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

Question

William Abdalla · Accepted Answer

#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)#

For functions with two variables, the Taylor expansion's general expression 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)#

Knowing that from is another way to accomplish that.

#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.

Managing multivariate series requires a great deal of work due to the necessary notation.

Aurora Alvarado · Answer

undefined To expand cos(x+h) in powers of x and h, we can use the Taylor series expansion. The Taylor series expansion for cos(x) is:

cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...

To expand cos(x+h), we substitute (x+h) in place of x in the above series:

cos(x+h) = 1 - ((x+h)^2)/2! + ((x+h)^4)/4! - ((x+h)^6)/6! + ...

Expanding each term using the binomial theorem, we simplify the expression:

cos(x+h) = 1 - (x^2 + 2xh + h^2)/2! + (x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4)/4! - ...

Simplifying further, we collect like terms and arrange them in powers of x and h:

cos(x+h) = 1 - (x^2)/2! + (x^4)/4! - ... + (-1)^n * (x^(2n))/((2n)!) + ... 
          - (xh)/2! + (4x^3h)/4! - ... + (-1)^n * ((2n-1)x^(2n-1)h)/((2n)!) + ...
          + (h^2)/2! - (6x^2h^2)/4! + ... + (-1)^n * ((2n-2)x^(2n-2)h^2)/((2n)!) + ...
          - (x^3h)/2! + (4xh^3)/4! - ... + (-1)^n * ((2n-1)x^(2n-1)h^3)/((2n)!) + ...
          + (h^4)/2! - (6x^2h^4)/4! + ... + (-1)^n * ((2n-2)x^(2n-2)h^4)/((2n)!) + ...

This expansion can be continued indefinitely, including terms involving higher powers of x and h.