I stumbled upon the equation lim n -> infinity (1+(1/n))^n= e. My textbook says that the larger n is, the closer it gets to e. In calculating interest rates, n is replaced by amount of times compounded in one year. Anyone know why the irrational number e is so special? In that (1 + (1 divided by any large number)) an all to the power of the same very large number brings you close to e? This is utterly nonsensical to me.

 

When this is applied to continuous compounding, where the compound amount A for a deposit of P dollars at an interest rate r per year compounded continuously for t years is given by A = Pe^rt. Does anyone know how did they derive the formula A = Pe^rt given the detail above?

 

Also, do you think using the calculator too much can affect your thinking? In that you think less, so you are not as smart? Is it worth the trade off for a faster calculation? Also what do you think about teaching math where the you plug in numbers and teaching math where the teacher teaches how an equation is formulated?

Edited November 2, 201114 yr by Vay

I stumbled upon the equation lim n -> infinity (1+(1/n))^n= e. My textbook says that the larger n is, the closer it gets to e. In calculating interest rates, n is replaced by amount of times compounded in one year. Anyone know why the irrational number e is so special? In that (1 + (1 divided by any large number)) an all to the power of the same very large number brings you close to e? This is utterly nonsensical to me.

 

When this is applied to continuous compounding, where the compound amount A for a deposit of P dollars at an interest rate r per year compounded continuously for t years is given by A = Pe^rt. Does anyone know how did they derive the formula A = Pe^rt given the detail above?

 

Also, do you think using the calculator too much can affect your thinking? In that you think less, so you are not as smart? Is it worth the trade off for a faster calculation? Also what do you think about teaching math where the you plug in numbers and teaching math where the teacher teaches how an equation is formulated?

 

 

What you mentioned above concerning [math] lim_{n \to \infty} (1 + \frac{1}{n})^{n} = e[/math] is the definition of [math] e[/math] . But if you'd like to know how, you should consider the sequence [math](1 + \frac{1}{n})^{n}[/math] and see if it converges, and which it does. Now, as to why [math] e[/math] is so special is because it is built in the definition of compounding.

 

[math]A = P(1 + \frac{j_{m}}{m})^{tm}[/math] where [math] j_{m}[/math] is nominal annual rate, let t = 1.

 

As [math] lim_{m \to \infty} (1 + \frac{j_{m}}{m})^{m} = e^{j_{m}}[/math].

 

[math] A = Pe^{rt}[/math] where [math] r[/math] is the continuous compounding that occurs.

 

Which can be related [math]e^r = (1 + \frac{j_{m}}{m})^{m} [/math].

 

Hope this helps.

What you mentioned above concerning [math] lim_{n \to \infty} (1 + \frac{1}{n})^{n} = e[/math] is the definition of [math] e[/math] .

 

You can, of course define e in any fashion that results in the number that is universally recognized as e.

 

However the usual definition comes in terms of the exponential function which is defined by the power series

 

[math] exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}[/math] and [math] e = exp(1)[/math]

You can, of course define e in any fashion that results in the number that is universally recognized as e.

 

However the usual definition comes in terms of the exponential function which is defined by the power series

 

[math] exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}[/math] and [math] e = exp(1)[/math]

 

Thanks!