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This is Maclaurin's series (Taylor series about zero) of
c) `e^x sec x`.
If function `f` is infinitely differentiable in neighborhood of `a` then
In case of Maclaurin's series `a=0.`
Let's now calculate first 3 derivatives of `f(x)=e^x sec x.`
`(e^xsec x)'=e^x sec x+ e^xsecxtanx`
Let's now calculate value of those derivatives in 0.
`f''(0)=e^0sec0+2e^0sec0tan0+e^0(sec^3 0+sec0tan^2 0)=2`
If we put that in formula (1) we get
which is what we were supposed to get.
Also Maclaurin's series for other two functions are:
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