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Re: Series Expansions in Mathematica
Posted:
Feb 20, 2013 10:26 PM


Hello, I am attempting to expand an equation, and then solve the 1st, 2nd, 3rd, 4th and 5th order expressions one at a time.
The equation is:
Exp[4 x]  c Exp[3 y] Exp[4 x] + c  1 = 0
With, x and y small variables to be expanded and c a constant. Once this is expanded, the variables x and y are in turn replaced with expansions written as:
x = x[1] + x[2]/2 + x[3]/6 + x[4]/24 + x[5]/120 (to 5th order)
By the time I have completed these expansions, I have a somewhat long expression. Does Mathematica have an intelligent way of deciding which terms would be 1st/2nd/etc order? For example, x[1] y[2] and x[1]^2 y[1] would both be third order.
Many thanks, Sam Young
Hi, Sam,
Yes, you may go for troubles of solving this using series. Though you did not write what are you looking for as the function of what, I assume that you are looking for x=x(y), or reversed: y=y(x).
To explicitly answer your question:
a) This is the expansion:
expr2 = Series[ Exp[4 x]  c* Exp[3 y] Exp[4 x] + c  1, {x, 0, 5}, {y, 0, 5}] // Normal // Simplify
1/600 (45 c y (40 + 60 y + 60 y^2 + 45 y^3 + 27 y^4)  60 x (40 + c (40 + 120 y + 180 y^2 + 180 y^3 + 135 y^4 + 81 y^5))  120 x^2 (40 + c (40 + 120 y + 180 y^2 + 180 y^3 + 135 y^4 + 81 y^5))  160 x^3 (40 + c (40 + 120 y + 180 y^2 + 180 y^3 + 135 y^4 + 81 y^5))  160 x^4 (40 + c (40 + 120 y + 180 y^2 + 180 y^3 + 135 y^4 + 81 y^5))  128 x^5 (40 + c (40 + 120 y + 180 y^2 + 180 y^3 + 135 y^4 + 81 y^5)))
b) Now, like this you get coefficients:
Coefficient[expr2, #] & /@ {x, x^2, x^3, x^4, x^5} /. y > 0
{1/10 (40  40 c), 1/5 (40  40 c), (4/15) (40 + 40 c), (4/15) (40 + 40 c), (16/ 75) (40 + 40 c)}
Coefficient[expr2, #] & /@ {xy, x^2 y^2, x^3 y^3, x^4 y^4, y^5 x^5} /. {y > 0, x > 0}
{0, 36 c, 48 c, 36 c, ((432 c)/25)}
I did not extract them all. It is your job to accurately do this. I only show the way.
However, your equation Mathematica can solve almost as it is, let alone that each of us is able to solve without can solve explicitly without any computer.
This slightly transforms your expression:
expr = Exp[4 x]  c* Exp[3 y] Exp[4 x] + c  1 /. c > Exp[b]
1 + E^b + E^(4 x)  E^(b + 4 x + 3 y)
And this gives a solution.
Solve[expr == 0, x]
Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>
{{x > Log[((1 + E^b)^(1/4)/(1 + E^(b + 3 y))^(1/4))]}, {x > Log[((I (1 + E^b)^(1/4))/(1 + E^(b + 3 y))^(1/4))]}, {x > Log[(I (1 + E^b)^(1/4))/(1 + E^(b + 3 y))^(1/4)]}, {x > Log[(1 + E^b)^(1/4)/(1 + E^(b + 3 y))^(1/4)]}}
Or another way around:
Solve[expr == 0, y]
{{y > ConditionalExpression[ 1/3 (b  4 x + 2 I \[Pi] C[1] + Log[1 + E^b + E^(4 x)]), C[1] \[Element] Integers]}}
Have fun, Alexei
Alexei BOULBITCH, Dr., habil. IEE S.A. ZAE Weiergewan, 11, rue Edmond Reuter, L5326 Contern, LUXEMBOURG
Office phone : +35224542566 Office fax: +35224543566 mobile phone: +49 151 52 40 66 44
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