MacLaurin exercice

[nestor95]

how can I begin this problem? Develop according MacLaurin's serie:

1/(b-ax)

Thanks!!!

 

[HallsofIvy]

Okay, first do you know what a "MacLaurin series" is?

A "MacLaurin series" for a given function, f(x), is an infinite sum of constants times powers of x, $\displaystyle \sum_{n=0}^\infty c_mx^n$, where the constants are constructed in a particular way. If you are asked to find the MacLaurin series of $\displaystyle f(x)= 1/(b- ax)$ then I have to assume that you have been taught that "particular way" of constructing those coefficients!
That way is $\displaystyle c_n= \frac{1}{n!}\frac{d^nf}{dx^n}(0)$. That is, the coefficient of $\displaystyle x^n$ is one over n factorial times the nth derivative of the function evaluated at x= 0 (the "zeroth" derivative being the value of the function itself).

Here, $\displaystyle f(x)= \frac{1}{a+ bx}= (a+ bx)^{-1}$. The value of the function at x= 0 is $\displaystyle c_0= f(0)= a^{-1}= \frac{1}{a}$. The first derivative is $\displaystyle f'(x)= -(a+ bx)^{-2}(b)= \frac{-b}{(a+ bx)^2}$ and its value at x= 0 is $\displaystyle c_1= \frac{-b}{a^2}$. The second derivative is $\displaystyle 2(a+ bx)^{-3}b^2= \frac{2b^2}{(a+ bx)^3}$. Its value at x= 0 is $\displaystyle \frac{2b^2}{a^3}$ and, since 2!= 2, the coefficient of $\displaystyle x^2$ is $\displaystyle \frac{b^2}{a^3}$.

The first three terms of the MacLaurin Series for $\displaystyle \frac{1}{a+ bx}$ are $\displaystyle \frac{1}{a}- \frac{b}{a^2}x+ \frac{b^2}{a^3}x^2$.

To get the entire MacLaurin series you will want to calculate enough terms explictiely so that you can "guess" the general formula for all n (so far we have $\displaystyle c_0= \frac{1}{a}$, $\displaystyle c_1= -\frac{b}{a^2}$, and $\displaystyle c_2= \frac{b^2}{a^3}$ so that shouldn't be too hard!)

 

[nestor95]

I think I found the solution for f(x)=1/(b-ax)

(a^n·x^n)/(b^(n-1))

Thank you very much!!

 

[HallsofIvy]

That doesn't match the values I got. I got that the first three coefficients are $\displaystyle c_0= \frac{1}{a}$, $\displaystyle c_2= -\frac{b}{a^2}$, and $\displaystyle c_3= \frac{b^2}{a^3}$ while your formula gives $\displaystyle c_0= \frac{a^0}{b^{-1}}= b$, $\displaystyle c_2= \frac{a^2}{b^0}= a^2$, and $\displaystyle c_3= \frac{a^3}{b^2}$.