Question

Question #421de

Answer 1

As this is a polynomial, its MacLaurin expansion coincides with the polynomial itself. Using the binomial theorem:

`(1-x)^5 = 1-5x+10x^2-10x^3+5x^4-x^5`

Answer 2

`(1-x)^5=1-5x+10x^2-10x^3+5x^4-x^5`

Explanation:

We can also try to begin constructing a Maclaurin expansion to see that we get the same result as with the binomial theorem:

We start by taking derivatives of `(1-x)^5`:

`d/dx((1-x)^5)=-5(1-x)^4`

`d^2/dx^2((1-x)^5)=5*4(1-x)^3`

`d^3/dx^3((1-x)^5)=-5*4*3(1-x)^2`

`d^4/dx^4((1-x)^5)=5*4*3*2(1-x)`

`d^5/dx^5((1-x)^5)=-5*4*3*2*1`

`d^6/dx^6((1-x)^5)=0`

`d^7/dx^7((1-x)^5)=0`

We see that all derivatives after the 5th will be zero. This means the Maclaurin expansion is only those first five derivatives.

If we plug in `0` to all our derivatives we see that the `1-x` terms become `1`, so we get that the derivatives are:

`f^0(0)=1`

`f^1(0)=-5=-(5!)/(4!)=-5`

`f^2(0)=5*4=(5!)/(3!)=20`

`f^3(0)=-5*4*3=-(5!)/(2!)=-60`

`f^4(0)=5*4*3*2=(5!)/(1!)=120`

`f^5(0)=-5*4*3*2*1=-(5!)/(0!)=-120`

This means our Maclaurin expansion is:
`1-5x+20/(2!)x^2-60/(3!)x^3+120/(4!)x^4-120/(5!)x^5=`

`=1-5x+10x^2-10x^3+5x^4-x^5`

which is the same result the Binomial Theorem gives us.

In fact, you can use this type of Maclaurin expansion to derive the Binomial Theorem itself. Have a look here if you're interested in seeing that:
https://bringholm.com/docs/Deriving%20the%20Binomial%20Theorem%20using%20Maclaurin%20expansions.pdf