What is this kind of number sequence called?

[jedidia]

So I was fooling around with some stuff, and I accidentally discovered a scaling rule when messing with one of my functions. This is not me wanting to toot my horn, I'm sure the rule is well understood for a couple hundred years, I just want to know if I can make my life easier and cut out some iteration.

The relation is this: Take the numbers from 1 to 10 (and beyond, but 1 to 10 works fine for an example) and map them in the following way:

2/1, 3/2, 4/3, 5/4, 6/5, 7/6, 8/7, 9/8, 10/9

This gives you the factors between step, i.e. the proportionality for how much magnitude is actually gained by each step. Put simply, the factor between each step can easily be expressed as x/(x-1).
If you'd multiply all factors in this sequence you'd get approximately 9.969
Now there's two questions:

• What is such a sequence called in mathematics (knowing that might enable me to conduct some further googling on my own, without that I'm pretty much lost)?
• Is there a way to calculate the product of all factors of any such sequence without iteration? (the product of factors of the sequence 1..10 is 9.969, the product of factors of the sequence x..y is ...?)

 

[Urwumpe]

You mean a \( \frac{n+1}{n} \) sequence? (Wait a few minutes, I consult the big Bronstein book from university)

No, no name found. Maybe this should be written as a chain of sums.

 

[BrianJ]

2/1, 3/2, 4/3, 5/4, 6/5, 7/6, 8/7, 9/8, 10/9

If you'd multiply all factors in this sequence you'd get approximately 9.969

• Is there a way to calculate the product of all factors of any such sequence without iteration? (the product of factors of the sequence 1..10 is 9.969, the product of factors of the sequence x..y is ...?)

Oh, I'm sorry to jump in here - I'm sure I have the "wrong end of the stick" ? - but its making my brain itch!

How did you get 9.969?
I think: (2/1) x (3/2) x (4/3) x (5/4) x (6/5) x (7/6) x (8/7) x (9/8) x (10/9) = 10
and in general for (n+1)/n, from x->y = (y+1)/x

Well, its late, I should be asleep, just ignore me!

 

[n72.75]

Superparticular sequence. Shows up a lot in music.

 

[jedidia]

Well, its late, I should be asleep, just ignore me!

Hmmm, no, you are right! Looks like I had quite a bit of rounding in there when I ran the thing. Like when you run 9*(1/9) you don't get 1, depending on where the result is truncated, but here it gets accumulated a couple of times over.
I guess calculating the product of factors becomes rather trivial then... ?

Superparticular sequence. Shows up a lot in music.

Huh... I have a solid background in music theory, but nowhere near that deep...

 

[Moach]

Various ways to integrate it, with 10 steps for a range of 0~10 it should be pretty much the same as summation, no?
I mean, assuming you're using the simple x += dx method for it and not some higher-order RK4 type deal, of course...

anyways - It doesn't seem to have a specific name, but apparently the answer is: 7.60208 (best approximation, Boole wins)

 

[RisingFury]

You asked for the product, to which the answer is pretty obvious:

Let's start with your sequence:

2/1 * 3/2 * 4/3 * ...

You can already see that 2's cancel out, because 2/1 * 3/2 has a 2 in the numerator and denominator. Same goes for 3's and 4's...
In this case, the only things that don't drop out are the /1 and 10/ at the end, leaving you with 10 / 1 = 10

If you instead start with some other term, like 4/3 * 5/4 * 6/5, then the answer is... first and last! = 6/3 = 2

Ok, you asked for the product, but the answer you gave gives a sum, so I'm assuming you actually wanted a sum.
First thing, an infinite series of this does not converge, so we're stuck with the finite series.

Let's take your series:
2/1 + 3/2 + 4/3 + 5/4 + 6/5 + ... and rewrite it just a little bit. I will express each term as 1 + r:
(1 + 1/1) + (1 + 1/2) + (1 + 1/3) + (1 + 1/4) + (1 + 1/5) + ...
n + 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... so we're getting warmer. This is called the harmonic series and... it's a bit of a mess from here on. There's no simple formula, but there is an approximation:

n + Euler's constant + ln[n]. Euler's constant = 0.5772

Where does the ln[n] come from? Simple, from integration! You know that Integral[1/x] from 1 to n = ln[n], so you know the series increases at the same rate as a logarithm.