`1+x+x^2+2/3x^3+cdots`is the expansion of:a) `(1-x)^-1`  b) `secx`  c) `e^xsecx`  d) None of these 

1 Answer

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tiburtius | High School Teacher | (Level 2) Educator

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This is Maclaurin's series (Taylor series about zero) of

c) `e^x sec x`.

Taylor series

If function `f` is infinitely differentiable in neighborhood of `a` then

`f(x)=f(a)+(f'(a))/(1!)(x-a)+(f''(a))/(2!)(x-a)^2+(f^(3)(a))/(3!)(x-a)^3+cdots`         (1)

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: