# Why Does The Taylor Remainder Formula Work?

I've been studying calculus on my own and have come across Taylor series. It is very intuitive until I came across the remainder part of the formula where things got fuzzy. I understand why the remainder exists but not the mathematical description. Why is the value being plugged into the derivative of the remainder some number between $x$ and $a$? What is the connection to the mean value theorem? Is the remainder used to bound the function? Lastly and my most important question is what is the intuitive (for me most likely a geometrical approach would be great) for how this remainder is derived. I've spent a long time trying to figure it out for myself and looking online but it seems I'm missing something for there are virtually no questions being asked about this.

Thanks for you time, Jackson

• do you understand the mean value theorem, where $f(x) - f(a)$ equals the derivative evaluated somewhere between $x$ and $a?$ – zhw. May 18 '15 at 1:11
• Yes, I do understand it. – Jackson H May 18 '15 at 1:13
• Well then, that takes care of the "value being plugged into the derivative" part of your question. – zhw. May 18 '15 at 2:15

Perhaps not quite the way you are looking for, but:

You can derive Taylor's theorem with the integral form of the remainder by repeated integration by parts: $$f(x)-f(a) = \int_a^x f'(t) \, dt = \left[-(x-t)f'(t) \right]_a^x + \int_a^x (x-t) f''(t) \, dt \\ = (x-a)f'(a) + \int_a^x (x-t) f''(t) \, dt,$$ and so on, integrating the $(x-t)$ and differentiating the $f$ each time, to arrive at $$R_N = \int_a^x \frac{(x-t)^N}{N!} f^{(N+1)}(t) \, dt. \tag{1}$$ Interpretation for this is simply that integrating by parts in the other direction will give you back precisely $f(x) - f(a) - \dotsb - \frac{1}{N!}(x-a)^N f^{(N)}(a)$.

Now, we can get from (1) to the Lagrange and Cauchy forms of the remainder by using the Mean Value Theorem for Integrals, in the form:

Let $g,h$ be continuous, and $g>0$ on $(a,b)$. Then $\exists c \in (a,b)$ such that $$\int_a^b h(t) g(t) \, dt = h(c) \int_a^b g(t) \, dt.$$

(this is easy if you think about weighted averages and the usual Mean Value Theorem).

Applying this to (1) with $h=f$, $g(t)=(x-t)^N/N!$ gives $$R_N = f^{(N+1)}(c)\frac{(x-a)^{N+1}}{(N+1)!},$$ which is the Lagrange form of the remainder; using $h(t)=f(t)(x-t)^N/N!$, $g(t)=1$ gives $$R_N = f^{(N+1)}(c')\frac{(x-c')^N}{N!}(x-a),$$ which is the Cauchy form of the remainder.

The weighted averages mentioned above are a way to think about what we did here: we take an average of $\frac{(x-t)^N}{N!} f^{(N+1)}(t)$ over $[a,x]$, and use the MVT to equate this to a value of the function at a specific point: how much of the function we count as in the weighting affects what answer we obtain.

• Where does the divided by x-a part of the mean theorem go? – Jackson H May 20 '15 at 0:12
• Where in particular are you asking about? – Chappers May 20 '15 at 0:44
• The first line of f(x) -f (a). – Jackson H May 20 '15 at 0:52
• I think the answer to your question is that I am not using the MVT there: the fundamental theorem of calculus implies that $\int_a^x f'(t) \, dt = f(x)-f(a)$. – Chappers May 20 '15 at 1:58
• You know that $t$ is an antiderivative for $1$. So is $t-b$ for constant $b$, and if I choose $b=x$, I get something that vanishes at the upper limit, which is helpful if I want a series that depends only on $(x-a)$ and derivatives of $f$ at $a$. Therefore $\int_a^x 1 \cdot f'(t) \, dt = [ (t-x)f'(t) ]_{t=a}^x - \int_a^x (t-x)f''(t) \, dt$, and then I just absorb some of the extra $-$ signs to get all positive terms. – Chappers May 22 '15 at 0:18