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Here is the question:

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Also, we will use the notation $p_{n}: \mathbb{R} \rightarrow \mathbb{R}$ given by $p_{n}(x) = x^n$ for $n \in \mathbb{N}$ where we take $0 \in \mathbb{N}.$ Now, since $p_{1} = id_{\mathbb{R}},$ where I worked this problem:

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My questions are:

1- Is, in the 6th line of problem 6, it should be $\mathcal{C}$(actually I do not know how to write this type of C that is in the picture) instead of $C$?

2-Could anyone help me in solving problem 6?

Emptymind
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3 Answers3

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As far as I can see, line six does say $\mathcal{C}$, the set of functions under discussion. Anything else would be strange after the intro.

Note that $p_1(x)=x$, and closedness under products gives that $p_1 \cdot p_1 \in \mathcal{C}$ and this map is just $x \to x \cdot x=x^2$ so $p_2 \in \mathcal{C}$.

By induction $p_n(x) \in \mathcal{C}$ as $p_{n+1}=p_1 \cdot p_n$ for all $n$. As $\text{const}_c$ is in $\mathcal{C}$ as well, all functions of the form $x \to cx^n$ are in $\mathcal{C}$ too, by the product requirement.

Also show that $\mathcal{C}$ is closed under all finite sums (not just sums of two), by induction; a standard argument.

Then any polynomial $f$ can be written as $f(x)=c_0 + c_1x + c_2x^2 + \ldots + c_n x^n$ a finite sum of functions that we've already shown to be in $\mathcal{C}$ and so is in $\mathcal{C}$ by the sum condition and its finite sum extension.

Henno Brandsma
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  • @OP Beware Above is not a rigorous proof (use of ellipses is always a red flag when it comes to rigor). One natural way to fix (and simplify) this is described in my answer. Here the essence of the matter is the relationship between closure properties and induction so one should be very precise to avoid circular or incomplete arguments. – Bill Dubuque Mar 09 '20 at 17:12
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  1. It is most definitely a typo and should be $\mathcal{C}$. It tells us that constant functions are a part of the collection.
  2. A polynomial can be written as $P(x)=\sum_{k=1}^n a_k x^k$, and since it is closed under sums, you have to show that each $a_k x^k$ is in $\mathcal{C}$. $a_k \in \mathcal{C}$ by assumption 2, and $x^n \in \mathcal{C}$ by assumption 1 and 4. Thus $a_k x^k \in \mathcal{C}$ again by assumption 4, then by assumption 3, $P(x) \in \mathcal{C}$.
Dan Calzone
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$(6)$ has a natural rigorous proof by induction on degree. For the base case all zero degree polynomials (constants) are in $\cal C$ by hypothesis. Else $\deg f \ge 1$ hence by the polynomial division algorithm $\,f = q\, x + c\,$, with $\deg q < \deg f\,$ so $\,q\in\cal C\,$ by induction, hence $\,x = p_1\in\cal C\,$ and constant $\,c\in\cal C\,$ implies $\,q\,x+c = f\in\cal C,\,$ by $\,\cal C\,$ closed under multiplication and addition.

Remark $ $ Unwinding the induction essentially writes $f(x)$ in Horner form

$$ f(x) \,=\, f_0 + x (f_1 + x(f_2 + x(\cdots)))$$

It deserves to be better known that exponentiation by repeated squaring is simply a multiplicative analog.

One more general perspective that may prove instructive is structural induction and recursion.

Bill Dubuque
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