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

##### 2 Answers
Mar 14, 2017

${f}^{\left(7\right)} \left(0\right) = 1008$

#### Explanation:

We can construct the Maclaurin sequence for $f \left(x\right) = {x}^{2} \ln \left(1 + x\right)$. Start with:

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$

Replace $x$ with $- x$:

$\frac{1}{1 + x} = {\sum}_{n = 0}^{\infty} {\left(- x\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{n}$

Integrate:

$\int \frac{\mathrm{dx}}{1 + x} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \int {x}^{n} \mathrm{dx}$

$\ln \left(1 + x\right) = C + {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{n + 1} / \left(n + 1\right)$

At $x = 0$ we see that $C = 0$:

$\ln \left(1 + x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(n + 1\right) {x}^{n + 1}$

Multiply by ${x}^{2}$:

${x}^{2} \ln \left(1 + x\right) = {x}^{2} {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(n + 1\right) {x}^{n + 1} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(n + 1\right) {x}^{n + 3}$

We now have the Maclaurin series for $f \left(x\right) = {x}^{2} \ln \left(1 + x\right)$.

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 ${x}^{2} \ln \left(1 + x\right)$, we see that the ${x}^{7}$ term will occur when $n = 4$, which gives a term of ${\left(- 1\right)}^{4} / \left(4 + 1\right) {x}^{4 + 3} = \frac{1}{5} {x}^{7}$.

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 ${f}^{\left(7\right)} \left(0\right)$ let us consider the Taylor Series for $f \left(x\right)$ 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 $\ln \left(1 + x\right)$ is given by:

$\ln \left(1 + x\right) = x - {x}^{2} / 2 + {x}^{3} / 3 - {x}^{4} / 4 + {x}^{5} / 5 \ldots$

And so it must be that the Maclaurin Series for ${x}^{2} \ln \left(1 + x\right)$ is given by:

${x}^{2} \ln \left(1 + x\right) = {x}^{2} \left(x - {x}^{2} / 2 + {x}^{3} / 3 - {x}^{4} / 4 + {x}^{5} / 5 + \ldots\right)$
$\text{ } = {x}^{3} - {x}^{4} / 2 + {x}^{5} / 3 - {x}^{6} / 4 + {x}^{7} / 5 + \ldots$

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)
$\text{ } = \frac{5040}{5}$
$\text{ } = 1008$