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## #1 2005-11-04 01:56:58

Mathfun
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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.

## #2 2005-11-04 03:17:25

mathfun
Guest

### Re: Help Please With Functions!!

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

## #3 2005-11-04 03:46:59

mathsyperson
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### Re: Help Please With Functions!!

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.

Why did the vector cross the road?
It wanted to be normal.

## #4 2005-11-04 04:40:15

kylekatarn
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### Re: Help Please With Functions!!

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.

Last edited by kylekatarn (2005-11-04 04:42:54)

## #5 2005-11-04 04:43:57

kylekatarn
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### Re: Help Please With Functions!!

#### mathsyperson wrote:

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

Btw, thanks! : )

## #6 2005-11-04 09:02:15

MathsIsFun

Online

### Re: Help Please With Functions!!

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

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #7 2005-11-04 11:18:38

kylekatarn
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### Re: Help Please With Functions!!

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

## #8 2005-11-05 01:46:31

Mathfun
Guest

### Re: Help Please With Functions!!

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

## #9 2005-11-06 03:44:51

kylekatarn
Power Member

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### Re: Help Please With Functions!!

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)²x

f( f(x) ) = x
(a+b)²x = x ⇔ (a+b)² = 1

Conclusion from H1.a)
∀x≥0 [ f( f(x) ) = x ] ⇔ (a+b)² = 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)²x

f( f(x) ) = x
(a-b)²x = x ⇔ (a-b)² = 1

Conclusion from H1.b)
∀x<0 [ f( f(x) ) = x ] ⇔ (a-b)² = 1

Conclusion from H1.a) and H1.b)
∀x≥0 [ f( f(x) ) = x ] ⇔ (a+b)² = 1
and
∀x<0 [ f( f(x) ) = x ] ⇔ (a-b)² = 1
therefore
∀x∊ℝ [ f( f(x) ) = x ] ⇔ [ (a+b)² = (a-b)² =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)²x

g(g(x)) = x
(a-b)²x = x ⇔ (a-b)² = 1

Conclusion from T1.a)
∀x≥0 [ g( g(x) ) = x ] ⇔ (a-b)² = 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)²x

g( g(x) ) = x
(a+b)²x = x ⇔ (a+b)² = 1

Conclusion from T1.b)
∀x<0 [ g( g(x) ) = x ] ⇔ (a+b)² = 1

Conclusion from T1.a) and T1.b)
∀x≥0 [ g( g(x) ) = x ] ⇔ (a-b)² = 1
and
∀x<0 [ g( g(x) ) = x ] ⇔ (a+b)² = 1
therefore
∀x∊ℝ [ g( g(x) ) = x ] ⇔ [ (a-b)² = (a+b)² =1 ]

----------------------------------------------------------------------------------------------------------
Conclusions from H1 and T1
∀x∊ℝ [ f( f(x) ) = x ] ⇔ [ (a+b)² = (a-b)² =1 ]
and
∀x∊ℝ [ g( g(x) ) = x ] ⇔ [ (a-b)² = (a+b)² =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

Last edited by kylekatarn (2005-11-06 03:46:17)

## #10 2005-11-06 09:56:21

Mathfun
Guest

### Re: Help Please With Functions!!

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

kylekatarn
Power Member

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no problem : )

MathsIsFun