What is the Taylor series expansion for (1 + x)^.5 and its derivatives?

[rjs123]

Homework Statement

a. Find the first four nonzero terms in the Taylor series expansion about x = 0 for f(x) = (1+x)^.5

b. Use the results found in part (a) to find the first four nonzero terms in the Taylor series expansion about x= 0 for g(x) = (1 + x^3)^.5

c. Find the first four nonzero terms in the Taylor series expansion about x = 0 for the function h such that h'(x) = (1 + x^3)^.5 and h(0) = 4

The Attempt at a Solution

a. Pn(x) = 1 + 1/2x - x^2/(8) + x^3/(16)

b. stuck...

 

[ojs]

since the taylor expansion for g(x) and f(x) are both about 0 and get the same value when you insert 0 into it then you can just put x^3 into the taylor expansion for f(x) instead of x and get

1 + 1/2x^3 - (x^3)^2/8 + (x^3)^3/16

which is the solution to part b.

Part c is then to integrate the taylor expansion for g(x) and use the h(0) = 4 to find the integration constant.

 

[rjs123]

since the taylor expansion for g(x) and f(x) are both about 0 and get the same value when you insert 0 into it then you can just put x^3 into the taylor expansion for f(x) instead of x and get

1 + 1/2x^3 - (x^3)^2/8 + (x^3)^3/16

which is the solution to part b.

Part c is then to integrate the taylor expansion for g(x) and use the h(0) = 4 to find the integration constant.

this is what i was thinking...but when i try and find the 1st 4 nonzero terms for (1 + x^3)^.5 by finding the derivatives you end up with just 1...since every term after the first is going to equal 0 when x = 0.

 

[Char. Limit]

this is what i was thinking...but when i try and find the 1st 4 nonzero terms for (1 + x^3)^.5 by finding the derivatives you end up with just 1...since every term after the first is going to equal 0 when x = 0.

But that's just not true. For example, the third derivative is nonzero at x=0.

 

[rjs123]

But that's just not true. For example, the third derivative is nonzero at x=0.

f(x) (1+x^3)^.5...... x= 0 .... 1
f'(x) 3/2x^2(1+x^3)^-.5... x = 0 ... 0
f''(x) -9/4x^4(1+x^3)^-1.5... x= 0 ... 0
f'''(x) -81/8x^6(1+x^3)^-2.5... x= 0 ... 0

im probably doing something wrong here

 

[Char. Limit]

Your first and second derivatives aren't correct. Are you using the product rule?

EDIT: Sorry, your first is right. Your second derivative isn't right, though. Neither is your third.

 

[rjs123]

Your first and second derivatives aren't correct. Are you using the product rule?

EDIT: Sorry, your first is right. Your second derivative isn't right, though. Neither is your third.

ah i see...i was just using the power rule and completely forgot about the product rule here...for f''(x) it would be 3/2x^2(dx)*(1+x^3)^-.5 + 3/2x^2 * (1+x^3)^-.5 (dx) it kinda gets messy after the first derivative

 

[Char. Limit]

Yeah, it does. I recommend using the Taylor series for sqrt(1+x) and then just plugging in x^3 wherever there is an x. It will get you the first four terms a lot faster and easier.

 

[rjs123]

Yeah, it does. I recommend using the Taylor series for sqrt(1+x) and then just plugging in x^3 wherever there is an x. It will get you the first four terms a lot faster and easier.

b. 1 + 1/2x^3 -x^6/8 + x^9/16


c. integrate b... h(0) = 4

x + x^4/8 - x^7/56 + x^10/160

h(0) = 4

therefore the integral constant is 4 since all the terms equal 0 when 0 is plugged into them...correct?

 

[Char. Limit]

Yes, that's correct. If you want, I believe you can discard the x^10 term at this point, as you have 4 non-zero terms without it.

 

[rjs123]

Yes, that's correct. If you want, I believe you can discard the x^10 term at this point, as you have 4 non-zero terms without it.

the integral constant would be appended to the end like this...then discard the x^10

x + x^4/8 - x^7/56 + 4

 

[Char. Limit]

Looks good.

 

[rjs123]

Looks good.

thanks for your help...appreciate it.