Math time!

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.

encrypted.google.com/#q=y=5-1/x

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?

lol why are you on this board?

>he doesn't know what a maclaurin expansion is at 25 and he's proud of it

Google is your friend, lazy shit. Maclaurin expansion is Taylor expansion about the point 0. If you don't know what Taylor expansion is, you didn't even learn Calculus, in which case gtfo. Or, again, google because Calculus is really fucking easy.

whew lad

>ln(x)
>not \ln(x)
>lim
>not \lim
>\epsilon
>not /varepsilon
anyway your problem is because you need to include a space before [/math]
This should work:
[eqn]\ln(x) = \lim_{\varepsilon \rightarrow 0}\frac{x^\varepsilon-1}{\varepsilon} [/eqn]

You want the a function which maps n to the nth harmonic number. See en.m.wikipedia.org/wiki/Harmonic_number.

Who the hell taught you math?

The remainder is [math] R_n(x)=(-1)^n \frac{ \xi^{n+1} (x-1)^{n+1} }{ n+1 } = \ln(x) - Taylor_n(x) [/math]
Where [math] \xi \in [1,x] [/math] makes the remainder exact which can then be bounded for error analysis by picking [math] \xi [/math] to be the value that maximizes it.

Is there a real convention for \varx vs \x? From a little bit of searching, it seems to be overloaded as fuck which is a little irritating. \varsigma is final-position sigma (an actual VARIANT!), \vartheta, \varphi, \varepsilon seem to be script form (arguable), and the rest I could find are fucking italic (that's a fucking typeface).

Here's an idea.
[math]\log(x+2^n) = n \log(2) + \log(x/2^n+1)[/math]
If x

Yes, yes. Would you please tell us what [math]\xi ^ {n+1} [/math] is? Because it simplifies to what I just said

No there isn't, but for most people \varepsilon and \varnothing look better than \epsilon and \emptyset

The Taylor series centered on 'a' will converge for all points in the open ball B_a(a). Do you even complex variables?

[math] \ln(x) = \ln(a) - \sum^{ \infty } _ {1} (-1)^{n} \frac{ (x-a)^n } { n a^n } [/math]

[math] radius = \lim_{n \to \infty} \frac { \frac{ 1 } { n a^n } } { \frac{ 1 } { (n+1) a^{n+1} } } = a [/math]

1. ln(x) has a vertical asymptote at x=0
2. The slope of ln(x) decreases as x->∞
A Taylor expansion can't satisfy the first requirement and can only satisfy the second over a limited range.

f(x)=sqrt(x)-1/sqrt(x) satisfies both of those properties, and provides a reasonable approximation for small x (and f(1)=0).

No, you're off by 1 and forgot the (x-1)^n term

You failed it.

This is taught in calc 2. Its first applicatiom is in diffeq with eulers equation for solving linear, second degree, constant coefficient, differential equations when the factorization of the auxillary equation is r=a+/-bi. Even then, you just produce another equation to memorize. You should know this or not comment about your ignorance.

I have wondered about the
following logarithmic functions:
L(x, 1) = ln(x)
L(x, 2) = ln(ln(x)+1)
L(x, 3) = ln(ln(ln(x)+1)+1)
...
L(x, n) = ln(L(x, n-1)+1)
... and how to represent
an intermediate function
L(x, r) where r = a/b
for whole number _a_ and _b_

should have stated
"posint _a_ and _b_"
late night, sorry

interpolation.