• Support PF! Buy your school textbooks, materials and every day products Here!

Simple generating function

  • Thread starter talolard
  • Start date
  • #1
125
0

Homework Statement



I am trying to show that [tex]\sum{ \frac{x^n}{n}} = -ln(1-x)[/tex]
But I am doing something wrong and I cant find my mistake.
Please find my mistake and let me know what it is.
Thanks

The Attempt at a Solution


set [tex] f(x)=\sum {\frac{x^n}{n}} [/tex]
then [tex] f'(x)= \sum {x^n-1} [/tex]
so [tex] xf'(x)=\sum{x^n} = \frac {1}{1-x} [/tex]
which means that [tex] f'(x)=\frac {1}{x} - \frac{1}{1-x} [/tex]
integrtang we get [tex] f(x)=ln|x|-ln|1-x| [/tex]
which is bad

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
478
32
The derivative of the function is:

f'(x)=Sum[x^(n-1),{n,1,Infinity}]

which is the same as:

f'(x) = Sum[x^(n),{n,0,Infinity}] = 1/(1-x)

Integrating:

Sum[x^(n+1)/(n+1),{n,0,Infinity}] = - ln(1-x)

which is the same as:

Sum[x^(n)/n,{n,1,Infinity}] = -ln(1-x).

The trick lies in repositioning the starting index.
 

Related Threads on Simple generating function

  • Last Post
Replies
4
Views
879
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
715
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
819
  • Last Post
Replies
2
Views
818
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
17
Views
3K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
4
Views
2K
Top