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

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Answer 1 · 2016-10-03T18:48:03

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

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.