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Post by mondo on Jun 24, 2023 3:10:47 GMT
This subject is a bit detached from the book but anyway I decided to post it here.
I did some interesting read about Taylor series derivation. Looks like it starts from the tangential approximation which is nothing else than a first derivative of a function: $f'(a) = \frac{f(x) - f(a)}{(x-a)}$. After rearranging the terms we are getting a first, crude approximation function $g(x) = f(a) + f'(a)(x-a)$. So far so good, all is clear. But then the author of the text I read says if the function is twice differentiable we can get an even better approximation by calculating a second derivative and here he comes with this: $\frac{f(x) - [f(a)+f'(a)(x-a)]}{(x-a)^2}$. At a first glance all seems to be clear, it looks like he calculates a second derivative but .. not quite. It looks like he did some trick here which allows him to fit it into a taylor polynomial I try to figure out what this trick is. The only explanation I can come up with at the moment is from the formula for a second derivative $f''(a) = \frac{f'(x) - f'(a)}{(x-a)}$ and if we plug in the expression for first derivative we will indeed get $(x-a)^2$ in the denominator. But what about the numerator?
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Post by Admin on Jun 24, 2023 17:58:44 GMT
Mondo
$\displaystyle\frac{f(x)-[f(a)+f'(a)(x-a)]}{(x-a)^2}=\frac{\frac{f(x)-f(a)}{x-a}-f'(a)}{x-a}=\frac{f'(x)-f'(a)}{x-a}=f''(x)$ as $x\rightarrow a$
Vasco
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Post by Admin on Jun 24, 2023 18:32:12 GMT
Mondo
Strictly speaking these equalities are only ultimately equal,(i.e. $x\rightarrow a$), so we should write $\asymp$ rather than $=$.
Or use limit notation.
Vasco
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Post by mondo on Jun 24, 2023 19:19:45 GMT
Mondo $\displaystyle\frac{f(x)-[f(a)+f'(a)(x-a)]}{(x-a)^2}=\frac{\frac{f(x)-f(a)}{x-a}-f'(a)}{x-a}=\frac{f'(x)-f'(a)}{x-a}=f''(x)$ Vasco Not exactly as in the preceding step it is defined as $f(x) \approx f(a) + f'(a)(x-a)$. Hence we can't say, as you did, $f(a) + f'(a)(x-a) = f'(a)$ |
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Post by Admin on Jun 24, 2023 21:40:50 GMT
Mondo
The correct maths is $\displaystyle f'(a) = \lim_{x\rightarrow a}\frac{f(x) - f(a)}{(x-a)}$
If $(x-a)$ is small then $f(x)\approx f(a)+(x-a)f'(a)$
which is a truncated form of the Taylor series,
Vasco
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Post by Admin on Jun 24, 2023 21:51:36 GMT
Mondo
I have updated my reply#1 above in red. It is now correct.
Vasco
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Post by mondo on Jun 25, 2023 4:12:50 GMT
Vasco, I agree with what you have said in all your responses. Nonetheless we still didn't fill the missing step between the derivative definition and the "mysterious" formula $\frac{f(x)-[f(a)+f'(a)(x-a)]}{(x-a)^2} = f''(a)$
I think the trick comes from the following fact which I overlooked at first "The function $f(x)$ and the approximating polynomial have the same value at $a$ and the same first derivative at $a$". Hence if that is true, we can indeed write $f(x)$ (as the approx polynomial) $- [{f(a) + f'(a)(x-a)}]$ (the aprrox. polynomial as $f'(a)$. When $a$ is an argument it simplifies to just $f(a)$ because $(a-a) = 0$. All of that divided by $(x-a)^2$ - and this part I can't explain. normally it should be just $(x-a)$ but they somehow know, probably base on a definition of a second derivative that the denominator shall be in a second power.
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Post by Admin on Jun 25, 2023 6:58:06 GMT
Mondo
It would be useful if you could post a photocopy of the text where you read this so that I can see it in context,
Vasco
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Post by Admin on Jun 25, 2023 7:23:35 GMT
Mondo
The derivative definition you quoted in your original post is incorrect.
Vasco
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Post by mondo on Jun 25, 2023 7:36:15 GMT
Mondo The derivative definition you quoted in your original post is incorrect. Vasco I looked again and it looks all fine for me, which part do you thing is wrong? Here are the two pages that we discuss attached. Attachments:taylor_doc.docx (292.18 KB)
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Post by Admin on Jun 25, 2023 10:30:07 GMT
Mondo
It's wrong because you've used an equals sign when it should be as in reply#4.
Vasco
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Post by Admin on Jun 25, 2023 14:29:22 GMT
Mondo
I've looked at the document you posted and it seems to me that in the expression (which you quoted in your first post about this)
$\displaystyle\frac{f(x)-\{f(a)+f'(a)(x-a)\}}{(x-a)^2}$,
the numerator is the Taylor series Remainder since it is the difference between $f(x)$ and the approximation to $f(x)$, namely $f(a)+f'(a)(x-a)$.
Vasco
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Post by Admin on Jun 25, 2023 19:23:53 GMT
Mondo
If you repeat the process over and over you will generate all the terms of the Taylor series. This method emphasises the approximation aspect of the Taylor series.
Vasco
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Post by mondo on Jun 25, 2023 19:33:10 GMT
Mondo If you repeat the process over and over you will generate all the terms of the Taylor series. This method emphasises the approximation aspect of the Taylor series. Vasco Yes, this is what I know, but the formula from the first post is still a bit mysterious or rather "guessed" instead of derived. I wish author added more information on how he got it instead of a bare "consider this expression". |
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Post by Admin on Jun 25, 2023 20:23:12 GMT
Mondo
I think I could produce a document which would make the formula more derived. Would you like me to do it or are you happy with things as they are?
Vasco
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