mooshupork34
Junior Member
- Joined
- Oct 29, 2006
- Messages
- 72
Okay, so I estimated the sq. root of 7.9 with a Taylor polynomial of degree three.
I based the polynomial at 8.
First I calculated the 1st-3rd derivatives of x^(1/2):
g'(x) = 1/2(x)^(-1/2)
g''(x) = -1/4(x)^(-3/2)
g'''(x) = 3/8(x)^(-5/2)
Then there's g(x), which is just the sq. root of 8.
When plugging in 8 for x, you get for the Taylor polynomial of degree three:
T3(x) = sqrt[8] + 1/2(8)^-1/2 (x-8) + 1/2 * -1/4(8)^(-3/2) (x-8)^2 + 1/6 * 3/8(8)^(-5/2) (x-8)^3
Plugging in 7.9 for x in the polynomial above, you get 2.810693867, which is not far off from what you get when you enter the square root of 7.9 into a calculcator. When you enter it into the calculator, you get 2.810693865.
And now for the question that is confusing me...
Apply Taylor's theorem to find an upper bound on the error in this approximation.
If someone could explain how this is done, I would REALLY appreciate it. Thanks!
I based the polynomial at 8.
First I calculated the 1st-3rd derivatives of x^(1/2):
g'(x) = 1/2(x)^(-1/2)
g''(x) = -1/4(x)^(-3/2)
g'''(x) = 3/8(x)^(-5/2)
Then there's g(x), which is just the sq. root of 8.
When plugging in 8 for x, you get for the Taylor polynomial of degree three:
T3(x) = sqrt[8] + 1/2(8)^-1/2 (x-8) + 1/2 * -1/4(8)^(-3/2) (x-8)^2 + 1/6 * 3/8(8)^(-5/2) (x-8)^3
Plugging in 7.9 for x in the polynomial above, you get 2.810693867, which is not far off from what you get when you enter the square root of 7.9 into a calculcator. When you enter it into the calculator, you get 2.810693865.
And now for the question that is confusing me...
Apply Taylor's theorem to find an upper bound on the error in this approximation.
If someone could explain how this is done, I would REALLY appreciate it. Thanks!