How to calculate the constant e?

Question

I specifically need to know how e is calculate from taking the limit of (1+1/N)^N as N goes to infinity
it's been bugging me in calculus.... my teacher is too lazy to show me how to calculate it
i want to know how that limit turns into, and don't just tell me because it is
plz should me some simple steps so i get it is calculated

Answer 1

Ok..

Define the limit as L. So you have



lim n-> inf (1+1/n)^n = L



take the natural logarithm. Since n>0 and ln is monotonically increasing, then you get



ln (lim n->inf (1+1/n)^n) = lim n->inf ln((1+1/n)^n) = ln L



redfine x=1/n. so you now get



lim x->0 ln(1+x)/x = ln L



you're taking the limit of a ratio of functions of the form 1/0, so use L'Hospital's rule



lim x-> 0 (1/(1+x))/(1) = ln L



now you can evaluate and get



1= ln L



solve for L

L=e^1



L=e

Answer 2

1

Answer 3

well, as a practical matter that limit isn't used computationally, since it doesn't converge as fast as other expressions for e.



some approximations for a few values of N:



N = 1:



(1 + 1/1)^1 = 2



N = 2:



(1 + 1/2)^2 = 9/4 = 2.25



N = 10:



(1 + 1/10)^10 = (11/10)^10 = (25,937,424,601)/(10,000,000,000) = 2.5937424601



N = 100:



(1 + 1/100)^100 = (101/100)^100 = 2.7048138294215260932671947108075 (or so)



as you can see, the exponents get pretty big, making calculation difficult.



a better way to find a value for e, is using the taylor series for e^x centered at 0:



e^x = 1 + x + x^2/2! + x^3/3! + ....+x^n/n! +......



using x = 1, this becomes:



e = 1 + 1 + 1/2 + 1/6 + 1/24 +......+ 1/n! +.......



so for n = 1, we get the approximation 2



for n = 2, we get 1 + 1 + 1/2 = 2.5



for n = 3, we get 1 + 1 + 1/2 + 1/6 = 8/3 = 2.66666.....



for n = 4, we get 1 + 1 + 1/2 + 1/6 = 65/24 = 2.708333......



(it took N = 100, to get this close with the other limit).



for n = 5, we get 326/120 = 2.7166666......



for n = 6, we get 1957/720 = 2.71805555......



which is accurate to 3 decimal places.



you can carry out the terms as much as you have patience for,



but factorials get big very fast (so their reciprocals get small very fast).

Answer 4

I believe e = (1 + 1/n) ^n just as you said



So let's take 2

(1 + 1/2) ^ 2 = (1.5) • (1.5) = 2.25

for 3

(1+1/3) ^3 = 2.37037



for 4

(1+1/4) ^4 = 2.4414



for 5

(1+1/5) ^5 = 2.48832



for 6 (1+1/6) ^6 = 2.52162

for 7 (1+1/7) ^7 = 2.54649970

for 8 (1+1/8) ^8 = 2.56578

for 9 (1+1/9) ^9 = 2.58117

for 10 (1+1/10) ^10 = 2.59374



Okay it is S L O W L Y approaching 2.71828 so let's choose larger numbers

20       10 (1+1/20) ^20 = 2.65329

50       10 (1+1/50) ^50 = 2.69158

100       10 (1+1/100) ^100 = 2.704813829422

Well we have e calculated to 2 significant figures

1,000      10 (1+1/1,000) ^1,000 = 2.71692

10,000      10 (1+1/10,000) ^10,000 = 2.7181459

Well. at 10,000 we get 5 significant figures



100,000      10 (1+1/100,000) ^100,000 = 2.71826823



1,000,000      10 (1+1/1,000,000) ^1,000,000 = 2.718280469



Okay, at n = 1,000,000 we see that e gets calculated to a "gigantic" 6 significant figures.



Anyway, I hope this helps you out.

Answer 5

Graph it, thays the easiest way to understand it graph y= (1+ 1/x)^x then trace the graph out to large values of x and compare that with the value of e^1 . my AP calc teacher always told me when in doubt graph it, and i scored a 5 on the AP exam.

Answer 6

e = lim(1+ (1/n)) ^ n

yes you are correct



read the wikipedia, it explains it very well



http://en.wikipedia.org/wiki/E_%28mathematical_constant%29

Answer 7

e61664285d54cffd863b610d1e6398c52does61664285d54cffd863b610d1e6398c52't61664285d54cffd863b610d1e6398c52have61664285d54cffd863b610d1e6398c52a61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52"exact61664285d54cffd863b610d1e6398c52a61664285d54cffd863b610d1e6398c52swer".61664285d54cffd863b610d1e6398c52The61664285d54cffd863b610d1e6398c52de61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52t61664285d54cffd863b610d1e6398c52o61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52o61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52e61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52s61664285d54cffd863b610d1e6398c52l61664285d54cffd863b610d1e6398c52m61664285d54cffd863b610d1e6398c52[61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52]n > inf(161664285d54cffd863b610d1e6398c52+61664285d54cffd863b610d1e6398c521/n > inf)^n > inf,61664285d54cffd863b610d1e6398c52wh61664285d54cffd863b610d1e6398c52ch61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52s61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52ou61664285d54cffd863b610d1e6398c52d61664285d54cffd863b610d1e6398c52through61664285d54cffd863b610d1e6398c52calculus.61664285d54cffd863b610d1e6398c52I61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52you61664285d54cffd863b610d1e6398c52have61664285d54cffd863b610d1e6398c52a61664285d54cffd863b610d1e6398c52quest61664285d54cffd863b610d1e6398c52o61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52that61664285d54cffd863b610d1e6398c52asks61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52or61664285d54cffd863b610d1e6398c52a61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52exact61664285d54cffd863b610d1e6398c52a61664285d54cffd863b610d1e6398c52swer,61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52t61664285d54cffd863b610d1e6398c52wa61664285d54cffd863b610d1e6398c52ts61664285d54cffd863b610d1e6398c52you61664285d54cffd863b610d1e6398c52to61664285d54cffd863b610d1e6398c52leave61664285d54cffd863b610d1e6398c52"e"61664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c5261664285d54cffd863b610d1e6398c52that61664285d54cffd863b610d1e6398c52a61664285d54cffd863b610d1e6398c52swer.