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Suppose f(x) = g(x) for all real x, and both f and g are sufficiently nice (perhaps we might limit them to be polynomials or analytic functions). Can we always manipulate one (with algebraic transformations) into the other? Is the same true for functions with multiple arguments?

To say more precisely what I mean let me quote Arthur's comment:

interested in the expressions of functions, like having the two functions $f(x)=\frac{4x+2}{2}$ and $g(x)=2x+1$ and use algebra to transform one into the other

NPS
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  • If $f=g$, there is no need for manipulations; they are the same. Are you asking for manipulations of their definining expressions maybe? – Marc van Leeuwen Mar 17 '14 at 14:23
  • @Marc I believe he is asking if this equality can be extended to the complex plane. – Joel Mar 17 '14 at 14:24
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    Note that functions are defined by what they do on their arguments. So if, for any $x$ we have $f(x) = g(x)$, then $f = g$. I believe, however, that you are more interested in the expressions of functions, like having the two functions $f(x) = \frac{4x + 2}{2}$ and $g(x) = 2x + 1$ and use algebra to transform one into the other (very easy in this case, but still). Is that what you're asking? – Arthur Mar 17 '14 at 14:25
  • Yes, Arthur, exactly! – NPS Mar 17 '14 at 14:32

2 Answers2

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That depends on what kind of restrictions you put on the expressions defining $f$ and $g$, and what your exact definition of algebraic transformations is.

For example, $$ \lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^n = e^x = \lim_{n\to\infty}\sum_{i=0}^n \frac{x^i}{i!} $$ yet simply applications of algbraic identities won't prove the equivalence, since $$ \left(1 + \frac{x}{n}\right)^n \neq \sum_{i=0}^n\frac{x^i}{i!} \text{.} $$

On the other hand, for polynomials algebraic identifies are sufficient to prove or disprove $p(x) = q(x)$ - it's sufficient to fully expand both sides, and compare the coefficients.

If you restrict yourself to functions definable in the first-order language of real closed fields, and $\forall x\, f(x) = g(x)$ is true over $\mathbb{R}$, then you can prove that by using just the axioms for RCF (real close fields) and the rules of deduction of first-order logic. In other words, if $$ \mathbb{R} \models \forall x\, f(x) = g(x) $$ then $$ \textrm{RCF} \vdash \forall x\, f(x) = g(x) \text{.} $$ This works because RCF is complete, i.e. for every possible logical statement $\varphi$, RCF either proves $\varphi$ or its negation $\lnot \varphi$.

fgp
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  • I'm no expert on what can be defined in first-order language, but I am afraid it could be not so much beyond polynomials, since it seems like a rather algebraic language. For instance can trigonometric functions be defined in that language? – Marc van Leeuwen Mar 17 '14 at 14:48
  • @MarcvanLeeuwen $f$ being definable doesn't require that there be a first-order expression representing $f$, it's sufficient for an first-order formula $\varphi(x,y)$ to exists such that $f(x)=y$ exactly if $\mathbb{R} \models \varphi(x,y)$. And yeah, that means I cheated a bit when writing $\mathbb{R} \models \forall x, f(x) = g(x)$... – fgp Mar 17 '14 at 15:05
  • So I rephrase my question: does there exists a first-order formula $\varphi(x,y)$ in the language of real closed fields such that $\sin(x)=y$ exactly if $\mathbb{R} \models \varphi(x,y)$ – Marc van Leeuwen Mar 17 '14 at 15:19
  • @MarcvanLeeuwen I knew you'd ask that ;-) Unfortunately, I don't know :-(. My guess is no, because if $\sin$ was definable, all analytic function would probably be, and then $f \equiv g$ would be decidable for all analytic function, but maybe I'm missing something there... Sadly, I'm no expert on this either... – fgp Mar 17 '14 at 17:17
  • @MarcvanLeeuwen Upon further thought, no, $\sin x$ isn't definable. If it were, you could define $\pi$ as the smallest zero of $\sin x$ larger than $0$, and then the natural numbers as the $x \geq 0$ for which $\sin(\pi x) = 0$. But since RCF is recursively axiomatizable and complete, that would make the peano axioms recursively axiomatizable and complete, which contradicts the incompleteness theorem. – fgp Mar 17 '14 at 18:01
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I am assuming you are asking if this equality can be extended to the entire complex plane. The answer for polynomials and analytic functions is yes.

Polynomials of degree $n$ are completely determined by $n+1$ points. So if two polynomials agree on the real line, they must be the same polynomial.

This same idea can be extended to analytic functions. If the zeros of an analytic function have an accumulation point inside of its domain, then the function must be the zero function. Since, $f-g$ is an analytic function (if $f$ and $g$ are) then $f(x)-g(x)=0$ for all real $x$. Thus the function $f-g$ has an accumulation of zeros at every real number. Therefore, $f-g=0$ on the complex plane, and $f=g$.

Joel
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  • Hmm... now that I read your question again, this might not be the answer you are looking for exactly. Haven't had my coffee yet, but I believe this does address part of your question. – Joel Mar 17 '14 at 14:31