1
$\begingroup$

Let $f(x)=\frac{e^{2x-1}}{1+e^{2x-1}}$ Then find $\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)$

How I proceed: $\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)=\int_{1}^{1233}\frac{e^{\frac{2x}{1234}-1}}{1+e^{\frac{2x}{1234}-1}}~dx$ then how I solve this integral. please help.

  • 0
    Then how I proceed?2012-12-22

3 Answers 3

4

Note that:

\begin{align} f(k/1234)+f((1234-k)/1234)=& \frac{e^{\frac{k}{617}-1}}{e^{\frac{k}{617}-1}+1}+\frac{e^{\frac{1234-k}{617}-1}}{e^{\frac{1234-k}{617}-1}+1}&=&\\ =&\frac{e^{k/617}}{e^{k/617}+e}+\frac{e}{e^{k/617}+e}&=&1 \end{align}

So your sum is equal to $616+f(1/2)$.

  • 0
    Thanks to $M$$a$thematica for simplifying the terms, I wouldn't have voluntarily done this.2012-12-22
3

Observe that $f(x)+f(1-x)=1$

Put $x=\frac k{1234}\implies f\left(\frac k{1234}\right)+ f\left(\frac {1234-k}{1234}\right)=1$

Then put $k=1,2,,\cdots ,\frac{1234}2=617$

so $ f\left(\frac 1{1234}\right)+ f\left(\frac {1234-1}{1234}\right)=1,$

$ f\left(\frac 2{1234}\right)+ f\left(\frac {1234-2}{1234}\right)=1,$

**

$ f\left(\frac {617}{1234}\right)+ f\left(\frac {1234-617}{1234}\right)=1\implies f\left(\frac 12\right)=\frac12$

Adding we get, $\sum_{k=1}^{1233}f\left(\frac{k}{1234}\right)=617(1)-f\left(\frac {617}{1234}\right)=617-\frac12$

  • 0
    I like how you can solve for $f(1/2)$.2012-12-22
1

I don't know if a closed form for your sum exists. However, we note that:

$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{k=1}^{N}f(\frac{k}{N})=\int_0^1f(x)dx$

Thus, for large values of $N$ you can estimate the sum using the integral.

To evaluate the integral, we note that $\frac{d}{dx}[e^{2x-1}+1]=2e^{2x-1}$, thus: $\int_0^1\frac{e^{2x-1}}{e^{2x-1}+1}dx=\frac{1}{2}\int_0^1\frac{2e^{2x-1}}{e^{2x-1}+1}dx=[\frac{1}{2}\log(1+e^{2x-1})]_{x=0}^{x=1}$

  • 0
    @Argha See akkkk's answer,way better than mine2012-12-22