Define the number e algebraically.?

Answer 1

Lot's of cool and correct responses. Although as several have stated, e is not the root of any polynomial with rational coefficients and finite degree, there are lots of neat ways to express e as infinite sums (as limits).

Here is my favorite:

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

where n! = n*(n-1)* ... * 1

Answer 2

Can't be defined algebraically but it can be described in a manner that will help to see where it comes from.
Think of compound interest. Let's say we've found a very generous bank that pays you 100 percent interest per year.
Amount earned = Principal X (1 + i/n)^(nt),
where n is the number of times per year the interest is computed
and t is the number of years you have the money invested.
year, and the amonut invested is also 1. So now we have
A = 1 X (1 + 1/n)^(nt) or simply (1 + 1/n)^(nt)
At the end of 1 year, if you invest 1 dollar, you'd have 2 dollars.
Now imagine compounding the interest more frequently during the year. For compounding monthly we have
A = (1 + (1/12))^12 ~= (1.0833)^12 ~= 2.6121
Wow! At this rate it seems our money could grow infinitely without bound! Let's see...
For compounding weekly you'd have
A = (1 + (1/52))^52 ~= (1.016)^52 ~=2.6926
For compounding daily you'd have
A = (1 + (1/365.25))^365.25 ~=2.7146
Looks like the number is not increasing without bound. Instead it seems to be levelling off at around 2.7
Let's see what would happen if the interest were compounded a hundred times per day...
A = (1 + (1/(100x365.25))^(100x365.25) ~= 2.717
So you can see that if you keep higher numbers of times interest is compounded, that is, let n go to infinity, your amount A would go to (drum roll...) e ~= 2.71828182846...
(I know it looks like it's going to repeat for a while in there, but trust me, it doesn't.
So there you have it.

Answer 3

a million +a million=?. we are able to apply some algebraic equations with a mushy touch of boolean. we are able to replace the ? with a 0. if we circulate the only to different area a million=0-a million . yet this potential we would desire to discover 2 numbers that provide one and nil. applying the and gate a million and a million = 0 applying the Xor gate a million + a million = 0. so the numbers we are able to apply are a million and a million in tis case we are able to alter the 0 to a million and upload the extra one to the equation so the undertaking will become a million=a million+a million+a million if we circulate the 1st a million to the different area it is going to become a ? returned ?+a million=a million+a million then ?+a million=2 what do you upload to a million to make 2. This once you get the calculator (medical calculator) type in a million+a million-a million/2 (sq. root - sin(2) = then click coach as graph and have been y intercepts x is the selection you like a million.987310. Then divided this selection by potential of cos(2) which = a million in accordance this the respond is 32

Answer 4

Because Euler's number e is a transcendental, it cannot be expressed as the root of any algebraic equation. However, we have Euler's equation, which is the closest to a simple "algebraic" equation:

e^(pi i) = -1, where i is Sqrt(-1).

Why this is possible in spite that e is a transcendental is because pi is also. In fact, this was how the transcendence of pi was first proven, by first proving the transcendence of e, and then using Euler's equation for a simliar proof of transcendence of pi. The PDF file is the german language proof of transcendence of pi.

Answer 5

As someone stated, e is transcendental, and by definition, transcendental numbers are those which doesn't have algebraic solutions.

It can, however, be expressed as a limit.

lim (1 + 1/n)^n
x -> infinity

If that qualifies algebraically enough for you, then that's your answer.

Answer 6

It is not an algebraic number.

"Certainly it was Hermite who proved that e is not an algebraic number in 1873."

Answer 7

e=sum i=0--->infinity 1/i! Remember 0! is defined as 1.

or

lim n---->infinity (1+1/n)^n

Answer 8

Another way to get it: e^x=(d/dx)e^x, x != 0

Answer 9

e= e^2 * 1/e

...sorry :)