Question

Find the Taylor degree 4 polynomial of ? (?) = −? ∗ ??? (?)
centered on 0 and find the interval

for which the approximation has a smaller error or than a.
???.

Answer #1

Find the degree 3 Taylor polynomial T3 (x) centered
at a = 4 of the function f(x) = (-7x+36)4/3

Find the Taylor polynomial of degree 3, centered at a=4 for the
function f(x)= sqrt (x+4)

Find the ?? (?) Taylor polynomial of ? (?) = (? - ?) ?? (? - ?) centered on ?. Using the Taylor inequality, find the error of the approximation if .? ≤ ? - ? ≤ ?. ?.

(1 point) Find the degree 3 Taylor polynomial T3(x) centered
at a=4 of the function f(x)=(7x−20)4/3.
T3(x)=
? True False Cannot be determined The function f(x)=(7x−20)4/3
equals its third degree Taylor polynomial T3(x) centered at a=4.
Hint: Graph both of them. If it looks like they are equal, then do
the algebra.

let f(x)=cos(x). Use the Taylor polynomial of degree 4
centered at a=0 to approximate f(pi/4)

1.
Find the Taylor polynomial, degree 4, T4, about 1/2 for f (x) = tan-inv (x) and use it to approximate tan-inv (1/16).
2.
Find the taylor polynomial, degree 4, S4, about 0 for f (x) = tan-inv (x) and use it to approximate tan-inv (1/16).
3.
who provides the best approximation, S4 or T4? Prove it.

1.
Use a deﬁnition of a Taylor polynomial to ﬁnd the Taylor
polynomial T2(x) for f(x) = x^3/2 centered at a = 4.
We use T1(3.98) to approximate (3.98)^3/2. Apply Taylor’s
inequality on the interval [3.98,4.02] to answer the following
question: can we guarantee that the error |(3.98)^3/2 −T1(3.98)| of
our approximation is less than 0.0001 ?

1- Please explain the difference between a Taylor Polynomial of
degree 4 and a Taylor Polynomial of degree 100. How would they be
different from each other? Assuming your computer has the specs to
handle either with ease, which is going to be the more desirable
one to utilize and why?

Find the second degree polynomial of Taylor series for f(x)=
1/(lnx)^3 centered at c=2. Write step by step.

Find the 4th degree, T4 taylor polynomial for f(x)=arctan (x)
centered at c=1/2 and use it to aproximate f(x)= arctan
(1/16)

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