How do you find f^7(0) where f(x)=x^2ln(1+x)?

2 Answers
Mar 14, 2017

f^((7))(0)=1008

Explanation:

We can construct the Maclaurin sequence for f(x)=x^2ln(1+x). Start with:

1/(1-x)=sum_(n=0)^oox^n

Replace x with -x:

1/(1+x)=sum_(n=0)^oo(-x)^n=sum_(n=0)^oo(-1)^nx^n

Integrate:

intdx/(1+x)=sum_(n=0)^oo(-1)^nintx^ndx

ln(1+x)=C+sum_(n=0)^oo(-1)^nx^(n+1)/(n+1)

At x=0 we see that C=0:

ln(1+x)=sum_(n=0)^oo(-1)^n/(n+1)x^(n+1)

Multiply by x^2:

x^2ln(1+x)=x^2sum_(n=0)^oo(-1)^n/(n+1)x^(n+1)=sum_(n=0)^oo(-1)^n/(n+1)x^(n+3)

We now have the Maclaurin series for f(x)=x^2ln(1+x).

The terms of a any general Maclaurin series are given by:

f(x)=sum_(n=0)^oof^((n))(0)/(n!)x^n

So the x^7 term of a general Maclaurin series is f^((7))(0)/(7!)x^7.

Using the Maclaurin series for x^2ln(1+x), we see that the x^7 term will occur when n=4, which gives a term of (-1)^4/(4+1)x^(4+3)=1/5x^7.

So the two coefficients of the x^7 terms must be equal:

1/5=f^((7))(0)/(7!)

f^((7))(0)=(7!)/5=1008

Mar 14, 2017

We know that the power series for a function is unique. It does not matter how we obtain the power series (it could from the Binomial Theorem, a Taylor Series, or a Maclaurin series).

As we are looking for f^((7))(0) let us consider the Taylor Series for f(x) about x=0, ie its Maclaurin series.

By definition:

f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f^((3))(0))/(3!)x^3 + ... + (f^((n))(0))/(n!)x^n + ...

The (well known) Maclaurin Series for ln(1+x) is given by:

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

And so it must be that the Maclaurin Series for x^2ln(1+x) is given by:

x^2ln(1+x) = x^2(x-x^2/2+x^3/3-x^4/4+x^5/5 +... )
" " = x^3-x^4/2+x^5/3-x^6/4+x^7/5+ ...

And if we equate the coefficients of x^7 from this derived series and the definition then we have:

\ \ \ (f^((7))(0))/(7!)= 1/5

:. f^((7))(0) = (7!)/(5)
" " = 5040/5
" " = 1008