MATLAB Answers

How can I simplify a symbolic expression?

1 view (last 30 days)
Omar B.
Omar B. on 3 Apr 2021
Commented: Omar B. on 5 Apr 2021
I have computed orthonormal polynomials using Gram-Schmidt process.
I got the following last polynomial,
How do I simplify it? I used simplifyFraction(P10), but still need to simplify.
P10= -(0.000000000000000000000000000000131072*(- 1.5154155839676911728909442039037e+34*x^10 + 8.2506771081142959211403787378745e+34*x^9 - 1.8992595448506783923507009086271e+35*x^8 + 2.3655495760953774286575822531395e+35*x^7 - 1.652095984482293770493859931794e+35*x^6 + 5.3305646291987002322965016172684e+34*x^5 + 7.0517607813983145020182411184835e+33*x^4 - 1.353163416021423788305989629406e+34*x^3 + 5.2851909224935648181523245483154e+33*x^2 - 950949168750141165645381083597780.0*x + 67958017824320084984093552226347.0))/(x^3*(5.0*x - 6.0)^2)

Accepted Answer

Walter Roberson
Walter Roberson on 3 Apr 2021
That is not a polynomial. Notice it has a division in it. It becomes infinite at 0 (triple root) and 1.2 (double root), which is not something that a polynomial with finite coefficients can have happen.
You can expand out the numerator, but that is not much of a change.
S = @(v) sym(v)
S = function_handle with value:
syms x
P10= -(S('0.000000000000000000000000000000131072')*(- S('1.5154155839676911728909442039037e+34')*x^10 + S('8.2506771081142959211403787378745e+34')*x^9 - S('1.8992595448506783923507009086271e+35')*x^8 + S('2.3655495760953774286575822531395e+35')*x^7 - S('1.652095984482293770493859931794e+35')*x^6 + S('5.3305646291987002322965016172684e+34')*x^5 + S('7.0517607813983145020182411184835e+33')*x^4 - S('1.353163416021423788305989629406e+34')*x^3 + S('5.2851909224935648181523245483154e+33')*x^2 - S('950949168750141165645381083597780.0')*x + S('67958017824320084984093552226347.0')))/(x^3*(S('5.0')*x - S('6.0'))^2)
P10 = 
[N,D] = numden(P10);
PS = simplify(expand(N))/D
PS = 

Sign in to comment.

More Answers (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!