Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

MathsIsFun
2005-11-06 20:30:16

Most impressive, kylekatarn.

kylekatarn
2005-11-06 14:04:47

no problem : )

Mathfun
2005-11-06 09:56:21

Big BIg BIg BIg BIg thanks!!!!!!!!!

kylekatarn
2005-11-06 03:44:51

I finnaly got it. I don't know if you still want the solution but here it is, anyway:

----------------------------------------------------------------------------------------------------------
We want to show that if f( f(x) ) = x then g( g(x) ) = x, for all x∊ℝ

H1 (Hipothesis):
∀x∊ℝ f( f(x) ) = x

T1 (Thesis):
∀x∊ℝ g( g(x) ) = x


This is an implication where
H1 ⇒ T1
∀x∊ℝ [ f(f(x)) = x ⇒ g(g(x)) = x ]


We will determine the conditions for the Hipothesis to be true and then do the same for the Thesis. The we find the conditions are the same for both Hipothesis and Thesis. This allows us to conclude that the inicial statement is more than an implication, its something "logically stronger" - an equivalence H1⇔T1
----------------------------------------------------------------------------------------------------------
L1 (Lemma 1):
∀x≥0 ⇒ |x| = x
∀x<0 ⇒ |x| = -x
(the case x=0 was included in the first implication, but it could be on the second one, too)

Proof: Its obvious, considering the definition of absolute value.
----------------------------------------------------------------------------------------------------------
Proof of H1:
∀x∊ℝ f( f(x) ) = x

H1.a) Case x≥0
f(x) = ax+b|x|
f(x) = ax+bx     //by L1
f(x) = (a+b)x
f( f(x) ) = f( (a+b)x ) = (a+b)x

f( f(x) ) = x
(a+b)x = x ⇔ (a+b) = 1

Conclusion from H1.a)
∀x≥0 [ f( f(x) ) = x ] ⇔ (a+b) = 1

H1.b) Case x<0
f(x) = ax+b|x|
f(x) = ax+b(-x)     //by L1
f(x) = ax-bx
f(x) = (a-b)x
f( f(x) ) = f( (a-b)x ) = (a-b)x

f( f(x) ) = x
(a-b)x = x ⇔ (a-b) = 1

Conclusion from H1.b)
∀x<0 [ f( f(x) ) = x ] ⇔ (a-b) = 1

Conclusion from H1.a) and H1.b)
∀x≥0 [ f( f(x) ) = x ] ⇔ (a+b) = 1
and
∀x<0 [ f( f(x) ) = x ] ⇔ (a-b) = 1
therefore
∀x∊ℝ [ f( f(x) ) = x ] ⇔ [ (a+b) = (a-b) =1 ]

----------------------------------------------------------------------------------------------------------
Proof of T1:
∀x∊ℝ g( g(x) ) = x

T1.a) Case x≥0
g(x) = gx-b|x| 
g(x) = ax-bx     //by L1
g(x) = (a-b)x
g(g(x)) = g((a-b)x) = (a-b)x

g(g(x)) = x
(a-b)x = x ⇔ (a-b) = 1

Conclusion from T1.a)
∀x≥0 [ g( g(x) ) = x ] ⇔ (a-b) = 1

H1.b) Case x<0
g(x) = ax-b|x|
g(x) = ax-b(-x)     //by L1
g(x) = ax+bx
g(x) = (a+b)x
g( g(x) ) = f( (a+b)x ) = (a+b)x

g( g(x) ) = x
(a+b)x = x ⇔ (a+b) = 1

Conclusion from T1.b)
∀x<0 [ g( g(x) ) = x ] ⇔ (a+b) = 1

Conclusion from T1.a) and T1.b)
∀x≥0 [ g( g(x) ) = x ] ⇔ (a-b) = 1
and
∀x<0 [ g( g(x) ) = x ] ⇔ (a+b) = 1
therefore
∀x∊ℝ [ g( g(x) ) = x ] ⇔ [ (a-b) = (a+b) =1 ]

----------------------------------------------------------------------------------------------------------
Conclusions from H1 and T1
∀x∊ℝ [ f( f(x) ) = x ] ⇔ [ (a+b) = (a-b) =1 ]
and
∀x∊ℝ [ g( g(x) ) = x ] ⇔ [ (a-b) = (a+b) =1 ]
therefore
∀x∊ℝ [ f( f(x) ) = x ] ⇔ ∀x∊ℝ [ g( g(x) ) = x ]     //by the transitivity property of '⇔'
∀x∊ℝ [ f( f(x) ) = x ⇔ g( g(x) ) = x ]
so we can say that
H1⇔ T1 ∴ H1⇒ T1     //this is what we wanted to prove
and more..
T1⇒ H1

Mathfun
2005-11-05 01:46:31

I still can't fix it..... ;(

kylekatarn
2005-11-04 11:18:38

...indeed:)
Again, thank you all:)

MathsIsFun
2005-11-04 09:02:15

Congrats kylekatarn, and thank you on behalf of everyone you have helped to get you there! (That was a curious sentence wasn't it?)

kylekatarn
2005-11-04 04:43:57

mathsyperson wrote:

And with that post, kylekatarn becomes a power member. Well done to you.

Btw, thanks! : )

kylekatarn
2005-11-04 04:40:15

f(f(x)) = x => g(g(x)) = x
g(g(x))=f(f(x))

f(x)=ax+b|x|
g(x)=ax-b|x|

----------------------------------
f(f(x))=a.f(x)+b|f(x)|
g(g(x))=a.g(x)-b|g(x)|
----------------------------------
a.f(x)+b|f(x)|=a.g(x)-b|g(x)|

:: k=b/a ::

f(x)+k|f(x)|=g(x)-k|g(x)|
f(x)-g(x)=-k|g(x)|-k|f(x)|
g(x)-f(x)=k|g(x)|+k|f(x)|
g(x)-f(x)=k(|g(x)|+|f(x)|)
g(x)-f(x)=k(|g(x)|+|f(x)|)

g(x)-f(x) = ax - b|x| - (ax + b|x|) = ax - b|x| - ax - b|x|) = -2b|x|

-2b|x|=k(|g(x)|+|f(x)|)
-2b(1/k)|x|=|g(x)|+|f(x)|

:: 1/k=a/b ::

-2b(a/b)|x|=|g(x)|+|f(x)|
-2a|x|=|g(x)|+|f(x)|

That's as far as I can gow for now.
You can try other things like: triangular inequality, proofing by expanding the module or proof by contradiction.
Best luck for you now.

mathsyperson
2005-11-04 03:46:59

And with that post, kylekatarn becomes a power member. Well done to you.

Back on topic, I tried it and got f(f(x)) = a (ax + b|x|) + b|(ax + b|x|)|.

For g(g(x)) to equal x when f(f(x)) is equal to x, then f(f(x) would have to be equal to g(g(x) and so b would have to be 0.

So now you have to prove that a (ax + b|x|) + b|(ax + b|x|)| ≠ x for all values of x when b ≠ 0. That's the tricky bit.

mathfun
2005-11-04 03:17:25

No one can help?? PLEASE It's veary important for me I have to present it tomorrow in class!!

Mathfun
2005-11-04 01:56:58

Let "a" and "b" be real numbers. We have 2 functions:

f(x) = ax + b|x|   and   g(x) = ax - b|x|

Show that if:  f(f(x)) = x    for every real x, then:

g(g(x)) = x   for every real x

Pleas help 'cause I haven't got any idea how to solve it.

Board footer

Powered by FluxBB