Home > Find The > Find The Order Of The Error Term For This Approximation# Find The Order Of The Error Term For This Approximation

## So because we know that p prime of a is equal to f prime of a when we evaluate the error function, the derivative of the error function at "a" that

take the **second derivative, you're going to get** a zero. Generated Sat, 15 Oct 2016 18:08:45 GMT by s_wx1094 (squid/3.5.20) The question wasn't about what the central difference approximation for $f''(x)$ is in terms of $f$, it was what the order of approximation is in the expression $f(t_0+h) \approx f(t_0) + Sep 29 '10 at 11:42 In your formula for $f''(x)$, you've forgotten to divide the remainder term by $h$; it should be $O(h^2)$ instead of $O(h^3)$. check over here

And that's the whole point of where I'm trying to go with this video, and probably the next video We're going to bound it so we know how good of an ProofBig O Truncation ErrorBig O Truncation Error Exploration. Now, what is the n+1th derivative of an nth degree polynomial? Is there any job that can't be automated?

And then plus go to the third derivative of f at a times x minus a to the third power, (I think you see where this is going) over three factorial, Probably the most popular **approximation formula that in** general behaves better than the ordinary $(f(x+h)-f(x))/h$ is $(f(x+h)-f(x-h))/2h$. And not even if I'm just evaluating at "a". derivatives approximation share|cite|improve this question edited Apr 12 '13 at 8:05 Cortizol 2,4601031 asked Apr 12 '13 at 6:59 Guest 211 1 Look up second order central difference for the

- And we already said that these are going to be equal to each other up to the nth derivative when we evaluate them at "a".
- Take the 3rd derivative of y equal x squared.
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Remark: Suppose that a numerical differentiation procedure has error behaviour of the type that you are looking for. What **is a** type system? Let $f(x)=x$. Your cache administrator is webmaster.

If we solve for $f^{\prime\prime}(x)$ like so: $$f^{\prime\prime}(x)=\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2h}+O(h^2)$$ and substitute in the first expression, $$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h^2}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2h}+O(h^2)\right)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$ we can take the $O(h^2)$ within the parentheses out as an $O(h^4)$ term: $$f(x+h)=f(x)+h f^{\prime}(x)+\frac{h}{2}\left(\frac{f^{\prime}(x+h)-f^{\prime}(x-h)}{2}\right)+\frac{h^3}{3!}f^{\prime\prime\prime}(x)+O(h^4)$$ from where our approximation is centered. share|cite|improve this answer edited Jan 31 '12 at 5:08 answered Jan 31 '12 at 4:21 André Nicolas 418k31358699 In general, symmetric differences have better numerical properties than unsymmetric differences. In it, you'll get: The week's top questions and answers Important community announcements Questions that need answers see an example newsletter By subscribing, you agree to the privacy policy and terms

this one already disappeared, and you're literally just left with p prime of a will equal to f prime of a. numerical-methods share|cite|improve this question edited Sep **28 '10 at 21:01** asked Sep 28 '10 at 20:46 jjkparker 1134 It looks like $h$ is being used differently in your Taylor Please try the request again. It has error behaviour of the kind you want, with I think $1/6$ instead of $1/3$.

more hot questions question feed about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation Science The n+1th derivative of our nth degree polynomial. Digital Diversity Are RingCT signatures malleable? It will help us bound it eventually, so let me write that.

But what I want to do in this video is think about, if we can bound how good it's fitting this function as we move away from "a". check my blog If we do know some type of bound like this over here, so I'll take that up in the next video.Finding taylor seriesProof: Bounding the error or remainder of a taylor but it's also going to be useful when we start to try to bound this error function. Let's think about what happens when we take the (n+1)th derivative.

How? Solution **1. **So the error at "a" is equal to f of a minus p of a, and once again I won't write the sub n and sub a, you can just assume this content So if you measure the error at a, it would actually be zero, because the polynomial and the function are the same there.

Is there any job that can't be automated? This article does not cite any sources. assist.

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There is just one such formula, namely $$f'(x)={1\over2h}\bigl(4 f(x+h)-f(x+2h)-3 f(x)\bigr)\ .$$ To obtain an error estimate fix $x$ and consider the auxiliary function $$g(h):=4f(x+h)-f(x+2h)-3f(x)-2h f'(x)\ .$$ Then $$g'(h)=4f'(x+h)-2f'(x+2h)-2f'(x), \quad g''(h)=4f''(x+h)-4f''(x+2h)\ .$$ A Scenarios and Animations related to this module. Few simplifying assumptions are made, and when a number is needed, an answer with two or more significant figures ("the town has 3.9×103 or thirty nine hundred residents") is generally given. http://a1computer.org/find-the/find-the-error-in-this.php And so when you evaluate it at "a" all the terms with an x minus a disappear because you have an a minus a on them...

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