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I have a very simple problem, the technique of which I want to apply to more difficult problems. Here is the simple example version:

Suppose we have four functions: $f_1(x)=\sin(x)$, $g_1(x)=0$, and $f_2(x)=\cos(x)$, $g_2(x)=0$. I want to show that there does not exist an $x\in\mathbb{R}$ for which $f_1(x)=g_1(x)$ and $f_2(x)=g_2(x)$ simultaneously. I proceed like this:

For a contradiction, suppose there exists an $x\in\mathbb{R}$ such that $f_1(x)=g_1(x)$ and $f_2(x)=g_2(x)$. Then multiplying the first expression by $\sin(x)$ and the second by $\cos(x)$ gives $$\sin^2(x)=0$$ and $$\cos^2(x)=0.$$ Now add the two above expressions to obtain $$\sin^2(x)+\cos^2(x)=0,$$ from which we conclude that $$1=0.$$ Since this is absurd then we have shown that no such $x\in\mathbb{R}$ exists and our proof is complete.

I want to be able to apply this method to more complicated functions $f_i\neq 0$ and $g_i\neq 0$. My question is this: (1) is the above technique correct for the simple example, and (2) what general tools might I use to solve such problems? Simultaneous equations spring to mind for linear $f$ and $g$, for example.

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So, your question is "how can I show that a given pair of functions has no simultaneous zeros ?" Is this fair? –  James S. Cook Sep 25 '12 at 16:56
    
No this is not my question. My first question asks whether my approach to the example problem is correct. My second question asks if anyone knows of any general tools or theories which might be of use to attack such problems. For example, for $f$ and $g$ linear I am aware we can frame the question as a simultaneous equation and solve that way. –  pbs Sep 25 '12 at 17:03
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2 Answers 2

up vote 4 down vote accepted

Your procedure was correct, and in a sense general. We have (in the reals) $f_1(x)=g_1(x)$ and $f_2(x)=g_2(x)$ for some $x$ if and only if there is an $x$ such that $H(x)=0$, where $$H(x)=(f_1(x)-g_1(x))^2+(f_2(x)-g_2(x))^2.$$ The same applies to any finite number of equations of the form $f_i(x)=g_i(x)$. This is because in general $a_1^2+a_2^2+\cdots+a_n^2=0$ iff all the $a_i$ are $0$.

Unfortunately, that leaves the problem of determining whether a possibly messy function $H(x)$ can ever be $0$.

And in certain cases at least, the strategy is not best possible. For example, if you have a set of simultaneous linear equations (in several variables), then the usual techniques of Linear Algebra can efficiently determine whether the system has a solution. Summing squares would be messy and unhelpful.

For a one variable but artificial example, consider the system $x^{66}+x^2+1=0$, $x^{17}-41x^{16}-2012x^{15}-3x-1=0$. It is clear if we glance at the system that there is no real solution, while summing squares and simplifying will leave us with an equation $H(x)=0$ that is not pleasant to analyze.

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This is very insightful, thanks. Is there a typo here... should we not have $\sum a_i^2=0$. –  pbs Sep 25 '12 at 17:14
    
@pbs: Thank you, this one is now fixed. Probably several more to go. –  André Nicolas Sep 25 '12 at 17:21
    
@pbs: To answer your question, no, there is no general theory. Sometimes we even replace a single equation by an equivalent system of "simpler" equations, in a sense going in the opposite direction to the one you described. By the way, the reduction you mentioned to sums of squares plays a modest but useful role in the "arithmetization" of logic. –  André Nicolas Sep 25 '12 at 17:28
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I think trying to prove that there are no common solutions to functions of any type might be a question that is too general.

In Regards to the second question:

You must have learned about solving linear equations, so maybe the next step is learning some techniques used to solved common zeros of polynomials. These can be done using a some pretty interesting tools like Groebner basis.

I reasonably accessible book is "Ideals, Varieties and Algorithms" by Cox, Little and O'shea.

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