Question about log/ln

  • Thread starter Physics197
  • Start date
  • #1
73
0
If you had something like ln(x+1) is there are way of breaking that term up?

I have a question where I have to find a two term approximation and i'm at a stage similar to this one and I don't know what I can do.
 

Answers and Replies

  • #2
mathman
Science Advisor
7,877
453
What do you mean by breaking it up? There is a power series for ln(1+x)

x-x2/2+x3/3+....
 
  • #3
73
0
I have y=ln(3u+4) and need a 2 term approximation (we are given a hint 4 = 3+1) so I assume that we must do: y=ln(3u+1+3) and the only way we know to approximate is y=ln(1+u) ~ u so in order to get that I need to somehow get rid of the 3 in y=ln(3u+1+3)
 
  • #4
lurflurf
Homework Helper
2,432
132
What kind of approximation.
We know log(x+1)~x
so expansion about x+1=1->x=0 would be nice
let f(x)=log(x+1)
f(x)~f(0)+f'(0)x+f''(0)x^2/2+f''(0)x^3/6+...
 
  • #5
73
0
The entire question is:

Find a two term approximation for the following function near t=1.

h = 5ln(3t+1)
and after substituting u+1 for t

I get:

h = 5ln(3u+4)
Now it gives us the hint: 4 = 3+1

But im stuck after this.
 
  • #6
mathman
Science Advisor
7,877
453
The entire question is:

Find a two term approximation for the following function near t=1.

h = 5ln(3t+1)
and after substituting u+1 for t

I get:

h = 5ln(3u+4)
Now it gives us the hint: 4 = 3+1

But im stuck after this.
Let v=3u+1, so h=5ln(1+v)~5[v - v2/2 + ...]
Take it from there.
 

Related Threads on Question about log/ln

  • Last Post
Replies
3
Views
5K
  • Poll
  • Last Post
Replies
17
Views
5K
  • Last Post
Replies
5
Views
798
  • Last Post
Replies
6
Views
7K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
2
Replies
25
Views
35K
  • Last Post
Replies
17
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
638
Top