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Use a computer algebra system to find the Taylor polynomials $ T_n $ centered at $ a $ for $ n $ = 2, 3, 4, 5. Then Graph these polynomials and $ f $ on the same screen.

$ f (x) = \sqrt [3]{1 + x^2,} $ $ a = 0 $

$\sqrt[3]{1+x^{2}} \approx T_{5}(x)=1+\frac{1}{3} x^{2}-\frac{1}{9} x^{4}$

Calculus 2 / BC

Chapter 11

Infinite Sequences and Series

Section 11

Applications of Taylor Polynomials

Sequences

Series

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

01:11

In mathematics, integratio…

06:55

In grammar, determiners ar…

01:32

Use a computer algebra sys…

02:02

Find the Taylor polynomial…

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03:23

02:41

05:14

04:59

03:53

01:51

Graphing Taylor polynomial…

03:04

indescribable. Where us? To find Taylor's polynomial off foot cracks around zero for orders, off to tree for five. Then, whereas to use a computer system, I used Wolfram Alpha. You're free to use any so afraid that you'd like. And, uh, this is what to be got 48 order approximation off this function around zero using Taylor's pull No meals here. What we observe is does this first term is our zeroth order perks mission. Um, this is our second order approximation. Since we have the power to this is the fourth order approximation. This is the sixth order, and this whole thing is that the eighth order here? We're missing the old terms. If you notice we're gonna say that tea for sort of approximation is, um, equal to want, just like t off zero. We're going to say that second order approximation teaching FX is excreted or three plus one. We know that Fort order approximation is that we know that third order approximation will be go to second order approximation. Also, we can see that Ford order approximation will be equal to fifth order approximation, and that is equal to x 34th or negative off exit. Afford over nine plus X corridor three plus one. Where else has to graft that again? You could use an alleged function plotter since you know the function itself and the approximations to balance function. Here we see that, Bobby Yeah, t not so zero order approximation as just a straight line. And the idea here is that as we increase the number of terms, as you can see that we're capturing the exact behavior that is our original function off effects is behaving as you can see, not off the functions captured a behavior that this Ford order function is capturing. If he one of the capture the behavior more accurately and for a longer period of you'd need to include more terms and use high order derivatives.

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