As it is not entirely clear whether the question is asking for the first five *nonzero* terms, or the first five terms produced from the sigma notation of a Taylor series, both will be given.

The Taylor series centered at #c# for a function #f(x)# is given by

#f(x) = sum_(n=0)^oo f^((n))(c)/(n!)(x-c)^n#

where #f^((n))(c)# is the #n^"th"# derivative of #f#, evaluated at #c#.

In our case, we have #f(x) = x^8+x^4+3#

As #f# is a polynomial, though, we can tell that #f^((n))(0)# will be nonzero only when #f^((n))(x)# contains a constant term, that is, for #n in {0, 4, 8}#. For all other derivatives, every term will have a factor of #x#, and thus will be #0# when evaluated at #0#.

As such, the Taylor series for #f(x)# will have only three terms:

#f(0)/(0!)(x-0)^0 = (0^8+0^4+3)/1*1 = 3#

#f^((4))(0)/(4!)(x-0)^4 = (8*7*6*5*0^4+4*3*2*1)/(4*3*2*1)x^4 = x^4#

#f^((8))(0)/(8!)(x-0)^8 = (8*7*6*5*4*3*2*1)/(8*7*6*5*4*3*2*1)x^8 = x^8#

Therefore, the Taylor series for #f(x)# is #3+x^4+x^8#, which is just #f(x)#. We should not be surprised by this result, as a Taylor series is, in effect, a polynomial representation of a function. When the function is a polynomial, then, they should be identical.

As for the terms produced from #n=0# to #n=4# in the series, we have, by the above reasoning, that #f^((n))(0)/(n!)(x-0)^n = 0# for #n in {1, 2, 3}#. Thus the first five terms the series produces are #3, 0, 0, 0, x^4#.