amoodarya
  • amoodarya
can someone prove this by geometry !
Mathematics
katieb
  • katieb
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amoodarya
  • amoodarya
\[x \rightarrow 0\\sinx \approx x-\frac{x^3}{3!}\]
ChillOut
  • ChillOut
Does a graphical approach for the taylor series work?
anonymous
  • anonymous
Not that it helps, but http://www.wolframalpha.com/input/?i=sinx%2C+x-x^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.

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amoodarya
  • amoodarya
"Concentrationalizing" thank you , but ...I am not at the beginning of studying math . I am phd candidate ....
Loser66
  • Loser66
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amoodarya
  • amoodarya
graphical approach for the tailor series rise from limit ,derivation not geometry
anonymous
  • anonymous
I have this feeling that you can modify the geometric argument that \(\dfrac{\sin x}{x}\to1\) as \(x\to0\)...
anonymous
  • anonymous
In 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
  • anonymous
Whether this modification is indeed possible, I'm not quite seeing it right away.
IrishBoy123
  • IrishBoy123
i think the answer is 'no'
Empty
  • Empty
I 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
  • anonymous
For \(x>0\), 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
anonymous
  • anonymous
Correction, \(x-\dfrac{x^3}{3!}\) isn't decreasing for \(0
anonymous
  • anonymous
*\(>\)
ganeshie8
  • ganeshie8
.

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