Error Bounds for derivative estimation

[renolovexoxo]

So we are estimated derivatives using a three point formula in class and the giving bounds for the error. I was given a table of x1=1.1, x2=1.2, x3=1.3, x4=1.4. I have to find an error bound to estimate the error for x=1.1 and 1.3 for f(x)=ex

In class we did x=1.1 as follows

error<=(h2/3)eg(x) for some g between (1.1, 1.3)
error<=(h2/3)e1.3
I understand all of this, except why it is between 1.1 and 1.3. I went to do the x3 error bound, and I wasn't sure what to choose for my upper bound. She added 2h to it, so would I use 1.5, even though it's not in my table?

 

[Ray Vickson]

So we are estimated derivatives using a three point formula in class and the giving bounds for the error. I was given a table of x1=1.1, x2=1.2, x3=1.3, x4=1.4. I have to find an error bound to estimate the error for x=1.1 and 1.3 for f(x)=ex

In class we did x=1.1 as follows

error<=(h2/3)eg(x) for some g between (1.1, 1.3)
error<=(h2/3)e1.3
I understand all of this, except why it is between 1.1 and 1.3. I went to do the x3 error bound, and I wasn't sure what to choose for my upper bound. She added 2h to it, so would I use 1.5, even though it's not in my table?

Do you mean e^x? You can type that either as e^x or use the "X²" button in the menu at the top of the input panel. Also: you write "error<=(h2/3)eg(x)". That could mean several things (and I truly do not know which you intend). It could mean:
(1) \frac{2}{3}h e g(x),\\<br /> (2) \frac{h^2}{3} e g(x),\\<br /> (3) \frac{2}{3} h e^{g x}.\\<br /> (4) \frac{h^2}{3} e^{g x}.<br />
In plain text you could write the first as (2/3)*h* e*g(x), the second as (h^2 /3)*e*g(x), the third as (2/3)*h*e^(g x) and the fourth as (h^2 /3) * e^(g x); or, you could use the "X²" button to get (2/3)*h*e*g(x), (h²/3)*e*g(x), (2/3)*h*e^gx and (h²/3)*e^gx.

RGV

 

[renolovexoxo]

I'm sorry, I meant e^x, I copied it from somewhere else and didn't catch the format change. I'm sorry about that!
I want to know (h^2)/3 * e^x, so the error was defined by h^2/3 * e^1.3

 

[Ray Vickson]

I'm sorry, I meant e^x, I copied it from somewhere else and didn't catch the format change. I'm sorry about that!
I want to know (h^2)/3 * e^x, so the error was defined by h^2/3 * e^1.3

The actual error will have the form (h^2)/3 * exp(y), for some y between 1.1 and 1.3, so--of course--the maximum possible error is obtained if we take the largest possible value of exp(y), which means taking the largest possible y; that is, y = 1.3. Is that what is bothering you?

RGV

 

[renolovexoxo]

No, It is when I move to evaluate it for x=1.3. I was given x values for 1.1, 1.2, 1.3, and 1.4. When I evaluate 1.3, what is the range of possible y values that I can consider? Does it stay between (1.1, 1.3) or do I move to include the points I used in my derivative estimation?

 

[Ray Vickson]

No, It is when I move to evaluate it for x=1.3. I was given x values for 1.1, 1.2, 1.3, and 1.4. When I evaluate 1.3, what is the range of possible y values that I can consider? Does it stay between (1.1, 1.3) or do I move to include the points I used in my derivative estimation?

It stays in the interval used in the derivative estimation. So, if you use a forward difference
f&#039;(x) \approx \frac{f(x+h) - f(x)}{h} you would have y in [x,x+h]. If you use a central difference
f&#039;(x) \approx \frac{f(x+h) - f(x-h)}{2h} you would have y in [x-h,x+h], etc. Didn't your textbook or course notes discuss this?

RGV