Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Infinity Solution


Date: 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/   
    
Associated Topics:
High School Number Theory

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/