Represent the function 5ln(4-x) as a power series (Maclaurin series) in the format: f(x) = ∞∑n=0 c (subscript) n x (superscript) n

Question

Represent the function 5ln(4-x) as a power series (Maclaurin series) in the format: f(x) = ∞∑n=0 c_nxⁿ

Answer 1

Hi Sam,

The Maclaurin series f(x) = ∞∑n=0 c (subscript) n x (superscript) n

is the Tailor series expansion of function f(x) near x=0 :

f(x)=[_n=0to∞]∑f⁽ⁿ⁾(0)xⁿ/n! (1)

where

f⁽ⁿ⁾(0)=dⁿf(x)/dxⁿ|_x=0 (2)

Now, using f(x)=5ln(4-x) (3) we'd find that

f⁽ⁿ⁾(0)≡f(0)=5ln4 (4.0) ;

f⁽¹⁾(0)=d¹f(x)/dx¹|_x=0=

5/(4-x)|_x=0=-5/4 (4.1)

f⁽²⁾(0)=d²f(x)/dx²|_x=0=

+5/(4-x)²|_x=0*(-1)=

-5/(4-x)²|_x=0=-5/4² (4.2) Same way, we'd find the 3rd and derivatives of n≥3 order:

f⁽³⁾(0)=d³f(x)/dx³|_x=0=

+2*5/(4-x)³|_x=0*(-1)=

-2*5/(4-x)³|_x=0=-2*5/4³ (4.3) ................

f⁽ⁿ⁾(0)=dⁿf(x)/dxⁿ|_x=0=

-5(n-1)!/4ⁿ (4.4)

Finally, substitute (4.0)-(4.4) into (1) and get

:f(x)=[_n=0to∞]∑f⁽ⁿ⁾(0)xⁿ/n!

=5ln4-

-[_n=1to∞]∑{5(n-1)!/4ⁿ}xⁿ/n!}= 5ln4-[_n=1to∞]∑5xⁿ/[n(4)ⁿ](4.5)

Hope it is useful. Best,

Dr.Ariel B.