to prove: if f ( a ) < 0 , f ( b ) > 0 , f ′ ( x ) > 0 , f ″ ( x ) > 0 {\displaystyle f(a)<0,f(b)>0,f'(x)>0,f''(x)>0\!} , then Newton's method x n + 1 = x n − f ( x n ) f ′ ( x n ) {\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}} converges; lim n → ∞ x n = r {\displaystyle \lim _{n\rightarrow \infty }x_{n}=r} such that f(r)=0.
g n ( x ) = f ′ ( x n ) ( x − x n ) + f ( x n ) g n ( x n + h ) = h f ′ ( x n ) + f ( x n ) f ( x n + h ) − f ( x n ) h = f ′ ( c ) , c > x n f ( x n + h ) = h f ′ ( c ) + f ( x n ) > g n ( x n + h ) {\displaystyle {\begin{aligned}g_{n}(x)&=f'(x_{n})(x-x_{n})+f(x_{n})\\g_{n}(x_{n}+h)&=hf'(x_{n})+f(x_{n})\\{\frac {f(x_{n}+h)-f(x_{n})}{h}}&=f'(c),c>x_{n}\\f(x_{n}+h)&=hf'(c)+f(x_{n})\\&>g_{n}(x_{n}+h)\end{aligned}}}
Now define h 1 {\displaystyle h_{1}} and h 2 {\displaystyle h_{2}} such that g n ( x n + h 1 ) = 0 , f ( x n + h 2 ) = 0 {\displaystyle g_{n}(x_{n}+h_{1})=0,f(x_{n}+h_{2})=0\!} . If h 1 ≤ h 2 {\displaystyle h_{1}\leq h_{2}} , then g n ( x n + h 1 ) = f ( x n + h 1 ) + k , k ≥ 0 {\displaystyle g_{n}(x_{n}+h_{1})=f(x_{n}+h_{1})+k,k\geq 0} , so g n ( x n + h 1 ) ≥ f ( x n + h 1 ) {\displaystyle g_{n}(x_{n}+h_{1})\geq f(x_{n}+h_{1})} , a contradiction. So h 1 > h 2 , x n + h > r {\displaystyle h_{1}>h_{2},x_{n}+h>r\!} .
g n ( x n + h ) = 0 {\displaystyle g_{n}(x_{n}+h)=0} implies x n + h = x n − f ( x n ) f ′ ( x n ) {\displaystyle x_{n}+h=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}} so f ( x n − f ( x n ) f ′ ( x n ) ) = f ( x n + 1 ) > f ( r ) > 0 {\displaystyle f\left(x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}\right)=f(x_{n+1})>f(r)>0} .
x n + 1 = x n − f ( x n ) f ′ ( x n ) {\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}} so x n + 1 < x n {\displaystyle x_{n+1}<x_{n}} for n>0 so f ( x n + 1 ) < f ( x n ) {\displaystyle f(x_{n+1})<f(x_{n})} so f ( x 1 ) > f ( x 2 ) > ⋯ > 0 {\displaystyle f(x_{1})>f(x2)>\cdots >0} so x 1 > x 2 > ⋯ > r {\displaystyle x_{1}>x2>\cdots >r} so by the monotone sequence theorem, x n {\displaystyle x_{n}} converges. So lim n → ∞ ( x n − x n + 1 ) = 0 {\displaystyle \lim _{n\rightarrow \infty }\left(x_{n}-x_{n+1}\right)=0} . But f ( x n ) = f ′ ( x n ) ( x n − x n − 1 ) {\displaystyle f(x_{n})=f'(x_{n})(x_{n}-x_{n-1})\!} and f ′ ( x n ) ≤ max { f ′ ( x ) : r ≤ x ≤ x 1 } {\displaystyle f'(x_{n})\leq \max \left\{f'(x):r\leq x\leq x_{1}\right\}} so lim n → ∞ f ( x n ) = 0 {\displaystyle \lim _{n\rightarrow \infty }f(x_{n})=0} so x n → r {\displaystyle x_{n}\rightarrow r} as required.