Question regarding "e"

[Projector]

I've been stumped by this question for a while now:

Find a value of n for which (1+ 1/n)^n gives the value of "e" correct to 9 decimal places.

[D H]

Have you tried any values of n?

[HallsofIvy]

Looks like a straightforward calculation problem to me!

[ice109]

lol, look up e on wikipedia and you'll know why we're giggling over here

[Jeff Reid]

Trick question? 9 significant digits or 9 digits after the decimal point?

[D H]

It is a trick question in a sense. Just throwing numbers at a typical computer implementation of (1+1/n)^n won't work. The problem lies in how computers represent real numbers.

Suppose one uses a program like Excel or writes a simple program to calculate (1+1/n)^n (I did just this with Excel, perl, and C on a Mac and on a Linux box.) The error shrinks as n increases up until n=10^7 or so. At this point the error stops shrinking. The problem is that the computer performs the calculation (1+1/n)^n as \exp(n*\log(1+1/n))[/tex]. For large n, [itex]log(1+1/n)\approx(1+1/n)-1, and this will err from 1/n using off-the-shelf double-precision numbers. For sufficiently large n (1016 or so), (1+1/n)-1=0 using off-the-shelf double-precision numbers!

A brute-force way to overcome this problem is to program using an extended precision package. An even better way to overcome the problem is to approach it analytically. A simple error analysis of (1+1/n)^n-e yields a value of n above which the error will be smaller than 1e-9. I am not revealing this value in the chance that this thread is asking us to solve a homework problem.

Since I have a "certain type of brain that's easily disabled (http://www.physicsforums.com/showthread.php?t=206096)", I of course had to double-check this limit using an extended precision math package. Ta-da, it works as predicted.

[mathwonk]

i had a homework assignment in 1960 to use the taylor series to calculate e to 9 digits, and found it too tedious.

In 2006 I felt professionally embarrassed after telling the story to my calc class so sat down and did it by hand, using a small $10 calculator for some arithmetic. (It took about 14 terms, but your sequence will have a different error term from the taylor series, so i doubt if i am giving away the answer.)

My former calc prof was still working in texas so i sent him the hw with an apology for being over 40 years late. he kindly wrote back his appreciation. maybe you too will eventually get credit for this one.

[arildno]

Psst, the 9th digit is...8! :smile:

Two point 7, two times Ibsen, 459045.

[Dodo]

Ibsen, or Verne, or Schubert... (google for the win).