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Closed form of $\int_0^2\frac{1}{2+\sqrt{3\,e^x+3\,e^{-x}-2}}dx$ Could you please help me to solve this integration problem?
$$\int_0^2\frac{1}{2+\sqrt{3\,e^x+3\,e^{-x}-2}}dx$$
Its approximate numeric value is $0.419197813818367...$, but I could not find an exact symbolic expression for it.
|
$\int_0^2\dfrac{1}{2+\sqrt{3e^x+3e^{-x}-2}}dx$
$=\int_1^{e^2}\dfrac{1}{2+\sqrt{3y+\dfrac{3}{y}-2}}d(\ln y)$
$=\int_1^{e^2}\dfrac{1}{y\left(2+\sqrt{3y+\dfrac{3}{y}-2}\right)}dy$
$=\int_1^{e^2}\dfrac{1}{2y+y\sqrt{3y+\dfrac{3}{y}-2}}dy$
$=\int_1^{e^2}\dfrac{1}{\sqrt{3y^3-2y^2+3y}+2y}dy$
$=\int_1^{e^2}\dfrac{\sqrt{3y^3-2y^2+3y}-2y}{\left(\sqrt{3y^3-2y^2+3y}+2y\right)\left(\sqrt{3y^3-2y^2+3y}-2y\right)}dy$
$=\int_1^{e^2}\dfrac{\sqrt{3y^3-2y^2+3y}-2y}{3y^3-2y^2+3y-4y^2}dy$
$=\int_1^{e^2}\dfrac{\sqrt{3y^3-2y^2+3y}}{3y^3-6y^2+3y}dy-\int_1^{e^2}\dfrac{2y}{3y^3-6y^2+3y}dy$
$=\int_1^{e^2}\dfrac{\sqrt{3y^3-2y^2+3y}}{3y(y-1)^2}dy-\int_1^{e^2}\dfrac{2}{3(y-1)^2}dy$
which can express in terms of Lauricella Functions
|
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|
How to express sum of even numbers in sigma notation Given a positive integer number n, how can I express the sum of all positive even numbers up to n in sigma notation?
|
Given a real number $x$ we can define $\lfloor x \rfloor$ to be the greatest integer less than or equal to $x$. For instance, $\lfloor 3 \rfloor = 3$ and $\lfloor 7.81 \rfloor = 7$.
Thus we can write the sum of even integers less than or equal to $n$ as
$$\sum_{k=0}^{\lfloor n/2 \rfloor} 2k$$
If you stare hard enough at this you'll see why it works: if $n$ is even then $\lfloor \frac{n}{2} \rfloor = \frac{n}{2}$, so the last term is $2 \cdot \frac{n}{2} = n$. If $n$ is odd then $\lfloor \frac{n}{2} \rfloor = \frac{n-1}{2}$, so the last term is $2 \cdot \frac{n-1}{2} = n-1$.
|
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|
Prove $\lim_{n\to\infty}{\frac{3n+1}{2n+5}=\frac{3}{2}}$ Problem Prove $$\lim_{n\to\infty}{\frac{3n+1}{2n+5}=\frac{3}{2}}$$ using a epsilon proof
Scratchwork
We see that $\mid{\frac{3n+1}{2n+5}-\frac{3}{2}}\mid=\mid{\frac{6n+2}{4n+10}-\frac{6n+15}{4n+10}}\mid= \mid\frac{-13}{4n+10}\mid=\frac{13}{4n+10}<\frac{13}{4n}<\frac{13}{n}<\epsilon$.
(I like to go all the way to $\frac{13}{n}$ which im not sure i can do). If we let $\epsilon >0$ then $\frac{1}{\epsilon}>0\implies \frac{13}{\epsilon}$ By the Archimdean property we can choose a $k\in\mathbb{N}$ such that $\frac{13}{\epsilon}<k\implies \frac{13}{k}<\epsilon$.
Proof:
Let let $\epsilon >0$ then $\frac{1}{\epsilon}>0\implies \frac{13}{\epsilon}>0$ Then by the Archimedean property there exists a $k\in\mathbb{N}$ such that $k>\frac{13}{\epsilon}\implies \frac{13}{k}<\epsilon$. If $n \geq k$ then $\frac{1}{n} \leq \frac{1}{k}\implies \frac{13}{n}\leq \frac{13}{k} <\epsilon$. Thus$\mid{\frac{3n+1}{2n+5}-\frac{3}{2}}\mid<\frac{13}{n}<\epsilon$. Thus $\lim{\frac{3n+1}{2n+5}=\frac{3}{2}}$. Would this suffice?
|
Yes, absolutely :)
Would post this as a comment, but then this question remains unanswered, so...
|
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|
Prove $\frac {1}{\cos 0^\circ \cdot \cos 1^\circ} + \ldots +\frac {1}{\cos 88^\circ \cdot \cos 89^\circ}= \frac{\cos 1^\circ}{\sin 1^\circ}$ Prove the following identity:
$$\frac {1}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {1}{\cos 88^{\circ} \cdot \cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$
After hours of trying, I wasn't able to make any significant progress, that worth mentioning. Then I finally decided to look into the solution, but it seems that author had the same problem, the listed solution is obviously wrong.
$$\sin 1^{\circ}\left(\frac {1}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {1}{\cos 88^{\circ} \cdot \cos 89^{\circ}}\right) = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$
$$\frac {\sin 1^{\circ}}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {\sin 1^{\circ}}{\cos 88^{\circ} \cdot \cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$
$$\frac {\sin (1^{\circ} - 0^{\circ})}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {\sin (89^{\circ} - 88^{\circ})}{\cos 88^{\circ} \cdot \cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$
$$\frac {\sin 1^{\circ}\cos0^{\circ} - \cos 1^{\circ}\sin 0^{\circ}}{\cos 0^{\circ} \cdot \cos 1^{\circ}} + \ldots +\frac {\sin 89^{\circ}\cos 88^{\circ} - \cos 89^{\circ}\sin 88^{\circ}}{\cos 88^{\circ} \cdot \cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$
$$\frac {\sin 1^{\circ}}{\cos 1^{\circ}} - \frac {\sin 0^{\circ}}{\cos 0^{\circ}} + \ldots +\frac {\sin 89^{\circ}}{\cos 89^{\circ}} - \frac {\sin 88^{\circ}}{\cos 88^{\circ}} + = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$
I hope I'm clear till now, beacuse everything I do was just elementary alegraic and trigonometric transformation. So now it's just telescopic series.
$$\frac {\sin 89^{\circ}}{\cos 89^{\circ}} - \frac {\sin 0^{\circ}}{\cos 0^{\circ}}= \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$
$$\frac {\sin 89^{\circ}}{\cos 89^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$
$$\frac {\cos 1^{\circ}}{\sin 1^{\circ}} = \frac{\cos 1^{\circ}}{\sin 1^{\circ}}$$
Q.E.D.
But I think this is actualy a counterproof, rather than a proof, because the identity that the author started the proof is different from the one we want. And obviously $\sin 1^{\circ} \neq 1$
My question is the identity wrong, is it a typo? Or am I missing something obvious?
|
I think it's a typo.
$$
\frac{1}{\cos k \cos (k+1)}=\frac{1}{\sin 1}\frac{\sin 1}{\cos k \cos (k+1)}=\frac{1}{\sin 1} \frac{\sin(k+1)\cos(k)-\sin(k)\cos(k+1)}{\cos k \cos (k+1)} \\ =\frac{1}{\sin 1} \left(\tan(k+1)-\tan (k)\right)
$$
Then
$$
\sin 1\; \sum_{k=0}^{88}\frac{1}{\cos k \cos (k+1)}=\tan(89)-\tan(0)=\frac{\cos(1)}{\sin(1)}
$$
|
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|
How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$ let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}
{5}\cdots (1)$$
I find sometimes,and I find this same problem:
let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$ \frac{1}{5a^2-4a+11}+\frac{1}{5b^2-4b+11}+\frac{1}{5c^2-4c+11}\leq\frac{1}{4} $$
and this problem have some methods,you can see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=223910&start=20
and I like this can_hang2007 methods and Honey_S methods,But for $(1)$ I can't prove it.Thank you
|
Here's what I've got so far.
It's not a complete solution,
but the ideas might be useful
to someone else.
$2x^2-6x+9
=x^2+(x-3)^2
$.
If
$f(x)
=2x^2-6x+9
$,
$f'(x)
=4x-6
$
is zero at $x = 3/2$.
Since $f(3/2)
=9/2
$,
$f(x) \ge 9/2$
for all $x$.
This means that
$\dfrac1{f(x)}
\le \dfrac{2}{9}
$
for all $x$,
so
$\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(c)}
\le \dfrac{6}{9}
= \dfrac23
$
without any hypothesis on $a, b, c$.
If $c = 3-a-b$,
$f(c)
=(3-a-b)^2+(a+b)^2
$.
If $a=b=3/2$,
$c = 0$
so
$f(c) = 9$
so
$\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(c)}
= \dfrac{5}{9}
< \dfrac{3}{5}
$.
If $a=b=c$,
$a=b=c=1$.
Since $f(1)
=1+4=5
$,
$\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(c)}
= \dfrac{3}{5}
$,
so this is the equality case.
If we can show that
$\dfrac1{f(a)}+\dfrac1{f(b)}+\dfrac1{f(3-a-b)}
$
is a maximum when
$a=b=1$,
that would do it.
But I don't know how right now,
so I'll leave it at this.
|
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|
Is $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$ Question is to check if $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$
we have $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\prod \limits_{n=2}^{\infty}(\frac{n^2-1}{n^2})=\prod \limits_{n=2}^{\infty}\frac{n+1}{n}\frac{n-1}{n}=(\frac{3}{2}.\frac{1}{2})(\frac{4}{3}.\frac{2}{3})(\frac{5}{4}.\frac{3}{4})...$
In above product we have for each term $\frac{a}{b}$ a term $\frac{b}{a}$ except for $\frac{1}{2}$.. So, all other terms gets cancelled and we left with $\frac{1}{2}$.
So, $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\frac{1}{2}$.
I would be thankful if some one can assure that this explanation is correct/wrong??
I am solving this kind of problems for the first time so, it would be helpful if some one can tell if there are any other ways to do this..
Thank you
|
One of the easiest ways to deal with infinite sums/products, is to stop at a finite value, say $N$, and look at what happens to the finite sum/product and then let $N \to \infty$. (In fact, this is the typical way infinite sums/products are to be understood/interpreted.)
Hence, in your case, let us look at
\begin{align}
S_N & = \prod_{n=2}^N \left(1-\dfrac1{n^2}\right) = \dfrac{1}{2}\cdot\dfrac{3}{2}\cdot \dfrac23 \cdot \dfrac43 \cdot \dfrac34 \cdot \dfrac54 \cdots \dfrac{N-2}{N-1} \cdot \dfrac{N}{N-1} \cdot \dfrac{N-1}{N} \cdot \dfrac{N+1}N\\
& = \dfrac12 \cdot \dfrac{N+1}N = \dfrac{N+1}{2N}
\end{align}
Now let $N \to \infty$ to conclude that
$$\prod_{n=2}^{\infty} \left(1-\dfrac1{n^2}\right) = \dfrac12$$
|
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|
Are there identities which show that every odd square is the sum of three squares? I am looking for algebraic identities of the form
$$
(2n+1)^2 = f(n)^2 + g(n)^2 + h(n)^2,
$$
where the functions are polynomials in $n$.
EDIT: Evidently $(6k)^2 = 36k^2$ is trivially the sum of three squares when $k$ is odd. We also have the identity
$$
(6k+3)^2 = (2(2k+1))^2 + (2(2k+1))^2 + (2k+1)^2.
$$
Are there similar identities for the other two odd residues modulo $6$, i.e., $6k+1$ and $6k-1$?
|
That's not really going to work, the highest degree terms do not cancel here, so all you get is $(2n+1)^2 = (2n+1)^2 + 0^2 + 0^2.$
What does work is due to Gordon Pall; every number is the sum of four squares, so write $$ 2n+1 = a^2 + b^2 + c^2 + d^2.$$ Then you get nontrivial expressions
$$ (2n+1)^2 = \left(a^2 + b^2 - c^2 - d^2 \right)^2 + (-2ad+2bc)^2 + (2ac+2bd)^2 $$
and similar things resulting from rearranging the letters $a,b,c,d$ and choosing many $\pm$ signs.
See Pall_Automorphs_1940.pdf at http://zakuski.math.utsa.edu/~kap/forms.html
Let's see, if $2n+1$ is already a square, perform the same task for $\sqrt {2n+1},$ so as to be assured of at least two nonzero summands. If $2n+1$ is a fourth power...
Given a specific integral expression such as $9 = 4 + 4 + 1,$ we can produce a rational expression
$$ \left( \frac{4n+2}{3}\right)^2 + \left( \frac{4n+2}{3}\right)^2 + \left( \frac{2n+1}{3}\right)^2 = (2n+1)^2. $$
While $49 = 36 + 9 + 4$ results in
$$ \left( \frac{12n+6}{7}\right)^2 + \left( \frac{6n+3}{7}\right)^2 + \left( \frac{4n+2}{7}\right)^2 = (2n+1)^2. $$
These are just three rational multiples of $(2n+1),$ that is the only way it can work. Plus, these only produce integer expressions for certain $n.$
|
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Any 'odd unit fraction' whose denominator is not $1$ can be represented as the sum of three different 'odd unit fractions'? Let us call a fraction whose denominator is odd 'odd fraction'. Also, let us call an odd fraction whose numerator is 1 'odd unit fraction'.
Then, here is my question.
Question : Is the following true?
"Any odd unit fraction whose denominator is not $1$ can be represented as the sum of three different odd unit fractions."
Motivation : I've been asking this question. Then, I reached the above expectation.
Examples :
$$\frac 13=\frac 15+\frac 19+\frac 1{45}$$
$$\frac 15=\frac 1{7}+\frac 1{21}+\frac 1{105}$$
$$\frac 17=\frac 19+\frac 1{33}+\frac 1{693}$$
$$\frac 19=\frac 1{11}+\frac 1{51}+\frac 1{1683}$$
$$\vdots$$
$$\frac 1{99}=\frac 1{101}+\frac 1{5001}+\frac 1{16668333}$$
$$\vdots$$
|
If $n$ is not a multiple of 3, then $$\frac{1}{n}=\frac{1}{n+2}+\frac{3}{(n+2)(n+4)}+\frac{1}{n(n+2)(n+4)}$$
Second try: You have the first fraction is $1/(n+2)$. Then you want to solve $$\frac{2}{n(n+2)}=\frac{1}{a}+\frac{1}{b}$$
Rearrange that into $(2a-n(n+2))(2b-n(n+2))=n^2(n+2)^2$ You can make the right-hand side equal to $pq$, where both $p$ and $q$ are $3\bmod 4$. (Take $p=n$ or $p=n+2$) Then $a$ and $b$ will be odd.
So $a=n(n+3)/2, b=n(n+2)(n+3)/2$ or $a=(n+1)(n+2)/2, b=n(n+2)(n+1)/2$
|
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|
finding range of function of three variables Three real numbers $x$, $y$, $z$ satisfy the following conditions.
$x^{2}+y^{2}+z^{2}=1~$, $~y+z=1$
Find the range of $~x^{3}+y^{3}+z^{3}~$ without calculus.
I solved this problem only with Lagrange-Multiplier and wonder if there exist other methods.
|
Let $z=1-y$ to get $x^2+y^2+(1-y)^2 = x^2 + 2 y^2 -2y +1 = 1$, or $x^2 + 2 y^2 -2y = 0$.
This gives $x = \pm\sqrt{2y(1-y)}$ and $y \in [0,1]$.
Substituting gives:
$x^3+y^3+z^3 = \pm (2y(1-y))^\frac{3}{2} +3 y(y-1) +1$, with $y \in [0,1]$.
Let $\phi_\pm(y) = \pm (2y(1-y))^\frac{3}{2} -3 y(1-y) +1$, with $y \in [0,1]$.
Let $\eta(y) = y(1-y)$, then for $y \in [0,1]$, we have $n(y) \in [0,1]$, $\eta(y)=\eta(1-y)$, $\eta$ is minimized at $y=\pm 1$, and maximized at $y=\frac{1}{2}$, and $\eta(\frac{1}{2}) = \frac{1}{4}$.
Hence $\eta(y) \in [0,\frac{1}{4}]$ for $y \in [0,1]$.
We see that $\phi_\pm(y) = \eta(y) (\pm 2\sqrt{2}\sqrt{\eta(y)}-3) +1$, and note that $(\pm 2\sqrt{2}\sqrt{\eta(y)}-3) \le 0$ for all $y \in [0,1]$.
It follows that both $\phi_+, \phi_-$ have a maximum value of $1$ for $y \in \{0,1\}$. Clearly $\phi_-(y) \le \phi_+(y)$, so to find a minimum value we need only consider $\phi_-$.
Since $\eta(y) \in [0,1]$, to find the minimum value, we need only consider of $x^2(-2\sqrt{2}x-3)+1$, for $x \in [0,\frac{1}{2}]$. This is easily seen to be decreasing for $ x \ge 0$, hence the minimum value occurs at $x=\frac{1}{2}$. and the minimum value is $\frac{1}{4}-\frac{1}{2 \sqrt{2}}$.
|
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|
Find the greatest common divisor (gcd) of $f(x) = x^2 + 1$ and $g(x) = x^6 + x^3 + x + 1$ Find the greatest common divisor (gcd) of $f(x) = x^2 + 1$ and $g(x) = x^6 + x^3 + x + 1$.
Since $x^6 + x^3 + x + 1 = (x^2 + 1)(x^4 - x^2 + x + 1)$, $\mathrm{gcd}[f(x),g(x)] = x^2 + 1$.
My question is how could I JUSTIFY that the answer is ACTUAL gcd of $f(x)$ and $g(x)$. thanks
|
To justify, you could say that it divides g(x) and f(x) to the rational polynomials such that they have no other common divisor, or common factor. So the answer is proved.
|
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|
Induction $(1+\frac{1}{x^n})(1+\frac{1}{y^n}) \geq (1+2^n)^2$ How to prove this inequality using Induction (or any simpler method):
Let (x,y) be real positive numbers, so that x+y=1; and n an integer:
Prove this:
$\begin{align}(1+\frac{1}{x^n})(1+\frac{1}{y^n}) \geq (1+2^n)^2\end{align}$
|
Replace $ y$ with $1-x$.
Hint: Show that $ \frac{1}{4} \geq x(1-x)$.
Hint: Show that $\frac{1}{x^n} \times \frac{1}{(1-x)^n} \geq 2^{2n} $.
Hint: Show that $ \frac{ 1}{x^n} + \frac{1}{(1-x)^n} \geq 2 \times 2^n$ using the technique AM-GM.
Expand and compare.
|
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|
How to find the orthogonal trajectories of the family of all the circles through the points $(1,1)$ and $(-1,-1)$? I'm trying to find the orthogonal trajectories of the family of circles through the points $(1,1)$ and $(-1, -1)$. Now such a family can be given by an equation of the form $$ x^2 + y^2 + 2g(x-y) - 2 = 0, $$ where $g$ is a parameter.
Now upon differentiation with respect to x, we obtain $$ 2x + 2y y^\prime + 2g (1 - y^\prime ) = 0, $$ where $y^\prime$ denotes the derivative of $y$ with respect to $x$. Upon dividing out by $2$, we arrive at $$ x + y y^\prime + g(1 - y^\prime ) = 0, $$ from which we get $$ g = \frac{x + y y^\prime}{y^\prime - 1}. $$ Putting this value of $g$ into the equation of the family of circles, we get $$ x^2 + y^2 +2 \frac{x + y y^\prime}{y^\prime - 1} ( x - y ) - 2 = 0, $$ so $$ (x^2 + y^2 -2 ) (y^\prime - 1 ) + 2 (x + y y^\prime ) (x - y) = 0$$ or $$ (x^2 + y^2 - 2 + 2xy - 2y^2 ) y^\prime + (2x^2 - 2xy - x^2 - y^2 + 2) = 0 $$ or $$ ( x^2 + 2xy - y^2 - 2) y^\prime + (x^2 - 2xy - y^2 + 2) = 0, $$ from which we get $$ y^\prime = - \frac{ x^2 - 2xy - y^2 + 2}{ x^2 + 2xy - y^2 - 2}. $$ Now for the orthogonal trajectories, we get $$ y^\prime = \frac{x^2 + 2xy - y^2 - 2}{x^2 - 2xy - y^2 + 2}. $$ How to solve this differential equation?
|
Assuming your calculations are correct up to the last part, I begin with:
$$ y^\prime = \frac{x^2 + 2xy - y^2 - 2}{x^2 - 2xy - y^2 + 2}. $$
Observe that:
$$ y^\prime = \frac{1 + 2y/x - (y/x)^2 - 2/x^2}{1 - 2y/x - (y/x)^2 + 2/x^2}. $$
Thus, let $v = y/x$ for which $y=vx$ and $y' = xv'+v$ and the given problem changes to:
$$ xv'+v = \frac{1 + 2v - v^2 - 2/x^2}{1 - 2v - v^2 + 2/x^2}. $$
I suspect this can be solved. But, I'll stop here for now. This is a standard substitution for ODEs which can be written as a function of $y/x$ which reflects a certain symmetry, namely that $x$ and $y$ can be scaled by the same factor.
|
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|
Ellipse: product of the distance from foci to a tangent is a constant I am supposed to determine what is the result of said product. Given $P(x_0,y_0)$, I need to calculate the distance from the foci of an ellipse to the tangent line that passes through $P$, and then multiply the distances.
In essence it is quite simply. We take:
$$
\frac{x_0}{a^2}x + \frac{y_0}{b^2}y = 1
$$
as the tangent line. Then we simply calculate its distance to each focus $(c,0)$ and $(-c,0)$, using the formula and then, multiplying.
$$
d=\frac{\frac{x_0c}{a^2}±1}{\sqrt{\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}}
$$
$$
\text{Some constant k}=\frac{\frac{x_0^2c^2}{a^4}-1}{\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4}}
$$
I'm having trouble getting things cancelled here. The constant k is $b^2$, but I can't get to it. Help?
|
$$
\frac{1-\frac{c^2x_0^2}{a^4}}{\frac{x_0^2}{a^4}+\frac{y_0^2}{b^4}}
=\frac{1-\frac{c^2x_0^2}{a^4}}{\frac{x_0^2}{a^4}+\frac1{b^2}\left(1-\frac{x_0^2}{a^2}\right)}
=\frac{1-\frac{c^2x_0^2}{a^4}}{\frac1{b^2}\left(1-(a^2-b^2)\frac{x_0^2}{a^4}\right)}
=b^2,
$$
where in the final step we used the identity $a^2-b^2=c^2$.
|
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|
Show that the vector field $X(x, y, z)=(xy-z^2, yz-x^2, x^2+z^2+xz-1)$ is tangent to the set $x^2 + y^2 + z^2 = 1$ I know I need to find functions $F(t)$, $G(t)$, and $H(t)$ such that $F(0)=x$, $G(0)=y$, and $H(0)=z$ and $F'(0)=xy-z^2$, $G'(0)=yz-x^2$, and $H'(0)=x^2+z^2+xz-1$. It's also necessary that $(F(t))^2 +(G(t))^2 + (H(t))^2 = 1$ for all $t$. Just not sure where to go from here...
|
We will show that the vector field is perpendicular to any point $(x,y,z)$ on sphere by considering the dot product of the vector field and the normal to the sphere at that point.
$(xy-z^2,yz-x^2,x^2+z^2+xz-1)\cdot(x,y,z)=x^2y-xz^2+zy^2-yx^2+zx^2+z^3+xz^2-z=zy^2+zx^2+z^3-z=z(x^2+y^2+z^2-1).$ We know on the surface of the sphere $x^2+y^2+z^2-1=0$. Therefore the result holds.
|
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|
How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form, like$(A+B)^{3}$.
The first thing I think is $(A+B)^{3}=A^3+3A^2B+3AB^2+B^3$, then try to make it become the the form of $A^3+3A^3B+3AB^3+B^3$. However, it it so difficult to obtain this form.
I need help.
Update :
Now I have $\left(x - \frac 4x\right)^3$ but how to find the $f^{-1}(x)$ of $f(x)=\left(x - \frac 4x\right)^3$?
Thank you for your attention
|
Hint: Try matching first and last terms: $A^3 = x^3$ and $B^3=-\frac{64}{x^3}$ and check if it fits the other terms.
|
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|
Limit of a Lebesgue integral What is the value of:
$$\lim_{n\to\infty}\sqrt{n}\int_0^{1}(1-t^2)^ndt$$
I think I have to use the Theorem of dominated convergence
|
From my comment:
$$\begin{align*}
A_n & = \int_0^1 (1-t^2)^n dt = \int_0^1 (1-t^2)^{n-1} dt - \int_0^1 t^2 (1-t^2)^{n-1} dt \\
& = A_{n-1} - \left( \underbrace{\frac{1}{n} \left[ t^2 (1-t^2)^n \right]_0^1}_{=0} - \frac{2}{n} \int_0^1 t (1-t^2)^n dt \right) \\
& = A_{n-1} + \frac{2}{n} \left( \underbrace{\frac 1 {n+1} \left[ t(1-t^2)^{n+1} \right]_0^1}_{=0} - \frac{1}{n+1} \int_0^1 (1-t^2)^{n+1} \right) \\
& = A_{n-1} - \frac 2 {n(n+1)} A_{n+1}
\end{align*}$$
So
$$A_{n+1} = \frac{n(n+1)} 2 (A_{n-1} - A_n)$$
With $A_0 = 1, A_1 = \frac{2}{3}$.
Does this get you any further?
|
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|
Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$ Prove that $1<\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.
By using the Mathematical induction. Suppose the statement holds for $n=k$.
Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}=(\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1})+(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$
we know $\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}>1$
What can we do for $(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$?
|
Another way: You can bound it by $$\frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{3n+1} \ge \int_{n+1}^{3n+1} \frac{1}{x} = \log \Big(\frac{3n+1}{n+1}\Big)$$ where $\frac{3n+1}{n+1}$ is monotone increasing in $n$, and $\log(\frac{3*7 + 1}{7+1}) > 1$ already. Then check the cases $n=1$ through $6$.
|
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|
Matrix multiplication - Express a column as a linear combination Let $A = \begin{bmatrix}
3 & -2 & 7\\
6 & 5 & 4\\
0 & 4 & 9
\end{bmatrix} $ and $B = \begin{bmatrix}
6 & -2 & 4\\
0 & 1 & 3\\
7 & 7 & 5
\end{bmatrix} $
Express the third column matrix of $AB$ as a linear combination of the column matrices of $A$
I don't get this... surely the 3rd column would be an expression of the row matrices of $A$ since the 3rd column of$AB$ would be $
\begin{bmatrix}
3(4) & -2(3) & 7(5)\\
6(4) & 5(3) & 4(5)\\
0(4) & 4(3) & 9(5)
\end{bmatrix} $
As I typed out the question I see my answer...
the 3rd column is $4\begin{bmatrix}
3\\
6\\
0
\end{bmatrix} 3 \begin{bmatrix}
-2\\
5\\
4
\end{bmatrix} 5 \begin{bmatrix}
7\\
4\\
9
\end{bmatrix}$
Is this correct?
|
The third column of $AB$ is $$A\begin{bmatrix}
4\\
3\\
5
\end{bmatrix}$$
and if we denote the columns of $A=[C_1 C_2 C_3]$ then the third column of $AB$ is
$$A\begin{bmatrix}
4\\
3\\
5
\end{bmatrix}=[C_1 C_2 C_3]\begin{bmatrix}
4\\
3\\
5
\end{bmatrix}=4C_1+3C_2+5C_3$$
|
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|
$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$? I am trying to show that
$$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$$
This question stems from the underlying homework problem, which asks to show
$$ \frac{\pi}{\sin(\pi z)} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}, $$
to which I am at my wits end. I have a couple of identities on hand, namely
$$ \pi \cot (\pi z) = \frac{1}{z} + \sum_{n \in \mathbb{Z}; n \neq 0} \frac{1}{z - n} + \frac{1}{n} $$
and
$$ \frac{\sin (\pi z)}{\pi} = z \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right) $$
and
$$ \frac{\pi^2}{\sin^2 (\pi z)} = \sum_{n \in \mathbb{Z}} \frac{1}{(z - n)^2} $$
I've tried fooling around with these identities and am getting nowhere. Any hints or suggestions would be greatly appreciated.
|
$$\sum_{n=1}^\infty\frac{(-1)^n}{z^2-n^2}=2\sum_{n=1}^\infty\frac{1}{z^2-(2n)^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$
$$=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(z/2)^2-n^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$
Now since,
$$\pi \cot(\pi z)=\frac{1}{z}+2z\sum_{n=1}^\infty\frac{1}{z^2-n^2}$$
We get that, $$\pi\coth(\frac{\pi z}{2})=\frac{2}{z}+2z\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(z/2)^2-n^2}$$
And so, $$\pi\cot(\frac{\pi z}{2})-\pi \cot(\pi z)=\frac{1}{z}+2z(\frac{1}{2}\sum_{n=1}^\infty\frac{1}{(z/2)^2-n^2}-\sum_{n=1}^\infty\frac{1}{z^2-n^2})
$$
$$=\frac{1}{z}+2z\sum_{n=1}^\infty \frac{(-1)^n}{z^2-n^2}$$
Now since $$\cot(\frac{\pi z}{2})-\cot(\pi z)=\frac{1}{\sin(\pi z)}$$
We get that:
$$\frac{\pi}{\sin(\pi z)}=\frac{1}{z}+2z\sum_{n=1}^\infty \frac{(-1)^n}{z^2-n^2}$$
As required
|
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|
How to prove the inequality $\sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2} $ for $n\in\mathbb{Z}^+$? I have to prove this inequality:
$$
\forall n \in Z^+, \sum_{i=1}^n \frac{\sqrt{i+1}}{2i} > \frac{\sqrt{n}}{2}
$$
So far, I have done the base cases and assumed the inequality is true for some integer k, and I have gotten to the point where I need to show that:
$$
\frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} < \frac{\sqrt{k+1}}{2}
$$
Can somebody point me in the right direction? I have tried squaring both sides, but because there is a + sign on the LHS, there are still square roots and it only makes things more complicated.
Is there a name for this inequality?
|
I am not a fan of using (in)equality that you have to prove, since you have to be really careful what you do with that. So, let's try this:
\begin{align*}
\frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+2} &> \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2k+4} = \frac{\sqrt{k}}{2} + \frac{\sqrt{k+2}}{2(k+2)} \\
&> \frac{\sqrt{k}}{2} + \frac{1}{2\sqrt{k+2}} = \frac{1}{2} \left(\sqrt{k} + \frac{1}{\sqrt{k+2}} \right) = \frac{1}{2} \sqrt{ k + 2\sqrt{\frac{k}{k+2}} + \frac{1}{k+2} } \\
&> \frac{1}{2} \sqrt{k + 1}
\end{align*}
You'll need to justify why
$$2\sqrt{\frac{k}{k+2}} + \frac{1}{k+2} > 1,$$
But that should be really easy.
|
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|
How find this inequality find the maximum $z_{5}$ let $z_{1},z_{2},z_{3},z_{4},z_{5}\in C$,such
$$\begin{cases}
|z_{1}|\le 1,|z_{2}|\le 1\\
|2z_{3}-(z_{1}+z_{2})|\le|z_{1}-z_{2}|\\
|2z_{4}-(z_{1}+z_{2})|\le|z_{1}-z_{2}|\\
|2z_{5}-(z_{3}+z_{4})|\le|z_{3}-z_{4}|
\end{cases}$$
Find the maximum $|z_{5}|$
My ugly solution:
note
$$|z_{1}-z_{2}|+|z_{1}+z_{2}|\le 2\sqrt{|z1_{1}|^2+|z_{2}|^2}$$
then we have
\begin{align*}&4|z_{5}|=2|2z_{5}-(z_{3}+z_{4})+(z_{3}+z_{4})|\le 2[|2z_{5}-(z_{3}+z_{4})|+|z_{3}+z_{4}|]\\
&\le 2(|z_{3}-z_{4}|+|z_{3}+z_{4}|)\\
&=|[2z_{3}-(z_{1}+z_{2})]-[2_{4}-(z_{1}+z_{2})]|+|[2z_{3}-(z_{1}+z_{2})]+[2z_{4}-(z_{1}+z_{2})]+2(z_{1}+z_{2})|\\
&\le|[2z_{3}-(z_{1}+z_{2})]-[2_{4}-(z_{1}+z_{2})]|+|[2z_{3}-(z_{1}+z_{2})]+[2z_{4}-(z_{1}+z_{2})]|+2|(z_{1}+z_{2})|\\
&\le 2|z_{1}+z_{2}|+2\sqrt{|2z_{3}-(z_{1}+z_{2})|^2+|2z_{4}-(z_{1}+z_{2})|^2}\\
&\le 2|z_{1}+z_{2}|+2\sqrt{2}|z_{1}-z_{2}|\\
&\le2\sqrt{[1^2+(\sqrt{2})^2]\cdot(|z_{1}+z_{2}|^2+|z_{1}-z_{2}|^2)}\\
&=2\sqrt{3}\cdot\sqrt{2(|z_{1}|^2+|z_{2}|^2)}\\
&\le2\sqrt{3}\cdot \sqrt{4}=4\sqrt{3}
\end{align*}
so
$$|z_{5}|\le \sqrt{3}$$
if and only if $z_{1}=e^{i\theta},z_{2}=e^{-\theta},z_{3}=\dfrac{1}{2}(z_{1}+z_{2})+\dfrac{\sqrt{6}}{3}e^{\frac{\pi}{4}i},z_{4}=\dfrac{1}{2}(z_{1}+z_{2})+\dfrac{\sqrt{6}}{3}e^{-\frac{\pi}{4}i},z_{5}=\sqrt{3},\theta=\arctan{\sqrt{2}}$
My question: Have other nice methods? or can someone can use geometric Explain and solve the problem?
|
It seems the following.
Your solution is nice but I shall use a geometric approach.
The points $z_1$ and $z_2$ belong to a disk $D_1$ of radius $R_1=1$ centered at the point $o_1=0$. The points $z_3$ and $z_4$ belong to a disk $D_2$ of radius $R_2=|z_1-z_2|/2$ centered at the point $o_2=(z_1+z_2)/2$. The point $z_5$ belongs to a disk of radius $R_3=|z_3-z_4|/2$ centered at the point $o_3=(z_3+z_4)/2$. Put $r_1=|o_1o_2|$ and $r_2=|o_2o_3|$. Since $z_1,z_2\in D_1$, then $$R_2^2=|z_1o_2||z_2o_2|\le (R_1+r_1)(R_1–r_1)=R_1^2-r_1^2$$ (here we use the theorem claiming that the product of lengths of the parts of the chords going through a fixed point is constant). Similarly, $$R_3^2=|z_3o_3||z_3o_4|\le (R_2+r_2)(R_2–r_2)=R_2^2-r_2^2=1-r_1^2-r_2^2.$$ Moreover, $|O_3Z_5|\le R_3$. Now $$|z_5|=|O_1Z_5|\le |o_1o_2|+|o_2o_3|+|o_3z_5|=r_1+r_2+\sqrt{1-r_1^2-r_2^2}\le$$ $$3\sqrt{\frac {r_1^2+r^2_2+1- r_1^2-r_2^2}{3}}=\sqrt{3},$$
and the equality is possible only if the points $o_2$ and $o_3$ consequently lie on a segment $[o_1z_5]$ and $r_1=r_2=\frac 1{\sqrt 3}$.
|
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|
Proof of $(\forall x)(x^2+4x+5 \geqslant 0)$ $(\forall x)(x^2+4x+5\geqslant 0)$ universe is $\Re$
I went about it this way
$x^2+4x \geqslant -5$
$x(x+4) \geqslant -5$
And then I deduce that if $x$ is positive, then $x(x+4)$ is positive, so it's $\geqslant 5$
If $ 0 \geqslant x \geqslant -4$, then $x(x+4)$ is also $\geqslant -5$.
If $ x < -4$, then $x(x+4)$ will be negative times negative = positive, so obviously $\geqslant -5$
I'm wondering if there's an easier to solve this problem? My way seems a little clunky.
|
\begin{align}
x^2+4x+5
&= x^2 + 2x(2) +(2)^2 +1\\
&= (x+2)^2 +1
\end{align}
$(x+2)^2\ge0$ for any $x\in\mathbb{R}$. Adding $1$ for both sides of inequalities, we have
$$
(x+2)^2+1\ge 1$$
for any $x\in \mathbb{R}$.
As $x^2+4x+5=(x+2)^2+1$,
$$
x^2+4x+5\ge 1>0$$
for any $x\in \mathbb{R}$.
Finally,
$$
x^2+4x+5>0$$
for any $x\in \mathbb{R}$.
Note
The quadratic expression $x^2+4x+5$ is always greater than zero, it never equals to zero. Be careful!
|
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|
Prove that $n^9 \equiv n \pmod{30}$ if $(30,n) > 1$ I'm looking at the following number theory problem:
Prove that $n^9 \equiv n \pmod{30}$ for all positive integers $n$ if $(30,n) > 1$.
It is easy to show that $n^9 \equiv n \pmod{30}$ if $(30,n)=1$ by using Euler's Theorem. Is there any way to prove the above easily when $(30,n)$ are not relatively prime? I'm looking for some easy realization, and not factoring $n^9-n=n(n^8-1)=n(n^4+1)(n^4-1)=n(n^4+1)(n^2+1)(n-1)(n+1)$ and then deducing divisibility. Is there a simple realization for this?
Thanks!
|
You know that $n^9-n=0\mod 30$ if, and only if $n^9-n=0\mod 2$, $n^9-n=0\mod 3$ and $n^9-n=0\mod 5$. But modulo $2$; we have $n^2=n$, so $n^9=n$ immediately. If $n\neq 0$, $n^2= 1$ modulo $3$, so $n^3=n$, so $n^9=n^3=n$ so $n^9=n$. Modulo $5$, if $n\neq 0$, $n^2=\pm 1$ so $n^4=1$ so $n^5=n$ so $n^9=n^5n^4=n^5=n$ and the proof is finished.
|
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|
How come when $2^{k} | (x-1)(x+1)$ one of the terms is divisible by $2$ and not by $4$ when $k \in \mathbb{N} $ and $3 \leq k$ So I'm reading Knuth's 'Discrete Mathematics' at the moment and there's a paragraph detailing how many solutions are there for $x^{2} \equiv 1 \pmod{p}$.
So other cases (when $p$ is an odd prime or when p is equal to $2$ or $4$) are clear to me, but I'm having trouble wrapping my head around the case when $p=2^{k}$ with $3 \leq k$.
Here's what's troubling me
If $2^{k}|(x - 1)(x + 1)$ then either $(x - 1)$
or $(x + 1)$ is divisible by $2$ but not by $4$, so the other one must be divisible
by $2^{k-1}$.
How come? I don't see that at all...
|
In general, there are cases in which neither $x+1$ nor $x-1$ is divisible by $2$ (take $x$ to be an even number).
It is given that $2^k | (x+1)(x-1) $.
Suppose neither $x+1$ nor $x-1$ is divided by $2$, then $2$ does not divide $(x+1)(x-1)$. So, there is no point of considering $2^k | (x+1)(x-1)$.
So, without loss of generality, suppose $2$ divides $x+1$
suppose both $(x+1)$ and $(x-1)$ are divisible by $4$ then, we should have $2^{2m} |(x+1)(x-1) $ which need not be the case always... (it is not specified that power of $2$ is even)
So, we can suppose $x+1$ is divisible by $2$ but not by $4$.
So, we should have $2^{k-1} | (x+1)$
|
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|
Evaluating $\sum_{n=0}^\infty \frac{n}{3^n}$ How can i prove that $ \sum\limits_{n=0}^\infty \frac{n}{3^n}=\frac{3}{4}$?
I tried to find some pattern but didn't succeed
|
Another approach: As the series converges, let L denote its limit. Now you have $$L=\sum_{n=0}^\infty \frac{n}{3^n}=\sum_{n=1}^\infty \frac{n}{3^n}= \frac{1}{3}\sum_{n=1}^\infty \frac{n}{3^{n-1}}= \frac{1}{3}\sum_{n=0}^\infty \frac{n+1}{3^{n}}= \frac{1}{3}\left[\sum_{n=0}^\infty \frac{n}{3^{n}}+\sum_{n=0}^\infty \frac{1}{3^{n}}\right] =\frac{1}{3}\left[L+ \sum_{n=0}^\infty \frac{1}{3^{n}}\right].$$
Now just use the usual formula for the limit of a geometric series, and solve algebraically for $L$.
|
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|
we need to find the points where $|f(z)|$ has maximum and minimum value $f(z)=(z+1)^2$ and $R$ be the triangle with vertices $(0,0),(0,1),(2,0)$, we need to find the points where $|f(z)|$ has maximum and minimum value
so here $z=2$ is the point of maximum and $z=0$ is minimmum (intuitively), as $f$ is analytic so it must attain maxima minima on boundary,
but my question is how can I show $f$ has no extrimum value on the other points of the boundary of the triangle? Thank you for help
|
On the lower edge of the triangle, we have $z = x$, and
$$|f(z)| = (x + 1)^2$$
This is a strictly increasing function of $x$, and so its minimum is at $0$ and maximum at $1$.
On the vertical edge, we have $z = iy$ and
$$|f(z)| = |(iy + 1)^2| = 1 + y^2$$
Again, this is strictly increasing in $y$ (take a derivative), and so maximized at $y = 1$, but the maximum value is only $2$.
The third edge is defined by the line $y = 1 - \frac{x}{2}$, so
\begin{align*}
|f(z)|^2 &= |1 + z|^2 \\
&=\left| 1 + x + i \left(1 - \frac{x}{2}\right)\right|^2 \\
&= (1 + x)^2 + \left(1 - \frac{x}{2}\right)^2 \\
&= 1 + 2x + x^2 + 1 - x + \frac{x^2}{4} \\
&= 2 + x + \frac{x^2}{4}
\end{align*}
This is strictly increasing in $x$ for $0 \le x \le 2$, and the maximum is obtained at the right endpoint, giving a maximum of $5$, as expected.
|
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|
$\cos x+\cos 3x+\cos 5x+\cos 7x=0$, Any quick methods? How to solve the following equation by a quick method?
\begin{eqnarray}
\\\cos x+\cos 3x+\cos 5x+\cos 7x=0\\
\end{eqnarray}
If I normally solve the equation, it takes so long time for me.
I have typed it into a solution generator to see the steps. One of the steps shows that :
\begin{eqnarray}
\\\cos x+\cos 3x+\cos 5x+\cos 7x&=&0\\
\\-4\cos x+40\cos ^3x-96\cos ^5x+64\cos ^7x&=&0\\
\end{eqnarray}
How can I obtain this form? It seems very quick. or this quick methd do not exist?
Thank you for your attention
|
In case you want to crack the question through solving polynomial equation, let $cos(x)=y$
You've found $$64y^7-96y^5+10y^2-4y=0$$
It can be factorized to
$$4y(2y^2-1)(8y^4-8y^2+1)=0$$
Furthermore
$$2y^2-1=(\sqrt{2}y+1)(\sqrt{2}y-1)$$
and
$$(8y^4-8y^2+1)=(2\sqrt{2}y^2+1)^2-(8+4\sqrt{2})y^2=(2\sqrt{2}y^2+\sqrt{8+4\sqrt{2}}y+1)(2\sqrt{2}y^2-\sqrt{8+4\sqrt{2}}y+1)$$
Using quadratic formula you'll get the same solutions other people post.
|
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|
Sum of squares of the quadratic nonresidues modulo $p$ is divisible by $p$ Let $p$ be a prime number with $p > 5$. Prove that the sum of the squares of the quadratic nonresidues modulo $p$ is divisible by $p$.
My idea is to use the fact that any quadratic residue is congruent modulo $p$ to one integer in the set $\{1^2,2^2,\ldots,\left( \frac{p-1}{2} \right)^2 \}$. And none of the quadratic residues are congruent to each other. So the quadratic nonresidues must all be in the set $\{ \left( \frac{p+1}{2} \right)^2 , \left( \frac{p+3}{2}\right)^2 \ldots, (p-1)^2\}$. So the sum of the quadratic nonresidues must be given by $$\sum_{k=1}^{p-1} k^2 - \sum_{k=1}^{\frac{p-1}{2}}k^2 = \frac{(p-1)p(2p-1)}{6} - \frac{\left( \frac{p-1}{2} \right) \left( \frac{p+1}{2} \right) p}{6} = \frac{p}{6} \left( (p-1)(2p-1) - \frac{1}{4}(p-1)(p+1) \right).$$
Is this the correct approach? Prove that the term in parenthesis is an integer divisible by $6$?
|
Using primitive roots finishes the problem quickly: Let $g$ be a primitive root, so that the sum of the squares of the quadratic non-residues is just
$$\sum_{i=1}^{\frac{p-1}{2}}{(g^{2i-1})^2} \equiv g^2\sum_{i=0}^{\frac{p-3}{2}}{g^{4i}} \equiv g^2\frac{1-(g^4)^{\frac{p-1}{2}}}{1-g^4} \equiv \frac{g^2(1-(g^{p-1})^2)}{1-g^4} \equiv 0 \pmod{p}$$
where since $p>5$, $p \nmid 1-g^4$ so the manipulations above are valid.
If you don't want to appeal to primitive roots, then something similar to what you tried can also be done. What we want is to take the sum of the squares of all non-zero elements and subtract the squares of the quadratic residues. So we get
\begin{align}
&\sum_{k=1}^{p-1}{k^2}-\sum_{k=1}^{\frac{p-1}{2}}{(k^2)^2} \\
&\equiv \frac{(p-1)p(2p-1)}{6}-\frac{(\frac{p-1}{2})(\frac{p+1}{2})(p)(3(\frac{p-1}{2})^2+3(\frac{p-1}{2})-1)}{30} \pmod{p}\\
& \equiv p[\frac{(p-1)(2p-1)}{6}-\frac{(\frac{p-1}{2})(\frac{p+1}{2})(3(\frac{p-1}{2})^2+3(\frac{p-1}{2})-1)}{30}] \pmod{p}\\
& \equiv 0 \pmod{p}
\end{align}
where we may treat $\frac{1}{30}=30^{-1}$ as an element in $\mathbb{Z}_p$ since $p>5$.
|
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|
$a^2-b^2=bc$ and $b^2-c^2=ac \Rightarrow a^2-c^2=ab$ Some weeks ago our math teacher asked the following question and gave us a week to solve it:
If $a^2-b^2=bc$ and $b^2-c^2=ac ,$ Prove $a^2-c^2=ab$, Where $a,b,c$ are non-zero real numbers.
This seemed really easy at the first, but when i tried to prove it i just failed every time. After a week, I only came up with this idea: Assume our case is true. $a^2-c^2=ab\Rightarrow a^2-b^2+b^2-c^2=ab\Rightarrow bc+ac=ab\Rightarrow\frac{1}{abc}(bc+ac)=\frac{1}{abc}(ab) \Rightarrow$ $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$. Now if we Prove this, our case will be proved as well.
In my opinion this seemed like a really nice question so i wanted to share it with everyone.
|
You can chug through this by eliminating the $a$ variable and following your nose.
$$b(b + c) = b^2 + bc = a^2 = {(b^2 - c^2)^2 \over c^2}$$
We can assume that $b \neq -c$, since if $b = -c$ the first equation implies $a = 0$, contradicting that $a,b,c$ are all nonzero. So we can divide by $b+c$ to get
$$b = {(b^2 - c^2)(b-c) \over c^2}$$
Multiplying by $c^2$ and expanding this becomes
$$bc^2 = b^3 - c^2b - cb^2 + c^3 \tag 1$$
You want to show $a^2 - c^2 = ab$. Substituting your first equation into $a^2$ and the second into $ab$ this is equivalent to showing
$$b^2 + bc - c^2= {b^2 - c^2 \over c}b$$
Multiplying by $c$ this is equivalent to
$$b^2c + bc^2 - c^3= b^3 - c^2b \tag 2$$
Rearranging terms, $(2)$ is the same as $(1)$.
|
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|
How find this $\int_{0}^{\pi}\frac{\cos{(nx)}}{\cos{x}+a}dx$ Fin the integral
$$I_{n}=\int_{0}^{\pi}\dfrac{\cos{(nx)}}{\cos{x}+a}dx$$
where $n\in {\mathbb N}\,,\ a>1$
My try:
let
$$I_{n}-I_{n-1}=\int_{0}^{\pi}\dfrac{\cos{(nx)}-\cos{(n-1)x}}{\cos{x}+a}dx$$
and note
$$\cos{x}-\cos{y}=-2\sin{\dfrac{x+y}{2}}\sin{\dfrac{x-y}{2}}$$
so
$$I_{n}-I_{n-1}=-2\int_{0}^{\pi}\dfrac{\sin{(nx-\dfrac{x}{2})}\sin{\dfrac{x}{2}}}{\cos{x}+a}dx$$
Then I can't ,Thank you for your help.
|
A little complex analysis helps. First, by the evenness of $\cos$, we have
$$I_n = \frac12 \int_{-\pi}^\pi \frac{\cos (nx)}{a+\cos x}\,dx,$$
and by Euler's formula, we have $\cos (nx) = \operatorname{Re} e^{inx}$, so
$$I_n = \frac12 \operatorname{Re} \int_{-\pi}^\pi \frac{e^{inx}}{a+\cos x}\,dx.$$
Since the imaginary part is odd, the integral is real, and we can omit the $\operatorname{Re}$.
Now we write $z = e^{ix}$, then we have
$$\cos x = \frac12(z+z^{-1});\quad e^{inx} = z^n;\quad dx = \frac{dz}{iz};$$
so we obtain
$$I_n = \int_{\lvert z\rvert = 1} \frac{z^n}{2a + z + z^{-1}}\, \frac{dz}{iz} = 2\pi\cdot\frac{1}{2\pi i}\int_{\lvert z\rvert = 1} \frac{z^n}{z^2 + 2az + 1}\,dz.$$
We need the zeros of the denominator $z^2 + 2az + 1 = (z+a-\sqrt{a^2-1})(z+a+\sqrt{a^2-1})$ to apply the Cauchy integral formula. Of the zeros, $-a+\sqrt{a^2-1}$ lies inside the unit disk, and $-a-\sqrt{a^2-1}$ outside the closed unit disk. With $f(z) = \frac{z^n}{z+a+\sqrt{a^2-1}}$, the Cauchy integral formula yields
$$\begin{align}
I_n &= 2\pi\cdot\frac{1}{2\pi i}\int_{\lvert z\rvert = 1} \frac{f(z)}{z+a-\sqrt{a^2-1}}\,dz\\
&= 2\pi\cdot f(-a+\sqrt{a^2-1})\\
&= \frac{\pi}{\sqrt{a^2-1}}(\sqrt{a^2-1}-a)^n.\tag{1}
\end{align}$$
|
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|
Finding a,b,c,d in a quartic expression Let $p(x)=x^4+ax^3+bx^2+cx+d$ where a,b,c,d are constants. If $p(1)=10$, $p(2)=20$, $p(3)=30$, compute $\frac {p(12)+p(-8)}{10}$. I have tried so far.
\begin{align}
a+b+c+d=&9\\8a+4b+2c+d=&4\\27a+9b+3c+d=&-51
\end{align}
Manipulating these, I got $6a+b=-25$. Now, $$\frac {p(12)+p(-8)}{10}=\frac{24832+1216a+208b+4c+2d}{10}$$ $$=\frac{24832+202(6a+b)+(4a+4b+4c)+2b+2d}{10}$$ $$=\frac{19782+(36-4d)+2b+2d}{10}$$ $$=\frac{19818+2b-2d}{10}$$ How do I get rid of the $2b-2d$?
|
(2012's answer is correct, but the algebraic manipulations doesn't reveal what is happening. This answer explains why we could calculate the expression, despite not having enough information.)
By the remainder factor theorem, since $p(x) - 10x$ is a monic quartic polynomial with roots 1, 2, and 3, hence
$$p(x) - 10x = ( x-1) (x-2) ( x-3) (x-k), $$
where $k$ is some constant.
Hence, $ p(12) - 120 = 11 \times 10 \times 9 \times (12-k)$ and $p(-8) - (-80) = (-9) \times (-10) \times (-11) \times (-8-k)$.
Note that the coefficients of $ (12-k) $ and $-(-8-k)$ are the same, namely $11\times 10 \times 9$, so we can add them up to get:
$$p(12) + p(-8) = 120 + (-80) + 11 \times 10 \times 9 \times (12+8) = 19840.$$
If you want a similar problem to practice what you learnt in this problem, try this math problem on Brilliant.
|
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|
How to solve the diophantine equation:$ xa^3+yb^3=c^3$ Let $a,b,c,x,y \in \mathbb{Z}> 1$. Any hint on how to solve of the diophantine equation $ xa^3+yb^3=c^3$?
|
Consider following relations:
$1^3+6^3+8^3=9^3$
$3^3+4^3+5^3=6^3$
$⇒1^3+3^3+4^3+5^3+8^3=9^3$
or: $1^3 + 3^3+5^3+ 4^3(1+2^3)=9^3$
or: $(1^3 + 3^3+5^3). 1^3+ 4^3(1+2^3)=9^3$
or: $153 . 1^3 +4^3 . 9 = 9^3$
Compare this with equation:
$x a^3 + y b^3=c^3$
you find: $x=153, a=1, y=9, b=4, c=9$
It can be seen that there are infinitely many relations like $a^3+b^3+c^3=d^3$. Now if two or more terms of LHS of relation, for example $b^3$ and $c^3$ have a common factor like k we have:
$b^3+c^3= k^3[(b/k)^3+(c/k)^3]$
That is relation $a^3+b^3+c^3=d^3$ reduces to an equation like $X.A^3 +YB^3=C^3$ which indicates this equation may have many forms and solutions.
|
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|
Prove $\frac{1}{b-a}\int_a^b\frac{x}{\sin x}dx\leqslant\frac{a+b}{\sin a+\sin b}$ $a,b\in(0,\frac{\pi}{2}),aFor $a,b\in(0,\frac{\pi}{2}),a<b$, prove
$$\frac{1}{b-a}\int_a^b\frac{x}{\sin x}dx\leqslant\frac{a+b}{\sin a+\sin b}.$$
By mean value theorem, there is $c\in(a,b)$ s.t.
$$\frac{c}{\sin c}=\frac{1}{b-a}\int_a^b\frac{x}{\sin x}dx$$
I want to prove that
$$\frac{c}{\sin c}\leqslant\frac{a+b}{\sin a+\sin b}$$
for any $c\in (a,b)$. I am not sure if I am on the right track
|
From the Hermite-Hadamard inequality, since $\frac{x}{\sin x}$ is convex on $[a,b]$, you have that $$\frac{1}{b-a}\int_a^b\frac{x}{\sin x}\,dx\leq\frac{1}{2}\left(\frac{a}{\sin a}+\frac{b}{\sin b}\right).$$ So it's enough to prove the following:
$$\frac{a\sin b+b\sin a}{2\sin a\sin a b}\leq\frac{a+b}{\sin a+\sin b}\Leftrightarrow(a+b)\sin a\sin b+b\sin^2a+a\sin^2 b\leq2(a+b)\sin a\sin b\Leftrightarrow b\sin^2a+a\sin^2b\leq(a+b)\sin a\sin b\Leftrightarrow$$$$b\sin a(\sin b-\sin a)+a\sin b(\sin a-\sin b)\geq 0\Leftrightarrow(\sin b-\sin a)(b\sin a-a\sin b)\geq 0.$$ But, $\sin b-\sin a>0$, so it is enough to show that $$b\sin a\geq a\sin b\Leftrightarrow\frac{\sin a}{a}\geq\frac{\sin b}{b}.$$ But, this holds, since $\frac{\sin x}{x}$ is decreasing in $[a,b]$.
|
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|
For what $(m, n)$, does $1+x+x^2 +\dots+x^m | 1 + x^n + x^{2n}+\dots+x^{mn}$? For what $(m, n)$, does $1+x+x^2 +\dots+x^m | 1 + x^n + x^{2n}+\dots+x^{mn}$?
Well,
$$\sum_{i = 0}^{m} x^i = \frac{x^{m+1} - 1}{x - 1}$$
and,
$$\sum_{i = 0}^m x^{in} = \frac{x^{n(m+1)} - 1}{x-1}$$
Notice that $x^{m+1} - 1|(x^{m+1})^n - 1$, therefore,
$$x^{n(m+1)} - 1 = q(x^{m+1} - 1)$$
Let
$$\frac{x^{m+1} - 1}{x - 1} = \alpha$$
Therefore,
$$\sum_{i = 0}^{m} x^i = \frac{x^{m+1} - 1}{x - 1} = \alpha$$
and,
$$\sum_{i = 0}^m x^{in} = \frac{x^{n(m+1)} - 1}{x-1} = \frac{q(x^{m+1} - 1)}{x-1} = \alpha q$$
Clearly, $\alpha|\alpha q$. Therefore, all $(m, n)$ such that $m, n \in \mathbb{N}$ should work. But the correct answer, according to the text is that $(m+1)$ and $n$ have to be relatively prime. Where did I go wrong?
|
In your second formula you need $x^n$ in denominator.
|
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|
Direct proof that $\sum_{n\geq 0}\frac{x^n}{n!}=\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n$ Is there a direct proof that $$\sum_{n\geq 0}\frac{x^n}{n!}=\lim_{n\rightarrow\infty}(1+\frac{x}{n})^n?$$
We dont know what logarithms or exponentials are.
|
Applying the Binomial theorem,
\begin{equation*}
\lim_{n\to\infty}(1+\frac{x}{n})^n = \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\frac{(n-2)(n-1)x^3}{6n^2}+\frac{(n-3)(n-2)(n-1)x^4}{24n^3}+\frac{(n-4)(n-3)(n-2)(n-1)x^5}{120n^4}+... \\
=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+... \\
=\sum_{n\geq 0}\frac{x^n}{n!}.
\end{equation*}
Note that we used L'Hopital's rule when taking the limits because we had the form $\frac{\infty}{\infty}$.
|
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|
Find the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $(9,3)$. Find the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $(9,3)$.
I did
$$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{9+h}-3}{h} \frac{\sqrt(9+h)+3}{\sqrt(9+h)+3} = \frac{9+h-3}{h\sqrt{9+h}+3}= \frac{6}{\sqrt{9+h}+3}= \frac{6}{6}=1.$$
|
let $f(x) = x^a$
$f'(x) = a*x^{a-1}$
for example:
$f(x) = x^2$ -> $f'(x) = 2*x^1$
$f(x) = x^3$ -> $f'(x) = 3*x^2$
So...
if $f(x)=x^\frac{1}{2}$ what is $f'(x)$?
|
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|
Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$ How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).
|
EDITED. Some simplifications were made.
Here is a solution.
1. Basic facts on the dilogarithm. Let $\mathrm{Li}_{2}(z)$ be the dilogarithm function defined by
$$ \operatorname{Li}_{2}(z) = \sum_{n=1}^{\infty} \frac{z^{n}}{n^{2}} = - \int_{0}^{z} \frac{\log(1-x)}{x} \, dx. $$
Here the branch cut of $\log $ is chosen to be $(-\infty, 0]$ so that $\operatorname{Li}_{2}$ defines a holomorphic function on the region $\Bbb{C} \setminus [1, \infty)$. Also, it is easy to check (by differentiating both sides) that the following identities hold
\begin{align*}
\operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right)
&= -\mathrm{Li}_{2}(z) - \tfrac{1}{2}\log^{2}(1-z); \quad z \notin [1, \infty) \tag{1} \\
\operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right)
&= \color{blue}{\boxed{\operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z)}} + \color{red}{\boxed{\log(-z)\log(1-z)}}; \quad z \notin [0, \infty) \tag{2}
\end{align*}
Notice that in (2), the blue-colored part is holomorphic on $|z| < 1$ while the red-colored part induces the branch cut $[-1, 0]$.
2. A useful power series. Now let us consider the power series
$$ f(z) = \sum_{n=0}^{\infty} \frac{H_n}{n} z^n. $$
Then $f(z)$ is automatically holomorphic inside the disc $|z| < 1$. Moreover, it is easy to check that
$$ \sum_{n=1}^{\infty} H_{n} z^{n-1}
= \frac{1}{z} \left( \sum_{n=1}^{\infty} \frac{z^{n}}{n} \right)\left( \sum_{n=0}^{\infty} z^{n}\right)
= -\frac{\log(1-z)}{z(1-z)}. $$
thus integrating both sides, together with the identity $\text{(1)}$, we obtain the following representation of $f(z)$.
$$f(z)
= \operatorname{Li}_{2}(z) + \tfrac{1}{2}\log^{2}(1-z)
= -\operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right). \tag{3}$$
3. Integral representation and the result. By the Parseval's identity, we have
$$ \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
= \frac{1}{2\pi} \int_{0}^{2\pi} f(e^{it})f(e^{-it}) \, dt
= \frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} f\left(\frac{1}{z}\right) \, dz \tag{4} $$
Since $\frac{1}{z}f(z)$ is holomorphic inside $|z| = 1$, the failure of holomorphy of the integrand stems from the branch cut of
\begin{align*}
f\left(\tfrac{1}{z}\right)
&= -\operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right) \\
&= -\color{blue}{\left( \operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z) \right)} - \color{red}{\log(-z)\log(1-z)},
\end{align*}
which is $[0, 1]$. To resolve this, we utilize the identity $\text{(2)}$. Note that the blue-colored portion does not contributes to the the integral $\text{(4)}$, since it remains holomorphic inside $|z| < 1$. That is, only the red-colored portion gives contribution to the integral. Consequently we have
\begin{align*}
\sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
&= -\frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} \color{red}{\log(-z)\log(1-z)} \, dz. \tag{5}
\end{align*}
Since the integrand is holomorphic on $\Bbb{C} \setminus [0, \infty)$, we can utilize the keyhole contour wrapping around $[0, 1]$ to reduce $\text{(5)}$ to
\begin{align*}
\sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
&=-\frac{1}{2\pi i} \Bigg\{ \int_{0^{-}i}^{1+0^{-}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \\
&\qquad \qquad + \int_{1+0^{+}i}^{+0^{+}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \Bigg\} \\
&=-\frac{1}{2\pi i} \Bigg\{ \int_{0}^{1} \frac{f(x)(\log x + i\pi)\log(1-x)}{x} \, dx \\
&\qquad \qquad - \int_{0}^{1} \frac{f(x)(\log x - i\pi)\log(1-x)}{x} \, dx \Bigg\} \\
&=-\int_{0}^{1} \frac{f(x)\log(1-x)}{x} \, dx. \tag{5}
\end{align*}
Plugging $\text{(3)}$ to the last integral and simplifying a little bit, we have
\begin{align*}
\sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
&= - \int_{0}^{1} \frac{\operatorname{Li}_2(x)\log(1-x)}{x} \, dx - \frac{1}{2}\int_{0}^{1} \frac{\log^{3}(1-x)}{x} \, dx \\
&= \left[ \frac{1}{2}\operatorname{Li}_2(x)^2 \right]_0^1 - \frac{1}{2} \int_{0}^{1} \frac{\log^3 x}{1-x} \, dx \\
&= \frac{1}{2}\zeta(2)^{2} + \frac{1}{2} \Gamma(4)\zeta(4) \\
&= \frac{17\pi^{4}}{360}
\end{align*}
as desired.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Laplace differential equation Can somebody help me work out $2y''+y'-y=\mathrm{e}^{3t}$, y(0)=2 and y'(0)=0 with the method of Laplace? I got
\begin{align*}
Y(s)&=\frac{1}{(s-3)(2s^2+s-1)}+ \frac{2+4s}{(s-3)(2s^2+s-1)}\\
&=-\frac{4}{15}\frac{1}{2s-1}+\frac{1}{12}\frac{1}{s+1}+\frac{1}{20}\frac{1}{s-3}-\frac{1}{6}\frac{1}{s+1}-\frac{16}{15}\frac{1}{2s-1}+\frac{7}{10}\frac{1}{s-3}\\
&=-\frac{2}{15}\mathrm{e}^{0.5t}+\frac{1}{12}\mathrm{e}^{-t}+\frac{1}{20}\mathrm{e}^{3t}-\frac{1}{6}\mathrm{e}^{-t}-\frac{8}{15}\mathrm{e}^{0.5t}+\frac{7}{10}\mathrm{e}^{3t}
\end{align*}
Is this correct?
|
Given:
$$\tag 1 2 y''+ y'- y = e^{3t},~ y(0)=2, ~ y'(0)=0$$
Taking the Laplace Transform of $(1)$, we have:
$$\mathcal{L}(2 y''+ y'- y) = 2(s^2y(s) - sy(0)-y'0) + (sy(s)-y(0)) - y(s) = \dfrac{1}{s-3}$$
Using the initial conditions from $(1)$, this reduces to:
$$y(s)(2s^2 + s - 1) -4s - 2 = \dfrac{1}{s-3}$$
Solving for $y(s)$ and doing partial fraction expansions/simplifications yields:
$$y(s) = \dfrac{1}{20(s-3)} + \dfrac{3}{4(s+1)} + \dfrac{12}{5(2s-1)}$$
Now we want to find $\mathcal{L^{-1}}(y(s))$, yielding:
$$y(t) = \dfrac{1}{20}e^{3t} + \dfrac{3}{4}e^{-t} + \dfrac{6}{5}e^{t/2}$$
Notes:
*
*You can always easily validate $y(t)$ by plugging it back in to the DEQ and making sure it satisfies it.
*You can solve this problems using other methods like undetermined coefficients or variation of parameters.
|
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|
Find the first three terms of the Maclaurin Series
Determine using multiplication/division of power series (and not via WolframAlpha!) the first three terms in the Maclaurin series for $y=\sec x$.
I tried to do it for $\tan(x)$ but then got kind of stuck. For our homework we have to do it for the $\sec(x)$. It is kind of tricky. Help would be awesome!
Thanks!
Taylor series for $\tan(x)$:
\begin{align*}
\tan (x)
&=\frac{\sin(x)}{\cos(x)}\\
&=\frac{x-\frac {x^3}6+\frac{x^5}{120}-\cdots}{1-\frac{x^2}2+\frac{x^4}{24}-\cdots}\\
&=x+\frac{x^3}3+\frac{2x^5}{15}+\cdots
\end{align*}
|
Let's write
$$1 = \cos{x} \sec{x} = \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots\right)(a_0 + a_1 x + a_2x^2 + \dots)$$
Expand the right hand side to find that
$$1 = 1 (a_0) + x (a_1) + x^2 \left(a_2 - a_0 \frac{1}{2!}\right) + x^3 \left(a_3 - a_1 \frac{1}{2!}\right) + x^4 \left(a_4 - a_2 \frac{1}{2!} + a_0 \frac{1}{4!}\right)+\dots$$
Equating coefficients gives (note that the left side is $1 + 0x + 0x^2 + 0x^3 + \dots$)
\begin{align*}
1 &= a_0 \\
0 &= a_1 \\
0 &= a_2 - \frac{a_0}{2} \\
0 &= a_3 - \frac{a_1}{2} \\
0 &= a_4 - \frac{a_2}{2} + \frac{a_0}{24}
\end{align*}
Solving this gives $a_0 = 1$, $a_1 = 0 = a_3$, $a_2 = \frac{1}{2}$ and $a_4 = \frac{5}{24}$, or
$$\boxed{\sec{x} \approx 1 + \frac{1}{2} x^2 + \frac{5}{24} x^4}$$
|
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|
Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $ Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $
Thanks in advance, my professor asked us to this a couple weeks ago, but I was enable to get to the right answer.
Good luck!
Here is what I got up to;
$\frac{(n+1)!}{(n-r)!(r+1)!} = \frac{(n)!}{(r)!(n-r)!} + \frac{(n)!}{(r+1)!(n-r-1)!} $
|
$$
\binom{n}{r} + \binom{n}{r+1} \\
\frac{n!}{(n-r)!r!} + \frac{n!}{(n-r-1)!(r+1)!} \\
\frac{n!}{(n-r)(n-r-1)!r!} + \frac{n!}{(n-r-1)!r!(r+1)} \\
\frac{n!}{(n-r-1)!r!}\left(\frac{1}{n-r} + \frac{1}{r+1}\right) \\
\frac{n!}{(n-r-1)!r!}\left(\frac{n+1}{(n-r)(r+1)}\right) \\
\frac{(n+1)!}{(n-r)!(r+1)!}\\
\binom{n+1}{r+1}
$$
|
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|
Uniform continuity of $f(x) = x \sin{\frac{1}{x}}$ for $x \neq 0$ and $f(0) = 0.$ For the $f(x) = x \sin{\frac{1}{x}}$ for $x \neq 0$ and $f(0) = 0,$ my text book asks the following questions.
(b) Why is $f$ uniformly continuous on any bounded subset of $\mathbb{R}$?
(c) Is $f$ uniformly continuous on $\mathbb{R}$??
The graph for the function is this.
For the question (b), if I take subset between $[0.2,0.6]$ or the subset where the slope is steep, I don't think the function is uniformly continuous because I think for a given $\epsilon>0$, there is no unique $\delta >0$ for the bounded subset. Therefore, it also cannot be uniformly continuous on $\mathbb{R}.$ However, the questions sounds like the function is uniformly continuous and the book says that it is uniformly continuous. The answer on the book says something but I need more explanation. Thanks.
|
Would anyone be so kind as to critique the following direct demonstration that $\displaystyle x \sin \frac{1}{x}$ is uniformly continuous on $(0,1)$ ?
Let $\epsilon > 0$ and let $x, y \in (0,1)$.
Then
\begin{align*}
x\sin\frac{1}{x} - y \sin\frac{1}{y}
&= x\sin\frac{1}{x} - y\sin\frac{1}{x} + y\sin\frac{1}{x} - y \sin\frac{1}{y} \\
&= (x-y)\sin\frac{1}{x} + y \left ( \sin\frac{1}{x} - \sin\frac{1}{y} \right ),
\end{align*}
so
\begin{equation}
\label{eq: star}
\left | x\sin\frac{1}{x} - y \sin\frac{1}{y} \right |
\leq |x - y| + y \left | \sin\frac{1}{x} - \sin\frac{1}{y} \right |.
\end{equation}
But
\begin{align}
\label{eq: starstar}
\left | \sin\frac{1}{x} - \sin\frac{1}{y} \right |
&= \left | 2 \cos \left ( \frac{1}{2} \left ( \frac{1}{x} + \frac{1}{y} \right ) \right ) \sin \left ( \frac{1}{2} \left ( \frac{1}{x} - \frac{1}{y} \right ) \right ) \right | \notag \\
&\leq 2 \left | \sin \frac{y-x}{2xy} \right | \notag \\
&\leq \frac{|y - x|}{|xy|} \notag \\
&= \frac{|x - y|}{xy}.
\end{align}
By \eqref{eq: star} and \eqref{eq: starstar} we have
\begin{equation}
\label{eq: potato}
\left | x \sin\frac{1}{x} - y \sin\frac{1}{y} \right |
\leq |x-y| + \frac{|x-y|}{x}
=|x-y| \left ( 1 + \frac{1}{x} \right ).
\end{equation}
By symmetry, we also have
\begin{equation}
\label{eq: banana}
\left | x \sin\frac{1}{x} - y \sin\frac{1}{y} \right |
\leq |x-y| \left ( 1 + \frac{1}{y} \right ).
\end{equation}
Now let $\gamma > 0$. \
\noindent
\textbf{Case I:}
Suppose $x \geq \gamma$ or $y \geq \gamma$.
Then \eqref{eq: potato} and \eqref{eq: banana} imply that
\begin{equation}
\left | x \sin\frac{1}{x} - y \sin\frac{1}{y} \right |
\leq |x-y| \left (1 + \frac{1}{\gamma} \right ),
\end{equation}
so we should let $\displaystyle \delta = \frac{\epsilon}{1 + \frac{1}{\gamma}}$. \
\noindent
\textbf{Case II:}
Suppose instead that $x < \gamma$ and $y < \gamma$.
Then $\displaystyle \left | x \sin\frac{1}{x} - y \sin\frac{1}{y} \right |$ equals either $\displaystyle x \sin\frac{1}{x} - y \sin\frac{1}{y}$ or $\displaystyle y \sin\frac{1}{y} - x \sin\frac{1}{x}$.
But
\begin{equation}
x \sin\frac{1}{x} - y \sin\frac{1}{y}
\leq x - (-y)
= x + y
< 2 \gamma,
\end{equation}
and similarly,
\begin{equation}
y \sin\frac{1}{y} - x \sin\frac{1}{x} < 2 \gamma,
\end{equation}
hence
\begin{equation}
\left | x \sin\frac{1}{x} - y \sin\frac{1}{y} \right | < 2 \gamma,
\end{equation}
so we should set $\displaystyle \gamma = \frac{\epsilon}{2}$.
\noindent
To summarise, let
\begin{equation}
\delta = \frac{\epsilon}{1 + \frac{1}{\epsilon/2}}
= \frac{\epsilon}{1 + \frac{2}{\epsilon}}
= \frac{\epsilon^2}{\epsilon + 2}.
\end{equation}
Then
\begin{equation}
|x-y| < \delta \Rightarrow \left | x \sin \frac{1}{x} - y \sin \frac{1}{y} \right | < \epsilon.
\end{equation}
\noindent
Therefore $\displaystyle h(x) = x \sin \frac{1}{x}$ is uniformly continuous on $(0,1)$.
|
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|
Check my solution: Find $\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x)$ Find $\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x)$
First I rationalized the numerator,
$$
\begin{align*}
&\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x) \cdot \frac{\sqrt{(x+a_1)(x+a_2)}+x}{\sqrt{(x+a_1)(x+a_2)}+x} \\
=&\lim_{x\to+\infty}\frac{x(a_1+a_2)+a_1a_2}{\sqrt{(x+a_1)(x+a_2)}+x} \\
=&\lim_{x\to+\infty}\frac{(a_1+a_2)+(a_1a_2)/x}{\sqrt{(\frac{1}{x}+\frac{a_1}{x^2})(\frac{1}{x}+\frac{a_2}{x^2})}+1} \\
=& a_1+a_2
\end{align*}
$$
Some calculations in my calculator tells me my answer is likely to be wrong.
I think the real answer is $\frac{a_1+a_2}{2}$.
Can anyone find my mistake or drop me a hint?
|
You do it right until the last line $\lim_{x\to+\infty}\frac {x(a_1+a_2)+a_1a_2}{\sqrt {(x+a_1)(x+a_2)}+x}=\lim_{x\to+\infty}\frac {x(a_1+a_2)+a_1a_2}{\sqrt {x^2+(a_1+a_2)x+a_1a_2} +x}$=$\lim_{x\to+\infty}\frac {x((a_1+a_2)+a_1a_2x^{-1})}{|x| \left( \sqrt {1+(a_1+a_2)x^{-1}+a_1a_2x^{-2}}+1 \right)} $
For $x>0$ we have $|x|=x$ and thus we have $\lim_{x\to+\infty}\frac {(a_1+a_2)+a_1a_2}{\sqrt {1+(a_1+a_2)x^{-1}+a_1a_2x^{-2}}+1}=
(a_1+a_2)/2$
|
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|
Express $\cos2\theta$ in terms of $\cos$ and $\sin$ (De Moivre's Theorem) Use De Moivre's to express $\cos2\theta$ in terms of powers of $\sin$ and $\cos$.
What I have is:
$\cos2\theta + i\sin2\theta\\
= (\cos\theta + i \sin\theta)^2\\
= \cos^2\theta + 2 \cos\theta ~i \sin\theta + (i \sin)^2\theta\\
= \cos^2\theta + i(2\cos\theta \sin\theta) - \sin^2\theta\\
= \cos^2\theta - \sin^2\theta + i(2\cos\theta \sin\theta)
$
so $\cos2\theta = \cos^2\theta - \sin^2\theta$
Is this correct?
|
using de moivres theorem we can generalise.$$\cos(n\theta)=\cos^n(\theta)-\frac{n(n-1)}{2!}\cos^{n-2}(\theta)\sin^2(\theta)+\frac{n(n-1)(n-2)(n-3)}{4!} \cos^{n-4}(\theta)\sin^4(\theta)........$$
|
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|
how to find the asymptotic expansion of the following sum: I need to determine an asymptotic expansion when $q \rightarrow 1$ of the sum
$$S(q)=\sum_{n=0}^{\infty} \frac{q^n}{ (q^n + 1)^2 }.$$
Numerical computations suggest that $S(q)\sim\frac{c}{|q-1|}$ with $c \approx 0.5$.
|
Suppose that $q>1$ and re-write the sum as
$$S(q) = \frac{1}{4} + \sum_{n\ge 1} \frac{q^n}{(q^n+1)^2}$$
The sum term is harmonic and may be re-written as
$$T(x) = \sum_{n\ge 1} \frac{e^{nx}}{(e^{nx}+1)^2}$$
which we will evaluate at $x=\log q.$
We will evaluate $T(x)$ by inverting its Mellin transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad
g(x) = \frac{e^x}{(e^x+1)^2}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \frac{e^x}{(e^x+1)^2} x^{s-1} dx
= \int_0^\infty \frac{e^{-x}}{(1+e^{-x})^2} x^{s-1} dx
= \int_0^\infty \sum_{m\ge 1} (-1)^{m+1} m e^{-mx} x^{s-1} dx\\
\sum_{m\ge 1} (-1)^{m+1} m \int_0^\infty e^{-mx} x^{s-1} dx
= \Gamma(s) \sum_{m\ge 1} \frac{(-1)^{m+1} m}{m^s}
= \Gamma(s) \sum_{m\ge 1} \frac{(-1)^{m+1}}{m^{s-1}}\\
= \Gamma(s)
\left(1-\frac{1}{2^{s-2}}\right) \zeta(s-1).$$
Therefore the Mellin transform $Q(s)$ of $T(x)$ is given by
$$Q(s) = \Gamma(s)
\left(1-\frac{1}{2^{s-2}}\right) \zeta(s-1) \zeta(s).$$
The Mellin inversion integral for $Q(s)$ is
$$\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds.$$
The two zeta function terms taken together cancel the poles of the gamma function, so we are left with just two poles/residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) = \frac{1}{2x}
\quad\text{and}\quad
\mathrm{Res}(Q(s)/x^s; s=0) = -\frac{1}{8}.$$
Therefore in a neighborhood of zero we have that
$$T(x) \sim \frac{1}{2x}-\frac{1}{8}.$$
Now as $q$ goes to one $\log q$ goes to zero, so this expansion at $x=\log q$ definitely applies. We get that $$S(q) = \frac{1}{4} + \frac{1}{2x}-\frac{1}{8}
= \frac{1}{8} + \frac{1}{2\log q}.$$
This approximation is better than the conjecture from the original question.
Now use that for $q$ close to one,
$$\frac{1}{2\log q} \sim \frac{1}{2(q-1)} + \frac{1}{4} - \frac{1}{24} (q-1)+\cdots$$
to conclude that
$$S(q) \sim \frac{3}{8} + \frac{1}{2(q-1)}.$$
|
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|
Arithmetic series relationship with difference of two consecutive cubes. Is this a thing? Excuse my dodgy notation and my write-up in general, this is the first proof I've done since leaving school a while back. Anywho, has anyone come across anything like this before? Read the whole thing, not just the arithmetic series stuff.
The sum of an ascending series, where a is the first integer, k is the common difference, and there are n terms.
$a+(a+k)+(a+2k)+⋯(a+(n-1)k)$
$$=na+[0+(n-1)k)+[1+(n-2)k]+[2+(n-3)k]+⋯$$
$$=na+\frac {n}{2}(n-1)k$$
$$=n\left[a+\frac{(n-1)k}{2}\right]$$
Difference between two consecutive cubes:
$2^3-1^3=7=(1×6)+1$
$3^3-2^3=19=(1+2)×6+1$
$4^3-3^3=37=(1+2+3)×6+1$
By observation, the difference between two consecutive cubes is equal to six times the sum of all integers between $0$ and the larger cubed integer, plus $1$.
Algebraically:
$$1+6\sum_{i=1}^x a_i=1+6{n\left[a+\frac{(n-1)k}{2}\right]}$$
Where $1$ is the first term and the common difference, and there are x terms.
$$=1+6{x\left[1+\frac{(x-1)}{2}\right]}$$
$$=1+6x+3x^2-3x$$
$$=3x^2+3x+1$$
$$=x^3+3x^2+3x+1-x^3$$
$$=(x+1)^3-x^3$$
Q.E.D
P.S. I always wanted to use Q.E.D but don't normally get the chance.
|
I think you can prove it directly :
$$(n+1)^3 - n^3 = 3 n^2 + 3 n + 1 = 3 n (n + 1) + 1 = 6 \frac{n (n + 1)} 2 + 1 $$
$\frac{n (n + 1)} 2$ being the sum of all integers from $1$ to $n$
|
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|
Approximating the square root of two with fractions I would like to prove that there exist only finitely many $m, n \in \mathbb{N}$ satisfying $$\left | \sqrt{2} - \frac{m}{n} \right | < \frac{1}{4n^2}.$$
Any thoughts?
Thank you for your help.
|
$$|\sqrt{2} - \frac{m}{n}| < \frac{1}{4n^2}\\\implies
|\sqrt{2}n-m|< \frac{1}{4n}\\\implies
\sqrt{2}n-\frac{1}{4n}<m<\sqrt{2}n+\frac{1}{4n}\\\implies
0<|(\sqrt{2}n-m)(\sqrt{2}n+m)|<\frac{1}{4n}(2\sqrt{2}n+\frac{1}{4n})=\frac{\sqrt{2}}{2}+\frac{1}{16n^2}<1.$$
But $|(\sqrt{2}n-m)(\sqrt{2}n+m)|=|2n^2-m^2|$ is an integer, a contradiction.
|
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|
If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this?
Prove that if $n$ is a natural number and $(3n+1)$ & $(4n+1)$ are both perfect squares, then $56$ will divide $n$.
Clearly we have to show that $7$ and $8$ both will divide $n$.
I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd perfect square.
- so we have $4n+1\equiv 1\pmod{8}$; from this $2|n$ so $3n+1$ is a odd perfect square.
- so $3n+1\equiv 1\pmod{8}$ so $8|n$ but I can't show $7|n$. How do I show this?
Thanks for the help.
|
Let $3n+1=K^2$ and $4n+1=P^2$, where $P$ and $K$ are integers. Therefore, we know that $4(K^2)-3(P^2)=1$. Rearranging terms gives us $K^2= \frac {1+3P^2}4$. By testing values and guesswork we know that $P=13$ and $K=15$. Therefore $3n+1=169$, and then solving for $n$ gives us $n=56$.
|
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|
Limit with Stolz-Cesàro theorem: $\lim\limits_{n\to \infty} \frac{1+2\sqrt2+3\sqrt3+\ldots+n\sqrt n}{n^2 \sqrt{n}}$ $$\lim_{n\to \infty} \frac{1+2\sqrt2+3\sqrt3+\ldots+n\sqrt n}{n^2 \sqrt{n}}= \text{?}$$
Book has no answers.
It's on Stolz-Cesàro theorem lesson, if that helps. Can't find a solution.
|
This is the direct application of the Stolz-Cesàro theorem
$$
\begin{align}
&\lim_{n\to \infty} \frac{1+2\sqrt{2}+...+n\sqrt{n}}{n^2 \sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(1+2\sqrt{2}+...+(n+1)\sqrt{n+1})-(1+2\sqrt{2}+...+n\sqrt{n})}{(n+1)^2 \sqrt{n+1}-n^2\sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(n+1)\sqrt{n+1}}{(n+1)^2 \sqrt{n+1}-n^2\sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(n+1)\sqrt{n+1}((n+1)^2 \sqrt{n+1}+n^2\sqrt{n})}{((n+1)^2 \sqrt{n+1}-n^2\sqrt{n})((n+1)^2 \sqrt{n+1}+n^2\sqrt{n})}\\
=&\lim_{n\to \infty} \frac{(n+1)\sqrt{n+1}((n+1)^2 \sqrt{n+1}+n^2\sqrt{n})}{(n+1)^5-n^5}\\
=&\lim_{n\to \infty} \frac{(n+1)^4+(n+1)n^2\sqrt{n(n+1)}}{5n^4+10n^3+10n^2+5n+1}\\
=&\lim_{n\to \infty} \frac{(1+n^{-1})^4+(1+n^{-1})\sqrt{1+n^{-1}}}{5+10n^{-1}+10n^{-2}+5n^{-3}+n^{-4}}\\
=&\frac{2}{5}
\end{align}
$$
|
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"url": "https://math.stackexchange.com/questions/575992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Finding determinant using properties of determinant without expanding show that determinant
$$\left|\matrix{
x^2+L & xy & xz \\
xy & y^2+L & yz \\
xz & yz & z^2+L \\
}\right|
= L^2(x^2+y^2+z^2+L)$$
without expanding by using the appropriate properties of determinant.
All i can do is LHS
$$x^2y^2z^2\left|\matrix{
1+L/x^2 & 1 & 1 \\
1 & 1+L/y^2 & 1 \\
1 & 1 & 1+L/z^2 \\
}\right|$$
Is it a must to relate to eigenvalue problem?
|
$$\left|\matrix{
x^2+L & xy & xz \\
xy & y^2+L & yz \\
xz & yz & z^2+L \\
}\right|
= $$
$$xyz\left|\matrix{
x+L/x & y & z \\
x & y+L/y & z \\
x & y & z+L/z \\
}\right|
= $$
$$=\left|\matrix{
x^2+L & y^2 & z^2 \\
x^2 & y^2+L & z^2 \\
x^2 & y^2 & z^2+L \\
}\right|
= $$
$$=\left|\matrix{
x^2+y^2+z^2+L & y^2 & z^2 \\
x^2+y^2+z^2+L & y^2+L & z^2 \\
x^2+y^2+z^2+L & y^2 & z^2+L \\
}\right|
= $$ $$=(x^2+y^2+z^2+L)\left|\matrix{
1 & y^2 & z^2 \\
0 & L & 0\\
0 & 0 & L \\
}\right|=L^2(x^2+y^2+z^2+L)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/578917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Determine one triple of positive integer Note that $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$ and $\frac{1}{6}-\frac{1}{7}=\frac{1}{6\times 7}$ and $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{42}=1$.
Determine one triple $(x,y,z)$ of positive integers with $1000<x<y<z<2000$ and $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{45}=1$.
A friend of mine asked me this question. I have only been able to guess one solution but $x,y,z$ doesn't fall into the range.
Note: This question is meant to be done within 10-15min and without a computer.
|
There are three solutions for
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} + \frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{45} = 1$
in the required range, I
found them by exhaustive search:
$$x=1771,\;y=1932,\;z=1980$$
$$x=1806,\;y=1892,\;z=1980$$
$$x=1830,\;y=1891,\;z=1953$$
|
{
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"url": "https://math.stackexchange.com/questions/580290",
"timestamp": "2023-03-29T00:00:00",
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|
Generalizing $\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\operatorname dx}{x^{2}+1} = \frac{5\pi^{2}}{96}$ The following integral
\begin{align*}
\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{dx}{x^{2}+1} = \frac{5\pi^{2}}{96}
\tag{1}
\end{align*}
is called the Ahmed's integral and became famous since its first discovery in 2002. Fascinated by this unbelievable closed form, I have been trying to generalize this result for many years, though not successful so far.
But suddenly it came to me that some degree of generalization may be possible. My conjecture is as follows: Define the (generalized) Ahmed integral of parameter $p$, $q$ and $r$ by
\begin{align*}
A(p, q, r) := \int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1}}{q \sqrt{p^{2}x^{2} + 1} } \, \frac{pqr \, dx}{(r^{2} + 1)p^{2} x^{2} + 1}.
\end{align*}
Now suppose that $p q r = 1$, and define its complementary parameters as
\begin{align*}
\tilde{p} = r \sqrt{\smash{q}^{2} + 1}, \quad
\tilde{q} = p \sqrt{\smash{r}^{2} + 1}, \quad \text{and} \quad
\tilde{r} = q \sqrt{\smash{p}^{2} + 1},
\tag{2}
\end{align*}
Then my guess is that
\begin{align*}
A(p, q, r) = \frac{\pi^{2}}{8} + \frac{1}{2} \left\{ \arctan^{2} (1 / \tilde{p} ) - \arctan^{2} ( \tilde{q} ) - \arctan^{2} ( \tilde{r} ) \right\}.
\end{align*}
Plugging the values $(p, q, r) = (1/\sqrt{2}, \sqrt{2}, 1)$, the corresponding complementary parameters become $(\tilde{p}, \tilde{q}, \tilde{r}) = (\sqrt{3}, 1, \sqrt{3})$. Then for these choices, the original Ahmed's integral $\text{(1)}$ is retrieved:
\begin{align*}
\int_{0}^{1} \frac{\arctan \sqrt{x^{2} + 2} }{\sqrt{x^{2} + 2} } \, \frac{dx}{x^{2} + 1}
&= \frac{\pi^{2}}{8} + \frac{1}{2} \left\{ \arctan^{2} \frac{1}{\sqrt{3}} - \arctan^{2} 1 - \arctan^{2} \sqrt{3} \right\} \\
&= \frac{5\pi^{2}}{96}.
\end{align*}
In fact, I have a more generalized conjecture involving dilogarithms depending on complementary parameters. But since this specialized version is sufficiently daunting, I won't deal with it here.
Unfortunately, proving this relation is not successful so far. I just heuristically calculated and made some ansatz to reach this form. Can you help me improve the situation by proving this or providing references to some known results?
EDIT. I finally succeeded in proving a general formula: let $k = pqr$ and complementary parameters as in $\text{(2)}$. Then whenever $k \leq 1$, we have
\begin{align*}
A(p, q, r)
&= 2\chi_{2}(k) - k \arctan (\tilde{p}) \arctan \left( \frac{k}{\tilde{p}} \right) \\
&\quad + \frac{k}{2} \int_{0}^{1} \frac{1}{1-k^{2}x^{2}} \log\left( \frac{1+\tilde{p}^{2}x^{2}}{1+\tilde{p}^{2}} \times \frac{1+\tilde{q}^{2}x^{2}}{1+\tilde{q}^{2}} \times \frac{1+\tilde{r}^{2}x^{2}}{1+\tilde{r}^{2}} \right) \, dx.
\end{align*}
Then the proposed conjecture follows as a corollary. I'm planning to gather materials related to the Ahmed's integrals and put into a combined one. You can find an ongoing proof of this formula here.
|
The following is only a partial answer, but it might be useful.
Assuming that all the parameters are positive, the integral $$ I(p,q,r) = \int_{0}^{1} \frac{\operatorname{arccot} q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \, dx $$ can be expressed in terms of $I \left(\frac{1}{q}, \frac{1}{p}, \frac{1}{r} \right)$.
$$ \small \begin{align} & \int_{0}^{1} \frac{\operatorname{arccot} q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \, dx \\ &= \int_{0}^{1} \int_{0}^{1} \frac{1}{t^{2}+p^{2}q^{2}x^{2}+q^{2}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \,dt \, dx \\&= \int_{0}^{1} \frac{(r^{2}+1)pqr}{q^{2}r^{2}+(r^{2}+1)t^{2}} \int_{0}^{1} \frac{1}{(r^{2}+1)p^{2}x^{2}+1} \, dx \, dt - \int_{0}^{1} \int_{0}^{1} \frac{pq^{3}r}{q^{2}r^{2}+(r^{2}+1)t^{2}} \frac{1}{t^{2}+p^{2}q^{2}x^{2}+q^{2}} \,dx \, dt \\ &= (r^{2}+1)pqr \, \frac{\arctan \left(\frac{\sqrt{r^{2}+1}}{qr} \right)}{qr \sqrt{r^{2}+1}} \frac{\arctan (p\sqrt{r^{2}+1})}{p \sqrt{r^{2}+1)}} - \int_{0}^{1} \frac{pq^{3}r}{q^{2}r^{2}+(r^{2}+1)t^{2}} \frac{\operatorname{arccot} \left(\frac{1}{p}\sqrt{\frac{t^{2}}{q^{2}}+1} \right)}{pq^{2}\sqrt{\frac{t^{2}}{q^{2}}+1}} \, dt \\ &= (r^{2}+1)pqr \, \frac{\arctan \left(\frac{\sqrt{r^{2}+1}}{qr} \right)}{qr \sqrt{r^{2}+1}} \frac{\arctan (p\sqrt{r^{2}+1})}{p \sqrt{r^{2}+1)}} - \int_{0}^{1} \frac{\frac{1}{pqr}}{\frac{1+r^{2}}{r^{2}} \frac{t^{2}}{q^{2}}+1}\frac{\operatorname{arccot} \left(\frac{1}{p}\sqrt{\frac{t^{2}}{q^{2}}+1} \right)}{p\sqrt{\frac{t^{2}}{q^{2}}+1}} \, dt \\ &= (r^{2}+1)pqr \, \frac{\arctan \left(\frac{\sqrt{r^{2}+1}}{qr} \right)}{qr \sqrt{r^{2}+1}} \frac{\arctan (p\sqrt{r^{2}+1})}{p \sqrt{r^{2}+1)}} - I \left(\frac{1}{q}, \frac{1}{p}, \frac{1}{r} \right).\end{align}$$
And by making the substitution $u= \frac{1}{x}$ followed by the substitution $w^{2}= p^{2}+u^{2}$, one can show that
$$ \begin{align} &\int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \,dx \\ &= \frac{\pi}{2} \int_{0}^{1} \frac{1}{ q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} dx - \int_{0}^{1} \frac{\text{arccot} \, q \sqrt{p^{2}x^{2}+1}}{q\sqrt{p^{2}x^{2}+1}} \frac{pqr}{(r^{2}+1)p^{2}x^{2}+1} \,dx \\ &= \frac{\pi}{2} \, \text{arctan} \left(\frac{pr}{\sqrt{p^{2}+1}} \right) -I(p,q,r). \end{align}$$
|
{
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"url": "https://math.stackexchange.com/questions/580521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "193",
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|
Compute a sum with partial fraction decomposition and generating functions I am trying to compute $$\sum_{k=1}^{n-1} \frac{1}{k(n-k)}$$ using a term by term partial fraction decomposition and also with generating functions, but I'm stuck.
|
Note that:
$$
\frac{1}{k (n - k)} = \frac{1}{k n} + \frac{1}{n (n - k)}
$$
Thus your sum is:
$\begin{align}
\sum_{1 \le k \le n - 1} \frac{1}{k (n - k)}
&= \sum_{1 \le k \le n - 1} \frac{1}{k n}
+ \sum_{1 \le k \le n - 1} \frac{1}{n (n - k)} \\
&= \frac{1}{n} \sum_{1 \le k \le n - 1} \frac{1}{k}
+ \frac{1}{n} \sum_{1 \le k \le n - 1} \frac{1}{n - k} \\
&= \frac{2}{n} \sum_{1 \le k \le n - 1} \frac{1}{k} \\
&= \frac{2}{n} H_{n - 1}
\end{align}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/580732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find all solutions for a complex equation: $(1+i)z^2 - (6+i)z + 9+7i=0$ There is this math assignment that we've been given to find all the answers for some diffrent math problems.
The problem is: $(1+i)z^2 - (6+i)z + 9+7i=0$, find all the solutions and answer in geometric form.
I've tried the following:
$(1+i)z^2 - (6+i)z + 9 + 7i = 0 $
$z^2 - ((6+i)*(1-i)z) + ((9+7i)*(1-i)) = 0$ //Leaves $z^2$ by itself.
$z^2 - (7-5i)z + (16-2i) = 0$
$z^2 - (7-5i)z + ((7-5i)/2)^2 - ((7-5i)/2)^2 + (16-2i) = 0$ //completing the square
$(z - ((7-5i)/2))^2 - ((7-5i)/2)^2 + (16-2i) = 0$
Have I done something wrong? And how should I proceed?
Alright, after some recalculations, I reach this:
$$(z-((7-5i)/2))^2 - ((7-5i)^2)/4 + (32-4i)/4 = 0$$
$$(z-((7-5i)/2))^2 - (56-4i)/4 = 0$$
|
$$(1+i)z^2-(6+i)z+9+7i=0\Longleftrightarrow$$
Use the abc-formula:
$$z=\frac{-\left(-(6+i)\right)\pm\sqrt{\left(-(6+i)\right)^2-4\cdot\left(1+i\right)\cdot\left(9+7i\right)}}{2\cdot\left(1+i\right)}\Longleftrightarrow$$
$$z=\frac{-\left(-6-i\right)\pm\sqrt{\left(-6-i\right)^2-4\cdot\left(1+i\right)\cdot\left(9+7i\right)}}{2+2i}\Longleftrightarrow$$
$$z=\frac{\left(6+i\right)\pm\sqrt{\left(35+12i\right)-4\cdot\left(1+i\right)\cdot\left(9+7i\right)}}{2+2i}\Longleftrightarrow$$
$$z=\frac{\left(6+i\right)\pm\sqrt{\left(35+12i\right)-4\cdot\left(2+16i\right)}}{2+2i}\Longleftrightarrow$$
$$z=\frac{\left(6+i\right)\pm\sqrt{\left(35+12i\right)-\left(8+64i\right)}}{2+2i}\Longleftrightarrow$$
$$z=\frac{\left(6+i\right)\pm\sqrt{27-52i}}{2+2i}\Longleftrightarrow$$
$$z=\frac{6+i}{2+2i}\pm\frac{1}{2+2i}\sqrt{27-52i}\Longleftrightarrow$$
$$z=\left(\frac{7}{4}-\frac{5}{4}i\right)\pm\left(\frac{1}{4}-\frac{1}{4}i\right)\sqrt{27-52i}$$
|
{
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"url": "https://math.stackexchange.com/questions/583213",
"timestamp": "2023-03-29T00:00:00",
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|
Find the residue of $\frac{e^{iz}}{(z^2+1)^5}$ at $z = i$ and evaluate $\int_0^{\infty} \cos x/(x^2+1)^5 dx$ I know the evaluation of $\int_0^{\infty} \cos x/(x^2+1)^5 dx$ requires that I solve the first part, but for some reason I'm stumped. I get that I should use $\lim_{z \to i}\frac{1}{24}\frac{d^4}{dz^4}((z-i)^5\frac{e^{iz}}{(z^2+1)^5})$, but I can't seem to evaluate that for the life of me.
|
$$\frac{d^4}{dz^4}\left((z-i)^5\frac{e^{iz}}{(z-i)^5(z+i)^5}\right)= \frac{d^4}{dz^4}\left(e^{iz}(z+i)^{-5}\right).$$
By Leibnitz Rule
$$ \frac{d^4}{dz^4}\left(e^{iz}(z+i)^{-5}\right) = \binom{4}{0} \left(\frac{d^4}{dz^4} e^{iz}\right)(z+i)^{-5}+\binom{4}{1} \left( \frac{d^3}{dz^3}e^{iz}\right) \left( \frac{d}{dz}(z+i)^{-5} \right) \\ +\binom{4}{2} \left( \frac{d^2}{dz^2}e^{iz}\right) \left( \frac{d^2}{dz^2}(z+i)^{-5} \right)+\binom{4}{3} \left( \frac{d}{dz}e^{iz}\right) \left( \frac{d^3}{dz^3}(z+i)^{-5} \right)\\+\binom{4}{4} \left( e^{iz}\right) \left( \frac{d^4}{dz^4}(z+i)^{-5} \right)$$
$$=e^{iz}(z+i)^{-5}+4 (-i)e^{iz} (-5) (z+i)^{-6} \\
-6e^{iz}(30)(z+i)^{-7} +4i e^{iz}(-210)(z+i)^{-8} +1680 e^{iz}(z+i)^{-9} $$
$$=\frac{e^{iz}}{(z+i)^{9}} \left( (z+i)^4+20i(z+i)^{3} -180(z+i)^2 -840i(z+i) +1680 \right)$$
|
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|
Prove using mathematical induction that $2^{3n}-1$ is divisible by $7$ So, i wanna prove $2^{3n}-1$ is divisible by $7$, so i made this:
$2^{3n}-1 = 7\cdot k$ -> for some $k$ value
$2^{3n+1} = 1+2\cdot1 - 2\cdot1 $
$2^{3n+1} - 1-2\cdot1 + 2\cdot1 $
$2^{3n}\cdot2 - 1-2\cdot1 + 2\cdot1$
$2(2^{3n}-1) -1 +2$
$2\cdot7k+1$ -> made this using the hypothesis.
so, i dont know if its right, or if its wrong, i dont know how to keep going from this, or if its the end.
Thanks.
|
$f(n) = 2^{3n}-1 $
$f(0) = 0$ and $7|0$
Suppose that $7|f(n)$, let's say $f(n) = 7k$, $\Rightarrow f(n+1) = 2^{3n+3}-1 = 8\cdot2^{3n}-1 = 8\cdot2^{3n}-1 + 8 - 8 = 8(2^{3n}-1) - 7 = 8\cdot(7k) - 7 = 7\cdot(8k-1)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/584686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Maximizing slope of a secant line Two points on the curve $$ y=\frac{x^3}{1+x^4}$$ have opposite $x$-values, $x$ and $-x$. Find the points making the slope of the line joining them greatest.
Wouldn't the maximum slope of the secant line be with the max/min of the curve?
So $x=3^{1/4}$ and $x=-3^{1/4}$?
|
Replace $x^2=\tan(\alpha)$. We have $$\text{Slope} = \frac{\frac{x^3}{1+x^4}-\frac{-x^3}{1+x^4}}{2x}= \frac{x^2}{1+x^4} = \frac{\sin(2\alpha)}{2}\le\frac{1}{2}$$ with equality exactly when $\alpha=n\pi+\frac{\pi}{4}, x^2=1$
|
{
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"url": "https://math.stackexchange.com/questions/586181",
"timestamp": "2023-03-29T00:00:00",
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|
logarithms equation Hello I have following problem: solve equation $\log{(x-5)^2}+\log{(x+6)^2}=2$
and I rewrited this equation as
$2\log{(x-5)}+2\log{(x+6)}=\log{100} \implies 2(\log{(x-5)(x+6))=\log{100}} \implies \log{x^2+x-30}=\log10 \implies x^2+x-40=0 $
and I solved this equation, but I obtained only two solutions and there should be four, so I wonder if it is necessary to create $\log{(x-5)^2(x+6)^2}=\log{100}$ or is a simplier way.
|
As the logarithm term contains squares, we can allow $x-5,x+6$ to be negative
In fact, $\log b^2=2\log |b|$ for real $b$
As $\log A+\log B=\log (AB),$
$$\implies \log_{10}\{(x-5)(x+6)\}^2=2$$
$$\implies \{(x-5)(x+6)\}^2=10^2=100$$
$$\implies (x-5)(x+6)=\pm10$$
|
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"url": "https://math.stackexchange.com/questions/588188",
"timestamp": "2023-03-29T00:00:00",
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|
Finding a Jordan canonical form $J$ and an invertible matrix $Q$ such that $J = Q^{-1}AQ$ $$
A = \begin{bmatrix}
0 & 1 & -1 \\
-4 & 4 & -2 \\
-2 & 1 & 1
\end{bmatrix}
$$
The characteristic polynomial can be found to be: $p(t)= -(t-1)(t-2)^2$.
I found $E_{\lambda_1} = N(A - I) = (1, 2, 1)$.
I'm having problems finding $E_{\lambda_2}$:
$$
(A - 2I) = \begin{bmatrix}
-2 & 1 & -1 \\
-4 & 2 & -2 \\
-2 & 1 & -1
\end{bmatrix}
$$
and
$$
(A - 2I)^2 = \begin{bmatrix}
2 & -1 & 1 \\
4 & -2 & 2 \\
2 & -1 & 1
\end{bmatrix}
$$
But there are numerous $\beta$'s that can be found for $(A - 2I)$, such as $\beta = (1, 1, -1)$ or $\beta = (1, 0, -2)$, etc., etc.
Then for $(A - 2I)^2$, these same $\beta$'s can be used, so I already know that I have messed up somewhere. Can anyone offer some suggestions?
|
We asked to find the Jordan Normal Form of the matrix:
$$
A = \begin{bmatrix}
0 & 1 & -1 \\
-4 & 4 & -2 \\
-2 & 1 & 1
\end{bmatrix}
$$
We start out by solving $|A - \lambda I| = 0$ to find the eigenvalues, which gives:
$$-(\lambda - 2)^2(\lambda-1) = 0 \rightarrow \lambda_1 = 1,~ \lambda_2 = 2, ~\lambda_3 = 2$$
We have one eigenvalue that has algebraic multiplicity one and a second one that has multiplicity two.
Next, we want to find the three linearly independent eigenvectors $[A - \lambda_i I]v_i = 0$.
For $\lambda_1 = 1$, we get:
$$\begin{bmatrix} -1 & 1 & -1 \\ -4 & 3 & -2 \\ -2 & 1 & 0 \end{bmatrix}v_1 = 0$$
The RREF yields:
$$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}v_1 = 0$$
This leads to the eigenvector $v_1 = (1,2,1)$.
For $\lambda_{2,3} = 2$, we get:
$$\begin{bmatrix} -2 & 1 & -1 \\ -4 & 2 & -2 \\ -2 & 1 & 1 \end{bmatrix}v_{2,3} = 0$$
The RREF yields:
$$\begin{bmatrix} 1 &-\dfrac{1}{2} & \dfrac{1}{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}v_{2,3} = 0$$
This fortunately leads to two independent eigenvectors (not always the case):
*
*$v_2 = (-1,0,2)$
*$v_3 = (1,2,0)$
We are now able to find the diagonal matrix as:
$$J = Q^{-1} \cdot A \cdot Q$$
$Q$ is made up of the column eigenvectors as:
$$Q = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix}$$
This should yield:
$$J = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$
Now, you can also do this the way you were trying and should get an identical result (can be different as eigenvectors are not unique).
|
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|
How to calculate such sums of Legendre symbols? How to calculate such sums as $\sum_{x\in\mathbb{F}_p} \left(\frac{x^2+ax+b}{p} \right)$
If $x^2+ax+b$has a root, $b$ may be eliminated and the sum is evaluated to be $0+\sum_{x\in\mathbb{F}_p^*} \left(\frac{1+ax^{-1}}{p} \right)=-1$. Otherwise to linearize it, you need to go into quadratic extensions. So I guess there may be a solution resorting to extensions of $\mathbb{F}_p$.
It is easy to extend the symbol since the extension is cyclic. But then I need to sum in an unfamiliar space. Can we fix this and eventually give a proof in this flavor?
Any reference is welcome.
|
We have in general for $p>2$ an odd prime
$$\sum_{x=0}^{p-1}{\left(\frac{x^2+ax+b}{p}\right)}=\begin{cases} -1 & \text{if} \; p\nmid a^2-4b \\ p-1 & \text{if} \; p\mid a^2-4b \end{cases}$$
We begin by completing the square, using that $(\frac{4}{p})=1$:
\begin{align}
\sum_{x=0}^{p-1}{\left(\frac{x^2+ax+b}{p}\right)}& =\sum_{x=0}^{p-1}{\left(\frac{4}{p}\right)\left(\frac{x^2+ax+b}{p}\right)} \\
& =\sum_{x=0}^{p-1}{\left(\frac{4x^2+4ax+4b}{p}\right)} \\
&=\sum_{x=0}^{p-1}{\left(\frac{(2x+a)^2-(a^2-4b)}{p}\right)} \\
&=\sum_{y=0}^{p-1}{\left(\frac{y^2-k}{p}\right)}
\end{align}
where in the last step $k=a^2-4b$ and we have used that $y=2x+a$ ranges over all values $\pmod{p}$ as $x$ ranges over all values $\pmod{p}$. Recall that we are only looking at $p>2$.
If $p \mid a^2-4b=k$, then we get
$$\sum_{y=0}^{p-1}{\left(\frac{y^2}{p}\right)}=p-1$$
If $p \nmid a^2-4b=k$, then from my answer to this question we get
$$\sum_{y=0}^{p-1}{\left(\frac{y^2-k}{p}\right)}=-1$$
I guess I should do $p=2$ as well:
$$\sum_{x=0}^{1}{\left(\frac{x^2+ax+b}{2}\right)}=\left(\frac{b}{2}\right)+\left(\frac{1+a+b}{2}\right)=\begin{cases} 1 & \text{if} \; (a, b) \equiv (0, 0) \pmod{2} \\ 1 & \text{if} \; (a, b) \equiv (0, 1) \pmod{2} \\ 0 & \text{if} \; (a, b) \equiv (1, 0) \pmod{2} \\ 2 & \text{if} \; (a, b) \equiv (1, 1) \pmod{2} \end{cases}$$
|
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|
How to prove that either $2^{500} + 15$ or $2^{500} + 16$ isn't a perfect square? How would I prove that either $2^{500} + 15$ or $2^{500} + 16$ isn't a perfect square?
|
Short solutions: $2^{1000} + 15 \equiv 3 \pmod{4}$, $2^{1000} + 16 \equiv 2 \pmod{3}$, neither of which are quadratic residues.
More elementary solution: $2^{1000} = (2^{500})^2$. The next perfect square is $$(2^{500} + 1)^2 = 2^{1000} + 2(2^{500}) + 1$$ which is clearly more than both $2^{1000} + 15$ and $2^{1000} + 16$. Therefore, these two can't possibly be perfect squares.
|
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|
Simpson Rule special case of Gauss quadrature Prove that the Simpson Rule for integrating the function $f$ on the interval $[a,b]$: $$I_2(f)=\frac{b-a}{6}(f(a)+4f(\frac{a+b}{2})+f(b))$$ is actually the Gauss quadrature for the weight $g=1$ and $3$ nodes.
I can not get why when we have $3$ nodes in the Gauss quadrature the nodes have to be the ones from the formula above! Can anyone give me some hints on how to prove this?
|
Start from the definition of a Gauss quadrature:
$$\int_{-1}^1 f(x)\ dx \approx \sum_{i=1}^3 g_i f(x_i).$$
We perform a change of variables to change the range of integration, so
$$\int_a^b f(x)\ dx = \frac{b-a}{2}\int_{-1}^1 f\left(\frac{b-a}{2}z+\frac{b+a}{2}\right)\ dz.$$
Apply gaussian quadrature to the new integral on the right.
$$\frac{b-a}{2}\int_{-1}^1 f\left(\frac{b-a}{2}z+\frac{b+a}{2}\right)\ dz \approx \frac{b-a}{2}\sum_{i=1}^3 g_i f\left(\frac{b-a}{2}z_i+\frac{b+a}{2}\right).$$
With a three-point quadrature, our points are $x_1 = a, x_2 = \frac{a+b}{2}, x_3 = b$. These correspond to $z_1 = -1, z_2 = 0, z_3 = 1$.
Plugging these in, we find
$$\begin{align*}
\int_a^b f(x)\ dx \approx \frac{b-a}{2}& \left[g_1 f\left(\frac{b-a}{2}\cdot -1
+ \frac{a+b}{2}\right)\right. \\
&+\left.g_2 f\left(\frac{b-a}{2}\cdot 0
+ \frac{a+b}{2}\right)\right. \\
&+ \left.g_3 f\left(\frac{b-a}{2}\cdot 1
+ \frac{a+b}{2}\right)\right].
\end{align*}$$
The right-hand side simplifies to $$\frac{b-a}{2}\left[ g_1 f(a) + g_2 f\left(\frac{a+b}{2}\right) + g_3 f(b)\right].$$
Now you can see the function arguments are in the right form, and now all you have to do is compute the weights, which is straightforward since $g(x) = 1$. You can use $g_i = \int_\alpha^\beta g(x) \prod_{\substack{1 \le j \le 3 \\ j \neq i}} \frac{x-x_j}{x_i-x_j}\ dx$. But note that we've already done the change of variables to shift $[a,b]$ to $[-1,1]$ in the quadrature, so $\alpha = -1$ and $\beta = 1$. It is very straightforward to compute the integrals.
For $g_1$, we have
$$g_1 = \int_{-1}^1 \frac{x-0}{-1-0}\cdot \frac{x-1}{-1 - 1}\ dx = \frac12\left[\left.\frac{x^3}{3}\right|_{-1}^1 - \left.\frac{x^2}{2}\right|_{-1}^1\right] = \frac12\left[\frac13+\frac13-\left(\frac12-\frac12\right)\right] = \frac13.$$
You should find that $g_2$ and $g_3$ are similar. The result follows.
|
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|
Don't understand simple logarithm problem with fractional base $$3\log_{\frac{4}{9}}\sqrt[4]{\frac{27}{8}}$$
$$\log_{\frac{3}{2}}\frac{16}{81}$$
I understand using the expansion property to expand the division into a subtraction but how do I proceed from there?
|
First observe that
$$
3\log_\frac{4}{9}\sqrt[4]{\frac{27}{8}} =\frac{3}{4}\log_\frac{4}{9}\frac{27}{8}
$$
So all we need now is to compute the log term. By definition $\log_ab=y$ means that $a^y=b$ so we need to find $y$ such that
$$
\left(\frac{4}{9}\right)^y=\frac{27}{8}
$$
Expanding this, we have
$$
\left(\frac{2^2}{3^2}\right)^y=\frac{3^3}{2^3}
$$
or
$$
\left(\frac{2}{3}\right)^{2y}=\left(\frac{3}{2}\right)^3
$$
and we're almost there, if we could make the left and right sides have the same base. Can you carry on from there?
|
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|
Find $x$ for $\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12}...$ $$\left(\frac1{1\times101} + \frac1{2\times102} + \dots +\frac1{10\times110}\right)x = \frac1{1\times11} + \frac1{2\times12} + \dots +\frac1{100\times110}$$
Find x
My younger sister in grade 5 had this question in a test. But I, a college student, still can't solve this. What a shame :<
|
$$\frac{1}{k(k+10)}=\frac{a}{k}+\frac{b}{k+10}$$
$$\begin{array}{l}
\frac{a}{k}+\frac{b}{k+10}=\frac{k(a+b)+10a}{k(k+10)}
\end{array}$$
$$k\leftarrow 0, a=\frac{1}{10}$$
$$k\leftarrow 1, b=-\frac{1}{10}$$
$$\boxed{\cfrac{1}{k(k+10)}=\cfrac{1}{10}\left(\cfrac{1}{k}-\cfrac{1}{k+10}\right)}$$
$$\begin{array}{l}
\sum_{k=1}^{100} \frac{1}{k(k+10)}&=\frac{1}{10}\sum_{k=1}^{100} \frac{1}{k}-\cfrac{1}{k+10}\\
&=\frac{1}{10}\left(\sum_{k=1}^{100} \frac{1}{k}-\sum_{k=1}^{100}\cfrac{1}{k+10}\right)\\
&=\frac{1}{10}\left(\sum_{k=1}^{100} \frac{1}{k}-\sum_{k=11}^{110}\cfrac{1}{k}\right)\\
&=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{1}{k}-\sum_{k=101}^{110}\cfrac{1}{k}\right)\\
&=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{1}{k}-\sum_{k=1}^{10}\cfrac{1}{k+100}\right)\\
&=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{1}{k}-\cfrac{1}{k+100}\right)\\
&=\frac{1}{10}\left(\sum_{k=1}^{10} \frac{100}{k(k+100)}\right)\\
\sum_{k=1}^{100} \frac{1}{k(k+10)}&=10\sum_{k=1}^{10} \frac{1}{k(k+100)}
\end{array}$$
|
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|
Expected value of a negative binomial that has finite $n: n \lt \infty$?
Cards from an ordinary deck are turned face up one at a time. Compute
the expected number of cards that need to be turned face up in order
to obtain
(a) 2 aces;
(c) all 13 hearts.
This is a homework problem straight from a chapter on Expectation from a probability textbook. The textbook has a section on finding the expected value of a negative binomial random variable, and says $E[X] = E[X_1]+ E[X_2[ \dots + E[X_r] = \frac{r}{p}$.
(The textbook defines the negative binomial distribution as the probability that $n$ trials are required until $r$ successes occur. It is assumed that $r$ is constant and $n$, the value of the negative binomial random variable, is unbounded i.e. may go to $\infty$.
$$P(X = n) = \binom{n-1}{r-1}p^r (1-p)^{n-r}\ \ \ \ \text{for}\ r \le n \lt \infty$$
The problem (a) here seems to want the expectation of a negative binomial random variable, however, the above equation for $E[X]$ assumes that $r \le n \lt \infty$, but in the case of this problem, only up to $50$ cards may actually be selected such that $2$ are aces (there are $48$ non-aces, so the next $49$th, $50$th cards must be aces). In other words, for this problem $2 \le n \le 50$.
So instead of following the textbook's equation for $E[X]$, I tried to find $E[X]$ for (a), given $r = 2, p = \frac{4}{52}$ as
$$E[X] = \sum_{n=2}^{50} n \binom{n-1}{2-1} \left(\frac{4}{52}\right)^2\left(\frac{48}{52}\right)^{n-2} \approx 19.8134 $$
If I follow the textbook, I get $E[X] = \frac{r}{p} = 2 \left(\frac{48}{52}\right)^{-1} = 26$.
Are either of these answers correct? And is problem (c) essentially the same as solving (a)?
|
Using the identity
$$
\sum_{k=a}^{n-b}\binom{k}{a}\binom{n-k}{b}=\binom{n+1}{a+b+1}\tag{1}
$$
here is how I would approach (c):
The number of arrangements which get all $13$ hearts in exactly $k$ draws is $\binom{k-1}{12}\binom{52-k}{0}$; that is, the number of arrangements, with the $k^\text{th}$ draw being a heart, to arrange the other $12$ hearts in the previous $k-1$ draws and none in the remaining $52-k$ draws. The total number of arrangements is therefore
$$
\sum_{k=13}^{52}\binom{k-1}{12}\binom{52-k}{0}=\binom{52}{13}\tag{2}
$$
and the expected number of draws would be
$$
\begin{align}
\frac1{\binom{52}{13}}\sum_{k=13}^{52}k\binom{k-1}{12}\binom{52-k}{0}
&=\frac1{\binom{52}{13}}\sum_{k=13}^{52}13\binom{k}{13}\binom{52-k}{0}\\
&=\frac{13\binom{53}{14}}{\binom{52}{13}}\\
&=\frac{13\cdot53}{14}\\[12pt]
&\doteq49.214\tag{3}
\end{align}
$$
Since André Nicolas has given a complete answer to both parts of this, I will show how to use $(1)$ to handle (a):
The number of arrangements which attains $2$ aces in exactly $k$ draws is $\binom{k-1}{1}\binom{52-k}{2}$; that is, the number of arrangements, with the $k^\text{th}$ draw an ace, to have one ace in the first $k-1$ draws and the other two aces in the remaining $52-k$ draws. The total number of arrangements is therefore
$$
\sum_{k=2}^{52}\binom{k-1}{1}\binom{52-k}{2}=\binom{52}{4}\tag{4}
$$
and the expected number of draws would be
$$
\begin{align}
\frac1{\binom{52}{4}}\sum_{k=2}^{50}k\binom{k-1}{1}\binom{52-k}{2}
&=\frac1{\binom{52}{13}}\sum_{k=2}^{52}2\binom{k}{2}\binom{52-k}{2}\\
&=\frac{2\binom{53}{5}}{\binom{52}{4}}\\
&=\frac{2\cdot53}{5}\\[12pt]
&=21.2\tag{5}
\end{align}
$$
Identity $(1)$ is proven using negative binomial coefficients in this answer. Let's give a generating function approach here.
$$
\begin{align}
\frac{(1+x)^{n+1}-(1+y)^{n+1}}{x-y}
&=\frac{(1+x)^{n+1}-(1+y)^{n+1}}{(1+x)-(1+y)}\\
&=\sum_{k=0}^n(1+x)^k(1+y)^{n-k}\\
&=\sum_{k=0}^n\sum_{i=0}^k\binom{k}{i}x^i\sum_{j=0}^{n-k}\binom{n-k}{j}y^j\\
&=\sum_{i=0}^k\sum_{j=0}^{n-k}x^iy^j\sum_{k=0}^n\binom{k}{i}\binom{n-k}{j}\tag{6}
\end{align}
$$
Thus, the sum in $(1)$ is the coefficient of $x^ay^b$ in the right hand side of $(4)$. Let's compute the left hand side of $(4)$ in a different manner.
$$
\begin{align}
\frac{(1+x)^{n+1}-(1+y)^{n+1}}{x-y}
&=\sum_{k=0}^{n+1}\binom{n+1}{k}\frac{x^k-y^k}{x-y}\\
&=\sum_{k=0}^{n+1}\binom{n+1}{k}\sum_{j=1}^kx^{j-1}y^{k-j}\tag{7}
\end{align}
$$
The sole occurrence of $x^ay^b$ in $(5)$ is when $j=a+1$ and $k-j=b$; that is, $k=a+b+1$. Therefore, the coefficient of $x^ay^b$ is $\binom{n+1}{a+b+1}$. Thus, combining $(6)$ and $(7)$ yields $(1)$.
|
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|
Asymptotic expansion of $J(t) = \int^{\infty}_{0}{\exp(-t(x + 4/(x+1)))}\, dx$ I want to derive an asymptotic expansion for the following Bessel function. I think I need to rewrite it in another form, from which I can integrate it by parts. I am interested in obtaining the expansion up to second order as $t$ approaches infinity.
$$J(t) = \int^{\infty}_{0}{\exp\left(-t\left(x + \dfrac{4}{x+1} \right) \right)}\, dx$$
|
First make the (cosmetic) change of variables $x = y+1$, so that
$$
\begin{align}
J(t) &= \int^{\infty}_{0} \exp\left[-t\left(x + \dfrac{4}{x+1} \right) \right]\, dx \\
&= \int_{-1}^{\infty} \exp\left[-t\left(y + 1 + \dfrac{4}{y+2} \right) \right]\, dy \\
&= \int_{-1}^{\infty} \exp\Bigl[-t f(y) \Bigr]\,dy.
\end{align}
$$
Now $f(y)$ has a minimum at $y=0$, and near there we have
$$
f(y) = 3 + \frac{y^2}{2} - \frac{y^3}{4} + \frac{y^4}{8} + \cdots.
$$
We would like to introduce a new variable by
$$
3 + \frac{z^2}{2} = 3 + \frac{y^2}{2} - \frac{y^3}{4} + \frac{y^4}{8} + \cdots
$$
or, taking the principal branch of the square root,
$$
z = y \sqrt{1 - \frac{y}{2} + \frac{y^2}{4} + \cdots},
$$
which of course would only hold in a small neighborhood of the origin. In general we can solve for $y$ in this equation using series reversion (the Lagrange formula, say), but in this case we can write down the answer explicitly as
$$
\begin{align}
y &= \frac{1}{4} \left(z^2 + z \sqrt{16+z^2}\right) \\
&= z + \frac{z^2}{4} + \frac{z^3}{32} - \frac{z^5}{2048} + O\left(z^7\right).
\end{align}
$$
Thus we have
$$
dy = \left[1+\frac{z}{2}+\frac{3 z^2}{32}-\frac{5 z^4}{2048} + O\left(z^6\right)\right]dz.
$$
The integrals over the tails of the intervals are exponentially small as $t \to \infty$, and by following the usual steps in the Laplace method we arrive at the expression
$$
J(t) = \int_{-\infty}^{\infty} \exp\left[-t \left(3 + \frac{z^2}{2}\right)\right] \left(1+\frac{z}{2}+\frac{3 z^2}{32}-\frac{5 z^4}{2048}\right)\,dz + O\left(\int_{-\infty}^{\infty} \exp\left[-t \left(3 + \frac{z^2}{2}\right)\right] z^6\,dz\right).
$$
We can integrate term-by-term and conclude that
$$
J(t) = e^{-3t} \sqrt{\frac{2\pi}{t}} \left[1 + \frac{3}{32} t^{-1} - \frac{15}{2048} t^{-2} + O\left(t^{-3}\right)\right]
$$
as $t \to \infty$.
|
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|
Proving $4(a^3 + b^3) \ge (a + b)^3$ and $9(a^3 + b^3 + c^3) \ge (a + b + c)^3$
Let $a$, $b$ and $c$ be positive real numbers.
$(\mathrm{i})$ Prove that $4(a^3 + b^3) \ge (a + b)^3$.
$(\mathrm{ii})$Prove that $9(a^3 + b^3 + c^3) \ge (a + b + c)^3.$
For the first one I tried expanding to get $a^3 + b^3 \ge a^2b+ab^2$ but I'm not sure how to prove it.
|
for (i);
notice that $(a^3+b^3)=(a+b)(a^2-ab+b^2)$. This is the standard factorization for the sum of cubes. Then $4(a^2-ab+b^2)\ge(a+b)^2$.
Thus,
$$4a^2-4ab+4b^2\ge (a+b)^2$$
$$a^2+2ab+b^2+3a^2-6ab+3b^2\ge (a+b)^2$$
$$(a+b)^2+3(a-b)^2\ge(a+b)^2$$
Since $3(a-b)^2$ is always positive or $0$, the inequality is proven.
|
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|
what is the limit of $\displaystyle \lim_{x\to 0} \frac{x}{a}\cdot \lfloor{\frac{b}{x}\rfloor}$
find the limit of:
$\displaystyle\lim_{x\to 0} \frac{x}{a}\cdot \lfloor{\frac{b}{x}\rfloor}$
$ a,b>0$
i know that:
$\lfloor\frac{b}{x}\rfloor\le \frac{b}{x} + 1$; and
$\lfloor\frac{b}{x}\rfloor\ge \frac{b}{x} -1$
but i cant seem to combine those using the squeezing theorem (i belive the limit is $\frac{b}{a}$)
|
Consider two cases: $x>0$ and $x<0$.
In each case multiply your inequalities by $\frac{x}{a}$.
Then, say for $x<0$, you get:
$\frac{b}{a} -\frac{x}{a} \geq \frac{x}{a} \lfloor \frac{b}{x} \rfloor \geq \frac{b}{a}+ \frac{x}{a}.$
Now use the squeeze theorem.
|
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|
Calculate the lim $\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}$ $$
\lim_{x\to 1} \frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1}
$$
I multiply it with:
$$
\frac{ \sqrt{x} + 1}{\sqrt{x} + 1}
$$
And I get :
$$
\lim_{x\to 1} \frac{ \sqrt{x}^2 - 1^2}{(\sqrt[3]{x} - 1) * (\sqrt{x} + 1)}
$$
But the solution is still division by 0 and not possible, so I think I've made a mistake somewhere..
|
Hint: $$\frac{ \sqrt{x} - 1}{\sqrt[3]{x} - 1} =\frac{3}{2} \cdot \frac{ e^\frac{\ln x}{2} - 1}{\frac{\ln x}{2}} \cdot \left( \frac{ e^\frac{\ln x}{3} - 1}{\frac{\ln x}{3}}\right)^{-1} $$
|
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|
Counting Combinations of a List of Numbers Let's say I have the following list of 9 (including repetitions) single-digit numbers: 7,7,7,5,5,3,2,2,2. I want to know, for an arbitrary list of natural numbers, how many ways I can select one, two,...,all of them without the obvious double counting. For example, for the list above I can select two of them in the following 9 ways:
77 75 55 53 32 22 73 72 52,
and again for the list above I can select three of them in the following 15 ways:
777 775 755 553 532 322 222 773 772 722 552 522 753 752 732
(let me know if I missed some).
I've tried thinking of it as a counting without replacement problem, e.g. thinking of each number as a colored marble in an urn but can't seem to get the proper formula.
|
Suppose we have $N$ balls colored $n$ different colors. For each color $k$, let $d_k$ be the number of balls of color $k$. Our goal is to find the number $a_r$ of ways to choose $r\leq N$ balls without repetition and ignoring order, where balls of the same color are considered identical. Now, any such choice of $r$ balls is completely determined by how many balls of each color it has. In other words, we're reduced to finding the number of $n$-tuples $(e_1,\ldots,e_n)$ of non-negative integers satisfying
*
*$e_k\leq d_k$ for all $k$, and
*$e_1+\cdots+e_k=r$.
Here's where we make use of a really nice algebraic trick: consider the product of polynomials
$$f(X)=\prod_{k=1}^n\left(1+X+X^2+\cdots+X^{d_k}\right).$$
Expanding this out, we get
$$\begin{align*}
f(X) &= \left(\sum_{j_1=0}^{d_1}X^{j_1}\right)\cdots\left(\sum_{j_n=0}^{d_n} X^{j_n}\right)\\
&= \sum_{j_1=0}^{d_1}\cdots\sum_{j_n=0}^{d_n} X^{j_1+\cdots +j_n},
\end{align*}$$
so the coefficient of $X^r$ is
$$\sum_{j_1+\cdots+j_n=r}1,$$
where $j_k$ is required to be less than or equal to $d_k$ for all $k$. But this is exactly $a_r$. Thus, all we need to do is find a nice representation of the coefficients of $f(X)$.
To do this, recall that $1+X+X^2+\cdots+X^m=(1-X^{m+1})/(1-X)$ for all $m$, and that $(1-X)^{-n}=\sum_{r=0}^\infty \binom{n+r-1}{n-1}X^r$. Then
$$\begin{align*}
f(X) &= \frac{(1-X^{d_1+1})\cdots(1-X^{d_n+1})}{(1-X)^n}\\
&= (1-X^{d_1+1})\cdots(1-X^{d_n+1})\left[\binom{n-1}{n-1}+\binom{n+1-1}{n-1}X + \binom{n+2-1}{n-1}X^2+\cdots\right].
\end{align*}$$
Now, $(1-X^{d_1+1})\cdots(1-X^{d_n+1})$ expands to
$$ 1- \sum_{k=1}^n X^{d_k+1} + \sum_{k_1,k_2=1}^n X^{d_{k_1}+d_{k_2}+2} - \cdots + (-1)^n\sum_{k_1,\ldots,k_n=1}^n X^{d_{k_1}+\cdots+d_{k_n}+n}, $$
where in each sum, the $k_i$'s are required to be distinct. To multiply this together with the series in brackets, let's figure out what happens when we multiply the series in brackets with a single guy of the form $X^s$, where $s$ is a positive integer. Well,
$$\begin{align*}
X^s\cdot\frac{1}{(1-X)^n} &= X^s\left[\binom{n-1}{n-1}+\binom{n+1-1}{n-1}X + \binom{n+2-1}{n-1}X^2+\cdots\right]\\
&= \binom{n-1}{n-1}X^s+\binom{n+1-1}{n-1}X^{s+1} + \binom{n+2-1}{n-1}X^{s+2}+\cdots\\
&= \binom{n+s-s-1}{n-1}X^s + \binom{n+s+1-s-1}{n-1}X^{s+1} + \binom{n+s+2-s-1}{n-1}X^{s+2}\cdots\\
&= \sum_{r=s}^\infty \binom{n+r-s-1}{n-1}X^r.
\end{align*}$$
But when $\binom{n+r-s-1}{n-1}=0$ whenever $r<s$ (by the definition of the binomial coefficient), so we can tack on all the $r<s$ terms to the above series expression and not change anything:
$$X^s\cdot\frac{1}{(1-X)^n} = \sum_{r=0}^\infty \binom{n+r-s-1}{n-1}X^r.$$
Finally, returning to our expansion of $(1-X^{d_1+1})\cdots(1-X^{d_n+1})$ and multiplying by the series, we get that the $r$th coefficient of $f(X)$, i.e. $a_r$, is
$$a_r= 1- \sum_{k=1}^n \tbinom{n+r-d_k-2}{n-1} + \sum_{k_1,k_2=1}^n \tbinom{n+r-d_{k_1}-d_{k_2}-3}{n-1} - \cdots + (-1)^n\sum_{k_1,\ldots,k_n=1}^n \tbinom{n+r-d_{k_1}-\cdots-d_{k_n}-n-1}{n-1},$$
where, as before, in each sum, the $k_i$'s are required to be distinct.
|
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|
Transforming from cartesian to cylindrical Here is the question:
Transform $\textbf{A} = \hat{\mathbf{x}} 2 - \hat{\mathbf{y}}5 + \hat{\mathbf{z}}3$ into cylindrical coordinates at point ($x=-2, y=3, z=1$).
What I have tried is this: consider the point the exercise has given and find $r$ and $\theta$ with it. This leads to
$$r = \sqrt{(-2)^2 + 3^2} = \sqrt{13}, \quad \cos \theta = \frac{-2}{\sqrt{13}}, \sin \theta = \frac{3}{\sqrt{13}}.$$
The book gives the relation
$$
\begin{bmatrix} A_r \\ A_{\theta} \\ A_z \end{bmatrix}
=
\begin{bmatrix} \cos \theta & \sin \theta & 0 \\ - \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix}
$$
as the relation of the coordinates. Using this, I plugged the values of $\cos \theta, \sin \theta$ I found and took $A_x = 2, A_y = -5, A_z = 3$. Therefore,
$$
\begin{bmatrix} A_r \\ A_{\theta} \\ A_z \end{bmatrix}
=
\begin{bmatrix} 2 \cos \theta - 5 \sin \theta \\ - 2 \sin \theta -5 \cos \theta \\ 3 \end{bmatrix}
=
\begin{bmatrix} \frac{2 \cdot (-2) - 5 \cdot (3)}{\sqrt{13}} \\ \frac{-2 \cdot (-2) -5 \cdot (3)}{\sqrt{13}} \\ 3 \end{bmatrix}
=
\begin{bmatrix} \frac{-19}{\sqrt{13}} \\ \frac{- 11}{\sqrt{13}} \\ 3 \end{bmatrix}.
$$
This gives approximately $(A_r, A_{\theta}, A_z) = (- 5.27, 3.05, 3)$, but the answer is
$$\textbf{A} = \hat{\mathbf{r}} 5.27 - \hat{\mathbf{\theta}}1.11 + \hat{\mathbf{z}}3.$$
Seemingly I get a wrong sign for the first coordinate and the second one is completely off. What am I doing wrong (if so)?
This comes from self-studying electromagnetism using this book, first chapter (vector algebra).
|
When you substituted in for the trig functions in the $A_{\theta}$ line, you put the values in wrong. You should have $-2\sin \theta - 5 \cos \theta=\frac 1{\sqrt{13}}(2(-2)-5(-2))=\frac 4{\sqrt{13}}\approx 1.11$
Now we have an overall sign error relative to the book. As I draw the picture, the book is wrong. The given point is in the second quadrant, while $\textbf{A}$ points down and toward the origin. $A_r$ is therefore negative. $\textbf{A}$ also points to the left of the radial vector from the point to the origin, so $A_{\theta}$ is positive.
|
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|
Showing that this IVP has a solution for all of $\mathbb{R}$ I'm preparing for an exam and came across this problem:
Given the IVP
$f'(x) = \frac{f(x)}{1+x^2+(f(x))^2}$
$f(x_0) = y_0$
-show this has an unique solution $f \in C(\mathbb{R})$ defined on all of $\mathbb{R}$.
-What is the solution if $y_0=0$?
We're using the Picard-Lindelof Theorem, so we need to show Lipschitz of $F(x,y)$ where $f'(x)=F(x,y)$ in y on the region on which the IVP is defined. Clearly $F(x,y)$ is continuous on $\mathbb{R}^2$, so I assume we'll just have to show Lipschitz and we're done. This is where I run into a snag, however - how do I show $|F(x,y)-F(x,y_1)| \leq C|y-y_1|$?
|
You have
$$
\left|\frac{y}{1+x^2+y^2}-\frac{y_1}{1+x^2+y_1^2}\right|
\leq
\left|\frac{y}{1+x^2+y^2}-\frac{y_1}{1+x^2+y^2}\right|
+\left|\frac{y_1}{1+x^2+y^2}-\frac{y_1}{1+x^2+y_1^2}\right|\\
=\frac{|y-y_1|}{1+x^2+y^2}+\left|\frac{|y_1|(1+x^2+y_1^2)-|y_1|(1+x^2+y^2)}{(1+x^2+y^2)(1+x^2+y_1^2)}\right|\\
=\frac{|y-y_1|}{1+x^2+y^2}+\frac{|y_1||y_1^2-y^2|}{(1+x^2+y^2)(1+x^2+y_1^2)}\\
=\frac{|y-y_1|}{1+x^2+y^2}+\frac{|y_1||y_1+y|\,|y_1-y|}{(1+x^2+y^2)(1+x^2+y_1^2)}\\
\leq|y-y_1|+|y-y_1|\,\frac{(y_1^2+|y|)}{(1+y_1^2)(1+y^2)}\\
\leq2|y_1-y|
$$
|
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|
squares which are not the sum of a square and twice a triangular number I'm trying to determine conditions on integer squares which cannot be written as a square and twice a triangular [all numbers positive], i.e. integers $n \ge 1$ where there are no integers $a,b \ge 1$ such that
$$ n^2 = a^2 + b^2+b.$$
For example, $$9 = 1^2 + 8 = 2^2 + 5,$$ so it is such a number; however, $16=2^2 + (2 \cdot 6)$, so it is not.
This paper http://math.nju.edu.cn/~zwsun/111o.pdf claims a proof about numbers which cannot be written as the sum of a square and two [not necessarily equal] triangular numbers — I will try to adapt their proof if I can't find another.
Any references or hints on how to approach the problem would be appreciated.
Thanks!
Kieren.
|
In fact every odd square can be written as $n^2= 4T_{k} + a^2 + (a+1)^2$. Every odd square can be decomposed into:
$n^2 = [(n-1)/2 + (n+1)/2]^2$ with $(n-1)/2=4T_{k}$ and $(n+1)/2 = a^2 + (a+1)^2$ or equivalently $n^2 = (a+b)^2 = a^2 +2ab +b^2$.
Your example $9= 2^2 +5$ can in fact be rewritten as:
$9=(9-1)/2 + (9+1)/2 = 4 + 5 = (1+2)^2 = 4T_{1} + 1^2 + 2^2$
If you lump the two squares $(a^2 +b^2)=m$, then you will end up with:
$n^2= m + 4T_{k}$. This is truly a general result. But lumping the two squares into an integer doesn't make the two squares disappear. They are just hidden.
Let's consider another example $n^2=9^2=(4+5)^2= 4^2 +2*4*5 + 5^2$. We can also rewrite$n^2=9^2=81= 2*4*5 +41$ which really doesn't change the fact that $41=4^2 + 5^2$.
One can convince oneself by decomposing any odd square as done above and look at the final result. $81=9^2= 16 + 2*20 +25$. $20$ is already twice the triangular number $T_{4}=10$. So $81$ is a square that definitely can be written as the sum of 4 times a triangular number and two consecutive squares. One needs only look at the multiplication table to see that $4^2+2*4*5+5^2 = 16 +20 + 20 +25$ add up to $81$.
The diagonals under and above the main diagonal that gives the squares give $2T_{k}$ but since we are adding $2T_{k} +2T_{k}$, we can claim that the odd square we considered is not the sum of a square and twice a triangular number.
So we can write $n^2=9^2=81$ as:
$81= 4^2 + 2*20 + 5^2$ or
$81= 41 + 4*10$
Even squares cannot be decomposed as done above.
|
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|
If the sum $\frac{1}{7} + \frac{1\cdot 3}{7\cdot 9} + \frac{1\cdot 3\cdot 5}{7\cdot 9\cdot 11} + ...$ to 20 terms is $\frac{m}{n}$, then $n-4m$?
If the sum $\frac{1}{7} + \frac{1\cdot 3}{7\cdot 9} + \frac{1\cdot 3\cdot 5}{7\cdot 9\cdot 11} + ...$ to 20 terms is $\frac{m}{n}$, reduced fraction, then what is $n-4m$?
This is a question I dug up from an old JEE Main paper (India). It's very intriguing to me because, although it looks simple, I can't seem to find he sum in this question.
The numerator of the nth term of the series seems to be the product of the first n terms of odd numbers.
The denominator is similarly made but the sequence starts from 7.
I have no idea how to find the sum to n terms in this situation. If it were the sum of odd numbers and not the product, I could have done it easily.
Please explain to me how you find the product of n terms of a sequence
and also the sum to n terms of the given sequence.
I have never been introduced to the product of n terms before, If you could give me a proper intuition for it, it would really make my christmas.
|
The following is not particularly clever but works.
Note that $$1\cdot 3 \cdot 5 \cdot \cdots \cdot (2N-1) = \frac{(2N)!}{2^N N!}$$
Using this (or just considering ratios of adjacent terms as suggested in comments), each term in the series is
$$\frac{120(2n)! (n+3)!}{n!(2n+6)!} = \frac{15}{(2n+1)(2n+3)(2n+5)} = \frac{15}{8}\left(\frac 1 {2n+1} - \frac 2 {2n+3} + \frac 1 {2n+5} \right)$$
Using telescopic sums, one finds the sum of $N$ terms to be
$$\frac 1 4 -\frac{15}8 \left( - \frac 1 {2N+3} + \frac 1 {2N+5}\right)$$
and the answer can be calculated.
Edit: As Cameron pointed out in the comments, the telescopic nature of the sum is made clearer by realizing you are summing the terms $$\left(\frac 1 3 - \frac 1 5\right) - \left(\frac 1 5 - \frac 1 7\right),\quad \left(\frac 1 5 - \frac 1 7\right) - \left(\frac 1 7 - \frac 1 9\right),\quad \cdots$$
|
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|
Find all postive integer numbers $x,y$,such $x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$ Find all postive integer $x$ and $y$ such that
$x+y+1$ divides $2xy$ and $x+y-1$ divides $x^2+y^2-1$
My try: since
$$(x+y)^2-2xy=x^2+y^2$$
I know this well know reslut:
$$xy|(x^2+y^2+1),\Longrightarrow \dfrac{x^2+y^2+1}{xy}=3,x,y\in N$$
and such condition $$(x,y)=(u_{n},u_{n+1})$$
where
$$u_{0}=u_{1}=1,u_{n+2}=3u_{n+1}-u_{n}$$
so I can't,Thank you
|
$$(x+y+1)(x+y-1)-1(x^2+y^2-1)=2xy$$
If we assume $x\equiv y\pmod 2$, then $\text{lcm}(x+y+1,x+y-1)=(x+y)^2-1$ since the expressions are coprime. If $x+y-1\mid x^2+y^2-1$, the above equality would imply that $x+y-1\mid 2xy$, but since $2xy<(x+y)^2-1$ ($x,y>0$), this implies $x+y+1\not \mid 2xy$.
If $x\not\equiv y\pmod2$, then $\text{lcm}(x+y+1,x+y-1)=\frac{(x+y)^2-1}2$. In order to construct a similar argument to before, we need $$\begin{align} \frac{(x+y)^2-1}{2}&>2xy\\
(x+y)^2-1&>4xy\\
(x+y)^2-4xy&>1\\
(x-y)^2&>1\\
|x-y|&>1 \end{align}$$
This leaves us with two cases: $x=y$ and $x=y+1$. We can dismiss the first since it implies $x\equiv y\pmod 2$. So we want $$(y+1)+y-1=2y\mid (y+1)^2+y^2-1=2y^2+2y$$ Which is true.
We also need $$(y+1)+y+1=2y+2\mid 2(y+1)y$$ Which is true for all $y$.
In conclusion, your solution set for the problem is $\{(x,y)\in\Bbb N^2\mid |x-y|=1\}$
|
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|
Help with derivative of $y=x^2\sin^5x+x\cos^{-5}x$
Find $y^{\prime}$ of $y=x^2\sin^5x+x\cos^{-5}x$
My try:
$\dfrac{d}{dx}(x^2\sin^5x)=x^2(-5\sin^4x)+(2x\sin^5x)$
$\dfrac{d}{dx}(x\cos^{-5}x)=x(-5\cos^{-6}x)+1(\cos^{-5}x)$
This doesn't seem right. Can you please show how to do it correctly?
The answer is $y^{\prime}=x\sin^4(x) ( 2 \sin(x)+5x\cos(x)) +(5x\tan(x)+1)\sec^5x$
|
$$y=x^2\sin^5x+x\cos^{-5}x$$
When taking the derivative of each summand of $y$, you left out a component of the chain rule for each of the trigonometric functions: $$[g(x)]^n = n[g(x)]^{n - 1} \cdot \color{blue}{g'(x)}$$ See the text highlighted in blue:
$$\dfrac{d}{dx}(x^2\sin^5x)=x^2(5\sin^4x){\bf\color{blue}{ \cos x}}+(2x\sin^5x)$$
$$\dfrac{d}{dx}(x\cos^{-5}x)=x(-5\cos^{-6}(x){\bf \color{blue}{(-\sin x)}})+1(\cos^{-5}x)$$
$$y' = 5x^2\sin^4x\cos x+ 2x\sin^5x +5x\cos^{-6}x\sin x + \cos^{-5}x$$
|
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|
Prove that if $10|A$ then $100|A$. For : $A=a^2+ab+b^2$ with $a,b \in \mathbb{N}$.
We known that $10|A$ prove that $100|A$.
|
Solution 1: Suppose $10|A$. If either $a$ or $b$ is odd, then $A$ is also odd, so $a$ and $b$ must be even and hence $4|A$. Now consider $A$ modulo $5$. If either $a$ or $b$ is divisible by $5$, the other must be as well, and so $25|A$. Otherwise, $a^2$ and $b^2$ can be either of $\pm 1$ modulo $5$. If one is $1$ and the other $-1$, then $A\equiv ab \pmod{5}$ is not divisible by $5$. If both are $1$, then $A\equiv ab+2$, and $a, b \equiv \pm 1 \pmod{5}$ so that $A \equiv 1, 3\pmod{5}$ and is not divisible by $5$. A similar argument shows that if $a^2\equiv b^2 \equiv -1\pmod{5}$, then $ab \equiv \pm 1\pmod{5}$ and hence $A$ is still not divisible by $5$. So $a$ and $b$ must both be divisible by $5$ and thus $25|A$. Therefore $100|A$.
Massively Overkill Solution 2: The numbers 2 and 5 are Eisenstein primes, so for any Eisenstein integer $a-b\omega$, if $5|N(a-b\omega)$, then $5|(a-b\omega)$ and thus $5|a$ and $5|b$. The same goes for $2$ in place of $5$ and the result follows since $N(a-b\omega) = a^2+ab+b^2$.
|
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|
An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$ $n$ is a positive integer, then
$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53.$$
please don't refer to the famous $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$.
I want to find a better proof.
My stupid method:
$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt \left(1+\frac1{2^2}+\dotsb+\frac1{10^2}\right)+\frac1{10\cdot11}+\dotsb+\frac1{n(n-1)}\\<1.549768...+\frac1{10}\lt\frac53$$
|
We show by induction that if $n\gt 1$ then
$$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2}\lt \frac{5}{3}-\frac{2}{2n+1}.$$
The result is true at $n=2$. Suppose that the result holds at $n=k$. We show it holds at $n=k+1$.
By the induction assumption,
$$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{k^2}+\frac{1}{(k+1)^2}\lt \frac{5}{3}-\frac{2}{2k+1}+\frac{1}{(k+1)^2}.\tag{1}$$
The right-hand side of (1) is equal to
$$\frac{5}{3}-\left(\frac{2}{2k+1}-\frac{1}{(k+1)^2}\right).$$
Now we need to show that $\frac{2}{2k+1}-\frac{1}{(k+1)^2}\gt \frac{1}{2k+3}$ or equivalently that $\frac{4}{(2k+1)(2k+3)}\gt \frac{1}{(k+1)^2}$. So we show that $4(k+1)^2\gt (2k+1)(2k+3)$. This is straightforward.
|
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|
Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
x-4=x-5-\sqrt{x-5}+1\\
x-4=x-4-\sqrt{x-5}\\
\text{substracting x, and adding 4 to both sides}\\
0=-\sqrt(x-5)\\
\text{switching both sides}\\
\sqrt{x-5}=0\\
\text{sqaring both sides}\\
x-5=0\\
x=5\\
\text{When I place 5 in the equation, I get:}\\
\sqrt{5-4}-\sqrt{5-5}+1=0\\
\sqrt{1}-\sqrt{0}+1=0\\
1-0+1=0\\
2=0\\
\text{this means that the equation dosent have any solution, right??}\\
$$
Any advice and suggestion is helpful.
Thanks!!!
|
Hint $\ \ \sqrt{z+1}+\sqrt{z}\, =:\, y\ $ times $\,(\sqrt{z+1}-\sqrt{z}\,=\,-1) \ \Rightarrow\ 1 = -y\,$ contra $\, y > 0\ \ \ $ QED
|
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|
Hermitian matrix such that $4M^5+2M^3+M=7I_n$ $n$ is a positive integer. Besides the identity matrix $I_n$, does there exist other $n\times n$ Hermitian matrix $M$, such that the following equality
$$4M^5+2M^3+M=7I_n $$
hold?
I try this: Since $4M^5+2M^3+M=7I_n$, then
$$(M-I_n)(4M^4+4M^3+6M^2+6M+7I_n)=0$$
but, What should I do next? Thanks!
|
Hermitian matrices are diagonalizable and they have only real eigenvalues. Let
$$
M=H^{-1}\mathrm{diag}(d_1,\ldots,d_n)H,
$$
where $D=\mathrm{diag}(d_1,\ldots,d_n)$ is the diagonal matrix which contains the eigenvalues of $M$. Note that
$$
4x^4+4x^3+6x^2+6x+7= (2x^2+1)^2+3(x+1)^2+2x^2+4 \ge 5,
$$
for all real $x$. This implies that
$$
L=4M^4+4M^3+6M^2+6M+7= H^{-1}\mathrm{diag}(r_1,\ldots,r_n)H,
$$
with $r_i\ge 5$, for all $i$. Thus $L$ is invertible. Hence
$$
(4M^4+4M^3+6M^2+6M+7)(M-I)=0,
$$
implies that $M=I$!
|
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Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
|
$\begin{eqnarray}{\bf Hint} &&\ \ \underbrace{2^{11}} \\
\,&=&\ \ \overbrace{2^{10} + \underbrace{2^{10}}}\\
\,&=&\ \ 2^{10} + \overbrace{{2^9+\underbrace{2^9}}}\\
\,&=&\ \ 2^{10} + 2^9+\overbrace{2^8+\underbrace{2^8}}\\
\,&=&\ \ 2^{10} + 2^9+2^8+\overbrace{2^7+2^7}\\
\,&&\qquad\qquad\ \ \vdots\qquad\qquad\qquad\ddots
\end{eqnarray}$
Alternatively we can write it in telescopic form
$\ \begin{eqnarray}
\color{#c00}{2^{11}-2^k} = \underbrace{\phantom{2^11 - 2^10}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\color{#c00}{2^{11}}}&&\overbrace{{-2^{10}} +\underbrace{\phantom{2^10 - 2^9}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!2^{10}}^{=\, 0}&&\overbrace{-2^9 +\underbrace{\phantom{2^9 - 2^8}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!2^9}^{=\,0}&&\overbrace{-2^8 + 2^8}^{=\,0} \cdots \overbrace{-2^{k+2} + \underbrace{\phantom{2^{k+2} - 2^{k+1}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!2^{k+2}}^{=\, 0}&&\overbrace{-2^{k+1}+\underbrace{\phantom{2^{k+1} - 2^{k}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!2^{k+1}}^{=\,0}&&\color{#c00}{-2^k}\\
&&\!\!2^{10}\quad\, + &&\!\!2^9\quad\, + &&\!\!2^8\quad +\quad\ \cdots\qquad + &&\!\!\!\!2^{k+1}\quad\ \ + &&\!\!\!2^k
\end{eqnarray}$
|
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|
How would you explain to a 9th grader the negative exponent rule? Let us assume that the students haven't been exposed to these two rules: $a^{x+y} = a^{x}a^{y}$ and $\frac{a^x}{a^y} = a^{x-y}$. They have just been introduced to the generalization: $a^{-x} = \frac{1}{a^x}$ from the pattern method: $2^2 = 4, 2^1 = 2, 2^0 = 1, 2^{-1} = \frac{1}{2}$ etc. However, some students confuse $2^{-3}$ to be $(-2)(-2)(-2)$ since they are familiar with $2^{3} = 2 \cdot 2 \cdot 2$. This is a low-income urban school and most kids in this algebra class struggle with math dealing with exponents, fractions and decimals. What would be the best approach to reach all 32 students?
|
*
*Help for confusing $$ 2^{-3}\neq (-2)(-2)(-2)$$
Advise them to watch out for the exact position of the - sign,
and tell them since it isn't next to 2, it simply does not belong to the 2, instead it belongs to the 3.
Example would be
$$2^{-3} = \frac{1}{2^3} = \frac{1}{2\times2\times2} = \frac{1}{8}$$
as already proposed by @DoktoroReichhard
*Learning the rule:
$$a^{x+y} = a^{x}a^{y}$$
First example (with x = 2, y = 3) would be
$$ 10^{2+3} = 10^{5} = 100 ~000= 100 \cdot 1000 = (10\cdot10) ~ \cdot ~ (10\cdot10\cdot10) = 10^{2}\cdot10^{3} $$
Now substitute 10 with a and you get
$$a^{2+3}= a^{5} = aaaaa =aa\cdot aaa =a^{2}\cdot a^{3}=a^{2+3} $$
Then use 2 instead of 10 and calculate each single step on the board together with class
*Learning the rule:
$$a^{x-y} = \frac{a^{x}}{a^{y}}$$
First example (with x = 2, y = 3) would be
$$ 10^{2-3} = 10^{-1} =\frac{1}{10}= \frac {100}{1000} = \frac {10^{2}}{10^{3}} $$
Now substitute 10 with a and you get
$$a^{2-3}=a^{-1} = \frac{1}{a}= \frac {aa}{aaa} = \frac {a^{2}}{a^{3}} $$
Then use 2 instead of 10 and calculate each single step on the board together with class
In the end feel free to point out (for the ambitious students) that these rules will also work with crazy numbers like
$$ 2^{1.45 - 2.45} =2^{-1} =\frac{1}{2} $$
It's great that you are so passionate about teaching your class.
|
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"timestamp": "2023-03-29T00:00:00",
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|
The equation $3a+5b=n$.
For a given number $n$, how can we find out whether we have any non-negative values for $a$ and $b$ for the equation
$$3a+5b=n,$$
where $1\le n\le 100,000$.
For example: If $n=5$, then $a=0, b=1$.
|
Your question is a special case of the Frobenius problem or coin problem. You want to know for which $n$ it is possible to find nonnegative integers $a,b$ such that $3a+5b=n$. Take a look at the two-variable case on the Wikipedia page: It is always possible if $n>3\cdot5-3-5$, that is, $n>7$.
For values smaller then or equal to $7$, it is known that exactly half of them can be written in the form $3a+5b$. All you have to do is check them manually:
$0=3\cdot0+5\cdot0$,
$3=3\cdot1+5\cdot0$,
$5=3\cdot0+5\cdot1$,
and $6=3\cdot2+5\cdot0$.
The other half ($1$, $2$, $4$ and $7$) indeed can't be written in the desired form.
These are explicit solutions:
If $n=3k$, then we have $n=3\cdot k+5\cdot0$.
If $n=3k+1$ with $k>2$, then we can write $n=3\cdot(k-3)+5\cdot2$.
If $n=3k+2$ with $k>1$, then $n=3\cdot(k-1)+5\cdot1$.
This covers all integers $n>7$.
All other solutions can now be derived from these using Bézout's identity.
|
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|
What's the explanation for why n^2+1 is never divisible by 3? What's the explanation for why $n^2+1$ is never divisible by $3$?
There are proofs on this site, but they are either wrong or overcomplicated.
It can be proved very easily by imagining 3 consecutive numbers, $n-1$, $n$, and $n+1$. We know that exactly one of these numbers must be divisible by 3.
$$(n-1)(n)(n+1)=(n)(n-1)(n+1)=(n)(n^2-1)$$
Since one of those first numbers had to have been divisible by $3$, this new product $(n)(n^2-1)$ must also be divisible by $3$. That means that either $n$ (and by extension $n^2$) or $n^2-1$ is divisible by $3$. If one of those has to be divisible by $3$, then $n^2+1$ cannot be.
So it is definitely true. My question is why is this true, what is inherent about $1$ more than a square number that makes it not divisible by $3$? Another way of saying this might be to explain it to me as if I don't know algebra.
|
Another approach is to write $n=3k+r$, with $r=0$, or $r=1$, or $r=2$, which is always possible: it's a simple division with the remainder. Now, the case $r=0$ is easily settled:
$$
n^2+1=9k^2+1
$$
which can't be divisible by $3$.
If $r=1$, then
$$
n^2+1=(3k+1)^2+1=9k^2+6k+1+1=3(3k^2+2k)+2
$$
which is not divisible by $3$.
If $r=2$, then
$$
n^2+1=(3k+2)^2+1=9k^2+12k+4+1=3(3k^2+4k)+3+2=3(3k^2+4k+1)+2
$$
and again this is not divisible by $3$.
|
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|
Prove that $\lim\limits_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$
Prove that:
$$\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)=0$$
Epsilon>0. According to the Archimedean Property of reals, we have n1 element N with epsilon*n1>11. (Why 11? Seems so random...)
n0:=max{3,n1} (What's the point of that? It doesn't appear anywhere in the proof...).
For each n element N with n>=n0 we have:
\begin{align}
0&\lt\sqrt{n^4+n^2+20n+7}-\sqrt{n^4+n^2+1} \\\,\\
&=\dfrac{\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}\right)\cdot\left(\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}\right)}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}} \\\,\\
&=\dfrac{20n+6}{\sqrt{n^4+n^2+20n+7}\,+\,\sqrt{n^4+n^2+1}}\leqslant\dfrac{20n+2n}{\sqrt{n^4}+\sqrt{n^4}}=\dfrac{11}n\leqslant\dfrac{11}{n_1}\lt\epsilon
\end{align}
(I don't get the circled part. Why 20n+2n, why "cut off" the roots? And why should it equal 11/n?)
From this we get
$$\left|\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right|\lt\epsilon$$
, thus the proof is complete.
Thanks for the clarifications!
|
You should, through looking at only the leading terms of
$$\dfrac{20 n + 6}{\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}}$$
see
$$\dfrac{20 n}{\sqrt{n^4}+\sqrt{n^4}}=\dfrac{20 n}{2 n^2}=\dfrac{10}{n}$$
which is clearly going to go to $0$ as $n\rightarrow \infty $. That's what we want to work towards.
So we try to get rid of the lower order terms but only by increasing the expression to preserve the inequality. So only to increase the denominator and reduce the numerator.
The numerator $\sqrt{n^4+n^2+20n+7}+\sqrt{n^4+n^2+1}$ can be replaced by the lesser $\sqrt{n^4}+\sqrt{n^4}$.
But we can only get rid of the $6$ term by increasing it up to a higher degree term. They could have just used $n$, but instead used $2n$, presumably so they could nicely cancel out the 2 in the denominator, and get the prettier $\dfrac{11}{n}$ rather than $\dfrac{21}{2n}$.
So the $11$ is arbitrary, and so is the $3$. It could have been $10.5$ and $6$ if we went with $6 \leq n$ (for $n\geq6$) rather than $6 \leq2n$ (for $n\geq3$).
|
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|
Derivative of trigonometric function help I am having trouble computing derivatives at $x=2k \pi$.
Let $f(x)=x \sqrt{1-\cos(x)}$.
Then $\displaystyle f'(x)=\frac{2(1-\cos x)+x\sin x}{2\sqrt{1-\cos x}}$.
Then $\displaystyle f'(2k\pi)=\lim_{x \rightarrow 2k\pi}\frac{2(1-\cos x)+x\sin x}{2\sqrt{1-\cos x}}$.
I know limits such as $\displaystyle \lim_{x \rightarrow o}\frac{1-\cos x}{x^{2}}=\frac{1}{2}$, but I don't know how to use it here correctly.
|
As $x$ approaches $2k\pi$ from the negative side, $\sin x$ is negative. From the positive side, $\sin x$ is positive. We have
$$\lim_{x\rightarrow2k\pi^-}\dfrac{2(1-\cos x)+x\sin x}{2\sqrt{1-\cos x}}=\lim_{x\rightarrow2k\pi^-}\dfrac{2(1-\cos x)-x\sqrt{1-\cos^2x}}{2\sqrt{1-\cos x}}=$$
$$\lim_{x\rightarrow2k\pi^-}\sqrt{1-\cos x}-\dfrac{x\sqrt{1+\cos x}}2=-k\pi\sqrt2$$
The positive side is similar except $\sin x=\sqrt{1-\cos^2x}$ instead of $-\sqrt{1-\cos^2x}$ yielding $k\pi\sqrt2$. The limit as $x$ approaches $2k\pi$ exists when these 2 are equal, when $k=0$.
|
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|
Integral of $\sqrt{16x^{2}+9}$ with respect to x Find the Integral of $\sqrt{16x^{2}+9}$ with respect to x
My try: $\int\sqrt{16x^{2}+9} dx =\int(16x^{2}+9)^{1/2}dx$ then make u substitution $u=16x^{2}+9$
but then have no idea how to continue
|
HINT:
$$I=\int\sqrt{x^2+a^2}dx=\int dx \sqrt{x^2+a^2}-\int\left(\int dx \frac{d \sqrt{x^2+a^2}}{dx}\right)dx$$
$$=x\sqrt{x^2+a^2}-\int \frac{x^2}{\sqrt{x^2+a^2}}dx$$
$$=x\sqrt{x^2+a^2}-\int \frac{x^2+a^2-a^2}{\sqrt{x^2+a^2}}dx$$
$$=x\sqrt{x^2+a^2}-I+a^2\int \frac1{\sqrt{x^2+a^2}}dx$$
For $\displaystyle \int \frac1{\sqrt{x^2+a^2}}dx,$ use Trigonometric substitution
Now $\displaystyle 16x^2+9=16\left(x^2+\left(\frac34\right)^2\right)$
|
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|
Find minimum value of this expression: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$ Let $a,b,c\in R$ and satisfying $a^2+b^2+c^2=1$
Find minimum value of this expression: $P=\sqrt{a^2+(1-bc)^2}+\sqrt{b^2+(1-ca)^2}+\sqrt{c^2+(1-ab)^2}$
|
We need to consider only non-negative $a, b, c$ as: $P(a, b, c)=P(-a, -b, -c)$ means if all three are negative we may replace them with positive numbers, if exactly two are negative then we can replace them with a triple having only one negative, and finally if only one is negative (say $a$), $P(a, b, c) \ge P(-a, b, c)$ gives us a better minimum in positive numbers. So WLOG, consider $a, b, c \in [0, 1]$.
By Cauchy-Schwarz, $\left(a^2+(1-bc)^2\right)(3+4) \ge \left(\sqrt3a+2(1-bc)\right)^2$
$\implies \sqrt{a^2+(1-bc)^2} \ge \frac1{\sqrt7} \left(2+\sqrt3a-2bc\right)$
$\implies \sqrt7P \ge 6+\sqrt3(a+b+c) -2(ab+bc+ca)$
Now if $x=a+b+c, \; 2(ab+bc+ca)=x^2-1$ so $\sqrt7P \ge 7+x(\sqrt3-x)$. As $x$ is non-negative and $x \le \sqrt{(a^2+b^2+c^2)(1+1+1)} = \sqrt3$, the minimum is in fact when $x=\sqrt3$, giving $P\ge \sqrt7$.
We may note that all inequalities used achieve equality simultaneously when $a=b=c=\frac1{\sqrt3}$.
|
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|
specify the limit of the following equation? What is the limit of the following $\lim_{n \to \infty}$ $\frac{(n^2-9)/(n+1)^3}{(n+3)/(3n+1)^2}$ and how to calculate it?
|
Write
$$\begin{array}{rcl}
\lim_{n \to \infty}\frac{(n^2-9)/(n+1)^3}{(n+3)/(3n+1)^2} &=& \lim_{n \to \infty}\frac{(n-3)\cdot(n+3)\cdot(3n+1)^2}{(n+3)\cdot(n+1)^3}\\
&=& \lim_{n \to \infty}\frac{(n-3)\cdot(3n+1)^2}{(n+1)^3}\\
&=& \lim_{n \to \infty}\frac{9n^3 -21n^2-17n-3}{n^3+3n^2+3n+1} \\
&=& \lim_{n \to \infty}\frac{9 -21n^{-1}-17n^{-2}-3n^{-3}}{1+3n^{-1}+3n^{-2}+1n^{-3}} \\
\end{array}$$
Thus the limit is $9$
Alternatively you can split up the fraction
$$\begin{array}{rcl}
\lim_{n \to \infty}\frac{(n^2-9)/(n+1)^3}{(n+3)/(3n+1)^2} &=& \lim_{n \to
\infty}\frac{(n-3)\cdot(n+3)\cdot(3n+1)^2}{(n+3)\cdot(n+1)^3}\\
&=& \lim_{n \to \infty}\frac{(n-3)\cdot(3n+1)^2}{(n+1)^3}\\
&=& \lim_{n \to \infty} 9 - \frac{48}{n+1}+\frac{52}{(n+1)^2}-\frac{16}{(n+1)^3}\\
&=& 9
\end{array}$$
|
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|
mixed numbers subtraction vertically In the following subtraction we are subtracting $2$ mixed numbers vertically. I know how it works except the last step.
$$ 7 \frac{1}{3} - 4 \frac{1}{2} = 3 + \frac{-1}{6} = 2 + \frac{5}{6} = 2 \frac{5}{6}$$
I am confused about this step: $ 3 + \frac{-1}{6} = 2 + \frac{5}{6}$
How does this work?
|
3+-1/6=2+(1-1/6)=2+(6/6-1/6)=2+5/6
|
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|
Fractions in convergence proof Sequence: $a_n = \sqrt{2+ \frac{3}{n}}$
To prove convergence, want to show that $\left|\sqrt{2+ \frac{3}{n}} - \sqrt{2}\right| \le \varepsilon$
Simplifying, we get that $\sqrt{2+ \frac{3}{n}} - \sqrt{2}= \frac{3/n}{\sqrt{2 + \frac{3}{n}} + \sqrt{2}}$.
I understand up to this part. But then, the textbook says we need to replace this fraction by a larger fraction, for example show that $\frac{3/n}{2 \sqrt{2}} \le \varepsilon$
Why must we replace it by a larger fraction? Why is it not sufficient to say that the simplified fraction is less than epsilon? Can we choose any larger fraction that we want?
|
$\sqrt{2 + \frac{3}{n}}$ is larger than $\sqrt{2}$; then $\sqrt{2 +\frac{3}{n}} + \sqrt{2}$ is larger than $2\sqrt{2}$.
So, $\frac{1}{\sqrt{2+\frac{3}{n}}+\sqrt{2}}$ is smaller than $\frac{1}{2\sqrt{2}}$ . So, now, you have an upper bound expressed in a simpler manner.
|
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|
Clarification Regarding evaluation of $\lim_{n\rightarrow \infty} n\sin(2\pi e n!)$- NBHM-$2009$ Question is to evaluate $$\lim_{n\rightarrow \infty} n\sin(2\pi e n!)$$
We have $e = 1 + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \cdots + \dfrac1{n!} + \dfrac1{(n+1)!} + \dfrac1{(n+2)!} + \cdots$
$$n!e=n!(1 + \dfrac1{1!} + \dfrac1{2!} + \dfrac1{3!} + \cdots + \dfrac1{n!} + \dfrac1{(n+1)!} + \dfrac1{(n+2)!} + \cdots)$$
$$=M+\dfrac1{n+1} + \dfrac1{(n+1)(n+2)} + \dfrac1{(n+1)(n+2)(n+3)} + \cdots$$
for some integer $M$.
Now, for $2\pi e n!$ we have :
$$2\pi e n!=2\pi (M+\dfrac1{n+1} + \dfrac1{(n+1)(n+2)} + \dfrac1{(n+1)(n+2)(n+3)} + \cdots)$$
$$=(2\pi M+\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$
For $\sin(2\pi e n!)$ We have :
$$\sin(2\pi e n!)=\sin(2\pi M+\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$
$$=\sin(\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$
For $n\sin(2\pi e n!)$ we have :
$$n\sin(2\pi e n!)=n\sin(\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$
For large $n$ we would have
$$\sin(\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$
$$=\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots$$
I hope I can say that for large $n$
$$n\sin(2\pi e n!)=n(\dfrac{2\pi}{n+1} + \dfrac{2\pi}{(n+1)(n+2)} + \dfrac{2\pi}{(n+1)(n+2)(n+3)} + \cdots)$$
$$=\frac{2\pi}{1+\frac{1}{n}}+\dfrac{2\pi}{(1+\frac{1}{n})(n+2)} + \dfrac{2\pi}{(1+\frac{1}{n})(n+2)(n+3)} + \cdots)$$
As $n\rightarrow \infty$ we would have :
$$\frac{2\pi}{1+0}+0+0+0+\dots=2\pi$$
So, $$\lim_{n\rightarrow \infty} n\sin(2\pi e n!)=2\pi$$
I would be thankful if some one can check what i have done is reasonably sufficient....
Thank you :)
|
your answer seems ok to me. my back-of-the-envelope calc using Landau's "big-O" was:
$$
n \sin (2\pi e n) = n \sin(2\pi E_n +2\pi e_n) = n \sin 2\pi e_n \\
= n \sin \left(\frac{2\pi}{n+1} + O(n^{-2})\right) = 2\pi (1+n^{-1})^{-1} + nO(n^{-2}) \\
= 2\pi + O(n^{-1})
$$
|
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|
Infinite Geometric Series Issue i have came across a series, i am trying to find its sum knowing the fact that, if it converges and its common ratio ex. r is: -1 < r < 1, then i can use the specified formula $\frac{a}{1-r}$ , which specifically means first term of series over 1 minus common ratio
here is the series
$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}$
i manipulated it this way to prove its convergence: $\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}(2n-1)\frac{1}{2^n}=\sum_{n=1}^{\infty}(2n-1)\left(\frac{1}{2}\right)^n$
$\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$
using it i get the result 1, which actually should be 3
|
$$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}\frac{2n}{2^n}-\sum_{n=1}^{\infty}\frac{1}{2^n}=\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}-1.$$
We use Maclaurin series for function $\frac{1}{(1-x)}$
$$\frac{1}{1-x}=1+x+x^2+x^3+\dots = \sum_{n=0}^{\infty} x^{n}$$
Differentiating both sides of this equation we get that
$$\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3+\dots = \sum_{n=1}^{\infty}n x^{n-1}$$
if $x=\frac 12$ then $\sum_{n=1}^{\infty}n (\frac 12)^{n-1}=\frac{1}{(1-\frac 12)^2}=4$
$$\sum_{n=1}^{\infty}\frac{2n-1}{2^n}=\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}-1=3$$
|
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|
How many zero elements are there in the inverse of the $n\times n$ matrix How many zero elements are there in the inverse of the $n\times n$ matrix
$A=\begin{bmatrix}
1&1&1&1&\cdots&1\\
1&2&2&2&\cdots&2\\
1&2&1&1&\cdots&1\\
1&2&1&2&\cdots&2\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
1&2&1&2&\cdots&\cdots
\end{bmatrix}$
My try: Denote by $a_{ij}$ and $b_{ij}$ the elements of $A$ and $A^{-1}$,respectively
then for $k\neq m$,we have
$$\sum_{i=0}^{n}a_{ki}b_{im}=0$$
Then I can't.Thank you very much
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Investigation with Maple indicates that for $n>1$, the inverse of $A$ has the following form:
(1) the main diagonal starts with $2$, finishes with $\pm1$, has zeros in between;
(2) the next diagonal each way has $1$ and $-1$ alternating;
(3) everything more than "one step away" from the main diagonal is $0$.
For example
$$\pmatrix{1&1&1&1\cr 1&2&2&2\cr 1&2&1&1\cr 1&2&1&2\cr}^{-1}
=\pmatrix{2&-1&0&0\cr -1&0&1&0\cr 0&1&0&-1\cr 0&0&-1&1\cr}$$
and
$$\pmatrix{1&1&1&1&1\cr 1&2&2&2&2\cr 1&2&1&1&1\cr 1&2&1&2&2\cr 1&2&1&2&1\cr}^{-1}
=\pmatrix{2&-1&0&0&0\cr -1&0&1&0&0\cr 0&1&0&-1&0\cr 0&0&-1&0&1\cr 0&0&0&1&-1\cr}\ .$$
I can't see a simple proof that this is true for all $n$, but if it is then the number of zeros is $n^2-2n$, provided $n>1$.
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|
Find the minimum value of $\frac{\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ac(c+a)}}{\sqrt{ab+bc+ca}}$ Let $a,b,c\ge0$ such that: $(a+b)(b+c)(c+a)=1$.
Find the minimum value of:
$$P=\frac{\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ac(c+a)}}{\sqrt{ab+bc+ca}}$$
I've tried many things but all failed. Please help. Thank you.
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If $c=0$ and $a=b=\frac{1}{\sqrt[3]2}$ then $P=\sqrt[3]2$.
We'll prove that it's a minimal value of $P$.
Indeed, we need to prove that $$\frac{\sum\limits_{cyc}\sqrt{ab(a+b)}}{\sqrt{ab+ac+bc}}\geq\sqrt[6]{4(a+b)(a+c)(b+c)}.$$
Now, by AM-GM
$$\sum\limits_{cyc}\sqrt{ab(a+b)}=\sqrt{\sum_{cyc}\left(a^2b+a^2c+2\sqrt{a^2bc(a+b)(a+c)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}\left(a^2b+a^2c+4abc\right)}.$$
Thus, it remains to prove that
$$\frac{\sum\limits_{cyc}(a^2b+a^2c+4abc)}{ab+ac+bc}\geq\sqrt[3]{4(a+b)(a+c)(b+c)}$$ or
$$a+b+c+\frac{9abc}{ab+ac+bc}\geq\sqrt[3]{4(a+b)(a+c)(b+c)}.$$
But by Schur
$$a+b+c+\frac{9abc}{ab+ac+bc}\geq\sqrt[3]{(a+b+c)^3+\frac{27abc(a+b+c)^2}{ab+ac+bc}}\geq$$
$$\geq\sqrt[3]{(a+b+c)^3+81abc}=\sqrt[3]{\sum_{cyc}(a^3-a^2b-a^2c+abc)+\sum_{cyc}(4a^2b+4a^2c+28abc)}\geq$$
$$\geq\sqrt[3]{\sum_{cyc}\left(4a^2b+4a^2c+\frac{8}{3}abc\right)}=\sqrt[3]{4(a+b)(a+c)(b+c)}$$
and we are done!
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|
Prove trigonometric relation Recently, I found this identities in a sheet of paper I was given as studying material:
$$\prod^n_{k=1}\sin\left(\frac{k\pi}{2n+1}\right)=\frac{\sqrt{2n+1}}{2^n}\tag1$$
$$\prod^n_{k=1}\cos\left(\frac{k\pi}{2n+1}\right)=\frac{1}{2^n}\tag2$$
$$\prod^n_{k=1}\tan\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}\tag3$$
They were there as auxiliary identites, meaning that we could use them if we needed to. But I found them really interesting, so I tried to prove them. After some time struggling with equations, I managed to prove $(2)$ for $n=1,2,3$, with increasing difficulty using the sum of cosines formula, the product of cosines, and the formula for the arithmetic progression of cosines. I couldn't finish the proof for $n=4$, because I needed that:
$$\left(2\cos\frac{\pi}{9}\right)\left(2\cos\frac{2\pi}{9}-1\right)=1\tag{4}$$
Equivalently, showing that $\cos\frac{\pi}{9}$ is a solution of
$$8x^3-2x-1=0$$
And no idea with $(1)$ and $(3)$.The best I could think of is that since the square root can't appear out of the blue, considering the square and doing a clever trick that results in a sum that yields $2n+1$ in some way would do it. But I have had no luck until now. Any ideas?
Edit: Actually, the proof for $(4)$ wasn't that hard:
$$2*2\cos\frac{\pi}{9}\cos\frac{2\pi}{9}-2\cos\frac{\pi}{9}=1$$
$$\iff 2*\left(\cos\frac{\pi}{3}+\cos\frac{\pi}{9}\right)-2\cos\frac{\pi}{9}=1$$
$$\iff 2*\left(\frac{1}{2}+\cos\frac{\pi}{9}\right)-2\cos\frac{\pi}{9}=1$$
The last one is clearly true.
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Consider
$$\prod_{k=1}^{2n} \sin \left(\frac{k\pi}{2n+1}\right).$$
By symmetry, it's the square of your first product, and expanding it with Euler's formula, we find
$$\begin{align}
\prod_{k=1}^{2n} \sin \left(\frac{k\pi}{2n+1}\right) &= \frac{1}{(2i)^{2n}}\prod_{k=1}^{2n} \left(\exp \frac{k\pi i}{2n+1} - \exp \frac{-k\pi i}{2n+1}\right)\\
&= \frac{1}{(2i)^{2n}} \exp\left(\frac{\pi i (2n)(2n+1)}{2(2n+1)}\right) \prod_{k=1}^{2n}\left(1-\exp \frac{-2k\pi i}{2n+1}\right)\\
&= \frac{1}{2^{2n}}\prod_{k=1}^{2n}\left(1-\exp \frac{-2k\pi i}{2n+1}\right).
\end{align}$$
Now consider the polynomial
$$P(X) = \prod_{k=1}^{2n} \left(X - \exp \frac{-2k\pi i}{2n+1}\right).$$
All its roots are $2n+1$-th roots of unity other than $1$, and they are all distinct, since
$$\frac{-k}{2n+1} - \frac{-j}{2n+1} \notin \mathbb{Z}$$
for $k\neq j$ and $1 \leqslant k,j \leqslant 2n$. So we have
$$P(X) = \frac{X^{2n+1}-1}{X-1} = X^{2n} + X^{2n-1} + \dotsc + X + 1,$$
and we see that $P(1) = 2n+1$, which proves the first equation.
The product of the cosines is similar, we get
$$\prod_{k=1}^{2n} \cos \frac{k\pi}{2n+1} = \frac{(-1)^n}{2^{2n}} \prod_{k=1}^{2n}\left(1 + \exp \frac{-2k\pi i}{2n+1}\right).$$
Considering
$$Q(X) = \prod_{k=1}^{2n}\left(X + \exp \frac{-2k\pi i}{2n+1}\right) = P(-X),$$
we find $Q(1) = 1$, so
$$\prod_{k=1}^{2n} \cos \frac{k\pi}{2n+1} = \frac{(-1)^n}{2^{2n}}.$$
Since the factor for $k = 2n+1-j$ is the negative of the factor for $j$, $1 \leqslant j \leqslant n$, we have
$$\prod_{k=1}^n \cos \frac{k\pi}{2n+1} = \frac{1}{2^n}.$$
The formula for the product of the tangents is a direct consequence.
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|
Solve the error in the simultaneous equation. The question is:
$$\begin{align}
\tag{1} 2x - 3y &= 3\\
\tag{2} 4x^2 - 9y^2 &= 3
\end{align}$$
From equation (1):
$$\tag{3} 2x = 3 - 3y.$$
Substitute equation (3) in (2):
$$\begin{align}
4x^2 - 9y^2 &= 3\\
(2x)^2 - 9y^2 &= 3\\
(3 - 3y)^2 - 9y^2 &= 3\\
[(3)^2 - 2\cdot 3\cdot 3y + (-3y)^2] - 9y^2 - 3 &= 0\\
9 +18y + 9y^2 - 9y^2 - 3 &= 0\\
-18y + 6 &= 0\\
-18y &= -6\\
y &= \frac 1 3
\end{align}$$
But the answer for y in my book is $ - \frac 1 3\ $
Which one is right?
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your equation (3) is wrong. it will be 2x=3+3y and then try.
the solving process is alternatively done by using the formula a^2-b^2 in the LHS of (2) and then use eq (1).
This particular type of problem follow always this type method.
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|
Proving a property of about the Fermat numbers
Show that the last digit in the decimal expansion of $F_n=2^{2^n}+1$ is $7$ for $n \geq 2$.
For our base step we let $n=2$. Now we have $2^{2^2}=16$. So the assertion holds for our base case. Then we assume it holds for $n$. For the $n+1$ case, is there a way to demonstrate this without resorting to this: $$(2^{2^n})^{{2^{n+1}}-2^n}.$$
That is the solution available in the back of the book. My attempt was to look at $2^{2^{n+1}}=2^{2^n}2^2.$
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$$ (2^{2^n}+1)^2 = 2^{2^{n+1}} + 2\cdot 2^{2^n}+1$$
Modulo 10 the left-hand side is 9 by the inductive hypothesis. Also
$$2\cdot 2^{2^n} = 2(2^{2^n}+1)-2$$
Modulo 10, this is 2.
What we have is:
$$(2^{2^n}+1)^2 - 2\cdot 2^{2^n}\,\,\, (\text{mod 10}) = 9-2=7$$
But that's the same as
$$2^{2^{n+1}} + 1 \,\,\, (\text{mod 10})$$
|
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|
Maximum and minimum distance from the origin
Find the maximum and minimum distances from the origin to the curve $5x^3+6xy+5y^2-8=0$
My attempt:
We have to maximise and minimise the following function $x^2+y^2$ with the constraint that $5x^3+6xy+5y^2-8=0$.
Let
$$F(x,y)=x^2+y^2+\lambda(5x^2+6xy+5y^2-8)$$
$$\frac{\delta F(x,y)}{\delta x}=2x+\lambda(10x+6y)$$
and
$$\frac{\delta F(x,y)}{\delta y}=2y+\lambda(6x+10y)$$
Multiplying the 2 equations by y,x respectively and subtracting I get
$$\lambda(y^2-x^2)=0$$
Hence $$y=x$$
Substituting $x=y$ in $5x^3+6xy+5y^2-8=0$, I get the $x=\pm \frac{1}{\sqrt2}$ and $y=\pm \frac{1}{\sqrt2}$ . Now I am stuck. Both the points corresponds to only one distance. Did I do something wrong?
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if the case is $5x^2+6xy+5y^2-8=0$, it could be solved in a easy way:
$5x^2+6xy+5y^2=8,2xy \le x^2+y^2 \implies 5x^2+5y^2 +3(x^2+y^2) \ge 8 \implies x^2+y^2 \ge 1$ when $x=y$
$2xy \ge -(x^2+y^2) \implies 2(x^2+y^2) \le 8 \implies x^2+y^2 \le 4$ when $x=-y$
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|
Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt:
$$
\begin{align}
\tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\
&= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\
&= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin(x)}\\
&= \frac{1}{\cos(x) \sin(x)}\\
&= \frac{1}{\frac{1}{\sec(x)}\frac{1}{\csc(x)}}\\
&=\frac{1}{\frac{1}{\sec \csc}}\\
&=\frac{1}{1}\cdot \frac{\sec(x) \csc(x)}{1}\\
&= \sec(x) \csc(x)
\end{align}
$$
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That is exactly correct! Just two things: First, $\tan,\sin,\cos,$ etc hold no meaning on their own, they need an argument. So just be sure to write $\tan x$, $\cos x$ etc rather than just $\tan$ or $\cos$.
Finally, you could save time on your proof by noticing on the fourth step that
$$
\frac{1}{\cos x\sin x}=\frac{1}{\cos x}\frac{1}{\sin x}=\sec x \csc x
$$
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|
How does $\frac1{n^4} =\frac1{(2n-1)^4} + \frac1{(2n)^4}$ I need to find the sum of $\displaystyle\sum_{n=1}^\infty \frac{1}{n^4}.$
In the solution document it written:
$$\displaystyle\sum_{n=1}^\infty \frac1{n^4} = \sum_{n=1}^\infty \frac1{(2n-1)^4} + \sum_{n=1}^\infty \frac1{(2n)^4}.$$
My question is, how did he get that $\displaystyle\sum_{n=1}^\infty \frac1{n^4} = \sum_{n=1}^\infty \frac1{(2n-1)^4} + \sum_{n=1}^\infty \frac1{(2n)^4} $?
Thanks in advance.
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Since the even and odd natural numbers form a complete residue system modulo two the author can separate the sum into two sums of even and odd terms. This could be done with any modulus for example instead of modulo two, we could separate the sum using a modulus of three:
$$\sum_{n=1}^\infty\frac{1}{n^4}=\sum_{n=1}^\infty\frac{1}{(3n)^4}+\sum_{n=1}^\infty\frac{1}{(3n-1)^4}+\sum_{n=1}^\infty\frac{1}{(3n-2)^4}$$
Or for any natural number $m$
$$\sum_{n=1}^\infty\frac{1}{n^4}=\sum_{k=0}^{m-1}\sum_{n=1}^\infty\frac{1}{(mn-k)^4}$$
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.