hay guyz

[I Took the Red Pill]

Use Taylor's Theorem to obtain an upper bound for the error of the approximation. Then calculate the exact value of the error.

arctan(0.4) ≈ 0.4 - (0.4)³/3

So it's screwing with my head.

[Flying Mullet]

I'm assuming that you're using a Taylor Polynomial of degree 3:
x - x³/3

To find the upper bound, or remainder, use the following formula:
( max | f(n+1)(z) | * (x – x0) n+1 ) / (n + 1)!

For you, f(x) is arctan(x), n is 3 (your Taylor Polynomial is up to x³) and x0 is 0, making your equation look like this:
( max | f(4)(z) | * (x)^4 ) / (4)!

You'll need to find the fourth derivative of arctan and then find the maximum value for the fourth derivative of arctan over the interval [0, 0.4].

The fourth derivative of arctan is x(6x^2+9)/(1-x^2)^(7/2). The max value on [0, 0.4] is 7.334.

So your final formula is:
((7.334)x^4) / (4)!

Substitute 0.4 in for x and you get:
((7.334)(0.4)^4) / (4)! = 0.007823

The forumla for the exact remainder is:
| f(x) - Pn(x)| or
| arctan(x) - x - x^3/3 |

So subtiitute 0.4 in for x again and you get:
| arctan(0.4) - (0.4) 0 (0.4)^3/3 | = 0.001833

Thus 0.007823 is the upper bound of the error and 0.001833 is the exact remainder.




You wouldn't happen to be doing problem #48 in section 9.7 of <a href="http://www.amazon.com/Calculus-Single-Variable-Transcendental-Functions/dp/0618606254/ref=sr_11_1?ie=UTF8&qid=1195139466&sr=11-1">this book</a>, would you?

[Kirobaito]

It amazes me that a year and a half ago I would have been able to do this problem. It's not hard at all. I remember knowing how to do Taylor polynomials. But it's all gone. I wouldn't have the first clue what to do.

[xXsarahXx]

what in the name of god is that?!?!

[I Took the Red Pill]

I'm assuming that you're using a Taylor Polynomial of degree 3:
x - x³/3

To find the upper bound, or remainder, use the following formula:
( max | f(n+1)(z) | * (x – x0) n+1 ) / (n + 1)!

For you, f(x) is arctan(x), n is 3 (your Taylor Polynomial is up to x³) and x0 is 0, making your equation look like this:
( max | f(4)(z) | * (x)^4 ) / (4)!

You'll need to find the fourth derivative of arctan and then find the maximum value for the fourth derivative of arctan over the interval [0, 0.4].

The fourth derivative of arctan is x(6x^2+9)/(1-x^2)^(7/2). The max value on [0, 0.4] is 7.334.

So your final formula is:
((7.334)x^4) / (4)!

Substitute 0.4 in for x and you get:
((7.334)(0.4)^4) / (4)! = 0.007823

The forumla for the exact remainder is:
| f(x) - Pn(x)| or
| arctan(x) - x - x^3/3 |

So subtiitute 0.4 in for x again and you get:
| arctan(0.4) - (0.4) 0 (0.4)^3/3 | = 0.001833

Thus 0.007823 is the upper bound of the error and 0.001833 is the exact remainder.




You wouldn't happen to be doing problem #48 in section 9.7 of <a href="http://www.amazon.com/Calculus-Single-Variable-Transcendental-Functions/dp/0618606254/ref=sr_11_1?ie=UTF8&qid=1195139466&sr=11-1">this book</a>, would you?

Thanks a bunch :)

And yeah, it's #47 in 9.7, but it's not Early Transcendental Functions, it's just Calculus of A Single Variable, Eighth Edition. But it's the same number and section, so I don't know how different they are.

[Flying Mullet]

If you run into any more problems, check out this site:
Calc Chat Free Solutions (http://calcchat.tdlc.com/free_solutions/main.html)

It's the online companion to the textbook and they worked out the solutions to the odd number problems there.

Hopefully if you're scratching your head on a problem it will help you.