At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

ummm I might be able to do this... we'll need to collaborate though

they gave the first part:
ln(2)+\[\sum_{n=1}^{\infty}\]

so whats in the sigma, thats important

thats what we have to try to find lol

does the question say exactly that? What is a taylor series for...

take the derivative of ln(x) like 4 or 5 times see if you can see a pattern

yes, What is a Taylor Series for f(x)=ln(2+2x)

of and at x=1/2 we know that ln(1) = 0

\[\sum_{?}^{?} (f^{(n)}(1/2))/n! * (2x-(1/2))^{n}\]

the general formula would be F^(n) (x) = (-1)^(n+1) (n-1) !
---------------
x^(n)

thats my best guess but I think I'm taking it too fat by kinda approxximating

>.<

ha, I have no idea

I dont know where I would stop using the general form...?