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noelevans
2012-11-09 12:29:58

LaTex input (not so pretty) vs LaTex output = Pretty math

anonimnystefy
2012-11-09 07:25:50

Pretty math?

noelevans
2012-11-09 06:25:25

Thanks all!

bobbym
2012-11-09 03:47:16

Hi noelevans;

Give this a try:

http://latex.codecogs.com/editor.php

perfect latex everytime!

anonimnystefy
2012-11-09 00:33:33

The forum also has built in tags which allows use of subscripts and superscripts withouth knowledge of latex:

For example an.

bob bundy
2012-11-08 19:21:42

hi noelevans

You could try Latex:

http://www.mathisfunforum.com/viewtopic.php?id=4397

Bob

noelevans
2012-11-08 18:22:40

Hi!

Suppose that a   is a constant and r is not zero or one and for each n in {1,2,3,...}
0
n
a   = r*a     .    Show that a  = a * r    for each n in {1,2,3,...}.
n         n-1                      n     0
1
For n=1 we have a  = r*a     = r*a  = a *1  .   So it works for n=1.
1        1-1        0     0

Now suppose that for some positive integer k,

k                                                               k               k          k+1
we have a  = a *r  .   Then  a      = r*a           = r*a   = r*a *r  = a * r * r   = a*r
k     0                   k+1        (k+1)-1        k         0         0                0

So given that it works for k we have shown that it works for k+1, which is the inductions step.
Therefore it works for all n in {1,2,3,...}.

Is there any way to put this page into a half-space mode?  Then the super and subscripts would
look pretty nice.

bob bundy
2012-11-07 18:53:11

hi hempy,

The following site has an explanation of proof by induction.  The final example is the proof for the geometric sequence.  The notation is a little different but you should be able to 'convert it'.

http://www.csee.umbc.edu/~stephens/203/PDF/4-2.pdf

Bob

hempy
2012-11-07 17:07:42

Let a(sub0) and r be fixed real numbers with r ≠ 0 and r ≠ 1, and suppose that for each n ∈ N, a(subn) = r*a(subn-1).

For every nonnegative integer n, a(subn) = a(sub0) * r^n.

Prove by induction.