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1018

  • one year ago

find the taylor series generated by f=x^3 -2x+1 at a=3

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  1. jtvatsim
    • one year ago
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    Are you familiar with the formula for Taylor series? What have you seen in class or your book? (I ask since there are different versions depending on the class). :)

  2. 1018
    • one year ago
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    i am only familiar with the one with polynomial order? like of 4 etc...

  3. jtvatsim
    • one year ago
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    OK, and what does that one look like? Could you type part of it?

  4. 1018
    • one year ago
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    f(a) + f(a)(x-a)... is that enough? the next one is with a factorial denominator

  5. jtvatsim
    • one year ago
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    OK, that's fine.

  6. jtvatsim
    • one year ago
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    So the equation we are using is \[f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3\] we won't need any more terms since the polynomial you have been given only goes up to the third power. Does that make sense so far?

  7. 1018
    • one year ago
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    yes :)

  8. 1018
    • one year ago
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    oh, so that's how i determine how long the series would be. ok thanks

  9. 1018
    • one year ago
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    then? :)

  10. jtvatsim
    • one year ago
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    Yes, if you didn't realize that, you could keep going but your derivatives would be 0 (the fourth derivative of an x^3 function will be zero, the fifth derivative will be zero, and so on)

  11. jtvatsim
    • one year ago
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    Alright, so now we need: 1) To find up to the third derivative of f. That is, f', f'', and f'''. 2) We already know that a = 3. 3) Then we just plug in a = 3 into the formulas for f, f', f'', and f''' and we will be done!

  12. jtvatsim
    • one year ago
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    So, can you find f' (the first derivative)?

  13. 1018
    • one year ago
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    3x^2 -2 ?

  14. jtvatsim
    • one year ago
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    Excellent!

  15. 1018
    • one year ago
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    ok hey i think i got it from here. haha. thanks a lot!

  16. jtvatsim
    • one year ago
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    No worries! Good luck!

  17. 1018
    • one year ago
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    i just follow the formula right?

  18. 1018
    • one year ago
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    thanks again!

  19. jtvatsim
    • one year ago
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    Yep, so for example... for f'(a) you will take f'(3) which is 3(3)^2 - 2 = 27 - 2 = 25. and put that in the formula.

  20. jtvatsim
    • one year ago
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    Good luck! You are getting to the fun part of math, where we finally figure out how calculators are able to "know" what sine, cosine and other nasty equation values are... (it's all because of Taylor Series). :)

  21. 1018
    • one year ago
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    hey i just saw this reply. haha. thanks! yeah, im starting to enjoy math more and more as i go further. thanks again!

  22. 1018
    • one year ago
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    thanks also for this nice info haha

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