I don't understand the initial set up of the theorem. It seems equivalent to stating that any continuous function on [-1,+1] which vanishes at the origin can be uniformly approximated by a polynomial of the form x^(n+1) * Q(x). What's the purpose of the a_1, a_2, a_n ? a_0 is there to kill the value at 0 but as soon as you fix the other coefficients you have essentially just modified the underlying f to become some alternative function g that is continuous and vanishes at 0. Also in your proof I am confused that you would use the name q for what seems to be a different q than from the theorem statement.
Great question! The idea is that if we have a continuous function on [-1,1] then for any choice of a_1,…, a_n we can find a polynomial of the form f(0)+a_1x +…+ a_nx^n + higher order terms that uniformly approximate f! The f(0) and symmetric interval are needed.
@@mikethemathematician Thanks, I understand that : ) What I'm saying is that I don't understand why one introduces the a_1x + a_2x^2 + ... a_nx^n in the first place. You could just aswell absorb them into f. The remarkable part (as I understood it) is that you can approximate a continuous function that vanishes at 0 with a polynomial of the form x^(n+1) * q(x) i.e. that you don't need lower order terms.
I don't understand the initial set up of the theorem. It seems equivalent to stating that any continuous function on [-1,+1] which vanishes at the origin can be uniformly approximated by a polynomial of the form x^(n+1) * Q(x). What's the purpose of the a_1, a_2, a_n ? a_0 is there to kill the value at 0 but as soon as you fix the other coefficients you have essentially just modified the underlying f to become some alternative function g that is continuous and vanishes at 0.
Also in your proof I am confused that you would use the name q for what seems to be a different q than from the theorem statement.
These remarks aside it's a very nice result and somewhat surprising actually.
Great question! The idea is that if we have a continuous function on [-1,1] then for any choice of a_1,…, a_n we can find a polynomial of the form f(0)+a_1x +…+ a_nx^n + higher order terms that uniformly approximate f! The f(0) and symmetric interval are needed.
@@mikethemathematician Thanks, I understand that : ) What I'm saying is that I don't understand why one introduces the a_1x + a_2x^2 + ... a_nx^n in the first place. You could just aswell absorb them into f. The remarkable part (as I understood it) is that you can approximate a continuous function that vanishes at 0 with a polynomial of the form x^(n+1) * q(x) i.e. that you don't need lower order terms.