# From the Taylor Series

Discussion in 'Pseudoscience' started by Anamitra Palit, Dec 19, 2020.

Messages:
102

3. ### BellsStaff Member

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24,120
Mod Note

Please provide a summary of the issue instead of expecting people to open a file that may not be safe for them to open.

If you have links (ie a URL) to articles for example, that's one thing.

Sharing files like this is not advisable.

exchemist likes this.

5. ### Anamitra PalitRegistered Senior Member

Messages:
102
The link relates to the google drive:URL to an article saved in the google drive.What makes Bells feel it to be of a suspicious nature?Is pasting links [especially from the google drive]disallowed in his forum?Is there any latex facility for writing rich text directly in the editor? Pl do let me know of it?

The paper incidentally is of a mathematical nature to be understood from the mathematics in the paper. Nevertheless a basic indication has been orovied in the initial posting
"The Taylor series has been analyzed in the enclosed paper in several ways to reveal discrepancies. The analysis is of course of a mathematical nature. Requesting the attention of the audience to the mathematics of the paper and the issues ensuing from it...."
The warning itself is of an unjustified nature

7. ### DaveC426913Valued Senior Member

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16,720
Because Bells' head is more than just a hat rack.

Not explicitly, no.

But since my head is also not just a hat rack, there's no way I'm opening the document of a stranger just cold. Expereinced users will feel likewise.

Write your discussion here so it's on a (harmless) webpage. Refer to the (questionably harmless) document as-needed. That's how it's done.

It is justified. Perhaps you are just new.

Last edited: Dec 20, 2020
8. ### BellsStaff Member

Messages:
24,120
Mod Note

No one is going to click on a link that looks absolutely dodgy (for lack of a better term). Nor should they feel required to.
We don't know you. Google drive is one of the most popular target among hackers. We don't know if you have malware in those files. This isn't a file sharing website.

You aren't posting a link to a magazine or study that is published on a university site, as one example.

You are free to post the information from the files (say, type the words, sums, etc) and/or provide a summary and even post the mathematical equations, etc (this is catered for here).

All I am asking is that you do not ask people to open links to files you are sharing online because as I noted above, we do not know you. We cannot guarantee the safety of those files.

No one should be made to open questionable links or files on this site as part of any discussion.

9. ### Anamitra PalitRegistered Senior Member

Messages:
102
Considering information on the link provided by Staff member Bells:
. Google Drive does the job for you

As we mentioned, Google Drive is fairly secure. The service has its own antivirus system, where it scans every single uploaded file for malicious content.

The scan is performed on absolutely every type of file users upload, so there’s no way a certain file is skipped. But there’s a catch: Google performs the scan only on smaller files, more precisely on files smaller than 25MB.

If you’re uploading and sharing a larger file, you’re going to be warned that file wasn’t scanned, and it might contain some malicious content. Here’s what Google’s support page says:

“Google Drive scans a file for viruses before the file is downloaded or shared. If a virus is detected, users can’t share the file with others, send the infected file via email, or convert it to a Google Doc, Sheet, or Slide, and they’ll receive a warning if they attempt these operations. The owner can download the virus-infected file, but only after acknowledging the risk of doing so.

Only files smaller than 25 MB can be scanned for viruses. For larger files, a warning is displayed saying that the file can’t be scanned.”

The file enclosed by me is much, much less than 25MB

10. ### originHeading towards oblivionValued Senior Member

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11,574
So instead of abiding by the rules you agreed to when you signed up, you're going to argue about them. Great plan...

11. ### exchemistValued Senior Member

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11,204
He's having exactly the same experience on the .net site. Naturally.

12. ### Anamitra PalitRegistered Senior Member

Messages:
102
Derivation 1 from the paper has been provided in Latex:

We expand $f(x)$ in Taylor series $^{[1]}$ about three distinct points $x_{1}, x_{2}$ and $x_{3}$ with the same increment $h$
assuming convergence for all three cases
$$\begin{array}{l} f\left(x_{1}+h\right)=f\left(x_{1}\right)+\frac{h}{1 !} f^{\prime}\left(x_{1}\right)+\frac{h^{2}}{2 !} f^{\prime \prime}\left(x_{1}\right)+\frac{h^{3}}{3 !} f^{\prime \prime \prime}\left(x_{1}\right)+\cdots \ldots(1) \\ f\left(x_{2}+h\right)=f\left(x_{2}\right)+\frac{h}{1 !} f^{\prime}\left(x_{2}\right)+\frac{h^{2}}{2 !} f^{\prime \prime}\left(x_{2}\right)+\frac{h^{3}}{3 !} f^{\prime \prime \prime}\left(x_{2}\right)+\cdots \ldots(2) \\ f\left(x_{3}+h\right)=f\left(x_{3}\right)+\frac{h}{1 !} f^{\prime}\left(x_{3}\right)+\frac{h^{2}}{2 !} f^{\prime \prime}\left(x_{3}\right)+\frac{h^{3}}{3 !} f^{\prime \prime \prime}\left(x_{3}\right)+\cdots \ldots(3) \end{array}$$
The increment 'h" in the above three equations may be arbitrary to the extent it does not upset
convergence.
Adding the last three equations we have
\begin{array}{l} f\left(x_{1}+h\right)+f\left(x_{2}+h\right)+f\left(x_{3}+h\right) \\ \qquad \begin{aligned} &=f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)+\frac{h}{1 !}\left[f^{\prime}\left(x_{1}\right)+f^{\prime}\left(x_{2}\right)+f^{\prime}\left(x_{3}\right)\right] \\ &+\frac{h^{2}}{2 !}\left[f^{\prime \prime}\left(x_{1}\right)+f^{\prime \prime}\left(x_{2}\right)+f^{\prime \prime}\left(x_{3}\right)\right]+\frac{h^{3}}{3 !}\left[f^{\prime \prime \prime}\left(x_{1}\right)+f^{\prime \prime \prime}\left(x_{2}\right)+f^{\prime \prime \prime}\left(x_{3}\right)\right] \ldots(4) \end{aligned} \end{array}
Let $x$ satisfy the following equation
$$f(x)=f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)(5)$$
Equation (39) does not involve 'h'explicitly
$$\begin{array}{c} \Rightarrow f\left(x_{1}+h\right)+f\left(x_{2}+h\right)+f\left(x_{2}+h\right)=f(x)+\frac{h}{1 !} f^{\prime}(x)+\frac{h^{2}}{2 !} f^{\prime \prime}(x)+\frac{h^{3}}{3 !} f^{\prime \prime \prime}(x)+\cdots \ldots \\ f\left(x_{1}+h\right)+f\left(x_{2}+h\right)+f\left(x_{2}+h\right)=f(x+h)(6) \end{array}$$
But $h$ could be arbitrary. In fact we may very it continuously over a small interval without upsetting convergence. On top of that there are many functions for which the corresponding Taylor expansion is convergent for any arbitrary increment for example $\sin x, \cos x, \exp (x)$
For the same $x_{1}, x_{2}, x_{3}$ and corresponding $x$ as given by (5) we do have an infinite number of equations of the type (6) for the various 'h' that comply with the convergence issue. . But (6) does not represent an identity. We do have a situation of a gross violation.
It follows from the Taylor's expansion(from what we have discussed in the preceding part):
If
$$f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{2}\right)=f(x)$$
then
$$f\left(x_{1}+h\right)+f\left(x_{2}+h\right)+f\left(x_{2}+h\right)=f(x+h)$$
subject to the issue $f$ convergence that the Taylor expansion has to converge for the function $f$ for $x_{1}+$ $h, x_{2}+h$ and $x_{3}+h$
The above inference ensuing from Taylor series does not hold in general[with the exception of the exponential function: $\left.f(x)=e^{x}\right]$
If
$$\sin \left(x_{1}\right)+\sin \left(x_{2}\right)+\sin \left(x_{3}\right)=\sin (x)$$
-------------------(7)
By our choice the absolute value of the left side of the above should be less than or equal to unity
Then do we have the following for an arbitrary 'h', [arbitrary to the extent the absolute value of the left
side of the following should be less than or equal to unity]?
$$\sin \left(x_{1}+h\right)+\sin \left(x_{2}+h\right)+\sin \left(x_{3}+h\right)=\sin (x+h)$$
----(8)
Any arbitrary value of 'h' has to satisfy equation (8) subject to equation (7)

The Convergence Condition:
If we consider the Taylor expansion of $\sin (x+h)$ for
$$\frac{t_{n+1}}{t_{n}}=\frac{h^{2}}{(n+2)(n+1)} \frac{f^{(n+2)}(x)}{f^{n}(x)}$$
We apply D'Alembert Test considering the absolute value of each term in the expansion. If this series converges then the original series also converges.
$$\lim _{n \rightarrow \infty} \frac{\left|t_{n+1}\right|}{\left|t_{n}\right|}=\operatorname{lm}_{n \rightarrow \infty}\left|\frac{f^{(n+2)}(x)}{f^{n}(x)} \frac{h^{2}}{(n+2)(n+1)}\right|=\left|\frac{f^{(n+2)}(x)}{f^{n}(x)} h^{2}\right| \lim _{n \rightarrow \infty} \frac{1}{(n+2)(n+1)}=0$$
The modified series consisting of positive terms converges. Hence the Taylor expansion also converges for any arbitrary 'h'[considering finite $\frac{f^{(n+1)}(x)}{f^{n}(x)}$ for all $n$ and $n+1$ :choosing $x$ in that way and this can be achieved for the sin function]. Those terms for which $f^{n}(x)=0$ simply drop out of the series. We may apply D'Alembert ratio test to the remaining terms.
We may alternatively think of Cauchy condition: if $\lim _{n \rightarrow \infty} u_{n}^{1 / n}<1,$ then the series converges.
Against the Taylor series we form a series comprising the absolute values of the corresponding terms of the Taylor series.
$$\begin{array}{c} \frac{|x|^{n}}{n !}\left|f^{(n)}(x)\right| \\ \lim _{n \rightarrow \infty} \frac{|x|^{n}}{n !}=0 \Rightarrow \lim _{n \rightarrow \infty}\left(\frac{|x|^{n}}{n !}\right)^{1 / n}=0 \Rightarrow \lim _{n \rightarrow \infty} \frac{|x|}{(n !)^{1 / n}}=0 \end{array}$$
Indeed $\lim _{n \rightarrow \infty} \frac{|x|^{n}}{n !}=0$ implies that for every preassigned $\epsilon>0,$ no matter how small, we have $N>0$ such that for $n>N$ we have, $\frac{|x|^{n}}{n !}<\epsilon \Rightarrow \frac{|x|}{(n !)^{1 / n}}<\epsilon^{1 / n}=\epsilon^{\prime}$
For any arbitrary $\epsilon^{\prime}>0$ no matter how small we can arrange for an $\epsilon=\epsilon^{\prime 1 / n}$ so that $\frac{|x|^{n}}{n !}<\epsilon \Rightarrow$ $\frac{|x|}{(n !)^{1 / n}}<\epsilon^{1 / n}=\epsilon^{\prime}$
If $\left|f^{(n)}(x)\right|$ is bounded for all 'n', $\lim _{n \rightarrow \infty} \frac{|x|^{n}}{n !}\left|f^{(n)}(x)\right|=0 \Rightarrow \frac{|x|}{(n !)^{1 / n}}\left|f^{(n)}(x)\right|^{1 / n}=0$

Last edited: Dec 26, 2020