Infinity SolutionDate: 04/12/2001 at 03:17:14 From: Mr. Chen Subject: Infinity puzzle I was reading a maths text on my way home today when I came across this section on quadratic equations: If you are to solve a quadratic equation in which the x^2 terms cancel out, you must include the solution (x = infinity) where necessary. E.g. Solve x^2 - 3x = x^2 + 1. Then in this case you obtain the answer x = -1/3 and x = infinity. So I was thinking: if this is really the case, what about the equation 1 + 2x = 3 + 2x? Technically speaking, the lines of the two equations don't intersect at all, so there shouldn't be any solution. But if we consider the question again, it appears that x = infinity is the solution. Of course a solution that is infinite is meaningless, but let's say x is replaced by tan x. The equation tan x = infinity can be solved, and gives the answer x = pi/2. So I extended this to reason that since this is possible, what about other equations? I tried considering the quadratic equation in which the x^2 terms don't cancel out. Let's say, x^2 + 2x = 3x^2 - x. The solution is obviously x = 0 and x = 3/2. But x = infinity can be the solution too, right? I mean, if you consider the graphs of the two functions on both sides of the equation, i.e. f(x) = x^2 + 2x and g(x)= 3x^2 - x, both f(x) and g(x) will tend to infinity when x tends to infinity as well. So technically speaking, does it mean that infinity is the universal solution of any equation?? I thought about it, but it appeared that such a speculation would bring about some problems. For example, let's say I'm to solve this quadratic equation: x^2 + 3x + 10 = 0. This equation has imaginary roots. However, if you adjust the equation a little to become x^2 = -3x - 10, the solution would be the point(s) of intersection between the graphs of the two functions f(x) = x^2 and g(x) = -3x - 10 but if you carried out the argument I did earlier (that both functions tend to infinity as x tends to infinity as well), you supposedly get the solution x = infinity. But looking back at the original equation, x^2 + 3x + 10 = 0, and substituting the value of x = infinity, I realised that I get the equation infinity = 0. What is going on? I'm sure there's a flaw in my logic somewhere, but I can't seem to figure it out. Date: 04/12/2001 at 12:36:51 From: Doctor Peterson Subject: Re: Infinity puzzle Hi, Mr. Chen. I don't agree with the text's statement in the first place: we don't consider infinity to be a solution to any equation, because it is not a valid number. For some purposes you can think of infinity as a solution (in the sense that, as the variable becomes larger, the equation comes closer to being true), but that is not a normal way to talk about solutions. But even if we accept this concept, you have to be careful about what it means. I think the claim that infinity is a solution of the given quadratic is based on the fact that, for large x, the squared terms will overwhelm the other terms, so that both sides are essentially equal. You can see that by dividing all terms by x^2: x^2 - 3x = x^2 + 1 1 - 3/x = 1 + 1/x^2 As x becomes large, this approaches 1 = 1, and becomes more nearly true. But the equation as given is really the same as -3x = 1 and this does NOT become true when x becomes infinite. The left side becomes infinite, and the right side does not. This is just one hint that working with infinity is not "safe"; things don't behave right. What's happening is that if x is infinite, we are subtracting infinity from both sides, and infinity minus infinity is indeterminate. Look at it one way and it's zero, another way and it's infinite. Now, take your next equation: 1 + 2x = 3 + 2x Here, as in the quadratic, the coefficients of the highest terms are the same, so when we divide by x we get an equation that remains finite at infinity, and becomes "true": 1/x + 2 = 3/x + 2 The fact that the lines are parallel, and therefore meet at the "point at infinity" (in some geometries) agrees with this. Now look at x^2 + 2x = 3x^2 - x Because the coefficients of x^2 are NOT the same, this does not become "true" when x approaches infinity: 1 + 2/x = 3 - 1/x becomes 1 = 3 when x is infinite. If you graph the two sides, the curves do not come close when x is large; one remains about 3 times the other. The two real solutions you get are the only solutions even if you include infinity. So your speculation that infinity is a solution of all equations is, as you found, nonsense. The whole claim that you can consider infinity to be a solution of a quadratic is highly questionable, and they should at least have (a) defined exactly what they mean, and (b) told you what they mean by "where necessary." I think they meant "when the coefficients of the highest order terms are the same, and infinity makes sense in the problem you are solving"; and even then I'm not sure this really makes sense. I'm very curious to know in what context this claim was made; perhaps it makes sense in connection with some particular type of problem, where it is not really the quadratic equation itself they are trying to solve. You'll want to look at our FAQ on dividing by zero, which raises some of the questions you have to consider if you try to treat infinity as a number: http://mathforum.org/dr.math/faq/faq.divideby0.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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