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I am trying to prove an exercise from 'Chapter 1, Calculus Vol. I' by Apostol: If $f(x)$ is a polynomial of degree $n$, then show that $f(x+a)$ is also a polynomial of degree $n$.

I have outlined two proofs for this:

  1. Assume $f(x) = \sum_{k = 0}^{n}c_{k}x^{k}$, which gives us $f(x+a) = \sum_{k = 0}^{n}c_{k}(x+a)^{k}$. Using the binomial expansion formula, we can expand each term and show: $$f(x+a)=\sum_{k=0}^{n}\left(\sum_{i=k}^{n}{i \choose k}a^{i-k}c_i \right) x^{k}$$
  2. Argue that $f(x+a)$ is a linear translation of $f(x)$ along the $x$-axis and, graphically, the functions have the same form.

I am looking for a more elegant proof not involving such heavy algebra or arguments involving higher-order objects such as graphs.

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    $\begingroup$ (2) isn't a proof at all, really. I think (1) is the canonical proof, and hints at the fact that we should practice this sort of algebraic manipulation with summations until it doesn't seem heavy. $\endgroup$ Jun 1, 2022 at 2:40
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    $\begingroup$ Heavy algebra??? You ain't seen nothing yet! $\endgroup$ Jun 1, 2022 at 2:41
  • $\begingroup$ How about that $\frac {d}{dx} f(x+a) = \frac {d}{dx} f(x)$ and $\frac {d^{n+1}}{dx^{n+1}} f(x+a) = \frac {d^{n+1}}{dx^{n+1}} f(x) = 0$ $\endgroup$
    – user317176
    Jun 1, 2022 at 2:41
  • $\begingroup$ @DougM that is an interesting approach. The $(n+1)^{\text{th}}$ derivative shows that the degree of the function is $<=n$. However, I don't think we can equate the lower order derivatives. $\endgroup$
    – ananta
    Jun 1, 2022 at 2:46
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    $\begingroup$ #1 is the standard argument, and arguably the more general one. #2 is basically just hand-waving it as written, but since this is tagged real-analysis and assuming you know the fundamental theorem of algebra you could perhaps argue that $x_k$ is a (complex) root of $f(x)$ iff $x_k-a$ is a root of $f(x+a)$ and the two have the same multiplicity. The degree of a complex polynomial equals the number of roots counting multiplicities, so $f(x)$ and $f(x+a)$ have the same degree. That said, using FTA here is quite heavy-handed. $\endgroup$
    – dxiv
    Jun 1, 2022 at 3:46

1 Answer 1

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$$f(x) = \sum_{n=0}^{k}c_{n}x^{n}$$ $$f(x+a) = \sum_{n=0}^{k}c_{n}(x+a)^{n}$$ $$(x+a)^n = \sum_{l=0}^{n}b_{l}a^{n-l}x^{l}$$ $$f(x+a) = \sum_{n=0}^{k}c_{n}\sum_{l=0}^{n}{\gamma}_{l}x^{l}$$ gamma is the new constant since $b_{l}a^{n-l}$ is constant

rhs

$$\sum_{n=0}^{k}c_{n}\sum_{l=0}^{n}{\gamma}_{l}x^{l} = \sum_{n=0}^{k-1}c_n\sum_{l=0}^n{\gamma}_{l}x^{l} + c_{k}\sum_{l=0}^{k}{\gamma}_{l}x^{l}$$ $$c_{k}\sum_{l=0}^{k}{\gamma}_{l}x^{l} = \sum_{l=0}^{k}c_{k}{\gamma}_{l}x^{l} = \sum_{l=0}^{k}{\beta}_{l}x^{l}$$

following the same recursive solution

$$f(x+a) = \sum_{l=0}^{k}{\beta}_{l}x^{l} + \sum_{l=0}^{k-1}{\beta}_{l}x^{l} + \sum_{l=0}^{k-2}{\beta}_{l}x^{l}.....$$ $$F(x+a) = \sum_{l=0}^{k}{\beta}_{l}x^{l}$$ which is a ploynomial

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  • $\begingroup$ don't hesitate to ask if you still have doubts $\endgroup$ Mar 11, 2023 at 15:27

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