2
$\begingroup$

I am attempting to use the LogicalExpand command to find an equation for each coefficient in a power series. The documentation gives the following example of this usage:

In[1]:= LogicalExpand[Series[g[x]^2,{x,0,2}]==2+x]

Out[1]= -2+g[0]^2==0&&-1+2 g[0] g'[0]==0&&g'[0]^2+g[0] g''[0]==0

The goal is to get a system of equations like the one above corresponding to the coefficients of each power of $x$ in the power series.

My series is more complicated than the one in the documentation, and it should be identically equal to $0$, so that each coefficient should be identically zero. When I apply the LogicalExpand function, however, Mathematica simply returns the original input rather than a system of equations. The expression to which I am trying to apply logical expand is: $$\frac{1}{6} x \left(f'(0) h''(0)+h'(0) \left(f'(0) \left(2 h'(0)+3\right)+3\right)+h''(0)+6\right)+\frac{1}{12} x^2 \left(f''(0) \left(h''(0)+h'(0) \left(2 h'(0)+3\right)\right)-2 f'(0)^2 \left(h''(0)+h'(0) \left(4 h'(0)+3\right)\right)-4 f'(0) \left(h''(0)+h'(0) \left(h'(0)+3\right)+3\right)\right)=0$$

In Mathematica:

In[2]:=LogicalExpand[
 1/6 x (6 + h'[0] (3 + f'[0] (3 + 2 h'[0])) + h''[0] + f'[0] h''[0]) + 
 1/12 x^2 (-4 f'[0] (3 + h'[0] (3 + h'[0]) + h''[0]) + f''[0] (h'[0] (3 + 2 h'[0]) + h''[0]) - 
      2 f'[0]^2 (h'[0] (3 + 4 h'[0]) + h''[0])) == 0]

Out[2]:=
 1/6 x (6 + h'[0] (3 + f'[0] (3 + 2 h'[0])) + h''[0] + f'[0] h''[0]) +
 1/12 x^2 (-4 f'[0] (3 + h'[0] (3 + h'[0]) + h''[0]) + f''[0] (h'[0] (3 + 2 h'[0]) + h''[0]) - 
     2 f'[0]^2 (h'[0] (3 + 4 h'[0]) + h''[0])) == 0

I would appreciate any thoughts on why this is failing. Thank you!

$\endgroup$

1 Answer 1

3
$\begingroup$

Try with ordersymbol O[x] :

LogicalExpand[1/6 x (6 + h'[0] (3 + f'[0] (3 + 2 h'[0])) + h''[0] +f'[0] h''[0]) +1/12 x^2 (-4 f'[0] (3 + h'[0] (3 + h'[0]) + h''[0]) +f''[0] (h'[0] (3 + 2 h'[0]) + h''[0]) -2 f'[0]^2 (h'[0] (3 + 4 h'[0]) + h''[0])) == O[x]^3] 

(*1/6 (6 + Derivative[1][h][0] (3 + Derivative[1][f][0] (3 + 2 Derivative[1][h][0])) 
+ (h^\[Prime]\[Prime])[0] + Derivative[1][f][0] (h^\[Prime]\[Prime])[0]) == 0 
&& 
1/12 (-4 Derivative[1][f][0] (3 + Derivative[1][h][0] (3 + Derivative[1][h][0])+ (h^\[Prime]\[Prime])[0]) + (f^\[Prime]\[Prime])[0] (Derivative[1][h][0] (3 + 2 Derivative[1][h][0]) 
+ (h^\[Prime]\[Prime])[0]) -2 Derivative[1][f][0]^2 (Derivative[1][h][0] (3 + 4 Derivative[1][h][0]) 
+ (h^\[Prime]\[Prime])[0])) == 0*)
$\endgroup$
1
  • $\begingroup$ Brilliant! That exactly solves it, thank you. $\endgroup$
    – mwalth
    Mar 11, 2021 at 17:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.