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How do you find the power series representation for the function #f(x)=ln(5-x)# ?

Answer 1

Emilia Aguilar

#f(x)=-sum_{n=1}^infty(x-4)^n/n#

Let us look at some details.

Since

#ln(1-x)=-sum_{n=1}^inftyx^n/n#,

by replacing #x# by #(x-4)#,

#ln(5-x)=ln[1-(x-4)]=-sum_{n=1}^infty(x-4)^n/n#.

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Answer 2

Luke Alvarez

To find the power series representation for the function f(x) = ln(5 - x), you can use the Taylor series expansion. The Taylor series expansion for ln(1 + x) is well-known:

ln(1 + x) = x - x^2/2 + x^3/3 - x^4/4 + ...

However, to use this expansion for ln(5 - x), we need to rewrite it in the form ln(1 + x). We can do this by noticing that 5 - x can be written as 1 - (-x/5). So:

ln(5 - x) = ln(1 - (-x/5)).

Now, we can use the Taylor series expansion for ln(1 + x):

ln(1 - (-x/5)) = -x/5 - (-x/5)^2/2 - (-x/5)^3/3 - (-x/5)^4/4 + ...

This can be simplified to:

= -x/5 - x^2/50 - x^3/750 - x^4/18750 + ...

Thus, the power series representation for f(x) = ln(5 - x) is:

ln(5 - x) = -x/5 - x^2/50 - x^3/750 - x^4/18750 + ...

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.