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I Took the Red Pill
11-15-2007, 12:13 AM
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
11-15-2007, 03:11 PM
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
11-15-2007, 03:17 PM
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
11-15-2007, 03:37 PM
what in the name of god is that?!?!
I Took the Red Pill
11-15-2007, 07:55 PM
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
11-15-2007, 08:06 PM
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.
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