Is this result related to the Taylor series?

Question

We have, $$f(b)-f(a)=\lim_{n\rightarrow \infty}\sum_{k=0}^{n-1} hf'(a+kh)\:\:\:\:\:..(1)$$

where $h=\frac{b-a}{n}$

Now, $$f'(a)=f'(a)$$

$$f'(a+h)=f'(a)+hf^2(a)$$ $f^2(a)$ meaning the second derivative at $a$

$$f'(a+2h)=f'(a+h)+hf^2(a+h)=f'(a)+hf^2(a)+h(f^2(a)+hf^3(a))$$ $$=f'(a)+2hf^2(a)+h^2f^3(a)$$

Similarly, I found $f'(a+3h)$ to be $$f'(a)+3hf^2(a)+3h^2f^3(a)+h^3f^4(a)$$

Observing the pattern,

$$f'(a+kh)=\sum_{r=0}^k \binom{k}{r} h^rf^{r+1}(a)$$

So, the expansion of $\sum_{k=0}^{n-1} f'(a+kh)$ will contain terms of the form $c_rh^r f^{r+1}(a)$ with $r$ ranging from $0$ to $n-1$ and the coefficient $c_r$ of $h^r f^{r+1}(a)$ being $\sum_{k=0}^{n-1}\binom{k}{r}$

Therefore, $$f(b)-f(a)=\lim_{n\rightarrow \infty}h\sum_{r=0}^{n-1} \left(h^{r}f^{r+1}(a)\sum_{k=0}^{n-1} \binom{k}{r}\right)$$

$$=\lim_{n\rightarrow \infty}\sum_{r=0}^{n-1} \left(\frac{b-a}{n}\right)^{r+1}f^{r+1}(a)\binom{n}{r+1}$$

Compare this with the Taylor series result where

$$f(b)-f(a)=\lim_{n\rightarrow \infty} \sum_{r=0}^n (b-a)^{r+1} f^{r+1}(a)\frac{1}{(r+1)!}$$

Is the result similar or equivalent to the Taylor series? Also, is there some formula in terms of $n$ and $r$ for $\sum_{k=0}^{n-1} \binom{k}{r}$?

EDIT:

I think

$$\lim_{n\rightarrow \infty}\sum_{r=0}^{n-1} \left(\frac{b-a}{n}\right)^{r+1}f^{r+1}(a)\binom{n}{r+1}$$

$$=\lim_{n\rightarrow \infty}\sum_{r=0}^{n-1} \left(\frac{b-a}{n}\right)^{r+1}f^{r+1}(a)\frac{n!}{(n-(r+1))!(r+1)!}$$ might be equivalent to

$$\lim_{n\rightarrow \infty} \sum_{r=0}^n (b-a)^{r+1} f^{r+1}(a)\frac{1}{(r+1)!}$$

because of the limit. Is this true?

EDIT: Would this post be more rigorous if I removed all the limits and used $\omega$ in place of $n$? $\omega$ is the infinity in hyperreal numbers. $\frac{1}{\omega}$ is the infinitesimal. Arithmetic with infinity and infinitesimals is allowed in hyper-reals. That way it'd be non-standard but pretty rigorous, I think? I was using limits just to make it look as much mainstream as possible.

@GuyFsone I just googled it. Yes, the formula at the top is also called Riemann sum. I modified it in the post to contain all the derivatives at $a$ instead of all the first derivatives at all the points in the interval $(a,b)$. — Ryder Rude, Jan 23 '18 at 09:59
Your equation $f'(a+h) =f'(a) +hf^2(a)$ is incorrect and so are the next set of equations. — Paramanand Singh, Jan 23 '18 at 10:32
The identity $(1)$ is valid for any Riemann integrable $f'$ and goes by the name Fundamental Theorem of Calculus. — Paramanand Singh, Jan 23 '18 at 10:33
@ParamanandSingh Yes, I know about the fundamental theorem of calculus. Just wasn't familiar with the 'Riemann Integrals' term. — Ryder Rude, Jan 23 '18 at 10:37
@ParamanandSingh We are working with $\lim_{h\rightarrow 0}$ in this context, so there's no error term in those equations. — , Jan 24 '18 at 01:26
@dove: I understand that. The problem is "not writing those error terms" and more importantly "not performing an analysis of those terms". It's like a world where people pay in cash and don't worry about the remaining loose change which is due. — Paramanand Singh, Jan 24 '18 at 02:36
@ParamanandSingh I think the equations are valid at least in non-standard analysis where $h$ is a number as close to zero as possible. — Ryder Rude, Jan 24 '18 at 04:33
You have to then use the proper terminology of non-standard analysis and use things like standard part function and transfer principle and non-standard analysis is nothing more than avoiding inequalities involving $\epsilon-\delta$. It does not change the substance of a typical analysis proof in any other manner. — Paramanand Singh, Jan 24 '18 at 06:09
@ParamanandSingh, if you were more knowledgeable about Robinson's framework for infinitesimal analysis, I would be more comfortable with your sweeping claims concerning it. — Mikhail Katz, Jan 24 '18 at 14:25
@MikhailKatz: those claims are primarily based on wiki articles and your own answer (https://math.stackexchange.com/a/1836900/72031) to my question. — Paramanand Singh, Jan 24 '18 at 15:03
@ParamanandSingh, I have just re-read my answer that you linked above but did not find anything even remotely resembling your sweeping claim to the effect that " non-standard analysis is nothing more than avoiding inequalities involving ϵ−δ". Relying on wiki is hazardous. If I were you I would actually learn something about this before expostulating about it; you could start with Keisler's book. — Mikhail Katz, Jan 24 '18 at 15:06
@MikhailKatz : your first two examples in the linked answer indicate that they provide an alternative to limiting processes and reduce the number of quantifiers. This is also given in wiki. Further I have commented on your answer that these things only add convenience and not generality and you don't seem to have any objections to that. So as far as the use of nsa in elementary calculus is concerned my claim is not far from reality. — Paramanand Singh, Jan 24 '18 at 15:19
Both of your claims are incorrect. Robinson's framework is a powerful research tool that has led to many new results. Similarly, in teaching infinitesimal calculus it is not merely a matter of convenience but makes calculus accessible to a vastly larger student audience; see e.g., this 2017 publication in Journal of Humanistic Mathematics. — Mikhail Katz, Jan 24 '18 at 15:31
@MikhailKatz: I can't comment about research level math and I am not involved directly in teaching math. But I do know a bit about studying math. Learning the construction of reals combined with learning $\epsilon, \delta$ definitions is far simpler than learning the construction of hyperreals from reals. The convenience offered by nsa and thereby wider accessibility is by hiding all the difficult aspects into the construction of heavy machinery of hyperreals. So if you really want all the details, nsa is significantly harder to understand than the usual standard route. — Paramanand Singh, Jan 24 '18 at 15:58
@ParamanandSingh, neither the construction of the reals nor the construction of the hyperreals is explained in a typical freshman calculus course, so this point is moot. You are confusing procedure and ontology; this point is explained in one of my recent articles; if you are interested in becoming unconfused I can indicate a reference. — Mikhail Katz, Jan 24 '18 at 16:02

score 1 · Answer 1 · answered Jan 23 '18 at 10:27

1

$\sum_{k=0}^{n-1} \binom{k}{r}=\binom {n}{r+1} $.

Note that $\binom {k}{r}:=0$ if $k<r$

answered Jan 23 '18 at 10:27

Lucio Tanzini

1,425

what is the $:=$ symbol? – Ryder Rude Jan 23 '18 at 10:29
But thanks for the formula. I'll edit the post. – Ryder Rude Jan 23 '18 at 10:31
@RyderRude $:=$ means "define to be" or "let it equal", as opposed to just the equal sign $=$ which just means "assert to be equal" or perhaps "equal for some value of the unknown". I believe it comes from some programming languages, where one distinguishes between assignment operators and boolean equality checking operators – ziggurism Jan 23 '18 at 14:59

Ryder Rude · Answer 2 · 2018-01-24T12:13:35.927

-1

This was my own question. I have figured it out.

$$\lim_{n\rightarrow \infty} \left(\frac{b-a}{n}\right)^{r+1}f^{r+1}(a)\frac{n!}{(n-(r+1))!(r+1)!}$$

$$(b-a)^{r+1} f^{r+1}(a) \frac{1}{(r+1)!} \cdot \lim_{n\rightarrow \infty} \frac{n\cdot(n-1)\cdot(n-2)....\cdot(n-r)}{n^{r+1}}$$

The limit $\lim_{n\rightarrow \infty} \frac{n\cdot(n-1)\cdot(n-2)....\cdot(n-r)}{n^{r+1}}$ evaluates to $1$.

So, this post is just another proof of Taylor Series, I guess. But this proof doesn't even use the power rule and is so different that any other proof I've ever seen.

edited Jan 24 '18 at 12:13

answered Jan 23 '18 at 13:26

Ryder Rude

1,531

@LucioTanzini: This is not a proof at all and this is probably the reason it hasn't been seen before. By definition a proof is rigorous. And there is another serious problem. There is no clear mention of either the hypotheses or the conclusion of the thing supposed to be proved. – Paramanand Singh Jan 24 '18 at 02:36
I think you should should learn the proper proofs of both the Fundamental Theorem of Calculus and Taylor Theorem. Both these theorems are ultimately dependent on the more basic result called Mean Value Theorem and that is the link between these results. You may later fix this proof by adding rigor and the downvote can be removed. – Paramanand Singh Jan 24 '18 at 02:46
@ParamanandSingh I've seen people use the term 'non-rigorous proof'. Maybe this isn't a proof in the strict sense of the word. – Jan 24 '18 at 05:05
2

@Dove: a "non-rigorous proof" is just an euphemism for "wrong proof". Btw an "informal proof" is something which is valid/rigorous but possibly does not use formal terminology and symbols. – Paramanand Singh Jan 24 '18 at 06:05
@ParamanandSingh Since the meaning of rigor isn't 'correct', so non-rigorous does not mean wrong. Mathematical proofs have varying degrees of rigor. Some proofs are perfectly rigorous while other proofs rely more on intuition. This proof is of the second kind. So, this is a proof but with less degree of rigor. – Jan 24 '18 at 07:09
@Dove: IMHO one should not try to make things grey in mathematics (like in other fields of human life). Either a proof is rigorous or it is non-rigorous. The varying degrees of rigor is a result of the fact that throughout history mathematical knowledge has improved and so some historical proofs were considered non-rigorous by modern authors. But to contrast, Euclid's proof of infinitude of primes was as rigorous in ancient times as it is today and so are the proofs in his Elements. The same goes for Conics by Apollonius. Cont'd. – Paramanand Singh Jan 24 '18 at 07:36
@Dove: the tradition of non-rigorous proofs became commonplace with the discovery of calculus, because there was no sound theory for it when these ideas were developed. Sadly the trend continues till this date (thanks to crappy textbook authors) even though the theoretical details have been sorted out with full rigor. – Paramanand Singh Jan 24 '18 at 07:40
@Dove : Also you will not find such style of intuitive proofs in other branches of mathematics like algebra, trigonometry, geometry or number theory. – Paramanand Singh Jan 24 '18 at 07:51
1

@ParamanandSingh Would this be rigorous if I removed all the limits and used $\omega$ in place of $n$? $\omega$ is the infinity in hyperreal numbers. $h=\frac{1}{\omega}$ is the infinitesimal. Arithmetic with infinity and infinitesimals is allowed in hyper-reals. That way it'd be non-standard but pretty rigorous, I think? I was using limits just to make it look as much mainstream as possible. – Ryder Rude Jan 24 '18 at 08:04
I can't comment on that because my knowledge of non-standard analysis is almost zero. – Paramanand Singh Jan 24 '18 at 08:34

Is this result related to the Taylor series?

2 Answers2