#### Mohammad Hammad

##### New member

- Joined
- Jul 11, 2019

- Messages
- 38

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Mohammad Hammad
- Start date

- Joined
- Jul 11, 2019

- Messages
- 38

- Joined
- Jun 18, 2007

- Messages
- 25,825

How did you arrive at the equality - please share your work/thoughts.

Have you learned Taylor series expansion?

- Joined
- Jul 11, 2019

- Messages
- 38

it came out with me in a long calculations but at the end I stucked on this level in the photo

- Joined
- Nov 12, 2017

- Messages
- 12,914

hi can I ask if this series right or wrong. I found it with my calculations thanks

View attachment 29222

Using a spreadsheet, it looks like it may be valid; for x=0 the RHS is a harmonic series, which in fact is divergent, so that isn't a problem. Of course, as written it is defined only for positive integer x. It's not hard to prove for x=1, and I see hints of how it may work in general.

But in order to be sure it's right, we need to see your derivation (or else do all the work of figuring it out for ourselves, unless one of us happens to have seen it).

- Joined
- Jul 11, 2019

- Messages
- 38

- Joined
- Jul 11, 2019

- Messages
- 38

I mean did you see an equation like this before ?!Using a spreadsheet, it looks like it may be valid; for x=0 the RHS is a harmonic series, which in fact is divergent, so that isn't a problem. Of course, as written it is defined only for positive integer x. It's not hard to prove for x=1, and I see hints of how it may work in general.

But in order to be sure it's right, we need to see your derivation (or else do all the work of figuring it out for ourselves, unless one of us happens to have seen it).

- Joined
- Jul 11, 2019

- Messages
- 38

it give youNote that 1/x does not exist for x = 0. But the RHS does exist for x = 0. What does this mean?

-Dan

1/0 = zeta(1)

- Joined
- Nov 12, 2017

- Messages
- 12,914

I know Taylor series but I didn't use it here.

it came from a long derivative equations to reach this but in the end it came from this one .

I mean did you see an equation like this before ?!

I'm not familiar with your series, but I don't spend much time with them. Someone else may recognize it.

What you write here is just a different way to state your series; can you show us your work to derive it?

- Joined
- Jul 11, 2019

- Messages
- 38

- Joined
- Jan 27, 2012

- Messages
- 7,770

How are you defining x! for non-integer x?

- Joined
- Jul 11, 2019

- Messages
- 38

- Joined
- Nov 12, 2017

- Messages
- 12,914

I don't understand what you mean by "the minus side of a series", or why you say an infinite series can't be written as an infinite series!I wrote how it supposed to be if we write it as an infinite summation.

we can't write this series as infinite series.

it's like this one in this photo , it's the minus side of this series and it's define for negative integers.

View attachment 29234

But I asked for your derivation, and this is not a derivation. If you're saying the series you asked about is derived from this one, show how, and also how you know this one is true. I don't think you're wrong, but you aren't communicating in a way that gives us something to discuss. We want to see what you called your "calculations".

- Joined
- Oct 29, 2019

- Messages
- 1,206

[math]\frac{1}{2}=\frac{0!}{(2-1)}-\frac{1!}{(2-1)\color{red}(2-2)\color{black}}+...[/math]

however the red expression means the second term isn't a number. But if we take the negative x, x=-2 then the new series appears to work

[math]\frac{1}{-2}=\frac{0!}{(-2-1)}-\frac{1!}{(-2-1)(-2-2)}+...≈-0.49997537 \text{ (adding the first 200 terms)}[/math]

- Joined
- Jul 11, 2019

- Messages
- 38

[math]\frac{1}{2}=\frac{0!}{(2-1)}-\frac{1!}{(2-1)\color{red}(2-2)\color{black}}+...[/math]

however the red expression means the second term isn't a number. But if we take the negative x, x=-2 then the new series appears to work

[math]\frac{1}{-2}=\frac{0!}{(-2-1)}-\frac{1!}{(-2-1)(-2-2)}+...≈-0.49997537 \text{ (adding the first 200 terms)}[/math]

exactly you are right

but minus side it mean the derivative way I made to reach this equation is different than the way I derived the equation for the positive integers.

both equations looked similar but in fact they have a different derivations

- Joined
- Jul 11, 2019

- Messages
- 38

it's a long story and need a lecture to explain it

- Joined
- Jul 11, 2019

- Messages
- 38

I asked how to check if this equation is right or wrong first to know if my theory is right or wrong

- Joined
- Nov 12, 2017

- Messages
- 12,914

theoryand I derived these equations from it.

it's a long story and need a lecture to explain it

The way to check if it is right is to derive it. So, unless one of us happens to be familiar with it, or find it somewhere, you are asking us to derive it ourselves. That may be a lot to ask. It's much more polite to ask people to check your work than to ask them to do a difficult problem themselves.I askedhow to checkif this equation is right or wrong first to know if mytheoryis right or wrong

It looks right to me, judging merely from examples; but I have not taken the time to try to derive it, because I expected you to respond in a more helpful way (or someone else to say it's well-known!). I may try if I have time. And it may turn out to be obvious once I try. I can't say.

But if your "theory" is something non-standard, and you are thinking that the fact that a valid result can be derived from it would prove that the theory itself is true, you are misusing logic. It is possible to derive a true result from a false premise.

- Joined
- Jul 11, 2019

- Messages
- 38

I am still working on it.

I didn't finish it yet.

I am a busy man with 2 jobs so I tried the short way to be sure as a first step.

now I will try to have more time to work on it.

you were very helpful people here and I appreciate it.

- Joined
- Jul 11, 2019

- Messages
- 38