Taylor

]]>Thanks — I’ve fixed it now.

]]>For the benefit of students who find this page, and as an acknowledgement, I wrote it out carefully here: https://brilliant.org/wiki/taylors-theorem-with-lagrange-remainder/

]]>Wow, thanks for sharing that!

]]>The blog post is already quite old, but I just found a very nice proof in my high school book ([1] from the ’80s in Flanders, Belgium)

It is clear that there exist a real M so that

f(b) = f(a) + f'(a) (b -a) + f”(a) (b-a)^2/2! + … + f^(n)(a) (b-a)^n / n! + M (b-a)^(n+1)

Now consider the function

g(x) = f(b) – [ f(x) + f'(x) (b -x) + f”(x) (b-x)^2/2! + … + f^(n)(x) (b-x)^n / n! + M (b-x)^(n+1) ]

Applying Rolle’s theorem on the function g(x) gives directly Lagrange’s form of the remainder:

g(a) = g(b) =0, and almost all terms cancel in the calculation of g'(x)

I think this is more transparent than the proof above. There is almost nothing to remember.

[1] Wiskundig Leerpakket, 5.1, p 261

]]>https://web.math.princeton.edu/~gunning/bk.ps. ]]>

The way I see it, is that the remainder is an integral over a simplex, and the “intuitive” Mean Value Theorem (for integrals) gives the value of the integrand at some point times the volume of the simplex. I admit that’s not a proof.

]]>That gives you the integral form of the remainder, which is not the Lagrange form.

]]>$f(x+h)=f(x)+\int_x^{x+h} f'(x_1) dx_1$

and iterate the process by writing $f'(x_1)$ in the same fashion. Maybe that’s what Lagrange himself did.

]]>You’re in good company. A number of real analysis textbooks have faulty “proofs” of the convergence of certain Taylor series (e.g. for $e^x$) which make the same mistake.

]]>Yes, you’re right. I didn’t notice that when I first wrote it down. But i later returned, discovered my lapse, and tried to correct it. It turns out that you can still prove it in the way I wrote down, though it becomes more technical and demands more attention.

Ultimately, I think it’s sufficiently annoying that I don’t advise teaching it at all. This last semester, I taught the way that Dr. Gowers mentions in this post.

]]>This is a good heuristic, but not a proof since the $c$ depends on $t$ and thus is not constant in the integration.

]]>To be precise, I think my old exam question (from about 14 years ago) went as follows.

Suppose that is continuous. Prove that there exist and such that .

Very few students appeared to understand what this question was asking. A lot of students assumed or tried to prove that the function actually **was** the function .

I think more students would have at least understood the question if I had used (or ) instead of . (Whether they would have then been able to solve the problem is another matter.)

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