Approximating the cosine by Taylor polynomialLet f := cos ⁡ ( x ) I'm asked to find for which values of x we can be sure the 4th degree Taylor polynomial will give an error lesser than 1 1000 .Now, cos ⁡ ( x ) = 1 − x 2 2 + x 4 24 + o ( x 4 )Taylor's 4th degree remainder is then R = f ( 5 ) ( cos ⁡ ( θ ) ) 5 ! x 5 = − sin ⁡ ( θ ) 5 ! x 5 Since I'm only asked about the error, I can get rid of the sign, and can do the following boundering: | R | ≤ | 1 5 ! x 5 | ≤ 1 1000 Since that is strictly increasing, its easy to solve that last inequation, by doing 1 120 x 5 = 1 1000 ⟶ x = 120 1000 5 ≈ 0.65Which ends up giving | R | ≤ 1 1000 ∀ x : | x | ≤ 120 1000 5 ( ≈ 0.65 )Now, I thought this was an already correct solution, but to my surprise, I checked plotted both the function and its Taylor polynomial, and the error bound that produces below 0.65 is not only lesser than 1 1000 , but actually lesser than 1 10000 e.g: cos ⁡ ( 0.65 ) = 0.796187... t a y l o r ( c o s ( 0.65 ) ) = 0.796083...Which means it is overly precise, what is going on here, I'm asuming it must be some arithmetic mistake I made somewhere, but I can't find it and its driving me crazy.

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Approximating the cosine by Taylor polynomialLet   f  :=  cos  ⁡  (  x  ) I&#039;m asked to find for which values of x we can be sure the 4th degree Taylor polynomial will give an error lesser than       1    1000  .Now,   cos  ⁡  (  x  )  =  1  −            x      2        2    +            x      4        24    +  o  (      x    4    )Taylor&#039;s 4th degree remainder is then   R  =                    f                  (          5          )                    (      cos      ⁡      (      θ      )      )              5      !            x    5    =            −      sin      ⁡      (      θ      )              5      !            x    5  Since I&#039;m only asked about the error, I can get rid of the sign, and can do the following boundering:      |    R      |    ≤      |        1          5      !            x    5        |    ≤      1    1000  Since that is strictly increasing, its easy to solve that last inequation, by doing       1    120        x    5    =      1    1000    ⟶  x  =            120      1000        5    ≈  0.65Which ends up giving       |    R      |    ≤      1    1000    ∀  x  :      |    x      |    ≤            120      1000        5    (  ≈  0.65  )Now, I thought this was an already correct solution, but to my surprise, I checked plotted both the function and its Taylor polynomial, and the error bound that produces below 0.65 is not only lesser than       1    1000  , but actually lesser than       1    10000  e.g:  cos  ⁡  (  0.65  )  =  0.796187...  t  a  y  l  o  r  (  c  o  s  (  0.65  )  )  =  0.796083...Which means it is overly precise, what is going on here, I&#039;m asuming it must be some arithmetic mistake I made somewhere, but I can&#039;t find it and its driving me crazy.

faux0101d · Accepted Answer

Step 1You made us of       |    sin  ⁡  θ      |    ≤  1 in your derivation, which is a very conservative estimate and fine. Once you notice that in fact       |    sin  ⁡  θ      |    ≤      |    θ      |    ≤      |    x      |  , you can try to find the bounds for x from       |                  x        6                    5        !              |    &amp;lt;      1    1000  , which will allow you to go up to 0.7023… instead of just 0.65.Step 2Still, the fact that Taylor guarantees you an error   &amp;lt;      1    1000   does not exclude the possibility that the error is indeed much smaller &quot;by chance&quot;.

Nathalie Case · Accepted Answer

Step 1You have given an upper bound for the error, so you shouldn&#039;t be surprised that the actual error is below it. Further, you gave away more than you could have by assuming that   sin  ⁡  (  θ  ) gets as large as 1 in your interval, which it doesn&#039;t.So that&#039;s to say that nothing is wrong, and the trivial upper bound is exactly that - trivial.Step 2On the other hand, there is a better trivial upper bound. The cosine expansion is alternating, the terms go to 0, are decreasing in size (which isn&#039;t immediately obvious, nor true for all ranges of x - but it is true here). So you can recognize that the error will be less than the first omitted term.The first omitted term is       x    6        /    6  !. At   x  =  .65, this is about .0001, or 1/10000, as you&#039;ve requested.Since the terms of the cosine expansion quickly decay, this alternating series error should be pretty accurate very often.

Let f:=cos(x) I'm asked to find for which values of x we can be sure the 4th degree Taylor polynomial will give an error lesser than 1/(1000).

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