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## #1 2013-04-16 02:38:38

mrpace
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### 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

## #2 2013-04-16 02:40:33

Agnishom
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### Re: Prove by induction.......

'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'
'The whole person changes, why can't a habit?' -Alokananda

## #3 2013-04-16 02:45:49

mrpace
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### Re: Prove by induction.......

#### Agnishom wrote:

it's not the same question!

## #4 2013-04-16 03:50:43

anonimnystefy
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### Re: Prove by induction.......

What are Xn and X0?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment