This question 16 could you please help me?

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2 Answers
Mar 10, 2018

The series (to three terms) is #1 + x^3 + x^6#.

Explanation:

The Taylor series expansion of #f(x)=1/(1-x)# is

#f(x)=1+x+x^2+x^3+...+x^n+...#

(as long as #abs(x)<1#).

As an example,

#f(1/2) = 1/(1-1/2) =2#

# color(white)(f(1/2))= 1 + 1/2 + (1/2)^2+(1/2)^3+...+(1/2)^n+...#

# color(white)(f(1/2))= 1 + 1/2 + 1/4+1/8+...+1/(2^n)+...#

All we did was replace each #x# with #1/2#. So, what if we replace #x# with another expression, instead of a number? For example, what if, instead of #f(1/2)#, we take #f(x^3)?#

We'd do the same thing as we did with the #1/2#'s: just replace every #x# with an #x^3#.

Thus:

#f(x^3)=1/(1-x^3)#

#color(white)(f(x^3))=1+x^3+(x^3)^2+(x^3)^3+...+(x^3)^n+...#

#color(white)(f(x^3))=1+x^3+x^6+x^9+...+x^(3n)+...#

Mar 10, 2018

#1+x^3+x^6#

Explanation:

#"we are given the power series for "f(x)=1/(1-x)#

#"that is "1/(1-x)=1+x+x^2+x^3+....#

#"for "f(x)=1/(1-x^3)#

#"substitute "x" = "x^3" into the series"#

#rArrf(x)=1/(1-x^3)=1+x^3+(x^3)^2#

#color(white)(xxxxxxxxxxxxx)=1+x^3+x^6#