We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
2039
8
avatar

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.

 Feb 16, 2015

Best Answer 

 #6
avatar+101405 
+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.......!!!

 

 Feb 16, 2015
 #1
avatar
+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

 Feb 16, 2015
 #2
avatar+101769 
0

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

 Feb 16, 2015
 #3
avatar+101405 
+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....!!!!

 



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/
 Feb 16, 2015
 #4
avatar+28029 
+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!):

 

continuous interest

.

 Feb 16, 2015
 #5
avatar+101769 
0

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

 Feb 16, 2015
 #6
avatar+101405 
+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
 #7
avatar+28029 
+5

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

 Melody "

 

Apologies Melody, I should have looked more carefully!

 Feb 16, 2015
 #8
avatar+101769 
+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.

 

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

.
 Feb 19, 2015

23 Online Users

avatar
avatar