Radius of convergence for (1+x)^1/2

[asif zaidi]

Problem Statement:
Compute the Taylor Series for (1+x)^1/2 and find the radius of convergence


Problem Solution:

The Taylor Series expansion I get is

T(x) = 1 + (0.5*x) - (0.25*x^2)/2 + (0.375*x^3)/3! - (0.9375*x^4)/4! +...-...

So to get radius of convergence I have to find a closed solution form of the above equation and I simply am not able to come up with anything. Any pointers on this will be appreciated.

But then I was looking at the equation and saw that at least I can represent the x and the factorial portions as Sum (from 0-inf) x^i/i! and not worry about the coefficient.
If I use ratio test on this closed form, I will get the interval of convergence to be -inf to +inf. Therefore radius of convergence is +inf

Am I correct - if not please advise

Thanks

Asif

 

[HallsofIvy]

?? A closed form for the series? It is (1+ x)^1/2 of course!

I have no idea how you get -inf to inf as interval of convergence. You wrote the coefficients as "0.5", (0.25", "0.375", etc. which makes me wonder if you are sure of the correct values.
The nth derivative of (1+x)^-1/2, at 0, is (2n-3)!/((n-2)!2^n-2) so the nth term in the sum is (2n-3)!xⁿ/((n-2)!n!2^n-2). Applying the ratio test to that does not give infinity as radius of convergence.

But you can use the "closed form", which, as I said, was exactly what you were given: (1+ x)^-1/2. In general the Taylor's series of a function will converge as long as the function is "nice". Where is there a "problem" with (1+ x)^-1/2? How far is that value of x from x=0? That is the radius of convergence.

 

[asif zaidi]

I am a bit confused with the response so I will give more detail as to what I have done

First part is to find Taylor Series for (1+x)$$^{1/2}$$

BTW, in your response you have given (1+x)$$^{-1/2}$$ so I think the 1st difference is there.

Anyway for Taylor series I do the following.
Find derivatives

f'(x) = 1/2 (1+x)$$^{-1/2}$$
f''(x) = (1/2)(-1/2)(1+x)$$^{-3/2}$$
f'''(x) = (1/2)(-1/2)(-3/2)(1+x)$$^{-5/2}$$
f''''(x) = (1/2)(-1/2)(-3/2)(-5/2)(1+x)$$^{-7/2}$$
f'''''(x) = (1/2)(-1/2)(-3/2)(-5/2)(-7/2)(1+x)$$^{-9/2}$$
f''''''(x) = (1/2)(-1/2)(-3/2)(-5/2)(-7/2)(-9/2)(1+x)$$^{-11/2}$$

Simplifing above gives and evaluating at 0 gives

f'(x) = 1/2(1+x)$$^{1/2}$$ = 1/2
f''(x) = -1/4(1+x)$$^{-3/2}$$ = -1/4
f'''(x) = 3/8(1+x)$$^{-5/2}$$ = 3/8
f''''(x) = -15/16(1+x)$$^{-7/2}$$ = -15/16
f'''''(x) = 105/32(1+x)$$^{-9/2}$$ = 105/32
f''''''(x) = -945/64(1+x)$$^{-11/2}$$ = -945/64

Thus Taylor Series:

T(x) = 1 + x/2 -x$$^{2}$$/2!4 + 3x$$^{3}$$/3!8 -15x$$^4}$$/4!16 + 105x$$^{5}$$/5132 - 945x$$^{6}$$/6!64 +...

Next Step is to find an expression using $$\sum$$ (from 0-inf). And this is what I meant by closed sum. My terminology is probably wrong according to what you say. However, whenever I have done exercises to find radius/interval of convergence, I have always had an expr of $$\sum$$. I then evaluate a(n+1)/a(n) and see it is < |1| and find interval of convergence

Any advice on how I can find the right expression and then radius of convergence.

Note: in your case if you give me (1+x)$$^{-1/2}$$ then the radius of convergence will be 1. But I don't think this this applies to (1+x)$$^{1/2}$$ case because this function is 'nice' everywhere.


Thanks

Asif

 

[HallsofIvy]

Now set x= 0:
f'(0): 1/2
f''(0): (1/2)(-1/2)
f'''(0): (1/2)(-1/2)(-3/2)
f''''(0): (1/2)(-1/2)(-3/2)(-5/2)
f'''''(0): (1/2)(-1/2)(-3/2)(-5/2)(-7/2)
f''''''(0): (1/2)(-1/2)(-3/2)(-5/2)(-7/2)(-9/2)

The sign is alternating: it's (-1)ⁿ. The denominator is a power of 2: it's 2ⁿ The numerator is 1(3)(5)(7)(9)... That would be an exponential except that it is "missing" the even integers. Okay, put them in: 1(2)(3)(4)(5)(6)(7)(8)(9)/(2)(4)(6)(8) where I have also, of course, put those same numbers in the denominator- I have multiplied numerator and denominator by the same thing in order to keep the fraction the same thing. The numerator is now obviously 9!. What about the denominator? Well, since those are all even numbers, we can factor a 2 out of each: (2)(4)(6)(8)= (2)(1)(2)(2)(2)(3)(2)(4)= 2^{4(1)(2)(3)(4)= 24(4!). The numerator in f''''''(0) is 9!/244!. It should not be difficult to see that the general term, for f(n)(0) is (2n+1)!/2nn! Since the denominator was already 2n and the sign was alternating, f(n)(0)= (2n+1)!/[(2n(n!)2] and so the nth term in the Taylor's series is (2n+1)!/[(2n(n!)3]xn}

 

[asif zaidi]

Now set x= 0:
The numerator is now obviously 9!. What about the denominator? Well, since those are all even numbers, we can factor a 2 out of each: (2)(4)(6)(8)= (2)(1)(2)(2)(2)(3)(2)(4)= 2^{4(1)(2)(3)(4)= 24(4!). The numerator in f''''''(0) is 9!/244!. It should not be difficult to see that the general term, for f(n)(0) is (2n+1)!/2nn! Since the denominator was already 2n and the sign was alternating, f(n)(0)= (2n+1)!/[(2n(n!)2] and so the nth term in the Taylor's series is (2n+1)!/[(2n(n!)3]xn}

<sup>OK - I see what you are trying to do by getting 9! etc... But if I use the formula you give, it does not match
You have given f(n)(0)= (2n+1)!/[(2n(n!)2]
So if I calculate it for n=6, I get 13!/(12*6!2 = 187.6875.
However, using Taylor Series expansion, I get 14.76

However, using info you provided I was able to come up with the following form

1 + 0.5x + Sum (2-inf) [ (2n-3)! / n!(n-2)! 22n-2 ].
I did this for a few n's and got the same coefficient as Taylor Series.

But the problem is to find the radius of convergence for this, how do I take into account the 1st 2 terms (1+0.5x).

Thanks

Asif</sup>