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Solve for function

This is a precalc concept I am having trouble understanding as there is only one example for me I can find. What if I have f(g(x))=h(x), and h(x) equals something? How do I find possible f(x) and g(x)?

The questions are specifically h(x)=(x²-1)³ and h(x)=(x-2)²-5 but anything will do.— [Mac Davis] (talk)

For those examples, you could split the first one into (x²-1) and (x)³ and the second into (x-2) and (x)²-5. StuRat 04:28, 22 August 2006 (UTC)

Pick any g(x) that satisfies "g(x₁) = g(x₂) implies h(x₁) = h(x₂)". Then use f(z) = h(g^-1(z)). Note that no matter what h(x) is, an injective g(x) can always be used. For example, take g(x) = ax+b, where a is non-zero. Then use f(z) = h((z-b)/a), so f(g(x)) = h((g(x)-b)/a) = h(x). -- Four Dog Night 03:59, 22 August 2006 (UTC)

Really the easiest is to take f = id and g = h, or f = h and g = id, where id is the identity function defined by id(x) = x. For something a bit more challenging, solve the functional equation f(f(x)) = h(x) for f, where h(x) = – 1 / x. Extra constraint: stay within the realm of real numbers. --Lambiam^Talk 04:56, 22 August 2006 (UTC)

This is called function composition, and it may be helpful to read our article. We have little interest in the trivial case where either f or g are the identity function. For polynomials, like your two examples, we have algorithms for symbolic computation of decompositions using ideals and other methods of abstract algebra. However, in differential and integral calculus we often encounter a broader range of possibilities, including exponential and trigonometric functions combined with polynomials.

Let's look at two examples.

1. $h_{1}(x)={\frac {e^{x}+e^{-x}}{2}}$
2. $h_{2}(x)=2\cos ^{2}x-1\,\!$

In the first example we immediately notice the e^x as a tempting target. However, to pull that out we must rewrite e^−x as 1/e^x. That done, the function becomes

$h_{1}(x)={\frac {g(x)+1/g(x)}{2}}\qquad ;g(x)=e^{x}$

Since x now only appears in g(x), we recognize

$h_{1}(x)=f(g(x))\qquad ;f(x)=(x+1/x)/2.\,\!$

In the second example, x only appears as cos x, so the natural choice for g is that.

$h_{2}(x)=2(g(x))^{2}-1\qquad ;g(x)=\cos x\,\!$

This, in turn, immediately suggests f.

$h_{2}(x)=f(g(x))\qquad ;f(x)=2x^{2}-1\,\!$

Decompositions are never unique. Sometimes the choice of decomposition is a matter of taste; more often, of utility. Nor is decomposition always the best transformation. (For example, we might recognize h₁(x) as the hyperbolic function cosh(x); and h₂(x) as cos(2x), via the Chebyshev polynomials.)

Experience with many examples is probably the best way to learn, especially since the known algorithms are more suitable for computers than for humans.

Looking ahead, in the context of calculus we use function decomposition (and composition) whenever we use the two valuable tools known as the "chain rule" for differentiation and the "change of variables" method for integration. --KSmrq^T 07:47, 22 August 2006 (UTC)

Thanks for your help guys, but I have no idea what you are talking about. KSmrq, did you pull the f(x) and g(x) out of your ass or what? Somebody can explain that to me :P :) If one of you first three could.. .erm... clarify more on what you said that would be good. I'm not down with the symbolic algothirmic decoposition or injectives or Chebyshev polynomials yet. Right after I grasp this basic concept I'd like to learn about them though. — [Mac Davis] (talk)

It's simpler than you think. Let's take h(x)=(x²-1)³ as an example. Given a value for x, you can find h(x) in 3 steps: (a) square x; (b) subtract 1 from the result of (a); (c) cube the result of (b). In function notation, this can be written as

h(x)=f₃(f₂(f₁(x)))

where f₁(x)= x²; f₂(x)= x-1; f₃(x)= x³. So we have expressed h(x) as the composition of 3 functions. To express h(x) as the composition of two functions, just combine f₁(x) and f₂(x) into one function as StuRat did in his first response. Alternatively, you can combine f₂(x) and f₃(x) into one function, so f(x)=(x-1)³ and g(x)=x². Gandalf61 14:39, 22 August 2006 (UTC)

Ok, thanks. I think I get it. — [Mac Davis] (talk)

Questions on power towers

The tetration article touches on power towers but I have a few questions.

What would the appearance of the graph of $\lim _{N\rightarrow \infty }$${\ f(x)= \atop {\ }}{{\underbrace {x^{x^{\cdot ^{\cdot ^{x}}}}} } \atop {\textrm {N}}\ {\textrm {number}}\ {\textrm {of}}\ x}$

Also, what about infinitely nested logarithms?

$\lim _{N\rightarrow \infty }$$f(x)=\underbrace {\log _{b}(log_{b}(log_{b}(log_{b}(log_{b}(log_{b}(log_{b}(log_{b}(log_{b}(x))))))))).\,\!} \atop {\textrm {N}}\ {\textrm {number}}\ {\textrm {of}}\ log_{b}$

Or even, a power tower of logarithms?

$\lim _{N\rightarrow \infty }$${\ f(x)= \atop {\ }}{{\underbrace {(log_{b}(x))^{(log_{b}(x))^{(log_{b}(x))^{(log_{b}(x))^{(log_{b}(x))}}}}} } \atop {\textrm {N}}\ {\textrm {number}}\ {\textrm {of}}\ log_{b}}$

Also, what about nested sine functions?

$\lim _{N\rightarrow \infty }$$f(x)=\underbrace {\ \sin(\sin(\sin(\sin(\sin(\sin(\sin(\sin(\sin(\theta ))))))))).\,\!} \atop {\textrm {N}}\ {\textrm {number}}\ {\textrm {of}}\ \sin$

Weird questions that I have been pondering upon for a while. Thank you so much --Ķĩřβȳ♥ŤįɱéØ 07:36, 22 August 2006 (UTC)

If y is the limit of the x power tower, then y = x^y. Putting z = 1/y, we have z^z = 1/x. In the article Lambert's W function we see a way of expressing a solution, giving for the limit y the formula –W(–log x)/log x. Using the programs given in that article it should not be hard to graph this. For the log to be defined, x must be positive. For the W to be defined, x must be at most exp(1/e), which is about 1.444667861. I have not looked at the other problems in detail, but think they can be solved by similar means. --Lambiam^Talk 08:43, 22 August 2006 (UTC)

Addendum. The x power tower only approaches a limit for exp(–e) ≤ x ≤ exp(1/e). Between 0 and that lower bound on x, the successive towers asymptotically approach a cycle of two values. Apparently, Euler has analyzed this problem.[1]  --Lambiam^Talk 17:26, 22 August 2006 (UTC)

The infinitely nested natural logarithm always converges to the constant 0.31813150... + (1.33723570...)i, unless x is 0, 1, e, or a tower of powers of e. The nested sine simply tends to 0 if x is a real number (I'm not sure for which complex numbers it converges). These are simple cases of fixed point iterations; the solutions to x = log(x) and x = sin(x) respectively (to solve the first of these, use the Lambert W function as Lambiam said). The power tower of logarithms, as you specified it, is just an ordinary power tower with a change of variables. Fredrik Johansson 08:58, 22 August 2006 (UTC)

Interestingly, cosine has a nontrivial fixed point. If I remember correctly sin's fixed point is zero. Dysprosia 02:39, 23 August 2006 (UTC)

"BooBoo" without repetitions

How many ways are there to arange the word BooBoo without repeating a certain order?

Please Explain how to do it. —The preceding unsigned comment was added by 70.22.207.77 (talk • contribs) 17:58, August 22, 2006 (UTC).

I do not understand what you mean by "repeating a certain order". Could you give some examples of arrangements of "BooBoo" that do have such repetitions, and some that do not? --Lambiam^Talk 21:08, 22 August 2006 (UTC)

It's impossible to arrange the letters of BooBoo without repetitions, because there are four os and only two Bs to separate them. Every possible arrangement has two adjacent os, so unless I've misinterpreted your quetsion, the answer is zero. —Keenan Pepper 21:18, 22 August 2006 (UTC)

The most obvious interpretation of this question is "how many UNIQUE ways are there to arrange all 6 letters in the word BooBoo ?" I would say that the first B can be in any one of 6 positions, and the second B can be in any of the remaining 5 positions, for a total of 6x5 = 30. However, this will include two copies of each case, so, we get 30/2 = 15. Here are all the possibilities:

 1) BBoooo
 2) BoBooo
 3) BooBoo
 4) BoooBo
 5) BooooB
 6) oBBooo
 7) oBoBoo
 8) oBooBo
 9) oBoooB
10) ooBBoo
11) ooBoBo
12) ooBooB
13) oooBBo
14) oooBoB
15) ooooBB

StuRat 23:33, 22 August 2006 (UTC)

If that is the problem, the next questions is: Is there a way to find the number without exhaustively listing all cases? Here is how to do it. First, pretend that all letters are distinct, as in <B₁ o₁ o₂ B₂ o₃ o₄>, in which they have been made different by appending a number. This is a sequence of 6 symbols, so there are 6! permutations, such as <o₃ B₁ o₂ B₂ o₄ o₁> and <o₃ B₂ o₂ B₁ o₄ o₁>. Actually, these two are the same if we forget the numbers on the B's. We can pair off the sequences of the form <... B₁ ... B₂ ...> and <... B₂ ... B₁ ...> that are transformed into each other by swapping B₁ and B₂. So, as far as the Bs are concerned, we have counted everything 2! times (where 2! is the number of ways to permute <B₁ B₂>). After identifying equal arrangements for the B's we have 6!/2! arrangements left, for example <o₃ B o₂ B o₄ o₁> and <o₂ B o₁ B o₄ o₃>. But these are also the same if we forget the numbers of the o's. This time the equal arrangements come in groups of 4! elements, the number of permutations of the four letters <o₁ o₂ o₃ o₄>. In the end, we have 6!/(2!4!) = 15 arrangements. Let's apply this to the word "abracadabra". It has 12 letters: 5 a's, 2 b's, 1 c, 1 d and 2 r's, so we find 11!/(5!2!1!1!2!) = 83160 arrangements in total. --Lambiam^Talk 01:26, 23 August 2006 (UTC)

I did find the answer without listing all possible cases. The list was only including as a way to prove that my answer was correct. With so few cases (15), it would seem silly to me to not bother checking my answer. StuRat 01:33, 23 August 2006 (UTC)

If you read on Combination with repetition in permutations and combinations, and also multiset, you can find a solutions to more general problems than this. In this case, the answer is ${4+3-1} \choose {4}$ (Igny 02:18, 23 August 2006 (UTC))

I can not see, why you need so complicated formula of permutations with repetitions. The problem can be expressed as simple combinations counting: Choose two-element subset of six-elements set (two positions for B out of six positions) - how many unique choices can you make? And the answer is ${6 \choose 2}$. --CiaPan 06:30, 23 August 2006 (UTC)

Did it not strike anyone answering as highly likely that this is a homework problem? We can help by asking questions like "What do you not understand?" rather than by displaying a worked-out answer. Homework or not, it's better instruction. --KSmrq^T 08:48, 23 August 2006 (UTC)

If homework, the assignment was phrased somewhat incomprehensibly. --Lambiam^Talk 02:38, 24 August 2006 (UTC)

Commonly, those who want others to do their homework also do not understand the assignment. Some post it verbatim, which tends to make it obvious that it's homework. Others mangle it. Consider:

A young guy in business school hears the following joke pass between two mathematicians, with great effect.

• Q. What is purple and commutes?
• A. An abelian grape!

Eager to impress the lovely mathematician of his dreams, he tries it on her.

• Q. What is purple and travels?

We are accustomed to using special language and stating things precisely. That's an acquired skill. --KSmrq^T 12:01, 24 August 2006 (UTC)

Category:Accuracy disputes

I know this is not the usual use of the reference desk, but I was looking over T-Z in this category and thought that people here could help clear up the accuracy of these specific articles.--Birgitte§β ʈ Talk 18:38, 22 August 2006 (UTC)

• Unsolved problems in mathematics
• Trajectory

You may want to take this to Wikipedia talk:WikiProject Mathematics. -- Meni Rosenfeld (talk) 20:26, 22 August 2006 (UTC)

Thanks for the idea. I will do that if I get to any further sorting.--Birgitte§β ʈ Talk 12:58, 23 August 2006 (UTC)