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)#.

COMMON 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 more than once, and it is best to do this before writing out the sum in full.

#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)#

Let us now examine what results: 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)#

OLDER METHOD; MORE TIME

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 actions are:

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 Wolfram Alpha appears to concur with.

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Answer 2

To find the Maclaurin series for ( \frac{1}{(1+x)^3} ), we can use the geometric series expansion. The geometric series expansion is given by:

[ \frac{1}{1 - u} = 1 + u + u^2 + u^3 + \ldots ]

Substitute ( u = -x ) into the geometric series expansion:

[ \frac{1}{1 + x} = 1 - x + x^2 - x^3 + \ldots ]

Now, raise both sides to the power of 3:

[ \left(\frac{1}{1 + x}\right)^3 = (1 - x + x^2 - x^3 + \ldots)^3 ]

Expanding the right side using the binomial theorem:

[ = 1 - 3x + 6x^2 - 10x^3 + \ldots ]

Thus, the Maclaurin series for ( \frac{1}{(1+x)^3} ) is:

[ 1 - 3x + 6x^2 - 10x^3 + \ldots ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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