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amoodarya
 one year ago
can someone prove this by geometry !
amoodarya
 one year ago
can someone prove this by geometry !

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amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1\[x \rightarrow 0\\sinx \approx x\frac{x^3}{3!}\]

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.0Does a graphical approach for the taylor series work?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not that it helps, but http://www.wolframalpha.com/input/?i=sinx%2C+xx^3%2F6 But the interval of convergence for the sinx expansion is all real numbers, so being centered at 0 would make this approximation very accurate, even if it's only the 2nd term in the expansion.

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1"Concentrationalizing" thank you , but ...I am not at the beginning of studying math . I am phd candidate ....

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1graphical approach for the tailor series rise from limit ,derivation not geometry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have this feeling that you can modify the geometric argument that \(\dfrac{\sin x}{x}\to1\) as \(x\to0\)...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In case you're not familiar, the argument I'm referring to goes like this: dw:1437863206383:dw where \(m\angle AOD=x\). It follows that \[A_{\Delta OAB}\le A_\text{sector}\le A_{\Delta OCD}~~\implies~~\frac{1}{\cos x}\le\frac{x}{\sin x}\le\cos x\]and hence \(\dfrac{\sin x}{x}\to1\) as \(x\to0\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whether this modification is indeed possible, I'm not quite seeing it right away.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0i think the answer is 'no'

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I tried two things, they might help to consider in constructing something.dw:1437864683464:dw I was playing with the geometric relationship here, \[a^2+(x\frac{x^3}{3!})^2 \approx 1\] The other thing I was playing around with is in finding a very crude approximation to \(\pi\) \[0=\sin x \approx x\frac{x^3}{3!}\] \[0 \approx x(1\frac{x^2}{3!})\] \[\pi \approx \sqrt{3!}\] I'm still thinking about this, but I thought it'd be fun to share some ideas.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For \(x>0\), we have that \(x>x\dfrac{x^3}{3!}\). dw:1437863856850:dw One problem is that this establishes that \[\cos\left(x\frac{x^3}{3!}\right)\le\frac{x\dfrac{x^3}{3!}}{\sin\left(x\dfrac{x^3}{3!}\right)}\le\frac{1}{\cos\left(x\dfrac{x^3}{3!}\right)}\] At least for \(x\) near \(0\), say \(0<x<1\), the function \(x\dfrac{x^3}{3!}\) is decreasing, so \(\sin x\ge\sin\left(x\dfrac{x^3}{3!}\right)\). This means \[\frac{x\dfrac{x^3}{3!}}{\sin x}\le\frac{x\dfrac{x^3}{3!}}{\sin \left(x\dfrac{x^3}{3!}\right)}\]so it would seem that we only need to establish that the LHS is larger than \(\cos\left(x\dfrac{x^3}{3!}\right)\). Graphically, this appears to be true: http://www.wolframalpha.com/input/?i=Plot%5B%7BCos%5Bxx%5E3%2F6%5D%2C+%28xx%5E3%2F6%29%2FSin%5Bx%5D%2C%28xx%5E3%2F6%29%2FSin%5Bxx%5E3%2F6%5D%7D%2C%7Bx%2C1%2C1%7D%5D Whether this is a valid adaptation of the other argument, I'm not sure, just spitballing here.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correction, \(x\dfrac{x^3}{3!}\) isn't decreasing for \(0<x<1\); rather I meant to say that \(x<x\dfrac{x^3}{3!}\).
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