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

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Homework Help Overview

The discussion revolves around finding the Taylor series expansion for the function f(x) = (1 + x)^0.5 and its derivatives, as well as related functions g(x) = (1 + x^3)^0.5 and h(x) defined by its derivative. Participants explore the derivation of the first four nonzero terms for these functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to derive the Taylor series for f(x) and g(x) by substituting x^3 into the series for f(x). There is discussion about the derivatives of g(x) and the implications of evaluating them at x=0. Some participants express confusion about the nonzero terms in the derivatives and question the application of the product rule versus the power rule.

Discussion Status

There is ongoing exploration of the Taylor series and its derivatives, with some participants providing guidance on using known series expansions to simplify the process. Multiple interpretations of the derivatives are being discussed, and there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants note the challenge of finding nonzero terms in the derivatives of g(x) at x=0 and the potential for confusion arising from the application of differentiation rules. The homework context emphasizes deriving series expansions without providing complete solutions.

rjs123
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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...
 
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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.
 
ojs said:
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.
 
rjs123 said:
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.
 
Char. Limit said:
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
 
Last edited:
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.
 
Char. Limit said:
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
 
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.
 
Char. Limit said:
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?
 
  • #10
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.
 
  • #11
Char. Limit said:
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
 
  • #12
Looks good.
 
  • #13
Char. Limit said:
Looks good.

thanks for your help...appreciate it.
 

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