Ignoring Higher Order Terms in the Taylor Expansion?

Question

I have consulted a few texts in error anaylsis and the general formulation of error propgation does a taylor expansion and ignores second order terms and up. Why? From my differetial equation days, the numerical methods usually ignored terms past three and up b/c they become zero. I would imgaine their contrabution would be small if not zero due to the repetition in dervatities.

Answer 1

I've found the following site to be the most useful in explaining the theory of Taylor Expansion:

http://pathfinder.scar.utoronto.ca/~dyer/csca57/book_P/node26.html

Answer 2

The formula says that: (u + v)^r = u^r + r u^(r-a million)v + [r(r-a million)/2] u^(r-2)v^2 + ... + [r(r-a million)...(r-ok+a million)/ok!] u^(r-ok)v^ok + ... you want the z to have effective integer exponents, so allow v=z^2, leaving u=-a million and r=a million/2 (-a million + z^2)^(a million/2) = (-a million)^(a million/2) + (a million/2) (-a million)^(-a million/2) (z^2)^a million + [(a million/2)(-a million/2) / 2! ] (-a million)^(-3/2) (z^2)^2 + ... each and every time period is composed of portion of (-a million)^(a million/2). i have were given faith that the answer is composed of figuring out on a similar root of (i) in each and every time period, yet i do now no longer understand a thanks to coach it...diverse than interior the first time period the position f(0) = i because of very actuality the right root of sqrt(-a million) is common. Factoring this out gives you: (z^2 - a million)^(a million/2) = i * [ a million - (a million/2) z^2 - (a million/8) z^2 - (a million/16) z^3 - (5/128) z^4 - ...] be unsleeping that alternation of alerts in (a million/2)(-a million/2)(-3/2)(-5/2)... cancels with the alternation of (-a million)^(a million/2 - ok).

Answer 3

Actually, you just answered your own question..
Consider:
when dx = .000001, dx^2 = .00000000001