can two lorenz curves (L(0)=0, L(1)=1, convex and increasing in (0,1) ) intersect at more than once in (0,1)?
450251
Yes, it is quite possible. First imagine such a curve and take a tangent line at an interior point. Now move that line up slightly so it intersects the original function at two nearby points. Finally let the line curve very slightly between those intersection points and join the resulting curve to (0,0) and (1,1) smoothly. Specific examples should be easy with this intuition.\n\nI'm guessing that any number of intersection points are possible. Just modify this proceedure between intersection points.
606971
Do they have the same slope?  If so:\n\n(Y2-B2)/X2 = (Y1-B1)/X1\n\nThis implies that they will either never intersect or be equal to each other. (ie, if B2=B1, they are the same, if B2=/=B1 they will never be equal.)\n\nOK, so you have updated the question.  Now if you multiply these two equations together you will notice that x will be 2nd degree and y will be second degree (ie, both will have a 2 as their largest exponent.)  Therefore, the graph will be one of a conic section.  Either and ellipse, a circle, a hyperbola, or a parabola depending upon the scalars a,b,c and A,B,C.  If you don't know how to tell when a conic section is a circle, ellipse, hyperbola, etc. then ask another question, or google it.  Hope this helps.
322514
m1*m2=-1\n\nhold for the slope of two intersecting lines
904043
One variable is related to another by a function involving constant terms and terms of 1st order or higher.
786224
Actually, kaksi_guy seems to be onto something. Unfortunately, not only does his proof have the flaw that you noticed, but even if that were rectified, he would still have only proved that f is zero at infinitely many points on the interval, not that f is zero on every point in the interval. Still, his post is the inspiration for what follows:\n\nFirst, we need to assume that g(x) is defined everywhere on the interval between the two points. We don't need to assume that g(x) is continuous, but if we allow a function that is undefined at even one point, such as g(x)=(x-3/x)/2 (which is undefined at zero), we may easily generate exceptions to the theorem (in this case, f(x)=x²-1).\n\nWith that out of the way, we can begin our proof. Suppose f(x) = 0 at two points a and b. By the extreme value theorem, f(x) has both a maximum and a minimum on the closed interval [a, b]. Let us denote the minimum value f(x) attains on [a, b] as min(f(x)) (and analogously for max(f(x))). Now, since f(a)=f(b)=0, min(f(x))≤0 and max(f(x))≥0. Suppose that min(f(x))<0. Then the minimum does not occur on the endpoints, so it must occur on a point c in the open interval (a, b). It is readily shown that this implies f'(c)=0. Therefore, from the original differential equation, f''(c) - f(c) = 0 and thus f''(c) = f(c). However, f(c)<0, so this implies the function is concave down at c, contradicting the fact that c is a minimum. Thus min(f(x))=0. Similarly, if max(f(x))>0, then there would be a point c on (a, b) which is a maximum, but is concave up, which is also a contradiction. Thus, min(f(x))=max(f(x))=0, and a function whose minimum and maximum are both the same is a constant function, in this case f(x)=0 on the interval [a, b]. Q.E.D.
1011579
since you claim both lines intersect,\n5+λ=2+μ      (in direction i, both lines reach the same point)\n3-2λ=-11-4μ (in direction j, both lines reach the same point)\nsolving the above,\nμ=-10 λ=-13\nWe can substitute λ and μ in the k direction, and verify whether both lines really intersect:\n-2+2(-13)=-28\n14+5(-10)=-36\n\nBoth lines do not intersect
93062
x^2 + x + a = 0 <--- This is an example of a second degree equation and I think you are asking how it would change with a?  If so the amount of curvature would change as the a changes.  (more open less open curve)
740766
It is not possble!\n\nConsider the Fibonacci sequence 0, 1, 1, 2, 3, 5, 8 ..... This is clearly unbounded as |Un| > |Un-1|\n\n1, x, 1 + x, 1 + 2x, 2 + 3x. 3 + 5x, 5 + 8x ......\n\nIf x = 0 this is the Fibonacci sequence starting at U1 and so is still unbounded\n\nThe general term of the above sequence is therefore Un+1 + x* Un (Un, Un+1 of the Fibonacci sequence)\nNow Un diverges and so xUn also diverges\nIf x > 0, the divergence is more rapid\nif x < 0 the sequence oscillates without bound\n\nNB 1, 0, 1, 0, 1, 0, 1, ... does not follow the rules set \n\nshould go 1, 0, 1, 1. 2, 3, 5, ......
187414
2^y = 64/y\n\n\nthis is the intersection of two curves:\nt=2^y\nand\nt=64/y\n\nthat is an exponential and an equilateral hyperbola.\nPlotting the graphs you can have an idea of the intersection (in this case you have the exact solution by plotting integer values, but in general you may need to solve by iterative numeric formulas, after finding graphically a rough estimate of the solution)
151877
Theorytical if the line are formed on a flat gemetrical plane they will not intersect.Or if two tangent in a in the same circle are parallel they will never intersect.\nThe fith postulate of Euclid tackled this question.Later mathematicians, like gauss proved that they could intersect. But this proof was based on a different idea=paralell line having curvatures! Of course they would intersect. So the geometry which based on curvatures was started and was called Non Euclid Geometry.\nActually this could have been proven alot fast with some common sense! In reality it is imposible to construct a perfect circle or a perfect straight line(this is my own postulate).   This is because we are not able to have a perfect resolution of measurement.\nNot matter how you try to measure two parallel line  the results is ether convergence or divergence of the two lines if you make the line long enough!So in practice the 5th postulate of Euclid did have a practical application.
187139
Yes (-1,0) and (1,-2) share the tangent line y=-x-1\n\nwhich y=-x-1 is tangent to the function y= f(x)=x^4-2x^2-x\n\nNote that (-1,0) and (1,-2) satisfy y=x^4-2x^2-x
991638
I don't believe it can.  This is an example of a unicursal problem and you can find some more details at the puzzle link in the reference below, you'll need to scroll down to puzzle number 51.\n\nFrom my days of O Level Maths I remember that we touched on these and I seem to recall that there was an equation to do with the number of vertices and nodes that a shape had which would determine if it was unicursal or not.\n\nBasically only unicursal shapes (those with a specific ratio of nodes and vertices) can be drawn without removing a pencil from the page and without tracing over an existing line.\n\nI'll leave further research to you.\n\nHave fun
150933
What do you call the line that intersects a circle in only 1 point?
1179245
This is an inevitable occurrence; there is actually a theorem by Paul Erdos (I forget who he was working with) that is applicable...  Erdos-Schekeres Theorem, perhaps.  It's the one where they proved that if you take a piece of paper, cover the bottom of a box with it, remove it, wad it up and throw it back into the box, at least one point will be exactly above the same place it was before...  It doesn't take a whole lot of imagination to see a definite isomorphism between the E-S theorem (or whichever one it is) and what you're talking about.\n\nANYHOW, what it boils down to, in your case, is that you will have an experience (whether in dream or in reality is not germain) and it will match a future one.  On the whole, you should expect that to happen once in a while--and if you have a wide enough breadth of experience, perhaps more often...\n\nI'm not sure that this addresses your question.  If it doesn't, here's a shorter answer: It is part of normal life; it isn't a problem; and we all share it to some degree or other.
503588
In theory it will never get there as every half-distance will take an hour, and the number of half-distances is infinite. This is Xeno's Paradox in disguise, I think.\n\nIn practice the train cannot change speed with  a few metres left to a crawl that would take an hour, and it would grind to a halt or be unable to maintain the regime.\n\nBut the series 1 + 1/2 + 1/4 + 1/8 + 1/16 ... converges on 2 but never actually makes it.
823012
Either one is possible. Consider y=x² (one critical value, which is not an inflection point) and y=x³ (one critical value, which IS an inflection point).
1177010
If the you have a local max/min at an interior point, , the partial derivatives must be zero.\nSince the sum of the second partials is zero, the point must be a saddle point, not a max/min.