Find Maclaurin Series of ...?

Question

Find Maclaurin Series of function f(x) = (1-cosx)/x^2 . And with using this serie find the x=1.2 value of the function with 5 term far from serie and 4 term far from comma ( I mean 0.1254 like this).

Answer 1

The above is the right technique but I believe that the result is wrong. Knowing that:

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

It follows that:

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

And dividing by x^2:

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

You can also do this by direct expansion but you will encounter derivatives that require multiple application of l'Hopital's rule to evaluate.

Answer 2

The Maclaurin series for cos x is

1 - x^2/2! + x^4/4! - x^6/6! + x^8/8! - x^10/10! + ...

so if you substitute this for cos x in (1-cos x)/x^2, you get

1 - x^2/4! + x^4/6! - x^6/8! + x^8/10! - ...

and you can plug in x=1.2.

Answer 3

note that sin x = Sum from n = 0 to INF {(-a million)^n * x^(2n+a million) / (2n+a million)! only replace x with 3x. Then sin 3x = Sum from n = 0 to INF {(-a million)^n * (3x)^(2n+a million) / (2n+a million)!} only multiply the sequence by skill of x. x sin 3x = Sum from n = 0 to INF {(-a million)^n * x * (3x)^(2n+a million) / (2n+a million)!} Simplify x sin 3x = Sum from n = 0 to INF {(-a million)^n * x * 3^(2n+a million) * x^(2n+a million) / (2n+a million)!} x sin 3x = Sum from n = 0 to INF {(-a million)^n * 3^(2n+a million) * x^(2n+2) / (2n+a million)!}