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Why polynomial functions f(x)+g(x) is the same notation as (f+g)(x)? I've seen the sum of polynomials as f(x)+g(x) before, but never seen a notation as with a operator in a prenthesis as (f+g)(x). And author puts (f+g)(x) at the first.

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Source: Linear Algebra and Its Applications, Gareth Williams

Definition 8. Let X and Y be sets. A function from X to Y is a triple (f, X, Y), where f is a relation from X to Y satisfying
(a) Dom(f) = X.
(b) If (x, y)$\in f$ and (x, z) $\in f$, then y=z.

"We shall adhere to the custom of writing f: $X\space \rightarrow Y$ instead of (f, X, Y) and $y=f(x)$ instead of $(x,\space y) \in f$."

Source: Set Theory You-Feng Lin, Shwu-Yeng T.Lin

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    $\begingroup$ (f+g)(x) = f(x)+g(x) is the definition of the function (f+g). With this definition, polynomials form a vector space. $\endgroup$ – Paul Apr 13 '16 at 13:07
  • $\begingroup$ If you have trouble understanding this, consider $(f+g)$ as a name, e.g. you could define $h(x) = f(x) + g(x)$, only that you use symbol $(f+g)$ instead of $h$. The upside of this is that you need a new definition for any new name like $h$, but it is obvious what $(f+f)$ means or $(g+h)$ or even $(f + (f + g))$. $\endgroup$ – dtldarek Apr 13 '16 at 18:28
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Why polynomial functions $f(x)+g(x)$ is the same notation as $\mathsf (f+g\mathsf)(x)$?

Why not?   It is convenient and brief.   It lets us talk about the polynomial $f+g$ .

I've seen the sum of polynomials as $f(x)+g(x)$ before, but never saw a notation as with a operator in a parenthesis as $\mathsf(f+g\mathsf)(x)$. And author puts $\mathsf(f+g\mathsf)(x)$ at the first.

Well, the author is defining the notation; right there.   It just introduced you to it.   Shake hands and get to know it.   Now you will know what the author means when it is used in future.

What do you suppose $\mathsf(f+g+h\mathsf)(x)$ means?

For that matter $\mathsf (f\cdot g\mathsf)(x)$ should be introduced soon too.   Can you anticipate what that shall mean?

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  • $\begingroup$ As a side note, if this is a first linear algebra text it probably doesn't get into algebras. $f\cdot g$ is unlikely to be defined and if it is defined, it might mean an inner product via an integral or something. In any other intro to rigor text I would agree with you. $\endgroup$ – Mark S. Apr 13 '16 at 13:50
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This is very practical, as we indicates that f and g are being applied to the same element and it is the usual amount defined in the vector space of continuous functions.

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