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anonymous
 5 years ago
find the bound on the magnitude of the error if we approximate sqrt 2 using the taylor approximation of degree three for sqrt 1+x about x=0
anonymous
 5 years ago
find the bound on the magnitude of the error if we approximate sqrt 2 using the taylor approximation of degree three for sqrt 1+x about x=0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I hate approximations >.<

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i just dont know where to start or what to do

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey anwara can u plz look at my problem too when u r done here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I am almost there :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ty u so much i really appreciate it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know how to find the Taylor expansion of a function?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you please showme

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm, Do you have the formula of finding the Taylor expansion?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We're just looking for the third degree expansion, that's the first four terms.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay my only question is what do i plug into x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's good. So, we have the first four terms of the expansion: \[\sqrt{1+x}=1+{x \over 2}{x^2 \over 8}+{x^3 \over 16}+....\] We are looking for the approximation of sqrt(2), that's sqrt(1+1). So, we will plug x=1, and then we get: \[\sqrt{1+1}=\sqrt{2}=1+{1 \over 2}{1 \over 8}+{1 \over 16}={23 \over 16}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey anwara can u plz look at my problem too when u r done here 14 minutes ago

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay so now what do i do with this value

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now this is the approximated value of sqrt(2), we're looking now for the error in this value.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh I don't really know what the bound on the magnitude of the error exactly is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the taylor series expansion did u use the binomial series to figure it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now can u plz look at my problme plz

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, I didn't. But, you could, it's actually easier to be found by the binomial series.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now can u plz look at my problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so basically i would plug in sqrt of 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay so u dont know how do the rest of the problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Not really. I don't know what the bound on the magnitude of the error is. If you could provide me with a formula or any information regarding that, I might be able to help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you click on this link you will see it http://math.jasonbhill.com/courses/fall2010math2300005/lectures/taylorpolynomialerrorbounds

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its under this topic Theorem 10.1 Lagrange Error Bound
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