Imum Coeli
Junior Member
- Joined
- Dec 3, 2012
- Messages
- 86
QUESTION:
1) Find the second order polynomial for x^(1/2) at a = 1/9
2) Use the answer to find an estimate for sqrt(1/10)
3) Use Taylor's theorem to find an upper bound on the error of your estimate
NOTES:
1) I don't think caused me any trouble. Answer is P_2(x) = 1/6 + 3/2*x - 27/8*(x-1/9)^2
2) P_2(1/10) = 1/6 + 3/20 + 1/2400 = .31625
3) This is the part I am having difficulty with:
So...
By Taylor's theorem
R_2(x) = f^(3)(z)/(3)! * (x-1/9)^3 for some z between x and 1/9, where f^(3) is the 3rd derivative of f(x) = sqrt(x)
Now as f^(3)(x)= 3/8 * x^-(5/2)
abs(R_2(x)) = abs(2/8 * z^-(5/2) * (x-1/9)^3/3!
From this point I start to get confused... I'm not too sure how I'm supposed to pick the interval for z.
I think that I want the maximum possible value of z^-(5/2) , x <= z <= 1/9
so I pick z = 1/9 which gives
z = 243
Thus
abs(R_2(1/10)) = abs(3/8 * 243 * -1/4374000) = 2.084 * 10^-5
which seems like a reasonable answer.
Never-the-less, I'm not too confident about my z interval.
Is it true that in the above context that x <= z <= 1/9 is really just 1/10<= z <= 1/9, and if that is the case, wouldn't it be better to pick z = 1/10?
Thanks.
1) Find the second order polynomial for x^(1/2) at a = 1/9
2) Use the answer to find an estimate for sqrt(1/10)
3) Use Taylor's theorem to find an upper bound on the error of your estimate
NOTES:
1) I don't think caused me any trouble. Answer is P_2(x) = 1/6 + 3/2*x - 27/8*(x-1/9)^2
2) P_2(1/10) = 1/6 + 3/20 + 1/2400 = .31625
3) This is the part I am having difficulty with:
So...
By Taylor's theorem
R_2(x) = f^(3)(z)/(3)! * (x-1/9)^3 for some z between x and 1/9, where f^(3) is the 3rd derivative of f(x) = sqrt(x)
Now as f^(3)(x)= 3/8 * x^-(5/2)
abs(R_2(x)) = abs(2/8 * z^-(5/2) * (x-1/9)^3/3!
From this point I start to get confused... I'm not too sure how I'm supposed to pick the interval for z.
I think that I want the maximum possible value of z^-(5/2) , x <= z <= 1/9
so I pick z = 1/9 which gives
z = 243
Thus
abs(R_2(1/10)) = abs(3/8 * 243 * -1/4374000) = 2.084 * 10^-5
which seems like a reasonable answer.
Never-the-less, I'm not too confident about my z interval.
Is it true that in the above context that x <= z <= 1/9 is really just 1/10<= z <= 1/9, and if that is the case, wouldn't it be better to pick z = 1/10?
Thanks.
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