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

##### 2 Answers

#### Explanation:

We can construct the Maclaurin sequence for

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

Replace

#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

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

Multiply by

#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

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

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

So the

Using the Maclaurin series for

So the two coefficients of the

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

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

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

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) = 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) = 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

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

# :. f^((7))(0) = (7!)/(5) #

# " " = 5040/5 #

# " " = 1008 #