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• Write A Program To Compute The Exponential Function E^x Using The Infinite Series
• Write A Program To Compute Exponential Series 1
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I'm using the BigDecimal/BigInteger classes to work with really huge numbers.

I've got a formula for calculating a compound-growth series.

Write A Program To Compute The Exponential Function E^x Using The Infinite Series

For each n, the value = initial * (coef ^ n).

I'm trying to discover a fast way to calculate the sum of a subset of values between n0 and n1.

So for example where n0 = 4 and n1 = 6,

returns: initial * (coef ^ 4) + initial * (coef ^ 5) + initial * (coef ^ 6)

I don't know much maths, but maybe there is a formulaic way of expressing this?

I'm basically adding up all the values, clumping some of them into powers of 10 by raising the coefficient.

As far as I know the function is accurate. I can return a value for

n0 = 1, n1 = 50000, initial = 100, coef = 1.05 in under a second.

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Although I may never use the function for values higher than ~20,000, it would be nice to know if there are more effective approaches to this.

Next problem is to work out the largest value of n1 where the sum(n0, initial, coef) <= x.

EDIT:

(initial * coef ^ n0 - coef ^ n1 + 1) / 1 - coef

Thanks wikipedia.

Write A Program To Compute Exponential Series 1

1 Answer

I will write some algorithmic thoughts.

First of all lets simplify your formula:

So you should calculate: S = a * (c ^ n0) + a * (c ^ (n0+1)) +..+ a * (c ^ n1) where initial = a and coef = c

Let S(n) be a function of following sum:S(n) = a + a * c + a * (c^2) +..+ a * (c ^ n)

We will get S = S(n1)-S(n0-1)

In the other hand S(n) is a sum of a geometric progression, hence S(n)=a * (1-c^n)/(1-c).

So we will get S = S(n1)-S(n0-1)=a*(1-c^n1)/(1-c)-a*(1-c^(n0-1))/(1-c)=a*(c^(n0-1)-c^n1)/(1-c).

So now we have to deal with calculating c^n exponents (of course BigDecimal class has pow method, we are doing it just to be able to calculate complexity of algorithm). The following algorithm has O(log(n)) complexity:

So we can conclude that the sum can be calculated in O(log(n)) complexity if we taking into account the fact that algebraic operation has O(1) complexity.

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