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

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

Mathematics
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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). :)
i am only familiar with the one with polynomial order? like of 4 etc...
OK, and what does that one look like? Could you type part of it?

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f(a) + f(a)(x-a)... is that enough? the next one is with a factorial denominator
OK, that's fine.
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?
yes :)
oh, so that's how i determine how long the series would be. ok thanks
then? :)
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)
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!
So, can you find f' (the first derivative)?
3x^2 -2 ?
Excellent!
ok hey i think i got it from here. haha. thanks a lot!
No worries! Good luck!
i just follow the formula right?
thanks again!
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.
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). :)
hey i just saw this reply. haha. thanks! yeah, im starting to enjoy math more and more as i go further. thanks again!
thanks also for this nice info haha

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