Question #421de

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Answer 1 · 2018-01-04T11:38:17

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 · 2018-01-04T13:31:04

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

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