anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I hate approximations >.<
anonymous
  • anonymous
Let me see.
anonymous
  • anonymous
k ty

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anonymous
  • anonymous
i just dont know where to start or what to do
anonymous
  • anonymous
hey anwara can u plz look at my problem too when u r done here
anonymous
  • anonymous
I think I am almost there :)
anonymous
  • anonymous
ty u so much i really appreciate it
anonymous
  • anonymous
Do you know how to find the Taylor expansion of a function?
anonymous
  • anonymous
can you please showme
anonymous
  • anonymous
Hmm, Do you have the formula of finding the Taylor expansion?
anonymous
  • anonymous
We're just looking for the third degree expansion, that's the first four terms.
anonymous
  • anonymous
okay my only question is what do i plug into x
anonymous
  • anonymous
That'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
  • anonymous
hey anwara can u plz look at my problem too when u r done here 14 minutes ago
anonymous
  • anonymous
okay so now what do i do with this value
anonymous
  • anonymous
Yeah sure rsaad2.
anonymous
  • anonymous
its on work done
anonymous
  • anonymous
Now this is the approximated value of sqrt(2), we're looking now for the error in this value.
anonymous
  • anonymous
Oh I don't really know what the bound on the magnitude of the error exactly is.
anonymous
  • anonymous
for the taylor series expansion did u use the binomial series to figure it out
anonymous
  • anonymous
now can u plz look at my problme plz
anonymous
  • anonymous
Well, I didn't. But, you could, it's actually easier to be found by the binomial series.
anonymous
  • anonymous
now can u plz look at my problem
anonymous
  • anonymous
so basically i would plug in sqrt of 1
anonymous
  • anonymous
Yep.
anonymous
  • anonymous
okay so u dont know how do the rest of the problem
anonymous
  • anonymous
Not 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
  • anonymous
if you click on this link you will see it http://math.jasonbhill.com/courses/fall-2010-math-2300-005/lectures/taylor-polynomial-error-bounds
anonymous
  • anonymous
its under this topic Theorem 10.1 Lagrange Error Bound

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