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

2) Show |a_{n+1}/a_{n}|, lim (n->infinity) |a_{n+1}/a| and the radius of convergence for (sum) (-1)^{n}(n+1)x^{2n-1}/n (times)3^{n.}

Stu
Sep 30, 2014

#1**+10 **

Maclaurin series of f(x) says:

f(x) = f(0) + f`(0)x + f``(0)x^{2}/2! + f```(0)x^{3}/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 - x^{2}/8 + ...

To first order this is f(x) ≈ 1 + x/2

Alan
Sep 30, 2014

#1**+10 **

Best Answer

Maclaurin series of f(x) says:

f(x) = f(0) + f`(0)x + f``(0)x^{2}/2! + f```(0)x^{3}/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 - x^{2}/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(a_{o}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+....) 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.

Stu
Oct 1, 2014

#3**0 **

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)/dx_{at x=0} = 1/2

d( df(x)/dx )/dx = -(1/4)(1 + x)^{-3/2 }so putting x = 0, d( df(x)/dx )/dx_{at x = 0} = -1/4

and so on.

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

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

Alan
Oct 1, 2014