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[edit] October 6

[edit] Solving polynomials by composition of functions

I have been searching on the internet for any information about the use of composition of functions in solving polynomials. I have independently discovered this technique and I doubt I'm the first. It probably goes by another name or notation I am unfamiliar with. Furthermore I can create several formulas depending on the degree of the polynomial that represent factorizations of the highest power of x. It is also possible to achieve linear translations this way, by using 1*n where n is the highest power of x. Quartic equations represent the first good use of this method since 4 is not prime. You can create a formula for all Quartics that can be rewritten as a composition of 2 Quadratics, and then test a specific Quartic to see if it fits. Though this method takes a fair amount of work it can allow you to use basic algebraic techniques to solve problems otherwise requiring advanced knowledge such as the Quartic formula or Chebychev radicals, etc. any help in locating equivalent work would be much appreciated thanks.

I don't recall having heard of such a technique. I am somewhat skeptic as this seems to imply the existence of solutions by radicals to every polynomial equation (which is impossible). Can you give an example of how this technique is used? -- Meni Rosenfeld (talk) 11:19, 6 October 2007 (UTC)

To solve a quartic this way won't I need one of the quadratics first? - which I don't see how to get.14:22, 6 October 2007 (UTC) —Preceding unsigned comment added by 87.102.23.214 (talk)

It can't solve all quartics, not a general solution, but it's benefit is that you only need high school level knowledge to do it. I haven't played around w/ the sandbox enough to give you an example yet, sorry. Does copy and paste from microsoft's equation editor work??? —Preceding unsigned comment added by 69.54.140.201 (talk) 22:12, 6 October 2007 (UTC)

I see where I must have confused you, "You can create a formula for all Quartics that can be rewritten as a composition of 2 Quadratics" I didn't actually mean ALL quartics, rather all of them that fit the form created by a composition of 2 quadratics. —Preceding unsigned comment added by 69.54.140.201 (talk) 22:19, 6 October 2007 (UTC)

Copy+Paste from Microsoft's equation editor wouldn't work. Help:Formula contains information regarding displaying formulae, but you can give your example with plain text if you are uncomfortable with TeX. In any case, if I understand your method correctly, solving for the coefficients of the quadratics would lead back to a quartic equation. -- Meni Rosenfeld (talk) 22:29, 6 October 2007 (UTC)

Well, do you know about Polynomial long division? If you know a binomial factor (e.g. (x-2)), you can divide it out and reduce the degree of your polynomial by one (or if you know a quadratic that is a factor, by two, etc.). The thing is, in order to do this division, you need to know the factor beforehand, which comes back to the same problem you started with. There are a great many polynomials that can be easily factored. However, this is not often useful in practical applications of polynomial-solvers, e.g. in computer simulations the polynomial coming in is not known beforehand and the coefficients may be emperical measurements; a technique that doesn't work for the general case is likely to fail in this situation. - Rainwarrior 23:10, 6 October 2007 (UTC)

I'm assuming here that by "composition" you mean product, and not a composite function. I could be mistaken. - Rainwarrior 23:13, 6 October 2007 (UTC)

You still don't fully understand what I am suggesting here, (I know for sure this method works, I have tested it). I just looked at the help file so I'll try and give you an example.

Consider the function $f (g (x))$ , if we let $f (x) = x 2 + j x + k$ and $g (x) = x 2 + m x + n$ then $f (g (x))$ is a quartic. Do you recall how to solve for x in a composition of functions?? Well it is done by setting $f (y) = 0$ and solving for $y$ , where $y = g (x)$ . then set $g (x)$ equal to the solutions of $f (y)$ and solve for $x$ . So lets calculate the formula that discribe the Quartics the fit this form. Note, I have assumed that you have divided the leading term out first.

$f (g (x)) = (x 2 + m x + n) 2 + j (x 2 + m x + n) + k$

$f (g (x)) = (x 4 + 2 m x 3 + (m 2 + 2 n) x 2 + 2 m n x + n 2) + (j x 2 + j m x + j n) + (k)$

Drop parentheses and add.

$f (g (x)) = x 4 + 2 m x 3 + (m 2 + 2 n + j) x 2 + (2 m n + j m) x + (n 2 + j n + k)$

See now where the quartic comes from?? and that quartic is what we want to solve. So we let $f (g (x)) = 0$ and apply the method I described above. As long as a Quartic in question can fit the $f (g (x))$ equation above it can be solved this way, not all Quartics can (I believe). So lets find the general solution for the 'fitting' Quartics.

$f (y) = 0$

$y 2 + j y + k = 0$

And by the Quadratic formula.

$y=\frac{-j \pm \sqrt{j^2-4k}}{2}$

Now before things get confusing let me clarify. Note that $f$ is the last function to act on x, for this reason it is the first equation to solve. And it's solutions are the required outputs of $g (x)$ therefore we must set $g (x)$ not equal to 0, but equal to the solutions of $f$ .

$x^2+mx+n=\frac{-j \pm \sqrt{j^2-4k}}{2}$

Keep in mind that the right side represents two distinct values in most cases, and certainly in the general case. This means that the $\pm$ 's are independent and you must account for all possible combinations, (++,+-,-+,--) Which yields (of course) 4 solutions. Also note that the right side is a constant and therefore can be subtracted from both sides and grouped with $n$ .

$x^2+mx+(n-\frac{-j \pm \sqrt{j^2-4k}}{2})=0$

Once again we will use the quadratic formula.

$x=\frac{-m \pm \sqrt{m^2-4n+\frac{-j \pm \sqrt{j^2-4k}}{2}}}{2}$ Preceding As you can see, it can produce radical solutions not unlike the Quartic formula but the lack the presence of any Cube roots indicating that this is clearly not a general solution. It does however allow someone with only high school Algebra knowledge to solve 4th degree polynomials in cases not otherwise possible. And you don't have to have any prior knowledge of any solutions of the Quartic in order to solve it. It should be noted that one must be able to solve a linear system in two varibles in order to use this method. This is because the $x 2$ and $x$ terms both have $j$ and $n$ which means to determine both of their values when 'fitting' a Quartic to the formula. The last is handled by $k$ . — unsigned comment added by 69.54.140.201 (talk) 23:52, 6 October 2007 (UTC)

There is an error in the coefficient for the linear term in the expanded quartic; the formula should be

f(g(x)) = x⁴ + 2mx³ + (m²+2n+j)x² + (2mn + jm)x + n² + jn + k.

Not all quartic functions can be expressed as the composition of two quadratic ones. To apply this to a quartic of the form x⁴ + ax³ + bx² + cx + d, we have to find unknowns j, k, m and n from the equations resulting from equating the coefficients:

a = 2m

b = m²+2n+j

c = 2mn + jm

d = n² + jn + k

This means that a, b and c have to satisfy the relationship:

a³ − 4ab + 8c = 0.

If so, the unknowns can be found (in fact, we can then always choose n = 0) and we can find the roots of the quartic by repeated application of quadratic root finding as shown above. For example, your method works beautifully on x⁴ + 10x³ + 35x² + 50x + 24. But if that relationship between a, b and c does not hold, the system of equations has no solutions, so a decomposition of the quartic function into quadratic functions is then not possible. --Lambiam 04:50, 7 October 2007 (UTC)

An addition to my post. The condition on the three coefficients a, b and c can be reformulated as a condition on the multiset of roots of the quartic. For the quartic to be expressible as a composition of quadratic functions, you need that r₁ + r₄ = r₂ + r₃ for some permutation [r₁, r₂, r₃, r₄] of the four roots. This condition is fulfilled, for example, when the quartic function is symmetric about some axis. The example I gave above is actually symmetric about x = −⁵/₂. --Lambiam 12:45, 7 October 2007 (UTC)

I want to comment that your phrasing gives the impression that you assume some incompetence on our part (e.g. "Do you recall how to solve for x in a composition of functions?", as if we need to be reminded). That was unnecessary; I was only expressing skepticism regarding a method which you have described only vaguely, and Rainwarrior only suggested a different interpretation to what would otherwise be a somewhat unusual method.

Finding the coefficients does turn out to be easier than I had originally suspected (I haven't done the actual calculations then), so as Lambiam has commented, the method works well when it does (which is only rarely). -- Meni Rosenfeld (talk) 12:26, 7 October 2007 (UTC)

By the way, the required condition can be expressed in a neat way in terms of the roots of the polynomial; the roots need to come in pairs which have the same difference (for example, it will work for

(x - 1)(x - 2)(x - 6)(x - 7)

, since

2 - 1 = 7 - 6

I wonder how well this method will do for a polynomial of degree, say, 6 (too lazy to carry out any calculations though). -- Meni Rosenfeld (talk) 12:36, 7 October 2007 (UTC)

It may be a little unfair to bring up the phrasology the poster used - this style of language is quite polite in some parts of the world - I'm sure no sarcasm was intended.

With regards to the original question - I think the title used "Solving (certain) quartics by composition of functions" would be a name for it - I wonder if the question asker wanted to know if the method had ever been given a name after the original discover of the method - which I imagine is not likely since the method looks (on the whole) technically useless.87.102.115.31 13:59, 7 October 2007 (UTC)

Never said it was impolite. -- Meni Rosenfeld (talk) 16:05, 7 October 2007 (UTC)

Wasn't trying to be insulting, just through in addressing your understandable clarification questions, and thanks for pointing out the error, I have edited in the necessary changes. In the case of a 6th degree polynomial there are two distinct orders of composition of 2nd degree and 3rd degree polynomials, resulting in two formulas. My original question remains open however, has anyone seen similar work to the kind I have now described in detail? If not, this does present the unlikely possibility that this is new work. (One of reasons I'm asking) I think the best feature of this method I have independently discovered is that in general elementry techniques can be used, at least on polynomials who's $n$ is a power of 2. The case of a 6th degree polynomial present the necessity that the Cubic formula be known for most fitting cases to be solvable, especially when the cubic is the inside function (1st to act on x). I should also point out that in the case of an 8th degree polynomial there are three useful substitutions, 4 of 2, 2 of 4, and 2 of 2 of 2. Yes this method can be extended to three or more composits. I think I mentioned the value of my method being that it allows the solving of Quartics and other higher degree polynomials without the need for more advanced methods in 'fitting' cases. In general it seems using composite functions to solve other functions has been rarely attempted or overlooked, at least from my mathematical education so far... I doubt this method is of any real value to applied mathematics but it should be kept in mind when takling theoretical problems, as the form of the solutions is substancially simpler than the Quartic formula. —Preceding unsigned comment added by 69.54.140.201 (talk) 13:54, 7 October 2007 (UTC)

It has occured to me that what we have discussed here is just a generalization of a biquadratic equation, of the form

x 4 + a x 2 + b = 0

, which is solved using

t = x 2

. The method will work (for

n = 4

) exactly when the equation has the more general form

(x - p) 4 + a (x - p) 2 + b = 0

; the substitution

t = (x - p) 2

then solves it. Finding the a, b and p requires steps similar to what we have discussed, but with a slightly different methodology. p is the point where the third derivative is 0; The method will work if the first derivative is also 0 at that point; b is the value of the function at that point, and a is half the second derivative at that point.

In short, a much more restricted version of the method (for

n = 4

and

p = 0

) is already well known. The more I think about it, the more it seems that a more general discussion deserves a mention in the literature; I personally have not encountered it. -- Meni Rosenfeld (talk) 16:05, 7 October 2007 (UTC)

Your right it is similar to a Biquadratic function. I would like to point out that if you write your expression above as $((x - p) 2) 2 + a ((x - p) 2) + b = 0$ it is almost exactly what I have here save that the inside quadratic need not be a perfect square. I agree that it should get a more through mention in the literature as well. By the way, I found and corrected another error in my work, when I subtracted the first solution from both sides of the second equation i solved I forgot to switch the sign. —Preceding unsigned comment added by 69.54.140.201 (talk) 23:29, 7 October 2007 (UTC)

I should note that the use of Composite Functions is much more useful with exponential functions with complicated exponents. A basic example would be

$Y=e^{x^2-5x-7}-3$

the appropiate functions being.

$f (x) = e x - 3$ and

$g (x) = x 2 - 5 x - 7$

$f (g (x)) = Y$

$f (y) = e y - 3$

$e y - 3 = 0$

$e y = 3$

$\operatorname{ln} e^y= \operatorname{ln} 3$

$y= \operatorname{ln} 3$

$x^2-5x-7= \operatorname{ln} 3$

$x^2-5x-(7+ \operatorname{ln} 3)$

$x=\frac{5 \pm \sqrt{25+28+4 \operatorname{ln} 3}}{2}$

$x=\frac{5 \pm \sqrt{53+4 \operatorname{ln} 3}}{2}$

You could also set the original problem equal to zero and move the constant over, then $\operatorname{ln}$ both sides and work from there. —Preceding unsigned comment added by 69.54.140.201 (talk) 05:53, 8 October 2007 (UTC)

Note that once again, it is not necessary to think in terms of composite functions, and I find that it just complicates the matter. If we want to solve $e^{x^2-5x-7}-3=0$ , we can do so in a straightforward way:

$e^{x^2-5x-7}-3=0$

$e^{x^2-5x-7}=3$

$x^2-5x-7=\ln3\;\!$

$x^2-5x-(7+\ln3)=0\;\!$

$x=\frac{5 \pm \sqrt{53+4\ln3}}{2}$

By the way, note the calculation error you had. -- Meni Rosenfeld (talk) 13:04, 8 October 2007 (UTC)

See Galois theory. Bo Jacoby 16:55, 10 October 2007 (UTC).

Come to think of it, the composition technique is identical to substitution only, it bothers to more formally define the substitution. The one useful property is that you can work out general substitution cases to create short cut formulas.

[edit] using integration to find the lateral surface area of a cone

I'm trying to prepare for SAT II Math level 2, but I find a lot of stuff I haven't studied for 2 years since I'm in calc ... so I figure it would be best to try to rely on calc rather than memorise everything again. I'm stuck on the lateral surface area of a cone, which I know is equal to pi times its slant height times the radius. It's also equal to a Riemann sum of its circumference integrated along its height, right? So if the height is known, we can relate the radius to the height, which can be our dependent variable. So the ratio of the radius to the height can be expressed as a constant k, and the radius is h*k. So the area of each slice is 2*pi*r, or pi*hk. We can let our origin be the tip of the cone, and the height 12 and the radius 5 (one of the example problems I was doing). So we integrate this this from 0 to 12, and end up with 60pi? According to the formula though I should end up with 65pi .... 74.65.173.180 10:23, 6 October 2007 (UTC)

comment removed, was totally wrong. Ceroklis 16:07, 6 October 2007 (UTC)

I'm not sure who is confused more than who. The questioner is entirely correct in stating that the lateral surface area is equal to pi times its slant height times the radius. Slant height is not the same as (just) height, the usual meaning of the symbol h in this context. The radius r, height h and slant height s of a cone are related by the formula of the Pythagorean theorem: r² + h² = s². If r = 5 and h = 12, we find a nice integral value for s.

Integrating the area of the slices along the height will give you the volume of the cone. Perhaps the questioner meant something like the "lateral area" of a slice; however, that is not a directly meaningful or helpful concept. Integrating the circumference of the slices, as is done above, does not result in the correct value for the area of the lateral surface. The problem is that this gives, as it were, the area as seen by an observer at a straight angle to the cone axis, and not at a straight angle to the surface. Therefore the image undergoes foreshortening, in fact by a factor of h/s.

The easiest way of deriving the formula for the lateral surface is to cut the cone open along a straight line to the top and to flatten it, giving a circle segment. This derivation involves a bit of hand waving, because, strictly speaking, it requires a proof that the area does not change as we flatten the shape. --Lambiam 14:55, 6 October 2007 (UTC)

Finding the surface area of something with integrals involves arc length:

$\int_a^b 2\pi f(x)\sqrt{1+f'(x)^2}dx$

We let the function start from the point (0,r) and end at (h,0). The distance between these two points is the slant height s. We are given r=5 and h=12. r/h also happens to be the negative of the slope of the line passing through the two points we gave. So we create a function

$f(x) = -\frac{r}{h}x + r$

This function, when graphed, goes through those two points we made. Now, here's where the fun begins.

Suppose we rotate f(x) about the x-axis. What do we get? a cone. Now let's find the surface area of revolution using the formula above:

$\int_0^h 2\pi (-\frac{r}{h}x + r)(\sqrt{1+(-\frac{r}{h})^2})dx$

Let's pull out the constants:

$2\pi (\sqrt{1+(-\frac{r}{h})^2}) \int_0^h (-\frac{r}{h}x + r)dx$

Now, to solve the definite integral:

$\int_0^h (-\frac{r}{h}x + r)dx = -\frac{r}{2h}h^2 + rh - 0$

What do we do now? Well, now we need to multiply this by the constant:

$2\pi (\sqrt{1+(-\frac{r}{h})^2})(-\frac{r}{2h}h^2 + rh)$

Let's simplify a bit:

$2\pi (\sqrt{1+\frac{r^2}{h^2}})(-\frac{rh}{2} + rh)$

How to simplify this? Well, first let's make that radical look nicer. Using the identity $1 = \frac{h^2}{h^2}$ , we can rewrite the inside of that radical as:

$\sqrt{\frac{h^2 + r^2}{h^2}}$

Now, $\frac{1}{h^2}$ is a perfect square. We can remove it from inside the radical, leaving us with:

$\frac{2\pi}{h} (\sqrt{h^2 + r^2})(-\frac{rh}{2} + rh)$

But what is that radical? Well, this is where the Pythagorean Theorem comes in. Remember that a^2 + b^2 = c^2. So take the square root of both sides:

$\sqrt{a^2 + b^2} = c$

Now we let a=h and b=r, now we need to find c. Well, looking up, remember the slant height s? Well, c=s. So let's replace it:

$\frac{2\pi s}{h}(-\frac{rh}{2} + rh)$

So let's now simplify the second half, rewriting it:

$(rh -\frac{1}{2}rh) = \frac{rh}{2}$

So now we have:

$\frac{2\pi s}{h}\frac{rh}{2}$

Simplifies to:

$\pi r s \!$

and there you have it.--Mostargue 19:02, 6 October 2007 (UTC)

The formula used above can be found (with a slightly different notation) in Surface of revolution. --Lambiam 20:40, 6 October 2007 (UTC)

[edit] ʃ 1/x dx

So, the power rule states:

$\frac{d}{dx} x^n = nx^{n-1}.$

This can be rewritten as:

$\int x^{n} dx = \frac{x^{n+1}}{n+1} + C$ for n ≠ -1

If n = -1, then it becomes ln x. So my question is, how come:

$\lim_{n\to-1} \int x^{n} dx \neq \frac{x^{n+1}}{n+1} + C$

also, what is the TEX for the inequal sign --Mostargue 13:24, 6 October 2007 (UTC)

You can use \neq or \not= for the unequal sign. The latter, putting \not before an operator, is generic; for example, \not< produces $\not<$ .

That the limit above can't equal the right-hand side should be obvious. The variable n is bound on the left hand side, so that we can equivalently replace the whole expression by:

$\lim_{i\to-1} \int x^i dx,$

while on the right-hand side the variable n is free.

However, the following equality holds for positive x:

$\lim_{n\to-1} \int_1^x z^{n} dz = \lim_{n\to-1} \frac{x^{n+1}-1}{n+1}.$

--Lambiam 14:30, 6 October 2007 (UTC)

Right, that's what I meant. There should be a lim on both sides. So $\lim_{n\to-1} \frac{{|x|}^{n+1}-1}{n+1} = \ln{|x|}$ ?--Mostargue 15:01, 6 October 2007 (UTC)

Both numerator and denominator of that expression tend to zero as n tends to -1.

L'Hôpital's rule says that in this case,

$\lim_{x\to c} \frac{f(x)}{g(x)} = \lim_{x\to c} \frac{f\prime(x)}{g\prime(x)}$ .

In this case, we have;

$f(n) = x^{(n+1)} - 1\,$ and $g(n) = n+1\,$

We can rewrite f(n) as;

$f(n) = \exp[(n+1) \ln x] = \exp[(n \ln x + \ln x)]\,$

$f'(n) = \exp [(n+1)\ln x].(\ln x)\,$ - and $f'(-1) = \exp(0).\ln x\,$ , i.e. $f'(-1) = \ln x\,$

Also, $g(n) = n+1\,$ , so $g'(n) = 1\,$

So we have the result;

$\lim_{n\to-1} \frac{f(n)}{g(n)} = \lim_{n\to-1} \frac{f\prime(n)}{g\prime(n)} = \ln x$ Richard B 15:45, 6 October 2007 (UTC)

The point is, yes, $\lim_{n\to-1} \frac{{|x|}^{n+1}-1}{n+1} = \ln{|x|}$ . Here's an alternate derivation:

Definition of the exponential function: $e^x = \lim_{m\to \infty}\left(1+\frac{x}{m}\right)^m$

Corresponding definition of the natural logarithm: $\ln x = \lim_{m\to\infty} m(\sqrt[m]{x} - 1)$ .

Substitute $p = 1/m\!\,$ to get $\ln x = \lim_{p\to 0} \frac{x^p - 1}{p}$ .

Substitute $n = p - 1\!\,$ to get $\ln x = \lim_{n\to -1} \frac{x^{n+1} - 1}{n+1}$ .

Jim 18:05, 6 October 2007 (UTC)

While one may choose to define the natural logarithm this way, it is a definition I've never seen before, and I think one should actually first prove that it is equivalent to a more common definition before using it. --Lambiam 23:36, 6 October 2007 (UTC)