+0

# 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. 0 1808 8 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
+93038
+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
#2
+94203
0

I'd like to see a proof of this :))

Melody  Feb 16, 2015
#3
+93038
+10

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....!!!!

CPhill  Feb 16, 2015
#4
+27251
+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
+94203
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
+93038
+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
#7
+27251
+5

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

Apologies Melody, I should have looked more carefully!

Alan  Feb 16, 2015
#8
+94203
+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