Question

How do you find the Maclaurin series for `1/(1+x)^3`?

Answer 1

I got

`1 - 3x + 6x^2 - 10x^3 + cdots`,

or

`1/2sum_(n=0)^(N) (-1)^(n) (n+2)(n+1)x^(n)`.

I did it the normal way with derivatives, but I also showed a way to do it using the old power series way where you start from `1/(1-x)`.

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NORMAL WAY

The normal/easier way to do it is to modify the Taylor series so that `a = 0`:

`stackrel("Taylor Series")overbrace(sum_(n=0)^(N) (f^((n))(a))/(n!)(x-a)^n) => stackrel("Maclaurin Series")overbrace(\mathbf(sum_(n=0)^(N) (f^((n))(0))/(n!)x^n))`

`= (f(0))/(0!) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + cdots`

We can take the derivative a few times. I'd recommend you do this first, instead of in the process of writing out the sum itself.

`color(green)(f^((0))(x)) = f(x) = color(green)(1/(1+x)^3)`

`color(green)(f'(x)) = -3(1+x)^(-4) = color(green)(-3/(1+x)^4)`

`color(green)(f''(x)) = -3*-4(1+x)^(-5) = color(green)(12/(1+x)^5)`

`color(green)(f'''(x)) = 12*-5(1+x)^(-6) = color(green)(-60/(1+x)^6)`

So let's see what we get. The Maclaurin series becomes:

`= (f(0))/(0!) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + cdots`

`= (1/(1+(0))^3)/1 + (-3/(1+(0))^4)/1 x + (12/(1+(0))^5)/2 x^2 + (-60/(1+(0))^6)/6 x^3 + cdots`

`= color(blue)(1 - 3x + 6x^2 - 10x^3 + cdots)`

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OLDER WAY; TAKES LONGER

Technically, the Maclaurin series for this can actually be derived from what you should have been taught as the power series based on `1/(1-x)`. Let's try it that way too so we can see the similarities.

We know that `1/(1-x) = sum_(n=0)^(N) x^n = 1 + x + x^2 + x^3 + cdots`.

If we modify `1/(1-x)` to look like `1/(1+x)^3`, we can similarly modify the power series itself to achieve the same result as we just got above in blue.

Our steps are:

1. Differentiate once to get `-1(1-x)^(-2)*-1 = 1/(1-x)^2`.
2. Differentiate again to get `-2*(1-x)^(-3)*-1 = 2*1/(1-x)^3`.
3. Multiply by `1/2`.
4. Change `1 - x` to `1 + (-x)`. That means we substituted `-x` in place of `x`!

Therefore, if we do the same thing to the power series, we get what you would call the "power series" for `1/(1+x)^3`.

Step 1:

`d/(dx)[1 + x + x^2 + x^3 + x^4 + x^5 + cdots]`

`= 1 + 2x + 3x^2 + 4x^3 + 5x^4 + cdots` for `n = 1` to `n = N`.

Step 2:

`d/(dx)[1 + 2x + 3x^2 + 4x^3 + 5x^4 + cdots]`

`= 2 + 6x + 12x^2 + 20x^3 + cdots` for `n = 2` to `n = N`.

Step 3:

`1/2(2 + 6x + 12x^2 + 20x^3 + cdots)`

`= 1 + 3x + 6x^2 + 10x^3 + cdots` for `n = 2` to `n = N`.

Step 4:

Finally, substitute `-x` to get:

`= 1 + 3(-x) + 6(-x)^2 + 10(-x)^3 + cdots`

`= 1 + 3x(-1)^1 + 6x^2(-1)^2 + 10x^3(-1)^3 + cdots`

`= color(blue)(1 - 3x + 6x^2 - 10x^3 + cdots)` for `n = 2` to `n = N`.

which is the same result as we got before! We just now know that the first term has shifted two indices relative to the power series for `1/(1-x)`.

So, the sum notation for the power series would end up having gone through the above steps as well. Note that for step 4, by replacing `x` with `-x`, you've turned it into an alternating series (`+,-,+,-, cdots`), which uses `(-1)^n`.

`sum_(n=0)^(N) x^n` (starting at `1/(1-x)`)

`=> d/(dx)[sum_(n=0)^(N) x^n] = sum_(n=1)^(N) nx^(n-1)` (Step 1)

`=> d/(dx)[sum_(n=1)^(N) nx^(n-1)] = sum_(n=2)^(N) n(n-1)x^(n-2)` (Step 2)

`=> 1/2sum_(n=2)^(N) n(n-1)x^(n-2)` (Step 3)

`=> 1/2sum_(n=2)^(N) (-1)^(n) n(n-1)x^(n-2)` (Step 4)

Finally, to bring it back to an index of `n=0` as for standard power series (instead of `n=2`), you know that `n` has to start `2` earlier than before.

So, add `2` to each term that involves `n` to compensate (except for `(-1)^n`, which should remain as-is to retain the alternating aspect).

`=> color(blue)(1/2sum_(n=0)^(N) (-1)^(n) (n+2)(n+1)x^(n) = 1/(1+x)^3)`

which you can see Wolfram Alpha agrees with.