Sinus
Hey everybody!
Anybody knows how to prove that sin(x) < x for 0 < x < 1?
Thanx!
Sometimes movement is a result of a kick in the ass!
Well, just look at the function f(x)=x-sin(x) on that interval.
You would like to show that f(x) is always non-negative. So, note that f(0)=0, and then all you need to show is that the derivative of f on that interval is always positive (thus f is increasing on that interval).
You would like to show that f(x) is always non-negative. So, note that f(0)=0, and then all you need to show is that the derivative of f on that interval is always positive (thus f is increasing on that interval).
August 01, 2003 04:53 AM
If this is homework, as I suspect, he is probably not allowed to use derivation in the proof. There is quite a simple geometrical proof though.
You can write sin(x) as the following infinite sequence:
sin(x) = x - (1/(3!))*x^3 + (1/(5!))*x^5 - (1/(7!))*x^7 + (etc..)
If x>0 and x<1 the term x^3 is greater or equal to x^5.
Additionally (1/(3!)) is greater than (1/(5!)).
That is: subtracting (from x) a bigger number between 0..1 and adding a smaller number between 0..1 forces x to
be smaller than before these operations.
Thus: Computing (x2 = x - (1/(3!))*x^3 + (1/(5!))*x^5) with x in [0,1] yields (x2
sequence (with its higher powers).
Cheers,
Oliver
sin(x) = x - (1/(3!))*x^3 + (1/(5!))*x^5 - (1/(7!))*x^7 + (etc..)
If x>0 and x<1 the term x^3 is greater or equal to x^5.
Additionally (1/(3!)) is greater than (1/(5!)).
That is: subtracting (from x) a bigger number between 0..1 and adding a smaller number between 0..1 forces x to
be smaller than before these operations.
Thus: Computing (x2 = x - (1/(3!))*x^3 + (1/(5!))*x^5) with x in [0,1] yields (x2
sequence (with its higher powers).
Cheers,
Oliver
Thanx for the replies..
Can you please explain me what derivation is. (I live in Israel and I just does not know the right term for it in Hebrew.. Sorry).
2pilarsor:
Can you please explain me how did you get the following equation:
sin(x) = x - (1/(3!))*x^3 + (1/(5!))*x^5 - (1/(7!))*x^7 + (etc..)
Can you please explain me what derivation is. (I live in Israel and I just does not know the right term for it in Hebrew.. Sorry).
2pilarsor:
Can you please explain me how did you get the following equation:
sin(x) = x - (1/(3!))*x^3 + (1/(5!))*x^5 - (1/(7!))*x^7 + (etc..)
Sometimes movement is a result of a kick in the ass!
i think I have understood what derivation of f is:
it''s lim h->0 [f(x + h) - f(x)] / h?
If it''s this I am not allowed to you this..
AnonymousPoster said that there is a simple geometric proof..
Can anyone explain it to me?
Thanx guys..
it''s lim h->0 [f(x + h) - f(x)] / h?
If it''s this I am not allowed to you this..
AnonymousPoster said that there is a simple geometric proof..
Can anyone explain it to me?
Thanx guys..
Sometimes movement is a result of a kick in the ass!
> Can you please explain me how did you get the following equation:
> sin(x) = x - (1/(3!))*x^3 + (1/(5!))*x^5 - (1/(7!))*x^7 + (etc..)
That is the power series, that defines the sine-function.
You may look this up in any (good) mathbook.
Cheers,
Oliver
> sin(x) = x - (1/(3!))*x^3 + (1/(5!))*x^5 - (1/(7!))*x^7 + (etc..)
That is the power series, that defines the sine-function.
You may look this up in any (good) mathbook.
Cheers,
Oliver
Well, I don''t know how rigorous you want the proof to be. The geometric argument works well too. I guess I won''t give it away since it seems that this is a homework assigment, but draw a unit circle and look at some angle x (keeping in mind that an angle is just the length of the arc that it subtends on a unit circle). Then, look at the sin of the angle. It should be pretty clear from that.
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