Issue with solving an expression using solve command

I am doing coding for my thesis in Cognitive Radio Systems. After a series of simplification and substitution I got following expression for R_new which is the function of t_new

R_new=29310*t_new - (740230637807590039*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1))/9007199254740992 + (553376302050334301*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(31*t_new^(1/2) - 2383690921325797/562949953421312))/2) - 1))/9007199254740992 + (2982482666894658895*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(26*t_new^(1/2) - 7691227197842525/2251799813685248))/2) - 1))/18014398509481984 + (6130259553882274789*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(32*t_new^(1/2) - 618966936424713/140737488355328))/2) - 1))/36028797018963968 + (2335679196965696725*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(25*t_new^(1/2) - 3661259950175195/1125899906842624))/2) - 1))/18014398509481984 + (495882660278870997*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(81*t_new^(1/2) - 1748195917190933/140737488355328))/2) - 1))/18014398509481984 + (380895376735944389*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(74*t_new^(1/2) - 6347298562227079/562949953421312))/2) - 1))/9007199254740992 + (4391076890295509843*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(35*t_new^(1/2) - 688099554704501/140737488355328))/2) - 1))/36028797018963968 + (2810001741580268983*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(37*t_new^(1/2) - 2936751867564147/562949953421312))/2) - 1))/36028797018963968 + (4247342785866851583*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(38*t_new^(1/2) - 6057857383874553/1125899906842624))/2) - 1))/18014398509481984 + (725857227364724213*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(27*t_new^(1/2) - 4029967247667347/1125899906842624))/2) - 1))/4503599627370496 + (4031741629223864193*2^(1/2)*pi^(1/2)*(erf((2^(1/2)*(39*t_new^(1/2) - 6242211032620389/1125899906842624))/2) - 1)*((10*t_new)/3 - 1))/36028797018963968 + (1660128906151002903*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(23*t_new^(1/2) - 3292552652683065/1125899906842624))/2) - 1))/9007199254740992 + (1573888443493807947*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(51*t_new^(1/2) - 8454454817717421/1125899906842624))/2) - 1))/36028797018963968 + (1617008674822405425*2^(1/2)*pi^(1/2)*((10*t_new)/3 - 1)*(erf((2^(1/2)*(24*t_new^(1/2) - 1738453150714565/562949953421312))/2) - 1))/9007199254740992

now I need to calculate the value of t_new, which can be solved by taking the derivative and setting it to zero for this I did

R_derivative= diff(R_new)

t=solve(R_derivative)

but I am getting following result

Warning: Explicit solution could not be found. In solve at 83 In algorithm_temp at 84

t =

[ empty sym ]

but when I solve R_new for t_new it will give exact value. why I can't solve R_derivative ?

thanks in advance !!

Answer 1

[Walter Roberson]

Are you getting a symbolic solution for t_new from the R_new expression? One that is not expressed in terms of RootOf() ? I do not seem to be able to eliminate the RootOf(). If you are substituting in a numeric value for R_new then the solver would do a numeric solution rather than an exact solution.

The numeric solution for the differential is approximately 0.04082971273

[Suwas Karki]

If

t=solve(R_new)

then I got the result as

t =

0.010311267847402536009590909191235*i - 0.012822346460427825769512705063589

if I further use eval as

t1=eval(t)

then I got the value as

t2 =

      1.0183

And how you got that value 0.04082971273 ?? Is there any mistake that I had made in coding during calculation ?

[Walter Roberson]

Never eval() a symbolic expression. Likely you had a variable named "i" with value 100 at the time you eval()'d. The "i" in the symbolic expression represents the square root of negative 1.

The solution you get is not exact: it is a numeric solution corresponding to R_new = 0. You cannot get a closed-form algebraic ("exact") solution for the answer.

The value I gave of 0.0408<etc> is for the solution of R_derivative = 0, not for the solution of R_new = 0. R_new = 0 has no real-valued solutions

For any given value of R_new between approximately 1551 and 8789.714457, there are two corresponding t_new, one between 0 and 0.0408<etc> and the other between 0.0408<etc> and 0.2985<etc> . For larger real values of R_new there is only a single t_new greater than 0.2985<etc>

[Suwas Karki]

ok I also want solution of R_derivative=0 not of R_new=0. So how can I get the solution through coding ? and which value should I consider as you wrote above there are two values of t_new and how can I know it ?

[Walter Roberson]

The solution of R_derivative = 0 will have to be found by numeric approaches. It is the 0.04082971273 I indicated.

Rdfun = matlabFunction(diff(R_new, t_new));

initialguess = 1;

fzero(Rdfun, initialguess)

[Suwas Karki]

I am getting following error with your code

??? Error using ==> erf
Input must be real and full.

Error in ==>

Error in ==> fzero at 352
        a = x - dx;  fa = FunFcn(a,varargin{:});

Error in ==> algorithm_temp at 86
fzero(Rdfun, initialguess)

[Walter Roberson]

Oh yes, the numeric erf does not handle complex inputs. You will need to use the interval form of fzero to prevent t_new from going negative.

fzero(Rdfun, [0 realmax])

[Suwas Karki]

Again getting following error

??? Error using ==> fzero at 246
Function values at interval endpoints must be finite and real.

Error in ==> algorithm_temp at 84
fzero(Rdfun, [0 realmax])

[Walter Roberson]

Sorry, it appears that instead of realmax, you will need to use a number on the order of pi * 10^303 (approximately.) If you need to deal with larger t_new values than that, we are going to have to change tactics.

[Suwas Karki]

you mean I have to write code like below

Rdfun = matlabFunction(diff(R_miss_t, t_new))
fzero(Rdfun, [0 pi*10^303])

if yes then again I am getting the same error as

??? Error using ==> fzero at 246
Function values at interval endpoints must be finite and real.

Error in ==> algorithm_temp at 84
fzero(Rdfun, [0 pi*10^303])

[Suwas Karki]

sorry instead of R_miss_t it is R_new in above my comment

[Walter Roberson]

Ah, it appears that the interval you should use for that is [realmin realmax]

[Suwas Karki]

No with

fzero(Rdfun, [realmin realmax])

again getting same error :(

??? Error using ==> fzero at 246
Function values at interval endpoints must be finite and real.

Error in ==> algorithm_temp at 84
fzero(Rdfun, [realmin realmax])

[Walter Roberson]

I don't know. Try using [0.0001 100] as the interval.