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Homework Statement
A sequence [tex]{u_n}[/tex] is defined by [itex]x_{n+1}=x_n^2 +\frac{1}{4} ,x_1<\frac{1}{2}[/itex]
Prove by mathematical induction or otherwise that:-
a)[itex]x_{n+1}-\frac{1}{2}<0[/itex]
b)[itex]x_{n+1}>x_n[/itex]
Homework Equations
The Attempt at a Solution
a)
Assume statement true for n=N
therefore
[tex]
x_{N+1}-\frac{1}{2}<0[/tex]
squaring both sides
[tex]x_{N+1}^2-x_{N+1} +\frac{1}{4}<0[/tex]
[tex]x_{N+2}-x_{N+1}<0[/tex]
[tex]x_{N+2}<x_{N+1}[/tex]
[tex]-\frac{1}{2}[/tex]
[tex]x_{N+2}-\frac{1}{2}<x_{N+1}-\frac{1}{2}[/tex]
By the inductive hypothesis[itex]x_{N+1}-\frac{1}{2}<0[/itex]
[tex]
x_{N+2}-\frac{1}{2}<0[/tex]
[tex]x_{N+2}<\frac{1}{2}[/tex]
Therefore true for n=N+1. After testing [tex]x_1,x_2,x_3,...[/tex] true for all Natural numbers...
b) Assume true for n=N
[tex] x_{N+1}>x_N[/tex]
sq.both sides
[tex]x_{N+1}^2>x_N^2[/tex]
[tex]+\frac{1}{4}[/tex]
[tex]x_{N+1}^2+\frac{1}{4}>x_N^2+\frac{1}{4}[/tex]
[tex]x_{N+2}>x_{N+1}[/tex]
hence true for n=N+1...etc etc..true for n an element of N
What would be the way to prove this in the "otherwise method"? (Not by exhaustion)
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