Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding all the sets of three real numbers that satisfy specific equations I am having trouble finding a set of three numbers of real numbers $(x, y, z)$ satisfying $x + y + z = xy + xz + yz = 3$. I have tried factoring the equations around but I'm not having any luck and I don't think I'm going on the right path.
| Given
$$a+b+c=3\tag1$$
$$ab+bc+ca=3\tag2$$
Hint:
$$(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)\tag3$$
Spoiler
From $(3)$ it can easily be shown that
$$a^2+b^2+c^2 = 3$$
Now
$$(a+b+c)^2 = 3^2$$
$$(a+b+c)^2 + a^2+b^2+c^2 - 4(ab+bc+ca) = 3^2 + 3 - 4*3$$
$$a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2 = 0$$
$$(a-b)^2+(b-c)^2+(c-a)^2=0$$
As $a,b,c$ are real numbers and squares of real numbers cannot be negative so
$a-b=b-c=c-a=0;a=b=c$
Replacing in $(3)$ we get
$$a+a+a=b+b+b=c+c+c=3$$
$$a=b=c=1$$
| {
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"url": "https://math.stackexchange.com/questions/336924",
"timestamp": "2023-03-29T00:00:00",
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Calculate:$\int \frac{1}{(x+1)^\frac{3}{4}(x+2)^{\frac{5}{4}}}\ dx$ Calculate following integration
$$\int \frac{1}{(x+1)^\frac{3}{4}(x+2)^{\frac{5}{4}}}\ dx$$
| $\dfrac{1}{(x+1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}=\dfrac{1}{(x+1)^{2}(\dfrac{x+2}{x+1})^{\frac{5}{4}}}=\dfrac{1}{(x+1)^{2}(1+\dfrac{1}{x+1})^{\frac{5}{4}}}$ So, make a $\dfrac{1}{x+1}$ substitution and we get the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Having trouble showing that these series are the same. $$\frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\tfrac{n(n+1)}{2}+1}\frac{(x-\pi/4)^n}{n!} $$
$$= \frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\tfrac{n(n-1)}{2}}\frac{(x-\pi/4)^{n+1}}{(n+1)!} + 1$$
*(added the +1) sorry didn't see this before, my answer guide is a poor quality photocopy
The second one is in the answer guide. The first one is my answer. The problem from the book is
Write the taylor series for $\sin x$ centered at $\frac{\pi}{4}$
my work:
$f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$f^{\prime}\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$f^{\prime\prime}\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
$f^{\prime\prime\prime}\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
$f^{\left(4\right)}\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$\sin x = \sum\limits_{n=0}^{\infty} \frac{f^{(n)}(\pi/4)}{n!}(x -\pi/4)^n$
how did the book get its answer (the second one listed at start of this question).
| Your two series I will call S1 and S2
$$n=m+1$$
$$S1 = \sum \limits_{n=0}^{\infty} (-1)^{\tfrac{n(n+1)}{2}+1}\frac{(x-\pi/4)^n}{n!}$$
$$=\sum \limits_{m=-1}^{\infty} (-1)^{\tfrac{(m+1)(m+2)}{2}+1}\frac{(x-\pi/4)^{(m+1)}}{(m+1)!}$$
$$=\sum \limits_{m=-1}^{\infty} (-1)^{\tfrac{(m-1)m}{2}}\frac{(x-\pi/4)^{(m+1)}}{(m+1)!}$$
note how this is very simular to S2
so,
$$S1 = S2 + (-1)^{\tfrac{(c-1)c}{2}}\frac{(x-\pi/4)^{(c+1)}}{(c+1)!}~where~m=-1$$
$$S1 = S2 - 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify : $\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a}$? Just out of curiosity, is $$\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a},a \gt0\quad?$$
Thanks
| $$\sqrt{\frac{a}{2}} + \sqrt{\frac{a}{2}} = 2\sqrt{\frac{a}{2}} = \frac{2}{\sqrt{2}}\sqrt{a}$$
| {
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Calculate the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\tan ^4 x}{\cos 2x} dx$ Calculate the integral
$$\int\limits_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}\dfrac{\tan ^4 x}{\cos 2x} dx$$
| I evaluate this integral via its Cauchy Principal Value. To wit, note that
$$\cos{2 x} = \frac{2}{\sec^2{x}}-1$$
so that the integral takes the form
$$\int_{\pi/3}^{\pi/6} dx \sec^2{x} \frac{\tan^4{x}}{1-\tan^2{x}}$$
We can then let $u=\tan{x}$ and substitute to get
$$\int_{1/\sqrt{3}}^{\sqrt{3}} du \frac{u^4}{1-u^2} = \int_{1/\sqrt{3}}^{\sqrt{3}} du \frac{1}{1-u^2} - \int_{1/\sqrt{3}}^{\sqrt{3}} du \: (1+u^2)$$
The first integral on the RHS blows up, but its Cauchy PV is zero by symmetry about the pole at $u=1$. The other integral is straightforward; the result is
$$\begin{align}PV \int_{\pi/3}^{\pi/6} dx \frac{\tan^4{x}}{\cos{2 x}} &= -\left (\sqrt{3}-\frac{1}{\sqrt{3}} \right) - \frac{1}{3}\left (3\sqrt{3}-\frac{1}{3\sqrt{3}} \right) \\ &= -\frac{44}{9 \sqrt{3}}\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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an odd perfect number cannot be a prime number or a product of two prime numbers or power of prime number. how to prove :
an odd perfect number cannot be a prime number or a product of two prime numbers or power of prime number.
| Let $\sigma$ denote the sum-of-divisors function. A positive integer $N$ is said to be perfect if $\frac{\sigma(N)}{N}=2$.
Let $p^\alpha$ be an odd prime power and suppose that $N=p^\alpha$ is an odd perfect number. Since $\sigma(p^\alpha)=1+p+p^2+\dots+p^\alpha=\frac{p^{\alpha+1}-1}{p-1}$, we can observe that
$$\sigma(p^\alpha) = \frac{p^{\alpha+1}-1}{p-1} < \frac{p^\alpha p}{p-1} \implies \frac{\sigma(p^\alpha)}{p^\alpha} < \frac{p}{p-1} \le \frac{3}{2} < 2$$
since $p \ge 3$, so $N$ must be deficient.
Now, let $p^\alpha$ and $q^\beta$ be two odd prime powers with $p<q$ and suppose that $N=p^\alpha q^\beta$ is an odd perfect number. Noting that $\sigma$ is multiplicative, we obtain
$$\sigma(p^\alpha q^\beta) = \frac{p^{\alpha+1}-1}{p-1} \cdot \frac{q^{\beta+1}-1}{q-1} < \frac{p^\alpha p}{p-1} \cdot \frac{q^\beta q}{q-1} \implies \frac{\sigma(p^\alpha q^\beta)}{p^\alpha q^\beta} < \frac{p}{p-1} \cdot \frac{q}{q-1} \le \frac{3}{2} \cdot \frac{5}{4} = \frac{15}{8} < 2$$
since $p \ge 3$ and $q \ge 5$, so again $N$ must be deficient.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating function as Rational Function Find the generating function for the sequence ${a_n}$ defined by:
$$a_n=4a_{n-1}-4a_{n-2}+{n\choose 2}2^n+1$$
for $n\leq 2$ and $a_0=a_1=1$. Write your answer as a rational function.
| To find the generating function, write the recurrence as
$$a_n-4 a_{n-1}+4 a_{n-2}=n(n-1)2^{n-1}+1$$
Multiply both sides by $x^n$ and sun from $n=2$ on:
$$\sum_{n=2}^{\infty} (a_n-4 a_{n-1}+4 a_{n-2}) x^n = \sum_{n=2}^{\infty} n(n-1)2^{n-1} x^n + \sum_{n=2}^{\infty} x^n$$
Define the generating function as
$$g(x) = \sum_{n=0}^{\infty} a_n x^n$$
We can manipulate the left hand side to produce $g$:
$$\sum_{n=0}^{\infty} a_n x^n - (a_0-a_1x ) - 4 x \left (\sum_{n=0}^{\infty} a_n x^n - a_0\right ) + 4 x^2 \sum_{n=0}^{\infty} a_n x^n \\= (1-4 x+4 x^2)g(x) - (1+x) + 4 x$$
On the RHS, we note that
$$\begin{align}\sum_{n=2}^{\infty} n(n-1)2^{n-1} x^n &= \frac{x^2}{2}\sum_{n=2}^{\infty} n(n-1)2^n x^{n-2} \\ &= \frac{x^2}{2} \frac{d^2}{dx^2} \frac{1}{1-2 x}\end{align}$$
Also
$$\sum_{n=2}^{\infty} x^n = \frac{1}{1-x}-(1+x)$$
Putting this all together:
$$(1-4 x+4 x^2)g(x) - (1+x) + 4 x = \frac{4 x^2}{(1-2 x)^3} + \frac{1}{1-x} - (1+x)$$
With some algebra, we get the rational function sought:
$$g(x) = \frac{1-10 x+44 x^2-84 x^3+80 x^4-32 x^5}{(1-x)(1-2 x)^3 (1-4 x+4 x^2)}$$
The first several terms in the series expansion for this function are
$$g(x) = 1+x+5 x^2+41 x^3+241 x^4+1121 x^5+4481 x^6+16129
x^7+53761 x^8+168961 x^9+506881
x^{10}+O\left(x^{11}\right)$$
I leave it to the reader to compare the coefficients to the terms generated in the original recurrence.
| {
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What does the square root of minus $i$ equal? Can you enter the rabbit hole recursively?
If the $ \sqrt{-1} = i $ then, what does $ \sqrt{-i} $ equal?
| Let us assume there exist a complex number $m$ such that $m * m = -i$, then $m$ will be the answer we are looking for $\sqrt{-i}$.
we can expand $m$ as $a + i.b$, then we have $(a + i.b)^2 = -i$; expanding we get
$a^2 - b^2 + i.2ab = -i$ $[i^2 = - 1] $; to make this true we must have
$a^2 - b^2 = 0$ --> (1)
and
$a * b = -1/2$ --> (2)
From (2) we have $b = -1/2a$ substituting in (1) we have
$a^2 - (-1/2a)^2 = 0$
rearranging we get the below
$a2 - (1/4a^2) = 0$
Multiplying both sides of equation with $a^2$ we get
$a^4 = 1/4$ and so $a=\pm1/\sqrt{2}$ we know $b=-1/2a$
So, when $a = 1/\sqrt{2}$ then $b = - 1/\sqrt{2}$
like wise when $a = -1/\sqrt{2}$ then $b = 1/\sqrt{2}$
Finally substituting the $a$ and $b$ values we get two roots for $\sqrt{-i}$
$$(-1/{\sqrt{2}}) + i.(1/\sqrt{2})$$ and $$(1/\sqrt{2}) - i.(1/\sqrt{2})$$
You'll see squaring any of the above solution you'll get $-i$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove: $\int_{0}^{\pi}\frac{x\tan x}{\sec x+\cos x}dx = \frac{\pi^{2}}{4}$ How to prove following?
$$\int_{0}^{\pi}\frac{x\tan x}{\sec x+\cos x}dx = \frac{\pi^{2}}{4}$$
| $$\dfrac{x \tan(x)}{\sec(x) + \cos(x)} = \dfrac{x\sin(x)}{\cos(x) \cdot \sec(x) + \cos^2(x)} = \dfrac{x \sin(x)}{1+\cos^2(x)} = \sum_{k=0}^{\infty} (-1)^kx \sin(x) \cos^{2k}(x)$$
Hence,
$$I = \int_0^{\pi} \dfrac{x \tan(x)}{\sec(x) + \cos(x)} dx = \sum_{k=0}^{\infty}(-1)^k \int_0^{\pi} x \sin(x) \cos^{2k}(x) dx$$
Let $I_k = \displaystyle \int_0^{\pi} x \sin(x) \cos^{2k}(x) dx$. Replacing $x$ by $\pi-x$, we get $$I_k = \int_{0}^{\pi} (\pi-x) \sin(x) \cos^{2k}(x)dx = \pi \int_0^{\pi} \sin(x) \cos^{2k}(x)dx - I_k$$
Hence,
$$2I_k = \pi \int_0^{\pi} \sin(x) \cos^{2k}(x)dx = \pi \cdot \dfrac2{2k+1} \implies I_k = \dfrac{\pi}{2k+1}$$
Hence,
$$I = \sum_{k=0}^{\infty}(-1)^k I_k = \pi \left(\sum_{k=0}^{\infty} \dfrac{(-1)^k}{2k+1}\right) = \pi \cdot \dfrac{\pi}4 = \dfrac{\pi^2}4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{AB}{A'B'}+\frac{BC}{B'C'}+\frac{CA}{C'A'} \geq 4 \left(\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}\right). $ Let be a circle inscribed in the triangle $\triangle ABC$ wiht the center $I$. The intersection of the circle with $AI$ is $A'$, with $BI$ is $B'$ and with $CI$ is $C'$.
Prove that:
$$\frac{AB}{A'B'}+\frac{BC}{B'C'}+\frac{CA}{C'A'} \geq 4 \left(\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}\right). $$
thanks. seems to be to hard for me.
| $AB=r\left(\cot\dfrac{A}{2}+\cot\dfrac{B}{2}\right)=r\dfrac{\sin \left(\dfrac{A+B}{2}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}=r\dfrac{\cos \dfrac{C}{2}}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}=r\dfrac{\sin \dfrac{π+C}{2}}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}},$
$A'B'=r\sqrt{2\left(1-\cos \left(\dfrac{π+C}{2}\right)\right)}=2r\sin \left(\dfrac{π+C}{4}\right)$
$\dfrac{AB}{A'B'}=\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}$ then the inequality become:
$\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}+\dfrac{\cos \left(\dfrac{π+B}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{C}{2}}+\dfrac{\cos \left(\dfrac{π+A}{4}\right)}{\sin \dfrac{C}{2}\sin \dfrac{B}{2}} \ge 4\left(\sin {\dfrac{A}{2}}+\sin {\dfrac{B}{2}}+\sin {\dfrac{C}{2}}\right) $ $\iff$
$\cos \left(\dfrac{π+C}{4}\right)\sin \dfrac{C}{2}+\cos \left(\dfrac{π+B}{4}\right)\sin \dfrac{B}{2}+\cos \left(\dfrac{π+A}{4}\right)\sin \dfrac{A}{2} \ge 4\sin {\dfrac{A}{2}}\sin {\dfrac{B}{2}}\sin {\dfrac{C}{2}}\left(\sin {\dfrac{A}{2}}+\sin {\dfrac{B}{2}}+\sin {\dfrac{C}{2}}\right)$
$x=\dfrac{A}{2},y=\dfrac{B}{2},z=\dfrac{C}{2} →$
$f=\cos {\dfrac{π+2z}{4}}\sin {z}+\cos {\dfrac{π+2y}{4}}\sin {y}+\cos {\dfrac{π+2x}{4}}\sin {x} - 4\sin {x}\sin {y}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)$
$g=x+y+z-\dfrac{π}{2}$ , $F=f-\lambda g$
$F_{x}=\cos {\dfrac{π+2x}{4}}\cos {x}-\dfrac{1}{2}\sin {\dfrac{π+2x}{4}}\sin {x}-4\cos {x}\sin {y}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)-4\sin {x}\sin {y}\sin {z}\cos {x})=\lambda$ <1>
$F_{y}=\cos {\dfrac{π+2y}{4}}\cos {y}-\dfrac{1}{2}\sin {\dfrac{π+2y}{4}}\sin {y}-4\cos {y}\sin {x}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)-4\sin {x}\sin {y}\sin {z}\cos {y})=\lambda$ <2>
$F_{z}=\cos {\dfrac{π+2z}{4}}\cos {z}-\dfrac{1}{2}\sin {\dfrac{π+2z}{4}}\sin {z}-4\cos {z}\sin {y}\sin {x}\left(\sin {x}+\sin {y}+\sin {z}\right)-4\sin {x}\sin {y}\sin {z}\cos {z})=\lambda$ <3>
<1>- <2> :
$ \sin {\dfrac{x-y}{4}}\left(\sin {\dfrac{π+2z}{8}}-3\sin {\dfrac{3z}{4}} \left(3-4\left(\sin {\dfrac{x-y}{4}}\right)^2 \right)+32\cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)-16\sin {x}\sin {y}\sin {z}\cos {\dfrac{x-y}{4}}\sin {\dfrac{x+y}{2}} \right)=0 $ $→ \sin {\dfrac{x-y}{4}}P_{1}\left(x,y,z\right)=0$ <4>
<2>-<3>: $\sin {\dfrac{y-z}{4}}P_{2}\left(x,y,z\right)=0$, $P_{2}\left(x,y,z\right)$ is similar as above.
$|x-y|<\dfrac{π}{2} → \cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}} > 0.65 $ and $ \sin x+\sin y+\sin z \ge \sin \left(x+y+z\right)=1 → 16\cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)> 9\sin {\dfrac{3z}{4}}$
it is trivial that:$16\cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)>16\sin {x}\sin {y}\sin {z}\cos {\dfrac{x-y}{4}}\sin {\dfrac{x+y}{2}}$
$→ P_{1}\left(x,y,z\right)>0$ \sin ce it is symmetric, we have $P_{2}\left(x,y,z\right)>0$ also.
so the only solution is $x=y=z=\dfrac{π}{6}$ ,we put in $ f, f=0$,put $x=y=\dfrac{π}{2},z=0, → f>0$ so $f_{min}=0$ QED.
| {
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Proving $\gcd(n^2(n^2+1),2n+1)=\gcd(2n+1,5)$ We suppose $\forall n \in \mathbb {N}\setminus{0}$.
How can I prove that $\gcd(n^2(n^2+1),2n+1)=\gcd(2n+1,5)$?
| Using the Euclidean Algorithm, we get
$$
16n^2(n^2+1)-(8n^3-4n^2+10n-5)(2n+1)=5\tag{1}
$$
Therefore, $(n^2(n^2+1),2n+1)\mid5$.
Furthermore, since $(5,16)=1$, we have $5\mid2n+1\Longleftrightarrow5\mid n^2(n^2+1)$. Thus,
$$
(n^2(n^2+1),2n+1)=(2n+1,5)\tag{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345154",
"timestamp": "2023-03-29T00:00:00",
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Solving $\sin \theta + \cos \theta=1$ in the interval $0^\circ\leq \theta\leq 360^\circ$
Solve in the interval $0^\circ\leq \theta\leq 360^\circ$ the equation $\sin \theta + \cos \theta=1$.
I've got the two angles in the interval to be $0^\circ$ and $90^\circ$, it's not an answer I'm after, I'd just like to see different approaches one could take with a problem like this.
Thank you!
Sorry, my approach:
$$\begin{align}
\frac{1}{\sqrt 2}\sin \theta + \frac{1}{\sqrt 2}\cos \theta &= \frac{1}{{\sqrt 2 }} \\
\cos 45^\circ\sin \theta + \sin 45^\circ\cos \theta &= \frac{1}{\sqrt 2} \\
\sin(\theta + 45^\circ) &= \frac{1}{\sqrt 2} \\
\theta + 45^\circ &= 45^\circ,\ 135^\circ \\
\theta &= 0^\circ, \ 90^\circ
\end{align}$$
| I'd write $\sin \theta + \cos \theta = \sqrt{2} \sin \left(\theta + \dfrac{\pi}{4}\right)$ and go from there.
A similar tactic works for all equations of the form $a \sin \theta + b \cos \theta = c$ for constant $a,b,c$.
| {
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$5 \frac {3}{*} \times 3 \frac {*}{2}=19$? One of my friends gave me this apparently easy-looking problem which I do not know how to crack. The problem is to find the values of "*" where
$$5 \frac {3}{*} \times 3 \frac {*}{2}=19\text{ ?}$$
I can rearrange the problem as $5 \frac {3}{x} \times 3 \frac {y}{2}=19$ and I have to find the values of $x,y.$ Now $5 \frac {3}{x} \times 3 \frac {y}{2}=19 \implies \frac {5x+3}{x} \times \frac {6+y}{2}=19.$ Now, I do not know which way to go?
Can someone point me in the right direction? Thanks in advance for your time.
EDIT: Here "*"-s are not same. Infact I know the answer but do not know how to get it. Here $x=7,y=1.$
| I assume $x, y$ are supposed to be positive integers (otherwise there are infinitely many real solutions for $x, y$). If $y \geq 2$, then $19=(5+\frac{3}{x})(3+\frac{y}{2})>5(3+\frac{2}{2})=20$, a contradiction. Thus $y=1$, so $x=7$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$2x^4 \le\sin^4x+\cos^6x -1 $ inequality Solve the inequality:
$2x^4 \le \sin^4x+\cos^6x -1 $
What I did:
Using trig. identities I simplified the RHS to: $3\sin^2(x)\cos^2(x) + (\sin^2(x))\cos^2(x)\sin^2(x)$ but still I don't know where to go from here. I realize that both sides are positive (at least if I move the 1 to the LHS). Wolfram says the solution is $x=0$. But I can't figure how to get there.
Thanks
| Hint:
Maximum value of $\sin^4(x) + \cos^6(x)$ is $1$
$$
\begin{cases}
f(x)=\sin^4(x) + \cos^6(x) \\
f(x)=(1-\cos^2(x))^2 + cos^6(x) \\
f(x)=1-2\cdot cos^2(x)+cos^4(x)+cos^6(x) \\
f(x)=cos^4(x)+cos^6(x) - 2\cdot cos^2(x) + 1\\
\text{As all the terms are in the same phase}\\
\max(f(x)) = max(cos^4(x)+cos^6(x) - 2\cdot cos^2(x)) + 1\\
\max(f(x)) =\max(\cos^4(x)) + \max(\cos^6(x)) -2\cdot \max(\cos^2(x)) +1\\
\max(f(x)) =\max(\cos(x))^4 + \max(\cos(x))^6 -2\cdot \max(\cos(x))^2 +1\\
\text{As } \max(\cos(x)) = 1 \\
\max(f(x)) = 1 + 1 - 2 + 1= 1
\end{cases}
$$
So, Maximum Value of $sin^4(x) + cos^6(x) - 1$ is $0$
if $x$ is real, minimum value of $x^4$ is $0$
Now its easy to solve the inequality
Its always good to visualize,
Here is the plot of $sin^4(x) + cos^6(x) - 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/346433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How find this maximum $f=\dfrac{8a^2-6ab+b^2}{4a^2-2ab+ac}$ The quadratic equation $ax^2+bx+c=0$ has two roots in the interval $[0,2]$,Find the maximum of
$$f=\dfrac{8a^2-6ab+b^2}{4a^2-2ab+ac}$$
my idea:we have
(1):if $a>0$,then let $g(x)=ax^2+bx+c$
$$\Delta=b^2-4ac>0,g(0)\ge 0,g(2)\ge 0$$
then we have $$b^2>4ac,c\ge0,4a+2b+c\ge0$$
(2):if $a<0$,then we have
$$b^2-4ac>0,c\le 0,4a+2b+c\le 0$$.
so I think this method is very ugly. can someone have nice methods? Thank you
by the @Yimin hint:we have
$x_{1}+x_{2}=-\dfrac{b}{a},x_{1}x_{2}=\dfrac{c}{a}$,then we have $$f=\dfrac{8a^2-6ab+b^2}{4a^2-2ab+ac}=\dfrac{8-\dfrac{b}{a}+\left(\dfrac{b}{a}\right)^2}{4-2\dfrac{b}{a}+\dfrac{c}{a}}=\dfrac{8+x_{1}+x_{2}+(x_{1}+x_{2})^2}{4+2(x_{1}+x_{2})+x_{1}x_{2}},0\le x_{1},x_{2}\le 2$$
| First of all, simultaneously multiplying $a$, $b$, and $c$ a constant does change the roots, and $f$, so we can set $a = 1$ without loss of generality. This yields
$$
f =
\frac{ 8 - 6 b + b^2 } { 4 - 2 b + c }.
$$
Next, let $b = -2p$, $c = p^2 - q^2$ ($q > 0$).
The target function is now
$$
f(p, q)
= \frac{
8 + 12 p + 4 p^2
}
{
4 + 4 p + p^2 - q^2
}
= \frac{ 4 (p + 1)(p + 2) }
{ (2 + p)^2 - q^2 }.
$$
The quadratic equation becomes,
$$ (x - p)^2 = q^2 $$
and the two roots are $x_1 = p-q$, and $x_2 = p+q$, with
\begin{align}
0 &\le p \le 2, \\
q &\le \min\{p, 2 - p\}, \\
\end{align}
to satisfy $0 \le x_1 \le x_2 \le 2$.
Now $f$ increases with $q$, so $q$ must take the maximal possible value.
If $p \le 1$, $q = p$ satisfies the above constraints, and
\begin{align}
f(p, q)
&\le
f(p, p)
\\
&=
\frac{4(p+1)(p+2)}
{(2+p)^2 - p^2}
\\
&=p+2 \le 3
\qquad
(p \le 1).
\end{align}
If $p \ge 1$, $q$ should assume the value of $2-p$ to maximize $f(p, q)$,
\begin{align}
f(p, q)
&\le
f(p, 2 - p)
\\
&=
\frac{4(p+1)(p+2)}
{(2+p)^2-(2-p)^2}
\\
&=
\frac{1}{2}
\left(
p+\frac{2}{p} + 3
\right)
\qquad
(p \ge 1).
\end{align}
The function has a minimum at $p = \sqrt 2 \approx 1.414$, so we only need to check the value at the two boundaries. We know that at $p = 1$, $f = 3$. For $p = 2$
$$
f \le
\frac{1}{2}
\left(
2+\frac{2}{2}+3
\right)
=3.
$$
So the maximum is $3$, which is achieved at $a = 1, b = -2, c = 0$ or $a = 1, b = -4, c = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/346949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum Is there a formula for the following sum?
$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots + n\cdot (n+1)$
| Divide each term of the series by $2$. The result is
$$\binom{2}{2}+\binom{3}{2}+\cdots+\binom{n+1}{2}.\tag{$1$}$$
We give a combinatorial argument that the sum $(1)$ is equal to $\binom{n+2}{3}$.
Now how many ways are there to choose three numbers from the numbers $1$ to $n+2$? The smallest number chosen could be $n$. Then there are $\binom{2}{2}$ ways to choose the other two. Or the smallest chosen number could be $n-1$, in which case there are $\binom{3}{2}$ ways to choose the other two. Or the smallest chosen number could be $n-2$, in which case there are $\binom{4}{2}$ ways to choose the other two. And so on, up to the smallest chosen number being $1$, in which case there are $\binom{n+1}{2}$ ways to choose the other two.
Thus half our sum is $\binom{n+2}{3}$, and we arrive at
$$1\cdot 2+2\cdot 3+\cdots+n\cdot(n+1)=2\binom{n+2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/347585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
If $A,B,C,D$ are complex numbers on the unit circle with $A+B+C+D=0$, then they form a rectangle
Let $A, B, C, D$ be points on a unit circle. Prove that if $A+B+C+D=0$, then $A,B,C,D$ make a rectangle. (Use complex numbers.)
How do I prove this? I tried to use the dot product of 2 adjacent sides, but I got an ugly trig expression.
| Well, if you really want a proof which uses complex numbers...
If $A+B \not =0, A+C \not =0$,
$$(\frac{A+B}{2}) \cdot (A-B)=0$$
$$(\frac{A+B}{2}) \cdot (C-D)=(-\frac{C+D}{2}) \cdot (C-D)=0$$
(Here we are using dot product)
Since $A+B \not =0$, then the vector represented by $(\frac{A+B}{2})$ is perpendicular to $AB,$ and $CD$, so $AB//CD$. Similarly $AC//BD$, since $A+C \not =0$. Thus $ABDC$ is a paralellogram, so $A-B=C-D$ (since $A-B \not =D-C$), giving $A+D=B+C=0$.
Thus either $A+B=0, A+C=0,$ or $A+D=0$.
By symmetry it suffices to consider when $A+B=0$, then $C+D=0$. $(A-C) \cdot (B-C)=(A-C) \cdot (-A-C)=0$ so $AC \perp BC$. Similarly the other 3 angles are also right angles, so we get a rectangle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/348758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 2
} |
Proofs of $\cos(x+y) = \cos x\cos y - \sin x \sin y$ Define $\sin x $ and $\cos x$ via their infinite series:
$$
\sin x = \sum_n (-1)^{n}\frac{x^{2n+1}}{(2n+1)!}, \qquad
\cos x = \sum_n (-1)^n \frac{x^{2n}}{(2n)!}.
$$
Is there a short, clever proof that $\cos(x+y) = \cos x \cos y - \sin x \sin y$ for all real $x,y$? I can prove it using product series, or by showing that both sides (with $y$ fixed) are solutions of $f''(x) = -f(x)$, $f(0) = \cos y$, $f'(0) = - \sin y$. Does anyone know other (preferably slick!) proofs?
| Suppose we know that $\sin(x+y) = \sin x \cos y + \cos x \sin y$, then we can do this to get a proof:
$\cos (x + y) \\ = \sin (x + y + \frac{\pi}{2}) \\ = \sin x \cos (y + \frac{\pi}{2}) + \cos x \sin (y + \frac{\pi}{2}) \\ = \sin x (-\sin y) + \cos x \cos y \\ = \cos x \cos y - \sin x \sin y$
By the way, both have a geometric proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 8
} |
Trigonometrical limit $\lim\limits_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}?$ Can you help me solve this without using de l'Hôpital's rule (just using Standard rules):
$$ \lim_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}? $$
| Even though I like the approach of Dominic Michaelis, here is another one using infinite products.
First note that
\begin{align}
\sin(x) &= x \prod_{n=1}^\infty (1-\frac{x^2}{\pi^2n^2}) \\
&= x(1-\sum_{n=1}^\infty \frac{x^2}{\pi^2 n^2} + O(x^4))
\end{align}
and
\begin{align}
\cos(x) &= \prod_{n=1}^\infty (1-\frac{x^2}{\pi^2(n-1/2)^2})
\\
\Leftrightarrow x\cos(x) &= x\prod_{n=1}^\infty (1-\frac{x^2}{\pi^2(n-1/2)^2})
\\ &= x(1-\sum_{n=1}^\infty \frac{x^2}{\pi^2 (n-1/2)^2} + O(x^4))
\end{align}
Therefore
\begin{align}
\frac{\sin(x)-x\cos(x)}{x^3}&= \frac{1}{x^2}\Bigl[\sum_{n=1}^\infty \frac{x^2}{\pi^2 (n-1/2)^2} - \sum_{n=1}^\infty \frac{x^2}{\pi^2 n^2} +O(x^4) \Bigr]
\\
&=\sum_{n=1}^\infty \frac{1}{\pi^2 (n-1/2)^2} - \sum_{n=1}^\infty \frac{1}{\pi^2 n^2} +O(x^2) \\
&=\frac{1}{\pi^2} \Bigl[\sum_{n=1}^\infty \frac{1}{ (n-1/2)^2} - \sum_{n=1}^\infty \frac{1}{n^2} \Bigr]+O(x^2)
\end{align}
Together with the fact that
\begin{align}
&\sum_{n=1}^\infty \frac{1}{ (n-1/2)^2} =\frac{\pi^2}{2} \quad \text{and} \\
&\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}
\end{align}
we get
\begin{align}
\frac{\sin(x)-x\cos(x)}{x^3}
&=\frac{1}{\pi^2} \Bigl[\sum_{n=1}^\infty \frac{1}{ (n-1/2)^2} - \sum_{n=1}^\infty \frac{1}{n^2} \Bigr]+O(x^2) \\
&= \frac{1}{\pi^2} (\frac{\pi^2}{2}-\frac{\pi^2}{6}) +O(x^2)
\\ &= \frac{1}{3} + O(x^2) = 0 \quad \text{for } x\rightarrow 0
\end{align}
Note: I am aware, that this is not as simple as requested by Maths'problem and even more complex than the answer of Dominic Michaelis. We also need to know the infinite products and the identities involving $\sum 1/n^2$ and $\sum 1/(n-1/2)^2$. However, I just wanted to provide another method to solve the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/351038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Simple Modulo Question $6x = 9 \pmod{11}$ I am trying to solve $6x = 9\pmod{11}$
The solution suggests notice that $2*6 = 12 = 1 \pmod {11}$
$$6x = 9 \pmod {11}\\12x = 18 \pmod {11}\\x = 7 \pmod{11}$$
I don't get the final step from $12x = 18\pmod{11}$ to the solution.
| $6x \equiv 9 \pmod{11} \Rightarrow 2x\equiv 3 \pmod{11} (\because (3,11)=1) \Rightarrow 2x\equiv 14 \pmod{11} (\because 14\equiv 3 \pmod{11}) \Rightarrow x\equiv 7 \pmod{11}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/353600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Show that function is strictly monotone increasing I want to show that $$ f(x)=\dfrac{x-\sin(x)}{1-\cos(x)} $$ is strictly increasing in $(0,2 \pi) $. Unforunately, this is not that easy for me , as the derivative is not very manageable and trigonometric identities appear to be not very helpful, too.$$$$
Are there any further ways to do this?
| let $f(x)=\dfrac{x-\sin(x)}{1-\cos(x)}$
EDIT: I am wrong in the formula for the last version and I correct here,sorry for it.
$f'(x)=\dfrac{(x-\sin x)(-\sin x)}{(1-\cos x)^2}+\dfrac{1-\cos x}{1-\cos x}=\dfrac{\sin^2x-x\sin x+1-2\cos x+\cos^2 x}{(1-\cos x)^2}=\dfrac{2(1-\cos x)-x\sin x}{(1-\cos x)^2}=\dfrac{2\cdot 2\sin^2 \dfrac{x}{2}-2x\sin\dfrac{x}{2}\cos{x}{2}}{(1-\cos x)^2}=\dfrac{2\cdot sin \dfrac{x}{2}}{(1-\cos x)^2}\cdot (2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}) $
$\dfrac{2\cdot \sin \dfrac{x}{2}}{(1-\cos x)^2}>0 $ when $0<x<2\pi$,
so we only check $(2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}) $.
When $0< x < \pi, \tan\dfrac{x}{2}>\dfrac{x}{2},$ we have $ 2\sin\dfrac{x}{2}>x\cos x \dfrac{x}{2}$, ie. $2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}>0$.
When $\pi \leq x < 2\pi$, $\cos\dfrac{x}{2} \leq 0, -x \cos\dfrac{x}{2} \ge 0,$ and $2\sin\dfrac{x}{2}>0$, ie. $2\sin\dfrac{x}{2}-x\cos x \dfrac{x}{2}>0$
so $f'(x)>0$ when $0<x<2\pi$, QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Solving the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$ $$\int_0^{\infty} \frac{dx}{1+x^3}$$
So far I have found the indefinite integral, which is:
$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$
Now what do I need to do in order to calculate the improper integral?
| Next, simplify
$$
F(x)=-\frac{1}{6}\ln|x^2-x+1|+\frac{1}{\sqrt{3}}\arctan{\frac{2x-1}{\sqrt{3}}}+\frac{1}{3}\ln|x+1|
$$
$$
=\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|-\frac{1}{3}\ln\sqrt{|x^2-x+1|}
$$
$$
=\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln\left(\frac{|x+1|}{\sqrt{|x^2-x+1|}}\right).
$$
Then
$$\int_0^\infty \frac{dx}{1+x^3}=\lim_{X\rightarrow\infty}F(X)-F(0).$$
Compute the limit, and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/358262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Integrate $2\int x^2\, \sec^2x \,\tan x\, dx$ $$
2\int x^2\, \sec^2x \,\tan x\, \mathrm{d}x
$$
How to solve this using integration by parts? WolframAlpha can solve it, but is unable to give a step-by-step solution, and has a different answer to the one in the back of my textbook. There is also a question/answer on yahoo answers, but yet again, that gives a different answer to the one given to me.
| Integrate by parts, differentiating $x^2$ and integrating $\sec ^{2}x\tan x$ by substitution$^1$:
$$\begin{eqnarray*}
I &=&\int x^{2}\sec ^{2}x\tan x\, dx=\frac{1}{2}x^{2}\sec ^{2}x-\int x\sec
^{2}x\,dx \\
&=&\frac{1}{2}x^{2}\sec ^{2}x-\left( x\tan x-\int \frac{\sin x
}{\cos x}\,dx\right)\qquad \text{by parts; note }^2 \\
&=&\frac{1}{2}x^{2}\sec ^{2}x-x\tan x-\ln |\cos
x|+C.
\end{eqnarray*}$$
So
$$
\begin{equation*}
2I=2\int x^{2}\sec ^{2}x\tan x dx=x^{2}\sec ^{2}x-2x\tan
x-2\ln | \cos x|+C.
\end{equation*}
$$
--
$^1$ Let $u=\sec x$. Then $du=\sec x\, \tan x$ and
$$\int \sec ^{2}x\tan xdx=\int u\,du=\frac{1}{2}u^{2}=\frac{1}{2}\sec ^{2}x.$$
$^2$ Differentiate $x$ and integrate $\sec^2 x$. Since $\dfrac{d}{dx}\tan x=1+\tan ^{2}x=\sec ^{2}x$, $\displaystyle\int \sec ^{2}xdx=\tan x
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Area of a Regular Polygon I was trying to find the area of a regular polygon in terms of n, the side length and s, the number of sides.
Because there are $s$ sides number of isosceles triangles in a regular polygon, I decided to work out the area of an isosceles triangle in terms of $A$, the unique angle and $a$, the unique side:
$$Area=\frac{1}{2}ab\sin{C}$$
$$b=\frac{\sin{C}\times a}{\sin{A}}$$
(Sine rule, $C=B$)
$$b=\frac{\sin{\frac{180-A}{2}}\times a}{\sin{A}}$$
$$Area=\frac{1}{2}a\times\frac{\sin{C}\times a}{\sin{A}}\times\sin{C}=\frac{{(\sin{C}\times{a})}^{2}}{\sin{A}}$$
$$C=\frac{180-A}{2}=90-\frac{A}{2}$$
$$Area=\frac{{(\sin({90-\frac{A}{2}})\times{a})}^{2}}{\sin{A}}$$
And that was where I got to in finding the area of an isosceles triangle. Then I tried to find the area of the whole regular polygon:
$$Area=s\times\frac{{(\sin({90-\frac{A}{2})}\times{n})}^{2}}{\sin{A}}$$
Where $s$ is the number of sides and $n$ replaces $a$
$$A=\frac{360}{s}$$
$$Area=s\times\frac{{(\sin({90-\frac{\frac{360}{s}}{2}})\times{n})}^{2}}{\sin{\frac{360}{s}}}=s\times\frac{\sin^{2}({90-\frac{180}{s}})\times{n}^{2}}{\sin{\frac{360}{s}}}$$
$$Area=\frac{s{n}^{2}\sin^{2}({90-\frac{180}{s}})}{\sin{\frac{360}{s}}}$$
However, when I tested this formula it was wrong. Can someone tell me where I've gone wrong?
|
$$Area=\frac{1}{2}a\times\frac{\sin{C}\times a}{\sin{A}}\times\sin{C}=\frac{{(\sin{C}\times{a})}^{2}}{\sin{A}}$$
It seems that you dropped $\frac{1}{2}$ here. :-)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/360300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Minimum value of inverse trigo function (cubic) The minimum value of $(\sin^{-1}x)^3+(\cos^{-1}x)^3$ is equal to ( following options)
a) $\displaystyle \frac{\pi^3}{32}$
b) $\displaystyle\frac{5\pi^3}{32}$
c) $\displaystyle\frac{9\pi^3}{32}$
d) $\displaystyle\frac{11\pi^3}{32}$
Can we go like this :
$ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2}$ Therefore minimum value of $(\sin^{-1}x)^3 = (\frac{-\pi}{2})^3$= -$\frac{\pi}{8}$
Please guide..
| Using $$\frac{(\arcsin x)^3+(\arccos x)^3}{2}\geq \bigg(\frac{\arcsin x+\arccos x}{2}\bigg)^3$$
So $$\displaystyle (\arcsin x)^3+(\arccos x)^3\geq \frac{1}{4}\cdot \frac{\pi^3}{8} = \frac{\pi^3}{32}$$
and equality hold when $\displaystyle \arcsin x = \arccos x\Rightarrow x = \frac{1}{\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/362185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Two concentric circles have radii 1 and 4
What is $k+m+n$?
Thank you for helping. Please give a solution for me, if you don't mind.
| Let $O$ be the center of the concentric circles, let $A$ be the center of the top light-gray circle, and let $B$ be the center of one of the upper dark-gray circles; let $r$ be the common radius of the smaller circles.
Clearly, symmetry dictates that $\angle AOB = 60^\circ$. Note that $|OA| = 4-r$, and $|OB| = 1+r$, and $|AB| = 2r$. Thus, by the Law of Cosines (and the assumption that $r$ must be postive):
$$\begin{align}
|AB|^2 &= |OA|^2 + |OB|^2 - 2 |OA| |OB| \cos 60^\circ \\
\implies \qquad (2r)^2 &= (4-r)^2 + (1+r)^2 - ( 4-r )( 1 + r ) \\
\implies \qquad 0 &= r^2 + 9 r - 13 \\
\implies \qquad r &= \frac{1}{2}\left( -9 + \sqrt{7\cdot 19} \right)
\end{align}$$
Thus, $k = -9$, $m = 133$, and $n = 2$, so that $k+m+n = 126$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/364244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding $n$ such that $\frac{n^4 + 1}{n^2 +n + 1}$ is an integer
Find all integers such that $\dfrac{n^4 + 1}{n^2 +n + 1}$ is an integer.
I have no idea how to solve things like this and what i tried to do didn't get me anywhere. I'd be grateful for any help!
| HINT:
$$n^4+n^2+1=(n^2)^2+1^2+n^2=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$$
$$\implies \frac{n^4+1}{n^2+n+1}=n^2-n+1-\frac{n^2}{n^2+n+1}$$
So, $n^2+n+1$ must divide $n^2$
But $(n^2,n^2+n+1)=1$ (Proof Below)
$\implies n^2+n+1=\pm1$ as $ n^2+n+1$ being denominator $\ne0$
If $ n^2+n+1=-1\implies n^2+n+2=0\implies n=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot2}}2=\frac{-1\pm\sqrt7i}2$ which is not real.
If $ n^2+n+1=1\implies n(n+1)=0\implies n=0,-1$
[
Proof :
If prime $p>1$ divides both $n^2, n^2+n+1$
$p$ will divide $(n^2+n+1)-n^2=n+1$
As $p$ is prime, $p$ divides $n^2\implies p$ divides $n$
Then $p$ divides $(n+1)-n=1$ clear contradiction as $p>1$
]
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
how to find the value of a function from the first and second derivative.
The function $f$ is twice differentiable, and the graph of $f$ has no points of inflection. if $f\left(6\right)=3,\, f^{\prime}\left(6\right) = -1/2,$ and $f^{\prime\prime}\left(6\right) = -2$ Which of the following could be the value of $f\left(7\right)$?
(A) 2 (B) 2.5 (C) 2.9 (D) 3 (E) 4
From the answer sheet, I know that the answer should be two; however, I am unable to figure out why.
I have tried $y=f\left(a\right)+f^{\prime}\left(a\right)\left(x-a\right)$ but that gives the wrong answer. I also tried to approximate with making a taylor series; but that failed horribly.
I know that the value must be less than $f\left(6\right)$ because the tangent is negative and the second derivative is negative as well.
copied the question exactly from the pdf.
| If we use a truncated taylor series, the answer does indeed work out:
$$
f(7) \approx f(6) + f^\prime (6)(7-6) + \frac{1}{2} f^{\prime\prime} (6) (7-6)^2 = 3 - \frac{1}{2} + \frac{1}{2} \cdot (-2) = \frac{3}{2}
$$
We now need a bound on our error, using Lagrange Error Bound:
$$
|E_n (x)| \leq \frac{M}{(n+1)!} |x-a_0|^{n+1} = \frac{M}{(3)!} |x-6|^{3}
$$
Where $M$ is some value satisfying $|f^{(n+1)}| = |f^{\prime\prime\prime} (x) | \leq M$ on the interval $(6,x)$.
Since we don't know the value of $f^{\prime\prime\prime}$, we can sloppily just assume the closest answer is correct.
Alternatively, we truncate earlier:
$$
f(7) \approx f(6) + f^\prime (6)(7-6) \pm \frac{M}{2!} |7-6|^{2} = 3 - \frac{1}{2} + \frac{M}{2} = \frac{5}{2} \pm \frac{M}{2}
$$
Recalling that $f^{\prime\prime}(6) = -2$, we know that $|M| \geq 2$, which creates a lower bound on $|M|$.
So, any value $y \in \mathbb{R}$ that satisfies this equation is a possible answer,
$$
\frac{3}{2} = \frac{5}{2} - 1 \leq y \leq \frac{5}{2} + 1 = \frac{7}{2}
$$
So, the best possible answer is indeed $2$, however, $2.5$ is also within the tolerance.
edit: Added more about definition of Lagrange's Error Bound
edit: I now see that I did not read fully the question, since there are no inflection points, we can modify our upper bound to reflect that the derivative is always decreasing, so $\frac{3}{2} \leq y \leq \frac{5}{2}$ since there is no way for the derivative to be greater than $\frac{-1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How can be done by the method of mathematical induction? We are given that $P(x+1)-P(x)=2x+1$
We also know that $P(0)=1$
We want to prove that $P(2004)=(2004)^2 +1$
Can someone explain how can be solved with mathematical induction?
Thank you in advance!
| Observe that $P(y)=y^2+1$ satisfies the conditions [Proof below]
So, if $P(x)=x^2+1, P(x+1)=P(x)+2x+1=x^2+2x+1+1=(x+1)^2+1$
[
Proof:
As $P(x+1)-P(x)=2x+1$
$P(x)$ can be at most Quadratic Polynomial
Let $P(x)=Ax^2+Bx+C$ as $P(0)=1, A\cdot0^2+B\cdot0+C+C\implies C=1$
So, $P(x)=Ax^2+Bx+1, P(x+1)=A(x+1)^2+B(x+1)+1$
So, $2x+1=P(x+1)-P(x)=A(2x+1)+B=2Ax+A+B$
Equating the coefficients of $x,2=2A\implies A=1$
Equating the constants $A+B=1\implies B=0$
]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding $\lim_{x \to +\infty} \left(1+\frac{\cos x}{2\sqrt{x}}\right)$ Let $f(x)=x+\sin(\sqrt{x})$. I want to find $\lim_{x \to +\infty} f'(x)$.
Attempt 1
We have $$f'(x)=1+\frac{\cos x}{2\sqrt{x}} \leq 1 + \left|\frac{\cos x}{2\sqrt{x}}\right| \leq 1 + \frac{1}{2\sqrt{x}}.$$ As $x \rightarrow \infty$, $\sqrt{x} \rightarrow \infty$, hence $\frac{1}{2\sqrt{x}} \rightarrow 0$. Then $1 + \frac{1}{2\sqrt{x}} \rightarrow 1$. Therefore by the Sandwich Theorem $f'(x) \rightarrow 1$.
Lemma
$\lim_{x \to \infty} g(x)=l$ if and only if $\lim_{n \to \infty} g(x_{n})=l$ for all sequences $(x_{n}) \subset E$ with $\lim_{n \to \infty} x_{n} = \infty$, where $E$ is the domain of $g$.
Attempt 2
$f'(x)=1+\frac{\cos x}{2\sqrt{x}}$. Now let $g(x)=f'(x)$ and $x_{n}=n^2$. Then $\lim_{n \to \infty} (x_{n})=\infty$. We have $g(x_{n})=1+\frac{\cos(n)}{2n}$. Then as $n \rightarrow \infty$, $g(x_{n}) \rightarrow 1.$ Therefore by the Lemma above, $\lim_{x \to +\infty} g(x)=1$.
Question
Are the attempts above correct?
Thank you for your time.
Edited Attempt 1
We have $$f'(x)=1+\frac{\cos x}{2\sqrt{x}} \leq 1 + \left|\frac{\cos x}{2\sqrt{x}}\right| \leq 1 + \frac{1}{2\sqrt{x}}.$$ For all $x \geq 0$, $1+\frac{\cos x}{2\sqrt{x}} \geq 0$. So $$0 \leq 1+\frac{\cos x}{2\sqrt{x}} \leq 1+\frac{1}{2\sqrt{x}}.$$ As $x \rightarrow \infty$, $\sqrt{x} \rightarrow \infty$, hence $\frac{1}{2\sqrt{x}} \rightarrow 0$. Then $1 + \frac{1}{2\sqrt{x}} \rightarrow 1$. Therefore by the Sandwich Theorem $f'(x) \rightarrow 1$.
Question
Is this correct? Also, I would like to know if it is necessary to show that $$1-\frac{1}{2\sqrt{x}} \leq 1 + \frac{\cos x}{2\sqrt{x}} \leq 1+\frac{1}{2\sqrt{x}} \tag{1}$$
instead of $$0 \leq 1+\frac{\cos x}{2\sqrt{x}} \leq 1+\frac{1}{2\sqrt{x}}. \tag{2}$$
In my attempt to show inequality $(1)$, I got as far as $$1-\left|\frac{\cos x}{2\sqrt{x}}\right| \le \left|1 -\left(-\frac{\cos x}{2\sqrt{x}}\right)\right| \leq 1 + \left|\frac{\cos x}{2\sqrt{x}}\right| \leq 1 + \frac{1}{2\sqrt{x}}.$$ Could you please help me show that $$1-\frac{1}{2\sqrt{x}} \leq 1 + \frac{\cos x}{2\sqrt{x}}.$$
Thank you.
| For the first attempt, I suppose to use the sandwich theorem or squeeze theorem, you want to find suitable functions $L(x)$ and $U(x)$ such that for all $x$, $$L(x) \leq f(x) \leq U(x)$$ and $$1 = \lim_{x \to \infty} L(x) \leq \lim_{x \to \infty} f(x) \leq \lim_{x \to \infty} U(x) = 1.$$ Right now I only see an upper bound $U(x)$ with limiting value $1$, but no lower bound $L(x)$.
For the second attempt, you want to prove that for all sequences $(x_n)$, the limit is $1$, and not just for one particular choice. So using this lemma in a useful way is not so easy, and if I were you I'd try to make the first attempt work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/372681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding quadratic residues in a finite field by using a primitive element Let $1+2x$ be a primitive element of the field $\mathbb F_9$ obtained via the irreducible polynomial
$$x^2 + 1$$
over the base field $\mathbb F_3$.
i) Make a list of the elements of $\mathbb F_9$ together with the primitive element $1+2x$ and all the powers of primitive element.
ii) Which powers are quadratic residues and which are quadratic non-residues? Why?
| $$\begin{array}{rcl}
\left(1+2x\right)^0 & = & 1 \\
\left(1+2x\right)^1 & = & 1+2x \\
\left(1+2x\right)^2 & = & x \\
\left(1+2x\right)^3 & = & 1+x \\
\left(1+2x\right)^4 & = & 2 \\
\left(1+2x\right)^5 & = & 2+x \\
\left(1+2x\right)^6 & = & 2x \\
\left(1+2x\right)^7 & = & 2x+2 \\
\end{array}$$
Since $1+2x$ is primitive, every element of $\mathbb{F}_9$ has the form $\left(1+2x\right)^\alpha$ for some $\alpha$. Thus $$\left(\left(1+2x\right)^\alpha\right)^2=\left(1+2x\right)^{2\alpha}$$ so the quadratic residues are $1,2,x,$ and $2x$ - the even powers of $1+2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/374870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral Question - $\int\frac{1}{x^2-6x}\,\mathrm dx$ How I can evaluate the indefinite integral? :
$$\int\frac{1}{x^2-6x}\,\mathrm dx$$
Do I need to bring it to this format? : $\displaystyle \int\frac{1}{x^2-a^2}\,\mathrm dx\;$?
Thanks!
| Use partial fraction decomposition to obtain a sum of easily integrable terms:
$$\frac{1}{x^2-6x}=\frac{1}{x(x-6)}=\frac{A}{x}+\frac{B}{x-6}$$
$$\text{multiply by }x \implies \frac{1}{x-6}=A+\frac{Bx}{x-6},\quad \text{set }x=0 \implies \color{blue}{A=-\frac{1}{6}}$$
$$\text{multiply by }x-6 \implies \frac{1}{x}=\frac{A(x-6)}{x}+B,\quad \text{set }x=6 \implies \color{red}{B=\frac{1}{6}}$$
Hence
$$\int\frac{1}{x^2-6x}dx=\int\left(\color{blue}-\frac{\color{blue}1}{\color{blue}6x}+\frac{\color{red}1}{\color{red}6(x-6)}\right)dx$$
Now, use the fact that
$$\int\frac{1}{x-a}dx=\ln|x-a|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/377410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
nasty exponentials While trying to find the fourier transform of $\Large \frac{1}{1 + x^4} $, using the definition and the residue theorem has required me to evaluate nasty looking expressions like
$$\large \rm e^{-ike^{i\frac{\pi}{4}}} .$$
Mathematica tells me this is the same as $$ \rm e^{\frac{k}{\sqrt{2}}} \cos(\frac{k}{\sqrt{2}}) - i \rm e^{\frac{k}{\sqrt{2}}} \sin(\frac{k}{\sqrt{2}}) $$
Now these two expressions aren't obviously equivalent to me, and my question is, how can I get from the first expression to the second expression?
| It's just Euler's formula $$
e^{\sigma + i\varphi} = e^\sigma\cos\varphi + ie^\sigma\sin\varphi
$$ applied twice. First to get $$
e^{i\frac{\pi}{4}}
= \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})
= \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \text{,}
$$ and then again to get $$
\begin{align}
e^{-ike^{i\frac{\pi}{4}}}
&= e^{-ik\left(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right)}
= e^{\frac{k}{\sqrt{2}} - i\frac{k}{\sqrt{2}}}
= e^{\frac{k}{\sqrt{2}}}\cos\left({-\frac{k}{\sqrt{2}}}\right)
+ ie^{\frac{k}{\sqrt{2}}}\sin\left({-\frac{k}{\sqrt{2}}}\right) \\
&= e^{\frac{k}{\sqrt{2}}}\cos\left({\frac{k}{\sqrt{2}}}\right)
- ie^{\frac{k}{\sqrt{2}}}\sin\left({\frac{k}{\sqrt{2}}}\right) \text{.}
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Tricky elementary integral $\int_{0}^{\frac{\pi}{2}}x\cot(x)dx$.
$$\int_{0}^{\frac{\pi }{2}}x\cot(x)dx$$
I tried integration by parts and got $\dfrac{1}{2}\int_{0}^{\frac{\pi }{2}}x^{2} \csc^{2}x dx$ which doesn't help at all. I don't really know what to do. Any help will be greatly appreciated.
| that is a great question but there is a splendid Q , I will post for knowlodge
$$I=∫_{0}^{\frac{π }{4}}x^2 \cot xdx\ \ \ \ \ \ \ \ \ by\ parts\ we\ have\\
\\
I=x^2\ln(\sin x)\tfrac{\frac{π }{4}}{0}-2∫_{0}^{\frac{π }{4}}x\ln(\sin(x))dx\\
\\
=-\frac{π ^{2}\ln(2)}{32}+∫_{0}^{\frac{π }{4}}(2\ln(2)+2∑_{k=1}^{\infty }\frac{\cos(2kx)}{k})xdx\\
\\
=\frac{-π ^{2}\ln(2)}{32}+2\ln(2)∫_{0}^{\frac{π }{4}}xdx+2∑_{k=1}^{\infty }\frac{1}{k}∫_{0}^{\frac{π }{4}}x\cos(2kx)dx\\
\\
\\
but\ ∫ x\cos(2kx)=\frac{x\sin(2kx)}{2k}+\frac{\cos(2kx)}{4k^2}+c\\
\\
\therefore ∫_{0}^{\frac{π }{4}}x\cos(2kx)dx=\frac{π \sin(\frac{kπ }{2})}{8k}+\frac{\cos(\frac{π k}{2})}{4k^2}-\frac{1}{4k^2}\\
\\$$
$$\therefore I=\frac{-π ^{2}\ln(2)}{32}+\ln(2)x^2\tfrac{\frac{π }{4}}{0}+2∑_{k=1}^{\infty }(\frac{π \sin(\frac{π k}{2})}{8k^2}+\frac{\cos(\frac{π k}{2})}{4k^3}-\frac{1}{4k^3})\\
\\
=\frac{-π ^{2}}{32}\ln(2)+\frac{π ^{2}}{16}\ln(2)-\frac{1}{2}∑_{k=1}^{\infty }\frac{1}{k^3}+\frac{1}{2}∑_{k=1}^{\infty }(\frac{\cos(π k)}{8k^3})+\frac{π }{4}∑_{k=1}^{\infty }\frac{(-1)^{k-1}}{(2k-1)^2}\\
\\
=\frac{π ^{2}}{32}\ln(2)+\frac{π G}{4}-\frac{1}{2}\zeta (3)-\frac{3}{64}\zeta (3)\\
\\
\\
\therefore I=∫_{0}^{\frac{π }{4}}x^2.\cot(xdx=\frac{π ^{2}}{32}.\ln(2)-\frac{35}{64}\zeta (3)+\frac{π G}{4}\\$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/381976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
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} |
A problem on range of a trigonometric function: what is the range of $\frac{\sqrt{3}\sin x}{2+\cos x}$? What is the range of the function
$$\frac{\sqrt{3}\sin x}{2+\cos x}$$
| Let $$y = \frac{\sqrt{3}\sin x}{2+\cos x}\Rightarrow 2y+y\cos x = \sqrt{3}\sin x$$
So $$\sqrt{3}\sin x-y\cos x = 2y$$
Now Using $\bf{Cauchy\; Schwarz \; Inequality}$
$$\left[(\sqrt{3})^2+(-1)^2\right]\cdot \left[\sin^2 x+\cos^2 x\right]\geq \left[\sqrt{3}\sin x-y\cos x\right]^2$$
So $$4\geq 4y^2\Rightarrow y^2\leq 1\Rightarrow y \in \left[-1,1\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/382136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
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Can you prove if $5^n \mid r$ and $r$ odd then $5^{n+1} \mid 2^r+3^r$? I've looked around at this, but been unable to find it. After generating prime factorizations in Maple of $2^r+3^r$ for $r>0$ it seems that this holds. It's easy enough to prove case by case. The case $n=0$ is trivial since for $r$ odd $x+y|x^r+y^r$ (I'm pretty sure it works by factoring by grouping).
For $n=1$ suppose $r>0$ is odd and divisible by 5. Then $r=20k+5$ or $r=20k+15$ for some $k\in\mathbf{N}$.
Case 1: $2^{20k+5}+3^{20k+5}=2^5(2^k)^{20}+3^5(3^k)^{20}\equiv 7(1)+18(1)$ (mod 25) $\equiv$ 0 (mod 25)
Case2: $2^{20k+15}+3^{20k+15}=2^{15}(2^k)^{20}+3^{15}(3^k)^{20}\equiv 18(1)+7(1)$ (mod 25) $\equiv$ 0 (mod 25)
This works nicely because 20 is $\phi(25)$. In fact for any power of 5, $\phi(5^{n+1})=5^n(4)$ so $r$ can be broken into two cases $5^{n}(4)k+5^{n}$ or $5^nk(4)+3(5^n)$. Since a power of 2 or 3 can't be congruent to a power of 5, raising $2^k$ and $3^k$ to $\phi(5^{n+1})$ gives 1 modulo $5^{n+1}$. But here's is where I'm stuck, showing that $2^{5^n}+3^{5^n}\equiv$ $0$ (mod $5^{n+1}$). Or maybe I'm just approaching it the wrong way? It seems that by looking at Maple data this holds for any $x$, $y$, and $x+y$ where $x+y$ is odd.
| I'll continue from where you stopped.
Hint: Sow by induction that $5^{n+1}|2^{5^n}+3^{5^n}$. The base $n=0$ is clear. To show the step write $2^{5^{n+1}}+3^{5^{n+1}}$ as:
$$(2^{5^n}+3^{5^n})(2^{4.5^n}-2^{3.5^n}3^{5^n}+2^{2.5^n}3^{2.5^n}-2^{5^n}3^{3.5^n}+3^{5^n})$$
By the induction hypothesis, $(2^{5^n}+3^{5^n})$ is divisibe by $5^{n+1}$. Now it suffices to show that $(2^{4.5^n}-2^{3.5^n}3^{5^n}+2^{2.5^n}3^{2.5^n}-2^{5^n}3^{3.5^n}+3^{5^n})$ is divisible by 5. To do this use the fact that $2^{5^n}\equiv -(3^{5^n})\,(mod\,5)$. This follows from the fact $2\equiv -3\,(mod\,5)$ and the fact that $5^n$ is odd.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An easier way to find the integral of: $\int {x\sqrt {2 + x} {\rm{ }}dx} $, where ${u^2} = 2 + x$ My attempt at the question:
$\eqalign{
& \int {x\sqrt {2 + x} {\rm{ }}dx} \cr
& {u^2} = 2 + x \cr
& 2u{{du} \over {dx}} = 1 \cr
& {{du} \over {dx}} = {1 \over {2u}} \cr
& u = \sqrt {2 + x} \cr
& x = {u^2} - 2 \cr
& so: \cr
& \int {x\sqrt {2 + x} {\rm{ }}dx} = \int {x\sqrt {2 + x} } {\rm{ }}{{dx} \over {du}}du \cr
& = \int {x\sqrt {2 + x} } {\rm{ }} \times 2\sqrt {2 + x} du \cr
& = \int {2x} (2 + x)du \cr
& = \int {4x} + 2{x^2}du \cr
& = \int {4({u^2} - 2)} + 2{({u^2} - 2)^2}du \cr
& = \int {4{u^2} - 8} + 2({u^4} - 4{u^2} + 4)du \cr
& = \int {2{u^4} - 4{u^2}} du \cr
& = {2 \over 5}{u^5} - {4 \over 3}{u^3} + C \cr
& = {2 \over 5}{(\sqrt {2 + x} )^5} - {4 \over 3}{(\sqrt {2 + x} )^3} + C \cr
& = {2 \over 5}{(2 + x)^{{5 \over 2}}} - {4 \over 3}{(2 + x)^{{3 \over 2}}} + C \cr} $
A few questions I have:
Given ${u^2} = 2 + x$, $u = \pm \sqrt {2 + x} $, so why is it that we only take the principal square root and not the negative one for substitution?
The second question I have is a general one; is there an easier way of finding the integral? Have I done things in a manner that isn't overly longwinded? If so please suggest ways that would allow me to reach an answer quicker.
I'm on shakey grounds with integration at the moment so I was wondering if I could integrate this part of my working out without expanding out:
$ = \int {4({u^2} - 2)} + 2{({u^2} - 2)^2}du$
Thank you for all your help!
| I think it is simpler by parts:
$$u=x\;,\;\;u'=1\\v'=\sqrt{x+2}\;,\;\;v=\frac23(x+2)^{3/2}$$
so
$$\int x\sqrt{x+2}\,dx=\frac23x(x+2)^{3/2}-\frac23\int(x+2)^{3/2}dx=\frac23x(x+2)^{3/2}-\frac4{15}(x+2)^{5/2}+C=$$
$$=\frac2{15}(x+2)^{3/2}\left(3x-4\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How do you calculate the modulo of a really high number with a large power, with a really high mod number? I need to work out $516489222^{22} \pmod{96899}$. I know there are easier ways of working this out, but am really struggling.
| Note that $516489222 \equiv 17552 \pmod{96899}$. This gives
$$ 516489222^{22} \equiv 17552^{22} \pmod{96899}.$$
Now it is easy to calculate
\begin{align*}
17552^{2^{1}} &\equiv 30783 \pmod{96899}\\
17552^{2^{2}} &\equiv 30783^{2} \equiv 17768 \pmod{96899}\\
17552^{2^{3}} &\equiv 17768^{2} \equiv 4882 \pmod{96899}\\
17552^{2^{4}} &\equiv 4882 ^{2} \equiv 93669 \pmod{96899}.
\end{align*}
Since $22 = 2^{4} + 2^{2} + 2^{1},$ it follows that
\begin{align*}
17552^{22}
&\equiv 17552^{2^{4}} \cdot 17552^{2^{2}} \cdot 17552^{2^{1}} \pmod{96899} \\
&\equiv 93669 \cdot 17768 \cdot 30783 \pmod{96899} \\
&\equiv 4647 \pmod{96899}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/386590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$ I just found this identity but without any proof, could you just give me an hint how I could prove it?
$$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$
I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me
| Inspired by robjohn's answer I show a more general statement: for $x\in \mathbb{R}$ and $n\in\mathbb{N}$,
$$\bbox[5px,border:2px solid #0000A0]{\sum_{k=0}^{n}\binom{n+k}{k}((1-x)^{n+1}x^k+x^{n+1}(1-x)^k)=1}$$
Such identity appears as equation 69 at this page (see hypergeometric's comment). OP's identity follows by setting $x=1/2$.
Let
$$f_n(x)=\sum_{k=0}^{n}\binom{n+k}{k}x^k$$
then we have to show by induction that
$$(1-x)^{n+1}f_n(x)+x^{n+1}f_n(x)=1.$$
Base case: for $n=0$ it is trivial.
\noindent Inductive step step. We note that
\begin{align*}
f_{n+1}(x)
&=\sum_{k=0}^{n+1}\left(\binom{n+k}{k}+\binom{n+1+k-1}{k-1}\right)x^k\\
&=f_n(x)+\binom{2n+1}{n+1}x^{n+1}+xf_{n+1}(x)-\binom{2n+2}{n+1}x^{n+2}\\
&=f_n(x)+\binom{2n+1}{n}x^{n+1}+xf_{n+1}(x)-2\binom{2n+1}{n}x^{n+2}
\end{align*}
Therefore
$$(1-x)f_{n+1}(x)=f_n(x)+\binom{2n+1}{n}x^{n+1}(1-2x)$$
Hence
$$(1-x)^{n+2}f_{n+1}(x)
=(1-x)^{n+1}f_n(x)+\binom{2n+1}{n}(1-x)^{n+1}x^{n+1}(1-2x)$$
and, by replacing $x$ with $(1-x)$, we get
$$x^{n+2}f_{n+1}(1-x)
=x^{n+1}f_n(1-x)-\binom{2n+1}{n}x^{n+1}(1-x)^{n+1}(1-2x).$$
Finally, by adding the last two equations we find
$$(1-x)^{n+2}f_{n+1}(x)+x^{n+2}f_{n+1}(1-x)=
x^{n+1}f_n(1-x)+(1-x)^{n+1}f_n(x)=1$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/389099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Proof of $\frac{1}{e^{\pi}+1}+\frac{3}{e^{3\pi}+1}+\frac{5}{e^{5\pi}+1}+\ldots=\frac{1}{24}$ I would like to prove that $\displaystyle\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{n}{e^{n\pi}+1}=\frac1{24}$.
I found a solution by myself 10 hours after I posted it, here it is:
$$f(x)=\sum_{\substack{n=1\\n\text{ odd}}}^{\infty}\frac{nx^n}{1+x^n},\quad\quad g(x)=\displaystyle\sum_{n=1}^{\infty}\frac{nx^n}{1-x^n},$$
then I must prove that $f(e^{-\pi})=\frac1{24}$. It was not hard to find the relation between $f(x)$ and $g(x)$, namely $f(x)=g(x)-4g(x^2)+4g(x^4)$.
Note that $g(x)$ is a Lambert series, so by expanding the Taylor series for the denominators and reversing the two sums, I get
$$g(x)=\sum_{n=1}^{\infty}\sigma(n)x^n$$
where $\sigma$ is the divisor function $\sigma(n)=\sum_{d\mid n}d$.
I then define for complex $\tau$ the function
$$G_2(\tau)=\frac{\pi^2}3\Bigl(1-24\sum_{n=1}^{\infty}\sigma(n)e^{2\pi in\tau}\Bigr)$$ so that
$$f(e^{-\pi})=g(e^{-\pi})-4g(e^{-2\pi})+4g(e^{-4\pi})=\frac1{24}+\frac{-G_2(\frac i2)+4G_2(i)-4G_2(2i)}{8\pi^2}.$$
But it is proven in Apostol "Modular forms and Dirichlet Series", page 69-71 that $G_2\bigl(-\frac1{\tau}\bigr)=\tau^2G_2(\tau)-2\pi i\tau$, which gives $\begin{cases}G_2(i)=-G_2(i)+2\pi\\ G_2(\frac i2)=-4G_2(2i)+4\pi\end{cases}\quad$. This is exactly was needed to get the desired result.
Hitoshigoto oshimai !
I find that sum fascinating. $e,\pi$ all together to finally get a rational. This is why mathematics is beautiful!
Thanks to everyone who contributed.
| Integrate the function $$f(z) = \frac{z e^{iz}}{\cosh (z) + \cos(z)} \, $$ around the contour $[-\sqrt{2} \pi N, \sqrt{2} \pi N] \cup \sqrt{2} \pi Ne^{i[0, \pi]}$, where $N$ is a positive integer.
The integral vanishes on the semicircle as $N \to \infty$ because $|f(z)|$ decays exponentially fast to zero along the entire semicircle.
We therefore have
$\begin{align} \int_{-\infty}^{\infty}f(x) \, \mathrm dx &= 2 \pi i \left(\sum_{n=0}^{\infty}\operatorname{Res} \left[f(z), \frac{(2n+1) \pi (1+i)}{2} \right] + \sum_{n=0}^{\infty}\operatorname{Res} \left[f(z), \frac{(2n+1) \pi (-1+i)}{2} \right]\right) \\ &= 2 \pi i \left(\sum_{n=0}^{\infty} \lim_{z \to \frac{(2n+1) \pi (1+i)}{2} }\frac{ze^{iz}}{\sinh(z) - \sin(z)} + \sum_{n=0}^{\infty} \lim_{z \to \frac{(2n+1) \pi (-1+i)}{2} }\frac{ze^{iz}}{\sinh(z) - \sin(z)} \right) \\ &= 2 \pi i \left(\frac{\pi}{2}\sum_{n=0}^{\infty} \frac{(2n+1)e^{-(2n+1) \pi /2}}{\cosh \left(\frac{(2n+1) \pi }{2} \right)} + \frac{\pi}{2}\sum_{n=0}^{\infty} \frac{(2n+1)e^{-(2n+1) \pi /2}}{\cosh \left(\frac{(2n+1) \pi }{2} \right)}\right) \\ &= 2 \pi^{2} i \sum_{n=0}^{\infty} \frac{(2n+1) e^{- (2n+1) \pi /2}}{\cosh \left(\frac{(2n+1)\pi }{2} \right)} \\ &= 4 \pi^{2}i \sum_{n=0}^{\infty} \frac{2n+1}{e^{(2n+1)\pi }+1}. \end{align}$
Equating the imaginary parts on both sides of the equation, we get $$ \begin{align} \sum_{n=0}^{\infty} \frac{2n+1}{e^{(2n+1) \pi }+1} &= \frac{1}{4 \pi^{2}} \int_{-\infty}^{\infty} f(x) \, \mathrm dx \\ &= \frac{1}{2 \pi^{2}}\int_{0}^{\infty} \frac{x \sin(x)}{\cosh (x) + \cos(x)} \, \mathrm dx \\ &= \frac{1}{\pi^{2}} \, \Im \int_{0}^{\infty} x \sum_{n=1}^{\infty} (-1)^{n-1} e^{(-1+i)nx} \, \mathrm dx \\ &= \frac{1}{\pi^{2}} \, \Im \sum_{n=1}^{\infty} (-1)^{n-1} \int_{0}^{\infty} x e^{(-1+i)nx} \, \mathrm dx \\ &= \frac{1}{\pi^{2}} \, \Im \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{(1-i)^{2}n^{2}} \\ &= \frac{1}{2 \pi^{2}} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}} \\ &= \frac{1}{24} . \end{align} $$
In general, integrating the function $$g(z) = \frac{z^{4k+1} e^{iz}}{\cosh (z) + \cos(z)} \, , \quad k \in \mathbb{N}_{\ge 0},$$ around the same contour shows that
$$ \begin{align} \sum_{n=0}^{\infty} \frac{(2n+1)^{4k+1}}{e^{(2n+1)\pi}+1} &= \frac{(-1)^{k}2^{2k-1}}{\pi^{4k+2}} \int_{0}^{\infty} \frac{x^{4k+1} \sin(x)}{\cosh(x) + \cos(x)} \, \mathrm dx \\ &= \frac{(-1)^{k}2^{2k-1}}{\pi^{4k+2}} \frac{1}{2^{2k}} (-1)^{k} (4k+1)! \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{4k+2}} \\ &\overset{(1)}{=} \frac{B_{4k+2}}{2 (4k+2)} \left(2^{4k+1}-1\right), \end{align}$$ where $B_{k}$ is the $k$th Bernoulli number.
$(1)$ https://en.wikipedia.org/wiki/Dirichlet_eta_function#Particular_values
| {
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"url": "https://math.stackexchange.com/questions/389146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "168",
"answer_count": 6,
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} |
Definite Integral of square root of polynomial I need to learn how to find the definite integral of the square root of a polynomial such as:
$$\sqrt{36x + 1}$$ or $$\sqrt{2x^2 + 3x + 7} $$
EDIT: It's not guaranteed to be of the same form. It could be any polynomial that can't be easily factored into squares.
This isn't homework, I'm studying for a final. And for context, I'm finding the arc length of a function.
| $$\int {\sqrt{36x+1}}\,dx$$
use this formula
$$\int {\sqrt{x}}\,dx=\frac {x^{\frac{3}{2}}}{\frac32}\implies \frac {2x^\frac32}{3}+C$$
so
$$\int {\sqrt{36x+1}}\,dx=\frac {2(36x+1)^\frac32}{(3)(36)}$$
$$\frac {(36x+1)^\frac32}{54}+C$$
Q2
$$\int\sqrt{2x^2 + 3x + 7}\,dx $$
$$\int\sqrt{2(x^2 + \frac{3x}{2} + \frac72)}\,dx $$
$$\int\sqrt{2(x^2 + \frac{3x}{2} + \frac72)}\,dx $$
$$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} + \frac72)}\,dx $$
$$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} + \frac72+\frac{3^2}{4^2}-\frac{3^2}{4^2})}\,dx $$
$$\sqrt2\int\sqrt{(x^2 + \frac{3x}{2} +\frac{3^2}{4^2}+ \frac72-\frac9{16})}\,dx $$
$$\sqrt2\int\sqrt{(x+ \frac{3}{4})^2 + \frac72-\frac9{16})}\,dx $$
$$\sqrt2\int\sqrt{(x+ \frac{3}{4})^2 + (\frac{\sqrt47}{4})^2}\,dx $$
now use this formula
$$\int{\sqrt{x^2+a^2}}\,dx=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2\log|x+\sqrt{x^2+a^2}|}{2}+C$$
I hope you can take it now on your own
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$ Please help me to solve this integral:
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$
I managed to calculate an indefinite integral of the left part:
$$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \Im\ \text{Li}_2(e^{2\ x\ i}),$$
where $\Im\ \text{Li}_2(z)$ denotes the imaginary part of the dilogarithm. The corresponding definite integral $$\int_0^\pi\frac{\cos x}{\sin x}x\ \mathrm dx$$ diverges. So, it looks like in the original integral summands compensate each other's singularities to avoid divergence.
I tried a numerical integration and it looks plausible that
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx\stackrel{?}{=}\pi \log 54,$$
but I have no idea how to prove it.
| Let
$$y=\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x},$$
then, solving this with respect to $x$, we get
$$x=\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}.$$
So,
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx=\int_0^\infty\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\mathrm dy.$$
The latter integral can be solved by Mathematica and yields $$\pi\log54.$$
Of course, we want to prove that the result returned by Mathematica is correct.
The following statement is provably true, that can be checked directly by taking derivatives of both sides:
$$\int\frac{6y(8+y^2)}{(4+y^2)(16+y^2)}\left(\frac{\pi}{2}+\text{arccot}\frac{6y}{8-y^2}\right)\mathrm dy =\\ \frac{1}{2} i \left(2 \text{Li}_2\left(\frac{iy}{8}+\frac{1}{2}\right)+\text{Li}_2\left(\frac{iy}{6}+\frac{1}{3}\right)+2\text{Li}_2\left(\frac{iy}{6}+\frac{2}{3}\right)+\text{Li}_2\left(\frac{iy}{4}+\frac{1}{2}\right)+\text{Li}_2\left(\frac{2i}{y-2 i}\right)-\text{Li}_2\left(-\frac{2 i}{y+2i}\right)-\text{Li}_2\left(-\frac{1}{6} i (y+2i)\right)-\text{Li}_2\left(-\frac{1}{4} i (y+2i)\right)-2 \left(-\text{Li}_2\left(-\frac{2i}{y-4 i}\right)+\text{Li}_2\left(\frac{2 i}{y+4i}\right)+\text{Li}_2\left(-\frac{1}{8} i (y+4i)\right)+\text{Li}_2\left(-\frac{1}{6} i (y+4i)\right)\right)\right)+\pi \left(\frac{1}{2}\log \left(3 \left(y^2+4\right)\right)+\log\left(\frac{3}{64}\left(y^2+16\right)\right)\right)+\log \left(4\left(y^2+4\right)\right) \arctan\left(\frac{y}{4}\right)-\left(\log576-2\log \left(y^2+16\right)\right) \arctan\left(\frac{4}{y}\right)+\log\left(y^2+4\right) \text{arccot}\left(\frac{6y}{8-y^2}\right)-\arctan\left(\frac{2}{y}\right)\log12 +\arctan\left(\frac{y}{2}\right)\log2$$
The remaining part is to calculate $\lim\limits_{y\to0}$ and $\lim\limits_{y\to\infty}$ of this expression, which I haven't done manually yet, but it looks like a doable task.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/390957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
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How to find the minimum of $x+y^2+z^3$? let $x,y,z>0$, and $x+3y+z=9$, find the minimum of
$$x+y^2+z^3$$
I think this problem is very interesting. I have found this
when $$x=\dfrac{9}{2}-\dfrac{1}{\sqrt{3}},y=\dfrac{3}{2},z=\dfrac{1}{\sqrt{3}}$$
I belive this inequality have $AM-GM$ methods,becasue I have see this same problem can use $AM-GM$ methods,and I think this methods is very very nice.
if $a,b,c>0$,and $ a+b^2+c^3=\dfrac{325}{9}$, prove that
$$a^2+b^3+c^4\ge\dfrac{2807}{27}$$
my methods: let $a=x,b=y,c=z$,and then
$$x+y^2+z^3=\dfrac{325}{9}$$
use $AM-GM$,we have
$$a^2+x^2\ge 2ax$$
$$b^3+b^3+y^3\ge 3yb^2$$
$$c^4+c^4+c^4+z^4\ge 4zc^3$$
then we have
$$a^2+b^3+c^4+x^2+\dfrac{y^3}{2}+\dfrac{z^4}{3}\ge 2ax+\dfrac{3}{2}b^2+\dfrac{4}{3}zc^3$$
| Since $x+y^2+a^2+z^3+b^3+b^3\ge x+2ay+3b^2z$, now let $2a=3, 3b^2=1$. You can determine $ y=a=\frac{3}{2}, z=b=\frac{1}{\sqrt{3}}$ and $x=9-3y-z$ if equality holds.
You do get an AM-GM approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/391301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Proving the quadratic formula (for dummies) I have looked at this question, and also at this one, but I don't understand how the quadratic formula can change from $ax^2+bx+c=0$ to $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$. I am not particularly good at maths, so can someone prove the quadratic formula in a simple way, with no complicated words? All help appreciated.
| Look at each step here:
$$
\begin{align*}
a x^2 + b x + c
&= 0 \\
a \left( x^2 + \frac{b}{a} x \right) + c
&= 0 \\
a \left( x^2 + \frac{b}{a} x + \frac{b^2}{4 a^2} \right) - \frac{b^2}{4 a} + c
&= 0 \\
a \left( x + \frac{b}{2 a} \right)^2
&= \frac{b^2}{4 a} - c \\
\left( x + \frac{b}{2 a} \right)^2
&= \frac{b^2 - 4 a c}{4 a^2} \\
x + \frac{b}{2 a}
&= \frac{\pm\sqrt{b^2 - 4 a c}}{2 a} \\
x &= \frac{-b \pm\sqrt{b^2 - 4 a c}}{2 a}
\end{align*}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$\int_0^{\pi/4}\!\frac{\mathrm dx}{2+\sin x}$ , $\int_0^{2\pi}\!\frac{\mathrm dx}{2+\sin x}$ Please help me integrate
$$\int_0^{\pi/4}\!\frac{\mathrm dx}{2+\sin x}$$
and
$$\int_0^{2\pi}\!\frac{\mathrm dx}{2+\sin x}$$
I've tried the standard $u = \tan \frac{x}{2}$ substitution but it looks horrible.
Thanks in advance!
| Let's give another try to your failed technique...
$$\displaystyle\int \frac{dx}{2+\sin x}$$
Let $u = \tan \frac{x}{2}$
$$\int \frac{du}{u^2+u+1} = \int \frac{du}{\left(u+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}$$
Let $s=u+\frac{1}{2}$
$$\begin{align}\int \frac{ds}{s^2 + \left(\frac{\sqrt{3}}{2}\right)^2} &= \frac{2}{\sqrt{3}}\arctan\left(\frac{2s}{\sqrt{3}}\right)\\ &=\frac{2}{\sqrt{3}}\arctan\left(\frac{2u+1}{\sqrt{3}}\right)\\&=\frac{2}{\sqrt{3}}\arctan\left(\frac{2\tan\frac{x}{2}+1}{\sqrt{3}}\right)\end{align}$$
Evaluating the above from $0$ to $\dfrac{\pi}{4}$ yields approximately $0.33355$ while $0$ to $2\pi$ gives $\dfrac{2\pi}{\sqrt{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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For what values of a the function $y=x^6+ax^3-2x^3-2x^2+1$ is even I want to know for what valuyes this function is even
I know that $f(x)=f(-x)$ to proove that function is even. how its helps me?$$y=x^6+ax^3-2x^3-2x^2+1$$
Thanks!
| Let us start by evaluationg $f(x)$ and $f(-x)$.
$f(x) = x^6 + ax^3 - 2x^3 -2x^2 +1$
and
$f(-x) = x^6 - ax^3 + 2x^3 -2x^2 +1$
If $f(x)$ should equal $f(-x)$ then $a = $...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Develop the next function:$f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0)$ Develop the next function:$\displaystyle f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0).$
For the first part:
$\displaystyle\frac {4x+53}{(x-6)(x-5)}\to\frac {4x+53}{(x-6)(x+5)}=\frac{A}{(x-6)}+\frac{B}{(x+5)}=\frac{7}{(x-6)}+\frac{-3}{(x+5)}=\frac{7}{-6(1-\frac x6)}+\frac{-3}{5\left(1-(-\frac x5)\right)}$
$\displaystyle=-\frac76\sum_{n=0}^\infty{\left(\frac x6\right)^n}-\frac35\sum_{n=0}^\infty(-1)^n\left(\frac x5\right)^n$
Now i'm having trouble with finding the radius of convergence and $f^{(20)}(0)$.
What does $f^{(20)}(0)$ means? 20 derivatives and $x=0$? That means i should find a some-kind of pattern and then place $x=0$?
| When you have
$$f(x) = \sum_{n=0}^{\infty} a_n x^n$$
then
$$f^{(k)}(0) = k! \, a_k$$
You know $a_{20}$:
$$a_{20} = -\frac{7}{6} \frac{1}{6^{20}} - \frac{3}{5} \frac{1}{5^{20}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/395312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A numerical coincidence with continued fractions My brother built a garage that measures 45 feet by 30 feet. To make sure the right angles were accurate, he measured the two diagonals of the rectangle to see that they were equal. In inches,
\begin{align}
& \sqrt{540^2+360^2} \approx 648.999229548\text{ inches} \\[6pt]
= {} & 54\text{ feet}+1\text{ inch} - \text{less than $0.001$ inches}.
\end{align}
It's a bit odd to come within a thousandth of an inch when rounding to the nearest inch, but there's more:
$$
\sqrt{540^2+360^2} = 649 - \cfrac{1}{1298+\cfrac{1}{24073+\cdots}}.
$$
One part in twenty-four thousand??
Did I just happen to be there when someone rolled boxcars two dozen times in a row, or is there more to be said?
(Maybe I should add that $540^2+360^2 = 649^2-1$.)
PS:
$$
\sqrt{540^2+360^2} = 648 + \cfrac{1}{1+\cfrac{1}{1297+\cfrac{1}{25700+\cdots}}} \\
\text{(This part is mistaken; see below.)}
$$
Later edit: A calculator gave me the results above repeatedly; later another calculator disagreed, just as persistently, and I figured out what the truth is.
| More generally, consider $\sqrt{n^2 - 1}$. The square root of $(n^2 - 1)$ should be just slightly less than the square root of $n^2$ (which is $n$), so letting $x = n - \sqrt{n^2 - 1}$, we have $(n-x)^2 = n^2 - 1$ and therefore $2nx - x^2 = 1$, or the distance from the square root to the integer $n$ is
$$x = \dfrac{1}{2n - x}
= \cfrac{1}{2n - \cfrac{1}{2n - x}}
= \cfrac{1}{2n - \cfrac{1}{2n - \cfrac{1}{2n - \cfrac{1}{\dots}}}}
$$
Anyway we don't have to go that far; just the fact that $x = \frac{1}{2n + x} \le \frac{1}{2n-1}$ suffices to show that $x$ must be really small. To come within a thousandth of an inch when rounding to the nearest inch isn't surprising considering this.
Similar to continued fractions, we can also use the binomial theorem:
$$\begin{align}
\sqrt{n^2 - 1}
&= n\sqrt{1 - 1/n^2} = n(1 - 1/n^2)^{1/2} \\
&= n\left(1 - \frac{1}{2n^2} - \frac{1}{8n^4} - \frac{1}{16n^6} - \frac{5}{128n^8} - \dots \right)
\end{align}$$
and so
$$x = n - \sqrt{n^2 - 1} = \frac{1}{2n} + \frac{1}{8n^3} + O\left(\frac{1}{n^6}\right)$$
to see that $x$, the offset from an integer, is itself very close to $\frac{1}{2n}$, which is probably related to what the first calculator (incorrectly) gave in the first case.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given certain conditions, prove a function G(x) is always equal to 4. Theres a question I've been having trouble with:
$G(x)$ is the function $|x+2|+|x-2|$. Show that if $-2<x<2$ then $G(x) = 4$.
Any help would be greatly appreciated. No calculus please :D
| For problems with absolute values, break into cases. Remember,
$$|y|=\begin{cases}\hphantom{-}y & \text{if }y\geq 0,\\ -y & \text{if }y<0.\end{cases}$$
Thus,
$$\begin{align*}
|x+2|&=\begin{cases}\hphantom{-(}x+2 & \text{if }x+2\geq 0,\\ -(x+2) & \text{if }x+2<0\end{cases}\\\\\\
&=\begin{cases}\hphantom{-}x+2 & \text{if }x\geq -2,\\ -x-2 & \text{if }x<-2\end{cases}
\end{align*}$$
and
$$\begin{align*}
|x-2|&=\begin{cases}\hphantom{-(}x-2 & \text{if }x-2\geq 0,\\ -(x-2) & \text{if }x-2<0\end{cases}\\\\\\
&=\begin{cases}\hphantom{-}x-2 & \text{if }x\geq 2,\\ -x+2 & \text{if }x<2.\end{cases}
\end{align*}$$
Putting it all together,
$$\begin{align*}
G(x)&=|x+2|+|x-2|\\\\\\
&=\begin{cases}
\hphantom{-}x+2+|x-2| &\text{if }x\geq -2=\begin{cases}\hphantom{-}x+2+x-2 &\text{if }x\geq -2\textbf{ and }x\geq 2,\\
\hphantom{-}x+2-x+2 &\text{if }x\geq -2\textbf{ and }x<2
\end{cases}\\\\\\\\\\\\\\\\
-x-2+|x-2| &\text{if }x< -2=\begin{cases}-x-2+x-2 &\text{if }x<-2\textbf{ and }x\geq 2,\\
-x-2-x+2 &\text{if }x<-2\textbf{ and }x<2
\end{cases}
\end{cases}\\\\\\
&=\begin{cases}
\hphantom{-}2x & \text{if }x\geq -2\textbf{ and }x\geq 2,\\
\hphantom{-}4 & \text{if }x\geq -2\textbf{ and }x<2,\\
-4 & \text{if }x<-2 \textbf{ and }x\geq 2\; (\color{red}{\text{impossible}}),\\
-2x & \text{if }x<-2\textbf{ and }x<2
\end{cases}\\\\\\\\\\
&=\begin{cases}
\hphantom{-}2x & \text{if }x\geq 2,\\
\hphantom{-}4 & \text{if }-2\leq x<2,\\
-2x & \text{if }x<-2.
\end{cases}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/397118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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How to prove a generalized integral identity $$
\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})}dt=-\frac{1}{4}+\frac{\gamma}{2}
$$
where $\gamma$ = Euler Gamma
$$
\int_{0}^{\infty }\frac{t}{( e^{2\pi t}-1)(1+t^{2}) ^{2}}dt=\frac{\pi^2}{24} -\frac{3}{8}
$$
$$
\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(
1+t^{2})^{3}}dt=\frac{\pi^2}{96} +\frac{\zeta(3)}{8} -\frac{7}{32}
$$
| By
$$
\int_{0}^{\infty }\frac{\arctan \frac{t}{z}}{e^{2\pi t}-1}dt=\frac{z-z\ln z}{2}-\frac{1}{4}\ln \frac{2\pi }{z}+\frac{1}{2}\ln \Gamma (z) [{Re}z>0]
$$
we have
$$
\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right) }dt=\frac{\ln z}{4}-\frac{1}{4z}-\frac{1}{2}\psi \left(
z\right),
$$
$$
\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right) ^{2}}dt=\frac{1}{4z}\psi ^{\prime }\left( z\right) -\frac{%
1}{4z^{2}}-\frac{1}{8z^{3}} ,
$$
$$
\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right) ^{3}}dt=\frac{1}{16z^{3}}\psi ^{\prime }\left( z\right) -
\frac{1}{16z^{2}}\psi ^{\prime \prime }\left( z\right) -\frac{1}{8z^{4}}-
\frac{3}{32z^{5}}
$$
and the recurrence formula
$$
I_{p}(z)=\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left(
z^{2}+t^{2}\right)^p }dt=\frac{-1}{2pz}I_{p-1}^{\prime }(z).
$$
For $\psi ^{(k)}(z)$ we have the following identities
\begin{equation}
\begin{array}{c}
\psi ^{(k)}(n+z)=\psi ^{(k)}(z)+k!(-1)^{k}\sum\limits_{l=1}^{n-1}\frac{1}{%
\left( {l+z}\right) ^{k+1}}, \\
\psi ^{(k)}(z-n)=\psi ^{(k)}(z)+k!\sum\limits_{l=1}^{n}\frac{1}{\left( {l-n-z%
}\right) ^{k+1}}.%
\end{array}
\end{equation}
and
\begin{equation}
\begin{array}{c}
\psi ^{(k)}(0)=\left\{
\begin{array}{c}
-\gamma ,k=0 \\
k{{!(-1)^{k+1}}\zeta (k+1)},k>0%
\end{array}%
\right. , \\
\psi ^{(k)}(\frac{1}{2})=\left\{
\begin{array}{c}
-\gamma -2\ln 2,k=0 \\
k{{!(-1)^{k+1}}}\left( 2^{k+1}-1\right) {\zeta (k+1)},k>0%
\end{array}
\right. , \\
\psi ^{(k)}(\frac{1}{4})=\left\{
\begin{array}{c}
-\gamma -2\ln 2,k=0 \\
-\frac{k{!}}{2\pi }\left( \left( k+2+4^{k+2}\right) {\zeta }%
(k+2)-2\sum\limits_{l=0}^{k/2-1}4^{k-2l}{{{\zeta }(k-2l)\zeta }}(2l+2)\right)
\\
-k{!}2^{k}(2^{k+1}-1){\zeta }(k+1),k\text{ is even}%
\end{array}
\right. , \\
\psi ^{(k)}(\frac{3}{4})=\left\{
\begin{array}{c}
-\gamma -2\ln 2,k=0 \\
\frac{k{!}}{2\pi }\left( \left( k+2+4^{k+2}\right) {\zeta }%
(k+2)-2\sum\limits_{l=0}^{k/2-1}4^{k-2l}{{{\zeta }(k-2l)\zeta }}(2l+2)\right)
\\
-k{!}2^{k}(2^{k+1}-1){\zeta }(k+1),k\text{ is even}%
\end{array}
\right.
\end{array}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How is "n+n/2+n/4....1" equal to "2n-1" using the formula for geometric series? I never knew not having good knowledge of basic maths will be so crippling!! So please help me out this time. I'll be working on my maths from today on.
I was discussing about complexity of an algorithm on StackOverflow and I was told that the series $n+n/2+n/4 + \dots + 1$ evaluates to $2n-1$ and I was linked to the following formula on Wikipedia:
Even after trying hard, I regret to say I still don't get it how using this formula I can conclude that my series evaluates to $2n-1$. Please help me out as I am sure it will take only a few seconds for you.
https://stackoverflow.com/questions/16748454/complexity-for-nested-loops-diving-by-2?noredirect=1#comment24124682_16748454
| Firstly, in the linked StackOverflow question, the program does integer division at each step, so "n/2" in that context actually means the greatest integer less than or equal to $\frac{n}{2}$: more correctly, it should be written as $\left\lfloor \frac{n}{2} \right\rfloor$ (where $\left\lfloor x \right\rfloor$ is the floor function, e.g. $\left\lfloor \frac{7}{2} \right\rfloor = \left\lfloor 3.5 \right\rfloor = 3$).
Secondly, you missed a clause mentioned at the linked StackOverflow question: the correct statement is that if $n$ is a power of $2$, then $\left\lfloor n \right\rfloor + \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{4} \right\rfloor + \dots$ (which in this case is the same as $n + \frac{n}2 + \frac{n}4 + \dots + 1$) is exactly $2n - 1$.
For this special case when $n$ is a power of $2$ (which is what it takes for all the numbers $\frac{n}2, \frac{n}4, \dots$ to be integers, all the way to $1$), this is easy to prove. When $n$ is a power of $2$, say $n = 2^k$, the sum is
$$2^k + \frac{2^k}{2} + \frac{2^k}{4} + \dots + 1
= 2^k + 2^{k-1} + 2^{k-2} + \dots + 1
= 2^{k+1} - 1
= 2n - 1
$$
For example, $16 + 8 + 4 + 2 + 1 = 31$.
This can be proved by induction, or it follows from the formula for the geometric series, which states that $$a + ar + ar^2 + \dots + ar^{m-1} = a\frac{1-r^m}{1-r}.$$
In this case, we have $a = n = 2^k$, $r = 1/2$, and $m$ (the number of terms) is $k+1$, so the left-hand side is the sum $n + \frac{n}{2} + \dots + 1$, and the right-hand side is $$n \frac{1 - (1/2)^{k+1}}{1 - (1/2)} = n\frac{2 - (1/2)^k}{1} = 2n - n/2^k = 2n - 1.$$
But more generally, when $n$ need not be a power of $2$, still we can upper-bound the sum by $n + \frac{n}{2} + \frac{n}{4} + \dots$ (all the way to infinitely many terms). The formula for the infinite geometric series (when $|r| < 1$) is
$$a + ar + ar^2 + \dots = a\frac{1}{1-r}.$$
Here with $a = n$ and $r = 1/2$, we have $$n + \frac{n}{2} + \dots = n\frac{1}{1-1/2} = 2n.$$
As our finite sum is an integer strictly less than this upper bound, we can say it's at most $2n - 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/401937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
Prove $\exists x_m + x_n + x_p \ge 100$
Let $x_i \in \mathbb {R^+}$ (with $i = \overline{1,2,3,\ldots,100}$)
such that: $$x_1^2+x_2^2+x_3^2+\ldots+x_{100}^2 > 10^4 \land x_1+x_2+x_3 + \ldots+x_{100}<300$$
Prove that there exists $x_m+x_n+x_p \ge 100$
(with $m,n,q \in \{1,2,3,\ldots,100\} \land (m-n)(m-p)(n-p) \ne 0\ $)
I see this problem on a math forum. It is nice but I don't know where to start, I can't solve it.
| Suppose $x_1\ge x_2 \ge x_3\ge \dots \ge x_{100}$, and $x_1+x_2+x_3=s < 100$. Then
$$
x_1^2+x_2^2+x_3^2+\dots +x_{100}^2\le x_1^2+x_2^2+x_3^2+\lfloor\frac{300-s}{x_3}\rfloor x_3^2+(300-\lfloor\frac{300-s}{x_3}\rfloor x_3)^2
$$
since it attains the maximal at $(x_1, x_2, x_3, \dots, x_3,x_k,0,\dots)$, and
$$
x_1^2+x_2^2+x_3^2+\lfloor\frac{300-s}{x_3}\rfloor x_3^2+(300-\lfloor\frac{300-s}{x_3}\rfloor x_3)^2 \\
\le (s-2x_3)^2+2x_3^2+ \lfloor\frac{300-s}{x_3}\rfloor x_3^2+(300-\lfloor\frac{300-s}{x_3}\rfloor x_3)^2,
$$
so the problem becomes
If $a\ge b\ge c \ge 0, n\le 98$, $a+2b<100$ and $a+nb+c<300$, then $a^2+n b^2+c^2<100^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/404513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Need help simplifying complicated rational expression. Studying for my final and I can't figure this out.
Simplify:
$$\large\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}$$
| $$\large\frac{\frac{3}{x^3y} + \frac{5}{xy^4}}{\frac{5}{x^3y} -\frac{4}{xy}}$$
$$\large\frac{\frac{3y^3+5x^2}{x^3y^4}}{\frac{5-4x^2}{x^3y}}$$
$$\dfrac{3y^3+5x^2}{x^3y^4}\cdot{\dfrac{x^3y}{5-4x^2}}$$
$$\dfrac{3y^3+5x^2}{{y^3}\cdot{(5-4x^2)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
For any integer $a$, there is an integer $k$ such that $a^2=3k$ or $a^2=3k+1$
Let $a$ be an integer. Prove that there exists an integer $k$ such that $a^2=3k$ or $a^2=3k+1$.
Here is what I did:
I said: $a\in\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}$. When I put 2 into $a^2$, I get $4$ but when I put $2=k$, $3k$, I get the number $4$ on the LHS and the number $6$ on the RHS. But $4$ is not equal to $6$. So how can I prove that $a^2=3k$ equivalent?
I've learned what the division algorithm is. However, when I tried to solve this problem, I got confused.
| Note that any integer can be written as one of $3j+0$, $3j+1$, or $3j+2$
$$
\begin{align}
(3j+0)^2&=9j^2&=3k+0&\text{for }k=3j^2\\
(3j+1)^2&=9j^2+6j+1&=3k+1&\text{for }k=3j^2+2j\\
(3j+2)^2&=9j^2+12j+4&=3k+1&\text{for }k=3j^2+4j+1\\
\end{align}
$$
Therefore, the square of any integer can be written as $3k$ or $3k+1$.
Since the tag modular-arithmetic appears in the question, it should be said that the preceding can be written as
$$
\begin{align}
0^2\equiv0\pmod{3}\\
1^2\equiv1\pmod{3}\\
2^2\equiv1\pmod{3}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/407181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Can I find the equation of an ellipse with these points? How can I solve an ellipse with its major axis on the $x$-axis, given one focus, and two points on the ellipse, one of which I know to be on the major axis?
| Let $F$, $V$, and $P$ be the given focus, the given vertex (point on the major axis), and the other point. We'll assume that $V$ is the "closer" vertex to $F$ (that is, $|VF| = a-c$, where $a$ is the semi-major axis, and $c > 0$ is the focal radius; otherwise, $|VF|=a+c$, and the argument proceeds similarly).
The (fraction-free) standard form of the ellipse is
$$b^2 ( x - h )^2 + a^2 ( y - k )^2 = a^2 b^2$$
where $(h,k)$ is the center, $a>0$ is the semi-major axis, and $b>0$ is the semi-minor axis. With the major axis given to be on the $x$-axis, we have $k=0$. Let us assume that the focus $F$ is positioned at the origin, and that the ellipse's center is to the left of that focus, so that $h = -c < 0$. The equation becomes
$$b^2 ( x + c )^2 + a^2 y^2 = a^2 b^2$$
We therefore need to determine the three quantities $a$, $b$, $c$.
Let us position focus $F$ at the origin, and assign coordinates to the other points thusly: $V(v,0)$ and $P(p,q)$ (with $v > 0$ and $v \ge p$). We have three equations:
$$\begin{align}
v &= a - c &(1) \\
a^2 &= b^2 + c^2 &(2) \\
b^2 ( p + c )^2 + a^2 q^2 &= a^2 b^2 &(3)
\end{align}$$
We use $(1)$ to eliminate $c$ from $(2)$ and $(3)$:
$$\begin{align}
b^2 = a^2 - c^2 = ( a - c )( a + c ) &= v ( 2a - v ) &(4) \\
b^2(p+a-v)^2+a^2q^2 &= a^2 b^2 &(5)
\end{align}$$
... and then use $(4)$ to eliminate $b^2$ from $(5)$:
$$v(2a-v)(p+a-v)^2+a^2q^2 = a^2 v(2a-v)$$
whence
$$a^2 \left(p^2 + q^2- (2v-p)^2\right) + 2 a v\left(v-p\right)\left(2v-p\right) - v^2 \left(v-p\right)^2 = 0$$
so that
$$a = \frac{v \left(v-p\right) \left(-\left( 2v - p \right) \pm \sqrt{p^2+q^2}\right)}{p^2 + q^2- (2v-p)^2} = \frac{v\left(v-p\right)}{2v-p \pm \sqrt{p^2+q^2}}$$
with the "$\pm$" chosen to make $a > v > 0$, and
$$b = \sqrt{\;v(2a-v)\;} \qquad\qquad c = a - v$$
(You should double-check my algebra.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/407317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
finding overlapping permutations I have a data set $3\; 4\; 5\; 6\; 7\; 8\; 9$
I want to find all the permutations that can be formed using this such that neither $7$ nor $8$ is adjacent to $9$.
| We count the complement, the bad permutations where $7$ or $8$ is adjacent to $9$.
How many permutations have $7$ adjacent to $9$? There are $6$ places for the pair to go, and they can be in either order, giving a total of $(6)(2)$ ways to place $7$ and $9$. The rest can be placed in $5!$ ways, for a total of $(6)(2)(5!)$.
Similarly, there are $(6)(2)(5!)$ bad permutations with $8$ adjacent to $9$.
If we add, we have double-counted the permutations where $7$ and $8$ are both adjacent to $9$. to count these, note that the $9$ can be put in $5$ places. For each of these, there are $2$ ways to place the $7$ and $8$, and $4!$ ways to place the rest. So the number of bad permutations is
$$(6)(2)(5!)+(6)(2)(5!)-(5)(2)(4!).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/415610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$1=2$ | Continued fraction fallacy It's easy to check that for any natural $n$
$$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$
Now,
$$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots
=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},$$
$$2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots
=\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.$$
Since the right hand sides are the same, hence $1=2$.
| This is a general feature of continued fractions.
What you are doing is iterating the function
$$ s(x) = \frac{1}{2-x} $$
... so, for instance, you have that
$$ 1 = s(1) = s(s(1)) = s(s(s(1))) = \cdots $$
and
$$ 2 = s(3/2) = s(s(4/3)) = s(s(s(5/4))) = \cdots .$$
The function $s$ has a single fixed point at 1 (first line). It is known that iterating $s$ from any starting point except $x=2$ converges to 1 (think of $s$ as a mapping of the Riemann sphere; it takes $2 \mapsto \infty$ and $\infty \mapsto 0$). For $x<1$, this is easy to check: $x<s(x)<1$.
So, if $x\neq 2$, then
$$ \lim_{n \to \infty} s^n(x) = 1 , $$
where $s^n(x)=s(s(\cdots s(x)))$ is the $n$th iterate of $s$.
But if you start with some $y$, it's always possible to find an $x_n$ so that $s^n(x_n) = y$ (including $\infty$, the map is one-to-one). You've shown that $$s^n\left(\frac{n+2}{n+1}\right) = 2.$$
Another nice thing to note here is that to end up with a limit above 1, we need the numbers at the end of the fractions (the $x_n$) to approach 1 from above: i.e., if for all $n$ that
$$ s^n(x_n) = y > 1 $$
then
$$ x_n \searrow 1 .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "89",
"answer_count": 7,
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How to solve this integral $\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx$? I will be grateful if you would write me a solution procedure for this integral
$$\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx. $$
I am sure that an antiderivative is
$$\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right).$$
Now if I put $+\infty $ instead of $x$ I get
\begin{align*}
\left[ \frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right) \right]^{\infty }_{0}
&= \frac{1}{4} \left( \frac{\infty}{\infty}-\infty \right)-\frac{1}{4} \left( \frac{2 \log 1}{1}-\log 1 \right) \\
&= \frac{1}{4} \left( \frac{\infty}{\infty}-\infty \right).
\end{align*}
As you can see, it is useless. Can you help me please? Thanks
Can I use this solution below?
Let $$I=\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right).$$
Now if I calculate the limit of I i get:
$$\lim_{x\to\infty}I=0$$
So the final result is
\begin{align*}
\left[ \frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right) \right]^{\infty }_{0}=0
\end{align*}
| Putting $x=\frac1y$
$$I=\int_0^{\infty}\frac{x\ln x}{(1+x^2)^2}dx$$
$$=\int_{\infty}^0\frac{(-\ln y)}{y\left(1+\frac1{y^2}\right)^2}\cdot\frac{(-dy)}{y^2}$$
$$=\int_{\infty}^0\frac{\ln ydy}{(1+y^2)^2}$$
$$=-\int_0^{\infty}\frac{\ln dy}{(1+y^2)^2}\text{ as }\int_a^bf(x)dx=-\int_b^af(x)dx$$
$$=-I$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Proving that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$ How am I suppose to prove that:
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$$
Do I use the way like how we count $1+2+ \cdots+100$ to estimate?
So $1/5050 \lt 20$, implying that it is indeed less than $20$?
| Mean Value Theorem can also be used,
Let $\displaystyle f(x)=\sqrt{x}$
$\displaystyle f'(x)=\frac{1}{2}\frac{1}{\sqrt{x}}$
Using mean value theorem we have:
$\displaystyle \frac{f(n+1)-f(n)}{(n+1)-n}=f'(c)$ for some $c\in(n,n+1)$
$\displaystyle \Rightarrow \frac{\sqrt{n+1}-\sqrt{n}}{1}=\frac{1}{2}\frac{1}{\sqrt{c}}$....(1)
$\displaystyle \frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{n}}$
Using the above ineq. in $(1)$ we have,
$\displaystyle \frac{1}{2\sqrt{n+1}}<\sqrt{n+1}-\sqrt{n}<\frac{1}{2\sqrt{n}}$
Adding the left part of the inequality we have,$\displaystyle\sum_{k=2}^{n}\frac{1}{2\sqrt{k}}<\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=\sqrt{n}-1$
$\Rightarrow \displaystyle\sum_{k=2}^{n}\frac{1}{\sqrt{k}}<2\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=2(\sqrt{n}-1)$
$\Rightarrow \displaystyle1+\sum_{k=2}^{n}\frac{1}{\sqrt{k}}<1+2\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=2\sqrt{n}-2+1=2\sqrt{n}-1$
$\Rightarrow \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}<2\sqrt{n}-1$
Similarly adding the right side of the inequality we have,
$\displaystyle\sum_{k=1}^{n}\frac{1}{2\sqrt{k}}>\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})=\sqrt{n+1}-1$
$\Rightarrow \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}>2(\sqrt{n+1}-1)$
Showing that,
$\displaystyle 2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1.\tag 1$
Put $n=100$ to prove,
$\displaystyle 2\sqrt{101}-2<\sum_{k=1}^{100}{\frac{1}{\sqrt{k}}}<2\sqrt{100}-1=19<20.$
| {
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"url": "https://math.stackexchange.com/questions/420733",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "15",
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$3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. Is my induction solution correct? Show using mathematical induction that $3^{3n+1} < 2^{5n+6} $ for all non-negative integers $n$. I'm not sure whether what I did at the last is valid?
Basis step:
for all non-negative integers
$$P(n) = 3^{3n+1} < 2^{5n+6} $$
$$P(0) = 3^{3(0) + 1} = 3 < 64 = 2^{5(0) + 6}$$
$$P(0) = T$$
Inductive Step:
Assume: $3^{3k+1} < 2^{5k+6}$
Show: $3^{3(k+1)+1} < 2^{5(k+1)+6}$
$$ 3^{3(k+1)+1} = 3^{3k+4} = 3^3 \cdot 3^{3k+1}$$
By inductive hypothesis~
$$3^3 \cdot 3^{3k+1} < 3^3 \cdot 2^{5n+6} $$
This is the part where I'm not sure if you can do this in induction but it seems logically correct.
$$3^3 \cdot 2^{5n+6} = 27 \cdot 2^{5n+6}$$
$$2^{5(k+1)+6} = 2^{5k+5+6}= 2^5 \cdot 2^{5n+6} = 32 \cdot 2^{5n+6}$$
I'm not sure whether it should be $\le$ or $<$ but I used '$<$' for $3^3 \cdot 2^{5n+6}<2^{5(k+1)+6} $
Therefore:
$$3^{3(k+1)+1} < 3^3 \cdot 2^{5n+6}<2^{5(k+1)+6} $$
| After your first invokation of the inductive hypothesis you reach
$3^3 \cdot 3^{3k+1} < 3^3 \cdot 2^{5k+6}$
So why not just direction show that $3^3 < 2^5$, completing the proof?
$3^3 \cdot 2^{5k+6} < 2^5 \cdot 2^{5k+6} = 2^{5(k+1)+6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/421362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the limit of a matrix Suppose that $A=\begin{pmatrix}4&1 & 5\\ 2& 7& 1\\ 2& 2& 6\end{pmatrix}.\;$ How can I find $\;\displaystyle \lim_{n\to\infty}A^n$?
What theorem(s) should I use to solve this?
It's harder than what it looks like.....
I know it's stochastic, right?
| Let $B=\det A.$ $B=100$.
If there exists $\lim\limits_{n\to \infty}A^n$, then exists $\lim\limits_{n\to \infty}B^n$.
Neccessary condition is $|B|\leqslant 1$.
I think there is reason to consider scaled matrix
$$\color{#AA0000}{A=\begin{pmatrix}
.4& .1& .5\\
.2& .7& .1\\
.2& .2& .6
\end{pmatrix}},\tag{1}$$
where sum of elements in each row equals to 1.
If there exists limit
$$
M = \lim_{n\to \infty} A^n =
\begin{pmatrix}
a& b& c\\
d& e& f\\
g& h& k
\end{pmatrix},
$$
then
$$
A\cdot M = M,\tag{2}
$$
$$
M\cdot A = M.\tag{3}
$$
(2) implies
$$
\left\{\begin{array}{c}
0.4a+0.1d+0.5g=a;\\
0.2a+0.7d+0.1g=d;\\
0.2a+0.2d+0.6g=g;\\
\end{array}
\right.
$$
$$
\left\{\begin{array}{c}
-0.6a+0.1d+0.5g=0;\\
0.2a-0.3d+0.1g=0;\\
0.2a+0.2d-0.4g=0;\\
\end{array}
\right.
$$
then $a=d=g$.
Same way we get $b=e=h$, $c=f=k$.
So,
$$
M =
\begin{pmatrix}
a& b& c\\
a& b& c\\
a& b& c
\end{pmatrix}.
$$
Now, (3) implies
$$
\left\{\begin{array}{c}
0.4a+0.2b+0.2c=a;\\
0.1a+0.7b+0.2c=b;\\
0.5a+0.1b+0.6c=c;\\
\end{array}
\right.
$$
$$
\left\{\begin{array}{c}
-0.6a+0.2b+0.2c=0;\\
0.1a-0.3b+0.2c=0;\\
0.5a+0.1b-0.4c=0;\\
\end{array}
\right.
$$
then $b=1.4a$, $c=1.6a$.
If we'll use the rule, that sum of elements in each row equals to 1, then
$a+1.4a+1.6a=1$ $\implies$ $a=0.25$ $\implies$ $(a,b,c)=(0.25,\; 0.35,\; 0.4)$.
Answer (with asumption (1)):
$$
\color{#AA0000}{
\lim_{n\to\infty} A^n =
\begin{pmatrix}
0.25& 0.35& 0.4\\
0.25& 0.35& 0.4\\
0.25& 0.35& 0.4
\end{pmatrix}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/422366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Finding the derivative of a relational problem I am self studying some calculus and I have gotten really stuck! I thought I had the right idea but I keep getting the answer totally wrong. I am sure I am missing something important. Here is the problem:
For the equation $6x^{\frac{1}{2}}+12y^{-\frac{1}{2}} = 3xy$, find an equation of the tangent line at the point $(1, 4)$.
Here is my work:
$$6x^{\frac{1}{2}}+12y^{-\frac{1}{2}} = 3xy$$
$$6(x^{\frac{1}{2}}+2y^{-\frac{1}{2}}) = 3xy$$
$$2(x^{\frac{1}{2}}+2y^{-\frac{1}{2}}) = xy$$
$$2x^{\frac{1}{2}}+4y^{-\frac{1}{2}} = xy$$
$$\ln(2x^{\frac{1}{2}})+\ln(4y^{-\frac{1}{2}}) = \ln(xy)$$
$$\ln2 + \ln x^{\frac{1}{2}}+\ln4 + \ln y^{-\frac{1}{2}} = \ln x + \ln y$$
$$\ln2 + \frac{1}{2}\ln x+\ln4 -\frac{1}{2} \ln y = \ln x + \ln y$$
$$\ln2 + \ln4 + \frac{1}{2}\ln x - \ln x = \ln y + \frac{1}{2} \ln y$$
$$\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x = \ln y$$
$$e^{\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x} = y$$
$$\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x = \ln y$$
$$\frac{1}{3x} = y'\frac{1}{y}$$
$$y' = \frac{y}{3x}$$
$$y' = \frac{e^{\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x}}{3x}$$
$$y - 4 = \frac{4}{3}(x-1)$$
$$y = \frac{4}{3}x-\frac{4}{3} + 4$$
$$y = \frac{4}{3}x+\frac{8}{3}$$
The Solution I found was: $y = \frac{4}{3}x+\frac{8}{3}$ but this is wrong! Can you tell me where I have gone wrong?
| From the fourth line to the fifth, there is a mistake. You assumed that $\ln(a+b)=\ln(a)+\ln(b)$. This is not true.
How to do it:
Start from $6x^{1/2}+12y^{-1/2}=3xy$. It is probably a good idea to do no algebraic manipulation. But let's divide through by $3$, getting $2x^{1/2}+4y^{-1/2}=xy$.
Don't wait, differentiate, using implicit differentiation.
We get
$$2\cdot\frac{1}{2}x^{-1/2}-4\cdot\frac{1}{2}y^{-3/2}\frac{dy}{dx}=x\frac{dy}{dx}+y.$$
Now substitute our values of $x$ and $y$ to find $\frac{dy}{dx}$ at the point $(1,4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the inverse Laplace transformation of $\dfrac{s+1}{(s^2 + 1)(s^2 +4s+13)}$ My question is : find the function $f(t)$ that has the following Laplace transform
$$ F(s) = \frac{s+1}{(s^2 + 1)(s^2 +4s+13)} $$
thanks
| The partial fraction decomposition is
$$\frac{1}{20} \cdot \left[ \frac{s+2}{s^2 + 1} \ - \ \frac{s+6}{[s+2]^2 + 3^2 }\right] \ , $$
omitting the somewhat tedious linear algebra $ ^* $ and completing the square in the denominator. Arranging this for interpretation produces
$$\frac{1}{20} \cdot \left[ \frac{s}{s^2 + 1} \ + \ \frac{2 \cdot 1}{s^2 + 1} \ - \ \frac{s+2}{[s+2]^2 + 3^2 } \ - \ \frac{\frac{4}{3} \cdot 3}{[s+2]^2 + 3^2 }\right] \ , $$
from which we can read off the inverse Laplace transform as
$$\frac{1}{20} \cdot \left[ \ \cos t \ + \ 2 \sin t \ - \ e^{-2t} \ \cdot \left( \cos 3t \ + \ \frac{4}{3} \cdot \sin 3t \right) \right] \ . $$
$ ^* $ all right, I did check this afterwards with WolframAlpha...
$$ \\ $$
ADDENDUM (6/19) --
As for setting up the decomposition, we note that the two quadratic factors in the denominator are "irreducible over the real numbers" (both have only complex conjugate zeroes). The sum of "fractions" (actually rational functions) that we set up involve linear functions divided by the irreducible quadratic ones thusly,
$$ \frac{As + B}{s^2 + 1} \ + \ \frac{Cs + D}{s^2 + 4s + 13} \ = \ \frac{s + 1}{(s^2 + 1 ) \cdot (s^2 + 4s + 13)} \ . $$
This leaves us to solve for the four coefficients satisfying
$$ (As + B) \cdot (s^2 + 4s + 13) \ + \ (Cs + D) \cdot (s^2 + 1) \ = \ s + 1 , $$
(so, for instance, the cubic and quadratic terms must "zero out"), which prove to be
$$ A = \frac{1}{20} \ , \ B = \frac{1}{10} \ , \ C = - \frac{1}{20} \ , \ D = - \frac{3}{10} \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve for non-linear system of equations without computer Find $a,b,c \in \mathbb R$ such that:
$$\left\{
\begin{align}
a+b+c &=4 \\
\left( a+b \right) \left( b+c \right) \left( c+a \right) &=18 \\
\frac 1 a+\frac 1b+\frac 1c &=\frac 52
\end{align}
\right.$$
How do I solve for $a,b$ and $c$ manually (without using a computer) ? Thanks !
All solutions are $(1,1,2)$ and permutations
| $$
\begin{align}
a+b+c &=4 \\
\left( a+b \right) \left( b+c \right) \left( c+a \right) &=18 \\
\frac 1 a+\frac 1b+\frac 1c &=\frac 52
\end{align}
$$
It is useful to notice that all expressions here are expressible using symmetric polynomials and, in turn, they are expressible using the elementary symmetric polynomials.
$S_1=a+b+c$, $S_2=ab+ac+bc$, $S_3=abc$
First equation is simply $S_1=4$.
From the second equation we get $(a+b)(b+c)(c+a)=2abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=(ab+ac+bc)(a+b+c)-abc=S_1S_2-S_3=18$.
The third one gives $\frac 1a+\frac 1b+\frac 1c =\frac 52$ $\Leftrightarrow$ $\frac{ab+ac+bc}{abc}=\frac{S_2}{S_3}=\frac52$.
So we have
$$\begin{align}
S_1&=4\\
S_1S_2-S_3&=18\\
S_2&=\frac52S_3
\end{align}$$
which has the solution $S_1=4$, $S_2=5$, $S_3=2$.
Now using Vieta's formulas we get that $a$, $b$, $c$ are solutions of the cubic equation
$$x^3-4x^2+5x-2=0.$$
Now we can apply rational roots test, which reveals that $x=1$ is a solution and then we get $x^3-4x^2+5x-2=(x-1)(x^2-3x+2)=(x-1)^2(x-2)$.
(Or we could employ one of the methods used for solving cubic equations.)
We can check the solution or WolframAlpha or simply verify that $a=b=1$, $c=2$ fulfill the original system. (Of course, the permutations are solutions of the original equations, too.)
| {
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"url": "https://math.stackexchange.com/questions/425907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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On the number of quadratic residues $\pmod{pq}$ where$p$ and $q$ are odd primes. I have read that the formula for the number of quadratic residues $\pmod{pq}$ for odd primes $p$ and $q$ is $\frac{(p-1)(q-1)}{4}$. Is this the case, and if it is, why is it the case and how would one going about proving it? More generally, is there a formula for the number of quadratic residues for any composite modulus $m$ (and if so, why is it so and how would I prove that the formula holds)?
| For any $x$ between $1$ and $pq-1$ which is relatively prime to $pq$, let $f(x)$ be the remainder when $x^2$ is divided by $pq$. We show that the function $f$ is $4$-to-$1$. That yields your formula.
To show that $f$ is $4$-to-$1$, note that $x^2\equiv y^2 \pmod{pq}$ if and only if $(zx)^2\equiv 1\pmod{pq}$, where $z$ is the multiplicative inverse of $y$.
Now all we need to do is to show that the congruence $w^2\equiv 1\pmod{pq}$ has precisely $4$ solutions. Note that $w$ is a solution of the congruence if and only if $w^2\equiv 1\pmod{p}$ and $w^2\equiv 1\pmod{q}$. Each of these congruences has $2$ solutions, so using the Chinese Remainder Theorem we find that the congruence $w^2\equiv 1\pmod{q}$ has $4$ solutions.
Essentially the same argument shows that if $n=p_1^{a_1}\cdots p_k^{a_k}$, where the $p_i$ are distinct odd primes, there are $\frac{\varphi(n)}{2^k}$ numbers $x$ in the interval $[1,n-1]$, relatively prime to $n$, which are squares modulo $n$. Here $\varphi$ is the Euler $\varphi$-function.
Remark: Even $n$ are a little more complicated, one has to treat separately the cases where the highest power of $2$ that divides $n$ is $2^1$ or $2^2$ and the cases where the highest power is $2^k$ for some $k\ge 3$.
Added: In case the reduction to the congruence $w^2\equiv 1 \pmod{pq}$ seems mysterious, here is a more elementary approach. Let $a$ be fixed. We want to find how many solutions there are to the congruence $x^2\equiv a^2\pmod{pq}$.
Rewrite this as $(x-a)(x+a)\equiv 0 \pmod{pq}$. The congruence holds if one of $x-a$ or $x+a$ is divisible by $p$, and one of $x-a$ or $x+a$ is divisible by $q$. The $4$ solutions are obtained by setting (1) $x\equiv a \pmod{p}$, $x\equiv a \pmod{q}$; (2) $x\equiv -a \pmod{p}$, $x\equiv -a \pmod{q}$; (3) $x\equiv a \pmod{p}$, $x\equiv -a \pmod{q}$; (4) $x\equiv -a \pmod{p}$, $x\equiv a \pmod{q}$. Each of these systems of $2$ congruences has a unique solution modulo $pq$ by the Chinese Remainder Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/427997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Roots of $8x^3-4x^2-4x+1$ It is known that the roots of polynomial $8x^3-4x^2-4x+1$ are $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$ and $\cos\frac{5\pi}{7}$.
However this is what Wolfram Alpha/Wolfram Mathematica gives:
$$x = \frac{1}{6}+\frac{7^{2/3}}{3 2^{2/3} \sqrt[3]{-1+3 i
\sqrt{3}}}+\frac{1}{6} \sqrt[3]{\frac{7}{2} \left(-1+3 i
\sqrt{3}\right)}$$
$$x= \frac{1}{6}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1+i
\sqrt{3}\right)}{6 \sqrt[3]{-1+3 i \sqrt{3}}}-\frac{1}{12} \left(1-i
\sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}$$
$$x= \frac{1}{6}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1-i
\sqrt{3}\right)}{6 \sqrt[3]{-1+3 i \sqrt{3}}}-\frac{1}{12} \left(1+i
\sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 i \sqrt{3}\right)}$$
These must be the same roots, but is there a way to show it? Or at least a way to show that these roots given by Wolfram are real?
Can you show that
$$\cos\frac{\pi}{7} = \frac{1}{6}+\frac{7^{2/3}}{3 2^{2/3} \sqrt[3]{-1+3 i
\sqrt{3}}}+\frac{1}{6} \sqrt[3]{\frac{7}{2} \left(-1+3 i
\sqrt{3}\right)}$$
| First, with regard to the appearance of $i$ in the solutions, it's very well known that, when an irreducible cubic has three real roots, then any expression of those roots in terms of radicals must involve the imaginary unit $i$. This is the so called casus irreducibilis and was perhaps the most fundamental historical motivation for the acceptance of complex numbers.
Second, with regard to using Mathematica to see that these expressions are the same, you can use the RootReduce command to put either of these into the same canonical form. Thus, for example, the following input yields zero:
RootReduce[(1+7^(2/3)/((-1+(3*I)*Sqrt[3])/2)^(1/3)+
((7*(-1+(3*I)*Sqrt[3]))/2)^(1/3))/6 - Cos[Pi/7]]
This works in WolframAlpha as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/429883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Evaluating $\int^1_0 \frac2{\sqrt{2-x^2}} dx$ $$\int^1_0 \frac2{\sqrt{2-x^2}} dx$$
using substitution $x=\sqrt 2 \sin \theta$
$$\int^{\pi/4}_0 \frac{2\cos \theta d\theta}{\sqrt{2-2\sin^2 \theta}} = \int^{\pi/4}_0 \frac{2\cos\theta d\theta}{\sqrt2 \cos\theta} = \int^{\pi/4}_0 \frac{2d\theta}{\sqrt2} = \int^{\pi/4}_0 \frac{2\cdot\sqrt2 d\theta}{\sqrt2\cdot\sqrt2} = \int^{\pi/4}_0 \sqrt2d\theta = \sqrt2 \theta = \sqrt2 (\frac\pi4 - 0) = \sqrt2 \frac\pi4 $$
The problem is that the answer is $\frac\pi2$. Where did I make a mistake?
UPDATE:
using substitution $x=\sqrt 2 \sin\theta \rightarrow dx=\sqrt2\cos\theta$
$$\int^{\pi/4}_0 \frac{2\sqrt2\cos\theta}{\sqrt{2-2\sin^2\theta}} = \int^{\pi/4}_0 \frac{2\sqrt2\cos\theta}{\sqrt2 cos\theta} = \int^{\pi/4}_0 \frac{2\sqrt2}{\sqrt2} = \int^{\pi/4}_0 2 = 2 \theta = 2 (\frac\pi4 - 0) = \frac\pi2 $$
| An approach:
$$\int\limits_0^1\frac2{\sqrt{2-x^2}}dx=\frac2{\sqrt2}\int\limits_0^1\frac{dx}{\sqrt{1-\left(\frac x{\sqrt2}\right)^2}}=2\int\limits_0^1\frac{\left(\frac x{\sqrt2}\right)'dx}{\sqrt{1-\left(\frac x{\sqrt2}\right)^2}}=$$
$$=\left.2\arcsin\frac x{\sqrt2}\right|_0^1=2\arcsin\frac1{\sqrt2}=2\left(\frac{\pi}4\right)=\frac\pi2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/432533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Sum of squares of binomial coefficients I came across the following sum in reference to this question
$$\sum_{n=0}^{\infty} \frac{1}{2^{5 n}} \binom{2 n}{n}^2 = \frac{\sqrt{\pi}}{\Gamma \left( \frac{3}{4}\right)^2}$$
The sum on the left was generated from expanding the square root in the integrand of the following elliptic integral:
$$K\left( \frac{1}{2}\right) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\frac12 \sin^2{\theta}}} $$
For the life of me, I cannot figure out how to evaluate this sum directly. Mathematica has no problem in doing so. Can someone point the way?
| I'm assuming you meant $K \left(\frac{1}{\sqrt{2}} \right)$ because that's what you have on the right side of the equation.
$$ \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\frac{1}{2} \sin^{2} x}} \ dx = \sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{2-\sin^{2}x}} \ dx = \sqrt{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1+ \cos^{2} x}} \ dx$$
$$= \sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1+u^{2}}} \frac{du}{\sqrt{1-u^{2}}} = \sqrt{2} \int_{0}^{1} \frac{1}{\sqrt{1-u^{4}}} du $$
$$= \frac{\sqrt{2}}{4} \int_{0}^{1} t^{-\frac{3}{4}} (1-t)^{\frac{-1}{2}} \ dt = \frac{\sqrt{2}}{4} B \left( \frac{1}{4}, \frac{1}{2} \right) = \frac{\sqrt{2}}{4} \frac{\sqrt{\pi} \ \Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/435340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Prove an inequality concerning $\sqrt[3]{4a^3+4b^3}+\sqrt[3]{4b^3+4c^3}+\sqrt[3]{4c^3+4a^3}$ Let $a,b,c$ be positive. I need to prove
$\sqrt[3]{4a^3+4b^3}+\sqrt[3]{4b^3+4c^3}+\sqrt[3]{4c^3+4a^3}\leq \dfrac{4a^2}{a+b}+\dfrac{4b^2}{b+c}+\dfrac{4c^2}{c+a}$
Thanks!
| Observe that
$$\sum \frac{4a^2 - 4b^2} { a+b} = \sum (4a-4b) = 0. $$
Hence, we have
$$\sum \frac{4a^2} { a+b} = \sum \frac{4b^2} {a+b}$$
Hence, we may rewrite the RHS and make it more symmetric, namely
$$\sum \sqrt[3]{4a^3+4b^3} \leq \sum \frac{4a^2} {a+b} = \sum \frac{ 2a^2+2b^2} { a+b}.$$
Now, we can break it up into the individual terms, and just show that
$$ \frac{ a^3+b^3} {2} \left( \frac{a+b}{2} \right)^3 \leq \left( \frac{a^2+b^2} {2} \right)^3$$
This is true because
[Newton] $$S_1 S_3 \leq S_2^2$$
[Maclaurin] $$S_1^2 \leq S_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Prove that: $a^2+b^2+(1-a-b)^2\ge \frac {1}{3}$ Where $a$ and $b$ are any given real number. I have tried solving it using partial derivative.
$$
s=a^2+b^2+(1-a-b)^2$$
$$\frac{\partial s}{\partial a}=2a-2(1-a-b) \tag{1}$$
$$\frac{\partial s}{\partial b}=2b-2(1-a-b) \tag{2}$$
for maxima both (1) and (2) are 0..from here we get two equations from where we get values of $a$ and $b$
$$2a-2(1-a-b)=0 \tag{3}$$
$$2b-2(1-a-b)=0 \tag{4}$$
putting the values of $a$ and $b$ we find from (3) and (4) in the function the maximum value of the function should be $\frac {1}{3}$.which in this case is not
| LHS$ =a^2+b^2+1-2(a+b)+(a+b)^2 \ge \dfrac{(a+b)^2}{2}+1-2(a+b)+(a+b)^2=\dfrac{3}{2}(a+b)^2-2(a+b)+1=\left(\sqrt{\dfrac{3}{2}}(a+b)-\sqrt{\dfrac{2}{3}}\right)^2+\dfrac{1}{3} \ge \dfrac{1}{3} $
first "=" is $a=b$,2nd "=" is $a+b=\dfrac{2}{3} \to a=b=\dfrac{1}{3}$ to get $\dfrac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to find the $x^2+y^2=?$ Suppose that $ x$ and $ y$ are complex numbers, such $x+y=1$ and $x^{20}+y^{20}=20$, Find the sum of all possible values $x^2+y^2$
I find this problem answer is $-90$, But my methods is very very ugly, and I think this Problem have some nice methods.Thank you everyone.
| Let $xy=c$
So, $x^2+y^2=(x+y)^2-2xy=1-2c$
$x^3+y^3=(x+y)^3-3xy(x+y)=1-3c$
On multiplication, $(x^2+y^2)(x^3+y^3)=(1-2c)(1-3c)$
$\implies x^5+y^5+x^2y^2(x+y)=1-5c+6c^2\implies x^5+y^5=1-5c+5c^2$
On squaring, $ x^{10}+y^{10}+2(xy)^5=(1-5c+5c^2)^2$
$\implies x^{10}+y^{10}=1-10c+35c^2-50c^3+25c^4-2c^5$
On squaring, $ x^{20}+y^{20}+2(xy)^{10}=(1-10c+35c^2-50c^3+25c^4-2c^5)^2$
$\implies x^{20}+y^{20}=4c^{10}-100c^9+\cdots+1-2c^{10}$
$\implies 2c^{10}-100c^9+\cdots+1=20\implies 2c^{10}-100c^9+\cdots-19=0$
This is a $10$ degree equation in $c=xy$
Using Vieta's formula, $\sum xy=\frac{100}2=50$
$\implies \sum x^2+y^2=10-2\sum xy=10-2\cdot 50=-90$
| {
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Does $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ converge or diverge? I have to show whether $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ is convergent or divergent. I tried using the squeeze theorem to prove it was convergent. So what I did was
bound ${y_n}$ in between $\frac{-1}{n}$ and $\frac{1}{n}$. That is $\frac{-1}{n} \leq \frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \leq \frac{1}{n}$ Since we also proved earlier that $lim\frac{1}{n}=0$ and $lim\frac{-1}{n}= -lim\frac{-1}{n}=0$ It follows by the squeeze theorem that $lim(y_n)=0$. Would this be correct?
Edit: This what I have now.
Well this what I get so far that $ \sum_{k=n+1}^{2n}\frac{1}{k}≥ \sum_{k=n+1}^{2n}\frac{1}{k+1}$ Hence its divergent. I think it works since $ \sum_{k=n+1}^{2n}\frac{1}{k+1}=\frac{1}{(n+1)+1}+\frac{1}{(n+2)+1}+...+\frac{1}{2n+1}$ ≤ $ \sum_{k=n+1}^{2n}\frac{1}{k}= \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} $.
| Here is how to proceed using the Riemann sum idea
$$ y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}=\sum_{i=1}^{n}\frac{1}{n+i}$$
$$=\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+\frac{i}{n}} \longrightarrow_{n\to \infty} \int_{0}^{1}\frac{dx}{1+x}=\dots.$$
| {
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Find the rectangle with the maximum area inside the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
Find the rectangle with the maximum area inside the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, whose edges are parallel to the axises. Hint: Find the function we need to maximize.
Well in the answers it says that the function we need to maximize is $4xy$ and the area is $2ab$. How do we solve this?
| Another solution is:
the area of rectangle with one vertex $(x,y)$ (in the first quadrant) is $4xy$. From the equation of ellipse we can express $y$ by terms of $x$, as $y=b\sqrt{1-\frac{x^2}{a^2}}$, so the function to maximize is
$$A(x)=4x\cdot b\sqrt{1-\frac{x^2}{a^2}}$$
We can then apply the inequality of geometric and arithmetic means for the positive numbers $\frac{x^2}{a^2}$ and $1-\frac{x^2}{a^2}$ to get the solution.
| {
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Closed form for $a_{n+1} = (a_n)^2+\frac{1}{4}$ I've been given the following sequence:
\begin{align*}
&a_0 = 0; \\
&a_{n+1} = (a_n)^2+\frac{1}{4}.
\end{align*}
I also have to prove that whatever I come up with is correct, but that will likely be the easy part.
Here are the first few values:
\begin{align}
&a_0 = 0 \\
&a_1 = \frac{1}{4}\\
&a_2 = \frac{5}{16}\\
&a_3 = \frac{89}{256} \\
&a_4 = \frac{24305}{65536}
\end{align}
I've managed to to determine that the denominators are of the form $2^{2^n}$. I've tested up to one million terms of this sequence and it appears that $\lim_{n\rightarrow\infty}a_n = \frac{1}{2}$. I spent a while trying to find something of the form $a_n = \frac{P(n)}{Q(n)}$. I haven't had any luck with this, so I started looking into some sums. I've found that
\begin{align*}
a_2 = \frac{1}{4} + \frac{1}{16} = \frac{5}{16}
\end{align*}
and,
\begin{align*}
a_3 = \frac{1}{4} + \frac{1}{16} + \frac{1}{32} + \frac{1}{256} = \frac{89}{256}
\end{align*}
But now,
\begin{align*}
a_4 = \frac{1}{4} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128} + \frac{1}{512} + \frac{1}{1024} + \frac{1}{2048} + \frac{1}{4096} + \frac{1}{65536}= \frac{24305}{65536}.
\end{align*}
So it seems that there is some type of sum involving negative powers of 2 going on, but it isn't clear to me that there is even a pattern here. Any hints/help would be appreciated!
| I just checked OEIS sequence A167424 for which $$f(1)=1/2 \\ f(n+1)=[f(n)^2+1]/2.$$ If this $f$ is divided by 2 you get your sequence $a_1,a_2,\cdots,$
since then it gives correctly $a_1=1/4$ and $a_{n+1}=a_n^2+1/4.$ The recursion on the $a_n=f(n)/2$ follows on dividing through the recursion $f(n+1)=[f(n)^2+1]/2$ by 2 to obtain
$$\frac{f(n+1)}{2}=\left( \frac{f(n)}{2}\right) ^2 +\frac{1}{4}.$$
So anything like closed forms etc. should be extractable from the OEIS page, if it's there.
Side note: The Mandlebrot set intersects the real axis in the interval $[-2,1/4]$. Since it consists of those points $c$ such that the orbit of $0$ under the iteration of $f(x)=x^2+c$ is bounded, we see that your sequence is precisely the (bounded) orbit of the rightmost real point of the Mandlebrot set. (In this sense it seems not surprising there isn't a closed form, as typically Mandlebrot iterations bounce around the set unpredictably. There are periodic points, but it seems $1/4$ is not one of them. (Just thought this connection might be of interest.)
| {
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prove：$\left|x+\frac{1}{x}\right|\geq 2$ prove：
$\left|x+\frac{1}{x}\right|\geq 2$
Can I just use
$\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$
and
$\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$?
| Without loss of genarality we can consider $x\ge 0$(Reason: $x$ and $\frac{1}{x}$ has same sign so if $x<0$ the in place of $x$ we will take $-x(>0)$ and it will not make any difference since $\left| x+\frac{1}{x}\right |=\left| -x+\frac{1}{-x}\right |$)
Then we have,
$\left| x+\frac{1}{x}\right |=x+\frac{1}{x}=(\frac{1}{\sqrt{x}}-\sqrt{x})^2+2\ge 2$
You can use(with its proof which is done by my above arguement) either one(not both ) of the following(if you want to handle only real stuff) according to $x$ is positive or negative,
$\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$ or $\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$
You cant use both since ,if $x$ is positive then $\sqrt{-x}$ is complex.And if $x<0$ then $\sqrt{x}$ is complex.
| {
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For a positive integer $n$, $p_n$ denotes the product of the digits of n, and $s_n$ denotes the sum of the digits of n. For a positive integer $n$, $p_n$ denotes the product of the digits of $n$, and $s_n$ denotes the sum of the digits of $n$. What is the number of integers between 10 and 1000 for which $p_n + s_n = n$ ?
Let $n = xy$ be the two digit number satisfying the given condition.
Given, the product of the digits of $n + \text{sum of the digits of $n$} = n$,
\begin{align}
p_n + s_n &= n\\
xy + x + y &= 10x + y\\
9x – xy &= 0\\
x (9 - y) &= 0.
\end{align}
But $x$ is not zero, because $xy$ is a two digit number.
So, $9 - y = 0 ⇒ y = 9$.
So, $xy$ can be 19, 29, 39, 49, 59, 69, 79, 89, and 99. i.e, 9 numbers.
Let $n = xyz$ be the three digit number satisfying the given condition.
Since $p_n + s_n = n$
\begin{align}
xyz + x + y + z &= 100x + 10y + z.\\
99x + 9y – xyz &= 0.\\
xyz &= 99x + 9y.
\end{align}
| Last equation modulo 11 is:
$xyz \equiv 9(11x + y) \equiv 9y $mode$(11)$ so $xz \equiv 9$mode$(11)$ or $z \equiv 9 x^-1$ mode$(11)$. because $x$ and $z$ are digits, we have pairs $(1,9)$, $(2,6)$, $(3,3)$, $(4,3)$, $(5,4)$, $(6,2)$, $(7,6)$, $(8,8)$, $(9,1)$ for $(x,z)$.
by putting in the original equation we see that there is no solution.
| {
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Why this Equivalence of integrals is true? $$\int_b^{\infty}(1+y^{2})e^{-y^2}\left[\frac{f(y)-f(x)}{y-x}\right]^{2}dy=O(1)\int_b^{\infty}e^{-y^2}\left[f^2(y)+f^2(x)\right]dy$$
enter link description here
go to pg 72 in the end this pg. I don't understand how to get to the product of $O(1)$ by the integral. One idea for my problem.
What happen with $\int_{-b}^{b}(1+y)^2e^{-y^2}\varphi^2(x,y)dy$? converges? why?
thanks
| The integrand can be rewritten as
$$
\begin{align}
&(1+y^{2})e^{-y^2}\left[\frac{f(y)-f(x)}{y-x}\right]^{2} \\
&\qquad = e^{-y^2}\left[f^2(y) - 2f(y)f(x) + f^2(x)\right] \frac{1+y^2}{(y-x)^2} \\
&\qquad = e^{-y^2}\left[f^2(y) + f^2(x)\right]\left[1 - \frac{2f(y)f(x)}{f^2(y)+f^2(x)}\right]\frac{1+y^2}{(y-x)^2}.
\end{align}
$$
Now
$$
\left|1 - \frac{2f(y)f(x)}{f^2(y)+f^2(x)}\right| \leq 1 + \left|\frac{2f(y)f(x)}{f^2(y)+f^2(x)}\right| \leq 2,
$$
which can be seen by maximizing the function
$$
x \mapsto \frac{2ax}{x^2+a^2}.
$$
Also
$$
\lim_{y\to\infty} \frac{1+y^2}{(y-x)^2} = 1
$$
if $x$ is fixed (as is the case in the paper), so the quantity $\frac{1+y^2}{(y-x)^2}$ is bounded for all $y \geq b > x$.
Thus
$$
\left[1 - \frac{2f(y)f(x)}{f^2(y)+f^2(x)}\right]\frac{1+y^2}{(y-x)^2} = O(1),
$$
and so
$$
\int_b^{\infty}(1+y^{2})e^{-y^2}\left[\frac{f(y)-f(x)}{y-x}\right]^{2}dy=O(1)\int_b^{\infty}e^{-y^2}\left[f^2(y)+f^2(x)\right]dy
$$
for fixed $x$ and $b > x$.
| {
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Solution to linear equations as parameterized matrices. I want to solve the following matrix equation:
$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 3 & 2 \\ 4 & 2 \end{pmatrix}=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix}
$$
and I thus get for equations:
$$
\\2a +4b -2c = 0
\\2a + b -2d= 0
\\-3a -c +4d = 0
\\-3b +2c -2d = 0
$$
which reduces to:
$$
\\a+2b-c=0
\\3b-2c+2d=0
$$
Now I want to parameterize this with parameters $s$ and $t$, and intuition tells me I should be able to express the solution as:
$$
T=\begin{pmatrix} a & b \\ c & d \end{pmatrix}=s\begin{pmatrix} x & x \\ x & x \end{pmatrix}+t\begin{pmatrix} x & x \\ x & x \end{pmatrix}
$$
(where $x_i$ is som natural number)
This should be similar to parameters with vectors, but I havn't been able to solve this.
So whats the trick here?
Thanks!
/Alexander
| Choose two variables to represent $s$ and $t$. I'll use $a=s$ and $b=t$.
$$\begin{align}c&=s+2t\\
\\
2d & = 2c-3t \\
2d & = 2(s+2t)-3t \\
2d & = 2s + t \\
d &= s + \frac{1}{2}t
\end{align}
$$
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} s & t \\ (s+2t) & (s+\frac{1}{2}t) \end{bmatrix}=\begin{bmatrix} s & 0 \\ s & s \end{bmatrix}+\begin{bmatrix} 0 & t \\ 2t & \frac{1}{2}t \end{bmatrix}$$
| {
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Sums $\sum_{k=1}^n \sin(2k-1)\theta$, $\sum_{k=1}^n \sin^2(2k-1)\theta $ To prove:
$1.$ $$\sum_{k=1}^n \sin(2k-1)\theta = \frac{\sin^2 n\theta}{\sin \theta}.$$
$2.$ $$\sum_{k=1}^n \sin^2(2k-1)\theta = \frac{n}{2} - \frac{\sin 4n\theta}{4\sin 2\theta}.$$
| Alternatively, from $\cos(a+b)=\cos a\cos b-\sin a\sin b$ we get $$\sin a\sin b=\frac{\cos(b-a)-\cos(b+a)}2$$
Giving appropriate values to $a,b$, obtain a telescopical sum on the left.
Let $a=x,b=(2k-1)x$. Then $$\eqalign{
& \sin x\sin \left( {2k - 1} \right)x = \frac{{\cos (2k - 2)x - \cos 2kx}}{2} \cr
& \sin x\sum\limits_{k = 1}^n {\sin \left( {2k - 1} \right)x} = \frac{{1 - \cos 2nx}}{2} \cr} $$
so $$\sum\limits_{k = 1}^n {\sin \left( {2k - 1} \right)x} = \frac{{{{\sin }^2}nx}}{{\sin x}}$$
For the second one, note that $${\sin ^2}(2k - 1)\theta = \frac{{1 - \cos \left( {2k - 1} \right)2\theta }}{2}$$ so it suffices to find $$\sum\limits_{k = 1}^n {\cos \left( {2k - 1} \right)x} $$
And again, we can use $$\frac{{\sin \left( {a + b} \right) - \sin \left( {a - b} \right)}}{2} = \sin b\cos a$$ to get $$\sin x\cos \left( {2k - 1} \right)x = \frac{{\sin 2kx - \sin \left( {2k - 2} \right)x}}{2}$$ $$\sum\limits_{k = 1}^n {\cos \left( {2k - 1} \right)x} = \frac{{\sin 2nx}}{{2\sin x}}$$
so $$\sum\limits_{k = 1}^n {{{\sin }^2}(2k - 1)x} = \frac{n}{2} - \frac{{\sin 4nx}}{{4\sin 2x}}$$
ADD Note that above gives $$\sum\limits_{k = 0}^{n - 1} {\frac{{\sin \left( {k + \frac{1}{2}} \right)t}}{{\sin \frac{t}{2}}}} = {\left( {\frac{{\sin \frac{{nt}}{2}}}{{\sin \frac{t}{2}}}} \right)^2}$$ which is fundamental when computing the Fejér Kernel as the Cesaro mean of the Dirichlet Kernel.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Series expansion of $\sqrt{\log(1+x)}$ at $x=0$ Mathematica gives the following series expansion of $\sqrt{\log(1+x)}$ at $x=0$.
$$
x^{1/2}-\frac{1}{4}x^{3/2}+\frac{13}{96}x^{5/2}-\cdots
$$
You can find it from Wolfram alpha too.
How can I obtain the expansion?
Obviously Taylor expansion is impossible because $\sqrt{\log(1+x)}$ is not analytic at $x=0$.
Taylor expansion of the $\log(1+x)$ at $x=1$ is possible. But I don't know how to take sequre root on the expanded series.
I think I have not learned about square root of a series from calculs or analysis course.
From what material can I study about such things?
| We have
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
and recall that
$$(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)x^2}{2}+\frac{\alpha(\alpha-1)(\alpha-2)x^3}{3}+\cdots$$
so for $\alpha=\frac{1}{2}$ we have
$$\sqrt{\log(1+x)}=\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4)\right)^{1/2}=\sqrt{x}\left(1+(\underbrace{-\frac{x}{2}+\frac{x^2}{3}+O(x^3))}_{=u}\right)^{1/2}\\=\sqrt{x}(1+\frac{1}{2}u-\frac{1}{8}u^2+O(u^3))=\sqrt{x}(1-\frac{x}{4}+\frac{x^2}{6}-\frac{1}{8}\frac{x^2}{4}+O(x^3))\\=\sqrt{x}(1-\frac{x}{4}+\frac{13x^2}{96}+O(x^3))$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find a point on a parabola that's closest to another point. Find the point on the parabola $3x^2+4x-8$ that is closest to the point $(-2,-3)$.
My plan for this problem was to use the distance formula and then that the derivative to get my answer. I'm having a little trouble along the way.
$$ d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$$
| $$ d = \sqrt{(x - (-2))^2 + ((3x^2+4x-8) - (-3))^2} = \sqrt{\left(x+2\right) ^2 + \left(3x^2+4x-5 \right)^2}$$
Instead of minimizing $d$, lets minimize $d^2$. It should be noted that the minimum of $d$ and $d^2$ are exactly the same as $d \geq 0$. This gets rid of the square root, thereby simplifying the problem.
$$d^2 = (x+2)^2 + (3x^2+4x-5)^2.$$
Now, we can use calculus to minimize $d^2$,
$$(d^2)' = 2(x+2) + 2(3x^2+4x-5)(6x+4) = 36x^3 + 72x^2 - 26x - 36$$
This has roots at $x \approx -2.118, -.631, .749$. We know that one of these must be the absolute minimum, so evaluate $d^2$ for each $x$-value and whichever is smallest is the $x$-value of the point on the parabola nearest to the given point. You should then plug that $x$-value into the equation for the parabola to determine the $y$-value where this occurs. I got,
Ans: $ (-2.118, -3.014) $
| {
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Show that the vertex is the point on a branch of a hyperbola that is closest to the focus associated with that branch Given the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,$$ show that the point closest to focus $F(c, 0)$ where $c^2=a^2+b^2$ is the vertex $V(a, 0)$
| Solving for $x$ and considering the right branch where $x>0$
$$x = a\sqrt{1+\frac{y^2}{b^2}}$$
The distance between $F(c,0)$ and $P\left(a\sqrt{1+\frac{y^2}{b^2}}\;,\; y\right)$ is
$$\sqrt{\left(\frac{a}{b}\sqrt{b^2+y^2} - c\right)^2 + y^2}$$
The derivative of the expression under the radical is
$$2y\left(\frac{a^2}{b^2} + \frac{ac}{b\sqrt{b^2+y^2}} + 1\right)$$
The derivative has only one solution, $y=0$ and by the derivative test, we can see that the minimum value for the distance occurs at $y=0$, $x=a$.
| {
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Absolute extrema of the function $f(x,y)=2xy-x-y$ Find the absolute extrema of the function $$f(x,y)=2xy-x-y$$ over the region of the $xy$-plane bounded by the parabola $y=x^2$ and the line $y=4.$
I was wondering if I needed to use Lagrange multipliers to solve this problem and if I do, how would I go about solving this problem? If someone could help me, that would be great! Thanks
| To return to this problem, the reason the Lagrange-multiplier method won't be enough (or possibly even necessary) to solve the problem is that it "looks" for points on a level curve or surface of the function to be extremized where the normal vector is parallel (or anti-parallel) to a normal vector to the constraint curve or surface. So it can help with the boundary of a region, but may not tell us much about the interior.
When a closed region is specified on which to locate extrema, to be thorough, we follow a procedure analogous to what we use in single-variable calculus for locating extrema on a closed interval: find critical points in the interior of the region using derivatives; find critical points on bounding curves of the region (often also using derivatives or Lagrange multipliers); evaluate the function at vertices; compare the values of the function among these results.
The first partial derivatives of the function $ \ f(x, \ y) \ = \ 2xy \ - \ x \ - y \ $ are $ \ f_x \ = 2y \ - \ 1 \ $ and $ \ f_x \ = 2x \ - \ 1 \ $ ; these are both equal to zero at $ \ ( \ \frac{1}{2} , \ \frac{1}{2} \ ) \ $ , so this is the only critical point within the parabolic region. The value of the function there is $ \ f( \ \frac{1}{2} , \ \frac{1}{2} \ ) \ = \ \frac{1}{2} \ - \ \frac{1}{2} \ - \ \frac{1}{2} \ = \ -\frac{1}{2} \ $ .
The two "bounding curves" of the region are the parabola $ \ y \ = \ x^2 \ $ and the "horizontal" line $ \ y \ = \ 4 \ $ . We can use these equations to reduce the extremization problem to one with single-variable functions:
$ \mathbf{on \ y = 4 :} \quad f(x, \ 4) \ = \ 8x \ - \ x \ - 4 \ = \ 7x \ - \ 4 \ \ \Rightarrow \ \ \frac{df}{dx} \ = \ 7 \ \ ; $
$ \mathbf{on \ y = x^2 :} \quad f(x, \ x^2) \ = \ 2x^3 \ - \ x \ - \ x^2 \ \ \Rightarrow \ \ \frac{df}{dx} \ = \ 6x^2 \ - \ 2x \ - \ 1 \ \ . $
There is no critical point on the line, since the derivative is never zero there; the fact that the function $ \ f(x, \ 4) \ $ increases linearly with $ \ x \ $ suggests that the endpoints of this boundary segment may be extrema. On the parabola, $ \frac{df}{dx} \ = \ 6x^2 \ - \ 2x \ - \ 1 \ = \ 0 \ $ for $ \ x \ = \ \frac{2 \ \pm \ \sqrt{28}}{12} \ = \ \frac{1 \ \pm \ \sqrt{7}}{6} \ $ . The parabola intersects the line at $ \ ( \pm 2, \ 4) \ $ , so both of these values of $ \ x \ $ do correspond to locations on the restricted parabolic boundary: $ \ \left( \frac{1 \ - \ \sqrt{7}}{6} , \ \frac{4 \ - \ \sqrt{7}}{18} \right) $ and $ \ \left( \frac{1 \ + \ \sqrt{7}}{6} , \ \frac{4 \ + \ \sqrt{7}}{18} \right) \ $ . At these points, the value of our function is
$$ f \left( \frac{1 \ - \ \sqrt{7}}{6} , \ \frac{4 \ - \ \sqrt{7}}{18} \right) \ = \ \frac{7 \sqrt{7} \ - \ 10}{54} \ \approx \ 0.158 $$
and
$$ f \left( \frac{1 \ + \ \sqrt{7}}{6} , \ \frac{4 \ + \ \sqrt{7}}{18} \right) \ = \ -\frac{7 \sqrt{7} \ + \ 10}{54} \ \approx \ -0.528 \ \ . $$
Finally, we need to evaluate the function at the vertices of the parabolic region:
$$ f ( -2 , \ 4 ) \ = \ -16 \ - \ (-2) \ - 4 \ = \ -18 \ \ \ \text{and} \ \ \ f ( 2 , \ 4 ) \ = \ 16 \ - \ 2 \ - 4 \ = \ 10 \ \ , $$
which are just the values that $ \ 7x \ - \ 4 \ $ gives us. These are by far the greatest and least values of the function, so they are the absolute extrema for the region.
Had we applied Lagrange multipliers, we would need to treat the two portions of the boundary separately. The gradient of our function is $ \ \nabla f \ = \ \langle \ 2y \ - \ 1 \ , \ 2x \ - \ 1 \ \rangle \ $ . The constraint function for the parabolic curve is $ \ g(x,y) \ = \ y \ - \ x^2 \ $ , so its gradient is $ \ \nabla g \ = \ \langle \ -2x \ , \ 1 \ \rangle \ $ . From the Lagrange equation $ \ \nabla f \ = \ \lambda \nabla g \ $ , we obtain
$$ 2y \ - \ 1 \ = \ \lambda \ \cdot \ ( \ -2x \ ) \ \ , \ \ 2x \ - \ 1 \ = \ \lambda \ \cdot \ 1 $$
$$ \Rightarrow \ \ \lambda \ = \ -\frac{2y \ - \ 1}{2x} \ = \ 2x \ - \ 1 \ \ \Rightarrow \ \ 2y \ = \ 1 \ + \ 2x \ - \ 4x^2 \ \ ; $$
applying the constraint $ \ y \ = \ x^2 \ $ gives us the quadratic equation we found earlier, $ \ 6x^2 \ - \ 2x \ - \ 1 \ = \ 0 \ $ , leading to the extrema on the curve that we determined above.
As for the horizontal line, $ \ y \ = \ 4 \ $ , Lagrange multipliers will not give us anything meaningful in this case. It would seem that we would use a constraint function $ \ h(x, \ y) \ = \ y \ - \ 4 \ $ , for which the gradient is $ \ \nabla h \ = \ \langle \ 0, \ 1 \ \rangle \ $ . The consequent equations from $ \ \nabla f \ = \ \lambda \nabla g \ $ are $ 2y \ - \ 1 \ = \ \lambda \ \cdot \ 0 \ \ , \ \ 2x \ - \ 1 \ = \ \lambda \ \cdot \ 1 $ . But since we require $ \ y \ = \ 4 \ $ , this set of equations has no solution, implying that there is no level curve of $ \ f(x, \ y) \ $ which has a normal in the "vertical direction" where it touches this line. While this confirms that there is no critical point for the function along that line segment, it doesn't do much else for us.
| {
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If $p =\frac{4\sin\theta \cos\theta}{\sin\theta +\cos\theta}$ Find the value of $\frac{p+2\sin\theta}{p-2\sin\theta}$ Problem :
If $\displaystyle p =\frac{4\sin\theta\cos\theta}{\sin\theta +\cos\theta}$, find the value of $\displaystyle \frac{p+2\sin\theta}{p-2\sin\theta} + \frac{p+2\cos\theta}{p-2\cos\theta}$.
Please help how to proceed in such problem..Thanks..
| Note that $$\frac{p}{2\cos\theta}=\frac{2\sin \theta}{\cos\theta+\sin\theta}$$ and
$$\frac{p}{2\sin\theta}=\frac{2\cos \theta}{\cos\theta+\sin\theta}$$
Now apply the componendo-dividendo method to the first and second equalities and sum them up, you'll get the answer. It'll be $2$.
| {
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Find the point of intersection of the straight line $\frac{X+1}{4}=\frac{Y-2}{-2}=\frac{Z+6}{7}$ and plane $3X+8Y-9Z=0$ Find the point of intersection of the straight line
$$\frac{X+1}{4}=\frac{Y-2}{-2}=\frac{Z+6}{7}$$ and plane $3X+8Y-9Z=0$
the point of the line is $M(-1,2,-6)$ and direction vector of the line is $A(4,-2,7)$
I would like to get some advice how to do that.
Thanks!
| $$
\frac {X+1}4 = \frac {Y-2}{-2} = \frac {Z+6}7 = k \\
X = 4k - 1 \\
Y = -2k + 2 \\
Z = 7k - 6
$$
Substitute it to the equation of plane
$$
3X+8Y-9Z = 0 \\
12k-3 - 16k + 16 - 63k + 54 = 0 \\
-67k + 67 = 0 \\
k = 1
$$
Since you know $k$, you can easily find $X,Y,Z$.
$$
X = 4k-1 = 3 \\
Y = -2k+2 = 0 \\
Z = 7k-6 = 1
$$
So, point of intersection is $(3,0,1)$.
| {
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what is$ \sqrt{8i}$ Very simple question with an answer that I cannot understand:
I have $\sqrt{8i}$, which, I suppose, is the same as $\sqrt{\sqrt{-64}}$.
How come that $2+2i$ is the same as $\sqrt{8i}$?
My understanding is that $\sqrt{8i}$ is the same as:
(a) $\sqrt{2^3i}$
OR
(b) $2\sqrt{2i}$
I'm pretty sure (a) is correct and (b) might also be correct, but how can you get from there to $2+2i$?
Thanks in advance
| let $$x+iy=\sqrt{8i}$$
$$(x+iy)^2={8i}$$
$$x^2-y^2+2xyi=0+8i$$
compare real and imaginary parts of both sides:
$x^2-y^2=0\;; 2xy=8$
$$x^2-y^2=0$$
square both sides
$$\implies (x^2-y^2)^2 =0\implies (x^2+y^2)^2-(2xy)^2=0$$
$$\implies (x^2+y^2)^2=64 \implies (x^2+y^2)=\pm 8 $$
since sum of square of two real no. can't be negative so
$$x^2+y^2=8\text{ and }x^2-y^2=0$$
solving these two eqn will give $x=y=\pm 2 $
so $$\sqrt {8i}=\pm 2(1+i)\implies 2+2i,-2-2i$$
| {
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How many zeroes at the end of $3^34^45^56^6 - 3^64^55^46^3$? The quantity $3^34^45^56^6 - 3^64^55^46^3$ will end in how many zeros?
This is a GRE Practice question and I have to do it without using a calculator. Can anyone help me on this?
| Using $a^m\cdot a^n=a^{m+n}; a^{mn}=(a^m)^n$ and $(a\cdot b)^m=a^m\cdot b^m$ where $a,b,m$ are real numbers
$$3^34^45^56^6 - 3^64^55^46^3$$
$$=3^3(2^2)^45^5(2\cdot3)^6 - 3^6(2^2)^55^4(2\cdot3)^3$$
$$= 2^{8+6}5^53^{6+3} - 3^{3+6}2^{10+3}5^4$$
$$=2^{13}5^43^9(2\cdot5-1)$$
As $10=2\cdot5,$ the highest power of $10$ will be min $(13,4)$
| {
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Computing Galois Group of $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ Galois Group of $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ is cumbersome to computing. Is easy to find the all four possible candidates but is cumberstone to show that they are automorphisms:
For multiplication is too long showing that
$f((a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a'+b'\sqrt{2}+c'\sqrt{3}+d'\sqrt{6}))=f(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})f(a'+b'\sqrt{2}+c'\sqrt{3}+d'\sqrt{6})$
For some non trivial candidate since the left hand side expands in $16$ terms. What about $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$? there are $7$ non trivial candidates and multiplication expands in $64$ terms!
How to do it in a tricker form?
Edit(Let me be more precise in my question):
I know that if $f\in\Gamma(\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q})$, then $f(\sqrt{2})=\pm\sqrt{2}$, $f(\sqrt{3})=\pm\sqrt{3}$ so there are $4$ candidates for $\mathbb{Q}$-Automorphisms. Candidates are given by:
$f_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$
$f_2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$
$f_3(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}$
$f_4(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}$
But I need to check that they are $\mathbb{Q}$-Automorphisms in $\mathbb Q(\sqrt{2},\sqrt{3})$. All candidates $f_i$ fixes $\mathbb{Q}$ and is obvious that all preserves sums, $f_1$ obviously preserves product also since is the identity, but is cumberstone to show that $f_i(xy)=f_i(x)_if(y)$ to conclude $f_i\in\Gamma(\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q})$.
| $\mathbb{Q}(\sqrt 2, \sqrt 3)$ is the splitting field of $(x^2 - 2)(x^2 - 3)$ over $\mathbb{Q}$. Hence, the extension $\mathbb{Q}(\sqrt 2, \sqrt 3) / \mathbb{Q}$ is Galois. Since $\sqrt 2$ and $\sqrt 3$ are not linearly dependent, we have $[\mathbb{Q}(\sqrt 2, \sqrt 3) : \mathbb{Q}] = 4$. Thus, the Galois group $G$ has order $4$.
We know that automorphisms in $G$ permute the roots of $x^2 - 2$ and $x^2 - 3$. It's easy to verify that
$$
\sigma = \begin{cases}
\sqrt 2 \mapsto -\sqrt 2 \\
\sqrt 3 \mapsto \sqrt 3
\end{cases}
$$
and
$$
\tau = \begin{cases}
\sqrt 2 \mapsto \sqrt 2 \\
\sqrt 3 \mapsto -\sqrt 3
\end{cases}
$$
are members of $G$. Thus, $G = \{1, \sigma, \tau, \sigma \tau\}$. I'll let you find its isomorphism type.
For $\mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt 5)$, follow the same method, as it's the splitting field of $(x^2 - 2)(x^2 - 3)(x^2 - 5)$. Consider the automorphisms:
$$
\begin{cases}
\sqrt 2 \mapsto -\sqrt 2 \\
\sqrt 3 \mapsto \sqrt 3 \\
\sqrt 5 \mapsto \sqrt 5
\end{cases}
$$
and
$$
\begin{cases}
\sqrt 2 \mapsto \sqrt 2 \\
\sqrt 3 \mapsto -\sqrt 3 \\
\sqrt 5 \mapsto \sqrt 5
\end{cases}
$$
and
$$
\begin{cases}
\sqrt 2 \mapsto \sqrt 2 \\
\sqrt 3 \mapsto \sqrt 3 \\
\sqrt 5 \mapsto -\sqrt 5
\end{cases}.
$$
| {
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Infinite solutions of Pell's equation $x^{2} - dy^{2} = 1$
Let $d > 1$ be a squarefree integer. Prove that the equation $x^{2} - dy^{2} = 1$ has infinitely many solutions in $\mathbb{Z} \times \mathbb{Z}$.
What I have done: let $ \ \mathbb{K} = \mathbb{Q}[\sqrt{d}]$. If $ d \not\equiv 1$ mod $4$, then $O_{\mathbb{K}} = \mathbb{Z}[\sqrt{d}]$ and the statement follows by the Dirichlet Unit Theorem. What about the case $ d \equiv 1 $ mod $4$ ?
| You can handle the case $d\equiv 1 \pmod{4}$ in essentially the same way. For any unit $u$, $u^6=a+b\sqrt{d}$ with $a,b\in\mathbb Z$ and $a^2-db^2=1$, then infinitude follows from Dirichlet's Unit Theorem.
If $u$ is a unit then $u^2=\frac{\alpha}{2}+\frac{\beta}{2}\sqrt{d}$ with $\alpha,\beta\in \mathbb Z$, $\alpha \equiv \beta \pmod{2}$ (by definition of $O_{\mathbb K}$) and
$$
N(u^2)=\frac{\alpha^2}{4}-\frac{d\beta^2}{4}=1
$$
Then either:
Case 1 $\alpha\equiv\beta\equiv 0 \pmod{2}$. Then $\frac{\alpha}{2},\frac{\beta}{2}\in\mathbb Z$ and it follows from multiplication in $\mathbb K$ that $\exists a,b\in \mathbb Z$ with $u^6 =a+b\sqrt{d}$.
Case 2 $\alpha\equiv\beta\equiv 1 \pmod{2}$. Since $\alpha^2-d\beta^2=4$ and $\alpha^2\equiv\beta^2\equiv 1 \pmod{8}$ this case cannot arise unless $d\equiv 5\pmod{8}$. Then
$$
u^6 = \frac{1}{8}\left(\alpha(\alpha^2+3d\beta^2)+\beta(3\alpha^2+d\beta^2)\sqrt{d}\right) \\
\alpha^2+3d\beta^2\equiv 3\alpha^2+d\beta^2\equiv 0 \pmod{8} \\
a=\frac{\alpha(\alpha^2+3d\beta^2)}{8},b=\frac{\beta(3\alpha^2+d\beta^2)}{8}
$$
and $a,b\in\mathbb Z$.
| {
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Simplify $\left({\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}\right)\left({\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}\right)^{-1}$ Simplify $$\frac{\displaystyle\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}{\displaystyle\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}$$
I don't have any good idea. I need your help.
| Here is my answer.
I've just got the following result:$$\frac{\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt k}}}}{\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt k}}}}=1+\sqrt2+\sqrt{4+2\sqrt2}=\cot\frac{\pi}{16}.$$
Proof: Suppose that $\sum$ represents $\sum_{k=1}^{2499}$. Let the numerator and the denominator be $A$ and $B$ respectively. Letting $a_k=\sqrt{10+\sqrt{50+\sqrt k}}, b_k=\sqrt{10-\sqrt{50+\sqrt k}}$, we can represent $A, B$ as $A=\sum a_k, B=\sum b_k.$ Letting $p_k=\sqrt{50+\sqrt k}$ and $q_k=\sqrt{50-\sqrt k}$, since ${p_k}^2+{q_k}^2=10^2$ and $p_k\gt0, q_k\gt0$, there exists a real number $0\lt x_k\lt \frac{\pi}{2}$ such that $p_k=10\cos x_k, q_k=10\sin x_k$. Then, we get $$a_k=\sqrt{10+10\cos x_k}=\sqrt{10+10\left(2{\cos^2{\frac{x_k}{2}}}-1\right)}=\sqrt{20}\cos \frac{x_k}{2},$$$$b_k=\sqrt{10-10\cos x_k}=\sqrt{10-10\left(1-2{\sin^2{\frac{x_k}{2}}}\right)}=\sqrt{20}\sin \frac{x_k}{2}.$$ Then, since $\sum a_k=\sum a_{2500-k}$, let's consider $a_{2500-k}$. $$\begin{align}a_{2500-k}&=\sqrt{10+\sqrt{50+\sqrt{(50+\sqrt k)(50-\sqrt k)}}}\\&=\sqrt{10+\sqrt{50+{p_kq_k}}}\\&=\sqrt{10+\sqrt{50+100\cos {x_k}\sin {x_k}}}\\&=\sqrt{10+\sqrt{50(\cos {x_k}+\sin {x_k})^2}}\\&=\sqrt{10+\sqrt{50}\cdot\sqrt2\sin \left(x_k+\frac{\pi}{4}\right)}\\&=\sqrt{10+10\cdot2\cos \left(\frac{x_k}{2}+\frac{\pi}{8}\right)\sin \left(\frac{x_k}{2}+\frac{\pi}{8}\right)}\\&=\sqrt{10\left(\cos \left(\frac{x_k}{2}+\frac{\pi}{8}\right)+\sin \left(\frac{x_k}{2}+\frac{\pi}{8}\right)\right)^2}\\&=\sqrt{10}\left(\cos \left(\frac{x_k}{2}+\frac{\pi}{8}\right)+\sin \left(\frac{x_k}{2}+\frac{\pi}{8}\right)\right)\\&=\frac{\left(\cos \left(\frac{\pi}{8}\right)+\sin \left(\frac{\pi}{8}\right)\right)a_k+\left(\cos \left(\frac{\pi}{8}\right)-\sin \left(\frac{\pi}{8}\right)\right)b_k}{\sqrt2}\\&=\sqrt{\frac{\sqrt2+1}{2\sqrt2}}a_k+\sqrt{\frac{\sqrt2-1}{2\sqrt2}}b_k.\end{align}$$ Hence, $$A=\sqrt{\frac{\sqrt2+1}{2\sqrt2}}A+\sqrt{\frac{\sqrt2-1}{2\sqrt2}}B.$$
So, the proof is completed with $$\frac AB=1+\sqrt2+\sqrt{4+2\sqrt2}. $$
After a while, I got the following theorem in the same way as above.
Theorem: For any natural number $n$, $$\frac{\sum_{k=1}^{n^2+2n}\sqrt{\sqrt{2n+2}+{\sqrt{n+1+\sqrt k}}}}{\sum_{k=1}^{n^2+2n}\sqrt{\sqrt{2n+2}-{\sqrt{n+1+\sqrt k}}}}=1+\sqrt2+\sqrt{4+2\sqrt2}=\cot {\frac{\pi}{16}}.$$
Note that the case $n=49$ in this theorem is the question at the top.
P.S. I think it's worth adding a link where user mercio provided a background why the theorem holds for any $n$. (it's a background, not a proof. The theorem has already been proved.)
| {
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How to find the value of $a$ for which $\;\tan^2x + (a+1)\tan x-(a-3)<0$ is true I wanted to know, how can I find the value of $a$ for which the inequality $\tan^2x + (a+1)\tan x-(a-3)<0$ is true for at least one $x\in(0,\pi/2)$.
I don't know how to proceed, any help is appreciated.
| Set $\tan x = y$. Note that $y$ can take only positive values since $x\in (0,\dfrac{\pi}{2})$.
Then $$y^2 + (a+1)y - (a-3) <0$$
$$\iff\left(y+\dfrac{a+1}2 \right)^2 - \dfrac{(a+1)^2}4- (a-3) <0\quad \text{(completing the square)}$$
$$\iff \left(y+\dfrac{a+1}2 \right)^2 < \dfrac{a^2+6a-11}4 = \dfrac{(a+3)^2-20}4$$
So,
$$ -\sqrt{\dfrac{(a+3)^2-20}4}<y+\dfrac{a+1}2 < \sqrt{\dfrac{(a+3)^2-20}4}$$
$$\iff -\sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2<y < \sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2$$
We just need to make sure that there exists a positive $y$ that satisfies this.
So we need to make sure that the upper bound is positive.
Thus, $\sqrt{\dfrac{(a+3)^2-20}4}-\dfrac{a+1}2>0\iff \sqrt{(a+3)^2-20}>a+1$
If $a+1$ negative then the inequality is true, and if $a+1\geq0$, we can square both sides and simplify to obtain $a>3$ which is consistent with $a+1\geq0$. Also, we need $(a+3)^2-20$ to be non-negative, so $(a+3)^2\geq20\iff a\geq\sqrt{20}-3 \text{ or } a\leq-\sqrt{20}-3$.
Thus,
$ a>3 \text{ or } a\leq-\sqrt{20}-3$.
| {
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Using the Legendre Symbol Which have solutions? $x^{2} \equiv7 \mod{53}$, $x^{2} \equiv53 \mod{7}$, $x^{2} \equiv 14 \mod{31}$, $x^{2} \equiv 25\mod{997}$?
I have all of these properties for this Legendre symbol and no idea what to do with it to find whether these have a soltion, let alone solve them if they do.
| For the first, we want $(7/53)$. Both of our numbers are primes, and one has form $4k+1$. So by Quadratic Reciprocity, $(7/53)=(53/7)$.
But $53\equiv 4\pmod 7$, so $(53/7)=(4/7)$. But $4$ is a perfect square, so $(4/7)=1$. It follows that $(7/53)=1$, so $x^2\equiv 7\pmod{53}$ has a solution.
While solving the first problem, we solved the second. We saw that $(53/7)=1$, so the congruence $x^2\equiv 53\pmod{7}$ has a solution. But we really need minimal machinery for that, since $53\equiv 4\pmod{7}$, so the congruence has the obvious solutions $x\equiv \pm 2\pmod{7}$.
Jumping ahead, the congruence $x^2\equiv 25\pmod{997}$ has at least one obvious solution, no Legendre symbol calculation needed.
For the congruence $x^2
\equiv 14\pmod{31}$, the Legendre symbol approach has us evaluate $(14/31)$. By one of the standard properties of Legendre symbols, this is $(2/31)(7/31)$.
Because $31$ is a prime of the shape $8k+1$, we have $(2/31)=1$. Now we need to evaluate $(7/31)$. Since each of $7$ and $31$ is a prime of the form $4k+3$, by Reciprocity argument we $(7/31)=-(31/7)$. But $31\equiv 3\pmod{7}$, so we want $-(3/7)$. By Reciprocity, this is the negative of the negative of $(7/3)$, so it is $(7/3)$. But $(7/3)=(1/3)=1$. It follows that the congruence $x^2\equiv 14\pmod{31}$ has a solution.
| {
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If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta $ in terms of $x$. Given $\cos^4 \theta −\sin^4 \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta $ .
Here is what I did:
$\cos^4 \theta −\sin^4 \theta = x$.
($\cos^2 \theta −\sin^2 \theta)(\cos^2 \theta +\sin^2 \theta) = x$
Thus
($\cos^2 \theta −\sin^2 \theta)=x$ ,
so $\cos 2\theta=x$ .
Now $x^3=(\cos^2 \theta −\sin^2 \theta)^3=\cos^6 \theta-\sin^6 \theta +3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $
So if I can find the value of $3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $ in terms of $x$ , the question is solved. But how to do that ?
| You have
$$\cos^2 \theta + \sin^2 \theta = 1$$
$$\cos^2 \theta - \sin^2 \theta = x$$
Adding and subtracting the two equations gives
$$\cos^2 \theta = {1 + x \over 2}$$
$$\sin^2 \theta = {1 - x \over 2}$$
Substituting you have
$$\cos^6 \theta - \sin^6 \theta = \bigg({1 + x \over 2}\bigg)^3 - \bigg({1 - x \over 2}\bigg)^3$$
$$ = {3 \over 4} x + {1 \over 4} x^3$$
| {
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Arc length $y = \frac{1}{4} x^2 - \frac{1}{2} \ln x$ $$y = \frac{1}{4} x^2 - \frac{1}{2} \ln x$$
$$\int_1^{2e} \sqrt{1 + (y')^2}$$
$$y' = \frac{x}{2} - \frac{1}{2x}$$
$$y' = \frac{2x^2-1}{2x}$$
$$\left(\frac{2x^2-1}{2x}\right)^2$$
$$\frac{4x^4-4x^2+1}{4x^2}$$
$$\int_1^{2e} \sqrt{1 + \frac{4x^4-4x^2+1}{4x^2} }$$
The 1 cancels out the negative term in the numerator
$$\int_1^{2e} \sqrt{ \frac{4x^4+1}{4x^2} }$$
So now if i have done this right I have now idea how to integrate this, subsitution doesn't seem to help. What is the trick here?
| $$1+y'^2=1+\frac14\left(x^2-2+\frac1{x^2}\right)=\frac14\left(x+\frac1x\right)^2\implies$$
$$\int\limits_1^{2e}\sqrt{1+y'^2}dx=\frac12\int\limits_1^{2e}\left(x+\frac1x\right)dx=\ldots$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with square roots: $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$
Let $x$ and $y$ be nonnegative real numbers such that $x+y=1$. How do I show that $\sqrt{x^2+1}+\sqrt{y^2+1}\ge \sqrt{5}$?
How do I deal with square roots inside the inequality?
| Another interpretation of the problem is this: $ \ y = \sqrt{x^2 + 1} \ \ \text{and} \ \ x = \sqrt{y^2 + 1} \ $ are the "positive" branches of the "vertical" and "horizontal" unit hyperbolas centered on the origin. They thus share the asymptotes $ \ y = \pm \ x \ $ .
It is specified that $ \ x \ge 0 \ \ \text{and} \ \ y \ge 0 \ $ and that they be related through $ \ x + y = 1 \ . $ So we in fact require that $ \ 0 \ \le \ x,y \ \le \ 1 \ . $ We can thus imagine a point in the first quadrant which slides along the line $ \ y = 1 - x \ $ in said quadrant and consider the lengths of the vertical and horizontal line segments extending from each coordinate axis to the appropriate hyperbolic branch, as depicted in the figure below.
It is clear that the behavior of this geometric arrangment is symmetrical about the asymptote $ \ y = x \ , $ so we only need look at the results for, say, $ \ 0 \ \le \ x \ \le \ \frac{1}{2} \ . $ For $ \ x = 0 \ , $ we attain the maximum for the sum sought,
$$ \sqrt{0^2 + 1} \ + \ \sqrt{1^2 + 1} \ = \ 1 + \sqrt{2} \ \approx \ 2.4142 \ , $$
while for $ \ x = \frac{1}{2} \ , $ the sum has its minimum,
$$ \sqrt{(\frac{1}{2})^2 + 1} \ + \ \sqrt{(\frac{1}{2})^2 + 1} \ = \ 2 \cdot \sqrt{\frac{5}{4}} \ = \ \sqrt{5} \ \approx \ 2.2361 \ . $$
The sum of radicals has a fairly narrow range of values for the specified domain. [This approach is most nearly similar to Michael Albanese's analysis. (It also seems a sort of "inside-out" version of peterwhy's elegant method.)]
| {
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"timestamp": "2023-03-29T00:00:00",
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Factor this quadratic expression I need to do the following:
Prove that a quadratic expression of the form $A(x^2-y^2) - (B-C)xy$ can be always factored into two linear factors.
It is easy enough to compare this with the standard representation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c$ and say that the condition $abc + 2fgh -af^2 - bg^2 - ch^2 = 0$ is satisfied.
However, what are those two linear factors?
At best the given expression can be written as $Ax^2 - Ay^2 - Bxy + Cxy$ or $x(Ax-By) -y(Ax-Cy)$. But this doesn't look like it can be always factored.
Second try:
Let $B-C = D$ for convenience.
Now the expression is:
$Ax^2 - Ay^2 - Dxy$
$= Ax^2 - (Dy)x + Ay^2$
If this be equated to zero,
$x = \dfrac{Dy \pm \sqrt{(Dy)^2 - 4A^2}}{2A}$
$\Rightarrow 2Ax - Dy = \pm \sqrt{(Dy)^2 - 4A^2}$
Squaring both sides and cancelling $D^2y^2$, we are left with
$Ax^2 - Dxy + A = 0$
Sigh! Looks like I'm just going in circles, and while it's easy to prove using the standard results, I can't seem to find the factors. How can this be done?
| You want to factor $A(x^2-y^2) - (B-C)xy$. To ease typing, write $E$ for $B-C$.
Rewrite our quadratic as
$$\frac{1}{4A}\left(4A^2 x^2-4AExy -4A^2y^2\right).$$
Complete the square. We get
$$\frac{1}{4A}\left((2Ax-Ey)^2 -(4A^2+E^2)y^2\right).$$
This is a difference of squares, and we get
$$\frac{1}{4A}\left(2Ax-Ey -\sqrt{4A^2+E^2}\,y\right)\left(2Ax-Ey +\sqrt{4A^2+E^2}\,y\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the remainder of $6^{17}$ divided by $17^6$? What is the general method to solve such questions ?
| Well, let's have a go by hand. We note that $6^2=36=2(17+1)$ so that $$6^{17}=6\cdot(6^2)^8=6\cdot 2^8 \cdot(17+1)^8$$
Now we use $2^4=17-1$ to give $$6\cdot(17-1)^2\cdot(17+1)^8=6\cdot(17^2-1)^2\cdot(17+1)^6$$
Let's put $x=17$ and reduce modulo $x^6=17x^5$ as we go (I prefer to keep the powers and $x$ works better for me than $17$).
$(x+1)^6=6x^5+15x^4+20x^3+15x^2+6x+1$
$(x^2-1)^2=x^4-2x^2+1$
So that $(x^2-1)^2(x+1)^6=$ (dropping $x^6$ and higher powers, all equalities now are modulo $17^6$)
$6x^5+x^4-40x^5-30x^4-12x^3-2x^2+6x^5+15x^4+20x^3+15x^2+6x+1 =$
$-28x^5-14x^4+8x^3+13x^2+6x+1 =$
$6x^5-14x^4+8x^3+13x^2+6x+1$
We now need to multiply this by 6 to obtain
$36x^5-84x^4+48x^3+78x^2+36x+6$
and using $x=17$ this becomes
$2x^5-5x^5+x^4+3x^4-3x^3+5x^3-7x^2+2x^2+2x+6$
$=14x^5+4x^4+2x^3-5x^2+2x+6$
The value of this expression with $x=17$ is $20220503$ - see Dan Shved's comment. It doesn't look like there is any much simpler way to do it.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum value of this expression. Let $x$ and $y$ be two real numbers satisfying $$(x+y)^3 +4xy \geqslant 2.$$
Find the minimum value of the expression
$$E=x^3+y^3 -2(x^2+y^2)-1.$$
I tried. We have
$$2\leqslant (x+y)^3+(x+y)^2.$$
Therefore, $$x+y \geqslant 1.$$
$$E = (x+y)^3-3xy(x+y)-2[(x+y)^2-2xy]-1.$$
Or
$$E = (x+y)^3-xy(3x+3y-4)-2(x+y)^2-1.$$
Using $$xy\leqslant \dfrac{(x+y)^2}{4}$$ and if $$3x + 3y -4 >0,$$
$$E \geqslant (x+y)^3-\dfrac{(x+y)^2}{4}(3x+3y-4).$$
Put $t = x + y$, and using derivative.
A problem is, if $$3x + 3y -4 <0,$$ how to prove?
| when $x,y$ are non negative. the case is very easy as $x^3-2x^2$ have min value when $x>0,\to E_{min}=-\dfrac{91}{27}$ .
if $x \le 0,y>0$, let $x=-u, u \ge 0$, let $y=3u+a,$, it is not difficult to prove
when $a \ge 1.3, (2u+a)^3-4u(3u+a)\ge 2$
$E=26u^3+(27a-20)u^2+(9a^2-12a)u+a^3-2a^2-1$
it is obvious that $E_{min}=a^3-2a^2-1$ when $u=0 \to E_{min}=-\dfrac{59}{27}$
so final answer is when $x=y=\dfrac{4}{3},$ min is $-\dfrac{91}{27}$
edit: I add the reason why $x,y$ can't be negative at same time.
if $x,y$ are both negative, let $x=-u, y=-v, \implies -(u+v)^3+4uv \ge 2 \iff (u+v)^3-4uv+2 \le 0$ with $u>0,v>0$
but $(u+v)^3 \ge (2\sqrt{uv})^3$, let $t=\sqrt{uv}$
LHS$\ge 8t^3-4t^2+2=(2t+1)((t-1)^2+1) > 0$ so $x,y$ can't be negative at same time.
edit 3: indeed, the last step is not necessary as op already show $x+y \ge 1$ which hints $x,y$ can't be negative at same time.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to factor $2x^2-x-3$? I know its:
$$(x+1)(2x-3)$$
But how do you come to that conclusion?
| Note that $2-3=-1,$ so that $$\begin{align}2x^2-x-3 &= 2x^2+(2-3)x-3\\ &= 2x^2+2x-3x-3\\ &= 2x(x+1)-3(x+1)\\ &= (x+1)(2x-3).\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the acceleration of $ghg^{-1}h^{-1}$ is $2[X, Y]$. This problem has been driving me up the wall!
Let $G$ be a matrix group and let $g(t)$ and $h(t)$ be paths in $G$ with $g(0) = I$, $g^\prime(0) = X$, $h(0) = I$, and $h^\prime(0) = Y$. Prove that the group commutator $g(t)h(t)g^{-1}(t)h^{-1}(t)$ is equal to $I + [X, Y]t^2 + \mathcal{O}(t^3)$.
My problem: My calculation keeps showing that the group commutator's acceleration vanishes...
To simplify notation, I will write $g$ instead of $g(t)$.
(PLEASE: Before leaving the correct solution, please tell me where my solution is going wrong..)
First, I take for granted that
\begin{equation} \tag{1}
(g^{-1})^\prime(0) = -g^\prime(0)
\end{equation}
and
\begin{equation} \tag{2}
\left(ghg^{-1}h^{-1}\right)^\prime(0) = 0.
\end{equation}
I now show that
\begin{equation} \tag{3}
(g^{-1})^{\prime \prime}(0) = 2(g^\prime)^2(0) - g^{\prime \prime}(0).
\end{equation}
$I = g^{-1}g$ implies that $0 = (g^{-1})^\prime g + g^{-1}g^\prime$, so that
\begin{equation}
0 = (g^{-1})^{\prime \prime} g + (g^{-1})^\prime g^\prime + (g^{-1})^\prime g^\prime + g^{-1} g^{\prime \prime}.
\end{equation}
Evaluating at $0$ and rearranging, we obtain the desired result for $(g^{-1})^{\prime \prime}(0)$.
From Eqs. $(1)$ and $(2)$, we conclude that
\begin{align}
\left( (hg)^{-1} \right)^{\prime \prime}(0)
&= \left\{ \left( \left( g^{-1} \right)^\prime h^{-1} + g^{-1} \left( h^{-1} \right)^\prime \right)^\prime \right\} \bigg|_{t=0}
\\ &= \left\{ \left( g^{-1} \right)^{\prime \prime} + 2 g^\prime h^\prime + \left( h^{-1} \right)^{\prime \prime} \right\} \bigg|_{t=0}
\\ &= \left\{ 2 \left( g^\prime \right)^2 - g^{\prime \prime} +2 g^\prime h^\prime + 2 \left( h^\prime \right)^2 - h^{\prime \prime} \right\} \bigg|_{t=0}. \tag{4}
\end{align}
Now, for the main result:
Taking the first derivative of the group commutator, we obtain
\begin{equation}
(ghg^{-1}h^{-1})^\prime = (gh)^\prime (hg)^{-1} + gh \left( (hg)^{-1} \right)^\prime.
\end{equation}
Taking the second derivative and evaluating at $0$, we have (using Eq. $(3)$)
\begin{align}
(ghg^{-1}h^{-1})^{\prime \prime}(0)
& = \left\{ (gh)^{\prime \prime}(hg)^{-1} + (gh)^\prime \left( (hg)^{-1} \right)^\prime + (gh)^\prime \left( (hg)^{-1} \right)^\prime + gh \left( (hg)^{-1} \right)^{\prime \prime} \right\} \bigg|_{t=0}
\\ & = \left\{ (gh)^{\prime \prime} - 2(gh)^\prime (hg)^\prime + \left( (hg)^{-1} \right)^{\prime \prime} \right\} \bigg|_{t=0}
\\ &= \left\{ g^{\prime \prime} + 2 g^\prime h^\prime + h^{\prime \prime} \right\} \bigg|_{t=0}
\\ &- \left\{ 2 \left( g^\prime + h^\prime \right) \left( h^\prime + g^\prime \right) \right\} \bigg|_{t=0}
\\ &+ \left\{ 2 \left( g^\prime \right)^2 - g^{\prime \prime} +2 g^\prime h^\prime + 2 \left( h^\prime \right)^2 - h^{\prime \prime} \right\} \bigg|_{t=0}
\\ &= 0. \left(\text{Edit as of }8/21/13: \text{This is not equal to } 0, \text{but }2[X,Y]! \right)
\end{align}
Obviously, something has gone terribly wrong...
Thanks for your help!
| I would do it this way: if
$g = I + X t + g''(0) t^2/2 + O(t^3)$ and
$h = I + Y t + h''(0) t^2/2 + O(t^3)$,
$$g h = I + (X + Y) t + (g''(0)/2 + X Y + h''(0)/2) t^2 + O(t^2)$$
Similarly, $g^{-1} = I - X t + (X^2 - g''(0)/2) t^2 + O(t^3)$ and $h^{-1} = I - Y t + (Y^2 - h''(0)/2) t^2 + O(t^3)$ so
$$g^{-1} h^{-1} = I - (X + Y) t + (X^2 - g''(0)/2 + X Y + Y^2 - h''(0)/2) t^2 + O(t^3)$$
Multiply $gh$ and $g^{-1} h^{-1}$:
$$ I + \left(\frac{g''(0)}{2} + X Y + \frac{h''(0)}{2} - (X+Y)^2 + X^2 - \frac{g''(0)}{2} + X Y + Y^2 - \frac{h''(0)}{2}\right) t^2 + O(t^3)$$
Now note that the $g''$ and $h''$ terms cancel and $(X+Y)^2 = X^2 + X Y + Y X + Y^2$.
| {
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