# `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

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### 1 Answer

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`

`(e^xsecx)''=(e^xsecx+e^xsecxtanx)'=`

`e^xsecx+2e^xsecxtanx+e^x(sec^3x+secxtan^2x)`

`(e^xsecx)^((3))=`

`e^xsecx+3e^xsecxtanx+3e^x(sec^3x+secxtan^2x)+`

`e^x(5sec^3xtanx+secxtan^3x)`

Let's now calculate value of those derivatives in 0.

`f(0)=e^0sec0=1cdot1=1`

`f'(0)=e^0sec0+e^0sec0tan0=1cdot1+1cdot1cdot0=1`

`f''(0)=e^0sec0+2e^0sec0tan0+e^0(sec^3 0+sec0tan^2 0)=2`

`f^((3))(0)=cdots=4`

If we put that in formula (1) we get

`f(x)=1+1/1x+2/2x^2+4/6x^3=1+x+x^2+2/3x^3+cdots`

which is what we were supposed to get.

Also Maclaurin's series for other two functions are:

`secx=1+x^2/2+(5x^4)/24+(61x^6)/720+cdots`

`(1-x)^(-1)=1+x+x^2+x^3+x^4+cdots`