I have a situation where the tempo (beats/minute) can be increasing or decreasing linearly as a function of beats. So B/M = a+bB and I know a and b. I need to find the total length in minutes it would take to reach a specified tempo. My math is very rusty but I think this would be a geometric progression. For any given tempo the number of minutes would be M = B / (a+bB). Thus the length in minutes to that point would be: 1/(a+b) + 2/(a+2b) +...+ n/(a+nb). Can a simple formula be derived for this sum?
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.
Music Math Help-Geometric Progression?
- Thread starter danika
- Start date
The closed form depends on the values of a and b. Are there specific values for a and b you have to work with?.
But, in general:
The series \(\displaystyle \sum_{k=1}^{n}\frac{k}{a+kb}\) has no nice closed form. It is not geometric.
A general closed form in terms of a and b can be found using what is known as the Psi Function, but it is rather complicated.
Just for kicks, I ran it through Maple and got:
\(\displaystyle \frac{n}{b}-\frac{a\cdot \Psi\left(n+1+\frac{a}{b}\right)}{b^{2}}+\frac{a\cdot \Psi\left(1+\frac{a}{b}\right)}{b^{2}}\)
A numerical value would depend on the values of a and b.
To answer your question, "can a SIMPLE formula be derived for this sum?". I am sorry, but no. At least, not without knowing what a and b are.
But, in general:
The series \(\displaystyle \sum_{k=1}^{n}\frac{k}{a+kb}\) has no nice closed form. It is not geometric.
A general closed form in terms of a and b can be found using what is known as the Psi Function, but it is rather complicated.
Just for kicks, I ran it through Maple and got:
\(\displaystyle \frac{n}{b}-\frac{a\cdot \Psi\left(n+1+\frac{a}{b}\right)}{b^{2}}+\frac{a\cdot \Psi\left(1+\frac{a}{b}\right)}{b^{2}}\)
A numerical value would depend on the values of a and b.
To answer your question, "can a SIMPLE formula be derived for this sum?". I am sorry, but no. At least, not without knowing what a and b are.
J
JeffM
Guest
This sort of problem can be solved to a high degree of precision using a spread sheet and a simple iterated formula.
Is this answer overly abbreviated, or can you take it from here?
Is this answer overly abbreviated, or can you take it from here?
JeffM said:This sort of problem can be solved to a high degree of precision using a spread sheet and a simple iterated formula.
Is this answer overly abbreviated, or can you take it from here?
I was thinking along the same line, but I couldn't find any formula in Excel that I thought would apply. Any suggestions?
Type a known values for a and b into, say, cells A1 and A2.
Let a=A1=1 and b=A2=2 for instance.
Then, type the formula into another cell. Say into B1.
=K1/($A$1+K1*$A$2)
Drag a list of consecutive numbers down from, say, K1 to K50
Then, drag the formula down to B50.
At the bottom in B51, type =SUM(B1:B50)
You have your sum.
There are other ways to go about it. Jeff may have an easier way.
Many calculators have summation as well. Look for the \(\displaystyle \sum\) symbol
Let a=A1=1 and b=A2=2 for instance.
Then, type the formula into another cell. Say into B1.
=K1/($A$1+K1*$A$2)
Drag a list of consecutive numbers down from, say, K1 to K50
Then, drag the formula down to B50.
At the bottom in B51, type =SUM(B1:B50)
You have your sum.
There are other ways to go about it. Jeff may have an easier way.
Many calculators have summation as well. Look for the \(\displaystyle \sum\) symbol
J
JeffM
Guest
No I doubt there are any canned formulas in excel to do this. I meant that you could write the formula that you specified in your original post.
Galactus has a perfectly decent way to implement what I was suggesting, provided you fix the a and b in your formula as $A$1 and $A$2.
What might work a little better is to calculate a running sum so that you can increase the precision by decreasing grain size and increasing iterations without rewriting. I do not have time right now to write that out in excel-speak, but if you do not understand what I mean, I'll be back in about 2 to 3 hours.
Galactus has a perfectly decent way to implement what I was suggesting, provided you fix the a and b in your formula as $A$1 and $A$2.
What might work a little better is to calculate a running sum so that you can increase the precision by decreasing grain size and increasing iterations without rewriting. I do not have time right now to write that out in excel-speak, but if you do not understand what I mean, I'll be back in about 2 to 3 hours.
galactus said:Type a known values for a and b into, say, cells A1 and A2.
Let a=A1=1 and b=A2=2 for instance.
Then, type the formula into another cell. Say into B1.
=K1/($A$1+K1*$A$2)
Drag a list of consecutive numbers down from, say, K1 to K50
Then, drag the formula down to B50.
At the bottom in B51, type =SUM(B1:B50)
You have your sum.
There are other ways to go about it. Jeff may have an easier way.
Many calculators have summation as well. Look for the \(\displaystyle \sum\) symbol
Thanks, but given the granularity I need, it would have to be a very large array. I'm working on a film cue calculator based on tempo and am trying to include a facility for a tempo ramp. So, for example, If you assume a max frame rate of 30 fps and a cue length of 2 minutes, that would be an array of 3600 cells.
J
JeffM
Guest
So what if the array is large?
The machine will do the work.
However it now may be good numeric practice to use a running sum to avoid round-off or truncation errors.
The machine will do the work.
However it now may be good numeric practice to use a running sum to avoid round-off or truncation errors.