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1018
 one year ago
find the taylor series generated by f=x^3 2x+1 at a=3
1018
 one year ago
find the taylor series generated by f=x^3 2x+1 at a=3

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jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Are 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). :)

1018
 one year ago
Best ResponseYou've already chosen the best response.0i am only familiar with the one with polynomial order? like of 4 etc...

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3OK, and what does that one look like? Could you type part of it?

1018
 one year ago
Best ResponseYou've already chosen the best response.0f(a) + f(a)(xa)... is that enough? the next one is with a factorial denominator

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3So the equation we are using is \[f(a) + f'(a)(xa) + \frac{f''(a)}{2!}(xa)^2 + \frac{f'''(a)}{3!}(xa)^3\] 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?

1018
 one year ago
Best ResponseYou've already chosen the best response.0oh, so that's how i determine how long the series would be. ok thanks

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Yes, 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)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Alright, 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!

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3So, can you find f' (the first derivative)?

1018
 one year ago
Best ResponseYou've already chosen the best response.0ok hey i think i got it from here. haha. thanks a lot!

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3No worries! Good luck!

1018
 one year ago
Best ResponseYou've already chosen the best response.0i just follow the formula right?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Yep, 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.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.3Good 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). :)

1018
 one year ago
Best ResponseYou've already chosen the best response.0hey i just saw this reply. haha. thanks! yeah, im starting to enjoy math more and more as i go further. thanks again!

1018
 one year ago
Best ResponseYou've already chosen the best response.0thanks also for this nice info haha
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