sqrt{x} ]^2 \le x < ([\sqrt{x}] + 1)^2 $.
(... | First, you need that $x \geq 0$. Write $x=a+ \alpha$ where $\left\lfloor \sqrt x \right \rfloor=a$ and $\alpha <1$. Take $n^2 \leq a < (n+1)^{2}$ . Now, you need to prove that $n= \left \lfloor \sqrt a \right \rfloor = \left \lfloor \sqrt {a+ \alpha} \right \rfloor$. $\left \lfloor \sqrt {a+ \alpha} \right \rfloor \leq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/705597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Area of the portion of the sphere Find the area of the portion of the sphere of radius 1 (centered at the origin) that is in the cone $$z > \sqrt{x^2 + y^2}.$$
I tried to find the formula by the integral but I still did not get it!
| For a sphere, we have $x^2+y^2+z^2=1 \Leftrightarrow z=\sqrt{1-x^2-y^2}$
$f_y=-2y(\frac{1}{2})(1-x^2-y^2)^{-1/2}$, $f_x=-2x(\frac{1}{2})(1-x^2-y^2)^{-1/2}$.
So by the formula for a surface area, we have
$$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{(-2y(\frac{1}{2})(1-x^2-y^2)^{-1/2})^2+(-2x(\frac{1}{2})(1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/706288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $\frac{x}{1-x}$ using definition of derivative I was trying to find the equation of the tangent line for this function. I solved this using the quotient rule and got $\frac{1}{(x-1)^2}$ but I can't produce the same result using definition of derivatives. Can someone show me how to do it? I tried looking it up o... | First we create a difference quotient for this function: $$\lim\limits_{h \rightarrow 0} \frac{(x+h)(1-(x+h))^{-1} - x(1-x)^{-1}}{h}$$
Now we try to simplify the numerator a bit by making it one fraction as opposed to the difference between two fractions:
$$
\begin{align}
\lim\limits_{h \rightarrow 0} \frac{(x+h)(1-(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Guessing root of polynomials
Given $p(x)=x^5+(1+2i)x^4-(1+3i)x^2+8+44i$ check with the Horner-scheme if $(-2-i)$ is a root of $p(x)$.
First I have to guess a root, then proceed with the Horner-method and if i factorized it, i can say if $(-2-i)$ is a root or not, but how can i guess the first root, are there any tri... | $p(x) = 8 + 44i + x^2(-(1+3i) + x^2(1+2i + x))$
Now, let $x=-2-i$, thus, $x^2 = 4-1-4i = 3+4i$
$\begin{align}p(x) &= 8 +44i + (3+4i)(-(1+3i) + (3+4i)(-1+i)))\\
&=8+44i + (3+4i)(-(1+3i) -3+3i-4i-4))\\
&=8+44i +(3+4i)(-1-3i-3+3i-4i-4)\\
&=8+44i + (3+4i)(-8 - 4i)\\
&=8+44i -24 -12i -32i +16\\
&=0
\end{align}
$
Thus $-2-i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/709380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\sum_{n=0}^{\infty}\left(\frac{4}{(-3)^n} - \frac{3}{3^n}\right)$? I'm currently trying to compute the following series (found on page 65 of this textbook):
$$\sum_{n=0}^{\infty}\left(\frac{4}{(-3)^n} - \frac{3}{3^n}\right)$$
I've tried to somehow transform it into a geometric series (which I'm fairly s... | $\begin{array}{l}
\left( {\frac{4}{{\left( { - 3} \right)^n }} - \frac{3}{{3^n }}} \right) = \frac{{\left( { - 1} \right)^n }}{{3^n }} + \frac{{3\left( { - 1} \right)^n }}{{3^n }} - \frac{3}{{3^n }} \\
= \left\{ \begin{array}{l}
\frac{1}{{3^{2p} }}\quad ;n = 2p \\
\frac{{ - 7}}{{3^{2p + 1} }}\quad ;n = 2p + 1 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/711573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $2^x \cdot 3^y - 5^z \cdot 7^w = 1$ has no solutions Prove that $$2^x \cdot 3^y - 5^z \cdot 7^w = 1$$ has no solutions in $\mathbb{Z}^+$, if $y\ge 3$.
| Consider any rational number $2^x 3^y 5^{-z} 7^{-w}$, where $x$, $y$, $z$, $w \in \mathbb{Z}^+$ . Størmer's theorem guarantees that there are a finite number of such fractions where the numerator and denominator are consecutive integers.
Here are all of the solutions:
$\frac{7}{6}, \frac{8}{7}, \frac{15}{14}, \frac{21... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Construct an endomorphism $f$ such that $f\circ f=-Id$
Let $E$ be a real vector space of finite dimension $n$ and $f$ an endomorphism such that
$$f\circ f=-Id_E$$
*
*Show that $n = \dim (E)$ is an even integer
*Assume $n$ is even, $n=2p$. Construct an endomorphism $f$ such that $f\circ f=-Id$
1) I have,... | Suppose $f:\mathbb R^2\to\mathbb R^2$ is a linear map such that $f^2=-I$. If $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is the matrix of $f$ with respect to the standard basis, and then squaring we see that $$
\begin{pmatrix}
a^2+b c & a b+b d \\
a c+c d & b c+d^2 \\
\end{pmatrix}
=
\begin{pmatrix}
-1 & 0 \\
0 & -1 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Factoring $x^5 + x^4 + x^3 + x^2 + x + 1$ without using $\frac{x^n - 1}{x-1}$? I was at a math team meet today and one of the problems was to factor $x^5 + x^4 + x^3 + x^2 + x + 1$. It also gave the hint that it decomposes into two trinomials and a binomial.
The solution they gave was based on the fact that $\frac{x^6 ... | You can still factor this. Notice that
\begin{align*}
x^5+x^4+x^3+x^2+x+1 &= x^3(x^2+x+1)+1(x^2+x+1) \\
&=(x^3+1)(x^2+x+1) \\
&=(x+1)(x^2-x+1)(x^2+x+1).
\end{align*}
Now you can go one step further and show that both $(x^2-x+1)$ and $(x^2+x+1)$ are irreducible in the real numbers just by using the discriminant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/716183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Series Expansion Of An Integral. I want to find the first 6 terms for the series expansion of this integral:
$$\int x^x~dx$$
My idea was to let: $$x^x=e^{x\ln x}$$
From that we have: $$\int e^{x\ln x}~dx$$
The series expansion of $e^x$ is: $$\sum\limits_{n=0}^\infty\frac{x^n}{n!}$$
Then we have:
$$\int e^{x\ln x}~dx=\i... | http://en.wikipedia.org/wiki/Exponential_formula
\begin{align}
f(x) & = a_1 x + \frac{a_2}{2} x^2 + \frac{a_3}{6} x^3 + \frac{a_4}{24} x^4 + \frac{a_5}{120} x^5 + \frac{a_6}{720} x^6 + \cdots \\[10pt]
e^{f(x)} & = 1 + a_1 x + \frac{a_2+a_1^2}{2} x^2 + \frac{a_3+3a_2a_1+ a_1^3}{6} x^3 + \frac{a_4 + 4a_3 a_1+ 3a_2^2 + 6a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Rationalization of $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$
Question:
$$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ equals:
My approach:
I tried to rationalize the denominator by multiplying it by $\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$. And got the result to be (after a long calculat... | Your "long calculation" was obviously wrong, since the denominator should be
$$(\sqrt 2 + \sqrt 3 + \sqrt 5)(\sqrt 2 - (\sqrt 3 + \sqrt 5))=\\=\sqrt2^2 - (\sqrt 3 + \sqrt 5)^2 = 2 - (3 + 5 + 2\sqrt{15}) = -6-2\sqrt{15}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
How to solve this elliptic integral ?? Can anyone explain to me how to find the integral ?
$$ \int_0^1\sqrt{9x^4+4x^2+1}dx =? $$
| Although this integral really belongs to an elliptic integral, but this does not means we can always express an elliptic integral to the elliptic integral of the three standard types conveniently, http://integrals.wolfram.com/index.jsp?expr=%289x%5E4%2B4x%5E2%2B1%29%5E%281%2F2%29&random=false can tell us why.
Besides, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/721070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Competition math geometry question The perimeter of triangle ABC is $36$, and its area is $36$. Compute $\tan\frac{A}2 \tan\frac{B}2 \tan\frac{C}2$.
I found that the answer is $1/9$, but I was not able to find a reason for this. Could someone please give me a good explanation as to why it is this?
| Let $r$ be the inradius of a triangle, classical geometry tell us the perimeter $\mathcal{P}$ and area $\mathcal{A}$ is related to $r$ through the relations:
$$
\mathcal{P} = 2r\left(\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2}\right)
\quad\text{ and }\quad
\mathcal{A} = \frac{r\mathcal{P}}{2}
$$
Eliminating $r$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/723690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Multivariable limit: $\lim\limits_{(x,y) \to (0,0)}\dfrac{(y^2-x)^3}{x^2+y^4}$ I need to solve the limit
$$\lim\limits_{(x,y) \to (0,0)}\dfrac{(y^2-x)^3}{x^2+y^4}$$
and I can't think on a possible upper bound for it.
Any ideas?
Thanks
|
$$\lim\limits_{(x,y) \to (0,0)}\dfrac{(y^2-x)^3}{x^2+y^4}$$
Let $$
f(x,y)=\dfrac{(y^2-x)^3}{x^2+y^4}
$$
Then $$
\lim_{(0,0)} f(x,y) = \lim_{(0,0)} f(x^2,y)\\
|f(x^2,y)|= \dfrac{|y^2-x^2|^3}{x^4+y^4}\le
\dfrac{(y^2+x^2)^3}{x^4+y^4} \\
= \dfrac{r^6}{r^4 (\cos^4\theta + \sin^4\theta)} = O(r^2)
$$using polar coordinate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding one sided limits algebraically I was wondering what the best method was for proving this limit algebraically:
$$\lim_{x \to 1}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}$$
I know the answer to this question is ;
$$\lim_{x \to 1^+}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={-\infty}$$
$$\lim_{x \to 1^-}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}... | We have
$$
\frac{3x^4-8x^3+5}{x^3-x^2-x+1} = 3x - 5 + \frac{4}{x+1} - \frac{6}{x-1},
$$
which can be found by using polynomial long division and partial fraction decomposition. From this form the limits $x\to 1^+$ and $x\to 1^-$ are easy to compute without looking at any plots.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving an inequality using induction Use induction to prove the following:
$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2^n}\geq1+\frac{n}{2}$
What would the base case be? Would it still be $n=0$
so $\frac{1}{1}+\frac{1}{2}\geq 1+\frac{0}{2}$, which holds true.
then how would you prove for $n$ and $n+1$ to pro... | Yes, the base case here would be $n = 0$, and your assessment is correct, although there is no $\frac{1}{2}$ term in the $n = 0$ case.
For our inductive step (proving $n+1$ from $n$), consider that we know that
$$\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{2^n} \ge 1 + \frac{n}{2}$$ and are trying to show:
$$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate $\int_0^1\frac{1+x^4}{1+x^6}\,dx$ $$\int_0^1\frac{1+x^4}{1+x^6}\,dx$$
Can anyone help me solve the question? I am struggling with this.
| RV's great answer notwithstanding, here's another way using the residue theorem. Note that
$$\int_0^1 dx \frac{1+x^4}{1+x^6} = \frac14 \int_{-\infty}^{\infty} dx \frac{1+x^4}{1+x^6}$$
The integral on the RHS is, by the residue theorem, $i 2 \pi$ times the sum of the residues of the poles of the integrand in the upper ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find characteristic polynomial of this matrix?
Let, $A=\begin{bmatrix} 4&0&1&0\\1&1&1&0\\0&1&1&0 \\0&0&0&4 \end{bmatrix}$. Knowing that $4$ is one of its eigenvalues, find the characteristic polynomial of $A$.
Well if $4$ is an eigenvalues of $A$, one should have $|A-4I_{4}|=0$ .
And so,
$\begin{vmatrix} 0&0&1... | Working across the top row....
$$\begin{align}\begin{vmatrix} -\lambda&0&1\\1&-3-\lambda&1\\0&1&-3-\lambda&\end{vmatrix} &= -\lambda\left[(-3 -\lambda)(-3 -\lambda) -1\right] -0[\,] + 1\left[1-0 \right]=0 \\
&= -\lambda\left[9 + 6\lambda +\lambda^2 -1 \right] + 1= 0 \\
&= -\lambda^3 -6\lambda^2 -8\lambda +1 = 0\\
&= \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/727972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A series from a physics problem: $\sum_{k = 0}^{\infty}\frac{2}{2k + 1}e^{-(2k + 1)\pi x/a}\sin\left( \frac{(2k + 1)\pi y}{a}\right)$ How might we show that $$\sum_{k = 0}^{\infty}\frac{2}{2k + 1}e^{-(2k + 1)\pi x/a}\sin\left( \frac{(2k + 1)\pi y}{a}\right) = \tan^{-1}\left( \frac{\sin(\pi y/a)}{\sinh(\pi x/a)} \right... | Expanding on Mhenni Benghorbal's answer, the Maclaurin series of the inverse hyperbolic tangent function is $$\sum_{k=0}^{\infty} \frac{z^{2k+1}}{2k+1} = \text{artanh} (z) \, , \ |z| <1.$$
So assuming $x, a >0$ and $y \in \mathbb{R}$, we get
$$ \begin{align} &\sum_{k = 0}^{\infty}\frac{2}{2k + 1}e^{-(2k + 1)\pi x/a}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/728523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why $\cos^3 x - 2 \cos (x) \sin^2(x) = {1\over4}(\cos(x) + 3\cos(3x))$? Wolfram Alpha says so, but step-by-step shown skips that step, and I couldn't find the relation that was used.
| Here's a detailed step by step:
$=\frac{1}{4}(\cos(x)+3\cos(3x))$
$=\frac{1}{4}(\cos(x)+3\cos(x+2x))$
$=\frac{1}{4}(\cos(x)+3[\cos(x)\cos(2x)-\sin(x)\sin(2x)])$
$=\frac{1}{4}(\cos(x)+3[\cos(x)(1-2\sin^2(x))-\sin(x)(2\sin(x)\cos(x))])$
$=\frac{1}{4}(\cos(x)+3[\cos(x)-2\cos(x)\sin^2(x))-2\sin^2(x)\cos(x)])$
$=\frac{1}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/729587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)$
Prove the following integral
$$I=\int\limits_{0}^{\frac{\pi}{2}}\dfrac{dx}{1+\sin^2{(\tan{x})}}=\dfrac{\pi}{2\sqrt{2}}\left(\dfrac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\right)$$
This in... | First note that
$$ \begin{align} \int_{0}^{\pi /2} \frac{1}{1+ \sin^{2} ( \tan x)} \ dx &= \int_{0}^{\infty} \frac{1}{(1+\sin^{2} t)(1+t^{2})} \ dt \\ &= 2 \int_{0}^{\infty} \frac{1}{(3 - \cos 2t)(1+t^{2})} \ dt \end{align}$$
Then using the identity $$\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/730876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 3,
"answer_id": 0
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How to calculate the diagonal matrix The question says: find the eigenvalues and corresponding eigenvectors of the matrix $A$. This I could do. But then it says: hence find a non-singular matrix $P$ and a diagonal matrix $D$ such that $A + A^2 + A^3 = PDP^{-1}$ , where
$$
A =\begin{pmatrix}
6 & 4 & 1 \... | Let $A,D,X \in \mathbb C^{n,n}$, $D$ diagonal, $X$ non-singular. Let $\vec x_i$ be i-th column of $X$ and $\lambda_i $ i-th diagonal entry of $D$. Hopefully you can see that if $A = XDX^{-1} \leftrightarrow AX=XD$, that is to say $A$ and $D$ are similar, $A\vec x_i = \lambda_i \vec x_i$ from the way matrix multiplicati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/731995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is a multiple of $9$. Please mark/grade. What do you think about my first induction proof? Please mark/grade.
Theorem
The sum of the cubes of three consecutive natural numbers is a multiple of 9.
Proof
First, introducing a predicate $P$ over $\mathbb{N}$, we rephra... | It's fine, here's a simpler proof without induction:
$n^3\equiv n\ (\text{mod }3)$, because it obviously holds for $n=-1,0,1$.
Therefore $3n^3\equiv3n\ (\text{mod 9})$ and
$$(n-1)^3+n^3+(n+1)^3\equiv3n^3+6n\equiv0\ (\text{mod }9)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/732445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 1
} |
Help in evaluating $\int_0^{\infty} \frac{2 \sin x \cos^2 x}{x e^{x \sqrt{3}}} dx$ I need some help in evaluating $ \displaystyle \int_0^{\infty} dx \frac{2 \sin x \cos^2 x}{x e^{x \sqrt{3}}}$
The original question: Evaluate $\displaystyle \int_0^{\infty} dx \frac{e^{- x \sqrt 3}}{x} (1 - \sin x)(1 + 2 \sin x - \cos 2x... | I would consider the Laplace transform
$$F(p)=\int_0^{\infty} dx \, \frac{\sin{x}}{x} \cos^2{x} \, e^{-p x}$$
Then
$$\begin{align}F'(p) &= -\int_0^{\infty} dx \, \sin{x} \cos^2{x} \, e^{-p x}\\ &= -\frac14 \int_0^{\infty} dx \,(\sin{3 x}+\sin{x}) e^{-p x}\\ &= -\frac14 \left (\frac{3}{p^2+9}+\frac1{p^2+1} \right )\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/737467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Optimize function on $x^2 + y^2 + z^2 \leq 1$ Optimize $f(x,y,z) = xyz + xy$ on $\mathbb{D} = \{ (x,y,z) \in \mathbb{R^3} : x,y,z \geq 0 \wedge x^2 + y^2 + z^2 \leq 1 \}$. The equation $\nabla f(x,y,z) = (0,0,0)$ yields $x = 0, y = 0, z \geq 0 $ and we can evaluate $f(0,0,z) = 0$.
Now studying the function on the boun... | Now $f(x,y,z) = xyz + xy=xy(1+z)$
Note that on $x^2 + y^2 + z^2 = 1$, $|z| \le 1$ so $1+z \ge 0$.
Hence, for a given $z$, you maximise $f(x,y,z)$ by maximising $xy$ on $x^2 + y^2 = 1-z^2$
$x^2 + y^2 = 1-z^2$ is a circle of radius $\sqrt{1-z^2}$, so we can parametrise $(x,y)$ by $(\sqrt{1-z^2} \;sin(\theta), \sqrt{1-z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/742031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Area bounded by two circles $x^2 + y^2 = 1, x^2 + (y-1)^2 = 1$ Consider the area enclosed by two circles: $x^2 + y^2 = 1, x^2 + (y-1)^2 = 1$
Calculate this area using double integrals:
I think I have determined the region to be $D = \{(x,y)| 0 \leq y \leq 1, \sqrt{1-y^2} \leq x \leq \sqrt{1 - (1-y)^2}\}$
Now I can't se... | Draw a picture. You will note that part of the region is in the second quadrant.
If you want to use rectangular coordinates, it will be necessary to see where circles meet. That is not at $x=1$. If we solve the system of two equations, pretty quickly we get $y=\frac{1}{2}$, which gives $x=\pm \frac{\sqrt{3}}{2}$. Your... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/742949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is limits $\lim_{x\to\infty}{(x-2)^2\over2x+1}=\dfrac{1}{2}$? $$\lim_{x\to\infty}{(x-2)^2\over2x+1}=\dfrac{1}{2}$$
I used an online calculator and it said it was actually $=\infty$
Here's how I calculate it:
$$\lim_{x\to\infty}\dfrac{x^2+4-4x}{2x+1}=\lim_{x\to\infty}\dfrac{({x\over x}-{2\over x})({x\over x}-{2\over x})... | The online calculator is correct. The value of the limit $\to\infty$. Here is the proof:
$$
\begin{align}
\lim_{x\to\infty}{(x-2)^2\over2x+1}&=\lim_{x\to\infty}\frac{x^2-4x+4}{2x+1}\\
&=\lim_{x\to\infty}\frac{x^2-4x+4}{2x+1}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\\
&=\lim_{x\to\infty}\frac{\frac{x^2}{x}-\frac{4x}{x}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/743368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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