# Prove that if $f(x)$ is a polynomial of degree $n$, then $f(x+a)$ is also a polynomial of degree $n$

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.

• (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. Jun 1, 2022 at 2:40
• Heavy algebra??? You ain't seen nothing yet! Jun 1, 2022 at 2:41
• How about that $\frac {d}{dx} f(x+a) = \frac {d}{dx} f(x)$ and $\frac {d^{n+1}}{dx^{n+1}} f(x+a) = \frac {d^{n+1}}{dx^{n+1}} f(x) = 0$ Jun 1, 2022 at 2:41
• @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. Jun 1, 2022 at 2:46
• #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.
– dxiv
Jun 1, 2022 at 3:46

$$f(x) = \sum_{n=0}^{k}c_{n}x^{n}$$ $$f(x+a) = \sum_{n=0}^{k}c_{n}(x+a)^{n}$$ $$(x+a)^n = \sum_{l=0}^{n}b_{l}a^{n-l}x^{l}$$ $$f(x+a) = \sum_{n=0}^{k}c_{n}\sum_{l=0}^{n}{\gamma}_{l}x^{l}$$ gamma is the new constant since $$b_{l}a^{n-l}$$ is constant
$$\sum_{n=0}^{k}c_{n}\sum_{l=0}^{n}{\gamma}_{l}x^{l} = \sum_{n=0}^{k-1}c_n\sum_{l=0}^n{\gamma}_{l}x^{l} + c_{k}\sum_{l=0}^{k}{\gamma}_{l}x^{l}$$ $$c_{k}\sum_{l=0}^{k}{\gamma}_{l}x^{l} = \sum_{l=0}^{k}c_{k}{\gamma}_{l}x^{l} = \sum_{l=0}^{k}{\beta}_{l}x^{l}$$
$$f(x+a) = \sum_{l=0}^{k}{\beta}_{l}x^{l} + \sum_{l=0}^{k-1}{\beta}_{l}x^{l} + \sum_{l=0}^{k-2}{\beta}_{l}x^{l}.....$$ $$F(x+a) = \sum_{l=0}^{k}{\beta}_{l}x^{l}$$ which is a ploynomial