Calculating Natural Number e: History & Methods

[Menthol]

Would you please tell me how to calculate natural number e ? How that number came into being ?
Thanks

 

[inha]

you can calculate it with a number of different ways. check http://mathworld.wolfram.com/e.html

 

[whozum]

So I just realized that all this doesn't get me to the numerical value of e, just one of the nice properties of it. So if anyone wants the limit derivation for e^x, let this be your guide:

\lim_{h\rightarrow 0} \frac{n^{x+h} - n^x}{h} \Rightarrow \frac{n^xn^h - n^x}{h} \Rightarrow \frac{n^x(n^h-1)}{h}


The limit of the product is the product of the limits:

\lim_{h\rightarrow 0} \frac{n^h}{h} \lim_{h\rightarrow 0} n^x

L'hospitals simplifies the left to n^hln(n)

\lim_{h\rightarrow 0} n^hln(n) \lim_{h\rightarrow 0} n^x = n^0ln(n)n^x

Result: n^x' = n^x(ln(n))

For some n, n^x' = n^x, and that is only when n = e.

 

[dextercioby]

The definition is with the limit of that sequence,but the calculation is done with the series of e^{x} for x=1.


Daniel.

 

[Jameson]

Or you could use the limit e = \lim_{x\rightarrow\infty} (1 + \frac{1}{x})^{x}

Jameson

 

[HallsofIvy]

As to "How that number came into being ?": it has the very nice property that the rate of change of the function e^x is just e^x itself- no other function has that property. That's why it was recognized as being important. I wouldn't begin to speculate as to how any number "came into being"!

 

[quetzalcoatl9]

yet another way would be to integrate the hyperbola, and then take the inverse of that function:

\int \frac{1}{x} dx = ln(x)

ln^{-1}(x) = e^x

this integration could be done numerically, and a table of ln values be built (ala Napier and friends) and then a column could be made of e^x by indexing backward into the table.

tables of logs were actually very popular at one time (and led to the slide rule) because multiplication and division could be reduced to addition and subtraction (think algebra of exponents). this greatly benefited fields like artillery and naval navigation since complicated calculations could be done quickly by indexing into a precomputed table (same for trig functions).

 

[mathwonk]

of course all functions of form ce^x have that property as well, and these are the only ones.

 

[josephcollins]

The way I was introduced to e was to try to find the exponentional which was its own derivative. So say we have a^x, we know that a^0 is 1, so try to find the value of a such that the gradient of a^x is 1 at the point (0,1). From this you get the definition of a, or as we know it e: Lim (1+(1/n))^n as n tends to infinity. I think this is how it was approached, I think Eli Maor has a book on the history of e if you're interested, it's on Amazon. Cheers, Joe

 

[James R]

You could always use:

e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ...

which is from the Taylor series for e^x.

 

[robert Ihnot]

James R: You could always use:

e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + ...

which is from the Taylor series for e^x.

This is a really effective way, that is if the student seeks a good approximation. As my professor once said, "e gets to its limit very fast."

After n terms, starting with 0, the error is less than 1/(n!) for n =1 or greater. This can be shown. Consider:

1/n! >1/(n+1)! + 1/(n+2)! +1/{(n+3)! ++++

Since 1>1/(n+1) +1/{(n+1)(n+2)} + 1/{(n+1)(n+2)(n+3)} ++++

Since for n=1, we have 1>=1/2 + 1/(2*3) + 1/(3*4) +1/(4*5) +1/(5*6) ++++++ =(number of terms)/(number of terms +1)> the series above.

So that even at n=1, we can conclude 2<e<3. At n=3 we find that 2+2/3<e<2+5/6. At n=7, we have 2+3620/5041<e<2+3621/5041.