How to Derive and Verify the Taylor Series Expansion for log(1+x)?

  • Thread starter Thread starter uppaladhadium
  • Start date Start date
  • Tags Tags
    Expansion
Click For Summary
SUMMARY

The Taylor series expansion for log(1+x) is derived as log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ... This series can be proven using the integral representation log(1+x) = ∫dx/(1+x), where the series for 1/(1+x) is expanded as 1 - x + x^2 - x^3 + ... Integrating this term by term from 0 to x yields the Taylor series. The discussion also raises questions about the derivatives of log(1+x) at a specific point a, confirming that f'(a) = 0 is not generally true for a > -1.

PREREQUISITES
  • Understanding of Taylor series and their general formula
  • Basic knowledge of calculus, specifically integration and differentiation
  • Familiarity with logarithmic functions and their properties
  • Concept of series expansion and convergence
NEXT STEPS
  • Study the derivation of Taylor series for various functions
  • Learn about the convergence criteria for Taylor series
  • Explore integral representations of other logarithmic functions
  • Investigate the implications of derivatives in Taylor series expansions
USEFUL FOR

Mathematicians, students studying calculus, and anyone interested in series expansions and their applications in mathematical analysis.

uppaladhadium
Messages
7
Reaction score
0
We know that log(1+x) = x+((x^2)/2)+((x^3)/3)+....((x^n)/n)+...
Could anybody please tell me the proof
 
Physics news on Phys.org
log(1+x)=x - x2/2 + x3/3 - x4/4 + ...
(Alternate signs)

The easiest way to see it is by using an integral representation.

log(1+x) = ∫dx/(1+x)

Since 1/(1+x) = 1 - x + x2 - x3 + ...,

integrating term by term gives the series for log(1+x), where the integration limits are [0,x].
 
  • Like
Likes Adhruth Ganesh and Aaron John Sabu
thankyou for the answers i am grateful to you
 
In taylor series if f(x)=log(1+x) then is f'(a)=0?
and is f(a)=log(1+a)
 
Last edited:
uppaladhadium said:
In taylor series if f(x)=log(1+x) then is f'(a)=0?
and is f(a)=log(1+a)

Assuming a > -1:

What do you get when you take the derivative of log(1+x)? What do you get when you plug a in?
 

Similar threads

Undergrad Taylor expansion of f(x)=arctan(x) at infinity
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad Taylor Expansion Question about this Series
  • · Replies 2 ·
Replies
2
Views
2K
Graduate Series expansion of ##(1-cx)^{1/x}##
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad How to Derive the Taylor Series for log(x)?
  • · Replies 5 ·
Replies
5
Views
16K
Undergrad Why prove Taylor's theorem with p(x)=q(x)−((b−a)^n/(b−x)^n)q(a)?
  • · Replies 9 ·
Replies
9
Views
3K
High School Why is Big-O about how rapidly the Taylor graph approaches that of f(x)?
  • · Replies 19 ·
Replies
19
Views
4K
Undergrad Understanding the Taylor Expansion of a Translated Function
  • · Replies 3 ·
Replies
3
Views
6K
Undergrad Why Is There No Maclaurin Expansion for Log(x) Compared to Log(x+1)?
  • · Replies 7 ·
Replies
7
Views
2K
Graduate Solving a power series
  • · Replies 3 ·
Replies
3
Views
3K
High School Why do I not understand what seems to be basic Calculus?
  • · Replies 11 ·
Replies
11
Views
2K