Welcome & about me
Helpful Resorces & such
Just for fun: Bernoulli's Inequality Proof
For a natural number n and a nonnegative number b, $$(1+b)^n \geq 1+n b .$$ Proof:Let n be a natural number and b a nonnegative number. Consider the expression \( (1+b)^n \) By the binomial formula, this expands to $$(1+b)^n = {n \choose 0}1^n + {n \choose 1}1^{n-1}b + {n \choose 2}1^{n-2}b^2 \cdots {n \choose n-1}1b^{n-1} + {n \choose n} b^n $$ We know that by the multiplicative identity, one to any power equals one, so this binomial expansion simplifies to $$(1+b)^n = {n \choose 0} + {n \choose 1}b + {n \choose 2}b^2 \cdots {n \choose n-1}b^{n-1} + {n \choose n} b^n $$ Additionally, \({n \choose 0}\) is equal to one and \({n \choose 1}\) is equal to n, $$(1+b)^n = 1+ nb + {n \choose 2}b^2 \cdots {n \choose n-1}b^{n-1} + {n \choose n} b^n $$ We also know that the sum of the terms \({n \choose 2}b^2 \cdots {n \choose n-1}b^{n-1} + {n \choose n} b^n\) is positive since b is nonnegative. It follows that $$(1+b)^n = 1+ nb + {n \choose 2}b^2 \cdots {n \choose n-1}b^{n-1} + {n \choose n} b^n \geq 1 + nb$$ Thus, $$(1+b)^n \geq 1+n b $$