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#1 2013-04-15 04:38:38

mrpace
Member
Registered: 2012-08-16
Posts: 72

Prove by induction.......

Let f be some given function for which we wish to find a such that
f(alpha)= 0. Suppose that the equation f(x)= 0 may be arranged into
the form x=g(x), such that for some interval (a,b) alpha is in (a,b)  and
g(x) belongs to (a,b) . Further, suppose that g is differentiable with |g'(x)|</= C
for x belonging to (a,b), where C is some positive number. (Note that x=g(x)
implies that alpha=g(alpha)

please note that when i say "belonging to" i mean that symbol that looks a bit like an 'E'

Prove by induction that
|alpha-Xn| </= C^n|alpha-Xo|

thanks

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#2 2013-04-15 04:40:33

Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 23,790
Website

Re: Prove by induction.......

That is sad.... sad


'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

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#3 2013-04-15 04:45:49

mrpace
Member
Registered: 2012-08-16
Posts: 72

Re: Prove by induction.......

Agnishom wrote:

That is sad.... sad

it's not the same question!

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#4 2013-04-15 05:50:43

anonimnystefy
Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,015

Re: Prove by induction.......

What are Xn and X0?


“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

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