How do you find the Taylor polynomial of degree 4 for cos(x), for x near 0?

Question

28191 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-05T15:25:39

#T_4(x) = 1 -x^2/2+x^4/24#

The development of #cosx# in Taylor series around #x=0# is:

[1]...#cosx = sum_0^oo 1/(n!)[(d^((n))cosx)/(dx^n)]_(x=0)x^n#

Now, note that:

#(d^((n))cosx)/(dx^n) = cos((npi)/2+x)#

So [1] becomes:

#cosx = sum_0^oo x^n/(n!)cos((npi)/2)#

We can see that only terms with #n# even are non null, as:

#cos((npi)/2)=0# if #n# is odd,

and furthermore:

#cos((npi)/2)=(-1)^(n/2)# if #n# is even, so that:

#cosx = sum_0^oo (-1)^kx^(2k)/(2k!)#

The Taylor polynmial of degree 4 is just the partial sum of this series for #k=2#:

#T_4(x) = 1 -x^2/2+x^4/24#