Maclaurin Series for f(x)=(1-x)^-2: Find Solution

[Physicsnoob90]

Homework Statement

Find the Maclaurin series for f(x)=(1-x)^-2 using the definition of a Macluarin series,[ Assume that f has a power series expansion. Also find

Homework Equations

Maclaurin Series

The Attempt at a Solution

Steps:

1) I found the first 3 derivative of the function

2) I plugged in 0 for x

3) try to use the Macclaurin Series
f(x)= f(0) + f'(0) x + f''(0) x^2 / 2! + f'''(0) x^3 / 3! +...From here, I'm am stuck.

 

[STEMucator]

Homework Statement

Find the Maclaurin series for f(x)=(1-x)^-2 using the definition of a Macluarin series,[ Assume that f has a power series expansion. Also find

Homework Equations

Maclaurin Series

The Attempt at a Solution

Steps:

1) I found the first 3 derivative of the function

2) I plugged in 0 for x

3) try to use the Macclaurin Series
f(x)= f(0) + f'(0) x + f''(0) x^2 / 2! + f'''(0) x^3 / 3! +...


From here, I'm am stuck.

Let $t = -x$, then $f(x) = (1+t)^{-2}$

You now have a case of the binomial theorem with $\alpha = -2$.

Do you know the binomial series?

 

[Dick]

Homework Statement

Find the Maclaurin series for f(x)=(1-x)^-2 using the definition of a Macluarin series,[ Assume that f has a power series expansion. Also find

Homework Equations

Maclaurin Series

The Attempt at a Solution

Steps:

1) I found the first 3 derivative of the function

2) I plugged in 0 for x

3) try to use the Macclaurin Series
f(x)= f(0) + f'(0) x + f''(0) x^2 / 2! + f'''(0) x^3 / 3! +...


From here, I'm am stuck.

Ok, so what did you get for the first three derivatives evaluated at x=0? What do you get for the first three terms of the series? Don't you see a pattern to continue?

 

[Physicsnoob90]

Ok, so what did you get for the first three derivatives evaluated at x=0? What do you get for the first three terms of the series? Don't you see a pattern to continue?

Here is my work so far:

f(0) = 1
f'(x) = 2(1-x)^-3 ; f'(0) = 2
f''(x) = 6(1-x)^-4 ; f''(0) = 6
f'''(x) = 24(1-x)^-5 ; f'''(0) = 24


f(x) = (1-x)^-2

(1-x)^-2 = 1+2x+6/2! (x^2)+24/3! (x^3)+...

 

[Dick]

Here is my work so far:

f(0) = 1
f'(x) = 2(1-x)^-3 ; f'(0) = 2
f''(x) = 6(1-x)^-4 ; f''(0) = 6
f'''(x) = 24(1-x)^-5 ; f'''(0) = 24


f(x) = (1-x)^-2

(1-x)^-2 = 1+2x+6/2! (x^2)+24/3! (x^3)+...

Good! Expand the factorials. So you've got 1+2x+3x^2+4x^3+... Do you see the pattern? Can you argue why it must continue?

 

[STEMucator]

Notice every time you take a derivative, it has a "general form". Finding the $n^{th}$ derivative in a closed form, evaluated at $a$ will give you almost all of the work. Then every power series has the form:

$\sum_{n=0}^{∞} \frac{f^{(n)}(a)}{n!} (x-a)^n, \quad |x-a| < R$

Alternatively, you could appeal to the binomial theorem:

$(1+x)^{\alpha} = \sum_{k=0}^{∞} \begin{pmatrix} \alpha \\ k \end{pmatrix} x^k$

Where $\begin{pmatrix} \alpha \\ k \end{pmatrix} = \frac{\alpha(\alpha-1)(\alpha-2) \cdots (\alpha - k + 1)}{k!}$

and

$\begin{pmatrix} \alpha \\ 0 \end{pmatrix} = 1$

 

[Physicsnoob90]

Notice every time you take a derivative, it has a "general form". Finding the $n^{th}$ derivative in a closed form, evaluated at $a$ will give you almost all of the work. Then every power series has the form:

$\sum_{n=0}^{∞} \frac{f^{(n)}(a)}{n!} (x-a)^n, \quad |x-a| < R$

Alternatively, you could appeal to the binomial theorem:

$(1+x)^{\alpha} = \sum_{k=0}^{∞} \begin{pmatrix} \alpha \\ k \end{pmatrix} x^k$

Where $\begin{pmatrix} \alpha \\ k \end{pmatrix} = \frac{\alpha(\alpha-1)(\alpha-2) \cdots (\alpha - k + 1)}{k!}$

and

$\begin{pmatrix} \alpha \\ 0 \end{pmatrix} = 1$

Thanks! This really made more sense than what i wrote on my class notes

 

[STEMucator]

Thanks! This really made more sense than what i wrote on my class notes

No problem. If you want an easy way to get $R$, let $c_n = \frac{f^{(n)}(a)}{n!}$. Then every power series has the form:

$\sum_{n=0}^{∞} c_n (x-a)^n, \quad |x-a| < R$

With: $R = \displaystyle \lim_{n→∞} |\frac{c_n}{c_{n+1}}|$