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

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Discussion Overview

The discussion centers on deriving and verifying the Taylor series expansion for the function log(1+x). Participants explore various approaches to understand the series representation, including integral representations and the general form of Taylor series.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant states that log(1+x) can be expressed as a series: x + (x^2)/2 + (x^3)/3 + ...
  • Another participant mentions the general form of the Taylor series expansion and references an external source for further details.
  • A different participant provides an alternate series representation for log(1+x) with alternating signs: x - x^2/2 + x^3/3 - x^4/4 + ..., suggesting an integral representation as a method of derivation.
  • Several participants inquire about the values of the function and its derivatives at a point 'a', questioning if f'(a) equals zero and what f(a) would be.
  • One participant assumes a > -1 and prompts others to consider the derivative of log(1+x) and its evaluation at 'a'.

Areas of Agreement / Disagreement

Participants express various viewpoints on the Taylor series expansion, with no consensus reached on a single method of derivation or the specific values of derivatives at point 'a'.

Contextual Notes

Some assumptions regarding the domain of 'x' and the point 'a' are mentioned, particularly the condition a > -1, but these are not fully resolved within the discussion.

uppaladhadium
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We know that log(1+x) = x+((x^2)/2)+((x^3)/3)+....((x^n)/n)+...
Could anybody please tell me the proof
 
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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].
 
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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?
 

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