anonymous
  • anonymous
f(x) = e^(2x) + e^(−2x) Find an upper bound for the error in using the second degree Maclaurin polynomial of f to approximate f(0.5).
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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 this expert

answer on brainly

SEE EXPERT ANSWER

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

anonymous
  • anonymous
I guess I have to use the Lagrange remainder formula, right? \[E _{n}(x) = f(x) - T _{n}(x) =\frac{ f ^{(n+1)}(c) }{ (n+1)! }(x-a)^{(n+1)}\] Where c lies between x and a
anonymous
  • anonymous
yes just plug the rest in
anonymous
  • anonymous
So here a=0 right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
i think so
anonymous
  • anonymous
Is my 3rd derivative correct? I got... \[8e ^{2x}-8e ^{-2x}\]
anonymous
  • anonymous
ya
anonymous
  • anonymous
Cool, so then I get \[E _{2}(0.5)=\frac{ e ^{2c} -e ^{-2c}}{ 6 }\]
anonymous
  • anonymous
yes! im surprised u asked for help cuz u get it
anonymous
  • anonymous
Lol I don't know what to do next :D
anonymous
  • anonymous
that is f(x)
anonymous
  • anonymous
i see what u do now
anonymous
  • anonymous
u have to
anonymous
  • anonymous
replace x with .5
anonymous
  • anonymous
What if I was trying to find the lower bound?

Looking for something else?

Not the answer you are looking for? Search for more explanations.