Find the accumulated value of an investment of $15,000 for 5 years at an interest rate of 7% if the money is compounded continuously.

Guest Feb 16, 2015

#6**+10 **

Why does Pe^(rt) work??

Look at

lim n →∞ (1 + r/n)^n

Let m = n/r So we have

lim m →∞ ( 1 + 1/m)^(r*m) =

lim m →∞ [(1 + 1/m)^m ] ^r but ... lim m →∞ (1 + 1/m)^m = e

So we have

e^r

And

Pe^[r(1)] = P(e^r) = would represent continuous compounding for one year

So....

Pe^(rt) =

P(e^r)(e^r)(e^r)........... where (e^r) is multiplied "t" times ....{representing "t" years (or periods)......}

And that's it.......!!!

CPhill Feb 16, 2015

#1**+10 **

Let there be n divisions of time in 1 year.

formula after 1 year 15000*(1+7/n)^n

after 5 years 15000*(1+7/n)^(n*5)

n approaches infinity:

lim n->infinity [15000*(1+7/n)^(n*5)

=15000*e^7*5

=15000*e^35

Guest Feb 16, 2015

#3**+10 **

Reference https://www.physicsforums.com/threads/how-was-the-number-e-2-718-originated.525005/

Reference https://www.physicsforums.com/threads/how-was-the-number-e-2-718-originated.525005/

Reference https://www.physicsforums.com/threads/how-was-the-number-e-2-718-originated.525005/

The proof is this, Melody.....it's in the way that "e" is defined...

e = lim as n →∞ (1 + 1/n)^n

Notice, that Anonymous is letting n, the number of compounding periods per year, grow larger and larger......if we theoretically make "n" "huge," then the number of compoundings per year ≈ "continuous" because the time intervals of compounding → 0........thus, we have an "infinite" number of compoundings per year of extremely short duration each....!!!!

Reference https://www.physicsforums.com/threads/how-was-the-number-e-2-718-originated.525005/

Reference https://www.physicsforums.com/threads/how-was-the-number-e-2-718-originated.525005/

Reference https://www.physicsforums.com/threads/how-was-the-number-e-2-718-originated.525005/

CPhill Feb 16, 2015

#4**+10 **

This graph doesn't constitute a proof, but simply suggests that it's likely to be true (though you should have put 0.07 rather than 7 Melody!):

.

Alan Feb 16, 2015

#5**0 **

It wasn't me Alan, I didn't put anything but I also noticed that error, I suppose I should have said so :))

Melody Feb 16, 2015

#6**+10 **

Best Answer

Why does Pe^(rt) work??

Look at

lim n →∞ (1 + r/n)^n

Let m = n/r So we have

lim m →∞ ( 1 + 1/m)^(r*m) =

lim m →∞ [(1 + 1/m)^m ] ^r but ... lim m →∞ (1 + 1/m)^m = e

So we have

e^r

And

Pe^[r(1)] = P(e^r) = would represent continuous compounding for one year

So....

Pe^(rt) =

P(e^r)(e^r)(e^r)........... where (e^r) is multiplied "t" times ....{representing "t" years (or periods)......}

And that's it.......!!!

CPhill Feb 16, 2015

#8**+5 **

Thank you Chris and Alan,

I have finally looked properly at what you are telling me, and hopefully I will remember.

I have always had problems with this because I could never see why it was true and I have a reall problem reproducing things when I don't really understand.

I did not know that e could be defined that way

I will try and remember.

$$\boxed{e=\displaystyle \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n}$$

.Melody Feb 19, 2015