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