### e - unifying two common definitions of e

Note: The concept of infinity would be used liberally here. For a brief introduction to the origins of the infinities found in today's context, read this post.

Perhaps the most common expression that gives the value of e would be the sum of the reciprocal of factorials, from zero to infinity.

Equation (1)

Another well known expression is:

Equation (2)

I will attempt to show that these two expressions are identical. Firstly, I will let a be a finite number before pushing it towards the great infinity.

Equations (3)

Here, a takes the value of 4. Expanding the brackets, we see that the terms are results of multiplication between 1 and ¼. In the last line above, the coefficients 1, 4, 6, 4, 1 are actually binomial coefficients resulting from 4C0, 4C1, 4C2… 4C4.

For example, 4C2=6 arises because (¼)^2 appears 6 times as a result of the 6 different ways the terms can combine to give (¼)^2. The combinations are:

If you have no idea what I am talking about, try expanding f(6) into individual terms. It should become clear after that.

f(a) can then be rewritten as

Equations (4)

Expressing the binomial number as a ratio of factorials gives the final line in Equations (4).

Substituting f(a) into the definition of e (equation 2):

Equations (5)

The equation is beginning to resemble the infinite sum of inverse factorials:

Equations (6)

To show that these equations are equivalent, the following expression needs to be true:

Equations (7)

When a approaches infinity, the terms in the brackets should converge to unity, and the expressions in equations (6) would be identical.

Expanding the terms in the bracket:

Equations (8)

Cancelling terms that appear in the numerator and denominator:

Equations (9)

Note that there are an equal number of terms in the numerator and denominator. For (a-n)!, the missing terms are replaced with the a^n.

Replacing this result into the previous expression:

Equations (10)

When taking the limit of a approaching infinity, (a-k)/a is unity regardless of the value of k. After all, infinity minus 5 is still infinity.

Thus, it summary

Equations (11)

One very valid concern is, “what happens when k ≈ a? When k ≈ a, (a-k)/a is no longer unity and equation (6) might not match term-for-term.”

Fortunately, this never happens. Observe how the equation is structured: first, let a approach infinity. Then, incrementally increase n and k.

n and k can be as humongous as one wishes to make it, but they would always be dominated by a. You can even make n and k as large as the second Skewe’s number, and they would still not make a dent in a. The second Skewe’s number is finite; a is not.

Perhaps the most common expression that gives the value of e would be the sum of the reciprocal of factorials, from zero to infinity.

Equation (1)

Another well known expression is:

Equation (2)

I will attempt to show that these two expressions are identical. Firstly, I will let a be a finite number before pushing it towards the great infinity.

Equations (3)

Here, a takes the value of 4. Expanding the brackets, we see that the terms are results of multiplication between 1 and ¼. In the last line above, the coefficients 1, 4, 6, 4, 1 are actually binomial coefficients resulting from 4C0, 4C1, 4C2… 4C4.

For example, 4C2=6 arises because (¼)^2 appears 6 times as a result of the 6 different ways the terms can combine to give (¼)^2. The combinations are:

Multiply ¼ from the 1st and 2nd brackets, and multiply with 1 from the 3rd and 4th brackets.

Multiply ¼ from the 1st and 3rd brackets, and multiply with 1 from the 2nd and 4th brackets.

Multiply ¼ from the 1st and 4th brackets, and multiply with 1 from the 2nd and 3rd brackets.

Multiply ¼ from the 2nd and 3rd brackets, and multiply with 1 from the 1st and 4th brackets.

Multiply ¼ from the 2nd and 4th brackets, and multiply with 1 from the 1st and 3rd brackets.

Multiply ¼ from the 3rd and 4th brackets, and multiply with 1 from the 1st and 2nd brackets.

If you have no idea what I am talking about, try expanding f(6) into individual terms. It should become clear after that.

f(a) can then be rewritten as

Equations (4)

Expressing the binomial number as a ratio of factorials gives the final line in Equations (4).

Substituting f(a) into the definition of e (equation 2):

Equations (5)

The equation is beginning to resemble the infinite sum of inverse factorials:

Equations (6)

To show that these equations are equivalent, the following expression needs to be true:

Equations (7)

When a approaches infinity, the terms in the brackets should converge to unity, and the expressions in equations (6) would be identical.

Expanding the terms in the bracket:

Equations (8)

Cancelling terms that appear in the numerator and denominator:

Equations (9)

Note that there are an equal number of terms in the numerator and denominator. For (a-n)!, the missing terms are replaced with the a^n.

Replacing this result into the previous expression:

Equations (10)

When taking the limit of a approaching infinity, (a-k)/a is unity regardless of the value of k. After all, infinity minus 5 is still infinity.

Thus, it summary

Equations (11)

One very valid concern is, “what happens when k ≈ a? When k ≈ a, (a-k)/a is no longer unity and equation (6) might not match term-for-term.”

Fortunately, this never happens. Observe how the equation is structured: first, let a approach infinity. Then, incrementally increase n and k.

n and k can be as humongous as one wishes to make it, but they would always be dominated by a. You can even make n and k as large as the second Skewe’s number, and they would still not make a dent in a. The second Skewe’s number is finite; a is not.

Labels: mathematics, real numbers

## 2 Comments:

Ahh.. i see you've caught the mathematic fever, anyway here's another identity involving e which you can try proving.

determinant M= exp(log(trace M))

M is an n by n matrix.

sorry, it should be

exp(trace(log M))

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