I need help understanding how the series of e derives into the exponential series using the binomial theorem.
Here is a link to a pic of a page in my book, regarding the exponential series:
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(remove parentheses and spaces)
A couple of questions:
Where does the [1 + (1/k)]^k come from and why is it used?
Could you clarify the expansion of [1+(1/k)]^k?
I don't understand how it gets to ... k(1/k) + k(k-1)/2! (1/k^2) + ...
How does it end up with a 1 + 1 + 1[1-(1/k)]/2! + ...
Why are you finding the limit of the series?
And finally how do you end up with exponential series
x^n /n! = 1 + x + x^2/2! + ... ?
I'm confused and just really don't understand why or how you end up with everything. Help is VERY appreciated.
To understand how the series of e derives into the exponential series using the binomial theorem, let's break it down step by step.
First, let's start with the expression [1 + (1/k)]^k. This expression arises from considering the limit as k approaches infinity of the expression (1 + 1/k)^k. This limit is a special number known as the mathematical constant "e." The reason it is used is because the natural logarithm function, ln(x), is defined as the inverse function of the exponential function e^x. So understanding the series representation of e helps us understand the exponential series.
Next, let's clarify the expansion of [1 + (1/k)]^k using the binomial theorem. The binomial theorem states that for any real number a and any positive integer n, the expression (a + b)^n can be expanded as a sum of terms. Applying this theorem to our expression, we have:
(a + b)^n = C(n, 0)a^nb^0 + C(n, 1)a^(n-1)b^1 + C(n, 2)a^(n-2)b^2 + ... + C(n, n-1)a^1b^(n-1) + C(n, n)a^0b^n,
where C(n, r) represents the binomial coefficient, also denoted as "n choose r." In our case, a = 1, b = 1/k, and n = k. Plugging these values into the formula, we get:
[1 + (1/k)]^k = C(k, 0)1^k(1/k)^0 + C(k, 1)1^(k-1)(1/k)^1 + C(k, 2)1^(k-2)(1/k)^2 + ... + C(k, k-1)1^1(1/k)^(k-1) + C(k, k)1^0(1/k)^k.
Now, the binomial coefficient C(k, r) is defined as k! / (r!(k-r)!), where "!" denotes the factorial operation. Therefore, simplifying the terms, we have:
[1 + (1/k)]^k = 1 + k(1/k) + k(k-1)/(2!)(1/k)^2 + k(k-1)(k-2)/(3!)(1/k)^3 + ...
This expansion shows how [1 + (1/k)]^k can be broken down into a series of terms.
Now, let's address the expression 1 + 1 + 1[1-(1/k)]/2! + ... in your question. This expression arises when we simplify the terms of the expansion. Notice that each term after the second term has a factor of (1/k) raised to a power. We can rewrite these terms as:
k(k-1)/(2!)(1/k)^2 = (k^2 - k)/(2!k^2) = (1 - 1/k)/(2!)
Similarly, you can simplify further terms to obtain expressions like (1 - 1/k)/(3!), (1 - 1/k)/(4!), and so on.
The reason for finding the limit of the series is to evaluate the expression [1 + (1/k)]^k as k approaches infinity, which corresponds to evaluating e. Taking the limit allows us to understand the behavior of the expression as k becomes larger and larger.
Finally, the exponential series x^n/n! = 1 + x + x^2/2! + ... emerges from the fact that the terms in the expansion of [1 + (1/k)]^k can be written in the form of x^n/n! as k approaches infinity. As we substitute x = 1/k into the terms and take the limit, we obtain x^n/n!, which is a series representation for the exponential function.
I hope this explanation helps clarify how the series of e derives into the exponential series using the binomial theorem. Let me know if you have any further questions!