An example need to be turned into an integral

[vemvare]

OK. We are producing something and then storing it. The first year, "1" is produced, the second year, "1,1" the third 1,1², so that the production capacity increases 10% per year. How do we convert this into a general formula for how much we have produced in total after year x?

It is obviously an integral, but all formulas I find for ∫ n^x give nonsense when I apply them to this example, so I think I'm wrong in turing the formula into n^x

How do I solve this?

 

[Mentallic]

What you're looking for is called a geometric sum.

$$S=1+r+r^2+...+r^{n-1}=\frac{r^n-1}{r-1}$$

r is the ratio, in your case 1.1, n is the amount of years you're counting.

 

[vemvare]

Ah, that gives reasonable numbers. So the formula resulting from the example is discrete? Is it the non-integer factor (1,1) that does that?

EDIT: No, I specified the increase as happening annually, if it isn't, if it is continous, the (r-1) is replaced with ln(r), which I found by experimenting, and it gives exactly the value my calculator does. Oddly I cannot seem to find any reference to this formula, though I'm most cenrtainly again screwing up annotations.

I need to refreshen my math-math, I've been in the world of molecules for too long...

 

[SteamKing]

If the increase in production was constantly changing, every minute of every day saw an incremental increase, then you might use an integral to find the increase after a certain period. The integral is the limit of a certain process which can often be crudely represented as a summation.

 

[CellCoree]

To solve this problem, we can use the concept of a geometric series. The production capacity is increasing at a rate of 10% per year, which can be represented by a common ratio of 1.1. Therefore, the amount produced in each year can be expressed as 1, 1.1, 1.1^2, 1.1^3, and so on.

To find the total amount produced after x years, we can use the formula for the sum of a geometric series:

Sx = a(1-r^x)/(1-r)

where Sx is the total amount produced after x years, a is the initial production of 1, and r is the common ratio of 1.1.

Substituting these values into the formula, we get:

Sx = 1(1-1.1^x)/(1-1.1)

Sx = 1(1-1.1^x)/(-0.1)

Sx = -10(1-1.1^x)

Therefore, the general formula for the total amount produced after x years is:

Sx = -10(1-1.1^x)

This can also be written as an integral:

Sx = ∫ 1.1^x dx

where the limits of integration are from 0 to x.

Thus, the integral represents the total amount produced after x years, and the general formula is -10(1-1.1^x).