Mohammad Hammad
New member
- Joined
- Jul 11, 2019
- Messages
- 38
How did you arrive at the equality - please share your work/thoughts.
hi can I ask if this series right or wrong. I found it with my calculations thanks
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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).
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
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 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.
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I think I know what OP means by "minus side of this series". If x=2 then the new series implies...
[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]
I have a theory and 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 asked how to check if this equation is right or wrong first to know if my theory is right or wrong