Linear Approximation of ln3?

Question

anonymous · Answer

I know that e^x will be the closest thing, but I am not sure how to work that into the problem.

amistre64 · Answer

a linear approx is the line equation defined by:

y = f'(a)(x-a) + f(a)

this simply defines the slope of the line at a (the derivative), using the point a,f(a) as its anchor

anonymous · Answer

yes, I see, but what point?

amistre64 · Answer

3,ln3   which may defeat the purpose since it is ln3 you are wanting to find :/

amistre64 · Answer

im thinking the taylor polynomial would be the next best thing

amistre64 · Answer

if we could make a polynomial that best fits this curve, we would want it to "move" in the same way as it .... movement is a rate of change, or derivative.

amistre64 · Answer

spose
$$f(x)=a_0+a_1x+a_2x^2+a_3x^3+...$$
$$f'(x)=a_1+2a_2x+3a_3x^2+...$$
$$f''(x)=2a_2+3.2a_3x+...$$
all we would have to do is find some way to define the constants .... and we would have a polynomial f(x) that fits the curve

amistre64 · Answer

one strategy we could do is to zero out all the x parts, that would leave us with the constants at the start of each derivative.  For x=3, we would want to modify each x to get (x-3) ... in order to zero it out

amistre64 · Answer

$$f(x)=a_0+a_1(x-3)+a_2(x-3)^2+a_3(x-3)^3+...$$
$$f'(x)=a_1+2a_2(x-3)+3a_3(x-3)^2+...$$
$$f''(x)=2a_2+3.2a_3(x-3)+...$$

now when x=3, we get

$$f(3)=a_0$$
$$f'(3)=a_1$$
$$f''(3)=2a_2$$
$$...$$
and we can solve for the coefficients

amistre64 · Answer

any of this making sense?