# Identifying and evaluating $$e$$

## Exponentials

### Defining $$e$$ as a binomial

#### Lemma

$$f(n,i)=\dfrac{n!}{n^i(n-i)!}$$

$$f(n,i)=\dfrac{(n-i)!\prod_{j=n-i+1}^nj}{n^i(n-i)!}$$

$$f(n,i)=\dfrac{\prod_{j=n-i+1}^nj}{n^i}$$

$$f(n,i)=\dfrac{\prod_{j=1}^i(j+n-i)}{n^i}$$

$$f(n,i)=\prod_{j=1}^i\dfrac{j+n-i}{n}$$

$$f(n,i)=\prod_{j=1}^i(\dfrac{n}{n}+\dfrac{j-i}{n})$$

$$f(n,i)=\prod_{j=1}^i(1+\dfrac{j-i}{n})$$

$$\lim_{n\rightarrow \infty }f(n,i)=\lim_{n\rightarrow \infty }\prod_{j=1}^i(1+\dfrac{j-i}{n})$$

$$\lim_{n\rightarrow \infty }f(n,i)=\prod_{j=1}^i1$$

$$\lim_{n\rightarrow \infty }f(n,i)=1$$

#### Defining $$e$$

We know that:

$$(a+b)^n=\sum^n_{i=0} a^i b^{n-i} \dfrac{n!}{i!(n-i)!}$$

Let’s set $$b=1$$

$$(a+1)^n=\sum^n_{i=0} a^i \dfrac{n!}{i!(n-i)!}$$

Let’s set $$a=\dfrac{1}{n}$$

$$(1+\dfrac{1}{n})^{n}=\sum^{n}_{i=0} \dfrac{1}{n^i} \dfrac{n!}{i!(n-i)!}$$

$$(1+\dfrac{1}{n})^{n}=\sum^{n}_{i=0} \dfrac{1}{i!} \dfrac{n!}{n^i(n-i)!}$$

$$\lim_{n\rightarrow \infty }(1+\dfrac{1}{n})^{n}=\lim_{n\rightarrow \infty }\sum^{n}_{i=0} \dfrac{1}{i!} \dfrac{n!}{n^i(n-i)!}$$

From the lemma above:

$$\lim_{n\rightarrow \infty }(1+\dfrac{1}{n})^{n}=\sum^{\infty }_{i=0} \dfrac{1}{i!}$$

$$e=\sum^{\infty }_{i=0} \dfrac{1}{i!}$$

#### Defining $$e^x$$

$$e=\lim_{n\rightarrow \infty }(1+\dfrac{1}{n})^{n}$$

$$e^x=\lim_{n\rightarrow \infty }(1+\dfrac{1}{n})^{nx}$$

$$e^x=\lim_{n\rightarrow \infty }\sum^{nx}_{i=0} \dfrac{1}{n^i} \dfrac{(nx)!}{i!(nx-i)!}$$

$$e^x=\lim_{n\rightarrow \infty }\sum^{nx}_{i=0} \dfrac{x^i}{i!} \dfrac{(nx)!}{(nx)^i(nx-i)!}$$

From the lemma:

$$e^x=\sum^{\infty }_{i=0} \dfrac{x^i}{i!}$$

### Differentiating $$e^x$$

#### Intro

We have $$e^x=\sum^{\infty }_{i=0} \dfrac{x^i}{i!}$$

$$\dfrac{\delta }{\delta x}e^x=\dfrac{\delta }{\delta x}\sum^{\infty }_{i=0} \dfrac{x^i}{i!}$$

$$\dfrac{\delta }{\delta x}e^x=\sum^{\infty }_{i=0} \dfrac{\delta }{\delta x}\dfrac{x^i}{i!}$$

$$\dfrac{\delta }{\delta x}e^x=\sum^{\infty }_{i=1} \dfrac{\delta }{\delta x}\dfrac{x^i}{i!}$$

$$\dfrac{\delta }{\delta x}e^x=\sum^{\infty }_{i=1} \dfrac{x^{i-1}}{(i-1)!}$$

$$\dfrac{\delta }{\delta x}e^x=\sum^{\infty }_{i=0} \dfrac{x^{i}}{i!}$$

$$\dfrac{\delta }{\delta x}e^x=e^x$$

### Differentiating exponents, logarithms and power functions

#### Differentiating the natural logarithm

$$\dfrac{\delta }{\delta x}\ln (x)=\lim_{\delta \rightarrow 0}\dfrac{\ln(x+\delta)-ln(x)}{\delta }$$

$$\dfrac{\delta }{\delta x}\ln (x)=\lim_{\delta \rightarrow 0}\dfrac{\ln\dfrac{x+\delta}{x}}{\delta }$$

$$\dfrac{\delta }{\delta x}\ln (x)=\lim_{\delta \rightarrow 0}\dfrac{\ln(1+\dfrac{\delta}{x})}{\delta }$$

$$\dfrac{\delta }{\delta x}\ln (x)=\dfrac{1}{x}\lim_{\delta \rightarrow 0}\dfrac{x}{\delta}\ln(1+\dfrac{\delta}{x})$$

$$\dfrac{\delta }{\delta x}\ln (x)=\dfrac{1}{x}\ln(\lim_{\delta \rightarrow 0}(1+\dfrac{\delta}{x})^{\dfrac{x}{\delta }})$$

$$\dfrac{\delta }{\delta x}\ln (x)=\dfrac{1}{x}\ln(e)$$

$$\dfrac{\delta }{\delta x}\ln (x)=\dfrac{1}{x}$$

#### Differentiating logarithms of other bases

$$log_a(x)=\dfrac{\log_b(x)}{\log_b(a)}$$

$$log_a(x)=\dfrac{\ln(x)}{\ln(a)}$$

$$\dfrac{\delta }{\delta x}\log_a(x)=\dfrac{\delta }{\delta x}\dfrac{\ln(x)}{ln(a)}$$

$$\dfrac{\delta }{\delta x}\log_a(x)=\dfrac{1}{xln(a)}$$

#### Exponents

$$y=a^x$$

$$\ln(y)=x\ln(a)$$

$$\dfrac{\delta }{\delta x}\ln(y)=\dfrac{\delta }{\delta x}x\ln(a)$$

$$\dfrac{\delta }{\delta x}\ln(y)=\ln(a)$$

$$\dfrac{1}{y}\dfrac{\delta }{\delta x}y=\ln(a)$$

$$\dfrac{\delta }{\delta x}a^x=a^x\ln(a)$$

#### Power functions

$$y=x^n$$

$$\dfrac{\delta }{\delta x}y=\dfrac{\delta }{\delta x}x^n$$

$$\dfrac{\delta }{\delta x}y=\dfrac{\delta }{\delta x}e^{n\ln(x)}$$

$$\dfrac{\delta }{\delta x}y=\dfrac{n}{x}e^{n\ln(x)}$$

$$\dfrac{\delta }{\delta x}y=nx^{n-1}$$