5 ms·

Loved the article, but there was this big jump between 1 - x^2/2! + x^4/4! - ... and cos x (and similarly with sin x). Why exactly are these equal?

• gohrt 13y ago
  https://en.wikipedia.org/wiki/Taylors_theorem https://en.wikipedia.org/wiki/Taylors_theorem
• Permit 13y ago
  gohrt linked an article, and although you could derive the identity for yourself using that information, the article doesn't contain the series you're looking for. You can find it here if you'd like: https://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions https://en.wikipedia.org/wiki/Taylor_series#List_of_Maclauri...
• lmm 13y ago
  In advanced mathematics it's common to define cos and sin by these series (and pi is defined as the smallest strictly positive x with sin x = 0). (Of course that just reduces the question to "why do certain geometrical identities match this sin function")
  • dnautics 13y ago
    or you could use MacLaurin polynomial series (Taylor series at zero)
    • dbaupp 13y ago
      I'm probably missing something, but that is the Taylor series at 0.
      • dnautics 13y ago
        Yeah I mean to say you can derive the sin and cosine as abstract functions from their geometrical properties, get the form of their derivatives, and then derive their Taylor series, instead of defining them as infinite series, and then showing that the function looks like their geometric equivalents.
• dnautics 13y ago
  I was also perturbed by the jump from the definition of e to the taylor expansion. I know how to get there the long way (define e first, derive properties of the exponential derivative, then construct the Taylor series), does anyone know a shortcut?
  • iTrollFreely 13y ago
    exactly. the dude has no idea what he's talking about, or didn't bother with even a cursory proofread. e^x as "x tends to infinity" is infinity.
  • lmkg 13y ago
    The Taylor series is actually the expansion of the limit in the line above. There's some trickery in proving that the limit converges, but you can derive one line from the other with some straightforward combinatorics.
  • betterunix 13y ago
    Here is another way (a bit informal): e = lim_{n->infinity} (1 + 1/n)^n Now, apply the binomial theorem: 1 + n * 1/n + n! / (2 (n-2)! n^2) + ... + n! / (m! (n - m!) n^m) + ... Now, for each m, we have this sequence: a_n = n! / (m! (n - m)! n^m) Which converges on 1/m!, so we are left with this: 1 + 1 + 1/2! + 1/3! + 1/4! + ...
    • dnautics 13y ago
      it's a little ugly, because you have to have some strong conditions to use associativity on infinite series (and I forget what they are off the top of my head). Of course, this is true for splitting up the e^x into cos(x) and sin(x) as well.
      • pmiller2 13y ago
        The series has to be absolutely convergent. That is, you can rearrange the terms of \sum_{i=0}^\infty a_n freely if and only if \sum_{i=0}^\infty |a_n| converges. See http://en.wikipedia.org/wiki/Riemann_series_theorem http://en.wikipedia.org/wiki/Riemann_series_theorem