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

\(y=x^n\)

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

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

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

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