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# Maclaurin Series

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1) Show that f(x) = Sqrt(1+x) is 1+ (x/2) using the Maclaurin Series.

2) Show |an+1/an|, lim (n->infinity) |an+1/a| and the radius of convergence for (sum) (-1)n(n+1)x2n-1/n (times)3n.

Sep 30, 2014

#1
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Maclaurin series of f(x) says:

f(x) = f(0) + f`(0)x + f``(0)x2/2! + f```(0)x3/3! + ...

If

f(x) = (1 + x)1/2 then  f(0) = 1

f`(x) = (1/2)(1+x)-1/2 and f`(0) = 1/2

f``(x) = -(1/4)(1+x)-3/2 and f``(0) = -1/4

... etc.

So f(x) = 1 + x/2 - x2/8 + ...

To first order this is f(x) ≈ 1 + x/2

Sep 30, 2014

#1
+10

Maclaurin series of f(x) says:

f(x) = f(0) + f`(0)x + f``(0)x2/2! + f```(0)x3/3! + ...

If

f(x) = (1 + x)1/2 then  f(0) = 1

f`(x) = (1/2)(1+x)-1/2 and f`(0) = 1/2

f``(x) = -(1/4)(1+x)-3/2 and f``(0) = -1/4

... etc.

So f(x) = 1 + x/2 - x2/8 + ...

To first order this is f(x) ≈ 1 + x/2

Alan Sep 30, 2014
#2
+5

I did not understand. What we have done to solve it is set f(x) d/dx = d/dx(ao+a1x+a2x2+a3x3+a4x4+....) Such that all terms with x become zero and non x terms become the values, then taking the derivative of both sides again to find the next term and so on. So i am confused by the above post. Thanks.

Oct 1, 2014
#3
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Is it the notation I've used that is confusing?

f(x) = (1 + x)1/2 is the square root of (1+x), so putting x = 0, f(x)at x = 0 = 1

df(x)/dx = (1/2)(1 + x)-1/2 so putting x = 0, df(x)/dxat x=0 = 1/2

d( df(x)/dx )/dx = -(1/4)(1 + x)-3/2 so putting x = 0,  d( df(x)/dx )/dxat x = 0 = -1/4

and so on.

Putting these back into the Maclaurin series f(x) = 1 + x/2 - x2/8 ...

Notice that f(x) is not exactly 1 + x/2.  This is just a first-order approximation.

Oct 1, 2014
#4
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thank you. did it today and made more sense now. cheers.

Oct 2, 2014