Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find a $2 \times 2$ matrix $A$ that satisfies $A^3 = \begin{pmatrix} -1 & -1 \\ 1 & -1 \\ \end{pmatrix}$
Find a $2 \times 2$ matrix $A$ such that
$$A^3 =
\begin{pmatrix}
-1 & -1 \\
1 & -1 \\
\end{pmatrix}$$
I'm trying to think about this geometrically. Could this have something to do with a rotation dilation?
| There are several ways of tackling this problem, a nice one of which you have already thought of. Since $\ \begin{pmatrix}-1&-1\\1&-1\end{pmatrix}\ $ is indeed a dilated rotation matrix, namely $\ \sqrt{2}\begin{pmatrix}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{pmatrix}\ $, where $\ R=\begin{pmatrix}\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&\frac{-1}{\sqrt{2}}\end{pmatrix}\ $ is a rotation matrix, then the matrix $\ 2^\frac{1}{6} S\ $ will be an answer for your question if $\ S\ $ is a rotation matrix which rotates vectors through one third of the angle that $\ R\ $ does. The angle through which $\ R\ $ rotates vectors can be found as the angle between the unit vectors $\ \begin{pmatrix}1\\0\end{pmatrix}\ $ and $\ R\begin{pmatrix}1\\0\end{pmatrix}\ $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Understanding the constraints to find a $2\times 2$ non-zero matrix $A$ such that $A^2=0$ Using the rules for matrix multiplication, I have found four algebraic equations as "constraints" for getting every element in the resulting matrix to be zero. Assuming that the elements in the resulting matrix are $a, b, c$ and $d$ ($a, b$ are the elements in the first row and $c, d$ are the elements in the second). The equations are therefore;
*
*$a^2 + bc = 0$
*$ab + bd = 0$
*$ac + cd = 0$
*$d^2 + bc = 0$
I have found from the four equations that $a=-d$ and that $bc=-a^2$ hence $bc$ must be a negative quantity thus $b$ and $c$ have opposite signs but that does not seem to be enough to guarantee that the resulting matrix is always zero. Now, I do not know how to think about this problem;
How do I find more constraints from these equations?
And how do I know that I have found all possible constraints?
Why is that when I square one equation, some how I get a new constraint?
Shouldn't the four equations be enough to determine the exact conditions for $a, b, c$ and $d$?
| With
$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \tag 1$
we have
$A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & (a + d)b \\ (a +d)c & d^2 + bc \end{bmatrix} = 0, \tag 2$
whence
$a^2 + bc = 0, \tag 3$
$d^2 + bc = 0, \tag 4$
$(a + d)b = 0, \tag 5$
$(a + d)c = 0; \tag 6$
suppose
$a + d \ne 0; \tag 7$
then via (5) and (6),
$b = c = 0; \tag 8$
then via (3) and (4),
$a^2 = d^2 = 0 \Longrightarrow a = d = 0 \Longrightarrow a + d = 0 \Rightarrow \Leftarrow a + d \ne 0; \tag 9$
this contradiction rules out the case (7), so
$a + d = 0 \Longrightarrow d = -a; \tag{10}$
now if
$a = d = 0, \tag{11}$
then (3)-(4) imply, assuming $A \ne 0$, exactly one of
$b = 0, \; c \ne 0, \tag{12}$
$b \ne 0, \; c = 0, \tag{13}$
holds. Thus the solutions in the event that (11) binds are one of the forms
$A = \begin{bmatrix} 0 & b \\ 0 & 0 \end{bmatrix}, \; \begin{bmatrix} 0 & 0 \\ c & 0 \end{bmatrix}; \tag{14}$
when
$d = -a \ne 0, \tag{15}$
we find that (3)-(4) imply
$b, c \ne 0 \tag{16}$
and
$c = -\dfrac{a^2}{b}; \tag{16}$
therefore
$A = \begin{bmatrix} a & b \\ -\dfrac{a^2}{b} & -a \end{bmatrix}. \tag{17}$
We see that (2) implies $A$ takes one of the three forms (14), (17); the former being comprised of two one-parameter families of matrices, and the latter comprised of one two-parameter family.
| {
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"url": "https://math.stackexchange.com/questions/3180064",
"timestamp": "2023-03-29T00:00:00",
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Use ratio test to determine convergence or divergence. Find general term first. I have this series:
$$\frac{2}{3} + \frac{2*5}{3*5} + \frac{2*5*8}{3*5*7} ...$$
I'm confused how to find the general term here to then apply the ratio test.
So it starts with a $$\frac{2}{3}$$ and the next term is $\frac{2*5}{3*5}$
So each time we multiply by what appears to be another term that is $3(n-1)$ bigger than the original 2 in the numerator, and $2(n-1)$ bigger than the original 3 in the denominator. But how do I capture this in a general formula?
But the general term is not:
$$\frac{2 + 3(n-1)}{3 + 2(n-1)}$$
because that would be:
$$\frac{2}{3} + \frac{2*5}{3*5} + \frac{2*8}{3*7}$$
and would leave out all the middle terms. What am I doing wrong?
| Find formulae for the product terms in the numerator and the denominator. The questions seems to suggest that they are arithmetic progressions. In particular, the $n$th term (not the $n$th partial product) in the product $2 \cdot 5 \cdot 8 \cdot \ldots$ is supposedly $2 + 3(n - 1)$. Similarly, the $n$th term in the product $3 \cdot 5 \cdot 7 \cdot \ldots$ is supposed to be $3 + 2(n - 1)$. Therefore, to get from the $n$th term to the $(n + 1)$th term, you must multiply by the $(n + 1)$th term of the former product, divided by the $(n + 1)$th term of the latter product. More succinctly, if $a_n$ is the $n$th series term, then
$$a_{n+1} = a_n \cdot \frac{2 + 3(n + 1 - 1)}{3 + 2(n + 1 - 1)} = a_n \cdot \frac{2 + 3n}{3 + 2n}.$$
The ratio you want is therefore
$$\frac{a_{n+1}}{a_n} = \frac{2 + 3n}{3 + 2n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Generating function of $\frac{h(x)}{(1-x)^2}$
If $h(x)$ is the generating function for $a_r$, what is the generating function of $$\frac{h(x)}{(1-x)^2}$$
Let $h(x)$ be written as
$$h(x) = \sum_{r} a_r x^r $$
Consider more simply
$$\frac{h(x)}{1-x} = \frac{1}{1-x} h(x) =\sum_{r} x^r \sum_{r} a_r x^r$$
I tried to expand this and see what I could get
$$(1+x+x^2+x^3+\dots+x^r+\dots)(a_0+a_1x+a_2x^2+a_3x^3+\dots+a_rx^r+\dots)$$
there are two ways to simplify the product, either
$$a_0(1+x+x^2+\dots)+a_1(x+x^2+x^3+\dots)+ a_2(x^2+x^3+x^4+\dots)+\dots= \sum_ra_r\sum_{k\ge r}x^k$$
or
$$a_0 + (a_0+a_1)x+(a_0+a_1+a_2)x^2+(a_0+a_1+a_2+a_3)x^3 = \sum_r \left(\sum_{k\le r}a_k \right)x^r$$
obviously this is only for one factor of $\frac{1}{1-x}$ but I assume If I can get help for this I can extend it to two factors.
I'm not sure what form the answer is expected to be in? Because I could say the generating function is
$$\frac{h(x)}{1-x} = h\left(\sum_{k \ge r}x^k\right)$$
but I'm not sure that makes any sense. I was expecting to say something like
$$\frac{h(x)}{1-x} \mapsto h(x^2)$$
(the $x^2$ is not intentional, just some idea of what I believe the answer could look like)
Any help for the case of $\frac{1}{1-x}$ would be great and then I could extend it to $\frac{1}{(1-x)^2}$
| Another way.
Since
$(1-x)^2 = 1-2x+x^2$,
if
$\dfrac{h(x)}{(1-x)^2}
=\sum_{n=0}^{\infty} a_nx^n$
then
$\begin{array}\\
h(x)
&=(1-x)^2\sum_{n=0}^{\infty} a_nx^n\\
&=(1-2x+x^2)\sum_{n=0}^{\infty} a_nx^n\\
&=\sum_{n=0}^{\infty} a_nx^n-2x\sum_{n=0}^{\infty} a_nx^n+x^2\sum_{n=0}^{\infty} a_nx^n\\
&=\sum_{n=0}^{\infty} a_nx^n-\sum_{n=0}^{\infty} 2a_nx^{n+1}+\sum_{n=0}^{\infty} a_nx^{n+2}\\
&=\sum_{n=0}^{\infty} a_nx^n-\sum_{n=1}^{\infty} 2a_{n-1}x^{n}+\sum_{n=2}^{\infty} a_{n-2}x^{n}\\
&=a_0+a_1x+\sum_{n=}^{\infty} a_nx^n-2a_0x-\sum_{n=2}^{\infty} 2a_{n-1}x^{n}+\sum_{n=2}^{\infty} a_{n-2}x^{n}\\
&=a_0+(a_1-2a_0)x+\sum_{n=}^{\infty} (a_n-2a_{n-1}+a_{n-2})x^n\\
\end{array}a_n
$
If
$h(x)
=\sum_{n=0}^{\infty} h_nx^n$
then,
equating coefficients,
$h_0 = a_0$,
$h_1 = a_1-2a_0$,
so
$a_1
= h_1+2a_0
= h_1+2h_0
$,
and,
for $n \ge 2$,
$h_n
=a_n-2a_{n-1}+a_{n-2}
$
so
$a_n
=h_n+2a_{n-1}-a_{n-2}
$.
The advantage of this method
is that getting each new $a_n$
takes only
3 operations
(multiply, add, subtract)
instead of $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that for all $x\in \mathbb R$ $, \arctan x=\frac{\pi}{2}-\arccos(\frac{x}{\sqrt{1+x^{2}}})$
Prove that for all $x\in \mathbb R$, $$\arctan x=\frac{\pi}{2}-\arccos \left(\frac{x}{\sqrt{1+x^{2}}}\right)$$
From Lagrange form of Taylor's theorem I have:$$\arctan x+\arccos\left(\frac{x}{\sqrt{1+x^{2}}}\right)=x-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}+\frac{\pi}{2}-x-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}=-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}+\frac{\pi}{2}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}$$I know that: $$\left(-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}\right) \rightarrow 0$$But I don't know how to show that: $$\left(-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}\right) = 0$$ Can you help me?
| Hint:
Let $\arctan x=y,-\pi/2<y<\pi/2,\cos y>0$
$\arccos\dfrac{\tan y}{\sec y}=\arccos(\sin y)=\dfrac\pi2-\arcsin(\sin y)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\sum^{20}_{r=0} \binom{20}{r}\cos r \theta$
Evaluate $$\sum^{20}_{r=0} \binom{20}{r}\cos r \theta$$
To do this my initial thought was to set up an alternative series, $S$, $$S=\sum^{20}_{r=0} \binom{20}{r}\sin r \theta$$ If yoy let the original series be $C$, then $$C+iS = \sum^{20}_{r=0} \binom{20}{r}\cos r \theta + i\sum^{20}_{r=0} \binom{20}{r}\sin r \theta $$
Nw, I'm not sure how to get this into a ncie obvious looking form, but i am assuming it is equal to, $$C+iS = 1+20(\cos \theta + i\sin\theta) + \binom{20}{2}(\cos 2\theta + i\sin2\theta) + \cdots (\cos20\theta + i\sin20\theta)$$
$$C+iS = 1+20(\cos \theta + i\sin\theta) + \binom{20}{2}(\cos \theta + i\sin\theta)^2 + \cdots (\cos\theta + i\sin\theta)^{20}$$
$$C+iS = (1+(\cos\theta + i\sin\theta))^{20}$$ This is the point I didn't know where to continue from. All I know is that I want to get to a point where I can just take the real part which would be $C$. I know the answer is $$C=2^{20} \cos^{20}(\frac12\theta) \cos10\theta $$
| You did very well. Now
$$
\eqalign{
& C + i\,S = \sum\limits_r {\left( \matrix{
20 \cr
r \cr} \right)e^{\,ir\,\theta } } = \sum\limits_r {\left( \matrix{
20 \cr
r \cr} \right)\left( {e^{\,i\,\theta } } \right)^{\,r} } = \cr
& = \left( {1 + e^{\,i\,\theta } } \right)^{\,20} \cr
& \left| {1 + e^{\,i\,\theta } } \right| = \sqrt {\left( {1 + \cos \theta } \right)^{\,2} + \sin ^{\,2} \theta } = \sqrt {2 + 2\cos \theta } = 2\cos \left( {\theta /2} \right) \cr
& \phi = \angle \left( {1 + e^{\,i\,\theta } } \right) = \arctan \left( {{{\sin \theta } \over {1 + \cos \theta }}} \right) = \cr
& = \arctan \left( {{{2\sin \left( {\theta /2} \right)\cos \left( {\theta /2} \right)} \over {1 + \cos ^{\,2} \left( {\theta /2} \right) - \sin ^{\,2} \left( {\theta /2} \right)}}} \right) = \cr
& = \arctan \left( {{{\sin \left( {\theta /2} \right)} \over {\cos \left( {\theta /2} \right)}}} \right) = \theta /2 \cr}
$$
and you know how to conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Occurance of Prime Factor 3 Let $P^{3}(a, b)$ return the sum of the frequency of Prime Factor 3 of all Integers in $[a, b]$, e.g: $P^{3}(5, 10) = 3, P^{3}(12, 20) = 4, ...$
Then
$\forall x>10 \in \mathbb N: P^{3}(2^{x}, 2^{x+1}) > P^{3}(1, 2^{x}) + 1$
?
| Partial
Notice that:
$$P^3(a,b)=P^3(b,1)-P^3(a,1)$$
So your question becomes:
$$P^3(2^{k+1},1)>2P^3(2^k,1)+1$$
There is a standard formula for $P^3(N,1)$:
$$ P^3(N,1)=\sum_{i=1}^{\lfloor \log_3(N) \rfloor} \lfloor \frac{N}{3^i} \rfloor $$
So:
$$\sum_{i=1}^{\lfloor \log_3(2^{k+1}) \rfloor} \lfloor \frac{2^{k+1}}{3^i} \rfloor \geq 2\sum_{i=1}^{\lfloor \log_3(2^{k}) \rfloor} \lfloor \frac{2^{k}}{3^i}\rfloor+1$$
Notice that $\lfloor 2 f(x) \rfloor \geq 2\lfloor f(x) \rfloor$ so:
$$\sum_{i=1}^{\lfloor \log_3(2^{k+1}) \rfloor} \lfloor \frac{2^{k+1}}{3^i} \rfloor \geq 2 \sum_{i=1}^{\lfloor \log_3(2^{k+1}) \rfloor} \lfloor \frac{2^{k}}{3^i} \rfloor \geq 2 \sum_{i=1}^{\lfloor \log_3(2^{k}) \rfloor} \lfloor \frac{2^{k}}{3^i}\rfloor $$
And this is a pretty good partial result.
:/
| {
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Proving $8 | x^2 + y^2$ iff x,y are both even is false Per the question above I am trying to prove this statement false. As such only one of two conditions have to be met both x and y being odd, or them both being odd. I've seen a lot of examples on this site regarding x^2 - y^2 for a similar example but none where they are added together so I was looking for a pointer on what I might be missing. This is what I have so far:
False via contraposition: a^2 + b^2 is not divisible by 8 iff a or b are not even
Assume both a and b are odd
A = 2k + 1 and b = 2n + 1 where k and n are both integers
a^2 + b^2 = (2k + 1)^2 + (2n + 1)^2
a^2 + b^2 = 4k^2 + 4k + 1 + 4n^2 + 4n + 1
a^2 + b^2 = 4(k^2 + k + n^2 + n) + 2 (not divisble by 8 so doesnt help the case)
Assume a is odd
A = 2k + 1 and b = 2n
x^2 = 4k(k + 1) + 1
k(k + 1) is even (product of 2 numbers)
x^2 is of form 8k + 1
a^2 + b^2 = 8x + 1 + 8y + 1
a^2 + b^2 = 8(x - y) + 2 (the extra 2 is stoping this case from being true as well)
| You want to disprove the statement
$$8\mid x^2+y^2\iff 2\mid x, y $$
As @J.W. Tanner pointed out in the comments, a simple counterexample - such as $2^2+4^2$ or its generalization $(2n)^2+(2n+2)^2=8n^2+8n+4\not\equiv 0\pmod 8$ - is enough. This proves that $$2\mid x, y\color{red}{\not\Rightarrow}8\mid x^2+y^2$$
However, you won't be able to disprove the other direction, since it is true $$8\mid x^2+y^2\color{red}{\implies}2\mid x,y$$ This follows from the fact that the only quadratic residues modulo $8$ are $\{0, 1, 4\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$
Prove that $\sin (\sin x -x) =\sin x - x + o(x^5)$
In the task which I do I need record $\sin (\sin x-x)$ in a way to have $ax^3$. So: $$\sin (\sin x -x)=\sin x -x +r(\sin x -x)=x-\frac{x^5}{6}+\frac{x^5}{120}+r(x)-x +r(\sin x -x)$$ I know that $r(x)=o(x^5)$ and it is easy. However how to prove that $r(\sin x -x)=o(x^5)$? I tried to do it but then I have: $$\frac{r(\sin x -x)}{\sin x -x}\cdot \frac{\sin x -x}{x^5}=\frac{r(\sin x -x)}{\sin x -x}\cdot(\frac{\sin x}{x}-1)\cdot \frac{1}{x^4} \rightarrow 0\cdot(1-1)\cdot(-\infty)$$Can you help me how can I prove it?
| You can note that $$\sin x - x=-\frac{x^3}{6}+o(x^3)\tag{1}$$ and replacing $x$ with $\sin x - x$ we get $$\sin(\sin x - x) - (\sin x - x) =-\frac{(\sin x - x)^3}{6}+o((\sin x - x) ^3)$$ Using $(1)$ we can see that both the terms in right hand side of above equation are $o(x^5) $ and we are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$, at $a = 0$ Find Taylor Polynomial order 5 of $f(x) = \frac{1}{(1-x)}$ , at $a = 0$
So I start of with:
$f(0) = \frac{1}{(1-0)}=1$
$f'(0) = \frac{1}{(1-0)^2 }= 1$
$f''(0) = \frac{2}{(1-0)^3 }= 2$
$f'''(0) = \frac{6}{(1-0)^4 }= 6$
$f^{(4)}(0) = \frac{24}{(1-0)^5 }= 24$
$f^{(5)}(0) = \frac{120}{(1-0)^6 }= 120$
and I get
$1 + x + \frac{2x^2}{2} + \frac{6x^3}{3} + \frac{24x^4}{4} + \frac{120x^5}{5} $
Answer = $1 + x + x^2 + 2x^3 + 5x^4 + 24x^5 $
The problem is that a calculator i used to check it up said the answer was
answer = $1 + x + x^2 + x^3 + x^4 + x^5 $
So im confused, im I right or is the calculator wrong? Thanks!
| Its easy to do, take the inverse of (1-x) and then expand
$(1-x)^{-1}$ = 1 + $x + x^{2} + x^{3}$ + .....
So coefficient of each term is 1.
From your method, you are skipping the factorial term in the denominator. Calculator is right.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many cricket teams can the manager choose if one particular batsman refuses to play when one particular bowler does?
A certain country has a cricket squad of 16 people, consisting of 7 batsmen, 5 bowlers, 2 all- rounders and 2 wicket-keepers. The manager chooses a team of 11 players consisting of 5 batsmen, 4 bowlers, 1 all-rounder and 1 wicket keeper. Find the number of different teams the manager can choose if one particular batsman refuses to be in the team when one particular bowler is in the team.
I did it
$6C5 \cdot 5C4 \cdot 2 \cdot 2 + 7C5 \cdot 4C3 \cdot 2 \cdot 2$. It is completely wrong. How to do it?
| There are $5$ from $7 = 21$ ways to choose batsmen. For everyone of those there are $4$ from $5 = 5$ bowlers and for every one of those there are $2$ choices of wicket keeper and $2$ choices of all-rounders. Subtracting combinations that contain a certain bowler and batter involves counting the number of ways a batter can be in the group of $5$ which is $4$ from $6 = 15$, and for those, the choic of bowlers is $4$ from $4 = 1$. So the calculation looks like........
$([\binom{7}{5} - \binom{6}{4}] \cdot \binom{5}{4} + \binom{6}{4})\cdot 2 \cdot 2$
$=(6\cdot 5 + 15)\cdot 2 \cdot 2 = 180$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $3x^2 - 4x -2 = 0$ by completing the square I can't understand the solution from the textbook (Stroud & Booth's "Engineering Mathematics" on a problem that involves solving a quadratic equation by completing the square.
The equation is this:
$$
\begin{align}
3x^2 - 4x -2 = 0 \\
3x^2 - 4x = 2
\end{align}
$$
Now, divide both sides by three:
$$x^2 - \frac{4}{3}x = \frac{2}{3}$$
Next, the authors add to both sides the square of the coefficient of $x$, completing the square on the LHS:
$$x^2 - \frac{4}{3}x + \left(\frac{2}{3}\right)^2 = \frac{2}{3} + \left(\frac{2}{3}\right)^2$$
Now, the next two steps (especially the second step) baffle me. I understand the right-hand side of the first quation (how they get the value of $\frac{10}{9}$), but the last step is a complete mystery to me:
$$
\begin{align}
x^2 - \frac{4}{3}x + \frac{4}{9} = \frac{10}{9} \\
\left(x - \frac{2}{3}\right)^2 = \frac{10}{9}
\end{align}
$$
Can anyone please explain how they went from the first step to the second step?
| We get from $$3x^2-4x-2=0$$ to $$3x^2-4x=2$$ by adding two on both sides. Then dividing by $3$ we obtain
$$x^2-\frac{4}{3}x=\frac{2}{3}$$ then we can write $$x^2-2\cdot \frac{2}{3}x+\frac{4}{9}=\frac{2}{3}+\frac{4}{9}$$
| {
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"answer_id": 4
} |
Integral $\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx$
Prove that$$I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$
I've found this integral in my notebook and perhaps I encountered it before since it looks quite familiar.
Anyway I thought it's quite a trivial integral so I'm gonna solve it quickly, but I am having some hard time to finish it. I went on with Feynman's trick:
$$I(a)=\int_0^\infty \frac{\ln((1+x^2)a+x)}{1+x^2}dx\Rightarrow I'(a)=\int_0^\infty \frac{dx}{a+x+ax^2}$$
$$=\frac1a\int_0^\infty \frac{dx}{\left(x+\frac{1}{2a}\right)^2+1-\frac{1}{4a^2}}=\frac{1}{a}\frac{1}{\sqrt{1-\frac{1}{4a^2}}}\arctan\left(\frac{x+\frac{1}{2a}}{\sqrt{1-\frac{1}{4a^2}}}\right)\bigg|_0^\infty$$$$=\frac{\pi}{\sqrt{4a^2-1}}-\frac{2}{\sqrt{4a^2-1}}\arctan\left(\frac{1}{\sqrt{4a^2-1}}\right)=\frac{2\arctan\left(\sqrt{4a^2-1}\right)}{\sqrt{4a^2-1}}$$
We can prove easily via the substitution $x\to \frac{1}{x}$ that $I(0)=0$ so we have that:
$$I=I(1)-I(0)=2\int_0^1 \frac{\arctan\left(\sqrt{4a^2-1}\right)}{\sqrt{4a^2-1}}da$$
Now I thought about two substitutions:
$$ \overset{a=\frac12\cosh x}=\int_{\operatorname{arccosh}(0)}^{\operatorname{arccosh}(2)} \arctan(\sinh x)dx$$
$$\overset{a=\frac12\sec x}=\int_{\operatorname{arcsec}(0)}^{\frac{\pi}{3}}\frac{x}{\cos x}dx$$
But in both cases the lower bound is annoying and I think I am missing something here (maybe obvious).
So I would love to get some help in order to finish this.
Edit: We can apply once again Feynman's trick. First consider: $$I(t)=\int_0^1 \frac{2\arctan(t\sqrt{4a^2-1})}{\sqrt{4a^2-1}}da\Rightarrow I'(t)=2\int_0^1 \frac{1}{1+t^2(4a^2-1)}da$$
$$=\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2at}{\sqrt{1-t^2}}\right)\bigg|_0^1=\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2t}{\sqrt{1-t^2}}\right)$$
So once again we have $I(0)=0$, so $I=I(1)-I(0)$.
$$\Rightarrow I=\int_0^1\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2t}{\sqrt{1-t^2}}\right)dt\overset{t=\sin x}=\int_0^\frac{\pi}{2}\frac{\arctan(2\tan x)}{\sin x}dx$$
At this point Mathematica can evaluate the integral to be:
$$I=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$
I didn't try the last integral yet, but I am thinking of Feynman again $\ddot \smile$.
Edit 2: Found that I already was on it some time ago, and actually posted it here, which means I have solved it before using Feynman's trick, but right now I can't remember how I did it.
So given the circumstances I am positive that it can be solved starting with my approach, but if you have any other ways then feel free to share it.
| Solution 3. Consider the following integral:
$$ I(a)=\int_0^\frac{\pi}{2}\frac{\arctan(a\tan x)}{\sin x}dx\Rightarrow I'(a)=\int_0^\frac{\pi}{2}\frac{\sec x}{1+a^2\tan^2 x}dx$$
$$ =\int_0^\frac{\pi}{2}\frac{\cos x}{\cos^2 x+a^2\sin^2 x}dx\overset{\sin x=y}=\int_0^1 \frac{dy}{1+(a^2-1)y^2}=\frac{\arctan\sqrt{a^2-1}}{\sqrt{a^2-1}}$$
$$\rm I=\underbrace{I(2)-I(1)}_{=J}+I(1), \quad I(1)=\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx$$
$$J=\int_1^2 \frac{\arctan\sqrt{a^2-1}}{\sqrt{a^2-1}}da\overset{a=\sec x}=\int_0^\frac{\pi}{3}\frac{x}{\cos x}dx\overset{x=\frac{\pi}{2}-t}=\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{\frac{\pi}{2}-t}{\sin t}dt$$
$$\rm=\frac{\pi}{2}\int_\frac{\pi}{6}^\frac{\pi}{2} \frac{1}{\sin t}dt- \int_0^\frac{\pi}{2} \frac{t}{\sin t}dt+\int_0^\frac{\pi}{6} \frac{t}{\sin t}dt$$
$$ \Rightarrow I=J+I(1)=\frac{\pi}{2}\ln\left(\tan \frac{x}{2}\right)\bigg|_\frac{\pi}{6}^\frac{\pi}{2}+\int_0^\frac{\pi}{6} \frac{t}{\sin t}dt$$
Finally, using the result from here, we get:
$$ I=\frac{\pi}{2}\ln(2+\sqrt 3)-\frac{\pi}{6}\ln(2+\sqrt 3)+\frac43G=\boxed{\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G}$$
Solution 4.
$$I=\int_0^2\frac{\arctan\sqrt{a^2-1}}{\sqrt{a^2-1}}da=\int_0^2\frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}da$$
$$\sf I=\int_0^1\frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}da+\int_1^2\frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}da$$
The second integral is like the one from above, and for the first integral we need to use the complex definition of $\sf \arccos z$, namely $\sf -i\ln\left(z+\sqrt{z^2-1}\right)$.
$$\sf \Rightarrow \frac{\operatorname{arcsec} a}{\sqrt{a^2-1}}=\frac{-\ln\left(\frac{1-\sqrt{1-a^2}}{a}\right)}{\sqrt{1-a^2}}$$
And now via the substitution $a=\sin y$ everything goes smoothly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12$ If $x,y\in\mathbb R^+$, $x+y=12$, what is the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$?
I got the question from a mathematical olympiad competition (of China) for secondary 2 student, so I don't expect an "analysis" answer.
The answer should be $15$, when $x=\frac{20}{3}$ and $ y=\frac{16}{3}$ (just answer, no solution :< ).
I try to use AM-GM inequality, but I couldn't manage to get the answer.
$$\sqrt{x^2+25}+\sqrt{y^2+16}=\sqrt{x^2+25}+\sqrt{(x-12)^2+16}=\sqrt{x^2+25}+\sqrt{x^2-24x+160}$$
Can this help?
I also tried to plot the graph, see here.
Any help would be appreciated. Thx!
| Let
$f = \sqrt{x^2 + 25} + \sqrt{y^2 + 16}$
$g = x+y -12 =0 $
By Lagrange's Undetermined multipliers method,
$F = f + g\lambda = \sqrt{x^2 + 25} + \sqrt{y^2 + 16} + \lambda(x+y -12)$
Differentiating partially w.r.t x and y and using $F_x = 0$ and $F_y = 0 $ at extremum,
$F_x = \frac{x}{\sqrt{x^2 + 25}} +\lambda = 0$
$F_y = \frac{y}{\sqrt{y^2 + 16}} +\lambda = 0$
So,
$\frac{x}{\sqrt{x^2 + 25}} = \frac{y}{\sqrt{y^2 + 16}}$
$\frac{x^2}{x^2 + 25} = \frac{y^2}{y^2 + 16}$
$\frac{x^2 + 25}{x^2} = \frac{y^2 + 16}{y^2}$
$1 + \frac{25}{x^2} = 1 + \frac{16}{y^2}$
On solving,
$5y = \pm 4x$
Using this in g,
$x \pm 4x/5 = 12$
$x = 20/3, 60$
Using these values,
$y = 12 - x $
$y = 12 -20/3 = 16/3$ and $y = 12 - 60 = -48$
Using these values of x and y in f we find its value is minimum at $(\frac{20}{3},\frac{16}{3})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Efficiently compute $f'(a)f'(b)f'(c)$ for the roots of a cubic polynomial Let $a,b, c$ be the zeroes of $f(x) = x^3+3x^2-7x+1$. Find $f'(a)f'(b)f'(c)$.
My idea involved substituting in $a,b,c$ into $f'(x)$ then using Vieta's but that would take far too long of a time, especially since I'm training for competitions. What would be a great way to solve this within a time limit?
| If $f(x)=(x-r)g(x)$, then $f'(x)=g(x)+(x-r)g'(x)$ and hence $f'(r)=g(r)$.
As $f(x)=(x-a)(x-b)(x-c)$, $f'(a)f'(b)f'(c)=(a-b)(a-c)(b-a)(b-c)(c-a)(c-b)$.
Let $g(x)=f(x-1)$ and $\alpha$, $\beta$ and $\gamma$ be the roots of $g(x)=0$. Then $\alpha=a+1$, $\beta=b+1$ and $\gamma=c+1$. Therefore, $g'(\alpha)g'(\beta)g'(\gamma)=f'(a)f'(b)f'(c)$.
$g(x)=(x-1)^3+3(x-1)^2-7(x-1)+1=x^3-10x+10$.
$g'(x)=3x^2-10$.
\begin{align*}
g'(\alpha)g'(\beta)g'(\gamma)&=
\frac{(3\alpha^3-10\alpha)(3\beta^3-10\beta)(3\gamma^3-10\gamma)}{\alpha\beta\gamma}\\
&=\frac{(20\alpha-30)(20\beta-30)(20\gamma-30)}{\alpha\beta\gamma}\\
&=\frac{-8000(\frac{3}{2}-\alpha)(\frac{3}{2}-\beta)(\frac{3}{2}-\gamma)}{\alpha\beta\gamma}\\
&=\frac{-8000}{\alpha\beta\gamma}g\left(\frac{3}{2}\right)\\
&=\frac{-8000}{-10}\left[\left(\frac{3}{2}\right)^3-10\left(\frac{3}{2}\right)+10\right]\\
&=-1300
\end{align*}
We can also use the discriminant of $g(x)=0$, which is $-4(-10)^3-27(10)^2=1300$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Problem with $\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}$ How to simplify $$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}?$$
Rationalise the denominator
$$\frac{\sqrt{6+4\sqrt{2}}}{4}(2-\sqrt{2})$$
This is still not simplify.
| We have $$6+4\sqrt{2}=4+2+4\sqrt{2}$$ so $$2\sqrt{6+4\sqrt{2}}=4+2\sqrt{2}$$ and we get
$$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Arc length of $f(t)=(cos^4(t),sin^4(t))$ from $t=0$, to $t=2\pi$. I'm trying to compute the arc length $L$ of $f(t)=(cos^4(t),sin^4(t))$ from $t=0$, to $t=2\pi$. Using the regular formula we have that:
$$L=\int_{0}^{2\pi}\sqrt{16\cos^2(t)\sin^2(t)(\cos^4(t)+\sin^4(t))}dt$$
Then after some (long) calculi it reduced to
$$L=4\int_0^{\pi/2}\sin(2t)\sqrt{3+cos(4t)}dt$$
and then this can be computed using integration by parts, which leads to the integral
$$I=\int\frac{\sin(4t)\cos(2t)}{\sqrt{3+\cos(4t)}}dt$$
Now, by making $z=cos(2t)$ we have that
$$I=\int\frac{z^2}{\sqrt{2z^2+2}}dz=-\frac{1}{\sqrt2}\int\tan^2\theta\sec\theta d\theta$$
taking $z=\tan\theta$. Finally, i can make the tangent a secant squared, which will make me compute the integral of a secant cube, which involves more integration by parts. The point is that this is a very long and tedious approach to compute this arc length. Can you suggest a simpler way of computing this number? Any idea will be helpful.
| It is better to write $$\sin 2t \sqrt{3 + \cos 4t} = \sin 2t \sqrt{2 + 2 \cos^2 2t},$$ hence with the substitution $$u = \cos 2t, \quad du = -2 \sin 2t \, dt,$$ we obtain $$\int_{t=0}^{\pi/2} \!\!\sin 2t \sqrt{2 + 2 \cos^2 2t} \, dt = -\frac{1}{2} \int_{u=1}^{-1} \!\!\!\!\sqrt{2 + 2u^2} \, du = \frac{1}{\sqrt{2}} \int_{u=-1}^1 \!\!\!\! \sqrt{1 + u^2} \, du \\ = \sqrt{2} \int_{u=0}^1 \!\!\sqrt{1+u^2} \, du.$$ Then integration by parts yields
$$\begin{align*}I = \int_{u=0}^1 \sqrt{1+u^2} \, du
&= \left[ u \sqrt{1+u^2} \right]_{u=0}^1 - \int_{u=0}^1 \frac{u^2}{\sqrt{1+u^2}} \, du \\
&= \sqrt{2} - \int_{u=0}^1 \frac{1+u^2}{\sqrt{1+u^2}} - \frac{1}{\sqrt{1+u^2}} \, du \\
&= \sqrt{2} - I + \left[ \sinh^{-1} u \right]_{u=0}^1,
\end{align*}$$
consequently $$I = \frac{\sqrt{2} + \sinh^{-1} 1}{2} = \frac{\sqrt{2} + \log(1+\sqrt{2})}{2},$$ and the rest is just bookkeeping of the other constants and symmetries.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Computing volume inside a ball and outside a cylinder I try to calculate the volume inside the ball $\ x^2 + y^2 + z ^2 = 4 $ the outside the cylinder $\ x^2+y^2=2x $ using double integral.
Since the shape is symmetric I chose $\ z = 0 $ as bottom limit and the ball is the upper limit. I try to convert to polar form and so I get
$$\ x^2 + y^2 + z ^2 = 4 \Rightarrow z = \sqrt{4-r^2} \\x^2 + y^2 = 2x \Rightarrow r = 2 \cos \theta $$
therefore the integral should be
$$\ 4 \cdot \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta$$
but this integral is way to messy for me to calculate. I mean first step of calculating integral for $\ r $ is okay but then the value I get and calculating integral of $\ \theta $ is beyond me and I believe it is beyond the scope of the course. I guess I'm missing something in the process here?
| You've correctly set up the integral representing the volume of the inside cylinder. The scalar $4$ on the left represents
*
*the reflectional symmetry about the $xy$-plane; and
*the reflectional symmetry about the $yz$-plane.
\begin{align}
& 4 \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \ r \ dr \ d \theta \\
=& 2 \int_0^{\pi/2} \int_0^{2\cos\theta} (\sqrt{4-r^2}) \, d(r^2) \ d \theta \\
=& 2 \int_0^{\pi/2} \left[- \frac{(4-r^2)^{3/2}}{3/2} \right]_0^{2\cos\theta} \, d(r^2) \ d \theta \\
=& \frac{32}{3} \int_0^{\pi/2} (1-\sin^3\theta) \, d \theta \\
=& \frac{32}{3} \cdot \frac{\pi}{2} - \frac{32}{3} \cdot \frac{2}{3} \\
=& \frac{16(3\pi - 4)}{9}
\end{align}
The second last equality is due to Wallis's integrals.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Exact value of $100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$ Problem
Find the exact value of $$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)$$
What I tried :
multiply $(n-1)$ to sumnation's numerator and denominator then it changed to
$$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^3-1)}{n(n-2)!}\right)$$
But this didn't give me any clues to solve this.
I just guess telescoping is the way to solve because this is finite-sum, but I don't have any idea.
| Want
$100!\times\left(1+\sum_{n=1}^{100} \frac{(-1)^n (n^2+n+1)}{n!}\right)
$
Going up to $m$,
$\begin{array}\\
\sum_{n=1}^{m} \dfrac{(-1)^n (n^2+n+1)}{n!}
&=\sum_{n=1}^{m} \dfrac{(-1)^n (n^2-n+2n+1)}{n!}\\
&=\sum_{n=1}^{m} \dfrac{(-1)^n (n^2-n+2n+1)}{n!}\\
&=\sum_{n=1}^{m} (-1)^n(\dfrac{ n^2-n}{n!}+\dfrac{2n}{n!}+\dfrac{ 1}{n!})\\
&=\sum_{n=1}^{m} (-1)^n\dfrac{ n^2-n}{n!}+2\sum_{n=1}^{m} (-1)^n\dfrac{ n}{n!}+\sum_{n=1}^{m} (-1)^n\dfrac{ 1}{n!}\\
&=\sum_{n=2}^{m} (-1)^n\dfrac{ n(n-1)}{n!}+2\sum_{n=1}^{m} (-1)^n\dfrac{ 1}{(n-1)!}+\sum_{n=1}^{m} (-1)^n\dfrac{ 1}{n!}\\
&=\sum_{n=2}^{m} (-1)^n\dfrac{ 1}{(n-2)!}+2\sum_{n=0}^{m-1} (-1)^{n+1}\dfrac{ 1}{n!}+\sum_{n=1}^{m} (-1)^n\dfrac{ 1}{n!}\\
&=\sum_{n=0}^{m-2} (-1)^n\dfrac{ 1}{n!}-2\sum_{n=0}^{m-1} (-1)^{n}\dfrac{ 1}{n!}+\sum_{n=1}^{m} (-1)^n\dfrac{ 1}{n!}\\
&=1+\sum_{n=1}^{m-2} (-1)^n\dfrac{ 1}{n!}-2-2\sum_{n=1}^{m-2} (-1)^{n}\dfrac{ 1}{n!}-2\dfrac{(-1)^{m-1}}{(m-1)!}\\
&\quad +\sum_{n=1}^{m-2} (-1)^n\dfrac{ 1}{n!}+\dfrac{(-1)^{m-1}}{(m-1)!}+\dfrac{(-1)^m}{m!}\\
&=-1-2\dfrac{(-1)^{m-1}}{(m-1)!}+\dfrac{(-1)^{m-1}}{(m-1)!}+\dfrac{(-1)^m}{m!}\\
&=-1+\dfrac{(-1)^{m}}{(m-1)!}+\dfrac{(-1)^m}{m!}\\
\end{array}
$
so
$\begin{array}\\
m!\left(1+\sum_{n=1}^{m} \frac{(-1)^n (n^2+n+1)}{n!}\right)
&=m!(1+(-1+\dfrac{(-1)^{m}}{(m-1)!}+\dfrac{(-1)^m}{m!})\\
&=(-1)^{m}m+(-1)^m)\\
&=(-1)^{m}(m+1)\\
\end{array}
$
For $m=100$,
this is
$101$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3221253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^4}$
Evaluate $\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$
The answer stated is $\displaystyle {1 \over 6}$.
What I've tried:
$$\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{\cos (\sin x) -1+1- \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{1- \cos x}{x^4} - \frac {1-\cos (\sin x)}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{2 \sin^2(\frac {x}{2})}{x^4} - \frac {2 \sin^2(\frac {\sin x}{2})}{x^4}$$
$$=\displaystyle \lim_{x\to0} \left(\frac{\sin(\frac {x}{2})}{x} \right)^2. \left( \dfrac{1}{2x^2} \right) - \frac {2 \sin^2(\frac {\sin x}{2})}{x^4}$$
I'm not sure how I can evaluate the limit by proceeding this way. All help will be appreciated.
P.S. I'd prefer not using L'Hôpital's rule, it can get really messy.
EDIT: I should have mentioned that I would prefer if the solution does not use taylor series approximations (or any approximations) for that matter.
| By using trigonometry identity and Taylor series,\begin{align} &\lim_{x\to0} \frac{\cos (\sin x)-\cos(x)}{x^4}\\
&=-2\lim_{x\to 0}\frac{\sin \left( \frac{\sin x - x}2\right) \sin\left(\frac{\sin x + x}2 \right)}{x^4}\\
&= -2\lim_{x \to 0 }\frac{\sin \left( \frac{-x^3}{2(6)}\right)\sin \left( \frac{x+x}{2}\right)}{x^4}\\
&= -2 \lim_{x \to 0}\frac{-\frac{x^3}{6(2)}\cdot x}{x^4}\\
&= \frac16\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find antiderivative $\int (2x^3+x)(\arctan x)^2dx $
Find antiderivative $$\int (2x^3+x)(\arctan x)^2dx $$
My try:
$$\int (2x^3+x)(\arctan x)^2dx =(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int \frac{2\arctan x}{1+x^2}(\frac{1}{2}x^4+\frac{1}{2}x^2)dx=(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int (\arctan x) (x^2)dx= (\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\arctan x\cdot \frac{1}{3}x^3-\int (\frac{1}{1+x^2}) (\frac{1}{3}x^3)dx$$And then I don't know how I can find $\int (\frac{1}{1+x^2}) (\frac{1}{3}x^3)dx$. Can you help me with it?
| Let $t=x^2$. Your integral is $\int \frac{1}{6}\frac{tdt}{t+1}=\int (1-\frac{1}{t+1})dt=\frac{t}{6}-\int\frac{1}{t+1}dt$. Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
$$ $$
Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x = a + \sqrt{b}, z = c$. We can find $y$ as
$$ y = \sqrt{c^{2}- (a + \sqrt{b})^{2}} $$
so we have
$$ \cos(18) = \frac{y}{z} = \frac{\sqrt{c^{2}- (a + \sqrt{b})^{2}}}{c}$$
I also found out that
$$b = (c \sin(18) - a)^{2} = c^{2} \sin^{2}(18) - 2ac \sin(18) + a^{2}$$
I got no clue after this.
The solution says that $$ \sin(18) = \frac{-1 + \sqrt{5}}{4} $$
I gotta intuition that we must find $A,B,C$ such that
$$ A \sin(18)^{2} + B \sin(18) + C = 0 $$
then $\sin(18)$ is a root iof $Ax^{2} + Bx + C$, and $a = -B, b = B^{2} - 4AC, c = 2A$.
Totally different. This question is not asking to prove that $sin(18)=(-1+\sqrt{5})/4$, that is just part of the solution.
|
In the figure, let $AB=AC=x$. Note that $AD=CD=2$. As $\triangle ABC\sim\triangle CDB$, $\dfrac x2=\dfrac 2{x-2}$
So, $x^2-2x-4=0$ and hence $x=1+\sqrt{5}$.
$\sin18^\circ=\dfrac 1x=\dfrac{\sqrt{5}-1}4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3230730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Why is this integration method not valid? Let $$I=\int \frac{\sin x}{\cos x + \sin x}\ dx \tag{1}$$
Now let $$u=\frac{\pi}{2} - x \tag{2}$$ so $$I=\int \frac{\sin (\frac{\pi}{2} - u)}{\cos (\frac{\pi}{2} - u)+\sin (\frac{\pi}{2} - u)}\ du \tag{3}$$
$$=\int\frac{-\cos u}{\sin u + \cos u} \ du \tag{4}$$
$$= \int\frac{-\cos x}{\sin x + \cos x} \ dx \tag{5}$$
and hence $$2I=\int\frac{\sin x - \cos x}{\sin x + \cos x} \ dx \tag{6}$$
$$=-\ln\ |\sin x + \cos x| + c \tag{7}$$
$\implies I=-\frac{1}{2}\ln|\sin x + \cos x| + c \tag{8}$
But the actual answer is $$I= \frac{1}{2}x -\frac{1}{2}\ln|\sin x + \cos x| + c \tag{9}$$
according to Wolfram Alpha and supported by a different method.
Why does my method not yield the correct result?
| $$\int_{a}^{b}\frac{\sin \ x}{\cos \ x + \sin \ x}dx\neq \int_{a}^{b}\frac{-\cos \ x}{\cos \ x + \sin \ x}dx$$
but
$$\int_{a}^{b}\frac{\sin \ x}{\cos \ x + \sin \ x}dx= \int_{\pi/2-a}^{\pi/2-b}\frac{-\cos \ x}{\cos \ x + \sin \ x}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3232250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 6,
"answer_id": 2
} |
integration of a gaussian with $x^2$ I need to integrate
$$\int_{-\infty}^{\infty} x^2 e^{-ax^2} \qquad \text{where } a\in R$$
The book does the following:
I don't understand what's happening. I tried solving the integral using integration by parts and this is what I got
$$
\begin{align}
\int_{-\infty}^{\infty} x^2 e^{-ax^2} &= x^2\sqrt{\frac{\pi}{a}} - \int_{-\infty}^{\infty} \sqrt{\frac{\pi}{a}} 2x dx && \text{as we are told } \int_{-\infty}^{\infty}dx e^{-ax^2}= \sqrt{\frac{\pi}{a}}\\
&=x^2\sqrt{\frac{\pi}{a}} - 2\sqrt{\frac{\pi}{a}}\frac{x^2}{2} && \text{as } \int xdx = \frac{x^2}{2} \\
&= 0
\end{align}
$$
What am I doing wrong?
Actually, I might have found a way of solving this
$$
\begin{align}
\int_{-\infty}^{\infty} x^2 e^{-ax^2} dx &= \int_{-\infty}^{\infty} x\cdot x e^{-ax^2} dx \\
&= - \frac{1}{2a} \int_{-\infty}^{\infty} x(-2ax)e^{-ax^2} dx \\
&= -\frac{1}{2a}\left(\left[e^{-ax^2}\right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} e^{-ax2} dx\right) \\
&= -\frac{1}{2a}\left(0 - \sqrt{\frac{\pi}{a}}\right) \\
&= \frac{1}{2a}\sqrt{\frac{\pi}{a}}
\end{align}
$$
| note that the derivative of $e^{-ax^2}$ is $-2ax e^{-ax^2}$, so
$$ x^2 e^{-ax^2} = \frac{-1}{2a}x \,(-2ax e^{-ax^2}) = \frac{-1}{2a}x\, \frac{d}{dx} e^{-ax^2}$$
Insert this into the integral and integrate by parts.
| {
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"url": "https://math.stackexchange.com/questions/3233400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Next Term Of Strange Sequence I tutored a 10th grader and I was asked this puzzle and I had spent nearly an hour with it and got “no where”. Any one can crack it? Please let me know. Thank you.
Question: Find the $14$ th term of the sequence: $$ \frac{1}{2}, \frac{3}{7}, \frac{1}{3}, \frac{5}{19}, \frac{3}{14}, .... $$.
| $\displaystyle\frac24\ ,\ \frac37\ ,\ \frac4{12}\ ,\ \frac5{19}\ ,\ \frac6{28}\ ,...$
differences between denominators are $3,5,7,9,...$
(and numerators are consecutive integers $2,3,4,5,6,...$ )
$\displaystyle\frac24,\frac37,\frac4{12},\frac5{19},\frac6{28}, \\
\displaystyle\frac7{39},\frac8{52},\frac9{67},\frac{10}{84},\frac{11}{103},\\
\displaystyle\frac{12}{124},\frac{13}{147},\frac{14}{172},\frac{15}{199},\frac{16}{128}$
elementary watson, $\displaystyle a_{14}=\frac{15}{199}$ .
More generally, $\displaystyle a_n = \frac{n+1}{n^2+3}$ .
Using the formula, $\displaystyle a_{14}=\frac{14+1}{14^2+3}=\frac{14+1}{196+3}=\frac{15}{199}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me please
| Upon completing the square the equation $$(x^2+6x+7)^2+6(x^2+6x+7)+7=x$$
$$(x^2+6x+7)^2+6(x^2+6x+7)+9=x+2$$
$$[(x^2+6x+7)+3]^2=x+2$$
Which simplifies to $$[(x+3)^2+1]^2=x+2$$ which has no real solutions.
Upon comparison of derivatives for $x>-2$, we see that function on the LHS is always above the one on the RHS so they do not meet at any real values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Rotation Matrix and Triple Angle Formulas? Define $R_{\theta}:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ as the rotation matrix by angle $\theta$, where
$$R_{\theta} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$
Observe that
$$
(R_\theta)^2 = \begin{pmatrix} \cos^2\theta-\sin^2\theta & -2\sin\theta\cos\theta \\ 2\sin\theta\cos\theta & \cos^2\theta-\sin^2\theta \end{pmatrix}
= \begin{pmatrix} \cos2\theta & -\sin2\theta \\ \sin2\theta & \cos2\theta \end{pmatrix}=R_{2\theta}
$$
This all makes sense of course since if you rotate a vector by $\theta$ twice, the net result should be a rotation by $2\theta$. The algebra of it all can be verified with the double angle formulas.
However, how do you prove that
$$
(R_\theta)^3 = R_{3\theta}
$$
or perhaps that
$$
(R_\theta)^n = R_{n\theta}
$$
Are there triple angle formulas that can be used to make the algebra work? n-tuple angle formulas?
| You may use these identities
$\cos x \cos y - \sin x \sin y = \cos(x+y)$
$\sin x \cos y + \cos x \sin y = \sin(x+y)$
and use $\theta $ and $2\theta$ in place of $x$ and $y$. For $R_{n\theta}$, try proving it by induction, assuming $(R_{\theta})^n = R_{n\theta}$ to be true and find $(R_{\theta})^{n+1}$
| {
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"url": "https://math.stackexchange.com/questions/3236455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
Evaluate the following integral:
$ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
I've tried to apply Chebyshev theorem on the integration of binomial differentials.
We have $ m=4,a=2,b=-1,n=2,p=-3/2$.
$\frac{m+1}{n}+p$ is integer then we do the substitution
$t^2=2x^{-2}-1$, $x^2=\frac{2}{t^2+1}$
It go me there: $\int\frac{2}{t^2+1}^2(2-\frac{2}{t^2+1})^{-3/2}-\sqrt{2}t(\frac{1}{t^2+1})^{3/2}dt$ which is just more complicated expression. Where I went wrong?
| Another way:
Integrate by parts
$$\int\dfrac{x^4}{(2-x^2)^{3/2}}dx=\int x^3\cdot\dfrac x{(2-x^2)^{3/2}}dx$$
$$=x^3\int\dfrac x{(2-x^2)^{3/2}}dx-\int\left[\dfrac{d(x^3)}{dx}\int\dfrac x{(2-x^2)^{3/2}}dx\right]dx$$
Finally, $$\dfrac{x^2}{\sqrt{2-x^2}}=\dfrac{x^2-2+2}{\sqrt{2-x^2}}=\dfrac2{\sqrt{2-x^2}}-\sqrt{2-x^2}$$
Use $\#1,\#8$ of INTEGRALS CONTAINING THE SQUARE ROOT OF $a^2-x^2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Identity of tan(x) I came across the following formulas for analytical expressions of fundamental modes of asymmetric dielectric waveguide. $$ \tan(x) = x\frac{\pi^2-x^2}{\pi^2-4x^2} $$
This approximation is not present in Abramowitz's handbook. It has the same poles and zeros as $\tan(x)$.
Can someone guide me on its merits or where to find it. This doesn't seem like an identity to me.
| There are several ways to think about approximating $\tan x$ as a rational function.
Taking the log-derivative of $\sin x=x\prod_{k\ge 1}\left(1-\frac{x^2}{k^2\pi^2}\right)$ gives $$\cot x=\frac{1}{x}-2\sum_{k\ge 1}\frac{x}{k^2\pi^2-x^2}.$$Keeping only the $k=1$ term,$$\tan x\approx\frac{x(\pi^2-x^2)}{\pi^2-3x^2}.$$This is a reasonable approximation if $|x|\ll\pi$, so the right-hand side approximates$$x\left(1-\frac{x^2}{\pi^2}\right)\left(1+\frac{3x^2}{\pi^2}\right)\approx x+\frac{2}{\pi^2}x^3.$$However, the $3\mapsto 4$ replacement changes the $x^3$ coefficient to $\frac{3}{\pi^2}$, a much closer approximation of the exact result $\frac13$ (the Taylor series of $\tan x$ of course continues past the $x^3$ term).
A simpler motive for $\frac{\tan x}{x}\approx\frac{\pi^2-x^2}{\pi^2-4x^2}$ is that the right-hand side is an odd function and approximates $1$ for small $x$, diverges at $x=\pm\frac{\pi}{2}$ and vanishes at $x=\pm\pi$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Calculate $\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}$ The question:\,
Calculate $$\int_{-\infty}^\infty{x^2\,dx\over (1+x^2)^2}.$$
Book's final solution: $\dfrac\pi 2$.
My mistaken solution: I don't see where is my mistake because my final solution is $-\dfrac{\pi i}2$:
$$\begin{align}
\text{A}:\int_{-\infty}^\infty {x^2\over (1+x^2)^2}dx &= 2\int_{0}^\infty {x^2\over (1+x^2)^2}dx \quad\text{as the function is even}\\&=
-2\left(\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,i\right)+\text{Res}\left({z^2\over (1+z^2)^2}\cdot \ln(z) ,-i\right)\right)\end{align}
$$
Calculate the residues: Since $\pm i$ are poles of order $2$, then
$$\begin{align}\text{B}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,i\right)&={(2z\ln z+z^2\cdot z^{-1})(z+i)^2-z^2\ln z\cdot 2(z+i)\over (z+i)^4} \\&= \frac{(2i\ln i+i)(-4)+\ln i\cdot 4i }{16} \\ &= {1\over 16} (-8i\ln i -4i+4i\ln i)={-1\over 4}i(\ln i+1)\end{align}
$$
whereas
$$\begin{align}\text{C}:\text{Res}\left({z^2\ln z\over (1+z^2)^2} ,-i\right)&={(2z\ln z+z^2\cdot z^{-1})(z-i)^2-z^2\ln z\cdot 2(z-i)\over (z-i)^4} \\ &=\frac{(-2i\ln(- i)-i)(-4)+\ln (-i)\cdot (-4i)}{16} \\ &= {1\over 16} (8i\ln (-i) +4i-4i\ln(- i))={1\over 4}i(\ln (-i)+1)\end{align}
$$
Combining $\text A$, $\text B$ and $\text C$, we get
$$
\int_{-\infty}^\infty {x^2\over (1+x^2)^2}\,dx=-2\left({i\over 4}(\ln(-i)+1)-{i\over 4}(\ln i+1)\right)={-i\over 2}\left({3\pi\over 2}-{\pi\over 2}\right)={-\pi i \over 2}
$$
Where is my mistake? Thanks in advance!
| If we take the contour $\sf C$ to be the upper half-plane (with $\sf{\Im z>0}$), then we can write the integrand as $$\sf{\frac{z^2}{(z^2+1)^2}=\frac{\left(\frac{z}{z+i}\right)^2}{(z-i)^2}}$$ since the singularity in $\sf C$ is at $\sf{z=i}$ (of order two). Thus, using Cauchy's Integral Formula, we obtain $$\sf{\int_{-\infty}^\infty{x^2\over (1+x^2)^2}\,dx=\oint_C\frac{\left(\frac{z}{z+i}\right)^2}{(z-i)^2}\,dz=2\pi i\frac d{dz}\left[\left(\frac{z}{z+i}\right)^2\right]_{z=i}}=2\pi i\cdot\frac{2i\cdot i}{(i+i)^3}=\frac\pi2$$ as required. In your solution, there is no reason to introduce the natural logarithm, as it is not even part of the integrand!
| {
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"url": "https://math.stackexchange.com/questions/3239096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimum value of $\frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$ where $a, b > 0$ and $a + b \le 1$.
$a$ and $b$ are two positives such that $a + b \le 1$. Find the minimum value of $$\large \frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$$
We have that $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} = \dfrac{a^2 + b^2 + a + b}{ab + a + b + 1} \ge \dfrac{\dfrac{(a + b)^2}{2} + (a + b)}{\dfrac{(a + b)^2}{4} + 2(a + b)} = \dfrac{2(a + b) + 4}{(a + b) + 8}$
$\implies \dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b} \ge \dfrac{2(a + b) + 4}{(a + b) + 8} + \dfrac{1}{a + b}$
Let $x = 1 - (a + b) \implies x \ge 0$.
Thus $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b} \ge \dfrac{2(1 - x) + 4}{(1 - x) + 8} + \dfrac{1}{1 - x} = \dfrac{2x^2 - 9x + 15}{x^2 - 10x + 9}$
$= \dfrac{x(x + 23)}{3(9 - x)(1 - x)} + \dfrac{5}{3} \ge \dfrac{5}{3}$
The equality sign happens when $a = b$ and $x = 0$ or $a + b = 1 \implies a = b = \dfrac{1}{2}$.
I want to ask if the above solution is correct and if there are any other solutions that are more rational and reasonable.
| $\large \dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b}$
Taking $a+b = c$ for some c; $0<c\leq 1$, and expressing $b=c-a$; the above expression becomes (after modification)
$-2 + \{2(c+1) +(c^2+c)\}(\dfrac{1}{-a^2+ac+(c+1)} ) +1/c$
here only variable is a ; c is assumed constant.
Differentiating it and equating it to zero we get
$0+\{2(c+1) +(c^2+c)\}(-\dfrac{-2a+c}{(-a^2+ac+(c+1))^2} ) +0 = 0$
From this we get $a = c/2 = b$, using $a=b$ from here we get
$\dfrac{2a}{a+1}+\dfrac{1}{2a} = \dfrac{4a^2+a+1}{2a^2+2a} = 2 - \dfrac{3}{2a}+\dfrac{2}{a^2+a}$ again differentiatig wrt a and setting it to zero we get
$3a^2-2a-1 = 0$, from this we get $a = 1,-1/3$,
Now let $a = 1-d$ then $\dfrac{2a}{a+1}+\dfrac{1}{2a} = \dfrac{2(1-d)}{2-d}+\dfrac{1}{2(1-d)} = 2 +\dfrac{3d-2}{2(1-d)(2-d)}$ it is very clear from here that above function is strictly increasing from d = 0 to d = 1; or strictly decreasing from a = 0 to a = 1, but the maximum value of "a" possible here is $a=b = 1/2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$. I need to show that $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$.
This happens if $(x+1)\ln\Big(1 + \frac 2 {2x+1}\Big) > 1$ so let's study the function $f(x) = (x+1)\ln\Big(1 + \frac 2 {2x+1}\Big)$
Edit
We have abandoned the previous method of only using a characterization of $e$ with inequalities since we thought it was impractical to do it. Let's return instead to analysis.
*
*$f(1) = 2 \ln(\frac 5 3) > 1$
*$f'(x) = \ln(1+ \frac 2 {2x+1}) - \frac {4(x+1)}{(4x^2+8x+3)} < 0$ on $[1,+\infty[$
*$\lim f(x) = 1$
However how do I show formally 2.?
| Consider the function $f(y):=(1+\frac{2}{y})^{y+1}$. Clearly, $\lim_{y\rightarrow\infty}f(y)=e^2$. Moreover, $f$ is monotone decreasing since
\begin{align*}
f'(y)<0 &\iff \left({1+\frac{2}{y}}\right)\log\left({1+\frac{2}{y}}\right) < \left({\frac{2}{y}}\right)+\frac{1}{2}\left({\frac{2}{y}}\right)^2 \\
& \iff (1+t)\log(1+t) < t+\frac{1}{2}t^2 \quad \text{for }t=\frac{2}{y},
\end{align*}
where the last inequality can be verified for all $t> 0$ since equality holds for $t=0$ and the right-hand side grows faster than the left-hand side (just look at the derivatives). Choosing $y=2x+1$, we can thus deduce
\begin{align*}
e^2<\Big(1+\frac{2}{2x+1}\Big)^{2x+2}.
\end{align*}
Taking the square root, we conclude
\begin{align*}
e<\Big(1+\frac{2}{2x+1}\Big)^{x+1}.
\end{align*}
Edit: Modified due to the oversight pointed out by GFauxPas.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating factor for $(x^2-y^2-y)dx-(x^2-y^2-x)dy=0$ I am having trouble finding the integrating factor for turning the below differential equation into an exact one (Tenenbaum and Pollard, exercise 10, problem 6). Any hints and suggestions would extremely helpful and lead me to the solution.
Solve the differential equation :
$$ (x^2-y^2-y)dx - (x^2-y^2-x)dy=0$$
My attempt:
The coefficients of $dx$ and $dy$ are not homogenous functions. Further,
$
\begin{align}
P(x,y) &= x^2 - y^2 - y \\
\frac{\partial P(x,y)}{\partial y}&=-2y-1 \\
Q(x,y) &= -(x^2-y^2-x)\\
\frac{\partial Q(x,y)}{\partial x}&=-(2x-1) \\
\therefore \frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}
\end{align}
$
The given differential equation is not exact. We have :
$\begin{align}
& \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \\
=& -2y-1+(2x-1)\\
=&(2x-2y-2)
\end{align}$.
Moreover,
$\begin{align}
& yQ-xP\\
= & -y(x^2-y^2-x)-x(x^2-y^2-y)\\
= & -x^2 y + y^3 + xy - x^3 + xy^2 + xy\\
= & y^3 - x^3 + xy - xy(x - y - 1)
\end{align}$
It doesn't look like $(\partial P / \partial y - \partial Q / \partial x)/(yQ-xP)$ will be a function of $u=xy$ alone.
Also,
$\begin{align}
& yQ+xP\\
= & -y(x^2-y^2-x)+x(x^2-y^2-y)\\
= & -x^2 y + y^3 + xy + x^3 - xy^2 - xy\\
= & y^3 + x^3 - x^2 y - x y^2 \\
= & (y + x)(y^2 + x^2 - xy) - xy(y + x) \\
= & (y + x)(y^2 + x^2 - 2xy)\\
= & (y + x)(y - x)^2
\end{align}$
It doesn't look like $y^2(\partial P / \partial y - \partial Q / \partial x)/(yQ+xP)$ will be a function of $u=x/y$ or $x^2(\partial P / \partial y - \partial Q / \partial x)/(yQ+xP)$ will be a function of $u=y/x$ alone.
| The given differential equation is not exact and I think you can't find the integrating factor by known way and an easier way, rather you can solve it as follows
${}$
$(x^2-y^2-y)dx - (x^2-y^2-x)dy=0$
$\implies (x^2-y^2)(dx-dy)+xdy-ydx=0$
$\implies (1-\frac{y^2}{x^2})(dx-dy)+\frac{xdy-ydx}{x^2}=0$
$\implies (1-\frac{y^2}{x^2})(dx-dy)+d(\frac{y}{x})=0$
$\implies (dx-dy)+ \frac{d(\frac{y}{x})}{1-\frac{y^2}{x^2}}=0$
Integrating,
$$x-y+\tanh^{-1} (\frac{y}{x}) =c \quad \text{when} \quad |y|\lt|x|$$, or in other form $$x-y+\log|\frac{1+\frac{y}{x}}{1-\frac{y}{x}}|=c$$ where $c$ is integrating constant.
| {
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"url": "https://math.stackexchange.com/questions/3247724",
"timestamp": "2023-03-29T00:00:00",
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Solve $x^2+5x+6 \equiv 0 \pmod{\!11\cdot 17}$ Solve $x^2+5x+6 \equiv 187 \mod 187$
Solution
$$x^2+5x+6 \equiv 187 \mod 187$$
$$ (x+\frac{5}{2})^2 \equiv \frac{1}{4}$$
$$ 4(x+\frac{5}{2})^2 \equiv 1$$
$$ y:= x+\frac{5}{2} $$
$$ 4y^2 \equiv 1 \mod 11 \wedge 4y^2 \equiv 1 \mod 17 $$
$$ ( 2y \equiv 1 \mod 11 \vee 2y \equiv 10 \mod 11 ) \wedge ( 2y \equiv 1 \mod 17 \vee 2y \equiv 13 \mod 17) $$
$$ ( y \equiv 6 \mod 11 \vee y \equiv 5 \mod 11 ) \wedge ( y \equiv 9 \mod 17 \vee y \equiv 15 \mod 17) $$
Combining that from CRT I got:
$$ y \in \left\{49, 60,83,94 \right\} $$
and for example:
$$ x+\frac{5}{2} \equiv 94 \mod 187$$
$$ 2x \equiv 183 \mod 187$$
some calculus and get...
$$x \equiv 185 $$
And the same thing for each other case.
Question
Is there any faster (or smarter) way to solve equations like that?
| $$x^2+5x+6\equiv187\equiv0 \pmod {187=11\times17}$$
$$(x+2)(x+3)\equiv 0 \pmod {11 , 17}$$
$$x\equiv-2 \text { or } -3 \pmod {11, 17}$$
Now use the Chinese Remainder Theorem.
| {
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Find the limit of a multiplying term function when n tends to infinity. How do I evaluate the limit,
$$\lim_{n \to \infty} \ \ \ \left(1-\frac1{2^2}\right)\left(1-\frac1{3^2}\right)\left(1-\frac1{4^2}\right)...\left(1-\frac1{n^2}\right)$$
I tried to break the $n^{\text{th}}$ term into $$\frac{(n+1)(n-1)}{n.n}$$ and then tried to do something but couldn't get anywhere. Please help as to how to solve such questions in general. Thanks.
| The following will not be of help for such questions "in general", but it is too long for a comment.
Following Euler, we have
$$
\begin{align}
\frac{\sin x}{x} &=
\left( 1 - \frac{x}{\pi} \right) \left( 1 + \frac{x}{\pi} \right)
\left( 1 - \frac{x}{2 \pi} \right) \left( 1 + \frac{x}{2 \pi} \right)
\left( 1 - \frac{x}{3 \pi} \right) \left( 1 + \frac{x}{3 \pi} \right)
\dots \\
&=
\left[ 1 - \left( \frac{x}{\pi} \right)^2 \right]
\left[ 1 - \frac{1}{2^2} \left( \frac{x}{\pi} \right)^2 \right]
\left[ 1 - \frac{1}{3^2} \left( \frac{x}{\pi} \right)^2 \right]
\dots,
\end{align}
$$
or
$$
\frac{\sin x}{x} \cdot \frac{1}{1 - (x/\pi)^2} =
\left[ 1 - \frac{1}{2^2} \left( \frac{x}{\pi} \right)^2 \right]
\left[ 1 - \frac{1}{3^2} \left( \frac{x}{\pi} \right)^2 \right]
\dots.
$$
Hence the required product is
$$
\begin{align}
\lim_{x \to \pi} \frac{\sin x}{x} \cdot \frac{1}{1 - (x/\pi)^2}
&= \lim_{\epsilon \to 0} \frac{\sin \epsilon}{\pi - \epsilon} \cdot \frac{1}{1 - ([\pi - \epsilon]/\pi)^2} \\
&= \lim_{\epsilon \to 0} \frac{\epsilon}{\pi} \cdot \frac{1}{2 \pi \epsilon / \pi^2} \\
&= \frac{1}{2}.
\end{align}
$$
| {
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"url": "https://math.stackexchange.com/questions/3250940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Calculate the maximum value of $\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$ where $3a + 4b + 5c = 12$
$a$, $b$ and $c$ are positives such that $3a + 4b + 5c = 12$. Calculate the maximum value of $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$$
I want to know if there are any other solutions that are more practical. This came from my homework my teacher gave today. I have given my solution down below if you want to check out.
| Also, we can use C-S and AM-GM:
$$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}\leq$$
$$\leq\frac{ab}{(1+2)^2}\left(\frac{1^2}{ab}+\frac{2^2}{a+b}\right)+\frac{2ca}{9}\left(\frac{1}{ca}+\frac{4}{c+a}\right)+\frac{3bc}{9}\left(\frac{1}{bc}+\frac{4}{b+c}\right)=$$
$$=\frac{2}{3}+\frac{4}{9}\cdot\frac{ab}{a+b}+\frac{8}{9}\cdot\frac{ca}{c+a}+\frac{4}{3}\cdot\frac{bc}{b+c}\leq$$
$$\leq\frac{2}{3}+\frac{4}{9}\cdot\frac{a+b}{4}+\frac{8}{9}\cdot\frac{c+a}{4}+\frac{4}{3}\cdot\frac{b+c}{4}=\frac{2}{3}+\frac{3a+4b+5c}{9}=2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If the HCF of the polynomials $x^3+px+q $ and $x^3+rx^2+lx+x$ is $x^2+ax+b$, then their LCM is? (provided that $r≠0$) I tried multiplying the two polynomials together and then dividing them by the HCF (as product=HCF*LCM), but reached nowhere.
Then, I used the factor theorem but also got stuck.
Can somebody help?
| If $b \neq 0$,
$$x^2+ax+b \mid x(x^2+rx+l+1) $$
This would imply that $x^2+ax+b = x^2+rx+l+1 $.
And the LCM would be,
$$\frac{x(x^2+rx+l+1)(x^3+px+q)}{x^2+ax+b}=x^4+px^2+q$$
If $b=0$,
$$x+a|x^2+rx+l+1$$
if you divide you'll find the quotient to be $x+(r-a)$ and remainder to be $l+1-a(r-a)=0$
LCM is,
$$(x^3+px+q)(x+r-a)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$ and $k \le 2m+1$.
My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}.$$ I need prove that for all real $x \ge 1$ and for all positive integers $m$ the inequality $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$$ holds.
| By the power mean inequality, with $x\geq 0, n \geq k$, we have
$$ \left( \frac{ x^n + 1 } {2} \right) ^\frac{1}{n} \geq \left( \frac{x^k + 1 } { 2} \right) ^ \frac{1}{k}. $$
Hence, all that we need to show is:
$$ \frac{ x^{m+1} + 1 } { x^m + 1} \geq \left( \frac{ x^{2m+1 } + 1 } {x^0 + 1 } \right) ^ \frac{1}{2m+1} \quad (1) . $$
We will show that
$$ \left(\frac{1+x^{m+1}}{1+x^m}\right)^2 \geq \frac{1+x^{m+1+i}}{1+x^{m+i}} \times \frac{1+x^{m+1-i}}{1+x^{m-i}} \quad (2) $$
By cross multiplying and expanding terms (Thanks Wolfram), this is equivalent to:
$$ x^{m-i} (x^i-1)^2(x-1)(x^{2m+1} - 1) \geq 0. $$
This is obviously true for $ x \geq 0$.
(Easier to see by splitting up into cases $x\geq 1, x < 1$, which might explain where that condition came from.)
Then, (1) follows by multiplying the chain of inequalities of (2), going from $i=0$ to $m$.
Note: The (pseudo-)reasoning behind why (2) should be true, is that when we write it as $\prod ( 1+ x^{a_i}) \geq \prod ( 1+ x^{b_i})$, with $a_i, b_i$ in decreasing order, we have $ \sum a_i = \sum b_i$. Since $a_1 = m+i < m+i+1 = b_1$, it follows that after cancelling the $x^{\sum a_i}$ term on both sides, the LHS has a larger exponent, so it will be true for large enough $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
This has become the norm now... This problem is adapted from a recent competition.
We have that $6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) \ge 4(x + y + z)(xy + yz + zx)$
Furthermore, $$(x + y + z)^3 - 6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 9xyz$$
$$ = x^3 + y^3 + z^3 - 3(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 15xyz$$
In addition, $x^3 + y^3 + z^3 + 3xyz \ge x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2$.
So now we need to prove that $6xyz \ge x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2$, which I don't even know if it is true or not.
| If we write $x+y+z=1$ (we can do that) then we have to prove $$1+9xyz\geq 4(xy+yz+zx)$$ Clearly one of $x,y,z$ is smaller than $4/9$, we can assume $y<4/9$.
If we put $z=1-x-y$ we get $$0\geq (9y-4)x^2+x(9y^2-13y+4)-(2y-1)^2=:f(x)$$
Since the discirminant for $f(x)$ is: $$\Delta= (9y-4)y(3y-1)^2$$
is also nonegative for all $y<4/9$ we are done.
| {
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"url": "https://math.stackexchange.com/questions/3254924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate $\iiint_\omega (x+y+z)^2\,dxdydz$. Problem: Calculate $$\displaystyle\iiint_\Omega (x+y+z)^2\,dxdydz,$$ where $\Omega$ is the domain bounded by the sphere $x^2+y^2+z^2\le 3a^2$ and the paraboloid $x^2+y^2\le 2az$, where $z\ge 0$.
Progress: So, by sketching the domain, one can see that the domain includes a part from sphere and a part from the paraboloid, so we need to calculate the integral on two seperate part, then add the to get the result.
To calculate the integral in the sphere part, I've tried using spherical coordinate. Using spherical coordinate, I made the following substitution:
$$\begin{cases} x = r\sin\theta\cos\varphi\\ y = r\sin\theta\sin\varphi\\ z = r\cos\theta \end{cases}$$ with $0\le \varphi\le 2\pi$, $0\le\theta\le\alpha$, where $\alpha =\arccos\left(\dfrac{1}{\sqrt{3}}\right)$ (one can get the bound for angle $\theta$ by working out the intersection of the paraboloid and the sphere) and $\dfrac{a}{\cos\theta}\le r\le\dfrac{a\sqrt 2}{\sin\theta}$. Now, the expression $(x+y+z)^2$ become
$$r^2(1+\sin^2\theta\sin 2\varphi + \sin 2\theta+\sin\varphi+\sin 2\theta\cos\varphi).$$
Now, we multiply the Jacobian determinant and the original integral become
$$\int^{2\pi}_0d\varphi\int^\alpha_0d\theta\int^\tfrac{a\sqrt 2}{\sin\theta}_\tfrac{a}{\cos\theta}r^4\sin^2\theta(1+\sin^2\theta\sin 2\varphi + \sin 2\theta+\sin\varphi+\sin 2\theta\cos\varphi)dr.$$
As you can see, this one is tedious and almost impossible to calculate it. Do you have any idea to solve this problem?
| Note that $(x+y+z)^2 = x^2+y^2+z^2 +2(xy+yz+zx)$ and, due to the symmetry of the integration region, the integral simplifies to
$$I=\iiint_\Omega (x+y+z)^2\,dxdydz
= \iiint_\Omega (x^2+y^2+z^2)\,dxdydz
$$
Then, in cylindrical coordinates, the integration region is the circular area of radius $r=\sqrt 2 a$ and the integral is given by
$$I= \int_0^{2\pi}\int_0^{\sqrt2 a}\int_{\frac{r^2}{2a}}^{\sqrt{3a^2-r^2}} (r^2+z^2) r\ dzdr d\theta
=\frac{2\pi a^5}5\left(9\sqrt3-\frac{97}{12}\right)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding solutions to a system of equations using elementary symmetric polynomials Find the value of a, b, and c given:
$a^2 + b^2 + c^2 = 129$
$ab + ac + bc = -4$
$(a^2)(b^2) + (a^2)(c^2) + (b^2)(c^2)$ = 984
I attempted this problem using elementary symmetrical polynomials and found that the second equation can be written as $e_2(a, b, c)$. However, the squares in the other equations made me quite confused and I was unable to write them in the previous from. If I was able to write it in the form of elementary symmetrical polynomials, I should have been able to use simple substitution to find the value of $e_1, e_2$, and $e_3$. This then would have allowed me to form a cubic equation, where I could find the solutions by solving it.
However, as I mentioned before, I'm a little bit stuck on the part where I need to find the values of $e_1, e_2, e_3$. Any help would be extremely appreciated :) Also, is there any other way to solve for a, b, and c using elementary symmetric polynomials and substitution, but not cubic equations?
Update:
Thanks to @Gerry_Myerson, I've worked out that the first equation is:
$(e_1)^2 - 2e_2 = 129$
and the second equation is:
$e_2 = -4$
However, I still don't know how to work out the last one.
| Let $P(t) = (t-a)(t-b)(t-c) = t^3-e_1t^2+e_2t-e_3$.
Trivially, $e_1 = a+b+c$, $e_2 = ab+bc+ca = -4$, and $e_3 = abc$.
As noted in the comments, $129 = a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) = e_1^2-2e_2 = e_1^2+8$, so $e_1^2 = 121$, and thus, $e_1 = \pm 11$.
Next, note that $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} = \left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2 - 2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right) = \left(\dfrac{ab+bc+ca}{abc}\right)^2 - 2\dfrac{a+b+c}{abc} = \left(\dfrac{e_2}{e_3}\right)^2 - \dfrac{2e_1}{e_3}$
Multiplying both sides by $a^2b^2c^2 = e_3^2$ yields
$984 = a^2b^2+b^2c^2+c^2a^2 = e_2^2-2e_1e_3 = (-4)^2-2(\pm 11)e_3$, and thus, $e_3 = \mp44$.
So, $a,b,c$ are the roots of either $t^3-11t^2-4t+44 = 0$ or $t^3+11t^2-4t-44 = 0$
Solving these equations yeilds $(a,b,c) = (2,-2,11)$ and $(2,-2,-11)$ and permutations as solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}$
Find limit of $ \underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}}$
How can I do that? It is interesting due to mathematica says that
$$\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{-\frac{x y}{2}+\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}} = 0 $$
but wolfram that limit doesn't exists. What is more I am not sure too about existance of limit due to
$$\underset{\{x,y\}\to \{0,0\}}{\text{lim}}\frac{\sqrt{x y+1}-1}{y \sqrt{x^2+y^2}} $$ doesn't exists too...
| Using the Maclaurin formula
$$
\sqrt {A + 1} - 1 - \frac {A}2 \sim -\frac 18 A^2 [A \to 0],
$$
we get
$$
\frac {\sqrt { xy+ 1} - 1 - \frac {xy}2}{ y \sqrt {x^2 + y^2}} \sim -\frac 18\cdot \frac {x^2 y^2} {y \sqrt {x^2 + y^2}}.
$$
Now let
$$
x = r \cos t, y =r \sin t,
$$
then
$$
\frac {x^2 y^2}{y \sqrt {x^2 + y^2}} = \frac {r^4 \sin^2 t \cos^2 t}{r^2 \sin t} = r^2 \sin t \cos^2 t \rightrightarrows 0 \, [r \to 0^+],
$$
hence the limit exists and equals $0$ [if the "equivalence" holds].
| {
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How to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$. Please tell if the problem can be solved using telescoping technique or not.
If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using that. It is given that $a,b \in \mathbb{R}{+},\, a\gt b,\, n \in \mathbb{N}.$
I tried as follows, but was unsuccessful to pursue:
$a^n − b^n = a^n+\sum_{i=1}^{n-1}(a^ib^{n-i}-a^ib^{n-i})-b^n=a^n+\sum_{i=1}^{n-1}a^ib^{n-i}-\sum_{i=1}^{n-1}a^ib^{n-i}-b^n$
Edit : based on the selected answer's comment.
Writing a few terms of the series, $\sum_{i=1}^n (a^{n+1-i}b^{i-1}-a^{n-i}b^i)$ get:
For $n =5$, get the terms as:
$i=1, \,\, a^{5+1-1}b^{1-1}-a^{5-1}b^1 = a^5-a^4b.$
$i=2, \,\, a^{5-1}b^{2-1}-a^{5-2}b^2 = a^4b-a^3b^2.$
$i=3, \,\, a^{5-2}b^{3-1}-a^{5-3}b^3 = a^3b^2-a^2b^3.$
$i=4, \,\, a^{5-3}b^{4-1}-a^{5-4}b^4 = a^2b^3-a^1b^4.$
$i=5, \,\, a^{5-4}b^{3-1}-a^{5-3}b^5 = a^1b^4-b^5.$
Adding all the terms, get:
$a^5-a^4b+ a^4b-a^3b^2+a^3b^2-a^2b^3+a^2b^3-a^1b^4+a^1b^4-b^5 = a^5 - b^5$
| For the equalty,
By telescoping: We have,
$$(a-b)\sum_{i=1}^na^{n-i}b^{i-1}=a\sum_{i=1}^na^{n-i}b^{i-1}-b\sum_{i=1}^na^{n-i}b^{i-1},$$
which can be rewritten
$$\sum_{i=1}^na^{n+1-i}b^{i-1}-\sum_{i=1}^na^{n-i}b^i=a^n+\sum_{i=2}^na^{n+1-i}b^{i-1}-b^n-\sum_{i=1}^{n-1}a^{n-i}b^i,$$
which on simplifying gives
$$a^n-b^n+\sum_{i=1}^{n-1}a^{n-i}b^i-\sum_{i=1}^{n-1}a^{n-i}b^i=a^n-b^n.$$
By induction: You wish to prove
$$a^n-b^n=(a-b)\sum_{i=1}^n a^{n-i}b^{i-1}.$$
Let's try induction. For the $n=1$ case we have
$$(a-b)a^0b^0=a^1-b^1,$$
so the base case holds.
Now suppose the general case is true and consider the $n+1$ case. We have
$$(a-b)\sum_{i=1}^{n+1}a^{n+1-i}b^{i-1}=(a-b)\sum_{i=1}^n a^{n+1-i}b^{i-1}+(a-b)a^0b^n,$$
which can be rewritten
$$a(a^n-b^n)+(a-b)b^n=a^{n+1}-ab^n+ab^n-b^{n+1}=a^{n+1}-b^{n+1},$$
so in fact the general case holds.
For the inequality, you can use induction too:
$$a^n-b^n\leq (a-b)na^{n-1}.\tag{*}$$
The base case clearly holds since $a^1-b^1\leq (a-b)a^0$. Now suppose (*) holds and consider the $n+1$ case,
$$a^{n+1}-b^{n+1}=a^na-b^nb=(a^n-b^n)(a+b)-a^nb+b^na\tag{1}$$
But since $a>b$, then
$$(1)\leq (a^n-b^n)(a+b)-a^nb+a^nb=(a^n-b^n)(a+b)-b(a^n-b^n),$$
which we can write
$$(a^n-b^n)a\leq a(a-b)na^{n-1}=(a-b)na^n\leq(a-b)(n+1)a^n,$$
as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the degree of a ODE $(y''')^{\frac{4}{3}}+(y')^{\frac{1}{5}}+ y = 0$ Find the degree of the differential equation.
$$(y''')^{\frac{4}{3}}+(y')^{\frac{1}{5}}+ y = 0$$
The answer is available (order $= 3$; degree $= 60$). I need help with the steps. I'm stuck in eliminating the radicals of the differential coefficients.
| Went through @Lutz's answer Find the degree of the differential equation $\left( \frac{d^3y}{dx^3} \right)^{\frac{4}{3}} + \left( \frac{dy}{dx} \right)^{\frac{1}{5}} + y = 0.$. But despite his elaborate explanations, I found it really hard to understand.
Then, I tried it the hard way. It's only a 2-step solution. Nothing fancy.
$$ (y''')^{4/3}+y=-(y')^{1/5}$$
First, take power $5$ both sides,
$$ 5y^4(y''')^{4/3}+10y^{3}(y''')^{8/3}+5y(y''')^{16/3}+(y''')^{20/3}=-y'-y^5-10y^2(y''')^{4}$$
$$ (y''')^{4/3}[5y^4+5y(y''')^{4}]+(y''')^{8/3}[10y^{3}+(y''')^{4}]=-y'-y^5-10y^2(y''')^{4}$$
This is equivalent to $A+B=C$. $A$ is $O(\frac{16}{3})$. $B$ is $O(\frac{20}{3})$. $C$ is $O(4)$.
Cubing both sides $A^3+B^3 +3AB(A+B)=C^3$, where $A+B=C$ can be substituted from the above equation.
$$A^3+B^3 +3ABC=C^3$$
$$ (y''')^{4}[5y^4+5y(y''')^{4}]^{3}+(y''')^{8}[10y^{3}+(y''')^{4}]^{3}+3(y''')^{4}[5y^4+5y(y''')^{4}][10y^{3}+(y''')^{4}][-y'-y^5-10y^2(y''')^{4}]=[-y'-y^5-10y^2(y''')^{4}]^{3}$$
$$ O(16)+O(20)+O(16)=O(12)$$
So, the degree comes out to be $20$. Evidently, the bookish answer of $60$ must be wrong.
| {
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I need to define the cubic equation $f(x)= ax^3 + bx^2 + cx + d$ It is given that if cubic function $f(x)$ is divided by $(x^2+4)$ and $(x+3)$, the remainders will be $7x-20$ and $-2$ respectively. Also it is said show that $f(x)$ is $x^3 + 6x^2 + 11x +4$. I tried but can i find the $a,b,c$, and $d$ without knowing them in from the start?
| $f(x) = (x^2 + 4)(ax +b) +7x - 20$
$f(-3)=-2 = 13(-3a+b)-41 $
therefore $b= 3a+3$
so, $f(x)= (x^2+4)(ax +3a+3)+ 7x-20 $ (non-zero $a$)
| {
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Linear quadratic cubic system of equation I have been trying to make something out of this system of equations for quite sometimes now
$x+y+z=4\\x^2+y^2+z^2=38\\x^3+y^3+z^3=106$
I have tried direct substitution but the equation keeps expanding, and keeps getting complicated. I have also tried multipling by $x,$ $y,$ $z$ in several ways. But it was all a futile effort.
Any insight to this??
| $$16=38+2(xy+xz+yz),$$ which gives
$$xy+xz+yz=-11.$$
Thus, since
$$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+xz+yz)+3xyz,$$ we obtain:
$$106=64-3\cdot4\cdot(-11)+3xyz,$$ which gives $$xyz=-30$$ and $x$, $y$ and $z$ are roots of the equation:
$$t^3-4t^2-11t+30=0.$$
Can you end it now?
I got $(x,y,z)=(2,-3,5)$ and all symmetric permutations of this.
| {
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Exercise 3 in Section 7.11 of Apostol's Calculus (Vol. 1) little o-notation The problem is stated as:
Find the polynomial $P(x)$ of minimal degree such that
$$
\sin(x-x^2)=P(x)+o(x^6)\quad \textrm{as }x\to 0.
$$
and the answer in the book is
$$
P(x)=x-x^2-\frac{x^3}{6}+\frac{x^4}{2}-\frac{59x^5}{120}+\frac{x^6}{8}
$$
I am wondering why the answer cannot be
$P(x) = x-x^2$
Since we can use the fact that $\sin(x-x^2) = x-x^2 + o((x-x^2)^2)$ as $x \to 0$
And we know that $o((x-x^2)^2) = o(x^5) = o(x^6)$ as $x$ goes to zero.
Then why is the answer from the back of the book not equal to $x-x^2$??
| What they are asking you is to find a polynomial $p(x)$ such that
$$
\lim_{x\rightarrow0}\frac{|\sin(x-x^2)-p(x)|}{x^6}=0
$$
No to your statements:
*
*$o((x-x^2)^2) =o(x^5)$ is in general false. For example $h(x)=x(x-x^2)^2=o((x-x^2)^2$ as $x\rightarrow0$ but $x^{-5}h(x)\nrightarrow0$ as $x\rightarrow0$.
*$o(x^5) = o(x^6)$ is false in general: $g(x):=x^6=o(x^5)$ but $g(x) \neq o(x^6)$.
To help you solve the problem, use Taylor expansion of $\sin z= z-\frac{z^3}{6}+\frac{z^5}{5!}-\frac{z^7}{7!}+...$. Notice that when $z=x-x^2$, $z^7$ is a polynomial on $x$ of degree 14 of the $z^7=x^7-7x^8+...-x^{14}$ This term is of course $o(x^6)$ since all powers involved are larger than $6$. So truncating the sine series up to the $7-th$ power gives the right decay of residual. The rest is to check that the
$$
(x-x^2)-\frac{(x-x^2)^3}{6}+\frac{(x-x^2)^5}{5!}
$$
is the polynomial in the statement of your problem. A little but tedious algebra.
| {
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Rotated Ellipse It is well known that the equation
$$\frac {(x\cos\alpha+y\sin\alpha)^2}{a^2}+\frac {(x\sin\alpha-y\cos\alpha)^2}{b^2}=1\tag{1}$$
(where $\beta\neq\alpha$) represents an ellipse centred at the origin with semimajor/minor axes $a,b$, and rotated by $\alpha$.
Question
The equation
$$\frac {(x\cos\alpha+y\sin\alpha)^2}{a^2}+\frac {(x\sin\beta-y\cos\beta)^2}{b^2}=1\tag{2}$$
represents a rotated ellipse centred at the origin, but its semimajor/minor axes are no longer $a,b$.
How can this be transformed into a form similar to $(1)$, such that the semimajor/minor axes and angle of rotation can be easily determined?
| If you rewrite your equation as
$$
ux^2 + vy^2 +2wxy = 1
$$
then the squared reciprocal lengths of the semimajor/semiminor axes of the ellipse are the eigenvalues of the matrix
$$
\begin{pmatrix} u & w \\ w & v \end{pmatrix}
$$
You get
$$
\begin{eqnarray*}
u & = & \frac{\cos^2 \alpha}{a^2} + \frac{\sin^2 \beta}{b^2} \\
v & = & \frac{\sin^2 \alpha}{a^2} + \frac{\cos^2 \beta}{b^2} \\
w & = & \frac{\cos\alpha\sin\alpha}{a^2} - \frac{\cos\beta\sin\beta}{b^2}
\end{eqnarray*}
$$
The eigenvalues of the matrix are
$$
\lambda_{1,2} = \frac{u+v}{2} \pm \sqrt{\left(\frac{u-v}{2}\right)^2+w^2}
$$
Using
$$
\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \\
\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \\
\cos^2\alpha+\sin^2\alpha = 1
$$
this simplifies a lot:
$$
\lambda_{1,2} = \frac{1}{2}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)
\pm\frac{1}{2}\sqrt{\frac{1}{a^4} - \frac{2\cos(2\alpha-2\beta)}{a^2b^2}
+\frac{1}{b^4}}
$$
The length of the semimajor axis is $1/\sqrt{\lambda_2},$ and the length of the semiminor axis is $1/\sqrt{\lambda_1}.$
| {
"language": "en",
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proof that $\frac{a_{4n}-a_2}{a_{2n+1}}$ : integer I would appreciate if somebody could help me with the following problem:
Q: How to proof?
If $\{a_n\}$ satisfy $a_{1}=a$, $a_2=b$, $a_{n+2}=a_{n+1}+a_{n}$($a,b$: positive integers)
then proof that
$\frac{a_{4n}-a_2}{a_{2n+1}}$ : integer
I try start by mathematical induction but....,
find $f(n)=\frac{a_{4n}-a_2}{a_{2n+1}}$ $f(1)=1$, $f(2)=4$, $f(3)=11$, $f(4)=29$,...
| $R=\frac{a_{4n}-a_2}{a_{2n+1}}$
$a_1=a$
$a_2=b$
$a_3=a+b$
$a_4=a+2b$
$a_5=2a+3b$
$a_6=3a+5b$
$a_7=5a+8b$
$a_8=8a+13b$
For n=2 we have:
$\frac{8a+13b-b}{2a+3b}=4$
By experimental induction if R is true for n=2 it must also be true for n+1=3, we check this:
$\frac{a_{12}-b}{a_7}=\frac{55a+89b-b}{5a+8b}=11$
This true for any value of a and b,for example $a_{12}=65$ for a=0, b=1 and n=3 :
$\frac{65-1}{8}=8$
| {
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Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)$$
I found $A=\frac{1}{3}$ by letting $x=1$.
Now in the book they let me know that $B=\frac{2}{3}$, $C=-\frac{1}{3}$, $D=-1$ and $E=0$. But I would really like to figure out how I can find the values for $B, C, D, E$.
| After clearing the denominators we have $$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)(*)$$
as you obtained.
Indeed, $A=\frac{1}{3}$ if we let $x=1$.
Differentiate $(*)$ to get that $$4x^3-3x^2+4x-1=\frac{4}{3}x(x^2+2)+B(4x^3-3x^2+4x-2)+C(3x^2-2x+2)+D(2x-1)+E(**)$$
Differentiate the above relation again to get that $$12x^2-6x+4=\frac{4}{3}(3x^2+2)+B(12x^2-6x+4)+C(6x-2)+2D(***)$$
Differentiate again to get that $$24x-6=8x+B(24x-6)+6C(****)$$
Now let $x=\frac{1}{4}$. It follows that $-2=6C$, so $C=-\frac{1}{3}$.
For $x=0$ in $(****)$ we have $-6=-6B-2$, so $B=\frac{2}{3}$.
For $x=0$ in $(***)$ we obtain that $4=6+2D$, so $D=-1$.
Finally, for $x=0$ in $(**)$ we have $-1=1+E$, so $E=0$.
| {
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Does any (right) triangle exist such that $a^3+b^3=c^3$?
Does any right triangle exist such that $a^3+b^3=c^3$? Does any triangle exist such that $a^3+b^3=c^3$?
I'm stuck on this problem; I tried applying the Pythagorean theorem in three dimensions, but in vain. Any tips?
| We need $a\ne0$ and $b\ne0$. If it holds, then we have
$(a^3+b^3)^2=c^6=(a^2+b^2)^3$
$a^6+2a^3b^3+b^6=a^6+3a^4b^2+3a^2b^4+b^6$
$2ab=3a^2+3b^2$
$(a-b)^2+2a^2+2b^2=0$
$a=b=0$
| {
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If one of the lines given by $6x^2 – xy + 4cy^2 = 0$ is $3x+4y = 0$ then c is equal to? If one of the lines given by $6x^2 – xy + 4cy^2 = 0$ is $3x+4y = 0,$ then $c$ is equal to?
I have tried solving it by putting $x = \frac{-4y}{3}$ inside the given equation, as $3x + 4y$ is a solution of $6x^2 – xy + 4cy^2 = 0$, but I am getting two variables, $c$ and $y$. Please tell me how to solve it.
| Your idea is good; plugging in $x=-\tfrac43y$ yields
$$0=\frac{96}{9}y^2+\frac43y^2+4cy^2=\frac49\left(36+12c\right)y^2,$$
which must hold for all $y$. Then $36+12c=0$ so $c=-3$.
Alternatively, you could note that the point $(x,y)=(4,-3)$ is on the line, and hence on the curve, so
$$6\cdot4^2-4\cdot(-3)+4\cdot c\cdot(-3)^2=0.$$
This shows that $24+3+9c=0$ so $c=-3$.
| {
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Find the base system, $x$, such that $\frac{1}{5}$ and $\overline{.17}$ are numerals for the same number. So far I know that $\frac{1}{5} = 0.2$ in base 10. However, I'm not really sure how to convert it into another base. Any suggestions would be greatly appreciated.
| $$\begin{align} (0.1717...)_b =& \left(\frac{1}{b}+\frac{7}{b^2}\right) +
\left(\frac{1}{b^3}+\frac{7}{b^4}\right) + \cdots \\
=& \frac{b+7}{b^2}+\frac{b+7}{b^4} +\cdots \\
=& \sum_{n=1}^{\infty} \frac{b+7}{b^{2n}} \\
=& (b+7)\sum_{n=1}^{\infty}(b^{-2})^n \\
=& \frac{b+7}{b^2-1}
\end{align}$$
and this is equal to $1/5$, so $b=9$.
| {
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Solve $\int_{-\pi}^{\pi}\frac{1}{1+\sin^{^{2}}t}dt$ I need to solve $\int_{-\pi}^{\pi}\frac{1}{1+\sin^{^{2}}t}dt$. This is what I did, but I think the answer should be $\sqrt{2}\pi$.
${\sin^{^{2}}t}=\frac{1-\cos(2t)}{2}$
$z=e^{2ti}=\cos(2t)+i\sin(2t)$
$-2\pi \leq 2t\leq 2\pi$
$\cos(2t)=\Re(z)=\frac{z+\bar{z}}{2}$
$z^{-1}=\frac{1}{z}=\frac{\bar{z}}{|z|^{2}}$ with $|z|=1$
$\frac{1}{1+\sin^{2}t}=\frac{4}{6-z-\frac{1}{z}}$
$dt=\frac{dz}{2iz}$
$\int_{-\pi}^{\pi}\frac{1}{1+\sin^{^{2}}t}dt=\oint_{|z|=1}\frac{4}{6-z-\frac{1}{z}}\frac{dz}{2iz}=\frac{2}{i}\oint_{|z|=1}\frac{1}{6z-z^{2}-1}dz$
$6z-z^{2}-1=0$
$z_{1}=3-2\sqrt{2}\in\Omega$
$z_{2}=3+2\sqrt{2}\notin\Omega$
$f(z)=\frac{1}{6z-z^{2}-1}=\frac{1}{[z-(3-2\sqrt{2})][z-(3+2\sqrt{2})]}$
$\operatorname{Res}(f,z_{1})=\lim_{z\to z_{1}}(z-z_{1})f(z)=-\frac{1}{4\sqrt{2}}$
Using the Residue theorem $\frac{2}{i} 2\pi i(-\frac{1}{4\sqrt{2}})=\pi (-\frac{1}{\sqrt{2}})$
I've solved problems like this before just fine, but the parametrizations were always $z=e^{ti}$ $-\pi \leq t\leq \pi$, so I think that's where my problem is, but I don't know how to do it right.
| Here is an alternative approach without using complex analysis.
The change of variables $\sin(t)=\dfrac{2u}{1+u^2}$ with $u=\tan\left(\dfrac{t}{2}\right)$ gives $\mbox{d}u=\dfrac{1}{2}\left(1+\tan\left(\dfrac{t}{2}\right)^2\right)\mbox{d}t$.
And therefore
\begin{align*}I&=\int_{-\pi}^{\pi}\dfrac{1}{1+\sin(t)^2}\mbox{d}t\\
&=2\int_{0}^{\pi}\dfrac{1}{1+\sin(t)^2}\mbox{d}t\\
&=4\int_{0}^{+\infty}\dfrac{1}{1+\left(\dfrac{2u}{1+u^2}\right)^2 }\dfrac{\mbox{d}u}{1+u^2}\\
&=4\int_{0}^{+\infty}\dfrac{1+u^2}{1+u^4+6u^2}\mbox{d}u.
\end{align*}
Now notice that, noting $\alpha=3-2\sqrt{2}$ and $\beta=3+2\sqrt{2}$, you have $1+u^4+6u^2=(u^2+\alpha)(u^2+\beta)$ and
\begin{align*}
\dfrac{1+u^2}{1+u^4+6u^2}&=\dfrac{1+u^2+\alpha-\alpha}{1+u^4+6u^2}\\
&=\dfrac{1-\alpha}{1+u^4+6u^2}+\dfrac{1}{u^2+\beta}\\
&=\dfrac{1}{u^2+\beta}+(1-\alpha)\dfrac{1}{4\sqrt{2}}\left(\dfrac{1}{u^2+\alpha}+\dfrac{-1}{u^2+\beta}\right).
\end{align*}
Thus, \begin{align*}
\int_{0}^{\infty}\dfrac{1+u^2}{1+u^4+6u^2}du=\dfrac{1}{\sqrt{\beta}}\dfrac{\pi}{2}+(1-\alpha)\dfrac{1}{4\sqrt{2}}\dfrac{\pi}{2}\left(-\dfrac{1}{\sqrt{\beta}}+\dfrac{1}{\sqrt{\alpha}}\right).
\end{align*}
Since $\sqrt{\alpha}=\sqrt{2}-1$ and $\sqrt{\beta}=1+\sqrt{2}$ we obtain
\begin{align*}
I&=4\int_{0}^{+\infty}\dfrac{1+u^2}{1+u^4+6u^2}\mbox{d}u\\
&=\dfrac{4}{\sqrt{\beta}}\dfrac{\pi}{2}+(1-\alpha)\dfrac{1}{\sqrt{2}}\dfrac{\pi}{2}\left(\dfrac{-1}{\sqrt{\beta}}+\dfrac{1}{\sqrt{\alpha}}\right)\\
&=\pi\left(\dfrac{2}{1+\sqrt{2}}+\dfrac{\sqrt{2}-1}{\sqrt{2}}\dfrac{-\sqrt{\alpha}+\sqrt{\beta}}{\sqrt{\alpha\beta}}\right)\\
&=\pi(2\sqrt{2}-2+\sqrt{2}(\sqrt{2}-1))\\
&=\pi(2\sqrt{2}-2+2-\sqrt{2})\\
&=\sqrt{2}\pi.
\end{align*}
| {
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What is the correct solution of $\sqrt[7]{(-\sqrt{3}-i)^5}$? $\sqrt[7]{(-\sqrt{3}-i)^5}=(-\sqrt{3}-i)^\frac{5}{7}=
2^\frac{5}{7}(\cos(\frac{5}{7}\alpha)+i\sin(\frac{5}{7}\alpha)=$
$\tan\alpha=\frac{-1}{-\sqrt{3}} \implies \alpha=\frac{\pi}{6}+2k\pi$
$=2^\frac{5}{7}(\cos(\frac{5\pi}{42}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{42}+\frac{2k\pi}{7})$
The problem I have currently is that I have no idea where I made an error. The correct solution is $2^\frac{5}{7}(\cos(\frac{5\pi}{6}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{6}+\frac{2k\pi}{7})$.
| What you want is the set of solutions to the equation
$$
x^7 = (-\sqrt3 -i)^5.
$$
Now, taking the polar form of $-\sqrt3 -i$ you get
$$
(-\sqrt3 -i)^5 = \left(2\, e^{-5\pi/6}\right)^5
= 2^5 e^{-25\pi/6}
= 2^5 e^{-\pi/6}
$$
therefore the solutions are
$$
\{
2^{5/7} e^{-\pi/6 + 2 k \pi/7}, k=0,1,...,6
\}.
$$
| {
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Three cevians in a triangle create four sub-triangles of area $1$. Find the area of a non-triangular region.
I'm having trouble proving that all the white and green areas have the same area, from there on we can obtain the answer $1+\sqrt5$ by proving that the inner red triangle points are midpoints.
| Let $AD$, $BE$ and $CF$ be our chevians of $\Delta ABC$.
Also, let $AD\cap BE=\{P\}$, $AD\cap CF=\{R\}$ and $BE\cap CF=\{Q\}.$
Thus, $$S_{\Delta APE}=S_{\Delta BQF}=S_{\Delta CRD}=S_{\Delta PQR}=1.$$
Let $S_{AFQP}=x,$ $S_{BDRQ}=y$ and $S_{CEPR}=z$.
Thus, $$S_{\Delta BDR}=\frac{S_{\Delta BDR}}{S_{\Delta CDR}}=\frac{BD}{DC}=\frac{x+y+2}{z+2},$$
which gives
$$S_{\Delta BQR}=y-\frac{x+y+2}{z+2}=\frac{yz+y-x-2}{z+2}.$$
Now, $$S_{\Delta AFQ}=\frac{S_{\Delta AFQ}}{S_{\Delta BFQ}}=\frac{AF}{BF}=\frac{x+z+2}{y+2},$$
which gives
$$S_{\Delta AQP}=x-\frac{x+z+2}{y+2}=\frac{xy+x-z-2}{y+2}.$$
Thus, $$\frac{\frac{x+z+2}{y+2}+1}{\frac{xy+x+z-2}{y+2}}=\frac{S_{\Delta AFQ}+S_{\Delta BQF}}{S_{\Delta AQP}}=\frac{S_{\Delta ABQ}}{S_{\Delta AQP}}=$$
$$=\frac{BQ}{PQ}=\frac{S_{\Delta BQR}}{S_{\Delta PQR}}=S_{\Delta BQR}=\frac{yz+y-x-2}{z+2}.$$
Id est, $$\frac{x+y+z+4}{xy+x-z-2}=\frac{yz+y-x-2}{z+2}$$ or
$$(y+1)(z^2+x^2-xy+2x+4z-xyz+4)=0,$$ which gives
$$z^2+x^2-xy+2x+4z=xyz-4.$$
By the same way we can obtain:
$$x^2+y^2-yz+2y+4x=xyz-4$$ and
$$y^2+z^2-zx+2z+4y=xyz-4.$$
Now, let $x=\max\{x,y,z\}$.
Thus, $$x^2+y^2-yz+2y+4x-(y^2+z^2-zx+2z+4y)=0$$ or
$$(x-z)(x+z)+z(x-y)+4x-2y-2z=0$$ or
$$(x-z)(x+z+2)+(x-y)(z+2)=0,$$ which gives $$x=y=z,$$
$$x^2+6x=x^3-4$$ or
$$x^3-x^2-6x-4=0$$ or
$$x^3+x^2-2x^2-2x-4x-4=0$$ or
$$(x+1)(x^2-2x-4)=0$$ or
$$x=1+\sqrt5$$ and we are done!
| {
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Show that $r^2 \cot(A/2) \cot(B/2) \cot(C/2) = [ABC]$ In triangle $\Delta~ ABC$,
$~r~$ is the in-radius and
$~[ABC]~$ is the area.
Please explain $$ r^2 \cot(A/2) \cot(B/2) \cot(C/2) = [ABC]$$ thanks.
|
\begin{align}
[ABC]&=[IAB]+[IBC]+[ICA]
.
\end{align}
\begin{align}
[IAB]&=\tfrac12\cdot|AB|\cdot|IC_t|
=\tfrac12(|AC_t|+|BC_t|)\cdot|IC_t|
=\tfrac12(r\cot\tfrac\alpha2+r\cot\tfrac\beta2)\cdot r
=r^2(\tfrac12\,\cot\tfrac\alpha2+\tfrac12\,\cot\tfrac\beta2)
.
\end{align}
Similarly,
\begin{align}
[IBC]&=
r^2(\tfrac12\,\cot\tfrac\beta2+\tfrac12\,\cot\tfrac\gamma2)
,\\
[ICA]&=
r^2(\tfrac12\,\cot\tfrac\gamma2+\tfrac12\,\cot\tfrac\alpha2)
.
\end{align}
Hence,
\begin{align}
[ABC]&=[IAB]+[IBC]+[ICA]
=r^2(\cot\tfrac\alpha2+\cot\tfrac\beta2+\cot\tfrac\gamma2)
.
\end{align}
And
\begin{align}
\cot\tfrac\alpha2+\cot\tfrac\beta2+\cot\tfrac\gamma2
&=
\cot\tfrac\alpha2\cot\tfrac\beta2\cot\tfrac\gamma2
\quad \text{ for }\alpha+\beta+\gamma=180^\circ
\end{align}
is a known triple cotangent identity,
which is easy to check using
substitution
\begin{align}
\cot\tfrac\gamma2&=
\frac{\cot\tfrac\alpha2+\cot\tfrac\beta2}{\cot\tfrac\alpha2\cot\tfrac\beta2-1}
.
\end{align}
| {
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Why $ \frac{n+2}{n-2}<(n+2)^{2/n}$ for $n\geq 7$. In some paper, the authors mentioned the following statement:
One can easily check that for $n\geq 7$,
$$ \frac{n+2}{n-2}<(n+2)^{2/n}.$$
This statement is correct, and their objective was to find an upper bound of $\frac{n+2}{n-2}$, eventually starting from some integer. Now my question is how we can see that $(n+2)^{2/n}$ is an upper bound for $\frac{n+2}{n-2}$ starting from some integer ( here it is $7$).
Thank you.
| (1) The inequality is true for $n=7$.
(2) Assume $(k+2)^{\frac{k-2}{k}} < k - 2$
(3) We must show $(k+3)^{\frac{k-1}{k+1}} < k - 1$.
Consider $f(x) = (x+3)^{\frac{x-1}{x+1}} - x + 1$. (We want to show $f(x) < 0$ for $x \geq 7$.)
Well, $(x+3)^{\frac{x-1}{x+1}} - x + 1 < 0$ iff $(x+3)^{\frac{x-1}{x+1}} < x - 1$ iff $\frac{x-1}{x+1}\ln(x+3) < \ln(x-1)$ iff $\frac{x-1}{x+1} < \frac{\ln(x-1)}{\ln(x+3)},$ which is true for $x=7$; and, since $\frac{d}{dx}\left(\frac{x- 1}{x+1}\right) < 1$ and $\frac{d}{dx}\left(\frac{\ln(x-1)}{\ln(x+3)}\right) > 1$ for all $x > 7$, the last inequality holds for all $x \geq 7$.
| {
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"question_score": "7",
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Change of variable in integrals I am trying to solve a definite integral of a positive function, but I keep getting 0.
| On using $ \int^a_{-a} f(x) dx = 2 \int^a_0 f(x) dx$ and $ \int^{2a}_0 f(x) dx = \int^a_0 f(x) dx + \int^a_0 f(2a - x) dx$
$$I = \int^{\pi}_{-\pi} \dfrac{1}{1 + \sin^2 (x)} dx = 4 \int^{\pi /2}_0 \dfrac{1}{1 + \sin^2 (x) } dx $$
Multiply, both denominator and numerator by $\sec^2 (x)$ , Use $\sec^2 (x) = \tan^2 (x) + 1$, and substitute $\sqrt{2} \tan x = u$, Following integral would be obtained
$$ I = 2 \sqrt{2} \int^{\infty}_0 \dfrac{du}{1 + u^2} = 2 \sqrt{2} \cfrac{\pi}{2} = \sqrt{2} \pi $$
The problem with your substition is that, you've took $du = \cos(x) dx = \color{red} {+} \sqrt{1 - \sin^2 x} dx $, but the right way is $du = \cos(x) dx = \color{red} {\pm} \sqrt{1 - \sin^2 x} dx $ and then substitute the required limits.
| {
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If I roll a dice 4 times then what is the probability of getting 20 total dots? So, here the total evets = $6^4$
If I get $a,b,c,d$ doing each rolling then, $$a+b+c+d=20$$
$$where\ a,b,c,d\in\mathbb{N} \ and\ 1\le a,b,c,d\le6$$
Now, how can I solve this by stars and bars method? Or there are other easier ways to solve this?
| Using generating functions: The number of ways to roll a total of $20$ on four dice is $$\begin{align}[x^{20}](x+x^2+x^3+x^4+x^5+x^6)^4 &= [x^{16}](1+x+x^2+x^3+x^4+x^5)^4 \\ &= [x^{16}]\left({1-x^6 \over 1-x}\right)^4 \\
&= [x^{16}]{\binom40-\binom41x^6+\binom42x^{12}-\cdots \over (1-x)^4} \\
&= [x^{16}]{\binom40 \over (1-x)^4}-[x^{10}]{\binom41 \over (1-x)^4}+[x^4]{\binom42 \over (1-x)^4} \\
&= \binom40\binom{19}3-\binom41\binom{13}3+\binom42\binom{7}3 \\ &= 35. \end{align}$$ The probability is therefore $\frac{35}{6^4} = \frac{35}{1296} \approx 2.7\%$.
| {
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Rearranging the formula Transpose this formula to make $y$ the subject.
$$x=\sqrt{x^2y^2+1-y}$$
My try:
$$x^2=x^2y^2+1-y$$
$$x^2-x^2y^2=1-y$$
$$x^2(1-y^2)=1-y$$
Here I got 2 $y$ terms, but I am not sure what to do next.
| You can write $$0=y^2-\frac{1}{x^2}y+\frac{1-x^2}{x^2}$$ using the quadratic formula we obtain
$$y_{1,2}=\frac{1}{2x^2}\pm\sqrt{\left(\frac{1}{2x^2}\right)^2-\frac{1-x^2}{x^2}}$$
| {
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Let $a,b,c$ be the sides of a triangle. The find maximum value $ \frac{A}{S^{2}}$ Let $a,b,c$ be the sides of a triangle, $A$ is the area and $S$ is the semi-perimeter $(a+b+c)/2$.
Find the maximum value $\frac{A}{S^{2}}$.
My Approach:
Method 1:
Applying AM-GM inequality on $S,S-a,S-b,S-c$
$$\frac{4S-2S}{4} \ge \sqrt[4]{S(S-a)(S-b)(S-c)}$$
$$\frac{1}{4}\ge \frac{A}{S^2}$$
Method 2:
Applying AM-GM inequality on $S-a,S-b,S-c$
$$\frac{3S-2S}{3} \ge \sqrt[3]{(S-a)(S-b)(S-c)}$$
$$\frac{1}{3\sqrt{3}}\ge \frac{A}{S^2}$$
For equality in method 1 $a=b=c=0$ which is not true hence we get maximum value from method 2. But not sure if $\frac{1}{3\sqrt{3}}$ is the maximum because some other method may give me some other maximum value. So is there a method which gives me the exact maximum value.
| Your Method 2 is fine and it gives the exact maximum value that you are looking for.
Since $A=\sqrt{S(S-a)(S-b)(S-c)}$, the AM-GM inequality in Method 2 yields
$$\frac{(S-a)+(S-b)+(S-c)}{3} \ge \sqrt[3]{(S-a)(S-b)(S-c)}$$
that is
$$\frac{S^3}{3^3}\geq(S-a)(S-b)(S-c)=\frac{A^2}{S},$$
and therefore
$$\frac{1}{3\sqrt{3}}\ge \frac{A}{S^2}.$$
Here equality holds when $S-a=S-b=S-c$ that is when $a=b=c$ and we may conclude that the maximum value of $A/S^2$ is just $\frac{1}{3\sqrt{3}}$.
| {
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How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$. I am getting x = 4 but the book gets x = 81. What am I doing wrong?
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-\log_{3}(3)} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-1} + 8\\
\log_{3}(x) & = -\log_{3}(x) + 8\\
2\log_{3}(x) & = 8\\
\log_{3}(x) & = 4\\
x & = 4
\end{align*}
What am I doing wrong?
| $\log_3 (x)=\log_{1/3} (x)+8 $
$\iff \log_3 (x)=-\log_3 (x)+8$
$\iff 2\log_3 (x)=8$
$\iff \log_3 (x)=4$
$\implies x=3^4=81$
| {
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Haar functions are basis in $L^2[0,1]$ Define the Haar functions as $e_0^0=1$ and for $n\ge 1$, $k=1,\ldots,2^n$
$$
e_n^k(t)=\left\{ \begin{array}{ll}
2^{\frac{n-1}{2}} & \mbox{if $x \in \big(\frac{K-1}{2^n},\frac{K}{2^n}\big)$};\\
-2^{\frac{n-1}{2}} & \mbox{if $x \in \big(\frac{K}{2^n},\frac{K+1}{2^n}\big)$};\\
0 &\mbox{otherwise}
\end{array} \right.
$$
As part of showing that this is an orthonormal basis in $L^2[0,1]$, I need to show first that if $\langle f,e^k\rangle=0$ for each $k$ and $n$ then $\langle f,\chi_{[0,x]}\rangle = 0$ for every dyadic $x$. This would mean $f=0$.
I was looking at this answer but I do not understand the construction he uses. Thanks in advance for your help.
| We can use $f_n^j := \frac1{2^{\frac{n-1}2}} e_n^j$ since they have the same linear span, denote it by $S$.
We shall prove by induction on $n$ that $$\chi_{\left(\frac{k-1}{2^n},\frac{k}{2^n}\right)} \in S, \quad \text{ for all } n \ge 0, 1 \le k \le 2^n.$$
Indeed, if $n=1$ then
$$\chi_{(0,1)} \stackrel{\text{a.e.}}{=} e_0^0$$
so $\chi_{(0,1)} \in S$.
Assume that for some $n \in \Bbb{N}$ we have
$$\chi_{\left(\frac{k-1}{2^n},\frac{k}{2^n}\right)} \in S, \quad \text{ for all } 1 \le k \le 2^n.$$
Pick an odd number $1 \le k \le 2^{n+1}-1$, or in other words $k=2r+1$ with $1 \le r \le 2^n-1$. We have
$$\chi_{\left(\frac{k-1}{2^{n+1}},\frac{k}{2^{n+1}}\right)} - \chi_{\left(\frac{k}{2^{n+1}},\frac{k+1}{2^{n+1}}\right)} = f_{n+1}^k\in S,$$
$$\chi_{\left(\frac{k-1}{2^{n+1}},\frac{k}{2^{n+1}}\right)} + \chi_{\left(\frac{k}{2^{n+1}},\frac{k+1}{2^{n+1}}\right)} \stackrel{\text{a.e.}}{=} \chi_{\left(\frac{k-1}{2^{n+1}},\frac{k+1}{2^{n+1}}\right)} = \chi_{\left(\frac{r}{2^n},\frac{r+1}{2^n}\right)} \in S$$
using the assumption. By taking $\frac12$ of the sum and difference, we get
$$\chi_{\left(\frac{k-1}{2^{n+1}},\frac{k}{2^{n+1}}\right)},\chi_{\left(\frac{k}{2^{n+1}},\frac{k+1}{2^{n+1}}\right)} \in S$$
which covers all of them so
$$\chi_{\left(\frac{k-1}{2^{n+1}},\frac{k}{2^{n+1}}\right)} \in S, \quad \text{ for all } 1 \le k \le 2^{n+1}$$
finishes the induction.
Now for any dyadic interval we have
$$\chi_{\left(0,\frac{k}{2^n}\right)} = \sum_{j=1}^{k-1} \chi_{\left(\frac{j-1}{2^n},\frac{j}{2^n}\right)} \in S$$
which proves your claim.
| {
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Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
Find the number of roots of the polynomial $P(z)=z^5+2z^3+3$ in the closed unit disk $\{z\colon |z|\le 1\}$.
My solution:
Set $f(z)=3$.
For every $0<\delta<1$ and $|z|=\delta$, we have $|P(z)-f(z)|\le \delta^5+2\delta^3<3\delta<3=|f(z)|$. Hence $P(z)$ and $f(z)=3$ has the same number of zeros, which is $0$, inside the open unit disk.
Now consider the behavior of $P(z)$ when $|z|=1$. If we have $P(z)=0$, then $z^5=-2z^3-3$, and $1=|z^5|=|-2z^3-3|$ when $|z|=1$ which forces $-2z^3=2$ and $z^3=-1$.
Clearly, $z=-1$ is a zero of $P(z)$.
For the other two roots of $z^3=-1$, note that: $$ P(z)=z^2\cdot z^3+2z^3+3=z^2-2+3=z^2+1=0\implies z=\pm 1 $$
Therefore we conclude that there is only one root, $z=-1$, of $P(z)$ in the closed unit disk.
However, I am asking for other solutions, such as applying the symmetric version of Rouché''s theorem like in this post. Thank you.
| Note that on $|z| < 1$ there are no roots as $|z^5+2z^3| < 3$ and so the roots of $z^5+2z^3+3$ on $|z| \leq 1$ lie on the boundary $|z| = 1$. Let $z = cos(\theta) + isin(\theta)$ satisfy the polynomial relation $P(z)$ stated above.
We must have $|(e^{i\theta})^5 + 2(e^{i\theta})^3|^2 = 9$ and hence $|e^{i2\theta}+2|^2 = 9$ which is exactly when $\theta \equiv 0$ $ mod(\pi)$ so there are atmost two roots of $P(z)$ in $|z| = 1$ at $z=1$ and $z=-1$; but only $z = - 1$ is a root. Your conclusion should be correct.
| {
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Prove That $3^n + 8^n$ is Not Divisible by $5$ (Using Induction) Prove that $3^n+8^n$ is not divisible by 5.
I know that this can be proved by using congruence and I am providing the proof by congruence below. But is there any way to Prove It By Induction.
The proof by congruence goes like this:
$3\equiv 3\pmod 5 \\ 3^2 \equiv 4\pmod 5 \\ 3^3\equiv 7\pmod 5 \\ 3^4\equiv 1\pmod 5 \\ 3^5\equiv 3\pmod 5$
Also,
$8\equiv 3\pmod 5 \\ 8^2 \equiv 4\pmod 5 \\ 8^3\equiv 7\pmod 5 \\ 8^4\equiv 1\pmod 5 \\ 8^5\equiv 3\pmod 5$
Adding the congruence up (since the same cycle repeats after the 4th power) none of them are divisible by 5 or equal to 0.
But I need a proof by Induction.
Any help will be appreciated.
| You can go like this. For the base case, let $n=1$. Then
$$3^n+8^n = 3+8 = 11,$$
which is not divisible by $5$.
For the induction step, let $n\geq 1$ be arbitrary and assume that $3^n+8^n$ is not divisible by $5$. Now
\begin{align*}
3^{n+1} + 8^{n+1} &= 3\cdot 3^n + 8\cdot 8^n\\
&= 3\cdot 3^n + (3+5)\cdot 8^n\\
&= 3\cdot 3^n + 3\cdot 8^n + 5\cdot 8^n\\
&= 3(3^n + 8^n) + 5\cdot 8^n.
\end{align*}
The first term is not divisible by $5$ by the induction hypothesis, and because $3$ and $5$ are relatively prime, so since $5$ divides the second term, it does not divide their sum.
| {
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Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$. $$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\} \tag{$\star$}$$
My proof of $(\star)$ is as follows:
\begin{align}
10^{2n+1}+1
&= 10\cdot10^{2n}+1 \\
&= (11-1)\cdot10^{2n}+1 \\
&= 11\cdot10^{2n}-10^{2n}+1 \\
&= 11\cdot10^{2n}-\left(10^{2n}-1\right) \\
&= 11\cdot10^{2n}-\left(100^{n}-1\right) \\
&= 11\cdot10^{2n}-\left((99+1)^{n}-1\right) \\
&= 11\cdot10^{2n}-\left(1+\binom{n}{1}99+\binom{n}{2}99^2+\cdots+\binom{n}{n-1}99^{n-1}+99^n-1\right) \\
&= 11\cdot10^{2n}-\underbrace{99}_{11\cdot9} \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right) \\
&= 11\left(10^{2n}-9 \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right)\right)
\end{align}
Is there an easier way to prove $(\star)$? The expansion of $(99+1)^n$ seems unnecessarily complicated, but I wasn't sure how else to go from there. Easier proofs are welcome!
| Because for $n\geq1$ we have: $$10^{2n+1}+1=(10+1)(10^{2n}-10^{2n-1}+...-10+1).$$
For $n=0$ it's obvious.
| {
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shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{25}- \frac{8}{5})}^2} = \frac{\sqrt{52}}{25}$
but my answer is wrong. Where am i wrong?
| Hint
$3x=12-4y=4(3-y)$
WLOG any point on the line $P(4m,3-3m)$
$\dfrac43=\dfrac{3-3m-0}{4m-0}$
$m=?$
Can you find the distance of $P$ from the origin
Subtract the length of the radius from the distance
| {
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When going from $(x+2)^2=5$ to $x+2=\pm \sqrt{5}$, why isn't there also a $\pm(x+2)$? Say I am solving the following equation:
$$(x+2)^2 = 5$$
$$x + 2 = \pm \sqrt{5}$$
$$x = -2 \pm \sqrt{5}$$
However, when I took the positive and negative square root of $5$ in the second line, I did not take the positive and negative square root of $(x+2)^2$, which would be $\pm (x+2)$. Why is this?
| When solving $(x+2)^2=5$, recall in general that for $x\in\mathbb{R}$ we have $\sqrt{x^2}=|x|$. And since clearly $x+2\in\mathbb{R}$, we have by the last identity that by taking the square root of both sides of $(x+2)^2=5$ $$\sqrt{(x+2)^2}=|x+2|=\sqrt{5}\tag1$$
Then we have reduced the problem to solving $$|x+2|=\sqrt{5}\tag2$$
Recall once more that in general $$|x|=b>0\implies x=b\text{ or }x=-b\tag3$$
Thus putting $(1)$ and $(3)$ together, $$|x+2|=\sqrt{5}\iff x+2=\sqrt{5}\text{ or } x+2=-\sqrt{5}$$
Futhermore, $$x=\sqrt{5}-2\text{ or } x=-\sqrt{5}-2\iff \boxed{x=-2\pm\sqrt{5}}$$
| {
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Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
I saw instantly that $$h(x - \frac{1}{x})= x^2 - \frac{1}{x^2} = \left( x -\frac{1}{x} \right)\left( x + \frac{1}{x} \right),$$ but I don't know how to proceed. I tried something like $$\left(x -\frac{1}{x} \right) = a$$ and $$\left( x + \frac{1}{x} \right) = a + \frac{2}{x},$$ trying to apply $h(a)$ but it doesn't seem to work.
Any hints?
| There is no function satisfying $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $ x \ne 0$.
For any $a \ne 0$ the equation $x - \frac 1x = a$ has two solutions $x_1, x_2$, which are related by $x_1 x_2 = -1$. Then
$$
x_1^2 - \frac{1}{x_1^2} = h(a) = x_2^2 - \frac{1}{x_2^2} = \frac{1}{x_1^2} - x_1^2 \\
\implies x_1^4 = 1 \implies x_1 = \pm 1 \\
\implies a= 0 \, ,
$$
a contradiction.
More concretely, for $x=2$ you would get
$$
h(\frac 32) = h(2 - \frac 12) = 4 - \frac 14 = \frac{15}{4}
$$
and for $x = -1/2$
$$
h(\frac 32) = h(-\frac 12 + 2) = \frac 14 - 4 = -\frac{15}{4}
$$
which is a clear contradiction.
| {
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Find partial fractions of $\frac{3x^2-2x+1}{x^2(1-x^2)}$ and then deduce partial fractions of $\frac{3x^2-8x+6}{x(x-1)^2(2-x)}$ Find partial fractions of, $$\frac{3x^2-2x+1}{x^2(1-x^2)}$$
Hence deduce partial fractions of $$\frac{3x^2-8x+6}{x(x-1)^2(2-x)}$$
My Try
I was able to do the first part,$$\frac{3x^2-2x+1}{x^2(1-x^2)}=\frac1{x^2}-\frac2x+\frac3{x+1}-\frac1{x-1}$$
But I find it difficult to do the second part. Should I use a substitute or any other method? Please give me a hint to work this out. Thank you very much!
| Let $X=x-1$ then
$$\frac{3x^2-8x+6}{x(x-1)^2(2-x)}=-\frac{3(X+1)^2-8(X+1)+6}{(X+1)X^2(X-1)}=
-\frac{3X^2-2X+1}{X^2(X^2-1)}$$
Now you can use the first part.
| {
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Knowing that $u_1 = 1, u_2 = 3$ and $u_{n + 2} = 2u_{n + 1} - u_n + 1, \forall n \in \mathbb Z^+$, prove that $4u_{n + 2}u_n + 1$ is a square number.
Knowing that $$\large \left\{ \begin{align*} u_1 = 1&, u_2 = 3\\ u_{n + 2} = 2u_{n + 1} - u_n + 1&, \forall n \in \mathbb Z^+ \end{align*} \right.$$, prove that $\large 4u_{n + 2}u_n + 1$ is a square number.
Here are my thoughts.
$f\colon \mathbb Z \to \mathbb Z^+, u_n \mapsto v_n = 2u_n + 1, n \in \mathbb Z^+$. The problem becomes
Knowing that $$\large \left\{ \begin{align*} v_1 = 3&, v_2 = 7\\ v_{n + 2} + v_n = 2(v_{n + 1} + 1)&, \forall n \in \mathbb Z^+ \end{align*} \right.$$, prove that $\large v_{n + 2}v_n - v_{n + 1}$ is a square number.
This is a problem using induction, but I've just started studying the basics of it and haven't had any experience on the implications of it.
| Use characteristic polynomials, like here, to get to the conclusion that
$$(x-1)(x^2-2x+1)=0$$
is the characteristic polynomial and $x=1$ is a root of multiplicity 3. Then
$$u_n=(an^2+bn+c)\cdot 1^n$$
and
$$u_1=1=a+b+c$$
$$u_2=3=4a+2b+c$$
$$u_3=6=9a+3b+c$$
which resolves into $a=\frac{1}{2}$, $b=\frac{1}{2}$ and $c=0$, thus
$$u_n=\frac{1}{2}(n+1)n$$
and
$$4u_{n+2}u_n +1=(n+3)(n+2)(n+1)n+1=\\
(n^2 + 3 n)^2 + 2 (n^2 + 3 n) + 1=\\
(n^2 + 3 n+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Cauchy Schwarz: Proving this inequality For positive reals a,b,x and y, prove that
$(5a^2 + 2ab + 3b^2)(5x^2 + 2xy + 3y^2) \ge (5ax + ay + bx + 3by)^2$
My attempt: By Cauchy we can show that the LHS $\ge(5ax + 2(abxy)^{1/2} + 3by)^2$
I can then show that $ay + bx \ge 2(abxy)^{1/2}$ using AM-GM, but this doesn't help.
How can I proceed?
Thank you.
| By C-S we obtain:
$$(5a^2+2ab+3b^2)(5x^2+2xy+3y^2)=\left(\frac{1}{5}(5a+b)^2+\frac{14}{5}b^2\right)\left(\frac{1}{5}(5x+y)^2+\frac{14}{5}y^2\right)\geq$$
$$\geq\left(\frac{1}{5}(5a+b)(5x+y)+\frac{14}{5}by\right)^2=(5ax+ay+bx+3by)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$ Find the approximation of $\sqrt{80}$ with an error $\lt 0.001$
I thought it could be good to use the function $f: ]-\infty, 81] \rightarrow \mathbb{R}$ given by $f(x) = \sqrt{81-x}$
Because this function is of class $C^{\infty}$, we can compute its Taylor expansion given by :
$$T^n_{0} = 9 - \frac{1}{2}(81)^{-1/2}(x) + \frac{1}{4}(81)^{-3/2}\frac{x^2}{2}-\frac{3}{8}(81)^{-5/2}\frac{x^3}{6}\ + \dotsm $$
By the Lagrange remainder, $\exists$ for each $n \in \mathbb{N}$, $c_n \in [0,1]$ such that :
$$R^n(1) = f^{n+1}(c) \frac{1-0^{n+1}}{(n+1)!} = f^{n+1}(c) \frac{1}{(n+1)!} \leq 9.\frac{1}{(n+1)!} \lt 10^{-3}$$
$=> n(+1)! \gt \frac{9}{10^{-3}} = 9000$
So we can take $n = 8$
The approximation seems a little bit tricky to calculate especially without a calculator. I'm wondering if everything above is correct ?
| Barry Cipra's answer does everything that is required with no fuss at all, so it gets my vote, but I was still curious to know how much work it would be to get the result using the binomial series. Not a lot, I think:
$$
1 - \sqrt{1 - x} = \tfrac{1}{2}\sum_{k=0}^\infty a_kx^{k+1} \quad (|x| < 1),
\qquad a_k = \frac{1\cdot3\cdots(2k-1)}{4\cdot6\cdots(2k+2)},
\qquad a_0 = 1, \ a_1 = \tfrac{1}{4}.
$$
Because $a_{k+1} < a_k$, we can estimate the remainder using the
geometric series:
$$
0 < 1 - \sqrt{1 - x} - \tfrac{1}{2}\sum_{k=0}^{p-1} a_kx^{k+1} <
\frac{a_px^{p+1}}{2(1 - x)} \quad (p \geqslant 0, \ 0 < x < 1),
$$
In the present case, it is enough to take $p = 1$:
$$
0 < 1 - \frac{x}2 - \sqrt{1 - x} < \frac{x^2}{8(1 - x)} \quad (0 < x < 1).
$$
Taking $x = \frac1{81},$ and multiplying throughout by $9,$ we get $x/(1 - x) = \frac1{80},$ and so
$$
0 < \frac{161}{18} - \sqrt{80} < \frac1{8 \times 9 \times 80} = \frac1{5760}.
$$
In fact,
\begin{align*}
\frac{161}{18} - \sqrt{80} & \bumpeq 0.0001725, \\
\frac1{5760} & \bumpeq 0.0001736.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 8
} |
Find the inverse in $F = \frac{\Bbb{Z}_{7}[x]}{g(x)}$ of a polynomial $p(x)$ $g(x) = x^{3}+2x^{2}+1$
$p(x) = x^{2}+x+2$
I don't know what I did wrong, hope you figured it out:
$g(x) = p(x)(x+1) + 4x + 6$
$p(x) = (4x+6)(2x+6)+1$
$1=p(x)-(2x+6)(4x+6)$
$1=p(x)-(2x+6)[g(x) - p(x)(x+1)]$
$1=p(x)+(5x+1)[g(x) + p(x)(6x+6)]$
$1=(5x+1)g(x) + 2(5x+1)(6x+6)p(x)]$
$1=(5x+1)g(x) + 2(2x^{2}+x+6)p(x)]$
$1=(5x+1)g(x) + (4x^{2}+2x+5)p(x)]$
So: my solution is $(4x^{2}+2x+5)$, but the text book report another result. Any suggestion?
Also: how can I say, without a doubt, that $F$ is a field?
| $F$ can be seen as the algebra generated by the matrix $M = \begin{pmatrix}
0 & 0 & -1 \\
1 & 0 & 0 \\
0 & 1 & -2 \\
\end{pmatrix}$( with entries in $GF(7)\; $), called the companion matrix of the polynomial $g(x) = x^3 + 2x^2 + 1$. This matrix describes the action of the multiplication with $x$ namely $1 \mapsto x$, $x \mapsto x^2$ and $x^2 \mapsto x^3=-1-2x^2$. Then the polynomial $p(x)=x^2+x+2$ is represented by the matrix $N = M^2+M+2M^0 = \begin{pmatrix}
2 & -1 & 1 \\
1 & 2 & -1 \\
1 & -1 & 4 \\
\end{pmatrix}$. The inverse of $p$ is then represented by the inverse $N^{-1} = \begin{pmatrix}
0 & -2 & 3 \\
1 & 0 & 5 \\
2 & 4 & -1 \\
\end{pmatrix} = 2M^2 + M $ (note that the coefficients of an element of the algebra in the basis $\{M^0, M, M^2\}$ always can be read from the first column) which corresponds to the polynomial $2x^2+x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I solve this Entrance Exam problem? This question is from NEST 2018.
Suppose ${2+ \sqrt{3}}$ and $1-i$ are roots of the equation $(x^2+px+1)(x^2-2x+q)=0$ where $p,q$ are integers and $i=\sqrt{-1}$. Then what is $p+q$ ?
| If you substitute ${2+ \sqrt{3}}$ and $1-i$ in $(x^2+px+1)(x^2-2x+q)=0$, I obtain:
$$\left\{\begin{matrix}
(({2+ \sqrt{3}})^2+p({2+ \sqrt{3}})+1)(({2+ \sqrt{3}})^2-2({2+ \sqrt{3}})+q)=0
\\ ((1-i)^2+p(1-i)+1)((1-i)^2-2(1-i)+q)=0
\end{matrix}\right.$$
From the first equation, I obtain: $(2+\sqrt3)(p+4)(3+\sqrt3+q)=0$, so $p=-4$. Now, plugging in the second equation, I have: $(2i-4)(q-2)=0$ and $q=2$.
In conclusion $p+q=-4+2=-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the min value of $\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$, if $x\ne y \ne z$ and $x,y,z\in {1,2,3,4,5}$ My answer is $\frac{5}{29}$, I just use logic to substitute numbers in the expression, but I can't prove my answer If this expression be minimum then the denominator should be the greatest, so I just let x=5 and I need to let $\frac{1}{y+\frac{1}{z}}$ be the greatest so just need this denominator be smallest so y=1 and z=4, this is what I thought.
| It's just
$$\frac{1}{5+\frac{1}{1+\frac{1}{4}}}=\frac{5}{29}.$$
Indeed, we need to prove that
$$\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\geq\frac{5}{29}$$ or
$$\frac{1}{x+\frac{z}{yz+1}}\geq\frac{5}{29}$$ or
$$\frac{yz+1}{xyz+x+z}\geq\frac{5}{29}$$ or
$$5yz(5-x)+5(5-x)+4z(y-1)+4-z\geq0.$$
Now, if $z<4$ so $$5yz(5-x)+5(5-x)+4z(y-1)+4-z>0,$$ which gives
$$\frac{1}{x+\frac{1}{y+\frac{1}{z}}}>\frac{5}{29}$$ and we'll get a greater value than $\frac{5}{29}.$
If $z=5$, so $x<5$ and we obtain $$25y(5-x)+5(5-x)+20(y-1)+4-5>0,$$ which gives again:
$$\frac{1}{x+\frac{1}{y+\frac{1}{z}}}>\frac{5}{29}.$$
For $z=4$ we obtain $$20y(5-x)+5(5-x)+16(y-1)\geq0,$$
where the equality occurs for $x=5$ and $y=1$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to find least square solution to Ax=b when columns of A are not linear independent? Let $A =\begin{bmatrix} 0 & -1 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \\\end{bmatrix}$ and $b =\begin{bmatrix} 1 \\2\\ 3 \\\end{bmatrix}$
I couldn't just solve $A^TAx=A^Tb$, since the columns of A are not independent, I was told by my professor that I must find another matrix with same $im(A)$
I found the image of A = $\text{span}\left\lbrace\begin{bmatrix} 1 \\0\\ 1 \\\end{bmatrix},\begin{bmatrix} 0 \\1\\ 1 \\\end{bmatrix}\right\rbrace$
How do we go from here to find another matrix and find the least square solution to the equation?
| I don’t understand either what you are having issues with, or what your professor told you to do.
The first thing to notice here is that the system has solutions! So a “least squares solution” really just means a regular solution. If you solve the system directly, you get $x=4-t$, $y=-1$, $z=t$. So those are “least squares solutions”.
The second thing to remember is that even if it didn’t have solutions, there is still no problem. If $A^TA\mathbf{x} = A^T\mathbf{b}$ has multiple solutions, then they all give you least squares solutions: they all end up in the same point when you apply $A$, and that point will be the one closest to $\mathbf{b}$ in the range of $A$. There is absolutely nothing else to be done.
Even if you don’t notice that the system actually has exact solutions, you can just proceed as usual.
You want to find a least squares solutions; you solve the system $A^TA\mathbf{x} = A^T\mathbf{b}$. Now, if $A^TA$ is invertible, then this system has a unique solution. When $A^TA$ is not invertible, then the system will have multiple solutions, and any such solution will be a least squares solution.
Here, we have
$$\begin{align*}
A^TA &= \left(\begin{array}{rrr}
0 & 1 & 1\\
-1 & 2 & 1\\
0 & 1 & 1
\end{array}\right) \left(\begin{array}{rrr}
0 & -1 & 0\\
1 & 2 & 1\\
1 & 1 & 1
\end{array}\right)= \left(\begin{array}{ccc}
2 & 3 & 2\\
3 & 6 & 3\\
2 & 3 & 2
\end{array}\right)\\
A^T\mathbf{b} &= \left(\begin{array}{rrr}
0 & 1 & 1\\
-1 & 2 & 1\\
0 & 1 & 1
\end{array}\right) \left(\begin{array}{c}1\\2\\3\end{array}\right) = \left(\begin{array}{c}5\\6\\5\end{array}\right).
\end{align*}$$
This system has solutions. Using Gaussian elimination, we have:
$$\begin{align*}
\left(\begin{array}{ccc|c}
2 & 3 & 2 & 5\\
3 & 6 & 3 & 6\\
2 & 3 & 2 & 5
\end{array}\right) &\to \left(\begin{array}{ccc|c}
2 & 3 & 2 & 5\\
3 & 6 & 3 & 6\\
0 & 0 & 0 & 0
\end{array}\right) \to \left(\begin{array}{ccc|c}
1 & 2 & 1 & 2\\
2 & 3 & 2 & 5\\
0 &0 & 0 & 0
\end{array}\right)\\
&\to \left(\begin{array}{rrr|r}
1 & 2 & 1 & 2\\
0 & -1 & 0 & 1\\
0 & 0 & 0 & 0
\end{array}\right) \to \left(\begin{array}{ccc|r}
1 & 0 & 1 & 4\\
0 & 1 & 0 & -1\\
0 & 0 & 0 & 0
\end{array}\right).
\end{align*}
$$
So the solutions are given by $y=-1$, $z=t$, and $x=4-t$. These are exact solution, but even if they were not, any such solution would have the same image under $A$ and that image would minimize the error, i.e., be a least squares solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Given $x, y$ that $xy-\frac{x}{y^2}-\frac{y}{x^2}=3$, work out $xy-x-y$. Today I had a competition in Xiamen, China. I know how to do the questions except this strange equation.
Given $x, y$ that $xy-\dfrac{x}{y^2}-\dfrac{y}{x^2}=3$, work out $xy-x-y$.
Such a strange question, right? I have found the integral solution are $0$ when $x=y=2$ and $3$ when $x=y=-1$, but I think there are infinitely many solutions but I can’t really prove it at that time. Can you guys help me?
| Making $y = \lambda x$ we have
$$
xy -\frac{x}{y^2}-\frac{y}{x^2}=3\Rightarrow \lambda x^3-3x=\lambda+\frac{1}{\lambda^2}
$$
and solving for $x$
$$
x = \left\{
\begin{array}{c}
\frac{\lambda +1}{\lambda } \\
-\frac{\lambda ^2+\lambda +\sqrt{3} \sqrt{-(\lambda -1)^2 \lambda ^2}}{2 \lambda ^2} \\
-\frac{\lambda ^2+\lambda -\sqrt{3} \sqrt{-(\lambda -1)^2 \lambda ^2}}{2 \lambda ^2} \\
\end{array}
\right.
$$
so the solutions for $x$ are all real if $\lambda = 1$ and in those circumstances
$$
x = y = \cases{2\\
-1}
$$
giving respectively
$$
x y - x - y = 0\\
x y - x - y = 3
$$
NOTE
$$
\lambda^3 x^3-3\lambda^2x-\lambda^3-1 = (\lambda x-\lambda-1)(\lambda^2x^2+(\lambda^2+\lambda) x+\lambda^2-\lambda+1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}$ Evaluate $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}$.
My approach is to do $\lim_{n\to\infty} \frac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}} = \lim_{n\to\infty} \frac{\left(\frac{9}{4}\right)^{n}}{\frac{3}{2}\left(\frac{2}{3}\right)^{n}+\frac{3}{2}\left(\frac{9}{4}\right)^{n}}$ but not sure what to do next.
How could I convert the $\left(\frac{2}{3}\right)^{n}$ term into $\left(\frac{9}{4}\right)^{n}$? Thanks.
| $$\dfrac{\left(\frac{3}{2}\right)^{2n}}{\left(\frac{2}{3}\right)^{n-1}+\left(\frac{3}{2}\right)^{2n+1}}
=\dfrac1{\left(\frac{2}{3}\right)^{3n-1}+\frac{3}{2}}$$
Since $\dfrac{2}{3} <1$ , its limit will be zero.
Thus the answer is $\dfrac{1}{0+\frac{3}{2}}=\dfrac{2}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to isolate $x$ when $\cos(x/2)\cos(3x)/2−3\sin(x/2)\sin(3x)= 0$ Asked to find all relative and absolute extrema of $f(x) = \sin\left(\frac{1}{2}x\right)\cos(3x)$ on the interval $[0,\pi]$. I've gotten the derivative, $$f'(x)=\frac{\cos(\frac{x}{2})\cos(3x)}{2} - 3\sin(\frac{x}{2})\sin(3x),$$ just fine, but how would I proceed to isolate x after this? I noticed that the situation is similar to the sine angle sum identity, but I don't know if there's an identity for an equation with the form $$h\cos(\alpha)\cos(\beta) - k\sin(\alpha)\sin(\beta)$$ and can't find anything online. I don't have much experience working with Euler's Formula in this context but know enough that I could probably follow an explanation based on it.
| Trying approximate solutions.
First, using trigonometric formulae, the derivative can write
$$f'(x)=\frac{1}{4} \left(7 \cos \left(\frac{7 x}{2}\right)-5 \cos \left(\frac{5
x}{2}\right)\right)$$ Forget the $\frac{1}{4}$ and plot the function. You should notice that, for the range of concern, the solution are "close" to $\frac \pi{12}$, $\frac \pi{3}$, $\frac {2\pi}{3}$ and the last one is exactly $\pi$.
So, around each of these values, build Taylor expansions to get successively
$$f'(x)=\left(7 \sin \left(\frac{5 \pi }{24}\right)-5 \cos \left(\frac{5 \pi
}{24}\right)\right)+\left(x-\frac{\pi }{12}\right) \left(\frac{25}{2} \sin
\left(\frac{5 \pi }{24}\right)-\frac{49}{2} \cos \left(\frac{5 \pi
}{24}\right)\right)+\left(x-\frac{\pi }{12}\right)^2 \left(\frac{125}{8} \cos
\left(\frac{5 \pi }{24}\right)-\frac{343}{8} \sin \left(\frac{5 \pi
}{24}\right)\right)+O\left(\left(x-\frac{\pi }{12}\right)^3\right)$$ and we know the values of the trigonometric functions of $\frac{n \pi }{24}$. Solving the quadratic will give $x_1=0.286024$ (while the "exact" solution would be $0.286061$).
$$f'(x)=-\sqrt{3}+\frac{37}{2} \left(x-\frac{\pi }{3}\right)+\frac{109}{8} \sqrt{3}
\left(x-\frac{\pi }{3}\right)^2+O\left(\left(x-\frac{\pi }{3}\right)^3\right)$$
Solving the quadratic will give $x_2=1.13170$ (while the "exact" solution would be $1.13265$).
$$f'(x)=1-\frac{37}{2} \sqrt{3} \left(x-\frac{2 \pi }{3}\right)-\frac{109}{8}
\left(x-\frac{2 \pi }{3}\right)^2+O\left(\left(x-\frac{2 \pi
}{3}\right)^3\right)$$
Solving the quadratic will give $x_3=2.12520$ (while the "exact" solution would be $2.12525$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
initial value problem $y'=-y^2, y(1)= \sqrt{2}$ hey I cant seem to solve this at all, not sure how you do it with no x values
$$y'=-y^2, y(1)= \sqrt{2}$$
| We have $y'=-y^2 \iff \frac{y'}{-y^2}=1 \iff \frac{d}{dx}(\frac{1}{y})=\frac{dx}{dx} \iff \frac{1}{y}=x+c$
$y(1)=\sqrt{2} \iff c=\frac{\sqrt{2}-2}{2}$
So $1/y=\frac{2x+\sqrt{2}-2}{2}\iff y=\frac{2}{2x+\sqrt{2}-2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
System of three non-linear equations Can anyone help me solve this system of equations?
$$a_1 a_3 -a_2 ^2=0$$
$$a_1+a_3-2a_2-16=0$$
$$a_1 a_3 +64a_1 - a_2 ^2 -16 a_2 -64 =0$$
After couple of steps I got $4a_1-a_2-4=0, (a_2-a_3)^2=16$. Then we have two cases $a_2-a_3=4$ and $a_2-a_3=-4$, but I couldn't finish this. Thank you for your time
| I'm going to use $x=a_1, \ y=a_2, \ z=a_3$
$$xz - y^2=0 \tag A$$
$$x + z - 2y - 16 = 0 \tag B$$
$$xz + 64x - y^2 - 16y - 64 = 0 \tag C$$
Substitute $xz = y^2$ from (A) into (C)
\begin{align}
xz + 64x - y^2 - 16y - 64 &= 0 \\
y^2 + 64x - y^2 - 16y - 64 &= 0 \\
64x - 16y - 64 &= 0 \\
y &= 4x - 4 \tag D
\end{align}
Substitute $y = (4x - 4)$ from (D) into (B)
\begin{align}
x + z - 2y - 16 = 0 \\
x + z - 2(4x - 4) - 16 = 0 \\
-7x + z - 8 &= 0 \\
z &= (7x + 8) \tag E
\end{align}
Substitute (D) and (E) into (A)
\begin{align}
xz - y^2 &= 0 \\
x(7x + 8) - (4x - 4)^2 &= 0 \\
7x^2 + 8x - 16x^2 + 32x - 16 &= 0 \\
-9x^2 + 40x - 16 &= 0 \\
-(9x - 4)(x - 4) &= 0 \\
x &\in \left\{ \dfrac 49, 4 \right\}
\end{align}
We get $$
(a_1,a_2,a_3) = \left(\dfrac 49, -\dfrac{20}{9}, \dfrac{100}{9} \right)
\ \text{and} \
(a_1,a_2,a_3) = (4, 12, 36)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Find the sum of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$ up to $n$ terms
Find the sum of first $n$ terms of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$
*
*When $n$ is even.
*When $n$ is odd.
This sum can be written as
$$\sum_{1}^n (2k-1)^{3} +3 \sum_{1}^n (2k)^{2} $$
I can handle the sum up to n terms when it is not specified that $n$ is even or odd.
In this problem I'm confused, what changes should be done to get sum for even or odd $n$.
In my textbook, $n$ is replaced by $2m$ and then they solved the problem for first $m$ terms and then substituted $m = n/2$ and same is done for odd case, by substituting $n=2m-1$.
I didn't get that solution.
Any suggestion would be helpful.
| Details for your comment above:
$n$ is even:
$$\sum_{1}^{n/2} (2k-1)^{3} +3 \sum_{1}^{n/2} (2k)^{2}$$
Example:
$$1^3 + 3\cdot 2^2 = \sum_{k=1}^{2/2}(2k-1)^3+3\sum_{k=1}^{2/2}(2k)^2$$
$n$ is odd:
$$\sum_{1}^{(n+1)/2} (2k-1)^{3} +3 \sum_{1}^{(n+1)/2-1} (2k)^{2}$$
Example:
$$1^3 + 3\cdot 2^2 + 3^3 = \sum_{k=1}^{(3+1)/2}(2k-1)^3+3\sum_{k=1}^{(3+1)/2-1}(2k)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Discrete Mathematics - Combinatorics proof $\\$ I need to prove $\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot\left(\begin{array}{c}n\\ k\end{array}\right)4^k=\frac{5^n+(-3)^n}{2}$
my Attempt below
$\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot \left(\begin{array}{c}n\\ k\end{array}\right)4^k=\frac{1^0+(-1)^0}{2}\cdot \frac{n!}{0!(n-0)!}\cdot4^0\\ +\frac{1^1+(-1)^1}{2}\cdot \frac{n!}{1!(n-1)!}\cdot4^1+\frac{1^2+(-1)^2}{2}\cdot \frac{n!}{2!(n-2)!}\cdot4^2+\frac{1^3+(-1)^3}{2}\cdot \frac{n!}{3!(n-3)!}\cdot4^3+...+\frac{1^n+(-1)^n}{2}\cdot \frac{n!}{n!(n-n)!}\cdot4^n=\\ =\frac{1+1}{2}\cdot \frac{n!}{1n!}\cdot1+\frac{1+1}{2}\cdot \frac{n!}{2(n-2)!}\cdot16+\dotso+\frac{1^n+(-1)^n}{2}\cdot \frac{n!}{n!(n-n)!}\cdot4^n=\\= \frac{n!}{n!}+ \frac{4n!}{(n-1)!}+1\cdot \frac{16n!}{2(n-2)!} +1\frac{64n!}{6(n-3)!}+1\dotsm\ \frac{4^nn!}{n!(n-n)!}+1=$
$\frac{1}{2}\left[\sum_{k = 0}^{n} \binom{n}{k}4^k + \sum_{k = 0}^{n} \binom{n}{k}(-4)^k\right]=\\$1/2$(1+x)^n=\frac{1}{2}((1+4)^n+(1-4)^n=\frac{5^n+(-3)^n}{2}$
| Observe that
$$\sum_{k = 0}^{n} \left(\frac{1^k + (-1)^k}{2}\right)\binom{n}{k}4^k = \frac{1}{2}\left[\sum_{k = 0}^{n} \binom{n}{k}4^k + \sum_{k = 0}^{n} \binom{n}{k}(-4)^k\right]$$
By the Binomial Theorem,
$$(1 + x)^n = \sum_{k = 0}^{n} \binom{n}{k}x^k$$
For the first summation, $x = 4$; for the second summation $x = -4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$ the same as $\frac{\frac{2-b}{2b}}{b-2}$? Have I subtracted 2 fractions correctly?
$\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$
Starting with the numerator which is a difference of fractions, the least common denominator is 2b? So:
$\frac{2(1)}{2b}-\frac{b(1)}{2b}$ = $\frac{2}{2b}-\frac{b}{2b}$ = $\frac{2-b}{2b}$
So my newly simplified expression is:
$\frac{\frac{2-b}{2b}}{b-2}$
My question - is this correct? If yes, great - now how would I go about simplifying further by taking the denominator $b-2$ into account? If no, where did I go wrong?
| It is correct, since $$\frac{1}{b}-\frac{1}{2}=\frac{2}{2b}-\frac{b}{2b}=\frac{2-b}{2b}$$
| {
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Quickly evaluating this limit: $\lim\limits_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$ I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.
$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right).$$
If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as
$$\frac{1}{x^2}-\frac{1}{\sin^2 x} = \frac{\sin^2 x - x^2}{x^2\sin^2 x} = \left(\frac{x^2}{\sin^2 x}\right)\left(\frac{\sin x + x}{x}\right)\left(\frac{\sin x - x}{x^{3}}\right),$$
where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.
Edit: Thanks to everyone's clear responses, I now understand:
*
*Taylor expansions of trig functions are immensely powerful in evaluating limits, in this case turning the problem into a limit of a rational function.
*If a factor of a limit exists and is nonzero, it can be factored out without affecting convergence of the product. Formally, suppose $\lim f(x)$ exists and $g(x)$ is a factor of $f(x)$.
Then if $\lim g(x)$ exists in $\mathbb{R}\backslash\{0\}$, $\lim f(x)/g(x)$ exists also. That is to say, one can be "opportunistic" about simplifying ones limits if a factor with a real nonzero limit is discovered.
| $$
\frac {1}{x^2}-\frac{1}{\sin^2x} = \left(\frac {1}{x}+\frac{1}{\sin x}\right)\left(\frac {1}{x}-\frac{1}{\sin x}\right) = \frac{1}{x^2}\left(\frac {x}{x}+\frac{x}{\sin x}\right)\left(\frac {x}{x}-\frac{x}{\sin x}\right)
$$
so
$$
\lim_{x\to 0}\left(\frac {1}{x^2}-\frac{1}{\sin^2x}\right) =2\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{x}{\sin x}\right) = 2\lim_{x\to 0}\frac{x}{\sin x}\frac{1}{x^2}\left(\frac{\sin x}{x}-1\right) = 2\lim_{x\to 0}\frac{1}{x^2}\left(\frac{\sin x}{x}-1\right) = 2\lim_{x\to 0}\frac{1}{x^2}\left(1-\frac{x^2}{3!}+o(x^4)-1\right) = -\frac 13
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a formula for $(1^3)\cdot(n)+(2^3)\cdot(n-1)+(3^3)\cdot(n-2)+ \cdots + (n^3)\cdot(1)$
I need a formula for this sum:
$$(1^3)\cdot(n)+(2^3)\cdot(n-1)+(3^3)\cdot(n-2)+ \cdots + (n^3)\cdot(1)$$
I have found this formula :
$$1\cdot n +2\cdot(n-1)+3\cdot(n-2)+ \cdots +(n-1)\cdot 2 +n\cdot1= \frac16n(n+1)(n+2)$$
but it is not exactly what I need.
Thanks in advance.
| I can rewrite the sum as $S_n=\sum\limits _{k=1}^{n}k^3(n+1-k)=(n+1)\sum\limits _{k=1}^{n}k^3 - \sum\limits _{k=1}^{n}k^4$. Now, we can show by induction that $A_n=\sum\limits _{k=1}^{n}k^3=\frac{n^4+2n^3+n^2}{4}$ and $B_n=\sum\limits _{k=1}^{n}k^4=\frac{6n^5+15n^4+10n^3-n}{30}$. Substituing this formulas in: $S_n=(n+1)A_n - B_n$, I have: $$S_n=(n+1)\cdot \frac{n^4+2n^3+n^2}{4}-\frac{6n^5+15n^4+10n^3-n}{30}=\frac{3n^5+15n^4+25n^3+15n^2+2n}{60}$$
This is the expansion of your sum.
| {
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Elementary number theory, 'concert-ticket-arithmetic' Four friends, call them A,B,C and D are planning to go to a concert, but they realize that they
are a few dollars short to buy tickets.(50 $ per ticket).
We know that each of them has an integer amount of dollars.
If B borrowed 1$ from A,then B would have 2/3 of A’s balance
If C borrowed 2$ from B,then C would have 3/5 of B’s balance
If D borrowed 3$ from C, then D would have 5/7 of C’s balance
At least how much more money do they need(in $) in order to afford 4 tickets?
| Let $a$ be the money of $A$.
$b$ be the money of $B$. $c$ the money of $C$ and $d$ the money of $D$. I have the following system:
$$\left\{\begin{matrix}
b+1=\frac{2}{3}(a-1)
\\c+2=\frac{3}{5}(b-2)
\\d+3=\frac{5}{7}(c-3)
\end{matrix}\right.$$
Solving for $a,b,c,d$, I obtain:
$$\left\{\begin{matrix}
b=\frac{2}{3}(a-1)-1
\\c=\frac{1}{5}(2a-1)-4
\\d=\frac{1}{7}(2a-1)-8
\end{matrix}\right.$$
From this I can say: $3|a-1
\land 7|2a-1 \land 5|2a-1$ or in other words: $3|a-1\land 35|2a-1$. Also:
$$\left\{\begin{matrix}
35\alpha=2a-1
\\3\beta=a-1
\end{matrix}\right.$$
With $\alpha,\beta \in Z$. Solving for $\alpha$ and $\beta$, I obtain:
$$\left\{\begin{matrix}
\alpha=6n+5
\\\beta=35n+29
\end{matrix}\right.$$
With $n \in N$, So, setting $n=0$, I obtain the solutions:
$$\left\{\begin{matrix}
a=88
\\b=57
\\c=31
\\d=17
\end{matrix}\right.$$
Now let $R$ the money they need: $R=200-a-b-c-d=200-88-57-31-17=7$.
| {
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How to determine algebraically whether an equation has an infinite solutions or not? I was learning for the first time about partial fraction decomposition. Whoever explains it, emphasises that the fraction should be proper in order to be able to decompose the fraction. I was curious about knowing what happens if I try to decompose an improper fraction, So I tried to do one:
$\frac{x^2 - 4}{(x + 5)(x - 3)}$
I got the equation:
$\frac{(Ax + B)(x - 3) + (Cx + D)(x + 5)}{(x + 5)(x - 3)} = \frac{x^2 - 4}{(x + 5)(x - 3)}$. I have 4 unknowns: A, B, C and D.
$\therefore (Ax + B)(x - 3) + (Cx + D)(x + 5) = x^2 - 4$
After expanding and regrouping the coefficients:
$(A + C) x^2 + (-3A + B + 5C + D)x + (-3B + 5D) = x^2 - 4$
Here the coefficient of the term $x^2$ is 1 therefore:
$(A+C) = 1$
similarly:
$(-3A + B + 5C + D) = 0$
$(-3B + 5D) = -4$
I still have to get one more equation to be able to solve this system so I substituted 1 for x and I got this equation:
$-2A - 2B + 6C + 6D = -3$
After getting four equations I used this site to solve the system of equations. Unfortinetly I got no soultion. Tried another site and also the same result.
I've tried to use different values for x and got another equaitons like:
for x = 2 : $-2A - B + 24C + 7D$
for x = -1 : $4A - 4B - 4C + 4D$
for x = -2 : $10A - 5B - 6C + 3D$
But also that didn't work. Always the system of equations have an infinite solutions.
After tring to figure out why this is happening, I've managed to prove logically that this equation:
$(Ax + B)(x - 3) + (Cx + D)(x + 5) = x^2 - 4$
has an infinite solutions and my approach was as follows:
After doing polynomial long division and decomposing the fraction using the traditional way, the result should be:
$\frac{5}{8(x - 3)} - \frac{21}{8(x - 5)} + 1$
Now I can add the last term (the one) to the first term and get the follows:
$\frac{8x-19}{8(x-3)} - \frac{21}{8(x+5)}$
From that solution I can see that $A = 0, B = \frac{-21}{8}, C = 1, D = \frac{-19}{8}$. After all these are just the coefficients of the terms. and this solution worked fine.
Alternatively I can add the one to the second term instead and get:
$\frac{5}{8(x-3)} + \frac{8x + 19}{8(x+5)}$
Now $A = 1, B = \frac{19}{8}, C = 0, D = \frac{5}{8}$
Generally, after adding the one to any of the terms, I can add any number to one of the terms and add its negative to the other term and the equation will remain the same, But the value of the 4 constants (A, B, C, and D) will change. And from that I got convinced that there are an infinite solutions to this equation.
But Algebraically? I'm not able to prove that it has an infinite solutions algebraically. And my questions is how to prove algebraically that this equation has an infinite solutions? Or generally how to know whether the equation has just one solution or an infinite?
| You are trying to represent $\frac{x^2-4}{(x+5)(x-3)}$ in the form $\frac{Ax+B}{x+5}+\frac{Cx+D}{x-3}$. As explained in other responses, you have three equations in four unknowns so there are an infinite set of solutions.
The reason why there are an infinite set of solutions is that if you have any one solution $(A,B,C,D)$ such that
$\displaystyle\frac{x^2-4}{(x+5)(x-3)}=\frac{Ax+B}{x+5}+\frac{Cx+D}{x-3}$
then we also have, for any number $t$,
$\displaystyle\frac{x^2-4}{(x+5)(x-3)}=(\frac{Ax+B}{x+5}+t)+(\frac{Cx+D}{x-3}-t)
\\\Rightarrow \frac{x^2-4}{(x+5)(x-3)}=(\frac{Ax+B}{x+5}+\frac{tx+5t}{x+5})+(\frac{Cx+D}{x-3}-\frac{tx-3t}{x-3})
\\\Rightarrow \frac{x^2-4}{(x+5)(x-3)}=\frac{(A+t)x+(B+5t)}{x+5}+\frac{(C-tx)x+(D+3t)}{x-3}$
and so $(A+t,B+5t,C-t,D+3t)$ is also a solution.
So from the solution $(1, \frac{19}{8}, 0, \frac{5}{8})$ we can generate an infinite family of solutions $(1+t, \frac{19}{8}+5t,-t,\frac{5}{8}+3t)$.
On the other hand, if you change your representation to
$\displaystyle\frac{x^2-4}{(x+5)(x-3)}=A + \frac{B}{x+5}+\frac{C}{x-3}$
then you will find you have three equations in three unknowns and these equations will have a unique solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Problem with proof by induction I am struggling with the following equation, which I need to proof by induction:
$$\sum_{k=1}^{2n}\frac{(-1)^{k+1}}{k}= \sum_{k=n+1}^{2n}\frac{1}{k}$$
$n\in \mathbb{N}$. I tried a few times and always got stuck.
Help would be appreciated.
| The proof is simple. By induction hypothesis, assume that the statement is true for $s < n$. Then, we have
\begin{align}
\sum\limits_{k = 1}^{2n} \dfrac{\left( 1 \right)^{k + 1}}{k} &= \sum\limits_{k = 1}^{2 \left( n - 1 \right)} \dfrac{\left( 1 \right)^{k + 1}}{k} + \dfrac{1}{2n - 1} - \dfrac{1}{2n} \\
&= \sum\limits_{k = n}^{2 \left( n - 1 \right)} \dfrac{1}{k} + \dfrac{1}{2n - 1} - \dfrac{1}{2n} \\
&= \sum\limits_{k = n + 1}^{2 \left( n - 1 \right)} \dfrac{1}{k} + \dfrac{1}{2n - 1} + \dfrac{1}{n} - \dfrac{1}{2n} \\
&= \sum\limits_{k = n + 1}^{2n} \dfrac{1}{k}
\end{align}
which completes the proof!
| {
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Non Linear Algebraic Equation If$$ abc = 4(4+a+b+c) , find $$ $$1/(2+a) +1/(2+b) + 1/(2+c).$$
My approach; put $$a=b=c=4 $$ and we are done. Looking at the symmetry of the equation, i think it is the only value that will satisfy the given equation or if otherwise, it is too difficult to get it by hit and trail.
Any better solution will be appreciated.
Thank you.
| This is basically an exercise in simplification.
Rewrite the target expression as (after adding and simplifying): $$\frac{12+4(a+b+c)+ab+bc+ca}{8+4(a+b+c)+2(ab+bc+ca)+abc}=\frac{4(4+a+b+c)+ab+bc+ca-4}{4(4+a+b+c)+2(ab+bc+ca)+abc-8}.$$
Substituting using the given relationship results in $$\frac{abc+ ab+bc+ca-4}{2abc+2(ab+bc+ca)-8}=\frac{1}{2}.$$
| {
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Evaluating $ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $
I need help with this limit:
$$ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $$
I've tried using the $$ \tan(a-b) = \frac{\tan a-\tan b}{1+\tan a\tan b} $$ with $(x+a)$ as $a$ and $(x)$ as $b$ but having difficulty getting beyond that.
| The best way to do this problem is to realize that the limit is, by definition, $\tan ' (x)$. Knowing that $\tan x = \frac{\sin x}{\cos x}$ and knowing $\sin '(x) = \cos x$ and $\cos ' (x) = -\sin x$, you can calculate the limit with the quotient rule.
As you've said in a comment, the key to doing this problem straight ahead without first computing $\sin'(x)$ and $\cos '(x)$ is to use the identity
$$
\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}
$$
Substituting, we have
\begin{align}
\frac{\tan(x + a) - \tan x}{a} &= \frac{\frac{\tan x + \tan a}{1 - \tan x \tan a} - \tan x}{a}
\\&=\frac{\tan x + \tan a - (1 - \tan x \tan a) \tan x}{a(1 - \tan x \tan a)}
\\&= \frac{\tan x + \tan a(1 + \tan^2 x) - \tan x}{a - a\tan x \tan a}
\\&= \frac{\tan a \sec^2 x}{a - a \tan x \tan a}
\\&= \frac{\tan a}{a} \cdot \frac{1}{1 - \tan x \tan a} \cdot \sec^2 x
\end{align}
So, if we can just show $\lim_{a \to 0} \frac{\tan a}{a} = 1$ and $\lim_{a \to 0} \frac{1}{1 - \tan x \tan a} = 1$, we'll have arrived at the expected result.
The second limit is easy.
$$
\lim_{a \to 0} \frac{1}{1 - \tan x \tan a} = \frac{1}{1 - \tan x \lim_{a \to 0} \tan a} = \frac{1}{1 - \tan x \cdot 0} = 1
$$
The first limit is a bit trickier. On geometric grounds, one can argue that
$$
\cos a \leq \frac{\sin a}{a} \leq 1
$$
Hence
$$
1 \leq \frac{\tan a}{a} \leq \sec a
$$
Since $\lim_{a \to 0} \sec a = 1$, the squeeze theorem tells us that $\lim_{a \to 0} \frac{\tan a}{a} = 1$.
Now all we need to do is pass the limit on the difference quotient calculation above, use the two limits we just "proved" (a more careful proof might be called for depending on the demands of the course), and conclude that the limit is $\sec^2 x$.
| {
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Proving limits of multivariate function with epsilon-delta definition I want to solve this problem using epsilon-delta definition :
$$\lim_{(x,y) \to (0,0)} \frac{e^{-\frac{1}{x^{2}+y^{2}}}}{x^{2}+y^{2}}$$
However, I have no idea where to start first - how to modify and apply some kind of inequality.
| For $r \ge 1$, you have $1 \le \frac{e^r}{r^2}$ hence $0 < e^{-r}\le \frac{1}{r^2}$.
Replace $r$ with $1/(x^2+y^2)$. You get
$$e^{-1/(x^2+y^2)} \le (x^2+y^2)^2$$ and therefore
$$0 < \frac{e^{-1/(x^2+y^2)}}{x^2+y^2}\le (x^2+y^2)$$ for
$\sqrt{x^2+y^2} \le 1$. Now, fix $\epsilon > 0$ and take $\delta = \min(1, \sqrt{\epsilon})$. If $\sqrt{x^2+y^2} \le \delta$, you get
$$0 < \frac{e^{-1/(x^2+y^2)}}{x^2+y^2}\le \epsilon$$
| {
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For the function $f(x,y,z)= xyz$ how close to the point $(0,0,0)$ should one take the point $(x,y,z)$ in order to make $|f(x,y,z)-f(0,0,0)|<0.008$?
For the function
$$f(x,y,z)= xyz$$
how close to the point $(0,0,0)$ should one take the point $(x,y,z)$ in order to make $|f(x,y,z)-f(0,0,0)|<0.008$?
here is my solution:
after substitution in the absolute value inequality we get
$|xyz|<0.008$ $\Rightarrow$ $\sqrt {x^{2}y^{2}z^{2}}<0.008$
but for $x,y,z <1$
$\sqrt {x^{2}y^{2}z^{2}} < \sqrt{x^{2}+y^{2}+z^{2}}$
And If we choosed the distance to be less than $\delta = 0.008$ the condition will hold.
the solution of the book is $\delta = 0.2\sqrt{3}$
my questions are :
Is my solution correct?
how to obtain the solution of the book?
| To get the solution of the book you may use GM-AM:
*
*$\sqrt[3]{abc}\leq \frac{a+b+c}{3}$
$$|xyz| = \sqrt{x^2y^2z^2} \stackrel{GM-AM}{\leq}\sqrt{\frac{(x^2+y^2+z^2)^3}{3^3}} \stackrel{!}{<}0.008 = 0.2^3$$
Now, isolate $\sqrt{x^2+y^2+z^2}$:
$$\sqrt{\frac{(x^2+y^2+z^2)^3}{3^3}} <0.2^3 \Leftrightarrow \sqrt{x^2+y^2+z^2} < \left(0.2^3\sqrt{3^3} \right)^{\frac{1}{3}}=0.2\sqrt{3}$$
So, $\delta =0.2\sqrt{3}$ does the job.
| {
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Observe that all natural numbers n^3 can be written as a sum of n odd integers that are revolving around n^2 For example,
$1^3$ = $1$,
$2^3$ = $(2^2-1) + (2^2+1)$
$3^3$ = $(3^2-2) + (3^2) + (3^2 + 2)$
$4^3$ = $(4^2-3) + (4^2 - 1) + (4^2 + 1) + (4^2 + 3)$
I have been able to deduce a general formula for this behavior, which would be:
$n^3 = (n^2 - (n-1)) + (n^2 - (n-3)) + ... + (n^2 + (n-3)) + (n^2 + (n-1))$
I am then asked to prove this general formula via induction and I believe applying the second principle of induction would be more prudent than the first.
I read an explanation about it here: What is the second principle of finite induction?
It tells me that I have to assume all $n$ from 1 to $k$ is true instead of just $n = k$, so I assume I have to factor multiple instances of $i$ (where $i$ is an integer from 1 to $k$) to prove this general formula, but I don't really know how to approach it..
| Note that:
In the expression
$n^3 = (n^2 - (n-1)) + (n^2 - (n-3)) + ... + (n^2 + (n-3)) + (n^2 + (n-1))$,
$n-1$ cancels from the first term and the last term.
$n-3$ cancels from the second term and the second to last term.
$n-5$ cancels from the thirst term and the third to last term.
And so on.
$n-a$ cancels from the $(\frac{a+1}{2})^\text{th}$ term and the $(\frac{a+1}{2})^\text{th}$ to last term, for any odd positive integer $a$.
We left with $n^3=n^2+n^2+n^2+\dots$ $[n \text{ times}]$. This means that $n^3=n\cdot n^2$, which is true for any $n$.
| {
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Finding $x^3 + y^3$ when $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $ $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $
$x^3 + y^3 =?$
my answer =
$(3 + \sqrt5)^3 = 47 + 32\sqrt5$
$(3 - \sqrt5)^3 = 47 - 32\sqrt5$
$x^3 + y^3 = \frac{47 + 32\sqrt5}{47 - 32\sqrt5} + \frac{47 - 32\sqrt5}{47 + 32\sqrt5} = 2*7329/-2911$
why my answer is wrong? please help me
| Someone has already pointed out where your mistake is. This is just to point to another alternative, for the fun of it.
You can factor $x^3+y^3$ as $$(x+y)(x^2-xy+y^2).$$ To compute the sum of squares, note that $2(x^2+y^2)=(x+y)^2+(x-y)^2.$ So, you only need to compute the sum, difference and product of your numbers, and combine these accordingly. This simplifies the calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
find minimum of maximum of two functions
I need to find the angle $\theta$ so that: $$\max(\cos^2(\theta),1-\cos^2(45-\theta))$$ is minimized.
OK, so I wrote \begin{align*}f(\theta)&=\max(\cos^2(\theta),1-\cos^2(45-\theta))\\
&=\frac{\cos^2(\theta)+1-\cos^2(45-\theta)+|\cos^2(\theta)-1+\cos^2(45-\theta)|}{2}\end{align*}
Then I'm stuck.
| Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$D=\cos^2t-(1-\cos^2(45^\circ-t))=\cos^2t-\sin^2(45^\circ-t)=\cos(45^\circ)\cos(2t-45^\circ)$$
Case $\#1:$
$\cos^2t$ will be maximum if $D\ge0$ if $\cos(2t-45^\circ)\ge0$
$\iff360^\circ n-90^\circ\le2t-45^\circ\le360^\circ n+90^\circ$
$\iff360^\circ n-45^\circ\le2t\le360^\circ n+135^\circ$
Now in the above range $\cos^2t=\dfrac{1+\cos2t}2$ will be minimum $\iff$ if $\cos2t$ is minimum
which happens if $2t=360^\circ n+135^\circ$
Case $\#2:$
Check when $$1-\cos^2(45^\circ-t)=\sin^2(45^\circ-t)$$ will be maximum
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
For positive real numbers $a,b,c$ prove that $ a^4 + b^4 + c^4 \ge abc(a+b+c)$ For positive reals $a,b,c$ prove that $$ a^4+b^4+c^4 \ge abc(a+b+c). $$
I tried to pls around trying to reorganize to get AM-GM but i couldn't
Thanks for the help in advance.
| Why should the mean inequalities not work? Geometric mean vs. mean of power 4
$$
abc\le\left(\frac{a^4+b^4+c^4}3\right)^{\frac34}
$$and arithmetic mean vs. mean of power 4$$
\frac{a+b+c}3\le\left(\frac{a^4+b^4+c^4}3\right)^{\frac14}
$$
implies
$$
\frac{abc(a+b+c)}3\le \frac{a^4+b^4+c^4}3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
A hard inequality indian olympiad problem If $x,y,z$ are positive real numbers, prove that:
$\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\le 3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)$.
I have been stuck in it. It is an Indian Olympiad problem. Can you guys help me out, please?
| We have that:
$$x^2+xy+y^2=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\geq \frac{3}{4}(x+y)^2$$
Therefore
$$(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)\geq \frac{27}{64}(x+y)^2(y+z)^2(z+x)^2$$
And it remains to prove
$$9(x+y)(y+z)(z+x)\geq 8(x+y+z)(xy+yz+zx)$$
We can prove this with AM-GM:
$$8(x+y+z)(xy+yz+zx)=8(x+y)(y+z)(z+x)+8xyz \leq 9(x+y)(y+z)(z+x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.