Taylor Polynomials, find a good upper bound limit to error?

[匿名]

2017-06-15 6:49 am

I vaguely remember the question, but it followed along the lines of:

Q: Given 3cos(0.1), give a good approximation for what the upper bound limit is for the second Taylor Polynomial?

f(x) = 3cos(x)
f'(x) = 3 * -sin(x)
f''(x) = 3 * -cos(x)
So x is about 0, so a = 0.

If the approximation is remainder, R(x)
Do I find R(x) such that:
R(x) = f(x) - f(a) - f'(a)(x-a) - f''(a)(x-a)^(2) / 2! ?

回答 (1)

Lopez

2017-06-16 3:54 pm

✔ 最佳答案

f(x) = f(a) + f ' (a)( x - a ) + f '' (a)( x - a )²/2! + ......
Take f(x) = 3*cos x , x = 0.1 , a = 0 , then

3*cos 0.1
= f( 0.1 )
≒ f(a) + f ' (a)( x - a ) + f '' (a)( x - a )²/2!
= f(0)+ f ' (0)*0.1 + f '' (0)*0.1²/2!
= 3 + 0 + (-3)*0.01/2
= 2.985

The Remainder Estimation Theorem
If there are positive constants M and r such that
｜f^(n+1) (t)｜≦ M * r^(n+1) , for all t between a and x, then
｜Rn(x)｜≦ M * r^(n+1) * ｜x - a｜^(n+1) / (n+1)!

For this question,
f(x) = 3*cos x
f ' (x) = - 3*sin x
f '' (x) = - 3*cos x
f(³) (x) = 3*sin x
f(⁴) (x) = 3*cos x
.........................

Since ｜sin x｜and ｜cos x｜are bounded by 1,
｜f^(n+1) (t)｜≦ 3 , for all t
So M = 3 , r = 1
｜R₂(0.1)｜≦ 3 * 0.1^3 / 3! = 0.0005

So we conclude that
3*cos 0.1 = 2.985 + R , where｜R｜≦ 0.0005
That is, an upper bound for error is 0.0005

👍 1 👎 0