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Find the Taylor series for the given function at the specific value x=a...

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wah24 | Student | eNoter

Posted August 16, 2012 at 6:13 PM via web

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Find the Taylor series for the given function at the specific value x=a

f(x)=1/(1-x)^2                 a=2

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 17, 2012 at 11:38 AM (Answer #1)

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You need to write the Taylor's seriesfor the given function, about `x = a = 2`  such that:

`f(x) = f(2) + f'(2)*x + (f''(2))/(2!)*x^2 + ....`

You need to evaluate `f(2)`  substituting 2 for x such that:

`f(2) = 1/(1-2)^2 => f(2) = 1/(-1)^2 => f(2) = 1/1 = 1`

You need to evaluate `f'(x)`  using the quotient rule such that:

`f'(x) = (1'*(1 - x)^2 - 1*((1 -x)^2)')/((1 - x)^4)`

`f'(x) = (-2(1 - x)*(1 - x)')/((1 - x)^4)`

Reducing by `1-x`  yields:

`f'(x) = (-2*(-1))/((1 - x)^3) => f'(x) = 2/((1 - x)^3)`

You need to evaluate `f'(2)`  such that:

`f'(2) = 2/((1 - 2)^3) => f'(2) = 2/((-1)^3) => f'(2) = -2`

You need to evaluate `f''(x)`  using the quotient rule such that:

`f''(x) = (2'*((1 - x)^3) - 2*((1 - x)^3)')/((1 - x)^6)`

Since 2' yields 0, the first product vanishes and it only remains `- 2*((1 - x)^3)'`  such that:

`f''(x) = (-2*3(1-x)^2*(1-x)')/((1 - x)^6)`

Reducing by `(1-x)^2`  yields:

`f''(x) = 2/((1 - x)^4)`

You need to evaluate `f''(2)`  such that:

`f''(2) = 2/((1 - 2)^4) = 2`

Hence, evaluating Taylor's series for the given function yields `f(x) = 1 - 2*x + 2/(2!)*x^2 + ...`

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lfryerda | High School Teacher | (Level 2) Educator

Posted August 30, 2012 at 2:25 AM (Answer #2)

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There is another way to find a Taylor series of a function like this about a different point.  If we look at the geometric series

`1/{1-x}=1+x+x^2+x^3+x^4+cdots`

we can differentiate it term-by-term to get

`1/{(1-x)^2}=1+2x+3x^2+4x^3+cdots=sum_{n=1}^infty nx^{n-1}`

which is the series we want, but unfortunately it's about the point x=0 instead of x=2.

So we go back to the original geometric series and rewrite it to the equivalent form

`1/{1-x}=1/{1-(x-2)+2}={-1}/{1+(x-2)}`

`=-1+(x-2)-(x-2)^2+(x-2)^3-cdots=sum_{n=0}^infty (-1)^{n+1}(x-2)^n`

which is the geometric series but now about the point x=2.

Again we differentiate term-by-term to get

`1/{(1-x)^2}=1/{(1+(x-2))^2}`

`=1-2(x-2)+3(x-2)^2-cdots=sum_{n=1}^infty (-1)^{n+1}n(x-2)^{n-1}`

which is the Taylor series about the point x=2.  Note that this is in the expect form about a point `sum_{n=0}^infty c_n(x-a)^n` instead of about x=0.

The Taylor series is `1-2(x-2)+3(x-2)^2-cdots=sum_{n=1}^infty(-1)^{n+1}n(x-2)^{n-1}` .

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