Best polynomial approximation of degree n is the orthogonal projection which can be rewritten as [eqn]\langle\,ln(x),e_1\rangle e_1 + ... + \langle\,ln(x),e_n\rangle e_n[/eqn]
for any orthonormal polynomial basis e_1, ..., e_n.
where [eqn]\langle\,f, g \rangle = \int_T f(x)g(x) dx[/eqn]
T is the target interval.
i'm too lazy to use gram schmidt and find an orthonormal polynomial basis and apply it.
I'm guessing they mean using PEMDAS? (what I always think of for "simple math")
Not really sure how you'd go about this
Great!
Is there any way to stretch it out? Make the hike up take longer but even it out?
x-(1/2)x^2+(1/6)x^3...
Taylor series you mong
>Take the taylor series
You can't take the taylor series of ln(x), you can take the taylor series of ln(x+a) where a ≠ 0, and even then its only valid around x = a±1
Up to what error, and between what numbers?
[math]ln(x) \approx (x-1)+\frac{(x-1)^2}{2}-\frac{(x-1)^3}{3}+\frac{(x-1)^4}{4}\cdots + (-1^n)\left (\frac{(x-1)^n}{n}\right )[/math]
That should give you the nth Taylor aproximation of grade N of ln(x) on point X0=1.
The error at x would be [math]R_{n}(x)=\pm \frac{(\varepsilon -1)^{n+1}}{n+1}[/math]
with [math]\varepsilon[/math] being a number between 1 and x, and its usually taken the number that will make the error the largest
>and even then its only valid around x = a±1
CS major detected. It's valid around x = a±a
No it isn't.
Idk wtf a Maclaurin expansion even is and I'm 25. Even had to go back to read your comment to make sure I spelled it right. Exactly how many nipples do you own anyway?