find the taylor's series expansion of ln x about x subscript 0 = 0

Question

the question is about taylor series

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer 1

(i) ln(x) is undefined for x=0, but is defined ln(1+X), where X=0, so if we work out the series for X we can replace X with x-1, because x=1+X. Let y=a₀+a₁X+a₂X²+a₃X³+... represented by ∑a_nXⁿ.

When X=0, y=ln(1)=0, so a₀=0.

y⁽¹⁾=1/(1+X), y⁽²⁾=-1/(1+X)², y⁽³⁾=2/(1+X)³, y⁽⁴⁾=-6/(1+X)⁴, ..., y⁽ⁿ⁾=(-1)ⁿ⁺¹n!/(1+X)ⁿ+...

Also, y⁽¹⁾(0)=1=a₁, so a₁=1; y⁽²⁾(0)=-1=2a₂, so a₂=-½; y⁽³⁾(0)=2=6a₃, so a₃=⅓; y⁽⁴⁾(0)=-6=4!a₄, so a₄=-¼; etc. Therefore, a_n=(-1)ⁿ⁺¹/n for n>0. The series is therefore:

ln(X)=X-X²/2+X³/3-X⁴/4+... ln(1+x)=x-x²/2+x³/3-x⁴/4+...

ln(x)=x-1-(x-1)²/2+(x-1)³/3-(x-1)⁴/4+... When x=0 ln(0)=-1-1/2-1/3-1/4-... which must be (and is) divergent.

(ii) Let y=cos(x), then:

y⁽¹⁾=-sin(x); y⁽²⁾=-cos(x); y⁽³⁾=sin(x); y⁽⁴⁾=cos(x)=y; y⁽⁵⁾=-sin(x)=y⁽¹⁾, ...

In particular, y⁽⁰⁾(0)=y(0)=1, y⁽¹⁾(0)=0, y⁽²⁾(0)=-1, y⁽³⁾(0)=0. The pattern (1,0-1,0) repeats indefinitely.

If y=∑a_nxⁿ as a series, then a₀=1 because y(0)=1=a₀.

All odd powers of x have a zero coefficient, so a_2k+1=0, where k is an integer≥0.

a_4k=1/(4k)! and a_4k+2=-1/(4k+2)! making the series 1-x²/2!+x⁴/4!-x⁶/6!+...