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How to derive sine series without Taylor's formula?

  1. Aug 5, 2008 #1
    1. The problem statement
    Derive the sine power series without using Taylor's formula.
     
  2. jcsd
  3. Aug 5, 2008 #2

    HallsofIvy

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  4. Aug 6, 2008 #3
    Well, this is what the question reads. Without using the calculus, is it possible to derive sine power series using geometry, etc.
     
  5. Aug 6, 2008 #4
    I think you need more information. Like what is given because I use the power series as a definition of what sine is. Is this a differential equation problem?
     
  6. Aug 6, 2008 #5

    HallsofIvy

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    My question was really, "why should I do it". As I am sure you are aware, you are expected to try it yourself first and show us what you have done.
     
  7. Aug 6, 2008 #6
    This is the only standalone-statement of the problem. Nothing else is given. No it's not a dfn eqn problem. The question's standard is upto 1st year undergraduate (including Real Analysis) only.
     
  8. Aug 6, 2008 #7
    Well problem is that I've no idea from where to start, my guesses are geometry, complex number or the geometric definition of sine. Actually it's a part of my report, my main subject was Calculus of Kerela(India), the ancient Indian people knew the series without Taylor's formula, so that's what I'm intended to write in my report but it's nowhere on the internet, any suggestions are welcome.
     
  9. Aug 6, 2008 #8

    Dick

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    You got me curious. You just aren't searching the internet widely enough. Add Madhava to your queries. There's plenty of stuff. It looks to me like he and his school invented the concept of a limit, yes, probably couched in a geometric language. But I don't think they did it in any fundamentally different way than it was done two centuries later in the West. As a side note, Archimedes was also doing "method of exhaustion" proofs (i.e. limits) circa 200BC. Interesting subject. Keep researching.
     
  10. Aug 8, 2008 #9

    Gib Z

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    Well I came up with this, but it uses some integral calculus. Should still apply though. It comes up with cosines series as well.

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    Start with the inequality [itex]\sin x \leq x[/itex] for all [itex]x\geq 0[/itex].
    Integrate both sides of the inequality from 0 to some positive value t. Then you should get an inequality for cos t. In turn integrate that between 0 and x, you get one for sine again. You should get what to do now. Formalize the proof with induction and the squeeze theorem to really impress your teacher.
     
    Last edited: Aug 8, 2008
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