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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: $$ 6(2x+5)^{2}(3x-1)^{3}(7x+9)$$
| 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 $$f'(x)=(2x+5)^3\cdot 4(3x-1)\cdot(3x-1)'+(3x-1)^4\cdot3(2x+5)^2\cdot(2x+5)',$$ or $$f'(x)=12(2x+5)^3(3x-1)^3+6(2x+5)^2(3x-1)^4.$$ Now, pull out the common factor $6(2x+5)^2(3x-1)^3,$ and see what happens.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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,
$$I = \int_0^1 \dfrac{1-x^2}{(1+x^2)^2}\log(x)dx = \int_0^1 (1-x^2)\left(\sum_{k=0}^{\infty} (k+1)(-1)^k x^{2k} \right)\log(x)dx$$
We now have
$$\int_0^1 x^{2n} \log(x) dx = -\dfrac1{(2n+1)^2}$$
Hence,
$$I = \sum_{k=0}^{\infty}(k+1)(-1)^k\left(-\dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2}\right)= -\dfrac{\pi}4$$
To get the last step, note that
$$(k+1)\left(\dfrac1{(2k+1)^2} - \dfrac1{(2k+3)^2}\right) = \dfrac14 \left(\dfrac1{2k+1} - \dfrac1{2k+3} + \dfrac1{(2k+1)^2} + \dfrac1{(2k+3)^2} \right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 induction that this equals $\left(2-\dfrac{1}{(n+1)}-\dfrac{h_{n+1}}{n} \right)$ for every $n\ge1$ but am not sure how to use this in solving my problem. Could anyone give me some tips?
| This just begs to be telescoped:
$$\sum_{k=2}^\infty\sum_{n=1}^k \frac{1}{k(k-1)n} = \sum_{k=2}^\infty \frac{1}{k(k-1)} + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty\frac{1}{k(k-1)}$$
$$= \sum_{k=2}^\infty (\frac{1}{k-1} -\frac{1}{k}) + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty(\frac{1}{k-1} -\frac{1}{k})$$
$$= 1 + \sum_{n=2}^\infty \frac{1}{n}\cdot\frac{1}{n-1} =1 + \sum_{n=2}^\infty(\frac{1}{n-1}-\frac{1}{n}) =2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $(2)$ and $(3)$
$$
a^xb^{1-x}+a^{1-x}b^x\le a+b\tag{4}
$$
Since $a+b=1$, set $x=a$ and $1-x=b$ in $(4)$
$$
a^ab^b+a^bb^a\le1\tag{5}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $P(x)=0$ are the reciprocals of the roots of $Q(x)=0$, then prove that $a$ and $b$ are integers. Also find the greatest common divisor of $P(2013!+1)$ and $Q(2013!+1)$.
Let the roots of $P(x)=0$ be $\alpha,\; \beta, \;and\;\gamma$. Then we have the following four relations. $$\alpha+\beta+\gamma=-a$$ $$\alpha\beta\gamma=-b$$$$\frac{1}{\alpha\beta}+\frac{1}{\beta\gamma}+\frac{1}{\alpha\gamma}=b$$$$\frac{1}{\alpha\beta\gamma}=-a$$
From these, we get $a=b=1$ So, $P(x)=x^3+x^2+1$ and $Q(x)=x^3+x+1$. Now, how to proceed further?
| 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
$$
R(x)=x^3+\frac abx+\frac1b=0\tag{3}
$$
Since $Q$ and $R$ have the same roots, so must their difference,
$$
L(x)=\left(a-\frac ab\right)x+\left(a-\frac1b\right)=0\tag{4}
$$
If $a-\frac ab\ne0$, $L$ has one root, and so must $Q$. If $Q$ has only one root, then because the coefficient of $x^2$ is $0$, that root must be $0$. However, since the roots of $P$ and $Q$ are reciprocals, $0$ can't be a root of either. Contradiction.
Therefore, $a-\frac ab=0$ and $a-\frac1b=0$, which means $a=b=1$.
Thus,
$$
\begin{align}
P(x)&=x^3+x^2+1\\
Q(x)&=x^3+x+1
\end{align}\tag{5}
$$
Furthermore,
$$
\gcd(P(x),Q(x))\mid (2x^2-x+1)P(x)-(2x^2+x-2)Q(x)=3\tag{6}
$$
So, considering $\text{mod }3$:
If $x\equiv0\pmod{3}$, then $P(x)\equiv1\pmod{3}$
If $x\equiv1\pmod{3}$, then $P(x)\equiv Q(x)\equiv0\pmod{3}$
If $x\equiv2\pmod{3}$, then $P(x)\equiv1\pmod{3}$
Therefore,
$$
\gcd(P(x),Q(x))=\left\{\begin{array}{l}
3&\text{if }x\equiv1\pmod{3}\\
1&\text{if }x\not\equiv1\pmod{3}
\end{array}\right.\tag{7}
$$
Since $2013!+1\equiv1\pmod{3}$,
$$
\gcd(P(2013!+1),Q(2013!+1))=3\tag{8}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\cos x-\cos(x+y)=0$, then $y=2k\pi-2x$ for some $k\in \Bbb Z$. If their sum is equal to $0$, $y=(2k+1)\pi-2x$.
Then
$$
\begin{align}
f(x)&=\left\{\begin{array}{l}|\sin x|+|\sin(2k\pi-2x)|+|\sin(2k\pi-x)|\\|\sin x|+|\sin((2k+1)\pi-2x)|+|\sin((2k+1)\pi-x)|\end{array}\right.\\
&=2|\sin x|+|\sin 2x|
\end{align}$$
Now, when is $f(x)$ maximized? Well, $f'(x)=2\left(\frac{\sin x}{|\sin x|}\cos x+\frac{\sin 2x}{|\sin 2x|}\cos 2x\right)$. So we either want $\cos x+\cos 2x=0$ or $\cos x-\cos 2x=0$ This gets $\cos x=\pm1,\pm\frac12$. When this is true, we get:
$$f(x)=0,\frac{3\sqrt3}2$$
| {
"language": "en",
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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 * -4 ) ) &= ( 1.5, -2 )\\
( ( 2 + 1.5 ) + ( 4 + -2 ) ) &= ( 3.5, 2 )
\end{align*}
| 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
$$2U = (2 * 1, 2 * 2) = (2, 4).$$
Likewise, you should write
$$\frac{1}{2}V = (0.5 * 3, 0.5 * (-4)) = (1.5, -2).$$
As for your final line, you haven't indicated what you are doing. You are adding $2U$ and $\frac{1}{2}V$ so you should write
$$2U + \frac{1}{2}V = (2 + 1.5, 4 + (-2)) = (3.5, 2).$$
In summary, I think it would be better to format your answer as follows:
\begin{align*}
2U &= (2 * 1, 2 * 2) = (2, 4)\\
&\\
\frac{1}{2}V &= (0.5 * 3, 0.5 * (-4)) = (1.5, -2)\\
&\\
2U + \frac{1}{2}V &= (2 + 1.5, 4 + (-2)) = (3.5, 2).
\end{align*}
Alternatively, you could do all of the calculations at the same time:
$$2U + \frac{1}{2}V = (2 * 1, 2 * 2) + (0.5 * 3, 0.5 * (-4)) = (2, 4) + (1.5, -2) = (3.5, 2).$$
These comments are less important than the ones made above, but I think still worth pointing out. They are more about style than content.
In is better to use $\cdot$ or $\times$ to denote multiplication rather than $*$.
You shouldn't swap between fractions and decimals unless you have a good reason. As the question involved a fraction, use fractions the whole way through.
With these points in mind, I would present the calculation as follows:
\begin{align*}
2U + \frac{1}{2}V &= 2(1, 2) + \frac{1}{2}(3, -4)\\
&= (2\times 1, 2\times 2) + \left(\frac{1}{2}\times 3, \frac{1}{2}\times (-4)\right)\\
&= (2, 4) + \left(\frac{3}{2}, -2\right)\\
&= \left(\frac{5}{2}, 2\right).
\end{align*}
| {
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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 largest block in the jordan form is $k$, because The exponent of the term corresponding to an eigenvalue in the minimal polynomial is the size of the largest block.
But there is also a thing we can say about the characteristic polynomial, but I don't understand it.
We know that "The algebraic multiplicity of an eigenvalue is the sum of the total sizes of the blocks." Could someone please explain and demonstrate what this means?
| The possibilities are
$$\begin{pmatrix}1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\end{pmatrix}\;\;,\;\;\begin{pmatrix}1&1&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1\end{pmatrix}$$
$$\begin{pmatrix}1&1&0&0\\
0&1&0&0\\
0&0&1&1\\
0&0&0&1\end{pmatrix}\;\;,\;\;\begin{pmatrix}1&1&0&0\\
0&1&1&0\\
0&0&1&0\\
0&0&0&1\end{pmatrix}$$
$$\begin{pmatrix}1&1&0&0\\
0&1&1&0\\
0&0&1&1\\
0&0&0&1\end{pmatrix}$$
and the corresponding minimal pol. in each case is
$$(x-1)\;,\;\;(x-1)^2\;,\;\;(x-1)^2\;,\;\;(x-1)^3\;,\;\;(x-1)^4$$
Cases $\;2-3\;$ depend on whether the geometry multiplicity of the eigenvalue (i.e., the eigenspace's dimension) is two or one
| {
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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",
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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 formula that I do not quite remember how to use it.
| 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 colorings is $$ \dfrac{n^4 + 11n^2}{12} .$$
Proof: If we use exactly $4$ colors, there are $2$ different colorings. Because, we apply one of the colors (only $1$ way) to the base. With the remaining $3$ colors, we can paint the faces with $(3-1)!=2$ ways that using circular permutation. The selection number of $4$ out of $n$ colors is $\dbinom{n}{4}$. With the multiplication principle, we get $2\cdot \dbinom{n}{4}$ colorings. Similarly, if we use exactly $3$ colors, there are $\dbinom{3}{1}\cdot \dbinom{n}{3}$ colorings. If we use exactly $2$ colors, there are $3\cdot \dbinom{n}{2}$ colorings. If we use $1$ color, there are $n$ colorings. In total, we find the number of different colorings is
$$ 2\cdot \dbinom{n}{4} + 3\cdot \dbinom{n}{3} + 3\cdot \dbinom{n}{2} + n = \color{red}{\dfrac{n^4 + 11n^2}{12}} .$$
Especially for $n=2$, we get $\dfrac{2^4 + 11\cdot 2^2}{12} = \dfrac{60}{12} = 5$.
Note: Moreover we can see that, a part of our proof is that it solves the problem. If we use exactly $2$ colors, there are $3\cdot \dbinom{2}{2}$ colorings. If we use $1$ color, there are $2$ colorings. In total, we get $3 + 2 = 5$ different colorings.
| {
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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 such that $r< 1$ and let $c_R$ be the upper semicircle with radius R such that $R>1$. Let $L_1$ be the interval $(r,R)$ on the reals and let $L_2$ be the interval $(-R,r)$ on the reals.
Then
$$\int_{c_r} f + \int_{c_R} f + \int_{L_1} f + \int_{L_2} f$$
taken counterclockwise is equal to
$$2\pi i\text{ res}(f,i) = -\frac{\pi^2 e^{\frac{\pi i}{6}}}{2}$$
The integrals along $c_r$ and $c_R$ should go to $0$ as $r$ goes to $0$ and $R$ goes to infinity.
So:
$$\int_{L_1} + \int_{L_2}= \int_0^{\infty}\frac{x^{1/3}\log(x)}{x^2 + 1}dx - \int_{0}^{\infty} \frac{x^{1/3}e^{i\pi/3}(ln(x) + i\pi) }{x^2 + 1}dx$$
where x is the polar radius.
Let $A = \int_0^\infty \frac{x^{1/3}\log(x)}{x^2 + 1}dx$
and let $B = \int_0^\infty \frac{x^{1/3}}{x^2 + 1}dx $.
Then we have $A*(1-e^{\frac{i\pi}{3}}) - \pi i e^{\frac{i\pi}{3}}B = $-$\frac{\pi^2 e^{\frac{\pi i}{6}}}{2}$.
This doesn't mesh with the given values for $A$ and $B$, so I'm wondering where I went wrong. Thanks in advance.
| 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, the contour integral is equal to
$$\int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} \log{x} - e^{i 2 \pi/3} \int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} (\log{x}+i 2 \pi) $$
which is equal to
$$\left (1-e^{i 2 \pi/3} \right )\int_0^{\infty} dx \frac{x^{1/3}}{1+x^2} \log{x} - i 2 \pi e^{i 2 \pi/3} \int_0^{\infty} dx \frac{x^{1/3}}{1+x^2}$$
By the residue theorem, the contour integral is $i 2 \pi$ times the sum of the residues at the poles $z=e^{i \pi/2}$ and $z=e^{i 3 \pi/2}$, or
$$i 2 \pi \frac{e^{i \pi/6}}{2 i} i \frac{\pi}{2} + i 2 \pi \frac{e^{i \pi/2}}{-2 i} i \frac{3 \pi}{2} = \frac{5 \pi^2}{4} + i \frac{\sqrt{3} \pi^2}{4}$$
As you did above, let $A$ be the first integral and $B$ be the second. Then
$$\left (1-e^{i 2 \pi/3} \right ) A - i 2 \pi e^{i 2 \pi/3} B = \frac{5 \pi^2}{4} + i \frac{\sqrt{3} \pi^2}{4}$$
So we equate real and imaginary parts. Rearranging, we get two equations and two unknowns:
$$\frac{3}{2} A + \sqrt{3} \pi B = \frac{5 \pi^2}{4} $$
$$-\frac{\sqrt{3}}{2} A + \pi B = \frac{\sqrt{3} \pi^2}{4}$$
From this, you may easily verify that
$$A = \int_0^{\infty}dx \frac{x^{1/3}}{1+x^2} \log{x} = \frac{\pi^2}{6}$$
$$B = \int_0^{\infty}dx \frac{x^{1/3}}{1+x^2} = \frac{\pi}{\sqrt{3}}$$
| {
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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 $$\begin{align} n^3=\sum_{k=0}^{n-1} (k+1)^3-k^3&=\sum_{k=0}^{n-1} 3k^2+3k+1\\ &=3\sum_{k=0}^{n-1}k^2+3\sum_{k=0}^{n-1}k+\sum_{k=0}^{n-1}1\\
&=3\sum_{k=0}^{n-1}k^2+3\frac{(n-1)n}{2}+n\\
\frac{n^3}{6}-\frac{n(n-1)}{2}-\frac n3&=\sum_{k=0}^{n-1}k^2\\
\frac{n(n-1)(2n-1)}{6}&=\sum_{k=0}^{n-1}k^2 \end{align}$$
So we get $$ \frac{n(n+1)(2n+1)}{6}=\sum_{k=1}^{n}k^2 $$
| {
"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
} |
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 that would produce critical damping.
My work:
The restoring force is $-kx$. Then
$$7.5 = -k(2.5) \\
-\frac{7.5}{2.5} = k \\
ma = -\frac{7.5x}{2.5} \\
my’’ + 9y’ + -3y = 0,\quad y(0) = 2.5, y(5) = 0 \\
\frac{-9 \pm \sqrt{81 + 4(m)(3)}}{2m} \\
-\frac{9}{2m} \pm \frac{\sqrt{81+12m}}{2m} \\
y = Ae^{-(9/2)x}\cos\left(\frac{\sqrt{81+12m}}{2m}x\right) + Be^{-(9/2)x}\sin\left(\frac{\sqrt{81+12m}}{2m}x\right) \\
2.5 = A + B\cdot 0 \\
0 = (2.5)e^{-45/2}\cos\left(\sqrt{81+12m}\frac{5}{2m}\right) + Be^{-45/2}\sin\left(\sqrt{81+12m}\frac{5}{2m}\right)$$
Any help would be appreciated
| 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",
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"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 other identities
$(1)\displaystyle\sin x(\sin x-\cos x+1)=\sin^2x+\sin x(1-\cos x)=1-\cos^2x+\sin x(1-\cos x)$
$\displaystyle\implies\sin x(\sin x-\cos x+1)=(1-\cos x)(\sin x+\cos x+1)$
$(2)\displaystyle\sin x(\sin x+\cos x-1)=\sin^2x-\sin x(1-\cos x)=(1-\cos x)(1+\cos x-\sin x)$
$(3)\displaystyle\cos x(\sin x+\cos x-1)=\cos^2x-\cos x(1-\sin x)=(1-\sin x)(1+\sin x-\cos x)$
and so on
| {
"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 : $x=\tan \theta$
\begin{align*}
\int\frac{x^3}{\sqrt{1+x^2}}\, dx&=\int \tan^3\theta\sec\theta\, d\theta\\
&=\int (\sec^2\theta-1)\sec\theta\tan\theta\, d\theta\\
&=\frac{\sec^3\theta}{3}-\sec\theta+C\\
&=\frac{(\sqrt{1+x^2})^3}{3}-\sqrt{1+x^2}+C
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/606431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
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 good example is something like
$$
M = \frac19 \begin{bmatrix}
1 & -4 & 8 \\
8 & 4 & 1 \\
-4 & 7 & 4
\end{bmatrix}
$$
Does anyone have any other examples, or a process for generating them?
One general formula for a rotation matrix is given here. So one possible approach would be to choose $u_x$, $u_y$, $u_z$ and $\theta$ so that you get something simple. Simple enough for hand calculations, but not trivial. Like the example given above.
| Here are some:
$$\left[ \begin {array}{ccc} 1/3&2/3&2/3\\ 2/3&-2/3&1/3\\ 2/3&1/3&-2/3\end {array} \right]
$$
$$\left[ \begin {array}{ccc} 2/7&3/7&6/7\\ 3/7&-6/7&2/7\\ 6/7&2/7&-3/7\end {array} \right]
$$
$$
\left[ \begin {array}{ccc} \frac{2}{11}&{\frac {6}{11}}&{\frac {9}{11}}\\ -{\frac {6}{11}}&-{\frac {7}{11}}&{\frac {6}{11}}
\\ {\frac {9}{11}}&-{\frac {6}{11}}&\frac{2}{11}\end {array}
\right]
$$
EDIT: here are the small positive integer solutions of $a^2 + b^2 + c^2 = d^2$ with
$a \le b \le c$ and $\gcd(a,b,c,d)=1$, in order of increasing $b^2 + c^2$:
$$\eqalign{1^2 + 2^2 + 2^2 &= 3^2\cr
2^2 + 3^2 + 6^2 &= 7^2\cr
4^2 + 4^2 + 7^2 &= 9^2\cr
1^2 + 4^2 + 8^2 &= 9^2\cr
6^2 + 6^2 + 7^2 &= 11^2\cr
2^2 + 6^2 + 9^2 &= 11^2\cr
3^2 + 4^2 + 12^2 &= 13^2\cr
2^2 + 10^2 + 11^2 &= 15^2\cr
2^2 + 5^2 + 14^2 &= 15^2\cr
8^2 + 9^2 + 12^2 &= 17^2\cr
1^2 + 12^2 + 12^2 &= 17^2\cr
}$$
| {
"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
} |
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. Does any know help me to show this?
| 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\[\sin2a\leqslant \sin2b\leqslant \sin 2c\iff\frac{1}{\sin2a}\geqslant\frac{1}{\sin2b}\geqslant\frac{1}{\sin2c}.\]By Chebyshev's inequality, we have\begin{align}&\frac{1-3\cos^2a}{\sin2a}+\frac{1-3\cos^2b}{\sin2b}+\frac{1-3\cos^2c}{\sin2c}\\\geqslant&\frac{1}{3}\left(\frac{1}{\sin2a}+\frac{1}{\sin2b}+\frac{1}{\sin2c}\right)\left((1-3\cos^2a)+(1-3\cos^2b)+(1-3\cos^2c)\right)=0\end{align}Note that\begin{align}&\tan{a}+\tan{b}+\tan{c}- 2(\cot{a}+\cot{b}+\cot{c})\\=&2\left(\frac{1-3\cos^2a}{\sin2a}+\frac{1-3\cos^2b}{\sin2b}+\frac{1-3\cos^2c}{\sin2c}\right).\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/609197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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
$\frac{d \log^2(2\sin(\pi x))}{dx} = \frac{1}{3}\log^3(2\sin(\pi x))\frac{1}{\frac{1}{2\sin(\pi x)}2cos(\pi x)\pi} = \frac{1}{3}\log^3(2\sin(\pi x))\frac{sin(\pi x)}{\pi cos(\pi x)} = \frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}$
$\int_0^{\frac{1}{2}}(2x-1)^2\log^2(2\sin(\pi x))\,dx =$
$[(2x-1)^2\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} - \int_0^{\frac{1}{2}}(4x-2)\log^2(2\sin(\pi x))\,dx =$
$[(2x-1)^2\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + [(2 - 4x)\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + 4\int_0^{\frac{1}{2}}\log^2(2\sin(\pi x))\,dx =$
$[(2x-1)^2\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + [(2 - 4x)\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}} + 4[\frac{\log^3(2\sin(\pi x))sin(\pi x)}{3\pi cos(\pi x)}]_0^{\frac{1}{2}}$
*
*of course it's possible I messed it up somewhere
*with $\cos(\frac{\pi}{2})$ you have a $0$ in the denominator which makes it rather hard to evaluate the integral.
| {
"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
} |
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
$$ \lim_{x \to 0}\frac{\sqrt{3+bx}-\sqrt{3}}{x}=\lim_{x \to 0}\frac{(3+bx)-3}{x(\sqrt{3+bx}+\sqrt{3})}=\lim_{x \to 0}\frac{b}{(\sqrt{3+bx}+\sqrt{3})}=3 $$
We can find that
$$ \frac{b}{2\sqrt{3}}=3 , \quad b=2\sqrt{3}. $$
This is the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/613546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
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 $OL:OM:ON \text{ is } 1:2:3$.
If $\ \dfrac{\text{area of }LONP}{\text{area of }\Delta PQR}=\dfrac{a}{b}, \quad$ where $a$
and $b$ are integers with no common factors,
what is the value of $a+b$ ?
All that I was able to do is:
Area $LONP=\frac{1}{2} |OL||PL|+\frac{1}{2} |NP||ON|=\frac{1}{2} |OL||PL|+\frac{1}{2} |NP|\ 3|OL|=\frac{1}{2} |OL|\ \left[\ |PL|+3|NP|\ \right]$
Area $PQR=\frac{1}{2} |PR||PQ|\sin 60^o=\frac{\sqrt 3}{4} |PR||PQ|=\frac{\sqrt 3}{4} |NP+RN||PL+LQ|=\frac{\sqrt 3}{4} \left[\ |NP|+|PL|+|RN|+|LQ| \ \right]$
Area $\Delta LON=\frac{1}{2} |ON||OL|\sin 120=\frac{\sqrt 3}{4} |OL|\ 3|OL|=\frac{3\sqrt 3}{4} |OL|^2$
$\mathbf{EDIT : }$Following Suraj M.S 's answer :
$$\begin{align}
\text{Area } \Delta PQR &=\dfrac{3x\ RN}{2}+\dfrac{3x\ PN}{2}+\dfrac{x\ PL}{2}+\dfrac{x\ QL}{2}+ {x\ QM}+{x\ MR} \\
\\
&=\dfrac{3x\ (PN+RN)}{2}+\dfrac{x\ (PL+QL)}{2}+{x\ (QM+MR)}\\
\\
&=\dfrac{3x\ (PR)}{2}+\dfrac{x\ (PQ)}{2}+{x\ (QR)}\\
\\
&=kx(\dfrac{3}{2}+\dfrac{1}{2}+{1})=3kx\\
\end{align}$$
Area $\Delta PQR=\dfrac{1}{2} k^2 \sin 60^o=\dfrac{k^2 \sqrt 3}{4} \implies x=\dfrac{k}{4\sqrt 3}$
$\mathbf{Question: }$How do I now find the area of $LONP$ in terms of $x'$s and/or $k'$s only ?
| 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 resultant area as
$$area(\triangle PQR)=3kx$$
equating it with the usual area
$$3kx=\frac{\sqrt{3}k^2}{2}$$
we get
$$x=\frac{\sqrt{3}k}{6}$$
with the help of $x$ you will get $OL, OM, ON$.
Now let $\angle NPO=\theta$ then $\angle OPL=60^\circ-\theta$. with the help of the sides find $\frac{\sin \theta}{\sin (60^\circ-\theta)}$
you get
$$ \frac{\sin\theta}{\sin (60^\circ-\theta)}=\frac{ON}{OL}=3$$
solving for $\theta$
$$\tan \theta=\frac{3\sqrt{3}}{5}$$
with the help of $\tan$ you get $NP=\frac{5}{6}k$
now further you can get $PL$ by interchanging the angles $\theta$ and $60^\circ-\theta$ and once more solving you get the new $\theta$ as
$$\tan \theta=\frac{\sqrt{3}}{7}$$
using the same method of finding $NP$ find $PL$.
$$PL=\frac{7}{6}k$$
now we have found all our needed unknowns.
$$a=area(\triangle NOP)+area(\triangle POL)$$
$$=\frac{1}{2}(\frac{\sqrt{3}k}{2}.\frac{5k}{6}+ \frac{\sqrt{3}k}{6}.\frac{7k}{6} )$$
$$a=\frac{11\sqrt{3}k^2}{72}$$
also
$$b=\frac{\sqrt{3}k^2}{2}$$
solving you get $$\frac{a}{b}=\frac{11}{36}$$
$$a+b=\left(\frac{a}{b}+1\right)b$$
which solves to
$$a+b=\frac{47\sqrt{3}k^2}{72}$$
where $k$ is the side of the equilateral triangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/615018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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:
$$\begin{align*}
\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}
=& \lim_{x \to \infty} \frac{(\sqrt[3]{x} - \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}{(\sqrt[3]{x} + \sqrt[5]{x})(\sqrt[3]{x} + \sqrt[5]{x})}\\
\\
=& \lim_{x \to \infty} \frac{(\sqrt[3]{x})^2 - (\sqrt[5]{x})^2}{(\sqrt[3]{x})^2+2\sqrt[3]{x}\sqrt[5]{x}+(\sqrt[3]{x})^2}\\
\\
=& \lim_{x \to \infty} \frac{x^{2/3}-x^{2/5}}{x^{2/3}+2x^{1/15}+x^{2/5}}\\
\\
=& \lim_{x \to \infty} \frac{x^{4/15}}{2x^{17/15}}
\end{align*}$$
Somehow I get stuck. I am sure I did something wrong somewhere.. Can someone please help me out?
| $$F=\lim_{x \to \infty} \frac{\sqrt[3]x - \sqrt[5]x}{\sqrt[3]x + \sqrt[5]x}=\lim_{x \to \infty} \frac{x^{\frac13} - x^{\frac15}}{x^{\frac13} + x^{\frac15}}$$
As lcm$(3,5)=15$ I will set $\displaystyle x^{\frac1{15}}=y\implies x^{\frac13}=y^5$ and $\displaystyle x^{\frac15}=y^3$
$$\implies F=\lim_{y\to\infty}\frac{y^5-y^3}{y^5+y^3}$$
Setting $\frac1y=h,$
$$F=\lim_{h\to0}\frac{(1-h^2)h^3}{(1+h^2)h^3}=\lim_{h\to0}\frac{1-h^2}{1+h^2}$$ the cancellation of $h$ is legal as $h\ne0$ as $h\to0$
| {
"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
} |
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 equation, but the $6f(n/4)$ confuses me.
Please help me find a solution.
Kind regards.
| 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 $n:$
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor}
[z^j] \frac{1}{1-\frac{5}{2}z+\frac{6}{4} z^2}
\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^k.$$
Observe that the roots of $$1-\frac{5}{2}z+\frac{6}{4} z^2
\quad\text{are}\quad\rho_0=1\quad\text{and}\quad\rho_1=\frac{2}{3}.$$
It follows that the coefficients of the rational term have the form
$$c_0 + c_1 \left(\frac{3}{2}\right)^j.$$
Actually solving for $c_{0,1}$ we obtain the formula
$$[z^j] \frac{1}{1-\frac{5}{2}z+\frac{6}{4} z^2}
= 3 \left(\frac{3}{2}\right)^j - 2.$$
This gives the exact formula for $T(n):$
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left(3 \left(\frac{3}{2}\right)^j - 2\right)
\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
Now for an upper bound on this consider a string of one digits, which gives
$$T(n)\le
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left(3 \left(\frac{3}{2}\right)^j - 2\right)
\sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^k
= \frac{27}{2} \times 3^{\lfloor \log_2 n \rfloor}
- (12+4\lfloor \log_2 n \rfloor) 2^{\lfloor \log_2 n \rfloor} -\frac{1}{2}.$$
For a lower bound consider a one followed by a string of zeros, which gives
$$T(n)\ge
\sum_{j=0}^{\lfloor \log_2 n \rfloor}
\left(3 \left(\frac{3}{2}\right)^j - 2\right) 2^{\lfloor \log_2 n \rfloor}
= 9 \times 3^{\lfloor \log_2 n \rfloor}
- (8+2\lfloor \log_2 n \rfloor) 2^{\lfloor \log_2 n \rfloor}.$$
Joining the two bounds it follows that $T(n)$ is asymptotic as follows:
$$T(n)\in\Theta\left(3^{\lfloor \log_2 n \rfloor}\right)
= \Theta(2^{\log_2 n \times \log_2 3}) = \Theta(n^{\log_2 3})$$
with the leading coefficient between $9/2$ and $9.$
Now we are supposed to have $f(0)=0$, $f(1)=2$, and $f(2)=1,$ so start with
$$T(n) + \left(3 \left(\frac{3}{2}\right)^{\lfloor \log_2 n \rfloor} - 2\right)
2^{\lfloor \log_2 n \rfloor}.$$
This is equal to two at $n=1$ but equal to twelve at $n=2$, so we need an additional contribution of
$$T(n) + \left(3 \left(\frac{3}{2}\right)^{\lfloor \log_2 n \rfloor} - 2\right)
2^{\lfloor \log_2 n \rfloor}
- 11 (1-d_{\lfloor \log_2 n \rfloor-1})\times
\left(3 \left(\frac{3}{2}\right)^{\lfloor \log_2 n \rfloor-1} - 2\right)
2^{\lfloor \log_2 n \rfloor-1}.$$
This simplifies to (for $n\ge 2$)
$$ T(n) + 3^{\lfloor \log_2 n \rfloor+1} -2^{\lfloor \log_2 n \rfloor+1}
- 11 (1-d_{\lfloor \log_2 n \rfloor-1})\times
\left(3^{\lfloor \log_2 n \rfloor} -2^{\lfloor \log_2 n \rfloor}\right).$$
The behavior here is highly erratic but the formula is exact and the order of growth remains the same. Here is another Master Theorem computation at MSE.
| {
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"url": "https://math.stackexchange.com/questions/615586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
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 $F(x,y,x_0) = 0$ by supposing $x = x(y)$ then:
$\frac{\mathrm{d}}{\mathrm{d}x} F(x,y,x_0) = \frac{\partial}{\partial x}F(x,y,x_0) + \frac{\partial}{\partial y} F(x,y,x_0) \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{\frac{\partial}{\partial x} F(x,y,x_0)}{\frac{\partial}{\partial y} F(x,y,x_0)}$
Then
$y' = \frac{- 2 (x - x_0)}{2 y} = \frac{x_0 -x}{y} $
Using
$F(x,y,x_0) = (x-x_0)^2 + y^2 - R^2 = 0 \implies x_0^2 - 2 x x_0 + x^2 + y^2 - R^2 = 0 \implies x_0 = \frac{2x \pm \sqrt{4 x^2 - 4 x^2 - 4 y^2 + 4 R^2}}{2} = x \pm \sqrt{R^2 - y^2}$
We write $y'$ as:
$y' = \frac{-y}{x_0 -x} = \frac{-y}{x \pm \sqrt{R^2 - y^2} - x} = \frac{-y}{\pm \sqrt{R^2 - y^2}} \implies \frac{\mp \sqrt{R^2 - y^2}}{y} \mathrm{d}y = \mathrm{d}x$
Solving the integral we get:
$R \log{\left(R - \sqrt{R^2 - y^2}\right)} - R \log{y} + \sqrt{R^2 - y^2} = \pm x + C_0$
But when I graph the solution, I can see I am missing some solutions, the complete solution set is
$R \log{\left(R - \sqrt{R^2 - y^2}\right)} - R \log{y} + \sqrt{R^2 - y^2} = \pm x + C_0$ and
$R \log{\left(R - \sqrt{R^2 - y^2}\right)} - R \log{(-y)} + \sqrt{R^2 - y^2} = \pm x + C_0$
When did I loose those solutions?
| Remember that $\displaystyle \int\frac{dy}y = \log|y|$. It seems to be that easy.
| {
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"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:$$\begin{cases}|a-b|\le2\\|b-c|\le2\\|c-a|\le2\end{cases}$$
If you think about it, we may as well prove that at least two of these are correct:$$\begin{cases}(a-b)^2\le4\\(b-c)^2\le4\\(c-a)^2\le4\end{cases}$$
We could square one of the inequalities (since both sides are positive):$$(a+b+c)^2\le16\Rightarrow a^2+b^2+c^2+2(ab+bc+ac)\le16$$
$$a^2+b^2+c^2+8\le a^2+b^2+c^2+2(ab+bc+ac)\le16\Rightarrow a^2+b^2+c^2\le 8$$
So now we know that: $$\begin{cases}a,b,c>0\\a+b+c\le4\\a^2+b^2+c^2\le8\\ab+bc+ac\ge4\end{cases}$$
Does anyone see how to solve this? Thanks.
| 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",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 $4$ from both sides: $$a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}\ge8.5$$
We could use AM-GM here ($a^2,\frac{1}{a^2},b^2,\frac{1}{b^2}$ are all positive), but obviously it wouldn't do anything useful. And we still have to use the fact that $a+b=4$ somehow.
I've tried substituting $b$ with $4-a$, but after clearing the denominators and simplifying we don't quite see anything useful, just a random 6th degree polynomial.
The polynomial is actually: $$2a^6-24 a^5+103.5 a^4-188 a^3+122 a^2-8 a+16\ge 0$$
How could I solve this? We can't use calculus by the way. Thanks.
| 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
$$
\frac 1{a^2}+\frac 1{b^2}\geq \frac{(\frac 1a+\frac 1b)^2}2
$$
We know that the minimum of $\frac 1a+\frac 1{4-a}$ is at $a=2$, with outcome $1$. (This can be done by differentiation, or multiply with $a(4-a)$ first.) Now we get
$$
\frac 1{a^2}+\frac 1{b^2}\geq \frac{(\frac 1a+\frac 1b)^2}2\geq \frac 12
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
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 show that $\lim\limits_{x\to\infty}\frac1x\sqrt{x+\sqrt{x+\dots}}=0$, show inductively that
$$
\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}\le\frac{1+\sqrt{1+4x}}{2}
$$
using $\sqrt{x}\le\frac{1+\sqrt{1+4x}}{2}$ and
$$
\left(\frac{1+\sqrt{1+4x}}{2}\right)^2=x+\frac{1+\sqrt{1+4x}}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 maximum of $\displaystyle \frac{4x-9y}{2}$, $x$ should be at its maximum and $y$ at its minimum. Solving the equations, I get that $x = 3$ and $\displaystyle y = -\frac{2}{3}$, but the answer i get is wrong. What am I doing wrong?
| 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 parallel lines $ 2x \ - \ \frac{9}{2}y \ = \ c \ \ \Rightarrow \ \ y \ = \ \frac{4}{9}x \ - \ \frac{2}{9}c \ \ , $ as mathlove describes. Thus, the value of $ \ c \ $ corresponds to a change in the $ \ y-$intercept of the line; there is a minimal value of $ \ c \ $ for a line which is just tangent to the ellipse and a maximum value of $ \ c \ $ for a line tangent elsewhere (we will discuss those tangent points shortly).
Looking at this in three dimensions, the equation $ \ x^2 - 4x + 9y^2 + 6y + 4 \ = \ 0 \ $ represents an elliptic cylinder, the cross-section of which is everywhere the same as the ellipse in the $ \ xy-$plane. The function $ \ z \ = \ 2x \ - \ \frac{9}{2}y \ $ represents a plane slicing obliquely through this cylinder. The problem at hand asks for the maximum "height" $ \ z \ $ above the $ \ xy-$plane at which there is an intersection point between the cylinder and the plane.
Returning to the problem in two dimensions, we see then that we are seeking the point(s) at which the lines $ \ y \ = \ \frac{4}{9}x \ - \ \frac{2}{9}c \ $ are tangent to the ellipse, which requires us to find the locations on the ellipse which match the slope of the lines. (It will quickly be evident that there is no point in the interior of the ellipse that maximizes or minimizes $ \ c \ $ . ) By using implicit differentiation, we can find the slope of the tangent line to a point on the ellipse to be
$$ \frac{d}{dx} \ [ x^2 - 4x + 9y^2 + 6y + 4 ] \ = \ \frac{d}{dx} \ [ 0 ] \ \ \Rightarrow \ \ 2x - 4 + 18yy' + 6y' \ = \ 0 $$
$$ \Rightarrow \ y' \ = \ \frac{4 - 2x}{18y + 6} \ = \ \frac{2 - x}{3 \ (3y + 1)} \ . $$
It is this slope that we wish to set equal to the slope of the level curve lines:
$$ \frac{2 - x}{3 \ (3y + 1)} \ = \ \frac{4}{9} \ \Rightarrow \ 3 ( 2 - x ) \ = \ 4 ( 3y + 1 ) \ \Rightarrow \ y \ = \ \frac{1}{6} \ - \ \frac{1}{4}x \ . $$
The significance of this new linear equation is that the tangent points we are looking for lie on this line; what is also of interest is that we can write the equation as
$$ y \ - \ \frac{1}{6} \ + \ \frac{1}{2} \ = \ - \frac{1}{4} x \ + \ \frac{1}{2} \ \Rightarrow \ y \ + \frac{1}{3} \ = \ - \frac{1}{4} ( x - 2 ) \ ; $$
this means that the tangent points lie at opposite ends of a line segment passing through the center of the ellipse, as found from the equation in standard form which user117638 noted. (We see this sort of result in similar problems for locating the extrema of linear functions on curves with four-fold symmetry.) The matching of slopes of the tangent lines to the local slope of the ellipse is the essence of the Lagrange-multiplier method Fly By Night describes (what we are "really" doing is matching the gradient of the linear function to the local normal lines to points on the ellipse).
Now substituting $ \ y \ = \ \frac{1}{6} - \frac{1}{4}x \ $ into the equation of the ellipse, $ \ (x - 2)^2 \ + \ 9(y+\frac{1}{3})^2 \ = \ 1 \ $ leads to the quadratic equation $ \ \frac{25}{16}x^2 \ - \ \frac{25}{4}x \ + \ \frac{21}{4} \ = \ 0 \ . $ Solving the two equations simultaneously leads to the coordinate pairs $ \ ( \frac{14}{5} \ , \ - \frac{8}{15}) \ \ \text{and} \ \ ( \frac{6}{5} \ , \ - \frac{2}{15}) \ . $ Inserting these now into the tangent line equation $ \ y \ = \ \frac{4}{9}x \ - \ \frac{2}{9}c \ $ gives us the two parallel lines
$$ y \ = \ \frac{4}{9}x \ - \ \frac{16}{9} \ \Rightarrow \ c = 8 $$
and
$$ y \ = \ \frac{4}{9}x \ - \ \frac{6}{9} \ \Rightarrow \ c = 3 \ \ , $$
thereby presenting us with the maximal and minimal values of the linear function on the ellipse, respectively.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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{1}{3x}\int_0^x\sum_{n=0}^\infty(-1)^nt^{2n}\,dt-\frac{1}{3x^4}\int_0^xt^3\sum_{n=0}^\infty(-1)^nt^{2n}\\
&=&\frac{1}{3x}\int_0^x\frac{1}{1+t^2}\,dt-\frac{1}{3x^4}\int_0^x\frac{t^3}{1+t^2}\,dt\\
&=&\frac{1}{3x}\arctan x-\frac{1}{3x^4}\int_0^x\left(t-\frac{t}{1+t^2}\right)\,dt\\
&=&\frac{1}{3x}\arctan x-\frac{1}{3x^4}\left[\frac{x^2}{2}-\frac12\ln(1+x^2)\right].
\end{eqnarray}
Thus
\begin{eqnarray}
\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)(2n+4)}\left(\frac13\right)^{n+2}&=&\frac19f\left(\frac{1}{\sqrt{3}}\right)=\frac{1}{9\sqrt{3}}\arctan\frac{1}{\sqrt{3}}-\frac13\left[\frac16-\frac12\ln\frac43\right]\\
&=&\frac{\pi}{54\sqrt{3}}-\frac{1}{18}+\frac16\ln\frac43.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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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 is only possible for $n\equiv7\bmod16$ and $n\equiv1\bmod5$ in other words $n\equiv71\bmod 80$.
Now we prove $n=71$ does the trick.
Check $2^{c+3}||71^{2^{c}}-1$ by lte. So $71^{2^{c}}\equiv {2^{c+2}}\bmod2^{y+1}$
Likewise $71^{5^c}\equiv 5^{c+1}+1\bmod5^{c+2}$
If there is a $w$ such that $2^w\equiv-9\bmod a^c$but $2^w\neq-9\bmod2^{c+1}$ then $2^w\equiv2^c-9$ which implies $2^{w+2^{c-2}}\equiv{-9}\bmod2^{c+1}$ so there is also a solution for $2^c+1$.
Likewise if there is $w$ so $5^w\equiv-9\bmod5^c$ then $ 71^w\equiv a5^c-9\pmod{5^{c+1}} $ for some a not divisible by 5. Then if $5|mb+a$ we have $ 71^{w+m5^{c-1}}\equiv(a5^c-9)(b5^c+1)^m\equiv(mb+a)5^c-9\equiv-9\pmod{5^{c+1}}$.
Now let $c=2012$. By the above there are $w_1$ and $w_2$ such that $71^{w_1}\equiv-9\bmod2^{2012}$ and $71^{w_2}\equiv -9 \bmod 5^{2012}$
By chinese remainder theorem there is a $w$ such that
$w\equiv w_1\bmod2^{c-3}$ and $w\equiv w_2 \bmod 5^{c-1}$
so $71^w\equiv -9\bmod10^c$ as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $n+0$. This can be done in $\binom{n+0}{n}$ ways.
Maybe the smallest number chosen is $n$. Then we must choose $n$ numbers from the remaining $n+1$ numbers. That can be done in $\binom{n+1}{n}$ ways.
Maybe the smallest number chosen is $n-1$. Then we must choose $n$ numbers from the remaining $n+2$ numbers. That can be done in $\binom{n+2}{n}$ ways.
And so on. Finally, the smallest number chosen might be $1$, in which case we must choose $n$ numbers from the remaining $n+n$. This can be done in $\binom{n+n}{n}$ ways.
More formally, let $C_k$ be the number of choices in which the smallest number chosen is $n+1-k$. Then we must choose the remaining $n$ numbers from the $n+k$ numbers greater than $n+1-k$. This can be done in $\binom{n+k}{n}=\binom{k+n}{k}$ ways. So $C_k=\binom{k+n}{k}$.
Add up from $k=0$ to $k=n$. We get that
$$\binom{2n+1}{n+1}=\sum_{k=0}^n C_k=\sum_{k=0}^n \binom{k+n}{k}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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}\left({\sqrt{\lfloor x^2+x \rfloor }-x}\right) = , $where $\lfloor x \rfloor = $ floor function of $x$
$\bf{My\; Try}::$ for (1) one :: We can write as $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln x-\lfloor x \rfloor }{\lfloor x \rfloor}$
and we can say that when $x\rightarrow \infty$, Then $\lfloor x\rfloor \rightarrow x$
So $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln(x)-x}{x} = n\lim_{x\rightarrow \infty}\frac{\ln (x)}{x}-1$
Now Let $\displaystyle L = \lim_{x\rightarrow \infty}\frac{\ln(x)}{x}{\Rightarrow}_{L.H.R} =\lim_{x\rightarrow \infty}\frac{1}{x} = 0$
So $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln x-\lfloor x \rfloor }{\lfloor x \rfloor} = n\cdot 0-1 =-1$
$\bf{My\; Try}::$ for (2)nd one::we can say that when $x\rightarrow \infty$, Then $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$
So $\displaystyle \lim_{x\rightarrow \infty}\left({\sqrt{x^2+x}-x}\right) = \lim_{x\rightarrow \infty}\frac{\left({\sqrt{x^2+x}-x}\right)\cdot \left({\sqrt{x^2+x}+x}\right)}{\left({\sqrt{x^2+x}+x}\right)}$
$\displaystyle \lim_{x\rightarrow \infty}\frac{x}{\left(\sqrt{x^2+x}+x\right)} = \frac{1}{2}$
Now my doubt is can we write when $x\rightarrow \infty$, Then $\lfloor x\rfloor \rightarrow x$
and when $x\rightarrow \infty$, Then $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$
please clear me
Thanks
| $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} =\lim_{x\rightarrow \infty}\frac{lnx^n}{\lfloor x\rfloor}-1=\lim_{x\rightarrow \infty}\frac{nlnx}{\lfloor x\rfloor}-1$
Now $\displaystyle\frac{nlnx}{x} \le\frac{nlnx}{\lfloor x \rfloor}\le \frac{nlnx}{x-1}$. Hence $\displaystyle \lim_{x\rightarrow\infty}\frac{nlnx}{x}-1\le\lim_{x\rightarrow\infty}\frac{nlnx}{\lfloor x \rfloor}-1\le\lim_{x\rightarrow\infty} \frac{nlnx}{x-1}-1$.
Since $\displaystyle\lim_{x\rightarrow\infty}\frac{nlnx}{x}\to0$ and so does $\displaystyle\lim_{x\rightarrow\infty}\frac{nlnx}{x-1}$. So the required limit $-1$
| {
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"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 resulting degree of the term is $8$.
i think you should get $0.002706025...$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $(x+1)^n=q(x)\cdot (x-1)^3+ax^2+bx+c........................(1)$
Now put $x=1$, we get $\boxed{a+b+c=2^n}$
Now differentitae $(1)$, we get $n(x+1)^{n-1} = q(x)\cdot 3(x-1)^2+(x-1)^3\cdot q^{'}(x)+2ax+b$
again put $x=1$, we get $\boxed{2a+b=n(2)^{n-1}}$
Now again differentitae $(1)$ and then put $x=1$, we get
$\displaystyle \boxed{2a=n(n-1)2^{n-2}\Rightarrow \displaystyle a=\frac{n(n-1)}{2}\cdot 2^{n-2}}$
Similarly we get $\displaystyle b = n(2)^{n-1}-n(n-1)\cdot 2^{n-2}$
Similarly we get $\displaystyle c= 2^{n}+n(n-1)\cdot 2^{n-2}-\frac{n(n-1)}{2}\cdot 2^{n-2}$
So Remainder
$\displaystyle r(x) = \frac{n(n-1)}{2}2^{n-2}x^2+\left\{n(2)^{n-1}-n(n-1) 2^{n-2}\right\}x+2^{n}+n(n-1)\cdot 2^{n-2}-\frac{n(n-1)}{2}2^{n-2}$
is my solution is right , if not then how can i calculate it
or If there is any better method , then plz explain here
Thanks
| 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 final answer if you so choose. Try it and you will be amazed at how simple the calculations are.
Once you have your $a$, $b$ and $c$ you can always write
$$ r=a x^2 + (b-2a) x + a - b + c$$
for the remainder $r$.
Try it, it is like magic!
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\displaystyle 2x-1=\sqrt3\tan\phi$ in the second part
| {
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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 $a^3=3+x$, $b^3=3-x$ ? I don't know how to continue. Thank you.
| 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 $$\frac3x=\frac6{2\sqrt5}\implies x=?$$
| {
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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 $K$
such that for all $n\gt K$, $$\left| \frac{23n+2}{4n+1} - \frac{23}{4}
\right| < \epsilon. $$
Note that $$ \displaystyle\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| = \left| \frac{-15}{16n+4} \right| $$ and for $ n > 1 $ $$ \displaystyle \left| \frac{-15}{16n+4} \right| = \frac{15}{16n+4} < \frac{1}{n}. $$
Suppose $ \epsilon \in \textbf{R} $ and $ \epsilon > 0 $. Consider $ K = \displaystyle \frac{1}{\epsilon} $. Allow that $ n > K $. Then $ n > \displaystyle \frac{1}{\epsilon} $. So $ \epsilon >\displaystyle \frac{1}{n} $.
Thus
$$ \displaystyle\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| = \left| \frac{-15}{16n+4} \right| = \frac{15}{16n+4} < \frac{1}{n} < \epsilon. $$ Thus $$ \displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4}. $$
Is this proof correct? What are some other ways of proving this? Thanks!
| $\displaystyle\lim_{n\to \infty} \frac{n(23 + \frac{2}{n})}{n(4+\frac{1}{n})} = \cdots$
Use the fact that $\frac{\alpha}{n}$ tends to $0$ when $n$ tends to infinity, and theorems of limits of sequences.
| {
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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:
$$2R\tan^*(B-C)= \frac{2abc(a^2+b^2-c^2-a^2-c^2+b^2)}{(a^2+c^2-b^2)(a^2+b^2-c^2)-16\Delta^2},$$
but Heron's formula gives $16\Delta^2 = 4b^2c^2-(b^2+c^2-a^2)^2$, from which
$$2R\tan^*(B-C) = \frac{4abc(b^2-c^2)}{a^4-(b^2-c^2)^2-4b^2c^2+(b^2+c^2-a^2)^2}$$ follows. This gives:
$$2R\tan^*(B-C) = \frac{2abc(b^2-c^2)}{a^2(a^2-b^2-c^2)}=\frac{c^2-b^2}{a \cos A}$$ and:
$$2Ra^2\tan^*(B-C) = \frac{a(c^2-b^2)}{\cos A} = 2R(c^2-b^2)\tan(A),$$
$$ a^2 \tan^*(B-C) = (c^2-b^2)\tan A = (c^2-b^2)\frac{4\Delta}{b^2+c^2-a^2}.\tag{1}$$
So we have, in terms of side lengths,
$$ \sum_{cyc}(c^2-b^2)(a^2+c^2-b^2)(a^2+b^2-c^2) = 0, \tag{2} $$
or:
$$ \sum_{cyc}(c^2-b^2)(a^4-b^4-c^4+2b^2c^2) = 0,$$
$$ \sum_{cyc}\left(b^6-c^6-a^4b^2+a^2b^4+3b^2c^4-3b^4c^2\right)=0, $$
$$ \sum_{cyc}\left(-a^4b^2+a^2b^4+3b^2c^4-3b^4c^2\right)=0, $$
$$ \sum_{cyc}\left(4b^2c^4-4b^4c^2\right)=0,\tag{3} $$
Now it is easy to notice that $a=\pm b,a=\pm c$ or $b=\pm c$ imply that the LHS is zero, so $(a^2-b^2)(b^2-c^2)(c^2-a^2)$ divides the LHS in $(3)$.
$$ \sum_{cyc}\left(4b^2c^4-4b^4c^2\right) = 4(a^2-b^2)(b^2-c^2)(c^2-a^2)\tag{4} $$
proves the (modified) claim.
| {
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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:
$$F(t)=\int_{0}^{\frac{\pi}{2}}\dfrac{2t+2\cos{x}}{t^2+2t\cos{x}+1}dx+\int_{\frac{\pi}{2}}^{\pi}\dfrac{2t+2\cos{x}}{t^2+2t\cos{x}+1}dx=I_{1}+I_{2}$$
where
$$I_{2}=\int_{0}^{\frac{\pi}{2}}\dfrac{2t-2\cos{x}}{t^2-2t\cos{x}+1}dx$$
then
\begin{align*}I_{1}+I_{2}&=2\int_{0}^{\frac{\pi}{2}}\dfrac{(t+\cos{x})(t^2-2t\cos{x}+1)+(t-\cos{x})(t^2+2t\cos{x}+1)}{(t^2+1)^2-(2t\cos{x})^2}dx\\
&=4t\int_{0}^{\frac{\pi}{2}}\dfrac{t^2-2\cos^2{x}+1}{-4t^2\cos^2{x}+(t^2+1)^2}dt
\end{align*}
then I fell very ugly, have other methods? Thank you
| 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.
$$\begin{align}
\int_0^\pi\frac{2t+2\cos x}{t^2+2t\cos(x)+1}\,dx
&=\int_0^\infty\frac{2t+2\frac{1-u^2}{1+u^2}}{t^2+2t\frac{1-u^2}{1+u^2}+1}\,\frac{1}{1+u^2}\,du\\
&=\int_0^\infty\frac{2t(1+u^2)+2(1-u^2)}{(t^2+1)(1+u^2)^2+2t(1-u^4)}\,du\\
&=\int_0^\infty\frac{(2t+2)+(2t-2)u^2}{(t^2+1)(1+u^2)^2+2t(1-u^4)}\,du\\
&=\int_0^\infty\frac{(2t+2)+(2t-2)u^2}{u^4(t^2-2t+1)+u^2(2t^2+2)+(t^2+2t+1)}\,du\end{align}$$
Since the denominator is quadratic in $u^2$, it's possible to factor it into linear factors in terms of $t$ and then use partial fraction decomposition to get an antiderivative.
| {
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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 idea what will be the last term
I did one same kind of problem $\prod (1-{1\over k^2}),k\ge2$ whose answer is $1\over 2$
here I think the answer will be $D$, none of this?
| 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:
$$
\prod_{n=2}^m\frac{n-1}{n+1}=\frac13\frac24\frac35\frac46\cdots\frac{m-3}{m-1}\frac{m-2}{m}\frac{m-1}{m+1}
$$
$$
\prod_{n=2}^m\frac{n^2+n+1}{n^2-n+1}=\frac73\frac{13}7\frac{21}{13}\frac{31}{21}\cdots\frac{m^2-m+1}{m^2-3m+3}\frac{m^2+m+1}{m^2-m+1}
$$
| {
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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 parallel How can I understand that.please help me.
| 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 solutions.
On the other hand reducing the augmented matrix
$$
\begin{pmatrix}
2 & 6 & -1 & 10\\
3 & 2 & 2 & 1\\
5 & 4 & 3 & 4
\end{pmatrix}
$$
$$
\longrightarrow
\begin{pmatrix}
2 & 6 & -1 & 10\\
0 & -2 & 1 & -7\\
0 & 0 & 0 & 35
\end{pmatrix}
$$. Seeing the last row you realize that can't be possible.
| {
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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 saying $2^{-n} = 5^n \times 10^{-n}$, and when we calculate that out, it's correct:
$\begin{align}
5^1 \times 10^{-1} &= 5 \times 0.1 = 0.5 \\
5^2 \times 10^{-2} &= 25 \times 0.01 = 0.25 \\
5^3 \times 10^{-3} &= 125 \times 0.001 = 0.125 \\
5^4 \times 10^{-4} &= 625 \times 0.0001 = 0.0625 \\
5^5 \times 10^{-5} &= 3125 \times 0.00001 = 0.03125 \\
...
\end{align}$
I calculated this out for $n = [0, 10]$ and it works out, but I have no idea how to prove it fully.
| $\dfrac{1}{2} = \dfrac{5}{10}$ (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 $a^{-n} = \dfrac{1}{a^n}$
| {
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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 $\displaystyle\frac{x^2}{(\sqrt{2})^2}+\displaystyle\frac{y^2}{2^2}=1$
Stewart define a region I if it is of the kind $\{(x,y,z):(x,y)\in D, u_1(x,y)\leq z \leq u_2(x,y)\}$.
I believe that the region I'm asked for cab be describe setting $D=\{(x,y): \displaystyle\frac{x^2}{(\sqrt{2})^2}+\displaystyle\frac{y^2}{2^2}=1\}$ and then $E=\{(x,y,z):(x,y)\in D, x^2\leq z \leq 4-x^2-y^2\}$.
Can the volume $V(E)$ be computed by $\displaystyle\int\displaystyle\int_D\displaystyle\int_{x^2}^{4-x^2-y^2}z\;dz$?
Or maybe considering that $0\leq \sqrt{x} \leq \sqrt{2}$ and $0 \leq y \leq \sqrt{4-x^2-y^2}$ could I compute the volume calculating $\displaystyle\int_0^{\sqrt{2}}\displaystyle\int_0^{\sqrt{4-2x^2}}\displaystyle\int_{x^2}^{4-x^2-y^2}z\;dzdydx ?$
The issue with the approach above is that doesn't seem too easy after the first two integrals, because:
$\displaystyle\int_0^{\sqrt{2}}\displaystyle\int_0^{\sqrt{4-2x^2}}\displaystyle\int_{x^2}^{4-x^2-y^2}z\;dz = \displaystyle\frac{1}{2} \displaystyle\int_0^{\sqrt{2}}\displaystyle\int_0^{\sqrt{4-2x^2}} [(4-x^2-y^2)^2-x^2]\; dydz \\
=\displaystyle\frac{1}{2} \displaystyle\int_0^{\sqrt{2}} \left(16y-7x^2y-\frac{8}{3}y^3+\frac{2}{3}x^2y^3+x^4y+\frac{1}{5}y^5\right)\rvert_0^{\sqrt{4-2x^2}}\;dx$ which after the substitution seems hard to evaluate.
Is there an easy way?
| 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, as you have. The other is using polar coordinates. Since $2x^2 +y^2 = 4$ we have
$$\frac{x^2}{(\sqrt{2})^2} + \frac{y^2}{2^2} = 1.$$
To use polar coordinates here we switch using
$$\begin{cases}
x & = \frac{r \cos \theta}{\sqrt{2}} \\
y & = \frac{r \sin \theta}{2},
\end{cases}$$
and the "area element" becomes
$$dA = \frac{r}{2 \sqrt{2}} \, dr \, d \theta.$$
This is obtained via the Jacobian. I can explicitly write it out if you need. Proceeding the computations yields
$$
\begin{align}
\iint 4 -2x^2 -y^2 \, dA & = \int_0^{2 \pi} \hspace{-5pt} \int_0^1 (4-r^2) \frac{r}{2\sqrt{2}} \, dr \, d \theta \\
& = \frac{2 \pi}{2 \sqrt{2}} \int_0^1 4r - r^3 \, dr \\
& = \frac{\pi}{\sqrt{2}} \left( 2r^2 \bigg\vert_0^1 - \frac{r^4}{4} \bigg\vert_0^1 \right) \\
& = \frac{\pi}{\sqrt{2}} \left( 2 - \frac{1}{4} \right) \\
& = \frac{\pi}{\sqrt{2}} \cdot \frac{7}{4} \\
& = \frac{7 \pi}{4 \sqrt{2}}.
\end{align}
$$
Best wishes. :)
| {
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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) = $
$\lim\limits_{x \to \infty}\frac{1}{\sqrt{x^2 +1}+x}+\lim\limits_{x \to \infty}\frac{1}{\sqrt{4x^2 + 1}+2x} -\lim\limits_{x \to \infty}\frac{1}{\sqrt{9x^2 + 1}+3x} = 0 + 0 + 0 = 0$
| {
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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:
*
*$(0,0) \rightarrow (1,1) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rightarrow (1,1) \rightarrow (2,0) \rightarrow (3,1) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rightarrow (1,1) \rightarrow (0,2) \rightarrow (1,3) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rightarrow (1,1) \rightarrow (0,2) \rightarrow (1,3) \rightarrow (2,2) \rightarrow (1,1) \rightarrow (2,0) \rightarrow (3,1) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rightarrow (1,1) \rightarrow (0,2) \rightarrow (1,3) \rightarrow (2,2) \rightarrow (3,1) \rightarrow (2,0) \rightarrow (1,1) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rightarrow (1,1) \rightarrow (2,2) \rightarrow (1,3) \rightarrow (0,2) \rightarrow (1,1) \rightarrow (2,0) \rightarrow (3,1) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rightarrow (1,1) \rightarrow (2,2) \rightarrow (3,1) \rightarrow (2,0) \rightarrow (1,1) \rightarrow (0,2) \rightarrow (1,3) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rightarrow (1,1) \rightarrow (2,0) \rightarrow (3,1) \rightarrow (2,2) \rightarrow (1,1) \rightarrow (0,2) \rightarrow (1,3) \rightarrow (2,2) \rightarrow (3,3)$
*$(0,0) \rightarrow (1,1) \rightarrow (2,0) \rightarrow (3,1) \rightarrow (2,2) \rightarrow (1,3) \rightarrow (0,2) \rightarrow (1,1) \rightarrow (2,2) \rightarrow (3,3)$
I hope I didn't mess that up...
| {
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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 go from here though.
| Observe that
$$ \left( \left( \tfrac{n}{2} \right)^2 + n + 1 \right) = \left( \frac{n^2 + 4n + 4}{4} \right) = \left( \frac{n+2}{2} \right)^2$$
| {
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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 these can be expressed in terms of $\alpha=3+2\sqrt 2$ and $\beta=3-2\sqrt 2$.
More precisely, we verify that
$$ a_n = \frac{1+\sqrt 2}{4}\alpha^n+\frac{1-\sqrt 2}4\beta^n-\frac12$$
$$ c_n = \frac{2+\sqrt 2}{4}\alpha^n+\frac{2-\sqrt 2}{4}\beta^n$$
fulfill both the initial conditions and the recursion hence describe our sequencs explcitly.
Then (using $\alpha\beta=1$)
$$ a_n^2= \frac{(1+\sqrt 2)^2}{16}\alpha^{2n}+\frac{(1-\sqrt 2)^2}{16}\beta^{2n}+\frac18-\frac{1+\sqrt 2}{4}\alpha^n-\frac{1-\sqrt 2}4\beta^n$$
$$ (a_n+1)^2= \frac{(1+\sqrt 2)^2}{16}\alpha^{2n}+\frac{(1-\sqrt 2)^2}{16}\beta^{2n}+\frac18+\frac{1+\sqrt 2}{4}\alpha^n+\frac{1-\sqrt 2}4\beta^n$$
$$ c_n^2=\frac{(2+\sqrt 2)^2}{16}\alpha^{2n}+\frac{(2-\sqrt 2)^2}{16}\beta^{2n}+\frac14$$
so that by a lot of cancellation and surd simplification $c_n^2=a_n^2+(a_n+1)^2$ for all $n$.
The first few solutins thus found are $0^1+1^2=1^2$, $3^2+4^2=5^2$, $20^2+2^ 2=29^2$, $119^2+120^2=169^2$, $696^2+697^2=985^2$, $4059^2+4060^2=5741^2$, and can be continued indefinitely.
As a matter of fact, the list thus obtained is complete, i.e. there are no other solutions. But to show that would involve more than the above straightforward veififcations.
| {
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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{(\cos\ x)(\sin\ x)(\cos\ x)}{(\sin\ x)(\cos\ x)} \\
&= \frac{(\cos^2 x)(\sin\ x)}{(\sin\ x)(\cos\ x)} \\
&= \cos\ x
\end{align*}
Thank you!
| 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?
| {
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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 monotonously to $\dfrac 1t$. Therefore, Dini's theorem guarantees uniform convergence on compact intervals, and it is not hard to extend this towards infinity, as all functions tend to zero there. However, near zero all the functions have a singularity, so it is unclear to me what happens to the difference at/near $0$.
| Let us write
$$d_n(t) = \frac{1}{t} - f_n(t).$$
Then we find
$$\begin{align}
f_{n+1}(t) &= \frac{3}{4t} + \frac{t^3}{4}f_n(t)^4\\
&= \frac{3}{4t} + \frac{t^3}{4}\left(\frac{1}{t} - d_n(t)\right)^4\\
&= \frac{3}{4t} + \frac{t^3}{4}\left(\frac{1}{t^4} - \frac{4}{t^3}d_n(t) + \frac{6}{t^2}d_n(t)^2 - \frac{4}{t}d_n(t)^3 + d_n(t)^4\right)\\
&= \frac{1}{t} - d_n(t) + \frac{3}{2}t\, d_n(t)^2 - t^2d_n(t)^3 + \frac{t^3}{4} d_n(t)^4,
\end{align}$$
so
$$d_{n+1}(t) = d_n(t) - \frac{3}{2}t\,d_n(t)^2 + t^2d_n(t)^3 - \frac{t^3}{4} d_n(t)^4.\tag{1}$$
Starting with
$$\begin{align}
d_0(t) &= \frac{1}{t} - \frac{1}{e^t-1}\\
&= \frac{1}{t} - \frac{1}{t\left(1 + \frac{t}{2} + \frac{t^2}{6} + O(t^3)\right)}\\
&= \frac{1}{t} - \frac{1}{t}\left(1 - \left(\frac{t}{2} + \frac{t^2}{6} + O(t^3)\right) + \left(\frac{t}{2}+\frac{t^2}{6}+O(t^3)\right)^2 + O(t^3)\right)\\
&= \frac{1}{t} - \frac{1}{t}\left(1 - \frac{t}{2} + \frac{t^2}{12} + O(t^3)\right)\\
&= \frac{1}{2} - \frac{t}{12} + O(t^2),
\end{align}$$
which is an entire meromorphic function with poles in $2\pi i k,\; k \in \mathbb{Z}\setminus\{0\}$ and nowhere else, the recursion $(1)$ shows that all $d_n$ are entire meromorphic functions with poles in $2\pi ik,\; k \in\mathbb{Z}\setminus\{0\}$ and nowhere else, and we have
$$d_{n+1}(0) = d_n(0) = d_0(0) = \frac{1}{2}$$
for all $n$. Thus
$$\sup_{t > 0} \left(\frac{1}{t} - f_n(t)\right) \geqslant \lim_{t\searrow 0} d_n(t) = \frac{1}{2},$$
and we see that the convergence is not uniform on any interval $(0,\varepsilon)$.
| {
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} |
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 follows that $abc$ is divisible by $2^{3+4+4}=2^{11}$.
But the highest power of $2$ that divides $abc$ is $2^{11}$.
So the highest power of $2$ that divides $b$ is $2^4$, and the same is true of $c$.
So the powers are $2^3$, $2^4$, $2^4$, which means that the exponent of $2$ in the LCM is $\max(3,4,4)$, which is $4$.
Continue, reasoning your way with $3$, with $5$, with $7$.
| {
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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 $x^3 - x^2 - x +1 = 0$?
| 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:
$$
\frac{T(z) - t(0) - t(1) z - t(2) z^2 - t(3) z^3}{z^4}
= \frac{T(z) - t(0) - t(z) z - t(2) z^2}{z^3}
+ \frac{T(z) - t(0)}{z} + T(z) \\
T(z) = - \frac{2 + z}{5 (1 + z^2)} + \frac{2 + 4 z}{5 (1 - z - z^2)}
$$
The first term is just:
$$
\begin{cases}
- \frac{2}{5} & \text{\(n\) is even} \\
- \frac{1}{5} & \text{\(n\) is odd}
\end{cases}
$$
The second term gives rise to Fibonacci numbers, which have generating function:
$$
F(z) = \sum_{n \ge 0} F_n z^n = \frac{z}{1 - z - z^2} \\
\frac{F(z) - F_0}{z} = \sum_{n \ge 0} F_{n + 1} z^n = \frac{1}{1 - z - z^2}
$$
Pulling all together:
$$
t(n) =
\begin{cases}
- \frac{2}{5} + \frac{2}{5} F_{n + 1} + \frac{4}{5} F_n & \text{\(n\) is even} \\
- \frac{1}{5} + \frac{2}{5} F_{n + 1} + \frac{4}{5} F_n & \text{\(n\) is odd}
\end{cases}
$$
| {
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How to prove $\frac{(a_1 a_2\cdots a_n)^2-1}{8}\equiv\sum_{i=1}^n\frac{a^2_i -1}{8}\pmod 8$ Let $a_1,a_2,\cdots,a_n$ be odd numbers, show that
$$\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}\pmod 8$$
$n=2$
$$\dfrac{a^2_{1}a^2_{2}-1}{8}\equiv\dfrac{(2k-1)^2(2m-1)^2-1}{8}\equiv\frac{4k^2-4k+4m^2-4m}{8}=\dfrac{a^2_{1}+a^2_{2}-2}{8}\pmod 8?$$
where $a_{1}=2k-1,a_{2}=2m-1$
because
$$(2k-1)(2m-1)^2-1=(4km-2k-2m+1)^2-1=(16k^2m^2-16k^2m-16km^2+8km+8km)+4k^2+4m^2-4k-4m+1-1$$
But in general I can't prove it.
| We prove the congruence by induction on the number of odd factors.
For $n = 1$, the congruence is even an equality, so the base case is settled.
Now assume that $n > 1$ and, as the induction hypothesis, that the congruence
$$\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}{8}\right) \pmod{8}$$
holds for products of $n-1$ odd integers $a_k$.
The induction step then goes:
\begin{align}
\frac{\left(\prod_{k=1}^n a_k\right)^2-1}{8} - \frac{a_n^2-1}{8}
&= \frac{\left(\prod_{k=1}^n a_k\right)^2 - a_n^2}{8}\\
&= a_n^2\frac{\left(\prod_{k=1}^{n-1} a_k\right)^2-1}{8}\\
&\equiv \left(a_n^2\sum_{k=1}^{n-1} \frac{a_k^2-1}{8}\right) \pmod{8}\tag{IH}\\
&\equiv \left(\sum_{k=1}^{n-1} \frac{a_k^2-1}{8}\right) \pmod{8}, \tag{S}
\end{align}
where IH denotes the induction hypothesis, and S the congruence $a_n^2\equiv 1 \pmod{8}$ holding for all odd integers. Rearranging yields the congruence in the desired form
$$\frac{\left(\prod_{k=1}^n a_k\right)^2-1}{8} \equiv \left(\sum_{k=1}^n \frac{a_k^2-1}{8}\right) \pmod{8}.$$
| {
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Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ It is known that
\begin{align}
\arcsin x + \arcsin y =\begin{cases}
\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 \le 1 &\text{or} &(x^2+y^2 > 1 &\text{and} &xy< 0);\\
\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &0< x,y \le 1;\\
-\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &-1\le x,y < 0.
\end{cases}
\end{align}
I tried to prove this myself, have no problem in getting the 'crux' $\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2})$ part of the RHS, but face trouble in checking the range of that 'crux' under the given conditions.
| Let $a=\sin^{-1}x,$ $b=\sin^{-1}y\implies\sin a=x,$ $\sin b=y$ and $a,b\in[-\pi/2,\pi/2]\implies a+b\in[-\pi,\pi]$
$$
\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(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
Case 1: $-\dfrac{\pi}{2}\leq\sin^{-1}x+\sin^{-1}y\leq\dfrac{\pi}{2}$
$$
\color{darkblue}{\sin^{-1}x+\sin^{-1}y=\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
$$
\cos(a+b)\geq0\implies \cos a\cos b-\sin a\sin b\geq0\implies \cos a\cos b\geq\sin a\sin b\\
\sqrt{1-x^2}\sqrt{1-y^2}\geq xy\implies\\
a)\ 1-x^2-y^2+x^2y^2-x^2y^2\geq0 \implies x^2+y^2\leq1\\
b)\ 1-x^2-y^2+x^2y^2-x^2y^2<0 \implies x^2+y^2>1, xy<0
$$
Case 2 & 3: $\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y\leq\pi$ and $-\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2}$
$$
\cos(a+b)<0\implies x^2+y^2>1
$$
Case 2-: $\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y\leq\pi$
$$
\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)=\pi-(\sin^{-1}x+\sin^{-1}y)\\
\implies \color{darkblue}{\sin^{-1}x+\sin^{-1}y=\pi-\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
$$
\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y\leq\pi \;\&\;-\dfrac{\pi}{2}\leq\sin^{-1}x,\;\sin^{-1}y\leq\dfrac{\pi}{2}\\
\implies 0<\sin^{-1}x,\sin^{-1}y\leq\dfrac{\pi}{2}\implies 0< x,y\leq1
$$
Case 3-: $-\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2}$
$$
\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)=-\pi-(\sin^{-1}x+\sin^{-1}y)\\
\implies \color{darkblue}{\sin^{-1}x+\sin^{-1}y=-\pi-\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}
$$
$$
-\pi\leq\sin^{-1}x+\sin^{-1}y<-\dfrac{\pi}{2} \;\&\;-\dfrac{\pi}{2}\leq\sin^{-1}x,\;\sin^{-1}y\leq\dfrac{\pi}{2}\\
\implies -\dfrac{\pi}{2}\leq\sin^{-1}x,\sin^{-1}y< 0\implies -1\leq x,y< 0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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How can I find all integers $x≠3$ such that $x−3|x^3−3$ How can I find all integers $x≠3$ such that $x−3|x^3−3$?
I tried expand $x^3−3$ as a sum but I couldn't find a way after that.
| $$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.$$
Also,
\begin{eqnarray}
a_n&=&\frac{(-1)^n}{2}(2-n)+\frac{3n}{2},\\
a_n&=&\frac{(-1)^n}{2}\cdot 2-\frac{(-1)^n}{2}\cdot n+\frac{3n}{2},\\
a_n&=&\frac{(-1)^n}{2}\cdot 2+\frac{3-(-1)^n}{2}\cdot n,\\
a_n&=&(-1)^n+\frac{3-(-1)^n}{2}\cdot n,
\end{eqnarray}
from where we get $a_{2k-1}=-1+2\cdot(2k-1)$ and $a_{2k}=1+2k$.
Therefore,
$$a_1+a_3\dots+a_{2n-1}=(-1+2\cdot 1)+(-1+2\cdot 3)+\dots+(-1+2\cdot (2n-1)),$$
$$a_1+a_3\dots+a_{2n-1}=n\cdot (-1)+2\cdot (1+3+\dots+2n-1)=-n+2 A_{2n-1}$$
and similarly
$$a_2+a_4\dots+a_{2n}=(1+2)+(1+4)+\dots+(1+2n),$$
$$a_2+a_4\dots+a_{2n}=n\cdot 1+(2+4+\dots+2n)=n+A_{2n}$$
Finally,
$$S_{2n}=(a_1+a_3\dots+a_{2n-1})+(a_2+a_4+\dots+a_{2n}),$$
$$S_{2n}=(n\cdot(-1)+2\cdot A_{2n-1})+(n\cdot 1+A_{2n}),$$
$$S_{2n}=(-n+2\cdot n^2)+(2n+n^2)=3n^2+n$$
Since $S_{2n+1}=S_{2n}+a_{2n+1}$, $$S_{2n+1}=3n^2+n+4n+1=3n^2+5n+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.
$$
For ease of notation, let $s_n = \frac{1}{n} (z_1 + z_2 + \ldots + z_n)$.
Attempt:
*
*Let $n \in \mathbb{N}$ be arbitrary and consider that
$$
\left| A - s_n \right| = \left|A - \frac{1}{n} (z_1 + \ldots + z_n) \right| = \left|A - \frac{z_1}{n} - \ldots - \frac{z_n}{n} \right|
$$
so that through repeated applications of the triangle inequality we have that
$$
\left| A - s_n \right| \le \left| A - \frac{z_n}{n} \right| + \left| - \frac{z_{n-1}}{n} - \ldots - \frac{z_{1}}{n} \right| \le \left| A - \frac{z_n}{n} \right| + \left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right|
$$
*Now as $n \rightarrow \infty$, we have that the numerator of the term $\left| -\frac{z_{n-1}}{n} \right|$ approaches $-A$ while the denominators of all of the terms in the sum $\left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right|$ approach infinity. \uline{[Gap]}. Then as $n \rightarrow \infty$, we have that $\left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right|$ approaches $0$.
*On the other hand, we have also that $z_n \rightarrow A$ (by hypothesis) so that the term $\left| A - \frac{z_n}{n} \right|$ can get as close to $\left| A - \frac{A}{n} \right|$ as we'd like. Yet since $\frac{A}{n} \rightarrow 0$, we have that $\left| A - \frac{z_n}{n} \right| \rightarrow \left| A - 0 \right| = \left| A \right|$.
*Then since
$$
\left( \left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right| \right) \rightarrow 0
$$
and
$$
\left| A - \frac{z_n}{n} \right| \rightarrow \left| A \right|
$$
we have that
$$
|A - s_n| \le \left| A - \frac{z_n}{n} \right| + \left( \left| -\frac{z_{n-1}}{n} \right| + \ldots + \left|- \frac{z_{1}}{n} \right| \right) \rightarrow |A| + 0 = |A|.
$$
Question: My argument doesn't quite work since I have shown only that $|A - s_n| \rightarrow |A|$ and yet we want $|A - s_n| \rightarrow |0|$. Is there a way to keep most of my argument in place and yet actually to prove the desired statement?
| I think the problem with your strategy appears in your first step, when you are trying to approximate $A - \frac{1}{n}z_1 - \cdots - \frac{1}{n}z_n$. The problem is that you really aren't ever considering the difference between $A$ and $z_n$ when you are coming up with your bounds; remember, the only information you have is that $z_n\to A$, so you have to use that the difference between $A$ and $z_n$ goes to $0$. I think it would be more fruitful to instead first rewrite as $$
A - \frac{z_1}{n} - \cdots - \frac{z_n}{n} = \frac{(A-z_1)}{n} + \cdots + \frac{(A-z_n)}{n},$$ and then apply the triangle inequality. Maybe you can finish the argument from there?
| {
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Incongruent solutions to $7x \equiv 3$ (mod $15$) I'm supposed to find all the incongruent solutions to the congruency $7x \equiv 3$ (mod $15$)
\begin{align*}
7x &\equiv 3 \mod{15} \\
7x - 3 &= 15k \hspace{1in} (k \in \mathbb{Z}) \\
7x &= 15k+3\\
x &= \dfrac{15k+3}{7}\\
\end{align*}
Since $x$ must be an integer, we must find a pattern for $k$ that grants this. We know that $\frac{k+3}{7}$ must be equal to some integer, say $m$. Solving for $k$, we have $k=4+7m$.
Substituting this into our value for $x$, we get:
\begin{align*}
x & = \dfrac{15(4+7m) + 3}{7}.\\
&= \dfrac{63}{7} + \frac{105m}{7}.\\
&= 9+15m.
\end{align*}
So, $x = 9+15m, m\in \mathbb{Z}.$
So, is this what I was looking for? I'm not exactly sure what is meant by incongruent solutions.
| $$x = 9+15m, m\in \mathbb{Z}.$$
is the same as
$$x \equiv 9 \pmod{15} \,.$$
Equations of the type $ax \equiv b \pmod{n}$ sometimes have no solution, sometimes have one solution $\pmod{n}$ and sometimes they have two or more (incongruent solutions) $\pmod{n}$.
In this case, there is only one, the one you found. A much easier way to find it is to multiply your equation by $2$ and use
$$14x \equiv -x \pmod{15}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/675867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this simple trignometric problem? So this is the question that was given in a textbook and i attempted to win from the book which was saying i was wrong?
If $$\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}$$
so i attempted to solve it this way$$\implies\frac{\tan^2\theta + 1}{\tan^2\theta - 1} = \frac{\frac{\sin^2\theta}{\cos^2\theta} + 1}{\frac{\sin^2\theta}{\cos^2\theta} - 1}$$
$$=\frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta - \cos^2\theta}$$$$=\frac{1}{\sin^2\theta - \cos^2\theta}$$$$=\frac{1}{(\sin\theta + \cos\theta)(\sin\theta - \cos\theta)}$$substituting the values from$$=\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{5}{4}$$we get$$=\frac{1}{20}$$ but this is wrong answer so where am i wrong and what is the mistake
Note = I am not looking for the solution, i want to know what is wrong with this approach and why
Thanks
Akash
| 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
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 correct. So, I was hoping maybe someone can help me out with this problem. Thank you so much!
| 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 solution, as seen by substitution: $0^2 \overset{\text{?}}{=} -5\cdot 0$.
When $x$ is between $-5$ and $0$, then $x^2+2$ is smaller than $-5x+2$, as seen by plugging in any number between $-5$ and $0$. So
$$
\begin{align}
& \phantom{{}=} \int_{-5}^0 (\text{bigger function minus smaller function}) \\[8pt]
& = \int_{-5}^0 (-5x+2)-(x^2+2) \,dx = \int_{-5}^0 -5x -x^2 \,dx \\[8pt]
& = \left.\frac{-5x^2}{2} -\frac{x^3}{3}\right|_{x=-5}^{x=0} = \frac{125}{2} -\frac{125}{3} = \frac{125}{6}.
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
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 infinity)
(A(X)-a0-a1X)-7X∈a(n-1)X^N-1+10x^2∈an-2x^n-2=0
I want to know and from where these are substituted?
I find it really hard to type the maths equations.So im attaching a picture.Please help
| 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 $\frac{F(x) - a_0}{x} = a_1 + a_2x + a_3x^2+\ldots =\sum_{n=0}^{\infty} a_{n+1}x^n$
$\frac{F(x)-a_0-a_1}{x^2}=a_2+a_3x+a_4x^2+\ldots=\sum_{n=0}^{\infty}a_{n+2}x^n$
We have:
$$
\sum_{n=0}^{\infty}a_{n+2}x^n= 7 \sum_{n=0}^{\infty} a_{n+1}x^n-10\sum_{n=0}^{\infty} a_nx^n
$$
Therefore:
$$
\frac{F(x)-a_0-a_1}{x^2}=7\left(\frac{F(x) - a_0}{x}\right)-10F(x) \longleftrightarrow
$$
$$
F(x)=\frac{(1-7x)a_0+a_1}{1-7x+10x^2}
$$
Using partial fractions, one obtains:
$$
F(x)=\frac{(2 a_0-5 a_1)}{3 (5 x-1)}+\frac{(2 a_1-5a_0)}{3 (2 x-1)}
$$
Expand the two terms into power series:
$$
F(x) = \frac{5a_1-2a_0}{3}\sum_{n=0}^{\infty} 5^nx^n + \frac{5a_0-2a_1}{3}\sum_{n=0}^{\infty} 2^nx^n
$$
So,
$$
a_{n+1} = \left(\frac{5a_1-2a_0}{3}5^n+\frac{5a_0-2a_1}{3}2^n\right)
$$
For example, in our case of $a_0 = 0, a_1=1$,
$$
a_{n+1} = \left(\frac{5}{3}5^n-\frac{2}{3}2^n\right)=\frac{5^{n+1}-2^{n+1}}{3}
$$
This is correct, as $a_2 = 7$, $a_3 = 7\cdot 7 - 10 = 39$. Verifying our formula in terms of $n$, we have:
$$
a_2 = \frac{5^2-2^2}{3} = \frac{21}{3}, a_3 = \frac{5^3-2^3}{3}=39, \text{etc..}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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
$$\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^2}=\int_{0}^{1}\dfrac{\arctan{x}}{x}dx=C$$
where the $C$ is Catalan constant
But for this$$\sum_{k=0}^{\infty}\dfrac{(-1)^{\frac{k(k+1)}{2}}}{(2k+1)^2}=\dfrac{\sqrt{2}\pi^2}{16}$$I can't,Thank you for you help
| 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
$$\sum_{k=0}^{\infty} \left [\frac1{(8 k+1)^2} - \frac1{(8 k+3)^2} - \frac1{(8 k+5)^2}+\frac1{(8 k+7)^2} \right ]$$
Note that this sum is symmetric in sign about $k$; thus we may write the sum as
$$\frac12 \sum_{k=-\infty}^{\infty} \left [\frac1{(8 k+1)^2} - \frac1{(8 k+3)^2} - \frac1{(8 k+5)^2}+\frac1{(8 k+7)^2} \right ]$$
We may thus evaluate this sum using the residue theorem. Omitting details, we use the result
$$\sum_{k=-\infty}^{\infty} f(k) = -\sum_n \operatorname*{Res}_{z=z_n} \pi \, \cot{(\pi z)} \, f(z)$$
where
$$f(z) = \frac12 \left [\frac1{(8 z+1)^2} - \frac1{(8 z+3)^2} - \frac1{(8 z+5)^2}+\frac1{(8 z+7)^2} \right ]$$
$f$ has (double) poles at $z_1=-1/8$, $z_2=-3/8$, $z_3=-5/8$, and $z_4=-7/8$. Thus, for instance,
$$\begin{align}\operatorname*{Res}_{z=-1/8} \pi \, \cot{(\pi z)} \, f(z) &= \frac12\pi \frac1{8^2}\left [\frac{d}{dz} \left ( \left [1- \frac{(8 z+1)^2}{(8 z+3)^2} - \frac{(8 z+1)^2}{(8 z+5)^2}+\frac{(8 z+1)^2}{(8 z+7)^2} \right ] \cot{(\pi z)} \right ) \right ]_{z=-1/8} \\ &= -\frac1{128} \pi^2 \csc^2{\frac{\pi}{8}}\end{align}$$
The factor of $1/8^2$ comes from the fact that the residue calculation has us multiply $f$ by $(z+1/8)^2$. Similarly,
$$\operatorname*{Res}_{z=-3/8} \pi \, \cot{(\pi z)} \, f(z) = \frac1{128} \pi^2 \csc^2{\frac{3 \pi}{8}}$$
$$\operatorname*{Res}_{z=-5/8} \pi \, \cot{(\pi z)} \, f(z) = \frac1{128} \pi^2 \csc^2{\frac{5 \pi}{8}}$$
$$\operatorname*{Res}_{z=-7/8} \pi \, \cot{(\pi z)} \, f(z) = -\frac1{128} \pi^2 \csc^2{\frac{7 \pi}{8}}$$
Using
$$\sin^2{\frac{\pi}{8}} = \sin^2{\frac{7\pi}{8}} = \frac{2-\sqrt{2}}{4} = \frac{1/2}{2+\sqrt{2}}$$
$$\sin^2{\frac{3\pi}{8}} = \sin^2{\frac{5\pi}{8}} = \frac{2+\sqrt{2}}{4} = \frac{1/2}{2-\sqrt{2}}$$
Then the sum is finally
$$\frac{\pi^2}{128} 2 \left (\frac{2+\sqrt{2}}{1/2} - \frac{2-\sqrt{2}}{1/2} \right ) = \frac{\sqrt{2} \pi^2}{16}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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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 know how to get a or be now.
*You're given $2x^2 + 4x + 2y^2 + 6y = 66$, how would you find the center and the radius? I was thinking maybe complete the square, but for some reason it doesn't seem to be working correctly?
If you could just give me some help on these two, that would be very much appreciated! Thank you so much.
| 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 half-dimensions of the box are $a$ and $b$, and we know that
$$
c^2 = a^2 + b^2.
$$
Now, call the slope of the (positively sloped) asymptote $m$. In this problem, $m = 4$. Then,
$$
\frac{b}{a} = m \qquad \text{so} \qquad b = ma.
$$
Substitute back into the pythagorean equation to obtain:
$$
\begin{align}
c^2 &= a^2 + (ma)^2 \\
&= a^2 + m^2 a^2 \\
&= (1 + m^2) a^2
\end{align}
$$
And so,
$$
a^2 = \frac{c^2}{1 + m^2} \qquad \Longrightarrow \qquad a = \frac{c}{\sqrt{1 + m^2}}
$$
and
$$
b = \frac{mc}{\sqrt{1 + m^2}}
$$
You can find the values of $a$ and $b$ for your particular hyperbola.
2. It will be easier to complete the square in $x$ and $y$ if you first make the coefficients of $x^2$ and $y^2$ equal to $1$. This can be accomplished by first dividing by $2$.
$$
\begin{align}
2x^2 + 4x + 2y^2 + 6y &= 66 \\
x^2 + 2x + y^2 + 3y &= 33 \\
x^2 + 2x + 1 + y^2 + 3y + \frac{9}{4} &= 33 + 1 + \frac{9}{4} \\
(x + 1)^2 + \left( y + \frac{3}{2} \right)^2 = \frac{145}{4}
\end{align}
$$
From there, you can read off the center and radius.
| {
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"timestamp": "2023-03-29T00:00:00",
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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 bounded as $x \to \infty$.
All this to show that $x\cdot \mathrm{tailend} = \mathcal O\sqrt{x}$
The purpose is to get the asymptotic formula for the distribution of square free integers.
p.269 Exercise 8 Stewart and Tall.
| 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:
$$S(x) = \sum_{k\ge 1} \frac{1}{k^2} \left(1-\frac{1}{(x/k+1)^2}\right).$$
The sum term is harmonic and may be evaluated by inverting its Mellin transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \frac{1}{k^2}, \quad \mu_k = \frac{1}{k} \quad \text{and} \quad
g(x) = 1-\frac{1}{(x+1)^2}.$$
We need the Mellin transform $h^*(s)$ of $h(x)=g(x)-1$ which is
$$\int_0^\infty -\frac{1}{(x+1)^2} x^{s-1} dx
= \left[\frac{1}{1+x} x^{s-1} \right]_0^\infty
- (s-1) \int_0^\infty \frac{1}{1+x} x^{s-2} dx.$$
The fundamental strips of these three components are $\langle 0, 2\rangle$, followed by $\langle 1,2 \rangle$ and finally, $\langle 1,2\rangle$.
The contribution in the square brackets disappears in the fundamental strip $\langle 1, 2\rangle.$
Now this last Mellin transform can be evaluated using a keyhole contour with the slot on the positive real axis, which gives
$$(1-e^{2\pi i (s-1)})\times
\int_0^\infty \frac{1}{1+x} x^{s-1} dx
= 2\pi i \times \mathrm{Res}\left(\frac{1}{1+x} x^{s-1}; x= -1\right)$$
which yields in turn
$$\int_0^\infty \frac{1}{1+x} x^{s-1} dx = 2\pi i
\frac{ e^{\pi i (s-1)}}{1-e^{2\pi i s}}
= - 2\pi i \frac{1}{e^{-\pi i s}-e^{\pi i s}}
= \frac{\pi}{\sin(\pi s)}.$$
It follows that the Mellin transform of $g(x)$ is
$$g^*(s) = - (s-1) \frac{\pi}{\sin(\pi (s-1))} = (s-1)\frac{\pi}{\sin(\pi s)}$$
with fundamental strip $\langle -1, 0\rangle$ as can be seen by expanding $g(x)$ about zero and about infinity and testing convergence of the integral.
Therefore the transform $Q(s)$ of $S(x)$ is
$$ Q(s) = (s-1)\frac{\pi}{\sin(\pi s)} \zeta(2-s)
\quad\text{because}\quad \sum_{k\ge 1}
\frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1}\frac{1}{k^2} k^s = \zeta(2-s).$$
The half plane of convergence of the zeta term is $\Re(s)<1.$
We thus obtain the Mellin inversion integral
$$ S(x) = \frac{1}{2\pi i} \int_{-1/2-i\infty}^{-1/2+i\infty} Q(s)/x^s ds$$
which we evaluate by shifting it to the right for an expansion about infinity.
We get $$\mathrm{Res}(Q(s)/x^s; s = 0) = -\frac{\pi^2}{6}$$
and
$$\mathrm{Res}(Q(s)/x^s; s = 1) = \frac{1}{x}$$
and
$$\mathrm{Res}(Q(s)/x^s; s = 2) = -\frac{1}{2x^2}.$$
For the remaining poles at $s=q$ where $q>2$ we obtain
$$\mathrm{Res}(Q(s)/x^s; s = q) = (q-1) (-1)^q \zeta(2-q)\frac{1}{x^q}
\\= (-1)^q (q-1) \zeta(-(q-2))\frac{1}{x^q}
= (-1)^{q-1} (q-1)\frac{B_{q-1}}{q-1} \frac{1}{x^q}
= (-1)^{q-1} B_{q-1}\frac{1}{x^q} .$$
Now this only contributes when $q$ is odd
so that we may simplify it to $B_{q-1}\frac{1}{x^q} .$
(Note that those zero values from the Bernoulli numbers at odd indices cancel the poles from the sine term, so it is in fact correct to let the summation for the expansion range over all residues at poles at $q>2.$)
This gives for the asymptotic expansion
$$S(x) \sim \frac{\pi^2}{6} - \frac{1}{x} + \frac{1}{2x^2}
- \sum_{q>2} \frac{B_{q-1}}{x^q}.$$
We finally have
$$\frac{\pi^2}{6} - S(N) \sim
\frac{1}{N} - \frac{1}{2N^2} + \sum_{q>2} \frac{B_{q-1}}{N^q}$$
as discovered by Ron Gordon in his excellent answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find permutation that solves $\;\tau \circ X = \sigma$ I need to find a permutation $X$ that solves $\;\tau \circ X = \sigma,\;$ given
$$\tau = \begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 4 & 5 & 2 & 1
\end{bmatrix} = (1,3,5)(2,4)$$
$$\sigma = \begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 2 & 1 & 5 & 4
\end{bmatrix} = (1,3)(2)(4,5)$$
Is there a trick on how to solve for $X$?
| 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
$$\sigma = \begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 2 & 1 & 5 & 4
\end{bmatrix} = (1,3)(2)(4,5)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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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,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=\dfrac12[(x-y)^2+(y-z)^2+(z-x)^2]\cdot\dfrac12[(a-b)^2+(b-c)^2+(c-a)^2]$
Now, I got into a mess. I have got two products with sum of three squares which I cannot manage nor can I show this to be a perfect square. Please help.
| 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 $$(z^2-xy)^3-(x^2-yz)(y^2-zx)(z^2-xy)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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",
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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^{-r^2}}{r}\rho^2\sin\phi dr d\phi d \theta \\
=& 2\pi \Biggl[-\cos\phi\Biggr]_0^{\pi} \int_0^2 re^{-r^2} dr \\
=& 2\pi (\underbrace{-\cos\pi}_{=1} +\underbrace{\cos0}_{=1}) \int_0^2 re^{-r^2} dr \\
=& 4\pi \Biggl[\frac{-e^{-r^2}}{2}\Biggr]_0^2 \\
=& 2\pi \left( -e^{-2^2}+1\right) \\
=& 2\pi -2\pi e^{-4}
\end{align*}
| 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 $t=r^2$
| {
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} |
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 hard to compute values. I am sure I am missing something obvious here.
| 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 $$\begin{bmatrix}3 \strut& -2\\1 & -1\end{bmatrix}-\begin{bmatrix}1+\sqrt{2} & 0\\0 & 1+\sqrt{2}\end{bmatrix}=\begin{bmatrix}2-\sqrt{2} & -2\\1 & -2-\sqrt{2}\end{bmatrix}$$
We row-reduce to simplify the matrix while keeping its null space the same:
$$\begin{bmatrix}2-\sqrt{2} & -2\\1 & -2-\sqrt{2}\end{bmatrix}\xrightarrow{\;\;\Large\mathsf{\text{row 2} \;-\; \left(\frac{1}{2-\sqrt{2}}\right)\,\text{row 1}}\;\;}\begin{bmatrix}2-\sqrt{2}\strut & -2\\0 & 0\end{bmatrix}$$
Obviously, for a non-zero vector $\Bigl[\begin{smallmatrix}x_1\\x_2\end{smallmatrix}\Bigr]$, we have that
$$\begin{align*}
\begin{bmatrix}2-\sqrt{2} & -2\\0 & 0\end{bmatrix}\begin{bmatrix}x_1\strut\\x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}&\iff (2-\sqrt{2})x_1+(-2)x_2=0\\\\\\
&\iff x_2=\frac{2-\sqrt{2}}{2}x_1
\end{align*}$$
and an easy choice of $x_1$ and $x_2$ that satisfy this is $x_1=2$ and $x_2=2-\sqrt{2}$.
Let's check that this worked:
$$\begin{bmatrix}3 \strut& -2\\1 & -1\end{bmatrix}\begin{bmatrix}2\strut\\2-\sqrt{2}\end{bmatrix}=\begin{bmatrix}3\cdot 2+(-2)\cdot(2-\sqrt{2})\strut\\1\cdot 2+(-1)\cdot(2-\sqrt{2})\end{bmatrix}=\begin{bmatrix}2+2\sqrt{2}\\2\sqrt{2}\end{bmatrix}=(1+\sqrt{2})\begin{bmatrix}2\strut\\2-\sqrt{2}\end{bmatrix}$$
$$\checkmark$$
I leave it to you to do this process for the eigenvalue $1-\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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
$$
A
=
\begin{pmatrix}
-2 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\quad
B=
\begin{pmatrix}
3 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\quad
C=
\begin{pmatrix}
-3 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$$
Then $\det (A)=-1$ and $\det( B)=\det( C)=3$. However,
$$
\det(A+B)=8
$$
while
$$
\det(A+C)=0
$$
Your problem is not solvable!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/697259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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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 the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$ and $\sin 45^{\circ} = \frac{\sqrt{2}}{2}$,
$$\begin{align*}\sin(x - 30^{\circ}) &= \sin 45^{\circ} \\ x - 30^{\circ} &= 45^{\circ} \\ x &= 75^{\circ}\end{align*}.$$
Also, we know that $\sin 135^{\circ} = \frac{\sqrt{2}}{2}$. Then,
$$\begin{align*}\sin(x - 30^{\circ}) &= \sin 135^{\circ} \\ x - 30^{\circ} &= 135^{\circ} \\ x &= 165^{\circ}\end{align*}.$$
Therefore, $x = 30^{\circ}$ and $x = 135^{\circ}$.
| {
"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 this problem. I haven't really ever had any experience with such equations.
The solution:
Let $t=\frac{x-3}{x+1}$. Then $$f(t)+f\left(\frac{t-3}{t+1}\right)=\frac{3+t}{1-t}$$
Now let $t=\frac{3+x}{1-x}$. Then $$f\left(\frac{3+t}{1-t}\right)+f(t)=\frac{t-3}{t+1}$$
Add both equalities: $$\frac{8t}{1-t^2}=2f(t)+f\left(\frac{t-3}{t+1}\right)+f\left(\frac{3+t}{1-t}\right)=2f(t)+t$$
Hence the answer is $$f(x)=\frac{4x}{1-x^2}-\frac{x}{2}$$
This is unclear to me. For instance, how come we can assign a different value to the same variable? Does anyone understand this? I'd appreciate any help.
| 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 $t_0\in\mathbb R\setminus\{1\}$.
Similarly if $x\ne1$ and $t_1\ne-1$, then $t_1=\frac{3+x}{1-x}\Longleftrightarrow(1-x)t_1=3+x\Longleftrightarrow t_1-3=x(1+t_1)\Longleftrightarrow x=\frac{t_1-3}{t_1+1}$
So the second equality holds for all $t_1\in\mathbb R\setminus\{-1\}$.
Now for any $x\in\mathbb R$, $|x|\ne1$ we know that $$f(x)+f\left(\frac{x-3}{x+1}\right)=\frac{3+x}{1-x}$$ by the first equation, substituting $x$ for $t_0$ and $$f\left(\frac{3+x}{1-x}\right)+f(x)=\frac{x-3}{x+1}$$ by the second equation, substituting $x$ for $t_1$.
Now add these to get the solution.
| {
"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
} |
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 example and come to a valid conclusion based on such example.
My work:
Come up with a square matrix A and B let both be a 2 by 2 matrix(rows and cols must be same).
Matrix $A$:
$A = \begin{array}{ccc}
3 & 5 \\
4 & 6 \\
\end{array} $
Matrix $B$:
$B = \begin{array}{ccc}
5 & 8 \\
9 & 4 \\
\end{array} $
$A + B = \begin{array}{ccc}
8 & 13 \\
13 & 10 \\
\end{array}$
$(A + B)^2 = \begin{array}{ccc}
233 & 234 \\
234 & 264 \\
\end{array}$
$A^2 = \begin{array}{ccc}
29 & 45 \\
36 & 56 \\
\end{array}$
$(AB) = \begin{array}{ccc}
60 & 44 \\
74 & 56 \\
\end{array}$
$2(AB) = \begin{array}{ccc}
120 & 88 \\
234 & 112 \\
\end{array}$
$B^2 = \begin{array}{ccc}
97 & 72 \\
81 & 88 \\
\end{array}$
$A^2 + 2AB + B^2 = \begin{array}{ccc}
246 & 205 \\
265 & 256 \\
\end{array}$
Based my above work, the answer is false.
Is there another way to approach the problem? It seems like a lot of work needed to be done for a true/false question which raised my suspicion about whether there is a better way to look at the problem.
| $$\begin{align*}
(A+B)^2 &= (A+B)(A+B) \\
&= AA+AB+BA+BB \\
&= A^2 + AB+BA+B^2.\end{align*}$$
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",
"answer_count": 2,
"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
$$
\left(\frac{6\alpha}{\alpha^3+1},\frac{6\alpha^2}{\alpha^3+1}\right)\tag{2}
$$
Since $x+y=\dfrac{6\alpha}{\alpha^2-\alpha+1}$, we have that
$$
-2\lt x+y\le6\tag{3}
$$
Note that $x+y=-2$ only happens when $\alpha=-1$, and that doesn't give a finite $(x,y)$ in $(2)$.
Thus, cubing $(3)$ yields
$$
-8\le x^3+3x^2y+3xy^2+y^3\le216\tag{4}
$$
and applying $x^3+y^3=6xy$,
$$
-\frac83\le(x+y+2)xy\le72\tag{5}
$$
Since $x+y+2\ge1$, inequality $(5)$ leaves only a finite number of $x,y\in\mathbb{Z}$ to check.
| {
"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 $.
(3) $[ \sqrt{x} ]^2 \le [x] \le x < ([\sqrt{x}] + 1)^2$
??
| 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 $n= \left \lfloor \sqrt a \right \rfloor = \left \lfloor \sqrt {a+ \alpha} \right \rfloor$. $\left \lfloor \sqrt {a+ \alpha} \right \rfloor \leq \left \lfloor \sqrt {(n+1)^2 -1 + \alpha} \right \rfloor <\left \lfloor \sqrt {(n+1)^2} \right \rfloor =n+1$. This implies $\left \lfloor \sqrt {a+ \alpha} \right \rfloor=n $ what you needed
| {
"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
$$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\sqrt{(-2y(\frac{1}{2})(1-x^2-y^2)^{-1/2})^2+(-2x(\frac{1}{2})(1-x^2-y^2)^{-1/2})^2+1}dydx.$$
$$=\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\left(\frac{x^2+y^2}{1-x^2-y^2}+1\right)^{\frac{1}{2}}dydx.$$
Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
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 on wolfram alpha but I can't get it to produce the result using definition of derivatives.
| 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:
$$
\begin{align}
\lim\limits_{h \rightarrow 0} \frac{(x+h)(1-(x+h))^{-1} - x(1-x)^{-1}}{h}&= \lim\limits_{h \rightarrow 0} \frac{1}{h} \frac{(x+h)(1-x)-x(1-x-h)}{(1-x-h)(1-x)}\\
&= \lim\limits_{h \rightarrow 0}\frac{1}{h}\frac{-x^2 +x - hx +h -x +x^2+hx}{1 - x -x + x^2 - h + hx}\\
&= \lim\limits_{h \rightarrow 0}\frac{1}{h}\frac{h}{x^2 + (h-2)x + 1-h }
\end{align}
$$
Now canceling the $\frac{1}{h}$ we get our answer :
$$\begin{align}
\lim\limits_{h \rightarrow 0}\frac{1}{h}\frac{h}{x^2 + (h-2)x + 1-h }
&= \lim\limits_{h \rightarrow 0}\frac{1}{x^2 + (h-2)x + 1-h }\\
\end{align}
$$
And now evaluating our limit we finally get
$$\begin{align}
\lim\limits_{h \rightarrow 0}\frac{1}{x^2 + (h-2)x + 1-h } &= \frac{1}{x^2 -2x + 1}\\
&= \frac{1}{(x-1)^2} \blacksquare
\end{align}
$$
| {
"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 tricks ?
| $p(x) = 8 + 44i + x^2(-(1+3i) + x^2(1+2i + x))$
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 $-2-i$ is a root.
| {
"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 sure is the strategy for this series), but I've been unable to. Any help in solving this would be appreciated (though I'd prefer a hint over a solid answer).
| $\begin{array}{l}
\left( {\frac{4}{{\left( { - 3} \right)^n }} - \frac{3}{{3^n }}} \right) = \frac{{\left( { - 1} \right)^n }}{{3^n }} + \frac{{3\left( { - 1} \right)^n }}{{3^n }} - \frac{3}{{3^n }} \\
= \left\{ \begin{array}{l}
\frac{1}{{3^{2p} }}\quad ;n = 2p \\
\frac{{ - 7}}{{3^{2p + 1} }}\quad ;n = 2p + 1 \\
\end{array} \right. \\
\Rightarrow \sum\limits_{n = 0}^{ + \infty } {\left( {\frac{4}{{\left( { - 3} \right)^n }} - \frac{3}{{3^n }}} \right)} = \left\{ \begin{array}{l}
\sum\limits_{p = 0}^{ + \infty } {3^{ - 2p} } \quad ;n = 2p \\
- 7\sum\limits_{p = 0}^{ + \infty } {3^{ - 2p + 1} } \quad ;n = 2p + 1 \\
\end{array} \right. \\
\end{array}$
| {
"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
} |
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:
$\frac{7}{6}, \frac{8}{7}, \frac{15}{14}, \frac{21}{20}, \frac{28}{27}, \frac{36}{35}, \frac{49}{48}, \frac{50}{49}, \frac{64}{63}, \frac{126}{125}, \frac{225}{224}, \frac{2401}{2400}, \frac{4375}{4374}$
We only care about solutions where the numerator is $2^x 3^y$, and where the denominator is $5^z 7^w$ (for positive $x$, $y$, $z$, $w$). The only fraction which satisfies this condition is $\frac{36}{35}$, or $(x, y, z, w) = (2, 2, 1, 1)$.
| {
"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
} |
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, $f^2 = -Id \implies \det{(f^2)} = (\det{f})^2 = (-1)^n$. So being on a real space, n is even.
2) I thought about rotation for $f$, but I have no ideas to construct it explicitly.
Thanks in advance,
| 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 $$
\begin{pmatrix}
a^2+b c & a b+b d \\
a c+c d & b c+d^2 \\
\end{pmatrix}
=
\begin{pmatrix}
-1 & 0 \\
0 & -1 \\
\end{pmatrix}.
$$
To find an example, then, we need to solve the system of equations
$$
\left\{
\begin{aligned}
a^2+b c &= -1 \\
(a + d)b &= 0 \\
(a + d) c &= 0 \\
b c+d^2 &= -1
\end{aligned}
\right.
$$
If $a+d\neq0$, the two middle equations show that $b=c=0$, and the first equation then says that $a^2=-1$, which is impossible. It follows that $d=-a$. Similarly, we cannot have $b=0$ because of the first equation, and we see that $c=(-1-a^2)/b$. The matrix of $f$ is then $$\begin{pmatrix}a&b\\-\frac{1+a^2}{b}&-a\end{pmatrix}$$ with $b\neq0$. We have obtained all $2$-dimensional examples in this way.
| {
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"url": "https://math.stackexchange.com/questions/713591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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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 $\frac{x^6 - 1}{x-1} = x^5 + x^4 + x^3 + x^2 + x + 1$ and from there the solution is pretty straightforward. However, I was not aware of that factorization. The only ones I have really learned are $x^2 - y^2 = (x-y)(x+y)$ and $x^3 \pm y^3 = (x \pm y)(x^2 \mp xy + y^2)$. Is there any other way I could have solved this factorization without using the ones they used?
| 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",
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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:
$$\int e^{x\ln x}~dx=\int\sum\limits_{n=0}^\infty\frac{(x\ln x)^n}{n!}dx$$
I integrated the first 3 terms (since it was the easiest the work with):
$$x+\frac{1}{4}x^2(2\ln x-1)+\frac{1}{54}x^3(9(\ln x)^2-6\ln x+2)+\cdots$$
Is there any other efficient way of getting the first 6 terms of the series expansion of the integral?
| 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_2a_1^2 + a_1^4}{24} x^4 \\[8pt]
& \phantom{{}={}} + \frac{a_5 + 5a_4 a_1 + 10a_3 a_2 + 10a_3 a_1^2 + 15 a_2^2 a_1 + 10 a_2 a_1^3 + a_1^5}{120} x^5 \\[8pt]
& \phantom{{}={}} + \frac{a_6 + 6a_5 a_1 + 15 a_4 a_2 + 15 a_4 a_1^2 + 10a_3^2 + 60 a_3 a_2 a_1 + 20 a_3 a_1^3 + 15 a_2^3 + 45 a_2 a_1^4+ a_1^6}{720} x^6 \\[8pt]
& \phantom{{}={}} + \cdots\cdots
\end{align}
The patterns in the indices correspond to integer partitions and the coefficients count set partitions. For example, here is how we get the coefficient of $x^6$:
$$
\begin{array}{l|c}
\text{integer partition} & \text{number of set partitions} \\
\hline
6 & 1 \\
5+1 & 6 \\
4+2 & 15 \\
4+1+1 & 15 \\
3+3 & 10 \\
3+2+1 & 60 \\
3+1+1+1 & 20 \\
2+2+2 & 15 \\
2+2+1+1 & 45 \\
2+1+1+1+1 & 15 \\
1+1+1+1+1+1 & 1
\end{array}
$$
| {
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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 calculation):
$$\frac{\sqrt{24}+\sqrt{40}-\sqrt{16}}{\sqrt{12}+\sqrt{5}}$$
which is totally not in accordance with the answer, $\sqrt{2}+\sqrt{3}-\sqrt{5}$.
Can someone please explain this/give hints to me.
| Your "long calculation" was obviously wrong, since the denominator should be
$$(\sqrt 2 + \sqrt 3 + \sqrt 5)(\sqrt 2 - (\sqrt 3 + \sqrt 5))=\\=\sqrt2^2 - (\sqrt 3 + \sqrt 5)^2 = 2 - (3 + 5 + 2\sqrt{15}) = -6-2\sqrt{15}$$
| {
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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, it is better to tackle this integral to this series approach:
Hint:
$\int_0^1\sqrt{9x^4+4x^2+1}~dx$
$=\int_0^\frac{\sqrt{\sqrt{13}-2}}{3}\sqrt{9x^4+4x^2+1}~dx+\int_\frac{\sqrt{\sqrt{13}-2}}{3}^1\sqrt{9x^4+4x^2+1}~dx$ (separation according to the root between $0$ and $1$ of $9x^4+4x^2=1$)
$=\int_0^\frac{\sqrt{\sqrt{13}-2}}{3}\sqrt{1+x^2(9x^2+4)}~dx+\int_\frac{\sqrt{\sqrt{13}-2}}{3}^1x\sqrt{9x^2+4}\sqrt{1+\dfrac{1}{x^2(9x^2+4)}}~dx$
$=\int_0^\frac{\sqrt{\sqrt{13}-2}}{3}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!(x^2(9x^2+4))^n}{4^n(n!)^2(1-2n)}dx+\int_\frac{\sqrt{\sqrt{13}-2}}{3}^1x\sqrt{9x^2+4}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)(x^2(9x^2+4))^n}dx$
$=\int_0^\frac{\sqrt{\sqrt{13}-2}}{3}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(2n)!C_k^n4^{n-k}9^kx^{2n+2k}}{4^n(n!)^2(1-2n)}dx+\int_\frac{\sqrt{\sqrt{13}-2}}{3}^1x\sqrt{9x^2+4}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)x^{2n}(9x^2+4)^n}dx$
$=\int_0^\frac{\sqrt{\sqrt{13}-2}}{3}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(2n)!9^kx^{2n+2k}}{4^kn!k!(n-k)!(1-2n)}dx+\int_\frac{\sqrt{\sqrt{13}-2}}{3}^1\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(1-2n)x^{2n-1}(9x^2+4)^{n-\frac{1}{2}}}dx$
| {
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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 $r$ gives us
$$\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2} = \frac{\mathcal{P}^2}{4\mathcal{A}}$$
Since $A + B + C = \pi$, we have
$$0 = \cot\frac{\pi}{2} = \cot\left(\frac{A}{2} + \frac{B}{2} + \frac{C}{2}\right)
= \frac{\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot \cot\frac{C}{2} - \left(\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2}\right) }{
\cot\frac{A}{2}\cdot\cot\frac{B}{2} + \cot\frac{B}{2}\cdot\cot\frac{C}{2} + \cot\frac{C}{2}\cdot\cot\frac{A}{2} - 1}$$
This implies (the triple cotangent identity)
$$\tan\frac{A}{2}\cdot\tan\frac{B}{2}\cdot\tan\frac{C}{2} = \frac{1}{\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2}} = \frac{4\mathcal{A}}{\mathcal{P}^2} = \frac{4\times36}{36^2} = \frac19$$
| {
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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 coordinates. Then the limit is 0.
| {
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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 ;
$$\lim_{x \to 1^+}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={-\infty}$$
$$\lim_{x \to 1^-}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={+\infty}$$
The only way I can solve this is using the graph of the function as the limit becomes very apparent. What is the best way to answer this algebraically?
| 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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"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 prove the proof with induction?
| 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:
$$\frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{2^{n+1}} \ge 1 + \frac{n+1}{2}.$$
We thus, subtracting, want to show that:
$$
\frac{1}{2^{n}+1} + \cdots + \frac{1}{2^{n+1}} \ge \frac{1}{2}
$$
Note that there are $2^{n+1} - (2^n + 1) + 1 = 2^n$ terms here, each of which is greater than or equal to $\frac{1}{2^{n+1}}$. Therefore we get that:
$$
\frac{1}{2^{n}+1} + \cdots + \frac{1}{2^{n+1}} \ge 2^n\left(\frac{1}{2^{n+1}}\right) = \frac{1}{2}.
$$
So we are done.
| {
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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 half plane, or
$$\begin{align}i \frac{\pi}{2} \left (\frac{1+e^{i 4 \pi/6}}{6 e^{i 5 \pi/6}} + \frac{2}{6 i}+\frac{1+e^{i 20 \pi/6}}{6 e^{i 25 \pi/6}}\right ) &= i\frac{\pi}{6} \left (-i + e^{-i \pi/6} + e^{-i 5 \pi/6} \right )\\ &= \frac{\pi}{6} \left ( 1+2 \sin{\frac{\pi}{6}}\right ) \\ &= \frac{\pi}{3}\end{align}$$
| {
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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,
$\begin{vmatrix} 0&0&1&0\\1&-3&1&0\\0&1&-3&0 \\0&0&0&0 \end{vmatrix}=0$
It's clear that the previous equation is true (the determinant of $(A-4I_{4})=0$). Now that the factor $(\lambda-4)$ was pull out, one gets a new matrix by removing the null row and null column.
$A'=\begin{bmatrix} 0&0&1\\1&-3&1\\0&1&-3&\end{bmatrix}$
The characteristic polynomial of $A'$ will be a $3^{th}$ degree polynomial, which product with $(\lambda-4)$ equals to a $4^{th}$ degree polynomial.
Now, in order of finding the characteristic polynomial of $A'$ one must to solve the characteristic equation:
$\begin{vmatrix} -\lambda&0&1\\1&-3-\lambda&1\\0&1&-3-\lambda&\end{vmatrix}=0$
My doubt is on finding this determinant. I already tryed Laplace's transformations in order to make null row or a column, but I couldn't do it.
Can you give me a clue? Thanks.
| Working across the top row....
$$\begin{align}\begin{vmatrix} -\lambda&0&1\\1&-3-\lambda&1\\0&1&-3-\lambda&\end{vmatrix} &= -\lambda\left[(-3 -\lambda)(-3 -\lambda) -1\right] -0[\,] + 1\left[1-0 \right]=0 \\
&= -\lambda\left[9 + 6\lambda +\lambda^2 -1 \right] + 1= 0 \\
&= -\lambda^3 -6\lambda^2 -8\lambda +1 = 0\\
&= \lambda^3 +6\lambda^2 +8\lambda -1 = 0
\end{align}$$
| {
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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 $$\sum_{k = 0}^{\infty}\frac{2}{2k + 1}e^{-(2k + 1)\pi x/a}\sin\left( \frac{(2k + 1)\pi y}{a}\right) = \tan^{-1}\left( \frac{\sin(\pi y/a)}{\sinh(\pi x/a)} \right) $$
where $x, y$ are independent variables and $a$ is a parameter?
This arose in the solution to Laplace's equation. My text asserted this without proof, so I'm curious.
| 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
$$ \begin{align} &\sum_{k = 0}^{\infty}\frac{2}{2k + 1}e^{-(2k + 1)\pi x/a}\sin\left( \frac{(2k + 1)\pi y}{a}\right) \\ &= \text{Im}\sum_{k = 0}^{\infty}\frac{2}{2k + 1}e^{-(2k + 1)\pi x/a} e^{i(2k+1) \pi y /a} \\ &= 2 \ \text{Im} \ \text{artanh} \left(e^{- \pi x /a} e^{- i \pi y/a} \right) \\ &= \ \text{Im} \left( \text{Log} (1+e^{- \pi x/a} e^{i \pi y/a}) - \text{Log}(1-e^{-\pi x /a} e^{i \pi y /a})\right) \\&= \arctan \left( \frac{e^{- \pi x/a}\sin (\frac{\pi y}{a})}{1 + e^{- \pi x/a} \cos (\frac{\pi y}{a})} \right) - \arctan\left( \frac{-e^{- \pi x /a} \sin (\frac{\pi y}{a})}{1 - e^{- \pi x /a}\cos (\frac{\pi y}{a})} \right) \tag{1} \\ &= \arctan \left( \frac{\sin (\frac{\pi y}{a})}{e^{\pi x /a} +\cos (\frac{\pi y}{a})} \right) + \arctan\left( \frac{\sin (\frac{\pi y}{a})}{e^{\pi x /a} - \cos (\frac{\pi y}{a})} \right) \\ &= \arctan \left(\frac{2 e^{\pi x/a} \sin (\frac{\pi y}{a})}{e^{2 \pi x/a} - \cos^{2} (\frac{\pi y}{a}) - \sin^{2} (\frac{\pi y}{a})} \right) \tag{2}\\ &= \arctan \left( \frac{2e^{\pi x/a} \sin (\frac{\pi y}{a})}{e^{2 \pi x/a}-1} \right) \\ &= \arctan \left( \frac{\sin (\frac{\pi y}{a})}{\sinh (\frac{\pi x}{a})}\right). \end{align}$$
$(1)$ The points $1+e^{- \pi x/a} e^{i \pi y/a}$ and $ 1-e^{- \pi x/a} e^{i \pi y/a}$ are in the right half-plane.
$(2)$ Since $$ \left|\frac{\sin (\frac{\pi y}{a})}{e^{\pi x/a} + \cos (\frac{\pi y}{a})} \left(\frac{\sin (\frac{\pi y}{a})}{e^{\pi x/a} - \cos (\frac{\pi y}{a})} \right) \right| = \frac{1-\cos^{2} (\frac{\pi y}{a})}{e^{2 \pi x/a} - \cos^{2} (\frac{\pi y}{a})} < 1, $$ we can use the formula $$\arctan(x+y) = \arctan \left(\frac{x+y}{1-xy} \right).$$
| {
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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:
$=\frac{1}{4}(\cos(x)+3\cos(3x))$
$=\frac{1}{4}(\cos(x)+3\cos(x+2x))$
$=\frac{1}{4}(\cos(x)+3[\cos(x)\cos(2x)-\sin(x)\sin(2x)])$
$=\frac{1}{4}(\cos(x)+3[\cos(x)(1-2\sin^2(x))-\sin(x)(2\sin(x)\cos(x))])$
$=\frac{1}{4}(\cos(x)+3[\cos(x)-2\cos(x)\sin^2(x))-2\sin^2(x)\cos(x)])$
$=\frac{1}{4}(\cos(x)+3[\cos(x)-4\cos(x)\sin^2(x))])$
$=\frac{1}{4}(\cos(x)+3\cos(x)-12\cos(x)\sin^2(x))$
$=\frac{1}{4}(4\cos(x)-12\cos(x)\sin^2(x))$
$=\cos(x)-3\cos(x)\sin^2(x))$
$=\cos(x)-\cos(x)\sin^2(x)-2\cos(x)\sin^2(x)$
$=\cos(x)(1-\sin^2(x))-2\cos(x)\sin^2(x)$
$=\cos^3(x)-2\cos(x)\sin^2(x)$
If your question is an assignment, please try to resolve it from scratch without looking at the answer. There are other ways to solve it. Also try that for practice.
| {
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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 integral result was calculated using Mathematica and
I like this integral. But I can't solve it.
My idea:
Let $$\tan{x}=t\Longrightarrow dx=\dfrac{1}{1+t^2}dt$$
so
$$I=\int\limits_{0}^{\infty}\dfrac{dt}{1+\sin^2{t}}\cdot \dfrac{1}{1+t^2}$$
then I can't proceed. Can you help me? Thank you.
| 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 $$\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x + a^{2}} , \ \ |a|<1$$
we have
$$1 + 2 \sum_{k=1}^{\infty} a^{k} \cos(kx) = 1 + 2 \left(\frac{1-a \cos x}{1-2a \cos x +a^{2}} -1\right) = \frac{1-a^{2}}{1-2 a \cos x + a^{2}}$$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{1}{(1-2a \cos 2x +a^{2})(1+x^{2})} \ dx &= \frac{1}{1-a^{2}} \int_{0}^{\infty} \Big(1+2 \sum_{k=1}^{\infty} a^{k} \cos(2kx) \Big) \ \frac{1}{1+x^{2}} \ dx \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{2}{1-a^{2}} \sum_{k=1}^{\infty} a^{k} \int_{0}^{\infty} \frac{\cos(2kx)}{1+x^{2}} \ dx \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{\pi}{1-a^{2}} \sum_{k=1}^{\infty} \Big(\frac{a}{e^{2}} \Big)^{k} \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} + \frac{\pi}{1-a^{2}} \frac{a/e^{2}}{1-a^/e^{2}} \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} \Big(1+ \frac{2a}{e^{2}-a} \Big) = \frac{\pi}{2} \frac{1}{1-a^{2}} \frac{e^{2}+a}{e^{2}-a} \end{align}$$
Now rewrite the integral as
$$ \frac{1}{1+a^{2}} \int_{0}^{\infty} \frac{1}{(1 - \frac{2a}{1+a^{2}} \cos 2x)(1+x^{2})} \ dx$$
and let $a= 3 - 2 \sqrt{2}$.
Then
$$ \begin{align} \frac{1}{1+a^{2}} \int_{0}^{\infty} \frac{1}{(1- \frac{1}{3} \cos 2x)(1+x^{2})} \ dx &= \frac{3}{1+a^{2}} \int_{0}^{\infty} \frac{1}{(3 - \cos 2x)(1+x^{2})} \ dx \\ &= \frac{\pi}{2} \frac{1}{1-a^{2}} \frac{e^{2} + 3 - 2 \sqrt{2}}{e^{2} - 3 + 2 \sqrt{2}} \end{align}$$
which implies
$$ \begin{align} \int_{0}^{\pi /2} \frac{1}{1+\sin^{2} (\tan x)} \ dx &= \frac{\pi}{3} \frac{1+a^{2}}{1-a^{2}}\frac{e^{2} + 3 - 2 \sqrt{2}}{e^{2} - 3 + 2 \sqrt{2}} \\ &= \frac{\pi}{3} \frac{3}{2 \sqrt{2}}\frac{e^{2} + 3 - 2 \sqrt{2}}{e^{2} - 3 + 2 \sqrt{2}} \\ &= \frac{\pi}{2 \sqrt{2}}\frac{e^{2} + 3 - 2 \sqrt{2}}{e^{2} - 3 + 2 \sqrt{2}} \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/730876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 3,
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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
$$
A =\begin{pmatrix}
6 & 4 & 1 \\
-6 & -1 & 3 \\
8 & 8 & 4 \\
\end{pmatrix}
$$
The eigenvalues are $-1, 2, 8$ with corresponding eigenvectors $\begin{pmatrix}
4 \\
-9 \\
8 \\
\end{pmatrix}$,
$\begin{pmatrix}
5 \\
-6 \\
4 \\
\end{pmatrix}$,
$\begin{pmatrix}
1 \\
0 \\
2 \\
\end{pmatrix}$
I cannot calculate $A + A^2 + A^3$ numerically, so how can I do this?
| 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 multiplication is defined. It follows that columns of $X$ are the eigenvectors of $A$ and diagonal entries of $D$ are the eigenvalues of $A$.
If you let $A+A^2+A^3 = B = PDP^{-1}$, then the eigenvectors of $B$ are the same as of $A$, eigenvalues of $B$ are therefore $-1, 14, 584$. So
$P = \begin{pmatrix}
4 & 5 & 1 \\
-9 & -6 & 0 \\
8 & 4 & 2 \\
\end{pmatrix}$ and $D = \begin{pmatrix}
-1 & 0 & 0 \\
0 & 14 & 0 \\
0 & 0 & 584 \\
\end{pmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/731995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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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 rephrase the theorem as follows.
$$\forall n \in \mathbb{N}, P(n)
\quad \text{where} \quad
P(n) \, := \, n^3 + (n + 1)^3 + (n + 2)^3 \text{ is a multiple of 9}$$
We prove the theorem by induction on $n$.
Basis
Below, we show that we have $P(n)$ for $n = 0$.
$$0^3 + 1^3 + 2^3 = 0 + 1 + 8 = 9 = 9 \cdot 1$$
Inductive step
Below, we show that for all $n \in \mathbb{N}$, $P(n) \Rightarrow P(n + 1)$.
Let $k \in \mathbb{N}$. We assume that $P(k)$ holds.
In the following, we use this assumption to show that $P(k + 1)$ holds.
By the assumption,
there is a $i \in \mathbb{N}$ such that
$i \cdot 9 = k^3 + (k + 1)^3 + (k + 2)^3$.
We use this fact in the following equivalent transformation.
The transformation turns the sum of cubes in the first line,
for which we need to show that it is a multiple of 9,
into a product of 9 and another natural number.
$(k + 1)^3 + (k + 2)^3 + (k + 3)^3 \\
= (k + 1)^3 + (k + 2)^3 + k^3 + 9k^2 + 27k + 27 \\
= k^3 + (k + 1)^3 + (k + 2)^3 + 9k^2 + 27k + 27 \quad | \text{ using the induction hypothesis} \\
= 9i + 9k^2 + 27k + 27 \\
= 9 \cdot i + 9 \cdot k^2 + 9 \cdot 3k + 9 \cdot 3 \\
= 9 \cdot (i + k^2 + 3k + 3)$
We see that the above product has precisely two factors: 9 and another natural number.
Thus the product is a multiple of 9.
This completes the induction.
| 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 $\displaystyle \int_0^{\infty} dx \frac{e^{- x \sqrt 3}}{x} (1 - \sin x)(1 + 2 \sin x - \cos 2x)$
Using $\cos 2x = 1 - 2\sin^2 x$ and $1 - \sin^2 x = \cos^2 x$ I was able to get it into the above form. However, I do not know how to proceed. I would like some guidance rather than a full answer, please.
| I would consider the Laplace transform
$$F(p)=\int_0^{\infty} dx \, \frac{\sin{x}}{x} \cos^2{x} \, e^{-p x}$$
Then
$$\begin{align}F'(p) &= -\int_0^{\infty} dx \, \sin{x} \cos^2{x} \, e^{-p x}\\ &= -\frac14 \int_0^{\infty} dx \,(\sin{3 x}+\sin{x}) e^{-p x}\\ &= -\frac14 \left (\frac{3}{p^2+9}+\frac1{p^2+1} \right )\end{align}$$
Thus
$$F(p) = -\frac14 \left (\arctan{\frac{p}{3}}+\arctan{p} \right ) +C$$
The integration constant is
$$\begin{align}C &= \int_0^{\infty} dx \frac{\sin{x}}{x} \cos^2{x}\\ &= \frac14 \int_0^{\infty} dx \frac{\sin{3 x}}{x} + \frac14 \int_0^{\infty} dx \frac{\sin{x}}{x} \\ &= \frac14 \left ( \frac{\pi}{2} + \frac{\pi}{2}\right )\\ &= \frac{\pi}{4}\end{align}$$
so that
$$F(p) = \frac{\pi}{4} - \frac14 \left (\arctan{\frac{p}{3}}+\arctan{p} \right ) $$
and your answer is $2 F(\sqrt{3})$ (which accounts for the factor of two in the original integral):
$$\begin{align}2 F(\sqrt{3}) &= \frac{\pi}{2} - \frac12 \left (\arctan{\frac{\sqrt{3}}{3}}+\arctan{\sqrt{3}} \right )\\ &= \frac{\pi}{2} - \frac12 \left (\frac{\pi}{6} + \frac{\pi}{3} \right ) \\ &= \frac{\pi}{4}\end{align}$$
| {
"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 boundary $x^2 + y^2 + z^2 = 1$ gets really hairy. I tried replacing $x$ with $\sqrt{1 - y^2 - z^2}$ in order to transform $f(x,y,z)$ into a two-variable function $g(y,z)$ and optimize it on $y^2 + z^2 \leq 1$ but $g(y,z)$ is a pain to differentiate. I then tried spherical coordinates which really did not make it any much easier.
Got any suggestions on how to tackle it?
| 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 $(\sqrt{1-z^2} \;sin(\theta), \sqrt{1-z^2} \;cos(\theta))$, for $0 \le \theta \lt 2\pi$
Hence we want to maximise $xy = \sqrt{1-z^2}*sin(\theta) * \sqrt{1-z^2} *cos(\theta) = (1-z^2) \dfrac{sin(2\theta)}{2}$.
$sin(2\theta)$ has a obvious maximum of $1$, so the maximum of $xy$ is $\dfrac{(1-z^2)}{2}$
Now we choose $z$ to maximise $\dfrac{(1-z^2)}{2}(1+z) = \dfrac{(1+z-z^2-z^3)}{2}$
Setting the derivative to zero gives $\dfrac{(1-2z-3z^2)}{2} \equiv 0$
$(3z-1)(z+1) \equiv 0$
i.e. $z=\dfrac{1}{3}$ or $z=-1$
$z=-1$ corresponds to $f(x,y,z)=0$ so this must be a minimum.
$z=\dfrac{1}{3}$ corresponds to $f(x,y,z)=\dfrac{(1 - \dfrac{1}{9})}{2}(1 + \dfrac{1}{3}) = \dfrac{16}{27}$
| {
"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 seem to integrate this. Is this region wrong? Should the integral just be $\int_0^1 \int_{\sqrt{1-y^2}}^{\sqrt{1- (1-y)^2}} dx dy$?
Do I need to convert this to polar form?
| 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 integral setup would be almost right if you had integrated from $x=-\frac{\sqrt{3}}{2}$ to $x=\frac{\sqrt{3}}{2}$. The $y$ should range from $1-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$.
However, it might be preferable to take advantage of the symmetry and integrate from $x=0$ to $x=\frac{\sqrt{3}}{2}$, and double the result.
But as you indicated, polar may be better. The two circles have polar equations $r=1$ and $r=2\sin\theta$. The curves meet where $2\sin\theta=1$, so in the first quadrant at $r=1$, $\theta=\frac{\pi}{6}$ and also at $r=1$, $\theta=\frac{5\pi}{6}$.
Let's find the first quadrant area and double. Up to $\frac{\pi}{6}$, our bounding curve is $r=2\sin\theta$, and then up to $\frac{\pi}{2}$ it is $r=1$. So half our area is
$$\int_0^{\pi/6} 2r\sin\theta \,dr\,d\theta+\int_{\pi/6}^{\pi/2}r\,dr\,d\theta.$$
(The second integral could be replaced by a simple geometric argument.)
Added: For the first integral, integrate first with respect to $r$. We get $\sin\theta$. Then integrate with respect to $\theta$. We get $1-\frac{\sqrt{3}}{2}$.
The second integral gives the area of a circular sector with angle $\frac{\pi}{2}-\frac{\pi}{6}$, which is $\frac{\pi}{3}$, one-sixth of a circle, so area $\frac{\pi}{6}$. Thus half our area is $1-\frac{\sqrt{3}}{2}+\frac{\pi}{6}$.
| {
"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:
$$\lim_{x\to\infty}\dfrac{x^2+4-4x}{2x+1}=\lim_{x\to\infty}\dfrac{({x\over x}-{2\over x})({x\over x}-{2\over x})}{({2x\over x}+{1\over x})}=\dfrac{(1-0)(1-0)}{(2+0)}={1 \over 2}$$
| The online calculator is correct. The value of the limit $\to\infty$. Here is the proof:
$$
\begin{align}
\lim_{x\to\infty}{(x-2)^2\over2x+1}&=\lim_{x\to\infty}\frac{x^2-4x+4}{2x+1}\\
&=\lim_{x\to\infty}\frac{x^2-4x+4}{2x+1}\cdot\frac{\frac{1}{x}}{\frac{1}{x}}\\
&=\lim_{x\to\infty}\frac{\frac{x^2}{x}-\frac{4x}{x}+\frac{4}{x}}{\frac{2x}{x}+\frac{1}{x}}\\
&=\lim_{x\to\infty}\frac{x-4+\frac{4}{x}}{2+\frac{1}{x}}\\
&=\frac{\left(\lim\limits_{x\to\infty}x\right)-4+0}{2+0}\to\boxed{\Large\color{blue}{\infty}}\\
\end{align}
$$
$$\\$$
$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
| {
"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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