Question

can someone prove this by geometry !

Answer 1

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

Answer 2

Does a graphical approach for the taylor series work?

Answer 3

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.

Answer 4

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

Answer 5

.

Answer 6

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

Answer 7

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

Answer 8

In case you're not familiar, the argument I'm referring to goes like this:

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 \(x\to0\).

Answer 9

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

Answer 10

i think the answer is 'no'

Answer 11

I tried two things, they might help to consider in constructing something. 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.