help finding sum of this series?

We are given S(x) = 1 + x^3/3! + x^6/6! + .... Observe that [ d^3/(dx)^3 ][ S ] = S, and also observe that if y = A⋅e^(Bx), then [ d^3/(dx)^3 ][ y ] = B^3⋅y. Consequently, if

B = 1,

C = (-1 + i√3) / 2,

D = (-1 - i√3) / 2,

i.e. A, B, C are the cube roots of unity and we let

y(x) = (A_1)⋅e^(Bx) + (A_2)⋅e^(Cx) + (A_3)⋅e^(Dx),

then

[ d^3/(dx)^3 ][ y ]

= (A_1)⋅B^3⋅e^(Bx) + (A_2)⋅C^3⋅e^(Cx) + (A_3)⋅D^3⋅e^(Dx)

= (A_1)⋅1⋅e^(Bx) + (A_2)⋅1⋅e^(Cx) + (A_3)⋅1⋅e^(Dx)

= y.

Suppose the coefficients A_1, A_2, A_3 are such that S(x) = y(x). Then

S(0) = y(0),

S'(0) = y'(0),

S"(0) = y"(0),

which, respectively, are

(1): 1 = A_1 + A_2 + A_3,

(2): 0 = (A_1)⋅B + (A_2)⋅C + (A_3)⋅D,

(3): 0 = (A_1)⋅B^2 + (A_2)⋅C^2 + (A_3)⋅D^2,

where (2) and (3) can be rewritten, respectively, as

(4): 0 = A_1 + (A_2)⋅C + (A_3)⋅D,

(5): 0 = A_1 + (A_2)⋅D + (A_3)⋅C,

because B^2 = B = 1, C^2 = D and D^2 = C. So (4) - (5) is 0 = (A_3 - A_2)(D - C). Thus

A_2 = A_3,

which turns (1), (4), respectively, into

1 = A_1 + 2⋅(A_2),

0 = A_1 + (D + C)(A_2).

Taking the difference of these, 1 = (2 - (D+C))(A_2) = 3⋅(A_2). Therefore

A_2 = 1/3,

A_3 = 1/3,

A_1 = 1/3.

In conclusion,

S(x)

= (1/3)⋅e^x + (1/3)⋅e^((-½ + i√3/2)⋅x) + (1/3)⋅e^((-½ - i√3/2)⋅x)

= (1/3)⋅e^x + (1/3)⋅e^(-x/2)⋅e^(x⋅i√3/2) + (1/3)⋅e^(-x/2)⋅e^(-x⋅i√3/2)

= (1/3)⋅e^x + (1/3)⋅e^(-x/2)⋅( e^(x⋅i√3/2) + e^(-xi√3/2) )

= (1/3)⋅e^x + (1/3)⋅e^(-x/2)⋅( cis(x⋅√3/2) + cis(-x⋅√3/2) )

= (1/3)⋅e^x + (1/3)⋅e^(-x/2)⋅( 2⋅cos(x⋅√3/2) ),

i.e.

S(x) = (1/3)⋅e^x + (2/3)⋅cos(x⋅√3/2)⋅e^(-x/2).

The Maclaurin's series for cosx, sinx are

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

sinx=x-(x^3)/3!+(x^5)/5!-...

cosx+isinx=

1+(ix)^2/2!+(ix)^4/4!+..+

ix+(ix)^3/3!+(ix)^5/5!+...

=

e^(ix).

What is the general term of

1+(x^3)/3!+(x^6)/6!+....?

If it is known and the series is

convergent, then its total sum

could be found approximately

for a given x.

see this looks like e^x . You just put omega*x in place of x why as icube of omega is 1 and square of omega and itself is an imaginary number . so find e^(iwx/i) and compare real and imaginary parts

. I think real part is cos(wx/i)