I often tell my students that Integration by Parts is Applied Math Trick #1.  I once had someone retort with “last year, you said Taylor Series was trick #1”.

Well, Taylor series is just really integration by parts… Forever! I was surprised that students hadn’t seen this before.  It’s remarkably simple, and yields the remainder theorem quite nicely.

Taylor Series via Integration by Parts

Beginning with the identity that

u(x) = u(0) + \int_{0}^x u'(s)\:ds, a quick integration by parts where dv = ds where we will let v = s - x will give

\displaystyle u(x) = u(0) + u'(s) (s-x)\bigg\vert_{0}^x -\int_{0}^x (s-x) u''(s)\:ds.

Repeating this process again, we find…

\displaystyle u(x) = u(0) + u'(0) x + \frac{1}{2}u''(0)x^2  + \int_{0}^x \frac{1}{2}(s-x)^2 u''(s)\:ds.

If you keep going, you can see precisely how this will all play out.  Furthermore, the error in truncating at some point is bounded by the integral!  A simple estimate of the integral gives you exactly what you expect in terms of bounds on the integral.


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