Question

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

Answer 1

`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`

Answer 2

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`