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I am trying to solve this equation for kss with mathematical but the system is failing and so I am hoping to get some pointers here.

EulSS= B * Css^(-G)*(1 - D + A* kss^(1-A)) == Css^(-G) 

When I try Solve[EulSS, kss, Reals] it fails. When I try Reduce[{EulSS && Css > 0 && G > 0 && A > 0 && A < 1 && B > 0 && B < 1 && G > 0}, kss, Reals]it just goes into a constant loop.

I will appreciate the help.

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  • $\begingroup$ I'm guessing it is too difficult for Reduce to sort out solutions in real space subject to the parameter constraints. You might instead solve with no restrictions. In[36]:= EulSS = B*Css^(-G)*(1 - D + A*kss^(1 - A)) == Css^(-G); Solve[EulSS, kss] During evaluation of In[36]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. Out[37]= {{kss -> (-((Css^G (-Css^-G + B Css^-G - B Css^-G D))/( A B)))^(1/(1 - A))}} $\endgroup$ Jan 5, 2023 at 23:36
  • $\begingroup$ First: divide both sides by Css^(-G). Then, replace the derivative operator D with d. $\endgroup$ Jan 5, 2023 at 23:43
  • $\begingroup$ Thanks a lot for the proposal. I initially figured that mathematical would be able to work out the division from the information I provided. $\endgroup$ Jan 6, 2023 at 1:15

1 Answer 1

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  1. Divide both sides by Css^(-G)
  2. Replace the derivative operator D by variable d
  3. Replace upper-case letters (which conflict with Mathematica's internal names) with lower-case symbols
  4. Ensure $0 < a < 1$

Then

Assuming[0<a<1,
Solve[k^(1 - a) == (1/b - 1 + d)/a, k,Reals]]

$$k = a^{\frac{1}{a-1}} \left(\frac{b}{b d-b+1}\right)^{\frac{1}{a-1}}$$

with constraints.

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  • $\begingroup$ Thanks a lot for the proposal. I initially figured that mathematical would be able to work out the division from the information I provided. $\endgroup$ Jan 6, 2023 at 1:14
  • $\begingroup$ But your D was as inappropriate as an isolated Sin or Factor or ... $\endgroup$ Jan 6, 2023 at 18:14
  • $\begingroup$ I reworded it with the A, B and so for. I was using greek letters in mathematica and couldn't copy and paste that here. But again, thanks a lot for the suggestion and clarifications. $\endgroup$ Jan 7, 2023 at 1:36

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