# How to convince myself that I can take n=1 here?

Hall
Homework Statement:
In the linear space ##P_n## of all real polynomials of degree ##\leq## ##n##, define
##
\langle f,g \rangle = \sum_{k=0}^{n} f \left(\frac{k}{n}\right) ~g\left( \frac{k}{n} \right)
##
If ##f(t)=t##, find all real polynomials ##g## orthogonal to ##f##.
Relevant Equations:
##\langle f(t)=t, g \rangle = \sum_{k=0}^{n} \frac{k}{n} ~g\left( \frac{k}{n} \right)##
The Homework Statement reads the question.

We have
$$\langle f,g \rangle = \sum_{k=0}^{n} f\left(\frac{k}{n}\right) ~g\left( \frac{k}{n} \right)$$
If ##f(t) = t##, we have degree of ##f## is ##1##, so, should I take ##n = 1## in the above inner product formula and proceed as follows
$$\langle f(t)=t, g \rangle = \sum_{k=0}^{1} \frac{k}{1} ~g\left( \frac{k}{1} \right)$$
$$\sum_{k=0}{1} \frac{k}{1} ~g\left( \frac{k}{1} \right) = 0$$
$$f(0) ~g(0) + f(1) ~g(1) = 0$$
$$g(1) = 0 ~~~\implies ~~ g(t) = a(1-t) ~~~~~~~~~~~~\text{for any real a} ?$$
I'm finding it hard to convince myself that I can take ##n=1## just because the degree of ##f## is 1, the ##n## given to us is the fixed degree upto which polynomials are allowed in our linear space ##P_n##.

If I don't take ##n=1## the thing would get a little messier:
$$\sum_{k=0}^{n} \frac{k}{n} g\left( \frac{k}{n} \right) = 0$$
$$\frac{1}{n} \sum_{k=1}^{n} k ~g\left(\frac{k}{n} \right) = 0$$
$$\sum_{k=1}^{n} k ~g\left(\frac{k}{n} \right) = 0$$
$$\text{One possible solution for g is}$$
$$g (t) = a~(t - 1/n) (t- 2/n) \cdots (t - n/n)$$

Do I have to take ##n=1## or not?

## Answers and Replies

Mentor
I'm finding it hard to convince myself that I can take n=1 just because the degree of f is 1, the n given to us is the fixed degree upto which polynomials are allowed in our linear space Pn.
I agree. It could be that n = 3, for instance, but the function f(t) = t would still belong to that space. I don't think you can assume that n = 1.