Math Puzzle

[Farmer]

What other sort of questions were there in that chapter in that textbook? What were you doing using in that chapter? There are ways to solve this, would be nice to know the way they are looking for.

OP states this is a first year calc text.

H(H(x)) = H(x) is linear in H(x), but H(x) cannot be linear in x since it is not monotonically increasing or decreasing.

 

[artikk]

If H(x) =0 or exp(x) it kind of fits.

 

[Farmer]

If H(x) =0 or exp(x) it kind of fits.

exp(x) monotonically increases, it doesn't fit. Neither does H(x) = 0.

 

[TuxDave]

Find a function H such that H(H(x)) = H(x) for all numbers x, and such that H(1) = 36, H(2) = pi/3 and H(13) = 47.

It's an optional question in one of my textbook's chapters, but it is quite interesting!

Can you double check for a typo in your question?

If H(1) = 36,

H(H(1)) = H(1)
H(36) = 36?

 

[silverpig]

H(x) = 0.999... on a treadmill.

 

[Born2bwire]

Can you double check for a typo in your question?

If H(1) = 36,

H(H(1)) = H(1)
H(36) = 36?

Heh, I did not catch that. When I first read the post earlier I just assumed that it meant:

H(1) = 36
H(36) = 1

Which would actually be:

H(H(x)) = x

 

[Numenorean]

Find a function H such that H(H(x)) = H(x) for all numbers x, and such that H(1) = 36, H(2) = pi/3 and H(13) = 47.

It's an optional question in one of my textbook's chapters, but it is quite interesting!

Public Function H()

Select Case x

Case 1

Return 36

Case 2

Return pi/3

Case 13

Return 47

Case Else

Return x

End Select

End Function

Something like that what you're looking for?

 

[iCyborg]

For what its worth, my answer was

Let f(x) = g(x) + Indicator_fn{x in range of f(x)}*[x - g(x)]

And then fit whatever you like to g(x), as long as it satisfies the three points. I'm not particularly happy with my answer, seems like a cop-out. The question was marked 'hard' in the textbook, so I don't think it is as easy as I make it out to be.

This doesn't look like a solution to me, it's more like the problem statement rephrased with symbols. Furthermore, it's not clear that the recursion that defines this Indicator_f is even well defined.
This property h(h(x))=h(x) is called idempotence, and there's a shi*load of solutions: for all except 1,2,36 etc you can have h(x)=x, h(x)=abs(x), h(x)=const, or even semiarbitrary constructions like h(x)=x for x non-integer, h(n)=n if n is a power of 2 and h(k)=2^k for all other integers...
There should be some additional restriction, this seems too general.

 

[videogames101]

Piecewise to the rescue? Although I always find that to be quite a copout.



This seems weird to me. It can be shown the input of H(x),which is H(x), must be equal to the output of H(x), yet the restrictions say H(1) = 36, which doesn't follow.

to demonstrate visually, say y = H(x)

obviously H(x) = H(x)

so y = y (duh)

H(H(x)) = H(x)

H(y) = y

yes y = H(x), but it still shows how H(a) must be equal to a.

Am I wrong?

 

[JTsyo]

Piecewise to the rescue? Although I always find that to be quite a copout.



This seems weird to me. It can be shown the input of H(x),which is H(x), must be equal to the output of H(x), yet the restrictions say H(1) = 36, which doesn't follow.

to demonstrate visually, say y = H(x)

obviously H(x) = H(x)

so y = y (duh)

H(H(x)) = H(x)

H(y) = y

yes y = H(x), but it still shows how H(a) must be equal to a.

Am I wrong?

x=1 and x=36 both give you 36 for an output. The difference between the 2 is 35. x=13 and 47 give you 47 but the difference between the two is 34. So the period varies since 35 and 34 don't share any common denominator.

 

[Farmer]

This doesn't look like a solution to me, it's more like the problem statement rephrased with symbols. Furthermore, it's not clear that the recursion that defines this Indicator_f is even well defined.
This property h(h(x))=h(x) is called idempotence, and there's a shi*load of solutions: for all except 1,2,36 etc you can have h(x)=x, h(x)=abs(x), h(x)=const, or even semiarbitrary constructions like h(x)=x for x non-integer, h(n)=n if n is a power of 2 and h(k)=2^k for all other integers...
There should be some additional restriction, this seems too general.

Thank you!

 

[TuxDave]

Piecewise to the rescue? Although I always find that to be quite a copout.



This seems weird to me. It can be shown the input of H(x),which is H(x), must be equal to the output of H(x), yet the restrictions say H(1) = 36, which doesn't follow.

to demonstrate visually, say y = H(x)

obviously H(x) = H(x)

so y = y (duh)

H(H(x)) = H(x)

H(y) = y

yes y = H(x), but it still shows how H(a) must be equal to a.

Am I wrong?

H(1) = 36 is possible if H(x) can NEVER equal 1 for all values of x.

So H(y) = y is true for all y values that H(x) can be equal to.

 

[Farmer]

Piecewise to the rescue? Although I always find that to be quite a copout.



This seems weird to me. It can be shown the input of H(x),which is H(x), must be equal to the output of H(x), yet the restrictions say H(1) = 36, which doesn't follow.

to demonstrate visually, say y = H(x)

obviously H(x) = H(x)

so y = y (duh)

H(H(x)) = H(x)

H(y) = y

yes y = H(x), but it still shows how H(a) must be equal to a.

Am I wrong?

What did you just show there?

H(H(x)) = H(x), this is perfectly valid. There is nothing wrong with the problem statement. I already posted a perfectly valid solution if you allow H(x) to be discontinuous.

The question is, what smooth function satisfies these constraints?

 

[iCyborg]

I think there can't be a continuous solution (and smooth assumes continuous). There's a theorem that says if f continuous and f(a)=x, f(b)=y, then for any z between x and y, there must be c between a and b such that f(c)=z. Can't remember its name, it's a straightforward generalization of the case where if f is negative at x1, and positive at x2, it must have a root (f(z)=0) somewhere between x1 and x2, it's used to prove odd-degree polynomials must have at least one real root.

Taking this into account: h(1)=36, h(2)=PI/3. Since 36<13<Pi/3, this means that there has to be s, 1<s<2 such that h(s)=13. Now h(h(s)) = h(s), h(13) = 13, and this is a contradiction with the original condition that h(13)=47.

Can anyone spot something wrong with this?

 

[TuxDave]

I think there can't be a continuous solution (and smooth assumes continuous). There's a theorem that says if f continuous and f(a)=x, f(b)=y, then for any z between x and y, there must be c between a and b such that f(c)=z. Can't remember its name, it's a straightforward generalization of the case where if f is negative at x1, and positive at x2, it must have a root (f(z)=0) somewhere between x1 and x2, it's used to prove odd-degree polynomials must have at least one real root.

Taking this into account: h(1)=36, h(2)=PI/3. Since 36<13<Pi/3, this means that there has to be s, 1<s<2 such that h(s)=13. Now h(h(s)) = h(s), h(13) = 13, and this is a contradiction with the original condition that h(13)=47.

Can anyone spot something wrong with this?

Sounds spot on to me. I'm still voting that there's a typo in the OP.

 

[Farmer]

I think there can't be a continuous solution (and smooth assumes continuous). There's a theorem that says if f continuous and f(a)=x, f(b)=y, then for any z between x and y, there must be c between a and b such that f(c)=z. Can't remember its name, it's a straightforward generalization of the case where if f is negative at x1, and positive at x2, it must have a root (f(z)=0) somewhere between x1 and x2, it's used to prove odd-degree polynomials must have at least one real root.

Taking this into account: h(1)=36, h(2)=PI/3. Since 36<13<Pi/3, this means that there has to be s, 1<s<2 such that h(s)=13. Now h(h(s)) = h(s), h(13) = 13, and this is a contradiction with the original condition that h(13)=47.

Can anyone spot something wrong with this?

Rolle's mean value theorem, or some kind of mean value theorem or intermediate value theorem. Probably even the definition of "differentiable." That's definitely a calc 1 topic.

Great reasoning, that proof is perfectly valid by contradiction. I had suspected there would be a trick solution to this, being a textbook problem, but I this is it. Now I feel stupid.

To the OP: I'm quite sure what iCyborg wrote was the solution intended by the author, as it is simple and totally relevant to a textbook topic.

Best OT thread of the day.

 

[DrPizza]

I realized exactly what iCyborg has mentioned. Since this isn't a continuous function, I see no reason not to use a piecewise defined function.