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Taylor series expansions

  • Thread starter rjs123
  • Start date
  • #1
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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...
 

Answers and Replies

  • #2
ojs
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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.
 
  • #3
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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.
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.
 
  • #4
Char. Limit
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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.
 
  • #5
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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:
  • #6
Char. Limit
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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.
 
  • #7
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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
 
  • #8
Char. Limit
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1,204
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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.
 
  • #9
90
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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
Char. Limit
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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
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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
Char. Limit
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Looks good.
 
  • #13
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