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

1. amoodarya

$x \rightarrow 0\\sinx \approx x-\frac{x^3}{3!}$

2. ChillOut

Does a graphical approach for the taylor series work?

3. 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.

4. amoodarya

"Concentrationalizing" thank you , but ...I am not at the beginning of studying math . I am phd candidate ....

5. Loser66

.

6. amoodarya

graphical approach for the tailor series rise from limit ,derivation not geometry

7. anonymous

I have this feeling that you can modify the geometric argument that $$\dfrac{\sin x}{x}\to1$$ as $$x\to0$$...

8. 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 $$x\to0$$.

9. anonymous

Whether this modification is indeed possible, I'm not quite seeing it right away.

10. IrishBoy123

i think the answer is 'no'

11. 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.

12. 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 $$0<x<1$$, 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

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

*$$>$$

15. ganeshie8

.