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amoodarya

  • one year ago

can someone prove this by geometry !

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  1. amoodarya
    • one year ago
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    \[x \rightarrow 0\\sinx \approx x-\frac{x^3}{3!}\]

  2. ChillOut
    • one year ago
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    Does a graphical approach for the taylor series work?

  3. anonymous
    • one year ago
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    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.

  4. amoodarya
    • one year ago
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    "Concentrationalizing" thank you , but ...I am not at the beginning of studying math . I am phd candidate ....

  5. Loser66
    • one year ago
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    .

  6. amoodarya
    • one year ago
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    graphical approach for the tailor series rise from limit ,derivation not geometry

  7. anonymous
    • one year ago
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    I have this feeling that you can modify the geometric argument that \(\dfrac{\sin x}{x}\to1\) as \(x\to0\)...

  8. anonymous
    • one year ago
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    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 \(x\to0\).

  9. anonymous
    • one year ago
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    Whether this modification is indeed possible, I'm not quite seeing it right away.

  10. IrishBoy123
    • one year ago
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    i think the answer is 'no'

  11. Empty
    • one year ago
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    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.

  12. anonymous
    • one year ago
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    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 \(0<x<1\), 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%5Bx-x%5E3%2F6%5D%2C+%28x-x%5E3%2F6%29%2FSin%5Bx%5D%2C%28x-x%5E3%2F6%29%2FSin%5Bx-x%5E3%2F6%5D%7D%2C%7Bx%2C-1%2C1%7D%5D Whether this is a valid adaptation of the other argument, I'm not sure, just spitballing here.

  13. anonymous
    • one year ago
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    Correction, \(x-\dfrac{x^3}{3!}\) isn't decreasing for \(0<x<1\); rather I meant to say that \(x<x-\dfrac{x^3}{3!}\).

  14. anonymous
    • one year ago
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    *\(>\)

  15. ganeshie8
    • one year ago
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    .

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