Question

how to generate e using Ae^T=A(1+I/n)^nT

Answer 1

how to generate exponential e, i have found that Ae^T = A(1+I/n)^nT

Answer 2

Ae^T is the formula for continuous compounding, in which growth occurs every moment, while A(1 + 1/n)^nT is the formula for compounding in specific intervals, such as compounding once every quarter year.

To make the two formulae equal, you need to set the limit of the second formula as n approaches infinity.

Answer 3

is there a way to find exponential e

Answer 4

how do i set the limit of the other equation to n approaches infinity

Answer 5

Can you clarify your question a bit? Do you want to find the decimal approximation for 'e' through this method ?

Answer 6

"investigate other mathematical techniques or methods to generate exponential e based on that An = A(1+I/n)^nT and An = Ae^T

Answer 7

i need a process to generate an approximation of e as the formula An=Ae^T has limitations

Answer 8

Okay. Did you by chance miss the r in both the equations?

Answer 9

what is r

Answer 10

this is the equation i got

Answer 11

okay

Answer 12

so first elminiate A:
e^T = (1 + 1/n)^nT

then T

e = (1 + 1/n)^n

Answer 13

can T be eliminated like that

Answer 14

yes because we have raised both sides to the power of 1/T, which means now e^T/T = e^1 and the same with the right side

Answer 15

ok thanks for the help i understand now

Answer 16

but it's not complete yet.. we didn't take the limit

Answer 17

oh can u finish then plz sorry about that

Answer 18

We have this equation:
\[e = \left(1 - \frac{1}{n}\right)^n\]
were n represents the number of compoundings done in a year

since the left side, that is, Ae^T represented continuous compounding, it means that there are infinite compoundings each year. Thus take the limit as n approaches infinity, and you have

\[e = \lim_{n \rightarrow \infty} \left( 1 - \frac{1}{n}\right)^n\]

This is the formula for 'e'

Answer 19

Hope that helps!

Answer 20

thank you very much