If \(\displaystyle e^{I+1}=\sum_{n=0}^{\infty }\left ( \frac{a}{b} \right )^n\) where \(\displaystyle a, b\) are coprime numbers, then calculate the sum \(\displaystyle S=a+b\).
Integral and power series
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Integral and power series
Let \(\displaystyle I=\int_{0}^{1}\left ( \left \lfloor \frac{2}{x} \right \rfloor-2\left \lfloor \frac{1}{x} \right \rfloor \right )\, {\rm d}x\).
If \(\displaystyle e^{I+1}=\sum_{n=0}^{\infty }\left ( \frac{a}{b} \right )^n\) where \(\displaystyle a, b\) are coprime numbers, then calculate the sum \(\displaystyle S=a+b\).
If \(\displaystyle e^{I+1}=\sum_{n=0}^{\infty }\left ( \frac{a}{b} \right )^n\) where \(\displaystyle a, b\) are coprime numbers, then calculate the sum \(\displaystyle S=a+b\).
Hidden message
Imagination is much more important than knowledge.
- Grigorios Kostakos
- Founder
- Posts: 461
- Joined: Mon Nov 09, 2015 1:36 am
- Location: Ioannina, Greece
Re: Integral and power series
We give a solution to some point:
\begin{align*}
I&=\int_{0}^{1}\left ( \left \lfloor \frac{2}{x} \right \rfloor-2\left \lfloor \frac{1}{x} \right \rfloor \right )dx\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\frac{1}{x}} \\
{dx\,=\,-\frac{1}{t^2}\,dt}
\end{subarray}}\,\int_{1}^{+\infty}\frac{1}{t^2}\bigl({\lfloor{2t}\rfloor-2\lfloor{t}\rfloor }\bigr)\,dt\\
&=\sum_{n=1}^{\infty }\int_{n+\frac{1}{2}}^{n+1}{\frac{1}{t^2}\,dt}\\
&=\sum_{n=1}^{\infty }\frac {1}{(n+1)(2n+1)}\\
&=2\log{2}-1\,.
\end{align*} So \(e^{I+1}=e^{\log4-1+1}=4\) and the geometric series must be converges to \(4\), i.e. \(a<b\) and \[\sum_{n=0}^{\infty }\left ( \frac{a}{b} \right )^n=\frac{b}{b-a}=4\,.\] So we have that \(b=\frac{4}{3}\,a\,.\)
Remains the evaluation of the number \(S=a+\frac{4}{3}a=\frac{7}{3}a\) with the assumption that \({\rm{gcd}}\bigl(a,\tfrac{4}{3}a\bigr)=1\,.\)
\(\bullet\) In the graph below appears that the integral \(\int_{1}^{+\infty}\frac{1}{t^2}\bigl({\lfloor{2t}\rfloor-2\lfloor{t}\rfloor }\bigr)\,dt\) is equal to the infinite sum of the integrals \(\int_{n+\frac{1}{2}}^{n+1}{\frac{1}{t^2}\,dt}\,,\; n\in\mathbb{N}\,.\)
[/centre]
\begin{align*}
I&=\int_{0}^{1}\left ( \left \lfloor \frac{2}{x} \right \rfloor-2\left \lfloor \frac{1}{x} \right \rfloor \right )dx\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\frac{1}{x}} \\
{dx\,=\,-\frac{1}{t^2}\,dt}
\end{subarray}}\,\int_{1}^{+\infty}\frac{1}{t^2}\bigl({\lfloor{2t}\rfloor-2\lfloor{t}\rfloor }\bigr)\,dt\\
&=\sum_{n=1}^{\infty }\int_{n+\frac{1}{2}}^{n+1}{\frac{1}{t^2}\,dt}\\
&=\sum_{n=1}^{\infty }\frac {1}{(n+1)(2n+1)}\\
&=2\log{2}-1\,.
\end{align*} So \(e^{I+1}=e^{\log4-1+1}=4\) and the geometric series must be converges to \(4\), i.e. \(a<b\) and \[\sum_{n=0}^{\infty }\left ( \frac{a}{b} \right )^n=\frac{b}{b-a}=4\,.\] So we have that \(b=\frac{4}{3}\,a\,.\)
Remains the evaluation of the number \(S=a+\frac{4}{3}a=\frac{7}{3}a\) with the assumption that \({\rm{gcd}}\bigl(a,\tfrac{4}{3}a\bigr)=1\,.\)
\(\bullet\) In the graph below appears that the integral \(\int_{1}^{+\infty}\frac{1}{t^2}\bigl({\lfloor{2t}\rfloor-2\lfloor{t}\rfloor }\bigr)\,dt\) is equal to the infinite sum of the integrals \(\int_{n+\frac{1}{2}}^{n+1}{\frac{1}{t^2}\,dt}\,,\; n\in\mathbb{N}\,.\)
[/centre]
Grigorios Kostakos
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 12 guests