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

Subhotosh Khan

Super Moderator
Staff member
How did you arrive at the equality - please share your work/thoughts.

Have you learned Taylor series expansion?

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

topsquark

Senior Member
Note that 1/x does not exist for x = 0. But the RHS does exist for x = 0. What does this mean?

-Dan

Dr.Peterson

Elite Member
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).

• topsquark

New member
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).
I mean did you see an equation like this before ?!

New member
Note that 1/x does not exist for x = 0. But the RHS does exist for x = 0. What does this mean?

-Dan
it give you
1/0 = zeta(1)

Dr.Peterson

Elite Member
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?

HallsofIvy

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

Dr.Peterson

Elite Member
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
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!

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".

Cubist

Senior Member
I think I know what OP means by "minus side of this series". If x=2 then the new series implies...
$\frac{1}{2}=\frac{0!}{(2-1)}-\frac{1!}{(2-1)\color{red}(2-2)\color{black}}+...$
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

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

• New member
the derivativ
I think I know what OP means by "minus side of this series". If x=2 then the new series implies...
$\frac{1}{2}=\frac{0!}{(2-1)}-\frac{1!}{(2-1)\color{red}(2-2)\color{black}}+...$
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

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

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

New member
I have a theory and I derived these equations from it.
it's a long story and need a lecture to explain it

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

Dr.Peterson

Elite Member
I have a theory and I derived these equations from it.
it's a long story and need a lecture to explain it
I asked how to check if this equation is right or wrong first to know if my theory is right or wrong
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.

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.