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This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$
let $0<a\le b\le c\le d$, and such $abcd=1$,show that
$$a+b^2+c^3+d^4\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}+\dfrac{1}{d^4}$$
it seems harder than This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
| Suppose that it is true for $n=3$:
$$
\forall 0 \le a \le b \le c \wedge abc = 1 :
a - \frac{1}{a}
+ b^2 - \frac{1}{b^2}
+ c^3 - \frac{1}{c^3} \ge 0.
$$
Let $d \ge c \ge 1$, then it is clear that
$$
\forall 0 \le a \le b \le c \le d \wedge abcd = d :
a - \frac{1}{a}
+ b^2 - \frac{1}{b^2}
+ c^3 - \frac{1}{c^3}
+ d^4 - \frac{1}{d^4} \ge 0.
$$
Let us now define
$$
a' = \frac{a}{\mu^{24}},
b' = \frac{b}{\mu^{12}},
c' = \frac{c}{\mu^{8}},
d' = \frac{d}{\mu^{6}},
$$
using
$$
24 = 4!, 12 = 4!/2, 8 = 4!/3, 6 = 4!/4.
$$
We get
$$
a' b' c' d' =
\frac{a}{\mu^{24}} \frac{b}{\mu^{12}} \frac{c}{\mu^{8}} \frac{d}{\mu^{6}} =
\frac{abcd}{\mu^{50}} = \frac{d}{\mu^{50}},
$$
so
$$
a' b' c' d' =1,
$$
if
$$
\mu = \sqrt[50]{d}.
$$
As $d \ge 1$, it is clear that $\mu \ge 1$.
The inequality can be written as
$$
\frac{a}{\mu^{24}} - \frac{\mu^{24}}{a}
+ \frac{b^2}{\mu^{24}} - \frac{\mu^{24}}{b^2}
+ \frac{c^3}{\mu^{24}} - \frac{\mu^{24}}{c^3}
+ \frac{d^4}{\mu^{24}} - \frac{\mu^{24}}{d^4} \ge 0.
$$
Therefore
$$
\mu^{24} \left(
\frac{a}{\mu^{48}} - \frac{1}{a}
+ \frac{b^2}{\mu^{48}} - \frac{1}{b^2}
+ \frac{c^3}{\mu^{48}} - \frac{1}{c^3}
+ \frac{d^4}{\mu^{48}} - \frac{1}{d^4}
\right) \ge 0.
$$
Whence
$$
\mu^{24} \left(
a - \frac{1}{a}
+ b^2 - \frac{1}{b^2}
+ c^3 - \frac{1}{c^3}
+ d^4 - \frac{1}{d^4}
\right) -
\left( \mu^{12} - \frac{1}{\mu^{12}} \right)\big( a + b^2 + c^3 + d^4 \big)
\ge 0.
$$
Which can be written as
$$
a - \frac{1}{a}
+ b^2 - \frac{1}{b^2}
+ c^3 - \frac{1}{c^3}
+ d^4 - \frac{1}{d^4}
\ge
\mu^{-24} \left( \mu^{12} - \frac{1}{\mu^{12}} \right)\big( a + b^2 + c^3 + d^4 \big).
$$
As $\mu \ge 1$, we obtain
$$
\forall 0 \le a \le b \le c \le d \wedge abcd = 1 :\\
a - \frac{1}{a}
+ b^2 - \frac{1}{b^2}
+ c^3 - \frac{1}{c^3}
+ d^4 - \frac{1}{d^4}
\ge 0.
$$
This method allows to show that the case $n+1$ is true, IF the case $n$ is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Evaluating the infinite product $ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $ How does one evaluate
$ \displaystyle\prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $?
Seems fairly straightforward, as I just plugged in some initial values $n = 1, 2, 3,\dotsc$
$n = 1$
$ \sin(y)= 2\sin(\frac{y}{2})\cos(\frac{y}{2})$
$ \cos(\frac{y}{2}) = \sin(y)/2\sin(\frac{y}{2}) $
$n = 2$
$\sin(\frac{y}{2})= 2\sin(\frac{y}{4})\cos(\frac{y}{4}) $
$\cos(\frac{y}{4}) = \sin(\frac{y}{2})/2\sin(\frac{y}{4}) $
Hence,
$ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) = \sin(y)/2\sin(\frac{y}{2}) \times \sin(\frac{y}{2})/2\sin(\frac{y}{4}) ... \times \sin(\frac{y}{2^{n-1}})/2\sin(\frac{y}{2^n}) $
Then some denominators/numerators cancel, reducing the expression into
$ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) = \sin(y)/2^n\sin(\frac{y}{2^n})$
Is this expression no longer reducible?
Because apparently, somehow, the denominator seems to reduce to $y$
How does that work?
| Hint: Use the Taylor series and let $n\rightarrow \infty:$
$$2^n\sin\left(\frac{y}{2^n}\right)= 2^n\left(\frac{y}{2^n}-\left(\frac{y}{2^n}\right)^3/3! + \dots\right) = y - \frac{y^3}{3!2^{2n}} + \dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Special Palindromic String A string of length $N$ can be made from $6$ characters $a$, $b$, $c$, $d$, $e$ and $f$. There are some rules to make such a string:
*
*$b$ can not come directly after $a$.
*$d$ can not come directly after $a$ or $f$
*$c$ can not come directly after $c$ or $e$
*The string is a palindrome.
How many strings of length $N$ can be made given such constraints?
For $N=1$, there are $6$ such strings: $a$, $b$, $c$, $d$, $e$ and $f$.
For $N=2$, there are $5$ such strings: $aa$, $bb$, $dd$, $ee$ and $ff$.
For $N=3$, there are $27$ such strings: $aaa$, $aca$, $aea$, $afa$, $bab$, $bbb$, $bcb$, $bdb$, $beb$, $bfb$, $cac$, $cbc$, $cdc$, $dbd$, $dcd$, $ddd$, $ded$, $eae$, $ebe$, $ede$, $eee$, $efe$, $faf$, $fbf$, $fcf$, $fef$ and $fff$.
| Letting $r=\lceil n/2 -1 \rceil$, we have
$$f(n)=\begin{cases}
\frac{2^{-r-1}}{\sqrt{17}} \left(\left(5 \sqrt{17}-21\right)
\left(5-\sqrt{17}\right)^r+\left(21+5 \sqrt{17}\right)
\left(5+\sqrt{17}\right)^r\right) & n \textrm{ even;}\\
\frac{3\ 2^{-r}}{\sqrt{17}} \left(\left(\sqrt{17}-4\right)
\left(5-\sqrt{17}\right)^r+\left(4+\sqrt{17}\right)
\left(5+\sqrt{17}\right)^r\right) & n \textrm{ odd.}
\end{cases}$$
This gives the sequence of values ${1, 6, 5, 27, 23, 123, 105, 561, 479, 2559, 2185\ldots}$ for $n\ge 0$.
To obtain this, define the $6\times6$ transition matrix $M$ to be $(m_{ij})$, with rows and columns labeled by the symbols a,$\ldots$, f, with $m_{ij}=1$ iff symbol $i$ can be followed by symbol $j$. Since the string must be a palindrome, this matrix is symmetric:
$$M=\left(
\begin{array}{cccccc}
1 & 0 & 1 & 0 & 1 & 1 \\
0 & 1 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 & 0 & 1 \\
0 & 1 & 1 & 1 & 1 & 0 \\
1 & 1 & 0 & 1 & 1 & 1 \\
1 & 1 & 1 & 0 & 1 & 1 \\
\end{array}
\right)$$
Define a string to be legal if it respects the transitions; the $ij$ entry of $M^r$ is the number of legal strings of length $r+1$. Now, a palindromic string of length $n$ corresponds precisely to a legal string of length $\lceil n/2\rceil = (n+1)/2$ if $n$ is odd. If $n$ is even, a palindromic string of length $n$ corresponds to a legal string of length $n/2$ which ends in a symbol which can be followed by itself (i.e. any symbol but "c"). Therefore, if we define the vectors $\textbf{j}=(1,\ldots,1)$, and $\textbf{k}=(1,1,0,1,1,1)$, we have
$$f(n)=\begin{cases}
\textbf{j} M^r \textbf{j}^T, & n \textrm{ even;}\\
\textbf{j} M^r \textbf{k}^T, & n \textrm{ odd,}
\end{cases}$$
where $r=\lceil n/2\rceil -1$.
This is an effective formula, but we can get a closed form by diagonalizing $M$. That is, the eigenvalues of $M$ are distinct, so $M=U^{-1}D U$ where $D$ is the diagonal matrix whose entries are the eigenvalues. Thus $M^r=U^{-1}D^r U$, which leads to the closed form claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the Integral: $\int\ e^{x}(1+e^x)^{\frac{1}{2}}\ dx$ Evaluate the Integral: $\int\ e^{x}(1+e^x)^{\frac{1}{2}}\ dx$
$u=1+e^x$
$du=e^x\ dx$
$\int\ u^{\frac{1}{2}}\ du$
$u+c=1+e^x+c$
| $$\int\ e^{3}(1+e^x)^{\frac{1}{2}}\ dx=$$
$$e^3\int\sqrt{e^x+1} dx=$$
(substitute $u=e^x$ and $du=e^xdx$):
$$e^3\int\frac{\sqrt{u+1}}{u} du=$$
(substitute $s=\sqrt{u+1}$ and $ds=\frac{1}{2\sqrt{u+1}}du$):
$$2e^3\int\frac{s^2}{s^2-1} ds=$$
$$2e^3\int\left(-\frac{1}{2(s+1)}+\frac{1}{2(s-1)}+1\right)ds=$$
$$-e^3\int \frac{1}{s+1}ds+e^3\int \frac{1}{s-1}ds+2e^3\int1ds=$$
(substitute $p=s+1$ and $dp=ds$):
$$-e^3\int \frac{1}{p}dp+e^3\int \frac{1}{s-1}ds+2e^3\int1ds=$$
$$-e^3\ln(p)+e^3\int \frac{1}{s-1}ds+2e^3\int1ds=$$
(substitute $q=s-1$ and $dw=ds$):
$$-e^3\ln(p)+e^3\int \frac{1}{w}dw+2e^3\int1ds=$$
$$-e^3\ln(p)+e^3\ln(w)+2e^3\int1ds=$$
$$-e^3\ln(p)+e^3\ln(w)+2e^3s+C=$$
$$-e^3\ln(p)+e^3\ln(s-1)+2e^3s+C=$$
$$-e^3\ln(\sqrt{u+1}+1)+e^3\ln(\sqrt{u+1}-1)+2e^3\sqrt{u+1}+C=$$
$$-e^3\ln(\sqrt{e^x+1}+1)+e^3\ln(\sqrt{e^x+1}-1)+2e^3\sqrt{e^x+1}+C=$$
$$2e^3\left(\sqrt{e^x+1}-\tanh^{-1}\left(\sqrt{e^x+1}\right)\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital's: $$\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{2}{x^3}- \frac{2\cos x}{\sin ^3x} = \frac{0}{0}\,\ldots$$ The terms are getting endless here. Any help? Thanks.
| Use Taylor-MacLaurin at order $2$:
\begin{align*}\frac1x - \frac{\cos x}{\sin x}&=\frac1x - \frac{1-\dfrac{x^2}2+o(x^2)}{x-\dfrac{x^3}6+o(x^3)}=\frac1x-\frac1x\left(\frac{1-\smash[t]{\dfrac{x^2}2}+o(x^2)}{1-\smash[b]{\dfrac{x^2}6}+o(x^2)}\right)\\&=\frac1x-\frac1x\Bigl(1-\frac{x^2}3+o(x^2)\Bigr)= \frac x3+o(x)\underset{x\to 0}{\longrightarrow} 0.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Prove by induction that $4$ divides $n^3+(n+1)^3+(n+2)^3+(n+3)^3$ Just looking for someone to check my work and for feedback, thanks!
Base case: $n=0$
$0+1+8+27 = 36$
$4$ divides $36.$
Inductive step: Assume $4$ divides $k^3+(k+1)^3+(k+2)^3+(k+3)^3$ for some number where $k$ is a natural number including zero. So $k^3+(k+1)^3+(k+2)^3+(k+3)^3 = 4b$ where $b$ is some integer. We need to show $4$ divides $(k+1)^3+(k+2)^3+(k+3)^3+(k+4)^3$.
\begin{align}
(k+1)^3 & +(k+2)^3+(k+3)^3+(k+4)^3\\
&= 4k^3+30k^2+90k+100\\
&=(k^3+(k+1)^3+(k+2)^3+(k+3)^3)+12k^2+48k+36\\
&=4b+12k^2+48k+36 \qquad \text{(by inductive hypothesis)}\\
&=4(b+3k^2+12k+9)
\end{align}
Since $b$ is an element of any integer this holds true for $(k+1)$. Hence proven.
| To go from
$n$ to $n+1$,
you subtract $n^3$
and add $(n+4)^3$.
This means that
the sum changes by
$(n+4)^3 - n^3$,
so if this is divisible by 4,
divisibility by 4 remains.
But
$(n+4)^3 - n^3
=(n^3+12n^2+48n+64)-n^3
=12n^2+48n+64
=4(3n^2+12n+16)
$
is divisible by 4.
Since the first sum
(for n=0)
is
$0^3+1^3+2^3+3^3
=1+8+27
=36
$
is divisible by 4,
all are divisible by 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Powers of a $2 \times 2$ matrix Let $A$ be a $2 \times 2$ matrix such that $$A = \begin{pmatrix} \sin\frac{\pi}{18} & -\sin\frac{4\pi}{9} \\ \sin\frac{4\pi}{9} & \sin\frac{\pi}{18} \end{pmatrix}$$ Find the smallest number $n \in N$ such that $A^n = I$, where $I$ is the identity matrix of order $2$.
| Using the identity $\sin\theta=\cos(\frac{\pi}{2}-\theta)$, you get $\sin\frac{\pi}{18}=\cos\frac{4\pi}{9}$.
Substitute this in and you get your standard rotation matrix:
$\begin{pmatrix}
\cos\frac{4}{9}\pi & -\sin\frac{4}{9}\pi \\
\sin\frac{4}{9}\pi & \cos\frac{4}{9}\pi
\end{pmatrix}$
This is equivalent to a counterclockwise rotation of $\frac{4}{9}\pi$. How many times do you need to apply this rotation so that you are back to where you started?
As $4$ and $9$ are coprime, you will need to do it $9$ times. Hence, $n=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\circ}))$$ However I don't seem to get anywhere with this.
Unfortunately I don't know how to solve this question. I would be really grateful for any help or suggestions. Many thanks in advance!
| As $4\cos^3x-3\cos x=\cos3x$
If $\cos3x=\cos3A, 3x=2n\pi\pm3A$ where $n$ is any integer
$\implies x=\dfrac{2n\pi}3\pm A$ where $n\equiv0,1,2\pmod3$
So, the roots of $4\cos^3x-3\cos x-\cos3A=0$ are $\cos\dfrac{2n\pi}3\pm A$ where $n\equiv0,1,2\pmod3$
$\implies\sum_{r=0}^2\cos\left(\dfrac{2n\pi}3+A\right)=0$
and $\sum_{r=0}^2\cos\left(\dfrac{2n\pi}3-A\right)=0$
Now use $\cos2A=2\cos^2A-1$
Generalization : Like Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $ we can prove for positive integer $m$,
$$\cos(mx)=2^{m-1}\cos^mx+0\cdot\cos^{m-1}x+\cdots=0$$
$\implies\sum_{r=0}^{m-1}\cos\left(\dfrac{2n\pi}m-A\right)=\sum_{r=0}^{m-1}\cos\left(\dfrac{2n\pi}m+A\right)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
} |
Show that $\left(1 + \frac{x}{n}\right)^{-n} \le 2^{-x}$, when $x,n \ge 0$, $x \le n$ Show that $\left(1 + \frac{x}{n}\right)^{-n} \le 2^{-x}$, when $x,n \ge 0$, $x \le n$. This is driving me crazy... I have plotted the graphs to be sure that the inequality is true, and it is, but I can't seem to show it. Here is what I have so far:
$$\left(1 + \frac{x}{n}\right)^{-n} \le \left(1 + \frac{x}{n}\right)^{-x}$$
since $n \ge x$. I want to then use the fact that $1 + \dfrac{x}{n} \le 2$, but that gives me a $\ge$, not a $\le$, so my first step must be wrong... I don't know what else to do, instead.
| I have already posted an answer where I considered asymptotic behavior of inequality as $n\to \infty$.
Below are more general conclusions.
Given $0\le x \le n$, show that $\left(1 + \dfrac{x}{n}\right)^{-n} \le 2^{-x}$.
First, note that
$$
0\le x \le n \implies 0 \le \frac{x}{n}\le 1
\implies 1\le \left(1 + \dfrac{x}{n}\right) \le 2
$$
Second, note that for any $\alpha \ge 1$ we have
$$
\alpha \ge 1, \quad x \le n \implies
\alpha^{x} \le \alpha^{n} \implies
\alpha^{-x} = \frac{1}{\alpha^{x}}\ge \frac{1}{\alpha^{n}} = \alpha^{-n}
$$
Third, for $1\le \alpha \le \beta$ and for $n \ge 0$ we have
$$
n\ge 0 \implies \alpha \le \beta \iff\alpha^{n} \le \beta^{n} \iff
\alpha^{-n} \ge \beta^{-n}
$$
Fourth, rewrite $2^{-x}$ as $\left(2^{-\frac{x}{n}}\right)^{n}$, get
$$
2^{-x} = \left(2^{-\frac{x}{n}}\right)^{n}
\implies
\left(1 + \dfrac{x}{n}\right)^{-n} \le 2^{-x} \iff
\left(1 + \dfrac{x}{n}\right)^{-n} \le \Big(2^{\frac{x}{n}}\Big)^{-n}
$$
Finally, we only have to show that
$ \left(1 + \dfrac{x}{n}\right) \ge 2^{\frac{x}{n}}$ whenever $0 \le x \le n$.
Let $y:= \frac{x}{n}\,\ln 2$, then
$$
\begin{aligned}
2^{\frac{x}{n}} = e^{\frac{x}{n}\cdot \ln 2} = e^{y}
& = \sum_{k=0}^{\infty}\frac{y^{k}}{k!}
= \sum_{k=0}^{\infty}\frac{\ln^{k}(2)}{n^{k}}\frac{x^{k}}{k!}
= 1 + \frac{\ln2 }{n}x + \frac{\ln^{2}(2)}{n^{2}}\frac{x^{2}}{2}
+ \frac{\ln^{3}(2)}{n^{3}}\frac{x^{3}}{6} + \cdots = \\
& = 1 + \frac{\ln2 }{n}x + \mathcal{O}\left(x^2\right)
\\
\left(1 + \dfrac{x}{n}\right) = 1 + \dfrac{1}{n}x
& = 1 + \dfrac{1}{n}x + \mathcal{O}\left(x^2\right).
\end{aligned}
$$
Since $\ln 2< 1$, we conclude that $ 2^{\frac{x}{n}} \le \left(1 + \dfrac{x}{n}\right) $, and thus establish inequality:
$$
\begin{aligned}
\ln 2< 1 &\implies
1 + \dfrac{\ln 2}{n}x + \mathcal{O}\left(x^2\right) \le 1 + \dfrac{1}{n}x + \mathcal{O}\left(x^2\right)\implies
\\ & \implies
2^{\frac{x}{n}} \le \left(1 + \dfrac{x}{n}\right) , \quad 0\le x \le n \implies
\\ & \implies
\boxed{\left(1 + \dfrac{x}{n}\right)^{-n} \le 2^{-x}}
\end{aligned}
$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find the values of $\cos(\alpha+\beta) $ if the roots of an equation are given in terms of tan It is given that $ \tan\frac{\alpha}{2} $ and $ \tan\frac{\beta}{2} $ are the zeroes of the equation $ 8x^2-26x+15=0$ then find the value of $\cos(\alpha+\beta$).
I attempted to solve this but I don't know if my solution is right.Can someone verify this.
sum of roots = $ \frac{-b}{a}$
so $ \tan\frac{\alpha}{2} + \tan\frac{\beta}{2} $ = $ \frac{26}{8}$
product of roots = $ \frac{c}{a}$
so $ \tan\frac{\alpha}{2} $ . $ \tan\frac{\beta}{2} $ = $ \frac{15}{8}$
$$ \tan(\frac{\alpha+\beta}{2} ) = \frac{\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}}{1-\tan\frac{\alpha}{2}.\tan\frac{\beta}{2}}$$
Puting the values in :-
$$ \tan(\frac{\alpha+\beta}{2} ) = -\frac{26}{7}$$
$$ \sec^2(\frac{\alpha+\beta}{2}) = \tan^2(\frac{\alpha+\beta}{2} ) + 1 $$
$$ \sec^2(\frac{\alpha+\beta}{2}) = \frac{(-26)^2 + 7^2}{7^2} $$
$$ \cos^2{\frac{\alpha+\beta}{2}} = \frac{7^2}{25^2}$$
$$ \frac{\cos(\alpha+\beta) + 1}{2} = \frac{49}{725}$$
$$ \cos(\alpha+\beta) = \frac{-627}{725}$$
| Yes, your answer and method, all are right. Just $\tan(\frac{a+b}{2})=\frac{\tan(\frac a2)+\tan(\frac b2)}{1-\tan(\frac a2)\cdot\tan(\frac b2)}$.
But your values are right. And this was the slight correction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question:
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\left(\frac{6^\circ+66^\circ}{2}\right)\sin\left(\frac{6^\circ-66^\circ}{2}\right) $$
$$ 2\cos(60^\circ)\sin(18^\circ) + 2\cos(36^\circ)\sin(-30^\circ) $$
$$ 2\frac{1}{2}\sin(18^\circ) - 2\cos(36^\circ)\cdot\frac{1}{2} $$
$$ \sin(18^\circ) - \cos(36^\circ) $$
At this point I had to use a calculator. Does anyone know a way to solve it without a calculator.Thanks in advance.
| Angles that are multple of $18$ degrees occur in the regular pentagon with its diagonals and their trigonometric ratios are related to the golden ratio.
To find them, take an isosceles triangle with angles $A=36$ and $B=C=72$ degrees. Draw the bisector of $B$ that intersects $AC$ in a point $D$. Note that the triangles $ABC$ and $BCD$ are similar. Now let $1=AB$, $x=AD$. Note now that $ABD$ is also isosceles, and $AD=BD=BC=x$. The mentioned similarity yields
$$\frac1x=\frac x{1-x}$$
which is precisely the equation that defines the golden ratio.
To find $\cos 36^\circ$ draw the altitude of the triangle $ABD$ from $D$, that bisects the side $AB$. Then
$$\cos 36^\circ=\frac{1/2}{x}$$
Now, with the classical trigonometric identities, you can easily find closed expressions for the sine and the cosine of $18$, $36$, $54$ and $72$ degrees. For the angle of $18$ degrees, don't use half-angle formulae, but
$$\sin 18^\circ=\sin(90^\circ-72^\circ)=\cos 72^\circ=\cos^2 36^\circ-\sin^2 36^\circ$$
Remark: Perhaps you are wandering what that pentagon had to do with all of this. If you choose a vertex $A$ of a regular pentagon and draw the two diagonals $AB$ and $AC$ from it, the triangle $ABC$ is similar to the one we have just used.
| {
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"url": "https://math.stackexchange.com/questions/1366530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Contradiction between integration by partial fractions and substitution Integration by substitution:
$$\int \frac {dx}{x^2-1}$$
Let $x=\sec\theta$ and $dx=\sec\theta\tan\theta \,d\theta$
$$\int \frac {dx}{x^2-1} = \int \frac{\sec\theta\tan\theta \,d\theta}{\sec^2\theta-1} = \int \frac {\sec\theta\tan\theta\, d\theta}{\tan^2\theta} = \int \frac {\sec\theta \,d\theta}{\tan\theta} $$
$$\int \frac {\sec\theta \,d\theta}{\tan\theta} = \int \frac {\cos \theta \,d\theta}{\cos\theta \sin \theta} = \int \csc\theta\, d\theta$$
$$=\ln|\csc\theta-\cot\theta|+C$$
$$=\ln| \frac{x}{\sqrt{x^2-1}}-\frac{1}{\sqrt{x^2-1}}|+C=\ln| \frac{x-1}{\sqrt{x^2-1}}|+C$$
Which is $Undefined$ for $|x|<1$
Integration by partial fractions:
$$\int \frac {dx}{x^2-1}$$
$$\int \frac {dx}{x^2-1}= \frac 12\int\frac{dx}{x-1}- \frac12\int \frac{dx}{x+1} = \frac12 \ln | x-1| - \frac12 \ln|x+1| +C$$
$$ = \frac12 \ln | \frac{x-1}{x+1}|+C$$
Which is $Defined$ for $|x|<1$ and this is right because the integrand is defined for $|x|<1$
What is the problem in the substitution method ?
| Let's try to clear out some of the confusion on what is going on here.
First of all, $x\mapsto \frac 1 {x^2 - 1}$ is continuous function everywhere except for $x=\pm 1$, so it is Riemann integrable on any segment not containing $\pm 1$. That said, we would very much like to find primitive function defined on $\mathbb R^2\setminus\{\pm 1\}$. Solution by partial fractions does just that, $$F_1(x) = \frac 12 \ln\left|\frac{x-1}{x+1}\right| + C$$ is defined everywhere except at $x=\pm 1$. So, if we wanted to calculate either $$I_1 = \int_{\frac 12}^{\frac 34} \frac {dx} {x^2 - 1}$$ or $$I_2 = \int_{2}^{3} \frac {dx} {x^2 - 1}$$ we can use $F_1$ with no worries.
Now, if we try to use substitution such as $x = \sec\theta$, as OP notices, we might run into some problems in the long run. The primitive function derived this way is $$ F_2(x) = \ln\left| \frac {x - 1}{\sqrt{x^2 -1}} \right| + C$$ which is defined only for $|x|>1$. Is this shocking? Well, no. As lulu points out in the comments, $|\sec\theta|\geq 1$ for any $\theta$, which is a simple consequence of the definition $\sec\theta = \frac 1 {\cos\theta}$. Thus, by substituting $x=\sec\theta$, we already gave up on $x\in\langle -1,1\rangle$, which is actually fine as long as we are trying to calculate $I_2$, but won't work for $I_1$.
So, the question is: are $F_1$ and $F_2$ both "good" solutions? More precisely, if we wanted to calculate $I_2$, can we use either of those two $F$'s?
Well, let's assume that $|x| > 1$. Then we have:
$$ \ln\left| \frac {x - 1}{\sqrt{x^2 -1}} \right| = \ln \frac {\left|x - 1\right|}{\sqrt{x^2 -1}} = \ln \frac {\sqrt{(x-1)^2}}{\sqrt{x^2 -1}} = \ln \sqrt{\frac {{(x-1)^2}}{{x^2 -1}}} = \frac 12 \ln\frac{x-1}{x+1}=\frac 12 \ln\left|\frac{x-1}{x+1}\right| $$ where the last equality holds because $x-1$ and $x+1$ have the same signs on $|x|>1$. Thus, we have shown that $F_2 = \left.F_1 \right|_{\mathbb R^2\setminus [-1,1]}$. I hope that clarifies the problem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How To Tackle Trigonometric Proofs involving $4$th and $6$th powers? How do I prove that
$\cos^4A - \sin^4A+1=2\cos^2A$
$\cos^6A + \sin^6A =1-3\sin^2A\cdot\cos^2A$
I was going through a very old and very rich book of Plane Trigonometry to build a nice foundation for calculus , when I had came across these two rather interesting equations.
Unfortunately the book i'm reading never mentions how to imperatively solve such a proof, and oddly enough I was never exposed on how to prove these types of equations when I took precalc last semester as well.
Hope you guys can help me out.
| 1: Let's take $$LHS=\cos^4 A-\sin^4 A+1$$ $$=\cos^4 A+(1-\sin^4 A)$$ $$=\cos^4 A+(1-\sin^2 A) (1+\sin^2 A)$$ $$=\cos^4 A+\cos^2 A (1+\sin^2 A)$$ $$=\cos^2 A(\cos^2A+\sin^2 A+1)$$ $$=\cos^2 A(1+1)$$ $$=\color{blue}{2\cos^2 A=RHS}$$
2: Again let's take $$LHS=\cos^6 A+\sin^6 A$$ $$=(\cos^2 A)^3+(\sin^2 A)^3$$ $$=(\cos^2A+\sin^2A)(\cos^4 A+\sin^4 A-\cos^2A\sin^2A)$$ $$=(1)(\cos^4 A+\sin^4 A+2\cos^2A\sin^2A-3\cos^2A\sin^2A)$$ $$=(\cos^2A+\sin^2 A)^2-3\sin^2A\cos^2A$$ $$=(1)^2-\sin^2A3\cos^2A$$ $$=\color{blue}{1-3\sin^2 A\cos^2A=RHS}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of an oscillating recursive sequence Define the recursive sequence
$ q_{n+1} = \dfrac{q_n+2}{q_n+1};\;q_0=1 $
If we knew that
$ q_n \to q;\;n\to \infty $
then it's easy to show what follows
$ q_{n+1}\left(q_n+1\right) = q_n+2 $
$ q_{n+1}\cdot q_n+q_{n+1}-q{n}=2 $
$ q^2 = 2 \rightarrow q = \sqrt 2;\;n\to\infty $
I don't know how to prove that $q_n$ converges.
The sequence is oscillating and I know, empirically, that the sequence
$a_n=q_n-q_{n+1}$
is an alternate sequence whose general term goes to zero, thus $a_n$ converges.
Any help will be greatly appreciated, thank you.
| Your series, when written as a sequence of fractions, is:
$1, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \frac{41}{29}...$
Notice that the denominator of term $n + 1$ is the sum of the denominator and numerator of term $n$, and the numerator of term $n + 1$ is twice the denominator of term $n$ plus its numerator.
So let's write two sequences, denominator $D(n)$ and numerator $N(n)$, mutually recursive:
$D(n + 1) = D(n) + N(n)$ and $N(n + 1) = 2D(n) + N(n)$ with $D(0) = N(0) = 1$
Now we just need to show that $q_n = \frac{N(n)}{D(n)}$. Well, it's true for the base case of $n = 0$. Then for the inductive case, well, we have your recurrence $q_{n + 1} = \frac{q_n + 2}{q_n + 1}$, and by inductive hypothesis: $q_{n + 1} = \frac{\frac{N(n)}{D(n)} + 2}{\frac{N(n)}{D(n)} + 1} = \frac{N(n) + 2D(n)}{N(n) + D(n)} = \frac{N(n + 1)}{D(n + 1)}$ as required.
Now, we can generate yet another recurrence. Recall $q_n = N(n)/D(n)$. Then we can show:
$q_{n + 1} = 1 + \frac{1}{1 + q_n}$
Here is why:
$q_{n + 1} = \frac{N(n) + 2D(n)}{N(n) + D(n)} = 1 + \frac{D(n)}{N(n) + D(n)} = 1 + \frac{1}{\frac{N(n) + D(n)}{D(n)}} = 1 + \frac{1}{1 + \frac{N(n)}{D(n)}} = 1 + \frac{1}{1 + q_n}$
Now you might notice the similarity to the continued fraction expansion of $\sqrt{2}$. Continued Fraction of root 2
To show convergence, we define the error term $e(n) = q_n - \sqrt{2}$ and show that the absolute value of this error term decreases faster than geometrically with $n$, which in turn shows that it converges to $0$. Start with our latest recurrence:
$q_{n + 1} = 1 + \frac{1}{1 + q_n}$ so
$\sqrt{2} - e(n + 1) = 1 + \frac{1}{1 + \sqrt{2} - e(n)}$
$(\sqrt{2} - e(n + 1))(1 + \sqrt{2} - e(n)) = 2 + \sqrt{2} - e(n)$
$\sqrt{2} + 2 - \sqrt{2}e(n) - e(n + 1)(1 + \sqrt{2} - e(n)) = 2 + \sqrt{2} - e(n)$
$\sqrt{2}e(n) + e(n + 1)(1 + \sqrt{2} - e(n)) = e(n)$
$e(n + 1)(1 + \sqrt{2} - e(n)) = e(n)(1 - \sqrt{2})$
$e(n + 1) = e(n)\frac{(1 - \sqrt{2})}{(1 + \sqrt{2} - e(n))}$
Now, since $f(0) = 1$, then $|e(0)| < 1$. Assuming then that $|e(n)| < 1$
we can show that $|\frac{(1 - \sqrt{2})}{(1 + \sqrt{2} - e(n))}| < \frac{\sqrt{2} - 1}{\sqrt{2}} < 1$. Proceed as follows.
$|e(n)| < 1$ so
$1-e(n) > 0$
$1 + \sqrt{2} - e(n) > \sqrt{2}$
$|\frac{(1 - \sqrt{2})}{(1 + \sqrt{2} - e(n))}| < \frac{\sqrt{2} - 1}{\sqrt{2}} < 1$
And the convergence is shown.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.
| First let
$$ x = \sqrt{\frac{a}{a+8}}, \,\, y = \sqrt{\frac{b}{b+8}}, \,\, z = \sqrt{\frac{c}{c+8}} \,\, $$
Then $1 > x,y,z > 0$ and
$$ a = \frac{8x^2}{1 - x^2}, \,\, b = \frac{8y^2}{1 - y^2}, \,\, c = \frac{8z^2}{1 - z^2},\,\, $$
So the question transforms to this:
Given that $1 > x,y,z > 0, \, \, \frac{512x^2y^2z^2}{(1 - x^2)(1 - y^2)(1 - z^2)} = 1$, prove that $x + y + z \geqslant 1$.
Prove this by contradiction. Suppose on the contrary that $x + y + z < 1$, then
$$ \begin{align}
(1 - x^2)(1 - y^2)(1 - z^2) &= (1 - x)(1 + x)(1 - y)(1 + y)(1 - z)(1 + z) \\
&>(x + x + y + z)(y + z)(x + y + y + z)(x + z)(z + x + y + z)(x + y) \\
&\geqslant 4x^{\frac12}y^{\frac14}z^{\frac14}\cdot 2y^{\frac12}z^{\frac12} \cdot 4y^{\frac12}x^{\frac14}z^{\frac14}\cdot 2x^{\frac12}z^{\frac12} \cdot 4z^{\frac12}y^{\frac14}x^{\frac14}\cdot 2y^{\frac12}x^{\frac12}\\
&=512 x^{\frac12 + \frac14 + \frac12 + \frac14 + \frac12}y^{\frac14 + \frac12 + \frac12 + \frac14 + \frac12}z^{\frac14 + \frac12 +\frac14 + \frac12 + \frac12} \\
&= 512x^2y^2z^2
\end{align}$$
And this is contradictory to the condition.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Find the sum: $\sum_{i=1}^{n}\dfrac{1}{4^i\cdot\cos^2\dfrac{a}{2^i}}$ Find the sum of the following :
$S=\dfrac{1}{4\cos^2\dfrac{a}{2}}+\dfrac{1}{4^2\cos^2\dfrac{a}{2^2}}+...+\dfrac{1}{4^n\cos^2\dfrac{a}{2^n}}$
| Hint:
$$\frac{1}{4^n\cos^2\frac{a}{2^n}}+\frac{1}{4^n\sin^2\frac{a}{2^n}}=
\frac{1}{4^{n-1}\sin^2\frac{a}{2^{n-1}}}.$$
Adding $\displaystyle \frac{1}{4^n\sin^2\frac{a}{2^n}}$ to the sum, the result thus telescopes to $\displaystyle\frac{1}{\sin^2a}$, and hence the initial sum is
$$\frac{1}{\sin^2a}-\frac{1}{4^n\sin^2\frac{a}{2^n}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Power serie of $f'/f$ It seems that I'm [censored] blind in searching the power series expansion of $$f(x):=\frac{2x-2}{x^2-2x+4}$$ in $x=0$.
I've tried a lot, e.g., partiell fraction decomposition, or regarding $f(x)=\left(\log((x+1)^2+3)\right)'$ -- without success.
I' sure that I'm overseeing a tiny little missing link; dear colleagues, please give me a hint.
| The given $f(x)$ is given by
\begin{align}
f(x) = \frac{2(x-1)}{x^2 - 2x + 4}
\end{align}
and can be seen as, where $a = 1 + \sqrt{3} i$ and $b = 1 - \sqrt{3} i$, such that $ab=4$,
\begin{align}
f(x) = \frac{-2 \, (1 - x)}{ab \, \left(1 - \frac{x}{a}\right) \left( 1 - \frac{x}{b}\right)}
\end{align}
for which
\begin{align}
\ln f(x) &= \ln\left(- \frac{4}{ab}\right) + \ln(1 - x) - \ln\left(1 - \frac{x}{a}\right) - \ln\left(1 - \frac{x}{b}\right) \\
&= \pi i - \ln 2 + \ln(1 - x) - \ln\left(1 - \frac{x}{a}\right) - \ln\left(1 - \frac{x}{b}\right) \\
&= \pi i - \ln 2 - \sum_{n=1}^{\infty} \left( 1 - \frac{1}{a^{n}} - \frac{1}{b^{n}} \right) \, \frac{x^{n}}{n} \\
&= \pi i - \ln 2 + \sum_{n=1}^{\infty} \left( \frac{a^{n} + b^{n}}{4^{n}} - 1 \right) \, \frac{x^{n}}{n} \\
&= \pi i - \ln 2 + \sum_{n=1}^{\infty} \frac{a^{n} + b^{n} - 4^{n} }{n} \, \left(\frac{x}{4}\right)^{n} \\
&= \pi i - \ln 2 + \sum_{n=1}^{\infty} \frac{ 2 \, \cos\left(\frac{n \pi }{3}\right) - 2^{n} }{n} \, \left(\frac{x}{2}\right)^{n}.
\end{align}
Now differentiating both sides leads to
\begin{align}
\frac{f'}{f} &= \sum_{n=1}^{\infty} \left( 2 \, \cos\left(\frac{n \pi }{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n} \\
&= \frac{1}{2} \, \sum_{n=1}^{\infty} \left(2 \, \cos\left(\frac{n \pi}{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n-1} \\
&= \sum_{n=0}^{\infty} \left( \cos\left(\frac{(n+1) \pi}{3}\right) - 2^{n} \right) \, \left(\frac{x}{2}\right)^{n}.
\end{align}
Expanding the first few terms provides
\begin{align}
- \frac{f'}{f} = \frac{1}{2} + \frac{5 \, x}{4} + \frac{5 \, x^{2}}{4} + \frac{17 \, x^{3}}{16} + \frac{31 \, x^{4}}{32} + \mathcal{O}(x^{5})
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does the Pythagorean Theorem have its simple form only in Euclidean geometry? Below are the right-angled forms of the Pythagorean Theorem in elliptic, Euclidean, and hyperbolic geometry, respectively.
$$\cos\left(\frac{c}{R}\right) = \cos\left(\frac{a}{R}\right)\cos\left(\frac{b}{R}\right)$$
$$c^2 = a^2 + b^2$$
$$\cosh c = \cosh a \; \cosh b$$
To me, it looks like the Euclidean version is the simplest. No trig functions (and only one trig function in the general case). Also, it is the only version to have anything squared; the other two have $f(c)=f(a)f(b)$ whereas the Euclidean version is $f(c)=f(a)+f(b)$. Furthermore, there are plenty of fairly simple proofs of the Pythagorean Theorem in Euclidean geometry, but I would guess that this is not the case for elliptic and hyperbolic geometries. Why is this so?
My hunch is that the Parallel Postulate is the key to this answer. However, I don't see the connection between having exactly one parallel line and whatever it is that makes the formula turn out so nicely.
| You can even generalize the law of cosines.
$$
\cos\left(\frac{z}{r}\right) =
\cos\left(\frac{x}{r}\right) \cos\left(\frac{y}{r}\right)
+ \cos(\phi) \sin\left(\frac{x}{r}\right) \sin\left(\frac{y}{r}\right).
\tag 1
$$
What we have is
$$
\begin{array}{rcl}
r^2 > 0 &\rightarrow& \textrm{spherical}\\
&& \cos\left(\frac{z}{r}\right) =
\cos\left(\frac{x}{r}\right) \cos\left(\frac{y}{r}\right)
+ \cos(\phi) \sin\left(\frac{x}{r}\right) \sin\left(\frac{y}{r}\right)\\\\
r^2 < 0 &\rightarrow& \textrm{hyperbolic}\\
&&
\cosh\left(\frac{z}{r}\right) =
\cosh\left(\frac{x}{r}\right) \cosh\left(\frac{y}{r}\right)
- \cos(\phi) \sinh\left(\frac{x}{r}\right) \sinh\left(\frac{y}{r}\right)\\\\
\lim_{\displaystyle r^2 \rightarrow \infty} &\rightarrow& \textrm{flat or Euclidean}\\
&& z^2 = x^2 + y^2 - 2 \cos(\phi) x y
\end{array}
$$
Write it out...
$$
\begin{array}{rcl}
\displaystyle
\left( 1 - \frac{1}{2} \left( \frac{z}{r} \right)^2 + \cdots \right)
&=&
\displaystyle
\left( 1 - \frac{1}{2} \left( \frac{x}{r} \right)^2 + \cdots \right)
\left( 1 - \frac{1}{2} \left( \frac{y}{r} \right)^2 + \cdots \right)\\
&& \displaystyle \hspace{2em} + \cos(\phi)
\left( \frac{x}{r} + \cdots \right)
\left( \frac{y}{r} + \cdots \right)
\end{array}
$$
So you end up with
$$
z^2 = x^2 + y^2 - 2 \cos(\phi) x y.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Why do remainders show cyclic pattern? Let us find the remainders of $\dfrac{6^n}{7}$,
Remainder of $6^0/7 = 1$
Remainder of $6/7 = 6$
Remainder of $36/7 = 1$
Remainder of $216/7 = 6$
Remainder of $1296/7 = 1$
This pattern of $1,6,1,6...$ keeps on repeating. Why is it so? I'm asking in general, that is for every case of type $a^n/b$'s remainder keeps on repeating as we increase $n$.
P.S: This is a follow up question of my previous question.
| Note that as you find more and more powers, you'll eventually run out of remainders. So you'll come back to a previous number, and since exponentiation is repeated multiplication, you'll generate the cycle again. As an example, let's look at powers of $2$ modulo $7$. We have
\begin{align*}
2^1 &\equiv 2 \pmod{7} \\
2^2 &\equiv 2\cdot 2 \equiv 4 \pmod{7} \\
2^3 &\equiv 4 \cdot 2 \equiv 1 \pmod{7} \\
2^4 &\equiv 1 \cdot 2 \equiv 2 \pmod{7} \\
2^5 &\equiv 2 \cdot 2 \equiv 4 \pmod{7} \\
2^6 &\equiv 4 \cdot 2 \equiv 1 \pmod{7}
\end{align*}
Hopefully, this clarifies how multiplication works in modular arithmetic too.
| {
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"timestamp": "2023-03-29T00:00:00",
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how can i prove this trigonometry equation I need help on proving the following:
$$\frac{\cos {7x} - \cos {x} + \sin {3x}}{ \sin {7x} + \sin {x} - \cos {3x} }= -\tan {3x}$$
So far I've only gotten to this step:
$$\frac{-2 \sin {4x} \sin {3x} + \sin {x}}{ 2 \sin {4x} \cos {3x} - \cos {x}}$$
Any help would be appreciated as trigonometry is not my forte.
| The LHS can seen as $$\frac{\cos 7x -\cos x + \sin 3x}{\sin 7x + \sin x - \cos 3x} = \frac{-2\sin 4x \sin 3x + \sin 3x}{2\sin 4x\cos 3x - \cos 3x} = \tan 3x \frac{1-2\sin 4x}{-1+2\sin 4x} = -\tan 3x.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove using mathematical induction that $x^{2n} - y^{2n}$ is divisible by $x+y$ Prove using mathematical induction that
$(x^{2n} - y^{2n})$ is divisible by $(x+y)$.
Step 1: Proving that the equation is true for $n=1 $
$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$
Step 2: Taking $n=k$
$(x^{2k} - y^{2k})$ is divisible by $(x+y)$
Step 3: proving that the above equation is also true for $(k+1)$
$(x^{2k+2} - y^{2k+2})$ is divisible by $(x+y)$.
Can anyone assist me what would be the next step? Thank You in advance!
| When $n=1$,
$$x^{2n} - y^{2n} =x^2-y^2=(x+y)(x-y)$$
which is divisible by $(x+y)$. Assume true for $n=k$. Then $x^{2k}-y^{2k}$ is divisible by $x+y$.
Putting $n=k+1$,
$$x^{2n}-y^{2n}=x^{2k+2}-y^{2k+2}=\left(x^{k+1}-y^{k+1}\right) \left(x^{k+1}+y^{k+1}\right)$$
When $k$ is an odd number, $x+y$ is a factor of $x^{k+1}-y^{k+1}$.
Similarly, when $k$ is an even number, $x+y$ is a factor of $x^{k+1} + y^{k+1}$.
Hence $x^{2n}-y^{2n}$ is divisible by $x+y$ for integers $n>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
} |
How to integrate ${x^3}/(x^2+1)^{3/2}$? How to integrate $$\frac{x^3}{(x^2+1)^{3/2}}\ \text{?}$$
I tried substituting $x^2+1$ as t, but it's not working
| Let $x^2+1=t\implies 2xdx=dt$ or $xdx=\frac{dt}{2}$ $$\int \frac{x^3 dx}{(x^2+1)^{3/2}}=\frac{1}{2}\int \frac{(t-1)dt}{(t)^{3/2}}$$
$$=\frac{1}{2}\int \frac{(t-1)dt}{t^{3/2}}$$ $$=\frac{1}{2}\int (t^{-1/2}-t^{-3/2})dt$$ $$=\frac{1}{2}\left(2t^{1/2}+2t^{-1/2}\right)$$ $$=\sqrt{x^2+1}+\frac{1}{\sqrt{x^2+1}}+C$$ $$=\frac{x^2+2}{\sqrt{x^2+1}}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int_{-\infty}^{\infty}e^{-\pi x^2}\cdot e^{-2\pi ix\xi}dx = e^{\pi\xi^2}$ Prove that for all $\xi \in \mathbb{C}$,
$$\int_{-\infty}^{\infty}e^{-\pi x^2}\cdot e^{-2\pi ix\xi}dx = e^{\pi\xi^2}$$
I don't really know how to compute this integral. Can you please help me?
| By noticing that
$$a x^{2} + b x = \left( \sqrt{a} x + \frac{b}{2 \sqrt{a}} \right)^{2} - \frac{b^{2}}{4 a}$$
then
\begin{align}
I &= \int_{-\infty}^{\infty} e^{-a x^{2} - b x} \, dx \\
&= e^{\frac{b^{2}}{4 a}} \, \int_{-\infty}^{\infty} e^{- \left( \sqrt{a} x + \frac{b}{2 \sqrt{a}} \right)^{2}} \, dx
\end{align}
Making the change $t = \sqrt{a} x + \frac{b}{2 \sqrt{a}}$ leads to
\begin{align}
I &= \frac{1}{\sqrt{a}} \, e^{\frac{b^{2}}{4 a}} \, \int_{-\infty}^{\infty} e^{-t^{2}} \, dt \\
&= \frac{2}{\sqrt{a}} \, e^{\frac{b^{2}}{4 a}} \, \int_{0}^{\infty} e^{-t^{2}} \, dt \\
&= \frac{1}{\sqrt{a}} \, e^{\frac{b^{2}}{4 a}} \, \int_{0}^{\infty} e^{- u} \, u^{-1/2} \, du \hspace{5mm} \mbox{ where } t = \sqrt{u} \\
&= \sqrt{\frac{\pi}{a}} \, e^{\frac{b^{2}}{4 a}}.
\end{align}
Hence
\begin{align}
\int_{-\infty}^{\infty} e^{-a x^{2} - b x} \, dx = \sqrt{\frac{\pi}{a}} \, e^{\frac{b^{2}}{4 a}}.
\end{align}
For the case $a = \pi$ and $b = 2 \pi i \, \eta$ the result becomes
\begin{align}
\int_{-\infty}^{\infty} e^{- \pi x^{2} - 2 \pi i \eta x} \, dx = e^{- \pi \eta^{2}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
How to factor intricate polynomial $ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $ I would like to know how to factor the following polynomial.
$$ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $$
What is the method I should use to factor it? If anyone could help.. Thanks in advance.
| Let
$$
f(a)=ab^3-a^3b+a^3c-ac^3+bc^3-b^3c.
$$
If $a=b$, then $f(a)=0$. Therefore $(a-b)\mid f(a)$. Analogously $(a-c)\mid f(a)$; so, $(a-b)(a-c)\mid f(a)$. But we can take it directly:
$$
ab^3-a^3b+a^3c-ac^3+bc^3-b^3c = ab(b^2-a^2) + c^3(b-a) - c(b^3-a^3) =\\=
ab(b-a)(b+a)+c^3(b-a)-c(b-a)(b^2+ba+a^2) =\\= (b-a)[ab(b+a)+c^3-cb^2-cab-ca^2]
$$
$$
ab(b+a)+c^3-cb^2-cab-ca^2=ab(a+b) + c(c^2-a^2) - cb(a+b) =\\= (a-c)b(a+b) - c(a-c)(a+c)=(a-c)[b(a+b)-c(a+c)]
$$
And then
$$
b(a+b)-c(a+c) = ba+b^2 - ca - c^2 = a(b-c) + (b^2-c^2) =\\= a(b-c)+(b+c)(b-c)=(b-c)(a+b+c).
$$
So,
$$
ab^3-a^3b+a^3c-ac^3+bc^3-b^3c=(b-a)(a-c)(b-c)(a+b+c)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Integration by Partial Fractions $\int\frac{1}{(x+1)^3(x+2)}dx$ I'm trying to do a problem regarding partial fractions and I'm not sure if I have gone about this right as my answer here doesn't compare to the answer provided by wolfram alpha. Is it that I can't Seperate things that are raised to powers on the denominator?
http://www.wolframalpha.com/input/?i=integrate+1%2F%28%28x%2B1%29%5E3%28x%2B2%29%29
My Work:
$$\int\frac{1}{(x+1)^3(x+2)}dx$$
$$\frac{1}{(x+1)^3(x+2)}=\frac{A}{(x+1)^3}+\frac{B}{(x+2)}$$
$$1=A(x+2)+B(x+1)^3$$
Plugging $x=-1$
$$1=A(1)+B(0)^3$$
$$A=1$$
Plugging $x=-2$
$$1=A(0)+B(-1)^3$$
$$B=-1$$
$$\int\frac{1}{(x+1)^3(x+2)}dx=\int\left(\frac{1}{(x+1)^3}-\frac{1}{(x+2)}\right)$$
$$\int\left(\frac{1}{(x+1)^3}-\frac{1}{(x+2)}\right)dx=\frac{-1}{2(x+1)^2}-ln|x+2|+C$$
| Observe that:
$$x+2 = (x+1)+1$$
Rewrite the numerator as:
$$\left( (x+1)^3+1^3 \right) - (x+1)$$
And cancel using the sum of cubes identity
$$a^3+b^3\equiv(a+b)(a^2-ab+b^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Maximum value of expression Let the maximum value of the expression $y=\frac{x^4-x^2}{x^6+2x^3-1}$ for $x>1$ is $\frac p q$,where $p$ and $q$ are relatively prime positive integers.Find $p+q$.
My attempt: $y=\frac{x^2(x^2-1)}{(x^3+1)^2-2}$, then I could not think of any way to find maximum value of $f(x)$. Can someone help me in solving for $p$ and $q$?
| Let $t:=x-\frac{1}{x}$, which can take any value in $\mathbb{R}_{>0}$ Then, $y=\frac{t}{t^3+3t+2}$. To maximize $y$, we have to minimize $\frac{1}{y}=t^2+3+\frac{2}{t}$. By the AM-GM Inequality,
$$\frac{1}{y}=\left(t^2+\frac{1}{t}+\frac{1}{t}\right)+3\geq 3\sqrt[3]{t^2\cdot\frac{1}{t}+\frac{1}{t}}+3=6\,.$$
That is, $y \leq \frac{1}{6}$. The equality holds if and only if $t^2=\frac{1}{t}$, or $t=1$. Hence, the maximum of $y$ is $\frac{1}{6}$, which is attained at $x=\frac{1+\sqrt{5}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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find the total differential of this equation $ xyz + \sqrt{ x^2 + y^2 + z^2} = \sqrt 2 $ How to calculate the total differential of $ z= z(x,y)$, which is $ xyz + \sqrt{ x^2 + y^2 + z^2} = \sqrt 2 $ at point (1, 0, -1)？
The evaluation of mine seems wrong,
$ dz= \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy = (yz+ \frac{x}{\sqrt{x^2 + y^2 + z^2}})dx + (xz+ \frac{y}{\sqrt{x^2 + y^2 + z^2}})dy = \frac{dx}{\sqrt2} - dy $
Appreciate any helps.
| I have done the following workings for the question. I believe the answer is ok but please forgive if it is incorrect (and point out my oversight if possible).
As mentioned in the comments $z$ is implicitly a function of both $x$ and $y$. To be specific $z=z(x,y)$
First let's calculate $\frac{\partial z}{\partial x}$ remembering the product and chain rules:
You have:
$$xyz+\left( x^2+y^2+z^2 \right)^{1/2}=\sqrt{2}\\
yz+xy\frac{\partial z}{\partial x}+\frac{1}{2}\left(2x+2z\frac{\partial z}{\partial x}\right)\left(x^2+y^2+z^2\right)^{-1/2}=0$$
Now substitute $(1,0,-1)$ which, after simplifying, leads to $$2-2\frac{\partial z}{\partial x}=0\implies\frac{\partial z}{\partial x}=1$$
The case for$ \frac{\partial z}{\partial y}$ plays out similarly: $$xz+xy\frac{\partial z}{\partial y}+\frac{1}{2} \left( 2y + 2z \frac{\partial z}{\partial y}\right) \left(x^2+y^2+z^2 \right)^{-1/2}=0$$
Now substitute $(1,0,-1)$ leading to $\frac{\partial z}{\partial y}=\sqrt{2}$
Now $dz=\frac {\partial z}{\partial x}dx+\frac{\partial z}{\partial y} dy \implies dz=dx+\sqrt{2}dy$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Rearrangements that never change the value of a sum Which bijections $f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}$ have the property that for every sequence $\{a_n\}_{n=1}^\infty$,
$$
\lim_{n\to\infty} \sum_{k=1}^n a_k = \lim_{n\to\infty} \sum_{k=1}^n a_{f(k)},
$$
where "$=$" is construed as meaning that if either limit exists then so does the other and in that case then they are equal?
It is clear that there are uncountably many of these.
Might it just be that $\{f(n)/n : n=1,2,3,\ldots\}$ is bounded away from both $0$ and $\infty$?
These bijections form a group. Can anything of interest be said about them as a group?
PS: Here's another moderately wild guess (the one above appears to be wrong): Might it be just the bijections whose every orbit is finite?
| The hypothesis is wrong.
Let $f$ be the following sequence:
$1,2 ; 3,5,4,6; 7,9,11,8,10,12; \cdots$
where each block has two more numbers than the previous block and goes through the odd numbers in order before the even ones.
Clearly $f(n) \in n + O(\sqrt{n})$ as $n \to \infty$.
Now consider the following series:
$\frac{1}{1} - \frac{1}{1} \quad + \quad \frac{1}{2} - \frac{1}{2} + \frac{1}{2} - \frac{1}{2} \quad + \quad \frac{1}{3} - \frac{1}{3} + \frac{1}{3} - \frac{1}{3} + \frac{1}{3} - \frac{1}{3} \quad + \quad \cdots$
split into blocks in the same manner where each block has an alternating sum of the same reciprocal of the block index. Clearly this series converges to $0$. However after applying $f$ to the sequence in the series we get:
$\frac{1}{1} - \frac{1}{1} \quad + \quad \frac{1}{2} + \frac{1}{2} - \frac{1}{2} - \frac{1}{2} \quad + \quad \frac{1}{3} + \frac{1}{3} + \frac{1}{3} - \frac{1}{3} - \frac{1}{3} - \frac{1}{3} \quad + \quad \cdots$
which clearly does not converge.
Also, it is easy to see that we can insert arbitrarily large blocks of identity function between the blocks in the permutation, and corresponding blocks of zeros in the series, and the behaviour is exactly the same. This means that there is a counterexample where $f(n) \in n + o(n)$ as $n \to \infty$, equivalently where $\lim_{n\to\infty} \frac{f(n)}{n}$ exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
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Solving simple mod equations Solve $3x^2 + 2x + 1 \equiv 0 \mod 11$
Additionally, I have an example problem, but a step in the middle has confused me:
$3x^2 + 5x - 7 \equiv 0 \mod 17$. Rearrange to get $3x^2 + 5x \equiv 7 \mod 17$.
$\implies 6\cdot 3x^2 + 6\cdot 5x \equiv 6*7 \mod 17$.
The next line reads $x^2 + 30x \equiv 8 \mod 17$, which I am confused about;
I understand how $8$ came to be ($42 - 34$, reduced to $8$) but what happened to the coefficient of $3x^2$? This example is from my notes in NT class, so perhaps I copied something incorrectly, or did I?
For the first problem, you don't need to finish the whole problem. Once you get to a point where it looks something like $(x+15)^2 \equiv 10 \mod 17$ (result of the example problem), I know how to proceed. It's the beginning parts that have muddled me a bit...
Help would be greatly appreciated.
| Note that $3 \equiv 5^2$, therefore $3x^2 +2x +1 \equiv 5^2x^2 + 2 \cdot 5 \cdot 5^{-1} x + 5^{-2} - 5^{-2} +1$ and since $5^{-1} \equiv 9$ this means $3x^2 +2x +1 \equiv 5^2x^2 + 2 \cdot 5 \cdot 9 x + 9^2 - 3 \equiv (5x + 9)^2 - 3$, so you want to solve $(5x + 9)^2 \equiv 3 \equiv 5^2$, therefore $(5x + 9) \equiv \pm 5$, giving $5x \equiv 7$ and $5x \equiv 8$, giving $x \equiv 8$ and $x \equiv 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the series has positive integer coefficients How can I show that the Maclaurin series for
$$
\mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4}
\\
= 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\,
{x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots
$$
has positive integer coefficients? (I have others to do, too, but this one will be a start.)
possibilities
(a) The coefficients $Q(n)$ satisfy the recurrence
$$
(n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0
$$
(b) $\mu(x)$ satisfies the differential equation
$$
(x^3+9x^2+7x-3)\mu(x)+(x^4+12x^3+14x^2-12x+1)\mu'(x)=0
$$
(c) factorization of $x^4+12x^3+14x^2-12x+1$ is
$$
\left( x-\sqrt {5}+3+\sqrt {15-6\,\sqrt {5}} \right) \left( x-\sqrt
{5}+3-\sqrt {15-6\,\sqrt {5}} \right) \left( x+\sqrt {5}+3-\sqrt {15+
6\,\sqrt {5}} \right) \left( x+\sqrt {5}+3+\sqrt {15+6\,\sqrt {5}}
\right)
$$
(d) $(1-8X)^{-1/4}$ has positive integer coefficients, But if $X$ is defined by $1-8X=x^4+12x^3+14x^2-12x+1$, then $X$ does not have integer coefficients.
(e) Can we compute the series for $\log\mu(x)$
$$3\,x+{\frac {29}{2}}{x}^{2}+99\,{x}^{3}+{\frac {3121}{4}}{x}^{4}+{
\frac {32943}{5}}{x}^{5}+{\frac {348029}{6}}{x}^{6}+\dots
$$
and then recognize that its exponential has integer coefficients?
| For the periodicity mod $5$, note that
$$(1 - x^4)^4 \equiv (1 + 3 x + 4 x^2 + 2 x^3)^4 (1-12 x+14 x^2+12 x^3+x^4) \mod 5 $$
and thus
in the field $\mathbb Z_5((x))$ of formal Laurent series in $x$ over the integers mod $5$ we have
$$ \eqalign{&\dfrac{1}{1-12 x+14 x^2+12 x^3+x^4} = \left(\dfrac{1+3x+4x^2+2x^3}{1-x^4}\right)^4\cr
&= \left(1+3x+4x^2+2x^3 + (1+3x+4x^2+2x^3) x^4 + (1+3x+4x^2+2x^3) x^8 + \ldots \right)^4\cr} $$
Given there is a solution over the integers, taking the coefficients mod $5$ gives us a solution in $\mathbb Z_5((x))$. Of course there are four fourth roots, but this is the only solution whose constant coefficient is $1$ (the others have $2, 3$ or $4$).
Similarly, to show the coefficients are odd,
$$(1-x)^4 \equiv 1 + x \equiv 1 -12 x + 14 x^2 + 12 x^3 + x^4 \mod 2$$
so that in $\mathbb Z_2((x))$
$$\dfrac{1}{1-12 x+14 x^2+12 x^3+x^4} = \dfrac{1}{(1-x)^4} = (1 + x + x^2 + x^3 + \ldots)^{4}$$
Combine the solutions mod $5$ and mod $2$ and we see that mod $10$ the coefficients repeat $1,3,9,7$.
The pattern mod $3$ also seems quite interesting (but not periodic): I'll leave that as an exercise. Solution tomorrow if nobody else posts it.
EDIT: OK, time's up. Mod $3$ we have
$$\eqalign{(1 - x^2 + x^4)^{-1/4} &\equiv 1+{x}^{2}+{x}^{6}+{x}^{8}+{x}^{18}+{x}^{20}+{x}^{24}+{x}^{26}
\ldots\cr
&= \prod_{j=0}^\infty \left(1 + x^{2\cdot 3^j}\right) = \sum_{k \in S} x^{k}\cr}$$
where $S$ is the set of positive integers whose base-3 representation contains no $1$'s. This is because
$$(1+x^{m})^3 \equiv 1 + x^{3m}$$ so that
$$ \dfrac{1+x^{2\cdot 3^{n+1}}}{1+x^2} \equiv \prod_{j=0}^n \dfrac{1+x^{2\cdot 3^{j+1}}}{1+x^{2\cdot 3^j}} \equiv \prod_{j=0}^n \left(1 + x^{2\cdot 3^j}\right)^2$$
Taking $n \to \infty$ we get
$$ \dfrac{1}{1+x^2} \equiv \prod_{j=0}^\infty \left(1 + x^{2\cdot 3^j}\right)^2$$
and then
$$ \dfrac{1}{1-x^2 + x^4} \equiv \dfrac{1}{(1+x^2)^2} \equiv
\prod_{j=0}^\infty \left(1 + x^{2\cdot 3^j}\right)^4 $$
| {
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"url": "https://math.stackexchange.com/questions/1386819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$ I have written the left side of the equation as $$\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right).$$ I don't know how to find the sums of these sequences. I know the sums for odd and even integers, but I can't figure this out.
| Denote $H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$. What you want to prove is exactly
$$
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1}-\frac{1}{2n} = H_{2n} - H_n
$$
i.e.,
$$
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n-1}-\frac{1}{2n} - H_{2n} = - H_n
$$
Note that the left side is equal to
$$
2(-\frac{1}{2} - \cdots - \frac{1}{2n}) = -H_n
$$
since those terms in odd positions have been removed and those in even positions have been doubled.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Solving algebraic equations with radicals: $x(x-\sqrt3)(x+1)+3-\sqrt3=0$ I have several problems requiring assistance.
Solve for $x$:
$x\left( x-\sqrt { 3 } \right) \left( x+1 \right) +3-\sqrt { 3 } \quad =\quad 0$
I've followed the suggestion to get x^2 - (√3 -1)x + (3-2√3) and shall proceed to the quadratic formula. Thanks.
Number 2:
The equation $2k{ x }^{ 2 }+(8k+1)x+8k\quad =\quad 0$ has 2 distinct real roots for $x$. Find the range of values for $k$.
I've used the discriminant and all that to find $k>\frac { -1 }{ 16 } $.
Am I missing anything?
| The originally posted equation appeared to be $(x - \sqrt{3}) x (x+1) + 3 - \sqrt{3} = 0$ for which this polynomial can be seen in the form
\begin{align}
& x(x+1)(x - \sqrt{3}) + 3 - \sqrt{3} = 0 \\
& x^{2} (x + 1 - \sqrt{3}) - \sqrt{3} (x + 1 - \sqrt{3}) = 0 \\
& (x - \sqrt[4]{3})(x + \sqrt[4]{3})(x + 1 - \sqrt{3}) = 0
\end{align}
of which the solutions are $x \in \{ \sqrt[4]{3}, - \sqrt[4]{3}, \sqrt{3}-1 \}$.
As it appears now the equation asked to be solved is $(x+1)(x-\sqrt{3}) + 3 - \sqrt{3} = 0$. Expanded and solved is as follows:
\begin{align}
& x^{2} + (1-\sqrt{3})x + 3 - 2 \sqrt{3} = 0 \\
& x= - \frac{1 - \sqrt{3}}{2} \pm \frac{1}{2} \sqrt{(1-\sqrt{3})^{2} - 4 (3 - 2 \sqrt{3})} \\
& x = \frac{\sqrt{3} -1}{2} \pm \frac{\sqrt{6 \sqrt{3} - 8}}{2}.
\end{align}
As to the additional question:
The equation given is $2 k \, x^{2} + (1+8k) \, x + 8k = 0$ for which
$$x^{2} + \left(4 + \frac{1}{2k}\right) \, x + 4 = 0$$
and
$$x = - 2 - \frac{1}{4k} \pm \frac{\sqrt{1 + 16 k}}{4 k}$$.
It is fairly evident that the case $k =0$ leads to an unbounded solution. If $k \to \infty$ then the equation becomes $(x+2)^{2}=0$ and has a double solution of $x = -2$. If $\sqrt{1 + 16 k} \geq 0$ is required then $k > - \frac{1}{16}$. The conclusion is $k \geq - \frac{1}{16}, k \neq 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A basic square numbers equation I couldn't solve this equation. Could you help me to solve this step by step.
Thank you.
$$\sqrt[4]{\frac{8^x}{8}}=\sqrt[3]{16}$$
| \begin{align}
\sqrt[4]{8^{x} \times \frac{1}{8}} &= \sqrt[3]{16} \\
\sqrt[4]{8^{x-1}} &= \sqrt[3]{16} \\
8^{x-1} &= \left( \sqrt[3]{16} \right)^{4} \\
8^{x-1} &= 16^{\frac{4}{3}} \\
2^{3x-3} &= 2^{\frac{16}{3}} \\
3x-3 &= \frac{16}{3} \quad \to \quad
3x = \frac{16}{3} +3 \quad \to \quad
x = 1+\frac{16}{9} \quad \to \quad
x = \frac{25}{9}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not relate$A_1A_3$ and $A_1A_4$ with the side length.Can someone assist me in solving this problem?
| I was casting about for a method that didn't require a lot of work with multiple-angles and trigonometric identities. Up to now, I had a different -- though ultimately related -- argument (without using a circumscribed circle) which led me to the same equation, $ \ \frac{1}{\sin\theta} \ = \ \frac{1}{\sin2\theta} \ + \ \frac{1}{\sin3\theta} $ , which Shailesh already produced. Here's something using more basic triangle geometry than what I'd had.
For convenience, we will call the lengths of the sides of the polygon $ \ s \ $ , making $ \ A_1A_2 \ = \ s \ $ and call the other lengths of interest $ \ A_1A_3 \ = \ t \ $ and $ \ A_1A_4 \ = \ u \ $ .
"Drop perpendiculars" to $ \ \overline{A_1A_4} \ $ from $ \ A_2 \ $ to produce point $ \ P \ $ and from $ \ A_3 \ $ to produce $ \ Q \ $ . Since we are working with a regular polygon, it is straightforward to show that $ \ A_1A_2A_3A_4 \ $ is a trapezoid and that $ \ PQ \ = \ s \ $ . Extend, say , $ \ A_4A_3 \ $ to a point $ \ R \ $ : since $ \ \angle A_2A_3R \ $ is an exterior angle of the polygon, $ \ m(\angle A_2A_3R) \ = \ \frac{2 \pi}{n} \ $ . We have $ \ \overline{A_2A_3} \ \ \Vert \ \ \overline{A_1A_4} \ $ , so corresponding angle $ \ \angle QA_4A_3 \ $ also has measure $ \ \frac{2 \pi}{n} \ $ .
Consequently, $ \ A_4Q \ = \ A_1P \ = \ s \ \cos \left(\frac{2 \pi}{n} \right) \ $ [it is clear that the two segments are congruent] and so
$$ u \ = \ s \ \left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right) \ \ . $$
Although $ \ \Delta A_1A_2A_3 \ $ is isosceles, and $ \ m(\angle A_1A_2A_3) \ = \ \pi \ - \ \frac{2 \pi}{n} \ $ by a familiar theorem (or because it is supplementary to an exterior angle), we will actually not exploit the Law of Cosines here to find $ \ t \ $ . Instead, we obtain that $ \ m(\angle A_1A_3A_2 ) \ = \ \frac{1}{2} \ ( \ \pi \ - \ [ \pi \ - \ \frac{2 \pi}{n} ] \ ) \ = \ \frac{\pi}{n} \ $ . From this, we find $ \ m(\angle A_1A_3A_4 ) \ = \ \left(\pi \ - \ \frac{2 \pi}{n} \right) \ - \frac{\pi}{n} $ $ \ = \ \left(\pi \ - \ \frac{3 \pi}{n} \right) \ $ .
By the Law of Sines,
$$ \frac{t}{\sin \left( \frac{2 \pi}{n} \right)} \ = \ \frac{u}{\sin \left( \ \pi \ - \ \frac{3 \pi}{n} \ \right) \ } \ \ \Rightarrow \ \ \frac{t}{\sin \left( \frac{2 \pi}{n} \right)} \ = \ \frac{s \ \left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)}{\sin \left( \frac{3 \pi}{n} \right) } $$
[using the identity "sine of an angle equals the sine of its supplement"].
At last coming to the equation under discussion, we obtain
$$ \frac{1}{s} \ = \ \frac{1}{t} \ + \ \frac{1}{u} \ = \ \frac{1}{s} \ \left[ \ \frac{1 }{\left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)} \ \left( \ \frac{\sin \left( \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right) \ \right] \ \left[ \ 1 \ + \ \left( \ \frac{\sin \left( \frac{2 \pi}{n} \right)}{\sin \left( \ \frac{3 \pi}{n} \ \right)} \ \right) \ \right] $$
$$ \Rightarrow \ \ 1 \ = \ \frac{1 }{\left( \ 1 \ + \ 2 \ \cos \left[ \frac{2 \pi}{n} \right] \ \right)} \ \ \left[ \ \left( \ \frac{\sin \left( \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right) \ + \ 1 \ \right] $$
$$ \Rightarrow \ \ 1 \ + \ 2 \ \cos \left( \frac{2 \pi}{n} \right) \ = \ 1 \ + \ \left[ \ \frac{\sin \left( \frac{3 \pi}{n} \right)}{\sin \left( \ \frac{2 \pi}{n} \ \right)} \ \right] $$ $$\Rightarrow \ \ 2 \ \sin \left( \frac{2 \pi}{n} \right) \ \cos \left( \frac{2 \pi}{n} \right) \ = \ \sin \left( \frac{3 \pi}{n} \right) \ \ \Rightarrow \ \ \sin \left( \frac{4 \pi}{n} \right) \ = \ \sin \left( \frac{3 \pi}{n} \right) \ \ , $$
applying the "double-angle formula" for sine at the end.
Since all of the angles discussed here have measure less than $ \ \pi \ $ , it is either the case that
$$ \frac{4 \pi}{n} \ = \frac{3 \pi}{n} \ \ , $$
which would require $ \ \frac{ \pi}{n} \ = \ 0 \ $ , or that
$$ \frac{4 \pi}{n} \ = \ \pi \ - \frac{3 \pi}{n} \ \ \Rightarrow \ \ \frac{7 \pi}{n} \ = \ \pi \ \ \Rightarrow \ \ n \ = \ 7 \ \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Give the equation of a line that passes through the point (5,1) that is perpendicular and parallel to line A. The equation of line $A$ is $3x + 6y - 1 = 0$. Give the equation of a line that passes through the point $(5,1)$ that is
*
*Perpendicular to line $A$.
*Parallel to line $A$.
Attempting to find the parallel,
I tried $$y = -\frac{1}{2}x + \frac{1}{6}$$
$$y - (1) = -\frac{1}{2}(x-5)$$
$$Y = -\frac{1}{2}x - \frac{1}{10} - \frac{1}{10}$$
$$y = -\frac{1}{2}x$$
| For the second case:
the line you want to find its equation is parallel to $A$, so it means they have the same slope$-\frac{1}{2}$, which yields:
$$
{y}_{1} = -\frac{1}{2}x+p
$$
Now you'll find $p$ from another condition you gave(the line passes from $(5,1)$), so we get :
$$
1 = \frac{-5}{2}+p\\
p = 1+\frac{5}{2} = \frac{7}{2}
$$
We finally get:
$$
{y}_{1} = -\frac{1}{2}x+\frac{7}{2}
$$
For the first question you should know that if $m$ is the slope of line $A$ and $m'$ is the slopeof line $B$, and $A$ and $B$ are perpendicular then we have:
$$
mm' = -1
$$
For our case we have $m = 1/2$, so $m' = 2$ and we get:
$$
{y}_{2} = 2x+p'
$$
Can you continue from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove for primes p $>2$ that $\sum_{k=1}^{p−1}{k^{2p−1}}\equiv\frac{1}{2}p(p+1)\pmod {p^2}$
Let $p$ be an odd prime. Prove that:
$$\sum_{k=1}^{p−1}{k^{2p−1}}\equiv\dfrac{p(p + 1)}{2}\pmod {p^2}$$
The problem is taken from the 2004 Canada National Olympiad.
I am only able to show that the sum is congruent to $0$ modulo $p$ (no congruence modulo $p^2$).
Since no value of $k$ is divisible by $p$, we have by Fermat's Little Theorem
$$k^{p-1}\equiv1\pmod p\implies k^{p-1}=np+1\text{ for an }n\in\mathbb{Z^+}$$
so then
$$\begin{align}
k^{2(p-1)}=1+2np+p^2&\equiv 1+p\pmod {2p} \\
&\equiv 1\pmod p
\end{align}$$
But this does no better than $$k^{2p-1}\equiv k\quad(\mod p)$$
Hence,
$$\begin{align}
\sum_{k=1}^{p−1}{k^{2p−1}}&\equiv\dfrac{p(p - 1)}{2}\pmod p \\
&\equiv 0\pmod p
\end{align}$$
which is trivial.
| In general: $$(a+b)^n \equiv a^n + nba^{n-1}\pmod{b^2}.$$ Letting $a=-k, b=p, n=2p-1$, you get that $$(p-k)^{2p-1} \equiv (2p-1)pk^{2p-2} - k^{2p-1}\pmod{p^2}$$
And $k^{2p-2}=(k^2)^{p-1}\equiv 1\pmod p$, we have $$(2p-1)pk^{2p-2}\equiv -pk^{2p-2}\equiv -p\pmod {p^2}.$$
So $$(p-k)^{2p-1} \equiv -p -k^{2p-1}\pmod{p^2}$$
So $$2\sum_{k=1}^{p-1} k^{2p-1} = \sum_{k=1}^{p-1} \left(k^{2p-1}+(p-k)^{2p-1}\right) \equiv -p(p-1)\equiv p\equiv p^2+p\pmod{p^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Perpendicular Bisector of Made from Two Points For a National Board Exam Review:
Find the equation of the perpendicular bisector of the line joining
(4,0) and (-6, -3)
Answer is 20x + 6y + 29 = 0
I dont know where I went wrong. This is supposed to be very easy:
Find slope between two points:
$${ m=\frac{y^2 - y^1}{ x^2 - x^1 } = \frac{-3-0}{-6-4} = \frac{3}{10}}$$
Obtain Negative Reciprocal:
$${ m'=\frac{-10}{3}}$$
Get Midpoint fox X
$${ \frac{-6-4}{2} = -5 }$$
Get Midpoint for Y
$${ \frac{-0--3}{2} = \frac{3}{2} }$$
Make Point Slope Form:
$${ y = m'x +b = \frac{-10}{3}x + b}$$
Plugin Midpoints in Point Slope Form
$${ \frac{3}{2} = \frac{-10}{3}(-5) + b}$$
Evaluate b
$${ b = \frac{109}{6}}$$
Get Equation and Simplify
$${ y = \frac{-10}{3}x + \frac{109}{6}}$$
$${ 6y + 20x - 109 = 0 }$$
Is the problem set wrong? What am I doing wrong?
| lol. a lot of confusion on this thread. When you are computing the midpoints, what you actually do is find the distance between them and divide by two. This isn't the coordinate of the midpoint; this is just the distance form one end to the midpoint.
What you meant to do is take the average of the x and y values, not the average of their difference.
You should get a midpoint of $(-1,-1.5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate the volume bounded by the surfaces Calculate the volume of the solid bounded by the surfaces $$\begin{aligned}z&=4x^2+4y^2, \\ z&=x^2+y^2, \\z&=4.\end{aligned}$$
I made an equation of $4x^2+4y^2=4-x^2+y^2$ and solved it to get $x^2+y^2=\dfrac{4}{5}$.
Then I did a double integration
$$\displaystyle
\iint_{x^2+y^2\leq \frac{4}{5}} \left(4x^2+4y^2\right)-\left(4-x^2-y^2\right)dA,\
$$
did the subtraction and the changed to polar
$$\displaystyle
\int_0^{2\pi} \int_0^{\frac{4}{5}} \left(3r^2-4\right) r \,dr \,d\theta,\
$$
got a result of
$$
\displaystyle\int_0^{ 2\pi}\Bigg[ \left.\left(\frac{3}{4}r^4-2r^2\right) \right\vert_{r=0}^{\frac{4}{5}} \Bigg]d\theta
$$
Obviously this is not correct. Can you tell me where I have gone wrong?
Thanks in advance :)
| Let
$(x,y,z) \in \mathbb{R}^{3}$ and let $S$ be the region enclosed by the surfaces $z = x^{2}+y^{2}$ and $z=4$. Then $(x,y,z) \in S$ if and only if $|x| \leq 4, |y| \leq \sqrt{4-x^{2}}, 0 \leq z \leq 4$. Thus
$$4\int_{0}^{4}\int_{0}^{\sqrt{4-x^{2}}} x^{2} + y^{2} dy dx = 4\int_{0}^{4} x^{2}\sqrt{4-x^{2}} + \frac{(4-x^{2})^{3/2}}{3} dx$$
is the content of $S$.
Let $T$ be the region enclosed by $z=4$ and $z = 4x^{2} + 4y^{2}$ and do the above for $T$. Then the absolute value of the difference of the resulting two contents is the desired content.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where is my error in solving $y'' + y' + y = 0, y(0) = 1, y'(0) = 0$ with Laplace transform? Im trying to solve a laplace transoform question, but i am stuck.
The question is $y′′(t) + 2ζy′(t) + y(t) = 0, y(0) = 1, y′(0) = 0$ and $ζ = 0.5$.
I have so far done:
Laplace transform which gives
$s^2Y(s)-sy(0)-y'(0)+2ζ(sY(s)-y(0))+Y(s)$
When $ζ = 0.5$
$S^2Y(s)-Sy(0)-y'(0)+sY(s)-y(0)+Y(s)$
$S^2Y(s)-s-sY(s)-1+Y(s)$
$s^2Y(s)-sY(s)+Y(s)=s+1$
$Y(s)[s^2-s+1]=(s+1)$
$Y(s)=(s+1)/(s^2-s+1)$
I'm stuck on this bit not sure what to do after this.
| Since the denominator is not easily factorable, we avoid partial fractions and instead complete the square. Recall the following inverse Laplace transforms:
$$
Y(s) = \frac{b}{(s - a)^2 + b^2} \implies y(t) = e^{at}\sin bt \\
Y(s) = \frac{s - a}{(s - a)^2 + b^2} \implies y(t) = e^{at}\cos bt
$$
Indeed, observe that:
\begin{align*}
\frac{s + 1}{s^2 - s + 1}
&= \frac{s + 1}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} \\
&= \frac{s - \frac{1}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} + \frac{\frac{3}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} \\
&= \frac{s - \frac{1}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2} + \frac{3}{\sqrt 3} \cdot \frac{\frac{\sqrt 3}{2}}{(s - \frac{1}{2})^2 + (\frac{\sqrt 3}{2})^2}
\end{align*}
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $ax^5 + by^5$ by matrix from $ \left\{ \begin{aligned} ax+by&=3\\ ax^2+by^2&=7\\ ax^3+by^3&=16\\ ax^4+by^4&=42 \end{aligned} \right.$
Find $ax^5 + by^5$ if the real numbers $a$, $b$, $x$, and $y$ satisfy the equations
$$
\left\{
\begin{aligned}
ax+by&=3 \\
ax^{2}+by^{2}&=7 \\
ax^{3}+by^{3}&=16 \\
ax^{4}+by^{4}&=42
\end{aligned}
\right.
$$
with matrix form!
Interestingly, when I googled the equations, the problem comes from 1990 AIME Problems/Problem 15. The link gives two methods for the solution.
Is there a fashioned way to find the solution in matrix form?
| The secuence $c_n=ax^n+by^n$ satisfies a linear two step recurrence relation of the form
$$c_{n+2}=Ac_{n+1}+Bc_n$$
for some constants $A,B$. This means that if we perform the row operations coming from that recurrence relation (subtract the second row multiplied by $A$ and the first row multiplied by $B$ from the third row) to the matrix
$$
\left(\begin{array}{ccc}c_1&c_2&c_3\\c_2&c_3&c_4\\c_3&c_4&c_5\end{array}\right)
$$
its last row becomes all zeros. This means that its determinant vanishes, i.e.
$$
0=\left\vert\begin{array}{ccc}c_1&c_2&c_3\\c_2&c_3&c_4\\c_3&c_4&c_5\end{array}\right\vert=\left\vert\begin{array}{ccc}3&7&16\\7&16&42\\16&42&c_5\end{array}\right\vert=20-c_5.
$$
Therefore $c_5=20$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integral of $\frac{1}{x^2+x+1}$ and$\frac{1}{x^2-x+1}$ How to integrate two very similar integrals. I am looking for the simplest approach to that, it cannot be sophisticated too much as level of the textbook this was taken from is not very high. $$\int \frac{1}{x^2+x+1} dx$$ and$$\int \frac{1}{x^2-x+1} dx$$
| For the first one:
$$
\int { \frac { 1 }{ 1+{ x }^{ 2 }+x } } dx\quad =\quad \quad \int { \frac { 1 }{ { x }^{ 2 }+x+\frac { 1 }{ 4 } +\frac { 3 }{ 4 } } dx } \\ \qquad \qquad \qquad \qquad =\quad \int { \frac { 1 }{ { \left( x+\frac { 1 }{ 2 } \right) }^{ 2 }+\frac { 3 }{ 4 } } dx } \\ \qquad \qquad \qquad \qquad \quad =\quad \frac { 4 }{ 3 } \int { \frac { 1 }{ 1+{ \left( \sqrt { \frac { 4 }{ 3 } } \left( x+\frac { 1 }{ 2 } \right) \right) }^{ 2 } } } dx\\ \qquad \qquad \qquad \qquad \quad =\quad \frac { 2 }{ \sqrt{3} } \arctan { \left( \frac { 2 }{ \sqrt{3} } \left( x+\frac { 1 }{ 2 } \right) \right) } +C
$$
The second one will be the same just a minus instead of a plus:
$$
\frac { 2 }{ \sqrt{3} } \arctan { \left( \frac { 2 }{ \sqrt{3} } \left( x-\frac { 1 }{ 2 } \right) \right) } +C
$$
Here's a general formula for $\int \frac{1}{ax^2+bx+c}dx$, when $b^2-4ac<0$:
$$
\int { \frac { 1 }{ a{ x }^{ 2 }+bx+c } } dx\quad =\frac { 1 }{ a } \quad \int { \frac { 1 }{ { x }^{ 2 }+\frac { b }{ a } x+\frac { c }{ a } } } dx\\ \qquad \qquad \qquad \qquad \quad =\quad \frac { 1 }{ a } \int { \frac { 1 }{ { x }^{ 2 }+\frac { b }{ a } x+\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } +\frac { c }{ a } -\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } } dx } \\ \qquad \qquad \qquad \qquad \quad =\quad \frac { 1 }{ a } \int { \frac { 1 }{ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }+\frac { c }{ a } -\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } } } dx\\ \qquad \qquad \qquad \qquad \quad =\quad \frac { 1 }{ a\left( \frac { c }{ a } -\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } \right) } \int { \frac { 1 }{ 1+{ \left( \frac { x+\frac { b }{ 2a } }{ \sqrt { \frac { c }{ a } -\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } } } \right) }^{ 2 } } dx } \\ \qquad \qquad \qquad \quad =\quad \frac { \sqrt { \frac { c }{ a } -\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } } }{ c-\frac { { b }^{ 2 } }{ 4a } } \arctan { \left( \frac { x+\frac { b }{ 2a } }{ \sqrt { \frac { c }{ a } -\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } } } \right) } +C
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $x$ and $y$ by Cramer from $\left\{\begin{aligned} \frac{2x+2}{2y-3}&=\frac{x-2}{y+3}\\ \frac{x+2}{3y-2}&=\frac{x-1}{3y+8} \end{aligned}\right.$
Find $x$ and $y$ with Cramer rule from these equations
$$\left\{
\begin{aligned}
\frac{2x+2}{2y-3}&=\frac{x-2}{y+3} \\
\frac{x+2}{3y-2}&=\frac{x-1}{3y+8}
\end{aligned}
\right.
$$
I'm getting stuck to make the $x$ and $y$ out for making system equation matrix. Any idea?
| I will show you the first one:
$$\frac{2x+2}{2y-3}=\frac{x-2}{y+3} \implies\\
(2x+2)(y+3)=(x-2)(2y-3)\implies\\
2xy+6x+2y+6=2xy-3x-4y+6$$
Can you see how you can cancel and combine the terms now?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find Sum of n terms of the Series: $ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \cdots $ I want to find the sum to n terms of the following series.
$$ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \frac{7}{4\times5\times6} + \cdots $$
I have found that the general term of the sum is
\begin{equation}
\sum\limits_{i=1}^n \frac{2i - 1}{i(i+1)(i+2)}
\end{equation}
so the nth term must be
\begin{equation}
\frac{2n - 1}{n(n+1)(n+2)}
\end{equation}
Now.. how should i evaluate the sum of the n terms? What is the method i must use in order to do so? Thank you in advance.
| $$\begin{align}
\sum\limits_{i=1}^n \frac{2i - 1}{i(i+1)(i+2)}
&=\sum_{i=1}^n\frac {Ai+B}{i(i+1)}-\frac{A(i+1)+B}{(i+1)(i+2)}\\
&=\sum_{i=1}^n\frac{2i-\frac12}{i(i+1)}-\frac{2(i+1)-\frac12}{(i+1)(i+2)}\\
&=\frac34-\frac{4n+3}{2(n+1)(n+2)}\qquad\text{by telescoping}\\
&=\frac{n(3n+1)}{4(n+1)(n+2)}\qquad\blacksquare
\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no roots hence proving that above polynomial has $4$ real roots.
But using online calculators I get zero real roots. Where am I wrong?
| First, rescale by $720$ to get integer coefficients: $$x^6+6x^5+30x^4+120x^3+360x^2+720x+720$$
Now repeated completion of binomial powers:
$$\begin{align}
&\phantom{{}={}}(x+1)^6+15x^4+100x^3+345x^2+714x+719\\
&=(x+1)^6+15(x+5/3)^4+95x^2+\frac{3926}{9}x+\frac{16288}{27}\\
\end{align}$$
You could complete the square again on this last quadratic and you will be left with a positive constant, or you can just compute its discriminant to see that the quadratic itself has no roots (and has a positive quadratic term, so is therefore positive). So $$(x+1)^6+(x+5/3)^4+q(x)$$ is positive.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To evaluate $\int_{0}^\pi \frac {\sin^2 x}{a^2-2ab \cos x + b^2}\mathrm dx$? How do we evaluate $\displaystyle\int_{0}^\pi \dfrac {\sin^2 x}{a^2-2ab \cos x + b^2}\mathrm dx$ ? I tried substitution and some other methods , but its not working ; please help .
| Let $I$ be the integral given by
$$I=\int_0^\pi \frac{\sin^2x}{a^2+b^2-2ab\cos x}\,dx$$
Exploiting the even symmetry of the integrand, we have
$$I=\frac12\int_{-\pi}^\pi \frac{\sin^2x}{a^2+b^2-2ab\cos x}\,dx$$
Next, using $\sin^2x=\frac12(1-\cos 2x)$ we obtain
$$\begin{align}
I&=\frac14\int_{-\pi}^\pi \frac{1-\cos 2x}{a^2+b^2-2ab\cos x}\,dx\\\\
&=\frac14\text{Re}\left(\int_{-\pi}^\pi \frac{1-e^{i2x}}{a^2+b^2-2ab\cos x}\,dx\right)
\end{align}$$
Now, we move to the complex plane. To that end, we let $z=e^{ix}$ so that $dx=\frac{1}{iz}dz$. Then,
$$I=\frac{1}{4ab}\text{Re}\left(i\oint_{|z|=1} \frac{1-z^2}{z^2-\frac{a^2+b^2}{ab}z+1}\,dz\right)$$
The singularities of the integrand are at $z=\frac{a^2+b^2}{2ab}\pm\frac{|a^2-b^2|}{2|ab|}$. We will analyze the case for which $0<b<a$ with the results from other cases following from symmetry arguments. We note that if $ab>0$, then $$\frac{a^2+b^2}{2ab}>1$$ and the only enclosed pole is at $z=\frac{a^2+b^2}{2ab}-\frac{|a^2-b^2|}{2|ab|}=b/a$ for $0<b<a$. Therefore, the residue is given by
$$\text{Res}\left(\frac{1-z^2}{z^2-\frac{a^2+b^2}{ab}z+1},z=b/a\right)=\frac{1-(b/a)^2}{-(a^2-b^2)/2ab}=-b/a$$
Putting it all together yields the final result
$$\bbox[5px,border:2px solid #C0A000]{I=\frac{\pi}{2a^2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 - 2\sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int -2 e^x \sin^2x \mathrm{d}x + 2 e^x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x -4 \int e^x \sin^2x \mathrm{d}x \end{align*}
First two steps use integration by parts. In the first step we differentiate $\sin^2 x$. In the second step we differentiate $\sin x \cos x$.
Using this, we reach $$5\int e^x \sin^2 x \mathrm{d}x = e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x$$
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$
I can't reach the form that most integral calculators give, which has terms $\cos(2x)$ and $\sin(2x)$ by just using trig identities, so I wonder whether the result is correct. I would also be interested in a method that immediately gives the form $$-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$
| We have $$\int e^{x}\sin^{2}\left(x\right)dx=\frac{1}{2}\int e^{x}dx-\frac{1}{2}\int e^{x}\cos\left(2x\right)dx
$$ and $$\int e^{x}\cos\left(2x\right)dx=\textrm{Re}\left(\int e^{x+2ix}dx\right)
$$ then $$\int e^{x+2ix}dx=\frac{e^{x+2ix}}{1+2i}
$$ and so $$\int e^{x}\sin^{2}\left(x\right)dx=\frac{1}{2}e^{x}-\frac{1}{2}e^{x}\left(\frac{\cos\left(2x\right)+2\sin\left(2x\right)}{5}\right)+C=$$ $$=-\frac{e^{x}\left(\cos\left(2x\right)+2\sin\left(2x\right)-5\right)}{10}+C.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$ My Calc 2 teacher wasn't able to solve this:
$$\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$$
Can someone help me solve this?
| $$
\int \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta} \,d\theta = \underbrace{\int \theta \, dx = \theta x - \int x\,d\theta}_{\text{integration by parts}}
$$
$$
dx = \frac { \cos \theta } { \sin \theta + \sin ^ 3 \theta} \,d\theta = \frac{du}{u+u^3} = \frac{du}{u(1+u^2)} = \left( \frac A u + \frac{Bu+C}{1+u^2} \right)\,du
$$
$$
\frac{Bu}{1+u^2} \, du = \frac B 2\cdot \frac{dw} w
$$
$$
\int \frac C {1+u^2} \, du = C \arctan u +\text{constant}
$$
You need to do a bit of algebra to find the three coefficients $A,B,C$, and then put everything together.
I'm getting $A=1$, $B=-1$, $C=0$, so
$$
\int \left( \frac 1 u - \frac u {1+u^2}\right)\, du = \log |u| - \frac 1 2 \log(1+u^2) + \text{constant}
$$
but we don't need the constant: in integration by parts we need only one antiderivative, not all of them.
This is
$$
\log|\sin\theta| - \frac 1 2 \log(1+\sin^2\theta).
$$
So the integral is
$$
\theta\left( \log|\sin\theta| - \frac 1 2 \log(1+\sin^2\theta) \right) - \int \left( \log|\sin\theta| - \frac 1 2 \log(1+\sin^2\theta) \right) \, d\theta + \text{constant}
$$
I haven't taken it beyond that yet, and I don't know whether it can be done is closed form.
However, sometimes it is possible to find a definite integral even if you can't find the indefinite integral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$ Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$
Since the area of ellipse is $A=\pi ab\Rightarrow A^2=\pi^2a^2b^2$ and the circle and the ellipse touch each other internally. This much I can visualise, but how to find out $a$ and $b$ from this?
| Means ellipse and circle touches each other. So $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ and $(x-1)^2+y^2 = 1$
So Ellipse and Circle touches each other. So we will solve these two equations.
$\displaystyle \frac{x^2}{a^2}+\frac{1-(x-1)^2}{b^2} = 1\Rightarrow b^2x^2+a^2\left[1-(x-1)^2\right] = a^2b^2$
so $\displaystyle b^2x^2+a^2\left[1-(x^2+1-2x)\right]=a^2b^2$
So $\displaystyle b^2x^2+a^2-a^2x^2-a^2+2a^2x-a^2b^2 =0$
So $(b^2-a^2)x^2+2a^2x-a^2b^2$
So Discriminat of above equation is $=0$(Bcz Courve touches each other)
So $\displaystyle 4a^4+4(b^2-a^2)\cdot (a^2b^2) =0$
So $\displaystyle a^2+(b^2-a^2)b^2=0\Rightarrow a^2+b^4-a^2b^2=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $(a+b)(b+c)(c+a) \ge8$
Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$,
Prove that $(a+b)(b+c)(c+a)\ge8$.
My attempt: By AM-GM inequality, we have
$$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$
and similarly
$$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$
$$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$
Multiplying $(1)$, $(2)$ and $(3)$ together we reach
$$(a+b)(b+c)(c+a) \ge8abc.$$
Now, I need to show that $abc = 1$.
Again by AM-GM inequality we have
$$\frac{a+b+c}{3}\ge\sqrt[3]{abc}
\implies \frac{\frac{3}{abc}}{3}\ge\sqrt[3]{abc}
\implies abc\le1$$
Now if we show somehow that $abc\ge1$, we are done and that's where I am stuck. Can someone please explain how to show that?
Other solutions to the above question are also welcomed.
| A geometric approach is to notice that $a,b,c>0$ and $abc(a+b+c)=3$ grant that there is a triangle with side lengths $a+b,a+c,b+c$ and area $\sqrt{3}$. Since the area of a triangle is given by the product of its side lengths divided by four times the length of the circumradius, the problem boils down to understanding what is the minimum circumradius for a given area, and it is pretty clear that the minimum is achieved by the equilateral triangle, since
$$ 2R = \frac{BC}{\sin\widehat{BAC}} $$
and if a variable $A$ point travels on a line parallel to a fixed $BC$ segment (so that the area of $ABC$ is constant), the maximum $\widehat{BAC}$ angle is achieved when $BA=CA$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Finding a perpendicular vector This is given: $P = (-1,1,0),\; Q = (1,5,6)\; \text{and} \; R = (3,-1,4).$
My question is:
Find the values of $x$ (where x is a real number) for which $PR + xQR\;$ is perpendicular to $PR$.
So far, what I have done is used the dot product of two vectors and equated
$\vec {u} \cdot \vec{v} = 0$ and my result was $x = -3$.
Am I missing any other values? Thank you.
| You are correct:
$$PR = (4, -2, 4)\\
QR = (2, -6, -2)\\
PR + xQR = (4 + 2x, -2-6x, 4-2x)\\
PR\cdot (PR + xQR) = 4(4+2x) - 2(-2 -6x) + 4(4-2x) = \\=16+8x+4+12x+16-8x=36+12x$$
So the solution is $28+12x=0$ or $x=-\frac{36}{12} = -3$.
Alternatively, you are solving the equation $$PR\cdot(PR-x\cdot QR) = 0$$
You can rewrite that into $$PR\cdot PR = -x\cdot (PR\cdot QR)$$
Now, since $PR\cdot QR$ is a real number, you get
$$x=-\frac{PR\cdot PR}{PR\cdot QR}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $(1+ \tan{A}\tan{2A})\sin{2A} = \tan{2A}$. Prove the following statement
$$ (1+ \tan{A}\tan{2A})\sin{2A} = \tan{2A}. $$
On the left hand side I have put the value of $\tan{2A}$ and have then taken the LCM.
I got $\sin{2A}\cos{2A}$. How do I proceed?
Thanks
| \begin{align}
& (1 + \tan A \tan(2A))\sin(2A) \\
= & \left(1 + \frac{\sin A}{\cos A}\frac{\sin(2A)}{\cos(2A)}\right)\sin(2A)\\
= & \left(1 + \frac{\sin A}{\cos A}\frac{2\sin A\cos A}
{\cos(2A)}\right)\sin(2A) \\
= & \left(1 + \frac{2\sin^2 A}{\cos(2A)}\right)\sin(2A) \\
= & \frac{\cos(2A) + 2\sin^2 A}{\cos(2A)}\sin(2A) \\
= & \frac{1 - 2\sin^2 A + 2\sin^2 A}{\cos(2A)}\sin(2A) \\
= & \frac{\sin(2A)}{\cos(2A)} \\
= & \tan(2A)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402684",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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An arctan integral $\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$ According to Mathematica, we have that
$$\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx=\pi \left(\frac{\log (2)}{2}-\frac{1321}{6144}\right)$$
that frankly speaking looks pretty nice.
However Mathematica shows that
$$\int \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$$
$$=-\frac{1}{2} i \text{Li}_2\left(e^{2 i \tan ^{-1}(x)}\right)-\frac{1}{2} i \tan ^{-1}(x)^2+\tan ^{-1}(x) \log \left(1-e^{2 i \tan ^{-1}(x)}\right)-\frac{65}{256} \sin \left(2 \tan ^{-1}(x)\right)-\frac{23 \sin \left(4 \tan ^{-1}(x)\right)}{1024}-\frac{5 \sin \left(6 \tan ^{-1}(x)\right)}{2304}-\frac{\sin \left(8 \tan ^{-1}(x)\right)}{8192}+\frac{65}{128} \tan ^{-1}(x) \cos \left(2 \tan ^{-1}(x)\right)+\frac{23}{256} \tan ^{-1}(x) \cos \left(4 \tan ^{-1}(x)\right)+\frac{5}{384} \tan ^{-1}(x) \cos \left(6 \tan ^{-1}(x)\right)+\frac{\tan ^{-1}(x) \cos \left(8 \tan ^{-1}(x)\right)}{1024}$$
and this form doesn't look that nice.
Having given the nice form of the closed form I wonder if we can find a very nice and simple way of getting the answer. What do you think?
A supplementary question:
$$\int_0^{\infty } \frac{\arctan^2(x)}{x \left(x^2+1\right)^5} \, dx=\frac{55}{108}-\frac{1321}{12288}\pi^2+\frac{\pi^2}{4} \log (2)-\frac{7 }{8}\zeta (3)$$
| I'm not sure if this is nice and simple, but here is one way:
Substitute $y = \arctan x$ to rewrite as $$\int_0^{\pi/2} \frac{y}{\tan y \sec^{8} y} \, dy$$
Now write this as an integral in $y$ and $\sec y$:
$$\int_0^{\pi/2} y \cdot \frac{\sec y \tan y}{(\sec^2 y - 1) \sec^{9} y} \, dy$$
You can integrate this by parts. Finding the antiderivative of $\dfrac{\sec y \tan y}{(\sec^2 y - 1) \sec^{9} y}$ amounts (after a change of variable) to finding the antiderivative of $$\frac{1}{t^{11} - t^{9}} = \frac{1}{t^{9}(t-1)(t+1)}.$$ This has an elementary partial fractions decomposition which I don't have the energy to carry out.
| {
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"url": "https://math.stackexchange.com/questions/1403038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$ $\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$
MyAttempt
$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|=\int_{0}^{\pi/2}\sqrt2\sin x+2\cos x dx+\int_{\pi/2}^{\pi}|\sqrt2\sin x+2\cos x| dx$
I could solve first integral but in second one,i could not judge the mod will take plus sign or minus sign or how to break it in further intervals?
| We Can Write $\displaystyle \sqrt{2}\sin x+2\cos x = \sqrt{\left(\sqrt{2}\right)^2+2^2}\left\{\frac{\sqrt{2}}{\sqrt{6}}\sin x+\frac{2}{\sqrt{6}}\cos x\right\}$
$\displaystyle = \sqrt{6}\left\{\sin x\cdot \frac{1}{\sqrt{3}}+\cos x\cdot \frac{2}{\sqrt{6}}\right\} = \sqrt{6}\left\{\sin x\cdot \cos \alpha+\cos x\cdot \sin \alpha\right\} = \sqrt{6}\cdot \sin (x+\alpha)$
Where $$\displaystyle \sin \alpha = \frac{2}{\sqrt{6}}$$ and $$\displaystyle \cos \alpha = \frac{1}{\sqrt{3}}$$ and $$\displaystyle \tan \alpha = \sqrt{2}$$ . So $\displaystyle \frac{\pi}{4}<\alpha <\frac{\pi}{3}$
So Integral Convert into $$\displaystyle \sqrt{6}\int_{0}^{\pi}\left|\sin (x+\alpha)\right|dx$$
Let $(x+\alpha) = t\;,$ Then $dx = dt$ and Changing Limit, we get
$$\displaystyle I = \sqrt{6}\int_{\alpha}^{\pi+\alpha}|\sin t|dt = \sqrt{6}\int_{\alpha}^{\pi}\sin tdt-\sqrt{6}\int_{\pi}^{\pi+\alpha}\sin tdt $$
So we get $$\displaystyle I = \sqrt{6}\left(1+\cos \alpha\right)+\sqrt{6}\left(-\cos \alpha +1\right) = 2\sqrt{6}$$
| {
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"url": "https://math.stackexchange.com/questions/1403526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\int_{0}^{2\pi}\frac{x^2\sin x}{8+\sin^2x}=\frac{2\pi^2}{3}\ln\frac{1}{2}$ Prove that $$\int_{0}^{2\pi}\dfrac{x^2\sin x}{8+\sin^2x}=\frac{2\pi^2}{3}\ln\frac{1}{2}$$
My Attempt:
$$\int_{0}^{2\pi}\dfrac{x^2\sin x}{8+\sin^2x}=\int_{0}^{2\pi}x^2\frac{\sin x}{8+\sin^2x}$$
I applied integration by parts,considering $x^2$ as first function and $\dfrac{\sin x}{8+\sin^2x}$ as second function
$$=x^2\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2x}dx-\int_{0}^{2\pi}2x\left(\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2x}dx\right)dx$$
but this calculates out to be zero because $$\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2x}dx=0$$
What has gone wrong?Please help me..
| Let $$\displaystyle I = \int_{0}^{2\pi}\frac{x^2\cdot \sin x}{8+\sin^2 x}dx ...................(1)$$
Now Replace $x\rightarrow (2\pi-x)\;,$ We get
$$\displaystyle I = -\int_{0}^{2\pi}\frac{\left(2\pi-x\right)^2\cdot \sin x}{8+\sin^2 x}dx...........(2)$$
Now Add these Two equations, We get
$$\displaystyle 2I = \int_{0}^{2\pi}\frac{2\pi(2\pi-2x)\cdot \sin x}{8+\sin^2 x}dx$$
So we get $$\displaystyle I = 2\pi\int_{0}^{2\pi}\frac{x\cdot \sin x}{9-\cos^2 x}dx-2\pi^2\underbrace{\int_{0}^{2\pi}\frac{\sin x}{8+\sin^2 x}dx}_{J}$$
Now value of $J=0\;,$ Using $\displaystyle \int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$
Now Using Integration By parts for $I\;,$ We get
$$\displaystyle I = 2\pi\left[-\frac{x}{6}\cdot \ln \left(\frac{3+\cos x}{3-\cos x}\right)_{0}^{2\pi}+\frac{2}{6}\underbrace{\int_{0}^{2\pi}\ln \left(\frac{3+\cos x}{3-\cos x}\right)}_{0}\right] = -\frac{2\pi^2}{3}\cdot \ln\left(2\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$
I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I want until the $x^4$ term. So what i found is the following:
$$ (1+e^x)^{-1} = 1 - e^x + e^{2x} + e^{3x} + e^{4x}+\cdots\tag1 $$
Then by expanding each $e^{nx}$ term individually:
$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+\cdots\tag2 $$
$$ e^{2x} = 1 + 2x + 2x^2 + \frac{3x^3}{2} + \frac{2x^4}{3}+\cdots\tag3 $$
$$ e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \frac{27x^4}{8}+\cdots\tag4 $$
$$ e^{4x} = 1 + 4x + 8x^2 + \frac{32x^3}{3} + \frac{32x^4}{3}+\cdots\tag5 $$
So substituting $(2),(3),(4),(5)$ into $(1)$ i get:
$$ (1+e^x)^{-1} = 1 +2x+5x^2+ \frac{22x^3}{3} + \frac{95x^4}{12} +\cdots$$
Which isn't the correct result. Am i not allowed to expand this series binomially? I've seen on this site that this is an asymptotic expansion. However, i don't know about these and i haven't been able to find much information on this matter to solve this. If someone could help me understand how to solve it and why my approach isn't correct, i would be VERY grateful. Thanks in advance.
| Just a sketch of a proof to be expanded later (sorry, I am in a hurry): by taking the logarithmic derivative of the Weierstrass product of the $\cosh$ function we have the Taylor series of the $\tanh$ function, with the coefficients depending on values of the $\zeta$ function at even integers. With a little maquillage it is not difficult to turn that into the Taylor series of $\frac{1}{1+e^x}$. Or we may just play a bit with the generating function of Bernoulli numbers.
So, I was saying:
$$\cosh z = \prod_{n\geq 0}\left(1+\frac{4z^2}{(2n+1)^2 \pi^2}\right)\tag{1}$$
hence:
$$ \tanh z = \sum_{n\geq 0}\frac{8z}{(2n+1)^2 \pi^2 + 4z^2}\tag{2} $$
leads to:
$$ \tanh z = \sum_{m\geq 0}(-1)^m z^{2m+1}\sum_{n\geq 0}\frac{2^{2m+3}}{\pi^{2m+2}(2n+1)^{2m+2}}\tag{3}$$
by expanding the general term of the RHS of $(2)$ as a geometric series. Computing the innermost sum in terms of the $\zeta$ function it follows that:
$$\tanh z = \sum_{m\geq 0}\frac{2(-1)^m\left(2^{2m+2}-1\right) }{\pi^{2m+2}}\zeta(2m+2)\, z^{2m+1}\tag{4}$$
but since $\frac{1}{1+e^{z}}=\frac{1-\tanh(z/2)}{2}$ it follows that:
$$\frac{1}{1+e^{z}}=\frac{1}{2}-\sum_{m\geq 0}\frac{(-1)^m\left(2^{2m+2}-1\right) }{2^{2m+1}\pi^{2m+2}}\zeta(2m+2)\, z^{2m+1}\tag{5}$$
so to solve the original problem it is enough to recall that $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$.
Moreover, through $(5)$ it is straightforward to compute the magnitude of the coefficients by simply approximating $\zeta(2m+2)$ with $1$. On the other hand, that is trivial also by considering that the radius of convergence of the Taylor series of $f(z)=\frac{1}{1+e^z}$ at $z=0$ is $\pi$, since $f(z)$ is a meromorphic function with two simple poles at $z=\pm \pi i$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/1404886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Find the range of the function,$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt$ Find the range of the function,$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt$
I tried to solve it,i got range $\frac{\pi}{2}$ but the answer is ${\frac{-\pi}{2},\frac{\pi}{2}}$
$f(x)=\int_{-1}^{1}\frac{\sin x}{1-2t\cos x+t^2}dt=\int_{-1}^{1}\frac{\sin x}{(t-\cos x)^2+\sin^2x}dt=\sin x\int_{-1}^{1}\frac{1}{(t-\cos x)^2+\sin^2x}dt$
$=\frac{\sin x}{\sin x}\left[\tan^{-1}\frac{t-\cos x}{\sin x}\right]_{-1}^{1}=\tan^{-1}\frac{1-\cos x}{\sin x}+\tan^{-1}\frac{1+\cos x}{\sin x}$
$=\tan^{-1}\tan\frac{x}{2}+\tan^{-1}\cot\frac{x}{2}=\frac{\pi}{2}$
But i am not getting $\frac{-\pi}{2}$.Have i done some wrong?Please enlighten me.
| Notice, second last line,
$$\frac{\sin x}{\sin x}\left[\tan^{-1}\left(\frac{t-cos x}{\sin x}\right)\right]_{-1}^{1}$$
$$=\tan^{-1}\left(\frac{1-cos x}{\sin x}\right)-\tan^{-1}\left(\frac{-1-cos x}{\sin x}\right)$$
$$=\tan^{-1}\left(\frac{1-cos x}{\sin x}\right)+\tan^{-1}\left(\frac{1+cos x}{\sin x}\right)$$
$$=\tan^{-1}\left(\frac{\frac{1-cos x}{\sin x}+\frac{1+cos x}{\sin x}}{1-\frac{1-cos x}{\sin x}\frac{1+cos x}{\sin x}}\right)$$
$$=\tan^{-1}\left(\frac{\frac{2}{\sin x}}{1-\frac{1-cos^2 x}{\sin^2 x}}\right)$$
$$=\tan^{-1}\left(\frac{\frac{2}{\sin x}}{1-\frac{sin^2 x}{\sin^2 x}}\right)$$
$$=\tan^{-1}\left(\frac{\frac{2}{\sin x}}{1-1}\right)=\tan^{-1}(\infty)=\frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integrating $\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$ Integrating $$\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$$
Using substitution of $x=\tan \theta$, I got the required answer. But is there a more elegant solution to the problem?
| Let $$ I = \int \sqrt{x+\sqrt{x^2+1}} \, dx\;,$$ Let $$x+\sqrt{x^2+1} = t^2 \tag 1$$
Then $$ \left(1+\frac{x}{\sqrt{x^2+1}}\right) \, dx = 2t \, dt\Rightarrow t^2 \, dx = 2t\sqrt{x^2+1} \, dt$$
Now using $$\bullet\; \left(\sqrt{x^2+1}+x\right)\cdot \left(\sqrt{x^2+1}-x\right) = 1$$
So we get $$ \sqrt{x^2+1}-x = \frac{1}{\sqrt{x^2+1}+x} = \frac{1}{t^2} \tag 2$$
Now add $(1)$ and $(2)\;,$ we get $\displaystyle \sqrt{x^2+1} = \frac{t^2+t^{-2}}{2}$
So we get $$ dx = \frac{\left(t^2+t^{-2}\right)\cdot 2t}{2t^2} \, dt$$
so Integral $$\displaystyle I = \int ( t^2+t^{-2}) \, dt = \frac{t^3}{3}-\frac{1}{t}+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 0
} |
Rotation and translation of coordinate axes I am studying rotation and translation of conical but have no doubt in basic concept (Sorry, I know this is a very stupid question but I'm really struggling to understand). Especially in this equation:
$$
9x^2 - 4y^2 - 18x - 16y - 7 = 0
$$
$$
\begin{cases}
x = u + h\\
y = v + k
\end{cases}\\
9u^2 - 4v^2 + (18h - 18)u + (-8k - 16)v + G(h, k) = 0
$$
Why is not this equation?
$$
9u^2 - 4v^2 - 18(u + h) - 16(v + k) + G(h, k) = 0
$$
| What happens when you substitute $u+h$ for $x$ in the equation
$9x^2 - 4y^2 - 18x - 16y - 7 = 0$?
First of all, $x^2 = (u + h)^2 = u^2 + 2hu + h^2$.
Therefore $$9x^2 = 9(u^2 + 2hu + h^2) = 9u^2 + 18hu + 9h^2.$$
And of course $-18x = -18u - 18h$. So
\begin{align}
9x^2 - 4y^2 - 18x - 16y - 7
& = 9u^2 + 18hu + 9h^2 - 18u - 18h - 4y^2 - 16y - 7 \\
& = 9u^2 + (18h - 18)u - 4y^2 - 16y + 9h^2 - 18h - 7.
\end{align}
Now substitute $v+k$ for $y$.
You should find that $-4y^2 = -4(v+k)^2 = -4v^2 - 8kv - 4k^2$
and $-16y = -16v - 16k$. The result is
\begin{align}
9x^2 - & 4y^2 - 18x - 16y - 7 \\
& = 9u^2 + (18h - 18)u - 4v^2 - 8kv - 4k^2 - 16v - 16k + 9h^2 - 18h - 7 \\
& = 9u^2 + (18h - 18)u - 4v^2 + (-8k - 16)v + (- 4k^2 - 16k + 9h^2 - 18h - 7).
\end{align}
Now, of the two equations that you wrote, which one could possibly be
equal to $9x^2 - 4y^2 - 18x - 16y - 7$?
As a hint, where is the $hu$ term in the second equation?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving $\cos^2{\theta}-\sin{\theta} = 1$ Can someone please help me solve this?
$$\cos^{2}{\theta}-\sin{\theta} = 1, \quad\theta\in[0^{\circ}, 360^{\circ}]$$
| Notice, we have
$$\cos^2\theta-\sin\theta=1$$
$$1-\sin^2\theta-\sin\theta=1$$
$$-\sin^2\theta-\sin\theta=0$$
$$\sin^2\theta+\sin\theta=0$$
$$\sin\theta(\sin\theta+1)=0$$ $$\sin\theta=0\iff \theta=n(180^\circ)$$
Where, $n$ is any integer.
But for the given interval $[0^\circ, 360^\circ]$, substituting $n=0, 1, 2$, we get
$$ \theta=0^\circ, 180^\circ, 360^\circ$$
Now, $$\sin\theta+1=0$$
$$\sin\theta=-1\iff \theta=2n(180^\circ)-90^\circ$$
But for given interval $[0^\circ, 360^\circ]$, substituting $n=1$ we get
$$\theta= 270^\circ$$
Hence, we have $$\color{red}{\theta}=\left\{\color{blue}{0^\circ, 180^\circ, 270^\circ, 360^\circ} \right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\frac{\cos x-\cos2x}{1-\cos x}$? I want to find $$\int\dfrac{\cos x-\cos2x}{1-\cos x}\ dx$$
I have tried solving the question using substitution. how do I solve it?
| Notice, we have $$\int\frac{\cos x-\cos 2x}{1-\cos x}dx$$
$$=\int\frac{\cos x-2\cos^2x+1}{1-\cos x}dx$$
$$=\int\frac{\cos x-2\cos^2x+1}{1-\cos x}dx$$
$$=\int\frac{\cos x-\cos^2x+1-\cos^2 x}{1-\cos x}dx$$
$$=\int\frac{\cos x(1-\cos x)+(1-\cos^2 x)}{1-\cos x}dx$$
$$=\int\frac{\cos x(1-\cos x)}{1-\cos x}dx+\int\frac{(1+\cos x)(1-\cos x)}{1-\cos x}dx$$
$$=\int \cos x dx+\int (1+\cos x)dx$$
$$= \sin x +x+\sin x+C$$
$$= 2\sin x+x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
| Let me try. You have: $$\cos x + \sin x \tan (\frac{x}{2}) = 1 - 2\sin^2 (\frac{x}{2}) + 2 \sin (\frac{x}{2}) \cos (\frac{x}{2}) \tan (\frac{x}{2}) = 1 -2\sin^2 (\frac{x}{2}) + 2\sin^2 (\frac{x}{2}) = 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 4
} |
I just need your approval Verify by susbtitution if the given functions are a solution to the next differential equations.
a)
$$x^2y''+xy'-y=\ln x \quad ,\quad y_p=x^{-1}-\ln x $$
simplifying:
$$ x^2y''+xy'-y=\ln x $$
$$ x^2y''=-xy'+\ln x+y $$
$$y''=\frac{-xy'+\ln x+y}{x^2} $$
$$y_p=x^{-1}-\ln x $$
$$y_p'= -\frac{1}{(x)^2}-\frac{1}{x} $$
$$y_p'= \frac{-x-1}{x^2} $$
$$y_p''=\frac{x^2\frac{d}{dx}(-x-1)-\frac{x}{dx}(x^2).(-x-1)}{(x^2)^2} $$
$$y_p''=\frac{(x^2)(-1)-(2x)(-x-1)}{x^4} $$
$$ y_p''=\frac{(-x^2)-(-2x^2-2x)}{x^4} $$
$$y_p''=\frac{x^2+2x}{x^4} $$
$$y_p''=\frac{x+2}{x^3} $$
replacing:
$$y''=\frac{-x(\frac{-x-1}{x^2})+\ln x+(x^{-1}-\ln x)}{x^2} $$
$$y''=\frac{\frac{x+1}{x}+\frac{1}{x}}{x^2} $$
$$y''=\frac{\frac{x+2}{x}}{x^2} $$
$$y''=\frac{x+2}{x^3} $$
The function is a solution for the diffential equation
b)
$$y'+2xy^2=0 \quad , \quad y_p=\frac{1}{1+x^2} $$
simplifying:
$$ y'=-2xy^2 $$
$$y_p'= -\frac{2x}{(x^2+1)^2} $$
replacing:
$$y'=-2x(\frac{1}{1+x^2})^2 $$
$$y'=-\frac{2x}{(x^2+1)^2} $$
The function is a solution for the diffential equation
c)
$$x^2y''-xy'+2y=0 \quad , \quad y_p=x\cos(\ln x)$$ simplifying:
$$y''=\frac{xy'-2y}{x^2} $$
$$y_p'=(\frac{d}{dx}(x))(\cos(\ln x))+x(\frac{d}{dx}(\cos(\ln x))) $$
$$y_p'=(1)(\cos(\ln x))+(x)(-\frac{\sin(\ln(x))}{x})$$
$$y_p'=\cos(\ln x)-\sin(\ln x) $$
$$y_p''=(\frac{d}{dx})\cos(\ln x)-(\frac{d}{dx})\sin(\ln x) $$
$$y_p''= -\sin(\ln x)-\cos(ln) $$
$$y_p''= -\sin(\ln x)((\frac{d}{dx})(\ln x))-\cos(\ln x)((\frac{d}{dx})(\ln x)) $$
$$y_p''= -\frac{1}{x}\sin(\ln x)-\frac{1}{x}\cos(\ln x) $$
$$y_p''= -\frac{\sin(\ln x)}{x}-\frac{\cos(\ln x)}{x} $$
$$y_p''= -\frac{\sin(\ln x)+\cos(\ln x)}{x} $$
replacing:
$$y''=\frac{x(\cos(\ln x)-\sin(\ln x))-2(x\cos(\ln x))}{x^2} $$
$$y''=\frac{x\cos(\ln x)-x\sin(\ln x)-2x\cos(\ln x)}{x^2} $$
$$y''=\frac{-x\sin(\ln x)-x\cos(\ln x)}{x^2} $$
$$y''=\frac{-\sin(\ln x)-\cos(\ln x)}{x} $$
$$y''= -\frac{\sin(\ln x)+\cos(\ln x)}{x} $$
The function is a solution for the diffential equation
| You first idea is correct,
but you could do it simpler without first simplification.
Just "plug" it in...
$$
\begin{array}{ccc}
x^2 y'' & + & x y' & - & y & = & \ln(x)\\\\
\hline\\
\displaystyle x^2 \left( \frac{2}{x^3} + \frac{1}{x^2} \right) &+&
\displaystyle x \left( - \frac{1}{x^2} - \frac{1}{x} \right) &-&
\displaystyle \left( \frac{1}{x} - \ln(x) \right) &=& \ln(x)\\\\
\displaystyle \color{red}{\frac{2}{x}} + \color{blue}{1} &+&
\displaystyle \color{red}{-\frac{1}{x}} - \color{blue}{1} &+&
\displaystyle \color{red}{-\frac{1}{x}} + \color{green}{\ln(x)} &=&
\color{green}{\ln(x)}
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the least positive residue of $10^{515}\pmod 7$. I tried it, but being a big number unable to calculate it.
| All answers here use Fermat's little theorem, which is cool, but there is another simpler way (simpler if you don't know Fermat's theorem, otherwise it's more complicated).
$$10^0\equiv 1\mod 7\\
10^1 \equiv 3\mod 7\\
10^2\equiv 10\cdot 10 \equiv 3\cdot 3\equiv 9\equiv 2\mod 7\\
10^3 \equiv 10\cdot 10^2\equiv 3\cdot 2\equiv 6\mod 7\\
10^4\equiv 10\cdot 10^3\equiv 3\cdot 6\equiv 18\equiv 4\mod7\\
10^5\equiv 10\cdot 10^4\equiv 3\cdot 4 \equiv 12\equiv 5\mod 7\\
10^6\equiv 10\cdot 10^5\equiv 3\cdot 5\equiv 15\equiv 1\mod 7$$
Therefore, you know that taking $10^n\mod 7$ will result in a sequence which has repeating values: $1,3,2,6,4,5$, i.e. every sixth (so the zeroth, sixth, twelfth,...) number is $1$. Since $510=6\cdot 85$, you know that the $510$-th number in the sequence will again be $1$, then $511$-th $3$ and so on, so the $515$-th will be $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $f: \Bbb{N} \to \Bbb{N}$ is strictly increasing and $f(f(n))=3n$, find $f(2001)$. I have this question which seems a little harder than I thought. It has been about an hour for me hitting aimless thoughts on this one. I can really use a hint here if some one knows how to tackle it.
Let $f: \Bbb{N} \to \Bbb{N}$ such that $f$ is strictly increasing and $f(f(n))=3n$ for all $n \in \Bbb{N}$. Find $f(2001)$.
| Part A: $f(1)=2$.
First $f(1)$ cannot be $1$. Otherwise $3=f(f(1)) = f(1) = 1$.
So $f(1)$ is at least $2$. But $3 = f(f(1)) > f(2)$ because $f$ is increasing and $f(1)>2$.
Then $f(2)$ can only be $2$ or $1$. But $f(1)$ is at least $2$ and $f(1)<f(2)$.
This is a contradiction. So $f(1)$ must be $2$.
Part B
Since $2 = f(1)$, then $f(2) = f(f(1)) = 3\cdot1 = 3$. Similarly $f(3) = f(f(2)) = 6$ and
$f(6) = f(f(3)) = 9$.
Since f is strictly increasing, then $f(4)$ and $f(5)$ has to be $7$ and $8$.
Part C
Claim: $f(3^n) = 2\cdot 3^n $
Why? Let $f(n) = x$. Then $f(f(n)) = f(x)$.
So $3n = f(x)$. And $f(3n) = f(f(x)) = 3x = 3f(n)$. So iteration follows:
$$f(3^n) = 3f(3^{n-1}) = \ldots = 3^n\cdot f(1) = 2\cdot3^n$$
From this result,
$$f(2\cdot3^n) = f(f(3^n)) = 3\cdot3^n$$
Now for $x\in[3^n, 2\cdot3^n]$, $f(x)\in[2\cdot3^n, 3\cdot3^n]$. Both $x$ and
$f(x)$ have exactly $3^n$ values. Note that $f(x)$ has to be positive integers
and strictly increasing. Thus if $x$ increases by $1$, $f(x)$ has to increase by
$1$ too. So
$$ f(3^n + m) = 2\cdot3^n + m ; m \in [0,3^n]$$
Part D
Claim:$ f(2\cdot3^n + m) = 3(3^n + m) ; m \in [0,3^n]$.
Why? From Part C, $f(2\cdot3^n) = 3\cdot3^n$ and $f(3\cdot3^n) = 2\cdot3^{n+1}$
For any $x\in[2\cdot3^n, 3\cdot3^n]$, based on result of C, there exists a z such that $x = f(z) = 2\cdot 3^n + m$ where $z = 3^n + m$, and $m\in[0,3^n]$. Hence, $f(x) = f(f(z)) = 3z$, i.e.
$$ f(2\cdot 3^n + m) = 3\cdot(3^n + m) ; m \in [0,3^n]$$
Part E
$f(2001) = f(2\cdot3^6 + 543) = 3(3^6 + 543) = 3\cdot(729+543) = 3816$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
$z=100^2-x^2$. Then, how many values of $x,z$ are divisible by $6$?
My approach:
For $x=1$, $z$ is not divisible by $6$.
For $x=2$, $z$ is divisible by $6$.
For $x=3$, $z$ is not divisible by $6$.
For $x=4$, $z$ is divisible by $6$.
For $x=5$, $z$ is not divisible by $6$.
For $x=6$, $z$ is not divisible by $6$.
For $x=7$, $z$ is divisible by $6$.
For $x=8$, $z$ is divisible by $6$.
I could not identify the pattern in these questions. Also can this problem be solved with a better approach?
| HINT:
$$z=100^2-x^2=(100-x)(100+x)$$
As $(100-x)+(100+x)=200,100\pm x$ have same parity and
if one is divisible by $3,$the other is not
So if $2|z,100\pm x$ must be even
If $3|z,3|(100-x)(100+x)\implies$
either $3|(100-x)\iff x\equiv1\pmod3$
or $3|(100+x)\iff x\equiv-1\pmod3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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How many terms required in $e =\sum^∞_{k=0}{1\over k!}$ to give $e$ with an error of at most ${6\over 10}$ unit in the $20$th decimal place? How many terms are required in the series $e =\sum^∞_{k=0}{1\over k!}$ to give $e$ with an error of at most ${6\over 10}$ unit in the $20$th decimal place?
Here is what I have:
$$e\approx \sum_{k=0}^n {1\over k!}$$ where the remainder term is $$R=\sum_{k=n+1}^\infty {1\over k!}={1\over (n+1)!} \left[1+{(n+1)!\over (n+2)!}+{(n+1)!\over (n+3)!}+\cdots\right] \leq {1\over(n + 1)!} \left[1 + {1\over 2} + {1\over 4}+{1\over 8} + \cdots \right] = {2\over (n + 1)!}.$$ I'm not sure how to do this and give the error of at most $6\over 10$ unit in the $20$th decimal place. Any solutions or hints are greatly appreciated.
| Every tail of this series is bounded above by a geometric series. For example:
\begin{align}
& \frac 1 {0!} + \frac 1 {1!} + \cdots + \frac 1 {6!} + \overbrace{\frac 1 {7!} + \frac 1 {8!} + \frac 1 {9!} + \frac 1 {10!} + \cdots} \\[10pt]
\le {} & \frac 1 {0!} + \frac 1 {1!} + \cdots + \frac 1 {6!} + \overbrace{\frac 1 {7!} + \frac 1 {7!\cdot 8} + \frac 1 {7!\cdot 8^2} + \frac 1 {7!\cdot 8^3} + \cdots} \\[10pt]
= {} & \frac 1 {0!} + \frac 1 {1!} + \cdots + \frac 1 {6!} + \frac{1/7!}{1 - 1/8} \\[10pt]
= {} & \frac{1957}{720} + \frac 1 {630}.
\end{align}
So the problem is: how many terms are needed to make the upper bound in place of $1/630$ as small as what you need?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Algebra Problem: Division Can someone help me with a problem involving the expression $$\frac{(2x^3-3x^2+b)}{(4-x^2)}?$$
The question is to find which values $b$ can be to simplify the expression, but I do not know how to begin.
Thanks for any help you can give me.
| $4-x^2=-(x-2)(x+2)$, so you are looking for values of $b$ that make $2x^3-3x^2+b$ divisible by either $(x-2)$ or $(x+2)$.
So if you substitute $x=2$ into $2x^3-3x^2+b$ it will be 0, and the same for $x=-2$. This give you two simple equations in $b$, which gives you two values for $b$.
For $x=2$, you get $b=-4$, so you would expect $2x^3-3x^2-4$ to be divisible by $x-2$. You can check this by doing the division:
$2x^3-3x^2-4=(x-2)(2x^2+x+2)$.
So the expression simplifies to
$-\frac{2x^2+x+2}{x+2}$
Try this with $2x^3-3x^2+28$ and $(x+2)$, and you'll see that it works out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"Trig Substitutions", I tried half- angle and trig indentity in this one, but doesn't work I´m really lost in this one.
$\int \sin^3 (2x) \cos^2 (2x) dx$
I know that the answer is:
$\frac{1}{10}cos^5(2x)-\frac{1}{6}cos^3(2x) + c$
Please help
| $$
I = \int sin^3 (2x) cos^2(2x) dx
$$
Substituting
$$
t = 2x
\\dt = 2 dx
\\dx = \frac{dt}{2}
$$
So,
$$
I = \frac{1}{2}\int sin^3(t) cos^2(t) dt
$$
I'm now going to factor out
$$
sin^3 t = sin(t) (1 - cos^2 t)
$$
Because the factor of $sin(t)$ will simplify the integral.
$$
I = \frac{1}{2}\int sin(t)\; (1 - cos^2 (t)) \;cos^2(t) dt
\\I = \frac{1}{2}\int sin(t) cos^2 (t) dt - \frac{1}{2}\int sin(t) cos^4(t) dt
\\I = \frac{1}{2}\int cos^2(t) (-dcos(t)) - \frac{1}{2}\int cos^4(t) (-dcos(t))
\\I = -\frac{1}{2}\int cos^2(t) (dcos(t)) + \frac{1}{2}\int cos^4(t) (-dcos(t))
\\I = -\frac{1}{6} cos^3(t) + c1 + \frac{1}{10} cos^5(t) + c2
\\I = \frac{1}{10} cos^5(2x) + -\frac{1}{6} cos^3(2x) + c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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eigenvalues of cycle graph and its complement graph I am trying to find the eigenvalue of cycle graph and its complement. How to simplify.Suppose $\omega^{1}+\omega^{n-1}=2\cos (2\pi/n) $, then, $\omega^{\frac{n-1}{2}}+\omega^{\frac{n+1}{2}}=\ ?$ Is it true that $\omega^{1}+\omega^{n-1}=\omega^{\frac{n-1}{2}}+\omega^{\frac{n+1}{2}},$ where $\omega=e^{\frac{2\pi i}{n}}$?
Similarly, what is the value of $\omega^{0}+0+\omega^{2}+\omega^{3}+\cdots+\omega^{n-2}+0$. Both the adjacency matrices $C_n$ and its complement $\bar{C_{n}}$ gives circulant matrix.
| We have $\omega$ is a primitive $n$th root of unity, $\omega^{k}$ and $\omega^{n-k}$ are conjugates. This means that $(1/2)(\omega^{k}+\omega^{n-k}) = \mathrm{Re}(\omega^{k}) = \cos{\frac{2\pi k}{n}}$ (and so $\omega^{\frac{n-1}{2}} + \omega^{\frac{n+1}{2}} = 2 \cos{\frac{(n-1)\pi}{n}}$).
Also, $\omega$ is a root of
$$z^n -1 = (z-1)(z^{n-1} + \ldots + z+1),$$ so
$$\omega^{n-1}+ \ldots + \omega+1 = 0,$$ therefore $$1 + 0 + \omega^{2} + \ldots + \omega^{n-2} + 0 = 0-\omega - \omega^{n-1} = -2\cos{\frac{2\pi}{n}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1418236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$? How to prove:
$$\sum _{k=1}^{\infty} \frac{k-1}{2 k (1+k) (1+2 k)}=\log_e 8-2$$
Is it possible to convert it into a finite integral?
| We can evaluate the series using partial fraction expansion. To that end, we have
$$\begin{align}
\sum_{k=1}^{\infty}\frac{k-1}{2k(k+1)(2k+1)}&=\frac12\sum_{k=1}^{\infty}\left(\frac{6}{2k+1}-\frac{3}{k+1}\right)+\frac12\sum_{k=1}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k}\right)\\\\
&=-\frac12+\frac32\sum_{k=1}^{\infty}\left(\frac{1}{k+1/2}-\frac{1}{k+1}\right)\end{align} \tag 1$$
Now, we recall from the definition of the digamma function that
$$\begin{align}
\sum_{k=1}^{\infty}\left(\frac{1}{k+1/2}-\frac{1}{k+1}\right)&=-1-\gamma-\psi^0(1/2)\\\\
&=2\log 2-1 \tag 2
\end{align}$$
where we used $\psi^0(1/2)=-\gamma-2\log 2$ to arrive at the right-hand side of $(2)$.
Finally, substituting $(2)$ into $(1)$ yields
$$\begin{align}
\sum_{k=1}^{\infty}\frac{k-1}{2k(k+1)(2k+1)}&=-\frac12+\frac32\left(2\log 2-1\right)\\\\
&=3\log 2-2
\end{align}$$
which was to be shown!
NOTE:
Here we show another way to evaluate the sum $S$ given in $(2)$ by
$$S=\sum_{k=1}^{\infty}\left(\frac{1}{k+1/2}-\frac{1}{k+1}\right)$$
To that end, we write
$$\begin{align}
S&=\sum_{k=1}^{\infty}\left(\int_0^1 (x^{k-1/2}-x^k)\,dx\right)\\\\
&=\int_0^1\sum_{k=1}^{\infty}\left(x^{k-1/2}-x^k\right)\,dx\\\\
&=\int_0^1\frac{x^{1/2}-x}{1-x}\,dx\\\\
&=\left.\left(x-2x^{1/2}+2\log(1+x^{1/2})\right)\right|_0^1\\\\
&=2\log(2)-1
\end{align}$$
as expected!
We remark that we tacitly used the Dominated Convergence Theorem to justify the interchange of the order of integration and summation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Proof by induction for "sum-of" Prove that for all $n \ge 1$:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$
What I have done currently:
Proved that theorem holds for the base case where n=1.
Then:
Assume that $P(n)$ is true. Now to prove that $P(n+1)$ is true:
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n}{n+1} + n+1$$
So:
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n+1}{n+2}$$
However, how do I proceed from here?
| This step is wrong:
$$\sum_{k=1}^{n+1} 1/k(k+1) = n/(n+1) + n+1$$
it should read:
$$\sum_{k=1}^{n+1} \dfrac{1}{k(k+1)} = \dfrac{n}{n+1} + \dfrac{1}{(n+1)(n+2)}$$
so we have:
$$\sum_{k=1}^{n+1} \dfrac{1}{k(k+1)} = \dfrac{n(n+2)+1}{(n+1)(n+2)}$$
$$= \dfrac{n^2+2n+1}{(n+1)(n+2)}$$
$$= \dfrac{(n+1)^2}{(n+1)(n+2)}$$
$$= \dfrac{n+1}{n+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1419749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $45^\frac {1-a-b}{2-2a}$ where $90^a=2$ and $90^b=5$ without using logarithm? Let $90^a=2$ and $90^b=5$, Evaluate
$45^\frac {1-a-b}{2-2a}$
I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.
| $$2=90^a=(2\cdot5\cdot3^2)^a\iff5^a3^{2a}=2^{1-a}$$
and similarly, $$5=90^b\iff5^{1-b}3^{-2b}=2^b$$
Equating the powers of $2,$
$$\implies(5^a3^{2a})^b=(5^{1-b}3^{-2b})^{1-a}\iff3^{2b}=5^{1-a-b}$$
$$45^{1-a-b}=(3^2\cdot5)^{1-a-b}=3^{2-2(a+b)}\cdot5^{1-a-b}=3^{2-2(a+b)}\cdot3^{2b}$$
$$\implies45^{1-a-b}=3^{2(1-a)}$$
$\implies$ one of the values of $$45^{\frac{1-a-b}{2(1-a)}}$$ is $3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluating $\int_{0}^{\frac{\pi}{2}} (\sin (x) +\cos (x))^4 - (\sin(x) - \cos (x))^4 dx$ I would like to evaluate $$\int_{0}^{\frac{\pi}{2}} (\sin (x) +\cos (x))^4 - (\sin(x) - \cos (x))^4 dx.$$ One approach I used was to split the integral into two pieces and evaluating each piece directly with the anti derivatives $$ \frac {1}{4} (\sin(x) - \cos (x))^4 $$ and $$ \frac {1}{4} (-\sin(x) - \cos (x))^4, $$ respectively. This gives me 0. However, one can use trig identities and manipulate the integrand so that the final integral is $$ 4\int_{0}^{ \frac{\pi}{2} }2\sin (x) \cos (x) dx.$$ This gives me 4. How is that I'm getting two answers?
| You didn't compute antiderivatives correctly. To check your work, simply differentiate your result:
$$\frac{\mathrm d}{\mathrm dx}\left[\frac {1}{4} (\sin(x) - \cos (x))^4 + C\right] = (\sin x + \cos x)(\sin x - \cos x)^3 \neq (\sin x + \cos x)^4$$
Please, observe that we used the chain rule to derive our result. The same argument can be used to show that the second antiderivative is wrong.
To evaluate the integral it's much easier to use the fact that
$$A^2 - B^2 = (A + B)(A - B),$$
which, in combination with the identity $\sin^2 x + \cos^2 x = 1$, yields
$$\begin{align}
I
&= \int_0^{\pi/2}\left((\sin x + \cos x)^2 + (\sin x - \cos x)^2\right)\left((\sin x + \cos x)^2 - (\sin x - \cos x)^2\right)\mathrm dx\\
&= \int_0^{\pi/2}8\sin x \cos x\,\mathrm dx\\
&= 4\int_0^{\pi/2}\sin 2x\,\mathrm dx\\
&= 4\text{.}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of the following series to n terms $\frac{1}{1\cdot3}+\frac{2^2}{3\cdot5}+\frac{3^2}{5\cdot7}+\dots$ Find the sum of the following series to n terms $$\frac{1}{1\cdot3}+\frac{2^2}{3\cdot5}+\frac{3^2}{5\cdot7}+\dots$$
My attempt:
$$T_{n}=\frac{n^2}{(2n-1)(2n+1)}$$
I am unable to represent to proceed further. Though I am sure that there will be some method of difference available to express the equation. Please explain the steps and comment on the technique to be used with such questions.
Thanks in advance !
| You can compute the partial fraction decomposition of $T_n$;
$$T_n = \frac18 \left( \frac{1}{2n-1} - \frac{1}{2n+1} + 2 \right)$$
Then, you can separate the sum into three sum :
$$\sum_{n=1}^N T_n = \frac18 \left( \sum_{n=1}^N \frac{1}{2n-1} - \sum_{n=1}^N\frac{1}{2n+1} + \sum_{n=1}^N2 \right)$$
$$ = \frac18 \left( \sum_{n=2}^N \frac{1}{2n-1} - \sum_{n=1}^{N-1}\frac{1}{2n+1} + \sum_{n=1}^N2 + 1 - \frac{1}{2N+1}\right)$$
$$ = \frac18 \left( \sum_{n=1}^N2 + 1 - \frac{1}{2N+1}\right)$$
$$ = \frac18 \left( 2N + 1 - \frac{1}{2N+1}\right)$$
$$ = \frac18 \left( \frac{4N^2 +4N }{2N+1}\right)$$
$$ = \frac{N^2 +N}{2(2N+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Use De Moivre's Theorem to express $\cos(4θ)$ in terms of sums and differences of powers of $\sin(θ)$ and $\cos(θ)$ So far I have gotten this far using the binomial theorem.
$(x + iy)^4 = x^4 + 4x^3y + 6i^2x^2y^2 + 4i^3xy^3 + i^4y^4$=
$x^4 - 6x^2y^2 + y^4 + i(4x^3y - 4xy^3)$
$i^2 = -1 $ let $x = cosθ$ and y = $sinθ$
$e^{i4θ} = cos4θ + isin4θ = (cosθ + i sinθ)^4 $
$cos4θ = cos^4θ - 6cos^2θsin^2θ + sin^4θ$
I am not quite sure where to go next from here? Have I expressed this correctly in terms of sinθ and cosθ?
Thanks for taking the time to review.
| ${(\cos{\theta}+i\sin{\theta})^4}\to$
${(c+is)^4}\to$ (for brevity)
$\begin{matrix}
1&4&6&4&1\\
c^4&c^3&c^2&c&1\\
1&is&-s^2&-is^3&s^4\\
\hline
c^4&4ic^3s&-6c^2s^2&-4ics^3&s^4
\end{matrix}$
The terms pertaining to $\cos(4\theta)$ are the real ones; the imaginaries, after dispensing with $i$, pertain to $\sin(4\theta)$ .
So, $\cos(4\theta)=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta$ .
And, taking advantage of the fundamental relation between sine and cosine, $\cos(4\theta)=8\cos^4-8\cos^2+1=8\sin^4\theta-8\sin^2\theta+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Computing $E(XY)$ for finding $Cov(X,Y)$ Consider tossing a cubic die once and let $n$ be the smallest number of dots that appear on top. Define two random variables $X$ and $Y$ such that:
*
*$X=1$ if $n \in \left \{1,2 \right \}$, $X=2$ if $n \in \left \{3,4 \right \}$ and $X=3$ if $n \in \left \{5,6 \right \}$ and
*$Y=0$ if $n \in \left \{3,6 \right \}$, $Y=1$ if $n \in \left \{1,4 \right \}$, and $Y=2$ if $n \in \left \{2,5 \right \}$
Find the $cov(X,Y)$.
The support of $X$ is $\left \{1,2,3 \right \}$ and the support of $Y$ is $\left \{0,1,2 \right \}$.
The marginal probabilities are $f_{x}(X=1)=\frac{1}{3}$, $f_{x}(X=2)=\frac{1}{3}$, and $f_{x}(X=3)=\frac{1}{3}$ and the same for $Y=\left \{ 0,1,2 \right \}$
I have found $E(Y)=\frac{1}{3} (0) + \frac{1}{3} (1) + \frac{1}{3} (2) = 1$ and $E(X)=2$.
The formula for $Cov(X,Y) = E[(X-E(X))(Y-E(Y))]$, so I have $$Cov(X,Y) = E[(X-2)(Y-1)]=E[XY - X - 2Y +2]=E[XY]-E[X]-2E[Y]+2= E[XY] -2-2+2= E[XY] - 2.$$
EDIT: I was stuck on the $E[XY]$ part but have an answer now.
| Tabulate:
$\boxed{\begin{array}{r|r}
n & 1& 2& 3& 4& 5& 6 \\ \hline
X & 1& 1& 2& 2& 3& 3 \\ \hline
Y & 1& 2& 0& 1& 2& 0 \\ \hline
(X{-}\mathsf E(X))(Y{-}\mathsf E(Y)) & 0 & -1 & 0 &0 &1 &-1
\end{array}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Simple limit of a sequence
Need to solve this very simple limit $$ \lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$
I know how to solve these limits: by using
$a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - very tedious, boring and tiring. I hope there is some artful and elegant solution. Thank you!
| You have $$f(x)=\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}=\sqrt[3]{3x^2}\left(\sqrt[3]{1+\frac{4}{3x}+\frac{1}{3x^2}}-\sqrt[3]{1+\frac{3}{x}+\frac{2}{3x^2}}\right)$$ Using Taylor expansion at order one of the cubic roots $\sqrt[3]{1+y}=1+\frac{y}{3}+o(y)$ at the neighborhood of $0$, you get: $$f(x)=\sqrt[3]{3x^2}\left(\frac{4}{9x}-\frac{1}{x}+o\left(\frac{1}{x}\right)\right)$$ hence $$\lim\limits_{x \to \infty} f(x)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Is there a closed-form of $\frac{\zeta (2)}{\pi ^2}+\frac{\zeta (4)}{\pi ^4}+\frac{\zeta (6)}{\pi ^6}+.....$ The value of $$\frac{\zeta (2)}{\pi ^2}-\frac{\zeta (4)}{\pi ^4}+\frac{\zeta (6)}{\pi ^6}-.....=\frac{1}{e^2-1}$$
Is there a closed-form of $$\frac{\zeta (2)}{\pi ^2}+\frac{\zeta (4)}{\pi ^4}+\frac{\zeta (6)}{\pi ^6}+.....$$
any help, thanks
| $$\begin{align}
\sum_{n=1}^{\infty}\frac{\zeta(2n)}{\pi^{2n}}
&=\sum_{n=1}^{\infty}\frac{1}{\pi^{2n}}\sum_{k=1}^{\infty}\frac{1}{k^{2n}}\tag{1}\\
&=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(\pi^2 k^2)^n}\tag{2}\\
&=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{(\pi^2 k^2)^n}\tag{3}\\
&=\sum_{k=1}^{\infty}\frac{1}{(\pi^2 k^2)-1}\tag{4}\\
&=\frac{1}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2-\frac1{\pi^2}}\tag{5}\\
&=\frac{1}{2\pi}\left[\pi-\pi\cot\left(\frac\pi\pi\right)\right]\tag{6}\\
&=\frac{1-\cot1}{2}\tag{7}
\end{align}$$
$$\sum_{n=1}^{\infty}\frac{\zeta(2n)}{\pi^{2n}}=\frac{1-\cot1}{2}$$
$\text{Explanation :}$
$\text{6}$ using
$$\sum_{k=1}^\infty\frac{1}{k^2-z^2}
=\frac{1}{2z^2}-\frac{\pi\cot(\pi z)}{2z}
=\frac{1}{2z}\left[\frac1z-\pi\cot(\pi z)\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Condition for quartic polynomial coefficients given at least one real root
Find the minimum possible value of $a^2+b^2$ where $a$, $b$ are two real numbers such that the polynomial
$$x^4+ax^3+bx^2+ax+1,$$
has at least one real root.
My attempt: Let p be a real root.
Therefore $p^4 + a(p^3) + b(p^2) + ap + 1 = 0$.
Divide both sides by $p^2$, noting that $p$ cannot be zero since $P(0) = 1$, we get
$$p^2 + ap + b + a/p + 1/(p^2) = 0.$$
I rearranged this into a quadratic; i.e.
$$(p+1/p)^2 + a(p+1/p) + (b-2) = 0.$$
If $p$ is real, $p+1/p$ is real, so discriminant is non-negative.
setting discriminant $\geq0$, I get:
$$a^2 - 4(b-2)\geq0,$$
$$a^2 \geq 4b-8,$$
$$a^2 + b^2 \geq b^2 + 4b - 8 = (b+2)^2 - 12 \geq -12$$
But this is useless because it is obvious $a^2 + b^2 \geq 0$.
I believe this is because I also need to use the fact that $|p+1/p| \geq 2$, however I am unsure how to use this inequality with the discriminant.
Thanks in advance
| Hint:
Put $t=x+\frac{1}{x}$ then:
$x^4+ax^3+bx^2+ax+1=0$ has a real root if and only if $t^2+at+(b-2)=0$ has a real root $t$ such that $|t| \ge 2$, which is equivalent to
\begin{equation}\tag{*}
\left[\begin{matrix}
\frac{-a-\sqrt{a^2-4(b-2)}}{2} \le -2 \\
\frac{-a+\sqrt{a^2-4(b-2)}}{2} \ge 2
\end{matrix}\right.
\end{equation}
Using this:
$$A \le \sqrt{B} \Longleftrightarrow
\left[\begin{matrix}
\left\{\begin{matrix}
B \ge 0 \\
A < 0
\end{matrix}\right. \\
\left\{\begin{matrix}
A \ge 0 \\
A^2 \le B
\end{matrix}\right.
\end{matrix}\right.
$$
After some operations we have $(*)$ is equivalent to
$$\left[\begin{matrix}
(1)\left\{\begin{matrix}
a \ge 4 \\
a^2-4(b-2) \ge 0
\end{matrix}\right. \\
(2)\left\{\begin{matrix}
a \le -4 \\
a^2-4(b-2) \ge 0
\end{matrix}\right. \\
(3)\left\{\begin{matrix}
a < 4 \\
2a -b \ge 2
\end{matrix}\right. \\
(4)\left\{\begin{matrix}
a > -4 \\
2a+b \le -2
\end{matrix}\right.
\end{matrix}\right.$$
For $(1)$ and $(2)$: $a^2+b^2 \ge 4^2+0 =16$.
For $(3)$: using $5(a^2+b^2) = (2a-b)^2+(a+2b)^2$ we have $a^2+b^2 \ge \frac{4}{5}$, attained when $a=\frac{4}{5},b=-\frac{2}{5}$.
For $(4)$: using $5(a^2+b^2) = (2a+b)^2+(a-2b)^2$ we have $a^2+b^2 \ge \frac{4}{5}$, attained when $a=-\frac{4}{5},b=-\frac{2}{5}$.
Conclusion: the minimum value of $a^2+b^2$ is $\frac{4}{5}$, attained when $(a,b)=\left(\pm\frac{4}{5},-\frac{2}{5}\right)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How many ways to show $\sum_{n=1}^{\infty}\ln \left|1-\frac{x^2}{n^2\pi^2}\right|$ is pointwise convergence? $\sum_{n=1}^{\infty}\ln \left|1-\frac{x^2}{n^2\pi^2}\right|$. where $x\not= k\pi, k\in \mathbb{Z}$ is pointwise convergence.
Assume $n$ is sufficiently large($\geq N$),for a fixed point $x_0$,$-\ln \left|1-\frac{x_0^2}{n^2\pi^2}\right|=\frac{x_0^2}{n^2\pi^2}+\frac{1}{2}\frac{x_0^4}{n^4\pi^4}+\cdots+\frac{1}{n}\left(\frac{x_0^2}{n^2\pi^2}\right)^n+\cdots <
\frac{x_0^2}{n^2\pi^2}+\left(\frac{x_0^2}{n^2\pi^2}\right)^2+\left(\frac{x_0^2}{n^2\pi^2}\right)^3+\cdots+\left(\frac{x_0^2}{n^2\pi^2}\right)^n+\cdots=\frac{x_0^2}{n^2\pi^2-x_0^2}$
then $\sum_{n\geq N}\ln \left|1-\frac{x^2}{n^2\pi^2}\right| \leq \sum_{n \geq N}\frac{x_0^2}{n^2\pi^2-x_0^2} \Rightarrow 0$
Is there any other way to show this? because my textbook omits this process.Maybe there is a straightforward way to solve this.
| Another way for math freaks: using the Euler formula for the sine
$$
\sum_{n=1}^\infty\ln\left|1-\frac{x^2}{n^2\pi^2}\right|=
\ln\prod_{k=1}^\infty \left|1-\frac{x^2}{n^2\pi^2}\right|=\ln\left|\frac{\sin x}{x}\right|.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ If the minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ when $x>0$ is $\frac{p}{q}$ where $p,q\in N$ then find the least value of $(p+q).$
I can find the minimum value of $\frac{4}{x}+2x$ using $AM- GM$ inequality but that too comes irrational and when i tried to find the minimum value of $\frac{3+x}{4x^2+1}$ using first derivative test,i am facing difficulty.Please help me in solving this problem.
| HINT: There is another approach using calculus
let $$y=\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$$
$$\frac{dy}{dx}=-\frac{4}{x^2}+2+\frac{(4x^2+1)(1)-(3+x)(8x)}{(4x^2+1)^2}$$
Now, for maxima or minima, put $$\frac{dy}{dx}=0$$
$$\frac{(4x^2+1)(1)-(3+x)(8x)}{(4x^2+1)^2}-\frac{4}{x^2}+2=0$$
Since, $x>0$ then find the positive roots of the equation for minima & proceed
| {
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How to find triangle vertices given midpoints? I have a task to find vertices if midpoints are given: $M1(2;1)$, $M2(5;3)$, $M3(3;-4)$. I know one way to solve it through making a system of equations with three variables.
My teacher says there is faster way by using the midline of a triangle, and I can`t find this way of solving it on the Internet. How can I do it?
| The matrix equation giving the midpoints from the vertices is
$$
\begin{bmatrix}
\frac12&\frac12&0\\
0&\frac12&\frac12\\
\frac12&0&\frac12
\end{bmatrix}
\begin{bmatrix}
a\vphantom{\frac12}\\b\vphantom{\frac12}\\c\vphantom{\frac12}
\end{bmatrix}
=
\begin{bmatrix}
\frac{a+b}2\\\frac{b+c}2\\\frac{c+a}2
\end{bmatrix}
$$
Inverting the equation above yields
$$
\begin{bmatrix}
1&-1&1\vphantom{\frac12}\\
1&1&-1\vphantom{\frac12}\\
-1&1&1\vphantom{\frac12}
\end{bmatrix}
\begin{bmatrix}
\frac{a+b}2\\\frac{b+c}2\\\frac{c+a}2
\end{bmatrix}
=
\begin{bmatrix}
a\vphantom{\frac12}\\b\vphantom{\frac12}\\c\vphantom{\frac12}
\end{bmatrix}
$$
The last equation simply says
$$
a=\frac{a+b}2+\frac{c+a}2-\frac{b+c}2\\
b=\frac{b+c}2+\frac{a+b}2-\frac{c+a}2\\
c=\frac{b+c}2+\frac{c+a}2-\frac{a+b}2
$$
This amounts to reflecting each midpoint across the midpoint of the line connecting the other two midpoints.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$?
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$
I tried this question.
$\frac{x^2+y^2}{x+y}=4\Rightarrow x+y-\frac{2xy}{x+y}=4\Rightarrow x+y=\frac{2xy}{x+y}+4$
$x-y=\sqrt{(\frac{2xy}{x+y}+4)^2-4xy}$, but I am not able to proceed. I am stuck here. Is my method wrong?
| $\frac{x^2+y^2}{x+y}=4$ is equivalent to $(x-2)^2+(y-2)^2 = 8$, hence $(x,y)$ lies on a circle centered at $(2,2)$ with radius $2\sqrt{2}$. The tangents at the points $(0,4)$ and $(4,0)$ are parallel to the $y=x$ line, so the right answer is $(C)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $\ x^2-19\lfloor x\rfloor+88=0 $ I have no clue on how to solve this. If you guys have, please show me your solution as well.
$$\ x^2-19\lfloor x\rfloor+88=0 $$
| $$\ x^2-19\lfloor x\rfloor+88=0 \tag{*}$$
Start by solving $x^2-19x+88=0$. This factors as $(x-8)(x-11)=0$ giving solutions $x\in\{8,11\}$. Since these are integers, they are solutions of (*) as well.
There are no more solutions to (*) for $8<x<9$ since 8 was an ordinary root and $x^2$ is increasing.
For $9\le x<10$, we have $\lfloor x \rfloor = 9$, so that (*) reduces to
$$x^2-19\cdot9+88=0 \implies x^2=171-88=83 \implies x=\sqrt{83}$$
For $10\le x<11$, we have $\lfloor x \rfloor = 10$, so that (*) reduces to
$$x^2-19\cdot10+88=0 \implies x^2=190-88=102 \implies x=\sqrt{102}$$
There are no more solutions to (*) for $11<x<12$ since 11 was an ordinary root and $x^2$ is increasing, no are there any for $x\ge12$.
So the solutions are:
$$x\in\{8,\sqrt{83},\sqrt{102},11\}$$
For all $x$, we have
$$\lfloor x \rfloor \le x < \lfloor x \rfloor + 1$$
So for $x>0$ and a fixed $\lfloor x \rfloor$, since $x^2$ is increasing in $x$, the expression in (*) is minimised when $x=\lfloor x \rfloor$, i.e. when $x$ is an integer. But then for the factorisation $(x-8)(x-11)$ we have
$$0\le x<8 \implies (x-8)(x-11)>0$$
as both factors are negative. So the LHS in (*) is at least this large, so is always positive for $0\le x<8$, so there are no solutions with $0\le x<8$.
It is trivial to see that there are no solutions for $x<0$, as all terms on the LHS in (*) are then positive.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$ $\int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}}$
I was given this question by my senior.I tried to solve it but could not reach the answer.
Let $I= \int\frac{x dx}{\sqrt{x^4+4x^3-6x^2+4x+1}} $
$I=\int\frac{dx}{\sqrt{x^2+4x-6+\frac{4}{x}+\frac{1}{x^2}}}$
Then after repeated attempts, i could not solve further.
I think this function is not integrable.Am i correct?If not,how should i move ahead.Please help.
| In case that there is a typo, as the comments suggest, and the function is:
$$\int\frac{x dx}{\sqrt{x^4+4x^3+6x^2+4x+1}}$$
Then the integral is fairly easy once noticing that:
$$x^4+4x^3+6x^2+4x+1=(x+1)^4$$
And thus, the integral can be simplified to:
$$\int\frac{x dx}{(x+1)^2}=\int\frac{(x+1) dx}{(x+1)^2}-\int\frac{1 dx}{(x+1)^2}=\ln(x+1)+\frac{1}{x+1}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}(1+\sqrt[3]{4-3x})}$$
Finally
$$\frac{3}{(1+\sqrt[3]{4-3x})}$$
But this evaluates to
$$\frac{3}{2}$$
When the answer should be
$$1$$
Where did I fail?
| Alternatively you may observe that, for any differentiable function $f$, you have
$$
\frac{f(x)-f(a)}{x-a} \to f'(a).
$$
Then use it with $a=1$ and
$$
f(x)=\sqrt[3]{4-3x},\quad \quad f'(x)=-\frac{1}{(4-3 x)^{2/3}}
$$ giving
$$
\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)=-f'(1)=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the limit as $n$ approaches infinity: $\lim_{n\to \infty} \sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{\frac{n}{2}}}$ $$\lim_{n\to \infty} \sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{\frac{n}{2}}}$$
I am taking calculus in university and this is the problem I have been given. I haven't even seen limits involving a variable in the exponent in the textbook, so I am really stuck.
I tried graphing and I can guess that the limit will probably be $0$. I've tried laws of exponents, limit laws, but nothing gives me a good answer.
Also, sorry about the formatting, but this is the best I could do - it's my first time on this website. The second part of the equation should also be under a square root, so very similar to the first square root, but with the second exponent at $\frac{n}{2}$ instead of $-n$.
Thank you so much for help solving this.
| Hint. You may write
$$
\begin{align}
\sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{0.5n}}&=\left(\sqrt{3^n + 3^{-n}} - \sqrt{3^n + 3^{0.5n}}\right)\dfrac{\sqrt{3^n + 3^{-n}} + \sqrt{3^n + 3^{0.5n}}}{\sqrt{3^n + 3^{-n}} + \sqrt{3^n + 3^{0.5n}}}\\\\
&=\dfrac{(3^n + 3^{-n})-(3^n + 3^{0.5n})}{\sqrt{3^n + 3^{-n}} + \sqrt{3^n + 3^{0.5n}}}\\\\
&=\dfrac{-3^{0.5n}+3^{-n}}{\sqrt{3^n}\sqrt{1 + 3^{-2n}} + \sqrt{3^n}\sqrt{1 + 3^{-0.5n}}}\\\\
&=\dfrac{3^{0.5n}\left(-1+3^{-1.5n}\right)}{\sqrt{3^n}\sqrt{1 + 3^{-2n}} + \sqrt{3^n}\sqrt{1 + 3^{-0.5n}}}\\\\
&=\dfrac{-1+ 3^{-1.5n}}{\sqrt{1 + 3^{-2n}} + \sqrt{1 + 3^{-0.5n}}}
\end{align}
$$ then it becomes easier to obtain your limit as $n \to +\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\lim_{a \to \infty} \frac{3a+2}{5a+4}= \frac35$ using definition of limit Prove the limit using definition of limit$$ \ \lim_{a \to \infty}\ \frac{3a+2}{5a+4}= \frac{3}{5} $$
Answer: Let $\varepsilon \ >0 $. We want to obtain the inequality $$\left|\frac{3a+2}{5a+4}- \frac{3}{5}\right|< {\varepsilon}$$
$$ \Rightarrow \left|\frac{3a+2}{5a+4} - \frac{3}{5}\right|\ =\left|\frac{5(3a+2)-3(5a+4)}{5(5a+4)}\right|\\= \left|\frac{-2}{5(5a+4)}\right|\le\frac{1}{a} $$
Therefore we choose $K \in N$ s.t $K> \frac{1}{\varepsilon} $
$$\Rightarrow\left|\frac{3a+2}{5a+4}- \frac{3}{5}\right| \le\frac{1}{a}\le\frac{1}{K} < \varepsilon $$ Is this correct?
| let $ζ>0$ be any arbitrary real number
consider,
$$\begin{align}
&|[(3a+2)/(5a+4)]-(3/5)| < ζ\\
\Leftrightarrow&|-2/5(5a+4)| < ζ\\
\Leftrightarrow& a > 1/5[(2/5ζ)-4]=∆
\end{align}$$
hence for every $ζ>0$ there exists $∆>0$ such that which satisfies the definition of convergence
hence $f(a)=[(3a+2)/(5a+4)]$ converges to $3/5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Three vertices of a triangle are (2,7),(4,-1),(0,Y).if the perimeter of the triangle is the least then express Y in the lowest rational form as (p/q). I don't know how to proceed in this question. What I know is that the perimeter is the sum of all the sides.
| Notice, the length of side with vertices $(2, 7)$ & $(4, -1)$, using distance formula, is $$=\sqrt{(2-4)^2+(7+1)^2}=\sqrt{68}$$
Similarly, the length of side with vertices $(2, 7)$ & $(0, y)$ is $$=\sqrt{(2-0)^2+(7-y)^2}=\sqrt{(y-7)^2+4}$$ the length of side with vertices $(4, -1)$ & $(0, y)$ is $$=\sqrt{(4-0)^2+(-1-y)^2}=\sqrt{(y+1)^2+16}$$
Hence, the perimeter $P$ is the sum of all three sides hence. we have $$P=\sqrt{68}+\sqrt{(y-7)^2+4}+\sqrt{(y+1)^2+16}$$
Differentiating w.r.t. $y$, we get $$\frac{dP}{dy}=\frac{y-7}{\sqrt{(y-7)^2+4}}+\frac{y+1}{\sqrt{(y+1)^2+16}}$$
$$\frac{d^2P}{dy^2}=\frac{4}{((y-7)^2+4)^{3/2}}+\frac{16}{((y-7)^2+4)^{3/2}}$$
It is clear that $\frac{d^2P}{dy^2}>0\ \ \forall\ \ y\in R$ Hence, the perimeter is minimum/least
Now, setting $\frac{dP}{dy}=0$ for minimum, we get $$\frac{y-7}{\sqrt{(y-7)^2+4}}+\frac{y+1}{\sqrt{(y+1)^2+16}}=0$$
$$(y+1)^2=4(y-7)^2$$ $$y+1=\pm 2(y-7)$$
$$\implies y=15, \ \ \frac{13}{3} $$
But, it is clear that at both the values $y=15$ & $y=\frac{13}{5}$, the perimeter $P$ of triangle is minimum/least.
But at $y=15$ the triangle does not exist hence $y=15$ is unacceptable.
Hence, we get $$\color{red}{y=\frac{13}{3}} $$
| {
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"url": "https://math.stackexchange.com/questions/1453278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complex numbers solving for all solutions
*
*Find all complex numbers $z$ such that $z + \frac{1}{z}$ is real
*Let $a \gt b \gt 0$ be integers. Find all the solutions in complex numbers to the equation $z^a = z^{-b}$
For the first one, I tried manipulating the equation $Re(z) = \frac{z + \bar{z}}{2}$ to equal $z + \frac{1}{z}$ but I'm not really getting anywhere.
Don't know where to start for the second.
| Question 1
Let $z=x+iy$ for real numbers $x,y$. Clearly, they may not be both $0$ given we are dividing by $z$. $$z + z^{-1} = (x+iy) + \left(\frac{x}{x^2+y^2} - \frac{iy}{x^2+y^2}\right).$$
In order for $z+z^{-1}$ to be real, its imaginary part must be $0$. Therefore,
$$y- \frac{y}{y^2+x^2}=\frac{y(x^2+y^2-1)}{x^2+y^2}=0.$$
So, $y=0$ or $x^2+y^2=1$. That is to say, the imaginary part of $z$ is $0$ or $z$ lies on the unit circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454551",
"timestamp": "2023-03-29T00:00:00",
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Recurrent integral formula $$ I_{m,n}= \int {\sin(x)^m}{\cos(x)^n} dx $$
Determine the recurrence formula for the given integral,where "$n$" and "$m$" are $2$ natural parameters.
EDIT: Can it be solved by using only definite integrals formulas or the integration by parts method?
| Use integration by parts on $I_{m,n}$ with:
$$\begin{array}{ll}
u=\sin^{m-1}x & dv=\sin x \cos^n x\,dx \\
du=(m-1)\sin^{m-2}x \cos x\,dx & v =\frac{-1}{n+1}\cos^{n+1}x
\end{array}$$
Then we have:
$$
I_{m,n} = \frac{-1}{n+1}\sin^{m-1}x\cos^{n+1}x + \frac{m-1}{n+1}\color{blue}{\int{\sin^{m-2}x\,\cos^{n+2}x\,dx}} \tag{1}$$
where the integral in blue is
$$\begin{align}
\int{\sin^{m-2}x\cos^{n+2}x\,dx}&=\int{\sin^{m-2}x(1-\sin^2x)\cos^{n}x\,dx} \\
&=\int{\sin^{m-2}x\,\cos^{n}x\,dx}-\int{\sin^{m}x\cos^{n}x\,dx} \\[1em]
&=I_{m-2,n}-I_{m,n}
\end{align}$$
The "trick" lies in replacing $\cos^2x$ by $1-\sin^2x$ in the integrand, so as to get the recursive relation.
So (1) reduces to
$$\begin{align}
I_{m,n} &= \frac{-1}{n+1}\sin^{m-1}x\,\cos^{n+1}x + \frac{m-1}{n+1}(I_{m-2,n}-I_{m,n}) \implies \\[1em]
\frac{m+n}{n+1}I_{m,n} &= \frac{-1}{n+1}\sin^{m-1}x\,\cos^{n+1}x + \frac{m-1}{n+1}I_{m-2,n} \implies \\[1em]
I_{m,n} &= \frac{-1}{m+n}\sin^{m-1}x\,\cos^{n+1}x + \frac{m-1}{m+n}I_{m-2,n} \tag{*}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Two numbers $a$ and $b$ are chosen at random from the set of first 30 natural numbers.Find the probability that $a^2-b^2$ is divisible by $3$ Two numbers $a$ and $b$ are chosen at random from the set of first 30 natural numbers.Find the probability that $a^2-b^2$ is divisible by $3$.
Total number of ways of choosing two numbers out of 1,2,3....30 is $\binom{30}{2}=435$.So total cases are 435.
But i could not count the favourable number of cases and hence the probability.Please help me.
| Note that $a^2-b^2$ is divisible by $3$ if and only if either (i) $a$ and $b$ are both divisible by $3$ or (ii) neither $a$ nor $b$ is divisible by $3$. This is because if $n$ is not divisible by $3$, then $n$ has remainder $1$ or $2$ on division by $3$. If $a$ and $b$ have the same remainder on division by $3$, then $3$ divides $a-b$. And if one has remainder $1$ and the other $2$, then $3$ divides $a+b$. Finally, $a^2-b^2=(a-b)(a+b)$.
There are $\binom{10}{2}$ choices of Type (i), and $\binom{20}{2}$ of Type (ii).
Alternately, we count the bad pairs, where one of the numbers is divisible by $3$ and the other is not. There are $(10)(20)$ bad pairs.
| {
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trigonometry equation $3\cos(x)^2 = \sin(x)^2$ I tried to solve this equation, but my solution is wrong and I don't understand why. the answer in the book is: $x = \pm60+180k$. my answer is: $x= \pm60+360k$.
please help :)
3cos(x)^2 = sin(x)^2
3cos(x)^2 = 1 - cos(x)^2
t = cos(x)^2
3t=1-t
4t=1
t=1/4
cos(x)^2 = 1/4
cos(x) = 1/2
cos(x) = cos(60)
x = +-60+ 360k
edited:
cos(x) = -1/5
cos(x) = cos(120)
x = +-120 + 360k
I still don't get the answer in the book
| \begin{align*}
3\cos^2 x & = \sin^2x\\
3 & = \frac{\sin^2x}{\cos^2x}\\
\tan^2x & = 3\\
\tan x & = \sqrt{3}\\
\tan x & = \tan 60
\end{align*}
From there, you can use the tangent equation:
$$x= \pm 60+180k$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Where am I going wrong while trying to solve this logarithmic equation? $$\log _{ 0.2 }{ x } +\log _{ \sqrt { 5 } }{ x } =\log _{ 25 }{ x } +1$$
Steps I took:
$$\log _{ \frac { 1 }{ 5 } }{ x } +\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ \log _{ \frac { 1 }{ 5 } }{ \sqrt { 5 } } } =\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ \log _{ \frac { 1 }{ 5 } }{ 25 } } +1$$
$$\log _{ \frac { 1 }{ 5 } }{ x } +-\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ 2 } =-\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ 2 } +1$$
I can keep going, but this doesn't seem to lead me to the correct answer. Where did I go wrong?
| at line 3: $\log _{ \frac { 1 }{ 5 } }{ \sqrt { 5 } }$ doesn't equal -2, but $-\frac {1}{2}$, so the third line need to be:
$\log _{ \frac { 1 }{ 5 } }{ x } +-2\log _{ \frac { 1 }{ 5 } }{ x } =-\frac { \log _{ \frac { 1 }{ 5 } }{ x } }{ 2 } +1$
| {
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How to sketch the subset of a complex plane? The question asks to sketch the subset of $\{z\ \epsilon\ C : |Z-1|+|Z+1|=4\}$
Here is my working:
$z=x+yi$
$|x+yi-1| + |x+yi+1|=4$
$\sqrt{ {(x-1)}^2 + y^2} + \sqrt{{(x+1)}^2+y^2}=4$
${ {(x-1)}^2 + y^2} + {{(x+1)}^2+y^2}=16$
$x^2 - 2x+1+y^2+x^2+2x+1+y^2=16$
$2x^2+2y^2+2=16$
$x^2+y^2=7$
$(x-0)^2+(y-0)^2=\sqrt7$
=This is a circle with center $0$ and radius $\sqrt7$
My answer is different from the correct answer given: "This is an ellipse with foci at $-1$ and $1$ passing through $2$"
I have no idea how to get to this answer. Could someone please help me here?
| Or in another but similar parametrization to Yves':
\begin{align}
\sqrt{a+b}+\sqrt{a-b}&=4\\
\text{after squaring: }a+\sqrt{a^2-b^2}&=8\\
\text{rearrange and square: }a^2-b^2&=(8-a)^2=64-16a+a^2\\
16a-b^2&=64
\end{align}
where $a=x^2+y^2+1$ and $b=2x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$ The number of solutions of the equation $4\sin^2x+\tan^2x+\cot^2x+\csc^2x=6$ in $[0,2\pi]$
$(A)1\hspace{1cm}(B)2\hspace{1cm}(C)3\hspace{1cm}(D)4$
I simplified the expression to $4\sin^6x-12\sin^4x+9\sin^2x-2=0$
But i could not solve it further.Please help me.Thanks.
| Subtract $6$ from both sides of the original equation.
$$4\sin^2x-4+\csc^2x+\tan^2x-2+\cot^2x=0$$
$$(2\sin x-\csc x)^2+(\tan x-\cot x)^2=0$$
Since the square of a real number is non-negative, the only solution is if both squares are $0$. So we have $\tan x=\cot x$ and
$$2\sin x=\csc x$$
$$\sin^2x=\frac12$$
$$\left\{\begin{eqnarray}\sin x=\pm\frac{\sqrt2}2\\
\cos x=\pm\frac{\sqrt2}2\end{eqnarray}\right.$$
Now all it takes is to verify which of the $4$ solutions to this equation make tangent and cotangent equal.
| {
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$\int_{-1}^{+1}\sin(\sqrt{1-x^2})\cos(x)\,dx$ =? I am trying to find the value of the definite integral:
$$\int_{-1}^{+1}\sin\left(\sqrt{1-x^2}\right)\cos(x)\,dx$$
The answer from WolframAlpha is $1.20949$ but I can't solve it analytically.
| $\int_{-1}^1\sin\sqrt{1-x^2}\cos x~dx$
$=2\int_0^1\sin\sqrt{1-x^2}\cos x~dx$
$=2\int_0^\frac{\pi}{2}\sin\cos\theta\cos\sin\theta\cos\theta~d\theta$
$=\int_0^\frac{\pi}{2}\sin(\cos\theta+\sin\theta)\cos\theta~d\theta+\int_0^\frac{\pi}{2}\sin(\cos\theta-\sin\theta)\cos\theta~d\theta$
$=\int_0^\frac{\pi}{2}\sin\left(\sqrt2\cos\left(\theta-\dfrac{\pi}{4}\right)\right)\cos\theta~d\theta-\int_0^\frac{\pi}{2}\sin\left(\sqrt2\sin\left(\theta-\dfrac{\pi}{4}\right)\right)\cos\theta~d\theta$
$=\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\cos\theta)\cos\left(\theta+\dfrac{\pi}{4}\right)d\theta-\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\cos\left(\theta+\dfrac{\pi}{4}\right)d\theta$
$=\dfrac{1}{\sqrt2}\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\cos\theta)\cos\theta~d\theta-\dfrac{1}{\sqrt2}\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\cos\theta)\sin\theta~d\theta-\dfrac{1}{\sqrt2}\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\cos\theta~d\theta+\dfrac{1}{\sqrt2}\int_{-\frac{\pi}{4}}^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\sin\theta~d\theta$
$=\sqrt2\int_0^\frac{\pi}{4}\sin(\sqrt2\cos\theta)\cos\theta~d\theta+\sqrt2\int_0^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\sin\theta~d\theta$
$=\sqrt2\int_\frac{\pi}{2}^\frac{\pi}{4}\sin\left(\sqrt2\cos\left(\dfrac{\pi}{2}-\theta\right)\right)\cos\left(\dfrac{\pi}{2}-\theta\right)d\left(\dfrac{\pi}{2}-\theta\right)+\sqrt2\int_0^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\sin\theta~d\theta$
$=\sqrt2\int_\frac{\pi}{4}^\frac{\pi}{2}\sin(\sqrt2\sin\theta)\sin\theta~d\theta+\sqrt2\int_0^\frac{\pi}{4}\sin(\sqrt2\sin\theta)\sin\theta~d\theta$
$=\sqrt2\int_0^\frac{\pi}{2}\sin(\sqrt2\sin\theta)\sin\theta~d\theta$
$=\dfrac{1}{\sqrt2}\int_0^\frac{\pi}{2}\cos(\theta-\sqrt2\sin\theta)~d\theta-\dfrac{1}{\sqrt2}\int_0^\frac{\pi}{2}\cos(\theta+\sqrt2\sin\theta)~d\theta$
$=\dfrac{1}{\sqrt2}\int_0^\frac{\pi}{2}\cos(\theta-\sqrt2\sin\theta)~d\theta-\dfrac{1}{\sqrt2}\int_\pi^\frac{\pi}{2}\cos(\pi-\theta+\sqrt2\sin(\pi-\theta))~d(\pi-\theta)$
$=\dfrac{1}{\sqrt2}\int_0^\frac{\pi}{2}\cos(\theta-\sqrt2\sin\theta)~d\theta+\dfrac{1}{\sqrt2}\int_\frac{\pi}{2}^\pi\cos(\theta-\sqrt2\sin\theta)~d\theta$
$=\dfrac{1}{\sqrt2}\int_0^\pi\cos(\theta-\sqrt2\sin\theta)~d\theta$
$=\dfrac{\pi J_1(\sqrt2)}{\sqrt2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A problem relating to triangles and progressions If $D,E,F$ are the points of contact of the inscribed circle with the sides $BC, CA, AB$ of a triangle $\triangle ABC$, we need to show that if the squares of $AD, BE, CF$ are in arithmetic progression, then the sides of a triangle are in harmonic progression.
I tried using cosine law:
$$AD\times AD = c^2 +(s-b)^2 -2c(s-b)\cos B\\
BE\times BE = a^2 + (s-c)^2 -2a(s-c)\cos C\\
CF\times CF = b^2 + (s-a)^2 -2b(s-a)\cos A\\$$
But I couldn't simplify further to prove that the sides of a triangle are in harmonic progression.
| $$AD^2=c^2+(s-b)^2-2c(s-b)cosB=(c-(s-b))^2+2c(s-b)(1-cosB)=(b+c-s)^2+4c(s-b)sin^2(\frac{B}{2})$$ But $$sin^2\left(\frac{B}{2}\right)=\frac{(s-a)(s-c)}{ac}$$ So
$$AD^2=(b+c-s)^2+4c(s-b)\frac{(s-a)(s-c)}{ac}=(s-a)^2+4\frac{\Delta^2}{as}=(s-a)^2+\frac{4r\Delta}{a}$$ Similarly
$$BE^2=(s-b)^2+\frac{4r\Delta}{b}$$ and
$$CF^2=(s-c)^2+\frac{4r\Delta}{c}$$ Now since $AD^2$,$BE^2$ and $CF^2$ are in A.P we have
$$BE^2-AD^2=CF^2-BE^2$$ $\implies$
$$(s-b)^2-(s-a)^2+4r\Delta\left(\frac{1}{b}-\frac{1}{a}\right)=(s-c)^2-(s-b)^2+4r\Delta\left(\frac{1}{c}-\frac{1}{b}\right)$$ $\implies$
$$(a-b)c+4r\Delta\left(\frac{1}{b}-\frac{1}{a}\right)=(b-c)a+4r\Delta\left(\frac{1}{c}-\frac{1}{b}\right)$$ $\implies$
$$2ac-b(a+c)=4r\Delta\left(\frac{1}{a}+\frac{1}{c}-\frac{2}{b}\right)$$ from that
$$2ac-b(a+c)=4r\Delta\left(\frac{(a+c)b-2ac}{abc}\right)$$ which is possible only if
$$2ac=b(a+c)$$ so $a$,$b$ and $c$ are in H.P
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do you factorize quadratics when the coefficient of $x^2 \gt 1$? So I've figured out how to factor quadratics with just $x^2$, but now I'm kind of stuck again at this problem:
$2x^2-x-3$
Can anyone help me?
| For quadratic polynomials of the form $f(x)=ax^2+bx+c$ with $a,b,c$ real numbers, the roots are known to be given by the following:
The Quadratic Formula: The roots of $f(x)=ax^2+bx+c$ are
$$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$
and
$$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$
Note that when $b^2-4ac=0$, then the root is "repeated" and when $b^2-4ac<0$ then the roots are non-real complex numbers.
The polynomial then can be factored as $ax^2+bx+c = a(x-x_1)(x-x_2)$
For your specific example, $2x^2-x-3$, this is of the form $ax^2+bx+c$ with $a=2,~b=-1,~c=-3$.
Applying the formula above gives the two roots as:
$x_1=\frac{-(-1)+\sqrt{(-1)^2-4(2)(-3)}}{2(2)} = \frac{1+\sqrt{1+24}}{4}=\frac{1+5}{4}=\frac{3}{2}$
$x_2=\frac{-(-1)-\sqrt{(-1)^2-4(2)(-3)}}{2(2)} = \frac{1-\sqrt{1+24}}{4}=\frac{1-5}{4}=-1$
You have then that $2x^2-x-3=2(x-\frac{3}{2})(x-(-1))$, which can be rewritten as $=(2x-3)(x+1)$
note: in some settings we are interested only in factorizations which involve only integers. It is possible in this setting then that no such factorizations exist and we say the polynomial is irreducible
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that
$$a^4+b^4+c^4>abc(a+b+c)$$
My attempt:
I used the inequality A.M>G.M to get two inequalities
First inequality
$$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$
or
$$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new --
first inequality
$$\frac{a^4+b^4+c^4}{3abc} > \sqrt[3]{abc}$$
second inequality:
$$\frac{a+b+c}{3} > \sqrt[3]{abc}$$
I am seeing the numbers in the required equation variables here but am not able to manipulate these to get the inequality I want?? Please direct me on which step should I take after this??
| Here is a slightly more general approach to construct the right AM-GM. Note, for non-negative $x,y$ s.t. $x+y\le 1$,
$$x a^4+yb^4+(1-x-y)c^4\ge a^{4x}b^{4y}c^{4-4x-4y}$$
Clearly we are done if we have $4x=2,4y=1\implies 4-4x-4y=1$, so $x=\frac12, y=\frac14$ and what you need is to sum
$$\frac12a^4+\frac14b^4+\frac14c^4\ge a^2bc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
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