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How do you call this type of math... - Printable Version

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How do you call this type of math... by .Roy on 11-08-2006 at 03:42 PM

I dont know the name in english so ill explain it.

Its when you have an equation and you have to prove its true for every n. for n=positive interger.

so u put in the equation n+1 and then u prove it that way...


So what is this type of math called??

----------

Its like geometry on a graph (x,y).... Also dont know the name.

RE: How do you call this type of math... by John Anderton on 11-08-2006 at 03:45 PM

quote:
Originally posted by .Roy
Its when you have an equation and you have to prove its true for every n. for n=positive interger.

so u put in the equation n+1 and then u prove it that way...
Induction.
RE: How do you call this type of math... by .Roy on 11-08-2006 at 03:54 PM

Also what do you call it in calculus when you make something from
f( x ) to f( x ) '

RE: RE: How do you call this type of math... by Jesus on 11-08-2006 at 04:20 PM

quote:
Originally posted by .Roy
Also what do you call it in calculus when you make something from
f( x ) to f( x ) '

quote:
derived function - the result of mathematical differentiation; the instantaneous change of one quantity relative to another; df(x)/dx
RE: How do you call this type of math... by qgroessl on 11-08-2006 at 05:10 PM

quote:
Originally posted by .Roy
f( x ) to f( x )

We just call Function of X or... F of X for short... (derived function is probably the mathematical term for it though.

:p... okay... so I didn't see the '  lol...
RE: How do you call this type of math... by CookieRevised on 11-08-2006 at 07:03 PM

quote:
Originally posted by .Roy
I dont know the name in english so ill explain it.

Its when you have an equation and you have to prove its true for every n. for n=positive interger.

so u put in the equation n+1 and then u prove it that way...


So what is this type of math called??
I dunno what it is called in English... Though I fail to see how putting n+1 in the equation would proof such an equation....

You just need to solve the equation in such a way that n is on the left of the equation and something else is on the right of the equation, where the equation is actually a "greater than" sign...

eg: solve it so you come to something like "n > 0".
eg: is n*8 > n always true if n=postive integer? yes, proof:
1) n*8 > n
2) (n*8)-n > n-n
3) n*7 > 0
4) n*7/7 > 0/7
5) n > 0

if that is what you meant


quote:
Originally posted by UTI
quote:
Originally posted by .Roy
f( x ) to f( x )'

We just call Function of X or... F of X for short... (derived function is probably the mathematical term for it though.
function of X is correct for f(x)... but note the ' after f(x), which means f(x)' is the derived function of x.

The ' wasn't a typo in his post...

;)
RE: How do you call this type of math... by .Roy on 11-08-2006 at 08:07 PM

quote:
Originally posted by CookieRevised
function of X is correct for f(x)... but note the ' after f(x), which means f(x)' is the derived function of x.

The ' wasn't a typo in his post...

;)

I didnt describe what i meant enough...

John anderton got it though :O.

I mean to prove for example an equation is true, you first "say its true" and then prove its true for n+1. And then you find a way to insert the true statement in the new equation and u get something that is true for example n^2>0 and u know n^2 is always positive so...
RE: How do you call this type of math... by Voldemort on 11-09-2006 at 12:00 AM

didnt f(x )' mean everything out of x?

RE: How do you call this type of math... by Rubber Stamp on 11-10-2006 at 01:35 PM

Voldemort, what you mean is compliment, it is used in set theory and related chapters like probabilty and stuff. it is also denoted by x^c or x bar. but in calculus, ' means derived function of the given function.
answer to first query : Mathematical Induction
answer to second query: Differentiation.

RE: How do you call this type of math... by Vilkku on 11-11-2006 at 12:40 PM

Cookie: In Dutch it's called "inductie" (used Wikipedia to get it).
It's a special kind of math, so I suggest you don't start doubting :P

Example:
Show that 1+2+3+...+n = (n(n+1))/2 is true when n is a positive integer.

You would solve that this way:
1. n=1, both sides equal 1, so it's true.
2. Assume that it is true for n=k, so if it is true for k+1 we have prooven it's always true.
3. We prooved that 1+2+3+...+k is (k(k+1))/2 is the first step, so we can replace the set of numbers with that expression. We get
(k(k+1))/2 + (k+1) = ((k+1)(k+2))/2
Then you solve the right and left side separately, in this both are
(1/2)k^2 +  (3/2)k + 1
We have now prooven it's true using induction :)