How to expand cos(x+h) in powers of x and h?

1 Answer
Oct 3, 2016

#cos(x+h) = sum_(n=0)^oo(1/(n!) sum_(k=0)^n((n),(k))cos((n+k)pi/2)x^(n-k)h^k)#

Explanation:

The general expression of the Taylor expansion for functions with two variables is

#f(x,h)=sum_(n=0)^oo(1/(n!) sum_(k=0)^n((n),(k))((partial^nf)/(partialx^(n-k)partialh^k))_({x_0,h_0})(x-x_0)^(n-k)(h-h_0)^k)#

In this case, considering #x_0=0,h_0=0#

#((partial^nf)/(partialx^(n-k)partialh^k))_(x=0,h=0) = cos((n+k)pi/2)#

so

#cos(x+h) = sum_(n=0)^oo(1/(n!) sum_(k=0)^n((n),(k))cos((n+k)pi/2)x^(n-k)h^k)#

but

#cos((n+k)pi/2) = i^(n+k)((1+(-1)^(n+k))/2)# and

#((n),(k)) = (n!)/((n-k)!k!)#

so

#cos(x+h) = sum_(n=0)^oosum_(k=0)^n ( i^(n+k)((1+(-1)^(n+k))/2))/((n-k)!k!)x^(n-k)h^k#

with #i = sqrt(-1)#

Another way to do that is knowing that from

#cosx = sum_(k=0)^oo (-1)^k(x^(2k))/(2k!)# follows

#cos(x+h) = sum_(k=0)^oo(-1)^k((x+h)^(2k))/(2k!)# but here the variables #x+h# appear added.

Multivariate series handling is very cumbersome because the required notation effort needed.