+1313 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.
+33615 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
+33615 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
+1313 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.
+33615 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.
