Wikipedia:Reference desk/Archives/Mathematics/2006 December 10

= December 10 =

cubic
What's the maximum value of a cubic function?


 * A cubic function has the form:


 * $$f(x)=ax^3+bx^2+cx+d\,$$


 * Therefore, the maxmum degree is 3, is that what you're asking ? StuRat 11:59, 10 December 2006 (UTC)

Well, the maximum value of, say,


 * $$f(x)=-(x^2-6x+8)\,$$

is 1 when x = 3, meaning the highest value of f is 1

I was wondering how to find the maximum value of a cubic. Or is it just positive infinity?


 * A cubic doesn't have a global maximum or minimum value - think about what happens when x is a very large positive number or a very large negative number. However, it may have local maximum or minimum values, which are stationary points. Sketch the graph of $$y=x^3-x$$ to see an example. Gandalf61 12:30, 10 December 2006 (UTC)


 * I posted a pic above, so you can see what we mean. And yes, the maximum is normally positive infinity (with possible exceptions for degenerate cases, such as where a = 0). StuRat 12:53, 10 December 2006 (UTC)


 * You can find the turning point - where the function changes direction from up to down or down to up (at about x=-3 and x=+1 on the pic) by setting the differential of the function to zero. Differential Equations has info on differentials. --h2g2bob 18:01, 10 December 2006 (UTC)
 * Should have read Gandalf61's post more carefully, that's more or less what was said already :-) A turning point is another name for a stationary point.
 * As for the very edges, what happens is that as x goes to infinity, y will also go to infinity; and as x goes to negative infinity, y will go to negative infinity. If the coefficient of x3 is negative, the sign will swap (ie: if y = -x3, then if x&rarr;+&infin; then y&rarr;-&infin; and if x&rarr;-&infin; then y&rarr;+&infin;) Technically you can't actually ever reach infinity, thats why it "goes to" or "tends to" infinity (see limit (mathematics)). If x ever reached infinity, y would also become infinity (substitute x=infinity and solve for y to prove this). The function will tend to a countable infinity.
 * The way to solve these is to think "what happens if x is really big" (or "really big negatively" for negative x). You only really have to worry about the highest order term - if there's an x3 term, you can ignore the x2 and x terms. Likewise, you only have to worry about the sign of the coefficient (if you go to infinity, then something times infinity is still infinity). --h2g2bob 18:49, 10 December 2006 (UTC)
 * What? Countability has nothing to do with this. Even if infinity is introduced to the number system, it is not much more than just an object called infinity. This is completely different from the cardinality of sets. And let's not forget that as long as we are working with real numbers, there's no such thing as substituting infinity in the equation. -- Meni Rosenfeld (talk) 21:21, 10 December 2006 (UTC)


 * The possibilities for a polynomial function are few and simple. As with any function, we first ask if we are given bounds, such as, "Find the maximum of f(x) for a &le; x &le; b." If so, then we might have a maximum at a boundary, or we might have one within the boundaries.
 * The unbounded behavior splits into three cases: constant, odd degree, even degree. The maximum of f(x) = c is c; done! Otherwise, let n be a positive integer, and let &lambda; be a non-zero real number. For odd degree, f(x) = &lambda;x2n−1+(lower-order terms), the behavior for large x is entirely controlled by &lambda;. If &lambda; is positive, then as x increases positively, eventually so will f(x), becoming as large as we like; and if &lambda; is negative, then this happens as x increases negatively. (An odd power of a number retains the sign.) Either way there is no maximum. For even degree, f(x) = &lambda;x2n+(lower-order terms), a maximum may exist, depending on the sign of &lambda;. If &lambda; is positive, then as x increases either positively or negatively, so does f(x), making a maximum impossible; but if &lambda; is negative, then a maximum is guaranteed. (An even power of a real number is never negative.)
 * As for where to find a maximum, that was discussed in a recent thread. And if the function is not polynomial, such as f(x) = 1&frasl;x, the possibilities are more delicate. It is even possible to have a function that cannot take on arbitrarily large values, yet that also does not achieve a maximum value; an example is f(x) = x2 /( x2+1 ). --KSmrqT 21:43, 10 December 2006 (UTC)

Bézier Curves
Hello, I've been looking at the Bézier curve article and have been drawing a few Bézier curves from points. (I've uploaded an image at [[Media:BezierCurve.gif]])

But is there a way, instead of getting the curve from the points, to get the points corresponding to the curve ? I mean, under what assumptions can any curve be represented as a Bezier curve ? I suppose for example that it's impossible to draw sharp points except at the end of the curve, but that you can simulate those points, maybe, with a higher degree curve and points placed quite far away... Is there any way I can get the points with a given parametric/implicit equation ? Otherwise, how can I get a good approximation ?

--Xedi 13:39, 10 December 2006 (UTC)


 * Hmm, I suppose stitching together multiple Bézier curves allows the sharp points. Any ideas to get close approximations ? --Xedi 17:55, 10 December 2006 (UTC)


 * Degree is a limiting factor. A parametric curve of degree n can be represented exactly by a Bezier curve of degree n. For a curve like y=sin(x) which has infinite degree then you will not be able to get an exact Bezier curve, but the aproximation will be as good as you need.
 * If you have a degree-n parametric representaion of the curve over a given interval (p(t),q(t)) t in [0,1], then you calculate the Bernstein polynomial for each polynomial. There is a faily well know algorithm to do this, see for example.
 * Alternatively if you are happy with a piecewise cubic aproximation, for a given segment, P0 and P3 will be the end points, P1 will lie along the tangent to the curve at P1 and it distance is related to the curvature of the curve at P0. Similar for P2. --Salix alba (talk) 11:46, 11 December 2006 (UTC)


 * That is a lot to answer, enough to fill a lecture series!


 * A Bézier curve is piece of a polynomial curve; that is x and y are polynomial functions of a parameter t. Such a function is infinitely differentiable, and will not have any exact sharp corners. We do have the flexibility to perfectly reproduce any polynomial curve, but that's more limiting that we like. If we allow rational Bézier curves, where we draw a curve in 3D then perspectively project to 2D, we can handle functions which are a ratio of polynomials. This allows us to handle any curve with a quadratic implicit equation, including circles, ellipses, and hyperbolas. However, algebraic geometry tells us that most implicit curves of higher degree have no parametric representation as a ratio of polynomial functions. This already applies to the so-called elliptic curves, cubics like y2 = x3−x+1&frasl;2. a
 * Thus we step up to B-spline curves (or sometimes non-uniform rational B-spline curves, NURBS). These piecewise (rational) polynomial curves allow us to join pieces either smoothly or sharply, and avoid a major problem of polynomial approximation, Runge's phenomenon. A practical illustration of the possibilities is visible in the letter shapes you see on this page. For example, the outline of the letter "a" is drawn with such curves. --KSmrqT 15:49, 11 December 2006 (UTC)


 * Okay, thanks a lot. --Xedi 21:14, 11 December 2006 (UTC)

rewrite using partial fractions
hello, I need help rewriting this

(s-1)/(s+3)

into this, possibly using partial fractions, but I can't figure it out.

(-1/3)/s + (4/9)/(s/3+1)

thanks, -Steve


 * Start from the second one and work back to the first. Then write it down as if you'd done it the right way round. That always worked for me... yandman  07:53, 11 December 2006 (UTC)


 * Or, if you're supposed to understand how to do those sorts of things ... first you need to write them down correctly because there's no way in hell what you've written there can be equal. I suspect the first fraction is meant to be (s-1)/s(s+3). To work this out, I'll just give you the first couple of steps (because I'm feeling just a tad too lazy to do it all here):

Let $$\frac{s-1}{s(s+3)} = \frac{a}{s} + \frac{b}{s+3}$$. Then a little rearranging gives $$s-1 = a(s+3) + b s\ $$, and it's just a case of finding a and b. Confusing Manifestation 10:29, 11 December 2006 (UTC)


 * yes you're rigyht, this is controls and there was a step change, hence the extra 1/s. Thanks for the help. -Steve

vector perpindicular
if A =5I+3J+2K AND B=4I+2J+K HOW DO I FIND TWO VECTORS PERPENDICULAR TO THEM —The preceding unsigned comment was added by Garrycp2p (talk • contribs) 14:03, 10 December 2006 (UTC).


 * Use the scalar product. Imagine the vector of coordinates xI+yJ+zK. You need 5x+3y+2z=0 and 4x+2y+z=0 (the vector must be perpendicular to both the vectors, so the scalar product must be 0). This gives you { y=-3x and z=2x } for example. Just pick two values of x, put them in the equations to get y and z, and write your name and class on the top of the sheet...  yandman  14:11, 10 December 2006 (UTC)


 * Well, that's one way of doing it. A far, far easier way would be to think of what other kind of product of vectors you may have been taught in class, and think about what properties the result of that product has in relation to the original two vectors. Confusing Manifestation 14:13, 10 December 2006 (UTC)
 * I was worried that he might not have seen that yet. I recall having studied vectorial products much (1 or 2 years) later than scalar ones. yandman  16:24, 10 December 2006 (UTC)


 * hi thank for the help both of ye. Con man will you explain your way better if its easier id like to know my brain is gone to mush for cramming for exams so any help is great - garrycp2p —The preceding unsigned comment was added by Garrycp2p (talk • contribs) 16:13, 10 December 2006 (UTC).


 * Well we're not supposed to answer homework questions here, but I'll give you a little more of a hint: do you know about the Cross product, aka the vector product? If you have two vectors in 3D Euclidean space (which you do), then what do you know about the cross product of the two, specifically, where does it point in relation to the two vectors? Confusing Manifestation 10:20, 11 December 2006 (UTC)


 * (As stated at the top of this page, please do not use ALL CAPS; also, sign your posts. Thanks.)
 * This question is awkward to answer for three reasons: (1) we don't give complete homework answers, (2) we don't know what you know, and (3) it's tricky to handle all the input possibilities. Most of us instantly think of the same well-known tool. However, you may not yet have learned that tool, and it has limitations.
 * The way you have presented your two vectors, we will assume you are working in three dimensions with real scalars. Therefore, at most three vectors can be mutually perpendicular. Since you want to extend a list of two to a list of four, each cannot be perpendicular to the other three; in fact, the four cannot be linearly independent.
 * A very general way to proceed is to construct an orthonormal basis for the whole space, partitioned into a basis for the span of the given vectors, and a basis for its orthogonal complement. Then we can make as many perpendicular vectors as we like by taking linear combinations of the complement basis. A robust algorithm for the construction is QR decomposition, with pivoting, of a matrix whose first columns are the given vectors and whose remaining columns form an identity matrix.
 * We do have special cases. In 2D, a perpendicular to nonzero vector (x,y) is (−y,x). In 3D, we typically use a cross product; however, it is dangerous to blindly trust it. If either input vector is zero, it fails. If the input vectors are not independent, it fails. And if the input vectors are nearly dependent, a numerical calculation may suffer "catastrophic cancellation" and fail. --KSmrqT 13:59, 11 December 2006 (UTC)

natural log integration
I can easily find the integral of ln(x) using int. by parts, but I am having trouble doing the same for integral of ln(x+2). Can anyone help? --68.126.3.68 21:47, 10 December 2006 (UTC)


 * Try substituting $$t=x+2$$. -- Meni Rosenfeld (talk) 22:00, 10 December 2006 (UTC)
 * If you know how to, use substitution : put t = x + 2. Otherwise you can just see that, as $$\frac{df(x+b)}{dx} = f'(x+b)$$ (because the derivative of a constant is 0), you can just say that if $$ F(x) = \int f(x) dx$$, then $$ F(x+b) = \int f(x+b) dx$$ --Xedi 22:01, 10 December 2006 (UTC)
 * Ok, Meni Rosenfeld was faster. --Xedi 22:02, 10 December 2006 (UTC)


 * To amplify slightly: Any time you see an integral with an expression in place of x, which otherwise looks like a known integral, a change of variables is a natural attack. Here you claim to know how to handle
 * $$ \int \ln x \, dx, $$
 * so it is reasonable to try the change of variables y = 3x−5 when presented with
 * $$ \int \ln (3x-5) \, dx . $$
 * (Note the different problem, to avoid doing homework.) If y is so defined, then x = (y+5)/3, and dx = 1&frasl;3dy. Thus the integral becomes
 * $$ \frac{1}{3}\int \ln y \, dy, $$
 * and the known answer does the rest. Three cautions: (1) Remember to reverse substitute to use x in the answer, (2) it is necessary to adjust the limits of integration in a definite integral, and (3) this attack may not give a known integral if the expression for y is complicated (because the complexity is moved to the dx substitution). --KSmrqT 14:33, 11 December 2006 (UTC)